Ordinary Differential Equations

Ordinary Differential Equations
Ordinary Differential Equations:
A Systems Approach
Bruce P. Conrad
November 24, 2010
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c
2010
Bruce P. Conrad
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Chapter 1
First-Order Equations
3
4
1.1
CHAPTER 1. FIRST-ORDER EQUATIONS
Introduction
A differential equation is a relation involving an unknown function and
some of its derivatives. For example,
dy
= y + et
dt
is a differential equation that asks for a function, y = f (t), whose derivative
is equal to the function plus et . By differentiating, you can verify that a
function such as y = tet meets this specification.
Differential equations are a source of fascinating mathematical problems, and they have numerous applications.
A mathematical model is a mathematical construction, such as a differential equation, that simulates a natural or engineering phenomenon. Most
applications of differential equations take the form of mathematical models. For example, consider the problem of determining the velocity v of a
falling object.
~
v
?
Newton’s second law of motion tells us that the net force on the object is
equal to the product of its mass, m, and its acceleration, dv
dt . This law is a
differential equation,
dv
m
= F,
dt
Ignoring air resistance, for an object falling close to the earth’s surface the
force is F = mg, directed downward, where g is approximately 9.80 meters
per second per second. Thus the differential equation
m
dv
= mg
dt
is a mathematical model corresponding to a falling object.
To solve the differential equation, cancel the mass and note that v is an
antiderivative of the constant g; thus v = gt + C, where C is an arbitrary
constant.
1.1. INTRODUCTION
5
We have just solved a differential equation: The solution is not a single
function, but a family of functions depending on an arbitrary constant C.
To determine the equation of motion of this particular falling object, we
need to refine our model.
For example, an initial condition, specifying the velocity when t = 0,
will determine the equation of motion. If we assume that the object was
falling from rest, so that v = 0 when t = 0, then we know that the solution
that we seek has the property v(0) = 0. We have knowledge of the starting,
or initial value of v. It is easy to infer from this initial condition that the
constant C is equal to 0, and v = gt.
What if we don’t ignore air resistance? It is known that the force of air resistance is directed oppositely to the motion, with magnitude proportional
to the square of the velocity (assuming the velocity is much less than the
speed of sound). In other words, air resistance = k v2 , where the constant
of proportionality k is the drag coefficient). Combining air resistance and
gravitation, we obtain the differential equation model
dv
k
= g − v2 .
dt
m
(1.1)
It may be tempting to integrate as we did before:
Z k 2
v=
g− v
dt.
m
Since the “unknown function” v appears in the integrand here, there is no
way to calculate this integral without first knowing the answer! Our objective is to manage this quandary.
Terminology. The order of a differential equation is the order of the highest derivative of the unknown function occurring in the equation. The differential equations describing the velocity of a falling object that we just
considered above were first order. In the related second order equation,
y00 = g, the unknown function represented by the variable y is the distance
the object has fallen. The velocity would be v = y0 . Including air resistance,
we get y00 = g − k (y0 )2 /m, another second order equation.
A differential equation involving only derivatives with respect to a single independent variable is called an ordinary differential equation, or
ODE. The falling body models that we just considered are ODEs, in which
the independent variable is t. A differential equation that involves partial
derivatives with respect to two or more independent variables is a partial
6
CHAPTER 1. FIRST-ORDER EQUATIONS
differential equation, or PDE. As an example, here is the second order PDE
that models the vibration of a guitar string:
2
∂2 y
2∂ y
−
c
= 0.
∂t2
∂x2
The unknown function y represents the displacement of a point on the
string x centimeters from the bridge at time t, and c is a constant related
to the tension and density of the string.
In any differential equation, a dependent variable is a variable that represents an unknown function. A solution of a differential equation is a
function that can be substituted for the dependent variable to produce an
identity.
Example 1.1.1 The ODE
dy
= ky
(1.2)
dt
can be as a model for population growth, compound interest on savings accounts,
etc. (see section 1.2). Show that y = Cekt , where C is an arbitrary constant, is a
solution of this ODE.
dy
S OLUTION . If y = Cekt then dt = Ckekt and ky = Ckekt as well. Hence
equation (1.2) becomes an identity.
Differential equations typically have infinite families of solutions, but
we often need just one solution from the family. We refer to a single solution
of a differential equation as a particular solution to emphasize that it is one
of a family.
The general solution of a differential equation is the family of all its
solutions. The general solution of an ODE on an interval ( a, b) is a family
of all solutions that are defined at every point of the interval ( a, b). Finding
the general solution of an ODE requires two steps: calculation and verification. The calculation step is exemplified by our solution of the falling body
equation v0 = g. Unless a mistake was made in the integration, the family
of solutions we found will satisfy the ODE.
The verification step is to show that all solutions of the ODE belong to
this family. The following theorem from calculus is useful for this purpose:
Theorem 1.1 (Equal derivatives theorem) Let f 1 (t) and f 2 (t) be defined and
differentiable on an interval ( a, b) (infinite endpoints are permitted), and assume
1.1. INTRODUCTION
7
that f 10 (t) = f 20 (t) for all t ∈ ( a, b). Then there is a constant C such that f 1 (t) =
f 2 (t) + C for all t ∈ ( a, b).
Example 1.1.2 Verify that the family of solutions v = gt + C for y0 = g is the
general solution.
S OLUTION . Let y0 (t) = gt. If y1 (t) is another solution of the differential
equation, then y10 (t) = y00 (t) = g. It follows from the equal derivatives
theorem that y1 (t) ≡ y0 (t) + C. In other words, every solution of the ODE
belongs to the family y = gt + C.
Because ODEs typically have families of solutions, they are frequently
coupled with additional information (called constraints) to single out a solution of interest. The constraint that we have already encountered, and
shall frequently encounter in the future is an initial condition, specifying
the value of the solution at an initial time. An ODE coupled with an initial
condition is called an initial value problem, or IVP. The motion of a body
falling from rest with air resistance would be modeled by the IVP,
dv
= g − kv2 ; v(0) = 0.
dt
Solution by computer There are many techniques available to find solutions of ODEs. However, many ODEs have solutions that can’t be expressed in terms of the familiar elementary functions you worked with in
calculus courses. Since the invention of the first computer, solutions of intractable ODEs and PDEs have been calculated by computers.
These calculations are done by numerical means; that is, the computer
does not work with a formula for a solution, but calculates a table of values
that give a close approximation of a solution of the differential equation
from the differential equation itself. As a simple example, consider the
ODE of the form y0 = f (t). The solution,
y=
Z
f (t) dt + C,
is helpful if a formula for the antiderivative of f (t) is available. Without
such a formula, we might turn to a numerical method of evaluating the
integral, such as the rectangle rule, the trapezoidal rule, or Simpson’s rule.
Numerical methods for solving ODEs of the form y0 = f (t, y), where the
right side involves y as well as t, are generalizations of these “rules.”
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CHAPTER 1. FIRST-ORDER EQUATIONS
A numerical method cannot compute a family of solutions; it can only
approximate one solution at a time. The user is expected to provide constraints to single out the solution to be approximated. A program that is
designed to approximate solutions of ODEs with initial conditions as the
constraints is called an IVP solver.
While there are programs that are primarily IVP solvers, computer algebra systems (CAS), such as Maple, Mathematica, and Matlab include subroutines that are IVP solvers, as well as the capability to find a formula for
the general solution of practically any differential equation for which there
is an established method of solution. These routines follow rules for manipulating formulas instead of performing numerical calculations. There
are advanced calculators that incorporate IVP solvers and CAS as well.
Qualitative study of an ODE Figure 1.1 displays the graphs of several
solutions of the ODE v0 = g − kv2 representing the motion of a falling object
with air resistance. These graphs appear to have a common
p asymptote,
v = 100. We can explain this feature as follows: Let v∞ = g/k, so that
g = kv2∞ . The ODE can be rewritten as
v0 = k (v2∞ − v2 )
For v < v∞ we see that v0 > 0. This means that v is increasing. When
v > v∞ , v0 < 0, and v is decreasing. In both cases, v tends toward v∞ , as in
figure 1.1.
An explanation of the behavior of solutions of a given ODE obtained by
analyzing the equation itself, without referring to a formula for the general
solution, is called a qualitative study. A qualitative study may result in a
better understanding of the physical phenomenon represented by the ODE
than a formula for the general solution could. For example, we see in figure 1.1 that regardless of the initial velocity, the velocity of a falling object
tends to a limiting velocity (known as the terminal velocity ).
Our qualitative study of the falling object correlates well with the physics.
For v < v∞ , the gravitational force is greater than the air resistance, so
the net force is directed downward and causes the object to accelerate. If
v > v∞ the air resistance is dominant, and the object decelerates. Finally,
if v = v∞ gravity and air resistance balance each other, and the speed is
constant.
Exercises
For each of the differential equations in problems 1 – 13,
1.1. INTRODUCTION
9
(i) determine the order,
(ii) determine whether the differential equation is an ODE or a PDE.
1.
y00 + y = 0.
See answer
2.
y 0 = ( x + y )4
3.
(y0 )2 + xy = e x .
See answer
4.
5.
6.
7.
∂y
∂t
+ 6y ∂y
∂t =
∂3 y
.
∂x3
y 0 = t2 + y2 .
See answer
y(100) − y = 0
y0 = sin y.
See answer
8.
y2 + (y0 )2 = 1.
9.
y3 + ( y 0 )3 = 1
See answer
10.
t2 y00 + ty0 + (t2 − 1)y = 0.
11.
∂2 u
∂x∂y
= 0.
See answer
12.
∂u
∂t
13.
d3 u
dt3
14.
=
∂2 u
.
∂x2
+ 3t du
dt − u cosh t =
See answer
1
.
t2 +1
Show that y = C sec(t), is a family of solutions of
y0 − tan(t)y = 0.
and find solutions that satisfy the following initial conditions.
(a) y(0) = 0.
(b) y(π/4) = 1.
10
15.
CHAPTER 1. FIRST-ORDER EQUATIONS
Show that y = C sin(t) + D cos(t) is a family of solutions of
y00 + y = 0,
and find all solutions that satisfy the following constraints
(a) y(0) = 0.
(b) y(0) = 0, y(π ) = 0.
(c) y(0) = 0, y(π/6) = 1.
(d) y(0) = 1, y0 (0) = −1.
See answer
16.
√
√
Show that y = 2( t − 1) + Ce− t is a solution of
y
y0 + √ = 1,
2 t
and use it to solve the IVP that couples this ODE with the initial condition y(1) = 0.
2
Show that y = Cet is a solution of the ODE y0 = 2ty, and solve
the IVP y0 = 2ty; y(0) = 3.
See answer
0 if t ≤ 0
18.
Let y(t) =
Show that y(t) is a solution of the IVP
t2 if t > 0.
√
y0 = 2 y; y(1) = 1. Also show that
17.
y = t2 for all real t
√
is not a solution! Hint: remember that t2 = |t|.
19.
Let y(t) be a differentiable function defined on the interval (−1, 1)
with the property that for all t,
t2 + y ( t )2 = 1
(there are two such functions). Show that y(t) is a solution of the ODE
yy0 + t = 0.
See answer
1.2. LINEAR GROWTH AND DECAY
11
The following problems require showing a given family of functions to
be the general solution of a given differential equation. This entails verifying that every member of the family is a solution, and that every solution
belongs to the family. In each case, the equal derivatives theorem can be used.
20.
Verify that y(t, C ) = (t + C )e−t is the general solution of the ODE
d t
(e y) = 1
dt
by showing that for every value of the constant C, y(t, C ) satisfies the
ODE, and that every solution has the form y(t, C ) for some C.
21.
Verify that y = t ln t + Ct is the general solution of
1
ty0 − y
=
2
t
t
on the interval (0, ∞). Hint: (ty0 − y)/t2 =
See answer
1.2
d
dt
y
t .
Linear Models for Growth and Decay
An ODE used to simulate a scientific or engineering phenomenon is a type
of mathematical model of the phenomenon. Developing mathematical models involves theorizing, computation, and testing. Theorizing starts by
identifying a variable, or variables that describe the phenomenon (for a
falling body, the variables might be the velocity, the distance fallen, or
both). The ODE then describes how the variable or variables change with
time. The ODE may include parameters such as masses, friction constants,
etc. These are constants in a given system, but can be varied to make the
model applicable to other systems.
Nature is complicated, so a model may not reflect every aspect of the
phenomenon under study. For example, a model used to compute the trajectory of a projectile might ignore air resistance, so that the only force to
be considered is gravitational.
The computation and testing stage of the modeling process involves
finding a solution of the ODE and comparing it with observed data. If the
computed solution and the data don’t agree, the model must be modified
or replaced.
The derivative of a quantity y with respect to time gives its absolute
rate of change. It often happens that the relative rate of change, rather than
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CHAPTER 1. FIRST-ORDER EQUATIONS
the absolute rate of change, fits naturally into a mathematical model . The
relative rate of change of y is defined to be the absolute rate of change,
expressed as a fraction of y. The distinction between absolute and relative
rates of change can be seen by comparing simple and compound interest.
Simple interest is calculated as a percentage of the original deposit, and
does not change as long as no further deposits are made. The absolute rate
of change of value of this account is the dollar amount of interest earned
per unit time. For example, if the account pays 4% interest and $4000 is
initially deposited, then interest payments of 0.04($4000) = $160 will be
paid each year for the life of the account. The value y of the account is
determined by the ODE
dy
= 160,
dt
with y(0) = 4000. This indicates a constant absolute rate of change. After t
years, the value will be
4000 +
Z t
0
160 dt = $(4000 + 160t).
With compound interest, the interest previously accumulated in the account is included in calculating the interest to be paid. The relative rate
of change is the absolute growth rate expressed as a percentage of the account’s current balance. Banks quote this rate as the Annual Percentage
Rate, and often use the abbreviation “APR.” The formula for the relative
rate of change of the balance y is
1 dy
.
y dt
Assuming that the bank uses “continuous compounding,” as most do, the
relative rate of change of the account balance is equal to the APR, which is
constant. If the APR is k, then the ODE
dy
= k · y,
dt
(1.3)
is an appropriate mathematical model . Equation (1.3) is called the linear
growth equation. The word “linear” refers to the ODE, not its solution.
The linear growth equation can be used as a mathematical model for the
balance on a bank account with continuously compounded interest, as well
as many other phenomena, including population growth and radioactive
decay.
1.2. LINEAR GROWTH AND DECAY
13
Example 1.2.1 The family of solutions y = Cekt of equation (1.3) was found in
example 1.1.1. Verify that it is the general solution.
S OLUTION . We must show that any solution y = φ(t) of equation (1.3)
belongs to the family y = Cekt , or, equivalently, that the product e−kt φ(t) is
constant.
By the product rule for differentiation,
d −kt
[e φ(t)] = −ke−kt φ(t) + e−kt φ0 (t).
dt
Since φ(t) is a solution of equation (1.3), we can replace φ0 (t) with kφ(t),
and then
d −kt
[e φ(t)] = −ke−kt φ(t) + e−kt kφ(t) = 0.
dt
By the equal derivatives theorem, e−kt φ(t) = C, where C is constant. Thus
φ(t) = Cekt .
Every solution of equation (1.3) will display exponential growth if k > 0,
and exponential decay if k < 0. Figures 1.2 and 1.3 show graphs of y = ekt for
k > 0 and k < 0, respectively.
Compound interest is a familiar example of exponential growth.
Example 1.2.2 The IVP
y0 = 0.05 y; y(0) = 1000
is a mathematical model for the balance of a savings account that earns 5% interest—
compounded continuously—where the initial principal is $1000. Solve this IVP to
find the balance as a function of time.
S OLUTION . The general solution of the ODE, y0 = k y is y = Cek t , where C
is constant. For this bank account, k = 0.05, so y = C e0.05 t . When we
substitute t = 0 and y = 1000 to incorporate the initial condition, we find
C = 1000. Hence y = 1000e0.05 t .
The following example shows how the relative growth rate, k, can be
calculated from two data points, y(0) and y(t1 ) for some t1 > 0.
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CHAPTER 1. FIRST-ORDER EQUATIONS
Example 1.2.3 An investment of $10,000, made on June 30, 2005 in the Standard and Poor’s 500 Stock Index would grow to $12,920 by December 31, 2007,
assuming all dividends were reinvested when received. What rate of compound
interest would be required to match this performance? If the investment continued
to increase at this rate, when would the investment be expected to double, thus
reaching a value of $20,000?
S OLUTION . Let k be the interest rate. Then if y(t) denotes the balance at
time t, the IVP y0 = ky; y(0) = 10,000 holds. The general solution of the
ODE is y = Cekt , and the initial condition specifies C = y(0) = 10,000. The
time elapsed is t = 2.5 years. Hence 12,920 = 10, 000e2.5k , so
1
k = 2.5
ln(1.292) = 0.1025 = 10.25%.
Under our assumption that the investment will continue to grow at this rate,
the date it would reach the value of $20,000 would be t years after June
30, 2005, where 10,000e0.1025t = 20,000. Solving for t by taking logarithms,
1
we have t = 0.1025
ln(2) = 6.76 years. Thus we would expect the
investment to reach $20,000 at the end of March, 2012. (The rate of
growth of stock prices in not constant, so we can’t “bank” on this
prediction.)
The doubling time of a solution of (1.3) is the time it takes
for a solution whose initial value is y(0) to reach the value
2y(0).
We saw in example 1.2.3 that the growth rate of the stock market for
the 18 months ending December 31, 2007 can be expressed as a doubling
time of about 6.75 years. The relation between the doubling time and the
relative growth rate can be expressed as follows:
Doubling time =
ln 2
relative growth rate
(1.4)
To verify equation (1.4) just note that if k is the relative growth rate and T
the doubling time, then
ekT = 2.
Taking logarithms, kT = ln 2, which is equivalent to (1.4).
A solution of the linear growth equation (1.3) will double over any time
interval of length T, quadruple in 2T units of time, and so on. When nT
units of time have elapsed, the initial value will be multiplied by 2n .
1.2. LINEAR GROWTH AND DECAY
15
Population growth. Let y(t) represent the population, at time t years after a reference time. Although the population at any time is an integer,
we will be using a continuous variable to represent it. This continuous approximation of a discrete variable restricts this mathematical model to large
populations. Under the assumption that the relative birth and death rates,
b and d, respectively (expressed per thousand population), are constant, the
b−d
relative rate of change for the population is 1000
y0 (t) =
b−d
· y ( t ),
1000
b−d
. In the past, birth
which is the linear growth equation (1.3), with k = 1000
and death rates for human populations have not remained constant for
long periods of time, so predictions based on the linear model are only
reliable over short time spans.
Example 1.2.4 According to the 1980 census, the population of the United States
was then 227 million. The birth rate in 1980 was 15.9 per thousand, and the death
rate was 8.7 per thousand. Use these data with the linear growth model to estimate
the population in 2000.
−8.7
S OLUTION . The relative growth rate is k = 15.9
1000 = 0.0072. Let y
represent the population, in millions. Our model takes the form of an IVP,
y0 = 0.0072y; y(1980) = 227. The general solution of the ODE is
y = Ce0.0072t . If we set t = 1980, and y = 227, we find
227 = Ce(0.0072)(1980)
and hence C = 227e−(0.0072)(1980) . The solution of the IVP is therefore
y = 227e−(0.0072)(1980) e0.0072t = 227e0.0072(t−1980) .
We therefore estimate that in 2000, the population will be
y(2000) = 227e0.0072(20) = 262 million (The 2000 census recorded a
considerably larger population, 281 million. The
discrepancy—approximately the population of New York State—shows the
linear growth model is an oversimplification.)
16
CHAPTER 1. FIRST-ORDER EQUATIONS
Radioactive decay. The linear model for the decay of a radioactive isotope agrees extremely well with experimental evidence. In this model, y(t)
denotes the mass (in grams) of the isotope that has not yet decayed at time
t. It is known that the rate at which atoms of the isotope disintegrate is proportional to the number of atoms present. Furthermore, all atoms of the
isotope have the same mass, so the rate of disintegration is proportional to
y(t). The situation is described by a differential equation, called the linear
decay model:
dy
= −ky.
dt
The parameter k is the decay constant. In this instance, the linear decay
model is a special case of the linear growth model (1.3), where the growth
rate is negative.
The general solution of the linear decay equation is y = C e−kt so we
can say that the mass of the isotope is undergoing exponential decay.
The analog of the doubling time in this case is the half life, which is the
time it takes for a solution of equation (1.3) whose initial value is y0 to reach
the value 12 y0 .
Example 1.2.5 The half life of the isotope 131 I of iodine is 8 days. A nuclear
accident releases 100 kilograms of 131 I into the environment. How much 131 I will
remain in the environment after a period of one year? After two years?
S OLUTION . A rough estimate is made by noting that a year is made up of
slightly more that 45 eight day periods. At the end of each eight day
period, there is half as much 131 I present as there was at the beginning.
Thus, after the first eight days, there are 50 kg of the isotope left, after the
first 16 days, 25 kg will be left, and so on. After a year, the amount will be
about 100 · 2−45 = 2.84 × 10−12 kilograms left. Two years make up more
than 91 eight day periods, so the amount left after two years will be about
100 · 2−91 = 4.04 × 10−26 kilograms left.
It is prudent to view answers involving extremely large or extremely
small numbers critically. An atom of 131 I has mass 2.17 × 10−25 kilograms,
and it is impossible to have less than an atom of any substance. Therefore, the estimate we have made of the amount of 131 I left after two years
is wrong. This is not surprising since the model that we are using is not
accurate when only a few atoms of isotope are present.
1.2. LINEAR GROWTH AND DECAY
17
Homogeneous linear equations
A first order linear ODE is an equation of the form
a1 ( t ) y 0 + a0 ( t ) y = b ( t ).
The functions a1 (t) and a0 (t) are called the coefficient functions , and b(t)
is the source term.
For example, equation (1.3), y0 − k y = 0 is linear, and the falling body
equation with air drag, v0 + kv2 = g, is not linear because of the quadratic
term in the variable v.
A linear ODE is homogeneous if the source term is 0. Equation (1.3) is
homogeneous, and every homogeneous first order linear ODE can be put
in the form
(1.5)
y0 = k (t) y,
where k (t) = − a0 (t)/a1 (t). Thus a homogeneous linear equation with variable coefficients can be used as a model for growth or decay when the relative growth rate is a function of time.
To solve a homogeneous first order linear ODE we can mimic the derivation of the solution of equation (1.3) as follows: substitute y = CeK(t) , where
K (t) is a function that we will attempt to determine and C is an arbitrary
constant. Using the chain rule, we find that
dy
= C K 0 ( t ) eK (t)
dt
Referring to equation (1.5),
C K 0 ( t ) eK (t) = k ( t ) C eK (t) .
Hence K 0 (t) = k (t). In other words, if K (t) is an antiderivative of k (t) then
y = CeK(t) is a family of solutions of equation (1.5). This generalizes what
we found in the constant coefficient case.
Proposition 1.2.1 Let k(t) be a function that is continuous on an interval (c, d),
and let K (t) be an antiderivative of k(t) on that interval. Then the family of solutions y = CeK(t) of y0 = k (t)y is the general solution.
P ROOF. Proceed as in Example 1.2.1. Let y = φ(t) be any solution of
equation (1.5). By the product rule for differentiation,
d −K (t)
[e
φ(t)] = −K 0 (t) e−K(t) φ(t) + e−K(t) φ0 (t)
dt
= − k ( t ) e−K (t) φ ( t ) + e−K (t) k ( t ) φ ( t )
= 0.
18
CHAPTER 1. FIRST-ORDER EQUATIONS
By the equal derivatives theorem, e−K(t) φ(t) is constant, and hence φ(t) =
CeK(t) .
The homogeneous case
An ODE that can be put into the form a1 (t)y0 + a0 (t)y = 0 is
called a homogeneous first order linear equation. If k (t) =
− a0 (t)/a1 (t) is continuous on an interval (c, d) then the general solution is y = CeK(t) , where K (t) is an antiderivative of
k (t) and C is a constant. Thus, the following formula for the
general solution holds:
y = Ce−
R
( a0 (t)/a1 (t)) dt
A linear ODE with variable coefficients may have singular points, where
k (t) = − a0 (t)/a1 (t) is discontinuous. The expression y = CeK(t) can represent the general solution only on an interval that does not contain any
singular points. This comment is applicable to the next example, since the
ODE there has a singular point at 0. In this example we will substitute
y = CeK(t) into the equation without first dividing through by the coefficient of y0 .
Example 1.2.6 Find the general solution of
ty0 + my = 0
where m is a constant, on the interval (0, ∞), and find the solution that satisfies
the initial condition y(1) = −1.
S OLUTION . Substitute y = CeK(t) , and y0 = CK 0 (t)eK(t) to obtain
tCK 0 (t)eK(t) + mCeK(t) = 0.
Solving for K 0 (t) we obtain K 0 (t) = −m/t. Hence
K (t) = −
Z
m
dt = −m ln |t|.
t
Because our solution is to be found on (0, ∞), |t| = t. Also note that the
integration constant does not appear, because we don’t need the most
1.2. LINEAR GROWTH AND DECAY
19
general antiderivative of m/t.) It follows that y = Ce−m ln t = C t−m is the
general solution. You should check this result by differentiation.
To satisfy the initial condition, set t = 1 and y = −1 to obtain −1 = C. The
solution of the IVP is y = −tm .
Exercises
1.
The doubling time for a certain population is 35 years. How long
does it take for this population to triple?
See answer
2.
The U.S. national debt amounted to $1 trillion in 1980. By 1988 it
had grown to $2 trillion, and by 1992, $4 trillion. Does this information indicate that the debt was growing exponentially between 1980
and 1992, or does it indicate otherwise?
3.
Thomas Robert Malthus1 was concerned that the population will
increase exponentially if it is not controlled by limitations of resources
or governmental regulation. In his words,
In the United States of America, where the means of subsistence have been more ample, the manners of the people
more pure, and consequently the checks to early marriages
fewer than in any of the modern states of Europe, the population has been found to double itself in twenty-five years.
By how much did the birth rate (per thousand) exceed the death rate
in the United States at the time of Malthus?
See answer
4.
If the birth and death rates of 1980 (see example 1.2.4) were sustained indefinitely, how long would it take for the population to double?
Answer: 96 years.
5.
Two countries, A and B, each had a population of 1 million in
1900. The population of A obeys the exponential growth equation (1.3)
1 Essay on the Principle of Population, 1798. The quotation was taken from page 22 of
Norton Critical Edition, edited by Philip Appleman, Norton, New York 1976
20
CHAPTER 1. FIRST-ORDER EQUATIONS
with relative growth rate k = .04, and it is observed that the population of B, in millions, is always equal to the square root of the population of A, in millions. Thus, for example, when the population of A
reached 4 million, the population of B was 2 million. Show that the
population of B also obeys equation (1.3), and find its relative growth
rate.
See answer
6.
An investor sold her stock in two companies, A and B, for $44,200
and $2800, respectively. Each stock had been bought for an initial
purchase price of $1000, with all dividends reinvested. The A stock
had been held for 13.26 years, while the B stock had been held for
5.75 years. For each stock, compute the compound interest rate which
would yield an equivalent return over the same time period. Which
was the more lucrative investment?
7.
A bond, dated 1700, promises to pay the bearer £1 sterling plus
accumulated interest, compounded continuously at 6% per annum.
What is the bond worth in 2000?
See answer
In Exercises 8–15, find the singular points (if any) and the general solution of the ODE on intervals not containing singular points. Sketch the
graph of the solution that satisfies the given initial condition.
8.
(a) y0 − 2y = 0; y(0) = 1
(b) y0 + 2y = 0; y(0) = 1.
9.
10.
11.
ty0 − 12y = 0; y(2) = 1.
See answer
y0 + sin(t)y = 0; y(0) = 1.
cos(t)y0 − sin(t)y = 0; y(0) = −1.
See answer
12.
y0 + ty = 0; y(1) = 1.
13.
ty0 + y = 0; y(1) = 1.
See answer
1.2. LINEAR GROWTH AND DECAY
21
14.
(a) ty0 + |t|y = 0; y(0) = 1.
(b) ty0 − |t|y = 0; y(0) = 1.
15.
(t2 + 1)y0 − ty = 0; y(3) = 0.
See answer
Carbon dating
A radioactive isotope of carbon, 14 C, has a half life of 5730 years. The fraction of 14 C found in atmospheric carbon dioxide is constant at about one
part per billion (abbreviated 1 ppb), because the decay rate is matched by
the rate at which new 14 C is created by the effect of cosmic radiation. A
living plant acquires all of its carbon by respiration of atmospheric carbon
dioxide; hence the fraction of a living plant’s carbon which is the isotope
14 C is also 1 ppb. When the plant dies, decaying 14 C is not replaced, so the
time elapsed since the death of the plant can be estimated by measuring
the fraction of 14 C which remains. This technique, known as 14 C radiometric dating, was discovered in 1947 by Willard F. Libby. Before his work,
no one had noticed that atmospheric carbon dioxide contained 14 C. Libby
received the 1960 Nobel Prize for Chemistry as a result of this discovery.
16.
Suppose that a tree dies at t = 0. Show that the fraction of 14 C in
wood from the tree, in ppb, is modelled by the IVP,
y0 = −0.000121y; y(0) = 1.
17.
King Tutankhamen died in 1325 B.C.E., and his tomb was discovered in 1922 by Howard Carter. What would the proportion of 14 C in
a wood sample taken from the tomb? (Of course, this wouldn’t have
been measured until after 1947).
See answer
18.
A sample of wood from an archaeological investigation contains
0.7 ppb 14 C. Assuming that the wood, when part of a living tree, had
1 ppb 14 C, estimate the age of the wood.
19.
A sample of wood taken from a bristlecone pine tree on White
Mountain is determined to be 4850 years old by counting growth
22
CHAPTER 1. FIRST-ORDER EQUATIONS
rings. What would the expected proportion of 14 C in the sample be?
See answer
20.
14 C
radiometric dating is considered useful to determine the age
of objects between 500 and 50,000 years old. Use this information to
estimate the following measures of precision of the device used to
determine the number of ppb 14 C, on the basis of this statement.
Sensitivity The least concentration that can be detected.
Relative error The measurement error, expressed as a percentage of
the true concentration.
Answer: sensitivity: .002 ppb, relative error, 6%.
1.3
Linear First Order Equations
A hot object in a cold environment will cool, and the temperature of a cold
object in a warm environment will increase. Newton’s Law of Cooling , is
a simple model of this phenomenon. Expressed as an ODE it is
T 0 (t) = −k [ T (t) − A(t)]
(1.6)
where T (t) is the temperature of the object at time t, and A(t) is the ambient temperature (the temperature of the environment). The coefficient k is
the transmission coefficient , which tells the rate of heat transfer between
the object and its environment.
Equation (1.6) can be written in the form
T 0 + kT = kA(t),
and this reveals that it is linear, but inhomogeneous, because of the presence of the source term, kA(t).
To solve an inhomogeneous equation, solve the associated homogeneous equation—obtained by removing the source term—first. The associated homogeneous equation for
is
a1 ( t ) y 0 + a0 ( t ) y = b ( t ),
(1.7)
a1 (t)y0 + a0 (t)y = 0.
(1.8)
Denote the particular solution eK(t) of (1.8) by yh (t). The subscript h
stands for “homogeneous;” thus the general solution of the homogeneous
1.3. LINEAR FIRST ORDER EQUATIONS
23
equation (1.8) is y = C yh (t). The following theorem tells how to express
the general solution of an inhomogeneous linear ODE.
Theorem 1.2 Suppose that the coefficient functions a0 (t) and a1 (t), and the source
term b(t) in the inhomogeneous linear ODE (1.7) are continuous on an interval
(c, d), and that a1 (t) 6= 0 on that interval. Let y p (t) be a particular solution
of (1.7), defined on (c, d), and let yh (t) be a particular solution of the associated
homogeneous equation (1.8). Then the general solution of (1.7) on (c, d) is
y = y p (t) + C yh (t)
(1.9)
P ROOF. To verify that every function of the form (1.9) is a solution, just
substitute y = y p (t) + C yh (t) in (1.7):
a1 (t)[y p (t) + C yh (t)]0 + a0 (t)[y p (t) + C yh (t)] =
a1 (t)y0p (t) + a0 (t)y p (t) +C [ a1 (t)y0h (t) + a0 (t)yh (t)]
|
{z
}
|
{z
}
=b(t)
=0
= b ( t ).
Now let’s verify that every solution of (1.7) belongs to the family (1.9).
Let y = φ(t) be any solution of (1.7) that is defined on (c, d), and put ψ(t) =
φ(t) − y p (t). Then
a1 ( t ) ψ 0 ( t ) + a0 ( t ) ψ ( t ) =
a1 (t)φ0 (t) + a0 (t)φ(t) −[ a1 (t)y0p (t) + a0 (t)y p (t)]
|
{z
} |
{z
}
=b(t)
=b(t)
= 0.
Therefore, ψ(t) is a solution of the associated homogeneous equation (1.8).
By proposition 1.2.1, there is a constant C such that ψ(t) = C yh (t). It follows that
φ ( t ) = y p ( t ) + C y h ( t ).
The technique that we will use to determine a particular solution y p (t)
of an inhomogeneous linear ODE is known as the method of variation of
constants 2 . We replace the constant C that appears in (1.9) with a new
dependent variable v, and substitute the resulting expression in the inhomogeneous equation. Thus, substitute y = vyh (t) in (1.7) and replace y0
2 Some
texts refer to the method as ”variation of parameters.”
24
CHAPTER 1. FIRST-ORDER EQUATIONS
with v0 yh (t) + vy0h (t) (obtained by the product rule for differentiation). This
yields the equation
a1 (t)(v0 yh (t) + vy0h (t)) + a0 (t)vyh (t) = b(t)
or, more simply,
a1 (t)v0 (t)yh (t) + v(t)[ a1 (t)y0h (t) + a0 (t)yh (t)] = b(t).
(1.10)
Since yh (t) is a solution of the homogeneous equation,
a1 (t)y0h (t) + a0 (t)yh (t) = 0,
so the expression in square brackets in equation (1.10) drops out. Now we
have
a1 ( t ) v 0 ( t ) y h ( t ) = b ( t ).
which we solve for v0 (t) to get
v0 (t) =
and hence
v(t) =
b(t)
,
a1 ( t ) y h ( t )
b(t)
dt + C,
a1 ( t ) y h ( t )
Z
where C is constant. Upon multiplying this v(t) by yh (t) we obtain the following expression for the general solution of equation (1.7) on the interval
(c, d):
y(t) = yh (t)
where
Z
b(t)
dt + Cyh (t)
a1 ( t ) y h ( t )
yh = Ce−
R
(1.11)
( a0 (t)/a1 (t)) dt
is the solution of the homogeneous equation
a1 (t)y0 + a0 (t)y = 0.
We will revisit the method of variation of constants when solving higher
order linear equations and systems of linear equations. In these cases, more
than one constant will be “varied” — that is the reason for the plural “constants” in “variation of constants.”
1.3. LINEAR FIRST ORDER EQUATIONS
25
Example 1.3.1 Find the general solution of
y0 + 3y = 2e−t
on the interval (−∞, ∞), and determine the solution that satisfies the initial condition y(0) = 0.
S OLUTION . The associated homogeneous equation is y0 + 3y = 0, and we
can take yh = e−3t .
Substitute y = ve−3t and y0 = v0 e−3t − 3ve−3t in the in the inhomogeneous
ODE to get
v0 e−3t − 3ve−3t + 3ve−3t = 2e−t .
When simplified, this reduces to v0 = 2e2t . Thus
v=
Z
2e2t dt = e2t + C
Since y = ve−3t , the general solution is the family y = e−t + Ce−3t .
To satisfy the initial condition, set y = 0 and t = 0. This yields 0 = 1 + C so
C = −1, and
y = e−t − e−3t
As expected, the general solution found in example 1.3.1 splits as the
sum of a particular solution, y p = e−t , and the general homogenous solution, Cyh = Ce−3t . Figure 1.5 displays the graphs of several solutions.
In the following example, we apply the method of variation of constants
to a cooling problem.
Example 1.3.2 The temperature in an oven is 200◦ C when the oven is turned
off. After 10 minutes, the temperature is 175◦ C. The temperature in the kitchen is
20◦ C. Find an expression for the temperature of the oven as a function of time.
S OLUTION . We will use the IVP
dT
= −k( T − A(t)); T (0) = 200
dt
26
CHAPTER 1. FIRST-ORDER EQUATIONS
as our model, where T is the temperature in ◦ C, and A(t) = 20 is the
ambient temperature. We can write the ODE simply as T 0 + kT = 20k. The
associated homogeneous equation is
T 0 + kT = 0,
and hence Th = e−kt . Substitute T = ve−kt in the inhomogeneous equation
and simplify to obtain v0 = 20kekt . Integrate and get
v = 20ekt + C.
The general solution is therefore T = e−kt v = 20 + Ce−kt . To evaluate the
parameters C and k, we need to use the data. Substituting the initial
condition T (0) = 200, we get
200 = 20 + Ce0 ,
and C = 180. Using the second data point, t = 10, T = 175 we get
175 = 20 + 180e−10k ,
1
so e−10k = 155/180, and, k = − 10
ln(155/180) ≈ 0.015. Thus
T (t) = 20 + 180e−0.015t .
The model predicts that the difference between the oven temperature
and the ambient temperature will decay exponentially.
If the object whose temperature is being modeled contains a source of
heat, equation (1.6) must be modified by inserting another source term. Let
H (t) denote the rate that heat is generated within the object (H (t) would
be negative in some cases, such as air conditioning). Then
T 0 (t) = −k [ T (t) − A(t)] + mH (t),
(1.12)
where m is a positive constant, inversely proportional to the heat capacity
of the object. For a small object, such as a toaster oven, m would be relatively large, while for a large object, such as a domed sports stadium, m
would be small. In practice, we do not need to know the individual values
of m or H (t); all we need is the product mH (t).
1.3. LINEAR FIRST ORDER EQUATIONS
27
Example 1.3.3 The insulation of a building has transmission coefficient k = 0.2
hour−1 . When the furnace of the building is operating continuously and the outdoor temperature is 0◦ C, the indoor temperature will be maintained at a constant
20◦ C. Find the temperature inside the building as a function of time, with the
furnace operating continuously, assuming that the outdoor temperature varies between −7◦ C and 13◦ C each day, and that at time t = 0 the initial temperature
indoors is 20◦ C, and outdoors it is 3◦ C and getting warmer.
S OLUTION . We are given that when the ambient temperature is 0, the
solution is T (t) ≡ 20. This specifies a constant solution of equation (1.12),
which can be used to determine the magnitude of the heat source term.
Substitute T (t) = 20, T 0 (t) = 0, and A(t) = 0 to get
0 = −0.2(20 − 0) + mH (t).
Hence mH (t) ≡ 4 degrees per hour.
We assume the ambient temperature A(t) varies sinusoidally with a
period of 24 hours, average temperature Ta = −72+13 = 3 degrees
centigrade, and amplitude V = 13 − Ta = 10 degrees. Hence
2πt
A(t) = 3 + 10 sin
,
24
where t = 0 is the time in the morning that A(t) = Ta . Hence the
temperature satisfies the IVP,
πt
0
T (t) = −0.2 T (t) − 3 + 10 sin
+ 4; T (0) = 20.
12
The ODE can be written more simply as
T 0 + 0.2T = 4.6 + 2 sin
πt
12
.
The homogeneous solution is e−0.2t so we substitute T = e−0.2t v and
simplify to find
v0 = e0.2t [4.6 + 2 sin (.26t)]
π
( 12
≈ 0.26). Now integrate (it is fair to use a table of integrals here) to find
v = 23e0.2t + 18.6e0.2t (0.2 sin(0.26t) − 0.26 cos(0.26t)) + C
28
CHAPTER 1. FIRST-ORDER EQUATIONS
Thus
T (t) = e−0.2t v = 23 + 3.7 sin(0.26t) − 4.8 cos(0.26t) + Ce−0.2t .
When t = 0 and T = 20, and we have 20 = 23 − 4.8 + C. It follows that
C ≈ 1.8.
Figure 1.4 displays a graph of the temperature function found in example 1.3.3. In our solution, we took the trouble to evaluate the constant C, using the initial condition. It is clear from figure 1.4 that this was unnecessary
for most purposes, because the temperature settles into a periodic regime
after some time has elapsed. The term Ce−0.2t , which is the homogeneous
solution, is called a transient because it decays to 0 with increasing time.
The periodic solution
T = 23 + 3.7 sin(0.26t) − 4.8 cos(0.26t))
of the inhomogeneous ODE is said to be stable because every solution is
the sum of this and a transient term. The keys to this example are that
• There is a stable periodic temperature over 24 hours.
• The mean temperature is 23◦ C.
• The temperature ranges between 17◦ and 29◦ C (the temperature range
was determined from the graph).
• The initial condition is practically irrelevant; it determines only the
transient.
The occupants of the building (or a thermostat) will turn off the heat
to prevent it from getting as warm as 29◦ C, which invalidates our model,
as the ODE would no longer be linear (see Exercise 22 at the end of this
section).
Example 1.3.4 Find the solution of the IVP,
ty0 − y = t2 ln(t); y(1) = 2.
(1.13)
on the interval (0, ∞).
S OLUTION . The general solution y = Ct−m of the homogeneous equation,
1.3. LINEAR FIRST ORDER EQUATIONS
29
ty0 + my = 0 was found in example 1.2.6. Here m = −1, so we will use the
homogeneous solution y = Ct.
Thus, substitute y = v t, and y0 = v0 t + v in equation (1.13). After
simplifying, we have v0 = ln(t). Thus
v=
Z
ln(t) dt = t ln(t) − t + C,
and the general solution of the ODE is
y = t2 ln(t) − t2 + Ct.
Now substitute initial conditions (t, y) = (1, 2) to obtain C = 3. The solution
of the IVP is
y = t2 (ln(t) − 1) + 3t.
The general solution in example 1.3.4 again has the form y p + Cyh , where
y p = t2 (ln(t) − 1) and yh = t.
Exercises
You may check your answers with a CAS, if you like; if an antiderivative is
hard to figure out, go ahead and use the CAS for that, too.
In problems 1 – 12, find the general solutions of the ODEs on the intervals indicated (if no interval is indicated, the solution should be valid for all
real t). Then find the particular solutions that satisfy the initial conditions.
In some of these problems, a pair of differential equations that differ only
in the sign of the coefficient is given. Use a CAS (or graphing calculator) to
compare the graphs of several solutions of each of these paired equations.
In cases where the solutions are defined for all real t, do the solutions seem
to converge to some value as t → ∞, or as t → −∞?
1.
2.
3.
4.
y0 = 3t − 4y on (−∞, ∞); y(0) = 0.
See Answer
ty0 − y = t3 − 2t on (0, ∞); y(1) = 0.
2y0 + y = t−1 e−t/2 on (0, ∞); y(2) = 0.
See Answer
ty0 + y = 1 on (0, ∞); y(1) = 0.
30
CHAPTER 1. FIRST-ORDER EQUATIONS
2
5.
y0 + 2ty = e−t on (−∞, ∞); y(0) = 0.
See Answer
6.
y0 + ty = t on (−∞, ∞); y(0) = 0.
7.
(a) y0 + 10y = 1 on (−∞, ∞).
(b) y0 − 10y = 1 on (−∞, ∞).
Initial condition for both: y(0) = 1.
See Answer
8.
(a) ty0 + 12y = t3 on (0, ∞).
(b) ty0 − 12y = t3 on (0, ∞).
Initial condition for both: y(1) = 0.
9.
(a) y0 + 4y = 2e−4t sin(2t) on (−∞, ∞).
(b) y0 − 4y = 2e−4t sin(2t) on (−∞, ∞).
Initial condition for both: y(0) = 0.
See Answer
10.
(a) y0 + tan(t)y = sec3 (t) on (−π/2, π/2).
(b) y0 − tan(t)y = sec3 (t) on (−π/2, π/2).
Initial condition for both: y(0) = 1.
11.
12.
13.
cos(t)y0 = (y − 1) sin(t); y(0) = 0 on (−π/2, π/2).
See Answer
y0
y −1
= t; y(0) = 1 − 10−6 on (−∞, ∞).
A penny is heated to 800◦ C and is then allowed to cool. The
temperature after a minute is 600◦ , and the room temperature is 20◦ .
When will it be safe to pocket the coin (the temperature should be less
than 50◦ C)?
See Answer
1.3. LINEAR FIRST ORDER EQUATIONS
31
14.
A roast has an internal temperature of 15◦ C when it is put in a
175◦ C oven. After 1 hour, a meat thermometer placed in the roast
registers 50◦ C. How much longer will it take for the roast to reach the
state of medium rare (65◦ C)?
15.
A turkey is at room temperature (20◦ C) when it is put into the
oven, and it is removed after 5 hours, when its internal temperature
has reached 85◦ C. The oven temperature is 160◦ C. The turkey is allowed to stand outside the oven for 30 minutes prior to carving. Estimate the internal temperature of the turkey when it is carved. What
assumptions are necessary?
See Answer
16.
The building described in example 1.3.3 is given additional insulation, reducing the transmission coefficient to k = 0.1 hour−1 . The
furnace output is reduced to half of its capacity to compensate for the
additional insulation. Find the new mean temperature of the building
and the amplitude of its variation. Ignore transients.
17.
At 3:00 am, the temperature inside a house is 15◦ C, and the heat
is turned off by a timer. When the heat is turned on again at 6:00
am, the temperature has fallen to 10◦ C. Throughout this period, the
temperature outdoors is −20◦ C. Find the rate constant for heat loss in
the building.
See Answer
18.
(continuation of problem 17) The outdoor temperature remains
at −20◦ C. Assume that the furnace produces heat at a constant rate,
and remains on until 9:00 am, when the temperature inside the house
reaches 20◦ C. How long would it take to warm the house from 10◦ C
to 20◦ C if the temperature outdoors were 0◦ C?
Answer: 2 hours 17.5 minutes
19.
A cabin has two identical wood stoves; only one is in operation.
The temperature outside is −15◦ C, and the indoor temperature has
stabilized at an uncomfortable 10◦ . Therefore, the second stove will
be put into use. At what temperature will the temperature now stabilize?
See Answer
32
20.
CHAPTER 1. FIRST-ORDER EQUATIONS
When the stove is lighted in a mountain cabin, the temperature
in the cabin is the same as the temperature outside: −20◦ C. After
an hour the temperature inside has reached 0◦ C, and after 2 hours,
the temperature is 10◦ . Find the temperature as a function of time,
assuming that the stove continues to operate. How warm will the
cabin eventually get?
Answer: 20◦ C.
21.
The insulation of a house has a transmission coefficient of k = .04
hour−1 . The furnace is capable of heating the house at 1◦ C per hour
(neglecting heat loss). Assume that the outdoor temperature is 0◦ C,
and that the heat is alternately turned on for 4 hours, then off for 1
hour, by an automatic timer. What will be the average temperature
inside the house, if this continues indefinitely?
See answer
22.
The building described in example 1.3.3 is equipped with a thermostat that turns off the furnace when the interior temperature T >
20◦ C and turns it on again when T ≤ 20◦ C.
(a) Let
H (T ) =
4 if T ≤ 20
0 if T > 20.
Show that the IVP T 0 + 0.2T = 0.6 + 2 sin πt
12 + H ( T ); T (0) =
20 is an appropriate model for the temperature in the building.
(b) Explain why the ODE in part (a) is not linear.
(c) Use an IVP solver to solve the IVP in part (a). Explain the dips in
the graph and the jagged appearance of the horizontal segments.
23.
For each of the following ODEs, find the periodic solution, if there
is one, and decide whether or not it is stable. In other words, express
the general solution as the sum of the periodic solution and a family
of exponential functions. If the exponential functions decay to 0 as
t → ∞, they represent transients and the periodic solution is stable.
(a) y0 + 5y = 5 cos 2t.
(b) y0 − y = 7 cos 4t.
(c) y0 + 2y = cos t − 3 sin t.
(d) y0 − 5y = 4 cos t + 3 sin t.
1.3. LINEAR FIRST ORDER EQUATIONS
33
(e) y0 + y = e−t sin t.
See Answer
24.
Find the periodic solution of each of the following ODEs and
show that it is stable.
(a) y0 + y = sin(2t)
(b) y0 + y = sin(2000t)
(c) y0 + 5y = cos(πt).
(d) y0 + 10000y = sin t.
(e) y0 + .0006y = sin(.0008t).
(f) y0 + py = cos(ωt), where p > 0 is a constant. What would
happen if p < 0?
The remaining problems in this section are designed to be done with a
CAS.
25.
Find the general solution of each of the following.
(a) y0 + 0.1y = sin(2t).
(b) y0 + 2ty = 1. Attempt to solve this equation first without the
computer’s help. How does the computer get around the probR 2
lem of evaluating et dt?
(c) ty0 + 12y = [ln(t)]3 .
√
(d) ty0 + y = t.
See Answer
26.
For each of the following linear differential equations, find the
general solution, and graph several solutions by substituting values
for the constant, using the domain −10 ≤ t ≤ 10, and range −10 ≤
y ≤ 10. Make note of any properties that are common to all solutions.
(a) y0 + y = 1.
(b) y0 − y = 1.
(c) y0 + y = t.
(d) y0 − y = −t.
(e) y0 + y = sin(t).
34
CHAPTER 1. FIRST-ORDER EQUATIONS
(f) y0 + 0.05y = sin(t).
sin(πt)y0 + cos(πt)y = 1
π
π
π
(h) cos 20
t y0 + 20
sin 20
t y = 1.
(g)
27.
1
π
Using the plot range −π ≤ t ≤ π, −10 ≤ y ≤ 10, graph the
solution of the differential equation
ty0 + t cot(t)y = 1
that has a finite value at t = 0. What is that value?
See Answer
28.
Let g(t) denote the solution of
ty0 − 2y = e−t
that has the property g(1) = g0 (1). Draw the graph of g(t) on the
interval 0 ≤ t ≤ 2.
29.
Plot the nonconstant solution of y0 + 2y = ty2 that has a relative
minimum at t = 41 . What is the value of y(1/4)?
See Answer
30.
Let φ(t) be the solution of y0 + 2ty = y2 that has a maximum at
t = 1. Plot the graph of φ(t) on the interval 0 ≤ t ≤ 2, and calculate
φ (0).
A NSWER : φ(0) = 1.074386372.
1.4
Mixture Problems
A typical mixture problem involves a tank that initially contains V0 liters
of a salt solution, with concentration C0 grams per liter. A salt solution
containing K grams of salt per liter is being poured into the tank at J liters
per second; simultaneously, solution is pumped out of the tank at L liters
per second. The problem is to find the concentration of salt in the tank as a
function of time. See figure 1.6.
The mathematical model of the mixture problem depends on a simplifying assumption, the uniformity hypothesis: the concentration of the solution
throughout the tank is uniform. Let C (t) denote the salt concentration in the
tank, in grams per liter.
1.4. *MIXTURE PROBLEMS
35
We will use the amount of salt in the tank, denoted x (t), rather than the
concentration C (t), as the dependent variable. We can determine the rate of
change of x (t) by subtracting the rate at which the salt removed (the output
rate) from the rate that salt enters the tank (the input rate). The input rate
is JK grams per minute, since J liters of solution, each containing K grams
of salt, enter the tank each minute. The output rate is LC (t), grams per
minute, where C (t) is the concentration of salt in the tank. Thus
x 0 (t) = JK − LC (t).
To complete this model, we have to express the concentration in terms
of x (t). Let V (t) be the volume of solution in the tank at that time. If V0 is
the initial volume in the tank, then
V (t) = V0 + ( J − L)t
since the tank gains J − L liters per minute. Therefore,
C (t) =
x (t)
x (t)
=
.
V (t)
V0 + ( J − L)t
It follows that x (t) satisfies the linear ODE
dx
x
= JK − L
,
dt
V0 + ( J − L)t
(1.14)
with the initial condition x (0) = C0 V0 .
In our first example, the volume V is constant, because the rate J at
which fluid flows in is the same as the rate L at which fluid is pumped out.
Example 1.4.1 An accident has caused 10 kg of potassium permanganate to spill
into a mountain lake, coloring its water purple. The lake covers an area of 100
square meters, and its average depth is 5 meters. A stream feeds the lake at 1000
liters per minute, and another stream takes water from the lake at the same rate.
Find the concentration of potassium permanganate in the lake as a function of
time, assuming that the uniformity hypothesis holds in this situation. How much
of the substance will remain in the lake after 24 hours have elapsed?
S OLUTION . The volume V of the lake is
5 m. deep × 100 m2 × 1000 liters per m3 = 5 × 105 liters.
36
CHAPTER 1. FIRST-ORDER EQUATIONS
Let x (t) denote the amount of potassium permanganate dissolved in the
x (t) kg
lake; the concentration is C (t) =
.
5 × 105 liters
The input of potassium permanganate zero (the stream feeding the lake
was not affected by the spill), and the output is the product of the
concentration and the rate of flow: C (t) kg/liter × 1000 liters per minute.
Thus
x (t)
x 0 (t) = −
× 1000 = −0.002 x (t) kg/minute
5 × 105
Thus, the amount of potassium permanganate in the lake satisfies the
linear decay equation, x 0 = −0.002 x. It follows that x (t) = Ae−0.002 t , where
A = 10, the initial value of x. The half-life of x (t) is (ln(2)/0.002) ≈ 347
minutes, about 6 hours. Since 24 hours amounts 4 half-lives, we expect
that 10 × 2−4 ≈ 0.6 kg will remain in the lake after a day.
When the volume of solution is not constant, we encounter variable
coefficients in the input-output equation, as in the following example.
Example 1.4.2 A tank initially contains 1000 liters of salt solution with 70 grams
of salt per liter. A solution containing 120 grams of salt per liter enters the tank
at the rate of 9 liters per minute, and the well mixed solution is pumped out at 10
liters per minute. Find the concentration of salt as a function of time.
S OLUTION . The salt input rate is
(120 grams/liter) × (9 liters/min.) = 1080 grams/min.
The volume of solution in the tank is decreasing at a rate of 1 liter/min., so
V (t) = 1000 − t liters. Thus, the output rate is
x
grams/liter × 10 liters/min.
1000 − t
The input-output model results in the ODE
dx
x
= 1080 − 10
,
dt
1000 − t
or, in standard form,
0
x +
10
1000 − t
x = 1080.
1.4. *MIXTURE PROBLEMS
37
The solution of the associated homogeneous equation is
Z
10
exp −
dt = e10 ln(1000−t) = (1000 − t)10 .
1000 − t
Now we substitute x = v(1000 − t)10 in the differential equation, and
simplify to get
v0 = 1080(1000 − t)−10 .
Integration yields v = 120(1000 − t)−9 + A, where A is a constant. Since
x (t) = v(1000 − t)10 , it follows that
x (t) = 120(1000 − t) + A(1000 − t)10
(1.15)
The initial value is x (0) = (70 grams/liter) × (1000 liters) = 70, 000 grams;
thus equation (1.15) with (t, x ) = (0, 70,000) yields
70, 000 = 120, 000 + A · 100010 , or A = −50 × 1000−9 . If we substitute this
value of A into equation (1.15), and simplify a little, we will have
9
t
x (t) = 120(1000 − t) − 50(1000 − t) 1 −
.
1000
To obtain the concentration C (t), divide by V (t) = 1000 − t:
C (t) = 120 − 50(1 − .001t)9 grams per liter.
This formula is valid for 0 ≤ t < 1000. When t = 1000 the tank becomes
empty, and the model is no longer applicable.
Input-output problems in personal finance
Many investment plans consider factors such as inflation and the effect of
regular withdrawals. Suppose, for example, that a person has a pension
account of P dollars upon retirement, and withdraws money at a rate of
E(t) dollars per year to meet living expenses. If y(t) denotes the account
balance t years after retirement, then the rate of change of y(t) is the rate at
which interest income is received minus E(t).
If r denotes the interest rate, then our model is an ODE,
y0 = ry − E(t).
Assuming that the expenses grow with inflation according to the exponential growth law E0 = k E, where k is the annual rate of inflation, the
expenses after t years will be E(t) = E0 ekt , where E0 is the initial rate of
withdrawal.
38
CHAPTER 1. FIRST-ORDER EQUATIONS
Example 1.4.3 Ms. Doe retired yesterday at age 65. Her IRA account has a
principal of $450,000, which invested with a guaranteed interest rate of 5.25%,
compounded continuously. Her budget calls for annual expenses of $20,000, with
projected inflation of 2.5%. Determine Ms. Doe’s savings account balance t years
after her retirement. How long will her money last?
S OLUTION . Let y(t) denote the balance t years after retirement, in dollars.
The input in this problem is interest amounting to 0.0525y(t), and the
output is expenses of 20, 000e0.025t dollars per year. Thus
y0 = 0.0525y − 20000e0.025t .
This can be put into the standard form for a linear ODE:
y0 − 0.0525y = −20000e0.025t .
The solution of the associated homogeneous equation is y = Ce0.0525t , so
we substitute y = ve0.0525t and simplify to obtain v0 = −20, 000e−0.0275t .
Therefore v = 20, 000e−0.017t /0.0275 + C, and since y = ve0.0525t ,
y=
20000 0.025t
e
+ Ce0.0525t
0.0275
Since y(0) = 450, 000, and 20, 000/0.0275 ≈ 730, 000,
450, 000 = 730, 000 + C, and thus C ≈ −280, 000. Ms. Doe’s account
balance in t years will be about
y(t) = 730, 000e0.025t − 280, 000e0.0525t .
This function is decreasing, and y(t) = 0 when 730, 000 = 280, 000e0.0275t ;
that is, when e0.0275t = 2.6. This will be when Ms. Doe reaches the age of
100.
Exercises
In Exercises 1–10, assume that the uniformity hypothesis holds.
1.
A tank initially contains 100 liters of water. A 16% salt solution
enters the tank at 2 liters per second. Brine is pumped out of the tank
at 2 liters per second. How long does it take for the concentration of
salt in the tank to reach: 8%? 12%? 15%?
See answer
1.4. *MIXTURE PROBLEMS
2.
39
A 4% salt solution is poured into a large tank at 2 liters per second.
Brine is removed from the tank at 1.9 liters per second. Given that
the tank initially contains 50 liters of 2% salt solution, what is the
concentration of salt in the tank after one minute has elapsed?
Answer: 3.79%
3.
4.
A tank contains 16 liters of 100 parts per million (ppm) potassium
iodide (KI), in which the iodide is the unstable isotope 131 I, with a
half life of 8 days. Solution is drained from the tank at the rate of 1
liter per day, and fresh 100 ppm solution is added at the rate of 1 liter
per day. Find the steady state concentration of 131 I in the tank.
See answer
(continuation of problem 3) Suppose fresh 100 ppm solution of
is added at the rate of 2 liters per day, while the well mixed solution drains out at 1 liter per day, as before. If the initial volume is 16
liters and the initial concentration is 100 ppm, find the concentration
after 8 days have elapsed.
K 131 I
Answer: 63 ppm
5.
A beverage bottling plant has had an accident. Someone poured
quinine (intended for the tonic) in the orange drink mixing tank, and
the resulting liquid is unpalatable. No one knows (or will say) when
this happened, but we must find out, because the orange drink bottled since the mishap must not leave the plant. The concentration of
quinine in the tank is now 0.01%. The tank now holds 10000 liters,
and orange drink is being transferred to the carbonation tank at 1000
liters per hour. Fresh ingredients have been added to the tank at the
rate of 900 liters per hour. It is known that the amount of quinine that
was put into the tank was not more than 4 kilograms, and that it was
introduced into the tank within the last 24 hours. Estimate how long
ago, at worst, that the quinine was introduced.
See answer
6.
A tank contains 4 kg of salt dissolved in 100 liters of water. A salt
solution with concentration 0.01 kilograms per liter enters the tank at
the rate of 2 liters per day. Evaporation removes 1 liter of water per
day (salt does not evaporate), and an additional liter of solution is
drained from the tank each day. Find the concentration of salt in the
tank as a function of time.
40
CHAPTER 1. FIRST-ORDER EQUATIONS
7.
When brought to the emergency room, an accident victim has 3
liters of blood, and is still losing blood at 0.25 liter per hour. He is immediately given continuous blood transfusions at 0.5 liter per hour,
and an antibiotic drug is administered intravenously at 0.5 gram per
hour. Four hours later, the bleeding is stopped, and the transfusions
stop in another two hours. Determine the concentration of the antibiotic in the patient’s blood at the time that the transfusions stop.
See answer
8.
Two mixing tanks each initially contain 2 liters of pure water. A
10% salt solution enters the first tank at 0.1 liter per minute. The well
mixed solution is pumped out of this tank at the same rate, and into
the second tank. Instantaneous and perfect mixing occurs in the second tank, and the resulting solution is removed from it, again at the
rate of 0.1 liters per minute. Determine the concentration of salt in the
second tank as a function of time.
9.
Consider two mixing tanks, labeled A and B. Initially, tank A contains 3 liters of 12% salt solution, and tank B contains 1 liter of water.
Both tanks are stirred constantly, and solution is pumped from tank
A to tank B at 0.2 liters per hour. Solution is also pumped from tank
B to tank A at the same rate. Find the salt concentration of each tank
as a function of time. Hint: Since this is a closed system, the total
amount of salt held by the tanks is constant.
See answer
10.
A 200 liter solar water heater absorbs heat at a rate proportional to
the sine of the angle of the sun over the horizon. The sun rises at 0600
and sets at 1800 on the 24hour scale. Assume that the heat absorption
π
rate is 1500 sin 12
(t − 6) kilocalories per hour, where t is hours after
midnight of the equinox. Of course, no heat is absorbed at night —
the absorption rate is 0 for 0 < t < 6, 18 < t < 30, etc.
The water heater loses heat by conduction to the environment, and by
hot water usage. The ambient temperature is 20◦ C, and the rate constant for heat conduction to the rest of the house is k = 0.01 hour−1 .
Hot water is removed from the tank at an average rate of 5 liters per
hour, and is replaced at the same rate by cold water with a temperature of 10◦ C with instant and perfect mixing. If the water in the tank
is at 60◦ C at 0600, find its temperature 24 hours later.
1.5. *BERNOULLI EQUATIONS
41
Hint: treat this as a mixture problem, and keep track of heat input
and output. One kilocalorie of heat will raise the temperature of one
liter by one centigrade degree.
11.
A retired citizen’s living expenses are $30,000 per year now, and
he wants to allow for inflation of 3% per year. How much money
should he invest, at 6% interest, so that he can meet his living expenses and not touch the principal, no matter how long he lives?
See answer
12.
An endowed chair in mathematics is to be funded with one million dollars. The endowment committee believes it can realize at least
an 8% return on the funds, and expects to offer the occupant of the
chair a 5% salary increase each year to keep pace with inflation. What
should the initial salary be, if the principal of the endowment fund is
never to be drawn upon?
13.
The age-specific death rate d(t) is the number of deaths of individuals who are t years old per 1000 individuals of that age. Assume that
d(20) = 2.8, d(50) = 17.6, and that d(t) satisfies the linear growth
equation d0 (t) = kd(t). Of a sample of 100,000 twenty year olds, how
many will survive to age 50?
See answer
1.5
Bernoulli Equations
It is sometimes possible to convert a nonlinear first order ODE to a linear
one by a clever change of variables, For example, substitute v = ym in the
nonlinear ODE,
y m −1 y 0 + p ( t ) y m = q ( t )
(1.16)
By the chain rule, dv/dt = mym−1 y0 . Multiply equation (1.16) by m and
make the substitution. The resulting ODE,
v0 + mp(t)v = mq(t),
is linear, and thus we can find its general solution. It is then a simple matter
to replace v with ym and obtain the general solution of (1.16).
A Bernoulli equation is a first order nonlinear ODE of the form
y0 + p(t)y = q(t)yn ,
(1.17)
42
CHAPTER 1. FIRST-ORDER EQUATIONS
where n 6= 0, 1. The power n to which y is raised on the right side is the
exponent. Observe that by dividing equation (1.17) by yn , it can be made
to assume the form of equation (1.16), in which m = 1 − n. Thus, the substitution v = ym also converts Bernoulli equation to a linear equation.
The first Bernoulli equation was encountered by Johann Bernoulli in
1695. Bernoulli was trying to determine the trajectory of a projectile subject
to the forces of gravity and air resistance. He knew that the force of air resistance would be proportional to the square of the speed v of the projectile,
and derived the ODE,
dv
= −kv2 − g sin(θ ),
dt
(1.18)
where g is the gravitational acceleration and θ denotes the inclination of the
trajectory from the horizontal. Of course, θ is not constant, and Bernoulli
derived an ODE that it would satisfy:
dθ
g
= − cos(θ )
dt
v
(1.19)
The two ODEs, (1.18) and (1.19), form a coupled system, a topic that we
will take up in chapter 2.
It may seem strange to do so, but Bernoulli decided to treat the speed v
as a function of the inclination θ. Then, by the chain rule,
dv
dv dθ
=
,
dt
dθ dt
and hence
dv
v0 (t)
= 0 .
dθ
θ (t)
Bernoulli divided (1.18) by (1.19) to obtain
k v3
dv
=
+ tan(θ ) v.
dθ
g cos(θ )
(1.20)
Example 1.5.1 Show that (1.20) is a Bernoulli equation and make the appropriate
substitution to convert it to a linear ODE.
S OLUTION . Rearrange (1.20) in the form
dv
k
− tan(θ ) v = sec(θ ) v3
dθ
g
(1.21)
1.5. *BERNOULLI EQUATIONS
43
This has the form of (1.17), where v is the dependent variable, θ is the
independent variable, p(θ ) = − tan(θ ), and q(θ ) = (k/g) sec(θ ). Thus it is
a Bernoulli equation, with exponent n = 3. To linearize, define a new
variable u = v1−n = v−2 . Then
du
dv
= −2v−3 ;
dθ
dθ
or
dv
v3 du
=−
.
dθ
2 dθ
Substitute this expression in (1.21) to obtain
−
k
v3 du
− tan(θ ) v = sec(θ ) v3 .
2 dθ
g
Divide through by v3 to obtain
−
1 du
k
− tan(θ ) v−2 = sec(θ ).
2 dθ
g
Finally, replace v−2 with u, and multiply through by −2 to obtain
2k
du
+ 2 tan(θ ) u. = − sec(θ ).
dθ
g
Exercises
Find the general solution of the following equations:
1.
2.
ty2 y0 + y3 = 1.
See answer
y0 + (tan t)y = y2 .
3.
y0 + 3y = y3 sin t.
See answer
4.
e2y y0 + e2y = e−2t .
5.
6.
7.
y
y0 + 2t+1 = 12[(2t + 1)/y]3 .
See answer
y0 + (1 − 3t )y =
y4/3
t .
Can your CAS handle Bernoulli equations? Attempt to find the
general solution of each of these equations with your computer.
44
1.6
CHAPTER 1. FIRST-ORDER EQUATIONS
Separable ODEs
A separable ODE has the form
dy
= g ( t ) h ( y ),
dt
(1.22)
in which the right side can be expressed as the product of two singlevariable functions. Here are some examples:
dy
= t2 sin y
dt
and
dy
= et+3y (= et e3y )
dt
are separable. Homogeneous linear ODEs are separable because a1 (t) y0 +
a0 (t)y = 0 is equivalent to
a0 ( t )
0
y =−
y = g ( t ) h ( y ),
a1 ( t )
a (t)
where g(t) = − a0 (t) and h(y) = y. However, inhomogeneous linear ODEs
1
are not separable (unless the coefficients and source term are constants).
For an example of a nonlinear ODE that is not separable, consider
dy
= 1 + t y2 .
dt
The reason this equation is not separable is that the right side can’t be factored as the product of a function of t and a function of y.
ODEs in terms of differentials Consider an ODE
dy
= F (y, t).
dt
(1.23)
If y = φ(t) is a solution of (1.23) then the differentials of t and y are related
by the equation dy = φ0 (t) dt. Also, because φ is a solution,
φ0 (t) = F (φ(t), t) = F (y, t).
It follows that the (1.23) is equivalent to
dy = F (y, t) dt.
(1.24)
1.6. SEPARABLE ODES
45
This form of an ODE is often convenient, and we will call (1.24) an ODE in
differential form.
Let’s put the generic separable ODE, (1.22), in differential form:
dy = g(t) h(y) dt,
(1.25)
and see how to approach solving it. We will want to divide through by
h(y), but we have to be careful, since this would be questionable if h(y) has
zeros. Thus, at the outset, find the zeros of h(y). Suppose that h(y0 ) = 0
and plug y = y0 (constant function) into either (1.22) or (1.25). Because
both sides (either equation) are 0, this y = y0 is a solution of the ODE.
To find the solutions that are not constant, divide (1.25) by h(y) to get
the following separated equation:
Because
dy
= g(t) dt.
h(y)
(1.26)
φ0 (t) dt
= g(t) dt
h(φ(t))
(1.27)
implies that φ0 (t) = g(t)h(φ(t)), every solution of (1.26) is also a solution
of the original ODE (1.22). It must be emphasized that constant solutions
of (1.22) are usually not solutions of the separated equation, (1.26).
R
Integrating the separated equation Let H (y) = h(1y) dy be an antiderivaR
tive of h(y)−1 and let G (t) = g(t) dt. If y = φ(t) is a solution of the
separated equation, then we can make the following substitutions in equation (1.26):
1/h(φ(t)) ; H 0 (φ(t))
dy ; φ0 (t) dt
g ( t ) ; G 0 ( t ).
This yields
H 0 (φ(t)) φ0 (t) dt = G 0 (t) dt.
By the chain rule,
d
H (φ(t)) = H 0 (φ(t)) φ0 (t).
dt
(1.28)
46
CHAPTER 1. FIRST-ORDER EQUATIONS
Thus, (1.28) tells us that H (φ(t)) and G (t) have equal derivatives. It follows from the equal derivatives theorem that on any interval upon which
H (φ(t)) and G (t) are both defined, there is a constant C such that
H (φ(t)) = G (t) + C.
(1.29)
Equation (1.29) can be expressed as H (y) = G (t) + C. This is the solution of (1.22) in implicit form.
If H (y) happens to be strictly increasing or strictly decreasing, there is
a unique inverse function H −1 , and
φ ( t ) = H −1 ( G ( t ) + C )
is the general solution of the separated equation (1.26).
Here’s a summary of what we’ve learned so far:
Solving a separable ODE
To solve the ODE
dy = g(t) h(y) dt
start by finding the constant solutions y ≡ yi . This is done
by determining the roots y1 , y2 , . . . of h(y) = 0. Then find
the nonconstant solutions by integrating both sides of the
separated equation:
Z
dy
=
h(y)
Z
g(t) dt.
As a first example, consider the ODE (1.1) that is a model for the velocity
of a falling object with a drag force.
Example 1.6.1 Solve the IVP v0 = g − k v2 ; v(0) = 0.
2
S OLUTION . In differential form we have
r dv = ( g − k v ) dt. The constant
g
solutions are v = ±v∞ , where v∞ =
k
1.6. SEPARABLE ODES
47
We have noted that the ODE, which—because g = kv2∞ —we can write as
dv = k (v2∞ − v2 ) dt,
is separable. Of the two constant solutions, v = ±v∞ , v = v∞ represents
the terminal velocity, while v = −v∞ is extraneous, as it indicates a
constant upward velocity. In fact, the ODE (1.1) is incorrect for v < 0,
because the motion and air resistance cannot be in the same direction.
See exercise 16. Divide through by (v∞ + v)(v∞ − v) to get the separated
equation:
1
dv = k dt
(v∞ + v)(v∞ − v)
Expanding in partial fractions, we get
1
1
1
+
dv = k dt
2v∞ v∞ − v v∞ + v
which we integrate and obtain
1
ln |v∞ + v| − ln |v∞ − v| = kt + C.
2v∞
The value of C can be determined at this point by referring to the initial
condition, t = 0 and v = 0. We get C = 0, and after making this
replacement and simplifying the logarithmic expression
v∞ + v = 2kv∞ t,
ln v∞ − v To solve for v, first exponentiate
100 + v
= ±e2kv∞ t .
100 − v
and choose the plus sign since it is in accord with the initial condition.
Thus
e2kv∞ t − 1
= v∞ tanh (kv∞ t)
v = v∞ 2kv t
e ∞ +1
The solution of example 1.6.1 has the expected properties: starting from
a velocity of 0, the object will accelerate and its velocity will approach the
terminal velocity v∞ .
To explore further the properties of the linear and quadratic models, see
Exercise 6.
48
CHAPTER 1. FIRST-ORDER EQUATIONS
Example 1.6.2 Solve the initial value problem
y 0 = t y −1 ; y (0 ) = − 1
S OLUTION . Since h(y) = y−1 6= 0 for all y, there are no constant solutions.
The separated equation is y dy = t dt, and integration of both sides yields
1 2
1 2
2 y = 2 t + C. To determine the value of C, substitute t = 0 and y = −1:
1
1
(−1)2 = (0)2 + C
2
2
so C = 12 . Thus 12 y2 = 21 t2 + 12 . Solving for y yields two solutions
y=±
p
t2 + 1
Only one of these solutions,
y=−
p
t2 + 1,
is valid, since the initial condition specifies that y(0) is negative. Figure 1.8
shows several solutions of the differential equation y0 = t y−1 .
It is sometimes impractical to find a formula for y as an explicit function of t and the constant C, and even if it is possible to do this, it may be
unnecessary. We will say that a function F (t, y) is an integral of an ODE
y0 = f (t, y) if, for all solutions y = φ(t) of the ODE, F (t, φ(t)) is a constant
function of t. Technically speaking, a constant function F (t, y) ≡ C will be
an integral for any ODE, but it does not define a solution implicitly. Integrals are usually obtained by integrating the ODE in some way, as we have
done in the process of solving separable equations.
If F (t, y) is a function of two variables, its level curves are the curves
defined by the equations F (t, y) = C, where C is constant. If we have found
an integral F (t, y) of an ODE, then the graphs of all solutions will be subsets
of level curves of F. For example, in the process of working out the solution
of example 1.6.2, we found that
F (t, y) =
1 2 1 2
y − t is an integral of y0 = t y−1 .
2
2
It follows that if y = φ(t) is a particular solution of y0 = t y−1 , then the
graph of φ(t) is a subset of a hyperbola F (t, y) = C.
1.6. SEPARABLE ODES
49
Example 1.6.3 Find the general solution of
y0 =
2y
.
t
(1.30)
S OLUTION . Since h(y) = y has a zero at y = 0, there is a constant
solution, y ≡ 0. The separated equation is
dy
2 dt
=
.
y
t
Integrating both sides, we have
ln |y| = 2 ln |t| + C,
so F (t, y) = ln |y| − 2 ln |t| is an integral. Simplifying, we have
y
F (t, y) = ln 2 .
t
Since F (t, y) is an integral, so is ±e F(t,y) = y/t2 . Thus, solutions take the
form y/t2 = C, or y = C t2 , for some constant C. It happens that the
constant solution y ≡ 0 that we found at the outset belongs to this family
(put C = 0).
In the following example we can again find a solution y ≡ 0 because
the right side vanishes when y = 0. This time, the constant solution does
not belong to the family obtained by integrating. When this happens, the
constant solution is called a singular solution.
Example 1.6.4 Find a family of solutions of the ODE
y0 = 3y2/3 .
(1.31)
S OLUTION . We see that there is the constant solution y ≡ 0. Let’s turn to
the separated equation
1 −2/3
y
dy = dt.
3
for y 6= 0. Integrating, we have y1/3 = t + C; thus
y = ( t + C )3 ,
which is not constant for any value of C. Thus our constant solution is
singular. The graphs of several solutions from the family are shown in
figure 1.9. Notice that the t-axis (the graph of the singular solution) is
tangent to the graph of each of these solutions.
50
CHAPTER 1. FIRST-ORDER EQUATIONS
Exercises
1.
Which of the following ODEs is separable?
(a)
dy
dt
=
t2 +5
ty
(b) y dt = t dy
(c)
(d)
(e)
dy
dt
dy
dt
dy
dt
= ety
= e−(t
2 + y2 )
= cot y
(f) y0 =
t+y
t−y
See answer
2.
Find an integral for each of the following ODEs, if it is separable.
The non-separable equations should be skipped!
(a) y0 = tan(ty).
(b) y0 = ln(ty ).
(c) y0 = sec y.
(d) y0 = ln(t + y)
3.
A ball whose mass is 1 kilogram rolls on a level surface, subject
only to the force of friction. The initial velocity is 1 meter per second;
after 1 second, the velocity is 0.8 meters per second. Assuming the
magnitude of the frictional force is proportional to the velocity,
(a) What is the friction constant?
(b) When will the velocity be .5 meters per second?
(c) How far did the ball roll in the first second?
(d) Evaluate an improper integral to determine how far the ball will
roll if given an infinite amount of time.
See answer
4.
Repeat problem 3, assuming the magnitude of the frictional force
is proportional to the square of the velocity. The friction constant is
now the ratio between the magnitude of the friction force and v2 .
Answer: (a) 0.25 kg/m. (c) 0.893 m.
1.6. SEPARABLE ODES
51
5.
An object with a mass of 1 kg in free fall is subject to gravitational
force of 9.8 newtons and a frictional force of −bv, where v is the velocity. Given that the terminal velocity is v∞ = 10 m/s, calculate b.
See answer
6.
An object falls from rest. The terminal velocity is observed to be
98 meters per second. Assuming that the force of air resistance is
(a) proportional to the velocity,
(b) proportional to the square of the velocity,
how long does it take for the object to reach a speed of 49 meters per
second? Plot graphs of the velocity as a function of time to make a
comparison of the two models.
Answer: (a) 6.93 sec., (b) 5.49 sec.
In problems 7 – 15, find a family of solutions for the ODE. Then find
the particular solution that satisfies the given initial condition (be sure to
specify its domain).
dy
dt
7.
= (t − 1)y2 ; y(1) = 0.
See answer
8.
y0 = 3t2 y2 ; y(0) = −1.
9.
dy
dt
= 2ty ; y(1) = 2.
See answer
10.
tyy0 = 1; y(e) = 2.
11.
y0 = yet ; y(0) = 1.
See answer
12.
dy = (y2 + 1) dt; y(0) = 1.
13.
dy
dt
14.
15.
= y2 ; y(0) = 1.
See answer
dy
dt
=
1+ y
√ ;
t
y(4) = 0.
y0 = t(y2 − 1); y(0) = 0.
See answer
52
16.
CHAPTER 1. FIRST-ORDER EQUATIONS
If an object is initially moving upward (we consider upward velocity to be negative) then the force of air resistance will be downward
(positive), and the quadratic model for air resistance implies the following ODE holds during the upward motion:
v0 = g + kv2 .
When v > 0, the sign of the quadratic term should be changed to
minus.
(a) Show that the ODE
v0 = g − kv |v|
is applicable to both upward and downward motion.
(b) Suppose that the terminal velocity of the object is 100 m/s. The
object is thrown upward with an initial velocity of 100 meters/second
(v(0) = −100). Give its velocity as a function of time for the duration of its upward motion.
(c) How far upward does the object go?
(d) Find the velocity as a function of time for downward motion.
(e) Sketch the graph of v(t) for t > 0.
17.
A basketball is dropped from a tower. There is no wind, and the
drag force is proportional to the square of the velocity. After 2 seconds
of fall, its velocity is 14.7 meters per second. Calculate its terminal
velocity. It will be necessary to use Newton’s method.
See answer
18.
In this exercise, we must modify our model for the motion of a
falling body to take into account a variable mass. Let m(t) denote the
mass of a falling body. Then the velocity of the body is a solution of
the following ODE, derived from Newton’s second law of motion:
m(t)v0 = −m(t) g − k|v|v.
A rocket has a mass of 6 kilograms, including 2 kilograms of fuel.
The fuel burns uniformly for 20 seconds, leaving no residue, and produces a constant thrust of 120 newtons. After 20 seconds, the fuel is
spent, and the only forces are gravity and air resistance. The force of
air resistance is proportional to the square of the velocity, with drag
coefficient k = 0.02 kilograms per meter. Write initial value problems
to determine the velocity of the rocket as a function of time for
1.7. EXACT FORM
53
(a) 0 ≤ t ≤ 20,
(b) 20 ≤ t ≤ T1 , where T1 is the time that the rocket reaches its
maximum altitude, and for
(c) T1 ≤ t ≤ T2 , where T2 is the time that the rocket hits the ground.
Solving these ODEs is optional, and is recommended only for readers
who are using a CAS. Use the convention that upward motion has
positive velocity.
19.
A ball is thrown upward with a velocity v0 (negative). Its terminal
velocity in free fall is v∞ , and the drag force is proportional to the
square of the velocity. Find formulas for the following as as functions
of v0 and v∞ .
(a) the maximum height attained by the ball,
(b) the time taken to reach that height,
(c) the time taken to return to the ground, and
(d) the velocity when the ball hits the ground.
|v |v
Answer: √ 02 ∞ 2
v0 + v ∞
What happens if the drag force is negligible (v∞ = ∞)?
See answer
20.
Does the ball in problem 19 spend more time on the way up or on
the way down?
Answer: On the way down.
21.
A ball falls from rest from a high tower, with drag force proportional to the square of its velocity. How long does it take for the ball
to attain a speed of half its terminal velocity?
See answer
22.
A ball is thrown downward from a high tower, with v(0) = 4v∞ .
How long does it take for the ball to reach a speed of 2v∞ ? Assume
the drag force is proportional to the square of the velocity.
Answer: t = (|v∞ | ln(1.8))/(2g)
1.7
Exact Form and Integrating Factors
If we start with a function F ( x, y) that has continuous partial derivatives,
we can ask, is F an integral for some ODE? Suppose that there is such an
54
CHAPTER 1. FIRST-ORDER EQUATIONS
ODE, and let y = φ( x ) be a solution. Thus
F ( x, y) = C
would be an implicit solution of that ODE.
Let’s apply implicit differentiation with respect to x:
∂F ∂F dy
+
·
= 0.
∂x
∂y dx
(1.32)
Equation (1.32) is an ODE that has F as an integral. The answer to our
question is yes.
The total differential of F ( x, y) is defined as
dF =
∂F
∂F
dx +
dy.
∂x
∂y
The ODE (1.32) is equivalent to the equation dF = 0.
One approach to integrating an ODE
P( x, y) dx + Q( x, y) dy = 0
(1.33)
would be to try to find a function F ( x, y) whose total differential is
dF = P( x, y) dx + Q( x, y) dy.
This function F ( x, y) would have to satisfy the requirements
∂F
∂x ( x, y )
= P( x, y) and
∂F
∂y ( x, y )
= Q( x, y).
(1.34)
Equations (1.34) overdetermine F—that is, usually there is no function that
satisfies both equations. If function F ( x, y) that satisfies (1.34) for all ( x, y)
in a domain D in the x, y-plane does exist, the ODE (1.33) is said to be exact
on D . The remainder of this section will be devoted to explaining how to
recognize exact ODEs, and how to find an integral for an exact ODE. We
will also see that—in some circumstances—we can replace an ODE that is
not exact with an equivalent ODE that is exact.
Theorem 1.3 Assume that the first partial derivatives of P( x, y) and Q( x, y) are
continuous on a rectangular domain D in the plane. Then the ODE (1.33) is exact
on D if and only if
∂P
∂Q
( x, y) =
( x, y)
(1.35)
∂y
∂x
for all ( x, y) ∈ D
1.7. EXACT FORM
55
We will call condition (1.35) the exactness condition.
P ROOF. First we will prove that if the ODE (1.33) is exact, then the
exactness condition holds (this establishes that the exactness condition is
necessary). Suppose that there is a function F ( x, y) such that P( x, y) =
∂F
∂F
∂x ( x, y ) and Q ( x, y ) = ∂y ( x, y ). Since P and Q have continuous first partial
derivatives, F has continuous second partial derivatives on the domain D .
We can now refer to a theorem on partial derivatives that says if a function F ( x, y) has continuous second partial derivatives on a domain in the
plane, then
∂2 F
∂2 F
=
.
∂x∂y
∂y∂x
Thus
∂P
∂
=
∂y
∂y
∂Q
∂
=
∂x
∂x
and
∂F
∂x
∂F
∂y
=
∂2 F
∂y∂x
=
∂2 F
∂x∂y
are equal: the exactness condition holds.
Now we will show that if P( x, y) and Q( x, y) satisfy the exactness condition, then there is a differentiable function F ( x, y) defined on D such that
dF = P( x, y) dx + Q( x, y) dy.
Let ( x0 , y0 ) be a point in D . Define a function
H (y) =
and
F ( x, y) =
Z y
y0
Z x
x0
Q( x0 , s) ds.
P(t, y) dt + H (y).
By the fundamental theorem of calculus,
∂F
∂
= P( x, y) + H (y) = P( x, y).
∂x
∂x
Since P has continuous partial derivatives, we can apply the Leibniz rule
when differentiating F with respect to y:
∂F
∂y
=
=
∂
∂y
Z x
x0
Z x
∂P
x0
∂y
P(t, y) dt + H 0 (y)
(t, y) dt + H 0 (y).
56
CHAPTER 1. FIRST-ORDER EQUATIONS
∂Q
Now we will use the exactness condition to replace ∂P
∂y ( t, y ) with ∂t ( t, y )
in the integral. This enables us to use another form of the fundamental
theorem of calculus:
Z x
∂Q
∂t
x0
(t, y) dt = Q( x, y) − Q( x0 , y).
Thus we have shown that
∂F
= Q( x, y) − Q( x0 , y) + H 0 (y).
∂y
(1.36)
Since H (y) is by definition an antiderivative of Q( x0 , y) the last two terms
in (1.36) cancel, and the proof is complete.
Example 1.7.1 Which of the following ODEs are exact?
(i) (y2 + x2 − 2x + 3) dx + (2xy − y2 + 10) dy = 0.
(ii) x (y2 + x2 − 2x + 3) dx + x (2xy − y2 + 10) dy = 0.
S OLUTION .
(i)
(ii)
∂P
∂y ( x, y )
= 2y and
∂Q
∂x ( x, y )
= 2y so the equation is exact.
2
= 2xy and ∂Q
∂x ( x, y ) = 4xy − y + 10x; therefore the
exactness condition does not hold.
∂P
∂y ( x, y )
Observe that equation (ii) in example 1.7.1 was obtained by multiplying
equation (i) by x, but only equation (i) is exact. This shows that even though
two equations may be equivalent, one may be exact while the other is not.
To integrate an exact equation, we will use a simplified version of the
method used in the proof of theorem 1.3. Start by computing one of the
following indefinite integrals, whichever is easier:
M ( x, y) =
Z
P( x, y) dx or N ( x, y) =
Z
Q( x, y) dy.
When integrating with respect to x to find M ( x, y), treat y as a constant,
and when computing N ( x, y), treat x as a constant. Let us assume that we
have computed M( x, y). Then any function of the form
F ( x, y) = M ( x, y) + H (y)
1.7. EXACT FORM
57
will satisfy the requirement ∂F
∂x = P ( x, y ). We can view the H ( y ) term here
as a “constant of integration,” which depends on y since y was a constant
when the antiderivative M( x, y) was computed.
Now set ∂F
∂y = Q ( x, y ). In other words, solve the equation
∂M
+ H 0 (y) = Q( x, y)
∂y
to determine H (y). This yields
Z H (y) =
∂M
Q( x, y) −
∂y
dy.
Of course, H (y) is not allowed to depend on x. The exactness condition
ensures that all terms involving x in the difference Q( x, y) − ∂M
∂y cancel out,
since
∂
∂2 M
∂M
∂Q
Q( x, y) −
=
−
∂x
∂y
∂x
∂x∂y
∂P ∂P
−
= 0.
=
∂y
∂y
Therefore F ( x, y) = M ( x, y) + H (y) is an integral of the ODE (1.33).
Example 1.7.2 Find an integral for the ODE
(y2 + x2 − 2x + 3) dx + (2xy − y2 + 10) dy = 0.
S OLUTION . According to example 1.7.1, part (i), this equation is exact.
The integral F ( x, y) will have the form
F ( x, y) =
Z
(y2 + x2 − 2x + 3) dx + H (y)
1
= xy2 + x3 − x2 + 3x + H (y).
3
To determine H (y), differentiate this expression with respect to y:
∂F
dH
= 2xy +
.
∂y
dy
Since
∂F
∂y
= Q( x, y) = 2xy − y2 + 10, it follows that
2xy +
dH
= 2xy − y2 + 10.
dy
58
CHAPTER 1. FIRST-ORDER EQUATIONS
After cancelling, we obtain
and the integral is
dH
dy
= −y2 + 10. Therefore H (y) = − 13 y3 + 10y,
1
1
F ( x, y) = xy2 + x3 − x2 + 3x − y3 + 10y.
3
3
To find an explicit solution, we would have to solve the cubic equation
F ( x, y) = C for y in terms of x.
Integrating factors
Equation (ii) of example 1.7.1 isn’t exact, but comparing it with the exact
equation (i) of the example shows that it can be put into exact form by multiplying by an appropriate function, namely x −1 . This inspires the following definition: A function m( x, y) is an integrating factor for the equation
P( x, y) dx + Q( x, y) dy = 0.
(1.37)
if
m( x, y) P( x, y) dx + m( x, y) Q( x, y) dy = 0
is exact.
If we can determine an integrating factor for a given ODE, then we have
an equivalent ODE in exact form, and we can find an integral. In special
circumstances, we can put this strategy into action.
One-variable integrating factors. Suppose that the ODE (1.37) has an integrating factor m. Then
m P( x, y) dx + m Q( x, y) dy = 0
is exact, so by the exactness condition,
∂
∂
[m P( x, y)] =
[m Q( x, y)]
∂y
∂x
(1.38)
Equation (1.38) is a PDE whose unknown function is the integrating
factor m. It is not always easy to solve for m, but if it happens that there is
an integrating factor m( x ) or m(y) that depends on only one of the variables
x or y, then equation (1.38) reduces to a separable ODE, and the integrating
factor can be found.
1.7. EXACT FORM
59
Assume that there is an integrating factor of the form m( x ). By applying
the product rule for differentiation to equation (1.38), we obtain
m( x )
∂Q
∂P
= m0 ( x ) Q( x, y) + m( x ) .
∂y
∂x
This equation can be rearranged as
m0 ( x )
1
=
m( x )
Q( x, y)
∂P ∂Q
−
∂y
∂x
(1.39)
Since its left side is independent of y, equation (1.39) only makes sense if its
right side is also independent of y. Thus, if the right side is not independent
of y, there is no integrating factor m( x ). If the right side is independent of
y it is straightforward to integrate both sides and calculate m( x ).
The same reasoning shows that there is an integrating factor m(y) if and
only if
1
∂Q ∂P
−
P( x, y) ∂x
∂y
is a function of y alone, and that in this case
m0 (y)
1
∂Q ∂P
=
−
m(y)
P( x, y) ∂x
∂y
(see exercise 21).
Do not memorize the above material. It is easy to derive at a moment’s
notice from the exactness condition (in the form of equation (1.38)) by making the assumption that m is a function of x alone or of y alone.
Example 1.7.3 If possible, find an integrating factor m( x ) for the ODE
[2( x + y2 ) cos x + sin x ] dx + 2y sin x dy = 0
(1.40)
and use it to determine an integral for the ODE.
S OLUTION . An integrating factor m( x ) must satisfy
∂
∂
[m( x )(2( x + y2 ) cos( x ) + sin( x ))] =
[m( x )(2y sin( x ))]
∂y
∂x
If you do the differentiation (remember to use the product rule on the right
side) you will obtain
m( x )(4y cos( x )) = m0 ( x )(2y sin( x )) + m( x )(2y cos( x )).
60
CHAPTER 1. FIRST-ORDER EQUATIONS
We can cancel a factor of y and simplify this equation to the form
m0 ( x )
cos( x )
=
.
m( x )
sin( x )
Now we can integrate: ln(|m( x )|) = ln(| sin( x )|) + C. Therefore,
m = sin( x ) is an integrating factor. After multiplying (1.40) by this
integrating factor, we have the exact equation
[2( x + y2 ) sin( x ) cos( x ) + sin2 ( x )] dx + 2y sin2 ( x ) dy = 0.
Since it’s easier to integrate Q( x, y) = 2y sin2 ( x ) with respect to y than it is
to integrate P( x, y) with respect to x, set
F ( x, y) =
Then
∂F
∂y
Z
2y sin2 ( x ) dy + K ( x ) = y2 sin2 ( x ) + K ( x ).
= 2y sin2 ( x ). Since
∂F
∂x
= 2y2 sin( x ) cos( x ) + K 0 ( x ),
2y2 sin( x ) cos( x ) + K 0 ( x ) = 2( x + y2 ) sin( x ) cos( x ) + sin2 ( x ),
and K 0 ( x ) = 2x sin( x ) cos( x ) + sin2 ( x ). A final integration shows that
K ( x ) = x sin2 ( x ) and so F ( x, y) = ( x + y2 ) sin2 ( x ) is the integral that we
have sought. Graphs of all solutions of the ODE are subsets of level
curves
( x + y2 ) sin2 ( x ) = C.
Example 1.7.4 Find, if possible, a one-variable integrating factor for the ODE
2
3
2
dy = 0,
(1.41)
2xy − 3 dx + 4x + 2
x
x y
and use it to integrate the ODE.
S OLUTION .
We’ll start by looking for an integrating factor m( x ). If there is one, it must
satisfy
2
∂
3
∂
2
m( x ) 2xy − 3
=
m( x ) 4x + 2
.
∂y
x
∂x
x y
1.7. EXACT FORM
61
After you carry out the differentiation and simplify, you should obtain
6
3
= m0 ( x ) 4x2 + 2
,
m( x ) −6x + 3
x y
x y
and then
m0 ( x )
6 − 6x4 y
= 5
m( x )
4x y + 3
Since the right side depends on y, this equation has no solution, and
hence there is no integrating factor m( x ).
To find if there is an integrating factor m(y) we must find solutions (if any)
of
∂
2
3
∂
2
=
m(y) 2xy − 3
m(y) 4x + 2
.
∂y
x
∂x
x y
Now we use the product rule in differentiating the left side of the equation:
2
m (y) 2xy − 3
x
0
6
+ m(y)(2x ) = m(y) 8x − 3
x y
.
You can simplify this equation and obtain
3
m0 (y)
= .
m(y)
y
Integrate both sides to get ln |m(y)|) = 3 ln |y| + C. Thus m(y) = y3 is an
integrating factor. Multiplying equation (1.41) by y3 results in the exact
equation
2y3
3y2
4
2 3
2xy − 3 dx + 4x y + 2 dy = 0.
x
x
Put
F ( x, y) =
Z
(2xy4 − 2x −3 y3 ) dx + H (y)
= x 2 y 4 + x −2 y 3 + H ( y ).
= 4x2 y3 + 3x −2 y2 + H 0 (y), and it follows that H 0 (y) ≡ 0. The
integral is F ( x, y) = x2 y4 + x −2 y3 .
Then
∂F
∂y
62
CHAPTER 1. FIRST-ORDER EQUATIONS
Exercises
General instruction for problems 1 – 14: decide if the ODE is exact; if it is,
find an integral, and if it isn’t, do nothing. You are encouraged to use a
computer algebra system to help find antiderivatives.
1.
2.
(2x + 5y + 3) dx + (5x − 4y + 2) dy = 0.
Answer
y dx + ( x + y) dy = 0.
3.
(y2 − 1) dx + (2xy − x ) dy = 0.
Answer
4.
(y2 − y) dx + (2xy − x ) dy = 0.
5.
6.
3e3x y(ln y − 1) dx + ln y(e3x − y) dy = 0.
Answer
( x2 + 2xy − y2 ) dx + ( x2 − 2xy − y2 ) dy = 0.
x ( x2 + y2 − 1) dx + y( x2 + y2 + 1) dy = 0.
Answer
p
p
8.
x (3 x2 + y2 − 2) dx + y(3 x2 + y2 + 2) dy = 0.
2 2
y
x
9.
1 + x +y
dx +
− 1 dy = 0.
x +y
7.
Answer
10.
11.
12.
y dx
x
+ ln x dy = 0.
(3x2 + 6xy + 9y2 ) dx + (3x2 + 18xy + 51y2 ) dy = 0.
Answer
2x
y
dx +
y2 − x 2
y2
dy = 0.
13.
2xy dx + (y2 − x2 ) dy = 0.
Answer
14.
2xy dx − (y2 − x2 ) dy = 0.
In problems 15 – 19, find an integrating factor for the given ODE, and
use it to determine an integral.
1.7. EXACT FORM
15.
16.
17.
18.
63
( x2 + xy2 + 1) dx + 2y dy = 0.
Answer
x dy − (y − x ) dx = 0.
( x2 + 2x + 2xy + 2y + 3y2 ) dx + (2x + 6y) dy = 0.
Answer
(5x + 20y + 28y3 ) dx + (5x + 21xy2 ) dy = 0.
19.
2xy dx + (3x2 + 2y) dy = 0.
Answer
20.
Show that the ODE (5xy − 2y2 ) dx + (5x2 − 3xy) dy = 0 does not
have an integrating factor m( x ) or m(y).
21.
Show that there is an integrating factor m(y) for the ODE P( x, y) dx +
Q( x, y) dy = 0 if and only if the expression
1
P( x, y)
∂Q ∂P
−
∂x
∂y
is independent of x.
Answer
22.
Show that if y0 = f ( x, y) is a separable ODE, with f ( x, y) =
g( x ) h(y) then [h(y)]−1 is an integrating factor for
dy − f ( x, y) dx = 0.
Thus, the method for solving separable ODEs is a special case of the
integrating factor method.
23.
Write the linear ODE, y0 + p( x )y = q( x ), in the equivalent form
dy + [ p( x )y − q( x )] dx = 0,
and find a one-variable integrating factor m( x ). (This integrating factor, discovered by Leibniz in 1692, has been memorized by ten generations of students in ODE courses.)
Answer
64
1.8
CHAPTER 1. FIRST-ORDER EQUATIONS
Graphical Analysis of ODEs
Imagine a sheet of graph paper representing a section of the t, y-plane. We
want to sketch some solutions of an ODE y0 = f (t, y) on this paper. There
are vertical grid lines t = ti and horizontal grid lines y = y j , and we will
call the points (ti , y j ) where the grid lines intersect grid points.
A direction field for the ODE y0 = f (t, y) can be drawn by placing at
each grid point (ti , y j ) a short line segment, called an element. The element
is centered at the grid point, and its slope is equal to f (ti , y j ). Figure 1.10
shows a direction field for the ODE y0 = t − y2 on graph paper covering
the region −1.5 ≤ t ≤ 1.5, −1 ≤ y ≤ 1.
The elements of the direction field can be drawn at only a finite set
of points, but we should imagine an element located at each point of the
plane. If y = φ(t) is a solution of the ODE, then at each point (t, φ(t)) on
the graph of φ the slope of the graph of φ is equal to f (t, φ(t)), the slope of
the direction field element at that point. Thus at each point, the graph of a
solution is tangent to the element at its location.
Figure 1.11 shows two typical direction fields, as drawn by a CAS. They
belong to the ODEs,
(1.42)
v0 = 10 − .001v2 ,
and
v0 = 10 − .001e−.01t v2 ,
(1.43)
respectively. Both equations represent the velocity of a falling body. In (1.42),
the drag coefficient is constant,
while in (1.43), it decays with time. The hor√
izontal elements at v = 1000 in the constant drag case indicate that solutions of (1.42) tend to a terminal velocity while the direction field of (1.43)
does not have this property, and hence there is no terminal velocity.
Graphing the solution of an IVP
To make a rough sketch of the solution of an IVP, y0 = f (t, y); y(t0 ) = y0 ,
draw the direction field element centered at (t0 , y0 ). Let (t1 , y1 ) be the point
at the right end of that element, and draw the right half of the direction field
element at (t1 , y1 ), extending to a point (t2 , y2 ). Continuing this process, we
obtain a polygonal curve (broken line graph) that approximates the graph
of the solution of the IVP. The accuracy of the approximation depends on
the on the length of the direction field elements being used. With shorter
elements the graph will be more accurate, but requires more effort to draw.
1.8. GRAPHICAL ANALYSIS OF ODES
65
The endpoints (t0 , y0 ), (t1 , y1 ), (t2 , y2 ), . . . of the direction field elements
yield two sequences tn and yn . We will stipulate that the direction field
elements are drawn so that the tn are evenly spaced, with tn = t0 + nh,
where h is a constant increment.
We will use “difference notation:” For any sequence u0 , u1 , u2 , . . . of
numbers, form a sequence of forward differences
∆u0 = u1 − u0 , ∆u1 = u2 − u1 , . . . , ∆un = un+1 − un , . . . .
The forward differences for the sequence tn are constant: ∆tn = h. We
can calculate the forward differences for the yn because the slope of the
direction field element at the point (tn , yn ) is
∆yn
= f ( t n , y n ).
∆tn
Thus,
∆yn = h f (tn , yn ).
(1.44)
Equation (1.44) is a difference equation. Difference equations are like
ODEs, and we can exploit them to learn more about ODEs (and vice versa).
The analogy between difference equations and differential equations is
illustrated by compound interest. Bank A pays r% interest, compounded
k times per year. If an initial deposit of $P is made, and the account is
forgotten, let’s determine how the balance grows. Let yn denote the balance
at the end of the nth compounding period. The interest payment to the
account will be
r
∆yn =
yn
100k
The account balance at any time is determined by solving this difference
equation with initial condition y0 = P.
Bank B uses continuous compounding, thus replacing the above difference equation with an ODE,
y0 =
r
y
100
with initial condition y(0) = P to track the balance. Both methods give approximately the same result, with the value due to discrete compounding
increasing as k → ∞ and approaching the value due to continuous compounding in the limit.
In the context of difference equations, an IVP takes the form
∆ym = F (m, ym ); y0 = A.
66
CHAPTER 1. FIRST-ORDER EQUATIONS
Provided that F (m, y) is defined for all (m, y) such that m is a nonnegative
integer, this IVP is guaranteed to have a unique solution. We don’t even
have to assume F is continuous. To see why this is so, rewrite the difference
equation as
ym+1 = ym + F (m, ym ).
(1.45)
Equation (1.45) is recursive, in the sense it gives a formula for computing the
next term in the sequence. Since we have a starting value for y0 (the initial
condition), y1 , y2 , and so on can be computed, one after the other. Many
well known sequences are defined recursively—for example the factorial
sequence n! is defined by 0! = 1 and n! = n · (n − 1)!.
Euler’s method
Leonhard Euler, one of the brightest stars in the history of mathematics,
showed how to use a difference equation IVP to approximate the solution
of a differential equation IVP. Let y = φ(t) denote the solution of the IVP,
y0 =
f (t, y);
y(t0 ) = A.
(1.46)
(1.47)
Choose a time step h (h must be nonzero, but is allowed to be negative)
and set tn = nh + t0 . Euler’s method uses the difference equation (1.44)
with initial condition y0 = A to determine a sequence y0 , y1 , y2 , . . . . In this
sequence, yn serves as the approximation of φ(tn ).
We have pictured the approximate solution of an IVP given by Euler’s
method by plotting the sequence of points (tm , ym ) and connecting adjacent
points with straight line segments to form a polygonal curve.
Example 1.8.1 Use Euler’s method with time step h = ±0.25 to approximate the
solution of the IVP y0 = t − y2 , y(0) = 0, for −1.5 ≤ t ≤ 1.5
S OLUTION . The solution must be propagated to the left and to the right
from the initial point (0, 0). To propagate leftward, we use h = −0.25; and
to propagate rightward, h = 0.25. We will start with h = −0.25. Euler’s
method then generates a sequence ym defined by the initial condition
y(0) = 0 and the difference equation ∆ym = −0.25(tm − y2m ), where t0 = 0,
t−1 = −0.25, t−2 = −0.5, and so on. Since t−6 = −1.5, we will need to
compute ym for −1 ≥ m ≥ −6.
1.8. GRAPHICAL ANALYSIS OF ODES
67
Write the difference equation in recursive form as
ym−1 = ym − 0.25(tm − y2m ).
Take a moment to calculate y−1 through y−6 , starting with y0 = 0. You
should get y−1 = 0, y−2 = 0.0625, y−3 ≈ 0.1885, and after three more
steps, y−6 ≈ 1.0972.
This calculation approximates values of the solution y = φ(t) of the IVP at
the points t = −.25, −.5. − .75, −1, −1.25, and −1.5. Due to the large step
size, these approximations poor. A CAS provides a solution of the IVP and
evaluation yields φ(−0.5) = 0.1266 and φ(−1.5) = 1.7857 (both rounded to
4 decimal places).
To propagate to the right, start over with h = +0.25. The numbers
t0 , t1 , t2 . . . are 0, 0.25, 0.5 . . . , Again, calculate y1 through y6 ; you should
get y1 = 0, y2 = 0.0625, and so on, with y6 = 0.8098. Again, the
approximations you will obtain are not accurate. The values of the solution
φ(t) computed by the CAS, rounded to 4 decimal places, are
φ(0.25) = .0312, φ(0.5) = 0.1235, φ(0.75) = 0.2700, and φ(1.5) = 0.8574.
Figure 1.12 shows the graph of the approximate solution that we have
produced, and the exact solution produced by the CAS, with the direction
field for the ODE in the background.
To show that Euler’s method really does produce an approximation of
the solution φ(t) of a given IVP, we need to refer to a special case of Taylor’s
theorem, the second mean value theorem, The proof can be found in any
calculus text.
Theorem 1.4 (Second mean value theorem) Suppose that a function φ(t) is
twice differentiable on an interval ( A, B), and that t0 and h are numbers such that
t0 , t0 + h ∈ ( A, B). Then there is a number c1 between t0 and t0 + h, such that
1
φ(t0 + h) = φ(t0 ) + h φ0 (t0 ) + h2 φ00 (c1 ).
2
(1.48)
For any differentiable function φ(t),
φ(t0 + h) ≈ φ(t0 ) + hφ0 (t0 )
is the linear approximation of φ. The second mean value theorem gives a way
to see how precise it is.
68
CHAPTER 1. FIRST-ORDER EQUATIONS
Euler’s method relies on linear approximation because each step uses it.
If φ(t) is the solution of the IVP (1.46), (1.47) then we know that φ(t0 ) = y0
and
h φ0 (t0 ) = h f (t0 , y0 ) = ∆y0
(∆y0 = y1 − y0 as in (1.44)). If φ is sufficiently differentiable then by theorem 1.4
1
φ(t0 + h) = y0 + ∆y0 + h2 φ00 (c1 ).
| {z } 2
y1
and hence φ(t0 + h) − y1 = 21 h2 φ00 (c1 ), where c1 is between t0 and t1 . As
h → 0, the error term 12 h2 φ00 (c1 ) → 0 at a faster rate, and that is why y1 is a
good approximation of φ(t0 + h) if h is sufficiently small.
We still need to justify the Euler approximation as it proceeds through
many steps. In the second step, Euler’s method follows the direction field
element starting at (t1 , y1 ). Let φ1 (t) be the solution of the IVP
y0 = f (t, y) y(t1 ) = y1
Applying theorem 1.4 to φ1 (t), we find that
1
φ1 (t1 + h) = φ1 (t2 ) = y1 + ∆y1 + h2 φ100 (c2 ).
| {z } 2
y2
with c2 between t1 and t2 . The approximation error after the second step is
φ(t2 ) − y2 = [φ(t2 ) − φ1 (t2 )] + [φ1 (t2 ) − y2 ]
1
= [φ(t2 ) − φ1 (t2 )] + h2 φ100 (c2 ).
2
The two terms of this expression are accumulated error, and the local
error, respectively. The local error is defined to be the error due to one step
of the approximation. In the step where y2 is computed, the local error,
which we denote LE2 , is equal to φ1 (t2 ) − y2 . The purpose of referring to
the second mean value theorem is to estimate local error, and it tells us that
LE2 = 12 h2 φ100 (c2 ).
The accumulated error, denoted AEm , is the error due to all previous
steps taken in propagating the solution by Euler’s method. Thus the first
accumulated error is equal to
AE1 = φ(t2 ) − φ1 (t2 ).
1.8. GRAPHICAL ANALYSIS OF ODES
69
This is the difference between two solutions of the same ODE at t = t2 ,
given that the two solutions are close when t = t1 .
As we continue, each step approximates the solution φm (t) of a new
IVP, whose initial point is the point (tm , ym ) reached in the previous step.
There is a new local error LEm+1 = 21 φ00 (cm+1 ), and a new accumulated
error,
AEm = φ(tm+1 ) − φm (tm+1 )
We can’t expect that AEm will be equal to the sum of the local errors LE1 ,
LE2 . . . , LEm . Solutions of ODEs with nearby initial conditions may diverge
away from each other (if this happens, the accumulated error will be larger
than the sum of the previous local errors), or they may converge toward
each other, causing the accumulated error to be less than the sum of the
local errors.
Example 1.8.2 Determine the solution of the initial value problem
1
y0 = y + ; y(0) = 0.
4
Using a time step h = 0.5, use Euler’s method to calculate
(i) y1 , y2 , y3 , and y4 .
(ii) LE1 , LE2 , LE3 and LE4 .
(iii) AE1 , AE2 and AE3 .
S OLUTION . We will start by computing y1 . . . y4 . The large time step will
make the approximation unreliable, but we are studying errors.
The difference equation prescribed by Euler’s method is
∆ym = 0.5(ym + 0.25), with initial condition y0 = 0. Solving this difference
equation IVP, we obtain the following (See details).
i
ti
yi
0
0.0
0.0
1
0.5
0.125
2
1.0
0.3125
3
1.5
0.59375
4
2.0
1.015625
To determine the errors precisely, we need to have the general solution of
the ODE (in practice, we will never have access to this, since the purpose
of Euler’s method is to approximate a solution that can’t be obtained
analytically). Since y0 = y + 14 is a linear ODE, you can derive the solution
70
CHAPTER 1. FIRST-ORDER EQUATIONS
y = Cet − 14 . The solution that we are trying to approximate is satisfies the
initial condition φ(0) = 0, so C = 41 .
Each solution φm (t), satisfies initial conditions φ(tm ) = ym , so
φm (t) = Cm et − 14 where the constant Cm is determined by
1
ym = Cm etm − .
4
Thus Cm = (ym + 0.25)e−tm and
φm (t) = (ym + 0.25)et−tm − 0.25.
In particular
φ1 (t) = 0.375et−0.5 − 0.25
φ2 (t) = 0.5625et−1 − 0.25
φ3 (t) = 0.84375et−1.5 − 0.25
Figure 1.13 displays the graphs of φ(t), and φm (t) for m = 1, 2, 3. It also
shows the broken line graph connecting (t0 , y0 ), (t1 , y1 ), (t2 , y2 ), (t3 , y3 ),
and (t4 , y4 ). Notice that the graph of φ(t) is tangent to the first segment of
the broken line graph, φ1 (t) is tangent to the second segment, and so on.
The local and accumulated errors are shown in the following table (See
details), in which LEi = φi−1 (ti ) − yi and AEi = φ(ti ) − φi−1 (ti ).
i
LEi
AEi
1
0.037
—
2
0.056
0.061
3
0.084
0.193
4
0.125
0.456
In figure 1.13 we can picture LE4 as the vertical distance between the
bottom two curves at t = 2 (at the right edge of the graph). AE3 is the
distance from the top curve to the second-to-the-bottom curve, and
represents the consequences of the local errors LE1 , LE2 , and LE3 .
Example 1.8.3 The velocity y(t) (in meters per second) of a certain rock sinking
in sea water is given by the differential equation
y0 = 8 − 2y2 .
(1.49)
Given that y(0) = 0, use Euler’s method with h = 0.1 to approximate the solution
for 0 ≤ t ≤ 2 seconds.
1.8. GRAPHICAL ANALYSIS OF ODES
71
S OLUTION . The difference equation given by Euler’s method with h = 0.1
is
∆ym = 0.1(8 − 2y2m ).
The solution of this difference equation is the sequence beginning
0,0.8,1.472,. . . . By connecting successive points (tm , ym ), (tm+1 , ym+1 )
with line segments we obtain a broken line graph that serves as an
approximation of the solution of our initial value problem. This graph is
shown in figure 1.14. Although Euler’s method can’t be expected to give a
very accurate approximation with such a large time step, it should be noted
that the graph shown levels off at a velocity of 2 meters per second, which
is the correct terminal velocity. This difference equation, like the differential
equation it approximates, has a constant solution, ym = 2, and our solution
converges to it. In other words, the qualitative behavior of the Euler
difference equation matches that of the differential equation. Unfortunately,
this does not always happen—see problem 21 at the end of this section.
How to implement Euler’s method with a spreadsheet
Euler’s method uses four columns. Column A lists t0 , t1 , t2 , t3 , . . . ; Column
B is for y0 , y1 , y2 , y3 , . . . ; column C is for ∆t (all entries will be the same);
and we put ∆y0 , ∆y1 , ∆y2 , ∆y3 , . . . in column D. Enter the initial values of t
and y in A1 and B1, respectively, and put ∆t in C1. In D1, type the formula
for ∆y. Thus, if the ODE is
y 0 = y (1 − y ) +
1
sin2 (t)
10
(1.50)
the formula
=C1 * (B1 * (1 - B1) + (SIN(A1)ˆ2)/10)
would appear in cell D1.
Cell A2 will contain the formula for t1 , =A1+C1, and cell B2 will have
the formula for y1 , =B1+D1 One could copy cell C1 to C2, but it is better to
place in cell C2 the formula =C1 because this makes it possible to change
the time step by altering cell C1. Copy D1 to the clipboard and paste to cell
D2. Because copying is relative, if you are working with equation (1.50),
the expression
=C2 * (B2 * (1 - B2) + (SIN(A2)ˆ2)/10)
72
CHAPTER 1. FIRST-ORDER EQUATIONS
will appear in D2.
Now copy row 2 of the spreadsheet to the clipboard, and paste to rows
3, . . . , n. Column B will contain the approximations of yi corresponding to
the ti in column A.
Spreadsheets are capable of producing a graph of the approximate solution. The menu of graph types contains many choices; the appropriate
one is the “xy-graph,” with x-axis values from column A and “first series”
values from column B. Although a polygonal graph is appropriate, commercial spreadsheets can draw smooth curves by employing sophisticated
interpolation techniques. Smooth curves are no more accurate than the
polygonal graphs, but they look nicer.
Figure 1.15 displays the spreadsheet for the approximate solution of
equation (1.50), with initial condition y(0) = 0.5 and ∆t = 0.25.
Exercises
1.
Exercise 16 in section 1.6 the ODE
v0 = g − kv |v|
introduced a model for the velocity of a ball that was initially thrown
upward, subject to gravitational and drag forces (in the model, the
positive direction is downward). Draw a direction field for this ODE.
Use as parameters g = 9.8 meters per second2 and k = 0.002 meters−1 .
Answer
Print exercises 2 – 7, and sketch graphs of solutions of the IVPs on the
given direction fields.
1.8. GRAPHICAL ANALYSIS OF ODES
2.
73
y0 = ty; initial conditions y(0) = 0, y(0) = 1, and y(0) = −1.
1.5
1
0.5
0
-0.5
-1
-1.5
3.
-2
-1
0
1
2
3
y0 = 3t − 2y; initial conditions y(0) = −1, − 43 , and 0.
1.5
1
0.5
0
-0.5
-1
-1.5
Answer
-2
-1
0
1
2
3
74
CHAPTER 1. FIRST-ORDER EQUATIONS
4.
y0 = 2te−y ; initial conditions y(0) = 0, y(1) = 0, and y(−1) = 0.
1.5
1
0.5
0
-0.5
-1
-1.5
5.
y0 =
-2
1
t−3y ;
-1
0
1
2
3
0
1
2
3
y(0) = 1.
1.5
1
0.5
0
-0.5
-1
-2
Answer
6.
y0 = 2t; y(0) = 0.
-1
1.8. GRAPHICAL ANALYSIS OF ODES
75
1.5
1
0.5
0
-0.5
-1
-1.5
7.
-2
-1
0
1
2
3
0
1
2
3
y0 = −2y; y(0) = −1.
1.5
1
0.5
0
-0.5
-1
-1.5
Answer
-2
-1
76
CHAPTER 1. FIRST-ORDER EQUATIONS
8.
Match the ODEs with their direction fields.
(a) y0 = sin y
(b) y0 = t2 − y2
(c) y0 = y2 − 2ty + t2
(d) y0 = y − y2
(e) y0 = t2 − y
1
0.5
0
-0.5
-1
-1.5
(I)
-2
-1
0
1
2
3
1.8. GRAPHICAL ANALYSIS OF ODES
77
1.5
1
0.5
0
-0.5
-1
-1.5
(II)
-2
-1
-2
-1
0
1
2
3
1.5
1
0.5
0
-0.5
-1
-1.5
(III)
0
1
2
3
78
CHAPTER 1. FIRST-ORDER EQUATIONS
1.5
1
0.5
0
-0.5
-1
-1.5
(IV)
-2
-1
-5
-2.5
0
1
2
3
4
2
0
-2
-4
-7.5
(V)
9.
0
2.5
5
7.5
Show that the graph of the solution y(t) of the IVP y0 = t − 3y;
y(0) = 1 crosses the line t − 3y = 0 exactly once, and has a relative
minimum at the crossing point. Do not use the explicit solution, y =
1
−3t ). Instead, consider what the slope of the solution
9 (3t − 1 + 10e
will be at a crossing point.
Answer
1.8. GRAPHICAL ANALYSIS OF ODES
10.
79
Find the sequence of forward differences for each of the following
sequences.
(a) 1,2,4,8,16,32,. . . ,2m ,. . .
(b) 1,3,9,27,81,243,. . . ,3m ,. . .
(c) 0,1,8,27,64,125,. . . ,m3 ,. . .
(d) 1, 12 , 31 , 41 , 15 ,. . . , m1 ,. . .
11.
For each difference equation, determine the first three or four
terms of the sequence ym and then find an expression for ym .
(a) ∆ym = 1, y1 = 0.
(b) ∆ym = ym , y1 = 1.
(c) ∆ym = kym , y1 = C, where k and C are constants.
(d) ∆ym = ym + 1, y1 = 0.
Answer
12.
13.
The bank offers you a choice of an account bearing an interest
rate of 6%, compounded continuously, and an account with 6.25%
interest, compounded annually. Both rates are guaranteed for five
years, and you don’t intend to make withdrawals before that time is
up. Which account should you choose?
Let am and bm be sequences. The difference equation
∆ym = am ym + bm
is called the linear difference equation with coefficients am and source
bm . You will want to compare this definition with that of a linear ODE.
If bm = 0 for all m, the difference equation is homogeneous.
Define a new sequence Am by A1 = 1, and
Am = (1 + a1 )(1 + a2 ) · · · (1 + am−1 ) for m > 1.
(a) Show that every solution of a homogeneous difference equation
∆ym = am ym has the form ym = CAm , where C is constant.
(b) Explain how to solve an initial value problem involving an inhomogeneous linear difference equation. Hint: the procedure
resembles the variation of constants method for solving an inhomogeneous linear ODE.
80
CHAPTER 1. FIRST-ORDER EQUATIONS
(c) Find the general solution of
∆ym =
1
ym .
m
(d) Use the method of part (b) to solve
∆ym =
1
ym + m + 1, y1 = 0.
m
Answer
14.
(a) Show that the solution of the IVP
y0 = f ( x ); y( a) = b,
is
y( x ) = b +
Z x
a
f (u) du.
(1.51)
(1.52)
(b) Show that approximating the solution of the IVP 1.51 by Euler’s
method amounts to calculating a Riemann sum for the integral
in equation (1.52).
Use Euler’s method and a spreadsheet or programmable calculator to
find approximate solutions of the differential equations in problems 15
and 16, over the intervals indicated, and with the prescribed step sizes.
Draw broken line graphs of the solutions.
15.
y0 = y2 ; y(0) = 1. h = .05, 0 ≤ t ≤ 1.
Answer
16.
y0 = √ 2t
t + y2
; y(0) = 1. h = ±0.1
for −1 ≤ t ≤ 1. Can you save time by exploiting the symmetry in the
difference equations for propagating to the left and to the right?
17.
Let y = φ(t) denote the solution of the IVP,
y0 = t + y; y(0) = 0.
Since the differential equation is linear, you can solve this IVP and
find a formula for φ(t). This problem asks you to calculate φ(1) and
1.8. GRAPHICAL ANALYSIS OF ODES
81
to approximate φ(1) by using Euler’s method. The purpose is to see
how fast the approximation improves when the time step is reduced.
Use Euler’s method with time steps h = 1, 0.5, 0.25, 0.1, 0.05, 0.02, and
0.01 to approximate φ(1), and let E(h) = |φ(1) − y N | (where Nh =
1) denote the approximation error obtained with time step h. Plot a
graph of E(h) as a function of h.
Answer
18.
The purpose of this problem is to trace the consequence of one
local error in Euler’s method. Calculate the approximation y1 of e−0.1
given by solving the IVP y0 = −y; y(0) = 1 by Euler’s method with
h = 0.1. Let y = φ1 (t) be the solution of y0 = −y with initial condition
y(0.1) = y1 . Calculate e−t − φ1 (t) for t = 10 and for t = 20.
Repeat this calculation for the IVP y0 = y, y(0) = 1 and its solution,
y = et .
19.
Solve the initial value problem y0 = y + sin(πt); y(0) = 0, and
approximate the solution by calculating y1 , y2 , y3 , and y4 by Euler’s
method, using time step h = 0.25. Determine all of the local and accumulated errors.
Answer
20.
Given that the general solution of the ODE y0 = 1 + y2 is y =
tan(t − C ) where C is an arbitrary constant, calculate the local and
accumulated errors in approximating the solution of the IVP with
initial condition y(0) = 0 with time step h = 0.5 for y1 , y2 , and y3 .
Remember to use radians, and draw a graph showing the solution
φ(t) = tan(t) as well as the functions φ1 (t) and φ2 (t).
Euler’s method can give an answer that is drastically wrong if too large
a time step is used. Exercises 21–21b demonstrate this and give a modification of Euler’s method to avoid this problem.
21.
The solution y = e−100t of the initial value problem
y0 = −100y; y(0) = 1
(1.53)
converges to 0 very rapidly. This is apparent even if we restrict our
attention to 0 ≤ t ≤ 1, since e−100×1 is of the order of magnitude of
10−43 .
82
CHAPTER 1. FIRST-ORDER EQUATIONS
(a) Use Euler’s method with h = 0.1 to approximate the solution
of the initial value problem (1.53) for 0 ≤ t ≤ 1. Does your
computed solution appear to converge to 0?
(b) How small should h be to ensure that the solution converges to
0?
Answer
22.
The Backward Euler Method. One approach that avoids erratic
behavior observed in the solution computed in exercise 21 is to use
backward differences instead of forward differences with Euler’s Method.
Thus, we use the difference equation3
∇ym+1 = h f (tm+1 , ym+1 ) (m ≥ 0),
where ∇ym+1 = ym+1 − ym , as a model for the differential equation
y0 = f (t, y). This difference scheme is said to be implicit because it
defines ym+1 implicitly in terms of ym by means of the relation
ym+1 − ym − h f (tm+1 , ym+1 ) = 0.
Test the Backward Euler Method by approximating the solution of the
initial value problem (1.53), again with h = 0.1, for 0 ≤ t ≤ 1.
23.
This problem uses the forward and backward versions of Euler’s
method
with time step h = 0.1 to approximate the solution of y0 =
p
1 − y2 with y(0) = 0 for 0 ≤ t ≤ 2.
(a) Explain why the forward version fails.
(b) The difference equation for the backward version is
q
∇ym+1 = 0.1 1 − y2m+1 .
Square both sides and use the quadratic formula to derive the
following recursive equation:
p
ym ± 0.1 1.01 − y2m
y m +1 =
1.01
Decide which of the signs is correct, and which is extraneous.
3∇
is pronounced “nabla.”
1.8. GRAPHICAL ANALYSIS OF ODES
83
(c) Calculate ym for 1 ≤ m ≤ 20.
(d) The solution of the IVP is
sin(t) if 0 ≤ t ≤ π/2
y=
1
if t > π/2.
Determine the errors in both the forward and backward versions. Which is the more accurate?
Answer
In problems 24 – 27, a separable equation and a corresponding direction
field are given. Find a family of solutions and note any singular solutions.
Print the given direction field and sketch on it several solutions, including
any singular solutions.
84
CHAPTER 1. FIRST-ORDER EQUATIONS
y0 =
24.
√
ty.
4
3
2
1
0
0
y0 =
25.
4
2
p
6
8
9 − y2
3
2
1
0
-1
-2
-3
-4
Answer
-2
0
2
4
1.8. GRAPHICAL ANALYSIS OF ODES
85
y0 = cos(t) tan(y)
26.
6
4
2
0
-2
-4
-7.5
27.
-5
-2.5
0
5
2.5
7.5
10
(2 − t) dy = (1 − y) dt.
3.5
3
2.5
2
1.5
1
0.5
0
0
Answer
1
2
3
4
5
6
86
1.9
CHAPTER 1. FIRST-ORDER EQUATIONS
Initial Value Problems
We can solve IVPs involving linear first order ODEs, and we can approximate solutions of IVP involving nonlinear first order ODE. Now we will
see why we should expect an IVP to have one and only one solution.
Let us consider an IVP
y0 =
f (t, y);
y ( t0 ) = y0 .
(1.54)
(1.55)
Let’s refer to the point in the t, y-plane with coordinates (t0 , y0 ) as the initial
point. For the IVP (1.54), (1.55) to be correctly posed, the initial point must
be in the domain of f (t, y).
A solution φ(t) of (1.54) will be considered to be a solution of the IVP if
the domain on which φ(t) is defined is an open interval that contains t0 , and
φ(t0 ) = y0 . It is important to be aware that the domain of the solution of any IVP
is an open interval. The following example is meant to illustrate this point.
Example 1.9.1 Solve the IVP
dy
y
=−
; y(0) = 1.
dt
t+1
(1.56)
S OLUTION . The ODE is linear and homogeneous, and so its general
solution is of the form y = CeK(t) . Substituting this and y0 = CK 0 (t)eK(t)
yields
CeK(t)
CK 0 (t)eK(t) = −
,
t+1
so that K 0 (t) = −1/(t + 1). Hence we will put K (t) = − ln(t + 1) and
eK(t) = 1/(t + 1). The general solution of the ODE is
y=
C
.
t+1
Substituting t = 0 and y = 1 yields C = 1. The solution of the IVP is
y=
1
,
t+1
defined on the largest interval containing t = 0 and not containing t = −1:
(−1, ∞).
1.9. INITIAL VALUE PROBLEMS
87
The graph of the solution of example 1.9.1 is the blue curve in figure 1.16. In the figure, the portion of the graph shown as a red curve is not
part of the solution, because the solution must be defined on an interval.
Intuitively speaking, it should be possible to trace the graph of a solution
of an IVP without lifting one’s pencil from the paper.
Existence and Uniqueness Theorems
Is there some property of a two-variable function f (t, y) that we can easily
verify and that guarantees the IVP (1.54), (1.55) has a solution? This question is answered by the following theorem, which is one of the foundations
of our subject. Briefly, it says that the property we are looking for is continuity of f (as a function of two variables). Like most theorems, it must be
stated carefully.
Theorem 1.5 (Existence theorem) Let f (t, y) be a function that is continuous
at every point of some rectangle
D = {(t, y) : a < t < b, c < y < d},
and let (t0 , y0 ) be a point4 in D .
Then there is a function φ(t), defined on an open interval (h, k ) that contains
t0 , such that φ(t0 ) = y0 and for all t ∈ (h, k ),
φ0 (t) = f (t, φ(t)).
The existence theorem does not tell us how to find a solution of the
IVP. Furthermore, it says nothing about the extent of the interval on which
the solution is defined. The following example shows that even if f (t, y)
is continuous on the entire t, y-plane, the solution may only exist on an
extremely short interval.
Example 1.9.2 Show that
y=
100
.
1 − 100t2
(1.57)
is a solution of the IVP y0 = 2ty2 ; y(0) = 100. What is the domain of this
solution?
4 To ensure that ( t , y ) is not on an edge of the domain where f ( t, y ) is continuous, our
0 0
rectangle is open: D contains the interior points, but not the edges.
88
CHAPTER 1. FIRST-ORDER EQUATIONS
S OLUTION . Setting t = 0 in 100/(1 − 100t2 ) yields y(0) = 100. This is the
initial condition. Differentiating both sides of (1.57),
dy
dt
−100(−200 t)
(1 − 100t2 )2
2t (100)2
=
(1 − 100t2 )2
= 2t y2 .
=
1
Therefore (1.57) satisfies the ODE. This solution is undefined at t = ± 10
,
1
:
and the domain is the largest interval containing t = 0 but not t = ± 10
1 1
− 10
, 10 .
Figure 1.17 displays several graphs of solutions of the ODE that was the
focus of example 1.9.2.
There are many proofs of the existence theorem, the most direct being
the proof given by the Italian mathematician Giuseppe Peano in 1890. This
proof shows that a solution of the IVP can be obtained as a limit of approximations given by Euler’s method as the time step h → 0. The details of the
proof require a good dose of advanced calculus, and are therefore omitted.
Uniqueness of solutions
It is easy to verify that y ≡ 0 and y = t3 are solutions of the IVP
y0 = 3y2/3 ; y(0) = 0
(see figure 1.18). If y = t3 , then y0 = 3t2 and y2/3 = t2 ; hence the first
solution is valid. If y ≡ 0, then the ODE is satisfied because both sides are
identically 0.
IVPs often serve as mathematical models for physical, biological, social, or engineering phenomena. Predictions will not be reliable unless we
know that the IVP doesn’t have multiple solutions. Fortunately, if the function f (t, y) is nice enough, there will be only one solution. In rough terms,
a nice function should not increase or decrease too rapidly. Requiring a
function to be continuous rules out functions that have sudden jumps (discontinuities), but not functions that increase infinitely fast as the function
1.9. INITIAL VALUE PROBLEMS
89
f (y) = 3y2/3 does at y = 0: f 0 (0) = 2(0−1/3 ) = ∞. We will show that multiple solutions are impossible if the function f (t, y) satisfies the following
condition, known as a Lipschitz condition :
Lipschitz condition
A function f (t, y) satisfies a Lipschitz condition with respect
to the variable y in a rectangle D in the plane if there is a constant K (called a Lipschitz constant) such that for any (t, y1 )
and (t, y2 ) ∈ D ,
| f (t, y2 ) − f (t, y1 )| ≤ K |y2 − y1 |.
Example 1.9.3 Show that f (t, y) = 3y2/3 does not satisfy a Lipschitz condition
on any domain that intersects the t-axis.
S OLUTION . If we take y1 = 0 then
| f (t, y2 ) − f (t, y1 )| = 3 y2/3
2 ,
while K |y2 − y1 | = K |y2 |. If the Lipschitz condition holds, then for all y2
3 y2/3
≤ K | y2 |.
2
However, this can only hold for |y2 | > (3/K )3 . No matter how large we
choose K to be, there will be values of y2 close to 0 for which the inequality
does not hold.
The following theorem provides an easy way to check if a function
f (t, y) satisfies a Lipschitz condition.
Theorem 1.6 Suppose that the a function f (t, y) is defined of a closed rectangle
D in the t, y-plane. If ∂∂yf is continuous at each point (t, y) of D then f satisfies a
Lipschitz condition on D .
90
CHAPTER 1. FIRST-ORDER EQUATIONS
∂f
∂y
is continuous on the closed, bounded set D , there is a
∂f finite maximum value of ∂y in D . Let K denote this maximum.
P ROOF Since
Suppose that (t0 , y1 ) and (t0 , y2 ) are points in D , and for convenience,
assume y1 < y2 . . Let g(y) = f (t0 , y); then g is differentiable on [y1 , y2 ] —
in fact,
∂f
g0 (y) =
( t0 , y ).
∂y
Thus, by the mean value theorem, there is a number c ∈ (y1 , y2 ) such that
g(y2 ) − g(y1 ) = g0 (c)(y2 − y1 ).
Let’s take absolute values of both sides, and put this equation in terms of
f:
∂f
| f (t0 , y2 ) − f (t0 , y1 )| = (t0 , c) |y2 − y1 |.
∂y
∂f
Since ∂y (t0 , c) ≤ K, it follows that | f (t0 , y2 ) − f (t0 , y1 )| ≤ K |y2 − y1 |.
There are functions f (t, y) that satisfy Lipschitz conditions even though
they are not differentiable. See problem 22 for an example.
Theorem 1.7 (Uniqueness theorem) Suppose that y = φ1 (t) and y = φ2 (t)
are two solutions of the IVP
y0 = f (t, y); y(t0 ) = y0 .
If the function f (t, y) is continuous and satisfies a Lipschitz condition with respect
to y on a rectangle
D = {(t, y) : a < t < b, c < y < d}
that contains the initial point (t0 , y0 ), then there is an open interval (h, k ) with
t0 ∈ (h, k ), such that φ1 (t) = φ2 (t) for all t ∈ (h, k ).
Suppose that y = φ1 (t) and y = φ2 (t) are solutions of an ODE, y0 =
f (t, y), that satisfy initial conditions φ1 (t0 ) = y1 and φ2 (t0 ) = y2 , respectively. The proof of the uniqueness theorem is based on estimating how fast
the graphs of φ1 (t) and φ2 (t) can diverge from one another. This estimate
is of practical importance, beyond the proof of the uniqueness theorem, because initial conditions are often derived from measurements that can be
subject to error. The true initial value might be y1 and the measured value
might be y2 . The estimate tells us how far the true solution can diverge from
1.9. INITIAL VALUE PROBLEMS
91
the one based on a measurement error in the initial condition, in the worst
case.
In the case of a linear ODE,
y 0 − k ( t ) y = q ( t ),
we can compute d(t) = φ1 (t) − φ2 (t) precisely — no estimation is needed.
Since
d0 (t) = φ10 (t) − φ20 (t)
= (k(t)φ1 (t) + q(t)) − (k(t)φ2 (t) + q(t))
= k ( t ) d ( t ),
d(t) satisfies the homogeneous linear ODE y0 − k (t)y = 0. It follows that
d(t) = C eK(t) , where C is a constant and K (t) is an antiderivative of k (t).
Since d(t0 ) = y1 − y2 , we can calculate that C = (y1 − y2 )e−K(t0 ) and hence
d(t) = (y1 − y2 ) eK(t)−K(t0 ) .
This proves the uniqueness theorem in the linear case, because if y1 =
y2 , then clearly d(t) ≡ 0. When y1 6= y2 we can also draw interesting conclusions. For example, suppose it is known that K (t) → ∞ as t → ∞. Then
solutions with unequal initial conditions will diverge from one another. On
the other hand, if K (t) → −∞ as t → ∞, solutions will approach each other
and become indistinguishable as t → ∞.
Proof of the Uniqueness Theorem5
To see how fast solutions of a nonlinear ODE can diverge from one another,
we need two lemmas. The first allows us to replace an IVP with an integral
equation.
Lemma 1.9.1 The function φ(t) satisfies the IVP (1.54), (1.55) if and only if
φ ( t ) = y0 +
Z t
t0
f (u, φ(u)) du
(1.58)
P ROOF. By the fundamental theorem of calculus, for any differentiable
function φ,
φ ( t ) = φ ( t0 ) +
5 This
proof can be skipped if necessary.
Z t
t0
φ0 (u) du.
92
CHAPTER 1. FIRST-ORDER EQUATIONS
If φ(t) satisfies the IVP (1.54), (1.55) then we can replace φ(t0 ) with y0 and
φ0 (u) with f (u, φ(u)). This shows that φ(t) satisfies equation (1.58).
If φ(t) is a solution of equation (1.58) then φ(t0 ) = y0 because when t =
t0 the integral vanishes. Furthermore, the rule for differentiating integrals
shows that φ0 (t) = f (t, φ(t)); that is, φ(t) satisfies the ODE (1.54).
The second lemma is due to a Swedish-American mathematician, Thomas
Gronwall.
Lemma 1.9.2 (Gronwall’s inequality) Suppose g(t) is a function that is continuous on an interval [ a, b) and that g(t) ≥ 0 for all t ∈ ( a, b). Suppose also that
there are constants M ≥ 0 and C ≥ 0 such that the inequality
g(t) ≤ C + M
Z t
a
g(u) du
(1.59)
holds for all t ∈ [ a, b). Then g(t) ≤ C e M (t−a) for all t ∈ ( a, b).
P ROOF. Put
F (t) = e− M t
Z t
a
g(u) du.
By the product rule for differentiation and the Leibniz rule for differentiating integrals,
Z t
F 0 (t) = − M e− M t
g(u) du + e− M t g(t)
a
Z t
−M t
g(t) − M
g(u) du
= e
a
Referring to the inequality (1.59), it follows that F 0 (t) ≤ C e− M t for all t ∈
[ a, b). Since F ( a) = 0,
F (t) =
≤
Z t
a
Z t
a
F 0 (u) du
Ce− M u du
C −M t
(e
− e− M a )
M
C
= e − M t ( e M ( t − a ) − 1)
M
= −
Hence
Z t
a
g(u) du = e M t F (t)
≤
C M(t− a)
(e
− 1).
M
1.9. INITIAL VALUE PROBLEMS
93
Referring to the inequality (1.59) again, it follows that
g(t) ≤ C + M
C M(t− a)
(e
− 1) = C e M ( t − a ) .
M
The uniqueness theorem now follows from the following proposition.
Proposition 1.9.3 Let f (t, y) be continuous and satisfy a Lipschitz condition
with respect to y on a rectangular domain
D = {(t, y) : a < t < b, c < y < d}.
Let φ1 (t) and φ2 (t) be solutions of the ODE (1.54) that are defined on an interval
( p, q) ⊂ ( a, b), and let t0 ∈ ( p, q). Put y1 = φ1 (t0 ) and y2 = φ2 (t0 ). If c <
y1 , y2 < d then there is an interval (h, k ) such that for all t ∈ (h, k ),
|φ1 (t) − φ2 (t)| ≤ |y1 − y2 |eK|t−t0 | ,
where K is a Lipschitz constant for f on D .
P ROOF. Since φ1 and φ2 are continuous and c < φ1 (t0 ), φ2 (t0 ) < d
there is an interval (h, k ) ⊂ ( a, b) such that for all t ∈ (h, k ) the inequalities
c < φ1 (t), φ2 (t) < d hold. By lemma 1.9.1,
φi (t) = yi +
Z t
t0
f (u, φi (u)) du
for i = 1, 2. By subtracting we find
φ1 (t) − φ2 (t) = y1 − y2 +
Z t
t0
[ f (u, φ1 (u)) − f (u, φ2 (u))] du.
Assume t ∈ (t0 , k ). By the triangle inequality,
|φ1 (t) − φ2 (t)| ≤ |y1 − y2 | +
Z t
t0
| f (u, φ1 (u)) − f (u, φ2 (u))| du.
Since (u, φi (u)) ∈ D for u ∈ [t0 , t] the Lipschitz condition allows us to
substitute K |φ1 (u) − φ2 (u)| for | f (u, φ1 (u)) − f (u, φ2 (u))| in the integral to
get
|φ1 (t) − φ2 (t)| ≤ |y1 − y2 | + K
Z t
t0
|φ1 (u) − φ2 (u)| du.
94
CHAPTER 1. FIRST-ORDER EQUATIONS
By Gronwall’s inequality (lemma 1.9.2) it follows that for t0 ≤ t < k,
|φ1 (t) − φ2 (t)| ≤ |y1 − y2 | eK(t−t0 ) .
For t < t0 a symmetrical argument completes the proof.
To complete the proof of the uniqueness theorem we only have to apply
proposition 1.9.3 with y1 = y2 since the φi satisfy the same initial condition.
The conclusion is that |φ1 (t) − φ2 (t)| = 0 for all t ∈ (h, k ).
Proposition 1.9.3 also proves that the solution of an IVP depends continuously on the initial conditions.
Theorem 1.8 Suppose that f (t, y) is continuous and satisfies a Lipschitz condition with respect to y on a rectangular domain
D = {(t, y) : a < t < b, c < y < d}.
Let t0 ∈ ( a, b), and denote by φ(t, v) the solution of the IVP
y0 = f (t, y); y(t0 ) = v.
Given an interval (C, D ) with c < C < D < d, there is an interval (h, k )
containing t0 such that φ(t, v) is defined and continuous on the rectangle
D 0 = {(t, v) : h < t < k, C < y < D } ⊂ D .
While the proof of theorem 1.8 will not be given in its entirety it is worth
noting where proposition 1.9.3 fits in. To show φ(t, v) is continuous, we
need to show that the difference |φ(t, v) − φ(s, w)| can be made arbitrarily
small by choosing (s, w) sufficiently close to (t, v). By the triangle inequality,
|φ(t, v) − φ(s, w)| ≤ P + Q
where P = |φ(t, v) − φ(t, w)| and Q = |φ(t, w) − φ(s, w)|. We can make
Q small by choosing s close enough to t because φ(t, w), as a solution of
an IVP (with initial condition y(t0 ) = w) is continuous. Proposition 1.9.3
is used to show that P can be made arbitrarily small by choosing w close
enough to v.
The Picard proof of existence and uniqueness
An alternate proof of the existence and uniqueness of solutions of an IVP
y0 = f (t, y); y(t0 = y0 was published by Émile Picard in 1890. The proof
1.9. INITIAL VALUE PROBLEMS
95
based on lemma 1.9.1, and it constructs a sequence {φn (t)}n≥0 of approximate solutions that converge to a solution of the original IVP.
The first function in the sequence is constant: φ0 (t) ≡ y0 . For n > 0 the
function φn (t) is defined recursively as
φn (t) = y0 +
Z t
t0
f (s, φn−1 (s)) ds.
We’ll call this sequence the Picard sequence for the IVP. To summarize the
definition of the Picard sequence, the following observation is useful. Let
[ a, b] be an interval such that a < t0 < b, and let X [ a, b] denote the set of all
functions φ(t) that are defined and continuous on [ a, b]. For any φ ∈ X [ a, b]
let φ∗ = I(φ) ∈ X [ a, b] be the function defined by
∗
φ ( t ) = y0 +
Z t
t0
f (s, φ(s)) ds.
Then the recursive part of the definition of the sequence {φn } is just φn =
I(φn−1 ).
One can view I as a mapping
I : X [ a, b] −→ X [ a, b];
then lemma 1.9.1 can be restated as follows:
φ ∈ X [ a, b] has the property
I(φ) = φ
(1.60)
if and only if φ is a solution of the IVP y0 = f (t, y), y(t0 ) = y0 .
The property expressed in (1.60) may be stated verbally as “φ is fixed by I .”
Thus the existence problem for IVP’s can be rephrased as “Find a function
that is fixed by the transformation I .”
The following theorem is a tool from advanced calculus that allows us
to find functions that are fixed by certain transformations. It is stated in
terms of Banach spaces; our space X [ a, b] qualifies as a Banach space if,
given φ, ψ ∈ X [ a, b], we define
kφ − ψk = max{|φ(t) − ψ(t)| : t ∈ [ a, b]
.
A definition is necessary to proceed. Let X be a Banach space. (If you
haven’t heard of Banach spaces, please remember that the only Banach
96
CHAPTER 1. FIRST-ORDER EQUATIONS
space that we have in mind is X [ a, b].) A transformation T : X −→ X
is contractive if there is a constant C, with C < 1 such that for all φ, ψ ∈ X
the following inequality holds:
kT (φ) − T (ψ)k ≤ C kφ − ψk.
Intuitively, the transformation T brings elements of X closer together.
Theorem 1.9 (Banach Fixed Point Theorem) Let X be a Banach space and let
T : X −→ X be a contractive transformation. Then there is a unique φ̂ ∈ X that
is fixed by T .
To apply the Banach Fixed Point Theorem, we just have to show the
transformation I is contractive; then it will have a unique fixed point—and
as we have noted—fixed points of I and solutions of the IVP are the same
things.
The Picard proof works only if we assume that the function f (t, y) satisfies a Lipschitz condition with respect to y. In that case, let φ, ψ ∈ X [ a, b]
and denote I(φ), I(ψ) by φ∗ and ψ∗ , respectively. By definition of I ,
Z t
Z t
∗
∗
φ (t) − ψ (t) =
y0 +
f (s, φ(s)) ds − y0 +
f (s, ψ(s)) ds
t0
=
Z t
t0
t0
( f (s, φ(s)) − f (s, ψ(s))) ds
If L denotes the Lipschitz constant for f then for s ∈ [ a, b],
| f (s, φ(s)) − f (s, ψ(s))| ≤ L|φ(s) − ψ(s)|
In turn, for all s ∈ [ a, b] we have |φ(s) − ψ(s)| ≤ kφ − ψk, because kφ − ψk
is defined to be the maximum value of |φ(s) − ψ(s)| for s ∈ [ a, b]. Thus, for
all s ∈ [ a, b],
| f (s, φ(s)) − f (s, ψ(s))| ≤ Lkφ − ψk.
But
|φ∗ (t) − ψ∗ (t)| ≤
Z t
t0
| f (s, φ(s)) − f (s, ψ(s))| ds.
We can put the above two inequalities together to get
Z t
∗
∗
|φ (t) − ψ (t)| ≤ Lkφ − ψk ds = |t − t0 | Lkφ − ψk.
t0
1.9. INITIAL VALUE PROBLEMS
97
Thus, if we choose the interval [ a, b] to be of the form [t0 − h, t0 + h], so that
|t − t0 | ≤ h for all t ∈ [ a, b] then
kφ∗ − ψ∗ k ≤ hLkφ − ψk.
Thus, provided h < L−1 , the transformation I is contractive, and this completes a combined proof of existence and uniqueness of solutions of the
IVP y0 = f (t, y), y(t0 ) = y0 under the assumption that f (t, y) satisfies a
Lipschitz condition.
Because this proof depends critically on the Banach Fixed Point Theorem, a sketch of the proof is in order. See Exercise 27.
Exercises
In problems 1–8, solve the IVP, and find the largest interval on which the
solution is defined.
1.
2.
3.
4.
y0 = 0; y(1) = −2.
Answer
y0 − 5y = 25t; y(0) = 10.
y0 + 2y = sin 5t; y(0) = 0.
Answer
ty0 + y = et−1 ; y(1) = 1.
ty0 − 2y = t3 ; y(1) = 0.
Answer
√ 0
6.
ty − y = −t; y(1) = 2
5.
7.
8.
9.
y0 − 2ty = t; y(0) =
Answer
1
2
y0 − (tan t)y = sec3 t; y( π6 ) = 5.
Suppose that f (t, y) is continuous and satisfies a Lipschitz condition with respect to y on a rectangular domain D . Show that if
y = φ1 (t) and y = φ2 (t) are solutions of the ODE y0 = f (t, y), then
their graphs do not intersect each other in D , unless they are identical
in D . Hint: use the uniqueness theorem.
Answer
98
CHAPTER 1. FIRST-ORDER EQUATIONS
Solve the the ODE cos(t)y0 + y = 1, with initial condition y(0) =
10.
1.
11.
Let f (t) be a function that is continuous on (−∞, ∞), with f (t) 6=
0 for all t. Solve the IVP f (t)y0 + y = 1; y(0) = 1. Hint: The IVP in
exercise 10 is of this type.
Answer
In problems 12 – 21, find the set of initial points (t0 , y0 ) for which the
ODE, with initial value y(t0 ) = y0 satisfies the hypotheses of
(i) the existence theorem, and
(ii) the uniqueness theorem.
√
12.
y0 = t − y.
13.
y0 =
Answer
t
.
y2 +1
14.
yy0 = t2 + y2 .
15.
sin(t)y0 + y = 0.
Answer
16.
17.
(y0 )3 = t + 2y.
y = ty0 +
Answer
1
y0 .
y = ty0 + (y0 )2 .
√
3 y −1
0
19.
y = √ty .
Answer
18.
y0 = ln(t2 + y2 − 1) ln(9 − t2 − y2 ).
√
21.
y0 = 3 t − y
Answer
20.
22.
Determine which of the following functions satisfy a Lipschitz
condition with respect to y on the domain
D = {(t, y) : −1 < t < 1, −1 < y < 1},
and find a Lipschitz constant for those that do.
1.9. INITIAL VALUE PROBLEMS
99
= t − y2 .
= 4 t y.
= | y |.
= byc (byc is the greatest integer ≤ y).
= b t c.
p
23.
Show that f (t, y) = 1 − y2 satisfies a Lipschitz condition on the
domain
D = {(t, y) : a < t < b, c < y < d}
(a)
(b)
(c)
(d)
(e)
f (t, y)
f (t, y)
f (t, y)
f (t, y)
f (t, y)
if −1 < c and d < 1, but not if c = −1 or d = 1
Answer
24.
Use the uniqueness theorem to prove the equal derivatives theorem. Hint: Consider the IVP satisfied by f 1 (t) − f 2 (t).
25.
CAS exercise. Find an initial value y0 such that the solution of the
IVP
cos( x )y0 + y = sin( x ) + 2 sin(2x ); y(0) = y0
satisfies y(1) = 1.
Answer
26.
CAS exercise. Find an initial value y0 such that the solution of the
IVP
y0 + 2xy = 1, y(0) = y0
satisfies y(1) = 0.
A NSWER : y0 = −1.462651746.
27.
Let T : X −→ X be a contractive mapping with contraction constant C < 1.
(a) Show that T can leave at most one φ ∈ X fixed.
(b) Starting with any ψ0 ∈ X , recursively define, for n ≥ 1, ψn =
T (ψn−1 ). Show that for n, k > 0,
Cn
kψ1 − ψ0 k
1−C
(standard properties of Banach spaces show that this inequality
allows us to conclude that the sequence {ψn } converges, in the
sense that there is an element ψ∞ ∈ X such that
kψn+k − ψn k ≤
lim kψ∞ − ψn k = 0.
n→∞
100
CHAPTER 1. FIRST-ORDER EQUATIONS
(c) Show that if the limiting element ψ∞ exists, in the sense expressed
by the comment in part (b), then ψ∞ is fixed by t.
(d) Parts (a)–(c) show that the solution of the IVP
y0 = f (t, y); y(t0 ) = y0
is the limit of the Picard sequence—provided, of course, that f
satisfies the Lipschitz condition. For each of the following IVP’s,
construct the first four terms of the Picard sequence. Can you
guess the limiting solution?
i. y0 = y, y(0) = 1
ii. y0 = −t y, y(0) = 1.
√
iii. y0 = 3 y, y(0) = 0
iv. y0 =
2y
t ,
y (1) = 1
Answer
1.10
Nonlinear Growth Models
The linear model for population growth presented in section 1.2 was based
on an assumption that the relative growth rate of the population is constant.
It leads to the linear homogeneous ODE,
y0 = r y,
dy
in which r = 1y dt is a constant relative growth rate.
It is reasonable to expect the relative growth rate to decrease as resources become scarce. To include resource limits in our model we can
put
1 dy
= r ( y ),
y dt
where r (y) is a decreasing function of the population. The model thus leads
to an ODE,
dy
= y r ( y ).
(1.61)
dt
This ODE is not linear because r (y) is not a constant function.
Both the linear and the nonlinear models can be generalized by allowing the relative growth rate to depend on time. This is necessary to model
1.10. *NONLINEAR GROWTH MODELS
101
a population whose reproduction is seasonal. In this section, we will only
consider models that are nonlinear and independent of time.
The simplest way to impose resource limitations on a growth model is
to stipulate a carrying capacity M, representing the largest population that
the environment can support. If for some reason the population exceeds
the carrying capacity, it will decrease; and as long as the population is less
than the carrying capacity, it will increase.
In 1838, P. F. Verhulst presented such a model for population growth. In
his model , the relative growth rate decreases linearly from a value k when
the population is a negligible fraction of the carrying capacity to 0 when
the carrying capacity M is reached. The relative growth rate in this model
is
y
r (y) = k 1 −
.
M
When we substitute this r (y) in equation (1.61), the following ODE, known
as the logistic equation results:
y
y0 = k y 1 −
.
(1.62)
M
The logistic equation has two constant solutions, y = 0, and y = M.
The graphs of these constant solutions divide the t, y-plane into three regions. Below the line y = 0 and above the line y = M the right side of
equation (1.62) is negative. Therefore, any solution of the logistic equation
with initial conditions in one of these two regions will be decreasing. The
same reasoning shows that a solution with initial condition between the
two constant solutions must be increasing. The red curve in figure 1.19 is
one such solution. A nonconstant solution cannot cross either of the lines
y = 0 or y = M, because at the crossing point, the uniqueness theorem for
initial value problems would be violated.
The population values y = 0 and y = M are called stationary points of
the logistic equation, because when the initial population has either of these
values, it will remain constant (that is, stationary). The population y = M
is a stable stationary point because every solution whose initial value is
sufficiently close to M (any positive initial value will do) will converge to
M as t → ∞. On the other hand, y = 0 is an unstable stationary point,
because there are solutions with initial values y(0) arbitrarily close to 0 and
which diverge away from 0 as t → ∞.
In the following example, we will derive the general solution of the
logistic equation. The solution involves three parameters: M, k, and the
initial population; hence three data points are required to determine the
102
CHAPTER 1. FIRST-ORDER EQUATIONS
solution. To make the calculation as easy as possible, the data have been
taken at evenly spaced time intervals.
Example 1.10.1 Find the general solution of the logistic equation (1.62), and find
values of the parameters to fit the data y(0) = 1, y(1) = 2, y(2) = 3.
S OLUTION . The logistic equation is separable, and in separated form it
appears as follows:
1
k
dy =
dt.
y( M − y)
M
Integration by partial fractions yields
Z y 1
1
1
1
= k t + B,
+
dy =
ln M
y
M−y
M
M − y
M
where B denotes the integration constant. Thus
y
= ±ekt+ MB
M−y
Let’s assume our solution satisfies 0 < y < M; then the plus sign applies,
and we can solve for y to obtain
y=
M
,
1 + Ae−kt
where A = e− MB . To simplify further, set v = e−k ; then
y=
M
.
1 + Avt
Since v < 1, we have, limt→∞ y(t) = M. This confirms that the population
will converge to the carrying capacity as t → ∞. The parameters to be
evaluated are M, A, and v. The three data points yield the following
equations for these parameters:
M
1+A
M
1 + Av
M
1 + Av2
= 1
= 2
= 3
1.10. *NONLINEAR GROWTH MODELS
103
By the first equation, M = 1 + A which we put in the second equation and
solve for A to get A = (1 − 2v)−1 . Now substitute this and
M = 1+A = 1+
1
1 − 2v
in the third equation. After simplifying, the result is a quadratic equation,
3v2 − 4v + 1 = 0,
whose solutions are v = 13 and v = 1. Since v < 1, the second solution is
extraneous and v = 13 . The remaining parameters are readily evaluated:
A = 3 and M = 4, and the solution is
y=
4
.
1 + 31−t
The graph of this solution is shown in figure 1.20.
The classic application of the logistic equation to demography is a 1920
study by R. Pearl and L. Reed6 , who used the censuses of 1790, 1850, and
1910 to determine parameters for the logistic equation. They calculated
that the carrying capacity of the United States was 197,274,000 (in fact, the
actual population exceeded this value by 1970). Their solution of the logistic equation did not display a significant deviation from the census data
of the years 1800 - 1840 and 1860 - 1900, and was considered to be a great
success. Their predictions continued to be amazingly accurate until 1950.
Figure 1.21 shows the Pearl and Reed logistic curve, and the census data
covering the years 1790 – 2000. It is interesting to consider the causes of
failure for this model in the years after 1950.
General nonlinear growth models.
The key property of the growth models where the relative growth rate depends on the population is that they are based on an ODE
dy
= y r (y)
dt
6 Raymond
Pearl and Lowell Reed, “On the rate of growth of the population of the
United States since 1790 and its mathematical representation,” Proceedings of the National
Academy of Sciences, Washington vol. 6 (1920), pp. 275 - 288, http://www.pnas.org/
content/vol6/issue6/
104
CHAPTER 1. FIRST-ORDER EQUATIONS
that is autonomous ; that is, the independent variable t does not appear
on the right side. In general, an autonomous first order ODE will have the
form
dy
= g(y)
(1.63)
dt
Every autonomous ODE is separable, but we should not rush to integrate. Properties of solutions such as whether they are increasing or decreasing functions, the location of any asymptotes, etc. can be determined
without calculating an integral. We will be assuming that
(i) g0 (y) is continuous.
(ii) the zeros of g(y) are isolated. This means that if g(y j ) = 0, then there
is an open interval (y j − h, y j + h) that contains no other zeros of g.
The purpose of assumption (i) is to enable us to use the Existence and
Uniqueness Theorems (recall that if g0 (y) is continuous, then g(y) is also
continuous and satisfies a Lipschitz condition).
The zeros yi of g(y) are the stationary points of the ODE (1.63). Each
stationary point represents a constant solution y ≡ yi . It follows from assumption (ii) that the stationary points separate the real line into a collection of open intervals of the form (−∞, y1 ), (yi , yi+1 ), or (yn , ∞), and on
each of these intervals, g is nonzero, with no sign changes. If g(y) has no
zeros, there is just one interval, (−∞, ∞).
We will call an interval on which g(y) is positive an up interval, and an
interval where g(y) is negative is a down interval. It will be seen that the
range of any solution of the ODE (1.63) will be one of these intervals. If the
range is an up interval, the solution will be increasing—because in this case
y0 = g(y) is positive—and if the range is a down interval, then the solution
is decreasing.
y The stationary points of the logistic equation, g(y) = k y 1 − M , are
y1 = 0 and y2 = M. There is one up interval, (0, M), and two down intervals, (−∞, 0) and ( M, ∞). Figure 1.19 displays typical solutions whose
ranges are these intervals: note that the solution y = φ(t) with φ(0) ∈
(0, M) is strictly increasing, with φ(t) → M as t → ∞ and φ(t) → 0 as
t → −∞. If φ(0) lies in one of the two down intervals, then φ(t) is strictly
decreasing, but its graph has only one asymptote.
Proposition 1.10.1 Let y = φ(t) be a solution of an autonomous ODE of the
form (1.63). Then φ(t) is either a strictly increasing function, a strictly decreasing
function, or a constant function.
1.10. *NONLINEAR GROWTH MODELS
105
P ROOF. If φ(t) is not strictly increasing or decreasing, then it must have
a relative minimum or maximum. Thus there is a number t0 such that
φ0 (t0 ) = 0.
Set y0 = φ(t0 ). Since g(y0 ) = φ0 (t0 ) = 0, y0 is a stationary point. It
follows that y ≡ y0 and φ(t) are solutions of the same initial value problem, y0 (t) = g(y); y(t0 ) = y0 . By the Uniqueness Theorem, φ(t) ≡ y0 . We
have thus shown that if φ is neither an increasing function nor a decreasing
function then it must be constant.
To emphasize that proposition 1.10.1 is only applicable to autonomous
equations whose right sides g(y) are nice, let us consider the following
example.
Example 1.10.2 Explain why the following two examples are consistent with proposition 1.10.1.
(i) The function
y(t) =
t2 for t > 0,
0 for t ≤ 0.
√
is a solution of the ODE y0 = 2 y, and it is not strictly monotone.
2
(ii) The function y(t) = et is a solution of the ODE y0 = 2ty, and it is not
monotone.
S OLUTION .
√
(i) g(y) = 2 y is not differentiable at y = 0.
(ii) The ODE y0 = 2ty is not autonomous.
Our next observation will be that if φ(t) is a nonconstant solution of
equation (1.63) then the range of φ(t) will be an entire up interval (if φ is
increasing) or an entire down interval (if φ is decreasing). We reason first
that the range cannot overlap into two such intervals. If it did so, then the
graph of φ(t) would cross the graph of a constant solution corresponding
to a stationary point that separates the two intervals—a violation of the
uniqueness theorem. To see that the range of φ cannot be a proper subset of
an up or down interval, we will use the following proposition.
106
CHAPTER 1. FIRST-ORDER EQUATIONS
Proposition 1.10.2 Let y(t) denote a solution of equation (1.63). If
lim y(t) = L,
t→∞
where L is finite, then L is a stationary point of equation (1.63); and if N =
limt→−∞ y(t) is finite, then N is a stationary point.
P ROOF. If L is finite,
lim y0 (t) =
t→∞
lim g(y(t))
t→∞
= lim g(y)
y→ L
= g ( L ).
Thus, if g( L) 6= 0, then y0 (t) would have a nonzero limit as t → ∞. For
example, suppose g( L) = c > 0. Then for some number m, y0 (t) > c/2
for all t > m. Integrating, we would have y(t) > ct/2 + d (where d is an
integration constant) for t > m. This contradicts the fact that y(t) converges
to a finite number L as t → ∞. We can reach the same sort of contradiction
if we assume that g( L) < 0; hence g( L) = 0.
The proof that N is a stationary point if it is finite is the same.
Propositions 1.10.1 and 1.10.2 are the key ingredients of the proof of the
following theorem.
Theorem 1.10 Let y = φ(t) be a solution of the autonomous equation y0 = g(y),
where g(y) has isolated zeros and g0 (y) is continuous everywhere. Then either φ is
constant, or φ is a strictly increasing function whose range is an entire up interval,
or φ is strictly decreasing, with range an entire down interval.
With the aid of theorem 1.10, we can sketch graphs of solutions of an
autonomous ODE like equation (1.63) without integrating. We start by
drawing on the y-axis a phase diagram for the ODE, as follows. Locate
the stationary points yi by solving the equation g(y) = 0. These will be the
endpoints of the up and down intervals. Mark the stationary points on the
y-axis and draw horizontal lines to show the constant solutions. Test the
sign of g(y) by evaluating g at a convenient point in each interval (yi , yi+1 )
on the y-axis to see if it is an up or down interval. Finally, mark each up
interval with an arrow directed upward, and each down interval with a
downward arrow.
When the phase diagram and stationary solutions have been drawn, it
is easy to sketch graphs of the nonconstant solutions. All solutions with
1.10. *NONLINEAR GROWTH MODELS
107
initial points in a given up or down interval will be strictly increasing or
decreasing, with ranges covering the entire interval. Stationary points at
the ends of the up and down intervals will be horizontal asymptotes of the
solution.
Example 1.10.3 Draw the phase diagram for the ODE
y0 = 4y − y3 ,
(1.64)
and make a sketch of the graphs of the solutions of the four initial value problems
with y(0) = ±1, ±3.
S OLUTION . The stationary points are determined by solving 4y − y3 = 0 to
obtain y1 = −2, y2 = 0, and y3 = 2. The up and down intervals are
therefore (−∞, −2), (−2, 0), (0, 2), and (2, ∞). To determine the direction
of (−∞, −2), evaluate g(y) at an arbitrarily chosen point in the interval:
f (−3) = 15. Since we have found a positive value, this is an up interval,
and is to be marked accordingly. The other three intervals are marked as
follows. Since f (−1) < 0, (−2, 0) is a down interval; f (1) > 0 so (0, 2) is
an up interval; and, f (3) < 0, indicating that (2, ∞) is a down interval. The
phase diagram is imposed on the y-axis of figure 1.22.
The stationary solutions are indicated by the horizontal lines in figure 1.22.
Each nonconstant solution is asymptotic to the constant solutions
corresponding to the boundary of its range. Hence the solution of the
initial value problem with y(0) = 1 is asymptotic to the t-axis as t → −∞
and to the line y = 2 as t → ∞. In addition, the solution is strictly
increasing and has a slope of 3 at its crossing of the y-axis. With this
information, it is not difficult to make a sketch of the solution.
Similarly, if y(0) = −1, then y(t) is decreasing, asymptotic to the t-axis as
t → −∞, and to y = −2 as t → ∞. If y(0) = 3, y(t) is again decreasing,
and asymptotic to y = 2 as t → ∞; if y(0) = −3, then y(t) is increasing
and asymptotic to y = −2.
Stability
Consider an isolated stationary point y1 of an autonomous ODE y0 = f (y).
Since y1 is isolated, there is an interval ( a, b) containing y1 and no other
stationary points. We will say that y1 is a stable stationary point if all solutions φ(t) with initial values in ( a, b) converge to y1 as t → ∞. If there
108
CHAPTER 1. FIRST-ORDER EQUATIONS
are solutions with initial values in ( a, b) that do not converge to y1 , then we
would say that y1 is an unstable stationary point.
Figure 1.23 displays three phase diagrams in which the points yi are stationary points of autonomous ODEs. The intervals immediately above and
below the stationary point y1 are both directed toward it, so that stationary
point is stable. In fact, the interval I consisting of the point y1 , the up interval below it, and the down interval above it has the required property
that every solution with initial point in I converges to y1 as t → ∞. If one
or both of the intervals adjacent to a stationary point yi are directed away
from yi , then yi is unstable. In these cases, if we only know that the initial
point y0 is near to an unstable stationary point yi , but do not know whether
y0 < y1 or y0 > y1 , it is impossible to predict the limit of the function y(t).
The following proposition offers an alternative way to determine the
stability of a stationary point.
Proposition 1.10.3 Let y1 be a stationary point of the autonomous ODE y0 =
g(y) such that g0 (y1 ) 6= 0. Then, y1 is stable if g0 (y1 ) < 0, and is unstable if
g0 (y1 ) > 0.
The proof of proposition 1.10.3 is left as an exercise. (See problem 25 at
the end of this section.) In the following example, we will use the proposition to determine the stability of each stationary point of an ODE, and then
use the stability information to draw a phase diagram.
Example 1.10.4 Determine the stability status of each stationary point of
y0 = sin(y) −
2
y,
π
and draw the phase diagram.
S OLUTION . Figure 1.24 shows the graphs of u = sin(v) and u = 2v/π.
They cross at v = 0, and v = ±π/2; therefore the set of stationary points
is {−π/2, 0, π/2}. Since f 0 (0) = cos(0) − 2/π is positive, 0 is unstable.
f 0 (±π/2) = cos(π/2) − 2/π < 0, so the other two stationary points are
stable.
It follows that the intervals (0, π/2) and (−π/2, 0) are directed away from
the stationary point at 0: (0, π/2) is up and (−π/2, 0) is down. Since the
equilibria at ±π/2 are stable, adjacent intervals are directed toward these
points. Therefore (−∞, −π/2) is an up interval and (π/2, ∞) is a down
1.10. *NONLINEAR GROWTH MODELS
109
interval. The phase diagram is drawn on the y−axis of figure 1.24 (that is
the horizontal axis in this case). When sketching graphs of solutions of
y0 = sin(y) − 2y/π, the y-axis would be vertical, and the phase diagram in
figure 1.24 would be rotated counterclockwise 90◦ before sketching.
Proposition 1.10.3 does not apply to any stationary point y1 where f 0 (y1 ) =
0. Exercise 26 shows that in this case, it is possible for a stationary point to
be either stable or unstable.
Dependence on Parameters
The population P of codfish in a certain marine fishery is modeled by a
modified logistic equation,
P0 = kP(1 − P/M ) − H.
(1.65)
The growth parameter k and the carrying capacity M are taken from the
logistic equation, while H is the rate at which fish are harvested. We are
interested in how the fate of the fish population depends on the parameter
H.
Let f ( P) = −kP2 /M + kP − H be the quadratic expression on the right
side of equation (1.65). Figure 1.25 shows graphs of f ( P) representing four
different harvest rates.
Solving f ( P) = 0 with the quadratic formula, we find that the stationary points are
"
#
r
1
4H
P1 , P2 = M 1 ± 1 −
,
(1.66)
2
kC
provided that these are real numbers. If there are no real stationary points,
then f ( P) is negative for all P, as in the graph corresponding to H = 12000
in figure 1.25. Thus (−∞, ∞) is a down interval, and the codfish will be
extinct when the population reaches 0. If there are two distinct stationary
points, as in the graphs with H = 0 and H = 5000 of the figure, P1 will be
located between 0 and 12 M, and P2 will be between 21 M and M. The interval ( P1 , P2 ) is an up interval, and (−∞, P1 ) and ( P2 , ∞) are down intervals.
Hence P2 is stable, and will represent the limiting population.
We can identify a critical situation when P1 and P2 merge together as one
stationary point, as in the graph corresponding to H = 10000 of figure 1.25.
The intervals above and below are both down intervals, and the stationary
point is unstable, with extinction looming. The quadratic equation f ( P) =
110
CHAPTER 1. FIRST-ORDER EQUATIONS
0 has a double root when
1−
4H
= 0;
kC
that is, when H = 4k M. This is the critical harvest rate.
Equation (1.65) is actually an infinite family of ODEs depending on a
parameter H. The solutions undergo a fundamental change, called a bifurcation at a critical value H0 of the parameter. It is best to describe this
situation with a two-dimensional diagram called a bifurcation diagram,
as shown in figure 1.26. The horizontal axis represents the value of the H,
which is called the bifurcation parameter in this context. The vertical axis represents the population. The stationary points are considered as functions
P1 ( H ) and P2 ( H ) and plotted on the diagram. By convention, stable equilibria are depicted by solid curves; unstable equilibria by dashed curves.
Vertical arrows indicate whether the population is increasing or decreasing.
Exercises
1.
A lake can support a population of 1000 fish. There are now 600
fish in the lake, and on the same date last year there were 300. Assuming that the logistic model determines the fish population in the
lake, how many fish will be in the lake a year from now?
Answer
2.
Suppose that the lake in problem 1 has been stocked, so that now
it contains 1200 fish. How many will be in the lake a year from now?
3.
Solve the initial value problem:
y0 = .02y(200 − y); y(0) = 10.
Answer
4.
The populations of two communities are governed by the logistic
equation,
u
u0 = k u 1 −
.
M
The second community has a carrying capacity twice as large as the
first, and the growth constant k for the each community is the same.
Show that if the initial population of the second community is twice
the initial population of the first, then the second community will
always have twice the population of the first.
1.10. *NONLINEAR GROWTH MODELS
111
Let’s update Pearl and Reed, whose predictions of the US population
were based on census data for the years 1790, 1850, and 1910. We will
move our data points to 1890, 1950, and 2010, respectively and answer the
same questions that they did a century ago.
Problems 5 – 8 are based on the following data concerning the population of the United States.
Year
1890
1950
2010
Population (millions)
63.0
151.3
309.4
5.
Find a solution of the logistic equation to fit the data.
Answer
6.
What is the predicted carrying capacity, according to the logistic
model?
7.
When will the population reach a level of 500 million?
Answer
8.
Use the logistic model to predict the population in 2040.
9.
A rumor that a top psychic has predicted the sky is falling is
spreading in Gossipville. Each citizen of that town calls ten others per
day at random to chat; if one of the parties has heard the rumor and
the other hasn’t, the rumor spreads. The population of Gossipville
is 10,000, and at this point 1000 have already heard the rumor. How
long will it be before 90% of the people in Gossipville have heard the
rumor?
Hint: Let N (t) be the number of people who have been warned about
the sky. Show that N can be approximated by a continuous variable
which satisfies the logistic equation. Of course, it is necessary to figure out the values of M and k.
Answer
10.
The growth of a certain population is limited by two resources:
food and water. The food resources will support a population of M,
and there is enough water for a population of 2M. A researcher postulates the following nonlinear growth model, which accounts for both
limitations
y0 = k ( M − y)(2M − y)y.
112
CHAPTER 1. FIRST-ORDER EQUATIONS
Draw a phase diagram and show that the population has three stationary values, but that only one is stable. Do all solutions converge
to the same stable population? What happens if the population initially exceeds 2M? Does this make sense?
Problems 11 – 14 refer to an alternative to the logistic model for population growth, the Gompertz model.
It is based on the ODE
M
0
y = ky ln
,
y
where M is the carrying capacity and k is constant.
11.
Show that the Gompertz model has, like the logistic model, two
stationary populations, y ≡ 0 and y ≡ M, where y ≡ 0 is unstable,
and that y ≡ M is stable.
Answer
12.
Show that the substitution v = ln(y) produces a linear ODE with
v as the dependent variable.
13.
Given the data y(0) = 1, y(1) = 2, and y(2) = 3, determine the
carrying capacity according to the Gompertz model. It is interesting
to compare this result with that of example 1.10.1.
Answer: 5.31261.
Details
14.
Use the following data to find the carrying capacity of the United
States, according to the Gompertz model (These are the data used by
Pearl and Reed in the study cited above)
Answer: 18.35 billion
Year
1790
1850
1910
Population (millions)
4
23
92
A chemical reaction proceeds at a rate determined by the concentration
of the reactants and catalysts. Specifically, if y(t) denotes the concentration of one of the products of the reaction, and u(t), v(t), etc. denote the
1.10. *NONLINEAR GROWTH MODELS
113
concentrations of the reactants and catalysts, then
dy
= k[u(t)]n [v(t)]m · · ·
dt
where the exponents n, m, etc. are called the orders of the reaction with
respect to the corresponding reactant or catalyst. The number k is called
the rate constant.
15.
16.
In an autocatalytic reaction, the product acts as a catalyst to further
the reaction. Assume that an autocatalytic reaction has order 1 with
respect to the reactant and the catalyst, and that for each molecule
of the reactant that is consumed, one molecule of the catalyst is produced. If the reaction is allowed to proceed in a closed system, show
that its rate is governed by the logistic equation.
Answer
The reaction of gasses
2NOBr → 2NO + Br2
is second order with respect to the concentration of NOBr. In an
experiment, a 1 liter flask initially contains 2.7 × 1022 molecules of
NOBr. After 2 minutes, 1022 molecules of the reactant are left. Determine the rate constant for this reaction. When was exactly half of the
NOBr that was originally present consumed?
Answer: 1.2 minutes.
17.
Which of the following equations are autonomous?
(a) y0 = 1 − y.
(b) y0 = sin(t) + y2 .
y for 0 < t < 1
0
(c) y =
0, otherwise.
y for 0 < y < 1
(d) y0 =
0, otherwise.
Answer
Find the stationary points of the ODEs in problems 18 – 22 and draw
their phase diagrams.
18.
y0 = 4 + 3y − y2 .
114
19.
20.
21.
22.
23.
CHAPTER 1. FIRST-ORDER EQUATIONS
y 0 = y2 .
Answer
y0 = sin(y) − 12 .
y0 = ey
Answer
y0 = y − cos(y)
Which of the following functions are solutions of first order autonomous ODEs? In the case of those that are solutions, where would
the stationary points be located?
(a) y = 1 + t. Domain, (−∞, ∞).
(b) y = 1 + t2 . Domain, (−∞, ∞).
(c) y = e−2t . Domain, (−∞, ∞).
(d) y = tan(t). Domain, (− π2 , π2 ).
(e) y = sin(t). Domain, (−∞, ∞).
Answer
24.
The set of stationary points for an ODE y0 = f (y) is {0, 1, 2, 3, 4}.
Suppose that f 0 (0) = 2, f 0 (1) = −5, f 0 (2) = 0, f 0 (3) = −1, and
f 0 (4) = 12. Determine the stability of each stationary point, and draw
the phase diagram.
25.
Prove proposition 1.10.3. [Hint: Show that the hypothesis implies
g(y) changes sign at y1 .]
Answer
26.
An stationary point y1 of the ODE y0 = g(y) is said to be degenerate if f 0 (y1 ) = 0. For example, every point is a degenerate stationary
point of the equation y0 = 0. Give an example of a ODE with a degenerate, isolated, stable stationary point, and an example of a ODE
with a degenerate, isolated, unstable stationary point.
27.
Show that an ODE whose phase diagram is identical to the phase
diagram for y0 = y2 has a degenerate stationary point at 0.
Answer
1.10. *NONLINEAR GROWTH MODELS
28.
115
Use proposition 1.10.3 to determine the stability of 0 as an stationary point of
y0 = sin(y) − ky,
in the case where the parameter k > 1. Also determine the stability
when k < 1. What can be said when k = 1? Draw phase diagrams
k = 1, for k slightly less than 1, and for k slightly greater than 1.
29.
For the growth equation with harvesting,
P 0 = P (1 − P ) − H
(a) What is the critical harvest rate?
(b) Suppose H = 0.1, and P(0) = 1. What will be the limiting population?
(c) Draw the bifurcation diagram.
Answer
30.
The following ODE depends on a parameter m:
y0 = y2 − m.
(a) For which values of m are there stationary points?
(b) Draw the bifurcation diagram.
31.
Fitting a logistic curve to data. In example 1.10.1 we found that
we could find a logistic curve passing through three points in the
plane. What if there are more than three points? In this case, we
cannot expect the logistic curve to pass through all of the points, but
we can try to find parameters that will give the best logistic curve for
the data. The following method works well in cases where the logistic
equation is an appropriate model for population growth, such as the
U.S. population, 1790 – 1950.
Write the logistic equation P0 = kP(1 − P/M ) in the form
1 dP
= mP + b,
P dt
where m = −k/M and b = k. If we know the population at times
t0 − h, t0 , and t0 + h, then we can approximate the relative growth
rate as follows:
P ( t0 + h ) − P ( t0 − h )
1 dP ≈
.
(1.67)
P dt
2hP(t0 )
t = t0
116
CHAPTER 1. FIRST-ORDER EQUATIONS
(a) Let P(t) denote the U.S. population (see table 1.3) in year t. Tabulate, for t = 1800, 1810, 1820, . . . , 1940, P(t) and ( P(t + 10) −
P(t − 10))/(20P(t)).
(b) Using a statistical calculator or a spreadsheet7 , find the least
squares line that best fits the data, with the population as the
independent variable and the approximate relative growth rate,
as calculated by equation (1.67), as the dependent variable.
(c) The result of part (b) will be the slope m and the intercept b of the
least squares line. Use these to determine the carrying capacity.
(d) Use a CAS (or a pencil and paper) to find the solution of
P0 = P(mP + b)
with initial condition P(1900) = 76.21 million. Plot a graph of
the solution curve extending from the years 1790 to 2010. Include the census data, as was done in figure 1.21.
Answer
1.11
Review Exercises
Find the general solution of each of the ODEs in problems 1 – 12.
1.
y0 = ty + t + y + 1.
See Answer
2.
y 0 = y − t2 .
See Answer
3.
y0 = y + et .
See Answer
4.
y0 + tan(t)y = cos2 t.
See Answer
5.
y0 = et − y.
See Answer
6.
ty0 + y = et sin 2t.
See Answer
7 Statistical
calculators and spreadsheets have least squares functions that you may use.
1.11. REVIEW EXERCISES
7.
ty0 + 12y = 5t2 + 3t − 2.
See Answer
8.
y0 + tan(t)y = 1.
See Answer
9.
y0 = ty + 1.
See Answer
5y
10.
y0 = t
See Answer
11.
y0 + y = e−t
See Answer
12.
y0 = y cos t.
See Answer
In problems 13 – 6, solve the initial value problem.
13.
t+y
y0 = t ; y(1) = 2.
See Answer
14.
y0 + y = e−t ; y(0) = 0.
See Answer
15.
y0 = 2t; y(2) = 4.
See answer
16.
y0 + 4y = 3e−4t sin 3t; y(0) = 0.
See answer
ty0 − 3y = 5t3 ; y(1) = 1.
See answer
√
18.
y0 = 3 y; y(1) = 1.
See answer
17.
19.
y0 = (t + 3)y + 2t + 6; y(0) = 0.
See answer
20.
y0 = (t + 3)y + 2t + 6; y(0) = −2.
See answer
117
118
CHAPTER 1. FIRST-ORDER EQUATIONS
2y
√
y(0) = 3 2.
21.
y 0 − 4− t2 =
See answer
22.
y0 + ty = 2t + t3 ; y(0) = 1.
See answer
23.
A the balance of bank account bearing continuous compound interest at a constant rate triples in 30 years, with no deposits or withdrawals. How long did it take to double?
See answer
24.
If the world’s population follows the linear growth model with a
relative growth rate of 1.3% (0.013), how long will it take the population to double?
See answer
25.
A piece of wood contains 0.8 ppb 14 C, and when it was part of a
living tree, it contained 1 ppb. How old is it? The half-life of 14 C is
5730 years.
See answer
26.
Find the general solution of y0 + y = cos(4t). Is there a stable
periodic solution?
See answer
27.
A tank initially contains 1000 liters of brine with a concentration
of 100 grams per liter. Brine, containing 50 grams salt per liter, is
pumped into the tank at 200 liters per hour, and the well-mixed solution is pumped out at 100 liters per hour. After 10 hours, the tank
overflows. What is the concentration of salt in the tank at that time?
See answer
28.
Print a copy of the direction field shown below, and sketch on
it the graphs of the solutions of y0 = t2 − y2 , with initial conditions
y(−2) = 0 and y(2) = 0, respectively.
√1 ;
2− t
1.11. REVIEW EXERCISES
119
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-1
0
1
2
3
Answer
29.
Use Euler’s method with time step h = 0.25 to compute an approximate solution of the IVP, y0 = t2 − y2 ; y(0) = 0. Draw the polygonal graph of the approximate solution on your copy of the direction
field accompanying exercise 28.
Answer
120
CHAPTER 1. FIRST-ORDER EQUATIONS
30.
Sketch the graph of the solution of the IVP, y0 = t −
a hard copy of the direction field shown below.
√
y; y(0) = 2 on
4
3
2
1
0
-2
0
2
4
Answer
31.
Use Euler’s method with time step h = 0.5 to approximate the so√
lution of the IVP y0 = t − y; y(−2) = 4. Draw the polygonal graph
of this solution on your copy of the direction field accompanying exercise 30.
Answer
1.11. REVIEW EXERCISES
32.
121
Match the direction fields shown below with the ODEs.
(a) y0 = 0.03(5t − y2 ).
(b) y0 = sin(y).
(c) y0 = sin(t + y).
(d) y0 = −0.01y
(e) y0 = 0.05y(π − y).
(f) y0 = 0.02(t2 + y2 ).
I
II
III
IV
V
VI
Answer
In exercises 33 — 40, find a family of solutions and note any singular
solutions.
33.
y0 = ty2 .
Answer
122
34.
CHAPTER 1. FIRST-ORDER EQUATIONS
y0 = ey .
Answer
y 0 = y (1 − y ).
Answer
q
1− y2
36.
y0 = 1−t2 ; −1 < t < 1.
Answer
q
y2 −1
37.
y0 = t2 −1 ; |t| > 1.
Answer
35.
y 0 = t (1 + y2 ) / (1 + t2 ).
Answer
q
y
39.
y0 = t
Answer
38.
40.
y0 = e4y tan(3t).
Answer
41.
An object that is sinking in water is subject to forces due to gravity
(downward), and buoyancy and drag (both upward). By the Principle of Archimedes, the buoyancy force is equal to the weight of water
displaced by the object. Thus, if σ denotes the specific gravity of the
mg
object, the buoyancy force is equal to σ . The drag force is proportional to the square of the velocity. A diamond (σ = 3.5) has a terminal velocity of 21 meters per second when sinking in water. If you
toss it gently into the middle of Lake Tanganiyika (1463 meters deep),
(a) how far will the diamond travel in the first 10 seconds?
(b) Estimate how long it will take for the diamond to get to the bottom of the lake.
(c) Suppose that you throw the diamond vertically into the lake so
that its initial velocity after it is immersed in water is 42 meters
per second. Answer questions (a) and (b).
Answer
42.
A certain population is growing according to the logistic equation.
In year A, the population is 100,000 and growing at 4500 per year, and
in year B the population is 200,000, growing at 8000 per year.
1.11. REVIEW EXERCISES
123
(a) Calculate the relative growth rate in each of the years.
(b) Find the carrying capacity.
(c) If year A is 1980, when was year B?
Answer
In problems 43 — 48, draw phase diagrams for the ODEs, identify all
stable equilibrium points, and and describe the limiting behavior of the
solution of the IVP with increasing time. Do not solve these ODEs.
43.
y0 = y (2y + 1); y(0) = −1.
Answer
44.
y0 = −y sin2 (y); y(0) = 30.
Answer
45.
y0 = y2 + 2y + 2; y(0) = 0.
Answer
46.
y0 = 5 − 3 y; y(0) = −2.
Answer
47.
y0 = y−1 − 1; y(0) = 3.
Answer
48.
y0 = y−1 − 1; y(0) = −3.
Answer
In problems 49 – 51, it will be useful to differentiate both sides of the
ODE to obtain an expression for y00 . You don’t need formulas for the solutions to do these problems!
49.
Show that the graph of any solution of y0 = y1 is either increasing
and concave down, or decreasing and concave up.
Answer
50.
Imagine that the set of points in the plane satisfying the inequality
t + y + 1 > 0 is colored red, and the set of points satisfying t + y + 1 <
0 is colored blue. Show that if y(t) is a solution of the ODE y0 = t + y,
then the graph of y(t) is concave up at red points, and concave down
at blue points.
Answer
124
CHAPTER 1. FIRST-ORDER EQUATIONS
51.
Show that the graph of any solution of y0 = ey is increasing and
concave up. What can be said about the graph of a solution of y0 =
e−y ?
Answer
52.
Richardson’s extrapolation. Let φ(t) be the solution of the IVP
y0 = t − y2 ; y(0) = 0.
(a) Using a spreadsheet or other computer version of Euler’s method
with h = 0.1, 0.05, and 0.025, compute approximate values of
φ (1).
(b) Richardson’s extrapolation works with the assumption that the
accumulated error in each of these approximations is approximately equal to Ch where C is an unknown constant. If Yh1
and Yh2 are approximations of φ(1) obtained using steps h1 and
h2 = h1 /2, respectively,
φ(1) ≈ Yh1 + Ch1
1
φ(1) ≈ Yh2 + Ch1 .
2
(1.68)
(1.69)
Eliminate the constant C from (1.68) and (1.69), to find a better approximation of φ(1). Let Zh1 denote the approximation of
φ(1) thus obtained. This Zh1 is called a first-order extrapolated
approximation.
(c) Calculate Z0.1 and Z0.05 .
(d) Further analysis shows that there is a constant D such that
φ(1) ≈ Zh + Dh2 .
Use the values of Z (0.1) and Z0.05 just calculated to eliminate the
constant D, thus obtaining a second-order extrapolated approximation.
A NSWER
SUMMARY:
The second-order extrapolation yields φ(1) ≈ 0.4555383.
This is considerably closer to the correct value (found by solving the initial
value problem using a CAS), φ(1) = 0.4555445 than Y0.025 = 0.4477666.
Full Answer
53.
For which values of y0 does the IVP
s
1+y+t
0
y = 3
; y (0) = y0
1−y+t
1.11. REVIEW EXERCISES
125
have a solution? A unique solution?
Answer
54.
The IVP y0 = y sin(t − y), y(0) = 0 has the solution y ≡ 0. How
can you be sure that it’s the only solution?
Answer
126
CHAPTER 1. FIRST-ORDER EQUATIONS
Figure 1.1: Several solutions of the ODE v0 = g − kv2 . Each graph represents the velocity
of a falling object subject to air resistance. The values g = 9.8 meters/second2 and k =
9.8 × 10−4 kilogram /meter were used.
120
100
80
60
40
20
5
10
15
20
FIGURES FOR CHAPTER 1
127
Figure 1.2: Exponential growth: y = ekt (k > 0). Vertical lines mark the doubling time.
y
10
8
6
4
2
t
2
4
6
8
10
12
128
CHAPTER 1. FIRST-ORDER EQUATIONS
Figure 1.3: Exponential decay: y = ekt (k < 0). Vertical lines are at intervals of one half
life.
y
1
0.8
0.6
0.4
0.2
t
2
4
6
8
10
12
FIGURES FOR CHAPTER 1
129
Figure 1.4: Temperature in a heated building (black curve), and the stable periodic temperature (blue curve). The outdoor temperature is indicated by the red curve. See example 1.3.3.
30
25
20
15
10
5
10
-5
20
30
40
130
CHAPTER 1. FIRST-ORDER EQUATIONS
Figure 1.5: Solutions of y0 + 3y = 2e−t . See example 1.3.1.
2
1.5
1
0.5
-2
-1
1
-0.5
-1
-1.5
-2
2
FIGURES FOR CHAPTER 1
131
Figure 1.6: A typical mixture problem.
INPUT:
K grams solute per
liter
@
@ J liters/second
@
@
@
@
R
@
MIXING TANK
x (t) grams solute in
V (t) liters
@
@ L liters/second
@
@
@
@
R
@
OUTPUT
132
CHAPTER 1. FIRST-ORDER EQUATIONS
Figure 1.7: The velocity of a falling object.
50
40
30
20
10
5
10
15
20
FIGURES FOR CHAPTER 1
133
Figure 1.8: Solutions of the differential equation y0 = t/y. The solution
with initial value y(0) = −1 is shown in blue.
2
1.5
1
0.5
-4
-2
2
-0.5
-1
-1.5
-2
4
134
CHAPTER 1. FIRST-ORDER EQUATIONS
Figure 1.9: Family of solutions of a separable equation y0 = 3y2/3 . See example 1.6.4.
1
0.75
0.5
0.25
-2
-1
1
-0.25
-0.5
-0.75
-1
2
FIGURES FOR CHAPTER 1
135
Figure 1.10: Direction field for y0 = t − y2
1
0.5
0
-0.5
-1
-1.5
-1
-0.5
0
0.5
1
1.5
136
CHAPTER 1. FIRST-ORDER EQUATIONS
Figure 1.11: Direction fields corresponding to models for the velocity of a
falling object. On the left, the drag coefficient is constant, and on the right,
the drag coefficient decreases with time.
40
40
35
35
30
30
25
25
0
1
2
3
4
5
6
0
1
2
3
4
5
6
FIGURES FOR CHAPTER 1
137
Figure 1.12: Approximate solution (in red) by Euler’s method, the exact
solution (in green), and the direction field for y0 = t − y2 .
1
0.5
0
-0.5
-1
-1.5
-1
-0.5
0
0.5
1
1.5
138
CHAPTER 1. FIRST-ORDER EQUATIONS
Figure 1.13: Approximate solution of y0 = y + 0.25; y(0) = 0 (in blue) by
Euler’s method. The actual solution φ(t) (in red), and the graphs of φm (t)
for m = 1, 2, 3 are also shown.
1.5
1.25
1
0.75
0.5
0.25
0.5
1
1.5
2
FIGURES FOR CHAPTER 1
139
Figure 1.14: Velocity of a rock sinking in seawater, determined by Euler’s
method. See example 1.8.3.
2
1.5
1
0.5
0.25
0.5
0.75
1
1.25
1.5
1.75
2
140
CHAPTER 1. FIRST-ORDER EQUATIONS
Figure 1.15: Spreadsheet solution of equation (1.50). Column A contains
the t values, column B, the y-values. ∆t appears in column C, and ∆y in
column D. The graph is an XY-chart with data series from column B.
FIGURES FOR CHAPTER 1
141
Figure 1.16: The blue curve is the solution of the IVP 1.56; the red curve
is part of the graph of the equation y = 1/(t + 1) and satisfies the ODE
y0 = −y/(t + 1), but is not considered to be part of the solution of the IVP
with y(0) = 1.
3
2
1
-3
-2
-1
1
-1
-2
-3
2
3
142
CHAPTER 1. FIRST-ORDER EQUATIONS
Figure 1.17: Solutions of y0 = ty2 .
200
175
150
125
100
75
50
25
-0.6
-0.4
-0.2
0.2
0.4
0.6
FIGURES FOR CHAPTER 1
143
Figure 1.18: Two solutions of the IVP y0 = 3y2/3 ; y(0) = 0.
2
1.5
1
0.5
0
-0.5
-1
-1.5
-4
-2
0
2
4
144
CHAPTER 1. FIRST-ORDER EQUATIONS
Figure 1.19: Solutions of the logistic equation satisfying the initial conditions y(0) = 2, y(0) = 41 , and y(0) = −1. The parameters k and M were
0.04 and 1, respectively.
y
3
2
1
t
-50
50
-1
-2
100
150
FIGURES FOR CHAPTER 1
145
Figure 1.20: Solution of the logistic equation with parameters chosen to fit
the data given in example 1.10.1.
y
4
3
2
1
-1
1
2
3
4
5
6
t
146
CHAPTER 1. FIRST-ORDER EQUATIONS
Figure 1.21: Population of the United States in millions, censuses of 1790 –
2000. The logistic curve resulting from Pearl and Reed’s study is superimposed, and the data points used in their study are shown with diamondshaped marks. The dashed line represents the carrying capacity that Pearl
and Reed projected.
250
200
150
100
50
1850
1900
1950
2000
FIGURES FOR CHAPTER 1
147
Figure 1.22: Solutions of the ODE y0 = 4y − y3 . The phase diagram is
superimposed on the y-axis.
4
2
-1.5
-1
-0.5
0.5
-2
-4
1
1.5
148
CHAPTER 1. FIRST-ORDER EQUATIONS
Figure 1.23: Stationary points: On the left, y1 is stable. In the middle and
right diagrams, y2 and y3 are unstable.
?
s y1
6
s y2
6
6
6
s y3
?
FIGURES FOR CHAPTER 1
149
Figure 1.24: Graphs of u = sin(v) and u = 2v/π.
3
2
1
-4
-2
2
-1
-2
-3
4
150
CHAPTER 1. FIRST-ORDER EQUATIONS
Figure 1.25: Population growth rates, as functions of the population, for
various rates of harvest. In each graph, the growth parameter has been set
at k = 0.4 per individual organism per year, and the carrying capacity at
M = 105 .
H=0
H=5000
10000
5000
5000
50000
50000
100000
-5000
H=10000
50000
-5000
100000
H=12000
100000
50000
-5000
100000
FIGURES FOR CHAPTER 1
151
Table 1.1: Propagation of the solution of y0 = t − y2 , y(0) = 0, to the left;
see Example 1.8.1.
y0 = 0
y−1 = 0 − 0.25(0 − 02 ) = 0
y−2 = 0 − 0.25(−0.25 − 02 ) = 0.0625
y−3 = 0.0625 − 0.25(−0.5 − 0.06252 ) ≈ 0.1885
y−4 ≈ 0.1885 − 0.25(−0.75 − 0.18852 ) ≈ 0.3849
y−5 ≈ 0.3849 − 0.25(−1 − 0.38492 ) ≈ 0.6719
y−6 ≈ 0.6719 − 0.25(−1.25 − 0.67192 ) ≈ 1.0972
Propagation to the right:
y1 = 0 + 0.25(0 − 02 )
y2 = 0 + 0.25(0.25 − 02 ) = 0.0625
y3 = 0.0625 + 0.25(0.5 − 0.06252 ) ≈ 0.1865
y4 ≈ 0.1865 + 0.25(0.75 − 0.18852 ) ≈ 0.3653
y5 ≈ 0.3653 + 0.25(1 − 0.36532 ) ≈ 0.5820
y6 ≈ 0.5820 + 0.25(1.25 − 0.58202 ) ≈ 0.8098
152
CHAPTER 1. FIRST-ORDER EQUATIONS
Table 1.2: Euler method solution of the IVP in Example 1.8.2
y1 = y0 + ∆y0 = 0 + 0.5(0 + .25) = 0.125
y2 = y1 + ∆y1 = 0.125 + 0.5(0.125 + 0.25) = 0.3125
y3 = y2 + ∆y2 = 0.3125 + 0.5(0.3125 + .25) = 0.59375
y4 = y3 + ∆y3 = 0.65625 + 0.5(0.65625 + .25) = 1.015625
Local errors:
LE1 = φ(0.5) − y1 = (0.25e0.5 − 0.25) − 0.125 ≈ 0.037
LE2 = φ1 (1) − y2 = (0.375e0.5 − 0.25) − 0.3125 ≈ 0.056
LE3 = φ2 (1.5) − y3 = (0.5625e0.5 − 0.25) − 0.59375 ≈ 0.084
LE4 = φ3 (2) − y4 = (0.84375e0.5 − 0.25) − 1.015625 ≈ 0.125
Accumulated errors
AE2 = φ(1) − φ1 (1) = (0.25e1 − 0.25) − (0.375e0.5 − 0.25) ≈ 0.061
AE3 = φ(1.5) − φ2 (1.5) = (0.25e1.5 − 0.25) − (0.5625e0.5 − 0.25) ≈ 0.193
AE4 = φ(2) − φ3 (2) = (0.25e2 − 0.25) − (0.84375e0.5 − 0.25) ≈ 0.456
FIGURES FOR CHAPTER 1
153
Figure 1.26: Bifurcation for the logistic equation with harvesting. The harvesting rate is the bifurcation parameter; the growth parameter and the
carrying capacity are the same as in figure 1.25.
P
100000
50000
5000
10000
H
154
CHAPTER 1. FIRST-ORDER EQUATIONS
Table 1.3: U.S. population in millions, censuses of 1790 – 2010.
t
1790
1800
1810
1820
1830
1840
1850
1860
P(t)
3.9
5.3
7.2
9.6
12.9
17.1
23.2
31.4
t
1870
1880
1890
1900
1910
1920
1930
1940
P(t)
38.6
50.2
63.0
76.2
92.2
106.0
123.2
132.2
t
1950
1960
1970
1980
1990
2000
2010
P(t)
151.3
179.3
203.3
225.6
248.7
281.4
309.4
Chapter 2
Systems of Differential
Equations
155
156
2.1
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
Introduction
A system of ODEs is a set of two or more ODEs to be treated simultaneously. Each ODE involves the same independent variable (we will always
use t, for elapsed time, as the independent variable), but there are as many
dependent variables as there are equations. The systems that we will study
have the form
x 0 = f (t, x, y)
y0 = g(t, x, y),
where x 0 and y0 denote the derivatives of x and y with respect to t.
Consider the velocity of a object in motion, subject to forces such as
gravity and air resistance. In the previous chapters the object was assumed
to fall in a vertical direction, and the velocity was represented by a single ODE. If we are determining the trajectory of a ball that has been hit
or thrown, the motion will not be in a straight line. There will be two dependent variables, u(t) and v(t) representing the horizontal and vertical
components of the velocity, respectively. The drag force vector is pointed
in the direction opposite to the velocity vector, and we will denote its horizontal and vertical components by −h(u, v) and −k (u, v), respectively. The
gravitational force, mg, is vertical. Our model is a system of ODEs,
m u0 = −h(u, v)
m v0 = mg − k (u, v)
Just as single ODEs have families of solutions, so do systems. To find
a particular solution of interest, one may specify initial values to create
an IVP. The values of all dependent variables in an IVP must be specified
at a common starting time t0 . For example, suppose we are determining
the motion of a ball thrown horizontally from a tower 100 m. high with a
velocity of 30 m/s. Take the moment of release to be t = 0; then the initial
conditions are u(0) = 30, and v(0) = 0. The above system of ODEs can be
embellished by including the initial conditions:
m u0 = −h(u, v);
u(0) = 30
0
m v = mg − k (u, v); v(0) = 0
The height of the tower is not relevant in this model, because it ignores the
position of the ball. We could include the position by introducing two additional dependent variables, x (t), the horizontal distance from the tower at
time t, and y(t), the vertical distance from the ground. Noting that x 0 = u
2.1. INTRODUCTION
157
and y0 = v the following system is a model for the position and the velocity
of the ball:
x0
= u;
x (0) = 0
y0
= v;
y(0) = 100
m u0 = −h(u, v);
u(0) = 30
0
m v = mg − k (u, v); v(0) = 0
While the system we just constructed is a nice model, it is not for beginners. Here is a simpler IVP.
Example 2.1.1 Show that x = cos(t), y = − sin(t) is a solution of the initial
value problem
x 0 = y;
x (0) = 1
(2.1)
y0 = − x; y(0) = 0
S OLUTION . Substituting x = cos(t) and y = − sin(t) in the first equation
yields the identity − sin(t) = − sin(t). The same substitution in the second
equation results in cos(t) = cos(t). Finally, cos(0) = 1 = x0 and
− sin(0) = 0 = y0 .
The system in the Example 2.1.1 occurs in the following application to
mechanics. The dependent variables are the position x (t) and the velocity
y(t) of an object in motion. (Air resistance is ignored.) The object is attached
to a spring that obeys Hooke’s law: when the spring is stretched x units
past its equilibrium position (if the spring is compressed, x is negative), the
net force on the object, including the spring and gravity, is F = −k x. As
usual, the positive direction is downward. The variable y is the velocity, so
that y = dx
dt . By Newton’s second law of motion, the product of the mass
and the acceleration is equal to the net force. Since the acceleration is equal
dy
to dt , it follows that y satisfies
m
dy
= −k x.
dt
In (2.1), units were chosen so that the mass m = 1, and the spring constant
k = 1 as well. The initial values represent initial position and velocity, and
indicate that at t = 0, the object was pulled 1 unit below its equilibrium
position and released.
A second order ODE can also serve as a model for the motion of an
object suspended from a spring. The dependent variable x (t) represents
158
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
position. The acceleration is the second derivative of the position with respect to time, and Newton’s second law of motion provides the ODE,
m
d2 x
= −kx.
dt2
By viewing this physical phenomenon first as a system we see that an IVP
involving a second order ODE must be requires the initial values of the
position x and the velocity dx
dt . Second order ODEs have their purposes,
but the for the present we will be content with learning how to get rid of
them (and higher-order ODEs) by replacing them with equivalent systems
of first-order ODEs.
Given an explicit second order ODE,
d2 x
dx
= f t, x,
,
(2.2)
dt2
dt
put y =
dx
dt .
The system of first-order ODEs,
x0 = y
y0 = f (t, x, y).
(2.3)
is equivalent to (2.2), because if x = φ(t), y = ψ(t) is a solution of the
system (2.3) then φ0 (t) = ψ(t) (by the first equation). Thus
φ00 (t) = ψ0 (t) = f (t, φ(t), φ0 (t));
Conversely, if x = φ(t) is a solution of equation (2.2), put ψ(t) = φ0 (t).
Then x = φ(t), y = ψ(t) is a solution of the system (2.3). We say that the
system (2.3) replaces the ODE (2.2).
Example 2.1.2 Find a system of first order ODEs that replaces the ODE
y00 + t y0 + sin(y) = 0
S OLUTION . The ODE can be written as
d2 y
dy
= − t + sin(y)
dt2
dt
Set v =
dy
dt
and v0 =
d2 y
.
dt2
Then the system
y0 = v
v0 = −(t v + sin(y))
2.1. INTRODUCTION
159
replaces the given ODE.
An explicit higher-order ODE can also be replaced by a system of firstorder ODEs, with the number of equations in the system equal to the order
of the original equation. Thus, given an ODE
dn x
= f
dtn
dx
d n −1 x
t, x, , . . . , n−1
dt
dt
,
define new dependent variables y1 (t), y2 (t), . . . , yn−1 (t) by y1 (t) = x 0 (t),
y2 (t) = x 00 (t), . . . , yn−1 (t) = x (n−1) (t). The nth order equation shown above
is replaced by a system
x 0 = y1
y10 = y2
..
.
y0n−2 = yn−1
y0n−1 = f (t, x, y1 , y2 , . . . , yn−1 )
Uncoupled Systems
A system of two ODEs is uncoupled, if one of the differential equations
does not involve one of the dependent variables. Thus, an uncoupled system has the form
x 0 = f (t, x )
y0 = g(t, x, y)
To find the general solution of this system, we start by solving the first
equation. This gives a formula x = φ(t, C ), where C is a constant. Substituting this formula for x in the second equation yields another first order
equation with one dependent variable, y0 = g(t, φ(t, C ), y). Thus, solving
an uncoupled system of two differential equations is accomplished by solving two single first order equations.
Example 2.1.3 Find the general solution of the system
x 0 = 2 t x2
y0 = 2 t x y
In addition, find the solutions that satisfy the following sets of initial conditions.
(a) x (0) = 1, y(0) = 1;
160
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
(b) x (0) = 0, y(0) = 1;
(c) x (0) = 1, y(0) = 0.
S OLUTION . The equation x 0 = 2x2 t is separable, and can be written
dx
= 2t dt.
x2
Integration of both sides yields − x −1 = t2 + C, or
x=
−1
.
t2 + C
(2.4)
There is also a singular solution, x ≡ 0. We now substitute formula (2.4)
for x in the second equation to get
−1
0
y = 2t 2
y
t +C
which is also separable, and can be written in the equivalent form
dy
2t dt
=− 2
.
y
t +C
Integrating again, we have ln |y| = − ln |t2 + C | + D, where D is a second
constant. Taking the exponential of each side of this equation, we obtain
2
|y| = e− ln |t +C|+ D
eD
= 2
t +C
Thus
y=
K
t2 + C
where K = ±e D . It is also possible that K = 0 (why?).
If we substitute the singular solution x ≡ 0 in the second equation (instead
of formula (2.4)), then y0 ≡ 0, so y would be constant.
The solution of the system will thus be one of the following function pairs:
(
)
1
x = − t2 +
x = 0
C
or
K
y = K
y = t2 +
C
where C and K are constants.
We now solve the initial value problems.
2.1. INTRODUCTION
161
(a) Set t = 0 and x = 1 in x = t2−+1C . This yields 1 = −1/C, so C = −1.
K
Now we set t = 0, y = 1, and C = −1 in y = t2 +
, to get
C
1 = K/(−1). Hence K = −1 and the solution is
x =
y =
−1
−1
−1
.
t2 − 1
t2
(b) Since x (0) = 0, the solution of the first equation is the singular
solution x ≡ 0. It follows that y is constant, and must be equal to its
initial value of 1. The solution is
x ≡ 0
y ≡ 1.
(c) As in part (a), x =
−1
.
t2 −1
Since y(0) = 0, y ≡ 0. The solution is
−1
t2 − 1
y ≡ 0.
x ≡
Example 2.1.4 Find the general solution of the second order ODE,
dx
d2 x
=
+ 1.
2
dt
dt
S OLUTION . Put x 0 = y, so that the ODE is replaced by the system,
x0 = y
y0 = y + 1
(2.5)
The second equation does not involve the variable x, and thus (2.5) is
uncoupled. That equation is linear, and its solution is y = Cet − 1.
Referring to the first equation of (2.5), we have x 0 = Cet − 1, which can be
solved by antidifferentiating. The solution is
x = Cet − t + D,
y = Cet − 1
where C and D are constants.
162
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
Exercises
1.
Show that for all real C, x = C et , y = C et is a solution of the
system
x0 = y
y0 = x,
and find all solutions of the form x = A e−t , y = B e−t , where A and
B are constants.
Answer
Show that x = C e−2t , y = −3C e−2t is a solution of
2.
x0 = x + y
y0 = 3x − y,
and find all solutions of the form x = A e2t , y = B e2t , where A and B
are constants.
Show that x = 2C cos(3t), y = C [cos(3t) + 3 sin(3t)] is a solution
3.
of
x 0 = x − 2y
y0 = 5x − y,
and find all solutions where x = 2A sin(3t).
Answer
4.
Show that x = sin(2t) + et , y = 2 cos(2t) − 4et is a solution of
x 0 = 5 et + y
y0 = −4x,
and find a family of solutions of this system.
5.
Show that x = e2t (t + 1), y = e2t (t − 1) is a solution of
x 0 = x + y + 3e2t
y0 = 2x − 3e2t .
Furthermore, show that x = y = C e2t is a family of solutions of the
“associated homogeneous system”
x0 = x + y
y0 = 2x,
and find a family of solutions of the inhomogeneous system.
Answer
2.1. INTRODUCTION
163
In problems 6 – 9 find a system of two first order equations that replaces
the given second order ODE.
6.
7.
8.
9.
d2 x
dt2
= x dx
dt + 1.
y00 + 3y0 + 4y = t2 .
Answer
z00 + (z2 − 1) z0 + z = 0.
u00 u0 u = 1 + t2
Answer
In problems 10 – 18 determine if the system is uncoupled. If the system
is uncoupled, find its general solution.
0
x = 4x + 1
10.
y0 = y
0
x = x2
11.
y0 = 1/x
Answer
0
x = xy
12.
y0 = −y
0
x = y2
13.
y0 = − x2
Answer
0
x = x − et
14.
y0 = x + y
0
x = t/x
15.
y0 = t x y
Answer
0
x = x + y2
16.
y0 = y/2
0
x = x+y
17.
y0 = x − y
Answer
0
x = et x
18.
y0 = e2t x + y
164
2.2
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
The Phase Plane
There are several ways to present a solution x = φ(t), y = ψ(t) of a system
of two ODEs graphically, the simplest of which is to draw separate graphs
of φ(t) and ψ(t). This is useful, but it does not show any relationship between x and y. For example, it is easily verified that x = cos(t), y = sin(t)
is a solution of the system
x0 = −y
y0 = x,
but it is not obvious from the graphs of the sine and cosine functions that
x2 + y2 = 1.
The x, y-plane, whose coordinates represent the dependent variables of
the system of ODEs, is called the phase plane. Let x = φ(t), y = ψ(t)
be a solution of a system of two differential equations. The curve in the
phase plane described by the parametric equations x = φ(t), y = ψ(t) is
called the orbit of the solution. It is usually marked to show the direction
of increasing t. For example, the orbit of the solution x = cos(t), y = sin(t)
is the circle x2 + y2 = 1, marked with an arrow in the counterclockwise
direction, since as t increases from 0 to π2 , (cos(t), sin(t)) moves along the
circle from (1, 0) to (0, 1).
While it is natural to think of an orbit as a curve in the phase plane,
it is possible for an orbit to be just a point. If ( x1 , y1 ) is a point such that
f ( x1 , y1 ) = 0 and g( x1 , y1 ) = 0 then ( x, y) ≡ ( x1 , y1 ) is a solution of the
system
x 0 = f ( x, y)
(2.6)
y0 = g( x, y),
corresponding to an orbit that is the single point, ( x1 , y1 ). In this case,
( x1 , y1 ) is a stationary point of the system (2.6).
Example 2.2.1 Describe the orbits of the system
x0 = x
y0 = y
S OLUTION . This system is readily solved because it is uncoupled. The
general solution of x 0 = x is x = C et and the solution of y0 = y is y = D et .
(The constants C and D are independent of each other.)
If C = D = 0, the orbit is simply the origin itself (no arrow is attached to
this “stationary” orbit). If C = 0 the orbit is the positive y-axis, directed
2.2. THE PHASE PLANE
165
upward, if D > 0 or the negative y-axis, directed downward, if D < 0.
Assuming C 6= 0, we can substitute et = x/C in the formula for y to obtain
y = ( D/C ) x. The orbit consists of all points on this straight line that lie on
the same side of the origin as (C, D ), and is directed away from the origin.
See figure 2.1.
Example 2.2.2 Find a system of ODEs that replaces
y00 = y0 + 2y
(2.7)
Given that y = e−t , y = e2t and y = e2t − e−t are solutions of equation (2.7) find
the corresponding solutions of the system and describe their orbits in the phase
plane.
S OLUTION . Let v = y0 . Then the system
y0 = v
v0 = v + 2y
replaces equation (2.7). The solution corresponding to y = e−t is y = e−t ,
v = −e−t . Notice that v = −y for all t, and that as t → ∞, (y, v) → (0, 0).
Furthermore y is positive and v is negative, which puts the orbit in the
fourth quadrant of the y, v plane. Therefore, the orbit is the part of the
straight line v = −y in the fourth quadrant, and is directed toward the
origin.
The orbit of the solution corresponding to y = e2t is (y, v) = (e2t , 2 e2t ).
Here we have v = 2y, and as t → ∞, both y and v → +∞. This orbit is
therefore the part of the line v = 2y in the first quadrant, directed away
from the origin.
The parametric equation of the orbit of the solution corresponding to
y = e2t − e−t is (y, v) = (e2t − e−t , 2 e2t + e−t ). This will be left in parametric
form, since it is not easy to eliminate t to obtain a relation between v and
y. However, we can note that (y(0), v(0)) = (0, 3), and that as t → −∞,
(y(t), v(t)) → (−∞, ∞), while (y(t), v(t)) → (∞, ∞) as t → ∞. The three
orbits are displayed in figure 2.2.
A system
x 0 = f ( x, y)
y0 = g( x, y)
(2.8)
166
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
with the property that the independent variable does not appear on the
right side of either equation is said to be autonomous. An autonomous
system describes a vector field that assigns to each point ( x, y) a vector
~v( x, y) = f ( x, y)~i + g( x, y)~j.
(Here, ~i and ~j denote the usual horizontal and vertical unit vectors in the
plane.) A vector field can be represented graphically by drawing the vectors ~v( xm , yn ) at grid points ( xm , yn ) on graph paper. Such drawings resemble direction fields for ODEs, but there is a distinction: the vectors of the
vector field have varying magnitude—displayed by giving them different
lengths—and each has a direction and so is drawn as an arrow. Figure 2.3
shows the vector field ~v( x, y) = x~i − y~j that corresponds to the uncoupled
system
x0 = x
y0 = −y.
The orbits are not shown, but you should verify that they consist of the rays
of the x-axis, directed away from the origin, the rays of the y-axis, directed
toward the origin, and components of the hyperbolas xy = C, directed
downward in the first and second quadrants, and directed upward in the
third and fourth quadrants.
We have seen that an autonomous system of two ODEs can be displayed as a vector field in the plane. It is also useful to interpret a vector
field
~v( x, y) = f ( x, y)~i + g( x, y)~j
as a system of ODEs,
x0 =
y
0
f ( x, y)
= g( x, y)
If we think of ~v( x, y) as the velocity vector for a particle located at ( x, y),
then the particle will follow an orbit of this system.
An integral curve of a vector field ~v( x, y) is a curve with parametric
equations ( x, y) = (φ(t), ψ(t)) with tangent vector φ0 (t)~i + ψ0 (t)~j equal to
the field vector ~v( x, y) at each point of the curve; that is, for all t,
~v(φ(t), ψ(t)) = φ0 (t)~i + ψ0 (t)~j.
The term stationary point is used in the context of a vector field ~v to
indicate a point ( x1 , y1 ) where ~v( x1 , y1 ) = ~0. A stationary point of ~v( x, y) =
f ( x, y)~i + g( x, y)~j is also a stationary point of the system x 0 = f ( x, y), y0 =
g( x, y).
2.2. THE PHASE PLANE
167
Displaying a solution of a system
In chapter 3, we will learn how to solve systems such as the one in the
following example.
Example 2.2.3 Verify that
x (t) = et cos(10t)
y(t) = et sin(10t)
satisfies the system
x 0 = x − 10y
y0 = 10x + y,
(2.9)
and display the solution graphically.
S OLUTION . By the product rule for differentiation,
x 0 (t) = et cos(10t) − 10et sin(10t) = x (t) − 10y(t),
and
y0 (t) = et sin(10t) + 10et cos(10t) = y(t) + 10x (t).
There are several ways to display graphs of solutions of systems. Our first
plot, Figure 2.4, shows the components x (t) and y(t) of the solution
of (2.9) plotted together.
We see that the solution functions are oscillating, with increasing
amplitude. The orbit, displayed with the vector field representation of the
system (2.9), is shown in figure 2.5. The orbit is an outward spiral, which is
certainly consistent with the graphs shown in figure 2.4.
Using a three dimensional plot, we can display the graph of ( x, y) as a
function of t. See figure 2.6, in which the t-axis, for the independent
variable, is vertical, and the x, y-plane is the “axis” for the dependent
variables.
Exercises
In exercises 1 – 8, show that the given pair of functions is a solution of the
system of differential equations, and sketch the orbit determined by the
solution. If a second order ODE is given, replace it with a system of two
first order equations and proceed.
168
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
x 0 = 2y2 /t
y0 = y2 /x
x = t2 , y = t.
Answer
1.
2.
y00 = y; draw the orbits corresponding to three solutions: y =
y = et , and y = et − e−t .
− e−t ,
x0 = y
y0 = x
−
x = e t , y = − e−t .
Answer
3.
x0 = y
y0 = x
x = cosh(t), y = sinh(t). Hint: Remember the identity cosh2 (t) −
sinh2 (t) = 1.
4.
5.
y00 = y−3 ; solution, y =
Answer
√
t2 + 1
x 0 = xy
y0 = x2
x = sec(t), y = tan(t).
6.
7.
y00 = 21 t y0 − y; solution, y = t2 − 2.
Answer
x 0 = x − 2y
y0 = 2x + y
t
x = e cos(2t), y = et sin(2t).
8.
In problems 9 – 12, find stationary points for the given vector fields.
Draw some of the integral curves. (You may want to print copies of the
vector fields and draw the integral curves on these copies.)
9.
v( x, y) = [4yi − xj]/16.
2.2. THE PHASE PLANE
169
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-1
0
1
2
3
0
1
2
3
Answer
10.
v( x, y) = [(y2 − y)i + xj]/4.
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-1
170
11.
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
v( x, y) = [( x2 + 1)yi − 2xj]/16.
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
Answer
12.
v( x, y) = [ xi + 2yj]/8.
-1
0
1
2
3
2.2. THE PHASE PLANE
171
2
1.5
1
0.5
0
-0.5
-1
-1.5
-1
-2
13.
0
1
2
3
Use a CAS to plot the graphs of the component functions, the twodimensional orbits and the three-dimensional graphs of the solutions
of the following systems.
(a)
x 0 = 5y
y0 = −5x.
with initial conditions x (0) = 0, y(0) = 1; 0 ≤ t ≤ 2π.
(b)
x 0 = 2x − 6y
y0 = x − 2y.
with initial conditions x (0) = 1, y(0) = 0; 0 ≤ t ≤ 4π.
(c)
x 0 = 2y + sin(t)
y0 = −2x − sin(2t).
with initial conditions x (0) = 0, y(0) = 0; −2π ≤ t ≤ 4π.
172
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
(d)
x0 = x + y
y0 = t − x − y.
with initial conditions x (0) = 0, y(0) = 0; −2 ≤ t ≤ 4.
Answer
14.
Draw graphs of the component functions of the solution
( x, y) = (sinh(t), cosh(t))
of
x 0 = y; x (0) = 0
y0 = x; y(0) = 1,
and plot its orbit.
2.3
A User’s Guide to IVP Solvers
An IVP solver is a numerical method for computing solutions of IVPs.
We already know of one: Euler’s method. Although it was presented in
the context of a single ODE, it is easily modified to approximate solutions
of IVPs involving systems of ODEs. Recall that Euler’s method works by
replacing the ODE by an easily solved difference equation. For a system,
x 0 = f (t, x, y); x (t0 ) = A
y0 = g(t, x, y); y(t0 ) = B,
we use a system of difference equations,
∆x j = h f (ti , xi , yi ); x0 = A
∆y j = hg(ti , xi , yi ); y0 = B,
where h is the time step, and ti is the time after i steps have been taken; that
is, ti = t0 + i · h.
Using this system, we obtain sequences { xi } and {yi }. If x = φ(t),
y = ψ(t) denotes the solution of the IVP, then xi is an approximation of
φ(ti ) and yi approximates ψ(ti ).
Example 2.3.1 Plot the orbit of the system
x 0 = 2x + 2y
y0 = − x + 4y.
with initial conditions x (0) = 1, y(0) = 0, for t ranging from −1 to 1.
2.3. A USER’S GUIDE TO IVP SOLVERS
173
S OLUTION . It is possible to solve this system analytically, or to use an IVP
solver incorporated in a CAS to produce a numerical solution. We will use
a spreadsheet, with a moderately large stepsize, h = 0.1. Our
spreadsheet, shown in figure 2.7, has five columns, representing t, x, y,
∆x, and ∆y — the latter two are labelled Dx and Dy. The calculations are
done in two blocks, the first for 0 ≤ t ≤ 1, and the second for 0 ≥ t ≥ −1.
The formulas shown in figure 2.7 were entered once in each block, and
then copied. Thus, the entry in cell D2, +0.1*(2*B2+2*C2) was typed,
copied onto the clipboard, and then pasted into the cells D3 through D12,
and so on. This formula says to multiply the entry in cell B2 (the value of
x0 ) by 2 and add the result to 2× the entry in cell C2 (the value of y0 ) and
then multiply this sum by 0.1. The result is 0.1(2x0 + 2y0 ), which is equal
to ∆x1 . The entry in cell B3 adds the entries in cells B2 and D2 to form
x1 = x0 + ∆x1 .
The spreadsheet in figure 2.7 shows the formulas instead of numerical
values. The numerical values, and the orbit of the computed solution are
shown in figure 2.8. The graph was made by selecting xy graph in the
graphics type menu, with x values from column B and series values
from column C.
To assess the precision of the approximation just computed, see figure 2.9.
The strength of Euler’s method is its simplicity. As we have seen, the accumulated error varies proportionately with the step size. Since the amount
of effort is inversely proportional to the time step, we can expect to use 10
times as much computer time to improve accuracy by 1 decimal place. It is
worthwhile to find a more efficient method.
A classical IVP solver is an IVP solver in which the user specifies a
time step that will not be altered when the calculation is in progress. The
order is a number n such that the accumulated error varies proportionately
with |h|n , where h is the time step. Euler’s method, for example, is a classical method of order 1. Classical solvers can be constructed with arbitrarily
high order. Runge-Kutta methods (the Euler method is one of these) operate by setting
∆ym = h × ( a weighted average of values of f (t, y)), with tm ≤ t ≤ tm+1 .
For example, the “improved Euler method” is typical. Given (tm , ym ), set
ȳm+1 = ym + h f (tm , ym ) and tm+1 = tm + h.
174
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
Then set
∆ym =
1
( f (tm , ym ) + f (tm+1 , ȳm+1 ))
2
This method is second order. It requires two function evaluations per step,
but the payoff is√that the accumulated error is proportional to h2 . Thus it
will take about 10 times as much computer time to improve accuracy by
one decimal place.
The most commonly used classical methods are of order 4. To obtain a
fourth order Runge-Kutta method, it turns out that four function evaluations are required per step. For the “RK4” method, we define
k1 =
f (tm , ym )
k2 =
f (tm + 0.5h, ym + 0.5k1 h)
k3 =
f (tm + 0.5h, ym + 0.5k2 h)
k4 =
f (tm + h, ym + k3 h),
and then
∆ym = h
1
1
1
1
k1 + k2 + k3 + k4 .
6
3
3
6
To obtain an additional
decimal place of accuracy with the RK4 method,
√
we expect to do 4 10 ≈ 1.8 times as much work.
The predictor methods work a little differently. The Adams-Bashforth
method (also a fourth order method) calculates ∆ym as follows: let p0 =
f ( t m , y m ), p 1 = f ( t m −1 , y m −1 ), p 2 = f ( t m −2 , y m −2 ), p 3 = f ( t m −3 , y m −3 ).
Then
h × ( a weighted average of p0 . . . p3 ).
To be precise,
∆ym = h
59
37
9
55
p0 − p1 + p2 − p3 .
24
24
24
24
The advantage of this method is that it recycles previously computed data,
and requires only one new function evaluation per step. This strength is
also a weakness, because the error in the computed value of ym is recycled
and can cause accuracy to degenerate. To control error propagation, the
Adams-Bashforth method can be coupled with a “correction stage.” The
resulting method is known as the Adams-Bashforth-Moulton method, and
it works as follows. Numerical analysts refer to the sequence of calculations
by the acronym “PECE”.
2.3. A USER’S GUIDE TO IVP SOLVERS
175
Predict Use the Adams-Bashforth method to find a tentative value for ym+1 :
55
59
37
9
ȳm+1 = ym + h
p0 − p1 + p2 − p3
24
24
24
24
Evaluate Set p−1 = f (tm+1 , ȳm+1 ).
Correct Recalculate ym+1 , as
y m +1 = y m + h
19
5
1
9
p −1 + p 0 − p 1 + p 2
24
24
24
24
Evaluate Calculate f (tm+1 , ym+1 ) to prepare for the next step.
The improved Euler method that was mentioned above is a predictor-corrector method as well as a Runge-Kutta method, since ȳm+1 is predicted
by the Euler method, and used in a correction phase to get an improved
approximation of the solution at t = tm+1 .
The classical methods are not used for high precision work, because
it is hard to assess the accuracy of a solution without recomputing the
whole thing with smaller time steps and making comparisons. Methods
that change the stepsize as the computation progresses are called dynamic
methods. The Runge-Kutta-Fehlberg (RKF45) method is widely used. With
this method, each value of ym+1 is calculated twice in each step, first using
the RK4 method, followed by a Runge-Kutta method of order 5. When the
two computed values differ by more than a specified tolerance, both are
rejected, the time step is reduced, and the process is repeated. There is a
provision to stop when a minimum time step is reached: this avoids having the program running forever if, for example, the solution has a vertical
asymptote.
Other dynamic methods are designed for either increased precision or
to address problems encountered with certain types of higher order equations.
It is not practical to use a spreadsheet with dynamic IVP solvers. Fortunately, the programs supplied with CAS systems are excellent and easy to
use.
It may be convenient to use an IVP solver independently of a CAS, and
several such programs are available. John Polking has made pair of Java
applets, DFIELD and PPLANE, available on the web site
http://math.rice.edu/∼dfield/dfpp.html1
1 You will have to type this URL, because the tilde symbol will not copy correctly in your
browser.
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CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
These applets will draw direction fields for first order ODEs, and vector
fields representing systems of two autonomous first-order ODEs, respectively. By clicking any point in the field with the mouse pointer, one can
see a graph (or orbit) of the solution, with the initial condition specified
by the point that was selected.There are several graphing calculators that
incorporate IVP solvers as well.
Exercises
1.
The solution of the IVP
x 0 = − y x (0) = 1
y0 = x
y (0) = 0
is ( x, y) = (cos(t), sin(t)). Use the RKF45 algorithm with this IVP to
estimate sin(1). Compare the answer you get to the value from your
calculator (be sure it is set to radians!).
Answer
2.
The solution of the IVP
x 0 = −4y x (0) = 1
y0 = x
y (0) = 0
is ( x, y) = cos(2t), 21 sin(2t) . Thus, the identity x2 + 4y2 = 1 is satisfied, and the orbit is an ellipse. Use an IVP solver to draw this ellipse.
If you use a crude method like Euler’s, do you get a closed curve? Repeat the experiment with a classical fourth order method, and again
with a dynamic method.
3.
Consider the following system, which depends on a parameter e:
x 0 = − y + e x ( x 2 + y2 − 1) x (0) = A
y 0 = x + e y ( x 2 + y2 − 1)
y (0) = 0
When e = 0 this system is the same as that of exercise 1, and if A =
1, the solution of the IVP is the same as in exercise 1. This exercise
investigates what happens when e 6= 0 and A 6= 1. Let r ( x, y) denote
the distance of the point ( x, y) to the origin. Differentiation of r2 =
x2 + y2 with respect to t yields
r r 0 = x x 0 + y y0
2.3. A USER’S GUIDE TO IVP SOLVERS
177
Furthermore, if θ = arg( x + i y) is the polar angle,
θ0 =
x y0 − y x 0
x 2 + y2
Show that if ( x (t), y(t)) is the solution of the IVP then
r 0 = e r (r 2 − 1); r (0) = A
and draw a phase diagram for this IVP. Furthermore, show that θ 0 =
1. Thus orbits rotate counterclockwise about the origin, spiraling outward if r is increasing and inward if r is decreasing.
(a) Put e = −.25 and A = .05. Use the RKF45 algorithm to draw the
orbit of the solution for 0 ≤ 30. Explain how the orbit shown is
consistent with the conclusions reached above.
(b) Repeat the experiment with the following data.
i. (e, A) = (−.25, 4).
ii. (e, A) = (.25, .95).
iii. (e, A) = (.25, 1.05).
Answer
4.
The solution φ(t) of the IVP
y 0 = t2 + y2 ; y (0) = 0
(2.10)
has two vertical asymptotes, t = c and t = −c. Estimate the value of
c.
5.
Produce an accurate plot of the solution φ(t) of the IVP (2.10), for
−c < t < c. ranging from the negative asymptote t = −c to the positive asymptote t = c. Change the time step to 0.05, and use Euler’s
method to generate an approximate solution. Reduce the time step
by a factor of 2 repeatedly. Describe precisely the effect that reducing
the time step has on the error. Repeat the experiment with a second
order method, and the classical RK4 method.
Answer
6.
Draw the phase diagram for the ODE y0 = 4y − y3 , and sketch the
graphs of four solutions with different asymptotic behavior. Use an
IVP solver to graph the four solutions with initial conditions that you
should select so that the graphs resemble your sketches.
178
7.
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
Problem 6 suggests a way to find real roots of a polynomial f (y)
by using an IVP solver. The roots will be horizontal asymptotes of
solutions of the differential equation y0 = f (y), so by plotting numerous solutions, they can be located. Hint: if you try to solve y0 = f (y),
where f (y) is a high degree polynomial, your computer will probably complain of overflow problems. In this case, try replacing f (y)
with f (y)/10 N . Here you would choose N to be a positive integer
large enough to eliminate the overflow, but small enough so that the
solution you are plotting “levels off” soon.
(a) Try the method on a polynomial with known roots: f (y) =
(2y − 1)(2y + 1)2 (2y + 3)3 (2y + 5). Can you locate all of the roots
(including the multiple ones)?
(b) The polynomial
f (y) = y7 − 49y6 + 882y5 − 7350y4 + 29400y3 − 52920y2 + 35280y − 5040
has seven real roots in the interval [0, 20]. Approximate them.
Answer
8.
The differential equation
y0 + .05y = 2 sin(2t) + sin(3t)
has a periodic solution. Use an IVP solver to plot the graph of this
solution.
9.
Use the computer to determine and graph several solutions of the
following differential equations, and note any periodic solutions.
(a) y0 + 0.1y = sin(2t).
(b) y0 − y = sin(20t).
(c) y0 + 10y = sin(100t).
Answer
10.
Let S(t) = +1 if the integer part of t is even, (i.e. t ∈ [0, 1), [2, 3),
etc., and S(t) = −1 otherwise. This function is sometimes called a
square wave. Graph the periodic solution of
y 0 + y = S ( t ).
2.4. *AUTONOMOUS SYSTEMS
179
To represent S(t) with a simple formula your CAS can handle, note
that S(t) = (−1)btc where btc is the “floor” function that rounds numbers down to their integer parts. It is also possible to use S(t) =
sign(sin t) if the floor function is not available.
2.4
Autonomous Systems
A phase portrait for an autonomous system of ODEs,
x 0 = f ( x, y)
y0 = g( x, y)
(2.11)
is a drawing of the phase plane, showing a representative collection of orbits of the system.
Example 2.4.1 Draw a phase portrait of the system
x 0 = 2x
y0 = y,
S OLUTION . The system is uncoupled, and each equation in the system is
a homogeneous linear ODE. The general solution of the first equation is
x = Ce2t , and the general solution of the second is y = Det , where C and
D are constants. An orbit in the phase plane is therefore described by
parametric equations
x = Ce2t
(2.12)
y = Det ,
To eliminate t from these equations, note that y2 = D2 e2t . Therefore, if
D 6= 0,
x
C
= 2 = constant.
2
y
D
Therefore, x = Ay2 , where A is a constant. The orbit follows a parabola
with a horizontal axis of symmetry, opening to the right if A > 0, and to the
left if A < 0. If A = 0 the orbit lies on the y-axis, rising upward if D > 0,
and downward if D < 0. Finally, if D = 0 then y = 0: the orbit follows the
x-axis, headed away from the origin to the right if C > 0, and to the left if
C < 0. If C = D = 0 we have a stationary orbit.
Thus, each parabola or axis actually consists of three orbits. The origin is
a stationary point, and an orbit by itself. The other two orbits follow the
180
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
parabola or axis, leading away from the origin. Figure 2.10 is a phase
portrait of the system (2.12). This figure includes the vector field, but
normally the vector field is not shown in a phase portrait.
Example 2.4.1 is unusual because we were able to determine the general
solution of the system. Often, we must approach the problem of drawing a
phase portrait by relying on qualitative analysis of the system and an IVP
solver. The qualitative analysis should be done first, as it will aid us in
choosing initial points for the IVP solver to use.
A set of points where f ( x, y) = 0 is called an x-nullcline of the system (2.11). If ( x, y) = (φ(t), ψ(t)) is a solution whose orbit crosses an xnullcline at a point ( x1 , y1 ) = (φ(t1 ), ψ(t1 )), then φ0 (t1 ) = f ( x1 , y1 ) = 0. If
ψ0 (t1 ) = g( x1 , y1 ) 6= 0, the orbit must have a vertical direction at ( x1 , y1 ),
which will be upward if g( x1 , y1 ) > 0, and downward if g( x1 , y1 ) < 0. The
x-nullcline should be marked with vertical arrows according to the sign
of g( x, y). Similarly, a y-nullcline is a set of points where g( x, y) = 0, and
non-constant orbits have horizontal tangents at points where they cross a
y-nullcline. We mark the y-nullcline with horizontal arrows, pointed to the
right where f ( x, y) > 0, and to the left where f ( x, y) < 0.
Stationary points are located at the intersections of the x- and y-nullclines.
They can be found by solving the pair of equations f ( x, y) = 0, g( x, y) = 0.
Example 2.4.2 Find the nullclines and stationary points of the system
x 0 = x ( x − 4y)
y0 = y(y2 − 2x − 9)
S OLUTION . The x-nullclines are given by the equation
x ( x − 4y) = 0.
Since this equation is satisfied if either x = 0 or if x = 4y, the x-nullclines
are the y-axis and the line x = 4y.
The y-nullclines are determined by the equation
y(y2 − 2x − 9) = 0,
and hence they are the x-axis and the parabola x = (y2 − 9)/2.
The stationary points lie at the intersection of x-nullclines with y-nullclines.
The axes meet at (0, 0), and the y-nullcline x = (y2 − 9)/2 meets the
2.4. *AUTONOMOUS SYSTEMS
181
y-axis at (0, ±3). Finally, the x-nullcline x = 4y meets the y-nullcline
x = (y2 − 9)/2 where y2 − 8y − 9 = 0. The solutions are y = −1 and
y = 9, so (−4, −1) and (36, 9) are stationary points. Figure 2.11 shows the
nullclines and stationary points.
Integrals
An integral of a system of autonomous ODEs is a function F ( x, y) such
that for every solution x = φ(t), y = ψ(t) of the system, F (φ(t), ψ(t)) is
a constant function of t on the interval where (φ(t), ψ(t)) is defined. This
definition can be compared to the definition of integral of a single ODE. For
example, it is known that all solutions of the system
x0 = −y
y0 = x
have the form x = R cos(t − C ), y = R sin(t − C ), where R and C are
arbitrary constants. Hence the trigonometric identity cos2 (θ ) + sin2 (θ ) = 1
implies that x2 + y2 = R2 . In other words, F ( x, y) = x2 + y2 is an integral
of this system.
Finding an integral of a system by using the general solution of the
system is not practical, and a fact about parametric equations leads to a
better approach. If x = φ(t), y = ψ(t) are parametric equations for a curve,
then the slope of the tangent to the curve at a point ( x0 , y0 ) = (φ(t0 ), ψ(t0 ))
is equal to
dy
dy/dt
ψ 0 ( t0 )
=
= 0
dx
dx/dt
φ ( t0 )
If x = φ(t), y = ψ(t) are solutions of the system, then φ0 (t) = f (φ(t), ψ(t))
and ψ0 (t) = g(φ(t), ψ(t)). Thus, the slope of the orbit passing through a
point ( x0 , y0 ) is
dy
ψ 0 ( t0 )
g ( x1 , y1 )
= 0
=
.
dx
φ ( t0 )
f ( x1 , y1 )
The orbit therefore satisfies the first order ODE,
dy
g( x, y)
=
dx
f ( x, y)
This ODE is often written in differential form, as g( x, y) dx − f ( x, y) dy = 0.
182
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
Finding an Integral of an Autonomous System
Let F ( x, y) be an integral for the ODE
g( x, y) dx − f ( x, y) dy = 0.
(2.13)
Then F ( x, y) is also an integral of the autonomous system
x 0 = f ( x, y)
y0 = g( x, y).
Example 2.4.3 Draw a phase portrait of the system
x 0 = − p2 y
y 0 = q2 x
where p and q are nonzero constants.
S OLUTION . We will find an integral of the system by finding an integral for
the corresponding ODE
q2 x dx + p2 y dy = 0,
which is easily done since it is exact (and separable as well). The solution
is
q2 x2 + p2 y2 = C.
(2.14)
We can assume C ≥ 0 since otherwise equation (2.14) represents the
empty set. When C = 0 we have the origin, which is a stationary point. If
C > 0, put a2 = C/q2 and b2 = C/p2 . Dividing equation (2.14) by C, we
have
x2
y2
+ 2 =1
2
a
b
which is the equation of an ellipse, with principal axes of lengths 2a and 2b
on the two coordinate axes. Since x 0 = − p2 y, x is decreasing (directed to
the left) on the upper half plane y > 0, and increasing (directed to the
right) on the lower half-plane. Hence the orbit is directed counterclockwise.
2.4. *AUTONOMOUS SYSTEMS
183
What do these orbits have in common? Each is an ellipse, for which the
ratio of length of the vertical axis to the length of the horizontal axis is
equal to
√
b
C/q
p
=√
= .
a
q
C/p
Thus the orbits are the origin, and a family of similar ellipses. The phase
portrait is shown in figure 2.12.
The second order ODE, y00 + y2 = 0, can be replaced by the system
y0 = v
v 0 = − y2 .
(2.15)
Example 2.4.4 Find an integral of the system (2.15) and draw its phase portrait.
S OLUTION . An integral F (y, v) of the ODE
y2 dy + v dv = 0
will also be an integral of the system (2.15). This ODE is exact; integrating
yields
1
1
F (y, v) = y3 + v2 .
3
2
The orbits lie on level curves of F (y, v). Before drawing the phase portrait,
we note that the y-nullcline is the v-axis, and the v-nullcline is the y-axis.
There is one stationary point, the origin. The phase portrait shown in
figure 2.13, was drawn as a contour plot of the level curves of F (y, v). The
vector field
~v(y, v) = v~i − y2~j
was superimposed on the contour plot.
Can orbits intersect?
We will see that for autonomous systems, the answer is no. Is there an orbit passing through each point of the plane? Here the answer is yes. The
proofs are based on the existence and uniqueness theorems for systems of
ODEs. These theorems are modifications of the single-equation existence
and uniqueness theorems. As in the case of the single equation, there is a
184
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
continuity requirement for f and g for the existence statement, and a Lipschitz condition is also needed for uniqueness.
Consider an IVP (where the ODEs are not necessarily autonomous),
x 0 = f (t, x, y); x (t0 ) = x0
y0 = g(t, x, y); y(t0 ) = y0
(2.16)
The functions f and g depend on three variables, and are required to be
continuous in a box in 3-dimensional space, R3 , rather than a rectangle in
the t, y-plane. We will denote by B the box
B = {(t, x, y) : |t − t0 | < k, | x − x0 | < l, |y − y0 | < m},
centered at the point (t0 , x0 , y0 ) ∈ R3 . Here k, l, and m are positive numbers
which specify the size of the box.
To formulate the Lipschitz condition that we need for a system, it is
helpful to consider (2.16) as a single IVP where the dependent variable is a
vector. Let ~v denote the vector whose components are the dependent variables x and y, and let ~F (t, ~v) be the vector function whose components are
f (t, x, y) and g(t, x, y). In vector terms, the IVP (2.16) is stated as follows:
d~v ~
= F (t, ~v); ~v(t0 ) = ~v0
dt
where ~v0 is the initial vector, with components ( x0 , y0 ).
We will use absolute value signs to denote distance in the plane: thus if
~v1 = ( x1 , y1 ) and ~v2 = ( x2 , y2 ),
|~v1 − ~v2 | =
q
( x1 − x2 )2 + ( y1 − y2 )2 .
With these definitions in mind, the function ~F (t, ~v) is said to satisfy a
Lipschitz condition with respect to ~v in the box B if there is a constant K
such that for all (t, ~v1 ), (t, ~v2 ) ∈ B ,
|~F (t, ~v1 ) − ~F (t, ~v2 )| ≤ K |~v1 − ~v2 |.
Our reason for using vector notation here is economy. The Lipschitz
condition can be expressed without it, but it looks more complicated that
way. See problem 18 at the end of this section.
Theorem 2.1
2.4. *AUTONOMOUS SYSTEMS
185
Existence Suppose that f (t, x, y) and g(t, x, y) are each continuous at every
point in the box B . Then there is a number h > 0 and there are functions φ(t) and ψ(t), defined on the interval t0 − h < t < t0 + h such
that x = φ(t), y = ψ(t) is a solution of the IVP (2.16).
Uniqueness If the vector function ~F (t, ~v) = ( f (t, x, y), g(t, x, y)) satisfies a
Lipschitz condition with respect to ~v in B , and x = φ̄(t), y = ψ̄(t) is
another solution of the IVP 2.16, then there is a positive number h1 such
that for all t with t0 − h1 < t < t0 + h1 , φ(t) = φ̄(t) and ψ(t) = ψ̄(t).
Continuous Dependence Assume that ~F (t, ~v) satisfies a Lipschitz condition
with respect to ~v ∈ B . If ~u = (u1 , u2 ) is a vector such that (t0 , ~u) ∈ B , let
φ(t, ~u) denote the solution of
d~v ~
= F (t, ~v); ~v(t0 ) = ~u.
dt
Then there is a box B ∗ ⊆ B such that φ(t, ~u) is continuous as a function of
the three variables (t, u1 , u2 ) with domain the box B ∗ .
The proof of this theorem involves the same ideas as those of the singleequation theorems 1.5, 1.7 and 1.8, and is omitted.
Now let’s adapt the uniqueness theorem to the question of intersecting
orbits. If a system is not autonomous, it is certainly possible for orbits to
intersect. If ( x, y) = (φ1 (t), ψ1 (t)) and ( x, y) = (φ2 (t), ψ2 (t)) are solutions
of a system,
x 0 = f (t, x, y)
y0 = g(t, x, y)
their orbits could intersect at a point
( x ∗ , y∗ ) = (φ1 (t1 ), ψ1 (t1 )) = (φ2 (t2 ), ψ2 (t2 ))
and as long as t1 6= t2 this would be consistent with the uniqueness theorem. In the case of autonomous systems, we can show that the orbits do
not intersect. To apply the uniqueness theorem, the following lemma is
needed.
Lemma 2.4.1 Let ( x, y) = (φ(t), ψ(t)) be a solution of the autonomous system
x 0 = f ( x, y)
y0 = g( x, y).
Then for any t0 , ( x, y) = (φ(t + t0 ), ψ(t + t0 )) is also a solution.
186
dφ̄
dt
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
P ROOF. Put φ̄(t) = φ(t + t0 ) and ψ̄(t) = ψ(t + t0 ). By the Chain Rule,
= φ0 (t + t0 ) and ddtψ̄ = ψ0 (t + t0 ). Since for any u,
φ0 (u) = f (φ(u), ψ(u)),
it follows that
dφ̄
= φ 0 ( t + t0 ) =
dt
=
f (φ(t + t0 ), ψ(t + t0 ))
f (φ̄(t), ψ̄(t)).
dψ̄
Similarly, dt = g(φ̄(t), ψ̄(t)), and therefore ( x, y) = (φ̄(t), ψ̄(t)) is a solution.
The proof of this lemma was rather simple, and you should consider
why it does not work for systems that are not autonomous.
Theorem 2.2 Suppose that the functions f ( x, y) and g( x, y), are continuous and
satisfy a Lipschitz condition in a rectangle D ⊂ R2 . Then for any point ( x ∗ , y∗ ) ∈
D , there is exactly one orbit of the system
x 0 = f ( x, y)
y0 = g( x, y)
(2.17)
that contains ( x ∗ , y∗ ).
P ROOF. The Existence and Uniqueness Theorems imply that there is
exactly one solution, which we will denote ( x, y) = (φ1 (t), ψ1 (t)), of the
system (2.17) with initial values ( x (0), y(0)) = ( x ∗ , y∗ ). However, for any
t0 6= 0, there is also a solution, which we will denote ( x, y) = (φ2 (t), ψ2 (t)),
with initial values ( x (t0 ), y(t0 )) = ( x ∗ , y∗ ). Both solutions will describe an
orbit that passes through ( x ∗ , y∗ ).
We have to show that the two solutions describe the same orbit. Put
φ̄2 (t) = φ2 (t + t0 ), ψ̄2 (t) = ψ2 (t + t0 ). By Lemma 2.4.1, ( x, y) = (φ̄2 (t), ψ̄2 (t))
is a solution of the system (2.17), and it describes the same orbit that (φ2 , ψ2 )
does. Since φ̄2 (0) = φ2 (t0 ) = x0 , and ψ̄2 (0) = y0 , it follows that (φ̄2 , ψ̄2 )
satisfies the same initial conditions as (φ1 , ψ1 ). It follows from the Uniqueness Theorem that φ1 (t) = φ2 (t + t0 ) and ψ1 (t) = ψ2 (t + t0 ) for all t such
that all four functions are defined.
2.4. *AUTONOMOUS SYSTEMS
187
Closed orbits
Theorem 2.2 does not preclude an orbit from intersecting itself. This would
happen if the system (2.11) had a solution ( x, y) = (φ(t), ψ(t)) such that for
some pair of numbers t1 6= t2 ,
(φ(t1 ), ψ(t1 )) = (φ(t2 ), ψ(t2 )).
(2.18)
For example, if an orbit is simply an stationary point, the functions φ and
ψ would be constant; it therefore intersects itself infinitely often.
Theorem 2.3 Let ( x, y) = (φ(t), ψ(t)) be a solution of the system (2.11) such
that equation (2.18) holds for distinct numbers t1 < t2 . Then φ and ψ are Tperiodic, where T = t2 − t1 , and the orbit described by (φ(t), ψ(t)) is a closed
curve.
P ROOF. By lemma 2.4.1, ( x, y) = (φ(t + T ), ψ(t + T )) is also a solution of the system (2.11), and since it satisfies the same initial conditions as
(φ(t), ψ(t)) at t = t1 , we can use the uniqueness theorem to conclude that
(φ(t + T ), ψ(t + T )) ≡ (φ(t), ψ(t)). Therefore the functions φ and ψ are Tperiodic. A curve that is described by periodic parametric equations must
be closed, so the corresponding orbit is closed.
Van der Pol’s equation.
To illustrate theorem 2.3, let’s consider the second order ODE,
x 00 + c( x2 − 1) x 0 + x = 0.
which is known as van der Pol’s equation.The number c is a constant that
we can vary to produce a variety of examples. Van der Pol’s equation was
developed in the 1920’s as a model for the behavior of a triode vacuum
tube. The dependent variable x represents the current, as a function of
time, in an electrical circuit that includes a triode tube. We can replace van
der Pol’s equation with a system
x0 = y
y0 = − x + c(1 − x2 ) y.
(2.19)
188
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
The case c = 0. If c = 0, the system (2.19) is identical to the system that
we examined in example 2.4.3, where both parameters p and q have the
value 1. We found there that all orbits, except the stationary point at the
origin, were closed ellipses (in our case, since p = q, the orbits are actually
circles). Therefore, theorem 2.3 tells us that all of the solutions except the
stationary solution x ≡ 0, y ≡ 0 are periodic. In fact, you can easily verify
that the orbit passing through the point ( a, 0) on the x-axis is traced by the
solution
x = sin(t)
y = cos(t)
and is thus 2π-periodic.
The case c > 0. When c = 0, we are dismissing the important middle
term of van der Pol’s equation. We’ll use a positive value for c and see
what happens.
Example 2.4.5 Draw the x- and y-nullclines of the van der Pol system (2.19),
with c = 21 .
S OLUTION . The x-nullcline is given by the equation y = 0, and is thus the
x-axis. This means that all orbits cross the x-axis in the vertical direction.
Further inspection of the system (2.19) shows that when y = 0 we have
y0 = − x; thus the orbits crossing the positive x-axis are directed
downward, and the orbits crossing the negative part of the x-axis are
directed upward.
The y-nullcline has the equation
− x + c(1 − x2 ) y = 0.
We will substitute c =
1
2
and solve for y to get
y=
2x
.
1 − x2
The orbits will be directed to the right as they cross the y-nullcline in the
upper half-plane, and in the lower half plane they are directed to the left.
Figure 2.14 shows the nullclines.
2.4. *AUTONOMOUS SYSTEMS
189
Figure 2.14 indicates that orbits of the van der Pol system travel around
the origin in a counterclockwise direction, but they may not be closed orbits
— it is possible that they are spirals. Figure 2.15 displays segments of two
orbits. I have chosen initial points ( x0 , y0 ) = (1, 0) and ( x0 , y0 ) = (3, 0)
for these orbits, and followed each of them until it returned to the positive
x-axis. You can see that neither orbit is closed, since neither returns to its
starting point.
Let
x = φ(t, u)
y = ψ(t, u)
denote the solution of the van der Pol system (2.19) with c = 12 and initial
point (φ(0), ψ(0)) = (u, 0). By the continuous dependence statement in theorem 2.1, the functions φ and ψ are continuous functions of (t, u). Let g(u)
be the function, defined for all u ∈ [1, 3], as follows: First let s(u) be the
least positive number such that ψ(s(u), u) = 0 and φ(s(u), u) > 0. Thus,
the orbit starting at (u, 0) on the positive x-axis will return to the positive
x-axis for the first time at t = s(u). Now let
g ( u ) = φ ( s ( u ), u )
be the x-coordinate of the solution with initial point at (u, 0), when it first
returns to the positive x-axis.
It follows from the continuity of φ and ψ (and the implicit function theorem) that g is continuous. Now refer to figure 2.15. You will see that
1 < g(1) < g(3) < 3. Thus the function h(u) = g(u) − u is continuous,
with h(1) > 0, and h(3) < 0. By the intermediate value theorem, there is
a number u∗ ∈ (1, 3) with h(u∗ ) = 0; that is, g(u∗ ) = u∗ . The solution
( x, y) = (φ(t, u∗ ), ψ(t, u∗ )) thus satisfies
(φ(0, u∗ ), ψ(0, u∗ )) = (u∗ , 0) = (φ(s(u∗ ), u∗ ), ψ(s(u∗ ), u∗ )).
Now it follows from theorem 2.3 that φ(t, u∗ ) and ψ(0, u∗ )) are periodic
functions of t, and the orbit that they describe is a closed curve. This orbit
is shown in figure 2.16.
Exercises
In problems 1 – 7, sketch the nullclines and find the stationary points of the
given system of ODEs. Then use an IVP solver to draw a few orbits.
190
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
x0
y0
Answer
0
x
2.
y0
= x 2 + y2
= x 2 − y2
x0
y0
Answer
0
x
4.
y0
= −x
= 4y − y2 ,
1.
3.
x0
y0
Answer
0
x
6.
y0
5.
= x ( x + y)
= y(2x − y)
= x ( x − y2 )
= y ( x − 4)
= x2
=y
= y+1
= y
x0 = x + y − 2
y0 = x − y
Answer
7.
Find integrals for the systems of ODEs in problems 8 – 13.
8.
x0 = x
y0 = y + x2
x0
y0
Answer
0
x
10.
y0
= 2x
= x+y
x0
y0
Answer
0
x
12.
y0
= 3x
= 5y
9.
11.
= x (1 − y )
= y ( x − 1)
= 3y
= 5x
2.4. *AUTONOMOUS SYSTEMS
191
x 0 = x 2 + y2
y0 = −2xy
Answer
13.
14.
Suppose that each orbit of an autonomous system of ODEs is an
ellipse of the form x2 + 4y2 = C, where C ≥ 0. Show that the origin is
a stationary point.
15.
Show that the graph of any solution of the non-autonomous first
order ODE,
dy
= f ( x, y),
dx
passing through the initial point ( x0 , y0 ), is the same as the orbit of
the autonomous system
dx
dt
dy
dt
= 1
=
f ( x, y)
passing through the same point.
Answer
16.
Problem 15 shows that a non-autonomous first order ODE can
be treated as a system of two autonomous ODEs. Find a way to
treat a non-autonomous system of two ODEs as a system of three
autonomous ODEs.
17.
Prove the existence and uniqueness theorem for second order
ODEs: Suppose that f (t, x, y) is continuous in a box B in R3 and satisfies a Lipschitz condition there with respect to the variables x, y (specify
this Lipschitz condition). Then, given (t0 , x0 , y0 ) ∈ B , there is a unique
solution of the IVP
d2 x
dt2
= f t, x,
dx
dt
x ( t0 ) = x0
dx dt t=t
= y0 .
0
Base your proof on theorem 2.1.
Answer
18.
Write out the Lipschitz condition without using vector notation.
192
19.
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
Consider the system
ax 0 + by0 = f (t, x, y)
cx 0 + dy0 = g(t, x, y)
with initial conditions x (t0 ) = x0 , y(t0 ) = y0 . Show that if the functions f (t, x, y), g(t, x, y) are continuous and satisfy a Lipschitz condition with respect to ( x, y) in a box centered at (t0 , x0 , y0 ) and if
ad − bc 6= 0 then this IVP has a unique solution.
Answer
20.
Here is a nice application of Green’s theorem from advanced calculus. (People
had advanced calculus are excused). Let
who haven’t
f ( x, y)
~v( x, y) =
be a vector field defined on a rectangular dog( x, y)
main R, and suppose that for all ( x, y) ∈ R, div ~v( x, y) > 0. Show
that the system
x0 =
y
0
f ( x, y)
= g( x, y)
has no closed orbits in R. This fact, first proved in 1900 by the Swedish
mathematician Ivar O. Bendixson, is called Bendixson’s criterion.
2.5
Populations of Interacting Species
Suppose that two species, A and B, occupy the same environment. It may
happen that A is a predator species and B is its prey, or that A and B compete for the same food supply. In either case, the population of one of the
species cannot be successfully modeled without considering the influence
of the other. We will consider simple models involving systems of two
ODEs.
Denote the populations of the two species by x and y. The derivatives
0
x (t) and y0 (t) represent the absolute growth rates of these populations,
and the relative growth rates are x 0 (t)/x (t) and y0 (t)/y(t), respectively. Our
model posits that each relative growth rate is a function of x and y. It can
thus be expressed as a system of two ODEs,
x 0 = x f ( x, y)
y0 = y g( x, y).
(2.20)
2.5. *POPULATIONS OF INTERACTING SPECIES
193
The logistic equation
x 0 = kx (1 − x/C )
is a one-species model similar to the system (2.20). We used it in section 1.10
to study the dynamics of the population x of a single species in an environment with a carrying capacity of C individuals. In this model, the relative
growth rate of x is a linear function of x, and in the two-species models that
we will consider, the relative growth rate for each species will be a linear
function of x and y.
The Lotka-Volterra Equations
The Lotka-Volterra equations,
x 0 = x ( a − by)
y0 = cy( x − d),
(2.21)
was the first differential equations model for the populations of more than
one species. It was developed in 1926 by the Italian mathematician Vito
Volterra as a model to explain the fluctuating populations of predator and
prey fish in the Adriatic Sea. The same system was proposed in 1920 by
an American biophysicist, Alfred Lotka, to determine the rate of a hypothetical chemical reaction (See problem 2 at the end of this section). Lotka’s
intention was to show that the concentrations of chemicals involved in a
reaction could vary periodically2 . In Volterra’s model, x is the prey population, and y represents the population of a predator whose sole source of
food is this prey. The relative growth rate of the prey species is set equal
to a − by, indicating that in the absence of the predator, the number of prey
will increase exponentially, but the relative rate of increase decreases linearly with the predator population. Of course, even in the absence of the
predator, the prey population would be limited by the availability of resources, but the predator prevents the prey population from approaching
the carrying capacity. The predator keeps the prey from starvation. The
second equation of the system (2.21) indicates that the predator will starve
without the prey: when x = 0 it reduces to y0 = −cdy; thus y = y0 e−cdt
(where y0 is the initial population). The parameter d represents the minimum prey population necessary to support predators.
2 Oscillating
chemical reactions are no longer hypothetical. The first observations of an
oscillating reaction were made in 1951 by B. P. Belousov, a chemist in the former Soviet
Union.
194
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
The system (2.21) has two stationary points. These are the solutions of
x ( a − by) = 0
cy( x − d) = 0,
( x, y) = (0, 0) and ( x, y) = (d, b/a). The stationary point at the origin is
not unexpected: if both species are extinct, they will stay extinct. The other
stationary point represents a situation in which the appetite of the predator
is exactly matched by the reproductive rate of the prey. Our experience
with the logistic equation suggests that the stationary point at the origin
is probably unstable, and the stationary point at ( x, y) = (d, b/a) is stable.
This is true, but we will see that it is an oversimplification.
We can apply the procedure introduced in section 2.4 to find an integral
for the system (2.21), by solving the ODE
cy( x − d) dx − x ( a − by) dy = 0.
This ODE is separable, and upon dividing through by − xy we can integrate
to obtain
Z Z d
a
c
− 1 dx +
− b dy = constant
x
y
and it follows that
F ( x, y) = cd ln( x ) + a ln(y) − cx − by
is an integral.
To keep the notation simple, we will use the notation Fx , Fy , and so on,
∂F
for the partial derivatives ∂F
∂x , ∂y , etc. The critical points of F ( x, y ) are found
by setting Fx and Fy equal to 0 and solving for x and y:
Fx ( x, y) = c(d/x − 1) = 0
Fy ( x, y) = a/y − b = 0.
The stationary point (d, a/b) is the only critical point, and we will apply the
second derivative test to it. Since Fxx (d, a/b) = −c/d, Fyy (d, a/b) = −b2 /a,
and Fxy ≡ 0,
2
c
b
Fxx (d, a/b) Fyy (d, a/b) − [ Fxy (d, a/b)] = −
−
> 0,
d
a
2
and we conclude that F ( x, y) has an absolute maximum at ( x, y) = (d, a/b).
2.5. *POPULATIONS OF INTERACTING SPECIES
195
It follows that the level curves of F ( x, y) in the first quadrant — which
are the orbits of the Lotka-Volterra equations—are closed curves (see figures 2.17 and 2.18) surrounding the stationary point at (d, a/b). Orbits do
not converge to this stationary point as t → ∞, but they don’t diverge away
from it either. Since the orbits are closed curves, we conclude that the solutions of the Lotka-Volterra equations are periodic functions of time, and
hence that the populations of predator and of prey will oscillate.
The Lotka-Volterra system is only a starting point in the study of populations of predators and their prey.
Competing Species
We will now consider an ecosystem in which two species A and B, with
populations x and y, compete for resources. The organisms may be plants,
competing for nutrients and sunlight, grazing animals competing for forage, or predators in competition for prey. The model is thus applicable to a
single level of the food chain.
If the species B were removed from the environment, the population of
the species A would grow according to the logistic equation, which we will
write in the form
dx
= ax (C − x ),
dt
where C is its carrying capacity, and a = k/C is the ratio of the initial relative growth rate to the carrying capacity. The species B will consume some
of the resources required by A. This is taken into account by subtracting
from the carrying capacity an amount to reflect the rate at which the resources needed by species A are consumed by species B. Thus the relative
growth rate for the species A is
a(C − ry − x ),
where r is the rate at which species B consumes the resources.
The relative growth rate of B is, by the same reasoning, proportional to
L − sx − y, where L denotes the carrying capacity for B and s is the rate at
which A consumes the resources that B needs. Putting these together, we
have the system
1 dx
x dt
1 dy
y dt
= a(C − x − ry)
= d( L − sx − y),
196
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
We have thus derived the competing species equations,
x 0 = x (k − ax − by)
y0 = y(l − cx − dy)
(2.22)
where k = aC, b = ar, l = dL, and c = ds.
The system (2.22) has four stationary points. Three of these occur when
one or both of the species are extinct. For example, if A is extinct (x = 0)
then the species B can have the stationary populations y = 0 (also extinct)
or L (its carrying capacity). Furthermore, in this case the orbits would just
be those of the phase diagram of the logistic equation.
The fourth stationary point, ( x1 , y1 ) is located at the intersection of the
lines
ax + by = k, and
(2.23)
cx + dy = l.
(2.24)
The line (2.23) is an x-nullcline, and the line (2.24) is a y-nullcline. If the
lines do not intersect in the first quadrant, this stationary point is not of
interest, because neither population can be negative.
Figure 2.19 shows four possible configurations of the nullclines, depending on the relative size of the parameters. The intercepts of the line (2.23)
are at (C, 0) and p = (0, k/b) , and since (C, 0) is a stationary point, it is
marked with a heavy dot. The intercepts of the line (2.24) are q = (l/c, 0)
and the stationary point (0, L). In configurations (a) and (d) the stationary
point E = ( x1 , y1 ) is also present. The arrows on the nullclines indicate the
direction orbits must take when they cross. Thus, the x-nullcline has vertical arrows. The part of the x-nullcline that lies above the y-nullcline is in
a region where y0 < 0, so the arrows point down. When the x-nullcline is
below the y-nullcline, the arrows point up, for y0 > 0. The directions of the
arrows marking the y-nullclines are explained in the same way.
It is not feasible to find an integral for the system (2.22), and to understand it we will need to use a different approach. A closed polygon
T in the phase plane is called a trap if every orbit that crosses an edge or
vertex of T is directed toward the interior. Thus, orbits can enter T , but
once they do so, they are trapped, since they cannot cross any edge from
the inside outward. Identifying traps is an important technique for studying autonomous systems of ODEs. For the system (2.22) we will use traps
whose edges are nullclines.
Configurations (b) and (d) are the easiest to analyze, and we will start
with (b). The quadrilateral LqCP is a trap, because all orbits are directed
2.5. *POPULATIONS OF INTERACTING SPECIES
197
downward as they cross the x-nullcline (the line pC), to the right as they
cross the y-nullcline (the line Lq), and no orbits cross the axes. Furthermore,
all orbits starting outside this quadrilateral must either enter it, or converge
to one of the stationary points C or L.
Inside the trap, x 0 is positive (it doesn’t become negative until the other
side of the x-nullcline is reached), and, since the quadrilateral lies above
the y-nullcline, Lq, y0 is negative. With orbits trapped inside, and directed
downward and to the right, it can be shown that they all converge to the
stationary point (C, 0). The conclusion: in configuration (b), species A is
dominant, and B will become extinct. The same reasoning shows that exactly the reverse is true in configuration (c): species B is dominant, and A
dies out. The phase portrait corresponding to configuration (b) is shown in
figure 2.20.
In configuration (a) of Figure 2.19 the quadrilateral OCEL (O stands for
the origin) is a trap. Within this quadrilateral, the triangles pEL and qCE
are also traps. In triangle pEL, orbits are directed upward and to the left;
and in qCE they are directed downward and to the right. Thus orbits starting near E will converge to L or C if they enter the interior of one of these
triangles. In particular, consider the situation when the initial population
is inside the quadrilateral OqEp. Its orbit must either converge to E, or enter one of the above-mentioned triangles, and hence converge to C or L. It
can be shown that there is one orbit in OqEp that does converge to E; it is
called the separatrix. Orbits that start above the separatrix will cross the
edge Ep, entering the triangle where all orbits converge to L; and orbits
starting below the separatrix will enter the triangle qCE and converge to
E. This situation is called competitive exclusion: the species with the initial population advantage dominates; the other faces extinction. The phase
portrait in this case is shown in figure 2.21.
Now let us consider a happier situation: configuration (d). The quadrilateral OqEp is a trap, and within it, the triangles LEp and CqE are also
traps. The distinction between this and configuration (a) is that in this case,
the orbits inside the triangles are directed toward E instead of away from
it. Since ever orbit either must converge directly to E or enter one of these
triangles, we conclude that all orbits converge to E. Here we have a stable
equilibrium in which the species share the resources. The phase portrait is
displayed as figure 2.22.
It is interesting to speculate on which of the above cases is applicable in
familiar situations, such as crabgrass in the lawn, benign versus pathogenic
bacteria in the body, introduction of new species of fish in a pond, and so
on. Conclusions reached by using this model are not to be trusted without
198
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
further study, because it is based on many simplifying assumptions.
Exercises
1.
A population of insect pests is to be controlled by introducing
predator insects. The objective is to keep the maximum number of
pests present at any time as small as possible. Assuming that the
Lotka-Volterra model is valid in this situation,
(a) If the growth parameters a, b, c, d of the Lotka Volterra equations on page 193 are known, what is the optimum number of
predators to release?
(b) Evaluate this strategy: Apply a pesticide to reduce the number
of pests as much as possible; when the pesticide has degraded,
introduce predators to keep the pests under control.
(c) If the number of pests in the environment is initially small, is it
preferable to introduce the predators immediately, or to wait for
a while?
Answer
2.
Given a chemical reaction Q + R → S, the Law of Mass Action
states that the rate at which the product S is produced is proportional
to the product of the concentrations of Q and R. Thus, if x, y, and z
denote the concentrations of Q, R, and S, respectively, then
dz
= kxy,
dt
where k is constant.
Lotka3 considered a hypothetical sequence of three reactions
P + Q → 2Q, Q + R → 2R, R → S.
Suppose that the concentration of P is held fixed, by replenishing that
reactant at the same rate that it is consumed. Let x and y denote the
concentrations of Q and R, and derive the Lotka-Volterra equations
as a model of this chemical system.
3 A. J. Lotka, “Undamped oscillations derived from the law of mass action,” J. Amer.
Chem. Soc. 42 (1920) pp. 1595 – 1599
2.5. *POPULATIONS OF INTERACTING SPECIES
199
3.
Two species of fish inhabit a pond; neither is the prey of the other.
The pond’s carrying capacity is 1000 fish of species A, or 1500 of
species B. Also, if there are 800 A-fish and 800 B-fish, the population will be stationary. Set up a system of ODEs to model the fish
populations, and draw a sketch of the phase plane, identifying traps.
How will the populations stabilize, assuming both populations are
initially positive?
Answer
4.
Repeat problem 3 under the assumption that the carrying capacity
of the pond for each individual species is unchanged, but the combination of A and B-fish in the pond for a stationary population is
(a) 800 A-fish and 250 B-fish.
(b) 600 fish of each species.
5.
Symbiosis. If the relationship between two species is mutually
beneficial, the species are said to be in symbiosis. Lichens are familiar
examples. While each lichen appears to be a single organism, in fact
it consists of two species, an alga and a fungus, in symbiosis. Draw a
the nullclines for the two species model,
x 0 = ax (K − x + By)
y0 = dy( L + Cx − y).
and identify any traps. You will need to consider the cases where
BC ≥ 1 and BC < 1 separately.
Answer
6.
This problem is for readers who have had an advanced calculus
course. Let ( x, y) = (φ(t), ψ(t)) be a solution of the Lotka-Volterra
equations, and let T denote its period. The average populations are
x̄ =
1
T
Z T
0
φ(t) dt and ȳ =
1
T
Z T
0
ψ(t) dt.
Prove that x̄ = d and ȳ = a/b. Hint: One proof is based on Green’s
Theorem.
7.
This problem does not require advanced calculus, but readers
who have not had this course will have to take the result of problem 6
on faith.
200
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
Volterra’s predator-prey model4 was based on data from fish catches
in the Adriatic Sea. Volterra observed that the ratio of predators to
prey varied periodically, and this is reflected by his mathematical
model. Let R denote the catch rate: that is, the fraction of the population (of both species) that is caught. Taking fishing into account,
the Lotka-Volterra equations become
x 0 = x ( a − by) − Rx
y0 = cy( x − d) − Ry.
Assuming that a > R, show that as the catch rate increases, the average population of prey increases, and the average predator population decreases. You may use the result of problem 6. Although the
Lotka-Volterra equations oversimplify, this effect is substantiated by
Volterra’s data. For the same reason, it is unwise to apply a pesticide
to “help” an insect predator control a pest species: the result will be
to increase the average number of pests.
Answer
2.6
Chapter Glossary
R3 Three-dimensional space.
Autonomous system A system of ODEs in which the independent variable does not appear on the right side.
Bendixson’s criterion If a vector field, defined of a rectangle R in the plane,
has nonvanishing divergence then there are no closed integral curves
in R.
Classical IVP solver An IVP solver that has a fixed time step.
Competing species equations A system of ODEs that models the dynamics of the populations of two species that compete for the same resources. See page 196.
Competitive exclusion A situation in which a species with an initial population advantage is able to overwhelm its competitor.
4 V.
Volterra, “Variazionie fluttuazioni del numero d’individui in specie animali conviventi,” Mem. Acad. Lincei 2 (1926) pp. 31 – 113. English translation: “Variations and
fluctuations in the population of animals living together,” in Animal Ecology, by R. N. Chapman (New York: McGraw-Hill, 1931), pp. 409 – 448.
2.6. CHAPTER GLOSSARY
201
Dynamic IVP solver An IVP solver that automatically adjusts its time step
to maintain precision.
IVP solver A numerical method for computing solutions of initial value
problems.
Integral (of a system of ODEs) A function F ( x, y) that is constant on all
orbits of the system.
Integral curve (of a vector field v( x, y) = f ( x, y)i + g( x, y)j) A curve given
parametrically by x = p(t), y = q(t), where for all t, the tangent
vector p0 (t)i + q0 (t)j is equal to the vector at ( p(t), q(t)), that is,
v( p(t), q(t)) = p0 (t)i + q0 (t)j.
Lipschitz condition See page 184.
Lotka-Volterra equations A system of ODEs that is a model for the population dynamics of a predator and its prey. See page 193.
Nullcline (for a system x 0 = f ( x, y), y0 = g( x, y)) The set of points where
f ( x, y) = 0 (the x-nullcline) of g( x, y) = 0 (the y-nullcline).
Order (of an IVP solver) A number n such that the accumulated error
varies proportionately with the nth power of the time step.
Orbit The graph in the phase plane of parametric equations given by a
solution of a system of two ODEs. It is customary to mark orbits to
indicate the direction of increasing t.
Phase plane The plane, in which the coordinates correspond to the values
of the dependent variables of a system of two ODEs.
Phase portrait A drawing of the phase plane, showing a representative
collection of orbits of a given system of two ODEs.
Replace an ODE The system
dy
dt
dv
dt
= v
= f (t, y, v)
)
is said to replace the second order ODE
d2 y
dy
= f t, y,
,
dt2
dt
(2.25)
(2.26)
202
CHAPTER 2. SYSTEMS OF DIFFERENTIAL EQUATIONS
since (y, v) = (φ(t), ψ(t)) is a solution of (2.25) if and only if y = φ(t)
satisfies (2.26) and ψ = φ0 .
Separatrix An orbit of an autonomous system of two ODEs that separates
orbits with different limiting behavior.
Stationary point (of a vector field v = f ( x, y)i + g( x, y)j) A point ( x1 , y1 )
such that v( x1 , y1 ) = 0. An integral curve containing a stationary
point consists of the stationary point alone.
(of a system of ODEs, x 0 = f ( x, y), y0 = g( x, y)) A point ( x1 , y1 ) such
that f ( x1 , y1 ) = g( x1 , y1 ) = 0
System of ODEs Two or more ODEs to be treated simultaneously. To be
well-posed, the system must have one equation for each dependent
variable, and only one independent variable.
Trap A closed polygon in the phase plane such that all orbits that cross its
edges or vertices are pointed inward.
Uncoupled system A system of two ODEs in which one of the dependent
variables does not appear in one of the equations.
Vector field A function, defined on a subset of the xy plane, that assigns to
each point ( x, y) in its domain a vector v( x, y) = f ( x, y)i + g( x, y)j in
the plane.
2.7. REVIEW EXERCISES
2.7
1.
203
Review Exercises
Solve the initial value problem
x 0 = x + e t ; x (0) = 0
y0 = xy2 ;
y (0) = 1
Answer
2.
Find the general solution of
x0 = − x
y0 = y.
Also find an integral for this system and draw its phase portrait.
Answer
3.
Repeat Exercise 2 for the system
x0 = x
y0 = 2y
Answer
4.
Suppose that F 0 (y) = f (y). Find an integral for the system that
replaces the ODE y00 + f (y) = 0. Apply the result to the following
ODEs, in which α denotes a constant parameter:
(a) y00 = −α2 sin(y) (the pendulum equation).
(b) y00 = −α2 y (the linearized pendulum).
(c) y00 = −α2 y−2 (a falling body).
Answer
5.
Find systems of first order ODEs to replace the given second order
ODEs.
(a) y00 + y = 0.
(b) y00 = y0 − t2 sin(y)
Answer
204
6.
CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS
Given that y = et cos(t) is a solution of the ODE
y00 = 2y0 − 2y
sketch the corresponding orbit of the system that replaces this ODE.
Answer
7.
Find an integral for the system
x 0 = − x ( x + 4y)
y0 = 2y( x + y)
Answer
8.
Find the stationary points of the following vector fields:
(a) ~v = ( x + y − 2) i + ( x − 3y + 2) j.
(b) ~v = (y − x ) i + (y − x3 ) j.
(c) ~v = x ( x + y + 4) i + y( x + 5y) j.
Answer
9.
10.
Write out the systems of ODEs corresponding to the vector fields
in Exercise 3 and sketch their nullclines.
Answer
Draw the nullclines of the system
x 0 = ( x + y)(y2 − 1)
y0 = (y − x )( x2 − 1)
and identify a trap.
Answer
11.
Use an IVP solver to draw several integral curves for each of the
vector fields in Exercise 3. Choose initial points near (but not at!) the
stationary points, and let the time variable range from −5 to 5.
Answer
12.
Use an IVP solver to plot the graph of the solution of the damped
pendulum equation, y00 = −.05y0 − sin(y) with initial conditions y(0) =
0.25, y0 (0) = 0. Repeat with the “linearized version” of the equation,
y00 = −.05y0 − y with the same initial conditions.
Answer
FIGURES FOR CHAPTER 3
Figure 2.1: Orbits of the system x0 = x, y0 = y.
205
206
CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS
Figure 2.2: Three orbits of the system that replaces the ODE y00 = y0 + 2y. The horizontal
axis is the y-axis; the vertical axis represents y0 .
20
15
10
5
-7.5
-5
-2.5
2.5
-5
5
7.5
10
FIGURES FOR CHAPTER 3
207
Figure 2.3: The vector field xi − yj
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-1
0
1
2
3
208
CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS
Figure 2.4: Component graphs of a solution ( x(t), y(t)) of the system (2.9).
2
y
1
-1
-0.5
1
0.5
-1
-2
x
t
FIGURES FOR CHAPTER 3
209
Figure 2.5: Orbit of a solution ( x(t), y(t)) of the system (2.9). The vector field representation of the system of ODEs is also displayed.
2
1.5
1
0.5
-3
-2
-1
1
-0.5
-1
-1.5
2
210
CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS
Figure 2.6: Graph of a solution ( x(t), y(t)) of the system (2.9).
2
1
0
-1
1
0
-1
-2
-2
-1
0
1
FIGURES FOR CHAPTER 3
211
Figure 2.7: Spreadsheet for approximating the solution of x0 = 2x + 2y, y0 = − x + 4y,
with initial conditions ( x (0), y(0)) = (1, 0), by Euler’s method with stepsize h = 0.1.
212
CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS
Figure 2.8: Spreadsheet calculation and orbit plot for the IVP, x0 = 2x + 2y, y0 = − x + 4y;
( x (0), y(0) = (1, 0),. The formulas in the spreadsheet are shown in figure 2.7.
FIGURES FOR CHAPTER 3
213
Figure 2.9: Orbit of x0 = 2x + 2y, y0 = − x + 4y, with initial condition x(0) = 1, y(0) = 0.
The upper curve was drawn from the analytic solution, and the lower one was drawn from
the approximation obtained by Euler’s method in example 2.3.1.
-6
-4
-2
2
-2.5
-5
-7.5
-10
-12.5
-15
214
CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS
Figure 2.10: : Vector field and phase portrait for the system of ODEs in example 2.12.
FIGURES FOR CHAPTER 3
215
Figure 2.11: : Stationary points and nullclines for the system of ODEs in example 2.4.2.
The x-nullcline is marked with vertical arrows indicating the direction of orbital crossing;
the y-nullcline is marked with horizontal arrows. Stationary points are located where the
two nullclines intersect.
10
7.5
5
2.5
-10
10
-2.5
-5
20
30
40
216
CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS
Figure 2.12: Orbits of x0 = − p2 y, y0 = q2 x.
3
2
1
-6
-4
-2
2
-1
-2
-3
4
6
FIGURES FOR CHAPTER 3
217
Figure 2.13: Phase portrait of y00 = −y2 : see example 2.4.4.
v
2
1.5
1
0.5
-3
-2
-1
1
-0.5
-1
-1.5
-2
2
3
y
218
CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS
Figure 2.14: Nullclines of the van der Pol system (2.19), where c = 1/2.
4
2
-4
-2
2
-2
-4
4
FIGURES FOR CHAPTER 3
219
Figure 2.15: Segments of two orbits of the van der Pol system (2.19), with c = 1/2.
2
1
-2
-1
1
-1
-2
2
3
220
CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS
Figure 2.16: A closed orbit of the van der Pol system (2.19), with c = 1/2.
2
1
-2
-1
1
-1
-2
2
FIGURES FOR CHAPTER 3
221
Figure 2.17: An integral for the Lotka-Volterra equations. The orbits are level curves.
The parameters are a = 0.1, b = 0.01, c = 0.0005, and d = 2000. Thus, the stationary point
is located at ( x, y) = (2000, 10).
6.5
40
6
30
5.5
20
1000
2000
10
3000
4000
5000
222
CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS
Figure 2.18: Integral curves of the Lotka-Volterra equations.
40
30
20
10
500
1000
1500
2000
2500
3000
FIGURES FOR CHAPTER 3
223
Figure 2.19: Four possible configurations for the nullclines of the system (2.22) The stationary points are marked with heavy dots; the x-nullcline is marked with vertical arrows,
and the y-nullcline is marked with horizontal arrows.
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224
CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS
Figure 2.20: In this phase portrait, the parameters correspond to the configuration shown
in figure 2.19(b). The species whose population is represented by x dominates the species
represented by y regardless of the initial conditions. The nullclines are represented by
dashed lines, and the parameters are (k, a, b, l, c, d) = (30, 5, 6, 6, 2, 3).
8
6
4
2
2
4
6
8
FIGURES FOR CHAPTER 3
225
Figure 2.21: Competitive exclusion: the phase portrait of the system 2.22, when the
nullclines (shown as dashed lines) are configured as in figure 2.19(a). The parameter values
are: k = 3, a = 1, b = 2, l = 5, c = 2, and d = 3.
2
1.5
1
0.5
0.5
1
1.5
2
2.5
3
226
CHAPTER 3. SYSTEMS OF DIFFERENTIAL EQUATIONS
Figure 2.22: When the population growth parameters correspond to the configuration
shown in figure 2.19(d), the two species will coexist with populations represented by the
stationary point in the first quadrant. The initial conditions have no effect on the outcome,
as long as both species are present. Nullclines are represented by dashed lines, and the
parameters are (k, a, b, c, l, c, d) = (15, 5, 3, 12, 2, 6).
6
5
4
3
2
1
1
2
3
4
5
6
Chapter 3
Linear Systems
227
228
3.1
CHAPTER 3. LINEAR SYSTEMS
The Initial Value Problem
A system of linear, first order ODEs,
x 0 = p1 ( t ) x + q1 ( t ) y + r1 ( t )
y 0 = p2 ( t ) x + q2 ( t ) y + r2 ( t ),
(3.1)
expresses the rate of change of each dependent variable as a function of the
independent variable and both dependent variables. Thus, although each
equation in the system (3.1) is first order, we cannot directly solve for x
without knowing y, or for y without having first determined x. The problem is that the two equations are coupled. This kind of coupling appears
in many applications, including mechanical systems, parallel electrical circuits, and in economic models involving supplies of various commodities.
This chapter is an introduction to the general properties of linear systems. If the coefficients pi (t) and qi (t) are constants, we will learn how to
decouple the two equations of the system (3.1) and thus to solve it. Systems
with variable coefficients rarely decouple.
Systems in matrix form.
We will use matrix notation for systems. A matrix is a rectangular array
of numbers or functions. If there are m rows and n columns, the matrix
is called an m × n matrix. For example, a 2 × 2 matrix would be a square
array
a11 a12
.
a21 a22
This matrix uses the standard notation: the number or function that is located in the horizontal row i and vertical column j is aij . These aij ’s are
called the entries of the matrix.
A matrix with just one column is called a column vector. We’ll use
arrow notation ~v,~a, etc. for column vectors, and capital letters A, B, etc. will
denote more general matrices. Scalars (that is, constants and real-valued
functions) will be denoted with lower case italic letters x, a, t, etc.
It is possible to add matrices and to multiply them, provided that the
dimensions of the matrices are compatible. Any two matrices of the same
dimensions can be added; each entry of C = A + B is the sum of the corresponding entries of A and B. Thus cij = aij + bij .
The matrix product P = AB of an m × n matrix A and a p × q matrix B
can be formed if and only if n = p; that is, the number of columns of A is
3.1. THE INITIAL VALUE PROBLEM
229
equal to the number of rows of B. When this compatibility condition holds,
P will be a matrix of dimension m × q, with as many rows as A has, and
as many columns as B. For example, if A is a 2 × 2 matrix and B is a 2dimensional column vector (a 2 × 1 matrix), AB would be a 2-dimensional
column vector. The formula for the product is
a11 a12
b1
a11 b1 + a12 b2
=
.
a21 a22
b2
a21 b1 + a22 b2
More generally, if A is an m × k matrix and B is a k × n matrix, the
entries of the product C = AB are computed as follows:
cij = ai1 b1j + ai2 b2j + · · · + aik bkj .
If you are familiar with the dot-product of vectors, the i, jth entry of C is the
dot-product of row i of A and column j of B. Matrix multiplication satisfies
the associative law: Given three matrices A, B, and C, of dimensions such
that the products P = AB and Q = BC are defined, it always holds that
PC = AQ; in other words,
( AB)C = A( BC ).
Matrix multiplication also satisfies the distributive laws: assume that A and
A0 are matrices of the same dimensions, and that B is a matrix of dimensions such that AB is defined. Let B0 be a matrix of the same dimensions as
B. Then
( A + A0 ) B = AB + A0 B and A( B + B0 ) = AB + AB0 .
Matrix algebra shares these properties with the arithmetic of real numbers.
However, the commutative law does not hold. If A is a p × q matrix and
B is q × p, then unless p = q, AB and BA are of different sizes (p × p and
q × q, respectively); thus AB 6= BA. Even it A and B are square matrices,
so that p = q, usually AB 6= BA.
To convert the system (3.1) to matrix form, let the column vector
x
~v =
y
represent the dependent variables, and the matrix
p1 ( t ) q1 ( t )
A(t) =
p2 ( t ) q2 ( t )
230
CHAPTER 3. LINEAR SYSTEMS
be the coefficient matrix. The vector function
r1 ( t )
~r (t) =
r2 ( t )
is the source vector. To see that the matrix ODE
~v0 = A(t)~v +~r (t)
is equivalent to the system (3.1), we have only to multiply the matrix A and
the column vector ~v to obtain the column vector
p1 ( t ) x + q1 ( t ) y
A(t)~v =
,
p2 ( t ) x + q2 ( t ) y
add the column vector ~r (t),
p1 ( t ) x + q1 ( t ) y
r1 ( t )
p1 ( t ) x + q1 ( t ) y + r1 ( t )
+
=
p2 ( t ) x + q2 ( t ) y
r2 ( t )
p2 ( t ) x + q2 ( t ) y + r2 ( t )
and write out the matrix equation as
0 x
p1 ( t ) x + q1 ( t ) y + r1 ( t )
=
y0
p2 ( t ) x + q2 ( t ) y + r2 ( t )
Example 3.1.1 Put the system
x0 = y
y0 = 4 x
in matrix form.
S OLUTION .We need to find a coefficient matrix
a11 a12
A=
a21 a22
such that
A
x
y
=
a11 x + a12 y
a21 x + a22 y
=
y
4x
.
It follows that a12 = 1, a21 = 4, and the other entries are 0. Hence
0 1
A=
,
4 0
3.1. THE INITIAL VALUE PROBLEM
231
and the matrix form of the system is ~v0 = A ~v.
It is not difficult to check that x = e2t , y = 2 e2t is a solution of the
system that we considered in example 3.1.1. We can put this solution in
vector notation as
2t e
~v =
.
2 e2t
It is possible to multiply any matrix by a scalar constant or function:
just multiply each entry by the scalar. Thus we can write the above vector
solution as
1
2t
~v = e
2
Example 3.1.2 Show that
~ =e
w
−2t
1
−2
is also a solution of the system in example 3.1.1.
S OLUTION . Let
~c =
1
−2
,
~ = e−2t ~c. We need to show that
so that w
~ 0 = A · (e−2t ~c)
w
(3.2)
By matrix multiplication,
0 1
1
(0)(1) + (1)(−2)
−2
A~c =
=
=
= −2~c.
4 0
−2
(4)(1) + (0)(−2)
4
Thus
d −2t
(e ~c) = e−2t (−2)~c = e−2t A ~c.
dt
We can now conclude that equation (3.2) holds, since
e−2t A ~c = A · (e−2t ~c).
~0 =
w
With vector notation, the existence and uniqueness theorem 2.1 for systems looks the same as the existence and uniqueness theorems for first order ODEs that we encountered in chapter 1. The following theorem is for
232
CHAPTER 3. LINEAR SYSTEMS
linear systems only; it differs from theorem 2.1 in that the domain of the
solution of an IVP involving a linear system can be specified. We will refer
to this theorem frequently.
Theorem 3.1 (Existence & Uniqueness) Suppose that the entries of the coefficient matrix A(t) and the components of the source vector ~f (t) are continuous on
an interval ( a, b) containing the initial point t0 . Then the IVP
~v0 = A(t)~v + ~f (t); ~v(t0 ) = ~v0
has a solution ~v(t) that is defined for all t ∈ ( a, b). If ~u(t) is another solution of
the same IVP, then ~u(t) = ~v(t) for all t ∈ ( a, b).
Homogeneous systems
A system of ODEs
~v0 = A(t)~v
with a zero source vector is said to be homogeneous . The following corollary of theorem 3.1 is a useful property of homogeneous systems.
Corollary 3.1.1 Let A(t) be an n × n matrix whose entries are continuous on the
interval ( a, b), and let ~v(t) be a solution of the homogeneous system ~v0 = A(t)~v.
If, for some t0 ∈ ( a, b), ~v(t0 ) = ~0, then ~v(t) ≡ ~0 on ( a, b).
P ROOF. Notice that ~v(t) and ~z(t) ≡ ~0 are solutions of the IVP
~v0 = A(t)~v; ~v(t0 ) = ~0.
The uniqueness assertion of theorem 3.1 implies that ~v(t) = ~z(t) for all
t ∈ ( a, b)
Definition: If ~v1 (t) and ~v2 (t) are vector functions that are defined on for
all t in an interval ( a, b) If c1 and c2 are scalar constants, the vector function
~v3 (t) = c1 ~v1 (t) + c2 ~v2 (t)
(3.3)
is called a linear combination of ~v1 and ~v2 . (It is possible to form linear
combinations of three or more vector functions in the same way.)
Now suppose that ~v1 (t) and ~v2 (t) in (3.3) are solutions of a homogeneous linear system of ODEs, ~v0 = A~v. Then
~v30 = c1 ~v10 + c2 ~v20
= c1 A ~v1 + c2 A ~v2
3.1. THE INITIAL VALUE PROBLEM
233
Therefore, by the distributive law for matrix multiplication,
~v30 = A (c1 ~v1 + c2 ~v2 )
= A(t)~v3 .
The result of this calculation can be summarized as the following theorem.
Theorem 3.2 Let ~v1 (t), ~v2 (t), . . . , ~vn (t) be solutions of a linear homogeneous
system, ~v0 = A(t)~v. Then every linear combination
c1~v1 + c2~v2 + · · · + cn~vn
of these solutions is also a solution.
If we have just one solution ~v1 (t), then theorem 3.2 tells us that the family
of functions c1 ~v1 (t) is an infinite family of solutions (unless v1 (t) ≡ 0, for
then the family would not be infinite). If there are two solutions available,
c1~v1 (t) + c2~v2 (t) is a larger family of solutions, unless ~v2 (t) happens to be
a constant multiple of ~v1 (t).
Definition: The linear span of a set of solutions S = {~v1 (t), ~v2 (t), . . . , ~vn (t)}
of a system of ODEs is the set of all vector functions that can be expressed
as linear combinations of the functions in S . We can also consider the linear
span of a set T of vectors {~b1 , . . . ,~bk } ⊂ Rn : it is the set of all possible linear
combinations of the vectors in T .
Theorem 3.2 can be rephrased as follows: Every vector function in the
linear span of any set of solutions of a homogeneous linear system of ODEs
is also a solution.
In example 3.1.2 we found that
1
1
2t
−2t
~v1 = e
and ~v2 = e
2
−2
are solutions of the system
0
~v =
0 1
4 0
~v.
The linear span of these solutions is the family
1
1
2t
−2t
~v = c1 e
+ c2 e
.
2
−2
234
CHAPTER 3. LINEAR SYSTEMS
Definition: The vector functions ~v1 (t) and ~v2 (t) are linearly independent
if neither function is equal to a constant scalar multiple of the other. More
generally, we will say that a set of vector functions
S = {~v1 (t), ~v2 (t), . . . , ~vm (t)}
is linearly independent if none of the ~vi (t) belongs to the linear span of the
other (m − 1) vector functions in S.
A similar definition applies to vectors in the plane (or n-dimensional
space Rn ). We say that a set of constant vectors, S = {~a1 ,~a2 , . . . ,~am } is
linearly independent if none of the vectors in S belongs to the linear span
of the other vectors in S.
Proposition 3.1.2 The set S (of vectors in Rn or vector functions) is linearly
independent if and only if the only linear combination of the elements of S that is
equal to ~0 is the combination in which all of the coefficients are zero.
P ROOF : (We will refer to the case of vector functions; the proof for the
case of constant vectors is the same.) Suppose that S is linearly independent, and that
c1~v1 (t) + c2~v2 (t) + · · · + cn~vn (t) ≡ 0.
We have to show that c1 = c2 = · · · cn = 0. We will do so by supposing
that this is not the case and reaching a contradiction. Thus suppose that at
least one of the coefficients ci is nonzero. We can select this ci to be the last
coefficient that is nonzero, so that c j = 0 for j > i. Thus
c1~v1 (t) + c2~v2 (t) + · · · + ci−1~vi−1 + ci~vi (t) ≡ 0.
Since ci 6= 0 we can divide through by ci and solve for vi (t), thus expressing
it as a linear combination of {~v1 (t), . . . , ~vi−1 (t)}:
~vi (t) =
−1
(c1~v1 (t) + c2~v2 (t) + · · · + ci−1~vi−1 ).
ci
This shows that ~vi (t) belongs to the linear span of {~v1 (t), ~v2 (t), . . . , ~vi−1 },
contradicting our hypothesis that S is linearly independent.
To prove the converse, assume that the only linear combination of S
that is zero is the combination where all coefficients are zero. Now we will
assume that one of the elements of S, ~vi (t) lies in the span of the other n − 1
elements of S. This means there are scalar constants c1 , . . . , ci−1 , ci+1 , . . . , cn
such that
~vi (t) = c1~v1 (t) + c2~v2 (t) + · · · + ci−1~vi−1 + ci+1~vi+1 + · · · + cn~vn (t).
3.1. THE INITIAL VALUE PROBLEM
235
Put ci = −1. Then
c1~v1 (t) + c2~v2 (t) + · · · + ci−1~vi−1 + ci~vi (t) + ci+1~vi+1 + · · · + cn~vn (t) = 0
is a linear combination of S in which not all of the coefficients are zero,
contradicting our hypothesis.
Theorem 3.3 Let A(t) be an n × n matrix whose entries are continuous functions
on an interval ( a, b), and let t0 be any point in ( a, b). Then solutions ~v1 (t), ~v2 (t)
. . . , ~vn (t) of the linear homogeneous system ~v0 = A(t)~v are linearly independent
if and only if, as vectors in Rn ,
~v1 (t0 ), ~v2 (t0 ), . . . , ~vn (t0 )
are linearly independent.
P ROOF. For simplicity, let n = 2. (The proof can be used for general
n with only slight modification.) If ~v1 (t0 ) and ~v2 (t0 ) are not linearly independent, then we can assume that there is a number c such that ~v2 (t0 ) =
c~v1 (t0 ).
~v3 (t) = c ~v1 (t) − ~v2 (t)
is a linear combination of solutions of ~v0 = A(t)~v, and is therefore also a
solution. Since ~v3 (t0 ) = ~0, it follows from corollary 3.1.1 that ~v3 (t) = ~0 for
all t, and hence ~v2 (t) = c~v1 (t) for all t ∈ ( a, b).
Thus, if ~v1 and ~v2 are linearly independent as vector functions on ( a, b),
the initial vectors ~v1 (t0 ) and ~v2 (t0 ) must also be linearly independent.
To complete the proof we have to show that if ~v1 (t0 ) and ~v2 (t0 ) are
linearly independent, then the vector functions ~v1 and ~v2 are linearly independent. If there were a scalars c such that
~v2 (t) = c ~v1 (t)
or all t ∈ ( a, b) then
~v2 (t0 ) = c ~v2 (t0 ).
This contradicts our assumption that ~v1 (t0 ) and ~v2 (t0 ) are linearly independent.
Theorem 3.3 implies that the solution vectors ~v1 (t), ~v2 (t) . . . , ~vn (t) are
either linearly independent for all t in the domain ( a, b) where A(t) is continuous, or they are linearly dependent for all t ∈ ( a, b). They cannot be
linearly independent at some points of the domain ( a, b), and linearly dependent at others.
236
CHAPTER 3. LINEAR SYSTEMS
Example 3.1.3 Show that the solutions
~v1 = e
2t
1
2
and ~v2 = e
−2t
1
−2
of the system
0
~v =
0 1
4 0
~v.
are linearly independent.
1
−2
S OLUTION . Since ~v2 (0) =
is not a scalar multiple of
1
~v2 (0) =
, the two solutions are independent.
2
We are now almost ready for a full description of the general solution
of a homogeneous linear system of ODEs. The following definition is from
linear algebra:
Definition: A basis for Rn is a set S of vectors that is linearly independent, and whose linear span is all of Rn .
There is a parallel definition in the realm of systems of ODEs:
Definition: Let A(t) be an n × n matrix whose entries are functions of t. A
fundamental set of solutions of the homogeneous linear system ~v0 = A(t)~v
is a set of vector functions that is linearly independent, and whose linear
span is the set of all solutions of the system.
Just one more fact from linear algebra is needed: Let S = {~a1 ,~a2 , . . . ,~an }
be a set of n vectors in Rn . Then the following are equivalent:
• S is a basis for Rn .
• S is linearly independent.
• The linear span of S is equal to Rn .
Theorem 3.4 Let A(t) be an n × n matrix of functions, all continuous on an
interval ( a, b), let t0 be any point in ( a, b), and let S = {~a1 ,~a2 . . . ,~an } be a basis
of Rn .
3.1. THE INITIAL VALUE PROBLEM
237
Then the solutions ~v1 (t), ~v2 (t), . . . , ~vn (t) of the system
~v0 = A(t)~v,
(3.4)
satisfying the initial conditions ~vi (t0 ) = ~ai , respectively, form a fundamental set
of solutions of (3.4).
P ROOF. To keep the notation as simple as possible, we take n = 2. (The
proof works in the same way for any n.)
We’ll start by showing that the linear span of T = {~v1 (t), ~v2 (t)} con~ (t) of (3.4). Because {~a1 ,~a2 } is a basis for R2
tains an arbitrary solution w
2
~ (t0 ) ∈ R there are scalars c1 , c2 such that w
~ (t0 ) = c1~a1 + c2~a2 .
and w
~ (t) − (c1~v1 (t) + c2~v2 (t)). As it is
Define the vector function ~u(t) = w
a linear combination of solutions of (3.4), u(t) is also a solution. Because
~vi (t0 ) = ~ai , it follows that ~u(t0 ) = ~0. Then, by corollary 3.1.1, ~u = ~0 identically. Thus,
~ (t) = c1~v1 (t) + c2~v2 (t)
w
for all t ∈ ( a, b). This establishes that the linear span of T comprises all
solutions of (3.4).
To complete the proof, we need to show that T is linearly independent.
This follows from theorem 3.3, because S = {~a1 , a~2 } is a basis for R2 , thus
linearly independent.
The general solution of a system of ODEs is a family of solutions that
encompasses all solutions of the system.
Corollary 3.1.3 Let A(t) be an n × n matrix of functions that are continuous on
the interval ( a, b). Let t0 ∈ ( a, b), and let ~v1 (t), ~v2 (t) . . . , ~vn (t) be solutions of
~v0 = A(t)~v such that the set of vectors
S = {~v1 (t0 ), ~v2 (t0 ), . . . , ~vn (t0 )}
is a basis of Rn . Then the general solution of ~v0 = A(t)~v is
~v = c1 ~v1 + c2 ~v2 + · · · + cn~vn .
Example 3.1.4 Solve the IVP,
x 0 = y; x (0) = 1
y0 = 4x; y(0) = 0
238
CHAPTER 3. LINEAR SYSTEMS
S OLUTION . In matrix form, this system is
0 x
0 1
x
=
.
0
y
4 0
y
In example 3.1.3 we identified linearly independent solutions
1
1
2t
−2t
~v1 (t) = e
and ~v2 (t) = e
.
2
−2
Therefore, the general solution is ~v = c1~v1 + c2~v2 , and we need to
determine values for the coefficients so that
1
~v(0) =
.
0
Thus we need to solve
c1
1
2
+ c2
1
−2
=
1
0
,
which is equivalent to the pair of equations
c1 + c2 = 1
2c1 − 2c2 = 0
Thus c1 = c2 = 21 . The solution is x = 12 e2t + 12 e−2t = cosh(2t) and
y = e2t − e−2t = 2 sinh(2t).
Vector Spaces
Definition: A vector space is a set V of objects, together with an additive
~ ∈ V then there is a “vector” ~u = ~v + w
~ ∈ V . It is
structure, so that if ~v, w
also required that if r is a scalar (that is, a real or complex constant) and
~v ∈ V then there is a “vector” ~x = r~v ∈ V . It is customary to call V a real
vector space if the scalars are real, or a complex vector space if complex
scalars are used.
We can summarize this definition as follows: a vector space is a set
of objects that can be combined by taking linear combinations. To avoid
~
ambiguities, some rules have to be imposed. These are as follows (~u, ~v, w
represent arbitrary elements of V , and r, s are arbitrary scalars):
3.1. THE INITIAL VALUE PROBLEM
239
~ and w
~ + ~x = ~y then
• Addition is associative: If ~u = ~v + w
~u + ~x = ~v + ~y.
~ ) +~x = ~v + (~
(This can also be expressed as (~v + w
w + ~x ).)
| {z }
| {z }
~u
~y
~ =w
~ + ~v.
• Addition is commutative: ~v + w
• There is an additive identity, denoted ~0, such that ~0 + ~v = ~v for each
vector ~v.
• Each vector ~v has a unique additive inverse −~v such that ~v + (−~v) =
~0.
• Multiplication by the scalar 1 sends each vector to itself: 1~v = ~v.
• Two distributive laws that govern the interaction of scalar multiplication and addition:
Left: (r + s)~v = r~v + s~v
~ ) = r~v + r w
~
Right: r (~v + w
Here are some exercises:
1. Show that 0~v = ~0 by using a distributive law and the familiar fact
0 + 0 = 0 for scalars.
2. Show that (−1)~v = −~v
In addition to Rn and Cn , there are other vector spaces that will attract
out attention. If ( a, b) is an interval, C n ( a, b) will denote the collection of
dn f
all functions f (t) with the property that n (t) is continuous on ( a, b). In
dt
particular C0 ( a, b) is the collection of continuous functions on ( a, b). If the
functions are real valued, C n ( a, b) is a real vector space, because the functions (which play the role of “vectors”) can be added and multiplied by
scalars. When working with complex-valued functions, C n ( a, b) becomes a
complex vector space.
240
CHAPTER 3. LINEAR SYSTEMS
Definition: If V is a vector space and W is a nonempty subset of V then
we will say W is a subspace of V if W is a vector space in its own right.
~ ∈ W and r, s are
To see if W is a subspace, you have to show that if ~v, w
~ ∈ W too. If r = s = 0 this would mean ~0 ∈ W , so the
scalars, then r~v + sw
first thing to check when verifying that W is a subspace is that ~0 ∈ W .
Vector spaces are the environment for taking linear combinations, so
given S = {~v1 , . . . , ~vn } ⊂ V we can form the span of S: the set of all possible
linear combinations of the vectors in S. If S is nonempty then span S is a
subspace of V . It also makes sense to ask if the set S is linearly independent.
If that is the case then S is a basis of the subspace W = span S.
Definition: Let W be a vector space that has a finite basis S. The number
of vectors that comprise S is the dimension of W . You will wonder: “What
if S and T are separate bases of W . Must they have the same cardinality?”
The answer, proved in all linear algebra courses, is yes.
A vector space may not have a finite basis. For example, let P be the set
of all polynomials with real coefficients. Addition and scalar multiplication
are included in the rules for manipulating polynomials, so P is a real vector
space. Let S = {1, x, x2 , x3 , . . .}. Every polynomial is a linear combination
of a finite number of elements of S, so P = span S. Furthermore, we can
show S is linearly independent by noting that if the polynomial
f ( x ) = a0 + a1 x + · · · + a n x n
is identically zero, then it has more than n zeros, which is all a nontrivial
polynomial of degree n can have—therefore a0 = a1 · · · = an = 0. It follows
that S is a basis for P , and is not finite. A vector space that does not have
a finite basis is infinite dimensional. Besides P , the spaces C n [ a, b] that we
just introduced are infinite dimensional vector spaces—indeed, all contain
P as a subspace.
The vector space concept is important in the context of differential equations. In fact, let A(t) be an n × n matrix whose entries are continuous on
an interval ( a, b). Solutions of the homogeneous system ~v0 = A(t)~v will
be vector functions that are differentiable on ( a, b). Each such function has
n components, each belonging to C1 ( a, b); thus all solutions will belong to
V = (C1 ( a, b))n . The principal results of this section can be written compactly in terms of vector spaces as follows:
• The solutions of ~v0 = A(t)~v form a subspace of V (Theorem 3.2). Let
W denote that subspace.
3.1. THE INITIAL VALUE PROBLEM
241
• If S = {~v1 (t), ~v2 (t) . . . ,~vn (t)} ⊂ W and t0 ∈ ( a, b) then S is linearly
independent if and only if the set of vectors
S(t0 ) = {~v1 (t0 ), ~v2 (t0 ) . . . , ~vn (t0 )}
is a linearly independent subset of Rn (Theorem 3.3).
• Let S = {~a1 ,~a2 . . . ,~an } be a basis of Rn . Then there exist solutions
~vi (t) ∈ W such that ~vi (t0 ) = ~ai , for i = 1, 2, . . . , n (Theorem 3.1).
• The set of solutions ~vi (t), whose existence was just asserted, is a fundamental set of solutions of ~v0 = A(t)~v (Theorem 3.4).
• The definition of a fundamental set of solutions is the same as the
definition of a basis for W . Thus S is a basis for W .
• The vector space W of solutions of the homogeneous system ~v0 =
A(t)~v is n-dimensional, where n is the size of the square matrix A(t).
Exercises
~ ∈ R2 .
Let A be a 2 × 3 matrix, B be a 3 × 2 matrix, ~v ∈ R3 , and w
~
Make a list of six products that can be formed with A, B, ~v, and w
using matrix multiplication.
Answer


1 −1
8
1 4 7
3 , and ~v =
. Form
2.
Let A =
,B =  2
7
2 0 5
−5 4
~ = B~v, C = AB, ~x = Aw
~ , and ~y = C~v. If
the following products: w
you multiplied correctly, you should get ~x = ~y. Why?
1 0
a b
3.
Let E =
, and A =
.
0 0
c d
1.
(a) Calculate EA.
(b) Calculate AE.
(c) What property must the matrix A have in order to commute
with E (that is,what conditions must the entries a, b, c, d satisfy
to ensure that AE = EA)?
Answer
242
CHAPTER 3. LINEAR SYSTEMS
4.
Let F =
1 1
1 0
, and let A be the same as in Exercise 3.
(a) Calculate FA.
(b) Calculate AF.
(c) What property must the matrix A have in order to commute
with F?
5.
Put the system
x 0 = 3x − 2y
y0 = x + 7y
in matrix form.
Answer
6.
Put the system that replaces the second order ODE
y00 + 2y0 + y = 0
in matrix form.
7.
Find the matrix form of the system that replaces the inhomogeneous third order ODE
x 000 = (sin t) x 00 − tx 0 + 2et x + tan t
Answer
8.
Put the inhomogeneous system
x 0 = x + t y + t2
y0 = t x − y + 1
in matrix form.
9.
Show that
e
3t
1
1
,e
−t
1
−1
is a fundamental set of solutions of the system
0 x
1 2
x
=
.
0
y
2 1
y
Find the general solution.
Answer
3.1. THE INITIAL VALUE PROBLEM
10.
Show that
x (t)
y(t)
243
=
3t − 5
−6t + 4
is a solution of the system
0 x
1 2
x
9t
=
+
.
y0
2 1
y
0
In Exercises 11 – 15, solve the IVP.
0 x
2 0
x
x (0)
1
11.
=
;
=
.
y0
0 3
y
y (0)
2
Answer
0 x
−1 0
x
x (0)
2
12.
=
;
=
.
0
y
0 1
y
y (0)
0
x
1 2
x
x (0)
2
d
=
;
=
. (See Exercise 9.)
13.
dt
y
2 1
y
y (0)
0
Answer
x
1 2
x
x (0)
0
d
14.
=
;
=
. (See Exercise 9.)
dt
y
2 1
y
y (0)
5
0 x
0 0
x
x (0)
1
15.
=
;
=
.
y0
−1 1
y
y (0)
1
Answer
In Exercises 16 – 19, write the IVP in matrix form. For each system, use
corollary 3.1.3 to verify that the given family of solutions is actually the
general solution. Finally, use the general solution to solve the IVP.
x 0 = 4x + 3y; x (0) = 2
y0 = 3x − 4y; y(0) = 4.
x
3
1
5t
−
5t
General solution:
= c1 e
+ c2 e
.
y
1
−3
0
x = 2x + 3y; x (0) = −1
17.
y0 = − x − 2y; y(0) = 1.
x
3
1
General solution:
= c1 e t
+ c2 e − t
.
y
−1
−1
Answer
16.
244
18.
CHAPTER 3. LINEAR SYSTEMS
x 0 = x − 2y; x (0) = 1
y0 = 2x − 4y; y(0) = −1
1 + 2t
x
−2t
−
t
=e
c1
+ c2
.
General solution
2t
y
1 − 2t
x 0 = 4x + 3y; x (0) = 7
y0 = −2x − y; y(0) = −2
1
3
x
t
2t
= c1 e
+ c2 e
.
General solution
−1
−2
y
Answer
3
1
5t
−
t
20.
Show that e
,e
is a fundamental set of so−1
−3
lutions of the system in Exercise 16.
19.
3
1
t
−
t
21.
Show that e
,e
is a fundamental set of so−1
−1
lutions of the system in Exercise 17.
Answer
22.
The system
x0 = x + y
y0 = y
is uncoupled. Write the system in vector form and find a fundamental
set of solutions.
In Exercises 23–26 solve the IVP. The systems are not homogeneous,
but their general solutions are given.
x 0 = 2x − y;
x (0) = 1
y0 = 4x − 2y + 2; y(0) = −1.
1
2t + 1
x
− t2
General solution:
=
+
c
+
c
.
2
1
y
2t − 2t2
2
4t
Answer
0
x = x + 2y + 2et ; x (0) = 0
24.
y0 = − x − y;
y(0) = 0.
General
solution:
x
2
2 cos(t)
2 sin(t)
t
=e
+ c1
+ c2
.
y
−1
− cos(t) − sin(t)
− sin(t) + cos(t)
23.
3.2. SYSTEMS WITH CONSTANT COEFFICIENTS
25.
26.
27.
245
x 0 = y + cosh t; x (0) = 2
y0 = x + sinh t; y(0) = 2
x
cosh t
sinh t
= ( c1 + t )
+ c2
General solution:
y
sinh t
cosh t
Answer
x 0 = 2x + 2y + et ; x (0) = 0
.
y0 = 3x + y;
y (0) = −2
2
1
−5
x
−
t
4t
t
General solution:
= c1 e
+ c2 e
+e
.
−3
1
3
y
A linear second-order ODE is an ODE that can be put in the form
y00 + p(t)y0 + q(t)y = r (t).
(3.5)
The functions p(t) and q(t) are the coefficients of the ODE, and r (t) is
the source. Show that a linear second-order ODE can be replaced with
a linear system of first-order ODEs, and write the system corresponding to equation (3.5) in matrix form.
Answer
28.
3.2
State and prove, as a corollary to theorem 3.1, an existence and
uniqueness theorem for linear second-order ODEs.
Systems with Constant Coefficients
The general solution of the linear homogeneous ODE,
dy
= a(t) y,
dt
is found by substituting y = C e f (t) . This yields
C e f (t) f 0 (t) = a(t)(C e f (t) ).
After dividing through by Ce f (t) , we obtain f 0 (t) = a(t). It follows that
y = Ce
R
a(t) dt
is a family of solutions. By proposition 1.2.1, it is the general solution.
This idea can be used to solve a system of linear homogeneous ODEs
if the coefficient matrix of the system is a constant matrix. Consider a homogeneous system with n dependent variables x1 , . . . xn forming a vector
~v,
~v0 = A~v,
(3.6)
246
CHAPTER 3. LINEAR SYSTEMS
where A is an n × n constant matrix. Substitute ~v = est~b, where ~b is a
constant vector, and s is a constant that will be determined. Then ~v0 = sest~b,
and A~v = est A~b, so with this substitution, (3.6) becomes
sest~b = est A~b.
We may divide by the nonzero factor est to obtain
s~b = A~b.
(3.7)
Equation (3.7) will hold only if the vector A~b is a scalar multiple of ~b
(and s is the scalar). A vector ~b 6= ~0 with this property is called a eigenvector of the matrix A, and s is a eigenvalue of that matrix. Since there is an
association between s and ~b, we say that ~b belongs to s, and s belongs to ~b.
Theorem 3.5 If ~b is an eigenvector of the matrix A, and s is an eigenvalue belonging to ~b, then
~v(t) = est~b
is a solution of the system (3.6).
Theorem 3.5 is motivation to find eigenvalues and eigenvectors of square
matrices—then we can solve systems of ODEs with constant coefficients.
The method presented here is in many linear algebra texts. It is placed in
the context of the 2 × 2 matrix, and although it works in principle for n × n
matrices, the effort required makes it impractical for n ≥ 3; then one should
resort to a CAS to find eigenvalues and eigenvectors.
The matrix
1 0
I=
0 1
is called the 2 × 2 identity matrix. There is an n × n identity matrix for any
n, with the same pattern of zeros and ones, and you can readily check that
matrix I has the property that I ~b = ~b for any vector ~b.
For any scalar r, the matrix
r 0
rI =
0 r
is called a scalar matrix. Now, if r is an eigenvalue of an n × n matrix A
and ~b is an eigenvector belonging to r, then
A~b = r I~b,
3.2. SYSTEMS WITH CONSTANT COEFFICIENTS
247
or, in more compact form,
( A − r I )~b = ~0.
(3.8)
A matrix C with the property that there is a nonzero vector d~ such that
C d~ = ~0 is said to be singular. Since ~b is a nonzero vector (Remember!
~0 is not permitted to be an eigenvector), equation (3.8) implies that that
( A − r I ) is singular.
Conversely, if r is given so that ( A − r I ) is singular, then there must
exist a nonzero vector~b such that ( A − r I )~b = ~0, and thus r is an eigenvalue
of A belonging to the eigenvector ~b.
We have proved the following result:
Proposition 3.2.1 Let A be a square matrix. A scalar r is an eigenvalue of A if
and only if ( A − r I ) is singular matrix.
To find eigenvalues, find all values of r such that ( A − r I ) is a singular
matrix. Let
a c
B=
.
b d
be a 2 × 2 matrix, whose entries can be constants or functions. The determinant of B is defined to be a scalar that is calculated by the formula
det( B) = a d − b c.
Example 3.2.1 Find the determinant of
9 7
3 5
S OLUTION .
det
9 7
3 5
= 9 · 5 − 3 · 7 = 24.
The determinant can be defined for square matrices of any size, but in
this text we will not consider determinants of matrices larger than 2 × 2.
The following proposition will be familiar to all who have studied linear
algebra, where it is shown to be valid for n × n matrices.
m n
Proposition 3.2.2 Let C =
be a 2 × 2 matrix. Then C is singular if
p q
and only if det(C ) = 0.
248
CHAPTER 3. LINEAR SYSTEMS
h
,
k
(where h and k are not both 0), such that C~b = ~0. By matrix multiplication,
P ROOF Assume that C is singular. Then there is a vector ~b =
C~b =
mh + nk
ph + qk
=
0
0
Assuming for definiteness that h 6= 0,
k
k
m = − n and p = − q.
h
h
It follows that
det(C ) = mq − pn =
k
k
nq − qn = 0.
h
h
Conversely, assume that det(C ) = 0. Since mq = np, you can easily
show that
n
q
C
=C
= ~0
−m
−p
n
If C has any nonzero entries, then at least one of the vectors
and
−m
q
is nonzero, and hence C is singular.
−p
If all of the entries of C are equal to 0, then C~b = ~0 for every vector ~b,
and hence C is singular.
By propositions 3.2.1 and 3.2.2, r is an eigenvalue of A if and only if
when s = r, the determinant of the matrix (entries are functions of s)
a−s
b
A − sI =
c
d−s
is equal to 0. Let’s calculate this determinant, as a function of s!
det
a−s
b
c
d−s
= ( a − s)(d − s) − bc.
Thus det( A − s I ) is a quadratic polynomial in the variable s, which can be
simplified as
det( A − sI ) = s2 − ( a + d)s + ( ad − bc).
3.2. SYSTEMS WITH CONSTANT COEFFICIENTS
249
The coefficients of this equation are significant. You will recognize the
constant term as det( A). The coefficient of s is equal to the sum of the diagonal entries of A, and is called the trace of A. (Notation: tr ( A).)
We have thus proved the following result:
Proposition 3.2.3 Let A be a 2 × 2 matrix. Then eigenvalues of A are the roots
of the quadratic equation
s2 − tr ( A) s + det( A) = 0.
(3.9)
Equation (3.9) is called the characteristic equation of A.
Once an eigenvalue of a matrix A is known, the eigenvectors belonging
to it are found by solving a system of linear equations.
Example 3.2.2 Find the eigenvalues and eigenvectors of
1 2
A=
.
3 2
(3.10)
S OLUTION . Since tr ( A) = 1 + 2 = 3 and det( A) = (1)(2) − (3)(2) = −4,
the characteristic equation of A is
s2 − 3s − 4 = 0 or (s + 1)(s − 4) = 0.
It follows that the eigenvalues are s = −1 and s = 4.
The eigenvectors belonging to s = −1 satisfy the equation A~b = −~b. Set
~b = h .
k
Then
A~b =
h + 2k
3h + 2k
=
−h
−k
Hence h + 2k = −h and 3h + 2k = −k. Both of these equations reduce to
k = −h, so any nonzero vector that is a scalar multiple of
1
~b1 =
−1
is an eigenvector belonging to −1.
To find an eigenvector belonging to s = 4, we have to solve
1 2
h
4h
~
Ab =
=
.
3 2
k
4k
250
CHAPTER 3. LINEAR SYSTEMS
This is equivalent to solving the equations
h + 2k = 4h
3h + 2k = 4k
These equations both reduce to 3h = 2k, so it follows that any nonzero
scalar multiple of
~b2 = 2
3
is an eigenvector belonging to 4.
To determine the general solution of a system of ODEs, it is helpful
to know that solutions corresponding to different eigenvalues are linearly
independent. The following theorem from linear algebra can be applied for
this purpose.
Theorem 3.6 Let s1 , s2 , . . . sm be distinct eigenvalues of a matrix A, and let ~b1 ,
~b2 . . . , ~bm be eigenvectors belonging to them. Then
{~b1 ,~b2 , . . . ,~bm } is linearly independent.
P ROOF FOR m = 2. Suppose that ~b2 = c~b1 . Then A~b2 = cAb~1 = cs1~b1 =
s1~b2 . However, ~b2 is an eigenvector belonging to s2 . Hence A~b2 = s2~b2 It
follows that s1 = s2 , a contradiction.
Corollary 3.2.4 Let A be a 2 × 2 matrix that has two distinct eigenvalues s1 and
s2 . and let ~b1 and ~b2 be eigenvectors belonging to them. Then the general solution
of ~v0 = A~v is
~v(t) = c1 es1 t~b1 + c2 es2 t~b2 .
(3.11)
P ROOF. Let ~v1 (t) = es1 t~b1 and ~v2 (t) = es2 t~b2 . Then by theorem 3.6
~v1 (0) = ~b1 and ~v2 (0) = ~b2 are linearly independent. By corollary 3.1.3 on
page 237, it follows that the general solution can be expressed as in (3.11).
Example 3.2.3 Find the general solution of the system
x 0 = 4x − 4y
y0 = 3x − 3y
3.2. SYSTEMS WITH CONSTANT COEFFICIENTS
S OLUTION . Let
~v =
x
y
and A =
4 −4
3 −3
251
so that our system is equivalent to ~v0 = A~v. Since tr ( A) = 1 and
det( A) = 0, the characteristic equation is s2 − s = 0 and the eigenvalues
are 0 and 1. Let
~b1 = h
k
be an eigenvector belonging to 0. Then
4h − 4k = 0
3h − 3k = 0.
This reduces to h = k, so we can take
~b1 = 1
1
Now suppose that
~b2 =
h
k
is an eigenvector belonging to 1. Then
4h − 4k = h
3h − 3k = k,
so that 3h = 4k. Hence we can set
~b2 =
4
3
.
The general solution to the matrix equation is
1
4
0t
1t
~v = c1 e
+ c2 e
.
1
3
Thus x = c1 + 4c2 et and y = c1 + 3c2 et .
Example 3.2.4 Solve the IVP
2 4
1
0
~v =
~v; ~v(0) =
.
3 −2
0
252
CHAPTER 3. LINEAR SYSTEMS
S OLUTION . The trace of the coefficient matrix is 0, the determinant is −16,
and the characteristic equation is s2 − 16 = 0. The eigenvalues are ±4. To
find an eigenvector belonging to 4, we solve
2 4
h
4h
=
,
3 −2
k
4k
or,
2h + 4k = 4h
3h − 2k = 4k.
Both of these equations reduce to 2k = h, so
~b1 = 2
1
is an eigenvector belonging to 4. Turning to −4,
~b2 = h
k
is an eigenvector if
2h + 4k = −4h
3h − 2k = −4k.
These equations reduce to 2k = −3h so we will put h = 2 and k = −3.
The general solution is
2
2
4t
−4t
~v(t) = c1 e
+ c2 e
1
−3
Thus
~v(0) =
2c1 + 2c2
c1 − 3c2
Since the initial condition specifies that ~v(0) =
c2 by solving
2c1 + 2c2 = 1
c1 − 3c2 = 0.
1
0
, we can find c1 and
3.2. SYSTEMS WITH CONSTANT COEFFICIENTS
253
The solutions are c1 = 3/8, c2 = 1/8. We can now assemble the solution
of the IVP:
3 4t 1 −4t e + e
~v = 34 4t 34 −4t .
8e − 8e
Double eigenvalues
If the characteristic equation of a matrix A has a double root r, and ~b1 is an
eigenvector, then the system ~v0 = A~v has the family of solutions ~v = c ert ~b1 .
If there is a second, independent eigenvector ~b2 belonging to r, then every
vector in the plane can be expressed as a linear combination of ~b1 and ~b2 .
Since
A(c1~b1 + c2~b2 ) = c1 A~b1 + c2 A~b2
= r (c1~b1 + c2~b2 ),
every vector in the plane is an eigenvector belonging to r; that is, A~v = rI~v
for all ~v. Hence A = rI is a scalar matrix, and the general solution of
~v0 = A~v is ~v = ert~c where ~c is an arbitrary constant vector.
Not all 2 × 2 matrices with double eigenvalues are scalar matrices, and
here is an example of such a matrix. The characteristic polynomial of
r 0
A=
1 r
is s2 − 2rs + r2 = (s − r )2 , so r is a double eigenvalue. A is not a scalar
matrix, so it cannot have two independent eigenvectors. To see how to find
the general solution of ~v0 = A~v, let us put
x
~v =
y
and write the matrix equation as the system
x 0 = rx
y0 = x + ry.
This system is uncoupled. The solution of the first equation is x = c ert , and
we substitute this into the second equation to obtain
y0 = ry + c ert
254
CHAPTER 3. LINEAR SYSTEMS
We will solve this inhomogeneous first order equation by the method of
variation of constants, starting with the homogeneous solution yh (t) = ert .
Set y = w ert , where w is a new dependent variable, to obtain
w0 ert + rw ert = rw ert + c ert
which can be simplified as w0 = c. Thus w = ct + d and we have y =
(ct + d) ert . In vector form,
1
0
rt
~v = e c
+d
.
t
1
This solution is a linear combination of two solutions,
1
~v1 (t) = ert
= ert (~i + t~j)
t
and
~v2 (t) = e
rt
0
1
= ert~j,
where ~i and ~j are the standard basis vectors,
~i = 1 and ~j = 0 .
0
1
The solution ~v2 derives from the eigenvalue r and its eigenvector, ~j.
Let us now turn to an arbitrary matrix A that has a double eigenvalue
r. We will assume that A is not a scalar matrix, since scalar matrices are
handled differently.
Following the example computed above, we will try to derive a solution
of the form ~v(t) = ert (t~b +~c). By the product rule for differentiation,
~v0 = ert [r (t~b +~c) + ~b].
Since ~v0 = A~v and
A~v = ert (tA~b + A~c),
it follows that ~v is a solution if and only if for all t,
r (t~b +~c) + ~b = tA~b + A~c.
(3.12)
Setting t = 0 in (3.12), we have
~b = ( A − rI )~c.
(3.13)
3.2. SYSTEMS WITH CONSTANT COEFFICIENTS
255
On the other hand, if we differentiate (3.12) with respect to t, we have
A~b = r~b.
(3.14)
Equations (3.13) and (3.14) must both be satisfied. The following theorem from linear algebra tells us that since r is a double eigenvalue of A,
equation (3.14) will be satisfied automatically if (3.13) holds.
Theorem 3.7 (Cayley - Hamilton Theorem) Let B be a 2 × 2 matrix, and let
s2 − tr ( B) s + det( B) = 0
be its characteristic equation. Let B2 be the matrix formed by multiplying the
matrix B by itself. Then
B2 − tr ( B) B + det( B) I = 0I
Although stated here in terms of 2 × 2 matrices, the Cayley - Hamilton
theorem holds for square matrices of any size. It can be summarized to say
that “every square matrix satisfies its own characteristic equation.”
a b
P ROOF OF THE C AYLEY - H AMILTON THEOREM . If B =
, then
c d
tr ( B) = a + d and det( B) = ad − bc. By matrix multiplication,
2
a + bc ab + bd
2
B =
ac + cd bc + d2
and
tr ( B) B =
a2 + ad b( a + d)
c( a + d) ad + d2
.
It follows that
2
B − tr ( B) B =
bc − ad
0
0
bc − ad
= − det( B) I.
Because r is a double eigenvalue, the characteristic equation of A is (s −
= 0. By the Cayley - Hamilton Theorem, ( A − rI )2 = 0I. Since A 6= rI,
we can choose a vector ~c 6= ~0 that is not an eigenvector: A~c 6= r~c. Put
~b = ( A − rI )~c; then (3.13) holds. Furthermore,
r )2
( A − rI )~b = ( A − rI )2~c = 0I~c = ~0.
256
CHAPTER 3. LINEAR SYSTEMS
It follows that A~b = r~b: thus (3.14) also holds. Hence ~v(t) = es0 t (t~b +~c) is a
solution of the system ~v0 = A~v.
Matrices with Double Eigenvalues
Let A 6= rI be a 2 × 2 matrix that has a double eigenvalue, r.
Choose a vector ~c such that A~c 6= r~c and define ~b = A~c − r~c.
Then ~v1 (t) = ert~b and ~v2 (t) = ert (t~b + ~c) are linearly independent solutions of the system
~v0 = A~v.
Example
3.2.5
Find the general solution of the system ~v0 = A~v, where A =
1 −2
.
2 −3
S OLUTION . The characteristic equation of A is s2 + 2s + 1 = (s + 1)2 = 0,
which has the double root −1. Since A 6= (−1) I, we need a vector ~c that is
not an eigenvector. For simplicity, take one of the standard basis vectors.
1
(They cannot both be eigenvectors; if they were, A = − I.) Put ~c =
;
0
~c is not an eigenvector, because A~c is the first column of A, not a scalar
multiple of ~c. Set
~b = A~c − (−1)~c = 1 + 1 = 2 .
2
0
2
Then ~v1 (t) = e−t~b and ~v2 (t) = e−t (t~b +~c) are independent solutions. The
general solution is
−t 2e
(1 + 2t)e−t
~v = c1
+ c2
2e−t
2te−t
3.2. SYSTEMS WITH CONSTANT COEFFICIENTS
257
Exercises
In Exercises 1– 5, find all real eigenvalues and eigenvectors of the given
matrix.
1 0
1.
0 −4
Answer
2 3
2.
1 −2
1 1
3.
0 −4
Answer
3
3
4.
−3 −3
2 3
5.
−3 4
Answer
1 −2
6.
Show that the matrix
has no real eigenvalues.
2 1
a b
7.
Let s = r be an eigenvalue of A =
.
c d
(a) Show that
b
r−a
is an eigenvector belonging to r unless b = 0
and r = a.
a−d
(b) Show that if b = 0 then s = a is an eigenvalue, and
is
c
an eigenvector belonging to it, unless a = d and c = 0 as well.
(c) How do you find the eigenvectors if a = d and b = c = 0?
Answer
In Exercises 8 – 13, find the general solution for each system.
8.
~v0
=
3
4
−2 −3
~v
258
9.
CHAPTER 3. LINEAR SYSTEMS
~v0 =
3 2
1 2
~v
Answer
0
x = −3x − y
10.
y0 = x − y
0 1
0
~v =
~v
11.
0 0
Answer
0
x = x−y
12.
y0 = x − y
0
x = x + 2y
13.
y0 = 2x + y
Answer
In Exercises 14 – 18 solve the IVP.
1 2
1
0
~v =
~v; ~v(0) =
14.
−1 4
0
−1 1
0
0
~v =
~v; ~v(0) =
15.
−1 −3
1
Answer
0
x = x + y; x (0) = 4
16.
y0 = x + 2y; y(0) = 2
0
x = 5x − 2y x (0) = −1
17.
y0 = 2x + y y(0) = −2
Answer
0
x = x + y; x (0) = 3
18.
y0 = x + y; y(0) = 1
19.
Show that the characteristic equation of the system that replaces
the second order linear ODE
y00 + py0 + qy = 0
is s2 + ps + q = 0
Answer
3.3. SYSTEMS WITH OSCILLATING SOLUTIONS
20.
3.3
259
Let A be a 2 × 2 matrix with distinct eigenvalues s1 and s2 , and
let ~c be a vector that is not characteristic. Show that ( A − s2 I )~c is
an eigenvector belonging to s1 , and that ( A − s1 I )~c is an eigenvector
belonging to s2 .
Systems with Oscillating Solutions
We have already observed that x = cos(t), y = sin(t) is a solution of the
linear system
x0 = −y
(3.15)
y0 = x.
This system can be converted to vector form as ~v0 = A~v, where
A=
0 −1
1 0
.
Since tr ( A) = 0 and det( A) = 1, the characteristic equation of A, s2 +
√1 =
0, has no real roots. It does have two imaginary roots, ±i, where i = −1.
The system (3.15) is typical of the case where there are no real eigenvalues:
all such systems have solutions that oscillate.
Euler’s Formula
To solve a systems that has no real eigenvalues, we need to work with complex numbers. The following review is for the convenience of readers who
lack experience with the complex number system.
The complex number system is an extension of the real number system formed by including “imaginary numbers” to serve as square roots of
negative real numbers. Every complex number z can be expressed as
z = x + iy
where x and y are real numbers. The operations of arithmetic, +, −, ×, ÷
are all defined for complex numbers, and they satisfy the same commutative, associative, distributive, and existence of inverse laws that the real
numbers do. In addition, complex arithmetic has one operation that real
arithmetic does not have: conjugation. The conjugate of z = x + iy is
z̄ = x − iy. (It is standard practice to use z̄ to denote the conjugate of z.)
For example, 3 + 4i = 3 − 4i. Here is how to do complex arithmetic:
260
CHAPTER 3. LINEAR SYSTEMS
Addition: Use vector addition:
( x1 + iy1 ) + ( x2 + iy2 ) = ( x1 + x2 ) + i (y1 + y2 )
Thus, (3 + 4i ) + (5 − i ) − 6 = 2 + 3i.
Multiplication: Use binomial multiplication, and i2 = −1. Thus
( x1 + iy1 )( x2 + iy2 ) = x1 x2 + i ( x1 y2 + x2 y1 ) + y1 y2 i2
= ( x1 x2 − y1 y2 ) + i ( x1 y2 + x2 y1 ).
Thus, (12 − i ) × (1 + i ) = (12 − i2 ) + 12i − i = 13 + 11i. Notice that
zz̄ = x2 + y2 is real and positive (unless z = 0). If z = 3 + 4i, then
zz̄ = 32 + 42 = 25.
Division: Multiply the numerator and the denominator by the conjugate
of the denominator. The resulting quotient will be the same, but with
a real denominator.
x1 + iy1
x2 − iy2
x1 + iy1
=
x2 + iy2
x2 + iy2
x2 − iy2
x1 x2 + y1 y2
− x1 y2 + x2 y1
=
+i
x22 + y22
x22 + y22
Thus,
2+i
(2 + i )(3 − 4i )
10 − 5i
=
=
= 0.4 − 0.2i
3 + 4i
25
25
Just as it is customary to visualize real numbers as points on a line,
complex numbers are visualized as points in a plane, called the complex
plane. The horizontal axis of the complex plane is identified with the real
line, and is called the real axis. The vertical axis is the imaginary axis.
Figure 3.1 displays the complex plane.
The distance from z = x + iy to 0 in the
p complex plane is, by the usual
distance formula from analytic geometry, x2 + y2 . This
√ distance is the absolute value of z, and is denoted |z|. Notice that |z| = z z̄. All of the usual
properties of absolute value for real numbers are still true. For example the
triangle inequality,
| z1 + z2 | ≤ | z1 | + | z2 |
can be interpreted as saying that one side of a triangle must be shorter than
the sum of the lengths of the other two sides: see figure 3.2.
The components of a complex number z = x + iy are the real and imaginary parts of z, Re(z) = x and Im(z) = y, respectively. By convention,
the imaginary part of a complex number is a real number. For example,
Im(3 + 4i ) = 4, not 4i.
3.3. SYSTEMS WITH OSCILLATING SOLUTIONS
261
Polar representation of complex numbers. The polar representation of a
complex number is useful for multiplying and dividing complex numbers,
and for raising them to powers. It is not useful for complex addition. The
polar representation of z = x + iy is simply a setpof polar coordinates (r, θ )
for the point ( x, y), in which r ≥ 0. Thus, r = x2 + y2 = |z|. The angle
θ is called the argument of z, and is denoted arg(z). The representation
z = x + iy can be recovered from the polar representation by the formulas
x = r cos(θ ) and y = r sin(θ ); hence
z = r (cos(θ ) + i sin(θ )).
To compute arg(z), use the inverse cosine function:

 cos−1 Re(z)
if Im(z) ≥ 0
|z| arg(z) =
 − cos−1 Re(z)
if Im(z) < 0
|z|
The range of the argument as defined here is −π < arg(z) ≤ π, but we will
follow the custom of considering arg(z√
) as a function with values taken
mod 2π. For example, if I say arg(1 − 3i ) = − π3 and you say arg(1 −
√
3i ) = 5π
the polar representation
3 then we agree. Figure 3.3 illustrates
√
(r, θ ) of −1√+ 2i, where r = | − 1 + 2i | = 5 and θ = arg(−1 + 2i ) =
cos−1 (−1/ 5), approximately 117◦ .
Euler’s formula (Leonhard Euler)
eiθ = cos θ + i sin θ
We will take Euler’s formula as the definition of the exponential of an
imaginary number. It can be justified by substituting x = iθ in the Maclaurin series for e x :
∞
(iθ )n
.
eiθ = ∑
n=0 n!
In this series, the even terms are real and the odd terms are imaginary,
because (iθ )2m = (−1)m θ 2m and (iθ )2m+1 = (−1)m iθ 2m+1 . Therefore,
eiθ =
∞
θ 2m
∞
θ 2m+1
∑ (−1)m (2m)! + i ∑ (−1)m (2m + 1)! .
m =0
m =0
(3.16)
262
CHAPTER 3. LINEAR SYSTEMS
The first sum on the right side of equation (3.16) is the Maclaurin series for
the cosine function, and the second is i times the Maclaurin series for the
sine.
The identity eu+v = eu ev extends the definition of the exponential to all
complex numbers. Thus
eλ+iω = eλ eiω
= eλ (cos(ω ) + i sin(ω )).
To make use of the complex exponential, we need to be able to differentiate it. The derivative of a complex-valued function f (t) = p(t) + iq(t)
is
f 0 ( t ) = p 0 ( t ) + i q 0 ( t ).
Proposition 3.3.1 Let a = λ + iω be a complex constant. Then
d at
e = a e at .
dt
P ROOF.
d at
e
=
dt
d λt
{e [cos(ωt) + i sin(ωt)]}
dt
d λt
d
=
[e cos(ωt)] + i [eλt sin(ωt)]
dt
dt
λt
λt
= [λe cos(ωt) − e ω sin(ωt)]
+ i [λeλt sin(ωt) + ωeλt cos(ωt)]
= eλt [λ cos(ωt) + iω cos(ωt) − ω sin(ωt) + iλ sin(ωt)]
= eλt [(λ + iω ) cos(ωt) + (iω + λ)i sin(ωt)]
= a eλt [cos(ωt) + i sin(ωt)] = a e at .
Let A be a square matrix with real entries. The characteristic polynomial
of A may have some complex roots, and these are considered to be complex
eigenvalues. For example, the characteristic polynomial of the matrix
1 −1
A=
1
1
is s2 − 2s + 2. The eigenvalues of A can be calculated by using the quadratic
formula:
p
2 ± (−2)2 − 8
s1 , s2 =
= 1 ± i.
2
3.3. SYSTEMS WITH OSCILLATING SOLUTIONS
263
Since complex arithmetic follows the same rules as real arithmetic, we can
work with matrices and vectors that have complex entries just as we do
h
belongwhen the entries are real. Thus, to find an eigenvector ~b1 =
k
ing to s1 = 1 + i, we solve ( A − s1 I )~b1 = ~0:
1 −1
1
1
−
1+i
0
0 1+i
h
k
=
0
0
After simplifying, this matrix equation reduces to two complex equations,
−i h − k = 0
h−ik = 0
Observe that the second of these equations can be obtained from the first
by multiplying through by i. Thus for any h 6= 0, k = −i h will satisfy both
equations. Let’s put h = 1; then
~b1 =
1
−i
is an eigenvector belonging to 1 + i.
Suppose that s is a complex eigenvalue of a square matrix A, and ~b
is an eigenvector belonging to s. Put ~v(t) = est ~v. By proposition 3.3.1,
~v0 (t) = sest~b = s~v. Since A~v(t) = est A~b = est s~b = s~v, it follows
that ~v(t) is a
1 −1
solution of ~v0 = A~v. Returning to our matrix A =
, we see that
1
1
~v(t) = e
(1+ i ) t
1
−i
=
et (cos(t) + i sin(t))
et (sin(t) − i cos(t))
is a solution of ~v0 = A~v.
This solution is complex since ~b is a complex vector, but we require a
real solution, since ~v0 = A~v is a real system of ODEs. By the following
theorem, we can exchange a complex solution of a system of linear, homogeneous ODEs with real coefficients for two real solutions.
Theorem 3.8 Let ~z(t) be a complex solution of a homogeneous linear ODE or system of ODEs with real coefficients. Then ~v1 (t) = Re[~z(t)] and ~v2 (t) = Im[~z(t)]
are solutions as well.
264
CHAPTER 3. LINEAR SYSTEMS
P ROOF. The vector form of the system is ~v0 = A(t)~v, where the entries
of the matrix A(t) are real-valued. It is given that ~z(t) = ~v1 (t) + i~v2 (t) is a
complex-valued solution. Expand ~z0 = A(t)~z as
~v10 + i~v20 = A(t)~v1 + iA(t)~v2 .
(3.17)
Equating real and imaginary parts of both sides of (3.17), we see that ~v10 =
A(t)~v1 , and ~v20 = A(t)~v2 . It follows that ~v1 (t) and~v2 (t) are solutions. et (cos(t) + i sin(t))
of
By theorem 3.8 the complex solution ~v(t) =
et (sin(t) − i cos(t))
1 −1
~v0 =
~v, yields two real solutions,
1
1
cos(t)
sin(t)
~v1 (t) = et
and ~v2 (t) = et
sin(t)
− cos(t)
1
0
Since ~v1 (0) =
and ~v2 (0) =
are linearly independent, the
0
−1
solutions ~v1 (t) and ~v2 (t) are linearly independent.
Example 3.3.1 Find the general solution of the system ~v0 = A~v, where
−5 −4
A=
.
5 −1
S OLUTION . Since tr ( A) = −6 and det( A) = 25, the characteristic
equation of A is s2 + 6s + 25 = 0. By the quadratic formula, the
eigenvalues are
√
√
−6 ± 62 − 4 × 25
−6 ± −64
s=
=
= −3 ± 4i.
2
2
To find a complex eigenvector
~b =
h
k
(h and k will be complex numbers) belonging to s = −3 + 4i, we need to
solve
( A − (−3 + 4i ) I )~b = ~0.
This matrix equation is
−5 −4
(−3 + 4i )
0
h
0
−
=
,
5 −1
0
(−3 + 4i )
k
0
3.3. SYSTEMS WITH OSCILLATING SOLUTIONS
265
or
(−2 − 4i )
−4
5
(2 − 4i )
h
k
=
0
0
.
We must choose h and k so that (−2 + 4i )h − 4k = 0 and
5h + (2 − 4i )k = 0. These equations are redundant: any (h, k ) that
satisfies one will also satisfy the other. Thus we will put
−2
~b =
.
1 + 2i
By theorem 3.8, the real and the imaginary parts of
−2(cos(4t) + i sin(4t))
e(−3+4i)t~b = e−3t
cos(4t) + i sin(4t) + 2i (cos(4t) + i sin(4t))
−2 cos(4t)
2 sin(4t)
−3t
= e
+i
cos(4t) − 2 sin(4t)
2 cos(4t) + sin(4t)
are solutions. These solutions are ~v1 (t) = e
−2 sin(4t)
−3t
~v2 (t) = e
. Since
2 cos(4t) + sin(4t)
~v1 (0), ~v2 (0) =
−2
1
−3t
−2 cos(4t)
cos(4t) − 2 sin(4t)
and
0
,
2
are linearly independent, corollary 3.1.3 tells us that the general solution is
~v(t) = c1~v1 (t) + c2~v2 (t)
If we exchange a complex solution of a linear system for two real solutions, are the real solutions always linearly independent? The answer is
yes:
Theorem 3.9 Let s = λ + iω, where ω 6= 0, be a complex eigenvalue of a real
matrix A, and ~b = ~h + i~k be an eigenvector belonging to s. Then ~h and ~k are
linearly independent.
For a proof, see Exercise 11.
266
CHAPTER 3. LINEAR SYSTEMS
Corollary 3.3.2 Let A be a 2 × 2 real, constant matrix that has an eigenvalue
λ + iω with ω 6= 0, and let ~h + i~k be an eigenvector belonging to that eigenvalue.
Then the general solution of
~v0 = A~v
is
~v(t) = c1 eλt cos(ωt)~h − sin(ωt)~k + c2 eλt sin(ωt)~h + cos(ωt)~k ,
(3.18)
where c1 and c2 are constants.
Exercises
1.
Find the reciprocal of 2 + i.
Answer
2.
Locate each of the complex numbers on the complex plane, calculate arg(z), |z|, and find λ + ωi such that z = eλ+ωi .
(a) z = 2.
(b) z = −2.
(c) z = 1 − i.
√
(d) z = −1 + i 3.
(e) z = −2i.
3.
4.
5.
6.
Find the sixth roots of 1 (there are 6 of them), and locate them on
the complex plane. Hint: 1 = e2nπi .
Answer
Show that es̄ = es .
Show that |eλ+iω | = eλ (λ and ω are real).
Answer
Find all solutions of
e2s + 2es + 2 = 0.
In Exercises 7 – 10, find the general solution
0
x = − 12 x − 21 y
7.
y0 = 12 x − 12 y
Answer
3.4. MATRIX SOLUTIONS AND VARIATION OF CONSTANTS
8.
x 0 = 2x + 5y
y0 = − x
x0
y0
Answer
0
x
10.
y0
9.
11.
267
= 4x − 8y
= 4x − 4y
= −4x − 41y
= 21 x − 3y
Let s = λ + iω, where ω 6= 0, be a complex eigenvalue of a real
matrix A, and ~b = ~h + i~k be a eigenvector belonging to s.
(a) Show that the conjugate of ~b is an eigenvector belonging to the
conjugate of s.
¯
(b) Using theorem 3.6 , show that ~b and ~b are linearly independent.
(c) Prove theorem 3.9.
(d) Prove corollary 3.3.2.
Answer
12.
Let A be a 2 × 2 matrix such that tr A = 0 and det A < 0. Show
that the orbits of the system ~v0 = A~v are hyperbolas, their asymptotes,
and the origin.
Hint: Imitate the case where det A > 0.
3.4
Matrix Solutions and Variation of Constants
Let A(t) be an n × n matrix whose entries are either constants or continuous
functions of t, all defined of an interval ( a, b). A homogeneous system of n
linear ODEs
~v0 = A(t)~v,
(3.19)
admits a family of solutions on ( a, b), which can be described completely
if a linearly independent set {~v1 (t), ~v2 (t), . . . , ~vn (t)} of solutions is known.
By Theorem 3.4, every solution is then uniquely expressed as a linear combination
c1~v1 (t) + c2~v2 (t) + · · · + cn~vn (t).
It will be convenient to use matrix notation to simplify this statement.
268
CHAPTER 3. LINEAR SYSTEMS
Given an n × n matrix X (t) whose entries are functions of t, let X 0 be
the matrix formed by differentiating each entry of X . If the equation
X 0 = A(t)X
(3.20)
holds on the interval ( a, b), then X (t) is called a matrix solution of equation (3.19).
Proposition 3.4.1 Let the columns of the matrix X (t) be denoted ~v1 (t), ~v2 (t),
. . . , ~vn (t). Then X (t) is a matrix solution of the homogeneous system (3.19) if
and only if the ~vi (t) are solutions of (3.19) as vector functions.
P ROOF. The columns of X 0 are ~v10 , ~v20 , . . . , ~v0n . Furthermore, the columns
of A(t)X are A(t)~vi , for i = 1, 2, . . . , n. It follows each column of a matrix
solution of (3.19) is itself a (vector) solution, and conversely that if each
~vi (t) is a solution, then X (t) is a matrix solution.
We can use a matrix solution to generate a family of solutions of (3.19)
as follows.
Proposition 3.4.2 Suppose that X (t) is a matrix solution of (3.19), and let ~c ∈
Rn be a constant vector. Then the vector function
~v(t) = X (t)~c
is a solution of the system (3.19).
If, in addition, the columns of X (t) are linearly independent, then given any
~ (t) of the system (3.19), there is a unique ~c ∈ Rn such that ~v = X (t)~c.
solution w
P ROOF. Let ~v1 , ~v2 , . . . , ~vn denote the columns of X . By the definition of
matrix multiplication,
X~c = c1~v1 + c2~v2 + · · · + cn~vn
is a linear combination of solutions of the system (3.19) and thus itself a
solution of the system.
If {~v1 , ~v2 , . . . , ~vn } is linearly independent, then by theorem 3.4 there are
unique constants c1 , . . . , cn such that
~ = c1~v1 + c2~v2 + · · · + cn~vn .
w
~ = X~c.
The constants ci are the components of the vector ~c such that w
D EFINITION A matrix solution X (t) of equation (3.19) is called a fundamental matrix solution if there is a point t0 ∈ ( a, b) such that the columns
of X (t0 ) are linearly independent.
3.4. MATRIX SOLUTIONS AND VARIATION OF CONSTANTS
269
An n × n matrix P is said to be nonsingular if it is not singular. In
Exercise 13 at the end of this section, you are asked to prove that a matrix
P is nonsingular if and only if, as vectors, the columns of P are linearly
independent.
Proposition 3.4.3 Let X (t) be a matrix solution of (3.19), defined on an interval ( a, b). Then X (t) is a fundamental matrix solution if and only if one of the
following equivalent conditions holds.
• X (t) is nonsingular for all ordinary points t ∈ ( a, b).
• X (t0 ) is nonsingular for some ordinary point t0 ∈ ( a, b).
• det(X (t)) 6= 0 for all ordinary points t ∈ ( a, b).
• det(X (t0 )) 6= 0 for some ordinary point t0 ∈ ( a, b).
This proposition implies that a matrix solution that is nonsingular at
one point t0 ∈ ( a, b) is automatically nonsingular for all t ∈ ( a, b). This is
a special property of matrix solutions of homogeneous linear systems of
ODEs: it is possible for matrix functions in general to be nonsingular in
some places, and not in others.
P ROOF OF P ROPOSITION 3.4.3: The first statement follows from Exercise 13. For the second, we will apply theorem 3.4. Let ~a1 = v~1 (t0 ), . . . a~n =
~vn (t0 ) be the columns of X (t0 ), which is then nonsingular if and only if
{~a1 , . . . , a~n } is linearly independent. By theorem 3.4, the columns ~v1 (t), . . . , ~v(t)
of X (t) are then also linearly independent—that is, X (t) is nonsingular—
for all t ∈ ( a, b).
The statements about determinants follow from proposition 3.2.2.
Example 3.4.1 Find a fundamental matrix solution of
0 1
0
~v =
~v.
1 0
(3.21)
S OLUTION . The characteristic equation of
0 1
A=
1 0
is s2 − 1 = 0; hence the eigenvalues are ±1. We can use eigenvectors
1
1
~e1 =
and ~e2 =
1
−1
270
CHAPTER 3. LINEAR SYSTEMS
e1
belonging to 1 and −1, respectively, to obtain the solutions ~v1 (t) = et ~
−
t
and ~v2 (t) = e ~e2 . Since
t −t e
e
~v1 (t) =
and v~2 (t) =
,
et
− e−t
t
e
e−t
X (t) =
.
et − e−t
is a matrix solution.
1
1
Then X (0) =
; since det X (0) = −2 if follows from
1 −1
proposition 3.4.3 that X (t) is a fundamental matrix solution.
Example 3.4.2 Find the general
solution of the system 3.21, and also the solution
3
with initial value ~v(0) =
.
5
S OLUTION . In example 3.4.1, we found a fundamental matrix solution,
t
e
e−t
X =
et − e−t
The general solution is
~v = X~
ct
e
e−t
c1
=
et − e−t
c2
c1 e t + c2 e − t
=
c1 e t − c2 e − t
1
1
t
−t
= c1 e
+ c2 e
.
1
−1
To find the solution of the IVP, we must find a vector ~c such that X (0)~c is
equal to the specified value of ~v(0). Thus we must solve the equation
1
1
c1
3
=
.
1 −1
c2
5
or, equivalently, (c1 , c2 ) must satisfy the system
c1 + c2 = 3
c1 − c2 = 5
3.4. MATRIX SOLUTIONS AND VARIATION OF CONSTANTS
271
which yields (c1 , c2 ) = (4, −1). The solution of the IVP is therefore
1
1
~v(t) = 4et
− e−t
1
−1
Inhomogeneous systems
The method for solving an inhomogeneous system of linear ODEs,
~v0 = A(t)~v + ~f (t),
(3.22)
where A(t) is an n × n matrix, is basically the same as the method we used
in section 1.3 to solve a single first order linear ODE. We start with the
general solution of the associated homogeneous system,
~v0 = A(t)~v,
which we can express in terms of a fundamental matrix solution,
~vh (t) = X (t)~c
and add a particular solution ~v p (t) of the inhomogeneous equation.
Theorem 3.10 Suppose that all entries of the coefficient matrix A(t) and source
vector ~f (t) in (3.22) are continuous on an interval ( a, b). Let ~v p (t) be a particular
solution of the system (3.22), and let X (t) be a fundamental matrix solution of the
associated homogeneous system. Then the general solution of the system (3.22) on
( a, b) is
~v(t) = ~v p (t) + X (t)~c.
P ROOF. Let ~v(t) be an arbitrary solution of the inhomogeneous system (3.22), and put ~y(t) = ~v(t) − ~v p (t). Then
~y0 = ~v0 − ~v0p
= [ A(t)~v + ~f (t)] − [ A(t)~v p + ~f (t)]
= A(t)~v − A(t)~v p = A~y.
It follows that ~y(t) is a solution of the associated homogeneous system.
Therefore there is a unique constant vector ~c such that
~y(t) = X (t)~c;
272
CHAPTER 3. LINEAR SYSTEMS
and it follows that ~v(t) = ~v p (t) + X (t)~c.
Conversely, suppose that ~v(t) = ~v p (t) + X (t)~c, where ~c is a constant
vector. Then
d
(~v(t)) = ~v0p (t) + X 0 (t)~c
dt
= A(t)~v p (t) + ~f (t) + A(t)X (t)~c
= A(t)(~v p (t) + X (t)~c) + ~f (t)
= A(t)~v(t) + ~f (t);
in other words, ~v(t) is a solution of (3.22).
Theorem 3.10 can best be deployed in conjunction with a method for
finding particular solutions of systems.
Variation of constants
The method of variation of constants is used to find a particular solution of
an inhomogeneous system. Recall from section 1.3, where the method was
introduced for the scalar case, that the general solution of the associated
homogeneous equation was a required input. For systems, it is convenient
to start with a fundamental matrix solution of the associated homogeneous
system.
The key to adapting the method for systems of linear ODEs is inversion
of a matrix solution. This replaces division by yh (t) in the scalar case (as in
equation (1.11)).
Let P be a square matrix. A square matrix Q is called the inverse matrix
of P if P Q = Q P = I, where I is the identity matrix. We will use the
customary notation P−1 for the inverse matrix of P. The following theorem
is from linear algebra. Although it is true for square matrices of any size,
our proof only works for 2 × 2 matrices.
Theorem 3.11 Let
P=
a c
b d
be a nonsingular matrix. Then P has an inverse matrix.
P ROOF By proposition 3.2.2, det( P) = ad − bc is equal to 0 if and only if
P is singular.
Thus, if P is nonsingular, det( P) 6= 0. Define a matrix Q as
1
d −b
Q=
.
a
det( P) −c
3.4. MATRIX SOLUTIONS AND VARIATION OF CONSTANTS
273
You can verify that P Q = I and Q P = I by matrix multiplication.
We will now see how to find a particular solution of an inhomogeneous
system of ODEs. Let X (t) be a fundamental matrix solution of a homogeneous system ~v0 = A(t)~v. To find a particular solution of an inhomogeneous system
~v0 = A(t)~v + ~f (t)
(3.23)
set
~v p (t) = X (t)~
w ( t ),
~ (t) is a vector function that will be determined. By the product rule
where w
for differentiation,
~v0p = X 0 (t)~
w(t) + X (t)~
w0 (t)
= A(t)X (t)~
w(t) + X (t)~
w0 (t)
= A(t)~v p (t) + X (t)~
w 0 ( t ).
It follows that ~v p (t) is a particular solution of the system (3.23) if and only
if
X (t)~
w0 (t) = ~f (t).
(3.24)
Since X (t) is nonsingular, it has an inverse X −1 (t). Multiplying (3.24) by
this, we have
~ 0 (t) = X −1 (t)~f (t).
w
If t0 ∈ ( a, b), we can put
~ (t) =
w
Z t
t0
X −1 (s)~f (s) ds.
~ (t) has been determined we can find ~v p (t) by multiplying by
Now that w
X (t). This can be summarized as follows:
274
CHAPTER 3. LINEAR SYSTEMS
Variation of Constants Formula
Let X (t) be a fundamental matrix solution of ~v0 = A(t)~v, where
A(t) is a matrix whose entries are all continuous on an interval
( a, b). If ~f (t) is a vector function, also continuous on ( a, b), and
t0 ∈ ( a, b), then
~v p (t) = X (t)
Z t
t0
X −1 (s)~f (s) ds
is a particular solution of the inhomogeneous equation
~v0 = A(t)~v + ~f (t).
It is not necessary to memorize this formula; you just need to remember
to substitute ~v = X (t)~
w in (3.23).
Example 3.4.3 Find a particular solution of
x10 = x2 − e−t
x20 = x1 + e−t
et e−t
S OLUTION . In example 3.4.1, it was shown that X (t) =
et − e−t
fundamental matrix solution of the associated homogeneous
system (3.21). Set ~v p (t) = X (t)~
w(t). Thus
x1
x2
=
et e−t
et − e−t
w1
w2
=
e t w1 + e − t w2
e t w1 − e − t w2
Differentiating, and using the product rule we have
t
0 t 0
e w1 + e−t w20
e w1 − e − t w2
x1
=
+
et w10 − e−t w20
e t w1 + e − t w2
x20
and the right side of our system is
t
x2 − e − t
e w1 − e − t w2
− e−t
=
+
x1 + e − t
e t w1 + e − t w2
e−t
is a
3.4. MATRIX SOLUTIONS AND VARIATION OF CONSTANTS
275
After canceling, we have
et w10 + e−t w20 = −e−t
et w10 − e−t w20 = e−t .
The solution of this equation is w10 = 0, w20 = −1, or
0
~ (t) =
w
~ (t) =
Thus w
0
−t
0
−1
.
, and we have the particular solution
~v p (t) =
et e−t
et − e−t
0
−t
=
−te−t
te−t
.
The general solution is ~v p (t) + ~vh (t), where ~vh (t) = X (t)~c denotes the
general solution of the associated homogeneous equation.
Exercises
In Exercises 1 – 4, find a fundamental matrix solution of the system.
x0
y0
Answer
0
x
2.
y0
0
x
3.
y0
Answer
0
x
4.
y0
1.
=
=
3
2x
1
2x
+ 21 y
+ 23 y;
= −3x + y
= −5x − y;
= 2x − 4y
= 5x − 2y;
= −x − y
= x − 3y;
In each of Exercises 5 – 8, use the method of variation of constants to
find a particular solution of the system of ODEs, and then write down the
general solution. Fundamental matrix solutions of the associated homogeneous systems were found in Exercises 1 – 4.
276
CHAPTER 3. LINEAR SYSTEMS
x 0 = 23 x + 12 y + b1 (t)
where
y0 = 21 x + 32 y + b2 (t),
t b1
e
(a)
=
.
b2
et
t b1
e
=
.
(b)
b2
−et
t b1
te
=
.
(c)
b2
0
5.
Answer
0
x + 3x − y = −3e−2t
6.
y0 + 5x + y = e−2t
0 x
2 −4
x
−32 cos(4t)
7.
=
+
.
y0
5 −2
y
0
Answer
0 x
1 1
x
b1 (t)
8.
+
=
, where
y0
−1 3
y
b2 (t)
b1 (t)
1
−
2t
(a)
=e
.
b2 (t)
3
−2t b1 (t)
2e
(b)
=
.
b2 (t)
0
9.
Show that
X =
2t2
t3
− t2 − t3
is a fundamental matrix solution of
tx 0 = x − 2y
ty0 = x + 4y.
Using variation of constants, find a particular solution of
tx 0 = x − 2y + b1 (t)
ty0 = x + 4y + b2 (t),
where
b1 (t)
t
(a)
=
.
b2 (t)
2t
3.5. A FUNDAMENTAL MATRIX SOLUTION OF INTEREST
(b)
b1 (t)
b2 (t)
=
t3
− t3
277
.
Warning: the variation of constants formula is based on the assumption that the system has the form ~v0 = A(t)~v +~b(t); the system in this
Exercise must be written in that form before the formula can be used.
However, if you use the method of variation of constants there are no
worries.
Answer
10.
Find the general solution of each of the following inhomogeneous
systems.
0
x = x + y + t −1
(a)
y0 = − x − y.
0
x = x + y + t −1
(b)
y0 = −( x + y + t−1 ).
0
x = 4x + 5y + 5 sin(3t)
(c)
y0 = −5x − 4y + 3 cos(3t) − 4 sin(3t)
0
x = sec3 (t) − y
(d)
y0 = x − sec3 (t)
11.
Let C be a constant matrix, and suppose that X (t) is a matrix
solution of ~v0 = A(t)v. Show that X C is also a matrix solution.
Answer
12.
Let X1 (t) and X2 (t) be fundamental matrix solutions of the linear
system of differential equations ~v0 = A(t)~v, defined on an interval I .
Show that X1−1 · X2 is a constant matrix.
13.
Let Q be an n × n matrix. Show that Q is nonsingular if and only
if the columns of q are linearly independent.
Answer
3.5
A Fundamental Matrix Solution of Interest
The Maclaurin series expansion for the function e at is
e at = 1 + a t +
a2 2 a3 3
an
t + t + · · · + tn + · · · ,
2!
3!
n!
278
CHAPTER 3. LINEAR SYSTEMS
and it converges for all values of t.
Now consider the series with the scalar constant a replaced by a constant matrix A. By convention, A0 = I is the identity matrix, A1 = A,
A2 = A × A, etc.
The Matrix Exponential Function
Let A be an m × m matrix. Then let
e At = I + A t +
A2 2
An n
t +···+
t +···.
2!
n!
(3.25)
The first thing to consider is convergence. Let pij,k denote the i,jth entry
of Ak . Since A0 = I, and A1 = A,
1 if i = j
pij,0 =
,
0 if i 6= j
pij,1 = aij , and so on. Equation (3.25) defines e At to be the m × m matrix
whose i,jth entry is
∞
1
pij (t) = ∑ pij,k tk ,
(3.26)
k!
k =0
provided that this series is convergent for each i, j.
Theorem 3.12 For any m × m matrix A, the series (3.26) converges for all i, j
and for all values of t.
P ROOF. We will show that for any i, j,
∞
1
k
∑ k! pij,k t k =0
(3.27)
is convergent. Choose a number B such that for each entry aij of A, | aij | ≤ B.
We’ll see that
| pij,k | ≤ mk−1 Bk for k ≥ 1.
(3.28)
The proof of (3.28) is by mathematical induction. If k = 1, (3.28) holds
because | pij,1 | = | aij | ≤ B. Let k ≥ 2, and assume that (3.28) holds for the
power k − 1; that is
| pij,k−1 | ≤ mk−2 Bk−1 .
3.5. A FUNDAMENTAL MATRIX SOLUTION OF INTEREST
279
Since Ak = A · Ak−1 ,
pij,k = ai1 p1j,k−1 + ai2 p2j,k−1 + · · · + aim pmj,k−1 .
Starting with the triangle inequality, we have
| pij,k | ≤ | ai1 || p1j,k−1 | + | ai2 || p2j,k−1 | + · · · + | aim || pmj,k−1 |
k −1
k −2 k −1
≤ |B · mk−2 Bk−1 + B · mk−2 B
{z + · · · + B · m B }
m terms
k −1 k
= m B.
Therefore, if (3.28) holds for the power k − 1, then it holds for the kth power
as well. By the principle of mathematical induction, (3.28) is valid for all
k ≥ 1. Then
1
m k −1 B k k
| pij,k tk | ≤
| t |,
k!
k!
which implies that the series (3.27) is dominated by the series
∞
m k −1 B k k
∑ k! |t| .
k =0
The latter series converges since it is the Maclaurin series expansion of
f (t) = m1 emB|t| . By the comparison test, the series (3.27) also converges.
This means that the series (3.26) defining pij converges absolutely. Since
every absolutely convergent series is convergent, the proof is complete.
As a simple example, suppose that Z denotes the zero matrix (all entries
equal 0). Then the only nonzero term in the series
1
I + Zt + Z2 t2 + · · ·
2
defining e Zt is I. It follows that e Zt = I. The next example is also fairly
simple, because the series has only a finite number of nonzero terms.
Example 3.5.1 Let


0 1 0
A =  0 0 1 .
0 0 0
Calculate e At .
280
CHAPTER 3. LINEAR SYSTEMS
S OLUTION . By matrix multiplication,


0 0 1
A2 =  0 0 0  ,
0 0 0
and Ak = 0 for k ≥ 3. Hence
1
e At = I + At + ( A)2 t2
2 

1 2
1 t 2t
=  0 1 t 
0 0 1
Example 3.5.2 Calculate e At , where
A=
S OLUTION . If
B=
a 0
0 b
1 0
0 2
.
and C =
c 0
0 d
are diagonal matrices, you can readily verify that
ac 0
BC =
.
0 bd
Thus, the product of two diagonal matrices is also a diagonal matrix. The
diagonal entries of the product matrix are obtained simply by multiplying
the respective diagonal entries of B and C. It follows that we can find the
nth power of a diagonal matrix by raising each diagonal entry to the nth
power:
n
1
0
n
A =
.
0 2n
Thus
e At =
∞
1
∑ k! Ak tk
k =0
3.5. A FUNDAMENTAL MATRIX SOLUTION OF INTEREST
"
=
=
∑∞
k =0
0
tk
k!
et 0
0 e2t
#
0
∑∞
k =0
281
(2t)k
k!
.
The matrix exponential shares many properties with the ordinary exponential function, but there are limitations. The proof that er+s = er es does
not emphasize the commutative law, rs = sr, but it uses that fact. Multiplication of square matrices is not always commutative, so the matrix version
of that identity is as follows.
Theorem 3.13 Let A and B be n × n matrices that commute: AB = BA. Then
e( A+ B)t = e At e Bt .
The proof is an exercise.
Corollary 3.5.1 If A is a square matrix, then for all t ∈ R, the matrix e At is
nonsingular.
P ROOF : The matrices A and − A commute; therefore e At e− At = e( A− A)t =
where Z is the zero matrix. Since e Zt = I it follows that e− At is the inverse matrix of e At . A matrix is nonsingular iff it is invertible; hence e At is
nonsingular.
e Zt ,
Theorem 3.14 Let A be a square matrix whose entries are constants. Then e At is
a fundamental matrix solution of ~x 0 = A~x.
The proof depends on the chain rule applied to matrix polynomials. If
f ( x ) = c0 + c1 x + c2 x2 + · · · ck x k is an ordinary polynomial, f ( At) denotes
the matrix
f ( At) = c0 I + c1 A t + c2 A2 t2 + · · · + ck Ak tk .
Lemma 3.5.2 Let f 0 ( x ) denote the derivative of f with respect to x of a polynomial
f ( x ). If A is a square matrix with constant entries, then
d[ f ( At)]/dt = A · f 0 ( At).
282
CHAPTER 3. LINEAR SYSTEMS
P ROOF. We only need to consider the special case where f ( x ) = c x k is
a monomial, because a general polynomial is a sum of monomials. f 0 ( x ) =
c k x k−1 , so
A · f 0 ( At) = A · (c k Ak−1 tk−1 ) = c k Ak tk−1 .
On the other hand,
d[ f ( At)]/dt =
d
( c A k t k ) = c A k ( k t k −1 ).
dt
Since these are equal, the proof is complete
P ROOF OF THEOREM 3.14. By Corollary 3.5.1 e At is nonsingular, so we only
need to show that de At /dt = Ae At . Let Fm ( At) denote the partial sum
m
Fm ( At) =
1
∑ k! ( At)k .
k =0
By Lemma 3.5.2, dFm ( At)/dt = A · Fm0 ( At). Since
Fm0 ( At) =
m
1
∑ k! k( At)k−1
k =0
= AFm−1 ( At),
it follows that limm→∞ Fm0 ( At) = limm→∞ Fm−1 ( At) = e At .
It is the definition of the sum of an infinite series that
∞
lim Fm ( At) =
m→∞
1
∑ k! ( At)k = e At .
k =0
Hence
lim d[ Fm ( At)]/dt = A · e At ,
m→∞
and it might appear that the proof is complete. However, it still should be
checked that
d
d
lim
[ Fm ( At)] = [ lim Fm (t)];
m→∞ dt
dt m→∞
that is, that the limit of the derivatives is the derivative of the limit. The
proof of this fact is omitted.
3.5. A FUNDAMENTAL MATRIX SOLUTION OF INTEREST
283
Calculation
The calculation of e At , where A is a constant square matrix, involves developments that are found in a full course in linear algebra. While all of the
necessary ingredients are available in any linear algebra text, it is not unusual for 14-week linear algebra courses to omit some of this material. In
this text, the procedure for calculating e At , where A is an arbitrary constant
matrix with real or complex entries, will be described in detail, but we will
depend on linear algebra texts for proofs.
There is a simple algorithm to calculate e At when A is a 2 × 2 matrix.
Although the algorithm does not extend to n > 2, it is a good warm up for
the more general case.
Suppose that B is a 2 × 2 matrix with the special property that
tr B = 0.
The characteristic equation of B is then s2 + d = 0, where d = det B. Here
is the key formula:
B2 = −dI
(3.29)
The proof is just a reference to the Cayley-Hamilton theorem!
It follows that B2k = (−d)k I, and B2k+1 = B B2k = (−d)k B. Now that
we know the powers of B we can figure out the exponential:
e Bt =
∞
k =0
∞
=
∞
1
1
∑ (2k)! B2k t2k + ∑ (2k + 1)! B2k+1 t2k+1
∑
k =0
(−d)k
(2k)!
t2k I +
k =0
∞
(−d)k
∑ (2k + 1)! t2k+1 B
(3.30)
k =0
We can simplify this formula if we let |d| = ω 2 . If d < 0, so that −d =
ω 2 then (3.30) becomes
e Bt =
∞
(ωt)2k
1
I+
(2k)!
ω
k =0
∑
∞
(ωt)2k+1
∑ (2k + 1)! B
k =0
1
= cosh(ωt) I + sinh(ωt) B
ω
(3.31)
As an exercise, you should prove that when d > 0,
e Bt = cos(ωt) I +
1
sin(ωt) B
ω
(3.32)
284
CHAPTER 3. LINEAR SYSTEMS
Now for a matrix A with tr A = s 6= 0, put B = A − 2s I, and note that
tr B = tr A − tr 2s I = s − 2s (1 + 1) = 0. Furthermore, because the scalar matrix 2s I commutes with any matrix, by Theorem 3.13 we have e At = est/2 e Bt .
Example 3.5.3 Find
2 1
(c) A =
.
0 2
e At ,
where (a) A =
1 2
4 3
, (b) A =
2 −2
4 6
, and
S OLUTION . In each case, we will put B = A − ( 12 tr A) I.
−1 2
(a) Since tr A = 4 we get B = A − 2I =
. Thus d = det B = −9,
4 1
p
and ω = | − 9| = 3. Hence
2
1
cosh(3t) − 13 sinh(3t)
3 sinh(3t )
e Bt = cosh(3t) I + sinh(3t) B =
,
4
cosh(3t) + 13 sinh(3t)
3
3 sinh(3t )
and
2
cosh(3t) − 13 sinh(3t)
sinh(3t)
3
= e B=e
e
4
cosh(3t) + 13 sinh(3t)
3 sinh(3t )
1 5t 2 −t 1 5t
−t
3e + 3e
3 (e − e )
=
.
2 5t
2 5t
1 −t
−t
3 (e − e )
3e + 3e
−2 −2
(b) Since tr A = 8 we get B = A − 4I =
, and d = det B = 4.
4
2
Thus
1
cos(2t) − sin(2t)
− sin(2t)
Bt
e = cos(2t) I + sin(2t) B =
,
2 sin(2t)
cos(2t) + sin(2t)
2
At
2t
2t
and
cos(2t) − sin(2t)
− sin(2t)
e =e e =e
.
2 sin(2t)
cos(2t) + sin(2t)
0 1
(c) Here tr A = 4 so B = A − 2I =
. Because d = 0, we can’t
0 0
2
plug into (3.31)or (3.32).
However, we see that B = 0 so
1 t
e2t te2t
Bt
At
2t
Bt
e = I + tB =
It follows that e = e e =
.
0 1
0 e2t
At
4t Bt
4t
3.5. A FUNDAMENTAL MATRIX SOLUTION OF INTEREST
285
Inhomogeneous equations.
When applying the variation of constants formula to a system
~v0 = A~v + ~f (t),
(3.33)
where A is an n × n constant matrix, and ~v(t) and ~f (t) are n-dimensional
vector functions, we can use the matrix exponential e At as the fundamental
matrix solution of the associated homogeneous system, ~v0 = A~v.
Thus, set ~v(t) = e At~z(t). By the product rule for differentiation,
~v0 (t) = A e At~z(t) + e At~z0 (t)
= A~v(t) + e At~z0 (t).
Substituting this expression for ~v0 (t) in (3.33),
A~v(t) + e At~z0 (t) = A~v(t) + ~f (t)
and hence e At~z0 (t) = ~f (t). The inverse matrix of e At is simply e− At , so
~z0 (t) = e− At ~f (t).
Any antiderivative of e− At ~f (t) can be used as ~z(t); we will take
~z(t) =
Z t
e− As ~f (s) ds.
0
Then ~v(t) = e At~z(t), or
~v(t) = e
At
Z t
e− As ~f (s) ds
0
Since e At does not depend on the variable s of integration,
~v(t) =
Z t
e At e− As ~f (s) ds.
0
Noting that At and As are commuting matrices, we can combine exponentials to obtain a result:
286
CHAPTER 3. LINEAR SYSTEMS
Variation of Constants Formula
Systems with constant coefficients
The general solution of
~v0 = A~v + ~f (t),
where ~v and ~f are n × 1 matrices and A is an n × n constant
matrix, is
Z t
~v(t) =
e A(t−s) ~f (s) ds + e At~c.
0
Example 3.4.3 was to illustrate the variation of constants method for
finding a particular solution of an inhomogeneous system. We will return
to the system of that example, to see how the variation of constants formula
works.
Example 3.5.4 Use the variation of constants formula to find a particular solution of ~v0 = A~v + ~f (t), where
0 1
−1
−t
~
A=
and f = e
.
1 0
1
S OLUTION . The first step is to calculate e At . Since tr ( A) = 0 and
det( A) = −1, the characteristic equation of A is s2 − 1 = 0, and therefore,
by the Cayley-Hamilton theorem, A2 − I = 0. Even powers of A are thus
equal to I, and odd powers of A are equal to A. Thus
!
!
∞
∞
1
1
At
2k +1
e
=
A
∑ (2k)t2k ! I + ∑ (2k + 1)! t
k =0
k =0
= (cosh t) I + (sinh t) A
cosh t sinh t
=
sinh t cosh t
We can now apply the variation of constants formula. A particular solution
of ~v0 = A~v + ~f (t), will be given by
Z t
cosh(t − s) sinh(t − s)
−1
~v(t) =
e−s ds
sinh(t − s) cosh(t − s)
1
0
3.5. A FUNDAMENTAL MATRIX SOLUTION OF INTEREST
287
− es−t−s
ds
es−t−s
0
t
−e−t = s
e − t s =0
−1
−t
= te
.
1
=
Z t
Exercises

1.
0
 0
Show that if A = 
 0
0
1
0
0
0
0
1
0
0

0
0 
 , then A4 = 0. Hence calculate
1 
0
e At .
Answer
2.
Show that exp
t 1
0 t
=
et et
0 et
.
3.
Prove theorem 3.13. Hint: review the proof that es+t = es et , based
1 k
on the formula et = ∑∞
k=0 k! t . Where is the commutative law used?
Answer
4.
Let A(t) be a square matrix whose entries are functions of t, and
let B(t) = dA/dt. Show that if A(t) and B(t) commute, then
e A(t) =
∞
1
∑ k! [ A(t)]k
k =0
is a fundamental matrix solution of the system ~x 0 = B(t)~x.
t 1
5.
Show that if A(t) =
, then A and dA/dt do not com0 2t
mute.
Answer




0 1 −1
1 1 0
2  , and B =  0 1 2  . Show that e A =
6.
Let A =  0 0
0 0
0
0 0 1
B.
288
7.
CHAPTER 3. LINEAR SYSTEMS
Compute all powers of


0 a b
A =  0 0 c ,
0 0 0
and hence compute
(a) e At .
(b) e( A+λI )t , where λ is a scalar constant.
Use the result of part (b) to solve the following IVP:

x 0 = −2x + 3y + z; x (0) = 3 
y0 =
−2y − 4z; y(0) = 0

0
z =
−2z; z(0) = 1
Answer
8.
Let K =
0 −1
1 0
. Determine all powers of K and hence com-
pute
(a) eKt .
(b) e(K+λI )t , where λ is a scalar constant.
Use the result of part (b) to solve the following IVP:
x 0 = − x − y; x (0) = 1
y0 =
x − y; y(0) = 0
In Exercises 9 – 15, calculate the matrix e At .
1
1
9.
A=
.
−2 −1
Answer
0 4
10.
A=
.
2 2
−1 2
11.
A=
.
−3 3
Answer
3.5. A FUNDAMENTAL MATRIX SOLUTION OF INTEREST
12.
A=
13.
A=
−3
4
−4 −1
−3 −2
2
1
.
.
Answer
14.
A=
2 −5
4 −2
.


0 1 2
15.
A =  0 0 1 .
0 0 0
Answer
In Exercises 16– 19, find the general solution.
16.
x 0 = x − 4y + 2e−t
y0 = 2x − 3y + e−t
17.
x 0 = x + et
y0 = x − y + cosh t
Answer
18.
x 0 = x − y + t −1
y 0 = − x + y + t −1
19.
x 0 = 2x − y + e2t
y0 = 4x − 2y + 2e2t
Answer
289
290
CHAPTER 3. LINEAR SYSTEMS
The following two Exercises are intended to be challenging.
20.
Let A(t) be an n × n matrix whose entries are functions of t. Show
that if the matrices A(t) and A(u) commute for all t and u, then A(t)
and A0 (t) = dA/dt commute. Hence find a fundamental matrix solution for ~x 0 = A0 (t)~x.
21.
Let P be the vector space of all polynomials, and let D : P −→ P
denote the differentiation operator. Show that if f ( x ) ∈ P , then
(etD f )( x ) = f ( x + t)
Answer
3.6
Exponentiating a Matrix
Every CAS has the matrix exponential function: if a square matrix M is
entered, one issues the command Exp(M) and the CAS returns e M . To perform manual calculations, there are two steps. First, we’ll develop a set of
matrices that we can handle easily. We will call these matrices E-T-E (stands
for Easy-To-Exponentiate). That done, we will see how to adapt any n × n
matrix to fit our scheme.
Here is our first E-T-E:
Definition: A matrix N is nilpotent if there is an integer p > 0 such that
N p = 0.
We have seen some nilpotent matrices already (Examples 3.5.1 and 3.5.3(c)).
If N p = 0 then
1
1
e Nt = I + t N + t2 N 2 + · · · +
t p −1 N p −1 .
2
( p − 1) !
There is a particular sequence of nilpotent matrices of interest:
0 1
N2 =
0 0


0 1 0
N3 =  0 0 1 
0 0 0
..
.
3.6. EXPONENTIATING A MATRIX

Np



= 


291
0 1 0
0 0 1
.. .. ..
. . .
0 0 0
0 0 0

··· 0
··· 0 

..  .
..
. . 

··· 1 
··· 0
Lemma 3.6.1 If k < p the matrix Npk can be obtained from I p−k by adding k
columns of zeros on the left, and k rows of zeros on the bottom. For k ≥ p,
Npk = Z p .
Lemma 3.6.1 leads to a formula for e Np t :
1
1
e Np t = I + t Np + t 2 ( Np )2 + · · · +
t p −1 ( N p ) p −1
2
( p − 1) !


1 t 21 t2 · · · ( p−1 1)! t p−1


 0 1 t · · · ( p−1 2)! t p−2 


.
. .
..
..
..
= 

 .. ..
.
.
.



 0 0 0 ···
t
0 0 0 ···
1
(3.34)
The proofs of lemma 3.6.1 and of formula (3.34) are asked for in Exercise 3 at the end of this section.
We can build more E-T-E’s by adding scalar matrices and nilpotent matrices. For example, if A = kI p + Np then—because the matrices kI p and Np
commute—Theorem 3.13 tells us that
e At = ekIp t e Np t = ekt e Np
By (3.34), then

e ( k I p + Np ) t



kt 
=e 


1
0
..
.
0
0
t
1
..
.
0
0
1 2
2t
t
..
.
0
0
···
···
..
.
···
···
1
t p −1
( p −1) !
1
t p −2
( p −2) !
..
.
t
1




.



Example 3.6.1 Display the matrix A = −3I4 + N4 and calculate e At .
(3.35)
292
CHAPTER 3. LINEAR SYSTEMS

 
−3 0
0
0
0 1 0
 0 −3 0
  0 0 1
0
+
S OLUTION . −3I4 + N4 = 
 0
0 −3 0   0 0 0
0
0
0 −3
0 0 0


−3 1
0
0
 0 −3 1
0 

 . By formula (3.35),
 0
0 −3 1 
0
0
0 −3
1 3 3
1 2 2
(−3I4 + N4 )t
−3t
e
= e
I4 + tN4 + t N4 + t Nk
2
3!


1 t 21 t2 61 t3
1 2 

0 1 t
2t 
= e−3t 
 0 0 1
t 
0 0 0
1

0
0 
=
1 
0
Diagonal matrices. In a square matrix A, with entries denoted aij , the diagonal entries are the entries aii . These run down the diagonal extending
from the upper left corner to the lower right corner. The off-diagonal entries are the aij with i 6= j. If all of the off-diagonal entries of a matrix are
zeros, the matrix is called a diagonal matrix. Examples include scalar matrices, where all of the diagonal entries are equal. The diagonal matrix with
diagonal entries d1 , d2 , . . . , dn is denoted diag (d1 , d2 , . . . dn ). Unlike scalar
matrices, an arbitrary diagonal matrix does not commute with all other
matrices, although it does commute with all other diagonal matrices.
Diagonal matrices qualify as E-T-E’s. Following Example 3.5.2, you can
derive this formula:
ediag (d1 ,d2 ,...,dn )t = diag (ed1 t , ed2 t , . . . , edn t ).
(3.36)
Direct sums To build still more E-T-E’s, we will use a new matrix operation.
Let P be a p × p matrix, and let Q be a q × q matrix, and let n = p + q.
The direct sum of these matrices is an n × n matrix P ⊕ Q that looks like
this:
P
0 p×q
,
0q × p
Q
3.6. EXPONENTIATING A MATRIX
293
where 0 p×q denotes a p × q matrix of zeros. Note that if p = q = 1 then
P ⊕ Q is just a diagonal 2 × 2 matrix. Similarly, an n × n diagonal matrix
can be expressed as the direct sum of n 1 × 1 matrices. Here is another
example:


9 0 0
0
0
 0 3 −1 0
0 


3 −1

0
0 
(3.37)
⊕ (− I2 + N2 ) =  0 2 1
[9] ⊕

2 1
 0 0 0 −1 1 
0 0 0
0 −1
In this example we have used the fact, which you can prove, that the ⊕
operation is associative.
Direct sums are easily added and multiplied, provided that the summands are of the same size. Let P and P0 be p × p matrices, and Q, Q0 be
q × q matrices. Then (see exercise 5)
( P ⊕ Q)( P0 ⊕ Q0 ) = PP0 ⊕ QQ0 and ( P ⊕ Q) + ( P0 ⊕ Q0 ) = ( P + P0 ) ⊕ ( Q + Q0 ).
(3.38)
We can use formula (3.38) to raise direct sums to powers, ( P ⊕ Q)n = Pn ⊕
Qn , and to take exponentials:
e( P⊕Q)t = e Pt ⊕ eQt .
(3.39)
Formula (3.39) provides a way to find the exponential of direct sums of
E-T-E’s; thus these too are E-T-E’s.
Example 3.6.2 Find e At where A is the matrix in equation (3.37).
3 −1
−1 1
S OLUTION . Let U = [9], V =
, and W =
, so that
2 1
0 −1
A = U ⊕ V ⊕ W. Then e At = eUt ⊕ eVt ⊕ eWt .
First, eUt = [e9t ]. The other two
matrices
are 2 × 2. Noting that tr V = 4, we
1 −1
have V = 2I + B, where B =
. Since tr B = 0 and det B = 1, the
2 −1
Cayley-Hamilton Theorem implies B2 = − I. Hence B2n = (−1)n I and
B2n+1 = (−1)n B. Reasoning as we have before, we get
e Bt = (cos t) I2 + (sin t) B, and
cos t + sin t
− sin t
Vt
2t Bt
2t
e
= e e =e
2 sin t
cos t − sin t
2t
2t
e (cos t + sin t)
−e sin t
=
2e2t sin t
e2t (cos t − sin t)
294
CHAPTER 3. LINEAR SYSTEMS
Finally, W = − I + N2 so
e
It follows that

e
At


=


Wt
−t
= e ( I + tN2 ) =
e−t te−t
0
e−t
.
e9t
0
0
0
0
2t
2t
0 e (cos t + sin t)
−e sin t
0
0
0
2e2t sin t
e2t (cos t − sin t) 0
0
0
0
0
e−t te−t
0
0
0
0
e−t



.


Matrix conjugation
Proposition 3.6.2 If B = P−1 AP, where A, B, and P are n × n matrices and P
is invertible, then
e Bt = P−1 e At P.
P ROOF : Notice that for any power p,
B p = ( P−1 AP)( P−1 AP) · · · ( P−1 AP) = P−1 A p P
|
{z
}
p factors
Therefore
1
1 m m
e Bt = I + Bt + B2 t2 + · · · +
B t +···
2
m!
1
= I + P−1 APt + P−1 A2 Pt2 + · · ·
2
1 2 2
−1
= P
I + At + A t + · · · P
2
= P−1 e At P
Definition: Let A and P be n × n matrices, P being invertible. The
matrix B = P−1 AP is called the conjugate of A by P.
Definition. An n × n matrix A is semisimple if A has n linearly independent eigenvectors.
3.6. EXPONENTIATING A MATRIX
295
Proposition 3.6.3 Let A be a semisimple n × n matrix. Then A is conjugate
to a diagonal matrix D = diag (s1 , s2 , . . . , sn ), where the s1 , s2 , . . . , sn are the
eigenvalues of A (multiple eigenvalues are repeated in the list).
There is a converse to proposition 3.6.3.
Proof: Let ~b1 ,~b2 , . . . ,~bn be n linearly independent eigenvectors of A,
with eigenvalues s1 , s2 , . . . , sn . We will take P to be the matrix whose columns
are the vectors ~bi . Because the columns of P are linearly independent, P is
nonsingular, and thus invertible. Set Q = P−1 .
The columns of AP are the vectors A~b1 , . . . A~bn . Since the vectors ~bi are
eigenvectors, the ith column of AP is si~bi .
Now consider the matrix PD. When any matrix M on the right by a
diagonal matrix, the ith column of M is multiplied by the ith diagonal entry. Thus the columns of PD are also s1~b1 , s2~b2 , . . . , sn~bn , and we have the
equality
AP = PD, and thus AQ−1 = Q−1 D.
Multiply on the right by Q to obtain
A = Q−1 DQ.
Proposition 3.6.3 shows how our program is to work: since all diagonal
matrices are E-T-E’s and all semisimple matrices are conjugate to diagonal
matrices, we have a method to compute e At —and thus to find a fundamental matrix solution of ~v0 = A~v—for any semisimple constant matrix A.

2 −3 −3
where A =  1 −6 −3 . The characteris−1 9
4

Example 3.6.3 Compute e At
tic polynomial of A is t3 − t.
S OLUTION . The eigenvalues are the roots of the characteristic polynomial,
−1, 0, 1. The eigenvectors satisfy equations ( A − si I )~bi = ~0. Thus ~b1 is
determined by

   
3 −3 −3
x
0
( A − (−1) I )~b1 =  1 −5 −3   y  =  0 
−1 9
5
z
0
296
CHAPTER 3. LINEAR SYSTEMS


2 0 −1
The matrix A + I is row equivalent to  0 2 1 . We can therefore
0 0 0


1
take ~b1 =  −1 . By similar means, eigenvectors belonging to 0 and 1
2


 
3
3
are determined to be ~b2 =  −1  and ~b3 =  0 . Thus
3
1
−
1
P AP = diag (−1, 0, 1), where P is the matrix with columns ~b1 ,~b2 ,~b3 .
Exponentiating and referring to proposition 3.6.2 yields
P−1 e At P = diag (e−t , 1, et )
Multiplying on the left by P and on the right by P−1 we have
e At = P (e−t , 1, et ) P−1

  −t

e
0 0
1 3 3
=  −1 0 −1   0 1 0  
0 0 et
2 1 3
 −t
e + 3 − 3et −6e−t − 9 + 15et
6e−t − 5et
=  − e−t + et
2e−t + 1 − et −12e−t − 3 + 15et

1 −6 −3
1 −3 −2 
−1 5
3

−3e−t − 6 + 9et

3e−t − 3et
−
t
t
−6e − 2 + 9e
If a matrix is not semisimple, there is a theorem from linear algebra that
we can use. It refers to the following definition:
Definition: An elementary Jordan matrix is a an m × m matrix E(s, m) =
sIm + Nm , where Nm is the standard m × m nilpotent matrix. (Note that
E(s, 1) = [s], because N1 = [0].)
We have encountered elementary Jordan matrices already; Equation (3.35)
shows that they are E-T-E’s. See also Example 3.6.1. The following theorem
includes the result of Proposition 3.6.3 as the special case when each mi = 1.
It can be paraphrased as “Every square matrix is conjugate to a direct sum
of elementary Jordan matrices.” We have seen (equation (3.39)) that a matrix that can be expressed as a direct sum of E-T-E’s is itself an E-T-E; hence
every square matrix is conjugate to an E-T-E.
3.6. EXPONENTIATING A MATRIX
297
Theorem 3.15 Let A be an n × n matrix. There exist elementary Jordan matrices
Ei = E(si , mi ) (i = 1, . . . , p), such that A is conjugate to the direct sum of the Ei .
In other words, there exists an invertible n × n matrix P such that
!
p
A = P −1
M
Ei
P.
i =1
Furthermore, the dimensions mi of the matrices Ei satisfy m1 + · · · + m p = n and
the si are the eigenvalues of A.
We will omit the proof of Theorem 3.15. It is a corollary of the existence
of a matrix called the Jordan canonical form of the matrix A, which is proved
in all worthy linear algebra texts. It is worth reminding you that some of
the eigenvalues si of A may be complex, even if all entries of A are real.
Then the corresponding matrices Ei , and the matrix P, would have some
complex entries.
The procedure for calculating e At is to find an E-T-E B that is conjugate
to A (theorem 3.15 guarantees that there is one); of course one also needs a
matrix P such that P−1 AP = B; then A = PBP−1 and by proposition 3.6.2,
e At = Pe Bt P−1 .
This reduces finding a fundamental matrix solution of a system of linear,
constant coefficient ODE’s to a routine problem in linear algebra. Because
the matrix computations can be strenuous, they are often left to a computer
running a CAS, or specialized software.
In Example 3.6.4, we will find the exponential of a matrix that is not
semisimple. Before proceeding, we will need to introduce some helpful
definitions from linear algebra.
Definitions: Let s be an eigenvalue of a matrix A. The set of all eigenvectors of A that belong to s is called the eigenspace of A belonging to s. The
generalized eigenspace belonging to s is the set of all vectors ~a ∈ Rn such
that for some m,
( A − sI )m~a = ~0
You can verify that the eigenspace, and the generalized eigenspace belonging to an eigenvalue of A are subspaces of Rn .
The algebraic multiplicity of an eigenvalue s of a matrix A is the multiplicity of s as a root of the characteristic polynomial of A. Contrast this
298
CHAPTER 3. LINEAR SYSTEMS
with the geometric multiplicity, which is the dimension of the eigenspace
belonging to s.
If ~a is an eigenvector of A belonging to an eigenvalue of s then ( A −
sI )1 = ~0; thus ~a is also a generalized eigenvector. It is shown in linear
algebra courses that the generalized eigenspace of an eigenvalue is a subspace of Rn of dimension equal to its algebraic multiplicity. The set of
eigenvectors belonging to the same eigenvalue also form a subspace of the
generalized eigenspace. If a matrix is semisimple, the the geometric and
algebraic multiplicities of each eigenvalue are the same. If a matrix is not
semisimple, then for at least one eigenvalue, the algebraic multiplicity is
greater than the geometric multiplicity.
Example 3.6.4 Find e At , where


−19 22 21
A =  −9 11 9  ,
−9 10 11
given that the characteristic polynomial of A is p(s) = (s + 1)(s − 2)2 .
This example could be restated as “Find a fundamental matrix solution
X (t) of the following system such that X (0) = I.”
x 0 = −19x + 22y + 21z
y0 = −9x + 11y + 9z
z0 = −9x + 10y + 11z
S OLUTION . The eigenvalues of A
are 2(double) and −1. The solutions of
1
( A − 2I )~b = ~0 are all multiples of  0 . Therefore the geometric
1
multiplicity of the eigenvalue 2 is 1, while the algebraic multiplicity of the
same eigenvalue is 2. It follows that A is not semisimple. We calculate


54 −54 −54
( A − 2I )2 =  27 −27 −27 
18 −18 −18
which is row equivalent to the matrix


1 −1 −1
0 .
C= 0 0
0 0
0
3.6. EXPONENTIATING A MATRIX
299
The solutions of ( A − 2I )2~a = ~0 are the same as the solutions of C~a = ~0,
and one can easily
of the latter equation are the
 see

 that
 the
solutions
1 
 1
linear span of  1  ,  0  . These two vectors will form the first


0
1
 
2

3 
two columns of the matrix P; the third will be the eigenvector
2
belonging to the eigenvalue −1. Thus


1 1 2
P =  1 0 3 ,
0 1 2
and by matrix multiplication we find that


2 0 0
P−1 AP =  1 2 0  ,
0 0 −1
0 0
which is E-T-E, as the direct sum of 2I2 + N, where N =
is
1 0
nilpotent, and a 1 × 1 matrix. We have e(2I2 + N )t = e2t e Nt = e2t ( I + Nt).
Thus, P−1 e At P = (e2t ( I + Nt) ⊕ [e−t ] F, and so
 2t

e
0
0
e At = P  te2t e2t 0  P−1
0
0 e−t
 2t

e (−3t − 5) + 2e−t 2(e2t − e−t ) e2t (3t + 2) − 2e−t
1

3(−e2t + e−t )
−3e−t
3(e2t − e−t )
=
3
2t
−
2t
2t
−
t
2t
−
t
e (−3t − 2) + 2e
2(e − e ) e (3t − 1) − 2e
Clearly, A is not an E-T-E.
Exercises


2 1
0
0 .
1.
Calculate e At , where A =  −1 0
0 0 −1
Answer
300
CHAPTER 3. LINEAR SYSTEMS
2.
Let M be a p × n matrix, and let Np be the standard p × p nilpotent
matrix. Then M0 = Np M is also a p × n matrix. Show that the bottom
row of M0 is a zero row, and for i < p the ith row of M0 is equal to the
(i + 1)st row of M.
3.
Prove lemma 3.6.1, and formula (3.34).
Answer
4.
Prove that the direct sum operation of matrices, ( P, Q) 7→ P ⊕ Q,
is associative. That is, show that for any 3 matrices P, Q, and R, if we
put A = P ⊕ Q and B = Q ⊕ R then A ⊕ R = P ⊕ B.
5.
Use the definitions of matrix addition and multiplication to verify
the formula (3.38) for the product of direct sums of matrices. Then
prove formula 3.39) for the matrix exponential of direct sums of matrices.
Answer
6.
Let the n × n matrix A be conjugate to a diagonal matrix D; in
other words, we are assuming that there exists an invertible matrix P
such that P−1 AP = D. Prove that A is semisimple.
7.
If B a conjugate of A by some matrix P, write B ∼ A. Prove that
∼ is an equivalence relation: you need to prove the following:
• A ∼ A (reflexive law),
• A ∼ B ⇒ B ∼ A (symmetric law), and
• ( A ∼ B & B ∼ C ) ⇒ A ∼ C (transitive law).
Answer
In Exercises 8–13, calculate e At . As a practical matter, you will probably want to use a CAS for routine matrix calculations. Please limit
your automated calculations to the following operations:
• Matrix multiplication,
• Matrix inversion,
8.
• Row reduction.

−2 1
0
 1 −2 0
A=
 0
0 −5
0
0 −2

0
0 
.
4 
1
3.6. EXPONENTIATING A MATRIX
301

−2 0 0 0
 0 0 −4 0 

A=
 0 1 0 0 .
0 0 0 1
Answer


−2 −4 −8
2
2  . The characteristic polynomial of A is
A =  2
0
1
3
3
2
f (s) = s − 3s + 2s.


−1 1
2
A =  1 −1 −2  . The characteristic polynomial of A is
−2 2
4
f (s) = s3 − 2s2 .
Answer


7 −4 −1
A =  11 −6 −2  . The characteristic polynomial of A is
6 −4 0
3
2
f (s) = s − s .


2 −8 −2 5
 2 −4 −6 6 

A=
 2 −2 −10 8  . The characteristic polynomial of A is
2 −2 −8 6
2
f (s) = (s + 2s + 2)(s + 2)2 .
Answer

9.
10.
11.
12.
13.
14.
Use the result of Exercise 8 to determine the solution of the IVP
x 0 = −2x + y,
0
y = x − 2y,
0
z = −5z + 4w,
0
w = −2z + w,
15.
x (0) = 1
y (0) = −1
z (0) = 0
w (0) = 3
Use the result of Exercise 11 and the method of variation of constants to determine the solution of the IVP
x 0 = − x + y + 2z + 1,
0
x (0) = 0
y = x − y − 2z − 1,
y (0) = 0
z = −2x + 2y + 4y + 1
z (0) = 0
0
302
CHAPTER 3. LINEAR SYSTEMS
Answer
3.7
Chapter Glossary
e At The matrix exponential.
I The identity matrix.
Im(z) The imaginary part of z = x + iy: Im( x + iy) = y.
Re(z) The real part of z = x + iy: Re( x + iy) = x.
Associated homogeneous system (of a system of ODEs, v0 = A(t)v +
b(t)) The system obtained by removing the source term: v0 = A(t)v.
Absolute value The distance
of a complex number z = x + iy to 0. The
p
2
formula is |z| = x + y2
Argument (of a complex number z) The angle made by a line connecting z
to the origin with the positive real axis in the complex plane.

 cos−1 Re(z)
if Im(z) ≥ 0
|z|
arg(z) =
 − cos−1 Re(z)
if Im(z) < 0
|z|
Belongs to Corresponding characteristic roots and characteristic vectors of
a matrix are said to belong to each other.
Cayley-Hamilton theorem Let f (s) = 0 be the characteristic equation of a
matrix A. Then f ( A) is equal to the zero matrix.
Characteristic equation A polynomial equation used to determine the characteristic roots of a matrix A. For an n × n matrix A, this is
det( A − sI ) = 0.
For a 2 × 2 matrix
A=
a c
b d
,
the characteristic equation is s2 − p s + q = 0, where p = trace ( A) =
a + d, and q = det( A).
3.7. CHAPTER GLOSSARY
303
Coefficient matrix The matrix A(t) in the system
v 0 = A ( t ) v + b ( t ).
Column vector A matrix that has only one column. We use boldface letters
for column vectors.
Complex plane The ordinary xy-plane, in which the point ( x, y) is made
to correspond the complex number z = x + iy.
Determinant A scalar-valued function of square matrices. For a 2 × 2 matrix
a c
A=
,
b d
det( A) = ad − bc. Determinants are defined for n × n matrices in
linear algebra courses.
Eigenvalue (of a matrix A) A scalar s such that for some characteristic vector b, Ab = sb.
Eigenvector (of a matrix A). A nonzero vector b such that Ab = sb for
some scalar s.
Entry A scalar that appears at a specified location in a matrix.
Exponential (of a matrix A) A matrix function, denoted e At , that is the sum
of the power series
1
1
e At = I + At + A2 t2 + · · · + An tn + · · · .
2
n!
If A is a constant matrix, e At is the fundamental matrix solution of
v0 = Av having initial value X (0) = I.
Fundamental matrix solution A matrix solution X (t) that is a nonsingular
square matrix.
Generalized eigenspace belonging to an eigenvalue of a matrix A. The set
of vectors ~a with the property that ( A − sL) p~a = ~0 for some p.
Homogeneous system A system of linear ODEs that is satisfied by the vector function v(t) ≡ 0. A homogeneous system always has a zero
source vector.
304
CHAPTER 3. LINEAR SYSTEMS
Imaginary axis The vertical axis of the complex plane, which corresponds
to complex numbers z = 0 + iy.
Imaginary part (of a complex number z = x + iy) Im(z) = y.
Identity matrix The matrix




I=


1 0 0 ···
0 1 0 ···
0 0 1 ···
.
0 0 0 ..
0 0 0 ···
0
0
0




.

0 
1
I has the property that for all b ∈ Rn , Ib = b.
Inverse matrix (of a square matrix A) A matrix B such that AB = BA = I.
Matrix A rectangular array of scalars.
Matrix product The result of multiplying an m × n matrix A with entries
aij and a p × q matrix B with entries bkl is defined if and only if n = p;
in this case it is an m × q matrix C whose entries are
crs = ar1 b1s + ar2 b2s + · · · + arn bns .
Matrix solution (of a homogeneous system of ODEs, v0 = A(t)v) A matrix
function X (t) such that dX (t)/dt = A(t)X (t).
Nonsingular A matrix that is not singular: A is nonsingular if and only if
Ab 6= 0 for all b 6= 0.
Nilpotent matrix A square matrix N with the property that some power
N p is equal to the zero matrix.
Real axis The horizontal axis of the complex plane, which corresponds to
numbers z = x + 0i.
Real part (of a complex number z = x + iy) Re(z) = x.
Scalar A real or complex number, or a real or complex-valued expression.
We use lower case italic letters for scalars.
Singular A matrix A such that Ab = 0 for some nonzero vector b.
3.7. CHAPTER GLOSSARY
305
Source vector The vector function b(t) in the system
v 0 = A ( t ) v + b ( t ).
Trace The sum of the diagonal entries of a square matrix.
Variation of constants A method for finding a particular solution of an inhomogeneous system. Given a fundamental matrix solution X of
v0 = Av, substitute v = X w in
v0 = Av + f(t),
where w is to be determined. The resulting equation is readily simplified and integrated to find w. If A is a constant matrix then we can
take X(t) = e At . Then the following formula for the solution results:
v(t) =
Z t
0
e A(t−s) f (s) ds + e At c.
306
3.8
1.
CHAPTER 3. LINEAR SYSTEMS
Review Exercises
Solve the initial value problem
x 0 = 2x + y; x (0) = 1
y0 = x + 2y; y(0) = 0.
Answer
2.
Find a fundamental matrix solution for each differential equation.
0
x = x − 2y
(a)
y0 = 5x − y
0
x = x−y
(b)
y0 = x + y
0 x
1 2
x
(c)
=
.
y
0 1
y
(d) ~x 0 = A~x, where


1 2 3
A =  0 2 3 .
0 0 3
Answer
3.
Find the general solution of
x 0 = x + 5y + tan 2t
y0 = − x − y,
2 sin 2t − cos 2t sin 2t + 2 cos 2t
is a fundacos 2t
− sin 2t
mental matrix solution of the associated homogeneous equation.
Answer
given that X (t) =
4.
Suppose A is an n × n matrix with independent characteristic vectors ~b1 , . . . , ~bn . Denote the corresponding characteristic roots by r1 ,
. . . , rn . Let E(t) denote the diagonal matrix whose diagonal entries
are er1 t , . . . , ern t and let B denote the matrix whose columns are the
characteristic vectors ~b1 , . . . , ~bn . Show that B · E(t) is a fundamental
matrix solution of the system ~x 0 = A~x. Find an expression for e At .
Answer
3.8. REVIEW EXERCISES
307
Show that a 2 × 2 matrix is nilpotent if and only if its trace and
determinant are both zero.
Answer
a b
6.
Let A =
and let B = A − 12 tr ( A) I. Find a formula for
c d
det( B).
Answer
5.
308
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Figure 3.1: The complex plane.
−1 + 2i
r
6
2i r
i r
−2r
−1r
−(2 + i ) r
−1 + 2i
r
−i r
r
−2i r
r2+i
0
1r
2r r2+i
FIGURES FOR CHAPTER 4
309
Figure 3.2: The triangle inequality: |z1 + z2 | ≤ |z1 | + |z2 |
| z2 |
z1
| z1 | r
((((
0 r((((
(
(((
(((
(
(
(
(
((((
|z + z |
1
2
z1 + z2
(((r
((((
310
CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Figure 3.3: Polar representation of a complex number. The length of the heavy line to the
point representing the complex number z = −1 + 2i is r = |z|, and the radian measure of
the directed arc shown is θ = arg(z).
z
2
1.5
1
0.5
-1
-0.5
0.5
1
1.5
2
Chapter 4
Stability Theory
311
312
CHAPTER 4. STABILITY THEORY
4.1
Phase Portraits of Linear Systems
Let A be a 2 × 2 constant matrix. We will see how to draw a phase portrait
the system of ODEs,
~v0 = A~v.
(4.1)
All linear systems have a stationary point at the origin, since ~00 = ~0 = A~0.
If A is singular then there is a vector ~b 6= ~0, such that A~b = ~0. In this case,
all scalar multiples of ~b are stationary points, and we say that the system is
degenerate. If A is nonsingular, then ~0 is the only stationary point, and we
say that (4.1) is nondegenerate.
Nodes
The simplest case is that of a scalar matrix. Let A = rI, where r 6= 0 is
constant. Then eAt = ert I so the general solution of the system (4.1) is ~v =
c1
ert~c, where ~c =
. Thus every solution has the equations x = c1 er t ,
c2
y = c2 er t . Since y/x = c2 /c1 is constant, the orbits are half-lines directed
toward the origin (if r < 0) or away from the origin (if r > 0). The phase
portrait, shown as figure 4.1, is called a proper node. We say that the node
is stable if r < 0, because the solutions approach the origin with increasing
time. If r > 0, the node is unstable.
Now let’s consider the case of a matrix A that has unequal real eigenvalues. In this case, A is conjugate to a diagonal matrix
D=
r 0
0 s
,
and the phase portrait of (4.1) can be obtained by distorting the phase portrait of
~ 0 = Dw
~
w
(4.2)
in a very specific way.
To determine the general solution of (4.2) we note that e Dt
=
er t 0
0 es t
c1 er t
.
c2 e s t
To be definite, let’s suppose that r = −2 and s = −1. Orbits where
c2 = 0 follow the x-axis toward the origin, and when c1 = 0 instead, the
~ = e Dt~c =
so w
,
4.1. PHASE PORTRAITS OF LINEAR SYSTEMS
313
orbit follows the y-axis toward the origin. If c1 and c2 are both nonzero,
then the orbit is given by the parametric equations
x = c1 e−2t
y = c2 e − t .
Since (e−t )2 = e−2t we can eliminate t from the parametric equations and
obtain
2
y
x
=
,
c1
c2
or x = a y2 where a = c1 /c22 . The orbits that do not follow either coordinate axis follow parabolas toward the origin. The phase portrait, shown
in figure 4.3, displays the orbits on the x and y-axes, and four other orbits. The reason that not all orbits are straight lines is that as t increases,
x = c1 e−2t decreases in magnitude faster than y = c2 e−t . This causes the
orbits to become squashed against the y-axis, unless c2 = 0. This phase portrait is called an improper node. It is stable, because all orbits converge to
the origin.
In general, the phase portrait of a linear system of ODEs is called a
stable node if every orbit is directed toward the origin as t → ∞. If all orbits
approach the origin as t → −∞, the phase portrait is an unstable node.
Stable and unstable nodes can be recognized by calculating the eigenvalues
of the coefficient matrix.
Proposition 4.1.1 Let A be a 2 × 2 matrix. The phase portrait of the system
~v0 = A~v is a stable node if and only if both eigenvalues of A are negative, or are
complex numbers with negative real parts, and it is an unstable node if and only if
both eigenvalues are positive, or are complex numbers that have positive real parts.
The proof is left to you; see exercise 29 at the end of this section.
The following example is to illustrate the way to draw a phase portrait
of an improper node.
Example 4.1.1 Draw a phase portrait of the system
x 0 = 2x − y
y0 = − x + 2y.
S OLUTION . The eigenvalues of the coefficient matrix,
2 −1
,
−1 2
314
CHAPTER 4. STABILITY THEORY
1
are 1 and 3, so the phase portrait is an unstable node. The vectors
1
1
, respectively, are eigenvectors belonging to these
and
−1
eigenvalues. The slopes of these eigenvectors are 1 and −1, respectively.
The general solution of the system is therefore
x
1
1
t
3t
= c1 e
+ c2 e
(4.3)
y
1
−1
It follows that the half lines with slopes ±1, directed away from the origin,
are orbits of the system. The other orbits are not straight. As t → −∞, the
second term of (4.3) is negligible; the orbit therefore recedes from the
origin with slope 1. As t becomes large, the first term of (4.3) becomes
negligible, so the slope approaches −1, the slope of the other eigenvector.
See figure 4.4.
If a 2 × 2 matrix A is not a scalar matrix and has a double eigenvalue
r 6= 0, then the phase portrait of
0 x
x
=
A
(4.4)
y0
y
is an improper node, stable if r < 0, and unstable if r > 0. Here is an
example:
Example 4.1.2 Draw the phase portrait of
x0 = r x
y0 = k x + r y
S OLUTION . The system has a double eigenvalue of r, and all eigenvectors
are vertical. Assuming that r > 0, the phase portrait is an unstable node,
with straight orbits emanating from the origin and following the y-axis.
The general solution of this system,
x = c ert
y = (kct + d)e
(4.5)
rt
(4.6)
was derived on page 254 (for k = 1). We can eliminate t by solving (4.5) to
get
1
1
(4.7)
t = ln( x/c) = (ln | x | − ln |c|)
r
r
4.1. PHASE PORTRAITS OF LINEAR SYSTEMS
315
(the absolute value signs are necessary because x may be negative — we
only know that x and c have the same sign). Now divide (4.6) by (4.5) to
obtain y/x = (k t + (d/c)), and substitute (4.7) for t. This yields
y
k
k
d
= ln | x | − ln |c|) +
x
r
r
c
Let b = − kr ln |c|) + dc . Then
y=x
k
ln | x | + b .
r
(4.8)
Each orbit (except for the stationary point at the origin and the orbits on
the positive and negative y-axis) follows the graph of (4.8) for some
particular value of b. The phase portrait can be drawn by sketching graphs
of equation (4.8) for various b. Figure 4.5 shows the phase portrait for
k = 2 and r = 1.
Saddles
Let A be a 2 × 2 matrix whose eigenvalues r and s have opposite signs:
r < 0 < s. Let ~e and ~f be eigenvectors belonging to r and s, respectively.
Five orbits of ~v0 = A~v are readily identified: the stationary point at the
origin; the pair of half-lines, directed toward the origin and parallel to ~e;
and the two half-lines directed away from the origin and parallel to ~f . The
lines through the origin in the ~e and ~f directions are called the stable line
and the unstable line, respectively.
The remaining nonstationary orbits resemble hyperbolas, with the stable and unstable lines as asymptotes. As t → ∞ the orbits approach the
unstable line, and as t → −∞ they approach the stable line.
Example 4.1.3 Draw a phase portrait of the system
x0 = − x
y0 = x + y.
−1 0
S OLUTION . The eigenvalues of A =
are −1 and 1,
1 1
2
0
~
corresponding to eigenvectors ~e =
and f =
These vectors
−1
1
316
CHAPTER 4. STABILITY THEORY
determine the half-line orbits, which are drawn first. The remaining orbits
are asymptotic to these. Figure 4.6 displays the phase portrait and the
vector field.
The phase portrait of a system having eigenvalues of opposite sign is
called a saddle. To see why, let us find an integral for the system presented
in example 4.1.3. An integral for this system is also an integral for the ODE
( x + y) dx + x dy = 0,
which happens to be in exact form. Knowing this, It is straightforward
to obtain the integral F ( x, y) = 12 x2 + xy. The orbits follow level curves
F ( x, y) = C. The origin is a critical point of F (that is, both partial derivatives of F vanish there), and by the second derivative test,
2 ∂2 F ∂2 F
∂ F
−
= (1)(0) − (1)2 < 0,
2
2
∂x ∂y
∂x∂y
this critical point is a saddle point of the function F ( x, y).
You can classify a linear system with constant coefficients as a saddle,
degenerate, or a node with very little effort. The following lemma from
linear algebra is the key. Although it is stated in terms of on 2 × 2 matrices
it holds for all square matrices.
Lemma 4.1.2 Let A be an 2 × 2 matrix, with eigenvalues r1 and r2 . Then the
trace of A is equal to the sum of the eigenvalues:
tr( A) = r1 + r2 ;
and the determinant is equal to the product of the eigenvalues:
det( A) = r1 · r2 .
P ROOF. The characteristic polynomial of A can be factored as (s −
r1 )(s − r2 ). Multiply this out to obtain
s2 − (r1 + r2 )s + r1 r2 = s2 − tr( A) s + det( A).
It follows that tr ( A) = r1 + r2 and det( A) = r1 r2 .
By the lemma, the eigenvalues of a 2 × 2 matrix A are of opposite sign
if and only if det( A) < 0, and 0 is a eigenvalue if and only if det( A) = 0.
4.1. PHASE PORTRAITS OF LINEAR SYSTEMS
317
Thus, the phase portrait of the system ~v0 = A~v is a saddle if and only if
det( A) < 0, and it is degenerate if and only if det( A) = 0.
If det( A) > 0, it is possible that A has no real eigenvalues at all. If we
assume that A has real eigenvalues, then the eigenvalues must be of the
same sign, and hence the phase portrait is a node. If the eigenvalues are
positive, the node is unstable, and if they are negative, the node is stable.
Since the trace of A is equal to the sum of the eigenvalues, it must be of the
same sign as the roots, and thus serves as a stability indicator.
Classifying phase portraits of 2 × 2 Linear Systems:
Real eigenvalues.
Let A be a 2 × 2 matrix with real orbits of ~v0 = A~v. The phase
portrait of
~v0 = A~v
is
• degenerate if det( A) = 0,
• A saddle if det( A) < 0,
• A stable node if tr ( A) < 0 and det( A) > 0, and
• An unstable node if tr ( A) > 0 and det( A) > 0.
Complex eigenvalues
Consider a system ~v0 = A~v, where A is a 2 × 2 matrix with real entries and
complex eigenvalues. We will denote the eigenvalues, which are conjugate
to each other, by λ ± iω, and the eigenvectors will be ~h ± i~k. The general
solution is
~v = eλt [(c1 cos(ωt) + c2 sin(t))~h + (−c1 sin(ωt) + c2 cos(ωt))~k],
where c1 and c2 are constants (see equation 3.18 on page 266). If λ > 0,
the exponential factor will cause all solutions to converge to 0 as t → −∞,
and the phase portrait will be an unstable node. Similarly, if λ < 0, the
318
CHAPTER 4. STABILITY THEORY
phase portrait is a stable node. These are called spiral nodes, since the
phase portraits turn out to be spirals, expanding outward in the unstable
case and inward in the stable case.
If λ = 0, then ~v(t) is periodic (the period is 2π/ω), and hence the orbits
are closed. A phase portrait in which all orbits are closed is called a center.
Referring to lemma 4.1.2, we see that the trace of A is the sum of the
eigenvalues:
tr A = (λ + iω ) + λ − iω ) = 2λ.
Therefore, if A has complex eigenvalues tr A = 0 if and only if the eigenvalues of A are pure imaginary, and if the eigenvalues are not pure imaginary,
their real part has the same sign as tr A.
These observations can be summarized as follows:
Classifying phase portraits of 2 × 2 Linear Systems:
Complex eigenvalues.
Let A be a 2 × 2 real matrix that has no real eigenvalues.
Then the phase portrait of ~v0 = A~v is
• a center if tr A = 0;
• a stable node if tr A < 0; and
• an unstable node if tr A > 0.
Centers
Let
A=
a b
c −a
be a matrix whose trace is 0. An integral for the system
x 0 = ax + by
y0 = cx − ay
(4.9)
(4.10)
can be found by solving the differential equation
(cx − ay) dx − ( ax + by) dy = 0.
(4.11)
4.1. PHASE PORTRAITS OF LINEAR SYSTEMS
319
This equation is in exact form:
−a =
∂
∂
(cx − ay) =
[−( ax + by)].
∂y
∂x
Integrating equation (4.11) yields F ( x, y) = C, where
F ( x, y) =
1
1 2
c x − a xy − b y2 .
2
2
Recall from analytic geometry that an equation
px2 + qxy + ry2 = C
represents an ellipse if its discriminant d = 4pr − q2 is positive, and a hyperbola if d < 0. The discriminant of F ( x, y) = C is equal to −bc − a2 =
det( A). Thus, if det A > 0, the level curves of F ( x, y) are ellipses, and
if det( A) < 0, the level curves are hyperbolas. Since the orbits of (4.10)
must follow the level curves, they are ellipses when det( A) > 0. We were
aware that when det( A) < 0 the phase portrait is a saddle. Now we know
that when tr A = 0 and det A < 0, the orbits follow hyperbolas and their
asymptotes.
To determine the direction of the orbits of a center, consider a point
(0, y0 ) on the positive y-axis. At that point dx
dt = by0 . Thus the orbit will be
directed to the right at (0, y0 ) if b > 0; this indicates the clockwise direction.
Similarly, if b < 0, the direction is counterclockwise. (b = 0 is impossible
because then det A = − a2 indicates the phase portrait would be a saddle
rather than a center.)
Example 4.1.4 Draw a phase portrait of the system
x 0 = 2x + 8y
y0 = −5x − 2y.
S OLUTION . The matrix A =
2
8
−5 −2
has trace 0 and determinant 36.
Therefore,
e
At
1
= (cos 6t) I + (sin 6t) A =
6
4
cos 6t + 13 sin 6t
3 sin 6t
5
− 6 sin 6t
cos 6t − 13 sin 6t
.
It follows that the orbits rotate clockwise(since b = 8 > 0) around the origin
with angular frequency is 6 radians per unit time.
320
CHAPTER 4. STABILITY THEORY
The phase portrait is a center, and it can be drawn with the aid of the
integral found by integrating the exact equation
(−5x − 2y) dx − (2x + 8y) dy = 0.
Thus, the orbits are the ellipses
5 2
x + 2xy + 4y2 = c.
2
(4.12)
See figure 4.7.
Details related to how the ellipses are drawn are optional.
Spiral nodes
Suppose that the eigenvalues of the system
x 0 = ax + by
y0 = cx + dy
(4.13)
are complex numbers λ ± iω with λ, ω 6= 0. Let A =
a b
c d
be the coef-
ficient matrix. Then λ = 21 tr A = 12 ( a + d) and
q
q
1
1
ω=
−( a + d)2 + 4( ad − bc) =
−( a − d)2 − 4bc
2
2
Let B = A − λI. A quick calculation shows tr B = 0 and det B = ω 2 Hence
e Bt = (cos ωt) I +
and e At = eλt e Bt .
The phase portrait of
1
(sin ωt) B
ω
~ 0 = Bw
~
w
(4.14)
is a center. Orbits cycle around the origin with angular frequency ω, clockwise if b > 0 and counterclockwise if b < 0. Again, b = 0 is impossible,
for in that case the eigenvalues of A would be a and d, which are real; our
assumption is that the eigenvalues of A are not real.
If we start at some point other than the origin, the orbit of (4.14) follows
an ellipse. An orbit of the system (4.13) starting at the same point will
revolve around the origin with the same angular frequency, but will move
outward (if λ > 0) or inward (if λ < 0). In either case, the orbit will
be a spiral. Thus this phase portrait is called a spiral node. Focus is an
alternative term used instead of spiral node by some authors.
4.1. PHASE PORTRAITS OF LINEAR SYSTEMS
321
Example 4.1.5 Draw a phase portrait for the system
x0 = x − y
y0 = x + y.
S OLUTION . Let A =
1 −1
1 1
(4.15)
be the coefficient matrix of the system.
0 −1
1
; then tr B = 0
Noting that tr A = 2, we let B = A − 2 (tr A) =
1 0
and det B = 1 so
e Bt = (cos t) I + (sin t) B,
and
e
At
t Bt
=ee
=e
t
cos t − sin t
sin t cos t
The orbits are oriented counterclockwise and expand outward, with an
angular frequency of 1. Thus the orbit that starts at (1,0) will cross the
negative x-axis at (−eπ , 0), then the positive x-axis at (e2π , 0), and so on.
These crossing points form a geometric sequences on the positive and
negative x-axis, with ratio e2π ≈ 535. Unless a very large or very detailed
drawing is made, it is unlikely that more than one crossing will be seen.
The phase portrait in figure 4.9 shows four orbits of the system.
Example 4.1.6 Draw a phase portrait of the system
x 0 = x + 8y
y0 = −5x − 3y
S OLUTION . Again, let A =
1
8
−5 −3
(4.16)
be the coefficient matrix. We
2
8
1
calculate tr A = −2 and put B = A − 2 (tr A) I =
. Then
−5 −2
tr B = 0 and det B = 62 , so
1
e Bt = (cos 6t) I + (sin 6t) B
6
~ 0 = Bw
~ was the subject of example 4.1.6. Its orbits are
The system w
clockwise ellipses (see figure 4.7), revolving around the origin with angular
frequency 6.
322
CHAPTER 4. STABILITY THEORY
The orbits of system (4.16) can be obtained by following the elliptical orbits
~ 0 = Bw
~ and contracting inward at the same time at a rate e−t .
of w
The angular frequency of 6 indicates that each orbit, each orbit crosses
the positive x-axis once every 2π/6 = π/3 units of time. The exponential
factor in the solution is e−t . It follows from this that the crossing points form
a geometric sequence with ratio e−π/3 ≈ 0.351. Figure 4.10 displays one
~ 0 = Bw
~ with the same
orbit of the system (4.16), and the elliptical orbit of w
starting point.
Degenerate systems
If A is a 2 × 2 matrix that is singular, then the system ~v0 = A~v is degenerate.
All points of the phase plane are stationary if A is the zero matrix, and if
A 6= 0 we still know that 0 is an eigenvalue since there must be a nonzero
vector ~b with
A~b = ~0 = 0 · ~b;
thus ~b is an eigenvector belonging to 0. The line through the origin with
direction ~b consists of stationary points. Here are two examples of degenerate systems: in the first, there is a nonzero eigenvalue; and in the second,
0 is a double root.
Example 4.1.7 Draw a phase portrait of the system
x0 = − x
y0 = 0
−1 0
S OLUTION . Let A =
be the coefficient matrix. The eigenvalues
0 0
are −1 and 0, and ~i and ~j are corresponding eigenvectors. The stationary
line is therefore the y-axis. The general solution is
c1 e − t
−t~
~
~v = c1 e i + c2 j =
c2
or x = c1 e−t , y = c2 . The eigenvectors indicate the direction of the orbits, in
this case horizontal, pointed toward the stationary line. See figure 4.11.
Example 4.1.7 is typical of a degenerate system ~v0 = A~v, where A is
a matrix whose eigenvalues are 0 and r 6= 0. Let ~e and ~f be eigenvectors
4.1. PHASE PORTRAITS OF LINEAR SYSTEMS
323
belonging to 0 and r, respectively. The stationary line has the direction
of ~e, and the nonstationary orbits will be half lines parallel to ~f , directed
toward or away from the stationary line, depending on the sign of r. In
our example, the nonstationary orbits were perpendicular to the stationary
line, but in many cases the angle is oblique.
Example 4.1.8 Draw a phase portrait of the system
x0 = 0
y0 = − x
S OLUTION . The coefficient matrix A =
e
At
= I + At =
(4.17)
0 0
−1 0
1 0
−t 1
is nilpotent. Thus
The solutions of the system (4.17) are therefore
x
x
At
e
=
y
y − tx
0
Thus all points
on they-axis are stationary. Orbits not on the y-axis
y
follow vertical lines with speed | x |, directed downward if x > 0 and upward
if x < 0. See figure 4.12.
Notice that in this example, the nonstationary orbits are lines parallel
to the stationary line, with opposite directions on either side of the stationary line, and with speed → 0 as we approach the stationary line. This is
typical of degenerate systems of two linear equations when 0 is a double
eigenvalue. The only special property of the system in example 4.1.8 is that
the stationary line is vertical — it could have been any direction.
324
CHAPTER 4. STABILITY THEORY
Exercises
In Exercises 1 – 12 select the phase portrait on the next page that most
closely corresponds to the system ~v0 = A~v, where A is the given matrix.
−1 1
1.
A=
−1 1
Answer
−3 1
2.
A=
1 −1
4 −3
3.
A=
6 −5
Answer
2 0
4.
A=
0 1
1 3
5.
A=
2 2
Answer
2 1
6.
A=
1 2
0 −1
7.
A=
−1 0
Answer
1 0
8.
A=
0 −1
2 0
9.
A=
0 2
Answer
−4 3
10.
A=
−3 2
−1 1
11.
A=
1 −1
Answer
3 −2
12.
A=
4 −3
4.1. PHASE PORTRAITS OF LINEAR SYSTEMS
325
C
A
B
D
E
F
G
I
H
K
J
L
326
CHAPTER 4. STABILITY THEORY
Draw a phase portrait for the system x 0 = y0 = x + y.
Answer
0
x = x+y
14.
Draw a phase portrait for the system
y0 = − x − y.
13.
15.
Show that the each orbit of the system in example 4.1.3 on page 315
is a branch of a hyperbola x2 + 2xy = c, a half-line with slope − 12 , or
a vertical half-line.
Answer
0
x = x
Show that
16.
Draw a phase portrait of the system
y0 = −3y.
the orbits are horizontal or vertical half lines, and curves (not hyperbolas) satisfying the equation x3 y = constant.
In problems 17 – 26, draw the phase portrait of the given system, and
classify it as degenerate, a saddle, a center, or a node (stable or unstable).
Drawings of saddles must clearly indicate the stable and unstable lines.
x0
y0
Answer
0
x
y0
0
x
y0
Answer
0
x
y0
0
x
y0
Answer
0
x
y0
0
x
y0
Answer
17.
18.
19.
20.
21.
22.
23.
= 2x + y
= 3x
= 5x + 5y
= 3x + 7y
= −3x + 5y
= − 52 x + 2y
= x+y
= −5x − y
= x−y
= 0
= 2x + y
= −x
= 2x − 5y
= 17x − 2y
4.1. PHASE PORTRAITS OF LINEAR SYSTEMS
24.
x 0 = 3x − 10y
y0 = 8x − 5y
x0
y0
Answer
0
x
26.
y0
25.
327
= 3x + y
= −5x − 3y
= 8x + 16y
= 4x + 8y
27.
Show that if the phase portrait of example 4.1.1 is rotated 45◦
clockwise, the nonvertical orbits will lie on the graphs of functions
y = kx3 .
Answer
28.
Let A denote a 2 × 2 matrix such that tr A = 0 and det A < 0.
Show that the orbits of the system ~v0 = A~v are hyperbolas, their
asymptotes, and the origin. Hint: imitate the case where det A > 0.
29.
Prove proposition 4.1.1.
Answer
30.
Construct if possible, linear systems of two ODEs that fulfill the
specifications. (Answers are not unique).
(a) A degenerate system; stationary line: y = 2x.
(b) A nondegenerate system; the positive and negative half lines on
the x-axis are orbits directed away from the origin; all other orbits converge to the origin.
(c) A system in which all orbits except the stationary point at the
origin approach ∞ as t → ∞.
(d) A system with exactly one nonstationary orbit that converges to
the origin.
(e) A system in which there are exactly two nonstationary orbits
that converge to the origin.
(f) A system in which there are exactly four nonstationary orbits
that converge to the origin.
31.
Consider a homogeneous linear system
x 0 = ax + by
y0 = cx + dy
(4.18)
328
CHAPTER 4. STABILITY THEORY
(a) Show that if we substitute y = vx, where v is a new independent
variable, the result is the following system:
x 0 = x ( a + bv)
(4.19)
xv0 + vx 0 = x (c + dv)
(b) Multiply the first equation in the system (4.19) by v and subtract
it from the second. After dividing through by x you should obtain
v0 = −bv2 + (d − a)v + c.
(4.20)
(c) Draw a phase diagram for equation (4.20). You will have to consider three cases, depending on the number of stationary points:
0, 1, or 2.
(d) Show that the stationary points of equation (4.20) are the slopes
of the eigenvectors of the system (4.18).
(e) Give an interpretation of the direction of the orbits in the phase
diagram for equation (4.20), and explain how it is related to the
phase portrait of the system (4.18).
(f) What does the phase diagram indicate in the case where the
quadratic equation −bv2 + (d − a)v + c = 0 has no real roots?
Answer
Match the systems in problems 32– 37 with the phase portraits show
below.
A
B
C
D
E
F
4.2. NONLINEAR SYSTEMS
32.
33.
34.
35.
36.
37.
4.2
x 0 = x + 3y
y0 = x − y
x0
y0
Answer
0
x
y0
0
x
y0
Answer
0
x
y0
0
x
y0
Answer
329
= 4y
= −x
= 2x − 4y
= 2x − 2y
= −3x + 10y
= −2x + y
= x+y
= −x + y
= −4x − 4y
= 2x
Stationary Points of Nonlinear Systems
Let us consider an autonomous system of ODEs
x 0 = f ( x, y)
y0 = g( x, y),
(4.21)
The system is nonlinear unless the functions f and g have the form f ( x, y) =
ax + by and g( x, y) = cx + dy (where a, b, c, d are constants). We will always
assume that the functions f and g have continuous first partial derivatives
on an open domain D ⊂ R2 . By a two-dimensional analogue of Theorem 1.6, this will ensure that the vector field
f ( x, y)
~v( x, y) =
g( x, y)
satisfies the Lipschitz condition. Therefore the existence and uniqueness,
and continuous dependence theorem 2.1 for solutions of IVPs applies.
A stationary point ( x1 , y1 ) of ~v( x, y) is isolated if a circle can be drawn
around it that contains no other stationary points. For example, the origin
is an isolated stationary point of a linear system if and only if the system is
nondegenerate, because a degenerate system would have a stationary line
330
CHAPTER 4. STABILITY THEORY
through the origin, in the direction of eigenvectors belonging to the eigenvalue 0. Stationary points arbitrarily close to the origin can be found on the
stationary line.
A nonlinear system cannot be classified as stable or unstable as a whole;
it must be analyzed locally. The one-dimensional case was covered in section 1.10. Each stationary point could be individually classified as stable
or unstable. If the two adjacent orbits were directed toward it, the stationary point was stable; and it was unstable if at least one of these orbits was
directed away from it. In dimensions ≥ 2, we will need three categories:
asymptotically stable, neutrally stable, and unstable.
Consider a nonstationary point ( x0 , y0 ), and let ~φ(t) be the solution
of (4.21) such that ~φ(0) = ( x0 , y0 )
Definitions: If limt→∞ ~φ(t) = ( x1 , y1 ) then we will say ( x0 , y0 ) is attracted
to ( x1 , y1 ), and if limt→−∞ ~φ(t) = ( x2 , y2 ) then we will say ( x0 , y0 ) is repelled from ( x2 , y2 ).
It is an Exercise to show that if a point ( x0 , y0 ) is either attracted to or
repelled from a point ( x1 , y1 ) then ( x1 , y1 ) is a stationary point. Perhaps it
is surprising, but we will soon see an example where a point is simultaneously attracted to and repelled from the same stationary point!
Definition: A stationary point ~p = ( x1 , y1 ) ∈ D of the system (4.21) is
asymptotically stable if there is a circle C ⊂ D with center ~p such that every
point inside C is attracted to ~p.
If A is a 2 × 2 constant matrix, the phase portrait of
~v0 = A~v
was defined to be a stable node if every orbit is attracted to origin. Thus,
the origin is an asymptotically stable stationary point of a stable node.
Before turning to a detailed analysis of stability for stationary points of
nonlinear systems, let’s look at a couple of pictures.
Example 4.2.1 Figure 4.13 is the phase portrait of
p
x 0 = −4y + x (1 −
x 2 + y2 )
p
y0 = 4x + y(1 − x2 + y2 )
(4.22)
Find a stationary point of the system, and draw a conclusion about its stability. In
what sense does the stability (or lack thereof) appear to be local?
4.2. NONLINEAR SYSTEMS
331
S OLUTION . You can see that the origin is a stationary point, and the unit
circle x2 + y2 = 1 is an orbit (in fact, it is easy to verify that ( x, y) = (0, 0),
and ( x, y) = (cos(4t), sin(4t)) are solutions). Since (4.22) is nonlinear, the
formula
x
cos(4t)
=C
y
sin(4t)
only yields a solution when |C | = 1. The orbits inside the unit circle appear
to spiral away from the origin, and the orbits outside the circle spiral inward
toward it. Therefore, the phase portrait indicates that the origin is not
stable, but the circular orbit is stable. Of course, these statements are
speculative: we can’t use a computer-generated sketch as a proof, and we
haven’t defined the notion of a stable orbit. We can look at the figure as a
scientist would view an experiment, and seek an explanation in the form of
a rigorous discussion.
Example 4.2.1 provides a preview of a topic we will explore in more
depth in section 4.5: a limit cycle. The
be simpler
p system looks as if it might
y
2
2
in terms of polar coordinates, r = x + y and θ = arctan x . Differentiate
y
r2 = x2 + y2 , tan θ =
x
2
xy0 − x 0 y
to get r r 0 = x x 0 + y y0 and sec2 θθ 0 = xr 2 θ 0 = x2 . We can simplify the
latter equation to get r2 θ 0 = xy0 − x 0 y. Now substitute the expressions for x 0
and y0 in the system (4.2.1) to get an equivalent system in polar coordinates
(the details are for you to fill in):
r 0 = r (1 − r )
(4.23)
θ 0 = 4.
This system is uncoupled. It says that orbits circulate counterclockwise
around the origin with an angular4 velocity of 4. The variable r satisfies
the logistic equation, which has a stable stationary point at r = 1. Thus
orbits all converge (in a sense we will be able to make precise later) to the
periodic orbit in the unit circle.
Example 4.2.2 Locate all stationary points of the system
x0 = y
y0 = 2x − 3x2 /2,
whose phase portrait is shown in figure 4.14, and discuss their stability.
(4.24)
332
CHAPTER 4. STABILITY THEORY
S OLUTION . If ( x1 , y1 ) is astationary point, then x 0 = 0 implies y1 = 0. Thus
all stationary points are located on the x-axis. Also, y0 = 0 implies
3
2x1 − x12 = 0.
2
This quadratic has two solutions: x1 = 0 and x1 = 43 . We conclude that
there are two stationary points: (0,0) and 43 , 0 . The origin looks like a
saddle point of a linear
system, and is not stable. The orbits appear to
circulate around 43 , 0 as they do in a linear system that is a
center(eigenvalues are located on the imaginary axis). We infer that it too
is not asymptotically stable, although here we are stretching. The
computer-drawn orbits circulating around 43 , 0 appear to be closed ovals,
but possibly they spiral inward (or outward) very gradually.
Linearly Stable Equilibria
Near the stationary points in the examples that we just considered, the
phase portraits all appear to resemble phase portraits of linear systems. To
relate a stationary point ( x1 , y1 ) of (4.21) to a specific linear system, we use
the matrix of partial derivatives of f and g, all evaluated at ( x1 , y1 ). We will
be generalizing the stability test (proposition 1.10.3) for stationary points of
a one-dimensional system. The following definition may be familiar if you
have taken an advanced calculus course:
Definition: Let f x ( x1 , y1 ), f y ( x1 , y1 ), gx ( x1 , y1 ), and gy ( x1 , y1 ) denote the
partial
derivatives
of f and g. The derivative of the vector function ~v( x, y) =
f ( x, y)
at the point ( x1 , y1 ) is defined to be the matrix
g( x, y)
∂( f , g)
( x1 , y1 ) =
∂( x, y)
f x ( x1 , y1 ) f y ( x1 , y1 )
g x ( x1 , y1 ) gy ( x1 , y1 )
.
The following proposition indicates that the derivative of a vector function
is the best linear approximation of the function. For simplicity, we will
assume that ~v( x1 , y1 ) = ~0; that is, ( x1 , y1 ) is a stationary point.
4.2. NONLINEAR SYSTEMS
333
Proposition 4.2.1 Let ( x1 , y1 ) be a point such that f ( x1 , y1 ) = g( x1 , y1 ) = 0.
Suppose that the second partial derivatives of f ( x, y) and g( x, y) are continuous
on D . Then
x − x1
~v( x, y) = A( x1 , y1 )
+ O(( x − x1 )2 + (y − y1 )2 ).
(4.25)
y − y1
as ( x, y) → ( x1 , y1 ).
Equation (4.25) uses “Big O” notation, which can be explained as fol~ ( x, y) be vector functions, and let h( x, y) be a reallows: let ~v( x, y) and w
valued function. The statement
~v( x, y) = w
~ ( x, y) + O(h( x, y)) as ( x, y) → ( x1 , y1 )
means there is a circle C centered at ( x1 , y1 ) and a constant K such that
~ ( x, y)| ≤ K h( x, y)
|~v( x, y) − w
for all ( x, y in the interior of C . Thus we can interpret (4.25) to say that there
is a constant K and a circle C centered at ( x1 , y1 ) such that
~v( x, y) − A( x1 , y1 ) x − x1 ≤ K (( x − x1 )2 + (y − y1 )2 )
(4.26)
y − y1 for all ( x, y) within C .
P ROOF OF PROPOSITION 4.2.1. Put x̄ = x − x1 and ȳ = y − y1 , and
define a function
p(t) = f ( x1 + t x̄, y1 + tȳ).
Since f has continuous second partial derivatives, it is easy to see that p has
continuous second derivatives on the interval [0, 1]. By Taylor’s theorem
there is a number c1 ∈ (0, 1) such that
1
p(1) = p(0) + p0 (0)(1 − 0) + p00 (c1 )(1 − 0)2 .
2
But, p(1) = f ( x1 + x̄, y1 + ȳ) = f ( x, y), and p(0) = f ( x1 , y1 ) = 0. Furthermore, by the chain rule for partial derivatives,
p 0 (0) =
and
∂f
∂f
( x1 , y1 ) x̄ + ( x1 , y1 )ȳ
∂x
∂y
p00 (c1 ) = B11 x̄2 + 2B12 x̄ ȳ + B13 ȳ2 ,
334
CHAPTER 4. STABILITY THEORY
where
B11 =
B12 =
B13 =
∂2 f
( x1 + c1 x̄, y1 + c1 ȳ),
∂x2
∂2 f
( x1 + c1 x̄, y1 + c1 ȳ), and
∂x∂y
∂2 f
( x1 + c1 x̄, y1 + c1 ȳ).
∂y2
Thus
f ( x, y) − [ f x ( x1 , y1 ) x̄ + f y ( x1 , y1 )ȳ] = B11 x̄2 + 2B12 x̄ ȳ + B13 ȳ2 .
(4.27)
Similarly,
g( x, y) − [ gx ( x1 , y1 ) x̄ + gy ( x1 , y1 )ȳ] = B21 x̄2 + 2B22 x̄ ȳ + B23 ȳ2 ,
(4.28)
where B21 , B22 , and B23 represent the second partial derivatives of g, evaluated at a point ( x1 + c2 x̄, y1 + c2 ȳ). The Bij are not constants: all depend on
( x̄, ȳ). Since any function that is continuous on D must be bounded within
any circle C ⊂ D , there is a constant M such that each second partial derivative of either f or g has absolute value less than M inside C . Since each Bij
is equal to a second partial derivative of either f or g, the bounds | Bij | < M
hold.
By combining (4.27) and (4.28), we obtain
2
2 ~v( x, y) − A( x1 , y1 ) x̄ = B11 x̄ + 2B12 x̄ ȳ + B13 ȳ (4.29)
ȳ B21 x̄2 + 2B22 x̄ ȳ + B23 ȳ2 Using the fact that ( x̄ ± ȳ)2 ≥ 0, you can show that
2| x̄ ȳ| ≤ ( x̄2 + ȳ2 ).
You can use this inequality, the bounds | Bij | < M, and (4.29) to show that
√
~v( x, y) − A( x1 , y1 ) x̄ < 2M( x̄2 + ȳ2 ) 1 = 2 2M( x̄2 + ȳ2 ).
ȳ
1
√
Thus (4.26) holds, with K = 2 2M.
Proposition 4.2.1 provides the motivation to explore similarities of the
nonlinear system
0 x
f ( x, y)
=
(4.30)
y0
g( x, y)
4.2. NONLINEAR SYSTEMS
335
and the linear system
x̄ 0
ȳ0
= A ( x1 , y1 ) ·
x̄
ȳ
.
(4.31)
Any comparison would only be valid when ( x, y) is close to the stationary
point ( x1 , y1 ). It turns out that the two systems have much in common,
unless the matrix A( x1 , y1 ) has an eigenvalue that is either zero or pure
imaginary.
The system (4.30) is said to be linearly stable at the stationary point
( x1 , y1 ) if all eigenvalues of A( x1 , y1 ) lie to the left of the imaginary axis of
the complex plane. The following theorem, which was published in 1929 by
the German mathematician Oskar Perron, is the promised two-dimensional
version of proposition 1.10.3. (It is actually valid in n dimensions.)
Theorem 4.1 Let ( x1 , y1 ) be a linearly stable stationary point of the system (4.21)
of differential equations, where f and g are assumed to have continuous second
partial derivatives in D . Then ( x1 , y1 ) is asymptotically stable.
You will be asked to prove this theorem in exercise 24 in section 4.5,
using a technique that will be developed there.
The set of all points that are attracted to a stationary point ~p is called
the attracted set of ~p. For example, consider a linear system ~v0 = A~v, where
A is a constant matrix. The attracted set of the origin is
• the origin alone if A has no eigenvalues in the left half-plane,
• the entire plane if both eigenvalues of A are in the left half-plane, and
• the stable line if the eigenvalues of A are real and of opposite sign.
Similarly, the repelled set of ~p defined to be the set of points repelled by
~p. For a linear system ~v0 = A~v, the repelled set of the origin is
• the origin alone if A has no eigenvalues in the right half-plane,
• the entire plane if both eigenvaluesof A are in the right half-plane,
and
• the unstable line if the eigenvalues of A are real and of opposite sign.
For another example, consider an asymptotically stable stationary point
~p of a nonlinear system. By definition, there is a circle C centered at ~v1 , (C
336
CHAPTER 4. STABILITY THEORY
may be very small), such that all points within C are attracted to ~p. Thus
we can say that the attracted set of ~p contains the points inside some circle
centered at ~p.
Advanced texts go beyond theorem 4.1 and show that the phase portrait
of the system (4.30) and its “linearization” at ~p, the system (4.31), share the
following properties:
(i) If the phase portrait of the linear system (4.31) is an unstable node,
then the repelled set of ~p in the nonlinear system (4.30) contains the
the set of points inside a circle C centered at ~p .
(ii) If the phase portrait of the linear system (4.31) is a saddle, then, in the
nonlinear system (4.30), the repelled set of ~p is a curve, whose tangent direction at ~p is parallel to the eigenvector of A( x1 , y1 ) belonging
to the positive eigenvalue. The attracted set is a curve with tangent
direction at ~p parallel to the eigenvector belonging to the negative
eigenvalue. In figure 4.14, the origin is a saddle for the linearized system. The nonlinear system is more complicated (there is a center on
the positive x-axis), but the stationary point at the origin is of the saddle type. The stable and repelled sets are no longer straight lines; they
consist of the orbits directed toward and away from the origin. Notice
that in the right half-plane, the attracted and repelled sets coincide.
(iii) If the phase portrait of the linear system (4.31) is a spiral node, there
will be a spiral node near at the stationary point ~p of the nonlinear
system (4.30). This is illustrated by figures 4.13, in which the node is
unstable, and by figure 4.15, where the node is stable, although orbits
not starting inside the circular orbit will not converge to the origin.
(iv) If the phase portrait of the linear system (4.31) is a center, the orbits
near ~v1 of the system (4.30) will swirl around ~v1 , but they may not be
closed. The second order term that was dropped when passing from
the nonlinear system to the linear system is small, but not too small
to disrupt a family of closed orbits. This is illustrated in figure 4.16.
Example 4.2.3 Find the stationary points of the system
x 0 = x (2x + 3y − 7)
y0 = y(3x − 4y − 2)
and determine which are stable.
(4.32)
4.2. NONLINEAR SYSTEMS
337
S OLUTION . The stationary points are the solutions of the equations
x (2x + 3y − 7) = 0
y(3x − 4y − 2) = 0.
The graphs of these equations are the nullclines of the system. When
drawing a phase portrait, it is often useful to
include them. The nullclines
intersect at the points (0, 0), 0, − 21 , 72 , 0 , and (2, 1). These are the
stationary points.
The derivative matrix is
#
"
∂
∂
x
(
2x
+
3y
−
7
)
x
(
2x
+
3y
−
7
)
∂x
∂y
A( x, y) =
∂
∂
y
(
3x
−
4y
−
2
)
∂x
∂y y (3x − 4y − 2)
4x + 3y − 7
3x
=
.
3y
3x − 8y − 2
This matrix is only of interest when ( x, y) is one of the four stationary
points of the system (4.32).
At the stationary point (0, 0)
A(0, 0) =
−7 0
0 −2
,
which has eigenvalues −2 and −7. Since both are negative, (0, 0) is a
linearly stable stationary point, and by theorem 4.1, asymptotically stable.
The eigenvalues of
17
1
−2 0
A 0, −
=
− 32 2
2
are − 17
is a saddle point at
2 and 2. Since these have opposite signs, there
7
0, − 12 , with attracted set tangent to the eigenvector
belonging to
1
0
− 17
belonging to 2.
2 and with repelled set tangent to the eigenvector
1
The eigenvalues of A 72 , 0 are both positive, and it follows that the
stationary point 27 , 0 is unstable. Finally,
A(2, 1) =
4
6
3 −4
,
338
CHAPTER 4. STABILITY THEORY
√
6
√
with eigenvalues ± 34 and corresponding eigenvectors
.
± 34 − 4
This indicates
that there is a saddle point at (2, 1); the repelled set has
√
slope 16 ( 34 − 4) ≈ 0.3 as it crosses the stationary point, and the
√
attracted set has slope − 16 ( 34 + 4) ≈ −1.6 there. The phase portrait is
shown in figure 4.17.
Invariant Sets
A subset S of the x, y-plane is an invariant set for a system (4.21) of differential equations if every orbit of the system that contains a point of S is a
subset of S . An equivalent way of stating this definition is that S is invariant if and only if S is the union of a collection of orbits of the system.
Question: It is obvious that the union of two or more invariant sets is also an invariant set.
What about the intersection?
A typical example of an invariant set would be a level curve of an integral F ( x, y) of the system (4.21).
Example 4.2.4 Show that the x- and y-axes, and the four quadrants of the coordinate plane are invariant sets for the system in example 4.2.3.
S OLUTION . The points on the y-axis have x-coordinate equal to 0. If we
set x = 0, the system reduces to
x0 = 0
y0 = y(−4y − 2).
In particular, x 0 = 0 assures x is constant. Thus, if an orbit meets the
y-axis, then x ≡ 0 on that orbit, so that it is a subset of the y-axis.
Furthermore, on the y-axis, the phase portrait is just the phase diagram of
y0 = −y(4y + 2). Similarly, the x-axis is invariant, and the the phase
portrait reduces to the phase diagram of x 0 = x (2x − 7). No orbit can
cross either of the coordinate axis, and thus every orbit is confined to the
quadrant of its initial point. Thus, each quadrant is invariant.
When drawing phase portraits, it is important to draw the attracted and
repelled sets of each saddle point. Each of these will be a smooth curve,
called a separatrix, passing through the saddle point, and they divide the
4.2. NONLINEAR SYSTEMS
339
region around the saddle point into four quadrants. In each of the quadrants, the behavior of the orbits is distinctive. To use an ODE solver to draw
a separatrix, follow these steps:
1. Find the eigenvectors of the linearized system at the saddle point.
Draw axes through the saddle point with these characteristic directions. These axes are tangent to the separatrices.
2. Select four points ~v1 , . . . , ~v4 on the axes, on opposite sides of the saddle point. These points should be close to the saddle point.
3. Use the ODE solver to draw the orbits passing through the points ~v1 ,
. . . , ~v4 . Each solution ~vi (t) should have initial point ~vi (0) = ~vi , and
should be drawn for a ≤ t ≤ b where a < 0 and b > 0.
4. If the above steps have been done correctly, the four orbits drawn
will seem to merge to form the two separatrices. This method was
employed in drawing the separatrices for the two saddle points in
figure 4.17.
Example 4.2.5 Find the attracted set of the origin for the system in example 4.2.3.
S OLUTION . In example 4.2.4 it was shown that each quadrant of the plane
is an invariant set. We will determine the portion to the attracted set of the
origin in each quadrant, starting with the second quadrant and proceeding
counterclockwise.
It is easy to see that at every point in the second quadrant, y0 < 0. Thus all
orbits will eventually cross the x-nullcline, y = (7 − 2x )/3, and after that
crossing, x 0 > 0. It follows that all orbits in the second quadrant are
attracted toward the origin. In other words, the attracted set of the origin
contains the second quadrant.
In the third quadrant, x 0 > 0, so all orbits in that quadrant are directed to
the right.
One of these orbits is part of a separatrix of the stationary point
1
0, − 2 . It separates the other orbits in that quadrant into two classes:
those above it are attracted to the origin, and those below it are asymptotic
to the negative y-axis. The portion of the third quadrant that lies above this
separatrix is also in the attracted set of the origin,
The entirefourth quadrant is contained in the repelled set of the stationary
point 27 , 0 on the positive x-axis. It also contains part of the separatrix of
the stationary point 0, − 12 . This is an orbit that goes from 27 , 0 to
340
CHAPTER 4. STABILITY THEORY
0, − 12 , and the wedge-shaped region in the fourth quadrant that lies
above this orbit is the portion of the attracted set of the origin in the fourth
quadrant. Outside this wedge orbits are asymptotic to the negative y-axis.
Finally, in the first quadrant, the attracted set of the saddle at (2, 1) forms a
boundary of the attracted set of the origin.
Figure 4.18 shows the attracted set.
Neutrally Stable stationary points
If the phase portrait of a linear system is a center, then the origin is not
asymptotically stable, because no orbit, other than the origin itself, actually
converges to the origin. In fact, the attracted and repelled sets of the origin
consist of the origin alone. However, it is desirable to have a definition of
stability that encompasses centers. For example, it is in the context of such
a definition that we can say that the solar system, as described by Newton’s
laws, is stable1 .
Definitions: A stationary point ~p ∈ D of the system (4.21) of differential
equations is said to be neutrally stable if, given any circle C1 with center ~p
there is a (perhaps smaller) circle C2 , also centered at ~p, such that for any
solution ~v(t) of (4.21) such that ~v(0) is inside C2 , ~v(t) lies inside C1 for all
t ∈ R.
On the other hand, the stationary point ~p is unstable if there exists a
circle C centered at ~p such that there are points ~q arbitrarily close to ~p such
that the solution ~φ(t) with φ(0) = ~q eventually gets outside C ; that is, for
some t1 > 0, the point φ(t1 ) is outside C .
A stationary point that is not unstable is, naturally, said to be stable
Examples of neutrally stable stationary points that are not asymptotically stable are not hard to find: see example 4.2.6 below. Surprisingly,
there are also examples of unstable stationary points that are asymptotically
stable. (See Exercise 17 at the end of this section.)
Example 4.2.6 Show that the stationary point (0, 0) of the system
x 0 = 36y
y0 = − x,
is neutrally stable.
1 This has never been proved, but we have several billion years of experimental evidence.
4.2. NONLINEAR SYSTEMS
341
S OLUTION .
The phase portrait of this system is a center; the orbits are ellipses; in fact,
the reader can show that each orbit has equation 36x2 + y2 = C, by
following the procedure established in example 4.1.4 on page 319. The
major axis of each ellipse is horizontal, and 6 times as long as the minor
axis. Figure 4.19 displays one of the orbits.
Let R > 0 be given, and put r = 16 R. Any orbit starting inside the circle of
radius r will be an ellipse whose minor semiaxis has length less than r; its
major semiaxis is therefore less than R units long. Hence the entire orbit is
inside the circle of radius R.
Example 4.2.7 Show that (0, 0) is not a stable stationary point of the system
x0 = y
y0 = x
S OLUTION . We will show that no matter how small r is taken to be, there
are soluitons with initial points inside the circle of radius r that leave the
circle of radius 1. In fact, the phase portrait is a saddle, so all solutions
starting at a point inside the circle of radius r, except for the orbits lying on
the attracted set (the line y = − x), are unbounded as t → ∞, and
eventually leave the circle of radius 1.
Exercises
1.
Find the repelled set to the stationary point of
in example 4.2.3.
Answer
7
2, 0
of the system
In problems 2 – 13, find the stationary points, and sketch a phase portrait. In each case, is the stationary point stable? Linearly stable? asymptotically stable? Draw the separatrices of each saddle point.
0
x = x−y
2.
y0 = 3x − 2y
0
x = x−y
3.
y0 = 5x − 3y
Answer
342
CHAPTER 4. STABILITY THEORY
4.
x0
y0
Answer
0
x
y0
0
x
y0
Answer
0
x
y0
0
x
y0
Answer
0
x
y0
0
x
y0
Answer
0
x
y0
0
x
y0
Answer
5.
6.
7.
8.
9.
10.
11.
12.
13.
x 0 = x (1 − x )
y 0 = y (2 − y )
= x + 4y
= −9x − y
= x + 2y
= 2x − y
= y (1 − x 2 − y2 )
= − x (1 − x 2 − y2 )
= y (1 − x 2 )
= − x (1 − x 2 )
= x (2 − x − y )
= y( x − y)
= x (1 + x 2 + y2 )
= y (2 + x − y )
= x ( x2 + y2 − 10)
= y( xy − 3)
= x ( x + y + 6)
= y(5x + y − 10)
= x (1 − y )
= y ( x − 1)
14.
Show that the origin is the only stationary point of (4.22).
15.
Analyze the system
x 0 = −y + kx (1 − x2 − y2 )
y0 = x + ky(1 − x2 − y2 ).
(a) Show that the circle x2 + y2 = 1 is always a closed orbit.
(b) Show that the stability of the stationary point at the origin depends on k.
4.3. *COMPETING SPECIES
343
(c) Show that if we set r2 = x2 + y2 then r satisfies the differential
equation r 0 = kr (1 − r2 ).
(d) Conclude that the phase portrait is a stable spiral node if k < 0,
a center if k = 0, and an unstable spiral node if k > 0.
(e) Draw typical phase portraits for the system corresponding to
k > 0, k = 0, and k < 0.
Answer
16.
Consider the system
x 0 = −y + kx ( x2 + y2 )
where k is a
y0 = x + ky( x2 + y2 ),
constant (see figure 4.16).
(a) Show that theorem 4.1 does not determine whether or not the
origin is a stable stationary point, and that the stability of the
stationary point at the origin depends on k.
(b) Show that if we set r2 = x2 + y2 then r satisfies the differential
equation r 0 = kr3 .
(c) Conclude that the phase portrait is a stable spiral node if k < 0,
a center if k = 0, and an unstable spiral node if k > 0.
17.
Challenging problem: Asymptotic stability does not imply stability. Show that all nonconstant orbits of
x 0 = − y + x (1 − x 2 − y2 ) + p
y 0 = x + y (1 − x 2 − y2 ) − p
xy
x 2 + y2
x2
x 2 + y2
converge to the stationary point (1, 0) as t → ∞. Furthermore, show
that (1, 0) is unstable (see figure 4.20). Hint: try polar coordinates!
Answer
18.
4.3
Modify the proof of proposition 1.10.2 to show that in the phase
space of the system (4.21), if ~p = ( x1 , y1 ) is either repels or attracts a
point ( x0 , y0 ) then ~p is a stationary point.
Competing Species
When we considered the population dynamics of two-species systems in
section 2.5, we had not yet developed the concept of stability. With the
344
CHAPTER 4. STABILITY THEORY
definitions of neutrally and asymptotically stable stationary points in place,
we will reexamine the competing species equations
x 0 = x (k − ax − by)
(4.33)
y0 = y(l − cx − dy)
that were derived in section 2.5.
The four stationary points of (4.33) are the origin, where both species
are extinct, (k/a, 0), where the species represented by y is extinct, (0, l/d),
where the x-species is extinct, and a point ( x1 , y1 ) at the intersection of the
nullclines ax + by = k and cx + dy = l. The latter stationary point has no
biological significance unless it is located in the first quadrant, but then it
represents a situation where the two species share the resources. Since neither nullcline enters the third quadrant, ( x1 , y1 ) must lie in the first, second,
or fourth quadrant. Figure 2.19 displays the possible configurations of the
nullclines. As in figure 2.19, let L = l/d be the coordinate of the stationary
point on the y-axis, and C = k/a be the coordinate of the stationary point
on the x-axis. Also put p = k/b; this is the point where the x-nullcline intersects the y-axis, and let q = l/c be the point where the y-nullcline crosses
the y-axis. The points p and q are not stationary points.
The derivative matrix for the system (4.33) is
k − 2ax − by
−bx
A( x, y) =
.
−cy
l − cx − 2dy
k 0
Thus, A(0, 0) =
has two positive eigenvalues; as expected, there
0 l
is an unstable node at the origin.
At the stationary point on the y-axis,
k − bL 0
A(0, L) =
.
−cL −l
The eigenvalues of A(0, L) are −l and k − bL = b( p − L). The second of
these eigenvalues will be negative if p < L, and positive if L < p. It follows
that the stationary point at (0, L) is asymptotically stable if p < L, and is
a saddle when L < p. Similar reasoning shows that the stationary point
(K, 0) is stable if q < C and a saddle if C < q.
The fourth stationary point ( x1 , y1 ) lies in the first quadrant if and only
if either
Case I. p < L and q < C, or
4.3. *COMPETING SPECIES
345
Case II. L < p and C < q.
Furthermore, you can calculate that
A ( x1 , y1 ) =
− ax1 −bx1
−cy1 −dy1
.
Since det A( x1 , y1 ) = x1 y1 ( ad − bc), the eigenvalues are real with opposite
sign if ad < bc and ( x1 , y1 ) is a saddle. In Case I,
l
k
k
l
> and > .
d
b
a
c
Since each of the parameters a, b, c, d, k, and l is positive, we can multiply
the inequalities to obtain
kl
kl
>
(4.34)
ad
bc
from which it follows that ad − bc < 0. Hence, in Case I the stationary point
at ( x1 , y1 ) is a saddle.
In Case II the inequality in (4.34) is reversed, so det( A( x1 , y1 )) > 0.
Hence the eigenvalues are either real with the same sign, or complex conjugate. The trace of A( x1 , y1 ) is negative, and thus the roots, if real, are negative, and if they are complex, the real parts are negative. The system (4.33)
is therefore linearly stable in Case II. By the theorem of Perron, the ( x1 , y1 )
is an asymptotically stable stationary point in Case II.
What are the biological implications? We will start with the situation
in Case I, where there is a saddle at ( x1 , y1 ). If the two populations started
at their exact equilibrium values of x1 and x2 respectively, that equilibrium
would be maintained (unless disturbed by external influences). Since the
stationary point at ( x1 , y1 ) is unstable, coexistence is extremely improbable in this situation. Usually competitive exclusion occurs: the species with
the initial population advantage overwhelms the other. The separatrix of
( x1 , y1 ) serves as a boundary between the attracted sets of (0, L) and (C, 0),
and the destiny of the two competing species is determined by which side
of the separatrix the initial population vector lies. Figure 4.21 is a phase
portrait displaying the phenomenon of competitive exclusion.
When the stationary point ( x1 , y1 ) does not lie in the first quadrant, the
two species cannot coexist: one species will dominate and eventually exclude the other without regard to the initial condition, unless all members
of the dominant species are removed from the environment. To see why
this is so, refer to figure 4.22, which is a phase portrait of the system for
346
CHAPTER 4. STABILITY THEORY
case where C > q and L < p. We have noted that when the parameters are
configured in this way, (0, L) is a saddle and (C, 0) is asymptotically stable.
Furthermore, the attracted set for (0, L) is the positive y-axis, and we can
infer that the repelled set is a curve extending into the first quadrant. We
saw in section 2.5 that the quadrilateral LqCp is a trap, in the sense that
every orbit in the first quadrant enters LqCp unless it converges directly
to (C, 0), and no orbit can leave the quadrilateral. An orbit that enters the
quadrilateral is easily seen to converge to (C, 0). Since every orbit in the
first quadrant must enter LqCp, the attracted set of (C, 0) contains the entire first quadrant.
When L < p and C < q, both stationary points (C, 0) and (0, L) are
saddles. As long as both species are represented in the initial population
vector, the populations will approach the stationary point ( x1 , y1 ), which
is asymptotically stable: see figure 4.23. In this case, the attracted set of
( x1 , y1 ) is the first quadrant.
Exercises
1.
In Exercise 5 in Section 2.5, the system
x 0 = ax (K − x + By)
y0 = dy( L + Cx − y).
was presented as model for the population dynamics of two species
in symbiosis. Draw a phase portrait for the system, considering the
cases where BC > 1 and BC < 1 separately.
Answer
2.
Suppose that the differential equations model of a predator-prey
system has a limit cycle. If the prey is a pest, how would you devise
a pest control strategy?
3.
Draw a phase portrait for the following differential equation model
for a predator-prey system, in which the prey population would have
logistic growth in the absence of the predator.
x 0 = x [ A(C − x ) − by]
y0 = cy( x − d)
(4.35)
Consider the following three cases, and draw conclusions about the
fate of the predator in each.
4.4. ENERGY INTEGRALS
347
(a) C − d < 0.
(b) 0 < C − d <
(c) C − d >
Ad
4c
Ad
4c
Answer
4.
In the system (4.35) (see problem 3), let a = AC. Then, as C → ∞,
with A = aC −1 so that a is constant, the right side of the system (4.35)
approaches the right side of the Lotka-Volterra equations (2.21). However,
show
that for any finite C > d, the stationary point ( x1 , y1 ) =
d, a−bAd of the system (4.35) is asymptotically stable, even though
the corresponding stationary point of the Lotka-Volterra equations
is neutrally stable. We express this failure of thephase portrait of
the Lotka-Volterra system to resemble the phase portrait of the system (4.35), no matter how large the carrying capacity is taken to be,
by saying that the Lotka-Volterra system is not structurally stable.
4.4
Energy Integrals
Suppose that a particle with unit mass is in motion with one degree of freedom. This is the case if the particle is constrained to move on an axis. If
the force on the particle is dependent only on its position, not on its velocity (this rules out friction), then Newton’s Second Law of Motion takes the
form
d2 x
= g ( x ),
(4.36)
dt2
where g( x ) denotes the force at position x. For example, the particle might
be attached to a nonlinear spring with restoring force g( x ).
Solutions of equation (4.36) can be represented on the phase plane as
orbits of the system
x0 = y
(4.37)
y 0 = g ( x ).
Proposition 4.4.1 Orbits of the system (4.37) are directed to the right at points
in the first and second quadrants, and to the left in the third and fourth quadrants.
An orbit has a vertical tangent only at points where it crosses the x-axis.
P ROOF. In the first and second quadrants, y > 0. Since y = x 0 , it follows that x is increasing in these quadrants; in other words, every orbit is
348
CHAPTER 4. STABILITY THEORY
directed to the right as it passes through them. Similarly, in the third and
fourth quadrants, x 0 < 0, and it follows that an orbit will be directed to the
left while in these quadrants. A vertical tangent can only appear where the
orbit crosses the x-nullcline, which is the x-axis.
We can find an integral for the system (4.37) by solving the differential
equation
y dy − g( x ) dx = 0.
The result is
1 2
y − G ( x ) = C,
2
where G ( x ) denotes an antiderivative of g( x ). It follows that
F ( x, y) =
(4.38)
1 2
y − G(x)
2
is an integral of the system (4.37). We call F an energy integral.
The variable y the system (4.37) in represents the velocity of the object,
and 12 y2 is the kinetic energy. The potential energy of the particle is defined to
be U ( x ) = − G ( x ). Since G ( x ) is an antiderivative, the potential energy is
determined only up to the addition of a constant. The integral F ( x, y) represents the total mechanical energy of the particle. In deriving this integral
we have established a special case of the Law of Conservation of Energy.
To draw a phase portrait of the system (4.37), it is helpful to sketch the
graph of the potential energy function first. Since on any orbit, the sum
of the kinetic and potential energy is constant, when the potential energy
increases, the kinetic energy must decrease, and vice versa. All stationary
points of the system lie at points ( x1 , 0) on the x-axis where g( x1 ) = 0; that
is, U 0 ( x1 ) = 0, so that x1 is a critical point of the potential energy. We will
say that x1 is an isolated critical point if there is an interval ( x1 − r, x1 + s)
containing no other critical points.
Proposition 4.4.2 Suppose that the potential energy U ( x ) has an isolated critical
point at x = x1 . If U ( x1 ) is a relative minimum, then ( x1 , 0) is a neutrally stable
stationary point of the system (4.37). Furthermore, there are numbers r, s > 0
such that every orbit that crosses the x-axis in the interval ( x1 − r, x1 + s) is a
closed orbit, traversed clockwise.
P ROOF. For convenience, assume that x1 = 0. Let r > 0 and s > 0 be
chosen so that U ( x ) has no other critical points in the interval I = (−r, s).
The potential energy is then decreasing on I− = (−r, 0) and increasing on
I+ = (0, s). We will assume that r and s are chosen so that U (−r ) = U (s).
4.4. ENERGY INTEGRALS
349
Suppose that an orbit crosses the x-axis at a point x0 ∈ I− . Since the kinetic
energy at the crossing is 0, the total energy of this orbit is then U ( x0 ). As
shown in figure 4.24, the orbit cannot proceed to the left, because then both
the kinetic and potential energy would increase, and the total energy would
not be constant.
The orbit must therefore proceed to the right. Since a right directed orbit
must go above the x-axis, our orbit enters the upper half-plane. It continues
to move to the right, reaching a maximum kinetic energy as it crosses the
y-axis, where the potential energy is minimum. Then the kinetic energy
(and y) will decrease, until the x-axis is crossed again. This crossing will
occur at the point x ∈ I+ where U ( x ) = U ( x0 ). Now the orbit cannot go
farther to the right, because the potential energy cannot exceed the value
U ( x0 ). In going to the left, it enters the lower half-plane. To describe the rest
of the orbit we can appeal to symmetry. The orbit is the locus of solutions
of (4.38), which is obviously symmetric with respect to the x-axis. Therefore
the orbit in the lower half-plane is just the reflection of the portion already
traversed in the upper half-plane, and is therefore a closed curve.
We have verified the second assertion of the proposition. It still must be
proved that the stationary point at the origin is neutrally stable. This means
that for any number R > 0, it must be possible to find a number e > 0 such
that every orbit that starts inside the circle of radius e centered at the origin
remains inside the circle of radius R.
We can assume that R < r, s. (If this is not the case, replace R by a
smaller number that satisfies the assumption.) Let C R denote the circle of
radius R, centered at the origin. Put m = min{ F ( x, y) : ( x, y) ∈ C R },
the minimum value of the energy integral on C R . (This is the minimum on
the perimeter of the circle.) If m = F ( x, y), where ( x, y) ∈ C R then m =
1 2
2 y + U ( x ). Our choice of r ensures that x ∈ I \ {0}, so F ( x, y ) ≥ U ( x ) >
U (0) = f (0, 0). In short, m > F (0, 0). Appealing to the continuity of F,
there is a number e > 0 such that if ( x, y) is inside the circle Se with radius
e, then F ( x, y) < m. Any orbit which starts inside Se has total energy less
than m, and since every point of C R has total energy at least m, the orbit
cannot cross C R .
Proposition 4.4.3 Let x1 be an isolated critical point of the potential U ( x ). If the
potential does not have a relative minimum at x1 , then ( x1 , 0) is a not a stable
stationary point of the system (4.37).
P ROOF. As in the proof of proposition 4.4.2, assume that x1 = 0, and let
I = (−r, s) denote an open interval containing no other critical points of
350
CHAPTER 4. STABILITY THEORY
U ( x ). Either U 0 ( x ) > 0 on I− or U 0 ( x ) < 0 on I+ since otherwise, by the
first derivative test, U (0) would be a relative minimum. To be definite, we
will suppose that U 0 ( x ) > 0 on I− .
Let Cr denote the circle with center at the origin and radius r. We will
show that for any x0 ∈ (−r, 0), the orbit of the solution ( x, y) = (φ(t), ψ(t))
of the system (4.37) with initial condition (φ(0), ψ(0)) = ( x0 , 0) crosses the
circle Cr . As shown in figure 4.25, there will be two crossing points, since
the orbit is symmetric with respect to the x-axis; the one in the lower halfplane is directed to the left and thus outward from the circle.
Since the initial point ( x0 , 0) can be made arbitrarily close to the origin,
this will demonstrate instability.
The orbit through the point ( x0 , 0) is on the level curve of the total energy given by the equation
1
U ( x ) + y2 = U ( x0 ).
2
(4.39)
x 2 + y2 = r 2 .
(4.40)
The circle Sr has equation
We will show that there is a point ( x ∗ , y∗ ), with −r < x ∗ < 0, satisfying
equations (4.39) and (4.40) simultaneously. Multiply equation (4.39) by 2
and subtract it from equation (4.40) to eliminate the variable y and get
x2 − 2U ( x ) = r2 − 2U ( x0 ).
(4.41)
Let P( x ) denote the expression on the left side of equation (4.41). Notice
that
P( x0 ) = x02 − 2U ( x0 ) < r2 − 2U ( x0 ).
Also, since U (−r ) < U ( x0 ),
P(−r ) = r2 − 2U (−r ) > r2 − 2U ( x0 )
Appealing to the continuity of P( x ) and the intermediate value theorem,
there is a solution x ∗ ∈ (−r, x0 ) of equation (4.41);
p the orbit through the
point ( x0 , 0) crosses Sr at ( x ∗ , ±y∗ ), where y∗ = 2[U ( x0 ) − U ( x ∗ )].
If x1 is an isolated critical point of U ( x ), not a relative minimum, then
( x1 , 0) lies on a level energy curve
1 2
y + U ( x ) = U ( x1 ).
2
4.4. ENERGY INTEGRALS
351
This level curve is the separatrix of the stationary point ( x1 , 0). If U ( x ) has
a relative maximum at x1 , then ( x1 , 0), is a saddle point, as shown in figure 4.26. If x1 is an inflection point of U ( x ), the separatrix has only one
branch; this is illustrated by figure 4.25.
Example 4.4.1 Draw a phase portrait for the differential equation
x 00 = 6x (1 − x ).
S OLUTION . We will draw the phase portrait of the system
x0 = y
y0 = 6x (1 − x ).
Rx
The potential is U ( x ) = − 0 6z(1 − z) dz = 2x3 − 3x2 , which has a relative
maximum at x = 0 and a minimum at x = 1. Hence the stationary point at
(0, 0) is unstable, while the stationary point (1, 0) is stable. The separatrix
corresponding to the stationary point at the origin is the cubic curve
1 2
y + 2x3 − 3x2 = 0.
2
This curve consists of three orbits in addition to the origin itself. The
portion of the curve in the second quadrant is an orbit directed toward the
origin, and forms part of the attracted set. The intersection of the curve
with the third quadrant is an orbit directed away from the origin: it is part of
the repelled set. The situation in the right half-plane seems paradoxical,
but actually represents a relatively common occurrence: the portion of the
separatrix in the right half-plane is a single orbit, which converges to the
origin as t → ∞ and as t → −∞; it therefore lies in the intersection of the
attracted and repelled sets.
The orbits inside the separatrix in the right half-plane are closed; all other
orbits were drawn by reference to the potential function. The phase
portrait is shown in figure 4.14 on page 387.
The Pendulum
Figure 4.27 shows a pendulum. L denotes the length of the pendulum arm,
θ the angle made by the arm with the vertical axis, and m the mass.
352
CHAPTER 4. STABILITY THEORY
The force of gravity on the bob is F = mg, directed downward. We will
use a moving coordinate system in which the origin is the center of the bob,
with the y-axis is aligned along the pendulum arm, and the x-axis tangent
to the trajectory of the pendulum. In this coordinate system,
F = −mg(sin(θ )i + cos(θ )j).
The net force on the pendulum bob is F + Tj, where Tj, the support provided by the pendulum arm, is exactly cancelled by the j-component of F.
Thus, the net force is equal to −mg sin(θ )i.
By Newton’s Second Law of Motion, the net force on the pendulum
bob is equal to its mass multiplied by its acceleration. The acceleration is
the product Lθ 00 of the angular acceleration θ 00 of the pendulum arm and
its length L. Therefore, the motion of the pendulum is governed by the
differential equation mLθ 00 = −mg sin θ, or
θ 00 = −
g
sin θ,
L
(4.42)
The potential energy of the pendulum, normalized so that U (0) = 0, is
U (θ ) =
Z θ
g
0
L
sin(z) dz =
g
(1 − cos θ ).
L
As we would expect, the stationary points at θ = 2nπ, when the pendulum
is at the bottom of its trajectory, correspond to minima of the potential energy, and are stable. The stationary points at odd multiples of π (where the
pendulum is inverted) correspond to maxima of the potential energy and
are unstable. The separatrix is the curve
1 2 g
g
ω − cos θ = ,
2
L
L
in the phase plane (the variable ω = θ 0 represents angular velocity). In this
case, the separatrix separates the closed (periodic) orbits from those that
are not periodic. See figure 4.28.
Exercises
1.
Let x1 denote an isolated critical point of U ( x ). Is it possible for
the stationary point ( x1 , 0) of the system (4.37) to be asymptotically
stable?
Answer
4.4. ENERGY INTEGRALS
353
2.
Suppose that U ( x ) has a non-degenerate critical point at ( x1 , 0); that
is, U 0 ( x1 ) = 0 and U 00 ( x1 ) 6= 0. Let A denote the derivative matrix of
the system (4.37) at ( x1 , 0). Show that if U ( x ) has a relative maximum
at x1 , then the eigenvalues of A have opposite sign, while if U ( x ) has
a relative minimum at x1 , then the eigenvalues are pure imaginary.
3.
Show that if g( x1 ) = 0 and g0 ( x1 ) < 0, then ( x1 , 0) is a neutrally
stable stationary point of the system (4.37).
Answer
In problems 4 – 19, find a potential energy function and sketch its graph.
Then draw a phase portrait for the differential equation.
4.
x 00 = ax, where a is a positive constant.
5.
x 00 = ax, where a is a negative constant.
Answer
6.
x 00 = 0.
7.
x 00 = 1.
Answer
8.
9.
10.
x 00 = x2 .
x 00 = − x2 .
Answer
x 00 = 4x3 .
11.
x 00 = −4x3 .
Answer
12.
x 00 = 4x − x3 . Show that the attracted and repelled sets of the
origin are the same.
13.
x 00 = 3x | x |.
Answer
14.
x 00 = e− x (cos x + sin x ).
15.
x 00 =
x 2 −1
. To save time
( x 2 +1)2 h
i
x 2 −1
1
=
Re
.
( x 2 +1)2
( x + i )2
that
Answer
in computing the potential energy, note
354
16.
CHAPTER 4. STABILITY THEORY
x 00 = x2 sin x − 2x cos x.
17.
x 00 = sin x − 12 .
Answer
18.
x 00 = sin x − 1.
19.
x 00 = sin x −
Answer
20.
Suppose that the potential energy function U ( x ) for the differential equation (4.36) has a relative maximum at x = x1 . Show that the
when the two branches of the separatrix
cross the x-axis of the phase
p
0
plane at x = x1 , their slopes are ± g ( x1 ).
4.5
2
π
x.
The Lyapunov Stability Tests
Let θ denote the angular displacement of a pendulum arm from the vertical
position. If the pendulum is subject to linear friction, then θ satisfies the
differential equation
d2 θ
dθ
= −b − k sin θ,
(4.43)
2
dt
dt
where b is the friction constant, and k = g/L.
Let ω denote the angular velocity of the pendulum. The phase portrait
of the differential equation (4.43) is that of the system
dθ
= ω
dt
(4.44)
dω
= −bω − k sin θ
dt
Mechanical energy is not conserved when this pendulum is in motion,
because the friction dissipates kinetic energy as heat. The total energy,
F (θ, ω ) =
1 2
ω + k (1 − cos θ ),
2
is an integral for the motion of the frictionless pendulum, but it cannot be
expected to serve as an integral when there is friction.
We can validate this physical intuition by differentiating F (θ, ω ).
d
dω
dθ
F (θ, ω ) = ω
+ k sin θ
dt
dt
dt
= ω [−bω − k sin θ ] + k(sin θ )ω
= −bω 2 .
(4.45)
4.5. LYAPUNOV STABILITY TESTS
355
It follows that F (θ, ω ) decreases at a rate proportional to ω 2 and is not an
integral.
When (θ, ω ) = (2nπ, 0) the pendulum is motionless at the bottom of
its trajectory. Physical intuition suggests that if the pendulum starts with
a modest initial velocity near this equilibrium position, it will eventually
stop there. In other words, we expect the stationary point (2nπ, 0) to be
asymptotically stable. It is possible to prove this by using theorem 4.1, but
we will explore a second method, due to Lyapunov, which focuses on the
mechanical energy F (θ, ω ).
Lyapunov Functions
Let L( x, y) be a function with L(0, 0) = 0. If there is a number R > 0
with L( x, y) > 0. for all points ( x, y) except the origin inside the circle
x2 + y2 = R2 , then L( x, y) is said to be positive definite near the origin.
If the inequality is not strict; and we only have L( x, y) ≥ 0 for x2 + y2 < R2 ,
then L( x, y) is positive semidefinite near the origin.
We also define a function N ( x, y), with N (0, 0) = 0, to be negative
definite or semidefinite if − N ( x, y) is positive definite of semidefinite, respectively. A function L( x, y) with L( x1 , y1 ) = 0 is positive or negative
(semi)definite near ( x1 , y1 ) if the function F ( x, y) = L( x + x1 , y + y1 ) has
the corresponding property near the origin.
Consider an autonomous system of differential equations,
x 0 = f ( x, y)
(4.46)
y0 = g( x, y).
Let L( x, y) be a function whose partial derivatives
ous, and define the function
L0 ( x, y) =
∂L
∂x
and
∂L
∂y
are continu-
∂L
∂L
f ( x, y) +
g( x, y).
∂x
∂y
Proposition 4.5.1 Let ( x, y) = (φ(t), ψ(t)) be a solution of (4.46). Then
d
L(φ(t), ψ(t)) = L0 (φ(t), ψ(t)).
dt
P ROOF. By the Chain Rule,
dL
( x, y) =
dt
∂L dx ∂L dy
+
∂x dt
∂y dt
356
CHAPTER 4. STABILITY THEORY
∂L
∂L
f ( x, y) +
g( x, y)
∂x
∂y
= L0 ( x, y).
=
Now set x = φ(t), y = ψ(t), and the proof is complete.
Definition: A Lyapunov function for a stationary point ( x1 , y1 ) of (4.46)
is a function L( x, y) that is positive definite near ( x1 , y1 ), while L0 ( x, y) is
either positive or negative semidefinite.
Example 4.5.1 The total energy, F (θ, ω ) = 21 ω 2 + k (1 − cos θ ), is a Lyapunov
function for the pendulum system (4.44), for each of the stationary points (2nπ, 0).
S OLUTION . We will confine our attention to the stationary point at the
origin. Let R = 2π. Then for 0 < |θ | < R, we have 1 − cos θ > 0, and
hence F (θ, ω ) > 0 for 0 < θ 2 + ω 2 < R2 .
∂F
Since ∂F
∂θ = k sin( θ ), and ∂ω = ω, we have
F 0 (θ, ω ) = (k sin θ )(ω ) + (ω )(−bω − k sin θ )
= −bω 2 .
Thus F 0 is negative semidefinite. Since it vanishes on the θ-axis, F 0 is not
negative definite.
The first stability test of Lyapunov is as follows.
Proposition 4.5.2 Suppose that L( x, y) is a Lyapunov function for a stationary
point ( x1 , y1 ) of the system (4.46), where L0 ( x, y) is negative semidefinite near
( x1 , y1 ). Then ( x1 , y1 ) is stable.
For simplicity, we will assume in the proof of this proposition, and of
subsequent results in this section, that the point ( x1 , y1 ) is the origin. The
set of points in the phase plane that lie inside the circle with radius R, centered at the origin, will be denoted B R . The boundary circle is C R .
P ROOF OF PROPOSITION 4.5.2. Choose R > 0 so that for all ( x, y) ∈
B R ∪ C R with ( x, y) 6= (0, 0), L( x, y) > 0 and L0 ( x, y) ≤ 0. Let m denote
the minimum value of L( x, y) for ( x, y) ∈ C R . Since the origin does not
lie on C R , this number m is positive. Furthermore, since L is continuous
and L(0, 0) = 0, there is a number r, with 0 < r ≤ R, such that for all
4.5. LYAPUNOV STABILITY TESTS
357
points ( x, y) ∈ Br , L( x, y) < m. Let ( x, y) = (φ(t), ψ(t)) be a solution of the
system (4.46) with (φ(0), ψ(0)) ∈ Br . By proposition 4.5.1,
d
L(φ(t), ψ(t)) = L0 (φ(t), ψ(t)) ≤ 0,
dt
provided that (φ(t), ψ(t)) ∈ B R . Therefore2 L(φ(t), ψ(t)) cannot increase
unless the orbit of (φ(t), ψ(t)) crosses C R , and the circle C R cannot be reached
unless L(φ(t), ψ(t)) increases to at least m.
Proposition 4.5.2 and example 4.5.1 together imply that the lower equilibrium position of the pendulum is stable. Since we expect asymptotic stability, this is disappointing, but we will see that the method of Lyapunov
can be refined to obtain the desired result.
Example 4.5.2 Show that L( x, y) = x2 + y2 is a Lyapunov function for the origin as a stationary point of the system
x0 = y
y0 = − x
and conclude that the origin is stable.
S OLUTION . Since L( x, y) > 0 for all ( x, y) 6= 0, L is positive definite.
Further, L0 ( x, y) = (2x ) · y + (2y) · (− x ) ≡ 0. Therefore, L0 ( x, y) ≤ 0 for all
( x, y), and, by proposition 4.5.2, the origin is stable.
The result of example 4.5.2 is not surprising, since the phase portrait is
a center, and the orbits are circles — level curves of L( x, y).
Limit Sets
A point ( x̄, ȳ) is said to be a forward limit point of a solution ( x, y) =
(φ(t), ψ(t)) of the system (4.46) if there is a sequence tn → ∞ such that
lim (φ(tn ), ψ(tn )) = ( x̄, ȳ).
n→∞
The set of all forward limit points is denoted lim+ (φ, ψ).
Here are some examples to consider. Proofs are omitted.
2 Readers
of Catch-22, by Joseph Heller, will recognize this line of thought.
358
CHAPTER 4. STABILITY THEORY
• If a stationary point is asymptotically stable, it is the only forward
limit point of each orbit whose initial point is sufficiently near to it.
• A saddle point of a linear system is the forward limit point of any
solution in its attracted set. Orbits which do not lie on the attracted
set may not have any forward limit points.
• Some solutions of nonlinear systems are are only defined on an interval ( a, b) with b < ∞; these automatically have empty forward limit
sets.
• It is possible for the forward limit set of an orbit to contain more than
one point. For example, the forward limit set of a periodic orbit is the
entire orbit.
• If an orbit converges to a limit cycle, as in figure 4.13, each point of
the limit cycle is a forward limit point.
The forward limit set of a solution
( x, y) = (φ(t), ψ(t))
of the system (4.46) is the set of all its forward limit points. Our notation
for this set is
lim + (φ, ψ)
The following two propositions are presented without proof. The first
says that an orbit passing close enough to a stable stationary point must
have forward limit points. These forward limit points are not necessarily
stationary points themselves. For example, consider the case where the
equilibrium is neutrally stable: the limit points may lie in a periodic orbit.
Of course, if the stationary point is asymptotically stable, all nearby orbits
converge to it.
Proposition 4.5.3 Suppose that ( x1 , y1 ) is a stable stationary point of the system (4.46). Then there is a circle C centered at ( x1 , y1 ) such that if ( x, y) =
(φ(t), ψ(t)) is a solution of the system (4.46) starting inside C then
lim + (φ, ψ) 6= ∅.
The forward limit set of an orbit may contain more than one point, but
according to the next proposition, it must be an invariant set. In other
words, if a point x1 belongs to the forward limit set of an orbit, then the
4.5. LYAPUNOV STABILITY TESTS
359
entire orbit determined by x1 is either equal to or a subset of the forward
limit set. The best example to visualize for this proposition is the limit cycle.
Proposition 4.5.4 Assume that the functions f ( x, y) and g( x, y) on the right
side of the system (4.46) have continuous partial derivatives everywhere in the
x, y-plane. Then the forward limit set lim+ (φ, ψ) of any solution (φ(t), ψ(t)) is
an invariant set.
The Test for Asymptotic Stability
We can now formulate a criterion for asymptotic stability by combining the
concepts of limit sets and Lyapunov functions.
Proposition 4.5.5 Let ( x1 , y1 ) be a stationary point of the system (4.46). Suppose
that there is a Lyapunov function L( x, y) for ( x1 , y1 ) such that L0 ( x, y) is negative
semidefinite near ( x1 , y1 ), and that there is a radius R > 0 such that the only
nonempty invariant subset of
A R = B R ∩ {( x, y) : L0 ( x, y) = 0}
is {( x1 , y1 )} itself (B R denotes the set of points inside the circle with center ( x1 , y1 )
and radius R). Then ( x1 , y1 ) is asymptotically stable.
The proof depends on the following lemma.
Lemma 4.5.6 Let g(t) be a continuous, nonincreasing function, defined for all
t > 0. Suppose that g(t) ≥ 0 for all t > 0. Then limt→∞ g(t) exists.
The proof of lemma 4.5.6 is omitted. It is based on a fundamental property of the real number system: every set that has a lower bound has a
greatest lower bound.
P ROOF OF PROPOSITION 4.5.5. Again, assume that ( x1 , y1 ) = (0, 0). Let
R be a radius such that A R , the set of points ( x, y) ∈ B R where L0 ( x, y) = 0,
contains no invariant set except (0, 0). By proposition 4.5.2, this stationary
point is stable; hence there is an r > 0 such that if ( x, y) = (φ(t), ψ(t)) is a
solution of the system (4.46) with
(φ(0), ψ(0)) ∈ Br
(4.47)
(φ(t), ψ(t)) ∈ B R
(4.48)
then
360
CHAPTER 4. STABILITY THEORY
for all t > 0.
We will show that any solution ( x, y) = (φ(t), ψ(t)) satisfying the condition (4.47) converges to the origin. Since condition (4.48) then holds for all
t > 0, L(φ(t), ψ(t)) ≥ 0 and L0 (φ(t), ψ(t)) ≤ 0. It follows that L(φ(t), ψ(t))
is a non-increasing non-negative function. Therefore limt→∞ L(φ(t), ψ(t))
exists, by lemma 4.5.6. Denote this limit by l.
If ( x̄, ȳ) is a forward limit point of the solution (φ(t), ψ(t)) then there
is a sequence tn → ∞ such that (φ(tn ), ψ(tn )) → ( x̄, ȳ). Therefore, for any
( x̄, ȳ) ∈ lim+ (φ, ψ)
L( x̄, ȳ) = lim L(φ(tn ), ψ(tn )) = l.
n→∞
Let ( x, y) = (φ̄(t), ψ̄(t)) be the solution of the system (4.46) with initial
condition (φ̄(0), ψ̄(0)) = ( x̄, ȳ). Since, by proposition 4.5.4, lim+ (φ, ψ) is
invariant, (φ̄(t), ψ̄(t)) ∈ lim+ (φ, ψ) for all t. It follows that L(φ̄(t), ψ̄(t)) ≡ l
for all t; differentiating, we have L0 (φ̄(t), ψ̄(t)) ≡ 0. It follows that L0 ( x̄, ȳ) =
0; thus ( x̄, ȳ) ∈ A R .
We have shown that the forward limit set of any orbit starting in Br
is an invariant subset of A R . By proposition (4.5.3), this forward limit set
cannot be empty; therefore it must be {(0, 0)}. It follows that the origin is
asymptotically stable.
Example 4.5.3 Show that the stationary points (θ, ω ) = (2nπ, 0) of the system (4.44), which describes the motion of a linearly damped pendulum, are asymptotically stable.
S OLUTION . We have seen in example 4.5.1 that the total energy
F (θ, ω ) = 21 ω 2 + k (1 − cos θ ) is a Lyapunov function for each of these
stationary points, and F 0 (θ, ω ) = −bω 2 is negative semidefinite. To apply
proposition 4.5.5, it is necessary to find a number R such that the only
nonempty invariant subset of A R is the stationary point (2nπ, 0).
The function F 0 vanishes only on the θ-axis. The invariant subsets of this
axis are composed of the equilibria (nπ, 0). Hence, if R < π, the set A R
will only contain one stationary point, and thus only one nonempty
invariant subset. The equilibrium point, (2nπ, 0) is therefore asymptotically
stable. Figure 4.29 is a phase portrait for equation (4.44).
There is a Lyapunov test for instability:
4.5. LYAPUNOV STABILITY TESTS
361
Proposition 4.5.7 Suppose that L( x, y) is a Lyapunov function for the stationary point ( x1 , y1 ) of the system (4.46), such that L0 ( x, y) is positive definite near
( x1 , y1 ). Then ( x1 , y1 ) is unstable.
P ROOF. As usual, we will take ( x1 , y1 ) to be the origin. Choose R > 0
such that L( x, y) > 0 and L0 ( x, y) > 0 for all ( x, y) ∈ B R or on its boundary
circle C R (except for ( x, y) = (0, 0)). We will show that every orbit that
starts at a point inside B R other than the origin eventually crosses the circle
CR .
Let M be the maximum value of L( x, y) for ( x, y) ∈ B R ∪ C R . Suppose
that ( x, y) = (φ(t), ψ(t)) is a solution of the system (4.46) with (φ(0), ψ(0)) =
( x0 , y0 ) ∈ B R . Put m = L( x0 , y0 ); assuming that ( x0 , y0 ) 6= (0, 0), m > 0.
Since L is continuous and L(0, 0) = 0, a number r > 0 can be chosen such
that for all ( x, y) ∈ Br , L( x, y) < m2 . Since
d
L(φ(t), ψ(t)) = L0 (φ(t), ψ(t)) > 0
dt
as long as L(φ(t), ψ(t)) ∈ B R , L(φ(t), ψ(t)) is increasing. Therefore
L(φ(t), ψ(t)) ≥ m
for all t ≥ 0, unless (φ(t), ψ(t)) escapes frompB R .
Let K denote the annulus {( x, y) : r ≤ x2 + y2 ≤ R}. We will show
that the orbit of (φ(t), ψ(t)) must leave K. Since it cannot cross the inner
boundary circle, it must cross the outer one, and the proof will be complete
(see figure 4.30).
Let k = L0 ( x ∗ , y∗ ) be the minimum value of L0 ( x, y) for all ( x, y) ∈ K.
Since
L0 ( x, y) > 0
for all ( x, y) ∈ K, L( x ∗ , y∗ ) > 0, and thus k > 0. Put T =
assume that (φ(t), ψ(t)) ∈ K for all t > 0, then
L(φ( T ), ψ( T )) = m +
Z T
0
M − m +1
.
k
If we
L0 (φ(t), ψ(t)) dt
≥ m + kT = M + 1.
This is impossible, since L( x, y) ≤ M for all ( x, y) ∈ K.
Example 4.5.4 Let q( x, y) be a function with continuous partial derivatives. Show
that L( x, y) = x2 + y2 is a Lyapunov function for the origin as a stationary point
of the system
x 0 = −y + xq( x, y)
(4.49)
y0 = x + yq( x, y)
362
CHAPTER 4. STABILITY THEORY
if either
(a) q is positive definite, or
(b) q is negative definite.
Use this Lyapunov function to determine whether or not the origin is stable, asymptotically stable, or unstable.
S OLUTION . The function L( x, y) = x2 + y2 is positive definite, and
L0 ( x, y) = 2x (y + xq( x, y)) + 2y(− x + yq( x, y))
= 2( x2 + y2 )q( x, y)
is positive (negative) definite if q( x, y) is positive (negative) definite.
Therefore, L( x, y) is a Lyapunov function in either case. By Lyapunov’s
stability tests, the origin is an unstable stationary point if q is positive
definite, and it is asymptotically stable if q is negative definite.
The origin is not a linearly stable stationary point of the
system (4.49). A
−y + xq( x, y)
brief calculation shows that the derivative of F( x, y) =
x + yq( x, y)
is
xq x ( x, y) + q( x, y)
1 + xqy ( x, y)
A( x, y) =
−1 + yq x ( x, y)
yqy ( x, y) + q( x, y)
where q x and qy denote the partial derivatives of q. Since q(0, 0) = 0,
0 1
A(0, 0) =
, with eigenvalues ±i. Since the eigenvalues are not in
−1 0
the left half-plane this does not imply linear stability. Indeed, A(0, 0) does
not involve q at all, whereas we have seen that the stability of the origin
depends on q.
Example 4.5.5 Show that L( x, y) = x2 + y2 is a Lyapunov function for the system
x 0 = −3kx + y + kx3
(4.50)
y0 = − x
and use it to draw conclusions about the stability of the equilibrium at the origin.
4.5. LYAPUNOV STABILITY TESTS
363
S OLUTION . L( x, y) is obviously positive definite, and
L0 ( x, y) = 2x (−3kx + y + kx3 ) − 2xy
= 2kx2 (−3 + x2 ).
√
If k > 0, L0 ( x, y) ≤ 0 for | x | < 3, with equality only if x = 0. Thus the
origin is stable by proposition 4.5.2. To apply proposition 4.5.5, the
invariant subsets of the y-axis, where L0 ( x, y) vanishes, must be
determined. If x = 0, the first equation reduces to x 0 = y. If y 6= 0, an orbit
starting at (0, y) cannot be a subset of the y-axis, since any orbit on the
y-axis must have x 0 = 0. Therefore, the origin is the only invariant subset
of the y-axis, and we conclude that if k > 0, then the origin is
asymptotically stable.
If k < 0, the origin is not stable. Unfortunately, L0 is not positive definite
near the origin (it is only positive semidefinite), so we cannot apply
proposition 4.5.7 to reach this conclusion. Instead, we will assume stability
and reach a contradiction. By proposition 4.5.4, there is a number r > 0
such that if ( x, y) = (φ(t), ψ(t)) is a solution of the system with
(φ(0), ψ(0)) ∈ Br (0, 0), then lim+ (φ, ψ) 6= ∅. Reasoning similar to the
proof of proposition 4.5.5 shows that lim+ (φ, ψ) must be an invariant
subset of the y-axis. The only such subset is the origin. However, if
(φ(0), ψ(0)) 6= (0, 0) then L(φ(t), ψ(t)) ≥ L(φ(0), ψ(0)) > 0, and it follows
that lim+ (φ, ψ) 6= {(0, 0)}, a contradiction.
Notice that if k = 0, the system (4.50) is the linear system x 0 = y, y0 =
− x, whose orbits are circles. This system is neutrally stable. The derivative
matrix of the system (4.50) is
A( x, y) =
3k ( x2 − 1) 1
−1
0
.
The characteristic polynomial of A(0, 0) is p(s) = s2 + 3ks + 1. The eigenvalues are complex for |k | < 23 , double for k = ± 23 , and a pair of real numbers for |k | > 32 . The roots have negative real parts for k > 0 and positive
real parts for k < 0. Therefore, theorem 4.1 is applicable in this case, and
the phase portrait will be an unstable node if k < 0 and a stable node if
k > 0, of spiral type if |k | < 32 . See figure 4.31.
364
CHAPTER 4. STABILITY THEORY
Finding Lyapunov Functions
The principal difficulty in using the Lyapunov tests for stability is in finding a Lyapunov function. There is no general procedure, but systems which
arise in applications are frequently accompanied by natural Lyapunov functions. For example, the total mechanical energy of the damped pendulum
is a Lyapunov function.
If the functions f ( x, y) and g( x, y) on the right side of the system (4.46)
are polynomials in x and y, there may be a Lyapunov function of the form
L( x, y) = a( x − x1 )2 + b( x − x1 )(y − y1 ) + c(y − y1 )2 . You can show that
this function is definite near ( x1 , y1 ) if and only if b2 − 4ac < 0. Of course,
L0 will not be quadratic (unless we are dealing with a linear system). The
following proposition makes it possible to use this test for definiteness for
some non-quadratic functions. A function P( x, y) is said to vanish to order 2
at the origin if P(0, 0) = 0, and
∂P
∂P
∂2 P
∂2 P
∂2 P
(0, 0) =
(0, 0) = 2 (0, 0) =
(0, 0) = 2 (0, 0) = 0.
∂x
∂y
∂ x
∂x∂y
∂ x
Proposition 4.5.8 Let L2 ( x, y) = ax2 + bxy + cy2 and let
L( x, y) = L2 ( x, y) + P( x, y),
where P( x, y) vanishes to order 2 at the origin. Then
• If L2 is positive or negative definite, then L has the same property near the
origin.
• If L2 is positive semidefinite, L is not necessarily positive semidefinite near
the origin.
The proof of the proposition is left to you; see problem 26 at the end
of this section for hints. To see that there can be no conclusion when L2
is only semidefinite, consider L( x, y) = x3 + y3 , which is neither positive
nor negative semidefinite. If we put L2 ( x, y) = 0 (this is both positive and
negative semidefinite, then L( x, y) = L2 ( x, y) + P( x, y) where P( x, y) =
x3 + y3 vanishes to order 2 at the origin.
Example 4.5.6 Find a Lyapunov function of the form L( x, y) = ax2 + cy2 for
the system
x 0 = − x + 4y + x3 + xy2
y0 = − x − 2y + 2y3 + 2x2 y
and determine the stability of the origin.
4.5. LYAPUNOV STABILITY TESTS
365
S OLUTION . Since the function L( x, y) = ax2 + cy2 is required to be
positive definite, a and c must be positive. We will now calculate L0 .
L0 ( x, y) = 2ax (− x + 4y + x3 + xy2 ) + 2by(− x − 2y + 2y3 + 2x2 y)
= −2ax2 + (8a − 2c) xy − 4cy2 + 2ax4 + (2a + 4c) x2 y2 + 4cy4 .
By proposition 4.5.8, L0 ( x, y) will be positive or negative definite if the
expression obtained by deleting the third and higher order terms is also
positive or negative definite. We will therefore try to adjust the parameters
a and c so that the expression
L20 ( x, y) = −2ax2 + (8a − 2c) xy − 4cy2
is negative definite (positive definite is out of the question, since the
coefficients of x2 and y2 are negative). Observe that if we put a = 1 and
c = 4, the coefficient of xy will vanish, giving the expression
L20 ( x, y) = −2x2 − 16y2 , which is negative definite. Therefore L0 ( x, y) is
also negative definite, and hence L( x, y) = x2 + 4y2 is a Lyapunov
function that shows the origin to be asymptotically stable.
Exercises
1.
Show that L( x, y) = ax2 + bxy + cy2 is positive definite near the
origin if and only if b2 − 4ac < 0 and a > 0.
Answer
2.
Suppose that L( x, y) is a Lyapunov function for a stationary point
( x1 , y1 ) of a system of differential equations. Show that if L0 ≡ 0
then L is an integral, and that ( x1 , y1 ) is stable but not asymptotically
stable.
In each of problems 3 – 8, show that L( x, y) = x2 + y2 is a Lyapunov
function for the given system at the origin, and draw a conclusion about
stability.
0
x = −y + x3
3.
y 0 = x + y3 .
Answer
0
x = −5x + 2y
4.
y0 = 4x − 3y.
366
CHAPTER 4. STABILITY THEORY
x0
y0
Answer
0
x
6.
y0
0
x
7.
y0
Answer
0
x
8.
y0
5.
= y − sin x
= − x − sin y
= 2xy2 − x3 + 4y3
= 4x3 − 8x2 y − y3 .
= x3 + x + y
= −2x + y
= −2y + y2
= 2x + y − xy − y3
In problems 9 – 17, try to find a Lyapunov function of the form L( x, y) =
ax2 + cy2 and determine the stability of the origin as a stationary point. It
will occasionally be necessary to take L( x, y) = ax2 + bxy + cy2 .
x0
y0
Answer
0
x
y0
0
x
y0
Answer
0
x
y0
0
x
y0
Answer
0
x
y0
0
x
y0
Answer
0
x
y0
9.
10.
11.
12.
13.
14.
15.
16.
= −x + y
= −x − y
= 100y
= −x
= 2x − y
= 40x + 2y
= − x − 5y
= 5x + 5y.
= x−y
= 50x − y
= − x3 − y
= x − 4y3
= 2x + 5(y2 − y)
= 5( x − x2 ) − 4y
= 4y − xy2
= − x − x2 y
4.5. LYAPUNOV STABILITY TESTS
367
x 0 = 3y − x3 − xy2
y0 = −4x − x2 y − y3
Answer
17.
18.
Prove Lyapunov’s Test for Asymptotic Stability: a stationary point is
asymptotically stable if there is a Lyapunov function L for it with the
property that L0 is negative definite.
19.
What happens if, in the system 4.49 in example 4.5.4, q( x, y) is
negative semidefinite?
Answer
20.
The Lotka-Volterra equations
x 0 = x ( a − by)
y0 = cy( x − d),
are a model for describing the dynamics of a predator-prey relationship, with x representing the prey population, and y the predator
population. The system was explored in some detail in section 2.5,
where we found that there are two stationary points: the origin ~0,
and ~p = (d, ba ).
Show that ~p is stable, but not asymptotically stable.
21.
Put q( x, y) = ( x2 + y2 )(1 − x2 − y2 ) in the system (4.49) in example 4.5.4
(a) Show that the origin is unstable.
(b) Show that x = cos t, y = sin t is a solution of the system (4.49),
whose orbit is the unit circle, traversed counterclockwise.
(c) Show that the orbit found in part (b) is a limit cycle.
(d) Computer laboratory problem. Draw the phase portrait of this
system.
Answer
22.
A certain nonlinear, damped mass spring apparatus is governed
by the differential equation
x 00 + b| x 0 | x 0 + k ( x − x3 ) = 0.
Show that the total mechanical energy is a Lyapunov function for this
spring, and that the solution x = x 0 ≡ 0 is asymptotically stable.
368
CHAPTER 4. STABILITY THEORY
23.
Computer laboratory problem. Figure 4.31 indicates that the system 4.50 has a limit cycle. Investigate what happens for other values
of k. Does the limit cycle persist?
Answer
24.
The following is a proof, in outline form, of theorem 4.1 on page 335.
Suppose that ~x1 is a linearly stable stationary point of the system
~x 0 = ~f (~x );
(4.51)
that is, ~f (~x1 ) = ~0, and all eigenvalues of A(~x1 ), the derivative matrix
of ~f at ~x1 , are either negative real numbers or complex numbers with
negative real parts. We will refer to this matrix simply as A.
A symmetric matrix S will be called positive definite if the function
L(~x ) = ~x T · S · ~x is positive definite. Here, we view ~x as a column
matrix; ~x T denotes its transpose, a row matrix, and L(~x ), which is the
product of the three matrices, is a 1 × 1 matrix.
(a) Show that if there is a nonsingular matrix P such that S = P · P T
then S is positive definite.
(b) Show that L(~x ) = ~x T · S · x is a Lyapunov function for the origin
as a stationary point of
~x 0 = A~x
if S is positive definite, and
A T · S + S · A is negative definite.
(4.52)
(c) Use problem 18 to conclude that if there exists a positive definite
symmetric matrix S that satisfies condition (4.52), then ~x1 is an
asymptotically stable stationary point of the system (4.51).
To prove theorem 4.1, we will show how to find a positive definite
symmetric matrix S that satisfies condition (4.52), provided that all
eigenvalues of A have negative real parts.
T
T
(d) Let Y (t) = e A t · (e A t )T . Show that if the eigenvalues of A are
either negative, or have negative real parts, then
Z ∞
Y (t) dt
0
converges, and let S denote the value of this integral.
4.6. CHAPTER GLOSSARY
369
(e) Show that S, as defined in part (d), is positive definite.
(f) Show that A T · S + S · A = −Y (0), and complete the proof. Hint:
differentiate Y (t), and note that
−Y (0) =
Z ∞
0
[dY (t)/dt] dt.
25.
26.
4.6
−1 1
. Calculate S by the method outlined in
0 −1
problem 24 and show that S is positive definite and satisfies condition (4.52).
Answer
Take A =
Prove proposition 4.5.8. Hints: (1) The origin is a critical point of
L. (2) Use the second derivative test.
Chapter Glossary
In this glossary, definitions apply to a system of n ODEs except when stated
to the contrary. Thus we refer to phase space rather than the phase plane. To
fix notation, the system will be denoted
v 0 = F ( v ),
if it is autonomous, and v0 = F(v, t) if it is not. Here,


x1
 x2 


v =  . ,
 .. 
xn
is the vector of dependent variables, and

f 1 (v, t)
 f 2 (v, t),

F(v, t) = 
..

.





f n (v, t)
denotes a vector function defined on a subset of Rn × R. Finally, if v∗ ∈ Rn
and r > 0, let Br (v∗ ) denote the set of points in Rn at a distance less than r
from v∗ . The set Br (v∗ ) is called the ball of radius r centered at v∗ .
370
CHAPTER 4. STABILITY THEORY
Asymptotically stable (stationary point) A stationary point v∗ such that
there is a ball Br (v∗ ) and every solution v(t) with v(t1 ) ∈ Br (v∗ ) at
some time t1 has the property
lim v(t) = v∗
t→∞
Attracted A point v1 is attracted to a stationary point v∗ if any solution
with v(t1 ) = v1 for some t1 converges to v∗ . The set of points that are
attracted to v∗ is called the attracted set of v∗ . More generally, if S is
an invariant set, then v1 is attracted to S if the forward limit set of
every solution with v(t1 ) = v1 for some t1 is contained in S .
Center A system of two linear ODEs in which the characteristic roots are
imaginary numbers. The orbits are ellipses, centered at the origin.
Degenerate system A system of ODEs v0 = Av in which the matrix A is
singular.
Forward limit point See Limit point.
Improper node A system of linear ODEs in which not all orbits are halflines, but all orbits are directed toward the origin as t → ∞ (the stable
case), or all orbits are directed toward the origin as t → −∞ (the
unstable case).
Invariant set (of an autonomous system) A subset S of phase space with
the property that every orbit of the system that contains a point of S
is a subset of S .
lim+ See Limit point.
Limit cycle A closed orbit that is the forward limit set of all nearby orbits.
Limit point (of a solution v(t)). There are forward and backward limit
points. v∗ is a forward limit point if there is an increasing, unbounded
sequence of real numbers (we use the suggestive notation tm ↑ ∞)
such that
lim v(tm ) = v∗ .
(4.53)
m→∞
Similarly,
is a backward limit point if there is a sequence tm ↓ −∞
such that (4.53) holds. Limit points can also be ascribed to the orbit
represented by v(t). The set of forward limit points of v(t) is denoted
lim+ (v) and the set of backward limit points is lim− (v). These sets
v∗
4.6. CHAPTER GLOSSARY
371
can be empty, consist of one point to which the orbit converges, or
can be infinite. Some texts refer to the backward and forward limit
sets as the alpha and omega sets, respectively, of the orbit. In these
texts, the notations α(v) and ω (v) are used in place of lim− (v) and
lim+ (v), respectively
Limit set See Limit point.
Linearly stable (stationary point of an autonomous system) A stationary
point v∗ such that all of the characteristic roots of the derivative A(v∗ )
of F at v∗ are situated to the left of the imaginary axis of the complex
plane. By Perron’s theorem (page 335), linearly stable implies asymptotically stable, provided that the second partial derivatives of the
components of F are continuous.
Node A system of linear ODEs in which all orbits are directed toward the
origin as t → ∞ (the stable case), or all orbits are directed toward the
origin as t → −∞ (the unstable case).
Positive definite A function L( x1 , x2 , . . . , xn ) such that L(0, 0, . . . , 0) = 0
and L( x1 , x2 , . . . , xn ) > 0 for ( x1 , x2 , . . . , xn ) 6= L(0, 0, . . . , 0). The function L is said to be positive semidefinite. if L( x1 , x2 , . . . , xn ) ≥ 0. If
one of the above inequalities holds only for all ( x1 , x2 , . . . , xn ) ∈ Br (0)
for some r > 0, then we say that L is positive definite or semidefinite
near the origin.
Proper node A linear system of ODEs in which the orbits are all half-lines.
The half-lines will all be directed toward the origin (the stable case)
or all directed away from the origin (the unstable case).
Repelled A point v1 is repelled by stationary point v∗ if any solution with
v(t1 ) = v1 for some t1 converges to v∗ as t → −∞. The set of points
that are repelled by v∗ is called the repelled set of v∗ . More generally,
if S is an invariant set, then v1 is repelled by S if the backward limit
set of every solution with v(t1 ) = v1 for some t1 is contained in S .
Saddle A linear system of two ODEs that has two real characteristic roots
of opposite sign, or a stationary point v∗ of a nonlinear autonomous
system of two equations v0 = F(v) such that the characteristic roots
of the derivative A(v∗ ) of F at v∗ are real and of opposite sign. Near
the stationary point, the phase portrait looks like a linear saddle.
372
CHAPTER 4. STABILITY THEORY
Semisimple characteristic root A characteristic root of a matrix with multiplicity (as a root of the characteristic equation) equal to the maximum number of linearly independent characteristic vectors belonging to it.
Separatrix (applies to an autonomous system of two equations) An orbit
that converges to a saddle. Orbits on either side of a separatrix have
different limiting behavior.
Spiral node A linear system of two ODEs in which the characteristic roots
are neither real nor imaginary. The orbits swirl around the origin in
spirals.
Stable (stationary point v∗ of an autonomous system) For any number R >
0, there is a number r > 0 such that every orbit that passes within a
distance r from v∗ will be entirely within the set of points at distance
less than R from v∗ .
Stable line A straight line through the origin that is the union of the origin,
and two orbits directed toward the origin. A linear system of two
ODEs that is a saddle has a unique stable line. In the case of a stable
proper node, all lines through the origin are stable.
Stable node A system of linear ODEs in which all orbits approach the origin as t → ∞.
Stationary point (of an autonomous system) A point v∗ such that F(v∗ ) =
0. A stationary point represents a constant solution of the system.
Unstable line A straight line through the origin that is the union of the
origin, and two orbits directed away from the origin. A linear system
of two ODEs that is a saddle has a unique unstable line. In the case of
an unstable proper node, all lines through the origin are unstable.
Suggestions for Further Reading
Many advanced texts give complete proofs of the relation between the
phase portrait of a system of differential equations near an equilibrium
point and the phase portrait of the linearized system. The books Ordinary
Differential Equations, by Philip Hartman3 and Ordinary Differential Equa3 Second
edition, New York: Wiley, 1973.
4.6. CHAPTER GLOSSARY
373
tions, by Jack Hale4 are graduate level texts which contain rather complete
treatments of this topic.
Applications of differential equations to biology is the subject of a great
deal of current research. An outstanding text in this area is Mathematical
Biology, by J. D. Murray5 . Murray’s book is comprehensive, has few prerequisites, and is well written. It has the additional advantage, unusual for
a mathematics text, of being suitable for browsing.
4 Second
5 New
edition, Malabar, Florida: Krieger, 1980
York: Springer-Verlag, 1989.
374
CHAPTER 4. STABILITY THEORY
Figure 4.1: A proper node: the phase portrait of v0 = Iv.
FIGURES FOR CHAPTER 4
375
Figure 4.2: Obtaining the phase portrait of a system by distortion of a the
phase portrait of a conjugate system with a diagonal coefficient matrix.
Let A be a 2 × 2 matrix with distinct real eigenvalues r and s, and let ~e and ~f be
eigenvectors belonging to these eigenvalues. Let P be the matrix with columns
~e, ~f , and let D be the diagonal matrix diag (r, s). Then A and D are conjugate:
A = PDP−1 .
The general solution of (4.2) is
~ (t) = e Dt~c = αer t~i + βes t~j,
w
α
. Therefore the phase portrait of (4.2) is made up of orbits with
β
parametric equations
x = αert , y = βest .
where ~c =
These will be half-lines on the x- and y-axes (directed toward the origin if the eigenvalues are positive, and away from it if the eigenvalues are negative) and curves
|y| = constant · | x |s/r
We want to see how to distort this phase portrait to obtain the phase portrait
of (4.1).
The general solution of (4.1) is
~v(t) = e At~k = αer t~e + βes t ~f ,
where we have expressed the vector ~k as a linear combination of ~e = P~i and ~f = P~j:
~k = α~e + β ~f = P(α~i + β~j).
Because e At = Pe Dt P−1 , then it follows that
~v(t) = Pe Dt P−1~k
= Pe Dt P−1 [ P(α~i + β~j)]
~ (t)
= Pe Dt (α~i + β~j) = Pw
~ 7→ ~v = Pw
~ distorts the phase portrait of (4.2) to
Therefore the transformation w
obtain the phase portrait of (4.1).
376
CHAPTER 4. STABILITY THEORY
Figure 4.3: A stable improper node: phase portrait of x0 = −2x, y0 = −y.
FIGURES FOR CHAPTER 4
377
Figure 4.4: A node with two distinct, positive characteristic roots: see example 4.1.1. The
arrows in this figure show the vector field corresponding to the system of ODEs.
378
CHAPTER 4. STABILITY THEORY
Figure 4.5: An unstable improper node corresponding to a double characteristic root:
the vector field and phase portrait of x 0 = x, y0 = 2x + y. See example 4.1.2.
FIGURES FOR CHAPTER 4
Figure 4.6: A saddle: see example 4.1.3.
379
380
CHAPTER 4. STABILITY THEORY
Figure 4.7: A center: see example 4.1.4.
FIGURES FOR CHAPTER 4
Figure 4.8: Optional: Drawing the elliptical orbits in example 4.1.4
Let R(t) = [ x (t)]2 + [y(t)]2 . Then
dR
= 2x · x 0 + 2y · y0 .
dt
Substituting x 0 = 2x + 8y and y0 = −5x − 2y (from the differential
equations), we find that
dR
dt
= 2x (2x + 8y) + 2y(−5x − 2y)
= 4x2 + 6xy − 4y2
= 2(2x − y)( x + 2y)
At the extreme points of the ellipse, dR
dt = 0; this will occur when
y = 2x and when y = − 21 x. These lines contain the axes of each
orbit. You may wish to show that the longer axis is on the line y =
2x and that the ratio of the lengths of the axes is 3.
381
382
CHAPTER 4. STABILITY THEORY
Figure 4.9: A spiral node: see example 4.1.5.
FIGURES FOR CHAPTER 4
383
Figure 4.10: A spiral node. See example 4.1.6. The dashed ellipse is an orbit of the related
system that was presented in Example 4.1.4.
384
CHAPTER 4. STABILITY THEORY
Figure 4.11: A degenerate system: see example 4.1.7
-
- FIGURES FOR CHAPTER 4
385
Figure 4.12: A degenerate system: 0 is a double characteristic root. See example 4.1.8
6 6
?
?
386
CHAPTER 4. STABILITY THEORY
Figure 4.13: ). Phase portrait of the nonlinear system (4.22 The unit circle is a limit cycle.
(See example 4.2.1.)
1.5
1
0.5
-1.5
-1
-0.5
0.5
-0.5
-1
-1.5
1
1.5
FIGURES FOR CHAPTER 4
387
Figure 4.14: The origin is a saddle point of the system 4.24
3
2
1
-3
-2
-1
1
-1
-2
-3
2
3
388
CHAPTER 4. STABILITY THEORY
Figure 4.15: The
origin is a stable stationary point of the system
p
p
x 0 = − y − x (1 −
x 2 + y2 ), y 0 = x − y (1 −
x 2 + y2 ).
2
1.5
1
0.5
-2
-1.5
-1
-0.5
0.5
-0.5
-1
-1.5
-2
1
1.5
2
FIGURES FOR CHAPTER 4
389
Figure 4.16: The system x0 = −y − x( x2 + y2 ), y0 = x − y( x2 + y2 ) has an asymptotically
stable stationary point at the origin. The characteristic roots of the linearization at the origin
are ±i, so the linearization is a center. One orbit is shown.
0.5
-1
1
-0.5
390
CHAPTER 4. STABILITY THEORY
Figure 4.17: Phase portrait of the nonlinear system in example 4.2.3. The dashed lines
are nullclines.
4
2
-4
-2
2
-2
-4
4
FIGURES FOR CHAPTER 4
391
Figure 4.18: Attracted set of the origin for the system in examples 4.2.3 – 4.2.5.
4
2
-4
-2
2
-2
-4
4
392
CHAPTER 4. STABILITY THEORY
Figure 4.19: A stable stationary point. See example 4.2.6.
1
0.5
r
-1
-0.5
0.5
R
-0.5
-1
1
FIGURES FOR CHAPTER 4
393
Figure 4.20: The stationary point at (1, 0) is unstable, although all orbits except x ≡
0, y ≡ 0 converge to it. See problem 17.
1
0.5
-1.5
-1
-0.5
0.5
-0.5
-1
1
1.5
394
CHAPTER 4. STABILITY THEORY
Figure 4.21: Competitive exclusion The attracted set of the stationary point ( x1 , y1 ) is the
separatrix. If the initial population vector lies below the separatrix, the x-species dominates
and the y-species becomes extinct. If the initial population is above the separatrix, the yspecies is dominant. The parameter configuration is p < L and q < C. The nullclines are
shown as dashed lines. In drawing this phase portrait, the parameter values (k, a, b, l, c, d) =
(3, 1, 2, 5, 2, 3) were used.
2
1.5
1
0.5
0.5
1
1.5
2
2.5
3
FIGURES FOR CHAPTER 4
395
Figure 4.22: Dominance. The species with population x will outcompete its competitor
unless it is entirely removed from the environment. The parameter configuration is C > q
and L < p. The nullclines are represented by dashed lines. In drawing the phase portrait,
the parameter values (k, a, b, l, c, d) = (30, 5, 6, 6, 2, 3) were used.
8
6
4
2
2
4
6
8
396
CHAPTER 4. STABILITY THEORY
Figure 4.23: Coexistence.When L < p and C < q the species will coexist at the stationary
point in the first quadrant. The initial population vector has no effect on the outcome, as
long as both species are represented. The nullclines are represented by dashed lines. In
drawing this phase portrait, the parameter values (k, a, b, c, l, c, d) = (15, 5, 3, 12, 2, 6) were
used.
6
5
4
3
2
1
1
2
3
4
5
6
FIGURES FOR CHAPTER 4
397
Figure 4.24: A potential well: the orbit cannot escape.
1
0.5
-4
-2
2
-0.5
-1
-1.5
4
398
CHAPTER 4. STABILITY THEORY
Figure 4.25: An unstable stationary point. The critical point of U ( x) at the origin is not a
relative minimum.
FIGURES FOR CHAPTER 4
Figure 4.26: A separatrix: U ( x) has a relative maximum at x1 .
399
400
CHAPTER 4. STABILITY THEORY
Figure 4.27: A simple pendulum
r
θ
y
HH
H,q
HHH
H
j
H
x
mg
?
FIGURES FOR CHAPTER 4
401
Figure 4.28: Phase portrait of the pendulum equation. The variables θ and ω represent
angular displacement and angular velocity, respectively.
402
CHAPTER 4. STABILITY THEORY
Figure 4.29: A phase portrait of the linearly damped pendulum equation (4.44).
FIGURES FOR CHAPTER 4
403
Figure 4.30: L and L0 are both positive definite. In the annulus K, L0 ( x, y) ≥ k, and
L( x, y) ≤ M.
y
K
x
404
CHAPTER 4. STABILITY THEORY
Figure 4.31: Phase portrait of the system (4.50). The parameter k = −0.2.
FIGURES FOR CHAPTER 4
Answers to Selected Exercises
1.
(i) order 2, (ii) ODE.
Return
405
406
ORDINARY DIFFERENTIAL EQUATIONS
3.
(i) order 1, (ii) ODE.
Return
ANSWERS TO SELECTED EXERCISES
5.
(i) order 1, (ii) ODE.
Return
407
408
ORDINARY DIFFERENTIAL EQUATIONS
7.
(i) order 1, (ii) ODE.
Return
ANSWERS TO SELECTED EXERCISES
9.
(i) order 1, (ii) ODE.
Return
409
410
ORDINARY DIFFERENTIAL EQUATIONS
11.
(i) order 2, (ii) PDE.
Return
ANSWERS TO SELECTED EXERCISES
13.
(i) order 3, (ii) ODE.
Return
411
412
ORDINARY DIFFERENTIAL EQUATIONS
15.
(a) y = C sin(t)
(b) y = C sin(t)
(c) y = 2 sin(t)
(d) y = − sin(t) + cos(t)
Return
Details
ANSWERS TO SELECTED EXERCISES
17.
2
y = 3et .
Return
Details
413
414
ORDINARY DIFFERENTIAL EQUATIONS
19.
Use implicit differentiation.
Return
Details
ANSWERS TO SELECTED EXERCISES
21.
d
dt
y
t
= 1t . Use the equal derivatives theorem.
Return
Details
415
416
ORDINARY DIFFERENTIAL EQUATIONS
1.
55.47 years.
Return
Details
ANSWERS TO SELECTED EXERCISES
3.
By 27.726 per thousand.
Details
Return
417
418
ORDINARY DIFFERENTIAL EQUATIONS
5.
The relative growth rate of B is 2% per year.
Details
Return
ANSWERS TO SELECTED EXERCISES
7.
£65,659,969.14
Details
Return
419
420
ORDINARY DIFFERENTIAL EQUATIONS
9.
y=
1 12
t .
212
14
12
10
8
6
4
2
0.5
Details
Return
1
1.5
2
2.5
ANSWERS TO SELECTED EXERCISES
421
11.
y = − cos1(t) for − π2 < t <
-1.5
-1
-0.5
0.5
-2
-4
-6
-8
-10
-12
-14
Details
Return
π
2
1
1.5
422
ORDINARY DIFFERENTIAL EQUATIONS
13.
y=
1
t
for t > 0.
4
3
2
1
2
Details
Return
4
6
ANSWERS TO SELECTED EXERCISES
15.
y = 0.
Details
Return
423
424
ORDINARY DIFFERENTIAL EQUATIONS
17.
0.675 ppb.
Details
Return
ANSWERS TO SELECTED EXERCISES
19.
0.556 ppb.
Details
Return
425
426
ORDINARY DIFFERENTIAL EQUATIONS
1.
y = 34 t −
Details
Return
3
16
+
3 −4t
.
16 e
ANSWERS TO SELECTED EXERCISES
3.
y = 12 ln(t)e−t/2 − 12 ln(2)e−t/2 .
Details
Return
427
428
ORDINARY DIFFERENTIAL EQUATIONS
5.
2
y = te−t .
Details
Return
ANSWERS TO SELECTED EXERCISES
429
7.
(a) y =
9 −10t
.
10 e
1
10t
+ 11
− 10
10 e .
1
10
+
(b) y =
Details
Return
1.5
1.25
1
0.75
0.5
0.25
-1
-0.5
0.5
1
430
ORDINARY DIFFERENTIAL EQUATIONS
9.
(a) y = − cos(2t)e−4t + e−4t .
1 −4t
(b) y = − 17
e (4 sin(2t) + cos(2t)) +
1 4t
17 e .
80
60
40
20
-2
-1
1
-20
-40
Details
Return
2
ANSWERS TO SELECTED EXERCISES
11.
y = 1 − sec(t).
Details
Return
431
432
ORDINARY DIFFERENTIAL EQUATIONS
13.
Wait 11 minutes.
Details
Return
ANSWERS TO SELECTED EXERCISES
433
15.
81◦ C. The assumptions are that the transmission coefficient in the
oven is the same as that outside the oven, and that there are no other
sources of heat.
Details
Return
434
ORDINARY DIFFERENTIAL EQUATIONS
17.
The heat loss rate is 0.0514 hour−1
Details
Return
ANSWERS TO SELECTED EXERCISES
19.
35◦ C.
Details
Return
435
436
ORDINARY DIFFERENTIAL EQUATIONS
21.
There are two approaches, with slightly different answers. For a quick
approximation, just assume the heat source mH (t) = 0.8, because the
heat is on four-fifths of the time. This yields an average temperature
of 20◦ C.
A more ambitious project is to determine the stable periodic temperature, which is not constant—the heat does cycle on and off. Taking
the average value of that function over a full period, [0, 5] one obtains
an average temperature of about 19.7◦ C.
Details
Return
ANSWERS TO SELECTED EXERCISES
23.
(a)
5
29
(5 cos 2t + 2 sin 2t); stable.
7
(b) y = − 17
(cos 4t − 4 sin 4t) ; not stable.
(c) y = cos t − sin t; stable.
1
(d) y = − 26
(23 cos t + 11 sin t) ; not stable.
(e) There is no periodic solution.
Details
Return
437
438
ORDINARY DIFFERENTIAL EQUATIONS
25.
(−0.498753) e0.1 t cos(2. t) + (0.0249377) e0.1 t sin(2. t)
e(0.1) t
e(0.1) t
(b) The solution is
Z
2
2
− t2
et dt + C e−t
y=e
C
(a)
+
The CAS found
√
C
π Erfi(t)
,
2 +
2
t
e
2 et
and seems to be using a special function of the form
2
Erfi(t) = √
π
12
t12 ln(t)
t12 ln(t)2
48
−t
+ 288 −
C
(c) 12 + 3456
t
t12
1 √
C
(d) − t + t + √
2
e2 t
Details
Return
+
Z
2
et dt.
t12 ln(t)3
12
.
ANSWERS TO SELECTED EXERCISES
27.
y(0) = 1.
Details
Return
439
440
ORDINARY DIFFERENTIAL EQUATIONS
29.
y(1/4) = 8.
Details
Return
ANSWERS TO SELECTED EXERCISES
441
In Exercises 1–10, assume that the uniformity hypothesis holds.
1.
35, 69, and 138 seconds, respectively. Notice that the time doubles for
each increment — can you say why?
Details
Return
442
ORDINARY DIFFERENTIAL EQUATIONS
3.
42 ppm
Details
Return
ANSWERS TO SELECTED EXERCISES
5.
No more than 13 hours ago.
Details
Return
443
444
ORDINARY DIFFERENTIAL EQUATIONS
7.
0.55 grams per liter.
Details
Return
ANSWERS TO SELECTED EXERCISES
445
9.
4
The concentration in tank A is 0.09 + 0.03e− 15 t , and the concentration
4
in tank B is 0.09 − 0.09e− 15 t .
Details
Return
446
ORDINARY DIFFERENTIAL EQUATIONS
11.
$1 million.
Details
Return
ANSWERS TO SELECTED EXERCISES
13.
About 79,000
Details
Return
447
448
ORDINARY DIFFERENTIAL EQUATIONS
1.
y = (1 + Ct−3 )1/3 .
Details
Return
ANSWERS TO SELECTED EXERCISES
3.
√
y = ±√
Details
Return
37
2 cos t + 12 sin t + Ce6t
.
449
450
ORDINARY DIFFERENTIAL EQUATIONS
5.
q
y = 4 4(2t + 1)4 + C (2t + 1)−2 .
Details
Return
ANSWERS TO SELECTED EXERCISES
1.
(a) Separable.
(b) Separable.
(c) Not separable.
(d) Separable.
(e) Separable.
(f) Not separable.
Details
Return
451
452
ORDINARY DIFFERENTIAL EQUATIONS
3.
(a) − ln(0.8) = 0.223.
ln(0.5)
(b)
= 3.106 seconds.
ln(0.8)
0.2
(c) −
= 0.8963 meters.
ln(0.8)
1
(d) −
= 4.4814 meters.
ln(0.8)
Details
Return
ANSWERS TO SELECTED EXERCISES
5.
b = 0.98 kg/s.
Details
Return
453
454
ORDINARY DIFFERENTIAL EQUATIONS
7.
y = −2
Details
Return
1 2
2t
−t+C
−1
; y ≡ 0.
ANSWERS TO SELECTED EXERCISES
9.
√
√
y = ±C t; y = 2 t.
Details
Return
455
456
ORDINARY DIFFERENTIAL EQUATIONS
11.
t
t
y = ±Cee ;y = ee −1 .
Details
Return
ANSWERS TO SELECTED EXERCISES
13.
y = − t+1C ; y = − t−1 1 .
Details
Return
457
458
ORDINARY DIFFERENTIAL EQUATIONS
15.
y −1
y +1
2
= ±Cet ; y =
Details
Return
2
− e t +1
2 .
1+ e t
ANSWERS TO SELECTED EXERCISES
17.
About 19 meters per second.
Details
Return
459
460
ORDINARY DIFFERENTIAL EQUATIONS
19.
(a)
v2∞
g
q
ln
v20 + v2∞
v∞
.
v∞
| v0 |
arctan
.
g
v∞
(c) Let T2 be the time taken to
√fall from the maximum height to the
(b) T1 =
ground. Then T2 =
v∞
g
ln
v20 +v2∞ +|v0 |
.
v∞
If the drag force is negligible then
(a) the maximum height attained by the ball is
(b) the time taken to reach that height is
v20
2g
v0
g
(c) the time taken to fall from the maximum height to the ground is
v0
g
(d) the velocity when the ball hits the ground is |v0 |
Details
Return
ANSWERS TO SELECTED EXERCISES
21.
t = (|v∞ | ln(3))/(2g)
Details
Return
461
462
ORDINARY DIFFERENTIAL EQUATIONS
1.
F ( x, y) = x2 + 5xy + 3x − 2y2 + 2y.
Details
Return
ANSWERS TO SELECTED EXERCISES
3.
The exactness condition does not hold.
Details
Return
463
464
ORDINARY DIFFERENTIAL EQUATIONS
5.
F ( x, y) = e3x y(ln y − 1) +
Details
Return
y2
4
−
y2 ln y
2 .
ANSWERS TO SELECTED EXERCISES
7.
F ( x, y) =
Details
Return
x4
4
+
x 2 y2
2
−
x2
2
+
y4
4
+
y2
2.
465
466
ORDINARY DIFFERENTIAL EQUATIONS
9.
F ( x, y) = x −
Details
Return
y2
x +y .
ANSWERS TO SELECTED EXERCISES
11.
F ( x, y) = x3 + 3yx2 + 9y2 x + 17y3 .
Details
Return
467
468
ORDINARY DIFFERENTIAL EQUATIONS
13.
The exactness condition does not hold.
Details
Return
ANSWERS TO SELECTED EXERCISES
469
15.
x2
x2
x2
m = e 2 is an integrating factor, and F ( x, y) = y2 e 2 + xe 2 is an
integral.
Details
Return
470
ORDINARY DIFFERENTIAL EQUATIONS
17.
m = e x is an integrating factor, and F ( x, y) = 2xye x + 3y2 e x + x2 e x is
an integral.
Details
Return
ANSWERS TO SELECTED EXERCISES
471
19.
m = y2 is an integrating factor, and F ( x, y) = x2 y3 +
Details
Return
y4
2
is an integral.
472
ORDINARY DIFFERENTIAL EQUATIONS
23.
R
m = e p(x) dx is an integrating factor.
Details
Return
ANSWERS TO SELECTED EXERCISES
473
1.
100
50
0
-50
-100
0
Return
2
4
6
8
10
474
ORDINARY DIFFERENTIAL EQUATIONS
3.
1.5
1
0.5
0
-0.5
-1
-2
Return
-1
0
1
2
3
ANSWERS TO SELECTED EXERCISES
475
5.
1.5
1
0.5
0
-0.5
-1
-1.5
Return
Details
-2
-1
0
1
2
3
476
ORDINARY DIFFERENTIAL EQUATIONS
7.
1.5
1
0.5
0
-0.5
-1
-2
Return
-1
0
1
2
3
ANSWERS TO SELECTED EXERCISES
477
8.
1.5
1
0.5
0
-0.5
-1
-2
Return
-1
0
1
2
3
478
ORDINARY DIFFERENTIAL EQUATIONS
9.
At a crossing point, y0 = 0 and y00 = 1 > 0; hence there is a relative
minimum by the second derivative test.
Details
Return
ANSWERS TO SELECTED EXERCISES
11.
(a) 0, 1, 2, 3, 4, . . .; ym = m − 1.
(b) 1, 2, 4, 8, . . . ; ym = 2m−1 .
(c) C, Ck + 1, C (k + 1)2 , C (k + 1)3 , . . .; ym = C (k + 1)m−1 .
(d) 0, 1, 3, 7, . . .; ym = 2m−1 − 1.
Details
Return
479
480
ORDINARY DIFFERENTIAL EQUATIONS
13.
(c) ym = C m, where C is a constant.
(d) ym = m(m − 1).
Details
Return
ANSWERS TO SELECTED EXERCISES
481
15.
y20 = 9.55. In this case, the final local error (LE20 ) is infinite, since the
analytic solution, y = 1/(1 − t), has a singularity at t = 1.
10
8
6
4
2
0.2
Details
Return
0.4
0.6
0.8
1
482
ORDINARY DIFFERENTIAL EQUATIONS
17.
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.2
Details
Return
0.4
0.6
0.8
1
ANSWERS TO SELECTED EXERCISES
19.
Local errors: 0.101691, 0.0771727, 0.013463, -0.0506145
Accumulated errors: 0.130573, 0.266751, 0.359802.
Details
Return
483
484
ORDINARY DIFFERENTIAL EQUATIONS
21.
(a) No.
(b) 0 < h < 0.02
Details
Return
ANSWERS TO SELECTED EXERCISES
485
23.
p
, ym even(a) In the forward version,using ym+1 = ym + 1 − y2mp
tually increases to a value bigger than 1, and then 1 − y2m is
undefined.
(b) The plus sign should be used.
(c) Forward version: -0.000166583, 0.199499, 0.297489, 0.392961, 0.484917,
0.572373, 0.654372, 0.729989, 0.798335, 0.858556, 0.909828, 0.951327,
0.982145, 1.00096 (terminates)
Backward version: 0.0995037, 0.00113599, 0.29314, 0.385415, 0.473494,
0.55657, 0.633914, 0.70485, 0.768799, 0.825273, 0.873886, 0.914373,
0.946611, 0.970657, 0.986832, 0.99589, 0.999388, 0.999982, 1, 1, . . .
(d) The backward version is the more accurate.
Details
Return
486
ORDINARY DIFFERENTIAL EQUATIONS
25.
y = ±3 (singular solutions); y = 3 sin(t + C ), where C is a constant,
is a solution for all values of t such that y0 = 3 cos(t + C ) is positive.
3
2
1
0
-1
-2
-3
-4
Details
Return
-2
0
2
4
ANSWERS TO SELECTED EXERCISES
487
27.
|1 − y| = C |2 − t|, where C is a constant.
3.5
3
2.5
2
1.5
1
0.5
1
Details
Return
2
3
4
5
6
488
ORDINARY DIFFERENTIAL EQUATIONS
1.
y = −2, defined on (−∞, +∞).
Details
Return
ANSWERS TO SELECTED EXERCISES
3.
1
(5 cos(5t) − 2 sin(5t) − 5e−2t ), defined on (−∞, +∞).
29
Details
Return
y=−
489
490
ORDINARY DIFFERENTIAL EQUATIONS
5.
y = t3 − t2 , defined on (−∞, +∞).
Details
Return
ANSWERS TO SELECTED EXERCISES
7.
2
1
y = − + et defined on (−∞, +∞).
2
Details
Return
491
492
ORDINARY DIFFERENTIAL EQUATIONS
9.
If the graphs intersect, then there would be two solutions to the IVP
with initial condition at the intersection point.
Details
Return
ANSWERS TO SELECTED EXERCISES
11.
y ≡ 1.
Details
Return
493
494
ORDINARY DIFFERENTIAL EQUATIONS
13.
(i) t0 , y0 could be any real numbers.
(ii) t0 , y0 could be any real numbers.
Details
Return
ANSWERS TO SELECTED EXERCISES
15.
(i) t0 6= kπ, where k is an integer.
(ii) t0 6= kπ, where k is an integer.
Details
Return
495
496
ORDINARY DIFFERENTIAL EQUATIONS
17.
(i) y20 > 4t0 , and t0 6= 0.
(ii) ∅.
Details
Return
ANSWERS TO SELECTED EXERCISES
19.
(i) t0 y0 > 0.
(ii) t0 y0 > 0 and y0 6= 1.
Details
Return
497
498
ORDINARY DIFFERENTIAL EQUATIONS
21.
(i) t0 , y0 could be any real numbers.
(ii) t0 − y0 6= 0.
Details
Return
ANSWERS TO SELECTED EXERCISES
499
23.
∂f
∂y ( t, y )
= −√ y
continuous if |y| < 1, but not continuous on |y| =
p
1 This implies that f (t, y) = 1 − y2 satisfies a Lipschitz condition
on the domain D if −1 < c and d < 1, but there is no implication if
either c = −1 or d = 1.
1− y2
In the case d = 1 let y2 = 1 and y1 < 1. Then
q
s
1 − y21
| f (t, y2 ) − f (t, y1 )|
1 + y1
=
=
.
y2 − y1
1 − y1
1 − y1
It follows
sfrom this calculation that if there is a Lipschitz constant
1 + y1
K, then
≤ K for all y1 ∈ (−1, 1), which is not possible:
1 − y1
s
1 + y1
lim
= ∞. Therefore f does not satisfy a Lipschitz condi1 − y1
y 1 → 1−
tion when d = 1. A similar argument works when c = −1.
Return
500
ORDINARY DIFFERENTIAL EQUATIONS
25.
y0 = −2.001386626.
Details
Return
ANSWERS TO SELECTED EXERCISES
501
27.
(a) If φ and ψ are fixed by T . Then
kφ − ψk ≤ C kφ − ψk
and hence kφ − ψk = 0.
(b) By the triangle inequality,
∞
n + k −1
kψn+k − ψn k ≤
∑
k ψ j +1 − ψ j k <
j=n
∑ k ψ j +1 − ψ j k
j=n
By the contractive property of T ,
kψj+1 − ψj k ≤ C j kψ1 − ψ0 k.
Using the formula for the sum of a geometric series, the formula
follows.
(c) Because T is contractive,
kT (ψ∞ ) − ψn k = kT (ψ∞ ) − T (ψn−1 )k ≤ C kψ∞ − ψn−1 k
. Thus
lim kT (ψ∞ ) − ψn k = 0
n→∞
It follows that the sequence {ψn } converges to both ψ∞ and to
T (ψ∞ ) and hence the two are equal.
i. φ0 = 1, φ1 (t) = 1 + t, φ2 (t) = 1 + t + 21 t2 , and φ3 (t)1 + t +
n
1 2
1 3
1 m
t
2 t + 6 t . Thus φn ( t ) = ∑m=0 m! t and φ∞ ( t ) = e .
ii. φ0 = 1, φ1 (t) = 1 − 12 t2 , φ2 (t) = 1 − 12 t2 + 81 t4 , and φ3 (t) =
1 6
1 − 12 t2 + 81 t4 − 48
t Thus φn (t) = ∑nm=0 2m1m! t2m and φ∞ (t) =
2
et /2 .
iii. We have φn ≡ 0 for 0 ≤ n ≤ ∞.
iv. φ0 = 1, φ1 (t) = 1 + 2 ln t, φ2 (t) = 1 + 2 ln t + 2(ln t)2 , and
1
φ3 (t) = 1 + 2 ln t + 2(ln t)2 + 43 (ln t)3 . Thus φn (t) = ∑nm=0 m!
(2 ln t)m
2
and φ∞ (t) = t .
Return
Details
502
ORDINARY DIFFERENTIAL EQUATIONS
1.
840
Details
Return
ANSWERS TO SELECTED EXERCISES
3.
y = 200/(19e−4t + 1).
Details
Return
503
504
ORDINARY DIFFERENTIAL EQUATIONS
5.
Let p(t) denote the population, in millions, 60t years after 1890. Then
p(t) =
Details
Return
772.6
million.
1 + (11.264) (0.3246t )
ANSWERS TO SELECTED EXERCISES
7.
In 2070.
Details
Return
505
506
ORDINARY DIFFERENTIAL EQUATIONS
9.
About 10 12 hours, assuming they chatter 24 hours per day.
Details
Return
ANSWERS TO SELECTED EXERCISES
507
11.
The function f (y) = ky ln M
is defined for y > 0 only, but limy→0 f (y) =
y
0. Thus we can say 0 is stationary. The other stationary point at M is
verified because ln(1) = 0. If 0 < y < M then ln( M/y) > 0 indicating the population is increasing; for y > M we have ln( M/y) < 0
and the population is decreasing. Thus M is a stable stationary point
and 0 is unstable.
Return
508
ORDINARY DIFFERENTIAL EQUATIONS
14.
dy
dt
= ku(t)y(t). In a closed system, the number of total molecules is a
dy
constant C. Therefore u(t) + y(t) = C, and dt = ky(t)(C − y(t)). It’s
a logistic equation.
Details
Return
ANSWERS TO SELECTED EXERCISES
16.
(a) and (d)
Details
Return
509
510
ORDINARY DIFFERENTIAL EQUATIONS
18.
Stationary point: 0.
- s
0
Details
Return
-
ANSWERS TO SELECTED EXERCISES
511
20.
No stationary points.
-
Details
Return
512
ORDINARY DIFFERENTIAL EQUATIONS
22.
(a) This is the solution of first order autonomous ODE with no stationary points.
(b) This is not a solution of a first order autonomous ODE.
(c) This is the solution of first order autonomous ODE with a stationary point y = 0.
(d) This is the solution of first order autonomous ODE with no stationary points.
(e) This is not a solution of a first order autonomous ODE.
Details
Return
ANSWERS TO SELECTED EXERCISES
513
24.
If g0 (y1 ) < 0, there exists ε > 0 such that g0 (y) < 0 in (y1 − ε, y1 + ε),
because g0 (y) is continuous. That means g(y) is strictly decreasing in
(y1 − ε, y1 + ε). Because g(y1 ) = 0, it follows that g(y) changes sign
from positive to negative at y = y1 Therefore (y1 − ε, y1 ) is an up
interval, and (y1 , y + ε) is a down interval; the phase diagram for the
ODE looks like this near y1 :
- s .
Thus y1 is stable.
If g0 (y1 ) > 0, the same reasoning shows that near y1 , the phase diagram looks like this:
s - . Thus y1 is an unstable stationary point.
Details
Return
514
ORDINARY DIFFERENTIAL EQUATIONS
26.
Suppose the ODE is y0 = f (y). Then f (y) has a same stationary point
as y0 = y2 at y ≡ 0, thus f (0) = 0. The two up intervals (−∞, 0),
(0, +∞) for y0 = y2 is also the up intervals for f (y). therefore f 0 (0) =
f (0+h)− f (0)
f (0)
limh→0
= limh→0 f (hh) ≥ 0, and f 0 (0) = limh→0 f (0−h−)−
=
h
h
f (−h)
limh→0 −h ≤ 0, We obtain f 0 (0) = 0 and y ≡ 0 is a degenerate stationary point.
Return
ANSWERS TO SELECTED EXERCISES
515
28.
(a) H = 0.25
(b) P2 = 0.816
P
1
0.25
(c)
Details
Return
H
516
ORDINARY DIFFERENTIAL EQUATIONS
30.
250
200
150
100
50
1850
Details
Return
1900
1950
2000
ANSWERS TO SELECTED EXERCISES
Find the general solution of each of the ODEs in problems 1 – 12.
1.
y = −1 + Cet
Details
Return
2 /2+ t
.
517
518
ORDINARY DIFFERENTIAL EQUATIONS
2.
y = (2 + 2t + t2 ) + Cet .
Details
Return
ANSWERS TO SELECTED EXERCISES
3.
y = (t + C )et .
Details
Return
519
520
ORDINARY DIFFERENTIAL EQUATIONS
4.
y = (sin t + C ) cos t.
Details
Return
ANSWERS TO SELECTED EXERCISES
5.
y = 12 et + Ce−t .
Details
Return
521
522
ORDINARY DIFFERENTIAL EQUATIONS
6.
y = − 5t1 et (2 cos(2t) − sin(2t)) + C/t.
Details
Return
ANSWERS TO SELECTED EXERCISES
7.
5 2
y = 14
t +
Details
Return
3
13 t
− 61 + Ct−12 .
523
524
ORDINARY DIFFERENTIAL EQUATIONS
8.
y = (ln | sec(t) + tan(t)| + C ) cos t.
Details
Return
ANSWERS TO SELECTED EXERCISES
9.
R
2
2
y = ( e−t /2 dt + C )et /2 .
Details
Return
525
526
ORDINARY DIFFERENTIAL EQUATIONS
10.
y = Ct5
Details
Return
ANSWERS TO SELECTED EXERCISES
11.
y = ( t + C ) e−t .
Details
Return
527
528
ORDINARY DIFFERENTIAL EQUATIONS
12.
y = Cesin t
Details
Return
ANSWERS TO SELECTED EXERCISES
13.
y = t ln(t) + 2t.
Details
Return
529
530
ORDINARY DIFFERENTIAL EQUATIONS
14.
y = te−t .
Details
Return
ANSWERS TO SELECTED EXERCISES
15.
y = t2 .
Return
Details
531
532
ORDINARY DIFFERENTIAL EQUATIONS
16.
y = − cos(3t)e−4t + e−4t .
Return
Details
ANSWERS TO SELECTED EXERCISES
17.
y = 5t3 ln t + t3 .
Return
Details
533
534
ORDINARY DIFFERENTIAL EQUATIONS
18.
y = (2/3t + 1/3)3/2 .
Return
Details
ANSWERS TO SELECTED EXERCISES
19.
y = −2 + 2et
Return
Details
2 /2+3t
.
535
536
ORDINARY DIFFERENTIAL EQUATIONS
20.
y = −2.
Return
Details
ANSWERS TO SELECTED EXERCISES
21.
y = 2 √2+t
(2− t )
Return
Details
+
q
t
2 22+
−t .
537
538
ORDINARY DIFFERENTIAL EQUATIONS
22.
t2
y = t2 + e − 2 .
Return
Details
ANSWERS TO SELECTED EXERCISES
23.
2
30 ln
ln 3 years.
Return
Details
539
540
ORDINARY DIFFERENTIAL EQUATIONS
24.
53 years.
Return
Details
ANSWERS TO SELECTED EXERCISES
25.
1844 years.
Return
Details
541
542
ORDINARY DIFFERENTIAL EQUATIONS
26.
1
y = 17
(cos(4t) + 4 sin(4t)) + Ce−t ; the solution with C = 0 is the
stable periodic solution.
Return
Details
ANSWERS TO SELECTED EXERCISES
27.
62.5 grams per liter.
Return
Details
543
544
ORDINARY DIFFERENTIAL EQUATIONS
28.
1.5
1
0.5
0
-0.5
-1
-1.5
-2
Return
-1
0
1
2
3
ANSWERS TO SELECTED EXERCISES
545
29.
1.5
1
0.5
0
-0.5
-1
-1.5
-2
Details
Return
-1
0
1
2
3
546
ORDINARY DIFFERENTIAL EQUATIONS
30.
5
4
3
2
1
-3
Return
-2
-1
0
1
2
3
4
ANSWERS TO SELECTED EXERCISES
547
31.
5
4
3
2
1
0
-3
Details
Return
-2
-1
0
1
2
3
4
548
ORDINARY DIFFERENTIAL EQUATIONS
32.
(a) y0 = 0.03(5t − y2 ) : Graph IV.
(b) y0 = sin(y) : Graph III.
(c) y0 = sin(t + y) : Graph II.
(d) y0 = −0.01y : Graph VI.
(e) y0 = 0.05y(π − y) : Graph V.
(f) y0 = 0.02(t2 + y2 ) : Graph I.
Details
Return
ANSWERS TO SELECTED EXERCISES
33.
y = C−2 t2 , with a singular solution, y ≡ 0.
Details
Return
549
550
ORDINARY DIFFERENTIAL EQUATIONS
34.
y = − ln(−t − C ).
Details
Return
ANSWERS TO SELECTED EXERCISES
35.
t
y = 1+CeCet , with a singular solution, y ≡ 1.
Details
Return
551
552
ORDINARY DIFFERENTIAL EQUATIONS
36.
y = sin(arcsin t + C ), with two singular solutions: y = ±1.
Details
Return
ANSWERS TO SELECTED EXERCISES
37.
y = cosh(cosh−1 (t) + C ), with two singular solutions: y = ±1.
Details
Return
553
554
ORDINARY DIFFERENTIAL EQUATIONS
38.
y = tan(ln
Details
Return
√
1 + t2 + C ).
ANSWERS TO SELECTED EXERCISES
39.
√
y = ( t + C )2 , with a singular solution Y ≡ 0.
Details
Return
555
556
ORDINARY DIFFERENTIAL EQUATIONS
40.
y = − 41 ln(ln(C (cos(3t))4/3 )).
Details
Return
ANSWERS TO SELECTED EXERCISES
41.
(a) 166.4 meters.
(b) 72 seconds.
(c) 236 meters; 68.5 seconds.
Details
Return
557
558
ORDINARY DIFFERENTIAL EQUATIONS
42.
(a) For year A, 4.5%. For year B, 4%
(b) 1,000,000
(c) 1996.
Details
Return
ANSWERS TO SELECTED EXERCISES
559
43.
- r
r
− 12
0
y = − 12 is stable. y(t) → − 21 .
Details
Return
-
560
ORDINARY DIFFERENTIAL EQUATIONS
44.
- r - r r r r r r r r r r r
−π 0
π 2π 3π 4π 5π 6π 7π 8π 9π 10π
y = 0 is stable. If y(0) = 30 then y(t) → 9π.
Details
Return
ANSWERS TO SELECTED EXERCISES
561
45.
-
No stationary points; y(t) → +∞.
Return
562
ORDINARY DIFFERENTIAL EQUATIONS
46.
- r
5
3
y ≡ 53 is stable, and y(t) → 53 .
Details
Return
ANSWERS TO SELECTED EXERCISES
563
47.
- r
0
y ≡ 1 is stable and y(t) → 1.
Details
Return
1
564
ORDINARY DIFFERENTIAL EQUATIONS
48.
- r
0
1
There are no stationary points in (−∞, 0); y(t) → −∞.
Details
Return
ANSWERS TO SELECTED EXERCISES
49.
y0
If y0 = 1y then y00 = − y2 . Thus the signs of y0 and y00 are opposite.
Details
Return
565
566
ORDINARY DIFFERENTIAL EQUATIONS
50.
Notice that y00 = 1 + y0 = 1 + t + y, and use the second derivative to
determine concavity.
Details
Return
ANSWERS TO SELECTED EXERCISES
567
51.
For y0 = ey , we have y00 = e2y . Both are positive. Solutions of y0 = e−y
are increasing and concave down.
Details
Return
568
ORDINARY DIFFERENTIAL EQUATIONS
52.
(a)
h
0.1
0.05
0.025
Y (h)
0.423
0.440
0.448
(b) Z (h1 ) = 2Yh2 − Yh1 .
(c) Z0.1 = 0.456638, and Z0.05 = 0.455814.
Details
Return
ANSWERS TO SELECTED EXERCISES
569
53.
There will be a solution unless y0 = 1, and a unique solution unless
y0 = ±1.
Details
Return
570
ORDINARY DIFFERENTIAL EQUATIONS
54.
The right side of the differential equation satisfies a Lipschitz condition; with respect to y with Lipschitz constant equal to 2.
Details
Return
ANSWERS TO SELECTED EXERCISES
55.
x = A e−t , y = − A e−t
Return
Details
571
572
ORDINARY DIFFERENTIAL EQUATIONS
57.
x = 2A sin(3t), y = A sin(3t) − 3A cos(3t).
Return
Details
ANSWERS TO SELECTED EXERCISES
59.
x = e2t (t + 1 + C ), y = e2t (t − 1 + C ).
Return
Details
573
574
ORDINARY DIFFERENTIAL EQUATIONS
61.
y0 = v
v0 = −3v − 4y + t2
Return
Details
ANSWERS TO SELECTED EXERCISES
63.
u0 = v
t2
v 0 = 1+
uv
Return
Details
575
576
ORDINARY DIFFERENTIAL EQUATIONS
65.
1
−1
, y = − t2 − Ct + D, where C, D are constants.
t+C
2
Return
Details
x=
ANSWERS TO SELECTED EXERCISES
67.
This system is not uncoupled.
Return
577
578
ORDINARY DIFFERENTIAL EQUATIONS
69.
√
1 2
3/2
x = ± t2 + C, y = De± 3 (t +C) , where C and D are constants.
Return
Details
ANSWERS TO SELECTED EXERCISES
71.
This system is not uncoupled.
Return
579
580
ORDINARY DIFFERENTIAL EQUATIONS
73.
2
1
1
-1
-2
Return
Details
2
3
4
ANSWERS TO SELECTED EXERCISES
581
75.
1
-1
-2
-3
-4
-5
-6
Return
Details
2
3
4
5
6
582
ORDINARY DIFFERENTIAL EQUATIONS
77.
1
0.5
1
-0.5
-1
Return
Details
2
3
4
ANSWERS TO SELECTED EXERCISES
583
79.
4
2
-2
-1
1
-2
-4
Return
Details
2
584
ORDINARY DIFFERENTIAL EQUATIONS
81.
Stationary point: (0, 0).
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
Return
Details
-1
0
1
2
3
ANSWERS TO SELECTED EXERCISES
585
83.
Stationary point (0, 0).
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
Return
Details
-1
0
1
2
3
586
ORDINARY DIFFERENTIAL EQUATIONS
85.
1
1
0.5
0.5
1
(a)
2
3
4
5
6
-1 -0.5
-1
-0.5
1
00.5
0.5
1
0
-0.5
-1
6
0.5
-0.5
-0.5
-1
-1
1
4
2
0
ANSWERS TO SELECTED EXERCISES
587
0 2 -5
-2
6
4
2
2
1
2
-2
-4
-6
(b)
4
6
8
10
12
-6
-4
-2
-1
-2
0
5
10
2
4
6
5
0
588
ORDINARY DIFFERENTIAL EQUATIONS
6
6
4
4
2
-5
5
-2.502.5
2.5
5
0
-2.5
10
2
5
-5 -2.5
-2
2.5 5 7.5 10 12.5
-4
-2
2
-2
-4
-4
(c)
4
6
0
-5
ANSWERS TO SELECTED EXERCISES
1.5
1.5
1
1
0.5
0.5
589
0
5
10
4
2
-2
-1
-0.5
-1
(d)
Return
Details
1
2
3
4
-1
1
-0.5
-1
2
3
0
-2
1
0
-1
590
ORDINARY DIFFERENTIAL EQUATIONS
1.
y(1) ≈ 0.841471076.
Return
Details
ANSWERS TO SELECTED EXERCISES
591
3.
The phase diagram depends on e:
- r
r - r
−1 0
1
Phase diagram for e < 0
r - r
r
-
−1 0
1
Phase diagram for e > 0
1
0.5
-1
-0.5
0.5
1
-0.5
-1
(a)
1.5
1
0.5
-1
1
-0.5
(b)
i.
-1
2
3
4
592
ORDINARY DIFFERENTIAL EQUATIONS
1
0.5
-1
-0.5
0.5
-0.5
-1
ii.
1
-1.5 -1 -0.5
0.5
-1
-2
-3
iii.
Return
Details
-4
1
1
ANSWERS TO SELECTED EXERCISES
593
5.
The blue curve in the following graph is the actual solution, and the
black curve is the Euler approximation using step size 0.05.
300
200
100
-2
-1
1
-100
-200
-300
Return
2
594
ORDINARY DIFFERENTIAL EQUATIONS
7.
(a) Solve the ODE y0 = f (y)/1000 with initial conditions y(0) =
−3, −2, −1, and 0, over the interval [−5, 5]. The zeros of f will
be the ordinates of the horizontal asymptotes of these solutions.
(b) The roots are approximately 0.193, 1.027, 2.568, 4.900, 8.182, 12.734,
and 19.396.
Return
Details
ANSWERS TO SELECTED EXERCISES
595
9.
0.4
0.2
92
94
96
98
100
-13.5
-13
-0.2
-0.4
(a)
0.04
0.02
-14.5
-0.02
-0.04
(b)
-14
596
ORDINARY DIFFERENTIAL EQUATIONS
0.01
0.005
0.8
0.85
0.9
0.95
-0.005
(c)
Return
-0.01
ANSWERS TO SELECTED EXERCISES
597
1.
x-nullcline: (0.0).
y-nullclines: y = ± x.
Stationary point: (0, 0).
10
7.5
5
2.5
-10 -7.5
-5
-2.5
-2.5
-5
-7.5
-10
Return
Details
2.5
5
7.5
10
598
ORDINARY DIFFERENTIAL EQUATIONS
3.
x-nullcline: the y-axis.
y-nullclines: y = 4 and y = 0.
Stationary points: (0, 0) and (0, 4).
6
4
2
-3
-2
-1
1
-2
Return
Details
2
3
ANSWERS TO SELECTED EXERCISES
599
5.
x-nullcline: the y-axis.
y-nullcline: the x-axis.
Stationary point: (0, 0).
4
3
2
1
-4
-3
-2
-1
1
-1
-2
-3
-4
Return
Details
2
3
4
600
ORDINARY DIFFERENTIAL EQUATIONS
7.
x-nullcline: y = 2 − x.
y-nullcline: y = x.
Stationary point: (1, 1).
4
3
2
1
-4
-3
-2
-1
1
-1
-2
-3
-4
Return
Details
2
3
4
ANSWERS TO SELECTED EXERCISES
9.
F ( x, y) = −2x −1/2 y + 2x1/2 .
Return
Details
601
602
ORDINARY DIFFERENTIAL EQUATIONS
11.
F ( x, y) = 5 ln x − 3 ln y.
Return
Details
ANSWERS TO SELECTED EXERCISES
13.
F ( x, y) = x2 y +
Return
Details
y3
3.
603
604
ORDINARY DIFFERENTIAL EQUATIONS
15.
The system is uncoupled. The solution of the first equation, x 0 = 1,
with x (0) = x0 is x = t + x0 . Thus the second equation can be written
dy
dx = f ( x, y ); the solution of the system that passes through ( x0 , y0 )
when t = 0 would satisfy the initial condition y(0) = y0 , and since
x = t + x0 the solution follows the graph of the solution of the IVP
dy
= f ( x, y); y( x0 ) = y0 .
dx
Return
ANSWERS TO SELECTED EXERCISES
605
17.
Set y =
dx
dt
and y0 =
d2 x
.
dt2
Then the system
x 0 = y; y(t0 ) = y0
y0 =
f (t, x, y); x (t0 ) = x0
replaces the given ODE. Let g(t, x, y) = y, then f (t, x, y) and g(t, x, y)
satisfy Lipschitz conditions. By the existence and uniqueness theorems there is a unique solution of the system. Thus there is a unique
solution of the IVP.
Return
606
ORDINARY DIFFERENTIAL EQUATIONS
19.
The system is equivalent to
x0 =
y0 =
1
( a g(t, x, y) − c f (t, x, y))
ad − bc
1
(d f (t, x, y) − b g(t, x, y))
ad − bc
If the functions f (t, x, y), g(t, x, y) are continuous and satisfy a Lips1
1
chitz condition then ad−
bc ( ag ( t, x, y ) − c f ( t, x, y )) and ad−bc ( d f ( t, x, y ) −
bg(t, x, y)) are continuous and satisfy a Lipschitz condition. Thus by
the existence and uniqueness theorems the system has a unique solution with the initial conditions.
Return
ANSWERS TO SELECTED EXERCISES
1.
(a)
a
b.
(b) The strategy is ineffective.
(c) Wait until there are ≈ d pests.
Return
Details
607
608
ORDINARY DIFFERENTIAL EQUATIONS
3.
The fish populations are governed by the system
x 0 = k1 x (4000 − 4x − y)
y0 = k2 y(12000 − 7x − 8y)
The phase portrait has the configuration (d) of Figure 2.19. The triangles ABC and CDE in the figure below are traps, and orbits within
these triangles converge to the point C = (800, 800).
4000
A
3000
2000
B
1000
C
D
500
Return
Details
1000
E
1500
2000
ANSWERS TO SELECTED EXERCISES
609
5.
The x-nullclines are the lines x = 0 and K − x + By = 0. The ynullclines are the lines y = 0 and L + Cx − y = 0. The system has
at most four stationary points. Three of these occur when one or both
of the species are extinct. The fourth stationary point, ( x1 , y1 ) exists if
and only if BC 6= 1. Then the stationary point is
BL + K CK + L
( x1 , y1 ) =
,
.
1 − BC 1 − BC
If BC > 1 the fourth stationary point is in the third quadrant. There
is no trap, and the species will increase in numbers without bound. If
BC < 1 we have the fourth stationary point in the first quadrant, and
it is stable. All orbits in the first quadrant converge to this point.
The drawing shows the configuration when BC < 1. The quadrilateral whose vertices are the four stationary points is a trap
2000
1500
1000
500
500
Return
Details
1000
1500
2000
610
ORDINARY DIFFERENTIAL EQUATIONS
7.
The system with fishing can be written as Lotka-Volterra equations:
x 0 = x [( a − R) − by]
R
0
y = cy x − d +
.
c
By the result of problem 6, the average populations will be x̄ = d + Rc
and ȳ = a−b R . Thus the average prey population increases proportionally to the catch rate, and the average predator population decreases.
Return
Details
ANSWERS TO SELECTED EXERCISES
1.
x = tet , y =
Return
Details
1
.
e t (1− t )
611
612
ORDINARY DIFFERENTIAL EQUATIONS
2.
General solution: x = Ce−t , y = Det .
Integral: F ( x, y) = xy.
Phase Portrait:
Return
Details
ANSWERS TO SELECTED EXERCISES
3.
General solution: x = Cet , y = De2t .
Integral: F ( x, y) = x2 /y.
Phase portrait:
Return
Details
613
614
ORDINARY DIFFERENTIAL EQUATIONS
4.
F (y, v) =
v2
2
+
R
f (y) dy is an integral for the system y0 = v, v0 = f (y).
(a) F (y, v) =
v2
2
− α2 cos(y).
(b) F (y, v) =
v2
2
+
(c) F (y, v) =
v2
2
− α 2 y −1 .
Return
Details
α2 y2
2 .
ANSWERS TO SELECTED EXERCISES
5.
(a) y0 = v, v0 = −y
(b) y0 = v, v0 = v − t2 sin(y)
Return
Details
615
616
ORDINARY DIFFERENTIAL EQUATIONS
6.
Return
Details
ANSWERS TO SELECTED EXERCISES
7.
F ( x, y) = yx2 + 2y2 x
Return
Details
617
618
ORDINARY DIFFERENTIAL EQUATIONS
8.
(a) (1, 1).
(b) (0, 0), (−1, −1), (1, 1).
(c) (0, 0), (−4, 0), (−5, 1).
Return
Details
ANSWERS TO SELECTED EXERCISES
9.
(a) x 0 = x + y − 2, y0 = x − 3y + 2.
(b) x 0 = y − x, y0 = y − x3 .
(c) x 0 = x ( x + y + 4), y0 = y( x + 5y).
619
620
ORDINARY DIFFERENTIAL EQUATIONS
Return
Details
ANSWERS TO SELECTED EXERCISES
10.
In the drawing, the trap is shown in blue.
Return
Details
621
622
ORDINARY DIFFERENTIAL EQUATIONS
11.
(a)
(b)
(c)
ANSWERS TO SELECTED EXERCISES
Return
623
624
ORDINARY DIFFERENTIAL EQUATIONS
12.
Both the linear and nonlinear versions are shown: the linear version
is in black; the nonlinear version in blue. The bottom figure is an
enlargement of the region inside the rectangle in the top figure, and
shows more detail.
0.2
0.1
5
-0.1
-0.2
Return
10
15
20
ANSWERS TO SELECTED EXERCISES
1.
~ , ABw
~ , BA~v
AB, BA, A~v, Bw
Return
Details
625
626
ORDINARY DIFFERENTIAL EQUATIONS
3.
(a) EA =
(b) AE =
a b
0 0
a 0
c 0
.
.
(c) b = c = 0. In other words, A must be a diagonal matrix.
Return
Details
ANSWERS TO SELECTED EXERCISES
627
5.
3 −2
Let A =
. Then the matrix equation ~v0 = A~v is equivalent
1 7
to the given system.
Return
Details
628
ORDINARY DIFFERENTIAL EQUATIONS
7.



0
1
0
x
0
1 , and ~b =
~v0 = A~v + ~b where ~v =  x 0 , A =  0
t
00
2e −t sin t
x


0
 0 .
tan t
Return
Details

ANSWERS TO SELECTED EXERCISES
9.
x
1
1
3t
−t
= C1 e
+ C2 e
, where C1 , C2 are constants.
y
1
−1
Return
Details
629
630
ORDINARY DIFFERENTIAL EQUATIONS
11.
x = e2t , t = 2e3t
Return
Details
ANSWERS TO SELECTED EXERCISES
13.
x = e3t + e−t , y = e3t − e−t .
Return
Details
631
632
ORDINARY DIFFERENTIAL EQUATIONS
15.
x ≡ 1, y ≡ 1.
Return
Details
ANSWERS TO SELECTED EXERCISES
17.
1
e
−1
Return
Details
−t
633
634
ORDINARY DIFFERENTIAL EQUATIONS
19.
x
1
3
t
2t
= −8e
+ 5e
.
y
−1
−2
Return
Details
ANSWERS TO SELECTED EXERCISES
635
21.
Put t = 0 and see that the vectors ~v1 (0) and ~v2 (0) are linearly independent.
Return
Details
636
ORDINARY DIFFERENTIAL EQUATIONS
23.
−t2 + 3t + 1
−2t2 + 8t + 1
Return
Details
.
ANSWERS TO SELECTED EXERCISES
25.
cosh(t)
t
sinh(t)
Return
Details
+2
et
et
637
638
ORDINARY DIFFERENTIAL EQUATIONS
27.
y0
0
1
y
0
=
+
v0
−q(t) − p(t)
v
r (t)
Return
Details
ANSWERS TO SELECTED EXERCISES
639
1.
Eigenvalues: s1 = −4 and s2 = 1. Eigenvalues: ~b1 =
Return
Details
0
1
and
1
0
640
ORDINARY DIFFERENTIAL EQUATIONS
3.
Eigenvalues: s1 = −4 and s2 = 1; eigenvectors: ~b1 =
~b2 = 1 .
0
Return
Details
1
−5
and
ANSWERS TO SELECTED EXERCISES
5.
This matrix has no real eigenvalues.
Return
Details
641
642
ORDINARY DIFFERENTIAL EQUATIONS
7.
(c) Every nonzero vector is an eigenvector in this case.
Return
Details
ANSWERS TO SELECTED EXERCISES
9.
−1
c1 e
1
Return
Details
t
+ c2 e
4t
2
1
643
644
ORDINARY DIFFERENTIAL EQUATIONS
11.
1
t
c1
+ c2
0
1
Return
Details
ANSWERS TO SELECTED EXERCISES
13.
c1 e
3t
Return
Details
1
1
+ c2 e
−t
1
−1
645
646
ORDINARY DIFFERENTIAL EQUATIONS
15.
~v = e
−2t
Return
Details
t
1−t
ANSWERS TO SELECTED EXERCISES
17.
x = (2t − 1)e3t , y = (2t − 2)e3t .
Return
Details
647
648
ORDINARY DIFFERENTIAL EQUATIONS
19.
The coefficient matrix of the system is A =
teristic equation is s2 + ps + q = 0.
Return
Details
0
1
−q − p
; its charac-
ANSWERS TO SELECTED EXERCISES
1.
1
(2 − i )
5
Details
Return
649
650
ORDINARY DIFFERENTIAL EQUATIONS
3.
√
√
√
√
1
3
1
3
1
3
1
3
1, + i
,− +i
, −1, − − i
, and + i
.
2
2
2
2
2
2
2
2
Details
Return
ANSWERS TO SELECTED EXERCISES
5.
|eλ+iω | = eλ |eiω | = eλ .
Return
Details
651
652
ORDINARY DIFFERENTIAL EQUATIONS
7.
e
−t/2
Details
Return
−c1 sin(t/2) + c2 cos(t/2)
c1 cos(t/2) + c2 sin(t/2)
ANSWERS TO SELECTED EXERCISES
653
9.
(c1 + c2 ) cos(4t) + (c2 − c1 ) sin(4t)
c1 cos(4t) + c2 sin(4t)
Details
Return
654
ORDINARY DIFFERENTIAL EQUATIONS
11.
(a) Verify that the conjugate of a product (or sum) of complex numbers is the product (or sum) of the conjugates of these numbers.
Apply this when conjugating the equation A~b = s~b, noting that
since A has real entries, Ā = A.
(b) Because s is complex, s 6= s̄.
¯
(c) Since ~b and ~b are linearly independent, ~h and ~k must also be linearly independent.
(d) The real part and the imaginary part of a complex-valued solution of v0 = Av are themselves solutions of the system when A
is a matrix with real entries.
Details
Return
ANSWERS TO SELECTED EXERCISES
1.
X (t) =
Return
Details
et e2t
−et e2t
655
656
ORDINARY DIFFERENTIAL EQUATIONS
3.
X (t) =
Return
Details
2 cos(4t)
2 sin(4t)
cos(4t) + 2 sin(4t) sin(4t) − 2 cos(4t)
ANSWERS TO SELECTED EXERCISES
657
5.
(a) ~v(t) =
(b) ~v(t) =
(c) ~v(t) =
Return
Details
et e2t
−et e2t
et e2t
−et e2t
et e2t
−et e2t
−et
−et
tet
−tet
(t2 /4 − t/2 − 1/2)et
(−t2 /4 − t/2 − 1/2)et
~c +
~c +
~c +
658
ORDINARY DIFFERENTIAL EQUATIONS
7.
2 cos(4t)
2 sin(4t)
~c
~v(t) =
cos(4t) + 2 sin(4t) sin(4t) − 2 cos(4t)
−16t cos(4t) − 8t sin(4t) − cos(4t) − 2 sin(4t)
+
−20t sin(4t) − 52 cos(4t)
Return
Details
ANSWERS TO SELECTED EXERCISES
9.
(a)
(b)
Return
Details
−7
2 t
.
1
2t
t3 ln |t|
−t3 ln |t|
.
659
660
ORDINARY DIFFERENTIAL EQUATIONS
11.
(X C )0 = X 0 C = A(t)(X C ).
Return
ANSWERS TO SELECTED EXERCISES
661
13.
Suppose Q is singular; then there is a nonzero vector d such that Qd~ =
~0. Let ~qi denote the i th column of Q. You can show that
Qd~ = d1~q1 + · · · + dn~qn
and hence that the columns of Q are linearly dependent.
The proof that a matrix with linearly dependent columns is singular
also follows from the above observation.
Return
Details
662
ORDINARY DIFFERENTIAL EQUATIONS
1.

1

0
e At = 
 0
0
Return
Details
t
1
0
0
1 2
2t
t
1
0
1 3
6t
1 2
2t



t 
1
ANSWERS TO SELECTED EXERCISES
663
3.
∞
1 i ∞ 1 j
A )( ∑ B )
i!
j!
i =0
j =0
e A eB = ( ∑
∞
=
∞
1
1
∑ ∑ i! Ai j! B j
i =0 j =0
∞
=
∞
1
∑ ∑ i!j! Ai B j
i =0 j =0
∞
=
k
1
∑ ∑ i!(k − i)! Ai Bk−i
k =0 i =0
∞
=
k
1
k!
∑ k! ∑ i!(k − i)! Ai Bk−i .
i =0
k =0
By the binomial theorem, if A and B commute, then
( A + B)k =
k
k!
∑ i!(k − i)! Ai Bk−i .
i =0
If A and B do not commute, this won’t work: for example ( A + B)2 =
A( A + B) + B( A + B) = A2 + AB + BA + B2 . If AB 6= BA, it would
not be possible to combine the two middle terms to get ( A + B)2 =
A2 + 2AB + B2 .
It follows that for commuting square matrices A and B,
e A eB =
∞
k =0
Return
1
∑ k! ( A + B)k = e A+B .
664
ORDINARY DIFFERENTIAL EQUATIONS
5.
A(t) =
t 1
0 2t
, dA/dt =
t 1
but dA/dt · A(t) =
.
0 4t
Return
1 0
0 2
. Thus A(t) · dA/dt =
t 2
0 4t
,
ANSWERS TO SELECTED EXERCISES
7.


0 0 ac
A2 =  0 0 0  , and Ak = 0 for k ≥ 3.
0 0 0

1 a t b t + 21 ac t2
(a) e At =  0 1
ct 
0 0
1
 λt

λt
e
ate
(bt + 21 act2 )eλt
(b) e( A+λI )t =  0
eλt
cteλt  .
0
0
eλt

x = e−2t (3 + t + 6t2 ), y = −4te−2t , and z = e−2t . Return
Details
665
666
ORDINARY DIFFERENTIAL EQUATIONS
9.
cos(t) + sin(t)
sin(t)
−2 sin(t)
cos(t) − sin(t)
Details
Return
ANSWERS TO SELECTED EXERCISES
667
11.
e At = et
cos
!
√
√
√
√
2t − 2√sin 2t
2
sin
2t
√
√
√
Return
− √3 sin 2t
cos 2t + 2 sin 2t
√
2
Details
668
ORDINARY DIFFERENTIAL EQUATIONS
13.
1 − 2t −2t
2t
1 + 2t
Return
Details
ANSWERS TO SELECTED EXERCISES
15.

1 t 2t + 12 t2

 0 1
t
0 0
1
Return
Details

669
670
ORDINARY DIFFERENTIAL EQUATIONS
17.
x = (cosh t + sinh t)(t + c1 )
y = sinh t(t + c1 ) + (cosh t − sinh t)(t + c2 )
Return
Details
ANSWERS TO SELECTED EXERCISES
19.
1 2t
e + c1 (1 + 2t) − c2 t
2
y = e2t + 4c1 + c2 (1 − 2t).
x =
Return
Details
671
672
ORDINARY DIFFERENTIAL EQUATIONS
21.
Let f (t) be a polynomial of degree N. Then for n > N, D n ( f ) ≡ 0.
Hence
t2 2
tn n
tD
(e f )( x ) =
[ I + tD + D + · · · + D + · · ·] f ( x )
2!
n!
2
t
tN (N)
= f ( x ) + t f 0 ( x ) + f 00 ( x ) + · · · +
f ( x ).
2!
N!
By Taylor’s theorem, for any function g that is ( N + 1) times differentiable on R, there is a number ξ between x and t such that
g( x + t) = g( x ) + t g0 ( x ) +
t2 00
tN (N)
t N +1
g (x) + · · · +
g (x) +
g ( N +1) ( ξ ).
2!
N!
( N + 1) !
Since f ( N +1) (ξ ) = 0 it follows that
(etD f )( x ) = f ( x + t).
Return
ANSWERS TO SELECTED EXERCISES
1.

( t + 1) e t
tet
0
 −tet
et − tet
0 
−
0
0
e t
Return
Details

673
674
ORDINARY DIFFERENTIAL EQUATIONS
3.
We can refer to the result of the previous exercise.
The formula asserted by lemma 3.6.1 is true if k = 1, because it
matches the actual definition of Np . the Assume it holds for powers of Np less than some number r. Then the rows of Npr = Np Npr−1
are obtained by moving the rows of Npr−1 up one level and adding
a row of zeros at the bottom. This results in a matrix an additional
row of zeros at the bottom, and lower p − 1 rows of Npr−1 above it. It
follows that Npr has r rows of zeros at the bottom, and r columns of
zeros at the left. The complement of these rows and columns is I p−r .
By the principle of mathematical induction, the proof of lemma 3.6.1
is complete.
The proof of (3.34) now follows directly from the lemma. We can see
that Npr is to be multiplied by tr /r!; the results are added to get e Np t .
Thus when r = 0 we have the identity matrix; as we move up from
the diagonal we encounter entries of t, 12 t2 , and so on, until we get to
the first row.
Return
ANSWERS TO SELECTED EXERCISES
675
5.
Let P and P0 be n × n matrices, and let Q, Q0 be m × m matrices, and
let A, A0 denote P ⊕ Q and P0 ⊕ Q0 , respectively. Also let R = PP0 and
S = QQ0 We will use the notation pij , pij0 , qij , qij0 , aij , aij0 for the entries
of these matrices.
Thus if B = AA0 then bij =
n+m
∑
aik a0 kj. If 1 ≤ i, j ≤ n then aik a0 kj =
k =1
pik p0 kj when k ≤ n; and aik a0 kj = 0 for k > n. Hence when 1 ≤ i, j ≤
n, we have bij = rij . Also when n + 1 ≤ i, j, k ≤ n + m, aik a0 kj =
q(i−n)(k−n) p0 (k − n)( j − n), and when n + 1 ≤ i, j ≤ n + m and k ≤ n
both aik = 0 and a0 kj = 0. It follows that bij = s(i−n)( j−n) as well.
Now suppose i ≤ n < j Choose any k between 1 and n + m. If k ≤ n
then a0kj = 0, while if k > n we have aik = 0. Thus in the sum bij =
+m
0
∑nk=
1 aik a kj one factor of each term is equal to zero, and it follows
that bij = 0. Identical reasoning shows that bij = 0 when j ≤ n < i..
We have shown that B = R ⊕ S, thus verifying (3.38).
It now follows that we can raise direct sums to powers by the rule
( P ⊕ Q)n = Pn ⊕ Qn . By definition,
e( P⊕ Q)t =
∞
1
( P ⊕ Q)n tn
n!
n =0
∑
∞
=
1 n
( P ⊕ Qn )tn
n!
n =0
=
∞
1 n n
1
P
t
⊕
∑ n!
∑ n! Qn tn
n =0
n =0
∑
∞
= e Pt ⊕ eQt ,
and thus we have verified 3.39).
Return
676
ORDINARY DIFFERENTIAL EQUATIONS
7.
Recall that B ∼ A means that there exists an invertible matrix P such
that B = P−1 AP.
Reflexive: To prove A ∼ A let P = I (the identity matrix): A is the
conjugate of itself by I.
Symmetric: If A ∼ B then A is the conjugate of B by some invertible
matrix P. It follows that B is the conjugate of A by P−1 ; thus
B ∼ A.
Transitive: If ( A ∼ B and B ∼ C ) then there are invertible matrices
P, Q such that A is conjugate to B by P and B is conjugate to C
by Q. Thus A = P−1 BP and B = Q−1 CQ. Substitute for B in the
first equation to get
A = P−1 ( Q−1 CQ) P = ( QP)−1 C ( QP);
in other words A ∼ C because A is conjugate to C by the invertible matrix QP.
Return
ANSWERS TO SELECTED EXERCISES
9.
e−2t
 0
=
 0
0

e At
Return
Details

0
0
0
cos 2t −2 sin 2t 0 
.
1
cos 2t
0 
2 sin 2t
0
0
et
677
678
ORDINARY DIFFERENTIAL EQUATIONS
11.

e At = 
Return
Details
1
2t
2 (3 − e )
1 2t
2 ( e − 1)
1 − e2t
1 2t
2 ( e − 1)
1
2t
2 (3 − e )
2t
e −1

e2t − 1
1 − e2t  .
2e2t − 1
ANSWERS TO SELECTED EXERCISES
679
13.
e At = 
cos t + 3 sin t

2 sin t
e−t 

2 sin t
2 sin t
3 cos t − 3e−t − 11 sin t
4 cos t − 3e−t − 6 sin t
4 cos t − 4e−t − 6 sin t
4 cos t − 4e−t − 6 sin t
4(e−t − cos t) + 8 sin t − 6te−t
4(e−t − cos t) − 4 sin t − 6te−t
5e−t − 4 cos t + 4 sin t − 8te−t
4(e−t − cos t) + 4 sin t − 8te−t
Return
Details

cos t − e−t + 6te−t
cos t − e−t + 6te−t 

cos t − e−t + 8te−t 
cos t + 8te−t
680
ORDINARY DIFFERENTIAL EQUATIONS
15.


1
~v = t  −1  .
1
Return
Details
ANSWERS TO SELECTED EXERCISES
1.
1
1
1 3t 1 t
e + e , y = e3t − et .
2
2
2
2
Details
Return
x=
681
682
ORDINARY DIFFERENTIAL EQUATIONS
2.
Please notice that answers are not unique; in fact none of the answers
below is equal to the matrix exponential. (It would be correct—and
easier—to compute e At in each case.)
cos(3t) − 3 sin(3t) sin(3t) + 3 cos(3t)
(a)
5 cos(3t)
5 sin(3t)
−et sin(t) et cos(t)
(b)
et cos(t) et sin(t)
t
2e 2tet
(c)
0
et
 t

e 2e2t 92 e3t
(d)  0 e2t 3e3t 
0
0
e3t
Details
Return
ANSWERS TO SELECTED EXERCISES
3.
1
1
1
1 + sin(2t)
x = − +
sin(2t) − cos(2t) ln
2
2
4
cos(2t)
+ C1 [2 sin(2t) − cos(2t)] + C2 [sin(2t) + 2 cos(2t)]
1
1 + sin(2t)
cos(2t) ln
+ C1 cos(2t) + C2 sin(2t)
y =
4
cos(2t)
Return
Details
683
684
ORDINARY DIFFERENTIAL EQUATIONS
4.
e At = V · E(t) · V −1 .
Return
Details
ANSWERS TO SELECTED EXERCISES
685
5.
Let r be an eigenvalue of a nilpotent matrix M, and let ~v be corresponding eigenvector. Since M is nilpotent, Mn is the zero matrix
for some n. Therefore Mn~v = ~0. But Mn~v = r n~v so it follows that
r = 0. Since all eigenvalues of M are equal to zero, and since the trace
of any matrix is the sum of its eigenvalues (counted with algebraic
multiplicity), and the determinant of any matrix is the product of its
eigenvalues, it follows that the determinant of M is also zero.
Conversely, if the trace and determinant of M are equal to zero, then
the characteristic equation of M is s2 = 0. The Cayley-Hamilton theorem says that every matrix a satisfies its own characteristic equation;
hence M2 is is equal to the zero matrix.
Return
686
ORDINARY DIFFERENTIAL EQUATIONS
6.
det( B) = det( A) − 14 tr ( A)2 .
Return
Details
ANSWERS TO SELECTED EXERCISES
1.
E.
Return
Details
687
688
ORDINARY DIFFERENTIAL EQUATIONS
3.
I.
Return
Details
ANSWERS TO SELECTED EXERCISES
5.
F.
Return
Details
689
690
ORDINARY DIFFERENTIAL EQUATIONS
7.
L.
Return
Details
ANSWERS TO SELECTED EXERCISES
9.
A.
Return
Details
691
692
ORDINARY DIFFERENTIAL EQUATIONS
11.
D.
Return
Details
ANSWERS TO SELECTED EXERCISES
693
C
A
B
D
E
F
G
I
H
K
J
L
694
ORDINARY DIFFERENTIAL EQUATIONS
13.
Degenerate system:
4
2
-4
-2
2
-2
-4
Return
Details
4
ANSWERS TO SELECTED EXERCISES
695
15.
F ( x, y) = x2 + 2xy is an integral of this system. Thus the orbits follow level curves F ( x, y) = c, which are hyperbolas (for c 6= 0). The
asymptotes x = 0 and x + 2y = 0 are invariant sets consisting of the
origin and two half-line orbits, directed respectively away from and
toward the origin.
Return
Details
696
ORDINARY DIFFERENTIAL EQUATIONS
17.
The phase portrait is a saddle:
3
2
1
-3
-2
-1
1
-1
-2
-3
Return
Details
2
3
ANSWERS TO SELECTED EXERCISES
19.
The phase portrait is a stable spiral node:
4
2
-3
-2
-1
1
-2
-4
Return
Details
2
3
697
698
ORDINARY DIFFERENTIAL EQUATIONS
21.
This degenerate system has y = x as the stationary line:
3
2
1
-3
-2
-1
1
-1
-2
-3
Return
Details
2
3
ANSWERS TO SELECTED EXERCISES
699
23.
The phase portrait is a center, with orbits directed counterclockwise
on ellipses 5y2 − 4xy + 17x2 = C
7.5
5
2.5
-4
-2
2
-2.5
-5
-7.5
Return
Details
4
700
ORDINARY DIFFERENTIAL EQUATIONS
25.
This phase portrait is a saddle; the stable line is y = −5x, and the
unstable line is y = − x.
3
2
1
-3
-2
-1
1
-1
-2
-3
Return
Details
2
3
ANSWERS TO SELECTED EXERCISES
27.
The rotated solution is
√ c1 e t
cos(π/4)
sin(π/4)
c1 et + c2 e3t
.
= 2
−c2 e3t
− sin(π/4) cos(π/4)
c1 et − c2 e3t
Return
Details
701
702
ORDINARY DIFFERENTIAL EQUATIONS
29.
If both eigenvalues are negative −r, −s, then the general solution is
~v(t) = c1 e−rt~b1 + c2 e−st~b2 , where ~b1 ,~b2 are corresponding eigenvectors, and c1 , c2 are constants. As t → ∞, ~v(t) → ~0. Thus all orbits
converge toward the origin; the phase portrait is therefore a stable
node.
Now suppose that the eigenvalues are complex numbers −r ± is,
with negative real parts. Let ~h + i~k be an eigenvector belonging to
−r + is. Then the general solution of the system is ~v(t) = c1 e−rt (cos(st)h −
sin(st)~k ) + c2 e−rt (sin(st)~h + cos(st)~k). Again, ~v(t) approaches the origin as t → ∞; thus the phase portrait is a stable node.
If the eigenvalues are positive real numbers r, s, then the general solution is ~v(t) = c1 ert~b1 + c2 est~b2 , where ~b1 ,~b2 are corresponding eigenvectors. Since limt→−∞ ekt = 0 when k > 0, it follows that as t → −∞,
~v(t) approaches the origin. Hence the phase portrait is an unstable
node.
In the case where the eigenvalues are complex numbers r ± is, with
positive real parts, let ~h + i~k be a eigenvector belonging to r + is. The
general solution of the system is ~v(t) = ert [c1 (cos(st)~h − sin(st)k) +
c2 (sin(st)~h + cos(st)~k )]. As t → −∞, ~v approach the origin. Thus the
phase portrait is an unstable node.
If one of the eigenvalues is 0, the system is degenerate and hence
there is a stationary line through the origin. Each point (other than
the origin) on the stationary line is an orbit that does not approach
the origin. If the eigenvalues are complex numbers with 0 as real part,
the phase portrait is a center. Each orbit is an ellipse, which does not
converge to the origin. If one of the eigenvalues is positive and the
other is negative, the phase portrait is a saddle. The stable line will
consist of orbits directed toward the origin as t → ∞; the unstable
line consists of orbits that converge to the origin as t → −∞, and all
other orbits go to ∞ as t → ±∞.
Return
ANSWERS TO SELECTED EXERCISES
Details
703
704
ORDINARY DIFFERENTIAL EQUATIONS
31.
(a) Use the product rule to differentiate y = vx and simplify.
(c) If there are 2 stationary points and b > 0, the phase diagram that
looks like this:
r
- r
Thus the greater of the two stationary points is stable; the lesser
is unstable. If b < 0 all arrows in the phase diagram reverse, and
the lesser of the two stationary points is stable.
If there is only one stationary point then the phase portrait is as
follows if b > 0 (arrows reversed if b < 0):
r
Finally, if there are no stationary points, the phase diagram is a
down arrow if b > 0, or an up arrow if b < 0.
p
(d) The eigenvalues of the coefficient matrix are 12 ( a + d ± ( a − d)2 + 4bc).
The slopes of the eigenvectors are
−a + d ±
p
( a − d)2 + 4bc
,
2b
the same as the the stationary points of the ODE (4.20).
(e) Assume b > 0. If the equation (4.20) has two stationary points,
s1 < s2 then the two lines l1 : cx + dy = s1 ( ax + by) and l2 :
cx + dy = s2 ( ax + by) are parallel to eigenvectors. The halfplane on one side of the line ax + by = 0 is divided into three
regions by l1 and l2 . The region of the phase plane where
s1 <
cx + dy
< s2
ax + by
corresponds to the middle section of the first phase diagram in
part (c). The slope v of an orbit in this region increases as t increases, converging to s2 as t → ∞ and to s1 as t → −∞.
ANSWERS TO SELECTED EXERCISES
705
cx + dy
< s1 the slope v → −∞ when the
ax + by
orbit crosses the line ax + by = 0 with a vertical tangent; then the
sign of v changes and the slope will decrease from ∞, converging
to s2 as t → ∞.
If the system (4.20) has just one stationary point s1 , then the system (4.19) has a double eigenvalue. The slope of each orbit is
decreasing, but changes sign as it crosses the line ax + by = 0
with a vertical tangent. Hence the slope converges to s1 both as
t → −∞ and as t → +∞!
In the region where
(f) Again we will assume that b > 0. If the equation (4.20) has no
stationary points, then the system (4.18) has no real eigenvalues.
The orbits will be either clockwise ellipses (if the eigenvalues are
purely imaginary) or clockwise spirals. In either case, the slope
will be decreasing until the orbit assumes a vertical direction,
and then, as the orbit turns farther, the slope will decrease again
until a vertical direction is assumed again. The slope changes
from +∞ to −∞ twice in each revolution about the origin.
Return
Details
706
ORDINARY DIFFERENTIAL EQUATIONS
33.
D.
Return
Details
ANSWERS TO SELECTED EXERCISES
35.
B.
Return
Details
707
708
ORDINARY DIFFERENTIAL EQUATIONS
37.
A.
Return
Details
ANSWERS TO SELECTED EXERCISES
1.
Stable, but not asymptotically stable: degenerate system.
709
710
ORDINARY DIFFERENTIAL EQUATIONS
3.
Unstable spiral: every nonstationary orbit
as
is unbounded
t → ∞.
x
cos(t)
Example of an unbounded solution:
= et
y
− sin(t)
ANSWERS TO SELECTED EXERCISES
711
5.


0
 0 

Unstable. Here is an unbounded solution: v = e3t 
 1 .
1
712
ORDINARY DIFFERENTIAL EQUATIONS
7.


0
 0 

Unstable. Here is an unbounded solution: v = et 
 1 .
1
ANSWERS TO SELECTED EXERCISES
713
9.


t cos(t)
 cos(t) + t sin(t) 
.
Unstable. Here is an unbounded solution: v = Rev1 (t) = 


0
0
714
ORDINARY DIFFERENTIAL EQUATIONS
13.
The probability is equal to 0.
ANSWERS TO SELECTED EXERCISES
715
1.
The repelled set consists of the fourth quadrant, and the points of the
first quadrant that lie below repelled set (part of the separatrix) of the
stationary point (2, 1).
Return
Details
716
ORDINARY DIFFERENTIAL EQUATIONS
3.
The origin is the only stationary point. It is asymptotically stable, and
the attracted set is the entire plane.
4
2
-4
-3
-2
-1
1
-2
-4
Return
Details
2
3
4
ANSWERS TO SELECTED EXERCISES
5.
The origin is the only stationary point, and is neutrally stable.
4
2
-4
-3
-2
-1
1
-2
-4
Return
Details
2
3
4
717
718
ORDINARY DIFFERENTIAL EQUATIONS
7.
The stationary points are the origin, and all points on the circle of
radius 1, centered at the origin. The origin is neutrally stable. The
points on the circle are unstable.
1.5
1
0.5
-1.5
-1
-0.5
0.5
-0.5
-1
-1.5
Return
Details
1
1.5
ANSWERS TO SELECTED EXERCISES
719
9.
The stationary points (0, 0), which is linearly degenerate (eigenvalues, 0 and −2) and unstable, and its attracted set is the positive yaxis; (2, 0), a saddle point with the x-axis as the stable separatrix; and
(1, 1), a stable spiral node. Its attracted set is the first quadrant.
3
2
1
-3
-2
-1
1
-1
-2
-3
Return
Details
2
3
720
ORDINARY DIFFERENTIAL EQUATIONS
11.
√
The seven stationary points are (0, 0) (asymptotically stable), (± 10, 0)
(saddle points with the x-axis as unstable separatrices), (±3, ±1) (unstable), and (±1, ±3) (saddle points).
4
3
2
1
-4
-2
2
-1
-2
-3
-4
Return
Details
4
ANSWERS TO SELECTED EXERCISES
721
13.
There are two stationary points, (0, 0) (a saddle with stable separatrix
on the y-axis and unstable separatrix on the y-axis) and (1, 1) (eigenvalues are ±i). There is an integral: F ( x, y) = x + y − ln | x y| defined
except on the axes; F has a local minimum at (1,1), confirming that
this stationary point is a center.
2
1.5
1
0.5
-1
-0.5
0.5
-0.5
-1
Return
Details
1
1.5
2
722
ORDINARY DIFFERENTIAL EQUATIONS
15.
The origin is asymptotically stable if k < 0 (a counterclockwise inward spiral), stable if k = 0 (a linear center, oriented counterclockwise), and unstable if k > 0 (a counterclockwise outward spiral).
k=0
k=-0.05
2
-2
2
-2
Return
Details
k=0.05
2
-2
2
-2
ANSWERS TO SELECTED EXERCISES
723
17.
The unit circle minus the point (1, 0) is an orbit contained in both the
attracted and repelled set of (1, 0). Because the repelled set contains
points near (1, 0), this stationary point is unstable. The attracted set
contains the entire plane, except the origin. Thus all points inside a
circle of radius 12 centered at (1, 0) are attracted to (1, 0); it is therefore
asymptotically stable.
Return
Details
724
ORDINARY DIFFERENTIAL EQUATIONS
1.
There are four stationary points, including the origin, (K, 0) on the xaxis, and (0, L) on the y-axis. If BC < 1 the fourth stationary point is
a stable node in the first quadrant, and if BC > 1 the fourth stationary
point is in the third quadrant, where it is not of interest—populations
can’t be negative.
1
In the phase portraits shown below, ( B, C ) = 2,
and ( B, C ) =
3
2
2,
, respectively. The other parameters are the same for both:
3
( a, d, K, L) = (0.05, 0.05, 2, 1).
8
8
6
6
4
4
2
2
2 4 6 8 10 12 14
Return
Details
2 4 6 8 10 12 14
ANSWERS TO SELECTED EXERCISES
725
3.
(a) (C − d < 0) The small carrying capacity for the prey means that
there will never be enough prey in this environment to support
the predators. In the phase portrait that follows, the parameters
have been set as to be ( A, C, b, c, d) = (0.5, 1500, 1, 0.125, 2000)
500
400
300
200
100
500
1000
1500
2000
(b) ( Ad
4c > C − d > 0) There is an asymptotically stable stationary
point in the first quadrant, representing equilibrium populations
of each species. Their populations will converge directly to these
numbers. In the phase portrait that follows, the parameters have
been set to be ( A, C, b, c, d) = (0.5, 1500, 1, 0.125, 1000)
500
400
300
200
100
500
(c) When C − d >
1000
Ad
4c
1500
2000
the stationary point in the first quadrant is a
726
ORDINARY DIFFERENTIAL EQUATIONS
stable spiral node, because the eigenvalues have nonzero imaginary parts, and their real part is negative. Orbits are directed
counterclockwise, indicating fluctuations as they converge to the
equilibrium. In the following phase portrait, the parameters
have been set to be ( A, C, b, c, d) = (0.5, 1500, 2, 0.125, 500).
500
400
300
200
100
500
Return
Details
1000
1500
2000
ANSWERS TO SELECTED EXERCISES
1.
No.
Return
Details
727
728
ORDINARY DIFFERENTIAL EQUATIONS
3.
Use the second derivative test to detect a relative minimum of the
potential at x1 . Apply proposition 4.4.2.
Return
Details
ANSWERS TO SELECTED EXERCISES
729
5.
a
U ( x ) = − x2 .
2
Potential
Return
Details
Phase Portrait
730
ORDINARY DIFFERENTIAL EQUATIONS
7.
U (x) = −x
Potential
Return
Details
Phase Portrait
ANSWERS TO SELECTED EXERCISES
731
9.
U (x) =
1 3
x
3
Potential
Return
Details
Phase Portrait
732
ORDINARY DIFFERENTIAL EQUATIONS
11.
U ( x ) = x4
Potential
Return
Details
Phase Portrait
ANSWERS TO SELECTED EXERCISES
733
13.
U ( x ) = −| x3 |
Potential
Return
Details
Phase Portrait
734
ORDINARY DIFFERENTIAL EQUATIONS
15.
U (x) =
x2
x
+1
Potential
Return
Details
Phase Portrait
ANSWERS TO SELECTED EXERCISES
735
17.
1
U ( x ) = cos( x ) + x
2
Potential
Return
Details
Phase Portrait
736
ORDINARY DIFFERENTIAL EQUATIONS
19.
1
U ( x ) = cos( x ) +
x2
π
Potential
Return
Details
Phase Portrait
ANSWERS TO SELECTED EXERCISES
737
1.
Complete the square to get
L( x, y) = a( x +
1
b 2
y) + (4ac − b2 )y2
2a
4a
If b2 − 4ac < 0, and a > 0, then L( x, y) > 0 for ( x, y) 6= (0, 0). Therefore L( x, y) is positive definite.
Converse: Assume L( x, y) is positive definite. Then L(e, 0) = ae2
must be positive; hence a > 0. Aslo
L(be, −2ae) = a(4ac − b2 )e2 ,
so 4ac − b2 > 0.
Return
Details
738
ORDINARY DIFFERENTIAL EQUATIONS
3.
The origin is unstable.
Return
Details
ANSWERS TO SELECTED EXERCISES
5.
The origin is asymptotically stable.
Return
Details
739
740
ORDINARY DIFFERENTIAL EQUATIONS
7.
The origin is unstable.
Return
Details
ANSWERS TO SELECTED EXERCISES
741
9.
The origin is asymptotically stable. Use L( x, y) = x2 + y2 as the Lyapunov function.
Return
Details
742
ORDINARY DIFFERENTIAL EQUATIONS
11.
Using the Lyapunov function L( x, y) = 40x2 + y2 , the origin can be
shown to be unstable.
Return
Details
ANSWERS TO SELECTED EXERCISES
743
13.
L( x, y) = 50x2 − 2xy + y2 is a Lyapunov function that shows the origin to be neutrally stable but not asymptotically stable.
Return
Details
744
ORDINARY DIFFERENTIAL EQUATIONS
15.
L( x, y) = 5x2 − 5xy + 4y2 is a Lyapunov function that shows the origin to be asymptotically stable.
Return
Details
ANSWERS TO SELECTED EXERCISES
745
17.
L( x, y) = 4x2 + 3y2 is a Lyapunov function that shows the origin to
be asymptotically stable.
Return
Details
746
ORDINARY DIFFERENTIAL EQUATIONS
19.
If q( x, y) is negative semidefinite, L( x, y) is still a Lyapunov function
for the system, and the origin is a stable stationary point that may or
may not be asymptotically stable.
Return
Details
ANSWERS TO SELECTED EXERCISES
747
21.
(a) L0 ( x, y) = 2( x2 + y2 )2 (1 − x2 − y2 ) is positive definite for on the
set x2 + y2 < 1. Therefore, the origin is unstable.
(b) Since q(cos t, sin t) = 0, this is obvious.
(c) In polar coordinates, this system decouples as
r 0 = r 3 (1 − r 2 )
θ0 = 1
The first equation has a stable stationary point at r = 1, which
shows solutions approach the unit circle as t → ∞. The second
equation says that every solution (except the stationary solution
at the origin) has angular velocity of 1.
(d)
2
1.5
1
0.5
-2
-1.5
-1
-0.5
0.5
-0.5
-1
-1.5
-2
Return
Details
1
1.5
2
748
ORDINARY DIFFERENTIAL EQUATIONS
23.
The system has a limit cycle for any negative value of k.
k=-1
4
k=-10
20
2
-2
-1
Return
Details
10
1
2
-2
-1
1
-2
-10
-4
-20
2
ANSWERS TO SELECTED EXERCISES
25.
S=
Return
Details
1
2
1
4
1
4
3
4
749
0
ORDINARY DIFFERENTIAL EQUATIONS
2.
(i) order 1
(ii) This equation is an ODE
SOLUTIONS MANUAL
4.
(i) order 1
(ii) This equation is an ODE
1
2
ORDINARY DIFFERENTIAL EQUATIONS
6.
(i) order 1
(ii) This equation is an ODE
SOLUTIONS MANUAL
8.
(i) order 1
(ii) This equation is an ODE
3
4
ORDINARY DIFFERENTIAL EQUATIONS
10.
(i) order 2
(ii) This equation is a PDE for there are two independent variables:
x and y.
SOLUTIONS MANUAL
12.
(i) order 3
(ii) This equation is an ODE
5
6
ORDINARY DIFFERENTIAL EQUATIONS
14.
If y = C sin(t) + D cos(t) then y0 = C cos(t) − D sin(t) and
y00 = −C sin(t) − D cos(t), therefore y00 + y = 0. So y = C sin(t) +
D cos(t)is a family of solutions of y00 + y = 0.
(a) Substitute y = 0, t = 0 in y = C sin(t) + D cos(t), we get D = 0,
therefore the solutions are y = C sin(t).
(b) Substitute y = 0, t = 0 in y = C sin(t) + D cos(t), we get D = 0.
Substitute y = 0, t = π in y = C sin(t), we get C could be any
number. Therefore the solutions are y = C sin(t).
(c) Substitute y = 0, t = 0 in y = C sin(t) + D cos(t), we get D =
0. Substitute y = 1, t = π/6 in y = C sin(t),we get C = 2.
Therefore the solution is y = 2 sin(t).
(d) Substitute y = 1, t = 0 in y = C sin(t) + D cos(t),we get D = 1.
Substitute y0 = −1, t = 0 in y0 = C cos(t) − D sin(t) we get
C = −1. So the solution is y = − sin(t) + cos(t).
Return
SOLUTIONS MANUAL
7
16.
2
2
2
2
If y = Cet then y0 = Cet · (2t) = 2t · Cet = 2ty, therefore y = Cet
2
is a solution of the ODE. Substitute t = 0, y = 3 in y = Cet : we get
2
C = 3,therefore the solution of the IVP is y = 3et .
Return
8
ORDINARY DIFFERENTIAL EQUATIONS
18.
Use implicit differentiation: if t2 + y2 = 0, then
2t + 2y
dy
= 0.
dt
Divide by 2 to obtain yy0 + t = 0. Return
SOLUTIONS MANUAL
20.
9
If y > 0 and y = t ln t + Ct for some constant C, then y/t =
ln(t) + C, so
ty0 − y
d y
1
=
= .
t2
dt t
t
Thus, y = t ln t + Ct satisfies the ODE for any constant C. Now we
will show that all solutions have this form. Suppose that y is a solution; that is,
ty0 − y
1
= .
2
t
t
Since
d y
d
1
(ty0 − y)/t2 =
and (ln t) = ,
dt t
dt
t
we have
d y
d
= ln(t).
dt t
dt
Therefore, by the equal derivatives theorem, there is a constant C such
that
y
= ln t + C
t
and it follows that y = t ln t + Ct. Therefore y = t ln t + Ct is the
ty0 −y
general solution of the ODE t2 = 1t on the interval (0, ∞). Return
10
ORDINARY DIFFERENTIAL EQUATIONS
1.
Substitute t = 0 in the formula y = Cekt : y(0) = Ce0 = C. Now
substitute t = D, where D denotes the doubling time, and y = 2C.
We find that 2C = CekD , and hence ekD = 2. Taking logarithms, kD =
ln(2)
ln(2), and therefore k = D . Now substitute t = T, in y = Cekt where
T denotes the tripling time, and y = 3C. We find that 3C = CekT ,
and hence ekT = 3. Taking logarithms, kT = ln(3), and therefore
ln(3)
ln(3)
T = k . = 35 · ln 2) = 55.47 years.
Return
SOLUTIONS MANUAL
3.
11
Let c denote the relative growth rate (the birth rate minus the
c
death rate). Our model is y0 = 1000
y; y(t + 25) = 2y(t), where t ≥ 0.
c
The general solution of the ODE is y = Ce 1000 t , so y(0) = C, 2C =
c
c
2y(0) = y(25) = Ce 1000 ·25 . Therefore, 2 = e 40 so c = 40 ln 2 = 27.726.
Return
12
ORDINARY DIFFERENTIAL EQUATIONS
5.
For population A our model is y10 = 0.04y1 ; y1 (1900) = 1 million. The general solution for this ODE is y1 = Ce0.04t . Substituting t = 1900, y1 = 1, we get C = e−1900×0.04 , so y1 = e0.04(t−1900) .
√
Thus the population B is y2 = y1 = e0.02(t−1900) , therefore y20 =
0.02e0.02(t−1900) = 0.02y2 obeys equation (1.3), and its relative growth
rate is 2% per year.
Return
SOLUTIONS MANUAL
7.
13
The solution for the IVP y0 = 0.06y; y(1700) = 1 is y = e0.06(t−1700) ,
so y(2000) = e0.06(2000−1700) = e18 = 65, 659, 969.1
Return
14
ORDINARY DIFFERENTIAL EQUATIONS
9.
Substitute y = CeK(t) , and y0 = CK 0 (t)eK(t) to obtain CtK 0 (t)eK(t) −
12CeK(t) = 0. Solving for K 0 (t) we obtain K 0 (t) = 12t . Hence K (t) =
R 12
t dt = 12 ln t. It follows that the general solution of this ODE is
y = Ce12 ln t = Ct12 . To satisfy the initial condition, set t = 2 and y = 1
to obtain C = 2112 . Therefore the solution of the IVP is y = 2112 t12 .
14
12
10
8
6
4
2
0.5
1
1.5
2
2.5
Return
SOLUTIONS MANUAL
11.
15
Substitute y = CeK(t) , and y0 = CK 0 (t)eK(t) to obtain C cos(t)K 0 (t)eK(t) −
C sin(t)eRK(t) = 0. Solving for K 0 (t) we obtain K 0 (t) = tan(t). Hence
K (t) = tan(t) dt = − ln cos(t). It follows that the general solution
ln
1
of this ODE is y = Ce− ln cos(t) = Ce cos(t) = cosC(t) . To satisfy the initial
condition, set t = 0 and y = −1 to obtain C = −1. Therefore the
solution of the IVP is y = − cos1(t) for − π2 < t < π2
-1.5
-1
-0.5
0.5
1
1.5
-2
-4
-6
-8
-10
-12
-14
Return
16
13.
ORDINARY DIFFERENTIAL EQUATIONS
Substitute y = CeK(t) , and y0 = CK 0 (t)eK(t) to obtain tCK 0 (t)eK(t) +
CeK(t) = 0. Solving for K 0 (t) we obtain K 0 (t) = − 1t . Hence K (t) =
R
− 1t dt = − ln t. It follows that the general solution of this ODE is
1
y = Ce− ln t = Celn t = C 1t . To satisfy the initial condition, set t = 1
and y = 1 to obtain C = 1. Therefore the solution of the IVP is y = 1t
for t > 0.
4
3
2
1
2
4
6
Return
SOLUTIONS MANUAL
15.
17
Substitute y = CeK(t) , and y0 = CK 0 (t)eK(t) to obtain (t2 + 1)CK 0 (t)eK(t) −
t
tCeK(t) = 0. Solving for K 0 (t) we obtain K 0 (t) = t2 +
. Hence K (t) =
1
R t
R
1
1
1
2
2
dt = 2 t2 +1 d(t + 1) = 2 ln(t + 1). It follows that the general
t2 +1
√ 2
p
1
2
solution of this ODE is y = Ce 2 ln(t +1) = Celn (t +1) = C (t2 + 1).
To satisfy the initial condition, set t = 3 and y = 0 to obtain C = 0.
Therefore the solution of the IVP is y = 0.
Return
18
17.
ORDINARY DIFFERENTIAL EQUATIONS
y = e−0.000121t = e−0.000121(1922+1325) = e−0.3929 = 0.675 ppb.
Return
SOLUTIONS MANUAL
19.
y = e−0.000121t = e−0.000121×4850 = e−0.58685 = 0.556 ppb.
Return
19
20
ORDINARY DIFFERENTIAL EQUATIONS
1.
The associated homogeneous equation is y0 + 4y = 0, and we can
take yh = e−4t . Substitute y = ve−4t and y0 = v0 e−4t − 4ve−4t in the
inhomogeneous ODE to get v0 e−4t − 4ve−4t + 4ve−4t = 3t. When simplified, this reduces to Rv0 e−4t = 3t. Multiplying
through byRe4t yields
R
v0 = 3te4t . Thus v = 3te4t dt = 34 t d(e4t ) = 34 te4t − 34 e4t dt =
3 4t
3 4t
−4t , the general solution is the family
4 te − 16 e + C Since y = ve
3
+ Ce−4t . To satisfy the initial condition, set y = 0 and
y = 43 t − 16
3
3
3 −4t
3
+ C so C = 16
, and y = 34 t − 16
+ 16
e .
t = 0. This yields 0 = − 16
Return
SOLUTIONS MANUAL
3.
21
The associated homogeneous equation is 2y0 + y = 0, and we can
take yh = e−t/2 . Substitute y = ve−t/2 and y0 = v0 e−t/2 − 21 ve−t/2 in
the inhomogeneous ODE to get 2v0 e−t/2 − ve−t/2 + ve−t/2 =
t−1 e−t/2
R
1
1
When simplified, this reduces to v0 = 2 t−1 Thus v = 2 t−1 dt =
1
−t/2 , the general solution is the family y =
2 ln t + C Since y = ve
1
−t/2 + Ce−t/2 . To satisfy the initial condition, set y = 0 and
2 ln te
t = 2. This yields 0 = 21 ln 2e−1 + Ce−1 so C = − 12 ln 2, and y =
1
−t/2 − 1 ln(2) e−t/2 .
2 ln( t ) e
2
Return
22
ORDINARY DIFFERENTIAL EQUATIONS
5.
The associated homogeneous equation is y0 + 2ty = 0, and we
2
2
2
2
can take yh = e−t . Substitute y = ve−t and y0 = v0 e−t − 2tve−t
2
2
2
2
in the inhomogeneous ODE to get v0 e−t − 2tve−t + 2tve−t = e−t .
2
2
When simplified, this reduces to v0 e−t = e−t . Multiplying through
R
2
2
dt = t + C Since y = ve−t , the
by et yields v0 = 1. Thus v =
2
2
general solution is the family y = te−t + Ce−t . To satisfy the initial
2
condition, set y = 0 and t = 0. This yields C = 0 so y = te−t .
Return
SOLUTIONS MANUAL
23
7.
(a) The associated homogeneous equation is y0 + 10y = 0, and we
can take yh = e−10t . Substitute y = ve−10t and y0 = v0 e−10t −
10ve−10t in the inhomogeneous ODE to get v0 e−10t − 10ve−10t +
10ve−10t = 1. When simplified, this reduces to v0 e−10tR= 1. Multiplying through by e10t yields v0 = e10t . Thus v = e10t dt =
1 10t
+ C Since y = ve−10t , the general solution is the family
10 e
1
y = 10 + Ce−10t . To satisfy the initial condition, set y = 1 and
1
9
1
9 −10t
t = 0. This yields 1 = 10
+ C so C = 10
, and y = 10
+ 10
e
.
(b) The associated homogeneous equation is y0 − 10y = 0, and we
can take yh = e10t . Substitute y = ve10t and y0 = v0 e10t + 10ve10t
in the inhomogeneous ODE to get v0 e10t + 10ve10t − 10ve10t = 1.
When simplified, this reduces to v0 e10t R= 1. Multiplying through
1 −10t
by e−10t yields v0 = e−10t . Thus v = e−10t dt = − 10
e
+C
1
10t
+
Since y = ve , the general solution is the family y = − 10
10t
Ce . To satisfy the initial condition, set y = 1 and t = 0. This
1
11
1
10t
yields 1 = − 10
+ C so C = 10
, and y = − 10
+ 11
10 e .
1.5
1.25
1
0.75
0.5
0.25
-1
Return
-0.5
0.5
1
24
ORDINARY DIFFERENTIAL EQUATIONS
9.
(a) The associated homogeneous equation is y0 + 4y = 0, and we
can take yh = e−4t . Substitute y = ve−4t and y0 = v0 e−4t − 4ve−4t
in the inhomogeneous ODE to get v0 e−4t − 4ve−4t + 4ve−4t =
2e−4t sin(2t). When simplified, this reduces to v0 e−4t = 2e−4tRsin(2t).
Multiplying through by e4t yields v0 = 2 sin(2t). Thus v = 2 sin(2t) dt =
− cos(2t) + C Since y = ve−4t , the general solution is the family y = − cos(2t)e−4t + Ce−4t . To satisfy the initial condition,
set y = 0 and t = 0. This yields 0 = −1 + C so C = 1, and
y = − cos(2t)e−4t + e−4t .
(b) The associated homogeneous equation is y0 − 4y = 0, and we
can take yh = e4t . Substitute y = ve4t and y0 = v0 e4t + 4ve4t in the
inhomogeneous ODE to get v0 e4t + 4ve4t − 4ve4t = 2e−4t sin(2t).
When simplified, this reduces to v0 e4t = 2e−4t sin(2t). MultiplyR
ing Rthrough by e−4t yields v0 = 2e−8t sin(R2t). Thus v = 2 e−8t sin(2t) dt =
− 28 sin(2t) de−8t = − 14 e−8t sin(2t) + 14 e−8t cos(2t)(2) dt = − 41 e−8t sin(2t) +
R −8t
R
1
1
e cos(2t) dt = − 14 e−8t sin(2t) − 16
cos(2t) de−8t = − 14 e−8t sin(2t) −
2
R
1 −8t
2
1 −8t
e−8t sin(2t) dt = − 14 e−8t sin(2t) − 16
cos(2t) − 16
e cos(2t) −
16 e
16
1 −8t
1 −8t
1
1
−8t sin(2t ) +
v,
so
v
=
(−
e
sin
(
2t
)
−
e
cos
(
2t
))
+
C
=
−
16
17
4
16
17 (4e
e−8t cos(2t)) + C. Since y = ve4t , the general solution is the fam1
ily y = − 17
(4e−4t sin(2t) + e−4t cos(2t)) + Ce4t . To satisfy the
1
initial condition, set y = 0 and t = 0. This yields 0 = − 17
+ C so
1 −4t
1 4t
1
C = 17 , and y = − 17 e (4 sin(2t) + cos(2t)) + 17 e .
80
60
40
20
-2
-1
1
-20
-40
Return
2
SOLUTIONS MANUAL
11.
25
The associated homogeneous equation is y0 = tan(t)y, and we
can take yh = e− ln cos(t) = sec(t). Substitute y = v sec(t) and y0 =
v0 sec(t) + v sec(t) tan(t) in the inhomogeneous ODE to get
v0 sec(t) cos(t) + v sec(t) tan(t) cos(t) = (v sec(t) − 1) sin(t).
R
When simplified, this reduces to v0 = − sin(t). Thus v = − sin(t) dt =
cos(t) + C, the general solution is the family y = cos(t) sec(t) + C sec(t) =
1 + C sec(t). To satisfy the initial condition, set y = 0 and t = 0. This
yields 0 = 1 + C so C = −1, and y = 1 − sec(t).
Return
26
13.
ORDINARY DIFFERENTIAL EQUATIONS
We will use the IVP dT
dt = − k ( T − A ( t )); T (0) = 800 as our
model, where T is the temperature in ◦ C, and A(t) = 20 is the ambient temperature. The first step is to find the general solution of the
0
ODE, dT
dt = − k ( T − 20), or T + kT = 20k. The homogeneous solution is Th = e−kt . Substitute T = ve−kt ; after cancelling this results
in e−kt v0 = 20k, or v0 = 20kekt . Integration yields v = 20ekt + C, and
hence T = e−kt v = 20 + Ce−kt . To evaluate the parameters C and k,
we need to use the data. Substituting the initial condition T (0) = 800,
800 = 20 + Ce0 , we find that C = 780. Now substitute the other constraint that was given: t = 1, and T = 600: 600 = 20 + 780e−k .
It follows that e−k = 0.7436. Taking logarithms, k ≈ 0.2963. Thus
T (t) = 20 + 780e−0.2963t . Let 50 = 20 + 780e−0.2963t , we get t = 11
minutes.
Return
SOLUTIONS MANUAL
15.
27
While the turkey is in the oven, we will use the IVP dT
dt = − k ( T −
A(t)); T (0) = 20 as our model, where T is the temperature in ◦ C, and
A(t) = 160 is the ambient temperature. The first step is to find the
0
general solution of the ODE, dT
dt = − k ( T − 160), or T + kT = 160k.
The homogeneous solution is Th = e−kt . Substitute T = ve−kt ; after cancelling this results in e−kt v0 = 160k, or v0 = 160kekt . Integration yields v = 160ekt + C, and hence T = e−kt v = 160 + Ce−kt . To
evaluate the parameters C and k, we need to use the data. Substituting the initial condition T (0) = 20, 20 = 160 + Ce0 , we find that
C = −140. Now substitute the other constraint that was given: t = 5,
and T = 85: 85 = 160 − 140e−5k . It follows that e−5k = 0.62415. Taking
logarithms, k = − ln(0.62415) ≈ 0.12483.
Once the turkey is removed from the oven, we will use the IVP dT
dt =
−k( T − A(t)) = −0.12483( T − A(t)); T (0) = 85 as our model, where
T is the temperature in ◦ C, and A(t) = 20 is the ambient temperature. The first step is to find the general solution of the ODE, dT
dt =
0
−0.12483( T − 20), or T + 0.12483T = 20k. The homogeneous solution is Th = e−0.12483t . Substitute T = ve−0.12483t ; after cancelling this
results in e−0.12483t v0 = 20 × 0.12483, or v0 = 20 × 0.12483 × e0.12483t .
Integration yields v = 20e0.12483t + C, and hence T = e−0.12483t v =
20 + Ce−0.12483t . To evaluate the parameters C, we need to use the
data. Substituting the initial condition T (0) = 85, 85 = 20 + Ce0 ,
we find that C = 65. Thus T (t) = 20 + 65e−0.12483t . So T (0.5) =
20 + 65e−0.12483×0.5 = 81◦ C. The assumptions are that the transmission coefficient in the oven is the same as that outside the oven, and
that there are no other sources of heat.
Return
28
17.
ORDINARY DIFFERENTIAL EQUATIONS
We will use the IVP dT
dt = − k ( T − A ( t )); T (0) = 15 as our model,
where T is the temperature in ◦ C, and A(t) = −20 is the ambient
temperature. The first step is to find the general solution of the ODE,
dT
0
dt = − k ( T + 20), or T + kT = −20k. The homogeneous solution
is Th = e−kt . Substitute T = ve−kt ; after cancelling this results in
e−kt v0 = −20k, or v0 = −20kekt . Integration yields v = −20ekt + C,
and hence T = e−kt v = −20 + Ce−kt . To evaluate the parameters
C and k, we need to use the data. Substituting the initial condition
T (0) = 15, 15 = −20 + Ce0 , we find that C = 35. Now substitute
the other constraint that was given: t = 6 − 3 = 3, and T = 10:
10 = −20 + 35e−3k . It follows that e−3k = 0.857143. Taking logarithms, k = ln(0.857143)/3 ≈ 0.0514 hour−1 .
Return
SOLUTIONS MANUAL
19.
29
We use the model: T 0 (t) = −k [ T (t) − A(t)] + mN, where N is
the number of stoves in use. We are given that when the ambient
temperature is −15 and N = 1, the solution is T (t) ≡ 10. Substitute
T (t) = 10, T 0 (t) = 0, A(t) = −15, and N = 1 to get 0 = −k (10 +
15) + m. Hence m ≡ 25k. Then when N = 2 and the temperature has
stabilized at T = T∞ , we have
0 = −k ( T∞ + 15) + 2m = −k ( T∞ + 15) + 50k = −k ( T∞ − 35)
Therefore the temperature will stabilize at 35◦ C.
Return
30
21.
ORDINARY DIFFERENTIAL EQUATIONS
We use the model: T 0 (t) = −k [ T (t) − A(t)] + mH (t), with A(t) =
0, k = 0.04, and

1 if 0 ≤ t < 4




0 if 4 ≤ t < 5
mH (t) =
1 if 5 ≤ t < 9



..

.
In simpler terms, T 0 + 0.04T = mH (t). Since the furnace is on 45 of the
time, an intuitive approach would be to use a different model, where
mH (t) = 45 = 0.8.
The homogeneous solution is Th = e−0.04t . Substitute T = ve−0.04t ;
after canceling this results in e−0.04t v0 = mH (t), or v0 = mH (t)e0.04t .
With mH (t) = 0.8, we can integrate to get v = 20e0.04t + C. Hence
T = e−0.04t v = 20 + Ce−0.04t . As t → ∞, this solution converges to
20◦ C.
The periodic solution Tp (t) of the ODE
T 0 + 0.04T = mH (t)
is not constant, because the furnace does cycle on and off. First note
that Tp is stable, because if T (t) is any other solution then y = T − Tp
satisfies the ODE y0 = −.04y; hence y = Ce−0.04t → 0.
Let f 1 (t) be the solution of the IVP
y0 = −.04y + 1; y(0) = a,
(54)
where a is a parameter that we will determine. Let f 2 (t) be the solution of the IVP
y0 = −.04y; y(4) = f 1 (4).
If Tp (0) = a, then
Tp (t) =
f 1 (t) if 0 ≤ t < 4
f 2 (t) if 4 ≤ t ≤ 5,
and, because Tp is 5-periodic, Tp (5) = Tp (0) = a. Since Tp (5) = f 2 (5),
we must find the value of A such that f 2 (5) = a.
Solving equation 54, we have
f 1 (t) = 25[(.04a − 1)e−.04t + 1]
SOLUTIONS MANUAL
31
. Thus, f 1 (4) = 25(1 − e−.16 ) + a e−.16 = 3.696 + 0.8521 a, where we
are using four-digit precision. Thus f 2 (t) = (3.696 + 0.8521 a)e−.04(t−1)
and hence
f 2 (5) = (3.686 + 0.8521 a)e−.04 = a
Using the approximation e−.04 = 0.9608, we can solve for a to get
a = 19.59◦ C. This is the temperature when the furnace cycles on.
Using our value of a, we have, for 0 ≤ t ≤ 5,
f 1 (t) = 25 − 5.411 e−.04t
and
f 2 (t) = 20.39 e−.04(t−1)
The average temperature is
Z 4
Z
Z 5
1 5
1
Tp (t) dt =
f 1 (t) dt +
f 2 (t) dt
5 0
5 0
4
1
1
1
=
100 +
5.411(e(−.04)4 − 1) −
20.39(e−.04 − 1)
5
.04
.04
◦
= 19.72 C.
The drawing shows the graph of the periodic temperature.
20
15
10
5
2
Return
4
6
8
10
12
14
32
ORDINARY DIFFERENTIAL EQUATIONS
23.
(a) The associated homogeneous equation is y0 + 5y = 0, and we can
take yh = e−5t . Substitute y = ve−5t yields v0 = 5e5t cos 2t. Thus
5 5t
v = 29
e (5 cos 2t + 2 sin 2t) + C Since y = ve−5t , the general
5
solution is the family y = 29
(5 cos 2t + 2 sin 2t) + Ce−5t . The pe5
riodic solution is 29
(5 cos 2t + 2 sin 2t) when t → ∞, Ce−5t → 0,
therefore it is stable.
(b) The associated homogeneous equation is y0 − y = 0, and we can
take yh = et . Substitute y = vet and y0 = v0 et + vet in the inhomogeneous ODE to get v0 et = 7 cos 4t. So v0 = 7e−t cos 4t. Thus
7 −t
v = − 17
e (cos 4t − 4 sin 4t) + C and the general solution is the
7
family y = − 17
(cos 4t − 4 sin 4t) + Cet . The periodic solution is
7
y = − 17
(cos 4t − 4 sin 4t) . When t → ∞, Cet → ∞, therefore it
is not stable.
(c) The associated homogeneous equation is y0 + 2y = 0, and we
can take yh = e−2t . Substitute y = ve−2t yields v0 = e2t (cos t −
3 sin t). Thus v = e2t (cos t − sin t) + C and the general solution
is the family y = cos t − sin t + Ce−2t . The periodic solution is
y = cos t − sin t when t → ∞, Ce−2t → 0, therefore it is stable.
(d) The associated homogeneous equation is y0 − 5y = 0, and we
can take yh = e5t . Substitute y = ve5t and y0 = v0 e5t + 5ve5t in the
inhomogeneous ODE to get v0 = e−5t (4 cos t + 3 sin t) Thus v =
1 −5t
− 26
e (23 cos t + 11 sin t) + C and the general solution is the
1
family y = − 26
(23 cos t + 11 sin t) + Ce5t .The periodic solution
1
is y = − 26 (23 cos t + 11 sin t) . When t → ∞, Ce5t → ∞,therefore
it is not stable.
(e) The associated homogeneous equation is y0 + y = 0, and we can
take yh = e−t . Substitute y = ve−t yields v0 = et e−t sin t = sin t.
Thus v = − cos t + C and the general solution is the family y =
−e−t cos t + Ce−t , and there is no periodic solution.
Return
SOLUTIONS MANUAL
33
25.
(−0.498753) e0.1 t cos(2. t) + (0.0249377) e0.1 t sin(2. t)
e(0.1) t
e(0.1) t
(b) Substitute y = eK(t) in the homogeneous equation y0 + 2ty = 0,
to get K 0 (t)eK(t) + 2teK(t) = 0. Cancel eK(t) to obtain k0 (t) = −2t,
2
so K (t) = −t2 . Thus y = e−t is a solution of the associated
2
homogeneous equation. Put y = ve−t in the original equation
to obtain
2
2
d
(v e−t ) + 2t ve−t = 1
dt
C
(a)
+
2
Use the product rule and simplify to obtain v0 e−t = 1, so v0 =
2
et . It follows that
Z
2
v = et dt.
Hence the solution is
y = e−t
2
Z
2
et dt + C e−t
The CAS found
2
√
π Erfi(t)
C
,
2 +
2
t
e
2 et
and seems to be using a special function of the form
2
Erfi(t) = √
π
12
t12 ln(t)
t12 ln(t)2
48
−t
+ 288 −
C
(c) 12 + 3456
t
t12
√
1
C
(d) − t + t + √
2
2
e t
Return
+
Z
2
et dt.
t12 ln(t)3
12
.
34
27.
ORDINARY DIFFERENTIAL EQUATIONS
Put t = 0 in the ODE; noting that limt→0 t cot(t) = 1, we obtain
= 1; thus y(0) = 1. Our CAS tells us that the general
solution of the ODE is
0 · y 0 (0) + 1 · y (0)
(Si(t) + C ) csc(t)
where Si(t) is
R 1a special function. You can see by solving this ODE
that Si(t) = t sin(t) dt. If C 6= 0, this solution would be undefined
at t = 0.
10
7.5
5
2.5
-3
-2
-1
1
-2.5
-5
-7.5
-10
Return
2
3
SOLUTIONS MANUAL
29.
35
If there is a critical point at t = 1/4 then y0 (1/4) = 0. Put t = 1/4
in the ODE; we obtain
0+2y
2
1
1
1
= y
4
4
4
which we solve to find y(1/4) is either 0 or 8. By differentiating the
ODE, we find
y00 + 2y0 = y2 + 2t y y0 .
Set t = 1/4. Then y0 = 0 so y00 = y2 . If y(1/4) = 8 then there is a
relative minimum by the second derivative test. The test is indecisive
when y(1/4) = 0, but the solution with initial value y(1/4) = 0 is the
constant solution y ≡ 0.
20
17.5
15
12.5
10
7.5
5
2.5
-0.2
Return
0.2
0.4
0.6
36
ORDINARY DIFFERENTIAL EQUATIONS
In Exercises 1–10, assume that the uniformity hypothesis holds.
1.
Let x be the amount of salt in the tank, in kilograms. We use the
ODE x 0 (t) = JK − LC (t). Substitute J = 2, K = 16%, L = 2, C (t) =
x (t)/100, x (0) = 0. We get x 0 (t) + 0.02x (t) = 0.32 The homogeneous
solution is e−0.02t . Now we substitute x = ve−0.02t in the differential
equation, and simplify to get v0 = 0.32e0.02t . Integration yields v =
16e0.02t + C, where C is a constant. Since x (t) = ve−0.02t , it follows that
x (t) = 16 + Ce−0.02t . Substitute x (0) = 0, we get C = −16. Therefore
x (t) = 16 − 16e−0.02t . Let x (t) = 8% × 100 = 8. Solving 8 = 16 −
16e−0.02t , we find it takes t = 35 seconds to reach a concentration of
8%; solving 12 = 16 − 16e−0.02t yields t = 69 seconds to reach 12%;
and solving 15 = 16 − 16e−0.02t yields t = 138 seconds. to reach a
concentration of 15%.
Return
SOLUTIONS MANUAL
3.
37
Let y(t) denote the amount, in milligrams, of KI131 in the tank at
time t. (If the concentration is 100 ppm, then there are 100 milligrams
per liter.) The input rate is thus 100 milligrams per day; the output
rate is ky (due to radioactive decay, where k is the decay rate of I131 )
1
plus 16
y for the solution that is drained from the tank. To find k,
solve y0 = −ky with initial condition y(0) = 1. Thus y = e−kt . Since
the half-life is 8 days, y(8) = 0.5 = e−8k , which can be solved for k to
obtain k = ln82 .
Taking into account the decay of the radioactive iodide and the removal and replacement of solution, we will use the ODE
y0 = −
ln 2
1
y − y + 100 = 100 − p y,
8
16
where p = 2 ln162+1 . The homogeneous solution is y = e− p t . Substip t + C, and
tuting y = ve− p t results in v0 = 100e p t . Hence v = 100
p e
− p t . The limit, as t → ∞, is y = 100 milligrams. Dividing
y = 100
p + Ce
p
by the volume, 16 liters, the limiting concentration is
100
100
=
≈ 42 ppm.
16 p
2 ln 2 + 1
Return
38
ORDINARY DIFFERENTIAL EQUATIONS
5.
Let y(t) denote the amount of quinine in the tank with time t.
Our model will be the ODE y0 = − 10000+(1000
y = − 10010+t y. The
1000−900)t
initial value will be the amount of quinine now in the tank: y(0) =
10000
× 0.01% = 1. The ODE is homogeneous and linear, so y =
R
− ( 10010+t dt)
Ce
= Ce−10 ln(100+t) = C (100 + t)−10 . Substituting y(0) = 1
t −10
) .
we find C = 10010 . Therefore y(t) = 10010 (100 + t)−10 = (1 + 100
To find when the accident occurred, set y(t) = 4. We get t = 100(4−0.1 −
1) = −12.9449, so at worst the quinine was introduced about 13 hours
ago.
Return
SOLUTIONS MANUAL
7.
39
Let y(t) denote the amount of antibiotic drug at time t. Our model
0.25
is the ODE y0 = 0.5 − 3+(0.5
y. The homogeneous solution is
−0.25)t
0.25
R
1
1
. Substitute y = v 3+0.25t
; after canceling this
y = e− 3+0.25t dt = 3+0.25t
1
0
0
results in 3+0.25t v = 0.5, or v = 0.5(3 + 0.25t) = 1.5 + 0.125t. Integra1.5t+
1 2
t +C
1 2
16
t + C, and hence y = 3+0.25t
. Substituting
tion yields v = 1.5t + 16
7
y(0) = 0, we find C = 0, so y(4) = 4 , and after the transfusions stop
the amount of drug is 74 + 0.5 × 2 = 11
4 . The concentration of drug is
11
4
3+4×(0.5−0.25)+0.5×2
Return
=
11
20
= 0.55 grams per liter.
40
ORDINARY DIFFERENTIAL EQUATIONS
9.
let x (t) and y(t) denote the amounts of salt in tanks A and B,
0.2
respectively, at time t. Then x 0 = − 0.2
3 x + 1 y. Since x ( t ) + y ( t ) =
x ()) = 0.36,, we can replace y with .36 − x. Then, after simplifying,
4
4
x 0 = − 15
x + 0.072. The homogeneous solution is x = e− 15 t . Substitute
4
4
x = ve− 15 t ; after cancelling this results in e− 15 t v0 = 0.072, or v0 =
4
4
4
15 t + C = 0.27e 15 t + C,
0.072e 15 t . Integration yields v = 0.072 × 15
4 e
4
and hence x (t) = 0.27 + Ce− 15 t . Substitute x (0) = 3.6, get C = 0.09.
4
Then y(t) = 0.36 − x (t) = 0.09 − 0.09e− 15 t . The concentration in tank
4
A is x (t)/3 = 0.09 + 0.03e− 15 t , and the concentration in tank B is
4
y(t) = 0.09 − 0.09e− 15 t .
Return
SOLUTIONS MANUAL
11.
41
Let y(t) denote the funds available in the retirement account
at time t. Then y0 = 6%y − 30000e3%t , The homogeneous solution
is yh = e0.06t . Substitute y = ve0.06t ; after canceling this results in
e0.06t v0 = −30000e0.03t , or v0 = −30000e−0.03t . Integration yields v =
1000000e−0.03t + C, and hence y = e0.06t v = 1000000e0.03t + Ce0.06t . If
C < 0, as t → ∞, y → −∞, so C ≥ 0, and y(0) = 1000000 + C ≥
1,000,000.
Return
42
13.
ORDINARY DIFFERENTIAL EQUATIONS
The solution of d0 (t) = kd(t), is d(t) = Cekt . To determine the
values of k and C, substitute d(20) = 2.8 and d(50) = 17.6. Thus
Ce20k = 2.8, and Ce50k = 17.6; dividing these equations results in
1
17.6
e30k = 17.6
≈ 0.061, and C = 2.8/e20k = 0.82.
2.8 . Thus k = 30 ln 2.8
Therefore d(t) = 0.82e0.061t .
Let N (t) be the number of survivors in the sample t years later. Determining this function is an input-output problem where there is no
input and the output is by death. Thus, N (0) = 100, 000, and we are
to determine N (30). We will use the ODE
N 0 (t) = −
d(t + 20)
N (t)
1000
since at time t everyone in the sample is t + 20 years old.
Substitute N (t) = C eK(t) and d(t + 20) = 0.82e0.061(t+20) in the ODE
and simplify to obtain
K 0 (t) = −0.00082e0.061(t+20) .
Let’s choose the constant of integration so that K (0) = 0; thus K (t) =
−.013(e0.061(t+20) − e0.061(20) ). Then N (0) = C = 100, 000, and we have
the formula
0.061(t+20)
e−.013e
N (t) = 100, 000
0.061(20)
e−.013e
The number who survive to age 50 is N (30) ≈ 79,000.
Return
SOLUTIONS MANUAL
1.
43
Divide through by ty2 to obtain y0 + t−1 y = t−1 y−2 . To linearize,define
dy
1 −2 dv
2 dy
a new variable v = y3 . Then dv
dt = 3y dt ; or dt = 3 y
dt . Substitute
3
3
0
this expression in the equation to obtain v + t v = t . The associated
homogeneous equation is v0 + 3t v = 0, and we can take vh =
R 3
e− t dt = e−3 ln t = t−3 . Substitute v = ut−3 yields u0 = t3 3t = 3t2 .
Thus u = t3 + C Since v = ut−3 , the general solution is the family
v = 1 + Ct−3 . Therefore y = (1 + Ct−3 )1/3 .
Return
44
ORDINARY DIFFERENTIAL EQUATIONS
3.
To linearize, define a new variable v = y−2 . Then v0 = −2y−3 y0 .
Divide the ODE through by y−3 ; we obtain
y−3 y0 + 3y−2 = sin(t),
or,
1
− v0 + 3 v = sin(t)
2
The associated homogeneous equation is v0 − 6v = 0, and we can take
vh = e6t . Substituting v = ue6t yields
1
− u0 e6t = sin(t),
2
so
u = −2
Z
e−6t sin(t) dt =
1 −6t
e [2 cos(t) + 12 sin(t)] + C
37
Since v = ue6t , the general solution is the family v =
12 sin t) + Ce6t . Therefore
1
(2 cos t + 12 sin t + Ce6t ))−1/2
37
√
37
= ±√
2 cos t + 12 sin t + Ce6t
y = ±(
Return
1
37 (2 cos t
+
SOLUTIONS MANUAL
5.
45
To linearize,define a new variable v = y4 . Multiply through by y3
to obtain
y4
y3 y 0 +
= 12(2t + 1)3 ,
2t + 1
or,
1 0
v
v +
= 12(2t + 1)3 ,
4
2t + 1
because v0 = 4y3 y0 . The associated homogeneous
equation is v0 +
R 4
− 2t+1 dt
4v
= e−2 ln(2t+1) = (2t +
2t+1 = 0, and we can take v h = e
1)−2 . Substitute v = u(2t + 1)−2 yields u0 = 48[(2t + 1)]5 . Thus u =
4(2t + 1)6 + C Since v = u(2t + 1)−2 , the general solution is the family
v = 4(2t + 1)4 + C (2t + 1)−2 . Therefore y = ±(4(2t + 1)4 + C (2t +
1)−2 )1/4 .
Return
46
ORDINARY DIFFERENTIAL EQUATIONS
1.
2
dy
2
+5
5
1
(a) The right side of dt = t ty
= t+
t × y can be expressed as the
product of two single-variable functions, therefore the ODE is
separable.
dy
y
(b) The right side of dt = t = y × 1t can be expressed as the product
of two single-variable functions,therefore the ODE is separable.
dy
(c) The ODE dt = ety , is not separable, because ety can’t be factored
as the product of a function of t and a function of y.
dy
2
2
2
2
(d) The right side of dt = e−(t +y ) = e−t × e−y can be expressed as
the product of two single-variable functions, therefore the ODE
is separable.
(e) The right side of
ODE is separable.
t+y
dy
dt
= cot y depends only on y, therefore the
t+y
(f) The ODE y0 = t−y is not separable, because t−y can’t be factored
as the product of a function of t and a function of y.
Return
SOLUTIONS MANUAL
3.
47
According to Newton’s second law of motion, we have mv0 =
−b v, m = 1, v(0) = 1, v(1) = 0.8, where b is the friction constant.
dv
So dv
dt = − b v, v 6 = 0, and the separated equation is v = − b dt, Integrating both sides, we have ln |v| = −bt + C1 , so v = Ce−bt , where
C = eC1 is a constant. Substituting v(0) = 1, we obtain 1 = C × e0 .
Therefore C = 1.
(a) Substitute v(1) = 0.8 in v = e−bt , and solve for b. Thus, 0.8 =
e−b , and therefore b = − ln(0.8) = 0.223 kg/sec.
(b) Substituting v = 0.5 in v = eln(0.8)t , we obtain t =
seconds.
ln(0.5)
ln(0.8)
(c) To obtain the distance s, integrate the velocity. Thus,
s=
Z 1
eln(0.8)t dt = −
0
0.2
= 0.8963 meters.
ln(0.8)
(d) In an infinite amount of time, the ball will roll
Z ∞
0
Return
eln(0.8)t dt = −
1
= 4.4814 meters.
ln(0.8)
= 3.106
48
ORDINARY DIFFERENTIAL EQUATIONS
5.
According to Newton’s second law of motion, mv0 = −b v + mg.
Put m − 9.8 kg and g = 9.8 m/s2 . Now put v = 10 m/s and v0 = 0.
(The object will move at constant speed when v = 10.) This yields
0 = −b(10) + 9.8 and hence b = 0.98 kg/s.
Return
SOLUTIONS MANUAL
7.
49
Since h(y) = y2 has a zero at y = 0, there is a constant solution,
dy
y ≡ 0. The separated equation is y2 = (t − 1) dt. Integrating both
sides, we get a family of solutions in implicit form, −y−1 = 12 t2 − t +
C. Solving for y,
−2
y= 2
.
t − 2t + 2C
The singular solution satisfies the initial condition
Return
50
ORDINARY DIFFERENTIAL EQUATIONS
9.
Since h(y) = y has a zero at y = 0, there is a constant solution,
dy
y ≡ 0. The separated equation is y = dt
2t . Integrating both sides, we
get a family of solutions in implicit form, ln |y| = 12 ln |t| + C1 . Solving
√
for y, y = ±C t, √
where C = eC1 is a constant. Substituting y(1) = 2,
we obtain
√ 2 = C 1. Therefore C = 2 and the particular solution is
y = 2 t.
Return
SOLUTIONS MANUAL
11.
51
Since h(y) = y has a zero at y = 0, there is a constant solution,
dy
y ≡ 0. The separated equation is y = et dt. Integrating both sides,
we get a family of implicit solutions, ln |y| = et + C1 . Solving for
t
y, y = ±Cee , C = eC1 is a constant, substitute y(0) = 1 we obtain
t
1 = Ce, therefore C = e−1 and the particular solution is y = ee −1 .
Return
52
13.
ORDINARY DIFFERENTIAL EQUATIONS
Since h(y) = y2 has a zero at y = 0, there is a constant solution,
dy
y ≡ 0. The separated equation is y2 = dt. Integrating both sides,
we get a family of implicit solutions, − 1y = t + C. Solving for y, we
obtain y = − t+1C , where C is a constant. Substitute y(0) = 1 to obtain
1 = − C1 ; therefore C = −1 and the particular solution is y = − t−1 1 .
Return
SOLUTIONS MANUAL
15.
53
Since h(y) = y2 − 1 has two zeros, y = ±1, there are two constant
dy
solutions y ≡ 1, and y ≡ −1. The separated equation is y2 −1 = t dt.
y −1 Integrating both sides, we get a family of implicit solutions, 12 ln y+1 =
1
t
+ C1 . Take the exponential of both sides to obtain yy−
+1 = ±Ce ,
where C is a constant, and choose the plus sign since it is in accord
with the initial condition. Substitute y(0) = 0 we obtain −1 = Ce0 ;
2
y −1
therefore C = −1 and the particular solution is y+1 = −et . Hence
2
1 2
2t
t2
y = −e +t2 1 .
1+ e
Return
54
ORDINARY DIFFERENTIAL EQUATIONS
According to Newton’sq
second law of motion, mv0 = −k v2 + mg,
17.
Set v0 = 0 to obtain v∞ =
written
mg
k .
Thus the separated equation can be
dv
= dt,
g − kv2 /m
or
dv
1−
v2
v2∞
= g dt.
Integrating by the method of partial fractions, we obtain
v∞
v∞ + v
= gt + C.
ln
2
v∞ − v
By substituting v(0) = 0, we find C = 0. Therefore
v∞
v∞ + v
ln
= gt.
2
v∞ − v
Now substitute v(2) = 14.7 to obtain the equation
v∞ ln(
v∞ + 14.7
) = 39.2.
v∞ − 14.7
The following graph of f (u) = 39.2 − u (ln(u + 14.7) − ln(u − 14.7))
shows that f has a zero at approximately u = 19. This zero represents
the terminal velocity.
4
2
17
-2
-4
-6
-8
-10
Return
18
19
20
21
22
SOLUTIONS MANUAL
19.
55
According to Newton’s second law of motion, mv0 = kv2 + mg,
when the object is moving upward, and mv0 = −kv2 + mg, when
the object falls downward. Substitute v = v∞ and v0 = 0 in mv0 =
−kv2 + mg, and obtain k = mg
. When the object is moving upward,
v2
∞
g 2
v + g,
v2∞
v0 =
and the separated equation is
dv
1+
v2
v2∞
= g dt,
Integrating both sides, we have v∞ arctan
v
v∞
= gt + C. Substitut-
ing v(0) = v0 , we find
C = v∞ arctan
v0
v∞
,
and therefore v∞ arctan vv∞ = gt + v∞ arctan vv∞0 . To find the time T1 to
reach the maximum height, set v = 0, and obtain gT1 + v∞ arctan vv∞0 =
0. Thus
v∞
| v0 |
T1 =
arctan
g
v∞
For 0 ≤ t ≤ T1 the velocity is negative and is given by the formula
v = v∞ tan(( g/v∞ )t + arctan(v0 /v∞ ))
v0 + v∞ tan( g t/v∞ )
= −v∞
v0 tan( g t/v∞ ) − v∞
(a) the maximum height attained by the ball is
H = v∞
Z T1
0
tan(( g/v∞ )t + arctan(v0 /v∞ )) dt
Notice that since the upward direction is negative, v0 < 0, and H
gt
will be negative as well. Substitute s = v∞ + arctan vv∞0 and ds =
( g/v∞ ) dt. The limits of integration for s corresponding to t − 0
and t = T1 are arctan vv∞0 (which is negative), and 0, respectively.
56
ORDINARY DIFFERENTIAL EQUATIONS
Thus
H = (v2∞ /g)
Z 0
v
0 )
arctan( v∞
tan(s) ds
v2∞
ln(sec(arctan(v0 /v∞ )))
g
q

2
2
2
v ∞  v0 + v ∞ 
= −
.
ln
g
v∞
= −
(b) We have seen that the time taken to reach that height is
T1 =
v∞
| v0 |
arctan
.
g
v∞
(c) When the ball falls downward, we have the initial value problem
mg
mv0 = − 2 v2 + mg; v(0) = 0.
v∞
Dividing through by m, we obtain
v2
0
v = g 1− 2 .
v∞
The separated equation is
dv
1−
v2
v2∞
= g dt,
Integrating both sides, we have
v∞
v∞ + v
ln
= gt + C.
2
v∞ − v
Substituting v( T1 ) = 0, we obtain C = 0, and then
!
2gt
e v∞ − 1
v = v∞
= v∞ tanh( gt/v∞ ).
2gt
e v∞ + 1
Let T2 be the time taken for the ball to return to the ground from
the maximum height. The distance fallen, which is positive now,
is equal to − H. Thus
q

2 + v2
Z T2
2
v
∞
v
0
,
v∞
tanh( gt/v∞ ) dt = − H = ∞ ln 
g
v∞
0
SOLUTIONS MANUAL
57
or
v2∞
g
ln cosh( gT2 /v∞ )) =
v2∞
g
q
ln 
v20 + v2∞
v∞


It follows that
q
T2 =
=

v20 + v2∞
v∞

cosh−1 
g
v∞
s

2
v∞
v0
cosh−1 
+ 1
g
v∞
q
sinh2 ( x ) + 1 can now be used to see
v∞
−1 | v 0 |
T2 =
sinh
g
v∞
The identity cosh( x ) =
that
The time taken to return the ground is T1 + T2
(d) The velocity when the ball hits the ground is
−1 | v 0 |
V = v∞ tanh( gT2 /v∞ ) = v∞ tanh sinh
v∞
Now refer to the hyperbolic identity
tanh(u) =
sinh(u)
sinh(u)
.
=q
cosh(u)
sinh2 (u) + 1
It follows that
V = v∞ p
|v0 |/v∞
(|v0 |/v∞ )2 + 1
If the drag force is negligible (v∞ = ∞)
(a)
2
The maximum height attained by the ball equal limv∞ →∞ ( vg∞
v20
2g
√
ln
v20 +v2∞
)
v∞
(b) the time taken to reach that height equal limv∞ →∞ ( vg∞ arctan |vv∞0 | ) =
v0
g
=
58
ORDINARY DIFFERENTIAL EQUATIONS
(c) the time taken to return to the ground equal limv∞ →∞ ( T1 + T2) =
2v0
g
(d) the velocity when the ball hits the ground is limv∞ →∞ ( √|v02|v∞ 2 ) =
| v0 |
Return
v0 + v ∞
SOLUTIONS MANUAL
21.
59
According to Newton’s second law of motion, we have mv0 =
−kv2 + mg. To determine k, substitute v = v∞ , v0 = 0. Thus k =
mg
g
. It follows that v0 = − v2 v2 + g, and the separated equation is
v2∞
∞
v+v∞ v∞
dv
ln
=
g
dt.
By
integrating
both
sides,
we
find
2
2
v−v∞ = gt + C.
1− v2
v∞
v∞ Substitute v(0) = 0, we find C = 0. Thus v2∞ ln vv+
−v∞ = gt, and t =
v∞ ln(3)
v∞
v+v∞
v∞
v∞
2g (ln v−v∞ ). Finally, with v = 2 we obtain t = 2g (ln(3)) =
2g .
Return
60
ORDINARY DIFFERENTIAL EQUATIONS
1.
= 5 and ∂Q
∂x ( x, y ) = 5 so theRequation is exact. The integral F ( x, y) will have the form F ( x, y) = (2x + 5y + 3) dx + H (y) =
x2 + 5xy + 3x + H (y). To determine H (y), differentiate this expresdH
∂F
sion with respect to y: ∂F
∂y = 5x + dy . Since ∂y = Q ( x, y ) = 5x −
∂P
∂y ( x, y )
4y + 2, it follows that 5x +
dH
dy
dH
dy
= 5x − 4y + 2. After cancelling, we
= −4y + 2. Therefore H (y) = −2y2 + 2y, and the integral
is F ( x, y) = x2 + 5xy + 3x − 2y2 + 2y.
Return
obtain
SOLUTIONS MANUAL
3.
∂P
∂y ( x, y )
= 2y and
dition does not hold.
Return
61
∂Q
∂x ( x, y )
= 2y − 1; therefore the exactness con-
62
ORDINARY DIFFERENTIAL EQUATIONS
5.
3x
= 3e3x (ln y − 1) + 3e3x = 3e3x ln y and ∂Q
∂x ( x, y ) = 3e ln y
so
equation is exact. The integral F ( x, y) will have the form F ( x, y) =
R the
3e3x y(ln y − 1) dx + H (y) = e3x y(ln y − 1) + H (y). To determine
3x
H (y), differentiate this expression with respect to y: ∂F
∂y = e (ln y −
∂P
∂y ( x, y )
1) + e3x +
dH
dy
= ln y(e3x ) +
it follows that ln y(e3x ) +
tain
dH
dy
dH
dy
dH
dy .
Since
F ( x, y) =
Return
− 1) +
y2
4
= Q( x, y) = ln y(e3x − y),
= ln y(e3x − y). After cancelling, we ob-
= −y ln y. Therefore H (y) =
e3x y(ln y
∂F
∂y
−
y2 ln y
2 .
y2
4
−
y2 ln y
2 ,
and the integral is
SOLUTIONS MANUAL
7.
63
= 2xy and ∂Q
is exact. The
∂x ( x, y ) = 2xy so the equation
R
2
integral F ( x, y) will have the form F ( x, y) = x ( x + y2 − 1) dx +
∂P
∂y ( x, y )
4
x 2 y2
2
H (y) = x4 + 2 − x2 + H (y). To determine H (y), differentiate this
dH
∂F
2
expression with respect to y: ∂F
∂y = yx + dy . Since ∂y = Q ( x, y ) =
y( x2 + y2 + 1), it follows that yx2 +
dH
dy
4
= x4
celling, we obtain
integral is F ( x, y)
Return
dH
dy
= y( x2 + y2 + 1). After can-
= y3 + y. Therefore H (y) =
+
x 2 y2
2
−
x2
2
+
y4
4
+
y2
2
.
y4
4
+
y2
2,
and the
64
ORDINARY DIFFERENTIAL EQUATIONS
9.
2xy
y
x +y−y
x +y− x
∂Q
∂P
x
∂y ( x, y ) = 2 x +y × ( x +y)2 = ( x +y)3 and ∂x ( x, y ) = 2 x +y × ( x +y)2 =
2xy
; so the equation is exact. The integral F ( x, y) will have the form
( x + y )3
2 R
y2
y
F ( x, y) =
1+
x +y
dx + H (y) = x − x+y + H (y). To determine
y2 +2xy
H (y), differentiate this expression with respect to y: ∂F
∂y = − ( x +y)2 +
2
2
dH
x
∂F
x
− 1 + dH
− 1, it follows
dy =
x +y
dy . Since ∂y = Q ( x, y ) =
x +y
2
2
x
that x+x y − 1 + dH
− 1. After cancelling, we obtain dH
dy = x +y
dy =
0. Therefore H (y) = 0, and the integral is F ( x, y) = x −
Return
y2
x +y .
SOLUTIONS MANUAL
65
11.
= 6x + 18y and ∂Q
is ex∂x ( x, y ) = 6x + 18y; so the equation
R
2
act. The integral F ( x, y) will have the form F ( x, y) = (3x + 6xy +
9y2 ) dx + H (y) = x3 + 3yx2 + 9y2 x + H (y). To determine H (y), differdH
2
entiate this expression with respect to y: ∂F
∂y = 3x + 18xy + dy . Since
∂P
∂y ( x, y )
∂F
∂y
= Q( x, y) = 3x2 + 18xy + 51y2 , it follows that 3x2 + 18xy +
3x2
51y2 .
dH
dy
51y2 .
dH
dy
=
+ 18xy +
After cancelling, we obtain
=
Therefore
3
3
2
2
H (y) = 17y , and the integral is F ( x, y) = x + 3yx + 9y x + 17y3 .
Return
66
13.
ORDINARY DIFFERENTIAL EQUATIONS
∂P
∂y ( x, y )
= 2x and
tion does not hold.
Return
∂Q
∂x ( x, y )
= −2x; therefore the exactness condi-
SOLUTIONS MANUAL
15.
67
Let’s seek an integrating factor m( x ). This must satisfy
∂
∂
[m( x )( x2 + xy2 + 1)] =
[m( x )(2y)]
∂y
∂x
Do the differentiation to obtain
m( x )(2xy) = m0 ( x )(2y).
Simplify this equation to the form
m0 ( x )
= x.
m( x )
x2
2
Now we can integrate: ln(|m( x )|) = x2 + C. Therefore m = e 2 is an
integrating factor. After multiplying the equation by this integrating
factor, we have the exact equation
x2
x2
e 2 ( x2 + xy2 + 1) dx + 2e 2 y dy = 0.
x2
Since it’s easier to integrate Q( x, y) = 2e 2 y with respect to y than it
is to integrate P( x, y) with respect to x, set
F ( x, y) =
Then
∂F
∂y
Z
x2
= 2e 2 y. Since
x2
x2
2e 2 y dy + K ( x ) = y2 e 2 + K ( x ).
∂F
∂x
x2
= xy2 e 2 + K 0 ( x ),
x2
x2
xy2 e 2 + K 0 ( x ) = e 2 ( x2 + xy2 + 1),
x2
x2
and K 0 ( x ) = e 2 ( x2 + 1). A final integration shows that K ( x ) = xe 2
x2
x2
and so F ( x, y) = y2 e 2 + xe 2 .
Return
68
17.
ORDINARY DIFFERENTIAL EQUATIONS
Let’s seek an integrating factor m( x ). This must satisfy
∂
∂
[m( x )( x2 + 2x + 2xy + 2y + 3y2 )] =
[m( x )(2x + 6y)]
∂y
∂x
Do the differentiation to obtain
m( x )(2x + 2 + 6y) = m0 ( x )(2x + 6y) + 2m( x ).
Simplify this equation to the form
2x + 6y
m0 ( x )
=
= 1.
m( x )
2x + 6y
Now we can integrate: ln(|m( x )|) = x + C. Therefore m = e x is an
integrating factor. After multiplying the equation by this integrating
factor, we have the exact equation
( x2 + 2x + 2xy + 2y + 3y2 )e x dx + (2x + 6y)e x dy = 0.
Since it’s easier to integrate Q( x, y) = (2x + 6y)e x with respect to y
than it is to integrate P( x, y) with respect to x, set
F ( x, y) =
Then
∂F
∂y
Z
(2x + 6y)e x dy + K ( x ) = 2xye x + 3y2 e x + K ( x ).
= (2x + 6y)e x . Since
∂F
∂x
= 2xye x + 2ye x + 3y2 e x + K 0 ( x ),
2xye x + 2ye x + 3y2 e x + K 0 ( x ) = ( x2 + 2x + 2xy + 2y + 3y2 )e x ,
and K 0 ( x ) = ( x2 + 2x )e x . A final integration shows that K ( x ) = x2 e x
and so F ( x, y) = 2xye x + 3y2 e x + x2 e x .
Return
SOLUTIONS MANUAL
19.
69
Let’s seek an integrating factor m( x ). This must satisfy
∂
∂
[2xym( x )] =
[m( x )(3x2 + 2y)]
∂y
∂x
Do the differentiation to obtain
2xm( x ) = m0 ( x )(3x2 + 2y) + 6xm( x ).
Simplify this equation to the form
m0 ( x )
4x
=− 2
.
m( x )
3x + 2y
The right side of this equation is not a function of x, so there is no integrating factor m( x ). It is still possible that there will be an integrating
factor m(y). This must satisfy
∂
∂
[m(y)2xy] =
[m(y)(3x2 + 2y)]
∂y
∂x
Do the differentiation to obtain
2xm(y) + 2xym0 (y) = 6xm(y).
Simplify this equation to the form
m0 (y)
4x
2
=
= .
m(y)
2xy
y
Now we can integrate: ln |m(y)| = 2 ln |y| = ln(y2 ); thus we can multiply by the integrating factor m(y) = y2 to obtain the exact equation
2xy3 dx + (3x2 + 2y)y2 dy = 0.
Since it’s easier to integrate P( x, y) = 2xy3 with respect to x than it is
to integrate Q( x, y) with respect to y, set
F ( x, y) =
Then
∂F
∂x
= 2xy3 . Since
Z
2xy3 dx + K (y) = x2 y3 + K (y).
∂F
∂y
= 3x2 y2 + K 0 (y),
3x2 y2 + K 0 (y) = (3x2 + 2y)y2 ,
and K 0 (y) = 2y3 . A final integration shows that K (y) =
F ( x, y) = x2 y3 +
Return
y4
2.
y4
2
and so
70
21.
ORDINARY DIFFERENTIAL EQUATIONS
(i) An integrating factor m(y) must satisfy
∂
∂
[m(y) P( x, y)] =
[m(y) Q( x, y)]
∂y
∂x
Do the differentiation to obtain
m(y)
∂P
∂Q
+ m0 (y) P( x, y) = m(y) .
∂y
∂x
Simplify this equation to the form
m0 (y)
1
∂Q ∂P
=
−
m(y)
P( x, y) ∂x
∂y
∂Q
1
∂P
Therefore P(x,y
−
is independent of x.
) ∂x
∂y
∂Q
∂P
1
(ii) If R(y) = P(x,y
−
∂y is independent of x, let
) ∂x
m(y) = e
R
R(y) dy
R
=e
1
P( x,y)
∂Q
∂P
∂x − ∂y
Then m(y) is an integrating factor for the ODE.
Return
dy
SOLUTIONS MANUAL
24.
71
Let’s seek the integrating factor m( x ). This must satisfy
∂
∂
[m( x )[ p( x )y − q( x )]] =
m( x )
∂y
∂x
Do the differentiation: we will obtain
m ( x ) p ( x ) = m 0 ( x ).
Simplify this equation to the form
m0 ( x )
= p ( x ).
m( x )
R
Now
we can integrate: ln(|m( x )|) = p( x ) dx + C. Therefore m =
R
e p(x) dx is an integrating factor.
Return
72
ORDINARY DIFFERENTIAL EQUATIONS
1.
SOLUTIONS MANUAL
7.
73
Let f (t) denote the solution of y0 = t − 3y; y(0) = 1, and let
g(t) = t/3. Then f (0) = 1 > g(0) = 0. Above the graph of g(t),
f 0 (t) = t − 3y < 0, so f (t) is decreasing above the line t − 3y = 0. But
g(t) is increasing, therefore f (t) and g(t) should intersect at some
point (t0 , f (t0 ))(existence). If the graphs of f (t) and g(t) intersect
at another point (t1 , f (t1 )),then f 0 (t1 ) > g0 (t1 ) > 0, On the other
hand (t1 , f (t1 )) is a point on the graph of g(t), so f 0 (t1 ) = 0. This is a
contradiction, showing that the point of intersection is unique. At the
crossing, f 0 (t0 ) = 0. By differentiating both sides of the ODE, we find
y00 = 1 − 3y0 . Hence f 00 (t0 ) = 1 − 3 f 0 (t0 ) = 1 > 0, and by the second
derivative test (t0 , f (t0 )) is a relative minimum point for f (t).
Return
74
ORDINARY DIFFERENTIAL EQUATIONS
9.
(a) y2 = ∆y1 + y1 = 1, y3 = ∆y2 + y2 = 2, y4 = ∆y3 + y3 = 3, and
ym = m − 1.
(b) y2 = ∆y1 + y1 = 2, y3 = ∆y2 + y2 = 4, y4 = ∆y3 + y3 = 8, and
y m = 2m −1 .
(c) y2 = ∆y1 + y1 = C + kC = (k + 1)C, y3 = ∆y2 + y2 = (k + 1)2 C,
y4 = ∆y3 + y3 = (k + 1)3 C, and ym = (k + 1)m−1 C.
(d) y2 = ∆y1 + y1 = 0 + 0 + 1 = 2 − 1, y3 = ∆y2 + y2 = 4 − 1,
y4 = ∆y3 + y3 = 8 − 1, and ym = 2ym−1 + 1 = 2m−1 − 1
Return
SOLUTIONS MANUAL
75
11.
(a) ∆ym = am ym can be written ym − ym−1 = am−1 ym−1 . Therefore ym = (1 + am−1 )ym−1 , and similarly we have ym−1 = (1 +
am−2 )ym−2 , ym−2 = (1 + am−3 )ym−3 , . . . , y2 = (1 + a1 )y1 , so
ym = (1 + a1 )(1 + a2 ) · · · (1 + am−1 )y1 = CAm , where C = y1
is a constant.
(b) Substitute ym = vm Am , where vm is a sequence to be determined, to obtain vm+1 Am+1 = (1 + am )vm Am + bm . Since Am+1 =
(1 + am ) Am , we get: vm+1 = vm + Abmm+1 . It follows that vm =
m −1
∑
k =1
bk
.
A k +1
(c)
Am
1
1
1
1
=
1+
1+
1+
··· 1+
1
2
3
m−1
( m − 1) + 1
1+1 2+1 3+1
·
·
···
=
1
2
3
m−1
= m
Therefore ym = Cm, where C is a constant.
(d) Put ym = mvm , where vm is a sequence to be determined. Then
ym+1 = (m + 1)vm+1 , Substitute these in ∆ym = m1 ym + m + 1 to
obtain (m + 1)vm+1 − mvm = vm + m + 1, which we can simplify
as vm+1 − vm = 1. Because v1 = y1 = 0, vm = m − 1, and ym =
m ( m − 1).
Return
76
13.
ORDINARY DIFFERENTIAL EQUATIONS
Write the difference equation in recursive form as ym+1 = ym +
0.05y2m . Starting with y0 = 1, we have
y1 = 1 + 0.05(1)2 = 1.05
y2 = 1.05 + 0.05(1.05)2 = 1.10513
y3 = 1.10513 + 0.05(1.10513)2 = 1.16619
y4 = 1.16619 + 0.05(1.16619)2 = 1.23419
y5 = 1.23419 + 0.05(1.23419)2 = 1.31035
y6 = 1.31035 + 0.05(1.31035)2 = 1.3962
y7 = 1.3962 + 0.05(1.3962)2 = 1.49367
y8 = 1.49367 + 0.05(1.49367)2 = 1.60522
y9 = 1.60522 + 0.05(1.60522)2 = 1.73406
y10 = 1.73406 + 0.05(1.73406)2 = 1.88441
y11 = 1.88441 + 0.05(1.88441)2 = 2.06196
y12 = 2.06196 + 0.05(2.06196)2 = 2.27454
y13 = 2.27454 + 0.05(2.27454)2 = 2.53322
y14 = 2.53322 + 0.05(2.53322)2 = 2.85408
y15 = 2.85408 + 0.05(2.85408)2 = 3.26137
y16 = 3.26137 + 0.05(3.26137)2 = 3.7932
y17 = 3.7932 + 0.05(3.7932)2 = 4.51261
y18 = 4.51261 + 0.05(4.51261)2 = 5.5308
y19 = 5.5308 + 0.05(5.5308)2 = 7.06029
y20 = 7.06029 + 0.05(7.06029)2 = 9.55267
10
8
6
4
2
0.2
Return
0.4
0.6
0.8
1
SOLUTIONS MANUAL
15.
77
The associated homogeneous equation is y0 − y = 0, and we can
take yh = et . Substitute y = vet to get v0 = te−t . Thus v = −(1 +
t)e−t + C, the general solution is the family y = −(1 + t) + Cet . To
satisfy the initial condition, set y = 0 and t = 0. This yields 1 = C so
φ(t) = et − t − 1. and φ(1) = e − 2.
Write the difference equation in recursive form as ym+1 = ym + h(tm +
ym ) with initial condition y0 = 0.
For h = 1,
y1 = 0 + 1 × (0 + 0) = 0
E(1) = |φ(1) − y1 | = 0.71828
For h = 0.5,
y1 = 0 + 0.5 × (0 + 0) = 0
y2 = 0 + 0.5 × (0.5 + 0) = 0.25
E(0.5) = |φ(1) − y2 | = 0.46828
For h = 0.25,
y1 = 0 + 0.25 × (0 + 0) = 0
y2 = 0 + 0.25 × (0.25 + 0) = 0.0625
y3 = 0.0625 + 0.25 × (0.5 + 0.0625) = 0.203125
y4 = 0.203125 + 0.25 × (0.75 + 0.203125) = 0.441406
E(0.25) = |φ(1) − y4 | = 0.276874.
Similarly, if h = 0.1, then the approximation of φ(1) is y10 = 0.457948 +
0.1 × (0.9 + 0.457948) = 0.593742, and hence E(0.1) = |φ(1) − y10 | =
0.124538.
If h = 0.05 then the approximation of φ(1) is y20 = 0.57695 + 0.05 ×
(0.95 + 0.57695) = 0.653298, and E(0.05) = |φ(1) − y20 | = 0.0649823.
if h = 0.02 then the approximation of φ(1) is y50 = 0.691588, and
E(0.02) = |φ(1) − y50 | = 0.026692.
Finally, if h = 0.01 then the approximation of φ(1) is y100 = 0.704814,
and E(0.01) = |φ(1) − y100 | = 0.0134662.
In the following graph, the horizontal axis represents the time step h
and the vertical axis gives the approximation error. It shows that the
error decreases as h decreases; in fact the error seems to be decreasing
78
ORDINARY DIFFERENTIAL EQUATIONS
almost as a linear function of h (follow the graph downward and to
the left as h decreases).
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.2
Return
0.4
0.6
0.8
1
SOLUTIONS MANUAL
17.
79
The associated homogeneous equation is y0 − y = 0, and we
can take yh = et . Substitute y = vet to get v0 = e−t sin(πt). Thus
e−t (π cos(πt)+sin(πt))
v=−
+ C , the general solution is the family y =
π 2 +1
)+sin(πt)
− π cos(πt
+ Cet . To satisfy the initial condition, set y = 0 and
π 2 +1
−π cos(πt)−sin(πt)+et π
. Subt = 0. This yields C = π2π+1 and y =
π 2 +1
stitute 0.25, 0.5, 0.75, 1 in we obtain:y(0) = 0, y(0.25) = 0.101691,
y(0.5) = 0.384523, y(0.75) = 0.751185, y(1) = 1.07468.
Write the difference equation in recursive form as ym+1 = ym + 0.25(sin(π ×
tm ) + ym ). Starting with y0 = 0 ,we have
y1 = 0 + 0.25 × (sin(π × 0) + 0) = 0
y2 = 0 + 0.25 × (sin(π × 0.25) + 0) = 0.176777
y3 = 0.176777 + 0.25 × (sin(π × 0.5) + 0.176777) = 0.470971
y4 = 0.470971 + 0.25 × (sin(π × 0.75) + 0.470971) = 0.76549
Each solution φm (t), which we will use to quantify the errors, satisfies
π cos(πt)+sin(πt)
initial conditions φ(tm ) = ym , so φm (t) = −
+ Cm et
π 2 +1
where the constant Cm is determined by
ym = −
Thus Cm = (ym +
π cos(πtm ) + sin(πtm )
+ Cm etm
π2 + 1
π cos(πtm )+sin(πtm ) −tm
)e
π 2 +1
and
π cos(πt) + sin(πt)
π cos(πtm ) + sin(πtm ) t−tm
+ (ym +
)e
2
π +1
π2 + 1
In particular
π cos(πt)+sin(πt)
φ1 (t) = −
+ 0.269425et−0.25 ,
π 2 +1
φm (t) = −
π cos(πt)+sin(πt)
π 2 +1
π cos(πt)+sin(πt)
−
π 2 +1
φ2 (t) = −
+ 0.268776et−0.5 ,
φ3 (t) =
+ 0.331653et−0.75 ,
Therefore we obtain:
LE1 = φ(0.25) − y1 = 0.101691
LE2 = φ1 (0.5) − y2 = 0.0771727
LE3 = φ2 (0.75) − y3 = 0.013463
LE4 = φ3 (1) − y4 = −0.0506145
and:
AE1 = φ(0.5) − φ1 (1) = 0.130573
AE2 = φ(0.75) − φ2 (1.5) = 0.266751
AE3 = φ(1) − φ3 (2) = 0.359802
Return
80
ORDINARY DIFFERENTIAL EQUATIONS
19.
(a) Write the difference equation in recursive form as ym+1 = ym +
0.1(−100ym ) = −9ym . Starting with y0 = 1 ,we have
y1 = −9 × 1 = −9
y2 = −9 × (−9) = 81
y3 = −9 × (81) = −729
y4 = −9 × (−729) = 6561
y5 = −9 × 6561 = −59049
y6 = −9 × (−59049) = 531441
y7 = −9 × 531441 = −4.78297e × 106
y8 = −9 × (−4.78297 × 106 ) = 4.30467 × 107
y9 = −9 × 4.30467 × 107 = −3.8742 × 108
y10 = −9 × (−3.8742 × 108 ) = 3.48678 × 109
Obviously it diverges.
(b) Write the difference equation in recursive form as ym+1 = ym +
h(−100ym ) = (1 − 100h)ym . When |1 − 100h| < 1, ym → 0. Thus,
the solution of the difference equation converges for 0 < h <
0.02.
Return
SOLUTIONS MANUAL
81
21.
p
, ym even(a) In the forward version, using ym+1 = ym + 1 − y2mp
tually increases to a value bigger than 1, and then 1 − y2m is
undefined which causes the calculation to terminate.
p
(b) y0 = 1 − y2 > 0 therefore y is increasing on (0,2),and ym+1 >
ym . therefore the ”+” sign is correct,and the ”-” sign is extraneous.
(c) Forward version: Write
p the difference equation in recursive form
as ym+1 = ym + 0.1 1 − y2m with initial condition y0 = 0.
n
Error: sin(tn ) − yn
√yn
2
1
0 + 0.1√
× 1 − 0 = 0.1
−0.000166583
2
2
0.1 + 0.1 × √1 − 0.1 = 0.199499
−0.000829413
3 0.199499 + 0.1 × √1 − 0.1994992 = 0.297489
−0.00196835
2
4 0.297489 + 0.1 × √1 − 0.297489 = 0.392961
−0.00354275
2
5 0.392961 + 0.1 × √1 − 0.392961 = 0.484917
−0.00549106
6 0.484917 + 0.1 × √1 − 0.4849172 = 0.572373
−0.00773017
2
7 0.572373 + 0.1 × √1 − 0.572373 = 0.654372
−0.0101543
8 0.654372 + 0.1 × √1 − 0.6543722 = 0.729989
−0.0126332
2
9 0.729989 + 0.1 × √1 − 0.729989 = 0.798335
−0.0150082
2
10 0.798335 + 0.1 × √1 − 0.798335 = 0.858556
−0.0170855
11 0.858556 + 0.1 × √1 − 0.8585562 = 0.909828
−0.018621
2
12 0.909828 + 0.1 × √1 − 0.909828 = 0.951327
−0.0192878
2
13 0.951327 + 0.1 × √1 − 0.951327 = 0.982145
−0.018587
14 0.982145 + 0.1 × 1 − 0.9821452 = 1.00096
−0.0155079
The calculation ends here: y15 is undefined because y14 > 1.
Backward version: Write
√ the difference equation in recursive
ym ±0.1 1.01−y2m
form as ym+1 =
with initial condition y0 = 0.
1.01
82
ORDINARY DIFFERENTIAL EQUATIONS
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Return
0+0.1
√
yn
1.01−02
= 0.0995037
1.01
√
0.0995037+0.1 1.01−0.09950372
= 0.197533
1.01
√
2
0.197533+0.1 1.01−0.197533
= 0.29314
1.01
√
2
0.29314+0.1 1.01−0.29314
= 0.385415
1.01
√
2
0.385415+0.1 1.01−0.385415
= 0.473494
1.01
√
2
0.473494+0.1 1.01−0.473494
= 0.556574
1.01
√
2
0.556574+0.1 1.01−0.556574
= 0.633914
1.01
√
2
0.633914+0.1 1.01−0.633914
= 0.70485
1.01
√
2
0.70485+0.1 1.01−0.70485
= 0.768799
1.01
√
2
0.768799+0.1 1.01−0.768799
= 0.825273
1.01
√
2
0.825273+0.1 1.01−0.825273
= 0.873886
1.01
√
2
0.873886+0.1 1.01−0.873886
= 0.914373
1.01
√
0.914373+0.1 1.01−0.9143732
= 0.946611
1.01
√
0.946611+0.1 1.01−0.9466112
= 0.970657
1.01
√
0.970657+0.1 1.01−0.9706572
= 0.986832
1.01
√
0.986832+0.1 1.01−0.9868322
= 0.99589
1.01
√
0.99589+0.1 1.01−0.995892
= 0.999388
1.01
√
0.999388+0.1 1.01−0.9993882
= 0.999982
1.01
√
0.999982+0.1 1.01−0.9999822
=1
1.01
√
1+0.1 1.01−12
=1
1.01
Error: sin(tn ) − yn
0.000329698
0.00113599
0.00237993
0.00400371
0.00593122
0.00806836
0.0103033
0.012506
0.0145278
0.0161984
0.0173217
0.0176661
0.0169473
0.0147923
0.0106628
0.00411033
0.000612076
1.76663 × 10−5
0
0
SOLUTIONS MANUAL
23.
83
The constant solutions are y = ±3. Notice that for −3 < y <
dy
3, the sign of y0 is positive. The separated equation is √ 2 = dt.
9− y
y
Integrating both sides we obtain arcsin( 3 ) = t + C. Therefore y =
3 sin(t + C ), where C is a constant is a solution for all values of t such
that y0 = 3 cos(t + C ) is positive. The solutions y = ±3, not belonging
to this family, are singular solutions.
3
2
1
0
-1
-2
-3
-4
Return
-2
0
2
4
84
ORDINARY DIFFERENTIAL EQUATIONS
The constant solution is y = 1. The separated equation is
25.
dt
2− t .
dy
1− y
=
Integrating both sides we obtain: − ln |1 − y| = − ln |2 − t| +
C1 , Therefore |1 − y| = C |2 − t|, where C = e−C1 is a constant. The
solution y = 1, belongs to this family (when C = 0). There are no
singular solutions.
3.5
3
2.5
2
1.5
1
0.5
1
Return
2
3
4
5
6
SOLUTIONS MANUAL
1.
85
The general solution of y0 = 0 is y = C, where C is a constant.
Substitute y(1) = −2 to obtain C = −2. Therefore the solution of the
IVP is y = −2, defined on (−∞, +∞).
Return
86
ORDINARY DIFFERENTIAL EQUATIONS
3.
The solution of associated homogeneous equation is yh = e−2t .
1
Substitute y = ve−2t to get v0 = sin(5t)e2t . Thus v = − 29
(5 cos(5t) −
1
2 sin(5t))e2t + C, and the general solution is the family y = − 29
(5 cos(5t) −
5
−
2t
2 sin(5t)) + Ce . Substituting y(0) = 0 yields C = 29 . Therefore the
1
solution of the IVP is y = − 29
(5 cos(5t) − 2 sin(5t) − 5e−2t ) defined
on (−∞, +∞).
Return
SOLUTIONS MANUAL
5.
87
The solution of associated homogeneous equation is yh = t2 .
Substitute y = vt2 to get v0 = 1. Thus v = t + C, and the general
solution is the family y = t3 + Ct2 . Substituting y(1) = 0 yields
C = −1. Therefore the solution of the IVP is y = t3 − t2 defined on
the largest interval containing t = 1 : (−∞, +∞).
Return
88
ORDINARY DIFFERENTIAL EQUATIONS
7.
2
The solution of associated homogeneous equation is yh = et .
2
2
2
Substitute y = vet to get v0 = te−t . Thus v = − 12 e−t + C, and the
2
general solution is the family y = − 12 + Cet . Substituting y(0) = 12
2
yields C = 1. Therefore the solution of the IVP is y = − 21 + et defined
on the largest interval containing t = 0 : (−∞, +∞).
Return
SOLUTIONS MANUAL
9.
89
If y = φ1 (t) and y = φ2 (t) intersect at (t0 , y0 ), then y = φ1 (t)
and y = φ2 (t) are two solutions of the IVP y0 = f (t, y); y(t0 ) = y0 .
Because f (t, y) is continuous and satisfies a Lipschitz condition with
respect to y on a rectangular domain D , they are identical in D by the
uniqueness theorem.
Return
90
11.
ORDINARY DIFFERENTIAL EQUATIONS
The given IVP has a constant solution y = 1, just as in the previous Exercise. We have to prove that it is unique.
Write the ODE in the form y0 =
[−ε, +ε], f (t) 6= 0. Let g(t, y) =
1
|, t
f (t)
1− y
.
f (t)
Let ε > 0. On the interval
1− y
, then | g(t, y1 ) − g(t, y2 )|
f (t)
=
| y1 − y2 |
| f (t)|
≤
K |y1 − y2 |, K = max {|
∈ [−ε, +ε]} Therefore g(t, y) is continuous and satisfies Lipschitz condition with respect to y on {(t, y) :
−ε < t < +ε, −∞ < y < +∞}. By the uniqueness theorem the IVP
has a unique solution.
Return
SOLUTIONS MANUAL
91
13.
t
(i) f (t, y) = y2 +
is continuous in the plane. Thus we obtain t0 , y0
1
could be any pair of real numbers.
(ii)
∂f
∂y ( t, y )
= − (1+2tyy2 )2 is continuous in the plane. Thus we obtain
t0 , y0 could be any pair of real numbers.
Return
92
ORDINARY DIFFERENTIAL EQUATIONS
15.
y
(i) f (t, y) = − sin t . is continuous if t 6= kπ. Thus we obtain t0 6= kπ,
where k is an integer.
(ii)
∂f
∂y ( t, y )
= − sin1 t is continuous if t 6= kπ. Thus we obtain t0 6= kπ.
k is an integer.
Return
SOLUTIONS MANUAL
93
17.
(i) Write the√equation in the form: √
ty02 − yy0 + 1 = 0, We obtain
2
y± y −4t
y± y2 −4t
Let
f
(
t,
y
)
=
, f (t, y) is continuous if
y0 =
2t
2t
y2 > 4t, and t 6= 0 Thus we obtain y20 > 4t0 , and t0 6= 0
(ii) There are two functions for y0 , therefore the set of initial points
(t0 , y0 ) for the uniqueness theorem is ∅.
Return
94
ORDINARY DIFFERENTIAL EQUATIONS
19.
√
3
(i) f (t, y) =
(ii)
y −1
√
. is continuous if ty
ty
1/3
> 0. Thus we obtain t0 y0 > 0.
1)
+ 3(y−1)12/3 √ty is continuous if ty > 0 and
= − t(2y(−
ty)3/2
y 6= 1 Thus we obtain t0 y0 > 0 and y0 6= 1.
Return
∂f
∂y ( t, y )
SOLUTIONS MANUAL
95
21.
(i) f (t, y) = (t − y)1/3 . is continuous in the plane. Thus we obtain
t0 , y0 could be any real numbers.
(ii)
∂f
∂y ( t, y )
= − 3(t−1y)2/3 is continuous if t − y 6= 0 Thus we obtain
t0 − y0 6= 0.
Return
96
23.
ORDINARY DIFFERENTIAL EQUATIONS
∂f
∂y ( t, y )
= −√ y
continuous if |y| < 1, but not continuous
p
on |y| = 1 This implies that f (t, y) = 1 − y2 satisfies a Lipschitz
condition on the domain D if −1 < c and d < 1, but there is no
implication if either c = −1 or d = 1.
1− y2
In the case d = 1 let y2 = 1 and y1 < 1. Then
q
s
1 − y21
| f (t, y2 ) − f (t, y1 )|
1 + y1
=
=
.
y2 − y1
1 − y1
1 − y1
It follows
sfrom this calculation that if there is a Lipschitz constant
1 + y1
K, then
≤ K for all y1 ∈ (−1, 1), which is not possible:
1 − y1
s
1 + y1
= ∞. Therefore f does not satisfy a Lipschitz condilim
1 − y1
y 1 → 1−
tion when d = 1. A similar argument works when c = −1.
Return
SOLUTIONS MANUAL
25.
97
Our CAS found the following expression for the general solution, where a denotes the initial value y(0), and w( x ) = sin( x/2) −
cos( x/2):
y=
w( x )[8 ln |w( x )| + 4 sin( x ) + x − a] + 2 sin( x/2)
sin( x/2) + cos( x/2)
Solving y(1) = 0 for a yields a = −2.001386626.
Return
98
ORDINARY DIFFERENTIAL EQUATIONS
27.
(a) Suppose φ and ψ are fixed by T . Then
kφ − ψk = kT (φ) − T (ψ)k ≤ C kφ − ψk
and because 0 < C < 1, this inequality is impossible unless
kφ − ψk = 0.
(b) By the triangle inequality,
kψn+k − ψn k ≤ kψn+k − ψn+k−1 k + · · · + kψn+1 − ψn k.
Therefore, by the contractive property of T ,
kψn+k − ψn k ≤ (C n+k + · · · + C n ){ψ1 − ψ0 k.
Using the formula for the sum of a geometric series,
∞
∑
C n+k + · · · + C n <
Cm =
m=n
Cn
1−C
and the formula follows.
(c) For n > 0, note that kT (ψ∞ ) − T (ψn−1 )k ≤ C kψ∞ − ψn−1 k. Also
note that T (ψn−1 ) = ψn . Since |ψ∞ − ψn−1 k → 0 as n → ∞, it
follows that
lim kT (ψ∞ ) − ψn k = 0
n→∞
It follows that the sequence {ψn } converges to both ψ∞ and to
T (ψ∞ ) and hence the two are equal: ψ∞ is fixed by T .
i. The initial condition tells us φ0 = 1. Then
φ1 (t) = 1 +
Z t
0
1 ds = 1 + t
Z t
1
(1 + s) ds = 1 + t + t2
2
0
Z t
1
1
1
φ3 (t) = 1 + (1 + s + s2 ) ds = 1 + t + t2 + t3
2
2
6
0
One can deduce that
φ2 (t) = 1 +
n
φn (t) =
and that φ∞ (t) = et .
1 m
t
m!
m =0
∑
SOLUTIONS MANUAL
99
ii. The initial condition yields φ0 = 1. Then
Z t
1
−s ds = 1 − t2
2
0
Z t
1
1
1
φ2 (t) = 1 +
−s 1 − s2 ds = 1 − t2 + t4
2
2
8
0
Z t
1
1
1
1
1
φ3 (t) = 1 +
−s 1 − s2 + s4 ds = 1 − t2 + t4 − t6
2
8
2
8
48
0
φ1 (t) = 1 +
One can deduce that
n
∑
φn (t) =
m =0
1
2m m!
t2m
2
and that φ∞ (t) = et /2 .
iii. We have φ0 = 0. Note that the sequence is thus given reRtp
cursively by φn+1 = 0 3 φn (s) ds. Thus, if φn = 0, then
φn+1 = 0, too. It follows that φ∞ = 0.
iv. The initial condition yields φ0 = 1. Then
φ1 (t) = 1 +
φ2 (t) = 1 +
φ3 (t) = 1 +
Z t
2
1
Z t
2 + 4 ln s
1
s
s
ds = 1 + 2 ln t
ds = 1 + 2 ln t + 2(ln t)2
Z t
2 + 4 ln s + 2(ln s)2
s
1
4
ds = 1 + 2 ln t + 2(ln t)2 + (ln t)3 .
3
One can deduce that
n
φn (t) =
1
(2 ln t)m
m!
m =0
∑
and that φ∞ (t) = e2 ln t = t2 .
Return
100
1.
ORDINARY DIFFERENTIAL EQUATIONS
y We use the equation: y0 = k y 1 − M , where M = 1000. Write
1
k
it in separated form:
dy =
dt. Integration by partial
y(1000
− y) 1000
1
y
k
fractions yields
ln
=
t + B, where B denotes
1000
1000 − y
1000
the integration constant. This equation, solved for y, is y = 1+1000
,
Ae−kt
−
1000B
where A = e
is a constant. Substituting y(0) = 300 for last year
and y(1) = 600 this year, we get 1 + A = 1000/300 and 1 + Ae−k =
1000/600. Solving these equations, we obtain A = 7/3, k = ln(3.5).
Thus y(t) = 1+ 71000
, and y(2) = 1+ 71000
= 840
(3.5)−t
(3.5)−2
Return
3
3
SOLUTIONS MANUAL
101
3.
Write the ODE in separated form: y(2001 −y) dy = 0.02 dt. Integration by
y
1
partial fractions yields 200
ln 200−y = 0.02t + B, where B denotes
the integration constant. This equation, solved for y, is y = 1+200
,
Ae−4t
−
200B
where A = e
is a constant. Substituting y(0) = 10, we get: 1 +
A = 200/10. Thus A = 19 and y(t) = 1+200
.
19e−4t
Return
102
5.
ORDINARY DIFFERENTIAL EQUATIONS
Let p0 = 63.0, p1 = 151.3, and p2 = 309.4 be the three population
data. As these measurements were taken at 60-year intervals, we will
use 60 years as our unit of time.
Following the calculations in Example 1.10.1, the population can be
expressed as
M
p(t) =
1 + Avt
where M is the carrying capacity, A is a constant we’ll need to determine, and v = e−k , (where k is the ratio of p0 (t) to p(t)(1 − p(t)/M )
and t is measured in 60-year increments after 1890, so when t = 1 the
year is 1850, etc. Thus
M
p0 =
,
1+A
so M = p0 (1 + A). Substitute for M in
p1 =
M
1 + Av
to get
p1 − p0
and M = p0 p1
A=
p0 − p1 v
Substitute these in
p2 =
1−v
p0 − p1 v
.
M
1 + Av2
and simplify to get
p0 p1
1−v
=
.
p2 = p0 p1
2
( p1 − p0 ) v − p1 v + p0
p0 − ( p1 − p0 ) v
(The second equality is due to the factorization of the denominator as
(1 − v)( p0 − ( p1 − p0 )v).) Now solve for v:
p0 p2 − p1
v=
p2 p1 − p0
Substituting our data, we get v = 0.3246, A = 11.264, and M = 772.6
million. Thus, 60t years after 1890, the population is
p(t) =
Return
772.6
million.
1 + (11.264) (0.3246t )
SOLUTIONS MANUAL
7.
103
Let P = 500 million. Solve
t=
M
1+ Avt
= P to obtain
ln( M − P) − ln( PA)
,
ln v
and plug in the numerical values to get t = 3.00. Remembering to
multiply by 60, we see that the logistic model predicts that the population will reach 500 million in the year 1890 + 180 = 2070.
Return
104
9.
ORDINARY DIFFERENTIAL EQUATIONS
Let N (t) be the number of people who have been warned about
the sky. Then the number of people who have not heard the rumor
is 10000 − N (t). The rumor will spread to another person in a given
call if one of the parties calling has not heard the rumor, and the other
has. The probability that caller A will spread the rumor to caller B
is equal to the product of the probability that caller A has heard the
N (t)
N (t)
rumor, 10000 , and that caller B hasn’t, 1 − 10000 . There are 100000 calls
per day; hence
N 0 = 100000
N (10000 − N )
N
= 10N (1 −
).
10000
10000
10000
Thus, the model that we will use is the logistic equation.
Solving we get N = 1+10000
, where A is a constant. Substituting
Ae−10t
N (0) = 1000, we obtain A = 9. Let N (t) = 9000, we have 1 + 9e−10t =
10
9 , 10t = − ln(1/81) = 4.39. Thus it takes 0.439 day or = 10.536
hours to warn 90% of the population.
Return
SOLUTIONS MANUAL
11.
The function f (y) = ky ln
105
M
y
is defined for y > 0 only, but
limy→0 f (y) = 0. Thus we can say 0 is stationary. The other stationary point at M is verified because ln(1) = 0. If 0 < y < M then
ln( M/y) > 0 indicating the population is increasing; for y > M we
have ln( M/y) < 0 and the population is decreasing. Thus M is a stable stationary point and 0 is unstable.
Return
106
13.
ORDINARY DIFFERENTIAL EQUATIONS
Let v = ln(y). In terms of v the data are v(0) = 0, v(1) = ln 2, and
v(2) = ln 3. The general solution the ODE v0 = k(ln( M) − v) is v =
ln M + Ce−kt . From the data we get 0 = ln M + C, ln 2 = ln M + Ce−k ,
and ln 3 = ln M + Ce−2k . By the first equation, C = − ln( M ); hence
the second and third equations are
ln( M )(1 − e−k ) = ln(2)
ln( M )(1 − e−2k ) = ln(3).
Divide the second of these equations by the first (cancelling ln( M)) to
obtain
1 − p2
=α
1− p
where p = e−k and α =
ln(3)
.
ln(2)
Since 1 − p2 = (1 − p)(1 + p) we have
1 + p = α,
or, p = α − 1. Now we can find M: Since ln(2) = ln( M)(1 − ek ) =
ln( M)(1 − p),
ln(2)
ln( M ) =
2−α
and
ln(2)2
M = exp
≈ 5.31261.
2 ln(2) − ln(3)
Return
SOLUTIONS MANUAL
15.
dy
dt
107
= ku(t)y(t). In a closed system, the number of total molecules
dy
is a constant C. Therefore u(t) + y(t) = C, and dt = ky(t)(C − y(t)).
It’s a logistic equation.
Return
108
ORDINARY DIFFERENTIAL EQUATIONS
17.
(a) This ODE is autonomous because the right side is independent
of t.
(b) This ODE is not autonomous because the right side is depends
on t.
(c) Again, the right side depends on t, so the ODE isn’t autonomous.
(d) The right side of this ODE is independent of t, so it is autonomous.
Return
SOLUTIONS MANUAL
19.
109
Solving f (y) = y2 = 0 we get one stationary point y1 = 0,
f 0 (y) = 2y. Thus f 0 (0) = 0 if y > 0 or y < 0, we all get f (y) > 0.
Therefore (−∞, 0) and (0, +∞) are up intervals, and the stationary
point is unstable.
- s
0
Return
-
110
21.
ORDINARY DIFFERENTIAL EQUATIONS
There is no solution for f (y) = ey = 0. Therefore there are no
stationary points. Because f (y) = ey > 0, (−∞, +∞) is up interval.
-
Return
SOLUTIONS MANUAL
111
23.
(a) This is the solution of first order autonomous ODE: y0 = 1. There
are no stationary points.
(b) y0 = 2t, y0 > 0 if t > 0; y0 < 0 if t < 0. Therefore y is not strictly
increasing, strictly decreasing or a constant. It is therefore not a
solution of a first order autonomous ODE.
(c) This is the solution of first order autonomous ODEs: y0 = −2e−2t =
−2y. There is a stationary point y = 0.
(d) This is the solution of first order autonomous ODEs: y0 = sec2 (t) =
1 + y2 There are no stationary points.
(e) y0 = cos(t), Thus y is not strictly increasing, strictly decreasing or constant. It is therefore not a solution of a first order autonomous ODE.
Return
112
25.
ORDINARY DIFFERENTIAL EQUATIONS
If g0 (y1 ) < 0, there exists ε > 0 such that g0 (y) < 0 in (y1 − ε, y1 +
ε), because g0 (y) is continuous. That means g(y) is strictly decreasing
in (y1 − ε, y1 + ε). Because g(y1 ) = 0, it follows that g(y) changes sign
from positive to negative at y = y1 Therefore (y1 − ε, y1 ) is an up
interval, and (y1 , y + ε) is a down interval; the phase diagram for the
ODE looks like this near y1 :
- s .
Thus y1 is stable.
If g0 (y1 ) > 0, the same reasoning shows that near y1 , the phase diagram looks like this:
s - . Thus y1 is an unstable stationary point.
Return
SOLUTIONS MANUAL
27.
113
Suppose the ODE is y0 = f (y). Then f (y) has a same stationary point as y0 = y2 at y ≡ 0, thus f (0) = 0. The two up intervals (−∞, 0), (0, +∞) for y0 = y2 is also the up intervals for f (y).
f (0+h)− f (0)
therefore f 0 (0) = limh→0
= limh→0 f (hh) ≥ 0, and f 0 (0) =
h
f (0−h)− f (0)
f (−h)
limh→0
= limh→0 −h ≤ 0, We obtain f 0 (0) = 0 and y ≡ 0
−h
is a degenerate stationary point.
Return
114
ORDINARY DIFFERENTIAL EQUATIONS
29.
(a) The critical harvest rate is the maximum harvest rate that the
population can sustain without becoming extinct. Let f ( P) =
P(1 − P) − H = − P2 + P − H be the growth rate. The stationary
points are the roots of f ( P) = 0. By the quadratic formula,
for
√
1± 1−4H
H < 0.25 there are two stationary points, P1 , P2 =
. The
2
phase diagram for this case is
s - s
P1
P2
As the harvest rate increases, the stationary point P1 moves to
the right, and P2 moves left, until H = 0.25, when there is only
one stationary point, P = 0.5, and the phase diagram looks like
this:
s
0.5
When H > 0.25 there is no stationary point, and the population
becomes extinct. Thus the critical harvest rate is 0.25.
Notice that 0 is not a stationary point. The model predicts that
populations will become negative if P(0) < P1 (in the case H <
0.25), and for P(0) < 0.5 in the critical case. If the harvest rate
exceeds this critical value, negative populations are predicted
for any starting population. Of course the model will cease to
apply when the resource is exhausted; the model actually tell us
that that the state of extinction will not be approached asymptotically, it will occur in a finite span of time.
√
(b) For H = 0.1 we have P1 , P2 = 1± 21−0.6 = 0.5 ± 0.316. P1 =
0.184, P2 = 0.816. Referring to the first phase diagram in part (a),
( P1 , P2 ) is an up interval, and (−∞, P1 ) and ( P2 , +∞) are down
intervals. Therefore y = P2 is stable. Because P(0) = 1 > P2 , the
limiting population will be P2 = 0.816
(c) The bifurcation diagram merges the phase diagrams (presented
vertically) in a graph, where the horizontal axis gives the harvest
rate.
SOLUTIONS MANUAL
115
P
1
0.25
Return
H
116
ORDINARY DIFFERENTIAL EQUATIONS
31.
(a) P(1800)
P(1810)
P(1820)
P(1830)
P(1840)
P(1850)
P(1860)
P(1870)
P(1880)
P(1890)
P(1900)
P(1910)
P(1920)
P(1930)
P(1940)
= 5.3, ( P(1810) − P(1790))/(20P(1800)) = 0.0311321
= 7.2, ( P(1820) − P(1800))/(20P(1810)) = 0.0298611
= 9.6, ( P(1830) − P(1810))/(20P(1820)) = 0.0296875
= 12.9, ( P(1840) − P(1820))/(20P(1830)) = 0.0290698
= 17.1, ( P(1850) − P(1830))/(20P(1840)) = 0.030117
= 23.2, ( P(1860) − P(1840))/(20P(1850)) = 0.030819
= 31.4, ( P(1870) − P(1850))/(20P(1860)) = 0.0245223
= 38.6, ( P(1880) − P(1860))/(20P(1870)) = 0.0243523
= 50.2, ( P(1890) − P(1870))/(20P(1880)) = 0.0243028
= 63, ( P(1900) − P(1880))/(20P(1890)) = 0.0206349
= 76.2, ( P(1910) − P(1890))/(20P(1900)) = 0.0191601
= 92.2, ( P(1920) − P(1900))/(20P(1910)) = 0.0161605
= 106, ( P(1930) − P(1910))/(20P(1920)) = 0.0146226
= 123.2, ( P(1940) − P(1920))/(20P(1930)) = 0.0106331
= 132.2, ( P(1950) − P(1930))/(20P(1940)) = 0.0106278
(b) m = −0.000165504 b = 0.0317447
0.0317447
(c) M = −b/m = − −0.000165504
= 191.806 million.
(d) P0 = P(mP + b) The separated form is
dP
P(mP+b)
= dt, Integrating
both sides, we have 1b ln( mPP+b ) = t + C1 Simplifying we obtain
P = Ce−btb −m , where C is a constant. Substitute the initial condition, P(0) = 76.21 to get C = 0.000251038. Thus
p(t) =
0.0317447
.
0.000251038e−0.0317447t + 0.000165504
SOLUTIONS MANUAL
117
250
200
150
100
50
1850
Return
1900
1950
2000
118
ORDINARY DIFFERENTIAL EQUATIONS
Find the general solution of each of the ODEs in problems 1 – 12.
1.
The associated homogeneous
equation is y0 − (t + 1)y = 0, and
R
2 /2+ t
2
(
t
+
1
)
dt
t
we can take yh = e
= e
. Substitute y = vet /2+t yields
2
2
2
v0 = (t + 1)e−t /2−t . Thus v = −e−t /2−t + C. Since y = vet /2+t , the
2
general solution is the family y = −1 + Cet /2+t .
Return
SOLUTIONS MANUAL
2.
119
The associated
homogeneous equation is y0 − y = 0, and we can
R
take yh = e dt = et . Substitute y = vet yields v0 = −t2 e−t . Thus
v = e−t (2 + 2t + t2 ) + C. Since y = vet , the general solution is the
family y = (2 + 2t + t2 ) + Cet .
Return
120
3.
ORDINARY DIFFERENTIAL EQUATIONS
The associated
homogeneous equation is y0 − y = 0, and we can
R
take yh = e dt = et . Substitute y = vet yields v0 = 1 .general solution
Return
SOLUTIONS MANUAL
4.
121
The associated homogeneous
equation is y0 + tan(t)y = 0, and
R
we can take yh = e− tan(t) dt = eln cos t = cos t. Substitute y = v cos t
yields v0 = cos t. Thus v = sin t + C. Since y = v cos t, the general
solution is the family y = (sin t + C ) cos t.
Return
122
5.
ORDINARY DIFFERENTIAL EQUATIONS
The associated
homogeneous equation is y0 + y = 0, and we can
R
take yh = e− dt = e−t . Substitute y = ve−t yields v0 = e2t . Thus
v = 12 e2t + C. Since y = ve−t , the general solution is the family y =
1 t
−t
2 e + Ce .
Return
SOLUTIONS MANUAL
6.
123
The associated homogeneous equation is ty0 + y = 0, and we can
take yh = e− ln t = 1/t. Substitute y = v/t and y0 = v0R/t − v/t2 in the
inhomogeneous ODE to get v0 = et sin 2t. Thus v = (et sin 2t) dt =
− 15 et (2 cos(2t) − sin(2t)) + C. Since y = v/t, the general solution is
the family y = − 5t1 et (2 cos(2t) − sin(2t)) + C/t.
Return
124
7.
ORDINARY DIFFERENTIAL EQUATIONS
The associated homogeneous equation is ty0 + 12y = 0, and we
can take yh = e−12 ln t = t−12 . Substitute y = vt−12 and y0 = v0 t−12 −
12vt−13 in the inhomogeneous ODE to get v0 = t11 (5t2 + 3t − 2) =
5 14
3 13
5t13 + 3t12 − 2t11 . Thus v = 14
t + 13
t − 61 t12 + C. Since y = vt−12 ,
3
5 2
t + 13
t − 16 + Ct−12 .
the general solution is the family y = 14
Return
SOLUTIONS MANUAL
8.
125
The associated homogeneous
equation is y0 + tan(t)y = 0, and
R
we can take yh = e− tan(t) dt = Reln cos t = cos t. Substituting y = v cos t
yields v0 cos t = 1. Thus v = sec(t) dt = ln | sec(t) + tan(t)| + C.
Since y = v cos t, the general solution is the family y = (ln | sec(t) +
tan(t)| + C ) cos t.
Return
126
9.
ORDINARY DIFFERENTIAL EQUATIONS
The associated homogeneous equation is y0 − ty = 0, and we can
2
2
take yh = et /2 . Substitute y = vet /2 in the inhomogeneous ODE to
R
2
2
2
get v0 = e−t /2 . Thus v = e−t /2 dt + C. Since y = vet /2 , the general
R −t2 /2
2
solution is the family y = ( e
dt + C )et /2 .
Return
SOLUTIONS MANUAL
10.
y = Ce
Return
R
5
t
dt
= Ce5 ln t = Ct5
127
128
11.
ORDINARY DIFFERENTIAL EQUATIONS
The associated homogeneous equation is y0 + y = 0, and we can
take yh = e−t . Substitute y = ve−t in the inhomogeneous ODE to get
v0 = 1. Thus v = t + C Since y = ve−t , the general solution is the
family y = (t + C )e−t .
Return
SOLUTIONS MANUAL
12.
y = Ce
Return
R
cos t dt
= Cesin t
129
130
ORDINARY DIFFERENTIAL EQUATIONS
13.
The associated homogeneous equation is ty0 − y = 0, and we can
take yh = eln t = t. Substitute y = vt and y0 = v0 t + v in the inhomogeneous ODE to get v0 = t−1 . Thus v = ln t + C Since y = vt, the
general solution is the family y = t ln t + Ct. To satisfy the initial condition, set y = 2 and t = 1. This yields C = 2, and y = t ln t + 2t.
Return
SOLUTIONS MANUAL
131
14.
The associated homogeneous equation is y0 + y = 0, and we can take
yh = e−t . Substitute y = ve−t in the inhomogeneous ODE to get v0 =
1. Thus v = t + C Since y = ve−t , the general solution is the family
y = (t + C )e−t . To satisfy the initial condition, set y = 0 and t = 0.
This yields C = 0, and y = te−t .
Return
132
15.
ORDINARY DIFFERENTIAL EQUATIONS
y = t2 + C, To satisfy the initial condition, set y = 4 and t = 2.
This yields C = 0, and y = t2 .
Return
SOLUTIONS MANUAL
16.
133
The associated homogeneous equation is y0 + 4y = 0, and we can
take yh = e−4t . Substitute y = ve−4t in the inhomogeneous ODE to
get v0 = 3 sin(3t). Thus − cos(3t) + C Since y = ve−4t , the general
solution is the family y = − cos(3t)e−4t + Ce−4t . To satisfy the initial
condition, set y = 0 and t = 0. This yields 0 = −1 + C so C = 1, and
y = − cos(3t)e−4t + e−4t .
Return
134
17.
ORDINARY DIFFERENTIAL EQUATIONS
The associated homogeneous equation is ty0 − 3y = 0, and we can
take yh = e3 ln t = t3 . Substitute y = vt3 to get v0 t4 = 5t3 , or v0 = 5t .
Thus v = 5 ln t + C Since y = vt3 , the general solution is the family
y = 5t3 ln t + Ct3 . To satisfy the initial condition, set y = 1 and t = 1.
This yields 1 = C, so y = 5t3 ln t + t3 .
Return
SOLUTIONS MANUAL
18.
135
To linearize,define a new variable v = y1−1/3 = y2/3 . Then
y = v3/2 , and y0 = 32 v1/2 v0 . Substitute this expression in the equation to obtain 32 v1/2 v0 = v1/2 ; or v0 = 2/3 the general solution is the
family v = 2/3t + C. Therefore y = (2/3t + C )3/2 . To satisfy the initial condition, set y = 1 and t = 1. This yields 1 = (2/3 + C )3/2 , so
c = 1/3 and y = (2/3t + 1/3)3/2 .
Return
136
19.
ORDINARY DIFFERENTIAL EQUATIONS
The associatedR homogeneous equation is y0 − (t + 3)y = 0, and
2
2
we can take yh = e (t+3) dt = et /2+3t . Substituting y = vet /2+3t yields
2
2
v0 = (2t + 6)e−t /2−3t . Thus v = −2e−t /2−3t + C, and the general
2
solution is y = −2 + Cet /2+3t . To satisfy the initial condition, set y =
2
0 and t = 0. This yields 0 = −2 + C, so C = 2 and y = −2 + 2et /2+3t .
Return
SOLUTIONS MANUAL
20.
137
The ODE is the same as in Exercise 19, so the general solution is
2
again y = −2 + Cet /2+3t . To satisfy the initial condition, set y = −2
and t = 0. This yields −2 = −2 + C, so C = 0 and y = −2.
Return
138
21.
ORDINARY DIFFERENTIAL EQUATIONS
2y
The associated homogeneous equation is y0 − 4−t2 = 0, and we
R 2
q
q
2+ t
dt
2+ t
2+ t
can take yh = e 4−t2 = e1/2 ln 2−t =
.
Substituting
y
=
v
2− t
2− t
q
√
2+ t
1
1
0
0
yields v 2−t = √2−t , and thus v = √2+t . Integrating, v = 2 2 + t +
q
t
√2+t
C. Since y = v 22+
−t , the general solution is the family y = 2 (2−t) +
q
√
t
C 22+
.
To
satisfy
the
initial
condition,
set
y
=
3
2 and t = 0. This
−t
q
√
√
√
t
yields 3 2 = 2 2 + C, so C = 2 and y = 2 √2+t + 2 22+
−t .
(2− t )
Return
SOLUTIONS MANUAL
139
22.
The associated homogeneous equation is y0 + ty = 0, and we can
t2
t2
t2
t2
take yh = e− 2 . Substitute y = ve− 2 and y0 = v0 e− 2 − tve− 2 in
R
t2
the inhomogeneous ODE to get v0 = (2t + t3 )e 2 . Thus v = (2t +
t2
t2
t2
t3 )e 2 dt = t2 e 2 + C Since y = ve− 2 , the general solution is the family
t2
y = t2 + Ce− 2 . To satisfy the initial condition, set y = 1 and t = 0.
t2
This yields 1 = C, so y = t2 + e− 2 .
Return
140
23.
ORDINARY DIFFERENTIAL EQUATIONS
Let y(t) denote the balance of the bank at time t, then y0 (t) = ky,
and y = Cekt (C is the balance at time 0). Substituting y(30) = 3C,
1
we obtain e30k = 3, and k = 30
ln 3. Thus y = C 3t/30 . To find the
2
doubling time, solve y(t) = 2C, or 3t/30 = 2. Thus t = 30 ln
ln 3 years.
Return
SOLUTIONS MANUAL
24.
141
Let y(t) denote the population at time t, We use the ODE, y0 =
0.013y, so y = Ce0.013t (C is the population at time 0) Solving y(t) =
ln 2
Ce0.013t = 2C, we obtain t = 0.013
= 53 years.
Return
142
25.
ORDINARY DIFFERENTIAL EQUATIONS
The fraction of C14 in wood from the tree, in ppb,is modelled
by the IVP y0 = −ky; y(0) = 1. The general solution for the ODE is
y = Ce−kt . Substituting t = 0, y = 1 in y = Ce−kt , we get C = 1.
Because C14 has a half life of 5730 years, e−5730k = 0.5, and therefore
ln(0.5)
k = −5730 = 0.000121. Solving y = e−0.000121t = 0.8 for t, we obtain
ln 0.8
t = −0.000121
= 1844 years.
Return
SOLUTIONS MANUAL
26.
143
The associated homogeneous equation is y0 + y = 0, and we can
take yh = e−t . Substitute y = ve−t yields v0 = et cos(4t). Thus v =
1 t
−t
17 e (cos(4t ) + 4 sin(4t )) + C. Since y = ve , the general solution is
1
the family y = 17
(cos(4t) + 4 sin(4t)) + Ce−t . The periodic solution
1
is y = 17 (cos(4t) + 4 sin(4t)), and when t → ∞, Ce−t → 0. Therefore
it is stable.
Return
144
27.
ORDINARY DIFFERENTIAL EQUATIONS
Let x (t) denote the amount (in kilograms) of salt in the tank. The
volume of brine in the tank starts at 1000 liters and increases at the
x (t)
rate of 100 liters per hour. Thus the concentration is C (t) = 1000+100t .
0
We will use the ODE, x (t) = JK − LC (t) as our model. Substituting J = 200, K = 0.050, L = 100, and using the initial condition
x (0) = 100, we get x 0 (t) + 101+t x (t) = 10, The homogeneous soluR
1
tion is e− 10+t dt = (10 + t)−1 . Now we substitute x = v(10 + t)−1
in the differential equation, and simplify to get v0 = 10(10 + t). Integration yields v = 100t + 5t2 + C, where C is a constant. Since
+5t2 +C
x (t) = v(10 + t)−1 , it follows that x (t) = 100t10
. Substituting
+t
x (0) = 100, we get C = 1000. Therefore x (t) =
100×10+5(10)2 +1000
x (10) =
10+10
125
the salt in the tank is 2000
per liter.
Return
100t+5t2 +1000
.
10+t
Then
= 125 kilograms, and the concentration of
= .0625 kilograms per liter, or 62.5 grams
SOLUTIONS MANUAL
145
28.
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-1
0
1
2
3
146
29.
ORDINARY DIFFERENTIAL EQUATIONS
Write the difference equation in recursive form as ym+1 = ym +
h × (t2m − y2m ) with y0 = 0, and h = −0.25 we have
y(−0.25) = 0 + (−0.25) × ((0)2 − (0)2 ) = 0
y(−0.5) = 0 + (−0.25) × ((−0.25)2 − (0)2 ) = −0.015625
y(−0.75) = −0.015625 + (−0.25) × ((−0.5)2 − (−0.015625)2 ) = −0.078064
y(−1) = −0.078064 + (−0.25) × ((−0.75)2 − (−0.078064)2 ) = −0.217165
y(−1.25) = −0.217165 + (−0.25) × ((−1)2 − (−0.217165)2 ) = −0.455375
y(−1.5) = −0.455375 + (−0.25) × ((−1.25)2 − (−0.455375)2 ) = −0.794159
y(−1.75) = −0.794159 + (−0.25) × ((−1.5)2 − (−0.794159)2 ) = −1.19899
y(−2) = −1.19899 + (−0.25) × ((−1.75)2 − (−1.19899)2 ) = −1.60522
y(−2.25) = −1.60522 + (−0.25) × ((−2)2 − (−1.60522)2 ) = −1.96104
y(−2.5) = −1.96104 + (−0.25) × ((−2.25)2 − (−1.96104)2 ) = −2.26525
y(−2.75) = −2.26525 + (−0.25) × ((−2.5)2 − (−2.26525)2 ) = −2.54491
y(−3) = −2.54491 + (−0.25) × ((−2.75)2 − (−2.54491)2 ) = −2.81639
with y0 = 0, and h = 0.25 we have
y(0.25) = 0 + (0.25) × ((0)2 − (0)2 ) = 0
y(0.5) = 0 + (0.25) × ((0.25)2 − (0)2 ) = 0.015625
y(0.75) = 0.015625 + (0.25) × ((0.5)2 − (0.015625)2 ) = 0.078064
y(1) = 0.078064 + (0.25) × ((0.75)2 − (0.078064)2 ) = 0.217165
y(1.25) = 0.217165 + (0.25) × ((1)2 − (0.217165)2 ) = 0.455375
y(1.5) = 0.455375 + (0.25) × ((1.25)2 − (0.455375)2 ) = 0.794159
y(1.75) = 0.794159 + (0.25) × ((1.5)2 − (0.794159)2 ) = 1.19899
y(2) = 1.19899 + (0.25) × ((1.75)2 − (1.19899)2 ) = 1.60522
y(2.25) = 1.60522 + (0.25) × ((2)2 − (1.60522)2 ) = 1.96104
y(2.5) = 1.96104 + (0.25) × ((2.25)2 − (1.96104)2 ) = 2.26525
y(2.75) = 2.26525 + (0.25) × ((2.5)2 − (2.26525)2 ) = 2.54491
y(3) = 2.54491 + (0.25) × ((2.75)2 − (2.54491)2 ) = 2.81639
SOLUTIONS MANUAL
147
1.5
1
0.5
0
-0.5
-1
-1.5
-2
Return
-1
0
1
2
3
148
30.
ORDINARY DIFFERENTIAL EQUATIONS
5
4
3
2
1
-3
-2
-1
0
1
2
3
4
SOLUTIONS MANUAL
31.
149
Write the difference equation in recursive form as ym+1 = ym +
√
h × (t − y) with y0 = 4, and h = 0.5 we have
√
y(−1.5) = 4 + (0.5) × (−2 − √4) = 2
y(−1) = 2 + (0.5) × (−1.5 − 2) = 0.542893
√
y(−0.5) = 0.542893 + (0.5) × (−1 − √0.542893) = −0.325513
y(0) = −0.325513 + (0.5) × (−0.5 − −0.325513) = ERROR
with y0 = 4, and h = −0.5 we have
√
y(−2.5) = 4 + (−0.5) × (−2 − √4) = 6
y(−3) = 6 + (−0.5) × (−2.5 − 6) =√8.47474
y(−3.5) = 8.47474 + (−0.5) × (−3 − √8.47474) = 11.4303
y(−4) = 11.4303 + (−0.5) × (−3.5 − 11.4303) = 14.8708
5
4
3
2
1
0
-3
Return
-2
-1
0
1
2
3
4
150
ORDINARY DIFFERENTIAL EQUATIONS
32.
(a) The direction field will have positive slope only when 5t > y2 .
Direction field IV has this property.
(b) This ODE has constant solutions y = kπ, for integers k. Direction
field III is the only one with this property.
(c) The direction field for this ODE must be periodic in t when y
is constant, and periodic in y when t is constant. It is therefore
direction field II.
(d) This autonomous ODE has the constant solution 0, which is stable. It therefore corresponds to direction field VI.
(e) This autonomous ODE has constant solutions y = 0 (unstable)
and y = π (stable). Therefore, the direction field is V.
(f) All solutions of this ODE are increasing; therefore all direction
field elements have nonnegative slope. Direction field I is the
only one with this property.
Return
SOLUTIONS MANUAL
33.
151
Since h(y) = y2 has a zero at y = 0, there is a singular solution,
dy
y ≡ 0. The separated equation is y2 = tdt. Integrating both sides, we
get the family of solutions satisfy − 1y =
2
,
C − t2
where C = −2C1 is a constant.
Return
t2
2
+ C1 , that is y = − t2 1
2
+C1
=
152
34.
ORDINARY DIFFERENTIAL EQUATIONS
The separated equation is e−y dy = dt. Integrating both sides, we
get the family of solutions satisfy −e−y = t + C, that is y = − ln(−t −
C ), where C is a constant.
Return
SOLUTIONS MANUAL
35.
153
Since h(y) = y(1 − y) has two zeros y = 0, y = 1, there are
two singular solutions: y ≡ 0 and y ≡ 1. The separated equation
dy
is y(1−y) = dt. Integrating both sides, we get the family of solutions
y in implicit form, ln y−1 = t + C1 . Solving for y yields the explicit
formula, y =
Return
Cet
1+Cet
where C = −eC1 is a constant.
154
36.
ORDINARY DIFFERENTIAL EQUATIONS
p
Since h(y) = 1 − y2 has two zeros y = ±1, there are two sindy
gular solutions: y = ±1. The separated equation is √ 2 = √ dt 2 .
1− y
1− t
Integrating both sides, we get the family of solutions in implicit form,
arcsin y = arcsin t + C. Solving for y yields the explicit formula, y =
sin(arcsin t + C ) where C is a constant.
Return
SOLUTIONS MANUAL
37.
155
p
Since h(y) = y2 − 1 has two zeros y = ±1, there are two sindy
gular solutions: y = ±1. The separated equation is √ 2
= √ dt2 .
y −1
t −1
Integrating
√ the family of solutionspin implicit form,
p both sides, we get
2 − 1| = ln | t +
ln
|
y
+
y
t2 − 1| + C1 . Thus y + y2 − 1 = C (t +
√
t2 − 1) where C = eC1 is a constant.
Return
156
38.
ORDINARY DIFFERENTIAL EQUATIONS
Since h(y) = (1 + y2 ) has no zeros. Thus there are no singudy
lar solutions. The separated equation is 1+y2 = 1tdt
. Integrating
+ t2
both
in implicit form, arctan y =
√ sides, we get the family of solutions
√
ln 1 + t2 + C. Thus y = tan(ln 1 + t2 + C ), where C is a constant.
Return
SOLUTIONS MANUAL
39.
Since h(y) =
157
√
y has a zero y = 0, there is a singular solution
y ≡ 0. The separated equation is
dy
√
y
=
dt
√
.
t
Integrating both sides, we
√
√
get the family of solutions in implicit form, 2 y = 2 t + C1 . Thus
√
y = ( t + C )2 , where C = C1 /2 is a constant.
Return
158
40.
ORDINARY DIFFERENTIAL EQUATIONS
Since h(y) = e4y has no zero, there are no singular solutions. The
separated equation is e−4y dy = tan(3t)dt. Integrating both sides, we
−4y
get the family of solutions in implicit form, − e 4 = − 31 ln | cos(3t)| +
C1 . Thus e−4y = 43 ln | cos(3t)| + C where C = −4C1 is a constant.
Return
SOLUTIONS MANUAL
159
Our model is the ODE mv0 = mg −
41.
m and simplify to obtain v0 =
mg
− kv2 . Divide through by
3.5
5g
kv2
−
. With g = 9.8 m/s we have
7
m
kv2
= 7, v0 = 7 −
. so By solving v0 = 0 for v we obtain the terminal
m
r
7m
m
= 21. Thus = 63.
velocity, v =
k
k
v2
0
With this parameter, our model is v = 7 1 − 2 . The separated
21
21 + v dv
=
= 7 dt. Integrating both sides, we have ln equation is
v2
21 − v 1 − 21
2
2
t + C. To evaluate C, set v(0) = 0: we find C = 0. In explicit form,
3
e2t/3 − 1
we have v = 21 2t/3
.
e
+1
5
7g
Z 10 2t/3
e
−1
dt = 210.08 − 43.6683 = 166.4.meters.
1 + e2t/3
(b) v(10) ≈ 21, thus T = 10 + (1463 − 166)/21 = 10 + 62 = 72
seconds.
(a) 21
0
(c) Suppose that you throw the diamond vertically into the lake so
that its initial velocity after it is immersed in water is 42 meters per second. Answer questions (a) and (b). We know that
21 + v
2
21 + v
ln |
| = t + C. Thus
= C1 e2t/3 , where C1 = ±eC .
21 − v
3
21 − v
(3e2t/3 + 1)
Substituting v(0) = 42, we obtain C1 = −3. Thus v = 21 2t/3
,
3e
−1
Z 10
2t/3
(3e
+ 1)
and the distance travelled in 10 seconds is 21
dt =
2t/3
3e
−1
0
236 meters. Also v(10) ≈ 21; thus T = 10 + (1463 − 236)/21 =
10 + 58.4 = 68.5 seconds.
Return
160
ORDINARY DIFFERENTIAL EQUATIONS
42.
(a) In year A the relative growth rate was
y0
= 4500/100,000 =
y
y0
= 8000/200,000 = 4%
y
y0
y
(b) According to the logistic equation,
= k 1−
. Substitute
y
M
100,000
y0
= .045 and y = 100,000 to obtain .045 = k(1 −
). With
y
M
y0
200,000
= 0.04 and y = 200,000 we get 0.04 = k(1 −
). Solving
y
M
these two equations, we obtain M = 1,000,000 and k = 0.05.
y
(c) The general solution of the logistic equation y0 = k y 1 −
M
M
1,000,000
is y =
. Thus y =
. Substituting y(0) =
1 + Ae−0.05t
1 + Ae−kt
1,000,000
100,000, we find A = 9. We can then solve 200,000 =
1 + 9e−0.05t
to obtain t = ln(9/4)/0.05 ≈ 16. Thus year B is 1980+16=1996.
4.5%, and in year B
Return
SOLUTIONS MANUAL
43.
161
The stationary points are y ≡ 0, y ≡ − 12 . The up intervals are
(−∞, − 12 ) and (0, +∞), and (− 21 , 0) is a down interval. Thus y = − 21
is stable. and y(t) → − 21 .
Return
- r
r
− 12
0
-
162
44.
ORDINARY DIFFERENTIAL EQUATIONS
The stationary points are y = kπ, where k is an integer. The up
intervals are (−(k + 1)π, −kπ ) for k ≥ 0, and the down intervals are
(kπ, (k + 1)π ). Thus y = 0 is stable. If y(0) = 30 then y(t) → 9π
because 30 belongs to the down interval (9π, 10π ).
- r - r r r r r r r r r r r
−π 0
Return
π 2π 3π 4π 5π 6π 7π 8π 9π 10π
SOLUTIONS MANUAL
45.
163
y0 = y2 + 2y + 2 = (y + 1)2 + 1 > 0 Thus (−∞, +∞) is up interval
and y(t) → +∞. There are no stationary points.
-
Return
164
46.
ORDINARY DIFFERENTIAL EQUATIONS
There is one stationary point, y ≡ 53 . (−∞, 35 ) is up interval, and
( 53 , +∞) is down interval. Thus y ≡ 35 is stable, and y(t) → 53 .
- r
5
3
Return
SOLUTIONS MANUAL
47.
165
There is one stationary point, y ≡ 1. (0, 1) is up interval, and
(−∞, 0) and (1, +∞) are down intervals. Thus y ≡ 1 is stable, and
y(t) → 1.
- r
0
Return
1
166
48.
ORDINARY DIFFERENTIAL EQUATIONS
The phase diagram is the same as in the previous exercise, but
now the initial point is in the down interval (−∞, 0). Therefore y(t) →
−∞.
Return
SOLUTIONS MANUAL
49.
Differentiate both sides of y0 =
167
1
y
y0
to obtain y00 = − y2 . Now replace
y0 with y−1 to obtain the equation y00 = −y−3 . Thus when y < 0 the
solution graphs are decreasing (y0 < 0) and concave up (y00 > 0), and
when y > 0 the solution graphs are increasing (y0 > 0) and concave
down (y00 < 0).
Return
168
50.
ORDINARY DIFFERENTIAL EQUATIONS
Differentiate both sides of y0 = t + y with respect to t to obtain
y00 = 1 + y0 . Replace y0 with t + y; thus y00 = 1 + t + y. In red region,
y00 > 0, and thus the solution graphs are concave up. In the blue
region: y00 < 0, and thus the solution graphs are concave down.
Return
SOLUTIONS MANUAL
51.
169
Since y0 = ey > 0 the graph of any solution is increasing. By
differentiation, we find y00 = y0 ey ; the substitution y0 = ey shows that
y00 = e2y > 0. Thus the graph of any solution is concave up.
Similarly, for the ODE y0 = e−y we have y0 > 0 and y00 = −e−y y0 =
−e−2y < 0. Thus all solution graphs are increasing and concave down.
Return
170
ORDINARY DIFFERENTIAL EQUATIONS
52.
(a) Write the difference equation in recursive form as ym+1 = ym +
h × (tm − y2m ). Thus, with y0 = 0, and h = 0.1 we have
y0.1 (0.1) = 0 + (0.1) × (0 − (0)2 ) = 0
y0.1 (0.2) = 0 + (0.1) × (0.1 − (0)2 ) = 0.01
y0.1 (0.3) = 0.01 + (0.1) × (0.2 − (0.01)2 ) = 0.02999
y0.1 (0.4) = 0.02999 + (0.1) × (0.3 − (0.02999)2 ) = 0.0599001
y0.1 (0.5) = 0.0599001 + (0.1) × (0.4 − (0.0599001)2 ) = 0.0995413
y0.1 (0.6) = 0.0995413 + (0.1) × (0.5 − (0.0995413)2 ) = 0.14855
y0.1 (0.7) = 0.14855 + (0.1) × (0.6 − (0.14855)2 ) = 0.206344
y0.1 (0.8) = 0.206344 + (0.1) × (0.7 − (0.206344)2 ) = 0.272086
y0.1 (0.9) = 0.272086 + (0.1) × (0.8 − (0.272086)2 ) = 0.344683
y0.1 (1) = 0.344683 + (0.1) × (0.9 − (0.344683)2 ) = 0.422802
Similarly, with y0 = 0, and h = 0.05 we have
y0.05 (0.05) = 0 + (0.05) × (0 − (0)2 ) = 0,
y0.05 (0.1) = 0 + (0.05) × (0.05 − (0)2 ) = 0.0025,
..
.
y0.05 (1) = 0.400229 + (0.05) × (0.95 − (0.400229)2 ) = 0.43972,
and with y0 = 0, and h = 0.025 we have
y0.025 (0.025) = 0 + (0.025) × (0 − (0)2 ) = 0
y0.025 (0.05) = 0 + (0.025) × (0.025 − (0)2 ) = 0.000625
y0.025 (0.075) = 0.000625 + (0.025) × (0.05 − (0.000625)2 ) = 0.00187499
y0.025 (0.1) = 0.00187499 + (0.025) × (0.075 − (0.00187499)2 ) =
0.0037499
..
.
y0.025 (1) = 0.427971 + (0.025) × (0.975 − (0.427971)2 ) = 0.447767.
(b) Multiply (2.46) by 2 and subtract from (2.45). Thus φ(1) ≈ 2Yh2 −
Yh1 . Thus Zh1 = 2Yh2 − Yh1 .
SOLUTIONS MANUAL
171
(c) Calculate Z0.1 and Z0.05 . Z0.1 = 2Y0.05 − Y0.1 = 2 × 0.43972 −
0.422802 = 0.456638.
Z0.05 = 2Y0.025 − Y0.05 = 2 × 0.447767 − 0.43972 = 0.455814.
(d) φ(1) ≈ Z0.1 + 0.01D, and φ(1) ≈ Z0.05 + 0.0025 D. Thus,
4( Z0.05 + 0.0025 D ) − ( Z0.1 + 0.01 D ) ≈ 3φ(1)
. Since the D terms cancel, we may divide through by 3 and
obtain
φ (1) ≈
Return
4 × 0.455814 − 0.456638
4Z0.05 − Z0.1
=
= 0.455539
3
3
172
53.
ORDINARY DIFFERENTIAL EQUATIONS
There will be a solution unless y0 = 1, since that is the only point
on the y-axis where the right side is not continuous. To apply the
uniqueness theorem, we need a Lipschitz condition; this precludes
y0 = −1.
Return
SOLUTIONS MANUAL
54.
173
By the uniqueness theorem, we only need to see that the right
side of the differential equation satisfies a Lipschitz condition; this in
turn can be shown by demonstrating that the partial derivative with
respect to y of the right side is continuous.
Return
174
55.
ORDINARY DIFFERENTIAL EQUATIONS
Substituting x = C et and y = C et in the first equation yields
the identity C et = C et . The same substitution in the second equation
results in C et = C et .
Substituting x = A e−t and y = B e−t in the first equation yields
− A e−t = B e−t . The same substitution in the second equation results in − B e−t = A e−t . Solve these equations for A and B to obtain
A = − B.
Return
SOLUTIONS MANUAL
57.
175
Substituting x = 2C cos(3t) and y = C [cos(3t) + 3 sin(3t)] in the
first equation yields the identity −6C sin(3t) = 2C cos(3t) − 2C [cos(3t) +
3 sin(3t)]. The same substitution in the second equation results in
C [−3 sin(3t) + 9 cos(3t)] = 10C cos(3t) − C [cos(3t) + 3 sin(3t)].
Substituting x = 2A sin(3t) in the first equation yields 6A cos(3t) =
2A sin(3t) − 2y. Thus if y = A sin(3t) − 3A cos(3t), the first equation is satisfied. Now substitute x = 2A sin(3t) and y = A sin(3t) −
3A cos(3t) in the second equation. This results in an identity, 3A cos(3t) +
9A sin(3t) = 10A sin(3t) − A sin(3t) + 3A cos(3t). Thus the solutions
have the form x = 2A sin(3t), y = A sin(3t) − 3A cos(3t).
Return
176
59.
ORDINARY DIFFERENTIAL EQUATIONS
Substituting x = e2t (t + 1) and y = e2t (t − 1) in the first equation
yields the identity 2e2t (t + 1) + e2t = e2t (t + 1) + e2t (t − 1) + 3e2t . The
same substitution in the second equation results in 2e2t (t − 1) + e2t =
2e2t (t + 1) − 3e2t .
Substituting x = C e2t and y = C e2t in the first equation of the “associated homogeneous system” yields the identity 2C e2t = C e2t +
C e2t . The same substitution in the second equation results in 2C e2t =
2C e2t .
A family of solutions of the inhomogeneous system is the sum of the
associated homogeneous solutions plus a particular solution. Thus
we obtain a family of solutions:5 x = e2t (t + 1 + C ), y = e2t (t − 1 + C )
Return
SOLUTIONS MANUAL
61.
177
The ODE can be written as
v0
=
d2 y
.
dt2
d2 y
dt2
2
= −3 dy
dt − 4y + t Set v =
Then the system
y0 = v
v0 = −3v − 4y + t2
replaces the given ODE.
Return
dy
dt
and
178
63.
ORDINARY DIFFERENTIAL EQUATIONS
The ODE can be written as
Then the system
d2 u
dt2
=
1+ t2
u du
dt
u0 = v
t2
v 0 = 1+
uv
replaces the given ODE.
Return
Set v =
du
dt
and v0 =
d2 u
.
dt2
SOLUTIONS MANUAL
65.
179
The first equation, x 0 = x2 , has a singular solution x ≡ 0, that
isn’t consistent with the second equation; thus it is extraneous.
Therefore we can separate the first equation and integrate to obtain
1
− x −1 = t + C, or explicitly, x = t−
+C . Substitute this in the second equation and obtain y0 = −t − C, which we integrate to get
y = − 12 t2 − Ct + D, where C, D are constants.
Return
180
67.
ORDINARY DIFFERENTIAL EQUATIONS
This system is not uncoupled.
SOLUTIONS MANUAL
69.
181
The first equation can be√
put in separated form as x dx = t dt.
2
2
Thus x = t + C, or x = ± √t2 + C. Substitute this in the second
equation to obtain y0 = ±y t t2 + C. This has a singular solution
p
dy
y ≡ 0, and the separated equation is
= ±t t2 + C. Integrating
y
1 2
3/2
both sides we obtain ln y = ± 13 (t2 + C )3/2 + D1 , or y = De± 3 (t +C) ,
where C and D = e D1 are constants. (The singular solution is in this
family.)
Return
182
71.
ORDINARY DIFFERENTIAL EQUATIONS
This system is not uncoupled.
SOLUTIONS MANUAL
73.
183
Substituting x = t2 and y = t in the first equation yields the identity 2t = 2t2 /t. The same substitution in the second equation results
in 1 = t2 /t2 .
Eliminate t in the solution x = t2 , y = t. we have x = y2 , and as
t → ∞, both x, y → ∞. This orbit is therefore the parabola x = y2
directed upward.
2
1
1
-1
-2
Return
2
3
4
184
75.
ORDINARY DIFFERENTIAL EQUATIONS
Substituting x = e−t and y = −e−t in the first equation yields the
identity −e−t = −e−t . The same substitution in the second equation
results in e−t = e−t .
Eliminate t in the solution x = e−t , y = −e−t to obtain y = − x. As
t → ∞, both x and y → 0. This orbit is therefore part of the line
y = − x in the fourth quadrant directed toward the origin.
1
-1
-2
-3
-4
-5
-6
Return
2
3
4
5
6
SOLUTIONS MANUAL
77.
185
d2 y
The ODE can be written as dt2 = y−3 Set v =
Then the system
y0 = v
v 0 = y −3
dy
dt
and v0 = y−3 .
replaces the given ODE.
√
√
t2 + 1 − t2 (t2 + 1)−1/2
t
If y = t2 + 1, then y0 = √
=
and y00 =
t2 + 1
t2 + 1
1
= y −3
2
(t + 1)3/2
√
The orbit of the solution corresponding to y = t2 + 1 is
p
t
2
t + 1, √
.
(y, v) =
t2 + 1
s
t2
y2 − 1
1
2
=
. Hence v = ± 1 − 2 .
To eliminate t, notice that v = 2
t +1
y2
y
As t → ∞, y → ∞ and v → 1, and when t → −∞, we have y → ∞
and v → −1. The orbit in the y, v-plane is therefore directed upward
and to the right in the first quadrant, and asymptotic to v = ±1.
1
0.5
1
-0.5
-1
Return
2
3
4
186
ORDINARY DIFFERENTIAL EQUATIONS
79.
d2 y
t dy
The ODE can be written as dt2 = 2dt y0 − y Set v =
Then the system
y0 = v
v0 = 12 tv − y
dy
dt
and v0 = y00 .
replaces the given ODE.
If y = t2 − 2, then y0 = 2t, and y00 = 2 = t2 − (t2 − 2) = 12 t y0 − y.
The orbit of the solution corresponding to y = t2 − 2 is (y, v) = (t2 −
2, 2t). Here we have v2 = 4y + 8, and as t → ∞, both y and v → ∞.
This orbit is therefore the parabola v2 = 4y + 8 directed upward.
4
2
-2
-1
1
-2
-4
Return
2
SOLUTIONS MANUAL
81.
187
By solving the equations 4y/16 = 0, − x/16 = 0 we get the stationary point (0, 0).
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
Return
-1
0
1
2
3
188
83.
ORDINARY DIFFERENTIAL EQUATIONS
By solving the equations ( x2 + 1)y/16 = 0, −2x/16 = 0 we get
the stationary point (0, 0).
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
Return
-1
0
1
2
3
SOLUTIONS MANUAL
85.
189
190
ORDINARY DIFFERENTIAL EQUATIONS
(a)
1
0.5
1
2
4
3
5
-0.5
-1
Graphs of x (t) = sin(5t) and y(t) = cos(5t).
1
0.5
-1
-0.5
0.5
1
-0.5
-1
Phase portrait
0.5
0
-1-0.5
1
0
-0.5
-1
6
4
2
0
Three-dimensional graph.
0.5
1
6
SOLUTIONS MANUAL
191
(b)
6
4
2
4
2
6
8
10
12
-2
-4
-6
√
√
√
√
√
Graphs of x (t) = 4(cos( 2t) + 2 sin( 2t)) and y(t) = 2 2 sin[( 2t).
2
1
-4
-6
-2
2
4
6
-1
-2
Phase portrait
2
0
-2
-5
0
5
10
5
0
Three-dimensional graph.
192
ORDINARY DIFFERENTIAL EQUATIONS
(c)
6
4
2
-5
-2.5
5
2.5
7.5
10
12.5
-2
-4
1
Graphs of x (t) = 12
(4 cos(t) + (−4 + 6t) cos(2t) − 3 sin(2t)) and
1
y(t) = − 3 ((2 + (−2 + 3 t) cos(t)) sin(t)).
6
4
2
-4
-2
2
4
6
-2
-4
Phase portrait
5 -5 -2.5
2.5
0
2.5
0
-2.5
10
5
0
-5
Three-dimensional graph.
5
SOLUTIONS MANUAL
193
(d)
1.5
1
0.5
-1
-2
1
2
3
-0.5
-1
Graphs of x (t) = 16 t3 and y(t) = 16 (3t2 − t3 ).
1.5
1
0.5
-1
1
2
3
-0.5
-1
Phase portrait
0
5
10
4
2
0
-2
1
0
-1
Three-dimensional graph.
Return
4
194
1.
ORDINARY DIFFERENTIAL EQUATIONS
Using MDEP with an initial stepsize of 0.1, I found that y(1) ≈
0.841471076. Your answer may be different, since other renditions of
the RKF45 algorithm may not have the same default error tolerance
settings that MDEP has. The correct value of sin(1), to nine decimal
places, is 0.8414709848.
Return
SOLUTIONS MANUAL
195
3.
(a) If we multiply the first equation by x, multiply the second by y
and add, we have
xx 0 + yy0 = e( x2 + y2 )( x2 + y2 − 1).
With r2 = x2 + y2 and rr 0 = xx 0 + yy0 we have
rr 0 = er2 (r2 − 1),
or r 0 = er (r2 − 1). This differential equation has three stationary
points, at 0 and ±1. The phase diagram depends on e:
- r
r - r
r - r
r
-
−1 0
1
−1 0
1
Phase diagram for e < 0
Phase diagram for e > 0
The stationary point 0 corresponds to the stationary point of the
system at the origin and the stationary point 1 corresponds to
the circular orbit r = 1. Since r ≥ 0, the stationary point −1 has
no significance. If e < 0, 0 is an unstable stationary point of
r 0 = er (r2 − 1) and 1 is stable; then the origin is unstable and the
the orbit converges to the unit circle as shown in the following
phase portrait.
1
0.5
-1
-0.5
0.5
1
-0.5
-1
If e > 0, the stability is reversed: the origin is stable, and the
circle is the limiting orbit for all non-constant orbits as t → −∞.
Now the two orbits look like this:
196
ORDINARY DIFFERENTIAL EQUATIONS
1.5
1
0.5
-2
-1.5
-1
-0.5
0.5
1
1.5
-0.5
-1
-1.5
(b)
i. With (e, A) = (−.25, 4), the orbit will spiral inward toward
and converge to the unit circle, which is stable because e <
0..
1.5
1
0.5
-1
1
2
3
-0.5
-1
ii. With (e, A) = (.25, .95) the orbit will spiral inward toward
and converge to the origin, which is stable, and away from
the circular orbit, which is unstable because e > 0.
4
SOLUTIONS MANUAL
197
1
0.5
-1
-0.5
0.5
1
-0.5
-1
iii. With (e, A) = (.25, 1.05) the orbit will spiral outward to infinity, and away from the circular orbit, which is unstable
because e > 0.
1
-1.5 -1 -0.5
0.5
-1
-2
-3
-4
Return
1
198
ORDINARY DIFFERENTIAL EQUATIONS
5.
300
200
100
-2
-1
1
-100
-200
-300
Return
2
SOLUTIONS MANUAL
199
7.
(a) Solve the ODE y0 = f (y)/100 with initial conditions y(0) =
−3, −2, −1, and 0, over the interval [−5, 5]. The zeros of f will
be the ordinates of the horizontal asymptotes of these solutions.
Notice that the convergence to the multiple roots is slow — can
you explain why?
-100
-50
50
100
-1
-2
-3
-4
(b) The roots are approximately 0.193, 1.027, 2.568, 4.900, 8.182, 12.734,
and 19.396. To locate them, start with a numerical solution of
y0 = f (y) × 10−4 with initial condition y(0) = 10. You will find
that the graph of the solution rapidly converges to two asymptotes: y → 8.18215 as t → −∞, and y → 12.7342 as t → +∞.
Now try the same experiment with initial condition y(0) = 4.
This locates two more asymptotes: y → 4.90035 as t → −∞,
and y → 2.56788 as t → +∞. Are there zeros between 4.90035
and 8.18215? To find out, find a solution with initial condition
between these values of y: we will take y(0) = 6. The solution
is a decreasing function converging to 8.19215 to the left, and
4.90035 to the right. Thus, there can be no zeroes between the
two values. Now try an initial condition y(0) = 1. This solution
converges to 1.02666 to the right, and 0.193044 to the left. Now
try a solution with y(0) = 0. The result is a decreasing function, converging to 0.193044 as t → −∞, and diverging to −∞
as t → ∞. We have located 6 of the seven zeros. To locate the
last one, try a solution with y(0) = 15. This solution will be decreasing, with convergence to 19.3957 to the left and 12.7342 to
200
ORDINARY DIFFERENTIAL EQUATIONS
the right. We have thus located the seventh zero.The following
graph shows the solutions just computed.
15
10
5
-4
Return
-2
2
4
SOLUTIONS MANUAL
9.
201
The graphs shown in (a) and (c) are stable periodic solutions; the
solution shown was drawn by an IVP solver using arbitrary initial
conditions y(0) = 0, and then displaying a segment of the graph after
the transients have decayed. There is a periodic solution in (b), but
it is not stable, so we have to go backwards: The solution shown is
part of the graph of a solution calculated on the interval −15 ≤ t ≤ 0,
with initial condition y(0) = 0. Only the portion for −15 ≤ t ≤ −13
is displayed.
(a)
0.4
0.2
92
94
96
98
100
-13.5
-13
-0.2
-0.4
(b)
0.04
0.02
-14.5
-0.02
-0.04
(c)
-14
202
ORDINARY DIFFERENTIAL EQUATIONS
0.01
0.005
0.8
0.85
0.9
0.95
-0.005
-0.01
Return
SOLUTIONS MANUAL
1.
203
The x-nullcline is given by the equation x2 + y2 = 0. Thus the
x-nullcline are the point (0, 0).
The y-nullclines are determined by the equation x2 − y2 = 0, and
hence they are the lines y = ± x.
The stationary points lie at the intersection of x-nullclines with ynullclines. Thus there is only one stationary point, (0, 0).
10
7.5
5
2.5
-10 -7.5
-5
-2.5
-2.5
-5
-7.5
-10
Return
2.5
5
7.5
10
204
3.
ORDINARY DIFFERENTIAL EQUATIONS
The x-nullcline is given by the equation − x = 0, hence it is the
y-axis.
The y-nullclines are determined by the equation 4y − y2 = 0, and
hence they are the line y = 4 and the x-axis.
The stationary points lie at the intersection of x-nullclines with ynullclines. Thus there are two stationary points, (0, 0) and (0, 4).
6
4
2
-3
-2
-1
1
-2
Return
2
3
SOLUTIONS MANUAL
5.
205
The x-nullcline is given by the equation x2 = 0; hence it is the
y-axis.
The y-nullcline is determined by the equation y = 0 and hence it is
the x-axis.
The stationary points lie at the intersection of x-nullclines with ynullclines. Thus the stationary points are: (0, 0).
4
3
2
1
-4
-3
-2
-1
1
-1
-2
-3
-4
Return
2
3
4
206
7.
ORDINARY DIFFERENTIAL EQUATIONS
The x-nullcline is given by the equation x + y − 2 = 0, hence it is
the line y = 2 − x.
The y-nullcline is determined by the equation x − y = 0 and hence it
is the line y = x.
The stationary points lie at the intersection of x-nullclines with ynullclines. Thus there is one stationary point, (1, 1).
4
3
2
1
-4
-3
-2
-1
1
-1
-2
-3
-4
Return
2
3
4
SOLUTIONS MANUAL
9.
207
We will find an integral of the system by finding an integral for
the corresponding ODE ( x + y) dx − 2x dy = 0,
This ODE can be written as
2x
dy
− y = x,
dx
and hence is linear. The general solution of the homogeneous equady
tion 2x
− y = 0 will have the form y = C eK(x) , where 2x K 0 ( x ) =
dx
1. It is easily seen that K ( x ) = 12 ln( x ) so y = C eln(x)/2 = C x1/2
satisfies the homogeneous equation. Let’s substitute y = x1/2 v( x )
dv
in the inhomogeneous equation; we obtain 2x3/2
= x, and thus
dx
1/2
v( x ) = x
+ C. It follows that the general solution of out ODE is
1/2
y = x + C x . Divide through by x1/2 to obtain an integral:
F ( x, y) = x −1/2 (y − x ).
Return
208
11.
ORDINARY DIFFERENTIAL EQUATIONS
We will find an integral of the system by finding an integral for
the corresponding ODE 5y dx − 3x dy = 0, which is separable. Divide through by xy to obtain (5/x ) dx − (3/y) dy = 0. The separated
equation is thus (5/x ) dx = (3/y) dy. Integrating both sides we get
5 ln x = 3 ln y + C, Thus F ( x, y) = 5 ln x − 3 ln y is an integral.
Return
SOLUTIONS MANUAL
13.
209
We will find an integral of the system by finding an integral for
the corresponding ODE 2xy dx + ( x2 + y2 ) dy = 0,
Because
∂P
∂y ( x, y )
= 2x =
F ( x, y) =
Z
∂Q
∂x ( x, y ),
the equation is exact. Set
2xy dx + K (y) = x2 y + K (y).
The function K (y) satisfies
∂F
∂y
= x 2 + K 0 ( y ),
Thus K 0 (y) = y2 . A final integration shows that K (y) =
F ( x, y) =
Return
x2 y
+
y3
3.
y3
3
and so
210
15.
ORDINARY DIFFERENTIAL EQUATIONS
The system is uncoupled. The solution of the first equation, x 0 =
1, with x (0) = x0 is x = t + x0 . Thus the second equation can be
dy
written dx = f ( x, y); the solution of the system that passes through
( x0 , y0 ) when t = 0 would satisfy the initial condition y(0) = y0 , and
since x = t + x0 the solution follows the graph of the solution of the
IVP
dy
= f ( x, y); y( x0 ) = y0 .
dx
Return
SOLUTIONS MANUAL
17.
Set y =
dx
dt
211
and y0 =
d2 x
.
dt2
Then the system
x 0 = y; y(t0 ) = y0
y0 =
f (t, x, y); x (t0 ) = x0
replaces the given ODE. Let g(t, x, y) = y, then f (t, x, y) and g(t, x, y)
satisfy Lipschitz conditions. By the existence and uniqueness theorems there is a unique solution of the system. Thus there is a unique
solution of the IVP.
Return
212
19.
ORDINARY DIFFERENTIAL EQUATIONS
The system is equivalent to
x0 =
y0 =
1
( ag(t, x, y) − c f (t, x, y))
ad − bc
1
(d f (t, x, y) − bg(t, x, y))
ad − bc
If the functions f (t, x, y), g(t, x, y) are continuous and satisfy a Lip1
schitz condition then, provided that ad − bc 6= 0, ad−
bc ( ag ( t, x, y ) −
1
c f (t, x, y)) and ad−bc (d f (t, x, y) − bg(t, x, y)) are continuous and satisfy a Lipschitz condition. Thus by the existence and uniqueness theorems the system has a unique solution with the initial conditions.
Return
SOLUTIONS MANUAL
213
1.
(a) For the Lotka Volterra equation the stable point is (d, ba ), Thus
the optimum number of predators to release is ba .
(b) The strategy is ineffective. For the solutions of the Lotka Volterra
equation are periodic functions of time with the same average
population d for pests and ba for predators. If the number of prey
is initially small, it will become even bigger at some other time.
(c) Wait until there are ≈ d pests.
Return
214
3.
ORDINARY DIFFERENTIAL EQUATIONS
Let x and y denote the populations of species A and B, respectively. The x-nullclines are given by the line x = 0, and line determined by the points (1000, 0) and (800, 800). The equation of this
x-nullcline is therefore 4x + y = 4000. It follows that x 0 = k1 x (4000 −
4x − y).
The y-nullclines are given by the line y = 0, and line determined by
the points (0, 1500) and (800, 800), which has the equation 7x + 8y =
12000. Hence y0 = k2 y(12000 − 7x − 8y).
Thus the fish populations are governed by the system
x 0 = k1 x (4000 − 4x − y)
y0 = k2 y(12000 − 7x − 8y)
The phase portrait has the configuration (d) of Figure 3.19. The triangles ABC and CDE in the figure below are traps, and orbits within
these triangles converge to the point C = (800, 800).
4000
A
3000
2000
B
1000
C
D
500
Return
1000
E
1500
2000
SOLUTIONS MANUAL
5.
215
The x-nullclines are the lines x = 0 and K − x + By = 0. The ynullclines are the lines y = 0 and L + Cx − y = 0. The system has at
most four stationary points. Three of these occur when one or both of
the species are extinct. The fourth stationary point, ( x1 , y1 ) if it exists,
is located at the intersection of the lines K − x + By = 0, and L +
Cx − y = 0. If BC = 1 then the two lines parallel, and there are only
three stationary points. If BC 6= 1 we solve the equations to obtain a
+K
CK + L
fourth intersection point ( x1 , y1 ), where x1 = BL
1− BC and y1 = 1− BC .
If BC > 1 ( x1 , y1 ) is in the third quadrant, and thus this stationary
point is extraneous. If BC < 1 we have the fourth stationary point in
the first quadrant. This configuration is shown in the drawing. The
quadrilateral whose vertices are the four stationary points is a trap,
and the dihedral with vertex at the stationary point interior to the
first quadrant, and which is opposite to the quadrilateral, is also a
trap. Orbits in these traps are directed toward that stationary point,
which is therefore stable.
Thus. if BC < 1, the model indicates a stable population for each
species. In the case BC ≥ 1 the populations will increase without
bound.
2000
1500
1000
500
500
Return
1000
1500
2000
216
7.
ORDINARY DIFFERENTIAL EQUATIONS
The system with fishing can be written as Lotka-Volterra equations:
x 0 = x [( a − R) − by]
R
y0 = cy x − d +
.
c
By the result of Exercise 6, the average populations will be x̄ = d + Rc
and ȳ = a−b R . Thus the average prey population increases proportionally to the catch rate, and the average predator population decreases.
Return
SOLUTIONS MANUAL
1.
217
This system is uncoupled, because the first equation does not
involve y. This equation s linear, and the general solution of its associated homogenous equation, x 0 = x, is x = Aet . Substitute x = vet
in the inhomogeneous equation to get v0 = 1. Thus v = t + C and
the general solution of the first equation is x = (t + C )et . To satisfy
the initial condition, substitute x (0) = 0. Solving, we get C = 0. Thus
x = tet . Substitute this expression for x in the second equation to
obtain y0 = tet y2 . This equation is separable. There is a singular solution y = 0, which does not satisfy the initial condition, y(0) = 1. The
separated equation is y−2 dy = tet dt. Integrating both sides we have
−y−1 = et (t − 1) + C. Substituting y(0) = 1, we get C = 0. Therefore
we get
1
x = tet , y = t
.
e (1 − t )
Return
218
2.
ORDINARY DIFFERENTIAL EQUATIONS
This system is uncoupled. The general solution of first equation
is x = Ce−t , and the general solution of the second is y = Det , where
C and D are constants.
We will find an integral of the system by finding an integral for the
corresponding ODE y dx + x dy = 0, which is separable. Divide through
dx
dy
by xy to get
= − . Integrating both sides we have ln | x | =
x
y
− ln |y| + C1 . An equivalent form of this equation is ln | xy| = C1 . This
shows that ln | xy| is an integral, but we can apply the exponential and
obtain a simpler integral, F ( x, y) = xy.
The orbits of this system will be hyperbolas asymptotic to the x- and
y-axes, directed so as to approach the y-axis as t → ∞, and to approach the X-axis as t → −∞.
Return
SOLUTIONS MANUAL
3.
219
This equation is also uncoupled. The general solution of the first
equation is x = Cet , and the general solution of the second is y =
De2t , where C, D are constants.
We will find an integral of the system by finding an integral for the
corresponding ODE 2y dx − x dy = 0, which is separable. Divide
dy
through by xy to get 2dx
x = y , and integrate both sides. Thus 2 ln | x | =
ln |y| + C1 , or ln | x2 /y| = C1 . Apply the exponential to obtain a simple integral, F ( x, y) = x2 /y.
orbits will be halves of parabolas y = Ax2 , directed away from the
stationary point at the origin.
Return
220
4.
ORDINARY DIFFERENTIAL EQUATIONS
Let v = y0 ; then the system y0 = v, v0 = − f (y) replaces the given
ODE. We will find an integral of the system by finding an integral for
the corresponding ODE, v dv + f (y) dy = 0, which is separable. We
v2
obtain F (y, v) =
+ F ( y ).
2
(a) Here f (y) = α2 sin(y), and we could take F (y) = −α2 cos(y).
2
Thus the integral is F (y, v) = v2 − α2 cos(y).
(b) Here f (y) = α2 y), and we could take F (y) =
integral is F (y, v) =
v2
2
+
α2 y2
2 .
α2 y2
2 .
Thus the
(c) Here f (y) = α2 y−2 , and we could take F (y) = −α2 y−1 . Thus the
2
integral is F (y, v) = v2 − α2 y−1 .
Return
SOLUTIONS MANUAL
221
5.
(a) Let y0 = v; then the system y0 = v, v0 = −y replaces the given
ODE.
(b) Let y0 = v; then the system y0 = v, v0 = v − t2 sin(y) replaces the
given ODE.
Return
222
6.
ORDINARY DIFFERENTIAL EQUATIONS
Let y0 = v then the system y0 = v, v0 = 2v − 2y replace the given
ODE. The corresponding orbit is (y, v) = (et cos(t), et cos t − et sin t).
Return
SOLUTIONS MANUAL
7.
223
We will find an integral of the system by finding an integral for
the corresponding ODE 2y( x + y) dx + x ( x + 4y) dy = 0,
= 2x + 4y and ∂Q
is exact.
∂x ( x, y ) = 2x + 4y so the equation
R
The integral F ( x, y) will have the form F ( x, y) = 2y( x + y) dx +
H (y) = yx2 + 2y2 x + H (y). To determine H (y), differentiate this exdH
∂F
2
pression with respect to y: ∂F
∂y = x + 4yx + dy . Since ∂y = Q ( x, y ) =
∂P
∂y ( x, y )
x2 + 4yx, it follows that
2y2 x.
Return
dH
dy
= 0. and the integral is F ( x, y) = yx2 +
224
ORDINARY DIFFERENTIAL EQUATIONS
8.
(a) Solve the equations: x + y − 2 = 0, x − 3y + 2 = 0 we get the
stationary point (1, 1).
(b) Solve the equations: y − x = 0, y − x3 = 0 we get the stationary
points (0, 0), (−1, −1), (1, 1).
(c) Solve the equations: x ( x + y + 4) = 0, y( x + 5y) = 0 we get the
stationary points (0, 0), (−4, 0), (−5, 1).
Return
SOLUTIONS MANUAL
225
9.
(a) The system of ODEs corresponding to the vector fields is x 0 =
x + y − 2, y0 = x − 3y + 2. The x-nullclines are the line x + y −
2 = 0, and the y-nullclines are the line x − 3y + 2 = 0,
(b) ~v = (y − x ) i + (y − x3 ) j. The system of ODEs corresponding to
the vector fields is x 0 = y − x, y0 = y − x3 . The x-nullclines are
the line y − x = 0, and the y-nullclines are the line y − x3 = 0,
(c) ~v = x ( x + y + 4) i + y( x + 5y) j. The system of ODEs corresponding to the vector fields is x 0 = x ( x + y + 4), y0 = y( x + 5y). The
x-nullclines are the lines x = 0, ( x + y + 4) = 0. The y-nullclines
are the line y = 0, ( x + 5y) = 0.
226
ORDINARY DIFFERENTIAL EQUATIONS
Return
SOLUTIONS MANUAL
10.
227
The x-nullclines are the lines x + y = 0 and y = ±1. The ynullclines are the line x − y = 0, and x = ±1.
Return
228
ORDINARY DIFFERENTIAL EQUATIONS
11.
(a)
(b)
(c)
SOLUTIONS MANUAL
Return
229
230
12.
ORDINARY DIFFERENTIAL EQUATIONS
Both the linear and nonlinear versions are shown: the linear version is in black; the nonlinear version in blue. The second figure is
an enlargement of the region inside the rectangle on left, and shows
more detail.
0.2
0.1
5
-0.1
-0.2
Return
10
15
20
SOLUTIONS MANUAL
1.
231
If P is a p × q matrix and M is an m × n matrix then PM is defined
iff q = m; then it is a p × n matrix. Thus BA is a 2 × 2 matrix, and
AB is a 3 × 3 matrix. A vector ~u ∈ Rd is a d × 1 matrix A~v is a 2 × 1
~ ∈ R3 . As AB is a 3 × 3
matrix (that is, a vector in R2 ; similarly Bw
~ ∈ R2 , and BA~v ∈ R3 .
matrix and BA is 2 × 2, the products ABw
Return
232
ORDINARY DIFFERENTIAL EQUATIONS
3.
(a) The matrix E, when multiplied by A on the right, operates on A
by multiplying the first row of A by 1 and the second row by 0.
(b) When E is multiplied by A on the left, it results in multiplying
column 1 of A by 1, and column 2 of A by 0.
(c) If EA = AE then the corresponding entries of each matrix must
be equal. In row 2, column 1, the entry of EA is 0; the corresponding entry in AE is c. Thus c = 0. The entries of row 1, column 2 of EA and AE are b and 0, respectively. Therefore b = 0
as well. The
matrices that commute with E are the diagonal
only a 0
matrices
.
0 d
Return
SOLUTIONS MANUAL
5.
233
x
3x − 2y
Let ~v =
. Then A~v =
. Thus the matrix equay
x + 7y
0 x
tion
= A~v is equivalent to the given system.
y0
Return
234
ORDINARY DIFFERENTIAL EQUATIONS
7.
The system that replaces the given ODE is
x0 = y
y0 = z
z0 = 2et x − ty + (sin t)z + tan t,
or
 

y
x0
.
 y0  = 
z
t
0
2e x − ty + (sin t)z + tan t
z




x
0
1
0
0
1 ,
We can recast this as ~v0 = A~v +~b where ~v =  x 0 , A =  0
t
00
2e −t sin t
x


0
and ~b =  0 .
tan t
Return

SOLUTIONS MANUAL
235
9.
~v10 (t)
e3t
=
3e3t
1+2
2+1
1
1
=
e3t
3
3
1 2
2 1
, and
~v1 =
e3t
1 2
2 1
1
1
=
; thus ~v1 (t) is a solution.
~v2 (t) is also a solution
1 2
~v2 (t).
to
2 1
because ~v20 (t)
=
− e−t
1
−1
, which is equal
We must verify that {~v1 (t), ~v2 (t)} is linearly independent. This is
1
done by noticing that ~v1 (0) =
is not a scalar multiple of ~v2 (0) =
1
1
. Therefore, the general solution of the system is of the form:
−1
x
y
= C1 e
where C1 , C2 are constants.
Return
3t
1
1
+ C2 e
−t
1
−1
236
11.
ORDINARY DIFFERENTIAL EQUATIONS
x 0 = 2x, thus x = Ce2t . Substituting x (0) = 1 in we have C = 1.
3t
y0 = 3y, thus
y(0) = 2 in we have D = 2.
y = De 2t. Substituting
x
e
=
.
Therefore,
y
2e3t
Return
SOLUTIONS MANUAL
237
13.
e3t
1
1
+
From Exercise 1 we know the general solution is ~v(t) = C1
1
C2 e−t
, and we need to determine values for the coefficients so
−1
2
1
1
2
that ~v(0) =
. We need to solve C1
+ C2
=
,
0
1
−1
0
which is equivalent to the pair of equations
C1 + C2 = 2
C1 − C2 = 0
3t
x (t)
e + e−t
Thus C1 = C2 = 1, and
=
y(t)
e3t − e−t
Return
238
15.
ORDINARY DIFFERENTIAL EQUATIONS
It follows from the first equation, x 0 = 0, that x = C. Since x (0) =
1, we have C = 1. Thus the second equation can be recast as y0 =
−1 + y. The solution of the associated homogeneous equation is yh =
Cet . Substituting y = vet in the inhomogeneous equation and using
variation of constants, we have v0 = −e−t . Thus v = e−t + D, and y =
1 + Det . The initial condition y(0) = 1 impliesthat D =
0. Therefore
x (t)
1
=
.
the solution of the IVP is the constant,
y(t)
1
Return
SOLUTIONS MANUAL
239
17. In matrix
form,this
IVPis:
x0
2
3
x
x (0)
−1
3
t
=
;
=
. Let ~v1 (t) = e
and ~v2 (t) =
y0
−1 −2
y
y (0)
1
−1
2 3
3
3
2 3
1
t
t
0
−
t
~v1
=e
=
then ~v1 (t) = e
e
−1 −2
−1
−1
−1 −2
−1
2 3
1
1
2 3
−
t
−
t
0
~v2
~v2 (t) = −e
=e
=
−1 −2
−1
−1
−1 −2
3
1
Since ~v1 (0) =
is not a scalar multiple of ~v2 (0) =
, it
−1
−1
follows that ~v1 , ~v2 are independent solutions of the system and the
given family of solutions in the general solution.
For the IVP we need to solve
3
1
−1
c1
+ c2
=
,
−1
−1
1
which is equivalent to the pair of equations
3c1 + c2 = −1
− c1 − c2 = 1
x
1
−
t
Thus c1 = 0, c2 = −1. The solution is
= −e
.
y
−1
Return
240
ORDINARY DIFFERENTIAL EQUATIONS
19. In matrix
form,this
IVPis:
4
3
x
x (0)
7
1
t
=
;
=
. Let ~v1 (t) = e
and ~v2 (t) =
−2 −1
y
y (0)
−2
−1
4 3
1
1
4 3
3
t
t
0
2t
~v1
=e
=
then ~v1 (t) = e
e
−2 −1
−1
−1
−2 −1
−2
4 3
3
3
4 3
−
5t
2t
0
~v2
~v2 (t) = 2e
=e
=
−2 −1
−2
−2
−2 −1
1
3
Since ~v1 (0) =
is not a scalar multiple of ~v2 (0) =
, it
−1
−2
follows that ~v1 , ~v2 are independent solutions of the system and the
given family of solutions in the general solution.
x0
y0
For the IVP we need to solve
1
3
7
c1
+ c2
=
,
−1
−2
−2
which is equivalent to the pair of equations
c1 + 3c2 = 7
−c1 − 2c2 = −2
x
1
3
t
2t
Thus c1 = −8, c2 = 5. The solution is
= −8e
+ 5e
.
y
−1
−2
Return
SOLUTIONS MANUAL
21.
241
3
1
and ~v2 (0) =
are linearly independent;
−1
−1
thus the solutions are linearly independent.
Return
~v1 (0) =
242
ORDINARY DIFFERENTIAL EQUATIONS
23. x0
y0
In matrix
form,this
IVP
is: 2 −1
x
0
x (0)
1
=
+
;
=
.
4 −2
y
2
y (0)
−1
For the IVP we need to solve
0
1
1
1
+ c1
+ c2
=
,
0
2
0
−1
which is equivalent to the pair of equations
c1 + c2 = 1
2c1 = −1
− t2
1
2t + 1
3
3
1
1
+2
=
−2
Thus c1 = − 2 , c2 = 2 . The solution is
2
4t
2t − 2t2
−t2 + 3t + 1
.
−2t2 + 8t + 1
Return
SOLUTIONS MANUAL
25. x0
y0
243
In matrix
form,this
IVP is:
0 1
x
cosh(t)
x (0)
2
=
+
;
=
.
1 0
y
sinh(t)
y (0)
2
For the IVP we need to solve
1
0
2
+ c2
=
,
( c1 + 0)
0
1
2
which is equivalent to the pair of equations
c1 = 2
c2 = 2
x
cosh t
Thus c1 = 2, c2 = 2. The solution is
= (2 + t )
+
y
sinh t
t sinh t
cosh(t)
e
2
=t
+2
.
cosh t
sinh(t)
et
Return
244
27.
ORDINARY DIFFERENTIAL EQUATIONS
Let v = y0 . Then the linear second-order ODE (4.6) can be replaced with a linear system of first-order ODEs: y0 = v, v0 = −q(t)y −
p(t)v + r (t). The system corresponding to (4.6) is
0 y
0
1
y
0
=
+
.
v0
−q(t) − p(t)
v
r (t)
Return
SOLUTIONS MANUAL
1.
245
Since tr ( A) = 1 + (−4) = −3 and det( A) = (1)(−4) − (0)(0) =
−4, the characteristic equation of A is s2 + 3s − 4 = 0 and it follows
that the eigenvalues are s = −4 and s = 1.
~
~
The eigenvectors
belongingto s = −4 satisfy
the
equation Ab = −4b.
h
h
−4h
. Then A~b =
=
. Hence h = −4h and
Set ~b =
k
−4k
−4k
−4k = −4k. these equations
reduce to h = 0, so any vector that is a
0
scalar multiple of ~b1 =
is an eigenvector belonging to −4.
1
To
to s = 1, we have to solve A~b =
find an eigenvector
belonging
1 0
h
h
=
. This is equivalent to solving the equations
0 −4
k
k
h = h, −4k = k These equations
reduce to k = 0, so it follows that any
1
scalar multiple of ~b2 =
is an eigenvector belonging to 1.
0
Return
246
3.
ORDINARY DIFFERENTIAL EQUATIONS
Since tr ( A) = 1 + (−4) = −3 and det( A) = (1)(−4) − (0)(1) =
−4, the characteristic equation of A is s2 + 3s − 4 = 0 and it follows
that the eigenvalues are s = −4 and s = 1.
~
~
The eigenvectors
belonging to s = −4 satisfy
theequation Ab = −4b.
h
h+k
−4h
. Then A~b =
=
. Hence 5h + k = 0
Set ~b =
k
−4k
−4k
and −4k = −4k. these equations reduce
to 5h + k = 0, so any vector
1
that is a scalar multiple of ~b1 =
is a eigenvector belonging to
−5
−4.
To
to s = 1, we have to solve A~b =
find an eigenvector
belonging
1 1
h
h
=
. This is equivalent to solving the equations
0 −4
k
k
h + k = h, −4k = k These equations
reduce to k = 0, so it follows that
1
any scalar multiple of ~b2 =
is an eigenvector belonging to 1.
0
Return
SOLUTIONS MANUAL
5.
247
Since tr ( A) = 1 + (−4) = 6 and det( A) = (2)(4) − (−3)(3) =
17, the characteristic equation of A is s2 − 6s + 17 = 0. The discriminant of this equation is equal to (−6)2 − 4 × 17 = −32 < 0, and thus
it has no real roots. The matrix has no real eigenvalues or vectors.
Return
248
ORDINARY DIFFERENTIAL EQUATIONS
7.
(a) If either b 6= 0 or r 6= a, then ~k =
b
r−a
is a nonzero vector,
and
A~k =
a b
c d
b
r−a
=
ab + rb − ab
bc + dr − ad
=
rb
bc + dr − ad
Since r is an eigenvalue of A, we have r2 − ( a + d)r + ad − bc = 0.
Thus bc + dr − ad = r2 − ar, and it follows that
rb
b
~
Ak =
=r
= r~k.
r2 − ra
r−a
Thus ~k is an eigenvector belonging to r.
(b) If b = 0 the characteristic equation is s2 − ( a + d)s + ad = 0,
and thus s = a, d are eigenvalues. If either c 6= 0 or a 6= d, then
~k = a − d is a nonzero vector, and
c
A~k =
a 0
c d
a−d
c
a2 − ad
=
ac − dc + dc
2
a − ad
a−d
=
=a
= a~k.
ac
c
Thus ~k is an eigenvector belonging to a.
(c) If a = d and b = c = 0, then A = aI is a scalar matrix. It has one
eigenvalue, s = a, and every nonzero vector is an eigenvector.
Return
SOLUTIONS MANUAL
9.
249
3 2
so that our system is equivalent to ~v0 = A~v.
1 2
2
Since tr ( A) = 5 and det( A) = 4, the characteristic equation
is s −
h
be an
5s + 4 = 0 and the eigenvalues are 1 and 4. Let ~b1 =
k
eigenvector belonging to 1. Then 3h + 2k = h, h+ 2k =
k. These equa−
1
tions reduce to h = −k, so we can take ~b1 =
Now suppose
1
h
that ~b2 =
is an eigenvector belonging to 4. Then 3h + 2k =
k
2
~
4h, h + 2k = 4k, so that h = 2k. Hence we can set b2 =
. The
1
general solution to the matrix equation is
−1
2
t
4t
~v = c1 e
+ c2 e
.
1
1
Let A =
Return
250
ORDINARY DIFFERENTIAL EQUATIONS
11.
0 1
so that our system is equivalent to ~v0 = A~v.
0 0
Since tr ( A) = 0 and det( A) = 0, the characteristic equation is s2 = 0,
which has the double root 0.
Let A =
Since A 6= 0I, the
A does not have two independent eigenvec matrix
0
tors. . Put ~c =
. It is easy to see that A~c 6= 0c; that is, ~c is not an
1
eigenvector. Set
~b = A~c − (0)~c = 1 .
0
Then ~v1 (t) = e0t~b and ~v2 (t) = e0t (t~b +~c) are independent solutions.
The general solution is
1
t
~v = c1
+ c2
0
1
Return
SOLUTIONS MANUAL
13.
251
x
1 2
and A =
so that our system is equivalent
y
2 1
to v0 = A~v. Since tr ( A) = 2 and det( A) = −3, the characteristic
2
equation
is
s − 2s − 3 = 0 and the eigenvalues are 3 and −1. Let
~b1 = h be an eigenvector belonging to 3. Then h + 2k = 3h, 2h +
k
1
~
k = 3k. These equations reduce to h = k, so we can take b1 =
1
h
Now suppose that ~b2 =
is an eigenvector belonging to −1.
k
Then h+ 2k = −h, 2h + k = −k, so that h = −k. Hence we can set
1
~b2 =
. The general solution to the matrix equation is ~v =
−1
1
1
3t
−
t
c1 e
+ c2 e
. or x = c1 e3t + c2 e−t , y = c1 e3t − c2 e−t .
1
−1
Return
Let ~v =
252
ORDINARY DIFFERENTIAL EQUATIONS
15.
−1 1
so that our system is equivalent to ~v0 =
−1 −3
A~v. Since tr ( A) = −4 and det( A) = 4, the characteristic equation is
s2 + 4s + 4 = 0 and which has the double root −2.
Let A =
Since A 6= −2I, the
A does not have two independent eigen matrix
1
. It is easy to see that A~c 6= −2c; that is, ~c is not
vectors. Put ~c =
0
an eigenvector. Set
−
1
2
1
~b = A~c − (−2)~c =
+
=
.
−1
0
−1
Then ~v1 (t) = e−2t~b and ~v2 (t) = e−2t (t~b +
solu ~c) are
independent
1
t
+
1
−
2t
−
2t
tions. The general solution is ~v = c1 e
+ c2 e
.
−1
−t
0
Substituting ~v(0) =
, we have c1 + c2 = 0, −c1 = 1. Thus
1
c1 = −1, c2 = 1. Therefore the solution of the IVP is
1
t+1
t
−2t
−2t
−2t
~v = −e
+e
=e
.
−1
−t
1−t
Return
SOLUTIONS MANUAL
253
17.
5 −2
so that our system is equivalent to ~v0 =
2 1
A~v. Since tr ( A) = 6 and det( A) = 9, the characteristic equation is
s2 − 6s + 9 = 0 and which has the double root 3.
Let A =
Since A 6= 3I, the
A does not have two independent eigenvec matrix
1
. It is easy to see that A~c 6= 3c; thus ~c is not an
tors. Put ~c =
0
eigenvector. Set
~b = A~c − (3)~c = 5 − 3 = 2 .
2
0
2
Then ~v1 (t) = e3t~b and ~v2 (t) = e3t (t~b + ~c) are independent
solu2
2t
+
1
tions. The general solution is ~v = c1 e3t
+ c2 e3t
. Sub2
2t
−1
stituting ~v(0) =
, we have 2c1 + c2 = −1, 2c1 = −2. Thus
−2
2
3t
c1 = −1, c2 = 1. Therefore the solution of the IVP is ~v = −e
+
2
2t + 1
2t − 1
e3t
= e3t
. Thus the solution of the IVP is x =
2t
2t − 2
(2t − 1)e3t , y = (2t − 2)e3t .
Return
254
19.
ORDINARY DIFFERENTIAL EQUATIONS
Let y0 = v. The system that replaces the ODE is
y0 = v
v0 = −q y − p v
0
1
The coefficient matrix of the system is A =
. Since the
−q − p
trace of A is equal to − p and the determinant is equal to q, thus the
characteristic equation is s2 + ps + q = 0.
Return
SOLUTIONS MANUAL
1.
1
2+ i
Return
=
2− i
(2+i )(2−i )
255
=
2− i
5
256
3.
ORDINARY DIFFERENTIAL EQUATIONS
2nπi
6
nπi
3
= cos nπi
+ i sin nπi
,
3
3
√
√
3
3
1
1
for n = 0, 1, 2, 3, 4, 5. The six sixth roots are 1, + i
, − +i
, −1,
2
2
2
2
√
√
3
3
1
1
− −i
, and + i
.
2
2
2
2
Suppose α6 = e2nπi , then α = e
Return
=e
SOLUTIONS MANUAL
5.
|eλ+iω | = eλ |eiω | = eλ | cos(ω ) + i sin(ω )|
q
λ
= e cos2 (ω ) + sin2 (ω ) = eλ .
Return
257
258
ORDINARY DIFFERENTIAL EQUATIONS
1
x
− 2 − 21
7.
Let ~v =
and A =
so that our system is equiv1
y
− 21
2
alent to ~v0 = A~v. Since tr ( A) = −1 and det( A) = 12 , the characteristic
−1 ± i
.
equation is s2 + s + 12 = 0 and the eigenvalues are s =
2
h
To find a eigenvector ~b =
(h and k will be complex numbers)
k
belonging to s = −1/2 + i/2, we need to solve
−1/2 −1/2
h
h
= (−1/2 + i/2)
,
1/2 −1/2
k
k
or −h − k = (−1 + i )h, h − k = (−1 + i )k.
Each
of these equations
i
reduces to h = ik. Thus we will put ~b =
. The corresponding
1
complex-valued solution is
i (cos(t/2) + i sin(t/2))
(−1/2+i/2)t~
−t/2
e
b = e
cos(t/2) + i sin(t/2)
− sin(t/2)
cos(t/2)
−t/2
= e
+i
cos(t/2)
sin(t/2)
The real part and the imaginary part of this solution are themselves
solutions of the system, and in fact form a fundamental set of solutions. Therefore, the general solution is
− sin(t/2)
cos(t/2)
−t/2
−t/2
~v = c1 e
+ c2 e
cos(t/2)
sin(t/2)
−c1 sin(t/2) + c2 cos(t/2)
= e−t/2
.
c1 cos(t/2) + c2 sin(t/2)
Return
SOLUTIONS MANUAL
259
9.
x
4 −8
and A =
so that our system is equivay
4 −4
lent to ~v0 = A~v. Since tr ( A) = 0 and det( A) = 16, the characteristic
equation is s2 + 16 = 0 and the eigenvalues are ±4i.
h
(h and k will be complex numbers)
To find a eigenvector ~b =
k
4 −8
h
h
belonging to s = 4i, we need to solve
= (4i )
,
4 −4
k
k
or 4h − 8k = (4i )h, 4h − 4k = (4i )k. Each
of these
equations reduces
1+i
to h = (1 + i )k. Thus we will put ~b =
. The corresponding
1
complex-valued solution is
(1 + i )(cos(4t) + i sin(4t))
(4i )t~
e
b =
cos(4t) + i sin(4t)
cos(4t) − sin(4t)
sin(4t) + cos(4t)
=
+i
cos(4t)
sin(4t)
Let ~v =
The real part and the imaginary part of this solution are themselves
solutions of the system, and in fact form a fundamental set of solutions. Therefore, the general solution is
cos(4t) − sin(4t)
sin(4t) + cos(4t)
~v = c1
+ c2
cos(4t)
sin(4t)
(c1 + c2 ) cos(4t) + (c2 − c1 ) sin(4t)
=
.
c1 cos(4t) + c2 sin(4t)
Return
260
ORDINARY DIFFERENTIAL EQUATIONS
11.
¯
¯
¯
(a) A~b = A~b = s~b = s̄~b. Thus ~b is a eigenvector corresponding to s̄.
¯
(b) ~b and ~b are eigenvectors corresponding to two distinct eigenvalues s and s̄ respectively, thus they are linearly independent.
¯
¯
(c) Suppose c1~h + c2~k = 0. Because ~h = 12 (~b + ~b) and ~k = −2i (~b − ~b),
it follows that
c1 − ic2~ c1 + ic2~¯
b+
b = 0.
2
2
¯
We know that ~b and ~b are independent; thus c1 − ic2 = 0, c1 +
ic2 = 0. Solving these equations, we have c1 = 0, c2 = 0. Therefore ~h and ~k are linearly independent.
(d) The complex-valued solution corresponding to the eigenvalue
λ + iω and eigenvector ~h + i~k is
e(λ+iω )t (~h + i~k ) = eλt (cos(ωt) + i sin(ωt))(~h − i~k )
= eλt (cos(ωt)~h − sin(ωt)~k + i (sin(ωt)~h + cos(ωt)~k)).
The real part and the imaginary part of this solution are themselves solutions of the system, and in fact form a fundamental
set of solutions. Therefore, the general solution is
~v(t) = c1 eλt cos(ωt)~h − sin(ωt)~k + c2 eλt sin(ωt)~h + cos(ωt)~k ,
where c1 and c2 are constants.
Return
SOLUTIONS MANUAL
261
1.
The characteristic equation of A =
3
2
1
2
1
2
3
2
is s2 − 3s + 2 = 0;
hence the characteristic
roots
are1, 2. We can use characteristic vec1
1
tors ~e1 =
and ~e2 =
belonging to 1 and 2, respectively,
−1
1
t e and ~
to obtain independent
v2 (t) = e2t~e2 . Thus
1
solutions ~v1 (t) = e ~
t
2t
e
e
is a fundamental matrix solution.
X (t) =
−et e2t
Return
262
ORDINARY DIFFERENTIAL EQUATIONS
3.
2 −4
is s2 + 16 = 0;
5 −2
hence the characteristic
roots are
±4i. A characteristic vector corre2
. The real and imaginary parts of
sponding to 4i is ~v =
1 − 2i
The characteristic equation of A =
2 cos(4t) + 2i sin(4t)
e v =
(1 − 2i )(cos(4t) + i sin(4t))
2 cos(4t)
2 sin(4t)
=
+i
cos(4t) + 2 sin(4t)
sin(4t) − 2 cos(4t)
4it
are two independent solutions. Thus we can take
2 cos(4t)
2 sin(4t)
X (t) =
cos(4t) + 2 sin(4t) sin(4t) − 2 cos(4t)
as a fundamental matrix solution.
Return
SOLUTIONS MANUAL
263
et e2t
is a fundamental matrix solution of the as−et e2t
sociated homogeneous system.
1
0
t
(a) Set ~v p (t) = X (t)~
w(t). It follows that X (t)~
w (t) = e
, or
1
5.
X (t) =
et w10 + e2t w20 = et
−et w10 + e2t w20 = et .
−t
0
0
~ 0 (t) =
The
of this equation
is
solution
w1 = 0, w2 = e , or w
0
0
~ (t) =
. Thus w
, and we have the particular soe−t
− e−t
lution
t
e
e2t
0
−et
~v p (t) =
=
.
−et e2t
− e−t
−et
The general solution is ~v p (t) + ~vh (t), where vh (t) = X (t)~c denotes the general solution of the associated homogeneous equation.
1
0
t
(b) Set ~v p (t) = X (t)~
w(t). It follows that X (t)~
w (t) = e
, or
−1
et w10 + e2t w20 = et
−et w10 + e2t w20 = −et .
The
is w10 = 1, w20 = 0, or w0 (t) =
solution of this equation
1
t
~ (t) =
. Thus w
, and we have the particular solution
0
0
t
e
e2t
t
tet
~v p (t) =
=
.
−et e2t
0
−tet
The general solution is ~v p (t) + ~vh (t), where vh (t) = X (t)~c denotes the general solution of the associated homogeneous equation.
t b1
te
(c)
=
.
b2
0
1
0
t
Set ~v p (t) = X (t)~
w(t). It follows that X (t)~
w (t) = te
, or
0
et w10 + e2t w20 = tet
−et w10 + e2t w20 = 0.
264
ORDINARY DIFFERENTIAL EQUATIONS
~ 0 (t) =
The solution of this equation is w10 = t/2, w20 = te−t /2, or w
t2 /4
t/2
~ (t) =
. Thus w
, and we have the
1 −t
−
t
te /2
− 2 e (1 + t )
particular solution
t
t2 /4
e
e2t
(t2 /4 − t/2 − 1/2)et
~v p (t) =
=
.
−et e2t
(−t2 /4 − t/2 − 1/2)et
− 12 e−t (1 + t)
The general solution is ~v p (t) + ~vh (t), where vh (t) = X (t)~c denotes the general solution of the associated homogeneous equation.
Return
SOLUTIONS MANUAL
265
7.
2 cos(4t)
2 sin(4t)
is a fundacos(4t) + 2 sin(4t) sin(4t) − 2 cos(4t)
mental matrix solution of the associated homogeneous
system.
Set
−32 cos(4t)
~v p (t) = X (t)~
, or
w(t). It follows that X (t)~
w0 (t) =
0
X (t) =
2 cos(4t)w10 + 2 sin(4t)w20 = −32 cos(4t)
(cos(4t) + 2 sin(4t))w10 + (sin(4t) − 2 cos(4t))w20 = 0.
The solution of this equation is w10 = −16 cos2 (4t) + 8 sin(4t) cos(4t),
w20 = 8 sin2 (4t) − 16 cos(4t) sin(4t) − 8, or
−16 cos2 (4t) + 8 sin(4t) cos(4t)
0
~ (t) =
w
.
8 sin2 (4t) − 16 cos(4t) sin(4t) − 8
− 12 (16t + cos(8t) + 2 sin(8t))
~ (t) =
Thus w
, and we have the par1
2 (−8t + 2 cos(8t ) − sin(8t ))
ticular solution
−16t cos(4t) − 8t sin(4t) − cos(4t) − 2 sin(4t)
~v p (t) = X (t)~
.
w(t) =
−20t sin(4t) − 52 cos(4t)
The general solution is ~v p (t) +~vh (t), where ~vh (t) = X (t)~c denotes the
general solution of the associated homogeneous equation.
Return
266
9.
ORDINARY DIFFERENTIAL EQUATIONS
Since det(X (1)) = (2)(−1) − (−1)(1) =−1 6=0, X (1) is nonsin 2
4t
3t2
1 −2
2t
t3
1
= t
gular. Furthermore, dX /dt =
.
−2t −3t2
1 4
− t2 − t3
Thus X (t) is a matrix solution. A nonsingular matrix solution is a
fundamental matrix solution.
t
0
(a) Set ~v p (t) = X (t)~
w(t). It follows that X (t)~
w (t)t =
, or
2t
2t2 w10 + t3 w20 = 1
−t2 w10 − t3 w20 = 2.
The solution of this equation is w10 = 3t−2 , w20 = −5t−3 , or
3t−2
~ 0 (t) =
w
.
−5t−3
−3t−1
~ (t) =
Thus w
, and we have the particular solution
5t−2 /2
2
−7 2t
t3
−3t−1
2 t
~v p (t) =
.
=
1
− t2 − t3
5t−2 /2
2t
3 t
0
(b) Set ~v p (t) = X (t)~
w(t). It follows that X (t)~
w (t)t =
, or
− t3
2t2 w10 + t3 w20 = t2
−t2 w10 − t3 w20 = −t2 .
The solution of this equation is w10 = 0, w20 = t−1 , or
0
0
~ (t) =
w
.
t −1
0
~ (t) =
Thus w
, and we have the particular solution
ln |t|
2
3
2t
t3
0
t ln |t|
~v p (t) =
=
.
− t2 − t3
ln |t|
−t3 ln |t|
Return
SOLUTIONS MANUAL
267

12.


~
Suppose Q = [~q1 , ~q2 , · · · , qn ] and d = 

d1
d2
..
.



.

dn



Then Qd = [~q1 , q2 , · · · , ~qn ] 

d1
d2
..
.



 = d1~q1 + d2 q2 + · · · + dn~qn . Thus

dn
~ where
any linear combination of columns of Q can be expresses as Qd,
the components of the vector d~ are the coefficients of the linear combination.
If Q is nonsingular, then Qd~ = ~0 if and only if d~ = ~0; in other words, if
and only if the only linear combination of columns that is equal to the
zero vector is the zero combination. Thus, the columns of a nonsingular matrix are linearly independent. Conversely, if the columns of
Q are linearly independent, and Qd~ = ~0, then the vector d must represent the coefficients of the zero combination; in other words, d~ = ~0.
Hence Q is nonsingular.
Return
268
ORDINARY DIFFERENTIAL EQUATIONS

0

0
1.
By matrix multiplication, A2 = 
 0
0
k
and A = 0 for k ≥ 4. Hence
0
0
0
0
1
0
0
0


0 0
0

1 
 , A3 =  0 0
 0 0
0 
0 0
0
1
1
e At = I + At + ( A)2 t2 + ( A)3 t3
2
6


1 t 12 t2 61 t3
1 2 
 0 1 t
2t 
= 
 0 0 1
t 
0 0 0
1
Return
0
0
0
0

1
0 
,
0 
0
SOLUTIONS MANUAL
269
3.
1 i ∞ 1 j
A )( ∑ B )
i!
j!
j =0
i =0
∞
e A eB = ( ∑
∞
=
∞
1
1
∑ ∑ i! Ai j! B j
i =0 j =0
∞
=
∞
1
∑ ∑ i!j! Ai B j
i =0 j =0
∞
=
k
1
∑ ∑ i!(k − i)! Ai Bk−i
k =0 i =0
∞
=
k!
1 k
∑ k! ∑ i!(k − i)! Ai Bk−i .
i =0
k =0
By the binomial theorem, if A and B commute, then
( A + B)k =
k
k!
∑ i!(k − i)! Ai Bk−i .
i =0
If A and B do not commute, this won’t work: for example ( A + B)2 =
A( A + B) + B( A + B) = A2 + AB + BA + B2 . If AB 6= BA, it would
not be possible to combine the two middle terms to get ( A + B)2 =
A2 + 2AB + B2 .
It follows that for commuting square matrices A and B,
e A eB =
∞
1
∑ k! ( A + B)k = e A+B .
k =0
270
5.
ORDINARY DIFFERENTIAL EQUATIONS
t 1
1 0
A(t) =
, dA/dt =
. Thus A(t) · dA/dt =
0 2t
0 2
t 2
t 1
, but dA/dt · A(t) =
.
0 4t
0 4t
SOLUTIONS MANUAL
271


0 0 ac
A2 =  0 0 0  ,and Ak = 0 for k ≥ 3.
0 0 0
7.
(a)
1
e At = I + At + ( A)2 t2
2


1 at bt + 12 act2
=  0 1
ct 
0 0
1

eλt ateλt (bt + 12 act2 )eλt
= 0
eλt
cteλt  .
0
0
eλt

(b) e( A+λI )t = eλIt e At = eλt e At

0 3 1
Now let A =  0 0 −4  Then the ODE is the same as ~v = ( A −
0 0 0.
2I
The solution is e( A−2I )t~c. Substitute the initial value we have ~c =
 )~v 
3
 0  . Thus
1



 
x
3
 y  = e( A−2I )t  0  .
z
1

 
1 3t t − 6t2
3
−4t   0 
= e−2t  0 1
0 0
1
1


2
3 + t − 6t

−4t
= e−2t 
1
Hence x = e−2t (3 + t − 6t2 ), y = −4te−2t , and z = e−2t .
Return
272
9.
ORDINARY DIFFERENTIAL EQUATIONS
Since tr A = 0 and det( A) = 1, the characteristic equation is s2 +
1 = 0 and by the Cayley-Hamilton Theorem it follows that A2 + I =
0. Therefore A2n = ( A2 )n = (− I )n = (−1)n I. Multiply this by A to
get the odd powers: A2n+1 = (−1)n A.
To calculate e At separate into two sums:
!
!
∞
∞
(−1)n 2n
(−1)n 2n+1
At
e
=
t
I+ ∑
t
A
∑
n=0 (2n ) !
n=0 (2n + 1) !
cos t + sin t
sin t
= cos tI + sin tA =
−2 sin t
cos t − sin t
Return
SOLUTIONS MANUAL
273
11.
−2 2
, and note that
−3 2
det A = 2. By the Cayley Hamilton theorem, B2 = −2I, and hence
Since tr A = 2 we’ll set B = A − I =
e Bt =
∞
1
1
B2n + ∑
B2n+1
(
2n
)
!
(
2n
+
1
)
!
n =0
n =0
∞
∑
∞
∞
(−2)n
(−2)n
I
+
B
∑
∑
n=0 (2n ) !
n=0 (2n + 1) !
√
√
∞
1 ∞ (−1)n ( 2)2n+1
(−1)n ( 2)2n
I+√ ∑
B
= ∑
(2n)!
(2n + 1)!
2 n =0
n =0
√
√
1
= (cos 2t) I + √ (sin 2t) B
2
!
√
√
√
√
√
cos 2t − 2√sin 2t
2
sin
2t
√
√
√
=
− √32 sin 2t
cos 2t + 2 sin 2t
=
Noting that A = I + B, and that I and B commute, we have
e At = et e Bt = et
cos
!
√
√
√
√
2t − 2√sin 2t
2
sin
2t
√
√
√
− √3 sin 2t
cos 2t + 2 sin 2t
√
2
Return
274
13.
ORDINARY DIFFERENTIAL EQUATIONS
Since tr A = −2, we’ll put
B = A+I =
−2 −2
2
2
.
Then tr B = 0 and det B = 0. It follows (from the Cayley-Hamilton
theorem) that B2 = Z, the zero matrix and hence
1 − 2t −2t
Bt
.
e = I + Bt =
2t
1 + 2t
Therefore,
e
Return
At
−t Bt
=e e
=e
−t
1 − 2t −2t
2t
1 + 2t
.
SOLUTIONS MANUAL
275

15.

0 0 1
A2 =  0 0 0  ,and Ak = 0 for k ≥ 3. Hence
0 0 0
1
e At = I + At + ( A)2 t2
2


1 t 2t + 21 t2

=  0 1
t
0 0
1
Return
276
17.
ORDINARY DIFFERENTIAL EQUATIONS
Write the system in the form ~v0 = A~v + ~f (t), where
x
1 0
et
~
~v =
, A=
, and f (t) =
.
y
1 −1
cosh t
Noting that tr A = 0 and det A = −1, we have A2 = (det A) I = I so
A2n = I and A2n+1 = A. Thus
!
!
∞
∞
t2n+1
t2n
At
I+ ∑
A
e
=
∑
n=0 (2n + 1) !
n=0 (2n ) !
= (cosh t) I + (sinh t) A
cosh t + sinh t
0
=
.
sinh t
cosh t − sinh t
~ . We get ~v0 = Ae At w
~+
Applying variation of constants, put ~v = e At w
At
0
At
0
~ so it follows that e w
~ = ~f . To find w
~ , multiply ~f by (e At )−1 =
e w
e− At and integrate:
Z cosh t − sinh t
0
et
~ =
w
dt
− sinh t
cosh t + sinh t
cosh t
Z Z et (cosh t − sinh t)
1
=
dt
=
dt
1
−et sinh t + cosh2 t + cosh t sinh t
1
= t
+~c
1
(To get the second-to-last equality, substitute sinh t = 12 (et − e−t ) and
cosh t = 21 (et + e−t ).) Finally, we have the general solution:
At
~v = e w
~ =
Return
(cosh t + sinh t)(t + c1 )
sinh t(t + c1 ) + (cosh t − sinh t)(t + c2 )
.
SOLUTIONS MANUAL
277
19.
2 −1
and note that A2 = 0. (The trace and the
4 −2
determinant of A are both 0.) Hence e At = I + At. Turning to the
~ . Our system, which is
method of variation of constants, set ~v = e At w
equivalent to
1
0
2t
~v = A~v + e
,
2
1
0
−
At
2t
~ =e e
, or
reduces to w
2
Let A =
~
w
0
2t
1
2
= e ( I − At)
1 − 2t
t
1
2t
= e
−4t 1 + 2t
2
1
.
= e2t
2
Integrating, we get
1
~ = e2t
w
2
1
2
+
c1
c2
The general solution is
~v = e At w
~ = ( I + At)~
w
1 2t
1 + 2t
−t
e + c1
2
=
4t
1 − 2t
e2t + c2
1 2t 1
1 + 2t
−t
=
e
+ c1
+ c2
2
4t
1 − 2t
2
Return
278
21.
ORDINARY DIFFERENTIAL EQUATIONS
SOLUTIONS MANUAL
279
1.
2 1
−1 0
. Thus e At = eCt ⊕ [e−t ]. To
1
1
1
Ct
. Since
determine e put B = C − 2 (tr C ) I = C − I =
−1 −1
tr B = det B = 0, it follows from the Cayley-Hamilton theorem that
B2 = 0 and hence e Bt = I + Bt. Therefore,
t
e (1 + t )
tet
Ct
t
e = e ( I + Bt) =
.
−tet
e t (1 − t )
A = C ⊕ [−1], where C =
and
e At
Return

e t (1 + t )
tet
0
e t (1 − t ) 0 
=  −tet
0
0
e−t

280
3.
ORDINARY DIFFERENTIAL EQUATIONS
SOLUTIONS MANUAL
5.
281
282
7.
ORDINARY DIFFERENTIAL EQUATIONS
SOLUTIONS MANUAL
283
9.
Decompose A as the direct sum [−2] ⊕ C ⊕ [1], where C =
0 −4
1 0
Since tr C = 0 and det C = 4, we have
e
Ct
1
= (cos 2t) + (sin 2t)C =
2
cos 2t −2 sin 2t
1
cos 2t
2 sin 2t
.
Therefore
e−2t
 0
= [e−2t ] ⊕ eCt ⊕ [et ] = 
 0
0

e At
Return

0
0
0
cos 2t −2 sin 2t 0 
.
1
cos 2t
0 
2 sin 2t
0
0
et
.
284
11.
ORDINARY DIFFERENTIAL EQUATIONS
The eigenvalues of A are the roots of f (s) = 0: s = 0 (a double
root) and s = 2. There
eigenvectors belonging
 are two independent
 
2
1
to s = 0: ~b1 =  0  and ~b2 =  1 . Thus the algebraic mul1
0
tiplicity of this eigenvalue matches its geometric

multiplicity, and
1
thus A is semisimple. We will take ~b1 =  −1  as an eigenvec2
tor belonging to 2. Let P be the matrix with columns ~b1 ,~b2 ,~b3 , and
let D = diag (0, 0, 2). Then AP = PD, so A = PDP−1 . Taking the
exponential,

e At = PDP−1 = 
Return
1
2t
2 (3 − e )
1 2t
2 ( e − 1)
1 − e2t
1 2t
2 ( e − 1)
1
2t
2 (3 − e )
2t
e −1

e2t − 1
1 − e2t  .
2e2t − 1
SOLUTIONS MANUAL
13.
285
The eigenvalues of A are the zeros of f (s): s = −1 ± i and s = −2
(a double root). Since the eigenvalue −2 has algebraic multiplicity 2,
we start by row-reducing A + 2I to get

1
 0

 0
0

3
4
3
4
0
1
0
0

0 − 34
0 − 34 
.
1 −1 
0 0




We find that ~b1 = 
 1  . is an eigenvector of A belonging to −2 and
1
that the geometric multiplicity of −2 is 1. Hence A is not semisimple.
Let


6 −22 16 −4
 4 −12 8 −2 

B = ( A + 2I )2 = 
 4 −12 8 −2  .
4 −12 8 −2
A vector ~c 6= ~0 is a belongs to the generalized eigenspace of A belonging to −2 if and only if B~c = ~0. Row reduction applied to B yields

1
 0
R=
 0
0

0 −1 41
1 −1 14 
.
0 0 0 
0 0 0


1
 1 

~
~
~
~
Let ~b2 = 
 1  , and note that Rb2 = 0. Thus Bb2 = 0 too, and since
0
~
~
S1 = {b1 , b2 } is linearly independent, S1 is a basis for the generalized
eigenspace of A belonging to −2. Notice that


−6
 −6 

~
( A + 2I )~b2 = 
 −8  = −8b1 .
−8
Thus A~b2 = 2~b2 − 8~b1 .
286
ORDINARY DIFFERENTIAL EQUATIONS

3−i
−8
−2
5
 2
−3 − i
−6
6 
 . Apply
Let C = A − (−1 + i ) I = 
 2
−2
−9 − i
8 
2
−2
−8
7−i
row reduction to C to get


1 0 0 − 32 − 12 i
 0 1 0

−1
.
R1 = 

 0 0 1
−1
0 0 0
0

To obtain an eigenvector of A we need a vector ~b3 such that C~b3 = ~0,
 3 1 
2 + 2i


1
. Let ~b4 be
or equivalently, R1~b3 = ~0. We will take b~3 = 


1
1
~
the complex conjugate of b3 ; it will be an eigenvector belonging to the
conjugate eigenvalue −1 − i. The matrix
 3
3
1
3
1 
4 1 2 + 2i 2 − 2i
 3 1

1
1

4
P=
 1 1

1
1
1 0
1
1
has as its columns the vectors ~b1 ,~b2 ,~b3 ,~b4 . Thus the columns of AP
are −2~b1 , −2~b2 − 8~b1 , (−1 + i )~b3 , (−1 − i )~b4 . Let


−2 −8
0
0
 0 −2

0
0
,
D=
 0

0 −1 + i
0
0
0
0
−1 − i
and consider what happens when we form a product MD. where M
~ 1, . . . , m
~ 4 . The four columns
is any four-column matrix, with columns m
~ 1 , −2m
~ 1 − 8m
~ 2 , (−1 + i )~
of MD will be −2m
m3 , and (−1 − i )~
m4 , respectively. Applying this rule to M = P we get PD = AP. Therefore
A = PDP−1 and we have accomplished the objective of finding an ET-E matrix D that is conjugate to A. (As a direct sum of a 2 × 2 matrix
and a diagonal matrix, D is E-T-E.)
3 1
−2 −8
+ 2i
0
2
Let H =
, and K =
, so that D = H ⊕
3
0 −2
0
− 1i
2 2 0 −8
K. We have H = −2I + N, where N =
is nilpotent, with
0 0
SOLUTIONS MANUAL
287
e−2t −8te−2t
.
0
e−2t
= diag (e(−1+i)t , e(−1−i)t ).
N 2 = 0. Thus e Ht = e−2t e Nt = e−2t ( I + Nt) =
Since K = diag (−1 + i, −1 − i ), we have eKt
Thus
 −2t
e
−8te−2t
0
0
−2t

0
e
0
0
e Dt = e Ht ⊕ eKt = 
 0
0
e(−1+i)t
0
0
0
0
e(−1−i)t


,

and
e At = PeDt P−1 =
cos t + 3 sin t

2 sin t
e−t 

2 sin t
2 sin t
3 cos t − 3e−t − 11 sin t
4 cos t − 3e−t − 6 sin t
4 cos t − 4e−t − 6 sin t
4 cos t − 4e−t − 6 sin t
4(e−t − cos t) + 8 sin t − 6te−t
4(e−t − cos t) − 4 sin t − 6te−t
5e−t − 4 cos t + 4 sin t − 8te−t
4(e−t − cos t) + 4 sin t − 8te−t
Return

cos t − e−t + 6te−t
cos t − e−t + 6te−t 

cos t − e−t + 8te−t 
cos t + 8te−t
288
ORDINARY DIFFERENTIAL EQUATIONS




x
−2 1 2
15.
Put ~v =  y , and let A =  1 −1 2  be the coefficient
z
−2 2 4
matrix, so that our system is


1
~v0 = A~v +  −1  .
1
In Exercise 11 we found that
 1
2t
2 (3 − e )
At
e =  12 (e2t − 1)
1 − e2t
1 2t
2 ( e − 1)
1
2t
2 (3 − e )
2t
e −1

e2t − 1
1 − e2t  .
2e2t − 1
~ to get
Substitute ~v = e At w


1
~ 0 + Ae At w
~ = Ae At w
~ +  −1  .
e At w
1
Thus
 
1
~ 0 = e− At  −1  = 
w
1

1
−2t )
2 (3 − e
1 −2t
− 1)
2 (e
−
1 − e 2t
1 −2t
− 1)
2 (e
1
−2t )
(
3
−
e
2
e−2t − 1

 

1
1
e−2t − 1
1 − e−2t   −1  =  −1  .
1
1
2e−2t − 1


t
~ =  −t  +~c, and hence
It follows that w
t


1
~v = e At w
~ = t  −1  + e At~c
1
Finally, set t = 0 for the initial condition to get ~v(0)= ~0 =e At~c. Since
1
e At is nonsingular it follows that ~c = ~0. Thus ~v = t  −1  .
1
Return
SOLUTIONS MANUAL
1.
2 1
and A =
so that the IVP is equivalent
1 2
1
0
to ~v = Av; ~v(0) =
. Since tr ( A) = 4 and det( A) = 3, the
0
2 − 4s + 3 = 0 and the characteristic roots
characteristic equation is
s h
be a characteristic vector belonging to
are 3 and 1. Let ~b1 =
k
3. Then 2h + k = 3h, h + 2k = 3k.Each of these equations reduces
to
1
h
h = k, so we can take ~b1 =
Now suppose that ~b2 =
is
1
k
a characteristic vector belonging to 1. Then 2h
+ k = h, h + 2k = k,
1
so that h = −k. Hence we can set ~b2 =
. The general solution
−1
1
1
3t
t
to the matrix equation is ~v = c1 e
+ c2 e
. Substituting
1
−1
1
~v(0) =
, we have c1 + c2 = 1, c1 − c2 = 0. Thus c1 = 1/2, c2 =
0
1
1
1 3t
1 t
1/2. Therefore the solution of the IVP is ~v = 2 e
+ 2e
,
1
−1
or x = 21 e3t + 12 et , y = 21 e3t − 12 et .
Return
Let ~v =
x
y
289
290
ORDINARY DIFFERENTIAL EQUATIONS
2.
x
1 −2
(a) Let ~v =
and A =
so that our system is equivy
5 −1
alent to ~v0 = A~v. Since tr ( A) = 0 and det( A) = 9, the characteristic equation is s2 + 9 = 0 and the characteristic roots are
±3i.
h
~
To find a characteristic vector b =
(h and k will be complex
k
1 −2
h
numbers) belonging to s = 3i, we need to solve
=
5 −1
k
h
(3i )
, That is h − 2k = 3ih, 5h − k = 3ik. These will all rek
3i + 1
~
duce to 5h = (3i + 1)k. Thus we will put b =
. The
5
corresponding complex solution of ~v0 = A~v is
(3i + 1)(cos(3t) + i sin(3t))
(3i )t~
e
b =
5 cos(3t) + 5i sin(3t)
cos(3t) − 3 sin(3t)
sin(3t) + 3 cos(3t)
=
+i
5 cos(3t)
5 sin(3t)
It follows that
cos(3t) − 3 sin(3t) sin(3t) + 3 cos(3t)
5 cos(3t)
5 sin(3t)
is a fundamental matrix solution.
x
1 −1
(b) Let ~v =
and A =
so that our system is equivay
1 1
lent to ~v0 = A~v. Since tr ( A) = 2 and det( A) = 2, the characteristic equation is s2 − 2s + 2 = 0 and the characteristic roots are
1 ± i.
h
~
To find a characteristic vector b =
(h and k will be complex
k
1 −1
h
numbers) belonging to s = 1 + i, we need to solve
=
1 1
k
h
(1 + i )
, That is h − k = (1 + i )h, h + k = (1 + i )k. Each if
k
i
~
these equations reduces to h = ik. Thus we will put b =
.
1
SOLUTIONS MANUAL
291
The corresponding complex solution of ~v0 = A~v is
i (cos(t) + i sin(t))
(1+i )t~
t
e
b = e
cos(t) + i sin(t)
− sin(t)
cos(t)
t
= e
+i
cos(t)
sin(t)
Thus a fundamental matrix solutions is
−et sin(t) et cos(t)
et cos(t) et sin(t)
1 2
0 1
so that our system is equivalent to ~v0 = A~v.
0 2
Notice that A = I + B, where B =
, that I and B com0 0
mute, and that B2 = 0. As with any homogeneous linear system
of ODEs of the form v0 = A~v, the matrix exponential e At is a
fundamental solution. We can use our observations about this
special matrix to compute e At as follows:
(c) Let A =
e At = e( I + B)t = et e Bt
because I and B commute.
e Bt = I + Bt
because B2 = 0. Thus
e
At
t
= e ( I + Bt) =
et 2tet
0 et
is a fundamental matrix solution.
(d) ~x 0 = A~x, where


1 2 3
A =  0 2 3 .
0 0 3
We will
 0
 use
 the fact that this system is actually
x
 x
y0
~x =  y ; then the system is the same as
 0
z
z
uncoupled. Let
= x + 2y + 3z
= 2y + 3z
= 3z
292
ORDINARY DIFFERENTIAL EQUATIONS
Solving the third equation (which does not involve x or y), we
get z = c3 e3t , where c3 is a constant. Let’s substitute this expression for z in the second equation to obtain y0 = 2y + 3c3 e3t .
This equation does not involve x or z; we can solve it and get
y = c2 e2t + 3c3 e3t ; c2 is another constant. Finally, substitute these
expressions for y and z in the first equation. The resulting equation, x 0 = x + 2(c2 e2t + 3c3 e3t ) + 3c3 e3t = x + 2c2 e2t + 9c3 e3t .
Solve it we obtain x = c1 et + 2c2 e2t + 92 c3 e3t , where c1 is a third
constant. The general solution is
   t


x
c1
e 2e2t 29 e3t
 y  =  0 e2t 3e3t   c2 
z
c3
0
0
e3t
and a fundamental matrix solution is
 t

e 2e2t 29 e3t
 0 e2t 3e3t 
0
0
e3t
Return
SOLUTIONS MANUAL
3.
293
First we need finda particular solution. Suppose X (t)v isa soluv1
x
. Substituting this expression for
in the
tion, where ~v =
v2
y
0
x
x
tan(2t)
system
+
, we have, after some cancel=A
y
y
0
tan(2t)
(2 sin 2t − cos 2t)v10 + (sin 2t + 2 cos 2t)v20 = tan(2t)
ing, X~v0 =
. That is
= 0
0
cos(2t)v10 − sin(2t)v20
2
1
sin (2t) 0
These equations can be solved to obtain v10 =
, v2 = − cos(2t).
2 cos(2t)
4
1 + sin(2t)
1
1
− sin(2t), v2 =
After integrating, we have v1 = ln
4
cos(2t)
4
− 41 cos(2t). and we get a particular solution:

sin(2t) 
− 21 + 12 sin(2t) − 14 cos(2t) ln 1+
cos(2t)
.
X (t)~v = 
1+sin(2t)
1
cos
(
2t
)
ln
4
cos(2t)
Therefore, the general solution is
1
1
1
1 + sin(2t)
x = − +
sin(2t) − cos(2t) ln
2
2
4
cos(2t)
+ C1 [2 sin(2t) − cos(2t)] + C2 [sin(2t) + 2 cos(2t)]
1 + sin(2t)
1
cos(2t) ln
+ C1 cos(2t) + C2 sin(2t)
y =
4
cos(2t)
Return
294
ORDINARY DIFFERENTIAL EQUATIONS
4.
X (t) = V · E(t)

e r1 t 0 · · · 0
 0 e r2 t · · · 0 

= [~v1 , ~v2 , · · · , vn ] · 
 ··· ··· ··· ··· 
0
0
0 ern t
rt
= e 1 ~v1 , er2 t~v2 , · · · , ern t vn

Then X (t) is the matrix for which column i is the vector eri t~vi ; we
have noted that such a matrix is a fundamental matrix solution. It
follows that
e At = X (t) · [X (0)]−1 = V · E(t) · V −1 .
Return
SOLUTIONS MANUAL
295
5.
Let r be an eigenvalue of a nilpotent matrix M, and let ~v be corresponding eigenvector. Since M is nilpotent, Mn is the zero matrix
for some n. Therefore Mn~v = ~0. But Mn~v = r n~v so it follows that
r = 0. Since all eigenvalues of M are equal to zero, and since the trace
of any matrix is the sum of its eigenvalues (counted with algebraic
multiplicity), and the determinant of any matrix is the product of its
eigenvalues, it follows that the determinant of M is also zero.
Conversely, if the trace and determinant of M are equal to zero, then
the characteristic equation of M is s2 = 0. The Cayley-Hamilton theorem says that every matrix a satisfies its own characteristic equation;
hence M2 is is equal to the zero matrix.
296
6.
ORDINARY DIFFERENTIAL EQUATIONS
a − ( a + d)/2
b
Since tr ( A) = a + d, B =
=
c
d − ( a + d)/2
( a − d)/2
b
. Thus
c
(d − a)/2
1
det( B) = − ( a − d)2 − bc
4
− a2 + 2ad − d2
=
− bc
4
a2 + 2add2
= −
+ ad − bc
4
1
= − tr ( A)2 + det( A)
4
Return
SOLUTIONS MANUAL
1.
297
(E). Because tr ( A) = 0 and det( A) = 0, the characteristic equation is s2 = 0 and which has a double root, 0. As this system is degenerate, there is a stationary line, which narrows the choices down to
(D) and (E). As the system corresponding to (D) has a negative eigenvalue (indicated by the orbits that approach the stationary line), the
only system that matches is (E).
1
is an eigenHere are more details about the system. The vector
1
vector belonging to the eigenvalue 0; thus the line y = x, which is
parallel to this vector, is the stationary line. The matrix A is nilpotent,
so
1−t
t
e At = I + At =
.
−t 1 + t
a
1−t
At
put ~c =
; then e ~c = a
, which traces a line of slope
0
−t
1 passing through the point ( a, 0), parallel to the stationary line. Its
direction is northeast if a < 0 and southwest if a > 0.
Return
298
3.
ORDINARY DIFFERENTIAL EQUATIONS
(I). Notice first that det( A) = −2, which indicates that the eigenvalues are of opposite sign. The phase portrait is therefore a saddle, which narrows the choices to (B), (F), (I), or (L). Noting that
tr ( A) = −1, the characteristic equation
is s2 +s − 2 = 0, which has
1
1
roots −2, 1. The eigenvectors,
and
, each have positive
2
1
slope. Thus the phase portrait is (I).The stable line is y = 2x, and the
unstable line is y = x.
Return
SOLUTIONS MANUAL
5.
299
(The determinant of A is equal to −4, negative. Hence the eigenvalues are opposite in sign and the phase portrait is a saddle. narrows
the choices to (B), (F), (I), or (L). Noting that tr ( A) = 3, the characteristic equation is s2 − 3s −4 = 0, which has roots λ1 = −1, λ2 = 4.
−3
1
The eigenvectors are
, belonging to λ1 , and
, belonging
2
1
to λ2 Thus the stable line, −3y = 2x, is of negative slope, and the
unstable line, y = x, has positive slope. The phase portrait is (F).
Return
300
7.
ORDINARY DIFFERENTIAL EQUATIONS
Here we have a saddle, because det( A) = −1 is negative, indicating eigenvalues of opposite sign. The saddles are choices (B),
(F), (I), and (L). Noting thst tr ( A) = 0, the characteristic equation is
s2 −1 =0, which has roots λ1 = −1, and
λ2 = 1. The eigenvectors
1
1
are
, belonging to λ1 ; and
, belonging to λ2 . Thus the
1
−1
stable line is y = x, the unstable line is y = − x, and hence the phase
portrait is (L).
Return
SOLUTIONS MANUAL
9.
301
(A). A = 2I. The phase portrait is an unstable proper node; (A) is
the only choice.
Return
302
11.
ORDINARY DIFFERENTIAL EQUATIONS
(D). The characteristic equation is s2 + 2s = 0, and hence the
eigenvalues are 0, −2. This is a degenerate system with single 0 root,
and hence there is a stationary line, with all orbits directed toward
it. The phase portrait (D) is the only one that is consistent with this
information.
1
, belonging
Here is the complete solution: The eigenvectors are
1
1
to the 0 root; and
belonging to −1. The stationary line is
−1
y = x, and all non-stationary
orbits
approach it from the direction
1
indicated by the eigenvector
.
−1
Return
SOLUTIONS MANUAL
303
C
A
B
D
E
F
G
I
H
K
J
L
304
ORDINARY DIFFERENTIAL EQUATIONS
13.
1 1
1 1
2
is s − 2s = 0, which has roots
0, 2.Hence the system is degenerate.
1
1
,
. The stationary line is y = − x.
The eigenvectors are
−1
1
The other orbits are half-lines, all with slope 1, directed away from
the stable line.
The characteristic equation of the coefficient matrix A =
4
2
-4
-2
2
-2
-4
Return
4
SOLUTIONS MANUAL
15.
305
Obtain an integral of the system by integrating the exact ODE
( x + y) dx + x dy = 0
to get x2 + 2xy = c, which we can solve for y to get
y=
1 c
−x .
2 x
The graph of this equation is a hyperbola with asymptotes x = 0 and
y = − 21 x. The asymptotes each contain two orbits and the stationary
point at the origin. On the y-axis, we have y0 = y and x 0 = 0; this
indicates that the orbits on this asymptote are directed away from the
origin. On the line y = − 21 x, we have x 0 = − x and y0 = 12 x; thus the
orbits on this asymptote are directed toward the origin.
Return
306
ORDINARY DIFFERENTIAL EQUATIONS
17.
2 1
, has eigenvalues of oppo3 0
site
signs:
−1, 3. Thus the system
us
a saddle. The eigenvectors are
1
1
, belonging to −1, and
, belonging to 3. Thus the stable
−3
1
line is y = −3x. The unstable line is y = x.
The coefficient matrix, A =
3
2
1
-3
-2
-1
1
-1
-2
-3
Return
2
3
SOLUTIONS MANUAL
307
−3 5
19.
The eigenvalues of the coefficient matrix, A =
, are
− 52 2
−1±5i
These have negative real part; hence the phase portrait is a sta2
ble spiral node.
4
2
-3
-2
-1
1
-2
-4
Return
2
3
308
ORDINARY DIFFERENTIAL EQUATIONS
21.
The coefficient matrix A =
1 −1
0 0
has characteristic equation,
1
2
~
s − s = 0, and eigenvalues 0, 1. The eigenvectors are b1 =
, be1
1
longing to 0, and ~b2 =
, belonging to 1. The stationary line is
0
y = x, in the direction of ~b1 , and the non-stationary orbits are horizontal half-lines directed away from the stationary line.
3
2
1
-3
-2
-1
1
-1
-2
-3
Return
2
3
SOLUTIONS MANUAL
309
23.
2 −5
is 0, and
17 −2
the determinant is 81 ¿ 0; hence the phase portrait is a center. The
direction is counterclockwise,
as the velocity vector at a point (0, b)
−5
on the positive y-axis is b
, pointed southwest. We can find an
−2
integral by integrating the exact equation
The trace of the coefficient matrix A =
(17x − 2y) dx − (2x − 5y) dy = 0;
we obtain 5y2 − 4xy + 17x2
7.5
5
2.5
-4
-2
2
-2.5
-5
-7.5
Return
4
310
ORDINARY DIFFERENTIAL EQUATIONS
25.
3
1
, are
−5 −3
−2, 2. Hence the phase portrait is a saddle. The stable and unstable
1
lines are determined by the eigenvectors,
, belonging to −1,
−5
1
and
, belonging to 1. Thus the stable line is y = −5x. The
−1
unstable line is y = − x.
The eigenvalues of the coefficient matrix, A =
3
2
1
-3
-2
-1
1
-1
-2
-3
Return
2
3
SOLUTIONS MANUAL
27.
311
The solution found in example 4.1.1 is x = c1 et + c2 e3t , y = c1 et −
c2 e3t . After rotation the corresponding solution will be
√ cos(π/4)
sin(π/4)
c1 et + c2 e3t
c1 e t
=
2
.
− sin(π/4) cos(π/4)
c1 et − c2 e3t
−c2 e3t
In other words, x =
Return
√
√
2c1 et , y = − 2c2 e3t = kx3 where k =
− c2
.
2c31
312
29.
ORDINARY DIFFERENTIAL EQUATIONS
If both eigenvalues of A are negative −r, −s, then the general
solution is ~v(t) = c1 e−rt~b1 + c2 e−st~b2 , As t → ∞, ~v(t) → ~0.
If the eigenvalues are −r ± is, with negative real parts, put B = A −
rI. Then tr B = 0 and det B = s2 . Therefore the phase portrait of
~ 0 = Bw
~ is a center. The general solution of ~v0 = A~v is ~v(t) = er t e Bt ,
w
and because r < 0, ~v(t) → ~0 as t → ∞.
In each of the above cases the phase portrait of ~v0 = a~v is a stable
node. The same reasoning shows that if the eigenvalues of A are positive (or have positive real parts) then ~v(t) → ~0 as t → −∞ so the
phase portrait is an unstable node.
If one of the eigenvalues is 0, the eigenvector belonging to 0 determines the stationary line. Each point (other than the origin) on the
stationary line is an orbit that does not approach the origin; hence the
phase portrait is not a stable or unstable node.
If the eigenvalues are pure imaginary numbers, the phase portrait
is a center. Each orbit is an ellipse, which does not converge to the
origin. If one of the eigenvalues is positive and the other is negative,
the phase portrait is a saddle. The stable line will consist of orbits
directed toward the origin as t → ∞; the unstable line consists of
orbits that converge to the origin as t → −∞, and all other orbits go
to ∞ as t → ±∞. Therefore the phase portrait is not a stable node in
these cases, either.
Return
SOLUTIONS MANUAL
313
31.
(a) If y = vx, then by the product rule, y0 = xv0 + vx 0 . The right
sides of the equations are obtained by factoring.
(c) To determine the stationary points, solve the equation −bv2 +
(d − a)v + c = 0. Thus, we have two stationary points if (d −
a)2 + 4bc > 0, one if (d − a)2 + 4bc = 0, and none if (d − a)2 +
4bc < 0. In the case of two stationary points, when b > 0 the sign
of −bv2 + (d − a)v + c changes from − to + to − as v increases.
This implies a phase diagram that looks like this:
r
- r
Thus the greater of the two stationary points is stable; the lesser
is unstable. If b < 0 all arrows in the phase diagram reverse, and
the lesser of the two stationary points is stable.
If there is only one stationary point then the quadratic expression −bv2 + (d − a)v + c does not change sign; the phase portrait
is as follows if b > 0 (arrows reversed if b < 0):
r
Finally, if there are no stationary points, the phase diagram is a
down arrow if b > 0, or an up arrow if b < 0.
(d) The characteristic equation
is s2 − ( a + d)s + ( ad − bc) =
p
p0. The
1
1
2
roots are 2 ( a + d ± ( a + d) − 4( ad − bc)) = 2 ( a + d ± ( a − d)2 + 4bc).
b
p
The corresponding eigenvectors are 1
.
( a − d)2 + 4bc)
2 (− a + d ±
The slopes of these vectors are
p
− a + d ± ( a − d)2 + 4bc
,
2b
the same as the the stationary points of the ODE (4.20).
(e) Assume b > 0. If the equation (4.20) has two stationary points,
s1 < s2 then the two lines l1 : cx + dy = s1 ( ax + by) and l2 :
cx + dy = s2 ( ax + by) are parallel to eigenvectors. Each line
consists of the origin and two nonstationary orbits converging
either toward or away from the origin. (If 0 is an eigenvalue, the
corresponding line is a stationary line.)
The half-plane on one side of the line ax + by = 0 is divided into
three regions by l1 and l2 . The region of the phase plane where
s1 <
cx + dy
< s2
ax + by
314
ORDINARY DIFFERENTIAL EQUATIONS
corresponds to the middle section of the first phase diagram in
part (c). The slope v of an orbit in this region increases as t increases, converging to s2 as t → ∞ and to s1 as t → −∞.
cx + dy
In the region where
< s1 the slope v → −∞ when the
ax + by
orbit crosses the line ax + by = 0 with a vertical tangent; then the
sign of v changes and the slope will decrease from ∞, converging
to s2 as t → ∞.
If the system (4.20) has just one stationary point s1 , then the system (4.19) has a double eigenvalue. The slope of each orbit is
decreasing, but changes sign as it crosses the line ax + by = 0
with a vertical tangent. Hence the slope converges to s1 both as
t → −∞ and as t → +∞!
(f) Again we will assume that b > 0. If the equation (4.20) has no
stationary points, then the system (4.18) has no real eigenvectors. The orbits will be either clockwise ellipses (if the eigenvalues are purely imaginary) or clockwise spirals. In either case, the
slope will be decreasing until the orbit assumes a vertical direction, and then, as the orbit turns farther, the slope will decrease
again until a vertical direction is assumed again. This is consistent with the picture we see from the phase diagram of (4.20),
consisting of one down arrow. The slope changes from +∞ to
−∞ twice in each revolution about the origin.
Return
SOLUTIONS MANUAL
33.
315
(D). The characteristic equation is s2 + 4 = 0. The eigenvalues are
±2i.The phase portrait is a center, directed clockwise, because orbits
are directed to the right as they cross the positive y-axis. The integral
is x2 + 4y2 . This confirms the choice, because the orbits are ellipses
with horizontal major axes and vertical minor axes.
Return
316
35.
ORDINARY DIFFERENTIAL EQUATIONS
(B). The characteristic equation is s2 + 2s + 17 = 0. The roots are
−1 ± 4i. It’s a stable spiral node, directed clockwise, because it crosses
the positive y-axis traveling to the right.
Let B = A + I. Then tr B = 0 and det B = 16 = 42 . Hence e Bt =
(cos 4t) I + 41 (sin 4t) B, and e At = e−t [(cos 4t) I + 14 (sin 4t) B]. This shows
that our spiral has angular velocity of 4, and crosses the positive axis
(or any raw emanating form the origin) in a geometric sequence with
ratio e−π/2 ≈ 5.8.
~ 0 = Bw
~ has an integral, F ( x, y) = x2 −
Note also that the system w
2
2xy + 5y . The level curves are ellipses which can be used to draw
the orbits of ~v0 = A~v.
Return
SOLUTIONS MANUAL
37.
317
(A). The characteristic equation is s2 + 4s + 8 = 0. The roots are
−2 ± 2i. It’s a stable spiral node, directed counterclockwise, because
it crosses the positive y-axis traveling to the left.
Return
318
1.
ORDINARY DIFFERENTIAL EQUATIONS
The characteristic equation of A is s2 + s = 0, or s(s + 1) = 0. The
eigenvalues are 0 and −1. According theorem 8.1, A is stable but not
asymptotically stable.
SOLUTIONS MANUAL
2.
319
The characteristic equation of A is s2 − 2s − 3 = 0, or (s − 3)(s +
1) = 0. The eigenvalues are 3 and −1. Accordingtheorem
8.1, A
1
. Thus we
is not stable. The eigenvector corresponding to 3 is
1
1
have an unbounded solution v = e3t
.
1
320
4.
ORDINARY DIFFERENTIAL EQUATIONS
The characteristic equation of A is s2 − 2s + 2 = 0. The eigenvalues are 1 ± i. According theorem 8.1, A is unstable.
1
. Thus we have an
The eigenvector corresponding to 1 + i is
i
1
cos(t)
(
1
+
i
)
t
t
unbounded solution v = Re e
=e
i
− sin(t)
SOLUTIONS MANUAL
5.
321
The characteristic equation of A is s2 = 0. It has double eigenvalue 0. According theorem
8.1, A is unstable. The eigenvector cor1
. Thus we have an unbounded solution
responding to 0 is
−1
t+1
v=
.
−t
322
ORDINARY DIFFERENTIAL EQUATIONS
7.
The characteristic equation of A is (s2 + s)(s2 − 2s − 3) = 0. It have
eigenvalues 0, −1, −1, 3. According
8.1, A is unstable. The

 theorem
0
 0 

eigenvector corresponding to 3 is 
 1  . Thus we have an unbounded
1
 
0
 0 

solution v = e3t 
 1 .
1
SOLUTIONS MANUAL
8.
323
The characteristic equation of A is s2 + 1 = 0. It have
eigenvalues ±i. According theorem 8.1, A is stable but not asymptotically stable.
324
10.
ORDINARY DIFFERENTIAL EQUATIONS
The characteristic equation of A is (s2 + 1)(s2 − 1) = 0. It
have eigenvalues ±i, ±1. According
8.1, A is unstable. The

 theorem
0
 0 

eigenvector corresponding to 1 is 
 1  . Thus we have an unbounded
1
 
0

0 

solution v = et 
 1 .
1
SOLUTIONS MANUAL
11.
325
The characteristic equation of A is (s + 1)(s + 2)(s + 3) = 0. It
have eigenvalues −1, −2, −3. According theorem 8.1, A is asymptotically stable.
326
ORDINARY DIFFERENTIAL EQUATIONS
13.
The characteristic equation of A is (s2 + 1)(s2 + 1) = 0. It has double
eigenvalues ±i.
 
0
 1 

It is easily checked that b = 
 0  is not an eigenvector, and we put
0


1
 −i 

c = ( A − i I )b = 
 0 , which is an eigenvector. Thus there is a
0
solution
v(t) = eit (tc + b)
This solution
 is unbounded.The same may be said for its real part,
t cos(t)
 cos(t) + t sin(t) 
.
Re(v) = 


0
0
SOLUTIONS MANUAL
14.
327
The characteristic equation of A is (s2 + 1)(s2 + 4) = 0. It have
eigenvalues ±i, ±2i. According theorem 8.1, A is stable, but not asymptotically stable.
328
16.
ORDINARY DIFFERENTIAL EQUATIONS
If s1 = a i where a 6= 0 is real, is an eigenvalue, then s2 = − a i
is also an eigenvalue, According theorem 8.1, A is stable, but not
asymptotically stable. Therefore the phase portrait of the system is
a center, and thus the solution is periodic.
SOLUTIONS MANUAL
18.
329
By theorem 8.1, A is stable, but not asymptotically stable. However, some solutions may not be periodic. Let

0
 −1
A=
 0
0
1
0
0
0

0 0
0 0 

0 −2 
1 0
The characteristic equation
of A is (s2 + 1)(s2 + 2) = 0 and hence the
√
eigenvalues are ±i, ± 2i. The vectors



0
1
 i 

0


b=
 0  and c = 
1
√
0
1/( 2i )




√
2i, respectively. The solution
are eigenvectors
belonging
to
i
and
√
it
i
2t
v(t) = e b + e
c is bounded but not periodic. To see this, note that
the real part of this solution is

cos(t)


− sin

√(t)
w=
 cos( 2t)  .
√
√
sin( 2t)/ 2

If this solution had a period T then T would have to be simultaneously a period of
2Nπ, where N is an integer; and
√cos(t) — thus T = √
2t) — thus T = M 2π, where M is also an integer.
a period of cos( √
Hence 2Nπ = M 2π, and it follows that
√
2M
= 2,
N
which is impossible, since
√
2 is an irrational number.
330
20.
ORDINARY DIFFERENTIAL EQUATIONS
If c could be expressed as the linear combination of the n − 1
eigenvectors corresponding the left half-plane eigenvalues, then e At c
will be bounded as t → ∞. However, the probability of randomly
selecting a vector with this property is equal to 0.
SOLUTIONS MANUAL
21.
331
If c can be expressed as a linear combination of eigenvectors belonging to pure imaginary eigenvalues of A, then e At c is periodic and
thus is bounded on R.
Assume c can not be expressed as a linear combination of eigenvectors belonging to pure imaginary eigenvalues of A, but that c can be
expressed as a linear combination of eigenvectors belonging to eigenvalues of A, then there are terms belonging to the eigenvalues of the
form a + bi, where a 6= 0. If a > 0, limt→∞ e At c → ∞, If a < 0,
limt→−∞ e At c → ∞. In either case e At c can not be bounded.
If c cannot be expressed as a linear combination of eigenvectors then
A has repeated eigenvalues and that e At c is unbounded.
332
1.
ORDINARY DIFFERENTIAL EQUATIONS
The repelled set consists of those points whose orbitslead back
to
as t → −∞. Because the eigenvalues of A 72 , 0 are both
positive, the repelled set contains a small disk centered at 27 , 0 .
7
2, 0
Inspection of figure 4.17 shows that the entire fourth quadrant is a
subset of the repelled set, which extends into the first quadrant as
well. The upper boundary is repelled set (part of the separatrix) of
the saddle point at (2,1).
Return
SOLUTIONS MANUAL
3.
333
This system is linear. The characteristic equation is s2 + 2s + 2 =
0, and the eigenvalues are −1 ± i, which are located in the left halfplane. Therefore the phase portrait is a stable spiral node. The origin
is the only stationary point; it is asyptotically stable and its attracted
4
2
-4
-3
-2
-1
1
2
3
4
-2
-4
set is the entire plane.
Return
.
334
5.
ORDINARY DIFFERENTIAL EQUATIONS
This system is √
linear; its characteristic equation is s2 + 35 = 0.
The eigenvalues, ± 35 i, are pure imaginary, so the phase portrait is
a center. The origin is the only stationary point, and it is neutrally
stable.
4
2
-4
-3
-2
-1
1
2
3
4
-2
-4
.
Return
SOLUTIONS MANUAL
7.
335
To obtain the stationary points, solve the equations
y (1 − x 2 − y2 ) = 0
− x (1 − x2 − y2 ) = 0.
We find that the origin is a isolated stationary point, and that every
point on the unit circle x2 + y2 = 1 is stationary. We can perform
analysis of the isolated stationary point by linear approximation, but
not of the non-isolated points on the circle.
The derivative matrix is
"
#
∂
∂
2
2
2
2
∂x ( y (1 − x − y ))
∂y ( y (1 − x − y ))
A( x, y) =
∂
∂
2
2
2
2
∂x (− x (1 − x − y )) ∂y (− x (1 − x − y ))
−2xy
1 − x2 − 3y2
=
.
−1 + 3x2 + y2
2xy
0 1
A(0, 0) =
has eigenvalues ±i. Thus linear analysis is in−1 0
conclusive for this stationary point, which could be a center or a spiral. Although the linear approximation is a center, the nonlinearity
could disrupt the closed orbits of the center, as we have seen in Figure 4.16.
To analyze this stationary point, consider the ODE that is satisfied by
the orbits:
x (1 − x2 − y2 ) dx + y(1 − x2 − y2 ) dy = 0.
Divide through by the factor (1 − x2 − y2 ) to obtain the separable
equation x dx + y dy = 0, and thus an integral x2 + y2 = C. It follows
that the stationary point at the origin is a center after all, and is thus
neutrally stable, for the non-stationary orbits are circles. The orientation is determined by the direction of the orbits as they cross the
positive y-axis, which is to the right when x2 + y2 < 1 (inside the stationary circle), and to the left when x2 + y2 > 1. Thus the orbits are
clockwise inside the stationary circle, and counterclockwise outside
it. The stationary points on the unit circle are unstable, because the
orbit of any point close to, but not on, the unit circle will be a circle
through that point with center at the origin. Hence it will not stay in
a circle of radius 1, centered at that point.
336
ORDINARY DIFFERENTIAL EQUATIONS
1.5
1
0.5
-1.5
-1
-0.5
0.5
1
1.5
-0.5
-1
-1.5
Return
.
SOLUTIONS MANUAL
9.
337
Solve the equations
x (2 − x − y ) = 0
y( x − y) = 0.
to obtain the stationary points (0, 0), (2, 0), (1, 1).
The derivative matrix is
"
A( x, y) =
=
∂
∂x x (2 − x − y )
∂
∂x y ( x − y )
∂
∂y x (2 − x − y )
∂
∂y y ( x − y )
2 − 2x − y
−x
y
x − 2y
#
.
2 0
A(0, 0) =
has eigenvalues 2, 0. Because A(0, 0) is singular,
0 0
linear approximation is not conclusive. However, we note that the
x- and y-axes are invariant sets. On the x-axis, the system reduces to
a single logistic ODE, x = x (2 − x ), found by setting y = 0 in the
first equation. This shows that the origin is unstable, since the orbits
on the x-axis lead away from it. Similarly, the system reduces to the
ODE y0 = −y2 on the y-axis; orbits are directed downward, toward
the origin on the positive y-axis, and away from it on the negative
y-axis. Therefore the origin is unstable.
−2 −2
A(2, 0) =
has eigenvalues ±2, and therefore (2, 0) is
0
2
a saddle point with attracted set tangent to the eigenvector (1, 0) belonging to −2 and with repelled set tangent to the eigenvector (1, −2)
belonging to 2. It is also unstable.
−1 −1
The characteristic equation of A(1, 1) =
is s2 + 2s + 2 =
1 −1
0. The roots, s = −1 ± i, are in the left half plane, and thus (1, 1) is
a stable and a asymptotically stable equilibrium point. Orbits spiral
toward (1,1); to determine the direction of the spiral we note that for
x = 1, we have x 0 = 1 − y. Hence orbits cross the vertical line x = 1
from right to left when y > 1 and from left to right when y < 1. The
spiral is counterclockwise. The attracted set is the first quadrant. To
see this, note that no orbit can cross either axis, since the positive axes
are orbits. Therefore the any orbit that leads to the point (1,1) must
be confined to the first quadrant. The portion of the x-nullcline in the
338
ORDINARY DIFFERENTIAL EQUATIONS
first quadrant is the line y = − x + 2. Above this line, x 0 > 0, and
below it x 0 < 0. The y-nullcline is the line y = x. Above it, y0 < 0,
and below it y0 < 0. Thus no point in the first quadrant is attracted
to either the origin or the point (2,0). Also, if y is large, then y0 < 0,
while if x is large, x 0 < 0 All orbits in the first quadrant thus spiral
toward (1,1).
3
2
1
-3
-2
-1
1
2
3
-1
-2
-3
Return
.
SOLUTIONS MANUAL
11.
339
Notice first that the x- and y-axes are invariant. On the x-axis, the
first equation reduces to x 0 = x ( x2 − 10); hence the following phase
diagram appears on the x-axis
r
- r
− 10
0
√
√
r
-
10
The second equation reduces to y0 = −3y on the y-axis, and hence
there is an upward arrow on the negative y-axis, and a downward
arrow on the positive y-axis.
To find all of the stationary points, solve the equations
x ( x2 + y2 − 10) = 0
y( xy − 3) = 0.
√
In addition to (0, 0) and (± 10, 0), we find the stationary points
(3, 1), (1, 3) (−1, −3), and (−3, −1).
The derivative matrix is
"
∂
2
2
∂x x ( x + y − 10)
∂
∂x y ( xy − 3)
A( x, y) =
=
∂
2
2
∂y x ( x + y − 10)
∂
∂y y ( xy − 3)
3x2 + y2 − 10
2xy
2
y
2xy − 3
#
.
As
we mightexpect, the origin is asymptotically stable, since A(0, 0) =
−10 0
has eigenvalues −10, −3.
0
−3
√
√
20 0
A(± 10, 0) =
has eigenvalues 20, −3. Therefore (± 10, 0)
0 −3
are saddle points. Their attracted sets are curves that is tangent to the
eigenvector (0, 1) belonging to −3, and their repelled sets are intervals on the x-axis, as shown in the phase diagram that we considered
earlier.
2 6
The characteristic equation of A(1, 3) = A(−1, −3) =
is
9 3
s2 − 5s − 48 = 0. Hence the √
stationary points ±(1, 3) are saddle points.
The eigenvalues are 12 (5 ± 217). The attracted set of each stationary
√
point is a√curve tangent to the eigenvector (1 − 217, 12) belonging
to 21 (5 − 217) and with repelled set is a curve tangent to the eigen√
√
vector (1 + 217, 12) belonging to 12 (5 + 217).
340
ORDINARY DIFFERENTIAL EQUATIONS
18 6
.
1 3
2
The characteristic
√ equation is s − 21s + 48 = 0, and hence the eigen1
values, 2 (21 ± 249), are positive. Therefore (3, 1), (−3, −1) are unstable points.
√
The incoming separatrices of the saddle points at (1,3) and ( 10, 0)
connect with the unstable node at (3,1). These separatrices form the
right boundary of the attracted set of the origin. The left boundary is
a symmetric curve in the second and third quadrants.
At the stationary points ±(3, 1), the derivative matrix is A =
4
3
2
1
-4
-2
2
4
-1
-2
-3
-4
Return
.
SOLUTIONS MANUAL
13.
341
Solve the equations
x (1 − y ) = 0
y( x − 1) = 0.
to obtain the stationary points (0, 0), (1, 1).
The axes are invariant sets of this system. On the x-axis, the system
reduces to the ODE x 0 = x, and on the y-axis, it reduces to y0 = −y.
Thus the nonstationary orbits on the x-axis are directed away from
the origin, and the nonstationary orbits on the y-axis are directed toward the origin. It follows that the origin is a saddle point, with the
y-axis as attracted set and the x-axis as repelled set.
The derivative matrix is
"
∂
∂x x (1 − y )
∂
∂x y ( x − 1)
∂
∂y x (1 − y )
∂
∂y y ( x − 1)
1 − y −x
y
x−1
A( x, y) =
=
#
.
1 0
has eigenvalues 1, −1, confirming the saddle
0 −1
point at the origin.
0 −1
A(1, 1) =
. The characteristic equation is s2 + 1 = 0, and
1 0
A(1, 1) thus has eigenvalues ±i. Because this is a nonlinear system,
we cannot be sure that (1, 1) is a center because the nonlinearity may
disrupt the closed orbits that the linear approximation has.
A(0, 0) =
The orbits of this system are curves that satisfy then ODE
(1 − x )y dx + (1 − y) x dy.
This ODE is separable. Divide through by xy to obtain ( x −1 − 1) dx +
(y−1 − 1) dy − 0. We can then integrate to obtain solutions
F ( x, y) = ln( x ) + ln(y) − ( x + y) = C.
The integral, F, has a critical point at (1, 1). The discriminant,
∂2 F ∂2 F
−
∂x2 ∂y2
∂2 F
∂x∂y
2
=
−1
x2
−1
,
y2
342
ORDINARY DIFFERENTIAL EQUATIONS
is positive at ( x, y) = (1, 1). It follows that F has a local maximum at
the stationary point (1, 1), and hence the level curves of F near that
stationary point are closed. This verifies that (1,1) is indeed a center.
The orbits are are oriented counterclockwise, to be consistent with the
saddle structure at the origin.
2
1.5
1
0.5
-1
-0.5
0.5
1
1.5
2
-0.5
-1
Return
.
SOLUTIONS MANUAL
343
15.
The derivative matrix is
#
"
∂
∂
2
2
2
2
∂x [− y + kx (1 − x − y )] ∂y [− y + kx (1 − x − y )]
A( x, y) =
∂
∂
2
2
2
2
∂x [ x + ky (1 − x − y )]
∂y [ x + ky (1 − x − y )]
k − 3x2 k − ky2
−1 − 2kxy
=
.
1 − 2kxy
k − kx2 − 3y2 k
k −1
are k ± i. Because they deThe eigenvalues of A(0, 0) =
1 k
pend on k, the stability also depends on k. If k < 0, the origin is a stable spiral node, and if k > 0 the origin is an unstable spiral node. In
either case, the spirals are counterclockwise, because the orbits will be
directed to the left as they cross the positive y-axis. If k = 0 the eigenvalues are pure imaginary, which means that linear approximation is
inconclusive. However, when k = 0 the system is linear! Hence the
phase portrait is a center when k = 0.
We have completed the local analysis at the stationary point of this
system. We now turn to polar coordinates to see the big picture. Set
r2 = x2 + y2 ; then
2rr 0 = 2xx 0 + 2yy0
= 2x [−y + kx (1 − x2 − y2 )] + 2y[ x + ky(1 − x2 − y2 )]
= 2k( x2 + y2 )(1 − x2 − y2 )
= 2kr2 (1 − r2 )
Therefore r satisfies the ODE r 0 = kr (1 − r2 ). For k < 0, phase diagram
for this ODE is s - s s -, where the stationary points
are at 0, and ±1. The orbits that start within the unit circle, so that
r < 1, will spiral toward the origin. If r 6= 0, 1, then as t → −∞ the
orbits spiral toward the unit circle. Thus the orbit r = 1 is an unstable
periodic orbit.
If k > 0 the phase diagram s - s s - for r 0 = kr (1 − r2 )
indicates that the periodic orbit r = 1 is stable, because r → 1 as
t → ∞, provided that r 6= 0, 1.
If k < 0 the attracted set of the origin is the set of points inside the
unit circle, and if k > 0 the repelled set of the origin is also the set of
points inside the unit circle.
344
ORDINARY DIFFERENTIAL EQUATIONS
k=0
k=-0.05
2
-2
2
-2
Return
k=0.05
2
-2
2
-2
.
SOLUTIONS MANUAL
17.
345
Use the identities
rr 0 = x x 0 + y y0 and r2 θ 0 = x y0 − y x 0
to convert the system to polar coordinates. We obtain an uncoupled
system
r 0 = r (1 − r 2 )
θ 0 = 1 − cos(θ ).
The stationary points are those with polar coordinates r = 0, 1 and
θ = 0. Thus the points with cartesian coordinates (0, 0) and (1, 0) are
the only stationary points. To determine the stability, we will consider the phase diagrams of the polar ODEs. The phase diagram for
r 0 = r (1 − r2 ) is s - s , indicating that the circle r = 1 is an invariant set, and that all nonstationary orbits converge to it. The phase
diagram for θ 0 = 1 − cos(θ ) is s - s - s - s - . . . , where the
stationary points are 0, 2π, 4π, . . .. We thus see that every orbit converges toward the unit circle. The positive x-axis is an invariant set,
and the orbits on that axis form the same phase diagram as the phase
diagram just drawn for the variable r. Orbits that do not start on that
axis follow a counterclockwise path that does not cross the positive
x-axis, but converges to the point (1, 0). Thus, (1,0) is asymptotically
stable. It is not stable, because orbits starting at points in the upper
half-plane, no matter how close to (1, 0), must navigate counterclockwise around the origin and approach (1,0) from the fourth quadrant.
See the phase portrait.
2
1.5
1
0.5
-2
-1.5
-1
-0.5
0.5
1
1.5
2
-0.5
-1
-1.5
-2
.
346
ORDINARY DIFFERENTIAL EQUATIONS
Return
SOLUTIONS MANUAL
1.
347
The stationary points of the system are (0,0), (K, 0), (0, L), and,
provided that BC 6= 1, a point ( x ∗ , y∗ ), where
x∗ =
K + BL
1− BC
and y∗ =
L+CK
1− BC .
The point ( x ∗ , y∗ ) is thus in the first quadrant if BC < 1, and in the
third quadrant (where it would represent negative populations, and
has no biological significance) if BC > 1.
The derivative matrix of the system is
a(K − 2x − By)
aBx
D ( x, y) =
dCy
d( L + Cx − 2y)
aK 0
Thus, D (0, 0) =
, which has positive eigenvalues, in0 dL
dicating that
node. On
the origin is an unstable
the axes, we have
−a K
aBK
a(K + BL)
0
D (K, 0) =
and D (0, L) =
,
0
d( L + CK )
dCL
−d L
indicating that both stationary points are saddles. If we remember
∗
∗
that K − x ∗ + B y∗ = 0 and L +
C x − y = 0, it is not hard to see that
∗
∗
−a x a B x
D ( x ∗ , y∗ ) =
. This matrix has a negative trace, and
d C y∗ −d y∗
in the relevant case where BC < 1, det D ( x ∗ , y∗ ) = a x ∗ y∗ (1 − B C ) >
0. It follows that both eigenvalues are negative, and that the stationary point in the first quadrantis a stable node. In the phase
por
1
2
traits shown below, ( B, C ) = 2,
and ( B, C ) = 2,
, respec3
3
tively. The other parameters are the same for both: ( a, d, K, L) =
(0.05, 0.05, 2, 1).
8
8
6
6
4
4
2
2
2 4 6 8 10 12 14
Return
2 4 6 8 10 12 14
348
3.
ORDINARY DIFFERENTIAL EQUATIONS
There are three stationary points: The origin, where both species
are extinct, the point (C, 0), where the predator is extinct, and the prey
population is equal to the carrying capacity, and (d, y∗ ), where y∗ =
A
b (C − d ). The axes are invariant sets, indicating that if one of the
species is extinct, it will stay extinct. If x = 0 (the prey is extinct) then
the ODE for the predator is y0 = −cdy, predicting that the predator
population will decay exponentially. If y = 0 (no predators) the prey
population follows the logistic growth equation, x 0 = Ax (C − x ). By
drawing the single-species phase diagram on each axis, we obtain a
framework on which to build the two-species phase portrait:
r?
- r
6
(C, 0)
We can see from this diagram that the origin is a saddle point, just as
it is in the Lotka-Volterra system. This is confirmed by the derivative
matrix,
A(C − 2x ) − b y
−b x
D ( x, y) =
.
cy
c( x − d)
Thus D (0, 0) is the diagonal matrix with entries A C and −c d indicating that there is an saddle point at the origin.
For the stationary point on the x-axis,
D (C, 0) =
− AC
−bC
0
c(C − d)
,
is upper triangular, so the eigenvalues are the diagonal entries, − AC
and c(C − d).
(a) If C − d < 0 then both eigenvalues at (C, 0) are negative, and
hence that stationary point is a stable node. Furthermore, y∗
is negative, so the stationary points on the x-axis are the only
ones with biological significance. It follows that y is decreasing
throughout the first quadrant, and that all nonstationary orbits
in the first quadrant converge to (C, 0). The small carrying capacity for the prey means that there will never be enough prey in
this environment to support the predators. In the phase portrait
SOLUTIONS MANUAL
349
that follows, the parameters have been set as to be ( A, C, b, c, d) =
(0.5, 1500, 1, 0.125, 2000)
500
400
300
200
100
500
1000
1500
2000
(b) When C > d, the third stationary point, (d, y∗ ), is in the first
quadrant. When the predator population is equal to y∗ , there
are just enough predators to limit the prey population to d, the
minimum number needed to support any predators at all.
The eigenvalues of derivative matrix D (C, 0) are of opposite sign,
indicating that there is a saddle point at (C, 0). Furthermore, the
characteristic polynomial of
− A d −b d
D (d, y∗ ) =
.
c y∗
0
is s2 + Ad s + bcdy∗ , which has positive real eigenvalues when
Ad
A2 d2 ≥ 4bcdy∗ = 4Acd(C − d); that is,
≥ C − d. Hence the
4c
stationary point (d, y∗ ) is a stable node. In the phase portrait
that follows, the parameters have been set to be ( A, C, b, c, d) =
(0.5, 1500, 1, 0.125, 1000)
500
400
300
200
100
500
1000
1500
2000
350
ORDINARY DIFFERENTIAL EQUATIONS
∗
(c) When C − d > Ad
4c the stationary point at ( d, y ) is a stable spiral
node, because the real part each eigenvalue, − 12 Ad, is negative.
Orbits are directed counterclockwise. In the following phase
portrait, the parameters have been set to be ( A, C, b, c, d) = (0.5, 1500, 2, 0.125, 500)
500
400
300
200
100
500
Return
1000
1500
2000
SOLUTIONS MANUAL
1.
351
Suppose that ( x1 , 0) is asymptotically stable—we will reach a contradiction. Then there is a circle C , centered at ( x1 , 0), in the phase
plane, such that every orbit starting at a point inside C is attracted to
( x1 , 0). Let ( x (t), y(t)) be a solution of (4.37) such that ( x (0), y(0)) lies
inside C . Because F is an integral, F ( x (t), y(t)) is constant. The orbit
of this solution is attracted to ( x1 , 0), so limt→∞ ( x (t), y(t)) = ( x1 , 0).
By continuity, limt→∞ F ( x (t), y(t)) = ( x1 , 0), and since F is constant
on this orbit, F ( x (0), y(0)) = F ( x1 , 0). But ( x (0), y(0)) was arbitrarily
chosen as a point within C ; therefore F ( x, y) = F ( x1 , 0) for all ( x, y)
inside C . In particular F ( x, 0) = F ( x1 , 0) if | x − x1 | < r, where r
denotes the radius of C . Since F ( x, 0) = U ( x ) it follows that U is constant on the interval ( x1 − r, x1 + r ), and hence U 0 ( x ) = 0 there. This
contradicts our assumption that x1 was an isolated critical point.
Return
352
3.
ORDINARY DIFFERENTIAL EQUATIONS
Since U 0 ( x1 ) = − g( x1 ) = 0, and U 00 ( x1 ) = − g0 ( x1 ) > 0, it follows that x1 is a critical point of U ( x ), and, by the second derivative
test, that U ( x1 ) is a relative minimum. Therefore by proposition 4.4.2,
( x1 , 0) is a neutrally stable stationary point of the system (4.37).
Return
SOLUTIONS MANUAL
353
5.
U (x) = −
Z x
0
Potential
Return
a
ax dx = − x2
2
Phase Portrait
354
ORDINARY DIFFERENTIAL EQUATIONS
7.
U (x) = −
Z x
0
Potential
Return
1 dx = − x
Phase Portrait
SOLUTIONS MANUAL
355
9.
U (x) = −
Z x
0
Potential
Return
1
(− x2 ) dx = x3
3
Phase Portrait
356
ORDINARY DIFFERENTIAL EQUATIONS
11.
U (x) = −
Z x
0
Potential
Return
(−4x3 ) dx = x4
Phase Portrait
SOLUTIONS MANUAL
13.
U (x) = −
357
Z x
0
(3s|s|) ds = − x2 | x |
There is one stationary point, located at (0,0). As U ( x ) has a local
maximum at X = 0, the stationary point is a saddle.
Potential
Return
Phase Portrait
358
ORDINARY DIFFERENTIAL EQUATIONS
15.
U (x) = −
Rx
x 2 −1
0 ( x 2 +1)2
dx = −Re
Rxh
0
1
( x + i )2
i
1
dx = Re x+
i =
x
x 2 +1
Because u( x ) has a local minimum at x = −1 and a local maximum at
x = 1, the phase portrait has a center at (−1, 0) and a saddle at (1, 0).
Potential
Return
Phase Portrait
SOLUTIONS MANUAL
17.
U (x) = −
359
Z x
0
1
1
(sin x − ) dx = cos( x ) + x
2
2
There is a doubly infinite sequence of stationary points ( xn , 0) where
7
1
5
sin( xn ) = 12 . Thus, xn = . . . − 11
6 π, − 6 π, 6 π, 6 π, . . . . These stationary
points alternate between saddles and centers, with a saddle at ( 16 π, 0),
a center at ( 56 π, 0), and so on.
Potential
Return
Phase Portrait
360
ORDINARY DIFFERENTIAL EQUATIONS
19.
U (x) = −
Z x
0
sin x −
2
1
x dx = cos( x ) +
x2
π
π
has critical points at x = 0, ± π2 , with a local maximum at 0 and local
minima at ± π2 . Thus the phase portrait has a saddle at (0,0) and centers at ( π2 , 0). The attracted set of the origin is the same as the repelled
set; all orbits except those attracted to the origin are closed.
Potential
Return
Phase Portrait
SOLUTIONS MANUAL
361
1.
L( x, y) = ax2 + bxy + cy2
b2
b 2
y ) + ( c − ) y2
2a
4a
b 2
1
= a( x + y) + (4ac − b2 )y2
2a
4a
= a( x +
Thus if b2 − 4ac < 0, and a > 0, then L( x, y) ≥ 0; equality holds if
b
1
and only if a( x + 2a
y)2 = 4a
(4ac − b2 )y2 = 0, and hence x = y = 0.
Therefore, L( x, y) is positive definite.
Conversely, if a ≤ 0, then L(1, 0) ≤ 0 so L is not positive definite. If
a > 0 and b2 − 4ac ≥ 0, then L(−b, 2a) = (4ac − b2 ) a ≤ 0, and again
L is not positive definite.
Return
362
3.
ORDINARY DIFFERENTIAL EQUATIONS
The function L( x, y) = x2 + y2 is positive definite, and
L0 ( x, y) = 2x (−y + x3 ) + 2y( x + y3 )
= 2( x 4 + y4 )
L0 ( x, y) is also positive definite, therefore L( x, y) is a Lyapunov function for the given system and the origin is unstable.
Return
SOLUTIONS MANUAL
363
5.
The function L( x, y) = x2 + y2 is positive definite, and
L0 ( x, y) = 2x (y − sin x ) + 2y(− x − sin y)
= −2x sin x − 2y sin y
When ( x, y) is in a small disk centered at the origin, L0 ( x, y) ≈ −2x2 −
2y2 ; thus L0 ( x, y) is negative definite. It follows that L( x, y) is a Lyapunov function for the given system, and that the origin is stable and
asymptotically stable.
Return
364
ORDINARY DIFFERENTIAL EQUATIONS
7.
The function L( x, y) = x2 + y2 is positive definite, and
L0 ( x, y) = 2x ( x3 + x + y) + 2y(−2x + y)
= 2x4 + 2x2 − 2xy + 2y2
= 2x4 + x2 + y2 + ( x − y)2
L0 ( x, y) is positive definite, therefore L( x, y) is a Lyapunov function
for the given system and the origin is unstable.
Return
SOLUTIONS MANUAL
9.
365
Since the function L( x, y) = ax2 + cy2 is required to be positive
definite, a and c must be positive. We will now calculate L0 .
L0 ( x, y) = 2ax (− x + y) + 2cy(− x − y)
= −2ax2 − 2cy2 + (2a − 2c) xy
If we take a = c = 1, then L0 ( x, y) is negative definite, and hence
L( x, y) = x2 + y2 is a Lyapunov function that shows the origin to be
asymptotically stable.
Return
366
11.
ORDINARY DIFFERENTIAL EQUATIONS
Since the function L( x, y) = ax2 + cy2 is required to be positive
definite, a and c must be positive. We will now calculate L0 .
L0 ( x, y) = 2ax (2x − y) + 2cy(40x + 2y)
= 4ax2 + (80c − 2a) xy + 4cy2
If we take a = 40, c = 1 then L0 ( x, y) is positive definite, and hence
L( x, y) = 40x2 + y2 is a Lyapunov function that shows the origin to
be unstable.
Return
SOLUTIONS MANUAL
13.
367
Since the function L( x, y) = ax2 + bxy + cy2 is required to be positive definite, a > 0 and b2 − 4ac < 0 . We will now calculate L0 .
L0 ( x, y) = (2ax + by)( x − y) + (2cy + bx )(50x − y)
= (2a + 50b) x2 + (−b − 2c)y2 + (b − 2a + 100c − b) xy
If we take a = 50, b = −2, c = 1 then L0 ( x, y) ≡ 0 is negative semidefinite, and hence L( x, y) = 50x2 − 2xy + y2 is a Lyapunov function that
shows the origin to be neutrally stable but not asymptotically stable.
Return
368
15.
ORDINARY DIFFERENTIAL EQUATIONS
Since the function L( x, y) = ax2 + bxy + cy2 is required to be positive definite, a > 0 and b2 − 4ac < 0 . We will now calculate L0 .
L0 ( x, y) = (2ax + by)[2x + 5(y2 − y)] + (2cy + bx )[5( x − x2 ) − 4y]
= (4a + 5b) x2 + (−5b − 8c)y2 + (2b − 10a + 10c − 4b) xy
+5y2 (2ax + by) − 5x2 (2cy + bx )
By proposition 4.5.8, L0 ( x, y) will be positive or negative definite if
the expression obtained by deleting the third and higher order terms
is also positive or negative definite. We will therefore consider the
expression
L20 ( x, y) = (4a + 5b) x2 + (−5b − 8c)y2 + (2b − 10a + 10c − 4b) xy.
If we put a = 5, b = −5 and c = 4, the coefficient of xy will vanish,
giving the expression L20 ( x, y) = −5x2 − 7y2 , which is negative definite. Therefore L0 ( x, y) is also negative definite, and hence L( x, y) =
5x2 − 5xy + 4y2 is a Lyapunov function that shows the origin to be
asymptotically stable.
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SOLUTIONS MANUAL
17.
369
Since the function L( x, y) = ax2 + cy2 is required to be positive
definite, a and c must be positive. We will now calculate L0 .
L0 ( x, y) = 2ax (3y − x3 − xy2 ) + 2cy(−4x − x2 y − y3 )
= (6a − 8c) xy − 2ax4 − 2ax2 y2 − 2cx2 y2 − 2cy4
If we take a = 4, c = 3 then L0 ( x, y) is negative definite, and hence
L( x, y) = 4x2 + 3y2 is a Lyapunov function that shows the origin to
be asymptotically stable.
Return
370
19.
ORDINARY DIFFERENTIAL EQUATIONS
If q( x, y) is negative semidefinite, then L0 ( x, y) is negative semidefinite, so L( x, y) is still a Lyapunov function for the system. If q( x, y) ≡
0, then L( x, y) is an integral of the system, and hence the origin is neutraly stable. On the other hand, if q( x, y) = −y2 then L0 ( x, y) vanishes
only on the x-axis, and is negative for y 6= 0. Because the system 4.49
reduces to x 0 = 0, y0 = x in this case, orbit are vertical as they intersect the x-axis, and hence the x-axis contains no invariant sets other
than the origin. It follows that the origin is a stable stationary point
in this case.
We conclude that if q( x, y) is only known to be negative semidefinite,
then the origin is a stable stationary point of (4.49), but may or may
not be asymptotically stable.
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SOLUTIONS MANUAL
371
21.
(a)
L0 ( x, y) = 2( x2 + y2 )2 (1 − x2 − y2 )
L0 ( x, y) is positive definite when ( x, y) is in the unit disk. Thus
the origin is unstable.
(b) Since q(cos t, sin t) = 0, substituting x = cos t, y = sin t in the
system (4.49) gives the identities
(cos t)0 ≡ − sin t + (cos(t))(0)
(sin t)0 ≡ cos t + (sin(t))(0)
Thus x = cos t, y = sin t is a solution of the system ((4.49). The
orbit of this solution is the unit circle, traversed, counterclockwise.
(c) Let r2 = x2 + y2 , and θ = arg( x + iy) be polar coordinates. Then
2rr 0 = 2xx 0 + 2yy0
=
=
=
=
2x [−y + xq( x, y)] + 2y[ x + yq( x, y)]
2( x2 + y2 )q( x, y)
2( x 2 + y2 )2 (1 − x 2 − y2 )
2r4 (1 − r2 )
Thus the radial coordinate satisfies the autonomous ODE r 0 =
r3 (1 − r2 ), which has stationary points 0, ±1. The phase diagram
of this ODE is
- s
s
- s
−1
0
1
It follows that if an orbit starts at a point other than the origin,
the radial coordinate converges to 1.
y
If we differentiate the relation tan(θ ) = x , we obtain
sec2 (θ )θ 0 =
x y0 − x 0 y
.
x2
Noting that q = r2 (1 − r2 ) in polar coordinates, we have
x y0 = r cos(θ )(−r cos(θ ) + r sin(θ )r2 (1 − r2 ))
x 0 y = (r sin(θ ) + r cos(θ )r2 (1 − r2 ))r sin(θ )
(55)
372
ORDINARY DIFFERENTIAL EQUATIONS
Combining these, we have
x y 0 − x 0 y = −r 2 .
Substitute this result in (55) to obtain
sec2 (θ )θ 0 = −
r2
= − sec2 (θ ),
x2
from which it follows that θ 0 = −1. We have thus established
that all orbits spiral clockwise toward the unit circle, which is
therefore a limit cycle.
(d)
2
1.5
1
0.5
-2
-1.5
-1
-0.5
0.5
-0.5
-1
-1.5
-2
Return
1
1.5
2
SOLUTIONS MANUAL
23.
373
The system has a limit cycle for any negative value of k. Here
are two phase portraits illustrating the limit cycle for k = −1 and for
k = −10.
k=-1
4
k=-10
20
2
-2
-1
Return
10
1
2
-2
-1
1
-2
-10
-4
-20
2
374
ORDINARY DIFFERENTIAL EQUATIONS
25.
0 0
−1 0
is nilpotent. There= − I + N, where N =
T
1 0
= I+tN =
and e A t = e−t e Nt . Now we calculate
t 1
T
T
1 0
1 t
Y (t) = e A t (e A t )T
= e−2t
t 1
0 1
1
t
= e−2t
t 1 + t2
The matrix
fore e Nt
AT
Integrating, we get
S=
Z ∞
Y (t) dt =
e
−2t
0
=
1
2
1
4
− 12
− 21 t −
1
4
.
3
1
4
− 12 t − 14
−12t2 − 12 t −
3
4
∞
0
4
We can directly verify that L is positive definite,
L(~x ) = ~x T S~x =
1 2 3 2 1
1
1
5
x + x + x1 x2 = ( x1 + x2 )2 + x22
2 1 4 2 2
2
2
8
is positive definite. Also,
−1 0
T
A S + SA =
1 −1
is negative definite.
Return
1
2
1
4
1
4
3
4
+
1
2
1
4
1
4
3
4
−1 1
0 −1
=
−1 0
0 −1
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