MAT244H1

MAT244H1
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Introduction
INTRODUCTION TO DIFFERENTIAL EQUATIONS
A differential equation is an equation involving some hypothetical function and its derivatives.
Example
y ′′ + 2 y ′ = x
is an differential equation. As such, the differential equation is a description of some function
(exists or not).
A solution to a differential equation is a function that satisfies the differential equation.
Example
y = x 3 + 5x
•
•
•
•
is a solution to
y ′′ = x + y − x 3
.
Some differential equations are famous/important:
y′ = y y = e x
,
.
ax
y ′ = ay y = e
,
.
y ′′ + y = 0 y = cos x
,
.
y ′′ + ay = 0 y = cos a x
,
.
Recall that a differential equation describes a phenomenon in terms of changes. For example, if P = mv , then
dP
=F
Fdt = dm ⋅ v or dt
.
Example
A pool contains V liters of water which contains M kg of salt. Pure water enters the pool at a constant rate of v
liters per minute, and after mixing, exits at the same rate. Write a differential equation that describes the
density of salt in the pool at an arbitrary time t.
M (t )
ρ (t ) =
ρ
(
t)
V .
• Let
be the density at time t. Then
• To model change in
ρ 2 − ρ1
v
V ≈ − ρ1
t 2 − t1
V
ρ (t ) , let ρ1 = ρ (t1 ) and ρ 2 = ρ (t 2 ) . Then (ρ 2 − ρ1 )V ≈ − ρ 1v(t 2 − t1 ) , so
ρ (t + ∆t ) − ρ (t )
v
V ≈ − ρ (t )
or
t + ∆t − t
dρ
v
v
= − ρ (t )
ρ ′ = −ρ
V or
V .
• Now, as t → 0 , dt
V .
• Without solving this equation, we can predict facts about this system. As t → ∞ , ρ → 0 .
dρ
v
= − dt
ρ
V
• To solve this differential equation, write
. Integrating both sides, we get
ln ρ (t ) = −
v
t +C
V
ρ (t ) = e
v
− t +C
V
= eC e
v
− t
V
= Ae
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v
− t
V
.
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ρ (0 ) = ρ to
• If we add
ρ ′ = −ρ
v
V , then we have an IVP (initial value problem).
ISSUES ABOUT THE USE OF DIFFERENTIAL EQUATIONS
1) How to translate a real problem to a differential equation. Keep your eyes open!
2) Some patterns of nature are ill-defined. Use different points of view and different mental differential equation
models to reformulate them.
3) Differential equations have infinitely many solutions. Which one is yours? The initial value are extremely
important.
4) There may be no analytic solution found.
• Is there a solution?
• Is this solution unique?
• If a numeric answer is required, i.e. the value of the solution at one particular point, then use numerical
approximations. It does not give any feelings for the pattern, nor does it give elbow room.
• Use theoretical analysis if you need the behavior of the solution. This does not give any values.
• To know the behavior locally/in a neighborhood, solve in series.
5) The data does not fit you solution. You need to repeat (as in a feedback/controlled system).
NOTATIONS WITH REGARD TO THE INPUT/OUTPUT SYSTEMS
Example
xy ′′ + 2 y ′ − sin x ⋅ y = tan x can be written as L[ y ] = tan x . Solve it, and the answer is the output.
•
L[ y ] is the “black box system”.
•
tan x is the ‘input”.
For theoretical purposes, mathematicians use these equivalents: y ′′′ =
y ′′′ = f ( x, y , y ′) or F (x, y, y ′, y ′′, y ′′′) = 0 .
LINEAR VS. NON-LINEAR DIFFERENTIAL EQUATIONS
1
1
y′ + e x y =
is linear.
x
1 + tan x
•
tan x ⋅ y ′′ +
•
y ′′′ + y ′ + y 2 = 0 is non-linear.
First Order Differential Equations
LINEAR EQUATIONS
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2 y′
tan x sin x
+
y−
is the same as
x
x
x
MAT244H1a.doc
First order linear equations have the form y ′ + p(t ) y = g (t ) .
Derivation
• Suppose I can find µ (t ) so that µ (t ) p(t ) = µ ′(t ) .
• Multiply both sides of y ′ + p(t ) y = g (t ) by µ (t ) : µ (t ) y ′ + µ (t ) p(t ) y = µ (t )g (t ) which is
µ (t ) y ′ + µ ′(t ) y = µ (t )g (t ) , i.e. (µ (t ) y )′ = µ (t )g (t ) .
• Integrate both sides: µ (t ) y = µ (t )g (t )dt + C , so y (t ) =
• But what is µ (t ) ? Since µ (t ) p(t ) = µ ′ ⇔
1
µ (t )g (t )dt + C .
µ (t )
µ ′(t )
= p(t ) , therefore ln µ (t ) =
µ (t )
p(t )dt . So µ (t ) = e p (t )dt .
General Solution
To solve y ′ + p(t ) y = g (t ) ,
1) Let µ (t ) = e p (t )dt (no constant needed).
1
2) The solution is y (t ) =
µ (t )g (t )dt + C .
µ (t )
[
]
Example
2
Solve y ′ + 2ty = 2te −t .
• Here, p(t ) = 2t , g (t ) = 2te −t .
2
• Let µ (t ) = e
2tdt
2
= et .
• The solution is y (t ) =
1
t2
[e
e
• Note that as t → ∞ , y → 0 .
t2
2
]
2te −t dt + C =
1
e
t2
[ 2tdt + C ]= e1 (t
t2
2
)
2
2
+ C = t 2 e −t + Ce −t .
Importance of Analysis
One needs to have an understand of the solution before (even after) solving it, with respect to:
1) Behavior of the solution as t → ∞ .
2) The nature and behavior of the solution within a family (depends on y0).
Variation of Parameter
Recall that the family of solutions of a first order linear differential equation y ′ + p (t ) y = g (t ) , µ (t ) = e p (t )dt ,
1
1
C
y (t ) =
µ (t )g (t )dt + C =
µ (t )g (t )dt +
. Notice that the family of solutions is generated by
µ (t )
µ (t )
µ (t )
µ (t ) . This leads to the technique of variation of parameter.
[
]
Recall a differential equation L[ y ] = g (x ) . If g (x ) = 0 , then we have a zero-input system, or a homogeneous
differential equation L[ y ] = 0 which describes the solutions to a great extent.
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1
1
For example, consider y ′ + y = 3 cos 2t . First solve the corresponding homogeneous equation y ′ + y = 0
t
t
1
dt
t
1
(0 + C ) = C . Then the general solution to
µ (t )
t
1
1
1
1
y ′ + y = 3 cos 2t looks like y (t ) = A(t ) . Then y ′ = A′(t ) − A(t ) 2 , so
t
t
t
t
1
1 1
1
A′(t ) − A(t ) 2 + A(t ) = 3 cos 2t
A′(t ) = 3t cos 2t
A(t ) = (3t cos 2t )dt . Thus,
t
t
t
t
1
(3t cos 2t )dt + C .
y (t ) =
t
to find y 0 (t ) . Since µ (t ) = e
[
= t , y 0 (t ) =
]
ASYMPTOTIC BEHAVIOR OF SOLUTIONS
Recall from calculus that f (x ) and g ( x ) are asymptotes of one another if lim ( f (x ) − g (x )) = 0 .
x →∞
Example
2t − 5 and 2t − 5 + ce −t and 2t − 5 +
c
are asymptotic to each other.
t
SEPARATION OF VARIABLES
Idea
A differential (not necessarily linear) may appear as
g ( y )dy =
dy f (x )
=
. Then, g ( y )dy = f (x )dx , and the solution is
dx g ( y )
f (x )dx .
Note
Other ways a separable differential equation can appear as: M ( y ) =
dy
dy
N (x ) or f (x )g ( y ) =
.
dx
dx
Example
dy
1
= x y . So
dy = xdx
dx
y
y
−
1
2 dy
1
= xdx
2y 2 =
x2
+C.
2
Note
The solution about is an implicit solution. When write a solution explicitly, be careful! Pay attention to the
domain and range.
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Example
dy
= x3 y
dx
y
2
1
3 dy
2
y3
3 3 x2 1
y =
−
2
2 2
x2 1
− >0
3 3
−
2
=
3
3 3 x2
y =
+ C . Now, suppose y (2) = 1 , then = 2 + C
2
2
2
xdx
x2 1
=
−
3 3
x 2 −1 > 0
x2 >1
x2 1
y=
−
3 3
3
2
C=−
1
. So
2
. We need
x > 1, x < −1 .
IMPLICIT VS. EXPLICIT SOLUTIONS
For separable (not linear in general), we have implicit solutions (not necessarily functions). So as the
solutions to non-linear equations are implicit, the analysis of the solution is very difficult, and we need to
know if such solutions have explicit forms or not (and where). We need to know an interval on which explicit
solutions exists.
Example
Consider 2 y
dy
= 1 . The solution is y 2 = x + c .
dx
• Now, if y (1) = 1 , then c = 0 and the solution is y 2 = x . An explicit solution y = x exists on the
interval (0, ∞) . This solution can’t be extended to y < 0 because it doesn’t pass the vertical line test.
dy
is undefined. This indicates the possibility of a problem with defining an
dx
explicit solution to the differential equation.
• At the point (0,0) ,
Theorem: Implicit Function Theorem
If F (x, y ) = 0 and (a, b ) is such that F (a, b ) = 0 and if F y (a, b ) ≠ 0 , then we have an explicit function
y = f (x ) on an interval containing (a, b ) .
Conclusion: On any interval as long as
dy
is defined, we will have an explicit solution.
dx
Example
dy x + 3 y
Solve
=
using separation of variables.
dx
x− y
• Let v =
y
x
y = vx
dy
dv
= v+ x
.
dx
dx
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dy x + 3 y
=
=
• So
dx
x− y
1− v
(1 + v )
2
dv =
y
x becomes v + x dv = 1 + 3v
y
dx 1 − v
1−
x
1+ 3
dv 1 + 3v
(1 + v )2 . So we have
=
−v =
dx 1 − v
1− v
1
dx .
x
• Let w = 1 + v , dv = dw . So
2
− − ln w = ln x + C
w
−
x
2−w
2
dw =
1
dx
x
2
2
w
w
2
−
− ln 1 + v = ln x + C
−
1+ v
x+ y
2x
− ln
= ln x + C
x+ y
x
−
1
dw = ln x + C
w
dw −
2
1+
y
x
− ln 1 +
2x
− ln x + y + ln x = ln x + C
x+ y
y
= ln x + C
x
−
2x
− ln x + y = C .
x+ y
• Our techniques (integration) limits us to 1 + v = w ≠ 0 , but are we excluding some solutions? Now
x + 3 y −2 x
y x+ y
dy
1+ v = 1+ =
= 0 means y = − x . Notice that if y = − x ,
= −1 and
=
= −1 , so
x− y
2x
x
x
dx
we can add y = − x to the family of solutions.
ISSUES ON MODELING
Example: Money Growth
• If we say that annual rate of interest is 5%, we mean that on $100, we get $5 in one year. So
P(1) − P(0) = 0.05P(0 ) .
• Equivalently,
dP(t )
= 0.05P (t ) . So P(t ) = Ae 0.05t
dt
P(1) = P (0)e 0.05 = P(0) 1 + 0.05 +
P(1) − P(0) = 0.05P(0) +
(0.05)2
2!
+
P(t ) = P(0)e 0.05t . So
= P (0) + 0.05P(0) +
(0.05)2 P(0) +
2!
, so
(0.05)2 P(0) +
> 0.05P(0) .
2!
• If we say the annual interest rate r is compounded semi-annually, we mean
P
r
r
r
r
1
r
1
r
1
− P(0) = P (0) ⇔ P
= P(0) 1 +
. So, P(1) = P
1+
= P(0) 1 +
1+
= P(0) 1 +
2
2
2
2
2
2
2
2
2
• In general, when we have an annual interest rate r compounding n times a year,
• Compounding continuously means n → ∞ . So P(1) = lim P(0 ) 1 +
n →∞
• If we contribute continuously to the account, then
r
n
n
r
P(1) = P (0) 1 +
n
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.
n
.
= P(0)e r . Similarly, P(1) = P(0)e rt .
dP(t )
= rP (t ) + k where k is the constant contribution.
dt
EXISTENCE OF A UNIQUE SOLUTION
2
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Example
• Recall that a differential equation can be built starting from the solution like
y= x
y=− x
. They both can be
“expressed” by y 2 = x and 2 yy ′ = 1 .
• Notice that the differential equation 2 yy ′ = 1 at (0,0) is a confused initial value problem.
Note
Recall that given a first order differential equation y ′ + p(t ) y = g (t ) , the solution is
[
]
1
µ (t )g (t )dt + C where µ (t ) = e
µ (t )
functions, then we have a good solution.
y=
p (t )dt
. So as long as p(t ) and g (t ) are integrable (continuous)
Theorem
If p(t ) and g (t ) are continuous on I = [α , β ] , then for any value y0 (that is already given), there is a unique
solution y = φ (t ) that on I it satisfies y ′ + p(t ) y = g (t ) , y 0 = y (t 0 ) where t 0 ∈ I .
Theorem
Let y ′ = f ( y, t ) . If there is an “open window” I × J on which f and
∂f
are continuous, then the initial value
∂y
problem y ′ = f ( y, t ) , y 0 = y (t 0 ) has an unique solution (for any t 0 ∈ I and y 0 ∈ J ⇔ (t 0 , y 0 ) ∈ I × J ).
Note
Notice that y ′ =
1
is not continuous at any neighborhood of (0,0) , so the theorem doesn’t apply.
2y
Example
(
)
3
For which initial values does y ′ = t 2 + y 2 2 = f ( y, t ) have a unique solution?
• f is continuous everywhere.
1
∂f 3 2
•
= t + y 2 2 2 y is also continuous everywhere.
∂y 2
(
)
(
)
• So y ′ = t 2 + y 2
3
2
has a unique solution for all initial values.
Example
For which initial values does y ′ =
cot y
= f ( y, t ) have a unique solution?
1+ y
• f is discontinuous at y = −1 and y = kπ .
•
∂f csc 2 y (1 + y ) − cot y
=
is also discontinuous at y = −1 and y = kπ .
∂y
(1 + y )2
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BERNOULLI
Bernoulli solved y ′ + p(t ) y = y n g (t ) as follows:
v = y 1− n , and v ′ =
y′
y
n
+ p(t )
dv
= (1 − n ) y 1− n −1 y ′ = (1 − n ) y − n y ′
dt
y
yn
= g (t )
y − n y ′ + y 1− n p(t ) = g (t ) . Let
1
v ′ = y − n y ′ . So now we get
1− n
1
v ′ + p(t )v = g (t ) .
1− n
AUTONOMOUS DIFFERENTIAL EQUATIONS
Autonomous differential equations look like y ′ = f ( y ) .
Examples
1) y ′ = 0 ; y = c .
2) y ′ = k ; y = kx + c .
3) y ′ = ry exponential growth.
4) y ′ = y (1 − y ) is know as logistic growth.
Example: Logistic Growth Model
Consider the spread of a disease. If y (t ) is the number of infected population, then
dy
≈ y (k − y ) . So
dt
dy
y
y
= ay(k − y ) = aky 1 −
= ry 1 −
, k is “environmental carrying capacity” or “saturation level” and r is
dt
k
k
the “intrinsic growth rate”.
To solve it,
dy
y 1−
ln
y
y
1−
k
= rt + C
y
k
= rdt
y
y
1−
k
= Ae rt
1
+
y
1
1−
y
k
dy = rdt
y = Ae rt − Ae rt
y
k
ln y − ln 1 −
y + Ae rt
y
= rt + C
k
y
= Ae rt
k
y 1+
A rt
e = Ae rt
k
y0
Ae rt
A
kA
k
=
= − rt
=A
. Now, y (0 ) = y 0 means
= A . So
y0
A
A ke + A
k − rt
1 + e rt e − rt +
e +1
1−
k
k
A
k
ky 0
. As t → ∞ , y → k .
y (t ) = =
y 0 + (k + y 0 )e − rt
y (t ) =
Example
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An important model:
dy
k
= ry (M − ln y ) = ry (ln k − ln y ) = ry ln .
dt
y
Example: Logistic Growth With Threshold
Some times we need enough of y0 to start the epidemic. Let
points at y = 0, k , T . Now we get
dy
= −ry (k − y )(T − y ) = f ( y ), r > 0 , with critical
dt
dy
y
y
= ry 1 −
1−
.
dt
k
T
CRITICAL POINTS OR EQUILIBRIUMS OF AN AUTONOMOUS DIFFERENTIAL EQUATION
Example
y ′ = y (1 − y ) = 0 gives constant solutions y (t ) = 0 and y (t ) = 1 . These are the critical points or the
equilibriums.
y
semi-stable
equilibrium
y (t ) = 1
unstable
equilibrium
t
y (t ) = 0
Example
y ′ = y (1 − y ) = 0 gives constant solutions y (t ) = 0 and y (t ) = 1 .
y
asymptotically
stable equilibrium
y (t ) = 1
y (t ) = 0
t
Example: Schaefer Model
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Let y (t ) be the fish population at time t, which follows the logistic growth model. Then
where E the rate of harvest which is proportional to the fish population. Rewriting
equilibriums are at y (t ) = 0 and r 1 −
is at y =
y (t )
−E =0
k
dy
y
= ry 1 −
− Ey ,
dt
k
dy
y
= y r 1−
− E , the
dt
k
k
y (t ) = (r − E ) . Now, if r > E , the stable equilibrium
r
k
(r − E ) . But if r < E , the stable equilibrium is at y = 0 .
r
PARAMETRIC DIFFERENTIAL EQUATIONS
Example
Consider y ′ = ay − y 2 = y (a − y ) . The equilibriums are at y = 0 and y = a .
• When a > 0 , we have a stable equilibrium around y = a .
• When a = 0 , y ′ = y 2 and we have a semi-stable solution around y = 0 .
• When a < 0 , we have a unstable solution around y = a .
Here, a = 0 is called a bifurcation point.
EXACT DIFFERENTIAL EQUATIONS
Suppose ψ ( x, y ) = c . It implicitly defines a function y ( x ) .
Now
dy
dy
∂ψ ∂ψ dy
d
ψ ( x, y ) = 0 =
+
:= M ( x, y ) + N (x, y )
. So M (x, y ) + N (x, y )
= 0 or
dx
∂x
∂y dx
dx
dx
M (x, y )dx + N (x, y )dy = 0 . This type of differential equations are called exact equations.
Existence of Solution
If we are given M (x, y )dx + N (x, y )dy = 0 , how would we know there is such a ψ (x, y ) corresponding to it?
Indeed the condition
Example
Consider
y
∂M 1
∂N 1
+ 6 x dx + (ln x − 2)dy = 0 = M (x, y ) + N (x, y ) . Since
= and
= , the equation is
x
∂y
x
∂x x
exact. Notice that
Now,
∂M ∂N
=
= is necessary and sufficient for the existence of such ψ (x, y ) .
∂y
∂x
y
∂ψ
= M (x, y ) = + 6 x . Integrating with respect to x, we get ψ (x, y ) = y ln x + 3x 2 + f ( y ) .
∂x
x
∂ψ
= N ( x, y ) = ln x + f ′( y ) = ln x − 2 , so f ′( y ) = −2
∂y
ψ (x, y ) = y ln x + 3x 2 − 2 y + c 0 . The solution is ψ (x, y ) = C
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f ( y ) = −2 y + c 0 . Hence
y ln x + 3 x 2 − 2 y = c .
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INTEGRATING FACTOR
• What if M y ≠ N x ? Does that mean ψ (x, y ) doesn’t exists? Yes, but we may be able to change the
differential equation M (x, y )dx + N ( x, y )dy = 0 to a new “better” one.
• If we multiply M (x, y )dx + N ( x, y )dy = 0 by µ (x, y ) , we may get (µM ) y = (µN ) x .
• How do we know if we should look for such µ? Finding µ is very difficult, unless µ is somehow a µ (x ) or
µ ( y ) only.
Example
• Suppose (µM )dx + (µN )dy = 0 . We check if it is exact, and if it is, that fact should lead us to the answer.
• To be exact, we need (µM ) y = (µN ) x
µ y M − µ x N = µ (N x − M y ) .
µ y M + µM y = µ x N + µN x
(
• Now if µ ( x, y ) = µ (x ) , then µ y = 0 . Then − µ x N = µ N x − M y
(N x − M y )
µx
=−
.
µ
N
)
Example
This tells us a criterion for finding integrating factors of different type: µ ( xy ) . If
there exists a µ ( xy ) .
Nx −M y
xM − yN
= R( xy ) , then
Example
Consider 3x +
Nx −M y
xM − yN
y dy
6
x2
+
+3
= 0 . It is obvious M y ≠ N x , but
y
y
x dx
2
=
y
6
x
−3 2 + 2
y
x
y
2
2
=
y
6
x
−3 2 + 2
y
x
y
2
y
y
x
x
− x2 −3
2x 2 + 6 − 3
y
x
y
x
makes this differential equation exact.
3x 2 + 6
=
1
xy
2x 2 − 3
2x 2 + 6
y2
x
+6
x
y
2
y
x
−3
y
x
=
1
. So there is µ ( xy ) that
xy
Note
We were looking for an integrating factor µ ( xy ) so that µM + µNy ′ = 0 is exact, i.e. (µM ) y = (µN ) x . This
means µ y M + µM y = µ x N + µN x . But since we are requiring that µ ( xy ) is a function of xy, then by letting
w = xy , µ ( xy ) becomes µ (w ) , and µ x =
becomes µ ′xM + µM y = µ ′yN + µN x
dµ ∂w
= µ ′y and µ y = µ ′x . Therefore the condition of exactness
dw ∂x
µ′ Nx − M y
µ ′(xM − yN ) = µ N x − M y
=
= R(w) = R(xy ) .
µ xM − yN
(
)
Now, we can easily solve for µ.
NUMERICAL APPROXIMATION SOLUTIONS TO DIFFERENTIAL EQUATIONS
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If g (t ) is a solution to the differential equation y ′ = f (t , y ) and we need to know y1 = g (t1 ) and an
approximate value is good enough, then we can use the tangent line instead of g (t ) . y ′ =
y1 − y 0 = y ′(t1 − t 0 )
y1 = y ′(t1 − t 0 ) + y 0
y1 = y 0 + f (t 0 , y 0 )(t1 − t 0 ) .
y1 − y 0
t1 − t 0
Euler suggested to evaluate y1 = y 0 + f (t 0 , y 0 )(t1 − t 0 ) , y 2 = y1 + f (t1 , y1 )(t 2 − t1 ) ,
y 3 = y 2 + f (t 2 , y 2 )(t 3 − t 2 ) .
So, to calculate g (T ) , we can take [t 0 , T ] instead of using y1 = y 0 + f (t 0 , y 0 )(t1 − t 0 ) and subdivide it into
t 0 , t1 ,
, t n −1 , T . This way, the answer is a lot closer.
Convergence of Euler’s Method
If we let h → 0 (step size), then the approximate answer equals the actual answer. So for first order
differential equations, it is a good idea to let h → 0 to get a better solution.
Example
Prove that Euler’s method converges for y ′ = y, y (0 ) = 1 . We know the solution is y (t ) = e t .
• Let h =
t
.
n
• Now, y1 = y 0 + y ′(0,1)h = 1 + h , y 2 = 1 + h + y ′(t 2 ,1 + h )h = 1 + h + (1 + h )h = (1 + h )2 , so y n = (1 + h )n .
• If n → ∞ , then (1 + h )n = 1 +
t
n
n
= e t = y (t ) .
EXISTENCE AND UNIQUENESS OF A SOLUTION TO AN INITIAL VALUE PROBLEM
Technique
An initial value problem y ′ = f (t , y ), y (0 ) = y 0 can be transformed into w ′ = g (t , w), w(0 ) = 0 .
Example
Consider y ′ = t 2 + y 2 , y (1) = 2 . Let t = s + 1 , then y (t ) = z (s ) , so the problem becomes
z ′ = (s − 1)2 + z 2 , z (0) = 2 . Now let w = z − 2 ⇔ z = w + 2 , then z ′ = w ′ , so w ′ = (s + 1)2 + (w + 2 )2 . Now,
w(0 ) = z (0 ) − 2 = 2 − 2 = 0 .
Theorem
If y ′ = f ( y, t ) and if f and
∂f
are continuous on the rectangle I × J , then for some h > 0 there is an unique
∂y
solution of the form y = φ (t ), ∀t 0 − h < t < t 0 + h .
Note
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To prove this theorem, we first showed it is alright to assume y 0 = 0 and t 0 = 0 , so we assume that our IVP
is of the form y ′ = f ( y, t ), y (0) = 0 . The method of the proof is Picard’s Method
Note
Observe that y = φ (t ) is a solution to y ′ = f ( y, t ), y (0) = 0 . So it also satisfies y ′ = φ ′(t ) = f (φ (t ), t ) and
t
t
0
0
y = φ (t ) = φ ′(s )ds =
f (φ (s ), s )ds . So the IVP is equivalent to an integral equation.
0
Notice that φ (0) = φ ′(s )ds = 0 , i.e. y (0) = 0 .
0
Picard’s Method
1) Find a function φ 0 (t ) that satisfies φ 0 (0) = 0 (ex: φ 0 (t ) = 0 ).
2) Define φ n (t ) such that φ k +1 =
t
0
f (φ k (s ), t )ds, k = 0,1,
.
So we create a sequence of {φ n (t ), n = 0,1, } . Now:
1) Is this an infinite sequence?
• If at some point we stop producing new functions when n = k , then we already have the solution
φ k (t ) of the integral equation.
• If the sequence is infinite, then does it converge? To converge, we need g n (t ) = φ n +1 (t ) − φ n (t ) to
converge, i.e. lim g n (t ) = 0, ∀t ∈ [t 0 − h, t 0 + h] , or lim φ n (t ) = φ (t ), ∀t ∈ [t 0 − h, t 0 + h] .
n →∞
•
n →∞
} is defined, and φ (t ) = lim φ n (t ) exists, then what properties does φ (t ) have?
2) Suppose {φ n (t ), n = 0,1,
n→ ∞
φ (0) = lim φ n (0) = lim 0 = 0 .
n →∞
t
•
0
f (φ (s ), s )ds =
n →∞
t
0
f lim φ n (s ), s ds =
So φ (t ) is a solution.
n →∞
t
lim f (φ n (s ), s )ds = lim
0n →∞
t
n →∞ 0
f (φ n (s ), s )ds = lim φ n −1 (t ) = φ (t ) .
n →∞
Steps of the Proof of the Existence and Uniqueness Theorem
We constructed a sequence of functions {φ n (t ), n = 0,1, } with the hopes that the sequence converges to the
solution φ (t ) . We saw that lim φ n (t ) satisfies the criteria for the solution.
n →∞
1) When does this sequence converge? If
f (t , y1 ) − f (t , y 2 ) ≤ K y1 − y 2 ⇔
2)
∂f
exits and is continuous in D, then there is a number K such that
∂y
f (t , y1 ) − f (t , y 2 )
y1 − y 2
≤K.
f (t , φ n (t )) − f (t , φ n −1 (t )) ≤ K φ n (t ) − φ n −1 (t ) .
3) Because f (t , y ) is continuous on D, then for some h, for all t 0 − h < t < t 0 + h , f (t , y ) ≤ M . Then
φ1 (t ) =
t
0
f (s, φ 0 (s ))ds ≤ M t , and because φ 0 = 0 , φ1 (t ) − φ 0 (t ) ≤ M t . Also,
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t
φ 2 (t ) − φ1 (t ) ≤
0
t
t
2
0
0
2
(MKs )ds = MK t
f (s, φ1 (s )) − f (s, φ 0 (s ))ds ≤ K φ1 (s ) − φ 0 (s )ds ≤
MK 2 t n +1
.
(n + 1)!
4) Now, notice that φ n (t ) − φ1 (t ) = (φ n (t ) − φ n −1 (t )) + (φ n −1 (t ) − φ n − 2 (t )) +
. So,
φ n (t ) − φ n −1 (t ) ≤
inequality, φ n (t ) − φ1 (t ) ≤ φ n (t ) − φ n −1 (t ) +
φ n (t ) ≤ φ n (t ) + φ n (t ) − φ n −1 (t ) +
+ φ 2 (t ) − φ1 (t ) . Finally,
+ φ 2 (t ) − φ1 (t ) ≤ M t + MK
t2
+
2
+ (φ 2 (t ) − φ1 (t )) , so by the triangle
=
(Kh )2 +
M
Kh +
K
2!
. As n → ∞ , it
M
M
, so all the φ n are bounded by e Kh
.
K
K
5) We need to prove the uniqueness of this solution. Suppose there are two solutions, φ (t ) and ψ (t ) . Then
converges to e Kh
φ (t ) −ψ (t ) =
U (t ) =
t
0
t
f (s, φ (s )) − f (s,ψ (s ))ds ≤
0
0
t
f (s, φ (s )) − f (s,ψ (s )) ds ≤ K φ (s ) −ψ (s ) ds . Now let
0
t
φ (s ) −ψ (s ) ds , where U (t ) ≥ 0 . Notice U ′(t ) = φ (t ) −ψ (t ) ≤ K φ (s ) −ψ (s ) ds = KU (t ) . So
e − kt U ′ − e − kt KU ≤ 0
U ′ − KU ≤ 0
But
t
U (t ) ≥ 0
U (t ) = 0
U (t ) ≤ 0
(e
− kt
U
) ≤ 0 , therefore, (e
′
t
0
− ks
0
)
′
U (s ) ds = e − kt U (t ) ≤ 0
U (t ) ≤ 0 .
φ (t ) = ψ (t ) .
First Order Difference Equations
Examples of famous discrete process are stochastic processes y n +1 = f (n, y n , y n −1 ,
, y 0 , b) .
However, we can only study simple ones y n +1 = f (n, y n ) . Compare it with y ′ = f (t , y ) . Notice that
y n +1 − y n
(n + 1) − n
=
f (n, y n ) − y n
= g (n, y n ) ≈ y n′ = g (n, y n ) .
(n + 1) − n
A solution to a difference equation is a sequence of numbers y = y1 ,
Notice that the sequence defines a function with domain {0,1,2,
} . So,
, y n that satisfies y n +1 = f (n, y n ) .
y 0 = y (0) , y1 = y (1) , etc.
Notice that if the increments become small, the difference equation becomes a differential equation.
Example
Solve y n +1 =
•
y1 =
n+3
y n with respect to y0.
n +1
0+3
3
1+ 3
y0 =
y0 , y2 =
y1 =
0 +1
1
1+1
• So, y n =
4×3
y0 , y3 =
2 ×1
(n + 2)(n + 1) (4)(3) y = (n + 2)(n + 1) y .
0
0
n(n − 1) (2 )(1)
2
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2+3
y2 =
2 +1
5× 4×3
y0 .
3 × 2 ×1
MAT244H1a.doc
Definitions
1) We say the annual rate is r compounding monthly to mean the period is one monthly and the rate on that
r
.
period is
12
2) Effective annual rate equivalent to rate r compounding monthly is the rate re (compounding annually) such
r
that the effect of re over a year is the same as the effect of
for 12 periods. That is, if Y1 = (1 + re ) y 0 and
12
y12 = 1 +
r
12
12
y 0 , then Y1 = y12 .
Note
To calculate the effective annual rate, 1 + re = 1 +
r
12
12
.
AUTONOMOUS DIFFERENCE EQUATIONS
The simplest of difference equations is autonomous linear y n +1 = f ( y n ) where f ( y n ) = ρy n + b .
Solution
To solve an autonomous linear difference equation, note that:
• y1 = f ( y 0 ) = ρy 0 + b .
•
•
y 2 = f ( y1 ) = ρy1 + b = ρ (ρy 0 + b ) + b = ρ 2 y 0 + ρb + b .
(
)
y 3 = f ( y 2 ) = ρ ρ 2 y 0 + ρb + b + b = ρ 3 y 0 + ρ 2 b + ρb + b .
So y n = ρ n y 0 + ρ n −1b +
(
)
+ ρb + b = ρ n y 0 + b ρ n −1 +
+ 1 = ρ n y0 + b
ρ n −1
.
ρ −1
Note
Notice that if ρ < 1 , then as n → ∞ , y n = ρ n y 0 + b
y n = ρ n y0 + b
ρ n −1
1
→b
. However, if ρ > 1, ρ > 0 , as n → ∞ ,
ρ −1
1− ρ
ρ n −1
1
→b
; but if ρ > 1, ρ < 0 , we are confused (the limit doesn’t exist)!
ρ −1
1− ρ
Example
Consider a 20 year mortgage at rate 10% (compounded semi-annually) with monthly payment of b = 1000 .
What will the maximum loan be under the these restrictions?
• Model: y n +1 = (1 + i ) y n − 1000 . At n = 12 × 20 = 240 , y 240 = 0 .
r
• What is the periodic rate i? 1 +
2
2
= (1 + i )12
1
(1.05) 6
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= 1 + i = 1.00816 .
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(1 − i )n − 1 . y = 0 (1 − i )n y − b (1 − i )n − 1 = 0
0
(1 − i ) − 1 240
(1 − i ) − 1
240
(1.00816) − 1 = 105121.37 .
y 0 = 1000
0.00816 × (1.00816)240
• So y n = (1 − i )n y 0 − b
• So the maximum loan is $105121.37.
LOGISTIC GROWTH EQUATION
Logistic growth equations have the form u n +1 = ρu n (1 − u n ) .
Equilibrium
Equilibrium solutions are obtained when u n +1 = u n , i.e. u n = ρu n (1 − u n )
ρu n − ρu n 2 − u n = 0
u n (ρ − ρu n − 1) = 0
u n = ρu n − ρu n 2
un = 0
ρ − 1 . So different values of ρ present different
ρ
equilibriums: if ρ = 1 , then the two solutions are the same; if ρ ≈ 1 , then the two equilibriums are very close.
un =
Stability of the Equilibriums
1) Suppose a system starts near the 0 equilibrium, and suppose for simplicity that it is linear, i.e. u n +1 = ρu n . If
u n ≈ 0 (very near the equilibrium 0), then we may assume any logistic growth would be u n +1 = ρu n − ρu n 2 .
Then u n 2 is closer to 0, so we can assume u n +1 ≈ ρu n . We know as n → ∞ , u n → ρ n u 0 , so u n → 0 if
ρ < 1 . Therefore the logistic growth difference equation u n +1 = ρu n (1 − u n ) will converge to 0 (the
equilibrium) if we start near 0 and ρ < 1 . More difficult analysis guarantees this situation is true for nonlinear ones.
2) The behavior near the
ρ −1
equilibrium. Let vn be an amount of perturbation that defines a new solution near
ρ
ρ −1
ρ −1
, that is u n =
+ v n , v n ≈ 0 . This is a solution and needs to be reformulated to a process. Notice that
ρ
ρ
ρ −1
ρ −1
ρ −1
u n +1 =
+ v n +1 , u n =
+ v n , and u n +1 = ρu n (1 − u n ) , so v n +1 = u n +1 −
ρ
ρ
ρ
ρ −1
v n +1 = ρu n (1 − u n ) −
v n +1 = (2 − ρ )v n − ρv n 2 . Notice that since v n ≈ 0 , v n +1 ≈ (2 − ρ )v n , so this new
ρ
process converges if 2 − ρ < 1 ⇔ 1 < ρ < 3 .
Second Order Differential Equations
A second order differential equation is of the form y ′′ = f (t , y, y ′) .
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Note
Because of y , y ′ in f, we have two degrees of freedom: at t = t 0 , we have
y 0 = y (t 0 )
.
y 0′ = y ′(t 0 )
Example
Consider y ′′ + y = 0 . Solutions to this are y = sin x , y = cos x , y = 0 ,…there are infinitely many.
Specifying y 0 = y (0) = 0 , we have y = sin x , y = 2 sin x , y = − sin x ,…there are still infinite number of
solutions.
To specify one solution, we need to fix y ′ also. If
y 0 = y (0) = 0
1
1 , then the solution is y = − sin x .
3
y 0′ = y ′(0) = −
3
Note
Sometimes, we can point to a solution by specifying two points on the plane.
y (0) = 0
′
′
, then y (t ) = 3 sin t is the solution. This is a boundary value problem.
• If y + y = 0,
π
y
=3
2
• If y ′′ + y = 0,
y (0) = y 0
, then this is a initial value problem.
y ′(0) = y 0′
Example
y (0 ) = 0
y (0 ) = 0
may have no solutions, but y ′′ + y = 0,
have infinitely many solutions. So
y (π ) = 2
y (π ) = 0
BVP are not guaranteed to give a unique solution, but IVP will.
y ′′ + y = 0,
SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS
Definitions
1) A second order linear differential equation occurs when y ′′ = f (t , y, y ′) = g (t ) − p(t ) y ′ − q(t ) y or
y ′′ + p(t ) y ′ + q(t ) y = g (t ) .
2) If p(t ) and q(t ) are constants, when we have a constant coefficient second order linear differential equation.
Note
Notice that if y (t ) = e rt is a solution to ay ′′ + by ′ + cy = 0 , then ar 2 e rt + bre rt + ce rt = 0
ar 2 + br + c = 0 .
So if r is a solution to ar 2 + br + c = 0 (characteristic equation of the differential equation), then surely
y (t ) = e rt is a solution to ay ′′ + by ′ + cy = 0 .
If the solution to ar 2 + br + c = 0 is imaginary, then e it = cos t + i sin t .
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Note
Notice that because the input is 0, any linear combination of y1 and y 2 is also a solution, i.e.
y ′(t ) = c1 r1e r1t + c 2 r2 e r2t
y (t ) = c1 y1 (t ) + c 2 y 2 (t ) is the general solution to the differential equation. Since
(
) (
) (
)
y ′′(t ) = c1 r1 2 e r1t + c 2 r2 2 e r2t
,
so ay ′′ + by ′ + cy = a c1 r1 2 e r1t + c 2 r2 2 e r2t + b c1 r1e r1t + c 2 r2 e r2t + c c1 e r1t + c 2 e r2t =
(
) (
)
c1 ar1 2 e r1t + br1e r1t + cre r1t + c 2 ar2 2 e r2t + br2 e r2t + ce r2t = y1 (t ) + y 2 (t ) = 0.
Example
The second order differential equation y ′′ − 2 y ′ − 2 y = 0 has the characteristic equation r 2 − 2r − 2 = 0 with
r1 = 1 + 3
and general solution y (t ) = c1 e r1t + c 2 e r2t (the solution space, since all possible vectors in
r2 = 1 − 3
a vector space can be written as a linear combination of basis vectors).
roots
y (0) = c1 + c 2 = 1
(
) (
)
y ′(t ) = 1 + 3 c1 + 1 + 3 c 2 = 2
1
c1 =
(
1
=
1− 3 −1− 3
=
3 +1
2 3
)
1
. The augmented matrix is
1− 3 − 2
)
3t
+ c 2 e (1−
y ′(t ) = 1 + 3 c1 e (1+
and c1 =
1+ 3 1− 3
)
3t
1
1
1+ 3 1− 3
1
2 1− 3
1
y (t ) = c1 e (1+
y (0 ) = 1
, we solve
y ′(0 ) = 2
To solve the IVP y ′′ − 2 y ′ − 2 y = 0,
1
1
1+ 3
2
1
1
=
)
3t
(
)
+ 1 + 3 c 2 e (1−
)
3t
, so by Cramer’s Rule,
2
2 −1− 3
1 − 3 −1 − 3
=
3 −1
2 3
.
1+ 3 1− 3
Example
The solution to y ′′ − 2 y ′ − 2 y = 0 is y (t ) = c1 e (1+ 3 )t + c 2 e (1− 3 )t . We can analyze the behavior of the family,
ex: where are the solutions positive, 0, negative; where are the max/min points; the behavior as t → ±∞ . To
answer these questions, it is better to factor.
To find y (t ) = 0 , let y (t ) = e (1+
As t → ∞ , y (t ) = e (1+
)
3t
)
3t
c1 + c 2 e − 2
c1 + c 2 e − 2
3t
3t
→ c1e (1+
=0
)
3t
t=−
1
2 3
ln
c2
.
c1
. As t → −∞ , y (t ) = e (1−
EQUILIBRIUM SOLUTIONS
Equilibrium solutions are the constant solutions, i.e. y = c .
Example
The equilibrium solution to y ′′ − 2 y ′ − 2 y = 3 is y e = −
3
.
2
Page 18 of 32
)
3t
c1 e 2
3t
+ c 2 → c 2 e (1−
).
3t
MAT244H1a.doc
We can transform this non-homogenous differential equation to a homogenous one. Let
Y = y − ye
Y ′ = y′
Y ′′ − 2Y ′ − 2Y = 0 . This transformation is shift.
Y ′′ = y ′′
Example
Consider y ′′ + y ′ = e −t . To transform it, let
v = f (t ) + c1
dy
= f (t ) + c1 , then y =
dt
v = y′
. Then v ′ + v = e −t . Now solve
v ′ = y ′′
f (t )dt + c1t + c 2 .
Example
Consider 2 y 2 y ′′ + 2 y ( y ′)2 = 1 (t is not involved). Let v = y ′ =
So we have 2 y 2 v
dy
d dy
dv dv dy
dv
, then y ′′ =
=
=
=v
.
dt dt
dt dy dt
dy
dt
[
]
dv
d 2 2
+ 2 yv 2 = 1 . Now y is independent, i.e.
y v = 1.
dy
dy
FUNDAMENTAL SOLUTIONS OF LINEAR HOMOGENEOUS EQUATIONS
Solutions to a second order differential equation are real valued functions which are at least twice
differentiable.
The collection of such functions is denoted by C 2 .
We know from linear algebra that C 2 and addition of functions ( f + g )(x ) = f (x ) + g ( x ) and scalar
multiplication (αf )(x ) = αf (x ) form a vector space.
Linear Transformations
• The derivative D( f ) = f ′ , D 2 ( f ) = D(D( f )) = f ′′ .
• p(t )D( f ) = p(t ) f ′ , p(t )D( f + g ) = p(t )( f + g )′ = p(t ) f ′ + p(t )g ′ = p(t )D ( f ) + p(t )D (g ) .
•
q(t ) : R 2 → R 2 , q(t ) = q(t )1 ⇔ q(t ) f = q(t )1 f .
Note
Any linear differential equation can be captured as a linear transformation. So y ′′ + p(t ) y ′ + q(t ) y = 0 can be
(
)
written as L[ y ] = D 2 + p(t )D + q(t ) ( y ) = D 2 ( y ) + p(t )D ( y ) + q(t ) y = 0 .
Note
Notice that if y1 and y 2 are solutions to L[ y ] , then so is y = c1 y1 + c 2 y 2 .
Proof:
L[ y1 ] = 0
L[ y 2 ] = 0
L[c1 y1 + c 2 y 2 ] = c1 L[c1 y1 ] + c 2 L[ y 2 ] = 0 + 0 = 0 .
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Theorem: Existence and Uniqueness Theorem
y (t 0 ) = y 0
The initial value problem L[ y ] = g (t ),
has a unique solution on any interval I on which t 0 ∈ I
y ′(t 0 ) = y 0′
and p(t ) , q(t ) , g (t ) are all continuous.
Example
Determine the largest interval I on which the IVP (x − 2 ) y ′′ + y ′ + ((x − 2 ) tan x ) y = 0,
y (3) = 1
has a unique
y ′(3) = 2
solution.
• At x = 2 , y ′ = 0 , so we have no solution at x = 2 .
• Since x ≠ 2 , the IVP is now y ′′ +
1
π
y ′ + (tan x ) y = 0 . Note that tan x is discontinuous at
and
2
x−2
3π
.
2
• So 2,
3π
2
is the largest interval on which we have a unique solution.
LINEAR INDEPENDENCE AND THE WRONSKIAN
Recall that in the case of a first order differential equation, whenever we found a solution, we added a
constant to it in order to included all possible solutions (the general/family of solutions).
In the case of second order differential equations, we find two solutions y1 and y 2 and claim that
c1 y1 + c 2 y 2 is the general solution.
How do we decide if two functions are linearly independent on an interval I? Use Wronskian of the two
functions.
Wronskian
If the Wronskian W ( y1 , y 2 ) =
y1
y1′
y2
is non zero for some t ∈ I , then y1 and y 2 are linearly independent.
y 2′
Theorem: Abel’s Theorem
If y1 and y 2 are solutions to L[ y ] = y ′′ + p(t ) y ′ + q(t ) y = 0 where p(t ) and q(t ) are continuous on an open
interval I, then the Wronskian (W ( y1 , y 2 ))(t ) =
y1
y1′
y2
− p (t )dt
(t ) = ce
where c is a constant that depends on
y ′2
y1 and y 2 but not on t.
COMPLEX ROOTS OF THE CHARACTERISTIC EQUATION
Recall: To find solutions to ay ′′ + by ′ + cy = 0 we decided to find the roots r1 and r2 of ar 2 + br + c = 0 ,
and y1 (t ) = e r1t , y 2 (t ) = e r2t .
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What if r1 = r2 (repeated root), or there are no real roots (complex roots)?
If ∆ = b 2 − 4ac < 0 , then r =
As usual we say
− b ± ∆ − b ± (− ∆ )(− 1) − b ±
=
=
2a
2a
(− 1) (− ∆ )
2a
=−
b i (− ∆ )
±
= λ ± iµ .
2a
2a
y1 (t ) = e = e (λ +iµ )t = e λt e iµt = e λt [cos(µt ) + i sin (µt )]
are the solutions.
y (t ) = e r2t = e (λ −iµ )t = e λt e −iµt = e λt [cos(µt ) − i sin (µt )]
r1t
2
Real-Valued Solutions
But we are looking for a real solution. Indeed,
y1 (t ) + y 2 (t ) = 2e λt cos(µt )
− iy1 (t ) + iy 2 (t ) = 2e λt sin (µt )
are solutions too. So the
general solution is y (t ) = c1 e λt cos(µt ) + c 2 e λt sin (µt ) = e λt [c1 cos(µt ) + c 2 sin (µt )] .
If λ = 0 i.e. −
b
=0
2a
b = 0 , then the solution becomes
y (t ) = c1 cos(µt ) + c 2 sin (µt ) = c 2
c
c1
sin α
cos(µt ) + sin (µt ) . By letting 1 = tan α =
, we get
c2
cos α
c2
c
sin α
cos(µt ) + sin (µt ) = 2 [(sin α )(cos(µt )) + (cos α )(sin (µt ))] = A sin (µt + α ) .
cos α
cos α
y (t ) = c 2
Note
Given y ′′ + p(t ) y ′ + q(t ) y = 0 we can change the variable t to x so y (t )
d2y
dt 2
=
d dy dx dy d dx
d dy
⋅ + ⋅
=
dt dx dt dx dt dt
dt dx
dy
dx
+
dt
dx
then y ′′ + p(t ) y ′ + q(t ) y = 0 becomes y ′′ + by ′ + c(t ) y = 0 .
Example
(
)
Consider ty ′′ + t 2 − 1 y ′ + t 3 y = 0 , or y ′′ +
x(t ) =
(y ′′t
2
1
q(t ) 2 dt = tdt =
)
+ y′ +
d 2x
dt 2
y (x ) . Now,
dy dy dx
=
,
dt dx dt
1
. It turns out that if we let x(t ) =
q(t ) 2 dt ,
t 2 −1
y ′ + t 2 y = 0 . So q(t ) = t 2 . Now let
t
t2
dx
d2x
t 2 −1
′′ +
, and so
=t,
=
1
.
Now,
y
y ′ + t 2 y = 0 becomes
2
dt
t
dt 2
t 2 −1
( y ′t ) + t 2 y = 0
t
y ′′t 2 + y ′ + t 2 y ′ − y ′ + t 2 y = 0
y ′′ + y ′ + y = 0 .
REPEATED ROOTS
b
− t
−b± ∆
b
Recall that r =
. If ∆ = b 2 − 4ac = 0 , then r = −
. So only one solution exists: y1 = e 2 a . But if
2a
2a
we let y (t ) = v(t ) y1 , then maybe we can come up with some other solutions as well.
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If y (t ) is to satisfy ay ′′ + by ′ + cy = 0 , then:
y (t ) = v(t ) y1
y ′(t ) = v ′(t ) y1 + v(t ) y1′
•
•
y ′′(t ) = v ′′(t ) y1 + v ′(t ) y ′ + v ′(t ) y ′ + v(t ) y1′′ = v ′′(t ) y1 + 2v ′(t ) y ′ + v(t ) y1′′
′
′
and ay + by ′ + cy = 0 becomes a(v ′′(t ) y1 + 2v ′(t ) y ′ + v(t ) y1′′ ) + b(v ′(t ) y1 + v(t ) y1′ ) + c(v(t ) y1 ) = 0
[av ′′(t )y1 + 2av ′(t )y ′ + bv ′(t )y1 ] + v(t )[ay1′′ + by1′ + cy1 ] = 0 av ′′(t ) y1 + 2av ′(t )y ′ + bv ′(t )y1 = 0
•
av ′′(t )e
−
b
t
2a
− bv ′(t )e
−
b
t
2a
b
− t
e 2a
y 2 (t ) = (c1t + c 2 )
+ bv ′(t )e
−
b
t
2a
=0
v ′(t ) = c1
v ′′(t ) = 0
v(t ) = c1t + c 2 . So another solution is
.
So the general solution is y (t ) = C1e
−
b
t
2a
+ C 2 (c1t + c 2 )e
−
b
t
2a
= (C1 + C 2 c1t + c 2 )e
−
b
t
2a
= (at + b )e
−
b
t
2a
.
NON-HOMOGENOUS DIFFERENTIAL EQUATIONS
A non-homogenous differential equation has the form L[ y ] = y ′′ + p(t ) y ′ + q(t ) y = g (t ) . L[ y ] = 0 decides the
nature of the system.
General Solution
Suppose y1 and y 2 are two solutions to L[ y ] = g (t ) , i.e. L[ y1 ] = g (t ) = L[ y 2 ] . Then
L[ y1 ] − L[ y 2 ] = g (t ) = g (t ) = 0 , so y1 − y 2 is a solution to L[ y ] = 0 . So if y (t ) = c1 y1 (t ) + c 2 y 2 (t ) is the
general solution to L[ y ] = 0 , then y1 − y 2 = c1 y1 + c 2 y 2 for some c1 and c 2 . Therefore,
y1 = y 2 + c1 y1 + c 2 y 2 , which means if one solution to L[ y ] = g (t ) is known ( y p a particular solution) then
the general solution to L[ y ] = g (t ) is y (t ) = y p (t ) + c1 y1 (t ) + c 2 y 2 (t ) .
So to find the generals solution y (t ) to L[ y ] = g (t ) , we need to first guess y p (t ) then solve L[ y ] = 0 .
How to Guess a Particular Solution?
Notice that a (linear) differential equation does not completely disfigure the input g (t ) (at least the
“reasonable” inputs will be recognizable). “Reasonable” inputs: exponential e rt , polynomial p(t ) ,
trigonometric functions sin (αt ) , cos(αt ) , tan (αt ) =
sin (αt )
, roots
cos(αt )
n
x.
Example
Consider L[ y ] = y ′′ + 2 y ′ + 5 y = 3 sin (2t ) . Guess a y p (t ) .
y p (t ) = A sin (2t ) + B cos(2t )
Let
y ′p (t ) = 2 A cos(2t ) − 2 B sin (2t )
y ′p′ (t ) = −4 A sin (2t ) − 4 B cos(2t ) = −4 y p (t )
[ ]
, then L y p = 3 sin (2t )
− 4 y p + 2[2 A cos(2t ) − 2 B sin (2t )] + 5 y p = 3 sin (2t )
Page 22 of 32
y p + 4 A cos(2t ) − 4 B sin (2t ) = 3 sin (2t )
MAT244H1a.doc
y p + 4 A cos(2t ) − 4 B sin (2t ) = 3 sin (2t )
A sin (2t ) + B cos(2t ) + 4 A cos(2t ) − 4 B sin (2t ) = 3 sin (2t )
( A − 4 B ) sin (2t ) + (4 A + B ) cos(2t ) = 3 sin (2t ) . Since
sin (2t ) and cos(2t ) are linearly independent,
3
17
.
12
B=−
17
1
So guess y p (t ) = [3 sin (2t ) − 12 cos(2t )] .
17
A − 4B = 3
4A + B = 0
A=
More Complicated Inputs
What if L[ y ] = g (t ) , but g (t ) = g 1 (t ) + g 2 (t ) + g 3 (t ) ? It can be reduced to finding particular solutions to
L[ y ] = g 1 (t )
L[ y ] = g 2 (t ) because L[ y1 + y 2 + y 3 ] = L[ y1 ] + L[ y 2 ] + L[ y 3 ] = g1 (t ) + g 2 (t ) + g 3 (t ) = g (t )
L[ y ] = g 3 (t )
y p (t ) = y1 (t ) + y 2 (t ) + y 3 (t ) .
Example
Consider y ′′ + y ′ = e −t . Let
y p (t ) = Ae −t
y ′p (t ) = − Ae −t
Ae −t − Ae −t = e −t
0 = e −t ! This problem happens
y ′p′ (t ) = Ae −t
[ ]
because y p is already a solution to L y p = 0 .
Try replacing y p with ty p .
Repeated Roots
[ ]
[y ′p + y ′p + ty ′p′ ]+ p(t )[y p + ty ′p ]+ q(t )t[y p ] = g (t )
(tp y )″ + p(t )(ty p )′ + q(t )(ty p ) = g (t )
t [ty ′p′ + p(t ) y ′p + q(t ) y p ]+ 2 y ′p + p(t ) y p = g (t )
If L[ y ] = y ′′ + p(t ) y ′ + q(t ) y = g (t ) , then L ty p = g (t )
2 y ′p + p(t ) y p = g (t ) . Now solve for y p (t ) .
METHOD OF UNDETERMINED COEFFICIENTS
Example
Solve the IVP y ′′ + 4 y = t 2 + 3e t ,
1) Solve y ′′ + 4 y = 0 . r 2 + 4 = 0
y (0) = 0
.
y ′(0) = 2
r = 0 ± 2i , so the fundamental solution is y c (t ) = c1 cos(2t ) + c 2 sin (2t ) .
Page 23 of 32
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y p (t ) = At 2 + Bt + C + De t
2) Look for a particular solution y p (t ) . Since g (t ) = t 2 + 3e t , let
y ′p′ + 4 y p = t 2 + 3e t
[2 A + De ]+ 4[At
t
2
]
y ′p (t ) = 2 At + B + De t
y ′p′ (t ) = 2 A + De t
+ Bt + C + De t = t 2 + 3e t
4A = 1
4B = 0
2 A + 4C = 0
5D = 3
4 At 2 + 4 Bt + (2 A + 4C ) + 5 De t = t 2 + 3e t
A=
(t ) = c1 cos(2t ) + c 2 sin (2t ) + 1 t 2 − 1 + 3 e t
4
y ′(t )
y (0) = 0
y ′(0) = 2
8
5
1
4
B=0
C = − 18
D=
3) The general solution is y (t ) = y c (t ) + y p (t ) = c1 cos(2t ) + c 2 sin (2t ) +
4) Now, y
. So
.
3
5
1 2 1 3 t
t − + e .
4
8 5
= −2c1 sin (2t ) + 2c 2 cos(2t ) +
1
3
t + e t , so
2
5
− 19
19
7
1
1 3
40
. So the solution to the IVP is y (t ) = −
cos(2t ) + sin (2t ) + t 2 − + e t .
7
40
10
4
8 5
c2 =
10
c1 =
Example
Consider y ′′ + 4 y = 3 sin 2t . We already know the fundamental solutions y c (t ) = c1 cos(2t ) + c 2 sin (2t ) . So
when we “guess” the particular solution, use y p (t ) = At sin (2t ) + Bt cos(2t ) . Then the general solution is
y (t ) = −
3
t cos(2t ) + c1 cos(2t ) + c 2 sin (2t ) .
4
VARIATION OF PARAMETERS
This technique works for general linear differential equations with arbitrary input g (t ) .
Idea
If y1 (t ) , y 2 (t ) are fundamental solutions to L[ y ] = y ′′ + p(t ) y ′ + q(t ) y = 0 , then the general solution to
L[ y ] = g (t ) “can” be y (t ) = u1 (t ) y1 (t ) + u 2 (t ) y 2 (t ) .
We want more restrictions on u1 and u 2 . If u1′ y1 + u 2′ y 2 = 0 , then (u1′ y1 + u 2′ y 2 )′ = 0
u1′′ y1 + u1′ y1′ + u 2′′ y 2 + u 2′ y ′2 = 0 . So
u1′ y1 + u 2′ y 2 = 0
. Now, we can solve for u1 (t ) and u 2 (t ) .
u1′ y1′ + u 2′ y ′2 = g (t )
Example
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Consider x 2 y ′′ − 3xy ′ + 4 y = x 2 ln x , and
y1 = x 2
y 2 = x 2 ln x
fundamental solutions. Rewrite as
u1′ x 2 + u 2′ x 2 ln x = 0
3
4
y ′′ − y ′ + 2 y = ln x . Let
x
u1′ (2 x ) + u 2′ (2 x ln x + x ) = ln x
x
u1′ x 2 + u 2′ x 2 ln x = 0
u1′ + u 2′ ln x +
1
ln x .
=
2
2x
Observation
0
′
′
g (t )
u y + u2 y2 = 0
To solve 1 1
, use Cramer’s rule. So u1′ =
u1′ y1′ + u ′2 y ′2 = g (t )
y1
y1′
u 2′ =
y1
y1′
g (t )
0
y1
y1′
y2
y ′2
=
y2
y 2′
=
y2
y ′2
− y 2 g (t )
and
W ( y1 , y 2 )
y1 g (t )
. Thus, the general solution is
W ( y1 , y 2 )
y (t ) = u1 (t ) y1 (t ) + u 2 (t ) y 2 (t ) = y1 (t )
t
t0
− y1 (s )g (s )
ds + c1 + y 2 (t )
W ( y1 , y 2 )(s )
t
t0
y 2 (s )g (s )
ds + c 2 .
W ( y1 , y 2 )(s )
REDUCTION OF ORDER
If only y1 (t ) is given as a solution to L[ y ] = y ′′ + p(t ) y ′ + q(t ) y = 0 , we claim that y (t ) = v(t ) y1 (t ) is a
solution to L[ y ] = g (t ) .
So v ′′y1 + 2v ′y1′ + p(t )v ′y1 = g (t ) or v ′′y1 + (2 y1′ + p(t ) y1 )v ′ = g (t ) , which we can solve!
SPRING VIBRATIONS
Hooke’s Law
“The more the spring stretches, the more force.” Fs = −kL .
Equilibrium
mg = −kL ⇔ mg + kL = 0 .
Example
Given a spring, a 100g mass is hanged on it. The spring stretches by 2cm. What is k?
mg
(0.1)(− 9.8) = 49 .
mg = −kL k = −
=−
L
0.02
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MAT244H1a.doc
General Solution
On the way down, the mass passes through u (t ) at time t. So
m(u ′′(t ) + g ) + k (L + u (t )) = 0 mu ′′(t ) + mg + kL + ku (t ) = 0 mu ′′(t ) + ku(t ) = 0 . Therefore, at any point u
(with respect to any reference point), we have mu ′′ + ku = 0 as the differential equation form of Hooke’s Law.
λ =0
k
Now, u ′′ + u = 0 , so
. The solution is u (t ) = c1 cos(ω 0 t ) + c 2 sin (ω 0 t ) =
k
m
µ=
= ω0
m
(A is the amplitude, δ is the phase). The period is T =
= A cos(ω 0 t − δ )
2π
m
= 2π
.
ω0
k
Fluid Resistance
The fluid that the mass is moving in causes resistance. Fr ~ v , so Fr = λu ′ . So now, the differential equation
becomes mu ′′ + λu ′ + ku = 0 . So r = −
λ
2m
±
λ 2 − 4mk
2m
=−
ωq =
λ
2m
2m
λ
±i
±
λ2
4m
2
−
k
.
m
k
.
m
If λ = 0 , we have undamped motion with frequency ω 0 =
If λ 2 < 4mk , then r = −
λ
−
k
λ2
2 m cos
−
.
So
u
(
t
)
=
Ae
m 4m 2
k
λ2
−
t − δ , where
m 4m 2
k
λ2
λ2
2
−
=
ω
−
< ω 0 is the quasi-frequency.
0
m 4m 2
4m 2
Change in Mass
k
λ2 k
λ2
λ2
k
−
>
.
As
m
increases,
ω
increases.
As
m
decreases,
increases faster than
,
,
q
2
2
2
m 4m m 4m
m
4m
and there will eventually be no oscillation.
ωq =
Pendulum
The motion of the pendulum is governed by θ ′′ +
ω0 =
T=
g
θ = 0.
L
g
is the number of oscillation per second.
L
2π
L
= 2π
is the period.
ω0
g
ELECTRIC CIRCUITS
Components
Page 26 of 32
θ
L
MAT244H1a.doc
dQ
dQ
1
V=
⇔ VR = R
, R is resistance in ohms.
R
dt
dt
1
2) Capacitor: V L = Q , C is capacity in Faraday.
C
1) Resistor: I =
d 2Q
dI
= L 2 , L is in Henry.
dt
dt
3) Coil: V L = L
LC Circuit
VC + V L = 0 , V L = LQ ′′ , VC =
So LQ ′′ +
1
Q = 0 with ω 0 =
C
1
Q.
C
1
.
LC
LCR Circuit
1
Q =0.
C
VC + V R + V L = 0 . So LQ ′′ + RQ ′ +
FORCED VIBRATION
Vibration with input is mu ′′ + γu ′ + ku = F (t ) .
Cases of Interest
1) F (t ) = F0 cos(ωt ) where ω ≠ ω 0 . Then u (t ) = Ae − rt cos(ω 0 t − δ ) + R cos(ωt − δ ) , where
R=
F0
(
m 2 ω0 2 − ω
)
2 2
+ γ 2ω 2
F0
then R ≈
=
m 2ω 4 + γ 2ω 2
. Since ω 0 2 =
F0
ω m 2ω 2 + γ 2
k
, so if ω ≈ 0 then R ≈
m
F0
2
m ω0
4
=
F0
mω 0
2
=
F0
; if ω >> ω 0
k
which approaches 0 as ω approaches infinity.
2) F (t ) = F0 cos(ωt ) where ω ≈ ω 0 . Then u (t ) = Ae − rt cos(ω 0 t − δ ) + R cos(ωt − δ ) , where
R=
2
(
F0
2
m ω0 − ω
)
2 2
2
+γ ω
2
. Since ω ≈ ω 0 , so R ≈
F0
γ 2ω 2
=
F0
γω
; if γ ≈ 0 then R is large (momentary
resonance).
3) F (t ) = F0 cos(ωt ) where γ = 0 and ω = ω 0 . Then u (t ) = c1 cos(ω 0 t ) + c 2 sin (ω 0 t ) +
u (0) = 0
, then we have resonance.
u ′(0) = 0
Page 27 of 32
F0
t sin (ω 0 t ) . If
2mω 0
MAT244H1a.doc
4) F (t ) = F0 cos(ωt ) where γ = 0 and ω ≠ ω 0 . Then u (t ) = c1 cos(ω 0 t ) + c 2 sin (ω 0 t ) +
u (0) = 0
c1 = −
u ′(0) = 0
c2 = 0
u (t ) =
(
F0
m ω0 2 − ω
)
(
F0
(
F0
m ω0 2 − ω 2
) cos(ωt ) . If
)
m ω 0 2 − ω 2 , then
[cos(ωt ) − cos(ω 0 t )] =
F0
(
m ω0 2 − ω
)
ω0 2 − ω 2
2 sin
2
t sin
ω0 2 + ω 2
2
t . We let the actual
frequency to be ω 0 2 + ω 2 , and then the amplitude keeps changing with time (with frequency
ω0 2 − ω 2 < ω0 2 + ω 2 .
Series Solution of Second Order Linear Equations
Definition: Regular Point
A point x 0 is said to be a regular point of the differential equation P( x ) y ′′ + Q(x ) y ′ + R( x ) y = 0 if P( x 0 ) ≠ 0 .
That is, if
Q( x )
R(x )
and
are continuous at x 0 , then the IVP has a solution defined near y (0 ) = x 0 .
P(x )
P(x )
Example
Consider the IVP (1 − x )2 y ′′ − 2 xy ′ + 12 y = 0,
∞
y=
∞
a n ( x − 0 )n =
n =0
a n x n . Then − 2 xy ′ = −2 x
n =0
(1 − x )y ′′ = (1 − x )
2
x(0) = 0
. Assuming y exists near x 0 = 0 , we have
x ′(0) = 1
2
∞
∞
n =0
∞
n(n − 1)a n x n − 2 =
n=0
∞
na n x n −1 =
− 2na n x n and
n =1
n(n − 1)a n x n − 2 −
n =2
∞
n(n − 1)a n x n .
n=2
So (1 − x )2 y ′′ − 2 xy ′ + 12 y = 0 becomes
∞
n(n − 1)a n x n − 2 +
n=2
∞
− n(n − 1)a n x n +
n= 2
∞
have
(n + 1)(n + 2 )a n + 2 x n +
n =0
∞
− 2na n x n +
n =1
∞
12a n x n = 0 . Making the powers look alike, we
n=0
− n(n − 1)a n x n +
n =2
∞
∞
− 2na n x n +
n =1
∞
12a n x n = 0 .
n=0
So the DE becomes
∞
[2a2 +12a0 ] + [6a3 x +10a1 x] + [(n +1)(n + 2)an+2 − n(n −1)an − 2nan +12an ]x n = 0
n=2
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a 2 = −6a 0
6a 0 + a 2 = 0
5a1 + 3a 3 = 0
(n + 1)(n + 2 )a n + 2 − n(n − 1)a n − 2na n + 12a n
that
=0
a2 = 0
y (0) = a 0 = 0
, so
(n + 3)(n − 3) a
a n+2 =
y ′(0) = a1 = 1
(n + 1)(n + 2) n
a3 = −
5
3
a 5 = a 3+ 2 = 0 a 3 = 0
a n+2 =
Now, if
(n + 3)(n − 3) a
(n + 1)(n + 2) n
a 2 n +1 = 0, ∀n = 2,3,
a1 = 0
y (0) = 1
, then
a3 = 0
y ′(0) = 0
5
. Notice
a 3 = − a1
3
n(n − 1) + 2n − 12
(n + 3)(n − 3) a
a n+2 =
a =
(n + 1)(n + 2) n (n + 1)(n + 2) n
a 2 n = 0, ∀n = 0,1,
and
. So the solution is y1 ( x ) = x −
5 3
x .
3
a0 = 1
a 2 n +1 = 0, ∀n = 0,1,
and a 2 = −6 . So y 2 ( x ) = 1 − 6 x 2 + 3x 4 +
.
a4 = 3
Therefore, the general solution is y ( x ) = c1 y1 + c 2 y 2 . Notice that if
y (0) = b1
, then y ( x ) = b2 y1 + b1 y 2 .
y ′(0) = b2
Definition: Regular Point
A regular point is a point where all the derivatives of the solution function are defined. In other words, the
Taylor series exists.
RESTRICTIONS ON POWER SERIES SOLUTION
Example: Radius of Convergence
Consider x 2 − 2 x − 3 y ′′ + xy ′ + 4 y = 0 at various points: x 0 = 4 , x 0 = −4 , and x 0 = 0 . Find a lower bound
for the radius of convergence of the given points.
• The issues arise when the coefficient of y ′′ becomes 0. x 2 − 2 x − 3 = ( x − 3)( x + 1) , so at x = 3 and
x = −1 , we have issues.
• The solution shall make sense on the intervals (− ∞,−1) , (− 1,3) , (3, ∞ ) .
(
)
• At x 0 = 4 , the radius of convergence is at least 1 ( ρ ≥ 1 ).
At x 0 = −4 , the radius of convergence is at least 3 ( ρ ≥ 3 ).
At x 0 = 0 , the radius of convergence is at least 1 ( ρ ≥ 1 ).
Another technique for solving differential equations (by series) is by only calculating a few terms of the
solution.
Example
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MAT244H1a.doc
Evaluate four terms of the solution to (cos x ) y ′′ + xy ′ − 2 y = 0 .
1−
Rewriting, we get
(
(2a
x2
+
2
2
+ 6a 3 x + 12a 4 x 2 + 20a 5 x 5 +
− 2 a 0 + a1 x + a 2 x 2 + a 3 x 3 +
[(2a + 6a x + 12a x
− [2a + 2a x + 2a x
2
3
0
4
1
2
2
2
5
+ 20a 5 x +
3
+ 2a 3 x +
)− (a
]= 0
2x
2
)= 0
+ 3a 3 x 3 +
)+ x(a
)]+ [a x + 2a
1
1
2x
2
+ 2a 2 x + 3a 3 x 2 + 4a 4 x 3 +
+ 3a 3 x 3 + 4a 4 x 4 +
)
]
− a0 + a2 = 0
(2a 2 − 2a 0 ) + (6a 3 − a1 )x + (12a 4 − a 2 )x 2 + (20a 5 − 2a 3 )x 3 +
− a1 + 6a 3 = 0
=0
− a 2 + 12a 4 = 0
. Solve in terms of
− a 3 + 10a 5 = 0
a 0 and a1 .
System of First Order Linear Equations
INTRODUCTION
x1′ = F1 (t , x1 , x 2 , x 3 )
A system of differential equation is like x ′2 = F2 (t , x1 , x 2 , x 3 ) .
x 3′ = F3 (t , x1 , x 2 , x 3 )
Example
Consider the system
x1′ = 2 x1 − x 2 + cos t
x′
2 − 1 x1
cos t
. Such a system is written as 1 =
+
or
x 2′ = 3x1
x 2′
3 0 x2
0
X ′ = AX + g (t ) , which is a first order differential equation.
Example
We can use a system approach to solve any differential equation of any order. Consider u ′′′ + 2u ′ − u = 1 . Let
x1 = u
x1′ = x 2
x1′
0 1 0 x1
0
x 2 = x1′ = u ′ , then x ′2 = x1
. So we have x 2′ = 0 0 1 x 2 + 0 or
x 3 = x 2′ = u ′′
x 3′ = u ′′′ = u − 2u ′ + 1 = x1 − x 2 + 1
0
X ′ = AX + 0 .
1
Example
Page 30 of 32
x 3′
1 − 2 0 x3
1
MAT244H1a.doc
Conversely, we one can translate a system to an ordinary differential equation. Consider
u = x1
. We have u ′ = x1′ = 2 x 2
x 2 (0) = 4
u ′′ = 2 x ′2 = −4 x1 = −4u
x1 (0) = 3
x1′ = 2 x 2
where
x ′2 = −2 x1
u (0) = 3
.
u ′(0) = 2 x 2 (0) = 8
u ′′ + 4u = 0 where
APPLICATIONS IN MODELING
Example: A Double Tank Mixing Problem
ρ g/min
x 2 (0) = 300
2 L/min
x1 (0) = 500
V1 = 20L
dx1
x (t )
x (t )
= 2ρ + 3 2 − 5 1
dt
V2
V1
dx 2
x (t )
x (t )
= 5 1 −5 2
dt
V1
V2
− 5
x1′
= 520
x ′2
20
3
50
5
− 50
x1
x2
+
3 L/min
V2 = 50L
5 L/min
x1′ = 2 ρ + −
, so we have
2ρ
0
⇔ X′=
2 L/min
5
3
x1 +
x2
V1
V2
5
5
x 2′ =
x1 −
x2
V1
V2
− 14
1
4
3
50
− 101
X+
Example: Double Mass-Spring Problem
F1(t)
k1
m1
The variables are x1 and x 2 . So
m2
d 2 x1
dt 2
d 2 x2
dt 2
2ρ
0
k2
where
where X (0) =
F2(t)
m1
m2
x1
x2
= F1 (t ) − k1 x1 − k 2 (x 2 − x1 )
.
= F2 (t ) − k 2 (x 2 − x1 )
EIGENVALUES AND EIGENVECTORS
Recall
Page 31 of 32
x1 (0) = 500
x 2 (0) = 300
500
300
.
. So
MAT244H1a.doc
Two solutions y1 and y 2 to a differential equation are independent on an interval I if
W ( y1 , y 2 ) =
y1
y1′
y2
≠ 0 on I (where I contains the initial values).
y 2′
x1(1)
x1(2 )
x1(3)
For systems X ′ = AX we say solution X (1) (t ) = x 2(1) , X (2 ) (t ) = x 2(2 ) , X (3) (t ) = x 2(3)
x 3(1)
x 3(2 )
x 3(3)
(
)
on I if W X (1) , X (2 ) , X (3) = X (1)
X (2 )
are independent
X (3) ≠ 0 .
Recall
If r is an eigenvalue of A and ξ is the corresponding eigenvectors, i.e. Aξ = rξ , then in particular
X ′ = AX = rX
Now, if
x1
x2
x1′ = rx1
x 2′ = rx 2
x1 (t ) = ae rt
x 2 (t ) = be
rt
or X = e rt
a
.
b
is an eigenvector then x 2 = αx1 , then be rt = αae rt .
Notice that we may have two independent eigenvectors for A. In this case, two solutions in two independent
directions can be found so that if ξ (1) and ξ (2 ) are independent eigenvectors of A corresponding to
eigenvalues r1 and r2 , then we have two independent solutions
X (t ) = c1 X (1) (t ) + c 2 X (2 ) (t ) .
Page 32 of 32
X (1) = e r1t ξ (1)
X (2 ) = e r2t ξ (2 )
and the general solution is
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