Ordinary Differential Equations: A Systems Approach Bruce P. Conrad November 24, 2010 1 c 2010 Bruce P. Conrad 2 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.” 8 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 12 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. 14 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. 176 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. 6 6 L sC [email protected] C C C C sE P p 6 P C PP PP C ?PPPss C q (a) @ ? @ @ L sJ J s J C 6 s [email protected] q p C (b) C 6 C @ @ J6 s Js @ ? @ @s - C @ @ pJ ? @ @ @ @ @@ - (c) q C? C sP C sE L P C PP C6 P PP PPs Cs C (d) q 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 = PeDt 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. Return 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. Return 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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