Nonlinear First-Order Equations I

Transcription

1 Chapter 3 Nonlinear First-Order Equations I Separable Equations Recall that a first-order differential equation is said to be separable if it can be written in the form ϕ(y) = ψ(x)dx. This form suggests the formal procedure of antidifferentiating ϕ(y) and ψ(x) with respect to y and x, respectively, and equating the results to arrive at an implicit general solution of the form ϕ(y) = ψ(x)dx + C, (1) where here ϕ(y) and ψ(x)dx represent any specific antiderivatives of ϕ(y) and ψ(x). (In specific cases, it may be possible to go further and solve explicitly for y.) Example 1 The equation dx = 1 + cos x 1 + 2y is separable, since it can be rewritten in the form (1 + 2y) = (1 + cos x) dx. Antidifferentiation of each side results in an implicit general solution: y + y 2 = x + sin x + C. A sampling of the resulting solution curves is plotted in Figure 1. Figure 1

2 52 Chapter 3. Nonlinear First-Order Equations I It is often convenient to solve an initial-value problem involving a separable equation and an initial condition y(x 0 ) = y 0 by integrating between limits. After writing (1) as ϕ(y(x))y (x) = ψ(x), we integrate each side over [x 0, x]: x x 0 ϕ(y(s))y (s)ds = x x 0 ψ(s)ds. With the substitution u = y(s) in the integral on the left side, this becomes y(x) y 0 ϕ(u)du = x Example 2 Consider the initial-value problem x 0 ψ(s)ds. dx = 3x2 y 2, y(2) = 1. Separating variables and preparing to integrate between limits, we obtain y 1 du u 2 = x 2 3s 2 ds. From this we get u 1 y 1 = s3 x 2, which leads to the implicit solution 1 y + 1 = x This is easily solved for y, yielding the explicit solution y = 1 x 3 7. Note that the maximal domain of this solution is ( 3 7, ). The Equation dx = g(y/x) There are many types of non-separable differential equations that can be made separable by an appropriate change of dependent variable. One such type consists of equations of the form = g(y/x), (2) dx i.e., equations dx = f(x, y) in which f(x, y) may be expressed solely in terms of the ratio y/x. The form of (2) suggests the substitution z = y/x, or y = xz. * A function f with this property is said to be homogeneous of degree 0. For that reason, many textbooks refer to these as homogeneous differential equations. To avoid confusion, we will not use this terminology.

3 3.1. Separable Equations 53 It follows by the product rule that dx dx = z + x dz dx. Thus the differential equation becomes whose separated form is Example 3 Consider the equation may be written in terms of z as z + x dz dx = g(z), dz g(z) z = dx x. dx = y2 xy x 2 + xy. We can express the right side here as a function of y/x by dividing its numerator and denominator by x 2. The result is dx = (y/x)2 y/x. 1 + y/x Now we substitute z for y/x and z + x dz dx for dx z + x dz dx = z2 z 1 + z, and obtain which after some rearrangement and simplification becomes x dz dx = 2z 1 + z. By separating variables and integrating, 1 + z 1 dz = 2 z x dx, we arrive at ln z + z = 2 ln x + C, which, after some manipulation, becomes x 2 ze z = C. Finally, replacing z with y/x, we obtain an implicit general solution: xye y/x = C. A sampling of the resulting solution curves is plotted in Figure 2.

4 54 Chapter 3. Nonlinear First-Order Equations I Figure 2 Problems In Problems 1 6, solve the initial-value problem by separation of variables, and state the maximal domain. 1. dx = dx 2xy2, y(0) = 1/4 2. dt = 1 x2, x(0) = 0 du 3. dt = eu, u(0) = 0 4. dx = 1 2 y3 sin x, y(π/3) = 2 5. y = y ln y, y(0) = e 6. dz ds = (1 + z2 ) tan 1 z, z(0) = 1 In Problems 7 and 8, obtain an implicit solution by separation of variables: 7. y cos x = cos y + sin y, y(0) = π 8. y = 9 + y 2 2, y(0) = 4 9. Consider the initial-value problem y = y, y(0) = 1. (a) Show that separation of variables produces a function that is defined for all t, yet satisfies the differential equation only on an interval of the form (, T ], where 0 < T <. (b) Find a piecewise-defined solution that is valid for all t. In Problems 10 12, obtain an implicit general solution of the equation by means of the substitution z = y/x. 10. y = x2 y x 3 + y y = x y x + y 12. y = 2x y x 2y In Problems 13 16, solve the initial-value problem and state the maximal domain of the solution. 13. dx = y2 + xy x 2, y(1) = dt = y + t2 y 2, y(1) = 1 t dt = t2 + ty + y 2 t 2, y(1) = dx = y + x e y/x, y(1) = 0 x

5 3.2. Torricelli s Law 55 In Problems 17 21, obtain a general solution by means of the suggested substitution. Solve for the solution explicitly when possible. 17. y = 1 t + y, z = t + y 18. y = sin 2 (t + y + 1), z = t + y y = t + y 3 t y 1, y = 1 + (t 2)z 20. y = 2y t + t3 sec 2 ( y t 2 ), y = t 2 z 21. y = y 2 sec 4 t 2y tan t, y = z cos 2 t 3.2 Torricelli s Law The situation with which we are concerned here is that of a tank containing a fluid that is draining through an orifice, we assume is located at the bottom of the tank. In the 1600s, Evangelista Torricelli discovered that, if the depth of the fluid is y, then the speed at which the fluid exits the orifice is the same as the speed of a stone after free fall through a distance y, which is given by 2gy. So, since the (volume) flow rate through the orifice is proportional to the exit speed, it is also proportional to y ; that is, dv dt = ρ y, (1) where ρ is a positive constant, V (t) is the volume of fluid in the tank, and y(t) is the depth of the fluid. This is Torricelli s law also known as the Torricelli- Borda law. The constant ρ naturally depends on the viscosity of the fluid as well as the size and type of orifice. In fact, if the orifice has cross-sectional area a, then ρ = ab 2g, where b is an empirical constant known as the Borda constant. For water through a simple hole with a sharp edge, b 0.6. For a less viscous fluid (such as alcohol or gasoline) or an orifice in the form of short exit pipe, b would be closer to 1. The first question to address here is how we can turn this law into a differential equation for the depth y. If the tank has a constant horizontal cross-sectional area a (i.e., it s some type of cylinder), then this is easy. The volume is always V = ay, and so the differential equation becomes a dt = ρ y. However, we may be interested in more general shapes, so we need to look at a variable horizontal cross-sectional area a(s), where we ll agree that s is the * Torricelli is best known for inventing the barometer in This follows from conservation of energy.

6 56 Chapter 3. Nonlinear First-Order Equations I distance from the bottom of the tank. Now we need to do an integration to find the volume of fluid in the tank when the depth is y: V = y 0 a(s)ds. By the Fundamental Theorem of Calculus and the chain rule, this gives us dv dt Therefore, Torricelli s law becomes = a(y) dt. a(y) dt = ρ y. (2) Note that this is a separable, nonlinear differential equation for the depth y. Example 1 Suppose that the tank in question is a paraboloid generated by revolving the arc y = x 2, 0 x 2, about the y-axis. Suppose further that the tank is initially full and that the fluid in it is draining through a hole located at the vertex. To determine the depth y(t) of the fluid in the tank at time t 0, we first note that the horizontal cross-sectional area of the tank is a(s) = πs at height s above the vertex. So the initial-value problem of interest is π y dt = ρ y, y(0) = 4. Separation of variables leads to y = ρ π dt. Next we introduce new dummy variables and use definite integration: This gives We solve for y and find that y = y 4 t u du = ρ π ds. 2 3 (y3/2 4 3/2 ) = ρ π t. ( 8 3ρ ) 2/3 2π t up to time t = 16π at which time the tank will be completely empty (i.e., y = 0). (We call t = 16π 3ρ as the tank s emptying time. ) Thus the solution of the problem on [0, ) is y = ( 8 3ρ 0 3ρ, if 0 t 16π 3ρ ; ) 2/3 2π t, 0, if t > 16π 3ρ,

7 3.2. Torricelli s Law 57 Example 2 A fuel tank is a cylinder with radius 3 feet, length 10 feet, and a horizontal central axis. It is initially full and begins to drain through a hole at the bottom at time t = 0. (a) Find the depth of the fuel in the tank as a function of t, and state the draining time. (b) Assume that the Borda constant here is Compute the emptying time for exit pipes of diameter 2 inches and 4 inches. Solution: The tank s horizontal cross-sections are rectangles, and the horizontal cross-sectional area at height y is So Torricelli s law says that a(y) = 2 9 (y 3) 2 10 = 20 6y y y y 2 dt = ρ y, y(0) = 6. After separating variables this becomes 6y y 2 = ρ y 20 dt, and simplification of the left side gives 6 y = ρ 20 dt. Next we introduce new dummy variables and use definite integration: y 6 6 u du = t 0 ρ 20 ds. By substituting w = 6 u, dw = du and on the left side, we get 6 y Carrying out the integration now gives Thus until the emptying time, which is 0 w dw = t 2 3 (6 y)3/2 = ρ 20 t. t = y = /2 3ρ 0 ( ) 3ρt 2/3 40 = 80 6 ρ. ρ 20 ds. Now, if the Borda constant associated with the fuel is.75, then with a 2-inchdiameter exit pipe the constant ρ would be.75π(1/12) 2 (2)(32) 0.13, and so

8 58 Chapter 3. Nonlinear First-Order Equations I the emptying time is approximately 1500 s = 25 min. If the diameter of the exit pipe is 4 inches, then ρ = 0.75π(1/6) 2 (2)(32) 0.52, and so the emptying time is approximately 380 s = 6.3 min. Problems 1. (a) A tank in the shape of a right circular cone with radius 5 ft and height 10 ft is initially full and drains through a hole at its bottom. The vertex of the tank points downward and the constant ρ is known to be approximately Solve the differential equation for the depth y of fluid in the tank. For what time interval is the solution valid? Write the solution in a form that is valid for all t 0, and plot its graph. (See Example 1.) (b) Consider the same problem with the vertex of the tank pointing upward. Obtain an implicit solution of the differential equation for the depth y, and determine the draining time without solving for y. Plot the graph of t as a function of y; then reflect the graph to obtain the graph of y versus t. 2. Consider a cylindrical tank with height H and radius R. Compute the time required for the tank, initially full, to drain completely through a hole at its bottom if the tank s central axis is (a) vertical and (b) horizontal. (c) Suppose that H = 2R, so that the initial depth is the same for each of the two positions (and the contest is fair). In which of the two positions does the tank drain faster? 3. A spherical water tank has a radius of 8 feet and is initally full of water, which then flows out of a hole at the bottom of the tank. (a) Set up the initial-value problem for the depth of water in the tank, and find an implicit solution. (b) Find the time T when the tank becomes empty. (c) Compute the tank s draining time T in minutes if the radius of the hole is 1 inch and the Borda constant is b = A spherical water tank has a diameter of 25 feet and rests atop a 50-foot tower. Water flows out of the bottom of the tank, straight down, through and out of a 48-foot length of pipe. Assume that the tank and the pipe are initially full. (a) Set up the initial-value problem for the depth of water in the tank, and find an implicit solution. (b) Find the time T when the tank (not the pipe) becomes empty. (c) Compute the tank s draining time T in minutes if the radius of the pipe is 2 inches and the Borda constant is b = Consider the draining tank in Example 2. (a) How long does it take for the tank to become half empty? (or half full?) (b) What fraction of the initial contents of the tank remains after one-half of the draining time?

9 3.3. Bernoulli and Riccati Equations Consider an upright cylindrical tank with radius R and a circular hole in its bottom with radius r. Assuming a Borda constant b = 1, show that, if the tank initially contains fluid with a depth of y 0, then the time it takes the tank to drain is T = 2y0 g 7. A water tank with horizontal cross-sectional area a(s) at height s drains through a hole in the bottom of the tank. At the same time, water is being pumped into the tank at a constant rate R. Write the differential equation for the depth y. Find the equilibrium solution of the differential equation. Under what condition will it be meaningful? 8. An upright cylindrical tank with radius R and height H has two identical spigots attached one at the very bottom and another half-way up the side. Both spigots are opened fully at time t = 0. Write down the differential equation governing the depth of fluid in the tank while (a) the depth is greater than H/2, and (b) the depth is less than H/2. 9. This problem refers to Example 1. Note that if the solution y = ( 8 3ρ 2π t) 2/3 were valid for all t > 0, then that would indicate that the tank begins to fill up again after it empties. Check that this function does not satisfy the differential equation for t > 16π 3ρ. 10. Can you design a tank, formed by revolving some arc about the y-axis, such that water draining through a hole at the bottom causes the water level to fall at a constant rate? Such a tank could be used as a time-keeping device if a uniform scale were marked down its side. This water clock is called a clepsydra. 11. Consider the initial-value problem dt = ρ y, y(t0 ) = 0 a for the depth of water in an empty cylindrical tank. Clearly, there should be only the trivial solution y(t) = 0 for t t 0. But what about for t t 0? The trivial solution is one solution, but should there be others? What do these backward solutions represent? Can you expect the problem to have just one solution? 12. This is not a draining tank problem. Suppose a water tank with height H has horizontal cross-sectional area a(s) at height s. Water evaporates from the tank at a rate that is proportional to the area of the water s surface. (a) Show that the depth of water in the tank decreases at a constant rate regardless of the shape of the tank. (b) Suppose that the tank is right circular cone with height H and radius R. Suppose further that the initial depth of the water is H/2 and that, in addition to the evaporation, water is also being pumped into the tank at a constant rate q. Set up the initial-value problem for the depth of water in the tank. Find the equilibrium solution of the differential equation. Under what condition will it be meaningful? ( R r ) 2.

10 60 Chapter 3. Nonlinear First-Order Equations I 3.3 Bernoulli and Riccati Equations This section is devoted to certain types of nonlinear first-order differential equations that become linear after an appropriate substitution. The most basic equations of this type are the so-called Bernoulli equations: y + p y = f y n, n 0, 1, (1) where p and f are continuous on an interval I. (Note that the equation is alrea linear if n has one of the exceptional values of 0 or 1.) An appropriate substitution for such an equation is y = u m, y = m u m 1 u, since a judicious choice of m produces a linear equation in u. (See Problem 4.) The process is illustrated in the following example. Example 1 Consider the equation y + 2y = y 3. Substituting y = u m, y = m u m 1 u results in Now dividing by u m 1 gives us m u m 1 u + 2u m = u 3m. m u + 2u = u 2m+1. The expression on the right side becomes 1 if we select m = 1/2. Then, after multiplying through by 2, we have the simple first-order linear equation u 4u = 2, which we solve by inspection, obtaining Now since y = u 1/2, we arrive at u(t) = C e4t = 1 2 (1 + C e4t ). y(t) = C e 4t. Bernoulli Equations with n = 2 Bernoulli equations that are quadratic in y (i.e., n = 2) are of particular interest, since they arise in numerous applications including population models. For the equation y + p y = f y 2, (2)

11 3.3. Bernoulli and Riccati Equations 61 the substitution y = u m, y = m u m 1 u results in which after division by u m 1 becomes So with m = 1, we have m u m 1 u + pu m = f u 2m, m u + pu = f u m+1. Example 2 Consider the initial-value problem u pu = f and y = 1 u. (3) y = (1 e t y) y, y(t 0 ) = y 0, in which the differential equation may rewritten in the form of (2) as y y = e t y 2. By substituting y = 1/u and assuming for the moment that y 0 0, we obtain a linear problem for u: u + u = e t, u(t 0 ) = 1 y 0. Next we multiply by the integrating factor e t, after which the differential equation becomes (e t u) = 1, and then integrate from t 0 to t: This now leads to and then, after solving for u, t t 0 (e s u(s)) ds = t t 0 ds. e t u e t 0 /y 0 = t t 0, u = et 0 + y 0 (t t 0 ) y 0 e t. Now since y = 1/u, the solution of our original initial-value problem, for any (t 0, y 0 ), is Riccati Equations y = y 0 e t e t 0 + y0 (t t 0 ). Another class of nonlinear first-order equations that become linear after an appropriate substitution are the Riccati equations: y + p y = f y 2 + g, (4)

12 62 Chapter 3. Nonlinear First-Order Equations I where the functions p, f, g are continuous on an interval I. Note that a Bernoulli equation with n = 2 is a special case of a Riccati equation. In fact, it turns out that shifting the independent variable by any particular solution results in such a Bernoulli equation. That is, if ỹ is any known particular solution, then the substitution z = y ỹ produces a Bernoulli equation for z. To see this, first note that the proposed substitution turns (4) into Rearrangement gives z + ỹ + p (z + ỹ) = f (z + ỹ) 2 + g. z + (p 2f ỹ)z = f z 2 ( ỹ + p ỹ f ỹ 2 g ). The quantity in parentheses on the right side is zero since ỹ is a solution. Thus z satisfies the Bernoulli equation z + (p 2f ỹ)z = f z 2, which can be solved by the substitution z = u 2. Example 3 For the Riccati equation y = y (5 y) 6, a pair of equilibrium solutions are y = 2 and y = 3. We can use either of those as ỹ in our substitution. Using ỹ = 2, we first make the substitution The differential equation then becomes In standard form that s y = z + 2, y = z. z = (z + 2)(3 z) 6 = z z 2. z z = z 2. Next we substitute z = 1/u, z = u /u 2. After multiplication of both sides by u 2, that gives us u + u = 1, whose general solution is u = C e t + 1. So and thus z = 1 u = 1 C e t , y = z + 2 = 1 C e t

13 3.3. Bernoulli and Riccati Equations 63 Example 4 Consider the equation y + y = t y 2 1 t 2. It is not difficult to see that ỹ = 1/t is a particular solution. So we begin with the substitution That gives which simplifies eventually to y = z + 1/t, y = z 1/t 2. z 1 t 2 + z + 1 ( t = t z + 1 ) 2 1 t t 2, z z = tz 2. Next we substitute z = 1/u, z = u /u 2. After simplification, we get the linear equation u + u = t. Using the integrating factor e t (or the particular solution 1 t), we find u = 1 t + C e t. So and finally z = 1 u = 1 1 t + C e t, y = z + 1 t = 1 1 t + C e t + 1 t = 1 + C e t (1 t + C e t ) t. Problems In Problems 1 3, find a general solution of the equation by substituting y = u m and determining the appropriate value of m. 1. y + y = y 2 2. y + y = y 3. y + y = 1 y 4. Find a general formula for the correct value of m (in terms of n) to use in solving y + p y = f y n, n 1, by means of the substitution y = u m. Observe that the transformed equation always takes the form m u + p u = f. For the equations in Problems 5 13, find a general solution by means of an appropriate substitution of the form y = u m.

14 64 Chapter 3. Nonlinear First-Order Equations I 5. y 1 e t t + e t y = y2 6. y y tan t = y 2 7. y y 2t ln t = ln t y 8. y + y t + 1 = y3 9. y t y t = 2t y y et y 1 + e t = et 3 y 11. y + t y = t y y 3 t y = ty 13. y y 2t 2 = t e 1/t y In 14 16, find the solution that satisfies y(0) = y 0 (as in Example 2, but with t 0 = 0). ( ) e 14. y = (1 e t y) y 15. y = (cot t y) y 16. y t = y 1 + e t e t y For the Riccati equations in Problems 17 22, (a) verify the indicated particular solution ỹ; (b) find a general solution; 17. y + y = y 2 2, ỹ = y + y = e t y 2 e t, ỹ = e t 19. y y tan t = y 2 cos t sec t, ỹ = sec t 20. y + (2e t 1)y = y 2 + e 2t, ỹ = e t 21. y t3 t y = t y 2 + t 3 + 2, ỹ = t 22. y + y cos t 1 + sin t = y 2 sin t, ỹ = cos t 1 + sin t 23. Let p and f be continuous on an interval I, and let g and g be continuous on an interval J, with g (x) 0 for all x in J. Show that the substitution u = g(y) transforms the equation into the linear equation dt = f(t) p(t) g(y) g (y) u + p u = f. Show that the Bernoulli equation (1) is of this form, where g(y) = y 1 n. Each of the equations in Problems can be written in the form of the equation in Problem 23. Find a general solution by means of an appropriate substitution. 24. y = e t sin y cos y 26. y = e y 1 t, 25. y = (e t + tan y) cos 2 y 27. y = e y t y = (1 + y 2 )(e t + tan 1 y) 29. y = t y y ln y t + 1

15 3.4. The Logistic Population Model The Logistic Population Model We are concerned here with modeling the growth of a single population in a closed, static environment. Let P (t) denote the size of the population at time t. In reality, the population size would take on only integer values, but we will assume that P (t) is a smooth function that reasonably approximates the discrete population size. An important quantity in modeling is the per capita growth rate of the population, which is defined as dp dt /P. The most basic model assumes that the per capita growth rate is a constant k. Thus the rate of growth of the population is proportional to the size of the population at any given time. This gives us the initial-value problem dp dt = kp, P (0) = P 0, which has the solution P (t) = P 0 e kt. This model, which predicts unlimited exponential growth (if k > 0), is known as the Malthusian model, so named after the English economist Thomas Malthus of the late eighteenth century. The Malthusian model is clearly not realistic for a population that is naturally limited by environmental factors such as finite space and finite food supply. The constant k can be viewed as the difference in the per capita birth and death rates of the population. Thus k will be negative if the death rate exceeds the birth rate. In this case, the model predicts that P (t) 0 as t. The key to making the model more realistic is to allow the per capita growth rate to depend on the population size itself. It makes sense that this growth rate should decrease as the population increases and to be negative if the population size is too large for the environment to sustain. A simple model that takes these considerations into account comes from assuming that the per capita growth rate depends linearly on P. We replace the constant k with the linear expression k(1 P/M), where k and M are positive constants. Thus the model becomes dp dt = k P (1 P/M), P (0) = P 0. (1) This is called the logistic model and is credited to Pierre Verhulst, a Belgian mathematician of the mid-nineteenth century. Note that the constant M in (1) is the population level at which there is no growth; that is, the population level at which birth and death rates are the same. It is called the carrying capacity of the environment, because it represents the maximum population that the environment can sustain. We will see shortly that M also represents the size that the population will approach after a long period of time. The coefficient k is called the intrinsic growth rate, because * The per capita growth rate may be thought of as the percent increase in the population per unit time.

16 66 Chapter 3. Nonlinear First-Order Equations I if P is very small relative to M, then the model becomes approximately one of simple exponential growth; that is, P kp. In other words, k would be the exponential growth rate if the carrying capacity were infinite. Thus, the constant k is influenced by natural tendencies of the population and not by environmental factors. The differential equation in (1) may be explicitly solved by either of two elementary techniques. It is both a separable equation and a Bernoulli equation. You will be asked to solve it both ways in Problem 1 at the end of this section. The result is P (t) = MP 0. (2) P 0 + (M P 0 )e kt Our purpose here, however, is to analyze the behavior of solutions using the ideas of the preceding section. Figure 1 shows a sampling of solution curves for P 0 0. Note that there are two equilibrium solutions, P (t) = 0 and P (t) = M. Since solutions with P 0 close to 0 increase and do not approach 0 as t, we say the equilibrium solution P (t) = 0 is unstable. On the other hand, we can see that all positive solutions approach M as t. So we say the equilibrium solution P (t) = M is asymptotically stable. Figure 1 Example 1 A fish population was estimated to number 700, 1150, and 1520 in November of three consecutive years, during which no fishing was allowed. Find the environmental carrying capacity M and the population s intrinsic growth rate k. Solution. Taking t = 0 to be time of the first annual census, we have P 0 = 700, P (1) = 1150, and P (2) = By equation (2), the population during those years is given by P (t) = 700M (M 700)e kt, and so we have two equations for k and M: 700M (M 700)e k = 1150 and 700M = (M 700)e2k

17 3.4. The Logistic Population Model 67 The first step in solving these by hand is to solve the first one for e k and the second one for e 2k. That gives us e k = 14 (M 1150) 23 (M 700) and e 2k = 35 (M 1520) 76 (M 700). The second of those is the square of the first, so we have ( 14 (M 1150) 23 (M 700) ) 2 = which after some manipulation becomes 35 (M 1520) 76 (M 700), and then eventually That gives us 2128 (M 1150) 2 = 2645 (M 1520)(M 700), ( M)M = 0. M = Now we return to the expression for e k : which gives us Harvesting e k = 14 (M 1150) 23 (M 700) k = ln 1890 fish. = 14 (740) 23 (1190) = , ( ) Consider a population that is governed by (1) when it is undisturbed by environmental changes or forces from outside the environment. We wish to consider the effect of harvesting on such a population. By harvesting, we mean continuous, deliberate removal of members of the population at some specified rate. We assume here proportional harvesting, meaning that the rate of harvesting is proportional to the size of the population. This has the advantage that a large harvest will be taken from a large population, while a small population perhaps won t be wiped out by harvesting. The disadvantage is that this requires knowledge of the population size on the part of the harvester. Nevertheless, if the harvesting rate is given by βp, where β > 0, then the problem becomes dp dt = k P (1 P/M) βp, P (0) = P 0. (3)

18 68 Chapter 3. Nonlinear First-Order Equations I We would like to examine the effect of this additional term on the solutions of the differential equation. Notice that some rearranging results in ( ) = (k β)p 1, dp dt P M(1 β/k) which has exactly the same form as the original logistic equation with k replaced by k β and M replaced by M(1 β/k), provided that β k. Thus, if β k, the solution can be written easily by modifying (2) as follows: P (t) = M(1 β/k)p 0 P 0 + (M(1 β/k) P 0 ) e (k β)t. In particular, the equilibrium solutions are now P (t) = 0 and P (t) = M(1 β/k). If 1 β/k > 0 (i.e., β < k), then the equilibrium solution P (t) = M(1 β/k) is positive and asymptotically stable, and so the primary effect of proportional harvesting is simply to decrease the carrying capacity of the environment by a factor of 1 β/k. With this modification, the solution curves look the same as those of the original logistic equation in Figure 1. If 1 β/k < 0 (i.e., β > k), then the equilibrium solution P (t) = M(1 β/k) is negative and therefore of no interest. All positive solutions become strictly decreasing and approach 0 as t. Thus P (t) = 0 is asymptotically stable. (See Figure 2.) The upshot of all of this is that if β < k, then any positive population will approach P = M(1 β/k), but if β k, then any positive population will approach zero as t. Thus, excessive proportional harvesting can result in an ever-dwindling (continuous) population, which will eventually lead to extinction of the actual discrete Figure 2 population. Finally, in the critical case β = k, the differential equation in (3) becomes dp dt = k P 2 /M, whose solutions do not differ in character from those seen in Figure 2. The harvesting rate βp is also called the yield. When β < k, we ve seen that the population will approach the equilibrium value M(1 β/k) as t. Thus the yield will approach a stea value of βm(1 β/k) as t. This stea yield is maximized by β = k/2, i.e., half the intrinsic growth rate of the population. The maximum value of the yield is then km/4. That quantity is called the maximum sustainable yield.

19 3.4. The Logistic Population Model 69 Example 2 Consider again the fish population in Example 1. (a) The harvesting of what fraction of the population annually will result in an equilibrium population 1200 fish? (b) The harvesting of what fraction of the population annually will result the maximum sustainable yield? And what is the maximum sustainable yield? Solution. (a) Using k = and M = 1890 that we found in Example 1, the logistic model with proportional harvesting becomes P = P (1 P/1890) βp. So P = 1200 will be an equilibrium solution if (1200) (1 1200/1890) 1200β = 0. Solving for β gives us β 0.14, or 14%. (b) Assume that β < k, the yield at the stable equilibrium P + is βp + = βm (1 β/k) = 1890β (1 β/0.379). This quadratic function of β is zero at β = 0 and β = and is therefore maximized when β = 0.379/2 0.19, or 19%. So the maximum sustainable yield is 1890(0.19)(0.5) 180 fish per year. Problems 1. Solve (1) to obtain the solution (2) by (a) separation of variables and partial fraction decomposition; (b) treating the differential equation as a Bernoulli equation. 2. Suppose that a lake was initially stocked many years ago with 100 trout. After two years the population had grown to about 300. In recent years the population has had a stea value of about 1000 trout. No fishing is allowed in the lake, and fairly constant environmental conditions have persisted since the lake was first stocked. (a) Assuming a logistic model, write down the solution of the appropriate initial-value problem in terms of the unknown coefficient k. (b) The observation P (2) = 300 leads to an equation for k. Solve for k. 3. Suppose that the number P (t) of mature trees per acre in a large hardwood forest obeys the logistic equation (1). For years, P (t) had maintained a stea value of about 400 trees. Ten years ago, a logging company began harvesting at a rate corresponding to one out of twenty mature trees per year. The population of mature trees is now estimated to be about 280 trees per acre. (a) Write down, in terms of the coefficient k, the solution of the appropriate initialvalue problem.

20 70 Chapter 3. Nonlinear First-Order Equations I (b) The observation P (10) = 280 leads to a transcendental equation for k. Estimate k by solving this equation numerically. (c) Predict the number of mature trees per acre after many years of such harvesting. 4. Records show that the average number of mature trees per acre in a large forest 150 years ago was 200, and 50 years later the number had reached 280. For the past 20 years the number has remained about 300. Based on this information, what percent of the trees in this forest can be harvested each year and still maintain a stea stand of 200 trees per acre in the long run? 5. Suppose that 20 years ago a deer population in a wildlife refuge was estimated to be 500 deer, and 10 years ago the population had reached 800 deer. The population has remained nearly constant at about 1000 deer for the past several years. (a) Find the environmental carrying capacity M and the population s intrinsic growth rate k. (b) Harvesting what fraction of the population continuously per year would result in a stea population of 700 deer? 6. Suppose that three consecutive annual observations of a deer population in a wildlife refuge showed populations of 100, 133, and 170 deer. (a) Predict the long-term stea population if the community remains undisturbed. (b) Redo part (a) if the third observation were 173 instead of 170. Comment on the reliability of predicting the long-term stea population in this manner. If a hunting policy were implemented based upon a bad prediction, what impact might it have upon the population? 3.5 Direction Fields We have so far learned a number of techniques for finding solutions of certain types of differential equations. However, such techniques apply to only a small set of nonlinear equations, and real-world problems rarely fall into these solvable categories. Therefore, it is important to be able to get qualitative information about solutions of a differential equation from the differential equation itself. Remember that our primary goal always is to understand the behavior of solutions, even when formulas for solutions are unobtainable. Consider an equation of the general form y = f(t, y), (1) where f is a given continuous function of two variables. From a geometric point of view, such an equation specifies the slope of the graph of any solution at each point through which it passes in the (t, y)-plane. It is therefore useful to associate with f a direction field that guides solution curves in the plane as t increases. We visualize this direction field as an array of arrows with uniform length, each located at a point (t, y) and having slope f(t, y).

21 3.5. Direction Fields 71 The following examples illustrate direction fields. Each deals with a simple, linear differential equation for the sake of simplicity and solvability; however, the construction of the direction field and the information gained from it depend in no way upon linearity. Example 1 The direction field for the simple linear equation y = y, along with one solution curve passing through (0, 1), is shown in Figure 1a. Note that this solution is simply y = e t. Figure 1b shows several solution curves, corresponding to various initial values, for the same equation. In both figures, the slope of each arrow is equal to the y-coordinate at its base, since the function specifying the slope is f(t, y) = y. Figure 1a Figure 1b Example 2 Figures 2a and 2b show the direction field for the equation y = y, first with one solution curve through (0, 1) (which is the graph of y = e t ) and then with several solution curves. In both figures, the slope of each arrow is equal to the negative of the y-coordinate at its base, since the function specifying the slope is f(t, y) = y. Figure 2a Figure 2b An important thing to notice about the direction fields in all four of the preceding figures is the effect of having the right side of the differential equation depend only on y. This causes all direction arrows on any given horizontal line to be parallel, since the slope depends only on y and not on t. These lines are simple examples of what are called isoclines. An isocline is a curve with the

22 72 Chapter 3. Nonlinear First-Order Equations I property that every solution curve that crosses it does so with the same slope. In Figures 1 and 2, horizontal lines have this property. In general, isoclines for solutions of y = f(t, y) are described by f(t, y) = m, where m is an arbitrary constant that represents the slope with which solution curves cross a particular isocline. Notice that these may be thought of as the level curves of the surface z = f(t, y) in tyz-space. Example 3 Figure 3 shows the direction field and several solution curves for y = t y. The direction field here depends in a simple way on both y and t. The isoclines (shown as dashed lines in the figure) are the lines t y = m, each of which has slope 1, crosses the t-axis at t = m, and crosses the y-axis at y = m. Notice, for example, that the line t y = 0 is an isocline along which tangents to solution curves are all horizontal. Figure 3 Example 4 The direction field and several solution curves for the equation y = t y are shown in Figure 4. Here the isoclines (the dashed curves in the figure) are the hyperbolas t y = m, m 0, together with the coordinate axes, corresponding to m = 0. Figure 4 The equations in Examples 3 and 4 are fairly simple to solve explicitly, and you ll be asked to do so in Problem 1. The point, however, is that an equation s direction field can help us understand the behavior of solutions directly from the equation itself.

23 3.5. Direction Fields 73 Problems 1. Sketch several solution curves over each direction field. (a) (b) Problems 2 5: Sketch several isoclines and solution curves over the given direction field. Identify the direction field with one of the equations y = t y 2, y = y t 2, y = y (y 2 4y + 4), y = y (y 2 4y + 3). What kind of curves are the isoclines in each case? For each of the differential equations in Problems 6 9, sketch isoclines corresponding to the indicated slopes. Then sketch the direction field and a few solution curves. 6. y = t 2 + y 2, m = 0, 1/4, 1, 4 7. y = e 1 t2 y 2, m = e 15/16, e 3/4, 1, e 5/4, e 3 8. y = y + sin t, m = 0, ± 1/2, ± 1

24 74 Chapter 3. Nonlinear First-Order Equations I 9. y = tan 1 (t y), m = 0, ± π/6, ± π/4, ± π/3, ± tan The following figure shows the graph and a contour plot of the function f(t, y) = 4y t + e1 t2 y t 2 + y 2. The contours are the curves f(t, y) = m for the indicated values of m. (Also notice the visible portions of the t-and y-axes in the surface plot.) Reproduce the contour plot, and sketch over it the direction field associated with f. Then sketch several solution curves of the equation y = f(t, y). Repeat Problem 10 for functions f with the contour plots in 11 and 12. The contours are labeled with their corresponding slope values

25 3.6. Numerical Approximation Sketch the direction field and several solution curves for the logistic equation y = ky (1 y/m), assuming that k and M are given positive numbers. 14. Sketch the direction field and several solution curves for the Ricatti equation y = r(a y)(b y), assuming that r, a, b are constant with r > 0 and 0 < a < b. 3.6 Numerical Approximation Solution curves such as those in depicted in Section 3.5 can be generated without explicitly solving the equations with the aid of a numerical approximation procedure. Here we will only scratch the surface of the subject by looking at two fairly simple methods, beginning with the most basic one. Euler s Method. Consider the equation y = f(t, y). By integrating each side of this equation over an interval [t, t + h], we obtain y(t + h) = y(t) + t+h t f(s, y(s))ds. A left-endpoint approximation of the integral on the right side gives y(t + h) y(t) + h f(t, y(t)). This formula describes a simple way of approximating the value of the solution at t + h from knowledge of the solution at t. The resulting iterative procedure is called Euler s method. We start with an initial value y 0 = y(t 0 ) and a small stepsize h. We then iteratively compute approximations using the formula y 1 y(t 0 + h), y 2 y(t 0 + 2h), y 3 y(t 0 + 3h),... y n+1 = y n + h f(t n, y n ), where t n = t 0 + nh. As one would expect, the size of h greatly affects the quality of the approximations. Generally speaking, the smaller h is, the better the approximations will be. However, smaller values of the stepsize require more steps to reach a given t-value from the starting point t 0, and with each step a new error can compound errors made at previous steps. A later section contains a more careful discussion of these ideas. For now we will simply proceed with the awareness that numerical approximations can be inaccurate and therefore should be viewed with a heathly skepticism.

26 76 Chapter 3. Nonlinear First-Order Equations I Example 1 Consider the initial-value problem y = t y, y(0) = 1. The general Euler step for this problem is y n+1 = y n + h ( t n y n ). Five steps with h = 0.1 are computed (to four decimal places) as follows: y(0.1) y 1 = 1 + (0.1)( 0)(1) = 1 y(0.2) y 2 = 1 + (0.1)( 0.1)(1) = 0.99 y(0.3) y 3 = (0.1)( 0.2)(0.99) = y(0.4) y 4 = (0.1)( 0.3)(0.9702) = y(0.5) y 5 = (0.1)( 0.4)(0.9411) = Example 2 Figure 5 shows approximate solutions obtained with Euler s method for the problem y = t 2 y, y(0) = 1, using stepsizes of h = 0.05, 0.5, and 1.0, respectively. The graphs are linear interpolants of the data points generated by the method. The nearly smooth curve corresponds to h = 0.05 and follows the exact solution curve fairly closely. The other two give a good graphical illustration of exactly what Euler s method does. It simply follows the direction field at the current point. Figure 5 It should be easy to imagine that errors can cause approximations to wander quite far away from the exact solution after many steps. In fact, rather chaotic behavior can arise easily, as illustrated in the next example. Example 3 Figure 6 shows approximate solutions of y = t y 2, y(0) = 1, obtained by Euler s method with stepsizes h = 0.06 and 0.6. The approximate solution obtained with h = 0.06 follows the exact solution reasonably well, but notice how the approximate solution obtained with h = 0.6 starts out poorly and Figure 6

27 3.6. Numerical Approximation 77 then wanders back close to the correct solution, where it remains for some time before beginning to oscillate wildly. This not surprising if we look closely at the direction field. Improved Euler. One of many methods that are more accurate than Euler s method at a given stepsize is the improved Euler s method, a.k.a., improved Euler. The method is based upon y(t + h) y(t) + h ( ) f(t, y(t)) + f(t + h, y(t + h)), 2 in which the integral t+h t f(s, y(s))ds is approximated by a simple two-point application the trapezoidal rule. In it, we approximate the y(t + h) on the right side with Euler s method, obtaining y(t + h) y(t) + h ( ) f(t, y(t)) + f(t + h, y(t) + hf(t, y(t))). 2 This gives rise to the iterative formula y n+1 = y n + h ( ) f(t n, y n ) + f(t n+1, y n + hf(t n, y n )). 2 A somewhat more efficient organization of the calculation is as follows: k 1 = f(t n, y n ), k 2 = f(t n + h, y n + h k 1 ), y n+1 = y n + h ) (k 1 + k 2. 2 Note here that k 1 and k 2 are scratch calculations that need not be kept track of from step to step. Example 3 Consider again the initial-value problem in Example 1: y = t y, y(0) = 1. For this particular problem, improved Euler assumes the form k 1 = t n y n, k 2 = (t n + h)(y n + h k 1 ), y n+1 = y n + h ) (k 1 + k 2. 2 Four steps with h = 0.1 are computed (rounding to five places) as follows: k 1 = (0)(1) = 0 k 2 = (0.1)(1 + (0.1)(0)) = 0.1 y(0.1) y 1 = 1 + (0.05)(0 + ( 0.1)) = * Improved Euler is a basic example of family of methods known as Runge-Kutta methods.

28 78 Chapter 3. Nonlinear First-Order Equations I k 1 = (0.1)(0.995) = k 2 = (0.2)( (0.1)( )) = y(0.2) y 2 = (0.05)( ( )) = k 1 = (0.2)( ) = k 2 = (0.3)( (0.1)( )) = y(0.3) y 3 = (0.05)( ( )) = k 1 = (0.3)( ) = k 2 = (0.4)( (0.1)( )) = y(0.4) y 4 = (0.05)( ( )) = The exact solution of this initial-value problem is y(t) = e t2 /2. Thus we can compare the preceding approximations to the correct values (again rounding to five places): y(0.1) = , y(0.2) = , y(0.3) = , y(0.4) = So we see that while Euler s method is giving roughly one-decimal-place accuracy (cf. Example 1), improved Euler gives four-decimal-place accuracy. Problems 1. Do this problem with a calculator. For the initial-value problem y = 1/y, y(0) = 1, perform enough steps of Euler s method to reach t = 0.5 using stepsize h = 0.1. Repeat with h = Then find the exact solution and compare your approximations of y(.5) to the exact value. 2. Rework Problem 1 using the improved Euler s method. 3. Find the result of one step of Euler s method in terms of the stepsize h, and then compute the error as a Taylor series in h. (a) y = y, y(0) = 1 (b) y = y 2, y(0) = 1 4. Rework Problem 3 using the improved Euler s method. 5. For the problem y = y cos t, y(0) = 1, compute an approximation of y(0.2) with (a) Euler s method and (b) improved Euler, using stepsize h = 0.1 in each case. Find the exact value of y(0.2) and compute the error in each approximation.

29 3.6. Numerical Approximation Consider the following logistic model with a seasonal carrying capacity: ( ) P P = 0.21P 1, P (0) = (3 + 2 sin(2πt)) Using a stepsize of h = 1/12 in improved Euler, approximate the population at monthly intervals for one year, and sketch the resulting graph. 7. Consider the following logistic model with a seasonal intrinsic growth rate: ( P = 0.03(7 + 6 sin(2πt))p 1 P ), P (0) = Using a stepsize of h = 1/12 in improved Euler, approximate the population at monthly intervals for one year, and sketch the resulting graph. 8. A method similar to improved Euler, but based on Simpson s rule, is k 1 = f(t n, y n ), k 2 = f (t n + h/2, y n + hk 1 /2), k 3 = f(t n + h, y n + hk 2 ), y n+1 = y n + h 6 (k 1 + 4k 2 + k 3 ). Rework Problem 1 using this method. 9. Rework Problem 3 using the method in Problem 8.