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Section 2.6 Exact Equations 85 Proof (a) If (My −Nx)/N is independent of y, then (2.6.8) holds with p = (My −Nx)/N and q ≡0. Therefore P (x) = ±e R p(x) dx and Q(y) = ±e R q(y) dy = ±e0 = ±1, so (2.6.10) is an integrating factor for (2.6.11) on R. (b) If (Nx −My)/M is independent of x then eqrefeq:2.6.8 holds with p ≡0 and q = (Nx −My)/M, and a similar argument shows that (2.6.12) is an integrating factor for (2.6.11) on R. The next two examples show how to apply Theorem 2.6.1. Example 2.6.1 Find an integrating factor for the equation (2xy3 −2x3y3 −4xy2 + 2x) dx + (3x2y2 + 4y) dy = 0 (2.6.13) and solve the equation. Solution In (2.6.13) M = 2xy3 −2x3y3 −4xy2 + 2x, N = 3x2y2 + 4y, and My −Nx = (6xy2 −6x3y2 −8xy) −6xy2 = −6x3y2 −8xy, so (2.6.13) isn’t exact. However, My −Nx N = −6x3y2 + 8xy 3x2y2 + 4y = −2x is independent of y, so Theorem 2.6.1(a) applies with p(x) = −2x. Since Z p(x) dx = − Z 2x dx = −x2, µ(x) = e−x2 is an integrating factor. Multiplying (2.6.13) by µ yields the exact equation e−x2(2xy3 −2x3y3 −4xy2 + 2x) dx + e−x2(3x2y2 + 4y) dy = 0. (2.6.14) To solve this equation, we must find a function F such that Fx(x, y) = e−x2(2xy3 −2x3y3 −4xy2 + 2x) (2.6.15) and Fy(x, y) = e−x2(3x2y2 + 4y). (2.6.16) Integrating (2.6.16) with respect to y yields F (x, y) = e−x2(x2y3 + 2y2) + ψ(x). (2.6.17) Differentiating this with respect to x yields Fx(x, y) = e−x2(2xy3 −2x3y3 −4xy2) + ψ′(x). Comparing this with (2.6.15) shows that ψ′(x) = 2xe−x2; therefore, we can let ψ(x) = −e−x2 in (2.6.17) and conclude that e−x2 | Elementary Differential Equations with Boundary Value Problems_Page_95_Chunk2501 |
86 Chapter 2 Integrating Factors −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −4 −3 −2 −1 0 1 2 3 4 y x Figure 2.6.1 A direction field and integral curves for (2xy3 −2x3y3 −4xy2 + 2x) dx + (3x2y2 + 4y) dy = 0 Example 2.6.2 Find an integrating factor for 2xy3 dx + (3x2y2 + x2y3 + 1) dy = 0 (2.6.18) and solve the equation. Solution In (2.6.18), M = 2xy3, N = 3x2y2 + x2y3 + 1, and My −Nx = 6xy2 −(6xy2 + 2xy3) = −2xy3, so (2.6.18) isn’t exact. Moreover, My −Nx N = − 2xy3 3x2y2 + x2y2 + 1 is not independent of y, so Theorem 2.6.1(a) does not apply. However, Theorem 2.6.1(b) does apply, since Nx −My M = 2xy3 2xy3 = 1 is independent of x, so we can take q(y) = 1. Since Z q(y) dy = Z dy = y, | Elementary Differential Equations with Boundary Value Problems_Page_96_Chunk2502 |
Section 2.6 Exact Equations 87 µ(y) = ey is an integrating factor. Multiplying (2.6.18) by µ yields the exact equation 2xy3ey dx + (3x2y2 + x2y3 + 1)ey dy = 0. (2.6.19) To solve this equation, we must find a function F such that Fx(x, y) = 2xy3ey (2.6.20) and Fy(x, y) = (3x2y2 + x2y3 + 1)ey. (2.6.21) Integrating (2.6.20) with respect to x yields F (x, y) = x2y3ey + φ(y). (2.6.22) Differentiating this with respect to y yields Fy = (3x2y2 + x2y3)ey + φ′(y), and comparing this with (2.6.21) shows that φ′(y) = ey. Therefore we set φ(y) = ey in (2.6.22) and conclude that (x2y3 + 1)ey = c is an implicit solution of (2.6.19). It is also an implicit solution of (2.6.18). Figure 2.6.2 shows a direction field and some integral curves for (2.6.18). −4 −3 −2 −1 0 1 2 3 4 −4 −3 −2 −1 0 1 2 3 4 y x Figure 2.6.2 A direction field and integral curves for 2xy3ey dx + (3x2y2 + x2y3 + 1)ey dy = 0 Theorem 2.6.1 does not apply in the next example, but the more general argument that led to Theo- rem 2.6.1 provides an integrating factor. | Elementary Differential Equations with Boundary Value Problems_Page_97_Chunk2503 |
88 Chapter 2 Integrating Factors Example 2.6.3 Find an integrating factor for (3xy + 6y2) dx + (2x2 + 9xy) dy = 0 (2.6.23) and solve the equation. Solution In (2.6.23) M = 3xy + 6y2, N = 2x2 + 9xy, and My −Nx = (3x + 12y) −(4x + 9y) = −x + 3y. Therefore My −Nx M = −x + 3y 3xy + 6y2 and Nx −My N = x −3y 2x2 + 9xy, so Theorem 2.6.1 does not apply. Following the more general argument that led to Theorem 2.6.1, we look for functions p = p(x) and q = q(y) such that My −Nx = p(x)N −q(y)M; that is, −x + 3y = p(x)(2x2 + 9xy) −q(y)(3xy + 6y2). Since the left side contains only first degree terms in x and y, we rewrite this equation as xp(x)(2x + 9y) −yq(y)(3x + 6y) = −x + 3y. This will be an identity if xp(x) = A and yq(y) = B, (2.6.24) where A and B are constants such that −x + 3y = A(2x + 9y) −B(3x + 6y), or, equivalently, −x + 3y = (2A −3B)x + (9A −6B)y. Equating the coefficients of x and y on both sides shows that the last equation holds for all (x, y) if 2A −3B = −1 9A −6B = 3, which has the solution A = 1, B = 1. Therefore (2.6.24) implies that p(x) = 1 x and q(y) = 1 y . Since Z p(x) dx = ln|x| and Z q(y) dy = ln |y|, we can let P (x) = x and Q(y) = y; hence, µ(x, y) = xy is an integrating factor. Multiplying (2.6.23) by µ yields the exact equation (3x2y2 + 6xy3) dx + (2x3y + 9x2y2) dy = 0. | Elementary Differential Equations with Boundary Value Problems_Page_98_Chunk2504 |
Section 2.6 Exact Equations 89 −4 −3 −2 −1 0 1 2 3 4 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 y x Figure 2.6.3 A direction field and integral curves for (3xy + 6y2) dx + (2x2 + 9xy) dy = 0 We leave it to you to use the method of Section 2.5 to show that this equation has the implicit solution x3y2 + 3x2y3 = c. (2.6.25) This is also an implicit solution of (2.6.23). Since x ≡0 and y ≡0 satisfy (2.6.25), you should check to see that x ≡0 and y ≡0 are also solutions of (2.6.23). (Why is it necesary to check this?) Figure 2.6.3 shows a direction field and integral curves for (2.6.23). See Exercise 28 for a general discussion of equations like (2.6.23). Example 2.6.4 The separable equation −y dx + (x + x6) dy = 0 (2.6.26) can be converted to the exact equation − dx x + x6 + dy y = 0 (2.6.27) by multiplying through by the integrating factor µ(x, y) = 1 y(x + x6). However, to solve (2.6.27) by the method of Section 2.5 we would have to evaluate the nasty integral Z dx x + x6 . Instead, we solve (2.6.26) explicitly for y by finding an integrating factor of the form µ(x, y) = xayb. | Elementary Differential Equations with Boundary Value Problems_Page_99_Chunk2505 |
90 Chapter 2 Integrating Factors −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 y x Figure 2.6.4 A direction field and integral curves for −y dx + (x + x6) dy = 0 Solution In (2.6.26) M = −y, N = x + x6, and My −Nx = −1 −(1 + 6x5) = −2 −6x5. We look for functions p = p(x) and q = q(y) such that My −Nx = p(x)N −q(y)M; that is, −2 −6x5 = p(x)(x + x6) + q(y)y. (2.6.28) The right side will contain the term −6x5 if p(x) = −6/x. Then (2.6.28) becomes −2 −6x5 = −6 −6x5 + q(y)y, so q(y) = 4/y. Since Z p(x) dx = − Z 6 x dx = −6 ln|x| = ln 1 x6 , and Z q(y) dy = Z 4 y dy = 4 ln|y| = lny4, we can take P (x) = x−6 and Q(y) = y4, which yields the integrating factor µ(x, y) = x−6y4. Multi- plying (2.6.26) by µ yields the exact equation −y5 x6 dx + y4 x5 + y4 dy = 0. | Elementary Differential Equations with Boundary Value Problems_Page_100_Chunk2506 |
Section 2.6 Exact Equations 91 We leave it to you to use the method of the Section 2.5 to show that this equation has the implicit solution y x 5 + y5 = k. Solving for y yields y = k1/5x(1 + x5)−1/5, which we rewrite as y = cx(1 + x5)−1/5 by renaming the arbitrary constant. This is also a solution of (2.6.26). Figure 2.6.4 shows a direction field and some integral curves for (2.6.26). 2.6 Exercises 1. (a) Verify that µ(x, y) = y is an integrating factor for y dx + 2x + 1 y dy = 0 (A) on any open rectangle that does not intersect the x axis or, equivalently, that y2 dx + (2xy + 1) dy = 0 (B) is exact on any such rectangle. (b) Verify that y ≡0 is a solution of (B), but not of (A). (c) Show that y(xy + 1) = c (C) is an implicit solution of (B), and explain why every differentiable function y = y(x) other than y ≡0 that satisfies (C) is also a solution of (A). 2. (a) Verify that µ(x, y) = 1/(x −y)2 is an integrating factor for −y2 dx + x2 dy = 0 (A) on any open rectangle that does not intersect the line y = x or, equivalently, that − y2 (x −y)2 dx + x2 (x −y)2 dy = 0 (B) is exact on any such rectangle. (b) Use Theorem 2.2.1 to show that xy (x −y) = c (C) is an implicit solution of (B), and explain why it’s also an implicit solution of (A) (c) Verify that y = x is a solution of (A), even though it can’t be obtained from (C). In Exercises 3–16 find an integrating factor; that is a function of only one variable, and solve the given equation. 3. y dx −x dy = 0 4. 3x2y dx + 2x3 dy = 0 5. 2y3 dx + 3y2 dy = 0 6. (5xy + 2y + 5) dx + 2x dy = 0 | Elementary Differential Equations with Boundary Value Problems_Page_101_Chunk2507 |
92 Chapter 2 Integrating Factors 7. (xy + x + 2y + 1) dx + (x + 1) dy = 0 8. (27xy2 + 8y3) dx + (18x2y + 12xy2) dy = 0 9. (6xy2 + 2y) dx + (12x2y + 6x + 3) dy = 0 10. y2 dx + xy2 + 3xy + 1 y dy = 0 11. (12x3y + 24x2y2) dx + (9x4 + 32x3y + 4y) dy = 0 12. (x2y + 4xy + 2y) dx + (x2 + x) dy = 0 13. −y dx + (x4 −x) dy = 0 14. cos x cos y dx + (sin x cos y −sin x siny + y) dy = 0 15. (2xy + y2) dx + (2xy + x2 −2x2y2 −2xy3) dy = 0 16. y sin y dx + x(sin y −y cos y) dy = 0 In Exercises 17–23 find an integrating factor of the form µ(x, y) = P (x)Q(y) and solve the given equation. 17. y(1 + 5 ln|x|) dx + 4x ln|x| dy = 0 18. (αy + γxy) dx + (βx + δxy) dy = 0 19. (3x2y3 −y2 + y) dx + (−xy + 2x) dy = 0 20. 2y dx + 3(x2 + x2y3) dy = 0 21. (a cos xy −y sin xy) dx + (b cos xy −x sin xy) dy = 0 22. x4y4 dx + x5y3 dy = 0 23. y(x cos x + 2 sin x) dx + x(y + 1) sin x dy = 0 In Exercises 24–27 find an integrating factor and solve the equation. Plot a direction field and some integral curves for the equation in the indicated rectangular region. 24. C/G (x4y3 + y) dx + (x5y2 −x) dy = 0; {−1 ≤x ≤1, −1 ≤y ≤1} 25. C/G (3xy + 2y2 + y) dx + (x2 + 2xy + x + 2y) dy = 0; {−2 ≤x ≤2, −2 ≤y ≤2} 26. C/G (12xy + 6y3) dx + (9x2 + 10xy2) dy = 0; {−2 ≤x ≤2, −2 ≤y ≤2} 27. C/G (3x2y2 + 2y) dx + 2x dy = 0; {−4 ≤x ≤4, −4 ≤y ≤4} 28. Suppose a, b, c, and d are constants such that ad −bc ̸= 0, and let m and n be arbitrary real numbers. Show that (axmy + byn+1) dx + (cxm+1 + dxyn) dy = 0 has an integrating factor µ(x, y) = xαyβ. 29. Suppose M, N, Mx, and Ny are continuous for all (x, y), and µ = µ(x, y) is an integrating factor for M(x, y) dx + N(x, y) dy = 0. (A) Assume that µx and µy are continuous for all (x, y), and suppose y = y(x) is a differentiable function such that µ(x, y(x)) = 0 and µx(x, y(x)) ̸= 0 for all x in some interval I. Show that y is a solution of (A) on I. | Elementary Differential Equations with Boundary Value Problems_Page_102_Chunk2508 |
Section 2.6 Exact Equations 93 30. According to Theorem 2.1.2, the general solution of the linear nonhomogeneous equation y′ + p(x)y = f(x) (A) is y = y1(x) c + Z f(x)/y1(x) dx , (B) where y1 is any nontrivial solution of the complementary equation y′ +p(x)y = 0. In this exercise we obtain this conclusion in a different way. You may find it instructive to apply the method suggested here to solve some of the exercises in Section 2.1. (a) Rewrite (A) as [p(x)y −f(x)] dx + dy = 0, (C) and show that µ = ±e R p(x) dx is an integrating factor for (C). (b) Multiply (A) through by µ = ±e R p(x) dx and verify that the resulting equation can be rewrit- ten as (µ(x)y)′ = µ(x)f(x). Then integrate both sides of this equation and solve for y to show that the general solution of (A) is y = 1 µ(x) c + Z f(x)µ(x) dx . Why is this form of the general solution equivalent to (B)? | Elementary Differential Equations with Boundary Value Problems_Page_103_Chunk2509 |
CHAPTER 3 Numerical Methods In this chapter we study numerical methods for solving a first order differential equation y′ = f(x, y). SECTION 3.1 deals with Euler’s method, which is really too crude to be of much use in practical appli- cations. However, its simplicity allows for an introduction to the ideas required to understand the better methods discussed in the other two sections. SECTION 3.2 discusses improvements on Euler’s method. SECTION 3.3 deals with the Runge-Kutta method, perhaps the most widely used method for numerical solution of differential equations. 95 | Elementary Differential Equations with Boundary Value Problems_Page_105_Chunk2510 |
96 Chapter 3 Numerical Methods 3.1 EULER’S METHOD If an initial value problem y′ = f(x, y), y(x0) = y0 (3.1.1) can’t be solved analytically, it’s necessary to resort to numerical methods to obtain useful approximations to a solution of (3.1.1). We’ll consider such methods in this chapter. We’re interested in computing approximate values of the solution of (3.1.1) at equally spaced points x0, x1, ..., xn = b in an interval [x0, b]. Thus, xi = x0 + ih, i = 0, 1, . . ., n, where h = b −x0 n . We’ll denote the approximate values of the solution at these points by y0, y1, ..., yn; thus, yi is an approximation to y(xi). We’ll call ei = y(xi) −yi the error at the ith step. Because of the initial condition y(x0) = y0, we’ll always have e0 = 0. However, in general ei ̸= 0 if i > 0. We encounter two sources of error in applying a numerical method to solve an initial value problem: • The formulas defining the method are based on some sort of approximation. Errors due to the inaccuracy of the approximation are called truncation errors. • Computers do arithmetic with a fixed number of digits, and therefore make errors in evaluating the formulas defining the numerical methods. Errors due to the computer’s inability to do exact arithmetic are called roundoff errors. Since a careful analysis of roundoff error is beyond the scope of this book, we’ll consider only trunca- tion errors. Euler’s Method The simplest numerical method for solving (3.1.1) is Euler’s method. This method is so crude that it is seldom used in practice; however, its simplicity makes it useful for illustrative purposes. Euler’s method is based on the assumption that the tangent line to the integral curve of (3.1.1) at (xi, y(xi)) approximates the integral curve over the interval [xi, xi+1]. Since the slope of the integral curve of (3.1.1) at (xi, y(xi)) is y′(xi) = f(xi, y(xi)), the equation of the tangent line to the integral curve at (xi, y(xi)) is y = y(xi) + f(xi, y(xi))(x −xi). (3.1.2) Setting x = xi+1 = xi + h in (3.1.2) yields yi+1 = y(xi) + hf(xi, y(xi)) (3.1.3) as an approximation to y(xi+1). Since y(x0) = y0 is known, we can use (3.1.3) with i = 0 to compute y1 = y0 + hf(x0, y0). However, setting i = 1 in (3.1.3) yields y2 = y(x1) + hf(x1, y(x1)), | Elementary Differential Equations with Boundary Value Problems_Page_106_Chunk2511 |
Section 3.1 Euler’s Method 97 which isn’t useful, since we don’t know y(x1). Therefore we replace y(x1) by its approximate value y1 and redefine y2 = y1 + hf(x1, y1). Having computed y2, we can compute y3 = y2 + hf(x2, y2). In general, Euler’s method starts with the known value y(x0) = y0 and computes y1, y2, ..., yn succes- sively by with the formula yi+1 = yi + hf(xi, yi), 0 ≤i ≤n −1. (3.1.4) The next example illustrates the computational procedure indicated in Euler’s method. Example 3.1.1 Use Euler’s method with h = 0.1 to find approximate values for the solution of the initial value problem y′ + 2y = x3e−2x, y(0) = 1 (3.1.5) at x = 0.1, 0.2, 0.3. Solution We rewrite (3.1.5) as y′ = −2y + x3e−2x, y(0) = 1, which is of the form (3.1.1), with f(x, y) = −2y + x3e−2x, x0 = 0, and y0 = 1. Euler’s method yields y1 = y0 + hf(x0, y0) = 1 + (.1)f(0, 1) = 1 + (.1)(−2) = .8, y2 = y1 + hf(x1, y1) = .8 + (.1)f(.1, .8) = .8 + (.1) | Elementary Differential Equations with Boundary Value Problems_Page_107_Chunk2512 |
98 Chapter 3 Numerical Methods Solution Table 3.1.1 shows the values of the exact solution (3.1.6) at the specified points, and the ap- proximate values of the solution at these points obtained by Euler’s method with step sizes h = 0.1, h = 0.05, and h = 0.025. In examining this table, keep in mind that the approximate values in the col- umn corresponding to h = .05 are actually the results of 20 steps with Euler’s method. We haven’t listed the estimates of the solution obtained for x = 0.05, 0.15, ..., since there’s nothing to compare them with in the column corresponding to h = 0.1. Similarly, the approximate values in the column corresponding to h = 0.025 are actually the results of 40 steps with Euler’s method. Table 3.1.1. Numerical solution of y′ + 2y = x3e−2x, y(0) = 1, by Euler’s method. x h = 0.1 h = 0.05 h = 0.025 Exact 0.0 1.000000000 1.000000000 1.000000000 1.000000000 0.1 0.800000000 0.810005655 0.814518349 0.818751221 0.2 0.640081873 0.656266437 0.663635953 0.670588174 0.3 0.512601754 0.532290981 0.541339495 0.549922980 0.4 0.411563195 0.432887056 0.442774766 0.452204669 0.5 0.332126261 0.353785015 0.363915597 0.373627557 0.6 0.270299502 0.291404256 0.301359885 0.310952904 0.7 0.222745397 0.242707257 0.252202935 0.261398947 0.8 0.186654593 0.205105754 0.213956311 0.222570721 0.9 0.159660776 0.176396883 0.184492463 0.192412038 1.0 0.139778910 0.154715925 0.162003293 0.169169104 You can see from Table 3.1.1 that decreasing the step size improves the accuracy of Euler’s method. For example, yexact(1) −yapprox(1) ≈ .0293 with h = 0.1, .0144 with h = 0.05, .0071 with h = 0.025. Based on this scanty evidence, you might guess that the error in approximating the exact solution at a fixed value of x by Euler’s method is roughly halved when the step size is halved. You can find more evidence to support this conjecture by examining Table 3.1.2, which lists the approximate values of yexact −yapprox at x = 0.1, 0.2, ..., 1.0. Table 3.1.2. Errors in approximate solutions of y′ + 2y = x3e−2x, y(0) = 1, obtained by Euler’s method. x h = 0.1 h = 0.05 h = 0.025 0.1 0.0187 0.0087 0.0042 0.2 0.0305 0.0143 0.0069 0.3 0.0373 0.0176 0.0085 0.4 0.0406 0.0193 0.0094 0.5 0.0415 0.0198 0.0097 0.6 0.0406 0.0195 0.0095 0.7 0.0386 0.0186 0.0091 0.8 0.0359 0.0174 0.0086 0.9 0.0327 0.0160 0.0079 1.0 0.0293 0.0144 0.0071 Example 3.1.3 Tables 3.1.3 and 3.1.4 show analogous results for the nonlinear initial value problem y′ = −2y2 + xy + x2, y(0) = 1, (3.1.7) | Elementary Differential Equations with Boundary Value Problems_Page_108_Chunk2513 |
Section 3.1 Euler’s Method 99 except in this case we can’t solve (3.1.7) exactly. The results in the “Exact” column were obtained by using a more accurate numerical method known as the Runge-Kutta method with a small step size. They are exact to eight decimal places. Since we think it’s important in evaluating the accuracy of the numerical methods that we’ll be studying in this chapter, we often include a column listing values of the exact solution of the initial value problem, even if the directions in the example or exercise don’t specifically call for it. If quotation marks are included in the heading, the values were obtained by applying the Runge-Kutta method in a way that’s explained in Section 3.3. If quotation marks are not included, the values were obtained from a known formula for the solution. In either case, the values are exact to eight places to the right of the decimal point. Table 3.1.3. Numerical solution of y′ = −2y2 + xy + x2, y(0) = 1, by Euler’s method. x h = 0.1 h = 0.05 h = 0.025 “Exact” 0.0 1.000000000 1.000000000 1.000000000 1.000000000 0.1 0.800000000 0.821375000 0.829977007 0.837584494 0.2 0.681000000 0.707795377 0.719226253 0.729641890 0.3 0.605867800 0.633776590 0.646115227 0.657580377 0.4 0.559628676 0.587454526 0.600045701 0.611901791 0.5 0.535376972 0.562906169 0.575556391 0.587575491 0.6 0.529820120 0.557143535 0.569824171 0.581942225 0.7 0.541467455 0.568716935 0.581435423 0.593629526 0.8 0.569732776 0.596951988 0.609684903 0.621907458 0.9 0.614392311 0.641457729 0.654110862 0.666250842 1.0 0.675192037 0.701764495 0.714151626 0.726015790 Table 3.1.4. Errors in approximate solutions of y′ = −2y2 + xy + x2, y(0) = 1, obtained by Euler’s method. x h = 0.1 h = 0.05 h = 0.025 0.1 0.0376 0.0162 0.0076 0.2 0.0486 0.0218 0.0104 0.3 0.0517 0.0238 0.0115 0.4 0.0523 0.0244 0.0119 0.5 0.0522 0.0247 0.0121 0.6 0.0521 0.0248 0.0121 0.7 0.0522 0.0249 0.0122 0.8 0.0522 0.0250 0.0122 0.9 0.0519 0.0248 0.0121 1.0 0.0508 0.0243 0.0119 Truncation Error in Euler’s Method Consistent with the results indicated in Tables 3.1.1–3.1.4, we’ll now show that under reasonable as- sumptions on f there’s a constant K such that the error in approximating the solution of the initial value problem y′ = f(x, y), y(x0) = y0, at a given point b > x0 by Euler’s method with step size h = (b −x0)/n satisfies the inequality |y(b) −yn| ≤Kh, | Elementary Differential Equations with Boundary Value Problems_Page_109_Chunk2514 |
100 Chapter 3 Numerical Methods where K is a constant independent of n. There are two sources of error (not counting roundoff) in Euler’s method: 1. The error committed in approximating the integral curve by the tangent line (3.1.2) over the interval [xi, xi+1]. 2. The error committed in replacing y(xi) by yi in (3.1.2) and using (3.1.4) rather than (3.1.2) to compute yi+1. Euler’s method assumes that yi+1 defined in (3.1.2) is an approximation to y(xi+1). We call the error in this approximation the local truncation error at the ith step, and denote it by Ti; thus, Ti = y(xi+1) −y(xi) −hf(xi, y(xi)). (3.1.8) We’ll now use Taylor’s theorem to estimate Ti, assuming for simplicity that f, fx, and fy are continuous and bounded for all (x, y). Then y′′ exists and is bounded on [x0, b]. To see this, we differentiate y′(x) = f(x, y(x)) to obtain y′′(x) = fx(x, y(x)) + fy(x, y(x))y′(x) = fx(x, y(x)) + fy(x, y(x))f(x, y(x)). Since we assumed that f, fx and fy are bounded, there’s a constant M such that |fx(x, y(x)) + fy(x, y(x))f(x, y(x))| ≤M, x0 < x < b, which implies that |y′′(x)| ≤M, x0 < x < b. (3.1.9) Since xi+1 = xi + h, Taylor’s theorem implies that y(xi+1) = y(xi) + hy′(xi) + h2 2 y′′(˜xi), where ˜xi is some number between xi and xi+1. Since y′(xi) = f(xi, y(xi)) this can be written as y(xi+1) = y(xi) + hf(xi, y(xi)) + h2 2 y′′(˜xi), or, equivalently, y(xi+1) −y(xi) −hf(xi, y(xi)) = h2 2 y′′(˜xi). Comparing this with (3.1.8) shows that Ti = h2 2 y′′(˜xi). Recalling (3.1.9), we can establish the bound |Ti| ≤Mh2 2 , 1 ≤i ≤n. (3.1.10) Although it may be difficult to determine the constant M, what is important is that there’s an M such that (3.1.10) holds. We say that the local truncation error of Euler’s method is of order h2, which we write as O(h2). | Elementary Differential Equations with Boundary Value Problems_Page_110_Chunk2515 |
Section 3.1 Euler’s Method 101 Note that the magnitude of the local truncation error in Euler’s method is determined by the second derivative y′′ of the solution of the initial value problem. Therefore the local truncation error will be larger where |y′′| is large, or smaller where |y′′| is small. Since the local truncation error for Euler’s method is O(h2), it’s reasonable to expect that halving h reduces the local truncation error by a factor of 4. This is true, but halving the step size also requires twice as many steps to approximate the solution at a given point. To analyze the overall effect of truncation error in Euler’s method, it’s useful to derive an equation relating the errors ei+1 = y(xi+1) −yi+1 and ei = y(xi) −yi. To this end, recall that y(xi+1) = y(xi) + hf(xi, y(xi)) + Ti (3.1.11) and yi+1 = yi + hf(xi, yi). (3.1.12) Subtracting (3.1.12) from (3.1.11) yields ei+1 = ei + h [f(xi, y(xi)) −f(xi, yi)] + Ti. (3.1.13) The last term on the right is the local truncation error at the ith step. The other terms reflect the way errors made at previous steps affect ei+1. Since |Ti| ≤Mh2/2, we see from (3.1.13) that |ei+1| ≤|ei| + h|f(xi, y(xi)) −f(xi, yi)| + Mh2 2 . (3.1.14) Since we assumed that fy is continuous and bounded, the mean value theorem implies that f(xi, y(xi)) −f(xi, yi) = fy(xi, y∗ i )(y(xi) −yi) = fy(xi, y∗ i )ei, where y∗ i is between yi and y(xi). Therefore |f(xi, y(xi)) −f(xi, yi)| ≤R|ei| for some constant R. From this and (3.1.14), |ei+1| ≤(1 + Rh)|ei| + Mh2 2 , 0 ≤i ≤n −1. (3.1.15) For convenience, let C = 1 + Rh. Since e0 = y(x0) −y0 = 0, applying (3.1.15) repeatedly yields |e1| ≤ Mh2 2 |e2| ≤ C|e1| + Mh2 2 ≤(1 + C)Mh2 2 |e3| ≤ C|e2| + Mh2 2 ≤(1 + C + C2)Mh2 2 ... |en| ≤ C|en−1| + Mh2 2 ≤(1 + C + · · · + Cn−1)Mh2 2 . (3.1.16) Recalling the formula for the sum of a geometric series, we see that 1 + C + · · · + Cn−1 = 1 −Cn 1 −C = (1 + Rh)n −1 Rh | Elementary Differential Equations with Boundary Value Problems_Page_111_Chunk2516 |
102 Chapter 3 Numerical Methods (since C = 1 + Rh). From this and (3.1.16), |y(b) −yn| = |en| ≤(1 + Rh)n −1 R Mh 2 . (3.1.17) Since Taylor’s theorem implies that 1 + Rh < eRh (verify), (1 + Rh)n < enRh = eR(b−x0) (since nh = b −x0). This and (3.1.17) imply that |y(b) −yn| ≤Kh, (3.1.18) with K = M eR(b−x0) −1 2R . Because of (3.1.18) we say that the global truncation error of Euler’s method is of order h, which we write as O(h). Semilinear Equations and Variation of Parameters An equation that can be written in the form y′ + p(x)y = h(x, y) (3.1.19) with p ̸≡0 is said to be semilinear. (Of course, (3.1.19) is linear if h is independent of y.) One way to apply Euler’s method to an initial value problem y′ + p(x)y = h(x, y), y(x0) = y0 (3.1.20) for (3.1.19) is to think of it as y′ = f(x, y), y(x0) = y0, where f(x, y) = −p(x)y + h(x, y). However, we can also start by applying variation of parameters to (3.1.20), as in Sections 2.1 and 2.4; thus, we write the solution of (3.1.20) as y = uy1, where y1 is a nontrivial solution of the complementary equation y′ + p(x)y = 0. Then y = uy1 is a solution of (3.1.20) if and only if u is a solution of the initial value problem u′ = h(x, uy1(x))/y1(x), u(x0) = y(x0)/y1(x0). (3.1.21) We can apply Euler’s method to obtain approximate values u0, u1, ..., un of this initial value problem, and then take yi = uiy1(xi) as approximate values of the solution of (3.1.20). We’ll call this procedure the Euler semilinear method. The next two examples show that the Euler and Euler semilinear methods may yield drastically different results. Example 3.1.4 In Example 2.1.7 we had to leave the solution of the initial value problem y′ −2xy = 1, y(0) = 3 (3.1.22) in the form y = ex2 3 + Z x 0 e−t2dt (3.1.23) | Elementary Differential Equations with Boundary Value Problems_Page_112_Chunk2517 |
Section 3.1 Euler’s Method 103 because it was impossible to evaluate this integral exactly in terms of elementary functions. Use step sizes h = 0.2, h = 0.1, and h = 0.05 to find approximate values of the solution of (3.1.22) at x = 0, 0.2, 0.4, 0.6, ..., 2.0 by (a) Euler’s method; (b) the Euler semilinear method. SOLUTION(a) Rewriting (3.1.22) as y′ = 1 + 2xy, y(0) = 3 (3.1.24) and applying Euler’s method with f(x, y) = 1 + 2xy yields the results shown in Table 3.1.5. Because of the large differences between the estimates obtained for the three values of h, it would be clear that these results are useless even if the “exact” values were not included in the table. Table 3.1.5. Numerical solution of y′ −2xy = 1, y(0) = 3, with Euler’s method. x h = 0.2 h = 0.1 h = 0.05 “Exact” 0.0 3.000000000 3.000000000 3.000000000 3.000000000 0.2 3.200000000 3.262000000 3.294348537 3.327851973 0.4 3.656000000 3.802028800 3.881421103 3.966059348 0.6 4.440960000 4.726810214 4.888870783 5.067039535 0.8 5.706790400 6.249191282 6.570796235 6.936700945 1.0 7.732963328 8.771893026 9.419105620 10.184923955 1.2 11.026148659 13.064051391 14.405772067 16.067111677 1.4 16.518700016 20.637273893 23.522935872 27.289392347 1.6 25.969172024 34.570423758 41.033441257 50.000377775 1.8 42.789442120 61.382165543 76.491018246 98.982969504 2.0 73.797840446 115.440048291 152.363866569 211.954462214 It’s easy to see why Euler’s method yields such poor results. Recall that the constant M in (3.1.10) – which plays an important role in determining the local truncation error in Euler’s method – must be an upper bound for the values of the second derivative y′′ of the solution of the initial value problem (3.1.22) on (0, 2). The problem is that y′′ assumes very large values on this interval. To see this, we differentiate (3.1.24) to obtain y′′(x) = 2y(x) + 2xy′(x) = 2y(x) + 2x(1 + 2xy(x)) = 2(1 + 2x2)y(x) + 2x, where the second equality follows again from (3.1.24). Since (3.1.23) implies that y(x) > 3ex2 if x > 0, y′′(x) > 6(1 + 2x2)ex2 + 2x, x > 0. For example, letting x = 2 shows that y′′(2) > 2952. SOLUTION(b) Since y1 = ex2 is a solution of the complementary equation y′ −2xy = 0, we can apply the Euler semilinear method to (3.1.22), with y = uex2 and u′ = e−x2, u(0) = 3. The results listed in Table 3.1.6 are clearly better than those obtained by Euler’s method. Table 3.1.6. Numerical solution of y′ −2xy = 1, y(0) = 3, by the Euler semilinear method. | Elementary Differential Equations with Boundary Value Problems_Page_113_Chunk2518 |
104 Chapter 3 Numerical Methods x h = 0.2 h = 0.1 h = 0.05 “Exact” 0.0 3.000000000 3.000000000 3.000000000 3.000000000 0.2 3.330594477 3.329558853 3.328788889 3.327851973 0.4 3.980734157 3.974067628 3.970230415 3.966059348 0.6 5.106360231 5.087705244 5.077622723 5.067039535 0.8 7.021003417 6.980190891 6.958779586 6.936700945 1.0 10.350076600 10.269170824 10.227464299 10.184923955 1.2 16.381180092 16.226146390 16.147129067 16.067111677 1.4 27.890003380 27.592026085 27.441292235 27.289392347 1.6 51.183323262 50.594503863 50.298106659 50.000377775 1.8 101.424397595 100.206659076 99.595562766 98.982969504 2.0 217.301032800 214.631041938 213.293582978 211.954462214 We can’t give a general procedure for determining in advance whether Euler’s method or the semilinear Euler method will produce better results for a given semilinear initial value problem (3.1.19). As a rule of thumb, the Euler semilinear method will yield better results than Euler’s method if |u′′| is small on [x0, b], while Euler’s method yields better results if |u′′| is large on [x0, b]. In many cases the results obtained by the two methods don’t differ appreciably. However, we propose the an intuitive way to decide which is the better method: Try both methods with multiple step sizes, as we did in Example 3.1.4, and accept the results obtained by the method for which the approximations change less as the step size decreases. Example 3.1.5 Applying Euler’s method with step sizes h = 0.1, h = 0.05, and h = 0.025 to the initial value problem y′ −2y = x 1 + y2 , y(1) = 7 (3.1.25) on [1, 2] yields the results in Table 3.1.7. Applying the Euler semilinear method with y = ue2x and u′ = xe−2x 1 + u2e4x, u(1) = 7e−2 yields the results in Table 3.1.8. Since the latter are clearly less dependent on step size than the former, we conclude that the Euler semilinear method is better than Euler’s method for (3.1.25). This conclusion is supported by comparing the approximate results obtained by the two methods with the “exact” values of the solution. Table 3.1.7. Numerical solution of y′ −2y = x/(1 + y2), y(1) = 7, by Euler’s method. x h = 0.1 h = 0.05 h = 0.025 “Exact” 1.0 7.000000000 7.000000000 7.000000000 7.000000000 1.1 8.402000000 8.471970569 8.510493955 8.551744786 1.2 10.083936450 10.252570169 10.346014101 10.446546230 1.3 12.101892354 12.406719381 12.576720827 12.760480158 1.4 14.523152445 15.012952416 15.287872104 15.586440425 1.5 17.428443554 18.166277405 18.583079406 19.037865752 1.6 20.914624471 21.981638487 22.588266217 23.253292359 1.7 25.097914310 26.598105180 27.456479695 28.401914416 1.8 30.117766627 32.183941340 33.373738944 34.690375086 1.9 36.141518172 38.942738252 40.566143158 42.371060528 2.0 43.369967155 47.120835251 49.308511126 51.752229656 | Elementary Differential Equations with Boundary Value Problems_Page_114_Chunk2519 |
Section 3.1 Euler’s Method 105 Table 3.1.8. Numerical solution of y′ −2y = x/(1 + y2), y(1) = 7, by the Euler semilinear method. x h = 0.1 h = 0.05 h = 0.025 “Exact” 1.0 7.000000000 7.000000000 7.000000000 7.000000000 1.1 8.552262113 8.551993978 8.551867007 8.551744786 1.2 10.447568674 10.447038547 10.446787646 10.446546230 1.3 12.762019799 12.761221313 12.760843543 12.760480158 1.4 15.588535141 15.587448600 15.586934680 15.586440425 1.5 19.040580614 19.039172241 19.038506211 19.037865752 1.6 23.256721636 23.254942517 23.254101253 23.253292359 1.7 28.406184597 28.403969107 28.402921581 28.401914416 1.8 34.695649222 34.692912768 34.691618979 34.690375086 1.9 42.377544138 42.374180090 42.372589624 42.371060528 2.0 51.760178446 51.756054133 51.754104262 51.752229656 Example 3.1.6 Applying Euler’s method with step sizes h = 0.1, h = 0.05, and h = 0.025 to the initial value problem y′ + 3x2y = 1 + y2, y(2) = 2 (3.1.26) on [2, 3] yields the results in Table 3.1.9. Applying the Euler semilinear method with y = ue−x3 and u′ = ex3(1 + u2e−2x3), u(2) = 2e8 yields the results in Table 3.1.10. Noting the close agreement among the three columns of Table 3.1.9 (at least for larger values of x) and the lack of any such agreement among the columns of Table 3.1.10, we conclude that Euler’s method is better than the Euler semilinear method for (3.1.26). Comparing the results with the exact values supports this conclusion. Table 3.1.9. Numerical solution of y′ + 3x2y = 1 + y2, y(2) = 2, by Euler’s method. x h = 0.1 h = 0.05 h = 0.025 “Exact” 2.0 2.000000000 2.000000000 2.000000000 2.000000000 2.1 0.100000000 0.493231250 0.609611171 0.701162906 2.2 0.068700000 0.122879586 0.180113445 0.236986800 2.3 0.069419569 0.070670890 0.083934459 0.103815729 2.4 0.059732621 0.061338956 0.063337561 0.068390786 2.5 0.056871451 0.056002363 0.056249670 0.057281091 2.6 0.050560917 0.051465256 0.051517501 0.051711676 2.7 0.048279018 0.047484716 0.047514202 0.047564141 2.8 0.042925892 0.043967002 0.043989239 0.044014438 2.9 0.042148458 0.040839683 0.040857109 0.040875333 3.0 0.035985548 0.038044692 0.038058536 0.038072838 Table 3.1.10. Numerical solution of y′ +3x2y = 1+y2, y(2) = 2, by the Euler semilinear method. | Elementary Differential Equations with Boundary Value Problems_Page_115_Chunk2520 |
106 Chapter 3 Numerical Methods x h = 0.1 h = 0.05 h = 0.025 “Exact” x h = 0.1 h = 0.05 h = 0.025 h = .0125 2.0 2.000000000 2.000000000 2.000000000 2.000000000 2.1 0.708426286 0.702568171 0.701214274 0.701162906 2.2 0.214501852 0.222599468 0.228942240 0.236986800 2.3 0.069861436 0.083620494 0.092852806 0.103815729 2.4 0.032487396 0.047079261 0.056825805 0.068390786 2.5 0.021895559 0.036030018 0.045683801 0.057281091 2.6 0.017332058 0.030750181 0.040189920 0.051711676 2.7 0.014271492 0.026931911 0.036134674 0.047564141 2.8 0.011819555 0.023720670 0.032679767 0.044014438 2.9 0.009776792 0.020925522 0.029636506 0.040875333 3.0 0.008065020 0.018472302 0.026931099 0.038072838 In the next two sections we’ll study other numerical methods for solving initial value problems, called the improved Euler method, the midpoint method, Heun’s method and the Runge-Kutta method. If the initial value problem is semilinear as in (3.1.19), we also have the option of using variation of parameters and then applying the given numerical method to the initial value problem (3.1.21) for u. By analogy with the terminology used here, we’ll call the resulting procedure the improved Euler semilinear method, the midpoint semilinear method, Heun’s semilinear method or the Runge-Kutta semilinear method, as the case may be. 3.1 Exercises You may want to save the results of these exercises, sincewe’ll revisit in the next two sections. In Exer- cises 1–5 use Euler’s method to find approximate values of the solution of the given initial value problem at the points xi = x0 + ih, where x0 is the point wher the initial condition is imposed and i = 1, 2, 3. The purpose of these exercises is to familiarize you with the computational procedure of Euler’s method. 1. C y′ = 2x2 + 3y2 −2, y(2) = 1; h = 0.05 2. C y′ = y + p x2 + y2, y(0) = 1; h = 0.1 3. C y′ + 3y = x2 −3xy + y2, y(0) = 2; h = 0.05 4. C y′ = 1 + x 1 −y2 , y(2) = 3; h = 0.1 5. C y′ + x2y = sinxy, y(1) = π; h = 0.2 6. C Use Euler’s method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + 3y = 7e4x, y(0) = 2 at x = 0, 0.1, 0.2, 0.3, ..., 1.0. Compare these approximate values with the values of the exact solution y = e4x+e−3x, which can be obtained by the method of Section 2.1. Present your results in a table like Table 3.1.1. 7. C Use Euler’s method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + 2 xy = 3 x3 + 1, y(1) = 1 | Elementary Differential Equations with Boundary Value Problems_Page_116_Chunk2521 |
Section 3.1 Euler’s Method 107 at x = 1.0, 1.1, 1.2, 1.3, ..., 2.0. Compare these approximate values with the values of the exact solution y = 1 3x2 (9 lnx + x3 + 2), which can be obtained by the method of Section 2.1. Present your results in a table like Table 3.1.1. 8. C Use Euler’s method with step sizes h = 0.05, h = 0.025, and h = 0.0125 to find approximate values of the solution of the initial value problem y′ = y2 + xy −x2 x2 , y(1) = 2 at x = 1.0, 1.05, 1.10, 1.15, ..., 1.5. Compare these approximate values with the values of the exact solution y = x(1 + x2/3) 1 −x2/3 obtained in Example 2.4.3. Present your results in a table like Table 3.1.1. 9. C In Example 2.2.3 it was shown that y5 + y = x2 + x −4 is an implicit solution of the initial value problem y′ = 2x + 1 5y4 + 1, y(2) = 1. (A) Use Euler’s method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of (A) at x = 2.0, 2.1, 2.2, 2.3, ..., 3.0. Present your results in tabular form. To check the error in these approximate values, construct another table of values of the residual R(x, y) = y5 + y −x2 −x + 4 for each value of (x, y) appearing in the first table. 10. C You can see from Example 2.5.1 that x4y3 + x2y5 + 2xy = 4 is an implicit solution of the initial value problem y′ = −4x3y3 + 2xy5 + 2y 3x4y2 + 5x2y4 + 2x, y(1) = 1. (A) Use Euler’s method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of (A) at x = 1.0, 1.1, 1.2, 1.3, ..., 2.0. Present your results in tabular form. To check the error in these approximate values, construct another table of values of the residual R(x, y) = x4y3 + x2y5 + 2xy −4 for each value of (x, y) appearing in the first table. 11. C Use Euler’s method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem (3y2 + 4y)y′ + 2x + cos x = 0, y(0) = 1; (Exercise 2.2.13) at x = 0, 0.1, 0.2, 0.3, ..., 1.0. | Elementary Differential Equations with Boundary Value Problems_Page_117_Chunk2522 |
108 Chapter 3 Numerical Methods 12. C Use Euler’s method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + (y + 1)(y −1)(y −2) x + 1 = 0, y(1) = 0 (Exercise 2.2.14) at x = 1.0, 1.1, 1.2, 1.3, ..., 2.0. 13. C Use Euler’s method and the Euler semilinear method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + 3y = 7e−3x, y(0) = 6 at x = 0, 0.1, 0.2, 0.3, ..., 1.0. Compare these approximate values with the values of the exact solution y = e−3x(7x + 6), which can be obtained by the method of Section 2.1. Do you notice anything special about the results? Explain. The linear initial value problems in Exercises 14–19 can’t be solved exactly in terms of known elementary functions. In each exercise, use Euler’s method and the Euler semilinear methods with the indicated step sizes to find approximate values of the solution of the given initial value problem at 11 equally spaced points (including the endpoints) in the interval. 14. C y′ −2y = 1 1 + x2 , y(2) = 2; h = 0.1, 0.05, 0.025 on [2, 3] 15. C y′ + 2xy = x2, y(0) = 3 (Exercise 2.1.38); h = 0.2, 0.1, 0.05 on [0, 2] 16. C y′ + 1 xy = sin x x2 , y(1) = 2; (Exercise 2.1.39); h = 0.2, 0.1, 0.05 on [1, 3] 17. C y′ + y = e−x tan x x , y(1) = 0; (Exercise 2.1.40); h = 0.05, 0.025, 0.0125 on [1, 1.5] 18. C y′ + 2x 1 + x2 y = ex (1 + x2)2 , y(0) = 1; (Exercise 2.1.41); h = 0.2, 0.1, 0.05 on [0, 2] 19. C xy′ + (x + 1)y = ex2, y(1) = 2; (Exercise 2.1.42); h = 0.05, 0.025, 0.0125 on [1, 1.5] In Exercises 20–22, use Euler’s method and the Euler semilinear method with the indicated step sizes to find approximate values of the solution of the given initial value problem at 11 equally spaced points (including the endpoints) in the interval. 20. C y′ + 3y = xy2(y + 1), y(0) = 1; h = 0.1, 0.05, 0.025 on [0, 1] 21. C y′ −4y = x y2(y + 1), y(0) = 1; h = 0.1, 0.05, 0.025 on [0, 1] 22. C y′ + 2y = x2 1 + y2 , y(2) = 1; h = 0.1, 0.05, 0.025 on [2, 3] 23. NUMERICAL QUADRATURE. The fundamental theorem of calculus says that if f is continuous on a closed interval [a, b] then it has an antiderivative F such that F ′(x) = f(x) on [a, b] and Z b a f(x) dx = F (b) −F (a). (A) This solves the problem of evaluating a definite integral if the integrand f has an antiderivative that can be found and evaluated easily. However, if f doesn’t have this property, (A) doesn’t provide | Elementary Differential Equations with Boundary Value Problems_Page_118_Chunk2523 |
Section 3.2 The Improved Euler Method and Related Methods 109 a useful way to evaluate the definite integral. In this case we must resort to approximate methods. There’s a class of such methods called numerical quadrature, where the approximation takes the form Z b a f(x) dx ≈ n X i=0 cif(xi), (B) where a = x0 < x1 < · · · < xn = b are suitably chosen points and c0, c1, ..., cn are suitably chosen constants. We call (B) a quadrature formula. (a) Derive the quadrature formula Z b a f(x) dx ≈h n−1 X i=0 f(a + ih) (where h = (b −a)/n) (C) by applying Euler’s method to the initial value problem y′ = f(x), y(a) = 0. (b) The quadrature formula (C) is sometimes called the left rectangle rule. Draw a figure that justifies this terminology. (c) L For several choices of a, b, and A, apply (C) to f(x) = A with n = 10, 20, 40, 80, 160, 320. Compare your results with the exact answers and explain what you find. (d) L For several choices of a, b, A, and B, apply (C) to f(x) = A + Bx with n = 10, 20, 40, 80, 160, 320. Compare your results with the exact answers and explain what you find. 3.2 THE IMPROVED EULER METHOD AND RELATED METHODS In Section 3.1 we saw that the global truncation error of Euler’s method is O(h), which would seem to imply that we can achieve arbitrarily accurate results with Euler’s method by simply choosing the step size sufficiently small. However, this isn’t a good idea, for two reasons. First, after a certain point decreasing the step size will increase roundoff errors to the point where the accuracy will deteriorate rather than improve. The second and more important reason is that in most applications of numerical methods to an initial value problem y′ = f(x, y), y(x0) = y0, (3.2.1) the expensive part of the computation is the evaluation of f. Therefore we want methods that give good results for a given number of such evaluations. This is what motivates us to look for numerical methods better than Euler’s. To clarify this point, suppose we want to approximate the value of e by applying Euler’s method to the initial value problem y′ = y, y(0) = 1, (with solution y = ex) on [0, 1], with h = 1/12, 1/24, and 1/48, respectively. Since each step in Euler’s method requires one evaluation of f, the number of evaluations of f in each of these attempts is n = 12, 24, and 48, respectively. In each case we accept yn as an approximation to e. The second column of Table 3.2.1 shows the results. The first column of the table indicates the number of evaluations of f required to obtain the approximation, and the last column contains the value of e rounded to ten significant figures. In this section we’ll study the improved Euler method, which requires two evaluations of f at each step. We’ve used this method with h = 1/6, 1/12, and 1/24. The required number of evaluations of f | Elementary Differential Equations with Boundary Value Problems_Page_119_Chunk2524 |
110 Chapter 3 Numerical Methods were 12, 24, and 48, as in the three applications of Euler’s method; however, you can see from the third column of Table 3.2.1 that the approximation to e obtained by the improved Euler method with only 12 evaluations of f is better than the approximation obtained by Euler’s method with 48 evaluations. In Section 3.1 we’ll study the Runge-Kutta method, which requires four evaluations of f at each step. We’ve used this method with h = 1/3, 1/6, and 1/12. The required number of evaluations of f were again 12, 24, and 48, as in the three applications of Euler’s method and the improved Euler method; however, you can see from the fourth column of Table 3.2.1 that the approximation to e obtained by the Runge-Kutta method with only 12 evaluations of f is better than the approximation obtained by the improved Euler method with 48 evaluations. Table 3.2.1. Approximations to e obtained by three numerical methods. n Euler Improved Euler Runge-Kutta Exact 12 2.613035290 2.707188994 2.718069764 2.718281828 24 2.663731258 2.715327371 2.718266612 2.718281828 48 2.690496599 2.717519565 2.718280809 2.718281828 The Improved Euler Method The improved Euler method for solving the initial value problem (3.2.1) is based on approximating the integral curve of (3.2.1) at (xi, y(xi)) by the line through (xi, y(xi)) with slope mi = f(xi, y(xi)) + f(xi+1, y(xi+1)) 2 ; that is, mi is the average of the slopes of the tangents to the integral curve at the endpoints of [xi, xi+1]. The equation of the approximating line is therefore y = y(xi) + f(xi, y(xi)) + f(xi+1, y(xi+1)) 2 (x −xi). (3.2.2) Setting x = xi+1 = xi + h in (3.2.2) yields yi+1 = y(xi) + h 2 (f(xi, y(xi)) + f(xi+1, y(xi+1))) (3.2.3) as an approximation to y(xi+1). As in our derivation of Euler’s method, we replace y(xi) (unknown if i > 0) by its approximate value yi; then (3.2.3) becomes yi+1 = yi + h 2 (f(xi, yi) + f(xi+1, y(xi+1)) . However, this still won’t work, because we don’t know y(xi+1), which appears on the right. We overcome this by replacing y(xi+1) by yi + hf(xi, yi), the value that the Euler method would assign to yi+1. Thus, the improved Euler method starts with the known value y(x0) = y0 and computes y1, y2, ..., yn successively with the formula yi+1 = yi + h 2 (f(xi, yi) + f(xi+1, yi + hf(xi, yi))) . (3.2.4) The computation indicated here can be conveniently organized as follows: given yi, compute k1i = f(xi, yi), k2i = f (xi + h, yi + hk1i) , yi+1 = yi + h 2 (k1i + k2i). | Elementary Differential Equations with Boundary Value Problems_Page_120_Chunk2525 |
Section 3.2 The Improved Euler Method and Related Methods 111 The improved Euler method requires two evaluations of f(x, y) per step, while Euler’s method requires only one. However, we’ll see at the end of this section that if f satisfies appropriate assumptions, the local truncation error with the improved Euler method is O(h3), rather than O(h2) as with Euler’s method. Therefore the global truncation error with the improved Euler method is O(h2); however, we won’t prove this. We note that the magnitude of the local truncation error in the improved Euler method and other methods discussed in this section is determined by the third derivative y′′′ of the solution of the initial value problem. Therefore the local truncation error will be larger where |y′′′| is large, or smaller where |y′′′| is small. The next example, which deals with the initial value problem considered in Example 3.1.1, illustrates the computational procedure indicated in the improved Euler method. Example 3.2.1 Use the improved Euler method with h = 0.1 to find approximate values of the solution of the initial value problem y′ + 2y = x3e−2x, y(0) = 1 (3.2.5) at x = 0.1, 0.2, 0.3. Solution As in Example 3.1.1, we rewrite (3.2.5) as y′ = −2y + x3e−2x, y(0) = 1, which is of the form (3.2.1), with f(x, y) = −2y + x3e−2x, x0 = 0, and y0 = 1. The improved Euler method yields k10 = f(x0, y0) = f(0, 1) = −2, k20 = f(x1, y0 + hk10) = f(.1, 1 + (.1)(−2)) = f(.1, .8) = −2(.8) + (.1)3e−.2 = −1.599181269, y1 = y0 + h 2 (k10 + k20), = 1 + (.05)(−2 −1.599181269) = .820040937, k11 = f(x1, y1) = f(.1, .820040937) = −2(.820040937) + (.1)3e−.2 = −1.639263142, k21 = f(x2, y1 + hk11) = f(.2, .820040937 + .1(−1.639263142)), = f(.2, .656114622) = −2(.656114622) + (.2)3e−.4 = −1.306866684, y2 = y1 + h 2 (k11 + k21), = .820040937 + (.05)(−1.639263142 −1.306866684) = .672734445, k12 = f(x2, y2) = f(.2, .672734445) = −2(.672734445) + (.2)3e−.4 = −1.340106330, k22 = f(x3, y2 + hk12) = f(.3, .672734445 + .1(−1.340106330)), = f(.3, .538723812) = −2(.538723812) + (.3)3e−.6 = −1.062629710, y3 = y2 + h 2 (k12 + k22) = .672734445 + (.05)(−1.340106330 −1.062629710) = .552597643. Example 3.2.2 Table 3.2.2 shows results of using the improved Euler method with step sizes h = 0.1 and h = 0.05 to find approximate values of the solution of the initial value problem y′ + 2y = x3e−2x, y(0) = 1 | Elementary Differential Equations with Boundary Value Problems_Page_121_Chunk2526 |
112 Chapter 3 Numerical Methods at x = 0, 0.1, 0.2, 0.3, ..., 1.0. For comparison, it also shows the corresponding approximate values obtained with Euler’s method in 3.1.2, and the values of the exact solution y = e−2x 4 (x4 + 4). The results obtained by the improved Euler method with h = 0.1 are better than those obtained by Euler’s method with h = 0.05. Table 3.2.2. Numerical solution of y′ + 2y = x3e−2x, y(0) = 1, by Euler’s method and the improved Euler method. x h = 0.1 h = 0.05 h = 0.1 h = 0.05 Exact 0.0 1.000000000 1.000000000 1.000000000 1.000000000 1.000000000 0.1 0.800000000 0.810005655 0.820040937 0.819050572 0.818751221 0.2 0.640081873 0.656266437 0.672734445 0.671086455 0.670588174 0.3 0.512601754 0.532290981 0.552597643 0.550543878 0.549922980 0.4 0.411563195 0.432887056 0.455160637 0.452890616 0.452204669 0.5 0.332126261 0.353785015 0.376681251 0.374335747 0.373627557 0.6 0.270299502 0.291404256 0.313970920 0.311652239 0.310952904 0.7 0.222745397 0.242707257 0.264287611 0.262067624 0.261398947 0.8 0.186654593 0.205105754 0.225267702 0.223194281 0.222570721 0.9 0.159660776 0.176396883 0.194879501 0.192981757 0.192412038 1.0 0.139778910 0.154715925 0.171388070 0.169680673 0.169169104 Euler Improved Euler Exact Example 3.2.3 Table 3.2.3 shows analogous results for the nonlinear initial value problem y′ = −2y2 + xy + x2, y(0) = 1. We applied Euler’s method to this problem in Example 3.1.3. Table 3.2.3. Numerical solution of y′ = −2y2 + xy + x2, y(0) = 1, by Euler’s method and the improved Euler method. x h = 0.1 h = 0.05 h = 0.1 h = 0.05 “Exact” 0.0 1.000000000 1.000000000 1.000000000 1.000000000 1.000000000 0.1 0.800000000 0.821375000 0.840500000 0.838288371 0.837584494 0.2 0.681000000 0.707795377 0.733430846 0.730556677 0.729641890 0.3 0.605867800 0.633776590 0.661600806 0.658552190 0.657580377 0.4 0.559628676 0.587454526 0.615961841 0.612884493 0.611901791 0.5 0.535376972 0.562906169 0.591634742 0.588558952 0.587575491 0.6 0.529820120 0.557143535 0.586006935 0.582927224 0.581942225 0.7 0.541467455 0.568716935 0.597712120 0.594618012 0.593629526 0.8 0.569732776 0.596951988 0.626008824 0.622898279 0.621907458 0.9 0.614392311 0.641457729 0.670351225 0.667237617 0.666250842 1.0 0.675192037 0.701764495 0.730069610 0.726985837 0.726015790 Euler Improved Euler “Exact” Example 3.2.4 Use step sizes h = 0.2, h = 0.1, and h = 0.05 to find approximate values of the solution of y′ −2xy = 1, y(0) = 3 (3.2.6) | Elementary Differential Equations with Boundary Value Problems_Page_122_Chunk2527 |
Section 3.2 The Improved Euler Method and Related Methods 113 at x = 0, 0.2, 0.4, 0.6, ..., 2.0 by (a) the improved Euler method; (b) the improved Euler semilinear method. (We used Euler’s method and the Euler semilinear method on this problem in 3.1.4.) SOLUTION(a) Rewriting (3.2.6) as y′ = 1 + 2xy, y(0) = 3 and applying the improved Euler method with f(x, y) = 1 + 2xy yields the results shown in Table 3.2.4. SOLUTION(b) Since y1 = ex2 is a solution of the complementary equation y′ −2xy = 0, we can apply the improved Euler semilinear method to (3.2.6), with y = uex2 and u′ = e−x2, u(0) = 3. The results listed in Table 3.2.5 are clearly better than those obtained by the improved Euler method. Table 3.2.4. Numerical solution of y′ −2xy = 1, y(0) = 3, by the improved Euler method. x h = 0.2 h = 0.1 h = 0.05 “Exact” 0.0 3.000000000 3.000000000 3.000000000 3.000000000 0.2 3.328000000 3.328182400 3.327973600 3.327851973 0.4 3.964659200 3.966340117 3.966216690 3.966059348 0.6 5.057712497 5.065700515 5.066848381 5.067039535 0.8 6.900088156 6.928648973 6.934862367 6.936700945 1.0 10.065725534 10.154872547 10.177430736 10.184923955 1.2 15.708954420 15.970033261 16.041904862 16.067111677 1.4 26.244894192 26.991620960 27.210001715 27.289392347 1.6 46.958915746 49.096125524 49.754131060 50.000377775 1.8 89.982312641 96.200506218 98.210577385 98.982969504 2.0 184.563776288 203.151922739 209.464744495 211.954462214 Table 3.2.5. Numerical solution of y′−2xy = 1, y(0) = 3, by the improved Euler semilinear method. x h = 0.2 h = 0.1 h = 0.05 “Exact” 0.0 3.000000000 3.000000000 3.000000000 3.000000000 0.2 3.326513400 3.327518315 3.327768620 3.327851973 0.4 3.963383070 3.965392084 3.965892644 3.966059348 0.6 5.063027290 5.066038774 5.066789487 5.067039535 0.8 6.931355329 6.935366847 6.936367564 6.936700945 1.0 10.178248417 10.183256733 10.184507253 10.184923955 1.2 16.059110511 16.065111599 16.066611672 16.067111677 1.4 27.280070674 27.287059732 27.288809058 27.289392347 1.6 49.989741531 49.997712997 49.999711226 50.000377775 1.8 98.971025420 98.979972988 98.982219722 98.982969504 2.0 211.941217796 211.951134436 211.953629228 211.954462214 | Elementary Differential Equations with Boundary Value Problems_Page_123_Chunk2528 |
114 Chapter 3 Numerical Methods A Family of Methods with O(h3) Local Truncation Error We’ll now derive a class of methods with O(h3) local truncation error for solving (3.2.1). For simplicity, we assume that f, fx, fy, fxx, fyy, and fxy are continuous and bounded for all (x, y). This implies that if y is the solution of (3.2.1 then y′′ and y′′′ are bounded (Exercise 31). We begin by approximating the integral curve of (3.2.1) at (xi, y(xi)) by the line through (xi, y(xi)) with slope mi = σy′(xi) + ρy′(xi + θh), where σ, ρ, and θ are constants that we’ll soon specify; however, we insist at the outset that 0 < θ ≤1, so that xi < xi + θh ≤xi+1. The equation of the approximating line is y = y(xi) + mi(x −xi) = y(xi) + [σy′(xi) + ρy′(xi + θh)] (x −xi). (3.2.7) Setting x = xi+1 = xi + h in (3.2.7) yields ˆyi+1 = y(xi) + h [σy′(xi) + ρy′(xi + θh)] as an approximation to y(xi+1). To determine σ, ρ, and θ so that the error Ei = y(xi+1) −ˆyi+1 = y(xi+1) −y(xi) −h [σy′(xi) + ρy′(xi + θh)] (3.2.8) in this approximation is O(h3), we begin by recalling from Taylor’s theorem that y(xi+1) = y(xi) + hy′(xi) + h2 2 y′′(xi) + h3 6 y′′′(ˆxi), where ˆxi is in (xi, xi+1). Since y′′′ is bounded this implies that y(xi+1) −y(xi) −hy′(xi) −h2 2 y′′(xi) = O(h3). Comparing this with (3.2.8) shows that Ei = O(h3) if σy′(xi) + ρy′(xi + θh) = y′(xi) + h 2 y′′(xi) + O(h2). (3.2.9) However, applying Taylor’s theorem to y′ shows that y′(xi + θh) = y′(xi) + θhy′′(xi) + (θh)2 2 y′′′(xi), where xi is in (xi, xi + θh). Since y′′′ is bounded, this implies that y′(xi + θh) = y′(xi) + θhy′′(xi) + O(h2). Substituting this into (3.2.9) and noting that the sum of two O(h2) terms is again O(h2) shows that Ei = O(h3) if (σ + ρ)y′(xi) + ρθhy′′(xi) = y′(xi) + h 2 y′′(xi), | Elementary Differential Equations with Boundary Value Problems_Page_124_Chunk2529 |
Section 3.2 The Improved Euler Method and Related Methods 115 which is true if σ + ρ = 1 and ρθ = 1 2. (3.2.10) Since y′ = f(x, y), we can now conclude from (3.2.8) that y(xi+1) = y(xi) + h [σf(xi, yi) + ρf(xi + θh, y(xi + θh))] + O(h3) (3.2.11) if σ, ρ, and θ satisfy (3.2.10). However, this formula would not be useful even if we knew y(xi) exactly (as we would for i = 0), since we still wouldn’t know y(xi +θh) exactly. To overcome this difficulty, we again use Taylor’s theorem to write y(xi + θh) = y(xi) + θhy′(xi) + h2 2 y′′(˜xi), where ˜xi is in (xi, xi + θh). Since y′(xi) = f(xi, y(xi)) and y′′ is bounded, this implies that |y(xi + θh) −y(xi) −θhf(xi, y(xi))| ≤Kh2 (3.2.12) for some constant K. Since fy is bounded, the mean value theorem implies that |f(xi + θh, u) −f(xi + θh, v)| ≤M|u −v| for some constant M. Letting u = y(xi + θh) and v = y(xi) + θhf(xi, y(xi)) and recalling (3.2.12) shows that f(xi + θh, y(xi + θh)) = f(xi + θh, y(xi) + θhf(xi, y(xi))) + O(h2). Substituting this into (3.2.11) yields y(xi+1) = y(xi) + h [σf(xi, y(xi))+ ρf(xi + θh, y(xi) + θhf(xi, y(xi)))] + O(h3). This implies that the formula yi+1 = yi + h [σf(xi, yi) + ρf(xi + θh, yi + θhf(xi, yi))] has O(h3) local truncation error if σ, ρ, and θ satisfy (3.2.10). Substituting σ = 1 −ρ and θ = 1/2ρ here yields yi+1 = yi + h (1 −ρ)f(xi, yi) + ρf xi + h 2ρ, yi + h 2ρf(xi, yi) . (3.2.13) The computation indicated here can be conveniently organized as follows: given yi, compute k1i = f(xi, yi), k2i = f xi + h 2ρ, yi + h 2ρk1i , yi+1 = yi + h[(1 −ρ)k1i + ρk2i]. Consistent with our requirement that 0 < θ < 1, we require that ρ ≥1/2. Letting ρ = 1/2 in (3.2.13) yields the improved Euler method (3.2.4). Letting ρ = 3/4 yields Heun’s method, yi+1 = yi + h 1 4f(xi, yi) + 3 4f xi + 2 3h, yi + 2 3hf(xi, yi) , | Elementary Differential Equations with Boundary Value Problems_Page_125_Chunk2530 |
116 Chapter 3 Numerical Methods which can be organized as k1i = f(xi, yi), k2i = f xi + 2h 3 , yi + 2h 3 k1i , yi+1 = yi + h 4(k1i + 3k2i). Letting ρ = 1 yields the midpoint method, yi+1 = yi + hf xi + h 2 , yi + h 2 f(xi, yi) , which can be organized as k1i = f(xi, yi), k2i = f xi + h 2 , yi + h 2 k1i , yi+1 = yi + hk2i. Examples involving the midpoint method and Heun’s method are given in Exercises 23-30. 3.2 Exercises Most of the following numerical exercises involve initial value problems considered in the exercises in Section 3.1. You’ll find it instructive to compare the results that you obtain here with the corresponding results that you obtained in Section 3.1. In Exercises 1–5 use the improved Euler method to find approximate values of the solution of the given initial value problem at the points xi = x0 + ih, where x0 is the point where the initial condition is imposed and i = 1, 2, 3. 1. C y′ = 2x2 + 3y2 −2, y(2) = 1; h = 0.05 2. C y′ = y + p x2 + y2, y(0) = 1; h = 0.1 3. C y′ + 3y = x2 −3xy + y2, y(0) = 2; h = 0.05 4. C y′ = 1 + x 1 −y2 , y(2) = 3; h = 0.1 5. C y′ + x2y = sinxy, y(1) = π; h = 0.2 6. C Use the improved Euler method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + 3y = 7e4x, y(0) = 2 at x = 0, 0.1, 0.2, 0.3, ..., 1.0. Compare these approximate values with the values of the exact solution y = e4x+e−3x, which can be obtained by the method of Section 2.1. Present your results in a table like Table 3.2.2. 7. C Use the improved Euler method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + 2 xy = 3 x3 + 1, y(1) = 1 | Elementary Differential Equations with Boundary Value Problems_Page_126_Chunk2531 |
Section 3.2 The Improved Euler Method and Related Methods 117 at x = 1.0, 1.1, 1.2, 1.3, ..., 2.0. Compare these approximate values with the values of the exact solution y = 1 3x2 (9 lnx + x3 + 2) which can be obtained by the method of Section 2.1. Present your results in a table like Table 3.2.2. 8. C Use the improved Euler method with step sizes h = 0.05, h = 0.025, and h = 0.0125 to find approximate values of the solution of the initial value problem y′ = y2 + xy −x2 x2 , y(1) = 2, at x = 1.0, 1.05, 1.10, 1.15, ..., 1.5. Compare these approximate values with the values of the exact solution y = x(1 + x2/3) 1 −x2/3 obtained in Example 2.4.3. Present your results in a table like Table 3.2.2. 9. C In Example 3.2.2 it was shown that y5 + y = x2 + x −4 is an implicit solution of the initial value problem y′ = 2x + 1 5y4 + 1, y(2) = 1. (A) Use the improved Euler method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of (A) at x = 2.0, 2.1, 2.2, 2.3, ..., 3.0. Present your results in tabular form. To check the error in these approximate values, construct another table of values of the residual R(x, y) = y5 + y −x2 −x + 4 for each value of (x, y) appearing in the first table. 10. C You can see from Example 2.5.1 that x4y3 + x2y5 + 2xy = 4 is an implicit solution of the initial value problem y′ = −4x3y3 + 2xy5 + 2y 3x4y2 + 5x2y4 + 2x, y(1) = 1. (A) Use the improved Euler method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of (A) at x = 1.0, 1.14, 1.2, 1.3, ..., 2.0. Present your results in tabular form. To check the error in these approximate values, construct another table of values of the residual R(x, y) = x4y3 + x2y5 + 2xy −4 for each value of (x, y) appearing in the first table. 11. C Use the improved Euler method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem (3y2 + 4y)y′ + 2x + cos x = 0, y(0) = 1 (Exercise 2.2.13) at x = 0, 0.1, 0.2, 0.3, ..., 1.0. | Elementary Differential Equations with Boundary Value Problems_Page_127_Chunk2532 |
118 Chapter 3 Numerical Methods 12. C Use the improved Euler method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + (y + 1)(y −1)(y −2) x + 1 = 0, y(1) = 0 (Exercise 2.2.14) at x = 1.0, 1.1, 1.2, 1.3, ..., 2.0. 13. C Use the improved Euler method and the improved Euler semilinear method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + 3y = e−3x(1 −2x), y(0) = 2, at x = 0, 0.1, 0.2, 0.3, ..., 1.0. Compare these approximate values with the values of the exact solution y = e−3x(2 + x −x2), which can be obtained by the method of Section 2.1. Do you notice anything special about the results? Explain. The linear initial value problems in Exercises 14–19 can’t be solved exactly in terms of known elementary functions. In each exercise use the improved Euler and improved Euler semilinear methods with the indicated step sizes to find approximate values of the solution of the given initial value problem at 11 equally spaced points (including the endpoints) in the interval. 14. C y′ −2y = 1 1 + x2 , y(2) = 2; h = 0.1, 0.05, 0.025 on [2, 3] 15. C y′ + 2xy = x2, y(0) = 3; h = 0.2, 0.1, 0.05 on [0, 2] (Exercise 2.1.38) 16. C y′ + 1 xy = sin x x2 , y(1) = 2, h = 0.2, 0.1, 0.05 on [1, 3] (Exercise 2.1.39) 17. C y′ + y = e−x tan x x , y(1) = 0; h = 0.05, 0.025, 0.0125 on [1, 1.5] (Exercise 2.1.40), 18. C y′ + 2x 1 + x2 y = ex (1 + x2)2 , y(0) = 1; h = 0.2, 0.1, 0.05 on [0, 2] (Exercise 2.1.41) 19. C xy′ + (x + 1)y = ex2, y(1) = 2; h = 0.05, 0.025, 0.0125 on [1, 1.5] (Exercise 2.1.42) In Exercises 20–22 use the improved Euler method and the improved Euler semilinear method with the indicated step sizes to find approximate values of the solution of the given initial value problem at 11 equally spaced points (including the endpoints) in the interval. 20. C y′ + 3y = xy2(y + 1), y(0) = 1; h = 0.1, 0.05, 0.025 on [0, 1] 21. C y′ −4y = x y2(y + 1), y(0) = 1; h = 0.1, 0.05, 0.025 on [0, 1] 22. C y′ + 2y = x2 1 + y2 , y(2) = 1; h = 0.1, 0.05, 0.025 on [2, 3] 23. C Do Exercise 7 with “improved Euler method” replaced by “midpoint method.” 24. C Do Exercise 7 with “improved Euler method” replaced by “Heun’s method.” 25. C Do Exercise 8 with “improved Euler method” replaced by “midpoint method.” 26. C Do Exercise 8 with “improved Euler method” replaced by “Heun’s method.” 27. C Do Exercise 11 with “improved Euler method” replaced by “midpoint method.” | Elementary Differential Equations with Boundary Value Problems_Page_128_Chunk2533 |
Section 3.3 The Runge-Kutta Method 119 28. C Do Exercise 11 with “improved Euler method” replaced by “Heun’s method.” 29. C Do Exercise 12 with “improved Euler method” replaced by “midpoint method.” 30. C Do Exercise 12 with “improved Euler method” replaced by “Heun’s method.” 31. Show that if f, fx, fy, fxx, fyy, and fxy are continuous and bounded for all (x, y) and y is the solution of the initial value problem y′ = f(x, y), y(x0) = y0, then y′′ and y′′′ are bounded. 32. NUMERICAL QUADRATURE (see Exercise 3.1.23). (a) Derive the quadrature formula Z b a f(x) dx ≈.5h(f(a) + f(b)) + h n−1 X i=1 f(a + ih) (where h = (b −a)/n) (A) by applying the improved Euler method to the initial value problem y′ = f(x), y(a) = 0. (b) The quadrature formula (A) is called the trapezoid rule. Draw a figure that justifies this terminology. (c) L For several choices of a, b, A, and B, apply (A) to f(x) = A + Bx, with n = 10, 20, 40, 80, 160, 320. Compare your results with the exact answers and explain what you find. (d) L For several choices of a, b, A, B, and C, apply (A) to f(x) = A + Bx + Cx2, with n = 10, 20, 40, 80, 160, 320. Compare your results with the exact answers and explain what you find. 3.3 THE RUNGE-KUTTA METHOD In general, if k is any positive integer and f satisfies appropriate assumptions, there are numerical methods with local truncation error O(hk+1) for solving an initial value problem y′ = f(x, y), y(x0) = y0. (3.3.1) Moreover, it can be shown that a method with local truncation error O(hk+1) has global truncation error O(hk). In Sections 3.1 and 3.2 we studied numerical methods where k = 1 and k = 2. We’ll skip methods for which k = 3 and proceed to the Runge-Kutta method, the most widely used method, for which k = 4. The magnitude of the local truncation error is determined by the fifth derivative y(5) of the solution of the initial value problem. Therefore the local truncation error will be larger where |y(5)| is large, or smaller where |y(5)| is small. The Runge-Kutta method computes approximate values y1, y2, ..., yn of the solution of (3.3.1) at x0, x0 + h, ..., x0 + nh as follows: Given yi, compute k1i = f(xi, yi), k2i = f xi + h 2 , yi + h 2 k1i , k3i = f xi + h 2 , yi + h 2 k2i , k4i = f(xi + h, yi + hk3i), and yi+1 = yi + h 6 (k1i + 2k2i + 2k3i + k4i). The next example, which deals with the initial value problem considered in Examples 3.1.1 and 3.2.1, illustrates the computational procedure indicated in the Runge-Kutta method. | Elementary Differential Equations with Boundary Value Problems_Page_129_Chunk2534 |
120 Chapter 3 Numerical Methods Example 3.3.1 Use the Runge-Kutta method with h = 0.1 to find approximate values for the solution of the initial value problem y′ + 2y = x3e−2x, y(0) = 1, (3.3.2) at x = 0.1, 0.2. Solution Again we rewrite (3.3.2) as y′ = −2y + x3e−2x, y(0) = 1, which is of the form (3.3.1), with f(x, y) = −2y + x3e−2x, x0 = 0, and y0 = 1. The Runge-Kutta method yields k10 = f(x0, y0) = f(0, 1) = −2, k20 = f(x0 + h/2, y0 + hk10/2) = f(.05, 1 + (.05)(−2)) = f(.05, .9) = −2(.9) + (.05)3e−.1 = −1.799886895, k30 = f(x0 + h/2, y0 + hk20/2) = f(.05, 1 + (.05)(−1.799886895)) = f(.05, .910005655) = −2(.910005655) + (.05)3e−.1 = −1.819898206, k40 = f(x0 + h, y0 + hk30) = f(.1, 1 + (.1)(−1.819898206)) = f(.1, .818010179) = −2(.818010179) + (.1)3e−.2 = −1.635201628, y1 = y0 + h 6 (k10 + 2k20 + 2k30 + k40), = 1 + .1 6 (−2 + 2(−1.799886895)+ 2(−1.819898206)−1.635201628) = .818753803, k11 = f(x1, y1) = f(.1, .818753803) = −2(.818753803)) + (.1)3e−.2 = −1.636688875, k21 = f(x1 + h/2, y1 + hk11/2) = f(.15, .818753803 + (.05)(−1.636688875)) = f(.15, .736919359) = −2(.736919359) + (.15)3e−.3 = −1.471338457, k31 = f(x1 + h/2, y1 + hk21/2) = f(.15, .818753803 + (.05)(−1.471338457)) = f(.15, .745186880) = −2(.745186880) + (.15)3e−.3 = −1.487873498, k41 = f(x1 + h, y1 + hk31) = f(.2, .818753803 + (.1)(−1.487873498)) = f(.2, .669966453) = −2(.669966453) + (.2)3e−.4 = −1.334570346, y2 = y1 + h 6 (k11 + 2k21 + 2k31 + k41), = .818753803 + .1 6 (−1.636688875 + 2(−1.471338457)+ 2(−1.487873498)−1.334570346) = .670592417. The Runge-Kutta method is sufficiently accurate for most applications. Example 3.3.2 Table 3.3.1 shows results of using the Runge-Kutta method with step sizes h = 0.1 and h = 0.05 to find approximate values of the solution of the initial value problem y′ + 2y = x3e−2x, y(0) = 1 | Elementary Differential Equations with Boundary Value Problems_Page_130_Chunk2535 |
Section 3.3 The Runge-Kutta Method 121 at x = 0, 0.1, 0.2, 0.3, ..., 1.0. For comparison, it also shows the corresponding approximate values obtained with the improved Euler method in Example 3.2.2, and the values of the exact solution y = e−2x 4 (x4 + 4). The results obtained by the Runge-Kutta method are clearly better than those obtained by the improved Euler method in fact; the results obtained by the Runge-Kutta method with h = 0.1 are better than those obtained by the improved Euler method with h = 0.05. Table 3.3.1. Numerical solution of y′ + 2y = x3e−2x, y(0) = 1, by the Runge-Kuttta method and the improved Euler method. x h = 0.1 h = 0.05 h = 0.1 h = 0.05 Exact 0.0 1.000000000 1.000000000 1.000000000 1.000000000 1.000000000 0.1 0.820040937 0.819050572 0.818753803 0.818751370 0.818751221 0.2 0.672734445 0.671086455 0.670592417 0.670588418 0.670588174 0.3 0.552597643 0.550543878 0.549928221 0.549923281 0.549922980 0.4 0.455160637 0.452890616 0.452210430 0.452205001 0.452204669 0.5 0.376681251 0.374335747 0.373633492 0.373627899 0.373627557 0.6 0.313970920 0.311652239 0.310958768 0.310953242 0.310952904 0.7 0.264287611 0.262067624 0.261404568 0.261399270 0.261398947 0.8 0.225267702 0.223194281 0.222575989 0.222571024 0.222570721 0.9 0.194879501 0.192981757 0.192416882 0.192412317 0.192412038 1.0 0.171388070 0.169680673 0.169173489 0.169169356 0.169169104 Improved Euler Runge-Kutta Exact Example 3.3.3 Table 3.3.2 shows analogous results for the nonlinear initial value problem y′ = −2y2 + xy + x2, y(0) = 1. We applied the improved Euler method to this problem in Example 3. Table 3.3.2. Numerical solution of y′ = −2y2 + xy + x2, y(0) = 1, by the Runge-Kuttta method and the improved Euler method. x h = 0.1 h = 0.05 h = 0.1 h = 0.05 “Exact” 0.0 1.000000000 1.000000000 1.000000000 1.000000000 1.000000000 0.1 0.840500000 0.838288371 0.837587192 0.837584759 0.837584494 0.2 0.733430846 0.730556677 0.729644487 0.729642155 0.729641890 0.3 0.661600806 0.658552190 0.657582449 0.657580598 0.657580377 0.4 0.615961841 0.612884493 0.611903380 0.611901969 0.611901791 0.5 0.591634742 0.588558952 0.587576716 0.587575635 0.587575491 0.6 0.586006935 0.582927224 0.581943210 0.581942342 0.581942225 0.7 0.597712120 0.594618012 0.593630403 0.593629627 0.593629526 0.8 0.626008824 0.622898279 0.621908378 0.621907553 0.621907458 0.9 0.670351225 0.667237617 0.666251988 0.666250942 0.666250842 1.0 0.730069610 0.726985837 0.726017378 0.726015908 0.726015790 Improved Euler Runge-Kutta “Exact” | Elementary Differential Equations with Boundary Value Problems_Page_131_Chunk2536 |
122 Chapter 3 Numerical Methods Example 3.3.4 Tables 3.3.3 and 3.3.4 show results obtained by applying the Runge-Kutta and Runge- Kutta semilinear methods to to the initial value problem y′ −2xy = 1, y(0) = 3, which we considered in Examples 3.1.4 and 3.2.4. Table 3.3.3. Numerical solution of y′ −2xy = 1, y(0) = 3, by the Runge-Kutta method. x h = 0.2 h = 0.1 h = 0.05 “Exact” 0.0 3.000000000 3.000000000 3.000000000 3.000000000 0.2 3.327846400 3.327851633 3.327851952 3.327851973 0.4 3.966044973 3.966058535 3.966059300 3.966059348 0.6 5.066996754 5.067037123 5.067039396 5.067039535 0.8 6.936534178 6.936690679 6.936700320 6.936700945 1.0 10.184232252 10.184877733 10.184920997 10.184923955 1.2 16.064344805 16.066915583 16.067098699 16.067111677 1.4 27.278771833 27.288605217 27.289338955 27.289392347 1.6 49.960553660 49.997313966 50.000165744 50.000377775 1.8 98.834337815 98.971146146 98.982136702 98.982969504 2.0 211.393800152 211.908445283 211.951167637 211.954462214 Table 3.3.4. Numerical solution of y′ −2xy = 1, y(0) = 3, by the Runge-Kutta semilinear method. x h = 0.2 h = 0.1 h = 0.05 “Exact” 0.0 3.000000000 3.000000000 3.000000000 3.000000000 0.2 3.327853286 3.327852055 3.327851978 3.327851973 0.4 3.966061755 3.966059497 3.966059357 3.966059348 0.6 5.067042602 5.067039725 5.067039547 5.067039535 0.8 6.936704019 6.936701137 6.936700957 6.936700945 1.0 10.184926171 10.184924093 10.184923963 10.184923955 1.2 16.067111961 16.067111696 16.067111678 16.067111677 1.4 27.289389418 27.289392167 27.289392335 27.289392347 1.6 50.000370152 50.000377302 50.000377745 50.000377775 1.8 98.982955511 98.982968633 98.982969450 98.982969504 2.0 211.954439983 211.954460825 211.954462127 211.954462214 The Case Where x0 Isn’t The Left Endpoint So far in this chapter we’ve considered numerical methods for solving an initial value problem y′ = f(x, y), y(x0) = y0 (3.3.3) on an interval [x0, b], for which x0 is the left endpoint. We haven’t discussed numerical methods for solving (3.3.3) on an interval [a, x0], for which x0 is the right endpoint. To be specific, how can we obtain approximate values y−1, y−2, ..., y−n of the solution of (3.3.3) at x0 −h, . . ., x0 −nh, where h = (x0 −a)/n? Here’s the answer to this question: Consider the initial value problem z′ = −f(−x, z), z(−x0) = y0, (3.3.4) on the interval [−x0, −a], for which −x0 is the left endpoint. Use a numerical method to obtain approxi- mate values z1, z2, ..., zn of the solution of (3.3.4) at −x0 + h, −x0 + 2h, ..., −x0 + nh = −a. Then | Elementary Differential Equations with Boundary Value Problems_Page_132_Chunk2537 |
Section 3.3 The Runge-Kutta Method 123 y−1 = z1, y−2 = z2, . . ., y−n = zn are approximate values of the solution of (3.3.3) at x0 −h, x0 −2h, ..., x0 −nh = a. The justification for this answer is sketched in Exercise 23. Note how easy it is to make the change the given problem (3.3.3) to the modified problem (3.3.4): first replace f by −f and then replace x, x0, and y by −x, −x0, and z, respectively. Example 3.3.5 Use the Runge-Kutta method with step size h = 0.1 to find approximate values of the solution of (y −1)2y′ = 2x + 3, y(1) = 4 (3.3.5) at x = 0, 0.1, 0.2, ..., 1. Solution We first rewrite (3.3.5) in the form (3.3.3) as y′ = 2x + 3 (y −1)2 , y(1) = 4. (3.3.6) Since the initial condition y(1) = 4 is imposed at the right endpoint of the interval [0, 1], we apply the Runge-Kutta method to the initial value problem z′ = 2x −3 (z −1)2 , z(−1) = 4 (3.3.7) on the interval [−1, 0]. (You should verify that (3.3.7) is related to (3.3.6) as (3.3.4) is related to (3.3.3).) Table 3.3.5 shows the results. Reversing the order of the rows in Table 3.3.5 and changing the signs of the values of x yields the first two columns of Table 3.3.6. The last column of Table 3.3.6 shows the exact values of y, which are given by y = 1 + (3x2 + 9x + 15)1/3. (Since the differential equation in (3.3.6) is separable, this formula can be obtained by the method of Section 2.2.) Table 3.3.5. Numerical solution of z′ = 2x −3 (z −1)2 , z(−1) = 4, on [−1, 0]. x z -1.0 4.000000000 -0.9 3.944536474 -0.8 3.889298649 -0.7 3.834355648 -0.6 3.779786399 -0.5 3.725680888 -0.4 3.672141529 -0.3 3.619284615 -0.2 3.567241862 -0.1 3.516161955 0.0 3.466212070 Table 3.3.6. Numerical solution of (y −1)2y′ = 2x + 3, y(1) = 4, on [0, 1]. | Elementary Differential Equations with Boundary Value Problems_Page_133_Chunk2538 |
124 Chapter 3 Numerical Methods x y Exact 0.00 3.466212070 3.466212074 0.10 3.516161955 3.516161958 0.20 3.567241862 3.567241864 0.30 3.619284615 3.619284617 0.40 3.672141529 3.672141530 0.50 3.725680888 3.725680889 0.60 3.779786399 3.779786399 0.70 3.834355648 3.834355648 0.80 3.889298649 3.889298649 0.90 3.944536474 3.944536474 1.00 4.000000000 4.000000000 We leave it to you to develop a procedure for handling the numerical solution of (3.3.3) on an interval [a, b] such that a < x0 < b (Exercises 26 and 27). 3.3 Exercises Most of the following numerical exercises involve initial value problems considered in the exercises in Sections 3.2. You’ll find it instructive to compare the results that you obtain here with the corresponding results that you obtained in those sections. In Exercises 1–5 use the Runge-Kutta method to find approximate values of the solution of the given initial value problem at the points xi = x0 + ih, where x0 is the point where the initial condition is imposed and i = 1, 2. 1. C y′ = 2x2 + 3y2 −2, y(2) = 1; h = 0.05 2. C y′ = y + p x2 + y2, y(0) = 1; h = 0.1 3. C y′ + 3y = x2 −3xy + y2, y(0) = 2; h = 0.05 4. C y′ = 1 + x 1 −y2 , y(2) = 3; h = 0.1 5. C y′ + x2y = sinxy, y(1) = π; h = 0.2 6. C Use the Runge-Kutta method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + 3y = 7e4x, y(0) = 2, at x = 0, 0.1, 0.2, 0.3, ..., 1.0. Compare these approximate values with the values of the exact solution y = e4x+e−3x, which can be obtained by the method of Section 2.1. Present your results in a table like Table 3.3.1. 7. C Use the Runge-Kutta method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + 2 xy = 3 x3 + 1, y(1) = 1 at x = 1.0, 1.1, 1.2, 1.3, ..., 2.0. Compare these approximate values with the values of the exact solution y = 1 3x2 (9 lnx + x3 + 2), which can be obtained by the method of Section 2.1. Present your results in a table like Table 3.3.1. | Elementary Differential Equations with Boundary Value Problems_Page_134_Chunk2539 |
Section 3.3 The Runge-Kutta Method 125 8. C Use the Runge-Kutta method with step sizes h = 0.05, h = 0.025, and h = 0.0125 to find approximate values of the solution of the initial value problem y′ = y2 + xy −x2 x2 , y(1) = 2 at x = 1.0, 1.05, 1.10, 1.15 ..., 1.5. Compare these approximate values with the values of the exact solution y = x(1 + x2/3) 1 −x2/3 , which was obtained in Example 2.2.3. Present your results in a table like Table 3.3.1. 9. C In Example 2.2.3 it was shown that y5 + y = x2 + x −4 is an implicit solution of the initial value problem y′ = 2x + 1 5y4 + 1, y(2) = 1. (A) Use the Runge-Kutta method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approxi- mate values of the solution of (A) at x = 2.0, 2.1, 2.2, 2.3, ..., 3.0. Present your results in tabular form. To check the error in these approximate values, construct another table of values of the residual R(x, y) = y5 + y −x2 −x + 4 for each value of (x, y) appearing in the first table. 10. C You can see from Example 2.5.1 that x4y3 + x2y5 + 2xy = 4 is an implicit solution of the initial value problem y′ = −4x3y3 + 2xy5 + 2y 3x4y2 + 5x2y4 + 2x, y(1) = 1. (A) Use the Runge-Kutta method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approxi- mate values of the solution of (A) at x = 1.0, 1.1, 1.2, 1.3, ..., 2.0. Present your results in tabular form. To check the error in these approximate values, construct another table of values of the residual R(x, y) = x4y3 + x2y5 + 2xy −4 for each value of (x, y) appearing in the first table. 11. C Use the Runge-Kutta method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem (3y2 + 4y)y′ + 2x + cos x = 0, y(0) = 1 (Exercise 2.2.13), at x = 0, 0.1, 0.2, 0.3, ..., 1.0. | Elementary Differential Equations with Boundary Value Problems_Page_135_Chunk2540 |
126 Chapter 3 Numerical Methods 12. C Use the Runge-Kutta method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + (y + 1)(y −1)(y −2) x + 1 = 0, y(1) = 0 (Exercise 2.2.14), at x = 1.0, 1.1, 1.2, 1.3, ..., 2.0. 13. C Use the Runge-Kutta method and the Runge-Kutta semilinear method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of the initial value problem y′ + 3y = e−3x(1 −4x + 3x2 −4x3), y(0) = −3 at x = 0, 0.1, 0.2, 0.3, ..., 1.0. Compare these approximate values with the values of the exact solution y = −e−3x(3 −x +2x2 −x3 +x4), which can be obtained by the method of Section 2.1. Do you notice anything special about the results? Explain. The linear initial value problems in Exercises 14–19 can’t be solved exactly in terms of known elementary functions. In each exercise use the Runge-Kutta and the Runge-Kutta semilinear methods with the indi- cated step sizes to find approximate values of the solution of the given initial value problem at 11 equally spaced points (including the endpoints) in the interval. 14. C y′ −2y = 1 1 + x2 , y(2) = 2; h = 0.1, 0.05, 0.025 on [2, 3] 15. C y′ + 2xy = x2, y(0) = 3; h = 0.2, 0.1, 0.05 on [0, 2] (Exercise 2.1.38) 16. C y′ + 1 xy = sin x x2 , y(1) = 2; h = 0.2, 0.1, 0.05 on [1, 3] (Exercise 2.1.39) 17. C y′ + y = e−x tan x x , y(1) = 0; h = 0.05, 0.025, 0.0125 on [1, 1.5] (Exercise 2.1.40) 18. C y′ + 2x 1 + x2 y = ex (1 + x2)2 , y(0) = 1; h = 0.2, 0.1, 0.05 on [0, 2] (Exercise 2.1,41) 19. C xy′ + (x + 1)y = ex2, y(1) = 2; h = 0.05, 0.025, 0.0125 on [1, 1.5] (Exercise 2.1.42) In Exercises 20–22 use the Runge-Kutta method and the Runge-Kutta semilinear method with the indi- cated step sizes to find approximate values of the solution of the given initial value problem at 11 equally spaced points (including the endpoints) in the interval. 20. C y′ + 3y = xy2(y + 1), y(0) = 1; h = 0.1, 0.05, 0.025 on [0, 1] 21. C y′ −4y = x y2(y + 1), y(0) = 1; h = 0.1, 0.05, 0.025 on [0, 1] 22. C y′ + 2y = x2 1 + y2 , y(2) = 1; h = 0.1, 0.05, 0.025 on [2, 3] 23. C Suppose a < x0, so that −x0 < −a. Use the chain rule to show that if z is a solution of z′ = −f(−x, z), z(−x0) = y0, on [−x0, −a], then y = z(−x) is a solution of y′ = f(x, y), y(x0) = y0, on [a, x0]. | Elementary Differential Equations with Boundary Value Problems_Page_136_Chunk2541 |
Section 3.3 The Runge-Kutta Method 127 24. C Use the Runge-Kutta method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of y′ = y2 + xy −x2 x2 , y(2) = −1 at x = 1.1, 1.2, 1.3, ...2.0. Compare these approximate values with the values of the exact solution y = x(4 −3x2) 4 + 3x2 , which can be obtained by referring to Example 2.4.3. 25. C Use the Runge-Kutta method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of y′ = −x2y −xy2, y(1) = 1 at x = 0, 0.1, 0.2, ..., 1. 26. C Use the Runge-Kutta method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of y′ + 1 xy = 7 x2 + 3, y(1) = 3 2 at x = 0.5, 0.6,..., 1.5. Compare these approximate values with the values of the exact solution y = 7 ln x x + 3x 2 , which can be obtained by the method discussed in Section 2.1. 27. C Use the Runge-Kutta method with step sizes h = 0.1, h = 0.05, and h = 0.025 to find approximate values of the solution of xy′ + 2y = 8x2, y(2) = 5 at x = 1.0, 1.1, 1.2, ..., 3.0. Compare these approximate values with the values of the exact solution y = 2x2 −12 x2 , which can be obtained by the method discussed in Section 2.1. 28. NUMERICAL QUADRATURE (see Exercise 3.1.23). (a) Derive the quadrature formula Z b a f(x) dx ≈h 6 (f(a) + f(b)) + h 3 n−1 X i=1 f(a + ih) + 2h 3 n X i=1 f (a + (2i −1)h/2) (A) (where h = (b −a)/n) by applying the Runge-Kutta method to the initial value problem y′ = f(x), y(a) = 0. This quadrature formula is called Simpson’s Rule. | Elementary Differential Equations with Boundary Value Problems_Page_137_Chunk2542 |
128 Chapter 3 Numerical Methods (b) L For several choices of a, b, A, B, C, and D apply (A) to f(x) = A + Bx + Cx + Dx3, with n = 10, 20, 40, 80, 160, 320. Compare your results with the exact answers and explain what you find. (c) L For several choices of a, b, A, B, C, D, and E apply (A) to f(x) = A + Bx + Cx2 + Dx3 + Ex4, with n = 10, 20, 40, 80, 160, 320. Compare your results with the exact answers and explain what you find. | Elementary Differential Equations with Boundary Value Problems_Page_138_Chunk2543 |
CHAPTER 4 Applications of First Order Equations IN THIS CHAPTER we consider applications of first order differential equations. SECTION 4.1 begins with a discussion of exponential growth and decay, which you have probably al- ready seen in calculus. We consider applications to radioactive decay, carbon dating, and compound interest. We also consider more complicated problems where the rate of change of a quantity is in part proportional to the magnitude of the quantity, but is also influenced by other other factors for example, a radioactive susbstance is manufactured at a certain rate, but decays at a rate proportional to its mass, or a saver makes regular deposits in a savings account that draws compound interest. SECTION 4.2 deals with applications of Newton’s law of cooling and with mixing problems. SECTION 4.3 discusses applications to elementary mechanics involving Newton’s second law of mo- tion. The problems considered include motion under the influence of gravity in a resistive medium, and determining the initial velocity required to launch a satellite. SECTION 4.4 deals with methods for dealing with a type of second order equation that often arises in applications of Newton’s second law of motion, by reformulating it as first order equation with a different independent variable. Although the method doesn’t usually lead to an explicit solution of the given equation, it does provide valuable insights into the behavior of the solutions. SECTION 4.5 deals with applications of differential equations to curves. 129 | Elementary Differential Equations with Boundary Value Problems_Page_139_Chunk2544 |
130 Chapter 4 Applications of First Order Equations 4.1 GROWTH AND DECAY Since the applications in this section deal with functions of time, we’ll denote the independent variable by t. If Q is a function of t, Q′ will denote the derivative of Q with respect to t; thus, Q′ = dQ dt . Exponential Growth and Decay One of the most common mathematical models for a physical process is the exponential model, where it’s assumed that the rate of change of a quantity Q is proportional to Q; thus Q′ = aQ, (4.1.1) where a is the constant of proportionality. From Example 3, the general solution of (4.1.1) is Q = ceat and the solution of the initial value problem Q′ = aQ, Q(t0) = Q0 is Q = Q0ea(t−t0). (4.1.2) Since the solutions of Q′ = aQ are exponential functions, we say that a quantity Q that satisfies this equation grows exponentially if a > 0, or decays exponentially if a < 0 (Figure 4.1.1). Radioactive Decay Experimental evidence shows that radioactive material decays at a rate proportional to the mass of the material present. According to this model the mass Q(t) of a radioactive material present at time t satisfies (4.1.1), where a is a negative constant whose value for any given material must be determined by experimental observation. For simplicity, we’ll replace the negative constant a by −k, where k is a positive number that we’ll call the decay constant of the material. Thus, (4.1.1) becomes Q′ = −kQ. If the mass of the material present at t = t0 is Q0, the mass present at time t is the solution of Q′ = −kQ, Q(t0) = Q0. From (4.1.2) with a = −k, the solution of this initial value problem is Q = Q0e−k(t−t0). (4.1.3) The half–life τ of a radioactive material is defined to be the time required for half of its mass to decay; that is, if Q(t0) = Q0, then Q(τ + t0) = Q0 2 . (4.1.4) From (4.1.3) with t = τ + t0, (4.1.4) is equivalent to Q0e−kτ = Q0 2 , | Elementary Differential Equations with Boundary Value Problems_Page_140_Chunk2545 |
Section 4.1 Growth and Decay 131 Q t a > 0 a < 0 Q0 Figure 4.1.1 Exponential growth and decay so e−kτ = 1 2. Taking logarithms yields −kτ = ln 1 2 = −ln 2, so the half-life is τ = 1 k ln2. (4.1.5) (Figure 4.1.2). The half-life is independent of t0 and Q0, since it’s determined by the properties of material, not by the amount of the material present at any particular time. Example 4.1.1 A radioactive substance has a half-life of 1620 years. (a) If its mass is now 4 g (grams), how much will be left 810 years from now? (b) Find the time t1 when 1.5 g of the substance remain. SOLUTION(a) From (4.1.3) with t0 = 0 and Q0 = 4, Q = 4e−kt, (4.1.6) where we determine k from (4.1.5), with τ= 1620 years: k = ln2 τ = ln 2 1620. Substituting this in (4.1.6) yields Q = 4e−(t ln 2)/1620. (4.1.7) | Elementary Differential Equations with Boundary Value Problems_Page_141_Chunk2546 |
132 Chapter 4 Applications of First Order Equations Q t τ Q0 .5Q0 Figure 4.1.2 Half-life of a radioactive substance Therefore the mass left after 810 years will be Q(810) = 4e−(810 ln 2)/1620 = 4e−(ln 2)/2 = 2 √ 2 g. SOLUTION(b) Setting t = t1 in (4.1.7) and requiring that Q(t1) = 1.5 yields 3 2 = 4e(−t1 ln 2)/1620. Dividing by 4 and taking logarithms yields ln 3 8 = −t1 ln2 1620 . Since ln3/8 = −ln 8/3, t1 = 1620ln8/3 ln 2 ≈2292.4 years. Interest Compounded Continuously Suppose we deposit an amount of money Q0 in an interest-bearing account and make no further deposits or withdrawals for t years, during which the account bears interest at a constant annual rate r. To calculate the value of the account at the end of t years, we need one more piece of information: how the interest is added to the account, or—as the bankers say—how it is compounded. If the interest is compounded annually, the value of the account is multiplied by 1 + r at the end of each year. This means that after t years the value of the account is Q(t) = Q0(1 + r)t. | Elementary Differential Equations with Boundary Value Problems_Page_142_Chunk2547 |
Section 4.1 Growth and Decay 133 If interest is compounded semiannually, the value of the account is multiplied by (1 + r/2) every 6 months. Since this occurs twice annually, the value of the account after t years is Q(t) = Q0 1 + r 2 2t . In general, if interest is compounded n times per year, the value of the account is multiplied n times per year by (1 + r/n); therefore, the value of the account after t years is Q(t) = Q0 1 + r n nt . (4.1.8) Thus, increasing the frequency of compounding increases the value of the account after a fixed period of time. Table 4.1.7 shows the effect of increasing the number of compoundings over t = 5 years on an initial deposit of Q0 = 100 (dollars), at an annual interest rate of 6%. Table 4.1.7. Table The effect of compound interest n $100 1 + .06 n 5n (number of compoundings (value in dollars per year) after 5 years) 1 $133.82 2 $134.39 4 $134.68 8 $134.83 364 $134.98 You can see from Table 4.1.7 that the value of the account after 5 years is an increasing function of n. Now suppose the maximum allowable rate of interest on savings accounts is restricted by law, but the time intervals between successive compoundings isn’t ; then competing banks can attract savers by compounding often. The ultimate step in this direction is to compound continuously, by which we mean that n →∞in (4.1.8). Since we know from calculus that lim n→∞ 1 + r n n = er, this yields Q(t) = lim n→∞Q0 1 + r n nt = Q0 h lim n→∞ 1 + r n nit = Q0ert. Observe that Q = Q0ert is the solution of the initial value problem Q′ = rQ, Q(0) = Q0; that is, with continuous compounding the value of the account grows exponentially. | Elementary Differential Equations with Boundary Value Problems_Page_143_Chunk2548 |
134 Chapter 4 Applications of First Order Equations Example 4.1.2 If $150 is deposited in a bank that pays 5 1 2% annual interest compounded continuously, the value of the account after t years is Q(t) = 150e.055t dollars. (Note that it’s necessary to write the interest rate as a decimal; thus, r = .055.) Therefore, after t = 10 years the value of the account is Q(10) = 150e.55 ≈$259.99. Example 4.1.3 We wish to accumulate $10,000 in 10 years by making a single deposit in a savings account bearing 5 1 2% annual interest compounded continuously. How much must we deposit in the account? Solution The value of the account at time t is Q(t) = Q0e.055t. (4.1.9) Since we want Q(10) to be $10,000, the initial deposit Q0 must satisfy the equation 10000 = Q0e.55, (4.1.10) obtained by setting t = 10 and Q(10) = 10000 in (4.1.9). Solving (4.1.10) for Q0 yields Q0 = 10000e−.55 ≈$5769.50. Mixed Growth and Decay Example 4.1.4 A radioactive substance with decay constant k is produced at a constant rate of a units of mass per unit time. (a) Assuming that Q(0) = Q0, find the mass Q(t) of the substance present at time t. (b) Find limt→∞Q(t). SOLUTION(a) Here Q′ = rate of increase of Q −rate of decrease of Q. The rate of increase is the constant a. Since Q is radioactive with decay constant k, the rate of decrease is kQ. Therefore Q′ = a −kQ. This is a linear first order differential equation. Rewriting it and imposing the initial condition shows that Q is the solution of the initial value problem Q′ + kQ = a, Q(0) = Q0. (4.1.11) Since e−kt is a solution of the complementary equation, the solutions of (4.1.11) are of the form Q = ue−kt, where u′e−kt = a, so u′ = aekt. Hence, u = a k ekt + c | Elementary Differential Equations with Boundary Value Problems_Page_144_Chunk2549 |
Section 4.1 Growth and Decay 135 Q t a/k Figure 4.1.3 Q(t) approaches the steady state value a k as t →∞ and Q = ue−kt = a k + ce−kt. Since Q(0) = Q0, setting t = 0 here yields Q0 = a k + c or c = Q0 −a k . Therefore Q = a k + Q0 −a k e−kt. (4.1.12) SOLUTION(b) Since k > 0, limt→∞e−kt = 0, so from (4.1.12) lim t→∞Q(t) = a k . This limit depends only on a and k, and not on Q0. We say that a/k is the steady state value of Q. From (4.1.12) we also see that Q approaches its steady state value from above if Q0 > a/k, or from below if Q0 < a/k. If Q0 = a/k, then Q remains constant (Figure 4.1.3). Carbon Dating The fact that Q approaches a steady state value in the situation discussed in Example 4 underlies the method of carbon dating, devised by the American chemist and Nobel Prize Winner W.S. Libby. Carbon 12 is stable, but carbon-14, which is produced by cosmic bombardment of nitrogen in the upper atmosphere, is radioactive with a half-life of about 5570 years. Libby assumed that the quantity of carbon- 12 in the atmosphere has been constant throughout time, and that the quantity of radioactive carbon-14 | Elementary Differential Equations with Boundary Value Problems_Page_145_Chunk2550 |
136 Chapter 4 Applications of First Order Equations achieved its steady state value long ago as a result of its creation and decomposition over millions of years. These assumptions led Libby to conclude that the ratio of carbon-14 to carbon-12 has been nearly constant for a long time. This constant, which we denote by R, has been determined experimentally. Living cells absorb both carbon-12 and carbon-14 in the proportion in which they are present in the environment. Therefore the ratio of carbon-14 to carbon-12 in a living cell is always R. However, when the cell dies it ceases to absorb carbon, and the ratio of carbon-14 to carbon-12 decreases exponentially as the radioactive carbon-14 decays. This is the basis for the method of carbon dating, as illustrated in the next example. Example 4.1.5 An archaeologist investigating the site of an ancient village finds a burial ground where the amount of carbon-14 present in individual remains is between 42 and 44% of the amount present in live individuals. Estimate the age of the village and the length of time for which it survived. Solution Let Q = Q(t) be the quantity of carbon-14 in an individual set of remains t years after death, and let Q0 be the quantity that would be present in live individuals. Since carbon-14 decays exponentially with half-life 5570 years, its decay constant is k = ln2 5570. Therefore Q = Q0e−t(ln 2)/5570 if we choose our time scale so that t0 = 0 is the time of death. If we know the present value of Q we can solve this equation for t, the number of years since death occurred. This yields t = −5570lnQ/Q0 ln 2 . It is given that Q = .42Q0 in the remains of individuals who died first. Therefore these deaths occurred about t1 = −5570ln.42 ln 2 ≈6971 years ago. For the most recent deaths, Q = .44Q0; hence, these deaths occurred about t2 = −5570ln.44 ln 2 ≈6597 years ago. Therefore it’s reasonable to conclude that the village was founded about 7000 years ago, and lasted for about 400 years. A Savings Program Example 4.1.6 A person opens a savings account with an initial deposit of $1000 and subsequently deposits $50 per week. Find the value Q(t) of the account at time t > 0, assuming that the bank pays 6% interest compounded continuously. Solution Observe that Q isn’t continuous, since there are 52 discrete deposits per year of $50 each. To construct a mathematical model for this problem in the form of a differential equation, we make the simplifying assumption that the deposits are made continuously at a rate of $2600 per year. This is essential, since solutions of differential equations are continuous functions. With this assumption, Q increases continuously at the rate Q′ = 2600 + .06Q | Elementary Differential Equations with Boundary Value Problems_Page_146_Chunk2551 |
Section 4.1 Growth and Decay 137 and therefore Q satisfies the differential equation Q′ −.06Q = 2600. (4.1.13) (Of course, we must recognize that the solution of this equation is an approximation to the true value of Q at any given time. We’ll discuss this further below.) Since e.06t is a solution of the complementary equation, the solutions of (4.1.13) are of the form Q = ue.06t, where u′e.06t = 2600. Hence, u′ = 2600e−.06t, u = −2600 .06 e−.06t + c and Q = ue.06t = −2600 .06 + ce.06t. (4.1.14) Setting t = 0 and Q = 1000 here yields c = 1000 + 2600 .06 , and substituting this into (4.1.14) yields Q = 1000e.06t + 2600 .06 (e.06t −1), (4.1.15) where the first term is the value due to the initial deposit and the second is due to the subsequent weekly deposits. Mathematical models must be tested for validity by comparing predictions based on them with the actual outcome of experiments. Example 6 is unusual in that we can compute the exact value of the account at any specified time and compare it with the approximate value predicted by (4.1.15) (See Exercise 21.). The follwing table gives a comparison for a ten year period. Each exact answer corresponds to the time of the year-end deposit, and each year is assumed to have exactly 52 weeks. Year Approximate Value of Q Exact Value of P Error Percentage Error (Example 4.1.6) (Exercise 21) Q −P (Q −P )/P 1 $ 3741.42 $ 3739.87 $ 1.55 .0413% 2 6652.36 6649.17 3.19 .0479 3 9743.30 9738.37 4.93 .0506 4 13,025.38 13,018.60 6.78 .0521 5 16,510.41 16,501.66 8.75 .0530 6 20,210.94 20,200.11 10.83 .0536 7 24,140.30 24,127.25 13.05 .0541 8 28,312.63 28,297.23 15.40 .0544 9 32,742.97 32,725.07 17.90 .0547 10 37,447.27 37,426.72 20.55 .0549 | Elementary Differential Equations with Boundary Value Problems_Page_147_Chunk2552 |
138 Chapter 4 Applications of First Order Equations 4.1 Exercises 1. The half-life of a radioactive substance is 3200 years. Find the quantity Q(t) of the substance left at time t > 0 if Q(0) = 20 g. 2. The half-life of a radioactive substance is 2 days. Find the time required for a given amount of the material to decay to 1/10 of its original mass. 3. A radioactive material loses 25% of its mass in 10 minutes. What is its half-life? 4. A tree contains a known percentage p0 of a radioactive substance with half-life τ. When the tree dies the substance decays and isn’t replaced. If the percentage of the substance in the fossilized remains of such a tree is found to be p1, how long has the tree been dead? 5. If tp and tq are the times required for a radioactive material to decay to 1/p and 1/q times its original mass (respectively), how are tp and tq related? 6. Find the decay constant k for a radioactive substance, given that the mass of the substance is Q1 at time t1 and Q2 at time t2. 7. A process creates a radioactive substance at the rate of 2 g/hr and the substance decays at a rate proportional to its mass, with constant of proportionality k = .1(hr)−1. If Q(t) is the mass of the substance at time t, find limt→∞Q(t). 8. A bank pays interest continuously at the rate of 6%. How long does it take for a deposit of Q0 to grow in value to 2Q0? 9. At what rate of interest, compounded continuously, will a bank deposit double in value in 8 years? 10. A savings account pays 5% per annum interest compounded continuously. The initial deposit is Q0 dollars. Assume that there are no subsequent withdrawals or deposits. (a) How long will it take for the value of the account to triple? (b) What is Q0 if the value of the account after 10 years is $100,000 dollars? 11. A candymaker makes 500 pounds of candy per week, while his large family eats the candy at a rate equal to Q(t)/10 pounds per week, where Q(t) is the amount of candy present at time t. (a) Find Q(t) for t > 0 if the candymaker has 250 pounds of candy at t = 0. (b) Find limt→∞Q(t). 12. Suppose a substance decays at a yearly rate equal to half the square of the mass of the substance present. If we start with 50 g of the substance, how long will it be until only 25 g remain? 13. A super bread dough increases in volume at a rate proportional to the volume V present. If V increases by a factor of 10 in 2 hours and V (0) = V0, find V at any time t. How long will it take for V to increase to 100V0? 14. A radioactive substance decays at a rate proportional to the amount present, and half the original quantity Q0 is left after 1500 years. In how many years would the original amount be reduced to 3Q0/4? How much will be left after 2000 years? 15. A wizard creates gold continuously at the rate of 1 ounce per hour, but an assistant steals it con- tinuously at the rate of 5% of however much is there per hour. Let W(t) be the number of ounces that the wizard has at time t. Find W(t) and limt→∞W(t) if W(0) = 1. 16. A process creates a radioactive substance at the rate of 1 g/hr, and the substance decays at an hourly rate equal to 1/10 of the mass present (expressed in grams). Assuming that there are initially 20 g, find the mass S(t) of the substance present at time t, and find limt→∞S(t). | Elementary Differential Equations with Boundary Value Problems_Page_148_Chunk2553 |
Section 4.1 Growth and Decay 139 17. A tank is empty at t = 0. Water is added to the tank at the rate of 10 gal/min, but it leaks out at a rate (in gallons per minute) equal to the number of gallons in the tank. What is the smallest capacity the tank can have if this process is to continue forever? 18. A person deposits $25,000 in a bank that pays 5% per year interest, compounded continuously. The person continuously withdraws from the account at the rate of $750 per year. Find V (t), the value of the account at time t after the initial deposit. 19. A person has a fortune that grows at rate proportional to the square root of its worth. Find the worth W of the fortune as a function of t if it was $1 million 6 months ago and is $4 million today. 20. Let p = p(t) be the quantity of a product present at time t. The product is manufactured continu- ously at a rate proportional to p, with proportionality constant 1/2, and it’s consumed continuously at a rate proportional to p2, with proportionality constant 1/8. Find p(t) if p(0) = 100. 21. (a) In the situation of Example 4.1.6 find the exact value P (t) of the person’s account after t years, where t is an integer. Assume that each year has exactly 52 weeks, and include the year-end deposit in the computation. HINT: At time t the initial $1000 has been on deposit for t years. There have been 52t deposits of $50 each. The first $50 has been on deposit for t −1/52 years, the second for t −2/52 years · · · in general, the jth $50 has been on deposit for t −j/52 years (1 ≤ j ≤52t). Find the present value of each $50 deposit assuming 6% interest compounded continuously, and use the formula 1 + x + x2 + · · · + xn = 1 −xn+1 1 −x (x ̸= 1) to find their total value. (b) Let p(t) = Q(t) −P (t) P (t) be the relative error after t years. Find p(∞) = lim t→∞p(t). 22. A homebuyer borrows P0 dollars at an annual interest rate r, agreeing to repay the loan with equal monthly payments of M dollars per month over N years. (a) Derive a differential equation for the loan principal (amount that the homebuyer owes) P (t) at time t > 0, making the simplifying assumption that the homebuyer repays the loan con- tinuously rather than in discrete steps. (See Example 4.1.6 .) (b) Solve the equation derived in (a). (c) Use the result of (b) to determine an approximate value for M assuming that each year has exactly 12 months of equal length. (d) It can be shown that the exact value of M is given by M = rP0 12 | Elementary Differential Equations with Boundary Value Problems_Page_149_Chunk2554 |
140 Chapter 4 Applications of First Order Equations (a) Determine the time T(α) when the loan will be paid off and the amount S(α) that the home- buyer will save. (b) Suppose P0 = $50, 000, r = 8%, and N = 15. Compute the savings realized by accelerated payments with α = 1.05, 1.10, and 1.15. 24. A benefactor wishes to establish a trust fund to pay a researcher’s salary for T years. The salary is to start at S0 dollars per year and increase at a fractional rate of a per year. Find the amount of money P0 that the benefactor must deposit in a trust fund paying interest at a rate r per year. Assume that the researcher’s salary is paid continuously, the interest is compounded continuously, and the salary increases are granted continuously. 25. L A radioactive substance with decay constant k is produced at the rate of at 1 + btQ(t) units of mass per unit time, where a and b are positive constants and Q(t) is the mass of the substance present at time t; thus, the rate of production is small at the start and tends to slow when Q is large. (a) Set up a differential equation for Q. (b) Choose your own positive values for a, b, k, and Q0 = Q(0). Use a numerical method to discover what happens to Q(t) as t →∞. (Be precise, expressing your conclusions in terms of a, b, k. However, no proof is required.) 26. L Follow the instructions of Exercise 25, assuming that the substance is produced at the rate of at/(1 + bt(Q(t))2) units of mass per unit of time. 27. L Follow the instructions of Exercise 25, assuming that the substance is produced at the rate of at/(1 + bt) units of mass per unit of time. 4.2 COOLING AND MIXING Newton’s Law of Cooling Newton’s law of cooling states that if an object with temperature T(t) at time t is in a medium with temperature Tm(t), the rate of change of T at time t is proportional to T(t) −Tm(t); thus, T satisfies a differential equation of the form T ′ = −k(T −Tm). (4.2.1) Here k > 0, since the temperature of the object must decrease if T > Tm, or increase if T < Tm. We’ll call k the temperature decay constant of the medium. For simplicity, in this section we’ll assume that the medium is maintained at a constant temperature Tm. This is another example of building a simple mathematical model for a physical phenomenon. Like most mathematical models it has its limitations. For example, it’s reasonable to assume that the temperature of a room remains approximately constant if the cooling object is a cup of coffee, but perhaps not if it’s a huge cauldron of molten metal. (For more on this see Exercise 17.) To solve (4.2.1), we rewrite it as T ′ + kT = kTm. Since e−kt is a solution of the complementary equation, the solutions of this equation are of the form T = ue−kt, where u′e−kt = kTm, so u′ = kTmekt. Hence, u = Tmekt + c, | Elementary Differential Equations with Boundary Value Problems_Page_150_Chunk2555 |
Section 4.2 Cooling and Mixing 141 so T = ue−kt = Tm + ce−kt. If T(0) = T0, setting t = 0 here yields c = T0 −Tm, so T = Tm + (T0 −Tm)e−kt. (4.2.2) Note that T −Tm decays exponentially, with decay constant k. Example 4.2.1 A ceramic insulator is baked at 400◦C and cooled in a room in which the temperature is 25◦C. After 4 minutes the temperature of the insulator is 200◦C. What is its temperature after 8 minutes? Solution Here T0 = 400 and Tm = 25, so (4.2.2) becomes T = 25 + 375e−kt. (4.2.3) We determine k from the stated condition that T(4) = 200; that is, 200 = 25 + 375e−4k; hence, e−4k = 175 375 = 7 15. Taking logarithms and solving for k yields k = −1 4 ln 7 15 = 1 4 ln 15 7 . Substituting this into (4.2.3) yields T = 25 + 375e−t 4 ln 15 7 (Figure 4.2.1). Therefore the temperature of the insulator after 8 minutes is T(8) = 25 + 375e−2 ln 15 7 = 25 + 375 7 15 2 ≈107◦C. Example 4.2.2 An object with temperature 72◦F is placed outside, where the temperature is −20◦F. At 11:05 the temperature of the object is 60◦F and at 11:07 its temperature is 50◦F. At what time was the object placed outside? Solution Let T(t) be the temperature of the object at time t. For convenience, we choose the origin t0 = 0 of the time scale to be 11:05 so that T0 = 60. We must determine the time τ when T(τ) = 72. Substituting T0 = 60 and Tm = −20 into (4.2.2) yields T = −20 + | Elementary Differential Equations with Boundary Value Problems_Page_151_Chunk2556 |
142 Chapter 4 Applications of First Order Equations t T 5 10 15 20 25 30 100 150 200 250 300 350 400 50 Figure 4.2.1 T = 25 + 375e−(t/4)ln 15/7 We obtain k from the stated condition that the temperature of the object is 50◦F at 11:07. Since 11:07 is t = 2 on our time scale, we can determine k by substituting T = 50 and t = 2 into (4.2.4) to obtain 50 = −20 + 80e−2k (Figure 4.2.2); hence, e−2k = 70 80 = 7 8. Taking logarithms and solving for k yields k = −1 2 ln 7 8 = 1 2 ln 8 7. Substituting this into (4.2.4) yields T = −20 + 80e−t 2 ln 8 7 , and the condition T(τ) = 72 implies that 72 = −20 + 80e−τ 2 ln 8 7 ; hence, e−τ 2 ln 8 7 = 92 80 = 23 20. Taking logarithms and solving for τ yields τ = −2 ln 23 20 ln 8 7 ≈−2.09 min. | Elementary Differential Equations with Boundary Value Problems_Page_152_Chunk2557 |
Section 4.2 Cooling and Mixing 143 T t −5 5 10 15 20 25 30 35 40 20 40 60 80 −20 100 T=72 Figure 4.2.2 T = −20 + 80e−t 2 ln 8 7 Therefore the object was placed outside about 2 minutes and 5 seconds before 11:05; that is, at 11:02:55. Mixing Problems In the next two examples a saltwater solution with a given concentration (weight of salt per unit volume of solution)is added at a specified rate to a tank that initially contains saltwater with a different concentra- tion. The problem is to determine the quantity of salt in the tank as a function of time. This is an example of a mixing problem. To construct a tractable mathematical model for mixing problems we assume in our examples (and most exercises) that the mixture is stirred instantly so that the salt is always uniformly distributed throughout the mixture. Exercises 22 and 23 deal with situations where this isn’t so, but the distribution of salt becomes approximately uniform as t →∞. Example 4.2.3 A tank initially contains 40 pounds of salt dissolved in 600 gallons of water. Starting at t0 = 0, water that contains 1/2 pound of salt per gallon is poured into the tank at the rate of 4 gal/min and the mixture is drained from the tank at the same rate (Figure 4.2.3). (a) Find a differential equation for the quantity Q(t) of salt in the tank at time t > 0, and solve the equation to determine Q(t). (b) Find limt→∞Q(t). SOLUTION(a) To find a differential equation for Q, we must use the given information to derive an expression for Q′. But Q′ is the rate of change of the quantity of salt in the tank changes with respect to time; thus, if rate in denotes the rate at which salt enters the tank and rate out denotes the rate by which it leaves, then Q′ = rate in −rate out. (4.2.5) | Elementary Differential Equations with Boundary Value Problems_Page_153_Chunk2558 |
144 Chapter 4 Applications of First Order Equations 600 gal 4 gal/min; .5 lb/gal 4 gal/min Figure 4.2.3 A mixing problem The rate in is 1 2 lb/gal × (4 gal/min) = 2 lb/min. Determining the rate out requires a little more thought. We’re removing 4 gallons of the mixture per minute, and there are always 600 gallons in the tank; that is, we’re removing 1/150 of the mixture per minute. Since the salt is evenly distributed in the mixture, we are also removing 1/150 of the salt per minute. Therefore, if there are Q(t) pounds of salt in the tank at time t, the rate out at any time t is Q(t)/150. Alternatively, we can arrive at this conclusion by arguing that rate out = (concentration) × (rate of flow out) = (lb/gal) × (gal/min) = Q(t) 600 × 4 = Q(t) 150 . We can now write (4.2.5) as Q′ = 2 −Q 150. This first order equation can be rewritten as Q′ + Q 150 = 2. Since e−t/150 is a solution of the complementary equation, the solutions of this equation are of the form Q = ue−t/150, where u′e−t/150 = 2, so u′ = 2et/150. Hence, u = 300et/150 + c, | Elementary Differential Equations with Boundary Value Problems_Page_154_Chunk2559 |
Section 4.2 Cooling and Mixing 145 100 200 300 400 500 600 700 800 900 50 100 150 200 250 300 t Q Figure 4.2.4 Q = 300 −260e−t/150 so Q = ue−t/150 = 300 + ce−t/150 (4.2.6) (Figure 4.2.4). Since Q(0) = 40, c = −260; therefore, Q = 300 −260e−t/150. SOLUTION(b) From (4.2.6), we see that that limt→∞Q(t) = 300 for any value of Q(0). This is intu- itively reasonable, since the incoming solution contains 1/2 pound of salt per gallon and there are always 600 gallons of water in the tank. Example 4.2.4 A 500-liter tank initially contains 10 g of salt dissolved in 200 liters of water. Starting at t0 = 0, water that contains 1/4 g of salt per liter is poured into the tank at the rate of 4 liters/min and the mixture is drained from the tank at the rate of 2 liters/min (Figure 4.2.5). Find a differential equation for the quantity Q(t) of salt in the tank at time t prior to the time when the tank overflows and find the concentration K(t) (g/liter ) of salt in the tank at any such time. Solution We first determine the amount W(t) of solution in the tank at any time t prior to overflow. Since W(0) = 200 and we’re adding 4 liters/min while removing only 2 liters/min, there’s a net gain of 2 liters/min in the tank; therefore, W(t) = 2t + 200. Since W(150) = 500 liters (capacity of the tank), this formula is valid for 0 ≤t ≤150. Now let Q(t) be the number of grams of salt in the tank at time t, where 0 ≤t ≤150. As in Example 4.2.3, Q′ = rate in −rate out. (4.2.7) | Elementary Differential Equations with Boundary Value Problems_Page_155_Chunk2560 |
146 Chapter 4 Applications of First Order Equations 2t+200 liters 4 liters/min; .25 g/liter Figure 4.2.5 Another mixing problem The rate in is 1 4 g/liter × (4 liters/min) = 1 g/min. (4.2.8) To determine the rate out, we observe that since the mixture is being removed from the tank at the constant rate of 2 liters/min and there are 2t + 200 liters in the tank at time t, the fraction of the mixture being removed per minute at time t is 2 2t + 200 = 1 t + 100. We’re removing this same fraction of the salt per minute. Therefore, since there are Q(t) grams of salt in the tank at time t, rate out = Q(t) t + 100. (4.2.9) Alternatively, we can arrive at this conclusion by arguing that rate out = (concentration) × (rate of flow out) = (g/liter) × (liters/min) = Q(t) 2t + 200 × 2 = Q(t) t + 100. Substituting (4.2.8) and (4.2.9) into (4.2.7) yields Q′ = 1 − Q t + 100, so Q′ + 1 t + 100Q = 1. (4.2.10) By separation of variables, 1/(t + 100) is a solution of the complementary equation, so the solutions of (4.2.10) are of the form Q = u t + 100, where u′ t + 100 = 1, so u′ = t + 100. | Elementary Differential Equations with Boundary Value Problems_Page_156_Chunk2561 |
Section 4.2 Cooling and Mixing 147 Hence, u = (t + 100)2 2 + c. (4.2.11) Since Q(0) = 10 and u = (t + 100)Q, (4.2.11) implies that (100)(10) = (100)2 2 + c, so c = 100(10) −(100)2 2 = −4000 and therefore u = (t + 100)2 2 −4000. Hence, Q = u t + 200 = t + 100 2 − 4000 t + 100. Now let K(t) be the concentration of salt at time t. Then K(t) = 1 4 − 2000 (t + 100)2 (Figure 4.2.6). 200 400 600 800 1000 t .05 .15 .25 .10 .20 K Figure 4.2.6 K(t) = 1 4 − 2000 (t + 100)2 | Elementary Differential Equations with Boundary Value Problems_Page_157_Chunk2562 |
148 Chapter 4 Applications of First Order Equations 4.2 Exercises 1. A thermometer is moved from a room where the temperature is 70◦F to a freezer where the tem- perature is 12◦F . After 30 seconds the thermometer reads 40◦F. What does it read after 2 minutes? 2. A fluid initially at 100◦C is placed outside on a day when the temperature is −10◦C, and the temperature of the fluid drops 20◦C in one minute. Find the temperature T(t) of the fluid for t > 0. 3. At 12:00 PM a thermometer reading 10◦F is placed in a room where the temperature is 70◦F. It reads 56◦when it’s placed outside, where the temperature is 5◦F, at 12:03. What does it read at 12:05 PM? 4. A thermometer initially reading 212◦F is placed in a room where the temperature is 70◦F. After 2 minutes the thermometer reads 125◦F. (a) What does the thermometer read after 4 minutes? (b) When will the thermometer read 72◦F? (c) When will the thermometer read 69◦F? 5. An object with initial temperature 150◦C is placed outside, where the temperature is 35◦C. Its temperatures at 12:15 and 12:20 are 120◦C and 90◦C, respectively. (a) At what time was the object placed outside? (b) When will its temperature be 40◦C? 6. An object is placed in a room where the temperature is 20◦C. The temperature of the object drops by 5◦C in 4 minutes and by 7◦C in 8 minutes. What was the temperature of the object when it was initially placed in the room? 7. A cup of boiling water is placed outside at 1:00 PM. One minute later the temperature of the water is 152◦F. After another minute its temperature is 112◦F. What is the outside temperature? 8. A tank initially contains 40 gallons of pure water. A solution with 1 gram of salt per gallon of water is added to the tank at 3 gal/min, and the resulting solution dranes out at the same rate. Find the quantity Q(t) of salt in the tank at time t > 0. 9. A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per gallon is added to the tank at 6 gal/min, and the resulting solution leaves at the same rate. Find the quantity Q(t) of salt in the tank at time t > 0. 10. A tank initially contains 100 liters of a salt solution with a concentration of .1 g/liter. A solution with a salt concentration of .3 g/liter is added to the tank at 5 liters/min, and the resulting mixture is drained out at the same rate. Find the concentration K(t) of salt in the tank as a function of t. 11. A 200 gallon tank initially contains 100 gallons of water with 20 pounds of salt. A salt solution with 1/4 pound of salt per gallon is added to the tank at 4 gal/min, and the resulting mixture is drained out at 2 gal/min. Find the quantity of salt in the tank as it’s about to overflow. 12. Suppose water is added to a tank at 10 gal/min, but leaks out at the rate of 1/5 gal/min for each gallon in the tank. What is the smallest capacity the tank can have if the process is to continue indefinitely? 13. A chemical reaction in a laboratory with volume V (in ft3) produces q1 ft3/min of a noxious gas as a byproduct. The gas is dangerous at concentrations greater than c, but harmless at concentrations ≤c. Intake fans at one end of the laboratory pull in fresh air at the rate of q2 ft3/min and exhaust fans at the other end exhaust the mixture of gas and air from the laboratory at the same rate. | Elementary Differential Equations with Boundary Value Problems_Page_158_Chunk2563 |
Section 4.2 Cooling and Mixing 149 Assuming that the gas is always uniformly distributed in the room and its initial concentration c0 is at a safe level, find the smallest value of q2 required to maintain safe conditions in the laboratory for all time. 14. A 1200-gallon tank initially contains 40 pounds of salt dissolved in 600 gallons of water. Starting at t0 = 0, water that contains 1/2 pound of salt per gallon is added to the tank at the rate of 6 gal/min and the resulting mixture is drained from the tank at 4 gal/min. Find the quantity Q(t) of salt in the tank at any time t > 0 prior to overflow. 15. Tank T1 initially contain 50 gallons of pure water. Starting at t0 = 0, water that contains 1 pound of salt per gallon is poured into T1 at the rate of 2 gal/min. The mixture is drained from T1 at the same rate into a second tank T2, which initially contains 50 gallons of pure water. Also starting at t0 = 0, a mixture from another source that contains 2 pounds of salt per gallon is poured into T2 at the rate of 2 gal/min. The mixture is drained from T2 at the rate of 4 gal/min. (a) Find a differential equation for the quantity Q(t) of salt in tank T2 at time t > 0. (b) Solve the equation derived in (a) to determine Q(t). (c) Find limt→∞Q(t). 16. Suppose an object with initial temperature T0 is placed in a sealed container, which is in turn placed in a medium with temperature Tm. Let the initial temperature of the container be S0. Assume that the temperature of the object does not affect the temperature of the container, which in turn does not affect the temperature of the medium. (These assumptions are reasonable, for example, if the object is a cup of coffee, the container is a house, and the medium is the atmosphere.) (a) Assuming that the container and the medium have distinct temperature decay constants k and km respectively, use Newton’s law of cooling to find the temperatures S(t) and T(t) of the container and object at time t. (b) Assuming that the container and the medium have the same temperature decay constant k, use Newton’s law of cooling to find the temperatures S(t) and T(t) of the container and object at time t. (c) Find lim .t→∞S(t) and limt→∞T(t) . 17. In our previous examples and exercises concerning Newton’s law of cooling we assumed that the temperature of the medium remains constant. This model is adequate if the heat lost or gained by the object is insignificant compared to the heat required to cause an appreciable change in the tem- perature of the medium. If this isn’t so, we must use a model that accounts for the heat exchanged between the object and the medium. Let T = T(t) and Tm = Tm(t) be the temperatures of the object and the medium, respectively, and let T0 and Tm0 be their initial values. Again, we assume that T and Tm are related by Newton’s law of cooling, T ′ = −k(T −Tm). (A) We also assume that the change in heat of the object as its temperature changes from T0 to T is a(T −T0) and that the change in heat of the medium as its temperature changes from Tm0 to Tm is am(Tm −Tm0), where a and am are positive constants depending upon the masses and thermal properties of the object and medium, respectively. If we assume that the total heat of the system consisting of the object and the medium remains constant (that is, energy is conserved), then a(T −T0) + am(Tm −Tm0) = 0. (B) (a) Equation (A) involves two unknown functions T and Tm. Use (A) and (B) to derive a differ- ential equation involving only T. | Elementary Differential Equations with Boundary Value Problems_Page_159_Chunk2564 |
150 Chapter 4 Applications of First Order Equations (b) Find T(t) and Tm(t) for t > 0. (c) Find limt→∞T(t) and limt→∞Tm(t). 18. Control mechanisms allow fluid to flow into a tank at a rate proportional to the volume V of fluid in the tank, and to flow out at a rate proportional to V 2. Suppose V (0) = V0 and the constants of proportionality are a and b, respectively. Find V (t) for t > 0 and find limt→∞V (t). 19. Identical tanks T1 and T2 initially contain W gallons each of pure water. Starting at t0 = 0, a salt solution with constant concentration c is pumped into T1 at r gal/min and drained from T1 into T2 at the same rate. The resulting mixture in T2 is also drained at the same rate. Find the concentrations c1(t) and c2(t) in tanks T1 and T2 for t > 0. 20. An infinite sequence of identical tanks T1, T2, ..., Tn, ..., initially contain W gallons each of pure water. They are hooked together so that fluid drains from Tn into Tn+1 (n = 1, 2, · · ·). A salt solution is circulated through the tanks so that it enters and leaves each tank at the constant rate of r gal/min. The solution has a concentration of c pounds of salt per gallon when it enters T1. (a) Find the concentration cn(t) in tank Tn for t > 0. (b) Find limt→∞cn(t) for each n. 21. Tanks T1 and T2 have capacities W1 and W2 liters, respectively. Initially they are both full of dye solutions with concentrations c1 and c2 grams per liter. Starting at t0 = 0, the solution from T1 is pumped into T2 at a rate of r liters per minute, and the solution from T2 is pumped into T1 at the same rate. (a) Find the concentrations c1(t) and c2(t) of the dye in T1 and T2 for t > 0. (b) Find limt→∞c1(t) and limt→∞c2(t). 22. L Consider the mixing problem of Example 4.2.3, but without the assumption that the mixture is stirred instantly so that the salt is always uniformly distributed throughout the mixture. Assume instead that the distribution approaches uniformity as t →∞. In this case the differential equation for Q is of the form Q′ + a(t) 150 Q = 2 where limt→∞a(t) = 1. (a) Assuming that Q(0) = Q0, can you guess the value of limt→∞Q(t)?. (b) Use numerical methods to confirm your guess in the these cases: (i) a(t) = t/(1 + t) (ii) a(t) = 1 −e−t2 (iii) a(t) = 1 −sin(e−t). 23. L Consider the mixing problem of Example 4.2.4 in a tank with infinite capacity, but without the assumption that the mixture is stirred instantly so that the salt is always uniformly distributed throughout the mixture. Assume instead that the distribution approaches uniformity as t →∞. In this case the differential equation for Q is of the form Q′ + a(t) t + 100Q = 1 where limt→∞a(t) = 1. (a) Let K(t) be the concentration of salt at time t. Assuming that Q(0) = Q0, can you guess the value of limt→∞K(t)? (b) Use numerical methods to confirm your guess in the these cases: (i) a(t) = t/(1 + t) (ii) a(t) = 1 −e−t2 (iii) a(t) = 1 + sin(e−t). | Elementary Differential Equations with Boundary Value Problems_Page_160_Chunk2565 |
Section 4.3 Elementary Mechanics 151 4.3 ELEMENTARY MECHANICS Newton’s Second Law of Motion In this section we consider an object with constant mass m moving along a line under a force F . Let y = y(t) be the displacement of the object from a reference point on the line at time t, and let v = v(t) and a = a(t) be the velocity and acceleration of the object at time t. Thus, v = y′ and a = v′ = y′′, where the prime denotes differentiation with respect to t. Newton’s second law of motion asserts that the force F and the acceleration a are related by the equation F = ma. (4.3.1) Units In applications there are three main sets of units in use for length, mass, force, and time: the cgs, mks, and British systems. All three use the second as the unit of time. Table 1 shows the other units. Consistent with (4.3.1), the unit of force in each system is defined to be the force required to impart an acceleration of (one unit of length)/s2 to one unit of mass. Length Force Mass cgs centimeter (cm) dyne (d) gram (g) mks meter (m) newton (N) kilogram (kg) British foot (ft) pound (lb) slug (sl) Table 1. If we assume that Earth is a perfect sphere with constant mass density, Newton’s law of gravitation (discussed later in this section) asserts that the force exerted on an object by Earth’s gravitational field is proportional to the mass of the object and inversely proportional to the square of its distance from the center of Earth. However, if the object remains sufficiently close to Earth’s surface, we may assume that the gravitational force is constant and equal to its value at the surface. The magnitude of this force is mg, where g is called the acceleration due to gravity. (To be completely accurate, g should be called the magnitude of the acceleration due to gravity at Earth’s surface.) This quantity has been determined experimentally. Approximate values of g are g = 980 cm/s2 (cgs) g = 9.8 m/s2 (mks) g = 32 ft/s2 (British). In general, the force F in (4.3.1) may depend upon t, y, and y′. Since a = y′′, (4.3.1) can be written in the form my′′ = F (t, y, y′), (4.3.2) which is a second order equation. We’ll consider this equation with restrictions on F later; however, since Chapter 2 dealt only with first order equations, we consider here only problems in which (4.3.2) can be recast as a first order equation. This is possible if F does not depend on y, so (4.3.2) is of the form my′′ = F (t, y′). Letting v = y′ and v′ = y′′ yields a first order equation for v: mv′ = F (t, v). (4.3.3) | Elementary Differential Equations with Boundary Value Problems_Page_161_Chunk2566 |
152 Chapter 4 Applications of First Order Equations Solving this equation yields v as a function of t. If we know y(t0) for some time t0, we can integrate v to obtain y as a function of t. Equations of the form (4.3.3) occur in problems involving motion through a resisting medium. Motion Through a Resisting Medium Under Constant Gravitational Force Now we consider an object moving vertically in some medium. We assume that the only forces acting on the object are gravity and resistance from the medium. We also assume that the motion takes place close to Earth’s surface and take the upward direction to be positive, so the gravitational force can be assumed to have the constant value −mg. We’ll see that, under reasonable assumptions on the resisting force, the velocity approaches a limit as t →∞. We call this limit the terminal velocity. Example 4.3.1 An object with mass m moves under constant gravitational force through a medium that exerts a resistance with magnitude proportional to the speed of the object. (Recall that the speed of an object is |v|, the absolute value of its velocity v.) Find the velocity of the object as a function of t, and find the terminal velocity. Assume that the initial velocity is v0. Solution The total force acting on the object is F = −mg + F1, (4.3.4) where −mg is the force due to gravity and F1 is the resisting force of the medium, which has magnitude k|v|, where k is a positive constant. If the object is moving downward (v ≤0), the resisting force is upward (Figure 4.3.1(a)), so F1 = k|v| = k(−v) = −kv. On the other hand, if the object is moving upward (v ≥0), the resisting force is downward (Fig- ure 4.3.1(b)), so F1 = −k|v| = −kv. Thus, (4.3.4) can be written as F = −mg −kv, (4.3.5) regardless of the sign of the velocity. From Newton’s second law of motion, F = ma = mv′, so (4.3.5) yields mv′ = −mg −kv, or v′ + k mv = −g. (4.3.6) Since e−kt/m is a solution of the complementary equation, the solutions of (4.3.6) are of the form v = ue−kt/m, where u′e−kt/m = −g, so u′ = −gekt/m. Hence, u = −mg k ekt/m + c, so v = ue−kt/m = −mg k + ce−kt/m. (4.3.7) Since v(0) = v0, v0 = −mg k + c, | Elementary Differential Equations with Boundary Value Problems_Page_162_Chunk2567 |
Section 4.3 Elementary Mechanics 153 v v F1 = − kv F1 = − kv (a) (b) Figure 4.3.1 Resistive forces so c = v0 + mg k and (4.3.7) becomes v = −mg k + v0 + mg k e−kt/m. Letting t →∞here shows that the terminal velocity is lim t→∞v(t) = −mg k , which is independent of the initial velocity v0 (Figure 4.3.2). Example 4.3.2 A 960-lb object is given an initial upward velocity of 60 ft/s near the surface of Earth. The atmosphere resists the motion with a force of 3 lb for each ft/s of speed. Assuming that the only other force acting on the object is constant gravity, find its velocity v as a function of t, and find its terminal velocity. Solution Since mg = 960 and g = 32, m = 960/32 = 30. The atmospheric resistance is −3v lb if v is expressed in feet per second. Therefore 30v′ = −960 −3v, which we rewrite as v′ + 1 10v = −32. | Elementary Differential Equations with Boundary Value Problems_Page_163_Chunk2568 |
154 Chapter 4 Applications of First Order Equations − mg/k t v Figure 4.3.2 Solutions of mv′ = −mg −kv Since e−t/10 is a solution of the complementary equation, the solutions of this equation are of the form v = ue−t/10, where u′e−t/10 = −32, so u′ = −32et/10. Hence, u = −320et/10 + c, so v = ue−t/10 = −320 + ce−t/10. (4.3.8) The initial velocity is 60 ft/s in the upward (positive) direction; hence, v0 = 60. Substituting t = 0 and v = 60 in (4.3.8) yields 60 = −320 + c, so c = 380, and (4.3.8) becomes v = −320 + 380e−t/10 ft/s The terminal velocity is lim t→∞v(t) = −320 ft/s. Example 4.3.3 A 10 kg mass is given an initial velocity v0 ≤0 near Earth’s surface. The only forces acting on it are gravity and atmospheric resistance proportional to the square of the speed. Assuming that the resistance is 8 N if the speed is 2 m/s, find the velocity of the object as a function of t, and find the terminal velocity. Solution Since the object is falling, the resistance is in the upward (positive) direction. Hence, mv′ = −mg + kv2, (4.3.9) | Elementary Differential Equations with Boundary Value Problems_Page_164_Chunk2569 |
Section 4.3 Elementary Mechanics 155 where k is a constant. Since the magnitude of the resistance is 8 N when v = 2 m/s, k(22) = 8, so k = 2 N-s2/m2. Since m = 10 and g = 9.8, (4.3.9) becomes 10v′ = −98 + 2v2 = 2(v2 −49). (4.3.10) If v0 = −7, then v ≡−7 for all t ≥0. If v0 ̸= −7, we separate variables to obtain 1 v2 −49v′ = 1 5, (4.3.11) which is convenient for the required partial fraction expansion 1 v2 −49 = 1 (v −7)(v + 7) = 1 14 1 v −7 − 1 v + 7 . (4.3.12) Substituting (4.3.12) into (4.3.11) yields 1 14 1 v −7 − 1 v + 7 v′ = 1 5, so 1 v −7 − 1 v + 7 v′ = 14 5 . Integrating this yields ln |v −7| −ln|v + 7| = 14t/5 + k. Therefore v −7 v + 7 = eke14t/5. Since Theorem 2.3.1 implies that (v −7)/(v + 7) can’t change sign (why?), we can rewrite the last equation as v −7 v + 7 = ce14t/5, (4.3.13) which is an implicit solution of (4.3.10). Solving this for v yields v = −7c + e−14t/5 c −e−14t/5. (4.3.14) Since v(0) = v0, it (4.3.13) implies that c = v0 −7 v0 + 7. Substituting this into (4.3.14) and simplifying yields v = −7v0(1 + e−14t/5) −7(1 −e−14t/5) v0(1 −e−14t/5) −7(1 + e−14t/5 . Since v0 ≤0, v is defined and negative for all t > 0. The terminal velocity is lim t→∞v(t) = −7 m/s, independent of v0. More generally, it can be shown (Exercise 11) that if v is any solution of (4.3.9) such that v0 ≤0 then lim t→∞v(t) = − rmg k (Figure 4.3.3). | Elementary Differential Equations with Boundary Value Problems_Page_165_Chunk2570 |
156 Chapter 4 Applications of First Order Equations t v v = − (mg/k)1/2 Figure 4.3.3 Solutions of mv′ = −mg + kv2, v(0) = v0 ≤0 Example 4.3.4 A 10-kg mass is launched vertically upward from Earth’s surface with an initial velocity of v0 m/s. The only forces acting on the mass are gravity and atmospheric resistance proportional to the square of the speed. Assuming that the atmospheric resistance is 8 N if the speed is 2 m/s, find the time T required for the mass to reach maximum altitude. Solution The mass will climb while v > 0 and reach its maximum altitude when v = 0. Therefore v > 0 for 0 ≤t < T and v(T) = 0. Although the mass of the object and our assumptions concerning the forces acting on it are the same as those in Example 3, (4.3.10) does not apply here, since the resisting force is negative if v > 0; therefore, we replace (4.3.10) by 10v′ = −98 −2v2. (4.3.15) Separating variables yields 5 v2 + 49v′ = −1, and integrating this yields 5 7 tan−1 v 7 = −t + c. (Recall that tan−1 u is the number θ such that −π/2 < θ < π/2 and tan θ = u.) Since v(0) = v0, c = 5 7 tan−1 v0 7 , so v is defined implicitly by 5 7 tan−1 v 7 = −t + 5 7 tan−1 v0 7 , 0 ≤t ≤T. (4.3.16) | Elementary Differential Equations with Boundary Value Problems_Page_166_Chunk2571 |
Section 4.3 Elementary Mechanics 157 0.2 0.4 0.6 0.8 1 10 20 30 40 50 t v Figure 4.3.4 Solutions of (4.3.15) for various v0 > 0 Solving this for v yields v = 7 tan −7t 5 + tan−1 v0 7 . (4.3.17) Using the identity tan(A −B) = tan A −tan B 1 + tan A tan B with A = tan−1(v0/7) and B = 7t/5, and noting that tan(tan−1 θ) = θ, we can simplify (4.3.17) to v = 7v0 −7 tan(7t/5) 7 + v0 tan(7t/5). Since v(T) = 0 and tan−1(0) = 0, (4.3.16) implies that −T + 5 7 tan−1 v0 7 = 0. Therefore T = 5 7 tan−1 v0 7 . Since tan−1(v0/7) < π/2 for all v0, the time required for the mass to reach its maximum altitude is less than 5π 14 ≈1.122 s regardless of the initial velocity. Figure 4.3.4 shows graphs of v over [0, T] for various values of v0. | Elementary Differential Equations with Boundary Value Problems_Page_167_Chunk2572 |
158 Chapter 4 Applications of First Order Equations y = − R y = 0 y = h y Figure 4.3.5 Escape velocity Escape Velocity Suppose a space vehicle is launched vertically and its fuel is exhausted when the vehicle reaches an altitude h above Earth, where h is sufficiently large so that resistance due to Earth’s atmosphere can be neglected. Let t = 0 be the time when burnout occurs. Assuming that the gravitational forces of all other celestial bodies can be neglected, the motion of the vehicle for t > 0 is that of an object with constant mass m under the influence of Earth’s gravitational force, which we now assume to vary inversely with the square of the distance from Earth’s center; thus, if we take the upward direction to be positive then gravitational force on the vehicle at an altitude y above Earth is F = − K (y + R)2 , (4.3.18) where R is Earth’s radius (Figure 4.3.5). Since F = −mg when y = 0, setting y = 0 in (4.3.18) yields −mg = −K R2 ; therefore K = mgR2 and (4.3.18) can be written more specifically as F = −mgR2 (y + R)2 . (4.3.19) From Newton’s second law of motion, F = md2y dt2 , | Elementary Differential Equations with Boundary Value Problems_Page_168_Chunk2573 |
Section 4.3 Elementary Mechanics 159 so (4.3.19) implies that d2y dt2 = − gR2 (y + R)2 . (4.3.20) We’ll show that there’s a number ve, called the escape velocity, with these properties: 1. If v0 ≥ve then v(t) > 0 for all t > 0, and the vehicle continues to climb for all t > 0; that is, it “escapes” Earth. (Is it really so obvious that limt→∞y(t) = ∞in this case? For a proof, see Exercise 20.) 2. If v0 < ve then v(t) decreases to zero and becomes negative. Therefore the vehicle attains a maximum altitude ym and falls back to Earth. Since (4.3.20) is second order, we can’t solve it by methods discussed so far. However, we’re concerned with v rather than y, and v is easier to find. Since v = y′ the chain rule implies that d2y dt2 = dv dt = dv dy dy dt = v dv dy . Substituting this into (4.3.20) yields the first order separable equation v dv dy = − gR2 (y + R)2 . (4.3.21) When t = 0, the velocity is v0 and the altitude is h. Therefore we can obtain v as a function of y by solving the initial value problem v dv dy = − gR2 (y + R)2 , v(h) = v0. Integrating (4.3.21) with respect to y yields v2 2 = gR2 y + R + c. (4.3.22) Since v(h) = v0, c = v2 0 2 −gR2 h + R, so (4.3.22) becomes v2 2 = gR2 y + R + v2 0 2 −gR2 h + R . (4.3.23) If v0 ≥ 2gR2 h + R 1/2 , the parenthetical expression in (4.3.23) is nonnegative, so v(y) > 0 for y > h. This proves that there’s an escape velocity ve. We’ll now prove that ve = 2gR2 h + R 1/2 by showing that the vehicle falls back to Earth if v0 < 2gR2 h + R 1/2 . (4.3.24) | Elementary Differential Equations with Boundary Value Problems_Page_169_Chunk2574 |
160 Chapter 4 Applications of First Order Equations If (4.3.24) holds then the parenthetical expression in (4.3.23) is negative and the vehicle will attain a maximum altitude ym > h that satisfies the equation 0 = gR2 ym + R + v2 0 2 −gR2 h + R . The velocity will be zero at the maximum altitude, and the object will then fall to Earth under the influence of gravity. 4.3 Exercises Except where directed otherwise, assume that the magnitude of the gravitational force on an object with mass m is constant and equal to mg. In exercises involving vertical motion take the upward direction to be positive. 1. A firefighter who weighs 192 lb slides down an infinitely long fire pole that exerts a frictional resistive force with magnitude proportional to his speed, with k = 2.5 lb-s/ft. Assuming that he starts from rest, find his velocity as a function of time and find his terminal velocity. 2. A firefighter who weighs 192 lb slides down an infinitely long fire pole that exerts a frictional resistive force with magnitude proportional to her speed, with constant of proportionality k. Find k, given that her terminal velocity is -16 ft/s, and then find her velocity v as a function of t. Assume that she starts from rest. 3. A boat weighs 64,000 lb. Its propellor produces a constant thrust of 50,000 lb and the water exerts a resistive force with magnitude proportional to the speed, with k = 2000 lb-s/ft. Assuming that the boat starts from rest, find its velocity as a function of time, and find its terminal velocity. 4. A constant horizontal force of 10 N pushes a 20 kg-mass through a medium that resists its motion with .5 N for every m/s of speed. The initial velocity of the mass is 7 m/s in the direction opposite to the direction of the applied force. Find the velocity of the mass for t > 0. 5. A stone weighing 1/2 lb is thrown upward from an initial height of 5 ft with an initial speed of 32 ft/s. Air resistance is proportional to speed, with k = 1/128 lb-s/ft. Find the maximum height attained by the stone. 6. A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exerts a resistive force with magnitude proportional to the square of the speed, with k = 1 lb-s2/ft2. Also find its terminal velocity. 7. A 96 lb weight is dropped from rest in a medium that exerts a resistive force with magnitude proportional to the speed. Find its velocity as a function of time if its terminal velocity is -128 ft/s. 8. An object with mass m moves vertically through a medium that exerts a resistive force with magni- tude proportional to the speed. Let y = y(t) be the altitude of the object at time t, with y(0) = y0. Use the results of Example 4.3.1 to show that y(t) = y0 + m k (v0 −v −gt). 9. An object with mass m is launched vertically upward with initial velocity v0 from Earth’s surface (y0 = 0) in a medium that exerts a resistive force with magnitude proportional to the speed. Find the time T when the object attains its maximum altitude ym. Then use the result of Exercise 8 to find ym. | Elementary Differential Equations with Boundary Value Problems_Page_170_Chunk2575 |
Section 4.3 Elementary Mechanics 161 10. An object weighing 256 lb is dropped from rest in a medium that exerts a resistive force with magnitude proportional to the square of the speed. The magnitude of the resisting force is 1 lb when |v| = 4 ft/s. Find v for t > 0, and find its terminal velocity. 11. An object with mass m is given an initial velocity v0 ≤0 in a medium that exerts a resistive force with magnitude proportional to the square of the speed. Find the velocity of the object for t > 0, and find its terminal velocity. 12. An object with mass m is launched vertically upward with initial velocity v0 in a medium that exerts a resistive force with magnitude proportional to the square of the speed. (a) Find the time T when the object reaches its maximum altitude. (b) Use the result of Exercise 11 to find the velocity of the object for t > T. 13. L An object with mass m is given an initial velocity v0 ≤0 in a medium that exerts a resistive force of the form a|v|/(1 + |v|), where a is positive constant. (a) Set up a differential equation for the speed of the object. (b) Use your favorite numerical method to solve the equation you found in (a), to convince your- self that there’s a unique number a0 such that limt→∞s(t) = ∞if a ≤a0 and limt→∞s(t) exists (finite) if a > a0. (We say that a0 is the bifurcation value of a.) Try to find a0 and limt→∞s(t) in the case where a > a0. HINT: See Exercise 14. 14. An object of mass m falls in a medium that exerts a resistive force f = f(s), where s = |v| is the speed of the object. Assume that f(0) = 0 and f is strictly increasing and differentiable on (0, ∞). (a) Write a differential equation for the speed s = s(t) of the object. Take it as given that all solutions of this equation with s(0) ≥0 are defined for all t > 0 (which makes good sense on physical grounds). (b) Show that if lims→∞f(s) ≤mg then limt→∞s(t) = ∞. (c) Show that if lims→∞f(s) > mg then limt→∞s(t) = sT (terminal speed), where f(sT ) = mg. HINT: Use Theorem 2.3.1. 15. A 100-g mass with initial velocity v0 ≤0 falls in a medium that exerts a resistive force proportional to the fourth power of the speed. The resistance is .1 N if the speed is 3 m/s. (a) Set up the initial value problem for the velocity v of the mass for t > 0. (b) Use Exercise 14(c) to determine the terminal velocity of the object. (c) C To confirm your answer to (b), use one of the numerical methods studied in Chapter 3 to compute approximate solutions on [0, 1] (seconds) of the initial value problem of (a), with initial values v0 = 0, −2, −4, ..., −12. Present your results in graphical form similar to Figure 4.3.3. 16. A 64-lb object with initial velocity v0 ≤0 falls through a dense fluid that exerts a resistive force proportional to the square root of the speed. The resistance is 64 lb if the speed is 16 ft/s. (a) Set up the initial value problem for the velocity v of the mass for t > 0. (b) Use Exercise 14(c) to determine the terminal velocity of the object. (c) C To confirm your answer to (b), use one of the numerical methods studied in Chapter 3 to compute approximate solutions on [0, 4] (seconds) of the initial value problem of (a), with initial values v0 = 0, −5, −10, ..., −30. Present your results in graphical form similar to Figure 4.3.3. In Exercises 17-20, assume that the force due to gravity is given by Newton’s law of gravitation. Take the upward direction to be positive. | Elementary Differential Equations with Boundary Value Problems_Page_171_Chunk2576 |
162 Chapter 4 Applications of First Order Equations 17. A space probe is to be launched from a space station 200 miles above Earth. Determine its escape velocity in miles/s. Take Earth’s radius to be 3960 miles. 18. A space vehicle is to be launched from the moon, which has a radius of about 1080 miles. The acceleration due to gravity at the surface of the moon is about 5.31 ft/s2. Find the escape velocity in miles/s. 19. (a) Show that Eqn. (4.3.23) can be rewritten as v2 = h −y y + Rv2 e + v2 0. (b) Show that if v0 = ρve with 0 ≤ρ < 1, then the maximum altitude ym attained by the space vehicle is ym = h + Rρ2 1 −ρ2 . (c) By requiring that v(ym) = 0, use Eqn. (4.3.22) to deduce that if v0 < ve then |v| = ve (1 −ρ2)(ym −y) y + R 1/2 , where ym and ρ are as defined in (b) and y ≥h. (d) Deduce from (c) that if v < ve, the vehicle takes equal times to climb from y = h to y = ym and to fall back from y = ym to y = h. 20. In the situation considered in the discussion of escape velocity, show that limt→∞y(t) = ∞if v(t) > 0 for all t > 0. HINT: Use a proof by contradiction. Assume that there’s a number ym such that y(t) ≤ym for all t > 0. Deduce from this that there’s positive number α such that y′′(t) ≤−α for all t ≥0. Show that this contradicts the assumption that v(t) > 0 for all t > 0. 4.4 AUTONOMOUS SECOND ORDER EQUATIONS A second order differential equation that can be written as y′′ = F (y, y′) (4.4.1) where F is independent of t, is said to be autonomous. An autonomous second order equation can be converted into a first order equation relating v = y′ and y. If we let v = y′, (4.4.1) becomes v′ = F (y, v). (4.4.2) Since v′ = dv dt = dv dy dy dt = v dv dy , (4.4.3) (4.4.2) can be rewritten as v dv dy = F (y, v). (4.4.4) The integral curves of (4.4.4) can be plotted in the (y, v) plane, which is called the Poincaré phase plane of (4.4.1). If y is a solution of (4.4.1) then y = y(t), v = y′(t) is a parametric equation for an integral | Elementary Differential Equations with Boundary Value Problems_Page_172_Chunk2577 |
Section 4.4 Autonomous Second Order Equations 163 curve of (4.4.4). We’ll call these integral curves trajectories of (4.4.1), and we’ll call (4.4.4) the phase plane equivalent of (4.4.1). In this section we’ll consider autonomous equations that can be written as y′′ + q(y, y′)y′ + p(y) = 0. (4.4.5) Equations of this form often arise in applications of Newton’s second law of motion. For example, suppose y is the displacement of a moving object with mass m. It’s reasonable to think of two types of time-independent forces acting on the object. One type - such as gravity - depends only on position. We could write such a force as −mp(y). The second type - such as atmospheric resistance or friction - may depend on position and velocity. (Forces that depend on velocity are called damping forces.) We write this force as −mq(y, y′)y′, where q(y, y′) is usually a positive function and we’ve put the factor y′ outside to make it explicit that the force is in the direction opposing the motion. In this case Newton’s, second law of motion leads to (4.4.5). The phase plane equivalent of (4.4.5) is v dv dy + q(y, v)v + p(y) = 0. (4.4.6) Some statements that we’ll be making about the properties of (4.4.5) and (4.4.6) are intuitively reasonable, but difficult to prove. Therefore our presentation in this section will be informal: we’ll just say things without proof, all of which are true if we assume that p = p(y) is continuously differentiable for all y and q = q(y, v) is continuously differentiable for all (y, v). We begin with the following statements: • Statement 1. If y0 and v0 are arbitrary real numbers then (4.4.5) has a unique solution on (−∞, ∞) such that y(0) = y0 and y′(0) = v0. • Statement 2.) If y = y(t) is a solution of (4.4.5) and τ is any constant then y1 = y(t −τ) is also a solution of (4.4.5), and y and y1 have the same trajectory. • Statement 3. If two solutions y and y1 of (4.4.5) have the same trajectory then y1(t) = y(t −τ) for some constant τ. • Statement 4. Distinct trajectories of (4.4.5) can’t intersect; that is, if two trajectories of (4.4.5) intersect, they are identical. • Statement 5. If the trajectory of a solution of (4.4.5) is a closed curve then (y(t), v(t)) traverses the trajectory in a finite time T, and the solution is periodic with period T; that is, y(t + T) = y(t) for all t in (−∞, ∞). If y is a constant such that p(y) = 0 then y ≡y is a constant solution of (4.4.5). We say that y is an equilibrium of (4.4.5) and (y, 0) is a critical point of the phase plane equivalent equation (4.4.6). We say that the equilibrium and the critical point are stable if, for any given ϵ > 0 no matter how small, there’s a δ > 0, sufficiently small, such that if q (y0 −y)2 + v2 0 < δ then the solution of the initial value problem y′′ + q(y, y′)y′ + p(y) = 0, y(0) = y0, y′(0) = v0 satisfies the inequality p (y(t) −y)2 + (v(t))2 < ϵ | Elementary Differential Equations with Boundary Value Problems_Page_173_Chunk2578 |
164 Chapter 4 Applications of First Order Equations for all t > 0. Figure 4.4.1 illustrates the geometrical interpretation of this definition in the Poincaré phase plane: if (y0, v0) is in the smaller shaded circle (with radius δ), then (y(t), v(t)) must be in in the larger circle (with radius ϵ) for all t > 0. y y v ε δ Figure 4.4.1 Stability: if (y0, v0) is in the smaller circle then (y(t), v(t)) is in the larger circle for all t > 0 If an equilibrium and the associated critical point are not stable, we say they are unstable. To see if you really understand what stable means, try to give a direct definition of unstable (Exercise 22). We’ll illustrate both definitions in the following examples. The Undamped Case We’ll begin with the case where q ≡0, so (4.4.5) reduces to y′′ + p(y) = 0. (4.4.7) We say that this equation - as well as any physical situation that it may model - is undamped. The phase plane equivalent of (4.4.7) is the separable equation v dv dy + p(y) = 0. Integrating this yields v2 2 + P (y) = c, (4.4.8) where c is a constant of integration and P (y) = R p(y) dy is an antiderivative of p. If (4.4.7) is the equation of motion of an object of mass m, then mv2/2 is the kinetic energy and mP (y) is the potential energy of the object; thus, (4.4.8) says that the total energy of the object remains | Elementary Differential Equations with Boundary Value Problems_Page_174_Chunk2579 |
Section 4.4 Autonomous Second Order Equations 165 constant, or is conserved. In particular, if a trajectory passes through a given point (y0, v0) then c = v2 0 2 + P (y0). Example 4.4.1 [The Undamped Spring - Mass System] Consider an object with mass m suspended from a spring and moving vertically. Let y be the displacement of the object from the position it occupies when suspended at rest from the spring (Figure 4.4.2). y (a) 0 (b) (c) Figure 4.4.2 (a) y > 0 (b) y = 0 (c) y < 0 Assume that if the length of the spring is changed by an amount ∆L (positive or negative), then the spring exerts an opposing force with magnitude k|∆L|, where k is a positive constant. In Section 6.1 it will be shown that if the mass of the spring is negligible compared to m and no other forces act on the object then Newton’s second law of motion implies that my′′ = −ky, (4.4.9) which can be written in the form (4.4.7) with p(y) = ky/m. This equation can be solved easily by a method that we’ll study in Section 5.2, but that method isn’t available here. Instead, we’ll consider the phase plane equivalent of (4.4.9). From (4.4.3), we can rewrite (4.4.9) as the separable equation mv dv dy = −ky. Integrating this yields mv2 2 = −ky2 2 + c, | Elementary Differential Equations with Boundary Value Problems_Page_175_Chunk2580 |
166 Chapter 4 Applications of First Order Equations y v Figure 4.4.3 Trajectories of my′′ + ky = 0 which implies that mv2 + ky2 = ρ (4.4.10) (ρ = 2c). This defines an ellipse in the Poincaré phase plane (Figure 4.4.3). We can identify ρ by setting t = 0 in (4.4.10); thus, ρ = mv2 0 + ky2 0, where y0 = y(0) and v0 = v(0). To determine the maximum and minimum values of y we set v = 0 in (4.4.10); thus, ymax = R and ymin = −R, with R = rρ k . (4.4.11) Equation (4.4.9) has exactly one equilibrium, y = 0, and it’s stable. You can see intuitively why this is so: if the object is displaced in either direction from equilibrium, the spring tries to bring it back. In this case we can find y explicitly as a function of t. (Don’t expect this to happen in more complicated problems!) If v > 0 on an interval I, (4.4.10) implies that dy dt = v = r ρ −ky2 m on I. This is equivalent to √ k p ρ −ky2 dy dt = ω0, where ω0 = r k m. (4.4.12) Since Z √ k dy p ρ −ky2 = sin−1 s k ρy ! + c = sin−1 y R + c | Elementary Differential Equations with Boundary Value Problems_Page_176_Chunk2581 |
Section 4.4 Autonomous Second Order Equations 167 y = R y = − R t y Figure 4.4.4 y = R sin(ω0t + φ) (see (4.4.11)), (4.4.12) implies that that there’s a constant φ such that sin−1 y R = ω0t + φ or y = R sin(ω0t + φ) for all t in I. Although we obtained this function by assuming that v > 0, you can easily verify that y satisfies (4.4.9) for all values of t. Thus, the displacement varies periodically between −R and R, with period T = 2π/ω0 (Figure 4.4.4). (If you’ve taken a course in elementary mechanics you may recognize this as simple harmonic motion.) Example 4.4.2 [The Undamped Pendulum] Now we consider the motion of a pendulum with mass m, attached to the end of a weightless rod with length L that rotates on a frictionless axle (Figure 4.4.5). We assume that there’s no air resistance. Let y be the angle measured from the rest position (vertically downward) of the pendulum, as shown in Figure 4.4.5. Newton’s second law of motion says that the product of m and the tangential acceleration equals the tangential component of the gravitational force; therefore, from Figure 4.4.5, mLy′′ = −mg sin y, or y′′ = −g L sin y. (4.4.13) Since sin nπ = 0 if n is any integer, (4.4.13) has infinitely many equilibria yn = nπ. If n is even, the mass is directly below the axle (Figure 4.4.6 (a)) and gravity opposes any deviation from the equilibrium. | Elementary Differential Equations with Boundary Value Problems_Page_177_Chunk2582 |
168 Chapter 4 Applications of First Order Equations m y L Figure 4.4.5 The undamped pendulum (a) Stable equilibrium (b) Unstable equilibrium Figure 4.4.6 (a) Stable equilibrium (b) Unstable equilibrium However, if n is odd, the mass is directly above the axle (Figure 4.4.6 (b)) and gravity increases any deviation from the equilibrium. Therefore we conclude on physical grounds that y2m = 2mπ is stable and y2m+1 = (2m + 1)π is unstable. The phase plane equivalent of (4.4.13) is v dv dy = −g L sin y, where v = y′ is the angular velocity of the pendulum. Integrating this yields v2 2 = g L cos y + c. (4.4.14) If v = v0 when y = 0, then c = v2 0 2 −g L, so (4.4.14) becomes v2 2 = v2 0 2 −g L(1 −cos y) = v2 0 2 −2g L sin2 y 2, which is equivalent to v2 = v2 0 −v2 c sin2 y 2, (4.4.15) where vc = 2 r g L. | Elementary Differential Equations with Boundary Value Problems_Page_178_Chunk2583 |
Section 4.4 Autonomous Second Order Equations 169 The curves defined by (4.4.15) are the trajectories of (4.4.13). They are periodic with period 2π in y, which isn’t surprising, since if y = y(t) is a solution of (4.4.13) then so is yn = y(t) + 2nπ for any integer n. Figure 4.4.7 shows trajectories over the interval [−π, π]. From (4.4.15), you can see that if |v0| > vc then v is nonzero for all t, which means that the object whirls in the same direction forever, as in Figure 4.4.8. The trajectories associated with this whirling motion are above the upper dashed curve and below the lower dashed curve in Figure 4.4.7. You can also see from (4.4.15) that if 0 < |v0| < vc,then v = 0 when y = ±ymax, where ymax = 2 sin−1(|v0|/vc). In this case the pendulum oscillates periodically between −ymax and ymax, as shown in Figure 4.4.9. The trajectories associated with this kind of motion are the ovals between the dashed curves in Figure 4.4.7. It can be shown (see Exercise 21 for a partial proof) that the period of the oscillation is T = 8 Z π/2 0 dθ q v2c −v2 0 sin2 θ . (4.4.16) Although this integral can’t be evaluated in terms of familiar elementary functions, you can see that it’s finite if |v0| < vc. The dashed curves in Figure 4.4.7 contain four trajectories. The critical points (π, 0) and (−π, 0) are the trajectories of the unstable equilibrium solutions y = ±π. The upper dashed curve connecting (but not including) them is obtained from initial conditions of the form y(t0) = 0, v(t0) = vc. If y is any solution with this trajectory then lim t→∞y(t) = π and lim t→−∞y(t) = −π. The lower dashed curve connecting (but not including) them is obtained from initial conditions of the form y(t0) = 0, v(t0) = −vc. If y is any solution with this trajectory then lim t→∞y(t) = −π and lim t→−∞y(t) = π. Consistent with this, the integral (4.4.16) diverges to ∞if v0 = ±vc. (Exercise 21) . Since the dashed curves separate trajectories of whirling solutions from trajectories of oscillating solu- tions, each of these curves is called a separatrix. In general, if (4.4.7) has both stable and unstable equilibria then the separatrices are the curves given by (4.4.8) that pass through unstable critical points. Thus, if (y, 0) is an unstable critical point, then v2 2 + P (y) = P (y) (4.4.17) defines a separatrix passing through (y, 0). | Elementary Differential Equations with Boundary Value Problems_Page_179_Chunk2584 |
170 Chapter 4 Applications of First Order Equations π − π x y Figure 4.4.7 Trajectories of the undamped pendulum Figure 4.4.8 The whirling undamped pendulum ymax − ymax Figure 4.4.9 The oscillating undamped pendulum Stability and Instability Conditions for y′′ + p(y) = 0 It can be shown (Exercise 23) that an equilibrium y of an undamped equation y′′ + p(y) = 0 (4.4.18) is stable if there’s an open interval (a, b) containing y such that p(y) < 0 if a < y < y and p(y) > 0 if y < y < b. (4.4.19) If we regard p(y) as a force acting on a unit mass, (4.4.19) means that the force resists all sufficiently small displacements from y. We’ve already seen examples illustrating this principle. The equation (4.4.9) for the undamped spring- mass system is of the form (4.4.18) with p(y) = ky/m, which has only the stable equilibrium y = 0. In this case (4.4.19) holds with a = −∞and b = ∞. The equation (4.4.13) for the undamped pendulum is of the form (4.4.18) with p(y) = (g/L) sin y. We’ve seen that y = 2mπ is a stable equilibrium if m is an integer. In this case p(y) = sin y < 0 if (2m −1)π < y < 2mπ and p(y) > 0 if 2mπ < y < (2m + 1)π. | Elementary Differential Equations with Boundary Value Problems_Page_180_Chunk2585 |
Section 4.4 Autonomous Second Order Equations 171 It can also be shown (Exercise 24) that y is unstable if there’s a b > y such that p(y) < 0 if y < y < b (4.4.20) or an a < y such that p(y) > 0 if a < y < y. (4.4.21) If we regard p(y) as a force acting on a unit mass, (4.4.20) means that the force tends to increase all sufficiently small positive displacements from y, while (4.4.21) means that the force tends to increase the magnitude of all sufficiently small negative displacements from y. The undamped pendulum also illustrates this principle. We’ve seen that y = (2m + 1)π is an unstable equilibrium if m is an integer. In this case sin y < 0 if (2m + 1)π < y < (2m + 2)π, so (4.4.20) holds with b = (2m + 2)π, and sin y > 0 if 2mπ < y < (2m + 1)π, so (4.4.21) holds with a = 2mπ. Example 4.4.3 The equation y′′ + y(y −1) = 0 (4.4.22) is of the form (4.4.18) with p(y) = y(y −1). Therefore y = 0 and y = 1 are the equilibria of (4.4.22). Since y(y −1) > 0 if y < 0 or y > 1, < 0 if 0 < y < 1, y = 0 is unstable and y = 1 is stable. The phase plane equivalent of (4.4.22) is the separable equation v dv dy + y(y −1) = 0. Integrating yields v2 2 + y3 3 −y2 2 = C, which we rewrite as v2 + 1 3y2(2y −3) = c (4.4.23) after renaming the constant of integration. These are the trajectories of (4.4.22). If y is any solution of (4.4.22), the point (y(t), v(t)) moves along the trajectory of y in the direction of increasing y in the upper half plane (v = y′ > 0), or in the direction of decreasing y in the lower half plane (v = y′ < 0). Figure 4.4.10 shows typical trajectories. The dashed curve through the critical point (0, 0), obtained by setting c = 0 in (4.4.23), separates the y-v plane into regions that contain different kinds of trajectories; again, we call this curve a separatrix. Trajectories in the region bounded by the closed loop (b) are closed curves, so solutions associated with them are periodic. Solutions associated with other trajectories are not periodic. If y is any such solution with trajectory not on the separatrix, then lim t→∞y(t) = −∞, lim t→−∞y(t) = −∞, lim t→∞v(t) = −∞, lim t→−∞v(t) = ∞. | Elementary Differential Equations with Boundary Value Problems_Page_181_Chunk2586 |
172 Chapter 4 Applications of First Order Equations 1 y v (a) (b) (c) (b) Figure 4.4.10 Trajectories of y′′ + y(y −1) = 0 The separatrix contains four trajectories of (4.4.22). One is the point (0, 0), the trajectory of the equi- librium y = 0. Since distinct trajectories can’t intersect, the segments of the separatrix marked (a), (b), and (c) – which don’t include (0, 0) – are distinct trajectories, none of which can be traversed in finite time. Solutions with these trajectories have the following asymptotic behavior: lim t→∞y(t) = 0, lim t→−∞y(t) = −∞, lim t→∞v(t) = 0, lim t→−∞v(t) = ∞ (on (a)) lim t→∞y(t) = 0, lim t→−∞y(t) = 0, lim t→∞v(t) = 0, lim t→−∞v(t) = 0 (on (b)) lim t→∞y(t) = −∞, lim t→−∞y(t) = 0, lim t→∞v(t) = −∞, lim t→−∞v(t) = 0 (on (c)). . The Damped Case The phase plane equivalent of the damped autonomous equation y′′ + q(y, y′)y′ + p(y) = 0 (4.4.24) is v dv dy + q(y, v)v + p(y) = 0. This equation isn’t separable, so we can’t solve it for v in terms of y, as we did in the undamped case, and conservation of energy doesn’t hold. (For example, energy expended in overcoming friction is lost.) However, we can study the qualitative behavior of its solutions by rewriting it as dv dy = −q(y, v) −p(y) v (4.4.25) | Elementary Differential Equations with Boundary Value Problems_Page_182_Chunk2587 |
Section 4.4 Autonomous Second Order Equations 173 and considering the direction fields for this equation. In the following examples we’ll also be showing computer generated trajectories of this equation, obtained by numerical methods. The exercises call for similar computations. The methods discussed in Chapter 3 are not suitable for this task, since p(y)/v in (4.4.25) is undefined on the y axis of the Poincaré phase plane. Therefore we’re forced to apply numerical methods briefly discussed in Section 10.1 to the system y′ = v v′ = −q(y, v)v −p(y), which is equivalent to (4.4.24) in the sense defined in Section 10.1. Fortunately, most differential equation software packages enable you to do this painlessly. In the text we’ll confine ourselves to the case where q is constant, so (4.4.24) and (4.4.25) reduce to y′′ + cy′ + p(y) = 0 (4.4.26) and dv dy = −c −p(y) v . (We’ll consider more general equations in the exercises.) The constant c is called the damping constant. In situations where (4.4.26) is the equation of motion of an object, c is positive; however, there are situations where c may be negative. The Damped Spring-Mass System Earlier we considered the spring - mass system under the assumption that the only forces acting on the object were gravity and the spring’s resistance to changes in its length. Now we’ll assume that some mechanism (for example, friction in the spring or atmospheric resistance) opposes the motion of the object with a force proportional to its velocity. In Section 6.1 it will be shown that in this case Newton’s second law of motion implies that my′′ + cy′ + ky = 0, (4.4.27) where c > 0 is the damping constant. Again, this equation can be solved easily by a method that we’ll study in Section 5.2, but that method isn’t available here. Instead, we’ll consider its phase plane equivalent, which can be written in the form (4.4.25) as dv dy = −c m −ky mv . (4.4.28) (A minor note: the c in (4.4.26) actually corresponds to c/m in this equation.) Figure 4.4.11 shows a typical direction field for an equation of this form. Recalling that motion along a trajectory must be in the direction of increasing y in the upper half plane (v > 0) and in the direction of decreasing y in the lower half plane (v < 0), you can infer that all trajectories approach the origin in clockwise fashion. To confirm this, Figure 4.4.12 shows the same direction field with some trajectories filled in. All the trajectories shown there correspond to solutions of the initial value problem my′′ + cy′ + ky = 0, y(0) = y0, y′(0) = v0, where mv2 0 + ky2 0 = ρ (a positive constant); thus, if there were no damping (c = 0), all the solutions would have the same dashed elliptic trajectory, shown in Figure 4.4.14. | Elementary Differential Equations with Boundary Value Problems_Page_183_Chunk2588 |
174 Chapter 4 Applications of First Order Equations −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 v y Figure 4.4.11 A typical direction field for my′′ + cy′ + ky = 0 with 0 < c < c1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 v y Figure 4.4.12 Figure 4.4.11 with some trajectories added Solutions corresponding to the trajectories in Figure 4.4.12 cross the y-axis infinitely many times. The corresponding solutions are said to be oscillatory (Figure 4.4.13) It is shown in Section 6.2 that there’s a number c1 such that if 0 ≤c < c1 then all solutions of (4.4.27) are oscillatory, while if c ≥c1, no solutions of (4.4.27) have this property. (In fact, no solution not identically zero can have more than two zeros in this case.) Figure 4.4.14 shows a direction field and some integral curves for (4.4.28) in this case. t y Figure 4.4.13 An oscillatory solution of my′′ + cy′ + ky = 0 Example 4.4.4 (The Damped Pendulum) Now we return to the pendulum. If we assume that some mechanism (for example, friction in the axle or atmospheric resistance) opposes the motion of the pen- dulum with a force proportional to its angular velocity, Newton’s second law of motion implies that mLy′′ = −cy′ −mg sin y, (4.4.29) where c > 0 is the damping constant. (Again, a minor note: the c in (4.4.26) actually corresponds to | Elementary Differential Equations with Boundary Value Problems_Page_184_Chunk2589 |
Section 4.4 Autonomous Second Order Equations 175 c/mL in this equation.) To plot a direction field for (4.4.29) we write its phase plane equivalent as dv dy = −c mL −g Lv sin y. Figure 4.4.15 shows trajectories of four solutions of (4.4.29), all satisfying y(0) = 0. For each m = 0, 1, 2, 3, imparting the initial velocity v(0) = vm causes the pendulum to make m complete revolutions and then settle into decaying oscillation about the stable equilibrium y = 2mπ. −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 v y Figure 4.4.14 A typical direction field for my′′ + cy′ + ky = 0 with c > c1 y v v3 v2 v1 v0 y = 2π y = 4π y = 6π Figure 4.4.15 Four trajectories of the damped pendulum | Elementary Differential Equations with Boundary Value Problems_Page_185_Chunk2590 |
176 Chapter 4 Applications of First Order Equations 4.4 Exercises In Exercises 1–4 find the equations of the trajectories of the given undamped equation. Identify the equilibrium solutions, determine whether they are stable or unstable, and plot some trajectories. HINT: Use Eqn. (4.4.8) to obtain the equations of the trajectories. 1. C/G y′′ + y3 = 0 2. C/G y′′ + y2 = 0 3. C/G y′′ + y|y| = 0 4. C/G y′′ + ye−y = 0 In Exercises 5–8 find the equations of the trajectories of the given undamped equation. Identify the equilibrium solutions, determine whether they are stable or unstable, and find the equations of the sepa- ratrices (that is, the curves through the unstable equilibria). Plot the separatrices and some trajectories in each of the regions of Poincaré plane determined by them. HINT: Use Eqn. (4.4.17) to determine the separatrices. 5. C/G y′′ −y3 + 4y = 0 6. C/G y′′ + y3 −4y = 0 7. C/G y′′ + y(y2 −1)(y2 −4) = 0 8. C/G y′′ + y(y −2)(y −1)(y + 2) = 0 In Exercises 9–12 plot some trajectories of the given equation for various values (positive, negative, zero) of the parameter a. Find the equilibria of the equation and classify them as stable or unstable. Explain why the phase plane plots corresponding to positive and negative values of a differ so markedly. Can you think of a reason why zero deserves to be called the critical value of a? 9. L y′′ + y2 −a = 0 10. L y′′ + y3 −ay = 0 11. L y′′ −y3 + ay = 0 12. L y′′ + y −ay3 = 0 In Exercises 13-18 plot trajectories of the given equation for c = 0 and small nonzero (positive and negative) values of c to observe the effects of damping. 13. L y′′ + cy′ + y3 = 0 14. L y′′ + cy′ −y = 0 15. L y′′ + cy′ + y3 = 0 16. L y′′ + cy′ + y2 = 0 17. L y′′ + cy′ + y|y| = 0 18. L y′′ + y(y −1) + cy = 0 19. L The van der Pol equation y′′ −µ(1 −y2)y′ + y = 0, (A) where µ is a positive constant and y is electrical current (Section 6.3), arises in the study of an electrical circuit whose resistive properties depend upon the current. The damping term −µ(1 −y2)y′ works to reduce |y| if |y| < 1 or to increase |y| if |y| > 1. It can be shown that van der Pol’s equation has exactly one closed trajectory, which is called a limit cycle. Trajectories | Elementary Differential Equations with Boundary Value Problems_Page_186_Chunk2591 |
Section 4.4 Autonomous Second Order Equations 177 inside the limit cycle spiral outward to it, while trajectories outside the limit cycle spiral inward to it (Figure 4.4.16). Use your favorite differential equations software to verify this for µ = .5, 1.1.5, 2. Use a grid with −4 < y < 4 and −4 < v < 4. y v Figure 4.4.16 Trajectories of van der Pol’s equation 20. L Rayleigh’s equation, y′′ −µ(1 −(y′)2/3)y′ + y = 0 also has a limit cycle. Follow the directions of Exercise 19 for this equation. 21. In connection with Eqn (4.4.15), suppose y(0) = 0 and y′(0) = v0, where 0 < v0 < vc. (a) Let T1 be the time required for y to increase from zero to ymax = 2 sin−1(v0/vc). Show that dy dt = q v2 0 −v2c sin2 y/2, 0 ≤t < T1. (A) (b) Separate variables in (A) and show that T1 = Z ymax 0 du q v2 0 −v2c sin2 u/2 (B) (c) Substitute sin u/2 = (v0/vc) sinθ in (B) to obtain T1 = 2 Z π/2 0 dθ q v2c −v2 0 sin2 θ . (C) | Elementary Differential Equations with Boundary Value Problems_Page_187_Chunk2592 |
178 Chapter 4 Applications of First Order Equations (d) Conclude from symmetry that the time required for (y(t), v(t)) to traverse the trajectory v2 = v2 0 −v2 c sin2 y/2 is T = 4T1, and that consequently y(t + T) = y(t) and v(t + T) = v(t); that is, the oscillation is periodic with period T. (e) Show that if v0 = vc, the integral in (C) is improper and diverges to ∞. Conclude from this that y(t) < π for all t and limt→∞y(t) = π. 22. Give a direct definition of an unstable equilibrium of y′′ + p(y) = 0. 23. Let p be continuous for all y and p(0) = 0. Suppose there’s a positive number ρ such that p(y) > 0 if 0 < y ≤ρ and p(y) < 0 if −ρ ≤y < 0. For 0 < r ≤ρ let α(r) = min Z r 0 p(x) dx, Z 0 −r |p(x)| dx and β(r) = max Z r 0 p(x) dx, Z 0 −r |p(x)| dx . Let y be the solution of the initial value problem y′′ + p(y) = 0, y(0) = v0, y′(0) = v0, and define c(y0, v0) = v2 0 + 2 R y0 0 p(x) dx. (a) Show that 0 < c(y0, v0) < v2 0 + 2β(|y0|) if 0 < |y0| ≤ρ. (b) Show that v2 + 2 Z y 0 p(x) dx = c(y0, v0), t > 0. (c) Conclude from (b) that if c(y0, v0) < 2α(r) then |y| < r, t > 0. (d) Given ϵ > 0, let δ > 0 be chosen so that δ2 + 2β(δ) < max n ϵ2/2, 2α(ϵ/ √ 2) o . Show that if p y2 0 + v2 0 < δ then p y2 + v2 < ϵ for t > 0, which implies that y = 0 is a stable equilibrium of y′′ + p(y) = 0. (e) Now let p be continuous for all y and p(y) = 0, where y is not necessarily zero. Suppose there’s a positive number ρ such that p(y) > 0 if y < y ≤y + ρ and p(y) < 0 if y −ρ ≤ y < y. Show that y is a stable equilibrium of y′′ + p(y) = 0. 24. Let p be continuous for all y. (a) Suppose p(0) = 0 and there’s a positive number ρ such that p(y) < 0 if 0 < y ≤ρ. Let ϵ be any number such that 0 < ϵ < ρ. Show that if y is the solution of the initial value problem y′′ + p(y) = 0, y(0) = y0, y′(0) = 0 with 0 < y0 < ϵ, then y(t) ≥ϵ for some t > 0. Conclude that y = 0 is an unstable equilibrium of y′′ + p(y) = 0. HINT: Let k = miny0≤x≤ϵ (−p(x)), which is positive. Show that if y(t) < ϵ for 0 ≤t < T then kT 2 < 2(ϵ −y0). (b) Now let p(y) = 0, where y isn’t necessarily zero. Suppose there’s a positive number ρ such that p(y) < 0 if y < y ≤y + ρ. Show that y is an unstable equilibrium of y′′ + p(y) = 0. (c) Modify your proofs of (a) and (b) to show that if there’s a positive number ρ such that p(y) > 0 if y −ρ ≤y < y, then y is an unstable equilibrium of y′′ + p(y) = 0. | Elementary Differential Equations with Boundary Value Problems_Page_188_Chunk2593 |
Section 4.5 Applications to Curves 179 4.5 APPLICATIONS TO CURVES One-Parameter Families of Curves We begin with two examples of families of curves generated by varying a parameter over a set of real numbers. Example 4.5.1 For each value of the parameter c, the equation y −cx2 = 0 (4.5.1) defines a curve in the xy-plane. If c ̸= 0, the curve is a parabola through the origin, opening upward if c > 0 or downward if c < 0. If c = 0, the curve is the x axis (Figure 4.5.1). x y Figure 4.5.1 A family of curves defined by y −cx2 = 0 Example 4.5.2 For each value of the parameter c the equation y = x + c (4.5.2) defines a line with slope 1(Figure 4.5.2). Definition 4.5.1 An equation that can be written in the form H(x, y, c) = 0 (4.5.3) is said to define a one-parameter family of curves if, for each value of c in in some nonempty set of real numbers, the set of points (x, y) that satisfy (4.5.3) forms a curve in the xy-plane. | Elementary Differential Equations with Boundary Value Problems_Page_189_Chunk2594 |
180 Chapter 4 Applications of First Order Equations x y Figure 4.5.2 A family of lines defined by y = x + c x y Figure 4.5.3 A family of circles defined by x2 + y2 −c2 = 0 Equations (4.5.1) and (4.5.2) define one–parameter families of curves. (Although (4.5.2) isn’t in the form (4.5.3), it can be written in this form as y −x −c = 0.) Example 4.5.3 If c > 0, the graph of the equation x2 + y2 −c = 0 (4.5.4) is a circle with center at (0, 0) and radius √c. If c = 0, the graph is the single point (0, 0). (We don’t regard a single point as a curve.) If c < 0, the equation has no graph. Hence, (4.5.4) defines a one– parameter family of curves for positive values of c. This family consists of all circles centered at (0, 0) (Figure 4.5.3). Example 4.5.4 The equation x2 + y2 + c2 = 0 does not define a one-parameter family of curves, since no (x, y) satisfies the equation if c ̸= 0, and only the single point (0, 0) satisfies it if c = 0. Recall from Section 1.2 that the graph of a solution of a differential equation is called an integral curve of the equation. Solving a first order differential equation usually produces a one–parameter family of integral curves of the equation. Here we are interested in the converse problem: given a one–parameter family of curves, is there a first order differential equation for which every member of the family is an integral curve. This suggests the next definition. Definition 4.5.2 If every curve in a one-parameter family defined by the equation H(x, y, c) = 0 (4.5.5) is an integral curve of the first order differential equation F (x, y, y′) = 0, (4.5.6) then (4.5.6) is said to be a differential equation for the family. | Elementary Differential Equations with Boundary Value Problems_Page_190_Chunk2595 |
Section 4.5 Applications to Curves 181 To find a differential equation for a one–parameter family we differentiate its defining equation (4.5.5) implicitly with respect to x, to obtain Hx(x, y, c) + Hy(x, y, c)y′ = 0. (4.5.7) If this equation doesn’t, then it’s a differential equation for the family. If it does contain c, it may be possible to obtain a differential equation for the family by eliminating c between (4.5.5) and (4.5.7). Example 4.5.5 Find a differential equation for the family of curves defined by y = cx2. (4.5.8) Solution Differentiating (4.5.8) with respect to x yields y′ = 2cx. Therefore c = y′/2x, and substituting this into (4.5.8) yields y = xy′ 2 as a differential equation for the family of curves defined by (4.5.8). The graph of any function of the form y = cx2 is an integral curve of this equation. The next example shows that members of a given family of curves may be obtained by joining integral curves for more than one differential equation. Example 4.5.6 (a) Try to find a differential equation for the family of lines tangent to the parabola y = x2. (b) Find two tangent lines to the parabola y = x2 that pass through (2, 3), and find the points of tangency. SOLUTION(a) The equation of the line through a given point (x0, y0) with slope m is y = y0 + m(x −x0). (4.5.9) If (x0, y0) is on the parabola, then y0 = x2 0 and the slope of the tangent line through (x0, x2 0) is m = 2x0; hence, (4.5.9) becomes y = x2 0 + 2x0(x −x0), or, equivalently, y = −x2 0 + 2x0x. (4.5.10) Here x0 plays the role of the constant c in Definition 4.5.1; that is, varying x0 over (−∞, ∞) produces the family of tangent lines to the parabola y = x2. Differentiating (4.5.10) with respect to x yields y′ = 2x0.. We can express x0 in terms of x and y by rewriting (4.5.10) as x2 0 −2x0x + y = 0 and using the quadratic formula to obtain x0 = x ± p x2 −y. (4.5.11) | Elementary Differential Equations with Boundary Value Problems_Page_191_Chunk2596 |
182 Chapter 4 Applications of First Order Equations We must choose the plus sign in (4.5.11) if x < x0 and the minus sign if x > x0; thus, x0 = x + p x2 −y if x < x0 and x0 = x − p x2 −y if x > x0. Since y′ = 2x0, this implies that y′ = 2 x + p x2 −y , if x < x0 (4.5.12) and y′ = 2 x − p x2 −y , if x > x0. (4.5.13) Neither (4.5.12) nor (4.5.13) is a differential equation for the family of tangent lines to the parabola y = x2. However, if each tangent line is regarded as consisting of two tangent half lines joined at the point of tangency, (4.5.12) is a differential equation for the family of tangent half lines on which x is less than the abscissa of the point of tangency (Figure 4.5.4(a)), while (4.5.13) is a differential equation for the family of tangent half lines on which x is greater than this abscissa (Figure 4.5.4(b)). The parabola y = x2 is also an integral curve of both (4.5.12) and (4.5.13). y y x x (a) (b) Figure 4.5.4 SOLUTION(b) From (4.5.10) the point (x, y) = (2, 3) is on the tangent line through (x0, x2 0) if and only if 3 = −x2 0 + 4x0, which is equivalent to x2 0 −4x0 + 3 = (x0 −3)(x0 −1) = 0. | Elementary Differential Equations with Boundary Value Problems_Page_192_Chunk2597 |
Section 4.5 Applications to Curves 183 Letting x0 = 3 in (4.5.10) shows that (2, 3) is on the line y = −9 + 6x, which is tangent to the parabola at (x0, x2 0) = (3, 9), as shown in Figure 4.5.5 Letting x0 = 1 in (4.5.10) shows that (2, 3) is on the line y = −1 + 2x, which is tangent to the parabola at (x0, x2 0) = (1, 1), as shown in Figure 4.5.5. y 1 2 3 4 5 6 7 8 9 10 11 x 1 2 3 y = x2 Figure 4.5.5 Geometric Problems We now consider some geometric problems that can be solved by means of differential equations. Example 4.5.7 Find curves y = y(x) such that every point (x0, y(x0)) on the curve is the midpoint of the line segment with endpoints on the coordinate axes and tangent to the curve at (x0, y(x0)) (Figure 4.5.6). Solution The equation of the line tangent to the curve at P = (x0, y(x0) is y = y(x0) + y′(x0)(x −x0). If we denote the x and y intercepts of the tangent line by xI and yI (Figure 4.5.6), then 0 = y(x0) + y′(x0)(xI −x0) (4.5.14) and yI = y(x0) −y′(x0)x0. (4.5.15) | Elementary Differential Equations with Boundary Value Problems_Page_193_Chunk2598 |
184 Chapter 4 Applications of First Order Equations From Figure 4.5.6, P is the midpoint of the line segment connecting (xI, 0) and (0, yI) if and only if xI = 2x0 and yI = 2y(x0). Substituting the first of these conditions into (4.5.14) or the second into (4.5.15) yields y(x0) + y′(x0)x0 = 0. Since x0 is arbitrary we drop the subscript and conclude that y = y(x) satisfies y + xy′ = 0, which can be rewritten as (xy)′ = 0. Integrating yields xy = c, or y = c x. If c = 0 this curve is the line y = 0, which does not satisfy the geometric requirements imposed by the problem; thus, c ̸= 0, and the solutions define a family of hyperbolas (Figure 4.5.7). x y xI .5 xI yI .5 yI Figure 4.5.6 x y Figure 4.5.7 Example 4.5.8 Find curves y = y(x) such that the tangent line to the curve at any point (x0, y(x0)) intersects the x-axis at (x2 0, 0). Figure 4.5.8 illustrates the situation in the case where the curve is in the first quadrant and 0 < x < 1. Solution The equation of the line tangent to the curve at (x0, y(x0)) is y = y(x0) + y′(x0)(x −x0). Since (x2 0, 0) is on the tangent line, 0 = y(x0) + y′(x0)(x2 0 −x0). Since x0 is arbitrary we drop the subscript and conclude that y = y(x) satisfies y + y′(x2 −x) = 0. | Elementary Differential Equations with Boundary Value Problems_Page_194_Chunk2599 |
Section 4.5 Applications to Curves 185 x0 x0 2 x y Figure 4.5.8 x = 1 x y Figure 4.5.9 Therefor y′ y = − 1 x2 −x = − 1 x(x −1) = 1 x − 1 x −1, so ln|y| = ln|x| −ln|x −1| + k = ln x x −1 + k, and y = cx x −1. If c = 0, the graph of this function is the x-axis. If c ̸= 0, it’s a hyperbola with vertical asymptote x = 1 and horizontal asymptote y = c. Figure 4.5.9 shows the graphs for c ̸= 0. Orthogonal Trajectories Two curves C1 and C2 are said to be orthogonal at a point of intersection (x0, y0) if they have perpen- dicular tangents at (x0, y0). (Figure 4.5.10). A curve is said to be an orthogonal trajectory of a given family of curves if it’s orthogonal to every curve in the family. For example, every line through the origin is an orthogonal trajectory of the family of circles centered at the origin. Conversely, any such circle is an orthogonal trajectory of the family of lines through the origin (Figure 4.5.11). Orthogonal trajectories occur in many physical applications. For example, if u = u(x, y) is the tem- perature at a point (x, y), the curves defined by u(x, y) = c (4.5.16) are called isothermal curves. The orthogonal trajectories of this family are called heat-flow lines, because at any given point the direction of maximum heat flow is perpendicular to the isothermal through the point. If u represents the potential energy of an object moving under a force that depends upon (x, y), the curves (4.5.16) are called equipotentials, and the orthogonal trajectories are called lines of force. From analytic geometry we know that two nonvertical lines L1 and L2 with slopes m1 and m2, re- spectively, are perpendicular if and only if m2 = −1/m1; therefore, the integral curves of the differential equation y′ = − 1 f(x, y) | Elementary Differential Equations with Boundary Value Problems_Page_195_Chunk2600 |
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