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588 Chapter 11 Boundary Value Problems and Fourier Expansions Proof Multiplying (11.2.2) by φn and integrating yields Z b a f(x)φn(x) dx = Z b a φn(x) ∞ X m=1 cmφm(x) ! dx. (11.2.4) It can be shown that the boundedness of the partial sums {fN}∞ N=1 and the integrability of f allow us to interchange the operations of integration and summation on the right of (11.2.4), and rewrite (11.2.4) as Z b a f(x)φn(x) dx = ∞ X m=1 cm Z b a φn(x)φm(x) dx. (11.2.5) (This isn’t easy to prove.) Since Z b a φn(x)φm(x) dx = 0 if m ̸= n, (11.2.5) reduces to Z b a f(x)φn(x) dx = cn Z b a φ2 n(x) dx. Now (11.2.1) implies (11.2.3). Theorem 11.2.1 motivates the next definition. Definition 11.2.2 Suppose φ1, φ2, ..., φn,...are orthogonal on [a, b] and R b a φ2 n(x) dx ̸= 0, n = 1, 2, 3, .... Let f be integrable on [a, b], and define cn = Z b a f(x)φn(x) dx Z b a φ2 n(x) dx , n = 1, 2, 3, . . .. (11.2.6) Then the infinite series P∞ n=1 cnφn(x) is called the Fourier expansion of f in terms of the orthogonal set {φn}∞ n=1, and c1, c2, ..., cn, ...are called the Fourier coefficients of f with respect to {φn}∞ n=1. We indicate the relationship between f and its Fourier expansion by f(x) ∼ ∞ X n=1 cnφn(x), a ≤x ≤b. (11.2.7) You may wonder why we don’t write f(x) = ∞ X n=1 cnφn(x), a ≤x ≤b, rather than (11.2.7). Unfortunately, this isn’t always true. The series on the right may diverge for some or all values of x in [a, b], or it may converge to f(x) for some values of x and not for others. So, for now, we’ll just think of the series as being associated with f because of the definition of the coefficients {cn}, and we’ll indicate this association informally as in (11.2.7). | Elementary Differential Equations with Boundary Value Problems_Page_598_Chunk3001 |
Section 11.2 Fourier Expansions I 589 Fourier Series We’ll now study Fourier expansions in terms of the eigenfunctions 1, cos πx L , sin πx L , cos 2πx L , sin 2πx L , . . ., cos nπx L , sin nπx L , . . .. of Problem 5. If f is integrable on [−L, L], its Fourier expansion in terms of these functions is called the Fourier series of f on [−L, L]. Since Z L −L 12 dx = 2L, Z L −L cos2 nπx L dx = 1 2 Z L −L 1 + cos 2nπx L dx = 1 2 x + L 2nπ sin 2nπx L L −L = L, and Z L −L sin2 nπx L dx = 1 2 Z L −L 1 −cos 2nπx L dx = 1 2 x − L 2nπ sin 2nπx L , L −L = L, we see from (11.2.6) that the Fourier series of f on [−L, L] is a0 + ∞ X n=1 an cos nπx L + bn sin nπx L , where a0 = 1 2L Z L −L f(x) dx, an = 1 L Z L −L f(x) cos nπx L dx, and bn = 1 L Z L −L f(x) sin nπx L dx, n ≥1. Note that a0 is the average value of f on [−L, L], while an and bn (for n ≥1) are twice the average values of f(x) cos nπx L and f(x) sin nπx L on [−L, L], respectively. Convergence of Fourier Series The question of convergence of Fourier series for arbitrary integrable functions is beyond the scope of this book. However, we can state a theorem that settles this question for most functions that arise in applications. Definition 11.2.3 A function f is said to be piecewise smooth on [a, b] if: (a) f has at most finitely many points of discontinuity in (a, b); (b) f′ exists and is continuous except possibly at finitely many points in (a, b); (c) f(x0+) = limx→x0+ f(x) and f′(x0+) = limx→x0+ f′(x) exist if a ≤x0 < b; (d) f(x0−) = limx→x0−f(x) and f′(x0−) = limx→x0−f′(x) exist if a < x0 ≤b. Since f and f′ are required to be continuous at all but finitely many points in [a, b], f(x0+) = f(x0−) and f′(x0+) = f′(x0−) for all but finitely many values of x0 in (a, b). Recall from Section 8.1 that f is said to have a jump discontinuity at x0 if f(x0+) ̸= f(x0−). The next theorem gives sufficient conditions for convergence of a Fourier series. The proof is beyond the scope of this book. | Elementary Differential Equations with Boundary Value Problems_Page_599_Chunk3002 |
590 Chapter 11 Boundary Value Problems and Fourier Expansions Theorem 11.2.4 If f is piecewise smooth on [−L, L], then the Fourier series F (x) = a0 + ∞ X n=1 an cos nπx L + bn sin nπx L (11.2.8) of f on [−L, L] converges for all x in [−L, L]; moreover, F (x) = f(x) if −L < x < L and f is continuous at x f(x−) + f(x+) 2 if −L < x < L and f is discontinuous at x f(−L+) + f(L−) 2 if x = L or x = −L. Since f(x+) = f(x−) if f is continuous at x, we can also say that F (x) = f(x+) + f(x−) 2 if −L < x < L, f(L−) + f(−L+) 2 if x = ±L. Note that F is itself piecewise smooth on [−L, L], and F (x) = f(x) at all points in the open interval (−L, L) where f is continuous. Since the series in (11.2.8) converges to F (x) for all x in [−L, L], you may be tempted to infer that the error EN(x) = F (x) −a0 − N X n=1 an cos nπx L + bn sin nπx L can be made as small as we please for all x in [−L, L] by choosing N sufficiently large. However, this isn’t true if f has a discontinuity somewhere in (−L, L), or if f(−L+) ̸= f(L−). Here’s the situation in this case. If f has a jump discontinuityat a point α in (−L, L), there will be sequences of points {uN} and {vN} in (−L, α) and (α, L), respectively, such that lim N→∞uN = lim N→∞vN = α and EN(uN) ≈.09|f(α−) −f(α+)| and EN(vN) ≈.09|f(α−) −f(α+)|. Thus, the maximum value of the error EN(x) near α does not approach zero as N →∞, but just occurs closer and closer to (and on both sides of ) α, and is essentially independent of N. If f(−L+) ̸= f(L−), then there will be sequences of points {uN} and {vN} in (−L, L) such that lim N→∞uN = −L, lim N→∞vN = L, EN(uN) ≈.09|f(−L+) −f(L−)| and EN(vN) ≈.09|f(−L+) −f(L−)|. This is the Gibbs phenomenon. Having been alerted to it, you may see it in Figures 11.2.2–11.2.4, below; however, we’ll give a specific example at the end of this section. Example 11.2.1 Find the Fourier series of the piecewise smooth function f(x) = −x, −2 < x < 0, 1 2, 0 < x < 2 on [−2, 2] (Figure 11.2.1). Determine the sum of the Fourier series for −2 ≤x ≤2. | Elementary Differential Equations with Boundary Value Problems_Page_600_Chunk3003 |
Section 11.2 Fourier Expansions I 591 1 2 1 2 −1 −2 x y Figure 11.2.1 Solution Note that wen’t bothered to define f(−2), f(0), and f(2). No matter how they may be defined, f is piecewise smooth on [−2, 2], and the coefficients in the Fourier series F (x) = a0 + ∞ X n=1 an cos nπx 2 + bn sin nπx 2 are not affected by them. In any case, Theorem 11.2.4 implies that F (x) = f(x) in (−2, 0) and (0, 2), where f is continuous, while F (−2) = F (2) = f(−2+) + f(2−) 2 = 1 2 2 + 1 2 = 5 4 and F (0) = f(0−) + f(0+) 2 = 1 2 0 + 1 2 = 1 4. To summarize, F (x) = 5 4, x = −2 −x, −2 < x < 0, 1 4, x = 0, 1 2, 0 < x < 2, 5 4, x = 2. We compute the Fourier coefficients as follows: a0 = 1 4 Z 2 −2 f(x) dx = 1 4 Z 0 −2 (−x) dx + Z 2 0 1 2 dx = 3 4. | Elementary Differential Equations with Boundary Value Problems_Page_601_Chunk3004 |
592 Chapter 11 Boundary Value Problems and Fourier Expansions If n ≥1, then an = 1 2 Z 2 −2 f(x) cos nπx 2 dx = 1 2 Z 0 −2 (−x) cos nπx 2 dx + Z 2 0 1 2 cos nπx 2 dx = 2 n2π2 (cos nπ −1), and bn = 1 2 Z 2 −2 f(x) sin nπx 2 dx = 1 2 Z 0 −2 (−x) sin nπx 2 dx + Z 2 0 1 2 sin nπx 2 dx = 1 2nπ(1 + 3 cos nπ). Therefore F (x) = 3 4 + 2 π2 ∞ X n=1 cos nπ −1 n2 cos nπx 2 + 1 2π ∞ X n=1 1 + 3 cos nπ n sin nπx 2 . Figure 11.2.2 shows how the partial sum Fm(x) = 3 4 + 2 π2 m X n=1 cos nπ −1 n2 cos nπx 2 + 1 2π m X n=1 1 + 3 cos nπ n sin nπx 2 approximates f(x) for m = 5 (dotted curve), m = 10 (dashed curve), and m = 15 (solid curve). 1 2 1 2 −1 −2 x y Figure 11.2.2 | Elementary Differential Equations with Boundary Value Problems_Page_602_Chunk3005 |
Section 11.2 Fourier Expansions I 593 Even and Odd Functions Computing the Fourier coefficients of a function f can be tedious; however, the computation can often be simplified by exploiting symmetries in f or some of its terms. To focus on this, we recall some concepts that you studied in calculus. Let u and v be defined on [−L, L] and suppose that u(−x) = u(x) and v(−x) = −v(x), −L ≤x ≤L. Then we say that u is an even function and v is an odd function. Note that: • The product of two even functions is even. • The product of two odd functions is even. • The product of an even function and an odd function is odd. Example 11.2.2 The functions u(x) = cos ωx and u(x) = x2 are even, while v(x) = sin ωx and v(x) = x3 are odd. The function w(x) = ex is neither even nor odd. You learned parts (a) and (b) of the next theorem in calculus, and the other parts follow from them (Exercise 1). Theorem 11.2.5 Suppose u is even and v is odd on [−L, L]. Then: (a) Z L −L u(x) dx = 2 Z L 0 u(x) dx, (b) Z L −L v(x) dx = 0, (c) Z L −L u(x) cos nπx L dx = 2 Z L 0 u(x) cos nπx L dx, (d) Z L −L v(x) sin nπx L dx = 2 Z L 0 v(x) sin nπx L dx, (e) Z L −L u(x) sin nπx L dx = 0 and (f) Z L −L v(x) cos nπx L dx = 0. Example 11.2.3 Find the Fourier series of f(x) = x2 −x on [−2, 2], and determine its sum for −2 ≤ x ≤2. Solution Since L = 2, F (x) = a0 + ∞ X n=1 an cos nπx 2 + bn sin nπx 2 where a0 = 1 4 Z 2 −2 (x2 −x) dx, (11.2.9) an = 1 2 Z 2 −2 (x2 −x) cos nπx 2 dx, n = 1, 2, 3, . . ., (11.2.10) and bn = 1 2 Z 2 −2 (x2 −x) sin nπx 2 dx, n = 1, 2, 3, . . .. (11.2.11) | Elementary Differential Equations with Boundary Value Problems_Page_603_Chunk3006 |
594 Chapter 11 Boundary Value Problems and Fourier Expansions We simplify the evaluation of these integrals by using Theorem 11.2.5 with u(x) = x2 and v(x) = x; thus, from (11.2.9), a0 = 1 2 Z 2 0 x2 dx = x3 6 2 0 = 4 3. From (11.2.10), an = Z 2 0 x2 cos nπx 2 dx = 2 nπ " x2 sin nπx 2 2 0 −2 Z 2 0 x sin nπx 2 dx # = 8 n2π2 " x cos nπx 2 2 0 − Z 2 0 cos nπx 2 dx # = 8 n2π2 " 2 cos nπ −2 nπ sin nπx 2 2 0 # = (−1)n 16 n2π2 . From (11.2.11), bn = − Z 2 0 x sin nπx 2 dx = 2 nπ " x cos nπx 2 2 0 − Z 2 0 cos nπx 2 dx # = 2 nπ " 2 cos nπ −2 nπ sin nπx 2 2 0 # = (−1)n 4 nπ . Therefore F (x) = 4 3 + 16 π2 ∞ X n=1 (−1)n n2 cos nπx 2 + 4 π ∞ X n=1 (−1)n n sin nπx 2 . Theorem 11.2.4 implies that F (x) = 4, x = −2, x2 −x, −2 < x < 2, 4, x = 2. Figure 11.2.3 shows how the partial sum Fm(x) = 4 3 + 16 π2 m X n=1 (−1)n n2 cos nπx 2 + 4 π m X n=1 (−1)n n sin nπx 2 approximates f(x) for m = 5 (dotted curve), m = 10 (dashed curve), and m = 15 (solid curve). Theorem 11.2.5 ilmplies the next theorem follows. Theorem 11.2.6 Suppose f is integrable on [−L, L]. (a) If f is even, the Fourier series of f on [−L, L] is F (x) = a0 + ∞ X n=1 an cos nπx L , where a0 = 1 L Z L 0 f(x) dx and an = 2 L Z L 0 f(x) cos nπx L dx, n ≥1. | Elementary Differential Equations with Boundary Value Problems_Page_604_Chunk3007 |
Section 11.2 Fourier Expansions I 595 1 −1 2 −2 1 2 3 4 5 6 x y Figure 11.2.3 Approximation of f(x) = x2 −x by partial sums of its Fourier series on [−2, 2] (b) If f is odd, the Fourier series of f on [−L, L] is F (x) = ∞ X n=1 bn sin nπx L , where bn = 2 L Z L 0 f(x) sin nπx L dx. Example 11.2.4 Find the Fourier series of f(x) = x on [−π, π], and determine its sum for −π ≤x ≤π. Solution Since f is odd and L = π, F (x) = ∞ X n=1 bn sin nx where bn = 2 π Z π 0 x sin nx dx = −2 nπ x cos nx π 0 − Z π 0 cos nx dx = −2 n cos nπ + 2 n2π sin nx π 0 = (−1)n+1 2 n. Therefore F (x) = −2 ∞ X n=1 (−1)n n sin nx. | Elementary Differential Equations with Boundary Value Problems_Page_605_Chunk3008 |
596 Chapter 11 Boundary Value Problems and Fourier Expansions 1 2 3 −1 −2 −3 1 2 3 −1 −2 −3 x y Figure 11.2.4 Approximation of f(x) = x by partial sums of its Fourier series on [−π, π] Theorem 11.2.4 implies that F (x) = 0, x = −π, x, −π < x < π, 0, x = π. Figure 11.2.4 shows how the partial sum Fm(x) = −2 m X n=1 (−1)n n sin nx approximates f(x) for m = 5 (dotted curve), m = 10 (dashed curve), and m = 15 (solid curve). Example 11.2.5 Find the Fourier series of f(x) = |x| on [−π, π] and determine its sum for −π ≤x ≤π. Solution Since f is even and L = π, F (x) = a0 + ∞ X n=1 an cos nx. Since f(x) = x if x ≥0, a0 = 1 π Z π 0 x dx = x2 2π π 0 = π 2 | Elementary Differential Equations with Boundary Value Problems_Page_606_Chunk3009 |
Section 11.2 Fourier Expansions I 597 and, if n ≥1, an = 2 π Z π 0 x cos nx dx = 2 nπ x sin nx π 0 − Z π 0 sin nx dx = 2 n2π cos nx π 0 = 2 n2π (cos nπ −1) = 2 n2π [(−1)n −1]. Therefore F (x) = π 2 + 2 π X n=0 (−1)n −1 n2 cos nx. (11.2.12) However, since (−1)n −1 = 0 if n = 2m, −2 if n = 2m + 1, the terms in (11.2.12) for which n = 2m are all zeros. Therefore we only to include the terms for which n = 2m + 1; that is, we can rewrite (11.2.12) as F (x) = π 2 −4 π ∞ X m=0 1 (2m + 1)2 cos(2m + 1)x. However, since the name of the index of summation doesn’t matter, we prefer to replace m by n, and write F (x) = π 2 −4 π ∞ X n=0 1 (2n + 1)2 cos(2n + 1)x. Since |x| is continuous for all x and | −π| = |π|, Theorem 11.2.4 implies that F (x) = |x| for all x in [−π, π]. Example 11.2.6 Find the Fourier series of f(x) = x(x2 −L2) on [−L, L], and determine its sum for −L ≤x ≤L. Solution Since f is odd, F (x) = ∞ X n=1 bn sin nπx L , where bn = 2 L Z L 0 x(x2 −L2) sin nπx L dx = −2 nπ " x(x2 −L2) cos nπx L L 0 − Z L 0 (3x2 −L2) cos nπx L dx # = 2L n2π2 " (3x2 −L2) sin nπx L L 0 −6 Z L 0 x sin nπx L dx # = 12L2 n3π3 " x cos nπx L L 0 − Z L 0 cos nπx L dx # = (−1)n 12L3 n3π3 . Therefore F (x) = 12L3 π3 ∞ X n=1 (−1)n n3 sin nπx L . Theorem 11.2.4 implies that F (x) = x(x2 −L2) for all x in [−L, L]. | Elementary Differential Equations with Boundary Value Problems_Page_607_Chunk3010 |
598 Chapter 11 Boundary Value Problems and Fourier Expansions Example 11.2.7 (Gibbs Phenomenon) The Fourier series of f(x) = 0, −1 < x < −1 2, 1, −1 2 < x < 1 2, 0, 1 2 < x < 1 on [−1, 1] is F (x) = 1 2 + 2 π ∞ X n=1 (−1)n−1 2n −1 cos(2n −1)πx. (Verify.) According to Theorem 11.2.4, F (x) = 0, −1 ≤x < −1 2, 1 2, x = −1 2, 1, −1 2 < x < 1 2, 1 2, x = 1 2, 0, 1 2 < x ≤1; thus, F (as well as f) has unit jump discontinuities at x = ± 1 2. Figures 11.2.6-11.2.7 show the graphs of y = f(x) and y = F2N−1(x) = 1 2 + 2 π N X n=1 (−1)n−1 2n −1 cos(2n −1)πx for N = 10, 20, and 30. You can see that although F2N−1 approximates F (and therefore f) well on larger intervals as N increases, the maximum absolute values of the errors remain approximately equal to .09, but occur closer to the discontinuities x = ± 1 2 as N increases. 1 −1 y = 1.00 y = 1.09 y = − 0.09 x y Figure 11.2.5 The Gibbs Phenomenon: Example 11.2.7, N = 10 1 −1 y = 1.00 y = 1.09 y = − 0.09 x y Figure 11.2.6 The Gibbs Phenomenon: Example 11.2.7, N = 20 | Elementary Differential Equations with Boundary Value Problems_Page_608_Chunk3011 |
Section 11.2 Fourier Expansions I 599 1 −1 y = 1.00 y = 1.09 y = − .09 x y Figure 11.2.7 The Gibbs Phenomenon: Example 11.2.7, N = 30 USING TECHNOLOGY The computation of Fourier coefficients will be tedious in many of the exercises in this chapter and the next. To learn the technique, we recommend that you do some exercises in each section “by hand,” perhaps using the table of integrals at the front of the book. However, we encourage you to use your favorite symbolic computation software in the more difficult problems. 11.2 Exercises 1. Prove Theorem 11.1.5. In Exercises 2-16 find the Fourier series of f on [−L, L] and determine its sum for −L ≤x ≤L. Where indicated by C , graph f and Fm(x) = a0 + m X n=1 an cos nπx L + bn sin nπx L on the same axes for various values of m. 2. C L = 1; f(x) = 2 −x 3. L = π; f(x) = 2x −3x2 4. L = 1; f(x) = 1 −3x2 5. L = π; f(x) = | sinx| | Elementary Differential Equations with Boundary Value Problems_Page_609_Chunk3012 |
600 Chapter 11 Boundary Value Problems and Fourier Expansions 6. C L = π; f(x) = x cos x 7. L = π; f(x) = |x| cosx 8. C L = π; f(x) = x sin x 9. L = π; f(x) = |x| sinx 10. L = 1; f(x) = 0, −1 < x < 1 2, cos πx, −1 2 < x < 1 2, 0, 1 2 < x < 1 11. L = 1; f(x) = 0, −1 < x < 1 2, x cos πx, −1 2 < x < 1 2, 0, 1 2 < x < 1 12. L = 1; f(x) = 0, −1 < x < 1 2, sin πx, −1 2 < x < 1 2, 0, 1 2 < x < 1 13. L = 1; f(x) = 0, −1 < x < 1 2, | sinπx|, −1 2 < x < 1 2, 0, 1 2 < x < 1 14. L = 1; f(x) = 0, −1 < x < 1 2, x sin πx, −1 2 < x < 1 2, 0, 1 2 < x < 1 15. C L = 4; f(x) = 0, −4 < x < 0, x, 0 < x < 4 16. C L = 1; f(x) = x2, −1 < x < 0, 1 −x2, 0 < x < 1 17. L Verify the Gibbs phenomenon for f(x) = 2, −2 < x < −1, 1, −1 < x < 1, −1, 1 < x < 2. 18. L Verify the Gibbs phenomenon for f(x) = 2, −3 < x < −2, 3, −2 < x < 2, 1, 2 < x < 3. 19. Deduce from Example 11.2.5 that ∞ X n=0 1 (2n + 1)2 = π2 8 . 20. (a) Find the Fourier series of f(x) = ex on [−π, π]. (b) Deduce from (a) that ∞ X n=0 1 n2 + 1 = π coth π −1 2 . | Elementary Differential Equations with Boundary Value Problems_Page_610_Chunk3013 |
Section 11.2 Fourier Expansions I 601 21. Find the Fourier series of f(x) = (x −π) cos x on [−π, π]. 22. Find the Fourier series of f(x) = (x −π) sin x on [−π, π]. 23. Find the Fourier series of f(x) = sin kx (k ̸= integer) on [−π, π]. 24. Find the Fourier series of f(x) = cos kx (k ̸= integer) on [−π, π]. 25. (a) Suppose g′ is continuous on [a, b] and ω ̸= 0. Use integration by parts to show that there’s a constant M such that Z b a g(x) cos ωx dx ≤M ω and Z b a g(x) sin ωx dx ≤M ω , ω > 0. (b) Show that the conclusion of (a) also holds if g is piecewise smooth on [a, b]. (This is a special case of Riemann’s Lemma. (c) We say that a sequence {αn}∞ n=1 is of order n−k and write αn = O(1/nk) if there’s a constant M such that |αn| < M nk , n = 1, 2, 3, . . .. Let {an}∞ n=1 and {bn}∞ n=1 be the Fourier coefficients of a piecewise smooth function. Con- clude from (b) that an = O(1/n) and bn = O(1/n). 26. (a) Suppose f(−L) = f(L), f′(−L) = f′(L), f′ is continuous, and f′′ is piecewise continuous on [−L, L]. Use Theorem 11.2.4 and integration by parts to show that f(x) = a0 + ∞ X n=1 an cos nπx L + bn sin nπx L , −L ≤x ≤L, with a0 = 1 2L Z L −L f(x) dx, an = − L n2π2 Z L −L f′′(x) cos nπx L dx, and bn = − L n2π2 Z L −L f′′(x) sin nπx L dx, n ≥1. (b) Show that if, in addition to the assumptions in (a), f′′ is continuous and f′′′ is piecewise continuous on [−L, L], then an = L2 n3π3 Z L −L f′′′(x) sin nπx L dx. 27. Show that if f is integrable on [−L, L] and f(x + L) = f(x), −L < x < 0 (Figure 11.2.8), then the Fourier series of f on [−L, L] has the form A0 + ∞ X n=1 An cos 2nπ L + Bn sin 2nπ L where A0 = 1 L Z L 0 f(x) dx, | Elementary Differential Equations with Boundary Value Problems_Page_611_Chunk3014 |
602 Chapter 11 Boundary Value Problems and Fourier Expansions and An = 2 L Z L 0 f(x) cos 2nπx L dx, Bn = 2 L Z L 0 f(x) sin 2nπx L dx, n = 1, 2, 3, . . .. L − L x y Figure 11.2.8 y = f(x), where f(x + L) = f(x), −L < x < 0 − L L x y Figure 11.2.9 y = f(x), where f(x + L) = −f(x), −L < x < 0 28. Show that if f is integrable on [−L, L] and f(x + L) = −f(x), −L < x < 0 (Figure 11.2.9), then the Fourier series of f on [−L, L] has the form ∞ X n=1 An cos (2n −1)πx L + Bn sin (2n −1)πx L , where An = 2 L Z L 0 f(x) cos (2n −1)πx L dx and Bn = 2 L Z L 0 f(x) sin (2n −1)πx L dx, n = 1, 2, 3, . . .. 29. Suppose φ1, φ2, ..., φm are orthogonal on [a, b] and Z b a φ2 n(x) dx ̸= 0, n = 1, 2, . . ., m. If a1, a2, ..., am are arbitrary real numbers, define Pm = a1φ1 + a2φ2 + · · · + amφm. Let Fm = c1φ1 + c2φ2 + · · · + cmφm, where cn = R b a f(x)φn(x) dx R b a φ2n(x) dx ; that is, c1, c2, ..., cm are Fourier coefficients of f. | Elementary Differential Equations with Boundary Value Problems_Page_612_Chunk3015 |
Section 11.3 Fourier Expansions II 603 (a) Show that Z b a (f(x) −Fm(x))φn(x) dx = 0, n = 1, 2, . . ., m. (b) Show that Z b a (f(x) −Fm(x))2 dx ≤ Z b a (f(x) −Pm(x))2 dx, with equality if and only if an = cn, n = 1, 2, . . ., m. (c) Show that Z b a (f(x) −Fm(x))2 dx = Z b a f2(x) dx − m X n=1 c2 n Z b a φ2 n dx. (d) Conclude from (c) that m X n=1 c2 n Z b a φ2 n(x) dx ≤ Z b a f2(x) dx. 30. If A0, A1, ..., Am and B1, B2, ..., Bm are arbitrary constants we say that Pm(x) = A0 + m X n=1 An cos nπx L + Bn sin nπx L is a trigonometric polynomial of degree ≤m. Now let a0 + ∞ X n=1 an cos nπx L + bn sin nπx L be the Fourier series of an integrable function f on [−L, L], and let Fm(x) = a0 + m X n=1 an cos nπx L + bn sin nπx L . (a) Conclude from Exercise 29(b) that Z L −L (f(x) −Fm(x))2 dx ≤ Z L −L (f(x) −Pm(x))2 dx, with equality if and only if An = an, n = 0, 1, ..., m, and Bn = bn, n = 1, 2, ..., m. (b) Conclude from Exercise 29(d) that 2a2 0 + m X n=1 (a2 n + b2 n) ≤1 L Z L −L f2(x) dx for every m ≥0. (c) Conclude from (b) that limn→∞an = limn→∞bn = 0. 11.3 FOURIER EXPANSIONS II | Elementary Differential Equations with Boundary Value Problems_Page_613_Chunk3016 |
604 Chapter 11 Boundary Value Problems and Fourier Expansions In this section we discuss Fourier expansions in terms of the eigenfunctions of Problems 1-4 for Sec- tion 11.1. Fourier Cosine Series From Exercise 11.1.20, the eigenfunctions 1, cos πx L , cos 2πx L , . . ., cos nπx L , . . . of the boundary value problem y′′ + λy = 0, y′(0) = 0, y′(L) = 0 (11.3.1) (Problem 2) are orthogonal on [0, L]. If f is integrable on [0, L] then the Fourier expansion of f in terms of these functions is called the Fourier cosine series of f on [0, L]. This series is a0 + ∞ X n=1 an cos nπx L , where a0 = Z L 0 f(x) dx Z L 0 dx = 1 L Z L 0 f(x) dx and an = Z L 0 f(x) cos nπx L dx Z L 0 cos2 nπx L dx = 2 L Z L 0 f(x) cos nπx L dx, n = 1, 2, 3, . . .. Comparing this definition with Theorem 6(a) shows that the Fourier cosine series of f on [0, L] is the Fourier series of the function f1(x) = f(−x), −L < x < 0, f(x), 0 ≤x ≤L, obtained by extending f over [−L, L] as an even function (Figure 11.3.1). Applying Theorem 11.2.4 to f1 yields the next theorem. Theorem 11.3.1 If f is piecewise smooth on [0, L], then the Fourier cosine series C(x) = a0 + ∞ X n=1 an cos nπx L of f on [0, L], with a0 = 1 L Z L 0 f(x) dx and an = 2 L Z L 0 f(x) cos nπx L dx, n = 1, 2, 3, . . ., converges for all x in [0, L]; moreover, C(x) = f(0+) if x = 0 f(x) if 0 < x < L and f is continuous at x f(x−) + f(x+) 2 if 0 < x < L and f is discontinuous at x f(L−) if x = L. | Elementary Differential Equations with Boundary Value Problems_Page_614_Chunk3017 |
Section 11.3 Fourier Expansions II 605 L − L y = f(x) y = f(−x) x y Figure 11.3.1 Example 11.3.1 Find the Fourier cosine series of f(x) = x on [0, L]. Solution The coefficients are a0 = 1 L Z L 0 x dx = 1 L x2 2 L 0 = L 2 and, if n ≥1 an = 2 L Z L 0 x cos nπx L dx = 2 nπ " x sin nπx L L 0 − Z L 0 sin nπx L dx # = −2 nπ Z L 0 sin nπx L dx = 2L n2π2 cos nπx L L 0 = 2L n2π2 [(−1)n −1] = − 4L (2m −1)2π2 if n = 2m −1, 0 if n = 2m. Therefore C(x) = L 2 −4L π2 ∞ X n=1 1 (2n −1)2 cos (2n −1)πx L . Theorem 11.3.1 implies that C(x) = x, 0 ≤x ≤L. Fourier Sine Series | Elementary Differential Equations with Boundary Value Problems_Page_615_Chunk3018 |
606 Chapter 11 Boundary Value Problems and Fourier Expansions From Exercise 11.1.19, the eigenfunctions sin πx L , sin 2πx L , . . ., sin nπx L , . . . of the boundary value problem y′′ + λy = 0, y(0) = 0, y(L) = 0 (Problem 1) are orthogonal on [0, L]. If f is integrable on [0, L] then the Fourier expansion of f in terms of these functions is called the Fourier sine series of f on [0, L]. This series is ∞ X n=1 bn sin nπx L , where bn = Z L 0 f(x) sin nπx L dx Z L 0 sin2 nπx L dx = 2 L Z L 0 f(x) sin nπx L dx, n = 1, 2, 3, . . .. Comparing this definition with Theorem 6(b) shows that the Fourier sine series of f on [0, L] is the Fourier series of the function f2(x) = −f(−x), −L < x < 0, f(x), 0 ≤x ≤L, obtained by extending f over [−L, L] as an odd function (Figure 11.3.2). Applying Theorem 11.2.4 to f2 yields the next theorem. Theorem 11.3.2 If f is piecewise smooth on [0, L], then the Fourier sine series S(x) = ∞ X n=1 bn sin nπx L of f on [0, L], with bn = 2 L Z L 0 f(x) sin nπx L dx, converges for all x in [0, L]; moreover, S(x) = 0 if x = 0 f(x) if 0 < x < L and f is continuous at x f(x−) + f(x+) 2 if 0 < x < L and f is discontinuous at x 0 if x = L. Example 11.3.2 Find the Fourier sine series of f(x) = x on [0, L]. | Elementary Differential Equations with Boundary Value Problems_Page_616_Chunk3019 |
Section 11.3 Fourier Expansions II 607 L − L y = f(x) y = − f(−x) x y Figure 11.3.2 Solution The coefficients are bn = 2 L Z L 0 x sin nπx L dx = −2 nπ " x cos nπx L L 0 − Z L 0 cos nπx L dx # = (−1)n+1 2L nπ + 2L n2π2 sin nπx L L 0 = (−1)n+1 2L nπ . Therefore S(x) = −2L π ∞ X n=1 (−1)n n sin nπx L . Theorem 11.3.2 implies that S(x) = x, 0 ≤x < L, 0, x = L. Mixed Fourier Cosine Series From Exercise 11.1.22, the eigenfunctions cos πx 2L, cos 3πx 2L , . . ., cos (2n −1)πx 2L , . . . of the boundary value problem y′′ + λy = 0, y′(0) = 0, y(L) = 0 (11.3.2) | Elementary Differential Equations with Boundary Value Problems_Page_617_Chunk3020 |
608 Chapter 11 Boundary Value Problems and Fourier Expansions (Problem 4) are orthogonal on [0, L]. If f is integrable on [0, L] then the Fourier expansion of f in terms of these functions is ∞ X n=1 cn cos (2n −1)πx 2L , where cn = Z L 0 f(x) cos (2n −1)πx 2L dx Z L 0 cos2 (2n −1)πx L dx = 2 L Z L 0 f(x) cos (2n −1)πx 2L dx. We’ll call this expansion the mixed Fourier cosine series of f on [0, L], because the boundary condi- tions of (11.3.2) are “mixed” in that they require y to be zero at one boundary point and y′ to be zero at the other. By contrast, the “ordinary” Fourier cosine series is associated with (11.3.1), where the boundary conditions require that y′ be zero at both endpoints. It can be shown (Exercise 57) that the mixed Fourier cosine series of f on [0, L] is simply the restriction to [0, L] of the Fourier cosine series of f3(x) = f(x), 0 ≤x ≤L, −f(2L −x), L < x ≤2L on [0, 2L] (Figure 11.3.3). y = f(x) L 2L y = − f(2L−x) x y Figure 11.3.3 Applying Theorem 11.3.1 with f replaced by f3 and L replaced by 2L yields the next theorem. | Elementary Differential Equations with Boundary Value Problems_Page_618_Chunk3021 |
Section 11.3 Fourier Expansions II 609 Theorem 11.3.3 If f is piecewise smooth on [0, L], then the mixed Fourier cosine series CM(x) = ∞ X n=1 cn cos (2n −1)πx 2L of f on [0, L], with cn = 2 L Z L 0 f(x) cos (2n −1)πx 2L dx, converges for all x in [0, L]; moreover, CM(x) = f(0+) if x = 0 f(x) if 0 < x < L and f is continuous at x f(x−) + f(x+) 2 if 0 < x < L and f is discontinuous at x 0 if x = L. Example 11.3.3 Find the mixed Fourier cosine series of f(x) = x −L on [0, L]. Solution The coefficients are cn = 2 L Z L 0 (x −L) cos (2n −1)πx 2L dx = 4 (2n −1)π " (x −L) sin (2n −1)πx 2L L 0 − Z L 0 sin (2n −1)πx 2L dx # = 8L (2n −1)2π2 cos (2n −1)πx 2L L 0 = − 8L (2n −1)2π2 . Therefore CM(x) = −8L π2 ∞ X n=1 1 (2n −1)2 cos (2n −1)πx 2L . Theorem 11.3.3 implies that CM(x) = x −L, 0 ≤x ≤L. Mixed Fourier Sine Series From Exercise 11.1.21, the eigenfunctions sin πx 2L, sin 3πx 2L , . . ., sin (2n −1)πx 2L , . . . of the boundary value problem y′′ + λy = 0, y(0) = 0, y′(L) = 0 (Problem 3) are orthogonal on [0, L]. If f is integrable on [0, L], then the Fourier expansion of f in terms of these functions is ∞ X n=1 dn sin (2n −1)πx 2L , | Elementary Differential Equations with Boundary Value Problems_Page_619_Chunk3022 |
610 Chapter 11 Boundary Value Problems and Fourier Expansions where dn = Z L 0 f(x) sin (2n −1)πx 2L dx Z L 0 sin2 (2n −1)πx 2L dx = 2 L Z L 0 f(x) sin (2n −1)πx 2L dx. We’ll call this expansion the mixed Fourier sine series of f on [0, L]. It can be shown (Exercise 58) that the mixed Fourier sine series of f on [0, L] is simply the restriction to [0, L] of the Fourier sine series of f4(x) = f(x), 0 ≤x ≤L, f(2L −x), L < x ≤2L, on [0, 2L] (Figure 11.3.4). L 2L y = f(x) y = f(2L−x) x y Figure 11.3.4 Applying Theorem 11.3.2 with f replaced by f4 and L replaced by 2L yields the next theorem. Theorem 11.3.4 If f is piecewise smooth on [0, L], then the mixed Fourier sine series SM (x) = ∞ X n=1 dn sin (2n −1)πx 2L of f on [0, L], with dn = 2 L Z L 0 f(x) sin (2n −1)πx 2L dx, | Elementary Differential Equations with Boundary Value Problems_Page_620_Chunk3023 |
Section 11.3 Fourier Expansions II 611 converges for all x in [0, L]; moreover, SM (x) = 0 if x = 0 f(x) if 0 < x < L and f is continuous at x f(x−) + f(x+) 2 if 0 < x < L and f is discontinuous at x f(L−) if x = L. Example 11.3.4 Find the mixed Fourier sine series of f(x) = x on [0, L]. Solution The coefficients are dn = 2 L Z L 0 x sin (2n −1)πx 2L dx = − 4 (2n −1)π " x cos (2n −1)πx 2L L 0 − Z L 0 cos (2n −1)πx 2L dx # = 4 (2n −1)π Z L 0 cos (2n −1)πx 2L dx = 8L (2n −1)2π2 sin (2n −1)πx 2L L 0 = (−1)n+1 8L (2n −1)2π2 . Therefore SM (x) = −8L π2 ∞ X n=1 (−1)n (2n −1)2 sin (2n −1)πx 2L . Theorem 11.3.4 implies that SM (x) = x, 0 ≤x ≤L. A Useful Observation In applications involving expansions in terms of the eigenfunctions of Problems 1-4, the functions being expanded are often polynomials that satisfy the boundary conditions of the problem under consideration. In this case the next theorem presents an efficient way to obtain the coefficients in the expansion. Theorem 11.3.5 (a) If f′(0) = f′(L) = 0, f′′ is continuous, and f′′′ is piecewise continuous on [0, L], then f(x) = a0 + ∞ X n=1 an cos nπx L , 0 ≤x ≤L, (11.3.3) with a0 = 1 L Z L 0 f(x) dx and an = 2L2 n3π3 Z L 0 f′′′(x) sin nπx L dx, n ≥1. (11.3.4) Now suppose f′ is continuous and f′′ is piecewise continuous on [0, L]. (b) If f(0) = f(L) = 0, then f(x) = ∞ X n=1 bn sin nπx L , 0 ≤x ≤L, | Elementary Differential Equations with Boundary Value Problems_Page_621_Chunk3024 |
612 Chapter 11 Boundary Value Problems and Fourier Expansions with bn = −2L n2π2 Z L 0 f′′(x) sin nπx L dx. (11.3.5) (c) If f′(0) = f(L) = 0, then f(x) = ∞ X n=1 cn cos (2n −1)πx 2L , 0 ≤x ≤L, with cn = − 8L (2n −1)2π2 Z L 0 f′′(x) cos (2n −1)πx 2L dx. (11.3.6) (d) If f(0) = f′(L) = 0, then f(x) = ∞ X n=1 dn sin (2n −1)πx 2L , 0 ≤x ≤L, with dn = − 8L (2n −1)2π2 Z L 0 f′′(x) sin (2n −1)πx 2L dx. (11.3.7) Proof We’ll prove (a) and leave the rest to you (Exercises 35, 42, and 50). Since f is continuous on [0, L], Theorem 11.3.1 implies (11.3.3) with a0, a1, a2,... as defined in Theorem 11.3.1. We already know that a0 is as in (11.3.4). If n ≥1, integrating twice by parts yields an = 2 L Z L 0 f(x) cos nπx L dx = 2 nπ " f(x) sin nπx L L 0 − Z L 0 f′(x) sin nπx L dx # = −2 nπ Z L 0 f′(x) sin nπx L dx (since sin 0 = sinnπ = 0) = 2L n2π2 " f′(x) cos nπx L L 0 − Z L 0 f′′(x) cos nπx L # dx = −2L n2π2 Z L 0 f′′(x) cos nπx L dx (since f′(0) = f′(L) = 0) = −2L2 n3π3 " f′′(x) sin nπx L L 0 − Z L 0 f′′′(x) sin nπx L dx # = 2L2 n3π3 Z L 0 f′′′(x) sin nπx L dx (since sin 0 = sin nπ = 0). (By an argument similar to one used in the proof of Theorem 8.3.1, the last integration by parts is le- gitimate in the case where f′′′ is undefined at finitely many points in [0, L], so long as it’s piecewise continuous on [0, L].) This completes the proof. Example 11.3.5 Find the Fourier cosine expansion of f(x) = x2(3L −2x) on [0, L]. | Elementary Differential Equations with Boundary Value Problems_Page_622_Chunk3025 |
Section 11.3 Fourier Expansions II 613 Solution Here a0 = 1 L Z L 0 (3Lx2 −2x3) dx = 1 L Lx3 −x4 2 L 0 = L3 2 and an = 2 L Z L 0 (3Lx2 −2x3) cos nπx L dx, n ≥1. Evaluating this integral directly is laborious. However, since f′(x) = 6Lx −6x2, we see that f′(0) = f′(L) = 0. Since f′′′(x) = −12, we see from (11.3.4) that if n ≥1 then an = −24L2 n3π3 Z L 0 sin nπx L dx = 24L3 n4π4 cos nπx L L 0 = 24L3 n4π4 [(−1)n −1] = − 48L3 (2m −1)4π4 if n = 2m −1, 0 if n = 2m. Therefore C(x) = L3 2 −48L3 π4 ∞ X n=1 1 (2n −1)4 cos (2n −1)πx L . Example 11.3.6 Find the Fourier sine expansion of f(x) = x(x2 −3Lx + 2L2) on [0, L]. Solution Since f(0) = f(L) = 0 and f′′(x) = 6(x −L), we see from (11.3.5) that bn = −12L n2π2 Z L 0 (x −L) sin nπx L dx = 12L2 n3π3 " (x −L) cos nπx L L 0 − Z L 0 cos nπx L dx # = 12L2 n3π3 " L −L nπ sin nπx L L 0 # = 12L3 n3π3 . Therefore S(x) = 12L3 π3 ∞ X n=1 1 n3 sin nπx L . Example 11.3.7 Find the mixed Fourier cosine expansion of f(x) = 3x3 −4Lx2 + L3 on [0, L]. Solution Since f′(0) = f(L) = 0 and f′′(x) = 2(9x −4L), we see from (11.3.6) that cn = − 16L (2n −1)2π2 Z L 0 (9x −4L) cos (2n −1)πx 2L dx = − 32L2 (2n −1)3π3 " (9x −4L) sin (2n −1)πx 2L L 0 −9 Z L 0 sin (2n −1)πx 2L # dx = − 32L2 (2n −1)3π3 " (−1)n+15L + 18L (2n −1)π cos (2n −1)πx 2L L 0 # = 32L3 (2n −1)3π3 (−1)n5 + 18 (2n −1)π . | Elementary Differential Equations with Boundary Value Problems_Page_623_Chunk3026 |
614 Chapter 11 Boundary Value Problems and Fourier Expansions Therefore CM(x) = 32L3 π3 ∞ X n=1 1 (2n −1)3 (−1)n5 + 18 (2n −1)π cos (2n −1)πx 2L . Example 11.3.8 Find the mixed Fourier sine expansion of f(x) = x(2x2 −9Lx + 12L2) on [0, L]. Solution Since f(0) = f′(L) = 0, and f′′(x) = 6(2x −3L), we see from (11.3.7) that dn = − 48L (2n −1)2π2 Z L 0 (2x −3L) sin (2n −1)πx 2L dx = 96L2 (2n −1)3π3 " (2x −3L) cos (2n −1)πx 2L L 0 −2 Z L 0 cos (2n −1)πx 2L dx # = 96L2 (2n −1)3π3 " 3L − 4L (2n −1)π sin (2n −1)πx 2L L 0 # = 96L3 (2n −1)3π3 3 + (−1)n 4 (2n −1)π . Therefore SM(x) = 96L3 π3 ∞ X n=1 1 (2n −1)3 3 + (−1)n 4 (2n −1)π sin (2n −1)πx 2L . 11.3 Exercises In exercises marked by C graph f and some partial sums of the required series. If the interval is [0, L], choose a specific value of L for the graph. In Exercises 1-10 find the Fourier cosine series. 1. f(x) = x2; [0, L] 2. C f(x) = 1 −x; [0, 1] 3. C f(x) = x2 −2Lx; [0, L] 4. f(x) = sin kx (k ̸= integer); [0, π] 5. C f(x) = 1, 0 ≤x ≤L 2 0, L 2 < x < L; [0, L] 6. f(x) = x2 −L2; [0, L] 7. f(x) = (x −1)2; [0, 1] 8. f(x) = ex; [0, π] 9. C f(x) = x(L −x); [0, L] 10. C f(x) = x(x −2L); [0, L] | Elementary Differential Equations with Boundary Value Problems_Page_624_Chunk3027 |
Section 11.3 Fourier Expansions II 615 In Exercises 11-17 find the Fourier sine series. 11. C f(x) = 1; [0, L] 12. C f(x) = 1 −x; [0, 1] 13. f(x) = cos kx (k ̸= integer); [0, π] 14. C f(x) = 1, 0 ≤x ≤L 2 0, L 2 < x < L; [0, L] 15. C f(x) = x, 0 ≤x ≤L 2 , L −x, L 2 ≤x ≤L; [0, L]. 16. C f(x) = x sinx; [0, π] 17. f(x) = ex; [0, π] In Exercises 18-24 find the mixed Fourier cosine series. 18. C f(x) = 1; [0, L] 19. f(x) = x2; [0, L] 20. C f(x) = x; [0, 1] 21. C f(x) = 1, 0 ≤x ≤L 2 0, L 2 < x < L; [0, L] 22. f(x) = cos x; [0, π] 23. f(x) = sin x; [0, π] 24. C f(x) = x(L −x); [0, L] In Exercises 25-30 find the mixed Fourier sine series. 25. C f(x) = 1; [0, L] 26. f(x) = x2; [0, L] 27. C f(x) = 1, 0 ≤x ≤L 2 0, L 2 < x < L; [0, L] 28. f(x) = cos x; [0, π] 29. f(x) = sin x; [0, π] 30. C f(x) = x(L −x); [0, L]. In Exercises 31-34 use Theorem 11.3.5(a) to find the Fourier cosine series of f on [0, L]. 31. f(x) = 3x2(x2 −2L2) 32. f(x) = x3(3x −4L) 33. f(x) = x2(3x2 −8Lx + 6L2) 34. f(x) = x2(x −L)2 35. (a) Prove Theorem 11.3.5(b). | Elementary Differential Equations with Boundary Value Problems_Page_625_Chunk3028 |
616 Chapter 11 Boundary Value Problems and Fourier Expansions (b) In addition to the assumptions of Theorem 11.3.5(b), suppose f′′(0) = f′′(L) = 0, f′′′ is continuous, and f(4) is piecewise continuous on [0, L]. Show that bn = 2L3 n4π4 Z L 0 f(4)(x) sin nπx L dx, n ≥1. In Exercises 36-41 use Theorem 11.3.5(b) or, where applicable, Exercise 11.1.35(b), to find the Fourier sine series of f on [0, L]. 36. C f(x) = x(L −x) 37. C f(x) = x2(L −x) 38. f(x) = x(L2 −x2) 39. f(x) = x(x3 −2Lx2 + L3) 40. f(x) = x(3x4 −10L2x2 + 7L4) 41. f(x) = x(3x4 −5Lx3 + 2L4) 42. (a) Prove Theorem 11.3.5(c). (b) In addition to the assumptions of Theorem 11.3.5(c), suppose f′′(L) = 0, f′′ is continuous, and f′′′ is piecewise continuous on [0, L]. Show that cn = 16L2 (2n −1)3π3 Z L 0 f′′′(x) sin (2n −1)πx 2L dx, n ≥1. In Exercises 43-49 use Theorem 11.3.5(c) or, where applicable, Exercise 11.1.42(b), to find the mixed Fourier cosine series of f on [0, L]. 43. C f(x) = x2(L −x) 44. f(x) = L2 −x2 45. f(x) = L3 −x3 46. f(x) = 2x3 + 3Lx2 −5L3 47. f(x) = 4x3 + 3Lx2 −7L3 48. f(x) = x4 −2Lx3 + L4 49. f(x) = x4 −4Lx3 + 6L2x2 −3L4 50. (a) Prove Theorem 11.3.5(d). (b) In addition to the assumptions of Theorem 11.3.5(d), suppose f′′(0) = 0, f′′ is continuous, and f′′′ is piecewise continuous on [0, L]. Show that dn = − 16L2 (2n −1)3π3 Z L 0 f′′′(x) cos (2n −1)πx 2L dx, n ≥1. In Exercises 51-56 use Theorem 11.3.5(d) or, where applicable, Exercise 50(b), to find the mixed Fourier sine series of the f on [0, L]. 51. f(x) = x(2L −x) 52. f(x) = x2(3L −2x) | Elementary Differential Equations with Boundary Value Problems_Page_626_Chunk3029 |
Section 11.3 Fourier Expansions II 617 53. f(x) = (x −L)3 + L3 54. f(x) = x(x2 −3L2) 55. f(x) = x3(3x −4L) 56. f(x) = x(x3 −2Lx2 + 2L3) 57. Show that the mixed Fourier cosine series of f on [0, L] is the restriction to [0, L] of the Fourier cosine series of f3(x) = f(x), 0 ≤x ≤L, −f(2L −x), L < x ≤2L on [0, 2L]. Use this to prove Theorem 11.3.3. 58. Show that the mixed Fourier sine series of f on [0, L] is the restriction to [0, L] of the Fourier sine series of f4(x) = f(x), 0 ≤x ≤L, f(2L −x), L < x ≤2L on [0, 2L]. Use this to prove Theorem 11.3.4. 59. Show that the Fourier sine series of f on [0, L] is the restriction to [0, L] of the Fourier sine series of f3(x) = f(x), 0 ≤x ≤L, −f(2L −x), L < x ≤2L on [0, 2L]. 60. Show that the Fourier cosine series of f on [0, L] is the restriction to [0, L] of the Fourier cosine series of f4(x) = f(x), 0 ≤x ≤L, f(2L −x), L < x ≤2L on [0, 2L]. | Elementary Differential Equations with Boundary Value Problems_Page_627_Chunk3030 |
CHAPTER 12 Fourier Solutions of Partial Differential IN THIS CHAPTER we use the series discussed in Chapter 11 to solve partial differential equations that arise in problems of mathematical physics. SECTION 12.1 deals with the partial differential equation ut = a2uxx, which arises in problems of conduction of heat. SECTION 12.2 deals with the partial differential equation utt = a2uxx, which arises in the problem of the vibrating string. SECTION 12.3 deals with the partial differential equation uxx + uyy = 0, which arises in steady state problems of heat conduction and potential theory. SECTION 12.4 deals with the partial differential equation urr + 1 r ur + 1 r2 uθθ = 0, which is the equivalent to the equation studied in Section 1.3 when the independent variables are polar coordinates. 619 | Elementary Differential Equations with Boundary Value Problems_Page_628_Chunk3031 |
Section 12.1 The Heat Equation 619 12.1 THE HEAT EQUATION We begin the study of partial differential equations with the problem of heat flow in a uniform bar of length L, situated on the x axis with one end at the origin and the other at x = L (Figure 12.1.1). We assume that the bar is perfectly insulated except possibly at its endpoints, and that the temperature is constant on each cross section and therefore depends only on x and t. We also assume that the thermal properties of the bar are independent of x and t. In this case, it can be shown that the temperature u = u(x, t) at time t at a point x units from the origin satisfies the partial differential equation ut = a2uxx, 0 < x < L, t > 0, where a is a positive constant determined by the thermal properties. This is the heat equation. x = 0 x = L x Figure 12.1.1 A uniform bar of length L To determine u, we must specify the temperature at every point in the bar when t = 0, say u(x, 0) = f(x), 0 ≤x ≤L. We call this the initial condition. We must also specify boundary conditions that u must satisfy at the ends of the bar for all t > 0. We’ll call this problem an initial-boundary value problem. We begin with the boundary conditions u(0, t) = u(L, t) = 0, and write the initial-boundary value problem as ut = a2uxx, 0 < x < L, t > 0, u(0, t) = 0, u(L, t) = 0, t > 0, u(x, 0) = f(x), 0 ≤x ≤L. (12.1.1) Our method of solving this problem is called separation of variables (not to be confused with method of separation of variables used in Section 2.2 for solving ordinary differential equations). We begin by looking for functions of the form v(x, t) = X(x)T(t) that are not identically zero and satisfy vt = a2vxx, v(0, t) = 0, v(L, t) = 0 | Elementary Differential Equations with Boundary Value Problems_Page_629_Chunk3032 |
620 Chapter 12 Fourier Solutions of Partial Differential for all (x, t). Since vt = XT ′ and vxx = X′′T, vt = a2vxx if and only if XT ′ = a2X′′T, which we rewrite as T ′ a2T = X′′ X . Since the expression on the left is independent of x while the one on the right is independent of t, this equation can hold for all (x, t) only if the two sides equal the same constant, which we call a separation constant, and write it as −λ; thus, X′′ X = T ′ a2T = −λ. This is equivalent to X′′ + λX = 0 and T ′ = −a2λT. (12.1.2) Since v(0, t) = X(0)T(t) = 0 and v(L, t) = X(L)T(t) = 0 and we don’t want T to be identically zero, X(0) = 0 and X(L) = 0. Therefore λ must be an eigenvalue of the boundary value problem X′′ + λX = 0, X(0) = 0, X(L) = 0, (12.1.3) and X must be a λ-eigenfunction. From Theorem 11.1.2, the eigenvalues of (12.1.3) are λn = n2π2/L2, with associated eigenfunctions Xn = sin nπx L , n = 1, 2, 3, . . .. Substituting λ = n2π2/L2 into (12.1.2) yields T ′ = −(n2π2a2/L2)T, which has the solution Tn = e−n2π2a2t/L2. Now let vn(x, t) = Xn(x)Tn(t) = e−n2π2a2t/L2 sin nπx L , n = 1, 2, 3, . . . Since vn(x, 0) = sin nπx L , vn satisfies (12.1.1) with f(x) = sin nπx/L. More generally, if α1, . . ., αm are constants and um(x, t) = m X n=1 αne−n2π2a2t/L2 sin nπx L , then um satisfies (12.1.1) with f(x) = m X n=1 αn sin nπx L . This motivates the next definition. | Elementary Differential Equations with Boundary Value Problems_Page_630_Chunk3033 |
Section 12.1 The Heat Equation 621 Definition 12.1.1 The formal solution of the initial-boundary value problem ut = a2uxx, 0 < x < L, t > 0, u(0, t) = 0, u(L, t) = 0, t > 0, u(x, 0) = f(x), 0 ≤x ≤L (12.1.4) is u(x, t) = ∞ X n=1 αne−n2π2a2t/L2 sin nπx L , (12.1.5) where S(x) = ∞ X n=1 αn sin nπx L is the Fourier sine series of f on [0, L]; that is, αn = 2 L Z L 0 f(x) sin nπx L dx. We use the term “formal solution” in this definition because it’s not in general true that the infinite series in (12.1.5) actually satisfies all the requirements of the initial-boundary value problem (12.1.4) when it does, we say that it’s an actual solution of (12.1.4). Because of the negative exponentials in (12.1.5), u converges for all (x, t) with t > 0 (Exercise 54). Since each term in (12.1.5) satisfies the heat equation and the boundary conditions in (12.1.4), u also has these properties if ut and uxx can be obtained by differentiating the series in (12.1.5) term by term once with respect to t and twice with respect to x, for t > 0. However, it’s not always legitimate to differentiate an infinite series term by term. The next theorem gives a useful sufficient condition for legitimacy of term by term differentiation of an infinite series. We omit the proof. Theorem 12.1.2 A convergent infinite series W(z) = ∞ X n=1 wn(z) can be differentiated term by term on a closed interval [z1, z2] to obtain W ′(z) = ∞ X n=1 w′ n(z) (where the derivatives at z = z1 and z = z2 are one-sided) provided that w′ n is continuous on [z1, z2] and |w′ n(z)| ≤Mn, z1 ≤z ≤z2, n = 1, 2, 3, . . ., where M1, M2, ..., Mn, ..., are constants such that the series P∞ n=1 Mn converges. Theorem 12.1.2, applied twice with z = x and once with z = t, shows that uxx and ut can be obtained by differentiating u term by term if t > 0 (Exercise 54). Therefore u satisfies the heat equation and the boundary conditions in (12.1.4) for t > 0. Therefore, since u(x, 0) = S(x) for 0 ≤x ≤L, u is an actual solution of (12.1.4) if and only if S(x) = f(x) for 0 ≤x ≤L. From Theorem 11.3.2, this is true if f is continuous and piecewise smooth on [0, L], and f(0) = f(L) = 0. In this chapter we’ll define formal solutions of several kinds of problems. When we ask you to solve such problems, we always mean that you should find a formal solution. | Elementary Differential Equations with Boundary Value Problems_Page_631_Chunk3034 |
622 Chapter 12 Fourier Solutions of Partial Differential Example 12.1.1 Solve (12.1.4) with f(x) = x(x2 −3Lx + 2L2). Solution From Example 11.3.6, the Fourier sine series of f on [0, L] is S(x) = 12L3 π3 ∞ X n=1 1 n3 sin nπx L . Therefore u(x, t) = 12L3 π3 ∞ X n=1 1 n3 e−n2π2a2t/L2 sin nπx L . If both ends of bar are insulated so that no heat can pass through them, then the boundary conditions are ux(0, t) = 0, ux(L, t) = 0, t > 0. We leave it to you (Exercise 1) to use the method of separation of variables and Theorem 11.1.3 to motivate the next definition. Definition 12.1.3 The formal solution of the initial-boundary value problem ut = a2uxx, 0 < x < L, t > 0, ux(0, t) = 0, ux(L, t) = 0, t > 0, u(x, 0) = f(x), 0 ≤x ≤L (12.1.6) is u(x, t) = α0 + ∞ X n=1 αne−n2π2a2t/L2 cos nπx L , where C(x) = α0 + ∞ X n=1 αn cos nπx L is the Fourier cosine series of f on [0, L]; that is, α0 = 1 L Z L 0 f(x) dx and αn = 2 L Z L 0 f(x) cos nπx L dx, n = 1, 2, 3, . . .. Example 12.1.2 Solve (12.1.6) with f(x) = x. Solution From Example 11.3.1, the Fourier cosine series of f on [0, L] is C(x) = L 2 −4L π2 ∞ X n=1 1 (2n −1)2 cos (2n −1)πx L . Therefore u(x, t) = L 2 −4L π2 ∞ X n=1 1 (2n −1)2 e−(2n−1)2π2a2t/L2 cos (2n −1)πx L . We leave it to you (Exercise 2) to use the method of separation of variables and Theorem 11.1.4 to motivate the next definition. | Elementary Differential Equations with Boundary Value Problems_Page_632_Chunk3035 |
Section 12.1 The Heat Equation 623 Definition 12.1.4 The formal solution of the initial-boundary value problem ut = a2uxx, 0 < x < L, t > 0, u(0, t) = 0, ux(L, t) = 0, t > 0, u(x, 0) = f(x), 0 ≤x ≤L (12.1.7) is u(x, t) = ∞ X n=1 αne−(2n−1)2π2a2t/4L2 sin (2n −1)πx 2L , where SM(x) = ∞ X n=1 αn sin (2n −1)πx 2L is the mixed Fourier sine series of f on [0, L]; that is, αn = 2 L Z L 0 f(x) sin (2n −1)πx 2L dx. Example 12.1.3 Solve (12.1.7) with f(x) = x. Solution From Example 11.3.4, the mixed Fourier sine series of f on [0, L] is SM (x) = −8L π2 ∞ X n=1 (−1)n (2n −1)2 sin (2n −1)πx 2L . Therefore u(x, t) = −8L π2 ∞ X n=1 (−1)n (2n −1)2 e−(2n−1)2π2a2t/4L2 sin (2n −1)πx 2L . Figure 12.1.2 shows a graph of u = u(x, t) plotted with respect to x for various values of t. The line y = x corresponds to t = 0. The other curves correspond to positive values of t. As t increases, the graphs approach the line u = 0. We leave it to you (Exercise 3) to use the method of separation of variables and Theorem 11.1.5 to motivate the next definition. Definition 12.1.5 The formal solution of the initial-boundary value problem ut = a2uxx, 0 < x < L, t > 0, ux(0, t) = 0, u(L, t) = 0, t > 0, u(x, 0) = f(x), 0 ≤x ≤L (12.1.8) is u(x, t) = ∞ X n=1 αne−(2n−1)2π2a2t/4L2 cos (2n −1)πx 2L , where CM(x) = ∞ X n=1 αn cos (2n −1)πx 2L is the mixed Fourier cosine series of f on [0, L]; that is, αn = 2 L Z L 0 f(x) cos (2n −1)πx 2L dx. | Elementary Differential Equations with Boundary Value Problems_Page_633_Chunk3036 |
624 Chapter 12 Fourier Solutions of Partial Differential L x u y = x Figure 12.1.2 Example 12.1.4 Solve (12.1.8) with f(x) = x −L. Solution From Example 11.3.3, the mixed Fourier cosine series of f on [0, L] is CM(x) = −8L π2 ∞ X n=1 1 (2n −1)2 cos (2n −1)πx 2L . Therefore u(x, t) = −8L π2 ∞ X n=1 1 (2n −1)2 e−(2n−1)2π2a2t/4L2 cos (2n −1)πx 2L . Nonhomogeneous Problems A problem of the form ut = a2uxx + h(x), 0 < x < L, t > 0, u(0, t) = u0, u(L, t) = uL, t > 0, u(x, 0) = f(x), 0 ≤x ≤L (12.1.9) can be transformed to a problem that can be solved by separation of variables. We write u(x, t) = v(x, t) + q(x), (12.1.10) where q is to be determined. Then ut = vt and uxx = vxx + q′′ | Elementary Differential Equations with Boundary Value Problems_Page_634_Chunk3037 |
Section 12.1 The Heat Equation 625 so u satisfies (12.1.9) if v satisfies vt = a2vxx + a2q′′(x) + h(x), 0 < x < L, t > 0, v(0, t) = u0 −q(0), v(L, t) = uL −q(L), t > 0, v(x, 0) = f(x) −q(x), 0 ≤x ≤L. This reduces to vt = a2vxx, 0 < x < L, t > 0, v(0, t) = 0, v(L, t) = 0, t > 0, v(x, 0) = f(x) −q(x), 0 ≤x ≤L (12.1.11) if a2q′′ + h(x) = 0, q(0) = u0, q(L) = uL. We can obtain q by integrating q′′ = −h/a2 twice and choosing the constants of integration so that q(0) = u0 and q(L) = uL. Then we can solve (12.1.11) for v by separation of variables, and (12.1.10) is the solution of (12.1.9). Example 12.1.5 Solve ut = uxx −2, 0 < x < 1, t > 0, u(0, t) = −1, u(1, t) = 1, t > 0, u(x, 0) = x3 −2x2 + 3x −1, 0 ≤x ≤1. Solution We leave it to you to show that q(x) = x2 + x −1 satisfies q′′ −2 = 0, q(0) = −1, q(1) = 1. Therefore u(x, t) = v(x, t) + x2 + x −1, where vt = vxx, 0 < x < 1, t > 0, v(0, t) = 0, v(1, t) = 0, t > 0, and v(x, 0) = x3 −2x2 + 3x −1 −x2 −x + 1 = x(x2 −3x + 2). From Example 12.1.1 with a = 1 and L = 1, v(x, t) = 12 π3 ∞ X n=1 1 n3 e−n2π2t sin nπx. Therefore u(x, t) = x2 + x −1 + 12 π3 ∞ X n=1 1 n3 e−n2π2t sinnπx. A similar procedure works if the boundary conditions in (12.1.11) are replaced by mixed boundary conditions ux(0, t) = u0, u(L, t) = uL, t > 0 | Elementary Differential Equations with Boundary Value Problems_Page_635_Chunk3038 |
626 Chapter 12 Fourier Solutions of Partial Differential or u(0, t) = u0, ux(L, t) = uL, t > 0; however, this isn’t true in general for the boundary conditions ux(0, t) = u0, ux(L, t) = uL, t > 0. (See Exercise 47.) USING TECHNOLOGY Numerical experiments can enhance your understanding of the solutionsof initial-boundaryvalue prob- lems. To be specific, consider the formal solution u(x, t) = ∞ X n=1 αne−n2π2a2t/L2 sin nπx L , of (12.1.4), where S(x) = ∞ X n=1 αn sin nπx L is the Fourier sine series of f on [0, L]. Consider the m-th partial sum um(x, t) = m X n=1 αne−n2π2a2t/L2 sin nπx L . (12.1.12) For several fixed values of t (including t = 0), graph um(x, t) versus t. In some cases it may be useful to graph the curves corresponding to the various values of t on the same axes in other cases you may want to graph the various curves sucessively (for increasing values of t), and create a primitive motion picture on your monitor. Repeat this experiment for several values of m, to compare how the results depend upon m for small and large values of t. However, keep in mind that the meanings of “small” and “large” in this case depend upon the constants a2 and L2. A good way to handle this is to rewrite (12.1.12) as um(x, t) = m X n=1 αne−n2τ sin nπx L , where τ = π2a2t L2 , (12.1.13) and graph um versus x for selected values of τ. These comments also apply to the situationsconsidered in Definitions12.1.3-12.1.5, except that (12.1.13) should be replaced by τ = π2a2t 4L2 , in Definitions 12.1.4 and 12.1.5. In some of the exercises we say “perform numerical experiments.” This means that you should perform the computations just described with the formal solution obtained in the exercise. | Elementary Differential Equations with Boundary Value Problems_Page_636_Chunk3039 |
Section 12.1 The Heat Equation 627 12.1 Exercises 1. Explain Definition 12.1.3. 2. Explain Definition 12.1.4. 3. Explain Definition 12.1.5. 4. C Perform numerical experiments with the formal solution obtained in Example 12.1.1. 5. C Perform numerical experiments with the formal solution obtained in Example 12.1.2. 6. C Perform numerical experiments with the formal solution obtained in Example 12.1.3. 7. C Perform numerical experiments with the formal solution obtained in Example 12.1.4. In Exercises 8-42 solve the initial-boundary value problem. Where indicated by C , perform numerical experiments. To simplify the computation of coefficients in some of these problems, check first to see if u(x, 0) is a polynomial that satisfies the boundary conditions. If it does, apply Theorem 11.3.5; also, see Exercises 11.3.35(b), 11.3.42(b), and 11.3.50(b). 8. ut = uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = x(1 −x), 0 ≤x ≤1 9. ut = 9uxx, 0 < x < 4, t > 0, u(0, t) = 0, u(4, t) = 0, t > 0, u(x, 0) = 1, 0 ≤x ≤4 10. ut = 3uxx, 0 < x < π, t > 0, u(0, t) = 0, u(π, t) = 0, t > 0, u(x, 0) = x sinx, 0 ≤x ≤π 11. C ut = 9uxx, 0 < x < 2, t > 0, u(0, t) = 0, u(2, t) = 0, t > 0, u(x, 0) = x2(2 −x), 0 ≤x ≤2 12. ut = 4uxx, 0 < x < 3, t > 0, u(0, t) = 0, u(3, t) = 0, t > 0, u(x, 0) = x(9 −x2), 0 ≤x ≤3 13. ut = 4uxx, 0 < x < 2, t > 0, u(0, t) = 0, u(2, t) = 0, t > 0, u(x, 0) = x, 0 ≤x ≤1, 2 −x, 1 ≤x ≤2. 14. ut = 7uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = x(3x4 −10x2 + 7), 0 ≤x ≤1 15. ut = 5uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = x(x3 −2x2 + 1), 0 ≤x ≤1 16. ut = 2uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = x(3x4 −5x3 + 2), 0 ≤x ≤1 | Elementary Differential Equations with Boundary Value Problems_Page_637_Chunk3040 |
628 Chapter 12 Fourier Solutions of Partial Differential 17. C ut = 9uxx, 0 < x < 4, t > 0, ux(0, t) = 0, ux(4, t) = 0, t > 0, u(x, 0) = x2, 0 ≤x ≤4 18. ut = 4uxx, 0 < x < 2, t > 0, ux(0, t) = 0, ux(2, t) = 0, t > 0, u(x, 0) = x(x −4), 0 ≤x ≤2 19. C ut = 9uxx, 0 < x < 1, t > 0, ux(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = x(1 −x), 0 ≤x ≤1 20. ut = 3uxx, 0 < x < 2, t > 0, ux(0, t) = 0, ux(2, t) = 0, t > 0, u(x, 0) = 2x2(3 −x), 0 ≤x ≤2 21. ut = 5uxx, 0 < x < √ 2, t > 0, ux(0, t) = 0, ux( √ 2, t) = 0, t > 0, u(x, 0) = 3x2(x2 −4), 0 ≤x ≤ √ 2 22. C ut = 3uxx, 0 < x < 1, t > 0, ux(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = x3(3x −4), 0 ≤x ≤1 23. ut = uxx, 0 < x < 1, t > 0, ux(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = x2(3x2 −8x + 6), 0 ≤x ≤1 24. ut = uxx, 0 < x < π, t > 0, ux(0, t) = 0, ux(π, t) = 0, t > 0, u(x, 0) = x2(x −π)2, 0 ≤x ≤π 25. ut = uxx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = sin πx, 0 ≤x ≤1 26. C ut = 3uxx, 0 < x < π, t > 0, u(0, t) = 0, ux(π, t) = 0, t > 0, u(x, 0) = x(π −x), 0 ≤x ≤π 27. ut = 5uxx, 0 < x < 2, t > 0, u(0, t) = 0, ux(2, t) = 0, t > 0, u(x, 0) = x(4 −x), 0 ≤x ≤2 28. ut = uxx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = x2(3 −2x), 0 ≤x ≤1 29. ut = uxx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = (x −1)3 + 1, 0 ≤x ≤1 30. C ut = uxx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = x(x2 −3), 0 ≤x ≤1 31. ut = uxx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = x3(3x −4), 0 ≤x ≤1 | Elementary Differential Equations with Boundary Value Problems_Page_638_Chunk3041 |
Section 12.1 The Heat Equation 629 32. ut = uxx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = x(x3 −2x2 + 2), 0 ≤x ≤1 33. ut = 3uxx, 0 < x < π, t > 0, ux(0, t) = 0, u(π, t) = 0, t > 0, u(x, 0) = x2(π −x), 0 ≤x ≤π 34. ut = 16uxx, 0 < x < 2π, t > 0, ux(0, t) = 0, u(2π, t) = 0, t > 0, u(x, 0) = 4, 0 ≤x ≤2π 35. ut = 9uxx, 0 < x < 4, t > 0, ux(0, t) = 0, u(4, t) = 0, t > 0, u(x, 0) = x2, 0 ≤x ≤4 36. C ut = 3uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 1 −x, 0 ≤x ≤1 37. ut = uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 1 −x3, 0 ≤x ≤1 38. ut = 7uxx, 0 < x < π, t > 0, ux(0, t) = 0, u(π, t) = 0, t > 0, u(x, 0) = π2 −x2, 0 ≤x ≤π 39. ut = uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 4x3 + 3x2 −7, 0 ≤x ≤1 40. ut = uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 2x3 + 3x2 −5, 0 ≤x ≤1 41. C ut = uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = x4 −4x3 + 6x2 −3, 0 ≤x ≤1 42. ut = uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = x4 −2x3 + 1, 0 ≤x ≤1 In Exercises 43-46 solve the initial-boundary value problem. Perform numerical experiments for specific values of L and a. 43. C ut = a2uxx, 0 < x < L, t > 0, ux(0, t) = 0, ux(L, t) = 0, t > 0, u(x, 0) = 1, 0 ≤x ≤L 2 , 0, L 2 < x < L 44. C ut = a2uxx, 0 < x < L, t > 0, u(0, t) = 0, u(L, t) = 0, t > 0, u(x, 0) = 1, 0 ≤x ≤L 2 , 0, L 2 < x < L | Elementary Differential Equations with Boundary Value Problems_Page_639_Chunk3042 |
630 Chapter 12 Fourier Solutions of Partial Differential 45. C ut = a2uxx, 0 < x < L, t > 0, ux(0, t) = 0, u(L, t) = 0, t > 0, u(x, 0) = 1, 0 ≤x ≤L 2 , 0, L 2 < x < L 46. C ut = a2uxx, 0 < x < L, t > 0, u(0, t) = 0, ux(L, t) = 0, t > 0, u(x, 0) = 1, 0 ≤x ≤L 2 , 0, L 2 < x < L 47. Let h be continuous on [0, L] and let u0, uL, and a be constants, with a > 0. Show that it’s always possible to find a function q that satisfies (a), (b), or (c), but that this isn’t so for (d). (a) a2q′′ + h = 0, q(0) = u0, q(L) = uL (b) a2q′′ + h = 0, q′(0) = u0, q(L) = uL (c) a2q′′ + h = 0, q(0) = u0, q′(L) = uL (d) a2q′′ + h = 0, q′(0) = u0, q′(L) = uL In Exercises 48-53 solve the nonhomogeneous initial-boundary value problem 48. ut = 9uxx −54x, 0 < x < 4, t > 0, u(0, t) = 1, u(4, t) = 61, t > 0, u(x, 0) = 2 −x + x3, 0 ≤x ≤4 49. ut = uxx −2, 0 < x < 1, t > 0, u(0, t) = 1, u(1, t) = 3, t > 0, u(x, 0) = 2x2 + 1, 0 ≤x ≤1 50. ut = 3uxx −18x, 0 < x < 1, t > 0, ux(0, t) = −1, u(1, t) = −1, t > 0, u(x, 0) = x3 −2x, 0 ≤x ≤1 51. ut = 9uxx −18, 0 < x < 4, t > 0, ux(0, t) = −1, u(4, t) = 10, t > 0, u(x, 0) = 2x2 −x −2, 0 ≤x ≤4 52. ut = uxx + π2 sin πx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = −π, t > 0, u(x, 0) = 2 sinπx, 0 ≤x ≤1 53. ut = uxx −6x, 0 < x < L, t > 0, u(0, t) = 3, ux(1, t) = 2, t > 0, u(x, 0) = x3 −x2 + x + 3, 0 ≤x ≤1 54. In this exercise take it as given that the infinite series P∞ n=1 npe−qn2 converges for all p if q > 0, and, where appropriate, use the comparison test for absolute convergence of an infinite series. Let u(x, t) = ∞ X n=1 αne−n2π2a2t/L2 sin nπx L where αn = 2 L Z L 0 f(x) sin nπx L dx and f is piecewise smooth on [0, L]. | Elementary Differential Equations with Boundary Value Problems_Page_640_Chunk3043 |
Section 12.2 The Wave Equation 631 (a) Show that u is defined for (x, t) such that t > 0. (b) For fixed t > 0, use Theorem 12.1.2 with z = x to show that ux(x, t) = π L ∞ X n=1 nαne−n2π2a2t/L2 cos nπx L , −∞< x < ∞. (c) Starting from the result of (a), use Theorem 12.1.2 with z = x to show that, for a fixed t > 0, uxx(x, t) = −π2 L2 ∞ X n=1 n2αne−n2π2a2t/L2 sin nπx L , −∞< x < ∞. (d) For fixed but arbitrary x, use Theorem 12.1.2 with z = t to show that ut(x, t) = −π2a2 L2 ∞ X n=1 n2αne−n2π2a2t/L2 sin nπx L , if t > t0 > 0, where t0 is an arbitrary positive number. Then argue that since t0 is arbitrary, the conclusion holds for all t > 0. (e) Conclude from (c) and (d) that ut = a2uxx, −∞< x < ∞, t > 0. By repeatedly applying the arguments in (a) and (c), it can be shown that u can be differentiated term by term any number of times with respect to x and/or t if t > 0. 12.2 THE WAVE EQUATION In this section we consider initial-boundary value problems of the form utt = a2uxx, 0 < x < L, t > 0, u(0, t) = 0, u(L, t) = 0, t > 0, u(x, 0) = f(x), ut(x, 0) = g(x), 0 ≤x ≤L, (12.2.1) where a is a constant and f and g are given functions of x. The partial differential equation utt = a2uxx is called the wave equation. It is necessary to specify both f and g because the wave equation is a second order equation in t for each fixed x. This equation and its generalizations utt = a2(uxx + uyy) and utt = a2(uxx + uyy + uzz) to two and three space dimensions have important applications to the the propagation of electromagnetic, sonic, and water waves. The Vibrating String We motivate the study of the wave equation by considering its application to the vibrations of a string – such as a violin string – tightly stretched in equilibrium along the x-axis in the xu-plane and tied to the points (0, 0) and (L, 0) (Figure 12.2.1). If the string is plucked in the vertical direction and released at time t = 0, it will oscillate in the xu- plane. Let u(x, t) denote the displacement of the point on the string above (or below) the abscissa x at time t. We’ll show that it’s reasonable to assume that u satisfies the wave equation under the following as- sumptions: | Elementary Differential Equations with Boundary Value Problems_Page_641_Chunk3044 |
632 Chapter 12 Fourier Solutions of Partial Differential Equations L x u Figure 12.2.1 A stretched string 1. The mass density (mass per unit length) ρ of the string is constant throughout the string. 2. The tension T induced by tightly stretching the string along the x-axis is so great that all other forces, such as gravity and air resistance, can be neglected. 3. The tension at any point on the string acts along the tangent to the string at that point, and the magnitude of its horizontal component is always equal to T, the tension in the string in equilibrium. 4. The slope of the string at every point remains sufficiently small so that we can make the approxima- tion p 1 + u2x ≈1. (12.2.2) Figure 12.2.2 shows a segment of the displaced string at a time t > 0. (Don’t think that the figure is necessarily inconsistent with Assumption 4; we exaggerated the slope for clarity.) The vectors T1 and T2 are the forces due to tension, acting along the tangents to the segment at its endpoints. From Newton’s second law of motion, T1 −T2 is equal to the mass times the acceleration of the center of mass of the segment. The horizontal and vertical components of T1 −T2 are |T2| cos θ2 −|T1| cos θ1 and |T2| sin θ2 −|T1| sinθ1, respectively. Since |T2| cos θ2 = |T1| cos θ1 = T (12.2.3) by assumption, the net horizontal force is zero, so there’s no horizontal acceleration. Since the initial horizontal velocity is zero, there’s no horizontal motion. Applying Newton’s second law of motion in the vertical direction yields |T2| sinθ2 −|T1| sin θ1 = ρ∆s utt(x, t), (12.2.4) | Elementary Differential Equations with Boundary Value Problems_Page_642_Chunk3045 |
Section 12.2 The Wave Equation 633 T1 θ2 T2 x u θ1 Figure 12.2.2 A segment of the displaced string where ∆s is the length of the segment and x is the abscissa of the center of mass; hence, x < x < x + ∆x. From calculus, we know that ∆s = Z x+∆x x p 1 + u2x(σ, t) dσ; however, because of (12.2.2), we make the approximation ∆s ≈ Z x+∆x x 1 dσ = ∆x, so (12.2.4) becomes |T2| sinθ2 −|T1| sin θ1 = ρ∆x utt(x, t). Therefore |T2| sin θ2 −|T1| sinθ1 ∆x = ρutt(x, t). Recalling (12.2.3), we divide by T to obtain tan θ2 −tan θ1 ∆x = ρ T utt(x, t). (12.2.5) Since tan θ1 = ux(x, t) and tan θ2 = ux(x + ∆x, t), (12.2.5) is equivalent to ux(x + ∆x) −ux(x, t) ∆x = ρ T utt(x, t). | Elementary Differential Equations with Boundary Value Problems_Page_643_Chunk3046 |
634 Chapter 12 Fourier Solutions of Partial Differential Equations Letting ∆x →0 yields uxx(x, t) = ρ T utt(x, t), which we rewrite as utt = a2uxx, with a2 = T/ρ. The Formal Solution As in Section 12.1, we use separation of variables to obtain a suitable definition for the formal solution of (12.2.1). We begin by looking for functions of the form v(x, t) = X(x)T(t) that are not identically zero and satisfy vtt = a2vxx, v(0, t) = 0, v(L, t) = 0 for all (x, t). Since vtt = XT ′′ and vxx = X′′T, vtt = a2vxx if and only if XT ′′ = a2X′′T, which we rewrite as T ′′ a2T = X′′ X . For this to hold for all (x, t), the two sides must equal the same constant; thus, X′′ X = T ′′ a2T = −λ, which is equivalent to X′′ + λX = 0 and T ′′ + a2λT = 0. (12.2.6) Since v(0, t) = X(0)T(t) = 0 and v(L, t) = X(L)T(t) = 0 and we don’t want T to be identically zero, X(0) = 0 and X(L) = 0. Therefore λ must be an eigenvalue of X′′ + λX = 0, X(0) = 0, X(L) = 0, (12.2.7) and X must be a λ-eigenfunction. From Theorem 11.1.2, the eigenvalues of (12.2.7) are λn = n2π2/L2, with associated eigenfunctions Xn = sin nπx L , n = 1, 2, 3, . . .. Substituting λ = n2π2/L2 into (12.2.6) yields T ′′ + (n2π2a2/L2)T = 0, which has the general solution Tn = αn cos nπat L + βnL nπa sin nπat L , where αn and βn are constants. Now let vn(x, t) = Xn(x)Tn(t) = αn cos nπat L + βnL nπa sin nπat L sin nπx L . | Elementary Differential Equations with Boundary Value Problems_Page_644_Chunk3047 |
Section 12.2 The Wave Equation 635 Then ∂vn ∂t (x, t) = −nπa L αn sin nπat L + βn cos nπat L sin nπx L , so vn(x, 0) = αn sin nπx L and ∂vn ∂t (x, 0) = βn sin nπx L . Therefore vn satisfies (12.2.1) with f(x) = αn sin nπx/L and g(x) = βn cos nπx/L. More generally, if α1, α2, ..., αm and β1, β2,..., βm are constants and um(x, t) = m X n=1 αn cos nπat L + βnL nπa sin nπat L sin nπx L , then um satisfies (12.2.1) with f(x) = m X n=1 αn sin nπx L and g(x) = m X n=1 βn sin nπx L . This motivates the next definition. Definition 12.2.1 If f and g are piecewise smooth of [0, L], then the formal solution of (12.2.1) is u(x, t) = ∞ X n=1 αn cos nπat L + βnL nπa sin nπat L sin nπx L , (12.2.8) where Sf(x) = ∞ X n=1 αn sin nπx L and Sg(x) = ∞ X n=1 βn sin nπx L are the Fourier sine series of f and g on [0, L]; that is, αn = 2 L Z L 0 f(x) sin nπx L dx and βn = 2 L Z L 0 g(x) sin nπx L dx. Since there are no convergence-producing factors in (12.2.8) like the negative exponentials in t that appear in formal solutions of initial-boundary value problems for the heat equation, it isn’t obvious that (12.2.8) even converges for any values of x and t, let alone that it can be differentiated term by term to show that utt = a2uxx. However, the next theorem guarantees that the series converges not only for 0 ≤x ≤L and t ≥0, but for −∞< x < ∞and −∞< t < ∞. Theorem 12.2.2 If f and g are pieceswise smooth on [0, L], then u in (12.2.1) converges for all (x, t), and can be written as u(x, t) = 1 2[Sf(x + at) + Sf(x −at)] + 1 2a Z x+at x−at Sg(τ) dτ. (12.2.9) Proof Setting A = nπx/L and B = nπat/L in the identities sin A cos B = 1 2[sin(A + B) + sin(A −B)] and sin A sin B = −1 2[cos(A + B) −cos(A −B)] | Elementary Differential Equations with Boundary Value Problems_Page_645_Chunk3048 |
636 Chapter 12 Fourier Solutions of Partial Differential Equations yields cos nπat L sin nπx L = 1 2 sin nπ(x + at) L + sin nπ(x −at) L (12.2.10) and sin nπat L sin nπx L = −1 2 cos nπ(x + at) L −cos nπ(x −at) L = nπ 2L Z x+at x−at sin nπτ L dτ. (12.2.11) From (12.2.10), ∞ X n=1 αn cos nπat L sin nπx L = 1 2 ∞ X n=1 αn sin nπ(x + at) L + sin nπ(x −at) L = 1 2[Sf(x + at) + Sf(x −at)]. (12.2.12) Since it can be shown that a Fourier sine series can be integrated term by term between any two limits, (12.2.11) implies that ∞ X n=1 βnL nπa sin nπat L sin nπx L = 1 2a ∞ X n=1 βn Z x+at x−at sin nπτ L dτ = 1 2a Z x+at x−at ∞ X n=1 βn sin nπτ L ! dτ = 1 2a Z x+at x−at Sg(τ) dτ. This and (12.2.12) imply (12.2.9), which completes the proof. As we’ll see below, if Sg is differentiable and Sf is twice differentiable on (−∞, ∞), then (12.2.9) satisfies utt = a2uxx for all (x, t). We need the next theorem to formulate conditions on f and g such that Sf and Sg to have these properties. Theorem 12.2.3 Suppose h is differentiable on [0, L]; that is, h′(x) exists for 0 < x < L, and the one-sided derivatives h′ +(0) = lim x→0+ h(x) −h(0) x and h′ −(L) = lim x→L− h(x) −h(L) x −L both exist. (a) Let p be the odd periodic extension of h to (−∞, ∞); that is, p(x) = h(x), 0 ≤x ≤L, −h(−x), −L < x < 0, and p(x + 2L) = p(x), −∞< x < ∞. Then p is differentiable on (−∞, ∞) if and only if h(0) = h(L) = 0. (12.2.13) (b) Let q be the even periodic extension of h to (−∞, ∞); that is, q(x) = h(x), 0 ≤x ≤L, h(−x), −L < x < 0, and q(x + 2L) = q(x), −∞< x < ∞. Then q is differentiable on (−∞, ∞) if and only if h′ +(0) = h′ −(L) = 0. (12.2.14) | Elementary Differential Equations with Boundary Value Problems_Page_646_Chunk3049 |
Section 12.2 The Wave Equation 637 Proof Throughout this proof, k denotes an integer. Since f is differentiable on the open interval (0, L), both p and q are differentiable on every open interval ((k −1)L, kL). Thus, we need only to determine whether p and q are differentiable at x = kL for every k. (a) From Figure 12.2.3, p is discontinuousat x = 2kL if h(0) ̸= 0 and discontinuous at x = (2k −1)L if h(L) ̸= 0. Therefore p is not differentiable on (−∞, ∞) unless h(0) = h(L) = 0. From Figure 12.2.4, if h(0) = h(L) = 0, then p′(2kL) = h′ +(0) and p′((2k −1)L) = h′ −(L) for every k; therefore, p is differentiable on (−∞, ∞). x = 0 x = L x = 2L x = − L x = − 2L x = − 3L x = 3L x y Figure 12.2.3 The odd extension of a function that does not satisfy (12.2.13) x = 0 x = L x = 2L x = − L x = − 2L x = − 3L x = 3L x y Figure 12.2.4 The odd extension of a function that satisfies (12.2.13) (b) From Figure 12.2.5, q′ −(2kL) = −h′ +(0) and q′ +(2kL) = h′ +(0), so q is differentiable at x = 2kL if and only if h′ +(0) = 0. Also, q′ −((2k −1)L) = h′ −(L) and q′ +((2k −1)L) = −h′ −(L), so q is differentiable at x = (2k −1)L if and only if h′ −(L) = 0. Therefore q is differentiable on (−∞, ∞) if and only if h′ +(0) = h′ −(L) = 0, as in Figure 12.2.6. This completes the proof. x = 0 x = L x = 2L x = − L x = − 2L x = − 3L x = 3L x y Figure 12.2.5 The even extension of a function that doesn’t satisfy (12.2.14) x = 0 x = L x = 2L x = − L x = − 2L x = − 3L x = 3L x y Figure 12.2.6 The even extension of a function that satisfies (12.2.14) | Elementary Differential Equations with Boundary Value Problems_Page_647_Chunk3050 |
638 Chapter 12 Fourier Solutions of Partial Differential Equations Theorem 12.2.4 The formal solution of (12.2.1) is an actual solution if g is differentiable on [0, L] and g(0) = g(L) = 0, (12.2.15) while f is twice differentiable on [0, L] and f(0) = f(L) = 0 (12.2.16) and f′′ +(0) = f′′ −(L) = 0. (12.2.17) Proof We first show that Sg is differentiable and Sf is twice differentiable on (−∞, ∞). We’ll then differentiate (12.2.9) twice with respect to x and t and verify that (12.2.9) is an actual solution of (12.2.1). Since f and g are continuous on (0, L), Theorem 11.3.2 implies that Sf(x) = f(x) and Sg(x) = g(x) on [0, L]. Therefore Sf and Sg are the odd periodic extensions of f and g. Since f and g are differen- tiable on [0, L], (12.2.15), (12.2.16), and Theorem 12.2.3(a) imply that Sf and Sg are differentiable on (−∞, ∞). Since S′ f(x) = f′(x) on [0, L] (one-sided derivatives at the endpoints), and S′ f is even (the derivative of an odd function is even), S′ f is the even periodic extension of f′. By assumption, f′ is differentiable on [0, L]. Because of (12.2.17), Theorem 12.2.3(b) with h = f′ and q = S′ f implies that S′′ f exists on (−∞, ∞). Now we can differentiate (12.2.9) twice with respect to x and t: ux(x, t) = 1 2[S′ f(x + at) + S′ f(x −at)] + 1 2a[Sg(x + at) −Sg(x −at)], uxx(x, t) = 1 2[S′′ f (x + at) + S′′ f (x −at)] + 1 2a[S′ g(x + at) −S′ g(x −at)], (12.2.18) ut(x, t) = a 2[S′ f(x + at) −S′ f(x −at)] + 1 2[Sg(x + at) + Sg(x −at)], (12.2.19) and utt(x, t) = a2 2 [S′′ f (x + at) −S′′ f (x −at)] + a 2[S′ g(x + at) −S′ g(x −at)]. (12.2.20) Comparing (12.2.18) and (12.2.20) shows that utt(x, t) = a2uxx(x, t) for all (x, t). From (12.2.8), u(0, t) = u(L, t) = 0 for all t. From (12.2.9), u(x, 0) = Sf(x) for all x, and therefore, in particular, u(x, 0) = f(x), 0 ≤x < L. From (12.2.19), ut(x, 0) = Sg(x) for all x, and therefore, in particular, ut(x, 0) = g(x), 0 ≤x < L. Therefore u is an actual solution of (12.2.1). This completes the proof. Eqn (12.2.9) is called d’Alembert’s solution of (12.2.1). Although d’Alembert’s solution was useful for proving Theorem 12.2.4 and is very useful in a slightly different context (Exercises 63-68), (12.2.8) is preferable for computational purposes. Example 12.2.1 Solve (12.2.1) with f(x) = x(x3 −2Lx2 + L2) and g(x) = x(L −x). | Elementary Differential Equations with Boundary Value Problems_Page_648_Chunk3051 |
Section 12.2 The Wave Equation 639 Solution We leave it to you to verify that f and g satisfy the assumptions of Theorem 12.2.4. From Exercise 11.3.39, Sf(x) = 96L4 π5 ∞ X n=1 1 (2n −1)5 sin (2n −1)πx L . From Exercise 11.3.36, Sg(x) = 8L2 π3 ∞ X n=1 1 (2n −1)3 sin (2n −1)πx L . From (12.2.8), u(x, t) = 96L4 π5 ∞ X n=1 1 (2n −1)5 cos (2n −1)πat L sin (2n −1)πx L +8L3 aπ4 ∞ X n=1 1 (2n −1)4 sin (2n −1)πat L sin (2n −1)πx L . Theorem 12.1.2 implies that uxx and utt can be obtained by term by term differentiation, for all (x, t), so utt = a2uxx for all (x, t) (Exercise 62). Moreover, Theorem 11.3.2 implies that Sf(x) = f(x) and Sg(x) = g(x) if 0 ≤x ≤L. Therefore u(x, 0) = f(x) and ut(x, 0) = g(x) if 0 ≤x ≤L. Hence, u is an actual solution of the initial-boundary value problem. REMARK: In solving a specific initial-boundary value problem (12.2.1), it’s convenient to solve the prob- lem with g ≡0, then with f ≡0, and add the solutions to obtain the solution of the given problem. Because of this, either f ≡0 or g ≡0 in all the specific initial-boundary value problems in the exercises. The Plucked String If f and g don’t satisfy the assumptions of Theorem 12.2.4, then (12.2.8) isn’t an actual solution of (12.2.1) in fact, it can be shown that (12.2.1) doesn’t have an actual solution in this case. Nevertheless, u is defined for all (x, t), and we can see from (12.2.18) and (12.2.20) that utt(x, t) = a2uxx(x, t) for all (x, t) such that S′′ f (x ± at) and S′ g(x ± at) exist. Moreover, u may still provide a useful approximation to the vibration of the string; a laboratory experiment can confirm or deny this. We’ll now consider the initial-boundary value problem (12.2.1) with f(x) = x, 0 ≤x ≤L 2 , L −x, L 2 ≤x ≤L (12.2.21) and g ≡0. Since f isn’t differentiable at x = L/2, it does’nt satisfy the assumptions of Theorem 12.2.4, so the formal solution of (12.2.1) can’t be an actual solution. Nevertheless, it’s instructive to investigate the properties of the formal solution. The graph of f is shown in Figure 12.2.7. Intuitively, we are plucking the string by half its length at the middle. You’re right if you think this is an extraordinarily large displacement; however, we could remove this objection by multiplying the function in Figure 12.2.7 by a small constant. Since this would just multiply the formal solution by the same constant, we’ll leave f as we’ve defined it. Similar comments apply to the exercises. From Exercise 11.3.15, the Fourier sine series of f on [0, L] is Sf(x) = 4L π2 ∞ X n=1 (−1)n+1 (2n −1)2 sin (2n −1)πx L , | Elementary Differential Equations with Boundary Value Problems_Page_649_Chunk3052 |
640 Chapter 12 Fourier Solutions of Partial Differential Equations L .5 L .5 L x y Figure 12.2.7 Graph of (12.2.21) which converges to f for all x in [0, L], by Theorem 11.3.2. Therefore u(x, t) = 4L π2 ∞ X n=1 (−1)n+1 (2n −1)2 cos (2n −1)πat L sin (2n −1)πx L . (12.2.22) This series converges absolutely for all (x, t) by the comparison test, since the series ∞ X n=1 1 (2n −1)2 converges. Moreover, (12.2.22) satisfies the boundary conditions u(0, t) = u(L, t) = 0, t > 0, and the initial condition u(x, 0) = f(x), 0 ≤x ≤L. However, we can’t justify differentiating (12.2.22) term by term even once, and formally differentiating it twice term by term produces a series that diverges for all (x, t). (Verify.). Therefore we use d’Alembert’s form u(x, t) = 1 2[Sf(x + at) + Sf(x −at)] (12.2.23) for u to study its derivatives. Figure 12.2.8 shows the graph of Sf, which is the odd periodic extension of f. You can see from the graph that Sf is differentiable at x (and S′ f(x) = ±1) if and only if x isn’t an odd multiple of L/2. | Elementary Differential Equations with Boundary Value Problems_Page_650_Chunk3053 |
Section 12.2 The Wave Equation 641 x = 0 x = L x = 2L x = − L x = − 2L x = − 3L x = 3L x y − .5L .5L Figure 12.2.8 The odd periodic extension of (12.2.21) x = 0 x = L x = 2L x = − L x = − 2L x = − 3L x = 3L x y .5L − .5L Figure 12.2.9 Graphs of y = Sf(x −at) (dashed) and y = Sf(x −at) (solid), with f as in (12.2.21) In Figure 12.2.9 the dashed and solid curves are the graphs of y = Sf(x −at) and y = Sf(x + at) respectively, for a fixed value of t. As t increases the dashed curve moves to the right and the solid curve moves to the left. For this reason, we say that the functions u1(x, t) = Sf(x + at) and u2(x, t) = Sf(x −at) are traveling waves. Note that u1 satisfies the wave equation at (x, t) if x + at isn’t an odd multiple of L/2 and u2 satisfies the wave equation at (x, t) if x −at isn’t an odd multiple of L/2. Therefore (12.2.23) (or, equivalently, (12.2.22)) satisfies utt(x, t) = a2uxx(x, t) = 0 for all (x, t) such that neither x −at nor x + at is an odd multiple of L/2. We conclude by finding an explicit formula for u(x, t) under the assumption that 0 ≤x ≤L and 0 ≤t ≤L/2a. (12.2.24) To see how this formula can be used to compute u(x, t) for 0 ≤x ≤L and arbitrary t, we refer you to Exercise 16. From Figure 12.2.10, Sf(x −at) = ( x −at, 0 ≤x ≤L 2 + at, L −x + at, L 2 + at ≤x ≤L and Sf(x + at) = ( x + at, 0 ≤x ≤L 2 −at, L −x −at, L 2 −at ≤x ≤L if (x, t) satisfies (12.2.24). Therefore, from (12.2.23), u(x, t) = x, 0 ≤x ≤L 2 −at, L 2 −at, L 2 −at ≤x ≤L 2 + at, L −x, L 2 −at ≤x ≤L if (x, t) satisfies (12.2.24). Figure 12.2.11 is the graph of this function on [0, L] for a fixed t in (0, L/2a). | Elementary Differential Equations with Boundary Value Problems_Page_651_Chunk3054 |
642 Chapter 12 Fourier Solutions of Partial Differential Equations L at .5L .5L x =.5L − at x =.5L + at x y Figure 12.2.10 The part of the graph from Figure 12.2.9 on [0, L] L .5L y = .5L − at x =.5L + at x = .5L − at x y Figure 12.2.11 The graph of (12.2.23) on [0, L] for a fixed t in (0, L/2a) USING TECHNOLOGY Although the formal solution u(x, t) = ∞ X n=1 αn cos nπat L + βnL nπa sin nπat L sin nπx L of (12.2.1) is defined for all (x, t), we’re mainly interested in its behavior for 0 ≤x ≤L and t ≥0. In fact, it’s sufficient to consider only values of t in the interval 0 ≤t < 2L/a, since u(x, t + 2kL/a) = u(x, t) for all (x, t) if k is an integer. (Verify.) You can create an animation of the motion of the string by performing the following numerical experi- ment. Let m and k be positive integers. Let tj = 2Lj ka , j = 0, 1, . . .k; thus, t0, t1, ...tk are equally spaced points in [0, 2L/a]. For each j = 0, 1 ,2, ...k, graph the partial sum um(x, tj) = m X n=1 αn cos nπatj L + βnL nπa sin nπatj L sin nπx L on [0, L] as a function of x. Write your program so that each graph remains displayed on the monitor for a short time, and is then deleted and replaced by the next. Repeat this procedure for various values of m and k. We suggest that you perform experiments of this kind in the exercises marked C , without other specific instructions. (These exercises were chosen arbitrarily; the experiment is worthwhile in all the exercises dealing with specific initial-boundary value problems.) In some of the exercises the formal solutions have other forms, defined in Exercises 17, 34, and 49; however, the idea of the experiment is the same. | Elementary Differential Equations with Boundary Value Problems_Page_652_Chunk3055 |
Section 12.2 The Wave Equation 643 12.2 Exercises In Exercises 1-15 solve the initial-boundary value problem. In some of these exercises, Theorem 11.3.5(b) or Exercise 11.3.35 will simplify the computation of the coefficients in the Fourier sine series. 1. utt = 9uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x, 0 ≤x ≤1 2, 1 −x, 1 2 ≤x ≤1 , 0 ≤x ≤1 2. utt = 9uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = x(1 −x), ut(x, 0) = 0, 0 ≤x ≤1 3. utt = 7uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = x2(1 −x), ut(x, 0) = 0, 0 ≤x ≤1 4. C utt = 9uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x(1 −x), 0 ≤x ≤1 5. utt = 7uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 0 ut(x, 0) = x2(1 −x), 0 ≤x ≤1 6. utt = 64uxx, 0 < x < 3, t > 0, u(0, t) = 0, u(3, t) = 0, t > 0, u(x, 0) = x(x2 −9), ut(x, 0) = 0, 0 ≤x ≤3 7. utt = 4uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = x(x3 −2x2 + 1), ut(x, 0) = 0, 0 ≤x ≤1 8. C utt = 64uxx, 0 < x < 3, t > 0, u(0, t) = 0, u(3, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x(x2 −9), 0 ≤x ≤3 9. utt = 4uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x(x3 −2x2 + 1), 0 ≤x ≤1 10. utt = 5uxx, 0 < x < π, t > 0, u(0, t) = 0, u(π, t) = 0, t > 0, u(x, 0) = x sinx, ut(x, 0) = 0, 0 ≤x ≤π 11. utt = uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = x(3x4 −5x3 + 2), ut(x, 0) = 0, 0 ≤x ≤1 12. C utt = 5uxx, 0 < x < π, t > 0, u(0, t) = 0, u(π, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x sinx, 0 ≤x ≤π 13. utt = uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x(3x4 −5x3 + 2), 0 ≤x ≤1 | Elementary Differential Equations with Boundary Value Problems_Page_653_Chunk3056 |
644 Chapter 12 Fourier Solutions of Partial Differential Equations 14. utt = 9uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = x(3x4 −10x2 + 7), ut(x, 0) = 0, 0 ≤x ≤1 15. C utt = 9uxx, 0 < x < 1, t > 0, u(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 0 ut(x, 0) = x(3x4 −10x2 + 7), 0 ≤x ≤1 16. We saw that the displacement of the plucked string is, on the one hand, u(x, t) = 4L π2 ∞ X n=1 (−1)n+1 (2n −1)2 cos (2n −1)πat L sin (2n −1)πx L , 0 ≤x ≤L, t ≥0, (A) and, on the other hand, u(x, τ) = x, 0 ≤x ≤L 2 −aτ, L 2 −aτ, L 2 −aτ ≤x ≤L 2 + aτ, L −x, L 2 −aτ ≤x ≤L. (B) if 0 ≤τ ≤L/2a. The first objective of this exercise is to show that (B) can be used to compute u(x, t) for 0 ≤x ≤L and all t > 0. (a) Show that if t > 0, there’s a nonnegative integer m such that either (i) t = mL a + τ or (ii) t = (m + 1)L a −τ, where 0 ≤τ ≤L/2a. (b) Use (A) to show that u(x, t) = (−1)mu(x, τ) if (i) holds, while u(x, t) = (−1)m+1u(x, τ) if (ii) holds. (c) L Perform the following experiment for specific values of L and a and various values of m and k: Let tj = Lj 2ka, j = 0, 1, . . .k; thus, t0, t1, ..., tk are equally spaced points in [0, L/2a]. For each j = 0, 1 , 2,..., k, graph the mth partial sum of (A) and u(x, tj) computed from (B) on the same axis. Create an animation, as described in the remarks on using technology at the end of the section. 17. If a string vibrates with the end at x = 0 free to move in a frictionless vertical track and the end at x = L fixed, then the initial-boundary value problem for its displacement takes the form utt = a2uxx, 0 < x < L, t > 0, ux(0, t) = 0, u(L, t) = 0, t > 0, u(x, 0) = f(x), ut(x, 0) = g(x), 0 ≤x ≤L. (A) Justify defining the formal solution of (A) to be u(x, t) = ∞ X n=1 αn cos (2n −1)πat 2L + 2Lβn (2n −1)πa sin (2n −1)πat 2L cos (2n −1)πx 2L , where CMf(x) = ∞ X n=1 αn cos (2n −1)πx 2L and CMg(x) = ∞ X n=1 βn cos (2n −1)πx 2L | Elementary Differential Equations with Boundary Value Problems_Page_654_Chunk3057 |
Section 12.2 The Wave Equation 645 are the mixed Fourier cosine series of f and g on [0, L]; that is, αn = 2 L Z L 0 f(x) cos (2n −1)πx 2L dx and βn = 2 L Z L 0 g(x) cos (2n −1)πx 2L dx. In Exercises 18-31, use Exercise 17 to solve the initial-boundaryvalue problem. In some of these exercises Theorem 11.3.5(c) or Exercise 11.3.42(b) will simplify the computation of the coefficients in the mixed Fourier cosine series. 18. utt = 9uxx, 0 < x < 2, t > 0, ux(0, t) = 0, u(2, t) = 0, t > 0, u(x, 0) = 4 −x2, ut(x, 0) = 0, 0 ≤x ≤2 19. utt = 4uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = x2(1 −x), ut(x, 0) = 0, 0 ≤x ≤1 20. utt = 9uxx, 0 < x < 2, t > 0, ux(0, t) = 0, u(2, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = 4 −x2, 0 ≤x ≤2 21. utt = 4uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x2(1 −x), 0 ≤x ≤1 22. C utt = 5uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 2x3 + 3x2 −5, ut(x, 0) = 0, 0 ≤x ≤1 23. utt = 3uxx, 0 < x < π, t > 0, ux(0, t) = 0, u(π, t) = 0, t > 0, u(x, 0) = π3 −x3, ut(x, 0) = 0, 0 ≤x ≤π 24. utt = 5uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = 2x3 + 3x2 −5, 0 ≤x ≤1 25. C utt = 3uxx, 0 < x < π, t > 0, ux(0, t) = 0, u(π, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = π3 −x3, 0 ≤x ≤π 26. utt = 9uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = x4 −2x3 + 1, ut(x, 0) = 0, 0 ≤x ≤1 27. utt = 7uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 4x3 + 3x2 −7, ut(x, 0) = 0, 0 ≤x ≤1 28. utt = 9uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x4 −2x3 + 1, 0 ≤x ≤1 29. C utt = 7uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = 4x3 + 3x2 −7, 0 ≤x ≤1 | Elementary Differential Equations with Boundary Value Problems_Page_655_Chunk3058 |
646 Chapter 12 Fourier Solutions of Partial Differential Equations 30. utt = uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = x4 −4x3 + 6x2 −3, ut(x, 0) = 0, 0 ≤x ≤1 31. utt = uxx, 0 < x < 1, t > 0, ux(0, t) = 0, u(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x4 −4x3 + 6x2 −3, 0 ≤x ≤1 32. Adapt the proof of Theorem 12.2.2 to find d’Alembert’s solution of the initial-boundary value problem in Exercise 17. 33. Use the result of Exercise 32 to show that the formal solution of the initial-boundary value problem in Exercise 17 is an actual solution if g is differentiable and f is twice differentiable on [0, L] and g′ +(0) = g(L) = f′ +(0) = f(L) = f′′ −(L) = 0. HINT: See Exercise 11.3.57, and apply Theorem 12.2.3 with L replaced by 2L. 34. Justify defining the formal solution of the initial-boundary value problem utt = a2uxx, 0 < x < L, t > 0, u(0, t) = 0, ux(L, t) = 0, t > 0, u(x, 0) = f(x), ut(x, 0) = g(x), 0 ≤x ≤L to be u(x, t) = ∞ X n=1 αn cos (2n −1)πat 2L + 2Lβn (2n −1)πa sin (2n −1)πat 2L sin (2n −1)πx 2L , where SMf(x) = ∞ X n=1 αn sin (2n −1)πx 2L and SMg(x) = ∞ X n=1 βn sin (2n −1)πx 2L are the mixed Fourier sine series of f and g on [0, L]; that is, αn = 2 L Z L 0 f(x) sin (2n −1)πx 2L dx and βn = 2 L Z L 0 g(x) sin (2n −1)πx 2L dx. In Exercises 35-46 use Exercise 34 to solve the initial-boundaryvalue problem. In some of these exercises Theorem 11.3.5(d) or Exercise 11.3.50(b) will simplify the computation of the coefficients in the mixed Fourier sine series. 35. utt = 64uxx, 0 < x < π, t > 0, u(0, t) = 0, ux(π, t) = 0, t > 0, u(x, 0) = x(2π −x), ut(x, 0) = 0, 0 ≤x ≤π 36. utt = 9uxx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = x2(3 −2x), ut(x, 0) = 0, 0 ≤x ≤1 37. utt = 64uxx, 0 < x < π, t > 0, u(0, t) = 0, ux(π, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x(2π −x), 0 ≤x ≤π | Elementary Differential Equations with Boundary Value Problems_Page_656_Chunk3059 |
Section 12.2 The Wave Equation 647 38. C utt = 9uxx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x2(3 −2x), 0 ≤x ≤1 39. utt = 9uxx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = (x −1)3 + 1, ut(x, 0) = 0, 0 ≤x ≤1 40. utt = 3uxx, 0 < x < π, t > 0, u(0, t) = 0, ux(π, t) = 0, t > 0, u(x, 0) = x(x2 −3π2), ut(x, 0) = 0, 0 ≤x ≤π 41. utt = 9uxx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = (x −1)3 + 1, 0 ≤x ≤1 42. utt = 3uxx, 0 < x < π, t > 0, u(0, t) = 0, ux(π, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x(x2 −3π2), 0 ≤x ≤π 43. utt = 5uxx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = x3(3x −4), ut(x, 0) = 0, 0 ≤x ≤1 44. C utt = 16uxx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = x(x3 −2x2 + 2), ut(x, 0) = 0, 0 ≤x ≤1 45. utt = 5uxx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x3(3x −4), 0 ≤x ≤1 46. C utt = 16uxx, 0 < x < 1, t > 0, u(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x(x3 −2x2 + 2), 0 ≤x ≤1 47. Adapt the proof of Theorem 12.2.2 to find d’Alembert’s solution of the initial-boundary value problem in Exercise 34. 48. Use the result of Exercise 47 to show that the formal solution of the initial-boundary value problem in Exercise 34 is an actual solution if g is differentiable and f is twice differentiable on [0, L] and f(0) = f′ −(L) = g(0) = g′ −(L) = f′′ +(0) = 0. HINT: See Exercise 11.3.58 and apply Theorem 12.2.3 with L replaced by 2L. 49. Justify defining the formal solution of the initial-boundary value problem utt = a2uxx, 0 < x < L, t > 0, ux(0, t) = 0, ux(L, t) = 0, t > 0, u(x, 0) = f(x), ut(x, 0) = g(x), 0 ≤x ≤L. to be u(x, t) = α0 + β0t + ∞ X n=1 αn cos nπat L + Lβn nπa sin nπat L cos nπx L , where Cf(x) = α0 + ∞ X n=1 αn cos nπx L and Cg(x) = β0 + ∞ X n=1 βn cos nπx L | Elementary Differential Equations with Boundary Value Problems_Page_657_Chunk3060 |
648 Chapter 12 Fourier Solutions of Partial Differential Equations are the Fourier cosine series of f and g on [0, L]; that is, α0 = 1 L Z L 0 f(x) dx, β0 = 1 L Z L 0 g(x) dx, αn = 2 L Z L 0 f(x) cos nπx L dx, and βn = 2 L Z L 0 g(x) cos nπx L dx, n = 1, 2, 3, . . .. In Exercises 50-59 use Exercise 49 to solve the initial-boundaryvalue problem. In some of these exercises Theorem 11.3.5(a) will simplify the computation of the coefficients in the Fourier cosine series. 50. utt = 5uxx, 0 < x < 2, t > 0, ux(0, t) = 0, ux(2, t) = 0, t > 0, u(x, 0) = 2x2(3 −x), ut(x, 0) = 0, 0 ≤x ≤2 51. utt = 5uxx, 0 < x < 2, t > 0, ux(0, t) = 0, ux(2, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = 2x2(3 −x), 0 ≤x ≤2 52. utt = 4uxx, 0 < x < π, t > 0, ux(0, t) = 0, ux(π, t) = 0, t > 0, u(x, 0) = x3(3x −4π), ut(x, 0) = 0, 0 ≤x ≤π 53. utt = 7uxx, 0 < x < 1, t > 0, ux(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = 3x2(x2 −2), ut(x, 0) = 0, 0 ≤x ≤1 54. C utt = 4uxx, 0 < x < π, t > 0, ux(0, t) = 0, ux(π, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x3(3x −4π), 0 ≤x ≤π 55. utt = 7uxx, 0 < x < 1, t > 0, ux(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = 3x2(x2 −2), 0 ≤x ≤1 56. utt = 16uxx, 0 < x < π, t > 0, ux(0, t) = 0, ux(π, t) = 0, t > 0, u(x, 0) = x2(x −π)2, ut(x, 0) = 0, 0 ≤x ≤π 57. C utt = uxx, 0 < x < 1, t > 0, ux(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = x2(3x2 −8x + 6), ut(x, 0) = 0, 0 ≤x ≤1 58. utt = 16uxx, 0 < x < π, t > 0, ux(0, t) = 0, ux(π, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x2(x −π)2, 0 ≤x ≤π 59. C utt = uxx, 0 < x < 1, t > 0, ux(0, t) = 0, ux(1, t) = 0, t > 0, u(x, 0) = 0, ut(x, 0) = x2(3x2 −8x + 6), 0 ≤x ≤1 60. Adapt the proof of Theorem 12.2.2 to find d’Alembert’s solution of the initial-boundary value problem in Exercise 49. 61. Use the result of Exercise 60 to show that the formal solution of the initial-boundary value problem in Exercise 49 is an actual solution if g is differentiable and f is twice differentiable on [0, L] and f′ +(0) = f′ −(L) = g′ +(0) = g′ −(L) = 0. | Elementary Differential Equations with Boundary Value Problems_Page_658_Chunk3061 |
Section 12.2 The Wave Equation 649 62. Suppose λ and µ are constants and either pn(x) = cos nλx or pn(x) = sin nλx, while either qn(t) = cos nµt or qn(t) = sin nµt for n = 1, 2, 3, .... Let u(x, t) = ∞ X n=1 knpn(x)qn(t), (A) where {kn}∞ n=1 are constants. (a) Show that if P∞ n=1 |kn| converges then u(x, t) converges for all (x, t). (b) Use Theorem 12.1.2 to show that if P∞ n=1 n|kn| converges then (A) can be differentiated term by term with respect to x and t for all (x, t); that is, ux(x, t) = ∞ X n=1 knp′ n(x)qn(t) and ut(x, t) = ∞ X n=1 knpn(x)q′ n(t). (c) Suppose P∞ n=1 n2|kn| converges. Show that uxx(x, y) = ∞ X n=1 knp′′ n(x)qn(t) and utt(x, y) = ∞ X n=1 knpn(x)q′′ n(t) (d) Suppose P∞ n=1 n2|αn| and P∞ n=1 n|βn| both converge. Show that the formal solution u(x, t) = ∞ X n=1 αn cos nπat L + βnL nπa sin nπat L sin nπx L of Equation 12.2.1 satisfies utt = a2uxx for all (x, t). This conclusion also applies to the formal solutions defined in Exercises 17, 34, and 49. 63. Suppose g is differentiable and f is twice differentiable on (−∞, ∞), and let u0(x, t) = f(x + at) + f(x −at) 2 and u1(x, t) = 1 2a Z x+at x−at g(u) du. (a) Show that ∂2u0 ∂t2 = a2 ∂2u0 ∂x2 , −∞< x < ∞, t > 0, and u0(x, 0) = f(x), ∂u0 ∂t (x, 0) = 0, −∞< x < ∞. (b) Show that ∂2u1 ∂t2 = a2 ∂2u1 ∂x2 , −∞< x < ∞, t > 0, and u1(x, 0) = 0, ∂u1 ∂t (x, 0) = g(x), −∞< x < ∞. | Elementary Differential Equations with Boundary Value Problems_Page_659_Chunk3062 |
650 Chapter 12 Fourier Solutions of Partial Differential Equations (c) Solve utt = a2uxx, −∞< t < ∞, t > 0, u(x, 0) = f(x), ut(x, 0) = g(x), −∞< x < ∞. In Exercises 64-68 use the result of Exercise 63 to find a solution of utt = a2uxx, −∞< x < ∞ that satisfies the given initial conditions. 64. u(x, 0) = x, ut(x, 0) = 4ax, −∞< x < ∞ 65. u(x, 0) = x2, ut(x, 0) = 1, −∞< x < ∞ 66. u(x, 0) = sin x, ut(x, 0) = a cos x, −∞< x < ∞ 67. u(x, 0) = x3, ut(x, 0) = 6x2, −∞< x < ∞ 68. u(x, 0) = x sinx, ut(x, 0) = sin x, −∞< x < ∞ 12.3 LAPLACE’S EQUATION IN RECTANGULAR COORDINATES The temperature u = u(x, y, t) in a two-dimensional plate satisfies the two-dimensional heat equation ut = a2(uxx + uyy), (12.3.1) where (x, y) varies over the interior of the plate and t > 0. To find a solution of (12.3.1), it’s necessary to specify the initial temperature u(x, y, 0) and conditions that must be satisfied on the boundary. However, as t →∞, the influence of the initial condition decays, so lim t→∞ut(x, y, t) = 0 and the temperature approaches a steady state distribution u = u(x, y) that satisfies uxx + uyy = 0. (12.3.2) This is Laplace’s equation. This equation also arises in applications to fluid mechanics and potential theory; in fact, it is also called the potential equation. We seek solutions of (12.3.2) in a region R that satisfy specified conditions – called boundary conditions – on the boundary of R. For example, we may require u to assume prescribed values on the boundary. This is called a Dirichlet condition, and the problem is called a Dirichlet problem. Or, we may require the normal derivative of u at each point (x, y) on the boundary to assume prescribed values. This is called a Neumann condition, and the problem is called a Neumann problem. In some problems we impose Dirichlet conditions on part of the boundary and Neumann conditions on the rest. Then we say that the boundary conditions and the problem are mixed. Solving boundary value problems for (12.3.2) over general regions is beyond the scope of this book, so we consider only very simple regions. We begin by considering the rectangular region shown in Figure 12.3.1. The possible boundary conditions for this region can be written as (1 −α)u(x, 0) + αuy(x, 0) = f0(x), 0 ≤x ≤a, (1 −β)u(x, b) + βuy(x, b) = f1(x), 0 ≤x ≤a, (1 −γ)u(0, y) + γux(0, y) = g0(y), 0 ≤y ≤b, (1 −δ)u(a, y) + δux(a, y) = g1(y), 0 ≤y ≤b, | Elementary Differential Equations with Boundary Value Problems_Page_660_Chunk3063 |
Section 12.3 Laplace’s Equation in Rectangular Coordinates 651 y x a b Figure 12.3.1 A rectangular region and its boundary where α, β, γ, and δ can each be either 0 or 1; thus, there are 16 possibilities. Let BVP(α, β, γ, δ)(f0, f1, g0, g1) denote the problem of finding a solution of (12.3.2) that satisfies these conditions. This is a Dirichlet problem if α = β = γ = δ = 0 (Figure 12.3.2), or a Neumann problem if α = β = γ = δ = 1 (Figure 12.3.3). The other 14 problems are mixed. y x a b uxx + uyy = 0 u(x,0) = f0(x) u(x,b) = f1(x) u(0,y) = g0(y) u(a,y) = g1(y) Figure 12.3.2 A Dirichlet problem y x a b uxx + uyy = 0 uy(x,0) = f0(x) uy(x,b) = f1(x) ux(0,y) = g0(y) ux(a,y) = g1(y) Figure 12.3.3 A Neumann problem For given (α, β, γ, δ), the sum of solutions of BVP(α, β, γ, δ)(f0, 0, 0, 0), BVP(α, β, γ, δ)(0, f1, 0, 0), | Elementary Differential Equations with Boundary Value Problems_Page_661_Chunk3064 |
652 Chapter 12 Fourier Solutions of Partial Differential Equations BVP(α, β, γ, δ)(0, 0, g0, 0), and BVP(α, β, γ, δ)(0, 0, 0, g1) is a solution of BVP(α, β, γ, δ)(f0, f1, g0, g1). Therefore we concentrate on problems where only one of the functions f0, f1, g0, g2 isn’t identically zero. There are 64 (count them!) problems of this form. Each has homogeneous boundary conditions on three sides of the rectangle, and a nonhomogeneous boundary condition on the fourth. We use separation of variables to find infinitely many functions that satisfy Laplace’s equation and the three homogeneous boundary conditions in the open rectangle. We then use these solutions as building blocks to construct a formal solution of Laplace’s equation that also satisfies the nonhomogeneous boundary condition. Since it’s not feasible to consider all 64 cases, we’ll restrict our attention in the text to just four. Others are discussed in the exercises. If v(x, y) = X(x)Y (y) then vxx + vyy = X′′Y + XY ′′ = 0 for all (x, y) if and only if X′′ X = −Y ′′ Y = k for all (x, y), where k is a separation constant. This equation is equivalent to X′′ −kX = 0, Y ′′ + kY = 0. (12.3.3) From here, the strategy depends upon the boundary conditions. We illustrate this by examples. Example 12.3.1 Define the formal solution of uxx + uyy = 0, 0 < x < a, 0 < y < b, u(x, 0) = f(x), u(x, b) = 0, 0 ≤x ≤a, u(0, y) = 0, u(a, y) = 0, 0 ≤y ≤b (12.3.4) (Figure 12.3.4). Solution The boundary conditions in (12.3.4) require products v(x, y) = X(x)Y (y) such that X(0) = X(a) = Y (b) = 0; hence, we let k = −λ in (12.3.3). Thus, X and Y must satisfy X′′ + λX = 0, X(0) = 0, X(a) = 0 (12.3.5) and Y ′′ −λY = 0, Y (b) = 0. (12.3.6) From Theorem 11.1.2, the eigenvalues of (12.3.5) are λn = n2π2/a2, with associated eigenfunctions Xn = sin nπx a , n = 1, 2, 3, . . .. Substituting λ = n2π2/a2 into (12.3.6) yields Y ′′ −(n2π2/a2)Y = 0, Y (b) = 0, so we could take Yn = sinh nπ(b −y) a ; (12.3.7) | Elementary Differential Equations with Boundary Value Problems_Page_662_Chunk3065 |
Section 12.3 Laplace’s Equation in Rectangular Coordinates 653 y x a b uxx + uyy = 0 u(x,0) = f(x) u(x,b) = 0 u(0,y) = 0 u(a,y) = 0 Figure 12.3.4 The boundary value problem (12.3.4) however, because of the nonhomogeneous Dirichlet condition at y = 0, it’s better to require that Yn(0) = 1, which can be achieved by dividing the right side of (12.3.7) by its value at y = 0; thus, we take Yn = sinhnπ(b −y)/a sinh nπb/a . Then vn(x, y) = Xn(x)Yn(y) = sinh nπ(b −y)/a sinh nπb/a sin nπx a , so vn(x, 0) = sin nπx/a and vn satisfies (12.3.4) with f(x) = sin nπx/a. More generally, if α1, ..., αm are arbitrary constants then um(x, y) = m X n=1 αn sinh nπ(b −y)/a sinh nπb/a sin nπx a satisfies (12.3.4) with f(x) = m X n=1 αn sin nπx L . Therefore, if f is an arbitrary piecewise smooth function on [0, a], we define the formal solution of (12.3.4) to be u(x, y) = ∞ X n=1 αn sinh nπ(b −y)/a sinh nπb/a sin nπx a , (12.3.8) where S(x) = ∞ X n=1 αn sin nπx a | Elementary Differential Equations with Boundary Value Problems_Page_663_Chunk3066 |
654 Chapter 12 Fourier Solutions of Partial Differential Equations is the Fourier sine series of f on [0, a]; that is, αn = 2 a Z a 0 f(x) sin nπx a dx, n = 1, 2, 3, . . .. If y < b then sinh nπ(b −y)/a sinh nπb/a ≈e−nπy/a (12.3.9) for large n, so the series in (12.3.8) converges if 0 < y < b; moreover, since also cosh nπ(b −y)/a sinh nπb/a ≈e−nπy/a for large n, Theorem 12.1.2 applied twice with z = x and twice with z = t, shows that uxx and uyy can be obtained by differentiating u term by term if 0 < y < b. (Exercise 37). Therefore u satisfies Laplace’s equation in the interior of the rectangle in Figure 12.3.4. Moreover, the series in (12.3.8) also converges on the boundary of the rectangle, and satisfies the three homogeneous boundary conditions conditions in (12.3.4). Therefore, since u(x, 0) = S(x) for 0 ≤x ≤L, u is an actual solution of (12.3.5) if and only if S(x) = f(x) for 0 ≤x ≤a. From Theorem 11.3.2, this is true if f is continuous and piecewise smooth on [0, L], and f(0) = f(L) = 0. Example 12.3.2 Solve (12.3.4) with f(x) = x(x2 −3ax + 2a2). Solution From Example 11.3.6, S(x) = 12a3 π3 ∞ X n=1 1 n3 sin nπx a . Therefore u(x, y) = 12a3 π3 ∞ X n=1 sinh nπ(b −y)/a n3 sinh nπb/a sin nπx a . (12.3.10) To compute approximate values of u(x, y), we must use partial sums of the form um(x, y) = 12a3 π3 m X n=1 sinh nπ(b −y)/a n3 sinh nπb/a sin nπx a . Because of (12.3.9), small values of m provide sufficient accuracy for most applications if 0 < y < b. Moreover, the n3 in the denominator in (12.3.10) ensures that this is also true for y = 0. For graphing purposes, we chose a = 2, b = 1, and m = 10. Figure 12.3.5 shows the surface u = u(x, y), 0 ≤x ≤2, 0 ≤y ≤1, while Figure 12.3.6 shows the curves u = u(x, 0.1k), 0 ≤x ≤2, k = 0, 1, . . ., 10. | Elementary Differential Equations with Boundary Value Problems_Page_664_Chunk3067 |
Section 12.3 Laplace’s Equation in Rectangular Coordinates 655 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 0 0.5 1 1.5 2 2.5 3 Figure 12.3.5 1 2 1 2 3 x y Figure 12.3.6 Example 12.3.3 Define the formal solution of uxx + uyy = 0, 0 < x < a, 0 < y < b, u(x, 0) = 0, uy(x, b) = f(x), 0 ≤x ≤a, ux(0, y) = 0, ux(a, y) = 0, 0 ≤y ≤b (12.3.11) (Figure 12.3.7). y x a b uxx + uyy = 0 u(x,0) = 0 uy(x,b) = f(x) ux(0,y) = 0 ux(a,y) = 0 Figure 12.3.7 The boundary value problem (12.3.11) Solution The boundary conditions in (12.3.11) require products v(x, y) = X(x)Y (y) such that X′(0) = | Elementary Differential Equations with Boundary Value Problems_Page_665_Chunk3068 |
656 Chapter 12 Fourier Solutions of Partial Differential Equations X′(a) = Y (0) = 0; hence, we let k = −λ in (12.3.3). Thus, X and Y must satisfy X′′ + λX = 0, X′(0) = 0, X′(a) = 0 (12.3.12) and Y ′′ −λY = 0, Y (0) = 0. (12.3.13) From Theorem 11.1.3, the eigenvalues of (12.3.12) are λ = 0, with associated eigenfunction X0 = 1, and λn = n2π2/a2, with associated eigenfunctions Xn = cos nπx a , n = 1, 2, 3, . . .. Since Y0 = y satisfies (12.3.13) with λ = 0, we take v0(x, y) = X0(x)Y0(y) = y. Substituting λ = n2π2/a2 into (12.3.13) yields Y ′′ −(n2π2/a2)Y = 0, Y (0) = 0, so we could take Yn = sinh nπy a . (12.3.14) However, because of the nonhomogeneous Neumann condition at y = b, it’s better to require that Y ′ n(b) = 1, which can be achieved by dividing the right side of (12.3.14) by the value of its derivative at y = b; thus, Yn = a sinh nπy/a nπ cosh nπb/a. Then vn(x, y) = Xn(x)Yn(y) = a sinhnπy/a nπ cosh nπb/a cos nπx a , so ∂vn ∂y (x, b) = cos nπx a . Therefore vn satisfies (12.3.11) with f(x) = cos nπx/a. More generally, if α0, ..., αm are arbitrary constants then um(x, y) = α0y + a π m X n=1 αn sinh nπy/a n cosh nπb/a cos nπx a satisfies (12.3.11) with f(x) = α0 + m X n=1 αn cos nπx L . Therefore, if f is an arbitrary piecewise smooth function on [0, a] we define the formal solution of (12.3.11) to be u(x, y) = α0y + a π ∞ X n=1 αn sinh nπy/a n cosh nπb/a cos nπx a , where C(x) = α0 + ∞ X n=1 αn cos nπx a is the Fourier cosine series of f on [0, a]; that is, α0 = 1 a Z a 0 f(x) dx and αn = 2 a Z a 0 f(x) cos nπx a dx, n = 1, 2, 3, . . .. | Elementary Differential Equations with Boundary Value Problems_Page_666_Chunk3069 |
Section 12.3 Laplace’s Equation in Rectangular Coordinates 657 Example 12.3.4 Solve (12.3.11) with f(x) = x. Solution From Example 11.3.1, C(x) = a 2 −4a π2 ∞ X n=1 1 (2n −1)2 cos (2n −1)πx a . Therefore u(x, y) = ay 2 −4a2 π3 ∞ X n=1 sinh(2n −1)πy/a (2n −1)3 cosh(2n −1)πb/a cos (2n −1)πx a . (12.3.15) For graphing purposes, we chose a = 2, b = 1, and retained the terms through n = 10 in (12.3.15). Figure 12.3.8 shows the surface u = u(x, y), 0 ≤x ≤2, 0 ≤y ≤1, while Figure 12.3.9 shows the curves u = u(x, .1k), 0 ≤x ≤2, k = 0, 1, . . ., 10. 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 0 0.5 1 1.5 Figure 12.3.8 1 2 1 x y Figure 12.3.9 Example 12.3.5 Define the formal solution of uxx + uyy = 0, 0 < x < a, 0 < y < b, u(x, 0) = 0, uy(x, b) = 0, 0 ≤x ≤a, u(0, y) = g(y), ux(a, y) = 0, 0 ≤y ≤b (12.3.16) (Figure 12.3.10). Solution The boundary conditions in (12.3.16) require products v(x, y) = X(x)Y (y) such that Y (0) = Y ′(b) = X′(a) = 0; hence, we let k = λ in (12.3.3). Thus, X and Y must satisfy X′′ −λX = 0, X′(a) = 0 (12.3.17) | Elementary Differential Equations with Boundary Value Problems_Page_667_Chunk3070 |
658 Chapter 12 Fourier Solutions of Partial Differential Equations y x a b uxx + uyy = 0 u(x,0) = 0 uy(x,b) = 0 u(0,y) = g(y) ux(a,y) = 0 Figure 12.3.10 The boundary value problem (12.3.16) and Y ′′ + λY = 0, Y (0) = 0, Y ′(b) = 0. (12.3.18) From Theorem 11.1.4, the eigenvalues of (12.3.18) are λn = (2n −1)2π2/4b2, with associated eigen- functions Yn = sin (2n −1)πy 2b , n = 1, 2, 3, . . .. Substituting λ = (2n −1)2π2/4b2 into (12.3.17) yields X′′ −((2n −1)2π2/4b2)X = 0, X′(a) = 0, so we could take Xn = cosh (2n −1)π(x −a) 2b . (12.3.19) However, because of the nonhomogeneousDirichlet condition at x = 0, it’s better to require that Xn(0) = 1, which can be achieved by dividing the right side of (12.3.19) by its value at x = 0; thus, Xn = cosh(2n −1)π(x −a)/2b cosh(2n −1)πa/2b . Then vn(x, y) = Xn(x)Yn(y) = cosh(2n −1)π(x −a)/2b cosh(2n −1)πa/2b sin (2n −1)πy 2b , so vn(0, y) = sin (2n −1)πy 2b . | Elementary Differential Equations with Boundary Value Problems_Page_668_Chunk3071 |
Section 12.3 Laplace’s Equation in Rectangular Coordinates 659 Therefore vn satisfies (12.3.16) with g(y) = sin(2n −1)πy/2b. More generally, if α1, ..., αm are arbitrary constants then um(x, y) = m X n=1 αn cosh(2n −1)π(x −a)/2b cosh(2n −1)πa/2b sin (2n −1)πy 2b satisfies (12.3.16) with g(y) = m X n=1 αn sin (2n −1)πy 2b . Thus, if g is an arbitrary piecewise smooth function on [0, b], we define the formal solution of (12.3.16) to be u(x, y) = ∞ X n=1 αn cosh(2n −1)π(x −a)/2b cosh(2n −1)πa/2b sin (2n −1)πy 2b , where SM(x) = ∞ X n=1 αn sin (2n −1)πy 2b is the mixed Fourier sine series of g on [0, b]; that is, αn = 2 b Z b 0 g(y) sin (2n −1)πy 2b dy. Example 12.3.6 Solve (12.3.16) with g(y) = y(2y2 −9by + 12b2). Solution From Example 11.3.8, SM(y) = 96b3 π3 ∞ X n=1 1 (2n −1)3 3 + (−1)n 4 (2n −1)π sin (2n −1)πy 2b . Therefore u(x, y) = 96b3 π3 ∞ X n=1 cosh(2n −1)π(x −a)/2b (2n −1)3 cosh(2n −1)πa/2b 3 + (−1)n 4 (2n −1)π sin (2n −1)πy 2b . Example 12.3.7 Define the formal solution of uxx + uyy = 0, 0 < x < a, 0 < y < b, uy(x, 0) = 0, u(x, b) = 0, 0 ≤x ≤a, ux(0, y) = 0, ux(a, y) = g(y), 0 ≤y ≤b (12.3.20) (Figure 12.3.11). Solution The boundary conditions in (12.3.20) require products v(x, y) = X(x)Y (y) such that Y ′(0) = Y (b) = X′(0) = 0; hence, we let k = λ in (12.3.3). Thus, X and Y must satisfy X′′ −λX = 0, X′(0) = 0 (12.3.21) and Y ′′ + λY = 0, Y ′(0) = 0, Y (b) = 0. (12.3.22) | Elementary Differential Equations with Boundary Value Problems_Page_669_Chunk3072 |
660 Chapter 12 Fourier Solutions of Partial Differential Equations y x a b uxx + uyy = 0 uy(x,0) = 0 u(x,b) = 0 ux(0,y) = 0 ux(a,y) = g(y) Figure 12.3.11 The boundary value problem (12.3.20) From Theorem 11.1.4, the eigenvalues of (12.3.22) are λn = (2n −1)2π2/4b2, with associated eigen- functions Yn = cos (2n −1)πy 2b , n = 1, 2, 3, . . .. Substituting λ = (2n −1)2π2/4b2 into (12.3.21) yields X′′ −((2n −1)2π2/4b2)X = 0, X′(0) = 0, so we could take Xn = cosh (2n −1)πx 2b . (12.3.23) However, because of the nonhomogeneous Neumann condition at x = a, it’s better to require that X′ n(a) = 1, which can be achieved by dividing the right side of (12.3.23) by the value of its deriva- tive at x = a; thus, Xn = 2b cosh(2n −1)πx/2b (2n −1)π sinh(2n −1)πa/2b. Then vn(x, y) = Xn(x)Yn(y) = 2b cosh(2n −1)πx/2b (2n −1)π sinh(2n −1)πa/2b cos (2n −1)πy 2b , so ∂vn ∂x (a, y) = cos (2n −1)πy 2b . Therefore vn satisfies (12.3.20) with g(y) = cos(2n −1)πy/2b. More generally, if α1, ..., αm are arbitrary constants then um(x, y) = 2b π m X n=1 αn cosh(2n −1)πx/2b (2n −1) sinh(2n −1)πa/2b cos (2n −1)πy 2b | Elementary Differential Equations with Boundary Value Problems_Page_670_Chunk3073 |
Section 12.3 Laplace’s Equation in Rectangular Coordinates 661 satisfies (12.3.20) with g(y) = ∞ X n=1 αn cos (2n −1)πy 2b . Therefore, if g is an arbitrary piecewise smooth function on [0, b], we define the formal solution of (12.3.20) to be u(x, y) = 2b π ∞ X n=1 αn cosh(2n −1)πx/2b (2n −1) sinh(2n −1)πa/2b cos (2n −1)πy 2b , where CM(y) = ∞ X n=1 αn cos (2n −1)πy 2b is the mixed Fourier cosine series of g on [0, b]; that is, αn = 2 b Z b 0 g(y) cos (2n −1)πy 2b dy. Example 12.3.8 Solve (12.3.20) with g(y) = y −b. Solution From Example 11.3.3, CM(y) = −8b π2 ∞ X n=1 1 (2n −1)2 cos (2n −1)πy 2b . Therefore u(x, y) = −16b2 π3 ∞ X n=1 cosh(2n −1)πx/2b (2n −1)3 sinh(2n −1)πa/2b cos (2n −1)πy 2b . Laplace’s Equation for a Semi-Infinite Strip We now seek solutions of Laplace’s equation on the semi-infinite strip S : {0 < x < a, y > 0} (Figure 12.3.12) that satisfy homogeneous boundary conditions at x = 0 and x = a, and a nonhomoge- neous Dirichlet or Neumann condition at y = 0. An example of such a problem is uxx + uyy = 0, 0 < x < a, y > 0, u(x, 0) = f(x), 0 ≤x ≤a, u(0, y) = 0, u(a, y) = 0, y > 0, (12.3.24) The boundary conditions in this problem are not sufficient to determine u, for if u0 = u0(x, y) is a solution and K is a constant then u1(x, y) = u0(x, y) + K sin πx a sinh πy a . is also a solution. (Verify.) However, if we also require — on physical grounds — that the solution remain bounded for all (x, y) in S then K = 0 and this difficulty is eliminated. | Elementary Differential Equations with Boundary Value Problems_Page_671_Chunk3074 |
662 Chapter 12 Fourier Solutions of Partial Differential Equations y x a uxx + uyy = 0 u(x,0) = f(x) u(0,y) = 0 u(a,y) = 0 Figure 12.3.12 A boundary value problem on a semi-infinite strip Example 12.3.9 Define the bounded formal solution of (12.3.24). Solution Proceeding as in the solution of Example 12.3.1, we find that the building block functions are of the form vn(x, y) = Yn(y) sin nπx a , where Y ′′ n −(n2π2/a2)Yn = 0. Therefore Yn = c1enπy/a + c2e−nπy/a where c1 and c2 are constants. Although the boundary conditions in (12.3.24) don’t restrict c1, and c2, we must set c1 = 0 to ensure that Yn is bounded. Letting c2 = 1 yields vn(x, y) = e−nπy/a sin nπx a , and we define the bounded formal solution of (12.3.24) to be u(x, y) = ∞ X n=1 bne−nπy/a sin nπx a , where S(x) = ∞ X n=1 bn sin nπx a is the Fourier sine series of f on [0, a]. | Elementary Differential Equations with Boundary Value Problems_Page_672_Chunk3075 |
Section 12.3 Laplace’s Equation in Rectangular Coordinates 663 See Exercises 29-34 for other boundary value problems on a semi-infinite strip. 12.3 Exercises In Exercises 1-16 apply the definition developed in Example 1 to solve the boundary value problem. (Use Theorem 11.3.5 where it applies.) Where indicated by C , graph the surface u = u(x, y), 0 ≤x ≤a, 0 ≤y ≤b. 1. uxx + uyy = 0, 0 < x < 1, 0 < y < 1, u(x, 0) = x(1 −x), u(x, 1) = 0, 0 ≤x ≤1, u(0, y) = 0, u(1, y) = 0, 0 ≤y ≤1 2. uxx + uyy = 0, 0 < x < 2, 0 < y < 3, u(x, 0) = x2(2 −x), u(x, 3) = 0, 0 ≤x ≤2, u(0, y) = 0, u(2, y) = 0, 0 ≤y ≤3 3. C uxx + uyy = 0, 0 < x < 2, 0 < y < 2, u(x, 0) = x, 0 ≤x ≤1, 2 −x, 1 ≤x ≤2, u(x, 2) = 0, 0 ≤x ≤2, u(0, y) = 0, u(2, y) = 0, 0 ≤y ≤2 4. uxx + uyy = 0, 0 < x < π, 0 < y < 1, u(x, 0) = x sinx, u(x, π) = 0, 0 ≤x ≤π, u(0, y) = 0, u(π, y) = 0, 0 ≤y ≤1 5. uxx + uyy = 0, 0 < x < 3, 0 < y < 2, u(x, 0) = 0, uy(x, 2) = x2, 0 ≤x ≤3, ux(0, y) = 0, ux(3, y) = 0, 0 ≤y ≤2 6. uxx + uyy = 0, 0 < x < 1, 0 < y < 2, u(x, 0) = 0, uy(x, 2) = 1 −x, 0 ≤x ≤1, ux(0, y) = 0, ux(1, y) = 0, 0 ≤y ≤2 7. uxx + uyy = 0, 0 < x < 2, 0 < y < 2, u(x, 0) = 0, uy(x, 2) = x2 −4, 0 ≤x ≤2, ux(0, y) = 0, ux(2, y) = 0, 0 ≤y ≤2 8. uxx + uyy = 0, 0 < x < 1, 0 < y < 1, u(x, 0) = 0, uy(x, 1) = (x −1)2, 0 ≤x ≤1, ux(0, y) = 0, ux(1, y) = 0, 0 ≤y ≤1 9. C uxx + uyy = 0, 0 < x < 3, 0 < y < 2, u(x, 0) = 0, uy(x, 2) = 0, 0 ≤x ≤3, u(0, y) = y(4 −y), ux(3, y) = 0, 0 ≤y ≤2 10. uxx + uyy = 0, 0 < x < 2, 0 < y < 1, u(x, 0) = 0, uy(x, 1) = 0, 0 ≤x ≤2, u(0, y) = y2(3 −2y), ux(2, y) = 0, 0 ≤y ≤1 11. uxx + uyy = 0, 0 < x < 2, 0 < y < 2, u(x, 0) = 0, uy(x, 2) = 0, 0 ≤x ≤2, u(0, y) = (y −2)3 + 8, ux(2, y) = 0, 0 ≤y ≤2 12. uxx + uyy = 0, 0 < x < 3, 0 < y < 1, u(x, 0) = 0, uy(x, 1) = 0, 0 ≤x ≤3, u(0, y) = y(2y2 −9y + 12), ux(3, y) = 0, 0 ≤y ≤1 | Elementary Differential Equations with Boundary Value Problems_Page_673_Chunk3076 |
664 Chapter 12 Fourier Solutions of Partial Differential Equations 13. C uxx + uyy = 0, 0 < x < 1, 0 < y < π, uy(x, 0) = 0, u(x, π) = 0, 0 ≤x ≤1, ux(0, y) = 0, ux(1, y) = sin y, 0 ≤y ≤π 14. uxx + uyy = 0, 0 < x < 2, 0 < y < 3, uy(x, 0) = 0, u(x, 3) = 0, 0 ≤x ≤2, ux(0, y) = 0, ux(2, y) = y(3 −y), 0 ≤y ≤3 15. uxx + uyy = 0, 0 < x < 1, 0 < y < π, uy(x, 0) = 0, u(x, π) = 0, 0 ≤x ≤1, ux(0, y) = 0, ux(1, y) = π2 −y2, 0 ≤y ≤π 16. uxx + uyy = 0, 0 < x < 1, 0 < y < 1, uy(x, 0) = 0, u(x, 1) = 0, 0 ≤x ≤1, ux(0, y) = 0, ux(1, y) = 1 −y3, 0 ≤y ≤1 In Exercises 17-28 define the formal solution of uxx + uyy = 0, 0 < x < a, 0 < y < b that satisfies the given boundary conditions for general a, b, and f or g. Then solve the boundary value problem for the specified a, b, and f or g. (Use Theorem 11.3.5 where it applies.) Where indicated by C , graph the surface u = u(x, y), 0 ≤x ≤a, 0 ≤y ≤b. 17. C u(x, 0) = 0, u(x, b) = f(x), 0 < x < a, u(0, y) = 0, u(a, y) = 0, 0 < y < b a = 3, b = 2, f(x) = x(3 −x) 18. u(x, 0) = f(x), u(x, b) = 0, 0 < x < a, ux(0, y) = 0, ux(a, y) = 0, 0 < y < b a = 2, b = 1, f(x) = x2(x −2)2 19. u(x, 0) = f(x), u(x, b) = 0, 0 < x < a, ux(0, y) = 0, u(a, y) = 0, 0 < y < b a = 1, b = 2, f(x) = 3x3 −4x2 + 1 20. u(x, 0) = f(x), u(x, b) = 0, 0 < x < a, u(0, y) = 0, ux(a, y) = 0, 0 < y < b a = 3, b = 2, f(x) = x(6 −x) 21. u(x, 0) = f(x), uy(x, b) = 0, 0 < x < a, u(0, y) = 0, u(a, y) = 0, 0 < y < b a = π, b = 2, f(x) = x(π2 −x2) 22. uy(x, 0) = 0, u(x, b) = f(x), 0 < x < a, ux(0, y) = 0, ux(a, y) = 0, 0 < y < b a = π, b = 1, f(x) = x2(x −π)2 23. C uy(x, 0) = f(x), u(x, b) = 0, 0 < x < a, u(0, y) = 0, u(a, y) = 0, 0 < y < b a = π, b = 1, f(x) = x, 0 ≤x ≤π 2 , π −x, π 2 ≤x ≤π 24. u(x, 0) = 0, u(x, b) = 0, 0 < x < a, ux(0, y) = 0, u(a, y) = g(y), 0 < y < b a = 1, b = 1, g(y) = y(y3 −2y2 + 1) | Elementary Differential Equations with Boundary Value Problems_Page_674_Chunk3077 |
Section 12.3 Laplace’s Equation in Rectangular Coordinates 665 25. C uy(x, 0) = 0, u(x, b) = 0, 0 < x < a, ux(0, y) = 0, u(a, y) = g(y), 0 < y < b a = 2, b = 2, g(y) = 4 −y2 26. u(x, 0) = 0, u(x, b) = 0, 0 < x < a, ux(0, y) = 0, ux(a, y) = g(y), 0 < y < b a = 1, b = 4, g(y) = y, 0 ≤y ≤2, 4 −y, 2 ≤y ≤4 27. u(x, 0) = 0, uy(x, b) = 0, 0 < x < a, ux(0, y) = g(y), ux(a, y) = 0, 0 < y < b a = 1, b = π, g(y) = y2(3π −2y) 28. uy(x, 0) = 0, uy(x, b) = 0, 0 < x < a, ux(0, y) = g(y), u(a, y) = 0, 0 < y < b a = 2, b = π, g(y) = y In Exercises 29-34 define the bounded formal solution of uxx + uyy = 0, 0 < x < a, y > 0 that satisfies the given boundary conditions for general a and f. Then solve the boundary value problem for the specified a and f. 29. u(x, 0) = f(x), 0 < x < a, ux(0, y) = 0, ux(a, y) = 0, y > 0 a = π f(x) = x2(3π −2x) 30. u(x, 0) = f(x), 0 < x < a, ux(0, y) = 0, u(a, y) = 0, y > 0 a = 3, f(x) = 9 −x2 31. u(x, 0) = f(x), 0 < x < a, u(0, y) = 0, ux(a, y) = 0, y > 0 a = π, f(x) = x(2π −x) 32. uy(x, 0) = f(x), 0 < x < a, u(0, y) = 0, u(a, y) = 0, y > 0 a = π, f(x) = x2(π −x) 33. uy(x, 0) = f(x), 0 < x < a, ux(0, y) = 0, u(a, y) = 0, y > 0 a = 7, f(x) = x(7 −x) 34. uy(x, 0) = f(x), 0 < x < a, u(0, y) = 0, ux(a, y) = 0, y > 0 a = 5, f(x) = x(5 −x) 35. Define the formal solution of the Dirichlet problem uxx + uyy = 0, 0 < x < a, 0 < y < b, u(x, 0) = f0(x), u(x, b) = f1(x), 0 ≤x ≤a, u(0, y) = g0(y), u(a, y) = g1(y), 0 ≤y ≤b 36. Show that the Neumann Problem uxx + uyy = 0, 0 < x < a, 0 < y < b, uy(x, 0) = f0(x), uy(x, b) = f1(x), 0 ≤x ≤a, ux(0, y) = g0(y), ux(a, y) = g1(y), 0 ≤y ≤b | Elementary Differential Equations with Boundary Value Problems_Page_675_Chunk3078 |
666 Chapter 12 Fourier Solutions of Partial Differential Equations has no solution unless Z a 0 f0(x) dx = Z a 0 f1(x) dx = Z b 0 g0(y) dy = Z b 0 g1(y) dy = 0. In this case it has infinitely many formal solutions. Find them. 37. In this exercise take it as given that the infinite series P∞ n=1 npe−qn converges for all p if q > 0, and, where appropriate, use the comparison test for absolute convergence of an infinite series. Let u(x, y) = ∞ X n=1 αn sinh nπ(b −y)/a sinh nπb/a sin nπx a , where αn = 2 a Z a 0 f(x) sin nπx a dx and f is piecewise smooth on [0, a]. (a) Verify the approximations sinhnπ(b −y)/a sinh nπb/a ≈e−nπy/a, y < b, (A) and cosh nπ(b −y)/a sinh nπb/a ≈e−nπy/a, y < b (B) for large n. (b) Use (A) to show that u is defined for (x, y) such that 0 < y < b. (c) For fixed y in (0, b), use (A) and Theorem 12.1.2 with z = x to show that ux(x, y) = π a ∞ X n=1 nαn sinh nπ(b −y)/a sinh nπb/a cos nπx a , −∞< x < ∞. (d) Starting from the result of (b), use (A) and Theorem 12.1.2 with z = x to show that, for a fixed y in (0, b), uxx(x, y) = −π2 a2 ∞ X n=1 n2αn sinh nπ(b −y)/a sinh nπb/a sin nπx a , −∞< x < ∞. (e) For fixed but arbitrary x, use (B) and Theorem 12.1.2 with z = y to show that uy(x, y) = −π a ∞ X n=1 nαn cosh nπ(b −y)/a sinh nπb/a sin nπx a if 0 < y0 < y < b, where y0 is an arbitrary number in (0, b). Then argue that since y0 can be chosen arbitrarily small, the conclusion holds for all y in (0, b). (f) Starting from the result of (e), use (A) and Theorem 12.1.2 to show that uyy(x, y) = π2 a2 ∞ X n=1 n2αn sinh nπ(b −y)/a sinh nπb/a sin nπx a , 0 < y < b. | Elementary Differential Equations with Boundary Value Problems_Page_676_Chunk3079 |
Section 12.4 Laplace’s Equation in Polar Coordinates 667 (g) Conclude that u satisfies Laplace’s equation for all (x, y) such that 0 < y < b. By repeatedly applying the arguments in (c)–(f), it can be shown that u can be differentiated term by term any number of times with respect to x and/or y if 0 < y < b. 12.4 LAPLACE’S EQUATION IN POLAR COORDINATES In Section 12.3 we solved boundary value problems for Laplace’s equation over a rectangle with sides parallel to the x, y-axes. Now we’ll consider boundary value problems for Laplace’s equation over regions with boundaries best described in terms of polar coordinates. In this case it’s appropriate to regard u as function of (r, θ) and write Laplace’s equation in polar form as urr + 1 r ur + 1 r2 uθθ = 0, (12.4.1) where r = p x2 + y2 and θ = cos−1 x r = sin−1 x r . We begin with the case where the region is a circular disk with radius ρ, centered at the origin; that is, we want to define a formal solution of the boundary value problem urr + 1 r ur + 1 r2 uθθ = 0, 0 < r < ρ, −π ≤θ < π, u(ρ, θ) = f(θ), −π ≤θ < π (12.4.2) (Figure 12.4.1). Note that (12.4.2) imposes no restriction on u(r, θ) when r = 0. We’ll address this question at the appropriate time. ur r + r−1 ur + r−2 uθ θ = 0 u(ρ,θ) = f(θ) x y Figure 12.4.1 The boundary value problem (12.4.2) | Elementary Differential Equations with Boundary Value Problems_Page_677_Chunk3080 |
668 Chapter 12 Fourier Solutions of Partial Differential Equations We first look for products v(r, θ) = R(r)Θ(θ) that satisfy (12.4.1). For this function, vrr + 1 r vr + 1 r2 vθθ = R′′Θ + 1 r R′Θ + 1 r2 RΘ′′ = 0 for all (r, θ) with r ̸= 0 if r2R′′ + rR′ R = −Θ′′ Θ = λ, where λ is a separation constant. (Verify.) This equation is equivalent to Θ′′ + λΘ = 0 and r2R′′ + rR′ −λR = 0. (12.4.3) Since (r, π) and (r, −π) are the polar coordinates of the same point, we impose periodic boundary con- ditions on Θ; that is, Θ′′ + λΘ = 0, Θ(−π) = Θ(π), Θ′(−π) = Θ′(π). (12.4.4) Since we don’t want RΘ to be identically zero, λ must be an eigenvalue of (12.4.4) and Θ must be an associated eigenfunction. From Theorem 11.1.6, the eigenvalues of (12.4.4) are λ0 = 0 with associated eigenfunctions Θ0 = 1 and, for n = 1, 2, 3, . . ., λn = n2, with associated eigenfunction cos nθ and sin nθ therefore, Θn = αn cos nθ + βn sin nθ where αn and βn are constants. Substituting λ = 0 into (12.4.3) yields the r2R′′ + rR′ = 0, so R′′ 0 R′ 0 = −1 r , R′ 0 = c1 r , and R0 = c2 + c1 ln r. (12.4.5) If c1 ̸= 0 then lim r→0+ |R0(r)| = ∞, which doesn’t make sense if we interpret u0(r, θ) = R0(r)Θ0(θ) = R0(r) as the steady state temperature distribution in a disk whose boundary is maintained at the constant temperature R0(ρ). Therefore we now require R0 to be bounded as r →0+. This implies that c1 = 0, and we take c2 = 1. Thus, R0 = 1 and v0(r, θ) = R0(r)Θ0(θ) = 1. Note that v0 satisfies (12.4.2) with f(θ) = 1. Substituting λ = n2 into (12.4.3) yields the Euler equation r2R′′ n + rR′ n −n2Rn = 0 (12.4.6) for Rn. The indicial polynomial of this equation is s(s −1) + s −n2 = (s −n)(s + n), | Elementary Differential Equations with Boundary Value Problems_Page_678_Chunk3081 |
Section 12.4 Laplace’s Equation in Polar Coordinates 669 so the general solution of (12.4.6) is Rn = c1rn + c2r−n, (12.4.7) by Theorem 7.4.3. Consistent with our previous assumption on R0, we now require Rn to be bounded as r →0+. This implies that c2 = 0, and we choose c1 = ρ−n. Then Rn(r) = rn/ρn, so vn(r, θ) = Rn(r)Θn(θ) = rn ρn (αn cos nθ + sin nθ). Now vn satisfies (12.4.2) with f(θ) = αn cos nθ + βn sin nθ. More generally, if α0, α1,..., αm and β1, β2, ..., βm are arbitrary constants then um(r, θ) = α0 + m X n=1 rn ρn (αn cos nθ + βn sin nθ) satisfies (12.4.2) with f(θ) = α0 + m X n=1 (αn cos nθ + βn sin nθ). This motivates the next definition. Definition 12.4.1 The bounded formal solution of the boundary value problem (12.4.2) is u(r, θ) = α0 + ∞ X n=1 rn ρn (αn cos nθ + βn sin nθ), (12.4.8) where F (θ) = α0 + ∞ X n=1 (αn cos nθ + βn sin nθ) is the Fourier series of f on [−π, π]; that is, α0 = 1 2π Z π −π f(θ) dθ, and αn = 1 π Z π −π f(θ) cos nθ dθ and βn = 1 π Z π −π f(θ) sin nθ dθ, n = 1, 2, 3, . . .. Since P∞ n=0 nk(r/ρ)n converges for every k if 0 < r < ρ, Theorem 12.1.2 can be used to show that if 0 < r < ρ then (12.4.8) can be differentiated term by term any number of times with respect to both r and θ. Since the terms in (12.4.8) satisfy Laplace’s equation if r > 0, (12.4.8) satisfies Laplace’s equation if 0 < r < ρ. Therefore, since u(ρ, θ) = F (θ), u is an actual solution of (12.4.2) if and only if F (θ) = f(θ), −π ≤θ < π. From Theorem 11.2.4, this is true if f is continuousand piecewise smooth on [−π, π] and f(−π) = f(π). Example 12.4.1 Find the bounded formal solution of (12.4.2) with f(θ) = θ(π2 −θ2). | Elementary Differential Equations with Boundary Value Problems_Page_679_Chunk3082 |
670 Chapter 12 Fourier Solutions of Partial Differential Equations Solution From Example 11.2.6, θ(π2 −θ2) = 12 ∞ X n=1 (−1)n n3 sin nθ, −π ≤θ ≤π, so u(r, θ) = 12 ∞ X n=1 rn ρn (−1)n n3 sin nθ, 0 ≤r ≤ρ, −π ≤θ ≤π. Example 12.4.2 Define the formal solution of urr + 1 r ur + 1 r2 uθθ = 0, ρ0 < r < ρ, −π ≤θ < π, u(ρ0, θ) = 0, u(ρ, θ) = f(θ), −π ≤θ < π, (12.4.9) where 0 < ρ0 < ρ (Figure 12.4.2). ur r + r−1 ur + r−2 uθ θ = 0 u(ρ,θ) = f(θ) u(ρ0,θ) = 0 x y Figure 12.4.2 The boundary value problem (12.4.9) Solution We use separation of variables exactly as before, except that now we choose the constants in (12.4.5) and (12.4.7) so that Rn(ρ0) = 0 for n = 0, 1, 2,.... In view of the nonhomogeneous Dirichlet condition on the boundary r = ρ, it’s also convenient to require that Rn(ρ) = 1 for n = 0, 1, 2,.... We leave it to you to verify that R0(r) = lnr/ρ0 lnρ/ρ0 and Rn = ρ−n 0 rn −ρn 0 r−n ρ−n 0 ρn −ρn 0 ρ−n , n = 1, 2, 3, . . . | Elementary Differential Equations with Boundary Value Problems_Page_680_Chunk3083 |
Section 12.4 Laplace’s Equation in Polar Coordinates 671 satisfy these requirements. Therefore v0(ρ, θ) = lnr/ρ0 ln ρ/ρ0 and vn(r, θ) = ρ−n 0 rn −ρn 0 r−n ρ−n 0 ρn −ρn 0 ρ−n (αn cos nθ + βn sin nθ), n = 1, 2, 3, . . ., where αn and βn are arbitrary constants. If α0, α1,..., αm and β1, β2, ..., βm are arbitrary constants then um(r, θ) = α0 ln r/ρ0 lnρ/ρ0 + m X n=1 ρ−n 0 rn −ρn 0 r−n ρ−n 0 ρn −ρn 0 ρ−n (αn cos nθ + βn sin nθ) satisfies (12.4.9), with f(θ) = α0 + m X n=1 (αn cos nθ + βn sin nθ). This motivates us to define the formal solution of (12.4.9) for general f to be u(r, θ) = α0 lnr/ρ0 ln ρ/ρ0 + ∞ X n=1 ρ−n 0 rn −ρn 0 r−n ρ−n 0 ρn −ρn 0 ρ−n (αn cos nθ + βn sin nθ), where F (θ) = α0 + ∞ X n=1 (αn cos nθ + βn sin nθ) is the Fourier series of f on [−π, π]. Example 12.4.3 Define the bounded formal solution of urr + 1 r ur + 1 r2 uθθ = 0, 0 < r < ρ, 0 < θ < γ, u(ρ, θ) = f(θ), 0 ≤θ ≤γ, u(r, 0) = 0, u(r, γ) = 0, 0 < r < ρ, (12.4.10) where 0 < γ < 2π (Figure 12.4.3). Solution Now v(r, θ) = R(r)Θ(θ), where r2R′′ + rR′ −λR = 0 (12.4.11) and Θ′′ + λΘ = 0, Θ(0) = 0, Θ(γ) = 0. (12.4.12) From Theorem 11.1.2, the eigenvalues of (12.4.12) are λn = n2π2/γ2, with associated eigenfunction Θn = sinnπθ/γ, n = 1, 2, 3,.... Substituting λ = n2π2/γ2 into (12.4.11) yields the Euler equation r2R′′ + rR′ n −n2π2 γ2 R = 0. | Elementary Differential Equations with Boundary Value Problems_Page_681_Chunk3084 |
672 Chapter 12 Fourier Solutions of Partial Differential Equations ur r + r−1 ur + r−2 uθ θ = 0 u(ρ,θ) = f(θ) u(r,γ) = 0 γ x y Figure 12.4.3 The boundary value problem (12.4.10) The indicial polynomial of this equation is s(s −1) + s −n2π2 γ2 = s −nπ γ s + nπ γ , so Rn = c1rnπ/γ + c2r−nπ/γ, by Theorem 7.4.3. To obtain a solution that remains bounded as r →0+ we let c2 = 0. Because of the Dirichlet condition at r = ρ, it’s convenient to have r(ρ) = 1; therefore we take c1 = ρ−nπ/γ, so Rn(r) = rnπ/γ ρnπ/γ . Now vn(r, θ) = Rn(r)Θn(θ) = rnπ/γ ρnπ/γ sin nπθ γ satisfies (12.4.10) with f(θ) = sin nπθ γ . More generally, if α1, α2, ..., αm and are arbitrary constants then um(r, θ) = m X n=1 αn rnπ/γ ρnπ/γ sin nπθ γ satisfies (12.4.10) with f(θ) = m X n=1 αn sin nπθ γ . | Elementary Differential Equations with Boundary Value Problems_Page_682_Chunk3085 |
Section 12.4 Laplace’s Equation in Polar Coordinates 673 This motivates us to define the bounded formal solution of (12.4.10) to be um(r, θ) = ∞ X n=1 αn rnπ/γ ρnπ/γ sin nπθ γ , where S(θ) = ∞ X n=1 αn sin nπθ γ is the Fourier sine expansion of f on [0, γ]; that is, αn = 2 γ Z γ 0 f(θ) sin nπθ γ dθ. 12.4 Exercises 1. Define the formal solution of urr + 1 r ur + 1 r2 uθθ = 0, ρ0 < r < ρ, −π ≤θ < π, u(ρ0, θ) = f(θ), u(ρ, θ) = 0, −π ≤θ < π, where 0 < ρ0 < ρ. 2. Define the formal solution of urr + 1 r ur + 1 r2 uθθ = 0, ρ0 < r < ρ, 0 < θ < γ, u(ρ0, θ) = 0, u(ρ, θ) = f(θ), 0 ≤θ ≤γ, u(r, 0) = 0, u(r, γ) = 0, ρ0 < r < ρ, where 0 < γ < 2π and 0 < ρ0 < ρ. 3. Define the formal solution of urr + 1 r ur + 1 r2 uθθ = 0, ρ0 < r < ρ, 0 < θ < γ, u(ρ0, θ) = 0, ur(ρ, θ) = g(θ), 0 ≤θ ≤γ, uθ(r, 0) = 0, uθ(r, γ) = 0, ρ0 < r < ρ, where 0 < γ < 2π and 0 < ρ0 < ρ. 4. Define the bounded formal solution of urr + 1 r ur + 1 r2 uθθ = 0, 0 < r < ρ, 0 < θ < γ, u(ρ, θ) = f(θ), 0 ≤θ ≤γ, uθ(r, 0) = 0, u(r, γ) = 0, 0 < r < ρ, where 0 < γ < 2π. | Elementary Differential Equations with Boundary Value Problems_Page_683_Chunk3086 |
674 Chapter 12 Fourier Solutions of Partial Differential Equations 5. Define the formal solution of urr + 1 r ur + 1 r2 uθθ = 0, ρ0 < r < ρ, 0 < θ < γ, ur(ρ0, θ) = g(θ), ur(ρ, θ) = 0, 0 ≤θ ≤γ, u(r, 0) = 0, uθ(r, γ) = 0, ρ0 < r < ρ, where 0 < γ < 2π and 0 < ρ0 < ρ. 6. Define the bounded formal solution of urr + 1 r ur + 1 r2 uθθ = 0, 0 < r < ρ, 0 < θ < γ, u(ρ, θ) = f(θ), 0 ≤θ ≤γ, uθ(r, 0) = 0, uθ(r, γ) = 0, 0 < r < ρ, where 0 < γ < 2π. 7. Show that the Neumann problem urr + 1 r ur + 1 r2 uθθ = 0, 0 < r < ρ, −π ≤θ < π, ur(ρ, θ) = f(θ), −π ≤θ < π has no bounded formal solution unless R π −π f(θ) dθ = 0. In this case it has infinitely many solu- tions. Find those solutions. | Elementary Differential Equations with Boundary Value Problems_Page_684_Chunk3087 |
CHAPTER 13 Boundary Value Problems for Second Order Ordinary Differential Equations IN THIS CHAPTER we discuss boundary value problems and eigenvalue problems for linear second order ordinary differential equations. Section 13.1 discusses point two-point boundary value problems for linear second order ordinary differ- ential equations. Section 13.2 deals with generalizations of the eigenvalue problems considered in Section 11.1 677 | Elementary Differential Equations with Boundary Value Problems_Page_686_Chunk3088 |
Section 13.1 Two-Point Boundary Value Problems 677 13.1 TWO-POINT BOUNDARY VALUE PROBLEMS In Section 5.3 we considered initial value problems for the linear second order equation P0(x)y′′ + P1(x)y′ + P2(x)y = F (x). (13.1.1) Suppose P0, P1, P2, and F are continuous and P0 has no zeros on an open interval (a, b). From Theo- rem 5.3.1, if x0 is in (a, b) and k1 and k2 are arbitrary real numbers then (13.1.1) has a unique solution on (a, b) such that y(x0) = k1 and y′(x0) = k2. Now we consider a different problem for (13.1.1). PROBLEM Suppose P0, P1, P2, and F are continous and P0 has no zeros on a closed interval [a, b]. Let α, β, ρ, and δ be real numbers such that α2 + β2 ̸= 0 and ρ2 + δ2 ̸= 0, (13.1.2) and let k1 and k2 be arbitrary real numbers. Find a solution of P0(x)y′′ + P1(x)y′ + P2(x)y = F (x) (13.1.3) on the closed interval [a, b] such that αy(a) + βy′(a) = k1 (13.1.4) and ρy(b) + δy′(b) = k2. (13.1.5) The assumptions stated in this problem apply throughout this section and won’t be repeated. Note that we imposed conditions on P0, P1, P2, and F on the closed interval [a, b], and we are interested in solutions of (13.1.3) on the closed interval. This is different from the situation considered in Chapter 5, where we imposed conditions on P0, P1, P2, and F on the open interval (a, b) and we were interested in solutions on the open interval. There is really no problem here; we can always extend P0, P1, P2, and F to an open interval (c, d) (for example, by defining them to be constant on (c, d] and [b, d)) so that they are continuous and P0 has no zeros on [c, d]. Then we can apply the theorems from Chapter 5 to the equation y′′ + P1(x) P0(x)y′ + P2(x) P0(x)y = F (x) P0(x) on (c, d) to draw conclusions about solutions of (13.1.3) on [a, b]. We call a and b boundary points. The conditions (13.1.4) and (13.1.5) are boundary conditions, and the problem is a two-point boundary value problem or, for simplicity, a boundary value problem. (We used similar terminology in Chapter 12 with a different meaning; both meanings are in common usage.) We require (13.1.2) to insure that we’re imposing a sensible condition at each boundary point. For example, if α2 + β2 = 0 then α = β = 0, so αy(a) + βy′(a) = 0 for all choices of y(a) and y′(a). Therefore (13.1.4) is an impossible condition if k1 ̸= 0, or no condition at all if k1 = 0. We abbreviate (13.1.1) as Ly = F , where Ly = P0(x)y′′ + P1(x)y′ + P0(x)y, and we denote B1(y) = αy(a) + βy′(a) and B2(y) = ρy(b) + δy′(b). We combine (13.1.3), (13.1.4), and (13.1.5) as Ly = F, B1(y) = k1, B2(y) = k2. (13.1.6) | Elementary Differential Equations with Boundary Value Problems_Page_687_Chunk3089 |
678 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations This boundary value problem is homogeneous if F = 0 and k1 = k2 = 0; otherwise it’s nonhomoge- neous. We leave it to you (Exercise 1) to verify that B1 and B2 are linear operators; that is, if c1 and c2 are constants then Bi(c1y1 + c2y2) = c1Bi(y1) + c2Bi(y2), i = 1, 2. (13.1.7) The next three examples show that the question of existence and uniqueness for solutions of boundary value problems is more complicated than for initial value problems. Example 13.1.1 Consider the boundary value problem y′′ + y = 1, y(0) = 0, y(π/2) = 0. The general solution of y′′ + y = 1 is y = 1 + c1 sin x + c2 cos x, so y(0) = 0 if and only if c2 = −1 and y(π/2) = 0 if and only if c1 = −1. Therefore y = 1 −sinx −cos x is the unique solution of the boundary value problem. Example 13.1.2 Consider the boundary value problem y′′ + y = 1, y(0) = 0, y(π) = 0. Again, the general solution of y′′ + y = 1 is y = 1 + c1 sin x + c2 cos x, so y(0) = 0 if and only if c2 = −1, but y(π) = 0 if and only if c2 = 1. Therefore the boundary value problem has no solution. Example 13.1.3 Consider the boundary value problem y′′ + y = sin 2x, y(0) = 0, y(π) = 0. You can use the method of undetermined coefficients (Section 5.5) to find that the general solution of y′′ + y = sin 2x is y = −sin 2x 3 + c1 sin x + c2 cos x. The boundary conditions y(0) = 0 and y(π) = 0 both require that c2 = 0, but they don’t restrict c1. Therefore the boundary value problem has infinitely many solutions y = −sin 2x 3 + c1 sin x, where c1 is arbitrary. Theorem 13.1.1 If z1 and z2 are solutionsof Ly = 0 such that either B1(z1) = B1(z2) = 0 or B2(z1) = B2(z2) = 0, then {z1, z2} is linearly dependent. Equivalently, if {z1, z2} is linearly independent, then B2 1(z1) + B2 1(z2) ̸= 0 and B2 2(z1) + B2 2(z2) ̸= 0. | Elementary Differential Equations with Boundary Value Problems_Page_688_Chunk3090 |
Section 13.1 Two-Point Boundary Value Problems 679 Proof Recall that B1(z) = αz(a) + βz′(a) and α2 + β2 ̸= 0. Therefore, if B1(z1) = B1(z2) = 0 then (α, β) is a nontrivial solution of the system αz1(a) + βz′ 1(a) = 0 αz2(a) + βz′(a) = 0. This implies that z1(a)z′ 2(a) −z′ 1(a)z2(a) = 0, so {z1, z2} is linearly dependent, by Theorem 5.1.6. We leave it to you to show that {z1, z2} is linearly dependent if B2(z1) = B2(z2) = 0. Theorem 13.1.2 The following statements are equivalent; that is, they are either all true or all false. (a) There’s a fundamental set {z1, z2} of solutions of Ly = 0 such that B1(z1)B2(z2) −B1(z2)B2(z1) ̸= 0. (13.1.8) (b) If {y1, y2} is a fundamental set of solutions of Ly = 0 then B1(y1)B2(y2) −B1(y2)B2(y1) ̸= 0. (13.1.9) (c) For each continuous F and pair of constants (k1, k2), the boundary value problem Ly = F, B1(y) = k1, B2(y) = k2 has a unique solution. (d) The homogeneous boundary value problem Ly = 0, B1(y) = 0, B2(y) = 0 (13.1.10) has only the trivial solution y = 0. (e) The homogeneousequation Ly = 0 has linearly independent solutionsz1 and z2 such that B1(z1) = 0 and B2(z2) = 0. Proof We’ll show that (a) =⇒(b) =⇒(c) =⇒(d) =⇒(e) =⇒(a). (a) =⇒(b): Since {z1, z2} is a fundamental set of solutions for Ly = 0, there are constants a1, a2, b1, and b2 such that y1 = a1z1 + a2z2 y2 = b1z1 + b2z2. (13.1.11) Moreover, a1 a2 b1 b2 ̸= 0. (13.1.12) because if this determinant were zero, its rows would be linearly dependent and therefore {y1, y2} would be linearly dependent, contrary to our assumption that {y1, y2} is a fundamental set of solutions of Ly = 0. From (13.1.7) and (13.1.11), B1(y1) B2(y1) B1(y2) B2(y2) = a1 a2 b1 b2 B1(z1) B2(z1) B1(z2) B2(z2) . | Elementary Differential Equations with Boundary Value Problems_Page_689_Chunk3091 |
680 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations Since the determinant of a product of matrices is the product of the determinants of the matrices, (13.1.8) and (13.1.12) imply (13.1.9). (b) =⇒ (c): Since {y1, y2} is a fundamental set of solutions of Ly = 0, the general solution of Ly = F is y = yp + c1y1 + c2y2, where c1 and c2 are arbitrary constants and yp is a particular solution of Ly = F . To satisfy the boundary conditions, we must choose c1 and c2 so that k1 = B1(yp) + c1B1(y1) + c2B1(y2) k2 = B2(yp) + c1B2(y1) + c2B2(y2), (recall (13.1.7)), which is equivalent to c1B1(y1) + c2B1(y2) = k1 −B1(yp) c1B2(y1) + c2B2(22) = k2 −B2(yp). From (13.1.9), this system always has a unique solution (c1, c2). (c) =⇒(d): Obviously, y = 0 is a solution of (13.1.10). From (c) with F = 0 and k1 = k2 = 0, it’s the only solution. (d) =⇒(e): Let {y1, y2} be a fundamental system for Ly = 0 and let z1 = B1(y2)y1 −B1(y1)y2 and z2 = B2(y2)y1 −B2(y1)y2. Then B1(z1) = 0 and B2(z2) = 0. To see that z1 and z2 are linearly independent, note that a1z1 + a2z2 = a1[B1(y2)y1 −B1(y1)y2] + a2[B2(y2)y1 −B2(y1)y2] = [B1(y2)a1 + B2(y2)a2]y1 −[B1(y1)a1 + B2(y1)a2]y2. Therefore, since y1 and y2 are linearly independent, a1z1 + a2z2 = 0 if and only if B1(y1) B2(y1) B1(y2) B2(y2) a1 a2 = 0 0 . If this system has a nontrivial solution then so does the system B1(y1) B1(y2) B2(y1) B2(y2) c1 c2 = 0 0 . This and (13.1.7) imply that y = c1z1 + c2z2 is a nontrivial solution of (13.1.10), which contradicts (d). (e) =⇒ (a). Theorem 13.1.1 implies that if B1(z1) = 0 and B2(z2) = 0 then B1(z2) ̸= 0 and B2(z1) ̸= 0. This implies (13.1.8), which completes the proof. Example 13.1.4 Solve the boundary value problem x2y′′ −2xy′ + 2y −2x3 = 0, y(1) = 4, y′(2) = 3, (13.1.13) given that {x, x2} is a fundamental set of solutions of the complementary equation Solution Using variation of parameters (Section 5.7), you can show that yp = x3 is a solution of the complementary equation x2y′′ −2xy′ + 2y = 2x3 = 0. | Elementary Differential Equations with Boundary Value Problems_Page_690_Chunk3092 |
Section 13.1 Two-Point Boundary Value Problems 681 Therefore the solution of (13.1.13) can be written as y = x3 + c1x + c2x2. Then y′ = 3x2 + c1 + 2c2x, and imposing the boundary conditions yields the system c1 + c2 = 3 c1 + 4c2 = −9, so c1 = 7 and c2 = −4. Therefore y = x3 + 7x −4x2 is the unique solution of (13.1.13) Example 13.1.5 Solve the boundary value problem y′′ −7y′ + 12y = 4e2x, y(0) = 3, y(1) = 5e2. Solution From Example 5.4.1, yp = 2e2x is a particular solution of y′′ −7y′ + 12y = 4e2x. (13.1.14) Since {e3x, e4x} is a fundamental set for the complementary equation, we could write the solution of (13.1.13) as y = 2e2x + c1e3x + c2e4x and determine c1 and c2 by imposing the boundary conditions. However, this would lead to some tedious algebra, and the form of the solution would be very unappealing. (Try it!) In this case it’s convenient to use the fundamental system {z1, z2} mentioned in Theorem 13.1.2(e); that is, we choose {z1, z2} so that B1(z1) = z1(0) = 0 and B2(z2) = z2(1) = 0. It is easy to see that z1 = e3x −e4x and z2 = e3(x−1) −e4(x−1) satisfy these requirements. Now we write the solution of (13.1.14) as y = 2e2x + c1 | Elementary Differential Equations with Boundary Value Problems_Page_691_Chunk3093 |
682 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations Theorem 13.1.3 Suppose the homogeneous boundary value problem Ly = 0, B1(y) = 0, B2(y) = 0 (13.1.15) has only the trivial solution. Let y1 and y2 be linearly independent solutions of Ly = 0 such that B1(y1) = 0 and B2(y2) = 0, and let W = y1y′ 2 −y′ 1y2. Then the unique solution of Ly = F, B1(y) = 0, B2(y) = 0 (13.1.16) is y(x) = y1(x) Z b x F (t)y2(t) P0(t)W(t) dt + y2(x) Z x a F (t)y1(t) P0(t)W(t) dt. (13.1.17) Proof In Section 5.7 we saw that if y = u1y1 + u2y2 (13.1.18) where u′ 1y1 + u′ 2y2 = 0 u′ 1y′ 1 + u′ 2y′ 2 = F, then Ly = F . Solving for u′ 1 and u′ 2 yields u′ 1 = −F y2 P0W and u′ 2 = F y1 P0W , which hold if u1(x) = Z b x F (t)y2(t) P0(t)W(t) dt and u2(x) = Z x a F (t)y1(t) P0(t)W(t) dt. This and (13.1.18) show that (13.1.17) is a solution of Ly = F . Differentiating (13.1.17) yields y′(x) = y′ 1(x) Z b x F (t)y2(t) P0(t)W(t) dt + y′ 2(x) Z x a F (t)y1(t) P0(t)W(t) dt. (13.1.19) (Verify.) From (13.1.17) and (13.1.19), B1(y) = B1(y1) Z b a F (t)y2(t) P0(t)W(t) dt = 0 because B1(y1) = 0, and B2(y) = B2(y2) Z b a F (t)y1(t) P0(t)W(t) dt = 0 because B2(y2) = 0. Hence, y satisfies (13.1.16). This completes the proof. We can rewrite (13.1.17) as y = Z b a G(x, t)F (t) dt, (13.1.20) where G(x, t) = y1(t)y2(x) P0(t)W(t) , a ≤t ≤x, y1(x)y2(t) P0(t)W(t) , x ≤t ≤b. . | Elementary Differential Equations with Boundary Value Problems_Page_692_Chunk3094 |
Section 13.1 Two-Point Boundary Value Problems 683 This is the Green’s function for the boundary value problem (13.1.16). The Green’s function is related to the boundary value problem (13.1.16) in much the same way that the inverse of a square matrix A is related to the linear algebraic system y = Ax; just as we substitute the given vector y into the formula x = A−1y to solve y = Ax, we substitute the given function F into the formula (13.1.20) to obtain the solution of (13.1.16). The analogy goes further: just as A−1 exists if and only if Ax = 0 has only the trivial solution, the boundary value problem (13.1.16) has a Green’s function if and only the homogeneous boundary value problem (13.1.15) has only the trivial solution. We leave it to you (Exercise 32) to show that the assumptions of Theorem 13.1.3 imply that the unique solution of the boundary value problem Ly = F, B1(y) = k1, B2(y) = k2 is y(x) = Z b a G(x, t)F (t) dt + k2 B2(y1)y1 + k1 B1(y2)y2. Example 13.1.6 Solve the boundary value problem y′′ + y = F (x). y(0) + y′(0) = 0, y(π) −y′(π) = 0, (13.1.21) and find the Green’s function for this problem. Solution Here B1(y) = y(0) + y′(0) and B2(y) = y(π) −y′(π). Let {z1, z2} = {cos x, sin x}, which is a fundamental set of solutions of y′′ + y = 0. Then B1(z1) = (cos x −sin x) x=0 = 1 B2(z1) = (cos x + sin x) x=π = −1 and B1(z2) = (sin x + cos x) x=0 = 1 B2(z2) = (sin x −cos x) x=π = 1. Therefore B1(z1)B2(z2) −B1(z2)B2(z1) = 2, so Theorem 13.1.2 implies that (13.1.21) has a unique solution. Let y1 = B1(z2)z1 −B1(z1)z2 = cos x −sinx and y2 = B2(z2)z1 −B2(z1)z2 = cos x + sin x. Then B1(y1) = 0, B2(y2) = 0, and the Wronskian of {y1, y2} is W(x) = cos x −sin x cos x + sin x −sin x −cos x −sin x + cos x = 2. Since P0 = 1, (13.1.17) yields the solution y(x) = cos x −sin x 2 Z π x F (t)(cos t + sin t) dt +cos x + sin x 2 Z x 0 F (t)(cos t −sint) dt. | Elementary Differential Equations with Boundary Value Problems_Page_693_Chunk3095 |
684 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations The Green’s function is G(x, t) = (cos t −sin t)(cos x + sin x) 2 , 0 ≤t ≤x, (cos x −sin x)(cos t + sin t) 2 , x ≤t ≤π. We’ll now consider the situation not covered by Theorem 13.1.3. Theorem 13.1.4 Suppose the homogeneous boundary value problem Ly = 0, B1(y) = 0, B2(y) = 0 (13.1.22) has a nontrivial solution y1, and let y2 be any solution of Ly = 0 that isn’t a constant multiple of y1. Let W = y1y′ 2 −y′ 1y2. If Z b a F (t)y1(t) P0(t)W(t) dt = 0, (13.1.23) then the homogeneous boundary value problem Ly = F, B1(y) = 0, B2(y) = 0 (13.1.24) has infinitely many solutions, all of the form y = yp + c1y1, where yp = y1(x) Z b x F (t)y2(t) P0(t)W(t) dt + y2(x) Z x a F (t)y1(t) P0(t)W(t) dt and c1 is a constant. If Z b a F (t)y1(t) P0(t)W(t) dt ̸= 0, then (13.1.24) has no solution. Proof From the proof of Theorem 13.1.3, yp is a particular solution of Ly = F , and y′ p(x) = y′ 1(x) Z b x F (t)y2(t) P0(t)W(t) dt + y′ 2(x) Z x a F (t)y1(t) P0(t)W(t) dt. Therefore the general solution of (13.1.22) is of the form y = yp + c1y1 + c2y2, where c1 and c2 are constants. Then B1(y) = B1(yp + c1y1 + c2y2) = B1(yp) + c1B1(y1) + c2B1y2 = B1(y1) Z b a F (t)y2(t) P0(t)W(t) dt + c1B1(y1) + c2B1(y2) = c2B1(y2) Since B1(y1) = 0, Theorem 13.1.1 implies that B1(y2) ̸= 0; hence, B1(y) = 0 if and only if c2 = 0. Therefore y = yp + c1y1 and B2(y) = B2(yp + c1y1) = B2(y2) Z b a F (t)y1(t) P0(t)W(t) dt + c1B2(y1) = B2(y2) Z b a F (t)y1(t) P0(t)W(t) dt, since B2(y1) = 0. From Theorem 13.1.1, B2(y2) ̸= 0 (since B2(y1 = 0). Therefore Ly = 0 if and only if (13.1.23) holds. This completes the proof. | Elementary Differential Equations with Boundary Value Problems_Page_694_Chunk3096 |
Section 13.1 Two-Point Boundary Value Problems 685 Example 13.1.7 Applying Theorem 13.1.4 to the boundary value problem y′′ + y = F (x), y(0) = 0, y(π) = 0 (13.1.25) explains the Examples 13.1.2 and 13.1.3. The complementary equation y′′ + y = 0 has the linear inde- pendent solutions y1 = sin x and y2 = cos x, and y1 satisfies both boundary conditions. Since P0 = 1 and W = sin x cos x cos x −sin x = −1, (13.1.23) reduces to Z π 0 F (x) sin x dx = 0. From Example 13.1.2, F (x) = 1 and Z π 0 F (x) sin x dx = Z π 0 sin x dx = 2, so Theorem 13.1.3 implies that (13.1.25) has no solution. In Example 13.1.3, F (x) = sin 2x = 2 sin x cos x and Z π 0 F (x) sinx dx = 2 Z π 0 sin2 x cos x dx = 2 3 sin3 x π 0 = 0, so Theorem 13.1.3 implies that (13.1.25) has infinitely many solutions, differing by constant multiples of y1(x) = sin x. 13.1 Exercises 1. Verify that B1 and B2 are linear operators; that is, if c1 and c2 are constants then Bi(c1y1 + c2y2) = c1Bi(y1) + c2Bi(y2), i = 1, 2. In Exercises 2–7 solve the boundary value problem. 2. y′′ −y = x, y(0) = −2, y(1) = 1 3. y′′ = 2 −3x, y(0) = 0, y(1) −y′(1) = 0 4. y′′ −y = x, y(0) + y′(0) = 3, y(1) −y′(1) = 2 5. y′′ + 4y = 1, y(0) = 3, y(π/2) + y′(π/2) = −7 6. y′′ −2y′ + y = 2ex, y(0) −2y′(0) = 3, y(1) + y′(1) = 6e 7. y′′ −7y′ + 12y = 4e2x, y(0) + y′(0) = 8, y(1) = −7e2 (see Example 13.1.5) 8. State a condition on F such that the boundary value problem y′′ = F (x), y(0) = 0, y(1) −y′(1) = 0 has a solution, and find all solutions. | Elementary Differential Equations with Boundary Value Problems_Page_695_Chunk3097 |
686 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations 9. (a) State a condition on a and b such that the boundary value problem y′′ + y = F (x), y(a) = 0, y(b) = 0 (A) has a unique solution for every continuous F , and find the solution by the method used to prove Theorem 13.1.3 (b) In the case where a and b don’t satisfy the condition you gave for (a), state necessary and sufficient on F such that (A) has a solution, and find all solutions by the method used to prove Theorem 13.1.4. 10. Follow the instructions in Exercise 9 for the boundary value problem y′′ + y = F (x), y(a) = 0, y′(b) = 0. 11. Follow the instructions in Exercise 9 for the boundary value problem y′′ + y = F (x), y′(a) = 0, y′(b) = 0. In Exercises 12–15 find a formula for the solution of the boundary problem by the method used to prove Theorem 13.1.3. Assume that a < b. 12. y′′ −y = F (x), y(a) = 0, y(b) = 0 13. y′′ −y = F (x), y(a) = 0, y′(b) = 0 14. y′′ −y = F (x), y′(a) = 0, y′(b) = 0 15. y′′ −y = F (x), y(a) −y′(a) = 0, y(b) + y′(b) = 0 In Exercises 16–19 find all values of ω such that boundary problem has a unique solution, and find the solution by the method used to prove Theorem 13.1.3. For other values of ω, find conditions on F such that the problem has a solution, and find all solutions by the method used to prove Theorem 13.1.4. 16. y′′ + ω2y = F (x), y(0) = 0, y(π) = 0 17. y′′ + ω2y = F (x), y(0) = 0, y′(π) = 0 18. y′′ + ω2y = F (x), y′(0) = 0, y(π) = 0 19. y′′ + ω2y = F (x), y′(0) = 0, y′(π) = 0 20. Let {z1, z2} be a fundamental set of solutions of Ly = 0. Given that the homogeneous boundary value problem Ly = 0, B1(y) = 0, B2(y) = 0 has a nontrivial solution, express it explicity in terms of z1 and z2. 21. If the boundary value problem has a solution for every continuousF , then find the Green’s function for the problem and use it to write an explicit formula for the solution. Otherwise, if the boundary value problem does not have a solution for every continuous F , find a necessary and sufficient condition on F for the problem to have a solution, and find all solutions. Assume that a < b. (a) y′′ = F (x), y(a) = 0, y(b) = 0 (b) y′′ = F (x), y(a) = 0, y′(b) = 0 (c) y′′ = F (x), y′(a) = 0, y(b) = 0 (d) y′′ = F (x), y′(a) = 0, y′(b) = 0 | Elementary Differential Equations with Boundary Value Problems_Page_696_Chunk3098 |
Section 13.1 Two-Point Boundary Value Problems 687 22. Find the Green’s function for the boundary value problem y′′ = F (x), y(0) −2y′(0) = 0, y(1) + 2y′(1) = 0. (A) Then use the Green’s function to solve (A) with (a) F (x) = 1, (b) F (x) = x, and (c) F (x) = x2. 23. Find the Green’s function for the boundary value problem x2y′′ + xy′ + (x2 −1/4)y = F (x), y(π/2) = 0, y(π) = 0, (A) given that y1(x) = cos x √x and y2(x) = sin x √x are solutions of the complementary equation. Then use the Green’s function to solve (A) with (a) F (x) = x3/2 and (b) F (x) = x5/2. 24. Find the Green’s function for the boundary value problem x2y′′ −2xy′ + 2y = F (x), y(1) = 0, y(2) = 0, (A) given that {x, x2} is a fundamental set of solutions of the complementary equation. Then use the Green’s function to solve (A) with (a) F (x) = 2x3 and (b) F (x) = 6x4. 25. Find the Green’s function for the boundary value problem x2y′′ + xy′ −y = F (x), y(1) −2y′(1) = 0, y′(2) = 0, (A) given that {x, 1/x} is a fundamental set of solutions of the complementary equation. Then use the Green’s function to solve (A) with (a) F (x) = 1, (b) F (x) = x2, and (c) F (x) = x3. In Exercises 26–30 find necessary and sufficient conditions on α, β, ρ, and δ for the boundary value problem to have a unique solution for every continuous F , and find the Green’s function. 26. y′′ = F (x), αy(0) + βy′(0) = 0, ρy(1) + δy′(1) = 0 27. y′′ + y = F (x), αy(0) + βy′(0) = 0, ρy(π) + δy′(π) = 0 28. y′′ + y = F (x), αy(0) + βy′(0) = 0, ρy(π/2) + δy′(π/2) = 0 29. y′′ −2y′ + 2y = F (x), αy(0) + βy′(0) = 0, ρy(π) + δy′(π) = 0 30. y′′ −2y′ + 2y = F (x), αy(0) + βy′(0) = 0, ρy(π/2) + δy′(π/2) = 0 31. Find necessary and sufficient conditions on α, β, ρ, and δ for the boundary value problem y′′ −y = F (x), αy(a) + βy′(a) = 0, ρy(b) + δy′(b) = 0 (A) to have a unique solution for every continuous F , and find the Green’s function for (A). Assume that a < b. 32. Show that the assumptions of Theorem 13.1.3 imply that the unique solution of Ly = F, B1(y) = k1, B2(y) = f2 is y = Z b a G(x, t)F (t) dt + k2 B2 (y1)y1 + k1 B1(y2)y2. | Elementary Differential Equations with Boundary Value Problems_Page_697_Chunk3099 |
688 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations 13.2 STURM-LIOUVILLE PROBLEMS In this section we consider eigenvalue problems of the form P0(x)y′′ + P1(x)y′ + P2(x)y + λR(x)y = 0, B1(y) = 0, B2(y) = 0, (13.2.1) where B1(y) = αy(a) + βy′(a) and B2(y) = ρy(b) + δy′(b). As in Section 13.1, α, β, ρ, and δ are real numbers, with α2 + β2 > 0 and ρ2 + δ2 > 0, P0, P1, P2, and R are continuous, and P0 and R are positive on [a, b]. We say that λ is an eigenvalue of (13.2.1) if (13.2.1) has a nontrivial solution y. In this case, y is an eigenfunction associated with λ, or a λ-eigenfunction. Solving the eigenvalue problem means finding all eigenvalues and associated eigenfunctions of (13.2.1). Example 13.2.1 Solve the eigenvalue problem y′ + 3y′ + 2y + λy = 0, y(0) = 0, y(1) = 0. (13.2.2) Solution The characteristic equation of (13.2.2) is r2 + 3r + 2 + λ = 0, with zeros r1 = −3 + √1 −4λ 2 and r2 = −3 −√1 −4λ 2 . If λ < 1/4 then r1 and r2 are real and distinct, so the general solution of the differential equation in (13.2.2) is y = c1er1t + c2er2t. The boundary conditions require that c1 + c2 = 0 c1er1 + c2er2 = 0. Since the determinant of this system is er2 −er1 ̸= 0, the system has only the trivial solution. Therefore λ isn’t an eigenvalue of (13.2.2). If λ = 1/4 then r1 = r2 = −3/2, so the general solution of (13.2.2) is y = e−3x/2(c1 + c2x). The boundary condition y(0) = 0 requires that c1 = 0, so y = c2xe−3x/2 and the boundary condition y(0) requires that c2 = 0. Therefore λ = 1/4 isn’t an eigenvalue of (13.2.2). If λ > 1/4 then r1 = −3 2 + iω and r2 = −3 2 −iω, with ω = √ 4λ −1 2 or, equivalently, λ = 1 + 4ω2 4 . (13.2.3) | Elementary Differential Equations with Boundary Value Problems_Page_698_Chunk3100 |
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