12.2E: The Wave Equation (Exercises)

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We suggest that you perform experiments of this kind in Exercises 12.2.4, 12.2.8, 12.2.12, 12.2.15, 12.2.22, 12.2.25, 12.2.29, 12.2.38, 12.2.44, 12.2.46, 12.2.54, 12.2.57, and 12.2.59, 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 12.2.17, 12.2.34, and 12.2.49; however, the idea of the experiment is the same.

Q12.2.1

In Exercises 12.2.1-12.2.15 solve the initial-boundary value problem. In some of these exercises, Theorem 11.3.5b or Exercise 11.3.35 will simplify the computation of the coefficients in the Fourier sine series.

1. $u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)= \left\{\begin{array}{cl} x,&0\le x\le{1\over2},\\1-x,&{1\over2}\le x\le1 \end{array}\right.,\quad0\le x\le1$

2. $u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=x(1-x),\quad u_t(x,0)=0,\quad0\le x\le 1$

3. $u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=x^2(1-x),\quad u_t(x,0)=0,\quad0\le x\le1$

4. $u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x(1-x),\quad0\le x\le 1$

5. $u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=0\quad u_t(x,0)=x^2(1-x),\quad0\le x\le1$

6. $u_{tt}=64u_{xx},\quad 0<x<3,\quad t>0$,
$u(0,t)=0,\quad u(3,t)=0,\quad t>0$,
$u(x,0)=x(x^2-9),\quad u_t(x,0)=0,\quad0\le x\le 3$

7. $u_{tt}=4u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=x(x^3-2x^2+1),\quad u_t(x,0)=0,\quad0\le x\le1$

8. $u_{tt}=64u_{xx},\quad 0<x<3,\quad t>0$,
$u(0,t)=0,\quad u(3,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x(x^2-9),\quad0\le x\le 3$

9. $u_{tt}=4u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x(x^3-2x^2+1),\quad0\le x\le1$

10. $u_{tt}=5u_{xx},\quad 0<x<\pi,\quad t>0$,
$u(0,t)=0,\quad u(\pi,t)=0,\quad t>0$,
$u(x,0)=x\sin x,\quad u_t(x,0)=0,\quad0\le x\le \pi$

11. $u_{tt}=u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=x(3x^4-5x^3+2),\quad u_t(x,0)=0,\quad0\le x\le1$

12. $u_{tt}=5u_{xx},\quad 0<x<\pi,\quad t>0$,
$u(0,t)=0,\quad u(\pi,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x\sin x,\quad0\le x\le \pi$

13. $u_{tt}=u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x(3x^4-5x^3+2),\quad0\le x\le1$

14. $u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=x(3x^4-10x^2+7),\quad u_t(x,0)=0,\quad0\le x\le1$

15. $u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=0\quad u_t(x,0)=x(3x^4-10x^2+7),\quad0\le x\le1$

Q12.2.2

16. We saw that the displacement of the plucked string is, on the one hand,

\[u(x,t)=\frac{4L}{\pi ^{2}}\sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{(2n-1)^{2}}\cos \frac{(2n-1)\pi at}{L}\sin\frac{(2n-1)\pi x}{L},\: 0\leq x\leq L,\:t\geq 0,\tag{A}\]

and, on the other hand,

\[u(x,\tau)= \left\{\begin{array}{cl} x,&0\le x\le{L\over2}-a\tau,\\[5pt] {L\over2}-a\tau,&{L\over2}-a\tau\le x\le{L\over2}+a\tau,\\[5pt] L-x,&{L\over2}-a\tau\le x\le L. \end{array}\right. \tag{B}\]

if $0\le \tau\le L/2a$. The first objective of this exercise is to show that (B) can be used to compute $u(x,t)$ for $0\le x\le L$ and all $t>0$.

Show that if $t>0$, there’s a nonnegative integer $m$ such that either \[{\bf(i)}\quad t={mL\over a}+\tau\quad \text{or} \quad {\bf(ii)}\quad t={(m+1)L\over a}-\tau,\nonumber \] where $0\le \tau\le L/2a$.
Use (A) to show that $u(x,t)=(-1)^mu(x,\tau)$ if (i) holds, while $u(x,t)=(-1)^{m+1}u(x,\tau)$ if (ii) holds.
Perform the following experiment for specific values of $L$ and $a$ and various values of $m$ and $k$: Let \[t_j={Lj\over 2ka},\quad j=0,1,\dots k;\nonumber \] thus, $t_0$, $t_1$, …, $t_k$ are equally spaced points in $[0,L/2a]$. For each $j=0$, $1$, $2$,…, $k$, graph the $m$th partial sum of (A) and $u(x,t_j)$ 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

\[\begin{array}{c} u_{tt}=a^2u_{xx},\quad 0<x<L,\quad t>0,\\ u_x(0,t)=0,\quad u(L,t)=0,\quad t>0,\\ u(x,0)=f(x),\quad u_t(x,0)=g(x),\quad 0\le x\le L. \end{array} \tag{A}\]

Justify defining the formal solution of (A) to be

\[u(x,t)=\sum_{n=1}^\infty \left(\alpha_n\cos{(2n-1)\pi a t\over2L}+{2L\beta_n\over(2n-1)\pi a}\sin{(2n-1)\pi at\over2L}\right) \cos{(2n-1)\pi x\over2L},\nonumber \]

where

\[C_{M\!f}(x)=\sum_{n=1}^\infty\alpha_n\cos{(2n-1)\pi x\over2L} \quad \text{and} \quad C_{M\!g}(x)=\sum_{n=1}^\infty\beta_n\cos{(2n-1)\pi x\over2L}\nonumber \]

are the mixed Fourier cosine series of $f$ and $g$ on $[0,L]$; that is,

\[\alpha_n={2\over L}\int_0^Lf(x)\cos{(2n-1)\pi x\over2L}\,dx \quad \text{and} \quad \beta_n={2\over L}\int_0^Lg(x)\cos{(2n-1)\pi x\over2L}\,dx.\nonumber \]

Q12.2.3

In Exercises 12.2.18-12.2.31, use Exercise 12.2.17 to solve the initial-boundary value problem. In some of these exercises Theorem 11.3.5c or Exercise 11.3.42b will simplify the computation of the coefficients in the mixed Fourier cosine series.

18. $u_{tt}=9u_{xx},\quad 0<x<2,\quad t>0$,
$u_x(0,t)=0,\quad u(2,t)=0,\quad t>0$,
$u(x,0)=4-x^2,\quad u_t(x,0)=0,\quad0\le x\le2$

19. $u_{tt}=4u_{xx},\quad 0<x<1,\quad t>0$,
$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=x^2(1-x),\quad u_t(x,0)=0,\quad0\le x\le 1$

20. $u_{tt}=9u_{xx},\quad 0<x<2,\quad t>0$,
$u_x(0,t)=0,\quad u(2,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=4-x^2,\quad0\le x\le2$

21. $u_{tt}=4u_{xx},\quad 0<x<1,\quad t>0$,
$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x^2(1-x),\quad0\le x\le 1$

22. $u_{tt}=5u_{xx},\quad 0<x<1,\quad t>0$,
$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=2x^3+3x^2-5,\quad u_t(x,0)=0,\quad0\le x\le1$

23. $u_{tt}=3u_{xx},\quad 0<x<\pi,\quad t>0$,
$u_x(0,t)=0,\quad u(\pi,t)=0,\quad t>0$,
$u(x,0)=\pi^3-x^3,\quad u_t(x,0)=0,\quad0\le x\le\pi$

24. $u_{tt}=5u_{xx},\quad 0<x<1,\quad t>0$,
$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=2x^3+3x^2-5,\quad0\le x\le1$

25. $u_{tt}=3u_{xx},\quad 0<x<\pi,\quad t>0$,
$u_x(0,t)=0,\quad u(\pi,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=\pi^3-x^3,\quad0\le x\le\pi$

26. $u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$,
$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=x^4-2x^3+1,\quad u_t(x,0)=0,\quad0\le x\le1$

27. $u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0$,
$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=4x^3+3x^2-7,\quad u_t(x,0)=0,\quad0\le x\le1$

28. $u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$,
$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x^4-2x^3+1,\quad0\le x\le1$

29. $u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0$,
$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=4x^3+3x^2-7,\quad0\le x\le1$

30. $u_{tt}=u_{xx},\quad 0<x<1,\quad t>0$,
$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=x^4-4x^3+6x^2-3,\quad u_t(x,0)=0,\quad0\le x\le1$

31. $u_{tt}=u_{xx},\quad 0<x<1,\quad t>0$,
$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x^4-4x^3+6x^2-3,\quad0\le x\le1$

Q12.2.4

32. Adapt the proof of Theorem 12.2.2 to find d’Alembert’s solution of the initial-boundary value problem in Exercise 12.2.17.

33. Use the result of Exercise 12.2.32 to show that the formal solution of the initial-boundary value problem in Exercise 12.2.17 is an actual solution if $g$ is differentiable and $f$ is twice differentiable on $[0,L]$ and

\[g'_+(0)=g(L)=f'_+(0)=f(L)=f''_-(L)=0.\nonumber \]

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

\[\begin{array}{c} u_{tt}=a^2u_{xx},\quad 0<x<L,\quad t>0,\\ u(0,t)=0,\quad u_x(L,t)=0,\quad t>0,\\ u(x,0)=f(x),\quad u_t(x,0)=g(x),\quad 0\le x\le L \end{array}\nonumber\]

to be

\[u(x,t)=\sum_{n=1}^\infty \left(\alpha_n\cos{(2n-1)\pi a t\over2L}+{2L\beta_n\over(2n-1)\pi a}\sin{(2n-1)\pi at\over2L}\right) \sin{(2n-1)\pi x\over2L},\nonumber \]

where

\[S_{M\!f}(x)=\sum_{n=1}^\infty\alpha_n\sin{(2n-1)\pi x\over2L} \quad \text{and} \quad S_{M\!g}(x)=\sum_{n=1}^\infty\beta_n\sin{(2n-1)\pi x\over2L}\nonumber \]

are the mixed Fourier sine series of $f$ and $g$ on $[0,L]$; that is,

\[\alpha_n={2\over L}\int_0^Lf(x)\sin{(2n-1)\pi x\over2L}\,dx \quad \text{and} \quad \beta_n={2\over L}\int_0^Lg(x)\sin{(2n-1)\pi x\over2L}\,dx.\nonumber \]

Q12.2.5

In Exercises 12.2.35-12.2.46 use Exercise 12.2.34 to solve the initial-boundary value problem. In some of these exercises Theorem 11.3.5d or Exercise 11.3.50b will simplify the computation of the coefficients in the mixed Fourier sine series.

35. $u_{tt}=64u_{xx},\quad 0<x<\pi,\quad t>0$,
$u(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$,
$u(x,0)=x(2\pi-x),\quad u_t(x,0)=0,\quad0\le x\le \pi$

36. $u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$,
$u(x,0)=x^2(3-2x),\quad u_t(x,0)=0,\quad0\le x\le 1$

37. $u_{tt}=64u_{xx},\quad 0<x<\pi,\quad t>0$,
$u(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x(2\pi-x),\quad0\le x\le \pi$

38. $u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x^2(3-2x),\quad0\le x\le 1$

39. $u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$,
$u(x,0)=(x-1)^3+1,\quad u_t(x,0)=0,\quad0\le x\le 1$

40. $u_{tt}=3u_{xx},\quad 0<x<\pi,\quad t>0$,
$u(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$,
$u(x,0)=x(x^2-3\pi^2),\quad u_t(x,0)=0,\quad0\le x\le\pi$

41. $u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=(x-1)^3+1,\quad0\le x\le 1$

42. $u_{tt}=3u_{xx},\quad 0<x<\pi,\quad t>0$,
$u(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x(x^2-3\pi^2),\quad0\le x\le\pi$

43. $u_{tt}=5u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$,
$u(x,0)=x^3(3x-4),\quad u_t(x,0)=0,\quad0\le x\le1$

44. $u_{tt}=16u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$,
$u(x,0)=x(x^3-2x^2+2),\quad u_t(x,0)=0,\quad0\le x\le1$

45. $u_{tt}=5u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x^3(3x-4),\quad0\le x\le1$

46. $u_{tt}=16u_{xx},\quad 0<x<1,\quad t>0$,
$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x(x^3-2x^2+2),\quad0\le x\le1$

Q12.2.6

47. Adapt the proof of Theorem 12.2.2 to find d’Alembert’s solution of the initial-boundary value problem in Exercise 12.2.34.

48. Use the result of Exercise 12.2.47 to show that the formal solution of the initial-boundary value problem in Exercise 12.2.34 is an actual solution if $g$ is differentiable and $f$ is twice differentiable on $[0,L]$ and

\[f(0)=f'_-(L)=g(0)=g_-'(L)=f''_+(0)=0.\nonumber\]

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

\[\begin{array}{c} u_{tt}=a^2u_{xx},\quad 0<x<L,\quad t>0,\\ u_x(0,t)=0,\quad u_x(L,t)=0,\quad t>0,\\ u(x,0)=f(x),\quad u_t(x,0)=g(x),\quad 0\le x\le L. \end{array}\nonumber\]

to be

\[u(x,t)=\alpha_0+\beta_0t+\sum_{n=1}^\infty \left(\alpha_n\cos{n\pi at\over L}+{L\beta_n\over n\pi a}\sin{n\pi at\over L}\right) \cos{n\pi x\over L},\nonumber\]

where

\[C_f(x)=\alpha_0+\sum_{n=1}^\infty\alpha_n\cos{n\pi x\over L} \quad \text{and} \quad C_g(x)=\beta_0+\sum_{n=1}^\infty\beta_n\cos{n\pi x\over L}\nonumber\]

are the Fourier cosine series of $f$ and $g$ on $[0,L]$; that is,

\[\alpha_0={1\over L}\int_0^Lf(x)\,dx,\quad \beta_0={1\over L}\int_0^Lg(x)\,dx,\nonumber\]

\[\alpha_n={2\over L}\int_0^Lf(x)\cos{n\pi x\over L}\,dx, \quad \text{and} \quad \beta_n={2\over L}\int_0^Lg(x)\cos{n\pi x\over L}\,dx,\quad n=1,2,3,\dots.\nonumber\]

Q12.2.7

In Exercises 12.2.50-12.2.59 use Exercise 12.2.49 to solve the initial-boundary value problem. In some of these exercises Theorem 11.3.5a will simplify the computation of the coefficients in the Fourier cosine series.

50. $u_{tt}=5u_{xx},\quad 0<x<2,\quad t>0$,
$u_x(0,t)=0,\quad u_x(2,t)=0,\quad t>0$,
$u(x,0)=2x^2(3-x),\quad u_t(x,0)=0,\quad0\le x\le 2$

51. $u_{tt}=5u_{xx},\quad 0<x<2,\quad t>0$,
$u_x(0,t)=0,\quad u_x(2,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=2x^2(3-x),\quad0\le x\le 2$

52. $u_{tt}=4u_{xx},\quad 0<x<\pi,\quad t>0$,
$u_x(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$,
$u(x,0)=x^3(3x-4\pi),\quad u_t(x,0)=0,\quad0\le x\le \pi$

53. $u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0$,
$u_x(0,t)=0,\quad u_x(1,t)=0,\quad t>0$,
$u(x,0)=3x^2(x^2-2),\quad u_t(x,0)=0,\quad0\le x\le 1$

54. $u_{tt}=4u_{xx},\quad 0<x<\pi,\quad t>0$,
$u_x(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x^3(3x-4\pi),\quad0\le x\le \pi$

55. $u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0$,
$u_x(0,t)=0,\quad u_x(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=3x^2(x^2-2),\quad0\le x\le 1$

56. $u_{tt}=16u_{xx},\quad 0<x<\pi,\quad t>0$,
$u_x(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$,
$u(x,0)=x^2(x-\pi)^2,\quad u_t(x,0)=0,\quad0\le x\le \pi$

57. $u_{tt}=u_{xx},\quad 0<x<1,\quad t>0$,
$u_x(0,t)=0,\quad u_x(1,t)=0,\quad t>0$,
$u(x,0)=x^2(3x^2-8x+6),\quad u_t(x,0)=0,\quad0\le x\le 1$

58. $u_{tt}=16u_{xx},\quad 0<x<\pi,\quad t>0$,
$u_x(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x^2(x-\pi)^2,\quad0\le x\le \pi$

59. $u_{tt}=u_{xx},\quad 0<x<1,\quad t>0$,
$u_x(0,t)=0,\quad u_x(1,t)=0,\quad t>0$,
$u(x,0)=0,\quad u_t(x,0)=x^2(3x^2-8x+6),\quad0\le x\le 1$

Q12.2.8

60. Adapt the proof of Theorem 12.2.2 to find d’Alembert’s solution of the initial-boundary value problem in Exercise 12.2.49.

61. Use the result of Exercise 12.2.60 to show that the formal solution of the initial-boundary value problem in Exercise 12.2.49 is an actual solution if $g$ is differentiable and $f$ is twice differentiable on $[0,L]$ and

\[f'_+(0)=f'_-(L)=g'_+(0)=g_-'(L)=0.\nonumber\]

62. Suppose $\lambda$ and $\mu$ are constants and either $p_n(x)= \cos n\lambda x$ or $p_n(x)=\sin n\lambda x$, while either $q_n(t)=\cos n\mu t$ or $q_n(t)=\sin n\mu t$ for $n=1$, $2$, $3$, …. Let

\[u(x,t)=\sum_{n=1}^\infty k_np_n(x)q_n(t), \tag{A}\]

where $\{k_n\}_{n=1}^\infty$ are constants.

Show that if $\sum_{n=1}^\infty |k_n|$ converges then $u(x,t)$ converges for all $(x,t)$.
Use Theorem 12.1.2 to show that if $\sum_{n=1}^\infty n|k_n|$ converges then (A) can be differentiated term by term with respect to $x$ and $t$ for all $(x,t)$; that is, \[u_x(x,t)= \sum_{n=1}^\infty k_np_n'(x)q_n(t)\nonumber \] and \[u_t(x,t)= \sum_{n=1}^\infty k_np_n(x)q_n'(t).\nonumber \]
Suppose $\sum_{n=1}^\infty n^2|k_n|$ converges. Show that \[u_{xx}(x,y)= \sum_{n=1}^\infty k_np_n''(x)q_n(t)\nonumber \] and \[u_{tt}(x,y)= \sum_{n=1}^\infty k_np_n(x)q_n''(t)\nonumber \]
Suppose $\sum_{n=1}^\infty n^2|\alpha_n|$ and $\sum_{n=1}^\infty n|\beta_n|$ both converge. Show that the formal solution \[u(x,t)=\sum_{n=1}^\infty\left(\alpha_n\cos{n\pi at\over L}+{\beta_nL\over n\pi a}\sin{n\pi at\over L}\right) \sin{n\pi x\over L}\nonumber \] of Equation 12.2.1 satisfies $u_{tt}=a^2u_{xx}$ for all $(x,t)$.
This conclusion also applies to the formal solutions defined in Exercises 12.2.17, 12.2.34, and 12.2.49.

63. Suppose $g$ is differentiable and $f$ is twice differentiable on $(-\infty,\infty)$, and let

\[u_0(x,t)={f(x+at)+f(x-at)\over2}\quad \text{and} \quad u_1(x,t)={1\over2a}\int_{x-at}^{x+at}g(u)\,du.\nonumber \]

Show that \[{\partial^2 u_0\over\partial t^2}=a^2{\partial^2u_0\over\partial x^2},\quad-\infty<x<\infty,\quad t>0,\nonumber \] and \[u_0(x,0)=f(x),\quad {\partial u_0\over\partial t}(x,0)=0,\quad -\infty<x<\infty.\nonumber \]
Show that \[{\partial^2 u_1\over\partial t^2}=a^2{\partial^2u_1\over\partial x^2},\quad-\infty<x<\infty,\quad t>0,\nonumber \] and \[u_1(x,0)=0,\quad {\partial u_1\over\partial t}(x,0)=g(x),\quad -\infty<x<\infty.\nonumber \]
Solve \[u_{tt}=a^2u_{xx},\quad-\infty<t<\infty,\quad t>0,\nonumber \] \[u(x,0)=f(x),\quad u_t(x,0)=g(x),\quad-\infty<x<\infty.\nonumber \]

Q12.2.9

In Exercises 12.2.64-12.2.68 use the result of Exercise 12.2.63 to find a solution of \[u_{tt}=a^{2}u_{xx},\quad -\infty <x<\infty \nonumber\] that satisfies the given initial conditions.

64. $u(x,0)=x$,$u_t(x,0)=4ax$,$-\infty<x<\infty$

65. $u(x,0)=x^2$,$u_t(x,0)=1$,$-\infty<x<\infty$

66. $u(x,0)=\sin x$,$u_t(x,0)=a\cos x$,$-\infty<x<\infty$

67. $u(x,0)=x^3$,$u_t(x,0)=6x^2$,$-\infty<x<\infty$

68. $u(x,0)=x\sin x$,$u_t(x,0)=\sin x$,$-\infty<x<\infty$