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# 12.2E: The Wave Equation (Exercises)

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$$\newcommand{\place}{\bigskip\hrule\bigskip\noindent} \newcommand{\threecol}[3]{\left[\begin{array}{r}#1\\#2\\#3\end{array}\right]} \newcommand{\threecolj}[3]{\left[\begin{array}{r}#1\$1\jot]#2\\[1\jot]#3\end{array}\right]} \newcommand{\lims}[2]{\,\bigg|_{#1}^{#2}} \newcommand{\twocol}[2]{\left[\begin{array}{l}#1\\#2\end{array}\right]} \newcommand{\ctwocol}[2]{\left[\begin{array}{c}#1\\#2\end{array}\right]} \newcommand{\cthreecol}[3]{\left[\begin{array}{c}#1\\#2\\#3\end{array}\right]} \newcommand{\eqline}[1]{\centerline{\hfill\displaystyle#1\hfill}} \newcommand{\twochar}[4]{\left|\begin{array}{cc} #1-\lambda\\#3-\lambda\end{array}\right|} \newcommand{\twobytwo}[4]{\left[\begin{array}{rr} #1\\#3\end{array}\right]} \newcommand{\threechar}[9]{\left[\begin{array}{ccc} #1-\lambda\\#4-\lambda\\#7 -\lambda\end{array}\right]} \newcommand{\threebythree}[9]{\left[\begin{array}{rrr} #1\\#4\\#7 \end{array}\right]} \newcommand{\solutionpart}[1]{\vskip10pt\noindent\underbar{\color{blue}\sc Solution({\bf #1})\ }} \newcommand{\Cex}{\fbox{\textcolor{red}{C}}\, } \newcommand{\CGex}{\fbox{\textcolor{red}{C/G}}\, } \newcommand{\Lex}{\fbox{\textcolor{red}{L}}\, } \newcommand{\matfunc}[3]{\left[\begin{array}{cccc}#1_{11}(t)_{12}(t)&\cdots _{1#3}(t)\\#1_{21}(t)_{22}(t)&\cdots_{2#3}(t)\\\vdots& \vdots&\ddots&\vdots\\#1_{#21}(t)_{#22}(t)&\cdots_{#2#3}(t) \end{array}\right]} \newcommand{\col}[2]{\left[\begin{array}{c}#1_1\\#1_2\\\vdots\\#1_#2\end{array}\right]} \newcommand{\colfunc}[2]{\left[\begin{array}{c}#1_1(t)\\#1_2(t)\\\vdots\\#1_#2(t)\end{array}\right]} \newcommand{\cthreebythree}[9]{\left[\begin{array}{ccc} #1\\#4\\#7 \end{array}\right]} 1 \ newcommand {\ dy} {\ ,\ mathrm {d}y} \ newcommand {\ dx} {\ ,\ mathrm {d}x} \ newcommand {\ dyx} {\ ,\ frac {\ mathrm {d}y}{\ mathrm {d}x}} \ newcommand {\ ds} {\ ,\ mathrm {d}s} \ newcommand {\ dt }{\ ,\ mathrm {d}t} \ newcommand {\dst} {\ ,\ frac {\ mathrm {d}s}{\ mathrm {d}t}}$$ We suggest that you perform experiments of this kind in the exercises marked , 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 [exer:12.2.17}, [exer:12.2.34}, and [exer:12.2.49}; however, the idea of the experiment is the same. [exer:12.2.1] $$u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$$, $$u(0,t)=0,\quad u(1,t)=0,\quad t>0$$, $$u(x,0)=0,\quad u_t(x,0)= \left\{\begin{array}{cl} x,&0\le x\le{1\over2},\\1-x,&{1\over2}\le x\le1 \end{array}\right.,\quad0\le x\le1$$ [exer:12.2.2] $$u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$$, $$u(0,t)=0,\quad u(1,t)=0,\quad t>0$$, $$u(x,0)=x(1-x),\quad u_t(x,0)=0,\quad0\le x\le 1$$ [exer:12.2.3] $$u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0$$, $$u(0,t)=0,\quad u(1,t)=0,\quad t>0$$, $$u(x,0)=x^2(1-x),\quad u_t(x,0)=0,\quad0\le x\le1$$ [exer:12.2.4] $$u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$$, $$u(0,t)=0,\quad u(1,t)=0,\quad t>0$$, $$u(x,0)=0,\quad u_t(x,0)=x(1-x),\quad0\le x\le 1$$ [exer:12.2.5] $$u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0$$, $$u(0,t)=0,\quad u(1,t)=0,\quad t>0$$, $$u(x,0)=0\quad u_t(x,0)=x^2(1-x),\quad0\le x\le1$$ [exer:12.2.6] $$u_{tt}=64u_{xx},\quad 0<x<3,\quad t>0$$, $$u(0,t)=0,\quad u(3,t)=0,\quad t>0$$, $$u(x,0)=x(x^2-9),\quad u_t(x,0)=0,\quad0\le x\le 3$$ [exer:12.2.7] $$u_{tt}=4u_{xx},\quad 0<x<1,\quad t>0$$, $$u(0,t)=0,\quad u(1,t)=0,\quad t>0$$, $$u(x,0)=x(x^3-2x^2+1),\quad u_t(x,0)=0,\quad0\le x\le1$$ [exer:12.2.8] $$u_{tt}=64u_{xx},\quad 0<x<3,\quad t>0$$, $$u(0,t)=0,\quad u(3,t)=0,\quad t>0$$, $$u(x,0)=0,\quad u_t(x,0)=x(x^2-9),\quad0\le x\le 3$$ [exer:12.2.9] $$u_{tt}=4u_{xx},\quad 0<x<1,\quad t>0$$, $$u(0,t)=0,\quad u(1,t)=0,\quad t>0$$, $$u(x,0)=0,\quad u_t(x,0)=x(x^3-2x^2+1),\quad0\le x\le1$$ [exer:12.2.10] $$u_{tt}=5u_{xx},\quad 0<x<\pi,\quad t>0$$, $$u(0,t)=0,\quad u(\pi,t)=0,\quad t>0$$, $$u(x,0)=x\sin x,\quad u_t(x,0)=0,\quad0\le x\le \pi$$ [exer:12.2.11] $$u_{tt}=u_{xx},\quad 0<x<1,\quad t>0$$, $$u(0,t)=0,\quad u(1,t)=0,\quad t>0$$, $$u(x,0)=x(3x^4-5x^3+2),\quad u_t(x,0)=0,\quad0\le x\le1$$ [exer:12.2.12] $$u_{tt}=5u_{xx},\quad 0<x<\pi,\quad t>0$$, $$u(0,t)=0,\quad u(\pi,t)=0,\quad t>0$$, $$u(x,0)=0,\quad u_t(x,0)=x\sin x,\quad0\le x\le \pi$$ [exer:12.2.13] $$u_{tt}=u_{xx},\quad 0<x<1,\quad t>0$$, $$u(0,t)=0,\quad u(1,t)=0,\quad t>0$$, $$u(x,0)=0,\quad u_t(x,0)=x(3x^4-5x^3+2),\quad0\le x\le1$$ [exer:12.2.14] $$u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$$, $$u(0,t)=0,\quad u(1,t)=0,\quad t>0$$, $$u(x,0)=x(3x^4-10x^2+7),\quad u_t(x,0)=0,\quad0\le x\le1$$ [exer:12.2.15] $$u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$$, $$u(0,t)=0,\quad u(1,t)=0,\quad t>0$$, $$u(x,0)=0\quad u_t(x,0)=x(3x^4-10x^2+7),\quad0\le x\le1$$ [exer:12.2.16] We saw that the displacement of the plucked string is, on the one hand, \[u(x,t)=\{4L\over\pi^2}\sum_{n=1}^\infty{(-1)^{n+1}\over(2n-1)^2} \cos{(2n-1)\pi at\over L}\sin{(2n-1)\pi x\over L},\; 0\le x\le L,\; t\ge 0, \eqno{\rm(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. \eqno{\rm(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

$\mbox{\bf(i)}\quad t={mL\over a}+\tau\quad \text{or} \quad \mbox{\bf(ii)}\quad t={(m+1)L\over a}-\tau,$

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

$t_j={Lj\over 2ka},\quad j=0,1,\dots k;$

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.

[exer:12.2.17] 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

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

Justify defining the formal solution of (A) to be

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

where

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

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

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

[exer:12.2.18] $$u_{tt}=9u_{xx},\quad 0<x<2,\quad t>0$$,
$$u_x(0,t)=0,\quad u(2,t)=0,\quad t>0$$,
$$u(x,0)=4-x^2,\quad u_t(x,0)=0,\quad0\le x\le2$$

[exer:12.2.19] $$u_{tt}=4u_{xx},\quad 0<x<1,\quad t>0$$,
$$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$$,
$$u(x,0)=x^2(1-x),\quad u_t(x,0)=0,\quad0\le x\le 1$$

[exer:12.2.20] $$u_{tt}=9u_{xx},\quad 0<x<2,\quad t>0$$,
$$u_x(0,t)=0,\quad u(2,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=4-x^2,\quad0\le x\le2$$

[exer:12.2.21] $$u_{tt}=4u_{xx},\quad 0<x<1,\quad t>0$$,
$$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=x^2(1-x),\quad0\le x\le 1$$

[exer:12.2.22] $$u_{tt}=5u_{xx},\quad 0<x<1,\quad t>0$$,
$$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$$,
$$u(x,0)=2x^3+3x^2-5,\quad u_t(x,0)=0,\quad0\le x\le1$$

[exer:12.2.23] $$u_{tt}=3u_{xx},\quad 0<x<\pi,\quad t>0$$,
$$u_x(0,t)=0,\quad u(\pi,t)=0,\quad t>0$$,
$$u(x,0)=\pi^3-x^3,\quad u_t(x,0)=0,\quad0\le x\le\pi$$

[exer:12.2.24] $$u_{tt}=5u_{xx},\quad 0<x<1,\quad t>0$$,
$$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=2x^3+3x^2-5,\quad0\le x\le1$$

[exer:12.2.25] $$u_{tt}=3u_{xx},\quad 0<x<\pi,\quad t>0$$,
$$u_x(0,t)=0,\quad u(\pi,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=\pi^3-x^3,\quad0\le x\le\pi$$

[exer:12.2.26] $$u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$$,
$$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$$,
$$u(x,0)=x^4-2x^3+1,\quad u_t(x,0)=0,\quad0\le x\le1$$

[exer:12.2.27] $$u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0$$,
$$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$$,
$$u(x,0)=4x^3+3x^2-7,\quad u_t(x,0)=0,\quad0\le x\le1$$

[exer:12.2.28] $$u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$$,
$$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=x^4-2x^3+1,\quad0\le x\le1$$

[exer:12.2.29] $$u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0$$,
$$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=4x^3+3x^2-7,\quad0\le x\le1$$

[exer:12.2.30] $$u_{tt}=u_{xx},\quad 0<x<1,\quad t>0$$,
$$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$$,
$$u(x,0)=x^4-4x^3+6x^2-3,\quad u_t(x,0)=0,\quad0\le x\le1$$

[exer:12.2.31] $$u_{tt}=u_{xx},\quad 0<x<1,\quad t>0$$,
$$u_x(0,t)=0,\quad u(1,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=x^4-4x^3+6x^2-3,\quad0\le x\le1$$

[exer:12.2.32] Adapt the proof of Theorem [thmtype:12.2.2} to find d’Alembert’s solution of the initial-boundary value problem in Exercise [exer:12.2.17}.

[exer:12.2.33] Use the result of Exercise [exer:12.2.32} to show that the formal solution of the initial-boundary value problem in Exercise [exer:12.2.17} is an actual solution if $$g$$ is differentiable and $$f$$ is twice differentiable on $$[0,L]$$ and

$g'_+(0)=g(L)=f'_+(0)=f(L)=f''_-(L)=0.$

[exer:12.2.34] Justify defining the formal solution of the initial-boundary value problem

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

to be

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

where

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

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

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

[exer:12.2.35] $$u_{tt}=64u_{xx},\quad 0<x<\pi,\quad t>0$$,
$$u(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$$,
$$u(x,0)=x(2\pi-x),\quad u_t(x,0)=0,\quad0\le x\le \pi$$

[exer:12.2.36] $$u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$$,
$$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$$,
$$u(x,0)=x^2(3-2x),\quad u_t(x,0)=0,\quad0\le x\le 1$$

[exer:12.2.37] $$u_{tt}=64u_{xx},\quad 0<x<\pi,\quad t>0$$,
$$u(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=x(2\pi-x),\quad0\le x\le \pi$$

[exer:12.2.38] $$u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$$,
$$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=x^2(3-2x),\quad0\le x\le 1$$

[exer:12.2.39] $$u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$$,
$$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$$,
$$u(x,0)=(x-1)^3+1,\quad u_t(x,0)=0,\quad0\le x\le 1$$

[exer:12.2.40] $$u_{tt}=3u_{xx},\quad 0<x<\pi,\quad t>0$$,
$$u(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$$,
$$u(x,0)=x(x^2-3\pi^2),\quad u_t(x,0)=0,\quad0\le x\le\pi$$

[exer:12.2.41] $$u_{tt}=9u_{xx},\quad 0<x<1,\quad t>0$$,
$$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=(x-1)^3+1,\quad0\le x\le 1$$

[exer:12.2.42] $$u_{tt}=3u_{xx},\quad 0<x<\pi,\quad t>0$$,
$$u(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=x(x^2-3\pi^2),\quad0\le x\le\pi$$

[exer:12.2.43] $$u_{tt}=5u_{xx},\quad 0<x<1,\quad t>0$$,
$$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$$,
$$u(x,0)=x^3(3x-4),\quad u_t(x,0)=0,\quad0\le x\le1$$

[exer:12.2.44] $$u_{tt}=16u_{xx},\quad 0<x<1,\quad t>0$$,
$$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$$,
$$u(x,0)=x(x^3-2x^2+2),\quad u_t(x,0)=0,\quad0\le x\le1$$

[exer:12.2.45] $$u_{tt}=5u_{xx},\quad 0<x<1,\quad t>0$$,
$$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=x^3(3x-4),\quad0\le x\le1$$

[exer:12.2.46] $$u_{tt}=16u_{xx},\quad 0<x<1,\quad t>0$$,
$$u(0,t)=0,\quad u_x(1,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=x(x^3-2x^2+2),\quad0\le x\le1$$

[exer:12.2.47] Adapt the proof of Theorem [thmtype:12.2.2} to find d’Alembert’s solution of the initial-boundary value problem in Exercise [exer:12.2.34}.

[exer:12.2.48] Use the result of Exercise [exer:12.2.47} to show that the formal solution of the initial-boundary value problem in Exercise [exer:12.2.34} is an actual solution if $$g$$ is differentiable and $$f$$ is twice differentiable on $$[0,L]$$ and

$f(0)=f'_-(L)=g(0)=g_-'(L)=f''_+(0)=0.$

[exer:12.2.49] Justify defining the formal solution of the initial-boundary value problem

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

to be

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

where

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

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

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

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

[exer:12.2.50] $$u_{tt}=5u_{xx},\quad 0<x<2,\quad t>0$$,
$$u_x(0,t)=0,\quad u_x(2,t)=0,\quad t>0$$,
$$u(x,0)=2x^2(3-x),\quad u_t(x,0)=0,\quad0\le x\le 2$$

[exer:12.2.51] $$u_{tt}=5u_{xx},\quad 0<x<2,\quad t>0$$,
$$u_x(0,t)=0,\quad u_x(2,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=2x^2(3-x),\quad0\le x\le 2$$

[exer:12.2.52] $$u_{tt}=4u_{xx},\quad 0<x<\pi,\quad t>0$$,
$$u_x(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$$,
$$u(x,0)=x^3(3x-4\pi),\quad u_t(x,0)=0,\quad0\le x\le \pi$$

[exer:12.2.53] $$u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0$$,
$$u_x(0,t)=0,\quad u_x(1,t)=0,\quad t>0$$,
$$u(x,0)=3x^2(x^2-2),\quad u_t(x,0)=0,\quad0\le x\le 1$$

[exer:12.2.54] $$u_{tt}=4u_{xx},\quad 0<x<\pi,\quad t>0$$,
$$u_x(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=x^3(3x-4\pi),\quad0\le x\le \pi$$

[exer:12.2.55] $$u_{tt}=7u_{xx},\quad 0<x<1,\quad t>0$$,
$$u_x(0,t)=0,\quad u_x(1,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=3x^2(x^2-2),\quad0\le x\le 1$$

[exer:12.2.56] $$u_{tt}=16u_{xx},\quad 0<x<\pi,\quad t>0$$,
$$u_x(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$$,
$$u(x,0)=x^2(x-\pi)^2,\quad u_t(x,0)=0,\quad0\le x\le \pi$$

[exer:12.2.57] $$u_{tt}=u_{xx},\quad 0<x<1,\quad t>0$$,
$$u_x(0,t)=0,\quad u_x(1,t)=0,\quad t>0$$,
$$u(x,0)=x^2(3x^2-8x+6),\quad u_t(x,0)=0,\quad0\le x\le 1$$

[exer:12.2.58] $$u_{tt}=16u_{xx},\quad 0<x<\pi,\quad t>0$$,
$$u_x(0,t)=0,\quad u_x(\pi,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=x^2(x-\pi)^2,\quad0\le x\le \pi$$

[exer:12.2.59] $$u_{tt}=u_{xx},\quad 0<x<1,\quad t>0$$,
$$u_x(0,t)=0,\quad u_x(1,t)=0,\quad t>0$$,
$$u(x,0)=0,\quad u_t(x,0)=x^2(3x^2-8x+6),\quad0\le x\le 1$$

[exer:12.2.60] Adapt the proof of Theorem [thmtype:12.2.2} to find d’Alembert’s solution of the initial-boundary value problem in Exercise [exer:12.2.49}.

[exer:12.2.61] Use the result of Exercise [exer:12.2.60} to show that the formal solution of the initial-boundary value problem in Exercise [exer:12.2.49} is an actual solution if $$g$$ is differentiable and $$f$$ is twice differentiable on $$[0,L]$$ and

$f'_+(0)=f'_-(L)=g'_+(0)=g_-'(L)=0.$

[exer:12.2.62] 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

$u(x,t)=\sum_{n=1}^\infty k_np_n(x)q_n(t), \eqno{\rm(A)}$

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,

$u_x(x,t)= \sum_{n=1}^\infty k_np_n'(x)q_n(t)$

and

$u_t(x,t)= \sum_{n=1}^\infty k_np_n(x)q_n'(t).$

Suppose $$\sum_{n=1}^\infty n^2|k_n|$$ converges. Show that

$u_{xx}(x,y)= \sum_{n=1}^\infty k_np_n''(x)q_n(t)$

and

$u_{tt}(x,y)= \sum_{n=1}^\infty k_np_n(x)q_n''(t)$

Suppose $$\sum_{n=1}^\infty n^2|\alpha_n|$$ and $$\sum_{n=1}^\infty n|\beta_n|$$ both converge. Show that the formal solution

$u(x,t)=\sum_{n=1}^\infty\left(\alpha_n\cos{n\pi at\over L}+{\beta_nL\over n\pi a}\sin{n\pi at\over L}\right) \sin{n\pi x\over L}$

of Equation Equation \ref{eq:12.2.1} satisfies $$u_{tt}=a^2u_{xx}$$ for all $$(x,t)$$.

This conclusion also applies to the formal solutions defined in Exercises [exer:12.2.17}, [exer:12.2.34}, and [exer:12.2.49}.

[exer:12.2.63] Suppose $$g$$ is differentiable and $$f$$ is twice differentiable on $$(-\infty,\infty)$$, and let

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

Show that

${\partial^2 u_0\over\partial t^2}=a^2{\partial^2u_0\over\partial x^2},\quad-\infty<x<\infty,\quad t>0,$

and

$u_0(x,0)=f(x),\quad {\partial u_0\over\partial t}(x,0)=0,\quad -\infty<x<\infty.$

Show that

${\partial^2 u_1\over\partial t^2}=a^2{\partial^2u_1\over\partial x^2},\quad-\infty<x<\infty,\quad t>0,$

and

$u_1(x,0)=0,\quad {\partial u_1\over\partial t}(x,0)=g(x),\quad -\infty<x<\infty.$

Solve

$u_{tt}=a^2u_{xx},\quad-\infty<t<\infty,\quad t>0,$

$u(x,0)=f(x),\quad u_t(x,0)=g(x),\quad-\infty<x<\infty.$

[exer:12.2.64] $$u(x,0)=x$$,$$u_t(x,0)=4ax$$,$$-\infty<x<\infty$$

[exer:12.2.65] $$u(x,0)=x^2$$,$$u_t(x,0)=1$$,$$-\infty<x<\infty$$

[exer:12.2.66] $$u(x,0)=\sin x$$,$$u_t(x,0)=a\cos x$$,$$-\infty<x<\infty$$

[exer:12.2.67] $$u(x,0)=x^3$$,$$u_t(x,0)=6x^2$$,$$-\infty<x<\infty$$

[exer:12.2.68] $$u(x,0)=x\sin x$$,$$u_t(x,0)=\sin x$$,$$-\infty<x<\infty$$