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Mathematics LibreTexts

12.2E: The Wave Equation (Exercises)

  • Page ID
    18304
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    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] \(u(x,0)=x\sin x\),\(u_t(x,0)=\sin x\),\(-\infty<x<\infty\)