Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 3.1

1. Solve the initial value problems:
   1. $u_{tt}-c^2u_{xx}=0$ with $u(x,0)=x^3$ and $u_t(x,0)=\sin x$.
   2. $u_{tt}-c^2u_{xx}=2t$ with $u(x,0)=x^2$ and $u_t(x,0)=1$.
Solution

1. Using d'Alembert's formula (6) with $g(x)=x^3$ and $h(x)=\sin x$, we have

   $\begin{aligned}u(x,t)&=\frac{(x+ct)^3+(x-ct)^3}2+\frac1{2c}\int_{x-ct}^{x+ct}\!\sin\xi\,\mathrm d\xi\\&=x^3+3c^2xt^2-\frac{\cos(x+ct)-\cos(x-ct)}{2c}=x^3+3c^2xt^2+\frac{\sin(x)\sin(ct)}c\quad\text{for}~(x,t)\in\mathbb R^2.\end{aligned}$

2. Since the wave equation is a linear equation, the solution $u$ can be decomposed as $u=v+w$, where

   $\left\{\begin{aligned} &v_{tt}-c^2v_{xx}=0,\\&v(x,0)=x^2,~~v_t(x,0)=1\end{aligned}\right.\quad\text{and}\quad\left\{\begin{aligned} &w_{tt}-c^2w_{xx}=2t,\\&w(x,0)=0,~~w_t(x,0)=0.\end{aligned}\right.$

   By d'Alembert's formula (6) with $g(x)=x^2$ and $h(x)=1$, we have

   $\displaystyle v(x,t)=\frac{(x+ct)^2+(x-ct)^2}2+\frac1{2c}\int_{x-ct}^{x+ct}\!1\,\mathrm d\xi=x^2+c^2t^2+t\quad\text{for}~(x,t)\in\mathbb R^2.$

   On the other hand, we use (19), which follows from Duhamel's principle and d'Alembert's formula, to get

   $\begin{aligned}w(x,t)&=\frac1{2c}\int_0^t\!\int_{x-c(t-s)}^{x+c(t-s)}\!2s\,\mathrm d\xi\,\mathrm ds=\frac1{c}\int_0^t\!s\cdot2c(t-s)\,\mathrm ds\\&=2\int_0^t\!(ts-s^2)\,\mathrm ds=\left.ts^2-\frac{2s^3}3\right|_0^t=\frac{t^3}3\quad\text{for}~(x,t)\in\mathbb R^2.\end{aligned}$

   Therefore, the solution is

   $\displaystyle u(x,t)=v(x,t)+w(x,t)=x^2+c^2t^2+t+\frac{t^3}3\quad\text{for}~(x,t)\in\mathbb R^2.$



2. Solve the initial/boundary value problem

   $\left\{\begin{aligned} &u_{tt}-u_{xx}=0\quad\text{for}~0<x<\pi~\text{and}~t>0,\\&u(x,0)=0,~~u_t(x,0)=1\quad\text{for}~0<x<\pi,\\&u(0,t)=0,~~u(\pi,t)=0\quad\text{for}~t\geq0\end{aligned}\right.$

   using a Fourier series. Using the parallelogram rule, find the values of the solution in the various regions. Is the resulting solution continuous? Is it $C^1$?
Solution

Using the separation of variables, we suppose that the solution $u(x,t)$ is of the form

$\displaystyle u(x,t)=\sum_{n=1}^\infty\!u_n(x,t)=\sum_{n=1}^\infty\!X_n(x)T_n(t)\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0$.

Due to the superposition principle, we firstly consider $u(x,t)=X(x)T(t)$ to fulfill the wave equation with the boundary condition, which gives

$X(x)T''(t)-X''(x)T(t)=0\quad\text{and}\quad X(0)T(t)=X(\pi)T(t)=0$.

Surely, we want to find a general solution so $T$ is not identically zero and $X(0)=X(\pi)=0$. From the differential equation, we have $X''(x)/X(x)=T''(t)/T(t)$ for $x\in(0,\pi)$ and $t>0$, which means there exists a constant $\lambda$ such that $X''(x)=\lambda X(x)$ and $T''(t)=\lambda T(t)$. To find nontrivial solution $X$, we divide three cases as follows.

• If $\lambda>0$, we may write $\lambda=\mu^2$ for some $\mu>0$. Then $X(x)=c_1e^{\mu x}+c_2e^{-\mu x}$. Then by $X(0)=X(\pi)=0$, we find $c_1=c_2=0$ and then $X\equiv0$, which is a trivial solution.
• If $\lambda=0$, we get $X(x)=c_1+c_2x$. By $X(0)=X(\pi)=0$, it is easy to see $c_1=c_2=0$ and hence $X\equiv0$, which is a trivial solution.
• If $\lambda<0$, we may write $\lambda=-\mu^2$ for some $\mu>0$. Then we have $X(x)=c_1\sin(\mu x)+c_2\cos(\mu x)$. By $X(0)=0$, we get $c_2=0$. From $X(\pi)=0$, we have $c_1\sin(\mu\pi)=0$. Since we want to seek the nontrivial solution, we hope $c_1\neq0$, which means $\sin(\mu\pi)$ should be zero and $\mu$ must be an integer.

Based on the above discussion, for $\lambda=-n^2$, we have $X_n(x)=\sin(nx)$. On the other hand, we have $T''(t)=-n^2T(t)$, which gives $T(t)=c_n\sin(nt)+d_n\cos(nt)$. Thus, the solution $u$ has the following form:

$\displaystyle u(x,t)=\sum_{n=1}^\infty(c_n\sin(nt)+d_n\cos(nt))\sin(nx)\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0$.

Since $u(x,0)\equiv0$ on $[0,\pi]$, we have

$\displaystyle0=u(x,0)=\sum_{n=1}^\infty d_n\sin(nx)\quad\text{for}~x\in[0,\pi]$,

which gives $d_n=0$ for $n\in\mathbb N$. Differentiating $u$ with respect to $t$ yields

$\displaystyle u_t(x,t)=\sum_{n=1}^\infty nc_n\cos(nt)\sin(nx)$.

From $u_t(x,0)=1$, we have

$\displaystyle1=u_t(x,0)=\sum_{n=1}^\infty nc_n\sin(nx)\quad\text{for}~x\in[0,\pi]$.

Multiplying it by $\sin(mx)$ and then integrating it over $[0,\pi]$, we obtain

$\begin{aligned}\frac{1-(-1)^m}{m}&=\int_0^\pi\!1\cdot\sin(mx)\,\mathrm dx=\int_0^\pi\!\sum_{n=1}^\infty nc_n\sin(nx)\sin(mx)\,\mathrm dx\\&=m\int_0^\pi\!c_m\sin^2(mx)\,\mathrm dx=\frac{mc_m}2\int_0^\pi\!(1-\cos(2mx))\,\mathrm dx=\frac{mc_m\pi}2.\end{aligned}$

This means $\displaystyle c_n=\frac{2(1-(-1)^n)}{\pi n^2}$ for $n\in\mathbb N$. Therefore, we obtain

$\displaystyle u(x,t)=\frac2\pi\sum_{n=1}^\infty\frac{1-(-1)^n}{n^2}\sin(nt)\sin(nx)=\frac4\pi\sum_{m=1}^\infty\frac{\sin[(2m-1)t]\sin[(2m-1)x]}{(2m-1)^2}\quad\text{for}~x\in[0,\pi],\,t\geq0$.

To use the parallelogram rule, we can firstly define the following regions.

• $R_0=\{(x,t)\in\mathbb R^2\,:\,t\leq x\leq\pi-t,\,0\leq t\leq\pi/2\}$.
• $R_{3k-2}=\{(x,t)\in\mathbb R^2\,:\,0\leq x\leq\pi/2,\,(k-1)\pi+x\leq t\leq k\pi-x\}$ for $k\in\mathbb N$.
• $R_{3k-1}=\{(x,t)\in\mathbb R^2\,:\,\pi/2\leq x\leq\pi,\,k\pi-x\leq t\leq(k-1)\pi+x\}$ for $k\in\mathbb N$.
• $R_{3k}=\{(x,t)\in\mathbb R^2\,:\,(k-1)\pi\leq t-x\leq k\pi,\,k\pi\leq t+x\leq(k+1)\pi\}$ for $k\in\mathbb N$.

In region $R_0$, by d'Alembert's formula, the solution is

$\displaystyle u(x,t)=\frac{u(x+t,0)+u(x-t,0)}{2}+\frac12\int_{x-t}^{x+t}\!u_t(s,0)\,\mathrm ds=t$.

In region $R_1$, by the parallelogram rule, the solution is

$\displaystyle u(x,t)=u(0,-x+t)+u\left(\frac{x+t}2,\frac{x+t}2\right)-u\left(\frac{-x+t}2,\frac{-x+t}2\right)=0+\frac{x+t}2-\frac{-x+t}2=x$.

In region $R_2$, by the parallelogram rule, the solution is

$\begin{aligned}u(x,t)&=u\left(\frac{x-t+\pi}2,\frac{-x+t+\pi}2\right)+u(\pi,x+t-\pi)-u\left(\frac{-x-t+3\pi}2,\frac{x+t-\pi}2\right)\\&=\frac{-x+t+\pi}2+0-\frac{x+t-\pi}2=\pi-x.\end{aligned}$

In region $R_3$, by the parallelogram rule, the solution is

$\begin{aligned}u(x,t)&=u\left(\frac{x-t+\pi}2,\frac{-x+t+\pi}2\right)+u\left(\frac{x+t}2,\frac{x+t}2\right)-u\left(\frac\pi2,\frac\pi2\right)\\&=\frac{x-t+\pi}2+\left(\pi-\frac{x+t}2\right)-\frac\pi2=\pi-t.\end{aligned}$

In region $R_4$, by the parallelogram rule, the solution is

$\begin{aligned}u(x,t)&=u(0,-x+t)+u\left(\frac{x+t-\pi}2,\frac{x+t+\pi}2\right)-u\left(\frac{-x+t-\pi}2,\frac{-x+t+\pi}2\right)\\&=0+\left(\pi-\frac{x+t+\pi}2\right)-\left(\pi-\frac{-x+t+\pi}2\right)=-x.\end{aligned}$

In region $R_5$, by the parallelogram rule, the solution is

$\begin{aligned}u(x,t)&=u\left(\frac{x-t+2\pi}2,\frac{-x+t+2\pi}2\right)+u(\pi,x+t-\pi)-u\left(\frac{-x-t+4\pi}2,\frac{x+t}2\right)\\&=\left(\pi-\frac{-x+t+2\pi}2\right)+0-\left(\pi-\frac{x+t}2\right)=x-\pi.\end{aligned}$

In region $R_6$, by the parallelogram rule, the solution is

$\begin{aligned}u(x,t)&=u\left(\frac{x-t+2\pi}2,\frac{-x+t+2\pi}2\right)+u\left(\frac{x+t-\pi}2,\frac{x+t+\pi}2\right)-u\left(\frac\pi2,\frac{3\pi}2\right)\\&=-\frac{x-t+2\pi}2+\left(\frac{x+t-\pi}2-\pi\right)-\left(\pi-\frac{3\pi}2\right)=t-2\pi.\end{aligned}$

Therefore, we may use the mathematical induction to show that $u(x,t+4\pi)=u(x,t)$ for $t\geq0$, i.e., we have

• $u(x,t)=(-1)^{k-1}x$ in $R_{3k-1}$ for $k\in\mathbb N$.
• $u(x,t)=(-1)^{k-1}(\pi-x)$ in $R_{3k-2}$ for $k\in\mathbb N$.
• $u(x,t)=(-1)^k(t-k\pi)$ in $R_{3k}$ for $k\in\mathbb N\cup\{0\}$.

Clearly, $u$ is continuous but not $C^1$.

Remark: We are able to sum the Fourier series solution in regions bounded by characteristics as follows. By the product to sum formula, the solution can be expressed as

$\displaystyle u(x,t)=\frac2\pi\sum_{m=1}^\infty\frac{\cos[(2m-1)(x-t)]-\cos[(2m-1)(x+t)]}{(2m-1)^2}\quad\text{for}~x\in[0,\pi],\,t\geq0.$

Note that $\displaystyle\sum_{m=1}^\infty\frac{\cos[(2m-1)x]}{(2m-1)^2}=\frac{\pi^2-2\pi x}8$ for $x\in[0,\pi]$ (see the details from Solution of Exercise 3).

In region $R_0$ ($0<x-t<\pi$ and $0<x+t<\pi$), we have

$\displaystyle u(x,t)=\frac2\pi\left[\frac{\pi^2-2\pi(x-t)}8-\frac{\pi^2-2\pi(x+t)}8\right]=t$.

In region $R_{3k-2}$ ($0<x-t+k\pi<\pi$ and $0<x+t-(k-1)\pi<\pi$), we have

$\begin{aligned}u(x,t)&=\frac{2(-1)^k}\pi\sum_{m=1}^\infty\frac{\cos[(2m-1)(x-t+k\pi)]+\cos[(2m-1)(x+t+(k-1)\pi)]}{(2m-1)^2}\\&=\frac{2(-1)^k}\pi\left[\frac{\pi^2-2\pi(x-t+k\pi)}8+\frac{\pi^2-2\pi(x+t-(k-1)\pi)}8\right]\\&=(-1)^{k-1}x.\end{aligned}$

In region $R_{3k-1}$ ($0<x-t+(k-1)\pi<\pi$ and $0<x+t-k\pi<\pi$), we have

$\begin{aligned}u(x,t)&=\frac{2(-1)^{k-1}}\pi\sum_{m=1}^\infty\frac{\cos[(2m-1)(x-t+(k-1)\pi)]+\cos[(2m-1)(x+t-k\pi)]}{(2m-1)^2}\\&=\frac{2(-1)^{k-1}}\pi\left[\frac{\pi^2-2\pi(x-t+(k-1)\pi)}8+\frac{\pi^2-2\pi(x+t-k\pi)}8\right]\\&=(-1)^{k-1}(\pi-x).\end{aligned}$

In region $R_{3k}$ ($0<x-t+k\pi<\pi$ and $0<x+t-k\pi<\pi$), we have

$\begin{aligned}u(x,t)&=\frac{2(-1)^k}\pi\sum_{m=1}^\infty\frac{\cos[(2m-1)(x-t+k\pi)]-\cos[(2m-1)(x+t-k\pi)]}{(2m-1)^2}\\&=\frac{2(-1)^k}\pi\left[\frac{\pi^2-2\pi(x-t+k\pi)}8-\frac{\pi^2-2\pi(x+t-k\pi)}8\right]\\&=(-1)^k(t-k\pi).\end{aligned}$



3. Consider the initial/boundary value problem

   $\left\{\begin{aligned} &u_{tt}-u_{xx}=0\quad\text{for}~0<x<\pi~\text{and}~t>0,\\&u(x,0)=x,~~u_t(x,0)=0\quad\text{for}~0<x<\pi,\\&u_x(0,t)=0,~~u_x(\pi,t)=0\quad\text{for}~t\geq0.\end{aligned}\right.$

   1. Find a Fourier series solution, and sum the series in regions bounded by characteristics. Do you think the solution is unique?
   2. Use the parallelogram rule to solve this problem; is the resulting solution unique? continuous? $C^1$?
Solution
1. Using the separation of variables, we suppose that the solution $u(x,t)$ is of the form

   $\displaystyle u(x,t)=\sum_{n=0}^\infty u_n(x,t)=\sum_{n=0}^\infty X_n(x)T_n(t)\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0.$

   Due to the superposition principle, we firstly consider $u(x,t)=X(x)T(t)$ to fulfill the wave equation with the boundary condition, which gives

   $X(x)T''(t)-X''(x)T(t)=0\quad\text{and}\quad X'(0)T(t)=X'(\pi)T(t)=0$.

   Surely, we want to find a general solution so $T$ is not identically zero and $X'(0)=X'(\pi)=0$. From the differential equation, we have $X''(x)/X(x)=T''(t)/T(t)$ for $x\in(0,\pi)$ and $t>0$, which means there exists a constant $\lambda$ such that $X''(x)=\lambda X(x)$ and $T''(t)=\lambda T(t)$. To find nontrivial solution $X$, we divide three cases as follows.
   • If $\lambda>0$, we may write $\lambda=\mu^2$ for some $\mu>0$. Then $X(x)=c_1e^{\mu x}+c_2e^{-\mu x}$. Then by $X'(0)=X'(\pi)=0$, we find $c_1=c_2=0$ and then $X\equiv0$, which is a trivial solution.
   • If $\lambda=0$, we get $X(x)=c_1+c_2x$. By $X'(0)=X'(\pi)=0$, it is easy to see $c_2=0$ and hence $X\equiv c_1$. This gives a nonzero solution, provided $c_1\neq0$.
   • If $\lambda<0$, we may write $\lambda=-\mu^2$ for some $\mu>0$. Then we have $X(x)=c_1\sin(\mu x)+c_2\cos(\mu x)$. By $X'(0)=0$, we get $c_1=0$. From $X'(\pi)=0$, we have $-c_2\mu\sin(\mu\pi)=0$. Since we want to seek the nontrivial solution, we hope $c_2\neq0$, which means $\sin(\mu\pi)$ should be zero and $\mu$ must be an integer.
   Based on the above discussion, for $\lambda=-n^2$ and $n\in\mathbb N\cup\{0\}$, we have $X_n(x)=\cos(nx)$. On the other hand, $T''(t)=-n^2T(t)$, we obtain $T(t)=c_n\sin(nt)+d_n\cos(nt)$ for $n>0$ and $T(t)=A+Bt$ for $n=0$. Thus, the solution $u$ has the following form:

   $\displaystyle u(x,t)=(A+Bt)+\sum_{n=1}^\infty\!(c_n\sin(nt)+d_n\cos(nt))\cos(nx)\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0.$

   Since $u_t(x,0)=0$ on $[0,\pi]$, we have

   $\displaystyle0=u_t(x,0)=B+\sum_{n=1}^\infty c_n\cos(nx)$,

   which gives $B=c_n=0$ for $n\in\mathbb N$. Hence the solution becomes

   $\displaystyle u(x,t)=A+\sum_{n=1}^\infty\!d_n\cos(nt)\cos(nx)\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0.$

   Since $u(x,0)=x$, we have

   $\displaystyle x=u(x,0)=A+\sum_{n=1}^\infty d_n\cos(nx)\quad\text{for}~x\in[0,\pi]$.

   Integrating it over $[0,\pi]$, we get $\displaystyle A=\frac\pi2$. Multiplying it by $\cos(mx)$ and then integrating it over $[0,\pi]$, we obtain

   $\displaystyle d_n=\frac{\displaystyle\int_0^\pi\!x\cos(nx)\,\mathrm dx}{\displaystyle\int_0^\pi\!\cos^2(nx)\,\mathrm dx}=\frac{\displaystyle\left.\frac{x\sin(nx)}n\right|_0^\pi-\frac1n\int_0^\pi\!\sin(nx)\,\mathrm dx}{\displaystyle\frac12\int_0^\pi\!(1-\cos(2nx))\,\mathrm dx}=\frac{2((-1)^n-1)}{n^2\pi}\quad\text{for}~n\in\mathbb N.$

   Thus, the Fourier solution must be

   $\begin{aligned}u(x,t)&=\frac\pi2+\frac2\pi\sum_{n=1}^\infty\frac{(-1)^n-1}{n^2}\cos(nt)\cos(nx)\\&=\frac\pi2-\frac4\pi\sum_{m=1}^\infty\frac{\cos[(2m-1)t]\cos[(2m-1)x]}{(2m-1)^2}\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0.\end{aligned}$

   Using the product to sum formula, the Fourier solution beomces

   $\displaystyle u(x,t)=\frac\pi2-\frac2\pi\sum_{m=1}^\infty\frac{\cos[(2m-1)(x+t)]+\cos[(2m-1)(x-t)]}{(2m-1)^2}\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0$.

   To sum the Fourier series in regions bounded by characteristics, we define
   • $R_0=\{(x,t)\in\mathbb R^2\,:\,t\leq x\leq\pi-t,\,0\leq t\leq\pi/2\}$.
   • $R_{3k-2}=\{(x,t)\in\mathbb R^2\,:\,0\leq x\leq\pi/2,\,(k-1)\pi+x\leq t\leq k\pi-x\}$ for $k\in\mathbb N$.
   • $R_{3k-1}=\{(x,t)\in\mathbb R^2\,:\,\pi/2\leq x\leq\pi,\,k\pi-x\leq t\leq(k-1)\pi+x\}$ for $k\in\mathbb N$.
   • $R_{3k}=\{(x,t)\in\mathbb R^2\,:\,(k-1)\pi \leq t-x\leq k\pi,\,k\pi\leq t+x\leq(k+1)\pi\}$ for $k\in\mathbb N$.
   From $u(x,0)=x$, we note that $\displaystyle\sum_{m=1}^\infty\frac{\cos[(2m-1)x]}{(2m-1)^2}=\frac{\pi^2-2\pi x}{8}$ for $x\in[0,\pi]$.
   In region $R_0$ ($0<x+t<\pi$ and $0<x-t<\pi$), we have

   $\displaystyle u(x,t)=\frac\pi2-\frac2\pi\left[\frac{\pi^2-2\pi(x+t)}8+\frac{\pi^2-2\pi(x-t)}8\right]=x$.

   In region $R_{3k-2}$ ($0<x+t-(k-1)\pi<\pi$ and $0<x-t+k\pi<\pi$), we have

   $\begin{aligned}u(x,t)&=\frac\pi2+\frac{2(-1)^k}\pi\sum_{m=1}^\infty\frac{\cos[(2m-1)(x+t-(k-1)\pi)]-\cos[(2m-1)(x-t+k\pi)]}{(2m-1)^2}\\&=\frac\pi2+\frac{2(-1)^k}{\pi}\left[\frac{\pi^2-2\pi(x+t-(k-1)\pi)}8-\frac{\pi^2-2\pi(x-t+k\pi)}8\right]\\&=\frac\pi2+(-1)^k\left(-t+k\pi-\frac\pi2\right).\end{aligned}$

   In region $R_{3k-1}$ ($0<x+t-k\pi<\pi$ and $0<x-t+(k-1)\pi<\pi$), we have

   $\begin{aligned}u(x,t)&=\frac\pi2-\frac{2(-1)^k}\pi\sum_{m=1}^\infty\left[\frac{\cos[(2m-1)(x+t-k\pi)]-\cos[(2m-1)(x-t+(k-1)\pi)]}{(2m-1)^2}\right]\\&=\frac\pi2-\frac{2(-1)^k}\pi\left[\frac{\pi^2-2\pi(x+t-k\pi)}8-\frac{\pi^2-2\pi(x-t+(k-1)\pi)}8\right]\\&=\frac\pi2+(-1)^k\left(t-k\pi+\frac\pi2\right).\end{aligned}$

   In region $R_{3k}$ ($0<x+t-k\pi<\pi$ and $0<x-t+k\pi<\pi$), we have

   $\begin{aligned}u(x,t)&=\frac\pi2-\frac{2(-1)^k}\pi\sum_{m=1}^\infty\frac{\cos[(2m-1)(x+t-k\pi)]+\cos[(2m-1)(x-t+k\pi)]}{(2m-1)^2}\\&=\frac\pi2-\frac{2(-1)^k}\pi\left[\frac{\pi^2-2\pi(x+t-k\pi)}8+\frac{\pi^2-2\pi(x-t+k\pi)}8\right]\\&=\frac\pi2-(-1)^k\left(\frac{\pi-2x}4\right)=\frac{2\pi-(-1)^k(\pi-2x)}4.\end{aligned}$

   Clearly, this solution is continuous but not differentiable along the characteristics, and such solution is unique.
2. To use the parallelogram rule, we can again employ the region $R_k$ for $k\in\mathbb N\cup\{0\}$ as in (a). Due to the lack of the boundary data $u_x$, it is impossible to determine the solution uniquely. But it is possible to obtain the unique continuous solution (which is equal to the Fourier series solution) and we cannot obtain $C^1$ solution. Surely, we may observe that $u_x(x,0)=1$ for $0<x<\pi$, which is discontinuous at $x=0$ because $u_x(0,t)=0$ for $t\geq0$. Therefore, the parallelogram rule only gives the solution in region $R_0$:

   $\displaystyle u(x,t)=\frac{u(x+t,0)+u(x-t,0)}2+\frac12\int_{x-t}^{x+t}\!u_t(s,0)\,\mathrm ds=\frac{(x+t)+(x-t)}2+\frac12\int_{x-t}^{x+t}\!0\,\mathrm ds=x.$

   
   Warning: It does not seem reasonable to extend $u_x$ continuously so we cannot use the parallelogram rule for $u_x$ due to the lack of information of $u_x$ in region $R_k$ for $k\in\mathbb N$.


4. Consider the initial boundary value problem

   $\left\{\begin{aligned} &u_{tt}-c^2u_{xx}=0\quad\text{for}~x,t>0,\\&u(x,0)=g(x),~~u_t(x,0)=h(x)\quad\text{for}~x>0,\\&u(0,t)=0\quad\text{for}~t\geq0,\end{aligned}\right.$

   where $g(0)=0=h(0)$. If we extend $g$ and $h$ as odd functions on $-\infty<x<\infty$, show that d'Alembert's formula (6) gives the solution.
Solution

Let $\tilde g:\mathbb R\to\mathbb R$ and $\tilde h:\mathbb R\to\mathbb R$ be defined by

$\tilde g(x)=\begin{cases}g(x)&\text{if}~x\geq0;\\-g(-x)&\text{if}~x<0,\end{cases}\quad\text{and}\quad\tilde h(x)=\begin{cases}h(x)&\text{if}~x\geq0;\\-h(-x)&\text{if}~x<0.\end{cases}$

Then we firstly consider the following initial value problem

$\left\{\begin{aligned} &\tilde u_{tt}-c^2\tilde u_{xx}=0\quad\text{for}~x\in\mathbb R~\text{and}~t>0,\\&\tilde u(x,0)=\tilde g(x),\quad\tilde u_t(x,0)=\tilde h(x)\quad\text{for}~x\in\mathbb R.\end{aligned}\right.$

By d'Alembert's formula, the solution $\tilde u$ is

$\displaystyle\tilde u(x,t)=\frac{\tilde g(x+ct)+\tilde g(x-ct)}2+\frac1{2c}\int_{x-ct}^{x+ct}\!\tilde h(s)\,\mathrm ds\quad\text{for}~(x,t)\in\mathbb R\times\bar{\mathbb R}_+$.

When $x=0$, it is easy to verify that

$\displaystyle\tilde u(0,t)=\frac{\tilde g(ct)+\tilde g(-ct)}2+\frac1{2c}\int_{-ct}^{ct}\!\tilde h(s)\,\mathrm ds=\frac{g(ct)-g(ct)}2+\frac1{2c}\int_{-ct}^0\![-h(-s)]\,\mathrm ds+\frac1{2c}\int_0^{ct}\!h(s)\,\mathrm ds=0$

Thus, function $u=\tilde u\Big|_{\bar{\mathbb R}_+^2}$ solves the initial boundary value problem. Moreover, the solution $u$ can be represented as

$u(x,t)=\begin{cases}\displaystyle\frac{g(x+ct)+g(x-ct)}2+\frac1{2c}\int_{x-ct}^{x+ct}\!h(s)\,\mathrm ds&\text{if}~x\geq ct;\\\displaystyle\frac{g(ct+x)-g(ct-x)}2+\frac1{2c}\int_{ct-x}^{ct+x}\!h(s)\,\mathrm ds&\text{if}~0\leq x<ct.\end{cases}$



5. Find in closed form (similar to d'Alembert's formula) the solution $u(x,t)$ of

   $\left\{\begin{aligned} &u_{tt}-c^2u_{xx}=0\quad\text{for}~x,t>0,\\&u(x,0)=g(x),~~u_t(x,0)=h(x)\quad\text{for}~x>0\\&u(0,t)=\alpha(t)\quad\text{for}~t\geq0,\end{aligned}\right.$

   where $g,h,\alpha\in C^2$ satisfy $\alpha(0)=g(0)$, $\alpha'(0)=h(0)$, and $\alpha''(0)=c^2g''(0)$. Verify that $u\in C^2$, even on the characteristic $x=ct$.
Solution 1

Let $v:\bar{\mathbb R}_+^2\to\mathbb R$ be defined by $v(x,t)=u(x,t)-\alpha(t)$. Then $v$ satisfies

$\left\{\begin{aligned} &v_{tt}-c^2v_{xx}=-\alpha''(t)\quad\text{for}~(x,t)\in\mathbb R_+^2,\\&v(x,0)=g(x)-\alpha(0),\quad v_t(x,0)=h(x)-\alpha'(0)\quad\text{for}~x\in\mathbb R_+,\\&v(0,t)=0\quad\text{for}~t\in\bar{\mathbb R}_+.\end{aligned}\right.$

Set $\tilde g(x)=g(x)-\alpha(0)$ and $\tilde h(x)=h(x)-\alpha'(0)$. By the assumption, we have $\tilde g(0)=0$ and $\tilde h(0)=0$. Now we consider the following two initival boundary value problems

$\left\{\begin{aligned} &\phi_{tt}-c^2\phi_{xx}=0\quad\text{for}~(x,t)\in\mathbb R_+^2,\\&\phi(x,0)=\tilde g(x),\quad\phi_t(x,0)=\tilde h(x)\quad\text{for}~x\in\mathbb R_+,\\&\phi(0,t)=0\quad\text{for}~t\in\bar{\mathbb R}_+\end{aligned}\right.\quad\text{and}\quad\left\{\begin{aligned} &\varphi_{tt}-c^2\varphi_{xx}=-\alpha''(t)\quad\text{for}~(x,t)\in\mathbb R_+^2,\\&\varphi(x,0)=\varphi_t(x,0)=0\quad\text{for}~x\in\mathbb R_+,\\&\varphi(0,t)=0\quad\text{for}~t\in\bar{\mathbb R}_+.\end{aligned}\right.$

Then the solution $u(x,t)=v(x,t)+\alpha(t)=\phi(x,t)+\varphi(x,t)+\alpha(t)$ for $(x,t)\in\bar{\mathbb R}_+^2$. By the result of problem 4, we get

$\begin{aligned}\phi(x,t)&=\begin{cases}\displaystyle\frac{\tilde g(x+ct)+\tilde g(x-ct)}2+\frac1{2c}\int_{x-ct}^{x+ct}\!\tilde h(s)\,\mathrm ds&\text{if}~x\geq ct;\\\displaystyle\frac{\tilde g(ct+x)-\tilde g(ct-x)}2+\frac1{2c}\int_{ct-x}^{ct+x}\!\tilde h(s)\,\mathrm ds&\text{if}~0\leq x<ct.\end{cases}\\&=\begin{cases}\displaystyle-\alpha(0)-t\alpha'(0)+\frac{g(x+ct)+g(x-ct)}2+\frac1{2c}\int_{x-ct}^{x+ct}\!h(s)\,\mathrm ds&\text{if}~x\geq ct;\\\displaystyle-\frac xc\alpha'(0)+\frac{g(ct+x)-g(ct-x)}2+\frac1{2c}\int_{ct-x}^{ct+x}\!h(s)\,\mathrm ds&\text{if}~0\leq x<ct.\end{cases}\end{aligned}$

On the other hand, we extend $\varphi$ to $\tilde\varphi:\mathbb R\times\bar{\mathbb R}_+\to\mathbb R$ by

$\tilde\varphi(x,t)=\begin{cases}\varphi(x,t)&\text{if}~x\geq0,\,t\geq0;\\-\varphi(-x,t)&\text{if}~x<0,\,t\geq0.\end{cases}$

Then $\tilde\varphi$ satisfies

$\left\{\begin{aligned} &\tilde\varphi_{tt}-c^2\tilde\varphi_{xx}=-\text{sgn}(x)\alpha''(t)\quad\text{for}~(x,t)\in\mathbb R\times\mathbb R_+,\\&\tilde\varphi(x,0)=\tilde\varphi_t(x,0)=0\quad\text{for}~x\in\mathbb R.\quad\end{aligned}\right.$

Now we can use (19), which follows from Duhamel's principle and d'Alembert's formula, to get

$\begin{aligned}\tilde\varphi(x,t)&=\frac1{2c}\int_0^t\!\int_{x-c(t-s)}^{x+c(t-s)}\![-\text{sgn}(\xi)\alpha''(s)]\,d\xi\,\mathrm ds\end{aligned}$

Since $\tilde\varphi(0,t)=0$, we arrive at $\varphi=\tilde\varphi\Big|_{\bar{\mathbb R}_+^2}$. When $x\geq ct\geq0$, we have $x-ct+cs\geq0$ for $0\leq s\leq t$ and hence

$\begin{aligned}\varphi(x,t)&=-\frac1{2c}\int_0^t\!\int_{x-c(t-s)}^{x+c(t-s)}\!\alpha''(s)\,\mathrm d\xi\,\mathrm ds=-\int_0^t\!(t-s)\alpha''(s)\,\mathrm ds\\&=-t\int_0^t\alpha''(s)\,\mathrm ds+\int_0^t\!s\alpha''(s)\,\mathrm ds=-t\alpha'(s)\Big|_0^t+s\alpha'(s)\Big|_0^t-\int_0^t\!\alpha'(s)\,\mathrm ds\\&=-t\alpha'(t)+t\alpha'(0)+t\alpha'(t)-0-\alpha(t)+\alpha(0)\\&=t\alpha'(0)-\alpha(t)+\alpha(0).\end{aligned}$

When $0\leq x<ct$, we have $x-ct+cs\geq0$ for $(ct-x)/c\leq s\leq t$ and $x-ct+cs<0$ for $0\leq s<(ct-x)/c$, and hence

$\begin{aligned}\varphi(x,t)&=\frac1{2c}\int_0^{(ct-x)/c}\!\int_{x-c(t-s)}^{x+c(t-s)}\![-\text{sgn}(\xi)\alpha''(s)]\,\mathrm d\xi\,\mathrm ds+\frac1{2c}\int_{(ct-x)/c}^t\!\int_{x-c(t-s)}^{x+c(t-s)}\![-\text{sgn}(\xi)\alpha''(s)]\,\mathrm d\xi\,\mathrm ds\\&=\frac1{2c}\int_0^{(ct-x)/c}\!\left[\int_{x-c(t-s)}^0\!\alpha''(s)\,\mathrm d\xi-\int_0^{x+c(t-s)}\!\alpha''(s)\,\mathrm d\xi\right]\,\mathrm ds-\int_{(ct-x)/c}^t\!(t-s)\alpha''(s)\,\mathrm ds\\&=-\frac xc\int_0^{(ct-x)/c}\!\alpha''(s)\,\mathrm ds-t\int_{(ct-x)/c}^t\!\alpha''(s)\,\mathrm ds+\int_{(ct-x)/c}^t\!s\alpha''(s)\,\mathrm ds\\&=\frac{x}c\left[\alpha'(0)-\alpha'\left(t-\frac xc\right)\right]+t\left[\alpha'\left(t-\frac xc\right)-\alpha'(t)\right]+\left[t\alpha'(t)-\left(t-\frac xc\right)\alpha'\left(t-\frac xc\right)\right]-\left[\alpha(t)-\alpha\left(t-\frac xc\right)\right]\\&=\frac xc\alpha'(0)-\alpha(t)+\alpha\left(t-\frac xc\right)\end{aligned}$

Hence the solution $u$ is

$u(x,t)=\begin{cases}\displaystyle\frac{g(x+ct)+g(x-ct)}2+\frac1{2c}\int_{x-ct}^{x+ct}\!h(s)\,\mathrm ds&\text{if}~x\geq ct\geq0;\\\displaystyle\alpha\left(t-\frac xc\right)+\frac{g(ct+x)-g(ct-x)}2+\frac1{2c}\int_{ct-x}^{ct+x}\!h(s)\,\mathrm dx&\text{if}~0\leq x<ct.\end{cases}$

Since $g,h,\alpha\in C^2$, it is clear that $u\in C^2$ in $\bar{\mathbb R}_+^2-\{(x,t)\in\mathbb R_+^2:x=ct,\,x\geq0\}$ and satisfies

$\begin{aligned} &u_x(x,t)=\begin{cases}\displaystyle\frac{g'(x+ct)+g'(x-ct)}2+\frac{h(x+ct)-h(x-ct)}{2c}&\text{if}~x>ct\geq0;\\\displaystyle-\frac1c\alpha'\left(t-\frac xc\right)+\frac{g'(ct+x)+g'(ct-x)}2+\frac{h(ct+x)+h(ct-x)}{2c}&\text{if}~0<x<ct,\end{cases}\\&u_t(x,t)=\begin{cases}\displaystyle\frac{c[g'(x+ct)-g'(x-ct)]}2+\frac{h(x+ct)+h(x-ct)}2&\text{if}~x>ct>0;\\\displaystyle\alpha'\left(t-\frac xc\right)+\frac{c[g'(ct+x)-g'(ct-x)]}2+\frac{h(ct+x)-h(ct-x)}2&\text{if}~0<x<ct,\end{cases}\\&u_{xx}(x,t)=\frac1{c^2}u_{tt}(x,t)=\begin{cases}\displaystyle\frac{g''(x+ct)+g''(x-ct)}{2}+\frac{h'(x+ct)-h'(x-ct)}{2c}&\text{if}~x>ct>0;\\\displaystyle\frac1{c^2}\alpha''\left(t-\frac xc\right)+\frac{g''(ct+x)-g''(ct-x)}2+\frac{h'(ct+x)-h'(ct-x)}{2c}&\text{if}~0<x<ct,\end{cases}\\&u_{xt}(x,t)=u_{tx}(x,t)=\begin{cases}\displaystyle\frac{c[g''(x+ct)-g''(x-ct)]+h'(x+ct)+h'(x-ct)}2&\text{if}~x>ct>0;\\\displaystyle-\frac1c\alpha''\left(t-\frac xc\right)+\frac{c[g''(ct+x)+g''(ct-x)]+h'(ct+x)+h'(ct-x)}2&\text{if}~0<x<ct.\end{cases}\end{aligned}$

Since $\alpha(0)=g(0)$, it follows that $u\in C^0(\bar{\mathbb R}_+^2)$. To show $u\in C^2$, even on the characteristic $x=ct$, we need to compute the second order derivatives and prove that they are continuous on the line $x=ct$. For any point $(x_0,t_0)$ satisfying $x_0=ct_0$ for $t_0>0$, we may use the definition of partial derivative with respect to $x$ to compute the right derivative and left derivative as follows.

$\begin{aligned}u_x^+(x_0,t_0)&=\lim_{s\to0^+}\frac{u(x_0+s,t_0)-u(x_0,t_0)}s\\&=\lim_{s\to0^+}\left[\frac{g(x_0+ct_0+s)-g(x_0+ct_0)}{2s}+\frac{g(s)-g(0)}{2s}+\frac1{2cs}\int_{x_0+ct_0}^{x_0+ct_0+s}\!h(\xi)\,\mathrm d\xi-\frac1{2cs}\int_0^s\!h(\xi)\,\mathrm d\xi\right]\\&=\frac{g'(x_0+ct_0)+g'(0)}2+\frac{h(x_0+ct_0)-h(0)}{2c},\\u_x^-(x_0,t_0)&=\lim_{s\to0^-}\frac{u(s_0+s,t_0)-u(x_0,t_0)}s\\&=\lim_{s\to0^-}\left[\frac{\alpha(-s/c)-\alpha(0)}s+\frac{g(ct_0+x_0+s)-g(ct_0+x_0)}{2s}-\frac{g(-s)-g(0)}{2s}+\frac1{2cs}\int_{ct_0+x_0}^{ct_0+x_0+s}\!h(\xi)\,\mathrm d\xi+\frac1{2cs}\int_{-s}^0\!h(\xi)\,\mathrm d\xi\right]\\&=-\frac1c\alpha'(0)+\frac{g'(ct_0+x_0)+g'(0)}2+\frac{h(ct_0+x_0)+h(0)}{2c}\\&=\frac{g'(ct_0+x_0)+g'(0)}2+\frac{h(ct_0+x_0)-h(0)}{2c}\quad\text{(by $\alpha'(0)=h(0)$)}.\end{aligned}$

Thus, $u_x^+(x_0,t_0)=u_x^-(x_0,t_0):=u_x(x_0,t_0)$ and $\displaystyle\lim_{(x,t)\to(x_0,t_0)}u_x(x,t)=u_x(x_0,t_0)$. Similarly, we get the partial derivative with repsect to $t$:

$\begin{aligned}u_t^+(x_0,t_0)&=\lim_{s\to0^+}\frac{u(x_0,t_0+s)-u(x_0,t_0)}s\\&=\lim_{s\to0^+}\left[\frac{g(x_0+ct_0+cs)-g(x_0+ct_0)}{2s}+\frac{g(-cs)-g(0)}{2s}+\frac1{2cs}\int_{x_0+ct_0}^{x_0+ct_0+cs}\!h(\xi)\,\mathrm d\xi+\frac1{2cs}\int_{-cs}^0\!h(\xi)\,\mathrm d\xi\right]\\&=\frac{c[g'(x_0+ct_0)-g'(0)]}2+\frac{h(x_0+ct_0)+h(0)}2,\\u_t^-(x_0,t_0)&=\lim_{s\to0^-}\frac{u(x_0,t_0+s)-u(x_0,t_0)}s\\&=\lim_{s\to0^-}\left[\frac{\alpha(s)-\alpha(0)}s+\frac{g(ct_0+x_0)-g(ct_0+x_0+cs)}{2s}-\frac{g(cs)-g(0)}{2s}+\frac1{2cs}\int_{ct_0+x_0}^{ct_0+x_0+cs}\!h(\xi)\,\mathrm d\xi-\frac1{2cs}\int_0^{cs}\!h(\xi)\,\mathrm d\xi\right]\\&=\alpha'(0)+\frac{c[g'(ct_0+x_0)-g'(0)]}2+\frac{h(ct_0+x_0)-h(0)}2\\&=\frac{c[g'(ct_0+x_0)-g'(0)]}2+\frac{h(ct_0+x_0)+h(0)}2\quad\text{(by $\alpha'(0)=h(0)$)}.\end{aligned}$

Thus, $u_t^+(x_0,t_0)=u_t^-(x_0,t_0):=u_t(x_0,t_0)$ and $\displaystyle\lim_{(x,t)\to(x_0,t_0)}u_t(x,t)=u_t(x_0,t_0)$. For the case that $x_0=t_0=0$, we can only use the right parital derivatives $u_x^+$ and $u_t^+$ to show the continuity of $u_x$ and $u_t$ at the connor. Therefore, $u\in C^1(\bar{\mathbb R}_+^2)$.

Now we compute the second order partial derivatives. According to the definition, for the point $(x_0,t_0)$ satisfying $x_0=ct_0$ for $t_0>0$, we have

$\begin{aligned}u_{xx}^+(x_0,t_0)&=\lim_{s\to0^+}\frac{u_x(x_0+s,t_0)-u_x(x_0,t_0)}s\\&=\lim_{s\to0^+}\left[\frac{g'(x_0+ct_0+s)-g'(x_0+ct_0)}{2s}+\frac{g'(s)-g'(0)}{2s}+\frac{h(x_0+ct_0+s)-h(x_0+ct_0)}{2cs}-\frac{h(s)-h(0)}{2cs}\right]\\&=\frac{g''(x_0+ct_0)+g''(0)}2+\frac{h'(x_0+ct_0)-h'(0)}{2c},\\u_{xx}^-(x_0,t_0)&=\lim_{s\to0^-}\frac{u_x(x_0+s,t_0)-u_x(x_0,t_0)}s\\&=\lim_{s\to0^-}\left[\frac{\alpha'(-s/c)-\alpha'(0)}{-cs}+\frac{g'(ct_0+x_0+s)-g'(ct_0+x_0)}{2s}+\frac{g'(-s)-g'(0)}{2s}+\frac{h(ct_0+x_0+s)-h(ct_0+x_0)}{2cs}+\frac{h(-s)-h(0)}{2cs}\right]\\&=\frac1{c^2}\alpha''(0)+\frac{g''(ct_0+x_0)-g''(0)}2+\frac{h'(ct_0+x_0)-h'(0)}{2c}\\&=\frac{g''(ct_0+x_0)+g''(0)}2+\frac{h'(ct_0+x_0)-h'(0)}{2c}\quad\text{(by $\alpha''(0)=c^2g''(0)$)}.\end{aligned}$

Thus, $u_{xx}^+(x_0,t_0)=u_{xx}^-(x_0,t_0)$ and $\displaystyle\lim_{(x,t)\to(x_0,t_0)}u_{xx}(x,t)=u_{xx}(x_0,t_0)$. Similarly, we observe that

$\begin{aligned}u_{xt}^+(x_0,t_0)&=\lim_{s\to0^+}\frac{u_x(x_0,t_0+s)-u_x(x_0,t_0)}s\\&=\lim_{s\to0^+}\left[\frac{\alpha'(s)-\alpha'(0)}{-cs}+\frac{g'(ct_0+x_0+cs)-g'(ct_0+x_0)}{2s}+\frac{g'(cs)-g'(0)}{2s}+\frac{h(ct_0+x_0+cs)-h(ct_0+x_0)}{2cs}+\frac{h(cs)-h(0)}{2cs}\right]\\&=-\frac1c\alpha''(0)+\frac{c[g''(ct_0+x_0)+g''(0)]}2+\frac{h'(ct_0+x_0)+h'(0)}{2}\\&=\frac{c[g''(ct_0+x_0)-g''(0)]+h'(ct_0+x_0)+h'(0)}2\quad\text{(by $\alpha''(0)=c^2g''(0)$)},\\u_{xt}^-(x_0,t_0)&=\lim_{s\to0^-}\frac{u_x(x_0,t_0+s)-u_x(x_0,t_0)}s\\&=\lim_{s\to0^-}\left[\frac{g'(x_0+ct_0+cs)-g'(x_0+ct_0)}{2s}+\frac{g'(-cs)-g'(0)}{2s}+\frac{h(x_0+ct_0+cs)-h(x_0+ct_0)}{2cs}-\frac{h(-cs)-h(0)}{2cs}\right]\\&=\frac{c[g''(x_0+ct_0)-g''(0)]+h'(x_0+ct_0)+h'(0)}2.\end{aligned}$

Thus, $u_{xt}^+(x_0,t_0)=u_{xt}^-(x_0,t_0):=u_{xt}(x_0,t_0)$ and $\displaystyle\lim_{(x,t)\to(x_0,t_0)}u_{xt}(x,t)=u_{xt}(x_0,t_0)$. One may verify that $u_{tx}^+(x_0,t_0)=u_{tx}^-(x_0,t_0):=u_{tx}(x_0,t_0)$, $u_{tt}^+(x_0,t_0)=u_{tt}^-(x_0,t_0):=u_{tt}(x_0,t_0)$, $\displaystyle\lim_{(x,t)\to(x_0,t_0)}u_{tx}(x,t)=u_{tx}(x_0,t_0)$ and $\displaystyle\lim_{(x,t)\to(x_0,t_0)}u_{tt}(x,t)=u_{tt}(x_0,t_0)$. Therefore, $u\in C^2(\bar{\mathbb R}_+^2)$ and we complete the proof.

Solution 2
When $x\geq ct$, the domain of dependence of $u(x,t)$ is the interval $[x-ct,x+ct]$ so we have
$\displaystyle u(x,t)=\frac{u(x-ct,0)+u(x+ct,0)}2+\frac1{2c}\int_{x-ct}^{x+ct}\!h(s)\,\mathrm ds\quad\text{for}~x\geq ct\geq0.$
For $0\leq x<ct$, we can employ the parallelogram rule to obtain
$\begin{aligned}u(x,t)&=u\left(0,t-\frac1cx\right)+u\left(\frac{ct+x}2,\frac{ct+x}{2c}\right)-u\left(\frac{ct-x}2,\frac{ct-x}{2c}\right)\\&=\alpha\left(t-\frac xc\right)+\left[\frac{u(0,0)+u(ct+x,0)}2+\frac1{2c}\int_0^{ct+x}\!h(s)\,\mathrm ds\right]-\left[\frac{u(0,0)+u(ct-x,0)}2+\frac1{2c}\int_0^{ct-x}\!h(s)\,\mathrm ds\right]\\&=\alpha\left(t-\frac xc\right)+\frac{g(ct+x)-g(ct-x)}2+\frac1{2c}\int_{ct-x}^{ct+x}\!h(s)\,\mathrm ds.\end{aligned}$
For the case that $x_0=t_0=0$, we can only use the right partial derivatives $u_{xx}^+$, $u_{xt}^+$, $u_{tx}^+$ and $u_{tt}^+$ to show the continuity of $u_{xx}$, $u_{xt}$, $u_{tx}$ and $u_{tt}$ at the connor. Therefore, we can obtain the same solution as in Solution 1. Following the same argument in Solution 1, $u\in C^2(\bar{\mathbb R}_+^2)$.


6. Solve the initial/boundary value problem

   $\left\{\begin{aligned} &u_{tt}-u_{xx}=1\quad\text{for}~0<x<\pi~\text{and}~t>0,\\&u(x,0)=0,~~u_t(x,0)=0\quad\text{for}~0<x<\pi,\\&u(0,t)=0,~~u(\pi,t)=-\pi^2/2\quad\text{for}~t\geq0.\end{aligned}\right.$

   Describe the singularities (i.e., is $u$ $C^2$? If not, where does it fail? Is $u$ $C^1$? etc.).
Solution

Let $v(x,t)=u(x,t)+x^2/2$ for $x\in[0,\pi]$ and $t\in[0,\infty)$. Then $v$ satisfies

$\left\{\begin{aligned} &v_{tt}-v_{xx}=0\quad\text{for}~0<x<\pi~\text{and}~t>0,\\&v(x,0)=x^2/2,\quad v_t(x,0)=0\quad\text{for}~0<x<\pi,\\&v(0,t)=0,\quad v(\pi,t)=0\quad\text{for}~t\geq0.\end{aligned}\right.$

Using the separation of variables, we suppose that the solution $v(x,t)$ is of the form

$\displaystyle v(x,t)=\sum_{n=1}^\infty v_n(x,t)=\sum_{n=1}^\infty X_n(x)T_n(t)\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0.$

Due to the superposition principle, we firstly consider $u(x,t)=X(x)T(t)$ to fulfill the wave equation with the boundary condition, which gives

$X(x)T''(t)-X''(x)T(t)=0\quad\text{and}\quad X(0)T(t)=X(\pi)T(t)=0$.

Surely, we want to find a general solution so $T$ is not identically zero and $X(0)=X(\pi)=0$. From the differential equation, we have $X''(x)/X(x)=T''(t)/T(t)$ for $x\in(0,\pi)$ and $t>0$, which means there exists a constant $\lambda$ such that $X''(x)=\lambda X(x)$ and $T''(t)=\lambda T(t)$. To find nontrivial solution $X$, we divide three cases as follows.

• If $\lambda>0$, we may write $\lambda=\mu^2$ for some $\mu>0$. Then $X(x)=c_1e^{\mu x}+c_2e^{-\mu x}$. Then by $X(0)=X(\pi)=0$, we find $c_1=c_2=0$ and then $X\equiv0$, which is a trivial solution.
• If $\lambda=0$, we get $X(x)=c_1+c_2x$. By $X(0)=X(\pi)=0$, it is easy to see $c_1=c_2=0$ and hence $X\equiv0$, which is a trivial solution.
• If $\lambda<0$, we may write $\lambda=-\mu^2$ for some $\mu>0$. Then we have $X(x)=c_1\sin(\mu x)+c_2\cos(\mu x)$. By $X(0)=0$, we get $c_2=0$. From $X(\pi)=0$, we have $c_1\sin(\mu\pi)=0$. Since we want to seek the nontrivial solution, we hope $c_1\neq0$, which means $\sin(\mu\pi)$ should be zero and $\mu$ must be an integer.

Based on the above discussion, for $\lambda=-n^2$, we have $X_n(x)=\sin(nx)$. On the other hand, $T''(t)=-n^2T(t)$, we obtain $T(t)=c_n\sin(nt)+d_n\cos(nt)$. Thus, the solution $u$ has the following form:

$\displaystyle v(x,t)=\sum_{n=1}^\infty(c_n\sin(nt)+d_n\cos(nt))\sin(nx)\quad\text{for}~x\in[0,\pi)~\text{and}~t\geq0$.

Since $v_t(x,0)\equiv0$ on $(0,\pi)$, we have

$\displaystyle0=v_t(x,0)=\sum_{n=1}^\infty nc_n\sin(nx)$ for $x\in(0,\pi)$,

which gives $c_n=0$ for $n\in\mathbb N$. On the other hand, from $v(x,0)=x^2/2$ on $[0,\pi)$, we have

$\displaystyle\frac{x^2}2=v(x,0)=\sum_{n=1}^\infty d_n\sin(nx)\quad\text{for}~x\in[0,\pi)$.

Multiplying it by $\sin(mx)$ and then integrating it over $[0,\pi]$, we obtain

$\begin{aligned}\frac{(-1)^m(2-m^2\pi^2)-2}{2m^3}&=(-1)^{m+1}\frac{\pi^2}{2m}+\frac{(-1)^m-1}{m^3}\\&=(-1)^{m+1}\frac{\pi^2}{2m}+\left.\frac{x\sin(mx)}{m^2}\right|_0^\pi-\frac1{m^2}\int_0^\pi\!\sin(mx)\,\mathrm dx\\&=\left.-\frac{x^2\cos(mx)}{2m}\right|_0^\pi+\frac1m\int_0^\pi\!x\cos(mx)\,\mathrm dx\\&=\int_0^\pi\!\frac{x^2}2\cdot\sin(mx)\,\mathrm dx=\int_0^\pi\!\sum_{n=1}^\infty d_n\sin(nx)\sin(mx)\,\mathrm dx\\&=d_m\int_0^\pi\!\sin^2(mx)\,\mathrm dx=\frac{d_m}2\int_0^\pi\!(1-\cos(2mx))\,\mathrm dx=\frac\pi2d_m.\end{aligned}$

This means $\displaystyle d_n=\frac{(-1)^n(2-n^2\pi^2)-2}{n^3\pi}$ for $n\in\mathbb N$. Therefore, we obtain

$\displaystyle v(x,t)=\frac1\pi\sum_{n=1}^\infty\frac{(-1)^n(2-n^2\pi^2)-2}{n^3}\cos(nt)\sin(nx)\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0~\text{except}~(x,t)=(\pi,0)$,

which gives the solution

$\displaystyle u(x,t)=-\frac{x^2}2+\frac1\pi\sum_{n=1}^\infty\frac{(-1)^n(2-n^2\pi^2)-2}{n^3}\cos(nt)\sin(nx)\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0~\text{except}~(x,t)=(\pi,0)$.

The regularity of $u$ is equivalent to that of $v$ because $x^2/2$ is analytic. Thus, it suffices to investigate the value of $v$ in the various regions bounded by characteristics. To sum the Fourier series of $v$ in regions bounded by characteristics, we define

• $R_0=\{(x,t)\in\mathbb R^2\,:\,t\leq x\leq\pi-t,\,0\leq t\leq\pi/2\}$.
• $R_{3k-2}=\{(x,t)\in\mathbb R^2\,:\,0\leq x\leq\pi/2,\,(k-1)\pi+x\leq t\leq k\pi-x\}$ for $k\in\mathbb N$.
• $R_{3k-1}=\{(x,t)\in\mathbb R^2\,:\,\pi/2\leq x\leq\pi,\,k\pi-x\leq t\leq(k-1)\pi+x\}$ for $k\in\mathbb N$.
• $R_{3k}=\{(x,t)\in\mathbb R^2\,:\,(k-1)\pi\leq t-x\leq k\pi,\,k\pi\leq t+x\leq(k+1)\pi\}$ for $k\in\mathbb N$.

Using the product to sum formula, we have

$\displaystyle v(x,t)=\frac1{2\pi}\sum_{n=1}^\infty\frac{(-1)^n(2-n^2\pi^2)-2}{n^3}[\sin(n(x+t))+\sin(n(x-t))]\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0~\text{except}~(x,t)=(\pi,0)$.

From $v(x,0)=x^2/2$ on $[0,\pi)$, we note that $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n(2-n^2\pi^2)-2}{n^3}\sin(nx)=\frac{\pi x^2}2$ for $x\in[0,\pi)$.

In region $R_0$ ($0<x+t<\pi$ and $0<x-t<\pi$), we have

$\displaystyle v(x,t)=\frac1{2\pi}\left[\frac{\pi(x+t)^2}2+\frac{\pi(x-t)^2}2\right]=\frac{x^2+t^2}2.$

In region $R_{3k-2}$ ($0<x+t-(k-1)\pi<\pi$ and $0<x-t+k\pi<\pi$), we have

$\begin{aligned}v(x,t)&=\frac1{2\pi}\sum_{n=1}^\infty\frac{(-1)^n(2-n^2\pi^2)-2}{n^3}[\sin(n(x+t-(k-1)\pi))+\sin(n(x-t+k\pi))]\\&=\frac1{2\pi}\left[\frac{\pi(x+t-(k-1)\pi)^2}2+\frac{\pi(x-t+k\pi)^2}2\right]\\&=\frac{(x+t-(k-1)\pi)^2+(x-t+k\pi)^2}4.\end{aligned}$

In region $R_{3k-1}$ ($0<x+t-k\pi<\pi$ and $0<x-t+(k-1)\pi<\pi$), we have

$\begin{aligned}v(x,t)&=\frac1{2\pi}\sum_{n=1}^\infty\frac{(-1)^n(2-n^2\pi^2)-2}{n^3}[\sin(n(x+t-k\pi))+\sin(n(x-t+(k-1)\pi))]\\&=\frac1{2\pi}\left[\frac{\pi(x+t-k\pi)^2}2+\frac{\pi(x-t+(k-1)\pi)^2}2\right]\\&=\frac{(x+t-k\pi)^2+(x-t+(k-1)\pi)^2}4.\end{aligned}$

In region $R_{3k}$ ($0<x+t-k\pi<\pi$ and $0<x-t+k\pi<\pi$), we have

$\begin{aligned}v(x,t)&=\frac1{2\pi}\sum_{n=1}^\infty\frac{(-1)^n(2-n^2\pi^2)-2}{n^3}[\sin(n(x+t-k\pi))+\sin(n(x-t+k\pi))]\\&=\frac1{2\pi}\left[\frac{\pi(x+t-k\pi)^2}2+\frac{\pi(x-t+k\pi)^2}2\right]\\&=\frac{x^2+(t-k\pi)^2}2.\end{aligned}$

It is easy to verify that $v$ is discontinuous along all characteristics, which means

$v\in C^\infty\left((0,\pi)\times[0,\infty)-\bigcup_{k=0}^\infty\{(x,t)\in\mathbb R^2\,:\,t\pm x=k\pi\}\right)$.



7. 1. Use Fourier series to solve the initial/boundary value problem for the Klein-Gordon equation

      $\left\{\begin{aligned} &u_{tt}-c^2u_{xx}+m^2u=0\quad\text{for}~0<x<\pi~\text{and}~t>0,\\&u(x,0)=g(x),~~u_t(x,0)=h(x)\quad\text{for}~0<x<\pi,\\&u(0,t)=0=u(\pi,t)\quad\text{for}~t>0.\end{aligned}\right.$

      Notice that the solution $u(x,t)$ is bounded as $t\to\infty$.
   2. Do the same for the equation $u_{tt}-c^2u_{xx}-m^2u=0$. If $c^2<m^2$, show that the solution could be unbounded as $t\to\infty$.
Solution
1. Using the separation of variables, we suppose that the solution $u(x,t)$ is of the form

   $\displaystyle u(x,t)=\sum_{n=1}^\infty u_n(x,t)=\sum_{n=1}^\infty X_n(x)T_n(x)\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0.$

   Due to the superposition principle, we firstly consider $u(x,t)=X(x)T(t)$ to fulfill the Klein-Gordon equation with the boundary condition, which gives

   $X(x)T''(t)-c^2X''(x)T(t)+m^2X(x)T(t)=0\quad\text{and}\quad X(0)T(t)=X(\pi)T(t)=0.$

   Surely, we want to find a general solution so $T$ is not identically zero and $X(0)=X(\pi)=0$. From the differential equation, we have

   $\displaystyle\frac{X''(x)}{X(x)}=\frac1{c^2}\frac{T''(t)}{T(t)}+\frac{m^2}{c^2}\quad\text{for}~x\in(0,\pi)~\text{and}~t\in(0,\infty)$,

   which means there exists a constant $\lambda$ such that $X''(x)=\lambda X(x)$ and $T''(t)+(m^2-\lambda c^2)T(t)=0$. To find nontrivial solution $X$, we divide three cases as follows.
   • If $\lambda>0$, we may write $\lambda=\mu^2$ for some $\mu>0$. Then $X(x)=c_1e^{\mu x}+c_2e^{-\mu x}$. Then by $X(0)=X(\pi)=0$, we find $c_1=c_2=0$ and then $X\equiv0$, which is a trivial solution.
   • If $\lambda=0$, we get $X(x)=c_1+c_2x$. By $X(0)=X(\pi)=0$, it is easy to see $c_1=c_2=0$ and hence $X\equiv0$, which is a trivial solution.
   • If $\lambda<0$, we may write $\lambda=-\mu^2$ for some $\mu>0$. Then we have $X(x)=c_1\sin(\mu x)+c_2\cos(\mu x)$. By $X(0)=0$, we get $c_2=0$. From $X(\pi)=0$, we have $c_1\sin(\mu\pi)=0$. Since we want to seek the nontrivial solution, we hope $c_1\neq0$, which means $\sin(\mu\pi)$ should be zero and $\mu$ must be an integer.
   Based on the above discussion, for $\lambda=-n^2$, we have $X_n(x)=\sin(nx)$. On the other hand, we have $T''(t)+(m^2+c^2n^2)T(t)=0$, which gives

   $T(t)=c_n\cos(t\sqrt{m^2+c^2n^2})+d_n\sin(t\sqrt{m^2+c^2n^2})\quad\text{for}~n\in\mathbb N$.

   Thus, the solution $u$ has the following form:

   $\displaystyle u(x,t)=\sum_{n=1}^\infty[c_n\cos(t\sqrt{m^2+c^2n^2})+d_n\sin(t\sqrt{m^2+c^2n^2})]\sin(nx)\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0.$

   From $u(x,0)=g(x)$ and $u_t(x,0)=h(x)$ for $x\in(0,\pi)$, we have

   $\left\{\begin{aligned} &g(x)=u(x,0)=\sum_{n=1}^\infty c_n\sin(nx),\\&h(x)=\sum_{n=1}^\infty d_n\sqrt{m^2+c^2n^2}\sin(nx).\end{aligned}\right.$

   Multiplying them by $\sin(kx)$ and then integrating them over $[0,\pi]$, we can obtain

   $\displaystyle c_n=\frac2\pi\int_0^\pi\!g(x)\sin(nx)\,\mathrm dx,\quad d_n=\frac2{\pi\sqrt{m^2+c^2n^2}}\int_0^\pi\!h(x)\sin(nx)\,\mathrm dx\quad\text{for}~n\in\mathbb N$.

   Therefore, the solution is

   $\displaystyle u(x,t)=\frac2\pi\sum_{n=1}^\infty\left[\cos(t\sqrt{m^2+c^2n^2})\int_0^\pi\!g(x)\sin(nx)\,\mathrm dx+\frac{\sin(t\sqrt{m^2+c^2n^2})}{\sqrt{m^2+c^2n^2}}\int_0^\pi\!h(x)\sin(nx)\,\mathrm dx\right]\sin(nx)$

   for $x\in(0,\pi)$ and $t>0$. Due to he Parseval's Theorem, we can estimate the sum of squares of $c_n$ and $d_n$, i.e., $\displaystyle\sum_{n=1}^\infty c_n^2$ and $\displaystyle\sum_{n=1}^\infty d_n^2$ is bounded, provided that $g$ and $h$ are $L^2$-functions. This implies $u$ is bounded as $t\to\infty$.
2. Using the separation of variables, we suppose that the solution $u(x,t)$ is of the form

   $\displaystyle u(x,t)=\sum_{n=1}^\infty u_n(x,t)=\sum_{n=1}^\infty X_n(x)T_n(x)\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0.$

   Due to the superposition principle, we firstly consider $u(x,t)=X(x)T(t)$ to fulfill the Klein-Gordon equation with the boundary condition, which gives

   $X(x)T''(t)-c^2X''(x)T(t)-m^2X(x)T(t)=0\quad\text{and}\quad X(0)T(t)=X(\pi)T(t)=0.$

   Surely, we want to find a general solution so $T$ is not identically zero and $X(0)=X(\pi)=0$. From the differential equation, we have

   $\displaystyle\frac{X''(x)}{X(x)}=\frac1{c^2}\frac{T''(t)}{T(t)}-\frac{m^2}{c^2}\quad\text{for}~x\in(0,\pi)~\text{and}~t\in(0,\infty)$,

   which means there exists a constant $\lambda$ such that $X''(x)=\lambda X(x)$ and $T''(t)-(m^2+\lambda c^2)T(t)=0$. To find nontrivial solution $X$, we divide three cases as follows.
   • If $\lambda>0$, we may write $\lambda=\mu^2$ for some $\mu>0$. Then $X(x)=c_1e^{\mu x}+c_2e^{-\mu x}$. Then by $X(0)=X(\pi)=0$, we find $c_1=c_2=0$ and then $X\equiv0$, which is a trivial solution.
   • If $\lambda=0$, we get $X(x)=c_1+c_2x$. By $X(0)=X(\pi)=0$, it is easy to see $c_1=c_2=0$ and hence $X\equiv0$, which is a trivial solution.
   • If $\lambda<0$, we may write $\lambda=-\mu^2$ for some $\mu>0$. Then we have $X(x)=c_1\sin(\mu x)+c_2\cos(\mu x)$. By $X(0)=0$, we get $c_2=0$. From $X(\pi)=0$, we have $c_1\sin(\mu\pi)=0$. Since we want to seek the nontrivial solution, we hope $c_1\neq0$, which means $\sin(\mu\pi)$ should be zero and $\mu$ must be an integer.
   Based on the above discussion, for $\lambda=-n^2$, we have $X_n(x)=\sin(nx)$. On the other hand, we have $T''(t)-(m^2-c^2n^2)T(t)=0$, which gives

   $T_n(t)=\begin{cases}c_n\cos(t\sqrt{m^2+c^2n^2})+d_n\sin(t\sqrt{m^2+c^2n^2})&\text{if}~n>m/c;\\c_n+d_nt&\text{if}~n=m/c;\\c_n\exp(t\sqrt{m^2-c^2n^2})+d_n\exp(-t\sqrt{m^2-c^2n^2})&\text{if}~1\leq n<m/c.\end{cases}$

   Thus, the solution $u$ has the following form:

   $\displaystyle u(x,t)=\sum_{n=1}^\infty T_n\sin(nx)\quad\text{for}~x\in[0,\pi]~\text{and}~t\geq0.$

   We can determine all coefficients of $c_n$ and $d_n$ as for part (a). When $c<m$, we may observe that $T_1$ includes the exponential function and becomes unbounded as $t$ tends to infinity.


8. Derive a formula similar to (6) for the pure initial value problem for the Klein-Gordon equation $u_{tt}-c^2u_{xx}+m^2u=0$.
Solution 1

This proof follows from the paper: "Moshinsky's shutter problem: an initial-value problem for the Klein-Gordon equation" (Author: P. A. Martin and F. V. Kowalski). Using the Fourier transform with repsect to $x$, we define

$\displaystyle U(\xi,t)=\int_{-\infty}^\infty\!e^{-i\xi x}u(x,t)\,\mathrm dx$.

Then $U$ satisfies

$\begin{aligned}U_{tt}(\xi,t)&=\int_{-\infty}^{\infty}\!e^{-i\xi x}u_{tt}(x,t)\,\mathrm dx=\int_{-\infty}^{\infty}\!e^{-i\xi x}[c^2u_{xx}(x,t)-m^2u(x,t)]\,\mathrm dx\\&=-\xi^2c^2U(\xi,t)-m^2U(\xi,t)=-(\xi^2c^2+m^2)U(\xi,t).\end{aligned}$

The initial condition $u(x,0)=f(x)$ and $u_t(x,0)=g(x)$ for $x\in\mathbb R$ can be transformed into

$\displaystyle U(\xi,0)=F(\xi):=\int_{-\infty}^{\infty}\!e^{-i\xi x}f(x)\,\mathrm dx,\quad U_t(\xi,0)=G(\xi):=\int_{-\infty}^{\infty}\!e^{-i\xi x}g(x)\,\mathrm dx$.

We can treat the differential equation of $U$ as ordinary differential equation with respect to $t$ and obtain

$\displaystyle U(\xi,t)=F(\xi)\cos\left(t\sqrt{\xi^2c^2+m^2}\right)+\frac{G(\xi)}{\sqrt{\xi^2c^2+m^2}}\sin\left(t\sqrt{\xi^2c^2+m^2}\right)$.

By the inverse Fourier transform, we have

$\displaystyle u(x,t)=\frac1{2\pi}\int_{-\infty}^{\infty}\!e^{i\xi x}F(\xi)\cos\left(t\sqrt{\xi^2c^2+m^2}\right)\,\mathrm d\xi+\frac1{2\pi}\int_{-\infty}^{\infty}\!e^{i\xi x}\frac{G(\xi)}{\sqrt{\xi^2c^2+m^2}}\sin\left(t\sqrt{\xi^2c^2+m^2}\right)\,\mathrm d\xi$.

By (20)-(22) in the reference paper, we have

$\displaystyle\frac1{2\pi}\int_{-\infty}^{\infty}\!e^{i\xi x}\frac{G(\xi)}{\sqrt{\xi^2c^2+m^2}}\sin(t\sqrt{\xi^2c^2+m^2})\,\mathrm d\xi=\frac1{2c}\int_{-ct}^{ct}\!g(x-y)J_0\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy$,

where $J_0$ is the Bessel function of first kind of order $0$. Moreover, we observe that

$\begin{aligned}\frac1{2\pi}\int_{-\infty}^{\infty}\!e^{i\xi x}F(\xi)\cos\left(t\sqrt{\xi^2c^2+m^2}\right)\,\mathrm d\xi&=\frac{\partial}{\partial t}\left[\frac1{2\pi}\int_{-\infty}^{\infty}\!e^{i\xi x}\frac{F(\xi)}{\sqrt{\xi^2c^2+m^2}}\sin\left(t\sqrt{\xi^2c^2+m^2}\right)\,\mathrm d\xi\right]\\&=\frac\partial{\partial t}\left[\frac1{2c}\int_{-ct}^{ct}\!f(x-y)J_0\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy\right]\\&=\frac{f(x-ct)+f(x+ct)}2+\frac{mt}2\int_{-ct}^{ct}\frac{f(x-y)}{\sqrt{c^2t^2-y^2}}J_0'\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy\\&=\frac{f(x-ct)+f(x+ct)}2-\frac{mt}2\int_{-ct}^{ct}\frac{f(x-y)}{\sqrt{c^2t^2-y^2}}J_1\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy.\end{aligned}$

Here we have used the fact that $J_0'(x)=-J_1(x)$ and $J_1$ is the Bessel function of first kind of order $1$. Therefore, the solution $u$ is given by

$\displaystyle u(x,t)=\frac{f(x-ct)+f(x+ct)}2-\frac{mt}2\int_{-ct}^{ct}\frac{f(x-y)}{\sqrt{c^2t^2-y^2}}J_1\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy+\frac1{2c}\int_{-ct}^{ct}g(x-y)J_0\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy$.

Solution 2

This proof follows from the paper: "Derivation of Solutions of the Klein-Gordon Equation from Solutions of the Wave Equation" (Author: LL. G. Chambers). Due to the linearity of Klein-Gordon equation, we can consider the following equations

$\left\{\begin{aligned} &v_{tt}-c^2v_{xx}+m^2v=0\quad\text{for}~(x,t)\in\mathbb R\times\mathbb R_+,\\&v(x,0)=f(x),\quad v_t(x,0)=0\quad\text{for}~x\in\mathbb R,\end{aligned}\right.\quad\text{and}\quad\left\{\begin{aligned} &w_{tt}-c^2w_{xx}+m^2w=0\quad\text{for}~(x,t)\in\mathbb R\times\mathbb R_+,\\&w(x,0)=0,\quad w_t(x,0)=g(x)\quad\text{for}~x\in\mathbb R.\end{aligned}\right.$

Using the Laplace transform over time, we set

$\displaystyle V(x,s)=\int_0^\infty\!e^{-st}v(x,t)\,\mathrm dt$.

The Laplace transforms of $v_{xx}$ and $v_{tt}$ are

$\begin{aligned}\int_0^\infty\!e^{-st}v_{xx}(x,t)\,\mathrm dt&=V_{xx}(x,s),\\\int_0^\infty\!e^{-st}v_{tt}(x,t)\,\mathrm dt&=e^{-st}v_t(x,t)\Big|_{t=0}^{t=\infty}+s\int_0^\infty\!e^{-st}v_t(x,t)\,\mathrm dt\\&=s\left[v(x,t)e^{-st}\Big|_{t=0}^{t=\infty}+s\int_0^\infty\!e^{-st}v(x,t)\,\mathrm dt\right]\\&=-sf(x)+s^2V(x,s).\end{aligned}$

Here we have used the initial condition $v(x,0)=f(x)$ and $v_t(x,0)=0$ for $x\in\mathbb R$. Then we have

$[s^2V(x,s)-sf(x)]-c^2V_{xx}(x,s)+m^2V(x,s)=0\quad\text{for}~x\in\mathbb R~\text{and}~s>0,$

which means

$(s^2+m^2)V(x,s)-c^2V_{xx}(x,s)=sf(x)\quad\text{for}~x\in\mathbb R~\text{and}~s>0.$

Let $\tilde v$ be the solution of

$\left\{\begin{aligned} &\tilde v_{tt}-c^2\tilde v_{xx}=0\quad\text{for}~x\in\mathbb R~\text{and}~t>0,\\&\tilde v(x,0)=f(x),\quad\tilde v_t(x,0)=0\quad\text{for}~x\in\mathbb R.\end{aligned}\right.$

Set $\displaystyle\tilde V(x,s)=\int_0^\infty\!e^{-st}\tilde v(x,t)\,\mathrm dt$. Then the wave equation with initial condition can be transformed into

$[s^2\tilde V(x,s)-sf(x)]-c^2\tilde V_{xx}(x,s)=0\quad\text{for}~x\in\mathbb R~\text{and}~s>0$.

Replacing $s$ with $\sqrt{s^2+m^2}$, we get

$(s^2+m^2)\tilde V(x,\sqrt{s^2+m^2})-c^2\tilde V_{xx}(x,\sqrt{s^2+m^2})=\sqrt{s^2+m^2}f(x)\quad\text{for}~x\in\mathbb R~\text{and}~s>0$.

Moreover, we may observe that

$\displaystyle(s^2+m^2)\left[\frac{s\tilde V(x,\sqrt{s^2+m^2})}{\sqrt{s^2+m^2}}\right]-c^2\left[\frac{s\tilde V(x,\sqrt{s^2+m^2})}{\sqrt{s^2+m^2}}\right]_{xx}=sf(x)$.

Thus, we arrive at

$\displaystyle V(x,s)=\frac{s}{\sqrt{s^2+m^2}}\tilde V(x,\sqrt{s^2+m^2})\quad\text{for}~x\in\mathbb R~\text{and}~s>0$.

To apply the inverse Laplace transform to $V$ with respect to $s$, we need to know the inverse Laplace transform of $\alpha(s):=s/\sqrt{s^2+m^2}$ and $\beta(s):=\tilde V(x,\sqrt{s^2+m^2})$. Suppose that the inverse Laplace transform of $\alpha(s)$ and $\beta(x,s)$ can be denoted by $A(t)$ and $B(x,t)$, respectively. Then by conventional theorem, we have

$\displaystyle v(x,t)=\int_0^t\!A(\tau)B(x,t-\tau)\,\mathrm d\tau$.

By the formula of inverse Laplace transform, one may get $$\begin{equation}\label{1}B(x,t)=\tilde v(x,t)-m\int_0^t\!\tilde v(x,\sqrt{t^2-\xi^2})J_1(m\xi)\,\mathrm d\xi,\end{equation}$$ where $J_1$ is the Bessel function of the first kind of order $1$. This formula for $m=1$ can be found at page 123 in the German book (Formeln und Sätze für die Speziellen Funktionen der Mathematischen Physik). However it seems extermely difficult to derive it directly. Here we may simplify the expression of $B$ by using $\tilde v(x,t)=\frac{f(x-ct)+f(x+ct)}2$ and the substitution for $y=c\sqrt{t^2-\xi^2}$:

$\begin{aligned}B(x,t)&=\frac{f(x-ct)+f(x+ct)}2-mc\int_0^{ct}\frac{y[f(x-y)+f(x+y)]}{2\sqrt{c^2t^2-y^2}}J_1\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy\\&=\frac{f(x-ct)+f(x+ct)}2-\frac{mc}2\int_{-ct}^{ct}\frac{yf(x-y)}{\sqrt{c^2t^2-y^2}}J_1\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy.\end{aligned}$

On the other hand, we can follow the derivation from paper or verify it to get

$A(t)=mJ_0'(mt)+\delta(t)=-mJ_1(mt)+\delta(t)$,

where $J_0$ is the Bessel function of first kind of order $0$ and $\delta$ is the Dirac delta functional. Thus, we obtain

$\begin{aligned}v(x,t)&=B(x,t)-m\int_0^tJ_1(m\tau)B(x,t-\tau)\,\mathrm d\tau.\end{aligned}$

It may become

$\displaystyle v(x,t)=\frac{f(x-ct)+f(x+ct)}2-\frac{mt}2\int_{-ct}^{ct}\frac{f(x-y)}{\sqrt{c^2t^2-y^2}}J_1\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy.$

To solve $w$, we use the Laplace transform over time again to set

$\displaystyle W(x,s)=\int_0^\infty\!e^{-st}w(x,t)\,\mathrm dt$.

The Laplace transforms of $w_{xx}$ and $w_{tt}$ are

$\begin{aligned}\int_0^\infty\!e^{-st}w_{xx}(x,t)\,\mathrm dt&=W_{xx}(x,s),\\\int_0^\infty\!e^{-st}w_{tt}(x,t)\,\mathrm dt&=e^{-st}w_t(x,t)\Big|_{t=0}^{t=\infty}+s\int_0^\infty\!e^{-st}w_t(x,t)\,\mathrm dt\\&=-g(x)+s\left[e^{-st}w(x,t)\Big|_{t=0}^{t=\infty}+s\int_0^\infty\!e^{-st}w(x,t)\,\mathrm dt\right]\\&=-g(x)+s^2W(x,s).\end{aligned}$

Here we have used the initial condition $w(x,0)=0$ and $w_t(x,0)=g(x)$ for $x\in\mathbb R$. Then we have

$(s^2+m^2)W(x,s)-c^2W_{xx}(x,s)=g(x)\quad\text{for}~x\in\mathbb R~\text{and}~s>0$.

Let $\tilde w$ be the solution of

$\tilde w_{tt}-c^2\tilde w_{xx}=0\quad\text{for}~x\in\mathbb R~\text{and}~t>0,\\\tilde w(x,0)=0,\quad\tilde w_t(x,0)=g(x)\quad\text{for}~x\in\mathbb R$.

Set $\displaystyle\tilde W(x,s)=\int_0^\infty\!e^{-st}\tilde w(x,t)\,\mathrm dt$. Then the wave equation with initial condition can be transformed into

$s^2\tilde W(x,s)-c^2\tilde W_{xx}(x,s)=g(x)\quad\text{for}~x\in\mathbb R~\text{and}~s>0$.

Replacing $s$ with $\sqrt{s^2+m^2}$, we may observe that

$W(x,s)=\tilde W(x,\sqrt{s^2+m^2})\quad\text{for}~x\in\mathbb R~\text{and}~s>0$.

Using the same formula as $\eqref{1}$, we have

$\begin{aligned}w(x,t)&=\tilde w(x,t)-m\int_0^t\!\tilde w\left(x,\sqrt{t^2-\xi^2}\right)J_1(m\xi)\,\mathrm d\xi\\&=\tilde w(x,t)+m\int_0^t\tilde w\left(x,\sqrt{t^2-\xi^2}\right)J_0'(m\xi)\,\mathrm d\xi\\&=\tilde w(x,t)+\left.J_0(m\xi)\tilde w\left(x,\sqrt{t^2-\xi^2}\right)\right|_{\xi=0}^{\xi=t}+\int_0^t\!\frac{\xi\tilde w_t\left(x,\sqrt{t^2-\xi^2}\right)}{\sqrt{t^2-\xi^2}}J_0(m\xi)\,\mathrm d\xi\\&=\int_0^t\!\frac{\xi\tilde w_t\left(x,\sqrt{t^2-\xi^2}\right)}{\sqrt{t^2-\xi^2}}J_0(m\xi)\,\mathrm d\xi\quad\text{(by $J_0(0)=1$ and $\tilde w(x,0)=0$)}\\&=-\frac1c\int_{ct}^0\!\tilde w_t(x,y/c)J_0\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy\quad\text{(by $y=c\sqrt{t^2-\xi^2}$)}\\&=\frac1{2c}\int_0^{ct}\![g(x+y)+g(x-y)]J_0\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy\quad\text{(by $\displaystyle w_t(x,t)=\frac{g(x+ct)+g(x-ct)}2$)}\\&=\frac1{2c}\int_{-ct}^{ct}\!g(x-y)J_0\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy.\end{aligned}$

Here the last equality is used the fact that

$\displaystyle\int_0^{ct}\!g(x+y)J_0\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy=\int_{-ct}^0\!g(x-y)J_0\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy.$

Finally, we obtain the solution $u$ is

$\displaystyle u(x,t)=B(x,t)+\tilde w(x,t)-m\int_0^tJ_1(m\tau)B(x,t-\tau)\,\mathrm d\tau-m\int_0^t\!\tilde w(x,\sqrt{t^2-\xi^2})J_1(m\xi)\,\mathrm d\xi.$

It could become

$\displaystyle u(x,t)=\frac{f(x-ct)+f(x+ct)}2-\frac{mt}2\int_{-ct}^{ct}\frac{f(x-y)}{\sqrt{c^2t^2-y^2}}J_1\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy+\frac1{2c}\int_{-ct}^{ct}g(x-y)J_0\left(m\sqrt{t^2-\frac{y^2}{c^2}}\right)\,\mathrm dy$.

Solution 3

The idea comes from the Exercise 5 of Section 3.2. Let $v(x,y,t)=\cos(my/c)u(x,t)$ for $(x,y,t)\in\mathbb R^2\times\bar{\mathbb R}_+$. Then it is easy to find that

$\begin{aligned}v_{tt}(x,y,t)&=\cos(my/c)u_{tt}(x,t)=\cos(my/c)[c^2u_{xx}(x,t)-m^2u(x,t)]\\&=c^2v_{xx}(x,y,t)+c^2v_{yy}(x,y).\end{aligned}$

This means $v$ satisfies the two-dimensional wave equation with the initial condition

$v(x,y,0)=\cos(my/c)u(x,0)=f(x)\cos(my/c),\quad v_t(x,y,0)=\cos(my/c)u_t(x,0)=g(x)\cos(my/c)\quad\text{for}~(x,y)\in\mathbb R^2.$

Then by the formula (39) in Section 3.2, we obtain

$\begin{aligned}v(x,y,t)&=\frac1{2\pi}\frac\partial{\partial t}\left(t\int_{\xi_1^2+\xi_2^2<1}\frac{f(x+ct\xi_1)\cos(m(y+ct\xi_2)/c)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\right)\\&\quad+\frac t{2\pi}\int_{\xi_1^2+\xi_2^2<1}\frac{g(x+ct\xi_1)\cos(m(y+ct\xi_2)/c)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2.\end{aligned}$

Hence the solution $u$ is

$\begin{aligned}u(x,t)&=v(x,0,t)\\&=\frac1{2\pi}\frac\partial{\partial t}\left(t\int_{\xi_1^2+\xi_2^2<1}\frac{f(x+ct\xi_1)\cos(mt\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\right)\\&\quad+\frac t{2\pi}\int_{\xi_1^2+\xi_2^2<1}\frac{g(x+ct\xi_1)\cos(mt\xi_2)}{\sqrt{1-\xi_1^2-\xi_2^2}}\,\mathrm d\xi_1\,\mathrm d\xi_2\\&=\frac1{2\pi}\frac\partial{\partial t}\left(t\int_0^{2\pi}\!\int_0^1\!\frac{r\cos\left(mtr\sin\theta\right)f(x+ctr\cos\theta)}{\sqrt{1-r^2}}\,\mathrm dr\,\mathrm d\theta\right)\\&\quad+\frac t{2\pi}\int_0^{2\pi}\!\int_0^1\!\frac{r\cos(mtr\sin\theta)g(x+ctr\cos\theta)}{\sqrt{1-r^2}}\,\mathrm dr\,\mathrm d\theta.\end{aligned}$