Method of reflections for the wave equation.

Question

For the Wave Equation:

$ \left\{\begin{matrix} u_{tt}-c^2u_{xx}=0 & x>0, t>0\\ u(x,0)=f(x) & x>0 & \hspace{0.5cm} (1) \\ u_{t}(x,0)=g(x) & x>0 & \hspace{0.5cm} (1) \\ u(0,t)=h(t) & t>0 \\ \end{matrix}\right. $

The general solution is:

$ u(x,t) = \frac{f(x+ct)-f(ct-x)}{2}+ \frac{1}{2c}\int_{x+ct}^{ct-x}g(s)ds + h(t-\frac{x}{c}) $

My notes explain how to obtain it. But then they say that when $h\equiv0$, we can use the method of reflections to solve the problem for $ x\in{\mathbb{R}} $ with the restrictions (1).

• Can someone post an example using method of reflections? (*)
• Also I don't get why we need to solve it for $x\in{\mathbb{R}}$... Isn't the problem defined for $x>0$ ?

(*) I have been searching for it but I found nothing. Maybe it is a problem with the name, I have translated the name method of reflections from spanish método de reflexiones.

Answer 1

• Here is an example of an application of the method of reflections to a PDE problem:

Suppose we have the following half-line interval problem \begin{align*} \left\{\begin{matrix} v_t-kv_{xx}, & ~\text{in}~0<x<\infty,~0<t<\infty, \\ v(x,0)=\phi(x), & ~\text{for}~t=0, \\ v(0,t)=0, & ~\text{for}~x=0, \end{matrix}\right. \end{align*} with Dirichlet boundary conditions.

Introduce an odd extension \begin{align*} \left\{\begin{matrix} \phi(x) & \text{for}~x>0, \\ -\phi(-x) & \text{for}~x<0, \\ 0 & \text{for}~x=0, \end{matrix}\right. \end{align*} to the whole line that is unique. Suppose $u(x,t)$ is a solution to the problem \begin{equation*} u_t-ku_{xx}=0,~u(x,0)=\phi_{\text{odd}}(x),~x\in (-\infty,\infty),~t\in (0,\infty) \end{equation*} of the form \begin{equation*} u(x,t)=\int^{\infty}_{-\infty}S(x-y,t)\phi_{\text{odd}}(y)dy. \end{equation*} Write the above as \begin{equation*} u(x,t)=\int^{\infty}_{0}S(x-y,t)\phi(y)dy-\int^{0}_{-\infty}S(x-y,t)\phi(-y)dy. \end{equation*} Doing a change of variables (i.e. transforming $y\rightarrow -y$ in the second integral ) \begin{equation*} u(x,t)=\int^{\infty}_{0}[S(x-y,t)-S(x+y,t)]\phi(y)dy \end{equation*} gives the solution \begin{equation*} v(x,t)=\frac{1}{\sqrt{4\pi kt}}\int^{\infty}_{0}[e^{-(x-y)^2/4kt}-e^{-(x+y)^2/4kt}]\phi(y)dy \end{equation*} where we have used \begin{equation*} u(x,t)=\frac{1}{\sqrt{4\pi kt}}\int^{\infty}_{-\infty}e^{-(x-y)^2/4kt}\phi(y)dy. \end{equation*}

• With the second point you raised, the solution we are searching for means that we can find a solution for any choice of $x$ or $t$ (as long as it is valid for the given conditions). In the question, we have results for $u$ for any $x$ as long as $t=0$. What about $t\neq 0$?

See "PDEs: An Introduction" by Walter Strauss.