Toronto Math Forum

\begin{equation}
u_{tt}-\frac{1}{2}u_{xx}=0
\end{equation}
\begin{equation}
u(x,t)=\phi(x+\frac{1}{2}t)+\psi(x-\frac{1}{2}t)
\end{equation}
by(2) and (3)
\begin{equation}
\phi(x)+\psi(x)=f(x)=2xe^{-2x^2}\\
\phi(x)-\psi(x)=g(x)=0
\end{equation}
Then
\begin{equation}
\phi(x)=xe^{-2x^2}\\
\psi(x)=xe^{-2x^2}
\end{equation}
Thus for $x>\frac{1}{2}t$
\begin{equation}
\mu(x,t)=(x+\frac{1}{2}t)e^{-2(x+\frac{1}{2}t)^2}+(x-\frac{1}{2}t)e^{-2(x-\frac{1}{2}t)^2}
\end{equation}
Now consider  $x<\frac{1}{2}t$
by (4)
\begin{equation}
\phi(\frac{1}{2}t)+\psi(-\frac{1}{2}t)=te^{-t^2/2}
\end{equation}
let $x=-\frac{1}{2}t$, then $t=-2x$
\begin{equation}
\phi(-x)+\psi(x)=-2xe^{-(-2x)^2/2}\\
\psi(x)=-2xe^{-2x^2}-\phi(-x)\\
\psi(x)=-2xe^{-2x^2}+xe^{-2x^2}\\
\psi(x)=-xe^{-2x^2}
\end{equation}
Thus for $x<\frac{1}{2}t$
\begin{equation}
\mu(x,t)=(x+\frac{1}{2}t)e^{-2(x+\frac{1}{2}t)^2}-(x-\frac{1}{2}t)e^{-2(x-\frac{1}{2}t)^2}
\end{equation}

\begin{equation}
\frac{dE(t)}{dt}= \frac{1}{2}\int_0^L \bigl( 2u_tu_{tt}+ 2u_{x}c^2u_{xt} + \omega^22 uu_t)\,dx +c^2\alpha u(0,t)u_t(0,t)+ c^2\beta u(L,t)u_t(L,t)
 \end{equation}
do integral by part on $\int_0^L \bigl(2u_{x}c^2u_{xt})\,dx$
\begin{equation}
\int_0^L \bigl(2u_{x}c^2u_{xt})\,dx=2c^2[(u_{x}u_{t})\Big|_{0}^L - \int _0^L u_{xx}u_{t}\,dx)]
\end{equation}
plugging these into $\frac{dE(t)}{dt}$
\begin{equation}
\frac{dE(t)}{dt}=c^2\int_0^L \bigl( u_tu_{tt}- u_{xx}u_{t} + \omega^2 uu_t)\,dx +c^2(u_{x}u_{t})\Big|_{0}^L  +c^2\alpha u(0,t)u_t(0,t)+ c^2\beta u(L,t)u_t(L,t)
\end{equation}
By condition $u_{tt}-c^2u_{xx} + \omega^2 u =0$
\begin{equation}
c^2\int_0^L \bigl( u_tu_{tt}- u_{xx}u_{t} + \omega^2 uu_t)\,dx=0
\end{equation}
\begin{equation}
\frac{dE(t)}{dt}=c^2(u_{x}u_{t})\Big|_{0}^L  +c^2\alpha u(0,t)u_t(0,t)+ c^2\beta u(L,t)u_t(L,t)
\end{equation}
by condition $(u_x -\alpha u)|_{x=0}=0$ and $(u_x +\beta u)|_{x=L}=0$, we have
\begin{equation}
\begin{split}
&\alpha u(0,t)=u_{x}(0,t)\\
&\beta u(L,t)=-u_{x}(L,t)
\end{split}
\end{equation}
plugging (6) into  $\frac{dE(t)}{dt}$, we have:
\begin{equation}
\begin{split}
\frac{dE(t)}{dt}&=c^2[(u_{x}u_{t})\Big|_{0}^L+u_{x}(0,t)u_t(0,t)-u_{x}(L,t)u_t(L,t)]\\
&=c^2[u_{x}(L,t)u_t(L,t)-u_{x}(0,t)u_t(0,t)+u_{x}(0,t)u_t(0,t)-u_{x}(L,t)u_t(L,t)]\\
&=c^2[0+0]\\
&=0
\end{split}
\end{equation}
Since $\frac{dE(t)}{dt}$=0, then E(t) does not depend on t.

Firstly, Let's use characteristic coordinates \begin{equation}\left\{\begin{aligned}&\xi=x+ct,\\&\eta=x-ct.\end{aligned}\right.\label{eq1}\end{equation} When using the characteristic coordinates, the problem become: \begin{gather}u_{\xi\eta}=0\qquad \text{as }\xi>0,\eta>0,\\[3pt]u|_{\xi=0}=g(t)\qquad \text{as }t<0 ,\\[3pt]u|_{\eta=0}=h(t)\qquad \text{as } t>0.\end{gather} Then by 2.4.1 equation(4), we have \begin{equation}u=\phi(\xi)+\psi(\eta)\label{5}\end{equation} is the general solution to (2). Then \begin{equation}u|_{\xi=0}=\phi(0)+\psi(\eta)=g(t)\end{equation} \begin{equation}u|_{\eta=0}=\phi(\xi)+\psi(0)=h(t)\end{equation} Thus by (5)(6)(7) \begin{equation}u=g(t)+h(t)-\phi(0)-\psi(0)\end{equation} Since we have \begin{equation}u(0,0)=\phi(0)+\psi(0)\end{equation} \begin{equation}u(0,0)=h(0)=g(0)\end{equation} Then \begin{equation}u=g(t)+h(t)-g(0)\end{equation} solves Goursat problem.