TITLE: Homogeneous Wave Equation with Dirichlet Conditions
QUESTION [2 upvotes]: Let $u(x,t)$ a solution  of $u_{xx} = \frac{1}{c^2}u_{tt}, a<x<b, t>0.$
The integral energy $u$ is given by $E(t)=\int^{b}_{a}[u_x(x,t)^2+\frac{1}{c^2}u_t(x,t)^2]dx, t>0$.
(i) 
Show that if $u\in C^2([a,b]\times (0,\infty))\cap C^1([a,b]\times [0,\infty))$ 
satisfies the above equation and the boundary conditions $u(a,t)=0=u(b,t),t\ge 0$, then $E(t)=E(0)$ whatever $t\ge 0$.
(ii) 
Use (i) to show that there is at most one solution of the problem
$u\in C^2([a,b]\times (0,\infty))\cap C^1([a,b]\times [0,\infty))$
$u_{xx}-\frac{1}{c^2}u_{tt}=q(x,t), a<x<b,\quad t>0,$
$u(x,0)=f(x),\quad a\le x\le b,$
$u_t(x,0)=g(x),\quad a\le x\le b,$
$u(a,t)=A(t),\quad t\ge 0,$
$u(b,t)=B(t),\quad t\ge 0$
where $q\in C([a,b]\times (0,\infty)),\quad f,g\in C([a,b]),\quad A,B \in C([0,\infty))$
First, I have difficult to find the eigenvalues, and for part (ii) what kind of suggestion you can give me?

REPLY [4 votes]: I'm not sure what you mean when you say you need to "find the eigenvalues of $u$."  A function doesn't have eigenvalues. Operators have eigenvalues. That being said, this problem doesn't require any consideration of eigenvalues.
For part (i), we see \begin{align*} \dot E(t) &= \frac{d}{dt} \int_a^b (u_x^2 + \frac{1}{c^2} u_t^2 ) \, dx \\ 
&= 2\int^b_a (u_xu_{xt} + \frac{1}{c^2} u_tu_{tt}) dx \\
&= 2\int_a^b(- u_{xx}u_t + \frac{1}{c^2} u_tu_{tt}) dx + \left. \frac{}{} u_xu_t \right|_{x=a}^{x=b} \\
&= 2\int_a^b(- u_{xx} + \frac{1}{c^2}u_{tt}) u_t dx + \left. \frac{}{} u_xu_t \right|_{x=a}^{x=b} 
\end{align*} The boundary term goes to zero since $u(a,t) = u(b,t) = 0$ (so also $u_t(a,t) = u_t(b,t) = 0$). The other term is zero since $u$ satisfies the heat equation. This shows that $\dot E(t) = 0$ for all $t$ so $E$ is constant; hence $E(t) = E(0)$ for all $t$. 
For part (ii), assume that $v,w$ are two solutions to the given equation and put $u = v- w$. Then $u$ satisfies $$u_{xx} - \frac{1}{c^2} u_{tt} = 0$$ with $u(a,t) = u(b,t) = 0$,  $u(x,0) = u_t(x,0)= 0.$ By part (i), we must have $E(t) = E(0)$ for all $t$. But since $u, u_t$ are zero at $t=0$, we have $E(0) = 0$. Thus $E(t) = 0$ for all $t$. Since the integrand in $E$ is continuous and non-negative, it must be identically zero. But then, $u_x^2 = 0 = u_t^2$ for all $t$ which shows that $u$ is constant. Since $u$ is zero on the boundaries, it must be zero everywhere. Hence $v = w$, so the solutions are the same. This proves uniqueness.
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