Let $\square u$ be $u_{tt}-c^2u_{xx}$ where $u$ is a function to times differentiable. Suppose that $\square u=0$, $\square v=0$ for $a<x<b$ and satisfying the initial value conditions $u(a,t)=u(b,t)=0$ (t>0). Prove that $$ \displaystyle\frac{d}{dt}\int _a^b\frac{1}{2}(u_tv_t+c^2u_xv_x)dx=0$$
I ask: this conditions are enough to prove the conclusion.
I did the following:
We can derive inside the sign of integral getting:
$$ \displaystyle\frac{d}{dt}\int _a^b\frac{1}{2}(u_tv_t+c^2u_xv_x)dx=\frac{1}{2}\int_a^b(u_{tt}v_t+u_tv_{tt}+c^2u_{tx}v_x+c^2u_xv_{xt})dx$$
Using that $u_{tt}=c^2 u_{xx}$ and $v_{tt}=c^2 v_{xx}$ we can re write the equation above as
$$\displaystyle \frac{d}{dt} \int _a^b\frac{1}{2}(u_tv_t+c^2u_xv_x)dx=\frac{c^2}{2}\int_a^b(u_{xx}v_t+u_tv_{xx}+u_{tx}v_x+u_xv_{xt})dx$$
Integrating by parts we got the formulas
$$\int_a^bu_{xx}v_{t}dx=v_tu_x\displaystyle\mid_a^b-\int_a^bu_xv_{tx}dx$$ $$\int_a^bu_{t}v_{xx}dx=u_tv_x\displaystyle\mid_a^b-\int_a^bu_{tx}v_xdx$$
getting
$$\displaystyle\frac{d}{dt}\int _a^b\frac{1}{2}(u_tv_t+c^2u_xv_x)dx= \displaystyle\frac{c^2}{2}[v_tu_x\displaystyle\mid_a^b+ u_tv_x\displaystyle\mid_a^b]$$
If this computation are right then I can not conclude that the equations given is zero.
There are hypothesis missing or am I missing something?
Thanks!
Hint. If $u(a,t) = 0$ for all $t > 0$, what can you say about $\partial_t u(a, t)$? (Similar for $b$ and $v$)