Originally Posted by **me**

Since $\displaystyle u(a,t)=v(a,t)=u(b,t)=v(b,t)=0$ for all $\displaystyle t>0$, we have $\displaystyle u_t(a,t)=v_t(a,t)=u_t(b,t)=v_t(b,t)=0$ for $\displaystyle t>0$. Recall also that $\displaystyle u_{tt}=c^2u_{xx}$ and $\displaystyle v_{tt}=c^2v_{xx}$. These facts allow us to see that

$\displaystyle \frac{d}{dt}\int_a^b\frac{1}{2}(u_tv_t+c^2u_xv_x)d x$

$\displaystyle =\int_a^b\frac{1}{2}(u_{tt}v_t+u_tv_{tt}+c^2(u_{xt }v_x+u_xv_{xt}))dx$

$\displaystyle =\frac{c^2}{2}\int_a^b(u_{xx}v_t+u_xv_{xt}+u_{xt}v _x+u_tv_{xx})dx$

$\displaystyle =\frac{c^2}{2}\left[u_xv_t+u_tv_x\right]_a^b=0$. $\displaystyle \blacksquare$