I have problems checking that the energy $E(t)=\frac{1}{2}\int_I(u_t^2+c^2u_x^2)dx$ on an open interval $I\subset \mathbb R$, such that $u(0,x)=0$ and $u_t(0,x)=0$ for $x\in\mathbb R\setminus I$ is monotonically decreasing.
I think the best way is to show $E'\le0$
Therefore $E'(t)=\int_I(u_tu_{tt}+c^2u_xu_{xt})dx=\int_I(u_tu_{xx}c^2+c^2u_xu_{xt})dx=c^2\int_I(u_tu_{xx}+u_xu_{xt})dx=c^2\int_I(\frac{\partial(u_tu_x)}{\partial x})dx$
Could you explain why this should be decreasing ?
You calculated $E'(t)=u_t(b) u_x(b)-u_t(a)u_x(a)$ where $[a,b]=I$. Using d'Alembert's formula you can find $$ u_tu_x=\frac{1}{4c}\left\{(cg'(x+ct)+h(x+ct))^2 - (cg'(x-ct)+h(x-ct))^2 \right\} $$ When $x=b$, the first square vanishes and we get $u_tu_x\le 0$. The opposite happens when $x=a$.