For wave equation with dissipation, the energy $E(t)=\frac{1}{2}\int_{0}^{l} (u_t^2+u_x^2) dx$ is a decreasing function of $t$

Question

Let $\phi, \psi$, be twice continuously differentiable functions. Let $u$ be the solution of the following initial bvp:

$u_{tt}-u_{xx}+u_{t}=0, 0\lt x\lt l,t \gt0$

$u(x,0)=\phi(x), 0 \le x\le l$

$u_t(x,0)=\psi(x), 0 \le x\le l$

$u(0,t)=u(l,t)=0, t \ge 0$

Prove that the energy $E(t)=\dfrac{1}{2}\int_{0}^{l} ({u_t}^2+{u_x}^2) dx$ is a decreasing function of $t$

I differentiated with respect to $t$, then I manipulated things and used the last two conditions to arrive at $$\frac{dE}{dt}=-\int_{0}^{l}u_t^2 dx+[u_x(l,t)u_t(l,t)-u_x(o,t)u_t(0,t)]$$

I need to show that the terms in the bracket vanish. I have not used the first two conditions yet. I cannot find ways to do them either.

Answer 1

Note that (if $u$ is supposed to be $C^1$ upto the boundary) from $u(0, t) = u(\ell, t) = 0$ for all $t \ge 0$, we get by differentiating that $$ u_t(0, t) = u_t(\ell, t) = 0, \quad t \ge 0. $$ Hence the terms in the bracket are both zero.