Damped wave equation does not increase the energy Consider the system: (functions are smooth) and $g$ has compact support.
$$\begin{cases} u_{tt}-u_{xx}+u_t = 0 \qquad x\in \mathbb{R}, t>0 \\
u(x,0)=f(x) \qquad \qquad x \in \mathbb{R}\\
u_t(x,0)=g(x) 
\end{cases}$$
Prove that $E(t)=\int_\mathbb{R} u_t^2+u^2_x dx$ is nonincreasing.
When we put derivative inside use equation and integrate by parts we get
$$E'(t)=u_tu_x |_{x=-\infty}^{x=+\infty}-2\int_\mathbb{R} u_t^2 dx$$
How to prove that the boundary term vanishes?
 A: The damped wave equation has finite propagation speed (see below), as the standard wave equation. Hence, for every fixed $t$ the function $x\mapsto u(x,t)$ has compact  support, which implies $u_tu_x$ vanishes at infinity.  
Finite propagation speed
I'll work in $\mathbb{R}^n$ since this proof is the same. The precise claim is that if $f,g$ vanish on the ball $\{x:|x-x_0|\le r\}$, then $u(x,t)$ vanishes on the cone $C=\{(x,t):|x-x_0|\le r-t\}$. 
Proof. (Following PDE by Evans, p. 86) Consider the energy within a time slice of the cone $C$: 
$$E(t) = \frac12\int_{|x-x_0|\le r-t} (u_t^2  +|\nabla_x u|^2)\,dx$$
Differentiation yields
$$
E'(t)=\int_{|x-x_0|\le r-t} (u_tu_{tt}  + \nabla_x u \nabla_{xt} u)\,dx
-\frac12\int_{|x-x_0| = r-t} (u_t^2  +|\nabla_x u|^2)\,dx
$$
The second term here comes from the fact that the ball is shrinking. Using the elementary inequality $a^2+b^2\ge 2ab$ we get 
$$
\frac12\int_{|x-x_0| = r-t} (u_t^2  +|\nabla_x u|^2)\,dx \ge 
\int_{|x-x_0| = r-t} |u_t| |\nabla_x u |\,dx
$$
which means that this subtracted term is enough to offset the boundary term coming from integration by parts: 
$$
\int_{|x-x_0|\le r-t} (u_tu_{tt}  + \nabla_x u \nabla_{xt} u)\,dx
 = \int_{|x-x_0|\le r-t} (u_tu_{tt} - \Delta u u_t)\,dx + \int_{|x-x_0| = r-t}  u_t \nabla_xu  \,dx
$$
In conclusion, 
$$
E'(t) \le \int_{|x-x_0|\le r-t} (u_tu_{tt} - \Delta u u_t)\,dx = 
\int_{|x-x_0|\le r-t} (-u_t^2 )\,dx \le 0
$$
This proves that $E(t)=0$ for $0\le t\le r$, hence $u$ vanishes in the cone $C$.
