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?

• you still can solve it first with the Fourier series $g(x) = \sum_n c_n e^{2 i \pi n x / T}$ where $T$ is greater than the size of $support(g)$, and $E(t)$ will be obtained from the Parseval theorem – reuns Mar 1 '16 at 2:16

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.
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 $C$.
• Thank you! I have a question regarding the formula for $E'(t)$, how to show rigorously that $- \frac{1}{h} \int_{r-t-h \leq |x| \leq r-t} f(x,t) dx$ converges to the boundary integral in your formula ($f$ being the integrand) as $h\rightarrow 0$? – user16015 Mar 6 '16 at 4:40