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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?

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  • $\begingroup$ 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 $\endgroup$ – reuns Mar 1 '16 at 2:16
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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 $C$.

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  • $\begingroup$ 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 $h\rightarrow 0$? $\endgroup$ – user16015 Mar 6 '16 at 4:40
  • $\begingroup$ Uniform continuity gives this. On each short radial segment, consider the maximum and minimum of f. These are functions of polar angle. The integral over the shell is between the integrals of those. But the maximum and minimum converge to boundary values uniformly. $\endgroup$ – user147263 Mar 6 '16 at 4:47

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