Consider the PDE $$ \begin{cases} u_{tt} - \Delta u = 0 & \{(x,t) | x \in \mathbb{R}^3, t \in \mathbb{R}^+, x_1 > \frac{t}{2}\}\\ u_t = 4u_{x_1} & \{x_1 = \frac{t}{2} \} \end{cases} $$ Show that $u = 0$ in $C_t := \{ |x| < R - t\} \cap \{ x > \frac{t}{2} \}$ when $u = u_t = 0$ at $t = 0$ in $C_0.$

Normally the proof of finite propagation speed goes along somewhat like \begin{align} e(t) &= \frac{1}{2} \int_{C_t} u_t^2 + |Du|^2 \, dx\\ \frac{de(t)}{dt} &= \int_{C_t} u_tu_{tt} + Du_t \cdot Du \, dx + \frac{1}{2}\int_{\partial C_t} u_t^2 + |Du|^2 \; dx\\ &= \underbrace{\int_{C_t} u_t(u_{tt} - \Delta u) \, dx}_{0} + \underbrace{\int_{\partial C_t} u_t \cdot Du \; dx}_{\text{from int. by parts}} + \underbrace{\frac{1}{2}\int_{\partial C_t} u_t^2 + |Du|^2 \; dx}_{\text{from diff. time-dept. domain}}\\ \end{align}

Now I feel like already something has gone wrong in the surface terms. First, I don't think the $\partial C_t$ is necessarily the same domain as I wrote it in the above steps. They arise either by integration by parts or by differentiating the time dependent domain. Does anyone see any problem with what has been done?

Proceeding along as if it were correct, all that is important to show is that when you do this time-domain differentiation, the surface terms spit out a negative sign. Then we could use the estimate, $$ |u_t Du| \leq \frac{1}{2}(u_t^2 + |Du|^2)$$ to show that the energy is decreasing, and the rest is trivial.

My question is, how valid is the second and third equality in the sequence above? It seems to take relatively no account of the weird shaped domain (the $\{x > t/2 \}$ part). Any comments, suggestions, calculations would all be helpful. Thanks!

  • $\begingroup$ FYI, I've figured it out, no need for a solution unless you feel so inclined :) $\endgroup$ – Merkh Sep 9 '17 at 21:42

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