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The question is the following. Let $u\in C^2(\mathbb{R}^n\times[0,+\infty))$ be a solution of the problem

\begin{cases} u_{tt}-\Delta u = 0\\ u(x,0) = \phi(x)\\ u_t(x,0)=\psi(x) \end{cases} where the initial data $\phi$ and $\psi$ have compact support (i.e. vanish identically outside a closed ball $\overline{B(0,R)}$ of $\mathbb{R}^n$). Prove that for every $t\geq0$, $u(x,t)$ vanish identically outside the closed ball $\overline{B(0,R+t)}$.

Probably it is a direct consequence of the finite propagation speed but I can't formalize it...

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  • $\begingroup$ What do you mean by finite propagation speed exactly? $\endgroup$ – user147263 Nov 22 '15 at 16:47
  • $\begingroup$ The statement which says that if the initial data are zero on $\overline{B(x^o,t^o)}\times\{t=0\}$, then $u$ is zero at all points of the cone $K(x^o,t^o)$ (i.e. the values of the initial data outside the ball $\overline{B(x^o,t^o)}$ do not affect the value of u in the cone $K(x^o,t^o)$. $\endgroup$ – Redeldio Nov 22 '15 at 16:56
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Pick a point $(x_0,t_0)$ with $|x_0|>R+t_0$. The initial values vanish on the ball $B(x_0,|x_0|-R)$. By finite speed of propagation, the solution vanishes on the cone $K$ built on top of this ball such that the $t$-time section of this cone is $B(x_0,|x_0|-R-t)\times \{t\}$ for $t<|x_0|-R$.

In particular, the $t_0$ section of $K$ is $B(x_0,|x_0|-R-t_0)\times \{t_0\}$ which contains the point $(x_0,t_0)$. This implies the claim.

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