Theorem: For a fixed $T > 0$, let $u\in\operatorname{C}^{2}(\{(x,t)\in\mathbb{R}^{N+1}:0\leq t\leq T\}$. such that $\partial_{t}^2u-\Delta_{x}u\equiv 0$. Suppose there exists $(x_0,t_0)\in \{(x,t)\in\mathbb{R}^{N+1}:0\leq t\leq T\}$ such that $u(\cdot,0)$ and $\partial_{t}u(\cdot,0)$ are zero on the ball $B = > \{x\in\mathbb{R}^N:|x-x_0|<t_0\}$; then $u\equiv 0$ in the cone

$$\Omega = \{(x,t)\in\mathbb{R}^{N+1}:0 \leq t \leq t_0\text{ and }|x-x_0| \leq t_0 - t\}$$

Let $u\in\operatorname{C}^{2}(\mathbb{R}^{N+1})$ be a solution of

$$ \begin{cases} \partial_{t}^{2}u(x,t) - \Delta_{x} u(x,t) = 0,\text{ for } (x,t)\in\mathbb{R}^{N+1}\\ u(\cdot, 0) = u_0\\ \partial_{t} u(\cdot,0) = u_1 \end{cases}$$ Suppose there exists a $t_0\in\mathbb{R}$ such that $u(\cdot,t_0)$ has compact support. How to show that, for each $t\in\mathbb R$, $u(\cdot,t)$ has compact support. This is exercise 2 of the Chapter 5 on Folland's Introduction to PDE.

I tried to solve the problem by considering the function $w(x,t) = u(x,t_0 - t)$. Then, $w$ is a solution of the problem

$$ \begin{cases} \partial_{t}^{2}w(x,t) - \Delta_{x} w(x,t) = 0,\text{ for } (x,t)\in\mathbb{R}^{N+1}\\ w(\cdot, 0) = u(\cdot, t_0)\\ \partial_{t} w(\cdot,0) = \partial_{t}u(\cdot, t_0) \end{cases}$$

Let $B\subset\mathbb{R}^{N}$ be a ball, whose center is $x_0$ and radius $R$, such that $\operatorname{supp}u(\cdot,t_0)\subseteq B$. Then, by the above theorem, $w$ is zero in the set $$\left\{(x,t)\in\mathbb{R}^{N+1}:t \leq \big||x-x_0| - R\big|\right\}$$ But I could not conclude for all $t$.


$w$ is zero in the set $\left\{(x,t)\in\mathbb{R}^{N+1}:t \leq \big||x-x_0| - R\big|\right\}$

This isn't correct. Try $t=0$ here to see why. The correct conclusion is

$w$ is zero in the set $\left\{(x,t)\in\mathbb{R}^{N+1}: |x-x_0| \ge |t|+R \right\}$

You are done. Just rephrase the conclusion: for every $t$, the support of the function $w(\cdot,t)$ is contained in the set $\left\{x\in\mathbb{R}^{N}: |x-x_0| \le |t|+ R\right\}$. So it is compact. Same holds for $u$, by translation.


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