# The solution of wave equation is compactly supported when the functions in initial conditions have compact support.

Theorem: Let $u\in C^2(\mathbb{R}\times[0,\infty))$ be a solution of the problem $$u_{tt}-u_{xx}=0\text{ in }\mathbb{R}\times[0,\infty).$$ Given $x_0\in\mathbb{R}$ and $t_0>0$ consider the cone $$C_{x_0,t_0}=\{(x,t)\in\mathbb{R}\times[0,t_0];\;0\leq t\leq t_0,\,|x-x_0|\leq t_0-t\}.$$ If $u(x,0)=u_t(x,0)=0$ for all $x\in(x_0-t_0,x_0 t_0)$, then $u=0$ in $C$.

Problem: let $u\in C^2(\mathbb{R}\times[0,\infty))$ be a solution of the problem $$\left\{\begin{matrix} u_{tt}-u_{xx}=0 &\text{in}&\mathbb{R}\times(0,\infty) \\ u=g,\,u_t=h &\text{on}&\mathbb{R}\times\{0\} \end{matrix}\right.$$ where $h,g$ have compact support in $\mathbb{R}$. Prove that, for each $t>0$, the function $u(\cdot,t)$ has compact support in $\mathbb{R}$.

Hint: theorem above.

My (not successful) approach: it's enough to prove that there exists a bounded set $K\subset\mathbb{R}$ such that $u(x,t)=0$ for all $(x,t)\in(\mathbb{R}\backslash K)\times(0,\infty)$. We know that there is a bounded set $K$ such that $$\text{supp}(g)\cup\text{supp}(h)\subset K.$$

Given $(y,s)\in(\mathbb{R}\backslash K)\times(0,\infty)$, there are two cases: $s<d(K,y)$ or $s\geq d(K,y)$. In the first case we can consider the cone $C_{y,s}$ and conclude (by theorem above) that $u(y,s)=0$ (because $u(y,0)=g(y)=0=h(y)=u_t(y,0)$ for all $y\in\mathbb{R}\backslash K\supset(y-s,y+s)$). This argument doesn't work for the second case. So, I need help.

Thanks.

There exists a compact set such that for all $t\in \mathbb{R}_+$, the function $u(\cdot,t)$ has support in that compact set.
For each $t$, there exists $K_t \Subset \mathbb{R}$ such that $\DeclareMathOperator{\supp}{supp} \supp u(\cdot, t) \subset K_t$.
And in this light the answer should be much more easily found. Hint: let $K_0 = \overline{ \supp g \cup \supp h}$. You can define $K_t$ as a set of points "sufficiently far" (depending on $t$) from $K_0$.