Global solutions of the wave equation with bounded initial condition Let $f,g$ be bounded compactly supported smooth functions, and assume $u$ is the solutions of the wave equation
$$u_{tt}-c^2(x)\Delta u=0 \ \ \hbox{on} \ \ \mathbb{R}^n \times (0,\infty)$$
$$u(x,0)=f, \ \ u_t(x,0)=g, \ \ x\in \mathbb{R}^n,$$
where $c(x)>c_0>0$ is also a bounded smooth functions on $\mathbb{R}^n.$
Does $u(x,t)$ remain bounded on $\mathbb{R}^n \times (0,\infty)$? It seems to me that this should follow from a standard result about hyperbolic equations but I can't find a relevant reference. 
 A: The answer is no, already for the wave equation $c(x)=1$.
Let me be more precise.
If the initial data belong to $H^s\times H^{s-1}$ with $s$ strictly larger than $n/2$, then of course energy estimates give you that the $H^s$ norm remains bounded and hence the solution remains bounded in $L^\infty$ for all finite times. You may still have that the sup norm is unbounded as $t\to\infty$, depending on the precise form of the equation; for the constant coefficient case you get a global bound. 
If the initial data are only bounded and $n\ge2$ then one can construct bounded $f$ (take $g=0$) such that $u$ is unbounded as $t\to1$. Consider for instance $n=3$: by Kirchhoff's representation of the solution we get
$$
u(0,t)=c\int_{\partial B(0,t)}[f(y)+t\partial _r f(y)]\ dS(y)
$$
where $\partial_r$ is the derivative in the radial direciion. If $f$ is bounded but the radial derivative is singular on the boundary of $B(0,1)$ you can easily make the integral go to infinity as $t$ approaches 1. 
If $n=1$ the answer is positive also with variable coefficients since the conservation of the $H^1$ energy controls the sup norm. You may have blow up at infinity of the sup norm depending on the precise form of the equation; if it is in divergence form you get a global bound.
