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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.

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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.

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  • $\begingroup$ Thank you Piero. How about the case when $1-c(x)$ is compactly supported? Can we also get a global bound for all time? $\endgroup$
    – A random mathematician
    Commented Mar 19, 2019 at 13:58
  • $\begingroup$ At what rate the $H^s$ norm could grow in time if it doesn't remain bounded? Could the growth be exponential? $\endgroup$
    – A random mathematician
    Commented Mar 19, 2019 at 14:13
  • $\begingroup$ An exponential bound is easy to prove with Gronwall. Anything better requires much more effort and some structure on the equation. Anyway, the problem of the growth of high order Sobolev norms is nontrivial $\endgroup$
    – Piero D'Ancona
    Commented Mar 19, 2019 at 15:41

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