# Proving that $\int u^2 dx < Ce^{-at}$ where $u$ solves a linear pde

Let the unit sphere in $\mathbb{R^n}$ be $B_1(0)$ and let $u$ be the smooth solution of $$\begin{cases} u_{tt} + a^2(x) u_t - \Delta u = 0 & B_1(0) \times (0,\infty)\\ u(x,t) = 0 & \partial B_1(0) \times (0,\infty) \\ u(x,0) = g, \; u_t(x,0) = h & B_1(0) \times \{0\} \end{cases}$$

Here, $g,h$ and $a(x)$ are smooth functions and $g,h$ vanish on the boundary of $B_1.$ Prove that $\int_{B_1(0)} u^2(x,t) dx \leq C e^{-At}$, where $A$ is the minimum of $a^2(x)$ and $C$ is a positive real number.

Usually the methods I have seen for problems like these has been to use some sort of an energy, so here is my twice updated attempt, now using Poincare's inequality in the first step:

Let \begin{align} \frac{d}{dt} e(t) &:= \frac{d}{dt}\frac{1}{2} \int_{\Omega} u^2 \,dx \\ &\leq \frac{d}{dt}\frac{C}{2} \int_{\Omega} u_t^2 + |\nabla u|^2 dx\\ &= C\int_{\Omega} u_tu_{tt} + \nabla u_t \cdot \nabla u \; dx\\ &= C\int_{\Omega} u_t( u_{tt} - \Delta u) \; dx + \underbrace{\int_{\partial\Omega} u_t\nabla u \cdot n \, dS}_{\text{0 Due to BCs}}\\ &= -C\int_{\Omega} a^2(x)u_t^2 \; dx \leq 0 \end{align}

So we have $\frac{d}{dt}e(t) \leq 0,$ and therefore we have $e(t) \leq e(0) = \int_{B_1(0)} g^2 \; dx.$ Still, this does not show the desired inequality.

EDIT: Another idea that comes to mind, although I realize this is a finite domain, is to look at the Fourier transform, and then show that $||{\hat u(\xi,t)}||_{L_2}^2 \leq C e^{-At},$ since the $L_2$ norm of the transform is more or less equivalent to the $L_2$ norm in the spatial domain. The finite domain aspect seems to ruin this though.

• If $u$ is a displacement of a medium, isn't the local energy proportional to $u_t^2$, i.e., the square of the velocity? So have you considered a net energy integral like $\frac{1}{2}\int_\Omega u_t^2\;dx$ ? Or I've sometimes seen energy proportional to $u_t^2 + u_x^2$ in these kinds of problems. – rajb245 Aug 8 '17 at 17:57
• That is to say, the sum of $u_t^2$ and $u_x^2$ is proportional to the total energy, kinetic plus potential, and if $a(x)=0$ everywhere then that would be the conserved quantity. So if $a$ isn't identically zero, I would expect the total energy to decay at a rate that has something to do with $a(x)$, and from that maybe you can say something about your integral. – rajb245 Aug 8 '17 at 18:06
• Perhaps using the symbol $e(t)$ for my integral $\int u^2 \; dx$ was too suggestive. Its just a value, that is bounded by what people normally refer to as the energy, i.e. $\int u^2 \; dx \leq \int u_t^2 + u^2 \; dx \leq C\int u_t^2 + |\nabla u|^2 dx$ by Poincare's inequality. Of course it makes sense that the presence of a damping term will make the traditional energy decay, but its not clear to me how this gets me the inequality I'm looking for. – Merkh Aug 8 '17 at 21:17
• And just to be clear, we know that the $a^2(x) u_t$ term is a damping term because $a^2(x) > 0,$ assuming $a \neq 0.$ – Merkh Aug 8 '17 at 21:18
• With $E(t) = \frac{1}{2}\int_\Omega u_t^2 + |\nabla u|^2 \;dx$, you can directly show $\frac{dE}{dt} \leq -\int_\Omega a^2(x) u_t^2\;dx$ by multiplying both sides of the PDE by $u_t$, integrating, and using the identities you used. So if you could bound that integral in terms of $E$, you would have something like $\frac{dE}{dt} \leq -A E$, with a solution $E(t) \leq C e^{-A t}$ directly, and the Poincare inequality finishes off the proof. I haven't been able to massage that integral into the correct form yet though. – rajb245 Aug 8 '17 at 22:02