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.
You are trying to show that there is damping. What is the equation satisfied by $v(x,t)=u(x,t)\exp(\lambda t)$? We find $$ v(x,0)=g,\quad v_t(x,0)=h+\lambda g,\quad v=0\text{ on }\partial B, $$ so no noticeable changes (just a factor). When it comes to the pde, it becomes $$ v_tt + v_t(a^2-2\lambda) + v\lambda (-\lambda +a^2) -\Delta v=0 $$ If $a^2-2\lambda\geq0$, your analysis above carries through without a change. So you have proved $$ \int_\Omega v^2 \text{d} x \leq \int_{B^1} g^2 \text{d} x, $$ for $\lambda =\frac{1}{2}\min a^2$. The conclusion follows by writing this inequality in terms of $u$.