Show:if $u=u_t=0$, then $u\equiv0$ on $\Omega$ for damped wave equation Let $\phi(x)$ be a function in $C_{0}^{\infty}(\mathbb R^3)$ and consider a damped wave equation: $$u_{tt} -\Delta u +\phi u=0$$
$u(x,0)=f$ and $u_t(x,0)=g$.
Now for fixed $(x_0, t_0) \in \mathbb R^3 \times(0, \infty)$, let $\Omega = \{(x,t): t \in [0,t_0] ~\text{and}~ |x-x_0|\leq|t-t_0| \}$ be the cone of dependence  and define $U_{\tau}=\Omega \cap \{t=\tau\}$.
Show:if $u=u_t=0$, then $u\equiv0$ on $\Omega$. I'm not sure how to use cone of dependence here for damped wave equation.
 A: The typical method of proof goes something like this (sorry that I have not finished the problem here, but perhaps someone could),
Consider the energy
$$ e(t) = \frac{1}{2}\int_{\Omega(t)} u_t^2 + |\nabla u|^2 \; dx.$$
Since we have that $e(0) = 0$ by hypothesis, the idea is to then show that ${\dot e}(t) \leq 0.$
Compute,
\begin{align}
{\dot e}(t) &= \int_{\Omega(t)} u_tu_{tt} + \nabla u_t \cdot \nabla u \; dx -\frac{1}{2}\int_{\partial\Omega} u_t^2 + |\nabla u|^2 \; dS \\
&= \int_{\Omega(t)} u_t(u_{tt} - \Delta u)\; dx + \int_{\partial\Omega(t)} u_t\frac{du}{dn} \; dS -\frac{1}{2}\int_{\partial\Omega} u_t^2 + |\nabla u|^2 \; dS\\
&= \int_{\Omega(t)} -\phi u u_t\; dx \int_{\partial\Omega(t)} u_t\frac{du}{dn} \; dS -\frac{1}{2}\int_{\partial\Omega} u_t^2 + |\nabla u|^2 \; dS.
\end{align}
Using the fact that $$ \bigg|\frac{du}{dn} u_t\bigg| \leq \frac{1}{2} u_t^2 + \frac{1}{2} |\nabla u|^2,$$
we have that $$\int_{\partial\Omega(t)} u_t\frac{du}{dn} \; dS -\frac{1}{2}\int_{\partial\Omega} u_t^2 + |\nabla u|^2 \; dS \leq 0$$
Thus,
\begin{align}
{\dot e}(t) &\leq \int_{\Omega(t)} -\phi u u_t\; dx\\
\end{align} 
Here is where I get stuck since $\phi$ can take both positive and negative values in $\Omega$ and not much can be said about $u$ either.  I was hoping for the inequality, ${\dot e}(t) \leq e(t)$ so then $e(t) \leq e(0) \implies e(t) = 0.$
