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.


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 $e(t) \leq e(0) \implies e(t) = 0.$


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