Vanishing of solution to $u_{tt} - \Delta u + F(x, t)u = 0$ Consider the PDE given by $$u_{tt} - \Delta u + F(x, t)u = 0$$ for $x \in \mathbb{R}$ and $t > 0$ with the initial conditions $u(x, 0) = \varphi(x)$ and $u_{t}(x, 0) = \psi(x)$. If $F$, $\varphi$, and $\psi$ are smooth functions which equal zero for $|x| > R$, the problem I am working on is to show that $u(x, t) = 0$ for $|x| > R + t$.
I first tried to find an explicit formula for $u(x, t)$ and maybe hope that the formula for $u$ might hint at where it's supported. To do this I tried the taking the Fourier transform of both sides. I got $$\widehat{u}(\xi, t)_{tt} = 4\pi\xi^{2}\widehat{u}(\xi, t) - \widehat{F}(\xi, t) \ast \widehat{u}(\xi, t)$$ which doesn't seem to be able to be easily solved. Does anyone else have other suggestions?
 A: Here is a proof for the $1$D case. For $|x|\geq R$ we have $u_{tt} - \Delta u = 0$ and d'Alembert's solution to the wave-equation tells us that we can write the solution as
$$u(x,t) = f(x-t) + g(x+t)~~~~~~~\text{for}~~~~~~~|x|\geq R$$
for some functions $f,g$ determined by the boundary conditions. Applying the boundary conditions
$$u(x,0) = f(x) + g(x) =  0~~~~~~~\text{for}~~~~~~~|x|\geq R$$
$$u_t(x,0) = -f'(x) + g'(x) =  0~~~~~~~\text{for}~~~~~~~|x|\geq R$$
gives us $f(x)\equiv g(x) \equiv 0$ for $|x|\geq R$. From the solution above we therefore have
$$u(x,t) = 0$$
if $|x-t| \geq R$ and $|x+t|\geq R$ both of which is satisfied if $|x| \geq R + t$. 
A: Let $R'>R\,.$ Using the energy method, we have that $\displaystyle \partial_t\left[\frac{1}{2}\int_{B(0,R'+t)-B(0,R+t)}u_t^2+|\nabla u|^2\right]=\\\int_{B(0,R'+t)-B(0,R+t)}u_tu_{tt}+\nabla u_t\cdot\nabla u-\frac{1}{2}\int_{\partial (B(0,R'+t)-B(0,R+t))}u_t^2+|\nabla u|^2\\ =\int_{B(0,R'+t)-B(0,R+t)}u_t\Delta u- u_t\Delta u+\int_{\partial(B(0,R'+t)-B(0,R+t))}u_t\partial_n u-\frac{1}{2}\int_{\partial (B(0,R'+t)-B(0,R+t))}u_t^2+|\nabla u|^2\\=\int_{\partial(B(0,R'+t)-B(0,R+t))}u_t\partial_n u-\frac{1}{2}\int_{\partial (B(0,R'+t)-B(0,R+t))}u_t^2+|\nabla u|^2\\
=\int_{\partial (B(0,R'+t)-B(0,R+t))}-\frac{u_t^2}{2}-\frac{|\nabla u|^2}{2}+u_t\partial_n u\\
\le \int_{\partial (B(0,R'+t)-B(0,R+t))}-\frac{u_t^2}{2}-\frac{|\nabla u|^2}{2}+\frac{u_t^2}{2}+\frac{|\nabla u|^2}{2}\le 0\,.$
So the derivative of $\displaystyle \frac{1}{2}\int_{B(0,R'+t)-B(0,R+t)}u_t^2+|\nabla u|^2\;\;$ is $\;\le 0\,,$ and since this function is always positive and is $0$ initially, it must always be zero, and since $R'>R$ was arbitrary, this implies the result.
