Prove domain of Dependence Inequality for the Wave Equation? Let $(x_0,t_0)\in R^{n+1}$ with $t_0>0$, and let $\Omega$ be the conical domain in $R^{n+1}$ bounded by the backward characteristic cone with apex at $(x_0,t_0)$ and by the plane $t=0$. Suppose $u\in C^2(\overline \Omega)$ and statisfies $$\Delta u -u_{tt}-q(x)u=0$$ in $\Omega$, where $q(x)>0$. Derive the domain of dependence inequality
$$\int_{B(x_0,t_0-T)}u_{x_1}^2+...+u_{x_n}^2+u_{t}^2+qu^2|_{t=T}dx\le \int_{B(x_0,t_0)}u_{x_1}^2+...+u_{x_n}^2+u_{t}^2+qu^2|_{t=0}dx$$
where $0\le T\le t_0$ and $B(x_0,r)$ denotes the ball ${x:|x-x_0|<r}$.
My attempt:
I have no clue about this problem. Maybe using energy's method? Can anyone give me some hints or references like lecture notes? Thanks so much!
 A: As in my answer to your related question, I am again referring to Evan's book, chapter 2.4, Theorem 6. I also use Evan's notation, so $Du$ for the gradient of $u$, and $\nu$ the outward pointing unit normal vector to the respective boundary. 
We proceed exactly as in the proof of the theorem and just use the different local energy functional
$$e(t):=\frac12\int_{B(x_0,t_0-t)}u_t^2(x,t)+|Du(x,t)|^2+q(x)u^2(x,t)dx.$$
As in the book, we show that the local energy is non-increasing, means that $\dot{e}(t)\leq 0$, for all $0\leq t\leq t_0$.
So we differentiate w.r.t. time and obtain
$$\dot e(t)=\int_{B(x_0,t_0-t)}u_tu_{tt}+Du\cdot Du_t + q\,uu_t \;dx - \frac12 \int_{\partial B(x_0,t_0-t)}u_t^2+|Du|^2+q\,u^2 dS.$$
Integrating by parts the divergence term gives
$$\dot e(t)=\int_{B(x_0,t_0-t)}u_t(u_{tt}-\Delta u + q\,u) + \int_{\partial B(x_0,t_0-t)}\frac{\partial u}{\partial \nu}u_t \;dx - \frac12 \int_{\partial B(x_0,t_0-t)}u_t^2+|Du|^2+qu^2 dS.$$
Since $u$ is a solution to the wave equation, the interior integral vanishes and we obtain
$$\dot e(t)=\int_{\partial B(x_0,t_0-t)}\frac{\partial u}{\partial \nu}u_t - \frac12 \left(u_t^2+|Du|^2+q\,u^2 \right)dS.$$
The normal derivative can be bounded using Cauchy-Schwarz and Cauchy inequalities (check the book for that), so we get
$$|\frac{\partial u}{\partial \nu}u_t|\leq\frac12 \left(u_t^2+|Du|^2\right),$$
which gives in total
$$\dot e(t)\leq 0$$
and therefore
$$e(t)\leq e(0).$$
