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!
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 $$e(t)\leq e(0).$$