# Flux estimate for semilinear wave equation

This is from §12.1.2 (pages 660-661) of PDE Evans, 2nd edition. My question is at the bottom of this post.

DEFINITION. Fix $x_0 \in \mathbb{R}^n$, $t_0 > 0$ and define the backwards wave cone *with apex $(x_0,t_0)$: $$K(x_0,t_0):=\{(x,t) \mid 0 \le t\ le t_0, |x-x_-| \le t_0-t\}.$$ The curved part of the boundary of $K(x_0,t_0)$ is $$\Gamma(x_0,t_0):=\{(x,t) \mid 0 \le t \le t_0, |x-x_0| = t_0-t\}.$$

THEOREM 2 (Flux estimate for the semilinear wave equation). Assume that $u$ is a smooth solution of the semilinear equation $$u_{tt}-\Delta u+f(u)=0.$$

$\quad$(i) For each point $(x_0,t_0) \in \mathbb{R}^n \times (0,\infty)$ we have the identity $$\frac 1{\sqrt{2}}\int_{\Gamma(x_0,t_0)} \frac 12|u_t \nu - Du|^2+F(u) \, dS = e(0), \tag{3}$$ where $\nu :=\frac{x-x_0}{|x-x_0|}$ and $$e(t) := \int_{B(x_0,t_0-t)} \frac 12(u_t^2+|Du|^2)+F(u) \, dx \quad (0\le t \le t_0).$$

$\quad$(ii) If $$F \ge 0$$ and $$u(\cdot,0), \, u_t(\cdot,0) \equiv 0 \quad \textit{within }B(x_0,t_0),$$ then $u \equiv 0$ within the cone $K(x_0,t_0)$.

Proof. 1. We compute that \begin{align} \dot{e}(t) &= \int_{B(x_0,t_0-t)} u_t u_{tt} + Du \cdot Du_t + f(u) u_t \, dx - \int_{\partial B(x_0,t_0-t)} \frac 12(u_t^2+ |Du|^2)+F(u) \, dS \\ &= \int_{\partial B(x_0,t_0-t)} \frac{\partial u}{\partial \nu} u_t - \frac 12(u_t^2+|Du|^2)-F(u) \, dS \\ &= -\int_{\partial B(x_0,t_0-t)} \frac 12 |u_t \nu - Du|^2 + F(u) \, dS, \end{align} since $$|u_t \nu - Du|^2 = u_t^2-2u_t \frac{\partial u}{\partial \nu} + |Du|^2.$$ Now integrate in time between $0$ and $t_0$ to derive $(3)$. Notice that the factor $\frac 1{\sqrt{3}}$ appears when we switch to integration over $\Gamma(x_0,t_0)$, since this surface is tilted at constant angle $\frac{\pi}4$ above $B(x_0,t_0) \times \{t=0\}$.

This is only a partial proof that establishes (i) of Theorem 2. A three-line proof that establishes (ii) is not printed or needed here.

I also understand all but the last three lines of the proof of (i). I am running into trouble trying to integrate in time between $0$ and $t_0$ to derive $(3)$. Should I say \begin{align} e(t_0)-e(0)&=\int_0^{t_0} \dot{e}(t) \, dt \\ &= -\int_0^{t_0} \int_{\partial B(x_0,t_0-t)} \frac 12 |u_t \nu - Du|^2 + F(u) \, dS \, dt \end{align} I am observing that at $t=t_0$, we would be integrating over a ball of radius $t-t_0=0$. Thus, $e(t_0)=0$. So we are left to having the relation \begin{align} e(0)&=\int_0^{t_0} \dot{e}(t) \, dt = \int_0^{t_0} \int_{\partial B(x_0,t_0-t)} \frac 12 |u_t \nu - Du|^2 + F(u) \, dS \, dt \end{align} But how can we use this to establish $(3)$?

You did the integration correctly. To show (3), first compute the surface integral, using the area formula (see Evans, L. C. and Gariepy, R. F.: Measure Theory and Fine Properties of Functions, 2015, Theorem 3.9): $$\int_{\Gamma(x_0,t_0)}\frac{1}{2}|u_t\nu-Du|^2+F(u)\;dS =\int_{B(x_0,t_0)}\left[\frac{1}{2}|(u_t\circ\varphi)\cdot(\nu\circ\varphi)-Du\circ\varphi|^2+F(u\circ\varphi)\right]\sqrt{2}r\;drd\theta,$$ where $$\varphi(r,\theta)=\begin{pmatrix}r\cos\theta\\r\sin\theta\\ t_0-r\end{pmatrix},\quad r\in(0,t_0),\; \theta\in(0,2\pi),$$ is a chart of the surface $$\Gamma(x_0,t_0)$$. The $$\sqrt{2}$$ comes from the Jacobian of $$\varphi$$, which is $$J\varphi(r,\theta)=\sqrt{\det([D\varphi(r,\theta)]^\top D\varphi(r,\theta))}=\sqrt{2}r.$$ The claim now follows, since by integration in polar coordinates (which parametrize the circle $$\partial B(x_0,t_0-t)$$ for any $$t$$), we have $$\int_0^{t_0}\int_0^{2\pi}\left[\frac{1}{2}|(u_t\circ\varphi)\cdot(\nu\circ\varphi)-Du\circ\varphi|^2+F(u\circ\varphi)\right]r\;d\theta dr\\ =\int_0^{t_0}\int_{\partial B(x_0,t_0-t)}\frac{1}{2}|u_t\nu-Du|^2+F(u)\;dS dt.$$ Note that it is immaterial if we integrate with respect to $$r$$ or $$t$$, as for our particular cone $$K(x_0,t_0)$$, the lateral surface $$\Gamma(x_0,t_0)$$ is tilted at constant angle $$\frac{\pi}{4}$$ above $$B(x_0,t_0)\times\{t=0\}$$ and the distances $$r$$ and $$t$$ coincide.