PDE, wave equation and energy function. Let $u = u(x,t)$ be a solution ot the wave equation $u_{tt} - \Delta u = 0$ in $\mathbb{R^2}$. 
Assuming $\nabla u$ goes to 0 as $|x|$ goes to $\infty$, prove that $$E(t) = \int_{\mathbb{R}^2} |u_t|^2 + |\nabla u|^2 dxdy$$
is a constant in $t$ for all $t$. 
I'm not exactly sure how to approach this question, maybe I got confused on the terms what $\Delta u$ and $\nabla u$ are. But to prove it's a constant in $t$, I think we can show that $dE/dt$ is $0$ by plugging $\nabla u$ into equation and some integration tricks. Have been stuck for a while, any suggestion? 
 A: We are given that
$u_{tt} - \Delta u = 0, \tag 1$
or
$u_{tt} = \Delta u = \nabla^2 u = \nabla \cdot \nabla u; \tag 2$
we define 
$E(t) = \displaystyle \int_{\Bbb R^2} (u_t^2 + \nabla u \cdot \nabla u) \; dx dy; \tag 3$
(note that $u_t^2 = \vert u_t \vert^2$).  
Then
$E_t(t) = \dfrac{dE(t)}{dt} = \displaystyle \int_{\Bbb R^2} (2u_t u_{tt} + \nabla u_t \cdot \nabla u + \nabla u \cdot \nabla u_t) \; dx dy$
$= 2 \displaystyle \int_{\Bbb R^2} (u_t u_{tt} + \nabla u \cdot \nabla u_t) \; dx dy; \tag 4$
via (2) this becomes
$E_t(t) = 2 \displaystyle \int_{\Bbb R^2} (u_t \nabla \cdot \nabla u + \nabla u \cdot \nabla u_t) \; dx dy; \tag 5$
we note that
$\nabla \cdot (u_t \nabla u) = \nabla u_t \cdot \nabla u + u_t \nabla \cdot \nabla u, \tag 6$
so that
$\nabla u_t \cdot \nabla u = \nabla \cdot (u_t \nabla u) - u_t \nabla \cdot \nabla u; \tag 7$
subtstituting (7) into (5) we find
$E_t(t)$
$= 2 \displaystyle \int_{\Bbb R^2} (u_t \nabla \cdot \nabla u + \nabla \cdot (u_t \nabla u) - u_t \nabla \cdot \nabla u) \; dx dy$
$= 2 \displaystyle \int_{\Bbb R^2} \nabla \cdot (u_t \nabla u) \; dx dy; \tag 8$
we compute the last integral in this equation via the divergence theorem applied to the circle $S_R$ centered at the origin which bounds the ball $B_R$ of radius $R$, also centered at $(0, 0)$:
$2 \displaystyle \int_{\Bbb R^2} \nabla \cdot (u_t \nabla u) \; dx dy = 2\lim_{R \to \infty} \int_{B_R} \nabla \cdot (u_t \nabla u) \; dx dy$
$= 2\displaystyle \lim_{R \to \infty} \int_{S_R} u_t \nabla u \cdot \vec n_R \; ds, \tag 9$
where $\vec n_R$ is the outward pointing unit normal on $S_R$ and $ds$ is the linear measure along $S_R$.
It now appears that, to complete the calculation of (9), a stronger assumption than the given
$\nabla u \to 0 \; \text{as} \; R \to \infty \tag{10}$
is required, for we need to ensure that
$\displaystyle \lim_{R \to \infty} \int_{S_R} u_t \nabla u \cdot \vec n_R \; ds = 0, \tag{11}$
which is not necessarily guaranteed by (10); for example, if $\vert \nabla u \vert$ falls off as, say, $R^{-1/2}$ as $R \to \infty$, and $u_t$ approaches a non-zero constant,
$\displaystyle \lim_{R \to \infty} u_t \to c \ne 0, \tag{12}$
then the integral occurring in (11) may in fact become unbounded as $R \to \infty$.  A more stringent set of assumptions is therefore needed.  For the present purposes we hypothesize that $\nabla u$ falls off as $1/R^{1 + \epsilon}$, $\epsilon > 0$ and $u_t$ is bounded for large $R$; then it is easy to see that (11) holds and thus that
$\displaystyle \int_{\Bbb R^2} \nabla \cdot (u_t \nabla u) \; dx dy = 0; \tag{13}$
from this in accord with (8)
$E_t(t) = 0, \tag{14}$
and hence that $E(t)$ is a constant for all $t$.
A: By directly differentiating the energy weobtain
$$
\dfrac{d}{dt}E(t)=\int u_tu_{tt}+\int \nabla u\cdot\nabla u_t
$$
Then, by using the equation on the first integral and the Divergence's Theorem on the second one (and the fact that $\nabla u$ decay sufficiently fast at $+\infty$) we obtain $$
\dfrac{d}{dt}E(t)=\int u_t\Delta u-\int u_t\Delta u=0. 
$$
