PDE, wave equation and energy function.

We are given that

$u_{tt} - \Delta u = 0, \tag 1$


$u_{tt} = \Delta u = \nabla^2 u = \nabla \cdot \nabla u; \tag 2$

we define

$E(t) = \displaystyle \int_{\Bbb R^2} (u_t^2 + \nabla u \cdot \nabla u) \; dx dy; \tag 3$

(note that $u_t^2 = \vert u_t \vert^2$).


$E_t(t) = \dfrac{dE(t)}{dt} = \displaystyle \int_{\Bbb R^2} (2u_t u_{tt} + \nabla u_t \cdot \nabla u + \nabla u \cdot \nabla u_t) \; dx dy$ $= 2 \displaystyle \int_{\Bbb R^2} (u_t u_{tt} + \nabla u \cdot \nabla u_t) \; dx dy; \tag 4$

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 $t$.

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 $$ \dfrac{d}{dt}E(t)=\int u_t\Delta u-\int u_t\Delta u=0. $$