Show that energy for the wave equation does not increase with time

Question

Define the energy $E(t)$ as $$E(t)= \frac{1}{2} \iiint u_t^2 + c^2(u_x^2 + u_y^2 + u_z^2)\,dx\,dy\,dz$$ where the integral extends over all of space. Show that the energy for the wave equation with dissipation $$u_{tt} = c^2(u_{xx} +u_{yy} +u_{zz} −u_t)$$ does not increase with time, assuming that the solution $u$ tends to zero as $x, y,$ and $z$ tend to infinity. Is it possible for $E(t)$ to be constant for a solution of this equation?

Answer 1

With

$u_{tt} = c^2(u_{xx} + u_{yy} + u_{zz} - u_t) = c^2 (\nabla^2 u - u_t), \tag 1$

and the energy defined as

$E(t) = \dfrac{1}{2} \displaystyle \int_{\Bbb R^3} (u_t^2 + c^2 u_x^2 + c^2 u_y^2 + c^2 u_z^2) \; dV = \int_{\Bbb R^3} (u_t^2 + c^2 \nabla u \cdot \nabla u) \; dV, \tag 2$

we have

$E_t = \dfrac{1}{2} \displaystyle \int_{\Bbb R^3} (2u_t u_{tt} + 2c^2 u_x u_{xt} + 2c^2 u_y u_{yt} + 2c^2 u_z u_{zt}) \; dV$ $= \displaystyle \int_{\Bbb R^3} (u_t u_{tt} + c^2 \nabla u \cdot \nabla u_t) \; dV; \tag 3$

we may substitute (1) into (3)

$E_t = \displaystyle \int_{\Bbb R^3} (c^2 u_t (\nabla^2u - u_t) + c^2 \nabla u \cdot \nabla u_t) \; dV = c^2 \int_{\Bbb R^3}(u_t \nabla^2 u + \nabla u \cdot \nabla u_t - u_t^2) \; dV$ $= \displaystyle c^2 \int_{\Bbb R^3}(u_t \nabla^2 u + \nabla u \cdot \nabla u_t) \; dV - c^2 \int_{\Bbb R^3} u_t^2 \; dV; \tag 4$

examining the first integral on the right-hand side, we recall the identity

$\nabla \cdot (u_t \nabla u) = u_t \nabla^2 u + \nabla u \cdot \nabla u_t; \tag 5$

in light of this, (4) may be written

$E_t = \displaystyle c^2 \int_{\Bbb R^3} \nabla \cdot (u_t \nabla u) \; dV - c^2 \int_{\Bbb R^3} u_t^2 \; dV. \tag 6$

Now let $B(R)$ be the closed ball of radius $R$ centered at the origin in $\Bbb R^3$, and let $S(R) = \partial B(R)$ be it's boundary, a $2$-sphere of radius $R$, also centered at $(0, 0, 0)$. The divergence theorem yields

$ \displaystyle c^2 \int_{B(R)} \nabla \cdot (u_t \nabla u) \; dV = \int_{S(R)} u_t \nabla u \cdot \vec n \; dS, \tag 7$

where $\vec n$ is the outward normal to, and $dS$ the area element of, $S(R)$; now if

$u_t \nabla u \to 0 \; \text{sufficiently rapidly as} \; R \to \infty, \tag 8$

then we have

$\displaystyle c^2 \int_{\Bbb R^3} \nabla \cdot (u_t \nabla u) \; dV = \lim_{R \to \infty} \int_{B(R)} \nabla \cdot (u_t \nabla u) \; dV = \lim_{R \to \infty} \int_{S(R)} u_t \nabla u \cdot \vec n \; dS = 0, \tag 9$

and thus,

$E_t = - c^2 \displaystyle \int_{\Bbb R^3} u_t^2 \; dV \le 0, \tag{10}$

which shows that $E(t)$ does not increase with time under these conditions (9); indeed, as long as $u_t \ne 0$ somewhere, $E(t)$ is a strictly decreasing function of $t$. (I assume here that $u$ and its derivatives are continuous functions of $x, y, z, t$.) If $E(t)$ is a constant, then according to (10),

$u_t = 0, \tag{11}$

and hence

$u_{tt} = 0; \tag{12}$

(1) thus becomes

$\nabla^2 u = 0, \tag{13}$

and $u$ is a harmonic function which does not depend on $t$ at all.

In closing we note that if $u(x,y, z, t)$ is compactly supported, that is, it vanishes outside of some bounded set, then (9) and hence (10) both apply.