Energy uniqueness of 3D wave equation help

Question

I am trying to use an energy argument for to show that the global Cauchy problem for the three-dimensional wave equation has a unique solution.

The wave equation is $$\partial^2u/\partial t^2=\nabla^2u$$

I looked up the energy functional and I want to use this $$E[u]=\frac1{2}\int_{R^3}((\partial u/\partial t)^2+\nabla u \nabla u))d^3 r$$

I have been reading what may be a similar question http://www.sgo.fi/~j/baylie/Partial%20Differential%20Equations%20in%20Action%20-%20From%20Modelling%20to%20Theory%20-%20S.%20Salsa%20(Springer,%202008)%20WW.pdf on page 265-266 but I can't figure out how to go to the 3D case.

I know the idea is the following:

1) define the energy integral as non negative non increasing function

2) for t =0 the energy is zero and so the energy is 0 for all t greater than or equal to 0. Then due to positivity of energy and the zero ic and bc we get solution as zero.

I just am struggling with how to find the derivative of the Energy integral above.

Thanks for your time.

Answer 1

The author uses the Leibniz integral rule, and this is possibly why he chose to expand the energy integral. For \begin{align*} e(t) & = \frac{1}{2}\int_0^{c(t_0 - t)}dr \int_{\partial B_r(x_0)}\left(u_t^2 + c^2|\nabla u|^2\right)d\sigma \\ & = \frac{1}{2}\int_0^{c(t_0 - t)}\left(\int_{\partial B_r(x_0)}\left(u_t^2 + c^2|\nabla u|^2\right)d\sigma\right)dr \end{align*} applying Leibniz integral rule and product rule gives \begin{align*} \dot{e}(t) & = \frac{1}{2}\left(\int_{\partial B_r(x_0)}\left(u_t^2 + c^2|\nabla u|^2\right)d\sigma\right)\bigg|_{r = c(t_0 - t)} \left(\frac{d}{dt}\left[c(t_0 - t)\right]\right) \\ & \qquad + \frac{1}{2}\int_0^{c(t_0 - t)}\frac{\partial}{\partial t}\left(\int_{\partial B_r(x_0)}\left(u_t^2 + c^2|\nabla u|^2\right)d\sigma\right)dr \\ & = -\frac{c}{2}\int_{\partial B_{c(t_0 - t)}(x_0)}\left(u_t^2 + c^2|\nabla u|^2\right)d\sigma + \frac{1}{2}\int_0^{c(t_0 - t)}\int_{\partial B_r(x_0)}\left(2u_tu_{tt} + 2c^2\nabla u\cdot\left(\nabla u\right)_t\right)d\sigma dr \\ & = -\frac{c}{2}\int_{\partial B_{c(t_0 - t)}(x_0)}\left(u_t^2 + c^2|\nabla u|^2\right)d\sigma + \int_{B_{c(t_0 - t)}(x_0)}\left(u_tu_{tt} + c^2\nabla u\cdot\nabla (u_t)\right)d\mathbf{x} \end{align*} The rest should be pretty straightforward I hope. The crucial difference compared to the case of bounded domain is that divergence theorem doesn't apply to infinite domain, so we need to consider the energy integral $e(t)$ over some bounded domain and do a limiting argument. In the case of wave equation, instead of considering $e(t)$ over the sphere we exploit the finite propagation property and consider the backward cone.

Answer 2

I first saw this problem on my 'droid and the picture was too small for me to make out what it was all about, so I drafted an answer based upon a somewhat different approach, viz. the derivative of the energy integral over $\Bbb R^3$ and not over the retrodgrade cone. But I thought it might be worth sharing here in any event. So here goes:

We start with the wave equation

$u_{tt} = \nabla^2 u = \nabla \cdot \nabla u; \tag 0$

as I wrote in my comment to the question, I think we need to take the energy as

$E = \dfrac{1}{2}\displaystyle \int_{\Bbb R^3} (u_t^2 + \nabla u \cdot \nabla u)\; d^3r; \tag 1$

then

$E_t = \dfrac{1}{2}\displaystyle \int_{\Bbb R^3} (2 u_t u_{tt} + 2\nabla u_t \cdot \nabla u)\; d^3r = \displaystyle \int_{\Bbb R^3} (u_t u_{tt} + \nabla u_t \cdot \nabla u)\; d^3r; \tag 2$

we note that

$\nabla \cdot (u_t \nabla u) = \nabla u_t \cdot \nabla u + u_t \nabla \cdot \nabla u, \tag 3$

which may be written

$\nabla \cdot (u_t \nabla u) - \nabla u_t \cdot \nabla u = u_t \nabla \cdot \nabla u. \tag 4$

We seek to integrate (4) over $\Bbb R^3$; to do so, we let $B(R)$ be the closed ball of radius $R$ in $\Bbb R^3$ centered at $(0, 0, 0)$; then

$\displaystyle \int _{B(R)} (\nabla \cdot (u_t \nabla u) - \nabla u_t \cdot \nabla u) \; d^3r = \int_{B(R)} u_t \nabla \cdot \nabla u \; d^3r; \tag 5$

by the divergence theorem we have

$\displaystyle \int _{B(R)} (\nabla \cdot (u_t \nabla u)) \; d^3r = \int_{S(R)} (u_t \nabla u) \cdot \vec n \; d\Sigma; \tag 6$

where $\vec n$ is the outward pointing unit vector field on $S(R)$, and $d\Sigma$ is the surface area element on $S(R)$ induced from $d^3r$ on $\Bbb R^3$. If we assume that $u_t \nabla u$ falls off sufficiently fast as $R \to \infty$, then we have

$\displaystyle \int_{\Bbb R^3} \nabla \cdot (u_t \nabla u) \; d^3r = \lim_{R \to \infty} \int_{B(R)} \nabla \cdot (u_t \nabla u) \; d^3r = \lim_{R \to \infty} \int_{S(R)} (u_t \nabla u) \cdot \vec n \; d\Sigma= 0; \tag 7$

we thus conclude that (5) implies

$\displaystyle \int_{\Bbb R^3} \nabla u_t \cdot \nabla u \; d^3r = \lim_{R \to \infty} \int_{B(R)} \nabla u_t \cdot \nabla u \; d^3r$ $= \displaystyle \lim_{R \to \infty} -\int_{B(R)} u_t \nabla \cdot \nabla u \; d^3r = -\int_{\Bbb R^3} u_t \nabla \cdot \nabla u \; d^3r; \tag 8$

it is now a straightforward matter to see that, with the aid of (8), (2) yields

$E_t = \displaystyle \int_{\Bbb R^3} (u_t u_{tt} - u_t \nabla \cdot \nabla u)\; d^3r = \int_{\Bbb R^3} u_t (u_{tt} - \nabla^2 u)\; d^3r = 0 \tag 9$

by virtue of (0). The conservation of the energy (1) is thus established.

We may use (9) to demonstrate the uniqueness of solutions to (0). For if $u$ and $v$ are solutions having the same initial conditions,

$u(x, 0) = v(x, 0) = f(x), \; u_t(x, 0) = v_t(x, 0) = g(x), \tag{10}$

where again, $f(x)$ and $g(x)$, as we have assumed of $u_t \nabla u$ in (7), decrease sufficiently rapidly for large $R = \Vert x \Vert$; then if we consider the function

$w(x, t) = u(x, t) - v(x, t), \tag{11}$

we see that $w(x, t)$ is also a solution to (0) by linearity, and its initial conditions are

$w(x, 0) = u(x, 0) - v(x, 0) = f(x) - f(x) = 0, \tag{12}$

$w_t(x, 0) = u_t(x, 0) - v_t(x, 0) = g(x) - g(x) = 0; \tag{13}$

it then follows that the energy $E[w]$ of the solution $w(x, t)$ satisfies

$E[w](0) = \dfrac{1}{2}\displaystyle \int_{\Bbb R^3} (w_t^2(x, 0) + \nabla w(x, 0) \cdot \nabla w(x, 0))\; d^3r = 0, \tag{14}$

and

$E_t[w](t) = 0, \; t \ge 0, \tag{15}$

which together imply

$E[w](t) = 0, \; t \ge 0; \tag{16}$

since both

$w_t^2(x, t), \nabla w(x, t) \cdot \nabla w(x, t) \ge 0, \tag{17}$

(14) and (16) force

$w_t(x, t) = 0, \; \nabla w(x, t) = 0 \; \forall x \in \Bbb R^3, t \ge 0; \tag{18}$

since the first derivatives of $w(x, t)$ all vanish it follows that $w(x, t)$ is constsnt for $x \in \Bbb R^3$, $t \ge 0$; since $w(x, 0) = 0$, we have

$w(x, t) = 0, \; x \in \Bbb R^3, t \ge 0; \tag{19}$

that

$u(x, t) = v(x, t), \; x \in \Bbb R^3, t \ge 0 \tag{20}$

now follows from (11), the definition of $w(x, t)$.