The problem at hand is to show that,

$\ \nabla^2u = m^2u $

has a unique solution where $\ u$ is specified on the boundary of some $\ V \subseteq \mathbb{R}^3 $.

Using the Divergence Theorem I managed to get to the following (which may be headed down the wrong path entirely):

$\ \int_V \! |\nabla w|^2 \, \mathrm{d}V = - \int_V \! w(\nabla^2 w) \, \mathrm{d}V $

where $\ w = u_1 - u_2 $ and $\ u_1 $ and $\ u_2 $ are assumed to be two distinct solutions to the original PDE on $\ \partial V $. It is here that I am at a standstill.

Any ideas/pointers would be appreciated. Cheers!


Here's a quick finish: note that, since the equation

$\nabla^2u = m^2u \tag{1}$

is linear, for solutions $u_1, u_2$, $w = u_2 - u_1$ is also a solution; if $u_1 = u_2$ on $\partial V$, then $w = 0$ on $\partial V$ as well. Now use the presented formula

$\int_V \! |\nabla w|^2 \, \mathrm{d}V = - \int_V \! w(\nabla^2 w) \, \mathrm{d}V, \tag{2}$

and set $\nabla^2 w = m^2w$ on the left-hand side, yielding:

$\int_V \! |\nabla w|^2 \, \mathrm{d}V = - m^2\int_V \! w^2 \, \mathrm{d}V \le 0; \tag{3}$

(3) shows that $\nabla w = 0$ on $V$, whence $w$ is constant. Since $w = 0$ on $\partial V$, $w = 0$ on $V$ as well, whence $u_1 = u_2$ on $V$.

Well, to really tighten this up we would need to address regularity of the solution and of the region $V$ and its boundary $\partial V$, make sure everything is sufficiently smooth for the arguments to fly, etc., but I believe most of that stuff is pretty typical for equations of the form (1), being a basic elliptic prototype. So I'll leave that discussion for another place.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

  • $\begingroup$ Thank you! I guess the ending step was right in front of my nose :) $\endgroup$ – S Valera Feb 17 '14 at 17:08
  • $\begingroup$ @ S Valera: indeed you were very close! Glad to help out, and thanks for the "acceptance"! $\endgroup$ – Robert Lewis Feb 17 '14 at 17:10

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