Proving Uniqueness of Solution to PDE on Boundary of Volume 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!
 A: 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!!!
