One of the basic properties of the Laplacian is that given a compact Riemannian manifold-with-boundary (to which all this ${\mathrm{div}, \mathrm{grad}, \Delta}$ business applies equally), then for ${u}$ vanishing on the boundary, the ${L^2}$ inner product ${(\Delta u, u)}$ is fairly large relative to ${u}$. As an immediate corollary, if ${u}$ satisfies the Laplace equation ${\Delta u= 0}$ and vanishes on the boundary, then ${u}$ is identically zero.

It turns out that the proof of this will require the divergence theorem. This is a familiar fact from multivariable calculus, but it generalizes to ${n}$-dimensions nicely as a corollary of Stokes theorem and some of the other machinery thus developed.

So, let’s choose an oriented Riemannian manifold ${M}$ of dimension ${n}$ with boundary ${\partial M}$. There is a volume form ${dV}$ because of the choice of orientation globally defined. On ${\partial M}$, there is an induced Riemannian metric and an induced orientation, with a corresponding volume form ${dS}$ on ${\partial M}$. If ${X}$ is a compactly supported vector field, the divergence theorem states that

$\displaystyle \boxed{ \int_M \mathrm{div} X dV = \int_{\partial M} dS ,}$ (more…)