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} <X, n> dS ,} (more…)