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

where ${n}$ is the outward-pointing normal on ${\partial M}$. The key idea is to use Cartan’s magic formula; since ${d(dV)=0}$, Stokes’ theorem implies $\displaystyle \int_M \mathrm{div} X dV = \int_M L_X(dV) = \int_{\partial M} i_X dV.$

Now ${i_X(dV)}$ can be seen to be ${dS}$ by definition of the volume forms. Indeed, if ${f_1, \dots, f_{n-1}}$ is an oriented basis of ${T_x(\partial M)}$ for ${x \in \partial M}$, then ${n, f_1, \dots, f_{n-1}}$ is an oriented basis for ${T_x(M)}$. This proves the divergence theorem.

A simple corollary of the divergence theorem occurs when we replace ${X}$ by ${fX}$, and use the identity ${\mathrm{div}(fX) = f\mathrm{div} X + Xf}$ proved earlier; we find $\displaystyle \int_{M} f \mathrm{div} X dV + \int_M (Xf) dV = \int_{\partial M} fdS.$

However, since we are interested in the Laplacian, we naturally enough try ${X = \mathrm{grad} g}$. In which case, since ${(\mathrm{grad} g)f = <\mathrm{grad} g, \mathrm{grad} f>}$: $\displaystyle \int_{M} f \Delta g dV + \int_M <\mathrm{grad} g,\mathrm{grad} f> dV = \int_{\partial M} f \frac{\partial g}{\partial n} dS ,$

where ${\frac{\partial g}{\partial n} := <\mathrm{grad} g, n>}$ is the outward normal derivative. Why is this interesting? Let’s consider the case when ${f,g}$ vanish on ${\partial M}$. Let’s also consider the vector spaces ${V_1, V_2}$ defined in the following way. ${V_1}$ is the vector space of smooth functions supported outside ${\partial M}$, and ${V_2}$ is the vector space of smooth vector fields supported outside ${\partial M}$. These are inner product spaces because of the volume element. For instance, given ${X,Y \in V_2}$, define ${(X,Y) = \int dV}$. We have operators: $\displaystyle \mathrm{grad}: V_1 \rightarrow V_2, \ \Delta: V_1 \rightarrow V_1.$

The identity above says that $\displaystyle (f,\Delta g) = (\mathrm{grad} f, \mathrm{grad} g).$

In particular ${(f, \Delta f) = (\mathrm{grad} f, \mathrm{grad} f)}$. What will ultimately turn out is that the right-hand-side of this will be approximately the Sobolev norm of ${f}$ squared, while the left side will be bounded by the Sobolev norm of ${\Delta f}$ times the Sobolev norm of ${f}$. So in particular, ${\Delta}$ will become a bounded-below operator between suitable Sobolev spaces. More on this later.

As another consequence, we find Green’s identity $\displaystyle \int_{M} ( f \Delta g - g \Delta f) dV = \int_{\partial M} \left( f \frac{\partial g}{\partial n} - g \frac{\partial f}{\partial n} \right) dS.$

One can apply this to prove the mean-value-property for harmonic functions on open regions of ${\mathbb{R}^2}$. Given ${f}$ harmonic on a region containing the closure of ${D_r(x)}$, choose ${g(y) = \log |y-x|}$, and apply Green’s identity to an annuluar region around ${x}$. I will cover this in detail at some point in the future.

The main final point I want to make here is that the use of an orientation on ${M}$ is actually irrelevant. As we saw, ${\mathrm{div} X}$ is defined without regards to an orientation, as is ${}$. The point is that we can, on any Riemannian manifold, define a volume element $\displaystyle dV = g^{1/2} | dx^1 \wedge \dots \wedge dx^n|;$

this is a map from ${T_x(M) \times \dots \times T_x(M) \rightarrow \mathbb{R}_{\geq 0}}$ with appropriate homogeneity properties. The point is that, in local coordinates, this volume element transforms kind of like an ${n}$-form, but with an absolute value sign being added to the determinant. In particular, it makes sense to integrate with respect to a volume element.

So to prove the divergence theorem and its corollaries here for an unoriented manifold, note that the claims are local (since one can use a partition of unity), and choose an orientation locally. Then piece everything together.