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 ,}

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 {<X,n>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} f<X, n>dS.

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 <X,Y> 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 {<X,n>}. 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.

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