I have now discussed what the Laplacian looks like in a general Riemannian manifold and can thus talk about the basic equations of mathematical physics in a more abstract context. Specifically, the key ones are the Laplace equation

\displaystyle \Delta u = 0

for {u} a smooth function on a Riemannian manifold. Since {\Delta = \mathrm{div} \mathrm{grad}}, this often comes up when {u} is the potential energy function of a field which is divergence free, e.g. in electromagnetism. The other major two are the heat equation

\displaystyle u_t - \Delta u = 0

for a smooth function {u} on the product manifold {\mathbb{R} \times M} for {M} a Riemannian manifold, and the wave equation

\displaystyle u_{tt} - \Delta u = 0

in the same setting. (I don’t know the physics behind these at all, but it’s probably in any number of textbooks.) We are often interested in solving these given some kind of boundary data. In the case of the Laplace equation, this is called the Dirichlet problem. In 2-dimensions for data given on a circle, the Dirichlet problem is solved using the Poisson integral, as already discussed. To go further, however, we would need to introduce the general theory of elliptic operators and Sobolev spaces. This will heavily rely on the material discussed earlier on the Fourier transform and distributions, and before plunging into it—if I do decide to plunge into it on this blog—I want to briefly discuss why Fourier transforms are so important in linear PDE. Specifically, I’ll discuss the solution of the heat equation on a half space. So, let’s say that we want to treat the case of {\mathbb{R}_{\geq 0} \times \mathbb{R}^n}. In detail, we have a function {u(x)=u(0,x)}, continuous on {\mathbb{R}^n}. We want to extend {u(0,x)} to a solution {u(t,x)} to the heat equation which is continuous on {0 \times \mathbb{R}^n} and smooth on {\mathbb{R}_+^{n+1}}. To start with, let’s say that {u(0,x) \in \mathcal{S}(\mathbb{R}^n)}. The big idea is that by the Fourier inversion formula, we can get an equivalent equation if we apply the Fourier transform to both sides; this converts the inconvenience of differentiation into much simpler multiplication. When we talk about the Fourier transform, this is as a function of {x}. So, assuming we have a solution {u(t,x)} as above:

\displaystyle \hat{u}_t = \widehat{\Delta u} = -4\pi^2 |x|^2 \hat{u}.

Also, we know what {\hat{u}(0,x)} looks like. So this is actually a linear differential equation in {\hat{u}( \cdot, x)} for each fixed {x} with initial conditions {\hat{u}(0,x)}. The solution is unique, and it is given by

\displaystyle \hat{u}(t,x) = e^{-4 \pi^2 |x|^2 t} \hat{u}(0,x).

Now recall that multiplication on the Fourier transform level corresponds to conveolution, and the Fourier transform of {K(t,x) = (4 \pi t)^{-n/2} e^{- |x|^2/ (4 t)}} is {e^{-4 \pi^2 |x|^2 t}}. As a result, given a putative solution {u(t,x)}, we have determined {u(t,x)} by

\displaystyle u(t,x) = (K(t, \cdot) \ast u) = (4 \pi t)^{-n/2} \int_{\mathbb{R}^n} e^{- |y-x|^2/ (4 t)} u(0, y) dy.

So we have a candidate for a solution. Conversely, if the boundary data {u \in L^1(\mathbb{R}^n)} alone, it is easy to check by differentiation under the integral (justified by the rapid decrease of the exponential) that we have something satisfying the heat equation in the upper half-space. Moreover {||u(t, \cdot) - u(0, \cdot)||_{L^1} \rightarrow 0} as {t \rightarrow 0} by general facts about approximation to the identity and a look at the definition of {K(t, x)}—note that {K(\sqrt{t}, x)} is just the orthodox version of an approximation to the idnetity. So, we have found a way to solve the heat equation on {\mathbb{R}^{n+1}}. It thus seems that the way to solve equations such as the heat equation by convolution with appropriate kernels. In fact, this is more generally true of nonhomogeneous constant-coefficient linear PDE on {\mathbb{R}^n} (we’re forgetting about boundary value problems). Suppose we are given a partial differential operator {P} with constant coefficients, i.e.

\displaystyle Pf = \sum_{|a| \leq k} c_a D^a f ,

where the {a}‘s are multi-indices. Then it is immediate that {P} extends to an operator on distributions. Moreover,

\displaystyle \boxed{ P(\phi \ast f) = P\phi \ast f = \phi \ast Pf }

whenever {\phi} is a distribution and {f \in \mathcal{S}}. (This is clear whenever {\phi \in \cal{S}}; in general any distribution can be approximated in the weak* sense by distributions by convolving with an approximation to the identity.) As a result, if we have a fundamental solution {\phi}, i.e. one with

\displaystyle P \phi = \delta

we can get a solution to any equation of the form {Pf = g} for {g \in \mathcal{S}} by taking

\displaystyle f = g \ast \phi,

which is not only a distribution but also a polynomially increasing {C^{\infty}} one. So we can solve any constant-coefficient PDE given a fundamental solution. There is a big theorem of Malgrange and Ehrenpreis that fundamental solutions always exist to constant-coefficient linear PDE. However, the above statement about solving PDEs can actually be proved in a more elementary fashion; perhaps this will be a future topic. For now, however, I want to show that the Gauss kernel {K(t,x)} is actually a fundamental solution to the heat equation, once it is extended to {\mathbb{R}^{n+1}} with {K(t,x) \equiv 0} for {t \leq 0}. (This is no longer smooth, but it is still a distribution.) We need to show that

\displaystyle \int_{\mathbb{R}^{n+1}} K(t,x) \left( - \frac{d}{dt} - \Delta \right)u(t,x) dt dx = u(0).

Let’s take the integral {I_{\epsilon}} where {t} is integrated over {[\epsilon, \infty)}; then by integration by parts

\displaystyle I_{\epsilon} = \int_{\epsilon}^{\infty} \int_{\mathbb{R}^n} u \left( \frac{d}{dt} - \Delta \right) K dt dx + \int_{\mathbb{R}^n} K(\epsilon,x) u(\epsilon,x).

Since {K} is a solution to the heat equation on {\mathbb{R}^{n+1}_+} (e.g. look at the Fourier transform) it is the second integral that is nonzero. We can write this as

\displaystyle \int_{\mathbb{R}^n} K(\epsilon, x)( u(\epsilon,x) - u(0,0)) dx + u(0)

and it is easy to see (the same approximation to the identity argument) that the former term tends to zero as {\epsilon \rightarrow 0}.

So we indeed have a fundamental solution to the heat equation. It thus seems fair that we get solutions to it by convolving with the Gauss kernel.