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

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