Apologies for the embarrassingly bad pun in the title.

Distributions in general

First, it’s necessary to talk about distributions on an arbitrary open set {\Omega \subset \mathbb{R}^n}, which are not necessarily tempered. In particular, they may “grow arbitrarily” as one approaches the boundary. So, instead of requiring a functional on a Schwarz space, we consider functionals on {C_0^{\infty}(\Omega),} the space of smooth functions compactly supported in {\Omega}. However, we need some notion of continuity, which would require a topology on {C_0^{\infty}(\Omega)}. There is now the tricky question of how we would require completeness of the topological vector space {C_0^{\infty}(\Omega)}, which we of course desire. We can get such a topology by talking about “strict inductive limits” and whatnot, but since I don’t really find that particularly fun, I’ll sidestep it (but not really—most of the ideas will still remain).

Anyway, the idea here will be to consider auxiliary spaces {C^{\infty}(K)} for {K \subset \Omega} compact. This is the space of smooth functions {f: \Omega \rightarrow \mathbb{R}} which are supported in {K}. We give the space a Frechet topology by the family of seminorms

\displaystyle ||f||_a := \sup_K |D^a f|. (more…)

So, we’re given a constant-coefficient operator

\displaystyle P(D) = \sum_{a: |a| \leq k} C_a D^a.

(Note that I’m using slightly different notations than before.)

Let {P} be the polynomial {P(\xi) := \sum_{a: |a| \leq k} C_a (2\pi i \xi)^a}; then the notation {P(D)} suggests the substitution of differential operators in a polynomial, as it should, even though technically {D} should be replaced with {D/2 \pi i}. This should not cause any confusion. We are interested in finding local solutions to an equation

\displaystyle P(D) g = f,

where {f} is smooth. This will follow when we prove the Malgrange-Ehrenpreis theorem and get a fundamental solution. For now, however, we’ll avoid that.

Informal motivation

Anyway, we’ll first proceed informally: the equation {P(D)g = f} is equivalent to

\displaystyle P \hat{g} = \hat{f},

so formally we write

\displaystyle \hat{g} = \frac{\hat{f}}{P}, \ i.e. \ g(x) = \int_{\mathbb{R}^n} e^{2 \pi i x.t} \frac{ \hat{f}(t) }{ P(t) } dt \ (*) .

This, of course, is utter nonsense. The polynomial {P} will generally have roots, in which case we need to find a way to make sense of this. (more…)

Back to elliptic regularity. We have a constant-coefficient partial differential operator {P = \sum_{a: |a| \leq k} C_a D^a} which is elliptic, i.e. the polynomial

\displaystyle Q(\xi) = \sum_{a: |a| \leq k} C_a \xi^a

satisfies {|Q(\xi)| \geq \epsilon |\xi|^k} for {|\xi|} large. We used this last property to find a near-fundamental solution to {P}. That is, we chose {E} such that {\hat{E} = (1-\varphi) Q^{-1}}, where {\varphi} was our arbitrary cut-off function equal to one in some neighborhood of the origin. The point of all this was that

\displaystyle P(E) = \delta - \hat{\varphi}.

In other words, {E} is near the fundamental solution. So given that {Pf = g}, we can use {E} to “almost” obtain {f} from {g} by convolution {E \ast g}—if this were exact, we’d have the fundamental solution itself.

We now want to show that {E} isn’t all that badly behaved.

The singular locus of the parametrix

We are going to show that {\mathrm{sing} E = \{0\}}. The basic lemma we need is the following. Fix {m}. Consider a smooth function {\phi} such that, for each {a}, there is a constant {M_a} with

\displaystyle |D^a \phi(x)| \leq M_a (1+|x|)^{m-|a|};

then this is a distribution, but it is not necessarily a Schwarz function. And {\hat{\phi}} cannot be expected to be one, thus. Nevertheless:

Lemma 1 {\hat{\phi}} is regular outside the origin. (more…)

The next application I want to talk about here of Fourier analysis is to (a basic case of) ellipic regularity. Later we will use refinements of these techniques to obtain all kinds of estimates. Anyway, for now, a partial differential operator

\displaystyle P = \sum_{a: |a| \leq k} C_a D^a

is called elliptic if the homogeneous polynomial

\displaystyle \sum_{a: |a| = k} C_a \xi^a, \quad \xi = (\xi_1, \dots, \xi_n)

has no zeros outside the origin. For instance, the Laplace operator is elliptic. Later I will discuss how this generalizes to other PDEs, and how this polynomial becomes the symbol of the operator. For the moment, though let’s define {Q(\xi) = \sum_{a: |a| \leq k} C_a (2 \pi i \xi)^a}. The definition of {Q} such that

\displaystyle \widehat{ Pf } = Q \hat{f},

and we know that {|Q(\xi)| \geq \epsilon |\xi|^k} for {|\xi|} large enough. This is a very important fact, because it shows that the Fourier transform of {Pf} exerts control on that of {f}. However, we cannot quite solve for {\hat{f}} by dividing {\widehat{Pf}} by {Q} because {Q} is going to have zeros. So define a smoothing function {\varphi} which vanishes outside a large disk {D_r(0)}. Outside this disk, an estimate {|Q(\xi)| \geq \epsilon |\xi|^k} will be assumed to hold. (more…)

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…)