I'm concerned with a generic uniformly elliptic operator $L$ on $\mathbb{R}^n$. If $L$ is uniformly elliptic and I am studying the equation $Lu=f$ then the way I can deduce regularity on $\mathbb{R}^n$ is via the Fourier transform: $\hat{Lu} = \hat{f}$ which leads to $P(\xi)\hat{u} = \hat{f}$. From this finally I use the assumption that $P(\xi) \geq c |\xi|^2$ to deduce along with Parseval that $\|u\|_{H^2} \lesssim \|f\|_{L^2} + \|u\|_{L^2}$. My question is, why does this become so complicated on bounded domains?

Question: Why can't we simply write $u = \sum_k \phi_k \hat{u}(k)$ as a Fourier series and deduce from the equation that $\|k\|^2|\hat{u}(k)|^2 = |f(k)|^2$ (the Fourier coefficients) and use this to deduce that $$\sum_k (1+|k|^2)^2 |\hat{u}(k)|^2 \lesssim \|u\|_{L^2} + \|f\|_{L^2}$$ where again I've used Parseval's identity. In other words, doesn't everything from the $\mathbb{R}^n$ case just get converted into statements about the Fourier series? (as opposed to tranform).

Hope this is clear! Thanks!

  • $\begingroup$ Ok but I don't want to think about this in terms of difference quotients and integration by parts (as is the approach done by Evans). I feel that one should be able to take advantage of the fact that the operators are constant coefficient to describe what's going on in frequency space. What do I mean by Fourier series? I mean expressing $u$ in an orthnormal basis of say $-Delta u$ with $0$ boundary conditions. $\endgroup$ – Dorian Aug 28 '10 at 17:31
  • $\begingroup$ The fact that operators are constant coefficients means that they are translation invariant and well Fourier transform works well in $\mathbb{R}^n$ because it, too, is translation invariant. It's not very useful on a general domain, especially if the boundary is complicated. $\endgroup$ – Victor Protsak Aug 28 '10 at 17:43
  • $\begingroup$ Sure but fourier series makes sense on any nice bounded domain (nice enough so that you can use as your eigenbasis the eigenfunctions of the laplacian operator with dirichlet boundary conditions for instance). $\endgroup$ – Dorian Aug 28 '10 at 18:05
  • $\begingroup$ The Fourier series makes sense, but the Fourier basis functions are no longer eigenfunctions for your operator, because they don't satisfy the boundary conditions. $\endgroup$ – Nate Eldredge Aug 28 '10 at 19:11
  • $\begingroup$ I'm presuming 0 dirichlet boundary conditions so boundary conditions are non problem... $\endgroup$ – Dorian Aug 28 '10 at 19:28

Dorian, aren't you messing up things a little? Surely you can expand any $L^2$ function in a series of eigenfunctions for the elliptic operator, but please notice that this simple fact already requires quite a detailed theory of elliptic operators on bounded domains. In order to prove the existence of eigenfunctions you must be able to solve the equation $Lu=\lambda u$, and if you want to use your expansion for regularity results, you need to study the properties of the eigenfunctions, their growth etc.

But actually you are right, in a sense. It is indeed possible to prove existence and regularity of solution at the interior of the domain by using essentially the same methods as on the whole space. However, if you want to control the properties of the solution at the boundary, then this requires new tools. As a minimum, in the lucky situation of a smooth boundary, you can reduce to the case of a half space, but no less than that. If you are not convinced, think of the fact that some results cease to be true if you drop the assumption that the boundary is Lipschitz or satisfies some suitable cone condition.

As you suspect, it is also possible to do frequency space analysis much in the same way as on the whole space, but I would not call this easy. There is a beautiful set of notes "Lectures on semiclassical analysis" available on the web, by Evans and Zworsky, see Theorem 3.17 there (they prove interior Schauder estimates using Paley-Littlewood, apparently following a suggestion of H.Smith). I repeat: this is interior regularity, the behaviour at the boundary is substantially more difficult.

  • $\begingroup$ That's a wonderful set of notes. Thank you for pointing it out. $\endgroup$ – Willie Wong Aug 28 '10 at 23:09

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