9
$\begingroup$

I'm seeking a simple example of where elliptic (preferably linear) boundary regularity fails due to a simple kink in the domain.

So far my gueses were to look at $-\Delta u = f$ on $[0,2\pi] \times [0,2\pi]$ with $0$ Dirichlet boundary conditions and choose an $f$ which was far from $0$. This hasn't seem to produce any results (I was checking regularity directly by the method of Fourier series).

So more precisely, I would like an example where

1) $Lu = f$ in $\Omega \subset \mathbb{R}^n$ with $f$ smooth

2) $L$ is elliptic and $u = 0$ on $\partial \Omega$

3) $\Omega$ is not smooth and consequently $u$ is not smooth up to the boundary.

$\endgroup$
7
$\begingroup$

This is the same idea as timur's answer but with more details and less generality. A frequent test problem in numerical analysis is the Poisson equation $-\Delta u = 1$ on the L-shaped domain

$\Omega = ([-1,1] \times [-1,1]) \setminus ([-1,0] \times [-1,0])$

with homogeneous Dirichlet boundary conditions: $u = 0$ on $\partial\Omega$. The solution has a singularity at the origin: it is continuous but not differentiable. More precisely, close to the origin we have

$u(r,\theta) \approx r^{2/3} \sin \frac{2\theta+\pi}{3}$

in polar coordinates, according to equation (1.6) in http://eprints.ma.man.ac.uk/894/02/covered/MIMS_ep2007_156_Sample_Chapter.pdf (sample chapter from Elman, Silvester and Wathen, Finite Elements and Fast Iterative Solvers, Oxford University Press, 2005).

Added: I don't know the details and I don't have time to do the necessary computations, but I think that you can solve the PDE by converting the Laplacian to polar coordinates and applying separation of variables. I imagine that you get that

$u(r,\theta) = r^{2n/3} \sin \frac{2n}{3} (\theta + \frac{1}{2}\pi)$

with $n$ a positive integer satisfies the boundary conditions at $r=0$ and $\theta=-\pi/2$ and $\theta=\pi$ (as Dorian comments below, these are all harmonic functions, so there must be something else). Then take a linear combination of those to match the conditions on the rest of the boundary of the L-shaped domain. Close to the origin, the $n=1$ term dominates. Perhaps somebody else can confirm / amend?

$\endgroup$
4
  • 1
    $\begingroup$ Hey this is very nice and explicit. Can you help me find a source or understand how one can say that $u(r,\theta) \sim r^{2/3} \sin \frac{2\theta + \pi}{3}$ near the origin? $\endgroup$
    – Dorian
    Sep 15 '10 at 14:34
  • $\begingroup$ I added a bit, but I'm afraid I don't know more than that. Also for references I can't do better than advising you to follow the reference I already gave, which points to the finite-element book of Strang & Fix. $\endgroup$ Sep 15 '10 at 21:02
  • $\begingroup$ THanks for the addition. However I don't think you have enough freedom in your linear combinations (only a constant) to get the right conditons on all the other parts of the domain. $\endgroup$
    – Dorian
    Sep 16 '10 at 1:01
  • 1
    $\begingroup$ My concern is that all of those functions you've written down are harmonic so I don't see how they can satisfy Poisson's equation. There must be some extra term hiding somewhere. $\endgroup$
    – Dorian
    Oct 4 '10 at 16:30
5
$\begingroup$

You might consider this cheating, but at some point this tripped me up: What is the first Dirichlet eigenvalue and eigenfunction for $\Delta$ on the ball minus the origin? Well, since points have measure $0$, from the min-max principle it is the same as the first eigenvalue and eigenfunction for $\Delta$ on the ball. However, the eigenfunction certainly doesn't vanish at the origin. What went wrong -> The boundary of the punctured ball is a sphere and a point, which is not smooth.

EDIT: I forgot to note that the dimension should be 2 or more for this to make sense (see comments below)

$\endgroup$
10
  • $\begingroup$ Hmm, but in one dimension if we consider your example on $[-\pi, \pi]$ then the first eigenfunction indeed does vanish (it's just $\sin(x)$) so is it so surprising that it should vanish at the origin? $\endgroup$
    – Dorian
    Sep 8 '10 at 17:40
  • 1
    $\begingroup$ In one dimension, the one-point boundaries are smooth. $\endgroup$ Sep 8 '10 at 17:53
  • 1
    $\begingroup$ No. Classic theorem in Fredholm theory states that the first eigenfunction of a self-adjoint, uniformly elliptic operator on some domain is bounded, continuous, and positive. $\endgroup$ Sep 8 '10 at 18:52
  • 1
    $\begingroup$ @Dorian No, as pointed out by Willie Wong, among other things it must be positive on $(-\pi,\pi)$. According to wwwmaths.anu.edu.au/~hassell/efns.colloq.pdf the first eigenfunction on $[0,1]$ is $u(x) = \sqrt{2}\sin(\pi x)$. So by scaling, up to a constant the eigenfunction on $[-\pi,\pi]$ is $\sin((1/2)(x+\pi))$. Note that this is strictly positive on $(-\pi,\pi)$. $\endgroup$ Sep 8 '10 at 20:05
  • 1
    $\begingroup$ Yes, that is correct $\endgroup$ Sep 8 '10 at 20:11
4
$\begingroup$

There is an explicit characterization of regularity loss on polygonal domains in e.g. Grisvard's book.

Consider $-\Delta u=f$ on a polygonal domain $\Omega$ with the homogeneous Dirichlet boundary condition and with $f\in L^2(\Omega)$. Now consider the set $X=(-\Delta)^{-1}L^2(\Omega)\supset H^2(\Omega)$. If $\Omega$ is reasonably smooth we know that $X=H^2(\Omega)$. Let us say a vertex of $\Omega$ is re-entrant if the internal angle is larger than $\pi$. Then the result I mentioned says that

$X=H^2(\Omega)\oplus\mathrm{span}(\phi_1,\ldots,\phi_m)$

where each $\phi_i\in H^1(\Omega)\setminus H^2(\Omega)$ corresponds to a re-entrant vertex of the polygon. The precise regularity of $\phi$ depends on the angle. So the regularity on polygon is almost as good as that on smooth domain; regularity loss is associated to only a finitely many singular functions. In 3 dimension it is not true since there can be "re-entrant edge" and infinitely many singular functions will be associated to it.

$\endgroup$
4
  • $\begingroup$ Would you mind saying more explicitly what this regularity loss is? I don't have a copy of this book. $\endgroup$
    – Dorian
    Sep 15 '10 at 5:24
  • $\begingroup$ Thanks. I don't understand enough numerics language to have a good sense of what you're saying but it seems to be in the right direction. $\endgroup$
    – Dorian
    Sep 15 '10 at 14:37
  • $\begingroup$ I don't think it's numerics language. $\endgroup$
    – timur
    Apr 28 '11 at 14:22
  • 1
    $\begingroup$ Timur, can you please indicate which theorems in Grisvard's book refer to the facts you state in your answer? $\endgroup$ Nov 24 '15 at 22:18

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .