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.


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?

  • 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. $\endgroup$ – Jitse Niesen 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

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)

  • $\begingroup$ 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. $\endgroup$ – Willie Wong 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. $\endgroup$ – Willie Wong 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 $(-\pi,\pi)$. $\endgroup$ – Yakov Shlapentokh-Rothman Sep 8 '10 at 20:05
  • 1
    $\begingroup$ Yes, that is correct $\endgroup$ – Yakov Shlapentokh-Rothman Sep 8 '10 at 20:11

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


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.

  • $\begingroup$ 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? $\endgroup$ – Beni Bogosel Nov 24 '15 at 22:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.