A simple example where elliptic boundary regularity fails due to a kink in the domain - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:45:53Z http://mathoverflow.net/feeds/question/38054 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38054/a-simple-example-where-elliptic-boundary-regularity-fails-due-to-a-kink-in-the-do A simple example where elliptic boundary regularity fails due to a kink in the domain Dorian 2010-09-08T13:44:47Z 2010-10-08T10:27:20Z <p>I'm seeking a simple example of where elliptic (preferably linear) boundary regularity fails due to a simple kink in the domain.</p> <p>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).</p> <p>So more precisely, I would like an example where </p> <p>1) $Lu = f$ in $\Omega \subset \mathbb{R}^n$ with $f$ smooth </p> <p>2) $L$ is elliptic and $u = 0$ on $\partial \Omega$ </p> <p>3) $\Omega$ is <em>not smooth</em> and consequently $u$ is not smooth up to the boundary.</p> http://mathoverflow.net/questions/38054/a-simple-example-where-elliptic-boundary-regularity-fails-due-to-a-kink-in-the-do/38075#38075 Answer by Yakov Shlapentokh-Rothman for A simple example where elliptic boundary regularity fails due to a kink in the domain Yakov Shlapentokh-Rothman 2010-09-08T16:48:50Z 2010-09-08T19:11:32Z <p>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.</p> <p>EDIT: I forgot to note that the dimension should be 2 or more for this to make sense (see comments below)</p> http://mathoverflow.net/questions/38054/a-simple-example-where-elliptic-boundary-regularity-fails-due-to-a-kink-in-the-do/38774#38774 Answer by timur for A simple example where elliptic boundary regularity fails due to a kink in the domain timur 2010-09-15T04:06:13Z 2010-09-15T06:35:27Z <p>There is an explicit characterization of regularity loss on polygonal domains in e.g. Grisvard's book.</p> <p>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 $X=(-\Delta)^{-1}L^2(\Omega)\subseteq H^2(\Omega)$. 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 </p> <p>$X=H^2(\Omega)\oplus\mathrm{span}(\phi_1,\ldots,\phi_m)$</p> <p>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.</p> http://mathoverflow.net/questions/38054/a-simple-example-where-elliptic-boundary-regularity-fails-due-to-a-kink-in-the-do/38803#38803 Answer by Jitse Niesen for A simple example where elliptic boundary regularity fails due to a kink in the domain Jitse Niesen 2010-09-15T10:38:09Z 2010-10-08T10:27:20Z <p>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 </p> <p>$\Omega = ([-1,1] \times [-1,1]) \setminus ([-1,0] \times [-1,0])$ </p> <p>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 </p> <p>$u(r,\theta) \approx r^{2/3} \sin \frac{2\theta+\pi}{3}$ </p> <p>in polar coordinates, according to equation (1.6) in <a href="http://eprints.ma.man.ac.uk/894/02/covered/MIMS_ep2007_156_Sample_Chapter.pdf" rel="nofollow">http://eprints.ma.man.ac.uk/894/02/covered/MIMS_ep2007_156_Sample_Chapter.pdf</a> (sample chapter from Elman, Silvester and Wathen, Finite Elements and Fast Iterative Solvers, Oxford University Press, 2005).</p> <p><strong>Added:</strong> 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</p> <p>$u(r,\theta) = r^{2n/3} \sin \frac{2n}{3} (\theta + \frac{1}{2}\pi)$</p> <p>with $n$ a positive integer satisfies the boundary conditions at $r=0$ and $\theta=-\pi/2$ and $\theta=\pi$ (<em>as Dorian comments below, these are all harmonic functions, so there must be something else</em>). 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?</p>