Divergence form Elliptic PDE Removable Singularity/Regularity Question - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T22:53:01Z http://mathoverflow.net/feeds/question/114531 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114531/divergence-form-elliptic-pde-removable-singularity-regularity-question Divergence form Elliptic PDE Removable Singularity/Regularity Question Spencer 2012-11-26T14:47:55Z 2012-11-28T14:33:24Z <p><strong>Idea</strong></p> <p>Given a $W^{1,2}$ solution to a linear divergence form uniformly elliptic pde with bounded coefficients, standard De Giorgi-Nash-Moser theory tells us that the solution is infact (Holder) continuous. If you have better regularity away from one isolated point, say you are $C^1$ on the puncutered ball, can the solution still fail to be differentiable at that isolated point?</p> <p>The situations I am most familiar with tend to be ones in which one can go from boundedness and continuity to smoothness via Schauder theory and bootstrapping. Here, we already have continuity: But can differentiability fail at a single point? TIt seems out of reach of all the theorems I've seen, which makes me suspect it is false, but I cannot be sure without a counterexample. Does anyone have any ideas?</p> <p><strong>Details</strong></p> <p>In the specific situation I am interested in, I know a bit more. I am considering the following:</p> <p>$u \in W^{1,\infty}(B_1(0)) \cap C^{1,1}_{loc}(B_1(0)\setminus${0}$)$ satisfies weakly the equation</p> <p>$D_i(A_{ij}(x)D_ju) = D_ig^i$ </p> <p>in $B_1(0)$, where $A_{ij},g^i \in L^{\infty}(B_1(0))\cap W^{1,2}_{loc}(B_1(0)\setminus${0}$)$. </p> <p><strong>Questions</strong></p> <p>1) Must $u$ in fact be a $C^1$ solution on $B_1(0)$? </p> <p>2) What about just being differentiable at 0? </p> <p>3) How about even just $u \in W^{2,p}(B_1(0))$ for some $p > 1$?</p> http://mathoverflow.net/questions/114531/divergence-form-elliptic-pde-removable-singularity-regularity-question/114767#114767 Answer by Daniel Spector for Divergence form Elliptic PDE Removable Singularity/Regularity Question Daniel Spector 2012-11-28T12:31:37Z 2012-11-28T12:41:07Z <p>I believe the following is a counterexample:</p> <p>Let $N=1$, $B_1(0)=(-1,1)$, $u(x)=|x|$, then $|u^\prime(x)| = 1$, $u^{\prime \prime}(x) = 2\delta_0$, </p> <p>and </p> <p>$u^\prime(x) = 2H(x)-1$,</p> <p>where $H$ is the Heaviside function, $H \in L^\infty \cap W^{1,2}_{loc}((-1,1)\setminus {0})$.</p> <p>Thus $u$ solves $(\frac{1}{2} u^\prime)^\prime = H^\prime$,</p> <p>which is the PDE in one dimension, with $H$ in the right space, $u$ is smooth away from the origin and $u \in W^{1,\infty}(-1,1)$, but $u^{\prime\prime}$ is only a measure and so not in $L^p(-1,1)$ for any $p>1$.</p> <p>In particular, $u$ is not $C^1$, $u$ is not smooth at the origin, and $u^{\prime\prime} \notin L^p(-1,1)$.</p> <p>More generally, the computation extends into more dimensions, as far as I can tell.</p> http://mathoverflow.net/questions/114531/divergence-form-elliptic-pde-removable-singularity-regularity-question/114771#114771 Answer by Denis Serre for Divergence form Elliptic PDE Removable Singularity/Regularity Question Denis Serre 2012-11-28T13:25:16Z 2012-11-28T13:25:16Z <p>The theory of removable singularities began with Laurent Véron. It was continued by H. Brézis and others. </p> <p>When the elliptic equation is linear, the important ingredient is the codimension of the subset where you are lacking information. Codimension one is too big (see the counter-example in Dan Spector' answer). Usually, codimension 2 is OK, because then the subset has zero capacity.</p> <p>When the equation is non-linear, for instance $\Delta u+|u|^{p-1}u=0$, the situation depends upon the exponent $p>1$ and the integrability that you <em>a priori</em> know about $\nabla u$ or $u$. Say that the equation is posed the punctured ball. It admits a radial solution $$u=\omega r^{-\alpha},\qquad \alpha=\frac2{p-1},\qquad\omega=(\alpha(\alpha+2-d))^{\frac1{p-1}}.$$ If you know (or assume) for instance that $u\in W^{1,2}$, then $\alpha&lt;\frac d2-1$, that is $\frac2{p-1}&lt;\frac d2-1$. In other words, if $p\le\frac{d+2}{d-2}$, then the singularity at origin is removable.</p> http://mathoverflow.net/questions/114531/divergence-form-elliptic-pde-removable-singularity-regularity-question/114774#114774 Answer by Deane Yang for Divergence form Elliptic PDE Removable Singularity/Regularity Question Deane Yang 2012-11-28T14:33:24Z 2012-11-28T14:33:24Z <p>With a linear elliptic PDE, there's no way to bootstrap. What you see is what you get. The regularity of $A^{ij}\partial_j u$ cannot be made any better than the regularity of $g^i$. So if all you assume about $g^i$ is that it is $L^\infty$, then that's all you get for $A^{ij}\partial_j u$. Once this observation is made, it's easy to find the simplest possible example (where $u$ is Lipschitz, $\partial u$ is $L^\infty$, and you just set $A^{ij} = \delta^{ij}$ and $g^i = \partial_iu$). And that's exactly the one found by Daniel, which works in all dimensions.</p>