Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\Omega\subset\mathbb{R}^2$ a bounded set with Lipshitz continuous boundary.

If $$z\in L_0^2(\Omega)=\{v\in L^2(\Omega):\int_\Omega v\,dx=0\},$$

it is true that exists $\phi\in H^{1+\delta}(\Omega)$, $\delta\in (\frac{1}{2},1]$, such that $\Delta \phi=\nabla\cdot (\nabla \phi)=z$ and $||\nabla \phi||_{\delta,\Omega}\leq c||z||_{0,\Omega}$?

I known that when $\Omega$ is a convex set, what is written above holds with $\delta=1$, but i'm not sure if it holds in a more general case. Thanks.

share|improve this question
    
No, I believe pure homogeneous Neumann problem does not have higher than $H^{1+\frac{1}{2}}$ regularity, assuming the worst case, in nonconvex domain. –  Shuhao Cao Apr 28 '13 at 20:19

1 Answer 1

As stated, the problem can be reduced to the convex case by letting $B$ be a ball containing $\Omega$, and extending $z$ by $0$ to $B\setminus \Omega$. Or, just convolve $z$ with the fundamental solution of the Laplacian on $\mathbb R^2$; the convolution will be in $H^2$ and the shape of $\Omega$ does not matter.

The comment by Shuhao Cao mentions the Neumann problem. If $\partial v/\partial n$ is required to vanish on $\partial\Omega$ then the story is different, but I don't see any such condition in the OP. (Maybe it's missing.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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