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We consider the boundary value problem for potential in the form: $$-\Delta u(\boldsymbol{x})=0,\quad \boldsymbol{x}\in \mathbb R^3\smallsetminus S,$$ with boundary conditions $$\nabla u(\boldsymbol{x}) = -\delta g(\boldsymbol{x}), \boldsymbol{x}\in \partial S, \\ u(\boldsymbol{x}) \rightarrow 0,\quad as \quad |\boldsymbol{x}|\rightarrow \infty,$$ where $S$ is a sphere. It is possible to prove the regularity of a modified problem on a bounded domain $\Omega$ in the form: $$-\Delta u(\boldsymbol{x})=0,\quad \boldsymbol{x}\in \Omega,$$ $$\nabla u(\boldsymbol{x}) = -\delta g(\boldsymbol{x}),\quad \boldsymbol{x}\in \partial S, \\ u(\boldsymbol{x}) = u_m,\quad \boldsymbol{x}\in \partial\Omega\smallsetminus \partial S,$$ where $u_m$ is measure value of $u$. In the presentation of Prof. Matthias Ehrhardt: A Review of Transparent and Artificial Boundary Conditions Techniques for Linear and Nonlinear Schrödinger Equations there is something similar for a nonlinear problem. When I asked him this question for the elliptic problem, I didn't get an answer. Is there any literature about this problem?

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  • $\begingroup$ what does $\delta g$ mean? $\endgroup$ Commented Nov 19, 2013 at 14:34
  • $\begingroup$ $\delta g$ is measured value of normal derivative on boundary $\partial S$ $\endgroup$ Commented Nov 19, 2013 at 16:37

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On the first problem, if you mean $$ -\Delta u=0\mbox{ in }\mathbb{R}^n\setminus \bar{B}_R $$ with $u\to 0$ as $|x|\to \infty$, that is on the exterior or a ball, not of a sphere, then regularity is standard. Given your solution, it is trivial to prove that it is $C^\infty$ away from $B_R$. Then, you are looking at a problem on a bounded domain, $\Omega=B_{2R}\setminus\bar{B}_R$, with a $C^\infty$ dirichlet boundary condition on $\partial B_{2R}$ ($u$ itself) and a Neumann boundary condition on $S=\partial B_R$. Since the boundary is $C^\infty$, the regularity of $g$ gives that of $u$.

I guess this is probably what you mean, looking at the second problem you suggest. The way you wrote it is confusing: $\mathbb{R}^n\setminus {S}$ has two connected components, inside the ball of boundary $S$ and outside it, so that makes it look like a transmission problem.

So, to summarise:

  • As far as the interior regularity is concerned, that is, at any positive distance from $B_R$, the solution is $C^\infty$
  • Up to the boundary, the solution has the regularity inherited from $g$: If $g\in W^{n-1-\frac{1}{p},p}(S)$ then $u\in W^{n,p}(\Omega)$.

Since this is a ball, it is easy to express this in simple computable terms for $p=2$. Say that $B_R$ is the ball centered at the origin. Compute the Spherical harmonics' moments of $g$ --the Fourier series of $g$ on the sphere. say $g_{n,p}$ with $n\in\mathbb{N}$ and $p\in\{-n,...,n\}$.

Then look at the function $r(s):=\sum_{n=0}^{\infty}\sum_{p=-n..n} (|n|^2+|p|^2)^{s}|g_{n,p}|^2$ with $s\in(-\infty,\infty)$ (see remark below). The biggest $s$ for which this series is convergent gives you the $W^{s,2}$ in which your solution is.

Remark: one needs to be careful with the normalisation of the spherical harmonics there could be an extra $f(n,p)$ factor in the sum depending on the one you choose.

A reference I would look at for this last part would be Nedelec's Acoustic and electromagnetic equations. I don't have it at hand, but I think I remember it is in it.

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