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Let us consider the unit ball $B^n$ of $\mathbf{R}^n$ and $K = B^{n-1}(0,1/2) \times \{0\}$ the ball of dimension $n-1$ and radius $1/2$ lying on the equator.

We wish to solve the Dirichlet problem (for harmonic functions) on $B^n \setminus K$ with boundary value $f = 0$ on $K$, and $f$ (let's say) is smooth on $\partial B^n$. Such a solution exists, is unique and we can find it within the class of Sobolev function.

My question is how this solution behaves near $K$. Is it Hölder continuous ? What is the best exponent of Hölder continuity one can expect ?

More generally, are there known criterias for the regularity at the boundary (not abstract continuity like with the Wiener criterion, but with a control of the modulus of continuity) ?

Thanks

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  • $\begingroup$ $K$ is not contained in the boundary of $B^n$; it does not even touch the boundary. $\endgroup$
    – Florian
    Commented Nov 4, 2011 at 9:38
  • $\begingroup$ @Florian : yes, the Dirichlet problem is solved in the domain $B^n \setminus K$. The solution is is harmonic on $B^n \setminus K$ and continuous on $\overline{B^n}$. $\endgroup$
    – vizietto
    Commented Nov 4, 2011 at 9:52

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Far from the rim of the inner disk one has a good regularity. At the rim though, I think you can get the solution behaviour from explicit constructions of potentials induced by a charged ellipsoid and collapsing one of axes, e.g., as given in Kellogg's book.

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    $\begingroup$ Marius Mitrea has a bunch of papers on the regularity of the Dirichlet problem on manifolds. $\endgroup$
    – Nilima Nigam
    Commented Nov 25, 2011 at 16:21
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Grisvard's book has an extensive discussion of 2d elliptic problems with corners. Your problem is singular at the rim, and the singularity there is essentially the same one as a 2d problem with a 360 degree corner.

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You can expect the solution to be at most $C^{1/2}$ at the rim of $K$, and $C^\infty$ at each side of $K$ (but only Lipschitz across).

At the rim of $K$, since the rim is smooth, you can directly apply the results in https://arxiv.org/pdf/1402.1098

The exponent coincides with the homeneity that you would have in 2D with a 360 degree corner, namely, $u = \operatorname{Re}(z^{1/2})$.

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