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I just want to know for a domain with smooth boundary whether exterior Newtonian potential has singular analytic extension into the domain or into a part of the domain.

Definition of Newtonian potential of domain $\Omega$ at x is defined to be $u(x)=\int_{R^{n}}\Gamma(|x-y|)\chi_{\Omega}(y)dy$. When $x\in \Omega$ we say it is the interior Newtonian potential, otherwise we call it exterior Newtonian potential. Here $\Gamma(x)=\frac{1}{|x|^{n-2}}$ is the fundamental solution of Laplace, $\chi_{\Omega}$ is the characterization fucntion of $\Omega$

For example, the exterior Newtonian potential of the ball $B(a,R)$ is just $\frac{R^{n-2}}{n(n-2)}\frac{1}{|x-a|^{n-2}}$ when $n>2$. Even though it is Newtonian potential of the ball out of it, we can regard it as a singular analytic function on $R^{n}$. So we can say the exterior newtonian potential of the ball have singular analytic extension in it.

Or in particular: what is the exterior Newtonian potential of the domain which is formed by moving the center of a ball along a arbitrary small smooth curve? Does this case allow singular analytic extension in the domain?

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  • $\begingroup$ Can someone explain her reasons for closing this question?? $\endgroup$
    – Alexandre Eremenko
    Commented Oct 16, 2014 at 19:57

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You must specify more exactly what do you mean by "singular analytic" (what singularities are allowed). Some version of this problem was investigated in arXiv:1309.5483, and in the literature mentioned there. Main results are in dimension 2 (logarithmic potential) but there are also results for higher dimension.

EDIT. With the definition you give in the comment to my answer, the domains you are asking for are the same as the "quadrature domains". See, for example, arXiv:1202.5013 and references there on the higher dimensional case. Very little is known about them in dimensions $\geq 3$. But in dimension $2$ (with log kernel) the understanding is much better. There is a nice book on them for dimension 2, Varchenko, Etingof, Why the boundary of a round drop becomes a curve of order four? MR1190012.

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  • $\begingroup$ Thanks for your concern. The "singular analytic" in my mind is the function that is analytic in the domain except several points and those points are unremovable, i.e you can not define the value at those points to make the function to be analytic in the whole domain. For exmaple, in the case of a ball the center is an unremovable singular point of analytic function $\frac{1}{|x-a|^{n-2}}$, which is the exterior potential of the ball. $\endgroup$
    – user60554
    Commented Oct 16, 2014 at 17:11
  • $\begingroup$ Thank you. What if I add the assumption that the boundary is analytic or piecewice analytic? I know I can extend the exterior potential analytically across the boundary, but I need to find some singular point in the extension. $\endgroup$
    – user60554
    Commented Oct 16, 2014 at 20:26
  • $\begingroup$ The boundary of these domains is analytic. Look in the literature I recommended. $\endgroup$
    – Alexandre Eremenko
    Commented Oct 16, 2014 at 22:07

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