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I need an explicit formula for the rotationally invariant solution of $\Delta u=0$ in cylindrical coordinate $(r,\theta,z)$ for a domain like $D=[0,2]\times [0,2\pi]\times [-2,2] - [0,1]\times [0,2\pi]\times [-1,1]$ (a cylinder from which a smaller cylinder is cut out), with zero boundary conditions on $[0,2]\times [0,2\pi]\times \{-2\}$ and $[0,2]\times [0,2\pi]\times \{2\}$ and $[0,1]\times [0,2\pi]\times \{-1\}$ and $[0,1]\times [0,2\pi]\times \{1\}$ and given boundary conditions $f_1(z)$ on $\{1\}\times [0,2\pi]\times [-1,1]$ and $f_2(z)$ on $\{2\}\times [0,2\pi]\times [-2,2]$.

Separation of variables doesn't work since the domain is not in product form.

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    $\begingroup$ Why the vote to close? Does the person who voted to close know such solution and thinks it is trivial, or knows that it does not exist, and also thinks it is trivial? $\endgroup$
    – Alexandre Eremenko
    May 24, 2014 at 12:16
  • $\begingroup$ "Analytical" is an unfortunate choice of words, as this is physics speak for what mathematicians would call "closed form" or "explicit" solutions, and (worse) it suggests something else. $\endgroup$
    – Christian Remling
    May 24, 2014 at 19:05
  • $\begingroup$ I agree, but this is not a reason to close. I suppose an answer with an integral or a series will be satisfactory. Like the Poisson forumla for the ball. $\endgroup$
    – Alexandre Eremenko
    May 24, 2014 at 19:37
  • $\begingroup$ @AlexandreEremenko: I wasn't referring to that (and it wasn't me, either); I was just suggesting a cosmetic edit. $\endgroup$
    – Christian Remling
    May 24, 2014 at 19:45
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    $\begingroup$ I am convinced that the term "analytical" was not appropriate. I changed the term to "explicit". $\endgroup$
    – Erfan Salavati
    May 25, 2014 at 19:21

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