I have a problem where the domain is like a box with a tube missing - e.g.

0< x<1,0< y<1, 0< z<1 less the region (x-0.5)^2+(y-0.5)^2 < 0.25

In order to solve Laplace's equation in this domain it was mapped to one which can be thought of as simply a box. This was done because the initial domain was not simply connected - and so Laplace's equation could not be solved in it. Is this correct? I haven't been able to find a simple clarification of this online.

In this link on wolframalpha, http://mathworld.wolfram.com/SimplyConnected.html, my initial domain is like the far right image - can Laplace's equation be solved in a domain such as that?

  • $\begingroup$ Can you add the image of what you talking about and the question verbatim? $\endgroup$
    – dustin
    Mar 19, 2015 at 15:15
  • $\begingroup$ Yes, if it is a tube (or any open domain). You can prove the existence and the uniqueness by constructing an auxiliary problem in that tube and glue using continuity and the boundary conditions of the original problem. If it's a point, then no. $\endgroup$
    – Shuhao Cao
    Mar 19, 2015 at 15:17
  • $\begingroup$ @ShuhaoCao Might you have a problem with that method when considering the problem which characterizes $Re(1/z)$ on an annulus centered at zero? (I'm not sure where you are extending to, exactly.) $\endgroup$
    – Ian
    Mar 19, 2015 at 15:22
  • $\begingroup$ I've not used the maths stack exchange before - can someone point me in the direction of how to add properly formatted equations to the question and then I'll update it with more detail. $\endgroup$
    – Kvothe
    Mar 19, 2015 at 15:34
  • $\begingroup$ @Kvothe Look up MathJaX. $\endgroup$
    – Ian
    Mar 19, 2015 at 16:34

2 Answers 2


You can solve the Laplace equation on some domains which are not simply connected. For instance, you can solve the Laplace equation on an annulus. However, you can't solve (the classical Dirichlet problem for) the Laplace equation on a domain whose boundary is "too thin". For instance, the Dirichlet problem for the punctured disk $B(0,1) \setminus \{ 0 \} \subset \mathbb{R}^2$ can't be solved. The general situation for this question is well understood, but complicated. It is one of the main topics of classical potential theory.

You can solve the Laplace equation directly in your particular situation. It might be convenient for calculations to write the solution as $u \circ \phi$, where $\phi$ "nicely" maps your domain to some nicer domain and $u$ is harmonic on the new domain. But this is not mathematically necessary.


There is no topological obstacle to solving Laplace's equation with Dirichlet data (i.e. if the value of the solution on the boundary is prescribed). The solution may however lose some regularity near corner singularities (in your case near $x^2 + y^2 = .25, z \in \{0,1\}$, and at the edges of the outer box).

In the case of Neumann data (where the normal derivative of the solution is prescribed), solutions are only unique up to a constant which may be different for each connected component, and the boundary data must satisfy compatibility conditions, one for each connected component. So the only topological constraint comes from the number of components of the domain, which is 1 in your case.


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