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The question arises from a paper on Schwarz's domain decomposition method (click here).
We consider a bounded domain in $\mathbb{R}^2$ and a curve splits it into two, see the figure below.

enter image description here

Now we solve an elliptic partial differential equation with zero source term: $\mathcal{L}u=0$ with the discontinuous Dirichlet boundary value shown in the figure. It was guessed that the upper limit of $u(\mathbf{x}),$ as $\mathbf{x}$ goes to the discontinuous point A along the cutting curve, is strictly between 0 and 1.

How can one prove this? In the original paper, it says for the Laplace equation one can
derive this by potential theory. I do not know the proof in this particular case either.

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    $\begingroup$ For $\mathcal L = \Delta$, you can just map to the unit disk or a half plane and use the explicit form of the Poisson kernel (and the fact that the conformal map will be very well behaved for a region like the one above). $\endgroup$
    – Christian Remling
    Commented May 17, 2014 at 18:34
  • $\begingroup$ @ChristianRemling This is an interesting idea that I have not thought about. The Laplace equation (with f=0) does not change under conformal mapping. Maybe the original paper means this way. Thanks very much! Then, for the general PDE I need only to compare with Laplace locally around the point `A'. $\endgroup$
    – Hui Zhang
    Commented May 17, 2014 at 19:13

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