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Let $\Omega=\mathbb{D}\cap\{ (x,y)\, \vert\, y>0\}$, $I=(-1,1)\times \{0\}$ and $A=\partial\Omega\setminus I$. Let $Q\in L^1(\Omega)$, and $R\in C^\infty_{loc}(I)$.

I am looking to the following problem

$$ \left\{ \begin{aligned} \Delta \psi = Q & \hbox{ in } \Omega \\ \psi = 0 & \hbox{ on } A \\ \partial_{\nu}\psi = R & \hbox{ on } I \\ \end{aligned} \right. $$

What is the best regularity I can expect on $\psi$? is it at least continuous? This case doesn't seem so classical, I found nothing precise in Gilbarg & Trudinger for instance, and perhaps the fact my domain get corners could be problematic.... any precise reference would be welcome.

Thanks in advance

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If Needed, but the main assumption are above:

I know a bit more on $Q$ and $R$, for instance I have $$ \vert Q(z)\vert \leq \frac{C}{(1-\vert z\vert)^2}$$ and $$ \vert R(z)\vert \leq \frac{C}{(1-\vert z\vert)}$$

but of cours no $L^\infty$ bound.

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  • $\begingroup$ If I remember correctly such problems with mixed Dirichlet-Neumann conditions are discussed in in the book P. Grisvard, Elliptic Problems in Nonsmooth Domains. $\endgroup$
    – Andrew
    Apr 11 '18 at 12:04
  • $\begingroup$ Unfortunately, there is nothing about this kind of condition, it is more about linear combination of Dirichlet and Neumann $\endgroup$
    – Paul
    Apr 11 '18 at 20:35
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The continuity of $\psi$ up to $\partial \Omega$ is false without more control on $R$.

Consider for example the harmonic function that is $1$ on the upper half-circle and $-1$ on the lower half-circle. A model for the behavior near e.g. the lower left corner is the (zero-homogeneous) angle function $\frac{2}{\pi} \Im\, \log(z + 1)$, which on $I$ has normal derivative blowing up like inverse of distance to the corner. One can compute the exact solution using complex analysis: $$\psi = \frac{2}{\pi}\Im\,\log\left(\frac{1+z}{1-z}\right) = \frac{2}{\pi}\tan^{-1}\left(\frac{2y}{1-|z|^2}\right),$$ which satisfies $$Q= 0, \quad R = \psi_y(x,\,0) = \frac{4}{\pi(1-x^2)} \leq \frac{C}{1-|x|}.$$

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  • $\begingroup$ It looks a good counter example for continuity. Can we expect boundeness? $\endgroup$
    – Paul
    Apr 11 '18 at 20:33

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