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


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.

  • $\begingroup$ 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, 2018 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, 2018 at 20:35

1 Answer 1


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 $$Q= 0, \quad R = \psi_y(x,\,0) = \frac{4}{\pi(1-x^2)} \leq \frac{C}{1-|x|}.$$

  • $\begingroup$ It looks a good counter example for continuity. Can we expect boundeness? $\endgroup$
    – Paul
    Apr 11, 2018 at 20:33

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