A have a question related to the boundary regularity of a solution of a Poisson equation on a bounded domain. But to make the question easier to pose I will state it on $ R_+^2:=\{ x \in R^2:x_2>0\}$ but now the statement may not be true, but keep in mind i am only concerned about local issues near the boundary.

So consider $$ \Delta \phi(x) = \frac{f(x)}{x_2^{2-\gamma}}$$ in $ R_+^2$ with $ \phi=0$ on $ x_2=0$ and where $f$ is bounded and where $ 0<\gamma$ is small. Using a scaling argument one can show that

$$ \sup_{R_+^2} x_2^{-\gamma} | \phi | + \sup_{R_+^2} x_2^{1-\gamma} | \nabla \phi | \le C \sup_{R_+^2} |f|$$. (This is the statement that may not be exactly correct since really i am on a bounded domain and $x_2=dist(x, \partial \Omega)$).

In any case my question is in obtaining more regularity for the gradient of $ \phi$ in the tangential direction. So suppose $ f$ is $C^1$ in $ x_1$ direction then we have something like $$ \Delta \phi_{x_1}= \frac{f_{x_1}(x)}{x_2^{2-\gamma}}$$ in $ R_+^2$ with $ \phi_{x_1}=0$ on $ x_2=0$ (at least morally). We can then see that we get added regularity in $x_1$ direction.

QUESTION. So assuming $f$ is only bounded i would like to obtain some added regularity in $x_1$ direction. I assume the proof is quite trivial and a non-research level question but I have attempted for a while and I am having difficulty.



1 Answer 1


Boundedness of $f$ isn't enough to get extra regularity for $\phi_1$. Take for example $\phi = r^{\gamma}\sin(\theta)$, which solves an equation of the desired form with $f = (\gamma^2-1)\sin(\theta)$, which is discontinuous at the origin. Then $$\phi_1(1/{\sqrt{2}},1/{\sqrt{2}}) = \frac{1}{2}(\gamma-1),$$ so by homogeneity $$\phi_1(\epsilon,\epsilon) \sim \epsilon^{\gamma-1}$$ and the estimate you got from scaling can't be improved.


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