Elliptic regularity - nonlinear case

Question

Let's say I have a weak solution $u \in H^1(\Omega)$, $\Omega \subset \mathbb{R}^n$ open, to the equation $$ \Delta u = e^u, $$ let's also assume that $e^u \in L^\infty(\Omega)$. Does it follow that $u \in C^\infty(\Omega)$? Also, does it work if we substitute the last requirement with $e^u \in L^1(\Omega)$?

I tried to look for this kind of result on Gilbarg-Trudinger, but it was not helpful. Also M.E. Taylor's book does not deal with this case. It might be that you could do this using the Calderon-Zygmund inequality.

If anyone could give me a hint or a reference I would be very grateful!

Answer 1

If $e^u\in L^\infty$, then you have that $u\in C^{1,\alpha}_{loc}(\Omega)$, which implies that $u\in C^1(\Omega)$. By taking any $\Omega'$ open in $\Omega$, with $\overline{\Omega}'\subset\Omega$, you have that $u,|\nabla u|\in L^\infty(\Omega')$, hence differentiating you equation, you can conclude that $\nabla u\in C^{1,\alpha}_{loc}(\Omega)$. You can continue this bootstrap argument to conclude that $u\in C^\infty(\Omega)$.

For the case $e^u\in L^1$, I dont have idea if you can get such regularity.

All I have said here can be found in the papers of Liebermann:

Boundary regularity for solutions of degenerate elliptic equations - 1988

The natural generalizationj of the natural conditions of ladyzhenskaya and uralľtseva for elliptic equations - 1991

Answer 2

$e^u\in L^\infty$ gives $u\in C^\alpha_{loc}$ (see Gilbarg-Trudinger, for instance). Then, $\Delta u\in C^\alpha_{loc}$. Thus, $u\in C^{2,\alpha}_{loc}$ (Gilbarg-Trudinger again). And now, the bootstrap argument gives you $u\in C^\infty$.