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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!

Arctic Char
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Second
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2 Answers2

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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

Tomás
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  • Nice. So, the first thing I wanted to know is that: RHS of equation $\in L^\infty$, then the solution is in $C^{1,\alpha}$, which seems to be the case. I will take a look at those papers, thanks. – Second Apr 29 '13 at 22:35
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    Also @second, if your domain is regular in the boundary, then you can conclude more, for instance, $u\in C^{1,\alpha}(\overline{\Omega})$. – Tomás Apr 29 '13 at 23:22
  • I think that Theorem 2, page 1, in here might be of some help in the case in which we have $L^2$-ness of the RHS, so that we conclude $L^\infty$-ness of the RHS. Does it help? Maybe there is a less complicated way to see it... And I'm really only concerned about interior regularity. – Second Apr 30 '13 at 09:26
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    @Second, you need more. You need that $e^u\in L^q$ with $q>\frac{N}{2}$. It depende on the dimesion. If your function satisfies it, I agree with you that you can use the same argument. Also take a look on this post, it is a generalization of the link you have send me. http://math.stackexchange.com/questions/262179/boundedness-of-solutions-to-p-laplace-equation – Tomás Apr 30 '13 at 10:46
  • yes, I have that (btw, I'm doing it in 2d) :D That's very nice. I think those two papers are a bit too much, do you know any simpler approach? My equation is uniformly elliptic, so I would expect some simpler proof to go from $L^\infty \to C^{1,\alpha}$... – Second Apr 30 '13 at 12:29
  • Do you have any boundary condition? – Tomás Apr 30 '13 at 12:45
  • No, I have no boundary condition. But we can restrict to the case, let's say, $\Omega$ is an open disk of radius 2, $u$ solves the equation, is in $L^\infty$, and I want to prove smoothness in the open disk (with the same center) of radius 1 (if that is of any help). – Second Apr 30 '13 at 13:28
  • To prove regularity is always a hard task. I think you could try to show that $u\in H^2$, which implies that $u\in C$ and then bootstrap. If you have Brezis book of functional Analysis, in chapter 9 there is an outline for what I am saying. – Tomás Apr 30 '13 at 13:45
  • In the last comment I have assumed that you are workin in 2d. – Tomás Apr 30 '13 at 13:51
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$e^u\in L^\infty$ 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 $u\in C^\infty$.

Noemi
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