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Let $\Omega\subset\mathbb{R}^N$ and suppose that $g\in C^{1,\alpha}(\mathbb{R},\mathbb{R})$, $f\in C^{0,\alpha}(U)$ for every open $U$ with $\overline{U}\subset\Omega$, $\alpha\in (0,1)$ and $g\geq a$, where $a$ is a positive constant. Assume that $$-\operatorname{div}(g(|\nabla u|^2)\nabla u)=f,\ \mbox{in}\ \Omega$$

in the classical sense, where $u\in C^{1,\alpha}(U)$ for every $U$. Can I conclde that $u\in C^2(\Omega)$?

I was trying to solve this problem, by using Schauder Estimates, but I could not find anything specific. For example, if we call $g(|\nabla u(x)|^2)=h(x)$, then we have that

$$-\operatorname{div}(h(x)\nabla u)=f,\ \mbox{in}\ \Omega$$

with $h\in C^{0,\alpha}(U)$. Now I would like to use Schauder Regularity. Does anyonw knows any result in this direction?

Thank You.

Tomás
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  • Is there anything on the sign of $g'$ or at least its magnitude relative to that of $g$? Because I'm not sure you have ellipticity if you do not .... – Ray Yang Apr 30 '13 at 17:43
  • @RayYang, suppose that we have ellipcity. I forgot to add this hypothesis. – Tomás Apr 30 '13 at 17:46

1 Answers1

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Under the ellipticity condition, take a look at Ladyzhenskaya and Uraltseva's book Linear and Quasilinear Elliptic Equations, Chapter 4, Theorem 6.3. It appears to provide what you need.

In case you do not have the book, basically your equation translates to $$g(|\nabla u|^2) \Delta u + 2 g'(|\nabla u|^2) \nabla u^T D^2 u \nabla u = 0$$ If you assume that $g$ is $C^{1,\alpha}$ and $u$ is as well, then the coefficients on every $u_{ij}$ are $C^{\alpha}$, treating the gradient terms as part of the coefficients, whence you can apply Schauder.

You do need to assume sufficient auxiliary conditions on $g'$ for the equation to be elliptic, however.

Ray Yang
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