Let $\Omega \subset \mathbb{R}^n$ be a domain and consider the PDE in divergence form

$$ D_i(a_{i,j}D_ju)=0 \tag{1}$$

where $a_{i,j}(x)$ are measurable and satisfly the uniform ellipticity condition:

$$\exists \lambda(x)>0 : \frac{1}{\lambda}|\psi|^2\leq \sum a_{i,j} \psi_i \psi_j\leq \lambda |\psi|^2\ \ \ \ \forall x,\ \forall \psi=(\psi_1,..,\psi_n)\in \mathbb{R}^n \tag 2 $$

then we have De Giorgi's Theorem, under the additional assuption that $A=(a_{i,j})$ is symmetric:

**Theorem:** Any weak solution of $(1)$
id est any $u\in H^1$such that $$\int_{\Omega} a_{i,j} \ D_i u \ D_j \phi \ dx =0 \ \ \ \ \ \forall \phi\in C^{\infty}_{c} $$ is Holder continuous.

Now, I was wondering if anyone extended his proof to the case where $A$ is not symmetric.

I can't see where this assumption is crucial in De Giorgi's proof.

In this paper, at the end of page 5, it is said that in Morrey's "Multiple integrals in the calculus of variation" there is something about this problem, infact, at page 126 there is a very general form of the elliptic PDE which does not include assuptions on the symmetry of A. The approach seems similar to De Giorgi's one, no special treatment of the fact that the matrix is not symmetric seems to be there (if anyone is interested it is at pages 134-140)

I am thinking, maybe the proof is the same but De Giorgi was only bothered with the symmetric case (he was motivated by a problem in calculus of variation for which the symmetry of the matrix could be assumed WLOG, I believe) could this be the case?

Thanks!

docommute. – Deane Yang Jul 4 '13 at 16:40