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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!

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Isn't this trivial? If you are considering a scalar equation, because partial derivatives commute, $D_{i,j}u = D_{j,i}u$, the antisymmetric part of $a_{i,j}$ simply drops out. So there is no loss of generality in assuming that the coefficient matrix is symmetric from the outset. –  Igor Khavkine Jul 4 '13 at 16:08
    
@IgorKhavkine scalar in the sense that $a_{i,j}$ are constant? they are just bounded and measurable. I think the issue is that all we know is that $u\in H^1 $ so partial derivatives do not commute! –  Moritzplatz Jul 4 '13 at 16:16
    
or maybe I am missing something obvious! –  Moritzplatz Jul 4 '13 at 16:21
5  
Weak derivatives do commute. –  Deane Yang Jul 4 '13 at 16:40
    
I guess that would be it. :-) –  Igor Khavkine Jul 4 '13 at 16:41
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2 Answers

You can look to Han&Lin's book(especially chapter 3 and 4) which is a clear exposition of Nash-De Giorgi result. Here the only hypothesis is that $u$ is in $H^1$, $a_{ij}$ are in $L^\infty$ and uniformly elliptic in the sense that $$ a_{ij}\xi^i \xi^j \geq \lambda \vert \xi\vert^2,$$ for some fixed $\lambda$.

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Something is not completely clear in your question. There are two possible ways to generalise the De Giorgi's statement

  1. $a \in L^1(\Omega)^{N\times N}$ and the bound you wrote on the symmetric part of $a$;
  2. $a$ satisfies an $L^\infty(\Omega)^{N\times N}$ bound and the bound you wrote on the symmetric part of $a$.

In the symmetric case, 1. = 2. but not in general. If you are happy with 2., then it is true and you can read it in Gilbarg & Trudinger's book, or in many other textbooks. If you really mean 1., then it is not clear that there is a solution in $H^1(\Omega)$ to a boundary value problem of that type in general. Since you assume existence, you question still makes sense, but that is not a textbook question anymore.

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