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Context: I am currently reading through the freely available lecture notes from Tristan Riviere (here) on the applicability of integration by compensation in the analysis of various geometrically motivated PDEs.

I have attempted to find something in the vast literature to the following effect: suppose $u:D^2\rightarrow\mathbb{R}^n$, $u\in W^{1,2}$, $\Delta u = f(u,\nabla u) = 2H\partial_xu\times\partial_yu$. Then

$$u\in C^{0,\alpha}(D') \rightarrow u\in C^\infty(D''),$$

where $D'' \subset\subset D' \subset\subset D$.

In other words, an interior estimate. With the little regularity on hand, it appears to be very difficult. I find this surprising because most of the time, proving the Hölder regularity of the solution is the 'most difficult' part. I have the feeling that I am missing an obvious reference or a well-known folklore argument.

Showing the Hölder continuity of the solution relies on deriving a Morrey-type estimate with the help of the Wente lemma. In the process of doing this, one also shows that

$$\sup_{\rho < 1/2, p\in B_{1/2}(0)} \rho^{-\alpha} \int_{B_\rho(p)} |\Delta u|$$

is bounded. This implies that $f\in\mathcal{H}^1$. I just include this extra detail in case this question fits into a general framework of optimal interior regularity for Poisson's equation on a disk when the right hand side is Hardy. (This is the reason for the earlier form of my question.)

Can anybody help?

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Why do you believe this to be true? –  Deane Yang Feb 10 '11 at 14:02
    
To be explicit, $u = (x_1)^p$ for an appropriate value of $p$ is perhaps the simplest counterexample. Also, you seem to imply that assuming $f$ is in the Hardy space is a weaker assumption than $L^1$, but it is a stronger assumption (since, as you say, the Hardy space is smaller). –  Deane Yang Feb 10 '11 at 15:32
    
Yes, I know the Hardy space is a stronger assumption. I meant to write 'must' not 'can'--- I will edit the question. I also failed to write that $f$ is a function of $u$ and $\nabla u$ with critical quadratic growth in $\nabla u$. –  Glen Wheeler Feb 10 '11 at 16:01
    
I'm still confused. Surely, you don't meant that $f$, as a function of $u$ and $\nabla u$ is only in a Hardy space? –  Deane Yang Feb 10 '11 at 16:07
    
$f(u,\nabla u) = 2H\partial_xu \times \partial_yu$, where $H$ is a constant. What I understand well is how to prove the Hölder continuity of the (weak) solution so long as we assume $u\in W^{1,2}$. To go from Hölder continuity to smoothness (in this situation) is the step which I do not understand. In the proof of the Hölder continuity, one also shows some growth of the laplacian of the solution (reminiscent of a Morrey-type estimate) which allows one to conclude that the right hand side is in $\mathcal{H}^1$. I am sorry to have confused you both. –  Glen Wheeler Feb 10 '11 at 16:14

1 Answer 1

up vote 3 down vote accepted

I'm too lazy to type-up the proof myself, so I'll send you to a reference.

Chang, S.-Y. A., Wang, L. and Yang, P. C. (1999), "Regularity of harmonic maps". CPAM has the proof in Section 3. Once you get $C^{1,\gamma}$ you immediately get RHS is in $C^\gamma$ and the rest follow by standard elliptic regularity.

Note that the structure of the equation (RHS being of the form $d(u\cdot du)$) is only used for Wente's lemma. For the upgrade of regularity one uses a Caccioppoli type inequality.

(BTW, the Chang-Wang-Yang result bypasses the Hardy space estimates. For that the result can be found in the original paper of Helein, though I'd guess the material is also in his book if you don't read French.)

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Thanks Willie. I will check out the reference. But in this question I only have Hoelder continuity of the solution, not $C^{1,\alpha}$. I am aware that once the RHS is $C^\alpha$ then the standard regularity argument works fine. Thanks for the reference though, if one needs to work harder and obtain $C^{1,\alpha}$ then that's still useful. On Helein: I have been through Helein's book. He obtains the Hoelder continuity of the solution and then refers to the elliptic book by Ladyzhenskaya and Uraltseva. I couldn't find the argument with the correct growth on the RHS in that book however. –  Glen Wheeler Feb 10 '11 at 18:48
    
The reference I gave is to obtain $C^{1,\alpha}$ from just Holder. I made that comment since in your question you ask for smooth, while the section title for section 3 of Chang, Wang, and Yang's paper is "$C^{1,\gamma}$ regularity", so I thought I'd remark that that is enough. –  Willie Wong Feb 10 '11 at 19:08
    
(In fact that paper also contains a proof of Holder regularity from just $W^{1,2}$...) –  Willie Wong Feb 10 '11 at 19:10
    
You're right. Thanks for the answer :). –  Glen Wheeler Feb 10 '11 at 22:04

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