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Consider the elliptic equation \begin{equation} D_i(a_{i,j}D_j u(x)) = 0 \end{equation} in some bounded $\Omega$ of $\mathbb{R}^{n}$ with \begin{equation} \lambda |\xi|^{2} \le a_{i,j}(c) \xi_i \xi_j \le \Lambda |\xi|^{2} \end{equation} for some $\lambda, \Lambda > 0$. I would like to understand a corollary of the theorem below:

Theorem: Let $p$ a real number $p>2$. Then there exist $\varepsilon = \varepsilon(p)>0$ such that if $I$ is the identity matrix in $\mathbb{R}^{n}$ and \begin{equation} \|I - a_{i,j}\|_{\infty} \le \varepsilon, \tag{1} \end{equation} then all solutions to (1) in $W^{1,2}$ satisfy $u \in W^{1,p}$.

Corollary: If we assume that the equation (1) has continuous coefficients, then each solution $u \in W^{1,2}$ verifies that $u \in W^{1,p}$ for all $p < \infty$.

Also I would like to know the estimate for Laplacian $$ \| \nabla u\|^{2}_{\infty} \le C \dfrac{1}{|Q|} \int_{Q} | \nabla u(y)|^2 dy.$$

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What norm are you using in (1)? –  Jose27 Jul 31 '12 at 0:35
    
Sorry $\|\cdot\|_{\infty}.$ –  user29999 Jul 31 '12 at 0:41
3  
Tip: You can use [stuff] \tag{1} to get $$[\text{stuff}] \tag{1}$$ –  Rahul Jul 31 '12 at 0:45
    
Dou you have the reference where you're getting this results from? Also, I'm assuming there's a Laplacian missing in your last inequality. –  Jose27 Jul 31 '12 at 1:41
    
yes in the page 7. [1]: karlin.mff.cuni.cz/~kaplicky/pages/pages/2010z/CaPe1998.pdf –  user29999 Jul 31 '12 at 1:44
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1 Answer 1

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The philosophy of this paper is: if the solutions of $\mathcal A_0u=0$ are nice, and $\mathcal {A}$ is an operator sufficiently close to $\mathcal A_0$, then the solutions of $\mathcal{A}u=0$ are okay. Theorem A is a precise formulation of this philosophy.

Theorem B follows from Theorem A by taking $\mathcal A_0=\Delta$, the Laplacian. First we verify that the solutions of Laplacian (harmonic functions) are nice, namely that (H1) holds. Indeed, if $u$ is harmonic, then $|\nabla u|^2$ is subharmonic (you can just do the computation, or argue by convexity). The sub-mean value property says that $$| \nabla u(x)|^{2} \le \frac{1}{|B|} \int_{B} | \nabla u(y)|^2 dy \tag{SUB}$$ for any ball $B$ centered at $x$. To see that (1.1) holds, apply (SUB) to any point $x\in Q$, choosing $B$ to be the ball of radius equal to the half the sidelength of $Q$. Then $B\subset 2Q$, and their measures are about the same, so we get $$\| \nabla u\|^{2}_{L^{\infty}(Q)} \le \frac{C}{|2Q|} \int_{2Q} | \nabla u(y)|^2 dy$$ Hopefully this explains the "For the Laplacian we have the classical estimate..." part.

Then we use Lemma 2.1 to make sure that (H2) holds when the coefficients $\mathcal A$ are sufficiently close to the identity. So much for the proof of Theorem B.

Next, the corollary. We have $\mathrm{div}\, A\nabla u=0$ where $A=A(x)$ has continuous coefficients. We want to prove that $u$ is locally in $W^{1,p}$ for all $p<\infty$. Given any point $x_0$ in the domain, let $A_0=A(x_0)$. Make the linear change of variable $v=u\circ A_0^t$. The chain rule says $\nabla v=A_0 \nabla u$. Hence, $v$ satisfies $\mathrm{div}\, \tilde A\nabla v=0$ where $\tilde A(x)=A(x)A_0^{-1}$.

Note that $\tilde A(x)$ also has continuous coefficients, and $\tilde A(x_0)=I$. By continuity, $\tilde A(x)$ is close to $I$ when $x$ is close to $x_0$. Therefore, Theorem B applies and says that $v$ is locally in $W^{1,p}$. The same holds for $u$, which is just the composition of $v$ with a linear map.

(Added)

If $u$ is harmonic, then each partial derivative such as $u_{x}$ is also harmonic because $\Delta (u_x)=(\Delta u)_{x}=0$. Furthermore, the square of any harmonic function $v$ is subharmonic because $$\Delta (v^2)= \mathrm{div}\,\nabla (v^2) = \mathrm{div}\, (2v \nabla v) = 2|\nabla v|^2+2v\, \mathrm{div}\, (\nabla v) =2|\nabla v|^2+2v\, \Delta v = 2|\nabla v|^2\ge 0 $$ Finally, $|\nabla u|^2$ is the sum of squares of partial derivatives and therefore is subharmonic.

The authors do not say that their "$u\in W^{1,p}$" results are local. But they are local, unless additional assumptions on boundary regularity are made. One cannot get global regularity in arbitrary bounded domains by cube-based estimates used in this paper. Here is a concrete example: in two dimensions let $\Omega$ be $\{(x,y):0<y<x^2, 0<x<1\}$. The function $u(x,y)=\log (x^2+y^2)$ belongs to $W^{1,2}(\Omega)$, which you can check by integrating $|\nabla u|^2\approx 1/x^2$ within $\Omega$. We have $\Delta u=0$, which is the nicest equation one could ask for. However, $u$ does not belong to $W^{1,3}(\Omega)$, which you also can check by integration. Therefore, the global version of Theorem B does not hold (without extra hypotheses on the smoothness of $\partial\Omega$).

By Hölder's inequality, $u\in W^{1,p}$ implies $u\in W^{1,s}$ for any $1\le s<p$, as long as the domain of integration has finite measure. Since only bounded domains are considered here, there is no need to worry about $p<2$.

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I would like to see the details or some reference that if $f$ is harmonic then $f$ is subharmonic and other two doubts. First, the article does not say that $u$ is locally in $ W^{1,p}$. Have you saw the comment of Jose27? The last comment. Did you agree with him? Second, I think that the theorem treats only the case $p>2$. Is possible to treat the case $p<2$? The corollary says $p<\infty$? –  user29999 Aug 2 '12 at 22:26
    
@user29999 I added answers to your questions. If you want to know what Jose27 meant by his comment, you should ask him. –  user31373 Aug 3 '12 at 1:35
    
First, thank you. I like a lot of yours answers. I would like to say that when I ask for details isn't because I don't believe in the solutions, really I want to understand the details. Now I have one question, a solution $u \in W^{1,2}$ would be in the weak sense, would not it? The argument $$\Delta (v^2)= \mathrm{div}\,\nabla (v^2) = \mathrm{div}\, (2v \nabla v) = 2|\nabla v|^2+2v\, \mathrm{div}\, (\nabla v) =2|\nabla v|^2+2v\, \Delta v = 2|\nabla v|^2\ge 0 $$ Clearly holds for classical solutions.Does it holds for weak solutions? –  user29999 Aug 4 '12 at 1:02
    
@user29999 This is not an issue due to Weyl's lemma which says that any weak solution of Laplace's equation is in fact classical. –  user31373 Aug 4 '12 at 1:05
    
I asked about the theorem in this article, as you read it and you undertood severals things there, maybe you can help me. The ask is here math.stackexchange.com/questions/179156/…. –  user29999 Aug 5 '12 at 20:24
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