Proof of L^p Elliptic Regularity - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T06:41:49Z http://mathoverflow.net/feeds/question/56829 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56829/proof-of-lp-elliptic-regularity Proof of L^p Elliptic Regularity Yakov Shlapentokh-Rothman 2011-02-27T15:51:42Z 2011-04-11T03:26:38Z <p>Let $L = \sum_{i,j=1}^n -\frac{\partial}{\partial x^i} (a^{ij}(x)\frac{\partial}{\partial x^j}) + \sum_{i=1}^n b^i(x) \frac{\partial}{\partial x^i} + c(x)$ be a second order elliptic operator with smooth coefficients, $\Omega$ a bounded open domain with smooth boundary in $\mathbb{R}^n$, and $f$ be a function in $L^p(\Omega)$. We say that $u \in W_0^{1,p}(\Omega)$ (one weak derivative in $L_p$ and vanishing boundary values) is a weak solution of $Lu = f$ if for all $g \in W_0^{1,q}(\Omega)$ $(q = p^*)$ we have </p> <p>$\int_{\Omega} \sum_{i,j=1}^n a^{ij}(x)\frac{\partial u}{\partial x^i}\frac{\partial g}{\partial x^j} + \sum_{i=1}^n b^i(x)\frac{\partial u}{\partial x^i}g(x) + c(x)u(x)g(x) = f(x)$.</p> <p>The standard result is of course that all such weak solutions $u$ actually belong to $W_0^{2,p}(\Omega)$.</p> <p>I am trying to complete the following proof of this statement:</p> <p>(1) First we establish an a priori estimate for strong solutions $v \in W_0^{2,p}(\Omega)$ of $Lv = f$:</p> <p>$\vert\vert v\vert \vert_{W_0^{2,p}(\Omega)} \leq C(\vert\vert f\vert\vert_{L^p(\Omega)} + \vert\vert v\vert\vert_{L^p(\Omega)})$</p> <p>This is non-trivial but can be established by proving the relevant estimate for the Laplacian with a Newton Potential argument and then using the freezing coefficients technique.</p> <p>(2) Next we observe that if $L$ is injective on $W_0^{1,p}$, then we are done. This is because $L$ injective implies</p> <p>$\vert\vert v\vert\vert_{L^p(\Omega)} \leq C\vert\vert Lv\vert\vert_{L^p(\Omega)}$</p> <p>One proves this by assuming it was false and then using Rellich compactness to produce a non-zero solution to $Lv = 0$.</p> <p>Having established this estimate, we consider the smooth mollifications $f_{\epsilon}$ of $f$. By $L^2$ theory we can find smooth $v_{\epsilon}$ strong solutions of $Lv_{\epsilon} = f_{\epsilon}$. Since we have</p> <p>$\vert\vert v_{\epsilon} - v_{\epsilon'}\vert\vert_{W_0^{2,p}} \leq C\vert\vert f_{\epsilon}-f_{\epsilon'}\vert\vert_{L^p(\Omega)}$</p> <p>The $v_{\epsilon}$ converge to some $v \in W_0^{2,p}(\Omega)$ which will solve $Lv = f$ strongly. Since strong solutions are clearly weak solutions, by the injectivity of $L$ on $W_0^{1,p}(\Omega)$ we conclude that $u = v \in W_0^{2,p}(\Omega)$ and we are done. </p> <p>(3) We have no way to guarantee that $L$ is injective, for example $0$ might be an $L^2$ eigenvalue of $L$. However, if $p=2$ then we could guarantee that $L_{\lambda} = L + \lambda I$ is injective for some large $\lambda$. If we could establish this fact in the general case we would be done since $L_{\lambda}u = f + \lambda u \in L^p$ and $L_{\lambda}$ injective imply that (2) is applicable.</p> <p><strong>Question</strong>: What is the simplest way to prove that $L_{\lambda} = L + \lambda I$ is injective on $W_0^{1,p}(\Omega)$ for large $\lambda$? Do there exist <strong>weak</strong> $L^P$ maximum principles?</p> <p>Of course, I would prefer that the proof not use $L^p$ regularity. </p> http://mathoverflow.net/questions/56829/proof-of-lp-elliptic-regularity/57148#57148 Answer by Dan Lee for Proof of L^p Elliptic Regularity Dan Lee 2011-03-02T18:19:10Z 2011-03-02T18:19:10Z <p>If you have $u\in W_0^{1,p}$ solving $L_\lambda u=0$, then by Sobolev embedding, $u$ is also in $W_0^{k,2}$ for some <em>negative</em> number $k$. (I've never seen this version of the Sobolev embedding theorem, but I'm assuming that one can prove it using Fourier analysis.) Once you have $u\in W_0^{k,2}$ solving $L_\lambda u=0$, where we now think of $L_\lambda:W_0^{k,2}\to W^{k-2,2}$, the $L^2$ theory takes over and tells us that $u$ is actually smooth. </p> http://mathoverflow.net/questions/56829/proof-of-lp-elliptic-regularity/61250#61250 Answer by Yakov Shlapentokh-Rothman for Proof of L^p Elliptic Regularity Yakov Shlapentokh-Rothman 2011-04-11T01:45:01Z 2011-04-11T03:26:38Z <p>I stumbled across the book <em>Second Oder Elliptic Equations and Elliptic Systems</em> by Yah-Ze Chen which appears to contain an answer to my question. It is available on google books here </p> <p><a href="http://books.google.com/books?id=eQcbiPQPweQC&amp;pg=PA49&amp;dq=strong+solution+dirichlet+problem+Lp&amp;hl=en&amp;ei=byKiTeXXO6GG0QGG7dGgBQ&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CCcQ6AEwADgK#v=onepage&amp;q=strong%20solution%20dirichlet%20problem%20Lp&amp;f=false" rel="nofollow">http://books.google.com/books?id=eQcbiPQPweQC&amp;pg=PA49&amp;dq=strong+solution+dirichlet+problem+Lp&amp;hl=en&amp;ei=byKiTeXXO6GG0QGG7dGgBQ&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CCcQ6AEwADgK#v=onepage&amp;q=strong%20solution%20dirichlet%20problem%20Lp&amp;f=false</a></p> <p>To save time for those who are interested, here is the relevant argument:</p> <p>For large $\lambda > 0$ we want to show that $L_{\lambda} = L - \lambda I$ is injective on $W_0^{1,p}$.</p> <p>Claim: Let $L^T_{\lambda}$ be the transpose of $L^{\lambda}$ with respect to the paring that defines weak solutions. Then we claim that $L^T_{\lambda}$ inective on $W_0^{2,p}$ implies that $L_{\lambda}$ is injective on $W_0^{1,p}$</p> <p>Proof: Suppose that $L^T_{\lambda}$ is injective on $W_0^{2,p}$. Then, by an argument contained in the original post above, for every $f \in L^p(\Omega)$ we can find $u \in W_0^{2,p}(\Omega)$ such that $L^T_{\lambda}u = f$. Now, suppose that $L_{\lambda}v = 0$ for some $v \in W_0^{1,p}$. After an integration by parts and the definition of weak solution, we see that $\varphi \in W_0^{2,q}$ implies that </p> <p>$\int_{\Omega}uL^T_{\lambda}\varphi = 0$. </p> <p>Now choose $\Omega'' \subset\subset \Omega' \subset\subset \Omega$ and a bump function $\rho$ identically one in $\Omega''$ with support in $\Omega'$. $\rho\text{sgn}(u)$ is in $L^q$, and we can find $g \in W_0^{2,q}$ such that $L^T_{\lambda}g = \rho\text{sgn}(u)$. Plugging this $g$ into the above equality gives</p> <p>$\int_{\Omega''}|u| = -\int_{\Omega\setminus\Omega''}\rho |u|$</p> <p>Due to the arbitrariness of $\Omega''$, this implies that $\int_{\Omega} |u| = 0$ and hence $u$ is $0$ a.e.</p> <p>Claim: For $\lambda$ large enough, $L_{\lambda}$ is injective on $W_0^{2,p}$. </p> <p>Proof: Suppose $L_{\lambda}u = 0$ for $u \in W_0^{2,p}$. Let $\tilde{\Omega} = \Omega \times (-1,1)$, and $\tilde{\Omega'} = \Omega \times (-1/2,1/2)$. Let $(x,t)$ be the coordinates on $\Omega \times (-1,1)$. Then define $v(x,t) = \cos(\sqrt{\lambda}t)u(x)$. Let $\hat{L_{\lambda}} = L_{\lambda} + \partial_t^2$. We have $\hat{L_{\lambda}}v = 0$. The strong solution estimates give</p> <p>$\vert\vert v\vert\vert_{W^{2,p}(\tilde{\Omega'})} \leq C\vert\vert v\vert\vert_{L^p(\tilde{\Omega})} \leq C\vert\vert u\vert\vert_{L^p(\Omega)} \Rightarrow$</p> <p>$\vert\vert \partial_t^2v\vert\vert_{L^p(\tilde{\Omega'})} \leq C\vert\vert u\vert\vert_{L^p(\Omega)} \Rightarrow$</p> <p>$\lambda\vert\vert u\vert\vert_{L^p(\Omega)}(\int_{-1/2}^{1/2}|\cos(\sqrt{\lambda}t|^p)^{1/p} \leq C\vert\vert u \vert\vert_{L^p(\Omega)} \Rightarrow$</p> <p>$\lambda^{1 - \frac{1}{2p}}\vert\vert u\vert\vert_{L^p(\Omega)}(\int_{-1/2}^{1/2}|\cos t|^p)^{1/p} \leq C\vert\vert u\vert\vert_{L^p(\Omega)}$</p> <p>Now taking $\lambda$ large enough implies that $u = 0$ almost everywhere.</p>