Proof of L^p Elliptic Regularity

Question

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

$\int_{\Omega} \sum_{i,j=1}^n a^{ij}(x)\frac{\partial u}{\partial x^i}\frac{\partial g}{\partial x^j} + \sum_{i=1}^n b^i(x)\frac{\partial u}{\partial x^i}g(x) + c(x)u(x)g(x) = f(x)$.

The standard result is of course that all such weak solutions $u$ actually belong to $W_0^{2,p}(\Omega)$.

I am trying to complete the following proof of this statement:

(1) First we establish an a priori estimate for strong solutions $v \in W_0^{2,p}(\Omega)$ of $Lv = f$:

$\vert\vert v\vert \vert_{W_0^{2,p}(\Omega)} \leq C(\vert\vert f\vert\vert_{L^p(\Omega)} + \vert\vert v\vert\vert_{L^p(\Omega)})$

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.

(2) Next we observe that if $L$ is injective on $W_0^{1,p}$, then we are done. This is because $L$ injective implies

$\vert\vert v\vert\vert_{L^p(\Omega)} \leq C\vert\vert Lv\vert\vert_{L^p(\Omega)}$

One proves this by assuming it was false and then using Rellich compactness to produce a non-zero solution to $Lv = 0$.

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

$\vert\vert v_{\epsilon} - v_{\epsilon'}\vert\vert_{W_0^{2,p}} \leq C\vert\vert f_{\epsilon}-f_{\epsilon'}\vert\vert_{L^p(\Omega)}$

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.

(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.

Question: 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 weak $L^P$ maximum principles?

Of course, I would prefer that the proof not use $L^p$ regularity.

Answer 1

I stumbled across the book Second Oder Elliptic Equations and Elliptic Systems by Yah-Ze Chen which appears to contain an answer to my question. It is available on google books here

http://books.google.com/books?id=eQcbiPQPweQC&pg=PA49&dq=strong+solution+dirichlet+problem+Lp&hl=en&ei=byKiTeXXO6GG0QGG7dGgBQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCcQ6AEwADgK#v=onepage&q=strong%20solution%20dirichlet%20problem%20Lp&f=false

To save time for those who are interested, here is the relevant argument:

For large $\lambda > 0$ we want to show that $L_{\lambda} = L - \lambda I$ is injective on $W_0^{1,p}$.

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 $W_0^{1,p}$

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

$\int_{\Omega}uL^T_{\lambda}\varphi = 0$.

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

$\int_{\Omega''}|u| = -\int_{\Omega\setminus\Omega''}\rho |u|$

Due to the arbitrariness of $\Omega''$, this implies that $\int_{\Omega} |u| = 0$ and hence $u$ is $0$ a.e.

Claim: For $\lambda$ large enough, $L_{\lambda}$ is injective on $W_0^{2,p}$.

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

$\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 $

$\vert\vert \partial_t^2v\vert\vert_{L^p(\tilde{\Omega'})} \leq C\vert\vert u\vert\vert_{L^p(\Omega)} \Rightarrow $

$\lambda\vert\vert u\vert\vert_{L^p(\Omega)}(\int_{-1/2}^{1/2}|\cos(\sqrt{\lambda}t|^p)^{1/p} \leq C\vert\vert u \vert\vert_{L^p(\Omega)} \Rightarrow$

$\lambda^{1 - \frac{1}{2p}}\vert\vert u\vert\vert_{L^p(\Omega)}(\int_{-1/2}^{1/2}|\cos t|^p)^{1/p} \leq C\vert\vert u\vert\vert_{L^p(\Omega)}$

Now taking $\lambda$ large enough implies that $u = 0$ almost everywhere.

Answer 2

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 negative 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.