A problem about convergence…

Question

I am studying an article of Berestychi-Caffarelli-Niremberg - Monotonicity for elliptic equations in unbounded Lipschitz domains, and I don't understand a convergence in the demonstration of the lemma 3.4.

Suppose that $u\in C^2(\Omega)\cap C(\overline\Omega)$, such that $$ \left\{ \begin{array}{rl} \Delta u+f(u)=0, & in \ \Omega,\\ u=0, & on \ \partial\Omega,\\ 0<u<1, & in \ \Omega, \end{array} \right. $$ where $\Omega=\{x=(x',x_n)\in\mathbb{R}^n\times\mathbb{R};x_n>\varphi(x')\}$, with $\varphi$ is a Lipschitz function.

LEMMA 3.4: For any $h>0$, the solution $u$ is bounded away from $1$ in $\Omega_h=\{x\in\Omega;\varphi(x')<x_n<\varphi(x')+h\}$.

Sketch of the proof: Suppose by contradiction that exists a sequence $(x'^j,x_n^j)_j=(x^j)_j\subset\Omega_h$ such that $u(x^j)\rightarrow1$. By the translation $T^j(x)=x-x^j$ we move the set $\Omega$ to $\Omega^j$, given by $$\Omega^j=\{z=(z',z_n)\in\mathbb{R}^n;z_n>\varphi^j(z')=\varphi(z'+x'^j)-x_n^j\}.$$ Is easy to verify that the functions $\varphi^j$ is Lipschitz continuous and uniformly bounded in compact sets, so by the Arzela-Ascoli Theorem, for a subsequence $\varphi^j$ tend to a function $\widehat\varphi$. For each set $\Omega^j$ you have a shifted solution $$u^j(z',z_n)=u(z'+x'^j,z_n+x_n^j),$$ satisfying $$ \left\{ \begin{array}{rl} \Delta u^j+f(u^j)=0, & in \ \Omega^j,\\ u^j=0, & on \ \partial\Omega^j,\\ 0<u^j<1, & in \ \Omega^j, \end{array} \right. $$

FINALLY, the doubt: In the article, he says that the shifted solutions converge uniformly in compact subsets of $$\widehat\Omega=\{x\in\mathbb{R}^n;x_n>\widehat\varphi(x')\},$$ to a solution $\widehat u$ that satisfies $$ \left\{ \begin{array}{rl} \Delta \widehat u+f(\widehat u)=0, & in \ \widehat\Omega,\\ \widehat u=0, & on \ \partial\widehat\Omega,\\ \end{array} \right. $$

If the initial sequence $(x^j)_j$ is bounded, I can to argument this implies (Because in compact sets, the shifted solutions and the first and second derivatives, would be uniformly continuous and uniformly bounded, then you could apply the Arzela-Ascoli theorem). But I think that the sequence could be unbounded, I don't know. Someone can help me in this argument?

Thank you.

Answer 1

First of all, you should think of the sequence $(x^j)$ as unbounded, because if it had a finite limit, we'd immediately hit a contradiction: $x^j\to x\in \partial \Omega$, $u\in C(\overline{\Omega})$, and $u(x)=0$.

The reason for convergence of shifted solutions on compact sets is also Arzela-Ascoli, but we need uniform continuity on an unbounded set, which does not come for free.

Claim: For any $C>1$ the solution $u$ is uniformly Lipschitz on the set $\Omega_C=\{x: C^{-1}\le d(x)\le C\}$ where $d(x)=\operatorname{dist}(x,\partial\Omega)$.

Proof. The function $u$ solves the Poisson equation $\Delta u = g$ where $g=f\circ u$. Notice that both $u$ and $g$ are bounded in $\Omega$. The standard inner regularity result for the Poisson equation (Theorem 3.9 in Gilbarg-Trudinger) implies that $\nabla u$ is uniformly bounded in $\Omega_C$. (Why? Because any point $x\in \Omega_C$ is the center of a ball of radius $C^{-1}$ contained in $\Omega$. Apply the theorem on this ball, and recall that $u$ and $\Delta u$ are uniformly bounded.) It follows that $u$ is Lipschitz on $\Omega_C$. $\Box$

When we fix a compactly contained subdomain $G\Subset \widehat{\Omega}$ and consider the restrictions $u^j$ to $G$, we are actually looking at the restriction of $u$ to a subset of $\Omega_C$. Hence $u^j$ are uniformly Lipschitz on $G$. Since $f$ is Lipschitz and $\Delta u^j=f(u^j)$, it follows that the sequence $\Delta u^j$ is uniformly Lipschitz as well. By Theorem 4.6 in Gilbarg-Trudinger, this implies uniform Hölder continuity of all second-order derivatives of $u^j$.

Therefore, we can choose a subsequence, still denoted $u^j$, such that both $u^j$ and $D^2 u^j $ converge uniformly on $G$. Say, $u^j\to \widehat{u}$ and $\Delta u^j \to g$. We should check that $\Delta \widehat{u}=g$. To that end, pick a test function $\psi$ supported in $G$, and observe that $$\int u^j\Delta\psi\to \int \widehat{u}\,\Delta\psi\tag{1}$$ $$\int \Delta u^j \,\psi\to \int g\,\psi\tag{2}$$ Since $\int u^j\Delta\psi = \int \Delta u^j \,\psi$ (integration by parts), the combination of (1) and (2) implies $\Delta \widehat{u}=g$.