Ngô Quốc Anh

September 18, 2009

The method of moving planes: An example inspired by Gidas, Ni, and Nirenberg

Filed under: Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 1:41

In this section we will use the moving plane method to discuss the symmetry of solutions. The following result was first proved by Gidas, Ni and Nirenberg.

Theorem. Suppose u \in C(\bar B_1) \cap C^2(B_1) is a positive solution of

-\Delta u = f(u) in B_1 and u=0 on \partial B_1

where f is locally Lipschitz in \mathbb R. Then u is radialy symmetric in B_1 and \frac{\partial u}{\partial r}(x)<0 for x \ne 0.

The original proof requires that solutions be C^2 up to the boundary. Here we give a method which does not depend on the smoothness of domains nor the smoothness of solutions up to the boundary.

Statement. Suppose that ­ is a bounded domain which is convex in x_1 direction and symmetric with respect to the plane \{x_1 = 0\}. Suppose u \in C^2(\Omega)\cap C(\bar\Omega) is a positive solution of

-\Delta u = f(u) in \Omega and u=0 on \partial \Omega

where f is locally Lipschitz in \mathbb R. Then u is radialy symmetric in x_1 and D_{x_1}u(x)<0 for x \ne 0.

Idea of proof.  Write x=(x_1, y)\in\Omega for y \in \mathbb R^{n-1}. We will prove that

u(x_1,y)<u(x_1^\star,y) for any x_1>0 and x_1^\star<x_1 with x_1^\star+x_1>0.

Then by letting x_1^\star \to -x_1, we get u(x_1,y)\leq u(-x_1,y) for any x_1. Then by changing the direction x_1 \to -x_1 we get the symmetry.

Proof. Let a=\sup x_1 for (x_1, y)\in\Omega. For 0<\lambda<a, define

\Sigma_\lambda=\{x \in \Omega: x_1>\lambda\}

T_\lambda = \{x_1=\lambda\}

\Sigma'_\lambda= the reflection of \Sigma_\lambda with the respect to T_\lambda

x_\lambda=(2\lambda-x_1,x_2,...,x_n) for x=(x_1,x_2,...,x_n).

In \Sigma_\lambda we define

w_\lambda(x)=u(x)-u(x_\lambda) for x\in \Sigma_\lambda.

Then we have by Mean Value Theorem

\Delta w_\lambda+c(x,\lambda)w_\lambda=0 in \Sigma_\lambda

w_\lambda \leq 0 and w_\lambda \not \equiv 0 on \partial \Sigma_\lambda.

where c(x,\lambda) is a bounded function in \Sigma_\lambda.

We need to show w_\lambda<0 in \Sigma_\lambda for any \lambda \in (0,a). This implies in particular that w_\lambda assumes along \partial \Sigma_\lambda \cap \Omega its maximum in \Sigma_\lambda. By Hopf Lemma we have for any such \lambda \in (0,a)

D_{x_1}w_\lambda\bigg|_{x_1=\lambda} = 2D_{x_1}u\bigg|_{x_1=\lambda}<0.

For any \lambda close to a, we have w_\lambda<0 by the maximum principle for narrow domain. Let (\lambda_0, a) be the largest interval of values of \lambda such that w_\lambda<0 in \Sigma_\lambda. We want to show that \lambda_0=0. If \lambda_0>0, by continuity, w_\lambda \leq 0 in \Sigma_{\lambda_0} and w_{\lambda_0} \not \equiv 0 on \partial \Sigma_{\lambda_0}. Then the Strong Maximum Principle implies w_{\lambda_0}<0 in \Sigma_{\lambda_0}. We will show that for any small \varepsilon>0

w_{\lambda_0-\varepsilon} <0 in \Sigma_{\lambda_0-\varepsilon}.

Fix \delta>0 (to be determined). Let K be a closed subset in \Sigma_{\lambda_0} such that |\Sigma_{\lambda_0} - K| <\frac{\delta}{2}. The fact that w_{\lambda_0}<0 in \Sigma_{\lambda_0} implies

w_{\lambda_0}(x)\leq -\eta <0 for any x \in K.

By continuity we have

w_{\lambda_0-\varepsilon}<0 in K.

For \varepsilon>0 small, |\Sigma_{\lambda_0-\varepsilon}-K|<\delta. We choose \delta in such a way that we may apply the Maximum Principle for domain with small volume to w_{\lambda_0-\varepsilon} in \Sigma_{\lambda_0-\varepsilon}-K. Hence we get

w_{\lambda_0-\varepsilon} \leq 0 in \Sigma_{\lambda_0-\varepsilon}-K

and then

w_{\lambda_0-\varepsilon}(x) <0 in \Sigma_{\lambda_0-\varepsilon} -K.

Therefore we obtain for any small \varepsilon>0

w_{\lambda_0-\varepsilon}(x) <0 in \Sigma_{\lambda_0-\varepsilon}.

This contradicts the choice of \lambda_0.

latex

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at WordPress.com.

%d bloggers like this: