# 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) for any $x_1>0$ and $x_1^\star 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, 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 $\lambda_0$.

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