# Ngô Quốc Anh

## May 1, 2011

### Method of moving planes: The cylindrically symmetric of solutions of critical Hardy–Sobolev operators

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 5:22

Recently, I have read a paper entitled “Classification of solutions of a critical Hardy-Sobolev operator” published in J. Differential Equations [here].

In that paper, the authors try to classify all positive solutions for the following equation

$\displaystyle -\Delta u(x)=\frac{u(x)^\frac{n}{n-2}}{|y|} \quad \text{ in } \mathbb R^n$

where $x=(y,z)\in \mathbb R^k \times \mathbb R^{n-k}$, and $u \in \mathcal D^{1,2}(\mathbb R^n)$.

The main result of the paper can be formulated as follows

Theorem. Let $u_0$ be the function given by

$\displaystyle u_0(x)=u_0(y,z)=c_{n,k}\left( (1+|y|)^2+|z|^2\right)^{-\frac{n-2}{2}}$

where

$\displaystyle c_{n,k}=\big((n-2)(k-1)\big)^\frac{n-2}{2}.$

Then $u$ is a solution to the equation above if and only if

$\displaystyle u(y,z)=\lambda^\frac{n-2}{2}u_0(\lambda y, \lambda z+z_0)$

for some $\lambda>0$ and some $z_0 \in \mathbb R^{n-k}$.

First, they use the method of moving planes to prove the cylindrically symmetric of solutions. Thanks to this symmetry, the equation reduces to an elliptic equation in the positive cone in $R^2$ which eventually leads to a complete identification of all the solutions of the equation. We skip the detailed discussion here and refer the reader to the original paper.

This result has recently been generalized by Cao and Li.

## June 11, 2010

### The method of moving planes: Elliptic systems in the whole space

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 5:13

Li Ma and Baiyu Liu recently announced a new result accepted in Advances in Mathematics journal [here]. In that elegant paper, they considered the following system of equations

$\displaystyle\left\{ \begin{gathered} - \Delta u = g(u,v), \hfill \\ - \Delta v = f(u,v), \hfill \\ u > 0,v > 0, \hfill \\ \end{gathered} \right.$

in the whole space $\mathbb R^n$ with $n\geqslant 3$ where $f$ and $g$ are two smooth functions in $\mathbb R^2_+$.

This work is closely related to works done in [here] and [here]. Precisely, in [here], the authors have proved that classical solution $(u, v)$ of the PDEs is symmetric about some points $x_1, x_2 \in \mathbb R^n$ respectively, under assumptions

1. $u(x) \to 0$, $v(x) \to 0$ as $|x| \to \infty$;

2. $\displaystyle\frac{{\partial g}}{{\partial v}}(u,v) \geqslant 0, \quad \frac{{\partial f}}{{\partial u}} (u,v)\geqslant 0,\quad\forall (u,v) \in \left[ {0, + \infty } \right) \times \left[ {0, + \infty } \right)$;

3. $\displaystyle\frac{{\partial g}}{{\partial u}}(0,0) < 0, \quad \frac{{\partial f}}{{\partial v}}(0,0) < 0$;

4. $\displaystyle\det \left( {\begin{array}{*{20}{c}} {\frac{{\partial g}}{{\partial u}}} & {\frac{{\partial g}}{{\partial v}}} \\ {\frac{{\partial f}}{{\partial u}}} & {\frac{{\partial f}}{{\partial v}}} \\ \end{array} } \right)(0,0) > 0$.

In this paper, the authors relax the hypothesis (4), for the price of supposing exact growth of the solutions at infinity, and of the nonlinearities at zero. Precisely, they proved

Theorem. Let $(u, v)$ be a classical solution of the system and $f,g \in C^1([0,\infty) \times [0,\infty), \mathbb R)$. Suppose

1. $\displaystyle u(x) \sim \frac{1}{{{{\left| x \right|}^\alpha }}}, \quad v(x) \sim \frac{1}{{{{\left| x \right|}^\beta }}}, \quad \left| x \right| \to \infty$;
2. $\displaystyle \frac{{\partial g}}{{\partial u}}\lesssim {u^{p - 1}},\quad\frac{{\partial f}}{{\partial v}} \lesssim {v^{q - 1}},\quad u \to {0^ + },\quad v \to {0^ + }$;
3. $\displaystyle \frac{{\partial g}}{{\partial v}}\lesssim {u^{a - 1}},\quad\frac{{\partial f}}{{\partial u}} \lesssim {v^{b - 1}},\quad u \to {0^ + },\quad v \to {0^ + }$;
4. $\displaystyle \frac{{\partial g}}{{\partial v}} > 0,\quad\frac{{\partial f}}{{\partial u}} > 0,\quad\forall (u,v) \in (0,\infty ) \times (0,\infty )$;

where positive constants $\alpha, \beta, p, q, a, b$ satisfy

$\displaystyle\alpha (p - 1) > 2,\beta (q - 1) > 2,\alpha (a - 1) > 2,\beta (b - 1) > 2$.

Then there exists a point in $x_0 \in \mathbb R^n$, such that

$u(x)=u(|x-x_0|), \quad v(x)=v(|x-x_0|)$.

The proof used is the method of moving planes. This kind of system is also a generalization of the Lane-Emden system given by

$\displaystyle\left\{ \begin{gathered} - \Delta u = {v^b }, \hfill \\ - \Delta v = {u^a }. \hfill \\ \end{gathered} \right.$

It can also be applied to the following system

$\displaystyle\left\{ \begin{gathered} - \Delta u = - {u^p} + {v^b}, \hfill \\ - \Delta v = {u^a} - {v^q}. \hfill \\ \end{gathered} \right.$

However, it is no longer able to apply to

$\displaystyle\left\{ \begin{gathered} - \Delta u = {u^p} - {v^b}, \hfill \\ - \Delta v = - {u^a} + {v^q}. \hfill \\ \end{gathered} \right.$

In fact, hypothesis (4) plays an important role in their argument. To our knowledge, the latter case is still open.

## May 7, 2010

### The method of moving planes: An integral form

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 0:41

Today we study a beautiful version of the method of moving planes. This method just invented around 2003 by W.X. Chen, C.M. Li and B. Ou and published in Comm. Pure Appl. Math. around 2006 (for the paper, please go here). In that paper, the author proved among other things that every positive regular solution $u(x)$ of the integral equation

$\displaystyle u(x) = \int_{{\mathbb{R}^n}} {\frac{1}{{{{\left| {x - y} \right|}^{n - \alpha }}}}u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}dy}$

is radially symmetric and monotone about some point and therefore assumes the form

$\displaystyle u(x) = c{\left( {\frac{t}{{{t^2} + {{\left| {x - {x_0}} \right|}^2}}}} \right)^{\frac{{n - \alpha }}{2}}}$

with some constant $c=c(n,\alpha)$ and for some $t>0$ and $x_0 \in \mathbb R^n$. These solutions in case $n=3$ and $\alpha=2$ have the following shape.

This integral equation is also closely related to the following family of semilinear PDEs

$\displaystyle {( - \Delta )^{\frac{\alpha }{2}}}u = {u^{\frac{{n + \alpha }}{{n - \alpha }}}}$.

## May 6, 2010

### The method of moving planes: The case of the whole space

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 15:06

The trick considered here can be applied successfully to any bounded set $\Omega$ that has appropriate symmetry under the reflection in a hyperplane. But when the PDE holds in the whole space $\mathbb R^n$, and no boundary condition is available, one must first locate a possible center of symmetry by means of information about $u(x)$ for $|x|\to \infty$ and then try to derive some estimates.

Definition. Let $u \in C^1(\mathbb R^n)$. We shall say that $u$ has admissible assymptotic behavior iff one of the following conditions holds outside some ball, say for $r=|x| \geqslant R_u$. The numbers $\kappa$, $m$, $\delta$ and $\gamma$ are all strictly positive constants, with $\delta \in (0,1]$ and $c$ is a constant vector.

(A) $u(x)=-\kappa r^m + (c \cdot x) r^{m-2} + h(x)$ where $|\nabla h(x)|=O(r^{m-2-\delta}$ and if $m-1<\delta$ then $h(x) \to 0$ as $r \to \infty$.

(B) $u(x)=\kappa\log\frac{\gamma}{r} + (c \cdot x) r^{-2} + h(x)$ where $|\nabla h(x)|=O(r^{-2-\delta})$ and $h(x)\to 0$ as $r \to \infty$.

(C) $u(x)=\kappa r^{-m} + (c \cdot x) r^{-m-2} + h(x)$ where $|\nabla h(x)|=O(r^{-m-2-\delta})$ and $h(x)\to 0$ as $r \to \infty$.

The essential features of the above definition are these.

• First, $u(x)$ decreases as $r \to \infty$.
• Second, if $c \ne 0$, then $c \cdot x$ does not depend only on $r$.

Therefore if $u$ should turn out to be spherically symmetric about some point $q$ then $u(x+q)$ will contain no term $(c\cdot x)r^\alpha$. This enables us to find a priori the only point $q$ that can be a center of symmetry of the function $u$.

In fact in the literature, we have the following result

Theorem. If solution $u \in C^1(\mathbb R^n)$ of the following PDE

$-\Delta u = f(u)$

has admissible asymptotic behavior and $u>0$ in the case (C). Then under some restriction on the nonlinearity $f$, the following function

$v(x) := u(x+q)$

is spherically symmetric where $q$ is given by

$\displaystyle q = \left\{ \begin{gathered} \frac{1}{{\kappa m}}c, {\rm \; for \; cases \; (\mathbf{A}) \; and \; (\mathbf{C}) } \hfill \\ \frac{1}{\kappa }c, {\rm \; for \; case \; (\mathbf{B})}. \hfill \\ \end{gathered} \right.$

## May 4, 2010

### The method of moving planes: An introduction

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 1:17

Let us go back to the following problem that we have discussed before in this topic. Denote by $B_a$ the ball centered at the origin with radius $a>0$. Here we just consider our problem within a ball. We will take care the case of whole space $\mathbb R^n$.

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

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

where $f$ is locally Lipschitz in $\mathbb R$. Then $u$ is radially symmetric in $B_a$.

Since we haven’t provided any solutions to this problem, here I just figure out the idea and several steps needed during the proof.

Definition. For each unit vector $k \in \mathbb R^n$ and each number $\mu \in \mathbb R$, define a hyperplane by

$T_\mu(k)=\{\xi \in \mathbb R^n : \xi \cdot k = \mu\}$.

The reflection in $T_\mu(k)$ of any point $x \in \mathbb R^n$ is

$x^{\mu, k}=x+2(\mu - x \cdot k)k$

and the reflection in $T_\mu(k)$ of any function $\varphi : \mathbb R^n \to \mathbb R$ is defined by

$\varphi_{\mu,k}(x)=\varphi(x^{\mu,k})$

for all $x \in \mathbb R^n$.

Obviously, $T_\mu(k)$ is parallel to $T_{\mu'}(k)$ for any $\mu \ne \mu'$. This is because vector $k$ is perpendicular to $T_\mu(k)$ for any $\mu$. Since vector $k$ is chosen to be a unit vector, it is easy to check that the distance from the origin to $T_\mu(k)$ exactly equals $\mu$. The following lemma plays an important role in our argument

Lemma. If $v : \mathbb R^n \to \mathbb R$ has the property that

$v(x^{0,k})=v(x)$

for each unit vector $k$ and for all $x \in \mathbb R^n$, then $v$ is spherically symmetric (i.e. depends only on $|x|$).

## 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 $\lambda_0$.

latex