# Ngô Quốc Anh

## April 30, 2010

### The Pohozaev identity: Toda systems and a priori estimates

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 10:15

In this topic we consider the analysis of solutions of the following system entitled Toda system

$\displaystyle\left\{ \begin{gathered} - \Delta {u_1} = 2{e^{{u_1}}} - {e^{{u_2}}}, \hfill \\ - \Delta {u_2} = - {e^{{u_1}}} + 2{e^{{u_2}}}. \hfill \\ \end{gathered} \right.$

Following is our main result

Lemma 1. The following identities

$\displaystyle\begin{gathered} \left( {1 - \frac{n}{2}} \right)\int_{{B_r}(0)} {|\nabla {u_1}{|^2}dx} + r\left[ {\frac{1}{2}\int_{\partial {B_r}(0)} {|\nabla {u_1}{|^2}d\sigma } - \int_{\partial {B_r}(0)} {{{\left| {\frac{{\partial {u_1}}}{{\partial \nu }}} \right|}^2}d\sigma } } \right] \hfill \\ \qquad\qquad= - 2n\int_{{B_r}(0)} {{e^{{u_1}}}dx} + 2\int_{\partial {B_r}(0)} {r{e^{{u_1}}}dx} - \int_{{B_r}(0)} {{e^{{u_2}}}(x\cdot\nabla {u_1})dx} , \hfill \\ \left( {1 - \frac{n}{2}} \right)\int_{{B_r}(0)} {|\nabla {u_2}{|^2}dx} + r\left[ {\frac{1}{2}\int_{\partial {B_r}(0)} {|\nabla {u_2}{|^2}d\sigma } - \int_{\partial {B_r}(0)} {{{\left| {\frac{{\partial {u_2}}}{{\partial \nu }}} \right|}^2}d\sigma } } \right] \hfill \\ \qquad\qquad= - 2n\int_{{B_r}(0)} {{e^{{u_2}}}dx} + 2\int_{\partial {B_r}(0)} {r{e^{{u_2}}}dx} - \int_{{B_r}(0)} {{e^{{u_1}}}(x\cdot\nabla {u_2})dx} , \hfill \\ \end{gathered}$

hold.

## April 18, 2010

### The Pohozaev identity: Elliptic problem with biharmonic operator

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 14:19

We now consider another kind of problem involving biharmonic operator. Let us assume $u>0$ a solution of the equation

$\displaystyle (\Delta^2u)(x)+Q(x)f(u(x))=0$

in $\mathbb R^n$. We shall prove the following result

Theorem. The following identity

$\displaystyle\begin{gathered} \frac{3}{2}\left[ {\int_{{B_r}(0)} {uQ(x)f(u)dx} - \int_{\partial {B_r}(0)} {u\frac{{\partial \Delta u}}{{\partial \nu }}d\sigma } + \int_{\partial {B_r}(0)} {\frac{{\partial u}}{{\partial \nu }}\Delta ud\sigma } } \right] + \hfill \\ \frac{1}{2}\int_{\partial {B_r}(0)} {r{{(\Delta u)}^2}d\sigma } + \int_{{B_r}(0)} {(x\cdot\nabla (\Delta u))\Delta udx} - \int_{\partial {B_r}(0)} {\left[ {\Delta u\frac{{\partial (x\cdot\nabla u)}}{{\partial \nu }} - (x\cdot\nabla u)\frac{{\partial \Delta u}}{{\partial \nu }}} \right]d\sigma } \hfill \\ \qquad\qquad= \int_{\partial {B_r}(0)} {ruQ(x)f(u)d\sigma } - \int_{{B_r}(0)} {u\left( {Q(x)f(u) + (x\cdot\nabla Q)f(u) + (x\cdot\nabla u)Q(x)f'(u)} \right)dx}\hfill \\ \end{gathered}$

holds.

## April 14, 2010

### The Pohozaev identity: Integral equation with exponential nonlinearity

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 1:15

We now consider the Pohozaev identity for some integral equations. We start with the following equation

$\displaystyle u(x) = \frac{2}{{{\omega _n}}}\int_{{\mathbb{R}^n}} {\log \frac{{|y|}}{{|x - y|}}K(y){e^{nu(y)}}dy} + {C_0}$

where $\omega_n$ is the volume of the unit sphere in $\mathbb R^{n+1}$ and $K$ and $C_0$ are a smooth function in $\mathbb R^n$ and a constant, respectively.

Theorem. Suppose $u$ is a $C^1$ solution of the above integral equation such that $K(x)e^{nu(x)}$ is absolutiely integrable over $\mathbb R^n$. And if one sets

$\displaystyle\alpha = \frac{2}{{{\omega _n}}}\int_{{\mathbb{R}^n}} {K(y){e^{nu(y)}}dy}$

then

$\displaystyle- \infty < \alpha < \infty$

and the following identity holds

$\displaystyle\alpha (\alpha - 2) = \frac{4}{{n{\omega _n}}}\int_{{\mathbb{R}^n}} {\left\langle {y,\nabla K(y)} \right\rangle {e^{nu(y)}}dy}$.

This theorem was due to Xu X.W. from the paper published in J. Funct. Anal. (2005). When $n=2$, it was due to Cheng and Lin from this paper published in Math. Ann. (1997).

Finiteness for $\alpha$ is just the assumption of the integrability of the function $K(x)e^{nu(x)}$. Here we mainly need to show the identity holds true.

## April 13, 2010

### The Pohozaev identity: Semilinear elliptic problem with exponential nonlinearity

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 2:07

We know consider another type of equation, precisely, we consider the positive solution to the following

$\displaystyle -\Delta u=Ke^{2u}$

in $\mathbb R^n$. Following is what we need to prove.

Theorem. The following identity

$\displaystyle\begin{gathered} \left( {1 + \frac{n}{2}} \right)\int_{{B_r}(0)} {uK{e^{2u}}dx} + \int_{{B_r}(0)} {u\left( {(x \cdot \nabla K){e^{2u}} + 2(x \cdot \nabla u)K{e^{2u}}} \right)dx} \hfill \\ \qquad= \int_{\partial {B_r}(0)} {ruK{e^{2u}}d\sigma } - \int_{\partial {B_r}(0)} {\left[ {\left( {1 - \frac{n}{2}} \right)u\frac{{\partial u}}{{\partial \nu }} + \frac{1}{2}r|\nabla u{|^2} - r{{\left| {\frac{{\partial u}}{{\partial \nu }}} \right|}^2}} \right]d\sigma } \hfill \\ \end{gathered}$

holds.

## April 11, 2010

### The Pohozaev identity: Semilinear elliptic problem with polygonal nonlinearity

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 12:15

Let us start with the Pohozaev identity for semilinear elliptic equation with polygonal nonlinear terms of the form

$\displaystyle -\Delta u = |u|^{p-1}u$

over an open, star-shaped domain $\Omega$. We also assume $u$ is identical to zero on the boundary $\partial \Omega$.

We multiply the PDE by $x \cdot \nabla u$ and integrate over $\Omega$ to find

$\displaystyle\int_\Omega {\left( { - \Delta u} \right)\left( {x \cdot \nabla u} \right)} dx = \int_\Omega {|u{|^{p - 1}}u\left( {x \cdot \nabla u} \right)} dx$.

The term on the left is just

$\displaystyle\begin{gathered} A = - \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {\int_\Omega {{u_{{x_i}{x_i}}}{x_j}{u_{{x_j}}}dx} } } \hfill \\ \quad= - \sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {\left[ { - \int_\Omega {{u_{{x_i}}}{{({x_j}u)}_{{x_i}}}dx} + \int_{\partial \Omega } {{u_{{x_i}}}{\nu ^i}{x_j}{u_{{x_j}}}d\sigma } } \right]} } \hfill \\ \quad= \underbrace {\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {\int_\Omega {{u_{{x_i}}}{{({x_j}u_{x_j})}_{{x_i}}}dx} } } }_{{A_1}} - \underbrace {\sum\limits_{j = 1}^n {\sum\limits_{i = 1}^n {\int_{\partial \Omega } {{u_{{x_i}}}{\nu ^i}{x_j}{u_{{x_j}}}d\sigma } } } }_{{A_2}}. \hfill \\ \end{gathered}$