Ngô Quốc Anh

April 16, 2020

Restriction of gradient, Laplacian, etc on level sets and applications

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 18:10

This topic is devoted to proofs of several interesting identities involving derivatives on level sets. First, we start with the case of gradient. We shall prove

The first identity

\displaystyle \partial_\nu f = \pm |\nabla f|

on the level set

\displaystyle \big\{ x \in \mathbf R^n : f(x) =0 \big\}.

The above identity shows that while the right hand side involves the value of f in a neighborhood, however, the left hand side indicates that only the normal direction is affected. Heuristically, any change of f along the level set does not contribute to any derivative of f, namely, on the boundary of the level set, the norm of \nabla f is actually the normal derivative \partial_\nu f. Therefore, the only direction taking into derivatives of f is in the normal direction and this should be true for higher-order derivatives of f.

Next we prove the following

The second identity

\displaystyle \partial_\nu \big(x \cdot \nabla f \big)=(\partial_\nu^2 f) (x \cdot \nu)

on the level set

\displaystyle \big\{ x \in \mathbf R^n : \partial_1 f(x) = \cdots = \partial_n f(x) = 0 \big\}.

Combining the above two identities, we can prove

The third identity

\displaystyle \partial_\nu^2 f=\Delta f

on the level set

\displaystyle \big\{ x \in \mathbf R^n : \partial_1 f(x) = \cdots = \partial_n f(x) = 0 \big\}

which basically tells us how to compute the restriction of Laplacian on level sets. This note is devoted to a rigorous proof of the above facts together with a simple application of all these identities.


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}



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}



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}


\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}



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}


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