Ngô Quốc Anh

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.

Since w is assumed to be C^1, \nabla w exists and is continuous. It is not hard to justify that we can take the derivative under the integral sign. First, differentiate the equation to get

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

which implies

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

Now one multiplies both sides by K(x)e^{nu(x)} and integrate the resulting equation both sides over the ball B_R(0) for any R>0, one obtains

\displaystyle\int_{{B_R}(0)} {K(x){e^{nu(x)}}\left[ { - \frac{2}{{{\omega _n}}}\int_{{\mathbb{R}^n}} {\frac{{\left\langle {x,x - y} \right\rangle }}{{|x - y{|^2}}}K(y){e^{nu(y)}}dy} } \right]dx} = \int_{{B_R}(0)} {K(x){e^{nu(x)}}\left\langle {x,\nabla u(x)} \right\rangle dx} .

Now on the right hand side, one uses the divergence theorem

\displaystyle\int_U {u{v_{{x_i}}}dx} = - \int_U {{u_{{x_i}}}vdx} + \int_{\partial U} {uv{\nu ^i}d\sigma }

to get

\displaystyle\begin{gathered} \int_{{B_R}(0)} {K(x){e^{nu(x)}}\left\langle {x,\nabla u(x)} \right\rangle dx} = \frac{1}{n}\int_{{B_R}(0)} {K(x)\left\langle {x,\nabla \left( {{e^{nu(x)}}} \right)} \right\rangle dx} \hfill \\ \qquad= \frac{1}{n}\sum\limits_{i = 1}^n {\int_{{B_R}(0)} {K(x){x_i}{{\left( {{e^{nu(x)}}} \right)}_{{x_i}}}dx} } \hfill \\ \qquad= \frac{1}{n}\sum\limits_{i = 1}^n {\left[ { - \int_{{B_R}(0)} {{{\left( {K(x){x_i}} \right)}_{{x_i}}}{e^{nu(x)}}dx} - \int_{{B_R}(0)} {K(x){x_i}{e^{nu(x)}}\frac{{{x_i}}}{R}d\sigma } } \right]} \hfill \\ \qquad= \frac{1}{n}\sum\limits_{i = 1}^n {\left[ { - \int_{{B_R}(0)} {\left( {K(x) + K{{(x)}_{{x_i}}}{x_i}} \right){e^{nu(x)}}dx} - \int_{\partial {B_R}(0)} {K(x){x_i}{e^{nu(x)}}\frac{{{x_i}}}{R}d\sigma } } \right]} \hfill \\ \qquad= - \int_{{B_R}(0)} {\left[ {K(x) + \frac{1}{n}\left\langle {x,\nabla K(x)} \right\rangle } \right]{e^{nu(x)}}dx} + \frac{1}{n}\int_{\partial {B_R}(0)} {K(x){e^{nu(x)}}Rd\sigma } . \hfill \\ \end{gathered}

Letting R \to \infty, one has

\displaystyle RHS = - \int_{\mathbb R^n} {\left[ {K(x) + \frac{1}{n}\left\langle {x,\nabla K(x)} \right\rangle } \right]{e^{nu(x)}}dx}

since \alpha is finite, the second term goes to zero as R \to \infty.

While on the left hand side, by using the fact that

\displaystyle\left\langle {x,x - y} \right\rangle = \frac{{|x - y{|^2} + \left\langle {x + y,x - y} \right\rangle }}{2}

one has the following identity

\displaystyle\begin{gathered} \int_{{B_R}(0)} {K(x){e^{nu(x)}}\left[ { - \frac{2}{{{\omega _n}}}\int_{{\mathbb{R}^n}} {\frac{{\left\langle {x,x - y} \right\rangle }}{{|x - y{|^2}}}K(y){e^{nu(y)}}dy} } \right]dx} \hfill \\ \quad= \frac{1}{2}\int_{{B_R}(0)} {K(x){e^{nu(x)}}\left[ { - \frac{2}{{{\omega _n}}}\int_{{\mathbb{R}^n}} {K(y){e^{nu(y)}}dy} } \right]dx} \hfill \\ \qquad+ \frac{1}{2}\int_{{B_R}(0)} {K(x){e^{nu(x)}}\left[ { - \frac{2}{{{\omega _n}}}\int_{{\mathbb{R}^n}} {\frac{{\left\langle {x + y,x - y} \right\rangle }}{{|x - y{|^2}}}K(y){e^{nu(y)}}dy} } \right]dx} . \hfill \\ \end{gathered}

Now it is easy to see that the last term will vanish when one takes the limit R \to \infty simply by changing variables x and y, the Fubini theorem, and the fact that

\displaystyle\left\langle {x + y,x - y} \right\rangle = - \left\langle {y + x,y - x} \right\rangle.


\displaystyle LHS = - \frac{1}{2}\alpha \int_{{\mathbb{R}^n}} {K(x){e^{nu(x)}}dx}.

The theorem follows.

The following is a very beautiful result.

Corollary. Positive solution to the following integral equation

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

verifies the following condition

\displaystyle 1= \frac{1}{{{\omega  _n}}}\int_{{\mathbb{R}^n}} {{e^{nu(y)}}dy}.

See also:

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