Ngô Quốc Anh

June 3, 2010

Lower bound for integral of exp(u)

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 21:51

As mentioned before, today I will derive a very short and beautiful proof concerning the lower bound of \int\exp(u(x))dx where u, a positive solution to the following PDE

\displaystyle -\Delta u =e^{u(x)}, \quad x \in \mathbb R^2.

This proof I firstly learned from a paper published in Duke Math. J. in 1991 by W. Cheng and C. Li [here].

We assume

\displaystyle \int_{\mathbb R^2} e^{u(x)}dx < \infty.

Denote by \Omega_t the following set

\Omega_t = \{ x \in \mathbb R^2 : u(x)>t\}.

It follows from this topic that

\displaystyle \int_{\Omega_t} e^{u(x)}dx =-\int_{\Omega_t}\Delta u dx = \int_{\partial \Omega_t} |\nabla u|d\sigma.

Also, it follows from this topic that

\displaystyle -\frac{d}{dt}\int_{\Omega_t}dx =\int_{\partial \Omega_t} \frac{1}{|\nabla u|}d\sigma.

Thus, by the Schwarz inequality and the isoperimetric inequality

\displaystyle \left(\int_{\partial\Omega_t} \frac{1}{|\nabla u|}d\sigma\right)\left(\int_{\partial\Omega_t}|\nabla u| d\sigma\right) \geqslant |\partial\Omega_t|^2 \geqslant 4\pi |\Omega_t|.


\displaystyle -\left(\frac{d}{dt}\int_{\Omega_t}dx\right) \left(\int_{\Omega_t} e^{u(x)}dx\right)\geqslant 4\pi |\Omega_t|.


\displaystyle \frac{d}{dt}\left(\int_{\Omega_t} e^{u(x)}dx\right)^2=2\left(\int_{\Omega_t} e^{u(x)}dx\right)\frac{d}{dt}\left(\int_{\Omega_t} e^{u(x)}dx\right).

It is worth noticing that

\displaystyle\frac{d}{dt}\left(\int_{\Omega_t} e^{u(x)}dx\right)=e^t\frac{d}{dt}\left(\int_{\Omega_t}dx\right)

which yields

\displaystyle \frac{d}{dt}\left(\int_{\Omega_t} e^{u(x)}dx\right)^2=2\left(\int_{\Omega_t} e^{u(x)}dx\right)e^t\frac{d}{dt}\left(\int_{\Omega_t}dx\right) \leqslant -8\pi e^t\int_{\Omega_t}dx.

Integrating from -\infty to \infty gives

\displaystyle -\left(\int_{\mathbb R^2} e^{u(x)}dx\right)^2\leqslant -8\pi\int_{\mathbb R^2}e^{u(x)}dx

which implies

\displaystyle \int_{\mathbb R^2} e^{u(x)}dx \geqslant 8\pi.


Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at

%d bloggers like this: