Ngô Quốc Anh

August 13, 2010

An application of the (Moser-)Trudinger inequality to the mean field equations

Let (M,g) be a compact Riemannian surface with the volume |M|. The simplest form of the mean field equation studied in the contexts of the prescribing Gaussian curvature, statistical mechanics of many vortex points in the perfect fluid and self-dual gauss theories is given by

\displaystyle - {\Delta _g}u = \lambda \left( {\frac{{{e^u}}}{{\int_M {{e^u}d{v_g}} }} - \frac{1}{{|M|}}} \right), \quad \text{ on } M

with

\displaystyle\int_M {ud{v_g}} = 0

where \lambda is a real number.

The mean field equation has a variational structure, and u is a solution if and only if it is a critical point of

\displaystyle {J_\lambda }(v) = \frac{1}{2}\int_M {{{\left| {\nabla v} \right|}^2}d{v_g}} - \lambda \log \int_M {{e^v}d{v_g}}

defined for v \in H^1(M) with

\displaystyle\int_M {vd{v_g}} = 0.

It is worth noticing from this entry that so far our Moser-Trudinger’s inequality is just for \mathbb S^2

Theorem (Moser-Trudinger’s inequality for \mathbb S^2). There are constants \eta>0 and c=c(g)>0 such that for each p \geqslant 2

\displaystyle \log \overline\int_{{\mathbb{S}^2}} {{e^{2u}}d{v_{{g_c}}}}  \leqslant \left[ {\frac{1}{{4\pi }}\int_{{\mathbb{S}^2}} {{{\left|  {\nabla u} \right|}^2}d{v_{{g_c}}}} + 2\overline u } \right] + {C_2}

for all u \in W^{1,2}(\mathbb S^2).

Thank to a work due to Luigi Fontana [here] we actually have Moser-Trudinger’s inequality for general manifold of a higher order gradient of function. Presicely,

Theorem (Moser-Trudinger’s inequality for general manifolds). Let (M,g) be a compact Riemannian manifold of dimension n and m a positive integer strictly smaller than n. There exists a constant C=C(m,M) such that for all u \in C^n(M) with

\displaystyle\int_M {ud{v_g}} = 0

and

\displaystyle\int_M {|\nabla^m u|^\frac{n}{m}d{v_g}} \leqslant 1

the following uniform inequality holds

\displaystyle \int_M {\exp \left( {\lambda (m,n){{\left| u \right|}^{\frac{n}{{n - m}}}}} \right)d{v_g}} \leqslant C

where the constant \lambda(m,n) is given below

\displaystyle\lambda (m,n) = \begin{cases} \dfrac{n}{{{\omega _{n - 1}}}}{\left[ {\dfrac{{{\displaystyle\pi ^{\frac{n}{2}}}{2^m}\Gamma \left( {\dfrac{{m + 1}}{2}} \right)}}{{\Gamma \left( {\dfrac{{n - m + 1}}{2}} \right)}}} \right]^{\frac{n}{{n - m}}}},&m \text{ odd, }\hfill \\ \dfrac{n}{{{\omega _{n - 1}}}}{\left[ {\frac{{{\displaystyle\pi ^{\frac{n}{2}}}{2^m}\Gamma \left( {\dfrac{m}{2}} \right)}}{{\Gamma \left( {\dfrac{{n - m}}{2}} \right)}}} \right]^{\frac{n}{{n - m}}}},&m \text{ even}. \hfill \\ \end{cases}

Consequently, for a general Riemannian surface, i.e. n=2, m=1 one gets \lambda(1,2)=\frac{1}{16\pi} and thus

Corollary. There holds

\displaystyle\int_M {{e^{\frac{1}{{16\pi }}{{\left| u \right|}^2}}}d{v_g}} \leqslant C.

For a general function u, we have

\displaystyle u - \overline u \leqslant \frac{1}{{16\pi }}\frac{{{{\left( {u - \overline u } \right)}^2}}}{{\left\| {\nabla (u - \overline u )} \right\|_{{L^2}(M)}^2}} + 4\pi \left\| {\nabla (u - \overline u )} \right\|_{{L^2}(M)}^2

which after integrating implies

\displaystyle\int_M {{e^{u - \overline u }}d{v_g}} \leqslant C{e^{4\pi \left\| {\nabla (u - \overline u )} \right\|_{{L^2}(M)}^2}}.

Hence, if \overline u=0 we deduce

\displaystyle\log \int_M {{e^u}d{v_g}} \leqslant 4\pi \int_M {{{\left| {\nabla u} \right|}^2}d{v_g}} + \log C.

Consequently, we get

\displaystyle - \frac{{\log C}}{{8\pi }} \leqslant \frac{1}{2}\int_M {{{\left| {\nabla u} \right|}^2}d{v_g}} - \frac{1}{{8\pi }}\log \int_M {{e^u}d{v_g}}.

The above inequality is about to say if \lambda=8\pi, the functional J is bounded from below, and thus has global minimizer which turns out to be a critical point, hence, a solution. When \lambda < 8\pi, the functional J is still bounded from below (by a different approach) but it is not when \lambda > 8\pi. The case of \lambda>8\pi is not completely solved. We refer the reader to works of Ohtsuka, Malchiodi, Jost, etc.

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:

WordPress.com Logo

You are commenting using your WordPress.com 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 WordPress.com.

%d bloggers like this: