The Moser-Trudinger inequality for domains with holes

August 16, 2010

The Moser-Trudinger inequality for domains with holes

Filed under: PDEs — Tags: Moser-Trudinger's inequality — Ngô Quốc Anh @ 2:42

In this entry, we are interested in the following result

Theorem (Moser-Trudinger’s inequality for domains with holes). Let $\Omega$ be a bounded smooth domain in $\mathbb R^2$ . Let $S_1$ and $S_2$ be two subsets of $\overline \Omega$ satisfying

${\rm dist}(S_1,S_2) \geqslant \delta_0>0$

and let $\gamma_0$ be a number satisfying $\gamma_0 \in \left(0,\frac{1}{2}\right)$ . Then for any $\varepsilon>0$ , there exists a constant $c=c(\varepsilon, \delta_0, \gamma_0)>0$ such that

$\displaystyle\int_\Omega {{e^u}} \leqslant C\exp \left[ {\frac{1}{{32\pi - \varepsilon }}\int_\Omega {{{\left| {\nabla u} \right|}^2}} + C} \right]$

holds for all $u \in H_0^1(\Omega)$ satisfying

$\displaystyle\frac{{\int_{{S_1}} {{e^u}} }}{{\int_\Omega {{e^u}} }} \geqslant {\gamma _0}, \quad \frac{{\int_{{S_2}} {{e^u}} }}{{\int_\Omega {{e^u}} }} \geqslant {\gamma _0}$ .

Proof. Let $g_1$ and $g_2$ be two smooth functions such that

$1 \geqslant g_i \geqslant 0, \quad g_i(x) \equiv 1 \quad \text{ for all } x \in \Omega_i, \quad i=1,2,$

and

${\rm supp}g_1 \cap {\rm supp}g_2 = \emptyset$ .

It suffices to show that for all $u \in H_0^1(\Omega)$ with

$\displaystyle \int_\Omega u=0$

we have

$\displaystyle\int_\Omega {{e^u}} \leqslant C\exp \left[ {\frac{1}{{32\pi - \varepsilon }}\int_\Omega {{{\left| {\nabla u} \right|}^2}} } \right]$ .

There are two possible cases

Case 1. If

$\displaystyle\int_\Omega {{{\left| {\nabla ({g_1}u)} \right|}^2}} \leqslant \int_\Omega {{{\left| {\nabla ({g_2}u)} \right|}^2}}$

then by our hypothesis

$\displaystyle\int_\Omega {{e^u}} \leqslant \frac{1}{{{\gamma _0}}}\int_{{S_1}} {{e^u}} \leqslant \frac{1}{{{\gamma _0}}}\int_\Omega {{e^{{g_1}u}}}$ .

It follows from the original Moser-Trudinger inequality that

$\displaystyle\int_\Omega {{e^{{g_1}u}}} \leqslant c\exp \left[ {\frac{1}{{32\pi }}\int_\Omega {{{\left| {\nabla ({g_1}u)} \right|}^2}} + \overline {{g_1}u} } \right]$ .

Precisely, from

$\displaystyle {g_1}u - \overline {{g_1}u} \leqslant 4\pi \frac{{{g_1}u - \overline {{g_1}u} }}{{\int_\Omega {{{\left| {\nabla ({g_1}u)} \right|}^2}} }} + \frac{1}{{16\pi }}\int_\Omega {{{\left| {\nabla ({g_1}u)} \right|}^2}}$

we have

$\displaystyle\int_\Omega {{e^{{g_1}u - \overline {{g_1}u} }}} \leqslant \exp \left[ {\frac{1}{{32\pi }}\int_\Omega {{{\left| {\nabla ({g_1}u)} \right|}^2}} } \right]\int_\Omega {\exp \left( {4\pi \frac{{{g_1}u - \overline {{g_1}u} }}{{\int_\Omega {{{\left| {\nabla ({g_1}u)} \right|}^2}} }}} \right)} \leqslant c\exp \left[ {\frac{1}{{16\pi }}\int_\Omega {{{\left| {\nabla ({g_1}u)} \right|}^2}} } \right]$ .

Thus

$\displaystyle\begin{gathered} \int_\Omega {{e^u}} \leqslant \frac{c}{{{\gamma _0}}}\exp \left[ {\frac{1}{{32\pi }}\int_\Omega {{{\left| {\nabla ({g_1}u)} \right|}^2}} + \overline {{g_1}u} } \right] \hfill \\ \qquad\leqslant \frac{c}{{{\gamma _0}}}\exp \left[ {\frac{1}{{32\pi }}\int_\Omega {{{\left| {\nabla ({g_1}u + {g_2}u)} \right|}^2}} + \overline {{g_1}u} } \right] \hfill \\ \qquad\leqslant \frac{{c(\varepsilon )}}{{{\gamma _0}}}\exp \left[ {\frac{1}{{32\pi - \varepsilon }}\int_\Omega {{{\left| {\nabla u} \right|}^2}} + {c_1}(\varepsilon )\int_\Omega {{{\left| u \right|}^2}} } \right] \hfill \\ \end{gathered}$

In order to get rid of the term $\|u\|_2$ on the right hand side of the above inequality, we employ the condition $\int_\Omega u=0$ .

Given any $\eta>0$ , choose $a$ such that

${\rm meas}\{x \in \Omega: u(x) \geqslant a\}=\eta$ .

We then have

$\displaystyle\int_\Omega {{e^u}} \leqslant {e^a}\int_\Omega {{e^{u - a}}} \leqslant {e^a}\int_\Omega {{e^{{{(u - a)}^ + }}}} \leqslant C\exp \left[ {\frac{1}{{32\pi - \varepsilon }}\int_\Omega {{{\left| {\nabla u} \right|}^2}} + {c_1}(\varepsilon )\int_\Omega {{{\left| {{{(u - a)}^ + }} \right|}^2}} + a} \right]$ .

By the Sobolev inequality,

$\displaystyle\int_\Omega {{{\left| {{{(u - a)}^ + }} \right|}^2}} \leqslant \sqrt \eta \sqrt {\int_\Omega {{{\left| {{{(u - a)}^ + }} \right|}^4}} } \leqslant c\sqrt \eta \int_\Omega {{{\left| {\nabla u} \right|}^2}}$ .

By the Poincare inequality,

$\displaystyle a\eta \leqslant \int_{\left\{ {u \geqslant a} \right\}} u \leqslant \int_\Omega {|u|} \leqslant c\int_\Omega {{{\left| {\nabla u} \right|}^2}}$ .

Hence for any $\delta>0$ ,

$\displaystyle a \leqslant \delta \int_\Omega {{{\left| {\nabla u} \right|}^2}} + \frac{{{c^2}}}{{4\delta {\eta ^2}}}$ .

The proof now follows.

Case 2. If

$\displaystyle\int_\Omega {{{\left| {\nabla ({g_2}u)} \right|}^2}} \leqslant \int_\Omega {{{\left| {\nabla ({g_1}u)} \right|}^2}}$

then by an argument similar to that in Case 1 we obtain the desired inequality.

Remark. This kind of inequality plays a central role in studying Riemannian surfaces with conical singularities. The proof above is adapted from a paper due to W.C and C.L [here] published in the Journal of Geometric Analysis in 1991.

Comments (5)

5 Comments »

This improved Moser Trudinger inequality is really crucial in the beautiful works of Malchiodi and others.
Do you Know this papers? I’m trying to understand a little bit of this stuff but it is very very hard (at least for me!).

http://people.sissa.it/~malchiod/degsumnewrevised.pdf
and all the related results.

Comment by Fab — February 14, 2012 @ 19:15

Reply
- Hi Fab, thanks for the paper. Please raise your question here then we can discuss.
  
  Comment by Ngô Quốc Anh — February 14, 2012 @ 19:17
  
  Reply
First of all, Malchiodi rewrite the inequality above in your notes for $l+1$ holes then he show a criterion which implies the situation described in your third inequalities. As a consequence, if $p$ belongs to $(8k\pi, 8(k+1)\pi)$ where $p$ is the real parameter in the mean field equation on compact surface and if the Euler Lagrange functional associated to this equation attains large negative values then $e^u$ has to concentrate near at most $k$ points of the surface.

Comment by Fab — February 14, 2012 @ 19:42

Reply
Hi Ngo,
why don’t you post something about the Lyapunov Schmidt reduction?

Comment by Fab — February 23, 2012 @ 17:07

Reply
- Dear Fab, you can read my honours project done when I was an undergraduate student here . I have used that method in that paper.
  
  Comment by Ngô Quốc Anh — February 23, 2012 @ 17:12
  
  Reply

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Ngô Quốc Anh

August 16, 2010