# Ngô Quốc Anh

## August 16, 2010

### The Moser-Trudinger inequality for domains with holes

Filed under: PDEs — Tags: — 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.

## 13 Comments »

1. 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!).

and all the related results.

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

• 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

2. 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

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

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

• 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

4. Dear Ngo,
in your opinion is it possible to extend this argument to all $u\in W_{0}^{1,N}$?

Comment by Fab — October 18, 2013 @ 23:51

5. Clearly for a bounded smooth domain in ${\mathbb R}^{N}$!

Comment by Fab — October 18, 2013 @ 23:54

6. In the second line of the proof there is a misprint, $\Omega_i$ are $S_i$.

Comment by Fab — October 19, 2013 @ 0:00

• Thanks Fab, yes you are right, it should be $1 \geqslant g_i \geqslant 0, \quad g_i(x) \equiv 1 \quad \text{ for all } x \in S_i, \quad i=1,2$

as you said.

Comment by Ngô Quốc Anh — October 19, 2013 @ 3:03

7. Sorry Ngo, what about my previous question?
I’m quite sure of this.

Comment by Fab — October 19, 2013 @ 17:00

• Dear Fab,

I still need time to think about your question. My first feeling is that it should be okay.

Comment by Ngô Quốc Anh — October 20, 2013 @ 23:56

8. This is my feeling too!

Comment by Fab — October 21, 2013 @ 20:46

• Dear Fab,

Except the singularities, the following result found in http://dx.doi.org/10.1016/j.na.2013.10.010 could be useful for you.

Lemma 2.1 (Trudinger–Moser inequality for bounded domains). Let $\Omega \subset \mathbb R^N$ ( $N \geqslant 2$) be a bounded domain. Given any $u\in W^{1,N}_0(\Omega)$, we have $\displaystyle \int_\Omega \exp\left( \alpha |u|^\frac{N}{N-1}\right) dx \leqslant \infty$

for every $\alpha>0$. Moreover, there exists a positive constant $C=C(N,|\Omega|)$ such that $\displaystyle \sup_{\|u\|_{W_0^{1,N}(\Omega)} \leqslant 1}\int_\Omega \exp\left( \alpha |u|^\frac{N}{N-1}\right) dx \leqslant C$

for all $\alpha \leqslant \alpha_N$ where $\alpha_N = N\omega_{N-1}^{1/(N-1)}>0$ and $\omega_{N-1}$ is the ( $N-1$)-dimensional measure of the ( $N-1$)-sphere.

This could help you to state a similar result with holes.

Comment by Ngô Quốc Anh — November 2, 2013 @ 0:09

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