# 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.

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!).

Click to access degsumnewrevised.pdf

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.