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!).

    http://people.sissa.it/~malchiod/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.

      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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