# Ngô Quốc Anh

## August 2, 2010

### The Brezis-Merle inequality

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 0:25

I am going to talk about uniform estimates and blow-up phenomena for solutions of

$\displaystyle -\Delta u=V(x)e^u$

in two dimensions done by Brezis and Merle around 1991 published in Comm. Partial Differential Equations [here]. As a first step, I am going to derive some inequality that we need later.

Assume $\Omega \subset \mathbb R^2$ is bounded domain and let $u$ be a solution of

$\displaystyle -\Delta u=f(x)$

together with Dirichlet boundary condition. Here function $f$ is assumed to be of class $L^1(\Omega)$.

Theorem (Brezis-Merle). For every $\delta \in (0,4\pi)$ we have

$\displaystyle\int_\Omega {\exp \left[ {\frac{{(4\pi - \delta )|u(x)|}}{{{{\left\| f \right\|}_1}}}} \right]dx} \leqslant \frac{{4{\pi ^2}}}{\delta }{\rm diam}{(\Omega )^2}$

where $\|\cdot\|_1$ denotes the $L^1$-norm and $u$ a solution to our PDE.

Proof. Let

$\displaystyle R=\frac{1}{2}{\rm diam}(\Omega)$

so that $\Omega \subset B_R$ for some ball of radius $R$. Extend $f$ to be zero outside $\Omega$ and set, for $x \in \mathbb R^2$,

$\displaystyle\overline u (x) = \frac{1}{{2\pi }}\int_{{B_R}} {\log \frac{{2R}}{{\left| {x - y} \right|}}|f(y)|dy}$

so that

$-\Delta \overline u = |f|$

on $\mathbb R^2$. Note that $\overline u \geqslant 0$ for $x \in B_R$ since $\frac{2R}{|x-y|} \geqslant 1$. Besides,

$-\Delta (\overline u-u) = |f|-f \geqslant 0$.

It follows from the maximum principle that $|u| \leqslant \overline u$ on $\Omega$ and thus

$\displaystyle\int_\Omega {\exp \left[ {\frac{{(4\pi - \delta )|u(x)|}}{{{{\left\| f \right\|}_1}}}} \right]dx} \leqslant \int_\Omega {\exp \left[ {\frac{{(4\pi - \delta )|\overline u (x)|}}{{{{\left\| f \right\|}_1}}}} \right]dx}$.

We now use the Jensen inequality

$\displaystyle F\left( {\int {w(y)\varphi (y)dy} } \right) \leqslant \int {w(y)F(\varphi (y))dy}$

with

$\displaystyle F(t) = {e^t}, \quad w(y) = \frac{{|f(y)|}}{\|f\|_1}, \quad \varphi (y) = \frac{{4\pi - \delta }}{{2\pi }}\log \frac{{2R}}{{\left| {x - y} \right|}}$

in order to estimate the RHS. We obtain

$\displaystyle\begin{gathered} \int_{{B_R}} {\exp \left[ {\frac{{(4\pi - \delta )|\overline u (x)|}}{{{{\left\| f \right\|}_1}}}} \right]dx} \leqslant \int_{{B_R}} {\left[ {\int_{{B_R}} {{{\left( {\frac{{2R}}{{\left| {x - y} \right|}}} \right)}^{2 - \frac{\delta }{{2\pi }}}}\frac{{|f(y)|}}{\|f\|_1}dy} } \right]dx} \hfill \\ \qquad\qquad= \int_{{B_R}} {\left[ {\int_{{B_R}} {{{\left( {\frac{{2R}}{{\left| {x - y} \right|}}} \right)}^{2 - \frac{\delta }{{2\pi }}}}dx} } \right]\frac{{|f(y)|}}{\|f\|_1}dy} . \hfill \\ \end{gathered}$

But for $y \in B_R$ we have

$\displaystyle\int_{{B_R}} {{{\left( {\frac{{2R}}{{\left| {x - y} \right|}}} \right)}^{2 - \frac{\delta }{{2\pi }}}}dx} \leqslant \int_{{B_R}} {{{\left( {\frac{{2R}}{{\left| x \right|}}} \right)}^{2 - \frac{\delta }{{2\pi }}}}dx} = \frac{{4{\pi ^2}}}{\delta }{\rm diam}{(\Omega )^2}$.

And thus the proof follows.

A simple consequence of the theorem is

Corollary. Let $u$ be a solution of PDE with $f \in L^1(\Omega)$. Then for every constant $k>0$

$\displaystyle e^{k|u|} \in L^1(\Omega)$.

Proof. Let

$0<\varepsilon<\frac{1}{k}$.

We may split $f$ as $f=f_1+f_2$ with $\|f_1\|_1 < \varepsilon$ and $f_2 \in L^\infty(\Omega)$. Write $u=u_1+u_2$ where $u_i$ are the solutions of

$-\Delta u_i=f_i$

with Dirichlet boundary condition. Choosing, for example, $\delta=(4\pi-1)$ in the theorem we find

$\displaystyle\int_\Omega {\exp \left[ {\frac{{|{u_1}(x)|}}{{{{\left\| {{f_1}} \right\|}_1}}}} \right]dx} < \infty$

and thus

$\displaystyle\int_\Omega {\exp \left[ {k|{u_1}(x)|} \right]dx} < \infty$.

The conclusion follows since $|u_1| \leqslant |u_1|+|u_2|$ and $u_2 \in L^\infty(\Omega)$.

This kind of result for biharmonic operator had also been done by C.S.L [here]. It is worth noticing that this result had also been extended to the $p$-Laplacian by Aguilar-Peral [here].

## 2 Comments »

1. Hi!
I think that the modulus in the left side of the 11th math line is unnecessary because in Br the used function is non negative.
Just for saving ink!

Comment by Fab — December 10, 2010 @ 21:12

• Thank Fab. You are right, thus it reads as the following

$\displaystyle\int_{{B_R}} {\exp \left[ {\frac{{(4\pi - \delta )\overline u (x)}}{{{{\left\| f \right\|}_1}}}} \right]dx} \leqslant \cdots$

Comment by Ngô Quốc Anh — December 10, 2010 @ 21:38