# Ngô Quốc Anh

## August 19, 2010

### L^infinity-boundedness for a single solution of -Delta u = Vexp(u)

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 7:00

The aim of this entry is to derive the $L^\infty$-boundedness for a single solution of the following PDE $\displaystyle -\Delta u = V(x) e^u$

over a domain $\Omega$. This elegant result had been done by Brezis and Merle around 1991 published in Comm. Partial Differential Equations [here].

There are two possible cases.

The case of bounded domain. Let us assume $u$ a solution of the following PDE $\displaystyle\begin{cases}- \Delta u = V(x){e^u}& \text{ in }\Omega , \hfill \\ u = 0&\text{ on }\partial \Omega ,\end{cases}$

where $\Omega \subset \mathbb R^2$ is a bounded domain and $V$ is a given function on $\Omega$.

Theorem. If $V \in L^p$ and $e^u \in L^{p'}$ for some $1 then $u \in L^\infty$.

Proof. It first follows from the Brezis-Meler inequality that $e^{ku} \in L^1, \quad \forall k>0$

which by the Holder inequality gives $e^{u} \in L^r, \quad \forall r<\infty$.

Therefore, if $p<\infty$ $Ve^u \in L^{p-\delta}, \quad \forall \delta>0$

while if $p=\infty$ $Ve^u \in L^r, \quad \forall r<\infty$.

Thus, a standard $L^p$-estimate argument from the elliptic theory implies that $u$ is bounded.