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<p\leqslant \infty 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.

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