# 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.

The conclusion of our theorem still holds for a solution $u$ of $\displaystyle\begin{cases}- \Delta u = V(x){e^u}+f(x)& \text{ in }\Omega , \hfill \\ u =g&\text{ on }\partial \Omega ,\end{cases}$

with $g \in L^\infty(\partial\Omega)$ and $f \in L^q$ for some $q>1$.

Remark. It is worth noticing that a local version of the corollary mentioned in this entry still holds. Taking into account this result still gives a local version of the theorem above.

Theorem (local version). If $V \in L_{loc}^p$ and $e^u \in L_{loc}^{p'}$ for some $1 then $u \in L_{loc}^\infty$.

The case of the whole space. Let us now turn to the case of $\Omega = \mathbb R^2$. Our PDE is simply $\displaystyle -\Delta u = V(x) e^u, \quad x \in \mathbb R^2$.

In this circumstance, we get

Theorem. Assume $u \in L_{loc}^1(\mathbb R^2)$. If $V \in L^p(\mathbb R^2)$ and $e^u \in L^{p'}(\mathbb R^2)$ for some $1 then $u \in L^\infty(\mathbb R^2)$.

Proof. Fix $0<\varepsilon<\frac{1}{p'}$ and split $Ve^u$ as $Ve^u=f_1+f_2$ with $\|f_1\|_{L^1}<\varepsilon, \quad f_2 \in L^\infty$.

Let $B_r$ be the unit ball of radius $r$ centered at $x_0$. We denote by $C$ various constants independent of $x_0$. Let $u_i$ be the solution of $\displaystyle\begin{cases}- \Delta u_i = f_i& \text{ in } B_1 , \hfill \\ u_i = 0&\text{ on }\partial B_1.\end{cases}$

By the Brezis-Meler inequality, $\displaystyle\int_{B_1}\exp\left[ \frac{1}{\varepsilon}|u_1|\right]\leqslant C$

and in particular $\|u_1\|_{L^1(B_1)} \leqslant C$.

We also have $\|u_2\|_{L^\infty(B_1)} \leqslant C$.

Let $u_2=u-u_1-u_2$ so that $\Delta u_3=0$ on $B_1$. The mean value theorem for harmonic functions implies that $\displaystyle {\left\| {u_3^ + } \right\|_{{L^\infty }({B_{1/2}})}} \leqslant C{\left\| {u_3^ + } \right\|_{{L^1}({B_1})}}$.

On the other hand we have $\displaystyle u_3^+ \leqslant u^++|u_1|+|u_2|$

and since $\displaystyle p'\int_{{\mathbb{R}^2}} {{u^ + }} \leqslant \int_{{\mathbb{R}^2}} {{e^{p'u}}} \leqslant C$

we see that $\displaystyle\|u_3^+\|_{L^1(B_1)} \leqslant C$.

Thus we find that $\displaystyle\|u_3^+\|_{L^\infty(B_{1/2})} \leqslant C$.

Finally we write $\displaystyle - \Delta u = V(x){e^u} = (V{e^{{u_1}}}){e^{{u_2} + {u_3}}} = g$

with $\displaystyle {\left\| g \right\|_{{L^{1 + \delta }}({B_{1/2}})}} \leqslant C$

for some $\delta >0$. Using once more the mean value theorem and standard elliptic estimates we deduce that $\displaystyle {\left\| {{u^ + }} \right\|_{{L^\infty }({B_{1/4}})}} \leqslant C{\left\| {{u^ + }} \right\|_{{L^1}({B_{1/2}})}} + C{\left\| g \right\|_{{L^{1 + \delta }}({B_{1/2}})}} \leqslant C$.

Since $C$ is independent of $x_0$, we conclude that $u^+\in L^\infty$.

Remark. The mean value theorem for harmonic functions says that if $u$ is harmonic then $\displaystyle u(x) = \frac{1}{n\omega_n r^{n-1}}\int_{\partial B(x,r)} u d\sigma = \frac{1}{\omega_n r^n}\int_{B (x,r)} u dy$

for any point $x$. Thus $\displaystyle u(x) = \frac{1}{{{\omega _2}{{\left( {\frac{1}{2}} \right)}^2}}}\int_{B\left( {x,\frac{1}{2}} \right)} {u(y)dy} \leqslant C\int_{B\left( {{x_0},1} \right)} {u(y)dy} = C{\left\| u \right\|_{{L^1}({B_1})}}, \quad \forall x \in {B_{\frac{1}{2}}}$.

Taking sup both sides we arrive at $\displaystyle {\left\| u \right\|_{{L^\infty }({B_{1/2}})}} \leqslant C{\left\| u \right\|_{{L^1}({B_1})}}$.

## 8 Comments »

1. Congratulations for the beautiful website, full of amazing purposes and tricks!
In particular, this file about L-infinity boundness argument is been to me so useful during the reading of the elegante Brezis-Merle paper.
A dumb question: in the whole paper BM use a “standar elliptic estimate” and (exactly in the proof above) a “standard Lp estimate” that I don’t know.
May I have some info about or book references?
Thanks in advance!
Fab

Comment by Fab — December 6, 2010 @ 18:38

• Hi Fab,

Thanks for your interest in my blog. Concerning to the standard $L^p$ estimate, I strongly recommend you to go through the book due to Gilbarg-Trudinger entitled “Elliptic partial differential equations of second order”.

Comment by Ngô Quốc Anh — December 7, 2010 @ 0:49

2. Thanks for your advice!

Comment by Fab — December 10, 2010 @ 20:56

3. However, if you use as boundary condition u<=0 instead of only u=0 you have no more modulus problem.

A suggestion: Why don't you post something on "the three alternative theorem" by Brezis Merle that is in the same elegant paper?

Fab

Comment by Fab — December 17, 2010 @ 22:35

• Thank Fab, I will think about it.

Comment by Ngô Quốc Anh — December 17, 2010 @ 22:37

4. The comment above, of course, is refered to Brezis Merle Inequality, sorry!

Comment by Fab — December 17, 2010 @ 22:38

5. Sorry for my english mistake, refer is a transitive verb, so I should use only “is referred” without the “to”, isn’t right?

Comment by Fab — December 17, 2010 @ 22:44

• Aha, you need to use “to”, so the latter was not correct but the first one.

Comment by Ngô Quốc Anh — December 17, 2010 @ 23:08

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