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.

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<p\leqslant \infty 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<p\leqslant \infty 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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