Ngô Quốc Anh

August 6, 2010

An upper bound for solutions via the maximum principle

Filed under: PDEs — Tags: , — Ngô Quốc Anh @ 1:29

It is known that [here] the following PDE

\displaystyle\begin{cases}\Delta u =e^u, & {\rm in } \, \mathbb R^2,\\\displaystyle\int_{\mathbb R^2}e^u<\infty.\end{cases}

has no C^2 solution. However, this is no longer true if we replace the whole space by a ball of radius R, say B_R(0). In this entry, we show that if u \in C^2(\overline B_R) is a solution of

\Delta u=e^u, \quad {\rm in }\; B_R


\displaystyle u(0) \leqslant \log 8- 2\log R.

To this purpose, let us recall the following

The Maximum Principle. Let assume U\subset \mathbb R^2 be open and bounded. We consider an elliptic operator L of the form

\displaystyle Lu = - \sum\limits_{i,j = 1}^2 {{a^{ij}}\frac{{{\partial ^2}u}}{{\partial {x_i}\partial {x_j}}} + \sum\limits_{k = 1}^2 {{b^k}\frac{{\partial u}}{{\partial {x_k}}}} } + cu

where coefficients are continuous and the standard uniform ellipticity condition holds.

Theorem (Weak MP for c\geqslant 0).

  • If

    Lu \leqslant 0 , \quad {\rm in }\; U


    \displaystyle\mathop {\max }\limits_{\overline U } u \leqslant \mathop {\max }\limits_{\partial U} {u^ + }

    where u^+=\max\{u,0\}.

  • If

    Lu \geqslant 0 , \quad {\rm in }\; U


    \displaystyle\mathop {\min }\limits_{\overline U } u \geqslant -\mathop {\max }\limits_{\partial U} {u^ - }

    where u^-=\max\{-u,0\}=-\min\{u,0\}.

A proof of this theorem can be found in, for example, Partial Differential Equation [chapter 6] by L.C. Evans.

Corollary. The sign of solutions to PDE Lu=0 depends on those signs on the boundary.

We are now in a position to solve the question above. In the proof, we will make use a so-called boundary blow-up argument.

Proof. Let

\displaystyle v(r) = \log \frac{{8{R^2}}}{{{{({R^2} - {r^2})}^2}}}

written in the polar coordinates. It follows from the presentation formula of Laplacian in the polar coordinate

\displaystyle\Delta f = {1 \over r} {\partial \over \partial r} \left( r {\partial f \over \partial r} \right) + {1 \over r^2} {\partial^2 f \over \partial \theta^2}


\Delta v=e^v \quad {\rm in }\; B_R.


\displaystyle\mathop {\lim }\limits_{x \to \partial {B_R}} v(x) = + \infty .



then w verifies

\displaystyle -\Delta w + \underbrace {\frac{{{e^v} - {e^u}}}{{v - u}}}_{ > 0}w = 0\quad {\rm in }\; B_R.

So by the maximum principle we have

w \geqslant 0\quad {\rm in }\; B_R.


\displaystyle u(0) \leqslant v(0)=\log 8- 2\log R.

Obviously, if we replace B_R by the whole space \mathbb R^2, then by taking the limit as R \to \infty we deduce that u(0)=-\infty which is impossible. This is why in the whole space, our PDE has no solution.


Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Create a free website or blog at

%d bloggers like this: