# 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

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

then

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

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

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

then

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

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

• If

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

then

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

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

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

that

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

Apparently,

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

Let

$w=v-u$

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

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

Consequently,

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

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.