# Ngô Quốc Anh

## June 28, 2010

### A non-existence result for PDE Delta u=exp(u) in R^2

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 0:28

We provide a proof of the following well known fact.

Theorem. There is no $C^2$ solution to

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

I found this proof in a paper due to Y.Y. Li published in Commun. Math. Phys. in 1999 [here]. Before deriving the proof, let us recall the following notation known as the sphere mean in the literature. In $\mathbb R^n$ we denote the integral

$\displaystyle\displaystyle\frac{1}{\omega _n}{r^{n - 1}}\int_{\partial B\left( {0,r} \right)} {f\left( x \right)dS_x}$

by $\overline f(r)$. We call $\overline f$ the average of $f$ on the sphere $S(0,r)$ of radius $r$, or sphere mean of a function around the origin. In this context, we simply have

$\displaystyle\displaystyle\frac{1}{2\pi r}\int_{\partial B\left( {0,r} \right)} {f\left( x \right)dS_x}$.

Proof. We derive from the Jensen inequality that

$\displaystyle\frac{1}{{2\pi r}}\int_{\partial B(0,r)} {{e^u}d{S_x}} \geqslant {e^{\overline u (r)}}$.

It follows that $\overline u$ satisfies

$\displaystyle\Delta \overline u \geqslant {e^{\overline u (r)}}$

in $\mathbb R^2$, namely,

$\displaystyle\frac{1}{r}{\left( {r{{\overline u }^\prime }(r)} \right)^\prime } \geqslant {e^{\overline u (r)}}$.

We derive from the above that

$\displaystyle r{\overline u ^\prime }(r) \geqslant \int_0^r {s{e^{\overline u (s)}}ds} \geqslant 0$

for all $r \geqslant 0$, i.e. $\overline u$ is monotone increasing. Consequently,

$\displaystyle r{\overline u ^\prime }(r) \geqslant \int_0^r {s{e^{\overline u (0)}}ds} \geqslant {e^{\overline u (0)}}\frac{{{r^2}}}{2}$

for all $r \geqslant 0$. In turn we have

$\displaystyle\overline u (r) \geqslant \overline u (0) + {e^{\overline u (0)}}\frac{{{r^2}}}{4}$

for all $r \geqslant 0$. It now follows that

$\displaystyle\int_{\mathbb R^2}e^udx=\infty$

since

$\displaystyle\int_{\partial B(0,r)} {{e^u}d{S_x}}$

1. Here u is real valued functions, isn’t it?

I think the result holds true also for all dimensions, and the $C^2$-assumption can be replaced by the continuity. In fact, we may simply use the maximum principle as follows:

The function $u$ is subharmonic since $\Delta u=e^{u}>0$. Thus $u(x)\le \limsup_{|y|\to \infty}u(y)$. One the other hand, since $u$ is continuous and $\int {e^u}<\infty$ we have $u(y)\to -\infty$ as $|y|\to \infty$. Thus $u(x)\le -\infty$ for a.e. $x$ which is impossible.

Comment by hoadai — June 29, 2010 @ 22:30

• Hello hoadai, thanks for your comment and sorry for this delay. This is because your comment has been moved to the spam category. You are right about solution $u$, we mean here function $u$ is real-valued. The $C^2$-regularity means we are talking about classical solutions. I am not so sure if this condition can be weakened. If this is the case, solutions should be though in the sense of distribution so that the following property

$u(x)\le \limsup_{|y|\to \infty}u(y)$

is not clear to see (in my opinion).

Comment by Ngô Quốc Anh — July 7, 2010 @ 5:21

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