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}}

diverges. A contradiction.

2 Comments »

  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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