A Liouville Theorem for Möbius Invariant Equations

Abstract

In this paper, we classify Möbius invariant differential operators of second order in two-dimensional Euclidean space, and establish a Liouville type theorem for general Möbius invariant elliptic equations. The equations are naturally associated with a continuous family of convex cones $\Gamma _p$ in $\mathbb R^2$, with parameter $p\in [1, 2]$, joining the half plane $\Gamma _1:=\{ (\lambda _1, \lambda _2):\lambda _1+\lambda _2>0\}$ and the first quadrant $\Gamma _2:=\{ (\lambda _1, \lambda _2):\lambda _1, \lambda _2>0\}$. Chen and C. M. Li established in 1991 a Liouville type theorem corresponding to $\Gamma _1$ under an integrability assumption on the solution. The uniqueness result does not hold without this assumption. The Liouville type theorem we establish in this paper for $\Gamma _p$, $1<p\le 2$, does not require any additional assumption on the solution as for $\Gamma _1$. This is reminiscent of the Liouville type theorems in dimensions $n\ge 3$ established by Caffarelli, Gidas and Spruck in 1989 and by A. B. Li and Y. Y. Li in 2003–2005, where no additional assumption was needed either. On the other hand, there is a striking new phenomena in dimension $n=2$ that $\Gamma _p$ for $p=1$ is a sharp dividing line for such uniqueness result to hold without any further assumption on the solution. In dimensions $n\ge 3$, there is no such dividing line.

Appendix A: Two Calculus Lemmas

We now state two calculus lemmas for the reader’s convenience.

Lemma A.1

Let $u\in C^1(\mathbb {R}^2)$ satisfy

$$\begin{aligned} u\bigg(x+\frac{\lambda ^2(y-x)}{|y-x|^2}\bigg)-4\ln \frac{|y-x|}{\lambda }\le u(y), \quad \forall\lambda >0, \ x\in \mathbb {R}^2, \ |y-x|\ge \lambda . \end{aligned}$$

Then u must be constant.

Proof

Let $f={\rm e}^u$, then we have

$$\begin{aligned} \bigg(\frac{\lambda }{|y-x|}\bigg)^4f\bigg(x+\frac{\lambda ^2(y-x)}{|y-x|^2}\bigg)\le f(y), \quad \forall \lambda >0, \ x\in \mathbb {R}^2, \ |y-x|\ge \lambda. \end{aligned}$$

By [43, Lemma 11.1] (see also [44, Lemma 3.3]), we conclude that f is a constant; hence, u is a constant. $\square$

Lemma A.2

Let $u\in C^1(\mathbb {R}^2)$. Suppose that for every $x\in \mathbb {R}^2$, there exists $\lambda (x)>0$ such that

$$\begin{aligned} u_{x,\lambda (x)}(y)=u(y),\quad y\in \mathbb {R}^2\backslash \{x\}. \end{aligned}$$

Then for some $a>0$, $b>0$, $\bar{x}\in \mathbb {R}^2$,

$$\begin{aligned} u(x)=2\ln \frac{8a}{8|x-\bar{x}|^2+b}. \end{aligned}$$

Proof

Let $f={\rm e}^u$, then we have, for every $x\in \mathbb {R}^2$, there exists $\lambda (x)>0$ such that

$$\begin{aligned} \bigg(\frac{\lambda }{|y-x|}\bigg)^4f\bigg(x+\frac{\lambda ^2(y-x)}{|y-x|^2}\bigg)=f(y),\quad y\in \mathbb {R}^2\backslash \{x\}. \end{aligned}$$

By [43, Lemma 11.1] (see also [44, Lemma 3.7]),

$$\begin{aligned} f(x)=\pm \bigg(\frac{a}{d+|x-\bar{x}|^2}\bigg)^2. \end{aligned}$$

Lemma A.2 follows. $\square$