# Ngô Quốc Anh

## August 22, 2010

### Liouville’s theorem and related problems

Filed under: Giải tích 7 (MA4247), PDEs — Tags: — Ngô Quốc Anh @ 6:35

The following theorem is well-known

Theorem (Liouville). Let $\Omega$ be a simply connected domain in $\mathbb R^2$. Then all real solutions of

$\displaystyle \Delta u +2Ke^u=0$

in $\Omega$ where $K$ a constant, are of the form

$\displaystyle u=\log\frac{|f'|^2}{\left(1+\frac{K}{4}|f|^2\right)^2}$

where $f$ is a locally univalent meromorphic function in $\Omega$.

In geometry, our PDE

$\displaystyle \Delta u +2Ke^u=0$

says that under the case $\Omega=\mathbb R^2$, it holds

$e^u|dz|^2=f^*g_K$

where $g_K$ denotes the standard metric on $\mathbb S^2$ with constant curvature $K$. Thus we have

Corollary. All solutions of the PDE in $\mathbb R^2$ with $K>0$ and

$\displaystyle \int_{\mathbb R^2} e^u<\infty$

are of the form

$\displaystyle u(x)=\log\frac{16\lambda^2}{\left(4+\lambda^2K|x-x_0|^2\right)^2},\quad \lambda>0, \quad x_0 \in \mathbb R^2$.

The proof of this corollary is quite simple.

Proof. By the integrability assumption $f$ cannot have an essential singularity at infinity, for otherwise, $f$ would conver $\mathbb S^2$ (possible except one point) infinitely many times near infinity, which is impossible.

Therefore

$\displaystyle\mathop {\lim }\limits_{z \to z_0 } f(z) = \infty$

for some $z_0 \in \mathbb C$. By composing with an inversion, we may assume the former case holds. Then $f$ maps $\mathbb S^2$ onto $\mathbb S^2$. Since $\mathbb C$ cannot cover $\mathbb S^2$ (notice that $f'(z) \ne 0$ for all $z \in \mathbb C$), $f$ does  not have poles in $\mathbb C$. This means $f : \mathbb C \to \mathbb C$ is a covering map and therefore it assumes the form

$f(z)=\alpha z+\beta$

for some $\alpha \ne 0$ and $\beta \in \mathbb C$. A substitution into our PDE gives the desired conclusion.

It is worth noticing that our Corollary was also proved by W. Chen and C. Li by the method of moving planes [here] and Y.Y. Li and M. Zhu by the method of moving spheres [here].

Besides, one can see that the intergrability condition is also necessary for asymptotic radial symmetry. All non-radial solutions, which arise from transcendental functions, satisfy

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

Let’s go back to the theorem, it does not hold for domains which are not simply connected. For instance, the function

$\displaystyle u=-\log\left[4r\left(1+\frac{K}{r}\right)^2\right]$

is a solution of our PDE in the punctured disc with an isolated singularity at the origin. Such a result for this domain was derived by K.Chou and Y. Wan [here].

A natural extension, but has its own interest, is the so-called Toda system SU(n). The simplest case, known as SU(2), is the following

$\displaystyle\left\{ \begin{gathered} - \Delta u = 2{e^u} - {e^v}, \hfill \\ - \Delta v = - {e^u} + 2{e^v}, \hfill \\ \end{gathered} \right.$

in $\mathbb R^2$.

The solutions of SU(2) are already known, precisely, they can be written as follows

$\displaystyle u(z) = \log \frac{{4\left( {a_1^2a_2^2 + a_1^2{{\left| {2z + c} \right|}^2} + a_2^2{{\left| {{z^2} + 2bz + bc - d} \right|}^2}} \right)}}{{{{\left( {a_1^2 + a_2^2{{\left| {z + b} \right|}^2} + \left| {{z^2} + cz + d} \right|} \right)}^2}}}$

and

$\displaystyle v(z) = \log \frac{{16a_1^2a_2^2\left( {a_1^2 + a_2^2{{\left| {z + b} \right|}^2} + \left| {{z^2} + cz + d} \right|} \right)}}{{{{\left( {a_1^2a_2^2 + a_1^2{{\left| {2z + c} \right|}^2} + a_2^2{{\left| {{z^2} + 2bz + bc - d} \right|}^2}} \right)}^2}}}$

where $a_1, a_2>0$ are real and $b,c,d$ are complex numbers. We refer the reader to a paper due to del Pino, M. Kowalczyk and J. Wei for details [here].

The following pictures demonstrate the shape of these solutions for a particular choice of eight parameters (i.e. $a_1=a_2=1$, $b=(1,1)$, $c=d=0$).

For

$\displaystyle u(z) = \log \frac{{4\left( {1 + 4{{\left| z \right|}^2} + {{\left| {{z^2} + 2z} \right|}^2}} \right)}}{{{{\left( {1 + {{\left| {z + 1} \right|}^2} + {{\left| {{z^2}} \right|}^2}} \right)}^2}}}$

For

$\displaystyle v(z) = \log \frac{{16\left( {1 + {{\left| {z + 1} \right|}^2} + {{\left| {{z^2}} \right|}^2}} \right)}}{{{{\left( {1 + 4{{\left| z \right|}^2} + {{\left| {{z^2} + 2z} \right|}^2}} \right)}^2}}}$.