# Ngô Quốc Anh

## March 20, 2011

### A note on the equation involving the prescribing Gaussian curvature problem

Filed under: Uncategorized — Ngô Quốc Anh @ 2:41

Let $M$ be a smooth and compact two dimensional Riemannian manifold. Let $g_o(x)$ be a metric on $M$ with the corresponding Laplace-Beltrami operator $\Delta$ and Gaussian curvature $K_o(x)$. Given a function $K(x)$ on $M$, can it be realized as the Gaussian curvature associated to the point-wise conformal
metric

$\displaystyle g(x) = e^{2u(x)}g_o(x)$

To answer this question, it is equivalent to solve the following semi-linear elliptic equation

$\displaystyle -\Delta u +K_0(x)=K(x)e^{2u}, \quad x \in M.$

In this entry, we summarize some basic steps in order to simplify the above PDE. We first let $\overline u=2u$, then our PDE becomes

$\displaystyle -\frac{1}{2}\Delta \overline u +K_0(x)=K(x)e^{\overline u}.$

Let $v$ be a solution of the following PDE

$\displaystyle -\Delta v=2K_0(x)-2\overline K_0$

where

$\displaystyle \overline K_0=\frac{1}{|M|}\int_M K_0(x)dv$

is nothing but the average of $K_0$ over $M$. The solvability of the foregoing PDE comes from the fact that

$\displaystyle\int_M (2K_0(x)-2\overline K_0)dv=0.$

We let $w=\overline u+v$. Then it is easy to verify that $w$ solves the following

$\displaystyle -\Delta w +2\overline K_0=2K(x)e^{-v}e^w.$

Finally, letting

$\alpha=2\overline K_0, \quad R(x)=2K(x)e^{-v(x)}$

we get

$\displaystyle -\Delta w +\alpha = R(x)e^w$

or by renaming $w$ by $u$

$\displaystyle -\Delta u +\alpha = R(x)e^u.$

The advantage of this equation is that here $\alpha$ is constant. To be precise, by the Gauss-Bonnet theorem, we have

$\displaystyle \alpha=\frac{4\pi}{|M|}\chi(M)$

where $\chi(M)$ is the characteristic of $M$.