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.

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