# Ngô Quốc Anh

## March 23, 2011

### A proof of the uniqueness of the solution of the prescribing Gaussian curvature problem

Filed under: Uncategorized — Ngô Quốc Anh @ 22:58

Let us continue the problem of prescribing Gaussian curvature. Our PDE reads as the follows

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

where $M$ is a compact manifold without the boundary. Today we show that if

$\displaystyle K(x) \leqslant 0$

then our PDE has unique solution.

Assume that $u_1$ and $u_2$ are solutions to the PDE, that is

$\displaystyle\begin{gathered} - \Delta {u_1} + {K_0}(x) = K(x){e^{2{u_1}}}, \hfill \\ - \Delta {u_2} + {K_0}(x) = K(x){e^{2{u_2}}}, \hfill \\ \end{gathered}$

By subtracting, we have

$\displaystyle - \Delta ({u_1} - {u_2}) = K(x)({e^{2{u_1}}} - {e^{2{u_2}}}).$

Multiplying both sides by $u_1-u_2$, integrating over $M$, and the using the integration by parts we arrive at

$\displaystyle\int_M {{{\left| {\nabla ({u_1} - {u_2})} \right|}^2}dv} = \int_M {K(x)\frac{{{e^{2{u_1}}} - {e^{2{u_2}}}}}{{{u_1} - {u_2}}}{{\left| {{u_1} - {u_2}} \right|}^2}dv} .$

Since $K(x) \leqslant 0$, it follows that

$\displaystyle\int_M {{{\left| {\nabla ({u_1} - {u_2})} \right|}^2}dv} \leqslant 0.$

In particular, $u_1 \equiv u_2$.