We approach the problem of finding a metric of constant curvature on a surface (i.e. a $C^\infty$ 2-manifold). Specifically, what we want to do is, given a surface $M$ and a metric $g_0$, show that there exists a new metric $g$ of the form $g=e^{2u}g_0$ for some $u\in C^\infty (M)$ such that $g$ has constant curvature. It is well known that if we let $K_0$ and $K$ be the curvatures associated to $g_0$ and $g$ respectively, then they must satisfy the curvature equation \begin{equation} K=(K_0-\triangle u)e^{-2u} \end{equation} or equivalently, \begin{equation} \triangle u-K_0+Ke^{2u}=0 \end{equation} where $\triangle$ stands for the Laplace-Beltrami operator associated to the original metric $g_0$. Thus, solving our problem amounts to solving the curvature equation (i.e. finding $u$) for different values of $K$. We have essentialy three cases (thanks to the Gauss-Bonnet theorem), which are $K=0$, $K=-1$, and $K=1$. The first to cases are fairly straightforward (simple solutions can be found in an article by Melvin S. Berger 1971, or the book on PDE's by Taylor), but the case when $K=1$ is more challenging. I have read solutions which use the Riemann-Roch theorem on one hand or Ricci flow on the other.

The question is: is there any other way to find a solution to the curvature equation (in the case $K=1$)? The reason I want to know this is because I am writing my thesis and I would like to give the simplest proof of this result to make it more accessible. Thanks!