Liouville's equation

Remarks

The metric $e^{2u}(dx^2 + dy^2)$ has Gaussian curvature $-e^{-2u} \Delta u$, so $u$ solves the equation exactly when this metric has constant curvature $K$. Example: for $K = 1$, $u(x, y) = \log 2 - \log(1 + r^2)$ with $r^2 = x^2 + y^2$ is a solution on $\R^2$. Indeed, $\Delta \log(1 + r^2) = 4/(1 + r^2)^2$, so $\Delta u = -4/(1 + r^2)^2 = -e^{2u}$. Non-example: for $K = 1$, the function $u = 0$ is not a solution, since $\Delta u + e^{2u} = 1$. Other normalizations are common: the substitution $u = v/2$ gives $\Delta v + 2K e^{v} = 0$, so the form $\Delta v + \lambda e^{v} = 0$ is equivalent. Locally, the solutions for $K > 0$ are $u = \log\big( 2 \abs{f'} / (1 + K \abs{f}^2) \big)$ with $f$ a holomorphic function of $z = x + iy$ whose derivative has no zeros.