Let suppose function $g(s,t)$ satisfies partial differential equations:
$g\_{ss} g\_{tt} - g\_{st}^2=0$. It may be treated as the surface has zero gaussian curvature. I am searching for general solution of this equation locally. Are there any results for it?



General description of surface with zero gaussian curvature
A surface of Gaussian curvature zero is locally isometric to the plane, and is said to be developable. A complete surface of Gaussian curvature zero in Euclidean three space is a cylinder (where a cylinder means the surface generated by the lines parallel to a given axis passing through a fixed curve in the subspace perpendicular to the axis; the plane is a cylinder in this sense when the curve is itself a line). This is due to Pogorelov, although it is usually attributed to Hartman-Nirenberg (they attribute it to Pogorelov). It can be found proved as Theorem 1 in W. Massey's [Surfaces of Gaussian curvature zero in Euclidean $3$-space](http://projecteuclid.org/euclid.tmj/1178244205) or in part II of Hartman and Nirenberg's *On spherical image maps whose Jacobians do not change sign* (Hartman and Nirenberg prove a more general result, for hypersurfaces in Euclidean space of arbitrary dimension). The papers of Massey and Hartman-Nirenberg contain more detailed results applicable to the incomplete case. Corollaries $3$ and $3^{\prime}$ of Hartman-Nirenberg show that every point of a $C^{2}$ surface of Gauss curvature zero has a neighborhood which admits a parameterization of the form $x = a(u)v + b(u)$ where $(u, v)$ varies over some simply-connected plane domain and $a$ and $b$ take values in $\mathbb{R}^{3}$. (see also chapter IX of the English translation *Extrinsic Geometry of Convex Surfaces* of a book of Pogorelov.)


