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?
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 HartmanNirenberg (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 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 HartmanNirenberg contain more detailed results applicable to the incomplete case. Corollaries $3$ and $3^{\prime}$ of HartmanNirenberg 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 simplyconnected 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.)

$\begingroup$ Thanks. Well, I know this paper. Actually I'm looking for solution locally. I can reduce this equation to first order equation $g_s = h(g_t)$ and then using technique of solving firstorder PDES, write in parametric form general solution. But it is rather huge and cumbersome... $\endgroup$ – user47116 Jan 3 '15 at 13:00

1$\begingroup$ @user12355: I added some remarks to the answer to address the local case. $\endgroup$ – Dan Fox Jan 3 '15 at 13:19

$\begingroup$ Thanks. The parametrization is very useful for me. But actually functions $a(u)$ and $b(u)$ cannot be arbitrary. Right? $\endgroup$ – user47116 Jan 16 '15 at 9:31
I think this article might be helpful (see section 3, http://arxiv.org/pdf/1402.4751v2.pdf), also see this one (http://arxiv.org/abs/1205.7018).
Let me explain you very briefly what is happening there.
Let us use the following notations: $g(x,y)=B(x,y)$ and $B$ satisfies $B_{11}B_{22}B_{12}^{2}=0$.
Take some suitable space curve $\gamma(t) = (f_{1}(t),f_{2}(t),f_{3}(t)) :I\to \mathbb{R}^{3}$ and lets require that $B(f_{1}(t),f_{2}(t))=f_{3}(t)$ (this is boundary condition for your function $B(x,y)$  this means that we are assuming that there is a domain $\Omega \subset \mathbb{R}^{2}$ where the function $B$ is given and $B$ has prescribed boundary data on $\partial \Omega$. By the way, in this case $(f_{1}(t),f_{2}(t))$ parametrizes $\partial \Omega$ and $f_{3}$ is your boundary data for $B$).
Then this already gives you one equation after differentiating $B(f_{1}(t),f_{2}(t))=f_{3}(t)$ in variable $t$: $$ B_{1}f'_{1}+B_{2}f'_{2}=f'_{3} $$ where $f'_{j}=\frac{df_{j}}{dt}$.
This information is nothing unless you use the fact that $B$ satisfies homogeneous MongeAmpere equation (i.e., the fact that it has zero Gaussian curvature). It means (thanks to Pogorelov) that you can draw some family of line segments close to the boundary $\Omega$ which start from the curve $(f_{1}(t),f_{2}(t))$ and go inside the $\Omega$. Moreover the function $B$ is linear along these segments and the gradient of $B$ is constant along these segments. The typical picture of these family of segments is given here (see the picture below)
This picture close to the boundary of $\Omega$ is true if the things are not degenerate (i.e., torsion of $\gamma$ does not vanish on any subinterval $I$) then the domains, where $B$ is linear, cannot touch $\partial\Omega$ on a thick interval. however they can touch $\partial \Omega$ at some finite number of points (or countable number of points if the torsion of $\gamma$ changes sign infinitely many times).
In other words this means that the gradient of $B$ in $\Omega$ you can parametrize by one parameter $s$ i.e., $\nabla B = (t_{1}(s),t_{2}(s))$ where $s \in I$
So our equation $B_{1}f'_{1}+B_{2}f'_{2}=f'_{3}$ can be rewritten as follows $$ t_{1}(s)f'_{1}(s)+t_{2}(s)f'_{2}(s)=f'_{3}(s). $$ Of course this information is not enough to find $(t_{1}(s),t_{2}(s))$. But there is one more equation which you can also obtain, namely: $$ t_{1}'(s)\cos(\alpha(s))+t'_{2}(s)\sin(\alpha(s))=0, $$ where $(\cos(\alpha(s)),\sin(\alpha(s)))$ is the direction of the line segment starting at point $(f_{1}(s),f_{2}(s))$ i.e., unit vector, starting at point $(f_{1}(s),f_{2}(s))$ and going inside $\Omega$ along the line segment, along which $B$ is linear.
These two equations $$ t_{1}(s)f'_{1}(s)+t_{2}(s)f'_{2}(s)=f'_{3}(s);\\ t_{1}'(s)\cos(\alpha(s))+t'_{2}(s)\sin(\alpha(s))=0, $$ allow you to find $(t_{1}(s),t_{2}(s))$ up to a constant $C$ which you still have to choose later in order to glue these local pieces and to get some global picture for $B$. Thus you find $B$ $$ B(x,y)=f_{3}(s)+t_{1}(s)(xf_{1}(s))+t_{2}(s)(yf_{2}(s)) \quad(*), $$ where $(x,y)$ belongs to the line segment starting at point $(f_{1}(s),f_{2}(s))$.
For example if $\gamma(t)=(t,g(t),f(t))$ then $$ t_{2}(s)=C\exp\left(\int_{s_{1}}^{s}\frac{g''(r)}{K(r)}\cos(\alpha(r))dr \right)+\frac{f''(r)}{g''(r)}\int_{s_{1}}^{s}\left[ \frac{f''(y)}{g''(y)}\right]'\exp\left(\int_{y}^{s}\frac{g''(r)}{K(r)}\cos(\alpha(r))dr \right)dy $$ where $K(s)=g'(s)\cos(\alpha(s))\sin \alpha(s)$, and you can also notice that the expression $\left[ \frac{f''(y)}{g''(y)}\right]'$ coincides up to a curvature factor of $\gamma$ with the torsion of $\gamma$ which further plays a crucial role.
By the way the equation $t_{1}'(s)\cos(\alpha(s))+t'_{2}(s)\sin(\alpha(s))$ also can be obtained by differentiating (*) with respect to $x$ and treating $s$ as a function of $s(x,y)$.
Now there are lot of questions left:
How do you find these family of segments (or it is the same as to ask how do you find directions $(\cos(\alpha(s)),\sin(\alpha(s)))$).
How do you glue a global picture?
Under what conditions this ``roughly speaking'' is justified?
Some partial answers are given in the articles that I mentioned above.