There is classical result of [Hartman and Nirenberg](http://www.jstor.org/discover/10.2307/2372995?sid=21105775249151&uid=2&uid=4):


**Theorem**. *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 $R^3$.*


It can be treated as general solution (in parametric form) of the partial differential equation for $g(s,t)$:
$$g\_{tt} g\_{ss} - g\_{st}^2 = 0.$$


I am solving system of partial differential equations. It contains two equations of the form:
$$g\_{tt} g\_{ss} - g\_{st}^2 = 0,$$
$$f\_{tt} f\_{ss} - f\_{st}^2 = 0,$$
and some other. Actually, I need concatenated parametrization for these two surfaces. 


Is there any idea how to extend Hartman result for this case?



Two surfaces with zero gaussian curvature
 **Revised version giving an explicit integration of the equations:**


 Consider the system of three PDE for two functions $f(s,t)$ and $g(s,t)$
$$
f\_{ss}f\_{tt}-{f\_{st}}^2 = g\_{ss}g\_{tt}-{g\_{st}}^2 = {f\_t}^2+{g\_t}^2-1 = 0.
$$
I am going to show that the general solution depends on two functions of one variable and give an explicit local parametrization of solutions on an open
domain in the $st$-plane.


 Say that a (local) solution on a simply-connected open domain in the $st$-plane is *nondegenerate* if, when one writes $(f\_t,g\_t) = (\cos u, \sin u)$, one has $\mathrm{d}u\not=0$. In this case, because the other equations imply
$$
0 = (f\_{ss}f\_{tt}-{f\_{st}}^2)\,\mathrm{d}s\wedge\mathrm{d}t = \mathrm{d}f\_s\wedge\mathrm{d}f\_t = -\sin u\, \mathrm{d}f\_s\wedge\mathrm{d}u
$$
and 
$$
0 = (g\_{ss}g\_{tt}-{g\_{st}}^2)\,\mathrm{d}s\wedge\mathrm{d}t = \mathrm{d}g\_s\wedge\mathrm{d}g\_t = \cos u\, \mathrm{d}g\_s\wedge\mathrm{d}u
$$
it follows that $f\_s$ and $g\_s$ must also be functions of $u$ 
(which is a function of $s$ and $t$ on the domain in question). 
Let us assume that we have restricted to a subdomain $U$ 
in the $st$-plane on which the
fibers of the submersion $u:U\to\mathbb{R}$ are connected.
Let $I = u(U)\subset\mathbb{R}$ be the image interval.


 I claim that the equations above imply
$$
\cos u\,\mathrm{d}f\_s + \sin u\,\mathrm{d}g\_s = 0.
$$
This follows because, if $\mathrm{d}u = p\,\mathrm{d}s + q\,\mathrm{d}t$,
then $p$ and $q$ do not simultaneously vanish (since $\mathrm{d}u\not=0$), 
and the equations $\mathrm{d}f\_t = f\_{st}\,\mathrm{d}s + f\_{tt}\,\mathrm{d}t$
and $\mathrm{d}g\_t = g\_{st}\,\mathrm{d}s + g\_{tt}\,\mathrm{d}t$
then imply 
$$ 
f\_{tt} = -q\,\sin u,\quad f\_{st} = -q\,\sin u,\quad
g\_{tt} = p\,\cos u,\quad g\_{st} = p\,\cos u.
$$
Then the equations $f\_{ss}f\_{tt}-{f\_{st}}^2 =g\_{ss}g\_{tt}-{g\_{st}}^2 =0$,
coupled with the fact that $\sin u$ and $\cos u$ only vanish on isolated
level curves of $u$, imply that
$$
q\,f\_{ss} = -p^2\,\sin u\qquad \text{and}\qquad q\,g\_{ss} = p^2\,\cos u.
$$ 
Since $p$ and $q$ do not simultaneously vanish, it follows that $q$
cannot vanish and, hence, that 
$$
f\_{ss} = -(p^2/q)\,\sin u\qquad \text{and}\qquad g\_{ss} = (p^2/q)\,\cos u.
$$
This establishes the claim. In particular, it follows (since the
fibers of $u$ are connected) that there is a function $h:I\to\mathbb{R}$ such that
$$
\mathrm{d} f\_s = -h(u)\,\sin u\,\mathrm{d}u
\qquad \text{and}\qquad
\mathrm{d} g\_s = h(u)\,\cos u\,\mathrm{d}u.
$$


 Now, because
$$
\begin{align}
\mathrm{d} f 
&= f\_s\,\mathrm{d}s + f\_t\,\mathrm{d}t= \mathrm{d}\bigl(sf\_s + t f\_t\bigr) 
 - \bigl(s\,\mathrm{d}f\_s + t\,\mathrm{d}f\_t\bigr)\\
&= \mathrm{d}\bigl(sf\_s + t f\_t\bigr) + (s\,h(u) + t)\,\sin u\,\mathrm{d}u,
\end{align}
$$
and, similarly,
$$
\begin{align}
\mathrm{d}g 
&= g\_s\,\mathrm{d}s + g\_t\,\mathrm{d}t= \mathrm{d}\bigl(sg\_s + t g\_t\bigr) 
 - \bigl(s\,\mathrm{d}g\_s + t\,\mathrm{d}g\_t\bigr)\\
&= \mathrm{d}\bigl(sg\_s + t g\_t\bigr) - (s\,h(u) + t)\,\cos u\,\mathrm{d}u,
\end{align}
$$
one sees, by applying $\mathrm{d}^2=0$, that one must have 
$$
\mathrm{d}(s\,h(u){+} t)\wedge \sin u\,\mathrm{d}u
= \mathrm{d}(s\,h(u){+} t)\wedge \cos u\,\mathrm{d}u = 0.
$$
Thus, $(s\,h(u){+} t)$ must be a function of $u$ alone, i.e., 
$s\,h(u){+} t = k(u)$ for some function $k:I\to\mathbb{R}$.


Now, set $v = t\,h(u) - s$, so that one can solve for $s$ and $t$ in
terms of $u$ and $v$ as
$$
s = \frac{k(u)\,h(u)+v}{\bigl(1+h(u)^2\bigr)}
\qquad\text{and}\qquad
t = \frac{k(u)-v\,h(u)}{\bigl(1+h(u)^2\bigr)}.
$$
It then follows, because $\mathrm{d}s\wedge\mathrm{d}t\not=0$, 
that $(u,v):U\to\mathbb{R}^2$ is a coordinate system on $U$.


To complete the integration, set $h(u) = a(u) + a''(u)$.
This determines $a:I\to\mathbb{R}$ uniquely up to the addition 
of a constant linear combination of $\cos u$ and $\sin u$, 
and, thus, one immediately sees that there is a unique choice of $a$ so that
$$
f\_s = a(u)\,\cos u - a'(u)\,\sin u
\qquad\text{and}\qquad
g\_s = a(u)\,\sin u + a'(u)\,\cos u.
$$
Similarly, choosing a function $b:I\to\mathbb{R}$ so that $k(u) = b(u)+b''(u)$,
one has
$$
\mathrm{d}\bigl(b(u)\,\cos u - b'(u)\,\sin u) 
= -\bigl(b(u)+b''(u)\bigr)\,\sin u\,\mathrm{d}u 
= -k(u)\,\sin u\,\mathrm{d}u
$$
and
$$
\mathrm{d}\bigl(b(u)\,\sin u + b'(u)\,\cos u) 
= \phantom{-}\bigl(b(u)+b''(u)\bigr)\,\cos u\,\mathrm{d}u
= \phantom{-}k(u)\,\cos u\,\mathrm{d}u.
$$
Thus, there is a unique choice of $b$ so that
$$
f = s\,f\_s + t\,f\_t - \bigl(b(u)\,\cos u - b'(u)\,\sin u\bigr)
$$
and 
$$
g = s\,g\_s + t\,g\_t - \bigl(b(u)\,\sin u + b'(u)\,\cos u\bigr).
$$
With the formulae for $f\_s$ and $g\_s$ above and the expanded formulae
$$
s = \frac{\bigl(b(u){+}b''(u)\bigr)\,\bigl(a(u){+}a''(u)\bigr)+v}
{\bigl(1+\bigl(a(u){+}a''(u)\bigr)^2\bigr)}
\qquad\text{and}\qquad
t = \frac{\bigl(b(u){+}b''(u)\bigr)-v\,\bigl(a(u){+}a''(u)\bigr)}
{\bigl(1+\bigl(a(u){+}a''(u)\bigr)^2\bigr)},
$$
one now has formulae for $s(u,v)$, $t(u,v)$, $f(u,v)$, and $g(u,v)$
in terms of two arbitrary functions $a$ and $b$ of $u$ and their first
two derivatives. 


Conversely,
given a choice of $C^2$ functions $a$ and $b$ 
on an interval $I\subset\mathbb{R}$, on the open set 
in $I\times\mathbb{R}$ on which $\mathrm{d}(s(u,v))\wedge\mathrm{d}(t(u,v))
\not=0$, the surface
$$
\bigl(s(u,v), t(u,v), f(u,v), g(u,v)\bigr)
$$
in $\mathbb{R}^4$ is (locally) a graph of the form
$$
\bigl(s, t, f(s,t), g(s,t)\bigr)
$$
where $f$ and $g$ satisfy the desired $3$ equations.  

Moreover, every nondegenerate solution is locally of the above form. 


Note that $s(u,v)$, $t(u,v)$, $f(u,v)$, and $g(u,v)$ are all
linear in $v$, which shows that the above surface in $\mathbb{R}^4$ 
is *ruled*, i.e., is a $1$-parameter family of lines, which, apparently, 
is what the OP meant by asking whether there was a 'concatenated parametrization'.


Finally, let us say that a solution $(f,g)$ is *totally degenerate* if $\mathrm{d}u\equiv0$. In this case, of course, $f\_t$ and $g\_t$ are constants and the above analysis does not apply. However, the general totally degenerate solution is 
$$
(f,g) = \bigl(a(s) + (\cos u)\,t, b(s) + (\sin u)\,t\bigr),
$$
where $a$ and $b$ are arbitrary functions of one variable and $u$ is a constant. Thus, this represents a different class of solutions that depend on two functions of one variable. 


Outside of a closed set with no interior, the graph of a general solution is a union of graphs of nondegenerate solutions and totally degenerate solutions. 














