# Two surfaces with zero gaussian curvature

There is classical result of Hartman and Nirenberg:

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?

• What do you mean by 'concatenated parametrization'? This is not a standard term; at least, I've never seen it before. – Robert Bryant May 6 '15 at 12:40
• Actually, I want to construct a parametrization for both surfaces, like: $g = A_1(u,v)$, $f = A_2(u,v)$, $s = A_3(u,v)$, $t = A_4(u,v)$ in explicit way. – user47116 May 6 '15 at 12:56
• You shouldn't, in general, expect such a parametrization to be linear in the variable $v$ in both $A_1$ and $A_2$. However, it could be that the extra equation ('some other') that you didn't mention would imply that such a parametrization exists. There's no way to know without knowing that other equation. – Robert Bryant May 6 '15 at 13:21
• $f_t^2+g_t^2 = 1$ – user47116 May 6 '15 at 15:00
• @AntonPetrunin: I don't see why this form for $f$ and $g$ is likely. This mixed-order system of 3 equations for 2 unknowns turns out to be involutive, with the general (local) solution depending on two functions of 1 variable. It's of Goursat-parabolic type, which means that the characteristics are double (as is usual in parabolic equations), but there is no 'diffusion' in the initial conditions, i.e., one gets to freely specify one 'arbitrary function of one variable, and then the (local) solution can be constructed using ODE only.(If there is interest, I can put in details when I have time.) – Robert Bryant May 6 '15 at 20:59

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.

• Thanks for detailed answer. By the way, I've just recently found that it is homogeneous Monge-Amper equation and it has a general solution in convenient form: $w = t x + \varphi (t) y + \psi (t),$ $x + \varphi' (t) y + \psi '(t) = 0.$ – user47116 May 8 '15 at 7:08
• @DmitryLyakhov: Yes, each single second-order equation (one for $f$ and one for $g$) is a homogeneous Monge-Ampère equation, but that doesn't help you show that the two solutions can be simultaneously ruled; in fact, this won't be true without the third equation. – Robert Bryant May 8 '15 at 8:56
• It would be nice, if we can get an explicit formula! – user47116 May 10 '15 at 10:21