Question

We say that a surface $f(x,y,z)=0$ is ruled if for each point $p$ in the surface there is a line that passes through $p$ and is contained in the surface. See http://en.wikipedia.org/wiki/Ruled_surface for more information.

Does anybody know if there is a partial differential equation whose solutions are all ruled surfaces and only them? And what is the equation? Some reference would be helpful, too (different from Salmon's old book about surfaces which I unfortunately find unreadable).

Answer 1

Here is a test for when a surface of the form $z = f(x,y)$, where $f$ is a sufficiently smooth function of two variables, is ruled.

To begin, set $II = f_{xx} dx^2 + 2f_{xy}dxdy + f_{yy}dy^2$. If $II$ vanishes identically, then the surface is a plane, so it is ruled.

Suppose that $II$ is nonzero. The discriminant of $II$ is defined to be $$ \Delta(II) = f_{xx}f_{yy}- {f_{xy}}^2. $$

If $\Delta(II) >0$, then the surface is locally strictly convex and so cannot be ruled.

If $\Delta(II) = 0$, then the surface is ruled. In fact, it has vanishing Gauss curvature. Moreover, $II = \pm \alpha^2$ for some nonzero $1$-form $\alpha$ on the domain of $f$, and the curves in this domain defined by $\alpha = 0$ (which turn out to be straight lines) lift to the graph $z = f(x,y)$ to be straight lines.

If $\Delta(II) < 0$, set $III = f_{xxx}\ dx^3+3f_{xxy}\ dx^2dy+3f_{xyy}\ dxdy^2+f_{yyy}\ dy^3$ and let $III_0$ be the $II$-trace-free part of $III$. Then the surface $z = f(x,y)$ is ruled if and only if the discriminant of $III_0$ vanishes.

(Added later: This latter condition turns out to be equivalent to the condition that $II$ and $III$ have a common linear factor, say, $\alpha$, and hence is equivalent to the vanishing of the resultant of $II$ and $III$, i.e., $\textrm{Reslt}(II,III) = 0$. When such an $\alpha$ exists, the leaves of $\alpha=0$ are lines on the surface.)

Notes:

1. The discriminant of a cubic form $C = p\ dx^3 + 3q\ dx^2dy + 3r\ dxdy^2 + s\ dy^3$ is, by definition, $$ \Delta(C) = s^2p^2 + 4r^3p + 4 q^3s - 3 r^2q^2 - 6 sqrp. $$ It is, up to a multiple, the unique polynomial of degree $4$ in the coefficients that vanishes if and only if $C$ has a multiple factor.

2. Given a quadratic form $Q = a\ dx^2 + 2b\ dxdy + c\ dy^2$ with nonvanishing discriminant $D$, the $Q$-trace of a form $C$ of degree $3$ is the linear form $$ tr_Q(C ) = \frac{(ar-2bq+cp)\ dx + (as-2br+cq)\ dy}{D}. $$ Any cubic form $C$ can be uniquely written in the form $$ C = C_0 + L\cdot Q $$ where $L$ is a linear form and $tr_Q(C_0) = 0$. (In fact, $L = \tfrac34 tr_Q(C)$.) The term $C_0$ is called the $Q$-trace-free part of $C$.

3. The resultant $\textrm{Reslt}(Q,C)$ of a quadratic form $Q$ and a cubic form $C$ is the (unique up to nonzero multiples) polynomial that is cubic in the coefficients of $Q$, quadratic in the coefficients of $C$, and vanishes exactly when they have a common linear divisor.

Answer 2

From the paper to which 'lowerbound' linked, "Symmetry groups and Lagrangians associated with Tzitzeica surfaces," by Nicoleta Bila (also arXiv:math/9910138v1), here is Theorem 1 and its preamble: Consider $D \subset \mathbb{R}^2$ and let
   
   

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I take no credit (or blame!) for this; just posting as a community service.

Answer 3

Does anybody know if there is a partial differential equation whose solutions are all ruled surfaces and only them?

An example is theorem 1 of http://www.kurims.kyoto-u.ac.jp/EMIS/journals/BJGA/10.1/bt-bil.pdf

Edit: I have made this a community wiki answer so that if anyone would care to type in the theorem then they are free to do so.

Answer 4

Isn't it just $K = 0$, where $K$ is the Gauss curvature?

Wrong. Although surface with zero Gauss curvature is necessarily ruled, the converse is not true. The catenoid is the best known counterexample. See comments below.