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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).

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    $\begingroup$ When you write "Does anybody know if there is a partial differential equation whose solutions are all ruled surfaces and only them?", do you mean to ask whether there is a partial differential equation that characterizes ruled surfaces? The natural PDE that does this is the one that sets equal to zero the product of the curvatures of the asymptotic curves. It's a 3rd order equation, and the general solution depends on 3 functions of 1 variable, as you'd expect. This equation is projectively invariant, so it is better to express it directly in terms of projective invariants. $\endgroup$ Oct 12, 2011 at 0:46
  • $\begingroup$ yes, i'm looking for a partial equation that characterizes the ruled surfaces, but preferably for surfaces given by $f(x,y,z)=0$ and not in the parametric form. so if we're given a surface $f(x,y,z)=0$ we just plug in $f$ in the partial equation and we know that it's ruled iff we get 0. $\endgroup$
    – filipm
    Oct 12, 2011 at 8:33
  • $\begingroup$ here faculty.fairfield.edu/jmac/rs/halftw.htm they mention an equation $x^2z_{xx}+2xyz_{xy}+y^2z_{yy}=0$, but it's not the right one, it misses e.g. $z=xy$ (probably this equation gives only a subclass of the ruled surfaces). $\endgroup$
    – filipm
    Oct 12, 2011 at 8:36
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    $\begingroup$ @filipm: In that case, set $II = f_{xx} dx^2 + 2f_{xy}dxdy + f_{yy}dy^2$. If the discriminant of $II$ vanishes, then the surface is ruled. If the discriminant is positive, it is not ruled. If the discriminant is negative, compute $III = f_{xxx}dx^3+3f_{xxy}dx^2dy+3f_{xyy}dxdy^2+f_{yyy}dy^3$ and let $III_0$ be the $II$-trace-free part of $III$. Then the surface is ruled if and only if the discriminant of $III_0$ vanishes. $\endgroup$ Oct 12, 2011 at 13:56
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    $\begingroup$ @Felipe: For $z=f(x,y)$, it's the equation ${f_{yy}}^3{f_{xxx}}^2+6{f_{yy}}{f_{xxx}}{f_{yyy}}{f_{xy}}{f_{xx}}$ $-6{f_{yy}}^2{f_{xxx}}{f_{xyy}}{f_{xx}}-6{f_{yyy}}{f_{xy}}{f_{xx}}^2{f_{xyy}}$ $+9{f_{yy}}{f_{xyy}}^2{f_{xx}}^2-6{f_{xy}}{f_{yy}}^2{f_{xxy}}{f_{xxx}}$ $+12{f_{xy}}^2{f_{xxy}}{f_{yyy}}{f_{xx}}-18{f_{xy}}{f_{yy}}{f_{xxy}}{f_{xyy}}{f_{xx}}$ $+12{f_{yy}}{f_{xyy}}{f_{xy}}^2{f_{xxx}}-8{f_{yyy}}{f_{xy}}^3{f_{xxx}}$ $+9{f_{xx}}{f_{yy}}^2{f_{xxy}}^2-6{f_{yy}}{f_{xxy}}{f_{yyy}}{f_{xx}}^2$ $+{f_{yyy}}^2{f_{xx}}^3 = 0$. $\endgroup$ Oct 13, 2011 at 18:45

6 Answers 6

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

Suppose that $I\!I$ is nonzero. The discriminant of $I\!I$ is defined (up to a factor of $(\mathrm{d}x\wedge\mathrm{d}y)^{\otimes 2}$) to be $$ \Delta(I\!I) = f_{xx}f_{yy}- {f_{xy}}^2. $$

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

If $\Delta(I\!I) = 0$, then the surface is ruled. In fact, it has vanishing Gauss curvature. Moreover, $I\!I = \pm \alpha^2$ 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 $\Delta(I\!I) < 0$, set $I\!I\!I = f_{xxx}\ dx^3+3f_{xxy}\ dx^2dy+3f_{xyy}\ dxdy^2+f_{yyy}\ dy^3$ and let $I\!I\!I_0$ be the $I\!I$-trace-free part of $I\!I\!I$. Then the surface $z = f(x,y)$ is ruled if and only if the discriminant of $I\!I\!I_0$ vanishes.

(Added later: This latter condition (i.e., the vanishing of the discriminant of $I\!I\!I_0$) turns out to be equivalent to the condition that $I\!I$ and $I\!I\!I$ have a common linear factor, say, $\alpha$ (which will necessarily be real when $\Delta(I\!I) < 0$), and hence is equivalent to the vanishing of the resultant of $I\!I$ and $I\!I\!I$, i.e., $\textrm{Reslt}(I\!I,I\!I\!I) = 0$. 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.

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  • $\begingroup$ It is easy to see that the condition that II and III have a common real linear factor is necessary for the surface to be ruled: just parametrize a line on the surface as $( x_0+at, y_0+bt, z_0+ct=f(x_0+at,y_0+bt) )$ and differentiate 3 times wrt $t$. Is there a good reference to the proof that the condition is sufficient? $\endgroup$ Mar 5, 2019 at 12:09
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    $\begingroup$ @MikhailSkopenkov: I don't know a reference off the top of my head, but I am sure that it is somewhere in the classical affine geometry literature. Maybe one should ask Udo Simon or Gary Jensen, both of whom have books on classical affine geometry of surfaces. Maybe Blaschke is a good reference; have a look at his book on the subject. The proof that I know is just the obvious one using moving frames. Actually, since it is a projective geometry fact, this ought to be in the classical projective geometry literature, but I'm not that familiar with that literature. Maybe Svec or Wilczinski? $\endgroup$ Mar 5, 2019 at 12:51
  • $\begingroup$ Many thanks! The exposition of this result in the Blaschke book is a bit cumbersome, thus let us try the other ones. $\endgroup$ Mar 5, 2019 at 12:55
  • $\begingroup$ @RobertBryant, why the curves in the domain defined by $\alpha=0$ must be straight lines? I know it is a comment fact that surface with vanishing Gaussian curvature everywhere must be ruled surfaces, but I cannot find a simple proof. Could you explain to me a little bit? Thanks! $\endgroup$
    – student
    Oct 5, 2019 at 19:53
  • $\begingroup$ @student This is not obviuos. We are currently writing a paper on that. If you give your e-mail address (say, via private message), we could send you a draft. $\endgroup$ Oct 22, 2019 at 8:32
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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
   Tz1
   alt text


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

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  • $\begingroup$ Joseph, how do you insert a pdf document in an MO answer box ? You have even managed to insert material from two different pages without showing the page break. I would like to use the trick... $\endgroup$ Oct 14, 2011 at 2:56
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    $\begingroup$ @Chandan: I converted the PDF to JPEG (actually, to two JPEGs) and then included each as images with the HTML img-tag. $\endgroup$ Oct 14, 2011 at 9:58
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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.

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  • $\begingroup$ Would you care to type in the theorem? There is no guarantee that this pdf file will stay at that location for the lifetime of MO. $\endgroup$
    – David Roberts
    Oct 11, 2011 at 21:43
  • $\begingroup$ @david: last time I did this for you the question was closed so I'm not going to spend the effort this time. see mathoverflow.net/questions/77643/pythagorean-analogue/77645 $\endgroup$
    – psd
    Oct 11, 2011 at 21:53
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    $\begingroup$ @David: If you look at the paper, you will see that typing in all the requisite notation is a bit painful. Perhaps including the math review pointer is a more practical solution. $\endgroup$
    – Igor Rivin
    Oct 11, 2011 at 21:54
  • $\begingroup$ @david: I see that in my 'pythagorean analogue' answer you asked me to summarize the important points while at the same time you voted to close the question. $\endgroup$
    – psd
    Oct 11, 2011 at 22:26
  • $\begingroup$ The cited theorem has nothing to do with characterization of ruled surfaces. It is on characterization of a very-restricted class of so-called Tzitzeica surfaces, which can be ruled or not. $\endgroup$ Nov 4, 2019 at 17:42
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First of all, for bivariate functions $z=u(x,y)$, let's write down the following notation

\begin{gather*} r=\dfrac{\partial^2u}{\partial\,\!x^2}\qquad\,s=\dfrac{\partial^2u}{\partial\,\!x\partial\,\!y}\qquad\,t=\dfrac{\partial^2u}{\partial\,\!y^2}\\ \\ \lambda_1=\dfrac{-s+\sqrt{s^2-rt}}{t}\qquad\,\lambda_2=\dfrac{-s-\sqrt{s^2-rt}}{t}\\ \\ \Diamond\,\!u=\dfrac{\partial\,\!\lambda_1}{\partial\,\!x}+\lambda_1\dfrac{\partial\,\!\lambda_1}{\partial\,\!y} \\\\ \bar\Diamond\,\!u=\dfrac{\partial\,\!\lambda_2}{\partial\,\!x}+\lambda_2\dfrac{\partial\,\!\lambda_2}{\partial\,\!y} \end{gather*}

  1. If the smooth surface $z=u(x,y)$ is ruled, also required $t\ne0$, then the bivariate function $z=u(x,y)$ satisfies the third order partial differential equation $\Diamond\,\!u=0$ or $\bar\Diamond\,\!u=0$;

  2. If the bivariate function $z=u(x,y)$ satisfies the third-order partial differential equation $\Diamond\,\!u=0$ or $\bar\Diamond\,\!u=0$, also satisfies the inequality $s^2-rt\ge0$ and $t\ne0$, the surface represented by it is ruled.

Cf. [Monge 1780] Gaspard Monge, “Mémoire sur les Propriétés de plusieurs genres de Surfaces courbes, particulièrement sur celles des Surfaces développables, avec une Application à la Théorie des Ombres et des Pénombres”, Savans Étrangers 9 (1780), pp.382-440.

https://archive.org/details/mmoiresdemath09acad/page/434

Cf. [J. Ockendon, S. Howison, A. Lacey, A. Movchan] Applied Partial Differential Equations (2003), pp.380-382.

Example: $u(x,y)=\dfrac{xy}{x^2+y^2}$, i.e. $z=\dfrac{xy}{x^2+y^2}$; Assume that $x>y>0$, then

\begin{gather*} \begin{split} r&=+\dfrac{2(x^2-3y^2)xy}{(x^2 +y^2)^3}\\\\ s&=-\dfrac{(x^2 - 2 x y - y^2) (x^2 + 2 x y - y^2)}{(x^2 +y^2)^3}\\\\ t&=-\dfrac{2(3x^2-y^2)xy}{(x^2 +y^2)^3}\\\\ \sqrt{s^2-rt}&=\dfrac{\left|x^2-y^2\right|}{(x^2 +y^2)^2}=\dfrac{x^2-y^2}{(x^2 +y^2)^2}>0\\ \end{split}\\ \\ \lambda_1=-\dfrac{(x^2-3y^2)x}{(3x^2-y^2)y}\qquad\,\lambda_2=\dfrac{y}{x}\\ \\ \begin{split} &\color{red}{\Diamond\,\!u=-\dfrac{3(x-y)(x+y)(x^2+y^2)^3}{(3x^2-y^2)^3y^3}}\\ &\color{red}{\phantom{\Diamond\,\!u}<0}\\ \\ &\color{blue}{\bar\Diamond\,\!u\equiv0} \end{split} \end{gather*}

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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.

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    $\begingroup$ A surface can still be ruled without $K = 0$. For example, one principal curvature could be positive and the other negative, so the second fundamental form still has null directions; if you're lucky enough, they line up in space. The catenoid in particular is ruled but curvy. $\endgroup$ Oct 11, 2011 at 22:19
  • $\begingroup$ The Wikipedia article en.wikipedia.org/wiki/Differential_geometry_of_surfaces shows a formula for the Gaussian curvature of a ruled surface, and gives conditions on when it vanishes. $\endgroup$ Oct 11, 2011 at 22:21
  • $\begingroup$ Oops. That's right. You just need a null direction that propagates along a straight line. Gotta downvote my own answer. $\endgroup$
    – Deane Yang
    Oct 11, 2011 at 22:49
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    $\begingroup$ I think it's useful to leave it. I will edit it. $\endgroup$
    – Deane Yang
    Oct 12, 2011 at 0:39
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    $\begingroup$ An obvious example of a ruled surface with $K<0$: a one-sheet hyperboloid. $\endgroup$ Oct 12, 2011 at 5:27
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The only chracterization of ruled surfaces I know is the following: A surface S in the 3-dimensional Euclidean space of negative Gaussian curvature is a ruled surface iff its (equi)affine Pick-invariant vanishes identically on S.

This theorem can be found in the books of W. Blaschke on Affine Differential Geometry.

For surfaces of positive Gaussian curvature the following theorem is valid: A surface S in the n-dimensional Euclidean space of positive Gaussian curvature is a quadric iff its (equi)affine Pick-invariant vanishes identically on S.

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  • $\begingroup$ How is this related to the question? $\endgroup$ Feb 19, 2014 at 19:24
  • $\begingroup$ @AlexDegtyarev Pick-invariant=0 is the required PDE (perhaps). Although seems very difficult to write down explicitly in terms of the original surface. $\endgroup$ Oct 22, 2019 at 8:35

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