TITLE: partial differential equation for ruled surfaces
QUESTION [14 upvotes]: 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).

REPLY [2 votes]: 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*}

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 $\bar\Diamond\,\!u=0$;
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*}