Characteristics of equations of the form $u_{xy}=f(u_x,u_y,u)$ In the usual treatment of hyperbolic differential equations, it is always assumed that there are two families of characteristics. That is, if the equation $L[u]-f(u_x,u_y,u)=au_{xx}+2bu_{xy}+cu_{yy}-f(u_x,u_y,u)=0$ is hyperbolic, by definition the roots $\zeta_{\pm}$ of the polynomial $q(\zeta)=a\zeta^2-2b\zeta+c$ are real, and then one defines the characteristic curves by means of the ODEs $dy/dx=\zeta_{\pm}(x,y)$. But for example for the equation $u_{xy}=u$, the only root is $\zeta_+=\zeta_-=0$, then we would have only one family of characteristics so the equation is not hyperbolic but parabolic, but this contradicts the fact that a equation is parabolic iff  $b^2-ac=0$ and hyperbolic iff $b^2-ac>0$.
What's the true nature of the equations of the form $u_{xy}=f(u_x,u_y,u)$,  and what are the implications of the fact that only one set of characteristics exists, even though it's supposed to be a hyperbolic equation?
 A: From an 2015 MSE answer called Indefinite double integral
we conclude that the general solution of the PDE
$$
\frac{\partial^2 u}{\partial p\,\partial q} = \frac{\partial^2 u}{\partial q\,\partial p} = 0
$$
is given by
$$
u(p,q) = F(p) + G(q)
$$
The wave equation has been considered as an example:
$$
\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} = 0
$$
With a lemma in the
abovementioned reference, this is converted to:
$$
\frac{\partial}{\partial (x-ct)}\frac{\partial}{\partial (x+ct)} u =
\frac{\partial}{\partial (x+ct)}\frac{\partial}{\partial (x-ct)} u = 0
$$
So we find that the general solution of our wave equation is given by:
$$
u(x,t) = F(p) + G(q) = F(x-ct) + G(x+ct)
$$
Here $\,x-ct=0\,$ and $\,x+ct=0\,$ are the characteristics.
Quite analogously, if the OP's simplified equation $\,u_{xy}=0\,$ is considered instead, then the solution must be:
$$
u(x,y) = F(x) + G(y)
$$
with $\,x=0\,$ and $\,y=0\,$ as the two characteristics. Indeed the equation is still hyperbolic, because $b^2-ac=b^2=1\gt 0$.

Let's try another hyperbolic argument :-)
Both the equations $\,xy=1\,$ and $\,x^2-y^2=1\,$ represent an orthogonal hyperbola, right?
They are equivalent up to a rotation over $45^o$.Let's try the same for the equation $\,u_{xy}=0\,$.
Consider the rotation transformation
$$
\begin{cases}
x_1 = \cos(\alpha)\,x-\sin(\alpha)\,y \\
y_1 = \sin(\alpha)\,x+\cos(\alpha)\,y
\end{cases}
$$
We want to know how the derivatives of the solution $\,u\,$ transform.
$$
\frac{\partial u}{\partial x} =
  \frac{\partial u}{\partial x_1}\frac{\partial x_1}{\partial x}
+ \frac{\partial u}{\partial y_1}\frac{\partial y_1}{\partial x} \\
\frac{\partial u}{\partial y} =
  \frac{\partial u}{\partial x_1}\frac{\partial x_1}{\partial y}
+ \frac{\partial u}{\partial y_1}\frac{\partial y_1}{\partial y}
$$
It follows, with operator notation:
$$
\frac{\partial}{\partial x} u =
  \left[ \cos(\alpha)\frac{\partial}{\partial x_1}
 +\sin(\alpha)\frac{\partial}{\partial y_1} \right] u\\
\frac{\partial}{\partial y} u =
 \left[ -\sin(\alpha)\frac{\partial}{\partial x_1}
 +\cos(\alpha)\frac{\partial}{\partial y_1} \right] u
$$
So we have:
$$
\frac{\partial^2}{\partial x \, \partial y} =
\left(\frac{\partial}{\partial x}\right)\left(\frac{\partial}{\partial y}\right) = \\
\left[ \cos(\alpha)\frac{\partial}{\partial x_1}
 +\sin(\alpha)\frac{\partial}{\partial y_1} \right]
 \left[ -\sin(\alpha)\frac{\partial}{\partial x_1}
 +\cos(\alpha)\frac{\partial}{\partial y_1} \right] = \\
-\sin(\alpha)\cos(\alpha)\left(\frac{\partial}{\partial x_1}\right)^2
+\left[\cos^2(\alpha)-\sin^2(\alpha)\right]\frac{\partial}{\partial x_1}\frac{\partial}{\partial y_1}
+\sin(\alpha)\cos(\alpha)\left(\frac{\partial}{\partial y_1}\right)^2 = \\
-\frac12 \sin(2\alpha)\frac{\partial^2}{\partial x_1^2}
+ \cos(2\alpha)\frac{\partial^2}{\partial x_1\,\partial y_1}
+ \frac12 \sin(2\alpha)\frac{\partial^2}{\partial y_1^2}
$$
Now choose $\,\alpha = 45^o = \pi/4\,$, just as with the orthogonal hyperbolas, then we have:
$$
\frac{\partial^2}{\partial x \, \partial y} = -\frac12 \frac{\partial^2}{\partial x_1^2} + \frac12 \frac{\partial^2}{\partial y_1^2}
$$
And so:
$$
\frac{\partial^2 u}{\partial x \, \partial y} = - \frac12 \left[ \frac{\partial^2 u}{\partial x_1^2} - \frac{\partial^2 u}{\partial y_1^2}\right] = 0
$$
Which is clearly a hyperbolic, not a parabolic, equation.
The characteristics $\,x=0\,$ and $\,y=0\,$ are likewise rotated, giving $\,x_1+y_1=0\,$ and $\,x_1-y_1=0\,$.
A: This question has already been answered by Han de Bruijn: however I feel I can make explicit some conceptual points and give a few references and insights.

*

*First of all let me point out the most important thing: Han has shown by an example that the following equation where $L$ is a linear hyperbolic operator
$$
L[u]-f(u_x,u_y,u)=au_{xx}+2bu_{xy}+cu_{yy}-f(u_x,u_y,u)=0 \label{a}\tag{HE}
$$
has two (almost, see the notes) equivalent normal form, i.e.
$$
\begin{align}
u_{\eta\xi} & = f_1(u_\eta,u_\xi,u) \label{1}\tag{1}\\
u_{\eta\eta}-u_{\xi\xi} & = f_2(u_\eta,u_\xi,u) \label{2}\tag{2}
\end{align}
$$
To see this, let's recall the basic effect of coordinate transformation applied to the linear part of \eqref{a}: given a one-to-one map
$$
\begin{cases}
\eta=\eta(x,y)\\
\xi = \xi(x,y)
\end{cases} \implies J
\begin{pmatrix}
\eta & \xi \\
x & y
\end{pmatrix}\neq 0
$$
we have that
$$
\begin{split}
L(u)=au_{xx}+2bu_{xy}+cu_{yy} &= [a\eta_x^2+2b\eta_x\eta_y+c\eta_y^2] u_{\eta\eta}\\ 
&\; + 2[a\eta_x\xi_x+b(\eta_x\xi_y+\eta_y\xi_x)+c\eta_y\xi_y] u_{\eta\xi}\\
&\; + [a\xi_x^2+2b\xi_x\xi_y+c\xi_y^2] u_{\xi\xi}\\
&\; + \text{ lower order linear terms in $\eta_x, \eta_y, \xi_x, \xi_y$}
\end{split}\label{3}\tag{3}
$$
Now consider the characteristic equation for $L$, i.e.
$$
a\varphi_x^2+2b\varphi_x\varphi_y+c\varphi_y^2 = 0.\label{b}\tag{CE}
$$
Providing proper condition are stated (see again the notes below), since $L$ is hyperbolic (i.e. $b^2- ac>0$) this equation has two distinct real solutions , say $\varphi_1$ and $\varphi_2$: it is possible to demonstrate that these functions define a one-to-one map. If in \eqref{3} we put
$$
\begin{align}
\eta (x,y) = \varphi_1 (x,y)\\
\xi (x,y) = \varphi_2 (x,y)
\end{align}\label{4}\tag{choice 1}
$$
we get the normal form \eqref{1}, while if we put
$$
\begin{align}
\eta (x,y) = \varphi_1 (x,y)+\varphi_2 (x,y)\\
\xi (x,y) =  \varphi_1 (x,y)-\varphi_2 (x,y)
\end{align}\label{5}\tag{choice 2}
$$
we get the \eqref{2} normal form.

*The two different change of variables generate two different lower order linear terms, are that are added to the transformed nonlinear right hand side of \eqref{1} i.e. $f(u_x, u_y, u)$ to give rise to the right hand sides $f_i(u_\eta,u_\xi, u)$, $i=1,2$ of respectively \eqref{1} and \eqref{2}.

*Note that \ref{5} corresponds to the $\alpha={\pi/ 4}$ choice in the example of Han. On the other hands, it is easy to see the reason for the apparent inconsistency of \eqref{2} respect to the definition of "hyperbolic PDO": the coodinate axes $\eta=0$ and $\xi=0$ generated by \ref{4} are images of the characteristic curves.

*Why the normal form \eqref{1} was classically more used than normal form \eqref{2}?: this is simply because the solutions of \eqref{1} need to be of class $C^2$, while the solution of \eqref{2} are "only" required to be of class $C^1$ with continuous mixed derivatives, i.e. $u_{xy}\in C^0$. This may seem silly, as today the concept of solution has evolved in order to include non-differentiable-at-all functions but in the classical context formulation \eqref{2}, requiring a somewhat lower regularity, may have seen as appealing in order to ease the study of the PDE.

*For this answer I heavily used reference [2] (see chapter II, §1-2 pp. 130-140)  since it deals comprehensively with this equation and many more in a classical setting. Of course the problem is dealt with in other wonderful references in English, like in [1] (see chapter I, §2.2, pp. 22-26) or [3] (see chapter III, §1.1, p. 155), however in a very concise fashion.

*The analysis of the homogeneous ultrahyperbolic equation, i.e.
$$
\frac{\partial^2}{\partial x\partial y}u=u_{xy}=0
$$ is describer in this MathOverflow answer. The references given therein bring to the original works of Gaetano Fichera and Mauro Picone who studied this equation thoroughly.

References
[1] Andreĭ Vasil’evich Bitsadze, Equations of Mathematical Physics (English),  Translated from the Russian by V. M. Volosov and I. G. Volosova, Moskva: Mir, p. 318 (1980), ISBN: 5-03-000533-1, MR1024787, Zbl 0499.35002.
[2] Silvio Cinquini, Maria Cinquini Cibrario, Equazioni a derivate parziali di tipo iperbolico. (Italian) Monografie Matematiche del Consiglio Nazionale delle Ricerche 12. Roma: Edizioni Cremonese, pp. VIII+552 (1964), MR0203199, Zbl 0145.35404.
[3] Richard Courant, David Hilbert, Methods of mathematical physics. Volume II: Partial differential equations. Translated and revised from the German Original. Reprint of the 1st Engl. ed. 1962. (English) Wiley Classics Edition. New York-London-Brisbane: John Wiley & Sons/Interscience Publishers, pp. xxii+830 (1989), ISBN: 0-471-50439-4, MR1013360, Zbl 0729.35001.
