Solution of the PDE which has Cauchy Data. In the case of the quasilinear PDE
$$
(y-z) \frac{\partial z}{\partial x} + (z-x) \frac{\partial z}{\partial y} = x-y \, ,
$$
the method of characteristics yields the following Lagrange-Charpit equations:
$$
\frac{dx}{y-z}=\frac{dy}{z-x}=\frac{dz}{x-y} .
$$ 
Find the solution of the Cauchy problem with data $z(x,y)=0$ on $xy=1$.
I found first charecteristic as $c_1=z+x+y$. But I can' t find the other characteristic.
I generally have difficulty finding the second one. What are your suggestions for finding easily second characteristic? What are the clues, signs?. I will be very happy if you say your clues.
 A: (see graphics below)
A clue for this particular case is to take "geometrical glasses" : the given system is equivalent to the fact that $\begin{pmatrix}dx\\dy\\dz\end{pmatrix}$ is proportional to cross product $\begin{pmatrix}x\\y\\z\end{pmatrix} \times  \begin{pmatrix}1\\1\\1\end{pmatrix}.$ Let us denote by $U$ this fixed vector.
Thus, the tangent vector will always be simultaneously :


*

*orthogonal to vector $\begin{pmatrix}1\\1\\1\end{pmatrix}$ which explains the fact that planes with equation $x+y+z=c_1$ ($c_1$ any constant) constitute a characteristic. As a consequence, if point $M_t$ is in plane $(P_1)$ with equation $x+y+z=c_1$ at $t=0$, it will remain all the time in $(P_1)$.

*orthogonal to vector $\begin{pmatrix}x\\y\\z\end{pmatrix}$. This gives the "clue" that $M_t$ has a circular "behavior" in its plane. More precisely, this orthogonality behavior is possible if and only if the circular loop made by $M_t$ is at the intersection of plane $(P_1)$ with the cone whose axis is directed by vector $U$ and that passes through point $P_0$.  The consequence of all this being that we have a preservation of the (square of the) distance from point $M_t$ to the center of the circle (situated on line $x=y=z$), i.e., 
$$(x-c_1)^2+(x-c_1)^2+(z-c_1)^2=c_2$$
We have thus obtained a second parameter $c_2$...
Here is a graphical representation displaying some integral curves (circles), all with initial points $M_0$ ("red stars") on the hyperbola $(H)$ with equation $xy=1$. 
It is in fact evident that this initial constraint corresponds to the intersection with the very specific cone, whose axis is directed by $U$ and whose intersection with $xOy$ plane is hyperbola $(H)$ ; it can be guessed on the graphics below. 
Remarks:
1) Initial points $(a,1/a)$ and $(1/a,a)$ generate the same integral curve.
2) we have kept our geometrical intuition till the end...

