Determine the shock regions of quasi-linear equation Assume the IVP: 
\begin{cases}
z^2 z_x + z_y = 0 \\
z(x,0) = f(x) \\
\end{cases}
The condition of existence of (locally) unique solution is:
$$
P(t_0) \frac{dy(t_0)}{dt} - Q(t_0) \frac{dx(t_0)}{dt} \neq 0
$$
on the curve $(t,0,f(t))$, which holds true since:
$$
0 - 1 \cdot 1 = -1 \neq 0
$$
If we let $f(x) = x$, the solution is:
$$
z = x - z^2 y \iff z(x,y) = \frac{-1 + \sqrt{1 + 4xy}}{2y}
$$
Shock waves:
By the Implicit Function Theorem, the solution exists and is implicitly defined as long as:
$$
1 + f'(x - z^2 y)2zy > 0
$$
So a shock wave forms when:
$$
1 + f'(x - z^2 y)2zy = 0 \iff z = \frac{-1}{2y} \iff 1 + 4xy = 0
$$
Thus, the solution is uniquely defined for $1 + 4xy > 0$ and the region of shock is $1 + 4xy \leq 0$.
Question: Is the aforementioned an adequate description when someone wants to determine the regions where the solution is uniquely defined and the shock regions?
 A: With the initial condition $z(x,0) = f(x) = x$, the solution deduced from the method of characteristics reads indeed $z = x-z^2 y$ in implicit form. Since this is a quadratic equation, we can easily analyze the number of solutions from the computation of the discriminant $\Delta = 1+4xy$, which we want to be non-negative for existence. From the two real roots $z = \frac12 (-1 \pm \sqrt{\Delta})/y$, only the '+' solution is compatible with the initial condition. Therefore the classical solution to the IVP is uniquely defined in the region $\Delta \geq 0$ where we have $z = \frac12 (-1 + \sqrt{\Delta})/y$, and there is no solution outside. A plot of the base characteristic lines in the $x$-$y$ plane is given below:

The characteristic lines are included in the region $\Delta \geq 0$, located between the hyperbola branches $y = -1/(4x)$. Those hyperbola branches appear to be the caustic envelope of the characteristic lines, and they mark the boundaries of the accessible region of the $x$-$y$ plane. The region $\Delta < 0$ is not a shock region, but a region of the $x$-$y$ plane which is inaccessible. To some extent, this problem is quite similar to this one.
