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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?

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1 Answers1

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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:

characteristics

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.

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