Solve $$F\frac{\partial F}{\partial x} - \frac{\partial F}{\partial y} = y$$
subject to $F(s,0) = s^2$.
This is the first time I am using the method of characteristics, so I would like to know if I have made any errors in my working. I have $$ \frac{dx}{dt} = z, \quad \frac{dy}{dt} = -1, \quad \frac{dF}{dt} = y $$
Solving these gives me $$ y = -t + y_0, \quad F =-\frac{t^2}{2} + y_0t + F_0, \quad x = -\frac{t^3}{6}+\frac{y_0 t^2}{2} + F_0t + x_0 $$
Then assume without loss of generality that $y_0 = 0$, so $$ y = -t, \quad F =-\frac{t^2}{2}+ F_0, \quad x = -\frac{t^3}{6} + F_0t + x_0 $$
The characteristic curves are therefore curves of the form $$ t \mapsto \left(-\frac{t^3}{6} + F_0t + x_0,\quad-t,\quad -\frac{t^2}{2}+ F_0 \right) $$
I can now use the initial condition $F(s,0) = s^2$. This means I substitute $x = s, y = 0$ and $F = s^2$ and solve for $x_0, F_0$ and $t$ . So $$ y = 0 \implies t = 0, \quad s = x_0, \quad s^2 = F_0 $$
So the parametrisation of the solution surface is $$ (s,t) \mapsto \left(-\frac{t^3}{6} + s^2t + s,\quad-t,\quad -\frac{t^2}{2}+ s^2 \right) $$
To get this in terms of $x$ and $y$ I know $-y = t$, hence $$ x = \frac{y^3}{6} - s^2y + s \implies s = \frac{1\pm\sqrt{1-4y\left(x-\frac{y^3}{6} \right)}}{2y} $$
Thus $$ F(x,y) = -\frac{y^2}{2} + \left( \frac{1\pm\sqrt{1-4y\left(x-\frac{y^3}{6} \right)}}{2y}\right)^2 $$
This function is not well defined when $y = 0$ or when the term inside the square root is negative.
Have I made a mistake somewhere or is the working fine?
