I have the nonlinear PDE $$p^2 + 2q = x$$ with the initial condition $u(0, y) = -y^2.$$u(0, y) = -y^2$, and $y > 0$.
Here's what I have done so far:
I defined the function $F$ to be equal $$F(x, y, p, q, u) = p^2 + 2q - x,$$
and got
$$(F_x, F_y, F_p, F_q, F_u) = (-1, 0, 2p, 2, 0).$$
Therefore I managed to write down the characteristic
$$\frac{dx}{-F_p} = \frac{dy}{-F_q} = \frac{dp}{F_x + pF_u} = \frac{dq}{F_y + qF_u} = \frac{du}{-pF_p - qF_q},$$
or
$$\frac{dx}{-2p} = \frac{dy}{-2} = \frac{dp}{-1} = \frac{dq}{0} = \frac{du}{-2p^2 - 2q}.$$
Then, clearly, $q = a$ for some constant $a$. Putting $q = a$ in the equation yields
$$p^2 + 2a = x,$$
or
$$p = \pm \sqrt{x - 2a}.$$
Since $du = pdx + qdy$, we have
$$du = \pm \sqrt{x - 2a}dx + ady.$$
Integrating both sides yields
$$u = \pm \frac{2}{3}(x - 2a)^{\frac{3}{2}} + ay + b,$$ where $b$ is a constant.
If this is a solution, then it must satisfy the initial condition. Putting $x = 0$ and $u = -y^2$ in the solution above gives us
$$-y^2 = \pm \frac{2}{3}(-2a)^{\frac{3}{2}} + ay + b.$$
However, there are no constants $a, b$ such that the equation above holds for all $y$. Then I concluded that there is no solution to the given PDE satisfying the given initial condition.
But I solved the given PDE with the given initial condition using another method (what I ended up having was a parametric solution), so I know there actually is a solution to the given PDE with the given initial condition.
I can't even guess what my mistake was. Any help would be highly appreciated.