Charpit's method and a nonlinear PDE

Question

I have the nonlinear PDE $$p^2 + 2q = x$$ with the initial condition $u(0, y) = -y^2$, 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.

Answer 1

If you are interested in only solutions on subdomains, near the set $\{x = 0, y > 0\}$, then your mistake is that in the method of characteristics, the fact that $q = a$ only holds along the characteristic. (Your supposition that $q = a$ everywhere already contradicts the initial data.)

Everything you did after that is not part of the method of characteristics.

Given $$ F = p^2 + 2q - x $$ The characteristic equations are $$ \dot{p} = 1, \dot{q} = 0, \dot{u} = 2p^2 + 2 q, \dot{x} = 2p, \dot{y} = 2 $$ You obtain that $$ p(s) = s + p(0), q(s) = q(0), u(s) = u(0) + \frac23 s^3 + 2 s^2 p(0) + 2p(0)^2 s + 2 q(0)s + u(0)$$ and $$ x(s) = s^2 + 2p(0) s, y(s) = 2 s + y(0) $$ The initial data are given in terms of $y(0)$: $$ q(0) = - 2y(0), p(0) = \pm 2\sqrt{ y(0)}, u(0) = - y(0)^2 $$