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

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 parametrics 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.

Zeyd Bahadır Kırçu
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  • What was that other solution that you obtained? – Hans Lundmark Apr 11 '23 at 12:56
  • It was $u = \frac{2}{3}t^3 + 4t^2\sqrt{s} + 4st - s^2$, where $x = t^2 + 4t\sqrt{s}$ and $y = 2t + s$. – Zeyd Bahadır Kırçu Apr 11 '23 at 14:14
  • That doesn't seem quite right. Unless I've made a mistake in my calculations, that means that $u_x = t+2\sqrt{s}$ and $u_y=-2s$, and then $u_x^2+2u_y = (t^2+4t \sqrt{s} + 4s) + (-2s) = x + 2s \neq x$, so the PDE isn't satisfied. – Hans Lundmark Apr 11 '23 at 17:42
  • You've made a mistake. $u_y = -2s$, so $2u_y = -4s$ – Zeyd Bahadır Kırçu Apr 11 '23 at 19:35
  • Oops, I messed up the simplest step! You're right, of course. So far, so good, then. But your solution only works for $s>0$ (because of $\sqrt{s}$). – Hans Lundmark Apr 11 '23 at 21:35
  • And actually I think that $s>0$ is the best that you can hope for. I suppose you got your parametric solution by extending the initial conditions to $(x_0,y_0,u_0,p_0,q_0)=(0,s,-s^2,?,?)$ where the missing entries must satisfy $p_0^2+2q_0=x_0=0$ and $q_0=\partial_s u_0 = -2s$, so that $p_0^2 = 4s$. But that's impossible if $s < 0$. So there is your solution, and one more solution for $s>0$ corresponding to the negative root $p_0 = -2\sqrt{s}$, but no solution for all real $s$, and hence no solution satisfying the original condition $u(0,y)=-y^2$ for all real $y$. – Hans Lundmark Apr 11 '23 at 21:55
  • Regarding your original calculation, $q$ is constant along each characteristic curve, but you can't say that $q$ is just constant, can you? Compare to your parametric solution, where $q=-2s$ depends on $s$ (although not on $t$), so that there's a different value of $q$ on each characteristic curve. – Hans Lundmark Apr 11 '23 at 22:26
  • You're right, but what I care about is the solution that is obtained by Charpit's method, if possible. – Zeyd Bahadır Kırçu Apr 12 '23 at 12:30
  • Well, you should get the same solution. Writing $\frac{dx}{F_p} = \cdots (= dt)$ is just a way of eliminating $t$ from the ODEs $dx/dt = F_p$ (etc.) that you use when solving it in the other way. – Hans Lundmark Apr 12 '23 at 15:21
  • Well, you're completely right again! Then, could you please help me solve the equation for u in terms of x and y, not only in s and t? – Zeyd Bahadır Kırçu Apr 12 '23 at 16:47
  • Hmm... You'd have to invert the relation $(x,y) = (t^2+4t \sqrt{s},2t+s)$ to get $(t,s)$ in terms of $(x,y)$ (and then substitute it into the formula for $u$), but that doesn't seem very easy. I'm getting $(x-t^2)^2 = 16t^2 (y-2t)$, which is a quartic equation for $t$. – Hans Lundmark Apr 12 '23 at 19:21

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