Charpit method for $u_x u_y - u = 0$

Question

Let $x = x(s), y= y(s), u= u(s), s \in \mathbb{R}$ be the characteristic curve of the PDE $$\bigg(\frac{\partial u}{\partial x}\bigg)\bigg(\frac{\partial u}{\partial y}\bigg) - u = 0 \tag{1}$$ passing through a given curve $x = 0, y = \tau, u = \tau^2, \tau \in \mathbb{R}$. Then the characteristic are given by

1. $x = 3\tau(e^s-1), y = \dfrac{\tau}{2}(e^{-s}+1), u = \tau^2e^{-2s}$

2. $x = 2\tau(e^s-1), y = {\tau}(2e^{2s}-1), u = \dfrac{\tau^2}{2}(1+e^{-2s})$

3. $x = 2\tau(e^s-1), y = \dfrac{\tau}{2}(e^{s}+1), u = \tau^2e^{2s}$

4. $x = \tau(e^{-s}-1), y = {-2\tau}\bigg(e^{-s}-\dfrac{3}{2}\bigg), u = \tau^2(2e^{-2s}-1)$

My attempt:- Applying Charpit equation for $$f(x,y,u,p,q)=pq-u$$ we have

$$\frac{dx}{q}=\frac{dy}{p}=\frac{du}{2pq}=\frac{dp}{p}=\frac{dq}{q} \tag{2}$$

Considering, $$\frac{dp}{p}=\frac{dq}{q}\implies p=aq$$

applying in (1), we get $aq^2=u \implies q=\sqrt{u/a}.$So, $p=\sqrt{au}$

(2) gives $pdx+qdy=du \implies ax+y=\sqrt{au}+b$

How do I get the characteristic curves? Can you prescribe some textbook for this?

Answer 1

Write $F(x,y,p,q,u) = pq - u \equiv 0$ and parametrise the solution as $(x,y,u) = (x(s,\tau), y(s,\tau), y(s,\tau))$, where $\tau$ corresponds to the parametrisation of the initial data, so that $(x(0,\tau),y(0,\tau),u(0,\tau))=(0,\tau, \tau^2)$. With overdot denoting differentiation by $s$, we have the following five equations by Charpit: \begin{equation} \dot{x} = F_p =q, \\ \dot{y} = F_q =p, \\ \dot{p} = -F_x-pF_u =p, \\ \dot{q} = -F_y-qF_u =q, \\ \dot{u} = pF_p + qF_q =2pq = 2u. \end{equation} Clearly the last three equations are the only ones which we can directly integrate (for now), to obtain $u = \tau^2 e^{2s}$, $p = p_0(\tau)e^s$, $q = q_0(\tau)e^s$. To determine $p_0,q_0$ we use the conditions from the data: \begin{equation} p_0(\tau)q_0(\tau) =u_0(\tau) = \tau^2\\ u_0'(\tau) = x_0'(\tau)p_0(\tau) + y_0'(\tau)q_0(\tau)= q_0(\tau). \end{equation} Thus $p_0 = \tau/2$, $q_0 = 2 \tau$. Returning to the first two equations from Charpit's method and utilising the data, we obtain: $x = 2\tau(e^s-1)$, $y = \frac{\tau}{2} \left( e^s +1 \right)$. So the correct answer is the third option.

A great source for these kinds of problems would be "Applied Partial Differential Equations" by Ockendon et al.