# Solve $u_{x_1}u_{x_2} = u$ with $u(0,x_2)=x_2^2$

I'm solving the non-linear IVP $$u_{x_1}u_{x_2} = u, \qquad u(0,x_2)=x_2^2$$

The general PDE has the form $$F(p,q,z)=pq-z=0$$. The characteristic differential equations are \begin{align*} \frac{dx_1}{dt} & = F_p = q \\ \frac{dx_2}{dt} & =F_q = p \\ \frac{dz}{dt} & = pF_p + qF_q = pq + qp = 2pq \\ \frac{dp}{dt} & = -F_x - pF_z = -p(-1) = p \\ \frac{dq}{dt} & = -F_y - qF_z = -q(-1) = q. \end{align*} On the initial curve $$\Gamma$$, which corresponds to the Cauchy-data for this problem, the initial values $$x_1 = f(s) = 0, \enspace x_2 = g(s) = s, \enspace z = h(s) = s^2$$ hold. Furthermore we have by the Chain Rule that \begin{align*} z' & = u_xf'(s) + u_yg'(s) \\ 2s & =\phi(s)\cdot 0 + \psi(s)\cdot 1, \end{align*} or that $$\psi(s)=2s$$, and this is the initial data for $$q$$. We can use this to find $$\phi(s)$$ as per $$pq - z = \phi(s)\cdot 2s - s^2 = 0,$$ so that $$\phi(s)=\frac{s}{2}$$. Next I begin to actually solve the system. I find that $$p=e^t\frac{s}{2},\enspace q=e^t2s,\enspace z=2pqt+s^2$$

and moreover that $$x_1=qt, \enspace x_2 = pt + s$$

Here is where I can't make any more progress. What remains to be done is to somehow solve $$t$$ and $$s$$ in terms of $$x_1$$ and $$x_2$$. How do I finish this one?

You need to substitute $$p$$ and $$q$$ into the DEs for $$x_1, x_2, z$$ before solving it: \begin{alignat*}{2} \frac{dx_1}{dt} & = q = 2se^t && \implies X(s, t) = 2se^t - 2s = 2s(e^t - 1) \\ \frac{dx_2}{dt} & = p = \frac{s}{2}e^t && \implies Y(s, t) = \frac{s}{2}e^t + \frac{s}{2} = \frac{s}{2}(e^t + 1) \\ \frac{dz}{dt} & = 2pg = 2s^2e^{2t} && \implies Z(s, t) = s^2e^{2t}, \end{alignat*} where the constants of integration are found with $$X(s, 0) = 0$$, $$Y(s, 0) = s$$, and $$Z(s, 0) = s^2$$. The last step is to solve the equations for $$X$$ and $$Y$$ in terms of $$s$$ and $$t$$ and then substitute into $$Z$$, which I will let you do it yourself.
• Is your equation $dz/dt$ correct? I think it should be $2s^2e^{2t}$. Jan 8, 2022 at 10:12