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?

## Best Answer

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.