$u_x u_y=1$ with initial data $u=0$ and $x+y=1$
Parametrise with $x=s$ and $y=1-s$
diffeq can be written as $F(p,q) = pq - 1 = 0$
Charpits equations give: $dx/dt = q$, $dy/dt = p$, $dp/dt = 0$, $dq/dt = 0$, $du/dt = 2pq$
Solving these we get $p=p_0(s)$ and $q=q_0(s)$ and $x = q_0(s)t + x_0 = q_0(s)t + s $ and $y = p_0(s)t + y_0 = p_0(s)t + 1-s $ and $u = 2p_0(s)q_0(s)t$
$F(p_0,q_0) = 0 $ implies that $p_0q_0 = 1$ and $du_0(s)/ds = p_0(dx_0/ds)+ q_0(dy_0/ds)$ implies $p_0 = q_0$
As a result $p_0 = q_0 = -1$ or $1 $ and so $u=2t$
How do i then go about stating where each solution exists and further proving that for a real solution to exist, $[du_0(s) / ds ]^2 \geq 4 [dx_0(s)/ds][dy_0(s)/ds]$
For general initial conditions $z_0(s)=u(x_0(s),y_0(s))$ you get the system of equations $$ p_0(s)q_0(s)=1\\ p_0(s)\dot x_0(s)+q_0(s)\dot y_0(s)=\dot z_0(s) $$ which looks almost like the Viete's equations for a quadratic equation. Multiply the first equation with $\dot x_0\dot y_0$ to get that $p_0(s)\dot x_0(s)$ and $q_0(s)\dot y_0(s)$ are the root pair of $$ r^2-\dot z_0\cdot r + \dot x_0\dot y_0=0. $$ Now check the discriminant for a condition to get real solutions.