I'm having trouble with solving the quasilinear PDE
$$\begin{cases} u_x+u_y&=2\sqrt{u}, \\ u(x,x)&=g(x) \end{cases}$$
via method of characteristics as in this paper.
My attempt:
First I started by formulating the ODEs
$$\begin{cases} \dot{x}&=1 \\ \dot{y}&=1 \\ \dot{z}&=2\sqrt{z}. \end{cases}$$
Solving those and applying the initial conditions gives me
$$\begin{cases} x&=t+x_0 \\ y&=t+x_0 \\ z&=\frac{1}{4}(2t+z_0)^2=\frac{1}{4}(2t+2\sqrt{g(x_0)})^2. \end{cases}$$
But how do I eliminate $t$ and $x_0$ if $x=y$?
Best Answer
The system $$\left\lbrace \begin{aligned} t+x_0 = x\\ t+x_0=y \end{aligned}\right. $$ has no solution $(t, x_0)$ if $x≠y$, and it has infinitely many solutions if $x=y$. Therefore, we can't express the solution uniquely in terms of $x$, $y$. Situation is somewhat similar to this post and related ones.
The Lagrange-Charpit equations $$ \frac{dx}{1} = \frac{dy}{1} = \frac{du}{2\sqrt u} $$ provides the characteristic families $x-y = c_1$ and $x - \sqrt{u} = c_2$. The second characteristic family might be rewritten $u = (x - c_2)^2$). General solutions read $$ u = \big(x - f(x-y)\big)^2 $$ where $c_2 = f(c_1)$ involves an arbitrary function $f$. If $x=y$, we find $u = (x - f(0))^2$ which needs to equal $g(x)$ according to the boundary condition. Two cases arise:
Again, we can't express the solution uniquely in terms of $x$, $y$.