Method of characteristics for quasilinear PDE $u_x+u_y=2\sqrt{u}$ 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 $x=y$?
 A: 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:


*

*if $g(x) = (x - c_3)^2$, then we obtain an infinity of solutions under the constraint $f(0) = c_3$;

*else, no function $f$ matches the boundary condition: there is no solution.


Again, we can't express the solution uniquely in terms of $x$, $y$.
