# 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 $$x=y$$?

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 $$x$$, $$y$$.
• Thanks! I got to that but computed $dz/dx=2\sqrt{z}$, which led me to the solution $u(x,y)=\frac{1}{4}(2x+f(y-x))^2$, with the condition $f(0)=2(\sqrt{g(x)}-x)$. Jun 2, 2020 at 5:55