So there is an example problem in the textbook and I really just don't understand what's going on, both mathematically and conceptually. The problem is solve $z_x + 2zz_y = 1$ with boundary conditions $z = 0$ on $y=0$ for $x \ge 0$ and $z=y$ on $x=0$ for $y \ge 0$.
I write the characteristic problem as $x_s = 1$, $y_s = 2z$, and $z_s = 1$. With the first boundary condition, we can write the line $\Gamma(\sigma) = (\sigma, 0, 0)$ therefore we get that $x = s + \sigma$, $y = s^2$, and $z = s$. Therefore we get characteristic curves $y = (x-\sigma)^2$. Then the textbook implies that the solution is only specified below the parabola $y = x^2$. I don't understand this. Also, given these characteristics, there will be infinitely many characteristic intersections below the parabola $y = x^2$, so this makes me think the solution is not specified at those points either.
Can anyone reconcile this or give me some guidance on what's going on here?
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$
$\dfrac{dz}{dt}=1$ , letting $z(0)=z_0$ , we have $z=z_0+t=z_0+x$
$\dfrac{dy}{dt}=2z=2z_0+2t$ , letting $y(0)=f(z_0)$ , we have $y=f(z_0)+2z_0t+t^2=f(z-x)+2(z-x)x+x^2=f(z-x)-x^2+2xz$ , i.e. $z=x+F(x^2-2xz+y)$
$z(0,y)=y$ :
$F(y)=y$
$\therefore z=x+x^2-2xz+y$
$2xz+z=x^2+x+y$
$(2x+1)z=x^2+x+y$
$z(x,y)=\dfrac{x^2+x+y}{2x+1}$
But this solution does not satisfy $z(x,0)=0$ ,
$\therefore$ the concept of piecewise solution should be introduced