Finding the uniquely determined region of a PDE (a) Solve the equation
$yu_x+xu_y=0$
 with the condition $u(0,y) =
e^{-y^2}$.
(b) In which region of the xy plane is the solution uniquely determined?   
I did the first part but I don't understand how to do part b.  I read this.
But I don't understand how the argument is based. A more clear explanation is appreciated
 A: This is Exercise 1.2.7 in Partial Differential Equations by Strauss.  
(a) The slope of characteristics is $x/y$. Solve 
$$
\frac{dy}{dx}=\frac{x}{y} \tag{1}
$$
by separating variables: $y\,dy=x\,dx$, hence $y^2/2=x^2/2+C$. Isolate $C$ here: 
$$y^2-x^2=C \tag{2}$$
The  general solution is $u(x,y)=f(y^2-x^2)$, an arbitrary function of the left side of (2). Plugging in the given condition $u(0,y) =
e^{-y^2}$, we get $$f(y^2)=e^{-y^2}\tag{3}$$ I prefer to use a new letter, such as $z$, for the argument of $f$. The function $f(z)=e^{-z}$ satisfies (3) and gives 
$$
{u(x,y)= e^{x^2-y^2}} \tag{4}
$$
(b) Looking at (3) more carefully, we see that it only involves the values of $f$ on $[0, \infty)$. Therefore, it is  satisfied by any function $f$ such that $f(z)=e^{-z}$ for $z\ge 0$. The values of $f$ for negative $z$ can be arbitrary. It follows that  the solution $u(x,y)=f(y^2-x^2)$ is uniquely determined in the region $y^2-x^2\ge 0$ (unpainted on the sketch below) but can be any function of the form $f(y^2-x^2)$ in the region $y^2-x^2  < 0$ (painted).

A: Another detailed explanation can be found in Exercise 4 here and a simple one in 7b) here.
