Method of Characteristics for nonlinear PDE I just started working on classical methods for nonlinear PDE's and I'm kind stucked in some questions. I would appreciate some help with the following Qing Hang question, so I could use it for trying to understand better:

Exercise 2.8. In the plane, find two solutions of the initial-value problem $$
\begin{aligned}
& xu_x+yu_y+\frac12 (u_x^2+u_y^2)=u, \\
& u(x,0) = \frac12 (1-x^2) .
\end{aligned}
$$

Here I put the deduction that he does in the book, as the way to solve such kind of problem:

thanks for any help.
 A: Observing the incomplete squares in the equation, one can easily complete them to
$$
(u_x+x)^2+(u_y+y)^2=2u+x^2+y^2
$$
strongly suggesting to consider $v(x,y)=u(x,y)+\frac12(x^2+y^2)$ so that then
$$
v_x^2+v_y^2=2v, ~~v(x,0)=\frac12
$$
Now with $p=v_x$ and $q=v_y$ establish the Lagrange-Charpit equations
$$
\frac{dx}{2p}=\frac{dy}{2q}=\frac{dv}{2(p^2+q^2)}=\frac{dv}{4v}=-\frac{dp}{-2p}
=-\frac{dq}{-2q}
$$
where one reads of the constants of the characteristics as $p=cq$, $x=p+a=cq+a$, $y=q+b$. From the given equation, $v=\frac12(1+c^2)q^2$. Now every is expressed in terms of $q$. 
Insert the initial conditions $x_0, y_0=0, v_0=\frac12$, $p_0=0$. As $1=2v_0=p_0^2+q_0^2=q_0^2$ we get $q_0=\pm 1$, which implies $c=0$. Next $a=x_0-p_0=x_0$ and $b=y_0-q_0=-q_0$.
$c=0$ gives $p=0$ or $v_x=0$ everywhere, so that $v(x,y)=f(y)$. $f'(y)^2=2f(y)$, $f(0)=\frac12$ is an implicit ODE with solutions $f(y)=\frac12\max(0,y+1)^2$ or $f(y)=\frac12\max(0,1-y)^2$. 
In total,
$$
u(x,y)=\frac12\max(0,y+1)^2-\frac12(x^2+y^2)
$$
or
$$
u(x,y)=\frac12\max(0,1-y)^2-\frac12(x^2+y^2)
$$
