Solve the First Order PDE equation 
Problem: For equation: 
  $$u=xu_x+yu_y+\frac{1}{2}(u_x^2+u_y^2) $$
   find the solution with 
  $$u(x,0)=\frac{1}{2}(1-x^2)$$

Here is what I have so far:
Let $$F(x,y,z,p,q)=z-xp-yq-\frac{1}{2}(p^2+q^2)$$ with $$p=u_x,\\ q=u_y$$
Characteristic Equations:
$$\begin{eqnarray}
\frac{dx}{dt}&=& -x-p \\ 
\frac{dy}{dt}&=&-y-q \\ 
\frac{dz}{dt}&=&-px-p^2-qy-q^2 \\
\frac{dp}{dt}&=&0 \\
\frac{dq}{dt}&=&0
\end{eqnarray}$$
I know that I need to get Characteristic strips( Solutions of Characteristic Equations) but I am stuck.
Please help. Thanks
 A: By inspection one find two solutions :
$$u(x,y)=\frac12(1-x^2)+y$$
$$u(x,y)=\frac12(1-x^2)-y$$
Both satisfy the PDE and the condition.
Of course this oversimplified approach doesn't say if they are more solutions or not.
$\underline {\text{HINT to solve with the characteristics method}}$ .
A preliminary transformation in order to simplify the job :
Change of function : $\quad u(x,y)=\frac12 (v^2-x^2-y^2)\quad $ with $v=v(x,y)$ .
$u_x=vv_x-x\quad$ and $\quad u_y=vv_y-y$
$x(vv_x-x)+y(vv_y-y)+\frac12(vv_x-x)^2+\frac12(vv_y-y)^2=\frac12(v^2-x^2-y^2)$
After simplification :
$$\boxed{(v_x)^2+(v_y)^2=1}$$
With condition $v(x,0)=2u(x,0)+x^2+0^2=(1-x^2)+x^2=1$
$$\boxed{v(x,0)=1}$$
I suppose that you can take it from here.
A: From last two equations: $$p=a, ~~q=b$$ where $a$, $b$ are arbitrary constants. First three equations can be written as $$dz=pdx+q dy=adx+bdy.$$ Integration gives$$z=ax+by+c$$ You can find $c={a^2+b^2\over 2}$ 
This is a special case in Charpit's method and the equation is called as Clairaut's equation.
