I am asked to solve the boundary value problem on a PDE about $u:\mathbb{R}_{\ge 0}^2\cap \{(x,y): x^2+y^2\ge 1\}\to\mathbb{R}$:$$\begin{cases}u_x^2+u_y^2=1 & x^2+y^2>1,\, x,y>0\\u(\cos s,\sin s)=s & s\in (0,\frac{\pi}{2})\end{cases}$$
I followed the way of solving characteristic ODE as is described on Evans's textbook.But I got the absurd result that $u(x,y)=\sqrt{x^2+y^2}+\arctan(\frac yx)-1$, which does not satisfy $u_x^2+u_y^2=1$.I do not know what is wrong.
My solution: Set $F(p_x,p_y,z,x,y)=p_x^2+p_y^2-1$. Then the characteristic ODE is\begin{align*}(p_x'(s),p_y'(s))&=-D_xF-D_zF\cdot \vec{p}=0.\\z'(s)&=D_pF\cdot\vec{p}=2(p_x^2+p_y^2)=2.\\(x'(s),y'(s))&=D_pF=(2p_x,2p_y).\end{align*}The first equation says $p_x,p_y$ are constant. Therefore we may set (since $p_x^2+p_y^2=1$)$$ p_x(s)\equiv \cos\theta, \, p_y(s)\equiv \sin\theta\, (\forall s). $$Then by the third equation we have $x(s)=2s\cos\theta$, $y(s)=2s\sin\theta$.
Finally, since $z'(s)=2$ we have $u(x(s),y(s))=z(s)=2s+z^0$.By letting $s=\frac 12$ we have $1+z^0=u(\cos \theta,\sin\theta)=\theta$.Hence $z^0=\theta-1=\arctan(\frac yx)-1$.Thus\begin{multline*}u(x,y)= u(2s\cos\theta,2s\sin\theta)=\\2s+z^0=2s+ \arctan(\frac yx)-1=\sqrt{x^2+y^2}+\arctan(\frac yx)-1.\end{multline*}But $u(x,y)=\sqrt{x^2+y^2}+\arctan(\frac yx)-1$ does not satisfy $u_x^2+u_y^2=1$. What is wrong with my solution?