Solving fully nonlinear 1st order PDE I am new to PDE and I am solving the following:
$$u_x^2+u_y^2=u^2$$
with the curve
$$\Gamma:\cos(s),\sin(s),1.$$
This is what I got so far. First, I set my PDE into the general first-order PDE:
$$F(x,y,z,p,q)=p^2+q^2-z^2=0.$$
Then I obtained the following:
$$\dfrac{dx}{dt}=F_p=2p$$
$$\dfrac{dy}{dt}=F_q=2q$$
$$\dfrac{dz}{dt}=pF_p+qF_q=2z^2$$
$$\dfrac{dp}{dt}=-F_x-pF_z=2pz$$
$$\dfrac{dq}{dt}=-F_y-qF_q=2qz.$$
By my initial curve $\Gamma,$ I need to find functions
$$p=\phi(s)\text{ and }q=\psi(s)$$
such that
$$F(f(s),g(s),h(s),\phi(s),\psi(s))=0$$
$$h'(s)=f'(s)\phi(s)+g'(s)\psi(s).$$
Using this relationship, I obtained the following functions:
$$\phi(s)=\pm\cos(s)$$
$$\psi(s)=\pm\sin(s)$$
where the signs for my functions have to be the same. First, I am checking when both $\phi(s)$ and $\psi(s)$ have positive sign. I obtained the following solutions for my functions when I solved by system of ODEs above:
$$x(s,t)=(2te^{\frac{2t}{1-2t}}+1)\cos(s)$$
$$y(s,t)=(2te^{\frac{2t}{1-2t}}+1)\sin(s)$$
$$z(s,t)=\dfrac{1}{1-2t}$$
$$p(s,t)=\cos(s)e^{\frac{2t}{1-2t}}$$
$$q(s,t)=\sin(s)e^{\frac{2t}{1-2t}}$$
I know that the next step is to solve for $s$ and $t$. Before I do this step, am I on the right track? Did I make any errors, and if so how can I make sure I avoid them when I solve first-order fully nonlinear PDEs in the future?
 A: To avoid the need of algebraic manipulation with $s,t$, you can use the folowing method. We have:
$$\frac{-dp}{2pz} = \frac{-dq}{2qz} \implies \frac{dp}{p} = \frac{dq}{q} \implies p = Cq$$
Now plugging into the equation we have that $C^2q^2 + q^2 = z^2 \implies q = \frac{z}{\sqrt{C^2+1}}$ and $p = \frac{Cz}{\sqrt{C^2+1}}$. Then plugging into $dz = pdx + qdy$ and integrating we have:
$$dz = \frac{Cz}{\sqrt{C^2+1}}dx + \frac{z}{\sqrt{C^2+1}}dy$$
$$\frac{dz}{z} = \frac{C}{\sqrt{C^2+1}}dx + \frac{1}{\sqrt{C^2+1}}dy$$
$$\ln|z| = \frac{Cx + y}{\sqrt{C^2+1}} + D$$
$$z = Ae^{\frac{Cx + y}{\sqrt{C^2+1}}}$$
This is the complete integral of the PDE.
Now for the particualr solution we use:
$$\frac{d}{dt}(1) = u_x\frac{dx}{dt} + u_y\frac{dy}{dt}$$
$$0 = C(-\sin t) + \cos t \implies C(t) = \cot t$$
Plugging this with the initial condition in the complete integral after some manipulation we have that
$$1 = A(t)e \implies A(t) = e^{-1}$$
Finally we get $u(x,y) = e^{\cos(t)x + \sin(t)y -1}$, where $t \in [0,2\pi]$. This is a family of solution that passes through the curve. To find a single solution that passes through the curve we need to find the envelope of the family of solutions given by $\phi(x,y,u,A(t),C(t)) = 0$. This happens when $\frac{d\phi}{dt} = 0$ and we get:
$$0 = \frac{d\phi}{dt} = \phi_A \frac{dA}{dt} + \phi_C \frac{dC}{dt} = \frac{x - Cy}{(1+C^2)^{\frac 32}} \implies x = yC$$
Plugging into the family of soltions we get the wanted solution:
$$\boxed{u(x,y) = e^{\sqrt{x^2+y^2} - 1}}$$
REMARK: From the solution we can conclude that if we write it in polar coordinates it will be independent of $\theta$. This prompts you to believe that if we introduce change of variables at the beginning then we would have an easier job. As an exercises maybe can check this.
A: Just as Stefan4024 remarked, we can find the radial solution without  resort to Characteristics method, by setting $u(x,y)=u(r), $ where $r=\sqrt{x^2+y^2}.$ Then the boundary problem is equivalent to
\begin{gather*}
\begin{cases}
u'(r)=\pm u(r),\\
u(1)=1.
\end{cases}
\end{gather*}
Solve it, and we have $u(x,y)=u(r)=e^{\pm(r-1)}=e^{\pm(\sqrt{x^2+y^2}-1)}.$
A: This problem has nice radial symmetry. We first work out
$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}=\frac{\partial u}{\partial r}\frac{x}{r}$$
So our PDE is
$$\left(\frac{\partial u}{\partial r}\frac{x}{r}\right)^2+\left(\frac{\partial u}{\partial r}\frac{y}{r}\right)^2=u^2$$
$$\frac{1}{r^2}\left(\frac{\partial u}{\partial r}\right)^2(x^2+y^2)=u^2$$
Since $r=\sqrt{x^2+y^2}$,
$$\frac{\partial u}{\partial r}=u$$
$$u(r)=Ae^r \text{ or equivalently }u(x,y)=Ae^{\sqrt{x^2+y^2}}.$$
Boundary conditions follow.
