Question
While solving a non-linear partial differential equation, i am stuck to a problem
$$z^2(x^2p^2+q^2)=1$$
And let $$F=z^2(x^2p^2+q^2)-1$$
Where $p=\frac{\partial z}{\partial x}$
And $q= \frac{\partial z}{\partial y}$
my approach
From Charpit method
$$\frac{dx}{F_p}=\frac{dy}{F_q}=\frac{dz}{pF_p+qF_q}= \frac{-dp}{F_x+pF_z}=\frac{-dq}{F_y+qF_q}$$
$$\frac{dx}{2px^2z^2}=\frac{dy}{2qz^2}=\frac{dz}{2p^2x^2z^2+2q^2z^2}=\frac{-dp}{2xp^2z^2+2pz(x^2p^2+q^2)}=\frac{-dq}{2zq(x^2p^2+q^2)}$$
The solution of the pde is given by
$$dz=pdx+qdy$$
My problem is, i am unable to seperate the functions for further solution.
Any subtle hints are highly apperciated.
Best Answer
$$z^2x^2\left(\frac{\partial z}{\partial x} \right)^2+z^2\left(\frac{\partial z}{\partial y}\right)^2=1$$
Below, this is not a direct answer to the question, but an hint, too long to be edited in the comments section.
The change of variables :$\quad\begin{cases} Z=\frac12 z^2\\ X=\frac12 x^2\\ Y=y \end{cases}\quad$ transforms the PDE into : $$\left(\frac{\partial Z}{\partial X} \right)^2+\left(\frac{\partial Z}{\partial Y}\right)^2=1$$ In the littérature one can find examples of solving the Eikonal equation with various boundary conditions :
Example 11, p.21 in http://www.ehu.eus/luis.escauriaza/apuntes_problemas_y_examene/method-of-characteristics.pdf
Example 2.26, p.21 in https://www.math.ualberta.ca/~xinweiyu/436.A1.12f/PDE_Meth_Characteristics.pdf
And on Math.StackExchange : Finding a complete integral solution for the Eikonal equation $u_x^2+u_y^2=1$