Solve the eikonal equation with radial data

Question

Consider the equation $$ \left(\frac{\partial f}{\partial r}\right)^{2} + \frac{1}{r^{2}}\left(\frac{\partial f}{\partial \varphi}\right)^{2} = g(r). $$ How to solve it? Here the left hand side is $|\nabla f|^2$ in polar coordinates.

After separating the variables, I get $$\sqrt{g(r)r^{2} - r^{2} (\partial_{r} f)^{2} } = \partial_{\varphi} f$$

This looks like an eikonal equation and has a very general solution. Do you have a context and/or a set of boundary conditions that could help guide a solution? — Ron Gordon, Feb 23 '13 at 13:51
No, I don't. I only have an equation in cartesian coordinate system, which I rewrite into an equation in a Polar system. — John Taylor, Feb 23 '13 at 13:55
Oh, thank you. After separation I can get
$$ \sqrt{g(r)r^{2} - r^{2} (\partial_{r} f)^{2} } = \partial_{\varphi} f. $$ So, the left and the right side is equal to some constant a. — John Taylor, Feb 24 '13 at 19:17

score 1 · Answer 1 · answered Mar 04 '16 at 23:53

An issue with separation of variables is that with a nonlinear PDE, there is no way to construct the general solution by summing separated solutions. But if the goal is to find just one solution, then looking for a separated one is a good idea. If both sides in $$\sqrt{g(r)r^{2} - r^{2} (\partial_{r} f)^{2} } = \partial_{\varphi} f$$ are constant, then the constant has to be zero because a nonzero constant value of $\partial_{\varphi} f$ would make $f$ multi-valued. From here $f$ can be found...

But it would be simpler to just assume from the outset that $f$ is radial (looking for a radial solution with radial data is natural). So $f=f(r)$, and the equation is simply $f'(r)=\sqrt{g(r)}$.

Solve the eikonal equation with radial data

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