 Inverse-square refractive-index profile

A transverse electric polarization with constant permeability and conserved transverse wavenumber $k_0\sin {\theta }_0$ satisfies $e^{\prime \prime }+[{\beta }^2-({\nu }^2-1/4)/(z-z_0{)}^2]e=0$, with ${\beta }^2=k_0^2(a^2-{\sin }^2{\theta }_0)$ and ${\nu }^2=1/4+k_0^2{b}^2$. Substitution $e=\sqrt{|z-z_0|}{Z}_{\nu }(\beta |z-z_0|)$ reduces it to the Bessel equation. The profile is singular at $z_0$, so each side requires its own boundary conditions.