= Inverse-square refractive-index profile
{title2=$k^2(z)=k_0^2[a^2-b^2/(z-z_0)^2]$}
A <transverse electric polarization> with constant permeability and conserved transverse wavenumber $k_0\sin\theta_0$ satisfies $e''+[\beta^2-(\nu^2-1/4)/(z-z_0)^2]e=0$, with $\beta^2=k_0^2(a^2-\sin^2\theta_0)$ and $\nu^2=1/4+k_0^2b^2$. 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.
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