The differential equation for a optical fiber with a refractive index $n(r)$ is given as $$\nabla^{2}_{\perp}A(r,\theta)+(k^{2}n(r)^2-\beta^2)A(r,\theta)=0.$$ which is separable in cylindrical coordinates, $A(r,\theta)=R(r)\Theta(\theta)$. The radial DE follows as $$ r^{2}R''+rR'+[r^{2}(k^{2}n(r)^{2}-\beta^{2})-m^{2}]R=0. $$ For a step function index ($n_{0}>n_{1}$, $\Delta \equiv n_{0}-n_{1}$), $$n(r)=\cases{n_{0} : r\leq a\\n_{1}:r>a} \implies R(r)=\cases{c_{1}J_{m}(r\sqrt{k^{2}n_{0}-\beta^{2}}) : r\leq a\\c_{2}K_{m}(r\sqrt{\beta^{2}-k^{2}n_{1}}):r>a}$$ the radial part becomes a Bessel function. In a similar way, if $n(r)$ is quadratic one obtains a Gaussian with a Laguerre profile (is very similar to the Bessel's).
$\textbf{My question is:}$ If one has a index function such as $n(r)=n_{1}+\Delta f(\alpha r)$ where $\alpha$ is a parameter of how smooth the transition is. By applying the approximation $n(r)^2\approx n_{1}^2-2\Delta n_{1}f(\alpha r)$ and setting $f(\alpha r)=(1-\tanh^2(\alpha r /a))$ one obtain the radial equation, $$ r^{2}R''+rR'+[r^{2}(k^{2}n_{1}^{2}-\beta^{2}+2\Delta n_{1}k^{2}(1-\tanh^{2}(\alpha r /a)))-m^{2}]R=0. $$ For simplicity I rewrite it as $$ r^{2}R''+rR'+[r^{2}(A-B\tanh^{2}(\xi r))-m^{2}]R=0, $$ where $A=k^{2}n_{0}^{2}-\beta^{2}+2\Delta n_{1}k^{2}$, $B=2\Delta n_{1}k^{2}$, $\xi=\alpha/a$.
$\textbf{Bonus:}$ I also tried with other index fuctions $$ f(r)=\left[ \tanh(\alpha (a-r)/a)+1 \right]/2 $$ which returns $$ r^2R''+rR'+[r^2(A+B\tanh(\xi (a-r)))-m^2]R=0 $$ where $A=k^{2}(n_{1}^2+\Delta)-\beta^2$, $ B=n_{1}k^{2}\Delta$, $\xi=\alpha /a$, but with no success.