The differential equation I am studying looks like this:
$$r(r-1)\psi_{,rr}+\psi_{,r}+\left[ \dfrac{w^2r^3}{r-1} -l(l+1)+\dfrac{\epsilon}{r} \right] \psi=0$$
I proceed like this: first I assume that the asymptotic behavoir is $e^{-S(r)}$ and susbstitute this into the differential equation, I obtain:
$$-l(l+1)+\dfrac{3}{r}+\dfrac{r^3w^2}{r-1} + S^\prime + (r-1)r (S^\prime)^2+r(r-1)S^{\prime\prime}=0$$
I assume that $S^{\prime\prime}\ll(S^\prime)^2$, so I can solve the above equation for $S^\prime$; I obtain a complicated result which simplifies, for large $r$, to $$S(r)=\pm i w r +C(r).$$ Now I substitute this result back into my differential equation in order to fix $C$; I get:
$$i w+\left( -l(l+1)+\dfrac{3}{r} + \dfrac{w^2r^3}{r-1} \right) (i r w +C) + C^\prime+r(r-1)C^{\prime\prime}=0$$
This is where I am stuck; the result should be $\psi\approx r^{\pm i w}e^{\pm i w r}$, so $C$ should be $\pm i w ln(r)$, but I do not understand how this can result from the last equation.
Can anyone please help?
You need to correctly insert your first approximation into the equation. As $S'=\pm iw+C'$ and $S''=C''$, you should get $$ 0=-l(l+1)+\dfrac{ϵ}{r}+\dfrac{r^3w^2}{r-1} + [\pm iw+C'] + (r-1)r ([\pm iw+C'])^2+r(r-1)C'' \\ =-l(l+1)+\dfrac{ϵ}{r}+\dfrac{(2r^2-r)w^2}{r-1} + [\pm iw+C'] + (r-1)r (\pm2iwC'+C'^2)+r(r-1)C'' $$ With $C'\ll 1$ and thus $C'^2\ll C'$ and still assuming $C''\ll C'$ for $r\to\infty$, balancing the leading terms give the equation $$ 2rw^2\pm2iwr^2C'=0\implies C'(r)=\pm\frac{iw}r $$ so that indeed $C(r)=\pm iw\ln(r)+D$.