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I want to solve the following ODE

$$x''(z)+ \frac{\frac{d}{dz} \left(\frac{f(z)}{z^2}\right)}{\frac{f(z)}{z^2}}x'(z)+\frac{\omega^2}{(f(z))^2}x(z)=0$$

where $$f(z) = 1- 4 \left(\frac{z}{z_*}\right)^3 + 3\left(\frac{z}{z_*} \right)^4.$$

Clearly this equation has irregular singularity at $z=z_*$. I was trying to extract the 'irregular singular' part of the solution using asymptotic analysis following Bender & Orszag (see pg. 76-88). The aim is to reduce the ODE into a regular singular one. That is the solution should look like

$$x(z) = (\text{singular part}) \times v(z)$$

where $v(z)$ satisfies an ODE which has only a regular singular point at $z=z_*$ and therefore can be solved by Frobenius method.

I could extract the leading and sub-leading singular part (which I believe are correct) of the solution. Solution so far looks like, $\displaystyle x(z) = e^{-i\frac{z_*^2 \omega}{6(z-z_*)} - i \frac{2}{9}z_* \omega \log(z-z_*)} v(z)$. But the ODE satisfied by $v(z)$ still has irregular singular point at $z=z_*$.

I could not fix that next order singular piece but but intuitively that term has to be less singular than $\log(z-z_*)$ at $z-z_*$ (may be $\log(\log(z-z_*))?$)

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    $\begingroup$ Note the proper use of \text{singular part} in MathJax; the way you did it is not proper. I'm not sure if you intended a mulitple-valued logarithm with a capital L rather than lower case in "log", but if so, it should be coded as \operatorname{Log} to get proper spacing and to prevent italicization. I changed it to \log with a backslash, which also has those two effects. $\qquad$ $\endgroup$ – Michael Hardy Jul 26 '16 at 0:10
  • $\begingroup$ Thanks for the edit. I am not very frequent at this site :-) $\endgroup$ – Physics Moron Jul 26 '16 at 0:27

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