Extended version of associated Legendre eq. Can I use the Frobenius method?

[omyojj]

Hello,

Let us assume we have a differential equation

$$( \frac{d^2}{dx^2}+ \frac{2}{1+1/n}\Theta_n^{n-1} - \nu^2 ) y = 0$$

where

$$\Theta_n = (1-\mu^2)^{1/(n+1)}$$

is a function of $$\mu$$ which is a function of $$x$$ :

$$\frac{d\mu}{dx} = \Theta_n^n = (1-\mu^2)^{n/(n+1)}$$

In terms of $$\mu$$ and $$\frac{d}{dx} = \frac{d\mu}{dx}\frac{d}{d\mu}$$, the above equation becomes

$$\left( \Theta_n^n\frac{d}{d\mu}\Theta_n^n\frac{d}{d\mu} + \frac{2}{1+1/n}\Theta_n^{n-1} - \nu^2 \right)y = 0$$

In the limit of $$n \rightarrow \infty$$, we have

$$\lim_{n\rightarrow \infty} \Theta_n^n = 1- \mu^2$$

$$\left( (1-\mu^2)\frac{d}{d\mu}(1-\mu^2)\frac{d}{d\mu} + 2(1-\mu^2) - \nu^2 \right) y = 0$$

and this is just associated Legendre equation.

$$y^{\prime\prime} - \frac{2\mu}{1-\mu^2}y^{\prime} + \left[ \frac{a(a+1)}{1-\mu^2}-\frac{\nu^2}{(1-x^2)^2}\right]y = 0$$

with a=1.

(The solutions go like this

$$P_{1}^{\pm\nu} \propto ( 1+ \mu )^{\pm \nu/2} \cdot ( 1 - \mu )^{\mp \nu/2} \cdot (\nu - \mu)$$

)

I think that solutions should show qualitatively similar behavior to that obtained from the associated Legendre's equation, But coudn't make any progress

Can I solve this equation utilizing a commonly-used technique like the Frobenius method for any rational number n > 0?

Can't I just have the form that the solutions might have? (like power series)

Thank you..

p.s. I should further note that the differential relation b/w x and mu can be directly integrated to give (x = 0 when mu = 0)

$$x = \mu {}_2 F_1 (1/2,\,\, n/(n+1);\,\,3/2;\,\,\mu^2)$$
In the limiting case (n -> ∞ )

$$x = \mu {}_2 F_1 (1/2,\,\,1;\,\,3/2;\,\,\mu^2) = \tanh^{-1}\mu$$

 

[gtoguy97]

So the differential equation can be rewritten in terms of x ( (1-\tanh^2x)\frac{d^2}{dx^2} + \frac{2}{1+1/n}\tanh x\,\Theta_n^{n-1} - \nu^2 )y = 0