# Solving for the recursion relation of a second order ODE

I want to solve for the asymptotic solution of the following differential equation

$$\left(y^2+1\right) R''(y)+y\left(2-q \left(b \sqrt{y^2+1}\right)^{-q}\right) R'(y)-l (l+1) R(y)=0$$

as $$y\rightarrow \infty$$, where $$q>0$$. I did the standard way by obtaining a series solution by the Frobenius method prescription in the form

$$R(y)=\sum_{n=0}^\infty \frac{a_{n}}{y^{n+k}}$$ where $$k=l+1$$ is the indicial exponent. I had difficulty finding for a recurrence relation for the coefficients $$a_n$$ for arbitrary value of the parameter $$q$$. Right now, I am just doing the brute force method of solving individual $$a_n$$ for every value of $$q$$. But I am just wondering whether the recurrence relation is possible to solve. Can this be solved analytically?

• It would help if you provide the recurrence relation you obtained – Yuriy S Aug 4 at 16:10
• I think you are supposed to use $y^2+1 \to y^2$ as $y \to \infty$, which should make for a nice explicit recurrence relation once you substitute the Frobenius series – Yuriy S Aug 4 at 16:43
• I havent solved yet the recurrence relation. Can it be solved? – user583893 Aug 5 at 0:12
• My problem is how to get rid with $q$. I dont know what to do – user583893 Aug 5 at 2:48
• Is $q$ an integer? If it's an integer, then I know what to do. If it's not an integer, the problem gets more complicated – Yuriy S Aug 5 at 8:15

First, we take the most complicated term and expand it in a series around $$y \to \infty$$:

$$\left(\sqrt{y^2+1}\right)^{-q}= \frac{1}{y^q} {_2 F_1} \left(\frac{q}{2},1;1;-\frac{1}{y^2} \right)= \sum_{k=0}^\infty \frac{(-1)^k (q/2)_k}{k!} \frac{1}{y^{2k+q}}$$

Where $$(p)_k=p(p+1)(p+2) \cdots (p+k-1)$$ is the Pochhammer symbol, and I'm using hypergeometric series instead of binomial series because it's more convenient.

Then we divide the equation by $$y^2$$ and obtain:

$$\left(1+\frac{1}{y^2}\right) R''(y)+\left(\frac{2}{y}-\frac{q}{b^q} \sum_{k=0}^\infty \frac{(-1)^k (q/2)_k}{k!} \frac{1}{y^{2k+q+1}} \right) R'(y)-\frac{l (l+1)}{y^2} R(y)=0$$

Now substitute:

$$R(y)=\sum_{n=0}^\infty \frac{a_{n}}{y^{n+\alpha}} \\ R'(y)=-\sum_{n=0}^\infty \frac{(n+\alpha) a_{n}}{y^{n+\alpha+1}} \\ R''(y)=\sum_{n=0}^\infty \frac{(n+\alpha)(n+\alpha+1) a_{n}}{y^{n+\alpha+2}}$$

We obtain:

$$\sum_{n=0}^\infty \frac{(n+\alpha)(n+\alpha+1) a_{n}}{y^{n+\alpha+2}}+\sum_{n=0}^\infty \frac{(n+\alpha)(n+\alpha+1) a_{n}}{y^{n+\alpha+4}}-2\sum_{n=0}^\infty \frac{(n+\alpha) a_{n}}{y^{n+\alpha+2}}+ \\ + \frac{q}{b^q} \sum_{n=0}^\infty \sum_{k=0}^\infty \frac{(-1)^k (q/2)_k}{k!} \frac{(n+\alpha) a_{n}}{y^{2k+n+q+\alpha+2}}-l (l+1) \sum_{n=0}^\infty \frac{a_{n}}{y^{n+\alpha+2}}=0$$

Where $$\alpha$$ is an unknown exponent.

Multiplying by $$y^{\alpha+2}$$, we have:

$$\sum_{n=0}^\infty \frac{(n+\alpha)(n+\alpha+1) a_{n}}{y^{n}}+\sum_{n=0}^\infty \frac{(n+\alpha)(n+\alpha+1) a_{n}}{y^{n+2}}-2\sum_{n=0}^\infty \frac{(n+\alpha) a_{n}}{y^{n}}+ \\ + \frac{q}{b^q} \sum_{n=0}^\infty \sum_{k=0}^\infty \frac{(-1)^k (q/2)_k}{k!} \frac{(n+\alpha) a_{n}}{y^{2k+n+q}}-l (l+1) \sum_{n=0}^\infty \frac{a_{n}}{y^{n}}=0$$

Let's assume $$q$$ is a positive integer, so $$q=1,2,3,\ldots$$. Then we can collect the powers of $$y$$.

$$\sum_{n=0}^\infty \frac{(n+\alpha)(n+\alpha+1) a_{n}}{y^n}+\sum_{n=2}^\infty \frac{(n+\alpha-2)(n+\alpha-1) a_{n-2}}{y^n}-2\sum_{n=0}^\infty \frac{(n+\alpha) a_{n}}{y^n}+ \\ + \frac{q}{b^q} \sum_{k=0}^\infty \sum_{n=2k+q}^\infty \frac{(-1)^k (q/2)_k}{k!} \frac{(n-2k-q+\alpha) a_{n-2k-q}}{y^n}-l (l+1) \sum_{n=0}^\infty \frac{a_{n}}{y^n}=0$$

Since $$q>0$$, it's clear that the indical equation reads:

$$\alpha(\alpha+1)-2\alpha-l(l+1)=0$$

$$\alpha(\alpha-1)=l(l+1)$$

So we have:

$$\alpha_{1,2}= \frac{1}{2} \pm \sqrt{l(l+1)+\frac{1}{4}}$$

The recurrence relation in general should probably look like this (check for a few values of $$q$$ and $$n$$ though, because I may have made a mistake in collecting the terms):

$$\left((n+\alpha-1)(n+\alpha)-\alpha(\alpha-1) \right) a_{n}+(n+\alpha-2)(n+\alpha-1) a_{n-2} + \\ + \frac{q}{b^q} \sum_{k=0}^{\lfloor \frac{n+q}{2} \rfloor} \frac{(-1)^k (q/2)_k}{k!} (n-2k-q+\alpha) a_{n-2k-q} =0$$

Where we obviously take $$a_{-1}=a_{-2}=a_{-3}= \dots=0$$.

I have replaced $$l(l+1)$$ by $$\alpha(\alpha-1)$$, which makes the expression work for the both roots of the indical equation.

For $$q$$ not an integer, I'm not sure what to do.

• Thanks Yuriy. I think i need to check this out. I'll get back to you. – user583893 Aug 5 at 14:01
• Hi @Yuriy S. How do I make sense with a recurrence relation containing three coefficients? – user583893 Aug 11 at 11:44