A Recurrence relations for series solution of differential equation

dim_d00m
Messages
1
Reaction score
0
TL;DR Summary
I'm looking for a three-term relation between the coefficients of a series solution to the spheroidal harmonics defining equation.
I am currently looking at section IIA of the following paper: https://arxiv.org/pdf/gr-qc/0511111.pdf. Eq. (2.5) proposes an ansatz to solve the spheroidal wave equation (2.1). This equation is
$$ \dfrac{d}{dx} \left((1-x^2) \dfrac{d}{dx}S_{lm} \right) + \left(c^2x^2 + A_{lm} - \dfrac{m^2}{1-x^2}\right)S_{lm}=0,$$

and the ansatz is

$$S_{lm}(x)= e^{cx}(1-x^2)^{|m|/2} \sum_{p=0}^{+\infty} a_p (1+x)^p.$$

Upon differentiating this ansatz and some algebra, I find the following equation:

$$ \sum_{p=0}^{+\infty} a_p (1+x)^p \left[ -2c(1+|m|)x(1+x^2) + c^2(1-x^2)^2 + |m|\left( (|m|+1)x^2 -1 \right) +(1-x^2)(c^2x^2+A_{lm}) -m^2 \right] + \sum_{p=0}^{+\infty} (p+1) a_{p+1} (1+x)^p \left[2x(1-x^2)^2 - 2(1+|m|)(1-x^2) \right] +\sum_{p=0}^{+\infty} (p+2)(p+1) a_{p+2} (1+x)^p (1-x^2)^2 =0. $$

Expanding and grouping powers of $x$, I get
$$\sum_{p=0}^{+ \infty} (1+x)^p \left[a_p[A_{lm} + c^2 - m(m+1) -2xc(m+1) + x^2(m(m+1)-c^2 -A_{lm}) + 2x^3 c (m+1)] \right]\\
+ 2(p+1) a_{p+1} [c- (m+1)x - 2cx^2 + (m+1)x^3 + cx^4] + (p+2)(p+1)a_{p+2}[1-2x^2+x^4]=0.$$

I have no idea how to proceed further. I've tried to look at equating the coefficients of the various powers of ##x## to 0, but I don't get anything like the recurrence relation outlined in Eqs. (2.6) and (2.7) of said paper. Any help would be much appreciated!
 
Physics news on Phys.org
Welcome to PF!

The basic idea with series solutions is to substitute into the DE, collect like powers of ##x## , and then equate the coefficient of each power of ##x## to zero. Doing this leads to the mess you've shown. Now, the authors have reduced this with perseverance (and or a computer algebra package) leading to the recurrence relations (2.6) and (2.7). Okay, starting with these recurrence relations, one needs to specify ##a_0## then use (2.6) to get ##a_1##. From here (2.7) may be used to find ##a_2## and so on. Sadly, this is where the real work starts.

The differential equation (2.1) has singular points at ##x = \pm 1##. The expansion, (2.5) is being taken around ##x=-1## so you will get an analytic solution valid in a region around ##-1##. The problem is the point ##x=1##. For almost all values of ##A_{lm}## the series will diverge at ##x=1##. The series will only converge for eigenvalues. Now, they solve this problem in equation (2.9). I have no clue how they arrived at this continued fraction. I'd be interested to find out.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top