# Solving associated Legendre equation

1. Mar 18, 2013

### ShayanJ

I'm trying to solve the associated Legendre differential equation:

$y''+\frac{2x}{x^2-1}y'+[ \lambda+\frac{m^2}{x^2-1}]y=0$

By series expansion around one of its regular singularities.(e.g. $x_0=1$)

This equation is of the form:

$y''+p(x)y'+q(x)y=0$

Which is solved by the following procedure:
1-Expand p(x) and q(x) in laurent series:

$p(x)=\sum_{-1}^{\infty} A_n (x-x_0)^n \\ \\ q(x)=\sum_{-2}^{\infty} B_n(x-x_0)^n$

2-Solve the equation $r^2+(A_{-1}-1)r+B_{-2}=0$ for r
3-Based on the value(s) of r, write y as a series and find the coefficients by substitution in the differential equation.

For the associated Legendre eqaution,I find $A_{-1}=1$ and $B_{-2}=0$
which means r=0.So the solutions have the forms:

$y_1=1+\sum_1^{\infty}a_n(x-1)^n \\ y_2=y_1 \ln{|x-1|}+\sum_0^{\infty}b_n(x-1)^n$

When I substitute $y_1$ in the equation,I get the following:

$\sum_1^{\infty} [ n(n+1)a_n + \large{\frac{2xna_n}{x+1} +\frac{m^2 a_{n-1}}{x+1} }](x-1)^n+[\lambda(x-1)+\frac{m^2}{x+1}](x-1)=0$

Now I have two questions:
1-Its obvious that the relation between $a_n$ and $a_{n-1}$ involves x.Is it acceptable?why?
2-How am I supposed to use the equation,which is gained by equating the terms outside the sum to zero?

Thanks

2. Mar 20, 2013

### ShayanJ

Well,I guess I should change my question a bit.
First,the differential equation I wrote in the last post isn't the associated legendre equation.
Second,I now understand its completely wrong to have x in the recursion relation.
So I should ask,what's wrong with my way of solving that equation?
Thanks