Solving associated Legendre equation

  • Thread starter ShayanJ
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  • #1
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Main Question or Discussion Point

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

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

By series expansion around one of its regular singularities.(e.g. [itex]x_0=1[/itex])

This equation is of the form:

[itex]
y''+p(x)y'+q(x)y=0
[/itex]

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

[itex]
p(x)=\sum_{-1}^{\infty} A_n (x-x_0)^n \\ \\
q(x)=\sum_{-2}^{\infty} B_n(x-x_0)^n
[/itex]

2-Solve the equation [itex] r^2+(A_{-1}-1)r+B_{-2}=0 [/itex] 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 [itex] A_{-1}=1 [/itex] and [itex] B_{-2}=0 [/itex]
which means r=0.So the solutions have the forms:

[itex]
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
[/itex]

When I substitute [itex] y_1 [/itex] in the equation,I get the following:

[itex]
\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
[/itex]

Now I have two questions:
1-Its obvious that the relation between [itex] a_n [/itex] and [itex] a_{n-1} [/itex] 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
 

Answers and Replies

  • #2
2,788
587
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
 

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