I'm trying to solve the associated Legendre differential equation:(adsbygoogle = window.adsbygoogle || []).push({});

[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

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# Solving associated Legendre equation

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