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Leading term of a power-series solution

  1. Jan 25, 2015 #1
    Hey! I have a question: how can you find the power-series solution and its leading term of the following ODE around the point x=+-(1-b) (where b is near x):

    $$-(1-x^2)\frac{\partial^2 f^m_l}{\partial x^2}+2x\frac{f^m_l}{\partial x}+\frac{m^2}{1-x^2}f^m_l=l(l+1)f^m_l ,$$

    and m and l are constants.

    EDIT: Let me ask a better question - how can you find the solution to the following ODE:

    $$ sin(\theta)\frac{\partial}{\partial \theta}(sin(\theta)\frac{\partial f(\theta)}{\partial \theta})+[l(l+1)sin^2(\theta)-m^2)]f(\theta)=0 $$

    Thank you.
     
    Last edited: Jan 25, 2015
  2. jcsd
  3. Jan 26, 2015 #2

    mfb

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    I didn't solve it, but defining f(θ)=g(sin(θ)) makes the equation easier. Maybe the more general f(θ)=g(sin(θ)) + h(cos(θ)) is even better.
     
  4. Jan 26, 2015 #3
    I already solved it: it isn't an homonegeous ODE, so to calculate its power-series solution you substitute x=cos(θ)=sqrt(1-(sin(θ))^2). The solution is:

    $$\Theta(\theta)=A_l^mP_l^m(x)$$

    where
    $$P_l^m(x)\equiv (1-x^2)^{\vert m \vert/2}\left(\frac{d}{dx}\right)P_l(x)$$.
    This is its associated Legendre function.
     
  5. Jan 27, 2015 #4
    What I meant was that the solution is:
    $$\Theta(\theta)=A_l^mP_l^m(cos\theta)$$

    where

    $$P_l^m(x)\equiv (1-x^2)^{\vert m \vert/2}\left(\frac{d}{dx}\right)P_l(x),$$

    where

    $$x=cos\theta.$$
     
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