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I Legendre Differential Equation

  1. Oct 12, 2016 #1
    I just started learning Legendre Differential Equation. From what I learn the solutions to it is the Legendre polynomial.

    For the legendre DE, what is the l in it? Is it like a variable like y and x, just a different variable instead?

    Legendre Differential Equation: $$(1-x^2) \frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + l(l+1)y = 0$$

    But what is the differences between the associated Legendre polynomial and unassociated Legendre polynomial? When do we use the associated one or the unassociated one, since both are solutions to the Legendre Differential Equation?

    Unassociated Legendre Polynomial: $$P_l(x) = \sum^M_{m=0}(-1)^m\frac{(2l-2m)!}{2^lm!(l-m)!(l-2m)!}x^{l-2m}$$

    Associated Legendre Polynomial: $$P^m_l(x) = (-1)^m(1-x^2)^{m/2}\frac{d^m}{dx^m}(P_l(x))$$

    Lastly, where did the m come from?
     
  2. jcsd
  3. Oct 12, 2016 #2

    DrClaude

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    Staff: Mentor

  4. Oct 17, 2016 #3
    How did you get from (8) to (12)?
    $$\sum^\infty_{n=0} n(n-1) a_nx^{x-2} ⇒ \sum^\infty_{n=0} (n+2)(n+1) a_{n+2} x^n$$

    Shouldnt it be $$\sum^\infty_{n=0} (n+2)(n+1) a_{n+2} x^n + 6a_{-2}x^{-4} +2a_{-1}x^{-3}$$ cause you replace n with n+2?

    Besides how do you find the value for a0 and a1 after getting (17)?
    Sorry for these questions as I am still new to this.
     
  5. Oct 17, 2016 #4

    Mark44

    Staff: Mentor

    The first three terms of the series on the left are ##0(-1)a_0x^{-2} + 1(0)a_1x^{-1} + 2(1)a_2x^0 ##, so you can discard the first two terms, as their coefficients are both zero. The first term of the series on the right is ##2(1)a_2x^0##, so now you can adjust the starting index of the series on the left, replacing n by n + 2.
     
  6. Oct 17, 2016 #5
    Thanks. But how to know the value of a0 and a1? Cause without knowing them, I can't use them to form the solutions (23)&(24).
     
  7. Oct 17, 2016 #6

    Mark44

    Staff: Mentor

    These constants are usually determined by the initial conditions (typically y(0) and y'(0)). If they're not given, you can't determine those coefficients of your series.
     
  8. Oct 19, 2016 #7
    So in this case, a0 and a1 each equals to 1 as the initial conditions right?

    As for eqn(23), "If Inline48.gif is an even integer, the series Inline49.gif reduces to a polynomial of degree Inline50.gif with only even powers of Inline51.gif and the series Inline52.gif diverges.", technically both are diverging right? Is just that one is in the even powers of x while the other is in the odd powers of x.

    I also don't understand the representation of (25) in (26)?
     
  9. Oct 19, 2016 #8

    Mark44

    Staff: Mentor

    No, I don't see how you can assume that.
    I assume you're talking about this page: http://mathworld.wolfram.com/LegendreDifferentialEquation.html
    One thing that bothers me about this page is their statement, right after equation (2), that "The above form is a special case of the so-called "associated Legendre differential equation" corresponding to the case [PLAIN]http://mathworld.wolfram.com/images/equations/LegendreDifferentialEquation/Inline1.gif." [Broken]
    I don't see m mentioned anywhere before that, so it's not clear to me what they're talking about here.

    Regarding equation (23), they assert that the series for ##y_2(x)## diverges if l is an even integer, and that ##y_1(x)## diverges if l is an odd integer. These statements omit a lot of work in showing why the two series diverge. I would want to confirm these for myself.
     
    Last edited by a moderator: May 8, 2017
  10. Oct 19, 2016 #9

    DrClaude

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    Staff: Mentor

    They're talking about the associated Legendre differential equation. It is that equation that has an ##m##, which for ##m=0## reduces to the
    Legendre differential equation.
     
    Last edited by a moderator: May 8, 2017
  11. Oct 21, 2016 #10
    Based on eqn(23) & eqn(24), a0 and a1 should be 1 right? If not then I don't know how to find the values of a0 and a1.
     
  12. Oct 21, 2016 #11
    Sir, do you mind elaborating on the part of equation (23) about y2(x)diverges if l is an even integer, and that y1(x) diverges if l is an odd integer.
    I am lost here cause I am new to this.
     
  13. Oct 21, 2016 #12

    Mark44

    Staff: Mentor

    In that Wolfram page, it appears that they set both a0 and a1 to 0. These constants normally come from initial conditions such as y(0) and y'(0), but there's no mention of initial conditions on that page.
     
  14. Oct 22, 2016 #13
    If a0 is 0, then won't a2, a4, ... equal to zero too? $$a_2=-\frac{l(l+1)}{1.2}a_0$$

    Likewise for a1, since the odd number of a depends on a1 too.
     
  15. Oct 23, 2016 #14

    Mark44

    Staff: Mentor

    Reconsidering, I think I misspoke. For the even solutions it appears that they set ##a_0=1## and ##a_1 = 0##. For the odd solutions, it appears they set ##a_0=0## and ##a_1 = 1##.
     
  16. Oct 23, 2016 #15
    Never mind I understand already. But I am not sure whether the equation (26) is correct on the site(http://mathworld.wolfram.com/LegendreDifferentialEquation.html) you suggested.

    Shouldn't it be $$_2F_1(-\frac{1}{2}.l, \frac{1}{2}(l+1); \frac{1}{2}; x^2)$$ instead of $$_2F_1(-\frac{1}{2}, \frac{1}{2}(l+1); \frac{1}{2}, x^2)$$ for l even?
     
  17. Oct 24, 2016 #16
    Thanks a lot for your help but do you know whether the equation (26) is correct on the site(http://mathworld.wolfram.com/LegendreDifferentialEquation.html)?

    Shouldn't it be $$_2F_1(-\frac{1}{2}.l, \frac{1}{2}(l+1); \frac{1}{2}; x^2)$$ instead of $$_2F_1(-\frac{1}{2}, \frac{1}{2}(l+1); \frac{1}{2}, x^2)$$ for l even?
     
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