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Eigenvalue problem using Bessel Functions

  1. Nov 29, 2008 #1
    1. The problem statement, all variables and given/known data

    Bessels equation of order n is given as the following:

    [tex] y'' + \frac{1}{x}y' + (1 - \frac{n^2}{x^2})y = 0 [/tex]

    In a previous question I proved that Bessels equation of order n=0 has the following property:

    [tex] J_0'(x) = -J_1(x) [/tex]

    Where J(x) are Bessel functions of order n=0 and order n=1 respectively.

    Use this property to solve the eigenvalue problem:

    [tex] \frac{1}{r}\frac{d}{dr}(r(\frac{d\phi}{dr}))+ \lambda^2\phi = 0 [/tex]

    where we have:

    [tex] 0 < r < a [/tex]

    [tex] \frac{d\phi}{dr}(a) = 0 [/tex]

    [tex] \phi(0) [/tex] bounded

    2. Relevant equations

    The question about the special property:
    The power series for the nth Bessel function is

    [tex] J_n(x) = (\frac{x}{n})^n\sum_{m=0}^\infty \frac{(-1)^m}{m!(n+m)!}(\frac{x}{2})^{2m} [/tex]

    Use this to show that:

    [tex] J_{n+1}(x) = -x^n\frac{d}{dx}(x^{-n}J_n(x)) [/tex]

    3. The attempt at a solution

    *Phew, that was alot of latex!*

    I've tried expanding:

    [tex] \frac{1}{r}\frac{d}{dr}(r(\frac{d\phi}{dr}))+ \lambda^2\phi = 0 [/tex]

    using the product rule, into something more familiar to see if I can make some sort of relation to the property I am supposed to use. Any help is appreciated, I've spent a lot of time rearranging terms and trying to relate things but can't quite put my finger on it.
  2. jcsd
  3. Dec 1, 2008 #2
    In case anyone is interested, I think this is right:

    Like I was trying before we expand:

    [tex] \frac{1}{r}\frac{d}{dr}(r(\frac{d\phi}{dr}))+ \lambda^2\phi = 0 [/tex]

    we get:

    [tex] \phi''+\frac{1}{r}\phi'+\lambda^2\phi = 0 [/tex]

    To get this into a more "familiar" form of bessels equation, multiply everything by [tex] r^2 [/tex]:

    [tex] r^2\phi"" + r\phi'+(r^2\lambda^2-0^2)\phi = 0 [/tex]

    Which is exactly Bessel's equation of order n=0.

    Digging into a textbook we can learn that the following is a solution to this equation:

    [tex] \phi(r) = C_1J_0(\lambda r) + C_2Y_0(\lambda r) [/tex]

    Since in our problem we are told that [tex] \phi(0) [/tex] is bounded. But looking into our text we find that [tex] Y_0(x) [/tex] is some ghastly function with [tex] ln(x) [/tex] inside of it. Since [tex] ln(x) [/tex] is not bounded at 0, we can conclude that [tex] C_2 = 0 [/tex].

    Our second condition says:

    [tex] \frac{d\phi}{dr}(a) = 0 [/tex]

    then we must have

    [tex] \phi'(a) = -\lambda C_1J_0(\lambda a) = 0 [/tex]

    So we'll throw out the case that [tex] C_1 [/tex] is zero because it isn't interesting. But if [tex] C_1 [/tex] is not equal to zero then we either require [tex] \lambda = 0 [/tex] or [tex] J_0(\lambda a) = 0 [/tex]

    Then if [tex] x_1, x_2, x_3, . . . [/tex] are roots of [tex] J_0(\lambda a) [/tex] we get [tex] \lambda = \frac{x_1}{a}, \frac{x_2}{a}, \frac{x_3}{a} . . . [/tex]

    If I'm missing something or this is completely wrong, someone let me know! :)
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