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Bessel Equation and Bessel fuctions

  1. Nov 24, 2013 #1
    We first express Bessel's Equation in Sturm-Liouville form through a substitution:

    Bessel1.png


    Next, we consider a series solution and replace v by m where m is an integer. We obtain a recurrence relation:


    bessel2.png

    Then, since all these terms must be = 0,

    Consider m = 0

    First term vanishes and second term =

    a1x = 0

    therefore, a1 = 0. Then by recurrence relation above, a1 = a3 = a5 = ..... = a2n-1 = 0

    Then only the even series give non-zero coefficients, so we start with a0 ≠ 0.

    Consider m = 1

    Second term vanishes, and first term =

    a0 = 0

    By recurrence relation, this implies a2n = 0

    So we start the series with a1 ≠ 0

    Book's Explanation
    bessel3.png

    I don't understand why a0 = a1 = .... = am-1 = 0

    Surely, for any value of m that is ≠ 1 we have:

    [tex]a_{1}x = \frac{m^2}{1 - m^2} a_0 [/tex]

    This doesn't imply anything above? Also, how did they get the resulting Bessel functions? It looks wildly different from mine.

    a0 = a1 = .... = am-1 = 0,

    [tex] u = \sum_{n=m}^\infty a_{n} x^n = \frac {1}{2^n n!} x^n [/tex]
     
    Last edited: Nov 24, 2013
  2. jcsd
  3. Nov 24, 2013 #2

    AlephZero

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    From 8.44 you have ##(k^2 - m^2)a_k = -a_{k-2}##

    When ##k = m##, that is ##0a_k = a_{k-2}##.

    Using ##a_{k-2} = 0## and the recurrence relation, ##a_{k-4} = a_{k-6} = \dots = 0##.
     
  4. Nov 24, 2013 #3

    pasmith

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    Start from
    [tex]
    -m^2 a_0 + (1 - m^2)a_1 x + \sum_{k= 2}^{\infty} ((k^2 - m^2)a_k + a_{k-2})x^k = 0
    [/tex]

    We need this equation to hold for all [itex]x[/itex]. Thus looking at the constant term we must have
    [tex]
    m^2 a_0 = 0
    [/tex]
    Thus if [itex]m \neq 0[/itex] we must have [itex]a_0 = 0[/itex]. On the other hand if [itex]m = 0[/itex] then [itex]m^2 a_0 = 0[/itex] whatever the value of [itex]a_0[/itex].

    Looking at the coefficient of [itex]x^1[/itex] we have
    [tex]
    (1 - m^2)a_1 = 0
    [/tex]
    so that if [itex]m \neq 1[/itex] we are forced to take [itex]a_1 = 0[/itex]. If [itex]m = 1[/itex] then [itex](1 - m^2)a_1 = 0[/itex] whatever the value of [itex]a_1[/itex].

    At order [itex]x^2[/itex] and higher we find that
    [tex](k^2 - m^2)a_k + a_{k-2} = 0[/tex]
    from which it follows that if [itex]a_{k-2} = 0[/itex] then we must have [itex]a_k = 0[/itex] unless [itex]k = m[/itex], in which case we can choose [itex]a_m[/itex] as we see fit. But this only happens if [itex]m \geq 2[/itex], so that [itex]m \neq 0[/itex] and [itex]m \neq 1[/itex], and by the above we must take [itex]a_0 = a_1 = 0[/itex]. The recurrence relation then tells us that [itex]a_0 = a_1 = \dots = a_{m - 1} = 0[/itex].

    We also have
    [tex]((m + 1)^2 - m^2)a_{m + 1} + a_{m - 1}
    = (2m + 1)a_{m+1} + 0 = 0[/tex]
    so that [itex]a_k = 0[/itex] for [itex]k = m + 2n + 1[/itex], [itex]n \geq 0[/itex].

    If [itex]k = m + 2n[/itex], [itex]n \geq 0[/itex], then
    [tex]
    ((m + 2n)^2 - m^2)a_{m + 2n} + a_{m + 2(n-1)} = 0
    [/tex]
    so that
    [tex]
    a_{m + 2n} = - \frac{a_{m + 2(n-1)}}{(m^2 + 4mn + 4n^2) - m^2}
    = - \frac{a_{m + 2(n-1)}}{4n(m + n)}
    [/tex]
    which can be solved to obtain
    [tex]
    a_{m + 2n} = \frac{(-1)^n}{4^n n!(m + 1) \times \dots \times (m + n) }a_m = \frac{(-1)^n m!}{2^{2n} n!(m + n)!}a_m
    [/tex]
    (looking at [itex]a_{m + 2}[/itex], [itex]a_{m + 4}[/itex], and [itex]a_{m + 6}[/itex] establishes the pattern which can then be proven by induction on [itex]n[/itex]) from which the given series for [itex]J_m(x)[/itex] follows.
     
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