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How is the Bessel function approximated by a ln function

  1. Aug 18, 2012 #1
    1. The problem statement, all variables and given/known data
    It is stated in "Mathematical methods of Physics" by J. Mathews, 2nd ed, p274, that the Bessel function of the second kind and of order zero, i.e. [itex]Y_0(x)[/itex] can be approximated by [itex]\frac{2}{\pi}\ln(x)[/itex]+constant as [itex]x \to 0[/itex], but no more details are given in the same text.


    2. Relevant equations
    Could you show me how that approximation comes about?


    3. The attempt at a solution
    Visually I see the resemblance between the functions, but I cannot derive the approximation myself.
     
    Last edited: Aug 19, 2012
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  3. Aug 18, 2012 #2

    gabbagabbahey

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    Hi sinbaski, welcome to PF!:smile:

    Do you mean the Bessel function of the second kind? (Bessel functions of the first kind are finite at the origin. and [itex]\frac{2}{\pi}\ln(x)[/itex] is not)
     
  4. Aug 19, 2012 #3
    Yes, gabbagabbahey, I meant the second kind. Sorry, that was a careless typo! :|
     
  5. Aug 19, 2012 #4

    gabbagabbahey

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    Okay, well there are a few different ways of defining the Bessel functions, so what definition are you using in your course?
     
  6. Aug 19, 2012 #5
    In the same book [itex]Y_m[/itex] is defined as

    [itex]
    Y_m(x) = \frac{J_m(x) \cos m\pi - J_{-m}(x)}{\sin m\pi}
    [/itex]

    while [itex]J_m(x)[/itex] is defined as
    [itex]
    J_m(x) = \sum_{r=0}^{\infty} \frac{(-1)^r}{r! \Gamma(r+m+1)}\left(\frac{x}{2}\right)^{2r+m}
    [/itex]
     
  7. Aug 19, 2012 #6

    vela

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    I think the answer will pop out if you go back to solving the Bessel differential equation for m=0. The method of Frobenius will yield one solution J0(x), and now you need to construct a second, independent one.
     
  8. Aug 19, 2012 #7
    I have to admit I can't solve by hand the Bessel equation
    [itex]
    x^2 y'' + x y' + x^2 y = 0
    [/itex]

    Matlab gives the solution to the equation as
    [itex]
    A J_0(x) + B Y_0(x)
    [/itex]

    Still I don't see how [itex]\frac{2}{\pi}\ln(x)[/itex] emerges from the equations :(
     
  9. Aug 19, 2012 #8

    vela

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    You want to learn about the Frobenius method. Depending on the roots so the indicial equation, sometimes it'll yield two independent solutions, but in this case, it doesn't. Wikipedia just tells you the form of the second solution.

    http://en.wikipedia.org/wiki/Frobenius_method#Double_roots

    If you look on page 16 of Mathews and Walker, there's a discussion of how to solve Bessel's equation using Frobenius, including how to find the second solution.
     
  10. Aug 19, 2012 #9

    vela

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    I did a bit more reading, and it looks like to get the constant factor of ##2/\pi##, you're going to have to look at the definition of Ym(x). It's an indeterminate form when m is an integer, so you want to evaluate it using L'Hopital's rule.
     
  11. Aug 19, 2012 #10

    gabbagabbahey

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    For integer [itex]m[/itex], this should really be

    [tex]Y_m(x) = \lim_{\alpha \to m} \frac{J_{\alpha}(x) \cos \alpha \pi - J_{-\alpha}(x)}{\sin \alpha \pi}[/tex]

    Okay, so use this definition, with the (corrected) one above to find an expression for [itex]Y_0(x)[/itex].

    Note: There's no need for you to actually solve Bessel's differential equation for this problem (the above equations are the two independent solutions), but it is a good exercise and one you may want to do after this problem (or before)
     
    Last edited: Aug 19, 2012
  12. Aug 19, 2012 #11
    Thank you for your kind reply! The wikipage is very comprehensive compared with the book. Now I think I have got something, following your advice:
    [itex]
    \lim_{m\to 0} \frac{\cos m\pi J_m(x) - J_{-m}(x)}{\sin m\pi}
    [/itex]
    [itex]
    = \lim_{m\to 0}\frac{
    -\pi\sin m\pi J_m(x) + \cos m\pi \frac{\partial}{\partial m}J_m(x) + \frac{\partial}{\partial (-m)}J_{-m}(x)
    }{\pi \cos m\pi}
    [/itex]
    [itex]
    = \frac{2}{\pi} \lim_{m\to 0} \frac{\partial}{\partial m}J_m(x)
    [/itex]

    Then
    [itex]
    \frac{\partial}{\partial m}J_m(x) = \frac{\partial}{\partial m}\sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(m+r+1)}\left(\frac{x}{2}\right)^{m+2r}
    [/itex]
    [itex]
    = \ln(x/2)\sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(m+r+1)}\left(\frac{x}{2}\right)^{m+2r} +
    \sum_{r=0}^{\infty} \left(\frac{x}{2}\right)^{m+2r} \frac{\partial}{\partial m}\frac{(-1)^r}{r!\Gamma(m+r+1)}
    [/itex]

    Taking the limit [itex]x \to 0, m\to 0[/itex], the first sum reduces to [itex]\ln(x/2)[/itex] while the second becomes a constant. Thus we get the desired result of [itex]\frac{2}{\pi}\ln(x)+C[/itex]

    Do you think this argument is correct?
     
    Last edited: Aug 20, 2012
  13. Aug 21, 2012 #12

    gabbagabbahey

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    I'm not sure it makes any sense to say that taking the limit [itex]x \to 0[/itex] and [itex]m \to 0[/itex] of [itex]\ln(x/2)\sum_{r=0}^{\infty} \frac{(-1)^r}{r!\Gamma(m+r+1)}\left(\frac{x}{2}\right)^{m+2r}[/itex] results in [itex]\ln(x/2)[/itex]. I think you should take the limit [itex]m \to 0[/itex] of the whole expression (obviously you will need to first calculate [itex]\frac{\partial}{\partial m}\frac{(-1)^r}{r!\Gamma(m+r+1)}[/itex] ), simplify it as much as possible, and then do a Taylor Series approximation for small [itex]x[/itex].
     
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