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Bessel function series expansion

  1. Dec 8, 2010 #1
    1. The problem statement, all variables and given/known data

    This is the how the question begins.

    1. Bessel's equation is [tex]z^{2}\frac{d^{2}y}{dz^{2}} + z\frac{dy}{dz} + \left(z^{2}- p^{2}\right)y = 0[/tex].

    For the case [tex]p^{2} = \frac{1}{4}[/tex], the equation has two series solutions which (unusually) may be expressed in terms of elementary functions:

    [tex]J_{1/2} = \left(\frac{2}{\pi z}\right)^{1/2} sin z[/tex]
    [tex]J_{-1/2} = \left(\frac{2}{\pi z}\right)^{1/2} cos z[/tex]

    [ The factors [tex]\left(\frac{2}{\pi z}\right)^{1/2}[/tex] are supefluous, but are included by convention, for reasons that are not relevant to the present purposes.]

    Clearly [tex]J_{-1/2}[/tex] is singular at z = 0. Show that [tex]J_{1/2}(0) = 0[/tex].

    2...................(for later)


    2. Relevant equations


    3. The attempt at a solution

    I am going to assume the solutions [tex]J_{1/2}[/tex] and [tex]J_{-1/2}[/tex] without worrying about why/how they come about.

    Obviously, when z = 0, [tex]cos z \neq 0 [/tex]. Therefore, [tex]\left(\frac{2}{\pi z}\right)^{1/2}[/tex] blows up and [tex]J_{-1/2}[/tex] is singular at z = 0.

    On the other hand, if we draw separately the graphs of [tex]\left(\frac{2}{\pi z}\right)^{1/2}[/tex] and [tex] sin z [/tex] and then combine the two in a single graph of [tex]J_{1/2}[/tex], we find that it is sinusoidal with an amplitude given by [tex]\left(\frac{2}{\pi z}\right)^{1/2}[/tex]. This means that the curve oscillates as it moves towards z = 0 with an amplitude that tends to infinity as z tends to 0. How do I conclude from this that [tex]J_{1/2}(0) = 0[/tex].
     
  2. jcsd
  3. Dec 8, 2010 #2
    Compare the order to which [tex]\sin z[/tex] vanishes at 0 to the order of growth of [tex]\sqrt{2/(\pi z)}[/tex] at 0. From that you should be able to prove an estimate of the form [tex]J_{1/2}(z) = O(f(z))[/tex] as [tex]z \to 0[/tex] for an appropriately chosen bound [tex]f[/tex].
     
  4. Dec 8, 2010 #3
    Thanks for your help!

    I don't understand what 'order' means in this context.
    How do we compare the orders?

    "an estimate of the form [tex]J_{1/2}(z) = O(f(z))[/tex] as [tex]z \to 0[/tex] for an appropriately chosen bound [tex]f[/tex].": I don't understand!
     
  5. Dec 8, 2010 #4
    "Order" is short for "order of growth", which means a simple function (usually a power, log, or exponential function or simple combination of those functions) which you can use to "measure" the speed with which a given function vanishes or grows near a point. These orders are measured up to a constant factor.

    For example, [tex]\csc x[/tex] grows like [tex]1/x[/tex] as [tex]x \to 0[/tex], that is, you can choose positive constants [tex]c, C[/tex] so that [tex]c|x^{-1}| < |\csc x| < C|x^{-1}|[/tex] for [tex]x[/tex] close enough to zero. One common way to try to find an order of growth for a function is to expand it in a Taylor or Laurent series.

    For the meaning of [tex]O(f(z))[/tex] look here: Wikipedia article on Big-O notation Using this notation you could express the example above as [tex]\csc x = \Theta(1/x)[/tex] as [tex]x \to 0[/tex].

    The point of my hint is that, to prove that [tex]J_{1/2}(0) = 0[/tex], you should examine the orders of growth near zero of the functions whose product is [tex]J_{1/2}[/tex], to prove that the product can be estimated by a function you know vanishes at 0.
     
  6. Dec 8, 2010 #5
    I will have a think about this method while we move on to the next part of the question.

    2. By means of the substitution [tex]z = kx[/tex], show that [tex]J_{1/2}(kx)[/tex] and [tex]J_{-1/2}(kx)[/tex] are solutions of the following equations:

    [tex]x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} + \left(k^{2}x^{2} - p^{2}\right)y = 0[/tex].

    3. Show that this equation, subject to boundary conditions [tex]y(0) = 0, y(1) = 0 [/tex], has eigenvalues [tex]k = n\pi, n = 1,2,3...[/tex] and eigenfunctions [tex]y_{n} = J_{1/2}(n\pi x)[/tex].

    I am fine with 2, but din't know how to begin 3.
     
  7. Dec 8, 2010 #6

    benorin

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    Homework Helper

    Try performing the limit calculation for a nore solid answer: show that [itex]\lim_{z\rightarrow 0} \left(\frac{2}{\pi z}\right)^{1/2} \sin z=1[/itex]
     
  8. Dec 8, 2010 #7
    [tex]\lim_{z\rightarrow 0} \left(\frac{\sin z}{z}\right)=1[/tex], so
    [tex]\lim_{z\rightarrow 0} \left(\frac{2}{\pi z}\right)^{1/2} (\left \sin z \right)[/tex]
    = [tex]\lim_{z\rightarrow 0} \left(\frac{2z}{\pi}\right)^{1/2}\left(\frac{\sin z}{z}\right) [/tex]
    = [tex]\lim_{z\rightarrow 0} \left(\frac{2z}{\pi}\right)^{1/2} \lim_{z\rightarrow 0} \left(\frac{\sin z}{z}\right)[/tex]

    The first limit is 0 and the second is 1, so the final answer is 0: a contradiction!!

    Thoughts?
     
  9. Dec 9, 2010 #8
    The function:

    [tex]J_{1/2}(z)=\sqrt{\frac{2}{\pi z}}\sin(z)[/tex]

    is undefined at z=0. However, it's limit there is zero by L'Hospital's rule so that if we explicitly define:

    [tex]J_{1/2}(z)=\begin{cases} \sqrt{\frac{2}{\pi z}}\sin(z) & z>0 \\
    0 & z=0
    \end{cases}
    [/tex]

    then [itex]J_{1/2}(z) [/itex] becomes continuous for real z greater than or equal to zero.

    and that thing about drawing separate graphs is not correct. If you plot the function, you should get a sinusoidal wave for positive real z.
     
    Last edited: Dec 9, 2010
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