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Sturm-Liouville Problem. Two eigenfunctions have the same number of zeros

  1. Jun 24, 2012 #1
    Hello,
    As I know, the "standard" Sturm-Liouville problem
    [tex]\frac{d}{dx}p\frac{d}{dx}y + \lambda\rho y = 0[/tex]
    with boundary conditions
    [tex]\begin{align}y^{\prime}(a)&=0 \\
    y^{\prime}(b)&=F(\lambda)\end{align}[/tex]
    has sequence of eigenvalues
    [tex]\lambda < \lambda_1 < \lambda_2 < ... ...[/tex]
    with corresponding eigenfunctions [itex]y_0, y_1, ... [/itex]
    The "Standard" (sorry I'm not a math guy) Sturm's oscillation theorem states that
    [itex]y_n[/itex] has exactly n zeros in (a,b).
    (In some cases, there is a skip in the counting of the zeros of the eigenfunction compared to the index of the eigenvalue, n-1, or n-2 ... depending on the form of [itex]F(\lambda)[/itex]. (reference: Sturm-Liouville theory: past and present page. 20)

    I solved the equation numerically and examined the number of zeros in [itex]y_n[/itex].
    I find that for [itex]\lambda_0[/itex], [itex]y_0[/itex] has no zeros in the interval.
    But for overtones, I find

    # of zeros in y_n = (n+1)/2 for odd n = 1, 3, 5, ...
    # of zeros in y_n = n/2 for even n = 2, 4, 6, ...

    that is to say, eigenfunctions [itex]y_1, y_2[/itex] with distinct eigenvalues [itex]\lambda_1, \lambda_2[/itex]
    have the same number of zeros (only one).

    I'm trying to figure out why this is so.
    Is there any theorem with regard to this?
     
    Last edited: Jun 24, 2012
  2. jcsd
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