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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?

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

Can you offer guidance or do you also need help?

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