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

1. Jun 24, 2012

### omyojj

Hello,
As I know, the "standard" Sturm-Liouville problem
$$\frac{d}{dx}p\frac{d}{dx}y + \lambda\rho y = 0$$
with boundary conditions
\begin{align}y^{\prime}(a)&=0 \\ y^{\prime}(b)&=F(\lambda)\end{align}
has sequence of eigenvalues
$$\lambda < \lambda_1 < \lambda_2 < ... ...$$
with corresponding eigenfunctions $y_0, y_1, ...$
The "Standard" (sorry I'm not a math guy) Sturm's oscillation theorem states that
$y_n$ 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 $F(\lambda)$. (reference: Sturm-Liouville theory: past and present page. 20)

I solved the equation numerically and examined the number of zeros in $y_n$.
I find that for $\lambda_0$, $y_0$ 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 $y_1, y_2$ with distinct eigenvalues $\lambda_1, \lambda_2$
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