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

[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?

 

[jvicens]

Hi there,

I can provide some insight into why this may be the case. The standard Sturm-Liouville problem is a widely studied topic in mathematics, and there are several theorems and results that can help explain why eigenfunctions with distinct eigenvalues have the same number of zeros.

One possible explanation is the Sturm oscillation theorem, which states that the number of zeros of an eigenfunction is equal to the index of the eigenvalue, plus or minus one depending on the form of the boundary conditions. In your case, the boundary conditions are such that there is a skip in the counting of zeros, which can lead to the eigenfunctions having the same number of zeros.

Another possible explanation is the Weyl's alternative theorem, which states that if two eigenfunctions have the same eigenvalue, then they must be linearly dependent. This means that the eigenfunctions cannot have a different number of zeros, as they would be identical in this case.

It is also worth mentioning that the number of zeros in an eigenfunction can be affected by the shape of the domain and the form of the boundary conditions. In some cases, the number of zeros may not follow a specific pattern, but in general, the number of zeros can provide valuable information about the eigenvalues and eigenfunctions of a Sturm-Liouville problem.

I hope this helps to shed some light on why eigenfunctions with distinct eigenvalues may have the same number of zeros. Keep exploring and researching, and you may come across other theorems and results that can provide further insights into this topic.