# Sturm-Liouville Problem

How could one prove that there are at most countably many eigenvalues of the Sturm-Liouville problem $-Lu = ju$, $j$ = eigenvalue, and $u$ is in $C^2[a,b]$? I have been attempting at this problem for a while and I am not sure how to proceed. Thank you very much.

#### Solutions Collecting From Web of "Sturm-Liouville Problem"

Suppose you have a regular Sturm-Liouville operator
$$Lf = \frac{1}{w(x)}\left[-\frac{d}{dx}\left(p(x)\frac{df}{dx}\right)+q(x)f\right]$$
For example, suppose $p \in C^1[a,b]$, $w,q\in C[a,b]$ and $p > 0$, $w > 0$ on $[a,b]$. The domain $\mathcal{D}(L)$ should include some real endpoint conditions such as
$$\cos\alpha f(a)+\sin\alpha f'(a)=0,\\ \cos\beta f(b)+\sin\beta f'(b) = 0.$$
(Assume that $\alpha,\beta$ are real so that the endpoint conditions are real.)

By standard existence theorems, there are unique $C^2[a,b]$ solutions of $Lf=\lambda f$ such that
$$\varphi_{\lambda}(a)=-\sin\alpha,\;\; \varphi_{\lambda}'(a)=\cos\alpha, \\ \psi_{\lambda}(b)=-\sin\beta,\;\;\psi_{\lambda}'(b)=\cos\beta.$$
The solution $\varphi_{\lambda}$ satisfies the left endpoint condition, while $\psi_{\lambda}$ satisfies the right endpoint condition. These solutions have power series expansions in $\lambda$ because of the fixed endpoint conditions. And you show that the following does not depend on $x$.
$$w(\lambda)=W_p(\varphi_{\lambda},\psi_{\lambda})(x)=p(x)\{\varphi_{\lambda}(x)\psi_{\lambda}'(x)-\varphi_{\lambda}'(x)\psi_{\lambda}(x)\}$$
The Wronskian $w(\lambda)$ is an entire function of $\lambda$, and you can show that this function of $\lambda$ vanishes for some $\lambda$ iff $\{ \varphi_{\lambda},\psi_{\lambda} \}$ is a linearly-dependent set of functions, which is equivalent to both functions satisfying both endpoint conditions. That, in turn, is equivalent to the existence of a non-trivial $C^2[a,b]$ solution of $Lf=\lambda f$ which satisfies both endpoint conditions. The zeros of the Wronskian $w$ are the eigenvalues, which cannot cluster because of the identity theorem for holomorphic functions.

If you want to avoid Complex Analysis, you can use Functional Analysis to work in the Banach space $C[a,b]$ and show that $(L-\lambda I)$ has a bounded inverse for $\lambda\notin\mathbb{R}$ that is given by
$$(L-\lambda I)^{-1}f= \frac{\psi_{\lambda}(x)}{w(\lambda)}\int_{a}^{x}f(x)\varphi_{\lambda}(x)dx + \frac{\varphi_{\lambda}(x)}{w(\lambda)}\int_{x}^{b}f(x)\psi_{\lambda}(x)dx.$$
This is a compact operator because of its smoothing properties, which is shown with a classical application of Arzela-Ascoli; this is done by showing that the integral operator maps bounded sets of functions to equicontinuous sets in $C[a,b]$. So you know a lot about its eigenfunctions and eigenvalues. And you can easily relate the eigenfunctions/eigenvalues of $(L-\lambda I)$ to those of $(L-\lambda I)^{-1}$. You can show that the eigenvalues of $L$ cannot cluster in the finite plane because of properties of the spectrum of compact operators.

You can approach the subject by looking at the zeros of $Lf=\lambda f$ for real $\lambda$. This is another approach, and one of the earliest taken by Sturm.

Dear Carlotto, for each eigenvalue, you may assume that $u$ is a real function. If it were not, you could take the real part of $u$ to make it real and it would still be an eigenstate.

Take an eigenvalue $\lambda$, find the corresponding eigenstate $u_\lambda$, and – the critical point is coming – count the number $n$ of zeros of $u_\lambda$ in the interval $[a,b]$. You’re finished if you can prove that for each number of zeros $n$, you can only find one eigenvalue (or two eigenvalues); the eigenvalues may be labeled by $n$ (and an extra bit if needed) and this map proves that the set of eigenvalues is countable.

Now, it’s not hard to see that the statement is right. The function $u$ has to satisfy the S-L boundary conditions at $x=a$. It is evolved, while staying real, in the direction towards $b$ where some S-L boundary conditions have to hold, too. The evolution depends on $\lambda$. As you adjust $\lambda$, you’re increasing the number of zeros and for each number of zeros, you will find just one value of $\lambda$ for which the boundary conditions are satisfied at $b$ again.

There are subtleties for a general S-L problem but I recommend you to start with the $-d^2/dx^2$ operator on an interval $[0,\pi]$ where the eigenstates are just standing waves $\sin(nx)$ and eigenvalues are $n^2$. The boundary conditions are $y=0$ at $x=0,\pi$. The general S-L problem is actually conceptually analogous although the detailed operators are shifted and the eigenvalues are changed. But you may continuously connect it to the standing waves case so the spectrum has to be countable – another way to see it.

There’s a minimization method to find the functions. The lowest-eigenvalue eigenstate with the minimum number of zeros minimizes the expectation value of $L$ among all states; the next-to-lowest eigenstate minimizes the eigenvalue of $L$ between all states that are perpendicular to the ground state; the next one minimizes it among all states that are orthogonal to both the previous ones; and so on.

You can show that the eigenvalues of that problem, with appropriate boundary conditions at the extremes of a finite interval (you omitted this very important piece of information), are precisely the zeros of a meromorphic function that doesn’t vanish identically. Thus those zeros cannot cluster at any finite point. Reference: Coddington-Levinson Theory of ordinary differential equations, chapter 7, theorem 2.1.

I know very little about differential equations, so I’ll turn to Wikipedia:
Taking this as my definition of “Sturmâ€“Liouville problem”, we abstract this to studying a self-adjoint (but maybe unbounded) operator $L$ acting on the Hilbert space $L^2[a,b]$. The general theory ensures that the eigenvectors form an orthonormal set (this follows easily from the fact that the operator $L$ is self-adjoint). So the collection of all eigenvectors is an orthonormal set in the Hilbert space $L^2[a,b]$. But this Hilbert space is separable, and all orthonormal bases of a Hilbert space have the same cardinality, so it’s immediate that you can have only countably many eigenvectors.
The question as stated only requires you to prove there are at most countably many eigenvalues, and since you refer to functional analysis I suppose the idea is to use that $L^2(0,1)$ is a separable Hilbert space (it appear you want to use the weight $w$ in the space). To do this, suppose you have two eigenvalues $\lambda$, $\mu$ with eigenfunctions $u$ and $v$. Then
$$\lambda\int_0^1u\overline{v}w=\int_0^1(-(pu’)’+qu)\overline{v} =\int_0^1u\overline{(-(pv’)’+qv)}=\overline{\mu}\int_0^1u\overline{v}w$$
where the middle equality comes from integrating by parts twice, using the boundary conditions. For $u=v$ this shows all eigenvalues are real, and for $\lambda\ne\mu$ that eigenfunctions to different eigenvalues are orthogonal. Thus, if there were uncountably many eigenvalues the space would not be separable.