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.
[Math] Sturm-Liouville Problem
functional-analysis
Related Solutions
The differential equation you seek is
$$y''+\lambda y = 0$$
which had the general solution
$$y(x) = A \cos{\sqrt{\lambda} x} + B \sin{\sqrt{\lambda} x}$$
The boundary value given by $y'(0)=y(0)$ implies that $A = B \sqrt{\lambda}$. The boundary value given by $y'(1) + y(1) = 0$ implies that
$$-A \sqrt{\lambda} \sin{\sqrt{\lambda}} + B \sqrt{\lambda} \cos{\sqrt{\lambda}} + A \cos{\sqrt{\lambda}} + B \sin{\sqrt{\lambda}}=0$$
Using the relation derived from the first boundary condition, we get
$$-\lambda \sin{\sqrt{\lambda}} + \sqrt{\lambda} \cos{\sqrt{\lambda}} + \sqrt{\lambda} \cos{\sqrt{\lambda}} + \sin{\sqrt{\lambda}}=0$$
or, after a little algebra,
$$\tan{\sqrt{\lambda}} = \frac{2 \sqrt{\lambda}}{\lambda-1}$$
Here is a plot of the LHS and RHS so you can see the intersections and hence, the square roots of the eigenvalues. Clearly, for the first few, they must be computed numerically. For the $k$th eigenvalue, where $k$ is large,
$$\lambda_k \sim k^2 \pi^2$$
The eigenfunction (unnormalized) corresponding to that eigenvalue is
$$y_k(x) = \sqrt{\lambda_k} \cos{\sqrt{\lambda_k} x} + \sin{\sqrt{\lambda_k} x}$$
The Green's function may then be found from the eigenvalues/eigenfunctions as follows:
$$G(x,x') = \sum_{k=1}^{\infty} \frac{y_k(x) y_k(x')}{\lambda-\lambda_k}$$
where $G$ satisfies
$$\frac{d^2}{dx^2} G(x,x') + \lambda G(x,x') = \delta(x-x')$$
Recall that I did find your eigenvalues and eigenfunctions, as well as the Green's function in terms of an expansion in those eigenfunctions.
To find the Green's function directly in the form you want, note that you are solving
$$\frac{d^2}{d x^2} G(x,x') = -\delta(x-x')$$
The general solution to this equation may be immediately written down as
$$G(x,x') = \begin{cases}\\A x + B & 0 \le x\le x'\\ C x+D & x' < x \le 1 \end{cases}$$
The constants are determined through boundary conditions, as well as conditions imposed on the Green's function itself through the differential equation. The boundary conditions at $x=0$ imply that $A=B$. The boundary conditions at $x=1$ imply that
$$C+C+D = 0 \implies D=-2 C$$
We need two other conditions to completely determine the Green's function. First, we require that $G$ be continuous at $x=x'$:
$$A x'+B = C x'+D$$
The last condition is a little tricky. We integrate the diff. equation with respect to $x$ in a small neighborhood about $x=-x'$; note that the integral of a delta function that goes through $x=x'$ is $1$. Thus, we have
$$\lim_{\epsilon \to 0}\left(\left[\frac{d}{dx} G(x,x')\right]_{x=x'+\epsilon}-\left[\frac{d}{dx} G(x,x')\right]_{x=x'-\epsilon}\right) = -1$$
This is saying that the derivative of $G$ is discontinuous at $x=x'$. What this means for our solution is that
$$C-A=-1$$
You then solve, four equations for four unknowns. Actually, these equations are simple enough, and you will find that
$$A = B = \frac13(2-x')$$ $$C=-\frac13(1+x')$$ $$D=\frac{2}{3}(1+x') $$
The sought result then follows:
$$G(x,x') = \begin{cases}\\\frac13 (2-x')(1+x) & 0 \le x\le x'\\ \frac13 (1+x')(2-x) & x' < x \le 1 \end{cases}$$
Best Answer
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.