Discrete Unbounded Sequence of Eigenvalues

With $x=b$ this ``Oscillation theorem'' tells us that $\theta (b,\lambda )$ is a function which increases without limit as $\lambda\to\infty$ . Consequently, as $\lambda$ increases from $\lambda =-\infty$ , there will be a first value, say $\lambda_0$ , for which the second boundary condition (the one at $x=b$ ), i.e.,

$$\theta (b,\lambda_0)=\delta$$

is satisfied. Moreover, as $\lambda$ increases beyond $\lambda_0$ , $\theta (b,\lambda )$ increases monotonically beyond $\delta$ until it reaches the value $\delta +\pi$ . This happens at a specific value of $\lambda$ , say $\lambda_1$ , which is larger than $\lambda_0$ ,

$$\lambda_0<\lambda_1 ~.$$

Continuing in this fashion, one finds that, regardless of how big an integer $n$ one picks, the equation

$$\theta (b,\lambda) =\delta +n\pi$$

always has a solution for $\lambda$ , which we shall call $\lambda_n$ . This yields an infinite discrete sequence of $\lambda$ 's which is monotonically increasing

$$\lambda_0<\lambda_1<\cdots <\lambda_n<\cdots ~~.$$

This sequence has no upper bound. Why? For any large $\Lambda >0$ consider $\theta (b,\Lambda)$ . This number lies between some pair of points, say,

$$\delta +N\pi \leq \theta (b,\Lambda) < \delta +(N+1)\pi$$

The Oscillation Theorem says that $\theta (b,\lambda )$ has the property of being a monotonic function of $\lambda$ whose range is the whole positive real line. The latter property guarantees that each of the two equations,

$$\theta (b,\lambda) =\delta +N\pi$$

and

$$\theta (b,\lambda) =\delta +(N+1)\pi~~,$$

has a solution. The former property guarantees that each of these two solutions is unique. Call them $\lambda _N$ and $\lambda _{N+1}$ . The former property also guarantees that

$$\lambda _N \leq \Lambda < \lambda _{N+1}~~.$$

Since $\Lambda >0$ can be as large as we please, the sequence of eigenvalues,

$$\lambda_0<\lambda_1<\cdots <\lambda_N <\lambda _{N+1} <\cdots ~~,$$

has no upper bound.

Corresponding to this sequence, there is the set of eigenfunctions

$$u_n(x)=r_n(x)\sin\theta (x;\lambda_n)~\quad~ n=0,1,2,\dots\,.$$

Each of these functions oscillates as a function of $x$ . How many times does each $u_n(x)$ pass through zero in the open interval $(a,b)$ ? Reference to Figure 3.13 shows that $u_n(x)$ has precisely $n$ zeroes inside $(a,b)$ ; zeroes at the endpoints, if any, do not count. Indeed, it must have at least $n$ zeroes because the graph of $\theta (x,\lambda )$ with $\lambda$ held fixed as in Figure 3.13, must cross at least $n$ multiples of $\pi$ (dotted horizontal lines in Figure 3.13 and 3.11). On the other hand, the function $u_n(x)$ cannot have more than $n$ zeroes because the graph of phase $\theta (x,\lambda )$ can cross each multiple of $\pi$ no more than once. This fact is guaranteed by Eq.(3.25) on page .

To summarize, we have the following

Theorem: Any regular S-L problem has an infinite number of solutions $u_n(x)$ which belong to the real eigenvalues

$$\lambda_0<\lambda_1<\lambda_2<\cdots~~~\textrm{with}~~~\lim_{n\to\infty} \lambda_n=\infty\,.$$

Furthermore, each eigenfunction $u_n(x)$

1. has exactly $n$ zeroes in the interval $a<x<b$ ,
2. is unique up to a constant multiplicative factor.

Lecture 25