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### Sturm's Comparison Theorem

When confronted with the regular boundary value problem

 0 (318)

 0 (319) 0

1. How does the oscillatory nature of , a solution to Eq. 3.18, depend on ?
2. Why do the values of permitted by (3.18) and (3.19) form a discrete and semi-infinite sequence

with a smallest eigenvalue and with as ?

The oscillatory nature'' of a solution is expressed qualitatively by the location of the zeroes of its graph. One could also inquire about its behaviour between successive zeroes. However, we shall see that such an inquiry always leads to the same answer: Provided is positive between a pair of successive zeroes, the graph of has only a single maximum (or minimum). This means that can not oscillate between two of its successive zeroes.

Thus the most important issue is the existence and location of the zeroes, which are controlled entirely by the phase of a given solution . This phase is a scalar function from which one directly constructs the solution. It is preferrable to discuss the behavior of the solution in terms of its phase because the key qualitative properties of the latter are very easy to come by. As we shall see, one only needs to solve a first order differential equation, not the second order S-L equation.

However, before establishing and solving this differential equation, let us use the second order S-L differential equation directly to determine how the zeroes of are affected if the parameter is changed. We express this behaviour in terms of the

Sturm Comparison Theorem:

Whenever , then between two zeroes of the nontrivial solution , there lies a zero of .

This theorem demands that one compare the two different solutions

of Eq. 3.18 corresponding two different constants and . The conclusion is obtained in three steps:

Step 1: Multiply these two equations respectively by and , and then form their difference. The result, after cancelling out the term, is

which is the familiar Lagrange identity, Eq.(3.15), in disguise. Upon integration one obtains

If both and satisfy the mixed Dirichlet-Neumann boundary conditions at , then

If is a singular point of the differential Eq.(3.18), is zero. Thus, if and are finite at , then the left hand side vanishes again at the lower limit . Thus both for a regular and for this singular Sturm-Liouville problem we have

Step 2: Now assume that, for some range of values , each of the corresponding solutions satisfying the boundary condition at , oscillates. In other words, as increases beyond , reaches a maximum, then decreases, passes through zero, reaches a minimum, increases and so on. That such a range of -values exists, we shall see later.

Let lie in this range, and let be the first zero of as in Figure 3.4. Consequently,

Step 3. One must now conclude that if , then the zeroes of are more closely spaced than those of . Why must this statement, the Sturm Comparison Theorem, be true?

a) Assume the contrary, i.e., assume that has no zero in . See Figure 3.4. In that case for all . This implies

But , and , so that . This is a contradiction. Hence our assumption was wrong; the function does have a zero in .

b) Now consider the circumstance where has two successive zeroes at and : . In that case one obtains

If does not change sign in , as in Figure 3.6, then we again have a contradiction because if , then , . In other words, the picture in Figure 3.6 is impossible. We conclude, therefore, that must have a zero inside the open interval .

Next: Phase Analysis of a Up: Basic Properties of a Previous: Orthogonality, Reality, and Uniqueness   Contents   Index
Ulrich Gerlach 2010-12-09