22.7 ESTIMATION OF EIGENVALUES AND EIGENFUNCTIONS

λmaxis finite, orλmax=∞andλminis finite. Notice that here we have departed

from direct consideration of the minimising problem and made a statement about

a calculation in which no actual minimisation is necessary.

Thus, as an example, for an equation with a finite lowest eigenvalueλ 0 any

evaluation ofI/Jprovides an upper bound onλ 0. Further, we will now show that

the estimateλobtained is a better estimate ofλ 0 than the estimated (guessed)

functionyis ofy 0 , the true eigenfunction corresponding toλ 0. The sense in which

‘better’ is used here will be clear from the final result.

Firstly, we expand the estimated ortrial functionyin terms of the complete

setyi:

y=y 0 +c 1 y 1 +c 2 y 2 +···,

where, if a good trial function has been guessed, theciwill be small. Using (22.25)

we have immediately thatJ=1+

∑

i|ci|

(^2). The other required integral is
I=
∫b
a
[
p
(
y′ 0 +
∑
i
ciy′i
) 2
−q
(
y 0 +
∑
i
ciyi
) 2 ]
dx.
On multiplying out the squared terms, all the cross terms vanish because of
(22.27) to leave
λ=
I
J

λ 0 +
∑
i|ci|
(^2) λ
i
1+
∑
j|cj|
2
=λ 0 +
∑
i
|ci|^2 (λi−λ 0 )+O(c^4 ).
Henceλdiffers fromλ 0 by a term second order in theci, even thoughydiffered
fromy 0 by a term first order in theci; this is what we aimed to show. We notice
incidentally that, sinceλ 0 <λifor alli,λis shown to be necessarily≥λ 0 , with
equality only if allci=0,i.e.ify≡y 0.
The method can be extended to the second and higher eigenvalues by imposing,
in addition to the original constraints and boundary conditions, a restriction
of the trial functions to only those that are orthogonal to the eigenfunctions
corresponding to lower eigenvalues. (Of course, this requires complete or nearly
complete knowledge of these latter eigenfunctions.) An example is given at the
end of the chapter (exercise 22.25).
We now illustrate the method we have discussed by considering a simple
example, one for which, as on previous occasions, the answer is obvious.