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Estimation of eigenvalues and eigenfunctions

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Estimation of eigenvalues and eigenfunctions
CALCULUS OF VARIATIONS
Show that
b
yj pyi − yj qyi dx = λi δij .
(22.27)
a
Let yi be an eigenfunction of (22.24), corresponding to a particular eigenvalue λi , so that
pyi + (q + λi ρ)yi = 0.
Multiplying this through by yj and integrating from a to b (the first term by parts) we
obtain
b
b
b
yj pyi
−
yj (pyi ) dx +
yj (q + λi ρ)yi dx = 0.
(22.28)
a
a
a
The first term vanishes by virtue of (22.26), and on rearranging the other terms and using
(22.25), we find the result (22.27). We see at once that, if the function y(x) minimises I/J, i.e. satisfies the Sturm–
Liouville equation, then putting yi = yj = y in (22.25) and (22.27) yields J and
I respectively on the left-hand sides; thus, as mentioned above, the minimised
value of I/J is just the eigenvalue λ, introduced originally as the undetermined
multiplier.
For a function y satisfying the Sturm–Liouville equation verify that, provided (22.26) is
satisfied, λ = I/J.
Firstly, we multiply (22.24) through by y to give
y(py ) + qy 2 + λρy 2 = 0.
Now integrating this expression by parts we have
b b 2
ypy −
py − qy 2 dx + λ
a
a
b
ρy 2 dx = 0.
a
The first term on the LHS is zero, the second is simply −I and the third is λJ. Thus
λ = I/J. 22.7 Estimation of eigenvalues and eigenfunctions
Since the eigenvalues λi of the Sturm–Liouville equation are the stationary values
of I/J (see above), it follows that any evaluation of I/J must yield a value that lies
between the lowest and highest eigenvalues of the corresponding Sturm–Liouville
equation, i.e.
λmin ≤
I
≤ λmax ,
J
where, depending on the equation under consideration, either λmin = −∞ and
792
22.7 ESTIMATION OF EIGENVALUES AND EIGENFUNCTIONS
λmax is finite, or λmax = ∞ and λmin is 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 eigenvalue λ0 any
evaluation of I/J provides an upper bound on λ0 . Further, we will now show that
the estimate λ obtained is a better estimate of λ0 than the estimated (guessed)
function y is of y0 , the true eigenfunction corresponding to λ0 . The sense in which
‘better’ is used here will be clear from the final result.
Firstly, we expand the estimated or trial function y in terms of the complete
set yi :
y = y0 + c1 y1 + c2 y2 + · · · ,
where, if a good trial function has been guessed, the ci will be small. Using (22.25)
we have immediately that J = 1 + i |ci |2 . The other required integral is
I=
a
b
2
2 ci yi − q y0 +
ci yi
p y0 +
dx.
i
i
On multiplying out the squared terms, all the cross terms vanish because of
(22.27) to leave
I
J
λ0 + i |ci |2 λi
=
1 + j |cj |2
= λ0 +
|ci |2 (λi − λ0 ) + O(c4 ).
λ=
i
Hence λ differs from λ0 by a term second order in the ci , even though y differed
from y0 by a term first order in the ci ; this is what we aimed to show. We notice
incidentally that, since λ0 < λi for all i, λ is shown to be necessarily ≥ λ0 , with
equality only if all ci = 0, i.e. if y ≡ y0 .
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.
793
CALCULUS OF VARIATIONS
y(x)
1
(c)
0.8
(b)
0.6
(a)
(d)
0.4
0.2
x
0.2
0.4
0.6
0.8
1
Figure 22.10 Trial solutions used to estimate the lowest eigenvalue λ of
−y = λy with y(0) = y (1) = 0. They are: (a) y = sin(πx/2), the exact result;
(b) y = 2x − x2 ; (c) y = x3 − 3x2 + 3x; (d) y = sin2 (πx/2).
Estimate the lowest eigenvalue of the equation
−
d2 y
= λy,
dx2
0 ≤ x ≤ 1,
(22.29)
with boundary conditions
y (1) = 0.
y(0) = 0,
(22.30)
We need to find the lowest value λ0 of λ for which (22.29) has a solution y(x) that satisfies
(22.30). The exact answer is of course y = A sin(xπ/2) and λ0 = π 2 /4 ≈ 2.47.
Firstly we note that the Sturm–Liouville equation reduces to (22.29) if we take p(x) = 1,
q(x) = 0 and ρ(x) = 1 and that the boundary conditions satisfy (22.26). Thus we are able
to apply the previous theory.
We will use three trial functions so that the effect on the estimate of λ0 of making better
or worse ‘guesses’ can be seen. One further preliminary remark is relevant, namely that the
estimate is independent of any constant multiplicative factor in the function used. This
is easily verified by looking at the form of I/J. We normalise each trial function so that
y(1) = 1, purely in order to facilitate comparison of the various function shapes.
Figure 22.10 illustrates the trial functions used, curve (a) being the exact solution
y = sin(πx/2). The other curves are (b) y(x) = 2x − x2 , (c) y(x) = x3 − 3x2 + 3x, and (d)
y(x) = sin2 (πx/2). The choice of trial function is governed by the following considerations:
(i) the boundary conditions (22.30) must be satisfied.
(ii) a ‘good’ trial function ought to mimic the correct solution as far as possible, but
it may not be easy to guess even the general shape of the correct solution in some
cases.
(iii) the evaluation of I/J should be as simple as possible.
794
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