Masujima M. Applied Mathematical Methods in Theoretical Physics
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166 5 Hilbert–Schmidt Theory of Symmetric Kernel
As t →∞, the asymptotic form of φ(x, t) is given either by
φ(x, t) = f (x) −
n
A
n
(0)φ
n
(x)ifallλ
n
< 0, (5.2.75)
or by
φ(x, t) = A
i
(0)φ
i
(x)exp
t
λ
i
if 0 <λ
i
< all other λ
n
. (5.2.76)
5.3
Bounds on the Eigenvalues
In the process of proving our lemma in the previous section, we managed to obtain
the upper bound on the lowest eigenvalue,
|
λ
1
|
≤
A
2
A
4
.
A better upper bound can be obtained as follows. If we call
R
2
= A
6
A
4
,
we note that
R
2
≥ R
1
,
or
1R
2
≤ 1R
1
.
Furthermore, we find
A
2
+ sA
4
+ s
2
A
6
+ s
3
A
8
+ s
4
A
10
+···
= A
2
+sA
4
[1 + s(A
6
A
4
) + s
2
(A
8
A
4
) +···]
≥ A
2
+sA
4
[1 + sR
2
+ s
2
R
2
2
+···],
which diverges if
sR
2
> 1.
Hencewehavesingularityfor
s ≤ 1R
2
≤ 1R
1
.
5.3 Bounds on the Eigenvalues 167
Thus we have an improved upper bound on λ
1
,
|
λ
1
|
≤
A
4
A
6
≤
A
2
A
4
.
So, we have the successively better upper bounds on
|
λ
1
|
,
A
2
A
4
,
A
4
A
6
,
A
6
A
8
, ...,
for the lowest eigenvalue, each better than the previous one, i.e.,
|
λ
1
|
≤···≤
A
6
A
8
≤
A
4
A
6
≤
A
2
A
4
. (5.3.1a)
Recall also that with a symmetric kernel, we have
A
2m
=
K
m
2
. (5.3.2)
The upper bounds (5.3.1a), in terms of the norm of the iterated kernels, become
|
λ
1
|
≤···≤
K
3
K
4
≤
K
2
K
3
≤
K
K
2
. (5.3.1b)
Now consider the question of finding the lower bounds for the lowest eigenvalue λ
1
.
Consider the expansion of the symmetric kernel,
K(x, y) ≈
n
φ
n
(x)φ
n
(y)λ
n
. (5.3.3)
This expression is an equation in the mean, and hence there is no guarantee that it
is true at any point as an exact equation. In particular, on the line y = x which has
zero measure in the square 0 ≤ x, y ≤ h, it need not be true. The equality
K(x, x) =
n
φ
n
(x)φ
n
(x)λ
n
need not be true. Hence
h
0
K(x, x)dx =
n
1λ
n
(5.3.4)
need not be true. The right-hand side of Eq. (5.3.4) may not converge.
However, for
K
2
(x, y) =
h
0
K(x, z)K(z, y)dz =
n
φ
n
(x)φ
n
(y)λ
2
n
,
168 5 Hilbert–Schmidt Theory of Symmetric Kernel
since we know K(x, y) to be square-integrable,
A
2
=
h
0
K
2
(x, x)dx =
n
1λ
2
n
(5.3.5)
must converge, and, in general, for m ≥ 2, we have
A
m
=
n
1λ
m
n
, m = 2, 3, ..., (5.3.6)
and we know that the right-hand side of Eq. (5.3.6) converges.
Consider now the expansion for A
2
,namely
A
2
=
1
λ
2
1
+
1
λ
2
2
+
1
λ
2
3
+···=
1
λ
2
1
1 +
λ
1
λ
2
2
+
λ
1
λ
3
2
+···
≥
1
λ
2
1
,
i.e.,
λ
2
1
≥ 1A
2
. (5.3.7)
Hence we have a lower bound for the eigenvalue λ
1
,
|
λ
1
|
≥ 1
A
2
,or
|
λ
1
|
≥ 1
K
. (5.3.8)
This is consistent with our early discussion of the series solution to the Fredholm
integral equation of the second kind for which we concluded that for
|
λ
|
< 1
K
, (5.3.9)
there are no singularities in λ, so that the first eigenvalue λ = λ
1
must satisfy
inequality (5.3.8).
We can obtain better lower bounds for the eigenvalue λ
1
.ConsiderA
4
,
A
4
=
1
λ
4
1
+
1
λ
4
2
+
1
λ
4
3
+···=
1
λ
4
1
1 +
λ
1
λ
2
4
+
λ
1
λ
3
4
+···
≥
1
λ
4
1
,
i.e.,
|
λ
1
|
≥
1
(
A
4
)
1/4
. (5.3.10)
This is an improvement over the previously established lower bound since we know
from Eqs. (5.3.1a) and (5.3.8) that
1A
1/2
2
≤
|
λ
1
|
≤ (A
2
A
4
)
1/2
,
5.4 Rayleigh Quotient 169
so that
A
4
≤ A
2
2
,
i.e.,
1
(
A
4
)
1/4
≥ 1/
(
A
2
)
1/2
.
Thus 1
(
A
4
)
1/4
is a better lower bound than 1
(
A
2
)
1/2
.
Proceeding in the same way with A
6
, A
8
,..., we get better and better lower
bounds,
1
(
A
2
)
1/2
≤ 1
(
A
4
)
1/4
≤ 1
(
A
6
)
1/6
≤···≤
|
λ
1
|
. (5.3.11)
Putting both the upper bounds (5.3.1a) and lower bounds (5.3.11) together, we have
for the smallest eigenvalue λ
1
,
1
(
A
2
)
1/2
≤
1
(
A
4
)
1/4
≤
1
(
A
6
)
1/6
≤···≤
|
λ
1
|
≤···
≤
A
6
A
8
1/2
≤
A
4
A
6
1/2
≤
A
2
A
4
1/2
. (5.3.12)
Strength permitting, we calculate A
6
, A
8
,..., to obtain better and better upper
bounds and lower bounds from Eq. (5.3.12).
5.4
Rayleigh Quotient
Another useful technique for finding the upper bounds for eigenvalues of self-
adjoint operators is based on the Rayleigh quotient. Consider the self-adjoint integral
operator,
˜
K =
h
0
dyK(x, y)withK(x , y) = K(y, x), (5.4.1)
with eigenvalues λ
n
and eigenfunctions φ
n
(x),
˜
Kφ
n
= (1λ
n
)φ
n
,
(φ
n
, φ
m
) = δ
nm
,
(5.4.2)
where the eigenvalues are ordered such that
|
λ
1
|
≤
|
λ
2
|
≤
|
λ
3
|
≤···.
170 5 Hilbert–Schmidt Theory of Symmetric Kernel
Consider any given function g(x)suchthat
g
= 0. (5.4.3)
Consider the series
n
b
n
φ
n
(
x
)
with b
n
= (φ
n
, g). (5.4.4)
This series expansion is the projection of g
(
x
)
onto the space spanned by the set
{φ
n
(x)}
n
, which may not be complete.
We can easily verify the Bessel inequality, which says
n
b
2
n
≤
g
2
. (5.4.5)
Proof of the Bessel inequality: Start with
g −
n
b
n
φ
n
2
≥ 0,
which implies
(g, g) −
m
b
m
(g, φ
m
) −
n
b
n
(φ
n
, g) +
n
m
b
n
b
m
(φ
n
, φ
m
) ≥ 0.
Thus we have (g, g) −
m
b
2
m
≥ 0, which states
n
b
2
n
≤
g
2
, completing the proof
of the Bessel inequality (5.4.5).
Now, consider the quadratic form (g,
˜
Kg),
(g,
˜
Kg) =
h
0
dxg(x)
˜
Kg(x) =
h
0
dx
h
0
dyg(x)K(x, y)g(y). (5.4.6)
Substituting the expansion
K(x, y) ≈
n
φ
n
(x)φ
n
(y)λ
n
into Eq. (5.4.6), we obtain
(g,
˜
Kg) =
h
0
dx
h
0
dyg(x)
n
(φ
n
(x)φ
n
(y)λ
n
)g(y) =
n
b
2
n
λ
n
. (5.4.7)
5.4 Rayleigh Quotient 171
Then, taking the absolute value of the above quadratic form (5.4.7), we obtain
(g,
˜
Kg)
=
n
b
2
n
λ
n
≤
b
2
1
|
λ
1
|
+
b
2
2
|
λ
2
|
+
b
2
3
|
λ
3
|
+···
=
1
|
λ
1
|
b
2
1
+
λ
1
λ
2
b
2
2
+
λ
1
λ
3
b
2
3
+···
≤
1
|
λ
1
|
b
2
1
+ b
2
2
+ b
2
3
+···
≤
1
|
λ
1
|
g
2
. (5.4.8)
Hence, from Eq. (5.4.8), the Rayleigh quotient Q,definedby
Q ≡ (g,
˜
Kg)(g, g), (5.4.9)
is bounded above by
Q
=
(g,
˜
Kg)
g
2
≤
1
|
λ
1
|
,
i.e., the absolute value of the lowest eigenvalue λ
1
is bounded above by
|
λ
1
|
≤ 1
Q
. (5.4.10)
To find a good upper bound on
|
λ
1
|
, choose a trial function g(x)withadjustable
parameters and obtain the minimum of 1
Q
. Namely, we have
|
λ
1
|
≤ min
1
Q
, (5.4.11)
with Q given by Eq. (5.4.9).
Example 5.2. Find an upper bound on the leading eigenvalue of the symmetric
kernel
K(x, y) =
(1 − x)y,0≤ y < x ≤ 1,
(1 − y )x,0≤ x < y ≤ 1,
using the Rayleigh quotient.
Solution. Consider the trial function g(x) = ax which probably is not very good.
We have
(g, g) =
1
0
dxa
2
x
2
= a
2
3,
172 5 Hilbert–Schmidt Theory of Symmetric Kernel
and
(g,
˜
Kg) =
1
0
dx
1
0
dyg(x)K(x, y)g(y)
=
1
0
dxax
x
0
dy(1 − x)yay +
1
x
dy(1 − y)xay
= a
2
1
0
dxx
2
(1 − x
2
)6 = a
2
30.
So, the Rayleigh quotient Q is given by Q = (g,
˜
Kg)(g, g) = 110, and we get
min(1
Q
) = 10. Thus, from Eq. (5.4.11), we obtain
|
λ
1
|
≤ 10,
which is a reasonable upper bound. The exact value of λ
1
turns out to be
λ
1
= π
2
≈ 9.8696,
so it is not too bad, considering that the eigenfunction for λ
1
turns out to be
A sin(π x), not well approximated by ax.
5.5
Completeness of Sturm–Liouville Eigenfunctions
Consider the Sturm–Liouville eigenvalue problem,
d
dx
[p(x)
d
dx
φ(x)] − q(x)φ(x) = λr(x)φ(x)on[0,h], (5.5.1)
with φ(0) = φ(h) = 0, and p(x) > 0, r(x) > 0, for x ∈ [0, h]. We proved earlier that
using Green’s function G(x, y)definedby
d
dx
p(x)
d
dx
G(x, y)
− q(x)G(x, y) = δ(x −y), (5.5.2)
with G(0, y) = G(h, y) = 0, Eq. (5.5.1) is equivalent to the integral equation
φ
(
x
)
= λ
h
0
G(y, x)r(y)φ(y)dy. (5.5.3)
Since the Sturm–Liouville operator is self-adjoint and symmetric, we have a
symmetric Green’s function, G(x, y) = G(y, x). Now, define
ψ
(
x
)
=
r(x)φ
(
x
)
, K(x, y) =
r(x)G(x, y)
r(y). (5.5.4)
5.5 Completeness of Sturm–Liouville Eigenfunctions 173
Then ψ
(
x
)
satisfies
ψ
(
x
)
= λ
h
0
K(x, y)ψ(y)dy, (5.5.5)
which has a symmetric kernel. Applying the Hilbert–Schmidt theorem, we know
that K(x, y) defined above is decomposable in the form
K(x, y) ≈
n
ψ
n
(x)ψ
n
(y)λ
n
=
n
r(x)r(y)φ
n
(x)φ
n
(y)λ
n
, (5.5.6)
with λ
n
real and discrete,andtheset{ψ
n
(x)}
n
orthonormal. Namely,
h
0
ψ
n
(x)ψ
m
(x)dx =
h
0
r(x)φ
n
(x)φ
m
(x)dx = δ
nm
. (5.5.7)
Note the appearance of the weight function r(x) in the middle equation of Eq. (5.5.7).
To prove the completeness, we will establish that any function f (x) can be expanded
in a series of {ψ
n
(x)}
n
or {φ
n
(x)}
n
.Letusdothisforthedifferentiable case (which
is stronger than square-integrable), i.e., assume that f (x) is differentiable. As such,
given any f (x), we can define g(x)by
g(x) ≡ Lf (x), (5.5.8)
where
L =
d
dx
p(x)
d
dx
− q(x). (5.5.9)
Taking the inner product of both sides of Eq. (5.5.8) with G(x, y), we get
(G(x, y), g(x)) = f (y), i.e.,
f (x) =
h
0
G(x, y)g(y)dy =
h
0
(K(x, y)
r(x)r(y))g(y)dy. (5.5.10)
Substituting expression (5.5.6) for K(x, y) into Eq. (5.5.10), we obtain
f (x) =
h
0
dy
n
(φ
n
(x)φ
n
(y)λ
n
)g(y)
=
n
(φ
n
(x)λ
n
)(φ
n
, g) =
n
(β
n
λ
n
)φ
n
(x), (5.5.11)
whereweset
β
n
≡ (φ
n
, g) =
h
0
φ
n
(y)g(y)dy. (5.5.12)
174 5 Hilbert–Schmidt Theory of Symmetric Kernel
Since the above expansion of f (x)intermsofφ
n
(x)istrueforanyf (x), this
demonstrates that the set {φ
n
(x)}
n
is complete.
Actually, in addition, we must require f (x )tosatisfythehomogeneous boundary
conditions in order to avoid boundary terms. Also, we must make sure that the
kernel for the Sturm–Liouville eigenvalue problem is square-integrable.Sincethe
set {φ
n
(x)}
n
is complete, we conclude that there must be an infinite number of
eigenvalues for Sturm–Liouville system. Also, it is possible to prove the asymptotic
results, λ
n
= O(n
2
)asn →∞.
5.6
Generalization of Hilbert–Schmidt Theory
In this section, we consider the generalization of Hilbert–Schmidt theory.
Direction 1: So f ar in our discussion of Hilbert–Schmidt theory, we assumed that
K(x, y) is real. It is straightforward to extend to the case when K(x, y)iscomplex.
We define the norm
K
of the kernel K(x, y)by
K
2
=
h
0
dx
h
0
dy
K(x, y)
2
. (5.6.1)
The iteration series solution to the Fredholm integral equation of the second
kind converges for
|
λ
|
< 1
K
. Also, the Fredholm theory still remains valid. If
K(x, y) is, in addition, self-adjoint, i.e., K(x, y) = K
∗
(y, x), then the Hilbert–Schmidt
expansion holds in the form
K(x, y) ≈
n
φ
n
(x)φ
∗
n
(y)λ
n
, (5.6.2)
where
h
0
φ
∗
n
(x)φ
m
(x)dx = δ
nm
and λ
n
= real, n integer.
Direction 2:Wenotethatinallthediscussionsofar,thevariablex is restricted to a
finite basic interval, x ∈ [0, h]. We extend the basic interval [0, h ]to[0,∞). We want
to solve the following integral equation:
φ
(
x
)
= f
(
x
)
+ λ
+∞
0
K(x, y)φ(y)dy, (5.6.3)
with
+∞
0
dx
+∞
0
dyK
2
(x, y) < ∞,
+∞
0
dxf
2
(x) < ∞. (5.6.4)
5.6 Generalization of Hilbert–Schmidt Theory 175
By a change of the independent variable x, it is always possible to transform the
interval [0, ∞)ofx into [0, h]oft, i.e., x ∈ [0, ∞) ⇒ t ∈ [0, h ]. For example, the
following transformation will do:
x = g(t) = t(h − t). (5.6.5)
Then, writing
˜
φ(t ) = φ(g(t)), etc.,
we have
˜
φ
(
t
)
=
˜
f
(
t
)
+ λ
h
0
˜
K(t, t
)
˜
φ
t
g
(t
)dt
.
On multiplying by
g
(t) on both sides of the above equation, we have
g
(t)
˜
φ(t) =
g
(t)
˜
f (t) + λ
h
0
g
(t)
˜
K(t, t
)
g
(t
)
g
(t
)
˜
φ(t
)dt
.
Defining ψ
(
t
)
by
ψ
(
t
)
=
g
(t)
˜
φ(t ),
we obtain
ψ
(
t
)
=
g
(t)
˜
f (t) +λ
h
0
g
(t)
˜
K(t, t
)
g
(t
)
ψ
t
dt
. (5.6.6)
If the original kernel K(x, y) is symmetric, then the transformed kernel is also
symmetric. Furthermore, the transformed kernel
g
(t)
˜
K(t, t
)
g
(t
) and the trans-
formed inhomogeneous term
g
(t)
˜
f (t) are square-integrable if K(x , y)andf (x)are
square-integrable, since
h
0
dt
h
0
dt
g
(t)
˜
K
2
(t, t
)g
(t
) =
+∞
0
dx
+∞
0
dyK
2
(x, y) < ∞, (5.6.7a)
and
h
0
dtg
(t)
˜
f
2
(t) =
+∞
0
dxf
2
(x) < ∞. (5.6.7b)
Thus, under appropriate conditions, the Fredholm theory and the Hilbert–Schmidt
theory both apply to Eq. (5.6.3). Similarly, we can extend these theories to the case
of infinite range.