Masujima M. Applied Mathematical Methods in Theoretical Physics

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156 5 Hilbert–Schmidt Theory of Symmetric Kernel

Then



1 −





(x)φ

(x)dx = 0. (5.2.7)

If λ

= λ

,thenwehave



(x)φ

(x)dx = 0forλ

= λ

. (5.2.8)

In the case of ﬁnite matrices, we know that the eigenvalues of a symmetric matrix

are real and that the matrix is diagonalizable. Also, the eigenvectors are orthogonal

to each other. We shall show that the same statements hold true in the case of

square-integrable symmetric kernels.

If K is symmetric, then ψ

(x) = φ

(x). Then, by Eq. (5.2.8), the eigenfunctions

of a symmetric kernel are orthogonal to each other,



(x)φ

(x)dx = 0forλ

= λ

. (5.2.9)

Furthermore, the eigenvalues of a symmetric kernel must be real. This is seen by

supposing that the eigenvalue λ

is complex. Then we have λ

= λ

∗

. The complex

conjugate of Eq. (5.2.4) is given by

∗

(x) = λ

∗



K(x, y)φ

∗

(y)dy, (5.2.10)

implying that λ

∗

and φ

∗

(x) are an eigenvalue and eigenfunction of the kernel

K(x, y), respectively. But, Eq. (5.2.9) with λ

= λ

∗

then requires that



(x)φ

∗

(x)dx =





(x)



dx = 0, (5.2.11)

implying then that φ

(x) ≡ 0, which is a contradiction. Thus the eigenvalue must

be real, λ

= λ

∗

, to avoid this contradiction. The eigenfunctions of a symmetric kernel

are orthogonal to each other and the eigenvalues are real.

We now rather boldly expand the symmetric kernel K(x, y)intermsofφ

(x),

K(x, y) =



(x). (5.2.12)

We then normalize the eigenfunctions such that



(x)φ

(x)dx = δ

(5.2.13)

5.2 Real and Symmetric Kernel 157

(even if there is more than one eigenfunction belonging to a certain eigenvalue,

we can choose linear combinations of these eigenfunctions to satisfy Eq. (5.2.13)).

From the orthogonality (5.2.13), we ﬁnd



dxφ

(x)K(x, y) =

(y), (5.2.14)

and thus obtain

K(x, y) =



(y)φ

(x)

. (5.2.15)

There is a problem though. We do not know if the eigenfunctions {φ

(x)}

are

complete. In fact, we are often sure that the set {φ

(x)}

is surely not complete. An

example is the kernel in the form of a ﬁnite sum of factorized terms. However, the

content of the Hilbert–Schmidt theorem (which will be proven shortly) is to claim

that Eq. (5.2.15) for K(x, y)isvalidwhether{φ

(x)}

is complete or not. The only

conditions are that K(x, y)besymmetric and square-integrable.

We calculate the iterated kernel,

(x, y) =



K(x, z)K(z, y)dz





(x)φ

(z)



(z)φ

(y)



(x)δ

(y) =



(x)φ

(y)

, (5.2.16)

and in general, we obtain

(x, y) =



(x)φ

(y)

, j = 2, 3, .... (5.2.17)

Now the deﬁnition for the resolvent kernel

H (x, y;λ) =−K(x, y) − λK

(x, y) −···−λ

j+1

(x, y) −···

becomes

H (x, y;λ) =−



(x)φ

(y)



1 +

+···+

+···



=−



(x)φ

(y)

1 −

158 5 Hilbert–Schmidt Theory of Symmetric Kernel

i.e.,

H (x, y;λ) =



(x)φ

(y)

λ − λ

. (5.2.18)

This elegant expression explicitly shows the analytic properties of H(x, y;λ)inthe

complex λ plane. We can use this resolvent to solve the inhomogeneous Fredholm

Integral Equation of the second kind with a symmetric and square-integrable kernel.

φ(x) = f (x) + λ



K(x, y)φ(y)dy = f (x) − λ



H (x, y;λ)f (y)dy (5.2.19)

= f (x) − λ



(x)

λ − λ



(y)f (y)dy .

Denoting

≡



(y)f (y)dy , (5.2.20)

we have the solution to the inhomogeneous equation (5.2.19),

φ(x) = f (x) − λ



(x)

λ − λ

. (5.2.21)

At λ = λ

, the solution does not exist unless f

= 0, as usual.

As an another application of the eigenfunction expansion (5.2.15), we consider

the Fredholm Integral Equation of the ﬁrst kind with a symmetric and square-integrable

kernel,

f (x) =



K(x, y)φ(y)dy. (5.2.22)

Denoting

≡



(y)φ(y)dy, (5.2.23)

we have

f (x) =



(x)

. (5.2.24)

Immediately we encounter the problem. Equation (5.2.24) states that f (x)isa

linear combination of φ

(x). In many cases, the set {φ

(x)}

is not complete, and

thus f (x) is not necessarily representable by a linear superposition of {φ

(x)}

and

5.2 Real and Symmetric Kernel 159

Eq. (5.2.22) has no solution. If f (x) is representable by a linear superposition of

{φ

(x)}

, it is easy to obtain φ

. From Eqs. (5.2.20) and (5.2.24),



(x)f (x)dx =

, (5.2.25)

= f

. (5.2.26)

A solution to Eq. (5.2.22) is then given by

φ(x) =



(x) =



(x). (5.2.27)

If the set {φ

(x)}

is not complete, solution (5.2.27) is not unique. We can add to it

any linear combination of {ψ

(x)}

that is orthogonal to {φ

(x)}

φ(x) =



(x) +



(x), (5.2.28)



(x)φ

(x)dx = 0foralli and n. (5.2.29)

If the set {φ

(x)}

is complete, solution (5.2.27) is the unique solution. It may,

however, still diverge since we have λ

in the numerator, unless f

vanishes

sufﬁciently rapidly as n →∞to ensure the convergence of the series (5.2.27).

We will now prove the Hilbert–Schmidt expansion, (5.2.15), to an extent that

everything beautiful about it is exhibited, but to avoid getting too mathematical, we

shall not be completely rigorous.

We will outline a plan of the proof. First, note the following lemma.

Lemma: For a nonzero normed symmetric kernel,

∞ >



> 0 and K(x, y) = K

(x, y), (5.2.30)

there exists at least one eigenvalue λ

and one eigenfunction φ

(x)(which we normalize

to unity).

Once this lemma is established, we can construct a new kernel

K(x, y)by

K(x, y) ≡ K(x, y) −

(x)φ

(y)

. (5.2.31)

Now φ

(x) cannot be an eigenfunction of K(x, y)becausewehave



K(x, y)φ

(y)dy =





K(x, y) −

(x)φ

(y)



(y)dy

(x)

−

(x)

= 0, (5.2.32)

160 5 Hilbert–Schmidt Theory of Symmetric Kernel

whichleavesustwopossibilities,

(A)



≡ 0.

We have an equality

K(x, y) =

(x)φ

(y)

, (5.2.33)

except over a set of points x whose measure is zero. The proof for this case is done.

(B)



= 0.

By the lemma, there exist at least one eigenvalue λ

and one eigenfunction φ

(x)of

akernel

K(x, y),



K(x, y)φ

(y)dy = φ

(x). (5.2.34)

Namely,





K(x, y) −

(x)φ

(y)



(y)dy = φ

(x). (5.2.35)

We then show that φ

(x)andλ

are an eigenfunction and eigenvalue of the original

kernel K(x, y) orthogonal to φ

(x), respectively.

To demonstrate the orthogonality of φ

(x)toφ

(x), multiply φ

(x) in Eq. (5.2.35)

and integrate over x:



(x)φ

(x)dx = λ



(x)dx





K(x, y) −

(x)φ

(y)



(y)dy

= λ





(y) −

(y)



(y)dy = 0. (5.2.36)

From Eq. (5.2.35), we then have



K(x, y)φ

(y)dy = φ

(x). (5.2.37)

Once we ﬁnd φ

(x), we construct a new kernel

K(x, y)by

K(x, y) ≡

K(x, y) −

(x)φ

(y)

= K(x, y) −



n=1

(x)φ

(y)

. (5.2.38)

We then repeat the argument for

K(x, y).

Ultimately either we ﬁnd after N steps,

K(x, y) =



n=1

(x)φ

(y)

, (5.2.39)

5.2 Real and Symmetric Kernel 161

or we ﬁnd the inﬁnite series,

K(x, y) ≈

∞



n=1

(x)φ

(y)

. (5.2.40)

We can show that the remainder R(x, y)deﬁnedby

R(x , y) ≡ K(x, y) −

∞



n=1

(x)φ

(y)

(5.2.41)

cannot have any eigenfunction.Ifψ (x) is the eigenfunction of R(x, y),



R(x , y)ψ(y)dy = ψ(x), (5.2.42)

we know that

(1): ψ(x)isdistinctfromall{φ

(x)}

ψ(x) = φ

(x)forn = 1, 2, ..., (5.2.43)

(2): ψ(x) is orthogonal to all {φ

(x)}



ψ(x)φ

(x)dx = 0forn = 1, 2, .... (5.2.44)

Substituting deﬁnition (5.2.41) of R(x, y) into Eq. (5.2.42) and noting the orthogo-

nality (5.2.44), we ﬁnd



K(x, y)ψ(y)dy = ψ(x), (5.2.45)

which is a contradiction to Eq. (5.2.43). Thus we must have





(x, y)dxdy = 0. (5.2.46)

Formula (5.2.15) holds in the sense of the mean square convergence.

So we only have to prove the Lemma stated with the condition (5.2.30) and the

proof of the Hilbert–Schmidt theorem will be ﬁnished. To do so, it is necessary to

work with the iterated kernel K

(x, y), which is also symmetric:

(x, y) =



K(x, z)K(z, y)dz =



K(x, z)K(y, z)dz. (5.2.47)

162 5 Hilbert–Schmidt Theory of Symmetric Kernel

This is because the trace of K

(x, y) is always positive,



(x, x)dx =



dzK

(x, z) =



> 0. (5.2.48)

First, we will prove that if K

(x, y) has an eigenvalue, then K(x, y) has at least one

eigenvalue equaling one of the square roots of the former.

Recall the deﬁnition of the resolvent kernel of K(x, y),

H (x, y;λ) =−K(x, y) − λK

(x, y) −···−λ

j+1

(x, y) −···, (5.2.49)

H (x, y;−λ) =−K(x, y) + λK

(x, y) −···−(−λ)

j+1

(x, y) −···. (5.2.50)

Taking the difference of Eqs. (5.2.49) and (5.2.50), we ﬁnd

[H(x, y;λ) − H(x, y;−λ)] =−λ[K

(x, y) + λ

(x, y) +···]

= λH

(x, y;λ

), (5.2.51)

which is the resolvent for K

(x, y). Equality (5.2.51), which is valid for sufﬁciently

small λ where the series expansion in λ is deﬁned, holds for all λ by analytic

continuation.

If c is an eigenvalue of K

(x, y), H

(x, y;λ

)hasapoleatλ

= c. From Eq. (5.2.51),

either H(x, y;λ)orH(x, y;−λ) must have a pole at λ =±

√

c. This means that at

least one of ±

√

c is an eigenvalue of K(x, y).

Now we prove that K

(x, y) has at least one eigenvalue. We have



(x, x;s)dxD

(s) =



(x, x;s)dx

=−(A

+ sA

+ s

+···) (5.2.52)

where



(x, x)dx, m = 2, 3, .... (5.2.53)

If K

(x, y) has no eigenvalues, then D

(s) has no zeros, and series (5.2.52) must be

convergent for all values of s. To this end, consider

m+n



dxK

m+n

(x, x) =



dzK

(x, z)K

(z, x)



dzK

(x, z)K

(x, z). (5.2.54)

5.2 Real and Symmetric Kernel 163

Applying the Schwarz inequality,

m+n

≤





dzK

(x, z)





dzK

(x, z)







dxK

(x, x)





dxK

(x, x)



= A

i.e., we have

m+n

≤ A

. (5.2.55)

Setting



m → n − 1,

n → n + 1,

in inequality (5.2.55), we have

≤ A

2n−2

2n+2

. (5.2.56)

Recalling that

> 0, (5.2.57)

which is precisely the reason why we consider K

(x, y), we have

2n−2

≤

2n+2

. (5.2.58)

Successively we have

2n+2

≥

2n−2

≥

2n−2

2n−4

≥···≥

≡ R

, (5.2.59)

so that

= R

, A

≥ R

, A

≥ R

and generally

≥ R

n−1

. (5.2.60)

164 5 Hilbert–Schmidt Theory of Symmetric Kernel

Thus we have

+ sA

+ s

+···≥A

(1 + sR

+ s

+···). (5.2.61)

The right-hand side of inequality (5.2.61) surely diverges for those s such that

s ≥

. (5.2.62)

Thus



(x, x;s)dx

is divergent for those s satisfying inequality (5.2.62). Then K

(x, y)hastheeigenvalue

s satisfying

s ≤

and K(x, y)hastheeigenvalueλ

satisfying

≤



. (5.2.63)

This completes the proof of the Lemma and ﬁnishes the proof of the

Hilbert–Schmidt theorem. 

The Hilbert–Schmidt expansion, (5.2.15), can be helpful in many problems

where a symmetric and square-integrable kernel is involved.

 Example 5.1. Solve the integro-differential equation,

∂

∂t

φ(x, t) =



K(x, y)φ(y, t)dy, (5.2.64a)

with the initial condition

φ(x,0)= f (x), (5.2.64b)

where K(x, y) is symmetric and square-integrable .

Solution. The Hilbert–Schmidt expansion, (5.2.15), can be applied giving

∂

∂t

φ(x, t) =



(x)



(y)φ(y, t)dy. (5.2.65)

5.2 Real and Symmetric Kernel 165

Deﬁning A

(t)by

(t) ≡



(y)φ(y, t)dy, (5.2.66)

and changing the dummy index of summation from n to m in Eq. (5.2.65), we have

∂

∂t

φ(x, t) =



(x)

(t). (5.2.67)

Taking the time derivative of Eq. (5.2.66) yields

(t) =



dyφ

(y)

∂

∂t

φ(y, t). (5.2.68)

Substituting Eq. (5.2.67) into Eq. (5.2.68) and, noting that the orthogonality of

{φ

(x)}

means that only the m = n term is left, we get

(t) =

(t)

. (5.2.69)

The solution to Eq. (5.2.69) is then

(t) = A

(0) exp





. (5.2.70)

Here we note that

(0) =



dxφ

(x)f (x). (5.2.71)

We can now integrate Eq. (5.2.67) from 0 to t,withA

(t) f rom Eq. (5.2.70),



∂

∂t

φ(x, t) =





(x)

(0) exp





. (5.2.72)

The left-hand side of Eq. (5.2.72) is now exact, and we obtain

φ(x, t) − φ(x,0)=



(x)

(0)



exp





− 1









. (5.2.73)

From the initial condition (5.2.64b) and Eq. (5.2.73), we ﬁnally get

φ(x, t) = f (x) +



(0)



exp





− 1



(x). (5.2.74)