## Mauro Fabrizio and Angelo Morro

Print publication date: 2003

Print ISBN-13: 9780198527008

Published to Oxford Scholarship Online: September 2007

DOI: 10.1093/acprof:oso/9780198527008.001.0001

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# (p.631) C Compact operators and eigenfunctions

Source:
Electromagnetism of Continuous Media
Publisher:
Oxford University Press

# C.1 Compact operators

Consider an abstract formulation of eigenvalue problems. Let

(C.1)
$Display mathematics$
where A : HH is a linear operator on a Hilbert space H. The nonzero solutions u to (C.1) and the corresponding values λ are called eigenvectors and eigenvalues.

Definition C.1 An operator A on H is said to be bounded if there is a positive constant c such that1

$Display mathematics$

The set of bounded linear operators is a linear space; the definition

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makes it a normed space.

Proposition C.1 A bounded linear operator is continuous, namely, if fnf in H then AfnAf in H. The converse also is true.

Proof. Since A is bounded, there is a constant M > 0 such that

$Display mathematics$

Hence, fnf implies AfnAf. The converse is proved by showing that unbounded operators are not continuous. Let fn be a sequence in H such that, for every integer n, we have ║Afn║ > nfn║. Hence, the sequence given by

$Display mathematics$
is such that ║gn║ is as small as we please, while ║Agn║ > 1. □

Proposition C.2 For every bounded linear operator A on H there is a bounded linear operator Asuch that 1

$Display mathematics$

Proof. By the Cauchy–Schwarz inequality, we have

$Display mathematics$
(p.632) which shows that, for every ν ∈ H, the inner product (Au, ν) is a bounded linear functional of u. By the Riesz theorem, there is a vector wH such that
$Display mathematics$

Accordingly, there is an operator A′ : HH and w = A′ν. The operator A′ proves to be linear. Moreover, because

$Display mathematics$
we have ║A′ν║ ≤ ║A║ ║ν║ and hence A′ is bounded. □

Definition C.2 The multiplicity of an eigenvalue λ is the number of independent eigenvectors u pertaining to that eigenvalue.

Definition C.3 An operator A on H is said to be symmetric if

$Display mathematics$

Two properties follow at once. First, the eigenvalues of a symmetric linear operator are real. For, the inner product of Au = λu with u, the symmetry of the inner product, and the symmetry of A yield

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whence λ = λ*. Second, the eigenvectors associated with distinct eigenvalues are mutually orthogonal. For, let
$Display mathematics$

Inner multiplication of the first equation by u 2 and use of the second one yield

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By

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and the assumption λ1 ≠ λ2, the conclusion (u 1, u 2) = 0 follows.

Definition C.4 An operator A on H is said to be compact (or completely continuous) if A transforms bounded sets into compact sets.

In fact, we use compactness through a consequence, namely, that if A is compact then, whenever {un} is a sequence in H with ║un║ < M, the sequence {Aun} contains a subsequence ${ A u n k }$ which converges to some vector in H and hence

$Display mathematics$

(p.633) Proposition C.3 A compact operator A is bounded and continuous.

Proof. Since A is linear, if it is bounded then it is continuous. We prove the boundedness by contradiction. If A is unbounded, there is a sequence {un} with ║un║ = 1 and ║Aun║ → ∞. We can then consider a subsequence ${ u n k }$ such that

$Display mathematics$

This sequence does not contain any Cauchy subsequence, which contradicts the compactness of A. □

Proposition C.4 A compact operator A on H transforms any bounded sequence {un} into a sequence {Aun}, which contains a subsequence ${ A u n k }$ such that

$Display mathematics$

Proof. To prove the result, we consider a bounded sequence, ║un║ < M, and a Cauchy subsequence ${ u n k }$. By means of the Cauchy–Schwarz inequality, we have

(C.2)
$Display mathematics$
whence the conclusion follows. □

Conversely, let A be a linear operator which transforms each bounded sequence {un}, unH, into a sequence {Aun} which contains a subsequence with the property (C.2). Hence (cf. [182]) the operator A is compact.

Proposition C.5 If A is a bounded symmetric operator then the quadratic form (Au, u) is bounded in that there exist two finite numbers m, M such that

$Display mathematics$

Proof. By the boundedness of A, there are two numbers m, M such that

(C.3)
$Display mathematics$
whence the desired result follows. □

(p.634) Proposition C.6 Let A be a bounded linear symmetric operator on H and M, m be given by (C.3). If uH, withu║ = 1, satisfies (Au, u) = M, or (Au, u) = m, then M, or m, is an eigenvalue of A.

Proof. Let u MH satisfy (Au M, u M) = M and hence

$Display mathematics$

By the definition of M, for every uH, we have

$Display mathematics$

Accordingly, the functional (Mu - Au, u) is minimal at u = u M. For every φ ∈ H let u = u M + tφ. We can write

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whence
$Display mathematics$

The arbitrariness of φ ∈ H implies that

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The proof for m proceeds along the same lines. □

Theorem C.1 For any compact symmetric operator A, the numbers m, M of (C.3) are eigenvalues of A.

Proof. By Proposition C.6, we only need to show that there is u MH such that (Mu M - Au M, u M) = Mu M2 -(Au M, u M) = 0. Assume, by contradiction, that no nonzero vector exists with such a property. Then

$Display mathematics$
(p.635) and the bilinear form (Mu - Au, u) is positive definite. We can then view (Mν - Aν, w) as an inner product in H; hence the Cauchy–Schwarz inequality yields
$Display mathematics$

By the definition of M, there is a sequence {un}, with ║un║ = 1, such that

$Display mathematics$
as n → ∞. Accordingly, letting ν = un and w = Mun - Aun, we have
$Display mathematics$
whence Mun - Aun → 0. Because A is compact, there is a subsequence ${ u n k }$ of {un} which converges to a vector, u say, of H, that is,
$Display mathematics$

Since A is bounded, and hence continuous,

$Display mathematics$

Accordingly, since Aun - Mun → 0, we have

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The analogous proof holds for m. □

Let λ0 be the eigenvalue satisfying

$Display mathematics$

Let u 0 be an eigenvector associated with λ0; for simplicity, let ║u 0║ = 1. Consider the space H 0 = {uH; (u, u 0) = 0}. The operator A, restricted to H 0, is still compact. We can then argue as with m and M. Thus, there exists an eigenvalue λ1 such that

$Display mathematics$

Of course, |λ0| ≥ |λ1|. Denote by u 1, ║u 1║ = 1, the eigenvector (or one of the eigenvectors) associated with λ1. By iterating the procedure, we set up a sequence of eigenvalues {λn} with

$Display mathematics$

(p.636) Lemma C.1 Let A be a symmetric compact operator and let0| = supuH |(Au, u)|/║u2 be the eigenvalue whose absolute value is the maximum in question. Then

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Proof. By the definition of ║A║ we have

(C.4)
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For any two vectors u 1, u 2, it follows that

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For any uH, let u 1 = |λ0|1/2 u, u 2 = |λ0|-1/2 Au. We have

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whence
$Display mathematics$

Comparison with (C.4) yields the conclusion. □

Proposition C.7 If H is an infinite-dimensional space then the sequencen} converges to zero.

Proof. We prove the result by contradiction. If λn → λ ≠ 0, then the sequence {unn} is bounded. Since A is compact, the sequence {Aunn} should contain a converging subsequence in H. Now, Aunn = un and

$Display mathematics$

Hence, no subsequence is convergent and the contradiction is found. □

Theorem C.2 The eigenvectors un of a compact operator A constitute an orthogonal system in H such that, for every wH,

$Display mathematics$

Further, if Aw=0 ⇔ w=0 then {un} is a basis for H.

(p.637) Proof. Given any wH, we consider the sequence

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By the orthonormality condition (uj, uk) = δjk, we have

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Hence, the function φn = w - wn is a vector in Hn = {φ ∈ H; (φ, u 1) = 0, …, (φ, un) = 0}. The operator A is well defined on Hn and the norm ║An in Hn equals |λn|. Hence, for any φnHn, we have

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In view of Proposition C.7, it follows that

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whence
$Display mathematics$

By use of Bessel's inequality, we have

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Hence, it follows that $∑ k = m n ( w , u k ) u k$ is a Cauchy sequence and there is a vector ν ∈ H such that

$Display mathematics$

Now suppose that

$Display mathematics$
since Aν = Aw, we have ν = w. Hence the eigenvectors {un} are a basis in that, for every wH, we have
$Display mathematics$

# (p.638) C.2 Eigenfunctions in spatially homogeneous resonators

Consider a resonator occupying a domain Ω ⊂ ℝ3 which is bounded and endowed with a Lipschitzian boundary. We look for time-harmonic solutions E = E 0(x) exp(iωt), H = H 0(x) exp(iωt). Hence, E 0, H 0 are required to satisfy the boundary-value problem

$Display mathematics$

The solutions for (E 0, H 0) : Ω → ℂ3 x ℂ3 are considered in the space $ℋ 0 1 ( Ω ) × ℋ ( Ω )$. Application of the curl operator and comparison yield

(C.5)
$Display mathematics$

(C.6)
$Display mathematics$

If the permittivity ε and the permeability μ are constant, eqns (C.5) and (C.6) take the form

$Display mathematics$

Now, since ▿ · E 0 = 0 (no free charges) and ▿ · H 0 = 0, the identity ▿ x (▿ x w) = ▿(▿ · w) - ▿w yields

$Display mathematics$
where
$Display mathematics$

$Display mathematics$
on the boundary according as w stands for E 0 or H 0. With this in mind, we revisit well-known eigenfunctions associated with various domains Ω (cf. [139–141, 148]).

## C.2.1 The one-dimensional interval

Let Ω = {x; x ∈ (0, a)}. The divergence-free condition implies that the x-component of w is constant, possibly zero. Hence, apart from an inessential constant, the vector w is transverse. Let w be the pertinent component and let the subscript x stand for d/dx; the differential equation reads

$Display mathematics$

(p.639) The boundary conditions become w = 0 or w′ = 0. Any function wL 2(0, a) has the expansion

$Display mathematics$
where
$Display mathematics$

If w = 0 at the end points then bn = 0, n = 0, 1, …. If, instead, w′ = 0 then cn = 0, n = 1, 2, …. In both cases, we have an infinity of eigenvalues kn, that is, $k n 2 = n 2 π 2 / a 2$, whence

$Display mathematics$

## C.2.2 The rectangle

Let Ω = {(x, y); x ∈ (0, a), y ∈ (0, b)}. Any component satisfies

$Display mathematics$

To fix ideas, let w = 0 on the boundary. We start from the expansion

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and, upon substitution, we solve for αm(y). We obtain that the admissible values of k are given by
$Display mathematics$

Hence, the eigenfunctions are sin(mπx/a) sin(nπy/b) and

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where
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(p.640) If ∂w/∂n = 0 on the boundary then constants, cos(nπx/a) and cos(mπy/b) are eigenfunctions, so that the solution is expressed as

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where
$Display mathematics$

If, instead, the boundary conditions are

$Display mathematics$
then
$Display mathematics$
where
$Display mathematics$

The solution for the case when w = 0 at y = 0, a and ∂w/∂x at x = 0, b is obtained by interchanging x and y.

In all cases

$Display mathematics$

The eigenvalues may be multiple; certainly each eigenvalue, with mn, is double if a = b because the pairs m, n and n, m correspond to the same eigenvalue.

The eigenvalues change if the boundary conditions are, for example, of the mixed type in the form

$Display mathematics$

The generalization to the rectangular parallelepiped is purely formal. Of course, the eigenfunctions depend on the boundary conditions. If, for (p.641) example, w vanishes on the boundary then the eigenfunctions take the form sin(mπx/a) sin(nπy/b) sin(pπz/c). The eigenvectors kmnp are given by

$Display mathematics$

## C.2.3 The circle

Let Ω be the circle of radius a and use the polar coordinates r, φ so that Ω = {(r, φ); r ∈ [0, a], φ ∈ [0, 2π]}. The unknown function w(r, φ) satisfies

$Display mathematics$
and is required to be bounded in Ω. Expand u(r, φ) in a Fourier series with respect to φ, parametrized by r,
$Display mathematics$
where
$Display mathematics$

In fact, substitution in the differential equation shows that αm satisfies the differential equation

$Display mathematics$

Letting ξ = kr, we have

$Display mathematics$
where a prime stands for d/dξ; this is Bessel's equation, of order m, for the unknown function αm in the variable kr. Hence,
$Display mathematics$

Since αm must be bounded when r = 0, we take dm = 0.

Let now w = 0 at r = a and let ξmn, n = 1, 2, …, be the zeroes of Jm(ξ), that is,

$Display mathematics$
(p.642) arranged in ascending order. Since ξ = kr, admissible values of k are given by kmna = ξmn. Hence, the eigenfunctions are exp(imφ)Jmmn r/a) and, if w belongs to L 2(Ω) then
$Display mathematics$
where
$Display mathematics$

The eigenvalues k 2 are then the squares of kmn = ξmn/a, namely, the squares of 1/a times the zeroes of Jm.

A series of the type $∑ n = 1 ∞ c n J m ( ξ m n r / a )$ where

$Display mathematics$
is called a Fourier–Bessel series. Since the expansion for w holds if wL 2(Ω) then, by identifying w with f(r) exp(imφ), it follows that the Fourier–Bessel expansion of a function f holds when $∫ 0 a | f | 2 r d r < ∞$.

If, instead, the boundary condition is ∂w/∂r = 0, at r = a, then the functions Jm(kr) must satisfy

$Display mathematics$

Hence the values kmn of k are allowed such that kmn = ηmn/a and ηmn are the zeroes of dJmn/dr; for most values of m, ηm1 = 0. We have

$Display mathematics$
where
$Display mathematics$
and d m1 = 0 unless m = 0. A series of the form $∑ m = 1 ∞ d n J m ( η m n r / a )$, where
$Display mathematics$
with d 1 = 0 when m ≠ 0, is called a Dini series. The Dini expansion of a function f holds when m is an integer and $∫ 0 a | f | 2 r d r$ exists.

## (p.643) C.2.4 The sphere

Let Ω be the sphere of radius a and describe Ω through the spherical polar coordinates r, θ, φ with r ∈ [0, a], θ ∈ [0, π], φ ∈ [0, 2π). The equation ▵w + k 2 w = 0 takes the form

$Display mathematics$

As with the circle, we start from a Fourier series in φ, namely,

$Display mathematics$
where
$Display mathematics$

Substitution shows that αm must satisfy the differential equation

$Display mathematics$

This shows that αm may conveniently be determined by separation of variables, αm(r, θ) = Rm(rm(θ), thus obtaining

$Display mathematics$

This equation holds if and only if there is a constant, λ say, such that

$Display mathematics$

Consider the last equation and observe first that d/d cos θ = -(1/sin θ)d/dθ hence, letting x = cos θ, we have the form

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The unknown function Θm is required to be bounded at x = ±1.

The Legendre polynomials

$Display mathematics$
(p.644) are the solutions to the equation
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bounded at x = ±1 and satisfying Pn(1) = 1, and hence they are eigenfunctions of
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with eigenvalue λ = n(n + 1), for every n ∈ N. We then recognize that
$Display mathematics$
is the solution for Θm, whereas the constant λ takes the values n(n + 1), n ∈ N. The functions $P n ( m )$ are the associated Legendre functions of order m; of course, $P n ( 0 ) ( x ) = P n ( x ) and P n ( m ) ( x ) ≠ 0$ only for mn.

Upon the substitution, Rm(r) = r -1/2 Ym(r), since λ = n(n + 1), we have

$Display mathematics$
where ′ ≡ d/dr. The boundedness at r = 0 implies that Ym is the Bessel function of the first kind of order n + 1/2, Jn+1/2(kr). Hence, we see that R is in fact parametrized by m through n, nm, and write
$Display mathematics$

In conclusion, the eigenfunctions have the form

$Display mathematics$

To determine k, we consider the boundary condition. Let w = 0 at r = a. If ζn+1/2,ν are the zeroes of jn, the admissible values of k are given by ka = ζn+1/2,ν and hence we have found the eigenvalues and the eigenfunctions of the Helmholtz equation for the sphere as

$Display mathematics$

## Notes:

(1) The operator A′ is called adjoint operator of A.