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Electromagnetism of Continuous MediaMathematical Modelling and Applications$

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

(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)
A u = λ u ,
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

A u c u u H .

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

A = sup f H A f f
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

A f n A f = A ( f n f ) M f n f .

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

g n = f n n f n
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

( A u , v ) = ( u , A v ) u , v H .

Proof. By the Cauchy–Schwarz inequality, we have

| ( A u , v ) | A u v M v u ,
(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
( A u , v ) = ( u , w ) u H .

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

A v 2 = ( A A v , v ) A A v v A A v v ,
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

( A u , v ) = ( u , A v ) u , v H .

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

λ ( u , u ) = ( A u , u ) = ( u , A u ) * = ( A u , u ) * = λ * ( u , u ) ,
whence λ = λ*. Second, the eigenvectors associated with distinct eigenvalues are mutually orthogonal. For, let
A u 1 = λ 1 u 1 , A u 2 = λ 2 u 2 , λ 1 λ 2 .

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

λ 1 ( u 1 , u 2 ) = ( A u 1 , u 2 ) = ( u 1 , A u 2 ) = λ 2 ( u 1 , u 2 ) ,

By

( λ 1 λ 2 ) ( u 1 , u 2 ) = 0
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

A u n k A u n h 0 a s n k , n h .

(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

A u n k + 1 > A u n k + 1

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

( A u n k A u n h , u n k u n h ) 0

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)
( A u n k A u n h , u n k u n h ) u n k u n h A u n k A u n h 2 M A u n k A u n h
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

m ( A u , u ) u 2 M .

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

(C.3)
m = inf u = 1 ( A u , u ) , M = sup u = 1 ( A u , u ) ,
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

( M u M A u M u M ) = 0.

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

0 M u 2 ( A u , u ) = ( M u A u , u ) .

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

0 = d ( M u A u , u ) d t | t = 0 = 2 M ( u M , φ ) 2 ( A u M , φ ) ,
whence
( M u M A u M , φ ) = 0

The arbitrariness of φ ∈ H implies that

A u M = M u M .

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

( M u A u , u ) > 0 u 0
(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
| ( M v A v , v ) | 2 ( M v A v , v ) ( M w A w , w ) .

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

( M u n A u n , u n ) 0
as n → ∞. Accordingly, letting ν = un and w = Mun - Aun, we have
M u n A u n 2 ( | M | + A ) ( M u n A u n , u n ) ,
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,
lim n k u n k = u .

Since A is bounded, and hence continuous,

lim n k A u n k = A u .

Accordingly, since Aun - Mun → 0, we have

A u = M u .

The analogous proof holds for m. □

Let λ0 be the eigenvalue satisfying

| λ 0 | = sup { | m | | M | } = sup u H | ( A u , u ) | u 2 .

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

| λ 1 | = sup u H 0 | ( A u , u ) | u 2 .

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

| λ 0 | | λ 1 | | λ n | .

(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

A = | λ 0 | .

Proof. By the definition of ║A║ we have

(C.4)
| λ 0 | sup h H A u u u 2 A .

For any two vectors u 1, u 2, it follows that

| ( A u 1 , u 2 ) | 1 4 ( A ( u 1 + u 2 ) , u 1 + u 2 ) ( A ( u 1 u 2 ) , u 1 u 2 ) 1 4 ( | λ 0 | u 1 + u 2 2 + u 1 u 2 2 ) 1 2 | λ 0 | ( u 1 2 + u 2 2 ) .

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

A u 2 1 2 | λ 0 | ( | λ 0 | u 1 2 + | λ 0 | 1 A u 2 ) ,
whence
A u | λ 0 | u .

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

u n u m 2 = 2 m n .

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,

A w = n = 1 λ n ( w , u n ) u n .

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

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

w n = i = 1 n ( w , u i ) u i .

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

( w w n , u i ) = 0 , i = 1 , ... , n .

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

A φ n λ n φ n .

In view of Proposition C.7, it follows that

lim n A φ n = 0 ,
whence
A w = n = 1 λ n ( w , u n ) u n .

By use of Bessel's inequality, we have

k = m n ( w , u k ) u k 2 = k = m n ( w , u k ) 2 w 2 .

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

v = n = 1 ( w , u n ) u n .

Now suppose that

A w = 0 w = 0 ;
since Aν = Aw, we have ν = w. Hence the eigenvectors {un} are a basis in that, for every wH, we have
w = n = 1 ( w , u n ) u n .

(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

i ω ε E 0 = × H 0 , i ω ε H 0 = × E 0 , x Ω , E 0 ( x ) × n ( x ) = 0 x Ω .

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

(C.5)
μ 1 × ( ε 1 × H 0 ) ω 2 H 0 = 0 ,

(C.6)
ε 1 × ( μ 1 × E 0 ) ω 2 E 0 = 0.

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

× ( × w ) ω 2 μ ε w = 0.

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

w + k 2 w = 0 ,
where
k 2 = ω 2 ε μ .

In addition,

w × n = 0 or ( × w ) × n = 0
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

w x x + k 2 w = 0.

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

w = 1 2 b 0 + n = 1 b n cos ( n π x / a ) + n = 1 c n sin ( n π x / a ) ,
where
b n = 2 a 0 a w cos ( n π x / a ) d x , c n = 2 a 0 a w sin ( n π x / a ) d x .

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

ω n 2 = n 2 π 2 / a 2 ε μ .

C.2.2 The rectangle

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

w x x + w y y + k 2 w = 0.

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

w ( x , y ) = m = 1 α m ( y ) sin ( m π x / a )
and, upon substitution, we solve for αm(y). We obtain that the admissible values of k are given by
k 2 = m 2 π 2 a 2 + n 2 π 2 b 2 , m = 1 , 2 , ... , n = 1 , 2 , ...

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

w = m = 1 n = 1 c m n sin ( m π x / a ) sin ( n π x / b ) ,
where
c m n = 4 a b 0 a 0 b w ( x , y ) sin ( m π x / a ) sin ( n π y / b ) d x d y .

(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

w = 1 4 b 00 + m = 1 { b m 0 cos ( m π x / a ) + b 0 m cos ( m π y / b ) } , + m = 1 n = 1 b m n cos ( m π x / a ) cos ( n π y / b ) ,
where
b m n = 4 a b 0 a 0 b w ( x , y ) cos ( m π x / a ) cos ( n π y / b ) d x d y .

If, instead, the boundary conditions are

w = 0 at x = 0 , a ; d w / d y = 0 at y = 0 , b
then
w = m = 1 n = 1 ( b 0 m + b n m ) sin ( m π x / a ) cos ( n π y / b ) ,
where
b m n = 4 a b 0 a 0 b w ( x , y ) cos ( n π y / b ) sin ( m π x / a ) d x d y , m = 1 , 2 , ... , n = 0 , 1 , ...

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

k m n 2 = m 2 π 2 a 2 + n 2 π 2 b 2 = ω n m 2 ε μ .

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

w ( 0 , y ) = 0 , d w ( a , y ) / d y = 0 ; w ( x , 0 ) = 0 , w ( w , b ) = 0.

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

k m n p 2 = m 2 π 2 a 2 + n 2 π 2 b 2 + p 2 π 2 c 2 , m , n , p .

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

1 r ( r w r ) r + 1 r 2 u φ φ + k 2 w = 0
and is required to be bounded in Ω. Expand u(r, φ) in a Fourier series with respect to φ, parametrized by r,
w ( r , φ ) = m = α m ( r ) exp ( i m φ ) ,
where
α m ( r ) = 1 2 π 0 2 π w ( r , φ ) exp ( i m φ ) d φ .

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

1 r d d r ( r d α m d r ) + ( k 2 m 2 r 2 ) α m = 0.

Letting ξ = kr, we have

ξ 2 α m + ξ α m + ( ξ 2 + m 2 ) α m = 0 ,
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,
α m = c m J m ( k r ) + d m Y m ( k r ) .

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,

J m ( ξ m n ) = 0 ,
(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
w = n = 1 c m n exp ( i m φ ) ( ξ m n τ / a ) ,
where
c m n = 1 a 2 π m + 1 2 ( ξ m n ) 0 a 0 2 π w ( r , φ ) exp ( i m φ ) J m ( ξ m n r / a ) r d r d φ .

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

c m = 2 a 2 J m + 1 2 ( ξ m n ) 0 a r f ( r ) J m ( ξ m n r / a ) d r ,
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

d d r J m ( k r ) = 0 at r = a .

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

w = m = n = 1 d m n exp ( i m φ ) J m ( η m n r / a ) ,
where
d m n = 1 π a 2 J m 2 ( ξ m n ) 0 a 0 2 π w ( r , φ ) exp ( i m φ ) J m ( η m n r / a ) r d r d φ
and d m1 = 0 unless m = 0. A series of the form m = 1 d n J m ( η m n r / a ) , where
d n = 2 π a 2 J m 2 ( η m n ) 0 a r f ( r ) J m ( η m n r / a ) d r ,
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

1 r 2 r ( r 2 w r ) + 1 r 2 sin θ θ ( sin θ w θ ) + 1 r 2 sin 2 θ 2 w φ 2 + k 2 w = 0.

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

w ( r , θ , φ ) = m = α m ( r , θ ) exp ( i m φ ) ,
where
α m ( r , θ ) = 1 2 π 0 2 π w ( r , θ , φ ) exp ( i m φ ) d φ

Substitution shows that αm must satisfy the differential equation

r ( r 2 α m r ) + 1 sin θ θ ( sin θ α m θ ) + ( k 2 r 2 m 2 sin 2 θ ) α m = 0.

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

1 R m r ( r R m r ) + k 2 r 2 + 1 Θ m sin θ θ ( sin θ Θ m θ ) m 2 sin 2 θ = 0.

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

1 R m r ( r R m r ) + k 2 r 2 = λ , + 1 Θ m sin θ θ ( sin θ Θ m θ ) m 2 sin 2 θ = λ .

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

d d x [ ( 1 x 2 ) d Θ m d x ] + ( λ m 2 1 x 2 ) Θ m = 0 , x ( 1 , 1 ) .

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

The Legendre polynomials

P n ( x ) = 1 2 n n ! d n d x n ( x 2 1 ) n
(p.644) are the solutions to the equation
d d x [ ( 1 x 2 ) dP n d x ] + n ( n + 1 ) P n = 0 ,
bounded at x = ±1 and satisfying Pn(1) = 1, and hence they are eigenfunctions of
d d x [ ( 1 x 2 ) d y d x ] + λ y
with eigenvalue λ = n(n + 1), for every n ∈ N. We then recognize that
P n ( m ) ( x ) = ( 1 x 2 ) m / 2 d m P n d x m
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

r 2 Y m + r Y m + [ k 2 r 2 ( n + 1 / 2 ) 2 ] Y m = 0 ,
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
R n ( r ) = j n ( k r ) : = ( π / 2 k r ) 1 / 2 J n + 1 / 2 ( k r ) .

In conclusion, the eigenfunctions have the form

j n ( k r ) P n m ( cos θ ) exp ( i m φ ) , n , m n .

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

k 2 = ( ς n + 1 / 2 , v / a ) 2 , w m n j n ( ς n + 1 / 2 , v r / a ) P n m ( cos θ ) exp ( i m φ ) , n , m n .

Notes:

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