The eigenvalue problem for a bianisotropic cavity

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1 The eigenvalue problem for a bianisotropic cavity Eftychia Argyropoulou, Andreas D.Ioannidis National and Kapodistrian University of Athens, Greece Linnaeus University, Sweden PIERS 2013, Stockholm, Sweden Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

2 Formulation Let Ω be filled with a bianisotropic material. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

3 Formulation Let Ω be filled with a bianisotropic material. Bianisotropic materials: optically active, meaning they can twist the polarization of light in either refraction or transmission. This is a consequence of the fact that electric and magnetic fields are coupled. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

4 Formulation Let Ω be filled with a bianisotropic material. Bianisotropic materials: optically active, meaning they can twist the polarization of light in either refraction or transmission. This is a consequence of the fact that electric and magnetic fields are coupled. We consider the most general case of a linear medium where the constitutive relations are D = εe + ξh, (1) B = ζe + µh. (2) Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

5 Formulation Let Ω be filled with a bianisotropic material. Bianisotropic materials: optically active, meaning they can twist the polarization of light in either refraction or transmission. This is a consequence of the fact that electric and magnetic fields are coupled. We consider the most general case of a linear medium where the constitutive relations are D = εe + ξh, (1) B = ζe + µh. (2) ε, ξ, ζ and µ: 3 3 matrices, having as entries complex functions of the position vector r and the angular frequency ω. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

6 Formulation In the six vector notation, the problem is stated as follows: [ ] [ ] ε ξ 0 curl iω e = e, (3) ζ µ curl 0 and can be considered as a generalized eigenvalue problem, where ω acts as the eigenvalue and e := (E, H) T as the eigenvector. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

7 Formulation In the six vector notation, the problem is stated as follows: [ ] [ ] ε ξ 0 curl iω e = e, (3) ζ µ curl 0 and can be considered as a generalized eigenvalue problem, where ω acts as the eigenvalue and e := (E, H) T as the eigenvector. Remark: The matrix in the left hand side depends on the eigenvalue ω. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

8 Formulation The formal eigenvalue problem under consideration is written as Qe = ωm(ω)e. ( ) Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

9 Formulation The formal eigenvalue problem under consideration is written as Qe = ωm(ω)e. ( ) where is the material matrix and M = M(ω) := [ ] ε ξ, ζ µ Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

10 Formulation The formal eigenvalue problem under consideration is written as Qe = ωm(ω)e. ( ) where M = M(ω) := [ ] ε ξ, ζ µ is the material matrix and [ ] 0 curl Q := i curl 0 is the (formally self-adjoint) Maxwell operator. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

11 The Mathematical Problem We are now going to formulate ( ) in a Hilbert space setting. We focus in the case of a cavity, and more precisely, Assumption 1 Ω is a bounded Lipschitz domain in R 3. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

12 The Mathematical Problem We are now going to formulate ( ) in a Hilbert space setting. We focus in the case of a cavity, and more precisely, Assumption 1 Ω is a bounded Lipschitz domain in R 3. We further assume that the wall Γ := Ω is perfect conducting, Assumption 2 ˆn E = 0 on Γ. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

13 The Mathematical Problem The domain of Q is D(Q) := H 0 (curl ; Ω) H(curl ; Ω). and X := L 2 (Ω; C 3 ) L 2 (Ω; C 2 ). Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

14 The Mathematical Problem The domain of Q is D(Q) := H 0 (curl ; Ω) H(curl ; Ω). and X := L 2 (Ω; C 3 ) L 2 (Ω; C 2 ). Proposition 1 Q is a selfadjoint operator. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

15 The Mathematical Problem The domain of Q is D(Q) := H 0 (curl ; Ω) H(curl ; Ω). and X := L 2 (Ω; C 3 ) L 2 (Ω; C 2 ). Proposition 1 Q is a selfadjoint operator. Assumption 3 The entries of ε, ξ, ζ and µ are L (Ω) functions (with respect to r). Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

16 The Mathematical Problem The domain of Q is D(Q) := H 0 (curl ; Ω) H(curl ; Ω). and X := L 2 (Ω; C 3 ) L 2 (Ω; C 2 ). Proposition 1 Q is a selfadjoint operator. Assumption 3 The entries of ε, ξ, ζ and µ are L (Ω) functions (with respect to r). Corollary 1 M defines a bounded multiplication operator on X. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

17 The Mathematical Problem The domain of Q is D(Q) := H 0 (curl ; Ω) H(curl ; Ω). and X := L 2 (Ω; C 3 ) L 2 (Ω; C 2 ). Proposition 1 Q is a selfadjoint operator. Assumption 3 The entries of ε, ξ, ζ and µ are L (Ω) functions (with respect to r). Corollary 1 M defines a bounded multiplication operator on X. Proposition 2 H 1 (Ω) L 2 (Ω) with a compact injection. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

18 The Mathematical Problem The spectrum of Q is discrete and consists of a sequence of eigenvalues with no accumulation point. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

19 The Mathematical Problem The spectrum of Q is discrete and consists of a sequence of eigenvalues with no accumulation point. Actually, the non-zero eigenvalues of Q, appear as a bilateral sequence (ω 0 n) n Z. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

20 The Mathematical Problem The spectrum of Q is discrete and consists of a sequence of eigenvalues with no accumulation point. Actually, the non-zero eigenvalues of Q, appear as a bilateral sequence (ω 0 n) n Z. (ω 0 n) n N is an increasing sequence of non-negative numbers, thus diverging at infinity, and ω 0 n = ω 0 n. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

21 The Mathematical Problem The spectrum of Q is discrete and consists of a sequence of eigenvalues with no accumulation point. Actually, the non-zero eigenvalues of Q, appear as a bilateral sequence (ω 0 n) n Z. (ω 0 n) n N is an increasing sequence of non-negative numbers, thus diverging at infinity, and ω 0 n = ω 0 n. Each ω 0 n, n 0, is counted as many times as its multiplicity. ω0 0 := 0 is always an eigenvalue but needs a special treatment, since the kernel ker Q is infinite dimensional. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

22 The Mathematical Problem Let us now focus on the corresponding eigenvectors. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

23 The Mathematical Problem Let us now focus on the corresponding eigenvectors. To each ω 0 n, n = 1, 2,..., corresponds one normalized eigenvector e n := (E n, H n ) T, which is obtained by solving an eigenvalue problem for the negative vector Laplacian. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

24 The Mathematical Problem Let us now focus on the corresponding eigenvectors. To each ω 0 n, n = 1, 2,..., corresponds one normalized eigenvector e n := (E n, H n ) T, which is obtained by solving an eigenvalue problem for the negative vector Laplacian. The eigenvector for ω 0 n = ω 0 n is then given by e 0 n = (E n, H n ) T. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

25 The Mathematical Problem Let us now focus on the corresponding eigenvectors. To each ω 0 n, n = 1, 2,..., corresponds one normalized eigenvector e n := (E n, H n ) T, which is obtained by solving an eigenvalue problem for the negative vector Laplacian. The eigenvector for ω 0 n = ω 0 n is then given by e 0 n = (E n, H n ) T. Remark: Eigenvectors corresponding to different eigenvalues are orthogonal, so (e 0 n) n Z can be chosen as an orthonormal sequence. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

26 The Mathematical Problem Let now H := [..., e n,..., e 2, e 1, e 1, e 2,..., e n,...]. The restriction of Q on H is denoted by Q H. Q H is both the restriction and the part of Q on H. Actually, it is the spectrum of Q H that is discrete and one can strictly prove that. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

27 The Mathematical Problem Let now H := [..., e n,..., e 2, e 1, e 1, e 2,..., e n,...]. The restriction of Q on H is denoted by Q H. Q H is both the restriction and the part of Q on H. Actually, it is the spectrum of Q H that is discrete and one can strictly prove that. Proposition 3 Q H is a self-adjoint operator in H and has compact inverse Q 1 H. The sequence of eigenvalues of Q H is (ωn) 0 n Z and the sequence of the corresponding eigenvectors is (en) 0 n Z. The latter is an orthonormal basis for H. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

28 The Mathematical Problem Then ( ) is restricted in H and is written e = ωq 1 H M(ω)e. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

29 The Mathematical Problem Then ( ) is restricted in H and is written e = ωq 1 H M(ω)e. Let F(ω) := ωq 1 H M(ω); we then conclude to the eigenvalue problem for a linear pencil (I F(ω))e = 0 (4) Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

30 The Mathematical Problem Then ( ) is restricted in H and is written e = ωq 1 H M(ω)e. Let F(ω) := ωq 1 H M(ω); we then conclude to the eigenvalue problem for a linear pencil (I F(ω))e = 0 (4) F(ω): is a compact operator in X. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

31 The abstract framework Consider an open and connected set D C and an operator-valued function F : D B(X ). Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

32 The abstract framework Consider an open and connected set D C and an operator-valued function F : D B(X ). Definition 2 An ω is called eigenvalue of the pencil I F ( ) if the equation F (ω)x = x has non trivial solutions. A non trivial solution of F (ω)x = x ω S, is called an eigenvector corresponding to ω and the linear span of the eigenvectors is called the eigenspace corresponding to ω. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

33 The abstract framework We have the following important result Proposition 4 (Analytic Fredholm Alternative) Let F be analytic and F (ω) K(X ) for all ω D. Then either a) I F (ω) is not injective for every ω D, or b) (I F (ω)) 1 B(X ) for all ω D\S, where S C is a countable set without any limit point. Remark: In case (b), the operator-valued function (I F ( )) 1 is analytic in D\S, meromorphic in D and the residues at the poles are finite rank operators. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

34 The abstract framework We now focus in the special case F (ω) := AB(ω), ω D, where A is compact self-adjoint and B(ω) B(X ). We treat the equation in an abstract sense in an arbitrary seperable Hilbert space X. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

35 The abstract framework We now focus in the special case F (ω) := AB(ω), ω D, where A is compact self-adjoint and B(ω) B(X ). We treat the equation in an abstract sense in an arbitrary seperable Hilbert space X. The spectral theorem ensures that A is represented as Ax = n λ n x, e 0 n e 0 n, (5) where (λ n ) is the sequence of (non-zero real) eigenvalues of A, in an absolutely descending order and counted as many times as their multiplicity, and (e 0 n) is the sequence of corresponding eigenvectors. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

36 The abstract framework We now focus in the special case F (ω) := AB(ω), ω D, where A is compact self-adjoint and B(ω) B(X ). We treat the equation in an abstract sense in an arbitrary seperable Hilbert space X. The spectral theorem ensures that A is represented as Ax = n λ n x, e 0 n e 0 n, (5) where (λ n ) is the sequence of (non-zero real) eigenvalues of A, in an absolutely descending order and counted as many times as their multiplicity, and (e 0 n) is the sequence of corresponding eigenvectors. The latter is an orthonormal basis for R(A). Here we assume that A has infinitely many eigenvalues and thus λ n 0. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

37 The abstract framework The problem F (ω)x = x, i.e., AB(ω)x = x, since x R(A), can now be written as Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

38 The abstract framework The problem F (ω)x = x, i.e., AB(ω)x = x, since x R(A), can now be written as x, e 0 n e 0 n, or, equivalently, as n λ n B(ω)x, e 0 n e 0 n = n Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

39 The abstract framework The problem F (ω)x = x, i.e., AB(ω)x = x, since x R(A), can now be written as or, equivalently, as λ n B(ω)x, e 0 n e 0 n = n n n x, e 0 n e 0 n, ( λ n x, B(ω) 1 ) I en 0 en 0 = 0. (6) λ n Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

40 The abstract framework The problem F (ω)x = x, i.e., AB(ω)x = x, since x R(A), can now be written as or, equivalently, as Let f n = f n (ω) := λ n B(ω)x, e 0 n e 0 n = n n (6) is a multiplier operator n x, e 0 n e 0 n, ( λ n x, B(ω) 1 ) I en 0 en 0 = 0. (6) λ n ( B(ω) 1 λ n I ) e 0 n = ( ) B(ω) 1 λ n I en. 0 The LHS of S = S(ω) := n λ n, f n e 0 n, corresponding to sequences (λ n ) R, (f n ) X, (e n ) X, where (λ n ) is bounded, (f n ) is a sequence and (e n ) is an orthonormal basis. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

41 The abstract framework Proposition 5 The following are equivalent: a) S is injective. b) (f n ) is a complete sequence, i.e., [f 1, f 2,..., f n,...] = X. c) x, f n = 0 for every n, implies x = 0. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

42 The abstract framework Proposition 5 The following are equivalent: a) S is injective. b) (f n ) is a complete sequence, i.e., [f 1, f 2,..., f n,...] = X. c) x, f n = 0 for every n, implies x = 0. Corollary 3 ω is an eigenvalue of I F ( ) if and only if the sequence (f n (ω)) is not complete. The corresponding eigenspace is ker S(ω). Moreover, x ker S(ω) if and only if x, f n (ω) = 0 for every n and, consequently, ker S(ω) = [f 1 (ω), f 2 (ω),..., f n (ω),...]. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

43 The inverse problem Let us now consider the inverse problem, that is to reconstruct the operator B( ) from the knowledge of the eigenelements of the problem F (ω)x = x. Actually, let us assume that ω is an eigenvalue with corresponding eigenspace ker S(ω). Then ker S(ω) T = [f 1 (ω), f 2 (ω),..., f n (ω),...], and assume that there is a way to calculate the sequence (f n (ω)). Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

44 The inverse problem Let us now consider the inverse problem, that is to reconstruct the operator B( ) from the knowledge of the eigenelements of the problem F (ω)x = x. Actually, let us assume that ω is an eigenvalue with corresponding eigenspace ker S(ω). Then ker S(ω) T = [f 1 (ω), f 2 (ω),..., f n (ω),...], and assume that there is a way to calculate the sequence (f n (ω)). The straightforward relation B(ω)e 0 n, e 0 m = e 0 n, f m (ω) + δ nm λ n, (7) (δ nm stands for the Kronecker delta) allows the recovery of the operator B(ω). Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

45 Eigenfrequencies and modes of the bianisotropic cavity We now return to problem where F(ω) := ωq 1 H M(ω). (I F(ω))e = 0 (8) Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

46 Eigenfrequencies and modes of the bianisotropic cavity We now return to problem where F(ω) := ωq 1 H M(ω). Assumption 4 (I F(ω))e = 0 (8) M is an analytic function D ω M(ω) B(X), where D is a domain in the complex plane, such that 0 D. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

47 Eigenfrequencies and modes of the bianisotropic cavity We now return to problem where F(ω) := ωq 1 H M(ω). Assumption 4 (I F(ω))e = 0 (8) M is an analytic function D ω M(ω) B(X), where D is a domain in the complex plane, such that 0 D. Consequently, F defines an analytic function D ω F(ω) K(X). Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

48 Eigenfrequencies and modes of the bianisotropic cavity We now return to problem where F(ω) := ωq 1 H M(ω). Assumption 4 (I F(ω))e = 0 (8) M is an analytic function D ω M(ω) B(X), where D is a domain in the complex plane, such that 0 D. Consequently, F defines an analytic function D ω F(ω) K(X). Moreover, for ω 0 = 0, F(ω 0 ) = 0 and thus I F(ω 0 ) is invertible. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

49 Eigenfrequencies and modes of the bianisotropic cavity The Analytic Fredholm Alternative then gives Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

50 Eigenfrequencies and modes of the bianisotropic cavity The Analytic Fredholm Alternative then gives Proposition 6 The pencil I F( ) has countably many eigenvalues with finite dimensional corresponding eigenspaces. They form a sequence (ω n ) of non-zero complex numbers, diverging at infinity. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

51 Eigenfrequencies and modes of the bianisotropic cavity The Analytic Fredholm Alternative then gives Proposition 6 The pencil I F( ) has countably many eigenvalues with finite dimensional corresponding eigenspaces. They form a sequence (ω n ) of non-zero complex numbers, diverging at infinity. ω n : an eigenfrequency of the cavity The above result reveals that such frequencies exist and are countably many. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

52 Eigenfrequencies and modes of the bianisotropic cavity The Analytic Fredholm Alternative then gives Proposition 6 The pencil I F( ) has countably many eigenvalues with finite dimensional corresponding eigenspaces. They form a sequence (ω n ) of non-zero complex numbers, diverging at infinity. ω n : an eigenfrequency of the cavity The above result reveals that such frequencies exist and are countably many. The eigenvectors corresponding to an eigenfrequency ω n are called the corresponding modes. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

53 Eigenfrequencies and modes of the bianisotropic cavity Let now A := Q 1 H, B(ω) := ωm(ω). The eigenvalues of A are calculated as follows λ n := 1 ω 0 n, n = ±1, ±2,..., and the sequence of the corresponding eigenvectors is again (e n ). Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

54 Eigenfrequencies and modes of the bianisotropic cavity Let now A := Q 1 H, B(ω) := ωm(ω). The eigenvalues of A are calculated as follows λ n := 1 ω 0 n, n = ±1, ±2,..., and the sequence of the corresponding eigenvectors is again (e n ). Consequently, one has the expansion Ax = λ n x, e 0 n e 0 n. n= Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

55 Eigenfrequencies and modes of the bianisotropic cavity Let now A := Q 1 H, B(ω) := ωm(ω). The eigenvalues of A are calculated as follows λ n := 1 ω 0 n, n = ±1, ±2,..., and the sequence of the corresponding eigenvectors is again (e n ). Consequently, one has the expansion Ax = λ n x, e 0 n e 0 n. n= Now F(ω) = AB(ω) and it is reformulated with the aim of the multiplier operator Sx = S(ω)x := λ n x, f n e 0 n, where f n = f n (ω) := n= ( B(ω) 1 ) I e 0 n = ( ωm(ω) ω λ ni 0 ) e 0 n, n = ±1, ±2,... n Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

56 Eigenfrequencies and modes of the bianisotropic cavity Proposition 7 ω 0 is an eigenfrequency of the cavity if and only if S(ω) is not an injective operator, if and only if (f n (ω) n Z is not complete. The corresponding subspace of modes is finite dimensional and is given by ker S(ω) = [,..., f n (ω),..., f 2 (ω), f 1 (ω), f 1 (ω), f 2 (ω),..., f n (ω),...]. Moreover, the following equality applies M(ω)e 0 n, e 0 m = ω 0 n e 0 n, f m (ω) + δ nm, from which the material matrix can be recovered. Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

57 Thank you! Eftychia Argyropoulou (UoA) The eigenvalue problem for a bianisotropic cavity August / 21

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