Computation of Maxwell's Transmission Eigenvalues and its Application in Inverse Medium Problems
Transcript of Computation of Maxwell's Transmission Eigenvalues and its Application in Inverse Medium Problems
Finite Element Method for TE Numerical Examples Application
Computation of Maxwell’s Transmission Eigenvalues andits Application in Inverse Medium Problems
Jiguang Sun
In collaboration with L. Xu, Chongqing University.
Novel Directions in Inverse Scattering, Jul. 29 - Aug. 2, 2013
Honoring David Colton
Funded in part by NSF under grant DMS-1016092
Finite Element Method for TE Numerical Examples Application
1 Finite Element Method for TE
2 Numerical Examples
3 Application
Finite Element Method for TE Numerical Examples Application
Maxwell’s transmission eigenvalues
In terms of electric fields, the transmission eigenvalue problem for theMaxwell’s equations can be formulated as the following (see [Kirsch 2009]).
Definition
A value of k2 6= 0 is called a transmission eigenvalue if there existreal-valued fields E ,E0 ∈ (L2(D))3 with E − E0 ∈ H0(curl2;D) such that
curl curl E − k2NE = 0, in D, (1a)
curl curl E0 − k2E0 = 0, in D, (1b)
ν × E = ν × E0, on ∂D, (1c)
ν × curl E = ν × curl E0, on ∂D. (1d)
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Researchers
Cakoni
Colton
Gintides
Haddar
Kirsch
Paivarinta
Monk
Sleeman
Sylvester
.......
Special Issue on Transmission Eigenvalues, Inverse Problems
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TE on balls
Now suppose that N = N0I for some constant N0. Then the solutions ofthe Maxwell’s equations on a ball are given by
Mu = curl xu, Nu =1
ikcurl Mu, n ≥ 1,
Mv = curl xv, Nv =1
ikcurl Mv, n ≥ 1,
where u = jn(kr)Ymn (x) and v = jn(kr
√N0)Ym
n (x), jn is the sphericalBessel’s function of order n and Ym
n is the spherical harmonic (see, e.g.,[Colton and Kress 1998]), and r = |x |.
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TE on balls (continued)
For TE mode, to satisfy the boundary conditions, the wave number k2’sneed to satisfy∣∣∣∣ jn(kr) jn(kr
√N0)
1r∂∂ρ (rjn(kr)) 1
r∂∂r
(rjn(kr
√N0)) ∣∣∣∣ = 0, n ≥ 1. (2)
For TM mode, the wave number k2’s need to satisfy [Monk and S. 2012]∣∣∣∣ 1r∂∂r (rjn(kr)) 1
r∂∂r
(rjn(kr
√N0))
k2jn(kr) k2N0jn(kr√N0)
∣∣∣∣ = 0, n ≥ 1. (3)
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Distribution of transmission eigenvalues
Real axis
Imagin
ary
axis
0 5 10 15 20 25 30 35 40 45 50−4
−3
−2
−1
0
1
2
3
4
0.5
1
1.5
2
2.5
Figure : The plot of determinant for n = 1 of the TM mode.
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The fourth order problem
The fourth order formulation [Paivarinta-Sylvester 08, Cakoni-Haddar 09]
(∇×∇×−k2N)(N − I )−1(∇×∇×−k2)(E − E0) = 0.
Setting τ := k2 and u = E − E0, we obtain a variation formulation for thetransmission eigenvalue problem: find τ ∈ C and u ∈ H0(curl2,D) suchthat
Aτ (u, v)− τB(u, v) = 0 ∀ v ∈ H0(curl2,D)
where
Aτ (u, v) =((N − I )−1(∇×∇× u − τu), (∇×∇× v − τv)
)+ τ2(u, v)
andB(u, v) = (∇× u,∇× v).
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An algebraic equation
If (N − I )−1 is a bounded positive definite matrix field on D, Aτ is acoercive Hermitian sesquilinear form on H0(curl2,D)× H0(curl2,D).Furthermore, the sesquilinear form B is Hermitian and non-negative. Thisleads us to consider the auxiliary eigenvalue problem for fixed τ
Aτ (u, v)− λ(τ)B(u, v) = 0 ∀ v ∈ H0(curl2,D). (4)
Note that the generalized eigenvalue λ(τ) depends on τ since Aτ dependson τ . Then the smallest transmission eigenvalue is the first positive root ofthe function
f (τ) := λ(τ)− τ (5)
where λ(τ) is the smallest generalized eigenvalue of (4).
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A mixed finite element method
The problem Aτ (u, v)− λ(τ)B(u, v) = 0 corresponds to
(∇×∇×−τ)(N − I )−1(∇×∇×−τ)w + τ2w = λ∇×∇× w . (6)
Letting u = w and v = (N − I )−1(∇×∇×−τ)u, we obtain
(∇×∇×−τ)v + τ2u = λ∇×∇× u, (7)
(∇×∇×−τ)u = (N − I )v . (8)
Find (λ, u, v) ∈ (R,H0(curl,D),H(curl,D)) such that
(∇× v ,∇× ξ)− τ(v , ξ) + τ2(u, ξ) = λ(∇× u,∇× ξ), (9)
(∇× u,∇× φ)− τ(u,φ) = ((N − I )v ,φ), (10)
for all ξ ∈ H0(curl,D) and φ ∈ H(curl,D).
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The edge element
Sh = the space of lowest order edge element on D,
S0h = Sh ∩ H0(curl,D)
= the subspace of functions in Sh that have vanishing DoF on ∂D,
where DoF stands for degree of freedom. Let ψ1, . . . , ψK be a basis for S0h
and ψ1, . . . , ψK , ψK+1, . . . , ψT be a basis for Sh. Let uh =∑K
i=1 uiψi and
vh =∑T
i=1 uiψi . Furthermore, let ~u = (u1, . . . , uK )T and~v = (v1, . . . , vT )T . Then the matrix form corresponding to the aboveproblem is
SK×T~v − τMK×T~v + τ2MK×K~u = λhSK×K~u, (11)
ST×K~u− τMT×K~u = MN−IT×T~v. (12)
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The matrix form
From (12) we obtain
~v =(MN−I
T×T
)−1(ST×K − τMT×K )~u.
A~u = λSK×K~u (13)
where
A =
((SK×T − τMK×T )
(MN−I
T×T
)−1(ST×K − τMT×K ) + τ2MK×K
).
Alternatively,(τ2MK×K SK×T − τMK×T
ST×K − τMT×K −MN−IT×T
)(~u~v
)= λ
(SK×K 0
0 0
)(~u~v
).
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AlgorithmS: (secant method)
generate a regular tetrahedra mesh for D
set it = 1 and δ =abs(x1 − x0)
compute the smallest generalized eigenvalue λA of (4) for τ = x0
compute the smallest generalized eigenvalue λB of (4) for τ = x1
while δ > tol and it < maxit
τ = x1 − λB x1−x0λB−λA
compute the smallest eigenvalue λτ of Ax = λBx
δ = abs(λτ − τ)
x0 = x1, x1 = τ, λA = λB , λB = λτ , it = it + 1.
end
Here x0 and x1 are initial values which are chosen close to zero.
Finite Element Method for TE Numerical Examples Application
1 Finite Element Method for TE
2 Numerical Examples
3 Application
Finite Element Method for TE Numerical Examples Application
Numerical examples:
Considering N1,N2 and N3 given by 16 0 00 16 00 0 16
,
16 1 01 16 00 0 14
,
16 x yx 16 zy z 14
.
The eigenvalues of N1 are 16 with multiplicity 3.
The eigenvalues of N2 are 14, 15, 17.
For the case of N3, N∗ = 13.5698 and N∗ = 17 for the unit ball andN∗ = 13.2679 and N∗ = 17.5616 for the unit cube.
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Plots of f (τ) = λ(τ)− τ v.s. τ
0.5 1 1.5 2 2.5 3 3.5 4 4.5 5−3
−2
−1
0
1
2
3
4
τ:=k2
f(τ):
=λ(τ
)−τ
ball, N1
ball, N2
ball, N3
cube, N1
cube, N2
cube, N3
Figure : The plot of f (τ) = λ(τ)− τ v.s. τ for D1 and D2 with N1, N2, and N3.
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Computed transmission eigenvalues
Table : The computed smallest Maxwell’s transmission eigenvalues of the unitball and the unit cube together with the number of iterations used in the secantmethod.
domain k1 number of iterations
unit ball N1 1.1837 4
unit ball N2 1.1702 4
unit ball N3 1.1952 4
unit cube N1 2.0595 4
unit cube N2 2.0411 4
unit cube N3 2.0527 4
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The convergence rate
−3.2 −3 −2.8 −2.6 −2.4 −2.2 −2 −1.8
−4.6
−4.4
−4.2
−4
−3.8
−3.6
−3.4
−3.2
−3
−2.8
−2.6
log(DoF−1/3
)
log(E
rror)
Figure : The plot of the error for the the computed smallest transmissioneigenvalue for the unit ball for N = 16I .
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More examples (A)
N = diag(16, 15, x) (14)
with x changing from 12 to 14.
14 14.5 15 15.5 16 16.5 17 17.5 181.26
1.28
1.3
1.32
1.34
1.36
1.38
N(3,3)
τ1 :
= k
12
12 12.2 12.4 12.6 12.8 13 13.2 13.4 13.6 13.8 1412
13
14
15
16
17
18
N(3,3)
λ
1
λ2
λ3
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More examples (B)
N :=
16 x zx 16 yz y 16
(15)
with x changing from 1 to 6, y from 6 to 1, and z from 2 to 4.
2 4 6 8 10 12 14 16 18 201.139
1.14
1.141
1.142
1.143
1.144
1.145
1.146
1.147
1.148
τ1 :
= k
12
1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 610
12
14
16
18
20
22
24
26
λ1
λ2
λ3
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1 Finite Element Method for TE
2 Numerical Examples
3 Application
Finite Element Method for TE Numerical Examples Application
The far field operator
We define the far field operator F : L2(Ω)→ L2(Ω):
(Fv)(x) =
∫Γu∞(y , x)v(y)ds(y). (16)
The linear sampling method is related to the following linear ill-posedintegral equation, for z in a sampling domain
(Fv)(x) = Φ∞(x , z). (17)
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Fundamental theorem for the linear sampling method
Assume that k2 is NOT a Transmission eigenvalue for D. Let F be thefar-field operator defined.
(a) If z ∈ D then there exist a sequence vn, such that
limn→∞
Fvn = Φ∞(·, z). (18)
Furthermore, vn converges in H1(D).
(b) If z ∈ Ω \ D then for every sequence vn, such that
limn→∞
Fvn = Φ∞(·, z) (19)
we have thatlimn→∞
‖vn‖H1(D) =∞.
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Determination of transmission eigenvalues
What happens if k2 is an eigenvalue? For Helmholtz case, Cakoni et al.show that [Cakoni-Colton-Haddar 2011]
Theorem
Let k be a transmission eigenvalue and assume that
limn→∞
F(u, vn) = Φ∞(·, z). (20)
Then for almost every z ∈ D, ‖vn‖H1(D) cannot be bounded as n→∞.
If we choose a point z inside D and plot the norms of the kernels of theregularized solutions against k , we would expect the norms are relativelylarge when k is a transmission eigenvalue and relatively small when k isnot a transmission eigenvalue.
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Transmission eigenvalues for a disk
1 1.5 2 2.5 3 3.5 40
0.5
1
1.5
2
2.5
3
3.5
wavenumber k
norm
of th
e H
erg
lotz
kern
el
Figure : The plot of ‖gz‖L2(S1) against k for a point (0.2, 0.2) inside the D. HereD is a disk with radius 1/2 and the index of refraction n(x) = 16. The exactlowest transmission eigenvalues are 1.99, 2.61, 3.22, . . ..
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Transmission eigenvalues for the unit square
1 1.5 2 2.5 3 3.5 40
0.5
1
1.5
2
2.5
3
wavenumber k
norm
of th
e H
erg
lotz
kern
el
Figure : The plot of ‖gz‖L2(S1) against k for a point inside the D. Here D is theunit square and the index of refraction n(x) = 16. The exact lowest transmissioneigenvalue is 1.89.
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An optimization scheme
Now we suppose that the first transmission eigenvalue kδ1 is reconstructedfrom scattering data.Let µD : L∞(D)→ R which maps a given index of refraction n to thelowest transmission eigenvalue of D, i.e.
µD(n) = k1(D). (21)
Assuming kδ1 (D) is obtained using Cauchy data, we seek a constant n0
minimizing the difference between µD(n) and kδ1 (D), i.e.,
n0 = argminn|µD(n)− kδ1 (D)|. (22)
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Lemma
(Cakoni et. al. 2010) The function µD is a differentiable function of n.Moreover, denoting τ := k2, if f (τ, n) := µ1(nτ2)− (n + 1)τ , then ∂f
∂τ < 0when τ < n+1
2n λ0(D) where λ0(D) is the first Dirichlet eigenvalue of thenegative Laplacian in D.
2 4 6 8 10 12 14 160
2
4
6
8
10
12
14
16
18
20
Index of refraction n
The low
est tr
ansm
issio
n e
igenvalu
e
disk with radius 1/2
unit square
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AlgorithmN n0 = algorithmN(D, kδ1 , tol)
generate a regular triangular mesh for D
estimate an interval [a, b]
compute ka1 and kb1while abs(a− b) > tol
c = (a + b)/2 and compute kc1if |kc1 − kδ1 | < |ka1 − kδ1 |
a = c
else
b=c
end
end
n0 = c
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Estimation of the index of refraction: 2D
Faber-Krahn type inequality:
k21 (D) >
λ0(D)
supD n(x), sup
Dn(x) >
λ0(D)
k21 (D)
(23)
where λ0(D) is the first Dirichlet eigenvalue.
Table : Estimation of the index of refraction. The last column is computed usingthe Faber-Krahn type inequality.
domain D exact n n0 lower bound
disk r = 1/2 centered at (0, 0) 16 16.40 5.80
(−1/2, 1/2)× (−1/2, 1/2) 16 18.30 6.37
disk r = 1/2 centered at (0, 0) 8 + 4|x | 9.33 3.00
(−1/2, 1/2)× (−1/2, 1/2) 8 + x1 − x2 7.87 2.35
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Estimation of the index of refraction: 3D
Look for N0 such that N0I gives kδ1 on D.
Table : The reconstructed index of refraction Ne .
domain Ne N∗ N∗
unit ball N2 14.66 14 17
unit ball N3 15.19 13.57 17
unit cube N2 15.38 14 17
unit cube N3 15.51 13.27 17.56
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The reconstructed N0 and the eigenvalues of N(x)
12 12.2 12.4 12.6 12.8 13 13.2 13.4 13.6 13.8 1412
13
14
15
16
17
18
N(3,3)
λ
1
λ2
λ3
N0
16 16.2 16.4 16.6 16.8 17 17.2 17.4 17.6 17.8 1814.5
15
15.5
16
16.5
17
17.5
18
N(3,3)
λ
1
λ2
λ3
N0
15 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9 1614.8
15
15.2
15.4
15.6
15.8
16
16.2
N(3,3)
λ
1
λ2
λ3
N0
2 4 6 8 10 12 14 16 18 2010
12
14
16
18
20
22
λ1
λ2
λ3
N0
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Thank you!