computation of maxwell's transmission eigenvalues and its application in inverse medium...

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


1 Finite Element Method for TE

2 Numerical Examples

3 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)


Researchers

Cakoni

Colton

Gintides

Haddar

Kirsch

Paivarinta

Monk

Sleeman

Sylvester

.......

Special Issue on Transmission Eigenvalues, Inverse Problems


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 |.


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)


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.


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).


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).


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).


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)


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

).


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.




3 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.


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.


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 cube N1 2.0595 4




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 .


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


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




3 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)


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) =∞.


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.


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, . . ..


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.


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)


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


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


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


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


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


Thank you!

computation of maxwell's transmission eigenvalues and its application in inverse medium...

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