A numerical method to compute electromagnetic ...

Introduction and motivation Solving the interior transmission eigenvalue problem Numerical results Summary and outlook

A numerical method to compute electromagnetic interior transmission eigenvalues Andreas Kleefeld Brandenburg Technical University Cottbus - Senftenberg Faculty 1 (Mathematics, Natural Sciences, Computer Science) P.O. Box 101344 03013 Cottbus Germany

September 18th , 2015 in Berlin, Germany Workshop Recent Developments in Inverse Problems

Kleefeld, Andreas

Transmission eigenvalues

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Table of Contents

1

Introduction and motivation

2

Solving the interior transmission eigenvalue problem

3

Numerical results

4

Summary and outlook

Kleefeld, Andreas


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Motivation

Is there an incident electric field that does not scatter?

Interior transmission eigenvalues κ1 , κ2 , κ3 , . . . for a homogeneous component are different from a component with an inhomogeneity.

Medical imaging

Nondestructive testing

Kleefeld, Andreas


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Setup R3 ν

D bounded open region in R3 . Boundary Γ consists of a finite number of disjoint, closed, bounded surfaces belonging to class C 2 .

D

¯ is connected. R3 \D κ given wave number.

ν denotes normal pointing in the exterior.

Γ

N is the relative electric permittivity (we assume N = nI with n > 1).

Kleefeld, Andreas


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Scattering by an inhomogeneous media Solve:

curl curl E s − κ 2 E s = 0

in R3 \D ,

curl curl E − κ 2 n E = 0 in D , s i on Γ , E +E ×ν = E ×ν s i curl E + E × ν = curl E × ν on Γ , r = |x | . lim (curl E s × x − iκ r E s ) = 0 ,

r →∞

Total electric field is E = E s + E i with incident electric field E i . Question: Is there an incident electric field that does not scatter?

Kleefeld, Andreas


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Interior transmission eigenvalue problem Question is related to the interior transmission eigenvalue problem (ITEP). If E i is given such that E s = 0, then setting E := E |D and E0 := E i |D yields the following problem: Find a solution (E , E0 ) 6= (0, 0) to the ITEP given by

curl curl E0 − κ 2 E0 = 0

curl curl E − κ n E = 0 2

in D , in D ,

on Γ , E × ν = E0 × ν curl E × ν = curl E0 × ν on Γ .

Then κ ∈ C will be an interior transmission eigenvalue (ITE). Kleefeld, Andreas


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Interior transmission eigenvalue problem

ITEP was introduced by Kirsch in 1986 and Colton & Monk in 1988.

Discreteness of ITEs: Colton-Kirsch-Päivärinta in 1989, Rynne-Sleeman in 1991, Cakoni-Haddar and Colton-Päivärinta-Sylvester in 2007, Kirsch, Cakoni-Haddar in 2009 and Hickmann in 2012. Existence of ITEs: Päivärinta-Sylvester and Kirsch in 2009, Cakoni-Gintides-Haddar, Cakoni-Haddar and Cakoni-Kirsch in 2011, and Bellis-Cakoni-Guzina and Cossonnière in 2011. And many more.

Kleefeld, Andreas


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Interior transmission eigenvalue problem

ITEs carry information of material properties ⇒ Can be used to say something about the presence of abnormalities inside an homogeneous media.

⇒ Can be used to test the integrity of a material (nondestructive testing).

Many open questions. ⇒ What does the first ITE tell us about the inhomogeneous media N (x )? ⇒ Do complex-valued ITEs exist? ⇒ How to compute ITEs efficiently?

Kleefeld, Andreas


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Solving the ITEP — Method I Integral operators in D:

(SLκ ψ ) (x ) (DLκ ψ ) (x )

= curl curl

Z

Γ

= curl

Z

Γ

ψ (y ) Φκ (x , y ) ds(y ) ,

ψ (y ) Φκ (x , y ) ds(y ) .

Integral operators on Γ:

(Sκ ψ ) (x ) (Kκ ψ ) (x )

= curl curl

Z

Γ

= curl

Z

Γ

ψ (y ) Φκ (x , y ) ds(y ) × ν (x ) ,

ψ (y ) Φκ (x , y ) ds(y ) × ν (x ) .

Here, Φκ (p, q ) = eiκ r /4π r with r = |p − q | and p 6= q.

Kleefeld, Andreas


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Solving the ITEP — Method I Use boundary integral equations (see Cossonnière & Haddar 2013, Cakoni & Haddar 2015). Using the Stratton-Chu formulas yields E = DLκ √n M + E0 = DLκ M +

1

κ 2n

1

κ2

SLκ √n J ,

in D ,

SLκ J ,

in D ,

with M = E × ν |Γ = E0 × ν |Γ and J = curl E × ν |Γ = curl E0 × ν |Γ . On the boundary holds

0 = E × ν − E0 × ν = Kκ √n − Kκ M −

1

κ 2n

Sκ √n −

1

κ

S 2 κ

J = 0.

Additionally,

0 = curl E × ν − curl E0 × ν = Sκ √n − Sκ M + Kκ √n − Kκ J = 0 . Kleefeld, Andreas


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Solving the ITEP — Method I

Leads to Z(κ ) with Z(κ ) =

M J

Sκ √n − Sκ Kκ √n − Kκ

=

0 0

on Γ

Kκ √n − Kκ 1 S √ − κ12 Sκ κ2n κ n

−1/2

1/ 2

Z(κ ) : Ht (Γ) × H −3/2,−1/2 (div, Γ) → Ht Fredholm of index zero.

.

(Γ) × H −1/2,1/2 (curl, Γ) is

Z(κ ) is analytic on C\R− . −1/2

Z(iκ ) is strictly coercive in Ht κ > 0.

Kleefeld, Andreas

−3/2,−1/2

(Γ) × H0


(div, Γ) for real

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Solving the ITEP — Method I Discretize Z(κ )X = 0. Compute eigenvalues of Z(κ ) ∈ Cm×m and look for κ for which the smallest eigenvalue is close to zero. Problem: Eigenvalues cluster around zero. Workaround: Solve generalized eigenvalue problem Z(κ )X = λ Z(iκ )X . Has to be calculated for a lot of wave numbers, say N = 200 (at least). Generalized eigenvalue problem has to be solved for large m.

⇒ Method is expensive if one is interested in many highly accurate ITE values.

Kleefeld, Andreas


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Solving the ITEP — Method I 50 45 40 35

1

1/|λ |

30 25 20 15 10 5 0 2.5 2.6 2.7 2.8 2.9

3

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 κ

4

Figure 1: Cossonnière’s method for an ellipsoidal-shaped obstacle using n = 4. Kleefeld, Andreas


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Solving the ITEP — Method II (new)

Consider the nonlinear eigenvalue problem of the form Z(κ )v = 0,

v ∈ Cm ,

v 6= 0,

κ ∈ Ω ⊂ C.

Assume large scale problem k m (k is number of eigenvalues including multiplicities). Problem can be reduced to a linear eigenvalue problem of dimension k (Keldysh’s theorem). One has to use complex-valued contour integrals. See article by W.-J. Beyn 2012.

Kleefeld, Andreas


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Solving the ITEP — Method II Choose a contour ∂ Ω in C which might contain eigenvalues. For example an ellipse ψ (t ) = µ + a cos(t ) + b sin(t )i . Then ψ 0 (t ) = −a sin(t ) + b cos(t )i Algorithm:

ˆ ∈ Cm×` randomly. 1. Choose an index ` ≤ m and V 2. Evaluate the contour integrals Z 1 ˆ dz A= M(z )−1 V 2π i ∂ Ω

and

B=

Z 1 ˆ dz zM(z )−1 V 2π i ∂ Ω

numerically with the trapezoidal rule. With tj = 2Nπ j , j = 0, . . . , N (ψ (t0 ) = ψ (TN )) we have AN =

1 N −1 ˆ ψ 0 (tj ) , M(ψ (tj ))−1 V iN j∑ =0

BN =

1 N −1 ˆ ψ (tj )ψ 0 (tj ) . M(ψ (tj ))−1 V iN j∑ =0

N = 50 is more than enough. Kleefeld, Andreas


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Solving the ITEP — Method II 3. Compute SVD AN = V ΣW H . 4. Perform a rank test for Σ, find 0 < k ≤ ` such that

σ1 ≥ . . . ≥ σk > tolrank > σk +1 ≈ 0 ≈ σ` ≈ 0 . If k = `, then increase ` and go to step 1. Otherwise let V0 = V (1 : m, 1 : k ), W0 = W (1 : `, 1 : k ), and Σ0 = diag(σ1 , . . . , σk ) . 1 k ×k . 5. Compute C = V0H BN W0 Σ− 0 ∈C

6. Solve the eigenvalue problem for C .

Eigenvalues are κi with corresponding eigenvectors si , i = 1, . . . , k. Eigenvectors for the nonlinear eigenvalue problem are vi = V0 si , i = 1, . . . , k.

Kleefeld, Andreas


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Numerical results — Unit sphere We choose n = 4 . Unit sphere (centered at the origin) has ITEs:

1

κ1 ≈ 3.141593 , κ2 ≈ 3.492822 , κ3 ≈ 3.592863 , κ4 ≈ 3.692445 , κ5 ≈ 3.902613 , κ6 ≈ 4.261683 , κ7 ≈ 4.430310 , κ8 ≈ 4.831855 , κ9 ≈ 4.979577 . Kleefeld, Andreas

z

0.5

0

−0.5

−1 −1

−1 −0.5

−0.5 0

0 0.5

0.5 1

1

x

y

Figure 2: Unit sphere: x = ρ sin(φ ) cos(θ ), y = ρ sin(φ ) sin(θ ), and z = ρ cos(φ ) with ρ = 1.


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Numerical results — Unit sphere

Im(κ)

0.2

0

−0.2

3

3.1

3.2

3.3

3.4

3.5 Re(κ)

3.6

3.7

3.8

3.9

4

Figure 3: The first five ITEs for the unit sphere. Reference values are κ1 ≈ 3.14159, κ2 ≈ 3.492822, and κ3 ≈ 3.592863.

ITE

κ1 κ2 κ3

calculated ITE (m = 1812) 3.163 3.517 3.617

relative error (m = 1812) 6.8· 10−3 6.9· 10−3 6.7· 10−3 Kleefeld, Andreas

calculated ITE (m = 7710) 3.146 3.498 3.598 Transmission eigenvalues

relative error (m = 7710) 1.4· 10−3 1.5· 10−3 1.4· 10−3 18 / 33


Numerical results — Ellipsoid

1.2 0.8

z

0.4 0 −0.4 −0.8 −1.2 −1

−1 −0.5

−0.5 0

0 0.5

0.5 1

1 x

y

Figure 4: Ellipsoid: x = sin(φ ) cos(θ ), y = sin(φ ) sin(θ ), and z = ρ cos(φ ) with ρ = 6/5. Kleefeld, Andreas


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Im(κ)

0.2

0

−0.2

2.5

2.6

2.7

2.8

2.9

3 Re(κ)

3.1

3.2

3.3

3.4

3.5

Figure 5: The first six ITEs for the ellipsoidal surface

Kleefeld, Andreas


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ITE

κ1 κ2 κ3 κ4 κ5 κ6

calculated ITE (m = 7710) 2.920 3.057 3.233 3.311 3.409 3.422

Old method: Previous results are not accurate, unsatisfactory, and time-consuming to calculate. My method (new): Numerical results are highly accurate and the computational cost is much smaller.

Kleefeld, Andreas


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Numerical results — Peanut

1.5 1

z

0.5 0 −0.5 −1 −1.5 −1

−0.5

0

0.5

1

1

0.5

−0.5

0

−1

x y

Figure 6: Peanut with ρ 2 = 9 cos2 (φ ) + sin2 (φ )/4 /4.

Kleefeld, Andreas


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Im(κ)

0.2

0

−0.2

2.5

2.6

2.7

2.8

2.9

3 Re(κ)

3.1

3.2

3.3

3.4

3.5

Figure 7: The first four ITEs for the peanut-shaped surface

Kleefeld, Andreas


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ITE

κ1 κ2 κ3 κ4

calculated ITE (m = 7710) 3.044 3.302 3.368 3.422

Kleefeld, Andreas


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Numerical results — Acorn

1.5 1

z

0.5 0 −0.5 −1 −1.5 −1.5

−1

−0.5

0

0.5

1

1.5

1.5

1

0.5

0

−0.5

−1

−1.5

x

y

Figure 8: Acorn with ρ 2 = 9 {17/4 + 2 cos(3φ )} /25. Kleefeld, Andreas


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Im(κ)

0.2

0

−0.2

2

2.1

2.2

2.3

2.4

2.5 Re(κ)

2.6

2.7

2.8

2.9

3

Figure 9: The first five ITEs for the acorn-shaped obstacle.

Kleefeld, Andreas


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ITE

κ1 κ2 κ3 κ4 κ5

calculated ITE (m = 7710) 2.634 2.720 2.768 2.842 2.930

Kleefeld, Andreas


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Complex-valued EV for the ellipsoid 0 −0.1 −0.2 −0.3

Im(κ)

−0.4 −0.5 −0.6 −0.7 −0.8 −0.9 −1

2

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 Re(κ)

3

Figure 10: Five complex-valued eigenvalues for the ellipsoid. Contour is a circle with center (5/2, −1/2) and radius 1/2. Values are from left to right 2.19 − 0.55i, 2.22 − 0.60i, 2.32 − 0.71i, 2.50 − 0.70i, and 2.64 − 0.64i. Kleefeld, Andreas


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Exterior transmission eigenvalues Are all complex-valued solution ITEs? No! The following problem (ETP): Find (E , E0 ) 6= (0, 0) such that

curl curl E0 − κ 2 E0 = 0

curl curl E − κ 2 n E = 0 E × ν = E0 × ν

in R3 \D , in R3 \D ,

on Γ ,

curl E × ν = curl E0 × ν on Γ ,

are satisfied. E, E0 satisfy the Silver-Müller radiation condition. Leads to the same system Z(κ )v = 0. Solutions are exterior transmission eigenvalues (ETEs).

Kleefeld, Andreas


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Complex-valued TE Assumption 1: ETEs lie in the lower half-plane.

1 0

Assumption 2: Complex-valued ITEs can be found in the upper half-plane. It holds κ ITE ⇐⇒ κ ITE.

Im(κ)

−1 −2 −3 −4 −5

Old method: Can only calculate real ITEs.

−6 0

1

2

3 Re(κ)

4

5

6

Figure 11: ETEs (red) and complex-valued ITEs (green) for the unit sphere (n = 4). Kleefeld, Andreas

My method (new): Can calculate complex-valued ITEs and ETEs.


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Interior Maxwell eigenvalues EVMax 1. 2. 3. 4.

Unit sphere 2.744 [3] 3.870 [5] 4.493 [3] 4.973 [7]

Ellipsoid 2.536 [2] 2.714 [1] 3.568 [2] 3.636 [1]

Peanut 2.568 [2] 3.039 [1] 3.255 [2] 3.368 [1]

Acorn 1.963 [1] 2.340 [2] 3.072 [2] 3.237 [2]

Table 1: The first four interior Maxwell eigenvalues (EVMax ).

ITE 1. 2. 3. 4.

Unit sphere 3.146 [3] 3.498 [5] 3.598 [3] 3.698 [5]

Ellipsoid 2.920 [2] 3.057 [1] 3.233 [3] 3.311 [2]

Peanut 3.044 [2] 3.302 [2] 3.368 [1] 3.422 [1]

Acorn 2.634 [2] 2.720 [1] 2.768 [1] 2.842 [2]

Table 2: The first four interior transmission eigenvalues in D (n = 4).

Kleefeld, Andreas


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Summary and outlook

Presented an alternative method to calculate ITEs for various surfaces. Results are very accurate with less computational costs. Complex-valued ITEs and ETEs can be calculated. Interior Maxwell eigenvalues can be calculated similarly.

Further investigation is needed for the complex-valued eigenvalues. This method might be useful for estimation of the relative electric permittivity knowing the first real ITE. Extension to the elastic case is also possible.

Kleefeld, Andreas


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References

A. K LEEFELD, The transmission problem for the Helmholtz equation in

R3 , Computational Methods in Applied Mathematics, 12 (2012), pp. 330–350. A. K LEEFELD, A numerical method to compute interior transmission eigenvalues, Inverse Problems, 29 (2013), 104012 (20pp). A. K LEEFELD, Numerical methods for acoustic and electromagnetic scattering: Transmission boundary-value problems, interior transmission eigenvalues, and the factorization method, Habilitation Thesis, May 2015.

Kleefeld, Andreas


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A numerical method to compute electromagnetic ...

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