Numerical methods for acoustic and electromagnetic ...

Brandenburg University of Technology Cottbus - Senftenberg Institute for Applied Mathematics and Scientific Computing Platz der Deutschen Einheit 1 03046 Cottbus, Germany

Habilitation Thesis

Numerical methods for acoustic and electromagnetic scattering: Transmission boundary-value problems, interior transmission eigenvalues, and the factorization method Andreas Kleefeld May 20, 2015

Accepted by Faculty 1 Mathematics, Natural Sciences and Computer Science

Opening of the habilitation procedure: Decision of the faculty council: Referees: Prof. Dr. rer. nat. habil. Georg Bader Prof. Dr. rer. nat. habil. Andreas Kirsch Prof. Dr. Fioralba Cakoni

November 12, 2014 May 13, 2015

Typeset with LATEX 2ǫ

Contents 1

General introduction

1.1 1.2 1.3 1.4 1.5 1.6 1.7 2

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The transmission problem for the Helmholtz equation . . . . . . . . . . . A numerical method to compute interior transmission eigenvalues . . . . The factorization method for the acoustic transmission problem . . . . . The exterior Maxwell problem . . . . . . . . . . . . . . . . . . . . . . . . . . Eigenvalues for the interior Maxwell problem . . . . . . . . . . . . . . . . . Interior transmission eigenvalues for electromagnetic scattering . . . . . Factorization for Maxwell’s equation and fractional Helmholtz equation

The transmission problem for the Helmholtz equation

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problem statement . . . . . . . . . . . . . . . . . . . . . . . . . . . Integral equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . Existence and uniqueness . . . . . . . . . . . . . . . . . . . . . . . Boundary element method for solving integral equations . . . . 2.5.1 Triangulation, interpolation, and numerical integration 2.5.2 Boundary element collocation method . . . . . . . . . . . 2.6 Superconvergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 Interpolation of even degree . . . . . . . . . . . . . . . . . 2.6.2 Interpolation of odd degree . . . . . . . . . . . . . . . . . 2.7 Numerical results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.1 Accuracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.2 Superconvergence . . . . . . . . . . . . . . . . . . . . . . . 2.7.2.1 Sphere . . . . . . . . . . . . . . . . . . . . . . . . 2.7.2.2 Peanut . . . . . . . . . . . . . . . . . . . . . . . . 2.7.2.3 Cushion . . . . . . . . . . . . . . . . . . . . . . . 2.8 Summary and outlook . . . . . . . . . . . . . . . . . . . . . . . . .

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A numerical method to compute interior transmission eigenvalues

3.1 3.2 3.3 3.4

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . The interior transmission problem . . . . . . . . . . . . . Surface integral equations . . . . . . . . . . . . . . . . . . Numerical calculation of the transmission eigenvalues 3.4.1 Method by Anne Cossonnière . . . . . . . . . . .

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2.1 2.2 2.3 2.4 2.5

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3.4.2 New method to calculate transmission eigenvalues . . . . . . . Generation of the matrices . . . . . . . . . . . . . . . . . . . . . . . . . . Numerical results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.1 Transmission eigenvalues for the unit sphere . . . . . . . . . . . 3.6.2 Transmission eigenvalues for an ellipsoidal obstacle . . . . . . 3.6.3 Transmission eigenvalues for a peanut-shaped obstacle . . . . 3.6.4 Transmission eigenvalues for an acorn-shaped obstacle . . . . 3.6.5 Transmission eigenvalues for a cushion-shaped obstacle . . . . 3.6.6 Transmission eigenvalues for a bumpy sphere-shaped obstacle 3.6.7 Transmission eigenvalues for a short cylinder-shaped obstacle 3.6.8 Transmission eigenvalues for a long cylinder-shaped obstacle 3.6.9 Additional remarks . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.10 Complex transmission eigenvalues . . . . . . . . . . . . . . . . . 3.6.11 The interior Dirichlet problem . . . . . . . . . . . . . . . . . . . . Estimation of the refraction index . . . . . . . . . . . . . . . . . . . . . . 3.7.1 Estimation of contrast for the unit sphere . . . . . . . . . . . . . 3.7.2 An algorithm to calculate an estimation for the contrast . . . . 3.7.3 Application of the proposed algorithm . . . . . . . . . . . . . . . Summary and conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The exterior Maxwell boundary-value problem . . . . . . . . . . . . . . . . Integral equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The direct scattering problem for time-harmonic electromagnetic waves The Boundary Element Collocation Method . . . . . . . . . . . . . . . . . . Superconvergence of the boundary element collocation method . . . . . 5.6.1 Interpolation of even degree . . . . . . . . . . . . . . . . . . . . . . . 5.6.2 Interpolation of odd degree . . . . . . . . . . . . . . . . . . . . . . . 5.7 Numerical results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Eigenvalues for the interior Maxwell problem

48 49 50 52 53 54 56 57 58 59 60 61 61 63 71 71 72 73 74 77

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The exterior Maxwell problem

5.1 5.2 5.3 5.4 5.5 5.6

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The factorization method for the acoustic transmission problem

4.1 Problem statement . . . . . . . . . . . . . . . . . 4.2 Numerical results . . . . . . . . . . . . . . . . . . 4.2.1 Generation of synthetic far-field data . 4.2.2 Series expansion for a sphere . . . . . . 4.2.3 Reconstruction of a variety of surfaces . 4.3 Summary . . . . . . . . . . . . . . . . . . . . . . . 5

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77 78 79 80 82 90 95

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95 96 96 98 99 103 109 115 119 125 127

6.1 Problem statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

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Contents 6.2 Numerical results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Eigenvalues for the unit sphere . . . . . . . . . . . . . . . . . 6.2.2 Eigenvalues for an ellipsoidal surface . . . . . . . . . . . . . 6.2.3 Eigenvalues for a peanut-shaped obstacle . . . . . . . . . . . 6.2.4 Eigenvalues for an acorn-shaped obstacle . . . . . . . . . . . 6.2.5 Eigenvalues for a cushion-shaped obstacle . . . . . . . . . . 6.2.6 Eigenvalues for a bumpy sphere-shaped obstacle . . . . . . 6.2.7 Eigenvalues for a short and long cylinder-shaped obstacle 6.2.8 Complex eigenvalues . . . . . . . . . . . . . . . . . . . . . . . 7

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Interior transmission eigenvalues for electromagnetic scattering

7.1 7.2 7.3 7.4 8

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Introduction . . . . . . Problem statement . . Numerical results . . . Summary and outlook

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Future research and open questions

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129 129 130 130 131 131 132 132 133 137

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137 138 141 144 145

8.1 The factorization method for Maxwell’s equation . . . . . . . . . . . . . . . . 145 8.2 Fractional Helmholtz equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 9

Conclusion and outlook

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List of Figures 2.1 Different surfaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Estimated order of convergence at six points for a unit sphere using constant interpolation with α = 1/3. The theoretical order of convergence is ˆ2 ln(δ ˆ−1 )) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . O(δ n n 2.3 Estimated order of convergence at six points for a unit sphere using linear ˆ3 ) . interpolation with α = 1/6. The theoretical order of convergence is O(δ n 2.4 Estimated order of convergence at six points for a unit sphere using quadratic interpolation with α = 1/10. The theoretical order of converˆ4 ln(δ ˆ−1 )) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . gence is O(δ n n 2.5 Estimated order of convergence at six points for a peanut using constant interpolation with α = 1/3. The theoretical order of convergence ˆ2 ln(δ ˆ−1 )) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . is O(δ n n 2.6 Estimated order of convergence at six points for a peanut using linear ˆ3 ) . interpolation with α = 1/6. The theoretical order of convergence is O(δ n 2.7 Estimated order of convergence at six points for a peanut using quadratic interpolation with α = 1/10. The theoretical order of convergence is ˆ4 ln(δ ˆ−1 )) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . O(δ n n 2.8 Estimated order of convergence at six points for a cushion using constant interpolation with α = 1/3. The theoretical order of convergence ˆ−1 )) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ˆ2 ln(δ is O(δ n n 2.9 Estimated order of convergence at six points for a cushion using linear ˆ3 ) . interpolation with α = 1/6. The theoretical order of convergence is O(δ n 2.10 Estimated order of convergence at six points for a cushion using quadratic interpolation with α = 1/10. The theoretical order of convergence is ˆ4 ln(δ ˆ−1 )) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . O(δ n n 3.1 Left to right: Surface of the unit sphere S2 , the ellipsoidal surface E , the peanut-shaped surface P , the acorn-shaped surface A , a cushion-shaped surface C , the surface of a bumpy sphere B, a (round) short cylindershaped surface S , and a (round) long cylinder-shaped surface L . . . . . . 3.2 The first two eigenvalues for the unit sphere S2 . Exact values are κ1,S2 ,4 ≈ 3.141 59 and κ2,S2 ,4 ≈ 3.692 45. . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 The first eight eigenvalues for the ellipsoidal obstacle E . . . . . . . . . . . . 3.4 The first seven eigenvalues for the peanut-shaped obstacle P . . . . . . . .

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List of Figures 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13

3.14 3.15 3.16 3.17

Cossonnière’s method for the peanut-shaped obstacle P . . . . . . . . . . . The first ten eigenvalues for the acorn-shaped obstacle A . . . . . . . . . . The first nine eigenvalues for the cushion-shaped obstacle C . . . . . . . . . The first five eigenvalues for the bumpy sphere-shaped obstacle B. . . . . The first thirteen eigenvalues for the short cylinder-shaped obstacle S . . . The first sixteen eigenvalues for the long cylinder-shaped obstacle L . . . . Eight complex-valued eigenvalues for the unit sphere S2 . . . . . . . . . . . . Fifty-three complex-valued eigenvalues for the ellipsoid E . . . . . . . . . . Left: Magnification of the previous domain. Thirteen complex-valued eigenvalues for the ellipsoid E forming two conglomerates. Right: Twelve complex-valued eigenvalues for the ellipsoid E forming three conglomerates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Real and imaginary part of the eigenfunctions vi , i = 1, . . . , 6 corresponding to κ j , j = 1, . . . , 3 for the unit sphere S2 . . . . . . . . . . . . . . . . . . . . Real and imaginary part of the eigenfunctions vi , i = 1, . . . , 6 corresponding to κ j , j = 1, . . . , 5 for the ellipsoidal surface E . . . . . . . . . . . . . . . . Real and imaginary part of the eigenfunctions vi , i = 1, . . . , 6 corresponding to κ j , j = 1, . . . , 5 for the peanut-shaped surface P . . . . . . . . . . . . Real and imaginary part of the eigenfunctions vi , i = 1, . . . , 6 corresponding to κ j , j = 1, . . . , 4 for the acorn-shaped surface A . . . . . . . . . . . . . (1) vi

63 66 67 68 69

(2) vi ,

3.18 Real and imaginary part of the eigenfunctions and i = 1, . . . , 3 corresponding to κ j,P ,4 , j = 1, . . . , 3. . . . . . . . . . . . . . . . . . . . . . . .

4.1 Different surfaces under consideration. . . . . . . . . . . . . . . . . . . . . . . 4.2 The slice W (x i , 0, 0) for i = 1, . . . , 55 of W for the far-field data of a sphere of radius one. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Nine reconstructed surfaces with the factorization method using γ = 6. Parameters are ke = 2, ki = 1, and τ = 1/2. . . . . . . . . . . . . . . . . . . . 4.4 The reconstructed acorn-shaped surface with the factorization method using γ = 6 for various noise levels. Parameters are ke = 2, ki = 1, and τ = 1/2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Reconstructed surfaces with the factorization method using various γ for the acorn-shaped surface. Parameters are ke = 2, ki = 1, and τ = 1/2. . . 4.6 Six reconstructed surfaces with the factorization method using γ = 4. Parameters are ke = 2, ki = 1, and τ = 2. . . . . . . . . . . . . . . . . . . . . 4.7 Six reconstructed surfaces with the factorization method using γ = 4 and F ∈ C258×258 . Parameters are ke = 2, ki = 1, and τ = 2. . . . . . . . . . . . . 4.8 Three reconstructed surfaces with the factorization method using γ = 6 for varying number of incidence and observation directions. Parameters are ke = 2, ki = 1, and τ = 1/2. . . . . . . . . . . . . . . . . . . . . . . . . . .

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List of Figures 4.8 Three reconstructed surfaces with the factorization method using γ = 6 for varying number of incidence and observation directions. Parameters are ke = 2, ki = 1, and τ = 1/2. . . . . . . . . . . . . . . . . . . . . . . . . . . 4.9 Two simultaneously reconstructed surfaces with the factorization method using γ = 6. Parameters are ke = 2, ki = 1, and τ = 1/2. . . . . . . . . . . . 4.10 Three simultaneously reconstructed surfaces with the factorization method using γ = 6. Parameters are ke = 2, ki = 1, and τ = 1/2. . . . . . 4.11 Reconstruction of the acorn-shaped surface using the unregularized and regularized version of the far-field equation with 66 incident waves and observation directions using δ = 1%. Parameters are ke = 2, ki = 1, and τ = 1/2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.12 Reconstruction of the acorn-shaped surface using the unregularized and regularized version of the far-field equation with 66 incident waves and observation directions using δ = 5%. Parameters are ke = 2, ki = 1, and τ = 1/2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

91 92 92

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5.1 Left to right: Surface of the unit sphere, the ellipsoidal surface, and the peanut-shaped surface. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 6.1 Complex-valued interior Maxwell and Dirichlet eigenvalues for the unit sphere enclosed by the rectangle [−0.2, 6] × [−6, 0.2]. . . . . . . . . . . . . 133 7.1 Complex-valued interior transmission eigenvalues for the unit sphere enclosed by the rectangle [−0.2, 6] × [−6, 0.2]. . . . . . . . . . . . . . . . . . . 141 7.2 Interior transmission eigenvalues for three different surfaces. . . . . . . . . 143 8.1 8.2 8.3 8.4 8.5

.Reconstructed unit sphere for various wave numbers. Reconstructed unit sphere for various q. . . . . . . . . . ˆ (z). . . . . . . . . . Four reconstructed surfaces using W Cossonnière’s method for S2 using various s. . . . . . . Cossonnière’s method for P using various s. . . . . . .

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List of Tables 2.1 Nodes and Lagrange basis functions over σ for constant, linear, and quadratic interpolation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Accuracy for constant, linear, and quadratic interpolation for various points situated in the exterior D1 and interior D2 with wave numbers κ1 = 2 and κ2 = 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Accuracy for constant, linear, and quadratic interpolation for various points situated in the exterior D1 and interior D2 with wave numbers κ1 = 3 and κ2 = 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Accuracy for constant, linear, and quadratic interpolation for various points situated in the exterior D1 and interior D2 with wave numbers κ1 = 4 and κ2 = 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Accuracy for constant, linear, and quadratic interpolation for various points situated in the exterior D1 and interior D2 with wave numbers κ1 = 5 and κ2 = 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Constant interpolation with α = 1/3 for a unit sphere. The theoretical ˆ2 ln(δ ˆ−1 )) . . . . . . . . . . . . . . . . . . . . . . . order of convergence is O(δ n n 3.1 Relative error for the first two eigenvalues (EV) with m = 768 and m = 3072 for S2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Values for the first four eigenvalues (EV) with m = 768 and m = 3072 for E . 3.3 Values for the first four eigenvalues (EV) with m = 768 and m = 3072 for P. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Values for the first seven eigenvalues (EV) with m = 3072, m = 6144, and m = 12 288 for P . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Values for the first four eigenvalues (EV) with m = 768 and m = 3072 for A. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Values for the first four eigenvalues (EV) with m = 768 and m = 3072 for C. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Values for the first four eigenvalues (EV) with m = 768 and m = 3072 for B. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8 Values for the first four eigenvalues (EV) with m = 768 and m = 3072 for S. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9 Values for the first four eigenvalues (EV) with m = 768 and m = 3072 for L. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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List of Tables 3.10 The first eigenvalues (EV) for the interior transmission problem in the ellipsoid E with semi axis (1, 1, a) for various choices of a for n = 4. . . . . 3.11 The first four eigenvalues (EVDir ) of −∆ in D. . . . . . . . . . . . . . . . . . . 3.12 The first four eigenvalues (EV) for the interior transmission problem in D for n = 4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.13 Estimated contrasts for various surfaces for m = 768. . . . . . . . . . . . . . 4.1 Far-field pattern errors for a sphere with R = 1 and the parameters ke = 2, ki = 1, and τ = 1/2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Nodes and Lagrange basis functions over σ for constant, linear, and quadratic interpolation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Accuracy for constant interpolation with α = 1/3, n = 1024, and n v = 1024 with wave number κ = 1 for the unit sphere. . . . . . . . . . . . . . . 5.3 Accuracy for linear interpolation with α = 1/6, n = 1024, and n v = 3072 with wave number κ = 1 for the unit sphere. . . . . . . . . . . . . . . . . . . 5.4 Accuracy for quadratic interpolation with α = 1/10, n = 256, and n v = 1536 with wave number κ = 1 for the unit sphere. . . . . . . . . . . . . . . 5.5 Constant, linear, and quadratic interpolation for an ellipsoid. . . . . . . . . 5.6 Constant, linear, and quadratic interpolation for a peanut. . . . . . . . . . 5.7 Generic test cast for checking the electric far-field pattern. . . . . . . . . . . 5.8 Third the electric and magnetic far-field for d = p component p p of T (1/ 3, 1/ 3, 1/ 3) , p = (1, 2, −3)T , κ = 1 at xˆ = (1, 0, 0)T . . . . . . . . . 6.1 6.2 6.3 6.4 6.5 6.6 6.7

Values for the four EV with m = 288, m = 1152, and m = 4608 for S2 . Values for seven EV with m = 1152 and m = 4608 for E . . . . . . . . . . Values for five EV with m = 1152 and m = 4608 for P . . . . . . . . . . . Values for nine EV with m = 1152 and m = 4608 for A . . . . . . . . . . Values for nine EV with m = 1152 and m = 4608 for C . . . . . . . . . . Values for six EV with m = 4608 and m = 9216 for B. . . . . . . . . . . Values for ten EV for S and ten EV for L with m = 4608. . . . . . . . .

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7.1 The four eigenvalues (EV) for the interior transmission problem for n = 4. 144

12

Acknowledgements First of all, I have to thank Professor Georg Bader for offering me the position at the Brandenburg University of Technology, for giving me the freedom to persue my own research fields, and for giving me the opportunity to teach right from the beginning various courses such as Numerical Analysis for Engineers, Numerical Analysis I, and Seminar Numerical Analysis. I am also very thankful to Professor Rembert Reemtsen for initiating the contact to his colleague Professor Andreas Kirsch. Next, I would like to thank Professor Andreas Kirsch for the invitation to Karlsruhe in April 2010 to present myself and the material of my Ph.D. Thesis to the Workgroup on Inverse Problems. There I had the opportunity to talk to him personally, which resulted in a joint project (The factorization method for a conductive boundary condition) and above all he introduced me to many open questions in the field of acoustic interior transmission problem (ITP for short). During my second visit in Karlsruhe in February 2012, we talked again about the ITP and he gave me Anne Cossonnière’s Ph.D. Thesis to read. From this, I had a breakthrough. I was able to attack the problem numerically, which I presented at Inverse Problems: Scattering, Tomography and Parameter Identification - Conference on the occasion of Andreas Kirsch’s 60th birthday in Bad Herrenalb in April 2013. Here, I would also like to thank the organizing committee consisting of Tilo Arens and Frank Hettlich for giving me the opportunity to present those results at this conference. There, I met Antonios Charalambopoulos and that was the starting point of a fruitful collaboration with him and his former student Konstantinos Anagnostopoulos (The factorization method for the acoustic transmission problem). I was able to include numerical achievements of considerable importance, applying the factorization Method to three dimensional reconstruction and incorporating a thorough regularization technique to the whole approach. I am thankful to both of them for this nice collaboration and the consent to include those numerical results in this thesis. Further, I would like to thank the organizing committee consisting of Andreas Kirsch, Tilo Arens and Frank Hettlich for giving me the opportunity to present those results at the thirteenth international conference on integral methods in science and engineering (IMSE 2014) in Karlsruhe in July 2014. Next, I would like to thank Professor K. E. Atkinson for his FORTRAN source code that solves the Laplace equation by a boundary element method (BIEPACK). I further thank Britta Kleefeld, Friedemann Kemm, and Professor Martin Reißel for reading this thesis for typing errors and grammar. I am also very thankful to the Faculty 1 (Mathematics, Natural Sciences and Computer Science) for its financial support. Special thanks goes to the thesis committee. Finally, I thank my wife Britta and my two children Joshua and Fabian for their loving support.

13

1 General introduction Every chapter can be read almost independently of the other ones. The content and the programs have been developed at the Brandenburg University of Technology starting in January 2010 up until now. Most of the content has been presented at a variety of conferences and is published in several journal articles.

1.1 The transmission problem for the Helmholtz equation In Chapter 2, the mathematical formulation of acoustic scattering by penetrable objects in three dimensions involving the Helmholtz equations with different wavenumbers for each media and transition conditions at the interface of the scatterer is considered. We solve this transmission problem mathematically using the boundary integral formulation, and solve the resulting system of boundary integral equations numerically with the boundary element method to high accuracy which is due to superconvergence. The presented numerical results agree with the theoretical results. Precisely, the problem under consideration is the three-dimensional transmission problem for time harmonic acoustic waves for two homogeneous media. A simply-connected and bounded region with sufficiently smooth boundary is immersed in an infinite medium. Each medium is characterized by the space independent wave number κ and the density µ. The system of boundary integral equations is reviewed as well as an existence and uniqueness result. The system is approximated by the boundary element collocation method and consistency, stability, and convergence is proved. In addition, superconvergence is proved and numerical results illustrate the agreement with these theoretical results. No numerical results have been reported for this method yet. The topic has been presented at the 17. Südostdeutsches Kolloquium zur Numerischen Mathematik at Ilmenau University of Technology, May 6th, 2011 and is published in Comput. Methods Appl. Math., 12 (2012), pp. 330–350, see [39]).

15

1 General introduction

1.2 A numerical method to compute interior transmission eigenvalues In Chapter 3 the calculation of transmission eigenvalues to high accuracy is presented. This is a challenging task, since the transmission eigenvalue problem is a non-selfadjoint and nonlinear eigenvalue problem. It appears in the study of scattering by inhomogeneous media and is closely related to non-scattering waves. The eigenvalues provide information about material properties of the scattering media and can be determined from scattering data. Hence, they can play an important role in a variety of inverse problems such as target identification and nondestructive testing. Precisely, the numerical calculation of eigenvalues of the interior transmission problem arising in acoustic scattering for constant contrast in three dimensions is considered. From the computational point of view existing methods are very expensive, and are able to show only the existence of such transmission eigenvalues. Furthermore, the algorithms have difficulties finding them if two or more eigenvalues are situated closely together. We present a new method based on complex-valued contour integrals and the boundary integral equation method which is able to calculate highly accurate transmission eigenvalues (a key ingredient is the solver used in Chapter 2). This chapter provides for the first time such accurate values for various surfaces different from a sphere in three dimensions. Additionally, the computational cost is even lower than that of existing methods. Furthermore, the algorithm is capable of finding complex-valued eigenvalues for which no numerical results have been reported yet. Until now, the proof of existence of such eigenvalues is still an open question. Finally, highly accurate eigenvalues of the interior Dirichlet problem are provided and might serve as test cases to check newly derived Faber-Krahn type inequalities for larger transmission eigenvalues which are not available yet. Additionally, we propose a method to estimate the contrast n from the knowledge of a few interior transmission eigenvalues for a given obstacle. The content of Chapter 3 has been presented at Inverse Problems: Scattering, Tomography and Parameter Identification - Conference on the occasion of Andreas Kirsch’s 60th birthday, Bad Herrenalb, April 8–11, 2013 and is published in Inverse Problems, 29 (2013), 104012 (20pp), see [40] with the exception of Section 3.7 and the eigenfunction plots.

1.3 The factorization method for the acoustic transmission problem In Chapter 4, the shape reconstruction problem of acoustically penetrable bodies from far-field data corresponding to time-harmonic plane wave incidents is considered within the framework of the factorization method. Recently, K. Anagnostopoulos et al. have established the necessary theoretical framework for the application of the factorization method to the inverse acoustic transmission

16

1.4 The exterior Maxwell problem problem rigorously. In this chapter, extended numerical examples in three dimensions are presented, where a variety of different surfaces are successfully reconstructed by the factorization method, thus complementing the method’s validation from the computational point of view. The material in Chapter 4 has been published in Inverse Problems, 29 (2013), 115015 (29pp), see [1, Section 4] and has been presented at the thirteenth international conference on integral methods in science and engineering (IMSE 2014) at Karlsruhe Institute of Technology, July 21–25, 2014.

1.4 The exterior Maxwell problem Electromagnetic scattering arises in science and engineering with applications such as radar imaging, biomedicine, environmental pollution, and geophysical prospecting. The direct electromagnetic scattering problem consists of finding the scattered field provided the geometry of the scatterer is known. In the inverse problem one tries to reconstruct the geometry provided the scattering amplitude also known as the far-field pattern is given. Such problem arises in military applications to identify objects by radar, in medical imaging to detect leukemia, or in non-destructive testing. There are several approaches to solve the inverse problem. However, each of these methods needs accurate far-field data for complex geometries in three dimensions, since exact solutions are only known for special geometries like a sphere. In Chapter 5, the scattering of electromagnetic waves impinging on a conductive scatterer is considered. The problem is formulated as a boundary integral equation which is solved numerically to high accuracy which is due to superconvergence. The presented numerical results confirm the derived theoretical results. The proofs are straightforward extensions of my Ph.D. Thesis, see [36], but the implemented solver is necessary for the next three chapters. The content of Chapter 5 has been presented at the 7th Workshop on Advanced Computational Electromagnetics at the Karlsruhe Institute of Technology, March 1st, 2012 and at the Colloquium of the Department of Mathematics Saarland University, Saarbrücken, November 23, 2012. This material has not been published yet.

1.5 Eigenvalues for the interior Maxwell problem In Chapter 6, we calculate — as a preliminary step — interior Maxwell eigenvalues with the integral equation method. Later, we aim at calculating interior transmission eigenvalues arising in electromagnetic scattering theory. The resulting integral equation is approximated with the boundary element collocation method which leads to highly accurate results due to superconvergence as shown in the previous chapter. The nonlinear eigenvalue problem is solved with the contour integral algorithm which we used to cal-

17

1 General introduction culate interior transmission eigenvalues arising in acoustic scattering theory. A variety of surfaces in three dimensions are considered and interior Maxwell eigenvalues including their multiplicities are calculated. Additionally, we are also able to calculate complexvalued eigenvalues. Note that this material has neither been published nor presented previously.

1.6 Interior transmission eigenvalues for electromagnetic scattering In Chapter 7, the numerical calculation of eigenvalues of the interior transmission problem arising in electromagnetic scattering for constant contrast in three dimensions is considered. As in Chapter 3 we use the method based on complex-valued contour integrals and the boundary integral equation method which is able to calculate accurate transmission eigenvalues, thus extending the acoustic to the electromagnetic interior transmission eigenvalue problem. This chapter provides for the first such accurate values for various surfaces different from a sphere and a cube in three dimensions. Additionally, the computational cost is even lower than the one of existing methods. Furthermore, the algorithm is capable of finding complex-valued eigenvalues for which no numerical results have been reported yet. Until now, the proof of existence of such eigenvalues is still an open question. Note also that this material has neither been published nor presented previously.

1.7 Factorization for Maxwell’s equation and fractional Helmholtz equation In Chapter 8, we report preliminary results regarding two different topics. In the first section of this chapter, we review the factorization method for Maxwell’s equation. In particular, a perfect conductive scatterer is considered. We show that the factorization method works from the computational point of view. Note that we generate highly accurate far-field data using the solver developed in Chapter 5. Note that currently the theoretical verification is still unknown and remains future research. In the second section, we report preliminary results for the fractional Helmholtz equation, a topic which might soon become of interest. Many questions arise and merit further consideration.

Finally, we give a short conclusion in Chapter 9 and illustrate and motivate possible future research.

18

2 The transmission problem for the Helmholtz equation The problem under consideration is the three-dimensional transmission problem for time harmonic acoustic waves for two homogeneous media. A simply-connected and bounded region with sufficiently smooth boundary is immersed in an infinite medium. Each medium is characterized by the space independent wave number κ and the density µ. The system of boundary integral equations is reviewed as well as an existence and uniqueness result. The system is approximated by the boundary element collocation method and consistency, stability, and convergence is proved. In addition, superconvergence is proved and numerical results illustrate the agreement with these theoretical results. No numerical results seem to be reported for this method yet.

2.1 Introduction In this chapter we present the three-dimensional transmission problem for time harmonic acoustic waves for two homogeneous media. A simply-connected and bounded region with sufficiently smooth boundary is immersed in an infinite medium. Each medium is characterized by the space independent wave number κ and the density µ. The problem at hand has been considered by Kress & Roach [43]. They formulate the problem statement, illustrate the system of boundary integral equations to solve the problem at hand and prove existence and uniqueness. For an excellent introduction we refer to the dissertation by Kittappa [30] and the article by him and his advisor Kleinman (see [31]). However, no analysis and numerical results have been reported yet for this problem in three dimensions. Therefore, we briefly summarize the problem statement in Section 2.2. In the following section, we review the system of boundary integral equations to solve the transmission problem. Next, the existence and uniqueness result of Kress & Roach [43] is reviewed. Section 2.5 is devoted to the boundary element collocation method which is used to solve the system of boundary integral equations numerically. We review consistency, stability, convergence, and order of convergence of this method which is an application of Atkinson’s results (see [2]). We will improve the order of convergence; that is, we prove superconvergence at the collocation nodes distinguishing the cases of even and odd interpolation. The two main results of this chapter are given in Theorem 8 and Theorem 10, respectively. The proofs are basically extensions to the author’s Ph.D. thesis (see [36]) — he considered the exterior Neumann and Robin problem

19

2 The transmission problem for the Helmholtz equation for Helmholtz’s equation — and are illustrated in Section 2.6. Note that some numerical analysis is already published in Kleefeld & Lin [42] and stated here for the sake of completeness. The numerical results are in agreement with the theoretical results as presented in Section 2.7. Note that the computations for small refinements are performed on a regular PC. Additionally, we parallelized the FORTRAN solver and performed computations on a huge cluster to demonstrate numerical results for large refinements. A summary including possible future work concludes this chapter.

2.2 Problem statement The mathematical description of the transmission problem considered as a boundaryvalue problem is given in [43] and [15], among others. Let D1 be an unbounded open region and D2 a bounded open region of R3 satisfying D1 ∩ D2 = ;

D 1 ∪ D 2 = D1 ∪ D 2 = R 3 .

and

The boundary between D1 and D2 is assumed to be a closed and smooth C 2 surface which is denoted by Γ. The normal drawn from D2 to D1 is denoted by ν. The transmission problem: Find two complex-valued functions u1 ∈ C 2 (D1 ) ∩ C 1 (D1 ) and u2 ∈ C 2 (D2 ) ∩ C 1 (D2 ) such that ∆u1 + κ21 u1 = 0 ,

in

D1 = R3 \D2 ,

∆u2 + κ22 u2 = 0 ,

in

D2 ,

µ1 u1 − µ2 u2 = f ,

on

Γ,

∂ν u1 − ∂ν u2 = g ,

on

Γ,

(2.1)

where u1 satisfies the Sommerfeld radiation condition ∂ν u1 − iκ1 u1 = O

1 |x|

,

|x| → ∞

uniformly for all directions x/|x|. For the sake of simplicity, the wave numbers κ1 and κ2 as well as the constants µ1 and µ2 are assumed to be positive real numbers. However, the results can be extended to the case of complex numbers (see [43, p. 1433] for details). The functions f ∈ C 1,α (Γ) and g ∈ C 0,α (Γ) are also given. Here C 0,α (Γ) denotes the space of all Hölder continuous functions on Γ and C 1,α (Γ) the space of all Hölder differentiable functions on Γ. The Banach space of all essentially bounded, complexvalued, and Lebesgue integrable functions on Γ with the essential supremum norm k· k∞ is denoted by (L ∞ (Γ), k· k∞ ) and the Banach space of all continuous complex-valued functions on Γ with the usual maximum norm k· k∞ is denoted by (C(Γ), k· k∞ ).

20

2.3 Integral equation

2.3 Integral equation The transmission problem can be solved with the aid of integral equations. Therefore, define the four compact boundary integral operators Z Z Lκ [σ](x) =

MκT [σ](x) =

Φκ (x, y)σ( y) ds( y) , Z

Γ

Γ

∂ν( y) Φκ (x, y)σ( y) ds( y) ,

Mκ [σ](x) =

∂ν(x) Φκ (x, y)σ( y) ds( y) ,

Γ

˜κ ,κ [σ](x) = Nκ [σ](x) − Nκ [σ](x) , N 1 2 1 2

for x ∈ Γ and a suitable smooth function σ, where Z ∂ν( y) Φκ (x, y)ψ( y) ds( y) ,

Nκ [ψ](x) = ∂ν(x) Γ

x ∈ Γ,

where ψ is a suitable smooth function (later we define the correct spaces). The function Φκ : R3 × R3 ∋ (x, y) → Φκ (x, y) ∈ C is the fundamental solution of the Helmholtz equation given by Φκ (x, y) =

eiκr 4πr

r = |x − y| ,

with

x 6= y ∈ R3 .

This function is C ∞ outside the diagonal of R3 × R3 . To solve the transmission problem we seek a solution in the form of a combined double- and single-layer function Z ¦ © u1 (x) = ∂ν( y) Φκ1 (x, y)φ( y) + µ1 Φκ1 (x, y)ψ( y) ds( y) , x ∈ D1 (2.2) Z

Γ

¦

u2 (x) =

© ∂ν( y) Φκ2 (x, y)φ( y) + µ2 Φκ2 (x, y)ψ( y) ds( y) ,

Γ

x ∈ D2

(2.3)

with unknown densities φ ∈ C 1,α (Γ) and ψ ∈ C 0,α (Γ). This ansatz solves (2.1) such that u1 ∈ C 2 (D1 ) ∩ C 1 (D1 ) and u2 ∈ C 2 (D2 ) ∩ C 1 (D2 ) which is due to the regularity of the single and double layer potentials (see [16]). Letting x approach the boundary, and using the jump relations (see [15, 16]) in (2.2) and (2.3) yields 1 u1 = Mκ1 φ + φ + µ1 Lκ1 ψ 2

and

1 u2 = Mκ2 φ− φ + µ2 Lκ2 ψ . 2

Combining the previous two equations and using the first boundary condition of (2.1) we obtain 1 (2.4) µ1 + µ2 φ + µ1 Mκ1 − µ2 Mκ2 φ + µ21 Lκ1 − µ22 Lκ2 ψ = f on Γ . 2 Taking the normal derivative, letting x approach the boundary, and using the jump relations in (2.2) and (2.3) yields 1 1 T T and ∂ν u2 = Nκ2 φ + µ2 Mκ2 ψ+ ψ . ∂ν u1 = Nκ1 φ + µ1 Mκ1 ψ − ψ 2 2

21

2 The transmission problem for the Helmholtz equation Combining the previous two equations and using the second boundary condition of (2.1) yields 1 2

µ1 + µ2 φ − Nκ1 − Nκ2 φ −

µ1 MκT1

− µ2 MκT2

ψ = −g

on

Γ.

(2.5)

Equations (2.4) and (2.5) can be written abstractly in the form (E + K ) χ = h , where K is given by K =

2 2 µ 1 Mκ 1 − µ 2 M κ 2 µ1 Lκ1 − µ2 Lκ2 − µ1 MκT1 − µ2 MκT2 − Nκ1 − Nκ2

The operator E reads as E=

1 2

µ1 + µ2 I 0

1 2

(2.6) !

=:

0 µ1 + µ2 I

K11 K12 K21 K22

.

(2.7)

,

where I denotes the identity operator, and χ, h are given by φ f χ= , h= . ψ −g Clearly, the operators Ki j : C(Γ) → C(Γ), i, j = 1, 2 are compact, since all operators have weakly singular kernels (refer also to [43, p. 1435]). Hence, the operators K and E map from C(Γ) × C → C(Γ) × C(Γ) and I from above is the identity operator on C(Γ). The product space C(Γ) × C(Γ) is equipped with the norm

φ

(2.8)

:= max{kφk∞ , kψk∞ } ,

ψ ∞

and similarly the product space L ∞ (Γ)× L ∞ (Γ). Note that for i, j = 1, 2 is also holds Ki j : C(Γ) → C 0,α (Γ) and Ki j : C 0,α (Γ) → C 1,α (Γ) which follows from the weak singularity and then by a standard argument (see [43, p. 1435]). Therefore, since we have f ∈ C 1,α (Γ) and g ∈ C 0,α (Γ), we automatically have that for any continuous solutions of (2.6), we obtain φ ∈ C 1,α (Γ) and ψ ∈ C 0,α (Γ). Note that, since Γ is compact, we have the relation C(Γ) ⊂ L ∞ (Γ); that is, for any f ∈ C(Γ) there is an equivalence class Θ ∈ L ∞ (Γ) with f ∈ Θ (see [27, p. 7]). Therefore, the integral operators can also be regarded as compact mappings from L ∞ (Γ) to C(Γ). We remark that the operator K is compact from L ∞ (Γ)× L ∞ (Γ) to C(Γ)×C(Γ) because each entry is a compact operator from L ∞ (Γ) to C(Γ), since all operators are weakly singular and the surface is assumed to be of class C 2 . In addition, K is also compact from C(Γ) × C(Γ) to C(Γ) × C(Γ) as explained above. Further note that µ1 and µ2

22

2.4 Existence and uniqueness are constants which are specified in the problem and that E is invertible because µ1 and µ2 are both positive. In summary, one has to solve the system (2.6) to obtain the continuous density functions φ and ψ that are defined on the surface. Then, one uses those functions to obtain the potentials u1 in D1 and u2 in D2 with the integral equations (2.2) and (2.3).

2.4 Existence and uniqueness In this section, we briefly mention the existence and uniqueness of a solution of the transmission problem. The following uniqueness result is taken from [43, Theorem 3.1, p. 1434]. Theorem 1. Let κ1 , κ2 ∈ R+ , and let µ1 , µ2 ∈ R be such that ̺=

µ2 κ22 µ1 κ21

> 0.

Then the only solution of the homogeneous transmission problem is the trivial solution. The existence of the solutions of the transmission problem can be established by Fredholm’s alternative (see [2, Theorem 1.3.1, p. 13]) since the operator is compact. The following results are stated for the sake of completeness and are taken from [43, Theorem 4.6, p. 1437]. Theorem 2. Either the homogeneous transmission problem has only the trivial solution, in which case the nonhomogeneous transmission problem has a unique solution for any inhomogeneous term, or the homogeneous transmission problem has a finite number of linearly independent solutions and the nonhomogeneous transmission problem is solvable if and only if the inhomogeneous term satisfies Z Γ

f ∂ν v1 − gµ1 v1 ds = 0

for all solutions v1 , v2 of the corresponding homogeneous problem.

2.5 Boundary element method for solving integral equations The boundary element method is discussed extensively in [2] and we assume that the reader is familiar with the used notation. We briefly summarize the important parts.

23

2 The transmission problem for the Helmholtz equation

2.5.1 Triangulation, interpolation, and numerical integration Assume that Γ is a connected smooth surface in R3 . Further, we assume that Γ can be written as Γ = Γ1 ∪ · · · ∪ Γ J , (2.9) where Γ j (pairwise disjoint) is a part of the surface. Each Γ j is divided into a triangular mesh and the collection of these is denoted by Tn = {∆k | 1 ≤ k ≤ n} .

(2.10)

Let the unit simplex in the st-plane be defined as σ = {(s, t) | 0 ≤ s, t, s + t ≤ 1} . For a given constant α with 0 < α < 1/3 let i + (r − 3i)α j + (r − 3 j)α (si , t j ) = , , r r

0 ≤ i, j, i + j ≤ r

(2.11)

be the uniform grid inside of σ with f r = (r + 1)(r + 2)/2 nodes. We use interior points to avoid the problem of defining the normal at the collocation points which are common to more than one face ∆k (see p. 280 in [3]). The ordering of this grid is denoted by the nodes {q1 , . . . , q f r }. For each ∆k , we assume there is a map 1−1

mk : σ −−−→ ∆k ,

(2.12)

onto

which is used for interpolation and integration on ∆k (the nodes and Lagrange basis functions for constant, linear, and quadratic interpolation are given in [3, pp. 266–267] and [46, p. 9]). Define the node points of ∆k by vk, j = mk (q j ) ,

j = 1, . . . , f r .

To obtain a triangulation (2.10) and the mapping (2.12), use a parametric representation for each region Γ j of (2.9). Assume that for each Γ j , there is a map 1−1

F j : R j −−−→ Γ j , onto

j = 1, . . . , J ,

(2.13)

where R j is a polygonal region in the plane and F j is sufficiently smooth. That is, a ˆ k, j be an element of the triangulation of R j is mapped onto a triangulation of Γ j . Let ∆ triangulation of R j with vertices ˆ vk,1 , ˆ vk,2 , and ˆ vk,3 . Then the map (2.12) is given by mk (s, t) = F j ((1 − s − t)ˆ vk,1 + t ˆ vk,2 + sˆ vk,3 ) = F j (Tk (s, t)) ,

(s, t) ∈ σ ,

(2.14)

with the obvious definition of the map Tk (see [36, Appendix A] for some surfaces). We collect all triangles of R j together for all j and denote the triangulation of the parametrization plane with ˆ k | 1 ≤ k ≤ n} Tˆn = {∆

24

(2.15)

2.5 Boundary element method for solving integral equations and mesh size ˆn = max diam(∆ ˆ k ) ≡ max hk δ 1≤k≤n

(2.16)

1≤k≤n

ˆn → 0 as n → ∞. satisfying δ Most smooth surfaces can be decomposed as in (2.9). We consider conforming triangulations satisfying the following condition: if two triangles in Tˆn have a nonempty intersection, then this intersection consists of a single common vertex or of a common ˆ k ∈ Tˆn is done by connecting the midpoints edge (see [2, p. 188]). The refinement of ∆ ˆ of the three sides of ∆k yielding four new triangles. This also leads to symmetry in the triangulation and cancellation of errors may occur (see [2, p. 173]). For a function g ∈ L ∞ (Γ), define the projection operator Pn : L ∞ (Γ) → L ∞ (Γ) by Pn g(q) = Pn g(mk (s, t)) =

fr X

g(mk (q j ))l j (s, t) ,

j=1

(s, t) ∈ σ ,

k = 1, . . . , n,

q ∈ ∆k ,

(2.17) with the nodes q j given by 2.11 and the Lagrange basis functions l j given in Table 2.1.

i 1

Constant qi Li (s, t) (α, α)

Linear

1

qi

Li (s, t)

(α, α)

u−α 1−3α t−α 1−3α s−α 1−3α

2

(α, 1 − 2α)

3

(1 − 2α, α)

4 5 6

Quadratic qi L (s, t) i u−α u−α (α, α) 2 − 1 1−3α 1−3α t−α t−α 2 (α, 1 − 2α) − 1 1−3α 1−3α s−α s−α 2 − 1 (1 − 2α, α) 1−3α 1−3α 1−α t−α u−α 4 1−3α α, 2 1−3α 1−α 1−α s−α t−α 4 , 2 2 1−3α 1−3α 1−α s−α u−α , α 4 2 1−3α 1−3α

Table 2.1: Nodes and Lagrange basis functions over σ for constant, linear, and quadratic interpolation.

This yields the piecewise polynomial of degree r interpolating g on the nodes of the mesh {∆k } for Γ (see [3, p. 265] and [46, p. 161] for its use and refer to Section 2 in [5] for more details on point evaluation in L ∞ ). Note that the interpolation error for g ∈ C r (Γ) is r. To reduce integration over ∆k to σ, use the map mk : σ → ∆k to obtain Z

Z

f (q) ds(q) = ∆k

with the Jacobian

f (mk (s, t))J(s, t) dσ σ

J(s, t) = ∂s mk × ∂ t mk (s, t) .

(2.18)

25


2.5.2 Boundary element collocation method We want to solve the integral equation (2.6) which fits into the format (λI − K )x = y

(2.19)

with a projection method — the collocation method. Note that the operator K is compact from L ∞ (Γ) × L ∞ (Γ) to C(Γ) × C(Γ) and that the equation (2.19) is assumed to be uniquely solvable. Define the projection operator Pn : L ∞ (Γ) × L ∞ (Γ) → L ∞ (Γ) × L ∞ (Γ) by Pn 0 , (2.20) Pn = 0 Pn where Pn is given by (2.17). Applying the projection operator Pn defined by (2.20) on both sides of (2.19) and using the collocation method, (2.19) yields λI − Pn K x n = Pn y , x n ∈ Xn × Xn , (2.21) where Xn ⊂ L ∞ (Γ) is a finite subspace with dimension n which equals m f r where m is the number of triangles and f r are the degrees of freedom for interpolation of degree r. Therefore, we obtain a linear system of size 2n × 2n in which all elements are surface integrals over a curved triangle. The pointwise convergence to the identity Pn x → x as n → ∞ for all x ∈ C(Γ) × C(Γ) ˆn → 0. From this and the compactness of is given in [2, Lemma 5.1.1, p. 165] provided δ K one can deduce consistency. More precisely, if {Pn } is a family of bounded projections on L ∞ (Γ) × L ∞ (Γ) with Pn x → x as n → ∞ for x ∈ C(Γ) × C(Γ), then kK − Pn K k → 0 as n → ∞ (see [2, Lemma 3.1.2, p. 57]). From this and the assumption that K : L ∞ (Γ) × L ∞ (Γ) → C(Γ) × C(Γ) is bounded and the assumption λI − K : L ∞ (Γ) × 1−1

L ∞ (Γ) −−−→ C(Γ) × C(Γ), we have stability and convergence. More precisely, we have onto

that the operator (λI − Pn K )−1 exists as a bounded operator from L ∞ (Γ) × L ∞ (Γ) to C(Γ) × C(Γ) for all sufficiently large n ≥ N . Moreover, it is uniformly bounded; i.e. supn≥N k(λI − Pn K )−1 k < ∞. With the help of (2.19) we have for the solution of |λ| (2.21): x − x n = λ(λI − Pn K )−1 (x − Pn x) and kλI −P K k kx − Pn xk∞ ≤ kx − x n k∞ ≤ n

|λ| k(λI − Pn K )−1 k· kx − Pn xk∞ . This leads to kx − x n k∞ converging to zero at exactly the same speed as kx − Pn xk∞ (see [2, Theorem 3.1.1, p. 55]). With this facts, one has the following theorem (see [2, Theorem 9.2.1] for a proof).

Theorem 3. Let Γ be a sufficiently smooth surface; that is, Γ can be parametrized as in (2.9) and (2.13), where each F j ∈ C r+2 . Let K be a compact integral operator from L ∞ (Γ) × L ∞ (Γ) to C(Γ) × C(Γ) and assume the equation (2.19) is uniquely solvable for all functions y ∈ C(Γ) × C(Γ). Let Pn be the projection operator (2.20) and consider the approximate solution of (λI −K )x = y by means of the collocation approximation (2.21). Then, we obtain Stability: The inverse operators (λI − Pn K )−1 exist and are uniformly bounded for all

26

2.6 Superconvergence sufficiently large n ≥ N . Convergence: The approximation x n has the error x − x n = λ(λI − Pn K )−1 (I − Pn )x

(2.22)

and therefore x n → x as n → ∞. Order of convergence: Assume x ∈ C r+1 (Γ) × C r+1 (Γ). Then ˆ r+1 , kx − x n k∞ ≤ c δ n

n≥N,

(2.23)

ˆn is the mesh size of the parametrization domain given by (2.16). where δ The result (2.23) can be improved at the collocation points for a symmetric triangulation and special values of α (see (2.11)), as we will show in the next section. That is, we will prove superconvergence at the collocation points.

2.6 Superconvergence We need some preliminaries and auxiliary results before we can prove superconvergence at the collocation points. We confine ourselves to triangles in the parametrization plane and then use the map F j . Therefore, let τ ⊂ R2 be an arbitrary triangle with vertices {ˆ v1 , ˆ v2 , ˆ v3 }. If f ∈ C(τ), then fr X Lτ f (x, y) = f (mτ (qi ))l i (s, t) , (x, y) = mτ (s, t) i=1

is a polynomial of degree r in the parametrization variables s and t that interpolates f at the nodes {mτ (q1 ), . . . , mτ (q f r )}, where qi and l i are the interpolation nodes and Lagrange basis functions, respectively, and mτ (s, t) = (1 − s − t)ˆ v1 + t ˆ v2 + sˆ v3 ,

(s, t) ∈ σ ,

(2.24)

which corresponds to the map Tk given in (2.14) by suppressing the index k. We will write explicitly Lτk and mτk if necessary. The integration formula over τ given by Z Z τ

f (x, y) dτ ≈

τ

Lτ f (x, y) dτ

(2.25)

has degree of precision at least r. If τ1 and τ2 are triangles for which τ1 ∪ τ2 is a parallelogram, then (2.25) has degree of precision r + 1 (see [4, p. 660]). For differentiable functions f , we define ∂ i f (x, y) i D i f (x, y) := max D i f (x, y) . and kD f k := max ∞ 0≤ j≤i ∂ x j ∂ y i− j (x, y)∈τ

27

2 The transmission problem for the Helmholtz equation In our case the kernel is given by K(P, Q) with points P = (ˆ x , ˆy ) and Q = (x, y). For simplicity we write K P (x, y) instead of K(P, Q) = K(P, (x, y)). Define h(τ) Υ(τ) := ∗ , h (τ) where h(τ) denotes the diameter of τ and h∗ (τ) denotes the radius of the circle inscribed ˆ n,k }, n ≥ 1 satisfies in τ and tangent to its sides. Our triangulation Tˆn = {∆

ˆ k) < ∞ ; sup max Υ(∆ n

ˆ k ∈Tˆn ∆

ˆ n,k from having i.e., it is uniformly bounded in n and therefore, it prevents the triangles ∆ angles which approach 0 as n → ∞ (see [3, p. 276]). For the sake of completeness we state the following lemma (this is a combination of Lemma 4.1 and Corollary 4.2 [42]). Lemma 4. Let τ be a planar triangle of diameter h, let f ∈ C r+1 (τ) and K P ∈ L 1 (τ). Then Z Z ¦ © K dτ · max D r+1 f , K P (x, y)(I − Lτ ) f (x, y) dτ ≤ c (Υ (τ)) hr+1 P τ τ τ (2.26) where c (Υ (τ)) is some multiple of a power of Υ(τ) and P ∈ / τ. Next, we need the following Lemma 5. Let i ∈ {0, 1, 2} and let Γ be a smooth C max{2,i+1} surface. Let K(P, Q) be the kernel of either K11 , K12 , K21 or K22 given in (2.7). Then c i , P 6= Q , (2.27) DQ K(P, Q) ≤ |P − Q|i+1 where c denotes a generic constant independent of P and Q. Proof. This is a straightforward but tedious calculation. The details of the kernel of MκT are given in [42] Lemma 4.7 and Lemma 4.14 or in [36] Lemma 3.4.12 and Lemma 3.4.42. Thus, equation (2.27) is satisfied for the kernel of K22 . By a symmetry argument we also have the result for Mκ and thus for K11 . In [36] Lemma 4.4.1 it is shown that equation (2.27) is satisfied for the kernel of Lκ and therefore we have the result for K12 . By the same arguments we can show that equation (2.27) is satisfied for the kernel of Nκ1 − Nκ2 and thus we also have the result for the kernel of K21 . The next lemma has already been stated and proved and is given for the sake of completeness (see [42, Corollary 4.4]).

28

2.6 Superconvergence Lemma 6. Let τ be a planar triangle of diameter h. Let f ∈ C r+1 (τ) and the kernel K satisfy Lemma 5 for i = 0. In addition, assume that there is a singularity inside of τ; that is, P = Q inside of τ. Then Z ¦ © (2.28) K P (x, y)(I − Lτ ) f (x, y) dτ ≤ c (Υ (τ)) hr+2 · max D r+1 f , τ τ where c (Υ (τ)) is some multiple of a power of Υ(τ).

2.6.1 Interpolation of even degree In this section, we assume that r is even. This implies whenever τ1 and τ2 are triangles for which τ1 ∪ τ2 is a parallelogram, then (2.25) has degree of precision r + 1. The next lemma is given by [46, Lemma 3.3.15, p. 64] and has been adapted to our notation. Lemma 7. Let τ1 and τ2 be two planar triangles with diameter h such that R = τ1 ∪ τ2 is a parallelogram. Let f ∈ C r+2 (R) and K P ∈ L 1 (R) be differentiable with first derivatives D x K P and D y K P belonging to L 1 (R). Then Z K P (x, y)(I − Lτ ) f (x, y) dτ R Z ¦ © 1 r+2 K + D K dτ · max D r+1 f , D r+2 f (2.29) ≤ c (Υ (R)) h P P R

R

/ R. with Υ(R) = maxi=1,2 Υ τi and c (Υ (R)) some multiple of a power of Υ(R) and P ∈ In the following, by f ∈ C k (Γ) we mean f ∈ C(Γ) and f ∈ C k (Γ j ) (that is, f ◦ F j ∈ C k (R j )), j = 1, . . . , J. In the sequel, we assume that the conforming triangulation Tˆn is ˆ k ) < ∞. symmetric and satisfies supn max∆ˆ k ∈Tˆn Υ(∆ We are now in position to prove the first main result of this chapter. Theorem 8. Assume the conditions of Theorem 3 with each parametrization function F j ∈ C r+3 (Γ) and χ ∈ C r+2 (Γ) × C r+2 (Γ). Assume that the kernel of K satisfies the conditions of Lemma 5. Then, for the integral equation (2.6) we have ˆ r+2 ln(δ ˆ−1 ) ; (2.30) max χ(vi ) − χn (vi ) ≤ c δ n n 1≤i≤ f r n

that is, max

ˆ r+2 ln(δ ˆ−1 ) . max φ(vi ) − φn (vi ) , max ψ(vi ) − ψn (vi ) ≤ c δ n n

1≤i≤ f r n

1≤i≤ f r n

29

2 The transmission problem for the Helmholtz equation Proof. We will bound

K (I − Pn )φ(vi ) + K12 (I − Pn )ψ(vi ) max K (I − Pn )χ(vi ) = max 11 1≤i≤ f r n K21 (I − Pn )φ(vi ) + K22 (I − Pn )ψ(vi ) 1≤i≤ f r n

to prove (2.30). It suffices to prove that ˆ r+2 ln(δ ˆ−1 ) max K11 (I − Pn )φ(vi ) ≤ c δ n n 1≤i≤ f r n

with kernel K11 (P, Q), since the proofs of K12 , K21 , and K22 are done by the same argument. Thus, we will bound Z max K11 (I − Pn )φ(vi ) = max K11 (vi , Q)(I − Pn )φ(Q) dΓQ 1≤i≤ f r n 1≤i≤ f r n T Z n K Pi (x, y)(I − Lτ ) f (x, y) dx d y , = max 1≤i≤ f r n ˆ T n

(2.31)

where we used a change of variables and the notation f (x, y) := φ(F j (x, y))| ∂ x F j × ∂ y F j (x, y) | , K Pi (x, y) := K11 (vi , F j (x, y)) ,

with

vi = F j (Pi ) .

For a given collocation point vi , denote ∆∗ the curved triangle containing this point ˆ ∗ the triangle in the parametrization plane containing the point Pi ∈ ∆ ˆ∗ and denote ∆ satisfying vi = F j (Pi ). Define ˆ∗ Tˆn∗ = Tˆn − ∆

and subdivide Tˆn∗ into two disjoint classes Tˆn(1) and Tˆn(2) such that Tˆn(1) ∪ Tˆn(2) = Tˆn∗ , where Tˆn(1) denotes the set of triangles making up parallelograms to the maximum extent possible and Tˆn(2) denotes the set of the remaining triangles. Hence, ˆ ∗ ∪ Tˆ(1) ∪ Tˆ(2) . Tˆn = ∆ n n

ˆ−2 ) and in Tˆ(2) is O(pn) = Recall that the number of triangles in Tˆn(1) is O(n) = O(δ n n ˆ−1 ). Moreover, all but a finite number of the triangles in Tˆ(2) , bounded independent O(δ n n of n, will be at a minimum distance d > 0 from Pi with d independent of n and i. Hence the function K Pi (x, y) is uniformly bounded for the point (x, y) being in a triangle in Tˆn(2) . Thus, we can split the integral in (2.31) into three parts Z Z Z

Tˆn

K Pi (x, y)(I − Lτ ) f (x, y) dx d y =

+ (1) Tˆn

K Pi (x, y)(I − Lτ ) f (x, y) dx d y +

We have the following three cases.

30

ˆ∗ ∆

Z

K Pi (x, y)(I − Lτ ) f (x, y) dx d y

(2) Tˆn

K Pi (x, y)(I − Lτ ) f (x, y) dx d y .

2.6 Superconvergence 1. By Lemma 6, the error in evaluating the integral Z K Pi (x, y)(I − Lτ ) f (x, y) dx d y ∆ˆ ∗ ˆ r+2 ) by the ˆ ∗ . Thus, we also have O(δ is O(hr+2 ), where h is the diameter of ∆ n definition (2.16).

2. Next, consider the error from triangles in Tˆn(1) . By Lemma 7 and (2.16) we have Z K Pi (x, y)(I − Lτ ) f (x, y) dx d y Tˆ(1) n Z X ≤ K Pi (x, y)(I − Lτk ) f (x, y) dτk R (1) k R k ∈Tˆn Z X r+2 (|K Pi | + |D1 K Pi |) dτk chk ≤ Rk

(1)

R k ∈Tˆn

ˆ r+2 ≤ cδ n

Z

X (1) R k ∈Tˆn

Z ˆ r+2 ≤ cδ n ≤

ˆ r+2 cδ n

Z

(1) Tˆn

Rk

(|K Pi | + |D1 K Pi |) dτk

(|K Pi | + |D1 K Pi |) dτ

ˆ∗ Tˆn −∆

(|K Pi | + |D1 K Pi |) dτ .

According to Lemma 5 the last quantity is bounded by

Z ˆ r+2 cδ n

ˆ∗ Tˆn −∆

1 ˆ |Pi − Q|

+

1 ˆ2 |Pi − Q|

dτ

(2.32)

ˆ = F −1 (Q) lying in the parametrization plane. with the points Pi = F j−1 (vi ) and Q j With the help of polar coordinates in this parametrization plane we can show that ˆ r+2 ln(δ ˆ−1 )). Therefore, the error arising the expression in (2.32) is of order O(δ n n ˆ r+2 ln(δ ˆ−1 )). from triangles in Tˆn(1) is O(δ n n

31

2 The transmission problem for the Helmholtz equation 3. Lastly, consider the error over each such triangle in Tn(2) . Applying Lemma 4 yields Z K Pi (x, y)(I − Lτ ) f (x, y) dx d y Tˆ(2) n Z X ≤ K Pi (x, y)(I − Lτk ) f (x, y) dτk τ (2) k τk ∈Tˆn Z X r+1 ≤ |K Pi | dτk chk τk

(2)

τk ∈Tˆn

≤

X

chr+1 h2k , k

(2)

τk ∈Tˆn

where the last step follows, since the area of the k-th triangle is O(h2k ) and since K Pi is uniformly bounded as mentioned above. Therefore, using (2.16) we have X X 2 ˆ r+3 ˆ r+2 , chr+1 h ≤ c δ 1 ≤ cδ k k n n (2)

(2)

τk ∈Tˆn

τk ∈Tˆn

ˆ−1 ) such triangles. Thus, the total error resulting from triangles since we have O(δ n ˆ r+2 ). in Tˆn(2) is O(δ n Combining the errors from the integrals over ∆∗ , Tˆn(1) , and Tˆn(2) yields the result (2.30).

2.6.2 Interpolation of odd degree Now assume r is odd. Then (2.25) has degree of precision r. Suppose we can find α = α0 such that (2.25) has degree of precision r + 1. Then (2.25) has degree of precision r + 2 over a parallelogram. The following lemma is taken from [46, Lemma 3.3.12, p. 62] and has been adapted to our notation. Lemma 9. Let τ1 and τ2 be two planar triangles with diameter h such that R = τ1 ∪ τ2 is a parallelogram. Let f ∈ C r+3 (R) and K P ∈ L 1 (R) be two times differentiable with derivatives of order 1 and 2 belonging to L 1 (R). In addition, assume α = α0 . Then Z K P (x, y)(I − Lτ ) f (x, y) dτ R Z 2 X ¦ © r+3 D i K dτ max D r+1 f , D r+2 f , D r+3 f ≤ c (Υ (R)) h (2.33) P

R i=0

R

/ R. with Υ(R) = maxi=1,2 Υ τi and c (Υ (R)) some multiple of a power of Υ(R) and P ∈

32

2.6 Superconvergence Now, we can prove the second main part of this chapter. Theorem 10. Assume the conditions of Theorem 3 with each parametrization function F j ∈ C r+4 (Γ) and χ ∈ C r+3 (Γ) × C r+3 (Γ). Assume α = α0 . Moreover, assume that the kernel of K satisfies the conditions Lemma 5. Then, for the integral equation (2.6) we have ˆ r+2 ; (2.34) max χ(vi ) − χn (vi ) ≤ c δ n 1≤i≤ f r n

that is,

max

ˆ r+2 . max φ(vi ) − φn (vi ) , max ψ(vi ) − ψn (vi ) ≤ c δ n 1≤i≤ f r n

1≤i≤ f r n

Proof. We will bound K (I − Pn )φ(vi ) + K12 (I − Pn )ψ(vi ) max K (I − Pn )χ(vi ) = max 11 1≤i≤ f r n K21 (I − Pn )φ(vi ) + K22 (I − Pn )φ(vi ) 1≤i≤ f r n

to prove (2.30). It suffices to prove that ˆ r+2 max K11 (I − Pn )φ(vi ) ≤ c δ n 1≤i≤ f r n

with kernel K11 (P, Q), since the proofs for K12 , K21 , and K22 are done by the same argument. Using the same notations as in Theorem 8 we have ˆ r+2 ). ˆ ∗ is O(δ 1. By Lemma 6, the error in evaluating the integral over ∆ n 2. Next, consider the error from triangles in Tˆn(1) . By Lemma 9 we have the bound Z ˆ r+3 (2.35) (|K P | + |D1 K P | + |D2 K P |) dτ . cδ n

ˆ∗ Tˆn −∆

i

i

i

According to Lemma 5 the quantity (2.35) is bounded by Z 1 1 1 r+3 ˆ cδ + + dτ . n ˆ ˆ 2 |Pi − Q| ˆ3 |Pi − Q| |Pi − Q| ˆ∗ Tˆn −∆

(2.36)

With the help of polar coordinates in this parametrization plane we can show that ˆ r+3 1 ). Therefore, the error arising from the expression in (2.36) is of order O(δ ˆn n δ r+2 (1) ˆ ˆ triangles in T is O(δ ). n

n

3. Lastly, consider the error over each such triangle in Tˆn(2) . As in Theorem (8) item 3, we apply Lemma 4. Thus, the total error resulting from triangles in Tˆn(2) is ˆ r+3 )). O(δ n ˆ ∗ , Tˆ(1) , and Tˆ(2) yields the result (2.34). Combining the errors from the integrals over ∆ n n

33


2.7 Numerical results In this section, we present the accuracy and the superconvergence of the system of boundary integral equations. Let the true solution be given by eiκ2 r2 eiκ1 r1 i i z and u2 (x, y, z) = 2 (z − 3) , u1 (x, y, z) = 2 1+ 1+ κ 1 r1 κ 2 r2 r1 r2 p p where r1 = x 2 + y 2 + z 2 and r2 = (x − 3)2 + ( y − 3)2 + (z − 3)2 . Let µ1 = 1 and µ2 = 2. We used NS = 128, NN S = 4 (see [2, pp. 457–459 and pp. 460–462], respectively). We denote the number of faces of the triangulation with n and the size of the linear system with n v . We consider three different surfaces which are illustrated in Fig. 2.1 (a) – (c) (for more surfaces refer to [36]). 1.5

1.0

1

1

0.5

0.5

0

z

z

z

0.5

0

−0.5

0

−0.5

−0.5 −1

−1 −1.5

−1.0 −1.0

−1.0 −0.5

−0.5 0

0 0.5

0.5 1.0

1.0

−1.5 −1

−0.5

0

0.5

x

y

(a) The unit sphere.

1

1

0.5

0

−0.5

−1

x y

(b) Surface of a peanut.

−1

−0.5

0

0.5

1

1.5

1.5

1

0

0.5

−0.5

−1

−1.5

x

y

(c) Surface of a cushion.

Figure 2.1: Different surfaces.

The first surface is the unit sphere, the second surface is a peanut which is given parametrically by x = ̺ sin(φ) ¦ © cos(θ ), y = ̺ sin(φ) sin(θ ), and z = ̺ cos(φ) where 2 2 2 ̺ = 9 cos (φ) + sin (φ)/4 /4. The third surface is a cushion represented in spherical coordinates parametrically by ̺ = 1 − cos(2φ)/2.

2.7.1 Accuracy In this section, we illustrate the accuracy of the integral equation (2.6) for constant, linear, and quadratic interpolation. We choose α = 1/3 for constant (n = 1024 and n v = 2048), α = 1/6 for linear (n = 256 and n v = 1536), and α = 1/10 for quadratic interpolation (n = 256 and n v = 3072). The surface under consideration is the unit sphere. Note that we consider four experiments. In the first case, we pick the wave numbers κ1 = 2 and κ2 = 1, for the second case κ1 = 3 and κ2 = 2, for the third case κ1 = 4 and κ2 = 3, and the fourth case κ1 = 5 and κ2 = 4. In each case we choose 66 points situated on a ball with radius R centered at the origin surrounding the object which we denote by B(R; 0) (see [34, Fig. 1] for a picture and [34, Appendix] for the generation of the 66 points lying on the unit sphere). We just have to multiply the coordinates by R, where we choose the ten radii R = 100, 10, 8, 6, 4, 2, 1.8, 1.6, 1.4, 1.2

34

2.7 Numerical results (the points are located in the exterior domain) and the five radii R = 0.8, 0.6, 0.4, 0.2, 0.1 (the points are located in the interior domain). In the following four tables, we report the relative error in percent using the true solutions u1 in D1 and u2 in D2 . Table 2.2: Accuracy for constant, linear, and quadratic interpolation for various points situated in the exterior D1 and interior D2 with wave numbers κ1 = 2 and κ2 = 1. Points in D1 located on B(100; 0) B(10; 0) B(8; 0) B(6; 0) B(4; 0) B(2; 0) B(1.8; 0) B(1.6; 0) B(1.4; 0) B(1.2; 0)

relative error in % constant linear quadratic 0.71382 0.05991 0.00114 0.73592 0.05807 0.00117 0.74099 0.05743 0.00117 0.74827 0.05622 0.00118 0.75775 0.05322 0.00120 0.79982 0.07149 0.00144 0.85830 0.08196 0.00149 0.92769 0.10159 0.00168 1.00636 0.14996 0.00333 1.06815 0.34166 0.01318

Points in D2 located on B(0.8; 0) B(0.6; 0) B(0.4; 0) B(0.2; 0) B(0.1; 0)

relative error in % constant linear quadratic 3.55241 1.53592 0.04021 1.85208 0.30779 0.00410 1.06858 0.10149 0.00209 0.50652 0.03105 0.00087 0.23437 0.01276 0.00040

Table 2.3: Accuracy for constant, linear, and quadratic interpolation for various points situated in the exterior D1 and interior D2 with wave numbers κ1 = 3 and κ2 = 2. Points in D1 located on B(100; 0) B(10; 0) B(8; 0) B(6; 0) B(4; 0) B(2; 0) B(1.8; 0) B(1.6; 0) B(1.4; 0) B(1.2; 0)




For the first experiment we are able to achieve an overall relative error of < 1.1 % in the exterior and < 3.6 % in the interior for constant interpolation as illustrated in Table 2.2. For linear interpolation we have < 0.4 % in the exterior and < 1.6 % in the interior whereas for quadratic interpolation we obtain < 0.02 % in the exterior and < 0.05 % in the interior. Note that in general the relative error decreases as the distance from the boundary increases. Interestingly, for linear interpolation the relative error is slightly increasing for points far away from the boundary. In sum, we can conclude that our results are very accurate. For the second experiment we obtain similar overall relative errors which are only slightly worse than those of the first experiment. We achieve < 1.2 % in the exterior and < 4.0 % in the interior for constant, < 0.4 % in the exterior and

35










< 1.8 % in the interior for linear, and < 0.02 % in the exterior and < 0.05 % in the interior for quadratic interpolation. In sum, we obtain very accurate results, although the wave numbers have been increased. The same observation is true for the third experiment. Interestingly, we get slightly better results compared to the first and second experiment although the wave numbers are larger. We achieve an overall relative error of < 0.82 % in the exterior and < 2.2 % in the interior for constant, < 0.3 % in the exterior and < 0.81 % in the interior for linear, and < 0.02 % in the exterior and < 0.04 % in the interior for quadratic interpolation. The relative error for the fourth experiment are only slightly worse than those of the other three experiments. We achieve an overall relative error of < 1.2 % in the exterior and < 4.1 % in the interior for constant, < 0.5 % in the exterior and < 1.1 % in the interior for linear, and < 0.03 % in the exterior and < 0.08 % in the interior for quadratic interpolation. In summary, we are able to obtain very accurate numerical results even for higher wave numbers. It is worthwhile mentioning that all computations presented above are done on a

36

2.7 Numerical results regular PC. To get better accuracy we have to use more refinements which implies that we would have to use the parallelized version of my program (see discussion in the next section).

2.7.2 Superconvergence In this section, we present numerical results to illustrate the superconvergence for smooth surfaces without surface approximation for the integral equation (2.6) and compare them with the theoretical results obtained in Section 2.6. Note that we do not know the density on the surface to illustrate the superconvergence. Therefore, we pick points close to the boundary and calculate the rates, since we know the true solution in the exterior domain. Denote the error between the calculated solution un and the true solution u at the point Pi by En (Pi ); that is, En (Pi ) = u(Pi ) − un (Pi ) . Define the estimated order of convergence (EOC) at the point Pi by EOC Pi = log2 En Pi /E4n Pi . Let the densities be given by µ1 = 1 and µ2 = 2 as well as the wave numbers by κ1 = 2 and κ2 = 1. 2.7.2.1 Sphere

As a first example consider domain p we p p the unit sphere. Let Pi be points in the exterior p given p by ωi p (1/ 3,p1/ 3, 1/ 3) with ω1 = 1.3, ω2 = 1.5, and ω3 = 3. Let Q i = υi (1/ 3, 1/ 3, 1/ 3) be points in the interior domain with υ1 = 0.7, υ2 = 0.5, and υ3 = 0.1. 0

absolute error

10

theoretical rate abs. error at P1

−2

10

abs. error at P

2

abs. error at P3 abs. error at Q1

−4

10

abs. error at Q

2

abs. error at Q3 −6

10

0

10

1

10

2

3

10

10

4

10

5

10

n

Figure 2.2: Estimated order of convergence at six points for a unit sphere using constant interpoˆ2 ln(δ ˆ−1 )) . lation with α = 1/3. The theoretical order of convergence is O(δ n n

37


Table 2.6: Constant interpolation with α = 1/3 for a unit sphere. The theoretical order of converˆ2 ln(δ ˆ−1 )) . gence is O(δ n n n (n v ) 4 (8) 16 (32) 64 (128) 256 (512) 1024 (2048) 4096 (8192) 16384 (32768) n (n v ) 4 (8) 16 (32) 64 (128) 256 (512) 1024 (2048) 4096 (8192) 16384 (32768)

En (P1 ) 1.9755−1 1.2953−1 6.6289−2 9.2130−3 3.2323−3 8.1902−4 2.0634−4 En (Q 1 ) 8.0202−2 7.6938−2 3.9308−2 6.5227−3 1.9854−3 5.0357−4 1.2685−4

En (P2 ) 1.5034−1 8.0251−2 3.5231−2 7.3618−3 1.9895−3 5.0539−4 1.2737−4 En (Q 2 ) 4.3632−2 3.3407−2 1.3669−2 3.4876−3 9.1512−4 2.3365−4 5.8990−5

EOC P1 0.61 0.97 2.85 1.51 1.98 1.99 EOCQ 1 0.06 0.97 2.59 1.72 1.98 1.99

EOC P2 0.91 1.19 2.26 1.89 1.98 1.99 EOCQ 2 0.39 1.29 1.97 1.93 1.97 1.99

En (P3 ) 1.1553−1 5.1855−2 1.9401−2 4.7803−3 1.2430−3 3.1606−4 7.9662−5 En (Q 3 ) 1.8261−3 3.7674−3 1.5987−3 4.8072−4 1.2911−4 3.3276−5 8.4343−6

EOC P3 1.16 1.42 2.02 1.94 1.98 1.99 EOCQ 3 -1.04 1.24 1.73 1.90 1.96 1.98

The numerical results for constant interpolation are in agreement with the theoretical results as presented in Table 2.6. Since it is difficult to see the rate of convergence clearly, a log-log plot of the absolute error against n is illustrated in Fig. 2.2. Recall that if we increase the number of faces n by a factor of four, the mesh size of the parametrization ˆn is halved. We see that we obtain the order of convergence of almost two. For domain δ points closer to the boundary we obtain a slightly worse rate of convergence. Note that we obtain similar results for different points located close to the boundary. 0

10


absolute error

−2

10

abs. error at P

2

abs. error at P3

−4

10

abs. error at Q1 abs. error at Q

−6

2

10


10

0

10

1

10

2

3

10

10

4

10

5

10

n

Figure 2.3: Estimated order of convergence at six points for a unit sphere using linear interpolaˆ3 ) . tion with α = 1/6. The theoretical order of convergence is O(δ n

As one can see in Fig. 2.3 the same is true for linear interpolation. We definitely obtain the theoretical order of three, and it seems like we obtain a higher order of convergence for interior points.

38

2.7 Numerical results 0

absolute error

10

theoretical rate abs. error at P1 abs. error at P

2

abs. error at P3

−5

10


2


10

0

1

10

2

10

3

10 n

4

10

10

Figure 2.4: Estimated order of convergence at six points for a unit sphere using quadratic interˆ4 ln(δ ˆ−1 )) . polation with α = 1/10. The theoretical order of convergence is O(δ n n

As presented in Fig. 2.4 the same is true for quadratic interpolation. We achieve the theoretical order of almost four.

2.7.2.2 Peanut

Next, a peanut shaped surface is considered. Let Pi = ωi (1, 1, 1) be points in the exterior domain with ω1 = 1.4, ω2 = 1.5, and ω3 = 1.6. Let Q i = υi (1, 1, 1) be points in the interior domain with υ1 = 0.3, υ2 = 0.2, and υ3 = 0.1. 0

absolute error

10


−2

10

abs. error at P2 abs. error at P

3

abs. error at Q1

−4

10


3

−6

10

0

10

1

10

2

3

10

10

4

10

5

10

n

Figure 2.5: Estimated order of convergence at six points for a peanut using constant interpolation ˆ−1 )) . ˆ2 ln(δ with α = 1/3. The theoretical order of convergence is O(δ n n

Using constant interpolation we obtain the theoretical order of almost two for all points as shown in Fig. 2.5. For linear interpolation we obtain the theoretical order of convergence of three as presented in Fig. 2.6. The rates however suggest a better order of convergence than three.

39

2 The transmission problem for the Helmholtz equation 0

10


absolute error

−2

10

abs. error at P

2

abs. error at P3

−4

10


−6

2

10


10

0

10

1

2

10

3

10

4

10

10

5

10

n

Figure 2.6: Estimated order of convergence at six points for a peanut using linear interpolation ˆ3 ) . with α = 1/6. The theoretical order of convergence is O(δ n 0

10

absolute error

theoretical rate abs. error at P

1


−5

10

3

abs. error at Q

1


3

−10

10

0

10

1

10

2

10 n

3

10

4

10

Figure 2.7: Estimated order of convergence at six points for a peanut using quadratic interpolaˆ4 ln(δ ˆ−1 )) . tion with α = 1/10. The theoretical order of convergence is O(δ n n

As we can see in Fig. 2.7 we obtain the theoretical order of convergence of almost four for the quadratic interpolation. Note that the rates are slightly better for interior points than for exterior points, since the rates are oscillating. 2.7.2.3 Cushion

A cushion surface is considered here. Let Pi = ωi (1, 1, 1) be points in the exterior domain with ω1 = 1.4, ω2 = 1.5, and ω3 = 1.6. Let Q i = υi (1, 1, 1) be points in the interior domain with υ1 = 0.3, υ2 = 0.2, and υ3 = 0.1. As presented in Fig. 2.8 we obtain the theoretical order of almost two for all the points using constant interpolation. The same is true for linear interpolation as one can see in Fig. 2.9. We definitely achieve the theoretical order of three. Figure 2.10 presents numerical results using quadratic interpolation. We obtain the theoretical order of almost four. For this example the rates are slightly better than the

40

2.7 Numerical results 0

absolute error

10


−2

10

abs. error at P

2

abs. error at P3 abs. error at Q1

−4

10

abs. error at Q

2


10

0

10

1

2

10

3

10

4

10

10

5

10

n

Figure 2.8: Estimated order of convergence at six points for a cushion using constant interpolaˆ2 ln(δ ˆ−1 )) . tion with α = 1/3. The theoretical order of convergence is O(δ n n 0

10


absolute error

−2

10


−4

10

3

abs. error at Q1 abs. error at Q2

−6

10

abs. error at Q

3

−8

10

0

10

1

2

10

3

10

4

10

10

5

10

n

Figure 2.9: Estimated order of convergence at six points for a cushion using linear interpolation ˆ3 ) . with α = 1/6. The theoretical order of convergence is O(δ n 0

10

absolute error

theoretical rate abs. error at P

1


−5

10

3

abs. error at Q

1


3

−10

10

0

10

1

10

2

10 n

3

10

4

10

Figure 2.10: Estimated order of convergence at six points for a cushion using quadratic interpoˆ4 ln(δ ˆ−1 )) . lation with α = 1/10. The theoretical order of convergence is O(δ n n

theoretical rate. In summary, the results for a sphere, peanut, and cushion are consistent with a su-

41

2 The transmission problem for the Helmholtz equation perconvergence rate as predicted by Theorem 8 and Theorem 10. Finally, note that the computation for n v ≤ 10000 are performed on a regular quad core computer with 2.33GHz and 4GB RAM memory. Due to the large memory requirement for larger n v , we parallelized the program and used 12 nodes of the BTU cluster (informatics) having 32 nodes each with 8 CPUs and 16 GB RAM memory.

2.8 Summary and outlook We have reviewed the system of boundary integral equations to solve the transmission problem in three dimensions for smooth obstacles as well as the existence and uniqueness results obtained by Kress & Roach [43]. We illustrated the consistency, stability, convergence, and order of convergence of the boundary element collocation method which is an application of results based on Atkinson [2]. We improve those results by proving superconvergence and verify this by numerical results in three dimensions. We have shown that the obtained numerical results are in agreement with the theoretical results. In addition, we present the accuracy of our method leading to highly accurate numerical results. Finally, we would like to motivate researchers in this area to consider also piecewise smooth surfaces. Ideas in this directions could be based on the article by Wendland [63].

42

3 A numerical method to compute interior transmission eigenvalues In this chapter the numerical calculation of eigenvalues of the interior transmission problem arising in acoustic scattering for constant contrast in three dimensions is considered. From the computational point of view existing methods are very expensive, and are only able to show the existence of such transmission eigenvalues. Furthermore, they have trouble finding them if two or more eigenvalues are situated closely together. We present a new method based on complex-valued contour integrals and the boundary integral equation method which is able to calculate highly accurate transmission eigenvalues. So far, this chapter provides for the first time such accurate values for various surfaces different from a sphere in three dimensions. Additionally, the computational cost is even lower than for existing methods. Furthermore, the algorithm is capable of finding complex-valued eigenvalues for which no numerical results have been reported yet. Until now, the proof of existence of such eigenvalues is still an open question. Finally, highly accurate eigenvalues of the interior Dirichlet problem are provided and might serve as test cases to check newly derived Faber-Krahn type inequalities for larger transmission eigenvalues that are not yet available. Additionally, a method to estimate the contrast n from the knowledge of a few interior transmission eigenvalues for a given obstacle is proposed.

3.1 Introduction The interior transmission problem arises in inverse acoustic scattering theory and has become a very important area of research. The so-called transmission eigenvalues carry information about material properties and thus serve for detecting abnormalities inside homogeneous media. Hence, they can be used to test the integrity of materials which is an often desired problem in many nondestructive testing scenarios. This topic has first been introduced by Kirsch [32] in 1986 and by Colton and Monk [18] in 1988. Based upon this, research has been mainly focused on the discreteness of transmission eigenvalues (see [10, 22, 23, 28, 32, 55] among others) and the existence (see [12,51] to mention a few), since the sampling method for reconstructing the support of an inhomogeneous medium fails for wave numbers that are transmission eigenvalues (see Cakoni and Colton (2006) [9] for the linear sampling method and Kirsch and Grinberg (2008) [33] for the factorization method). Note that this overview is by far not complete, since this topic is one of today’s central research subjects in inverse scattering

43

3 A numerical method to compute interior transmission eigenvalues theory (see Cakoni and Haddar (2012) [11] for a recent historical overview). It is noteworthy to mention that there are still many open questions (see [11, page 574]) which merit further investigation. So far, there are only a few methods available to calculate transmission eigenvalues (see for example [24]). But these methods still have many drawbacks. First, these methods are only able to show the existence of a transmission eigenvalue in a prescribed interval. Second, they miss transmission eigenvalues completely if they are closely situated together. Thus, one cannot infer how many eigenvalues are located closely together. Third, the computed transmission eigenvalues lack accuracy. Fourth, the computational cost is high. Fifth, existing methods sometimes miss transmission eigenvalues completely. Lastly, they still need further theoretical justification. In summary, currently existing methods are neither able to calculate transmission eigenvalues to high accuracy with low computational cost nor can they find all transmission eigenvalues. In this chapter, we present a new method which is able to find all transmission eigenvalues in a prescribed region to high accuracy with low computational cost. Briefly summarizing, we use Cossonnière and Haddar’s [25, 2013] recently derived system of surface integral equations and attack the nonlinear eigenvalue problem by a recent algorithm invented by Beyn [6] published in 2012 which is based on complex-valued contour integrals integrating the resolvent. The nonlinear eigenvalue problem can be reduced to a linear eigenvalue problem of small size. We refer the reader to [6] for a discussion of other existing methods that are able to solve nonlinear eigenvalue problems. Therefore, this chapter provides highly accurate transmission eigenvalues for a variety of obstacles in three dimensions for the first time. Those values might be used to check not yet existing Faber-Krahn type inequalities for larger transmission eigenvalues or to get an idea of how such inequalities might look like. Additionally, the new algorithm can find complex-valued transmission eigenvalues. Until now, the proof of existence of such complex-valued eigenvalues is still open. The rest of this chapter is organized as follows. In Section 3.2, we explain the interior transmission problem. In Section 3.3, the integral equation method is reviewed. The interior transmission problem can be solved with surface integrals since we assume constant contrasts. This will lead to a nonlinear eigenvalue problem. Its numerical calculation is discussed in Section 3.4. We first present an existing method that is able to find such eigenvalues. However, this method has many drawbacks and is very expensive from the computational point of view. Then, we explain a new method that is able to find highly accurate values with lower computational cost. Further, in Section 3.5 we briefly explain how to generate the needed matrices. Precisely, we explain how to discretize the integral equation of the first kind numerically. In Section 3.6, we provide highly accurate transmission eigenvalues for a variety of surfaces in three dimensions. For a special case, we compare the calculated solution with an exact solution. Further, we also calculate complex-valued eigenvalues. In addition, we provide values for the interior Dirichlet problem which might be needed to check not yet existing Faber-Krahn

44

3.2 The interior transmission problem type inequalities for larger transmission eigenvalues. Finally, we propose a method to estimate the contrast n from the knowledge of a few interior transmission eigenvalues for a given obstacle. A short summary with conclusions and some remarks are given in Section 3.8.

3.2 The interior transmission problem Let D be a domain in R3 with a connected boundary Γ belonging to class C 2 . Further, assume that the complement of R3 is connected. In what follows κ denotes the wave number, ν is the normal pointing in the exterior, and n is the contrast of the medium in D. We assume that the contrasts are constant, since we aim to apply the integral equation method. The interior transmission problem consists of finding non-trivial solutions (v, w) to ∆v + κ2 v

=0

in

D,

∆w + κ nw

=0

in

D,

v

=w

on

Γ,

on

Γ,

2

∂ν v = ∂ν w

where κ2 will be a transmission eigenvalue. To solve this partial differential equation, we will use the boundary integral equation method.

3.3 Surface integral equations The interior transmission problem for constant contrast can be solved with the integral equation method. This has first been shown by Cossonnière and Haddar in 2013 (see [25]). In this section, we review in detail the derivation of the system of integral equations. p First, we set κ0 = κ and κ1 = κ n for simplicity. Using Green’s representation theorem in D (see for example [16, Theorem 2.1]) yields v(P)

= Vκ0 (∂ν v)(P) − Wκ0 (v)(P) ,

w(P) = Vκ1 (∂ν w)(P) − Wκ1 (w)(P) ,

P ∈ D,

P ∈ D,

(3.1) (3.2)

where Vκ ( f )(P) Wκ ( f )(P) =

= R

R

Γ

Φκ (P, q) f (q) ds(q) ,

P ∈ D,

∂ν(q) Φκ (P, q) f (q) ds(q) ,

P ∈ D,

Γ

denote the single and double layer function defined for points in the domain D, respectively. The function Φκ (p, q) is the fundamental solution of the Helmholtz equation

45

3 A numerical method to compute interior transmission eigenvalues with wave number κ given by exp (iκr) /4πr with r = |p − q| and p 6= q. Letting D ∋ P → p ∈ Γ and using the jump relations of the single and double layer function (see for example [16, Theorem 3.1]) yields v(p)

= Lκ0 (∂ν v)(p) − Mκ0 (v)(p) + 21 v(p) ,

w(p) = Lκ1 (∂ν w)(p) − Mκ1 (w)(p) +

1 w(p) , 2

p ∈ Γ,

(3.3)

p ∈ Γ,

(3.4)

where Lκ ( f )(p) Mκ ( f )(p) =

= R

R

Γ

Φκ (p, q) f (q) ds(q) ,

p ∈ Γ,

∂ν(q) Φκ (p, q) f (q) ds(q) ,

p ∈ Γ,

Γ

denote the single and double layer function defined for points on the surface Γ, respectively. Note that after applying the boundary condition equation (3.4) reads v(p) = Lκ1 (∂ν v)(p) − Mκ1 (v)(p) +

1 2

v(p) ,

Taking the difference of (3.5) and (3.3) gives 0 = Lκ1 − Lκ0 (∂ν v)(p) + −Mκ1 + Mκ0 (v)(p) ,

p ∈ Γ.

(3.5)

p ∈ Γ.

(3.6)

Next, we take the normal derivative of (3.1) and (3.2), let D ∋ P → p ∈ Γ, and use the jump relations of the normal derivative of the single and double layer function (see for example [16, Theorem 3.1]) which yields ∂ν v(p)

= MκT0 (∂ν v)(p) + 12 ∂ν v(p) − Nκ0 (v)(p) ,

∂ν w(p) = MκT1 (∂ν w)(p) +

1 ∂ w(p) − 2 ν

Nκ1 (w)(p) ,

p ∈ Γ,

(3.7)

p ∈ Γ,

(3.8)

where MκT ( f )(p)

R

∂ν(p) Φκ (p, q) f (q) ds(q) , R Nκ ( f )(p) = ∂ν(p) Γ ∂ν(q) Φκ (p, q) f (q) ds(q) , =

Γ

p ∈ Γ, p ∈ Γ,

denote the normal derivative of the single and double layer function defined for points on the surface Γ, respectively. Note that after applying the boundary condition, (3.8) reads 1 ∂ν v(p) = MκT1 (∂ν v)(p) + ∂ν v(p) − Nκ1 (v)(p) , 2 Taking the difference of (3.9) and (3.7) gives 0 = MκT1 − MκT0 (∂ν v)(p) + −Nκ1 + Nκ0 (v)(p) ,

46

p ∈ Γ.

p ∈ Γ.

(3.9)

(3.10)

3.4 Numerical calculation of the transmission eigenvalues With the notation α = ∂ν v(p) and β = v(p), p ∈ Γ (3.6) and (3.10) can be written abstractly as Lκ1 − Lκ0 −Mκ1 + Mκ0 α 0 = . (3.11) MκT1 − MκT0 −Nκ1 + Nκ0 β 0 Finally, we define the surface operator A which depends on the wave number κ by Lκpn − Lκ −Mκpn + Mκ A(κ) = (3.12) MκTpn − MκT −Nκpn + Nκ p where we replace κ1 with κ n and κ0 with κ, respectively. Then, (3.11) can be written as A(κ)X = 0

(3.13)

with the obvious definition of X . Note that the operator A(κ) : H −3/2 (Γ) × H −1/2 (Γ) → H 3/2 (Γ)×H 1/2 (Γ) is of Fredholm type with index zero, and it is analytic on the upper halfplane of C (see [24, Lemma 5.3.9] or [25, Theorem 3.8]). Thus, the theory of eigenvalue problems for holomorphic Fredholm operator-valued functions applies to A(κ). In the next section, we will concentrate on A(κ).

3.4 Numerical calculation of the transmission eigenvalues In this section, we focus on the numerical calculation of highly accurate transmission eigenvalues in three dimensions. In addition, the computational cost has to be low.

3.4.1 Method by Anne Cossonnière To our knowledge the only available procedure so far for the calculation of transmission eigenvalues in three dimensions is the following. First, one has to discretize (3.13). Then A(κ) in (3.12) will become a matrix, say A(κ) ∈ Cm×m with m large. Next, one computes the eigenvalues of A(κ) for many values of κ and checks for which the smallest eigenvalue is close to zero. However, note that A(κ) : H −1/2 (Γ) × H 1/2 (Γ) → H 1/2 (Γ) × H −1/2 (Γ) is a compact operator and hence the eigenvalues of A(κ) accumulate at zero. Therefore, it is difficult to distinguish between the eigenvalue zero and the other eigenvalues close to zero due to numerical errors. The workaround to this problem would be to consider a generalized eigenvalue problem A(κ)X = λB(κ)X with preconditioner B(κ) = A(iκ). Since A(iκ) is injective (see [24, Lemma 5.3.10]), the accumulation point zero is shifted away to minus one. After computing the eigenvalues λi , i = 1, . . . , m such that 0 < |λ1 | < |λ2 | < . . . for a fixed κ, one plots the inverse of λ1 against κ. The

47

3 A numerical method to compute interior transmission eigenvalues transmission eigenvalues will be located at peaks of this curve (see for example Figure 3.5 on page 55). The problems are that one has to generate matrices A(κ) for many κ, say N = 1000, and then one has to solve 1000 generalized eigenvalue problems of size m with m very large, say 3000. Hence, from the computational point of view this method is very expensive. Moreover, if one is interested in very accurate transmission eigenvalues, the cost would be even higher. In summary, the method is good for showing the existence of transmission eigenvalues in a certain interval, but is not able to provide accurate values for them. If two eigenvalues are very close to each other, one will only see a single peak where both eigenvalues are merged together. It is noteworthy that the linear sampling method has the same drawbacks and thus does not serve as a possible alternative to provide highly accurate transmission eigenvalues either. Next, we will provide an alternative method that is able to calculate highly accurate transmission eigenvalues with even lower computational cost. Precisely, one only has to generate at most N = 50 matrices and has to solve one eigenvalue problem of size much smaller than m.

3.4.2 New method to calculate transmission eigenvalues The new method is based on the recent algorithm given by [6] in 2012. Without going into too much detail we briefly explain the idea and then present the algorithm. In the sequel, we consider the nonlinear eigenvalue problem of the form A(κ)v = 0,

v ∈ Cm ,

v 6= 0,

κ ∈ Ω ⊂ C,

where Ω is some open domain in the complex plane. Its boundary is denoted by ∂ Ω and we assume that there are k eigenvalues inside the contour including multiplicities. We further assume that we deal with a large scale problem k ≪ m. Using Keldysh’s theorem one can reduce the original nonlinear eigenvalue problem to a linear eigenvalue problem of size k. Then, this new method involves evaluating a complex-valued integral of the resolvent operator which has to be approximated numerically by the trapezoidal rule. The new method works as follows (see [6, integral algorithm 1]): Initially, one has to choose a contour ∂ Ω in C which might contain k eigenvalues. We will choose a 2πperiodic function, for example the ellipse ψ(t) = µ + a cos(t) + b sin(t)i with derivative ψ′ (t) = −a sin(t) + b cos(t)i. First, choose an index l ≤ m and Vˆ ∈ Cm×l randomly. Second, evaluate the contour integrals A=

48

1 2πi

Z −1 ˆ

A(κ) V dκ ∂Ω

and

B=

1 2πi

Z κA(κ)−1 Vˆ dκ ∂Ω

3.5 Generation of the matrices numerically with the trapezoidal rule. With the change of variable κ = ψ(t) and t j = 2π j/N , j = 0, . . . , N (ψ(t 0 ) = ψ(TN )) we have AN =

−1 1 NX

iN

j=0

−1 ˆ

′

A(ψ(t j )) V ψ (t j ) ,

BN =

−1 1 NX

iN

A(ψ(t j ))−1 Vˆ ψ(t j )ψ′ (t j ) .

j=0

Note that the choice N = 50 suffices, since we have an exponential rate of convergence of the trapezoidal rule. This is a standard result on the trapezoidal rule for holomorphic periodic integrands. Third, compute a singular value decomposition of AN = V ΣW H . Fourth, perform a rank test for Σ and find 0 < k ≤ l such that σ1 ≥ . . . ≥ σk > tolrank > σk+1 ≈ 0 ≈ σl ≈ 0 . If k = l, then increase l and go to step 1. Otherwise let V0 = V (1 : m, 1 : k), W0 = W (1 : l, 1 : k), and Σ0 = diag(σ0 , . . . , σk ) . Fifth, compute C = V0H BN W0 Σ−1 ∈ Ck×k . 0

(3.14)

Finally, solve the eigenvalue problem for C. In summary, one has to generate at most N = 50 matrices A(κ) and has to solve one eigenvalue problem of size k with k much smaller than m. The additional cost is the factorization of the N matrices for finding the inverse.

3.5 Generation of the matrices In this section, we briefly outline how we generate the N = 50 matrices A(κ) for κ ∈ ∂ Ω ⊂ C. We will use the boundary element method to discretize (3.13) with A(κ) given by (3.12) containing surface integral operators to obtain the matrices. A boundary integral equation package (BIEPACK) for the solution of Fredholm integral equation of the second kind arising from the Laplace equation has been given by Atkinson (see [2] and the reference therein). Kleefeld [36] has extended this package to solve the Helmholtz equation. Precisely, the operators Vκ , Wκ , and the surface integral operators Lκ , Mκ , MκT , and Nκ1 − Nκ2 can be approximated numerically for any complex wave number κ. It has been used for a variety of problems dealing with the Helmholtz equation (see for example [34, 37–39, 41, 42] to mention a few). Note that there the discretization resulted from the collocation method applied to a Fredholm integral equation of the second kind. We use the same approximation and integration technique for (3.13). We briefly demonstrate the discretization for the single layer function Lκ on Γ, since for the other integral operators onlySthe kernel has to be exchanged. Precisely, we generate a n conforming triangulation Tn = k=1 ∆k of the surface Γ. We use quadratic interpolation on the reference triangle σ = {(s, t) | 0 ≤ s, t, s + t ≤ 1} with the parameter α = 1/10 and

49

3 A numerical method to compute interior transmission eigenvalues we refer the reader to [42, page 119] for the six nodes qi inside σ and the six Lagrange 1−1

basis functions L i . We assume that there is a map mk : σ −−−→ ∆k (see [36, Appendix onto

A] for a variety of surfaces). Then the (i, l), (k, j)-th entry of L ≡ Lκ ∈ C6n×6n is given by Z L = Φ v , m (s, t) L (s, t) ∂ m × ∂ m (s, t) d(s, t) , κ

(i,l),(k, j)

i,ℓ

k

j

s

k

t

k

σ

where i, k = 1, . . . , n and l, j = 1, . . . , 6 and vi,l = mi (ql ). Additionally, the integrals are approximated numerically. Note that the kernel is weakly singular. Hence, we have to use a different integration routine provided vi = mk (s, t) for some (s, t) ∈ σ. Applying the Duffy transformation removes the singularity. Then one has to integrate over the unit square. This double integral is approximated by a two-dimensional Gaussian quadrature with NS quadrature points. The T2 :5–1 rule from Stroud [59, p. 314] is used for the nonsingular integrals (see pp. 457–459 and pp. 460–462 in [2], respectively). Depending on how close the point is located at the singularity, we use a local refinement process. We refine by connecting the midpoints leading to four new triangles. The number of refinements is denoted by NN S . We use the parameters NS = 128 and NN S = 4. Interestingly, the operator approximations lead to matrices that have small condition numbers (for all surfaces under consideration) and get bigger if we increase the number of unknowns, but they are not ill-conditioned for m ≤ 12288 (see also [2, p. 468] for the treatment of integral equations of the first kind). Otherwise, one would have to think of some kind of regularization to ensure the stability of the inverse.

3.6 Numerical results In this section, we provide transmission eigenvalues with high accuracy for various obstacles in three dimensions. Precisely, we calculate the eigenvalues κ and by squaring them we get the transmission eigenvalues κ2 . In the sequel, we will choose, without loss of generality, the contrast n = 4. Note that we use an ellipse as the domain Ω ∈ C. Its boundary ∂ Ω is given by ψ(t) = 3 + cos(t) + sin(t)i/4 and thus serves as the contour for all obstacles under consideration with the exception of the seventh obstacle. For it, we use the contour ψ(t) = 5/2 + cos(t) + sin(t)i/4. We will consider eight surfaces as depicted in Figure 3.1. The first surface is the unit sphere centered at the origin denoted by S2 . The second surface is an ellipsoidal surface with semi axes (1, 1, 6/5) which we denote by E . The third surface is a peanut-shaped surface denoted by P given parametrically ¦ by x = ̺ sin(φ) cos(θ © ), y = ̺ sin(φ) sin(θ ), and z = ̺ cos(φ) where 2 2 2 ̺ = 9 cos (φ) + sin (φ)/4 /4. The fourth surface is an acorn given by ̺ 2 = 9 17/4 + 2 cos(3φ) /25 and is denoted by A . The fifth surface is a cushion given by ̺ = 1 − cos(2φ)/2 and is denoted by C . The last three surfaces are the bumpy

50

3.6 Numerical results

1.5

1.2

1

0.8

1

0.4

0.5

0.5

0

z

0

z

z

0

−0.4

−0.5

−0.5 −0.8

−1 −1 −0.5

−0.5 0

−0.5 0

0.5 1

0 0.5

−0.5

0.5 1

1

−1.5 −1

−1 −0.5

0 0.5

−1

−1.2 −1

−1

0

0.5

1

1

0.5

1

x

x

−1

x

y

y

−0.5

0

y

1.5

1

1

1

0.5

0.5

0

z

z

z

0.5

0

0

−0.5 −0.5

−0.5

−1 −1

−1 −1.5 −1.5

−1 −1.5 −1

−0.5

0

0.5

1

1.5

1.5

1

0.5

−0.5

0

−1

−1.5

−1

−0.5

0

x

0.5

1

1.5

1.5

1

−0.5

0

0.5

−1

−0.5

−1.5

−1 0 0.5

x

y

y

−0.5 0

0.5 y

1 1

x

1.5

1

1 0.5

0

z

z

0.5

0 −0.5

−0.5

−1

−1 −1.5

−1

−0.5

0

0.5 y

1

1.5

1.5

1

0

0.5

−0.5

−1

−1.5 −1.5 −1

−0.5

0

0.5

x

1

1

0.5

−0.5

0

−1

x y

Figure 3.1: Left to right: Surface of the unit sphere S2 , the ellipsoidal surface E , the peanutshaped surface P , the acorn-shaped surface A , a cushion-shaped surface C , the surface of a bumpy sphere B, a (round) short cylinder-shaped surface S , and a (round) long cylinder-shaped surface L .

sphere B, the (round) short cylinder S , and the (round) long cylinder L given by ̺ = 1 + sin(3φ) sin(3θ )/5, ̺ 10 = 1/((2 sin(φ)/3)10 + cos10 (φ)), and ̺ 10 = 1/((2 cos(φ)/3)10 + sin10 (φ)), respectively. It is easy to check that S2 , E , P , A , C , B, S , and L are contained in the balls B(0, 1), B(0, 6/5), B(0, 3/2), B(0, 3/2), B(0, 3/2), B(0, 6/5), B(0, η), and B(0, η), respectively, where the ball centered at p B(0, r)2/5denotes 9/10 the origin with radius r. Here, the value of η is (9 6 + 8) 4 /8 ≈ 1.697 755 890. All computations are performed on a regular quad core computer with 2.33GHz and 4GB RAM memory.

51

3 A numerical method to compute interior transmission eigenvalues

3.6.1 Transmission eigenvalues for the unit sphere First, we consider the unit sphere S2 , since in this case the transmission eigenvalues can easily be calculated numerically by equation (3.15) which is given below. Since we assume that the unit sphere is centered at the origin, we have the following expansion of v and w (see for example [16, Chapter 2]) v(x) =

p ∞ X X

αm j (κ|x|) Ypm (x/|x|) , p p

p=0 m=−p

w(x) =

p ∞ X X

p β pm jp κ n|x| Ypm (x/|x|) ,

p=0 m=−p

where jp denotes the spherical Bessel function of order p and Ypm the spherical harmonics of order p. Using the boundary condition v = w and ∂ν v = ∂ν w on the surface of the unit sphere; i.e. |x| = 1 yields the two equations p m αm j (κ) − β j κ n =0 p p p p p p καm j ′ (κ) − κ n jp′ κ n = 0 p p which are equivalent to

jp (κ) − jp (2κ) jp′ (κ) −2 jp′ (2κ)

αm p β pm

=

0 0

,

where we used the contrast n = 4. The problem reduces to calculating the zeros of jp (κ) − jp (2κ) =0 (3.15) det jp′ (κ) −2 jp′ (2κ) for p ≥ 0 numerically. The first four zeros are κ1,S2 ,4 ≈ 3.141 59 ,

κ4,S2 ,4 ≈ 4.831 86 ,

κ2,S2 ,4 ≈ 3.692 45 ,

κ3,S2 ,4 ≈ 4.261 68 ,

where the first index denotes the number of the eigenvalue κ. The second index represents the surface under consideration and the third index is the value of the contrast n. Next, we calculate the eigenvalues κ in Ω for the unit sphere. There should be two eigenvalues located in Ω not counting multiplicities. Note that the order of the zero κ of A(κ)v(κ) is called the multiplicity of v at κ. As we can see in Figure 3.2, we obtain the two eigenvalues in the domain Ω with the new method. The first and second have multiplicities three and five, respectively. We can also see the chosen contour ∂ Ω and the N = 50 points chosen for the numerical integration of the contour integral. Next, we will provide the relative error for the first two eigenvalues, where we use the exact values provided above. In the first case, we have m = 768 and in the second case we have m = 3072. As can be seen in Table 3.1 we obtain highly accurate values for the eigenvalues (EV).

52

3.6 Numerical results 0.6

0.4

Im( )

0.2

0

0.2

0.4

0.6 1.8

2

2.2

2.4

2.6

2.8

3 Re( )

3.2

3.4

3.6

3.8

4

4.2

Figure 3.2: The first two eigenvalues for the unit sphere S2 . Exact values are κ1,S2 ,4 ≈ 3.141 59 and κ2,S2 ,4 ≈ 3.692 45.

EV κ1,S2 ,4 κ2,S2 ,4

calculated EV (m = 768) 3.141 56 3.692 21

relative error (m = 768) 9.3458· 10−6 6.3376· 10−5

calculated EV (m = 3072) 3.141 59 3.692 44

relative error (m = 3072) 1.8516· 10−7 2.1206· 10−6

Table 3.1: Relative error for the first two eigenvalues (EV) with m = 768 and m = 3072 for S2 .

3.6.2 Transmission eigenvalues for an ellipsoidal obstacle Now, we consider the ellipsoidal obstacle E . Looking at Figure 3.3 we are able to see eight eigenvalues that are located inside the contour ∂ Ω not counting multiplicities. In the next Table 3.2, we list the values of the first four eigenvalues. As one can see, we are able to obtain at least three digit accuracy for m = 768. More accurate values are obtained for m = 3072, however we do not have an exact solution to compare with. EV κ1,E ,4 κ2,E ,4 κ3,E ,4 κ4,E ,4

calculated EV (m = 768) 2.854 732 2.931 834 3.052 080 3.451 557

calculated EV (m = 3072) 2.854 706 2.932 387 3.052 206 3.451 519

Table 3.2: Values for the first four eigenvalues (EV) with m = 768 and m = 3072 for E .

The last four eigenvalues enclosed by ∂ Ω are 3.529 335, 3.614 770, 3.881 648, and 3.928 384 which are calculated using m = 3072.

53

3 A numerical method to compute interior transmission eigenvalues 0.6

0.4

Im(κ)

0.2

−0

−0.2

−0.4

−0.6 1.8

2

2.2

2.4

2.6

2.8

3 Re(κ)

3.2

3.4

3.6

3.8

4

4.2

Figure 3.3: The first eight eigenvalues for the ellipsoidal obstacle E .

3.6.3 Transmission eigenvalues for a peanut-shaped obstacle Next, we consider the peanut-shaped obstacle P . As we can see in Figure 3.4 there are seven eigenvalues located in the chosen contour ∂ Ω not counting multiplicities. They are reasonably separated from each other. 0.6

0.4

Im( )

0.2

0

0.2

0.4

0.6 1.8

2

2.2

2.4

2.6

2.8

3 Re( )

3.2

3.4

3.6

3.8

4

4.2

Figure 3.4: The first seven eigenvalues for the peanut-shaped obstacle P .

In Table 3.3, we provide the values of the first four eigenvalues. As one can see, we are able to obtain at least four digit accuracy for m = 768. More accurate values are obtained for m = 3072.

54

3.6 Numerical results EV κ1,P ,4 κ2,P ,4 κ3,P ,4 κ4,P ,4

calculated EV (m = 768) 2.825 074 3.044 954 3.514 955 3.574 801

calculated EV (m = 3072) 2.825 465 3.044 714 3.515 142 3.574 896

Table 3.3: Values for the first four eigenvalues (EV) with m = 768 and m = 3072 for P .

The last three eigenvalues enclosed by ∂ Ω are 3.625 503, 3.826 828, and 3.844 950 which are calculated using m = 3072. Additionally, we used Cossonnière’s method to calculate the eigenvalues. We used the interval [2, 4] with N = 201 equidistant points; i.e. we need to calculate 402 matrices and then solve 201 nonlinear eigenvalue problems. Thus, the computational cost is eight times higher than our new method and at the same time we have much higher accuracy. We are able to identify the existence of the first five eigenvalues and can roughly tell its value as shown in Figure 3.5. 180

160

140

120

1

1/|λ |

100

80

60

40

20

0

2

2.2

2.4

2.6

2.8

3 κ

3.2

3.4

3.6

3.8

4

Figure 3.5: Cossonnière’s method for the peanut-shaped obstacle P .

Beside mentioning the low accuracy, we have one single peak at the sixth and seventh eigenvalue. Hence, one is not able to identify the two very closely situated eigenvalues in the figure. Since we know that there are two eigenvalues one could use the interval [3.8, 3.9] and run the costly procedure again with a finer grid. Finally, note that we are not necessarily restricted to m = 3072. In Table 3.4 we additionally provide the first seven eigenvalues of the peanut-shaped obstacle P for m = 6144 and m = 12 288, where the last column has been calculated on a regular quad core computer with 3.4GHz and 16GB RAM memory.

55

3 A numerical method to compute interior transmission eigenvalues EV κ1,P ,4 κ2,P ,4 κ3,P ,4 κ4,P ,4 κ5,P ,4 κ6,P ,4 κ7,P ,4

calc. EV (m = 3072) 2.825 465 3.044 714 3.515 142 3.574 896 3.625 503 3.826 828 3.844 950

calc. EV (m = 6144) 2.825 457 3.044 766 3.515 129 3.574 903 3.627 455 3.827 094 3.844 734

calc. EV (m = 12 288) 2.825 456 3.044 765 3.515 130 3.574 902 3.627 453 3.827 094 3.844 736

Table 3.4: Values for the first seven eigenvalues (EV) with m = 3072, m = 6144, and m = 12 288 for P .

As we can see in Table 3.4, we are able to obtain at least three digits and at least five digits accuracy for m = 3072 and m = 6144, respectively, and even higher accuracy for m = 12 288. Note that the third, fourth, and sixth eigenvalue has multiplicity two whereas the others are simple.

3.6.4 Transmission eigenvalues for an acorn-shaped obstacle In this section, we calculate the eigenvalues for an acorn-shaped obstacle. We see in Figure 3.6 that we definitely have nine eigenvalues not counting multiplicities for m = 768. Between the fifth and sixth eigenvalues it looks like we additionally have two complex-valued eigenvalues located closely to the real axis. Note that our method is also able to identify complex-valued eigenvalues. Further investigation for complex-valued eigenvalues is done in Section 3.6.10.

0.5

0.4

0.3

0.2

Im( )

0.1

0

0.1

0.2

0.3

0.4

0.5

1.8

2

2.2

2.4

2.6

2.8

3 Re( )

3.2

3.4

3.6

3.8

4

Figure 3.6: The first ten eigenvalues for the acorn-shaped obstacle A .

56

4.2

3.6 Numerical results But looking closely at the two complex-valued eigenvalues for m = 3072 reveals the eigenvalue 3.357 039 of multiplicity two. The previous listed eigenvalue has value 3.344 123 and the next eigenvalue has value 3.367 874. In summary, we have ten eigenvalues located in ∂ Ω not counting multiplicities. In Table 3.5 we provide values for the first four eigenvalues which are obtained with the new method. Again we see that we obtain very accurate values for them. For m = 768 we obtain two digits accuracy. The accuracy of m = 3072 is even higher. calculated EV (m = 768) 2.694 649 2.711 716 2.910 972 2.986 754

EV κ1,A ,4 κ2,A ,4 κ3,A ,4 κ4,A ,4

calculated EV (m = 3072) 2.706 295 2.718 191 2.940 516 2.994 077

Table 3.5: Values for the first four eigenvalues (EV) with m = 768 and m = 3072 for A .

The eighth, ninth, and tenth eigenvalue for m = 3072 are 3.719 806, 3.736 285, and 3.795 809, respectively. The multiplicities are 1, 2, 1, 2, 2, 2, 1, 2, 2, and 2.

3.6.5 Transmission eigenvalues for a cushion-shaped obstacle The next surface under consideration is the cushion-shaped obstacle. We calculate the eigenvalues for it in Ω. As we can see in Figure 3.7, we have nine eigenvalues enclosed by ∂ Ω not counting multiplicities. 0.6

0.4

Im( )

0.2

0

0.2

0.4

0.6 1.8

2

2.2

2.4

2.6

2.8

3 Re( )

3.2

3.4

3.6

3.8

4

4.2

Figure 3.7: The first nine eigenvalues for the cushion-shaped obstacle C .

57

3 A numerical method to compute interior transmission eigenvalues In Table 3.6 we provide values for the first four eigenvalues for m = 768 and m = 3072. Again, we are able to obtain two digits for m = 768 and even higher accuracy for m = 3072. EV κ1,C ,4 κ2,C ,4 κ3,C ,4 κ4,C ,4

calculated EV (m = 768) 2.939 813 2.960 403 3.191 512 3.228 344

calculated EV (m = 3072) 2.941 084 2.962 924 3.192 652 3.234727

Table 3.6: Values for the first four eigenvalues (EV) with m = 768 and m = 3072 for C .

The fifth, sixth, seventh, eighth, and ninth eigenvalue for m = 3072 are 3.508 462, 3.848 378, 3.892 142, 3.941 219, and 3.991 618, respectively.

3.6.6 Transmission eigenvalues for a bumpy sphere-shaped obstacle The next surface under consideration is a bumpy sphere-shaped obstacle. We calculate its eigenvalues in Ω. As we can see in Figure 3.8, we have five eigenvalues enclosed by ∂ Ω not counting multiplicities. 0.6

0.4

Im(κ)

0.2

0

−0.2

−0.4

−0.6 1.6

1.8

2

2.2

2.4

2.6

2.8

3

3.2

3.4

3.6

3.8

4

4.2

Re(κ)

Figure 3.8: The first five eigenvalues for the bumpy sphere-shaped obstacle B.

In Table 3.7 we provide values for the first four eigenvalues for m = 768 and m = 3072. We are able to obtain two digits for m = 768 and even higher accuracy for m = 3072. The fifth eigenvalue for m = 3072 is 3.978 853.

58

3.6 Numerical results EV κ1,B,4 κ2,B,4 κ3,B,4 κ4,B,4

calculated EV (m = 768) 3.260 959 3.364 243 3.454 247 3.862 864

calculated EV (m = 3072) 3.290 592 3.372 266 3.518 822 3.870 518

Table 3.7: Values for the first four eigenvalues (EV) with m = 768 and m = 3072 for B.

3.6.7 Transmission eigenvalues for a short cylinder-shaped obstacle The next surface under consideration is a short cylinder-shaped obstacle. We calculate the eigenvalues for this obstacle in Ω. As we can see in Figure 3.9, we have thirteen eigenvalues enclosed by ∂ Ω not counting multiplicities. 0.6

0.4

Im(κ)

0.2

0

−0.2

−0.4

−0.6 1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

3

3.2

3.4

3.6

Re(κ)

Figure 3.9: The first thirteen eigenvalues for the short cylinder-shaped obstacle S .

In Table 3.8 we provide values for the first four eigenvalues for m = 768 and m = 3072. Again, we are able to obtain two to three digits for m = 768 and even higher accuracy for m = 3072. EV κ1,S ,4 κ2,S ,4 κ3,S ,4 κ4,S ,4

calculated EV (m = 768) 2.187 215 2.337 717 2.468 408 2.645 202

calculated EV (m = 3072) 2.188 338 2.341 010 2.470 157 2.645 873

Table 3.8: Values for the first four eigenvalues (EV) with m = 768 and m = 3072 for S .

59

3 A numerical method to compute interior transmission eigenvalues The next nine eigenvalues for m = 3072 are 2.749 359, 2.807 625, 2.959 770, 3.108 373, 3.162 479, 3.181 600, 3.226 746, 3.288 880, and 3.454 109, respectively.

3.6.8 Transmission eigenvalues for a long cylinder-shaped obstacle The last surface under consideration is a long cylinder-shaped obstacle. We calculate the eigenvalues for this obstacle in Ω. As we can see in Figure 3.10, we have sixteen eigenvalues enclosed by ∂ Ω not counting multiplicities. 0.6

0.4

Im(κ)

0.2

0

−0.2

−0.4

−0.6 1.8

2

2.2

2.4

2.6

2.8

3 Re(κ)

3.2

3.4

3.6

3.8

4

4.2

Figure 3.10: The first sixteen eigenvalues for the long cylinder-shaped obstacle L .

In Table 3.9 we provide values for the first four eigenvalues for m = 768 and m = 3072. Again, we are able to obtain three digits for m = 768 and even higher accuracy for m = 3072. EV κ1,L ,4 κ2,L ,4 κ3,L ,4 κ4,L ,4

calculated EV (m = 768) 2.397 055 2.524 415 2.893 946 2.959 708

calculated EV (m = 3072) 2.397 621 2.527 255 2.894 152 2.962 155

Table 3.9: Values for the first four eigenvalues (EV) with m = 768 and m = 3072 for L .

The next twelve eigenvalues for m = 3072 are 3.167 520, 3.244 287, 3.344 557, 3.437 854, 3.476 033, 3.528 842, 3.552 710, 3.657 617, 3.702 147, 3.860 546, 3.939 621, and 3.956 027, respectively.

60


3.6.9 Additional remarks In this section, we summarize some of the above results. In the sequel, we assume constant contrast n. Let D be a domain in R3 containing the origin with a connected boundary Γ belonging to class C 2 . Define the set D which contains all obstacles satisfying the aforementioned assumption. All eight obstacles that we have previously considered are contained in B(0, 1.7). Using n = 4, the smallest and largest eigenvalues for those are 2.188 338 for the (round) short cylinder S and 3.290 592 for the bumpy sphere B, respectively. Let D ∈ D and n be fixed. What is the smallest or largest possible real-valued eigenvalue κ1,D,n corresponding to an obstacle D that is contained in a prescribed ball B(0, r) with radius r centered at zero? Another question might be the following. For fixed n are there two obstacles D1 and D2 which have the same first transmission eigenvalue? Note that we have κ1,P ,4 ≈ 2.825 465. Next, we use the ellipsoid with semi axis (1, 1, a) for some given value a. We are able to obtain the following values as listed in Table 3.10. A choice of the parameter a between 1.2 and 1.3 might give the desired a κ1,E ,4

0.8 3.33

0.9 3.22

1.0 3.14

1.1 2.97

1.2 2.85

1.3 2.75

1.4 2.67

1.5 2.61

Table 3.10: The first eigenvalues (EV) for the interior transmission problem in the ellipsoid E with semi axis (1, 1, a) for various choices of a for n = 4.

equality. Note that monotonicity of the first transmission eigenvalue with respect to the contrast is known (see [10, 12]). However, in Table 3.10 monotonicity of the first transmission eigenvalue with respect to the surface parameter a is shown.

3.6.10 Complex transmission eigenvalues As previously mentioned our method is also able to calculate complex-valued eigenvalues. No numerical results have yet been reported for such complex-valued eigenvalues in three dimensions. The domain Ω we chose is a 12◦ clockwise rotated ellipse with center (3.0, −1.5) and (a, b) = (2.0, 0.5). As we can see in Figure 3.11 we have eight complex-valued eigenvalues located in the domain Ω not counting multiplicities for the unit sphere. Note that there are more complex-valued eigenvalues. To find others one has to choose an appropriate contour in the complex plane. The values are from left to right 1.288−1.017i, 1.740−1.130i, 2.198−1.228i, 2.659− 1.315i, 3.123 − 1.393i, 3.591 − 1.465i, 4.061 − 1.531i, and 4.533 − 1.593i. Interestingly, things look quite different for the above ellipsoid E . For the same contour as chosen for the unit sphere, we have now 53 complex-valued eigenvalues located in the contour that form an interesting pattern. See Figure 3.12 for it. First, we have

61


0.8

Im( )

1.2

1.6

2

2.4 0.8

1.2

1.6

2

2.4

2.8

3.2

3.6

4

4.4

4.8

5.2

Re( )

Figure 3.11: Eight complex-valued eigenvalues for the unit sphere S2 .

three eigenvalues. The next conglomerate consists of 7 eigenvalues. The remaining conglomerates from left to right consist of 8, 9, 10, 11, 12, and 13 eigenvalues, respectively. −0.8

−1

−1.2

Im(κ)

−1.4

−1.6

−1.8

−2

−2.2

−2.4 0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

3 Re(κ)

3.2

3.4

3.6

3.8

4

4.2

4.4

4.6

4.8

5

5.2

Figure 3.12: Fifty-three complex-valued eigenvalues for the ellipsoid E .

Additionally, we calculate some of the eigenvalues again. We choose the domain enclosed by the circle centered at (1.5, −1) with radius 1/2 which shows a magnification of the previous domain to verify that the pattern does not result from the numerical algorithm. The left part of Figure 3.13 provides the 13 eigenvalues enclosed by the circle which can be separated into two different conglomerates of size six and seven. The six values of the first conglomerate are from left to right 1.171 − 0.922i, 1.175 − 0.926i, 1.186−0.936i, 1.205−0.952i, 1.233−0.975i, and 1.269−1.003i. For the second conglomerate we have the seven values 1.582 − 1.024i, 1.585 − 1.027i, 1.596 − 1.035i, 1.615 − 1.049i, 1.641 − 1.067i, 1.676 − 1.090i, and 1.718 − 1.116i. Note that each eigenvalue from one of the conglomerates has multiplicity two with the exception of the first eigenvalue which has multiplicity one.

62


−0.25

−0.7

−0.45

−0.9

−0.65 Im(κ)

Im(κ)

−0.5

−1.1

−0.85

−1.3

−1.05

−1.5

−1.25

1

1.2

1.4

1.6

1.8

2

0.25

0.45

Re(κ)

0.65

0.85

1.05

1.25

Re(κ)

Figure 3.13: Left: Magnification of the previous domain. Thirteen complex-valued eigenvalues for the ellipsoid E forming two conglomerates. Right: Twelve complex-valued eigenvalues for the ellipsoid E forming three conglomerates.

The right part of Figure 3.13 shows the two previous conglomerates of size four and five, respectively. Also here each eigenvalue from one of the conglomerates has multiplicity two with the exception of the first eigenvalue which has multiplicity one. The values of the first conglomerate are from left to right 0.374−0.644i, 0.376−0.649i, 0.385 − 0.666i, and 0.40 − 0.694i. For the second conglomerate we have the five values 0.767 − 0.800i, 0.771 − 0.804i, 0.782 − 0.817i, 0.801 − 0.838i, and 0.829 − 0.867i. Of course, we can also calculate complex-valued eigenvalues for the other surfaces, but our main focus in this chapter is the calculation of real-valued transmission eigenvalues. We just wanted to show that the algorithm is capable of doing more.

3.6.11 The interior Dirichlet problem In this last section of the numerical results, we want to emphasize that this algorithm is also applicable to the interior Dirichlet problem. Precisely, we look for non-trivial solutions to ∆u + κ2 u = 0

in

D,

u =0

on

Γ,

where κ2 is the interior Dirichlet eigenvalue. This is solved with the single layer ansatz u(P) = Vκ (φ)(P), P ∈ D. Using the jump relations and the boundary condition leads to the investigation of Lκ (φ)(p) = 0, p ∈ Γ. The numerical approximation results in the matrix Lκ as explained in Section 3.5.

63

3 A numerical method to compute interior transmission eigenvalues Highly accurate results are given in Table 3.11 for the first four eigenvalues using m = 1536. Note that we used a contour of an ellipse centered at (9/2, 0) with semi axes (2, 1/4) for the sphere and the bumpy sphere. For the short and long cylinder we used a contour of an ellipse centered at (3, 0) with semi axes (1, 1/4) and semi axes (3/2, 1/4), respectively. For the remaining obstacles a contour of an ellipse centered at (4, 0) with p semi axes (3/2, 1/4) has been used. Note that we report κ = λi (D), where λi (D) denotes the i-th interior Dirichlet eigenvalue. Additionally, we report in square brackets the multiplicity of the eigenvalue. We also included a summary for the first four eigenvalues and their multiplicities for the interior transmission problem given in Table 3.12 using m = 3072. It is easy to check that the Faber-Krahn type inequality for the first transmission eigenvalue is satisfied for all eight obstacles. It is given by nκ21,D,n > λ1 (D) , where κ21,D,n is the smallest transmission eigenvalue and λ1 (D) is the first interior Dirichlet eigenvalue of −∆ on D (see [22, page 24] for a proof). EVDir 1. 2. 3. 4. EVDir 1. 2. 3. 4.

Unit sphere S2 3.141 593 [1] 4.493 408 [3] 5.763 441 [5] 6.283 185 [1] Cushion C 2.950 220 [1] 3.570 869 [2] 4.305 271 [2] 4.628 224 [1]

Ellipsoid E 2.975 506 [1] 4.056 326 [1] 4.351 067 [2] 5.164 684 [1] Bumpy sphere B 3.302 096 [1] 4.617 400 [2] 4.770 460 [1] 5.854 904 [2]

Peanut P 3.189 591 [1] 3.704 819 [1] 4.581 649 [1] 4.873 977 [2] Short cylinder S 2.246 704 [1] 3.002 660 [2] 3.532 333 [1] 3.772 009 [2]

Acorn A 2.714 524 [1] 3.652 853 [2] 3.854 271 [1] 4.528 995 [2] Long cylinder L 2.624 895 [1] 3.194 858 [1] 3.965 593 [1] 3.975 298 [2]

Table 3.11: The first four eigenvalues (EVDir ) of −∆ in D.

Note that there is still the open question of how Faber-Krahn type inequalities would look like for higher transmission eigenvalues (see [11, page 574] for a discussion). The values provided in Tables 3.11 and 3.12 might help to get ideas or to check newly derived Faber-Krahn type inequalities for higher transmission eigenvalues. p It is also noteworthy that we have κ1,D,n = λ1 (D) for the unit sphere with n = 4 as one can see in Tables 3.11 and 3.12. Now, let n be arbitrary but fixed. Assume that p (3.16) κ1,D,n = λ1 (D) , holds, where we assume real eigenvalues. Is the number of obstacles that satisfy (3.16) finite? Note that we are also able to calculate the eigenfunctions vi , i = 1, . . . , k that correspond to the eigenvalues κi by vi = V0 si , where si are the eigenvectors of the matrix

64

3.6 Numerical results EV 1. 2. 3. 4. EV 1. 2. 3. 4.

Unit sphere S2 3.141 593 [3] 3.692 445 [5] 4.261 683 [7] 4.831 855 [9] Cushion C 2.941 084 [2] 2.962 924 [2] 3.192 652 [2] 3.234 727 [1]

Ellipsoid E 2.854 706 [1] 2.932 387 [1] 3.052 206 [2] 3.451 519 [2] Bumpy sphere B 3.290 592 [1] 3.372 266 [2] 3.518 822 [1] 3.870 518 [2]

Peanut P 2.825 465 [1] 3.044 714 [1] 3.515 142 [2] 3.574 896 [2] Short cylinder S 2.188 338 [2] 2.341 010 [1] 2.470 157 [2] 2.645 873 [1]

Acorn A 2.706 295 [1] 2.718 191 [2] 2.940 516 [1] 2.994 077 [2] Long cylinder L 2.397 621 [1] 2.527 255 [1] 2.894 152 [2] 2.962 155 [2]

Table 3.12: The first four eigenvalues (EV) for the interior transmission problem in D for n = 4.

C ∈ Ck×k given by (3.14) on page 49. V0 is also defined on page 49. In Figure 3.14, 3.15, 3.16, and 3.17 we plot the first few eigenfunctions (both its real and imaginary part alternating in the subfigures) of the unit sphere, the ellipsoidal surface, the peanut-shaped surface, and the acorn-shaped surface, respectively. Note that we can also plot the eigenfunctions for the interior transmission eigenvalues, but for each eigenvalue we would have to plot four pictures, since we have a pair of functions defined on the surface each having a real and imaginary part. Precisely, we (1) (2) (1) (2) have vi := (vi , vi )T . In Figure 3.18 we show the two functions vi and vi defined on the surface for the first three interior transmission eigenvalues obtained for the peanutshaped surface.

65


0.04

1

1

0.03

0.02

0.02

0.5

0.5

0.01

z

0

0.01

0.01

0

0

z

0

−0.5

−0.5

−0.5

−1 −1

−0.03

−1 −0.5

0 0.5

−1 −1

−0.03

−0.5 0

−0.04

0.5

1

−1 −0.5

0

0.5 1

−0.02

−0.02

−0.5 0

0

−0.01

−0.5 −0.02

−1

0

−0.01

−0.01

−1 −1

0.03

0.02

0.5

z

0.04

0.04

0.03

1

1

0 0.5

1

1

x

x

y

(a) Re(v1 ) for κ1 .

y

(b) Re(v2 ) for κ2 .

(c) Re(v3 ) for κ2 . −3

−4

−3

x 10

x 10

x 10 5

4

4

1

−0.04

0.5 1

x y

−0.03

−0.5 0

−0.04

0.5

1

1

1

3

3 2 2

0.5

0

z

z

0

0

0

−1 −0.5

0.5

0.5 1

1

z

0.5

0

0

−1 −0.5

−2

−0.5

−0.5

−2 −3 −1 −1

−1 −0.5

−4

−0.5 0 1

−0.5 0

−5

0.5

−1 −1

−1 −0.5

0 0.5

−3

−1 −1

1

−1 −0.5

−4

0 0.5 1

0 0.5

1

0.5 1

x

1

x

y

x

y

(d) Im(v1 ) for κ1 .

−1

−0.5 0

0.5

y

(e) Im(v2 ) for κ2 .

(f) Im(v3 ) for κ2 .

0.04 0.04 0.04

0.03 1

1

1

0.03

0.03

0.02 0.02

0.02

0.01

0.5

0.5

0.5 0.01 0.01 0

0

z

0

z

z

0 0

0

−0.01 −0.01 −0.02

−0.5

−0.5

−0.01 −0.5

−0.02

−0.02

−0.03 −0.03 −1 −1

−1 −0.5

−1 −1

−0.04

−0.5 0 0.5

−1 −0.5

0

−0.05

−1 −0.5

0

0.5 1

−0.04

−0.5 0

−0.03

−1 −1

0.5

1

0.5 1

−0.5 0 0.5

−0.05

1

0.5 1

x

1

x

y

x

y

(g) Re(v4 ) for κ2 .

−0.04

0

y

(h) Re(v5 ) for κ3 .

(i) Re(v6 ) for κ3 . −3

−3

−5

x 10

x 10

x 10

4

2

1

3

3

1

1 2

2 1 0.5

0.5

0.5 1

1

0

0

−1 −0.5

−0.5 −2

−2

−1 −1

−1 −0.5

−0.5 0

0 0.5

−3

−3

−1 −1

−1 −0.5

−0.5 0

0

0.5 1

0.5

1 x

(j) Im(v4 ) for κ2 .

−1 −1 −1

−1 −0.5

−4

−0.5 0 0.5

1

0.5 1

x y

(k) Im(v5 ) for κ3 .

−2

0

0.5 1

y

0 0

−1 −0.5

z

z

z

0 0

1 x

y

(l) Im(v6 ) for κ3 .

Figure 3.14: Real and imaginary part of the eigenfunctions vi , i = 1, . . . , 6 corresponding to κ j , j = 1, . . . , 3 for the unit sphere S2 .

66


−0.019 0.03 −0.02

1.2

0.04

1.2

1.2 0.03

0.02

−0.021 0.8

0.8

0.8 0.02

−0.022 0.4

0.01

0.4

0.4

0.01

−0.023 0

0

−0.024 −0.4

z

z

z

0

−0.4

−0.025

0

0

−0.01

−0.4 −0.01

−0.026

−0.8

−0.8 −0.02

−0.027 −1.2 −1

−1.2 −1

−1 −0.5

−0.028

−0.5 0

−0.5 0 0.5

1

−1 −0.5

−0.03

0

0.5 1

−0.03 −1.2 −1

−1 −0.5

0 0.5

−0.02

−0.8

0

0.5 1

0.5

1

0.5 1

x

1

x

y

x

y


−0.04

−0.5 0

y


(c) Re(v3 ) for κ3 .

−8

−6

−6

x 10

x 10

x 10

4

8

8

1.2

1.2

1.2 3

7 0.8

6

0.8

0.8 2

0

z

z

0.4

4 0.4

1

0

0

z

6

0.4

2

0

0

5 −0.4

−1

−0.4

4

−0.8

−1.2 −1

−1 −0.5 0.5

−0.5 0 0.5

1

−1 −0.5

−4

0

0.5 1

−6 −1.2 −1

−1 −0.5

0

−4

−0.8

−3

−1.2 −1

3

−0.5 0

−2

−0.8

−2

−0.4

0

0.5 1

0.5

1

0.5 1

x

1

x

y

x

y


−8

−0.5 0

y



0.02 0.04

0.04 1.2

1.2

1.2

0.01

0.03

0.03 0.8

0.8

0.8

0.02 0.4

0.02

0 0.4

0.01

0.4

0.01

0

−0.4

−0.02

−0.8

0

0

−0.02

−0.01

−0.4

0

z

0

z

z

−0.01

−0.01

−0.4

−0.8

−0.03

−0.8

−1.2 −1

−0.04

−1.2 −1

−0.02

−0.03 −1.2 −1

−0.03

−1 −0.5

−0.04

−0.5 0

−1 −0.5

0 0.5

−0.5 0

0

0.5 1

−1 −0.5

0.5

1

0

0.5 1

0 0.5

−0.05

1

0.5 1

x

1

x

y

x

y


−0.04

−0.5

y


(i) Re(v6 ) for κ5 .

−6

−6

x 10

x 10

5

2

−6

x 10 8

4

1.2

1.2

1.2

1.5

6 3 0.8

0.8

0.8

1

4

2 0.4

0.4

−1

−0.4

0

2 z

0

z

z

0

0.4

0.5

1

0

−0.4

0 0 −0.4

−0.5

−2

−2 −0.8

−0.8

−0.8 −1

−3 −1.2 −1

−1 −0.5

−4

−0.5 0

0 0.5

−1.2 −1

−1 −0.5

−5

0 0.5

1

0.5 1

x y


−4 −1.2 −1

−1 −0.5

0

0.5 1

−1.5

−0.5

0

0 0.5

−2

1

0.5 1

x y


−6

−0.5

1 x

y


Figure 3.15: Real and imaginary part of the eigenfunctions vi , i = 1, . . . , 6 corresponding to κ j , j = 1, . . . , 5 for the ellipsoidal surface E .

67


0.04 −0.015

0.025

1.5

1.5

1.5 0.03

1

0.02

1

−0.02

1

0.015

0.02

0.01 0.5

0.01

−0.025 z

z

0

0.5

0

0.005 z

0.5

0

0

0

−0.03

−0.005

−0.5

−0.5

−0.01

−0.5

−1

−0.02

−1

−0.03

−1.5 −1

−0.01 −1

−0.035

−0.015 −0.02

−1.5 −1

−1 −0.5

−0.5 0

−1.5 −1

−0.04

−1 −0.5

0 0.5

−0.5 0

0.5 1

−1 −0.5

0 0.5

1

1

−0.5 0 0.5

1

0.5 1

x

1

x

y

x

y


−0.025

0

−0.04

0.5

y


(c) Re(v3 ) for κ3 . −5

−6

−7

x 10

x 10

x 10 4.5

4 4

1.5

1.5

1.5

4

0.5

3

0

2.5

−0.5

2

0

0

−1

−2

−1

1.5

0.5

1

−0.5

2

−1

1 2

z

z

0.5

3

3 1

3.5

z

1

1

0

0

−0.5

−1

−1

−2

−3 −1.5 −1

−1.5 −1

1 −1 −0.5

−0.5 0

−1 −0.5

0 0.5

−0.5 0 0.5

1

−3 −1 −0.5

−4

0

0.5

0.5 1

−1.5 −1 −0.5 0

0.5 1

0 0.5

1

x

1

x

y

x

y


−4

0.5 1

y



0.06 0.04

0.04

1.5

1.5

1.5

0.03

0.03

1

1 0.02

0.5

0.01

0

z

0

−0.01

−0.5

0

−0.01

−0.5

0.02

0.5

0.01

0

z

0.5

z

0.04

1 0.02

0

0

−0.5 −0.02

−0.02

−1

−0.02

−1

−0.03 −1.5 −1

−0.03 −1.5 −1

−1 −0.5

−0.5 0

−0.5 0

0.5 1

−0.04 −1.5 −1

−1 −0.5

−0.04

0 0.5

−1

1

−1 −0.5

−0.04

0 0.5

−0.5 0

0

0.5 1

0.5

1

0.5 1

x

1

x

y

x

y


y


(i) Re(v6 ) for κ5 . −3

−3

−4

x 10

x 10

x 10 1.5

4

2.5 1.5

1.5

1.5

2 1

3

1

1

1.5

1 2

1

0.5 0.5

0

z

z

0

0

0

−0.5 −0.5

0.5

1

0.5

−1

−0.5

z

0.5

0

0

−0.5 −0.5

−1 −1

−1.5

−2

−1

−1 −1

−3

−2 −1.5 −1

−1 −0.5

−0.5 0

0 0.5

−1.5 −1 −0.5

−0.5 0 0.5

1


−4

−1 −0.5

−0.5 0

0.5 1

x y

−1.5 −1

−1 0

0.5 1

−2.5

0 0.5

1


−1.5

0.5 1

x y

1 x

y


Figure 3.16: Real and imaginary part of the eigenfunctions vi , i = 1, . . . , 6 corresponding to κ j , j = 1, . . . , 5 for the peanut-shaped surface P .

68


−0.01

0.05

0.05

1.5

−0.02

1.5

1

−0.03

1

0.04

0.04

1.5

0.03

0.03 1

0.02

0.02 −0.04

0.5

0

−0.05

0

−0.5

−0.06

−0.5

−1

−0.07

−1

0.5 0.01

0.01 0

z

z

z

0.5

−0.01

0

0 −0.01

−0.5

−0.02

−0.02 −1

−0.03

−0.03 −1.5 −1.5

−1.5 −1

−1.5 −1.5

−0.08

−1 −0.5

−1.5 −1

−0.5 0

−0.5

0 0.5 1

1

1.5

−0.5 0

−0.05

0.5

0 0.5

1 1.5

1

1 1.5

1.5

x

y

x

y


−0.05

0.5

1.5

x

−0.04

−1 −0.5

0 0.5

1 1.5

−1.5 −1

−0.5 0

−0.09

0.5

−1.5 −1.5

−0.04

−1

y


(c) Re(v3 ) for κ2 . −5

−5

−6

x 10 3

x 10 3

x 10

1 1.5

1.5

1

1.5

2

1

0

2

1 1

0.5

1

0.5

0.5

−0.5

0

0

z

0

z

z

−1

−0.5

−1

−1

−3 −1 −0.5

−1

−2

−1.5 −1.5

−1.5 −1

−1 −0.5

0 0.5 1

1

1.5

0 0.5

−3

1 1.5

−0.5 0

0.5

0.5 1

1.5

1 1.5

x

−3

1.5

x

y

x

y


y



0.04

0.04 0.04

1.5

1.5

1.5

0.03

0.03 1

0.5

0

0

z

z

0.5

−0.5

1

0.02

0.02

0

0

0.02

0.5

0.01

z

1

−2

−1 −0.5

0 0.5

−4

1 1.5

−1.5 −1

−0.5 0

0.5

−1.5 −1.5

−1.5 −1

−0.5 0

0

−1

−1

−1.5 −1.5

0

−0.5

−2

0.01

0

0

−0.5

−0.01

−0.5

−0.01

−1

−0.02

−1

−0.02

−1.5 −1.5

−0.03

−1.5 −1.5

−0.02 −1

−0.04

−1.5 −1.5

−1.5 −1

−1 −0.5

−1.5 −1

−0.5 0

−1 −0.5

0 0.5 1

−0.06

1 1.5

−0.03 −1 −0.5

−0.04

0 0.5

−1.5 −1

−0.5 0

0.5

−0.5 0

0.5 1

1.5

1 1.5

0.5 1

1.5

1 1.5

x

1.5

x

y

x

y


−0.04

0 0.5

y


(i) Re(v6 ) for κ4 .

−6

−5

x 10

−5

x 10

x 10 8

8

6 1.5

1

4

0.5

6

1.5

6

1

4

1

0.5

2

0.5

2

0

0

0

0

1.5 4

z

z

z

2 0 0 −0.5

−0.5

−1

−2

−1.5 −1.5

−1.5 −1

−1 −0.5

−0.5 0

0 0.5

0.5 1

−4

−2

−1

−4

−1.5 −1.5

−1.5 −1 0

1.5

0.5 1

1.5


−1.5 −1 0

−8

0 0.5

0.5 1

1.5


−8

1 1.5

x y

−6

−0.5

1 1.5

x

−4 −1.5 −1.5 −0.5

0 0.5

−2

−1

−1

−0.5

1

y

−6

−1 −0.5

−0.5

1.5 x

y


Figure 3.17: Real and imaginary part of the eigenfunctions vi , i = 1, . . . , 6 corresponding to κ j , j = 1, . . . , 4 for the acorn-shaped surface A .

69


0.08 0.04 1.5

0.08 1.5

0.03 0.02

1

0.01

0.5

0.06

1

0.5

0.06

1

0.04

0.5

0.02

0.04

0 0

z

0

z

z

1.5

0 0

−0.01 −0.5

0.02

−0.5

−0.5

−0.02 −1

−0.02 −1

−0.03

−1.5 −1.5

−1.5 −1.5

−0.04 −1.5 −1

−1 −0.5

−1 −0.5

−0.05

0 0.5

0.5 1

1.5

1 1.5

x

1.5

x

x

y

(1) Re(v1 )

0 0.5

1 1.5

for κ1,P ,4 .

(b)

−0.06

−0.5 0

0.5 1

1.5

y

(a)

−1 −0.5

−0.02

0 0.5

1 1.5

−1.5 −1

−0.5 0

0.5 1

−0.04 −1.5 −1.5

−1.5 −1

−0.5 0

−1 0

y

(1) Re(v2 )

for κ2,P ,4 .

(c)

(1) Re(v3 )

for κ3,P ,4 . −4

−4

−4

x 10 4

1.5

3

1.5

1

2

1

x 10

x 10

6

1.5 1.5

5

1 1

4 0.5

0.5 1

3 0

z

z

0

z

0.5

0.5

0

0

2

0 −0.5

−0.5

−0.5 1

−0.5

−1 −1

−1

−1 0 −1

−2

−1.5 −1.5

−1.5 −1.5

−1.5 −1

−1 −0.5

−1.5 −1

−0.5 0 0.5

0.5 1

−1 −0.5

0 0.5

1 1.5

−1.5 −1

−0.5 0

−3

−1.5 −1.5

−1

−1 −0.5

0

0.5 1

1.5

0.5 1

1.5

1 1.5

x

1.5

x

y

x

y

(1)

y

(1)

(d) Im(v1 ) for κ1,P ,4 .

−1.5

0 0.5

−2

1 1.5

−0.5 0

(1)

(e) Im(v2 ) for κ2,P ,4 .

(f) Im(v3 ) for κ3,P ,4 .

−3

x 10 0.01

16 0.015

0.008 1.5

14

1.5 0.006

0.01

1

1

0.004

10 0.5

0.002

0

z

0

0

0

4

−0.002

−0.5

6

0

0.005

0.5

8

z

0.5

z

1.5

12

1

−0.5

−0.5 −0.005 2

−0.004 −1

−1

−1 0

−0.01

−0.006 −1.5 −1.5

−1.5 −1

−1 −0.5

−1.5 −1.5

−0.008

0

−1 −0.5

0 0.5

0 0.5 1

1.5

1 1.5

1.5

x

y

x

y

(2) Re(v1 )

0 0.5

1 1.5

for κ1,P ,4 .

(h)

−0.015

−0.5

0.5 1

1.5 x

(g)

−1 −0.5

−4

0 0.5

−0.01

1 1.5

−1.5 −1

−0.5 0

0.5 1

−1.5 −1.5

−2

−1.5 −1

−0.5

y

(2) Re(v2 )

(2)

for κ2,P ,4 .

−5

(i) Re(v3 ) for κ3,P ,4 . −5

−5

x 10

x 10 4

x 10 12

6 1.5

1.5

3

1.5

10

4 1

2

0

0

−0.5

0.5

z

z

0.5

8

6

0

4

−0.5

−1 −0.5

−0.5 0

0 0.5

0.5 1

−1.5 −1.5 −6

−1.5 −1

−2

−1 −0.5

−0.5 0

−2 −1.5 −1.5

−1.5 −1

−1 −0.5

(2)

0.5 1

1.5

−4

1 1.5

x

1.5 x

y

(j) Im(v1 ) for κ1,P ,4 .

−3

0 0.5

−4

1 1.5

−0.5 0

0.5 1

x

y

(2)

(2)

(k) Im(v2 ) for κ2,P ,4 .

(l) Im(v3 ) for κ3,P ,4 . (1)

Figure 3.18: Real and imaginary part of the eigenfunctions vi ing to κ j,P ,4 , j = 1, . . . , 3.

70

−1

0 0.5

1.5

y

0

−1

1 1.5

1

0

0

−4 −1.5 −1

2

0.5

−0.5

−1

−1.5 −1.5

1

2

−2 −1

z

1

(2)

and vi , i = 1, . . . , 3 correspond-

3.7 Estimation of the refraction index

3.7 Estimation of the refraction index In this section, we propose an algorithm to estimate the constant refraction index n from the knowledge of the obstacle and the first transmission eigenvalues (which is a desired procedure in nondestructive testing). Note that the only available algorithms so far use either the Faber-Krahn-type inequality for the first transmission eigenvalue (see [22]) or a simple bisection method (see [60]). The first method is not acceptable due to large discrepancies of the estimation and the true n. The second is expensive due to many evaluations of the direct problem (given n, compute the first transmission eigenvalue). We assume that we know a few interior transmission eigenvalues for a given obstacle. As a first step, we assume that we know one interior transmission eigenvalue, say 2 κ1 . Then, we consider equation (3.12) with the surface operator A defined by equation (3.13) which now depends on n rather than κ and any κ is replaced by the known κ1 . Hence, we have to solve the nonlinear eigenvalue problem A(n)X = 0 , where

A(n) :=

Lκ1 pn − Lκ1 −Mκ1 pn + Mκ1 MκT pn − MκT1 −Nκ1 pn + Nκ1

(3.17)

1

with the obvious definition of X (see (3.13) on page 47). Note that the operator A(n) is again Fredholm of index zero, and it is analytic on C\R− . After discretizing (3.17) as illustrated in Section 3.5, we again solve the nonlinear eigenvalue problem of the form A(n)v = 0,

v ∈ Cm ,

v 6= 0,

κ∈Ω⊂C

(3.18) p

˜. After with the algorithm as explained in Section 3.4.2. Note that we replace n by n 2 ˜, we have n = n ˜ . However, as one might expect the result will not be calculating n unique.

3.7.1 Estimation of contrast for the unit sphere The first eigenvalue for the unit sphere S2 is given by 3.141 593, where we used the ˜ leads contrast n = 4 (˜ n = 2). Using the new proposed way of calculating the contrast n ˜ including their multiplicities, where we used the circle with to the following result for n center (1.85, 0) and radius 1/2 as a contour and m = 192: 1.999 240[3] ,

2.000 000[1] ,

2.299 149[2] ,

2.305 344[3] . ˜ from the knowledge of the second Since, the value is not unique, we can calculate n eigenvalue κ2 = 3.692 445. Using m = 768, we get 1.800 427[3] ,

1.937 814[1] ,

1.995 092[2] ,

1.999 376[3] ,

2.258 103[3] ,

2.259 799[1] ,

2.272 973[3] .

71

3 A numerical method to compute interior transmission eigenvalues The next step would be to take the intersection of the two results. Since we are dealing with floating point numbers, we can compare any two elements only by using the convention that they are equal up to some difference. We say two elements E1 and E2 ˜ = 2.00 and are considered the same if |E1 − E2 | < 1%. Then, we get (after rounding) n hence n = 4.00, which was the contrast we started with. Interestingly, it is enough to consider only two eigenvalues to obtain a unique contrast n. If it would not be unique, one would continue using more eigenvalues until a uniqe element is found through the described intersection method. Note also that we used a very small m. Using m = 768, ˜ we get the contrasts n 1.999 986[3] ,

2.000 000[1] ,

2.306 060[2] ,

1.801 241[3] ,

1.937 814[1] ,

1.999 893[2] ,

1.999 910[3] ,

2.278 325[3] ,

2.279 122[3] ,

2.306 081[3] . for the first and

2.279 518[1] . ˜ = 2.00. It follows n = 4.00. Using larger for the second eigenvalue and hence the uniqe n ˜. We have two, four, and four possibilities for eigenvalue leads to more possibilities for n κ1 , κ2 , and κ3 , respectively (if we consider two elements the same if the distance is smaller than 1% for m = 768). We get the following values for m = 768 1.721816[3] ,

1.780663[1] ,

1.783978[5] ,

1.999187[3] ,

1.999881[3] ,

2.000219[1] ,

2.244925[2] ,

2.245649[1] ,

2.246163[3] ,

2.247330[3] . This specific example already gives inside of how an algorithm that is able to calculate an estimation for the constrast given a few eigenvalues and the obstacle might look like. In the next subsection, we summarize the procedure.

3.7.2 An algorithm to calculate an estimation for the contrast Now, we are in position to explain a procedure to calculate an estimation for the constant contrast. Assume that we have eigenvalues κi , i = 1, 2, 3, . . . for a known obstacle D with surface Γ.

72

3.7 Estimation of the refraction index Then the algorithm consists of the following steps: Step 0: Specify an appropriate contour in the complex plane, for example a circle. Initialize i with 1. Step 1: Use the discretization of (3.17) where κ is replaced by the known eigenvalue κi . Step 2: p ˜ := n using the algorithm from Solve the nonlinear eigenvalue problem (3.18) for n ˜ pi } containing pi ≥ 0 elements. Section 3.4.2. The result will be the set N˜i := {˜ n1i , . . . , n Step 3: If pi = 0, then go to Step 0 and choose another contour. Step 4: ˜21 will be the contrast n. If pi = 1, stop the algorithm. n Otherwise if i is 1, increase i by 1 and go to Step 1. Step 5: If i > 1 take an (appropriate) intersection of N˜i and N˜i−1 . If the intersection contains one element, then stop. Otherwise, set N˜i = N˜i ∩ N˜i−1 and go to Step 1. Note that the crucial part of this algorithm is the choice of an appropriate intersection function; i.e., the right threshold value. The question remains if one can directly combine solving two nonlinear eigenvalue problems for n1 and n2 under the constraint n1 = n2 (or even p nonlinear eigenvalue problems under the constraint n1 = n2 = . . . = n p ), which will be considered future research.

3.7.3 Application of the proposed algorithm In this subsection, we will apply the proposed algorithm from the previous subsection to the ellipsoidal, peanut-shaped, acorn-shaped, and cushion-shaped surfaces as well. Our findings are listed in Table 3.13. Here, we report the sets N˜i for i = 1, 2 for different surfaces using κi , i = 1, 2. Additionally, we use an oval box to mark the unique estimated ˜. Any of these values squared and rounded gives the contrast n. contrast n As we can see, we can recover the constant contrast n = 4.00 for all five surfaces under consideration. Interestingly, two to three eigenvalues were enough to ensure the uniqueness. We need three eigenvalues for the ellipsoidal, acorn, and cushion-shaped surface, respectively. The reason might be that the first two eigenvalues are very close to each other. Thus, our advise is to use two eigenvalues that are distinct (the distance is

73

3 A numerical method to compute interior transmission eigenvalues S2 N˜1 N˜2 E N˜1 N˜2 N˜3 P N˜1 N˜2 A N˜1 N˜2 N˜3 C N˜1 N˜2 N˜3

1.999 986 [3] , 1.801 241 [3] , 2.278 325 [3] ,

2.000 000 [1] , 1.937 814 [1] , 2.279 122 [3] ,

2.000 014 [1] , 1.961 259 [1] , 2.308 214 [2] , 1.907 732 [1] , 2.226 327 [2] ,

1.999 828 [1] , 1.916 773 [1] , 2.254 307 [2] ,

2.004 688 1.999 968

[1] , [1] ,

2.306 081 [3] , 1.999 910 [3] ,

2.106 339 [2] , 2.062 082 [2] ,

2.309 152 [1] , 2.259 299 [1] ,

1.991 011 [1] , 2.321 543 [2]

1.999 938 [2] ,

2.124 571 [1] , 2.000 124 [1] , 2.260 086 [2] ,

2.201 110 [1] , 2.176 429 [1] ,

2.227 661 [1] ,

2.098 831 [1] , 2.092 468 [1] , 1.987 280 [1] , 2.238 764 [1] ,

2.174 251 [2] , 2.165 971 [2] , 2.025 899 [2] , 2.243 261 [1] ,

2.148 752 [1] ,

2.148 908 [1] ,

2.134 612 [1] , 2.348 646 [2] , 1.999 386 [1] , 2.233 806 [2] ,

2.134 766 [1] ,

1.99 5548 [1] , 1.99 0991 [1] , 1.90 0786 [2] , 2.22 5917 [2] ,

2.00 2731 [2] , 1.99 6762 [2] , 1.90 4686 [1] , 2.22 8727 [1] ,

1.999 334 [2] , 2.151 459 [1] , 1.987 971 [2] , 2.138 575 [1] , 1.881 800 [2] , 2.016 798 [1] ,

2.306 060 [2] , 1.999 893 [2] , 2.279 518 [1] ,

2.006 331 [2] , 1.999 188 [2] , 2.335 564 [2] , 1.937 152 [2] , 2.178 440 [2] ,

2.191 287 [1] ,

1.999 511 [1] , 2.341 157 [2] ,

Table 3.13: Estimated contrasts for various surfaces for m = 768.

larger than the prescribed tolerance), but they should be as small as possible, since for ˜ is obtained. larger eigenvalues more possibilities for n

3.8 Summary and conclusions We provide a method to calculate transmission eigenvalues within high accuracy for many different obstacles in three dimensions. To our knowledge the new algorithm described in this chapter is the first that provides such highly accurate values for obstacles different from a sphere. Note that the computational cost is dramatically reduced compared to existing methods. We further provide eigenvalues for the interior Dirichlet

74

3.8 Summary and conclusions problem and thus those eigenvalues might serve for checking newly derived Faber-Krahn type inequalities. Faber-Krahn type inequalities for larger transmission eigenvalues are currently not available and remain a topic of future research. Moreover, we are able to calculate complex-valued transmission eigenvalues. For this problem no numerical values have yet been reported and the proof of existence still remains. The question remains whether all calculated complex-valued eigenvalues are really interior transmission eigenvalues or not. Finally note that one can also use this new algorithm to compute interior transmission eigenvalues for the electromagnetic scattering problem. Additionally, we are able to plot the corresponding eigenfunctions. Many open questions remain which merit further investigation. For example, the following inverse problem: compute the contrast n given the obstacle and the first few transmission eigenvalues (we already proposed a new method).

75

4 The factorization method for the acoustic transmission problem Recently, K. Anagnostopoulos et al. have rigorously established the necessary theoretical framework for the application of the factorization method to the inverse acoustic transmission problem (see [1, Section 3]). In this chapter, extended numerical examples in three dimensions are presented, where a variety of different surfaces are successfully reconstructed by the factorization method, thus complementing the method’s validation from the computational point of view.

4.1 Problem statement We refer the reader to [1, Section 2] for the precise formulation of the direct scattering problem and the physical background which lead to the so-called classical acoustic transmission scattering problem. To shorten things, we have the following problem at hand: In mathematical terms, the secondary fields uint (x) and usct (x) satisfy the classical acoustic transmission scattering problem: Find a pair of functions (uint , usct ) with uint ∈ C 2 (D) ∩ C 1 (D) and usct ∈ C 2 (R3 \D) ∩ C 1 (R3 \D) such that ∆uint (x) + ki2 uint (x) = 0 ,

x ∈ D,

(4.1)

∆usct (x) + ke2 usct (x) = 0 ,

x ∈ R3 \D ,

(4.2)

uint (x) − usct (x) = f (x) ,

x ∈ ∂ D,

(4.3)

τ−1 ∂ν uint (x) − ∂ν usct (x) = h(x) , lim r ∂ r usct (x) − ike usct (x) = 0 ,

x ∈ ∂ D,

(4.4)

r = |x| ,

(4.5)

r→∞

with specific data f (x) := uinc (x) and h(x) := (∂ν uinc )(x) for x ∈ ∂ D. Here, the incident plane wave is given by ˆ uinc (x) = eike x·d , x ∈ R3 , (4.6) where ke is the wave number of the acoustic waves in the host medium and ki is the wave number of the acoustic medium filling the interior of D. The ratio τ := ρi /ρe ∈ R+ is the mass density ratio of the two media. Note that this problem is a special case of (2.1) using u2 = τ−1 uint and u1 = usct .

77

4 The factorization method for the acoustic transmission problem The linear integral operator F : L 2 (S2 ) → L 2 (S2 ) is defined by Z

ˆ d) ˆ ds(d) ˆ , u∞ (ˆ x ; d)g(

(F g)(ˆ x) =

xˆ ∈ S2 ,

S2

whose integral kernel is formed by the forward data of the inverse problem. Here u∞ is the far-field of usct . We denote with Φ∞ (ˆ x , y) := exp{−ike xˆ · y}/(4π) ke the far-field pattern of the fundamental solution with wave number ke . The main theoretical result of K. Anagnostopoulos et al. is the following theorem (see [1, Theorem 3.8]): Theorem 11. Assume that ω is not an interior transmission eigenvalue and ki2 is neither a Dirichlet eigenvalue nor a Neumann eigenvalue of −∆ in D. Then the scatterer D can be explicitly characterized as follows: (·, z) ∈ R (F ∗ F )1/4 z ∈ D ⇐⇒ Φ∞ ke ⇐⇒ [W (z)]−1 :=

(z) ∞ X |β j |2 j=1

|λ j |

(4.7) < ∞,

(4.8)

(z) (ˆ x , z), xˆ ∈ S2 , with where β j := Φ∞ (·, z), ψ j L 2 (S2 ) are the expansion coefficients of Φ∞ ke ke respect to the complete orthonormal system of eigenfunctions {ψ j : j ∈ N} of the compact, normal and injective operator F in L 2 (S2 ) and λ j ∈ C are the corresponding eigenvalues. In the next section, we will be concerned with the task of reconstructing the shape of the scatterer D from the knowledge of the given far-field patterns using the theoretical foundation given in Theorem 11.

4.2 Numerical results In what follows, we briefly describe how to generate synthetic far-field data for a variety of surfaces in three dimensions and, additionally, its numerical approximation using the boundary element collocation method. Next, we will derive a series expansion for a sphere of radius R > 0 centered at the origin which we will use as a testing scenario to verify the correctness of the numerical approximation and show that we are able to obtain highly accurate far-field data due to superconvergence. Finally, we shortly explain the implementation of the factorization method and report the successful reconstruction for a variety of surfaces from the knowledge of the far-field data for different parameter settings.

78


4.2.1 Generation of synthetic far-field data First, we derive the system of boundary integral equations to solve the problem (4.1)– (4.5). We make the ansatz of a combination of an acoustic double and single layer potential of the form usct (x) =

Wke φ(x) + Vke ψ(x) ,

uint (x) = τ Wki φ(x) + τ Vki ψ(x) ,

x ∈ R3 \D , x ∈ D,

(4.9) (4.10)

where φ and ψ are two unknown density functions defined on the surface ∂ D. Letting the point x approach the boundary in (4.9) and (4.10) and using the jump relations leads to 1 usct = Mke φ + L ke ψ + φ on ∂ D , 2 τ uint = Mki φ + τ L ki ψ − φ on ∂ D . 2 Using the boundary condition usct − uint = −uinc , with uinc defined by (4.6), yields 1 2

(1 + τ) φ + Mke − τ Mki φ + L ke − τ L ki ψ = −uinc .

(4.11)

Taking the normal derivative in (4.9) and (4.10), letting the point x approach the boundary, and using the jump relations leads to 1 Nke φ + MkTe ψ − ψ on ∂ D , 2 τ = τ Nki φ + τMkTi ψ + ψ on ∂ D . 2

∂ν usct = ∂ν uint

Using the boundary condition ∂ν usct − τ−1 ∂ν uint = −∂ν uinc yields T T − ψ + Nke − Nki φ + Mke − Mki ψ = −∂ν uinc .

(4.12)

Equations (4.11) and (4.12) can be written in the form of a 2 × 2 system of boundary integral equations as follows 1 inc M − τ M L − τ L (1 + τ)I 0 φ −u k k k k e i e i 2 + = , Nke − Nki MkTe − MkTi ψ −∂ν uinc 0 −I (4.13)

which has to be solved for the unknown density functions φ and ψ. The far-field pattern of (4.9) is given by Z 1 ∞ xˆ ∈ S2 . (4.14) ∂ν( y) e−ike xˆ · y φ( y) + e−ike xˆ · y ψ( y) ds( y), u (ˆ x) = 4π ∂ D

79

4 The factorization method for the acoustic transmission problem Note that we used the layer ansatz, because all entries in (4.13) are integral operators with weakly singular kernel for which numerical approximations can be constructed. Furthermore, the derivation is similar to the one presented in [39]. We use the boundary element collocation method to solve (4.13) numerically (see [39, Chapter 5] for a detailed description). After obtaining the unknown densities at the collocation nodes, we can numerically evaluate (4.14) to obtain the far-field pattern. In a similar fashion as in [39, Chapter 6], we will be able to show a rate of convergence of almost four using quadratic basis functions, thus leading to superconvergence. Note that the numerical program developed by myself (see [36]) is an extension of BIEPACK (boundary integral equation package for the solution of integral equations of the second kind arising from the Laplace equation) developed by Atkinson (see [2]) which has been recently used in a variety of problems dealing with the Helmholtz equation (see [34, 37–39, 41, 42] among others).

4.2.2 Series expansion for a sphere In this section, we derive a series expansion for a sphere of radius R > 0 centered at the origin. Note that jn denotes the spherical Bessel function of the first kind of order n, h(1) n the spherical Hankel function of the first kind of order n, Pn is the Legendre polynomial of order n, and Ynm is the spherical harmonic. With the notation x = r xˆ , r > 0 and xˆ ∈ S2 we have sct

u (r xˆ ) =

∞ X n X

x) , (ke r)Ynm (ˆ anm h(1) n

r >R

n=0 m=−n

in the exterior (see [16, Theorem 2.14]). In the interior we have uint (r xˆ ) =

∞ X n X

x) , bnm jn (ki r)Ynm (ˆ

r < R.

n=0 m=−n

Using the Jacobi-Anger expansion of the incident field yields ˆ = u (r xˆ ; d) inc

∞ X n X

ˆ . 4πin jn (ke r)Ynm (ˆ x )Ynm (d)

n=0 m=−n

Using the boundary condition on the sphere r = R yields the following system of two equations for the unknown coefficients anm and bnm :   (1) m n m ˆ − jn (ki R) hn (ke R) an −4πi jn (ke R)Yn (d)  = . m (1)′ −1 ′ bn ke hn (ke R) −τ ki jn (ki R) ˆ −4πin j ′ (k R)Y m (d)k n

e

e

n

Solving this system for anm gives anm

80

= −4πi

n

ˆ Ynm (d)

ki jn′ (ki R) jn (ke R) − τke jn′ (ke R) jn (ki R) ′

(ke R) − τke h(1) ki jn′ (ki R)h(1) (ke R) jn (ki R) n n

.

4.2 Numerical results The far-field pattern of usct is given by (see [16, Theorem 2.15]) ∞

u (ˆ x) = =

∞ 1 1 X

ke i ke

n=0 ∞ X

i

m X

n+1

anm Ynm (ˆ x)

m=−n

(2n + 1)

n=0

ki jn′ (ki R) jn (ke R) − τke jn′ (ke R) jn (ki R) ′

ki jn′ (ki R)h(1) (ke R) − τke h(1) (ke R) jn (ki R) n n

ˆ , Pn (ˆ x · d) (4.15)

where the last step follows from the addition theorem for spherical harmonics (see [16, Theorem 2.8]). In the following, we create 66 incident waves and measure the far-field pattern in the same directions (see [34, Appendix A.1] for the generation of the 66 waves). This yields our set of data f i j with i, j = 1, . . . , 66, which we collect in a matrix, say F ∈ C66×66 . Due to the series expansion (4.15), we are able to compare our numerical approximation of the far-field pattern. The error between the calculated solution f i j and the ‘true’ far-field pattern u∞ (ˆ x i , dˆj ) at the point xˆi ∈ S2 corresponding to direction of 2 incidence dˆj ∈ S is denoted by En (ˆ x i , dˆj ). More precisely, we have x i , dˆj ) − f i j . En (ˆ x i , dˆj ) = u∞ (ˆ The estimated order of convergence is defined by ¦ © ¦ © x i , dˆj ) / log(2) . x i , dˆj ) / max E4n (ˆ EOC = log max En (ˆ i, j

i, j

In Table 4.1 we report the maximal error and the estimated order of convergence (EOC) for ke = 2, ki = 1, τ = 1/2, and R = 1, where n denotes the number of faces of the triangulation and the number of collocation points is denoted by n v . Using quadratic n (n v ) 4 (24) 16 (96) 64 (384) 256 (1536)

maxi, j En (ˆ x i , dˆj ) 2.3883D-01 3.0120D-02 6.2361D-04 1.3656D-05

Quadratic interpolation EOC n (n v ) maxi, j En (ˆ x i , dˆj ) 8 (48) 3.4775D-02 2.99 32 (192) 1.2316D-03 5.59 128 (768) 1.2885D-05 5.51 512 (3072) 1.0405D-06

EOC 4.82 6.58 3.63

Table 4.1: Far-field pattern errors for a sphere with R = 1 and the parameters ke = 2, ki = 1, and τ = 1/2.

ˆ3 ) (see [39, interpolation usually leads to a rate of convergence of order three; i.e., O (δ n ˆn is the mesh size of the parametrization domain. More precisely, Theorem 5.1]) where δ ˆn is the maximum diameter of the n elements that triangulate the parametrization δ domain (see [39, p. 335] for the definition). As shown in [39, Theorem 6.1], we are ˆ4 log δ ˆ−1 ). able to prove theoretically a rate of convergence of almost four; i.e., O (δ n n Although the estimated rate of convergence is slightly varying, one can observe that we

81

4 The factorization method for the acoustic transmission problem achieve numerically a rate of convergence of almost four which in turn shows that our boundary element collocation solver is able to correctly produce highly accurate far-field data. It is noted that other parameter choices lead to similar results.

4.2.3 Reconstruction of a variety of surfaces We proceed by demonstrating that one is able to reconstruct a variety of different surfaces with the factorization method from the knowledge of the far-field pattern successfully. The surfaces under consideration are shown in Figure 4.1; specifically, their triangulation is depicted. The first surface is a unit sphere. The second surface is an ellipsoid with semiaxes (1, 1, 6/5). The third surface is peanut-shaped and given in spherical coordi2 nates via x = ̺ sin(φ) ¦ © cos(θ ), y = ̺ sin(φ) sin(θ ), and z = ̺ cos(φ) with ̺ = 9 cos2 (φ) + sin2 (φ)/4 /4. The fourth surface is acorn-shaped and parametrically given by ̺ 2 = 9 17/4 + 2 cos(3φ) /25. The fifth surface is a cushion-shaped surface and represented by ̺ = 1 − cos(2φ)/2. The sixth and seventh surface are a round short and long cylinder given by ̺ 10 = 1/((2 sin(φ)/3)10 + cos10 (φ)) and ̺ 10 = 1/((2 cos(φ)/3)10 + sin10 (φ)), respectively. The eighth surface is a bumpy sphere arising in tumor growth modeling, which is given by ̺ = 1 + sin(3φ) sin(3θ )/5. The last surface is a cube centered at the origin with edge length two. Note that we use 3072 collocation nodes for all surfaces with the exception of the cube, where we use 4608 collocation nodes and the parameters NS = 128 and NN S = 4 to generate the synthetic far-field data. The implementation of the factorization method for reconstructing the surfaces is rather straightforward and easy (see, also, [33] for a description). First, define a grid G , say a cube C = [−t, +t]3 with some parameter t > 0, with N equidistant points in each of the three directions. Second, compute a singular value decomposition of the given matrix F = UΛV ∗ ∈ C66×66 , which contains the far-field data. Note that we are not necessarily restricted to the size of 66. Third, for each point z from the grid (N × N × N values) we have to compute the expansion coefficients of the expression rz = (exp(−ike z· dˆj )) j=1,...,66 ∈ C66 with respect to the columns of V by (z) βl

=

66 X

ˆ

Vj,l e−ike z·d j ,

l = 1, . . . , 66 ,

j=1

which is a simple matrix-vector multiplication ρ (z) := V T rz of V T and rz . Finally, we calculate  −1 (z) 66 X |βl |2  W (z) =  |λ | l l=1 for the indicator function in (4.8). Then, the isosurface of the map z 7→ W (z) will be plotted for an appropriately selected cut-off value, say γ, using the unit sphere as the ‘calibration’ scatterer.

82

4.2 Numerical results 1.5

1.2

1

0.8

1

0.4

0.5

0.5

0

z

0

z

z

0

−0.4

−0.5

−0.5 −0.8

−1 −1 −0.5

−0.5 0

−0.5 0

0

0.5 1

0.5

−0.5

0

0.5 1

1

−1.5 −1

−1 −0.5

0 0.5

−1

−1.2 −1

−1

0.5

1

1

0.5

1

x

x

(a) Triangulation of a unit sphere with 512 faces.

−1

x

y

y

−0.5

0

y

(b) Triangulation of an ellipsoidal surface with 512 faces.

(c) Triangulation of a peanut-shaped surface with 512 faces.

1.5

1

1

0.5

0.5

1

0

z

0

z

z

0.5

0

−0.5

−0.5

−0.5

−1

−1 −1.5

−1.5 −1.5

−1

−0.5

0

0.5

1

1.5

1.5

1

0.5

0

−0.5

−1

−1.5

−1

−0.5

0

0.5

x

1

1.5

1.5

0

0.5

1

−0.5

−1

−1.5

−1 −1.5

−1

−0.5

0

(d) Triangulation of an acorn-shaped surface with 512 faces.

1

1.5

x

y

y

0.5

1.5

1

−0.5

0

0.5

−1

−1.5

x

y

(e) Triangulation of a cushionshaped surface with 512 faces.

(f) Triangulation of a round short cylinder-shaped surface with 512 faces.

1.5

1.5

1

1

1

0.5 0.5 0

z

z

z

0.5

0 0

−0.5 −0.5

−0.5

−1 −1

−1

−1

−0.5 −1.5 −1

−1 −0.5

0

0.5

1

1

0.5

−0.5

0

0

−1

−0.5

−1.5 −1.5

−1

−0.5

0

0.5

0

0.5

0.5 x

y

y

(g) Triangulation of a round long cylindershaped surface with 512 faces.

1 1

x

(h) Triangulation of a bumpy sphere with 512 faces.

1

1.5

1.5

1

0.5

0

−0.5

−1

−1.5

x

y

(i) Triangulation of a cube with 768 faces.

Figure 4.1: Different surfaces under consideration.

As a first parameter choice, we select ke = 2, ki = 1, and τ = 1/2. Recall that the values of W (z) should be much larger for z ∈ D than those in the exterior. We plot the slice W (x i , 0, 0) for i = 1, . . . , 55 for the unit sphere to get an idea of how to choose the threshold parameter γ. As we can see in Figure 4.2, the threshold parameter should be chosen between 4 and 7; we choose γ = 6. The reconstructions of all nine surfaces are shown in Figure 4.3. Note that we have used t = 1.5 and N = 55 for all surfaces with the exception of the cylinders, where we have used t = 2 and N = 73, since the obstacles are slightly larger.

83

4 The factorization method for the acoustic transmission problem 2

10

1

W(P)

10

0

10

−1

10

−2

10

1.5

1

0.5

0 P(x,0,0)

−0.5

−1

−1.5

Figure 4.2: The slice W (x i , 0, 0) for i = 1, . . . , 55 of W for the far-field data of a sphere of radius one.

As we can observe, the reconstruction of all nine surfaces are quiet accurate. Interestingly, the method also works for piecewise smooth surfaces although this is not justified from the theoretical point of view. The reconstruction of the bumpy sphere is very smooth, but the trend is clear. Hence, we have a clear indication that the (F ∗ F )1/4 factorization method is justified not only from the theoretical but also from the computational point of view. Needless to say that the reconstruction’s quality depends on the amount of noise in the far-field data. Due to the generation of synthetic data by the forward solver, we actually, already, deal with noisy far-field data. To quantify this, one may employ the ‘normality’ criterion. More precisely, after measuring kF ∗ F − F F ∗ k2 using the matrix spectral norm, we acquire the following values for the previous nine surfaces: 0.5833, 1.3546, 2.9651, 5.8632, 7.0939, 21.6075, 11.0284, 0.8023, 4.2491, respectively. Next, we again consider the acorn-shaped surface and add Gaussian white noise to the real and imaginary part of F of various levels. The obtained reconstructions for various white noise scenarios are shown in Figure 4.4, where we used t = 2 and N = 73. As one might expect, an increase in the noise level on the far-field data results in a deterioration of the reconstruction’s quality. It should be noted, however, that the region containing the unknown scatterer is clearly identified by the method in all cases. We now turn our attention to examining the influence of the choice of the threshold parameter γ on the method’s performance. By virtue of Figure 4.5, where the reconstruction of the acorn-shaped surface for different values of γ is depicted, we clearly deduce that the choice of the threshold parameter is crucial, since different choices lead to good or less accurate reconstructions. In the same spirit, for a different set of parameters, namely ke = 2, ki = 1, and τ = 2, we again look at the slice W (x i , 0, 0) for the unit sphere. The suggested choice of γ should now be 4. Figure 4.6 illustrates the reconstruction of a selected choice of obstacles for the parameters ke = 2, ki = 1, and τ = 2. As

84


(a) Reconstructed surface of a unit sphere.

(d) Reconstructed shaped surface.

(b) Reconstructed surface.

acorn- (e) Reconstructed shaped surface.

(g) Reconstructed round long cylinder-shaped surface.

ellipsoidal

(c) Reconstructed shaped surface.

peanut-

cushion- (f) Reconstructed round short cylinder-shaped surface.

(h) Reconstructed surface of a bumpy sphere.

(i) Reconstructed surface of a cube.

Figure 4.3: Nine reconstructed surfaces with the factorization method using γ = 6. Parameters are ke = 2, ki = 1, and τ = 1/2.

one can observe, the first three surfaces are reconstructed satisfactorily whereas the last three surfaces are not that well reconstructed. Of course, one can, usually, achieve an increase of the reconstruction’s quality by employing a greater number of incidence and observation directions. This is revealed in Figure 4.7, where we show the reconstruction of the previous six surfaces (see Figure 4.6 for a comparison) for 258 incident waves and the threshold parameter chosen as γ = 4. As one easily confirms, the reconstructions are of higher accuracy for increasing number of incidence and observation directions compared to the previous reconstructions. This is evident for the acorn-shaped and the round long-cylinder shaped surfaces. However, the reconstruction of the bumpy sphere, although certainly better, is not that good. Finally, the surface of the cube appears, also, to be more accurately reconstructed.

85

4 The factorization method for the acoustic transmission problem

(a) Reconstructed acorn- (b) Reconstructed acorn- (c) Reconstructed acornshaped surface with no noise. shaped surface 0.5% noise. shaped surface 1% noise.

(d) Reconstructed acorn- (e) Reconstructed acorn- (f) Reconstructed acorn-shaped shaped surface 1.5% noise. shaped surface 2% noise. surface 5% noise.

Figure 4.4: The reconstructed acorn-shaped surface with the factorization method using γ = 6 for various noise levels. Parameters are ke = 2, ki = 1, and τ = 1/2.

Next, we further investigate the quality of the reconstructions for the peanut-shaped, acorn-shaped, and cushion-shaped surfaces for varying number of incidence and observation directions, aiming at identifying a ‘lower bound’ for this number. In Figure 4.8, we provide reconstructions for those three surfaces, where we have used the parameters ke = 2, ki = 1, τ = 1/2, t = 1.5, and N = 55. We observe that we need more than 34 incident waves and observation directions to obtain a good reconstruction for all three surfaces under consideration. A descent reconstruction is obtained for 34 incident waves and observation directions. Using 18 incident waves and observation directions results in poor reconstructions, even for the increased values t = 2 and N = 73. Of course, our criterion for assessing the reconstructions’ quality is the human visual perception, since there is no quantitative criterion to measure the quality of the reconstructions. In a next step, we provide some numerical examples, which aim at demonstrating the ability of the factorization method to deliver reconstructions with no a priori information about the number of connected components of the unknown scatterer. First, we consider an example with two scatterers. To create the far-field, we use the following heuristic approach. We separately compute the far-field for a cube centered at (0, 0, 3) with edge length two and for a unit sphere centered at (0, 0, −3) using 66 directions of incidence and observation and the parameters ke = 2, ki = 1, and τ = 1/2. We subsequently form the superposition of the two far-field patterns and use it as the forward data for the factorization method algorithm. Although a heuristic and mathematically unjustified

86


(a) Reconstructed acorn- (b) Reconstructed acorn- (c) Reconstructed acornshaped surface for γ = 3. shaped surface for γ = 3.5. shaped surface for γ = 4.

(d) Reconstructed acorn- (e) Reconstructed acorn- (f) Reconstructed acorn-shaped shaped surface for γ = 4.5. shaped surface for γ = 5. surface for γ = 5.5.

Figure 4.5: Reconstructed surfaces with the factorization method using various γ for the acornshaped surface. Parameters are ke = 2, ki = 1, and τ = 1/2.

approach which, of course, ignores both multiple scattering effects and all interactions among the members of the cluster, we have decided to adopt it as an ‘ultimate proof’ that the method indeed works with no a priori information1 . Indeed, this is the case in Figure 4.9(a). Additionally, we provide a second example in Figure 4.9(b) with an acorn centered at the origin and a unit sphere centered at (0, 0, −5). In both cases, we are able to reconstruct satisfactorily the two distinct obstacles. However, one observes that the reconstructions of the sphere and the cube, depicted in Figure 4.9(a), seem to ‘attract’ each other, which is also true for the reconstructions in Figure 4.9(b). To interpret this behaviour, one should take into account that the ‘real’ distance between the two objects (lower cube face and top of the sphere) is 4 units, which is very close to the wavelength λe = π, in the host medium. Not much can be expected for distances smaller than a wavelength. This is actually revealed in Figure 4.10, where we have used the first example (Figure 4.9(a)) and inserted an acorn-shaped obstacle between the cube and the sphere. Therein, the three obstacles, being too close to each other, tend to get merged. (Note that for the reconstruction of multiple scatterers, we have used t = 2 with N = 73 in x- and y-direction and N = 163 in the z-direction.) All the numerical reconstructions presented so far, have been obtained by solving a 1

It is noted that such an approach has already been used by pioneers in the field (see, for example, [35, p. 104] for generating far-field data for two open arcs).

87




ellipsoidal

(d) Reconstructed round long cylinder-shaped surface.

(e) Reconstructed surface of a bumpy sphere.


acorn-

(f) Reconstructed surface of a cube.

Figure 4.6: Six reconstructed surfaces with the factorization method using γ = 4. Parameters are ke = 2, ki = 1, and τ = 2.

discretized version of the far-field equation x ) = Φ∞ (ˆ x , z) , (F ∗ F )1/4 gz (ˆ ke

xˆ ∈ S2 ,

which is, of course, inherently ill-posed due to the compactness of the far-field operator. Alternatively, one may obtain an approximate solution of the continuous problem by employing an appropriate regularization scheme. In what follows, we investigate the use of a regularized version of the far-field equation according to the Tikhonov-Morozov regularization scheme [49, 62]. Thus, at the discrete level, instead of considering 2

kgz k =

(z) 66 X |β |2 l

l=1

|λl |

:=

1 W (z)

,

we compute kgz,α k2 =

66 X l=1

|λl | α + |λl |

1

(z)

2 2 |βl | :=

Wα (z)

,

where the regularization parameter α is determined for each point z on the selected grid as the unique zero of the discrepancy function f (α) =

88

66 X α2 − δ2 |λl | (z) 2 2 |βl | , α + |λ | l l=1




ellipsoidal

(d) Reconstructed round long cylinder-shaped surface.

(e) Reconstructed surface of a bumpy sphere.


acorn-

(f) Reconstructed surface of a cube.

Figure 4.7: Six reconstructed surfaces with the factorization method using γ = 4 and F ∈ C258×258 . Parameters are ke = 2, ki = 1, and τ = 2.

for some known estimate δ of the error in the far-field matrix. We use the following procedure to corrupt the far-field matrix in a way such that kF − Fδ k2 ¶ δ can be ensured: Let NR and NI be two random matrices with entries uniformly distributed in the interval [−1, 1]. For a given ε, we define the perturbed matrix Fδ by Fδ := F + ε NR + iNI F. Hence, we can always choose ε so that kF − Fδ k2 ¶ δ is guaranteed for some δ. With this δ at hand, we can then compute the regularization parameter α for each point z in the grid. It is to be noted, however, that this does not include the error in the far-field data due to the forward solver, which could be much larger than the error due to the artificially introduced random noise. In other words, the whole approach is very likely to underestimate the ‘true’ error in the far-field data. In figures 4.11 and 4.12, we show the reconstruction of the acorn-shaped surface for δ = 1% and δ = 5%, respectively, using both the unregularized and the regularized version of the far-field equation (i.e., plots of W (z) and Wα (z), respectively) for the parameters ke = 2, ki = 1, and τ = 1/2. For completeness, we also include isosurface plots for the map z 7→ α(z), after adjusting appropriately the threshold parameter γ. Figures 4.11 and 4.12 clearly demonstrate that a stable reconstruction can be obtained upon using the regularized far-field equation. Interestingly, we are also able to exploit the mapping z 7→ α(z) to get a reconstruction of the surface of the unknown scatterer

89


(a) Peanut-shaped surface with F ∈ C18×18 .

(b) Acorn-shaped surface with F ∈ C18×18 .

(c) Cushion-shaped with F ∈ C18×18 .

surface

(d) Peanut-shaped surface with F ∈ C34×34 .

(e) Acorn-shaped surface with F ∈ C34×34 .

(f) Cushion-shaped with F ∈ C34×34 .

surface

Figure 4.8: Three reconstructed surfaces with the factorization method using γ = 6 for varying number of incidence and observation directions. Parameters are ke = 2, ki = 1, and τ = 1/2.

(see, e.g., Figure 4.11(e)), which sometimes appears even better than both the regularized and unregularized reconstructions. However, note that the employment of a regularization scheme for solving the far-field equation does not lead to superior reconstructions compared to those obtained without regularization. This has also been observed in [19]. In fact, the regularized reconstruction appears to be, overall speaking, a little smoother than the unregularized one.

4.3 Summary The theoretical investigation of the method is complemented by numerical examples which exploit generated synthetic far-field data for a variety of surfaces in three dimensions. The elaborated numerical reconstructions reveal that the factorization method is capable of effectively identifying both connected and disconnected scatterers in terms of their location and geometry for various selections of the physical parameters involved.

90

4.3 Summary

(g) Peanut-shaped surface with F ∈ C66×66 .

(h) Acorn-shaped surface with F ∈ C66×66 .

(i) Cushion-shaped with F ∈ C66×66 .

surface

(j) Peanut-shaped surface with F ∈ C130×130 .

(k) Acorn-shaped surface with F ∈ C130×130 .

(l) Cushion-shaped with F ∈ C130×130 .

surface

(m) Peanut-shaped with F ∈ C258×258 .

(n) Acorn-shaped surface with F ∈ C258×258 .

(o) Cushion-shaped with F ∈ C258×258 .

surface

surface

Figure 4.8: Three reconstructed surfaces with the factorization method using γ = 6 for varying number of incidence and observation directions. Parameters are ke = 2, ki = 1, and τ = 1/2.

91


(a) Reconstructed cube of edge length two and unit sphere with F ∈ C66×66 .

(b) Reconstructed acorn-shaped surface and unit sphere with F ∈ C66×66 .

Figure 4.9: Two simultaneously reconstructed surfaces with the factorization method using γ = 6. Parameters are ke = 2, ki = 1, and τ = 1/2.

(a) Reconstructed cube of edge length two, acorn-shaped surface, and unit sphere with F ∈ C66×66 .

Figure 4.10: Three simultaneously reconstructed surfaces with the factorization method using γ = 6. Parameters are ke = 2, ki = 1, and τ = 1/2.

92

4.3 Summary

(a) Isosurface of W (z) with threshold γ = 6.

(b) Isosurface of Wα (z) with threshold γ = 6.

(c) Isosurface of α(z) with threshold γ = 0.01.

(d) Isosurface of α(z) with threshold γ = 0.005.

(e) Isosurface of α(z) with threshold γ = 0.001.

(f) Isosurface of α(z) with threshold γ = 0.0005.

Figure 4.11: Reconstruction of the acorn-shaped surface using the unregularized and regularized version of the far-field equation with 66 incident waves and observation directions using δ = 1%. Parameters are ke = 2, ki = 1, and τ = 1/2.

93


(a) Isosurface of W (z) with threshold γ = 6.

(b) Isosurface of Wα (z) with threshold γ = 6.

(c) Isosurface of α(z) with threshold γ = 0.1.

(d) Isosurface of α(z) with threshold γ = 0.05.

(e) Isosurface of α(z) with threshold γ = 0.01.

(f) Isosurface of α(z) with threshold γ = 0.005.

Figure 4.12: Reconstruction of the acorn-shaped surface using the unregularized and regularized version of the far-field equation with 66 incident waves and observation directions using δ = 5%. Parameters are ke = 2, ki = 1, and τ = 1/2.

94

5 The exterior Maxwell problem A boundary integral equation is described that solves the exterior Maxwell boundary-value problem in three dimensions. The solution is found by approximating the Fredholm integral equation of the second kind with the boundary element collocation method. We prove superconvergence at the collocation nodes distinguishing the cases of even and odd interpolation. Numerical results demonstrate the performance of the method and confirm the superconvergence.

5.1 Introduction Electromagnetic scattering arises in science and engineering with applications such as radar imaging, biomedicine, environmental pollution, and geophysical prospecting. The direct electromagnetic scattering problem consists of finding the scattered field provided the geometry of the scatterer is known. In the inverse problem one tries to reconstruct the geometry from knowledge of the scattering amplitude also known as the far-field pattern. Such problem arises in military applications to identify objects by radar, in medical imaging to detect leukemia, or in non-destructive testing (see Colton et al. [21]). There are several approaches to solve the inverse problem as presented in [21]. However, each of these methods needs accurate far-field data for complex geometries in three dimensions, since exact solutions are only known for special geometries like a sphere. Therefore, we present a boundary element collocation solver that is able to calculate accurate far-field patterns for complex smooth surfaces with perfect conductor boundary condition in three dimensions. The high accuracy is due to the obtained superconvergence which is in agreement with the theoretical rates of superconvengence as stated in Theorem 18 and Theorem 22, where we distinguish the cases of even and odd interpolation. The outline of this chapter is as follows. In Section 5.2, the exterior Maxwell boundaryvalue problem is presented. The next section discusses the integral equation to solve the problem at hand. A special case of the exterior Mawell boundary-value problem is the direct scattering problem for time-harmonic electromagnetic waves which is reviewed in Section 5.4. Section 5.5 explains the boundary element collocation method, and we give a convergence and error analysis; that is, we review the consistency, stability, and convergence order of the boundary element collocation method. In the next section, we prove superconvergence at the collocation nodes distinguishing the cases of even and odd interpolation. The two main results of this chapter are stated in Theorem 18 and

95

5 The exterior Maxwell problem Theorem 22. In Section 5.7, numerical results for several smooth surfaces are presented which are in agreement with the theoretical results. A short summary concludes this chapter.

5.2 The exterior Maxwell boundary-value problem Let D be a bounded open region in R3 . The boundary of D is denoted by Γ and is assumed to consist of a finite number of disjoint, closed, bounded surfaces belonging to ¯ is connected (see [15, p. 32]). class C 2 , and we assume that the complement R3 \ D The exterior boundary-value problem for the Maxwell equations consists of finding a ¯ ; C3 ) ∩ C(R3 \D; C3 ) to radiating solution E = (E1 , E2 , E3 )T , H = (H1 , H2 , H3 )T ∈ C 1 (R3 \ D the reduced Maxwell equations ∇ × E − iκH = 0 ,

∇ × H + iκE = 0

in

¯, R3 \ D

Im(κ) ≥ 0 ,

(5.1)

that satisfy the boundary condition ν×E= f

on

Γ,

(5.2)

0,α (Γ; C3 ) = {a ∈ C t0,α (Γ; C3 ) : Div a ∈ C 0,α (Γ; C)} is where the tangential field f ∈ C t,d given. Here, C t0,α (Γ; C3 ) denotes the space of Hölder continuous tangential fields with the Hölder norm. In addition, the wave number κ is given, and the solution has to satisfy one of the Silver-Müller radiation conditions

lim (H × x − r E) = 0

r→∞

or

lim (E × x + r H) = 0 ,

r→∞

where r = |x| and the limits hold uniformly for all directions x/|x|. We also need the space C t,d (Γ; C3 ) = {a ∈ C t (Γ; C3 ) : Div a ∈ C(Γ; C)}, where C t (Γ; C3 ) denotes the space of continuous tangential fields with the supremum norm.

5.3 Integral equations In this section, we derive an integral equation to solve the problem at hand. First, define the vector potential Z A(x) =

Φκ (x, y)a( y) ds( y) , Γ

x ∈ R3 \Γ

with density a = (a1 , a2 , a3 )T which is a given integrable vector field. The matrix kernel is given by Φκ (x, y) = Φκ (x, y)I3×3 , where Φκ (x, y) = exp(iκr)/4πr with r = |x − y| for x, y ∈ R3 , x 6= y is the fundamental solution of the Helmholtz equation. One can show that the fields Em = ∇ × A and H m = iκ1 ∇ × Em solve the Maxwell equations in

96

5.3 Integral equations the interior and exterior and, in addition, satisfy the Silver-Müller radiation conditions. They represent the electromagnetic field of a magnetic dipole distribution on Γ. Thus, we assume that the unknown electric and magnetic fields in the exterior are given by Z 1 E(x) = ∇ × Φκ (x, y)a( y) ds( y) , (5.3) H(x) = ∇ × E(x) , iκ Γ ¯ with unknown tangential field a ∈ C t,d (Γ; C3 ). Calculating respectively, for x ∈ R3 \ D ν(x)×E(x) of E(x) in (5.3), letting x approach the boundary and using the jump relation (see [16, Theorem 6.11]), we obtain Z ¦ © 1 ν(x) × E(x) = ν(x) × ∇ x × Φκ (x, y)a( y) ds( y) + a(x) , x ∈ Γ. 2 Γ Applying the boundary condition (5.2) yields Z ¦ © 1 a(x) + ν(x) × ∇ x × Φκ (x, y)a( y) ds( y) = f (x) , 2 Γ

x ∈ Γ,

(5.4)

0,α where the tangential field f = ( f1 , f2 , f3 )T ∈ C t,d (Γ; C3 ) is given. After obtaining the tangential vector field a on Γ, we insert a in (5.3) to obtain the electric and magnetic field in the exterior, respectively. To solve the Fredholm integral equation of the second kind (5.4), we first rewrite (5.4) in the following form Z 1 a(x) + K1 (x, y)a( y) ds( y) = f (x) , x ∈ Γ, (5.5) 2 Γ

where the vectors are a(x) = (a1 (x), a2 (x), a3 (x))T , f (x) = ( f1 (x), f2 (x), f3 (x))T ,

ν(x) = (v1 (x), v2 (x), v3 (x))T , ν( y) = (w1 ( y), w2 ( y), w3 ( y))T ,

and the matrix kernel K1 (x, y) is given by K1 (x, y) = −∂ x 1 Φκ v1 − ∂ x 2 Φκ v2 − ∂ x 3 Φκ v3 I   ∂ x 1 Φκ (v1 − w1 ) ∂ x 1 Φκ (v2 − w2 ) ∂ x 1 Φκ (v3 − w2 )   +  ∂ x 2 Φκ (v1 − w1 ) ∂ x 2 Φκ (v2 − w2 ) ∂ x 2 Φκ (v3 − w3 )  . ∂ x 3 Φκ (v1 − w1 ) ∂ x 3 Φκ (v2 − w2 ) ∂ x 3 Φκ (v3 − w3 )

(5.6) (5.7)

¯ the equations in (5.3) can be written as For x ∈ R3 \ D Z Z E(x) =

K2 (x, y)a( y) ds( y) , Γ

H(x) =

K3 (x, y)a( y) ds( y) ,

(5.8)

Γ

97

5 The exterior Maxwell problem respectively, where the matrix kernels K2 (x, y) and K3 (x, y) are given by   0 −∂ x 3 Φκ ∂ x 2 Φκ   0 −∂ x 1 Φκ  K2 (x, y) =  ∂ x 3 Φκ −∂ x 2 Φκ ∂ x 1 Φκ 0

(5.9)

and K3 (x, y)  −∂ x22 x 2 Φκ − ∂ x23 x 3 Φκ ∂ x21 x 2 Φκ ∂ x21 x 3 Φκ 1   ∂ x21 x 2 Φκ −∂ x21 x 1 Φκ − ∂ x23 x 3 Φκ ∂ x22 x 3 Φκ = iκ  ∂ x21 x 3 Φκ ∂ x22 x 3 Φκ −∂ x21 x 1 Φκ − ∂ x22 x 2 Φκ

  ,  (5.10)

respectively. The entries in K1 , K2 and K3 can be calculated explicitly. For j, k ∈ {1, 2, 3}, we have ∂ x j Φκ = ∂ x2j x k Φκ = with r = delta.

p

eiκr 4πr 3 eiκr 4πr

(iκr − 1)(x j − y j ) ,

(−r 2 κ2 − 3iκr + 3)(x j − y j )(x k − yk ) + δ jk 5

eiκr 4πr 3

(iκr − 1)

(x 1 − y1 )2 + (x 2 − y2 )2 + (x 3 − y3 )2 and where δ jk denotes the Kronecker

5.4 The direct scattering problem for time-harmonic electromagnetic waves In this section, we consider the direct scattering problem for time-harmonic electromagnetic waves. For two real-valued given vectors d and p with |d| = 1 and p ⊥ d, the incident plane waves are E i (x) = p eiκx·d ,

H i (x) = d × p eiκx·d

and solve the Maxwell equations in R3 . The direct problem consists of finding the scattered field E s and H s as a radiating solution to the Maxwell equations in the exterior of the scatterer D, where E i , H i , and the obstacle D are given. That is, we have to solve (5.1) with boundary condition (5.2) f = −ν × E i . The radiating solutions E s and H s have the asymptotic expansion 1 1 eiκ|x| eiκ|x| ∞ ∞ E (ˆ x) + O , H(x) = H (ˆ x) + O E(x) = |x| |x| |x| |x| 98

5.5 The Boundary Element Collocation Method for |x| → ∞ uniformly in all directions xˆ = x/|x|. The vector fields E ∞ and H ∞ are defined on the unit sphere S2 = {x ∈ R3 : |x| = 1} and are called the electric and magnetic far-field pattern. It can be shown that they satisfy H ∞ = ν × E ∞ and ν · E ∞ = ν · H ∞ = 0. The far-field patterns E ∞ and H ∞ of (5.3) are given by Z iκ ∞ E (ˆ x) = xˆ × e−iκ xˆ · y a( y) ds( y) , xˆ ∈ S2 (5.11) 4π Γ Z iκ ∞ xˆ × e−iκˆx · y a( y) × xˆ ds( y) , xˆ ∈ S2 (5.12) H (ˆ x) = − 4π Γ which follows directly from the two estimates given in [16, Equations (6.25) and (6.26)]. For xˆ ∈ S2 equations (5.11) and (5.12) can be written as Z Z E ∞ (ˆ x) =

K4 (ˆ x , y)a( y) ds( y) ,

H ∞ (ˆ x) =

K5 (ˆ x , y)a( y) ds( y) , Γ

Γ

respectively, where the matrix kernels K4 (ˆ x , y) and K5 (ˆ x , y) are given by   0 −ˆ x 3 xˆ2 iκ −iκ xˆ · y   0 −ˆ x1  , K4 (ˆ x , y) = e  xˆ3 4π −ˆ x 2 xˆ1 0  xˆ1 xˆ3 − xˆ22 + xˆ32 xˆ1 xˆ2 iκ −iκ xˆ · y  2 2  K5 (ˆ x , y) = e xˆ1 xˆ2 − xˆ1 + xˆ3  xˆ2 xˆ3 4π xˆ1 xˆ3 xˆ2 xˆ3 − xˆ12 + xˆ22

(5.13)    . (5.14) 

5.5 The Boundary Element Collocation Method In this section, we describe the boundary element collocation method to solve the integral equation of the second kind given by (5.5) numerically. Note that this projection method is also discussed in [3]. First, we assume that Γ is a connected smooth surface of class C 2 ; that is, Γ can be written as Γ = Γ1 ∪ · · · ∪ Γ J , (5.15) where every Γ j is divided into a triangular mesh and the collection of those is denoted by Tn = {∆k | 1 ≤ k ≤ n} . (5.16) Let the unit simplex in the st-plane be defined as σ = {(s, t) | 0 ≤ s, t, s + t ≤ 1} . For a given constant α with 0 < α < 1/3, let i + (2 − 3i)α j + (2 − 3 j)α , , (si , t j ) = 2 2

0 ≤ i, j, i + j ≤ r

(5.17)

99

5 The exterior Maxwell problem be the uniform grid inside of σ with f r = (r + 1)(r + 2)/2 nodes. The ordering of this grid is denoted by the nodes {q1 , . . . , q f r }. For each ∆k , we assume that there is a map 1−1

mk : σ −−−→ ∆k ,

(5.18)

onto

which is used for interpolation and integration on ∆k (see [36, Appendix A] for some surfaces). Define the node points of ∆k by vk, j = mk (q j ),

j = 1, . . . , f r .

To obtain a triangulation (5.16) and the mapping (5.18), we use a parametric representation for each region Γ j of (5.15). Assume that for each Γ j , there is a map 1−1

F j : R j −−−→ Γ j , onto

j = 1, . . . , J ,

(5.19)

where R j is a polygonal region in the plane and F j is sufficiently smooth. That means, ˆ k, j be an element of the a triangulation of R j is mapped onto a triangulation Γ j . Let ∆ triangulation of R j with vertices ˆ vk,1 , ˆ vk,2 , and ˆ vk,3 . Then the map (5.18) is given by mk (s, t) = F j ((1 − s − t)ˆ vk,1 + t ˆ vk,2 + sˆ vk,3 ) = F j (Tk (s, t)) ,

(s, t) ∈ σ ,

(5.20)

with the obvious definition of the map Tk . We collect all triangles of R j together for all j and denote the triangulation of the parametrization plane with ˆ k | 1 ≤ k ≤ n} Tˆn = {∆

(5.21)

ˆn = max diam(∆ ˆ k ) = max hk δ

(5.22)

and mesh size 1≤k≤n

1≤k≤n

ˆn → 0 as n → ∞. satisfying δ Most smooth surfaces can be decomposed as in (5.15). In the sequel, we consider conforming triangulations satisfying T3 (see [2, p. 188]). That is, if two triangles in Tˆ have a nonempty intersection, then that intersection consists either of (i) a single vertex, or (ii) all of a common edge. Note that T1 and T2 (see [2, p. 188] for the definition) are automatically satisfied, since our surface is assumed to be smooth. The refinement ˆ k ∈ Tˆn is done by connecting the midpoints of the three sides of ∆ ˆ k yielding four of ∆ new triangles. Thus, T3 is automatically satisfied and this also leads to symmetry in the triangulation, and cancellation of errors occur (see [2, p. 173]). The interpolation polynomial of degree r over σ and the corresponding Lagrange basis functions are obtained by the usual condition Li (qi ) = 1 ,

100

and

Li (q j ) = 0 ,

if

i 6= j .

(5.23)

5.5 The Boundary Element Collocation Method

i 1

Constant qi Li (s, t) (α, α)

1

Linear qi

Li (s, t)

(α, α)

u−α 1−3α t−α 1−3α s−α 1−3α

2

(α, 1 − 2α)

3

(1 − 2α, α)

4 5 6

Quadratic qi L (s, t) i u−α u−α (α, α) 2 − 1 1−3α 1−3α t−α t−α (α, 1 − 2α) 2 −1 1−3α 1−3α s−α s−α 2 − 1 (1 − 2α, α) 1−3α 1−3α 1−α t−α u−α 4 1−3α 1−3α α, 2 1−α 1−α s−α t−α 4 1−3α , 2 2 1−3α 1−α s−α u−α , α 4 2 1−3α 1−3α

Table 5.1: Nodes and Lagrange basis functions over σ for constant, linear, and quadratic interpolation.

In Table 5.1, we state the nodes and Lagrange basis functions over σ for constant, linear, and quadratic interpolation, where u = 1 − s − t. The interpolation operator of degree r over ∆k is given by an (mk (s, t)) := Pn a(mk (s, t)) =

fr X

a(mk (q j ))L j (s, t) ,

j=1

(s, t) ∈ σ ,

(5.24)

where k = 1, . . . , n. Recall that we have to solve the Fredholm integral equation of the second kind Z 1 a(x) + K1 (x, y)a( y) ds( y) = f (x) , x ∈ Γ, (5.25) 2 Γ written abstractly as 1 2

a + K a = f on Γ.

(5.26)

Using the map (5.20), equation (5.25) is equivalent to n Z X 1 K1 (x, mk (s, t))a(mk (s, t)) ∂s mk × ∂ t mk (s, t) dσ = f (x) , a(x) + 2 k=1 σ where x ∈ Γ. Next, define the collocation nodes by {vk, j } = {mk (q j ), k = 1, . . . , n, j = 1, . . . , f r }.

(5.27)

Then substitute the approximated solution an given by (5.24) in (5.25) and force the residual n Z X 1 rn (x) = K1 (x, mk (s, t))an (mk (s, t)) ∂s mk × ∂ t mk (s, t) dσ an (x) + 2 k=1 σ − f (x),

x ∈Γ

101

5 The exterior Maxwell problem to be zero at the collocation nodes. Thus, we have to solve the linear system of size 3n· f r × 3n· f r given by fr Z n X X 1 K1 (vi,l , mk (s, t))L j (s, t) ∂s mk × ∂ t mk (s, t) dσ a(mk (q j )) a(vi,l ) + 2 k=1 j=1 σ = f (vi,l ), (5.28) where i = 1, . . . , n and l = 1, . . . , f r . This can be written in matrix form as 1 I+A x = g, 2 where the (i, l)th element of x ∈ C3n· f r and g ∈ C3n· f r are given by x i,l = a(vi,l ) ∈ C3 and g i,l = f (vi,l ) ∈ C3 , respectively for i = 1, . . . , n and l = 1, . . . , f r . The (i, l), ( j, k)th element of A ∈ C3n· f r ×3n· f r is given by Z A(i,l),(k, j) = K1 (vi,l , mk (s, t))L j (s, t) ∂s mk × ∂ t mk (s, t) dσ ∈ C3×3 σ

for i, k = 1, . . . , n and l, j = 1, . . . , f r . Of course, the integrals over σ have to be approximated numerically. Since the kernel is weakly singular, we have to use a different integration routine if vi = mk (s, t) for some (s, t) ∈ σ. We use the Duffy transformation to remove the singularity. This results in integrating over a unit square. This double integral is approximated with a two-dimensional Gaussian quadrature with NS quadrature points. The T2 :5–1 rule from Stroud [59, p. 314] is used for the nonsingular integrals (see pp. 457–459 and pp. 460–462 in [2], respectively). In this case each triangle is refined by connecting the midpoints leading to four new triangles. The number of refinements is denoted by NN S . ¯ given by (5.8) can be written as The electric field for an arbitrary point x ∈ R3 \ D Z E(x) = =

K2 (x, Γ n Z X σ

k=1

y)a( y) ds( y)

K2 (x, mk (s, t))a(mk (s, t)) ∂s mk × ∂ t mk (s, t) dσ .

To approximate the electric field, replace a with an yielding fr Z n X X K2 (x, mk (s, t))L j (s, t) ∂s mk × ∂ t mk (s, t) dσ a(mk (q j )) , En (x) = k=1 j=1

σ

which can be written as fr n X X Ωk, j (x)· a(vk, j ) , En (x) = k=1 j=1

Z

Ωk, j (x) = σ

102

K2 (x, mk (s, t))L j (s, t) ∂s mk × ∂ t mk (s, t) dσ ∈ C3×3 .

5.6 Superconvergence of the boundary element collocation method In the same manner, we can obtain the approximated magnetic field H n (x) =

fr n X X

Υk, j (x)· a(vk, j ) ,

k=1 j=1

Z

Υk, j (x) = σ

K3 (x, mk (s, t))L j (s, t) ∂s mk × ∂ t mk (s, t) dσ ∈ C3×3 .

Similarly, we approximate the far-field patterns E ∞ and H ∞ . Note that the integrals are approximated numerically with the T2 :5–1 rule from Stroud [59, p. 314] and the a(mk (q j )) = a(vk, j ) is the solution of the linear system (5.28). We can establish the following result (this is an easy extension of [2, Theorem 9.2.1]). Theorem 12. Let Γ be a smooth surface. Further assume that Γ is parametrized as in (5.15) and (5.18), where each F j ∈ C r+2 . Let K be a compact integral operator from L ∞ (Γ; C3 ) t,d 0,α to C t,d (Γ; C3 ) (see equation (5.26)) and assume the equation (5.5) is uniquely solvable for 0,α all functions f ∈ C t,d (Γ; C3 ). Let Pn be the interpolation operator of degree r and consider the approximate solution of 12 I + K a = f by means of the collocation approximation. Then we obtain −1 Stability: The inverse operators 21 I + Pn K exist and are uniformly bounded for all sufficiently large n ≥ N . Convergence: The approximation an has error a − an =

1 2

1 2

I + Pn K

−1

I − Pn a

(5.29)

and therefore an → a as n → ∞. r+1 Order of convergence: Assume a ∈ C t,d (Γ). Then ˆ r+1 , ka − an k∞ ≤ c δ n

n≥N,

(5.30)

ˆn is the mesh size of the parametrization domain given by (5.22). where δ Next, we will show that we can derive a better result; that is, superconvergence at the collocation nodes.

5.6 Superconvergence of the boundary element collocation method In this section, we will show that we can improve the result (5.30) at the collocation nodes distinguishing the cases of even and odd interpolation.

103

5 The exterior Maxwell problem We confine ourselves to triangles in the parametrization plane and then use the map F j . Therefore, let τ ⊂ R2 be an arbitrary triangle with vertices {ˆ v1 , ˆ v2 , ˆ v3 }. If T 3 f = ( f1 , f2 , f3 ) ∈ C(τ; C ), then Lτ f i (x, y) =

fr X

f i (mτ (q j ))L j (s, t),

(x, y) = mτ (s, t),

i = 1, 2, 3

j=1

are polynomials of degree r in the parametrization variables s and t that interpolate f i at the nodes {mτ (q1 ), . . . , mτ (q f r )}, where q j and L j are given by (5.17) and (5.23), respectively. We have mτ (s, t) = (1 − s − t)ˆ v1 + t ˆ v2 + sˆ v3 ,

(s, t) ∈ σ ,

(5.31)

which corresponds to the map Tk given in (5.20) by suppressing the index k. We will write explicitly Lτk and mτk if necessary. The operator norm is given by kLτ k = max

(s,t)∈σ

fr X j=1

|L j (s, t)| .

(5.32)

The integration formula over τ given by Z Z τ

f i (x, y) dτ ≈

τ

Lτ f i (x, y) dτ ,

i = 1, 2, 3

(5.33)

has degree of precision of at least r. For differentiable functions f i , we define ∂ j f (x, y) i j ∂ j f (x, y) = max ∂ j f (x, y) , i = 1, 2, 3 , k∂ f k = max i i i ∞ 0≤k≤ j ∂ x k ∂ y j−k (x, y)∈τ and ∂ j f = max k∂ j f k . i ∞ 1≤i≤3

In our case, the kernel is given by   ˆ K12 ( P ˆ K13 ( P ˆ ˆ , Q) ˆ , Q) ˆ , Q) K11 ( P ˆ K22 ( P ˆ K23 ( P ˆ  ˆ , Q) ˆ , Q) ˆ , Q) ˆ = ˆ , Q) K( P   K21 ( P ˆ ˆ ˆ ˆ ˆ ˆ K31 ( P , Q) K32 ( P , Q) K33 ( P , Q) ˆ = (x, y). For simplicity, we write ˆ = (ˆ with points P x , ˆy ) and Q   K11, Pˆ (x, y) K12, Pˆ (x, y) K13, Pˆ (x, y)   K Pˆ (x, y) =  K21, Pˆ (x, y) K22, Pˆ (x, y) K23, Pˆ (x, y)  K31, Pˆ (x, y) K32, Pˆ (x, y) K33, Pˆ (x, y)

104

5.6 Superconvergence of the boundary element collocation method ˆ = K( P ˆ , Q) ˆ , (x, y)). Thus, the ith component of instead of K( P Z

Z K Pˆ (x, y) f (x, y) dτ

3 X

is given by

τ

Ki j, Pˆ (x, y) f j (x, y) dτ .

τ j=1

With this notation we are able to prove the following Lemma 13. Let τ be a planar right triangle. Further, assume that the two sides which form the right angle have length h. Let f ∈ C r+1 (τ; C3 ) and K Pˆ ∈ L 1 (τ; C3×3 ). Then Z 3 X Ki j, Pˆ (x, y)(I − Lτ ) f j (x, y) dτ τ j=1 Z 3 X K ˆ (x, y) dτ ∂ r+1 f , i = 1, 2, 3, (5.34) ≤ chr+1 i j, P

τ j=1

ˆ∈ where P / τ. Proof. Let i ∈ {1, 2, 3} be arbitrary. Let p j,r (x, y) be a Taylor polynomial of f j with degree r over τ. We have k f j − p j,r k∞ ≤ c j hr+1 k∂ r+1 f j k∞ ,

f j ∈ C r+1 (τ)

(5.35)

where c j is a suitable constant. For simplicity define Lˆ = I − Lτ and Ki j = Ki j, Pˆ (x, y). Then write Z

3 X τ j=1

Z Ki j Lˆ f j dτ =

3 X τ j=1

Ki j Lˆ f j − p j,r

Z dτ +

3 X τ k=1

Ki j Lˆp j,r dτ .

(5.36)

Because of (5.35), the first term on the right hand side of (5.36) is bounded Z Z 3 3 X X Ki j Lˆ f j − p j,r dτ ≤ |Ki j | kLˆk k f j − p j,r k∞ dτ τ j=1 τ j=1 Z 3 Z 3 X X r+1 r+1 r+1 |Ki j | k∂ r+1 f j k∞ dτ |Ki j |c j h k∂ f j k∞ dτ ≤ ch ≤ τ j=1

τ j=1

Z

≤ chr+1

3 X τ j=1

|Ki j | dτ ∂ r+1 f ,

where the bound for kLˆk is given by (5.32) and c = max j∈{1,2,3} c j . The second term on the right hand side of (5.36) is zero since Lτ p j,r = p j,r if and only if Lˆp j,r = 0. Thus, we obtain the assertion for each i ∈ {1, 2, 3}.

105

5 The exterior Maxwell problem The result of Lemma 13 can be extended to general triangles. However, the derivatives of f and K Pˆ will involve the mapping mτ . The bound of (5.34) will depend on a term proportional to some power of Υ(τ) = h(τ)/h∗ (τ), where h(τ) denotes the diameter of τ and h∗ (τ) denotes the radius of the circle inscribed ˆ n,k }, n ≥ 1 satisfies in τ and tangent to its sides. Our triangulation Tˆn = {∆

ˆ k) < ∞ ; sup max Υ(∆ n

ˆ k ∈Tˆn ∆

ˆ n,k from having i.e., it is uniformly bounded in n and therefore, it prevents the triangles ∆ angles which approach 0 as n → ∞ (see [3, p. 276]). Hence, we have the following Corollary 1. Let τ be a planar triangle of diameter h, let f ∈ C r+1 (τ; C3 ) and K Pˆ ∈ L 1 (τ; C3×3 ). Then Z 3 X Ki j, Pˆ (x, y)(I − Lτ ) f j (x, y) dτ τ j=1 Z 3 X r+1 K ˆ (x, y) dτ ∂ r+1 f , i = 1, 2, 3, (5.37) ≤ c (Υ (τ)) h i j, P τ j=1

ˆ∈ where c (Υ (τ)) is some multiple of a power of Υ(τ) and P / τ. Proof. Let i ∈ {1, 2, 3} be arbitrary. Let τ be a planar triangle of diameter h with vertices ˆ be a planar right triangle with vertices ˆ v1 , v2 , and v3 and let τ v1 , ˆ v2 , and ˆ v3 . Further, assume that the two sides which form the right angle have length h. The bijective map ˆ → τ is given by mτˆ : τ (x, y) = mτˆ xˆ , ˆy = T ◦ Tˆ −1 , ˆ and T : σ → τ are given by where the bijective maps Tˆ : σ → τ xˆ , ˆy x, y

= Tˆ (s, t) = (1 − s − t)ˆ v1 + sˆ v2 + t ˆ v3 ,

= T (s, t) = (1 − s − t)v1 + sv2 + t v3 ,

respectively. Thus, using a change of variables, Z Z =

3 X τ j=1 3 X ˆ j=1 τ

106

Ki j, Pˆ (x, y)(I − Lτ ) f j (x, y) dτ Ki j, Pˆ mτˆ xˆ , ˆy

(I − Lτ ) f j mτˆ

τ. xˆ , ˆy ∂ xˆ mτˆ × ∂ ˆy mτˆ xˆ , ˆy dˆ

5.6 Superconvergence of the boundary element collocation method One can easily check that the Jacobian of this transformation is simply Area(τ)/Area(ˆ τ), ˆ since the Jacobian of the maps T and T are 2· Area(τ) and 2· Area(ˆ τ), respectively. That is, the Jacobian is a constant. Thus, Z

3 X

Ki j, Pˆ mτˆ xˆ , ˆy

ˆ j=1 τ

=

Area(τ) Area(ˆ τ)

Z

3 X

τ xˆ , ˆy ∂ xˆ mτˆ × ∂ ˆy mτˆ xˆ , ˆy dˆ

(I − Lτ ) f j mτˆ

Ki j, Pˆ mτˆ xˆ , ˆy

ˆ j=1 τ

(I − Lτ ) f j mτˆ xˆ , ˆy

dˆ τ,

and, hence, this case can be reduced to the right triangle case. However, Ki j, Pˆ and f j , as well as their derivatives, will depend on the mapping mτˆ . In addition, the constant c (Υ (τ)) is some multiple of a power of Υ(τ). Before we prove the next lemma, we need the following Assumption 1. Let k = 0, 1 be an integer and let Γ be a smooth C 2 surface. Let each entry Ki j (P, Q) of the kernel matrix K(P, Q) of our integral operator be given in the form Ki j (P, Q) =

Ψi j (P, Q) |P − Q|

,

(5.38)

where each Ψi j is smooth and bounded. Assume k K (P, Q) ∂Q i j ≤

c |P − Q|k+1

,

P 6= Q ,

(5.39)

where c denotes a generic constant independent of P and Q. The next proof is based on the Duffy transformation (see [26]). Lemma 14. Let τ be a planar triangle of diameter h. Let f ∈ C r+1 (τ; C3 ) and the kernel K satisfy Assumption 1 for k = 0. In addition, assume that there is a singularity inside of τ; that is, P = Q inside of τ. Then Z 3 X i = 1, 2, 3, Ki j, Pˆ (x, y)(I − Lτ ) f j (x, y) dτ ≤ c (Υ (τ)) hr+2 ∂ r+1 f , τ j=1 (5.40) where c (Υ (τ)) is some multiple of a power of Υ(τ). Proof. We can assume without loss of generality the right triangular case, otherwise ˆ. proceed as in Corollary 1. Assume that the singularity occurs inside of τ, say at P ˆ . We obtain three triangles τ1 , τ2 , and τ3 . The singularity Connect the vertices of τ with P occurs at one of the vertices of each triangle. Without loss of generality we can assume

107

5 The exterior Maxwell problem ˆ = (0, 0) (use a that we deal with τ where the singularity sits at the origin; that is, P linear transformation for each triangle τ1 , τ2 , and τ3 ). Let i ∈ {1, 2, 3} be arbitrary. Let p j,r (x, y) be a Taylor polynomial of f j with degree r over τ. Using (5.35), Lτ p j,r = p j,r and the notation Lˆ = I − Lτ and Ki j = Ki j, Pˆ , we have Z 3 Z 3 X X ˆ (5.41) Ki j Lˆ f j − p j,r dτ . Ki j L f j dτ = τ j=1

τ j=1

Next, define the transformation from τ to the unit square = [0, 1] × [0, 1] by x = (1 − v)uh ,

y = uvh

with Jacobian

uh2 .

(5.42)

Note that this is the composition of the map τ to σ given by (x, y) = (sh, th) with Jacobian h2 and the Duffy transformation mapping σ to given by (s, t) = ((1 − v)u, uv) with Jacobian u. Hence, Z 3 X ˆ Ki j, Pˆ (x, y)L f j (x, y) dτ τ j=1 Z 3 X (5.41) ˆ Ki j ((0, 0), x, y)L f j (x, y) − p j,r (x, y) dτ = τ j=1 Z 3 X ≤ kLˆk k f j − p j,r k∞ Ki j ((0, 0), x, y) dτ τ j=1

(5.35)

≤

(5.42)

=

(5.38)

≤

=

ch

ch

ch

r+1

r+1

r+2

∂

∂

∂

r+1

r+1

r+1

f f f

chr+2 ∂ r+1 f

Z Z Z Z

3 X K ((0, 0), x, y) dτ ij τ j=1 3 X K ((0, 0), (1 − v)uh, uvh)uh2 du dv ij

j=1

3 X Ψi j ((0, 0), (1 − v)uh, uvh) · |u| · h du dv p (1 − v)2 u2 h2 + u2 v 2 h2 j=1

p

1

du dv . (1 − v)2 + v 2 −1/2 It is readily seen that g(u, v) = (1 − v)2 + v 2 is nonsingular together with all its ∞ derivatives on . Therefore g(u, v) ∈ C (), and hence, it is bounded on the compact set . Thus, Z 1 hr+2 ∂ r+1 f du dv ≤ chr+2 ∂ r+1 f . p (1 − v)2 + v 2

This can be done for all triangles τ j , and thus, we obtain the assertion (5.40) for each i ∈ {1, 2, 3}.

108

5.6 Superconvergence of the boundary element collocation method

5.6.1 Interpolation of even degree In this section, we assume that r is even. This implies whenever τ1 and τ2 are triangles for which τ1 ∪ τ2 is a parallelogram, then (5.33) has degree of precision r + 1. Now, we can state the next lemma. Lemma 15. Let τ1 and τ2 be two planar triangles with diameter h such that R = τ1 ∪ τ2 is a parallelogram. Let f ∈ C r+2 (R; C3 ) and let K Pˆ ∈ L 1 (R; C3×3 ) be differentiable with first derivatives ∂ x K Pˆ and ∂ y K Pˆ belonging to L 1 (R; C3×3 ). Then Z 3 X Ki j, Pˆ (x, y)(I − Lτ ) f j (x, y) dτ R j=1 Z 3 X r+2 K ˆ (x, y) + ∂ ˆ K ˆ (x, y) dτ ≤ c (Υ (R)) h i j, P

Q

i j, P

R j=1

¦ © i = 1, 2, 3 (5.43) × max ∂ r+1 f , ∂ r+2 f , / R. with Υ(R) = maxi=1,2 Υ τi and c (Υ (R)) some multiple of a power of Υ(R) and P ∈ Proof. Without loss of generality, we can assume that R is a square with side length h, otherwise proceed as in Lemma 1. Let i ∈ {1, 2, 3} be arbitrary. Let p j,r (x, y) and p j,r+1 (x, y) be Taylor polynomials of f j with degree r and r + 1 over R, respectively. We have k f j − p j,r k∞ ≤ c j hr+1 k∂ r+1 f j k∞ ,

k f j − p j,r+1 k∞ ≤ ˜c j h

r+2

k∂

r+2

f j k∞ ,

f j ∈ C r+1 (R) , fj ∈ C

r+2

(R) ,

(5.44) (5.45)

for a suitable constant c j and ˜c j , respectively. For simplicity, define Ki j = Ki j, Pˆ (x, y). In addition, we can find constants Ci j such that kKi j − Ci j k1 ≤ chk∂Qˆ Ki j k1 ,

j = 1, 2, 3,

(5.46)

where k· k1 denotes the norm on L 1 (R). Define Lˆ = I − Lτ and write Z 3 Z 3 Z 3 X X X ˆ ˆ Ki j L f j dτ = Ki j L ( f j − p j,r+1 ) dτ + (Ki j − Ci j )Lˆp j,r+1 dτ R j=1

Z +

R j=1

3 X R j=1

R j=1

Ci j Lˆp j,r+1 dτ .

(5.47)

Using (5.45), one obtains that the first term on the right hand side of (5.47) is bounded Z Z 3 3 X X ˆ ˆ k( f j − p j,r+1 )k∞ |Ki j | dτ K L ( f j − p j,r+1 ) dτ ≤ kL k R j=1 i j R j=1 Z 3 Z 3 X X (5.48) |K | dτ ∂ r+2 f . k∂ r+2 f k |K | dτ ≤ chr+2 ≤ chr+2 R j=1

j ∞

ij

ij

R j=1

109

5 The exterior Maxwell problem Since we have Lˆp j,r = 0, we obtain Lˆp j,r+1 = Lˆ(p j,r+1 − p j,r ) = Lˆ

p j,r+1 − f j − p j,r − f j

and, therefore, using (5.44) and (5.45), we have kLˆp j,r+1 k∞ ≤ chr+1 hk∂ r+2 f j k∞ + k∂ r+1 f j k∞

(5.49)

with c = max j∈{1,2,3} {c j , ˜c j }. Hence, the second term on the right hand side of (5.47) is bounded Z 3 X (Ki j − Ci j )Lˆp j,r+1 dτ R j=1 Z 3 X ≤ |Ki j − Ci j | kLˆp j,r+1 k∞ dτ R j=1

(5.49)

≤

Z

ch

r+1

Z ≤ (5.46)

≤

ch

r+1

Z chr+2

3 X R j=1 3 X R j=1 3 X R j=1

|Ki j − Ci j | hk∂ r+2 f j k∞ + k∂ r+1 f j k∞ dτ ¦ © |Ki j − Ci j | dτ max ∂ r+1 f , ∂ r+2 f ¦ © |∂Qˆ Ki j | dτ max ∂ r+1 f , ∂ r+2 f .

(5.50)

The third term on the right hand side of (5.47) is zero, since formula (5.33) has degree of precision r + 1 and Ci j is a constant. Combining (5.48) and (5.50) yields the result for each i ∈ {1, 2, 3}. Note that we will now consider a symmetric triangulation. We can construct symmetry in the triangulation Tˆn by the following procedure. We refine a triangle in the parametrization plane by dividing it into four new smaller triangles. This is accomplished by connecting the midpoints of the three sides by straight lines. Hence, the number of triangles of Tˆn increases by a factor of 4. In addition, most of the triangles ˆ−2 ). The can be grouped as parallelograms. The number of such triangles is O(n) = O(δ n p ˆ−1 ). The symmetric pairs can be chosen number of remaining triangles is O( n) = O(δ n ˆ (independent of n) such that the remaining elements are at a bounded distance from P (see [2, 4]). We want to apply these results to the individual subintegrals Z (K a) (vl ) =

K(vl , Q)a(Q) ds(Q) . Tn

110

(5.51)

5.6 Superconvergence of the boundary element collocation method The ith component of (K a) (vl ) is given by Z Z =

3 X

Ki j (vl , Q)a j (Q) ds(Q)

Tn j=1

3 X

Tˆn j=1

Ki j (vl , F j (x, y))a j (F j (x, y)) | ∂ x F j × ∂ y F j (x, y)| dx d y ,

(5.52)

where l = 1, . . . , n· f r and is now defined in the parametrization plane. We write F instead of F j , j = 1, . . . , J, to avoid confusion with the summation index j. We define f j (x, y) := a j (F (x, y)) ∂ x F × ∂ y F (x, y) , Ki j, Pˆl (x, y) := Ki j (vl , F (x, y)),

with

ˆl ) , vl = F ( P

(5.53)

which are defined in the parametrization plane. k k In the following, by a ∈ C t,d (Γ; C3 ), we mean that a ∈ C t,d (Γ; C3 ) and a ∈ C t,d (Γ j ; C3 ) holds (that is, a ◦ F j ∈ C k (R j ; C3 )), j = 1, . . . , J. In the sequel, weassume that the conforming triangulation Tˆn is symmetric and satisˆ k ) < ∞. fies supn max∆ˆ k ∈Tˆn Υ(∆ Now, we are in position to state the following Theorem 16. Assume the conditions of Theorem 12 with each parametrization function F ∈ r+2 (Γ; C3 ). Assume that the kernel K satisfies Assumption 1. C r+3 and a = (a1 , a2 , a3 )T ∈ C t,d Then ˆ−1 ) , ˆ r+2 ln(δ i = 1, 2, 3 . (5.54) max ai (vl ) − ai,n (vl ) ≤ c δ n n 1≤l≤n· f r

Proof. We will bound Z max K(vl , Q)(I − Pn )a(Q) ds(Q) 1≤l≤ f r n T Z n 3 X = max Ki j, Pˆl (x, y)(I − Lτ ) f j (x, y) dx d y 1≤l≤ f r n ˆ T j=1

max K (I − Pn )a(vl ) =

1≤l≤ f r n

n

(5.55) to prove (5.54), where we used (5.51) and (5.52) in the last step. By assumption f belongs to C r+2 in the parametrization plane. Let i ∈ {1, 2, 3} be arbitrary. For a given collocation point vl , denote by ∆∗ the curved ˆ ∗ the triangle in the parametrization plane triangle containing this point, and denote by ∆ ˆl ∈ ∆ ˆl ) (see (5.52)). Define ˆ ∗ satisfying vl = F ( P containing the point P ˆ∗ Tˆn∗ = Tˆn − ∆

111

5 The exterior Maxwell problem and subdivide Tˆn∗ into two disjoint classes Tˆn(1) and Tˆn(2) such that Tˆn(1) ∪ Tˆn(2) = Tˆn∗ , where Tˆn(1) denotes the set of triangles making up parallelograms to the maximum extent possible, and Tˆn(2) denotes the set of the remaining triangles; i.e. ˆ ∗ ∪ Tˆ(1) ∪ Tˆ(2) . Tˆn = ∆ n n ˆ−2 ) and in Tˆ(2) is O(pn) = Recall that the number of triangles in Tˆn(1) is O(n) = O(δ n n ˆ−1 ). Moreover, all but a finite number of the triangles in Tˆ(2) , bounded independent O(δ n n ˆl with d independent of n and i. Hence, of n, will be at a minimum distance d > 0 from P the function Ki j, Pˆl (x, y) is uniformly bounded for the point (x, y) being in a triangle in Tˆn(2) . Thus, we can split the integral in (5.55) into three parts Z

3 X Tˆn j=1

Z Ki j, Pˆl (x, y)(I − Lτ ) f j (x, y) dx d y =

Z

Z

+ ˆ∗ ∆

+

.

(1) Tˆn

(2) Tˆn

1. By Lemma 14, the error in evaluating the integral Z 3 X Ki j, Pˆl (x, y)(I − Lτ ) f j (x, y) dx d y , ∆ˆ ∗ j=1

∀i

ˆ r+2 ) by the definition ˆ ∗ . Thus, we have O(δ is O(hr+2 ) where h is the diameter of ∆ n (5.22). 2. Next, consider the error from triangles in Tˆn(1) . By Lemma 15 we have

≤ ≤ (5.22)

Z 3 X Ki j, Pˆl (x, y)(I − Lτ ) f j (x, y) dx d y Tˆ(1) j=1 n Z 3 X X ) f (x, y) dτ (x, y)(I − L K ˆl j k τk i j, P (1) R j=1

R k ∈Tˆn

X

(1) R k ∈Tˆn

≤

ˆ r+2 cδ n

≤

ˆ r+2 cδ n

k

Z

112

Z

X (1) R k ∈Tˆn

ˆ r+2 cδ n

3 X R k j=1

Z

≤

Z

chr+2 k

(|Ki j, Pˆl (x, y)| + |∂Qˆ Ki j, Pˆl (x, y)|) dτk

3 X R k j=1

3 X

(1) Tˆn j=1

(|Ki j, Pˆl (x, y)| + |∂Qˆ Ki j, Pˆl (x, y)|) dτk

(|Ki j, Pˆl (x, y)| + |∂Qˆ Ki j, Pˆl (x, y)|) dx d y

3 X

ˆ ∗ j=1 Tˆn −∆

(|Ki j, Pˆl (x, y)| + |∂Qˆ Ki j, Pˆl (x, y)|) dx d y .

5.6 Superconvergence of the boundary element collocation method According to Assumption 1, the last quantity is bounded by Z 3 X 1 1 r+2 ˆ cδ + dx d y, n ˆ ˆ2 ˆl − Q| ˆl − Q| | P | P ∗ ˆ ˆ Tn −∆ j=1

∀ i.

(5.56)

ˆ r+2 ln(δ ˆ−1 )), which can be obtained by The expression in (5.56) is of order O(δ n n using polar coordinates. Therefore, the error arising from triangles in Tˆn(1) is ˆ r+2 ln(δ ˆ−1 )). O(δ n n 3. Lastly, consider the error over each such triangle in Tˆn(2) . Applying Corollary 1 yields Z 3 X Ki j, Pˆl (x, y)(I − Lτ ) f j (x, y) dx d y Tˆ(2) j=1 n Z 3 X X Ki j, Pˆl (x, y)(I − Lτk ) f j (x, y) dτk ≤ τ j=1 (2) τk ∈Tˆn

≤

X

k

Z

3 X

chr+1 k

τk j=1

(2)

τk ∈Tˆn

|Ki j, Pˆl (x, y)| dτk ≤

X

chr+1 h2k , k

(2)

τk ∈Tˆn

where the last step follows, since the area of the kth triangle is O(h2k ) and since Ki j, Pˆl is uniformly bounded as mentioned above. Therefore, we have X (2) τk ∈Tˆn

(5.22)

ˆ r+3 chr+1 h2k ≤ c δ k n

X (2) τk ∈Tˆn

ˆ r+2 , 1 ≤ cδ n

ˆ−1 ) such triangles. Thus, the total error arising from the triangles since we have O(δ n ˆ r+2 ). in Tˆ(2) is O(δ n

n

ˆ ∗ , Tˆ(1) , and Tˆ(2) yields the result (5.54). Combining the errors from the integrals over ∆ n n We now show that Assumption 1 is satisfied for the kernel matrix K1 . Lemma 17. Assumption 1 is satisfied for the kernel matrix K1 . Proof. Without loss of generality, we only have to show that the (1, 1)-entry of K1 satisfies Assumption 1; that is, we have to show that K(P, Q) = −∂ν(P) Φκ (P, Q) +

eiκ|P−Q| 4π|P − Q|3

(iκ|P − Q| − 1)(P1 − Q 1 )(v1 (P) − w1 (Q))

satisfies Assumption 1. It suffices to show that Assumption 1 is satisfied for ˆ (P, Q) = K

eiκ|P−Q| 4π|P − Q|3

(iκ|P − Q| − 1)(P1 − Q 1 )(v1 (P) − w1 (Q)) ,

113

5 The exterior Maxwell problem since ∂ν(P) Φκ (P, Q) has already been considered in [42, Lemma 4.7]. We have to show this statement for a given fixed point P only for Q in a neighborhood Γ ∩ B(P; R) of P with 0 < |P − Q| < R, since, for Q ∈ Γ not belonging to Γ ∩ B(P; R), we 1 can always use the estimates |P − Q| ≤ d = diam(Γ), |(ν P , Q − P)| ≤ d and |P−Q| ≤ R1 , because |P − Q| ≥ R. According to the assumptions on the surface, there exists an R > 0 such that Γ∩B(P; R) can be projected onto the tangent plane in a one-one fashion. Thus, we can assume that the surface Γ can be represented locally by z = f (x, y) with f ∈ C 2 . Without loss of generality, we let P be the origin of the coordinate system and Q be an arbitrary point in Γ ∩ B(P; R). Therefore, we obtain P = (0, 0, 0),

Q = (x, y, f (x, y))

ν(P) = (0, 0, 1),

ν(Q) = (− f x (x, y), − f y (x, y), 1)

and implicitly we have f (0, 0) = f x (0, 0) = f y (0, 0) = 0. Clearly, this yields eiκ|Q| ˆ |P − Q| K (P, Q) = |Q| (iκ|Q| − 1) |x| | f x | . 3 4π|Q|

Using |x| ≤ 4.23]) yields

p

x 2 + y 2 + f 2 = |Q| and | f x | ≤

p

(5.57)

x 2 + y 2 + f 2 = |Q| (see [42, Equation

eiκ|Q| eiκ|Q| 2 ˆ (P, Q) ≤ |Q| |Q| = |P − Q| K (iκ|Q| − 1) (iκ|Q| − 1) ≤ c. 4π 4π|Q|3

Thus, K ˆ (P, Q) ≤

c

,

P 6= Q .

,

P 6= Q .

|P − Q|

Next, we claim that

ˆ (P, Q)| ≤ |∂Q K

114

c |P − Q|2

(5.58)

5.6 Superconvergence of the boundary element collocation method First, consider

eiκ|Q| ˆ (P, Q)| = |Q|2 (iκ|Q| − 1)(−x f x ) |P − Q|2 |∂ x K 3 4π|Q| x eiκ|Q| iκ Q· Q (iκ|Q| − 1)(−x f ) ≤ x x | {z } 4π|Q|2 ≤c | {z } ≤c by (5.57)

eiκ|Q| + 3 (iκ|Q| − 1)(−x f ) x 4π|Q|2 | {z } ≤c by (5.57)

Q· Q x |Q| | {z } ≤c by [42, Eq. 4.24]

eiκ|Q| Q· Q x + (−iκ x f x ) |Q| 4π|Q| | {z } {z } | ≤c by [42, Eq. 4.24] ≤c eiκ|Q| + (iκ|Q| − 1)(− f x − x f x x ) . 4π|Q| | {z } ≤c

c ≤ ∂ K ˆ ˆ (P, Q) ≤ (P, Q) , P = 6 Q . Similarly, we obtain Hence, we have ∂ x K y |P−Q|2 c , P 6= Q and thus, we have (5.58). |P−Q|2 Now, we state the following superconvergence theorem for the integral equation (5.25). Theorem 18. Let each parametrization function F j ∈ C r+3 and the tangential field a = r+2 (a1 , a2 , a3 )T ∈ C t,d (Γ; C3 ). Then we have ˆ r+2 ln(δ ˆ−1 ) , i = 1, 2, 3 max ai (vl ) − ai,n (vl ) ≤ c δ n n 1≤l≤n· f r

for the integral equation (5.25). Proof. Since the matrix kernel K1 satisfies Assumption 1 as shown in Lemma 17, we have ˆ r+2 ln(δ ˆ−1 ), max K (I − Pn )ai (vl ) ≤ c δ i = 1, 2, 3 n n 1≤l≤n· f r

by Theorem 16. This completes the proof.

5.6.2 Interpolation of odd degree We now assume that r is odd. Then, (5.33) has degree of precision r. Suppose we can find α = α0 such that (5.33) has degree of precision r + 1. Then, (5.33) has degree of precision r + 2 over a parallelogram. With this observation we have the following

115

5 The exterior Maxwell problem Lemma 19. Let τ1 and τ2 be two planar triangles with diameter h such that R = τ1 ∪ τ2 is a parallelogram. Let f ∈ C r+3 (R; C3 ) and K Pˆ ∈ L 1 (R; C3×3 ) be two times differentiable with derivatives of order 1 and 2 belonging to L 1 (R; C3×3 ). In addition, assume α = α0 . Then Z 3 X Ki j, Pˆ (x, y)(I − Lτ ) f j (x, y) dτ R j=1 Z 3 X r+3 |Ki j, Pˆ (x, y)| + |∂Qˆ Ki j, Pˆ (x, y)| + |∂Qˆ2 Ki j, Pˆ (x, y)| dτ ≤ c (Υ (R)) h R j=1

¦ © × max ∂ r+1 f , ∂ r+2 f , ∂ r+3 f ,

i = 1, 2, 3 (5.59)

ˆ∈ / R. with Υ(R) = maxi=1,2 Υ τi and c (Υ (R)) some multiple of a power of Υ(R) and P

Proof. Without loss of generality, we can assume that R is a square with side length h, otherwise proceed as in Lemma 1. Let i ∈ {1, 2, 3} be arbitrary. Let p j,r (x, y), p j,r+1 (x, y), and p j,r+2 (x, y) be Taylor polynomials of f j with degree r, r + 1, and r + 2 over R, respectively. We have

k f j − p j,r k∞ ≤ chr+1 k∂ r+1 f j k∞ ,

k f j − p j,r+1 k∞ ≤ ch

r+2

k∂

r+2

f j k∞ ,

k f j − p j,r+2 k∞ ≤ chr+3 k∂ r+3 f j k∞ ,

f j ∈ C r+1 (R)

fj ∈ C

r+2

(5.60)

(R)

(5.61)

f j ∈ C r+3 (R)

(5.62)

for a suitable constant c. It easily follows from (5.60), (5.61), and (5.62) that kp j,r+1 − p j,r k∞ ≤ chr+1 hk∂ r+2 f j k∞ + k∂ r+1 f j k∞ kp j,r+2 − p j,r+1 k∞ ≤ chr+2 hk∂ r+3 f j k∞ + k∂ r+2 f j k∞ .

(5.63) (5.64)

For simplicity, define Ki j = Ki j, Pˆ (x, y). In addition, we can find polynomials of degree 0 (0) (1) and 1 over R, denoted by Ci j and Ci j , such that

116

kKi j − Ci j k1 ≤ ch k∂Qˆ Ki j k1 ,

(0)

(5.65)

kKi j − Ci j k1 ≤ ch2 k∂Qˆ2 Ki j k1 .

(1)

(5.66)

5.6 Superconvergence of the boundary element collocation method Define Lˆ = I − Lτ and write Z 3 X Ki j Lˆ f j dτ Z

R j=1

3 X

= Z

R j=1 3 X

+ Z

R j=1 3 X

+

R j=1

Z Ki j Lˆ( f j − p j,r+2 ) dτ + (Ki j −

(1) Ci j )Lˆ(p j,r+1

Z (0) Ci j Lˆp j,r+2

dτ +

3 X R j=1

(0)

(Ki j − Ci j )Lˆ(p j,r+2 − p j,r+1 ) dτ Z

− p j,r ) dτ + 3 X

R j=1

3 X R j=1

Ki j Lˆp j,r dτ −

(1) (0) (Ci j − Ci j )Lˆp j,r+1 dτ

Z

3 X R j=1

(1) Ci j Lˆp j,r dτ .

(5.67) The last three integrals vanish, since the degree of precision is r + 2 and Lˆp j,r = 0. The integral Z 3 X (1) (0) (Ci j − Ci j )Lˆp j,r+1 dτ R j=1

is zero, since (1)

(0)

(1)

(0)

Lτ [(Ci j − Ci j )p j,r+1 ] = Lτ [(Ci j − Ci j )Lτ p j,r+1 ] and, therefore, Z 3 X

Z (1) (Ci j

R j=1

−

(0) Ci j )Lˆp j,r+1

dτ =

3 X R j=1

(1) (0) Lˆ[(Ci j − Ci j )p j,r+1 ] dτ = 0 .

Taking bounds in (5.67) and using (5.62) for the first integral, (5.64) and (5.65) for the second integral, and (5.63) and (5.66) for the third integral yields the estimate Z 3 X ˆ K L f dτ j R j=1 i j Z 3 X r+3 ≤ ch kLˆk k∂ r+3 f j k∞ |Ki j | dτ R j=1

Z

+ ch

r+2

kLˆk· ch

3 X R j=1

Z + chr+1 kLˆk· ch2 Z ≤ chr+3

3 X R j=1

hk∂ r+3 f k∞ + k∂ r+2 f j k∞ |∂Qˆ Ki j | dτ

3 X R j=1

hk∂ r+2 f k∞ + k∂ r+1 f j k∞ |∂Qˆ2 Ki j | dτ

|Ki j, Pˆ (x, y)| + |∂Qˆ Ki j, Pˆ (x, y)| + |∂Qˆ2 Ki j, Pˆ (x, y)|

dτ

¦ © × max ∂ r+1 f , ∂ r+2 f , ∂ r+3 f

117

5 The exterior Maxwell problem and, thus, the assertion (5.59). Before we prove the next theorem we need the following Assumption 2. Let Γ be a smooth C 3 surface. Let K(P, Q) be the kernel of our integral operator. Assume that each entry of K satisfies c 2 , P= 6 Q, ∂Q Ki j (P, Q) ≤ |P − Q|3 where c denotes a generic constant independent of P and Q. Now, we are in position to state the following Theorem 20. Assume the conditions of Theorem 12 with each parametrization function r+3 F ∈ C r+4 and a = (a1 , a2 , a3 )T ∈ C t,d (Γ; C3 ). Assume α = α0 . Moreover, assume that the kernel K satisfies Assumptions 1 and 2. Then ˆ r+2 , i = 1, 2, 3 . (5.68) max ai (vl ) − ai,n (vl ) ≤ c δ n 1≤l≤n· f r

Proof. We will bound Z K(vl , Q)(I − Pn )a(Q) ds(Q) max K (I − Pn )a(vl ) = max 1≤l≤ f r n 1≤l≤ f r n T n

to prove (5.68). Using the same notations as in Theorem 16, we have ˆ r+2 ). ˆ ∗ is O(δ 1. By Lemma 14 the error in evaluating the integral over ∆ n 2. Next, consider the error arising from triangles in Tˆn(1) . By Lemma 19 and using the same argument as in item 2 of Theorem 16, we have Z ˆ r+3 cδ n

3 X K (x, y) + ∂ ˆ K (x, y) + ∂ 2 K (x, y) dx d y . Q ij ij ˆ ij Q

ˆ ∗ j=1 Tˆn −∆

(5.69) According to Assumption 1 and Assumption 2, the quantity (5.69) is bounded by   Z 3 X 1 1 1 ˆ r+3  + δ (5.70) 2 + 3  dx d y . n ˆ ˆ ˆ ˆ ˆ ˆ P − Q ∗ ˆ j=1 P −Q P − Q l Tˆ −∆ n

l

l

ˆn ), which can be obtained by using ˆ r+3 /δ The expression in (5.70) is of order O(δ n ˆ r+2 ). polar coordinates. Therefore, the error arising from triangles in Tˆn(1) is O(δ n 3. Lastly, consider the error over each such triangle in Tˆn(2) . As in Theorem 16 item 3, ˆ r+2 ). we apply Lemma 1. Thus, the total error coming from triangles in Tˆn(2) is O(δ n

118

5.7 Numerical results ˆ ∗ , Tˆ(1) , and Tˆ(2) yields the result (5.68). Combining the errors from the integrals over ∆ n n We now show that Assumption 2 is satisfied for the matrix kernel K1 . Lemma 21. Assumption 2 is satisfied for the matrix kernel K1 . Proof. This involves a simple but lengthy and tedious calculation similar to Lemma 17.

Now, we state the following superconvergence theorem for the integral equation (5.25). Theorem 22. Let each parametrization function F ∈ C r+4 and the tangential field a = r+3 (a1 , a2 , a3 )T ∈ C t,d (Γ; C3 ). Assume α = α0 . Then we have ˆ r+2 , max ai (vl ) − ai,n (vl ) ≤ c δ n

1≤l≤n· f r

i = 1, 2, 3

for the integral equation (5.25). Proof. Since the matrix kernel K1 satisfies Assumption 1 and Assumption 2 as shown in Lemma 17 and 21, we have by Theorem 20 ˆ r+2 . max K (I − Pn )a(vl ) ≤ c δ n

1≤l≤n· f r

5.7 Numerical results In order to verify the superconvergence, we need an example that satisfies the exterior Maxwell problem. One can easily show that the columns of K2 and K3 given in (5.9) and (5.10), respectively, satisfy the Maxwell equations with respect to x. Therefore, let the electric and magnetic field be given by  0 e   E(x) = (iκr − 1)  x 3  3 4πr −x 2 

iκr

(5.71)

and    1 −x 2 x 2 − x 3 x 3 e     1 2 2 x1 x2 H(x) =  − 2(iκr − 1)  0  ,  (−r κ − 3iκr + 3)  4πiκr 3 r 2 0 x1 x3 



iκr

119

5 The exterior Maxwell problem where r =

p

x 12 + x 22 + x 32 . The boundary condition ν(x) × E(x) is given by  −x 2 v2 − x 3 v3 e   x 2 v1 (iκr − 1)  ν(x) × E(x) = . 3 4πr x 3 v1 

iκr

∞ To calculate the far-fieldppatterns and H ∞ , we use the incident plane wave E i with p E p incident direction d = (1/ 3, 1/ 3, 1/ 3)T and polarization p = (1, 2, −3)T . Thus, the boundary condition is given by   v2 p3 − v3 p2   −ν(x) × E i (x) = −eiκ x·d  v3 p1 − v1 p3  . v1 p2 − v2 p1

First, we illustrate the accuracy of the integral equation (5.25) for constant, linear, and quadratic interpolation. The surface under consideration is the unit sphere. The wave number κ equals 1. We use NS = 128 and NN S = 4. We denote the number of faces of the triangulation with n and the number of node points of the triangulation with n v . Point (10.0,11.0,12.0)

   

( 5.0, 6.0, 7.0)

  

( 1.0, 2.0, 3.0)

  

( 1.0, 1.0, 1.0)

Point (10.0,11.0,12.0)

 

   

( 5.0, 6.0, 7.0)

  

( 1.0, 2.0, 3.0)

  

( 1.0, 1.0, 1.0)

 

Approximated solution of E = (E1 , E2 , E3 )T real part imag. part   +4.8520D-07 −6.4488D-08   −7.9418D-04   +2.4975D-03 +7.2760D-04   −2.2893D-03 −7.0018D-07 −9.0763D-07   +4.6626D-03   −2.0386D-03 −3.9960D-03   +1.7480D-03 −2.4298D-06 −6.0809D-06   +1.3400D-02   −1.1496D-02 −8.9327D-03   +7.6660D-03 −1.9102D-05 −9.8969D-07   −2.3747D-02   −1.9395D-02 +2.3757D-02 +1.9396D-02 Approximated solution of H = (H1 , H2 , H3 )T real part imag. part   +8.0168D-04 −2.9152D-03   −5.0506D-04   +1.1549D-03 −5.5180D-04   +1.2600D-03 −5.2194D-03 +2.6376D-03   +2.0465D-03   −4.6134D-04 +2.3887D-03   −5.3668D-04 −1.4176D-02 +1.2790D-02   +3.3555D-03   −5.9656D-04 +5.0351D-03   −8.8572D-04 +3.0243D-02 +4.9782D-03   +4.2635D-03   −2.6197D-02 +4.2662D-03 −2.6171D-02





 

 





 

 





 

 





 

 





 

 





 

 





 

 





 

 

Absolute error 4.8947D-07 1.0100D-06 6.7897D-07 1.1463D-06 2.1445D-06 1.3840D-06 6.5484D-06 9.6292D-06 5.6296D-06 1.9127D-05 3.1084D-05 3.9924D-05 Absolute error 1.1587D-06 2.3306D-07 6.4908D-07 2.6290D-06 5.2856D-07 1.3005D-06 1.1991D-05 3.6450D-06 6.6562D-06 6.1403D-05 1.6144D-05 1.2782D-05

           

           

Table 5.2: Accuracy for constant interpolation with α = 1/3, n = 1024, and n v = 1024 with wave number κ = 1 for the unit sphere.

As we can see in Tables 5.2, 5.3, and 5.4, the closer the point to the boundary the worse the error. However, using constant interpolation, we obtain two to three digits accuracy

120


Point (10.0,11.0,12.0)

   

( 5.0, 6.0, 7.0)

  

( 1.0, 2.0, 3.0)

  

( 1.0, 1.0, 1.0)

Point (10.0,11.0,12.0)

 

   

( 5.0, 6.0, 7.0)

  

( 1.0, 2.0, 3.0)

  

( 1.0, 1.0, 1.0)

 

Approximated solution of E = (E1 , E2 , E3 )T real part imag. part   −5.7893D-09 +5.7082D-10   −7.9351D-04   +2.4968D-03 +7.2739D-04   −2.2887D-03 +8.1917D-09 +1.1154D-08   +4.6605D-03   −2.0390D-03 −3.9947D-03   +1.7477D-03 +2.9608D-08 +7.5732D-08   +1.3391D-02   −1.1500D-02 −8.9271D-03   +7.6664D-03 +2.8093D-07 +1.2367D-08   −2.3722D-02   −1.9375D-02 +2.3722D-02 +1.9375D-02 Approximated solution of H = (H1 , H2 , H3 )T real part imag. part   +8.0112D-04 −2.9142D-03   −5.0528D-04   +1.1548D-03 −5.5120D-04   +1.2598D-03 −5.2168D-03 +2.6374D-03   +2.0469D-03   −4.6089D-04 +2.3880D-03   −5.3772D-04 −1.4164D-02 +1.2792D-02   +3.3586D-03   −5.9449D-04 +5.0379D-03   −8.9185D-04 +3.0232D-02 +4.9174D-03   +4.2591D-03   −2.6181D-02 +4.2591D-03 −2.6182D-02





 

 





 

 





 

 





 

 





 

 





 

 





 

 





 

 


           

           

Table 5.3: Accuracy for linear interpolation with α = 1/6, n = 1024, and n v = 3072 with wave number κ = 1 for the unit sphere.

for both the electric and magnetic field. With the linear interpolation, we obtain four to five digits accuracy, whereas for the quadratic interpolation, the accuracy is five to six digits for the electric field. For the magnetic field, we obtain five to six digits for points far away and three to four digits for points close to the boundary. The reason is that the calculation of the magnetic field involves a hypersingular kernel and, thus, for points close to the boundary, the approximation is worse. Next, we present numerical results to illustrate the superconvergence of the collocation method for smooth surfaces without surface approximation (see [36, Appendix A]) for the integral equation (5.25) and compare them with the theoretical results. Let P1 (3, 3, 3), P2 (2, 5, 6), and P3 (10, 11, 12) be points in the exterior domain. The true solution is given by (5.71). Without loss of generality, we present numerical results for the second component of E = (E1 , E2 , E3 )T , since we obtain similar results for the first and last component. Denote the error between the calculated solution E2,n and the true solution E2 at the point Pi by En (Pi ); that is, En (Pi ) = E2 (Pi ) − E2,n (Pi ) . Define the estimated order of convergence (EOC) at the point Pi by EOC Pi = log2 En Pi /E4n Pi .

121

5 The exterior Maxwell problem

Point

  

(10.0,11.0,12.0)

  

( 5.0, 6.0, 7.0)

  

( 1.0, 2.0, 3.0)

  

( 1.0, 1.0, 1.0)

Point

  

(10.0,11.0,12.0)

  

( 5.0, 6.0, 7.0)

  

( 1.0, 2.0, 3.0)

  

( 1.0, 1.0, 1.0)

Approximated solution of E = (E1 , E2 , E3 )T real part imag. part   −1.2237D-09 −8.1840D-11   −7.9349D-04   +2.4968D-03 +7.2737D-04   −2.2887D-03 +1.5256D-09 +2.6973D-09   +4.6605D-03   −2.0390D-03 −3.9947D-03   +1.7477D-03 +5.3876D-09 +1.8217D-08   +1.3391D-02   −1.1500D-02 −8.9271D-03   +7.6665D-03 +9.7348D-08 +3.0493D-09   −2.3723D-02   −1.9375D-02 +2.3723D-02 +1.9375D-02 Approximated solution of H = (H1 , H2 , H3 )T real part imag. part   +8.0109D-04 −2.9142D-03   −5.0526D-04   +1.1548D-03 −5.5119D-04   +1.2598D-03 −5.2168D-03 +2.6375D-03   +2.0468D-03   −4.6091D-04 +2.3880D-03   −5.3773D-04 −1.4164D-02 +1.2792D-02   +3.3585D-03   −5.9454D-04 +5.0378D-03   −8.9183D-04 +3.0232D-02 +4.9173D-03   +4.2589D-03   −2.6182D-02 +4.2589D-03 −2.6183D-02





 

 





 

 





 

 





 

 





 

 





 

 





 

 





 

 


           

           

Table 5.4: Accuracy for quadratic interpolation with α = 1/10, n = 256, and n v = 1536 with wave number κ = 1 for the unit sphere.

We consider two different smooth surfaces. Precisely, we consider an ellipsoidal surface and a peanut-shaped surface which are the last two surfaces shown in Figure 5.1. As a 1.5

1.5

1

1

0.5

0.5

1

0

0

z

z

z

0.5

−0.5

0 −0.5

−0.5 −1

−1 −1

−1 −0.5

−0.5 0

0 0.5

0.5 1

y

1

x

−1.5 −1.2

−1

−0.8

−0.4

0

0.4 y

0.8

1.2

1

0.5

−0.5

0 x

−1

−1.5 −1

−0.5

0

0.5

1

1

0.5

−0.5

0

−1

x y

Figure 5.1: Left to right: Surface of the unit sphere, the ellipsoidal surface, and the peanut-shaped surface.

first example, we consider an ellipsoid with a = 1.0, b = 1.2, and c = 1.5. The second surface is a peanut. Its surface in spherical coordinates cos(θ ), ¦ is given by x = ̺ sin(φ) © 2 2 2 y = ̺ sin(φ) sin(θ ), and z = ̺ cos(φ), where ̺ = 9 cos (φ) + sin (φ)/4 /4. The estimated error of convergence illustrated in Table 5.5 is in agreement with the superconvergence rates predicted by Theorems 18 and 22; that is, we obtain the theoretical

122


n (n v ) 4 (4) 16 (16) 64 (64) 256 (256) 1024 (1024)

En (P1 ) 2.9762D-03 4.2113D-04 1.4572D-04 3.5525D-05 8.3149D-06

n (n v ) 4 (12) 16 (48) 64 (192) 256 (768) 1024 (3072)

En (P1 ) 3.6697D-03 5.1907D-04 6.8839D-05 4.7378D-06 2.9846D-07

n (n v ) 4 (24) 16 (96) 64 (384) 256 (1536)

En (P1 ) 7.5495D-04 1.1338D-04 1.8000D-06 5.7030D-08

Constant interpolation with α = 1/3 EOC(P1 ) En (P2 ) EOC(P2 ) En (P3 ) 3.4662D-04 6.4424D-04 2.82 2.79 4.9908D-05 4.8419D-05 1.53 -0.24 5.8948D-05 3.0700D-05 2.04 1.90 1.5819D-05 8.3320D-06 2.10 2.04 3.8457D-06 2.0005D-06 Linear interpolation with α = 1/6 EOC(P1 ) En (P2 ) EOC(P2 ) En (P3 ) 1.4628D-03 8.9289D-04 2.82 2.32 2.9260D-04 1.1519D-04 2.91 3.48 2.6186D-05 1.5273D-05 3.86 3.79 1.8917D-06 1.0339D-06 3.99 3.92 1.2494D-07 6.7074D-08 Quadratic interpolation with α = 1/10 EOC(P1 ) En (P2 ) EOC(P2 ) En (P3 ) 5.2098D-04 1.9630D-04 2.74 3.75 3.8600D-05 2.7001D-05 5.98 5.06 1.1606D-06 4.5041D-07 4.98 5.19 3.1850D-08 1.5686D-08

EOC(P3 ) 3.73 0.66 1.88 2.06 EOC(P3 ) 2.95 2.91 3.88 3.95 EOC(P3 ) 2.86 5.91 4.84

Table 5.5: Constant, linear, and quadratic interpolation for an ellipsoid.

ˆ2 ln(δ ˆ−1 )), O(δ ˆ3 ), and O(δ ˆ4 ln(δ ˆ−1 )), respectively. order of convergence O(δ n n n n n For the linear and quadratic interpolation case, we almost obtain an additional order of convergence compared to the theoretical results predicted by Theorems 18 and 22. This is possibly due to the smoothing effect of the integral given by equation (5.8). We get similar result for points closer to the boundary and for different wave numbers. The numerical results shown in Table 5.6 for a peanut are in agreement with the theoretical results predicted by Theorems 18 and 22. Again, for the linear case we obtain an additional order. This is not the case for the quadratic case. The rates for the constant and quadratic case are oscillating. Note that different wave numbers give similar rates of convergence. In the next paragraph, we consider a generic test case to calculate the electric far-field pattern for a unit sphere with wave number κ = 1. We use the solution      0 1 iκ|x|    iκ|x|   e es (x) = curl  ∂ x 3 e4πx  0  E =    4π|x|   iκ|x| 0 −∂ x 2 e4πx es . that satisfies the Maxwell equation. Then, the boundary condition is given by f = ν × E Its electric far-field is given by   0 iκ   e∞ (ˆ E x) =  xˆ3  . 4π −ˆ x2

123

5 The exterior Maxwell problem

n (n v ) 4 (4) 16 (16) 64 (64) 256 (256) 1024 (1024)

En (P1 ) 3.7835D-03 1.6565D-03 1.3018D-04 7.5900D-05 2.3346D-05

n (n v ) 4 (12) 16 (48) 64 (192) 256 (768) 1024 (3072)

En (P1 ) 2.4689D-03 1.2144D-03 1.9006D-04 1.2459D-05 4.2458D-07

n (n v ) 4 (24) 16 (96) 64 (384) 256 (1536)

En (P1 ) 2.1374D-03 1.0866D-03 6.7943D-06 6.4447D-07

Constant interpolation with α = 1/3 EOC(P1 ) En (P2 ) EOC(P2 ) En (P3 ) 1.9354D-03 1.1771D-03 1.19 0.72 1.1768D-03 4.2583D-04 3.67 3.24 1.2415D-04 3.9936D-05 0.78 1.49 4.4085D-05 2.0581D-05 1.70 1.61 1.4471D-05 6.2849D-06 Linear interpolation with α = 1/6 EOC(P1 ) En (P2 ) EOC(P2 ) En (P3 ) 9.1315D-04 6.2993D-04 1.02 0.02 9.0210D-04 3.0978D-04 2.68 2.49 1.6042D-04 5.3916D-05 3.93 4.02 9.9017D-06 3.3738D-06 4.88 4.83 3.4925D-07 1.1887D-07 Quadratic interpolation with α = 1/10 EOC(P1 ) En (P2 ) EOC(P2 ) En (P3 ) 1.6845D-03 5.7970D-04 0.98 0.95 8.7324D-04 2.9811D-04 7.32 8.43 2.5369D-06 1.4860D-06 3.40 2.42 4.7401D-07 1.6770D-07

EOC(P3 ) 1.47 3.41 0.96 1.71 EOC(P3 ) 1.02 2.52 4.00 4.83 EOC(P3 ) 0.96 7.65 3.15

Table 5.6: Constant, linear, and quadratic interpolation for a peanut.

Without loss of generality, we present the absolute error of the electric far-field for the second component for xˆ1 = (1, 0, 0)T and the third component for xˆ2 = (0, 1, 0)T , where we use quadratic interpolation with α = 1/10. The results are shown in Table 5.7. As n (n v ) 4 (24) 16 (96) 64 (384) 256 (1536) ∞

e∞ (ˆ Approx. solution E 2 x1) real part imag. part +2.0436D-05 –9.3209D-04 –1.2195D-04 +1.1576D-04 +3.0085D-06 –9.1620D-08 –1.1416D-07 –9.5627D-09 0 0

EOC(ˆ x1) 2.47 5.80 4.72

x2) Approx. solution Ee3∞ (ˆ real part imag. part +1.7821D-03 –8.1302D-02 +1.5468D-04 –7.9758D-02 +3.9575D-06 –7.9581D-02 +3.9411D-08 –7.9578D-02 0 –7.9577D-02

EOC(ˆ x2) 3.38 5.43 6.48

Table 5.7: Generic test cast for checking the electric far-field pattern.

we can see the estimated rate of convergence is in agreement with the theoretical results. Note that we obtain similar estimated rates of convergence for different wave numbers and points xˆ . Finally, we calculate the electric and magnetic far-field pattern for a sphere with radius r p= 1/2 wave number κ = 1 at xˆ = (1, 0, 0)T with incident direction p and p d = (1/ 3, 1/ 3, 1/ 3)T and polarization p = (1, 2, −3)T , where we use quadratic interpolation with α = 1/10. Precisely, we use f = −(ν × p)eiκx·d as boundary function. Without loss of generality, we present numerical results for the third component of E ∞ = (E1∞ , E2∞ , E3∞ )T and H ∞ = (H1∞ , H2∞ , H3∞ )T , since we obtain similar results for the other two components. The exact solution is given by an infinite series (see for ex-

124

5.8 Summary ample [29]). As we can see in Table 5.8, the calculated solution is accurate, and the n (n v ) 4 (24) 16 (96) 64 (384) 256 (1536) ∞

Approximated solution E3∞ real part imag. part –0.219756 –0.028676 –0.268501 –0.038828 –0.271561 –0.039328 –0.271647 –0.039338 –0.271649 –0.039339

EOC(ˆ x) 4.1 5.2 5.5

Approximated solution H3∞ real part imag. part +0.202135 +0.020138 +0.235726 +0.024156 +0.237765 +0.024310 +0.237821 +0.024320 +0.237823 +0.024320

EOC(ˆ x) 4.1 5.2 4.8

p p p Table 5.8: Third component of the electric and magnetic far-field for d = (1/ 3, 1/ 3, 1/ 3)T , p = (1, 2, −3)T , κ = 1 at xˆ = (1, 0, 0)T .

estimated rate of convergence agrees with the theoretical results. Note that different points on the unit sphere, different incident directions, and polarizations yield similar results.

5.8 Summary We use the boundary element collocation method to solve a Fredholm integral equation of the second kind, where we use interpolation at interior points. We prove superconvergence at the collocation nodes distinguishing the cases of even and odd interpolation and illustrate it with numerical results for several smooth surfaces.

125

6 Eigenvalues for the interior Maxwell problem In this chapter, we calculate — as a preliminary step — interior Maxwell eigenvalues with the integral equation method. Later, we aim to calculate interior transmission eigenvalues arising in electromagnetic scattering theory. The resulting integral equation is approximated with the boundary element collocation method which leads to highly accurate results due to superconvergence as shown in the previous chapter. The nonlinear eigenvalue problem is solved with the contour integral algorithm which we used to calculate interior transmission eigenvalues arising in acoustic scattering theory. A variety of surfaces in three dimensions are considered, and interior Maxwell eigenvalues including their multiplicities are calculated. Additionally, we are also able to calculate complex-valued eigenvalues which are not interior Maxwell eigenvalues. They correspond to scattering poles for the exterior problem.

6.1 Problem statement In this section, we consider the reduced Maxwell equations curl E − iκ H = 0 ,

curl H + iκ E = 0 in D ,

where κ is the wavenumber and E and H are the electric and magnetic field, respectively. The boundary condition is given by E × ν = 0 on Γ, where ν denotes the normal directed into the exterior domain (see [16, Chapter 6] for more details). The above equations can equivalently be written as curl curl E − κ2 E = 0 E×ν = 0

in D ,

(6.1)

on Γ .

The number κ2 will be called an interior Maxwell eigenvalue for the domain D, if there exists a non trivial solution E ∈ H(curl, D) := {u ∈ L 2 (D)3 : curl u ∈ L 2 (D)3 } to (6.1). In this section, we will calculate such eigenvalues for a variety of surfaces. One is interested in such eigenvalues, since the linear sampling method (LSM) or the Factorization method (FM) from the theoretical point of view fail at such values. The reason is that the far-field operator F : L 2t S2 → L 2t S2 defined by Z F g (ˆ x) = E∞ xˆ , d, g(d) ds(d), xˆ ∈ S2 S2

127

6 Eigenvalues for the interior Maxwell problem x , d, p) is the farfor all g ∈ L 2t S2 fails to be injective at such eigenvalues, where E∞ (ˆ i field pattern generated by the incident fields E (x, d, p) := iκ d × p × d eiκx·d , Hi := iκ d ×p eiκx·d for all d, x ∈ S2 for the scattering problem by a perfect conductor D (see [33, Chapter 5] for more details regarding the FM). For well-known properties of real-valued Maxwell eigenvalues we refer the reader to [47]. We use the boundary integral equation technique to solve the problem at hand. We make the ansatz E(x) = curl x A− (x) , x ∈ D, where

Z A− (x) :=

Φκ (x, y)a( y) ds( y) , Γ

x∈D

with a being a continuous tangential field. Calculating E × ν, letting x approach the boundary, and using the jump relation (see [16, p. 165] for the jump of ν(x) × curl x A− (x)), yields Z 1 curl x A− (x) × ν(x) = − ν(x) × curl x Φκ (x, y)a( y) ds( y) + a(x) = 0 on Γ , 2 Γ which is equivalent to −

1 2

Z a(x) + Γ

ν(x) × curl x Φκ (x, y)a( y) ds( y) = 0 ,

(6.2)

which can abstractly be written as E(κ)X = 0

(6.3)

with the obvious definition of X and E(κ). The operator E(κ) defined by (6.3) is a sum of −1/2 −1/2 divΓ , Γ divΓ , Γ into H× an invertible and a compact operator mapping from H× −1/2 divΓ , Γ , and [7, section 5 and 6] for the (see [7, section 2] for the definition of H× definition of the boundary integral operators and there compactness properties). Con −1/2 −1/2 divΓ , Γ is Fredholm of index divΓ , Γ → H× sequently, the operator E(κ) : H× zero. Since the kernel is analytic, we also have the analyticity of the operator E(κ) on C\R− . Hence again, the theory of eigenvalue problems for holomorphic Fredholm operator-valued functions applies to E(κ). Since the boundary integral equation (6.2) can be easily solved numerically by the boundary element collocation method with high accuracy as shown in the previous chapter (compare equations (5.4) and (5.5)) and since we can again apply algorithm 1 from Section 3.4.2, we are now in position to calculate highly accurate interior Maxwell eigenvalues for a variety of surfaces. Finally note that we could also use the ansatz E(x) = curl x curl x A− (x) ,

128

x ∈ D.

6.2 Numerical results Calculating E × ν, letting x approach the boundary, and using the jump relation (see [16, p. 165] for the jump of ν(x) × curl x curl x A− (x)), yields Z curl x curl x A− (x) × ν(x) = −ν(x) × curl x curl x

Φκ (x, y)a( y) ds( y) = 0 on Γ . Γ

6.2 Numerical results In this section, we calculate the eigenvalues κ of E(κ)X = 0 for a variety of surfaces. Note that κ2 will then be an interior Maxwell eigenvalue.

6.2.1 Eigenvalues for the unit sphere As a first test scenario, we calculate the eigenvalues for the unit sphere. It is known that the zeros of the spherical Bessel functions jn (κ) for n ≥ 1 are such eigenvalues. Note that such zeros are also eigenvalues of the interior Dirichlet problem (see Section 3.6.11) provided n ≥ 0. Some of those zeros are 3.141 593 ,

4.493 408 ,

5.763 441 .

Additionally, the zeros of jn (κ)+κ jn′ (κ) for n ≥ 1 are also eigenvalues (see [16, p. 189]). Some of those zeros are 2.743 707 ,

3.870 239 ,

4.973 420 .

Now, we choose the contour ∂ Ω as an ellipse with center (4, 0), semi axis (1.5, 0.25), and N = 50. We are able to find four eigenvalues not counting multiplicity. In Table 6.1 we report the four eigenvalues (EV) for a unit sphere including their multiplicity using m = 288, m = 1152, and m = 4608. EV calc. EV (m = 288) κ1,S2 2.744013 [3] κ2,S2 3.874976 [5] κ3,S2 4.490728 [3]

κ4,S2

4.987188 [7]

calc. EV (m = 1152) 2.743 714 [3] 3.870 289 [5] 4.493 355 [3]

calc. EV (m = 4608) 2.743 707 [3] 3.870 239 [5] 4.493 409 [3]

4.973 404 [7]

4.973 421 [7]

Table 6.1: Values for the four EV with m = 288, m = 1152, and m = 4608 for S2 .

Comparing the results with the listed zeros of the spherical Bessel functions, it can be seen that we have been able to compute highly accurate eigenvalues for m = 4608. Further comparing the entries in Table 6.1 with those of Table 3.11 on page 64, one can see that we have indeed some of the interior Dirichlet eigenvalues. We have put an oval box around those eigenvalue which are also interior Dirichlet eigenvalues.

129

6 Eigenvalues for the interior Maxwell problem

6.2.2 Eigenvalues for an ellipsoidal surface Next, we choose an ellipsoid E with semi axis (1, 1, 6/5). Picking an ellipse with center (3.5, 0) and semi axis (1, 0.25) as the contour ∂ Ω yields seven eigenvalues not counting multiplicities. In Table 6.2 we list the seven eigenvalues including their multiplicities using m = 1152 and m = 4608. Again, the results are highly accurate for m = 4608. EV calculated EV (m = 1152) κ1,E 2.535 577 [2] κ2,E 2.714 178 [1] 3.568 470 [2] κ3,E κ4,E 3.635 950 [1] 3.684 361 [2] κ5,E κ 4.274 289 [2] 6,E κ7,E 4.350 977 [1]

calculated EV (m = 4608) 2.535 563 [2] 2.714 185 [1] 3.568 412 [2] 3.635 930 [1] 3.684 265 [2] 4.274 299 [2] 4.351 068 [1]

Table 6.2: Values for seven EV with m = 1152 and m = 4608 for E .

6.2.3 Eigenvalues for a peanut-shaped obstacle In this subsection, we consider the peanut-shaped obstacle P . We choose an ellipse with center (3, 0) and semi axis (1, 0.25) as the contour. We have five eigenvalues enclosed in this contour not counting multiplicities. The five eigenvalues including their multiplicity are listed in Table 6.3. EV κ1,E κ2,E κ3,E κ4,E κ5,E

calculated EV (m = 1152) 2.568 018 [2] 3.039 661 [1] 3.254 149 [2] 3.368 310 [1] 3.979 029 [2]

calculated EV (m = 4608) 2.567 757 [2] 3.039 228 [1] 3.254 557 [2] 3.368 244 [1] 3.979 412 [2]

Table 6.3: Values for five EV with m = 1152 and m = 4608 for P .

130


6.2.4 Eigenvalues for an acorn-shaped obstacle Next, we consider calculating the eigenvalues for an acorn-shaped obstacle. The chosen contour is an ellipse centered at (2.5, 0) with semi axis (1.5, 0.25). We have nine eigenvalues situated in the domain enclosed by the contour. The eigenvalues are listed in Table 6.4 including their multiplicities. EV calculated EV (m = 1152) κ1,A 1.964 462 [1] κ2,A 2.342 690 [2] 3.072 173 [2] κ3,A κ4,A 3.237 901 [2] 3.290 775 [1] κ5,A κ 3.378 679 [2] 6,A κ7,A 3.652 106 [1]

κ8,A κ9,A

3.920 158 [1] 3.921 582 [2]

calculated EV (m = 4608) 1.962 936 [1] 2.340 299 [2] 3.071 505 [2] 3.237 245 [2] 3.289 999 [1] 3.385 842 [2] 3.652 890 [1] 3.919 095 [1] 3.922 311 [2]

Table 6.4: Values for nine EV with m = 1152 and m = 4608 for A .

6.2.5 Eigenvalues for a cushion-shaped obstacle The next surface under consideration is the cushion-shaped obstacle. We calculate nine eigenvalues not counting multiplicities which are enclosed by the ellipse with center (2.5, 0) and semi axis (1.5, 0.25). The eigenvalues and their multiplicities are listed in Table 6.5. EV calculated EV (m = 1152) κ1,C 1.723 257 [1] κ2,C 2.775 711 [2] 2.807 512 [2] κ3,C κ4,C 3.094 625 [1] κ 3.292 432 [2] 5,C κ6,C 3.570 760 [1]

κ7,C κ8,C κ9,C

3.660 673 [2] 3.699 290 [1] 3.918 955 [2]

calculated EV (m = 4608) 1.722 891 [1] 2.775 666 [2] 2.806 562 [2] 3.101 040 [1] 3.292 832 [2] 3.570 877 [1] 3.659 709 [2] 3.697 111 [1] 3.919 714 [2]

Table 6.5: Values for nine EV with m = 1152 and m = 4608 for C .

131

6 Eigenvalues for the interior Maxwell problem

6.2.6 Eigenvalues for a bumpy sphere-shaped obstacle The next surface under consideration is a bumpy sphere-shaped obstacle. We pick the contour generated by an ellipse centered at (3, 0) and semi axis (1, 0.25). We calculate six eigenvalues inside the contour. In Table 6.6 we list the eigenvalues including their multiplicities. We have to use a larger m to get good accuracy, since this surface is more complicated. EV κ1,B κ2,B κ3,B κ4,B κ5,B κ6,B

calculated EV (m = 4608) 1.985 628 [1] 2.735 292 [1] 2.746 863 [2] 3.708 920 [2] 3.755 871 [1] 3.786 961 [2]

calculated EV (m = 9216) 1.986 123 [1] 2.736 407 [1] 2.754 356 [2] 3.723 780 [2] 3.759 440 [1] 3.792 084 [2]

Table 6.6: Values for six EV with m = 4608 and m = 9216 for B.

6.2.7 Eigenvalues for a short and long cylinder-shaped obstacle The next two surfaces under consideration are a short and a long cylinder-shaped obstacle. We choose the ellipse with center (2.5, 0) and semi axis (1, 0.25) and the ellipse with center (3, 0) and semi axis (1, 0.25), respectively. As we can see in Table 6.7 we obtain in both cases ten eigenvalues. Additionally, we include their multiplicities. EV calculated EV (m = 4608) κ1,S 1.625 347 [1] 2.008 884 [2] κ2,S κ3,S 2.304 119 [1] κ4,S 2.584 565 [2] 2.593 090 [2] κ 5,S κ6,S 3.002 149 [1]

κ7,S κ8,S κ9,S κ10,S

3.073 223 [2] 3.234 881 [2] 3.403 848 [2] 3.455 438 [2]

calculated EV (m = 4608) 2.125 425 [2] 2.435 480 [1] 2.691 855 [1] 2.807 816 [2] 3.237 656 [2] 3.270 355 [1]

EV κ1,L κ2,L κ3,L κ4,L κ5,L κ6,L κ7,L κ8,L κ 9,L κ10,L

3.664 179 [2] 3.730 623 [2] 3.870 694 [2] 3.975 267 [1]

Table 6.7: Values for ten EV for S and ten EV for L with m = 4608.

132


6.2.8 Complex eigenvalues As we have seen in Section 3.6.10, we are also able to calculate complex eigenvalues easily. In Figure 6.1(a) we present complex eigenvalues enclosed by the rectangle [−0.2, 6] × [−6, 0.2] for the unit sphere. In Figure 6.1(b) we additionally show the eigenvalues for the interior Dirichlet problem. Hence, the reader can see that the interior Dirichlet eigenvalues are a subset of the interior Maxwell eigenvalues for the unit sphere.

0

−1

Im(κ)

−2

−3

−4

−5

−6 0

1

2

3 Re(κ)

4

5

6

(a) Complex-valued interior Maxwell eigenvalues.

0

−1

Im(κ)

−2

−3

−4

−5

−6 0

1

2

3 Re(κ)

4

5

6

(b) Complex-valued interior Dirichlet eigenvalues.

Figure 6.1: Complex-valued interior Maxwell and Dirichlet eigenvalues for the unit sphere enclosed by the rectangle [−0.2, 6] × [−6, 0.2].

One can see nice patterns in both cases. The fifty-two values are (from left to right, top

133

6 Eigenvalues for the interior Maxwell problem to bottom) 2.743707 + 0.000000i , 4.973420 + 0.000000i , 1.807339 − 0.701964i , 0.000000 − 1.000000i , 0.866025 − 1.500000i , 2.657418 − 2.103789i , 3.571023 − 2.324674i , 5.420694 − 2.685677i , 0.000000 − 2.948742i , 4.452565 − 3.447592i , 5.366222 − 3.673889i , 3.517174 − 4.070139i , 4.414443 − 4.368289i , 1.739286 − 4.758291i , 0.000000 − 4.971787i , 1.739811 − 5.426269i , 3.498157 − 5.604422i , 4.384947 − 5.967528i ,

3.870239 + 0.000000i , 5.763459 + 0.000000i , 2.757856 − 0.842862i , 4.676410 − 1.047673i , 0.000000 − 1.596072i , 0.870569 − 2.157138i , 4.492673 − 2.515932i , 0.867234 − 2.896211i , 3.544886 − 3.195236i , 0.868926 − 3.544265i , 2.626272 − 3.735708i , 0.867510 − 4.248359i , 2.623305 − 4.454008i , 3.509503 − 4.825424i , 4.401147 − 5.159794i , 5.297720 − 5.465000i , 0.000000 − 5.615557i ,

4.493409 + 0.000000i , 0.866025 − 0.500000i , 3.714784 − 0.954230i , 5.641635 − 1.128906i , 1.754381 − 1.838907i , 0.000000 − 2.322185i , 1.752303 − 2.571399i , 2.644316 − 2.908062i , 1.742661 − 3.351956i , 0.000000 − 3.646739i , 1.743040 − 4.033561i , 0.000000 − 4.284595i , 5.317272 − 4.638440i , 0.868391 − 4.897196i , 2.616175 − 5.204841i , 0.867614 − 5.587886i , 2.615285 − 5.895441i ,

for the interior Maxwell eigenvalues. The twenty-five values are (from left to right, top to bottom) 3.141593 + 0.000000i , 0.000000 − 1.000000i , 2.657418 − 2.103789i , 4.492673 − 2.515932i , 1.742661 − 3.351956i , 3.517174 − 4.070139i , 5.317272 − 4.638440i , 2.616175 − 5.204841i , 4.384947 − 5.967528i ,

4.493409 + 0.000000i , 0.866025 − 1.500000i , 0.000000 − 2.322185i , 5.420694 − 2.685677i , 0.000000 − 3.646739i , 0.867510 − 4.248359i , 1.739286 − 4.758291i , 0.867614 − 5.587886i ,

5.763459 + 0.000000i , 1.754381 − 1.838907i , 3.571023 − 2.324674i , 0.867234 − 2.896211i , 2.626272 − 3.735708i , 4.414443 − 4.368289i , 0.000000 − 4.971787i , 3.498157 − 5.604422i ,

for the interior Dirichlet eigenvalues. Note that we are also able to calculate complex eigenvalues for the other surfaces under consideration. Finally, we say a few things to the complex-valued eigenvalues. First it is noted that neither the interior Dirichlet nor the interior Maxwell problem have complex-valued eigenvalues which can easily be explained by using Greens first formula. The complexvalued numbers are so called scattering poles which correspond to the exterior problem. We refer the reader to [61, Chapter 9, p. 218 and p. 224] for the definition of scattering poles and the proof in the case of the interior Dirichlet problem. Solving the interior or

134

6.2 Numerical results exterior Dirichlet problem with a single layer ansatz leads to the same nonlinear eigenvalue problem Lκ φ = 0 which is then solved with the method of Beyn. One should then use a binary criterion if the found κ and function φ lead to a potential of zero in the interior or exterior, respectively to determine whether the found κ is an interior eigenvalue or a scattering pole.

135

7 Interior transmission eigenvalues for electromagnetic scattering In this chapter, the numerical calculation of eigenvalues of the interior transmission problem arising in electromagnetic scattering for constant contrast in three dimensions is considered. As in Chapter 3 we use a method based on complex-valued contour integrals and the boundary integral equation method which is able to calculate accurate transmission eigenvalues, thus extending the acoustic to the electromagnetic interior transmission eigenvalue problem. Up to our knowledge, this is the first time that such values for various surfaces different from a sphere and a cube in three dimensions are calculated numerically. Additionally, the computational cost is even lower than those of existing methods. Furthermore, the algorithm is capable of finding complex-valued eigenvalues for which no numerical results have been reported yet. Until now, the proof of existence of such eigenvalues is still open.

7.1 Introduction Transmission eigenvalues play an important role in acoustic and electromagnetic scattering (refer to Chapter 3 for an exhausting discussion). Therefore, we extent the results from Chapter 3 for the acoustic transmission eigenvalues to the electromagnetic transmission eigenvalues. Currently, Monk and Sun [48] report transmission eigenvalues arising in electromagnetic scattering for a unit ball and a unit cube using the finite element method. Cossonnière [24, Section 6.2.5] calculates transmission eigenvalues for a unit sphere, a unit cube, a sphere containing a box, and a drop, where she uses the boundary integral equations method. We use her method to calculate transmission eigenvalues for a sphere, an ellipsoidal surface, and a peanut-shaped surface. Additionally, we use the new method based on complex-valued contour integrals and the boundary integral equation method to report transmission eigenvalues. Note however that the results are not very accurate, since we currently approximate the unknown density function and the surface in the boundary element collocation method by linear elements. The reason is that we have to approximate the surface divergence of the unknown density function, which is easy to handle using linear elements. Note that other boundary element packages such as BEM++ [58] also use elements of low order. That means large linear system arise if one is interested in high accurate transmission eigenvalues. The extension to higher order elements (if possible) remains future research.

137

7 Interior transmission eigenvalues for electromagnetic scattering In Section 7.2, we illustrate the interior transmission eigenvalue problem arising in electromagnetic scattering. We solve the problem at hand with the boundary integral equation method. Additionally, we explain how to rewrite the strong singular boundary integral operators as boundary integral operators with a weakly singular kernel. Finally, numerical results are reported in Section 7.3. A short summary and directions for possible future research are given in Section 7.4.

7.2 Problem statement In this section, we describe the interior transmission eigenvalue problem for electromagnetic scattering. We denote the wavenumber by κ, with N = n I the constant contrast, and E and H are the electric and magnetic field, respectively. We have to find a non-trivial solution (E, E0 ) to curl curl E − κ2 N E = 0 2

curl curl E0 − κ E0 = 0

E × ν − E0 × ν = 0

curl E × ν − curl E0 × ν = 0

in D ,

(7.1)

in D ,

(7.2)

on Γ ,

(7.3)

on Γ .

(7.4)

Transmission eigenvalues are values of κ for which this problem has non-trivial solutions (see for example [24, 48] among others). Next, we derive the system of boundary integral equations to solve the problem at hand, which is an alternative approach compared to the original fourth order formulation. We closely follow the work of Cossonnière [24, Section 6.2]. p We set κ1 = nκ and κ0 = κ. Then, by rewriting the first Stratton-Chu formula (see [16, Theorem 6.2]) we get for x ∈ D Z E(x) = curl Γ

+

1 κ21

Γ

Z Γ

+

κ20

Z

curl curl

E0 (x) = curl 1

E( y) × ν( y) Φκ1 (x, y) ds( y) curl E( y) × ν( y) Φκ1 (x, y) ds( y) ,

(7.5)

E0 ( y) × ν( y) Φκ0 (x, y) ds( y) Z

curl curl Γ

curl E0 ( y) × ν( y) Φκ0 (x, y) ds( y) .

(7.6)

First, we let x approach the boundary in (7.5) and (7.6) each cross multiplied by ν(x), then use the jump relations (see [16, Theorem 6.11]), and the boundary condition (7.3).

138

7.2 Problem statement Taking the difference of the results leads to Z 0 = curl Z − curl + −

1 κ21 1 κ20

Γ

Γ

M ( y) Φκ1 (x, y) ds( y) × ν(x) M ( y) Φκ0 (x, y) ds( y) × ν(x) Z

curl curl Z

Γ

curl curl Γ

J( y) Φκ1 (x, y) ds( y) × ν(x) J( y) Φκ0 (x, y) ds( y) × ν(x) ,

where we set −1/2

M = E × ν = E0 × ν ∈ H×

divΓ , Γ −1/2

(7.7)

and J = curl E × ν = curl E0 × ν ∈ H×

x ∈Γ

divΓ , Γ

−1/2

on the surface Γ. Using the two boundary integral operators Kκ , Tκ : H× −1/2 divΓ , Γ defined as H×

divΓ , Γ →

Z

Kκ ψ (x) := curl Γ

ψ( y) Φκ (x, y) ds( y) × ν(x)

(7.8)

Z

Tκ ψ (x) := curl curl Γ

ψ( y) Φκ (x, y) ds( y) × ν(x)

(7.9)

equation (7.7) can be written as

Kκ 1 − Kκ 0 M −

1 κ21

Tκ 1 −

1 κ20

Tκ 0

J = 0,

on Γ .

(7.10)

Next, we rewrite the second Stratton-Chu formula (see [16, Theorem 6.2]). For x ∈ D we obtain Z curl E(x) = curl

J( y) Φκ1 (x, y) ds( y) Γ

Z

+ curl curl

M ( y) Φκ1 (x, y) ds( y) ,

(7.11)

Γ

Z

J( y) Φκ0 (x, y) ds( y)

curl E0 (x) = curl Γ

Z

+ curl curl

M ( y) Φκ0 (x, y) ds( y) .

(7.12)

Γ

139

7 Interior transmission eigenvalues for electromagnetic scattering First, we let x approach the boundary in (7.11) and (7.12) each cross multiplied by ν(x), then use the jump relations (see [16, Theorem 6.11]) and the boundary condition (7.4). Taking the difference of the results leads to T κ 1 − Tκ 0 M + K κ 1 − K κ 0 J = 0 , on Γ . (7.13) Combining equations (7.10) and (7.13), we obtain the following system of boundary integral equations Kκ 1 − Kκ 0 T κ 1 − Tκ 0 M 0 = , Kκ1 − Kκ0 κ12 Tκ1 − κ12 Tκ0 J 0 1

0

which can be written abstractly as Z(κ)X = 0 with

K κ p n − Kκ M Z(κ) := and X := , 1 T p − κ12 Tκ J κ2 n κ n p where we replaced κ1 and κ0 with nκ and κ, respectively. Again, we can use the contour integral algorithm to solve the nonlinear eigenvalue problem. Finally, we shortly explain how to rewrite the two boundary integral operators for which an approximation has to be found numerically. The boundary integral operator Mκ has already been considered previously (see the matrix kernel (5.7)) and has in fact a weakly singular matrix kernel. The boundary integral operator Tκ contains a strong singularity, but can be rewritten as an operator with weakly singular kernel. We have Z Z Tκ ψ (x) = κ2 ψ( y) Φκ (x, y) ds( y) + grad div ψ( y) Φκ (x, y) ds( y) × ν(x) Tκ p n − Tκ Kκ p n − Kκ

ZΓ

κ

=

2

= κ2 Γ

Z − −

ZΓ

ψ( y) Φκ (x, y) ds( y) + grad Γ

Z

Γ

Z

Div ψ( y) Φκ (x, y) ds( y) Γ

× ν(x)

ψ( y) Φκ (x, y) ds( y) × ν(x)

Div ψ( y) grad x Φκ (x, y) × ν( y) − ν(x) ds( y) −−→ Curl Div ψ( y)Φκ (x, y) ds( y) ,

Γ

where the first, second, and third step follow by using curl curl E = −∆E + grad divE = κ2 E + grad divE , [16, p. 170], and [50, p. 240], respectively. Here, Div and Grad denote the surface −−→ divergence and surface gradient, respectively. Curlψ is the vectorial surface curl defined by −−→ Curlψ = Gradψ × ν .

140


7.3 Numerical results In this section, we present some calculations of transmission eigenvalues, where the system of integral equations is approximated with the BEM++ package, since the results are currently slightly better than the ones obtained with my boundary element collocation package. The electromagnetic interior transmission eigenvalues of a unit sphere centered at the origin can alternatively be calculated by finding the zeros of the function p − jp (κ) n jp ( nκ) p f (κ) = −ψ′p (κ) ψ′p ( nκ) for p ≥ 1, where ψ p (κ) := κ jp (κ) is the first Ricatti-Bessel function of order p (see [24, Appendix F]) and additionally the zeros of (3.15) are such eigenvalues for p ≥ 1. In Figure 7.1, we show the transmission eigenvalues for the unit sphere centered at the origin which are located in the rectangle [−0.2, 6] × [−6, 0.2] using n = 4. As we can see, we obtain 243 complex-valued transmission eigenvalues, where 11 are real-valued transmission eigenvalues. The 243 values are (from left to right, top to bottom)

0

−1

Im(κ)

−2

−3

−4

−5

−6 0

1

2

3 Re(κ)

4

5

6

Figure 7.1: Complex-valued interior transmission eigenvalues for the unit sphere enclosed by the rectangle [−0.2, 6] × [−6, 0.2].

141

7 Interior transmission eigenvalues for electromagnetic scattering 3.141593 + 0.000000i , 3.902613 + 0.000000i , 4.979577 + 0.000000i , 0.000000 − 0.417047i , 5.637521 − 0.597791i , 0.390890 − 0.668177i , 3.552124 − 0.756501i , 5.831277 − 0.853925i , 5.173387 − 1.021924i , 1.740112 − 1.129857i , 2.658589 − 1.314448i , 0.421137 − 1.403659i , 4.009841 − 1.506344i , 0.833974 − 1.595360i , 5.424279 − 1.682644i , 5.951417 − 1.755198i , 1.286135 − 1.826640i , 0.418460 − 2.041027i , 2.589219 − 2.250085i , 0.954713 − 2.362152i , 3.071219 − 2.407960i , 3.525003 − 2.523913i , 2.701558 − 2.654138i , 4.438996 − 2.733621i , 4.859411 − 2.801172i , 5.319546 − 2.892683i , 5.781223 − 2.979732i , 0.000000 − 3.065664i , 0.000000 − 3.102584i , 3.026091 − 3.233350i , 0.424473 − 3.378611i , 0.428418 − 3.415234i , 5.501441 − 3.525555i , 2.144286 − 3.655131i , 0.858617 − 3.696717i , 4.820109 − 3.770659i , 1.774076 − 3.869973i , 5.306675 − 3.918364i , 5.762349 − 4.030047i , 0.000000 − 4.152064i , 3.486308 − 4.226093i , 3.906734 − 4.351623i , 0.000000 − 4.391602i , 4.352117 − 4.502664i , 3.558893 − 4.603719i , 4.799118 − 4.646625i , 0.881599 − 4.755777i , 5.276411 − 4.815913i , 5.697519 − 4.916194i , 0.855546 − 5.003837i , 2.147820 − 5.083172i , 3.899977 − 5.140505i , 1.285397 − 5.278403i , 4.573867 − 5.335974i , 0.430402 − 5.409904i , 3.018936 − 5.512350i , 5.232004 − 5.622784i , 0.856920 − 5.672460i , 2.647091 − 5.722024i , 2.149407 − 5.779657i , 3.896076 − 5.900616i ,

3.492822 + 0.000000i , 4.261683 + 0.000000i , 5.399436 + 0.000000i , 5.960285 − 0.459414i , 4.547834 − 0.650982i , 2.401019 − 0.676785i , 4.063658 − 0.797912i , 0.843734 − 0.881498i , 0.000000 − 1.075179i , 2.155065 − 1.197712i , 0.411176 − 1.364377i , 3.542004 − 1.439076i , 4.059075 − 1.530351i , 4.951239 − 1.627487i , 5.476354 − 1.703790i , 0.000000 − 1.777772i , 1.701748 − 1.960599i , 0.425062 − 2.078920i , 2.619980 − 2.282725i , 3.038258 − 2.376332i , 0.000000 − 2.440085i , 1.287674 − 2.559110i , 4.400989 − 2.704607i , 0.427135 − 2.748341i , 0.000000 − 2.827879i , 2.142873 − 2.908039i , 0.848020 − 2.983046i , 2.583108 − 3.076769i , 2.606925 − 3.110729i , 1.289334 − 3.263339i , 3.471517 − 3.379701i , 1.709363 − 3.449919i , 3.948559 − 3.549272i , 0.851389 − 3.660441i , 0.000000 − 3.728673i , 4.852403 − 3.801451i , 2.601238 − 3.880397i , 1.290745 − 3.953770i , 0.426007 − 4.044518i , 1.712315 − 4.155522i , 1.833072 − 4.232172i , 0.859868 − 4.369509i , 2.160744 − 4.411903i , 4.378351 − 4.535214i , 2.598472 − 4.618877i , 4.826681 − 4.678721i , 3.019507 − 4.779562i , 1.714716 − 4.849575i , 5.727690 − 4.948380i , 3.656804 − 5.010560i , 0.000000 − 5.090459i , 3.921463 − 5.174102i , 2.582586 − 5.303530i , 2.597068 − 5.338400i , 4.786610 − 5.468810i , 1.716688 − 5.536238i , 1.806995 − 5.626579i , 0.861565 − 5.708173i , 3.475287 − 5.745059i , 5.705085 − 5.802971i , 1.287033 − 5.952914i ,

3.592863 + 0.000000i 4.430310 + 0.000000i , 5.534295 + 0.000000i , 0.000000 − 0.459958i , 1.570796 − 0.658479i , 2.694533 − 0.709160i , 4.694144 − 0.812029i , 1.252868 − 0.982134i , 1.700798 − 1.097716i , 2.197347 − 1.227689i , 3.076553 − 1.366158i , 3.589963 − 1.464281i , 4.479683 − 1.568907i , 0.850468 − 1.633053i , 5.898619 − 1.734824i , 1.264820 − 1.790393i , 1.726846 − 1.995534i , 2.143504 − 2.112450i , 0.842876 − 2.297708i , 1.797722 − 2.376802i , 3.490138 − 2.493221i , 3.944483 − 2.602260i , 0.422196 − 2.711229i , 1.724443 − 2.760037i , 4.898734 − 2.829430i , 5.360051 − 2.920233i , 5.822798 − 3.006618i , 4.532361 − 3.080571i , 0.000000 − 3.157458i , 3.051998 − 3.266618i , 0.887134 − 3.406315i , 1.724316 − 3.485456i , 4.706938 − 3.593336i , 4.399653 − 3.678675i , 0.920435 − 3.749004i , 5.651110 − 3.804384i , 5.273133 − 3.888126i , 5.727662 − 4.000339i , 3.042833 − 4.058469i , 1.724807 − 4.191044i , 2.664325 − 4.261856i , 2.146076 − 4.376770i , 0.000000 − 4.427799i , 2.581851 − 4.584133i , 1.291903 − 4.636360i , 0.427111 − 4.709557i , 5.247620 − 4.784258i , 1.725455 − 4.885048i , 3.458921 − 4.964604i , 0.860818 − 5.039694i , 0.905741 − 5.101037i , 5.361425 − 5.184108i , 4.342572 − 5.308290i , 4.365426 − 5.341515i , 0.000000 − 5.476857i , 3.035047 − 5.546913i , 5.492308 − 5.633180i , 3.456792 − 5.711002i , 0.000000 − 5.753138i , 2.160616 − 5.814835i , 1.293645 − 5.988458i ,

for the interior transmission eigenvalue problem.

142

3.692445 + 0.000000i , 4.831855 + 0.000000i , 5.963690 + 0.000000i , 1.226850 − 0.541476i , 4.712389 − 0.658479i , 0.410947 − 0.709259i , 0.814202 − 0.843758i , 1.288183 − 1.016844i , 0.000000 − 1.116098i , 2.614001 − 1.286305i , 3.122991 − 1.392728i , 0.000000 − 1.500000i , 4.530003 − 1.591853i , 5.002496 − 1.649486i , 0.000000 − 1.739062i , 0.000000 − 1.810077i , 0.900703 − 2.020021i , 2.171685 − 2.146188i , 0.854402 − 2.334760i , 0.000000 − 2.402509i , 1.272231 − 2.522879i , 3.981014 − 2.632083i , 1.705760 − 2.724596i , 1.891885 − 2.762732i , 3.613329 − 2.883492i , 2.164308 − 2.942724i , 0.856900 − 3.019643i , 2.827762 − 3.084464i , 1.277179 − 3.227272i , 3.765840 − 3.356331i , 3.499281 − 3.412307i , 3.919132 − 3.517297i , 4.368723 − 3.647305i , 2.161676 − 3.690142i , 0.000000 − 3.765168i , 2.581693 − 3.845903i , 1.280707 − 3.917865i , 3.021362 − 4.024480i , 0.429291 − 4.080807i , 3.463099 − 4.192596i , 0.853771 − 4.333470i , 3.931522 − 4.384639i , 0.000000 − 4.490554i , 1.283348 − 4.600596i , 2.744844 − 4.645563i , 0.429923 − 4.745603i , 3.037897 − 4.813920i , 4.457968 − 4.908372i , 3.478924 − 4.998579i , 0.000000 − 5.054482i , 2.160521 − 5.118346i , 1.763260 − 5.270208i , 1.292855 − 5.314047i , 0.427943 − 5.374043i , 4.810733 − 5.501670i , 1.726112 − 5.571655i , 5.257309 − 5.655285i , 0.000000 − 5.717331i , 5.678676 − 5.770821i , 0.000000 − 5.820010i ,


50

50

45

45

40

40

35

35

30

30 1/|λ1|

1/|λ1|

Next, we use the method of Cossonnière to calculate real-valued interior transmission eigenvalues in the interval [3, 4] for a unit sphere centered at the origin, an ellipsoidal surface with semi-axis (1, 1, 6/5) in the interval [5/2, 4], and a peanut-shaped surface in the interval [5/2, 4] using n = 4.

25

25

20

20

15

15

10

10

5 0

5

3

3.1

3.2

3.3

3.4

3.5 κ

3.6

3.7

3.8

3.9

0 2.5 2.6 2.7 2.8 2.9

4

(a) Five peaks representing interior transmission eigenvalues for a unit sphere.

3

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 κ

4

(b) Eleven peaks representing interior transmission eigenvalues for an ellipsoidal surface.

50 45 40 35

1/|λ1|

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

(c) Nine peaks representing interior transmission eigenvalues for a peanut-shaped surface.

Figure 7.2: Interior transmission eigenvalues for three different surfaces.

As we can see in Figure 7.2 (a), we are able to see the existence of the first five real-valued transmission eigenvalues (refer to the previous table). We detect eleven and nine interior transmission eigenvalues for the ellipsoidal surface and the peanut-shaped surface, respectively. Refer to Figures 7.2 (b) and (c). In Table 7.1, we report the first four eigenvalues including their multiplicities using the new method, where we used the contours of an ellipse centered at (7/2, 0), (3, 0), and (3, 0) with semi-axis (1/2, 1/4) for the unit sphere, the ellipsoidal surface, and the peanut-shaped surface, respectively. Finally, we use the new method to calculate complex-valued eigenvalues. We use the contour of an ellipse centered at (5/2, −1/2) with semi-axis (1/2, 1/2) for the first two surfaces under consideration. We obtain the two eigenvalues 2.69−0.71i and 2.40−0.68i with multiplicities three and five for the unit sphere (also refer to the values on page

143

7 Interior transmission eigenvalues for electromagnetic scattering EV 1. 2. 3. 4.

Unit sphere S2 3.14 [3] 3.49 [5] 3.59 [3] 3.69 [5]

Ellipsoid E 2.92 [2] 3.06 [1] 3.23 [2] 3.31 [3]

Peanut P 3.00 [2] 3.04 [2] 3.37 [1] 3.42 [1]

Table 7.1: The four eigenvalues (EV) for the interior transmission problem for n = 4.

142). For the ellipsoidal surface, we get the five eigenvalues 2.19 − 0.55i, 2.22 − 0.60i, 2.64−0.64i, 2.50−0.70i, and 2.32−0.71i with multiplicities one, two, one, two, and two, respectively. The question remains whether all calculated complex-valued eigenvalues are really interior transmission eigenvalues or not.

7.4 Summary and outlook We calculate interior transmission eigenvalues for the electromagnetic scattering problem using the contour integral method. We are able to calculate both real-valued and complex-valued eigenvalues for a variety of surfaces similar to the acoustic scattering problem, thus extending those results. In future work, we will try to extend the program to higher order elements to be able to achieve higher accuracy with less elements. Currently, we are using 7710 elements.

144

8 Future research and open questions In this chapter, possible future research is presented. Some generalizations are possible, but lead to intrinsic problems and open questions which are now illustrated in detail.

8.1 The factorization method for Maxwell’s equation In this section, we show that the factorization method for scattering of a perfect conductor works from the computational point of view. That is, we present for the first time numerical reconstructions using the factorization method. Note that the theoretical justification is still open. We will now explain how to apply the factorization method for scattering of a perfect conductor and outline the open problem. The far-field operator F : T 2 S2 → T 2 S2 is defined by

Z

E ∞ xˆ ; d, g(d) ds(d) ,

F g (ˆ x) = S2

xˆ ∈ S2 ,

where E ∞ is the far-field pattern corresponding to the incident plane wave E i (x) = peiκx·d (see Section 5.4) and the point z ∈ R3 and the vector q ∈ R3 are fixed. The tangential space T 2 S2 is defined by T 2 S2 = {g : S2 → C3 | g ∈ L 2 S2 ; g· xˆ = 0 for xˆ ∈ S2 } .

Let xˆ , e1 (ˆ x ), e2 (ˆ x ) be an orthonormal basis of R3 associated with xˆ ∈ S2 . Note that such a basis can be found easily. Given xˆ = xˆ (θ , φ), θ ∈ [0, π], φ ∈ [0, 2π] which can be written in spherical coordinates as  sin(θ ) cos(φ)   xˆ =  sin(θ ) sin(φ)  ≡ e r cos(θ ) 

we deduce

(8.1)

 cos(θ ) cos(φ)   e1 (ˆ x ) =  cos(θ ) sin(φ)  ≡ eθ − sin(θ ) 

145

8 Future research and open questions and

 − sin(φ)   e2 (ˆ x ) =  cos(φ)  ≡ eφ . 0 

It is easily verified that (eφ , eθ , e r ) form a right-handed coordinate system. We have e r · eθ = e r · eφ = eθ · eφ = 0 and |e r | = |eθ | = |eφ | = 1. Obviously, e1 (ˆ x ) and e2 (ˆ x ) span the ˆ tangential plane at x . In the linear sampling method (see for example [14, 20]), one wants to solve the equation F g(· ; z, q) (ˆ x ) = Φz (ˆ x ; q) , where Φz (ˆ x ; q) =

iκ 4π

xˆ × q e−iκˆx ·z

(8.2)

(see [16, p. 164] for the derivation and [33, Theorem 5.11] for the definition) which is equivalent to (R x )· E ∞ xˆ ; d, g(d; z, q) ds(d) = e1 (ˆ x )· Φz (ˆ x ; q) 2 e1 (ˆ S R ∞ e (ˆ x )· E xˆ ; d, g(d; z, q) ds(d) = e2 (ˆ x )· Φz (ˆ x ; q) S2 2 since E ∞ (ˆ x ; d, q)· xˆ = 0 for all xˆ , d ∈ S2 and for all q ∈ R3 . Hence, we now consider Z e1 (ˆ x )· E ∞ xˆ ; d, g(d; z, q) ds(d) . (8.3) ∞ ˆ x ; d, g(d; z, q) e (ˆ x )· E 2 2 S Using the reciprocity relation (see [16, Theorem 6.28]) ei (ˆ x )· E ∞ xˆ ; d, g(d; z, q) = g(d; z, q)· E ∞ −d; −ˆ x , ei (ˆ x) for i = 1, 2, we can rewrite (8.3) as Z E ∞ −d; −ˆ x , e1 (ˆ x ) · g(d; z, q) ds(d) . ∞ −d; −ˆ x , e (ˆ x ) · g(d; z, q) E 1 2 S

(8.4)

The function g(d; z, q) is also a tangential field; that is, g(d; z, q)· d = 0. Hence, using the orthonormal basis of R3 ; i.e., (d, e1 (d), e2 (d)) one can write g(d; z, q) = g1 (d; z, q)e1 (d) + g2 (d; z, q)· e2 (d) . Then, (8.4) can be written as Z g1 (d; z, q) E ∞ −d; −ˆ x , e1 (ˆ x ) · e1 (d) E ∞ −d; −ˆ x , e1 (ˆ x ) · e2 (d) ds(d) . ∞ ∞ −d; −ˆ x , e (ˆ x ) · e2 (d) g2 (d; z, q) −d; −ˆ x , e (ˆ x ) · e (d) E E 2 2 1 2 S

146

8.1 The factorization method for Maxwell’s equation A variant of the linear sampling method is Kirsch’s factorization method, where one considers the equation

(F ∗ F )

1/4

g(· ; z, q) (ˆ x ) = Φz (ˆ x ; q) ,

where the operator F is normal (see [33, Theorem 5.7]). Precisely, the factorization method works as follows (see [33, p. 37]): 1. Given N and M , calculate θi = π(i − 1)/(N − 1), i = 1, . . . , N and φ j = 2π( j − 1)/(M − 1), j = 1, . . . , M . 2. Calculate xˆk = xˆ (θi , φ j ), k = j + M (i − 1) according to (8.1). Similarly, calculate e1 (ˆ x k ) and e2 (ˆ x k ), (k = 1, . . . , N · M ). 3. Analogously, calculate dk = xˆ (θi , φ j ) according to (8.1), (k = 1, . . . , N · M ). 4. Create the matrix A ∈ C2N M ×2N M , where each 2 × 2 block matrix Ak,l , k, l = 1, . . . , N · M is calculated by

Ak,l =

E ∞ −dl ; −ˆ x k , e1 (ˆ x k ) · e1 (dl ) E ∞ −dl ; −ˆ x k , e1 (ˆ x k ) · e2 (dl ) . E ∞ −dl ; −ˆ x k , e2 (ˆ x k ) · e1 (dl ) E ∞ −dl ; −ˆ x k , e2 (ˆ x k ) · e2 (dl )

Compare with (8.4). 5. Compute a singular value decomposition of A; i.e., A = UΛV ∗ . 6. Construct a grid G , say a cube [−1.5, 1.5]3 with equidistant spacing of size T . 7. For each point z ∈ R3 in the grid, we a) construct the vector rz =

Φz (ˆ x j ; q)· e1 (ˆ x j) Φz (ˆ x j , q)· e2 (ˆ x j)

, j=1,...,N ·M

where q ∈ R3 is fixed and Φz is given by (8.2), b) compute the matrix-vector product ̺ (z) = V ∗ rz , and c) evaluate

 W (z) = 

2N M X j=1

(z)

|̺ j |2 |λ j |

−1 

,

where λ j are the diagonal entries of Λ. 8. Finally, we plot an isosurface of z 7→ W (z) for a given threshold ε.

147

8 Future research and open questions In Figure 8.1, we show the reconstruction of a sphere of radius r = 1 centered at the origin for κ = 1, κ = 2, and κ = 3, where we used N = 10, M = 10, q = (1, 1, 1)T and ε = 0.001 (we skipped the scaling factor 1/(4π) in the calculation of rz ). The far-field data for the sphere are generated by a finite approximation of the series expansion (see for example [29])

=

+

E ∞ (ˆ x ; d, p) ∞ 4πi X 1 κ

n=1

(

n(n + 1)

n h jn (κR) X

(κR) m=−n h(1) n

jn (κR) + κR jn′ (κR)

n h X

′

(κR) + κRh(1) h(1) (κR) m=−n n n i

i x) p· GradYnm (d) GradYnm (ˆ

x ) × xˆ p· GradYnm (d) × d GradYnm (ˆ

) .

As we can see in Figure 8.1 (a) the reconstruction is very accurate for κ = 1, but for

(a) Reconstructed unit sphere, κ = 1, q = (1, 1, 1)T , and ε = 0.001.

(b) Reconstructed unit sphere, κ = 2, q = (1, 1, 1)T , and ε = 0.001.

(c) Reconstructed unit sphere, κ = 3, q = (1, 1, 1)T , and ε = 0.001.

Figure 8.1: .Reconstructed unit sphere for various wave numbers.

κ = 2 and κ = 3 the reconstructions are not satisfying as shown in Figures 8.1 (b) and

148

8.1 The factorization method for Maxwell’s equation (c). Using different vectors q yields the same unsatisfying result, which is illustrated in Figure 8.2 (a)–(c). The key to obtain a good result is to calculate W (z) for three linear

(a) Reconstructed unit sphere, κ = 2, q = (1, 0, 0)T , and ε = 0.003.

(b) Reconstructed unit sphere, κ = 2, q = (0, 1, 0)T , and ε = 0.003.

(c) Reconstructed unit sphere, κ = 2, q = (0, 0, 1)T , and ε = 0.003.

(d) Reconstructed unit sphere, κ = 2 ˆ (z). and ε = 0.003 using W

Figure 8.2: Reconstructed unit sphere for various q. T T independent vectors q,say q1 = (1, 0, 0)T , q2 = (0, 1, 0) , and q3 = (0, 0, 1) and plot an ˆ (z) := Wq (z) + Wq (z) + Wq (z) /3. The accurate reconstruction using isosurface of W 1 1 1 this idea is shown in Figure 8.2 (d). As expected, increasing N and M also improves the reconstructions. Note that it is possible to create synthetic far-field data with my programm for varying surfaces (see Chapter 5). We generate far-field data for a unit sphere, an ellipsoid with semi-axis (1, 1.2, 1), an acorn-shaped surface, and a cushion-shaped surface using the parameters α = 0.1 (quadratic interpolation) and n v = 4608 with wave number κ = 2, N = 10, and M = 10 (see Chapter 5 for details). In Figure 8.3, we show the reconˆ (z) for the threshold struction of the four surfaces, where we plot an isosurface of W ε = 0.004. The true surfaces are given in Figure 3.1 on page 51.

149

8 Future research and open questions

(a) Reconstructed unit sphere, κ = 2 ˆ (z). and ε = 0.004 using W

(b) Reconstructed ellipsoidal surface, κ = 2 ˆ (z). and ε = 0.004 using W

(c) Reconstructed acorn-shaped surface, κ = 2 ˆ (z). and ε = 0.004 using W

(d) Reconstructed cushion-shaped surface, κ = 2 ˆ (z). and ε = 0.004 using W

ˆ (z). Figure 8.3: Four reconstructed surfaces using W

As we can see, we are able to reconstruct different surfaces for varying wavenumbers and radii using three independent vectors q. Hence, the factorization method works from the computational point of view. From the theoretical point of view, it remains to show that the ranges of the operators (F ∗ F )1/4 and A coincide, where A is the operator that maps the boundary data onto the electric far-field pattern E ∞ . Precisely, one has to show that the electric dipole operator Tκ (see (7.9) for the definition) is coercive for κ = i (see [17, p. 260] for a recent discussion)1 . This is the missing ingredient to validate the factorization method for the electromagnetic case (the easiest to be considered here is scattering by a perfect conductor) from the theoretical point of view, but it is still an open problem.

1

According to Andreas Kirsch (private communication) the coercivity is known, but showing the compactness of Tκ − Ti is still open.

150

8.2 Fractional Helmholtz equation

8.2 Fractional Helmholtz equation + Dz1+α with 0 < α < 1, + D1+α The α-Laplacian in three dimensions is given by D1+α y x where D1+α is an α-fractional derivative with respect to t. The reader is referred to [45, t Appendix A] for a good introduction to fractional differentiation and integration, as well as the definition of the Riemann-Liouville and the Riesz/Weyl fractional operator (note that there are other fractional derivatives such as the Caputo fractional derivative). Recently, the α-Laplacian has been well-studied even in the context of boundary integral operators (see for example [13], [44] and [54]). As pointed out by William Rundell’s talk “Is the Universe Gaussian?” at the conference in Bad Herrenalb, April 9th, 2013 it is a short step to define the fractional Helmholtz equation. There he raised three questions, one of which was: “Transmission eigenvalues?”. I tried to attack this interesting idea/question and will now present my preliminary results. Note that there is (so far) no literature available regarding this issue. The first paper considering the fractional Helmholtz equation in two dimensions has been carried out by Samuel and Thomas (see [56]). According to Saxena et al. the fractional wave equation has been used in the theory of vibrations of smart materials in media, where the memory effects should not be ignored (see [57] for a discussion and the reference cited therein). Another application might be to use the operators arising for the fractional Helmholtz equation for stabilization purpose (we refer the reader to the article by Buffa and Sauter [8]). Here, stabilization means to avoid critical wave numbers in a boundary integral formulation. First, the layer potential of a fractional Laplacian ∆s with 1/2 < s < 1 for a function ψ ∈ L 2 (∂ Ω) is given by (see [13, p. 345]) Z Vs ψ(x) =

Ks (x, y)ψ( y) ds( y) , ∂Ω

where the fundamental solution Ks (x, y) is given by (see [44, p. 3037]) Ks (x, y) =

Γ(3/2 − s) 2s

2 π

3/2

1

Γ(s) |x − y|3−2s

.

It satisfies ∆s v(x) = 0 for x ∈ R3 \∂ Ω. The boundary integral operator SLs ψ is given by SLs ψ = (Vs ψ)|∂ Ω . To find this fundamental solution, one can use for example the Fourier transform, since the fractional Laplacian can also be defined as a pseudo-differential operator R Øs v(η) = (2π|η|)2s b (−∆) v (η), where b v (η) = R3 e−2πiη·x v(x) dx, η ∈ R3 is the Fourier transform of v in three dimensions (see [13, p. 346]). Alternatively, one can use the

151

8 Future research and open questions Riesz potential. Note that we get the same fundamental solution if we use the Weyl fractional derivative, since the Fourier transform of it is the same as above (see [45, p. 59]). If we now consider the fractional Helmholtz equation, the following question arises: Given 1/2 < s < 1, what is the fundamental solution for the single layer potential satisfying ∆s v(x) + κ2 v(x) = 0 for x ∈ R3 \∂ Ω, where κ is the wave number? With this at hand, one could define the single layer boundary integral operator and could then calculate the interior Dirichlet eigenvalues as in Section 3.6.11 for varying s, where the choice s = 1 yields the results presented in Section 3.6.11. From this, one could then try to attack the interior transmission problem, specifically the interior transmission eigenvalues. To illustrate the encountered problem, we first calculate the fundamental solution of the Helmholtz equation, which is a function f (~r) that depends only on r = |~r|. The 3D-Fourier transform (non-unitary, angular frequency) of −∆u − κ2 u = δ is given by (see above and any Fourier transform table) b(ω) b(ω) ω2 u ~ − κ2 u ~ = 1,

ω = |ω| ~

which implies 1

b(ω) u ~ =

. ω − κ2 The 3D-inverse Fourier transform of a radial symmetric function can easily be calculated by the 1D integral u(r) =

Z

1 2πr 2

∞

2

b(ω)ω sin(ωr) dω . u

(8.5)

0

R To see this, convert the 3D-inverse Fourier integral R3 e−i~r·ω~ f (ω) dω ~ to spherical coordinates and perform integration with respect to θ and φ. Precisely, we have to calculate u(r) =

1

Z

∞

1

2π r 0 ω − κ2 1 1 iκr = e π 2π2 r 2 eiκr = . 4πr

Note that we have

R∞

2

1 ω sin(ωr) dω 0R ω2 −κ2 ∞ 1

2

=

π iκr e 2

ω sin(ωr) dω

(see [52, p. 395 Formula 9]). Also

note that the integral 0 ω2 −κ2 ω sin(ωr) dω has to be understood as a Cauchy principal R∞ 1 value; i.e. limε→0 0 ω2 −(κ+iε) 2 ω sin(ωr) dω, because of the apparent singularity. Hence,

152

8.2 Fractional Helmholtz equation the fundamental solution is given by Gκ (x, y) =

eiκ|x− y| 4π|x − y|

,

x, y ∈ R3 ,

x 6= y .

To find now the fundamental solution of the fractional Helmholtz equation, we would proceed as above. The 3D-Fourier transform of (−∆u)s − κ2 u = δ for 1/2 < s < 1 is given by b(ω) b(ω) ω2s u ~ − κ2 u ~ = 1, ω = |ω| ~ which implies b(ω) u ~ = Using (8.5), we have to evaluate u(r) =

Z

1

∞

2

2π r

1 ω2s − κ2 1

2s

ω − κ2

0

.

ω sin(ωr) dω ,

(8.6)

which is only possible to be evaluated analytically (with Maple or Mathematica) for special values of s (for example s = 1) and even then, the result is a Meijer-G-function (see [53, Chapter 8.2] for the definition and notation of the Meijer-G-function). We will now use the extension of the Meijer-G-function, the Fox-H-functions to solve the integral in (8.6). Refer to [53, Chapter 8.3] for the definition and notation of the Fox-H-function. It is worth mentioning that neither Mathematica nor Maple can handle Fox-H-functions yet. We have the well-known relation of the inverse hyperbolic tangent −z

d dz

arctanh(z) =

z z2 − 1

and the Meijer-G-representation arctanh(z) = −

i 2

1,2 G2,2

2 −z

1 2 1 2

1

!

0

and hence we have z z2 − 1

=

i 2

1,4 2 G4,4

0 1 2 −z 1 2 2 0

1 2 1 2

1 1

! ,

(8.7)

where we used the differentiation formula [53, Formula 40, p. 621] ! ! 1 0 ap a d m,n p m,n+2 = 2G p+2,q+2 −z 2 21 z G p,q −z 2 bq bq dz 1 2

153

8 Future research and open questions with k = 1, ω = −1,r = 2, l = 1, c ∗ = 1, and µ = 0 (see [53, p. 798] for the definition of ∆(k, a) and ∆(k, (a p ))). Note that (8.7) can be simplified due to the formula of lowering the order and degeneracy (see [53, Formula 8, p. 619]). We have z

2 −z

1,1 = iG1,1

z2 − 1

!

1 2 1 2

.

Setting z = w s for 1/2 < s < 1 yields ws 2s

w −1

1,1 = iG1,1

−w 2s

1 2 1 2

! .

(8.8)

b s , we have to evaluate (see Equation (8.6)) Setting κ = κ u(R) =

Z

1

∞

2π R

1 b 2s ω −κ

2

2s

0

ω sin(ωR) dω ,

which after the change of variable z = ω/b κ is given by u(R) =

b 2−2s κ 2π2 R

Z

∞ 0

z 1−s

zs z 2s − 1

sin(b κRz) dz

and using (8.8) yields the expression

u(R) =

b 2−2s i κ 2π2 R

Z

∞

1,1 z 2−s−1 sin(b κRz)G1,1

0

−z 2s

1 2 1 2

! dz .

Since the Meijer-G-function can be written as a Fox-H-function (see [53, Formula 21, p. 629]); i.e. ! ! 1 1, 1 1,1 1,1 G1,1 −z 2s 21 = H1,1 −z 2s 12 2 ,1 2 b R, ω = −1, r = 2s to obtain we can use [53, Formula 4, p. 355] with α = 2 − s, σ = κ for 1/2 < s < 1

u(R) =

b 2−2s i 21−s p κ 2

2π R (b κR)

2−s

1,2 πH3,1

−

2s s 1 1 s , 1 , 2, s 2− 2 ,s 2 1 bR κ ,1 2 2

! .

(8.9)

Note that a Fox-H-function can be represented as a series using the residue theorem (see [53, Formula 4, p. 628] for the formula and the conditions). After some simplifications,

154

8.2 Fractional Helmholtz equation we obtain 2s s 1 1 s ! , 1 , 2, s 2 2− 2 ,s 1,2 2 H3,1 − 1 bR κ ,1 2 2s −( 3 − s +k) 1 2 2 s 2 ∞ Γ 1 − 3 − k Γ 3 + k (−1)k − X bR κ 2s s 2s s 3 = k!s Γ + k k=0 2 2s −( 1 +k) 3 2 k ∞ − κb2R (−1) X Γ 2 − s − ks Γ (1 + k) . + Γ (s + ks) k! k=0

In sum, we can simplify (8.9) to obtain the following expression for the fundamental solution:  2s −( 3 − s +k) 1 ∞ 2 2 s 2 i π(−1)k X u(R) = −  bR κ b s k=0 sin π 3 + k Γ 3 + k k!s 2s R3−s π3/2 κ s 2 2  2s −( 1 +k) ∞ k 2 X 2 π(−1)  − + , 1 1 b κ R Γ (s + ks) Γ s + ks + sin π s + ks + k=0 2

2

(8.10) where we used the reflection formula of the Gamma-function (see for example [53, p. 759]). Note that we assumed that κ is real. The formula is also valid for complex-valued κ with Im(κ) ≥ 0. How does the formula change if we use a complex-valued κ with Im(κ) < 0? Is it possible to write down a simplified closed-form solution? Intuitively, it should be possible, since Ks (x, y) from above is a special case for κ = 0. I would suggest something similar to Γ(3/2 − s) eh(r,s,κ) , 22s π3/2 Γ(s) |x − y|3−2s where the function h(r, s, κ) still has to be determined. The desired function might be a generalized Mittag-Leffler function Ea,b (z) =

∞ X j=0

(−z) j Γ(b + a j)

,

which is a natural generalization of the exponential function (see [45, Appendix B] and the references therein for the definition and some properties). Note that by using s = 1, the expression in (8.10) reduces to u(R) =

sin(κR) 4πR

+

i cos(κR) 4πR

=

eiκR 4πR

.

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8 Future research and open questions Finally, we present some numerical results using the method of Cossonnière (is it theoretically justified?), since our formula above is only valid for Im(κ) ≥ 0 (compare with Section 3.4). We use 300 equidistant points within the interval [2, 5] for the wave number κ. We calculate the eigenvalues for the unit sphere using s = 1; i.e., using the standard Laplacian (see Table 3.11 for the same results), s = 0.95, and s = 0.9. As we can see in Figure 8.4 (a), we are able to identify the two eigenvalues 3.141 593 and 4.493 408 that are located in the interval [2, 5], where we used s = 1 . Interestingly, 10

10

9

9

8

8

7

7

6

6

5

5

4

4

3

3

2

2

1

1

0

2

2.5

3

3.5

4

4.5

5

(a) Cossonnière’s method for S2 using s = 1.

0

10

9

9

8

8

7

7

6

6

5

5

4

4

3

3

2

2

1

1

2

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5

(c) Cossonnière’s method for S2 using s = 0.95.

2.5

3

3.5

4

4.5

5

(b) Cossonnière’s method for S2 using s = 0.975.

10

0

2

0

2

2.5

3

3.5

4

4.5

5

(d) Cossonnière’s method for S2 using s = 0.9.

Figure 8.4: Cossonnière’s method for S2 using various s.

we get seven peaks around the original eigenvalue 3.141 593 for s = 0.975 and no eigenvalue around the original eigenvalue 4.493 408 as shown in Figure 8.4 (b). I would have expected a slight shift of the two original eigenvalues. For s = 0.95, we have only three peaks as one can see in Figure 8.4 (c). Using s = 0.9, we get no peak as shown in Figure 8.4 (d). Additionally, we present the same plots for the peanut-shaped surface. As we can see in Figure 8.5 (a), we are able to identify the four eigenvalues 3.189 591, 3.704 819, 4.581 649, and 4.873 977 that are located in the interval [2, 5]. Interestingly, we get twenty peaks distributed in the interval [2, 5] for s = 0.975 as shown in Figure

156

8.2 Fractional Helmholtz equation 10

10

9

9

8

8

7

7

6

6

5

5

4

4

3

3

2

2

1

1

0

2

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3

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4

4.5

5

(a) Cossonnière’s method for P using s = 1.

0

2

2.5

3

3.5

4

4.5

5

(b) Cossonnière’s method for P using s = 0.975.

10 9 8 7 6 5 4 3 2 1 0

2

2.5

3

3.5

4

4.5

5

(c) Cossonnière’s method for P using s = 0.95.

Figure 8.5: Cossonnière’s method for P using various s.

8.5 (b). For s = 0.95, we get no peak as shown in Figure 8.5 (c). In sum, many questions arise and merit further investigation.

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9 Conclusion and outlook First, we give a general introduction in Chapter 1. In Chapter 2, the mathematical formulation of acoustic scattering by penetrable objects in three dimensions involving the Helmholtz equations with different wavenumbers for each media and transition conditions at the interface of the scatterer is considered. The transmission problem is solved with the boundary integral equation method. The resulting system of boundary integral equations is approximated by the boundary element collocation method. Superconvergence is proved theoretically and validated by numerical results. No numerical results seem to be reported for this method yet. A solver has been implemented to solve the acoustic transmission problem numerically in three dimensions for smooth surfaces to high accuracy due to superconvergence which is needed later for a variety of tasks (in Chapters 3 and 4). A possible theoretical and practical extension for piecewise smooth surfaces remains future research. In Chapter 3, the interior transmission eigenvalue problem is considered. Acoustic interior transmission eigenvalues play an important role in nondestructive testing and the calculation of those is a challenging task in this area. We present a new way based on complex-valued contour integrals and boundary integral equations to actually calculate interior transmission eigenvalues numerically to high accuracy for a variety of obstacles. The implemented algorithm outperforms existing algorithms and at the same time it is possible to determine multiplicities of an eigenvalue. Additionally, the algorithm is able to find complex-valued transmission eigenvalues. No such values have been reported yet and the proof of existence still remains. Above all, a new method is proposed to estimate the index of refraction from the knowledge of a few interior transmission eigenvalues. Chapter 4 provides extended numerical results for the application of Kirsch’s factorization method to solve the inverse acoustic transmission problem, since the theoretical framework has been established recently. It is shown that a variety of different surfaces are succesfully reconstructed by the factorization method, and hence we complement the method’s validation from the computational point of view. A possible extension of the previous topics is to consider electromagnetic instead of acoustic scattering. In Chapter 5, the exterior Maxwell problem for a conductive scatterer in three dimensions is solved numerically with the boundary element collocation method. Superconvergence is shown theoretically and validated numerically. Hence, it is possible to produce highly accurate approximations. An extension to other boundary conditions remains future research. With this solver at hand, we are able to numerically calculate interior Maxwell eigenvalues for a variety of surfaces to high accuracy using the approach from Chapter 3 with

159

9 Conclusion and outlook the aim to solve the electromagnetic interior transmission problem later on. Additionally, the multiplicities of the eigenvalues are reported and complex-valued eigenvalues are calculated. The interior transmission problem for electromagnetic scattering is considered in Chapter 7. Transmission eigenvalues including their multiplicities are calculated numerically using the new method presented in Chapter 3 for a variety of scatterers. This is the first time that results different from a sphere and a cube are presented. Additionally, complex-valued eigenvalues are reported. Until now, the proof of existence is still open. However, note that the boundary element solver has to be improved to create highly accurate approximations to calulate highly accurate transmission eigenvalues as in Chapter 3 numerically. This remains to be analyzed. Finally, two more topics of current interest are considered in Chapter 8, but merit future investigation. First, the factorization method for scattering of a perfect conductor is shown for the first time to work from the computational point of view. The highly accurate far-field data are calculated numerically with the solver of Chapter 5. The validation from the theoretical point of view is still open and remains future research. Second, a possible extension of the material considered in Chapters 2 to 4 is the use of the fractional Helmholtz equation (recently W. Rundell has inspired me). Preliminary results are presented here and merit possible future research. Note that this topic opens a complete new area of research.

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