Reduce the problem to one periodic cell Introduce Γ b : = { ( x 1 , b ) : 0 ≤ x 1 , ≤ 2 π } , Ω b : = { x ∈ Ω Λ : 0 < x 1 < 2 π , x 2 < b } α ( Ω b ) 2 : u satisfies one of the boundary conditions on Λ } . V α = V α ( Ω b ) : = { u ∈ H 1 D-to-N map 0 Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 9 (47)
The Dirichlet-to-Neumann map T T v = − ∑ v n exp ( i α n x 1 ) , v = ∑ v n exp ( i α n x 1 ) ∈ H 1 / 2 ( Γ b ) 2 W n ˆ ˆ α n ∈ ℤ n ∈ ℤ where ( ) ω 2 β n / d n 2 µα n − ω 2 α n / d n W n : = 1 , d n : = α 2 n + β n γ n . − 2 µα n + ω 2 α n / d n ω 2 γ n / d n i Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 10 (47)
The Dirichlet-to-Neumann map T T v = − ∑ v n exp ( i α n x 1 ) , v = ∑ v n exp ( i α n x 1 ) ∈ H 1 / 2 ( Γ b ) 2 W n ˆ ˆ α n ∈ ℤ n ∈ ℤ where ( ) ω 2 β n / d n 2 µα n − ω 2 α n / d n W n : = 1 , d n : = α 2 n + β n γ n . − 2 µα n + ω 2 α n / d n ω 2 γ n / d n i Lemma (i) The map T is a bounded linear map from H 1 / 2 ( Γ b ) 2 to H − 1 / 2 ( Γ b ) 2 . α α Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 10 (47)
The Dirichlet-to-Neumann map T T v = − ∑ v n exp ( i α n x 1 ) , v = ∑ v n exp ( i α n x 1 ) ∈ H 1 / 2 ( Γ b ) 2 W n ˆ ˆ α n ∈ ℤ n ∈ ℤ where ( ) ω 2 β n / d n 2 µα n − ω 2 α n / d n W n : = 1 , d n : = α 2 n + β n γ n . − 2 µα n + ω 2 α n / d n ω 2 γ n / d n i Lemma (i) The map T is a bounded linear map from H 1 / 2 ( Γ b ) 2 to H − 1 / 2 ( Γ b ) 2 . α α (ii) Re W n > 0 for all sufficiently large ∣ n ∣ . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 10 (47)
The Dirichlet-to-Neumann map T T v = − ∑ v n exp ( i α n x 1 ) , v = ∑ v n exp ( i α n x 1 ) ∈ H 1 / 2 ( Γ b ) 2 W n ˆ ˆ α n ∈ ℤ n ∈ ℤ where ( ) ω 2 β n / d n 2 µα n − ω 2 α n / d n W n : = 1 , d n : = α 2 n + β n γ n . − 2 µα n + ω 2 α n / d n ω 2 γ n / d n i Lemma (i) The map T is a bounded linear map from H 1 / 2 ( Γ b ) 2 to H − 1 / 2 ( Γ b ) 2 . α α (ii) Re W n > 0 for all sufficiently large ∣ n ∣ . (iii) T = T 1 + T 2 , ∫ − ∑ ∀ u ∈ H 1 α ( Ω b ) 2 . T 1 v = v n exp ( i α n x 1 ) , Re { − T 1 u ⋅ uds } ≥ 0 , W n ˆ Γ b ∣ n ∣≥ M − ∑ = v n exp ( i α n x 1 ) , T 2 v W n ˆ ∣ n ∣ < M Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 10 (47)
Variational formulation Variational formulation: find u ∈ V α such that ∫ ∫ ∫ ( Tu in − T u in ) ⋅ ϕ ds E ( u , ϕ ) − ω 2 u ⋅ ϕ dx − B ( u , ϕ ) : = ϕ ⋅ T uds = Ω b Γ b Γ b for all ϕ ∈ V α , where E ( u , ϕ ) = ( λ + 2 µ )( ∂ 1 u 1 ∂ 1 ϕ 1 + ∂ 2 u 2 ∂ 2 ϕ 2 )+ µ ( ∂ 2 u 1 ∂ 2 ϕ 1 + ∂ 1 u 2 ∂ 1 ϕ 2 ) + λ ( ∂ 1 u 1 ∂ 2 ϕ 2 + ∂ 2 u 2 ∂ 1 ϕ 1 )+ µ ( ∂ 2 u 1 ∂ 1 ϕ 2 + ∂ 1 u 2 ∂ 2 ϕ 1 ) . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 11 (47)
Variational formulation Variational formulation: find u ∈ V α such that ∫ ∫ ∫ ( Tu in − T u in ) ⋅ ϕ ds E ( u , ϕ ) − ω 2 u ⋅ ϕ dx − B ( u , ϕ ) : = ϕ ⋅ T uds = Ω b Γ b Γ b for all ϕ ∈ V α , where E ( u , ϕ ) = ( λ + 2 µ )( ∂ 1 u 1 ∂ 1 ϕ 1 + ∂ 2 u 2 ∂ 2 ϕ 2 )+ µ ( ∂ 2 u 1 ∂ 2 ϕ 1 + ∂ 1 u 2 ∂ 1 ϕ 2 ) + λ ( ∂ 1 u 1 ∂ 2 ϕ 2 + ∂ 2 u 2 ∂ 1 ϕ 1 )+ µ ( ∂ 2 u 1 ∂ 1 ϕ 2 + ∂ 1 u 2 ∂ 2 ϕ 1 ) . F ∈ V ∗ B u = F , α , where ∫ ( Tu in − T u in ) ⋅ ϕ ds , ( B u , ϕ ) Ω b : = B ( u , v ) , ( F , ϕ ) Ω b : = Γ b and ( ⋅ , ⋅ ) Ω b denotes the duality between V α and V ∗ α . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 11 (47)
Fredholm alternative Theorem Let A be a compact operator from a Hilbert space H 1 into a Hilbert space H 2 , then ■ The equation ( I + A ) u = g , g ∈ H 2 (1) admits a unique solution, if the homogeneous equation ( I + A ) u = 0 admits uniquely the trivial solution u = 0 . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 12 (47)
Fredholm alternative Theorem Let A be a compact operator from a Hilbert space H 1 into a Hilbert space H 2 , then ■ The equation ( I + A ) u = g , g ∈ H 2 (1) admits a unique solution, if the homogeneous equation ( I + A ) u = 0 admits uniquely the trivial solution u = 0 . ■ The equation (1) is solvable if and only if < g , v > = 0 holds for all v ∈ H ∗ 2 satisfying ( I + A ∗ ) v = 0 . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 12 (47)
Strong ellipticity of the sesquilinear form B ∫ ∫ E ( u , ϕ ) − ω 2 u ⋅ ϕ dx − B ( u , ϕ ) : = ϕ ⋅ T uds Ω b Γ b Korn’s inequality: ∫ E ( u , u ) dx ≥ C ∣∣ u ∣∣ 2 H 1 ( Ω α ) 2 Ω b Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 13 (47)
Strong ellipticity of the sesquilinear form B ∫ ∫ E ( u , ϕ ) − ω 2 u ⋅ ϕ dx − B ( u , ϕ ) : = ϕ ⋅ T uds Ω b Γ b Korn’s inequality: ∫ E ( u , u ) dx ≥ C ∣∣ u ∣∣ 2 H 1 ( Ω α ) 2 Ω b Properties of the Dirichlet-to-Neumann map T : T = T 1 + T 2 , where ∫ ∀ u ∈ H 1 α ( Ω b ) 2 , Re { − T 1 u ⋅ uds } ≥ 0 , Γ b T 2 : = T − T 1 is a finite dimensional operator . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 13 (47)
Strong ellipticity of the sesquilinear form B ∫ ∫ E ( u , ϕ ) − ω 2 u ⋅ ϕ dx − B ( u , ϕ ) : = ϕ ⋅ T uds Ω b Γ b Korn’s inequality: ∫ E ( u , u ) dx ≥ C ∣∣ u ∣∣ 2 H 1 ( Ω α ) 2 Ω b Properties of the Dirichlet-to-Neumann map T : T = T 1 + T 2 , where ∫ ∀ u ∈ H 1 α ( Ω b ) 2 , Re { − T 1 u ⋅ uds } ≥ 0 , Γ b T 2 : = T − T 1 is a finite dimensional operator . Conclusion : B is a Fredholm operator with index zero. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 13 (47)
Existence of a solution at arbitrary frequency Consider the operator equation F ∈ V ∗ B u = F , α , Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 14 (47)
Existence of a solution at arbitrary frequency Consider the operator equation F ∈ V ∗ B u = F , α , where ∫ ∫ ( Tu in − T u in ) ⋅ ϕ ds = ( F , ϕ ) Ω b = f 0 ⋅ ϕ ds Γ b Γ b { 2 i β 0 k p ( λ + 2 µ ) ( − α , γ 0 ) ⊤ e i α x 1 − i β 0 b incident plane pressure wave = d 0 f 0 − 2 i γ 0 k s µ ( β 0 , α , ) ⊤ e i α x 1 − i γ 0 b incident plane shear wave d 0 Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 14 (47)
Existence of a solution at arbitrary frequency Consider the operator equation F ∈ V ∗ B u = F , α , where ∫ ∫ ( Tu in − T u in ) ⋅ ϕ ds = ( F , ϕ ) Ω b = f 0 ⋅ ϕ ds Γ b Γ b { 2 i β 0 k p ( λ + 2 µ ) ( − α , γ 0 ) ⊤ e i α x 1 − i β 0 b incident plane pressure wave = d 0 f 0 − 2 i γ 0 k s µ ( β 0 , α , ) ⊤ e i α x 1 − i γ 0 b incident plane shear wave d 0 Lemma If ϕ ∈ V α satisfies B ∗ ϕ = 0 , then A p , n = 0 for ∣ α n ∣ < k p and A s , n = 0 for ∣ α n ∣ < k s . (2) Here A p , n and A s , n are Rayleigh coefficients for ϕ ∈ V α . In particular, A p , 0 = A s , 0 = 0 . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 14 (47)
Existence of a solution at arbitrary frequency Consider the operator equation F ∈ V ∗ B u = F , α , where ∫ ∫ ( Tu in − T u in ) ⋅ ϕ ds = ( F , ϕ ) Ω b = f 0 ⋅ ϕ ds Γ b Γ b { 2 i β 0 k p ( λ + 2 µ ) ( − α , γ 0 ) ⊤ e i α x 1 − i β 0 b incident plane pressure wave = d 0 f 0 − 2 i γ 0 k s µ ( β 0 , α , ) ⊤ e i α x 1 − i γ 0 b incident plane shear wave d 0 Lemma If ϕ ∈ V α satisfies B ∗ ϕ = 0 , then A p , n = 0 for ∣ α n ∣ < k p and A s , n = 0 for ∣ α n ∣ < k s . (2) Here A p , n and A s , n are Rayleigh coefficients for ϕ ∈ V α . In particular, A p , 0 = A s , 0 = 0 . Conclusion: B ∗ ϕ = 0 , ( F , ϕ ) Ω b = 0 if = ⇒ Existence of a solution ! Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 14 (47)
Uniqueness Lemma There exists small frequency ω 0 > 0 such that Re ( B u , u ) Ω b = Re B ( u , u ) ≥ C ∣∣ u ∣∣ 2 H 1 ( Ω b ) , ω ∈ ( 0 , ω 0 ] . Conclusion: Uniqueness and existence hold for small frequencies ! Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 15 (47)
Uniqueness Lemma There exists small frequency ω 0 > 0 such that Re ( B u , u ) Ω b = Re B ( u , u ) ≥ C ∣∣ u ∣∣ 2 H 1 ( Ω b ) , ω ∈ ( 0 , ω 0 ] . Conclusion: Uniqueness and existence hold for small frequencies ! According to the Analytic Fredholm theory : either B − 1 does not exist for any w ∈ ℝ + , or B − 1 exists for all w ∈ ℝ + ∖ D where D is a discrete subset of ℝ + . Conclusion: Uniqueness and existence hold for all frequencies excluding a discrete set with the only accumulating point at infinity. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 15 (47)
Uniqueness under the Dirichlet boundary condition If Λ is the graph of a smooth function f , then using a periodic Rellich identity: ∫ ( ∆ ∗ + ω ) u ⋅ ∂ 2 udx = − 2 Re 0 Ω b ( ∫ ) ∫ 2 Re ( Tu ⋅ ∂ 2 u ) − E ( u , u ) n 2 + ω 2 ∣ u ∣ 2 n 2 ds = Λ − Γ b ∫ µ ∣ ∂ n u ∣ 2 +( λ + µ ) ∣ div u ∣ 2 ) ( = n 2 ds , Λ Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 16 (47)
Uniqueness under the Dirichlet boundary condition If Λ is the graph of a smooth function f , then using a periodic Rellich identity: ∫ ( ∆ ∗ + ω ) u ⋅ ∂ 2 udx = − 2 Re 0 Ω b ( ∫ ) ∫ 2 Re ( Tu ⋅ ∂ 2 u ) − E ( u , u ) n 2 + ω 2 ∣ u ∣ 2 n 2 ds = Λ − Γ b ∫ µ ∣ ∂ n u ∣ 2 +( λ + µ ) ∣ div u ∣ 2 ) ( = n 2 ds , Λ Since √ 1 + ∣ f ′ ∣ 2 ≥ 0 n 2 = 1 / Λ , on we have ∂ n u = u = 0 Λ . on Applying Holmgren’s theorem leads to uniqueness at arbitrary frequency. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 16 (47)
Uniqueness under the Dirichlet boundary condition If Λ is the graph of a smooth function f , then using a periodic Rellich identity: ∫ ( ∆ ∗ + ω ) u ⋅ ∂ 2 udx = − 2 Re 0 Ω b ( ∫ ) ∫ 2 Re ( Tu ⋅ ∂ 2 u ) − E ( u , u ) n 2 + ω 2 ∣ u ∣ 2 n 2 ds = Λ − Γ b ∫ µ ∣ ∂ n u ∣ 2 +( λ + µ ) ∣ div u ∣ 2 ) ( = n 2 ds , Λ Since √ 1 + ∣ f ′ ∣ 2 ≥ 0 n 2 = 1 / Λ , on we have ∂ n u = u = 0 Λ . on Applying Holmgren’s theorem leads to uniqueness at arbitrary frequency. Remark Uniqueness holds even if Λ is given by the graph of a Lipschitz function or Λ is a binary grating profile. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 16 (47)
Ongoing work on direct scattering problems Scattering of elastic waves by rough surfaces: ( ∆ ∗ + ω 2 ) u ∆ ∗ : = µ ∆ +( λ + µ ) grad div , = Ω Λ , g in = Λ u 0 on O Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 17 (47)
Ongoing work on direct scattering problems Scattering of elastic waves by rough surfaces: ( ∆ ∗ + ω 2 ) u ∆ ∗ : = µ ∆ +( λ + µ ) grad div , = Ω Λ , g in = Λ u 0 on O If the source term g ∈ L 2 ( Ω Λ ) 2 has a compact support and Λ is the graph of a Lipschitz function, then existence and uniqueness of solutions in H 1 ( Ω Λ ) 2 can be proved. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 17 (47)
Inverse problem Inverse Problem (IP) Given incident field u in ( x ; θ ) and the near-field data u ( x 1 , b ; θ ) , determine the unknown scattering surface Λ . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 18 (47)
Inverse problem Inverse Problem (IP) Given incident field u in ( x ; θ ) and the near-field data u ( x 1 , b ; θ ) , determine the unknown scattering surface Λ . Helmholtz equation: ■ K. Ito and F. Reitich, 1999, conjugate gradient algorithm based on analytic continuation; ■ F. Hettlich, 2002, iterative regularization; ■ G. Bruckner, J. Elschner, G. C. Hsiao, A. Rathsfeld, 2002-2004, optimization method; ■ T. Arens, N. Grinberg, A. Kirsch, 2003,2005, factorization method; Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 18 (47)
Inverse problem Inverse Problem (IP) Given incident field u in ( x ; θ ) and the near-field data u ( x 1 , b ; θ ) , determine the unknown scattering surface Λ . Helmholtz equation: ■ K. Ito and F. Reitich, 1999, conjugate gradient algorithm based on analytic continuation; ■ F. Hettlich, 2002, iterative regularization; ■ G. Bruckner, J. Elschner, G. C. Hsiao, A. Rathsfeld, 2002-2004, optimization method; ■ T. Arens, N. Grinberg, A. Kirsch, 2003,2005, factorization method; Uniqueness for the Navier equation (using one incident plane wave): ■ Antonios, C., Drossos, G. and Kiriakie, K. 2001, smooth gratings with a small height, Dirichlet boundary condition, n=2. ■ J. Elschner and G. Hu, 2011, polygonal or polyhedral gratings, the third and fourth kind boundary conditions, n=2,3. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 18 (47)
Uniqueness of (IP) for polygonal grating profiles { } f ( x 1 ) is a continuous piecewise linear function of A = Λ f : period 2 π , and is not a straight line parallel to ox 1 Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 19 (47)
Uniqueness of (IP) for polygonal grating profiles { } f ( x 1 ) is a continuous piecewise linear function of A = Λ f : period 2 π , and is not a straight line parallel to ox 1 Assume 1. Λ 1 , Λ 2 ∈ A , and one of them has a corner point at the origin. 2. u 1 resp. u 2 satisfies the third (or forth) kind boundary conditions on Λ 1 resp. Λ 2 . 3. u 1 ( x 1 , b ; θ ) = u 2 ( x 1 , b ; θ ) holds for one incident plane elastic wave with the incident angle θ . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 19 (47)
Uniqueness of (IP) for polygonal grating profiles { } f ( x 1 ) is a continuous piecewise linear function of A = Λ f : period 2 π , and is not a straight line parallel to ox 1 Assume 1. Λ 1 , Λ 2 ∈ A , and one of them has a corner point at the origin. 2. u 1 resp. u 2 satisfies the third (or forth) kind boundary conditions on Λ 1 resp. Λ 2 . 3. u 1 ( x 1 , b ; θ ) = u 2 ( x 1 , b ; θ ) holds for one incident plane elastic wave with the incident angle θ . Questions: 1. Can we obtain Λ 1 = Λ 2 ? (uniqueness) Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 19 (47)
Uniqueness of (IP) for polygonal grating profiles { } f ( x 1 ) is a continuous piecewise linear function of A = Λ f : period 2 π , and is not a straight line parallel to ox 1 Assume 1. Λ 1 , Λ 2 ∈ A , and one of them has a corner point at the origin. 2. u 1 resp. u 2 satisfies the third (or forth) kind boundary conditions on Λ 1 resp. Λ 2 . 3. u 1 ( x 1 , b ; θ ) = u 2 ( x 1 , b ; θ ) holds for one incident plane elastic wave with the incident angle θ . Questions: 1. Can we obtain Λ 1 = Λ 2 ? (uniqueness) 2. If not, what kind of geometric characteristics do Λ 1 and Λ 2 share so as to generate the same near field on x 2 = b ? Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 19 (47)
Uniqueness of (IP) for polygonal grating profiles { } f ( x 1 ) is a continuous piecewise linear function of A = Λ f : period 2 π , and is not a straight line parallel to ox 1 Assume 1. Λ 1 , Λ 2 ∈ A , and one of them has a corner point at the origin. 2. u 1 resp. u 2 satisfies the third (or forth) kind boundary conditions on Λ 1 resp. Λ 2 . 3. u 1 ( x 1 , b ; θ ) = u 2 ( x 1 , b ; θ ) holds for one incident plane elastic wave with the incident angle θ . Questions: 1. Can we obtain Λ 1 = Λ 2 ? (uniqueness) 2. If not, what kind of geometric characteristics do Λ 1 and Λ 2 share so as to generate the same near field on x 2 = b ? 3. How many incident elastic waves are sufficient to uniquely determine an arbitrary grating profile Λ ∈ A ? Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 19 (47)
Uniqueness of (IP) for polygonal grating profiles { } f ( x 1 ) is a continuous piecewise linear function of A = Λ f : period 2 π , and is not a straight line parallel to ox 1 Assume 1. Λ 1 , Λ 2 ∈ A , and one of them has a corner point at the origin. 2. u 1 resp. u 2 satisfies the third (or forth) kind boundary conditions on Λ 1 resp. Λ 2 . 3. u 1 ( x 1 , b ; θ ) = u 2 ( x 1 , b ; θ ) holds for one incident plane elastic wave with the incident angle θ . Questions: 1. Can we obtain Λ 1 = Λ 2 ? (uniqueness) 2. If not, what kind of geometric characteristics do Λ 1 and Λ 2 share so as to generate the same near field on x 2 = b ? 3. How many incident elastic waves are sufficient to uniquely determine an arbitrary grating profile Λ ∈ A ? Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 19 (47)
Uniqueness under the fourth kind boundary conditions Theorem Assume that the fourth kind boundary conditions are imposed on Λ 1 and Λ 2 . if u 1 ( x 1 , b ; θ ) = u 2 ( x 1 , b ; θ ) , ∀ x 1 ∈ ( 0 , 2 π ) holds for one incident pressure wave, then either Λ 1 = Λ 2 or Λ 1 , Λ 2 ∈ D 2 ( θ , k p ) . In the latter case, the total field takes the form u = ˆ θ exp ( ik p x ⋅ ˆ θ ) − ˆ θ exp ( − ik p x ⋅ ˆ θ ) − e 1 exp ( ik p x 1 )+ e 1 exp ( − ik p x 1 ) in ℝ 2 , where e 1 = ( 1 , 0 ) ⊤ . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 20 (47)
Uniqueness under the fourth kind boundary conditions Theorem Assume that the fourth kind boundary conditions are imposed on Λ 1 and Λ 2 . if u 1 ( x 1 , b ; θ ) = u 2 ( x 1 , b ; θ ) , ∀ x 1 ∈ ( 0 , 2 π ) holds for one incident pressure wave, then either Λ 1 = Λ 2 or Λ 1 , Λ 2 ∈ D 2 ( θ , k p ) . In the latter case, the total field takes the form u = ˆ θ exp ( ik p x ⋅ ˆ θ ) − ˆ θ exp ( − ik p x ⋅ ˆ θ ) − e 1 exp ( ik p x 1 )+ e 1 exp ( − ik p x 1 ) in ℝ 2 , where e 1 = ( 1 , 0 ) ⊤ . Corollary If Λ ∈ A and Λ / ∈ D 2 ( θ , k p ) , then the near-field data u ( x 1 , b ; θ ) corresponding to one incident plane pressure wave can uniquely determine Λ , under the boundary conditions of the fourth kind. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 20 (47)
The unidentifiable class D 2 ( θ , k p ) . − 10 − 5 0 5 10 − 4*pi − 2*pi 0 2*pi 4*pi ⎧ ⎫ Each line segment of Λ lies on the straight line ⎨ ⎬ 2 π x 2 = x 1 tan ϕ + k p cos θ n for some n ∈ ℤ with D 2 ( θ , k p ) : = Λ ∈ A : . ϕ ∈ { θ 2 + π 4 , θ 2 − π 4 } , and k p ( 1 ± sin θ ) ∈ ℤ . ⎩ ⎭ Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 21 (47)
The unidentifiable class D 2 ( θ , k p ) , θ = π / 6 , k p = 2 . Features: (1) Λ ∈ D 2 ( π / 6 , 2 ) provided Λ lies on the − 10 grids. − 5 (2) The angle between two neighboring line 0 segments of Λ is fixed ( π / 2 ). 5 (3) The set D 2 ( θ , k p ) is uniquely determined 10 by the incident pressure wave. − 4*pi − 2*pi 0 2*pi 4*pi ⎧ ⎫ Each line segment of Λ lies on the straight line ⎨ ⎬ 2 π D 2 ( θ , k p ) : = Λ ∈ A : x 2 = x 1 tan ϕ + k p cos θ n for some n ∈ ℤ with . ϕ ∈ { θ 2 + π 4 , θ 2 − π 4 } , and k p ( 1 ± sin θ ) ∈ ℤ . ⎩ ⎭ Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 22 (47)
Uniqueness under the third kind boundary conditions Theorem If u 1 ( x 1 , b ; θ ) = u 2 ( x 1 , b ; θ ) , ∀ x 1 ∈ ( 0 , 2 π ) holds for one incident pressure wave with the incident angle θ , then either Λ 1 = Λ 2 or one of the following cases occurs: 1. (a) Λ 1 , Λ 2 ∈ N 2 ( θ , k p ) , and the total field takes the form u = ˆ θ exp ( ik p x ⋅ ˆ θ ) − ˆ θ exp ( − ik p x ⋅ ˆ θ ) . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 23 (47)
Uniqueness under the third kind boundary conditions Theorem If u 1 ( x 1 , b ; θ ) = u 2 ( x 1 , b ; θ ) , ∀ x 1 ∈ ( 0 , 2 π ) holds for one incident pressure wave with the incident angle θ , then either Λ 1 = Λ 2 or one of the following cases occurs: 1. (a) Λ 1 , Λ 2 ∈ N 2 ( θ , k p ) , and the total field takes the form u = ˆ θ exp ( ik p x ⋅ ˆ θ ) − ˆ θ exp ( − ik p x ⋅ ˆ θ ) . (b) Λ 1 , Λ 2 ∈ D 2 ( θ , k p ) , and the total field takes the form u = ˆ θ exp ( ik p x ⋅ ˆ θ ) − ˆ θ exp ( − ik p x ⋅ ˆ θ )+ e 1 exp ( ik p x 1 ) − e 1 exp ( − ik p x 1 ) . 2. 3. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 23 (47)
Uniqueness under the third kind boundary conditions Theorem If u 1 ( x 1 , b ; θ ) = u 2 ( x 1 , b ; θ ) , ∀ x 1 ∈ ( 0 , 2 π ) holds for one incident pressure wave with the incident angle θ , then either Λ 1 = Λ 2 or one of the following cases occurs: 1. (a) Λ 1 , Λ 2 ∈ N 2 ( θ , k p ) , and the total field takes the form u = ˆ θ exp ( ik p x ⋅ ˆ θ ) − ˆ θ exp ( − ik p x ⋅ ˆ θ ) . (b) Λ 1 , Λ 2 ∈ D 2 ( θ , k p ) , and the total field takes the form u = ˆ θ exp ( ik p x ⋅ ˆ θ ) − ˆ θ exp ( − ik p x ⋅ ˆ θ )+ e 1 exp ( ik p x 1 ) − e 1 exp ( − ik p x 1 ) . 2. Λ 1 , Λ 2 ∈ N 3 ( θ , k p ) with θ ∈ [ − π 6 , π 6 ] . The total field takes the form u = ˆ θ exp ( ik p x ⋅ ˆ ˆ ˆ ˆ ˆ θ )+ Rot 2 π θ exp ( ik p x ⋅ Rot 2 π θ )+ Rot 4 π θ exp ( ik p x ⋅ Rot 4 π θ ) . 3 3 3 3 3. Λ 1 , Λ 2 ∈ N 4 ( 0 , k p ) , θ = 0 , and the total field takes the form u = − e 2 exp ( − ik p x 2 )+ e 2 exp ( ik p x 2 )+ e 1 exp ( ik p x 1 ) − e 1 exp ( − ik p x 1 ) . Here N 2 , D 2 , N 3 , N 4 ∈ A are four classes of unidentifiable grating profiles. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 24 (47)
The unidentifiable classes N 3 and N 4 Figure: N 4 ( 0 , 4 ) Figure: N 3 ( π / 6 , 2 ) 6 6 4 4 2 2 0 0 − 2 − 2 − 4 − 4 − 6 − 2*pi 0 2*pi − 6 − 2*pi 0 2*pi Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 25 (47)
Uniqueness with minimal number of incident elastic waves Theorem ■ Under the fourth kind boundary conditions, two incident pressure waves or four incident shear waves are enough to uniquely determine a grating Λ ∈ A . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 26 (47)
Uniqueness with minimal number of incident elastic waves Theorem ■ Under the fourth kind boundary conditions, two incident pressure waves or four incident shear waves are enough to uniquely determine a grating Λ ∈ A . If Rayleigh frequencies of the compressional resp. shear part are excluded, the minimal number is one incident pressure wave resp. three incident shear waves. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 26 (47)
Uniqueness with minimal number of incident elastic waves Theorem ■ Under the fourth kind boundary conditions, two incident pressure waves or four incident shear waves are enough to uniquely determine a grating Λ ∈ A . If Rayleigh frequencies of the compressional resp. shear part are excluded, the minimal number is one incident pressure wave resp. three incident shear waves. ■ Under the third kind boundary conditions, four incident pressure waves or two incident shear waves are enough to uniquely determine a grating Λ ∈ A . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 26 (47)
Uniqueness with minimal number of incident elastic waves Theorem ■ Under the fourth kind boundary conditions, two incident pressure waves or four incident shear waves are enough to uniquely determine a grating Λ ∈ A . If Rayleigh frequencies of the compressional resp. shear part are excluded, the minimal number is one incident pressure wave resp. three incident shear waves. ■ Under the third kind boundary conditions, four incident pressure waves or two incident shear waves are enough to uniquely determine a grating Λ ∈ A . If Rayleigh frequencies of the compressional resp. shear part are excluded, the minimal number is three incident pressure waves resp. one incident shear wave. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 26 (47)
Uniqueness with minimal number of incident elastic waves Theorem ■ Under the fourth kind boundary conditions, two incident pressure waves or four incident shear waves are enough to uniquely determine a grating Λ ∈ A . If Rayleigh frequencies of the compressional resp. shear part are excluded, the minimal number is one incident pressure wave resp. three incident shear waves. ■ Under the third kind boundary conditions, four incident pressure waves or two incident shear waves are enough to uniquely determine a grating Λ ∈ A . If Rayleigh frequencies of the compressional resp. shear part are excluded, the minimal number is three incident pressure waves resp. one incident shear wave. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 26 (47)
Basic tool: reflection principle (Elschner & Yamamoto 2010) Suppose that ( ∆ ∗ + ω 2 ) u = 0 in ℝ 2 and u satisfies the third (resp. fourth) kind boundary conditions on both lines l 0 and l 1 (with the angle α ). Then ■ u satisfies the same boundary condition on l 2 , Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 27 (47)
Basic tool: reflection principle (Elschner & Yamamoto 2010) Suppose that ( ∆ ∗ + ω 2 ) u = 0 in ℝ 2 and u satisfies the third (resp. fourth) kind boundary conditions on both lines l 0 and l 1 (with the angle α ). Then ■ u satisfies the same boundary condition on l 2 , ■ Rot 2 α u ( x ) = u ( Rot 2 α x ) . Here Rot is the rotation operator around the origin. o Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 27 (47)
Other reflection principles Reflection principle for the Helmholtz equation ■ Cheng J., Yamamoto M., Alessandrini G., Rondi L., Liu H., Zou J., Elschner J. ( scattering by bounded obstacles) ■ Elschner J., Schmidt G. , Yamamoto M., Hu G. (scattering by periodic structures) Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 28 (47)
Other reflection principles Reflection principle for the Helmholtz equation ■ Cheng J., Yamamoto M., Alessandrini G., Rondi L., Liu H., Zou J., Elschner J. ( scattering by bounded obstacles) ■ Elschner J., Schmidt G. , Yamamoto M., Hu G. (scattering by periodic structures) Reflection principle for the Maxwell equation ■ Liu H., Yamamoto M. and Zou J. 2007 Reflection principle for the Maxwell equations and its application to inverse electromagnetic scattering ■ Bao G., Zhang H. and Zou J. 2010 Unique determination of periodic polyhedral gratings to appear in Trans. Amer. Math. Soc. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 28 (47)
Remarks 1. Our proofs mainly rely on the reflection principle for the Navier equation applied to polygonal periodic structures, under the third or fourth kind boundary conditions. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 29 (47)
Remarks 1. Our proofs mainly rely on the reflection principle for the Navier equation applied to polygonal periodic structures, under the third or fourth kind boundary conditions. 2. The uniqueness results can be extended to the case of bi-periodic structures. In ℝ 3 , there exist seven unidentifiable sets corresponding to one incident elastic wave. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 29 (47)
Remarks 1. Our proofs mainly rely on the reflection principle for the Navier equation applied to polygonal periodic structures, under the third or fourth kind boundary conditions. 2. The uniqueness results can be extended to the case of bi-periodic structures. In ℝ 3 , there exist seven unidentifiable sets corresponding to one incident elastic wave. 3. One incident quasi-periodic point source wave is enough to uniquely determine a polygonal or polyhedral grating profile under the third or fourth kind boundary conditions. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 29 (47)
Remarks 1. Our proofs mainly rely on the reflection principle for the Navier equation applied to polygonal periodic structures, under the third or fourth kind boundary conditions. 2. The uniqueness results can be extended to the case of bi-periodic structures. In ℝ 3 , there exist seven unidentifiable sets corresponding to one incident elastic wave. 3. One incident quasi-periodic point source wave is enough to uniquely determine a polygonal or polyhedral grating profile under the third or fourth kind boundary conditions. 4. Our method cannot apply to the Dirichlet boundary condition, because we do not know the corresponding reflection principle. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 29 (47)
Remarks 1. Our proofs mainly rely on the reflection principle for the Navier equation applied to polygonal periodic structures, under the third or fourth kind boundary conditions. 2. The uniqueness results can be extended to the case of bi-periodic structures. In ℝ 3 , there exist seven unidentifiable sets corresponding to one incident elastic wave. 3. One incident quasi-periodic point source wave is enough to uniquely determine a polygonal or polyhedral grating profile under the third or fourth kind boundary conditions. 4. Our method cannot apply to the Dirichlet boundary condition, because we do not know the corresponding reflection principle. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 29 (47)
Open problem Reflection principle under the Dirichlet boundary condition: Suppose that ( ∆ ∗ + ω 2 ) u = 0 in ℝ 2 and u = 0 on both l 0 and l 1 (with the angle α ). ■ Can we obtain u = 0 on l 2 ? ■ Does u exhibit any symmetry ? o Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 30 (47)
Open problem Reflection principle under the Dirichlet boundary condition: Suppose that ( ∆ ∗ + ω 2 ) u = 0 in ℝ 2 and u = 0 on both l 0 and l 1 (with the angle α ). ■ Can we obtain u = 0 on l 2 ? ■ Does u exhibit any symmetry ? o If α = π / 2 , ω = 0 , then one solution is u = ( − xy 2 , x 2 y ) ⊤ . Our conjecture : u is uniquely determined (up to a constant). Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 30 (47)
Numerical algorithm for (IP): Dirichlet boundary condition Inverse problem: Given an incident plane wave u in ( x ) and the near-field data u sc ( x 1 , b ) = ∑ A n = A p , n ( α n , β n ) ⊤ e i β n b + A s , n ( − γ n , α n ) ⊤ e i γ n b , A n exp ( i α n x 1 ) , (3) n ∈ ℤ find a grating profile Λ : = { ( x 1 , f ( x 1 )) : 0 < x 1 < 2 π } such that u in + u sc = 0 on Λ . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 31 (47)
Numerical algorithm for (IP): Dirichlet boundary condition Inverse problem: Given an incident plane wave u in ( x ) and the near-field data u sc ( x 1 , b ) = ∑ A n = A p , n ( α n , β n ) ⊤ e i β n b + A s , n ( − γ n , α n ) ⊤ e i γ n b , A n exp ( i α n x 1 ) , (3) n ∈ ℤ find a grating profile Λ : = { ( x 1 , f ( x 1 )) : 0 < x 1 < 2 π } such that u in + u sc = 0 on Λ . Inverse problem: Find f from the knowledge of finite number of Fourier coefficients A n , n = 1 , 2 , ⋅⋅⋅ , K . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 31 (47)
Numerical algorithm for (IP): Dirichlet boundary condition Inverse problem: Given an incident plane wave u in ( x ) and the near-field data u sc ( x 1 , b ) = ∑ A n = A p , n ( α n , β n ) ⊤ e i β n b + A s , n ( − γ n , α n ) ⊤ e i γ n b , A n exp ( i α n x 1 ) , (3) n ∈ ℤ find a grating profile Λ : = { ( x 1 , f ( x 1 )) : 0 < x 1 < 2 π } such that u in + u sc = 0 on Λ . Inverse problem: Find f from the knowledge of finite number of Fourier coefficients A n , n = 1 , 2 , ⋅⋅⋅ , K . K : the number of propagating modes involved in computation. Far field data : { A n : ∣ α n ∣ < k p or ∣ α n ∣ < k s } . We always assume that 1. β n ∕ = 0 , γ n ∕ = 0 for all n ∈ ℤ . (Rayleigh frequencies are excluded) 2. b > f ( t ) > 0 , t ∈ ( 0 , 2 π ) . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 31 (47)
Fundamental solutions Fundamental solution to the Helmholtz equation ( ∆ + k 2 ) u = 0 : Φ k ( x , y ) = i 4 H ( 1 ) x = ( x 1 , x 2 ) , y = ( y 1 , y 2 ) ∈ ℝ 2 . 0 ( k ∣ x − y ∣ ) , Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 32 (47)
Fundamental solutions Fundamental solution to the Helmholtz equation ( ∆ + k 2 ) u = 0 : Φ k ( x , y ) = i 4 H ( 1 ) x = ( x 1 , x 2 ) , y = ( y 1 , y 2 ) ∈ ℝ 2 . 0 ( k ∣ x − y ∣ ) , Fundamental solution to the Navier equation µ ∆ u +( λ + µ ) grad div u + ω 2 u = 0 : Γ ( x , y ) i i 4 µ H ( 1 ) [ H ( 1 ) 0 ( k s ∣ x − y ∣ ) − H ( 1 ) ] 4 ω 2 grad x grad T = 0 ( k s ∣ x − y ∣ ) I + 0 ( k p ∣ x − y ∣ ) x µ Φ k s ( x , y ) I + 1 1 [ ] ω 2 grad x grad T = Φ k s ( x , y ) − Φ k p ( x , y ) . x Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 32 (47)
Quasiperiodic fundamental solutions The α -quasiperiodic fundamental solution to the Helmholtz equation ∑ G k ( x , y ) = exp ( − i α 2 π n ) Φ k ( x + n ( 2 π , 0 ) , y ) x − y ∕ = n ( 2 π , 0 ) n ∈ ℤ i 1 4 π ∑ = exp ( i α n ( x 1 − y 1 )+ i β n ∣ x 2 − y 2 ∣ ) . β n n ∈ ℤ Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 33 (47)
Quasiperiodic fundamental solutions The α -quasiperiodic fundamental solution to the Helmholtz equation ∑ G k ( x , y ) = exp ( − i α 2 π n ) Φ k ( x + n ( 2 π , 0 ) , y ) x − y ∕ = n ( 2 π , 0 ) n ∈ ℤ i 1 4 π ∑ = exp ( i α n ( x 1 − y 1 )+ i β n ∣ x 2 − y 2 ∣ ) . β n n ∈ ℤ The α -quasiperiodic fundamental solution (Green’s tensor) to the Navier equation: Π ( x , y ) = ∑ exp ( − i α 2 π n ) Γ ( x + n ( 2 π , 0 ) , y ) n ∈ ℤ µ G k s ( x , y ) I + 1 1 [ ] ω 2 grad x grad T = G k s ( x , y ) − G k p ( x , y ) x ( ) ( )[ ∂ 2 1 G k s ( x , y ) 0 + 1 ∂ x 1 ∂ x 2 ] x 1 = G k s ( x , y ) − G k p ( x , y ) . ω 2 ∂ 2 µ 0 G k s ( x , y ) ∂ x 2 ∂ x 1 x 2 Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 33 (47)
A two-step algorithm Step 1 Reconstruct the scattered field u sc . Making the ansatz for u sc in the form ∫ 2 π u sc = 1 Π ( x 1 , x 2 ; t , 0 ) ϕ ( t ) dt , x 2 ≥ f ( x 1 ) , 2 π 0 we only need to solve the first kind integral equation ∫ 2 π T ϕ ( x 1 ) : = 1 Π ( x 1 , b ; t , 0 ) ϕ ( t ) dt = u sc ( x 1 , b ) . 2 π 0 Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 34 (47)
A two-step algorithm Step 1 Reconstruct the scattered field u sc . Making the ansatz for u sc in the form ∫ 2 π u sc = 1 Π ( x 1 , x 2 ; t , 0 ) ϕ ( t ) dt , x 2 ≥ f ( x 1 ) , 2 π 0 we only need to solve the first kind integral equation ∫ 2 π T ϕ ( x 1 ) : = 1 Π ( x 1 , b ; t , 0 ) ϕ ( t ) dt = u sc ( x 1 , b ) . 2 π 0 Step 2 Find f by minimizing the defect ∫ 2 π ∣∣ u in ( x 1 , f ( x 1 ))+ 1 Π ( x 1 , f ( x 1 ) t , 0 ) ϕ ( t ) dt ∣∣ 2 L 2 ( 0 , 2 π ) → inf f ∈ M , 2 π 0 where M ∑ M = { f ( x 1 ) = a 0 + a m cos ( mx 1 )+ a M + m sin ( mx 1 ) , b j ∈ ℝ , j = 0 , 1 , ⋅⋅⋅ 2 M } . m = 1 Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 34 (47)
Step 1: Solve the first kind integral equation T ϕ ( x 1 ) = u b Tikhonov regularization: γϕ + T ∗ T ϕ = T ∗ u b , (4) γ > 0 : regularization parameter, T ∗ : the adjoint operator of T . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 35 (47)
Step 1: Solve the first kind integral equation T ϕ ( x 1 ) = u b Tikhonov regularization: γϕ + T ∗ T ϕ = T ∗ u b , (4) γ > 0 : regularization parameter, T ∗ : the adjoint operator of T . σ ( n ) 2 ( ) Solution : ϕ γ = ∑ ϕ ( n ) ϕ ( n ) j A n , U ( n ) V ( n ) ∑ exp ( i α n t ) , : = , γ γ j j ( σ ( n ) ) 2 + γ j = 1 ∣ n ∣≤ K j Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 35 (47)
Step 1: Solve the first kind integral equation T ϕ ( x 1 ) = u b Tikhonov regularization: γϕ + T ∗ T ϕ = T ∗ u b , (4) γ > 0 : regularization parameter, T ∗ : the adjoint operator of T . σ ( n ) 2 ( ) Solution : ϕ γ = ∑ ϕ ( n ) ϕ ( n ) j A n , U ( n ) V ( n ) ∑ exp ( i α n t ) , : = , γ γ j j ( σ ( n ) ) 2 + γ j = 1 ∣ n ∣≤ K j where U ( n ) = ( U ( n ) 1 , U ( n ) 2 ) , V ( n ) = ( V ( n ) , V ( n ) ) , Σ ( n ) = diag ( σ ( n ) 1 , σ ( n ) 2 ) . 1 2 is the singular value decomposition of M ( n ) given by ( ) { ( )} e i β n b α 2 α 2 α n β n α n γ n i i M ( n ) : = n n e i γ n b . + I − β 2 γ 2 4 πω 2 β n α n β n 4 πµγ n 4 πω 2 γ n α n γ n n n Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 35 (47)
Step 1: Solve the first kind integral equation T ϕ ( x 1 ) = u b Tikhonov regularization: γϕ + T ∗ T ϕ = T ∗ u b , (4) γ > 0 : regularization parameter, T ∗ : the adjoint operator of T . σ ( n ) 2 ( ) Solution : ϕ γ = ∑ ϕ ( n ) ϕ ( n ) j A n , U ( n ) V ( n ) ∑ exp ( i α n t ) , : = , γ γ j j ( σ ( n ) ) 2 + γ j = 1 ∣ n ∣≤ K j where U ( n ) = ( U ( n ) 1 , U ( n ) 2 ) , V ( n ) = ( V ( n ) , V ( n ) ) , Σ ( n ) = diag ( σ ( n ) 1 , σ ( n ) 2 ) . 1 2 is the singular value decomposition of M ( n ) given by ( ) { ( )} e i β n b α 2 α 2 α n β n α n γ n i i M ( n ) : = n n e i γ n b . + I − β 2 γ 2 4 πω 2 β n α n β n 4 πµγ n 4 πω 2 γ n α n γ n n n Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 35 (47)
Step 2: Find f by minimizing ∣∣ u in ( x 1 , f ( x 1 ))+ u sc ( x 1 , f ( x 1 )) ∣∣ L 2 Suppose that a = ( a 0 , a 1 , ⋅⋅⋅ , a M , ⋅⋅⋅ , a 2 M ) ∈ ℝ 2 M + 1 such that M ∑ f ( t ) = a 0 + a m cos ( mt )+ a M + m sin ( mt ) . m = 1 for some M ∈ ℕ . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 36 (47)
Step 2: Find f by minimizing ∣∣ u in ( x 1 , f ( x 1 ))+ u sc ( x 1 , f ( x 1 )) ∣∣ L 2 Suppose that a = ( a 0 , a 1 , ⋅⋅⋅ , a M , ⋅⋅⋅ , a 2 M ) ∈ ℝ 2 M + 1 such that M ∑ f ( t ) = a 0 + a m cos ( mt )+ a M + m sin ( mt ) . m = 1 for some M ∈ ℕ . Then N f ∈ M ∣∣ u in ( x 1 , f ( x 1 ))+ u sc ( x 1 , f ( x 1 )) ∣∣ L 2 ⇔ r 2 j ( a ) , K ∈ ℕ + ∑ inf inf a ∈ ℝ 2 M + 1 j = 1 Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 36 (47)
Step 2: Find f by minimizing ∣∣ u in ( x 1 , f ( x 1 ))+ u sc ( x 1 , f ( x 1 )) ∣∣ L 2 Suppose that a = ( a 0 , a 1 , ⋅⋅⋅ , a M , ⋅⋅⋅ , a 2 M ) ∈ ℝ 2 M + 1 such that M ∑ f ( t ) = a 0 + a m cos ( mt )+ a M + m sin ( mt ) . m = 1 for some M ∈ ℕ . Then N f ∈ M ∣∣ u in ( x 1 , f ( x 1 ))+ u sc ( x 1 , f ( x 1 )) ∣∣ L 2 ⇔ r 2 j ( a ) , K ∈ ℕ + ∑ inf inf a ∈ ℝ 2 M + 1 j = 1 where, for s j = 2 π ( j − 1 ) / N , r j ( a ) 2 = 1 θ e − i β f ( s j ) + ∑ e i β n f ( s j ) + S ( n ) ϕ ( n ) ( P ( n ) ϕ ( n ) N ∣ ˆ e i γ n f ( s j ) ) ∣ 2 γ γ ∣ n ∣≤ N Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 36 (47)
Step 2: Find f by minimizing ∣∣ u in ( x 1 , f ( x 1 ))+ u sc ( x 1 , f ( x 1 )) ∣∣ L 2 Suppose that a = ( a 0 , a 1 , ⋅⋅⋅ , a M , ⋅⋅⋅ , a 2 M ) ∈ ℝ 2 M + 1 such that M ∑ f ( t ) = a 0 + a m cos ( mt )+ a M + m sin ( mt ) . m = 1 for some M ∈ ℕ . Then N f ∈ M ∣∣ u in ( x 1 , f ( x 1 ))+ u sc ( x 1 , f ( x 1 )) ∣∣ L 2 ⇔ r 2 j ( a ) , K ∈ ℕ + ∑ inf inf a ∈ ℝ 2 M + 1 j = 1 where, for s j = 2 π ( j − 1 ) / N , r j ( a ) 2 = 1 θ e − i β f ( s j ) + ∑ e i β n f ( s j ) + S ( n ) ϕ ( n ) ( P ( n ) ϕ ( n ) N ∣ ˆ e i γ n f ( s j ) ) ∣ 2 γ γ ∣ n ∣≤ N To solve the least-squares problem N r 2 ∑ F ( a ) = j ( a ) → a ∈ ℝ 2 M + 1 , inf j = 1 we may use Gauss-Newton Method, Levenberg-Marquardt Method or Trust-Region Reflective Algorithm. Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 36 (47)
Numerical examples ■ u in : incident pressure wave. ■ θ = 0 , k p = 4 . 2 , k s = 4 . 5 , ω = 5 . ■ γ : Tikhonov regularization parameter. ■ δ : noise level of the measurement u ( x 1 , b ) . ■ K : the number of propagating modes involved in computation ■ K < 4 : partial far-field data ■ K = 4 : far-field data ■ K > 4 : far-field data + partial evanescent modes Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 37 (47)
Example 1: Fourier gratings ( K = 7 , γ = 10 − 12 ) Example 1 Suppose that f ( t ) = 2 + ζ ( cos ( t )+ cos ( 2 t )+ cos ( 3 t )) , ζ = 0 . 05 π . f ∗ ( t ) = a 0 + a 1 cos ( t )+ a 2 cos ( 2 t )+ a 3 cos ( 3 t )+ a 4 sin ( t )+ a 5 sin ( 2 t )+ a 6 sin ( 3 t ) . Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 38 (47)
Example 1: Fourier gratings ( K = 7 , γ = 10 − 12 ) Example 1 Suppose that f ( t ) = 2 + ζ ( cos ( t )+ cos ( 2 t )+ cos ( 3 t )) , ζ = 0 . 05 π . f ∗ ( t ) = a 0 + a 1 cos ( t )+ a 2 cos ( 2 t )+ a 3 cos ( 3 t )+ a 4 sin ( t )+ a 5 sin ( 2 t )+ a 6 sin ( 3 t ) . a 0 a 1 a 2 a 3 a 4 a 5 a 6 Target 2 0.157 0.157 0.157 0 0 0 Initial 0 0 0 0 0 0 0 LB -5 0 0 0 0 0 0 UB 5 5 5 5 5 5 5 δ = 0 2.0026 0.1590 0.1610 0.1596 0 0 0 δ = 5% 2.0029 0.1585 0.1592 0.1609 0.0004 0.0001 0.0001 δ = 8% 2.0021 0.1594 0.1610 0.1595 0.0003 -0.0010 -0.0005 δ = 10% 2.0029 0.1609 0.1601 0.1600 0.0023 -0.0019 0.0018 Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 38 (47)
Sensitivity analysis to the parameter K ( δ = 0 , γ = 10 − 12 ) Figure: K = 1 , 2 , 3 , 4 , 5 , 6 . K=1 K=2 K=3 2.5 2.5 computed computed computed 2.6 2.4 target 2.4 target target initial initial initial 2.3 2.3 2.4 2.2 2.2 2.2 2.1 2.1 2 2 2 1.8 1.9 1.9 1.6 1.8 1.8 1.4 1.7 1.7 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 K=4 K=5 K=6 2.5 2.5 2.5 computed computed computed 2.4 target 2.4 target 2.4 target initial initial initial 2.3 2.3 2.3 2.2 2.2 2.2 2.1 2.1 2.1 2 2 2 1.9 1.9 1.9 1.8 1.8 1.8 1.7 1.7 1.7 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 39 (47)
Example 2: smooth gratings ( K = 7 , γ = 10 − 12 , δ = 0 ) Suppose that f ( t ) = 1 . 5 + 0 . 2exp ( sin ( 3 t ))+ 0 . 3exp ( sin ( 3 t )) , M f ∗ ( t ) = a 0 + ∑ a m cos ( mt )+ a M + m sin ( mt ) , M = 8 . m = 1 Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 40 (47)
Example 2: smooth gratings ( K = 7 , γ = 10 − 12 , δ = 0 ) Suppose that f ( t ) = 1 . 5 + 0 . 2exp ( sin ( 3 t ))+ 0 . 3exp ( sin ( 3 t )) , M f ∗ ( t ) = a 0 + ∑ a m cos ( mt )+ a M + m sin ( mt ) , M = 8 . m = 1 3 computed ( δ =0) target 2.8 initial 2.6 2.4 2.2 2 1.8 1.6 0 1 2 3 4 5 6 Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 40 (47)
Example 2: smooth gratings ( K = 7 , γ = 10 − 12 , δ ∕ = 0 ) Suppose that f ( t ) = 1 . 5 + 0 . 2exp ( sin ( 3 t ))+ 0 . 3exp ( sin ( 3 t )) , M f ∗ ( t ) = a 0 + ∑ a m cos ( mt )+ a M + m sin ( mt ) , M = 8 . m = 1 Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 41 (47)
Example 2: smooth gratings ( K = 7 , γ = 10 − 12 , δ ∕ = 0 ) Suppose that f ( t ) = 1 . 5 + 0 . 2exp ( sin ( 3 t ))+ 0 . 3exp ( sin ( 3 t )) , M f ∗ ( t ) = a 0 + ∑ a m cos ( mt )+ a M + m sin ( mt ) , M = 8 . m = 1 3 computed ( δ =0.05) computed ( δ =0.1) 2.8 target initial 2.6 2.4 2.2 2 1.8 1.6 0 1 2 3 4 5 6 Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 41 (47)
Example 2: different initial guesses ( K = 4 , γ = 10 − 12 , δ = 0 ) K=4,M=8 3 computed 2.8 target initial 2.6 2.4 2.2 2 1.8 1.6 0 1 2 3 4 5 6 K=4,M=8 computed 3.2 target initial 3 2.8 2.6 2.4 2.2 2 1.8 1.6 0 1 2 3 4 5 6 Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 42 (47)
Example 3: binary gratings ( K = 4 , δ = 0 , γ = 10 − 4 ) A prior information : the unknown surface is a binary grating with a finite number of corner points. 3.6 3.6 target target 3.4 3.4 initial initial computed computed 3.2 3.2 3 3 2.8 2.8 2.6 2.6 2.4 2.4 2.2 2.2 2 2 1.8 1.8 1.6 1.6 0 1 2 3 4 5 6 0 1 2 3 4 5 6 Reconstruct a binary grating profile from the far-field data corresponding to three incident angles θ = − π / 4 , 0 , π / 4 (left) or one single incident angle θ = 0 (right). Direct and Inverse Elastic Scattering Problems for Diffraction Gratings ⋅ Workshop 3, Linz, Nov. 24, 2011 ⋅ Page 43 (47)
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