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A Reduced Basis Method for Multiple Electromagnetic Scattering in Three Dimensions "Numerical methods for high-dimensional problems Ecole des Ponts Paristech, April 14th-18th 2014 Benjamin Stamm LJLL, Paris 6 and CNRS Thursday, April


  1. A Reduced Basis Method for Multiple Electromagnetic Scattering in Three Dimensions "Numerical methods for high-dimensional problems” Ecole des Ponts Paristech, April 14th-18th 2014 Benjamin Stamm LJLL, Paris 6 and CNRS Thursday, April 24, 14

  2. Outline • Problem setting • Single scatterer RBM • RBM for multiple scattering problem • Numerical results In collaboration with • M. Fares (CERFACS, Toulouse) • M. Ganesh (Colorado School of Mines) • J. Hesthaven (EPFL) • Y. Maday (Paris 6 and Brown University) • S. Zhang (City University of Hong Kong) Thursday, April 24, 14

  3. Problem setting Thursday, April 24, 14

  4. Physical problem/Geometrical Configuration [in 3D] E i ( x ) D i D J D 1 Incident plane wave impinging onto collection of J perfectly conducting obstacles D 1 , . . . , D J . Thursday, April 24, 14

  5. Physical problem/Geometrical Configuration [in 3D] E i ( x ) D i D J D 1 ⇒ Scattered field E s ( x ) Incident plane wave impinging onto collection of J perfectly conducting obstacles D 1 , . . . , D J . Thursday, April 24, 14

  6. Parametrization The system is parametrized by: • The wave number k , • The angle and polarization of the incident wave E i ( x ; k, p , ˆ k ) = − p e ik x · ˆ k ( θ , φ ) , • The location and shape of the obstacles: Thursday, April 24, 14

  7. Parametrization The system is parametrized by: • The wave number k , • The angle and polarization of the incident wave E i ( x ; k, p , ˆ k ) = − p e ik x · ˆ k ( θ , φ ) , • The location and shape of the obstacles: T i (ˆ x ) = γ i B i ˆ x + b i Reference shape: D i ˆ y y ˆ D ˆ x x Thursday, April 24, 14

  8. Parametrization The system is parametrized by: • The wave number k , • The angle and polarization of the incident wave E i ( x ; k, p , ˆ k ) = − p e ik x · ˆ k ( θ , φ ) , • The location and shape of the obstacles: T i (ˆ x ) = γ i B i ˆ x + b i Reference shape: D i ˆ y y ˆ D ˆ x The a ffi ne transformation T i includes: x B i ∈ SO (3): rotation γ i ∈ R + : stretching b i ∈ R 3 : translation Thursday, April 24, 14

  9. Parametrization The system is parametrized by: • The wave number k , • The angle and polarization of the incident wave E i ( x ; k, p , ˆ k ) = − p e ik x · ˆ k ( θ , φ ) , • The location and shape of the obstacles: T i (ˆ x ) = γ i B i ˆ x + b i Reference shape: D i ˆ y y ˆ D ˆ x The a ffi ne transformation T i includes: x B i ∈ SO (3): rotation γ i ∈ R + : stretching b i ∈ R 3 : translation Parameter: µ = ( k, ˆ � ) ∈ P ⊂ R 5+7 J k , p , b 1 , B 1 , γ 1 � , . . . , b J , B J , γ J � �� � �� Thursday, April 24, 14

  10. Governing equations (time-harmonic ansatz) Assume that the free space is a homogenous media with magnetic permeability µ and electrical permittivity ε . The total electric field E = E i + E s ∈ H (curl , Ω ) satisfies curl curl E − k 2 E = 0 in Ω , Maxwell E × n = 0 on Γ , boundary condition � � � � � curl E s ( x ) × | x | − ik E s ( x ) 1 � = O as | x | → ∞ . Silver-M¨ uller radiation condition x � � | x | Ω = R 3 \ ∪ J i =1 D i . Γ is the collection of all surfaces: Γ = ∪ J i =1 ∂ D i . see book of [Colton,Kress], [Nedelec] Thursday, April 24, 14

  11. Variational formulation of the Electric Field Integral Equation (EFIE) Change the unknown to be u : Electric currant on collection of surfaces. For any fixed µ ∈ P , find u ( µ ) ∈ V s.t. a [ u ( µ ) , v ; µ ] = f [ v ; µ ] , ∀ v ∈ V with � � � � u ( y ) · v ( x ) − 1 a [ u , v ; µ ] = ikZ G k ( x , y ) k 2 div y u ( y )div x v ( x ) d y d x Γ ( µ ) Γ ( µ ) � E i ( y ; k, p , ˆ f [ v ; µ ] = − k ) · v ( y ) d y Γ ( µ ) The scattered electric field E s is then uniquely determined by the electric currant u . Thursday, April 24, 14

  12. Variational formulation of the Electric Field Integral Equation (EFIE) Change the unknown to be u : Electric currant on collection of surfaces. For any fixed µ ∈ P , find u ( µ ) ∈ V s.t. a [ u ( µ ) , v ; µ ] = f [ v ; µ ] , ∀ v ∈ V with � � � � u ( y ) · v ( x ) − 1 a [ u , v ; µ ] = ikZ G k ( x , y ) k 2 div y u ( y )div x v ( x ) d y d x Γ ( µ ) Γ ( µ ) � E i ( y ; k, p , ˆ f [ v ; µ ] = − k ) · v ( y ) d y Γ ( µ ) The kernel function is given by e ik | x − y | G k ( x , y ) = 4 π | x − y | The scattered electric field E s is then uniquely determined by the electric currant u . Thursday, April 24, 14

  13. Output of interest: Radar Cross Section (RCS) • Describes pattern/energy of electrical field at infinity • Functional of the current on body d ] = ikZ � d ) e − ik x · ˆ A ∞ [ u ; µ , ˆ d × ( u ( x ) × ˆ ˆ d d x 4 π Γ � � | A ∞ [ u ; µ , ˆ d ] | 2 RCS[ u ; µ , ˆ d ] = 10 log 10 | E i ( x ; k, p , ˆ k ) | 2 where u : current on surface ˆ d : given directional unit vector Thursday, April 24, 14

  14. Output of interest: Radar Cross Section (RCS) • Describes pattern/energy of electrical field at infinity • Functional of the current on body d ] = ikZ � d ) e − ik x · ˆ A ∞ [ u ; µ , ˆ d × ( u ( x ) × ˆ ˆ d d x 4 π Γ � � | A ∞ [ u ; µ , ˆ d ] | 2 RCS[ u ; µ , ˆ d ] = 10 log 10 | E i ( x ; k, p , ˆ k ) | 2 where u : current on surface 40 ˆ d : given directional unit vector 30 20 RCS 10 0 -10 -20 0 1 2 3 4 5 6 φ rcs Thursday, April 24, 14

  15. Single obstacle scattering Thursday, April 24, 14

  16. Reduced Basis Method Reduced Basis Ansatz: V N = span { u δ ( µ 1 ) , . . . , u δ ( µ N ) } for some well-chosen sample points µ 1 , . . . , µ N . Thursday, April 24, 14

  17. Reduced Basis Method Reduced Basis Ansatz: V N = span { u δ ( µ 1 ) , . . . , u δ ( µ N ) } for some well-chosen sample points µ 1 , . . . , µ N . Example: 1 parameter: wavenumber k Di ff erent snapshots illustrated Thursday, April 24, 14

  18. Reduced Basis Method Reduced Basis Ansatz: V N = span { u δ ( µ 1 ) , . . . , u δ ( µ N ) } for some well-chosen sample points µ 1 , . . . , µ N . Example: 1 parameter: wavenumber k Di ff erent snapshots illustrated Question: How to find the sample points µ 1 , . . . , µ N such that V N ≈ M = { u δ ( µ ) : ∀ µ ∈ P } Thursday, April 24, 14

  19. Reduced Basis Method Reduced Basis Ansatz: V N = span { u δ ( µ 1 ) , . . . , u δ ( µ N ) } for some well-chosen sample points µ 1 , . . . , µ N . Example: 1 parameter: wavenumber k Di ff erent snapshots illustrated Question: How to find the sample points µ 1 , . . . , µ N such that V N ≈ M = { u δ ( µ ) : ∀ µ ∈ P } Answer: Greedy-algorithm Thursday, April 24, 14

  20. Affine decomposition for EFIE For any parameter value µ ∈ P , find u ( µ ) ∈ V s.t. a ( u ( µ ) , v ; µ ) = f ( v ; µ ) , ∀ v ∈ V with � � � � e ik | x − y | u ( x ) · v ( y ) − 1 a ( u , v ; µ ) = ikZ k 2 div Γ , x u ( x ) div Γ , y v ( y ) d x d y 4 π | x − y | Γ Γ � e ik x · ˆ k ( θ , φ ) v ( x ) d x f ( v ; µ ) = − p · Γ Thursday, April 24, 14

  21. Affine decomposition for EFIE For any parameter value µ ∈ P , find u ( µ ) ∈ V s.t. a ( u ( µ ) , v ; µ ) = f ( v ; µ ) , ∀ v ∈ V with � � � � e ik | x − y | u ( x ) · v ( y ) − 1 a ( u , v ; µ ) = ikZ k 2 div Γ , x u ( x ) div Γ , y v ( y ) d x d y 4 π | x − y | Γ Γ � e ik x · ˆ k ( θ , φ ) v ( x ) d x f ( v ; µ ) = − p · Γ Solution : Empirical Interpolation Method (EIM) (also based on a greedy algorithm) Given: A parametrized function g ( x ; µ ). Output: { µ q } Q q =1 such that Q S i m i l a r p r o b l e m � g ( x ; µ ) ≈ I Q ( g )( x ; µ ) = α g q ( µ ) g ( x ; µ q ) . formulation as for the q =1 RBM, but solutions are explicitly known (not solution to PDE) [Maday et al. 2004] (happy birthday!) Thursday, April 24, 14

  22. Affine decomposition for the EFIE Approximating Q e ik | x − y | q ( k ) e ikq | x − y | � α a 4 π | x − y | ≈ 4 π | x − y | q =1 Q e ik x · ˆ q ( µ ) e ik q x · ˆ � k ( θ , φ ) ≈ k ( θ q, φ q ) α f q =1 red: parameter-dependent, blue: parameter-independent. Thursday, April 24, 14

  23. Affine decomposition for the EFIE Approximating Q e ik | x − y | q ( k ) e ikq | x − y | � α a 4 π | x − y | ≈ 4 π | x − y | q =1 Q e ik x · ˆ q ( µ ) e ik q x · ˆ � k ( θ , φ ) ≈ k ( θ q, φ q ) α f q =1 results in � � � � e ik | x − y | e ik | x − y | 4 π | x − y | u ( x ) · v ( y ) d x d y − iZ a ( v , w ; µ ) = ikZ 4 π | x − y | div Γ , x u ( x ) div Γ , y v ( y ) d x d y k Γ Γ Γ Γ Q � � e ikq | x − y | � q ( k ) 4 π | x − y | u ( x ) · v ( y ) d x d y ikZ α a ≈ Γ Γ q =1 Q � � iZ α a q ( k ) e ikq | x − y | � 4 π | x − y | div Γ , x u ( x ) div Γ , y v ( y ) d x d y − k Γ Γ q =1 red: parameter-dependent, blue: parameter-independent. Thursday, April 24, 14

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