Acoustic and electromagnetic transmission problems: - PowerPoint PPT Presentation

R OMA , 14 M AY 2019 Acoustic and electromagnetic transmission problems: wavenumber-explicit bounds and resonance-free regions Andrea Moiola Joint work with E.A. Spence (Bath)

Part I Helmholtz equation

Helmholtz equation Acoustic waves in free space governed by wave eq. ∂ 2 U ∂ t 2 − ∆ U = 0 . Time-harmonic regime: assume U ( x , t )= ℜ{ u ( x ) e − i kt } and look for u . u satisfies Helmholtz equation ∆ u + k 2 u = 0 , with wavenumber k > 0 . Wavelength: λ = 2 π k , distance between two crests of a plane wave. 2

Helmholtz equation Acoustic waves in free space governed by wave eq. ∂ 2 U ∂ t 2 − ∆ U = 0 . Time-harmonic regime: assume U ( x , t )= ℜ{ u ( x ) e − i kt } and look for u . u satisfies Helmholtz equation ∆ u + k 2 u = 0 , with wavenumber k > 0 . Wavelength: λ = 2 π k , distance between two crests of a plane wave. 2

Helmholtz equation Acoustic waves in free space governed by wave eq. ∂ 2 U ∂ t 2 − ∆ U = 0 . Time-harmonic regime: assume U ( x , t )= ℜ{ u ( x ) e − i kt } and look for u . u satisfies Helmholtz equation ∆ u + k 2 u = 0 , with wavenumber k > 0 . Typical Helmholtz scattering problem: plane wave u Inc ( x ) = e i k x · d hitting a sound-soft (i.e. Dirichlet) obstacle Wavelength: λ = 2 π k , distance between two crests of a plane wave. 2

Helmholtz transmission problem Single penetrable homogeneous obstacle Ω i : � Sommerfeld u o = u i + g D radiation condition ∂ Ω i √ ∂ n u o = A N ∂ n u i + g N ∂ r u o − i ku o = o ( r 1 − d ) Ω i ⊂ R d ∆ u i + k 2 n i u i = f i Ω o = R d \ Ω i ∆ u o + k 2 u o = f o Data: f i ∈ L 2 (Ω i ) , comp (Ω o ) , g D ∈ H 1 ( ∂ Ω i ) , g N ∈ L 2 ( ∂ Ω i ) , f o ∈ L 2 refractive index 2 n i > 0 , wavenumber k > 0 , A N > 0 , scatterer Ω i ⊂ R d (Lipschitz bounded). � 1 in Ω o , What is A N ? E.g. in TE modes εµ = u = H z : A N = ε o ε i . n i in Ω i , In TM modes, u = E z : A N = µ o µ i . In acoustics A N = ρ o ρ i . Solution exists and is unique for Ω i Lipschitz and k ∈ C \ { 0 } , ℑ k ≥ 0 T ORRES , W ELLAND 1999. 3

Helmholtz transmission problem Single penetrable homogeneous obstacle Ω i : � Sommerfeld u o = u i + g D radiation condition ∂ Ω i √ ∂ n u o = A N ∂ n u i + g N ∂ r u o − i ku o = o ( r 1 − d ) Ω i ⊂ R d ∆ u i + k 2 n i u i = f i Ω o = R d \ Ω i ∆ u o + k 2 u o = f o Data: f i ∈ L 2 (Ω i ) , comp (Ω o ) , g D ∈ H 1 ( ∂ Ω i ) , g N ∈ L 2 ( ∂ Ω i ) , f o ∈ L 2 refractive index 2 n i > 0 , wavenumber k > 0 , A N > 0 , scatterer Ω i ⊂ R d (Lipschitz bounded). � 1 in Ω o , What is A N ? E.g. in TE modes εµ = u = H z : A N = ε o ε i . n i in Ω i , In TM modes, u = E z : A N = µ o µ i . In acoustics A N = ρ o ρ i . Solution exists and is unique for Ω i Lipschitz and k ∈ C \ { 0 } , ℑ k ≥ 0 T ORRES , W ELLAND 1999. 3

Helmholtz transmission problem Single penetrable homogeneous obstacle Ω i : � Sommerfeld u o = u i + g D radiation condition ∂ Ω i √ ∂ n u o = A N ∂ n u i + g N ∂ r u o − i ku o = o ( r 1 − d ) Ω i ⊂ R d ∆ u i + k 2 n i u i = f i Ω o = R d \ Ω i ∆ u o + k 2 u o = f o Data: f i ∈ L 2 (Ω i ) , comp (Ω o ) , g D ∈ H 1 ( ∂ Ω i ) , g N ∈ L 2 ( ∂ Ω i ) , f o ∈ L 2 refractive index 2 n i > 0 , wavenumber k > 0 , A N > 0 , scatterer Ω i ⊂ R d (Lipschitz bounded). � 1 in Ω o , What is A N ? E.g. in TE modes εµ = u = H z : A N = ε o ε i . n i in Ω i , In TM modes, u = E z : A N = µ o µ i . In acoustics A N = ρ o ρ i . Solution exists and is unique for Ω i Lipschitz and k ∈ C \ { 0 } , ℑ k ≥ 0 T ORRES , W ELLAND 1999. 3

Wave scattering The example we have in mind is scattering of incoming wave u Inc : f i = k 2 ( 1 − n i ) u Inc , f o = 0 , g D = 0 , g N = ( A N − 1 ) ∂ n u Inc . Scattered field Total field Incoming field u Inc = e i k x · d (datum) u = ( u i , u o ) u + u Inc √ 4 , A N = 1 , 2 ) , k = 20 , λ = 0 . 314 , 3 × 3 box, n i = 1 d = ( 1 3 2 , − figures represent real parts of fields. → U ( x , t ) = ℜ{ u ( x ) e − i kt } 4

Goal and motivation From Fredholm theory we have � u i � f i � g D � �� � �� � �� � � � � � � ≤ C 1 + C 2 � � � � � � u o f o g N � � � � � � Ω i / o Ω i / o ∂ Ω i Goal: find out how C 1 and C 2 depend on k , n i , A N , and Ω i and deduce results about resonances. Motivation: NA of Helmholtz problems with variable wavenumber: ◮ B ARUCQ , C HAUMONT -F RELET , G OUT (2016) ◮ O HLBERGER , V ERFÜRTH (2016) ◮ B ROWN , G ALLISTL , P ETERSEIM (2017) ◮ S AUTER , T ORRES (2017) ◮ G RAHAM , P EMBERY , S PENCE (2019) ◮ G RAHAM , S AUTER (2018) and with random parameters (from UQ perspective): ◮ F ENG , L IN , L ORTON (2015) ◮ H IPTMAIR , S CARABOSIO , S CHILLINGS , S CHWAB (2018) ◮ P EMBERY , S PENCE (2018). . . 5

Goal and motivation From Fredholm theory we have � u i � f i � g D � �� � �� � �� � � � � � � ≤ C 1 + C 2 � � � � � � u o f o g N � � � � � � Ω i / o Ω i / o ∂ Ω i Goal: find out how C 1 and C 2 depend on k , n i , A N , and Ω i and deduce results about resonances. Motivation: NA of Helmholtz problems with variable wavenumber: ◮ B ARUCQ , C HAUMONT -F RELET , G OUT (2016) ◮ O HLBERGER , V ERFÜRTH (2016) ◮ B ROWN , G ALLISTL , P ETERSEIM (2017) ◮ S AUTER , T ORRES (2017) ◮ G RAHAM , P EMBERY , S PENCE (2019) ◮ G RAHAM , S AUTER (2018) and with random parameters (from UQ perspective): ◮ F ENG , L IN , L ORTON (2015) ◮ H IPTMAIR , S CARABOSIO , S CHILLINGS , S CHWAB (2018) ◮ P EMBERY , S PENCE (2018). . . 5

Who cares? L AFONTAINE , S PENCE , W UNSCH , arXiv 2019: Allow to control: ◮ Quasi-optimality & pollution effect ◮ Gmres iteration # ◮ Matrix compression ◮ hp -FEM&BEM (Melenk–Sauter) ◮ . . . 6

Who cares? L AFONTAINE , S PENCE , W UNSCH , arXiv 2019: Allow to control: ◮ Quasi-optimality & pollution effect ◮ Gmres iteration # ◮ Matrix compression ◮ hp -FEM&BEM (Melenk–Sauter) ◮ . . . 6

“Cut-off resolvent”: R χ ( k ) Assume g D = g N = 0 (no jumps/boundary data). � f i � u i � � Solution operator: R ( k ) = R ( k , n i , A N , Ω i ) : �→ . f o u o Let χ 1 , χ 2 ∈ C ∞ 0 ( R d ) s.t. χ j ≡ 1 in a neighbourhood of Ω i . Then L 2 (Ω i ) ⊕ L 2 (Ω o ) H 1 (Ω i ) ⊕ H 1 (Ω o ) R χ ( k ) := χ 1 R ( k ) χ 2 : → ( f i , f o ) �→ ( u i , u o χ 1 ) . Well-known that R χ ( k ) is holomorphic on ℑ k > 0 . Resonances: poles of meromorphic continuation of R χ ( k ) to ℑ k < 0 . We want to bound the norm of R χ ( k ) , k ∈ R . Consider separately cases n i < 1 and n i > 1 : very different! 7

“Cut-off resolvent”: R χ ( k ) Assume g D = g N = 0 (no jumps/boundary data). � f i � u i � � Solution operator: R ( k ) = R ( k , n i , A N , Ω i ) : �→ . f o u o Let χ 1 , χ 2 ∈ C ∞ 0 ( R d ) s.t. χ j ≡ 1 in a neighbourhood of Ω i . Then L 2 (Ω i ) ⊕ L 2 (Ω o ) H 1 (Ω i ) ⊕ H 1 (Ω o ) R χ ( k ) := χ 1 R ( k ) χ 2 : → ( f i , f o ) �→ ( u i , u o χ 1 ) . Well-known that R χ ( k ) is holomorphic on ℑ k > 0 . Resonances: poles of meromorphic continuation of R χ ( k ) to ℑ k < 0 . We want to bound the norm of R χ ( k ) , k ∈ R . Consider separately cases n i < 1 and n i > 1 : very different! 7

“Cut-off resolvent”: R χ ( k ) Assume g D = g N = 0 (no jumps/boundary data). � f i � u i � � Solution operator: R ( k ) = R ( k , n i , A N , Ω i ) : �→ . f o u o Let χ 1 , χ 2 ∈ C ∞ 0 ( R d ) s.t. χ j ≡ 1 in a neighbourhood of Ω i . Then L 2 (Ω i ) ⊕ L 2 (Ω o ) H 1 (Ω i ) ⊕ H 1 (Ω o ) R χ ( k ) := χ 1 R ( k ) χ 2 : → ( f i , f o ) �→ ( u i , u o χ 1 ) . Well-known that R χ ( k ) is holomorphic on ℑ k > 0 . Resonances: poles of meromorphic continuation of R χ ( k ) to ℑ k < 0 . We want to bound the norm of R χ ( k ) , k ∈ R . Consider separately cases n i < 1 and n i > 1 : very different! 7

“Cut-off resolvent”: R χ ( k ) Assume g D = g N = 0 (no jumps/boundary data). � f i � u i � � Solution operator: R ( k ) = R ( k , n i , A N , Ω i ) : �→ . f o u o Let χ 1 , χ 2 ∈ C ∞ 0 ( R d ) s.t. χ j ≡ 1 in a neighbourhood of Ω i . Then L 2 (Ω i ) ⊕ L 2 (Ω o ) H 1 (Ω i ) ⊕ H 1 (Ω o ) R χ ( k ) := χ 1 R ( k ) χ 2 : → ( f i , f o ) �→ ( u i , u o χ 1 ) . Well-known that R χ ( k ) is holomorphic on ℑ k > 0 . Resonances: poles of meromorphic continuation of R χ ( k ) to ℑ k < 0 . We want to bound the norm of R χ ( k ) , k ∈ R . Consider separately cases n i < 1 and n i > 1 : very different! 7