The Half-Space Matching Method for scattering problems in - PowerPoint PPT Presentation

The idea of the Half-Space Matching formulation (in 2D) We split the infinite domain into 5 overlapping subdomains; A rectangle containing the defects 4 half-spaces which do not contain any defect For instance for the plate (with 2D pictures): A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 7 / 41

The idea of the Half-Space Matching formulation (in 2D) We split the infinite domain into 5 overlapping subdomains; A rectangle containing the defects 4 half-spaces which do not contain any defect And for the junction of waveguides: A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 7 / 41

The idea of the Half-Space Matching formulation (in 2D) We split the infinite domain into 5 overlapping subdomains; A rectangle containing the defects 4 half-spaces which do not contain any defect And for the junction of waveguides: A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 7 / 41

The idea of the Half-Space Matching formulation (in 2D) We split the infinite domain into 5 overlapping subdomains; A rectangle containing the defects 4 half-spaces which do not contain any defect And for the junction of waveguides: A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 7 / 41

The idea of the Half-Space Matching formulation (in 2D) We split the infinite domain into 5 overlapping subdomains; A rectangle containing the defects 4 half-spaces which do not contain any defect And for the junction of waveguides: A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 7 / 41

The idea of the Half-Space Matching formulation (in 2D) We split the infinite domain into 5 overlapping subdomains; A rectangle containing the defects 4 half-spaces which do not contain any defect The key point In each half-space, we can use a separation of variables. A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 7 / 41

The simple case of the 2D dissipative Helmholtz equation f is a compactly supported source term ω ε = ω + i ε with ε > 0 The initial problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 ε u = f in R 2 . You can add any bounded obstacle with usual BC. The problem is well-posed by Lax-Milgram theorem. The numerical method works for ε = 0 but not yet the theory... A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 8 / 41

Derivation of the half-space matching formulation The initial problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 ε u = f in R 2 . We introduce the splitting of R 2 : A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 9 / 41

Derivation of the half-space matching formulation The initial problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 ε u = f in R 2 . We introduce the splitting of R 2 and the notations: A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 9 / 41

Derivation of the half-space matching formulation The initial problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 ε u = f in R 2 . A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 10 / 41

Derivation of the half-space matching formulation The initial problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 ε u = f in R 2 . Unknowns of the new formulation The restriction of u to Ω b : u b The Dirichlet traces of u on the infinite lines Σ j a : ϕ j A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 10 / 41

Fourier representation in the half-spaces Consider for instance the half-space Ω 0 a = { x > a } . We know that: � ∆ u + ω 2 ε u = 0 ( x > a ) u ( a , y ) = ϕ 0 ( y ) ( y ∈ R ) A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 11 / 41

Fourier representation in the half-spaces Consider for instance the half-space Ω 0 a = { x > a } . We know that: � ∆ u + ω 2 ε u = 0 ( x > a ) u ( a , y ) = ϕ 0 ( y ) ( y ∈ R ) Applying a Fourier transform in y and solving the ODE in x , we get: ϕ 0 ( ξ ) e − √ ε ( x − a ) for x ≥ a ξ 2 − ω 2 u ( x , ξ ) = ˆ ˆ with ℜ ( √ z ) > 0, (the limit ε → 0 gives the outgoing solution) A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 11 / 41

Fourier representation in the half-spaces Consider for instance the half-space Ω 0 a = { x > a } . We know that: � ∆ u + ω 2 ε u = 0 ( x > a ) u ( a , y ) = ϕ 0 ( y ) ( y ∈ R ) Applying a Fourier transform in y and solving the ODE in x , we get: ϕ 0 ( ξ ) e − √ ε ( x − a ) for x ≥ a ξ 2 − ω 2 u ( x , ξ ) = ˆ ˆ and then ϕ 0 ( ξ ) e − √ � 1 ξ 2 − ω 2 ε ( x − a ) e iy ξ d ξ for x ≥ a √ u ( x , y ) = ˆ 2 π R A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 11 / 41

Fourier representation in the half-spaces Consider for instance the half-space Ω 0 a = { x > a } . We know that: � ∆ u + ω 2 ε u = 0 ( x > a ) u ( a , y ) = ϕ 0 ( y ) ( y ∈ R ) Applying a Fourier transform in y and solving the ODE in x , we get: ϕ 0 ( ξ ) e − √ � 1 ξ 2 − ω 2 ε ( x − a ) e iy ξ d ξ := U 0 ( ϕ 0 ) in Ω 0 u ( x , y ) = √ ˆ a 2 π R Proceeding in the same way for all half-planes, we get: u = U j ( ϕ j ) in Ω j a for j = 0 , 1 , 2 , 3 A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 11 / 41

Compatibility in the quarter-planes A first equation on the ϕ j is obtained by imposing the compatibility of the half-plane representations in the overlapping areas. A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 12 / 41

Compatibility in the quarter-planes Consider for instance the quarter-plane Ω 0 a ∩ Ω 1 a . We must ensure: U 0 ( ϕ 0 ) = U 1 ( ϕ 1 ) in Ω 0 a ∩ Ω 1 a which is equivalent to U 0 ( ϕ 0 ) = U 1 ( ϕ 1 ) in ∂ (Ω 0 a ∩ Ω 1 a ) A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 12 / 41

Compatibility in the quarter-planes Consider for instance the quarter-plane Ω 0 a ∩ Ω 1 a . We must ensure: U 0 ( ϕ 0 ) = U 1 ( ϕ 1 ) in Ω 0 a ∩ Ω 1 a which is equivalent to U 0 ( ϕ 0 ) = U 1 ( ϕ 1 ) in ∂ (Ω 0 a ∩ Ω 1 a ) Proof Take v = U 0 ( ϕ 0 ) − U 1 ( ϕ 1 ) and use the well-posedness in H 1 of problem: � ∆ v + ω 2 in Ω 0 a ∩ Ω 1 ε v = 0 a on ∂ (Ω 0 a ∩ Ω 1 v = 0 a ) A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 12 / 41

Compatibility in the quarter-planes Consider for instance the quarter-plane Ω 0 a ∩ Ω 1 a . We must ensure: U 0 ( ϕ 0 ) = U 1 ( ϕ 1 ) in Ω 0 a ∩ Ω 1 a which is equivalent to U 0 ( ϕ 0 ) = U 1 ( ϕ 1 ) in ∂ (Ω 0 a ∩ Ω 1 a ) Proof Take v = U 0 ( ϕ 0 ) − U 1 ( ϕ 1 ) and use the Open question ! well-posedness in H 1 of problem: � Prove this result when ε = 0 ∆ v + ω 2 in Ω 0 a ∩ Ω 1 ε v = 0 a on ∂ (Ω 0 a ∩ Ω 1 v = 0 a ) A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 12 / 41

Compatibility in the quarter-planes (2) Consider for instance the quarter-plane Ω 0 a ∩ Ω 1 a . We must ensure: U 0 ( ϕ 0 ) = U 1 ( ϕ 1 ) on ∂ (Ω 0 a ∩ Ω 1 a ) which can be rewritten: U 0 ( ϕ 0 ) = ϕ 1 on Ω 0 a ∩ Σ 1 a U 1 ( ϕ 1 ) = ϕ 0 on Ω 1 a ∩ Σ 0 a A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 13 / 41

Compatibility in the quarter-planes (2) Consider for instance the quarter-plane Ω 0 a ∩ Ω 1 a . We must ensure: U 0 ( ϕ 0 ) = U 1 ( ϕ 1 ) on ∂ (Ω 0 a ∩ Ω 1 a ) which can be rewritten: U 0 ( ϕ 0 ) = ϕ 1 on Ω 0 a ∩ Σ 1 a U 1 ( ϕ 1 ) = ϕ 0 on Ω 1 a ∩ Σ 0 a This leads to the following integral equation: ϕ 0 ( ξ ) e − √ � 1 ξ 2 − ω 2 ϕ 1 ( x ) = ε ( x − a ) e ia ξ d ξ := D ( ϕ 0 ) for x > a √ ˆ 2 π R A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 13 / 41

Compatibility in the quarter-planes (2) Consider for instance the quarter-plane Ω 0 a ∩ Ω 1 a . We must ensure: U 0 ( ϕ 0 ) = U 1 ( ϕ 1 ) on ∂ (Ω 0 a ∩ Ω 1 a ) which can be rewritten: U 0 ( ϕ 0 ) = ϕ 1 on Ω 0 a ∩ Σ 1 a U 1 ( ϕ 1 ) = ϕ 0 on Ω 1 a ∩ Σ 0 a This leads to the following integral equation: ϕ 0 ( ξ ) e − √ � 1 ξ 2 − ω 2 ϕ 1 ( x ) = ε ( x − a ) e ia ξ d ξ := D ( ϕ 0 ) for x > a √ ˆ 2 π R Proceeding in the same way for all half-planes, we get 8 equations: ϕ j = D ( ϕ j ± 1 ) on Ω j ± 1 ∩ Σ j j ∈ Z / 4 Z a a A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 13 / 41

Compatibility between the half-planes and the square Finally, we have to ensure the compatibility between the interior solution u b and the half-plane representations U j ( ϕ j ) in Ω b \ Ω a . A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 14 / 41

Compatibility between the half-planes and the square Finally, we have to ensure the compatibility between the interior solution u b and the half-plane representations U j ( ϕ j ) in Ω b \ Ω a . We match Dirichlet traces on the small square ∂ Ω a and Neumann traces on the large square ∂ Ω b (OK since ε > 0). A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 14 / 41

Compatibility between the half-planes and the square Finally, we have to ensure the compatibility between the interior solution u b and the half-plane representations U j ( ϕ j ) in Ω b \ Ω a . We match Dirichlet traces on the small square ∂ Ω a and Neumann traces on the large square ∂ Ω b (OK since ε > 0). More precisely, we impose: u b = ϕ 0 on Σ 0 a ∩ ∂ Ω a A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 14 / 41

Compatibility between the half-planes and the square Finally, we have to ensure the compatibility between the interior solution u b and the half-plane representations U j ( ϕ j ) in Ω b \ Ω a . We match Dirichlet traces on the small square ∂ Ω a and Neumann traces on the large square ∂ Ω b (OK since ε > 0). More precisely, we impose: u b = ϕ 0 on Σ 0 a ∩ ∂ Ω a ∂ u b ∂ n = ∂ ∂ n ( U 0 ( ϕ 0 )) on ∂ Ω b ∩ Ω 0 a A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 14 / 41

Compatibility between the half-planes and the square Finally, we have to ensure the compatibility between the interior solution u b and the half-plane representations U j ( ϕ j ) in Ω b \ Ω a . We match Dirichlet traces on the small square ∂ Ω a and Neumann traces on the large square ∂ Ω b (OK since ε > 0). More precisely, we impose: u b = ϕ 0 on Σ 0 a ∩ ∂ Ω a ∂ u b ∂ n = ∂ ∂ n ( U 0 ( ϕ 0 )) on ∂ Ω b ∩ Ω 0 a This leads to the following integral equation: ϕ 0 ( ξ ) e − √ � � ∂ u b ∂ n = − 1 ξ 2 − ω 2 ξ 2 − ω 2 ε ( b − a ) e iy ξ d ξ := Λ( ϕ 0 ) for | y | < a √ ε ˆ 2 π R A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 14 / 41

Compatibility between the half-planes and the square We match Dirichlet traces on the small square ∂ Ω a and Neumann traces on the large square ∂ Ω b (OK since ε > 0). More precisely, we impose: u b = ϕ 0 on Σ 0 a ∩ ∂ Ω a ∂ u b ∂ n = ∂ ∂ n ( U 0 ( ϕ 0 )) on ∂ Ω b ∩ Ω 0 a And we do the same for all half-planes: u b = ϕ j on Σ j a ∩ ∂ Ω a for j = 0 , 1 , 2 , 3 ∂ u b ∂ n = Λ( ϕ j ) on ∂ Ω b ∩ Ω j a for j = 0 , 1 , 2 , 3 A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 14 / 41

The half-space matching formulation The initial problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 ε u = f in R 2 . The new unknowns The restriction of u to Ω b : u b The traces of u on Σ j a : ϕ j A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 15 / 41

The half-space matching formulation The initial problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 ε u = f in R 2 . The new unknowns The restriction of u to Ω b : u b The traces of u on Σ j a : ϕ j The new formulation  ∆ u b + ω 2 ε u b = f in Ω b    Σ j  u b = ϕ j on a ∩ ∂ Ω a for j = 0 , 1 , 2 , 3   ∂ u b ∂ Ω b ∩ Ω j ∂ n = Λ( ϕ j ) on a for j = 0 , 1 , 2 , 3      ϕ j = D ( ϕ j ± 1 )  Ω j ± 1 ∩ Σ j on j ∈ Z / 4 Z a a A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 15 / 41

Several generalizations The power of this approach is that it applies to 1 Anisotropic scalar equations like: div ( A ∇ u ) + ω 2 ε u = f where A is a positive definite matrix. With 4 different integral operators Λ j and 8 operators D j ! A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 16 / 41

Several generalizations The power of this approach is that it applies to 1 Anisotropic scalar equations like: div ( A ∇ u ) + ω 2 ε u = f where A is a positive definite matrix. 2 Any number of half-planes (Internship of P. Merino (2014)). A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 16 / 41

Several generalizations The power of this approach is that it applies to 1 Anisotropic scalar equations like: div ( A ∇ u ) + ω 2 ε u = f where A is a positive definite matrix. 2 Any number of half-planes (Internship of P. Merino (2014)). 3 Open waveguides problems: use the generalized Fourier transform instead of the usual Fourier transform. A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 16 / 41

Several generalizations The power of this approach is that it applies to 1 Anisotropic scalar equations like: div ( A ∇ u ) + ω 2 ε u = f where A is a positive definite matrix. 2 Any number of half-planes (Internship of P. Merino (2014)). 3 Open waveguides problems: use the generalized Fourier transform instead of the usual Fourier transform. 4 Isotropic and anisotropic elastodynamics. A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 16 / 41

Outline Motivations and objectives 1 The new formulation 2 Mathematical analysis 3 Discretization and numerical results 4 Conclusion, future works and open questions 5 A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 17 / 41

Variational formulation  ∆ u b + ω 2 ε u b = f in Ω b     Σ j u b = ϕ j on a ∩ ∂ Ω a   ∂ u b ∂ n = Λ( ϕ j ) ∂ Ω b ∩ Ω j on  a     ϕ j = D ( ϕ j ± 1 )  Ω j ± 1 ∩ Σ j on a a A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 18 / 41

Variational formulation  ∆ u b + ω 2 ε u b = f in Ω b     Σ j u b = ϕ j on a ∩ ∂ Ω a   ∂ u b ∂ n = Λ( ϕ j ) ∂ Ω b ∩ Ω j on  a     ϕ j = D ( ϕ j ± 1 )  Ω j ± 1 ∩ Σ j on a a The appropriate functional space is: V = { ( u b , ϕ j ); u b ∈ H 1 (Ω b ) , ϕ j ∈ L 2 (Σ j a ) , u b = ϕ j on Σ j a ∩ ∂ Ω a } A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 18 / 41

Variational formulation  ∆ u b + ω 2 ε u b = f in Ω b     Σ j u b = ϕ j on a ∩ ∂ Ω a   ∂ u b ∂ n = Λ( ϕ j ) ∂ Ω b ∩ Ω j on  a     ϕ j = D ( ϕ j ± 1 )  Ω j ± 1 ∩ Σ j on a a The appropriate functional space is: V = { ( u b , ϕ j ); u b ∈ H 1 (Ω b ) , ϕ j ∈ L 2 (Σ j a ) , u b = ϕ j on Σ j a ∩ ∂ Ω a } A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 18 / 41

Variational formulation  ∆ u b + ω 2 ε u b = f in Ω b     Σ j u b = ϕ j on a ∩ ∂ Ω a   ∂ u b ∂ n = Λ( ϕ j ) ∂ Ω b ∩ Ω j on  a     ϕ j = D ( ϕ j ± 1 )  Ω j ± 1 ∩ Σ j on a a The appropriate functional space is: V = { ( u b , ϕ j ); u b ∈ H 1 (Ω b ) , ϕ j ∈ L 2 (Σ j a ) , u b = ϕ j on Σ j a ∩ ∂ Ω a } Then the weak form of the half-space matching formulation is:  � � � ∇ u b · ∇ v b − ω 2 � Λ( ϕ j ) , v b � ∀ ( v b , ψ j ) ∈ V  ε u b v b − = fv b    Ω b Ω b j � � � ϕ j ψ j − � D ( ϕ j ± 1 ) , ψ j � − u b ψ j ∀ ( v b , ψ j ) ∈ V  = 0    Σ j Σ j a ∩ ∂ Ω a a ± A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 18 / 41

Fredholm property V = { ( u b , ϕ j ); u b ∈ H 1 (Ω b ) , ϕ j ∈ L 2 (Σ j a ) , u b = ϕ j on Σ j a ∩ ∂ Ω a } ( u b , ϕ j ) ∈ V such that  � � � ∇ u b · ∇ v b − ω 2 � Λ( ϕ j ) , v b �  ε u b v b − = fv b    Ω b Ω b j ∀ ( v b , ψ j ) ∈ V � � � ϕ j ψ j − � D ( ϕ j ± 1 ) , ψ j � − u b ψ j  = 0    Σ j Σ j a ∩ ∂ Ω a a ± GOOD NEWS : Λ is a compact operator on L 2 (Σ j a ) BAD NEWS : D is not a compact operator on L 2 (Σ j a )... A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 19 / 41

Compactness of Λ Remember that Λ( ϕ 0 ) = ∂ ∂ x U 0 ( ϕ 0 ) on x = b , | y | < a . � Λ( ϕ 0 )( y ) = ϕ 0 ( ξ ) d ξ K Λ ( ξ, y ) ˆ R ε e − √ � K Λ ( ξ, y ) = − 1 ξ 2 − ω 2 ξ 2 − ω 2 ε ( b − a ) e iy ξ √ 2 π A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 20 / 41

Compactness of Λ Remember that Λ( ϕ 0 ) = ∂ ∂ x U 0 ( ϕ 0 ) on x = b , | y | < a . � Λ( ϕ 0 )( y ) = ϕ 0 ( ξ ) d ξ K Λ ( ξ, y ) ˆ R ε e − √ � K Λ ( ξ, y ) = − 1 ξ 2 − ω 2 ξ 2 − ω 2 ε ( b − a ) e iy ξ √ 2 π Direct proof of compactness For b > a and ε > 0, K Λ ∈ L 2 ( R × ] a , a [) ϕ 0 ⇒ Λ is compact ⇒ Λ is a Hilbert-Schmidt operator acting on ˆ A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 20 / 41

Compactness of Λ Remember that Λ( ϕ 0 ) = ∂ ∂ x U 0 ( ϕ 0 ) on x = b , | y | < a . � Λ( ϕ 0 )( y ) = ϕ 0 ( ξ ) d ξ K Λ ( ξ, y ) ˆ R ε e − √ � K Λ ( ξ, y ) = − 1 ξ 2 − ω 2 ξ 2 − ω 2 ε ( b − a ) e iy ξ √ 2 π Direct proof of compactness For b > a and ε > 0, K Λ ∈ L 2 ( R × ] a , a [) ϕ 0 ⇒ Λ is compact ⇒ Λ is a Hilbert-Schmidt operator acting on ˆ Alternative proof of compactness Interior regularity of U 0 ( ϕ 0 ) and compact embedding ⇒ Λ is compact A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 20 / 41

Fredholm property (2) V = { ( u b , ϕ j ); u b ∈ H 1 (Ω b ) , ϕ j ∈ L 2 (Σ j a ) , u b = ϕ j on Σ j a ∩ ∂ Ω a } ( u b , ϕ j ) ∈ V such that � �  � ∇ u b · ∇ v b − ω 2 � Λ( ϕ j ) , v b � ∀ ( v b , ψ j ) ∈ V ε u b v b − = fv b     Ω b Ω b  j  � �� �   � �� � coercive compact � � �  ϕ j ψ j � D ( ϕ j ± 1 ) , ψ j � u b ψ j ∀ ( v b , ψ j ) ∈ V − − = 0     Σ j Σ j � �� � a ∩ ∂ Ω a  a ±   � �� � ?? � �� � compact coercive A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 21 / 41

Fredholm property (2) V = { ( u b , ϕ j ); u b ∈ H 1 (Ω b ) , ϕ j ∈ L 2 (Σ j a ) , u b = ϕ j on Σ j a ∩ ∂ Ω a } ( u b , ϕ j ) ∈ V such that � �  � ∇ u b · ∇ v b − ω 2 � Λ( ϕ j ) , v b � ∀ ( v b , ψ j ) ∈ V ε u b v b − = fv b     Ω b Ω b  j  � �� �   � �� � coercive compact � � �  ϕ j ψ j � D ( ϕ j ± 1 ) , ψ j � u b ψ j ∀ ( v b , ψ j ) ∈ V − − = 0     Σ j Σ j � �� � a ∩ ∂ Ω a  a ±   � �� � ?? � �� � compact coercive Let us consider for instance the term � D ( ϕ 0 ) , ψ 1 � A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 21 / 41

Properties of operator D Remember that D ( ϕ 0 ) = U 0 ( ϕ 0 ) on x > a , y = a . � D ( ϕ 0 )( x ) = ϕ 0 ( ξ ) d ξ K D ( ξ, x ) ˆ R e − √ K D ( ξ, x ) = − 1 ξ 2 − ω 2 ε ( x − a ) e ia ξ √ 2 π Lack of compactness (due to the cross-point) ∈ L 2 ( R × ] a , a [) and D is not compact. Even for b > a and ε > 0, K D / A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 22 / 41

Properties of operator D Remember that D ( ϕ 0 ) = U 0 ( ϕ 0 ) on x > a , y = a . � D ( ϕ 0 )( x ) = ϕ 0 ( ξ ) d ξ K D ( ξ, x ) ˆ R e − √ K D ( ξ, x ) = − 1 ξ 2 − ω 2 ε ( x − a ) e ia ξ √ 2 π Lack of compactness (due to the cross-point) ∈ L 2 ( R × ] a , a [) and D is not compact. Even for b > a and ε > 0, K D / Partial result of compactness By the interior regularity of U 0 ( ϕ 0 ), ϕ 0 − → D ( ϕ 0 − ) is compact By definition of V , ϕ 0 a → D ( ϕ 0 a ) is compact A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 22 / 41

Properties of operator D (2) What about ϕ 0 + → D ( ϕ 0 + )? � D ( ϕ 0 )( x ) = ϕ 0 ( ξ ) d ξ K D ( ξ, x ) ˆ R e − √ K D ( ξ, x ) = − 1 ξ 2 − ω 2 ε ( x − a ) e ia ξ √ 2 π A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 23 / 41

Properties of operator D (2) What about ϕ 0 + → D ( ϕ 0 + )? � D ( ϕ 0 )( x ) = ϕ 0 ( ξ ) d ξ K D ( ξ, x ) ˆ R e − √ K D ( ξ, x ) = − 1 ξ 2 − ω 2 ε ( x − a ) e ia ξ √ 2 π It reminds the Laplace operator  L 2 ( R + ) → L 2 ( R + )  �L� = √ π � L : R + e − ξ x f ( ξ ) d ξ f →  A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 23 / 41

Properties of operator D (2) What about ϕ 0 + → D ( ϕ 0 + )? � D ( ϕ 0 )( x ) = ϕ 0 ( ξ ) d ξ K D ( ξ, x ) ˆ R e − √ K D ( ξ, x ) = − 1 ξ 2 − ω 2 ε ( x − a ) e ia ξ √ 2 π From the Laplace operator, we deduce that  L 2 ([ a , + ∞ )) → L 2 ([ a , + ∞ ))   � ˜ 1 ˜ e − √ D � ≤ � √ D : ξ 2 +1( x − a ) − e ia ξ 2 ϕ → √ ϕ ( ξ ) d ξ ˆ   2 π R A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 23 / 41

Properties of operator D (2) What about ϕ 0 + → D ( ϕ 0 + )? � D ( ϕ 0 )( x ) = ϕ 0 ( ξ ) d ξ K D ( ξ, x ) ˆ R e − √ K D ( ξ, x ) = − 1 ξ 2 − ω 2 ε ( x − a ) e ia ξ √ 2 π e − √ ξ 2 +1( x − a ) − e − √ ε ( x − a ) � e ia ξ � � ξ 2 − ω 2 D ( ϕ )( y ) = ˜ D ( ϕ )( y ) + √ ϕ ( ξ ) d ξ ˆ 2 π R � �� � Hilbert-Schmidt A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 23 / 41

Properties of operator D (2) What about ϕ 0 + → D ( ϕ 0 + )? � D ( ϕ 0 )( x ) = ϕ 0 ( ξ ) d ξ K D ( ξ, x ) ˆ R e − √ K D ( ξ, x ) = − 1 ξ 2 − ω 2 ε ( x − a ) e ia ξ √ 2 π e − √ ξ 2 +1( x − a ) − e − √ ε ( x − a ) � e ia ξ � � D ( ϕ )( y ) = ˜ ξ 2 − ω 2 D ( ϕ )( y ) + √ ϕ ( ξ ) d ξ ˆ 2 π R � �� � Hilbert-Schmidt Final result + ) is a sum ˜ D + ( D − ˜ D ) with � ˜ The operator ϕ 0 + → D ( ϕ 0 D � < 1 and D − ˜ D compact. A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 23 / 41

Fredholm property (3) V = { ( u b , ϕ j ); u b ∈ H 1 (Ω b ) , ϕ j ∈ L 2 (Σ j a ) , u b = ϕ j on Σ j a ∩ ∂ Ω a } ( u b , ϕ j ) ∈ V such that � �  � ∇ u b · ∇ v b − ω 2 � Λ( ϕ j ) , v b � ∀ ( v b , ψ j ) ∈ V ε u b v b − = fv b     Ω b Ω b  j  � �� �   � �� � coercive compact � � �  ϕ j ψ j � D ( ϕ j ± 1 ) , ψ j � u b ψ j ∀ ( v b , ψ j ) ∈ V − − = 0     Σ j Σ j � �� � a ∩ ∂ Ω a  a ±   � �� � ?? � �� � compact coercive A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 24 / 41

Fredholm property (3) V = { ( u b , ϕ j ); u b ∈ H 1 (Ω b ) , ϕ j ∈ L 2 (Σ j a ) , u b = ϕ j on Σ j a ∩ ∂ Ω a } ( u b , ϕ j ) ∈ V such that � �  � ∇ u b · ∇ v b − ω 2 � Λ( ϕ j ) , v b � ∀ ( v b , ψ j ) ∈ V ε u b v b − = fv b     Ω b Ω b  j  � �� �   � �� � coercive compact �  ϕ, � ψ ) − (˜ ϕ, � ϕ, � u b ψ j ∀ ( v b , ψ j ) ∈ V ( � D � ψ ) + ( K � ψ ) − = 0     Σ j � �� � a ∩ ∂ Ω a    coercive � �� � compact A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 24 / 41

Fredholm property (3) V = { ( u b , ϕ j ); u b ∈ H 1 (Ω b ) , ϕ j ∈ L 2 (Σ j a ) , u b = ϕ j on Σ j a ∩ ∂ Ω a } ( u b , ϕ j ) ∈ V such that � �  � ∇ u b · ∇ v b − ω 2 � Λ( ϕ j ) , v b � ∀ ( v b , ψ j ) ∈ V ε u b v b − = fv b     Ω b Ω b  j  � �� �   � �� � coercive compact �  ϕ, � ψ ) − (˜ ϕ, � ϕ, � u b ψ j ∀ ( v b , ψ j ) ∈ V ( � D � ψ ) + ( K � ψ ) − = 0     Σ j � �� � a ∩ ∂ Ω a    coercive � �� � compact Theorem The problem is of Fredholm type. It is well-posed (because of the equivalence with the initial problem). A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 24 / 41

Does this result hold for the generalizations ? 1 Anisotropic scalar equations like div ( A ∇ u ) + ω 2 ε u = f we did not succeed in proving � ˜ D � < 1 for some ˜ Open question: D . A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 25 / 41

Does this result hold for the generalizations ? 1 Anisotropic scalar equations like div ( A ∇ u ) + ω 2 ε u = f we did not succeed in proving � ˜ D � < 1 for some ˜ Open question: D . 2 Any number of half-planes OK: we prove � ˜ D � < 1 by using Mellin calculus. A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 25 / 41

Does this result hold for the generalizations ? 1 Anisotropic scalar equations like div ( A ∇ u ) + ω 2 ε u = f we did not succeed in proving � ˜ D � < 1 for some ˜ Open question: D . 2 Any number of half-planes OK: we prove � ˜ D � < 1 by using Mellin calculus. 3 Open waveguides problems: OK: ˜ D is the same as in the homogeneous case since quarter-planes are homogeneous ! A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 25 / 41

Does this result hold for the generalizations ? 1 Anisotropic scalar equations like div ( A ∇ u ) + ω 2 ε u = f we did not succeed in proving � ˜ D � < 1 for some ˜ Open question: D . 2 Any number of half-planes OK: we prove � ˜ D � < 1 by using Mellin calculus. 3 Open waveguides problems: OK: ˜ D is the same as in the homogeneous case since quarter-planes are homogeneous ! 4 Isotropic elastodynamics. Still to do. A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 25 / 41

Does this result hold for the generalizations ? 1 Anisotropic scalar equations like div ( A ∇ u ) + ω 2 ε u = f we did not succeed in proving � ˜ D � < 1 for some ˜ Open question: D . 2 Any number of half-planes OK: we prove � ˜ D � < 1 by using Mellin calculus. 3 Open waveguides problems: OK: ˜ D is the same as in the homogeneous case since quarter-planes are homogeneous ! 4 Isotropic elastodynamics. Still to do. 5 Anisotropic elastodynamics. Open question. A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 25 / 41

An alternative formulation We impose: u b = ϕ j on the blue lines ϕ j = D ( ϕ j ± 1 ) on the red lines First formulation: Alternative formulation: Equivalence: YES Equivalence: ?? Compactness of D : NO Compactness of D : YES but Fredholmness OK and Fredholmness OK A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 26 / 41

Outline Motivations and objectives 1 The new formulation 2 Mathematical analysis 3 Discretization and numerical results 4 Conclusion, future works and open questions 5 A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 27 / 41

Discrete problem For the discretization: We truncate the lines Σ j a (parameter T ) We use continuous Lagrange Finite Elements, 1D for ϕ j and 2D for u b b , ϕ j New unknown ( u h h ) A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 28 / 41

Discrete problem For the discretization: We truncate the lines Σ j a (parameter T ) We use continuous Lagrange Finite Elements, 1D for ϕ j and 2D for u b b , ϕ j New unknown ( u h h ) For the computation of integral terms: We truncate the Fourier integrals We use quadrature formulae A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 28 / 41

Discrete problem For the discretization: We truncate the lines Σ j a (parameter T ) We use continuous Lagrange Finite Elements, 1D for ϕ j and 2D for u b b , ϕ j New unknown ( u h h ) For the computation of integral terms: We truncate the Fourier integrals We use quadrature formulae Example: Exact formula � � Λ( ϕ 0 h ) , u h ϕ 0 u h b � = k ( ξ ) ˆ h ( ξ )ˆ b ( b , ξ ) d ξ R ε e − √ � ξ 2 − ω 2 ξ 2 − ω 2 ε ( b − a ) , ˆ ϕ 0 where k ( ξ ) = − h ( ξ ) and � b 1 u h u h b ( b , y ) e − i ξ y dy are computed analytically. ˆ b ( b , ξ ) = √ 2 π − b A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 28 / 41

Discrete problem For the discretization: We truncate the lines Σ j a (parameter T ) We use continuous Lagrange Finite Elements, 1D for ϕ j and 2D for u b b , ϕ j New unknown ( u h h ) For the computation of integral terms: We truncate the Fourier integrals We use quadrature formulae Example: Exact formula and approximate formula � � � Λ( ϕ 0 ϕ 0 ϕ 0 h ) , u h u h u h b � = k ( ξ ) ˆ h ( ξ )ˆ b ( b , ξ ) d ξ ≈ q n k ( ξ n ) ˆ h ( ξ n )ˆ b ( b , ξ n ) R ξ n ∈Q ε e − √ � ξ 2 − ω 2 ξ 2 − ω 2 ε ( b − a ) , ˆ ϕ 0 where k ( ξ ) = − h ( ξ ) and � b 1 u h u h b ( b , y ) e − i ξ y dy are computed analytically. ˆ b ( b , ξ ) = √ 2 π − b A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 28 / 41

Discrete problem For the discretization: We truncate the lines Σ j a (parameter T ) We use continuous Lagrange Finite Elements, 1D for ϕ j and 2D for u b b , ϕ j New unknown ( u h h ) For the computation of integral terms: We truncate the Fourier integrals We use quadrature formulae GOOD NEWS Convergence of FE discretization ensured by the Fredholm property. Error estimates for the semi-discretization in ξ have been obtained. A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 28 / 41

Numerical validation The problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 ε u = f in R 2 where f ≈ δ A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 29 / 41

Numerical validation The problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 ε u = f in R 2 where f ≈ δ The dissipative case: ω ε = 10 + 0 . 5 i and T = 24 a err rel < 0 . 06% A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 29 / 41

Numerical validation The problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 ε u = f in R 2 where f ≈ δ A power of our method: the solu- tion can be reconstructed outside A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 29 / 41

Numerical validation The problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 ε u = f in R 2 where f ≈ δ A power of our method: the solu- tion can be reconstructed outside A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 29 / 41

Numerical validation (2) The problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 u = f in R 2 where f ≈ δ A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 30 / 41

Numerical validation (2) The problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 u = f in R 2 where f ≈ δ The non-dissipative case: ω = 10 and T = 24 a A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 30 / 41

Numerical validation (2) The problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 u = f in R 2 where f ≈ δ refine the dis- The price to pay: cretization in ξ A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 30 / 41

Numerical validation (2) The problem Find u ∈ H 1 ( R 2 ) such that ∆ u + ω 2 u = f in R 2 where f ≈ δ A bad discretization in ξ is easily de- tectable a posteriori... A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 30 / 41

Application to an anisotropic equation The problem � div ( A ∇ u ) + ω 2 R 2 \O ε u = f in A ∇ u · n = 0 on O � � 1 0 . 5 where A = 0 . 5 1 A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 31 / 41