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The Half-Space Matching Method for scattering problems in anisotropic media Anne-Sophie Bonnet-Ben Dhia 1 Sonia Fliss 1 Antoine Tonnoir 2 with some results of Johanes Tjandrawidjadja 1 and Julian Ott 3 1 POEMS (UMR CNRS-ENSTA-INRIA), Palaiseau,

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SLIDE 1

The Half-Space Matching Method for scattering problems in anisotropic media

Anne-Sophie Bonnet-Ben Dhia1 Sonia Fliss1 Antoine Tonnoir2 with some results of Johanes Tjandrawidjadja1 and Julian Ott3

1 POEMS (UMR CNRS-ENSTA-INRIA), Palaiseau, France. 2 INSA, Rouen, France. 3 KIT, Karlsruhe, Germany.

RICAM, LINZ 2016

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 1 / 41

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SLIDE 2

Outline

1

Motivations and objectives

2

The new formulation

3

Mathematical analysis

4

Discretization and numerical results

5

Conclusion, future works and open questions

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 2 / 41

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SLIDE 3

Motivations

We consider time-harmonic scattering problems in infinite domains for which usual methods do not work.

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 3 / 41

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SLIDE 4

Motivations

We consider time-harmonic scattering problems in infinite domains for which usual methods do not work. TWO EXAMPLES AND MAIN MOTIVATIONS:

SHM experiment - INSA of Lyon Photonic waveguides - UPSaclay

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 3 / 41

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SLIDE 5

Motivations

We consider time-harmonic scattering problems in infinite domains for which usual methods do not work. TWO EXAMPLES AND MAIN MOTIVATIONS: The diffraction of an ultrasonic wave by a defect in an anisotropic elastic plate (Application to Non Destructive Testing)

SHM experiment - INSA of Lyon Photonic waveguides - UPSaclay

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 3 / 41

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SLIDE 6

Motivations

We consider time-harmonic scattering problems in infinite domains for which usual methods do not work. TWO EXAMPLES AND MAIN MOTIVATIONS: The diffraction of an ultrasonic wave by a defect in an anisotropic elastic plate (Application to Non Destructive Testing) The diffraction by a junction between open optical waveguides (Application to integrated optics and nanophotonics)

SHM experiment - INSA of Lyon Photonic waveguides - UPSaclay

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 3 / 41

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SLIDE 7

Motivations

We consider time-harmonic scattering problems in infinite domains for which usual methods do not work. TWO EXAMPLES AND MAIN MOTIVATIONS: The diffraction of an ultrasonic wave by a defect in an anisotropic elastic plate (Application to Non Destructive Testing) The diffraction by a junction between open optical waveguides (Application to integrated optics and nanophotonics) DIFFICULTIES: For the theory: Radiation condition ? Well-posedness ? For the numerics: Transparent boundary condition ?

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 3 / 41

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SLIDE 8

The anisotropic elastic plate

The objective is to formulate and compute the scattering of a Lamb mode by an arbitrary defect (crack, heterogeneity, corrosion zone etc...)

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 4 / 41

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SLIDE 9

The anisotropic elastic plate

The objective is to formulate and compute the scattering of a Lamb mode by an arbitrary defect (crack, heterogeneity, corrosion zone etc...) Usual approaches are not adapted because:

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 4 / 41

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SLIDE 10

The anisotropic elastic plate

The objective is to formulate and compute the scattering of a Lamb mode by an arbitrary defect (crack, heterogeneity, corrosion zone etc...) Usual approaches are not adapted because: Perfectly Matched Layers are ”unstable”

B´ ecache et al., Stability of perfectly matched layers, group velocities and anisotropic waves, JCP 2003 Skelton et al., Guided elastic waves and perfectly matched layers, Wave Motion 2007 Bonnet-Ben Dhia et al., On the use of perfectly matched layers in the presence of long or backward propagating guided elastic waves, Wave Motion 2014

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 4 / 41

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SLIDE 11

The anisotropic elastic plate

The objective is to formulate and compute the scattering of a Lamb mode by an arbitrary defect (crack, heterogeneity, corrosion zone etc...) Usual approaches are not adapted because: Perfectly Matched Layers are ”unstable” The computation of the Green tensor is too expensive.

Wang and Achenbach, Three-Dimensional Time-Harmonic Elastodynamic Green’s Functions for Anisotropic Solids, Proceeding of the Royal Society, 1937

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 4 / 41

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SLIDE 12

The anisotropic elastic plate

The objective is to formulate and compute the scattering of a Lamb mode by an arbitrary defect (crack, heterogeneity, corrosion zone etc...) Usual approaches are not adapted because: Perfectly Matched Layers are ”unstable” The computation of the Green tensor is too expensive. The derivation of a DtN condition on a circular boundary is impossible, because cylindrical coordinates are not compatible with anisotropy

Givoli et al., Non reflecting boundary condition for elastic waves, Wave Motion, 1990

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 4 / 41

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SLIDE 13

The anisotropic elastic plate

The objective is to formulate and compute the scattering of a Lamb mode by an arbitrary defect (crack, heterogeneity, corrosion zone etc...) Usual approaches are not adapted because: Perfectly Matched Layers are ”unstable” The computation of the Green tensor is too expensive. The derivation of a DtN condition on a circular boundary is impossible, because cylindrical coordinates are not compatible with anisotropy

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 4 / 41

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SLIDE 14

The junction of open waveguides

The objective is to modelize and simulate the scattering of a guided mode by a junction of open waveguides. ∆u + ω2n2(x, y)u = 0 where n(x, y) = n∞(y) outside the junction The problem is difficult because the exterior domain is not homogeneous:

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 5 / 41

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SLIDE 15

The junction of open waveguides

The objective is to modelize and simulate the scattering of a guided mode by a junction of open waveguides. ∆u + ω2n2(x, y)u = 0 where n(x, y) = n∞(y) outside the junction The problem is difficult because the exterior domain is not homogeneous: Radiation conditions are difficult to formulate.

Bonnet-Ben Dhia, Goursaud and Hazard, Mathematical analysis of the junction of two acoustic open waveguides, SIAM J. Appl. Math., 2011 Jerez-Hanckes and N´ ed´ elec, Asymptotics for Helmholtz and Maxwell solutions in 3-D open waveguides, Communications in Computational Physics, 2012

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 5 / 41

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SLIDE 16

The junction of open waveguides

The objective is to modelize and simulate the scattering of a guided mode by a junction of open waveguides. ∆u + ω2n2(x, y)u = 0 where n(x, y) = n∞(y) outside the junction The problem is difficult because the exterior domain is not homogeneous: Radiation conditions are difficult to formulate. Perfectly Matched Layers are not accurate because guided modes decay exponentially in the transverse directions.

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 5 / 41

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SLIDE 17

The junction of open waveguides

The objective is to modelize and simulate the scattering of a guided mode by a junction of open waveguides. ∆u + ω2n2(x, y)u = 0 where n(x, y) = n∞(y) outside the junction The problem is difficult because the exterior domain is not homogeneous: Radiation conditions are difficult to formulate. Perfectly Matched Layers are not accurate because guided modes decay exponentially in the transverse directions.

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 5 / 41

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SLIDE 18

Outline

1

Motivations and objectives

2

The new formulation

3

Mathematical analysis

4

Discretization and numerical results

5

Conclusion, future works and open questions

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 6 / 41

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SLIDE 19

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

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 7 / 41

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SLIDE 20

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 We will use Finite elements in the rectangle Semi-analytical representations in the half-spaces This approach has been first introduced by Sonia Fliss and Patrick Joly for periodic media:

Sonia Fliss and Patrick Joly, Exact boundary conditions for time-harmonic wave propagation in locally perturbed periodic media, Applied Numerical Mathematics, 2009

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 7 / 41

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SLIDE 21

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

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SLIDE 22

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

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SLIDE 23

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

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SLIDE 24

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

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SLIDE 25

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

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SLIDE 26

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

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SLIDE 27

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

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SLIDE 28

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

slide-29
SLIDE 29

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

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SLIDE 30

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

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SLIDE 31

The simple case of the 2D dissipative Helmholtz equation

f is a compactly supported source term ωε = ω + iε with ε > 0

The initial problem

Find u ∈ H1(R2) such that ∆u + ω2

εu = f in R2.

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

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SLIDE 32

Derivation of the half-space matching formulation

The initial problem

Find u ∈ H1(R2) such that ∆u + ω2

εu = f in R2.

We introduce the splitting of R2:

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 9 / 41

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SLIDE 33

Derivation of the half-space matching formulation

The initial problem

Find u ∈ H1(R2) such that ∆u + ω2

εu = f in R2.

We introduce the splitting of R2 and the notations:

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 9 / 41

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SLIDE 34

Derivation of the half-space matching formulation

The initial problem

Find u ∈ H1(R2) such that ∆u + ω2

εu = f in R2.

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 10 / 41

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SLIDE 35

Derivation of the half-space matching formulation

The initial problem

Find u ∈ H1(R2) such that ∆u + ω2

εu = f in R2.

Unknowns of the new formulation

The restriction of u to Ωb: ub The Dirichlet traces of u on the infinite lines Σj

a: ϕj

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 10 / 41

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SLIDE 36

Fourier representation in the half-spaces

Consider for instance the half-space Ω0

a = {x > a}.

We know that:

  • ∆u + ω2

εu = 0

(x > a) u(a, y) = ϕ0(y) (y ∈ R)

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 11 / 41

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SLIDE 37

Fourier representation in the half-spaces

Consider for instance the half-space Ω0

a = {x > a}.

We know that:

  • ∆u + ω2

εu = 0

(x > a) u(a, y) = ϕ0(y) (y ∈ R) Applying a Fourier transform in y and solving the ODE in x, we get: ˆ u(x, ξ) = ˆ ϕ0(ξ)e−√

ξ2−ω2

ε(x−a) for x ≥ a

with ℜ(√z) > 0, (the limit ε → 0 gives the outgoing solution)

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 11 / 41

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SLIDE 38

Fourier representation in the half-spaces

Consider for instance the half-space Ω0

a = {x > a}.

We know that:

  • ∆u + ω2

εu = 0

(x > a) u(a, y) = ϕ0(y) (y ∈ R) Applying a Fourier transform in y and solving the ODE in x, we get: ˆ u(x, ξ) = ˆ ϕ0(ξ)e−√

ξ2−ω2

ε(x−a) for x ≥ a

and then u(x, y) = 1 √ 2π

  • R

ˆ ϕ0(ξ)e−√

ξ2−ω2

ε(x−a)eiyξdξ for x ≥ a A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 11 / 41

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SLIDE 39

Fourier representation in the half-spaces

Consider for instance the half-space Ω0

a = {x > a}.

We know that:

  • ∆u + ω2

εu = 0

(x > a) u(a, y) = ϕ0(y) (y ∈ R) Applying a Fourier transform in y and solving the ODE in x, we get: u(x, y) = 1 √ 2π

  • R

ˆ ϕ0(ξ)e−√

ξ2−ω2

ε(x−a)eiyξdξ := U0(ϕ0) in Ω0

a

Proceeding in the same way for all half-planes, we get: u = Uj(ϕj) in Ωj

a for j = 0, 1, 2, 3

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 11 / 41

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SLIDE 40

Compatibility in the quarter-planes

A first equation on the ϕj 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

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SLIDE 41

Compatibility in the quarter-planes

Consider for instance the quarter-plane Ω0

a ∩ Ω1 a.

We must ensure: U0(ϕ0) = U1(ϕ1) in Ω0

a ∩ Ω1 a

which is equivalent to U0(ϕ0) = U1(ϕ1) in ∂(Ω0

a ∩ Ω1 a)

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 12 / 41

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SLIDE 42

Compatibility in the quarter-planes

Consider for instance the quarter-plane Ω0

a ∩ Ω1 a.

We must ensure: U0(ϕ0) = U1(ϕ1) in Ω0

a ∩ Ω1 a

which is equivalent to U0(ϕ0) = U1(ϕ1) in ∂(Ω0

a ∩ Ω1 a)

Proof

Take v = U0(ϕ0) − U1(ϕ1) and use the well-posedness in H1 of problem:

  • ∆v + ω2

εv = 0

in Ω0

a ∩ Ω1 a

v = 0

  • n ∂(Ω0

a ∩ Ω1 a)

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 12 / 41

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SLIDE 43

Compatibility in the quarter-planes

Consider for instance the quarter-plane Ω0

a ∩ Ω1 a.

We must ensure: U0(ϕ0) = U1(ϕ1) in Ω0

a ∩ Ω1 a

which is equivalent to U0(ϕ0) = U1(ϕ1) in ∂(Ω0

a ∩ Ω1 a)

Proof

Take v = U0(ϕ0) − U1(ϕ1) and use the well-posedness in H1 of problem:

  • ∆v + ω2

εv = 0

in Ω0

a ∩ Ω1 a

v = 0

  • n ∂(Ω0

a ∩ Ω1 a)

Open question !

Prove this result when ε = 0

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 12 / 41

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SLIDE 44

Compatibility in the quarter-planes (2)

Consider for instance the quarter-plane Ω0

a ∩ Ω1 a.

We must ensure: U0(ϕ0) = U1(ϕ1) on ∂(Ω0

a ∩ Ω1 a)

which can be rewritten: U0(ϕ0) = ϕ1 on Ω0

a ∩ Σ1 a

U1(ϕ1) = ϕ0 on Ω1

a ∩ Σ0 a

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 13 / 41

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SLIDE 45

Compatibility in the quarter-planes (2)

Consider for instance the quarter-plane Ω0

a ∩ Ω1 a.

We must ensure: U0(ϕ0) = U1(ϕ1) on ∂(Ω0

a ∩ Ω1 a)

which can be rewritten: U0(ϕ0) = ϕ1 on Ω0

a ∩ Σ1 a

U1(ϕ1) = ϕ0 on Ω1

a ∩ Σ0 a

This leads to the following integral equation: ϕ1(x) = 1 √ 2π

  • R

ˆ ϕ0(ξ)e−√

ξ2−ω2

ε(x−a)eiaξdξ := D(ϕ0) for x > a A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 13 / 41

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SLIDE 46

Compatibility in the quarter-planes (2)

Consider for instance the quarter-plane Ω0

a ∩ Ω1 a.

We must ensure: U0(ϕ0) = U1(ϕ1) on ∂(Ω0

a ∩ Ω1 a)

which can be rewritten: U0(ϕ0) = ϕ1 on Ω0

a ∩ Σ1 a

U1(ϕ1) = ϕ0 on Ω1

a ∩ Σ0 a

This leads to the following integral equation: ϕ1(x) = 1 √ 2π

  • R

ˆ ϕ0(ξ)e−√

ξ2−ω2

ε(x−a)eiaξdξ := D(ϕ0) for x > a

Proceeding in the same way for all half-planes, we get 8 equations: ϕj = D(ϕj±1) on Ωj±1

a

∩ Σj

a

j ∈ Z/4Z

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 13 / 41

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SLIDE 47

Compatibility between the half-planes and the square

Finally, we have to ensure the compatibility between the interior solution ub and the half-plane representations Uj(ϕj) in Ωb\Ωa.

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 14 / 41

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SLIDE 48

Compatibility between the half-planes and the square

Finally, we have to ensure the compatibility between the interior solution ub and the half-plane representations Uj(ϕj) in Ωb\Ωa. We match Dirichlet traces on the small square ∂Ωa and Neumann traces

  • n the large square ∂Ωb (OK since ε > 0).

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 14 / 41

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SLIDE 49

Compatibility between the half-planes and the square

Finally, we have to ensure the compatibility between the interior solution ub and the half-plane representations Uj(ϕj) in Ωb\Ωa. We match Dirichlet traces on the small square ∂Ωa and Neumann traces

  • n the large square ∂Ωb (OK since ε > 0).

More precisely, we impose: ub = ϕ0 on Σ0

a ∩ ∂Ωa

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 14 / 41

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SLIDE 50

Compatibility between the half-planes and the square

Finally, we have to ensure the compatibility between the interior solution ub and the half-plane representations Uj(ϕj) in Ωb\Ωa. We match Dirichlet traces on the small square ∂Ωa and Neumann traces

  • n the large square ∂Ωb (OK since ε > 0).

More precisely, we impose: ub = ϕ0 on Σ0

a ∩ ∂Ωa

∂ub ∂n = ∂ ∂n(U0(ϕ0)) on ∂Ωb ∩ Ω0

a

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 14 / 41

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SLIDE 51

Compatibility between the half-planes and the square

Finally, we have to ensure the compatibility between the interior solution ub and the half-plane representations Uj(ϕj) in Ωb\Ωa. We match Dirichlet traces on the small square ∂Ωa and Neumann traces

  • n the large square ∂Ωb (OK since ε > 0).

More precisely, we impose: ub = ϕ0 on Σ0

a ∩ ∂Ωa

∂ub ∂n = ∂ ∂n(U0(ϕ0)) on ∂Ωb ∩ Ω0

a

This leads to the following integral equation: ∂ub ∂n = −1 √ 2π

  • R
  • ξ2 − ω2

ε ˆ

ϕ0(ξ)e−√

ξ2−ω2

ε(b−a)eiyξdξ := Λ(ϕ0) for |y| < a A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 14 / 41

slide-52
SLIDE 52

Compatibility between the half-planes and the square

We match Dirichlet traces on the small square ∂Ωa and Neumann traces

  • n the large square ∂Ωb (OK since ε > 0).

More precisely, we impose: ub = ϕ0 on Σ0

a ∩ ∂Ωa

∂ub ∂n = ∂ ∂n(U0(ϕ0)) on ∂Ωb ∩ Ω0

a

And we do the same for all half-planes: ub = ϕj on Σj

a ∩ ∂Ωa for j = 0, 1, 2, 3

∂ub ∂n = Λ(ϕj) on ∂Ωb ∩ Ωj

a for j = 0, 1, 2, 3

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 14 / 41

slide-53
SLIDE 53

The half-space matching formulation

The initial problem

Find u ∈ H1(R2) such that ∆u + ω2

εu = f in R2.

The new unknowns

The restriction of u to Ωb: ub The traces of u on Σj

a: ϕj

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 15 / 41

slide-54
SLIDE 54

The half-space matching formulation

The initial problem

Find u ∈ H1(R2) such that ∆u + ω2

εu = f in R2.

The new unknowns

The restriction of u to Ωb: ub The traces of u on Σj

a: ϕj

The new formulation

             ∆ub + ω2

εub = f

in Ωb ub = ϕj

  • n

Σj

a ∩ ∂Ωa for j = 0, 1, 2, 3

∂ub ∂n = Λ(ϕj)

  • n

∂Ωb ∩ Ωj

a for j = 0, 1, 2, 3

ϕj = D(ϕj±1)

  • n

Ωj±1

a

∩ Σj

a

j ∈ Z/4Z

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 15 / 41

slide-55
SLIDE 55

Several generalizations

The power of this approach is that it applies to

1 Anisotropic scalar equations like:

div (A∇u) + ω2

εu = f

where A is a positive definite matrix. With 4 different integral operators Λj and 8 operators Dj !

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 16 / 41

slide-56
SLIDE 56

Several generalizations

The power of this approach is that it applies to

1 Anisotropic scalar equations like:

div (A∇u) + ω2

εu = 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

slide-57
SLIDE 57

Several generalizations

The power of this approach is that it applies to

1 Anisotropic scalar equations like:

div (A∇u) + ω2

εu = 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

slide-58
SLIDE 58

Several generalizations

The power of this approach is that it applies to

1 Anisotropic scalar equations like:

div (A∇u) + ω2

εu = 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

slide-59
SLIDE 59

Outline

1

Motivations and objectives

2

The new formulation

3

Mathematical analysis

4

Discretization and numerical results

5

Conclusion, future works and open questions

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 17 / 41

slide-60
SLIDE 60

Variational formulation

             ∆ub + ω2

εub = f

in Ωb ub = ϕj

  • n

Σj

a ∩ ∂Ωa

∂ub ∂n = Λ(ϕj)

  • n

∂Ωb ∩ Ωj

a

ϕj = D(ϕj±1)

  • n

Ωj±1

a

∩ Σj

a

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 18 / 41

slide-61
SLIDE 61

Variational formulation

             ∆ub + ω2

εub = f

in Ωb ub = ϕj

  • n

Σj

a ∩ ∂Ωa

∂ub ∂n = Λ(ϕj)

  • n

∂Ωb ∩ Ωj

a

ϕj = D(ϕj±1)

  • n

Ωj±1

a

∩ Σj

a

The appropriate functional space is: V = {(ub, ϕj); ub ∈ H1(Ωb), ϕj ∈ L2(Σj

a), ub = ϕj on Σj a ∩ ∂Ωa}

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 18 / 41

slide-62
SLIDE 62

Variational formulation

             ∆ub + ω2

εub = f

in Ωb ub = ϕj

  • n

Σj

a ∩ ∂Ωa

∂ub ∂n = Λ(ϕj)

  • n

∂Ωb ∩ Ωj

a

ϕj = D(ϕj±1)

  • n

Ωj±1

a

∩ Σj

a

The appropriate functional space is: V = {(ub, ϕj); ub ∈ H1(Ωb), ϕj ∈ L2(Σj

a), ub = ϕj on Σj a ∩ ∂Ωa}

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 18 / 41

slide-63
SLIDE 63

Variational formulation

             ∆ub + ω2

εub = f

in Ωb ub = ϕj

  • n

Σj

a ∩ ∂Ωa

∂ub ∂n = Λ(ϕj)

  • n

∂Ωb ∩ Ωj

a

ϕj = D(ϕj±1)

  • n

Ωj±1

a

∩ Σj

a

The appropriate functional space is: V = {(ub, ϕj); ub ∈ H1(Ωb), ϕj ∈ L2(Σj

a), ub = ϕj on Σj a ∩ ∂Ωa}

Then the weak form of the half-space matching formulation is:         

  • Ωb

∇ub · ∇vb − ω2

εubvb −

  • j

Λ(ϕj), vb =

  • Ωb

fvb ∀(vb, ψj) ∈ V

  • Σj

a

ϕjψj −

  • ±

D(ϕj±1), ψj −

  • Σj

a∩∂Ωa

ubψj = 0 ∀(vb, ψj) ∈ V

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 18 / 41

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SLIDE 64

Fredholm property

V = {(ub, ϕj); ub ∈ H1(Ωb), ϕj ∈ L2(Σj

a), ub = ϕj on Σj a ∩ ∂Ωa}

(ub, ϕj) ∈ V such that         

  • Ωb

∇ub · ∇vb − ω2

εubvb −

  • j

Λ(ϕj), vb =

  • Ωb

fvb

  • Σj

a

ϕjψj −

  • ±

D(ϕj±1), ψj −

  • Σj

a∩∂Ωa

ubψj = 0 ∀(vb, ψj) ∈ V GOOD NEWS : Λ is a compact operator on L2(Σj

a)

BAD NEWS : D is not a compact operator on L2(Σj

a)...

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 19 / 41

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SLIDE 65

Compactness of Λ

Remember that Λ(ϕ0) = ∂ ∂x U0(ϕ0) on x = b, |y| < a. Λ(ϕ0)(y) =

  • R

KΛ(ξ, y) ˆ ϕ0(ξ)dξ KΛ(ξ, y) = −1 √ 2π

  • ξ2 − ω2

εe−√ ξ2−ω2

ε(b−a)eiyξ A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 20 / 41

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SLIDE 66

Compactness of Λ

Remember that Λ(ϕ0) = ∂ ∂x U0(ϕ0) on x = b, |y| < a. Λ(ϕ0)(y) =

  • R

KΛ(ξ, y) ˆ ϕ0(ξ)dξ KΛ(ξ, y) = −1 √ 2π

  • ξ2 − ω2

εe−√ ξ2−ω2

ε(b−a)eiyξ

Direct proof of compactness

For b > a and ε > 0, KΛ ∈ L2(R×]a, a[) ⇒ Λ is a Hilbert-Schmidt operator acting on ˆ ϕ0 ⇒ Λ is compact

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 20 / 41

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SLIDE 67

Compactness of Λ

Remember that Λ(ϕ0) = ∂ ∂x U0(ϕ0) on x = b, |y| < a. Λ(ϕ0)(y) =

  • R

KΛ(ξ, y) ˆ ϕ0(ξ)dξ KΛ(ξ, y) = −1 √ 2π

  • ξ2 − ω2

εe−√ ξ2−ω2

ε(b−a)eiyξ

Direct proof of compactness

For b > a and ε > 0, KΛ ∈ L2(R×]a, a[) ⇒ Λ is a Hilbert-Schmidt operator acting on ˆ ϕ0 ⇒ Λ is compact

Alternative proof of compactness

Interior regularity of U0(ϕ0) and compact embedding ⇒ Λ is compact

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 20 / 41

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SLIDE 68

Fredholm property (2)

V = {(ub, ϕj); ub ∈ H1(Ωb), ϕj ∈ L2(Σj

a), ub = ϕj on Σj a ∩ ∂Ωa}

(ub, ϕj) ∈ V such that                 

  • Ωb

∇ub · ∇vb − ω2

εubvb

  • coercive

−

  • j

Λ(ϕj), vb

  • compact

=

  • Ωb

fvb ∀(vb, ψj) ∈ V

  • Σj

a

ϕjψj

  • coercive

−

  • ±

D(ϕj±1), ψj

  • ??

−

  • Σj

a∩∂Ωa

ubψj

  • compact

= 0 ∀(vb, ψj) ∈ V

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 21 / 41

slide-69
SLIDE 69

Fredholm property (2)

V = {(ub, ϕj); ub ∈ H1(Ωb), ϕj ∈ L2(Σj

a), ub = ϕj on Σj a ∩ ∂Ωa}

(ub, ϕj) ∈ V such that                 

  • Ωb

∇ub · ∇vb − ω2

εubvb

  • coercive

−

  • j

Λ(ϕj), vb

  • compact

=

  • Ωb

fvb ∀(vb, ψj) ∈ V

  • Σj

a

ϕjψj

  • coercive

−

  • ±

D(ϕj±1), ψj

  • ??

−

  • Σj

a∩∂Ωa

ubψj

  • compact

= 0 ∀(vb, ψj) ∈ V Let us consider for instance the term D(ϕ0), ψ1

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 21 / 41

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SLIDE 70

Properties of operator D

Remember that D(ϕ0) = U0(ϕ0) on x > a, y = a. D(ϕ0)(x) =

  • R

KD(ξ, x) ˆ ϕ0(ξ)dξ KD(ξ, x) = −1 √ 2π e−√

ξ2−ω2

ε(x−a)eiaξ

Lack of compactness (due to the cross-point)

Even for b > a and ε > 0, KD / ∈ L2(R×]a, a[) and D is not compact.

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 22 / 41

slide-71
SLIDE 71

Properties of operator D

Remember that D(ϕ0) = U0(ϕ0) on x > a, y = a. D(ϕ0)(x) =

  • R

KD(ξ, x) ˆ ϕ0(ξ)dξ KD(ξ, x) = −1 √ 2π e−√

ξ2−ω2

ε(x−a)eiaξ

Lack of compactness (due to the cross-point)

Even for b > a and ε > 0, KD / ∈ L2(R×]a, a[) and D is not compact.

Partial result of compactness

By the interior regularity of U0(ϕ0), ϕ0

− → D(ϕ0 −) is compact

By definition of V , ϕ0

a → D(ϕ0 a) is compact

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 22 / 41

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SLIDE 72

Properties of operator D (2)

What about ϕ0

+ → D(ϕ0 +)?

D(ϕ0)(x) =

  • R

KD(ξ, x) ˆ ϕ0(ξ)dξ KD(ξ, x) = −1 √ 2π e−√

ξ2−ω2

ε(x−a)eiaξ A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 23 / 41

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SLIDE 73

Properties of operator D (2)

What about ϕ0

+ → D(ϕ0 +)?

D(ϕ0)(x) =

  • R

KD(ξ, x) ˆ ϕ0(ξ)dξ KD(ξ, x) = −1 √ 2π e−√

ξ2−ω2

ε(x−a)eiaξ

It reminds the Laplace operator L :    L2(R+) → L2(R+) f →

  • R+ e−ξxf (ξ)dξ

L = √π

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 23 / 41

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SLIDE 74

Properties of operator D (2)

What about ϕ0

+ → D(ϕ0 +)?

D(ϕ0)(x) =

  • R

KD(ξ, x) ˆ ϕ0(ξ)dξ KD(ξ, x) = −1 √ 2π e−√

ξ2−ω2

ε(x−a)eiaξ

From the Laplace operator, we deduce that ˜ D :      L2([a, +∞)) → L2([a, +∞)) ϕ →

  • R

e−√

ξ2+1(x−a) −eiaξ

√ 2π ˆ ϕ(ξ)dξ ˜ D ≤

1 √ 2

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 23 / 41

slide-75
SLIDE 75

Properties of operator D (2)

What about ϕ0

+ → D(ϕ0 +)?

D(ϕ0)(x) =

  • R

KD(ξ, x) ˆ ϕ0(ξ)dξ KD(ξ, x) = −1 √ 2π e−√

ξ2−ω2

ε(x−a)eiaξ

D(ϕ)(y) = ˜ D(ϕ)(y) +

  • R
  • e−√

ξ2+1(x−a) − e−√ ξ2−ω2

ε(x−a) eiaξ

√ 2π ˆ ϕ(ξ)dξ

  • Hilbert-Schmidt

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 23 / 41

slide-76
SLIDE 76

Properties of operator D (2)

What about ϕ0

+ → D(ϕ0 +)?

D(ϕ0)(x) =

  • R

KD(ξ, x) ˆ ϕ0(ξ)dξ KD(ξ, x) = −1 √ 2π e−√

ξ2−ω2

ε(x−a)eiaξ

D(ϕ)(y) = ˜ D(ϕ)(y) +

  • R
  • e−√

ξ2+1(x−a) − e−√ ξ2−ω2

ε(x−a) eiaξ

√ 2π ˆ ϕ(ξ)dξ

  • Hilbert-Schmidt

Final result

The operator ϕ0

+ → D(ϕ0 +) is a sum ˜

D + (D − ˜ D) with ˜ D < 1 and D − ˜ D compact.

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 23 / 41

slide-77
SLIDE 77

Fredholm property (3)

V = {(ub, ϕj); ub ∈ H1(Ωb), ϕj ∈ L2(Σj

a), ub = ϕj on Σj a ∩ ∂Ωa}

(ub, ϕj) ∈ V such that                 

  • Ωb

∇ub · ∇vb − ω2

εubvb

  • coercive

−

  • j

Λ(ϕj), vb

  • compact

=

  • Ωb

fvb ∀(vb, ψj) ∈ V

  • Σj

a

ϕjψj

  • coercive

−

  • ±

D(ϕj±1), ψj

  • ??

−

  • Σj

a∩∂Ωa

ubψj

  • compact

= 0 ∀(vb, ψj) ∈ V

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 24 / 41

slide-78
SLIDE 78

Fredholm property (3)

V = {(ub, ϕj); ub ∈ H1(Ωb), ϕj ∈ L2(Σj

a), ub = ϕj on Σj a ∩ ∂Ωa}

(ub, ϕj) ∈ V such that                 

  • Ωb

∇ub · ∇vb − ω2

εubvb

  • coercive

−

  • j

Λ(ϕj), vb

  • compact

=

  • Ωb

fvb ∀(vb, ψj) ∈ V ( ϕ, ψ) − (˜ D ϕ, ψ)

  • coercive

+ (K ϕ, ψ) −

  • Σj

a∩∂Ωa

ubψj

  • compact

= 0 ∀(vb, ψj) ∈ V

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 24 / 41

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SLIDE 79

Fredholm property (3)

V = {(ub, ϕj); ub ∈ H1(Ωb), ϕj ∈ L2(Σj

a), ub = ϕj on Σj a ∩ ∂Ωa}

(ub, ϕj) ∈ V such that                 

  • Ωb

∇ub · ∇vb − ω2

εubvb

  • coercive

−

  • j

Λ(ϕj), vb

  • compact

=

  • Ωb

fvb ∀(vb, ψj) ∈ V ( ϕ, ψ) − (˜ D ϕ, ψ)

  • coercive

+ (K ϕ, ψ) −

  • Σj

a∩∂Ωa

ubψj

  • compact

= 0 ∀(vb, ψj) ∈ V

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

slide-80
SLIDE 80

Does this result hold for the generalizations ?

1 Anisotropic scalar equations like div (A∇u) + ω2

εu = f

Open question: we did not succeed in proving ˜ D < 1 for some ˜ D.

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 25 / 41

slide-81
SLIDE 81

Does this result hold for the generalizations ?

1 Anisotropic scalar equations like div (A∇u) + ω2

εu = f

Open question: we did not succeed in proving ˜ D < 1 for some ˜ 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

slide-82
SLIDE 82

Does this result hold for the generalizations ?

1 Anisotropic scalar equations like div (A∇u) + ω2

εu = f

Open question: we did not succeed in proving ˜ D < 1 for some ˜ 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

slide-83
SLIDE 83

Does this result hold for the generalizations ?

1 Anisotropic scalar equations like div (A∇u) + ω2

εu = f

Open question: we did not succeed in proving ˜ D < 1 for some ˜ 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

slide-84
SLIDE 84

Does this result hold for the generalizations ?

1 Anisotropic scalar equations like div (A∇u) + ω2

εu = f

Open question: we did not succeed in proving ˜ D < 1 for some ˜ 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

slide-85
SLIDE 85

An alternative formulation

We impose: ub = ϕj on the blue lines ϕj = D(ϕj±1) on the red lines First formulation: Equivalence: YES Compactness of D: NO but Fredholmness OK Alternative formulation: Equivalence: ?? Compactness of D: YES and Fredholmness OK

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 26 / 41

slide-86
SLIDE 86

Outline

1

Motivations and objectives

2

The new formulation

3

Mathematical analysis

4

Discretization and numerical results

5

Conclusion, future works and open questions

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 27 / 41

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SLIDE 87

Discrete problem

For the discretization:

We truncate the lines Σj

a (parameter T)

We use continuous Lagrange Finite Elements, 1D for ϕj and 2D for ub New unknown (uh

b, ϕj h)

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 28 / 41

slide-88
SLIDE 88

Discrete problem

For the discretization:

We truncate the lines Σj

a (parameter T)

We use continuous Lagrange Finite Elements, 1D for ϕj and 2D for ub New unknown (uh

b, ϕj 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

slide-89
SLIDE 89

Discrete problem

For the discretization:

We truncate the lines Σj

a (parameter T)

We use continuous Lagrange Finite Elements, 1D for ϕj and 2D for ub New unknown (uh

b, ϕj h)

For the computation of integral terms:

We truncate the Fourier integrals We use quadrature formulae

Example:

Exact formula

Λ(ϕ0

h), uh b =

  • R

k(ξ) ˆ ϕ0

h(ξ)ˆ

uh

b(b, ξ)dξ

where k(ξ) = −

  • ξ2 − ω2

εe−√ ξ2−ω2

ε(b−a), ˆ

ϕ0

h(ξ) and

ˆ uh

b(b, ξ) =

1 √ 2π b

−b

uh

b(b, y)e−iξydy are computed analytically.

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 28 / 41

slide-90
SLIDE 90

Discrete problem

For the discretization:

We truncate the lines Σj

a (parameter T)

We use continuous Lagrange Finite Elements, 1D for ϕj and 2D for ub New unknown (uh

b, ϕj h)

For the computation of integral terms:

We truncate the Fourier integrals We use quadrature formulae

Example:

Exact formula and approximate formula

Λ(ϕ0

h), uh b =

  • R

k(ξ) ˆ ϕ0

h(ξ)ˆ

uh

b(b, ξ)dξ ≈

  • ξn∈Q

qn k(ξn) ˆ ϕ0

h(ξn)ˆ

uh

b(b, ξn)

where k(ξ) = −

  • ξ2 − ω2

εe−√ ξ2−ω2

ε(b−a), ˆ

ϕ0

h(ξ) and

ˆ uh

b(b, ξ) =

1 √ 2π b

−b

uh

b(b, y)e−iξydy are computed analytically.

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 28 / 41

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SLIDE 91

Discrete problem

For the discretization:

We truncate the lines Σj

a (parameter T)

We use continuous Lagrange Finite Elements, 1D for ϕj and 2D for ub New unknown (uh

b, ϕj 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

slide-92
SLIDE 92

Numerical validation

The problem

Find u ∈ H1(R2) such that ∆u + ω2

εu = f in R2

where f ≈ δ

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 29 / 41

slide-93
SLIDE 93

Numerical validation

The problem

Find u ∈ H1(R2) such that ∆u + ω2

εu = f in R2

where f ≈ δ The dissipative case: ωε = 10 + 0.5i and T = 24a errrel < 0.06%

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 29 / 41

slide-94
SLIDE 94

Numerical validation

The problem

Find u ∈ H1(R2) such that ∆u + ω2

εu = f in R2

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

slide-95
SLIDE 95

Numerical validation

The problem

Find u ∈ H1(R2) such that ∆u + ω2

εu = f in R2

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

slide-96
SLIDE 96

Numerical validation (2)

The problem

Find u ∈ H1(R2) such that ∆u + ω2u = f in R2 where f ≈ δ

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 30 / 41

slide-97
SLIDE 97

Numerical validation (2)

The problem

Find u ∈ H1(R2) such that ∆u + ω2u = f in R2 where f ≈ δ The non-dissipative case: ω = 10 and T = 24a

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 30 / 41

slide-98
SLIDE 98

Numerical validation (2)

The problem

Find u ∈ H1(R2) such that ∆u + ω2u = f in R2 where f ≈ δ The price to pay: refine the dis- cretization in ξ

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 30 / 41

slide-99
SLIDE 99

Numerical validation (2)

The problem

Find u ∈ H1(R2) such that ∆u + ω2u = f in R2 where f ≈ δ A bad discretization in ξ is easily de- tectable a posteriori...

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 30 / 41

slide-100
SLIDE 100

Application to an anisotropic equation

The problem

  • div (A∇u) + ω2

εu = f

in R2\O A∇u · n = 0

  • n

O where A =

  • 1

0.5 0.5 1

  • A.-S. Bonnet-Ben Dhia (POEMS)

RICAM 2016 31 / 41

slide-101
SLIDE 101

Application to an anisotropic equation

The problem

  • div (A∇u) + ω2

εu = f

in R2\O A∇u · n = 0

  • n

O where A =

  • 1

0.5 0.5 1

  • ωε = 20 + 0.01i and T = 12a

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 31 / 41

slide-102
SLIDE 102

Application to the diffraction problem

Diffraction of a plane wave by a triangular obstacle: Here, the problem as been solved

with 3 half-planes instead of 4 without interior unknown

and with the ϕj chosen as the traces of the scattered field Here, the problem as been solved with the ϕj chosen as the traces of the total field

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 32 / 41

slide-103
SLIDE 103

Application to the diffraction problem

Diffraction of a plane wave by a triangular obstacle: Here, the problem as been solved

with 3 half-planes instead of 4 without interior unknown

and with the ϕj chosen as the traces of the scattered field Here, the problem as been solved with the ϕj chosen as the traces of the total field Surprisingly, the plane wave behavior does not alter the convergence !

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 32 / 41

slide-104
SLIDE 104

Application to the open waveguides problem (J. Ott)

As mentioned before, the Fourier transform has to be replaced by the generalized Fourier transform adapted to each stratified medium as in

Bonnet-Ben Dhia, Goursaud and Hazard, Mathematical analysis of the junction of two acoustic open waveguides, SIAM J. Appl. Math., 2011

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SLIDE 105

Application to the open waveguides problem (J. Ott)

The Half Space Matching method has been used in an optimization loop: Initial junction: almost no transmission Optimized junction: good transmission

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 33 / 41

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SLIDE 106

Application to the open waveguides problem (J. Ott)

The Half Space Matching method has been used to modelize a Y junction: again, 3 half-spaces are used instead of 4

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 33 / 41

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SLIDE 107

Application to the open waveguides problem (J. Ott)

The Half Space Matching method has been used to modelize a Y junction: Note that the angle between the branches has to be greater than π/2!

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 33 / 41

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SLIDE 108

Application to 2D elastic media (isotropic case)

The unknowns ub and the ϕj are vector fields (2 components).

The problem

  • div (σ(u) + ω2ρu = f

in R2\O σ(u) · n = 0

  • n

O where σij(u) = Cijkl ∂uk ∂xl

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 34 / 41

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SLIDE 109

Application to 2D elastic media (isotropic case)

The unknowns ub and the ϕj are vector fields (2 components).

The problem

  • div (σ(u) + ω2ρu = f

in R2\O σ(u) · n = 0

  • n

O where σij(u) = Cijkl ∂uk ∂xl P-wave S-wave S scattered wave

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 34 / 41

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SLIDE 110

Application to 2D elastic media (anisotropic case)

The unknowns ub and the ϕj are vector fields (2 components).

The problem

  • div (σ(u) + ω2ρu = f

in R2\O σ(u) · n = 0

  • n

O where σij(u) = Cijkl ∂uk ∂xl Slowness diagram

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 35 / 41

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SLIDE 111

Application to 2D elastic media (anisotropic case)

The unknowns ub and the ϕj are vector fields (2 components).

The problem

  • div (σ(u) + ω2ρu = f

in R2\O σ(u) · n = 0

  • n

O where σij(u) = Cijkl ∂uk ∂xl Slowness diagram QP-wave QS-wave QS scattered wave

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SLIDE 112

Comparison with PMLs

Isotropic case: PMLs Our method Anisotropic case: PMLs Our method

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SLIDE 113

Outline

1

Motivations and objectives

2

The new formulation

3

Mathematical analysis

4

Discretization and numerical results

5

Conclusion, future works and open questions

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SLIDE 114

Conclusion for 2D problems

Numerically, everything (we tried) works!

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SLIDE 115

Conclusion for 2D problems

Numerically, everything (we tried) works! But the theory is restricted to dissipative and weakly anisotropic media...

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SLIDE 116

Conclusion for 2D problems

Numerically, everything (we tried) works! But the theory is restricted to dissipative and weakly anisotropic media...

Main (difficult) objective

extend the theory to the non-dissipative case We know that in general ϕj / ∈ L2(Σj

a) but

ϕj ∈

  • ϕ ;
  • R

| ˆ ϕ(ξ|2

  • ξ2 − ω2
  • dξ < ∞
  • Bonnet-BenDhia and Tillequin, A limiting absorption principle for scattering

problems with unbounded obstacles, M2AS, 2001.

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 38 / 41

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SLIDE 117

Conclusion for 2D problems

Numerically, everything (we tried) works! But the theory is restricted to dissipative and weakly anisotropic media...

Main (difficult) objective

extend the theory to the non-dissipative case We know that in general ϕj / ∈ L2(Σj

a) but

ϕj ∈

  • ϕ ;
  • R

| ˆ ϕ(ξ|2

  • ξ2 − ω2
  • dξ < ∞
  • 1) Can we prove Fredholmness in this new framework ?

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 38 / 41

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SLIDE 118

Conclusion for 2D problems

Numerically, everything (we tried) works! But the theory is restricted to dissipative and weakly anisotropic media...

Main (difficult) objective

extend the theory to the non-dissipative case 1) Can we prove Fredholmness in this new framework ? 2) And I remind the question of equivalence: Is U0(ϕ0) = U1(ϕ1) in Ω0

a ∩ Ω1 a

equivalent to U0(ϕ0) = U1(ϕ1) in ∂(Ω0

a ∩ Ω1 a)?

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SLIDE 119

Ongoing work

A second main objective

Extend the method to 3D anisotropic elastic plates

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SLIDE 120

Ongoing work

A second main objective

Extend the method to 3D anisotropic elastic plates Lack of orthogonality of Lamb modes ⇒ No analytic representations in half- plates for Dirichlet data. Instead, we consider as unknowns both Cauchy data u and σ(u) · n on the infinite strips Σj

a.

This raises another open question concerning equivalence.

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SLIDE 121

Future directions

Main (difficult) objective

extend the theory to the non-dissipative case

A second main objective

Extend the method to 3D anisotropic elastic plates

And also...

Handle 3D Maxwell equations (ansitropic, open waveguides etc..) Introduce periocidity (junction of photonic waveguides) Justify the method without overlapping (a = b) ....

A.-S. Bonnet-Ben Dhia (POEMS) RICAM 2016 40 / 41

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SLIDE 122

Danke sch¨

  • n!

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