A factorization method for support characterization of an obstacle - PowerPoint PPT Presentation

A factorization method for support characterization of an obstacle with a generalized impedance boundary condition Mathieu Chamaillard , Nicolas Chaulet and Houssem Haddar POEMS (Propagation d’Ondes : Etude Mathématique et Simulation) CNRS / ENSTA / INRIA, Palaiseau October 25, 2013

Outline 1 The GIBC forward problem 2 The inverse GIBC problem 3 Numerical examples

The Generalized Impedance Boundary Conditions in acoustic scattering Context: • Imperfectly conducting obstacles • Periodic coatings (homogenized model) • Thin layers • Thin periodic coatings • ... Inverse problem: recover D from the scattered field.

General notions in inverse scattering Γ ∆ us + k 2 us = 0  D   ∂us � ∂ui �  + Z ui   + Z u = − on Γ  ∂ν ∂ν ∂us 2 � �  � − ikus  � �  lim ds = 0  � �   R →∞ � � | x | = R ∂r � � � 1 eikr u s ( x ) = u ∞ (ˆ x � x ) + O �� r − → + ∞ where ˆ x := | x | . r ( d − 1) / 2 r

General notions in inverse scattering Γ ∆ us + k 2 us = 0  D   ∂us � ∂ui �  + Z ui   + Z u = − on Γ  ∂ν ∂ν ∂us 2 � �  � − ikus  � �  lim ds = 0  � �   R →∞ � � | x | = R ∂r � � � 1 eikr u s ( x ) = u ∞ (ˆ x � x ) + O �� r − → + ∞ where ˆ x := | x | . r ( d − 1) / 2 r θ ) = e ik ˆ θ · z we define For incident plane waves u i ( z, ˆ u ∞ (ˆ x, ˆ θ ) ∈ L 2 ( S d , S d ) ,

General notions in inverse scattering Γ ∆ us + k 2 us = 0  D   ∂us � ∂ui �  + Z ui   + Z u = − on Γ  ∂ν ∂ν ∂us 2 � �  � − ikus  � �  lim ds = 0  � �   R →∞ � � | x | = R ∂r � � � 1 eikr u s ( x ) = u ∞ (ˆ x � x ) + O �� r − → + ∞ where ˆ x := | x | . r ( d − 1) / 2 r θ ) = e ik ˆ θ · z we define For incident plane waves u i ( z, ˆ u ∞ (ˆ x, ˆ θ ) ∈ L 2 ( S d , S d ) , Under minimal assumptions on Z design a method to recover D from u ∞ for all (ˆ x, ˆ θ ) .

The factorization method: a sampling method θ ) = e ik ˆ θ · x define For u i ( x, ˆ → u ∞ (ˆ x, ˆ ( Z , D ) − θ ) where u ∞ associated with u s ( Z , D ) is defined in dimension d by � 1 e ikr � �� u s ( x ) = u ∞ (ˆ x ) + O r − → + ∞ . r ( d − 1) / 2 r L 2 ( S d ) − → L 2 ( S d ) F : � Sd u ∞ (ˆ x, ˆ θ ) g (ˆ θ ) d ˆ g �− → θ Define the self-adjoint positive operator F # := |ℜ e ( F ) | + ℑ m ( F ) θ · z ∈ R ( F 1 / 2 ⇒ e − ik ˆ z ∈ D ⇐ ) #

The factorization method: a sampling method θ ) = e ik ˆ θ · x define For u i ( x, ˆ → u ∞ (ˆ x, ˆ ( Z , D ) − θ ) where u ∞ associated with u s ( Z , D ) is defined in dimension d by � 1 e ikr � �� u s ( x ) = u ∞ (ˆ x ) + O r − → + ∞ . r ( d − 1) / 2 r L 2 ( S d ) − → L 2 ( S d ) F : D � Sd u ∞ (ˆ x, ˆ θ ) g (ˆ θ ) d ˆ g �− → θ z Define the self-adjoint positive operator F # := |ℜ e ( F ) | + ℑ m ( F ) θ · z ∈ R ( F 1 / 2 ⇒ e − ik ˆ z ∈ D ⇐ ) # ∃ g s. t. F 1 / 2 g = e − ik ˆ θ · z #

The factorization method: a sampling method θ ) = e ik ˆ θ · x define For u i ( x, ˆ → u ∞ (ˆ x, ˆ ( Z , D ) − θ ) where u ∞ associated with u s ( Z , D ) is defined in dimension d by � 1 e ikr � �� u s ( x ) = u ∞ (ˆ x ) + O r − → + ∞ . r ( d − 1) / 2 r L 2 ( S d ) − → L 2 ( S d ) F : D � Sd u ∞ (ˆ x, ˆ θ ) g (ˆ θ ) d ˆ g �− → θ Define the self-adjoint positive operator F # := |ℜ e ( F ) | + ℑ m ( F ) z θ · z ∈ R ( F 1 / 2 ⇒ e − ik ˆ z ∈ D ⇐ ) # No solution!

State of the art • Factorization method for impenetrable scatterers: • Dirichlet and Neumann boundary condition: Kirsch 1998, • Impedance boundary condition ( Z = λ ): Kirsch & Grinberg 2002, • Inverse iterative methods with GIBC: Bourgeois, Chaulet & Haddar 2011–2012.

Outline 1 The GIBC forward problem 2 The inverse GIBC problem 3 Numerical examples

The GIBC forward problem A volume formulation • V an Hilbert space such that C ∞ (Γ) ⊂ V ⊂ H 1 / 2 (Γ) → V ∗ is linear and continuous and • Z : V − Z ∗ u = Z u For example for complex functions ( λ, µ ) ∈ ( L ∞ (Γ)) 2 Z = div Γ µ ∇ Γ + λ V = H 1 (Γ)

The GIBC forward problem A volume formulation • V an Hilbert space such that C ∞ (Γ) ⊂ V ⊂ H 1 / 2 (Γ) → V ∗ is linear and continuous and • Z : V − Z ∗ u = Z u • ℑ m � Z u, u � V ∗ ,V ≥ 0 for uniqueness reasons The GIBC problem writes: Find u s ∈ � � v ∈ D ′ (Ω ext ) , ϕv ∈ H 1 (Ω ext ) ∀ ϕ ∈ D ( R d ); v | Γ ∈ V ∆ u s + k 2 u s = 0 in Ω ext ,    ∂u s � f = − ∂u i �   ∂ν + Z u s = f on Γ , ∂ν − Z u i   ( P vol )  �  | ∂ r u s − iku s | 2 = 0 .  lim    R →∞ | x | = R

The GIBC forward problem A volume formulation • V an Hilbert space such that C ∞ (Γ) ⊂ V ⊂ H 1 / 2 (Γ) → V ∗ is linear and continuous and • Z : V − Z ∗ u = Z u • ℑ m � Z u, u � V ∗ ,V ≥ 0 for uniqueness reasons The GIBC problem writes: Find u s ∈ � � v ∈ D ′ (Ω ext ) , ϕv ∈ H 1 (Ω ext ) ∀ ϕ ∈ D ( R d ); v | Γ ∈ V ∆ u s + k 2 u s = 0 in Ω ext ,    ∂u s � f = − ∂u i �   ∂ν + Z u s = f on Γ , ∂ν − Z u i   ( P vol )  �  | ∂ r u s − iku s | 2 = 0 .  lim    R →∞ | x | = R Proof of uniqueness Assume f = 0 then on Γ :

The GIBC forward problem A volume formulation • V an Hilbert space such that C ∞ (Γ) ⊂ V ⊂ H 1 / 2 (Γ) → V ∗ is linear and continuous and • Z : V − Z ∗ u = Z u • ℑ m � Z u, u � V ∗ ,V ≥ 0 for uniqueness reasons The GIBC problem writes: Find u s ∈ � � v ∈ D ′ (Ω ext ) , ϕv ∈ H 1 (Ω ext ) ∀ ϕ ∈ D ( R d ); v | Γ ∈ V ∆ u s + k 2 u s = 0 in Ω ext ,    ∂u s � f = − ∂u i �   ∂ν + Z u s = f on Γ , ∂ν − Z u i   ( P vol )  �  | ∂ r u s − iku s | 2 = 0 .  lim    R →∞ | x | = R Proof of uniqueness Assume f = 0 then on Γ : ∂u s ∂ν + Z u s = 0 ,

The GIBC forward problem A volume formulation • V an Hilbert space such that C ∞ (Γ) ⊂ V ⊂ H 1 / 2 (Γ) → V ∗ is linear and continuous and • Z : V − Z ∗ u = Z u • ℑ m � Z u, u � V ∗ ,V ≥ 0 for uniqueness reasons The GIBC problem writes: Find u s ∈ � � v ∈ D ′ (Ω ext ) , ϕv ∈ H 1 (Ω ext ) ∀ ϕ ∈ D ( R d ); v | Γ ∈ V ∆ u s + k 2 u s = 0 in Ω ext ,    ∂u s � f = − ∂u i �   ∂ν + Z u s = f on Γ , ∂ν − Z u i   ( P vol )  �  | ∂ r u s − iku s | 2 = 0 .  lim    R →∞ | x | = R Proof of uniqueness Assume f = 0 then on Γ : ∂u s ∂ν + Z u s = 0 , ℑ m < ∂u s ∂ν , u s > + ℑ m < Z u s , u s > = 0 .

The GIBC forward problem A volume formulation • V an Hilbert space such that C ∞ (Γ) ⊂ V ⊂ H 1 / 2 (Γ) → V ∗ is linear and continuous and • Z : V − Z ∗ u = Z u • ℑ m � Z u, u � V ∗ ,V ≥ 0 for uniqueness reasons The GIBC problem writes: Find u s ∈ � � v ∈ D ′ (Ω ext ) , ϕv ∈ H 1 (Ω ext ) ∀ ϕ ∈ D ( R d ); v | Γ ∈ V ∆ u s + k 2 u s = 0 in Ω ext ,    ∂u s � f = − ∂u i �   ∂ν + Z u s = f on Γ , ∂ν − Z u i   ( P vol )  �  | ∂ r u s − iku s | 2 = 0 .  lim    R →∞ | x | = R Proof of uniqueness Assume f = 0 then on Γ : ∂u s ∂ν + Z u s = 0 , ℑ m < ∂u s ∂ν , u s > + ℑ m < Z u s , u s > = 0 . Or ℑ m < Z u s , u s > ≥ 0 ⇒ ℑ m < ∂u s ∂ν , u s > ≤ 0 and Rellich lemma ⇒ u s = 0 .

Variational formulation: Reduced problem Find u s ∈ { u s ∈ H 1 ( S R \ D ) , u s ∈ V } such that for all v ∈ { u s ∈ H 1 ( B R \ D ) , u s ∈ V } , a ( u s , v ) = < f, v > where � ∇ u ∇ v − k 2 uv d D + < Zu, v > V ∗ ,V + < Dtn S R u, v > a ( u, v ) := 2 ( S r ) , 1 H B R \ D Dtn S R exterior Dirichlet to Neumann operator on S R The sign of the real part of the impedance operator is imposed by the volume equation!

Well posedness of the forward problem A surface equivalent formulation Find u s ∈ � � v ∈ D ′ (Ω ext ) , ϕv ∈ H 1 (Ω ext ) ∀ ϕ ∈ D ( R d ); v | Γ ∈ V ∆ u s + k 2 u s = 0 in Ω ext ,    ∂u s � f = − ∂u i �   ∂ν + Z u s = f on Γ ,  ∂ν − Z u i  ( P vol )  �  | ∂ r u s − iku s | 2 = 0 .  lim    R →∞ | x | = R • n e : H 1 / 2 (Γ) − → H − 1 / 2 (Γ) the exterior DtN map → ∂u f f �− ∂ν where ∆ u f + k 2 u f = 0 in Ω ext ,     u f = f on Γ , � | ∂ r u f − iku s | 2 = 0 . lim    R →∞ | x | = R