Investigation of some transmission problems with sign changing - PowerPoint PPT Presentation

Outline of the talk 1 The coerciveness issue for the scalar case We develop a T -coercivity method based on geometrical transforma- tions to study div( µ − 1 ∇· ) : H 1 0 (Ω) → H − 1 (Ω) (improvement over Bonnet-Ben Dhia et al. 10 , Zwölf 08 ). 2 A new functional framework in the critical interval We propose a new functional framework when div( µ − 1 ∇· ) : X → Y is not Fredholm for X = H 1 0 (Ω) and Y = H − 1 (Ω) (extension of Dauge, Texier 97 , Ramdani 99 ). 3 Study of Maxwell’s equations We develop a T -coercivity method based on potentials to study curl ( ε − 1 curl · ) : V T ( µ ; Ω) → V T ( µ ; Ω) ∗ . 4 The T -coercivity method for the Interior Transmission Problem We study ∆( σ ∆ · ) : H 2 0 (Ω) → H − 2 (Ω) . 5 / 34

1 The coerciveness issue for the scalar case 2 A new functional framework in the critical interval 3 Study of Maxwell’s equations 4 The T -coercivity method for the Interior Transmission Problem 6 / 34

A scalar model problem Problem for E z in 2D in case of an invariance with respect to z : Find E z ∈ H 1 0 (Ω) such that: div( µ − 1 ∇ E z ) + ω 2 ε E z = − f in Ω . 7 / 34

A scalar model problem Problem for E z in 2D in case of an invariance with respect to z : Find E z ∈ H 1 0 (Ω) such that: div( µ − 1 ∇ E z ) + ω 2 ε E z = − f in Ω . H 1 0 (Ω) = { v ∈ L 2 (Ω) | ∇ v ∈ L 2 (Ω); v | ∂ Ω = 0 } f is the source term in H − 1 (Ω) 7 / 34

A scalar model problem Problem for E z in 2D in case of an invariance with respect to z : Find E z ∈ H 1 0 (Ω) such that: div( µ − 1 ∇ E z ) + ω 2 ε E z = − f in Ω . H 1 0 (Ω) = { v ∈ L 2 (Ω) | ∇ v ∈ L 2 (Ω); v | ∂ Ω = 0 } f is the source term in H − 1 (Ω) Since H 1 0 (Ω) ⊂⊂ L 2 (Ω), we focus on the principal part. Find u ∈ H 1 0 (Ω) s.t.: ( P ) div( µ − 1 ∇ u ) = − f in Ω . 7 / 34

A scalar model problem Problem for E z in 2D in case of an invariance with respect to z : Find E z ∈ H 1 0 (Ω) such that: Σ div( µ − 1 ∇ E z ) + ω 2 ε E z = − f in Ω . Ω 2 Ω 1 H 1 0 (Ω) = { v ∈ L 2 (Ω) | ∇ v ∈ L 2 (Ω); v | ∂ Ω = 0 } f is the source term in H − 1 (Ω) Since H 1 0 (Ω) ⊂⊂ L 2 (Ω), we focus on the principal part. Find u ∈ H 1 0 (Ω) s.t.: ( P ) div( µ − 1 ∇ u ) = − f in Ω . 7 / 34

A scalar model problem Problem for E z in 2D in case of an invariance with respect to z : Find E z ∈ H 1 0 (Ω) such that: Σ div( µ − 1 ∇ E z ) + ω 2 ε E z = − f in Ω . Ω 2 Ω 1 H 1 0 (Ω) = { v ∈ L 2 (Ω) | ∇ v ∈ L 2 (Ω); v | ∂ Ω = 0 } µ 1 = µ | Ω 1 > 0 f is the source term in H − 1 (Ω) µ 2 = µ | Ω 2 < 0 (constant) Since H 1 0 (Ω) ⊂⊂ L 2 (Ω), we focus on the principal part. Find u ∈ H 1 0 (Ω) s.t.: ( P ) div( µ − 1 ∇ u ) = − f in Ω . 7 / 34

A scalar model problem Problem for E z in 2D in case of an invariance with respect to z : Find E z ∈ H 1 0 (Ω) such that: Σ div( µ − 1 ∇ E z ) + ω 2 ε E z = − f in Ω . Ω 2 Ω 1 H 1 0 (Ω) = { v ∈ L 2 (Ω) | ∇ v ∈ L 2 (Ω); v | ∂ Ω = 0 } µ 1 = µ | Ω 1 > 0 f is the source term in H − 1 (Ω) µ 2 = µ | Ω 2 < 0 (constant) Since H 1 0 (Ω) ⊂⊂ L 2 (Ω), we focus on the principal part. Find u ∈ H 1 Find u ∈ H 1 0 (Ω) s.t.: 0 (Ω) s.t.: ( P ) ⇔ ( P V ) div( µ − 1 ∇ u ) = − f in Ω . a ( u , v ) = l ( v ) , ∀ v ∈ H 1 0 (Ω) . � µ − 1 ∇ u · ∇ v with a ( u , v ) = and l ( v ) = � f , v � Ω . Ω 7 / 34

A scalar model problem Problem for E z in 2D in case of an invariance with respect to z : Find E z ∈ H 1 0 (Ω) such that: Σ div( µ − 1 ∇ E z ) + ω 2 ε E z = − f in Ω . Ω 2 Ω 1 H 1 0 (Ω) = { v ∈ L 2 (Ω) | ∇ v ∈ L 2 (Ω); v | ∂ Ω = 0 } µ 1 = µ | Ω 1 > 0 f is the source term in H − 1 (Ω) µ 2 = µ | Ω 2 < 0 (constant) Since H 1 0 (Ω) ⊂⊂ L 2 (Ω), we focus on the principal part. Find u ∈ H 1 Find u ∈ H 1 0 (Ω) s.t.: 0 (Ω) s.t.: ( P ) ⇔ ( P V ) div( µ − 1 ∇ u ) = − f in Ω . a ( u , v ) = l ( v ) , ∀ v ∈ H 1 0 (Ω) . � µ − 1 ∇ u · ∇ v with a ( u , v ) = and l ( v ) = � f , v � Ω . Ω Definition. We will say that the problem ( P ) is well-posed if the operator A = div ( µ − 1 ∇· ) is an isomorphism from H 1 0 (Ω) to H − 1 (Ω). 7 / 34

Mathematical difficulty Classical case µ > 0 everywhere: � µ − 1 |∇ u | 2 ≥ min( µ − 1 ) � u � 2 a ( u , u ) = coercivity H 1 0 (Ω) Ω Lax-Milgram theorem ⇒ ( P ) well-posed. 8 / 34

Mathematical difficulty Classical case µ > 0 everywhere: � µ − 1 |∇ u | 2 ≥ min( µ − 1 ) � u � 2 a ( u , u ) = coercivity H 1 0 (Ω) Ω Lax-Milgram theorem ⇒ ( P ) well-posed. VS. The case µ changes sign: � µ − 1 |∇ u | 2 ≥ C � u � 2 loss of coercivity a ( u , u ) = H 1 0 (Ω) Ω 8 / 34

Mathematical difficulty Classical case µ > 0 everywhere: � µ − 1 |∇ u | 2 ≥ min( µ − 1 ) � u � 2 a ( u , u ) = coercivity H 1 0 (Ω) Ω Lax-Milgram theorem ⇒ ( P ) well-posed. VS. The case µ changes sign: � µ − 1 |∇ u | 2 ≥ C � u � 2 loss of coercivity a ( u , u ) = H 1 0 (Ω) Ω ◮ When µ 2 = − µ 1 , ( P ) is always ill-posed ( Costabel-Stephan 85 ). For a symmetric domain (w.r.t. Σ) we can build a kernel of infinite dimension. 8 / 34

Idea of the T -coercivity 1/2 Let T be an isomorphism of H 1 0 (Ω). Find u ∈ H 1 0 (Ω) such that: ( P ) ⇔ ( P V ) a ( u , v ) = l ( v ) , ∀ v ∈ H 1 0 (Ω) . 9 / 34

Idea of the T -coercivity 1/2 Let T be an isomorphism of H 1 0 (Ω). Find u ∈ H 1 0 (Ω) such that: ( P ) ⇔ ( P V ) ⇔ ( P T V ) a ( u , T v ) = l ( T v ) , ∀ v ∈ H 1 0 (Ω) . 9 / 34

Idea of the T -coercivity 1/2 Let T be an isomorphism of H 1 0 (Ω). Find u ∈ H 1 0 (Ω) such that: ( P ) ⇔ ( P V ) ⇔ ( P T V ) a ( u , T v ) = l ( T v ) , ∀ v ∈ H 1 0 (Ω) . � µ − 1 ∇ u · ∇ ( T u ) ≥ C � u � 2 Goal: Find T such that a is T -coercive: 0 (Ω) . H 1 Ω In this case, Lax-Milgram ⇒ ( P T V ) (and so ( P V )) is well-posed. 9 / 34

Idea of the T -coercivity 1/2 Let T be an isomorphism of H 1 0 (Ω). Find u ∈ H 1 0 (Ω) such that: ( P ) ⇔ ( P V ) ⇔ ( P T V ) a ( u , T v ) = l ( T v ) , ∀ v ∈ H 1 0 (Ω) . � µ − 1 ∇ u · ∇ ( T u ) ≥ C � u � 2 Goal: Find T such that a is T -coercive: 0 (Ω) . H 1 Ω In this case, Lax-Milgram ⇒ ( P T V ) (and so ( P V )) is well-posed. u 1 in Ω 1 1 Define T 1 u = − u 2 + ... in Ω 2 9 / 34

Idea of the T -coercivity 1/2 Let T be an isomorphism of H 1 0 (Ω). Find u ∈ H 1 0 (Ω) such that: ( P ) ⇔ ( P V ) ⇔ ( P T V ) a ( u , T v ) = l ( T v ) , ∀ v ∈ H 1 0 (Ω) . � µ − 1 ∇ u · ∇ ( T u ) ≥ C � u � 2 Goal: Find T such that a is T -coercive: 0 (Ω) . H 1 Ω In this case, Lax-Milgram ⇒ ( P T V ) (and so ( P V )) is well-posed. u 1 in Ω 1 1 Define T 1 u = in Ω 2 , with − u 2 + 2 R 1 u 1 R 1 transfer/extension operator R 1 Ω 1 Ω 2 Σ 9 / 34

Idea of the T -coercivity 1/2 Let T be an isomorphism of H 1 0 (Ω). Find u ∈ H 1 0 (Ω) such that: ( P ) ⇔ ( P V ) ⇔ ( P T V ) a ( u , T v ) = l ( T v ) , ∀ v ∈ H 1 0 (Ω) . � µ − 1 ∇ u · ∇ ( T u ) ≥ C � u � 2 Goal: Find T such that a is T -coercive: 0 (Ω) . H 1 Ω In this case, Lax-Milgram ⇒ ( P T V ) (and so ( P V )) is well-posed. u 1 in Ω 1 1 Define T 1 u = in Ω 2 , with − u 2 + 2 R 1 u 1 R 1 transfer/extension operator continuous from Ω 1 to Ω 2 R 1 R 1 u 1 = u 1 on Σ Ω 1 Ω 2 Σ R 1 u 1 = 0 on ∂ Ω 2 \ Σ 9 / 34

Idea of the T -coercivity 1/2 Let T be an isomorphism of H 1 0 (Ω). Find u ∈ H 1 0 (Ω) such that: ( P ) ⇔ ( P V ) ⇔ ( P T V ) a ( u , T v ) = l ( T v ) , ∀ v ∈ H 1 0 (Ω) . � µ − 1 ∇ u · ∇ ( T u ) ≥ C � u � 2 Goal: Find T such that a is T -coercive: 0 (Ω) . H 1 Ω In this case, Lax-Milgram ⇒ ( P T V ) (and so ( P V )) is well-posed. u 1 in Ω 1 1 Define T 1 u = in Ω 2 , with − u 2 + 2 R 1 u 1 R 1 transfer/extension operator continuous from Ω 1 to Ω 2 R 1 R 1 u 1 = u 1 on Σ Ω 1 Ω 2 Σ R 1 u 1 = 0 on ∂ Ω 2 \ Σ On Σ, we have − u 2 + 2 R 1 u 1 = − u 2 + 2 u 1 = u 1 ⇒ T 1 u ∈ H 1 0 (Ω). 9 / 34

Idea of the T -coercivity 1/2 Let T be an isomorphism of H 1 0 (Ω). Find u ∈ H 1 0 (Ω) such that: ( P ) ⇔ ( P V ) ⇔ ( P T V ) a ( u , T v ) = l ( T v ) , ∀ v ∈ H 1 0 (Ω) . � µ − 1 ∇ u · ∇ ( T u ) ≥ C � u � 2 Goal: Find T such that a is T -coercive: 0 (Ω) . H 1 Ω In this case, Lax-Milgram ⇒ ( P T V ) (and so ( P V )) is well-posed. u 1 in Ω 1 1 Define T 1 u = in Ω 2 , with − u 2 + 2 R 1 u 1 R 1 transfer/extension operator continuous from Ω 1 to Ω 2 R 1 R 1 u 1 = u 1 on Σ Ω 1 Ω 2 Σ R 1 u 1 = 0 on ∂ Ω 2 \ Σ T 1 is an isomorphism of H 1 2 T 1 ◦ T 1 = Id so 0 (Ω) 9 / 34

Idea of the T -coercivity 2/2 � � | µ | − 1 |∇ u | 2 − 2 µ − 1 3 One has a ( u , T 1 u ) = ∇ u · ∇ ( R 1 u 1 ) 2 Ω Ω 2 10 / 34

Idea of the T -coercivity 2/2 � � | µ | − 1 |∇ u | 2 − 2 µ − 1 3 One has a ( u , T 1 u ) = ∇ u · ∇ ( R 1 u 1 ) 2 Ω Ω 2 a is T -coercive when | µ 2 | > � R 1 � 2 µ 1 . Young’s inequality ⇒ 10 / 34

Idea of the T -coercivity 2/2 � � | µ | − 1 |∇ u | 2 − 2 µ − 1 3 One has a ( u , T 1 u ) = ∇ u · ∇ ( R 1 u 1 ) 2 Ω Ω 2 a is T -coercive when | µ 2 | > � R 1 � 2 µ 1 . Young’s inequality ⇒ u 1 − 2 R 2 u 2 in Ω 1 4 Working with T 2 u = in Ω 2 , where R 2 : Ω 2 → Ω 1 , one − u 2 a is T -coercive when µ 1 > � R 2 � 2 | µ 2 | . proves that 10 / 34

Idea of the T -coercivity 2/2 � � | µ | − 1 |∇ u | 2 − 2 µ − 1 3 One has a ( u , T 1 u ) = ∇ u · ∇ ( R 1 u 1 ) 2 Ω Ω 2 a is T -coercive when | µ 2 | > � R 1 � 2 µ 1 . Young’s inequality ⇒ u 1 − 2 R 2 u 2 in Ω 1 4 Working with T 2 u = in Ω 2 , where R 2 : Ω 2 → Ω 1 , one − u 2 a is T -coercive when µ 1 > � R 2 � 2 | µ 2 | . proves that 5 Conclusion: ∈ [ −� R 1 � 2 ; − 1 / � R 2 � 2 ], then the Theorem. If the contrast κ µ = µ 2 /µ 1 / operator div ( µ − 1 ∇· ) is an isomorphism from H 1 0 (Ω) to H − 1 (Ω). 10 / 34

Idea of the T -coercivity 2/2 � � | µ | − 1 |∇ u | 2 − 2 µ − 1 3 One has a ( u , T 1 u ) = ∇ u · ∇ ( R 1 u 1 ) 2 Ω Ω 2 a is T -coercive when | µ 2 | > � R 1 � 2 µ 1 . Young’s inequality ⇒ u 1 − 2 R 2 u 2 in Ω 1 4 Working with T 2 u = in Ω 2 , where R 2 : Ω 2 → Ω 1 , one − u 2 a is T -coercive when µ 1 > � R 2 � 2 | µ 2 | . proves that The interval depends on the 5 Conclusion: norms of the transfer operators ∈ [ −� R 1 � 2 ; − 1 / � R 2 � 2 ], then the [ −� R 1 � 2 ; − 1 / � R 2 � 2 ] Theorem. If the contrast κ µ = µ 2 /µ 1 / operator div ( µ − 1 ∇· ) is an isomorphism from H 1 0 (Ω) to H − 1 (Ω). 10 / 34

Choice of R 1 , R 2 ? A simple case: symmetric domain ◮ Ω 1 Σ Ω 2 11 / 34

Choice of R 1 , R 2 ? A simple case: symmetric domain ◮ R 1 = R 2 = S Σ Ω 1 Σ so that � R 1 � = � R 2 � = 1 Ω 2 ( P ) well-posed ⇔ κ µ � = − 1 11 / 34

Choice of R 1 , R 2 ? A simple case: symmetric domain ◮ R 1 = R 2 = S Σ Ω 1 Σ so that � R 1 � = � R 2 � = 1 Ω 2 ( P ) well-posed ⇔ κ µ � = − 1 Interface with a 2D corner ◮ Σ Ω 1 σ O Ω 2 11 / 34

Choice of R 1 , R 2 ? A simple case: symmetric domain ◮ R 1 = R 2 = S Σ Ω 1 Σ so that � R 1 � = � R 2 � = 1 Ω 2 ( P ) well-posed ⇔ κ µ � = − 1 Interface with a 2D corner ◮ Σ Action of R 1 : Ω 1 σ O Ω 2 11 / 34

Choice of R 1 , R 2 ? A simple case: symmetric domain ◮ R 1 = R 2 = S Σ Ω 1 Σ so that � R 1 � = � R 2 � = 1 Ω 2 ( P ) well-posed ⇔ κ µ � = − 1 Interface with a 2D corner ◮ Σ Action of R 1 : Ω 1 Ω 2 11 / 34

Choice of R 1 , R 2 ? A simple case: symmetric domain ◮ R 1 = R 2 = S Σ Ω 1 Σ so that � R 1 � = � R 2 � = 1 Ω 2 ( P ) well-posed ⇔ κ µ � = − 1 Interface with a 2D corner ◮ Σ Action of R 1 : symmetry w.r.t θ Ω 1 Ω 2 11 / 34

Choice of R 1 , R 2 ? A simple case: symmetric domain ◮ R 1 = R 2 = S Σ Ω 1 Σ so that � R 1 � = � R 2 � = 1 Ω 2 ( P ) well-posed ⇔ κ µ � = − 1 Interface with a 2D corner ◮ Σ Action of R 1 : symmetry + dilatation w.r.t θ Ω 1 Ω 2 11 / 34

Choice of R 1 , R 2 ? A simple case: symmetric domain ◮ R 1 = R 2 = S Σ Ω 1 Σ so that � R 1 � = � R 2 � = 1 Ω 2 ( P ) well-posed ⇔ κ µ � = − 1 Interface with a 2D corner ◮ Σ Action of R 1 : symmetry + dilatation w.r.t θ Ω 1 σ � R 1 � 2 O = R σ := (2 π − σ ) /σ Ω 2 11 / 34

Choice of R 1 , R 2 ? A simple case: symmetric domain ◮ R 1 = R 2 = S Σ Ω 1 Σ so that � R 1 � = � R 2 � = 1 Ω 2 ( P ) well-posed ⇔ κ µ � = − 1 Interface with a 2D corner ◮ Σ Action of R 1 : symmetry + dilatation w.r.t θ Ω 1 σ Action of R 2 : symmetry + dilatation w.r.t θ � R 1 � 2 = � R 2 � 2 = R σ := (2 π − σ ) /σ O Ω 2 11 / 34

Choice of R 1 , R 2 ? A simple case: symmetric domain ◮ R 1 = R 2 = S Σ Ω 1 Σ so that � R 1 � = � R 2 � = 1 Ω 2 ( P ) well-posed ⇔ κ µ � = − 1 Interface with a 2D corner ◮ Σ Action of R 1 : symmetry + dilatation w.r.t θ Ω 1 σ Action of R 2 : symmetry + dilatation w.r.t θ � R 1 � 2 = � R 2 � 2 = R σ := (2 π − σ ) /σ O Ω 2 ( P ) well-posed ⇐ κ µ / ∈ [ −R σ ; − 1 / R σ ] 11 / 34

Choice of R 1 , R 2 ? A simple case: symmetric domain ◮ R 1 = R 2 = S Σ Ω 1 Σ so that � R 1 � = � R 2 � = 1 Ω 2 ( P ) well-posed ⇔ κ µ � = − 1 Interface with a 2D corner ◮ Σ Action of R 1 : symmetry + dilatation w.r.t θ Ω 1 σ Action of R 2 : symmetry + dilatation w.r.t θ � R 1 � 2 = � R 2 � 2 = R σ := (2 π − σ ) /σ O Ω 2 ( P ) well-posed ⇔ κ µ / ∈ [ −R σ ; − 1 / R σ ] 11 / 34

Choice of R 1 , R 2 ? A simple case: symmetric domain ◮ R 1 = R 2 = S Σ Ω 1 Σ so that � R 1 � = � R 2 � = 1 Ω 2 ( P ) well-posed ⇔ κ µ � = − 1 Interface with a 2D corner ◮ Σ Σ Ω 2 Action of R 1 : symmetry + dilatation w.r.t θ Ω 1 σ σ Action of R 2 : symmetry + dilatation w.r.t θ Ω 1 � R 1 � 2 = � R 2 � 2 = R σ := (2 π − σ ) /σ O Ω 2 ( P ) well-posed ⇔ κ µ / ∈ [ −R σ ; − 1 / R σ ] 11 / 34

Choice of R 1 , R 2 ? A simple case: symmetric domain ◮ R 1 = R 2 = S Σ Ω 1 Σ so that � R 1 � = � R 2 � = 1 Ω 2 ( P ) well-posed ⇔ κ µ � = − 1 Interface with a 2D corner ◮ Σ Σ Ω 2 Action of R 1 : symmetry + dilatation w.r.t θ Ω 1 σ σ Action of R 2 : symmetry + dilatation w.r.t θ Ω 1 � R 1 � 2 = � R 2 � 2 = R σ := (2 π − σ ) /σ O Ω 2 ( P ) well-posed ⇔ κ µ / ∈ [ −R σ ; − 1 / R σ ] By localization techniques, we prove ◮ Proposition. ( P ) is well-posed in the Fredholm sense for a curvilinear polygonal interface iff κ µ / ∈ [ −R σ ; − 1 / R σ ] where σ is the smallest angle. ⇒ When Σ is smooth, ( P ) is well-posed in the Fredholm sense iff κ µ � = − 1. 11 / 34

Extensions for the scalar case The T -coercivity approach can be used to deal with non constant µ 1 , µ 2 ◮ and with the Neumann problem. 12 / 34

Extensions for the scalar case The T -coercivity approach can be used to deal with non constant µ 1 , µ 2 ◮ and with the Neumann problem. 3D geometries can be handled in the same way. ◮ The T -coercivity technique ◮ allows to justify convergence of standard finite element method for simple meshes ( Bonnet-Ben Dhia et al. 10, Nicaise, Venel 11, Chesnel, Ciarlet 12 ). 12 / 34

Transition: from variational methods to Fourier/Mellin techniques For the corner case, what happens when the contrast lies inside the criticial interval, i.e. when κ µ ∈ [ −R σ ; − 1 / R σ ]??? Σ Ω 1 σ O Ω 2 13 / 34

Transition: from variational methods to Fourier/Mellin techniques For the corner case, what happens when the contrast lies inside the criticial interval, i.e. when κ µ ∈ [ −R σ ; − 1 / R σ ]??? Σ Ω 1 σ O Ω 2 Idea: we will study precisely the regularity of the “solutions” using the Kondratiev’s tools, i.e. the Fourier/Mellin transform ( Dauge, Texier 97, Nazarov, Plamenevsky 94 ). 13 / 34

1 The coerciveness issue for the scalar case 2 A new functional framework in the critical interval ⇒ collaboration with X. Claeys (LJLL Paris VI). 3 Study of Maxwell’s equations 4 The T -coercivity method for the Interior Transmission Problem 14 / 34

Problem considered in this section We recall the problem under consideration ◮ Find u ∈ H 1 0 (Ω) such that: ( P ) − div( µ − 1 ∇ u ) = f in Ω . To simplify the presentation, we work on a particular configuration. ◮ Σ Ω 1 Ω 2 O µ 1 > 0 µ 2 < 0 15 / 34

Problem considered in this section We recall the problem under consideration ◮ Find u ∈ H 1 0 (Ω) such that: ( P ) − div( µ − 1 ∇ u ) = f in Ω . To simplify the presentation, we work on a particular configuration. ◮ Σ 15 / 34

Problem considered in this section We recall the problem under consideration ◮ Find u ∈ H 1 0 (Ω) such that: ( P ) − div( µ − 1 ∇ u ) = f in Ω . To simplify the presentation, we work on a particular configuration. ◮ Σ Ω 1 Ω 2 µ 1 > 0 µ 2 < 0 π 4 O O 15 / 34

Problem considered in this section We recall the problem under consideration ◮ Find u ∈ H 1 0 (Ω) such that: ( P ) − div( µ − 1 ∇ u ) = f in Ω . To simplify the presentation, we work on a particular configuration. ◮ Σ Ω 1 Ω 2 µ 1 > 0 µ 2 < 0 π 4 O O Using the variational method of the previous section, we prove the ◮ Proposition. The problem ( P ) is well-posed as soon as the contrast κ µ = µ 2 /µ 1 satisfies κ µ / ∈ [ − 3; − 1]. 15 / 34

Problem considered in this section We recall the problem under consideration ◮ Find u ∈ H 1 0 (Ω) such that: ( P ) − div( µ − 1 ∇ u ) = f in Ω . To simplify the presentation, we work on a particular configuration. ◮ Σ Ω 1 Ω 2 µ 1 > 0 µ 2 < 0 π 4 O O Using the variational method of the previous section, we prove the ◮ Proposition. The problem ( P ) is well-posed as soon as the contrast κ µ = µ 2 /µ 1 satisfies κ µ / ∈ [ − 3; − 1]. What happens when κ µ ∈ [ − 3; − 1)? 15 / 34

Analogy with a waveguide problem • Bounded sector Ω π/ 4 Σ Ω 1 Ω 2 O ( r , θ ) • Equation: − div( µ − 1 ∇ u ) = f � �� � − r − 2 ( µ − 1 ( r ∂ r ) 2 + ∂ θ µ − 1 ∂ θ ) u 16 / 34

Analogy with a waveguide problem • Bounded sector Ω π/ 4 Σ Ω 1 Ω 2 O ( r , θ ) • Equation: − div( µ − 1 ∇ u ) = f � �� � − r − 2 ( µ − 1 ( r ∂ r ) 2 + ∂ θ µ − 1 ∂ θ ) u • Singularities in the sector s ( r , θ ) = r λ ϕ ( θ ) 16 / 34

Analogy with a waveguide problem • Bounded sector Ω We compute the singularities s ( r , θ ) = r λ ϕ ( θ ) and we observe two cases: Outside the critical interval ◮ r �→ r λ 1 π/ 4 1 κ µ = − 4 1 Σ Ω 1 Ω 2 − λ 2 − λ 1 λ 1 λ 2 r O ( r , θ ) 0 -2 -1 1 2 • Equation: -1 not H 1 H 1 − 1 − div( µ − 1 ∇ u ) = f � �� � − r − 2 ( µ − 1 ( r ∂ r ) 2 + ∂ θ µ − 1 ∂ θ ) u • Singularities in the sector s ( r , θ ) = r λ ϕ ( θ ) 16 / 34

Analogy with a waveguide problem • Bounded sector Ω We compute the singularities s ( r , θ ) = r λ ϕ ( θ ) and we observe two cases: Outside the critical interval ◮ r �→ r λ 1 π/ 4 1 κ µ = − 4 1 Σ Ω 1 Ω 2 − λ 2 − λ 1 λ 1 λ 2 r O ( r , θ ) 0 -2 -1 1 2 • Equation: -1 not H 1 H 1 − 1 − div( µ − 1 ∇ u ) = f � �� � − r − 2 ( µ − 1 ( r ∂ r ) 2 + ∂ θ µ − 1 ∂ θ ) u Inside the critical interval ◮ r �→ ℜ e r λ 1 1 κ µ = − 2 • Singularities in the sector 1 λ 1 − λ 2 λ 2 s ( r , θ ) = r λ ϕ ( θ ) r 0 -2 -1 1 2 − λ 1 not H 1 -1 not H 1 H 1 − 1 16 / 34

Analogy with a waveguide problem • Bounded sector Ω We compute the singularities s ( r , θ ) = r λ ϕ ( θ ) and we observe two cases: Outside the critical interval ◮ r �→ r λ 1 π/ 4 1 κ µ = − 4 1 Σ Ω 1 Ω 2 − λ 2 − λ 1 λ 1 λ 2 r O ( r , θ ) 0 -2 -1 1 2 • Equation: -1 not H 1 H 1 − 1 − div( µ − 1 ∇ u ) = f � �� � − r − 2 ( µ − 1 ( r ∂ r ) 2 + ∂ θ µ − 1 ∂ θ ) u Inside the critical interval ◮ r �→ ℜ e r λ 1 1 κ µ = − 2 • Singularities in the sector 1 λ 1 − λ 2 λ 2 s ( r , θ ) = r λ ϕ ( θ ) r 0 -2 -1 1 2 − λ 1 not H 1 -1 not H 1 H 1 − 1 How to deal with the propagative singularities inside the critical interval? 16 / 34

Analogy with a waveguide problem • Bounded sector Ω π/ 4 Σ Ω 1 Ω 2 O ( r , θ ) • Equation: − div( µ − 1 ∇ u ) = f � �� � − r − 2 ( µ − 1 ( r ∂ r ) 2 + ∂ θ µ − 1 ∂ θ ) u • Singularities in the sector s ( r , θ ) = r λ ϕ ( θ ) 16 / 34

Analogy with a waveguide problem • Bounded sector Ω • Half-strip B θ ( z , θ ) = ( − ln r , θ ) π/ 4 B 1 Σ Σ Ω 1 Ω 2 θ = π/ 4 B 2 ( r , θ ) = ( e − z , θ ) z O ( r , θ ) • Equation: − div( µ − 1 ∇ u ) = f � �� � − r − 2 ( µ − 1 ( r ∂ r ) 2 + ∂ θ µ − 1 ∂ θ ) u • Singularities in the sector s ( r , θ ) = r λ ϕ ( θ ) 16 / 34

Analogy with a waveguide problem • Bounded sector Ω • Half-strip B θ ( z , θ ) = ( − ln r , θ ) π/ 4 B 1 Σ Σ Ω 1 Ω 2 θ = π/ 4 B 2 ( r , θ ) = ( e − z , θ ) z O ( r , θ ) • Equation: • Equation: − div( µ − 1 ∇ u ) = e − 2 z f − div( µ − 1 ∇ u ) = f � �� � � �� � − r − 2 ( µ − 1 ( r ∂ r ) 2 + ∂ θ µ − 1 ∂ θ ) u − ( µ − 1 ∂ 2 z + ∂ θ µ − 1 ∂ θ ) u • Singularities in the sector s ( r , θ ) = r λ ϕ ( θ ) 16 / 34

Analogy with a waveguide problem • Bounded sector Ω • Half-strip B θ ( z , θ ) = ( − ln r , θ ) π/ 4 B 1 Σ Σ Ω 1 Ω 2 θ = π/ 4 B 2 ( r , θ ) = ( e − z , θ ) z O ( r , θ ) • Equation: • Equation: − div( µ − 1 ∇ u ) = e − 2 z f − div( µ − 1 ∇ u ) = f � �� � � �� � − r − 2 ( µ − 1 ( r ∂ r ) 2 + ∂ θ µ − 1 ∂ θ ) u − ( µ − 1 ∂ 2 z + ∂ θ µ − 1 ∂ θ ) u • Singularities in the sector • Modes in the strip s ( r , θ ) = r λ ϕ ( θ ) m ( z , θ ) = e − λ z ϕ ( θ ) 16 / 34

Analogy with a waveguide problem • Bounded sector Ω • Half-strip B θ ( z , θ ) = ( − ln r , θ ) π/ 4 B 1 Σ Σ Ω 1 Ω 2 θ = π/ 4 B 2 ( r , θ ) = ( e − z , θ ) z r �→ ℜ e r λ z �→ ℜ e e − λ z O ( r , θ ) 1 1 • Equation: • Equation: r z − div( µ − 1 ∇ u ) = e − 2 z f 0 − div( µ − 1 ∇ u ) 0 = f � �� � � �� � − 1 − 1 − r − 2 ( µ − 1 ( r ∂ r ) 2 + ∂ θ µ − 1 ∂ θ ) u − ( µ − 1 ∂ 2 z + ∂ θ µ − 1 ∂ θ ) u • Singularities in the sector • Modes in the strip s ( r , θ ) = r λ ϕ ( θ ) m ( z , θ ) = e − λ z ϕ ( θ ) s ∈ H 1 (Ω) ℜ e λ > 0 m is evanescent 16 / 34

Analogy with a waveguide problem • Bounded sector Ω • Half-strip B θ ( z , θ ) = ( − ln r , θ ) π/ 4 B 1 Σ Σ Ω 1 Ω 2 θ = π/ 4 B 2 ( r , θ ) = ( e − z , θ ) z r �→ ℜ e r λ z �→ ℜ e e − λ z O ( r , θ ) 1 1 • Equation: • Equation: r z − div( µ − 1 ∇ u ) = e − 2 z f 0 − div( µ − 1 ∇ u ) 0 = f � �� � � �� � − 1 − 1 − r − 2 ( µ − 1 ( r ∂ r ) 2 + ∂ θ µ − 1 ∂ θ ) u − ( µ − 1 ∂ 2 z + ∂ θ µ − 1 ∂ θ ) u • Singularities in the sector • Modes in the strip s ( r , θ ) = r λ ϕ ( θ ) m ( z , θ ) = e − λ z ϕ ( θ ) s ( r , θ ) = r a (cos b ln r + i sin b ln r ) ϕ ( θ ) m ( z , θ ) = e − az (cos bz − i sin bz ) ϕ ( θ ) ( ℜ e λ = a , ℑ m λ = b ) s ∈ H 1 (Ω) ℜ e λ > 0 m is evanescent ∈ H 1 (Ω) s / ℜ e λ = 0 m is propagative 16 / 34

Analogy with a waveguide problem • Bounded sector Ω • Half-strip B θ ( z , θ ) = ( − ln r , θ ) π/ 4 B 1 Σ Σ Ω 1 Ω 2 θ = π/ 4 B 2 ( r , θ ) = ( e − z , θ ) z O ( r , θ ) • Equation: • Equation: − div( µ − 1 ∇ u ) = e − 2 z f − div( µ − 1 ∇ u ) = f � �� � � �� � − r − 2 ( µ − 1 ( r ∂ r ) 2 + ∂ θ µ − 1 ∂ θ ) u − ( µ − 1 ∂ 2 z + ∂ θ µ − 1 ∂ θ ) u • Singularities in the sector • Modes in the strip s ( r , θ ) = r λ ϕ ( θ ) m ( z , θ ) = e − λ z ϕ ( θ ) s ( r , θ ) = r a (cos b ln r + i sin b ln r ) ϕ ( θ ) m ( z , θ ) = e − az (cos bz − i sin bz ) ϕ ( θ ) ( ℜ e λ = a , ℑ m λ = b ) s ∈ H 1 (Ω) ℜ e λ > 0 m is evanescent ∈ H 1 (Ω) s / ℜ e λ = 0 m is propagative This encourages us to use modal decomposition in the half-strip. ◮ 16 / 34

Modal analysis in the waveguide κ µ = − 4 1 Outside the critical interval . All the ◮ modes are exponentially growing or decaying. − λ 2 − λ 1 λ 1 λ 2 → We look for an exponentially decaying -2 -1 1 2 H 1 framework solution. -1 17 / 34

Modal analysis in the waveguide κ µ = − 4 1 Outside the critical interval . All the ◮ modes are exponentially growing or decaying. − λ 2 − λ 1 λ 1 λ 2 → We look for an exponentially decaying -2 -1 1 2 H 1 framework solution. -1 Inside the critical interval . There are ◮ exactly two propagative modes. κ µ = − 2 1 λ 1 − λ 2 λ 2 -2 -1 1 2 − λ 1 -1 17 / 34

Modal analysis in the waveguide κ µ = − 4 1 Outside the critical interval . All the ◮ modes are exponentially growing or decaying. − λ 2 − λ 1 λ 1 λ 2 → We look for an exponentially decaying -2 -1 1 2 H 1 framework solution. -1 Inside the critical interval . There are ◮ exactly two propagative modes. κ µ = − 2 1 λ 1 → The decomposition on the outgoing modes − λ 2 λ 2 leads to look for a solution of the form -2 -1 1 2 c 1 ϕ 1 e λ 1 z − λ 1 u = + u e . -1 � �� � ���� propagative part evanescent part non H 1 framework 17 / 34

Modal analysis in the waveguide κ µ = − 4 1 Outside the critical interval . All the ◮ modes are exponentially growing or decaying. − λ 2 − λ 1 λ 1 λ 2 → We look for an exponentially decaying -2 -1 1 2 H 1 framework solution. -1 Inside the critical interval . There are ◮ exactly two propagative modes. κ µ = − 2 1 λ 1 → The decomposition on the outgoing modes − λ 2 λ 2 leads to look for a solution of the form -2 -1 1 2 c 1 ϕ 1 e λ 1 z − λ 1 u = + u e . -1 � �� � ���� propagative part evanescent part non H 1 framework ... but the modal decomposition is not easy to justify because two sign- changing appear in the transverse problem: ∂ θ σ∂ θ ϕ = − σλ 2 ϕ. 17 / 34

The new functional framework Consider 0 < β < 2, ζ a cut-off function (equal to 1 in + ∞ ) and define W − β = { v | e β z v ∈ H 1 0 ( B ) } space of exponentially decaying functions 18 / 34

The new functional framework Consider 0 < β < 2, ζ a cut-off function (equal to 1 in + ∞ ) and define W − β = { v | e β z v ∈ H 1 0 ( B ) } space of exponentially decaying functions = { v | e − β z v ∈ H 1 W β 0 ( B ) } space of exponentially growing functions 18 / 34

The new functional framework Consider 0 < β < 2, ζ a cut-off function (equal to 1 in + ∞ ) and define W − β = { v | e β z v ∈ H 1 0 ( B ) } space of exponentially decaying functions W + = span( ζϕ 1 e λ 1 z ) ⊕ W − β propagative part + evanescent part = { v | e − β z v ∈ H 1 W β 0 ( B ) } space of exponentially growing functions 18 / 34

The new functional framework Consider 0 < β < 2, ζ a cut-off function (equal to 1 in + ∞ ) and define W − β = { v | e β z v ∈ H 1 0 ( B ) } space of exponentially decaying functions ∩ W + = span( ζϕ 1 e λ 1 z ) ⊕ W − β propagative part + evanescent part ∩ = { v | e − β z v ∈ H 1 W β 0 ( B ) } space of exponentially growing functions 18 / 34

The new functional framework Consider 0 < β < 2, ζ a cut-off function (equal to 1 in + ∞ ) and define W − β = { v | e β z v ∈ H 1 0 ( B ) } space of exponentially decaying functions ∩ W + = span( ζϕ 1 e λ 1 z ) ⊕ W − β propagative part + evanescent part ∩ = { v | e − β z v ∈ H 1 W β 0 ( B ) } space of exponentially growing functions The operator A + : Let κ µ ∈ ( − 3; − 1) and 0 < β < 2. Theorem. div( µ − 1 ∇· ) from W + to W ∗ β is an isomorphism. 18 / 34

The new functional framework Consider 0 < β < 2, ζ a cut-off function (equal to 1 in + ∞ ) and define W − β = { v | e β z v ∈ H 1 0 ( B ) } space of exponentially decaying functions ∩ W + = span( ζϕ 1 e λ 1 z ) ⊕ W − β propagative part + evanescent part ∩ = { v | e − β z v ∈ H 1 W β 0 ( B ) } space of exponentially growing functions The operator A + : Let κ µ ∈ ( − 3; − 1) and 0 < β < 2. Theorem. div( µ − 1 ∇· ) from W + to W ∗ β is an isomorphism. Ideas of the proof: 1 A − β : div( µ − 1 ∇· ) from W − β to W ∗ β is injective but not surjective. 18 / 34

The new functional framework Consider 0 < β < 2, ζ a cut-off function (equal to 1 in + ∞ ) and define W − β = { v | e β z v ∈ H 1 0 ( B ) } space of exponentially decaying functions ∩ W + = span( ζϕ 1 e λ 1 z ) ⊕ W − β propagative part + evanescent part ∩ = { v | e − β z v ∈ H 1 W β 0 ( B ) } space of exponentially growing functions The operator A + : Let κ µ ∈ ( − 3; − 1) and 0 < β < 2. Theorem. div( µ − 1 ∇· ) from W + to W ∗ β is an isomorphism. Ideas of the proof: 1 A − β : div( µ − 1 ∇· ) from W − β to W ∗ β is injective but not surjective. 2 A β : div( µ − 1 ∇· ) from W β to W ∗ − β is surjective but not injective. 18 / 34

The new functional framework Consider 0 < β < 2, ζ a cut-off function (equal to 1 in + ∞ ) and define W − β = { v | e β z v ∈ H 1 0 ( B ) } space of exponentially decaying functions ∩ W + = span( ζϕ 1 e λ 1 z ) ⊕ W − β propagative part + evanescent part ∩ = { v | e − β z v ∈ H 1 W β 0 ( B ) } space of exponentially growing functions The operator A + : Let κ µ ∈ ( − 3; − 1) and 0 < β < 2. Theorem. div( µ − 1 ∇· ) from W + to W ∗ β is an isomorphism. Ideas of the proof: 1 A − β : div( µ − 1 ∇· ) from W − β to W ∗ β is injective but not surjective. 2 A β : div( µ − 1 ∇· ) from W β to W ∗ − β is surjective but not injective. 3 The intermediate operator A + : W + → W ∗ β is injective (energy integral) and surjective (residue theorem). 18 / 34

The new functional framework Consider 0 < β < 2, ζ a cut-off function (equal to 1 in + ∞ ) and define W − β = { v | e β z v ∈ H 1 0 ( B ) } space of exponentially decaying functions ∩ W + = span( ζϕ 1 e λ 1 z ) ⊕ W − β propagative part + evanescent part ∩ = { v | e − β z v ∈ H 1 W β 0 ( B ) } space of exponentially growing functions The operator A + : Let κ µ ∈ ( − 3; − 1) and 0 < β < 2. Theorem. div( µ − 1 ∇· ) from W + to W ∗ β is an isomorphism. Ideas of the proof: 1 A − β : div( µ − 1 ∇· ) from W − β to W ∗ β is injective but not surjective. 2 A β : div( µ − 1 ∇· ) from W β to W ∗ − β is surjective but not injective. 3 The intermediate operator A + : W + → W ∗ β is injective (energy integral) and surjective (residue theorem). 4 Limiting absorption principle to select the outgoing mode. 18 / 34

A funny use of PMLs We use a PML ( Perfectly Matched Layer ) to bound the domain B ◮ + finite elements in the truncated strip PML PML Contrast κ µ = − 1 . 001 ∈ ( − 3; − 1). 19 / 34

A black hole phenomenon The same phenomenon occurs for the Helmholtz equation. ◮ ( x , t ) �→ ℜ e ( u ( x ) e − i ω t ) for κ µ = − 1 . 3 ∈ ( − 3; − 1) (. . . ) (. . . ) Analogous phenomena occur in cuspidal domains in the theory of ◮ water-waves and in elasticity ( Cardone, Nazarov, Taskinen ). On going work for a general domain ( C. Carvalho ). ◮ 20 / 34

Summary of the results for the scalar problem Problem Σ Find u ∈ H 1 0 (Ω) s.t.: ( P ) − div ( µ − 1 ∇ u ) = f Ω 1 Ω 2 in Ω . µ 1 > 0 µ 2 < 0 π 4 O O 21 / 34

Summary of the results for the scalar problem Problem Σ Find u ∈ H 1 0 (Ω) s.t.: ( P ) − div ( µ − 1 ∇ u ) = f Ω 1 Ω 2 in Ω . µ 1 > 0 µ 2 < 0 π 4 O O Results ℑ m κ µ For κ µ ∈ C \ R − , ( P ) well-posed in H 1 0 (Ω) (Lax-Milgram) ℜ e κ µ − 3 − 1 21 / 34

Summary of the results for the scalar problem Problem Σ Find u ∈ H 1 0 (Ω) s.t.: ( P ) − div ( µ − 1 ∇ u ) = f Ω 1 Ω 2 in Ω . µ 1 > 0 µ 2 < 0 π 4 O O Results ℑ m κ µ For κ µ ∈ C \ R − , ( P ) well-posed in H 1 0 (Ω) (Lax-Milgram) For κ µ ∈ R ∗ − \ [ − 3; − 1], ( P ) well-posed in H 1 0 (Ω) ( T -coercivity) ℜ e κ µ − 3 − 1 21 / 34