Functional Analysis for Helmholtz Equation in the Framework of Domain Decomposition
aa r X i v : . [ m a t h . A P ] M a y Functional Analysis for Helmholtz Equation inthe Framework of Domain Decomposition
Mikhael Balabane ∗ Sept. 2008
Abstract
This paper gives a geometric description of functional spaces re-lated to Domain Decomposition techniques for computing solutionsof Laplace and Helmholtz equations. Understanding the geometricstructure of these spaces leads to algorithms for solving the equations.It leads also to a new interpretation of classical algorithms, enhanc-ing convergence. The algorithms are given and convergence is proved.This is done by building tools enabling geometric interpretations of theoperators related to Domain Decomposition technique. The Despresoperators, expressing conservation of energy for Helmholtz equation,are defined on the fictitious boundary and their spectral propertiesproved.It turns to be the key for proving convergence of the givenalgorithm for Helmholtz equation in a non-dissipating cavity.Using these tools, one can prove that the Domain Decomposition set-ting for the Helmholtz equation leads to an ill-posed problem. Never-theless, one can prove that if a solution exists, it is unique. And thatthe algorithm do converge to the solution. In the framework of domain decomposition, given a bounded open set Ω =Ω ∪ Ω ∪ Γ where the two open sets Ω and Ω are not overlapping, and Γ a ∗ University Paris 13 - France - email: [email protected] CMS 65M12, 35P10, 35P15, 35P25 u of the Helmholtz equation∆ u + k u = f ∈ L (Ω) and u ∈ H (Ω)the aim of this paper is to understand the dynamics of the sequence ( v n , v n ) n ∈ N solving separatly the Helmholtz equations on Ω and Ω , when equating thefluxes through Γ: ( m = 1 , m ′ = 2 , ∂v nm ∂n m − iγv nm = − ∂v n − m ′ ∂n m ′ − iγv n − m ′ on ΓIts ultimate aim is to prove convergence to ( u | Ω , u | Ω ) of the sequence ( u n , u n ) n ∈ N solving the Helmholtz equations on Ω and Ω with a penalization on theboundary Γ is added, namely: ∂u nm ∂n m − iγu nm = θ [ ∂u n − m ∂n m − iγu n − ] − (1 − θ )[ ∂u n − ∂n m ′ + iγu n − m ′ ] on ΓFor this sake, the geometry of the set of solutions of the Helmholtz equationon Ω × Ω with equated energy fluxes is studied, through the study of thecoupling operator defined on L (Γ) × L (Γ) which intertwins the fluxes. Itturns out that the key for understanding the convergence of the sequence( u n , u n ) n ∈ N is the analysis of the spectral properties of the intertwinnig op-erator.Using these tools, one can prove that the Domain Decomposition settingfor the Helmholtz equation leads to an ill-posed problem. Nevertheless, onecan prove that if a solution exists, it is unique. And that the algorithm doconverge to the solution.Convergence of the penalized algorithm is proven and numerical tests forsolving the Helmholtz equation through this domain decomposition algorithmare given.The geometric analysis given here provides the theoretical background foranother numerical algorithm for computing the global solution u , by a specificspectral method. A forthcoming paper describes and gives the numericalanalysis of this algorithm.This domain decomposition algorithm (in a dissipating cavity case, i.e. witha Sommerfeld-like radiation condition on part of the boundary), was first2nitiated and studied by B.Despres in [D1] [D2] [BD] , and computationalresults given by J.D.Benamou [B] [BD] , F.Collino and P.Joly [CGJ] .In order to perform the geometric analysis of the set of solutions of theHelmholtz equation on Ω × Ω , one has first to make a complete descriptionof the geometry of the set of solutions of the Laplace equation on Ω × Ω .Geometric properties of this set proven below makes it possible to revisitthe classical penalized Dirichlet/Neumann domain decomposition algorithm(with penalization) for solving the Laplace equation. A new version of thisalgorithm is given here, and proved to converge to the global solution, en-hancing the usual assumption on the penalization parameter.This completes classical results by O.Widlund [PW] , P.L.Lions [L] , or A.Quar-teroni and A.Valli [ FMQT] [FQZ] [QV] .The paper is organized as follows: in section 2 basic facts are revisited,although classical, and completed in order to set the geometric frameworkneeded. (It also makes the paper self contained). A precise study of duality,and the link with the Poincare-Steklov operators, is performed, which turnsto be central for the remainder of the paper. In section 3 a new version ofthe Dirichlet/Neumann algorithm for the Laplace equation is given, and con-vergence is proved. In section 4 geometric tools for the Helmholtz equation,and related domain decomposition algorithm, are given. Despres operatorsare studied and their spectral properties investigated. As is the intertwinnigoperator. In section 5 , convergence of the domain decomposition algorithmfor Helmholtz equation is proved. In section 6 numerical tests are given.Throughout this paper, when dealing with the Helmholtz equation, the fre-quency k is assumed to be non-resonnant for the Dirichlet boundary condi-tion. More precisely we shall always make the following Assumption (A) − k is not an eigenvalue of the Laplace operator on Ωwith Dirichlet boundary condition, i.e. the following problem is well posedfor f ∈ L (Ω): ∆ u + k u = f and u ∈ H (Ω)We shall also adopt the following Notation (N) normal derivatives at the boundary of an open set are alwaysmeant as the derivative along the outward unit normal vector3
Basics
Let Ω ⊂ R d be a bounded open set whose boundary ∂ Ω is a C -submanifoldof R d . Let Γ be an open C ∞ -submanifold of R d , such that:Ω = Ω ∪ Ω ∪ Γ , ∂ Ω = ( ∂ Ω ∩ ∂ Ω ) ∪ Γ , ∂ Ω = ( ∂ Ω ∩ ∂ Ω ) ∪ Γwhere Ω and Ω are open sets in R d . We assume that Ω and Ω fulfill thestrict cone property (see [Ag] for instance) and that Γ is transverse to ∂ Ω inthe following sense: Γ is a C -submanifold of R d with boundary, and thereexists a < σ ∈ ∂ Ω ∩ Γ, we have: − a ≤ n Γ ( σ ) .n ∂ Ω ( σ ) ≤ a (1)where n Γ ( σ ) ∈ C (Γ) is a unit vector normal to Γ at σ and n ∂ Ω ( σ ) ∈ C ( ∂ Ω)a unit vector normal to ∂ Ω at σ . Γ Let H (Ω) be endowed with the scalar product( u, v ) H (Ω) = Z Ω ∇ u ∇ vdx For m = 1 , H m = { u ∈ H (Ω m ); u | ∂ Ω ∩ ∂ Ω m = 0 } . Boundedness of thetrace operators from H (Ω) to H ( ∂ Ω ∩ ∂ Ω m ) imply that these are Hilbertspaces when endowed with the scalar products:( u, v ) H m = Z Ω m ∇ u ∇ vdx Let ρ Γ (resp. ρ Γ m for m = 1 ,
2) be the trace operator on Γ, i.e. the boundedlinear operator from H (Ω) (resp. H m ) to H / (Γ) which maps u to u | Γ .Let Λ = { u | Γ ; u ∈ H (Ω) } = H (Ω) /Kerρ Γ ≃ ( Kerρ Γ ) ⊥ and for m = 1 , m = { u | Γ ; u ∈ H m } = H m /Kerρ Γ m ≃ ( Kerρ Γ m ) ⊥ emark 1 Obviously
Kerρ Γ m = H (Ω m ) , Λ m ⊂ H / (Γ) , Λ ⊂ H / (Γ)Because ρ Γ and ρ Γ m are bounded, Λ and Λ m are Hilbert spaces when endowedwith the following norms: ∀ λ ∈ Λ , k λ k Λ = inf { u ; u | Γ = λ } k u k H (Ω) and ∀ λ ∈ Λ m , k λ k Λ m = inf { u ; u | Γ = λ } k u k H m Remark 2
Obviously, for any v ∈ H (Ω) and w ∈ H m k ρ Γ ( v ) k Λ ≤ k v k H (Ω) and k ρ Γ m ( w ) k Λ ≤ k w k H m Proposition 1
Let m = 1 , . For any λ ∈ Λ m , µ ∈ Λ m :1- There exists a unique u λm ∈ H m such that ∆ u λm = 0 in Ω m and ρ Γ m ( u λm ) = λ
2- One has: k λ k Λ m = k u λm k H m and ( λ, µ ) Λ m = ( u λm , u µm ) H m
3- For any λ ∈ Λ let u λ = u λm on Ω m , m = 1 , . Then k λ k = k u λ k H + k u λ k H
4- Using the previous notation, for any λ ∈ Λ , µ ∈ Λ( λ, µ ) Λ = ( u λ , u µ ) H (Ω) proof:
1- Uniqueness follows well posedness of the Laplace problem in H (Ω m ).In order to prove existence, let u ∈ H m be such that λ = ρ Γ m ( u ). Then∆ u ∈ H − (Ω m ). Let v be the unique solution in H (Ω m ) of ∆ v = ∆ u . Then u λm = u − v fulfills the property.2- Because of remark 1, one has to show: w ∈ H (Ω m ) ⇒ ( u λm , w ) H m = 0which follows from the Green formula.3- Because ρ Γ1 ( u λ ) = ρ Γ2 ( u λ ) = λ one has u λ ∈ H (Ω). Obviously it isorthogonal to Kerρ Γ . so k λ k = k u λ k H (Ω) = k u λ k H + k u λ k H .2 Λ = Λ = Λ By symmetry, it is enough to prove Λ = Λ . In order to prove this alge-braic and topological equality, two key tools are needed. The first tool isthe Calderon extension theorem [Ag] , which applies here because Ω hasthe strict cone property, by assumption, and which gives a bounded linearoperator E from H (Ω ) to H ( R d ) such that: ∀ w ∈ H (Ω ) , Ew | Ω = w The second key tool is:
Theorem 1
There exists a bounded linear operator τ in H ( R d ) such that: ∀ v ∈ H ( R d ) , v | ∂ Ω ∩ ∂ Ω = 0 ⇒ ( τ v | ∂ Ω = 0 and τ v | Ω = v | Ω ) proof: Assumption (1) gives a finite open covering ( ω j ) j of Γ ∩ ∂ Ω and a change ofvariables ( a j ) j such that V j = a j ( ω j ) is a neighbourhood of zero in R d and: a j (Γ ∩ ω j ) = { z j ∈ V j ; z j = 0 , z j > } and a j ( ∂ Ω ∩ ω j ) = { z j ∈ V j ; z j = 0 } a j (Ω ∩ ω j ) = { z j ∈ V j ; z j < , z j > } a j (Ω ∩ ω j ) = { z j ∈ V j ; z j > , z j > } Regularity of the submanifold Γ gives an open covering ( ω ′ k ) k of Γ and changeof variables ( b k ) k such that W k = b k ( ω ′ k ) is a neighbourhood of zero in R d and: b k (Γ ∩ ω ′ k ) = { z k ∈ W k ; z k = 0 } b k (Ω ∩ ω ′ k ) = { z k ∈ W k ; z k < } and b k (Ω ∩ ω ′ k ) = { z k ∈ W k ; z k > } Compactness of Γ enables to select a finite subcovering of Γ still denoted by( ω j ) j ∪ ( ω ′ k ) k having the previous properties.Let ω = R d \ Ω and ω = R d \ Ω , so: R d = ω ∪ ω ∪ ( ∪ j ω j ) ∪ ( ∪ k ω ′ k )Let ( α , α , ( α j ) j , ( α ′ k ) k )be a C ∞ -partition of unity associated with this open covering of R d .6et ε > ∀ z ∈ ( ∪ j V j ) ∪ ( ∪ k W k ) 0 < z j < ε ⇒ z ∈ Ω Let ψ ( s ) ∈ C ∞ ( R ) be equal to one for s < s > ε .Let ϕ ( θ ) ∈ C ∞ ( R ) be equal to zero for θ < θ > π For any v ∈ H ( R d ), v = α v + α v + X j α j v + X k α ′ k v we define τ v as: τ ( v ) = τ ( α v ) + τ ( α v ) + X j τ ( α j v ) + X k τ ( α ′ k v )with: τ ( α v ) ≡ τ ( α v ) = α vτ ( α ′ k v )( z k ) = ψ ( z k ) α ′ k ( z k ) v ( z k ) in the local coordinates.These three quantities are multiplication of v by C ∞ functions, which arebounded as well as all their derivatives. It is linear and bounded in H ( R d )with respect to v ∈ H ( R d ).In order to define τ ( α j v ), we first write α j v in the cylindrical coordinates asfollows: α j v ( z j , z j , z j , .., z jd ) = g α j v ( r j , θ j , z j , .., z jd ) with z j = r j cos ( θ j ) , z j = r j sin ( θ j )and define τ ( α j v ) in these coordinates as: g τ ( α j v )( r j , θ j , z j , .., z jd ) = ϕ ( θ j ) g α j v ( r j , θ j , z j , .., z jd )This quantity is linear with respect to v , and we prove its boundedness in H ( R d ) with respect to v ∈ H ( R d ) as follows (we omit the index j anddenote the measure dz ...dz d by dz ): k τ ( αv ) k H ( R d ) = Z V | g τ ( αv ) | rdrdθdz + Z V | ∂∂r g τ ( αv ) | rdrdθdz + Z V r | ∂∂θ g τ ( αv ) | rdrdθdz + d X Z V | ∂∂z j g τ ( αv ) | rdrdθdz = Z V | ϕ ( θ ) f αv | rdrdθdz + Z V | ϕ ( θ ) ∂∂r f αv | rdrdθdz Z V r | ϕ ( θ ) ∂∂θ f αv + ϕ ′ ( θ ) f αv | rdrdθdz + d X Z V | ϕ ( θ ) ∂∂z j f αv | rdrdθdz ≤ sup | ϕ | [ Z V | f αv | rdrdθdz + Z V | ∂∂r f αv | rdrdθdz + Z V r | ∂∂θ f αv | rdrdθdz + d X Z V | ∂∂z j f αv | rdrdθdz ]+2 sup | ϕ ′ | Z V r | f αv | rdrdθdz ≤ sup | ϕ | k αv k H ( R d ) + 2 sup | ϕ ′ | Z V r | f αv | rdrdθdz In order to estimate this last quantity we use the assumption v | ∂ Ω ∪ ∂ Ω = 0to have: f αv ( r, θ, z ) = − Z πθ ∂∂θ f αv ( r, s, z ) ds which gives (with ε ′ the radius of the support of α in the r variable, and B a ball containing the support of α in the z variable): Z V r | f αv | rdrdθdz ≤ Z B Z ε ′ Z π − π r | Z πθ ∂∂θ f αv ( r, s, z ) ds | rdrdθdz ≤ π Z B Z ε ′ Z π − π r | ∂∂θ f αv ( r, s, z ) | rdrdsdz ≤ π k αv k H ( R d ) we summarize to have: k τ ( αv ) k H ( R d ) ≤ C k αv k H ( R d ) ≤ C ′ k v k H ( R d ) and this ends the proof of the boundedness of τ in H ( R d ).We end the proof of theorem 1 using the following obvious observations: τ v | Ω = v | Ω because ψ ≡ R − and ϕ ≡ θ ≥ π τ v | ∂ Ω ∪ ∂ Ω = 0 by assumption τ v | ∂ Ω ∪ ∂ Ω = 0 because ϕ (0) = 0 and ψ ≡ z j > ε . Corollary 1
For m = 1 , , there exists a bounded linear map E m from H m to H (Ω) such that ∀ u ∈ H m ( E m u ) | Ω m ≡ u roof: for m = 1 for instance let E u be the restriction to Ω of τ Eu .Boundednessfollows from theorem 1. Corollary 2 Λ = Λ = Λ and the three norms k . k Λ , k . k Λ and k . k Λ areequivalent.proof: obviously Λ ⊂ Λ m and the previous corollary gives the converse inclu-sion. Moreover we have k . k Λ ≥ k . k Λ m and the previous corollary gives theconverse inequality. Corollary 3 D (Γ) is a dense subspace of Λ for any of the three norms.proof: D (Γ) is the set of traces on Γ of functions in D (Ω) because Γ is a C ∞ submanifold. Density of D (Γ) for the k . k Λ norm follows density of D (Ω)in H (Ω). Equivalence of the three norms ends the proof. Remark 3
If we denote as usual by H (Γ) the closure of D (Ω) in H (Γ) (which exists because Γ is C ∞ ), then the previous corollary asserts that Λ ⊂ H (Γ) . Boundedness of the trace operators gives constants C, C m such that: ∀ λ ∈ Λ , k λ k H (Γ) ≤ C k λ k Λ and ∀ λ ∈ Λ m , , k λ k H (Γ) ≤ C m k λ k Λ m H − (Ω m ) × Λ Theorem 2
For m = 1 ,
1- for any ( f, λ ) ∈ H − (Ω m ) × Λ there exists a unique u ∈ H such that ∆ u = f in Ω m and ρ Γ m ( u ) = λ
2- we have the estimate k u k H m ≤ k f k H − (Ω m ) + k λ k Λ m proof: Let u λm be given by Proposition 1. Let v = u − u λm . The problem isequivalent to v ∈ H (Ω m ) and ∆ v = f This is a well-posed problem and we have, because the Riesz representationoperator is isometric, k v k H (Ω m ) = k f k H − (Ω m ) . So k u k H m ≤ k u λm k H m + k v k H m ≤ k f k H − (Ω m ) + k λ k Λ m .4 Duality and the Poincare-Steklov operators Let Λ ′ denote the dual space to Λ, endowed with one of the three equivalentnorms associated with the equivalent norms on Λ defined previously.We denote by ( ., . ) ΛΛ ′ the duality product, and by ( ., . ) DD ′ the duality productin D ′ (Γ).Because of Corollary 3 and remark 3, we have the usual injections: D ⊂ Λ ⊂ H (Γ) ⊂ L (Γ) ⊂ H − (Γ) ⊂ Λ ′ ⊂ D ′ and for any λ ∈ D (Γ) and ν ∈ Λ ′ : ( ν, λ ) ΛΛ ′ = ( ν, λ ) DD ′ = ( ν, λ ) L (Γ) where η denotes the complex conjugate of functions or distributions η .For any of the three scalar products on Λ we have ( λ, η ) = ( λ, η ) so for allnorms: ∀ λ ∈ Λ , k λ k = k λ k and ∀ ν ∈ Λ ′ , k ν k = k ν k Notation 1
Let ˜ S denote the antilinear Riesz representation operator for Λ ′ in the Λ -scalar product, and ˜ S m ( m = 1 , ) this representation in the Λ m -scalar product, i.e. ∀ λ ∈ Λ , ∀ ν ∈ Λ ′ ( ν, λ ) ΛΛ ′ = ( λ, ˜ S − ν ) Λ = ( λ, ˜ S − ν ) Λ = ( λ, ˜ S − ν ) Λ Let S (resp. S m ) be the linear isometric bijections from Λ to Λ ′ defined by ∀ Λ ∈ Λ ˜ Sλ = Sλ, ˜ S λ = S λ, ˜ S λ = S λ We denote by n m the normal unit vector on Γ pointing outward with respectto Ω m . We denote by ∂ϕ∂n m the normal derivative on Γ of ϕ ∈ D (Ω), and by ∂∂n m bounded extensions of this operator to any functional space.We use Proposition 1 to have: Proposition 2 ∀ λ ∈ Λ S m λ = ∂u λm ∂n m and S = S + S proof: by Green formula ∀ λ ∈ Λ , η ∈ Λ ( ˜ S m η, λ ) Λ m Λ ′ m = ( λ, η ) Λ m =10 u λm , u ηm ) H (Ω m ) = Z Ω m ∇ u λm ∇ u ηm dx = Z Γ λ ∂u ηm ∂n m dσ this proves that the distribution ∂u ηm ∂n m is bounded in the Λ m norm, so ∂u ηm ∂n m ∈ Λ ′ and ˜ S m η = ∂u ηm ∂n m = ∂u ηm ∂n m = ∂u ηm ∂n m Corollary 4 Sλ = Sλ S m λ = S m λ Remark 4
For any v ∈ H m such that ∆ v ∈ L (Ω m ) , we have ∂v∂n m ∈ Λ ′ and k ∂v∂n m k Λ ′ ≤ C ( k v k H m + k ∆ v k L (Ω m ) )This is because for any ϕ ∈ D (Ω) we have:( ∂v∂n m , ϕ | Γ ) DD ′ = Z Ω m ∇ ϕ ∇ vdx + Z Ω m ϕ ∆ vdx and this formula shows that the distribution ∂v∂n m is bounded on Λ. Notation 2 :1- For a bounded linear operator T from Λ to Λ ′ , we denote by T ′ its adjointfor the (Λ , Λ ′ ) duality, i.e. ∀ λ ∈ Λ , ∀ η ∈ Λ (
T η, λ ) ΛΛ ′ = ( T ′ λ, η ) ΛΛ ′
2- For a bounded linear operator T from Λ to Λ , we denote by T ∗ the adjointoperator in Λ , i.e. ∀ λ ∈ Λ , ∀ η ∈ Λ (
T η, λ ) Λ = ( η, T ∗ λ ) Λ Proposition 3
For m = 1 , S ′ = S S ′ m = S m roof: by definition of S we have: ∀ λ ∈ Λ , ∀ η ∈ Λ( Sλ, η ) ΛΛ ′ = ( ˜ Sλ, η ) ΛΛ ′ = ( η, λ ) Λ = ( λ, η ) Λ = ( Sη, λ ) Λ , Λ ′ = ( Sη, λ ) Λ , Λ ′ Theorem 3
For m = 1 , let m ′ = 2 , . For all λ ∈ Λ , η ∈ Λ ( S − m S m ′ η, λ ) Λ m = ( η, λ ) Λ m ′ ( S − m S m ′ η, λ ) Λ m ′ = ( S m ′ η, S m ′ λ ) Λ ′ m S − m S m ′ is self adjoint in Λ m and in Λ m ′ ( S − S + S − S ) is selfadjoint in Λ proof:
1- ( S − m S m ′ η, λ ) Λ m = ( λ, S − m S m ′ η ) Λ m = ( S m ′ η, λ ) ΛΛ ′ = ( λ, η ) Λ m ′ = ( η, λ ) Λ m ′
2- We use Proposition 3 to have( S − m S m ′ η, λ ) Λ m ′ = ( S m ′ λ, S − m S m ′ η ) ΛΛ ′ = ( S m ′ η, S − m S m ′ λ ) ΛΛ ′ =( S m S − m S m ′ η, S − m S m ′ λ ) ΛΛ ′ = ( S − m S m ′ λ, S − m S m ′ η ) Λ m = ( S m ′ η, S m ′ λ ) Λ ′ m
3- We use Proposition 3 to have( S − m S m ′ η, λ ) Λ m = ( S m ′ η, λ ) ΛΛ ′ = ( S m ′ λ, η ) ΛΛ ′ = ( η, S − m S m ′ λ ) Λ m = ( η, S − m S m ′ λ ) Λ m On the other hand( S − m S m ′ η, λ ) Λ m ′ = ( S m ′ λ, S − m S m ′ η ) ΛΛ ′ = ( S m ′ η, S − m S m ′ λ ) ΛΛ ′ = ( S − m S m ′ λ, η ) Λ m ′ =( η, S − m S m ′ λ ) Λ m ′
4- follows 3
Corollary 5
Coerciveness: For m = 1 , let m ′ = 2 , . There exists C > such that for all λ ∈ Λ :1- ( S − m S m ′ λ, λ ) Λ m = k λ k m ′ ≥ C k λ k m ( S − m S m ′ λ, λ ) Λ m ′ = k S m ′ λ k ′ m ≥ C k λ k m ′ (( S − S + S − S ) λ, λ ) Λ ≥ (1 + C ) k λ k Proposition 4
For m = 1 , and for any ν ∈ Λ ′ there exists a unique u m ∈ H m such that ∆ u m = 0 in Ω m and ∂u m ∂n m = ν on Γ Moreover k u m k H m = k ν k Λ ′ roof: uniqueness is straightforward, and existence is provided by proposition1 and u m = u S − m ν Moreover isometry of the Riesz representation gives: k u m k H m = k S − m ν k Λ = k ν k Λ ′ Proposition 5
For m = 1 , and for any f ∈ L (Ω m ) there exists a unique u m ∈ H m such that ∆ u m = f and ∂u m ∂n m = 0 on Γ Moreover k u m k H m ≤ C k f k L (Ω m ) proof: Uniqueness is straightforward. For existence let v m ∈ H (Ω m ) be theunique solution of ∆ v m = f . In remark 4 we have shown that ν = ∂v m ∂n m ∈ Λ ′ and k ν k Λ ′ m ≤ C ( k f k L (Ω m ) + k v m k H m ) ≤ C ( k f k L (Ω m ) + k f k H − (Ω m ) ) ≤ C k f k L (Ω m ) The function u m = v m − u S − m ν solves the problem, and we have: k u m k H m ≤ k v m k H m + k u S − m ν k H m ≤ k f k L (Ω m ) + k S − m ν k Λ m ≤ k f k L (Ω m ) + k ν k Λ ′ m ≤ C k f k L (Ω m ) Proposition 6
Let f ∈ L (Ω) and u ∈ H (Ω) be the unique solution of ∆ u = f . For m = 1 , let f m = f | Ω m and g m ∈ H (Ω m ) be the uniquesolution of ∆ g m = f m . Let η m = g m | Γ λ = u | Γ ⇐⇒ ( S + S ) λ = − ( S η + S η )13 roof: the direct implication is stating continuity of u and its normal deriva-tives through Γ. The converse implication states that taking u | Ω m = u λm solves the global problem.We use the same notation as in the previous proposition to state: Theorem 4
Let < θ < with k S − S + S − S k L (Λ) < − θ ) θ . Any sequence ( λ n ) n ⊂ Λ which fulfills λ n +1 = ((1 − θ ) Id − θ S − S + S − S )) λ n − θ S − + S − ))( S η + S η ) do converge in Λ (with geometric rate − θ at least) and its limit is u | Γ .proof: theorem 3 states selfadjointness of S − S + S − S in Λ and theorem5 states coerciveness of S − S + S − S so: k (1 − θ ) Id − θ S − S + S − S ) k L (Λ) =sup Λ | (((1 − θ ) Id − θ ( S − S + S − S )) λ, λ ) Λ |k λ k =sup Λ (((1 − θ ) Id − θ ( S − S + S − S )) λ, λ ) Λ k λ k ≤ − θ − θ C ) < − θ < In the sequel we shall assume connnectedness of the open sets Ω m , m = 1 , j operator We define the linear bounded operator j from Λ to Λ ′ as the composition ofthe bounded linear injectionsΛ ⊂ H (Γ) ⊂ L (Γ) ⊂ H − (Γ) ⊂ Λ ′ It will play a key role for Helmholtz equations. Its properties are summarizedby: 14 roposition 7 .1- j is a compact and one-to-one operator2- For any λ ∈ Λ , j ( λ ) = j ( λ )
3- For m = 1 , ∀ λ ∈ Λ , σ ∈ Λ , ( j ( λ ) , σ ) ΛΛ ′ = ( σ, S − m j ( λ )) Λ m = ( σ, λ ) L (Γ) ∀ λ ∈ Λ , ( j ( λ ) , λ ) Λ , Λ ′ = k λ k L (Γ) j ′ = j
5- For m = 1 , the operator S − m j is selfadjoint in Λ m
6- For m = 1 , the operator jS − m is selfadjoint in Λ ′ m proof: Item 1 comes from the Rellich compactness of the injection from H (Γ)to L (Γ).Item 3 comes from:( σ, S − m j ( λ )) Λ m = R Ω m ∇ u σ ∇ u S − m j ( λ ) = R Γ σ ∂u S − m j ( λ ) ∂n m = R Γ σλ = ( σ, λ ) L (Ω m ) Item 4 follows item 3 because ( j ( λ ) , σ ) ΛΛ ′ = R Γ σλ = ( j ( σ ) , λ ) ΛΛ ′ Item 5 follows items 3 and 4 because( σ, S − m j ( λ )) Λ m = ( j ( λ ) , σ ) ΛΛ ′ = ( j ( σ ) , λ ) ΛΛ ′ = ( S − m j ( σ ) , λ ) Λ m Item 6 follows item 5 because( jS − m µ, ν ) Λ ′ m = ( S − m jS − m µ, S − m ν ) Λ m = ( S − m µ, S − m jS − m ν ) Λ m = ( µ, jS − m ν ) Λ ′ m This paragraph is devoted to the study of the operators m = 1 , S m + iγj ) : Λ −→ Λ ′ related to the Laplace equation, and to the like ( S km + iγj ) operators relatedto the Helmholtz equation. Let γ denote a real number. Proposition 8 .1- ∀ λ ∈ Λ (( S m + iγj ) λ, λ ) ΛΛ ′ = k λ k m + iγ k λ k L (Γ) ( S m + iγj ) has a bouded inverse.3- ∀ λ ∈ Λ ( S m + iγj ) λ = ( S m − iγj ) λ and T γm λ = T − γm ( λ ) proof:
1- is straightforward applying proposition 7 and notation 215- item 1 shows that
Ker ( S m + iγj ) = { } and Im ( S m + iγj ) is closed inΛ ′ . It remains to show that Im ( S m + iγj ) is everywhere dense in Λ ′ . Bypropositions 3 and 7 ∀ λ ∈ Λ , (( S m + iγj ) λ, η ) ΛΛ ′ = 0 = ⇒∀ λ, (( S m + iγj ) η, λ ) ΛΛ ′ = 0 = ⇒ ( S m + iγj ) η = 0 = ⇒ η = 0because ( S m + iγj ) is one to one.3- is straightforward. Proposition 9
For m = 1 , and any ν ∈ Λ ′ there exists a unique u ∈ H m such that ∆ u = 0 in Ω m and ∂u∂n m + iγjρ Γ m u = ν In fact u = u ( S m + iγj ) − ν and k u k H m ≤ C k ν k Λ ′ proof: Uniqueness is straightforward by the Green formula. For existence weapply the previous proposition to get λ ∈ Λ such that ( S m + iγj ) λ = ν , andcheck that u = u λ solves the problem.The following remark will be crucial to prove convergence of domain decom-position algorithms for the Helmholtz equation: Remark 5 with the notation of the preceeding proposition, if ν ∈ L (Γ) then ∂u∂n m ∈ L (Γ) and k ∂u∂n m k L (Γ) ≤ C k ν k L (Γ) proof: ∂u∂n m = ν − iγjρ Γ m u ∈ L (Γ) + Λ ⊂ L (Γ) and k ∂u∂n m k L (Γ) ≤ k ν k L (Γ) + | γ |k ρ Γ m u k L (Γ) ≤ k ν k L (Γ) + | γ |k u k H m ≤ k ν k L (Γ) + C | γ |k ν k Λ ′ ≤ C k ν k L (Γ) roposition 10 For m = 1 , and any f ∈ L (Ω m ) there exists a unique u ∈ H m such that ∆ u = f in Ω m and ∂u∂n m + iγjρ Γ m u = 0 and we have the estimate k u k H m ≤ C k f k L (Ω m ) proof: let v ∈ H (Ω m ) solve ∆ v = f . Let w = u − v . The function w fulfills ∆ w = 0 and ∂w∂n m + iγjρ Γ m w = ∂v∂n m . Remark 4 shows that ∂v∂n m ∈ Λ ′ .The result follows from the previous proposition and the estimate followsestimates in remark 4 and proposition 9. Remark 6
Using the notations of proposition 10 we have, as in remark 5, ∂u∂n m ∈ L (Γ) and the estimate: k ∂u∂n m k L (Γ) ≤ C k f k L (Ω m ) We can now proceed to compute the eigenfrequencies of the local Helmholtzproblems involved in the Domain Decomposition algorithm. For that sake,we will use the following
Notation 3
For m = 1 , we denote by D γm : L (Ω m ) −→ L (Ω m ) f −→ u where u is given by proposition 10. This map has the following properties:
Proposition 11 .1- D γm is a compact operator in L (Ω m )
2- the adjoint map of D γm for the L (Ω m ) scalar product is D − γm
3- we have D γm g = D − γm g and D γm D − γm f = D − γm D γm f ImD γm ⊂ H m ∂∂n m D γm f ∈ L (Γ) 17 roof:
1- translates Rellich compactness of the imbedding of H m in L (Ω m )2- If u = D γm f and v = D − γm g then Green formula gives Z Ω m ugdx − Z Ω m vf dx = Z Γ ∂v∂n m udσ − Z Γ ∂u∂n m vdσ = − iγ Z Γ uvdσ + iγ Z Γ uvdσ = 03- is straightforward4- follows proposition 105- follows remark 6We collect the spectral properties of D γm in the following Proposition 12
We denote by σ the spectrum of an operator and by σ p theset of its eigenvalues. We have for m = 1 , :1- σ ( D γm ) = { } ∪ σ p ( D γm ) µ ∈ σ ( D γm ) ⇐⇒ µ ∈ σ ( D − γm )
3- For any f ∈ L (Ω m ) , if u = D γm f then ( D γm f, f ) L (Ω m ) = − Z Ω m |∇ u | dx + iγ Z Γ | ρ Γ m u | dσ
4- there exists a constant c such that If µ ∈ σ ( D γm ) , µ = 0 , then Re µ ≤ , γIm µ ∈ R + , | Im µ | ≤ c | γ | | Re µ |
5- If γ = 0 then σ ( D γm ) ∩ R = { } proof:
1- follows compactness of D γm asserted in proposition 11.2- is obvious by taking the complex conjugate of the eigenfunction associatedwith µ .3- ( D γm f, f ) L (Ω m ) = Z Ω m uf dx = Z Ω m u ∆ udx = − Z Ω m |∇ u | dx + Z Γ ρ Γ m u ∂u∂n m | Γ dσ = − Z Ω m |∇ u | dx + iγ Z Γ | ρ Γ m u | dσ
4- if µ is an eigenvalue of D γm with associated eigenfunction f , and u = D γm f ,then µ Z L (Ω m ) | f | dx = − Z Ω m | u | dx + iγ Z Γ | ρ Γ m u | dσ c the constant of continuity of the trace operator ρ Γ m from H m to L (Γ).5- If µ = 0 is a real eigenvalue of D γm with associated eigenfunction f , then theprevious formula shows that ρ Γ m u = 0 and, because u ∈ Im ( D γm ), ∂u∂n m | Γ = − iγρ Γ m u = 0. On the other hand, the equality D γm f = µf translates to∆ u = µ u . Because the Laplace operator is hyperbolic in the direction n m ,and both data on Γ are zero, this implies u = 0 on a neighbourhood of Γ.Solutions of elliptic equations being analytic, and Ω m being connected, thisimplies u = 0 on Ω m . Then f = 0, which contradicts the assumption on f as an eigenfunction. Theorem 5
Let k ∈ R, k = 0 . Let γ ∈ R, γ = 0 . For m = 1 , :1- for any f in L (Ω m ) and ν ∈ Λ ′ , there exits a unique u ∈ H m such that: ∆ u + k u = f ∂u∂n m − iγjρ Γ m u = ν
2- we have the estimate k u k H m ≤ C ( k f k L (Ω m ) + k ν k Λ ′ )
3- if ν ∈ L (Γ) then ∂u∂n m ∈ L (Γ) and we have the estimate: k ∂u∂n m k L (Γ) ≤ C ( k f k L (Ω m ) + k ν k L (Γ) ) proof:
1- Let λ = ( S m − iγj ) − ν and let v = u − u λ . Then:∆ u + k u = f and ∂u∂n m − iγjρ Γ m u = ν ⇐⇒ ∆ v = f − k u λ − k v and ∂v∂n m − iγjρ Γ m v = 0 ⇐⇒ v = D γm ( f − k u λ − k v ) ⇐⇒ D γm v + k − v = k − D γm ( f − k u λ )The previous proposition 12 shows that − k − / ∈ σ ( D γm ) so this problem iswell-posed2- we have the estimate: k v k L (Ω) ≤ C ( k − k D γm f k L (Ω) + k D γm u λ k L (Ω) ) ≤ C ( k − k f k L (Ω) + k u λ k L (Ω) )19o k u k L (Ω) ≤ C ′ ( k − k f k L (Ω) + k u λ k L (Ω) ) ≤ C ′ ( k − k f k L (Ω) + k u λ k H m ) ≤ C ′ ( k − k f k L (Ω) + k λ k Λ ) ≤ C ′ ( k − k f k L (Ω) + k ν k Λ ′ )We write ∆ u = f − k u ∂u∂n m | Γ − iγjρ Γ m u = ν and the H m estimate follows proposition 9 and proposition 10.3- If ν ∈ L ( γ ) we write again∆ u = f − k u ∂u∂n m | Γ − iγjρ Γ m u = ν and the estimate follows remark 5 and remark 6. We now define the building blocks of the intertwinning operator on the ficti-tious boundary Γ: the Despres operators.
Definition 1 : Let k ∈ R , γ = 0 , γ ∈ R . For any ν ∈ Λ ′ , let u ∈ H m bethe unique solution, given by Theorem 5, of the following equation on Ω m , m = 1 , : ∆ u + k u = 0 , ∂u∂n m − iγjρ Γ m u = ν on Γ Let ˜ P γm be the linear bounded operator in Λ ′ defined by: ˜ P γm ν = ∂u∂n m + iγjρ Γ m u on Γ Remark 7
Boundedness of ˜ P γm follows Remark 3 and Proposition 7. Proposition 13 : We obviously have ˜ P γm ˜ P − γm = ˜ P − γm ˜ P γm = Id Λ ′ otation 4 Let ˜ A γ denote the linear bounded operator in Λ ′ × Λ ′ given by: ˜ A γ = − ˜ P γ − ˜ P γ ! Remark 8 : The inverse of ˜ A γ in Λ ′ × Λ ′ is the bounded linear operatorgiven by: ( ˜ A γ ) − = − ˜ P − γ − ˜ P − γ ! In order to use conservation of energy, and to gain compactness, we useTheorem 5 to introduce:
Notation 5
For γ = 0 and k ∈ R , let the bounded operator in L (Γ) denotedby P γm be the restriction of ˜ P γm to L (Γ) . Let the bounded operator in L (Γ) × L (Γ) denoted by A γ be the restriction of ˜ A γ to L (Γ) × L (Γ) . Conservation of energy fluxes through Γ reads:
Proposition 14
Let γ = 0 and m = 1 , .(i) P γm is an isometry in L (Γ) : ∀ ν ∈ L (Γ) , k ν k L (Γ) = k P γm ν k L (Γ) (ii) A γ is an isometry in L (Γ) × L (Γ) : ∀ ( ν, η ) ∈ L (Γ) × L (Γ) , k ( ν, η ) k L (Γ × L (Γ)) = k A γ ( ν, η ) k L (Γ) × L (Γ) proof: For ν ∈ L (Γ) let u ∈ H m solve by Theorem 5 the following equation:∆ u + k u = 0 in Ω m , ∂u∂n m − iγjρ γm u = ν ΓMultiplying by u the equation fulfilled by u and integrating on Ω m gives Z Ω m |∇ u | dx − k Z Ω m | u | dx = Z Γ ∂u∂n m udσ Taking the imaginary part gives I m Z Γ ∂u∂n m udσ = 021he result follows integration of the following identity on Γ: | ∂u∂n m + iγjρ γm u | − | ∂u∂n m − iγjρ γm u | = 4 iγ I m u ∂u∂n m An important consequence of this property will be crucial in the next section:
Corollary 6 : For m = 1 , and γ = 0 ,(i) P γm is a normal operator in L (Γ) (ii) A γ is a normal operator in L (Γ) × L (Γ) In the preceeding section we proved that the Despres operator P γm , ( m = 1 , γ = 0) is a bijective isometry in L (Γ), and consequently a normaloperator in L (Γ). It follows that its spectrum is a subset of the unit circlein the complex plane. We now investigate this spectrum more accurately. Definition 2
Let γ = 0 and m = 1 , . Let C γm be the operator in L (Γ) given by: ∀ ν ∈ L (Γ) , C γm ν = j ρ Γ m u where u ∈ Λ is the solution given by Theorem 5 of the equation: (∆ + k ) u = 0 in Ω m , ∂u∂n m − iγρ Γ m u = ν on ΓCompactness of the injection j from Λ to L ( γ ) gives: Proposition 15 :(i) For m = 1 , and γ = 0 , P γm = I + 2 iγC γm (ii) C γm is a normal and compact operator in L ( γ ) Notation 6
Let(i) Σ γm denote the spectrum of P γm (ii) Σ Dirm denote the sequence of eigenvalues of the Laplace operator on Ω m with Dirichlet boundary condition on ∂ Ω m , i.e. − k / ∈ Σ Dirm if and only if the following problem is well posed: (∆ + k ) u = f ∈ L (Ω m ) , u ∈ H (Ω m )22e have: Proposition 16 : Let γ = 0 , and m = 1 , .(i) belongs to Σ γm (ii) is an eigenvalue of P γm if and only if − k ∈ Σ Dirm proof: (i) Because C γm is a compact normal operator (Proposition 15) in L (Ω m ), itsspectrum is a sequence of eigenvalues and its limit zero. This implies that 1is in the closure of Σ γm , which is closed.(ii) If − k ∈ Σ Dirm , then there exist an eigenfunction ϕ k of the Laplace oper-ator such that ∆ ϕ k + k ϕ k = 0 and ϕ k | Γ = 0Let ν k = ∂ϕ k ∂n m | Γ . We have ν k = 0 or else ϕ k solving an elliptic equation,condition ϕ k | Γ = 0 and connectedness of Ω m would imply ϕ k = 0 on Ω m ,which contradicts the fact that ϕ k is an eigenfunction. Obviously ν k is aneigenvector of P γm for the eigenvalue 1.Conversely, if 1 is an eigenvalue of P γm , with eigenvector ν = 0, then thereexist u in H m such that∆ u + k u = 0 and ν = ∂u∂n m | Γ − iγjρ Γ m u = ∂u∂n m | Γ + iγjρ Γ m u This implies ρ Γ m u = 0, which added to the fact that u ∈ H m implies u ∈ H (Ω m ). So u is an eigenfunction of the Laplace operator for the eigenvalue − k , provided u is not identically zero. And this is ruled out because ∂u∂n m | Γ = ν = 0The following Theorem gives a complete spectral description of the Despresoperators: Theorem 6 : Assume that − k / ∈ Σ Dirm . For γ = 0 and m = 1 , :(i) Σ γm = { } ∪ ( e iσ nm ) n ∈ N where ( σ nm ) n ∈ N is a sequence of numbers with σ nm ∈ R , σ nm = 0 , and σ nm −→ when n −→ ∞ .(ii) For each n ∈ N , e iσ nm is an eigenvalue of P γm , with finite multiplicity.(iii) L (Γ) is the Hilbert direct sum of the eigenspaces associated with theeigenvalues e iσ nm roof: C γm is a normal and compact operator in L (Γ) (proposition 15).Its kernel is trivial (proposition 16). The diagonalization theorem gives asequence ( λ nm ) n ∈ N , ( λ nm = 0), for its eigenvalues (they have finite multiplic-ity), with limit zero, and L (Γ) is the Hilbert direct sum of the associatedeigenspaces. The diagonalization of P γm = I + 2 iγC γm follows, with eigen-values (1 + 2 iγλ nm ) n ∈ N , (1 + 2 iγλ nm = 1). Proposition 14 implies that theseeigenvalues have modulus one: we set 1 + 2 iγλ nm = e iσ nm . P γ P γ and P γ P γ Properties of the intertwinning operator A γ rely heavily on the spectral prop-erties of P γ P γ and P γ P γ that we investigate now.We first list obvious properties which follow from the previous section: Proposition 17 :(i) P γ P γ and P γ P γ are isometric bijections in L ( γ ) .(ii) P γ P γ and P γ P γ are normal operators in L ( γ ) .(iii) P γ P γ − I and P γ P γ − I are compact operators in L ( γ ) .Proof: (i) follows the fact that P γ and P γ are isometric bijections in L (Γ).(ii) follows (i)(iii) follows proposition 15 through: P γ P γ = ( I + 2 iγC γ )( I + 2 iγC γ ) = I + 2 iγC γ + 2 iγC γ − γ C γ C γ An important property that we sall need is the spectral status of 1:
Proposition 18 : 1 is not an eigenvalue of P γ P γ or P γ P γ in L (Γ) .Proof: By symmetry, it is enough to prove it for P γ P γ . Let ν ∈ L (Γ) be suchthat P γ P γ ν = ν . This translates to the existence of u ∈ H and u ∈ H satisfying:∆ u + k u = 0 in Ω ; ∂u ∂n − iku = ν, ∂u ∂n + iku = P γ ν on Γ24 u + k u = 0 in Ω ; ∂u ∂n − iku = P γ ν, ∂u ∂n + iku = ν on ΓThis implies that on Γ these functions fulfill: ∂u ∂n − iku − ∂u ∂n − iku = 0; ∂u ∂n + iku − ∂u ∂n + iku = 0Adding and substracting gives: u = − u and ∂u ∂n = ∂u ∂n We define u on Ω as u | Ω = u and u | Ω = − u . It solves the Helmholtzequation on Ω and Ω , its has no jump accross Γ, neither has its normalderivative. So it solves Helmholtz equation on Ω. Moreover u | ∂ Ω = 0. As-sumption (A) gives u = 0, so u = 0 and u = 0; and ν = 0 follows. Proposition 19 :(i) The spectrum of P γ P γ in L (Γ) is { } ∪ ( e iτ n ) n ∈ N , where ( τ n ) n ∈ N is aninfinite sequence of real numbers, τ n = 0 , and τ n −→ when n −→ ∞ .(ii) ( e iτ n ) n ∈ N is the set of eigenvalues of P γ P γ . They have finite multiplicity.If we denote by E n the eigenspace associated with e iτ n , then L (Γ) is theHilbert direct sum of the subspaces ( E n ) n ∈ N (iii) P γ P γ has the same properties, and we set the obvious notations: ( e iτ n ) n ∈ N for eigenvalues and ( E k ) n ∈ N for eigenspaces.Proof: The operator P γ P γ − I is a normal compact operator (proposition17). So by the diagonalization theorem its spectrum is the union of { } andan infinite sequence of eigenvalues with finite multiplicity ( t n ) n ∈ N , ( t n = 0).Zero is not an eigenvalue of P γ P γ − I (proposition 18). So the whole set ofeigenvalues is ( t n ) n ∈ N . If E n denotes the eigenspace associated with t n , then L (Γ) is the Hilbert direct sum of ( E n ) n ∈ N . By proposition 17 we know that P γ P γ is an isometry in L (Γ), so | t n | = 1, and we write it: t n = e iτ n .The theorem translates proven properties of t n into properties of τ n .In order to study the relationship between ( τ n ) n ∈ N and ( τ n ) n ∈ N , and be-tween ( E n ) n ∈ N and ( E n ) n ∈ N , we prove the following lemmi:25 emma 1 Let γ = 0 . For m = 1 , and any ν ∈ L (Γ) : P γm ν = P − γm ν Proof:
Let m = 1 or 2. By definition of P γm there exists u m ∈ H m with:∆ u m + k u m = 0; ∂u m ∂n m | Γ − iγjρ Γ m u m = ν ; ∂u m ∂n m | Γ + iγjρ Γ m u m = P γm ν Taking the complex conjugate of these equalities gives u m ∈ H m such that∆ u m + k u m = 0; ∂u m ∂n m | Γ + iγjρ Γ m u m = ν ; ∂u m ∂n m | Γ − iγjρ Γ m u m = P γm ν which by definition of P γm writes P − γm ν = P γm ν Lemma 2
Let γ = 0 (i) If λ is an eigenvalue of P γ P γ (resp P γ P γ ) for the eigenvector ν then itis an eigenvalue of P γ P γ (resp P γ P γ ) with associated eigenvector ν (ii) For all n ∈ N , τ n = τ n ( mod π ) ; we denote it by τ n (iii) If we denote by C the set of complex conjugates of distributions in a set C , then, for any n ∈ N , E n = E n and E n = E n Proof: (i) If P γ P γ ν = λν then P γ P γ ν = λν which by lemma 1 writes P − γ P − γ ν = λν , which implies, by proposition 13, P γ P γ ν = λ ν = λν because λ = λ byproposition 19.(ii) follows (i) and a renumbering.(iii) follows (i) because it gives: E n ⊂ E n and E n ⊂ E n but then E n ⊂ E n ⊂ E k and E n ⊂ E n ⊂ E n which gives E n = E n and E n = E n Lemma 3 : Let γ = 0 . For any n ∈ N , P γ E n = E n and P γ E n = E n roof: Let ν = 0 ∈ E n then P γ P γ ν = e iτ n ν so P γ P γ P γ ν = e iτ n P γ ν whichproves that P γ ν ∈ E n ( P γ ν = 0 because P γ is bijective (proposition 13)).This writes P γ E n ⊂ E n . Proposition 19, lemma 2 and invertibility of P γ (proposition 13) give dimE n = dimE n = dimP γ E n so P γ E n = E n .The following algebraic property and its consequences on the eigenprojectors(next theorem) are a key for understanding the geometric properties of theintertwinning operator: Lemma 4 : Let γ = 0 . For any µ / ∈ { } ∪ ( e iτ n ) n ∈ N : ( P γ P γ − µI ) − P γ = P γ ( P γ P γ − µI ) − ( P γ P γ − µI ) − P γ = P γ ( P γ P γ − µI ) − Proof: we prove the second assertion, using resolvant identity:( P γ P γ − µI ) − P γ = ( P γ P γ − µI ) − P γ ( P γ P γ − µI )( P γ P γ − µI ) − =[[ I + µ ( P γ P γ − µI ) − ] P γ − µ ( P γ P γ − µI ) − P γ ]( P γ P γ − µI ) − = P γ ( P γ P γ − µI ) − Lemma 5
Let γ = 0 . For any n ∈ N , if Π n (resp. Π n )denotes the spectralprojector of the operator P γ P γ (resp. P γ P γ ) on the eigenspace E n (resp. E n ) then we have P γ Π n = Π n P γ and P γ Π n = Π n P γ Proof:
By symmetry it is enough to prove the first formula. Let C n denote apositively oriented curve in the complex plane, which winds one time aroundthe eigenvalue e iτ n , and none around any other eigenvalue, then the Dunfordintegral representation formula gives:Π n = − iπ Z C n ( P γ P γ − µI ) − dµ and Π n = − iπ Z C n ( P γ P γ − µI ) − dµ The previous lemma gives the following: P γ Π n = − iπ Z C n P γ ( P γ P γ − µI ) − dµ = − iπ Z C n ( P γ P γ − µI ) − P γ dµ = Π n P γ .6 Spectral Properties of A γ We recall that A γ = − P γ − P γ ! and that this operator is a bijective isometry in L (Γ) × L (Γ) Theorem 7 : Let γ = 0 .(i) If λ / ∈ {± } ∪ ( ± e i τn ) n ∈ N then λ belongs to the resolvant set of A γ and ( A γ − λI ) − = λ ( P γ P γ − λ I ) − − ( P γ P γ − λ I ) − P γ − ( P γ P γ − λ I ) − P γ λ ( P γ P γ − λ I ) − ! (ii) For any n ∈ N , ± e i τn is an eigenvalue of A γ with associated eigenspace: F ± n = { ( µ , ∓ e − iτn P γ µ ); µ ∈ E n } (2) and associated eigenprojector : P ± n = Π n ∓ e − i τn P γ Π n ∓ e − i τn P γ Π n
12 12 Π n ! (3) (iii) {± } belong to the spectrum of A γ and are not eigenvalues of A γ (iv) ( F ± n ) n ∈ N ; ± is an orthogonal family of subspaces and we have the Hilbertdecomposition L (Γ) × L (Γ) = ( ⊕ ∞ F + n ) ⊕ ( ⊕ ∞ F − n ) (v) The following series are strongly convergent in L ( L (Γ) × L (Γ)) : I = ∞ X P + n + ∞ X P − n A γ = ∞ X e i τn P + n − ∞ X e i τn P − n Proof: (i) let λ / ∈ {± } ∪ ( ± e i τn ) n ∈ N ( A γ − λI ) is injective: let ( ϕ, ψ ) ∈ L (Γ) × L (Γ) be such that( A γ − λI ) ϕψ ! = ! P γ ψ + λϕ = 0 and P γ ϕ + λψ = 0which implies P γ P γ ψ − λ ψ = 0 and P γ P γ ϕ − λ ϕ = 0This implies ϕ = ψ = 0 by proposition 19 and lemma 2.( A γ − λI ) is surjective: For any ( ξ, η ) ∈ L (Γ) × L (Γ) let: ϕ = ( P γ P γ − λ I ) − ( λξ − P γ η ) and ψ = ( P γ P γ − λ I ) − ( λη − P γ ξ )We have, by lemma 4 P γ ϕ + λψ = P γ ( P γ P γ − λ I ) − ( λξ − P γ η ) + λ ( P γ P γ − λ I ) − ( λη − P γ ξ )= ( P γ P γ − λ I ) − ( λP γ ξ − P γ P γ η ) + λ ( P γ P γ − λ I ) − ( λη − P γ ξ )= ( P γ P γ − λ I ) − ( − P γ P γ η + λ η ) = − η Similarly P γ ψ + λϕ = − ξ These two equalities write( A γ − λI ) ϕψ ! = ξη ! So surjectivity is proven. These expressions for ( ϕ, ψ ) give the formula forthe resolvent of A .(ii) By definition of E n we have for any µ ∈ E n : A γ ( µ , ∓ e − i τn P γ µ ) = ( ± e − i τn P γ P γ µ , − P γ µ ) =( ± e i τn µ , − P γ µ ) = ± e i τn ( µ , ∓ e − i τn P γ µ )Because E n = { } , this proves that ± e i τk is an eigenvalue of A γ . Thisproves moreover that F ± n is a subset of the eigenspace of A γ associated withthe eigenvalue ± e i τn .On the other hand, if ( ξ, η ) is an eigenvector of A γ for the eigenvalue ± e i τn then − P γ η = ± e i τn ξ and − P γ ξ = ± e i τn η P γ P γ ξ = ∓ e i τn P γ η = e iτ n ξ so ξ ∈ E n and η = ∓ e − i τn P γ ξ This completes the caracterisation of the eigenspace.We compute now the eigenprojector: for this sake, we make a choice of abranch for √ z . We take a positively oriented curve C ± n in the complex planewhich winds one time around ± e i τn and not around ∓ e i τn nor does it windaround any ± e i τn ′ for n ′ = n . Let D n be the image of C ± n by the function z → z . D n winds one time around e iτ n and does not wind around e iτ n ′ for n ′ = n . Let D ′ n wind one time around e iτ n , lying in the interior set delimitedby D n .The eigenprojector is given by the Dunford formula: P ± n = − iπ Z C ± n ( A γ − λI ) − dλ Using the representation formula given by (i) for ( A γ − λI ) − leads to computeintegrals of two different types:For the first type it is straightforward and gives: − iπ Z C ± n λ ( P γ P γ − λ I ) − dλ = − iπ Z D n ( P γ P γ − λI ) − dλ = 12 Π n For the second type, we first use the resolvant identity to have: − iπ Z D n ( P γ P γ − λI ) − Π n dλ √ λ = − iπ Z D n ( P γ P γ − λI ) − dλ √ λ − iπ Z D ′ n ( P γ P γ − µI ) − dµ = − iπ − iπ Z D n Z D ′ n ( P γ P γ − λI ) − ( P γ P γ − µI ) − dλdµ √ λ = − iπ − iπ Z D n ( P γ P γ − λI ) − Z D ′ n dµλ − µ ! dλ √ λ − − iπ − iπ Z D ′ n ( P γ P γ − µI ) − Z D n dλ √ λ ( λ − µ ) ! dµ =30 iπ Z D ′ n ( P γ P γ − µI ) − dµ √ µ We compute now the second type of integral using this equality, propertiesof Π n , and lemma 5 to have: − − iπ Z C ± n ( P γ P γ − λ I ) − P γ dλ = − − iπ Z D n ( P γ P γ − λI ) − dλ √ λ ! P γ = − − iπ Z D n ( P γ P γ − λI ) − Π n dλ √ λ ! P γ = − − iπ Z D n e iτ n − λI ) Π n dλ √ λ ! P γ = ∓ e − i τn Π n P γ = ∓ e − i τn P γ Π n (iii) ± ± e − iτn ) n ∈ N so they belongto the spectrum of A . These values are not eigenvalues, or else 1 is aneigenvalue of P γ P γ and P γ P γ , which is ruled out by proposition 18(iv) and (v) Assertions (i), (ii) and (iii) prove that the spectrum of A γ is {± } ∪ ( ± e i τn ) n ∈ N . Normality of A γ (corollary 6) implies orthogonality ofthe family ( F ± n ) n ∈ N , and gives the decomposition of I and A γ as series of theeigenprojectors ( P ± n ) Remark 9 : Notice that the expression of F ± n in (ii) of the previous propo-sition is symmetric: in fact we have { ( µ , ∓ e − i τn P γ µ ); µ ∈ E n } = { ( ∓ e − i τn P γ µ ′ , µ ′ ); µ ′ ∈ E n } this is because P γ E n = E n following lemma 3, so if we set µ = ∓ e − i τn P γ µ ′ it ensures µ ∈ E n if µ ′ ∈ E n . Moreover, by definition of µ ′ we have: ∓ e − i τn P γ µ = e − iτ n P γ P γ µ ′ = µ ′ Domain Decomposition algorithm for theHelmholtz equation
Proposition 20 : Let k ∈ R , k = 0 . For any f ∈ L (Ω) , let u ∈ H (Ω) bethe unique solution of the Helmholtz equation ∆ u + k u = f Let γ ∈ R, γ = 0 . For m = 1 , , let v m ∈ H m solve the equation ∆ v m + k v m = f | Ω m and ∂v m ∂n m − iγjρ Γ v m = 0 on Γ Let ν m = 2 iγjρ Γ v m and η = ( P γ ν , P γ ν ) Then the equation in L (Γ) × L (Γ)( A γ − Id ) π = η has a unique solution: π = ( π , π ) with π m = ∂u∂n m + iγjρ Γ u − iγjρ Γ v m proof: First notice that the assumption f ∈ L (Ω) and assumption (A) on k imply u ∈ L (Ω), so ∆ u ∈ L (Ω). Regularity of ∂ Ω enables the use of classicalregularity results ( [Ag] ) for solutions of elliptic boundary problems to have u ∈ H (Ω), hence ∂u∂n m ∈ H (Γ) ⊂ L (Γ). This proves that π m ∈ L (Γ).Let w m = u | Ω m − v m . Then π m = ∂u∂n m + iγjρ Γ u − iγjρ Γ v m = ∂w m ∂n m + iγjρ Γ w m Because the function w m fulfills∆ w m + k w m = 0 , w m ∈ H m , ∂w m ∂n m − iγjρ Γ w m = ∂u∂n m − iγjρ Γ u on Γ32ne has P γm ( ∂u∂n m − iγjρ Γ u ) = π m so, if m ′ = 2 , m = 1 , π m = − P γm ( ∂u∂n m ′ + iγjρ Γ u ) = − P γm ( π m ′ + ν m ′ )This writes π = A γ π − η Uniqueness follows from theorem 7 (iii).
Proposition 21 : Let k ∈ R , k = 0 , γ ∈ R, γ = 0 . For f ∈ L (Ω) and for m = 1 , , let v m ∈ H m solve the equation ∆ v m + k v m = f | Ω m and ∂v m ∂n m − iγjρ Γ v m = 0 on Γ Let ν m = 2 iγjρ Γ v m and η = ( P γ ν , P γ ν ) Let π = ( π , π ) ∈ L (Γ) × L (Γ) solve the equation: ( A γ − Id ) π = η and let u m ∈ H m solve the equation ∆ u m + k u m = f | Ω m and ∂u m ∂n m + iγjρ Γ u m = π m + ν m on Γ Then u given by u | Ω m = u m solve the Helmholtz equation ∆ u + k u = f u ∈ H (Ω) proof: By definition of u m and v m one has:∆( u m − v m ) + k ( u m − v m ) = 0 , u m − v m ∈ H m and ∂ ( u m − v m ) ∂n m + iγjρ Γ ( u m − v m ) = π m on Γ33his implies through proposition 13:( P γm ) − π m = P − γm π m = ∂ ( u m − v m ) ∂n m − iγjρ Γ ( u m − v m ) = ∂u m ∂n m − iγjρ Γ u m Because π = A γ π − η this implies (with m ′ = 2 , m = 1 , ∂u m ∂n m − iγjρ Γ u m = − π m ′ − ν m ′ = − ∂u m ′ ∂n m ′ − iγjρ Γ u m ′ Adding and substracting these equalities gives: ∂u m ∂n m = − ∂u m ′ ∂n m ′ and ρ Γ u m = ρ Γ u m ′ These jump conditions through Γ imply that ∆ u + k u = f on Ω, and u fulfills the Dirichlet boundary condition on ∂ Ω because u m ∈ H m . Remark 10
Theorem 7 shows that the problem ( ˜ A − Id ) π = η is ill-posed for η ∈ L (Γ) × L (Γ) . Proposition 20 shows that if η has the spe-cific form given through the domain decomposition setting for the Helmholtzequation, the equation ( ˜ A − Id ) π = η do have a solution, (and this solutionis unique by Theorem 7). Proposition 21 shows that this solution providesthe solution of the Helmholtz equation. θ -algorithm for Helmholtzequation Let f ∈ L (Ω). Let u ∈ H fulfill the non-dissipating Helmholtz equation∆ u + k u = f in Ω. The classical algorithm used, (for dissipating cavities withSommerfeld-like boundary condition), to solve by a domain decompositiontechnique the Helmholtz equation ( [B],[BD],[D1],[D2],[CGJ] ) writes, inthe non-dissipating case that discussed here, as follows: for any π = ( π , π )given in L (Γ) × L (Γ) let π p +1 = θπ p + (1 − θ ) A γ π p − (1 − θ ) η η = ( P γ ν , P γ ν ) with ν m = 2 iγjρ Γ v m for v m ∈ H m solving∆ v m + k v m = f | Ω m and ∂v m ∂n m − iγjρ Γ v m = 0 on ΓIt is straightforward to translate the θ -algorithm in a PDE setting: one usetheorem 5 to get the (unique) function w nm ∈ H m such that∆ w pm + k w pm = 0 and π pm = ∂w pm ∂n m + iγjρ Γ w pm In these w p = ( w p , w p ) variables the θ -algorithm becomes: for m = 1 and 2(resp. m ′ = 2 and 1)∆ w p +1 m + k w p +1 m = 0 , w p +1 m ∈ H m ∂w p +1 m ∂n m − iγjρ Γ w p +1 m = θ [ ∂w pm ∂n m − iγjρ Γ w pm ] − (1 − θ )[ ∂w pm ′ ∂p m ′ + iγjρ Γ w pm ′ + ν m ′ ] on ΓFor practical use in computing codes, one writes this algorithm in the u n =( u n , u n ) variables with u pm = w pm + v m and gets:∆ u p +1 m + k u p +1 m = f | Ω m , u p +1 m ∈ H m ∂u p +1 m ∂n m − iγjρ Γ u p +1 m = θ [ ∂u pm ∂n m − iγjρ Γ u pm ] − (1 − θ )[ ∂u pm ′ ∂n m ′ + iγjρ Γ u pm ′ ] on Γ θ -algorithm Notice that if the sequence ( π p ) p ∈ N has a limit π ∞ in L (Γ) × L (Γ) thencontinuity of A γ gives: ( A γ − Id ) π ∞ = η and π ∞ provides the solution of the Helmholtz equation on Ω as stated inproposition 21.Alternatively, a way to solve the Helmholtz equation on Ω through solvingHelmholtz equations on Ω m , ( m = 1 , π p ) p ∈ N in L (Γ) × L (Γ) implies convergence in Λ ′ × Λ ′ , and theorem 5 showsthat the sequence ( w p ) p ∈ N = ( ( w p , w p ) ) p ∈ N has a limit in H × H , whichimplies convergence of the sequence ( u p ) p ∈ N = ( ( u p , u p ) ) p ∈ N in H × H .35roposition 21 shows that its limit u ∞ = ( u ∞ , u ∞ ) provides the solution ofthe Helmholtz equation, through.( u ∞ , u ∞ ) = ( u | Ω , u | Ω )Here are the convergence results for the θ -algorithm. We begin with a nega-tive result: Proposition 22 If θ = 0 then the sequence ( π p ) p ∈ N has no limit in L (Γ) × L (Γ) unless if its initial value fulfills ( A γ − Id ) π = η . Written for the ( u p ) p ∈ N sequence, this translates to u = u | Ω and u = u | Ω where u ∈ H (Ω) solves ∆ u + k u = f in Ω proof: for θ = 0 one has π p − π p − = A γ ( π p − − π p − ) and proposition 14gives ∀ p k π p − π p − k L (Γ) × L (Γ) = k π p − − π p − k L (Γ) × L (Γ) This prevents convergence unless if π = π , i.e. π fulfills π = A γ π − η i.e. unless u = u | Ω and u = u | Ω by proposition 21. Remark 11 If θ = 0 the sequence ( u p ) p ∈ N may have a limit in H × H even if u = u | Ω or u = u | Ω . This is because convergence of ( u p ) p ∈ N in H × H implies convergence of ( ∂u p ∂n + iγjρ Γ u p , ∂u p ∂n + iγjρ Γ u p ) in Λ ′ × Λ ′ , i.e.convergence of ( ∂w p ∂n + iγjρ Γ u p , ∂w p ∂n + iγjρ Γ u p ) in Λ ′ × Λ ′ , i.e. convergence of ( π p ) p ∈ N in Λ ′ × Λ ′ , which do not contradict divergence in L (Γ) × L (Γ) . We now turn to the main result:
Theorem 8 : For f ∈ L (Ω) and k ∈ R let u ∈ H (Ω) solve ∆ u + k u = f .Let γ = 0 , γ ∈ R . Then for any < θ < :(i) the sequence ( π p ) p ∈ N given by the θ -algorithm converge in L (Γ) × L (Γ) to π u = ( π u , π u ) with: π u = ∂u∂n + iγjρ Γ u and π u = ∂u∂n + iγjρ Γ u (ii) the sequence ( u p ) p ∈ N given by the θ -algorithm converge in H × H to ( u | Ω , u | Ω ) (iii) There is no uniform geometric rate of convergence. roof: (i) let π u = ( π u , π u ) with: π u = ∂u∂n + iγjρ Γ u and π u = ∂u∂n + iγjρ Γ u Nullity of jumps of u and its normal derivatives through Γ gives π u = Aπ u + η This implies π p − π u = θ [ π p − − π u ] + (1 − θ ) A [ π p − − π u ]We use eigenprojectors of A γ given by theorem 7 and denote by: δ pn, ± = P ± n ( π p − π u )Completeness of the set of orthogonal eigenprojectors ( P ± n ) n, ± proved intheorem 7 gives: π p − π u = X n δ pn, + + X n δ pn, − Decomposition of L (Γ) × L (Γ) by eigenspaces of A γ writes for successiveterms of the θ − algorithm sequence as follows: δ pn, ± = [ θ ± (1 − θ ) e i τn ] δ p − n, ± This implies: k δ pn, ± k L ( γ ) × L ( γ ) = [1 − θ (1 − θ )(1 ∓ cos τ n p k δ n, ± k L ( γ ) × L ( γ ) and orthogonality of the eigenprojectors writes: k π p − π u k L ( γ ) × L ( γ ) = X n, ± [1 − θ (1 − θ )(1 ∓ cos τ n p k δ n, ± k L ( γ ) × L ( γ ) Assumption 0 < θ < τ n = 0mod 2 π ) imply 0 < − θ (1 − θ )(1 ∓ cos τ n < π p L (Γ) × L (Γ) −→ π u π p to π u in L (Γ) × L (Γ)implies its convergence in Λ ′ × Λ ′ which implies convergence of the relatedsequence w p in H × H , and accordingly convergence of u p to ( u | Ω , u | Ω )Assertion (iii) is obvious by taking initial data for π p in the n − th eigenspaceof A γ and notice that τ n −→ [AG] Agmon S.. Lectures on elliptic boundary value problems. Van Nos-trand - Math Studies n2 - 1965 [B] Benamou J.D.
A domain decomposition method for the optimal controlof system governed by the Helmholtz equation. -Cannes-Mandelieu -(1995) [BD] Benamou J.D. -Despres B.
A domain decomposition method forthe Helmholtz equation and related optimal control problems.
J. of Comp.Physics vol 136 (1995) pp 68-82 [CGJ] Collino F. - Ghanemi S. - Joly P.
Domain decomposition methodfor harmonic wave propagation: a general presentation
Rapport de rechercheINRIA 3473 (1998) [D1] Despres B.
Domain decomposition method and the Helmholtz prob-lem
Math. and num. aspects of waves propagation phenomena. SIAM ed (1993) [D2] Despres B.
Domain decomposition method and the Helmholtz prob-lem (part ii) (1995) [FMQT] Faccioli E. - Maggio F. - Quarteroni A. - Tagliani A.
Spec-tral domain decomposition methods for the solution of elastice and acousticwave equation.
Geophysics 61 (1996) pp 1160-1174 [FQZ] Funaro D. - Quarteroni A. - Zanolli P.
An iterative procedurewith interface relaxation for domain decomposition methods
SIAM J. Numer.Anal.
25 (1988) pp 1213-1236 38
L] Lions P.L.
On the Schwarz alternating method I.
First InternationalSymposium on domain decomposition methods for PDE (1988) R.Glowinskiet als eds. SIAM.
Philadelphia. pp 1-42 [PW] Proskwowski W. - Widlund O.
On the numerical solutions ofHelmholtz’s equation by the capacitance matrix method
Math. Comp. (1976)vol 30 pp 433-468 [QV] Quarteroni A. - Valli A.
Domain decomposition methods for PDE.