Numerical computation of elastoacoustic vibrations

Numerical computation of elastoacoustic vibrations with interface damping Alfredo Berm´ udez ∗ and Rodolfo Rodr´ıguez †

Abstract. In this paper we introduce and analize a finite element method to compute the vibrations of a coupled fluid-solid system subjected to an external harmonic excitation. The effect of a thin layer of noise damping material between the fluid and the solid is taken into account by relaxing the interface kinematic condition of perfect contact. A complete analysis is included: the continuous problem is shown to be well posed and optimal error estimates are proved for the numerical method. Numerical results showing the response of the system with respect to the excitation frequency are presented. Key Words: Fluid-structure, damped vibrations, displacement formulation. 0

Introduction

In this paper we analize a displacement finite element method to compute the linear oscillations of an elastic structure containing an inviscid, compressible, barotropic fluid, subjected to a harmonic excitation. We are interested in studying the effect of introducing on the fluid-solid interface, a thin layer of damping material having a given frequency dependent impedance. In recent years large attention has being paid to this kind of problems, mostly related to the goal of decreasing the level of noise in aircraft or cars (cf. the EU program BRAIN[12]). Let us mention, for instance, the paper by Kher-Kandille and Ohayon[13] where a pressure/potential formulation for the fluid has been introduced and numerically solved by finite element and modal reduction methods. In the present paper we use a displacement formulation for both media discretized by piecewise linear Lagrangian finite elements in the solid and RaviartThomas edge finite elements in the fluid. This kind of approximation has been introduced in [2] and [5] for the numerical computation of elastoacoustic eigenmodes and has been extended to compute hydroelastic and sloshing eigenmodes in [3], [4] and [6]. Partially supported by Programa de Cooperación Cient´ıfica con Iberoamérica, Ministerio de Educaci´ on y Ciencias, Spain. † Partially supported by FONDECYT (Chile) through grant No. 1.960.615 and FONDAP-CONICYT (Chile) through Program A on Numerical Analysis. ∗

2

A. Berm´ udez and R. Rodr´ıguez

Here we are concerned with the source problem associated with an external harmonic excitation. The solution of such problem allows us to know the response of the fluid-solid system for different frequencies of the external source. This is the objective in many practical situations in which the level of noise in a cavity enclosed by an elastic solid due to an external source is the magnitude to be computed. Let us mention, for instance, the case in aeronautical engineering where the noise produced by propellers inside an aircraft needs to be reduced by introducing thin layers of viscoelastic material. The outline of the paper is as follows: after setting the mathematical model we give a weak formulation and characterize the frequencies of the external excitation for which this problem has a unique solution. Then we consider the above mentioned finite element discretization and obtain optimal error estimates by using some classical results from Babuˇska[1] for indefinite variational problems. Finally numerical results are given for some real test examples. In particular the response curves for an actually used damping material are shown. 1

The model problem

Let us consider an elastic (two-dimensional) vessel enclosing an inviscid compressible barotropic fluid (see Figure 1). ΩS . . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . . Ω. F . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

. . . . . . . . .

ΓN ΓI

- n ν

ΓD

Figure 1. Fluid and solid domains. We denote by ΩF and ΩS the domains occupied by the fluid and the solid, respectively. They are supposed to be polygonal but not necessarily convex or simply connected; even interior angles of 2π are allowed.

3

Elastoacoustic vibrations with interface damping

Let ΓI be the interface between both media and ν its unit normal vector pointing outwards ΩF . We denote by Γj , j := 1, . . . , J, the edges of this polygonal S interface (hence, ΓI = Jj=1 Γj ). We assume that the external boundary of the solid is the union of two parts, ΓD and ΓN and that the structure is fixed along ΓD (meas (ΓD ) > 0). Finally n denotes the unit outward normal vector along ΓN . We are interested in the steady state response of this fluid-solid system to external harmonic forces. In order to damp the elastoacoustic energy originated from them, a thin layer of viscoelastic material is introduced between the solid and the fluid. Following [13] the effect of this material is modelled by introducing an appropriate interface condition on ΓI to be made precise below. We use the following notations for the physical magnitudes in the fluid: U: displacement vector, ρF : density, c: acoustic speed, and in the solid: V: displacement vector, ρS : density, λS and µS : Lamé coefficients, 1 ε(V): strain tensor defined by εij (V) := 2

!

∂Vi ∂Vj , + ∂xj ∂xi

i, j = 1, 2,

σ(V): stress tensor, which is assumed to be related to the strain tensor by Hooke’s law: σij (V) = λS

2 X

εkk (V)δij + 2µS εij (V),

i, j = 1, 2.

k=1

Given an external harmonic force F(x, t) := Re e−iωt f (x) acting on ΓN , with f : ΓN → C2 and ω > 0, the response of the damped coupled system is given by displacement fields U : ΩF → R2 and V : ΩS → R2 such that (see [13]):

(1.1) (1.2) (1.3) (1.4) (1.5) (1.6)

∂2U − ρF c2 ∇( div U) = 0, ∂t2 ∂2V ρS 2 − div [σ(V)] = 0, ∂t ∂U ∂V σ(V)ν = − α (U · ν − V · ν) + β ·ν − · ν ν, ∂t ∂t ∂U ∂V −ρF c2 div U = α (U · ν − V · ν) + β ·ν − ·ν , ∂t ∂t σ(V)n = F, V = 0, ρF

in ΩF , in ΩS , on ΓI , on ΓI , on ΓN , on ΓD .

4


The coupling between the fluid and the structure is taken into account by equations (1.3) and (1.4). Notice that the usual kinematic interface condition U · ν = V · ν has been relaxed to take into account the effect of the viscoelastic material. Conditions (1.3) and (1.4) mean that forces on the interface are normal and continuous; furthermore they consist of two terms: the first one is proportional to the jump of the normal component of the displacements and accounts for the elastic response of the interface material, whereas the second one is proportional to the jump in the normal velocity and it models the viscous damping introduced by this material. The steady state solution of this evolution problem consists of harmonic in time displacement fields which can be found by solving the following stationary problem: Find u : ΩF → C2 and v : ΩS → C2 such that: −ρF ω 2 u − ρF c2 ∇( div u) = 0, −ρS ω 2 v − div [σ(v)] = 0, σ(v)ν = −(α − iωβ)(u · ν − v · ν)ν, −ρF c2 div u = (α − iωβ)(u · ν − v · ν), σ(v)n = f , v = 0,

(1.7) (1.8) (1.9) (1.10) (1.11) (1.12)

in ΩF , in ΩS , on ΓI , on ΓI , on ΓN , on ΓD .

In fact, clearly h

i

h

U(x, t) = Re e−iωt u(x) ,

i

V(x, t) = Re e−iωt v(x)

are harmonic in time solutions of problem (1.1)-(1.6). The specific impedance of a viscoelastic material for a particular frequency ω is defined by α Z(ω) := β + i. ω Notice that the coefficient in the interface equations (1.9) and (1.10) can then be written α − iωβ = −iωZ(ω). For a particular material, the real and imaginary parts of its impedance can be experimentally measured for any given frequency ω; both parts actually depend on ω. Because of this, we consider in this paper a general complex-valued function Z(ω) having strictly positive real and imaginary parts for all ω > 0. 2

Weak formulation

We introduce a weak formulation of (1.7)-(1.12) in order to analyze existence, uniqueness and regularity of solution for this problem. We use the standard notation for Sobolev spaces, norms and seminorms. We also denote H(div, ΩF ) := u ∈ L2 (ΩF )2 : div u ∈ L2 (ΩF ) and kukH(div,Ω

F

)

h

:= kuk2L2 (Ω

F

)2

+ k div uk2L2 (Ω

F

)

i1/2

.

5


Throughout the paper all functional spaces are complex valued and we denote by C a generic constant not necessarily the same at each occurrence. Let H := L2 (ΩF )2 × L2 (ΩS )2 and let n

o

V := (u, v) ∈ H(div, ΩF ) × H1Γ (ΩS )2 : u · ν ∈ L2 (ΓI ) , D

where H1Γ (ΩS ) is the subspace of functions in H1 (ΩS ) vanishing on ΓD . Endowed D with the norm h

k(u, v)k = kuk2H(div,Ω

F

2 ) + kvkH1 (Ω

S

)2

+ ku · νk2L2 (Γ

I

)

i1/2

V is a Hilbert space continuously included in H. We introduce the following sesquilinear forms, defined on H × H −→ C and V × V −→ C, respectively:

d (u, v), (φ, ψ)

aω (u, v), (φ, ψ)

:=

Z

ΩF

:=

Z

¯+ ρF u · φ

Z

ΩS

¯ + σ(v) : ε(ψ)

¯ ρS v · ψ, Z

ΩF

ΩS

Z

− iωZ(ω)

ΓI

¯ ρF c2 div u div φ

¯ ·ν −ψ ¯ · ν) (u · ν − v · ν)(φ

+ d (u, v), (φ, ψ) . The sesquilinear form d(·, ·) is an inner product on H inducing a norm equivalent to the L2 one. From now on we consider H endowed with this norm which we denote by | · |. On the other hand, notice that aω is not Hermitian but, because of the added mass term d(·, ·), it will be shown below that it is coercive. It is easy to check that the following is a variational formulation of the stationary problem (1.7)-(1.12): Given ω > 0 and f ∈ H−1/2 (ΓN )2 , find (u, v) ∈ V such that:

(2.1) aω (u, v), (φ, ψ) − (ω 2 + 1) d (u, v), (φ, ψ) = hf , ψi ,

∀(φ, ψ) ∈ V,

1/2

where h·, ·i stands for the duality pairing in H−1/2 (ΓN )2 × H00 (ΓN )2 (the latter, 1/2 H00 (ΓN )2 , being the space of traces on ΓN of functions in H1Γ (ΩS )2 ). D Our first goal is to characterize the values of ω for which this problem has a unique solution for any f ∈ H−1/2 (ΓN )2 . First we prove that aω is coercive: Proposition 2.1 For each ω > 0 there exists a positive constant γ(ω) such that h

i

Re aω (u, v), (u, v)

≥ γ(ω)k(u, v)k2 .

6


Proof : The definition of aω yields

h

i


(2.2)

Z

=

σ(v) : ε(¯ v) +

ΩS

ρS |v|2

ΩS

Z

+

Z

2

ΩF

Z

2

ρF c | div u| +

+ ω Im[Z(ω)]

Z

ΓI

ΩF

ρF |u|2

|u · ν − v · ν|2 .

Now, because of Korn’s inequality, there exists a constant c1 > 0 such that Z

Z

ΩS

ρS |v|2 ≥ c1 kvk2H1 (Ω

Z

ΩF

ρF |u|2 ≥ c2 kukH(div,ΩF ) ,

σ(v) : ε(¯ v) +

ΩS

S

)2 .

On the other hand, clearly, Z

ΩF

ρF c2 | div u|2 +

with c2 > 0. Finally, for any ε ∈ (0, 1) it holds Z

ΓI

2

|u · ν − v · ν| ≥ ε

Z

ε |u · ν| − 1−ε 2

ΓI

Z

ΓI

|v · ν|2 .

So, since ω Im[Z(ω)] is strictly positive, we may take ε small enough so as to satε c1 isfy 0 < ω Im[Z(ω)] C 2 ≤ , where C is the constant in the Trace Theorem: 1−ε 2 kv · νkL2 (ΓI ) ≤ CkvkH1 (ΩS )2 . Then, by substituting all these inequalities in (2.2) we have h

i


≥ c1 kvk2H1 (Ω

S

)2

+ c2 kuk2H(div,Ω

+ ω Im[Z(ω)] ε

Z

)

ε |u · ν| − 1−ε 2

ΓI

≥

F

c1 kvk2H1 (Ω )2 + c2 kuk2H(div,Ω ) F S 2 Z

+ εω Im[Z(ω)]

ΓI

Z

ΓI

|v · ν|

2

!

|u · ν|2

2

≥ γ(ω)k(u, v)k , with γ(ω) := min

c1 , c2 , εω Im[Z(ω)] > 0. 2

QED


7

Now, by identifying H with its dual space we have the classical continuous inclusions V ֒→ H ֒→ V′ . Then we may introduce the bounded linear operator Tω : V′ −→ V′ as follows: for G ∈ V′ , Tω G := (u, v), with (u, v) ∈ V being the unique solution of the problem

D

E

aω (u, v), (φ, ψ) = G, (φ, ψ)

V′ ×V

,

∀(φ, ψ) ∈ V,

Because of the coerciveness of aω and Lax-Milgram’s Theorem this problem has a unique solution for each G ∈ V′ . Problem (2.1) can be rewritten in terms of Tω as follows: find (u, v) ∈ V satisfying (u, v) = Tω (ω 2 + 1)(u, v) + F , or, equivalently,

1 1 I (u, v) = 2 Tω F, Tω − 2 ω +1 ω +1

where F the linear functional in V′ defined for f ∈ H−1/2 (ΓN )2 by D

E

F, (φ, ψ)

V′ ×V

:= hf, ψi ,

∀(φ, ψ) ∈ V.

This shows that problem (2.1) attains a unique solution for any f ∈ H−1/2 (ΓN )2 if and only if ω21+1 does not belong to the spectrum of Tω (which we denote by σ(Tω )). Our next goal is to characterize this spectrum. In particular, since for ω > 0, 0 < 1/(ω 2 + 1) < 1, we will show that only in very exceptional situations σ(Tω ) ∩ (0, 1) 6= ∅. Firstly notice that since Tω (V′ ) ⊂ V ⊂ H, then σ(Tω ) = σ(Tω |H ) ∪ {0} (see Lemma 4.1 in [2]). Hence, it is enough to analize σ(Tω |H ). Secondly, for all ω > 0, Tω coincides with the identity on the infinite dimensional subspace of H: K := {(u, 0) ∈ V : div u = 0 in ΩF and u · ν = 0 on ΓI } . This subspace is also characterized by (see [10]) K = {( curl ξ, 0) : ξ ∈ H1 (ΩF ) and ξ is constant on each connected component of ΓI }. Let G := K⊥H (i.e., the orthogonal complement of K in H). Then G is given by (see [10]) G = {(∇ϕ, v) : ϕ ∈ H1 (ΩF ) and v ∈ L2 (ΩS )2 }. The following result shows that G is an invariant subspace for Tω and that Tω |G is compact in H norm:

8


Proposition 2.2 It holds: i) Tω (G) ⊂ G;

ii) Tω : G −→ H is compact.

Proof : Let (z, w) ∈ G and (u, v) := Tω (z, w). For ( curl ξ, 0) ∈ K it holds Z

ΩF

D

E

ρF u · curl ξ = aω (u, v), ( curl ξ, 0) = (z, w), ( curl ξ, 0) =

Z

ΩF

V′ ×V

ρF z · curl ξ = 0,

Hence (u, v) ∈ K⊥H = G proving (i). On the other hand, Tω : H −→ V is continuous; then u ∈ H(div, ΩF ), u · ν ∈ L2 (ΓI ) and v ∈ H1 (ΩS )2 with kukH(div,ΩF ) + ku · νkL2 (ΓI ) + kvkH1 (ΩS )2 ≤ C|(z, w)|. Furthermore, because of (i), (u, v) ∈ G and hence there exists ϕ ∈ H1 (ΩF ) such that u = ∇ϕ; then ϕ satisfies ∆ϕ = div u, in ΩF , ∂ϕ = u · ν, on ΓI . ∂ν Therefore, by applying classical regularity results for the Laplace operator (see [11]) we know that u = ∇ϕ ∈ Hs/2 (ΩF )2 , with s = 1 if ΩF is convex or s = πθ otherwise (θ being the largest reentrant corner of ΩF ) and kukHs/2 (Ω

F

)2

h

i

≤ C kukH(div,ΩF ) + ku · νkL2 (ΓI ) ≤ C|(z, w)|.

So Tω : G −→ Hs/2 (ΩF )2 ×H1 (ΩS )2 is continuous and since the latter is compactly QED included in H, Tω : G −→ H is compact. Now we are able to characterize the spectrum of Tω : Proposition 2.3 Apart from µ = 0 and µ = 1, the spectrum of Tω consists of a sequence of (complex) eigenvalues of finite multiplicity, having µ = 0 as the only accumulation point.

9


Proof : It is an consequence of the analysis above and the spectral theorem for compact QED operators. The following proposition characterize the real eigenvalues of Tω and show that they are rather exceptional. Proposition 2.4 Let µ be a real eigenvalue of Tω different from 0 and 1 and (u, v) ∈ V be an associated eigenfunction. Then it holds: i) u · ν = v · ν on ΓI ; 1 µ

ii) λ =

and (0, 0) 6= (u, v) ∈ V must satisfy:

Z

(2.3)

¯ + σ(v) : ε(ψ)

ΩS

(2.4)

Z

ΩS

Z

¯+ ρF c2 div u div φ

ΩF

Z

ΩF

¯ =λ ρS v · ψ ¯ =λ ρF u · φ

Z

ΩS

¯ ∀ψ ∈ H1 (Ω )2 , ρS v · ψ, Γ S

ΩF

¯ ∀φ ∈ H(div, Ω ). ρF u · φ, F

Z

D

Proof : ¿From the assumptions on λ and (u, v) the following equality holds: (2.5)

Z

¯ + σ(v) : ε(ψ)

ΩS

Z

ΩF

− iωZ(ω)

Z

ΓI

¯+ ρF c2 div u div φ

Z

ΩS

¯+ ρS v · ψ

Z

ΩF

¯ ρF u · φ

¯ ·ν −ψ ¯ · ν) (u · ν − v · ν)(φ

= λ

Z

!

¯+ ρS v · ψ

Z

¯ , ρF u · φ

ρF c2 | div u|2 +

Z

ρS |v|2 +

ΩS

ΩF

∀(φ, ψ) ∈ V.

By taking (φ, ψ) = (u, v) we get Z

σ(v) : ε(¯ v) +

ΩS

Z

ΩF

− iωZ(ω)

Z

2

ΓI

ΩS

|u · ν − v · ν| = λ

Z

ΩS

Z

ΩF

2

ρS |v| +

ρF |u|2 Z

ΩF

ρF |u|

2

!

.

In the previous equality all terms are real except for the last one in the left hand side. Therefore we must have Im [iωZ(ω)]

Z

ΓI

|u · ν − v · ν|2 = 0

and then u · ν = v · ν, since we have assumed Re[Z(ω)] > 0. Now, by taking succesively ψ = 0 and φ = 0 in (2.5) we immediately deduce QED (2.3) and (2.4).

10


Notice that (2.3) and (2.4) are two uncoupled spectral problems. The former corresponds to the solid with vacuum inside while the latter corresponds to the fluid with free boundary ΓI (i.e., the fluid pressure p := ρF c2 div u is null on ΓI ). Moreover, according to the previous proposition, either u or v could be equal to 0 (only one of them), in which case condition (i) would imply v · ν = 0 or u · ν = 0, respectively. Hence, any λ satisfying (ii) and such that the corresponding eigenfunctions satisfy (i) should belong to one of the following sets: 1. the set of eigenvalues of the solid in vacuo whose corresponding eigenfunctions have, furthermore, null normal component on ΓI ; 2. the set of eigenvalues of the fluid with free boundary ΓI such that the corresponding eigenfunctions have, furthermore, null normal component on ΓI ; 3. the set of simultaneous eigenvalues of the two above problems such that the corresponding eigenfunctions have, furthermore, coinciding normal components on the interface ΓI . Any of these conditions are extremely rare to be fulfilled in practice. The previous proposition has a reciprocal. Indeed, it is easy to prove that if λ and (u, v) 6= (0, 0) satisfy the two conditions therein, then λ should be real and µ = λ1 would be an eigenvalue of the operator Tω , ∀ω > 0. In such case λ > 1 √ and ω = λ − 1 ∈ R. Furthermore the pair of eigenfunctions (u, v) is a solution of the implicit spectral problem for the fluid-solid system with interface damping associated with the source problem (2.1); i.e., ω > 0 and (0, 0) 6= (u, v) ∈ V satisfy

aω (u, v), (φ, ψ) = (ω 2 + 1) d (u, v), (φ, ψ) ,

∀(φ, ψ) ∈ V.

Now we are able to state the main result of this section: Theorem 2.5 Let ω ∈ R be such that λ := ω 2 + 1 does not fulfill condition (ii) in Proposition 2.4 for any (0, 0) 6= (u, v) ∈ V satisfying u · ν = v · ν. Then −1/2 (Γ )2 . Furthermore, there problem (2.1) N i a unique solution for each f ∈ H h has exists s ∈ 12 , 1 and t0 ∈ (0, 1] such that if f ∈ H−1/2+t (ΓN )2 with 0 ≤ t ≤ t0 , then u ∈ Hs (ΩF )2 , v ∈ H1+t (ΩS )2 , div u ∈ H1+s (ΩF ), u · ν ∈ H1/2+r (ΓI ) with r := min{s, t} and kukHs (ΩF )2 + kvkH1+t (ΩS )2 + k div ukH1+s (ΩF ) + ku · νkH1/2+r (Γ

I

)

≤ Ckf kH−1/2+t (Γ

N

)2 .

11


Proof : The existence of a unique solution is a consequence of the above discussion. Concerning its regularity, notice that by considering adequate test functions (φ, ψ) in (2.1) it can be shown that u and v satisfy (1.7)-(1.12). From (1.7) we obtain that div u ∈ H1 (ΩF ) and from this fact and (1.10) we have that u · ν ∈ H1/2 (ΓI ), in both cases with the respective norms bounded by Ck(u, v)k. Now, the arguments in the proof of (ii) in Proposition 2.2 can be repeated; however, because of the further regularity h ofi u · ν, we may improve that estimate and s 2 show that u ∈ H (ΩF ) with s ∈ 12 , 1 as defined therein. Consequently, (1.7) yields now div u ∈ H1+s (ΩF ). On the other hand, from (1.8), (1.9), (1.11), (1.12) and the regularity assumption on f , by applying classical regularity results for the elasticity equations (see [11]) we have that v ∈ H1+t (ΩS )2 and h

kvkH1+t (ΩS )2 ≤ C k(u, v)k + kf kH−1/2+t (Γ

N

)2

i

,

with t depending on the reentrant corners of ΩS , on the angles between ΓD and ΓN and on the Lamé coefficients λS and µS . Finally, equation (1.10) now yields u · ν ∈ H1/2+r (ΓI ) with r := min{s, t}. All these estimates and the continuity of Tω : V′ −→ V allow us to conclude the QED theorem.

3

Finite element discretization

Solid and fluid displacements belong to different spaces, H1 (ΩS )2 and H(div, ΩF ), respectively; hence, different type of finite elements should be used for each of them to discretize the variational problem (2.1). For the corresponding undamped eigenvalue problem (i.e., without any damping material on the fluid-solid interface) a discretization avoiding the spurious modes typical of displacement formulations and yielding optimal order approximations of eigenvalues and eigenfunctions has been introduced and analized in [2, 5]. In this section we show that the same discretization can be used for our problem yielding also optimal order results. Let {Th } be a family of regular triangulations of ΩF ∪ ΩS such that every triangle is completely contained either in ΩF or in ΩS and such that the end points of ΓD and ΓN coincide with nodes of the triangulation. For each component of the displacements in the solid we use the standard piecewise linear finite element space n

Lh (ΩS ) := v ∈ H1 (ΩS ) : v|T ∈ P1 (T ), ∀T ∈ Th , T ⊂ ΩS and, for the fluid, the Raviart-Thomas space (see [14]) n

o o

Rh (ΩF ) := u ∈ H(div, ΩF ) : u|T ∈ R0 (T ), ∀T ∈ Th , T ⊂ ΩF ,

12


where n

o

R0 (T ) := u ∈ P1 (T )2 : u(x, y) = (a + cx, b + cy), a, b, c ∈ C . The discrete analogue of V is n

o

Vh := (u, v) ∈ Rh (ΩF ) × Lh (ΩS )2 : v|Γ = 0 . D

With this finite element space we define an approximate problem to (2.1): Find (uh , vh ) ∈ Vh such that:

(3.1) aω (uh , vh ), (φ, ψ) − (ω 2 + 1) d (uh , vh ), (φ, ψ)

= hf , ψi , ∀(φ, ψ) ∈ Vh .

Let Rh : H(div, ΩF ) ∩ Hǫ (ΩF )2 −→ Rh (ΩF ) denote the Raviart-Thomas interpolant operator, which is well defined for any ǫ > 0 (see [7]). We also denote by Ih : H1 (ΩS )2 −→ Lh (ΩS )2 a Clement interpolant operator (see [8]). The following lemma shows that these interpolations provide approximations for (φ, ψ) ∈ G∩V from the discrete space Vh : Lemma 3.1 For all (φ, ψ) ∈ G ∩ V there exists (φh , ψ h ) := (Rh φ, Ih ψ) ∈ Vh such that k(φ, ψ) − (φh , ψ h )k → 0 as h → 0. Proof : Since (φ, ψ) ∈ G ∩ V, by proceeding as in the proof of Proposition 2.2 we have that φ ∈ Hs/2 (ΩF )2 (with s > 0 as in that proof). On the other hand, φ · ν ∈ L2 (ΓI ) and ψ ∈ H1 (ΩS )2 . Hence (see [7] and [8]) Rh φ → φ in H(div, ΩF ), Rh φ · ν → φ · ν in L2 (ΓI ), Ih ψ → ψ in H1Γ (ΩS )2 , D

and, consequently, (Rh φ, Ih ψ) ∈ Vh converges to (φ, ψ) in V. Let Kh :=

n

QED

( curl ξ, 0) : ξ ∈ Lh (ΩF ) and ξ is constant on each connected o

component of ΓI ⊂ Vh . Let Gh be the orthogonal complement of Kh in Vh . For any (φh , ψ h ) ∈ Vh we may write the orthogonal decomposition: (φh , ψ h ) = ( curl ξh , 0) + (χh , ψ h ), with ( curl ξh , 0) ∈ Kh and (χh , ψ h ) ∈ Gh . Notice that these subspaces are also orthogonal in H.

13


Let bω : V × V → C be the sesquilinear form defined by bω := aω − (1 + ω 2 ) d. The theorem below shows that bω satisfies a discrete inf-sup condition. This property will allow us to get optimal error estimates for the discrete source problem (3.1) by using a classical theorem from Babuˇska[1]. A similar analysis has been done by Demkowicz[9] for the Helmholtz equation, but his results cannot be directly applied to our situation. Proposition 3.2 For any real ω satisfying the assumptions of Theorem 2.5, the form bω satisfies the following property: there exists h0 > 0 and γ > 0, such that, for h ≤ h0 , bω (uh , vh ), (φ, ψ) ≥ γk(uh , vh )k. sup k(φ, ψ)k (φ,ψ )∈Vh (φ,ψ )6=(0,0)

Proof : We argue by contradiction. Let us suppose that ∀n ∈ N, there exists hn > 0 such that hn → 0 and (uhn , vhn ) ∈ Vhn , with k(uhn , vhn )k = 1, satisfying

bω (uhn , vhn ), (φ, ψ)

sup (φ,ψ )∈Vhn (φ,ψ )6=(0,0)

k(φ, ψ)k