A refined mixed finite-element method for the ... - Oxford Journals

IMA Journal of Numerical Analysis (2008) 28, 25−45 doi:10.1093/imanum/drm005 Advance Access publication on April 3, 2007

A refined mixed finite-element method for the stationary Navier–Stokes equations with mixed boundary conditions M OHAMED FARHLOUL† Département de Mathématiques et de Statistique, Université de Moncton, Moncton, New Brunswick E1A 3E9, Canada AND

S ERGE N ICAISE‡ AND L UC PAQUET§ Université de Valenciennes et du Hainaut Cambrésis, MACS, ISTV, F-59313 Valenciennes Cedex 9, France [Received on 15 October 2004; revised on 23 May 2006] This paper is concerned with the mixed formulation of the Navier–Stokes equations with mixed boundary conditions in 2D polygonal domains and its numerical approximation. We first describe the regularity of any solution. The problem is then approximated by a mixed finite-element method where the strain tensor and the antisymmetric gradient tensor, quantities of practical importance, are introduced as new unknowns. An existence result for the finite-element solution and convergence results are proved near a nonsingular solution. Quasi-optimal error estimates are finally presented. Keywords: Navier–Stokes system; corner singularity; mixed finite-element method; refined mesh.

1. Introduction Solutions of the Navier–Stokes equations in polygonal domains have in general corner singularities (Dauge, 1989; Orlt & Sändig, 1994; Farhloul et al., 2001). Hence, standard numerical methods lose accuracy on quasi-uniform meshes, and locally refined meshes are necessary to restore the optimal order of convergence. Standard finite-element methods for the Stokes or Navier–Stokes system with corner singularities have been analysed in Becker & Rannacher (1995), Bernardi & Raugel (1981), El Bouzid & Nicaise (1998) and Orlt & Sändig (1994), where it is shown that the use of appropriately refined meshes near the singular points allows one to restore the optimal order of convergence. Mixed methods for the Stokes and Navier–Stokes equations with Dirichlet boundary condition were initiated in Farhloul & Manouzi (1996) and Farhloul & Fortin (1993, 1997). Similarly, a mixed method for the Boussinesq equations (coupling between the Navier–Stokes equations and the heat equation) with Dirichlet boundary condition on the velocity field in the presence of corner singularities is analysed in Farhloul et al. (2001) (some geometrical restrictions were made due to the mixed boundary condition on the temperature). In these papers, only ∇u is introduced as a new unknown and therefore the natural Neumann boundary condition cannot be treated. Mixed methods for the linear elasticity equations with † Email: [email protected] ‡ Email: [email protected] § Email: [email protected] c The author 2007. Published by Oxford University Press on behalf of the Institute of Mathematics and its Applications. All rights reserved.

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M. FARHLOUL ET AL.

mixed boundary conditions (with or without singularities) were considered in Farhloul & Fortin (1993, 1997) and Farhloul & Paquet (2002) where the authors introduced the strain tensor and the vorticity as new unknowns. Finally, let us recall that interest in mixed methods is twofold. First, the Neumann boundary condition becomes an essential boundary condition, and second, its discretization gives directly the approximation to the strain tensor and the vorticity without any postprocessing. Our goal is then to consider the stationary Navier–Stokes equations with mixed boundary conditions (Dirichlet boundary condition on a part of the boundary and Neumann boundary condition on the remainder) in a 2D polygonal domain and to use a mixed finite-element method. Since Neumann boundary conditions are imposed on part of the boundary, our method uses the following new unknowns of physical significance: the rate of strain tensor ε(u) = 12 (∇u + (∇u)T ) and the vorticity ω(u) = 12 (∇u − (∇u)T ), u being the velocity field. We describe the regularity of a solution in terms of weighted Sobolev spaces. The poor regularity of any solution forces us to use appropriate Banach spaces and then to introduce an appropriate mixed formulation of the problem. We further state the equivalence between the classical variational formulation of the problem and this new mixed formulation. We then consider a discretization of our mixed formulation by using the mixed finite elements developed by Farhloul & Fortin (1997), namely piecewise constants for the velocity, piecewise P1 for the pressure, piecewise [P1 ]2 + (curl of the element bubbles) for each line of the strain tensor and piecewise [P1 ]2×2 for the vorticity. We then prove that the discrete mixed formulation has at least one solution near any nonsingular solution of the Navier–Stokes equations. Furthermore, using appropriately refined meshes near the singular points, we show a quasi-optimal error estimate. Our approach then combines some ideas from Farhloul & Fortin (1993, 1997) and Farhloul & Paquet (2002) developed for the linear elasticity equations with the techniques from Farhloul & Manouzi (1996), Farhloul & Fortin (1993, 1997) and Farhloul et al. (2001) used for the Navier–Stokes equations. The outline of the paper is as follows: In Section 2, we state the incompressible Navier–Stokes equations with mixed boundary conditions, introduce its classical variational formulation and prove some regularity results in terms of weighted Sobolev spaces that will be useful for our further analysis. We then state our new mixed formulation and show its equivalence with the variational formulation. We finally give some useful properties of linear operators in certain function spaces related to nonsingular solutions. In Section 3, we describe and analyse the discretization of the problem. Finally, section 4 is devoted to the proof of the quasi-optimal error estimate. 2. Mixed formulation of the Navier–Stokes equations with Dirichlet–Neumann boundary conditions Let Ω be a plane domain with a polygonal boundary. More precisely, we assume that Ω is a simply connected domain and that its boundary Γ is the union of a finite number of linear segments Γ j , 1 6 j 6 n e (Γ j is assumed to be an open segment). Let us further fix a partition {D, N } of the set { j ∈ N; 1 6 j 6 n e }. In Ω, we consider the following boundary-value problem for the Navier– Stokes equations: −2μdiv ε(u) + (u ∙ ∇)u + ∇ p = f

div u = 0

in Ω, in Ω,

u=0

on Γ D =

(2με(u) − pδ) ∙ n = 0

on Γ N =

(2.1) [

(2.2) Γj,

(2.3)

Γj,

(2.4)

j∈D

[

j∈N

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A REFINED MIXED FINITE-ELEMENT METHOD

where μ is the viscosity parameter, u the velocity field, p the pressure, f the density force which is supposed to be in [L 2 (Ω)]2 , δ is the 2 × 2 identity matrix, ε(u) = 12 (∇u + (∇u)T ) the strain tensor field and n the unit outward normal vector field along the boundary of Ω. For a tensor τ = (τ )16i, j 62 , the P2 P2 normal trace τ ∙ n is defined by τ ∙ n = j=1 τ1 j n j , j=1 τ2 j n j . In the sequel, we make the natural assumption that Γ D 6= ∅. The following variational formulation of problem (2.1)–(2.4) is quite standard (Orlt & Sändig, 1994). Let us introduce the space o n (2.5) W = u ∈ [H 1 (Ω)]2 ; u |Γ D = 0 . We then look for a pair (u, p) ∈ W × L 2 (Ω) that satisfies Z Z Z Z 2μ ε(u): ε(v)dx + [(u ∙ ∇)u] ∙ v dx − pdiv v dx = Ω

Ω

Z

Ω

Ω

Ω

f ∙ v dx

qdiv u dx = 0 ∀ q ∈ L 2 (Ω).

∀ v ∈ W,

(2.6) (2.7)

This problem possesses at least one solution (u, p) (not necessarily unique). Orlt & Sändig (1994) have proved that if f is sufficiently small, namely 2

k f kW ∗
2 and 1t < r11 + 12 .

L EMMA 2.2 Let v ∈ [L s (Ω)]2 such that div v ∈ L t (Ω). Let us suppose that s > 1 and that t > 1 1 satisfies 1t < 1s + 12 . Then, v ∙ n belongs to W − s ,s (Γ ) and the following Green’s formula holds: hv ∙ n, ϕi = (v, ∇ϕ) + (div v, ϕ)

0

∀ ϕ ∈ W 1,s (Ω),

(2.14) 1

where s 0 > 1 is the conjugate of s, i.e. 1s + s10 = 1 and h∙, ∙i means the duality pairing between W − s ,s (Γ ) 1

0

and W s ,s (Γ ).

29


Proof. Let us first note that (v, ∇ϕ) = 0

0

R

Ω

v ∙ ∇ϕ dx is meaningful since v ∈ [L s (Ω)]2 and ∇ϕ ∈ 2s 0

[L s (Ω)]2 . As ϕ ∈ W 1,s (Ω), the Sobolev embedding theorem leads to ϕ ∈ L 2−s 0 (Ω). Recalling that div v ∈ L t (Ω) and remarking that 1 1 1 1 + 2s 0 < + + t s 2 0 2−s

2 s0

−1 1 1 2 − 2s − 1 = + + = 1, s 2 2 2

we conclude that (div v)ϕ is integrable over Ω. ˉ 2 is dense in the Banach As in Theorem I.2.4 of Girault & Raviart (1986), we can show that [C ∞ (Ω)] space W s,t (div, Ω) = {v ∈ [L s (Ω)]2 : div v ∈ L t (Ω)}, equipped with its natural norm. ˉ 2 by this density and continuity, (2.14) holds for any Since (2.14) clearly holds for v ∈ [C ∞ (Ω)] s,t v ∈ W (div, Ω) and the conclusion follows. Now, the mixed formulation of our problem (2.6), (2.7) reads as follows: find σ˜ = (σ, p) ∈ Σ and u˜ = (u, ω) ∈ M such that 1 (σ, τ ) + (div(τ − qδ), u) + (τ, ω) = 0 ∀ τ˜ = (τ, q) ∈ Σ, 2μ (div(σ − pδ), v) + (σ, η) −

1 (σ ∙ u, v) − (ω ∙ u, v) + ( f, v) = 0 2μ

∀ v˜ = (v, η) ∈ M.

(2.15) (2.16)

R The first step is to verify that all of the above terms are well-defined. First, the terms (σ, τ ) = Ω σ : τ dx, R R (τ, ω) = Ω τ : ω dx and (σ, η) = Ω σ : η dx are meaningful because σ, τ, ω, η ∈ [L r1 (Ω)]2×2 and 1 1 2 2 1 2+α 1 1 1 r1 + r1 = r1 < 1. On the other hand, the condition r1 < α implies that t = 4 < 2 + 2r1 = 1− r2 and thus 1t + r12 < 1. This implies that the two terms (div(τ − qδ), u) and (div(σ − pδ), v) are well-defined.

The two nonlinear terms (σ ∙ u, v) and (ω ∙ u, v) make sense due to the fact that r11 + r22 = 1. The next result concerns the equivalence between our mixed formulation (2.15), (2.16) with the classical mixed formulation (2.6), (2.7). We skip its proof since it follows essentially the same lines as those of Theorem 3.2 of Farhloul & Paquet (2002). T HEOREM 2.3 (u, p) ∈ W × L 2 (Ω) is a solution of the variational formulation (2.6), (2.7) if, and only if, (σ˜ , u) ˜ = ((σ, p), (u, ω)) ∈ Σ × M is a solution of (2.15), (2.16). Moreover, we have the relations σ = 2με(u)

and ω =

1 (∇u − (∇u)T ). 2

(2.17)

In the sequel, we will analyse approximations to nonsingular solutions of the mixed formulation (2.15), (2.16) of the Navier–Stokes system. For this purpose, we introduce the linear operator S which associates to any function f ∗ ∈ [L t (Ω)]2 the solution (σ˜ ∗ = (σ ∗ , p ∗ ), u˜ ∗ = (u ∗ , ω∗ )) ∈ Σ × M of the problem 1 (σ ∗ , τ ) + (div(τ − qδ), u ∗ ) + (τ, ω∗ ) = 0 2μ (div(σ ∗ − p ∗ δ), v) + (σ ∗ , η) + ( f ∗ , v) = 0

∀ τ˜ = (τ, q) ∈ Σ,

(2.18)

∀ v˜ = (v, η) ∈ M.

(2.19)

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Problem (2.18), (2.19) is nothing else than a mixed formulation of the Stokes problem with mixed boundary conditions. We first note that the linear operator S is well-defined, i.e. if (u ∗ , p ∗ ) ∈ W × L 2 (Ω) denotes a solution to the classical variational formulation of the Stokes problem with datum f ∗ in the sense that it satisfies (2.6), (2.7) without the nonlinear term, then setting σ ∗ = 2με(u ∗ ) and ω∗ = 12 (∇u ∗ −(∇u ∗ )T ), the pair (σ˜ ∗ = (σ ∗ , p ∗ ), u˜ ∗ = (u ∗ , ω∗ )) belongs to Σ × M and is a solution of (2.18), (2.19). This second assertion simply follows from the same arguments as those used to prove Theorem 2.3 (without 4 2 > s := 1+α and then f ∗ ∈ the nonlinear terms). For the first assertion, we remark that t = 2+α s 2 ∗ ∗ [L (Ω)] with k f k0,s,Ω 6 Ck f k0,t,Ω . As the exponent s satisfies the assumptions of Corollary 4.2 of Orlt & Sändig (1994), we deduce by this corollary that (u ∗ , p ∗ ) ∈ [W 2,s (Ω)]2 × W 1,s (Ω) and that ku ∗ k2,s,Ω + k p ∗ k1,s,Ω 6 Ck f ∗ k0,s,Ω 6 Ck f ∗ k0,t,Ω .

(2.20) 2

By the Sobolev embedding theorem and the fact that 2 < r1 < α2 , W 1,s (Ω) ,→ L α (Ω) ,→ L r1 (Ω). Consequently, σ ∗ = 2με(u ∗ ) and ω∗ = 12 (∇u ∗ −(∇u ∗ )T ) belong to [L r1 (Ω)]2×2 . For the same reasons, the pressure p ∗ belongs to L r1 (Ω). Consequently, we get σ˜ ∗ = (σ ∗ , p ∗ ) ∈ Σ. Since the Sobolev embedding theorem guarantees that H 1 (Ω) ,→ L r2 (Ω), we directly deduce that u ∗ ∈ [L r2 (Ω)]2 as u ∗ ∈ [H 1 (Ω)]2 and thus u˜ ∗ = (u ∗ , ω∗ ) ∈ M. Note further that these arguments and the estimate (2.20) imply the following a priori estimate: ∗ k(σ˜ ∗ , u˜ ∗ )kΣ× ˆ Mˆ 6 Ck f k0,t,Ω ,

(2.21)

Σˆ = [L r (Ω)]2×2 × L r (Ω),

(2.22)

Mˆ = {v˜ = (v, η): v ∈ [L r (Ω)]2 , η ∈ [L r (Ω)]2×2 with η + ηT = 0},

(2.23)

where

with its natural norm ∗ ∗ ∗ ∗ k(σ˜ ∗ , u˜ ∗ )kΣ× ˆ Mˆ := kσ k0,r,Ω + k p k0,r,Ω + ku k0,r,Ω + kω k0,r,Ω ,

r being an arbitrary real number chosen but fixed in the open interval (2, min{r1 , r2 }). The a priori estimate (2.21) may be rewritten as ∗ kS f ∗ kΣ× ˆ Mˆ 6 Ck f k0,t,Ω ,

ˆ which means that S is a continuous operator from [L t (Ω)]2 into Σˆ × M. Next, we define the mapping H from Σ × M into itself by 1 τ ∙v−η∙v , H (τ˜ , v) ˜ := (τ˜ , v) ˜ −S f − 2μ

(2.24)

(2.25)

for every (τ˜ , v) ˜ = ((τ, q), (v, η)) ∈ Σ × M. With these notations, (σ˜ , u) ˜ = ((σ, p), (u, ω)) ∈ Σ × M is a solution of the mixed formulation (2.15), (2.16) for the Navier–Stokes equations if, and only if, (σ˜ , u) ˜ satisfies H (σ˜ , u) ˜ = 0.

(2.26)

31


In the sequel, we shall be concerned with nonsingular solutions of (2.26). Let us recall that a solution (σ˜ , u) ˜ ∈ Σ × M is called nonsingular if the Fréchet derivative of H at the point (σ˜ , u): ˜ H 0 (σ˜ , u): ˜ Σ×M →Σ×M (τ˜ , v) ˜ 7→ (τ˜ , v) ˜ +S

1 (σ ∙ v + τ ∙ u) + ω ∙ v + η ∙ u 2μ

(2.27)

is an isomorphism. Equivalently, (σ˜ , u) ˜ is a nonsingular solution of (2.26) if, and only if, the linearized mixed formulation of the Navier–Stokes problem at the point (σ˜ , u) ˜ = ((σ, p), (u, ω)) given by 1 ∗ (σ , τ ) + (div(τ − qδ), u ∗ ) + (τ, ω∗ ) = 0 2μ (div(σ ∗ − p ∗ δ), v) + (σ ∗ , η) −

∀ τ˜ = (τ, q) ∈ Σ,

(2.28)

1 (σ ∙ u ∗ + σ ∗ ∙ u, v) 2μ

− (ω ∙ u ∗ + ω∗ ∙ u, v) = 0 ∀ v˜ = (v, η) ∈ M

(2.29)

has a unique solution (σ˜ ∗ , u˜ ∗ ) = ((σ ∗ , p ∗ ), (u ∗ , ω∗ )) = 0 in Σ × M. Indeed, this amounts to saying that 0 H (σ˜ , u) ˜ is injective. To prove the above equivalence, we first show that the operator H 0 (σ˜ , u) ˜ defined by (2.27) from Σ ∗ × M → Σ ∗ × M is a Fredholm operator of index 0, where Σ ∗ denotes the space [L r1 (Ω)]2×2 × L r1 (Ω) endowed with the natural norm. P ROPOSITION 2.4 The operator K (σ˜ , u): ˜ Σ∗ × M → Σ∗ × M (τ˜ = (τ, q), v˜ = (v, η)) 7→ S

1 (σ ∙ v + τ ∙ u) + ω ∙ v + η ∙ u 2μ

˜ = I + K (σ˜ , u) ˜ is a Fredholm operator of index 0 is a linear compact operator. Consequently, H 0 (σ˜ , u) in the same space. Proof. We first show that the operator K (σ˜ , u) ˜ is well-defined. Indeed, as H 1,α (Ω) ,→ L s (Ω) for 2 all s < α , by Theorem 2.1 σ and ω belong to [L s (Ω)]2×2 for all s < α2 . Choosing s from the relation 1 1 1 4 1 1 α 1 1 1 s + r2 = t , recalling that t = 2+α and remarking that r2 < 2 − 4 , we see that s = t − r2 >

− 12 + α4 = α2 which is equivalent to s < α2 . As v ∈ [L r2 (Ω)]2 , by Hölder’s inequality we conclude that σ ∙ v, ω ∙ v ∈ [L t (Ω)]2 . For the products τ ∙ u and η ∙ u, by Hölder’s inequality we 2 have H 2,α (Ω) ,→ W 2, p (Ω) for p < 1+α , combined with the Sobolev embedding theorem, we get 2,α ˉ Since by Theorem 2.1 u belongs to [H 2,α (Ω)]2 , we obtain u ∈ [L ∞ (Ω)]2 . Again, H (Ω) ,→ C(Ω). 4 Hölder’s inequality allows us to conclude that τ ∙ u, η ∙ u ∈ [L r1 (Ω)]2 . As r1 > 2 > t = 2+α , this 1 t 2 ∗ immediately implies τ ∙ u, η ∙ u ∈ [L (Ω)] . Summarizing, f := 2μ (σ ∙ v + τ ∙ u) + ω ∙ v + η ∙ u belongs to [L t (Ω)]2 and therefore S f ∗ is in Σ × M ⊂ Σ ∗ × M. Let us now consider a bounded sequence (τñ , vñ ) = ((τn , qn ), (vn , ηn )) in Σ ∗ × M. Then, by 1 Hölder’s inequality or the Sobolev embedding theorem, we obtain that the sequence f n∗ := 2μ (σ ∙ t 2 ∗ ∗ ∗ vn + τn ∙ u) + ω ∙ vn + ηn ∙ u is a bounded sequence in [L (Ω)] . Now, setting (σ˜ n = (σn , pn ), u˜ ∗n = 2 (u ∗n , ωn∗ )) := S f n∗ and s := 1+α , it follows from our preceding study of the linear operator S that ∗ ∗ 2,s 2 1,s (u n , pn ) ∈ [W (Ω)] ×W (Ω) and is bounded in that space thanks to (2.20). Consequently, (σ˜ n∗ , u˜ ∗n ) 2+α 4

32

M. FARHLOUL ET AL.

belongs to ([W 1,s (Ω)]2×2 × W 1,s (Ω)) × ([W 2,s (Ω)]2 × [W 1,s (Ω)]2×2 ) and is a bounded sequence in that space. By the compact embeddings W 1,s (Ω) ,→c L r1 (Ω) and W 2,s (Ω) ,→c L r2 (Ω), there exists a ∗ ∗ subsequence σ˜ n k , u˜ n k k∈N that converges in Σ ∗ × M. This proves the compactness of the linear operator K (σ˜ , u): ˜ Σ ∗ × M → Σ ∗ × M.

C OROLLARY 2.5 If the operator H 0 (σ˜ , u) ˜ is injective from Σ × M into itself, then it is also an isomorphism in the same spaces.

Proof. We first show the injectivity of H 0 (σ˜ , u) ˜ from Σ ∗ × M into itself. Indeed, let (τ˜ , v) ˜ ∈ Σ∗ × M 0 be such that H (σ˜ , u)( ˜ τ˜ , v) ˜ = 0. From the properties of S, we deduce that (τ˜ , v) ˜ belongs to Σ × M, from which it follows that (τ˜ , v) ˜ = (0, 0). ˜ being injective from Σ ∗ × M into itself, by Proposition 2.4, it is an isomorphism from H 0 (σ˜ , u) ∗ Σ × M into itself. It remains to show that the mapping H 0 (σ˜ , u) ˜ is surjective from Σ × M into itself. Given (τ˜ ∗ , v˜ ∗ ) ∗ in Σ × M, since it also belongs to Σ × M, by the surjectivity of H 0 (σ˜ , u) ˜ from Σ ∗ × M into itself, ∗ there exists (τ˜ , v) ˜ in Σ × M such that ˜ τ˜ , v) ˜ = (τ˜ ∗ , v˜ ∗ ), H 0 (σ˜ , u)( or equivalently ∗

∗

(τ˜ , v) ˜ = (τ˜ , v˜ ) − S

1 (σ ∙ v + τ ∙ u) + ω ∙ v + η ∙ u . 2μ

Since this right-hand side belongs to Σ × M, we are done.

˜ from Σ × M into itself is not compact since we can R EMARK 2.6 We suspect that the operator K (σ˜ , u) ˜ τñ , vñ )) is only only show that for a bounded sequence (τñ , vñ ) in Σ × M, the sequence div(K (σ˜ , u)( bounded in [L t (Ω)]2 and so has no reason to be convergent in [L t (Ω)]2 . ˆ For further use, we also need the compactness of the operator K (σ˜ , u) ˜ in the larger space Σˆ × M.

˜ is compact in the Banach space Σˆ × Mˆ defined in (2.22) and P ROPOSITION 2.7 The operator K (σ˜ , u) (2.23). ˆ Fix (τ˜ , v) Proof. We first show that the operator K (σ˜ , u) ˜ is well-defined in the Banach space Σˆ × M. ˜ = 2 s 2×2 ˆ ˆ ((τ, q), (v, η)) ∈ Σ × M. Recall that σ, ω ∈ [L (Ω)] for all s < α . Choosing 1 < s < α2 such that 2 2 1 1 1+α 1+α (Ω) because v ∈ [L r (Ω)]2 . s + r = 2 (meaningful since r > 2), it follows that σ ∙ v, ω ∙ v ∈ L As u is bounded on Ω, it follows immediately that τ ∙ u and η ∙ u belong to [L r (Ω)]2 and thus to 2 2 2 1 2 L 1+α (Ω) as r > 2 > 1+α . Hence, we get 2μ (σ ∙ v + τ ∙ u) + ω ∙ v + η ∙ u ∈ [L s (Ω)]2 with s = 1+α . ∗ By the proof of the preceding proposition, we deduce that K (σ˜ , u)( ˜ τ˜ , v) ˜ ∈ Σ × M and therefore it ˆ also belongs to Σˆ × M. ∗ ∗ ˆ ˆ Let us now consider a bounded sequence (τñ , vñ ) = ((τn , qn ), (vn , ηn )) in Σ × M. Setting (σ˜ n , u˜ n ) 1 := S 2μ (σ ∙ vn + τn ∙ u) + ω ∙ vn + ηn ∙ u and arguing as in the proof of Proposition 2.4, we show that the sequence (σ˜ n∗ , u˜ ∗n )n∈N is bounded in ([W 1,s (Ω)]2×2 × W 1,s (Ω)) × ([W 2,s (Ω)]2 × [W 1,s (Ω)]2×2 ), 1 using the fact that the sequence 2μ (σ ∙ vn + τn ∙ u) + ω ∙ vn + ηn ∙ u is bounded in [L s (Ω)]2 . By 1,s the compact embedding W (Ω) ,→c L r (Ω), there exists a subsequence σ˜ n∗k , u˜ ∗n k k∈N converging in ˆ which proves the compactness result. Σˆ × M,


33

3. The discrete problem Let us now introduce the discrete version of (2.15), (2.16) by using one of the mixed finite elements developed by Farhloul & Fortin (1997). Since the singularities of our problem have a local character, the meshes have to be refined in a neighbourhood of each singular corner (a corner S j is called singular if ξ S (ω j ) < 1 implying that u does not belong to H 2 in a neighbourhood of S j ). So without loss of generality, we may suppose that only one corner is singular. Moreover, by an eventual translation, we may suppose that it is situated at the origin 0. Let (Th )h>0 be a family of triangulations of Ω satisfying the following conditions: there exist two positive constants γ and γ˜ such that (Th 1) hK 6γ ρK

∀ K ∈ Th , ∀ h > 0,

(Th 2) h K 6 γ h β for every K ∈ Th such that one of the corners of K is at 0, (Th 3) hK 6 γ

inf r 1−1/β (x) h for every K ∈ Th with no corner at 0,

x∈K

(Th 4) h K > γ˜ h β

∀ K ∈ Th , ∀ h > 0,

where h K denotes the diameter of K , ρ K denotes the supremum of the diameters of the inscribed circles in K and β > 1/(1 − α). R EMARK 3.1 The condition (Th 1) means that the family of triangulations (Th )h>0 is regular (see Ciarlet, 1978). The other conditions are, in particular, true for Raugel’s families of triangulations (see Raugel, 1978). For a triangle K ⊂ R2 , let Pk , k > 0, denote the restrictions of polynomials of total degree 6k to K . We set P = [P1 ]2 + R curl b K , where b K denotes the ‘bubble function’ defined by b K (x) = λ1 (x)λ2 (x)λ3 (x), with λ1 , λ2 and λ3 the ∂b K K barycentric coordinates on K , and curl b K = ∂b ∂ x2 , − ∂ x1 . We define the following finite-dimensional spaces: Σh = {τ˜h = (τh , qh ) ∈ Σ: qh |K ∈ P1 , τh |K ∈ [P]2 , ∀ K ∈ Th },

(3.1)

Mh = {v˜ h = (vh , ηh ) ∈ M: vh |K ∈ [P0 ]2 , ηh = θh χ with θh |K ∈ P1 , ∀ K ∈ Th },

(3.2)

where χ = ( 10 −1 0 ).

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M. FARHLOUL ET AL.

The discretization of (2.15) and (2.16) reads as follows: find σ˜ h = (σh , ph ) ∈ Σh and u˜ h = (u h , ωh ) ∈ Mh such that 1 (σh , τh ) + (div(τh − qh δ), u h ) + (τh , ωh ) = 0 ∀ τ˜h = (τh , qh ) ∈ Σh , 2μ (div(σh − ph δ), vh ) + (σh , ηh ) −

(3.3)

1 (σh ∙ u h , vh ) 2μ

− (ωh ∙ u h , vh ) + ( f, vh ) = 0 ∀ v˜ h = (vh , ηh ) ∈ Mh .

(3.4)

In order to analyse the discrete problem (3.3), (3.4) and prove the error estimates, we have to prove some intermediate results. We start by defining interpolation operators and proving their properties. P ROPOSITION 3.2 Let X h = {vh ∈ H (div, Ω): vh |K ∈ [P1 ]2 , ∀ K ∈ Th } and Yh = {qh ∈ L 2 (Ω): qh |K ∈ P1 , ∀ K ∈ Th }. Then, there exist two interpolation operators ρh ∈ L ([H 1,α (Ω)]2 , X h ) and Ph1 ∈ L (H 1,α (Ω), Yh ) such that for all s ∈ [2, 2/α), kρh v − vk0,s,Ω 6 Ch 1+β(2/s−1) |v|1,α,Ω ,

(3.5)

kPh1 q − qk0,s,Ω 6 Ch 1+β(2/s−1) |q|1,α,Ω ,

(3.6)

where C is, here and below, a positive constant independent of h. Proof. Let v ∈ [H 1,α (Ω)]2 . We define ρh v by (ρh v)|K = ρ K v ∈ [P1 (K )]2 and Z Z ρ K v ∙ np1 ds = v ∙ np1 ds ∀ p1 ∈ P1 (e), ∀ e ⊂ ∂ K . e

e

Observe that ρh v is nothing else but the lowest degree Brezzi–Douglas–Marini interpolant of v (see Brezzi et al., 1985). Now, using similar arguments as in the proof of Proposition 4.2 in Farhloul et al. (2001), we obtain 2/s−1 1−α h K |v|1,α,K

kv − ρh vk0,s,K 6 Ch K

∀ K ∈ Th ,

for all v ∈ [H 1,α (Ω)]2 and all s ∈ [2, 2/α). To conclude, we have to consider two cases: First case: The triangle K has one of its vertices at the origin. Owing to (Th 2), (Th 4) and the fact that 2/s−1 β > 1/(1 − α), we have h 1−α 6 γ 1−α h and h K 6 γ˜ 2/s−1 h β(2/s−1) . Thus, K kv − ρh vk0,s,K 6 Ch 1+β(2/s−1) |v|1,α,K . Second case: None of the vertices of the triangle K are at the origin. Owing to (Th 3), (Th 4) and the fact that β > 1/(1 − α), we have 2/s−1

kv − ρh vk0,s,K 6 Ch K

h K |v|1,K

hK |v|1,α,K infx∈K r α (x) 6 Ch 1+β(2/s−1) |v|1,α,K . 6 Ch β(2/s−1)

The estimate (3.5) follows from the two estimates above.

35


The operator Ph1 is defined as the L 2 -orthogonal projection from L 2 (Ω) onto q ∈ H 1,α (Ω). Then, it is easy to show that for all s ∈ [2, 2/α), ∀ K ∈ Th ,

Q

K ∈T h

P1 (K ). Let

kq − Ph1 qk0,s,K 6 C|J |1/s |q| ˆ 1,α,K , where qˆ = q ◦ FK . Therefore, using the estimate |q| ˆ 1,α, Kˆ 6 Ch −α K |q|1,α,K

and considering the two cases above, we get the estimate (3.6). R EMARK 3.3 One can prove that, under the hypothesis β > 2/(2 − α), ∀ s ∈ [2, 2/α), kρh v − vk0,s,Ω 6 Ch 2+β(2/s−1) |v|2,α,Ω

∀ v ∈ [H 2,α (Ω)]2 ,

kPh1 q − qk0,s,Ω 6 Ch 2+β(2/s−1) |q|2,α,Ω

∀ q ∈ H 2,α (Ω).

However, the operators ρh and Ph1 will serve to construct an interpolant of (σ, p) (see the the next proposition) but the solution (σ, p) of problem (2.15), (2.16) is only in the space [H 1,α (Ω)]2×2 × H 1,α (Ω). 1,s 2 L EMMA 3.4 For any γ ∈ L s (Ω) with s > 2, there exists w ∈ W0,Γ (Ω) := {v ∈ [W 1,s (Ω)]2 : N v = 0 on Γ N } satisfying div w = γ

in Ω,

kwk1,s,Ω 6 Ckγ k0,s,Ω ,

(3.7) (3.8)

for some positive constant C (independent of γ and w). ˉ Γ N ⊂ ∂ O and O\Ωˉ 6= ∅ (this is always Proof. Consider O a bounded domain of R2 such that Ωˉ ⊂ O, possible since Γ D is not empty). Now, consider an extension γ˜ defined as follows: ( γ on Ω, γ˜ = ˉ c on O\Ω, for a constant c chosen so that

Z

O

γ˜ (x)dx = 0,

which means that c is given by ˉ −1 c = −|O\Ω|

Z

γ (x)dx, Ω

ˉ means the area of O\Ω. ˉ Note that this identity implies that where |O\Ω| ˉ −1 |Ω|1/2 kγ k0,2,Ω 6 C1 kγ k0,s,Ω , |c| 6 |O\Ω|

(3.9)

ˉ this last estimate comes from the continuous for some C1 > 0 depending only on |Ω| and |O\Ω|; s 2 embedding L (Ω) ,→ L (Ω).

36

M. FARHLOUL ET AL.

Since γ˜ has a zero mean on O, adapting the arguments of Corollary I.2.4 of Girault & Raviart (1986) to the L s setting, there exists w˜ ∈ [W01,s (O)]2 such that div w˜ = γ˜

in O,

kwk ˜ 1,s,O 6 C2 kγ˜ k0,s,O ,

(3.10) (3.11)

for some positive constant C2 (independent of γ˜ and w). ˜ 2 1,s (Ω) since Γ N is Taking w as the restriction of w˜ to Ω, we directly see that it belongs to W0,Γ N a part of the boundary of O and that it satisfies (3.7) as a consequence of (3.10). To check the estimate (3.8), we remark that ˉ kγ˜ ks0,s,O = kγ ks0,s,Ω + |c|s |O\Ω|, and therefore by the estimate (3.9), we get kγ˜ k0,s,O 6 C3 kγ k0,s,Ω , ˉ Using this last estimate in (3.11) leads to (3.8). for some C3 > 0 depending only on |Ω| and |O\Ω|. P ROPOSITION 3.5 There exists an operator Πh : Σ ∩ ([H 1,α (Ω)]2×2 × H 1,α (Ω)) → Σh τ˜ = (τ, q) 7→ Πh τ˜ = (τh , qh ) such that for all v˜ h = (vh , ηh ) ∈ Mh , (div[(τ − qδ) − (τh − qh δ)], vh ) + (τ − τh , ηh ) = 0

(3.12)

kτ˜ − Πh τ˜ k0,s,Ω 6 Ch 1+β(2/s−1) |τ˜ |1,α,Ω ,

(3.13)

and for all s ∈ [2, 2/α), where |τ˜ |1,α,Ω = |τ |1,α,Ω + |q|1,α,Ω and C is a positive constant independent of h.

Proof. Let τ˜ = (τ, q) ∈ Σ ∩ ([H 1,α (Ω)]2×2 × H 1,α (Ω)) and set qh = Ph1 q. We define τh∗ such that for all K ∈ Th , (τh∗ − qh δ)|K ∈ [P1 (K )]2×2 , Z [(τh∗ − qh δ) − (τ − qδ)] ∙ n ∙ p1 ds = 0 e

∀ p1 ∈ [P1 (e)]2 , ∀ e ⊂ ∂ K .

1,s 2 Let γ = as(τ − τh∗ ), where as(τ ) = τ21 − τ12 . Due to Lemma 3.4, there exists w ∈ W0,Γ (Ω) such N that div w = γ and kwk1,s,Ω 6 Ckγ k0,s,Ω 6 Ckτ − τh∗ k0,s,Ω . Using the fact that the discretization of the Stokes problem (in primitive variables) by [P2 ⊕Rb K ]2 − P1 (i.e. the Crouzeix–Raviart element P2+ − P1 ; see Crouzeix & Raviart, 1973) is stable, we get h i2 1,s 2 wh ∈ v ∈ W0,Γ N (Ω) : v|K ∈ [P2 (K ) ⊕ Rb K ] , ∀ K ∈ Th ,


37

so that (div(w − wh ), μh ) = 0 for all μh ∈ {μ ∈ L 2 (Ω): μ|K ∈ P1 (K ), ∀ K ∈ Th }, and kwh k1,s,Ω 6 Ckwk1,s,Ω . Finally, set Πh (τ, q) = (τh , qh ), where τh = τh∗ +

curl wh,1 curl wh,2

with wh = (wh,1 , wh,2 ). Thus, we have Πh (τ, q) ∈ Σh , div(τh − qh δ) = div(τh∗ − qh δ) and as(τh ) = as(τh∗ ) + div wh . Now, using all these properties and ηh = θh χ , we get (div[(τ − qδ) − (τh − qh δ)], vh ) + (τ − τh , ηh ) = (div[(τ − qδ) − (τh∗ − qh δ)], vh ) + (as(τ − τh ), θh ) = (as(τ − τh∗ ) − div wh , θh ) = (as(τ − τh∗ ) − div w, θh ) = 0, which proves (3.12). To prove (3.13), first observe that each line of τh∗ − qh δ is nothing else but the lowest degree Brezzi–Douglas–Marini interpolant of the corresponding line of the tensor τ − qδ. Thus, using (3.5) and (3.6), we get kτ˜ − Πh τ˜ k0,s,Ω 6 C(k(τ − qδ) − (τh − qh δ)k0,s,Ω + kq − qh k0,s,Ω ) 6 C(k(τ − qδ) − (τh∗ − qh δ)k0,s,Ω + kwh k1,s,Ω + kq − qh k0,s,Ω ) 6 C(k(τ − qδ) − (τh∗ − qh δ)k0,s,Ω + kq − qh k0,s,Ω ) 6 Ch 1+β(2/s−1) (|τ |1,α,Ω + |q|1,α,Ω ). R EMARK 3.6 Let X h = {ψ ∈ L 2 (Ω): ψ|K ∈ P1 (K ), ∀ K ∈ Th }. Applying the results to the operators ρh and Ph1 (cf. Brezzi et al., 1985; Brezzi & Fortin, 1991; Ciarlet, 1978) and the fact that (Th )h>0 is regular and proceeding as in the proofs of Propositions 3.2 and 3.5, we have the following estimate: 2×2 0 0 ∀ t 0 ∈ (1, 2) and for all s such that 2 6 s < 2t 0 /(2−t 0 ), Πh ∈ L Σ ∩ W 1,t (Ω) ×W 1,t (Ω) , Σh , 0 Ph1 ∈ L W 1,t (Ω), X h and 0

kΠh τ˜ − τ˜ k0,s,Ω 6 Ch 1+2/s−2/t |τ˜ |1,t 0 ,Ω , 0

kPh1 ψ − ψk0,s,Ω 6 Ch 1+2/s−2/t |ψ|1,t 0 ,Ω ,

(3.14)

(3.15)

where |τ˜ |1,t 0 ,Ω = |τ |1,t 0 ,Ω + |q|1,t 0 ,Ω and C is a positive constant independent of h. Now, in order to express the discrete problem (3.3), (3.4) in the same form as the continuous problem (2.26), we introduce the discrete analogue Sh of the operator S. For fˆ ∈ [L t (Ω)]2 ,

38

M. FARHLOUL ET AL.

Sh fˆ = ((σˆ h , pˆ h ), (uˆ h , ωˆ h )) ∈ Σh × Mh is the solution of the problem 1 (σˆ h , τh ) + (div(τh − qh δ), uˆ h ) + (τh , ωˆ h ) = 0 2μ

∀ (τh , qh ) ∈ Σh ,

(div(σˆ h − pˆ h δ), vh ) + (σˆ h , ηh ) + ( fˆ, vh ) = 0 ∀ (vh , ηh ) ∈ Mh .

(3.16) (3.17)

The well-posedness of the problem (3.16), (3.17) is a consequence of the following results. L EMMA 3.7 There exists a positive constant C, independent of h, such that kτh k20,Ω > Ckτ˜h k20,Ω

∀ τ˜h = (τh , qh ) ∈ Vh ,

(3.18)

where Vh = {τ˜h ∈ Σh : (div(τh − qh δ), vh ) + (τh , ηh ) = 0 ∀ (vh , ηh ) ∈ Mh }.

(3.19)

Proof. The proof of this result is similar to the one of Proposition 3.2 in Farhloul & Fortin (1993).

L EMMA 3.8 There exists a positive constant C, independent of h, such that sup τ˜h ∈Σh

(div(τh − qh δ), vh ) + (τh , ηh ) > Ckv˜ h k0,Ω kτ˜h k0,Ω

∀ v˜ h ∈ Mh .

(3.20)

Proof. The proof of this result is similar to the one of Proposition 4.2 in Farhloul & Paquet (2002). Next, we define the mapping Hh from Σh × Mh into itself by 1 τh ∙ vh − ηh ∙ vh (3.21) Hh (τ˜h , v˜ h ) := (τ˜h , v˜ h ) − Sh f − 2μ for every (τ˜h , v˜ h ) = ((τh , qh ), (vh , ηh )) ∈ Σh × Mh . Then, the discrete problem (3.3), (3.4) can be written as follows: find (σ˜ h , u˜ h ) = ((σh , ph ), (u h , ωh )) ∈ Σh × Mh such that Hh (σ˜ h , u˜ h ) = 0.

(3.22)

Finally, we introduce the discrete operator K h of K . We denote by Ph0 the L 2 -orthogonal projection Q 2 ˜ , u) ˜ = ((σ, p), (u, ω)) ∈ Σ × M and set (σ˜ h∗ , u˜ ∗h ) = from [L 2 (Ω)]2 onto K ∈Th P0 (K ) . Let (σ 0 1 (Πh σ˜ , Ph u), ˜ where Ph u˜ = (Ph u, (Ph θ)χ) with ω = θ χ . The operator K h (σ˜ h∗ , u˜ ∗h ) ∈ L (Σh × Mh , Σh × Mh ) is defined by 1 ∗ ∗ ∗ ∗ ∗ ∗ (σ ∙ vh + τh ∙ u h ) + ωh ∙ vh + ηh ∙ u h . (3.23) K h (σ˜ h , u˜ h )(τ˜h , v˜ h ) = Sh 2μ h R EMARK 3.9 Similarly to (3.6), we have kPh0 v − vk0,s,Ω 6 Ch 1+β(2/s−1) |v|1,α,Ω

(3.24)

for all v ∈ [H 1,α (Ω)]2 and all s ∈ [2, 2/α). 2 0 We also have ∀ t 0 ∈ (1, 2), for all s such that 2 6 s < 2t 0 /(2 − t 0 ) and for all v ∈ W 1,t (Ω , 0

kPh0 v − vk0,s,Ω 6 Ch 1+2/s−2/t |v|1,t 0 ,Ω .

(3.25)


39

4. Error estimates To prove error estimates, we follow the procedure used in Farhloul et al. (2001) for the Boussinesq equations but adapted to our setting. L EMMA 4.1 Let fˆ ∈ [L 2 (Ω)]2 . Then, we have the following estimate:

1+β(2/r −1) kS fˆk1,α,Ω . k(Sh − S) fˆkΣ× ˆ Mˆ 6 Ch

(4.1)

ˆ and Proof. Let ((σˆ , p), ˆ (u, ˆ ω)) ˆ = S fˆ and ((σˆ h , pˆ h ), (uˆ h , ωˆ h )) = Sh fˆ and set (σˆ h∗ , pˆ h∗ ) = Πh (σˆ , p) ∗ ∗ ˆ ω). ˆ Then, similarly to Theorem 4.16 in Farhloul & Paquet (2002), using the theory (uˆ h , ωˆ h ) = Ph (u, of mixed finite-element methods (cf. Brezzi & Fortin, 1991) and (3.18), we have kσˆ h∗ − σˆ h k0,Ω + k pˆ h∗ − pˆ h k0,Ω + kuˆ ∗h − uˆ h k0,Ω + kωˆ h∗ − ωˆ h k0,Ω 6 C(kσˆ − σˆ h∗ k0,Ω + kωˆ − ωˆ h∗ k0,Ω ). This last inequality, with (3.6) and (3.13), gives us kσˆ h∗ − σˆ h k0,Ω + k pˆ h∗ − pˆ h k0,Ω + kuˆ ∗h − uˆ h k0,Ω + kωˆ h∗ − ωˆ h k0,Ω 6 Ch(|σˆ |1,α,Ω + | p| ˆ 1,α,Ω + |ω| ˆ 1,α,Ω ).

(4.2)

Now, using the inverse estimate (owing to (Th 4)) kψh k0,r,Ω 6 Ch β(2/r −1) kψh k0,Ω , the estimate (4.2) leads to kσˆ h∗ − σˆ h k0,r,Ω + k pˆ h∗ − pˆ h k0,r,Ω + kuˆ ∗h − uˆ h k0,r,Ω + kωˆ h∗ − ωˆ h k0,r,Ω 6 Ch 1+β(2/r −1) (|σˆ |1,α,Ω + | p| ˆ 1,α,Ω + |ω| ˆ 1,α,Ω ). Therefore, (4.1) is a consequence of (4.3), (3.6), (3.13), (3.24) and the triangle inequality. 2 0 L EMMA 4.2 Let fˆ ∈ L t (Ω) ∩ L t (Ω) , where t = 4/(2 + α) and

(4.3)

2 2r < t0 < . 2+r 2 − min j ξ S (ω j )

Then, we have the following estimate: 2−2/t 0 +β(2/r −1) k(Sh − S) fˆkΣ× kS fˆk1,t 0 ,Ω . ˆ Mˆ 6 Ch

(4.4)

2 0 Proof. First, from Corollary 4.2 in Orlt & Sändig (1994), we have that for fˆ ∈ L t (Ω) , with t 0 < 2 0 2/(2 − min ξ S (ω j )), the solution of the Stokes problem (2.18), (2.19) satisfies uˆ ∈ W 2,t (Ω) , pˆ ∈ 0 ˆ 2,t 0 ,Ω + k pk ˆ 1,t 0 ,Ω 6 Ck fˆk0,t 0 ,Ω . W 1,t (Ω) and kuk Now, following the proof of Lemma 4.1 and using the estimates (3.14) and (3.15), we get kσˆ h∗ − σˆ h k0,r,Ω + k pˆ h∗ − pˆ h k0,r,Ω + kuˆ ∗h − uˆ h k0,r,Ω + kωˆ h∗ − ωˆ h k0,r,Ω 0

6 Ch 2−2/t +β(2/r −1) (|σˆ |1,t 0 ,Ω + | p| ˆ 1,t 0 ,Ω + |ω| ˆ 1,t 0 ,Ω ),

where (σˆ h∗ , pˆ h∗ ) = Πh (σˆ , p) ˆ and (uˆ ∗h , ωˆ h∗ ) = Ph (u, ˆ ω). ˆ This last estimate, with (3.14), (3.15) and (3.25), gives us (4.4) for all t 0 such that 2r/(2 + r ) < t 0 < 2/(2 − min j ξ S (ω j )).

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M. FARHLOUL ET AL.

C OROLLARY 4.3 Assume that fˆ ∈ [L t (Ω)]2 , where we recall that t = 4/(2 + α). Let t 0 = 2/(1 + α). Then, for all r such that 2β 2 < r < min r1 , r2 , , β − (1 − α) we have ε ˆ k(Sh − S) fˆkΣ× ˆ Mˆ 6 Ch k f k0,t 0 ,Ω ,

(4.5)

for some ε > 0. Proof. First observe that t = 4/(2 + α) > t 0 = 2/(1 + α) and t 0 = 2/(1 + α) satisfy the conditions of Lemma 4.2. Thus, the estimate (4.4) holds. On the other hand, since t 0 = 2/(1+α) and r < 2β/(β −(1− α)), we have 2 − 2/t 0 + β(2/r − 1) > 0. Therefore, (4.4), with the fact that kS fˆk1,t 0 ,Ω 6 Ck fˆk0,t 0 ,Ω , yields (4.5). Now, we take β=

2/r , 2/r − α

(4.6)

with r < 4/(1 + α).

L EMMA 4.4 Assume that (σ˜ , u) ˜ is a nonsingular solution of (2.26). Then, for all r such that 2(2 − α) 2 < r < min r1 , r2 , , 1 + α(1 − α)

we have lim kK − K h kL (Σ× ˆ Σ× ˆ = 0. ˆ M, ˆ M)

(4.7)

2/r 2/r −α

and r > 2. On the other hand, the

h→0

Proof. First, let us point out that β < 1/(2/r − α) since β = condition r
0. We also have 1 + β(2/r − 1) > 0 since β < (1 − α)/(1 − 2/r ). Similar arguments as above, with (3.6) and (3.24), lead to 1+β(2/s−1) |u|1,α,Ω kτ k0,r,Ω , kS(τ ∙ (u − u ∗h )kΣ× ˆ Mˆ 6 Ch 1+β(2/s−1) kS((ω − ωh∗ ) ∙ v)kΣ× |ω|1,α,Ω kvk0,r,Ω , ˆ Mˆ 6 Ch 1+β(2/s−1) kS(η ∙ (u − u ∗h )kΣ× |u|1,α,Ω kηk0,r,Ω . ˆ Mˆ 6 Ch

Now, using (4.5) (since r satisfies the conditions of Corollary 4.3) and (3.13), we have ε ∗ k(S − Sh )(σh∗ ∙ v)kΣ× ˆ Mˆ 6 Ch kσh ∙ vk0,t0,Ω

6 Ch ε kσh∗ k0,s,Ω kvk0,r,Ω

6 Ch ε k(σ, p)k1,α,Ω kvk0,r,Ω , for some ε > 0. Similar arguments as above, with (3.6) and (3.24), lead to ε k(S − Sh )(τ ∙ u ∗h )kΣ× ˆ Mˆ 6 Ch kuk1,α,Ω kτ k0,r,Ω ,

ε k(S − Sh )(ωh∗ ∙ v)kΣ× ˆ Mˆ 6 Ch kωk1,α,Ω kvk0,r,Ω ,

ε k(S − Sh )(η ∙ u ∗h )kΣ× ˆ Mˆ 6 Ch kuk1,α,Ω kηk0,r,Ω .

Finally, using all the estimates above, we get γ k(K − K h )(τ˜ , v)k ˜ Σ× ˜ Σ× ˆ Mˆ 6 Ch k(τ˜ , v)k ˆ Mˆ ,

where γ = min{1 + β(2/s − 1), ε}. Therefore, the result follows immediately.

L EMMA 4.5 For small enough h, the operator (I + K h ) is an isomorphism from Σˆ × Mˆ into itself. Moreover, the operator (I + K h )−1 maps Σh × Mh into itself, and its norm is bounded, independently of h. Proof. First, since the operator K is compact from Σˆ × Mˆ into itself (cf. Proposition 2.7), I + K is a Fredholm operator of index 0. On the other hand, I + K is an isomorphism from Σ × M into itself since (σ˜ , u) ˜ is a nonsingular solution of (2.26). This implies that the range of I + K as an operator from

42

M. FARHLOUL ET AL.

Σˆ × Mˆ into itself contains the dense subset Σ × M. Consequently, its range is the whole space Σˆ × Mˆ and then I + K is an isomorphism from Σˆ × Mˆ into itself. This fact, with Lemma 4.4 and a classical perturbation argument (cf. Girault & Raviart, 1986), gives us the desired result. ˜ is a nonsingular solution of (2.26) and L EMMA 4.6 Assume that (σ˜ , u) 2(2 − α) 2 < r < min r1 , r2 , . 1 + α(1 − α) Then, there exists a constant C > 0 such that 1+β(2/r −1) , kHh (σ˜ h∗ , u˜ ∗h )kΣ× ˆ Mˆ 6 Ch

(4.8)

where σ˜ h∗ = (σh∗ , ph∗ ) = Πh (σ, p) and u˜ ∗h = (u ∗h , ωh∗ ) = Ph (u, ω).

Proof. Since (σ˜ , u) ˜ is a solution of (2.26), we may write

1 σ ∙u−ω∙u − u) ˜ + (S − Sh ) f − 2μ 1 ∗ ∗ ∗ ∗ + Sh (σ ∙ u − σ ∙ u) + (ωh ∙ u h − ω ∙ u) . 2μ h h

Hh (σ˜ h∗ , u˜ ∗h ) = (σ˜ h∗

− σ˜ , u˜ ∗h

(4.9)

From (3.6), (3.13), (3.24) and (4.1), we have 1+β(2/r −1) k(σ˜ h∗ − σ˜ , u˜ ∗h − u)k ˜ Σ× , ˆ Mˆ 6 Ch

1+β(2/r −1)

(S − Sh ) f − 1 σ ∙ u − ω ∙ u .

ˆ ˆ 6 Ch

2μ Σ× M

(4.10) (4.11)

The term Sh (σh∗ ∙ u ∗h − σ ∙ u) may be written as

Sh (σh∗ ∙ u ∗h − σ ∙ u) = Sh (σh∗ ∙ (u ∗h − u)) + Sh ((σh∗ − σ ) ∙ u). t 2 ˆ ˆ Using the fact that (owing to (4.5)) kSh fˆkΣ× ˆ Mˆ 6 Ck f k0,t 0 ,Ω for all f ∈ [L (Ω)] (with t = 4/(2 + α) 0 and t = 2/(1 + α)), (3.13) and (3.24), we get ∗ ∗ kSh (σh∗ ∙ (u ∗h − u))kΣ× ˆ Mˆ 6 Ckσh ∙ (u h − u)k0,t 0 ,Ω

6 Ckσh∗ k0,s,Ω ku ∗h − uk0,r,Ω

6 Ch 1+β(2/r −1) kσ˜ k1,α,Ω |u|1,α,Ω , ∗ kSh ((σh∗ − σ ) ∙ u)kΣ× ˆ Mˆ 6 Ck(σh − σ ) ∙ uk0,t 0 ,Ω

6 Ckσh∗ − σ k0,r,Ω kuk0,s,Ω

6 Ch 1+β(2/r −1) |σ˜ |1,α,Ω kuk1,α,Ω . Thus, 1+β(2/r −1) . kSh (σh∗ ∙ u ∗h − σ ∙ u)kΣ× ˆ Mˆ 6 Ch

(4.12)

43


Finally, a similar procedure leads to 1+β(2/r −1) , kSh (ωh∗ ∙ u ∗h − ω ∙ u)kΣ× ˆ Mˆ 6 Ch

and from (4.9)–(4.12) and the last estimate, we obtain (4.8). We are now able to prove an error estimate for nonsingular solutions of (2.26).

˜ is a nonsingular solution of (2.26). Then, for h small enough, T HEOREM 4.7 Suppose that (σ˜ , u) problem (3.22) has at least one solution (σ˜ h , u˜ h ) such that 2

r k(σ˜ , u) ˜ − (σ˜ h , u˜ h )kΣ× ˆ Mˆ 6 Ch

where

2 < r < min r1 , r2 ,

(2/r −1) + 2/r2/r −α

,

(4.13)

2(2 − α) . 1 + α(1 − α)

Proof. We define the following map R from Σh × Mh into itself:

R(τ˜h , v˜ h ) = (τ˜h , v˜ h ) − (I + K h )−1 Hh (τ˜h , v˜ h )

and prove that it has a fixed point in a neighbourhood of (σ˜ h∗ , u˜ ∗h ), where σ˜ h∗ = (σh∗ , ph∗ ) = Πh (σ, p) and u˜ ∗h = (u ∗h , ωh∗ ) = Ph (u, ω). To this end, we start by estimating kR(τ˜h , v˜ h ) − (σ˜ h∗ , u˜ ∗h )kΣ× ˆ Mˆ in terms of k(τ˜h , v˜ h ) − (σ˜ h∗ , u˜ ∗h )kΣ× ˆ Mˆ . Using Lemma 4.5, we may write R(τ˜h , v˜ h ) − (σ˜ h∗ , u˜ ∗h ) = (I + K h )−1 (I + K h )(R(τ˜h , v˜ h ) − (σ˜ h∗ , u˜ ∗h )) and then kR(τ˜h , v˜ h ) − (σ˜ h∗ , u˜ ∗h )kΣ× ˜ h∗ , u˜ ∗h ))kΣ× ˆ Mˆ 6 Ck(I + K h )(R(τ˜h , v˜ h ) − (σ ˆ Mˆ .

(4.14)

On the other hand, (I + K h )(R(τ˜h , v˜ h ) − (σ˜ h∗ , u˜ ∗h ))

= (I + K h )((τ˜h , v˜ h ) − (σ˜ h∗ , u˜ ∗h )) − Hh (τ˜h , v˜ h )

= Hh (σ˜ h∗ , u˜ ∗h ) − Hh (τ˜h , v˜ h ) + (I + K h )((τ˜h , v˜ h ) − (σ˜ h∗ , u˜ ∗h )) − Hh (σ˜ h∗ , u˜ ∗h ) 1 ∗ (σh − τh ) ∙ (u ∗h − vh ) + (ωh∗ − ηh ) ∙ (u ∗h − vh ) − Hh (σ˜ h∗ , u˜ ∗h ). = −Sh 2μ

(4.15)

t 2 ˆ ˆ Now, using the fact that kSh fˆkΣ× ˆ Mˆ 6 Ck f k0,t 0 ,Ω for all f ∈ [L (Ω)] (with t = 4/(2 + α) and t 0 = 2/(1 + α)) and the inverse inequality kvh k0,s,Ω 6 Ch β(2/s−2/r ) kvh k0,r,Ω , we have ∗ ∗ kSh ((σh∗ − τh ) ∙ (u ∗h − vh ))kΣ× ˆ Mˆ 6 Ck(σh − τh ) ∙ (u h − vh )k0,t 0 ,Ω

6 Ckσh∗ − τh k0,r,Ω ku ∗h − vh k0,s,Ω 2 2 6 Ch β s − r kσh∗ − τh k0,r,Ω ku ∗h − vh k0,r,Ω 2 2 6 Ch β s − r (kσh∗ − τh k0,r,Ω + ku ∗h − vh k0,r,Ω )2 ,

where 1/s = 1/t 0 − 1/r = (1 + α)/2 − 1/r .

44

M. FARHLOUL ET AL.

Similar arguments lead to β kSh ((ωh∗ − ηh ) ∙ (u ∗h − vh ))kΣ× ˆ Mˆ 6 Ch

2 2 s −r

Therefore, by (4.14), (4.15) and (4.8), we have β kR(τ˜h , v˜ h ) − (σ˜ h∗ , u˜ ∗h )kΣ× ˆ Mˆ 6 C 1 h

1+α− r4

Thus, if

(kωh∗ − ηh k0,r,Ω + ku ∗h − vh k0,r,Ω )2 .

1+β k(τ˜h , v˜ h ) − (σ˜ h∗ , u˜ ∗h )k2Σ× ˆ Mˆ + C 2 h

2 r −1

.

k(τ˜h , v˜ h ) − (σ˜ h∗ , u˜ ∗h )kΣ× ˆ Mˆ 6 ρ(h) with ρ(h) satisfying C1 h β

1+α− r4

we have

ρ 2 + C2 h 1+β

2 r −1

6 ρ(h),

kR(τ˜h , v˜ h ) − (σ˜ h∗ , u˜ ∗h )kΣ× ˆ Mˆ 6 ρ(h).

(4.16)

(4.17)

Now, owing to the definition of β given in (4.6) and the fact that r > 2, we have that, if h is small enough, the mean value ρ0 (h) of the roots of the equation 4 2 C1 h β 1+α− r ρ 2 − ρ + C2 h 1+β r −1 = 0 is

ρ0 (h) =

1 β h 2C1

4 r −1−α

.

Therefore, R has at least one fixed point (σ˜ h , u˜ h ) in the ball Bh = {(τ˜h , v˜ h ) ∈ Σh × Mh , k(τ˜h , v˜ h ) − (σ˜ h∗ , u˜ ∗h )kΣ× ˆ Mˆ 6 ρ0 (h)} and such a fixed point is a solution of Hh (σ˜ h , u˜ h ) = 0. 4 Since (σ˜ h , u˜ h ) ∈ Bh , ρ0 (h) 6 Ch β r −1−α and using (4.10), we have β r4 −1−α 6 Ch . k(σ˜ , u) ˜ − (σ˜ h , u˜ h )kΣ× ˆ Mˆ

Thus, since β =

2/r 2/r −α ,

we get 2

r k(σ˜ , u) ˜ − (σ˜ h , u˜ h )kΣ× ˆ Mˆ 6 Ch

(2/r −1) + 2/r2/r −α

,

which is the desired result. We close this paper by pointing out that the error estimate (4.13) is quasi-optimal when r → 2, 2 2/r (2/r −1) + 2/r −α → h owing to the refinement conditions on the regular family of triangulation. hr


45

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