Weak Imposition Of Boundary Conditions For. The Navier-Stokes Equations

Transcription

1 Weak Imposton Of Boundary Condtons For The Naver-Stokes Equatons Atfe Çağlar Ph.D. Student Department of Mathematcs Unversty of Pttsburgh Pttsburgh, PA 15260, USA November 23, 2002 Abstract We prove the convergence of a fnte element method for the Naver- Stokes equatons n whch the no-slp condton, u τ = 0 on Γ for = 1, 2 s mposed by a penalty method and the no-penetraton condton, u n = 0 on Γ, s mposed by Lagrange multplers. Ths approach has been studed for the Stokes problem n [2]. In most flows the Reynolds number s not Ths work partally supported by NSF grants DMS , INT , and INT atcst11@ptt.edu, atfe 1

2 neglgable so the u u nertal effects are mportant. Thus the extenson beyond the Stokes problem to the Naver-Stokes equatons s crtcal. We show exstence and unqueness of the approxmate soluton and optmal order of convergence can be acheved f the computatonal mesh follows the real boundary. Our results for the (nonlnear) Naver-Stokes equatons mprove known results for ths approach for the Stokes problem. Keywords. Naver-Stokes equatons, Lagrange multpler, slp wth frcton. AMS Subject classfcatons. 65N30. 1 Introducton The problem of predctng the equlbrum flow of a vscous ncompressble flud s one of solvng approxmately the statonary, ncompressble Naver-Stokes equatons: 2Re 1 (D(u)) u u p = f n Ω, u = 0 on Γ = Ω, (1.1) u = 0 n Ω, n a bounded polyhedral doman Ω R d d=2,3, where D s the deformaton tensor gven by D j (u) = 1 2 ( u u ) j x j x for 1, j d. The boundary Γ s assumed to be the unon of k flat parts Γ j : Γ = k Γ j. 2

3 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 3 The boundary condton u = 0 s decomposed of two separate condtons: no penetraton : u n = 0 on Γ, no slp : u τ = 0 on Γ, where n be the unt normal vector on Γ and τ, = 1,..., d 1, a system of orthonormal tangental vectors. There are physcal and computatonal reasons why these two boundary condtons should sometmes be separated and mposed usng dfferent technques see, e.g. the dscussons n [2]. For example, the Lagrange multpler mplementaton of the no-penetraton condton allows slght, but locally balancng n and out flow capturng some aspects of surface roughness. For the no-slp condton, the penalty formulaton (as we shall see) s equvalent to the Naver slp law. Thus, t s natural to use t for problems n whch large tangental stress mght occur. The numercal analyss of technques for mposng essental boundary condtons weakly was begun n [3], see also [4], [5, 6], [7], [2] and the references theren. In many flows nertal effects are mportant and thus the extenson beyond the Stokes problem, of [7,10] to the Naver-Stokes equatons s mportant. The purpose of ths report s to begn ths extenson. Our analyss both extends the work of [2] and [7] from the (lnear) Stokes problem to nonlnear Naver-Stokes problem and mproves the basc results of [2] even n the lnear case.

4 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 4 2 The Contnuous Problem For our mathematcal formulaton we ntroduce the followng spaces: X = [H 1 (Ω)] d, X 0 = [H 1 0 (Ω)] d, V = {v X v, q Ω = 0, q L 2 0(Ω)}, V 0 = { v X 0 v, q Ω = 0, q L 2 0 (Ω)}, Y = L 2 0 (Ω) = { q L 2 (Ω) q, 1 Ω = 0 }, Z = k H 1/2 (Γ j ), where k s the number of edges n 2d or faces n 3d of the boundary Γ..,. Ω s the usual L 2 nner product, H k (Ω) the usual W k,2 (Ω) Sobolev space wth norm. k,ω, and the space H k (Ω) s the dual of H0 k (Ω), the space of functons n H k (Ω) that vansh on Γ. The spaces H k 1/2 (Γ) consst of the traces of all functons n H k (Ω). Analogously, we denote by H (k 1/2) (Γ) the dual space of H k 1/2 (Γ) wth.,. Γ beng the dualty parng. A norm. Γ of a functon ϕ k H1/2 (Γ j ) s defned by ϕ Γ = ϕ 2 1/2,Γ j 1/2 wth the dual norm. Γ =:. Z. In addton, we set. Γj =. H 1/2 (Γ j) and. Γ j =. H 1/2 (Γ j). The most common formulaton of the Naver-Stokes equatons n (X 0, Y ) s

5 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 5 gven by (see e.g. [8], [9]) : Fnd (u, p) (X 0, Y ) such that: a 0 (u, v) a 1 (u; u, v) a 2 (v, p) = f, v Ω, v X 0, a 2 (u, q) = 0, q Y, where a 0 (u, v) = 2Re 1 D(u) : D(v) dx, Ω a 1 (u; v, w) = (u v) w dx, Ω b(u; v, w) = 1 2 [a 1(u; v, w) a 1 (u; w, v)], a 2 (u, p) = p( u) dx. Ω Lemma 2.1 For u V 0 and v, w X 0, b(u; v, w) = u v, w Ω. Snce a 1 (u; v, w) = a 1 (u; w, v) then b(u; v, v) = 0. Proof It follows from ntegral by parts. If we mpose the boundary condtons weakly we must seek a formulaton wth veloctes n X rather than X 0. Usng Green s formula, the defnton of deformaton and stress tensor we arrve at the followng weak formulaton of (1.1) n (X, Y ): Fnd (u, p) (X, Y ) such that: a 0 (u, v) b(u; u, v) a 2 (v, p) n j I(u, p), v Γj = f, v Ω, v X, a 2 (u, q) = 0, q Y. (2.1) Here, the tensor I(.,.) s defned by I k (u, p) = pδ k 2Re 1 D k (u) 1 2 u u k, for 1, k d,

6 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 6 and for all v X, q Y. Remark: I(.,.) s a modfcaton of the stress tensor where the pressure p s replaced by the Bernoull pressure pδ k 1 2 u u k Splt the fourth term of the left hand sde of equaton (2.1) to ts normal and tangental parts to obtan Lagrange multplers: ρ Γj := n j I(u, p) n j, λ 1 Γj := n j I(u, p) τ (j) 1, λ 2 Γj := n j I(u, p) τ (j) 2, Thus, ρ, λ 1, λ 2 are the ndvdual components of the normal stress on Γ j and 1 j k, where k s the number of boundary peces of Γ. Then we get 2Re 1 Ω Ω D(u) : D(v) dx 1 2 p( v) dx Ω(u u)v dx 1 2 Γ j ρ (v n j ) ds Defne c(.;.,.,.,.) on X Y Z Z Z as: Ω (u v)u dx Γ j λ 1 (v τ (j) 1 ) ds λ 2 (v τ (j) 2 ) ds = f v dx, (2.2) Γ j Ω q( u) dx = 0, Ω c(u; p, ρ, λ 1, λ 2 ) := p, u Ω ρ, u n j Γj λ, u τ (j) Γj. The correspondng weak formulaton of (2.1) s : a(u; u, v) c(v; p, ρ, λ 1, λ 2 ) = f, v Ω, c(u; q, σ, χ 1, χ 2 ) = 0, (2.3)

7 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 7 for all v X, q Y, σ, χ 1, χ 2 Z, where a(u; u, v) = a 0 (u; v) b(u; u, v). Defne the space K by K = {u X c(u; p, ρ, λ 1, λ 2 ) = 0 p Y, ρ, λ 1, λ 2 Z}. Lemma 2.2 The mult-lnear form c( ;,,, ) s contnuous on X Y Z Z. Proof Defne. as: u := [ u 2 u n j 2 Γ u τ 1 2 Γ u τ 2 2 ] 1/2 Γ whch s equal to. 1 on X, and apply Cauchy-Schwartz nequalty to c(. :.,.,.,.) we obtan: c(u; p, ρ, λ 1, λ 2 ) p u ρ Z u n j Γj λ Z u τ j 1 Γ j p u ρ 2 Z λ 2 Z 1/2 1/2 u n j 2 Γ j u τ j 2 Γ j 1/2 1/2 = p u ρ Z u n j Γ λ Z u τ Γ = ( p, ρ Z, λ 1 Z, λ 2 Z ), ( u, u n j Γ, u τ 1 Γ, u τ 2 Γ ) [ p 2 ρ 2 Z [ = p 2 ρ 2 Z ] 1/2 [ λ 2 Z u 2 u n j 2 Γ 1/2 λ Z] 2 u. u τ 2 Γ ] 1/2

8 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 8 Lemma 2.3 The space K s a closed subspace of the Hlbert space X. Proof Ths follows provded c(.;.,.,.) s contnuous on X Y Z Z Z whch was proven n Lemma 2.2 Lemma 2.4 For u K, Korn s nequalty u C K (Ω) D(u) and the Poncaré nequalty u C P (Ω) u hold. Proof If u K then c(v, q, ρ, λ 1, λ 2 ) = 0 for all q Y, ρ, λ 1, λ 2 Z. If we pck ρ = 0, q = 0, then u satsfes 2 k λ, u τ Γj = 0 for λ 1, λ 1 Z. Ths means that u τ = 0 n H 1/2 (Γ j ) for all j = 1,..., k, and thus u τ j = 0 a.e on Γ j. Smlarly, u n = 0 a.e on each Γ j. These mply for all 1 j k so that u = 0 a.e on Γ j wth meas(γ j ) s strctly postve. By Korn s nequaltes see [10], [11], [9] or [12] there exst C K > 0 such that u C K (Ω)(γ(u) D(u) ) for all u X. In partcular, for all u K, where γ(u) s a semnorm on L 2 (Ω) whch s a norm on the constants. Snce measure of Γ j are strctly postve, defne γ(u) = u Γj then, we get γ(u) = 0 f u s constant on Ω. The result thus follows. By the same argument, for the same γ(u) we get: u C P (Ω) (u)

9 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 9 for all u K. Corollary 2.1 The blnear form a 0 (, ) s coercve on K: α u 2 1 a 0(u, u) wth α = Re 1 mn{c 2 K (Ω), C 2 K (Ω)C 2 P (Ω)}. Proof Ths follows from the Korn and Poncaré nequaltes on K. Also note that the problem assocated wth problem (2.3) has the followng form n K. Fnd u K such that: a(u; u, v) = f, v Ω, v K. (2.4) Remark: The problem (2.3) s dfferent than the standard weak formulaton of the Naver Stokes equatons. Thus, we need to guarantee that the weak formulaton s correct. Lemma 2.5 (Exstence) There exsts a soluton to the formulaton (2.4) n K. Proof By the abstract theory developed n [8], the formulaton (2.4) has a soluton n K provded: 1) The form a(u; u, v) s coercve n K but snce b(u; u, u) = 0 t s suffcent to show that a 0 (u, v) s coercve. Ths was done n the Corollary ) The space K s separable and u a(u; u, v) s weakly contnuous.the space K s separable snce t s a closed subset of the Hlbert space X. Next, we prove that u a(u; u, v) s weakly contnuous n K. Let u be a functon n K

10 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 10 and let u m be a sequence n K so that u m u weakly as m. It s known by Rellch s theorem that H 1 (Ω) can be embedded compactly n L 2 (Ω). Then u m converges strongly to u L 2 (Ω) d as m. Let v be n a dense subset L of K, L = {w D(Ω) V : c(v, q, ρ, λ 1, λ 2 ) = 0, for all λ 1, λ 2, ρ Z}. b(u m ; u m, v) = 1 2 u m u m, v Ω 1 2 u m v, u m Ω. Snce u m K so s n V. Thus, we have u m u m, v Ω = u m v, u m Ω. Hence we can actually wrte b(.;.,.) as: d b(u m ; u m, v) = u m v, u m Ω = u m u mj v,j dx., Ω v,j L (Ω) snce v s nfntely many tmes dfferentable. Ths mples that lm u mu mj = u u j L 1 (Ω). m Thus, lm b(u m; u m, v) = m d, = b(u; v, u). Ω u u j v,j dx Snce u K V, b(u; v, u) = u; u, v Ω. The form a 0 (.,.) s contnuous and u m u n L 2 (Ω) mples that So we get lm a 0(u m, v) = a 0 (u, v) m lm a(u m; u m, v) = a(u; u, v) m

11 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 11 for all v L. By densty of L n K and contnuty of the forms a 0 (.,.) and b(.;.,.) the result follows. In order to prove unqueness we follow the approach of [8]. We defne the fnte constant as N := sup u,v,w K b(u; v, w) u 1 v 1 w 1. Lemma 2.6 (Unqueness) Suppose that the form a 0 (.,.) s coercve and the form b(.;.,.) s locally Lpschtz contnuous n K. Then, under the condton 1 α 2 N f < 1 where f f, v Ω := sup v K v 1 problem (2.4) has a unque soluton u K. Proof Suppose there are two solutons u 1 and u 2 to the problem (2.4) then a 0 (u 1, v) b(u 1 ; u 1, v) = f, v Ω, v K and a 0 (u 2, v) b(u 2 ; u 2, v) = f, v Ω, v K. Addng, subtractng the term b(u 1 ; u 2, v) and settng v = u 1 u 2 we obtan: a 0 (u 1 u 2, u 1 u 2 )b(u 1 ; u 1 u 2, u 1 u 2 )b(u 1 u 2 ; u 2, u 1 u 2 ) = 0. (2.7) The second term n (2.7) vanshes by the skew-symmetrc property of the form b(.;.,.). Thus a 0 (u 1 u 2, u 1 u 2 ) b(u 1 u 2 ; u 2, u 1 u 2 ). (2.8) Coercvty of the form a 0 (.,.) and the defnton of N mply that α u 1 u N u 2 1 u 1 u

12 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 12 Snce the form s coercve and contnuous n K by Lax-Mlgram lemma we have u α f. Substtutng (2.8) n (2.7) we obtan: α u 1 u N α f u 1 u Ths mples that: (α N α 2 f ) u 1 u Thus, u 1 = u 2 f ( 1 N α 2 f ) > 0. We have shown that the problem (2.4) has a unque soluton n K. Theorem 2.1 Assume that the multlnear form c(.;.,.,.) satsfes the nf-sup condton. There s a constant β > 0 such that: nf χ 1,χ 2,σ Z p Y c(v; q, σ, χ 1, χ 2 ) sup v X v ( p 2 0 χ 1 2 Z χ 2 2 Z β. (2.9) σ 2 Z )1/2 Then, for each soluton u of problem (2.4), there exst a unque p Y, ρ, λ 1, λ 2 Z such that (u, p, ρ, λ 1, λ 2 ) s a soluton of problem (2.3). Proof It follows from [13]. For error analyss we need the followng lemma. Lemma 2.7 There s a constant β > 0 such that: ) k nf sup ( χ 1, v τ (j) 1 Γ j χ 2, v τ (j) 2 Γ j σ, v n j Γj σ,χ 1,χ 2 Z v V v 1 ( χ 1 2 Z χ 2 2 Z β. σ 2 Z )1/2 Proof See [2].

13 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 13 3 Penalty-Lagrange Multpler Method It has been observed by [14] that slp wth frcton boundary condtons match the expermental behavor of real fluds at hgher Reynolds number better than no-slp condton. Consequently, we propose to mpose u τ (j) Γj = 0 for = 1, 2 usng a penalty technque that s equvalent to such a condton. We choose small, postve penalty parameters ɛ 1, ɛ 2 whose selecton wll be guded by the followng analyss. We consder the followng problem: Fnd u ɛ V and ρ ɛ Z such that: 2 a 0 (u ɛ, v) b(u ɛ ; u ɛ, v) 1 u ɛ, v τ (j) 2 Γ j u ɛ, v τ (j) 1 Γ j ρ ɛ, v n j Γj = f, v, v V, (3.1) σ, u ɛ n j Γj = 0, σ Z. Suppose that nstead of consderng penalty methods to weaken the condton u τ (j) Γj = 0, we consder slp wth non-lnear resstance to slp,.e.: u ɛ τ (j) 2Re 1 D(u ɛ ) u ɛ u ɛ p ɛ = f n Ω, (3.2) u ɛ = 0 n Ω, Γ j u ɛ n j ds = 0 on Γ = Ω, ɛ n j (D(u ɛ ) 1 2 u ɛ u ɛ ) τ (j) = 0 on Γ j, = 1, 2, 1 j k. Lemma 3.1 The varatonal formulaton of equaton (3.2) s the same as equaton (3.1).

14 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 14 Proof We multply equaton n (3.2) by v X, ζ Y, ρ Z, respectvely, and ntegrate by parts yeld: 2Re 1 D(u ɛ ) : D(v) Ω 1/2 u ɛ u ɛ, v Ω 1/2 u ɛ v, u ɛ Ω D(u ɛ ) n j, v Γj p ɛ, v Ω u ɛ τ (j) ρ ɛ, v n j Γj = f, v Ω, v X. σ, u ɛ n j Γj = 0, σ Z. ζ, u ɛ Ω = 0, ζ Y., v τ (j) Γj The lemma mples that the penalty method captures the dea of slp wth nonlnear resstance to slp.to smplfy the notaton we defne: λ ɛ 1 Γ j : = 1 u ɛ τ (j) 1, λ ɛ 2 Γ j : = 2 u ɛ τ (j) 2 ; where λ ɛ 1, λɛ 2 Z. Then equaton (3.1) takes the followng form: λ ɛ 2, v τ (j) 2 Γ j a 0 (u ɛ, v) b(u ɛ ; u ɛ, v) λ ɛ 1, v τ (j) 1 Γ j ρ ɛ, v n j Γj = f, v, v V, σ, u ɛ n j Γj = 0, σ Z, (3.3) u ɛ τ (j) 1, χ 1 Γj = ɛ 1 u ɛ τ (j) 2, χ 2 Γj = ɛ 2 λ ɛ 1, χ 1 Γj, χ 1 Z, λ ɛ 2, χ 2 Γj, χ 2 Z.

15 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 15 Proposton 3.1 Let (u, λ 1, λ 2, ρ) be the soluton of the Naver-Stokes equatons (2.1) and let (u ɛ, λ ɛ 1, λɛ 2, ρ ɛ) be the soluton to (3.2) Then: e ɛ (λ 1, λ 2, ρ) u u ɛ 1,Ω C[ɛ 2 1 λ 1 2 Γ ɛ 2 2 λ 2 2 Γ] 1/2, (3.4) where e ɛ (λ 1, λ 2, ρ) := [ λ 1 λ ɛ 1 2 Z λ 2 λ ɛ 2 2 Z ρ ρ ɛ 2 Z ]1/2 and C depends on Re, α, β, and f. Proof Subtractng equaton (3.3) from equaton (2.3) yelds: a 0 (u u ɛ, v) b(u u ɛ ; u ɛ, v) b(u; u u ɛ, v) λ λ ɛ, v τ (j) Γj ρ ρ ɛ, v n j Γj = 0, v V, (3.5) σ, (u u ɛ ) n j Γj = 0, σ Z, (u u ɛ ) τ (j), χ Γj = ɛ λ ɛ, χ Γj, χ Z, for = 1, 2. From Lemma 2.7 gven λ 1, λ 2, ρ Z we have: β e ɛ (λ 1, λ 2, ρ) sup 0 v V a 0 (u u ɛ, v) b(u u ɛ ; u ɛ, v) b(u; u u ɛ, v) v 1 2Re 1 u u ɛ 1 v 1 M u u ɛ 1 u ɛ 1 v 1 M u u ɛ 1 v 1 u 1 v 1 u u ɛ 1 (2Re 1 2Mα 1 f ) 2(Re 1 α) u u ɛ 1 (3.6)

16 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 16 where M := sup 0 u,v,w V b(u; v, w) u 1 v 1 w 1. In equaton (3.5) we take v = u u ɛ. Then, Thus, a 0 (u u ɛ, u u ɛ ) b(u u ɛ ; u ɛ, u u ɛ ) ɛ λ ɛ, λ λ ɛ Γ j ρ ρ ɛ, (u u ɛ ) n j Γj = 0. (3.7) a 0 (u u ɛ, u u ɛ ) = b(u u ɛ ; u u ɛ, u ɛ ) ɛ λ ɛ, λ λ ɛ Γ j. Add and subtract the terms 2 ɛ k λ, λ λ ɛ Γ j, and droppng the postve term 2 ɛ k λ λ ɛ, λ λ ɛ Γ j we get: a 0 (u u ɛ, u u ɛ ) b(u u ɛ ; u u ɛ, u ɛ ) Applyng Cauchy-Schwarz nequalty and defnton of M we get: a 0 (u u ɛ, u u ɛ ) b(u u ɛ ; u ɛ, u u ɛ ) ɛ λ, λ λ ɛ Γ j. ɛ λ, λ λ ɛ Γj Mα 1 f u u ɛ 2 1 ɛ λ λ ɛ Z λ Γ. By the arthmetc geometrc mean nequalty for n = 2, we have a 0 (u u ɛ, u u ɛ ) Mα 1 f u u ɛ 2 1 ( ) 1/2 ( 1/2 λ λ ɛ 2 Γ ɛ 2 λ Γ) 2. Usng the coercvty of a 0 (.,.) and nequalty (3.6) we get [ 2(Re 1 ] ( 1/2 α) u u ɛ 1 β α(1 Mα 2 f ɛ 2 ) λ Γ) 2.

17 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 17 Thus e ɛ (λ 1, λ 2, ρ) u u ɛ 1 2(Re 1 α) β u u ɛ 1 u u ɛ 1. Combnng the latter nequaltes we obtan: ( 1/2 e ɛ (λ 1, λ 2, ρ) u u ɛ 1 C ɛ 2 λ Γ) 2. (3.8) 4 Fnte Element Spaces The polyhedral doman Ω s subdvded nto d-smplces wth sdes of length less than h wth T h beng the famly of parttons. We wll assume that T h satsfes the usual regularty assumptons, see e.g. [15] that: 1. Each vertex of Ω s a vertex of a T T h, 2. Each T T h has at least one vertex n the nteror of Ω, 3. Any two d-smplces T, T T h may meet n a vertex, a whole edge, or a whole face, The constants c 0, c 1 denote dfferent constants whch are ndependent of h. 1. Each T T h contans a ball wth radus c 0 h and s contaned n a ball wth radus c 1 h. Denote by O h j the partton of Γ j whch s nduced by T h. Let X h X, Y h Y, Z h Z. Also let X h 0 = { u h X h u h = 0 on Γ }.

18 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 18 The spaces X h and Y h are assumed to satsfy the followng propertes: I. There s a constant β > 0 ndependent of h for whch: II. nf 0 p h Y h sup 0 u h X h 0 Ω ph dvu h dx p h 0,Ω u h 1,Ω β, (4.1) nf p p h 0,Ω ch p 1,Ω, p h Y h p H 1 (Ω), whch: III. There exsts a contnuous lnear operator Π h : H 1 (Ω) d X h for Π h (H 1 0 (Ω) d ) X h 0, u Π h u s,ω ch t s u t,ω, u H t (Ω) wth s = 0, 1 and t = 1, 2, u Π h u 0,Γ ch 1/2 u 1,Ω, where. 0,Γ = ( k. 0,Γ j ) 1/2. Assumpton I balances the nfluence of the constrant dvu = 0 and also mples that the space: V h 0 = { v h X h 0 qh, v h = 0, q h Y h}, V h = { v h X h q h, v h = 0, q h Y h} V h 0, s also not empty. As usual V h s not a subset of V and n partcular, the functons of V h are not dvergence free. Hence we ntroduce ant-symmetrc form b(.;.,.,.), N h = b(u h ; v h, w h ) sup 0 u h,v h,w h X u h h 1 v h 1 w h (4.2) 1 and f h = sup v h V h f, v h Ω v h 1.

19 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 19 Wth the above notaton, the dscrete analogue of problem (2.3) s: Fnd u h X h, p h Y h, ρ h, λ h 1, λh 2 Zh such that: a(u h ; u h, v h ) c(v h ; p h, ρ h, λ h 1, λ h 2 ) = f, v h Ω (4.3) c(u h ; q h, σ h, χ h 1, χ h 2) = 0 for all v h X h, q h Y h, σ h, χ h 1, χh 2 Zh. By assumpton I ths s equvalent to the followng problem n V h : Fnd u h X h, p h Y h, ρ h, λ h 1, λ h 2 Z h such that: a(u h, u h, v h ) = f, v h Ω (4.4) By the abstract theory developed n [8] the dscrete problem wll have unque soluton provded: α 2 N h f h < 1 (4.5) and, there s a constant β > 0, ndependent of h such that nf ρ h,λ h 1,λh 2 Zh p h Y h c(v h ; p h, ρ h, λ h 1, λ h sup 2) ] 1/2 0 v h X h v h 1 [ λ β. (4.6) h 1 2 Z λh 2 2 Z ρh 2 Z ph 2 0,Ω Now we need to fnd rght spaces X h, Y h, Z h so that they hold (4.6). An example of spaces X h, Y h, Z h satsfyng (4.6) and the classcal nf-sup condton are gven n [7]. Lemma 4.1 There exsts a constant β > 0 ndependent of h, such that: ) k ( λ h 1, vh τ (j) 1 Γ j λ h 2, vh τ (j) 2 Γ j ρ h, v h n j Γj nf ρ h Z h sup 0 v h X h λ h 1,λh 2 Zh [ v h 1 λ h 1 2 Z λh 2 2 Z ] ρh 2 1/2 β. Z (4.7)

20 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 20 Lemma 4.2 There s a constant β > 0, ndependent of h such that: nf ρ h,λ h 1,λh 2 Zh p h Y h c(v h ; p h, ρ h, λ h 1 sup, λh 2 ) ] 1/2 0 v h X h v h 1 [ λ β. (4.8) h 1 2 Z λh 2 2 Z ρh 2 Z ph 2 0,Ω Proof The last two lemmas are proven n [2]. 5 The Dscrete Penalty-Lagrange Multpler Method We can now wrte the dscrete analogue of the equaton (3.1) as: Fnd u h ɛ Xh satsfyng: a 0 (u h ɛ, vh ) b(u h ɛ ; uh ɛ, vh ) u h ɛ τ (j), v h τ (j) Γj ρ h ɛ, vh n j Γj p h ɛ, vh Ω = f, v h, (5.1) q h, u h ɛ Ω = 0, σ h, u h ɛ n j Γj = 0, for all v h X h, q h Y h, σ h Z h. Ths s equvalent to fndng u h ɛ V h such that: a 0 (u h ɛ, v h ) b(u h ɛ ; u h ɛ, v h ) for all v h V h, σ h Z h. u h ɛ τ (j), v h τ (j) Γj (5.2) ρ h ɛ, vh n j Γj = f, v h, σ h, u h ɛ n j Γj = 0,

21 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 21 Snce λ ɛ,h for = 1, 2 are not necessarly n Z h equaton (5.2) s not equvalent to (5.3). In fact λ ɛ,h Z h := X h τ (j) Γj, = 1, 2, 1 j k. a 0 (u h ɛ, v h ) b(u h ɛ ; u h ɛ, v h ) λ ɛ,h, v h τ (j) Γj ρ h ɛ, vh n j Γj = f, v h, v h V h, σ h, u h ɛ n j Γj = 0, σ h Z h, (5.3) u h ɛ τ (j), χ h Γ j = ɛ λ ɛ,h, χ h Γ j, χ h Zh, for = 1, 2. The latter problem wll be equvalent f we make the followng assumpton: There exsts β > 0 nf λ h Z h ρ h Z h sup 0 v h X h k λɛ,h 1, vh τ (j) 1 Γ j λ ɛ,h 2, vh τ (j) v h 1 [ λ ɛ,h 1 2 Z λɛ,h 2 2 Z ρh 2 Z 2 Γ j ρ h, v h n j Γj ] 1/2 β. (5.4) We shall now study the error u u h ɛ n 1 norm where (u, λ 1, λ 2, p, ρ) s the soluton of (2.3) and (u h ɛ, λ ɛ,h 1, λ ɛ,h 2, p h, ρ h ɛ ) s the soluton of the fnte element problem (5.3). We estmate the error n two cases. In the frst case the computatonal boundary Γ h does not follow the flud boundary Γ. For example, suppose the doman Ω has smooth boundary and we use quadratc elements for our approxmaton. Then by approxmatng the flud velocty u by penalzed velocty u ɛ and u ɛ by dscrete penalzed velocty u h ɛ. We wll conclude that the error n

22 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 22 1 norm has order h convergence. Ths requres that the penalty parameter ɛ should be scaled by h. In the second case, Γ h follows Γ exactly. Our doman s polygonal (n 2d) wth the choce of quadratc elements we approxmate u by dscrete velocty u h and u h by u h ɛ. Ths leads order h 2 convergence n the error 1 norm. Ths mples scalng ɛ by h 2. Thus, t has mproved the basc results of [2] even n the lnear case. CASE I: In ths case, the approxmaton u h ɛ satsfes (5.3). We need smlar equatons for the contnous u ɛ n the velocty space. For the reason we multply by v V to obtan (3.1), nstead we multply by v h V h. It gves: a 0 (u ɛ, v h ) b(u u ɛ ; u ɛ, v h ) p ɛ q h, v h Ω u ɛ τ (j), v h τ (j) Γj ρ ɛ, v h n j Γj = f, v h, (5.5) σ h, u ɛ n j Γj = 0, for all v h V h and σ h Z h, snce Z h Z. Subtractng (5.2) from (5.5) yelds: a 0 (u ɛ u h ɛ, v h ) b(u h ɛ ; u ɛ u h ɛ, v h ) b(u ɛ u h ɛ ; u ɛ, v h ) p ɛ q h, v h Ω (u ɛ u h ɛ ) τ (j), v h τ (j) Γj for all v h V h. We wrte: ρ ɛ ρ h ɛ, vh n j Γj = 0, (5.6) u ɛ u h ɛ = (u ɛ v h ) (u h ɛ v h ), (5.7) wth v h beng the best approxmaton of u ɛ n V h. Let η := u ɛ v h, φ h := u h ɛ vh.

23 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 23 Then equaton (5.6) becomes: a 0 (η φ h, v h ) b(η φ h ; u ɛ, v h ) b(u h ɛ ; η φh, v h ) p ɛ q h, v h Ω (η φ h ) τ (j), v h τ Γj ρ ɛ ρ h ɛ, v h n j Γj = 0, (5.8) for all v h V h. Equaton (5.8) can be rewrtten as: a 0 (φ h, v h ) b(φ h ; u ɛ, v h ) b(u h ɛ ; φ h, v h ) φ h τ, v h τ (j) = a 0 (η, v h ) b(η; u ɛ, v h ) b(u h ɛ ; η, v h ) p ɛ q h, v h Ω (5.9) ρ ɛ ρ h ɛ, vh n j Γj Set v h = φ h. Then: 2Re 1 D(φ h ) 2 0 b(φh ; u h ɛ, φh ) η τ (j), v h τ (j) Γj φ h τ 2 = 0,Γ = a 0 (η, φ h ) b(u h ɛ ; η, φh ) b(η; u ɛ, φ h ) p ɛ q h, φ h Ω η τ (j), φ h τ (j) Γj Γj ρ ɛ ρ h ɛ, φ h n j Γj. (5.10) But for each j, φ h n j Z h snce Z h Z. Thus we can replace ρ h ɛ n (5.10) wth any test functon, say σ h Z h. Then we have 2Re 1 D(φ h ) 2 φ h τ 2 0,Γ 2Re 1 η 1 φ h 1 N h u ɛ h 1 φ h 1 η 1 N h u ɛ φ h 1 η 1 2 η τ 2 0,Γ 2 φh τ 2 0,Γ ρ ɛ σ h Z φ h 1 p ɛ q h φ h 1 (5.11)

24 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 24 Re 1 D(φ h ) 2 2 φh τ 2 0,Γ Re 1 η 2 1 N h φ h 1 η 1 ( u h ɛ 1 u 1 ) p ɛ q h 2 φh η τ 2 0,Γ ρ ɛ σ h 2 Z φh (5.12) By Korn s nequalty we have: C 1 φ h 2 1 Re 1 D(φ h ) 2 2 φh τ 2 0,Γ Re 1 η 2 1 (N h f α 1 u ɛ 1 )( η 2 1 φh ) 1 p ɛ q h 2 C 1 2 η τ 2 0,Γ 1 ρ ɛ σ h 2 Z C C φ h (5.13) where C 1 depends on Re, α, and ɛ Snce N f α 2 < 1, we can choose h suffcently small so that: N h f h α 1 < α. Usng ths we obtan: Thus we get: C 1 φ h 2 1 Re 1 η 2 1 (α u ɛ 1 ) φh p ɛ q h 2 4 C 1 2 η τ 2 0,Γ 1 ρ ɛ σ h 2 0 C C φ h (5.14) C 2 φ h 2 1 C 3 η C 1 ( p ɛ q h 2 ρ ɛ σ h 2 ) where C 2 and C 3 are depend on Re, ɛ, α and u ɛ but not on h. 2 η τ 2 0,Γ (5.15)

25 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 25 Addng and subtractng η n every norm of the left hand sde of equaton (5.15) yelds: η φ h 2 1 C 4 η 2 1 C 5 η τ 2 0,Γ C 6 ( ρ ɛ σ h 2 Z p ɛ q h 2 ). Applyng the trangle nequalty and takng nfma we have: u ɛ u h ɛ 2 1 C 4 nf u ɛ v h 2 0 v h V h 1 C 5 C 6 ( nf ρ ɛ σ h 2 0 σ h Z h Z nf p ɛ q h 2 ), 0 q h Y h nf (u ɛ v h ) τ 2 0,Γ 0 v h V h where, for ɛ 1, ɛ 2 small. Now we have to derve a bound for the error n the Lagrange multpler. From Lemma 4.1 we take λ h 1 = λh 2 = 0 : nf sup 0 ρ h Z h 0 v h X h nf ρ h Z h sup 0 v h X h λ h 1,λh 2 Zh Equaton (5.16) mples that: k ρh, v h n j Γj v h 1 ρ h = Z k λh 1, v h τ (j) 1 Γ j λ h 2, v h τ (j) 2 Γ j ρ h, v h n j Γj [ v h 1 λ h 1 2 Z λh 2 2 Z ] ρh 2 1/2 β. Z (5.16) β ρ h ɛ σ h Z sup v h X h k ρh ɛ σ h, v h n j Γj v h 1. (5.17)

26 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 26 From (5.6) we have: ρ h ɛ σ h, v h n j Γj = a 0 (u ɛ u h ɛ, v h ) b(u ɛ u h ɛ ; u ɛ, v h ) b(u h ɛ ; u ɛ u h ɛ, vh ) p ɛ q h, v h Ω (u ɛ u h ɛ ) τ (j), v h τ (j) Γj ρ ɛ σ h, v h n j Γj 2Re 1 uɛ u h ɛ v h 1 1 2α v h uɛ 1 u h ɛ 1 pɛ q h v h 0 1 (u ɛ u h ɛ ) τ 0,Γ v h 1 ρ ɛ σ h Z v h 1 (C u ɛ u h ɛ 1 p ɛ q h where C := 2(α Re 1 ). (u ɛ u h ɛ ) τ 0,Γ ρ ɛ σ h Z ) v h 1 (5.18) Thus dvdng (5.18) by v h 1, combnng t wth (5.17), applyng the trangle nequalty and takng nfma yelds: ρɛ ρ h ɛ Z C uɛ u h ɛ β 1 1 β nf pɛ q h q h Y h β (uɛ u h ɛ ) τ 0,Γ 1 β β nf ρɛ σ h σ h Z h Z. Squarng both sdes of (5.19) and boundng the mxed terms yelds: ρ ɛ ρ h ɛ 2 Z 5C2 β 2 u ɛ u h 2 ɛ 5 1 5ɛ 2 β 2 β 2 nf q h Y h p ɛ q h 2 (5.19) (u ɛ u h ɛ ) τ 2 10 ( 0,Γ β 2 nf ρɛ σ h ) 2 Z. σ h Z h (5.20)

27 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 27 Combnng (5.20) wth (??) yelds the followng theorem for the total error. Theorem 5.1 Assume that the dscrete spaces satsfy condton (4.7). Then the error n the dscrete soluton of the penalty-lagrange multpler method s as follows: u ɛ u h ɛ 2 1 C 1 nf u ɛ v h 2 C 0 v h V h 1 2 ( nf uɛ v h) τ 2 0 v h V h 0,Γ ( C 3 nf ρ ɛ σ h 2 0 σ h Z h Z nf p ɛ q h ) 2. (5.21) 0 q h Y h 0 Incorporatng (3.6) nto (5.21) yelds the followng corollary for the total error for the penalty-lagrange multpler method: Corollary 5.1 Assume that the dscrete spaces satsfy condton (4.7). Then the error n the dscrete soluton of the penalty-lagrange multpler method s as follows: u u h ɛ 2 1 C 1 nf v h V h u v h 2 1,Ω C 2 ɛ ( C 3 nf ρ σ h 2 σ h Z h Z nf p q h 2 q h Y h 0 C 1 u u ɛ 2 1,Ω C 3 ρ ρ ɛ 2 Z p p ɛ 2 0 C 4 nf (u v h ) τ 2 v h V h 0,Γ ) C 2 ɛ (u u ɛ ) τ 2 0,Γ (5.22) ɛ 2 λ 2 Γ. Proof Squarng both sdes of (3.4) yelds the followng: u u ɛ 2 1,Ω λ λ ɛ 2 Z ρ ρ ɛ 2 Z C 1 ɛ 2 λ 2 Γ. (5.23) In (5.21) we replace the. 0,Γ norm on the left hand sde by. Z, add ( u u), ( p p), ( ρ ρ) n the approprate terms of the rght hand sde and add t to (5.22) to get the stated result.

28 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 28 Remark: The error bound that ncludes the dfference n the value of the stress vector between the true and the dscrete penalty-lagrange multpler solutons s as follows: u u h ɛ 2 1 λ λ ɛ,h 2 Z ρ ρh ɛ 2 Z { 1 C 1 nf } u v h 2 mn ɛ v h V h 1 C 2 nf (u v h ) τ 2 ɛ v h V h 0,Γ ( ) ( C 3 mn ɛ C 0 nf ρ σ h 2 σ h Z h Z nf ) p q h 2 q h Y h 0 ) ) C 1 u u ɛ 2 1 (C 3 mn ɛ C 0 ( ρ ρ ɛ 2 Z p p ɛ 2 0 C 2 (u u ɛ ) τ 2 0,Γ ɛ C 4 ɛ 2 λ 2 Γ. β To obtan a bound on p p ɛ 0 we use (1.12) wth χ = λ λ ɛ, σ = ρ ρ ɛ: ( ρ ρ ɛ 2 Z p p ɛ 2 0 λ λ ɛ 2 Z p p ɛ, v Ω 2 sup v X ) 1/2 a 0 (u u ɛ, v) b(u u ɛ ; u ɛ, v) sup C u u ɛ v X v 1, where C depends on Re and α. Thus: k λ λ ɛ, v τ (j) Γj k ρ ρ ɛ, v n j Γj v p p ɛ 2 0 C β 2 u u ɛ 2 1,Ω. (5.24) To get a bound on (u u ɛ ) τ 2 0,Γ we set χ = (u u ɛ ) τ (j) n (2.8): (u u ɛ ) τ 0,Γ ɛ λ ɛ 0,Γ. (5.25)

29 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 29 Add and subtract λ n the norm n the rght hand sde of (5.25) to yeld: (u u ɛ ) τ 0,Γ ɛ ( λ 0,Γ λ λ ɛ 0,Γ ( ) ɛ λ Γ λ λ ɛ 0,Γ, ) or ) (u u ɛ ) τ 2 0,Γ ɛ C 2 ( λ 2 Γ λ λ ɛ 2 0,Γ. (5.26) C 2 Now we must bound λ λ ɛ 2 0,Γ. Take v = u u ɛ n (3.7): a 0 (u u ɛ, u u ɛ ) b(u u ɛ ; u ɛ, u u ɛ ) Add 2 ɛ k λ λ ɛ, λ Γj to both sdes: ɛ λ λ ɛ 2 0,Γ ɛ λ λ ɛ, λɛ Γ j = 0. ɛ λ Γ λ λ ɛ Z a 0 (u u ɛ, u u ɛ )b(u u ɛ ; u ɛ, u u ɛ ). Defnton of N and the Cauchy-Schwarz nequalty: Thus overall: ɛ λ λ ɛ 2 0,Γ ɛ ( ) λ 2 1/2,Γ 2 λ λ ɛ 2 Z. C 2 ) (u u ɛ ) τ 2 0,Γ ɛ C 2 ɛ ( λ 2 Γ λ λ ɛ 2 Z. (5.27) Under the approxmaton assumpton: nf ( u v h v h V h 1 nf p q h 2 q h Y h 0 ) 1/2 nf ρ ρ h 2 ρ h Z h Z Ch k max { f 1, u 2 }, (5.28) where k s the degree of the polynomal space we are usng. If we let ɛ = ɛ 1 = ɛ 2 then corollary 5.1 ndcates that proper choce of ɛ n (5.22) s ɛ = h k/2.

30 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 30 Theorem 5.2 Assume that the dscrete spaces satsfy condton (4.7) and let ɛ = h k/2. Then the error n the soluton of the penalty-lagrange multpler method and the dscrete soluton of the penalty-lagrange multpler method s as follows: u u h ɛ 1 Cɛ. CASE II: Let (u, λ 1, λ 2, ρ, p) be the soluton of (2.3) and (u h, λ h 1, λ h 2, ρ h, p h ) be the fnte element approxmaton of (2.3), that s (4.3) then substractng (4.3) from (2.3) we get: a 0 (u u h, v h ) b(u; u u h, v h ) b(u u h ; u h, v h ) p p h, v h Ω (u u h ) τ (j), v h τ (j) Γj for all v h V h. We wrte: ρ ρ h, v h n j Γj = 0, (5.29) u u h = (u v h ) (u h v h ), wth v h beng the best approxmaton of u n V h. Let η := u v h, φ h := u h v h. Snce Γ h = Γ error terms whch come from approxmatng boundary vansh, then equaton (5.29) becomes: a 0 (η φ h, v h ) b(u; η φ h, v h ) b(η φ h ; u h, v h ) p p h, v h Ω = 0, for all v h V h. (5.31)

31 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 31 Equaton (5.31) can be rewrtten as: a 0 (φ h, v h ) b(u; φ h, v h ) b(φ h ; u h, v h ) = a 0 (η, v h ) b(u; η, v h ) b(η; u h, v h ) p p h, v h Ω (5.32) Settng v h = φ h n (5.32), applyng Cauchy-Schwartz nequalty and defnton of N h. We obtan: 2Re 1 D(φ h ) 2 2Re 1 D(η) D(φ h ) 2N h f h α 1 φ h 1 η 1 N h f h α 1 φ h 2 1 p ph φ h 1 (5.33) By Korn s nequalty we have: 2Re 1 D(φ h ) 2Re 1 D(η) 2α D(η) α D(φ h ) p p h Thus we get: C 1 D(φ h ) C 2 D(η) p p h. (5.34) where C 1 and C 2 are depend on Re, α,and f. Addng and substractng η from the left hand sde of (5.34) we get: D(u u h ) (1 C 1 /C 2 ) D(u v h ) 1/C 1 p p h. Takng nfma we get: D(u u h ) (1 C 1 /C 2 ) nf D(u v h ) 1/C 1 nf p p h. 0 v h V h 0 p h Y h (5.35) Under the approxmaton assumpton (5.30) we get u u h 1 h k. In order to bound u u h ɛ 1 besdes (5.35) we also need a bound on u h u h ɛ 1. For the

32 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 32 reason we substract equaton (5.3) from (4.3) then we set v h = u h u h ɛ and we obtan: a 0 (u h u h ɛ, u h u h ɛ ) b(u h u h ɛ ; u h ɛ, u h u h ɛ ) ɛ λ h, λ λ ɛ,h ρ h ρ h ɛ, (u h u h ɛ ) n j Γj = 0. (5.36) From Lemma 3.1 we have: where e h ɛ (λ ɛ,h 1, λɛ,h 2, ρ ɛ h ) 2(Re 1 α) u h u h ɛ β 1 (5.37) [ 1/2 e h ɛ (λ ɛ,h 1, λɛ,h 2, ρh ɛ ) := λ h 1 λ ɛ,h 1 2 Z λ h 2 λ ɛ,h 2 2 Z ρ h ρ h ɛ Z] 2. Γj Snce Γ h = Γ addng and substractng ɛ λ, λ λ ɛ,h Γj, and ρ, (u h u h ɛ ) n j Γj n (5.36) gves: a 0 (u h u h ɛ, u h u h ɛ ) b(u h u h ɛ ; u h ɛ, u h u h ɛ ) (5.38) ɛ λ, λ λ ɛ,h Γj ρ ρ h ɛ, (uh u h ɛ ) n j Γj = 0. By the same argument as n the proof of Proposton 2.1 we obtan: u h u h ɛ 1 C 1 [ ɛ 2 1 λ 1 2 Γ ɛ2 2 λ 2 2 Γ] 1/2, (5.39) where C 1 depends on Re, α, β, and f. Then combnng (5.37) and (5.39). We get: u h u ɛ h 1 ɛ

33 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 33 Thus from (5.35) and (5.38) u u ɛ h 1 h k ɛ. (5.40) where C depends on C 1 and λ Γ for = 1, 2. Thus (5.40) ndcates that the proper choce of ɛ n (5.40) s ɛ = h k. Theorem 5.3 Assume that the dscrete spaces satsfy condton (4.7) and let ɛ = h k. Then the error n the soluton of the penalty-lagrange multpler method and the dscrete soluton of the penalty-lagrange multpler method s as follows: u u h ɛ 1 Cɛ. Concluson. Weak mposton of essental boundary condtons by penalty- Lagrange multplers method lessens the ll-effects of domans wth non-smooth boundares. In ths paper we have shown that the optmal order of convergence can be acheved f the computatonal boundary follows the real flow boundary exactly. We dd not nclude numercal results because of the dffculty n mplementaton of the Lagrange multplers method. Tradtonally, Lagrange multplers method s used as a model problem. However, we have done smlar analyss by usng the penalty- penalty method and verfed our results numercally n [20]. Acknowledgments. I would lke to thank Prof. W. J. Layton and Dr. T. Lakos for ther valuable contrbutons to ths paper.

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