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Stokes coupling method for the exterior flow Part IV: stabilized finite element approximation
Applied Mathematics and Computation 172 (2006) 1225–1238
www.elsevier.com/locate/amc
Stokes coupling method for the exterior flow Part IV: stabilized finite element approximation q Yinnian He Faculty of Science, Xi’an Jiaotong University, Xi’an 710049, China
Abstract Based on the Stokes coupling method for solving the two dimensional exterior unsteady Navier–Stokes equations, the stabilized finite element approximation combining the boundary element with the stabilized finite element for solving the reduced Stokes coupling equations is presented. Finally, the optimal error estimates for the finite element solution are derived. Ó 2005 Elsevier Inc. All rights reserved. Keywords: Navier–Stokes equations; Stokes coupling method; Stabilized finite element
Let X0 be a simple connected bounded open set of R2 with smooth boundary C and let X denote the complement of X0 [ C where the origin of coordinates is located in the interior of X0. In X Rþ , we consider the following initial-boundary value problem of the Navier–Stokes equations:
q
Subsidized by the NSF of China 10371095 and the NSF of Shaanxi Province. E-mail address: [email protected]
0096-3003/$ - see front matter Ó 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2005.03.019
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Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
8 ou m Du þ ðu rÞu þ rp ¼ fðx; tÞ; ðx; tÞ 2 X Rþ ; > ot > > > < div u ¼ 0; ðx; tÞ 2 X Rþ ; ðN–SÞ uðx; tÞjC ¼ 0; lim uðx; tÞ ¼ 0; 8t 2 Rþ ; > > jxj!1 > > : x 2 X; uðx; 0Þ ¼ u0 ðxÞ; where uðx; tÞ and pðx; tÞ denote the unknown velocity vector and the pressure of the fluid at point ðx; tÞ 2 X Rþ , respectively, while m > 0 is the dynamic viscosity, fðx; tÞ represents a density of external forces and u0 ðxÞ is a initial velocity vector satisfying div u0 ¼ 0 in X with u0 njC ¼ 0, limjxj!1 u0 ðxÞ ¼ 0. For a fixed small number e > 0, we introduced an artificial smooth boundary C2 ¼ fx 2 X; jxj ¼ R1 g, separating an outer region X2 from an inner region X1 such that juðx; tÞj 6 e; 8ðx; tÞ 2 ðX2 [ C2 Þ Rþ .
ð1Þ
For the justification of Eq. (1) which is uniform with respect to time t, the readers can refer to the work of He and Xin [4]. For the reasons of the computational economy and the inertial term ðu rÞu being small in X2, we consider the following Stokes coupling problem: 8 ou m Du þ X X1 ðu rÞu þ rp ¼ f; ðx; tÞ 2 X Rþ ; > > > ot > < div u ¼ 0; ðx; tÞ 2 X Rþ ; 0 ðN–S Þ uðx; tÞjC ¼ 0; lim uðx; tÞ ¼ 0; 8t 2 Rþ ; > > jxj!1 > > : x 2 X; uðx; 0Þ ¼ u0 ðxÞ; where
( X X1 ðxÞ ¼
1; x 2 X1 ; 0; x 2 X1 .
Here, we also assume that the solution ðu; pÞ of (N–S 0 ) satisfies Eq. (1). The boundary element method (BEM) or the coupling method of BEM with finite element method (FEM)for solving the stationary Stokes problem on bounded domain and unbounded domain had been discussed by Yu [19,20], Zhu [21,22] and Sequeira [16,17]. The coupling method of BEM with FEM for solving the non-stationary Stokes problem had been discussed by He et al. [5] and Wang [18]. While coupling method of BEM with FEM for solving the stationary Navier–Stokes problem on exterior domain had been discussed by Li et al. [11,12]. Hereafter will we mainly consider the coupling method of BEM with FEM for solving the non-stationary Navier–Stokes problem (N–S) on exterior domain. In Ref. [7], we have proven that the difference of solutions between problems (N–S) and (N–S 0 ) is small; in Refs. [6,8,13], we presented the reduced Stokes coupling equations corresponding to problem (N–S 0 ) and discussed their well-posedness and the finite element approximation.
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
1227
In this paper, by using the stabilized finite element method presented by Hughes et al. [10], we will present the stabilized finite element approximation combining the boundary element with the stabilized finite element for solving the reduced Stokes coupling equations and provide the optimal error estimates for the finite element solution. 1. Reduced stokes coupling problem In order to derive the reduced Stokes coupling problem corresponding to problem (N–S 0 ), we take the equivalent form of problem (N–S 0 ): 8 ou m Du þ ðu rÞu þ rp ¼ f; ðx; tÞ 2 X1 Rþ ; > ot > > < div u ¼ 0; ðx; tÞ 2 X1 Rþ ; ðS1 Þ þ þ > > > uðx; tÞjC ¼ 0; rðu; pÞ njC2 ¼ k ; 8t 2 R ; : uðx; 0Þ ¼ u0 ðxÞ; x 2 X1 ; and 8 ou m Du þ rp ¼ f; ðx; tÞ 2 X2 Rþ ; > ot > > > < div u ¼ 0; ðx; tÞ 2 X2 Rþ ; ðS2 Þ uðx; tÞjC2 ¼ u ; lim uðx; tÞ ¼ 0; 8t 2 Rþ ; > > jxj!1 > > : x 2 X2 ; uðx; 0Þ ¼ u0 ðxÞ; where n denotes the unit normal vector of C and C2, corresponding to the exterior to X0 and X2, respectively, and ðu ; p Þ ¼ lim ðu; pÞjX1 ; x!C2
k ¼ rðu ; p Þ njC2 ; rðu; pÞ ¼
ðuþ ; pþ Þ ¼ lim ðu; pÞjX2 ; x!C2
kþ ¼ rðuþ ; pþ Þ njC2 ;
oui ouj þ ; dij p þ m oxj oxi 22
i; j ¼ 1; 2.
Now, we consider the equivalent system S1–S2 of the Stokes coupling equations (N–S 0 ). Applying the Greeen formula to S2, we derive the reduced Stokes coupling equations (see Refs. [6,8,13]):
ðR–S0 Þ
8 ou mDu þ ðu rÞu þ rp ¼ f; ðx; tÞ 2 X1 Rþ ; > ot > > > > ðx; tÞ 2 X1 Rþ ; > < div u ¼ 0; þ ujC ¼ 0; rðu; pÞ njC2 ¼ k ; 8t 2 Rþ ; > > > uðx; 0Þ ¼ u0 ðxÞ; x 2 X1 ; > > > : 1 Bðt; kÞ ¼ 2 c0 u þ Gðt; c0 uÞ þ Kðt; cX2 u0 ; cX2 fÞ; ðx; tÞ 2 C2 Rþ ;
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Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
where c0 is the trace operator defined by c0 u ¼ ujC2 and cXi is the restrict operator defined by cXi u ¼ ujXi , and the linear boundary integral operators B, G and K are defined as follows: Bi ðt; kÞ ¼
Z tZ 0
Gi ðt; c0 uÞ ¼
Ui ðx y; t sÞ kðy; sÞ dsy ds;
C2
Z tZ 0
uðy; sÞ rðUi ; P i Þðx y; t sÞ nðyÞ dsy ds;
C2
K i ðt; cX2 u0 ; cX2 fÞ ¼
Z
Ui ðx y; tÞ u0 ðyÞ dy
Z tZ 0
X2
Ui ðx y; t sÞ
X2
fðy; sÞ dy ds. Here ðUi ðx; tÞ; P i ðx; tÞÞ; i ¼ 1; 2 is the fundamental solution of the Stokes operator which satisfies oUi mDUi þ rP i ¼ dðxÞdðtÞei ; ot
ð2Þ
div Ui ¼ 0.
ð3Þ
For the detail case of the fundamental solution ðUi ; P i Þ, the readers can refer to Refs. [6,8,13]. In order to obtain the coupling variational formulation of the reduced Stokes coupling problem, let us introduce some Hilbert spaces. For the bounded domain X1, we define 2
H ¼ fv 2 L2 ðX1 Þ ; divv ¼ 0 in X1 with v njC ¼ 0g; with the norm j j0;X1 ; 2
W ¼ H 1C ¼ fv 2 H 1 ðX1 Þ ; vjC ¼ 0g;
with the norm j j1;X1 ;
V ¼ fv 2 W ; divv ¼ 0 in X1 g;
with the norm j j1;X1 ;
2
H 2C ¼ fv 2 H 2 ðX1 Þ ; vjC ¼ 0g; R M ¼ fq 2 L2 ðX1 Þ; X1 qdx ¼ 0g; 1
1
H 20 ðC2 Þ2 ¼ fw 2 H 2 ðC2 Þ2 ; 1 2
R C2
w ndsx ¼ 0g;
with the norm j j2;X1 ; with the norm j j0;X1 ; with the norm k k1;C2 ; 2
K ¼ the dual space of H 0 ðC2 Þ2 , with the norm k k1;C2 . 2 From the GreenÕs formula and problem (R–S 0 ), we obtain the following cou2 2 pling variational formulation: Given u0 2 L ðXÞ with div u0 ¼ 0 and 2 1 þ 1 f 2 L ðR ; W ðXÞ Þ find ðuðtÞ; kðtÞ; pðtÞÞ such that for t P 0,
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
8 ðut ; vÞ þ a0 ðu;vÞ þ a1 ðu;u;vÞ ðp; divvÞ þ hv;ki ¼ ðf; vÞ; > > > < ðq;divuÞ ¼ 0; ðQÞ > bðt;k; lÞ ¼ 12 c0 u þ Gðt; c0 uÞ þ Kðt;cX2 u0 ;cX2 fÞ;l ; > > : uðx;0Þ ¼ u0 ðxÞ; R where ut ¼ ou ; ðu; vÞ ¼ X1 u v dx and ot 2 Z mX oui ouj ovi ovj þ þ a0 ðu; vÞ ¼ dx; 2 i;j¼1 X1 oxj oxi oxj oxi a1 ðu; v; wÞ ¼
Z
1229
8v 2 W ; 8q 2 L20 ðX1 Þ; 8l 2 K; 8x 2 X1 ;
ðu rÞv w dx; bðt; k; lÞ ¼ hBðt; kÞ; li.
X1
By using a simply modified argument in Ref. [5], we have proven the following results. R Lemma 1. Let u 2 L2 ð0; T ; W Þ with C2 u n ds ¼ 0, u0 2 H ðXÞ; f 2 L1 ðRþ ; W 1 ðXÞ2 Þ. Then the variation formulation: Find k 2 L2 ð0; T ; KÞ such that for any l 2 L2 ð0; T ; KÞ, Z T Z T 1 c u þ Gðt; c0 uÞ þ Kðt; cX2 u0 ; cX2 fÞ; l dt; bðt; k; lÞ dt ¼ ð4Þ 2 0 0 0 admits a unique solution k ¼ kðc0 uÞ satisfying Z T Z T 2 2 2 2 kkk1;C2 dt 6 c ju0 j0;X2 þ ðjfj1;X2 þ juj1;X1 Þ dt ; 2
0
ð5Þ
0
Z t 1 1 2 2 hc0 u; kðc0 uÞi ds P ju0 j0;X2 jfj1;X2 ds 8t 2 ½0; T ; ð6Þ 2 4m 0 0 where we always assume that T > 0 is any fixed finite time, c and c0 ; c1 ; . . . appeared hereafter denotes some generic positive constants which depends only on the data ðm; X1 Þ. Z
t
Moreover, by recalling Ref. [8], we have the following well-posedness and regularity. Theorem 2. Let u0 2 V ðXÞ; f 2 L1 ðRþ ; W 1 ðXÞÞ \ L1 ðRþ ; L2 ðXÞ2 Þ. Then problem (R–S 0 ) admits a unique solution ðu; k; pÞ 2 L1 ð0; T ; V Þ \ L2 ð0; T ; H 2C Þ 1 L2 ð0; T ; H 2 ðC2 Þ2 Þ L2 ð0; T ; H 1 ðX1 Þ \ L20 ðX1 ÞÞ for all T > 0. Moreover; ðu; k; pÞ satisfies the following estimates: ! Z T 2 ou 2 2 2 2 juðtÞj1;X1 þ þ kuk2;X1 þ kkk1;C2 þ kpk1;X1 ds 6 cðX1 ; u0 ; f; T Þ; ot 2 0 0;X1 ð7Þ
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for all t 2 ½0; T , where cðm; X1 ; u0 ; f; T Þ > 0 denotes constant depending on the data ðm; X1 ; u0 ; f; T Þ. Finally, we conclude this section by recalling some estimates of the trilinear form a1 ð; ; Þ and bilinear forms a0 ð; Þ and bðt; ; Þ given in Refs. [5,8]: 1
ja1 ðu; v; wÞj 6 cðjuj0;X1 juj1;X1 jwj0;X1 jwj1;X1 Þ2 jvj1;X1 ; 8u; v; w 2 W ;
ð8Þ
ja1 ðu; v; wÞj 6 cjuj1;X1 jvj1;X1 jwj1;X1 ; 8u; v; w 2 W ;
ð9Þ
1
1
1
1
ja1 ðu; v; wÞj 6 cjuj20;X1 juj22;X1 jvj1;X1 jwj0;X1 ; 8u 2 H 2C ; v; w 2 W ; 1
ð10Þ
1
ja1 ðu; v; wÞj 6 cjuj20;X1 juj21;X1 jvj21;X1 jvj22;X1 jwj0;X1 ; 8u; w 2 W ; v 2 H 2C ; Z
Z
T
T
bðt; kðtÞ; lðtÞÞ dt 6 c 0
0
2 kkk1 ; C2 dt 2
1=2 Z
1=2
T 0
2 klk1;C2 2
dt
8k; l 2 L2 ð0; T ; KÞ; Z
bðt; kðtÞ; kðtÞÞ dt P c1 0
; ð12Þ
Z
T
ð11Þ
0
T
kkk21 ; C2 dt; 8k 2 L2 ð0; T ; KÞ; 2
akuk2 6 a0 ðu; uÞ; ja0 ðu; vÞj 6 mjuj1;X1 jvj1;X1 ; 8u; v 2 W ;
ð13Þ ð14Þ
where constant a > 0 depends on m.
2. Stabilized finite element approximation Let h > 0 be a real positive parameter. Finite element subspace ðW h ; Kh ; M h Þ of ðW ; K; MÞ is characterized by sh ¼ sh ðXÞ, a partitioning of X into triangles or quadrilaterals, assumed to be regular in the usual sense, i.e., for some r and x with r > 1 and 0 < x < 1, hK 6 rqK ; 8K 2 sh ;
ð15Þ
j cos hiK j 6 x;
ð16Þ
i ¼ 1; 2; 3; 4; 8K 2 sh ;
where hK is the diameter of element K, qK is the diameter of the inscribed circle of element K, and hiK are the angles of K in the case of a quadrilateral partitioning. The mesh parameter h is given by h ¼ maxfhK g, and the set of all interelement boundaries will be denoted by Ch. Let us denote by si, 1 6 i 6 m the finite number of segments of a line composing the boundary C2.
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
1231
The finite element subspaces used in this paper are defined by setting P 1 ðKÞ; if K is triangular; R1 ðKÞ ¼ ð17Þ Q1 ðKÞ; if K is quadrilateral; giving the continuous piecewise (bi)linear velocity subspace 2
W h ¼ fvh 2 C 0 ðX1 Þ \ W ; hvh ; ni ¼ 0; vh jK 2 R1 ðKÞ; 8K 2 sh g; and the piecewise constant stress subspace 2
Kh ¼ flh 2 L2 ðC2 Þ \ K; lh jsi 2 P 0 ðsi Þ; 1 6 i 6 mg; and the piecewise constant pressure subspace M h ¼ fqh 2 L2 ðX1 Þ \ M; qh jK 2 P 0 ðKÞ; 8K 2 sh g. Note that for finite element space pair ðW h ; M h Þ, neither of these methods are stable in standard Babuska-Brezzi sense; ðP 1 ; P 0 Þ triangle ‘‘locks’’ on regular grids (since there are more discrete incompressibility constrains than velocity degrees of freedom), and the ðQ1 ; P 0 Þ quadrilateral is the most infamous example of an unstable mixed method. The L2-orthogonal projection operators P h : L2 ðX1 Þ2 ! W h is defined as follows: 2
ðP h v; vh Þ ¼ ðv; vh Þ; 8v 2 L2 ðX1 Þ ; vh 2 W h . With these additional definitions a stabilized discrete formulation of the time-dependent reduced Stokes coupling problem (Qh) can be stated as follows. Definition 3. Stabilized finite element ðW h ; Kh ; M h Þ such that for all t > 0
formulation:
Find
ðuh ; kh ; ph Þ 2
8 ðu ; vÞ þ Bh ððuh ; ph Þ; ðv; qÞÞ þ hv; kh i þ a1 ðuh ; uh ; vÞ ¼ ðf ; vÞ; 8ðv; qÞ 2 ðW h ; M h Þ; > < ht ðQh Þ bðt; k; lÞ ¼ 12 c0 u þ Gðt; c0 uÞ þ Kðt; cX2 u0 ; cX2 fÞ; l ; 8l 2 Kh ; > : uh ð0Þ ¼ u0h ¼ P h u0 ;
where Bh ððuh ; ph Þ; ðvh ; qh ÞÞ ¼ aðuh ; vh Þ dðvh ; ph Þ þ dðuh ; qh Þ þ bCh ðph ; qh Þ; Ch ðph ; qh Þ ¼
X e2Ch
Z he e
½ph e ½qh e ds þ b
m X i¼1
Z h si
ph qh ds; 8ph ; qh 2 M h ;
si
½e is the jump operator across e 2 Ch and b > 0 is the local stabilization parameter (refer to Refs. [10,14]). For the above finite element spaces Wh, Kh and Mh, there exist the operators ph, sh and qh such that
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Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
jv ph vj0;X1 þ hjv ph vj1;X1 6 ch2 kvk2;X1 ; 8v 2 H 2 ðX1 Þ \ W ; 1=2
ð18Þ
2
kl sh lk1;C2 6 chklk1;C2 ; 8l 2 H 0 ðC2 Þ ;
ð19Þ
jq qh qj0;X1 6 chkqk1;X1 ; 8q 2 H 1 ðXÞ \ M;
ð20Þ
2
2
and the inverse inequality jvh j1;X1 6 ch1 jvh j0;X1 ; 8vh 2 W h ;
ð21Þ
hold (see Refs. [2,3,16]). Moreover, the following properties which are classical consequences of Eqs. (18)–(21) (see Refs. [2,3,9]) will be very useful jP h vj1;X1 6 cjvj1;X1 ; 8v 2 W ;
ð22Þ 2
jv P h vj0;X1 þ hjv P h vj1;X1 6 ch2 kvk2;X1 ; 8v 2 H 2 ðX1 Þ \ W ;
ð23Þ
jv P h vj0;X1 6 chjv P h vj1;X1 ; 8v 2 W .
ð24Þ
The following stability results of these mixed methods defined above are established by Hughes et al. [10] and Kechkar et al. [14]. Theorem 4. Given a stabilization parameter b P b0 > 0 and above finite element spaces Wh and Mh, there hold the following estimates: 8ðvh ; qh Þ 2 ðW h ; M h Þ, jBh ððuh ; ph Þ; ðvh ; qh ÞÞj 6 cðjuh j1;X1 þ jqh j0;X1 Þðjvh j1;X1 þ jqh j0;X1 Þ; 8ðuh ; ph Þ 2 ðW h ; M h Þ;
ð25Þ
Ch ðp qh p; p qh pÞ 6 ch2 kpk21;X1 ; 8p 2 H 1 ðXÞ \ M;
ð26Þ
cðjvh j21;X1 þ jqh j20;X1 þ Ch ðqh ; qh ÞÞ 6 Bh ððvh ; qh Þ; ðvh ; qh ÞÞ; 8ðvh ; qh Þ 2 ðW h ; M h Þ;
ð27Þ
where c > 0 is a constant dependent of m and b0, and b0 is some fixed positive constant. With above finite element spaces and the stabilized theorem for bilinear stabilized finite element form Bh ððuh ; ph Þ; ðvh ; qh ÞÞ, we can obtain the wellposedness of stabilized finite element formulation (Qh). 2
Lemma 5. Let uh 2 L2 ð0; T ; W h Þ, u0 2 H ðXÞ; f 2 L1 ðRþ ; W 1 ðXÞ Þ. Then the variation formulation: Find kh 2 L2 ð0; T ; Kh Þ such that for any l 2 L2 ð0; T ; Kh Þ,
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
Z
T
bðt; kh ; lÞ dt ¼
Z
0
T
0
1233
1 c u þ Gðt; c0 uh Þ þ Kðt; cX2 u0 ; cX2 fÞ; l dt; 2 0
ð28Þ
admits a unique solution kh ¼ kh ðc0 uh Þ satisfying Z
T 2 kkh k1;C2 2
0
Z
t 0
dt 6 c
2 ju0 j0;X2
þ
Z
T 2 ðjfj1;X2
0
1 1 2 hc0 u; kðc0 uÞ > ds P ju0 j0;X2 2 4m
þ
Z
2 juh j1;X1 Þ dt
;
ð29Þ
t
0
2
jfj1;X2 ds; 8t 2 ½0; T :
ð30Þ
Theorem 6. Under the assumptions of Theorems 2 and 4, problem (Qh) admits a unique solution ðuh ; kh ; ph Þ 2 L1 ð0; T ; W h Þ \ L2 ð0; T ; W h Þ L2 ð0; T ; Kh Þ L2 ð0; T ; M h Þ for all T > 0. Moreover, ðuh ; kh ; ph Þ satisfies the following estimates: Z T juh ðtÞj21;X1 þ ðjuh j21;X1 þ kkk21;C2 þ jph j20;X1 Þ dt 6 cðm; X1 ; u0 ; f; T Þ; ð31Þ 2
0
for all t 2 ½0; T .
4. Error estimates In this section, we mainly consider the error estimates of finite element solution ðuh ; kh ; ph Þ. Lemma 7. Under the assumptions of Theorems 2 and 4, Z
T
kh k21;C2 2
kk 0
2
dt 6 h cðm; X1 ; u0 ; f; T Þ þ c
Z
T 0
ju uh j21;X1 dt.
ð32Þ
Proof. From problem (Q) and problem (Qh), we have Z T bðt; sh k kh ; lÞ dt 0
1 ¼ 2 þ
Z
T
hc0 ðu uh Þ; li dt þ 0
Z 0
Z
T
hGðt; c0 ðu uh ÞÞ; li dt
0 T
bðt; sh k k; lÞdt; 8l 2 L2 ð0; T ; Kh Þ.
ð33Þ
1234
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
Due to the definition of the fundamental tension ðU k ðx; tÞ; P k ðx; tÞÞ, we can prove that for all l 2 L2 ð0; T ; Kh Þ, Z T Z T 1=2 Z T 1=2 2 2 < ðGðt;c ðu u ÞÞ;lÞdt 6 c kc ðuu Þk dt klk dt . 1 1 h h 0 0 ;C2 ;C2 0
2
0
2
0
ð34Þ Next, by taking lh ¼ sh k kh in Eq. (33) and using Eqs. (13),(14) and (34), we derive Z T Z T
2 2 2 ksh k kh k1;C2 dt 6 c kc0 ðu uh Þk1;C2 þ ksh k kk1;C2 dt. ð35Þ 2
0
2
0
2
Combining Eq. (35) with the triangle inequality gives Z T Z T
kk kh k21;C2 dt 6 c ku uh k21;C2 þ ksh k kk21 ; C2 dt.
ð36Þ
Using again the trace theorem [1,15] and Eq. (19), we derive Z T Z T
2 2 2 kk kh k1;C2 dt 6 c ju uh j1;X1 þ h2 kkk1;C2 dt.
ð37Þ
2
0
2
0
2
0
2
2
0
h
Combining Eq. (37) with regularity estimates Eq. (7) yields Eq. (32). Lemma 8. Under the assumptions of Theorems 2 and 4, Z T 2 kk kh ðc0 P h uÞk1 ; C2 dt 6 h2 cðm; X1 ; u0 ; f; T Þ; Z
ð38Þ
2
0
t
hc0 eh ; kh ðc0 P h uÞ kh i ds P 0; 8t 2 ½0; T ;
ð39Þ
0
where eh ¼ P h u uh . Proof. From the definition of k and kh ðc0 P h uÞ, we have Z T bðt; sh k kh ðc0 P h uÞ; lÞdt 0 Z T Z T 1 c0 ðu P h uÞ þ Gðt; c0 ðu P h uÞÞ; l dt; bðt; sh k k; lÞdt þ ¼ 2 0 0 2 8l 2 L ð0; T ; Kh Þ. ð40Þ A simply modified argument to Eq. (34) can yield Z
0
T
Z < ðGðt; c0 ðu P h uÞÞ; lÞdt 6 c
0
T
kc0 ðu P h uÞk21;C2 dt 2
1=2 Z 0
T
klk21;C2 dt 2
1=2 ;
ð41Þ
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
1235
for all l 2 L2 ð0; T ; Kh Þ. By taking l ¼ sh k kh ðc0 P h uÞ in Eq. (40) and using Eqs. (13),(14) and (41), we have Z T Z T Z T ksh k kh ðc0 P h uÞk21;C2 dt 6 c ksh k kk21;C2 dt þ c kc0 ðu P h uÞk21;C2 dt. 2
0
2
0
2
0
ð42Þ Combining Eq. (42) with the triangle inequality and the trace theorem [1,15] yields Z T Z T
2 kk kh ðc0 P h uÞk1;C2 dt 6 c ksh k kk21;C2 þ ju P h uj21;X1 dt. 2
0
2
0
ð43Þ Using again estimates Eqs. (19) and (23) in Eq. (43), we can derive Z
T
kk 0
2 kh ðc0 P h uÞk1;C2 2
dt 6 ch
2
Z
T
2 2 kkk1;C2 þ kuk2;X1 dt. 2
0
ð44Þ
Combining Eq. (44) with regularity estimates Eq. (7) yields Eq. (38). Moreover, by using the definition of kh ðc0 P h uÞ and kh, one finds that for all l 2 L2 ð0; T ; Kh Þ Z T Z T 1 c0 eh þ Gðt; c0 eh Þ; l dt. bðt; kh ðc0 P h uÞ kh ; lÞ dt ¼ 2 0 0 Applying Lemma 5 with u0 ¼ 0; f ¼ 0, to the above variational formulation, we know that kh ðc0 P h uÞ kh ¼ kh ðc0 eh Þ satisfies Eq. (39). h Lemma 9. Under the assumptions of Theorems 2 and 4, 2
juðtÞ uh ðtÞj0;X1 þ
Z
T 2
0
2
ðjuðtÞ uh ðtÞj1;X1 þ jpðtÞ ph ðtÞj0;X1 Þ dt 6 cðm; X1 ; u0 ; f; T Þh2 ; 8t 2 ½0; T . ð45Þ
Proof. From problems (Q) and (Qh), we obtain ðeht ; vÞ þ Bh ððeh ; gh Þ; ðv; qÞÞ þ a0 ðu P h u; vÞ dðv; p qh pÞ þ dðu P h u; qÞ þ bCh ðp qh p; qÞ þ hc0 v; k kh i þ a1 ðu uh ; u; vÞ þ a1 ðuh ; u uh ; vÞ ¼ 0; 8ðv; qÞ 2 ðW h ; M h Þ;
ð46Þ
where gh ¼ qh p ph . Then, by taking ðv; qÞ ¼ ðeh ; gh Þ in Eq. (46) and using Theorem 4, it follows that:
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Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
1 d jeh j20;X1 þ cðjeh j21;X1 þ jgh j20;X1 þ Ch ðgh ; gh ÞÞ þ a0 ðu P h u; eh Þ 2 dt dðeh ; p qh pÞ þ dðu P h u; gh Þ þ bCh ðp qh p; gh Þ þ hc0 eh ; k kh i þ a1 ðeh ; u; eh Þ þ a1 ðuh ; eh ; eh Þ þ a1 ðu P h u; u; eh Þ þ a1 ðuh ; u P h u; eh Þ ¼ 0. ð47Þ Due to Eqs. (8)–(11), the Poincare inequality and the trace theorem [1,15], one finds c ja0 ðu P h u; eh Þj 6 jeh j21;X1 þ cju P h uj21;X1 16 c 2 2 jdðeh ; p qh pÞj 6 jeh j1;X1 þ cjp qh pj0;X1 16 c 2 2 jdðu P h u; gh Þj 6 jgh j0;X1 þ cju P h uj1;X1 8 c bjCh ðp qh p; gh Þj 6 Ch ðgh ; gh Þ þ cCh ðp qh p; p qh pÞ 8 hc0 eh ; k kh i ¼ hc0 eh ; kh ðc0 P h uÞ kh i þ hc0 eh ; k kh ðc0 P h uÞi c 2 2 jhc0 eh ; k kh ðc0 P h uÞij 6 jeh j1;X1 þ ckk kh ðc0 P h uÞk1;C2 2 16 c 2 2 2 ja1 ðeh ; u; eh Þj 6 cjuj1;X1 jeh j1;X1 jeh j0;X1 6 jeh j1;X1 þ cjuj1;X1 jeh j0;X1 ; 16 1=2
1=2
3=2
1=2
ja1 ðuh ; eh ; eh Þj 6 cjuh j1;X1 juh j0;X1 jeh j1;X1 jeh j0;X1 c 6 jeh j21;X1 þ cjuh j20;X1 juh j21;X1 jeh j20;X1 ; 16 ja1 ðu P h u; u; eh Þjþja1 ðuh ; u P h u; eh Þj 6 cjeh j1;X1 ju P h uj1;X1 ðjuj1;X1 þ juh j1;X1 Þ c 2 2 2 2 6 jeh j1;X1 þ cðjuj1;X1 þ juh j1;X1 Þju P h uj1;X1 . 16 Combining the above estimates with Eq. (47) and applying Theorems 4 and 6 yields d 2 2 2 jeh j0;X1 þ cðjeh j1;X1 þ jgh j0;X1 Þ þ 2hc0 eh ; kh ðc0 P h uÞ kh i dt 6 cðm; X1 ; u0 ; f; T Þðju P h uj21;X1 þ jp qh pj20;X1 þ h2 kpk21;X1 Þ þ ckk kh ðc0 P h uÞk21;C2 þ cðm; X1 ; u0 ; f; T Þjeh j20;X1 . 2
ð48Þ
Integrating Eq. (48) from 0 to t and applying Lemma 8 and noting eh ð0Þ ¼ 0, we derive
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
jeh ðtÞj20;X1
þc
Z
1237
t 0
ðjeh j21;X1 þ jgh j20;X1 Þ ds
6 cðm; X1 ; u0 ; f; T Þ
Z
t 2
0
2
2
ðju P h uj1;X1 þ jp qh pj0;X1 þ h2 kpk1;X1 Þ ds
2
þ h cðm; X1 ; u0 ; f; T Þ þ cðm; X1 ; u0 ; f; T Þ
Z
t 2
0
jeh j0;X1 ds.
ð49Þ
Applying the Gronwall lemma [3] to Eq. (49) and using Eqs. (20), (23) and Theorems 2 and 6, we derive Z t 2 jeh ðtÞj0;X1 þ ðjeh ðsÞj21;X1 þ jgh ðsÞj20;X1 Þ ds 6 cðm; X1 ; u0 ; f; T Þh2 . ð50Þ 0
Combining Eq. (50) with the triangle inequality, noting Z t 2 2 2 ðju P h uj0;X1 þ jp qh pj0;X1 Þ ds juðtÞ P h uðtÞj0;X1 þ 0
6
ch2 juðtÞj21;X1
2
þ ch
Z 0
t
ðkuk22;X1 þ kpk21;X1 Þ ds;
ð51Þ
and applying Theorem 2, one finds Z t 2 2 2 ðju uh j1;X1 þ jp ph j0;X1 Þ ds juðtÞ uh ðtÞj0;X1 þ 0
6 cðm; X1 ; u0 ; f; T Þh2 .
ð52Þ
Finally, by combining Lemma 7 with Lemma 9, we obtain the following main optimal error estimates. Theorem 10. Under the assumptions of Theorems 2 and 4, Z T juðtÞ uh ðtÞj20;X1 þ ðjuðtÞ uh ðtÞj21;X1 þ kkðtÞ kh ðtÞk21;C2 0 2 þ jpðtÞ ph ðtÞj0;X1 Þ dt 6 cðm; X1 ; u0 ; f; T Þh2 ;
2
ð53Þ
for all t 2 ½0; T .
References [1] R.A. Adams, Sobolev Spaces, Academic Press, New York, 1975. [2] P.G. Ciarlet, The Finite Element Method for Elliptic Problems, Amsterdam, North-Holland, 1978.
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[3] V. Girault, P.A. Raviart, Finite Element Approximation of the Navier–Stokes Equations, Springer-Verlag, Berlin, 1979. [4] Chen He, Zhouping Xin, Weighted estimates for nonstationary Navier–Stokes equations in exterior domains, Meth. Appl. Anal. 7 (2000) 443–458. [5] Yinnian He, Kaitai Li, The coupling of boundary element and finite element methods for the exterior nonstationary Navier–Stokes equations, Acta Math. Sci. 11 (2) (1991) 190–207. [6] Yinnian He, Kaitai Li, The coupling of boundary Integral and finite element methods for the non-stationary flow, Numer. Meth. PDEs 14 (1998) 549–565. [7] Yinnian He, Kaitai Li, Stokes coupling method for the exterior flow Part I: Approximate accuracy, Acta Math. Appl. Sin. 15 (3) (1999) 333–336. [8] Yinnian He, Kaitai Li, Stokes coupling method for the exterior flow Part III: Regularity, Progr. Nat. Sci. 11 (2001) 734–745. [9] J.G. Heywood, R. Rannacher, Finite element approximation of the nonstationary Navier– Stokes problem, I: Regularity of solutions and second-order error estimates for spatial discretization 19 (1982) 275–311. [10] T.J.R. Hughes, L.P. Franca, A new finite element formulation for CFD: VII. The Stokes problem with various well-posed boundary conditions: Symmetric formulations that converge for all velocity/pressure space, Comput. Meth. Appl. Mech. Engrg. 65 (1987) 85–96. [11] Kaitai Li, Yinnian He, The coupling of boundary integral and finite element methods for the Navier–Stokes equations in an exterior domain, J. Comp. Math. 7 (2) (1989) 157–173. [12] Kaitai Li, Yinnian He, Coupling method for the exterior stationary Navier–Stokes equations, Numer. Meth. PDEs 9 (1993) 35–49. [13] Kaitai Li, Yinnian He, Stokes coupling method for the exterior flow Part II: Well-posedness analysis, Acta Math. Sci. 16 (4) (1996) 442–457. [14] N. Kechkar, D. Silvester, Analysis of locally stabilized mixed finite element method for the Stokes problem, Math. Comp. 58 (1992) 1–10. [15] J.L. Lions, E. Magenes, Non-homogeneous Boundary Value Problem and Applications, Springer-Verlag, Berlin, 1972. [16] A. Sequeira, The coupling of boundary integral and finite element methods for the bidimensional exterior steady Stokes problem, Math. Meth. Appl. Sci. 5 (1983) 356–375. [17] A. Sequeira, On the computer implementation of a coupled boundary and finite element methods for the bidimensional exterior steady Stokes problem, Math. Meth. Appl. Sci. 8 (1986) 117–133. [18] H. Wang, Boundary element methods for the nonstationary Stokes problem in three dimensions and their convergence analysis, in: C.A. Brebbia, G.S. Gipson (Eds.), Boundary Element, XIII, Computational Mechanics Publications Elsevier Applied Science, 1991, p. 161. [19] Dehao Yu, The coupling of natural BEM and FEM for the Stokes problem on unbounded domain, Chin. J. Numer. Math. Appl. 14 (4) (1992) 111–120. [20] Dehao Yu, Mathematical Theory of Natural Boundary Element Method, Science Press, Beijing, 1993. [21] Jialin Zhu, The boundary integral equation method for the 3-D stationary Stokes problem, Math. Numer. Sin. 7 (1) (1985) 40–49. [22] Jialin Zhu, A boundary integral equation method for solving stationary Stokes problem, Math. Numer. Sin. 8 (3) (1986) 281–289.
www.elsevier.com/locate/amc
Stokes coupling method for the exterior flow Part IV: stabilized finite element approximation q Yinnian He Faculty of Science, Xi’an Jiaotong University, Xi’an 710049, China
Abstract Based on the Stokes coupling method for solving the two dimensional exterior unsteady Navier–Stokes equations, the stabilized finite element approximation combining the boundary element with the stabilized finite element for solving the reduced Stokes coupling equations is presented. Finally, the optimal error estimates for the finite element solution are derived. Ó 2005 Elsevier Inc. All rights reserved. Keywords: Navier–Stokes equations; Stokes coupling method; Stabilized finite element
Let X0 be a simple connected bounded open set of R2 with smooth boundary C and let X denote the complement of X0 [ C where the origin of coordinates is located in the interior of X0. In X Rþ , we consider the following initial-boundary value problem of the Navier–Stokes equations:
q
Subsidized by the NSF of China 10371095 and the NSF of Shaanxi Province. E-mail address: [email protected]
0096-3003/$ - see front matter Ó 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2005.03.019
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Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
8 ou m Du þ ðu rÞu þ rp ¼ fðx; tÞ; ðx; tÞ 2 X Rþ ; > ot > > > < div u ¼ 0; ðx; tÞ 2 X Rþ ; ðN–SÞ uðx; tÞjC ¼ 0; lim uðx; tÞ ¼ 0; 8t 2 Rþ ; > > jxj!1 > > : x 2 X; uðx; 0Þ ¼ u0 ðxÞ; where uðx; tÞ and pðx; tÞ denote the unknown velocity vector and the pressure of the fluid at point ðx; tÞ 2 X Rþ , respectively, while m > 0 is the dynamic viscosity, fðx; tÞ represents a density of external forces and u0 ðxÞ is a initial velocity vector satisfying div u0 ¼ 0 in X with u0 njC ¼ 0, limjxj!1 u0 ðxÞ ¼ 0. For a fixed small number e > 0, we introduced an artificial smooth boundary C2 ¼ fx 2 X; jxj ¼ R1 g, separating an outer region X2 from an inner region X1 such that juðx; tÞj 6 e; 8ðx; tÞ 2 ðX2 [ C2 Þ Rþ .
ð1Þ
For the justification of Eq. (1) which is uniform with respect to time t, the readers can refer to the work of He and Xin [4]. For the reasons of the computational economy and the inertial term ðu rÞu being small in X2, we consider the following Stokes coupling problem: 8 ou m Du þ X X1 ðu rÞu þ rp ¼ f; ðx; tÞ 2 X Rþ ; > > > ot > < div u ¼ 0; ðx; tÞ 2 X Rþ ; 0 ðN–S Þ uðx; tÞjC ¼ 0; lim uðx; tÞ ¼ 0; 8t 2 Rþ ; > > jxj!1 > > : x 2 X; uðx; 0Þ ¼ u0 ðxÞ; where
( X X1 ðxÞ ¼
1; x 2 X1 ; 0; x 2 X1 .
Here, we also assume that the solution ðu; pÞ of (N–S 0 ) satisfies Eq. (1). The boundary element method (BEM) or the coupling method of BEM with finite element method (FEM)for solving the stationary Stokes problem on bounded domain and unbounded domain had been discussed by Yu [19,20], Zhu [21,22] and Sequeira [16,17]. The coupling method of BEM with FEM for solving the non-stationary Stokes problem had been discussed by He et al. [5] and Wang [18]. While coupling method of BEM with FEM for solving the stationary Navier–Stokes problem on exterior domain had been discussed by Li et al. [11,12]. Hereafter will we mainly consider the coupling method of BEM with FEM for solving the non-stationary Navier–Stokes problem (N–S) on exterior domain. In Ref. [7], we have proven that the difference of solutions between problems (N–S) and (N–S 0 ) is small; in Refs. [6,8,13], we presented the reduced Stokes coupling equations corresponding to problem (N–S 0 ) and discussed their well-posedness and the finite element approximation.
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
1227
In this paper, by using the stabilized finite element method presented by Hughes et al. [10], we will present the stabilized finite element approximation combining the boundary element with the stabilized finite element for solving the reduced Stokes coupling equations and provide the optimal error estimates for the finite element solution. 1. Reduced stokes coupling problem In order to derive the reduced Stokes coupling problem corresponding to problem (N–S 0 ), we take the equivalent form of problem (N–S 0 ): 8 ou m Du þ ðu rÞu þ rp ¼ f; ðx; tÞ 2 X1 Rþ ; > ot > > < div u ¼ 0; ðx; tÞ 2 X1 Rþ ; ðS1 Þ þ þ > > > uðx; tÞjC ¼ 0; rðu; pÞ njC2 ¼ k ; 8t 2 R ; : uðx; 0Þ ¼ u0 ðxÞ; x 2 X1 ; and 8 ou m Du þ rp ¼ f; ðx; tÞ 2 X2 Rþ ; > ot > > > < div u ¼ 0; ðx; tÞ 2 X2 Rþ ; ðS2 Þ uðx; tÞjC2 ¼ u ; lim uðx; tÞ ¼ 0; 8t 2 Rþ ; > > jxj!1 > > : x 2 X2 ; uðx; 0Þ ¼ u0 ðxÞ; where n denotes the unit normal vector of C and C2, corresponding to the exterior to X0 and X2, respectively, and ðu ; p Þ ¼ lim ðu; pÞjX1 ; x!C2
k ¼ rðu ; p Þ njC2 ; rðu; pÞ ¼
ðuþ ; pþ Þ ¼ lim ðu; pÞjX2 ; x!C2
kþ ¼ rðuþ ; pþ Þ njC2 ;
oui ouj þ ; dij p þ m oxj oxi 22
i; j ¼ 1; 2.
Now, we consider the equivalent system S1–S2 of the Stokes coupling equations (N–S 0 ). Applying the Greeen formula to S2, we derive the reduced Stokes coupling equations (see Refs. [6,8,13]):
ðR–S0 Þ
8 ou mDu þ ðu rÞu þ rp ¼ f; ðx; tÞ 2 X1 Rþ ; > ot > > > > ðx; tÞ 2 X1 Rþ ; > < div u ¼ 0; þ ujC ¼ 0; rðu; pÞ njC2 ¼ k ; 8t 2 Rþ ; > > > uðx; 0Þ ¼ u0 ðxÞ; x 2 X1 ; > > > : 1 Bðt; kÞ ¼ 2 c0 u þ Gðt; c0 uÞ þ Kðt; cX2 u0 ; cX2 fÞ; ðx; tÞ 2 C2 Rþ ;
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Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
where c0 is the trace operator defined by c0 u ¼ ujC2 and cXi is the restrict operator defined by cXi u ¼ ujXi , and the linear boundary integral operators B, G and K are defined as follows: Bi ðt; kÞ ¼
Z tZ 0
Gi ðt; c0 uÞ ¼
Ui ðx y; t sÞ kðy; sÞ dsy ds;
C2
Z tZ 0
uðy; sÞ rðUi ; P i Þðx y; t sÞ nðyÞ dsy ds;
C2
K i ðt; cX2 u0 ; cX2 fÞ ¼
Z
Ui ðx y; tÞ u0 ðyÞ dy
Z tZ 0
X2
Ui ðx y; t sÞ
X2
fðy; sÞ dy ds. Here ðUi ðx; tÞ; P i ðx; tÞÞ; i ¼ 1; 2 is the fundamental solution of the Stokes operator which satisfies oUi mDUi þ rP i ¼ dðxÞdðtÞei ; ot
ð2Þ
div Ui ¼ 0.
ð3Þ
For the detail case of the fundamental solution ðUi ; P i Þ, the readers can refer to Refs. [6,8,13]. In order to obtain the coupling variational formulation of the reduced Stokes coupling problem, let us introduce some Hilbert spaces. For the bounded domain X1, we define 2
H ¼ fv 2 L2 ðX1 Þ ; divv ¼ 0 in X1 with v njC ¼ 0g; with the norm j j0;X1 ; 2
W ¼ H 1C ¼ fv 2 H 1 ðX1 Þ ; vjC ¼ 0g;
with the norm j j1;X1 ;
V ¼ fv 2 W ; divv ¼ 0 in X1 g;
with the norm j j1;X1 ;
2
H 2C ¼ fv 2 H 2 ðX1 Þ ; vjC ¼ 0g; R M ¼ fq 2 L2 ðX1 Þ; X1 qdx ¼ 0g; 1
1
H 20 ðC2 Þ2 ¼ fw 2 H 2 ðC2 Þ2 ; 1 2
R C2
w ndsx ¼ 0g;
with the norm j j2;X1 ; with the norm j j0;X1 ; with the norm k k1;C2 ; 2
K ¼ the dual space of H 0 ðC2 Þ2 , with the norm k k1;C2 . 2 From the GreenÕs formula and problem (R–S 0 ), we obtain the following cou2 2 pling variational formulation: Given u0 2 L ðXÞ with div u0 ¼ 0 and 2 1 þ 1 f 2 L ðR ; W ðXÞ Þ find ðuðtÞ; kðtÞ; pðtÞÞ such that for t P 0,
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
8 ðut ; vÞ þ a0 ðu;vÞ þ a1 ðu;u;vÞ ðp; divvÞ þ hv;ki ¼ ðf; vÞ; > > > < ðq;divuÞ ¼ 0; ðQÞ > bðt;k; lÞ ¼ 12 c0 u þ Gðt; c0 uÞ þ Kðt;cX2 u0 ;cX2 fÞ;l ; > > : uðx;0Þ ¼ u0 ðxÞ; R where ut ¼ ou ; ðu; vÞ ¼ X1 u v dx and ot 2 Z mX oui ouj ovi ovj þ þ a0 ðu; vÞ ¼ dx; 2 i;j¼1 X1 oxj oxi oxj oxi a1 ðu; v; wÞ ¼
Z
1229
8v 2 W ; 8q 2 L20 ðX1 Þ; 8l 2 K; 8x 2 X1 ;
ðu rÞv w dx; bðt; k; lÞ ¼ hBðt; kÞ; li.
X1
By using a simply modified argument in Ref. [5], we have proven the following results. R Lemma 1. Let u 2 L2 ð0; T ; W Þ with C2 u n ds ¼ 0, u0 2 H ðXÞ; f 2 L1 ðRþ ; W 1 ðXÞ2 Þ. Then the variation formulation: Find k 2 L2 ð0; T ; KÞ such that for any l 2 L2 ð0; T ; KÞ, Z T Z T 1 c u þ Gðt; c0 uÞ þ Kðt; cX2 u0 ; cX2 fÞ; l dt; bðt; k; lÞ dt ¼ ð4Þ 2 0 0 0 admits a unique solution k ¼ kðc0 uÞ satisfying Z T Z T 2 2 2 2 kkk1;C2 dt 6 c ju0 j0;X2 þ ðjfj1;X2 þ juj1;X1 Þ dt ; 2
0
ð5Þ
0
Z t 1 1 2 2 hc0 u; kðc0 uÞi ds P ju0 j0;X2 jfj1;X2 ds 8t 2 ½0; T ; ð6Þ 2 4m 0 0 where we always assume that T > 0 is any fixed finite time, c and c0 ; c1 ; . . . appeared hereafter denotes some generic positive constants which depends only on the data ðm; X1 Þ. Z
t
Moreover, by recalling Ref. [8], we have the following well-posedness and regularity. Theorem 2. Let u0 2 V ðXÞ; f 2 L1 ðRþ ; W 1 ðXÞÞ \ L1 ðRþ ; L2 ðXÞ2 Þ. Then problem (R–S 0 ) admits a unique solution ðu; k; pÞ 2 L1 ð0; T ; V Þ \ L2 ð0; T ; H 2C Þ 1 L2 ð0; T ; H 2 ðC2 Þ2 Þ L2 ð0; T ; H 1 ðX1 Þ \ L20 ðX1 ÞÞ for all T > 0. Moreover; ðu; k; pÞ satisfies the following estimates: ! Z T 2 ou 2 2 2 2 juðtÞj1;X1 þ þ kuk2;X1 þ kkk1;C2 þ kpk1;X1 ds 6 cðX1 ; u0 ; f; T Þ; ot 2 0 0;X1 ð7Þ
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Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
for all t 2 ½0; T , where cðm; X1 ; u0 ; f; T Þ > 0 denotes constant depending on the data ðm; X1 ; u0 ; f; T Þ. Finally, we conclude this section by recalling some estimates of the trilinear form a1 ð; ; Þ and bilinear forms a0 ð; Þ and bðt; ; Þ given in Refs. [5,8]: 1
ja1 ðu; v; wÞj 6 cðjuj0;X1 juj1;X1 jwj0;X1 jwj1;X1 Þ2 jvj1;X1 ; 8u; v; w 2 W ;
ð8Þ
ja1 ðu; v; wÞj 6 cjuj1;X1 jvj1;X1 jwj1;X1 ; 8u; v; w 2 W ;
ð9Þ
1
1
1
1
ja1 ðu; v; wÞj 6 cjuj20;X1 juj22;X1 jvj1;X1 jwj0;X1 ; 8u 2 H 2C ; v; w 2 W ; 1
ð10Þ
1
ja1 ðu; v; wÞj 6 cjuj20;X1 juj21;X1 jvj21;X1 jvj22;X1 jwj0;X1 ; 8u; w 2 W ; v 2 H 2C ; Z
Z
T
T
bðt; kðtÞ; lðtÞÞ dt 6 c 0
0
2 kkk1 ; C2 dt 2
1=2 Z
1=2
T 0
2 klk1;C2 2
dt
8k; l 2 L2 ð0; T ; KÞ; Z
bðt; kðtÞ; kðtÞÞ dt P c1 0
; ð12Þ
Z
T
ð11Þ
0
T
kkk21 ; C2 dt; 8k 2 L2 ð0; T ; KÞ; 2
akuk2 6 a0 ðu; uÞ; ja0 ðu; vÞj 6 mjuj1;X1 jvj1;X1 ; 8u; v 2 W ;
ð13Þ ð14Þ
where constant a > 0 depends on m.
2. Stabilized finite element approximation Let h > 0 be a real positive parameter. Finite element subspace ðW h ; Kh ; M h Þ of ðW ; K; MÞ is characterized by sh ¼ sh ðXÞ, a partitioning of X into triangles or quadrilaterals, assumed to be regular in the usual sense, i.e., for some r and x with r > 1 and 0 < x < 1, hK 6 rqK ; 8K 2 sh ;
ð15Þ
j cos hiK j 6 x;
ð16Þ
i ¼ 1; 2; 3; 4; 8K 2 sh ;
where hK is the diameter of element K, qK is the diameter of the inscribed circle of element K, and hiK are the angles of K in the case of a quadrilateral partitioning. The mesh parameter h is given by h ¼ maxfhK g, and the set of all interelement boundaries will be denoted by Ch. Let us denote by si, 1 6 i 6 m the finite number of segments of a line composing the boundary C2.
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
1231
The finite element subspaces used in this paper are defined by setting P 1 ðKÞ; if K is triangular; R1 ðKÞ ¼ ð17Þ Q1 ðKÞ; if K is quadrilateral; giving the continuous piecewise (bi)linear velocity subspace 2
W h ¼ fvh 2 C 0 ðX1 Þ \ W ; hvh ; ni ¼ 0; vh jK 2 R1 ðKÞ; 8K 2 sh g; and the piecewise constant stress subspace 2
Kh ¼ flh 2 L2 ðC2 Þ \ K; lh jsi 2 P 0 ðsi Þ; 1 6 i 6 mg; and the piecewise constant pressure subspace M h ¼ fqh 2 L2 ðX1 Þ \ M; qh jK 2 P 0 ðKÞ; 8K 2 sh g. Note that for finite element space pair ðW h ; M h Þ, neither of these methods are stable in standard Babuska-Brezzi sense; ðP 1 ; P 0 Þ triangle ‘‘locks’’ on regular grids (since there are more discrete incompressibility constrains than velocity degrees of freedom), and the ðQ1 ; P 0 Þ quadrilateral is the most infamous example of an unstable mixed method. The L2-orthogonal projection operators P h : L2 ðX1 Þ2 ! W h is defined as follows: 2
ðP h v; vh Þ ¼ ðv; vh Þ; 8v 2 L2 ðX1 Þ ; vh 2 W h . With these additional definitions a stabilized discrete formulation of the time-dependent reduced Stokes coupling problem (Qh) can be stated as follows. Definition 3. Stabilized finite element ðW h ; Kh ; M h Þ such that for all t > 0
formulation:
Find
ðuh ; kh ; ph Þ 2
8 ðu ; vÞ þ Bh ððuh ; ph Þ; ðv; qÞÞ þ hv; kh i þ a1 ðuh ; uh ; vÞ ¼ ðf ; vÞ; 8ðv; qÞ 2 ðW h ; M h Þ; > < ht ðQh Þ bðt; k; lÞ ¼ 12 c0 u þ Gðt; c0 uÞ þ Kðt; cX2 u0 ; cX2 fÞ; l ; 8l 2 Kh ; > : uh ð0Þ ¼ u0h ¼ P h u0 ;
where Bh ððuh ; ph Þ; ðvh ; qh ÞÞ ¼ aðuh ; vh Þ dðvh ; ph Þ þ dðuh ; qh Þ þ bCh ðph ; qh Þ; Ch ðph ; qh Þ ¼
X e2Ch
Z he e
½ph e ½qh e ds þ b
m X i¼1
Z h si
ph qh ds; 8ph ; qh 2 M h ;
si
½e is the jump operator across e 2 Ch and b > 0 is the local stabilization parameter (refer to Refs. [10,14]). For the above finite element spaces Wh, Kh and Mh, there exist the operators ph, sh and qh such that
1232
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
jv ph vj0;X1 þ hjv ph vj1;X1 6 ch2 kvk2;X1 ; 8v 2 H 2 ðX1 Þ \ W ; 1=2
ð18Þ
2
kl sh lk1;C2 6 chklk1;C2 ; 8l 2 H 0 ðC2 Þ ;
ð19Þ
jq qh qj0;X1 6 chkqk1;X1 ; 8q 2 H 1 ðXÞ \ M;
ð20Þ
2
2
and the inverse inequality jvh j1;X1 6 ch1 jvh j0;X1 ; 8vh 2 W h ;
ð21Þ
hold (see Refs. [2,3,16]). Moreover, the following properties which are classical consequences of Eqs. (18)–(21) (see Refs. [2,3,9]) will be very useful jP h vj1;X1 6 cjvj1;X1 ; 8v 2 W ;
ð22Þ 2
jv P h vj0;X1 þ hjv P h vj1;X1 6 ch2 kvk2;X1 ; 8v 2 H 2 ðX1 Þ \ W ;
ð23Þ
jv P h vj0;X1 6 chjv P h vj1;X1 ; 8v 2 W .
ð24Þ
The following stability results of these mixed methods defined above are established by Hughes et al. [10] and Kechkar et al. [14]. Theorem 4. Given a stabilization parameter b P b0 > 0 and above finite element spaces Wh and Mh, there hold the following estimates: 8ðvh ; qh Þ 2 ðW h ; M h Þ, jBh ððuh ; ph Þ; ðvh ; qh ÞÞj 6 cðjuh j1;X1 þ jqh j0;X1 Þðjvh j1;X1 þ jqh j0;X1 Þ; 8ðuh ; ph Þ 2 ðW h ; M h Þ;
ð25Þ
Ch ðp qh p; p qh pÞ 6 ch2 kpk21;X1 ; 8p 2 H 1 ðXÞ \ M;
ð26Þ
cðjvh j21;X1 þ jqh j20;X1 þ Ch ðqh ; qh ÞÞ 6 Bh ððvh ; qh Þ; ðvh ; qh ÞÞ; 8ðvh ; qh Þ 2 ðW h ; M h Þ;
ð27Þ
where c > 0 is a constant dependent of m and b0, and b0 is some fixed positive constant. With above finite element spaces and the stabilized theorem for bilinear stabilized finite element form Bh ððuh ; ph Þ; ðvh ; qh ÞÞ, we can obtain the wellposedness of stabilized finite element formulation (Qh). 2
Lemma 5. Let uh 2 L2 ð0; T ; W h Þ, u0 2 H ðXÞ; f 2 L1 ðRþ ; W 1 ðXÞ Þ. Then the variation formulation: Find kh 2 L2 ð0; T ; Kh Þ such that for any l 2 L2 ð0; T ; Kh Þ,
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
Z
T
bðt; kh ; lÞ dt ¼
Z
0
T
0
1233
1 c u þ Gðt; c0 uh Þ þ Kðt; cX2 u0 ; cX2 fÞ; l dt; 2 0
ð28Þ
admits a unique solution kh ¼ kh ðc0 uh Þ satisfying Z
T 2 kkh k1;C2 2
0
Z
t 0
dt 6 c
2 ju0 j0;X2
þ
Z
T 2 ðjfj1;X2
0
1 1 2 hc0 u; kðc0 uÞ > ds P ju0 j0;X2 2 4m
þ
Z
2 juh j1;X1 Þ dt
;
ð29Þ
t
0
2
jfj1;X2 ds; 8t 2 ½0; T :
ð30Þ
Theorem 6. Under the assumptions of Theorems 2 and 4, problem (Qh) admits a unique solution ðuh ; kh ; ph Þ 2 L1 ð0; T ; W h Þ \ L2 ð0; T ; W h Þ L2 ð0; T ; Kh Þ L2 ð0; T ; M h Þ for all T > 0. Moreover, ðuh ; kh ; ph Þ satisfies the following estimates: Z T juh ðtÞj21;X1 þ ðjuh j21;X1 þ kkk21;C2 þ jph j20;X1 Þ dt 6 cðm; X1 ; u0 ; f; T Þ; ð31Þ 2
0
for all t 2 ½0; T .
4. Error estimates In this section, we mainly consider the error estimates of finite element solution ðuh ; kh ; ph Þ. Lemma 7. Under the assumptions of Theorems 2 and 4, Z
T
kh k21;C2 2
kk 0
2
dt 6 h cðm; X1 ; u0 ; f; T Þ þ c
Z
T 0
ju uh j21;X1 dt.
ð32Þ
Proof. From problem (Q) and problem (Qh), we have Z T bðt; sh k kh ; lÞ dt 0
1 ¼ 2 þ
Z
T
hc0 ðu uh Þ; li dt þ 0
Z 0
Z
T
hGðt; c0 ðu uh ÞÞ; li dt
0 T
bðt; sh k k; lÞdt; 8l 2 L2 ð0; T ; Kh Þ.
ð33Þ
1234
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
Due to the definition of the fundamental tension ðU k ðx; tÞ; P k ðx; tÞÞ, we can prove that for all l 2 L2 ð0; T ; Kh Þ, Z T Z T 1=2 Z T 1=2 2 2 < ðGðt;c ðu u ÞÞ;lÞdt 6 c kc ðuu Þk dt klk dt . 1 1 h h 0 0 ;C2 ;C2 0
2
0
2
0
ð34Þ Next, by taking lh ¼ sh k kh in Eq. (33) and using Eqs. (13),(14) and (34), we derive Z T Z T
2 2 2 ksh k kh k1;C2 dt 6 c kc0 ðu uh Þk1;C2 þ ksh k kk1;C2 dt. ð35Þ 2
0
2
0
2
Combining Eq. (35) with the triangle inequality gives Z T Z T
kk kh k21;C2 dt 6 c ku uh k21;C2 þ ksh k kk21 ; C2 dt.
ð36Þ
Using again the trace theorem [1,15] and Eq. (19), we derive Z T Z T
2 2 2 kk kh k1;C2 dt 6 c ju uh j1;X1 þ h2 kkk1;C2 dt.
ð37Þ
2
0
2
0
2
0
2
2
0
h
Combining Eq. (37) with regularity estimates Eq. (7) yields Eq. (32). Lemma 8. Under the assumptions of Theorems 2 and 4, Z T 2 kk kh ðc0 P h uÞk1 ; C2 dt 6 h2 cðm; X1 ; u0 ; f; T Þ; Z
ð38Þ
2
0
t
hc0 eh ; kh ðc0 P h uÞ kh i ds P 0; 8t 2 ½0; T ;
ð39Þ
0
where eh ¼ P h u uh . Proof. From the definition of k and kh ðc0 P h uÞ, we have Z T bðt; sh k kh ðc0 P h uÞ; lÞdt 0 Z T Z T 1 c0 ðu P h uÞ þ Gðt; c0 ðu P h uÞÞ; l dt; bðt; sh k k; lÞdt þ ¼ 2 0 0 2 8l 2 L ð0; T ; Kh Þ. ð40Þ A simply modified argument to Eq. (34) can yield Z
0
T
Z < ðGðt; c0 ðu P h uÞÞ; lÞdt 6 c
0
T
kc0 ðu P h uÞk21;C2 dt 2
1=2 Z 0
T
klk21;C2 dt 2
1=2 ;
ð41Þ
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
1235
for all l 2 L2 ð0; T ; Kh Þ. By taking l ¼ sh k kh ðc0 P h uÞ in Eq. (40) and using Eqs. (13),(14) and (41), we have Z T Z T Z T ksh k kh ðc0 P h uÞk21;C2 dt 6 c ksh k kk21;C2 dt þ c kc0 ðu P h uÞk21;C2 dt. 2
0
2
0
2
0
ð42Þ Combining Eq. (42) with the triangle inequality and the trace theorem [1,15] yields Z T Z T
2 kk kh ðc0 P h uÞk1;C2 dt 6 c ksh k kk21;C2 þ ju P h uj21;X1 dt. 2
0
2
0
ð43Þ Using again estimates Eqs. (19) and (23) in Eq. (43), we can derive Z
T
kk 0
2 kh ðc0 P h uÞk1;C2 2
dt 6 ch
2
Z
T
2 2 kkk1;C2 þ kuk2;X1 dt. 2
0
ð44Þ
Combining Eq. (44) with regularity estimates Eq. (7) yields Eq. (38). Moreover, by using the definition of kh ðc0 P h uÞ and kh, one finds that for all l 2 L2 ð0; T ; Kh Þ Z T Z T 1 c0 eh þ Gðt; c0 eh Þ; l dt. bðt; kh ðc0 P h uÞ kh ; lÞ dt ¼ 2 0 0 Applying Lemma 5 with u0 ¼ 0; f ¼ 0, to the above variational formulation, we know that kh ðc0 P h uÞ kh ¼ kh ðc0 eh Þ satisfies Eq. (39). h Lemma 9. Under the assumptions of Theorems 2 and 4, 2
juðtÞ uh ðtÞj0;X1 þ
Z
T 2
0
2
ðjuðtÞ uh ðtÞj1;X1 þ jpðtÞ ph ðtÞj0;X1 Þ dt 6 cðm; X1 ; u0 ; f; T Þh2 ; 8t 2 ½0; T . ð45Þ
Proof. From problems (Q) and (Qh), we obtain ðeht ; vÞ þ Bh ððeh ; gh Þ; ðv; qÞÞ þ a0 ðu P h u; vÞ dðv; p qh pÞ þ dðu P h u; qÞ þ bCh ðp qh p; qÞ þ hc0 v; k kh i þ a1 ðu uh ; u; vÞ þ a1 ðuh ; u uh ; vÞ ¼ 0; 8ðv; qÞ 2 ðW h ; M h Þ;
ð46Þ
where gh ¼ qh p ph . Then, by taking ðv; qÞ ¼ ðeh ; gh Þ in Eq. (46) and using Theorem 4, it follows that:
1236
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
1 d jeh j20;X1 þ cðjeh j21;X1 þ jgh j20;X1 þ Ch ðgh ; gh ÞÞ þ a0 ðu P h u; eh Þ 2 dt dðeh ; p qh pÞ þ dðu P h u; gh Þ þ bCh ðp qh p; gh Þ þ hc0 eh ; k kh i þ a1 ðeh ; u; eh Þ þ a1 ðuh ; eh ; eh Þ þ a1 ðu P h u; u; eh Þ þ a1 ðuh ; u P h u; eh Þ ¼ 0. ð47Þ Due to Eqs. (8)–(11), the Poincare inequality and the trace theorem [1,15], one finds c ja0 ðu P h u; eh Þj 6 jeh j21;X1 þ cju P h uj21;X1 16 c 2 2 jdðeh ; p qh pÞj 6 jeh j1;X1 þ cjp qh pj0;X1 16 c 2 2 jdðu P h u; gh Þj 6 jgh j0;X1 þ cju P h uj1;X1 8 c bjCh ðp qh p; gh Þj 6 Ch ðgh ; gh Þ þ cCh ðp qh p; p qh pÞ 8 hc0 eh ; k kh i ¼ hc0 eh ; kh ðc0 P h uÞ kh i þ hc0 eh ; k kh ðc0 P h uÞi c 2 2 jhc0 eh ; k kh ðc0 P h uÞij 6 jeh j1;X1 þ ckk kh ðc0 P h uÞk1;C2 2 16 c 2 2 2 ja1 ðeh ; u; eh Þj 6 cjuj1;X1 jeh j1;X1 jeh j0;X1 6 jeh j1;X1 þ cjuj1;X1 jeh j0;X1 ; 16 1=2
1=2
3=2
1=2
ja1 ðuh ; eh ; eh Þj 6 cjuh j1;X1 juh j0;X1 jeh j1;X1 jeh j0;X1 c 6 jeh j21;X1 þ cjuh j20;X1 juh j21;X1 jeh j20;X1 ; 16 ja1 ðu P h u; u; eh Þjþja1 ðuh ; u P h u; eh Þj 6 cjeh j1;X1 ju P h uj1;X1 ðjuj1;X1 þ juh j1;X1 Þ c 2 2 2 2 6 jeh j1;X1 þ cðjuj1;X1 þ juh j1;X1 Þju P h uj1;X1 . 16 Combining the above estimates with Eq. (47) and applying Theorems 4 and 6 yields d 2 2 2 jeh j0;X1 þ cðjeh j1;X1 þ jgh j0;X1 Þ þ 2hc0 eh ; kh ðc0 P h uÞ kh i dt 6 cðm; X1 ; u0 ; f; T Þðju P h uj21;X1 þ jp qh pj20;X1 þ h2 kpk21;X1 Þ þ ckk kh ðc0 P h uÞk21;C2 þ cðm; X1 ; u0 ; f; T Þjeh j20;X1 . 2
ð48Þ
Integrating Eq. (48) from 0 to t and applying Lemma 8 and noting eh ð0Þ ¼ 0, we derive
Y. He / Appl. Math. Comput. 172 (2006) 1225–1238
jeh ðtÞj20;X1
þc
Z
1237
t 0
ðjeh j21;X1 þ jgh j20;X1 Þ ds
6 cðm; X1 ; u0 ; f; T Þ
Z
t 2
0
2
2
ðju P h uj1;X1 þ jp qh pj0;X1 þ h2 kpk1;X1 Þ ds
2
þ h cðm; X1 ; u0 ; f; T Þ þ cðm; X1 ; u0 ; f; T Þ
Z
t 2
0
jeh j0;X1 ds.
ð49Þ
Applying the Gronwall lemma [3] to Eq. (49) and using Eqs. (20), (23) and Theorems 2 and 6, we derive Z t 2 jeh ðtÞj0;X1 þ ðjeh ðsÞj21;X1 þ jgh ðsÞj20;X1 Þ ds 6 cðm; X1 ; u0 ; f; T Þh2 . ð50Þ 0
Combining Eq. (50) with the triangle inequality, noting Z t 2 2 2 ðju P h uj0;X1 þ jp qh pj0;X1 Þ ds juðtÞ P h uðtÞj0;X1 þ 0
6
ch2 juðtÞj21;X1
2
þ ch
Z 0
t
ðkuk22;X1 þ kpk21;X1 Þ ds;
ð51Þ
and applying Theorem 2, one finds Z t 2 2 2 ðju uh j1;X1 þ jp ph j0;X1 Þ ds juðtÞ uh ðtÞj0;X1 þ 0
6 cðm; X1 ; u0 ; f; T Þh2 .
ð52Þ
Finally, by combining Lemma 7 with Lemma 9, we obtain the following main optimal error estimates. Theorem 10. Under the assumptions of Theorems 2 and 4, Z T juðtÞ uh ðtÞj20;X1 þ ðjuðtÞ uh ðtÞj21;X1 þ kkðtÞ kh ðtÞk21;C2 0 2 þ jpðtÞ ph ðtÞj0;X1 Þ dt 6 cðm; X1 ; u0 ; f; T Þh2 ;
2
ð53Þ
for all t 2 ½0; T .
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