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Multiplicative Schwarz methods for parabolic problems
Applied Mathematics and Computation 136 (2003) 593–610 www.elsevier.com/locate/amc
Multiplicative Schwarz methods for parabolic problems q Hongxing Rui School of Mathematics and System Science, Shandong University, Jinan 250100, China
Abstract Based on domain decomposition, we give a multiplicative Schwarz domain decomposition method for semi-linear parabolic problems. We consider the relation between the convergence rate and discretization parameters, including the diameter of the subdomain. We give the error estimate, which tells us that the convergence of the approximate solution is independent of the iteration number at each time level. Finally we give a numerical example. Ó 2002 Elsevier Science Inc. All rights reserved. Keywords: Multiplicative Schwarz method; Parabolic equation; Error estimate
1. Introduction Multiplicative Schwarz method, based on domain decomposition, is a powerful iteration method for solving elliptic equations and other stationary problems. A systematic theory has been developed for elliptic finite element problems in the past few years, see [2,5]. In [9,10], Lions presented a kind of Schwarz alternating algorithm in two sub-domain case for heat equations and gives a convergence result but does not give any error estimate. In [7,8] the authors presented a non-overlapping domain decomposition method for parabolic equations, but since they use explicit schemes at intersection points, the stability condition Dt 6 12H 2 is needed. In [3,4] Cai constructed a kind of
q
This work is supported by National Natural Science Foundation of China (No. 10071044). E-mail address: [email protected] (H. Rui).
0096-3003/02/$ - see front matter Ó 2002 Elsevier Science Inc. All rights reserved. PII: S 0 0 9 6 - 3 0 0 3 ( 0 2 ) 0 0 0 8 5 - 1
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H. Rui / Appl. Math. Comput. 136 (2003) 593–610
additive Schwarz algorithms and multiplicative Schwarz method and prove that the convergence rate is smaller than one for parabolic equations, there the author did not consider the dependence of the convergence and the discretization parameters. In this paper we are interesting in solving the parabolic problems using domain decomposition method. We use the Euler backward time discretization and Galerkin approximation in the space variables. At a fixed time level, the resulting equation is equivalent to an elliptic problem which depends on a time step increment and the approximate solution at the last time level. Therefore we apply the multiplicative Schwarz method, originally proposed for elliptic equations to the parabolic equations at each time level. The key point of this paper is to consider how the convergence and the error estimate depend on diameter of the sub-domain, the spacial mesh-size, the time step increment and the number of iterations at each time level. The outline of this paper is as follows. In Section 2 we present the time discrete parabolic equations and give a kind of multiplicative Schwarz algorithms with Euler backward time discretization. We also give the error estimate result, which tells us that the approximate solution converges after one cycle of iteration at each time level. In Section 3 we give some lemmas. In Section 4 we give the proof of the theorem mentioned in Section 2. Finally in Section 5 we give a numerical example. Throughout this paper, c or C, with or without subscripts, denotes a generic, strictly positive constant. Its value may be different at different occurrences, but is independent of the spacial mesh-size h and the time increment Dt, which will be introduced later.
2. Schwarz algorithms and convergence results Without loss of generality we consider the following model problem in a bounded polygonial domain X R2 : 2 ou X o ou aij ¼ f ðuÞ in X; ð1Þ ot i;j¼1 oxj oxi u¼0
ð2Þ
on oX; 0
uðx; 0Þ ¼ u ðxÞ
ð3Þ
in X;
where J ¼ ð0; T denote the time interval, aij ¼ aji and there exists a positive constant c such that 2 X i;j¼1
aij ni nj P cjnj
2
>
8n ¼ ðn1 ; n2 Þ 2 R2 :
ð4Þ
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
595
The standard variational formulation of the above problem is: Find uðtÞ 2 L2 ðJ ; H01 ðXÞÞ such that ou ; v þ aðu; vÞ ¼ ðf ðuÞ; vÞ; v 2 H01 ðXÞ; ot ð5Þ 0 ðuð0Þ; vÞ ¼ ðu ; vÞ; where aðu; vÞ ¼
Z X 2 X i;j¼1
ðf ðuÞ; vÞ ¼
Z
aij
ou ov dx; oxi oxj
ð6Þ
f ðuÞv dx:
X
Let Dt denote the time increment, tn ¼ nDt, un ¼ uðtn Þ, we have 2 un un1 ou ou þO ¼ Dt ; ot ot2 Dt Then Eq. (1) can be approximated by n u un1 ; v þ aðu; vÞ ¼ ðf ðun1 Þ; vÞ þ ðqn ; vÞ; Dt
ð7Þ
ð8Þ
where 2 un un1 oun ou n n1 þ f ðu Þ f ðu Þ ¼ O Dt : q ¼ ot2 Dt ot n
M
Let fX0i gi¼1 be a shape regular finite element triangulation, or a coarse mesh, M of X and fX0i gi¼1 has diameter of order H . To obtain an overlapping domain decomposition, we extend each subregion X0i to a larger region Xi such that X0i Xi X. We suppose that there exists a constant C such that distanceðoX0i \ X; Xi \ XÞ P CH for all i. Moreover we suppose that the decomposition satisfy the following condition (A). M
Definition 1. We say the decomposition fXi gi¼1 satisfy Condition (A) if the following conditions hold: 1. The subregion Xi ð1 6 i 6 MÞ can be divided into four parts: X Xi ; j ¼ 1; 2; 3; 4; r0 ¼ 0; r4 ¼ M: ð9Þ Dj ¼ rj1 þ1 6 i 6 rj
2. Sub-domains in Dj are disjoint. 3. There exists a constant C0 such that distðoD1 \ D12 ; oD2 \ D12 Þ P C0 H ;
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H. Rui / Appl. Math. Comput. 136 (2003) 593–610
distðoD3 \ D34 ; oD4 \ D34 Þ P C0 H ; distðoD12 \ X; D34 \ XÞ P C0 H ; where D12 ¼ D1 [ D2 , D34 ¼ D3 [ D4 . Now we give an example of domain decomposition satisfying Condition (A). We divide the following box into 16 subregions. Condition (A) holds for the subregion order listed in Figs. 1(a)–(d), where M ¼ 16, r1 ¼ 4, r2 ¼ 8, r3 ¼ 12, r4 ¼ 16. Extending the elements in H01 ðXj Þ to X by zero, and still let H01 ðXj Þ denote the set of the element. Let Th denote a quasi-uniform triangulation of X which is obtained by reM fining the above domain decomposition fXi gi¼1 , h is the mesh-size. Mh H01 ðXÞ denote the standard finite element space with piecewise polynomial of degree k, k P 1, such that inf ðku uk þ hku uk1 Þ 6 Ckukkþ1 hkþ1
u2Mh
8u 2 H01 ðXÞ \ H kþ1 ðXÞ: ð10Þ
Fig. 1.
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
597
Let Ti;h denote the restriction of Th on Xi , Mh ðXi Þ denote the restriction of Mh on Xi , Mh0 ðXi Þ ¼ Mh ðXi Þ \ H01 ðXi Þ. Set the initial approximation Wh0 2 Mh satisfying aðWh0 ; vÞ ¼ aðu0 ; vÞ
8v 2 Mh :
ð11Þ
We give the following multiplicative Schwarz algorithm. Scheme I. For n P 1 find W n 2 Mh by three steps: 1. Set W0n ¼ W n1 . n 2. Find WjMþi ðj ¼ 0; 1; . . . ; m 1; i ¼ 1; 2; . . . ; MÞ satisfying ! n WjMþi W n1 n ; v þ aðWjMþi ; vÞ ¼ ðf ðW n1 Þ; vÞ v 2 Mh0 ðXi Þ; Dt n n ¼ WjMþi1 WjMþi
x 2 X n Xi :
ð12Þ ð13Þ
n , x 2 X. 3. Let W n ¼ WmM Here m denotes the iteration number at each time level.
It is clear that the solution of the above scheme is defined uniquely. Remark 1. When Condition (A) holds the above method can be parallelized by coloring the sub-domains and solving simultaneously on disjoint sub-domains of the same color. Let Aðu; vÞ ¼ ðu; vÞ þ Dtaðu; vÞ kuka ¼ ðaðu; uÞÞ
1=2
kukA ¼ ðAðu; uÞÞ
8v 2 H01 ðXÞ;
ð14Þ ð15Þ
;
1=2
¼
2 ðkuk0
þ
2 1=2 kuka Þ :
ð16Þ
In Section 4 we can prove the following main result. Theorem 2.1. Suppose that the solution of (1) is sufficiently smooth and that Condition (A) holds for the above domain decomposition. Let W be the solution of Scheme I. When Dt and h are sufficiently small, if Dt ¼ oðH 2 Þ;
h ¼ oðH Þ;
ð17Þ
we have kun W n kA 6 C hkþ1 þ Dt þ
Dt h þ H2 H
m=2 ! ;
where C denotes a generic constant independent of Dt, h and H .
ð18Þ
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H. Rui / Appl. Math. Comput. 136 (2003) 593–610
Theorem 2.1 tells us that when using Scheme I the error caused by domain m=2 decomposing is ðDt=H 2 þ h=H Þ .
3. Some lemmas In this section we give some lemmas under Condition (A). Lemma 3.1. For u 2 H01 ðD12 Þ H01 ðXÞ, there exists a decomposition u ¼ P 2 1 l¼1 ul , ul 2 H0 ðDl Þ, such that 2 X l¼1
C1 Dt kul k2A 6 1 þ 2 kuk2 þ Dtkuk2a ; H
ð19Þ
where C1 denotes a constant independent of Dt, H and u. 2
Proof. From the condition we know that there exists a open covering fOl gl¼1 of X such that Ol \ X Xl . Using the decomposition theorem of unity P2 [1] we know that there exist smooth functions al ðl ¼ 1; 2Þ such that l¼1 al ¼ 1; 0 6 al 6 1; suppðal Þ Ol and for any positive integer r, we have kal kW r;1 ðXl Þ 6 CH r :
ð20Þ
Let ul ¼ al u. Then ul 2 H10 ðDl Þ, u ¼ fact, 2 X
Aðul ; ul Þ ¼
Z
2 X
a2l u2 dx
þ 2Dt
þ Dt
þ Dt
2 X 2 Z X
a2l 6
aij al u aij
D12
a2l
2 X
aij
ij¼1
ou ou dx oxi oxj
ou oal dx oxi oxj
oal oal 2 u dx: oxi oxj
2 X
D12 l¼1
ð21Þ
P2
l¼1
D12 l¼1
Z
2 X
al ¼ 1 we have that Z X 2 2 X a2l u2 dx 6 u2 dx;
l¼1
Z
Z
D12
2 X 2 Z X l¼1 ij¼1
ul . Next we prove (19) holds. In
D12 l¼1
l¼1 ij¼1
P2
l¼1
D12 l¼1
l¼1
Since
P2
ð22Þ
D12 l¼1
a2l
2 X
ou ou aij dx 6 ox i oxj ij¼1
Z
2 X D12 ij¼1
aij
ou ou dx ¼ aðu; uÞ: oxi oxj
ð23Þ
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
599
Integrating by parts we have equality Z Z ou oal 1 ou2 oa2l aij ual dx ¼ aij dx oxi oxj 4 Dl oxi oxj D12 Z Z 1 oa2l 1 o a2l 2 aij ni u dx aij ¼ u2 dx; ð24Þ 4 oDl 4 Dl oxi oxj oxj where ðn1 ; n2 Þ denote the unit vector on the outer normal direction. Since u 2 H01 ðD12 Þ and Suppðal Þ Ol , we have that Z oa2l aij ni u2 dx ¼ 0: oxj oDl Then there exists a constant C11 such that 2 Z 2 X X ou oal aij al u dx oxi oxj Dl ij¼1 l¼1 2 Z 2 X 1X o oa2l aij u2 dx ¼ 4 l¼1 Dl ij¼1 oxi oxj Z C11 6 2 u2 dx: H D12
ð25Þ
It is clear that exists a constant C12 such that Z 2 X 2 Z X oal oal 2 C12 aij u2 dx: u dx 6 2 oxi oxj H D12 D12 l¼1 ij¼1
ð26Þ
From (22), (23), (25), (26) and (21) we derive that 2 X DtðC11 þ C12 Þ 2 2 Aðul ; ul Þ 6 1 þ kuk þ Dtkuka : 2 H l¼1 Let C1 ¼ C11 þ C12 . Then we complete the proof.
ð27Þ
Let Mh ðDÞ be the restriction of Mh on D, Mh0 ðDÞ ¼ Mh ðDÞ \ H01 ðDÞ, D should be one of the sub-domains D1 ; D2 ; D3 ; D4 ; D12 and D34 . Lemma 3.2. For u 2 Mh0 ðD12 Þ, there exists a decomposition u ¼ ul 2 Mh0 ðDl Þ, such that 2 X Dt h 2 2 kul kA 6 1 þ C2 kukA ; þ 2 H H l¼1 where C2 denotes a constant independent of h, Dt, H and u.
P2
l¼1
ul ,
ð28Þ
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H. Rui / Appl. Math. Comput. 136 (2003) 593–610
Proof. Let e u l ¼ al u. From Lemma 3.1 we have that u ¼ and 2 X C1 Dt 2 2 2 ke u l kA 6 1 þ 2 kuk þ Dtkuka : H l¼1
P2
l¼1
e u l 2 H01 ðDl Þ, ul, e
ð29Þ
Let ul ¼ Ih e u l ¼ Ih ðal uÞ, where Ih : H01 ðD12 Þ ! Mh0 ðD12 Þ denotes the interpolation operator. Then u¼
2 X
ð30Þ
ul :
l¼1
Let s denote any of the triangles of the triangulation Th including in D12 . By interpolation theory we have that there exists a constant C independent of h; u such that 1=2 Z 2 ðul e 6 Chkþ1 je u l jH kþ1 ðsÞ ; ð31Þ u l Þ dx s
therefore Z
u2l dx
1=2
Z 6
s
s
¼
e u 2l dx
Z Z
e u 2l dx
s
6 s
1=2
e u 2l dx
1=2
þ Chkþ1 je u l jH kþ1 ðsÞ þ Chkþ1 kal ukH kþ1 ðsÞ
1=2 þC
k X
hkþ1 H ðkþ1rÞ kukH r ðsÞ ;
ð32Þ
r¼0
where in the last inequality we have used (20) and the fact that the function u is a polynomial of degree k on s. By inversive inequality [6] we have that for any u in Mh0 ðD12 Þ, kukH r ðsÞ 6 Chr kukL2 ðsÞ :
ð33Þ
Using (32), (33) and noticing hH 1 6 1, we have that " 1=2 Z 1=2 #2 Z XZ X Z h 2 2 2 2 e ul dx ¼ ul dx 6 u dx u l dx þC H D12 s s s s s Z XZ h2 X h e ¼ u2 dx þ 2C u 2l dx þ C 2 2 H H s s s s Z Z 1=2 1=2 X e u2 dx u 2l dx s
s
s
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
Z 6 D12
e u 2l dx
h2 þC 2 H
h H
l¼1
u2l dx 6
D12
D12
2 Z X l¼1
D12
2 Z X l¼1
Z D12
2
ðal uÞ dx
D12
1=2 Z
u2 dx
!1=2
XZ s
s
e u 2l dx þ C 2
D12
e u 2l dx
Z
h2 H2
u2 dx
D12
1=2 ð34Þ
;
D12
e u 2l dx þ 2C 2 2
h2 H2
ðal uÞ dx
Z
u2 dx þ 2C
D12
1=2 Z
D12
h H
1=2 u dx 2
D12
e u 2l dx
2 Z X l¼1
6
¼
u dx
2 Z X l¼1
6
2
Z
2 Z X l¼1
h u dx þ 2C H D12 2
s
s
þ 2C
Z
!1=2
XZ
2 Z X
2
601
h2 þ 2C 2 H 2
Z D12
!1=2 2
ðal uÞ dx
D12
e u 2l dx
u2 dx þ 2C 2 X
h H
jsignðal Þj
!1=2
Z
l¼1
2
u dx D12
Z h 2 h 3=2 þ 2C u2 dx: þ2 C H H D12
ð35Þ
Noticing the equivalence between kuka and kuk1 , we have that aðul ; ul Þ ¼ aðe ul; e u l Þ þ 2aðe u l ; ul e u l Þ þ aðul e u l ; ul e ulÞ 2 u l Þ þ Cje u l j1 jul e u l j1 þ Cjul e u l j1 : 6 aðe ul; e
Noticing k P 1 similarly to (32) and (34), we have that 1 kukL2 ðXl Þ þ jujH 1 ðXl Þ ; je u l j1 6 C H u l j1 6 C jul e
h juj 2 : H 2 L ðXl Þ
ð36Þ
ð37Þ ð38Þ
From (36)–(38) we have that aðul ; ul Þ 6 aðe ul; e ulÞ þ
CDt Ch 2 2 kuka : kukL2 ðXl Þ þ 2 H H
ð39Þ
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H. Rui / Appl. Math. Comput. 136 (2003) 593–610
From (35) and (39) we have that 2 X
2
kul kA ¼
l¼1
2 X
2
kul k þ Dt
2 X
l¼1 2 X
2
kul ka
l¼1
CDt Ch 2 2 Dtkuka kuk þ 2 H H l¼1 Dt h 2 6 1þC kukA : þ H2 H 6
2
ke u l kA þ
We complete the proof.
ð40Þ
Defining an operator Ri : Mh ! Mh0 ðXi Þ such that AðRi u u; vÞ ¼ 0
8v 2 Mh0 ðXi Þ; i ¼ 1; 2; . . . ; M:
ð41Þ
It is clear that Ri is bounded, kRi ukA 6 kukA
u 2 H01 ðXi Þ u 2 Mh0 ðXi Þ:
ð42Þ
Using the above lemmas we can prove the following lemma. Lemma 3.3. Suppose Condition (A) holds for the domain decomposition. Then there exists a constant C3 independent of u, h, Dt and H such that 1=2 Dt h kðI RM Þ ðI R2 ÞðI R1 ÞukA 6 C3 kukA 8u 2 Mh : þ H2 H ð43Þ e j : Mh ! M 0 ðDj Þ such that Proof. Define R h e j u u; vÞ ¼ 0 Að R
8v 2 Mh0 ðDj Þ; j ¼ 1; 2; 3; 4:
ð44Þ
Define R12 : Mh ! Mh0 ðD12 Þ and R34 : Mh ! Mh0 ðD34 Þ such that AðR12 u u; vÞ ¼ 0
8v 2 Mh0 ðD12 Þ;
ð45Þ
AðR34 u u; vÞ ¼ 0
Mh0 ðD34 Þ:
ð46Þ
8v 2
From Condition (A) we have that Mh0 ðDj Þ ¼ Mh0 ðXrj1 þ1 Þ Mh0 ðXrj1 þ2 Þ Mh0 ðXrj Þ;
ð47Þ
e j ¼ Rr þ1 þ Rr þ2 þ þ Rrj ; Ri Rj ¼ 0 if i 6¼ j, ij 2 frj1 þ 1; therefore R j1 j1 . . . ; rj g. Then we have that e j Þ; ðI Rrj Þ ðI Rrj1 þ1 Þ ¼ ðI ðRrj þ þ Rrj1 þ1 ÞÞ ¼ ðI R
ð48Þ
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
603
where I denote the unit operator. So for u 2 Mh we have that ðI RM ÞðI Rp1 Þ ðI R1 Þu e 3 ÞðI R e 2 ÞðI R e 1 Þu e 4 ÞðI R ¼ ðI R e 3 ÞðI R e 2 ÞðI R e 1 ÞR12 u þ ðI R e 4Þ e 4 ÞðI R ¼ ðI R e 3 ÞR34 ðI R e 2 ÞðI R e 1 ÞðI R e 12 Þu þ ðI R e 4 ÞðI R e 3Þ ðI R e 1 ÞðI R12 Þu: e 34 ÞðI R2 ÞðI R ðI R
ð49Þ
Using Lemma 3.2 we know that: For v 2 Mh0 ðD12 Þ there exists a decomposition v ¼ v1 þ v2 ; vj 2 Mh0 ðDj Þ; j ¼ 1; 2, such that 2 X Dt h Aðvi ; vi Þ 6 1 þ C2 Aðv; vÞ: ð50Þ þ H2 H i¼1 Using Lions’s result [9, p. 7], we have that e 2 ÞðI R e 1 Þvk 6 1 kðI R A 6 C31
1 þ C2
1
Dt
Dt h þ 2 2 H H
H2
!1=2 þ Hh2
1=2 kvkA
kvkA 8v 2 Mh0 ðD12 Þ:
ð51Þ
8v 2 Mh0 ðD34 Þ:
ð52Þ
Similarly e 4 ÞðI R e 3 Þvk 6 C32 kðI R A
Dt h þ H2 H2
1=2 kvkA
e1 ¼ R e 2, R e 1 we have R e 1 R12 ¼ R12 R e 1, R e 2 R12 ¼ Using the definition of R12 , R e2 ¼ R e 2 , therefore R12 R e 2 ÞðI R e 1 ÞðI R12 Þu ðI R34 ÞðI R e 2 ÞðI R e 1 Þu: ¼ ðI R34 ÞðI R12 ÞðI R
ð53Þ
Similarly to (51) we have that kðI R34 ÞðI R12 Þuk 6 C33
Dt h þ 2 2 H H
1=2 kukA
8v 2 Mh ;
ð54Þ
therefore e 2 ÞðI R e 1 ÞukA kðI R34 ÞðI R12 ÞðI R 6 C33
Dt1=2 e 2 ÞðI R e 1 ÞukA : kðI R H
ð55Þ
604
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
Using (49), (51), (52) and (55) and the bound of the operators, we have that kðI RM ÞðI RM1 Þ ðI R1 ÞukA 1=2 1=2 Dt h Dt h 6 CC31 kR12 ukA þ C32 þ þ H2 H2 H2 H2 1=2 e 2 ÞðI R e 1 ÞðI R12 ÞukA þ CC33 Dt þ h kR34 ðI R H2 H2 e 2 ÞðI R e 1 Þuk kðI R A 1=2 Dt h 6 C3 kukA ; þ H2 H2 where C denote a generic constant. The proof is completed.
ð56Þ
Lemma 3.3 tells us that the convergence rate for Scheme I is C3 ðDt= 1=2 H 2 þ h=H Þ , which becomes small as the discretization parameters become small when Dt ¼ oðH 2 Þ and h ¼ oðH Þ. 4. Convergence analyses and error estimates In this section we give the error estimates of Scheme I, that is the proof of Theorem 2.1. Define the auxiliary project uh 2 Mh such that aðuh u; vÞ ¼ 0 8v 2 Mh : Let g ¼ uh u. Then og n kþ1s kg ks þ ; ot 6 Ch s
ð57Þ
s ¼ 0; 1:
From (8) we derive that n uh un1 h ; v þ aðunh ; vÞ ¼ ðot gn þ qn ; vÞ þ ðf ðun1 Þ; vÞ 8v 2 Mh : Dt
ð58Þ
ð59Þ
Let en ¼ W n unh , eni ¼ Wi n unh . Then en ¼ enmp and combining (12) and (59) we have that for any v 2 Mh0 ðXi Þ, ! enjMþi en1 ; v þ aðenjMþi ; vÞ ¼ ðf 0 ðen1 þ gn1 Þ ot gn qn ; vÞ; ð60Þ Dt enjMþi enjMþi1 ¼ 0;
x 2 X Xi ;
ð61Þ
where f 0 denotes the first order derivative of f at a point between W n1 and un1 , f 0 ðen1 þ gn1 Þ ¼ f ðW n1 Þ f ðun1 Þ. This equation can be changed to
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
605
AðenjMþi ; vÞ ¼ ðenjMþi ; vÞ þ DtaðenjMþi ; vÞ ¼ ðen1 ; vÞ þ Dtðf 0 ðen1 þ gn1 Þ; vÞ Dtðot gn þ qn ; vÞ 8v 2 Mh0 ðXi Þ:
ð62Þ
Let En 2 Mh be the solution of the equation AðEn ; vÞ ¼ ðEn ; vÞ þ DtaðEn ; vÞ ¼ ðen1 ; vÞ þ Dtðf 0 ðen1 þ gn1 Þ; vÞ Dtðot gn þ qn ; vÞ ð63Þ
8v 2 Mh : From (62) and (63) we have that AðenjMþi En ðenjMþi1 En Þ; vÞ ¼ AðenjMþi1 En ; vÞ 8v 2 Mh0 ðXi Þ:
ð64Þ
Since enjMþi En ðenjMþi1 En Þ 2 Mh0 ðXi Þ, we have enjMþi En ðenjMþi1 En Þ ¼ Ri ðenjMþi1 En Þ; enjMþi En ¼ ðI Ri Þðenjpþi1 En Þ;
ð65Þ
i ¼ 1; 2; . . . ; M:
Therefore when Condition (A) holds we have en En ¼ ðenmM En Þ ¼ ððI RM Þ ðI R1 ÞÞm ðen0 En Þ;
ð66Þ
m
ken En kA 6 kððI Rh;M Þ ð1 Rh;1 ÞÞ ðen0 En ÞkA m=2 Dt h 6C kW n1 un1 þ uhn1 unh En kA ; þ h H2 H m=2 Dt h n n ken1 En þ un1 unh kA : þ ke kA 6 kE kA þ C h H2 H Noticing kunh
2 uhn1 kA
! ouh 2 ¼ O Dt ot dt ; tn1 A Z
tn
we have that ken kA 6 kEn kA þ C
Dt h þ H2 H
m=2
ken1 En kA þ Dt1=2
Z
tn
tn1
k
ouh 2 1=2 kA dt : ot ð67Þ
Now we estimate the right-hand side terms of (67). Let v ¼ En in (63). Then we have that
606
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
kEn k2A ¼ ðen1 ; En Þ þ Dt f 0 ðen1 þ gn1 Þ; En Dtðot gn þ qn ; En Þ
6 ken1 kkEn k þ CDtken1 kkEn k þ Dt kot gn k þ kqn k þ kgn1 k kEn k; ð68Þ therefore kEn kA 6 ð1 þ CDtÞken1 k þ CDtðkot gn k þ kqn k þ kgn1 kÞ:
ð69Þ
From (63) we have that ðEn en1 ; vÞ þ DtaðEn en1 ; vÞ ¼ Dtðf 0 ðen1 þ qn1 Þ; vÞ Dtðot gn þ qn ; vÞ Dtaðen1 ; En en1 Þ 8v 2 Mh :
ð70Þ
Let v ¼ En en1 . Using the same method as (69) we have that kEn en1 kA 6 Dt1=2 ken1 ka þ CDtðken1 k þ kot gn k þ kqn k þ kgn1 kÞ: ð71Þ Using (67), (69) and (71) we have that m=2 Dt h n n1 ke kA 6 ð1 þ CDtÞke k þ C Dt1=2 ken1 ka þ H2 H # " m=2 Dt h n n n1 þ CDt þ kot g k þ kq k þ kg k : þ H2 H
ð72Þ
By 2ab 6 a2 þ b2 we have that !2 m=2 Dt h 1=2 n1 Dt ke ka þ ke k þ C H2 H m=2 Dt h 2 ¼ ken1 k þ 2C Dt1=2 ken1 k ken1 ka þ H2 H m Dt h 2 þ C2 Dtken1 ka þ H2 H m Dt h 2 2 6 ken1 k þ Dtken1 ka þ C 2 ken1 k þ H2 H m m Dt h Dt h n1 2 2 þ C2 Dtke k 6 1 þ C þ þ ken1 k2A : a H2 H H2 H n1
ð73Þ
It is clear that there exists a constant C5 such that m m m 2 Dt h Dt h Dt h 1 þ C2 6 1 þ C ; þ 6 1 þ 2C þ þ 5 5 H2 H H2 H H2 H
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
607
therefore m 1=2 m Dt h Dt h 2 1þC 6 1 þ C5 ; þ þ H2 H H2 H so we have n1
ke
Dt h kþC þ H2 H
m=2 Dt
1=2
ke
n1
ka 6 1 þ C5
Dt h þ H2 H
m
ken1 kA :
Using (72) and (73) we have that m m=2 Dt h Dt h n n1 ke kA 6 1 þ C5 þ ke kA þ CDt þ H2 H H2 H ! þ kot gn k þ kqn k þ kgn1 k ;
ð74Þ
where the constant C is independent of Dt, h and H . Make summation for n we have " Z T n X n n Dtke ka 6 C ke0 k þ Dtke0 ka þ hkþ1 ke k þ 0
j¼1
dt1=2 þ Dt
Dt h þ H2 H
2 ! ou þ ot kþ1 H # 2 ouh 1=2 : ot dt A
2 kukH kþ1
m=2 Z
T
0
Since e0 ¼ 0, W n un ¼ W n unh þ unh un ¼ en þ gn : We have that n
n
n
n
kW u k 6 ke k þ kg k 6 C Dt þ h
kþ1
þ
Dt h þ 2 H H
m=2 ! :
ð75Þ
We complete the proof of Theorem 2.1. 5. Numerical example Consider the following parabolic equation: ou rðDruÞ ¼ f ot
in X J ;
ð76Þ
where X ¼ ð0; 1Þ ð0; 1Þ, J ¼ ð0; 1Þ, D ¼ dI, I is the unit matrix, the coefficient d > 0 is a constant. The right-hand side term, boundary condition and initial
608
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
Fig. 2. The sub-domains of X.
condition are selected in such a way that the exact solution is u ¼ exp ðx þ y tÞ. We divide X into four sub-domains: X1 ¼ ð0; 0:6Þ ð0; 0:6Þ, X2 ¼ ð0:4; 1Þ ð0; 0:6Þ, X3 ¼ ð0; 0:6Þ ð0:4; 1Þ, X4 ¼ ð0:4; 1Þ ð0:4; 1Þ, see Fig. 2. Based on this domain decomposition we give the triangulation. We first divide X into uniform squares with mesh-size h, here we let h ¼ 0:1; 0:05; 0:025, then we obtain the triangulation by dividing each square into two triangles. The time increment is Dt. The iterative number in each time step is m. Let kEkl2 ¼ max1 6 n 6 T =Dt kunh;ij unij kl2 , where kunh;ij unij kl2 be the approximation to the L2 norm of the error at the nth time level by using the 6-point Table 1 Dt
h
m
kEkl1
kEkl2
kEkl1
kEkl2
kEkl1
kEkl2
0.10 0.10 0.10 0.10 0.10
0.10 0.10 0.10 0.10 0.10
1 2 3 4 *
1.745E ) 1 4.977E ) 2 1.664E ) 2 7.349E ) 3 4.432E ) 3
7.423E ) 2 1.985E ) 2 8.630E ) 3 5.861E ) 3 4.892E ) 3
1.766E ) 1 4.552E ) 2 1.582E ) 2 1.017E ) 2 9.205E ) 3
7.663E ) 2 1.943E ) 2 9.614E ) 3 7.711E ) 3 7.220E ) 3
1.600E ) 1 5.234E ) 2 4.844E ) 2 4.829E ) 2 4.828E ) 2
7.719E ) 2 3.054E ) 2 2.809E ) 2 2.798E ) 2 2.797E ) 2
0.05 0.05 0.05 0.05 0.05
0.05 0.05 0.05 0.05 0.05
1 2 3 4 *
7.395E ) 2 1.769E ) 2 5.079E ) 3 2.538E ) 3 2.157E ) 3
3.151E ) 2 6.929E ) 2 2.715E ) 3 1.932E ) 3 1.746E ) 3
7.122E ) 2 1.391E ) 2 5.250E ) 3 4.622E ) 3 4.539E ) 3
3.107E ) 2 6.195E ) 3 3.360E ) 3 3.032E ) 3 2.987E ) 3
5.448E ) 2 2.510E ) 2 2.476E ) 2 2.476E ) 2 2.475E ) 2
2.798E ) 2 1.414E ) 2 1.395E ) 2 1.394E ) 2 1.394E ) 2
0.025 0.025 0.025 0.025 0.025
0.025 0.025 0.025 0.025 0.025
1 2 3 4
3.146E ) 2 5.480E ) 3 1.429E ) 3 1.076E ) 3 1.044E ) 3
1.338E ) 2 2.162E ) 3 8.648E ) 3 7.198E ) 4 7.009E ) 3
2.747E ) 2 3.835E ) 3 2.273E ) 3 2.225E ) 3 2.222E ) 3
1.203E ) 2 1.928E ) 3 1.365E ) 3 1.336E ) 3 1.335E ) 3
1.803E ) 2 1.257E ) 2 1.255E ) 2 1.255E ) 2 1.255E ) 2
9.863E ) 3 6.943E ) 3 6.933E ) 3 6.933E ) 3 6.933E ) 3
d ¼ 1:0
d ¼ 0:5
d ¼ 0:1
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
609
Table 2 Dt
h
m
l1
l2
l1
l2
l1
l2
0.10 0.10 0.10 0.10
0.10 0.10 0.10 0.10
1 2 3 4
1.701E ) 1 4.545E ) 2 1.232E ) 2 3.029E ) 3
6.934E ) 2 1.553E ) 2 3.378E ) 3 9.690E ) 4
1.674E ) 1 3.632E ) 2 6.615E ) 3 9.653E ) 4
6.941E ) 2 1.221E ) 2 1.920E ) 3 4.914E ) 4
1.117E ) 1 4.060E ) 2 1.613E ) 4 7.15E ) 6
4.922E ) 2 2.568E ) 3 1.168E ) 4 5.26E ) 6
0.05 0.05 0.05 0.05
0.05 0.05 0.05 0.05
1 2 3 4
7.179E ) 2 1.553E ) 2 2.922E ) 3 3.810E ) 4
2.976E ) 2 5.183E ) 2 9.690E ) 4 1.860E ) 4
6.668E ) 2 9.371E ) 3 7.100E ) 4 8.261E ) 5
2.808E ) 2 3.208E ) 3 3.716E ) 4 4.423E ) 5
2.973E ) 2 3.471E ) 4 3.62E ) 6 2.2E ) 7
1.404E ) 2 1.933E ) 4 1.99E ) 6 3.E ) 8
0.025 0.025 0.025 0.025
0.025 0.025 0.025 0.025
1 2 3 4
3.042E ) 2 4.436E ) 3 3.850E ) 4 3.195E ) 5
1.268E ) 2 1.462E ) 3 1.639E ) 4 1.874E ) 5
2.525E ) 2 1.613E ) 3 5.043E ) 5 2.385E ) 6
1.069E ) 2 5.931E ) 4 3.008E ) 5 1.513E ) 6
5.473E ) 3 1.144E ) 5
2.931E ) 3 7.39E ) 6
d ¼ 1:0
d ¼ 0:5
formulation on each triangle follows: kEkl1 ¼ max max 1 6 n 6 T =Dt
0
d ¼ 0:1
and define the discrete L1 norm error as n n uh;ij uij ;
where unij ; unh;ij denote the exact solution and approximate solution at point ðih; jh; nDtÞ, respectively. Setting d ¼ 1:, 0.5, 0.1, using different values of parameters Dt, h and M, the errors kEkl2 , kEkl1 are depicted in Table 1. Here ÔÕ denote the errors without using domain decomposition methods. The errors caused by domain decomposing are listed in Table 2. From the numerical results we can see that using the multiplicative Schwarz method we can get a good result for parabolic problem, even iterating only one or two cycle at each time level. From Table 2 we can see that the errors caused by decomposing domain become small as the discretization parameters h and Dt become small. Acknowledgement The author was supported by the National Natural Science Foundation of China. References [1] R.A. Adamas, Sobolev Space, Academic Press, New York, 1975. [2] J.H. Bramble, J.E. Pasciak, J. Wang, J. Xu, Convergence estimates for product iterative methods with application to domain decomposition, Math. Comp. 57 (195) (1991) 1–21.
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[3] X.C. Cai, Additive Schwarz algorithms for parabolic convection diffusion equations, Numer. Math. 60 (1991) 41–61. [4] X.C. Cai, Multiplicative Schwarz methods for parabolic problem, SIAM J. Sci. Comput. 15 (1994) 587–603. [5] X.C. Cai, Multiplicative Schwarz algorithms for nonsymmetric and indefinite problem, SIAM J. Numer. Anal. 30 (1993) 936–952. [6] P.G. Ciarlet, The Finite Element Method for Elliptic Problems, North-Holland, Amsterdam, 1978. [7] C.N. Dawson, T.F. Dupont, Explicit/implicit conservative Galerkin domain decomposition procedures for parabolic problems, Math. Comp. 58 (197) (1992) 21–34. [8] C.N. Dawson, Q. Du, T.F. Dupont, A finite difference domain decomposition algorithm for numerical solution of heat equation, Math. Comp. 57 (195) (1991) 63–71. [9] P.L. Lions, On the Scharz alternating method I, in: R. Glowinski, G.H. Golub, G.A. Meurant, J. Periaux (Eds.), First International Symposium on Domain Decomposition Methods for Partial Differential Equations, SIAM, Philadelphia, PA, 1988. [10] P.L. Lions, On the Schwarz alternating method II, in: T. Chan, R. Glowinski, J. Periaux, O.B. Widlund (Eds.), Second International Symposium on Domain Decomposition Methods for Partial Differential Equations, SIAM, Philadelphia, PA, 1989.
Multiplicative Schwarz methods for parabolic problems q Hongxing Rui School of Mathematics and System Science, Shandong University, Jinan 250100, China
Abstract Based on domain decomposition, we give a multiplicative Schwarz domain decomposition method for semi-linear parabolic problems. We consider the relation between the convergence rate and discretization parameters, including the diameter of the subdomain. We give the error estimate, which tells us that the convergence of the approximate solution is independent of the iteration number at each time level. Finally we give a numerical example. Ó 2002 Elsevier Science Inc. All rights reserved. Keywords: Multiplicative Schwarz method; Parabolic equation; Error estimate
1. Introduction Multiplicative Schwarz method, based on domain decomposition, is a powerful iteration method for solving elliptic equations and other stationary problems. A systematic theory has been developed for elliptic finite element problems in the past few years, see [2,5]. In [9,10], Lions presented a kind of Schwarz alternating algorithm in two sub-domain case for heat equations and gives a convergence result but does not give any error estimate. In [7,8] the authors presented a non-overlapping domain decomposition method for parabolic equations, but since they use explicit schemes at intersection points, the stability condition Dt 6 12H 2 is needed. In [3,4] Cai constructed a kind of
q
This work is supported by National Natural Science Foundation of China (No. 10071044). E-mail address: [email protected] (H. Rui).
0096-3003/02/$ - see front matter Ó 2002 Elsevier Science Inc. All rights reserved. PII: S 0 0 9 6 - 3 0 0 3 ( 0 2 ) 0 0 0 8 5 - 1
594
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
additive Schwarz algorithms and multiplicative Schwarz method and prove that the convergence rate is smaller than one for parabolic equations, there the author did not consider the dependence of the convergence and the discretization parameters. In this paper we are interesting in solving the parabolic problems using domain decomposition method. We use the Euler backward time discretization and Galerkin approximation in the space variables. At a fixed time level, the resulting equation is equivalent to an elliptic problem which depends on a time step increment and the approximate solution at the last time level. Therefore we apply the multiplicative Schwarz method, originally proposed for elliptic equations to the parabolic equations at each time level. The key point of this paper is to consider how the convergence and the error estimate depend on diameter of the sub-domain, the spacial mesh-size, the time step increment and the number of iterations at each time level. The outline of this paper is as follows. In Section 2 we present the time discrete parabolic equations and give a kind of multiplicative Schwarz algorithms with Euler backward time discretization. We also give the error estimate result, which tells us that the approximate solution converges after one cycle of iteration at each time level. In Section 3 we give some lemmas. In Section 4 we give the proof of the theorem mentioned in Section 2. Finally in Section 5 we give a numerical example. Throughout this paper, c or C, with or without subscripts, denotes a generic, strictly positive constant. Its value may be different at different occurrences, but is independent of the spacial mesh-size h and the time increment Dt, which will be introduced later.
2. Schwarz algorithms and convergence results Without loss of generality we consider the following model problem in a bounded polygonial domain X R2 : 2 ou X o ou aij ¼ f ðuÞ in X; ð1Þ ot i;j¼1 oxj oxi u¼0
ð2Þ
on oX; 0
uðx; 0Þ ¼ u ðxÞ
ð3Þ
in X;
where J ¼ ð0; T denote the time interval, aij ¼ aji and there exists a positive constant c such that 2 X i;j¼1
aij ni nj P cjnj
2
>
8n ¼ ðn1 ; n2 Þ 2 R2 :
ð4Þ
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
595
The standard variational formulation of the above problem is: Find uðtÞ 2 L2 ðJ ; H01 ðXÞÞ such that ou ; v þ aðu; vÞ ¼ ðf ðuÞ; vÞ; v 2 H01 ðXÞ; ot ð5Þ 0 ðuð0Þ; vÞ ¼ ðu ; vÞ; where aðu; vÞ ¼
Z X 2 X i;j¼1
ðf ðuÞ; vÞ ¼
Z
aij
ou ov dx; oxi oxj
ð6Þ
f ðuÞv dx:
X
Let Dt denote the time increment, tn ¼ nDt, un ¼ uðtn Þ, we have 2 un un1 ou ou þO ¼ Dt ; ot ot2 Dt Then Eq. (1) can be approximated by n u un1 ; v þ aðu; vÞ ¼ ðf ðun1 Þ; vÞ þ ðqn ; vÞ; Dt
ð7Þ
ð8Þ
where 2 un un1 oun ou n n1 þ f ðu Þ f ðu Þ ¼ O Dt : q ¼ ot2 Dt ot n
M
Let fX0i gi¼1 be a shape regular finite element triangulation, or a coarse mesh, M of X and fX0i gi¼1 has diameter of order H . To obtain an overlapping domain decomposition, we extend each subregion X0i to a larger region Xi such that X0i Xi X. We suppose that there exists a constant C such that distanceðoX0i \ X; Xi \ XÞ P CH for all i. Moreover we suppose that the decomposition satisfy the following condition (A). M
Definition 1. We say the decomposition fXi gi¼1 satisfy Condition (A) if the following conditions hold: 1. The subregion Xi ð1 6 i 6 MÞ can be divided into four parts: X Xi ; j ¼ 1; 2; 3; 4; r0 ¼ 0; r4 ¼ M: ð9Þ Dj ¼ rj1 þ1 6 i 6 rj
2. Sub-domains in Dj are disjoint. 3. There exists a constant C0 such that distðoD1 \ D12 ; oD2 \ D12 Þ P C0 H ;
596
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
distðoD3 \ D34 ; oD4 \ D34 Þ P C0 H ; distðoD12 \ X; D34 \ XÞ P C0 H ; where D12 ¼ D1 [ D2 , D34 ¼ D3 [ D4 . Now we give an example of domain decomposition satisfying Condition (A). We divide the following box into 16 subregions. Condition (A) holds for the subregion order listed in Figs. 1(a)–(d), where M ¼ 16, r1 ¼ 4, r2 ¼ 8, r3 ¼ 12, r4 ¼ 16. Extending the elements in H01 ðXj Þ to X by zero, and still let H01 ðXj Þ denote the set of the element. Let Th denote a quasi-uniform triangulation of X which is obtained by reM fining the above domain decomposition fXi gi¼1 , h is the mesh-size. Mh H01 ðXÞ denote the standard finite element space with piecewise polynomial of degree k, k P 1, such that inf ðku uk þ hku uk1 Þ 6 Ckukkþ1 hkþ1
u2Mh
8u 2 H01 ðXÞ \ H kþ1 ðXÞ: ð10Þ
Fig. 1.
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
597
Let Ti;h denote the restriction of Th on Xi , Mh ðXi Þ denote the restriction of Mh on Xi , Mh0 ðXi Þ ¼ Mh ðXi Þ \ H01 ðXi Þ. Set the initial approximation Wh0 2 Mh satisfying aðWh0 ; vÞ ¼ aðu0 ; vÞ
8v 2 Mh :
ð11Þ
We give the following multiplicative Schwarz algorithm. Scheme I. For n P 1 find W n 2 Mh by three steps: 1. Set W0n ¼ W n1 . n 2. Find WjMþi ðj ¼ 0; 1; . . . ; m 1; i ¼ 1; 2; . . . ; MÞ satisfying ! n WjMþi W n1 n ; v þ aðWjMþi ; vÞ ¼ ðf ðW n1 Þ; vÞ v 2 Mh0 ðXi Þ; Dt n n ¼ WjMþi1 WjMþi
x 2 X n Xi :
ð12Þ ð13Þ
n , x 2 X. 3. Let W n ¼ WmM Here m denotes the iteration number at each time level.
It is clear that the solution of the above scheme is defined uniquely. Remark 1. When Condition (A) holds the above method can be parallelized by coloring the sub-domains and solving simultaneously on disjoint sub-domains of the same color. Let Aðu; vÞ ¼ ðu; vÞ þ Dtaðu; vÞ kuka ¼ ðaðu; uÞÞ
1=2
kukA ¼ ðAðu; uÞÞ
8v 2 H01 ðXÞ;
ð14Þ ð15Þ
;
1=2
¼
2 ðkuk0
þ
2 1=2 kuka Þ :
ð16Þ
In Section 4 we can prove the following main result. Theorem 2.1. Suppose that the solution of (1) is sufficiently smooth and that Condition (A) holds for the above domain decomposition. Let W be the solution of Scheme I. When Dt and h are sufficiently small, if Dt ¼ oðH 2 Þ;
h ¼ oðH Þ;
ð17Þ
we have kun W n kA 6 C hkþ1 þ Dt þ
Dt h þ H2 H
m=2 ! ;
where C denotes a generic constant independent of Dt, h and H .
ð18Þ
598
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
Theorem 2.1 tells us that when using Scheme I the error caused by domain m=2 decomposing is ðDt=H 2 þ h=H Þ .
3. Some lemmas In this section we give some lemmas under Condition (A). Lemma 3.1. For u 2 H01 ðD12 Þ H01 ðXÞ, there exists a decomposition u ¼ P 2 1 l¼1 ul , ul 2 H0 ðDl Þ, such that 2 X l¼1
C1 Dt kul k2A 6 1 þ 2 kuk2 þ Dtkuk2a ; H
ð19Þ
where C1 denotes a constant independent of Dt, H and u. 2
Proof. From the condition we know that there exists a open covering fOl gl¼1 of X such that Ol \ X Xl . Using the decomposition theorem of unity P2 [1] we know that there exist smooth functions al ðl ¼ 1; 2Þ such that l¼1 al ¼ 1; 0 6 al 6 1; suppðal Þ Ol and for any positive integer r, we have kal kW r;1 ðXl Þ 6 CH r :
ð20Þ
Let ul ¼ al u. Then ul 2 H10 ðDl Þ, u ¼ fact, 2 X
Aðul ; ul Þ ¼
Z
2 X
a2l u2 dx
þ 2Dt
þ Dt
þ Dt
2 X 2 Z X
a2l 6
aij al u aij
D12
a2l
2 X
aij
ij¼1
ou ou dx oxi oxj
ou oal dx oxi oxj
oal oal 2 u dx: oxi oxj
2 X
D12 l¼1
ð21Þ
P2
l¼1
D12 l¼1
Z
2 X
al ¼ 1 we have that Z X 2 2 X a2l u2 dx 6 u2 dx;
l¼1
Z
Z
D12
2 X 2 Z X l¼1 ij¼1
ul . Next we prove (19) holds. In
D12 l¼1
l¼1 ij¼1
P2
l¼1
D12 l¼1
l¼1
Since
P2
ð22Þ
D12 l¼1
a2l
2 X
ou ou aij dx 6 ox i oxj ij¼1
Z
2 X D12 ij¼1
aij
ou ou dx ¼ aðu; uÞ: oxi oxj
ð23Þ
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
599
Integrating by parts we have equality Z Z ou oal 1 ou2 oa2l aij ual dx ¼ aij dx oxi oxj 4 Dl oxi oxj D12 Z Z 1 oa2l 1 o a2l 2 aij ni u dx aij ¼ u2 dx; ð24Þ 4 oDl 4 Dl oxi oxj oxj where ðn1 ; n2 Þ denote the unit vector on the outer normal direction. Since u 2 H01 ðD12 Þ and Suppðal Þ Ol , we have that Z oa2l aij ni u2 dx ¼ 0: oxj oDl Then there exists a constant C11 such that 2 Z 2 X X ou oal aij al u dx oxi oxj Dl ij¼1 l¼1 2 Z 2 X 1X o oa2l aij u2 dx ¼ 4 l¼1 Dl ij¼1 oxi oxj Z C11 6 2 u2 dx: H D12
ð25Þ
It is clear that exists a constant C12 such that Z 2 X 2 Z X oal oal 2 C12 aij u2 dx: u dx 6 2 oxi oxj H D12 D12 l¼1 ij¼1
ð26Þ
From (22), (23), (25), (26) and (21) we derive that 2 X DtðC11 þ C12 Þ 2 2 Aðul ; ul Þ 6 1 þ kuk þ Dtkuka : 2 H l¼1 Let C1 ¼ C11 þ C12 . Then we complete the proof.
ð27Þ
Let Mh ðDÞ be the restriction of Mh on D, Mh0 ðDÞ ¼ Mh ðDÞ \ H01 ðDÞ, D should be one of the sub-domains D1 ; D2 ; D3 ; D4 ; D12 and D34 . Lemma 3.2. For u 2 Mh0 ðD12 Þ, there exists a decomposition u ¼ ul 2 Mh0 ðDl Þ, such that 2 X Dt h 2 2 kul kA 6 1 þ C2 kukA ; þ 2 H H l¼1 where C2 denotes a constant independent of h, Dt, H and u.
P2
l¼1
ul ,
ð28Þ
600
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
Proof. Let e u l ¼ al u. From Lemma 3.1 we have that u ¼ and 2 X C1 Dt 2 2 2 ke u l kA 6 1 þ 2 kuk þ Dtkuka : H l¼1
P2
l¼1
e u l 2 H01 ðDl Þ, ul, e
ð29Þ
Let ul ¼ Ih e u l ¼ Ih ðal uÞ, where Ih : H01 ðD12 Þ ! Mh0 ðD12 Þ denotes the interpolation operator. Then u¼
2 X
ð30Þ
ul :
l¼1
Let s denote any of the triangles of the triangulation Th including in D12 . By interpolation theory we have that there exists a constant C independent of h; u such that 1=2 Z 2 ðul e 6 Chkþ1 je u l jH kþ1 ðsÞ ; ð31Þ u l Þ dx s
therefore Z
u2l dx
1=2
Z 6
s
s
¼
e u 2l dx
Z Z
e u 2l dx
s
6 s
1=2
e u 2l dx
1=2
þ Chkþ1 je u l jH kþ1 ðsÞ þ Chkþ1 kal ukH kþ1 ðsÞ
1=2 þC
k X
hkþ1 H ðkþ1rÞ kukH r ðsÞ ;
ð32Þ
r¼0
where in the last inequality we have used (20) and the fact that the function u is a polynomial of degree k on s. By inversive inequality [6] we have that for any u in Mh0 ðD12 Þ, kukH r ðsÞ 6 Chr kukL2 ðsÞ :
ð33Þ
Using (32), (33) and noticing hH 1 6 1, we have that " 1=2 Z 1=2 #2 Z XZ X Z h 2 2 2 2 e ul dx ¼ ul dx 6 u dx u l dx þC H D12 s s s s s Z XZ h2 X h e ¼ u2 dx þ 2C u 2l dx þ C 2 2 H H s s s s Z Z 1=2 1=2 X e u2 dx u 2l dx s
s
s
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
Z 6 D12
e u 2l dx
h2 þC 2 H
h H
l¼1
u2l dx 6
D12
D12
2 Z X l¼1
D12
2 Z X l¼1
Z D12
2
ðal uÞ dx
D12
1=2 Z
u2 dx
!1=2
XZ s
s
e u 2l dx þ C 2
D12
e u 2l dx
Z
h2 H2
u2 dx
D12
1=2 ð34Þ
;
D12
e u 2l dx þ 2C 2 2
h2 H2
ðal uÞ dx
Z
u2 dx þ 2C
D12
1=2 Z
D12
h H
1=2 u dx 2
D12
e u 2l dx
2 Z X l¼1
6
¼
u dx
2 Z X l¼1
6
2
Z
2 Z X l¼1
h u dx þ 2C H D12 2
s
s
þ 2C
Z
!1=2
XZ
2 Z X
2
601
h2 þ 2C 2 H 2
Z D12
!1=2 2
ðal uÞ dx
D12
e u 2l dx
u2 dx þ 2C 2 X
h H
jsignðal Þj
!1=2
Z
l¼1
2
u dx D12
Z h 2 h 3=2 þ 2C u2 dx: þ2 C H H D12
ð35Þ
Noticing the equivalence between kuka and kuk1 , we have that aðul ; ul Þ ¼ aðe ul; e u l Þ þ 2aðe u l ; ul e u l Þ þ aðul e u l ; ul e ulÞ 2 u l Þ þ Cje u l j1 jul e u l j1 þ Cjul e u l j1 : 6 aðe ul; e
Noticing k P 1 similarly to (32) and (34), we have that 1 kukL2 ðXl Þ þ jujH 1 ðXl Þ ; je u l j1 6 C H u l j1 6 C jul e
h juj 2 : H 2 L ðXl Þ
ð36Þ
ð37Þ ð38Þ
From (36)–(38) we have that aðul ; ul Þ 6 aðe ul; e ulÞ þ
CDt Ch 2 2 kuka : kukL2 ðXl Þ þ 2 H H
ð39Þ
602
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
From (35) and (39) we have that 2 X
2
kul kA ¼
l¼1
2 X
2
kul k þ Dt
2 X
l¼1 2 X
2
kul ka
l¼1
CDt Ch 2 2 Dtkuka kuk þ 2 H H l¼1 Dt h 2 6 1þC kukA : þ H2 H 6
2
ke u l kA þ
We complete the proof.
ð40Þ
Defining an operator Ri : Mh ! Mh0 ðXi Þ such that AðRi u u; vÞ ¼ 0
8v 2 Mh0 ðXi Þ; i ¼ 1; 2; . . . ; M:
ð41Þ
It is clear that Ri is bounded, kRi ukA 6 kukA
u 2 H01 ðXi Þ u 2 Mh0 ðXi Þ:
ð42Þ
Using the above lemmas we can prove the following lemma. Lemma 3.3. Suppose Condition (A) holds for the domain decomposition. Then there exists a constant C3 independent of u, h, Dt and H such that 1=2 Dt h kðI RM Þ ðI R2 ÞðI R1 ÞukA 6 C3 kukA 8u 2 Mh : þ H2 H ð43Þ e j : Mh ! M 0 ðDj Þ such that Proof. Define R h e j u u; vÞ ¼ 0 Að R
8v 2 Mh0 ðDj Þ; j ¼ 1; 2; 3; 4:
ð44Þ
Define R12 : Mh ! Mh0 ðD12 Þ and R34 : Mh ! Mh0 ðD34 Þ such that AðR12 u u; vÞ ¼ 0
8v 2 Mh0 ðD12 Þ;
ð45Þ
AðR34 u u; vÞ ¼ 0
Mh0 ðD34 Þ:
ð46Þ
8v 2
From Condition (A) we have that Mh0 ðDj Þ ¼ Mh0 ðXrj1 þ1 Þ Mh0 ðXrj1 þ2 Þ Mh0 ðXrj Þ;
ð47Þ
e j ¼ Rr þ1 þ Rr þ2 þ þ Rrj ; Ri Rj ¼ 0 if i 6¼ j, ij 2 frj1 þ 1; therefore R j1 j1 . . . ; rj g. Then we have that e j Þ; ðI Rrj Þ ðI Rrj1 þ1 Þ ¼ ðI ðRrj þ þ Rrj1 þ1 ÞÞ ¼ ðI R
ð48Þ
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
603
where I denote the unit operator. So for u 2 Mh we have that ðI RM ÞðI Rp1 Þ ðI R1 Þu e 3 ÞðI R e 2 ÞðI R e 1 Þu e 4 ÞðI R ¼ ðI R e 3 ÞðI R e 2 ÞðI R e 1 ÞR12 u þ ðI R e 4Þ e 4 ÞðI R ¼ ðI R e 3 ÞR34 ðI R e 2 ÞðI R e 1 ÞðI R e 12 Þu þ ðI R e 4 ÞðI R e 3Þ ðI R e 1 ÞðI R12 Þu: e 34 ÞðI R2 ÞðI R ðI R
ð49Þ
Using Lemma 3.2 we know that: For v 2 Mh0 ðD12 Þ there exists a decomposition v ¼ v1 þ v2 ; vj 2 Mh0 ðDj Þ; j ¼ 1; 2, such that 2 X Dt h Aðvi ; vi Þ 6 1 þ C2 Aðv; vÞ: ð50Þ þ H2 H i¼1 Using Lions’s result [9, p. 7], we have that e 2 ÞðI R e 1 Þvk 6 1 kðI R A 6 C31
1 þ C2
1
Dt
Dt h þ 2 2 H H
H2
!1=2 þ Hh2
1=2 kvkA
kvkA 8v 2 Mh0 ðD12 Þ:
ð51Þ
8v 2 Mh0 ðD34 Þ:
ð52Þ
Similarly e 4 ÞðI R e 3 Þvk 6 C32 kðI R A
Dt h þ H2 H2
1=2 kvkA
e1 ¼ R e 2, R e 1 we have R e 1 R12 ¼ R12 R e 1, R e 2 R12 ¼ Using the definition of R12 , R e2 ¼ R e 2 , therefore R12 R e 2 ÞðI R e 1 ÞðI R12 Þu ðI R34 ÞðI R e 2 ÞðI R e 1 Þu: ¼ ðI R34 ÞðI R12 ÞðI R
ð53Þ
Similarly to (51) we have that kðI R34 ÞðI R12 Þuk 6 C33
Dt h þ 2 2 H H
1=2 kukA
8v 2 Mh ;
ð54Þ
therefore e 2 ÞðI R e 1 ÞukA kðI R34 ÞðI R12 ÞðI R 6 C33
Dt1=2 e 2 ÞðI R e 1 ÞukA : kðI R H
ð55Þ
604
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
Using (49), (51), (52) and (55) and the bound of the operators, we have that kðI RM ÞðI RM1 Þ ðI R1 ÞukA 1=2 1=2 Dt h Dt h 6 CC31 kR12 ukA þ C32 þ þ H2 H2 H2 H2 1=2 e 2 ÞðI R e 1 ÞðI R12 ÞukA þ CC33 Dt þ h kR34 ðI R H2 H2 e 2 ÞðI R e 1 Þuk kðI R A 1=2 Dt h 6 C3 kukA ; þ H2 H2 where C denote a generic constant. The proof is completed.
ð56Þ
Lemma 3.3 tells us that the convergence rate for Scheme I is C3 ðDt= 1=2 H 2 þ h=H Þ , which becomes small as the discretization parameters become small when Dt ¼ oðH 2 Þ and h ¼ oðH Þ. 4. Convergence analyses and error estimates In this section we give the error estimates of Scheme I, that is the proof of Theorem 2.1. Define the auxiliary project uh 2 Mh such that aðuh u; vÞ ¼ 0 8v 2 Mh : Let g ¼ uh u. Then og n kþ1s kg ks þ ; ot 6 Ch s
ð57Þ
s ¼ 0; 1:
From (8) we derive that n uh un1 h ; v þ aðunh ; vÞ ¼ ðot gn þ qn ; vÞ þ ðf ðun1 Þ; vÞ 8v 2 Mh : Dt
ð58Þ
ð59Þ
Let en ¼ W n unh , eni ¼ Wi n unh . Then en ¼ enmp and combining (12) and (59) we have that for any v 2 Mh0 ðXi Þ, ! enjMþi en1 ; v þ aðenjMþi ; vÞ ¼ ðf 0 ðen1 þ gn1 Þ ot gn qn ; vÞ; ð60Þ Dt enjMþi enjMþi1 ¼ 0;
x 2 X Xi ;
ð61Þ
where f 0 denotes the first order derivative of f at a point between W n1 and un1 , f 0 ðen1 þ gn1 Þ ¼ f ðW n1 Þ f ðun1 Þ. This equation can be changed to
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
605
AðenjMþi ; vÞ ¼ ðenjMþi ; vÞ þ DtaðenjMþi ; vÞ ¼ ðen1 ; vÞ þ Dtðf 0 ðen1 þ gn1 Þ; vÞ Dtðot gn þ qn ; vÞ 8v 2 Mh0 ðXi Þ:
ð62Þ
Let En 2 Mh be the solution of the equation AðEn ; vÞ ¼ ðEn ; vÞ þ DtaðEn ; vÞ ¼ ðen1 ; vÞ þ Dtðf 0 ðen1 þ gn1 Þ; vÞ Dtðot gn þ qn ; vÞ ð63Þ
8v 2 Mh : From (62) and (63) we have that AðenjMþi En ðenjMþi1 En Þ; vÞ ¼ AðenjMþi1 En ; vÞ 8v 2 Mh0 ðXi Þ:
ð64Þ
Since enjMþi En ðenjMþi1 En Þ 2 Mh0 ðXi Þ, we have enjMþi En ðenjMþi1 En Þ ¼ Ri ðenjMþi1 En Þ; enjMþi En ¼ ðI Ri Þðenjpþi1 En Þ;
ð65Þ
i ¼ 1; 2; . . . ; M:
Therefore when Condition (A) holds we have en En ¼ ðenmM En Þ ¼ ððI RM Þ ðI R1 ÞÞm ðen0 En Þ;
ð66Þ
m
ken En kA 6 kððI Rh;M Þ ð1 Rh;1 ÞÞ ðen0 En ÞkA m=2 Dt h 6C kW n1 un1 þ uhn1 unh En kA ; þ h H2 H m=2 Dt h n n ken1 En þ un1 unh kA : þ ke kA 6 kE kA þ C h H2 H Noticing kunh
2 uhn1 kA
! ouh 2 ¼ O Dt ot dt ; tn1 A Z
tn
we have that ken kA 6 kEn kA þ C
Dt h þ H2 H
m=2
ken1 En kA þ Dt1=2
Z
tn
tn1
k
ouh 2 1=2 kA dt : ot ð67Þ
Now we estimate the right-hand side terms of (67). Let v ¼ En in (63). Then we have that
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H. Rui / Appl. Math. Comput. 136 (2003) 593–610
kEn k2A ¼ ðen1 ; En Þ þ Dt f 0 ðen1 þ gn1 Þ; En Dtðot gn þ qn ; En Þ
6 ken1 kkEn k þ CDtken1 kkEn k þ Dt kot gn k þ kqn k þ kgn1 k kEn k; ð68Þ therefore kEn kA 6 ð1 þ CDtÞken1 k þ CDtðkot gn k þ kqn k þ kgn1 kÞ:
ð69Þ
From (63) we have that ðEn en1 ; vÞ þ DtaðEn en1 ; vÞ ¼ Dtðf 0 ðen1 þ qn1 Þ; vÞ Dtðot gn þ qn ; vÞ Dtaðen1 ; En en1 Þ 8v 2 Mh :
ð70Þ
Let v ¼ En en1 . Using the same method as (69) we have that kEn en1 kA 6 Dt1=2 ken1 ka þ CDtðken1 k þ kot gn k þ kqn k þ kgn1 kÞ: ð71Þ Using (67), (69) and (71) we have that m=2 Dt h n n1 ke kA 6 ð1 þ CDtÞke k þ C Dt1=2 ken1 ka þ H2 H # " m=2 Dt h n n n1 þ CDt þ kot g k þ kq k þ kg k : þ H2 H
ð72Þ
By 2ab 6 a2 þ b2 we have that !2 m=2 Dt h 1=2 n1 Dt ke ka þ ke k þ C H2 H m=2 Dt h 2 ¼ ken1 k þ 2C Dt1=2 ken1 k ken1 ka þ H2 H m Dt h 2 þ C2 Dtken1 ka þ H2 H m Dt h 2 2 6 ken1 k þ Dtken1 ka þ C 2 ken1 k þ H2 H m m Dt h Dt h n1 2 2 þ C2 Dtke k 6 1 þ C þ þ ken1 k2A : a H2 H H2 H n1
ð73Þ
It is clear that there exists a constant C5 such that m m m 2 Dt h Dt h Dt h 1 þ C2 6 1 þ C ; þ 6 1 þ 2C þ þ 5 5 H2 H H2 H H2 H
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
607
therefore m 1=2 m Dt h Dt h 2 1þC 6 1 þ C5 ; þ þ H2 H H2 H so we have n1
ke
Dt h kþC þ H2 H
m=2 Dt
1=2
ke
n1
ka 6 1 þ C5
Dt h þ H2 H
m
ken1 kA :
Using (72) and (73) we have that m m=2 Dt h Dt h n n1 ke kA 6 1 þ C5 þ ke kA þ CDt þ H2 H H2 H ! þ kot gn k þ kqn k þ kgn1 k ;
ð74Þ
where the constant C is independent of Dt, h and H . Make summation for n we have " Z T n X n n Dtke ka 6 C ke0 k þ Dtke0 ka þ hkþ1 ke k þ 0
j¼1
dt1=2 þ Dt
Dt h þ H2 H
2 ! ou þ ot kþ1 H # 2 ouh 1=2 : ot dt A
2 kukH kþ1
m=2 Z
T
0
Since e0 ¼ 0, W n un ¼ W n unh þ unh un ¼ en þ gn : We have that n
n
n
n
kW u k 6 ke k þ kg k 6 C Dt þ h
kþ1
þ
Dt h þ 2 H H
m=2 ! :
ð75Þ
We complete the proof of Theorem 2.1. 5. Numerical example Consider the following parabolic equation: ou rðDruÞ ¼ f ot
in X J ;
ð76Þ
where X ¼ ð0; 1Þ ð0; 1Þ, J ¼ ð0; 1Þ, D ¼ dI, I is the unit matrix, the coefficient d > 0 is a constant. The right-hand side term, boundary condition and initial
608
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
Fig. 2. The sub-domains of X.
condition are selected in such a way that the exact solution is u ¼ exp ðx þ y tÞ. We divide X into four sub-domains: X1 ¼ ð0; 0:6Þ ð0; 0:6Þ, X2 ¼ ð0:4; 1Þ ð0; 0:6Þ, X3 ¼ ð0; 0:6Þ ð0:4; 1Þ, X4 ¼ ð0:4; 1Þ ð0:4; 1Þ, see Fig. 2. Based on this domain decomposition we give the triangulation. We first divide X into uniform squares with mesh-size h, here we let h ¼ 0:1; 0:05; 0:025, then we obtain the triangulation by dividing each square into two triangles. The time increment is Dt. The iterative number in each time step is m. Let kEkl2 ¼ max1 6 n 6 T =Dt kunh;ij unij kl2 , where kunh;ij unij kl2 be the approximation to the L2 norm of the error at the nth time level by using the 6-point Table 1 Dt
h
m
kEkl1
kEkl2
kEkl1
kEkl2
kEkl1
kEkl2
0.10 0.10 0.10 0.10 0.10
0.10 0.10 0.10 0.10 0.10
1 2 3 4 *
1.745E ) 1 4.977E ) 2 1.664E ) 2 7.349E ) 3 4.432E ) 3
7.423E ) 2 1.985E ) 2 8.630E ) 3 5.861E ) 3 4.892E ) 3
1.766E ) 1 4.552E ) 2 1.582E ) 2 1.017E ) 2 9.205E ) 3
7.663E ) 2 1.943E ) 2 9.614E ) 3 7.711E ) 3 7.220E ) 3
1.600E ) 1 5.234E ) 2 4.844E ) 2 4.829E ) 2 4.828E ) 2
7.719E ) 2 3.054E ) 2 2.809E ) 2 2.798E ) 2 2.797E ) 2
0.05 0.05 0.05 0.05 0.05
0.05 0.05 0.05 0.05 0.05
1 2 3 4 *
7.395E ) 2 1.769E ) 2 5.079E ) 3 2.538E ) 3 2.157E ) 3
3.151E ) 2 6.929E ) 2 2.715E ) 3 1.932E ) 3 1.746E ) 3
7.122E ) 2 1.391E ) 2 5.250E ) 3 4.622E ) 3 4.539E ) 3
3.107E ) 2 6.195E ) 3 3.360E ) 3 3.032E ) 3 2.987E ) 3
5.448E ) 2 2.510E ) 2 2.476E ) 2 2.476E ) 2 2.475E ) 2
2.798E ) 2 1.414E ) 2 1.395E ) 2 1.394E ) 2 1.394E ) 2
0.025 0.025 0.025 0.025 0.025
0.025 0.025 0.025 0.025 0.025
1 2 3 4
3.146E ) 2 5.480E ) 3 1.429E ) 3 1.076E ) 3 1.044E ) 3
1.338E ) 2 2.162E ) 3 8.648E ) 3 7.198E ) 4 7.009E ) 3
2.747E ) 2 3.835E ) 3 2.273E ) 3 2.225E ) 3 2.222E ) 3
1.203E ) 2 1.928E ) 3 1.365E ) 3 1.336E ) 3 1.335E ) 3
1.803E ) 2 1.257E ) 2 1.255E ) 2 1.255E ) 2 1.255E ) 2
9.863E ) 3 6.943E ) 3 6.933E ) 3 6.933E ) 3 6.933E ) 3
d ¼ 1:0
d ¼ 0:5
d ¼ 0:1
H. Rui / Appl. Math. Comput. 136 (2003) 593–610
609
Table 2 Dt
h
m
l1
l2
l1
l2
l1
l2
0.10 0.10 0.10 0.10
0.10 0.10 0.10 0.10
1 2 3 4
1.701E ) 1 4.545E ) 2 1.232E ) 2 3.029E ) 3
6.934E ) 2 1.553E ) 2 3.378E ) 3 9.690E ) 4
1.674E ) 1 3.632E ) 2 6.615E ) 3 9.653E ) 4
6.941E ) 2 1.221E ) 2 1.920E ) 3 4.914E ) 4
1.117E ) 1 4.060E ) 2 1.613E ) 4 7.15E ) 6
4.922E ) 2 2.568E ) 3 1.168E ) 4 5.26E ) 6
0.05 0.05 0.05 0.05
0.05 0.05 0.05 0.05
1 2 3 4
7.179E ) 2 1.553E ) 2 2.922E ) 3 3.810E ) 4
2.976E ) 2 5.183E ) 2 9.690E ) 4 1.860E ) 4
6.668E ) 2 9.371E ) 3 7.100E ) 4 8.261E ) 5
2.808E ) 2 3.208E ) 3 3.716E ) 4 4.423E ) 5
2.973E ) 2 3.471E ) 4 3.62E ) 6 2.2E ) 7
1.404E ) 2 1.933E ) 4 1.99E ) 6 3.E ) 8
0.025 0.025 0.025 0.025
0.025 0.025 0.025 0.025
1 2 3 4
3.042E ) 2 4.436E ) 3 3.850E ) 4 3.195E ) 5
1.268E ) 2 1.462E ) 3 1.639E ) 4 1.874E ) 5
2.525E ) 2 1.613E ) 3 5.043E ) 5 2.385E ) 6
1.069E ) 2 5.931E ) 4 3.008E ) 5 1.513E ) 6
5.473E ) 3 1.144E ) 5
2.931E ) 3 7.39E ) 6
d ¼ 1:0
d ¼ 0:5
formulation on each triangle follows: kEkl1 ¼ max max 1 6 n 6 T =Dt
0
d ¼ 0:1
and define the discrete L1 norm error as n n uh;ij uij ;
where unij ; unh;ij denote the exact solution and approximate solution at point ðih; jh; nDtÞ, respectively. Setting d ¼ 1:, 0.5, 0.1, using different values of parameters Dt, h and M, the errors kEkl2 , kEkl1 are depicted in Table 1. Here ÔÕ denote the errors without using domain decomposition methods. The errors caused by domain decomposing are listed in Table 2. From the numerical results we can see that using the multiplicative Schwarz method we can get a good result for parabolic problem, even iterating only one or two cycle at each time level. From Table 2 we can see that the errors caused by decomposing domain become small as the discretization parameters h and Dt become small. Acknowledgement The author was supported by the National Natural Science Foundation of China. References [1] R.A. Adamas, Sobolev Space, Academic Press, New York, 1975. [2] J.H. Bramble, J.E. Pasciak, J. Wang, J. Xu, Convergence estimates for product iterative methods with application to domain decomposition, Math. Comp. 57 (195) (1991) 1–21.
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[3] X.C. Cai, Additive Schwarz algorithms for parabolic convection diffusion equations, Numer. Math. 60 (1991) 41–61. [4] X.C. Cai, Multiplicative Schwarz methods for parabolic problem, SIAM J. Sci. Comput. 15 (1994) 587–603. [5] X.C. Cai, Multiplicative Schwarz algorithms for nonsymmetric and indefinite problem, SIAM J. Numer. Anal. 30 (1993) 936–952. [6] P.G. Ciarlet, The Finite Element Method for Elliptic Problems, North-Holland, Amsterdam, 1978. [7] C.N. Dawson, T.F. Dupont, Explicit/implicit conservative Galerkin domain decomposition procedures for parabolic problems, Math. Comp. 58 (197) (1992) 21–34. [8] C.N. Dawson, Q. Du, T.F. Dupont, A finite difference domain decomposition algorithm for numerical solution of heat equation, Math. Comp. 57 (195) (1991) 63–71. [9] P.L. Lions, On the Scharz alternating method I, in: R. Glowinski, G.H. Golub, G.A. Meurant, J. Periaux (Eds.), First International Symposium on Domain Decomposition Methods for Partial Differential Equations, SIAM, Philadelphia, PA, 1988. [10] P.L. Lions, On the Schwarz alternating method II, in: T. Chan, R. Glowinski, J. Periaux, O.B. Widlund (Eds.), Second International Symposium on Domain Decomposition Methods for Partial Differential Equations, SIAM, Philadelphia, PA, 1989.