Applied Mathematics and Computation 168 (2005) 1004–1019
www.elsevier.com/locate/amc
Non-conforming finite element
discretisations of a class of
non-Newtonian flows
A. Agouzal
a
a,*
, K. Allali b, K. Najib
c
U.M.R. 5585, Equipe de Mathe´matiques Applique´es de Lyon, Universite´ de Lyon 1,
Baˆt. 101, 69622 Villeurbanne cedex, France
b
De´partement de Mathe´matiques, Faculte´ des Sciences et Techniques, B.P. 146,
Route de Rabat, 20650 Mohammedia, Morocco
c
Ecole Nationale de lIndustrie Mine´rale, B.P. 753, Agdal, Rabat, Morocco
Abstract
We study an approximation for a class of non-Newtonian model, based on the general Ladyzhenskaya law for continuous and non-linear kinematic viscosity functions of
the fluid. We introduce a non-conforming finite element approximation of non-linear
Stokes problem; we give existence, uniqueness results and optimal error estimates under
minimal regularity assumptions. Finally, we introduce a post-processing to obtain conforming approximation of stress, which allow to derive various equivalence with two or
three field mixed dual approximations of non-linear Stokes problem.
2004 Elsevier Inc. All rights reserved.
Keywords: Non-conforming finite elements; Mixed finite elements; Non-Newtonian model
*
Corresponding author.
E-mail address: [email protected] (A. Agouzal).
0096-3003/$ - see front matter 2004 Elsevier Inc. All rights reserved.
doi:10.1016/j.amc.2004.09.042
A. Agouzal et al. / Appl. Math. Comput. 168 (2005) 1004–1019
1005
1. Introduction
In [9], a model for the motion of ideal incompressible viscous flow has been
proposed. Finite element approximation of this model has been studied, in the
case of conforming finite elements methods discretisation in primal formulation [5], and more recently in dual formulation in [6]. In this paper, we propose
and study the non-conforming finite element approximation method for such
model. We obtain optimal a priori error estimate in the quasi-norm for the
velocity, and obtain by post-processing a conforming approximation of the
stress as in the mixed dual formulation.
The model we work with is as follows: consider the motion of stationary
ideal incompressible viscous fluids in a bounded open connected, polyhedral
domain X Rd with Lipschitz boundary C (d = 2 or 3), and let u denote the
velocity field, p the pressure, and f the body force per unit mass. The model
is given by
8
Find ðu; pÞ such that;
>
>
>
>
>
>
< l0 div ðAðruÞruÞ l1 divðkrukr2 ruÞ þ rp ¼ f ; in X;
>
>
div u ¼ 0;
>
>
>
>
:
u ¼ 0;
in X;
on C ¼ oX;
2
d
where l0, l1 P 0 with l0 + l1 > 0, r 2 ]1, 2], f 2 (L (X)) , and A : Rd
d ! Rþ is
Lipschitz continuous function satisfying, there are positives constants c1 and c2
such that: for all a, b 2 Rd
d ,
c1 ka bk2 6 ðAðaÞa AðbÞbÞ : ða bÞ
ð1:1Þ
kAðaÞa AðbÞbk 6 c2 ka bk;
ð1:2Þ
and
(Colon denotes the scalar product in Rd
d ).
This kind of non-linear Stokes problem appears in the modeling of large
class of non-Newtonian fluids. In particular, the Carreau law for viscoplastic
flows (see, e.g., [10]) reads, with l1 = 0,
8a 2 R;
2
AðaÞ ¼ k 0 þ k 1 ð1 þ kak Þ
b2
2
with k0 P 0, k1 > 0 and b P 1. It is easy to check that the Carreau law satisfies
(1.1) and (1.2) for all k0 > 0 and b 2 [1, 2]. In particular, with b = 2 we recover
the usual linear Stokes model.
Let us recall that this problem has unique weak solution ðu; pÞ 2 H 10 ðXÞ
2
r0
0
r
L0 ðXÞ if l0 5 0 and ðu; pÞ 2 W 1;r
0 ðXÞ L ðXÞ with r ¼ r1, if l0 = 0.
1006
A. Agouzal et al. / Appl. Math. Comput. 168 (2005) 1004–1019
In the sequel, we denote by W s,p(X) and Ws,p(C), 0 6 s and 1 6 p 6 +1, the
usual Sobolev spaces (see e.g., [1]), endowed with the norms jj Æ jjs,p,X and
jj Æ jjs,p,C respectively. For a non-integer s we use the notations j Æ js,p,X and
j Æ js,p,C, where
ZZ
p
kD½s vðxÞ D½s vðyÞk
if p < þ1; jvjps;p;X ¼
dx dy
kx ykdþpr
X
X
and
jvjps;p;C ¼
ZZ
kD½s vðxÞ D½s vðyÞk
dx dy;
kx ykd1þpr
C
C
if p ¼ 1;
p
jvjs;þ1;X ¼ sup
X
X
kD½s vðxÞ D½s vðyÞkp
;
kx ykr
[s] being the integer part of s and r = s [s]. Hs(X) is the usual space Ws,2(X)
and H s0 ðXÞ the closure of DðXÞ in Hs(X).
2. Finite element approximation
In order to state the precise form of our discrete problem, we specify the
hypothesis on the class of finite elements spaces under questions. Let Th be
a family of regular triangulation by triangles or tetrahedrons of X in the sense
of Ciarlet [4]. We denoted by E the set of all edges (faces) of Th and by EI the
set of all interior edges (faces) of Th .
Let us introduce the following spaces:
1;p
W 0 ðTh Þ :¼ v 2 Lp ðXÞ; 8T 2 Th ; vhjT 2 H p ðT Þ; 8e 2 E;
Z
½v dc ¼ 0; 8e edgeðfaceÞ C
e
Z
v dc ¼ 0
e
with p 2 [1, 1[; and we set H 10 ðTh Þ ¼ W 1;2 ðTh Þ.
d
d
V h ¼ fvh 2 ðH 10 ðTh ÞÞ ; 8T 2 Th ; vhjT 2 ðP 1 ðT ÞÞ g;
M h ¼ fqh 2 L20 ðXÞ; 8T 2 Th ; qhjT 2 P 0 ðT Þg;
and
W h ¼ fr :¼ ðri;j Þi;j¼1;...;d 2 H ðdiv; XÞ; 8T 2 Th ;
d
d
d
r 2 ðRT 0 ðT ÞÞ :¼ ððP 0 ðT ÞÞ þ xP 0 ðT ÞÞ g:
A. Agouzal et al. / Appl. Math. Comput. 168 (2005) 1004–1019
We consider the following discrete problem (2.1):
8
Find ðuh ; ph Þ 2 V h M h such that;
>
>
P R
>
r2
>
>
8vh 2 V h ;
ðl0 Aðruh Þ þ l1 kruh k Þruh rvh
>
T
>
<
T 2Th
R
P R
p div vh dx ¼ X fh vh dx;
>
T h
>
>
T 2Th
>
>
P R
>
>
q div uh dx ¼ 0;
: 8q 2 M h ;
h
where fh is defined by :
1
f h :¼
measd ðT Þ
8T 2 Th ;
ð2:1Þ
h
T
T 2Th
1007
Z
f dx;
on T :
T
It is clear that this problem has unique solution (uh, ph) 2 Vh · Mh (see e.g.
[3,8]), moreover
div uh ¼ 0
8T 2 Th ;
ð2:2Þ
in T :
In the sequel, for abbreviation, we frequently write
(
)12
X
2
j j1;h ¼
l0 j j1;T
T 2Th
and
(
8q 21; 1½;
j j1;q;h ¼
X
)1q
j
q
j1;q;T
T 2Th
and we introduce the following form defined by
d 2
8ðu; vÞ 2 ððW 1;q
0 ðTh ÞÞ Þ ;
X Z
r2
ðl0 AðruÞ þ l1 kruk Þru rv dx
aðu; vÞ :¼
T 2Th
T
with q = 2 if l0 5 0 and q = r otherwise.
In the sequel, we denoted by C, C0, C1, . . . various positive generic constants
not dependent of fhT gT 2 Th and not necessarily the same in any place.
d
Let P : ðW 1;1
0 ðTh ÞÞ ! V h the linear operator defined by:
Z
d
8v 2 ðW 1;1
ðT
ÞÞ
;
8E
2
E;
ðPv vÞ dr ¼ 0:
h
0
E
First, we have
d
Lemma 2.1. The linear operator P satisfies, for all v 2 ðW 1;1
0 ðTh ÞÞ and for all
T 2 Th :
1008
A. Agouzal et al. / Appl. Math. Comput. 168 (2005) 1004–1019
Z
8sh 2 ðP 0 ðT ÞÞd
d ;
sh rðv PvÞ dx ¼ 0;
T
Z
8qh 2 P 0 ðT Þ
qh div ðv PvÞ dx ¼ 0:
T
Moreover, if v 2 W
1;q
ðTh Þ with q 2 ]1, 1[, then
(
) 10
q
X Z
X q0
d
q0
q0
8R 2 ðL ðXÞÞ ; Rðv PvÞ dx 6 Cjvj1;q;h
hT kRk0;q0 T ;
T 2T T
T 2T
h
0
where q ¼
h
q
.
q1
d
d·d
Proof. First, let v 2 ðW 1;1
, by elementwise integration
0 ðXÞÞ and sh 2 (P0(T))
by part, we infer
Z
Z
Z
sh rðv PvÞ dx ¼ ðv PvÞ div sh dx þ
ðv PvÞsh n dr
T
T
oT
Z
¼
ðv PvÞsh n dr ¼ 0:
oT
Again, by elementwise integration by part, we infer
Z
8qh 2 P 0 ðT Þ;
qh div ðv PvÞ dx
T
Z
Z
¼ rqh ðv PvÞ dx þ
ðv PvÞ nT dr ¼ 0:
oT
T
q0
1,q
q
. Since
Let q 2 ]1, 1[, v 2 W (X) and R 2 L ðXÞ with q0 ¼ q1
kv Pvk0;q;T 6 ChT jvj1;q;T ;
we have
(
) 10
q
X Z
X q0
q0
Rðv PvÞ dx 6 Cjvj1;q;h
hT kRk0;q0 T :
T 2T T
T 2T
h
h
Lemma 2.2. We have the generalise Green formulae, for all q 2 [1, 1[:
X Z
d
8v 2 ðW 1;q
ðT
ÞÞ
;
8r
2
W
;
ðrh re þ e div rh Þ dx ¼ 0:
h
h
h
0
T 2Th
d
T
Proof. Let ðv; rh Þ 2 ðW 1;q
0 ðTh ÞÞ W h , since
Z
d
8E 2 EI ;
½vdc ¼ 0 and rh nE 2 ðP 0 ðEÞÞ ;
E
A. Agouzal et al. / Appl. Math. Comput. 168 (2005) 1004–1019
we have
X Z
T 2Th
X Z
ðrh re þ e div rh Þ dx ¼
T
T 2Th
1009
r nT v dc
oT
Z
1X
¼
r nE ½v dc ¼ 0:
2 E2EI E
We have also (see e.g., [10]).
Lemma 2.3. For all s0, s1, r 2 (Lr(X))d ·d, we have
Z
ðks0 kr2 s0 ks1 kr2 s1 Þ r dx 6 Ckrk0;r;X
X
Z
12
r2
fks0 k0;r;X þ ks1 k0;r;X g 2
ðks0 k þ ks0 s1 kÞr2 ks0 s1 k2 dx
X
and
Z
r
r2
2
ks0 s1 k1;r;X þ ðks0 k þ ks0 s1 kÞ ks0 s1 k dx
X
Z
r2
r2
6 C ðks0 k s0 ks1 k s1 Þ ðs0 s1 Þ dx:
X
One of our main result in this section is the following.
Theorem 2.1. Let (u, p) the weak solution of the model problem and
(uh, ph) 2 Vh · Mh the solution of the discrete problem. If l0 5 0, then we have
X Z
r2
2
l1
ðkruk þ krðu uh ÞkÞ krðu uh Þk dx
T 2Th T
X Z
r
2
kru ruh k dx þ ju uh j1;h
þ l1
6C
T 2Th
(
X
T
)
h2T kf k20;T
T 2Th
and
(
kp 2
ph k0;X
þ
inf
rh 2W h ;div rh ¼fh
6 C l0 þ l1 kruk0;r;X þ
(
C
X
rh k20;X
kr X
r
kruh k0;r;T
T 2Th
1r r2
)
h2T kf k20;T
T 2Th
with r :¼ l0 AðruÞru þ l1 kruk
r2
þ
inf
rh 2W h ;div rh ¼fh
ru pId d .
kr rh k20;X
)
1010
A. Agouzal et al. / Appl. Math. Comput. 168 (2005) 1004–1019
Proof. We set e = u uh. Recall that
X Z
l1
ðkruk þ krðu uh ÞkÞr2 krðu uh Þk2 dx
T 2Th
T
þ l1
X Z
T 2Th
kru ruh kr dx þ ju uh j21;h
T
6 Cðaðu; eÞ aðuh ; eÞÞ ¼ C aðu; eÞ
Z
Z
fh e dx þ fh e dx aðuh ; eÞ ¼ I þ II:
X
X
On one hand, since 8T 2 Th , div e = 0 on T, and
X Z
8rh 2 W h ;
ðrh re þ e div rh Þ dx ¼ 0;
T
T 2Th
we have, "rh 2 Wh, div rh = fh:
Z
Z
X Z
aðu; eÞ fh e ¼ aðu; eÞ p div e dx þ e div rh dx
X
T 2Th
X Z
¼
T 2Th
X
T
ðr rh Þ re dx;
X
which implies that
jIj 6 jej1;h inf
rh 2W h ;div rh ¼fh
kr rh k0;X :
On the other hand
Z
Z
Z
aðuh ; eÞ fh e dx ¼ aðuh ; PeÞ fh e ¼
fh ðe PeÞ dx;
X
X
X
where Pe is the non-conforming interpolation operator. This last equality
implies
!12
X
2
jIIj 6 Cjej1;h
h2T kf k0;T :
T 2Th
Using the last inequalities and Young inequality, we obtain
X Z
r2
2
l1
ðkruk þ krðu uh ÞkÞ krðu uh Þk dx
T 2Th
T
þ l1
X Z
T 2Th
6C
X
2
r
kru ruh k dx þ ju uh j1;h
T
2
h2 kf k0;T
T 2Th T
þ
inf
rh 2W h ;div rh ¼fh
kr 2
rh k0;X
:
A. Agouzal et al. / Appl. Math. Comput. 168 (2005) 1004–1019
Finally, since
R
X
kp ph k0;X 6 C
X
v2ðH 10 ðXÞÞd
and Z
ðp ph Þ div v ¼
X
Z
1011
ðp ph Þ div v dx
jvj1;X
p div v X
X Z
T 2Th
ph div Pv dx
T
Z
fv dx aðuh ; PvÞ þ fh Pv dx
X
X
Z
Z
¼ aðu; vÞ aðuh ; vÞ þ fh ðPv vÞ dx þ ðfh f Þv dx
¼ aðu; vÞ Z
X
X
and using the fact that (see Lemma 2.3)
8
>
< XZ
jaðu;vÞ aðuh ;vÞj 6 Cjvj1;X
ðkruk
>
: T 2Th T
þkrðu uh ÞkÞ
r2
!12
2
krðu uh Þk dx
9
!1r 1r2
2
>
=
r
l1 @kruk0;r;X þ
kruh k0;r;T A þ ju uh j1;h ;
>
;
T 2Th
0
we have
8
>
<
X
9
!1r 1r2 >
=
X
r
kruh k0;r;T A
>
;
T 2Th
)
0
kp ph k0;X 6 C l0 þ l1 @kruk0;r;X þ
>
:
(
X
2
h2T kf k0;T þ
inf
2
T 2Th
rh 2W h ;div rh ¼fh
2
kr rh k0;X :
Using the similar arguments, we have also
Theorem 2.2. Let (u, p) the weak solution of the model problem and
(uh, ph) 2 Vh · Mh the solution of the discrete problem. If l0 = 0, then we have
X Z
r2
2
l1
ðkruk þ krðu uh ÞkÞ krðu uh Þk dx
T 2Th T
X Z
r
þ l1
kru ruh k dx
T
( T 2Th
)
X 0
r0
r0
r
6C
hT kf k0;r0 ;T þ
inf
kr rh k0;r0 ;X
T 2Th
rh 2W h ;div rh ¼fh
1012
A. Agouzal et al. / Appl. Math. Comput. 168 (2005) 1004–1019
and
X
1r r2
r
kru
kp ph k20;X 6 C kruk0;r;X þ
h k0;r;T
T 2Th
(
)
X 0
0
0
r
r
C
hrT kf k0;r0 ;T þ
inf
kr rh k0;r0 ;X
rh 2W h ;div rh ¼fh
T 2Th
r
.
with r: = l1k$ukr2$u pIdd and r0 ¼ r1
Using Theorems 2.1 and 2.2, we have unified a priori error estimate.
Theorem 2.3. Let (u, p) the weak solution of the model problem and
(uh, ph) 2 Vh · Mh the solution of the discrete problem. We have
X Z
r2
2
ðkruk þ krðu uh ÞkÞ krðu uh Þk dx
l1
T
T 2Th
þ l1
X Z
T 2Th
(
6C
X
2
r
kru ruh k dx þ ju uh j1;h
T
0
0
hqT kf kq0;q0 ;T
T 2Th
þ
inf
rh 2W h ;div rh ¼fh
kr 0
rh kq0;q0 ;X
)
and
(
kp 2
ph k0;X
X
1r r2
r
6 C l0 þ l1 kruk0;r;X þ
kru
k
h 0;r;T
T 2T
h
(
C
X
0
q0
hqT kf k0;r0 ;T
T 2Th
with r :¼ l0 AðruÞru þ l1 kruk
q
and q0 ¼ q1
.
r2
þ
inf
rh 2W h ;div rh ¼fh
kr q0
rh k0;q0 ;X
)
ru pId d , q = 2 if l0 5 0, q = r otherwise,
We are now in position to state the error estimates for our problem, more
precisely, using the classical approximation results for equilibrium finite element method [2], we have
Theorem 2.4. Let (u, p) the weak solution of the model problem and (uh, ph) 2
Vh · Mh the solution of the discrete problem. If
0
r :¼ l0 AðruÞru þ l1 krukr2 ru pId d 2 ðW s;q ðXÞÞd
d ;
s 20; 1:
A. Agouzal et al. / Appl. Math. Comput. 168 (2005) 1004–1019
1013
Then we have
X Z
ðkruk þ krðu uh ÞkÞr2 krðu uh Þk2 dx
l1
T 2Th
T
þ l1
X Z
T 2Th
(
X
6C
2
r
kru ruh k dx þ ku uh k1;h
T
0
q0
hqT kf k0;q0 ;T
þh
sq0
T 2Th
q0
krkW s;q0 ðXÞ
)
and
kp 2
ph k0;X
1
q1
(
6 Ckf k0;X
X
0
q0
hqT kf k0;q0 ;T
þh
sq0
T 2Th
q0
krkW s;q0 ðXÞÞ
)
q
.
with q = 2 if l0 5 0 and q = r otherwise, and q0 ¼ q1
3. Post-processing and mixed finite elements methods
First, we have the more general results.
Theorem 3.1. Let (uh,ph) 2 Vh · Mh the solution of discrete problem. The tensor
field rh defined by
8T 2 Th ;
rh :¼ l0 Aðruh Þruh þ l1 kruh kr2 ruh fh ðx xg Þ
;
d
where xg is the barycenter of T and by using the notations:
8f ; g 2 Rd ; ðf gÞij ¼ fi gj ;
i; j ¼ 1; . . . ; d;
satisfies
rh ph Id d 2 H ðdiv; XÞ
and
div ðrh ph Id d Þ ¼ fh
Proof. Remark that
8T 2 Th ;
d
rh ph Id d 2 ðRT 0 ðT ÞÞ :
Let e ¼ oT 1 \ oT 2 2 EI and vh 2 (Vh)d such that:
Z
8f 2 E;
vh dc ¼ dfe :
f
on X:
on T ;
1014
A. Agouzal et al. / Appl. Math. Comput. 168 (2005) 1004–1019
Since
8T 2 Th ;
Z
8vh 2 V h ;
T
fh ðx xg Þ
rvh dx ¼ 0
d
and using Green formulae, we have:
Z
½ðrh ph IdÞ ne ¼ ½ðrh ph Id d Þ:nvh dc
e
¼
(Z
2
X
)
ðrh ph IdÞ rvh þ vh div ðrh ph IdÞ dx
Ti
i¼1
2 Z X
¼
l0 Aðruh Þ þ l1 kruh kr2 ruh
Ti
i¼1
ph Id d ¼
2 Z X
l0 Aðruh Þ þ l1 kruh kr2 ruh rvh
Ti
i¼1
Z
fh ðx xg Þ
fh vh dx
rvh d
Ti
Z
ph div vh dx Ti
Z
fh vh dx
Ti
¼ 0;
then rh phIdd 2 H(div,X). Finally, it is clear that div(rh phId) = fh.
In the case where l1 = 0, we recovered the classical mixed dual formulations
of stokes problem (see e.g., [6,7]). First let
d
Y h ¼ fvh 2 ðL2 ðXÞÞ ;
8T 2 Th ;
d
fvh gjT 2 ðP 0 ðT ÞÞ g
and
qh 2 L20 ðXÞ;
X h ¼ fðsh ; qh Þ;
and
sh qh Id d 2 W h g:
8T 2 Th ;
fqh gjT 2 P 0 ðT Þ;
To obtain optimal a priori error estimate we need the following regularity
result.
Lemma 3.1. Let ðu; pÞ 2 ðH 1 ðXÞÞd L20 ðXÞ such that div u = 0 on X. Then
r ¼ l0 ru pId d 2 ðH s ðXÞÞd
d
d
u 2 ðH 1þs ðXÞÞ :
is equivalent to p 2 H s ðXÞ
and
A. Agouzal et al. / Appl. Math. Comput. 168 (2005) 1004–1019
1015
Proof. On one hand, it is clear that if p 2 H(s(X)) and u 2 (H1+s(X))d, then
$u 2 (Hs(X))d·d which imply
r ¼ l0 ru pId d 2 ðH s ðXÞÞd
d :
On the other hand, if r = l0$u pIdd 2 (Hs(X))d·d, then, since div u = 0, we
have
traceðrÞ ¼ l0 div u d p ¼ d p 2 H s ðXÞ;
which implies
d
d
l0 ru ¼ r þ pId d 2 ðH s ðXÞÞ
and then
p 2 H s ðXÞ
and
d
u 2 ðH 1þs ðXÞÞ :
We have the following a priori error estimates.
Theorem 3.2. Let (uh, ph) the solution of discrete problem with l1 = 0 and
A ¼ Id d . We set
8T 2 Th ;
rh :¼ l0 ruh 8T 2 Th ;
1
^uh ¼
measd ðT Þ
and
fh ðx xg Þ
;
d
Z
on T
1
uh dx þ
2
l0 d measd ðT Þ
T
Z
2
fh kx xg k dx:
T
Then ððrh ; ph Þ; ^
uh Þ 2 X h Y h is the unique solution of the following problem:
uh Þ 2 X h Y h ; such that;
Find ððrh ; ph Þ; ^
Z
Z
1
rh sh dx þ ^
uh div ðsh qh Id d Þ dx ¼ 0;
8ðsh ; qh Þ 2 X h ;
l0 X
X
Z
Z
8vh 2 Y h ;
vh div ðrh ph Id d Þ dx ¼ fvh dx:
X
X
s
Moreover, if r = l0$u pIdd 2 (H (X))
d ·d
, s 2 ]0,1], then
kr ðrh ph IdÞk0;X þ kp ph k0;X þ ku ^
uh k0;X
6 Cfhs krkH s ðXÞ þ hkf k0;X g:
Proof. Let us recall that rh phIdd 2 Xh and div(rh phIdd) = fh, which
implies that
Z
Z
vh div ðrh ph Id d Þ dx ¼ fvh dx:
X
X
1016
A. Agouzal et al. / Appl. Math. Comput. 168 (2005) 1004–1019
Let (sh, qh) 2 Xh, since div uh = 0 on T, for all T 2 Th and for all ðsh ; qh Þ 2
X h ; 8T 2 Th ,
Z
Z
fh ðx xg Þ
1
2
: sh ¼ 2
kx xg k dx
d
meas
ðT
Þ
d
T
T
d
Z
fh div ðsh ph Id d Þ dx;
T
we have
1
l0
Z
Z
1
rh sh dx ¼
ruh sh dx fh ðx xg Þ:sh dx
dl0 X
X
T 2T
Z
X
1
ruh ðsh qh Id d Þ dx fh ðx xg Þ
¼
dl0 X
T 2T
X
ðsh qh Id d Þ dx
X Z uh þ
¼
T 2Th
div ðsh qh Id d Þ dx ¼ Z
T
1
2
l0 d measd ðT Þ
Z
2
kx xg k dx fh
T
^
uh div ðsh qh Id d Þ dx:
X
Finally, it is clear that
8T 2 Th ; krh ph Id d rk0;T
6 Cfkp ph k0;T þ kru ruh k0;T þ hT kf k0;T g
and
8T 2 Th ;
ku ^
uh k0;T 6 CfkP 0 u uk0;T þ ku uh k0;T þ hT kf k0;T g;
where
8T 2 Th ;
1
P 0u ¼
measd ðT Þ
Z
u dx:
T
Using Theorem 2.3, and the fact that u 2 (Hs(X))d, we obtain
uh k0;X
kr ðrh ph IdÞk0;X þ kp ph k0;X þ ku ^
6 Cfhs krkH s ðXÞ þ hkf k0;X g:
In the general cases, we have the following:
Theorem 3.3. Let (uh, ph) the solution of discrete problem. We set
8T 2 Th ; rh :¼ l0 Aðruh Þruh þ l1 kruh k
r2
ruh fh ðx xg Þ
; on T ;
d
A. Agouzal et al. / Appl. Math. Comput. 168 (2005) 1004–1019
8T 2 Th ;
uh ¼
1
measd ðT Þ
Z
uh dx;
th ¼ ruh ;
1017
on T :
T
d
Then ððrh ; ph Þ; th ; uh Þ 2 X h ðY h Þ Y h is the unique solution of two-fold dual
mixed formulation
8
d
>
Find ððrh ; ph Þ; th ; uh Þ 2 X h ðY h Þ Y h such that;
>
>
R
>
<
8sh 2 ðY h Þd ;
ðrh l0 Aðth Þth l1 kth kr2 th Þsh dx ¼ 0;
X
R
>
8ðsh ; qh Þ 2 X h ;
ðt sh þ uh div ðsh qh Id d Þ dx ¼ 0;
>
X h
>
>
R
R
:
8vh 2 Y h ; X vh div ðrh ph Id d Þ dx ¼ X vh f dx:
Proof. Let (uh, ph) 2 Vh · Mh the solution
ððrh ; ph Þ; th ; uh Þ 2 X h ðY h Þd Y h defined by:
8T 2 Th ; rh :¼ l0 Aðruh Þruh þ l1 kruh k
8T 2 Th ;
uh ¼
1
measd ðT Þ
Z
uh dx;
of
r2
discrete
ruh th ¼ ruh ;
problem,
fh ðx xg Þ
; on T ;
d
on T :
T
On one hand, it is clear that
div ðrh ph Id d Þ ¼ fh
and
8sh 2 ðY h Þd ;
Z
ðrh l0 Aðrth Þrth l1 kth kr2 th Þsh dx
X
¼
X Z fh ðx xg Þ
sh dx ¼ 0:
d
T 2Th T
On the other hand, since div uh = 0 on T, for all T 2 Th , we have
Z
8ðsh ; qh Þ 2 X h ;
th : qh Id d dx ¼ 0;
X
which implies, using the generalise Green formula:
Z
ðth : sh þ uh div ðsh qh Id d ÞÞ dx
8ðsh ; qh Þ 2 X h ;
X
Z
¼ ðth : ðsh qh Id d Þ þ uh div ðsh qh Id d ÞÞ dx
ZX
¼ ðth : ðsh qh Id d Þ þ uh div ðsh qh Id d ÞÞ dx
ZX
¼ ðruh : ðsh qh Id d Þ þ uh div ðsh qh Id d ÞÞ dx ¼ 0:
X
and
1018
A. Agouzal et al. / Appl. Math. Comput. 168 (2005) 1004–1019
We deduce that ððrh ; ph Þ; th ; uh Þ 2 X h ðY h Þd Y h is solution of two-fold dual
mixed formulation
8
>
Find ððrh ; ph Þ; th ; uh Þ 2 X h ðY h Þd Y h such that;
>
>
R
>
<
d
r2
8sh 2 ðY h Þ ;
ðrh l0 Aðth Þth l1 kth k th Þsh dx ¼ 0;
X
R
> 8ðsh ; qh Þ 2 X h ;
ðth sh þ uh div ðsh qh Id d Þ dx ¼ 0;
>
>
>
R X
R
:
8vh 2 Y h ; X vh div ðrh ph Id d Þ dx ¼ X vh f dx:
To prove the uniqueness of the solution of this two fold dual mixed problem,
let
d
ððrih ; pih Þ; tih ; uih Þ 2 X h ðY h Þ Y h ;
i ¼ 1; 2;
two solution of the two field dual mixed problem, and set
rh ¼ r1h r2h ;
ph ¼ p1h p2h ;
uh ¼ u1h u2h
and
th ¼ t1h t2h :
It is clear that
div ðrh ph Id d Þ ¼ 0
then
rh 2 ðY h Þ
d
on X;
Z
and
th : rh dx ¼ 0;
X
which implies
aðt1h ; th Þ aðt2h ; th Þ ¼ 0;
then th = 0, and since rh 2 (Yh)d, we have
Z
rh : rh dx ¼ 0;
X
which is equivalent to rh = 0. Finally, since (see e.g., [3]), div(Xh) = Yh, we obtain uh ¼ 0, which implies that the two-fold mixed problem has at least one
solution. h
References
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