Analysis o a Non-o e lapping Domain Decomposi ion Me hod
o S okes Equa ions
Tom´as Chac´on Rebollo, Eliseo Chac´on Ve a
Dep o de Ecuaciones Di e enciales y An´alisis Num´e ico. Uni e sidad de Se illa
chacon@nume .us.es, eliseo@nume .us.es
Monog a ´ıas del Semin. Ma em. Ga c´ıa de Galdeano. 27: 201–208, (2003).
Abs ac
In his no e we ex end he analysis o ellip ic p oblems pe o med in [1] o saddle
poin p oblems like he S okes equa ions. We use a non o e lapping domain decom-
posi ion and he in oduc ion o a penal y e m. In a simply connec ed bounded
domain Ω ⊂R2wi h Lipschi z bounda y, we decompose Ω in o wo non-o e lapping
Lipschi z subdomains Ω1and Ω2wi h Ω1∩Ω2=∅, and suppose ha ∂Ωi= Γi∪Γ
whe e Γiis he common bounda y wi h Ω, Γi=∂Ω∩∂Ωiand Γ is he in e ace wi h
Ωj, Γ = ∂Ω1∩∂Ω2. The S okes equa ions on Ω a e sol ed ia he ollowing pa allel
p ocess: Fo n= 0,1,2, ..., gi en un
i,pn
iwe compu e un+1
iand pn+1
i(i= 1,2) such
ha 














−∆un+1
i+∇pn+1
i= in Ωi
∇ · un+1
i= 0 in Ωi
un+1
i= 0 on Γi
∂un+1
i
∂nij
−pn+1
inij =−1
(un+1
i−un
j) on Γ
whe e nij is he ou wa d no mal ec o on Γ poin ing om Ωiin o Ωj, > 0 is a
pa ame e ha ends o ce o and in o ce he ansmision condi ions on he in e ace
Γ and we s ess ha he p essu es pido no longe ha e ce o mean a e age. We
p esen he con e gence analysis o his echnique and some nume ical es s. An
amplia ion o his wo k will appea in [2].
Keywo ds: S okes P oblem, Pa allel echnique, Non O e lapping
AMS CLASSIFICATION: 65L50, 65L60, 65L70
1 In oduc ion
In a simply connec ed bounded domain Ω ⊂Rd(d= 2,3) wi h a Lipschi z bounda y
∂Ω and wi h ∈[L2(Ω)]d, we sea ch o a eloci y ield u∈[H1
0(Ω)]dand a p essu e
201
p∈L2
0(Ω) such ha
(−∆u+∇p= ,∇·u= 0 in Ω,
u= 0 on ∂Ω.
In he classical mixed o mula ion o his p oblem we look o (u, p)∈[H1
0(Ω)]d×L2
0(Ω)
wi h
(∇u,∇ )Ω−(p, ∇· )Ω−(∇·u, q)Ω= ( , )Ω,(1)
o all ( , q)∈[H1
0(Ω)]d×L2
0(Ω). Now we decompose Ω in o wo non-o e lapping Lipschi z
subdomains Ω1and Ω2wi h Ω1∩Ω2=∅( his choice is made o ease he exposi ion o
he main ideas, bu hese can be ex ended o mo e han wo subdomains). Suppose ha
∂Ωi= Γi∪Γ whe e Γiis he common bounda y wi h Ω, Γi=∂Ω∩∂Ωiand Γ is he
in e ace wi h Ωj, Γ = ∂Ω1∩∂Ω2, all o hese bounda ies a e Lipschi z (d−1)-dimensional
mani olds. Nex , we conside he Sobole spaces
Xi=H1
0(Ωi; Γi)d={ ∈[H1(Ωi)]ds. . |Γi= 0}
no med by | |2
1,Ωi= (∇ ,∇ )Ωiand he Hilbe spaces Mi=L2(Ωi) no med as usual.
Now o  > 0 we conside he p oblem (P):
Find (ui, pi)∈Xi×Miwi h
(P)






(∇u1,∇ 1)Ω1−(p1,∇· 1)Ω1−(q1,∇·u1)Ω1+1
(u1−u2, 1)0,Γ= ( , 1)Ω1,
(∇u2,∇ 2)Ω2−(p2,∇· 2)Ω2−(q2,∇·u2)Ω2+1
(u2−u1, 2)0,Γ= ( , 2)Ω2,
o all ( i, qi)∈Xi×Mi,i= 1,2. This p oblem is he a ia ional o mula ion o he
ollowing coupled pa ial di e en ial equa ions















−∆u1+∇p1= in Ω1
∇·u1= 0 in Ω1
u1= 0 on Γ1
∂u1
∂n12 −p1n12 =−1
(u1−u2) on Γ















−∆u2+∇p2= in Ω2
∇·u2= 0 in Ω2
u2= 0 on Γ2
∂u2
∂n21 −p2n21 =−1
(u2−u1) on Γ
whe e nij is he ou wa d no mal ec o on Γ poin ing om Ωiin o Ωjand we s ess ha
he p essu es pido no longe ha e ce o mean a e age. The app p ia ed ansmission
condi ions a e en o ced when −→ 0 because we show ha ku1−u−2k0,Γ=O().
The i e a ion p ocess ha we p oposse is he ollowing: Fo n= 0,1,2, ..., gi en un
1,un
2
202
we compu e un+1
1,un+1
2and pn+1
1, pn+1
2such ha he ollowing p oblems a e sa is ied















−∆un+1
1+∇pn+1
1= in Ω1,
∇·un+1
1= 0 in Ω1,
un+1
1= 0 on Γ1,
∂un+1
1
∂n12 −pn+1
1n12 =−1
(un+1
1−un
2) on Γ,















−∆un+1
2+∇pn+1
2= in Ω2,
∇·un+1
2= 0 in Ω2,
un+1
2= 0 on Γ2,
∂un+1
2
∂n21 −pn+1
2n21 =−1
(un+1
2−un
1) on Γ.
We ema k ha ou me hod may be iewed as a a ia ion o he Robin
2 Analysis o p oblem (P)
Le us in oduce he p oduc spaces X=X1×X2,M=M1×M2and deno e by capi al
le e s he elemen s (pai s) o Xand M. Then we no m Mwi h kPk2
M=P2
i=1 kpik2
0,Ωi
and X ia ((U,V))=P2
i=1(∇ui,∇ i)Ωi+1
(u1−u2, 1− 2)0,Γi.e., he no m in Xis gi en
by kUk= ((U,U)). Nex we de ine he o ms b(P,V) = −P2
i=1(pi,∇· i)Ωi, F(V) =
P2
i=1( , i)Ωiand w i e p oblem (P) in e ms o he a ia ional p oblem:
(Find (U,P)∈X×Msuch ha
((U,V))+b(P,V) + b(Q,U) = F(V),∀(V,Q)∈X×M.
Now we conside he ollowing symme ic and con inuous, acco ding o k·kand k·kM,
bilinea o m on X×Mgi en by
B(U,P;V,Q) = ((U,V))+b(P,V) + b(Q,U)
o all pai s (U,P),(V,Q)∈X×M. We ha e
Lemma 1 The e exis s a posi i e cons an γindependen o  > 0such ha o all
(U,P)∈X×M
S= sup
(V,Q)∈XxM
|B(U,P;V,Q)|
kVk+kQkM≥ γ (kUk+kPkM).
As a consequence, gi en ∈[L2(Ω)]dand o each  > 0p oblem (P) has a unique
solu ion (U,P)∈X×M.
We in oduce nex he consis ency e o o p oblem (P) as an app oxima ion o he S okes
equa ions in a ia ional o m. This e o is he esul o plugging he solu ion o (1) in o
(P).
203
Lemma 2 Le (u, p)be he solu ion o he S okes p oblem and U= (u|Ω1,u|Ω2),P=
(p|Ω1, p|Ω2). Then, we conside he consis ency e o o p oblem (P) ia
G(V) = ((U,V))+b(P,V)−F(V)
=
2
X
i=1
(∇u,∇ i)Ωi−
2
X
i=1
(p, ∇· i)Ωi−
2
X
i=1
( , i)Ωi
o all V= ( 1, 2)∈X. Then, assuming u∈[H2(Ω) ∩H1
0(Ω)]dand p∈H1(Ω), le
n1,2=n, we ha e
G(V) = ZΓ
(∂nu−pn)·( 1− 2)dσ
and he e o e
|G(V)| ≤ k∂nu−pnk0,Γk 1− 2k0,Γ.
Now we can es ima e he e o in app oxima ing he a ia ional o mula ion o he S okes
Equa ions wi h p oblem (P)
Lemma 3 Suppose ha u∈[H2(Ω) ∩H1
0(Ω)]dand p∈H1(Ω) is he solu ion o he
S okes p oblem. Fo each  > 0le (U,P)∈X×Mbe he unique solu ion o p oblem
(P), wi h U= (u
1,u
2)and P= (p
1, p
2). Le c(u, p) = k∂nu−pnk0,Γ,U= (u|Ω1,u|Ω2)
and cons uc
π=p
1χΩ1+p
2χΩ2−1
|Ω|(ZΩ1
p
1+ZΩ2
p
2).
Then
kU−Uk≤c(u, p)√and kp−πk0,Ω≤c(u, p)√.
As a consequence we ha e
2
X
i=1 |u−u
i|1,Ωi≤c(u, p)√and ku
1−u
2k0,Γ≤c(u, p).
3 Disc e e p oblem and e o es ima es
We suppose ha he domain Ω is polygonal and ake o h > 0 an admissible and egula
iangula ion Tho Ω o med by polygons (d= 2) o polyhed a (d= 3) elemen s such ha
Γ is o med by aces o sides o elemen s Kin Th. Then we use Ti
h=Th∩Ωi, o i= 1,2.
These iangula ions o Ωia e compa ible on Γ, i.e., hey sha e he same edges on Γ. Fo
he iangula ion Thwe conside ini e elemen subspaces (Vh, Ph) o ([H1
0(Ω)]d, L2
0(Ω))
sa is ying he disc e e in -sup condi ion o Ladyzhenskya-B ezzi-Babuˇska on Ω. Now we
conside he disc e e solu ion (uh, ph)∈Vh×Pho he disc e e e sion o he S okes
204
p oblem posed on Vh×Phand assume ha , when he solu ion (u, p) o he con inuous
S okes p oblem in Ω sa is ies u∈Hk+1(Ω) ∩H1
0(Ω)dand p∈Hk(Ω) (k≥1), hen
|uh−u|1,Ω+kph−pk0,Ω≤C0hk(2)
o some cons an C0=C0(u, p). Now, based on Ti
h, use ini e elemen subspaces o
(Xi, Mi), deno ed by (Xi,h, Mi,h), such ha each pai (Yi,h, Ni,h), whe e Yi,h =Xi,h ∩
[H1
0(Ωi)]dand Ni,h =Mi,h ∩L2
0(Ωi) also sa is ies he disc e e in -sup condi ion on Ωi. Fo
ins ance we could use he es ic ion o he spaces Vhand Ph o each o he Ωi. Se now
Xh=X1,h ×X2,h and Mh=M1,h ×M2,h and pose he disc e e e sion o (P), ha we
deno e by (P,h):
(Find (U
h,P
h)∈Xh×Mhsuch ha
((U
h,Vh))+b(P
h,Vh) + b(Qh,U
h) = F(Vh),∀(Vh,Qh)∈Xh×Mh.
The exis ence and uniqueness o solu ion o (P,h) is ca ied ou as o (P) and we ha e
he es ima es
Theo em 4 Le u∈Hk+1(Ω) ∩H1
0(Ω)dand p∈Hk(Ω) (k≥1) be he solu ion o he
S okes p oblem in Ωand o each h > 0le (uh, ph)∈Vh×Phsol e he disc e e S okes
p oblem on Vh×Ph. Now conside Uh= (uh|Ω1,uh|Ω2)∈Xh,Ph= (ph|Ω1, ph|Ω2)∈Mh.
Fo each  > 0le (U
h,P
h)∈Xh×Mhsol e (P,h)and w i e U
h= (u
1,h,u
2,h)and
P
h= (p
1,h, p
2,h). Now cons uc
π
h=p
1,hχΩ1+p
2,hχΩ2−1
|Ω|(ZΩ1
p
1,h +ZΩ2
p
2,h) (3)
hen, he ollowing e o es ima e hold
kUh−U
hk≤C(hk+√) (4)
kph−π
hk0,Ω≤C(hk+√) (5)
whe e C=C(u, p)is a posi i e cons an jus depending on (u, p). As a consequence o
(4) we ha e
2
X
i=1 |uh−u
i,h|1,Ωi≤C(hk+√)and ku
1,h −u
2,hk0,Γ≤C(√ hk+).
Via he iangula inequali y, we gi e he main esul o his sec ion
Theo em 5 Le u∈Hk+1(Ω) ∩H1
0(Ω)dand p∈Hk(Ω),(k≥1) be he solu ion o he
S okes p oblem in Ω. Fo each h > 0and  > 0le (U
h,P
h)∈Xh×Mhsol e (P,h)wi h
ini e dimensional spaces o accu acy k≥1and w i e P
h= (p
1,h, p
2,h). Then cons uc
π
has in (3). The ollowing bounds hold
2
X
i=1 |u−u
i,h|1,Ωi+1
√ku
1,h −u
2,hk0,Γ≤C(hk+√) (6)
kp−π
hk0,Ω≤C(hk+√) (7)
205

whe e C=C(u, p, )is a posi i e cons an jus depending on he da a. When =O(h2k)
we ha e
2
X
i=1 |u−u
i,h|1,Ωi+kp−π
hk0,Ω≤C hkand ku
1,h −u
2,hk0,Γ≤C h2k.
4 I e a ion p ocess
We sea ch o he solu ion o (P,h) ia he ollowing pa allelizable echnique: Fo n=
0,1,2, ..., gi en un
1=u,n
1,h and un
2=u,n
2,h we compu e un+1
1∈X1,h,un+1
2∈X2,h and
pn+1
1∈M1,h, pn+1
2∈M2,h such ha he ollowing p oblem (Pn
,h) is sa is ied











(∇un+1
1,∇ 1)Ω1−(pn+1
1,∇· 1)Ω1−(q1,∇·un+1
1)Ω1+1
(un+1
1−un
2, 1)0,Γ= ( , 1)Ω1,
(∇un+1
2,∇ 2)Ω2−(pn+1
2,∇· 2)Ω2−(q2,∇·un+1
2)Ω2+1
(un+1
2−un
1, 2)0,Γ= ( , 2)Ω2
o all ( 1, 2)∈Xh(we d op he indices and hwhen no needed). We ob ain he
ollowing geome ic a e o con e gence
Theo em 6 Le U
h= (u
1,h,u
2,h)∈Xhand P
h= (p
1,h, p
2,h)∈Mhbe he solu ion o
(P,h)and U,n
h= (u,n
1,h,u,n
2,h)∈Xh,P,n
h= (p,n
1,h, p,n
2,h)∈Mhbe he solu ion o (Pn
,h). Le
us de ine π
hand π,n
has in (3). Then, s a ing o he i e a i e p ocess, o ins ance, wi h
u0,
i,h = 0, he e exis s a posi i e cons an C0such ha o each , h > 0and all n≥0
2
X
i=1 |u,n+1
i,h −u
i,h|1,Ωi≤P k k0,Ω
√(1 + 2 C0)n/2,
kπ,n+1
h−π
hk0,Ω≤Pk k0,Ω
(1 + 2 C0)n/2
o some cons an Pp opo ional o he cons an in Poinca e’s Inequali y.
Via he iangula inequali y we ob ain he inal bound
Theo em 7 Le u∈Hk+1(Ω) ∩H1
0(Ω)dand p∈Hk(Ω), o k≥1, be he solu ion o
he S okes p oblem in Ω. Fo each h > 0and  > 0le (U,n
h,P,n
h)∈Xh×Mh(n≥1)
sol e he i e a ion p oblem (Pn
,h)s a ing o he i e a ion wi h U,0
h= 0, and using ini e
elemen spaces o accu acy k≥1. Then he ollowing bounds hold o all n≥0
2
X
i=1 |u−un+1,
i,h |1,Ωi≤C(hk+√+1
√(1 + 2 C0)n/2) (8)
kp−πn+1,
hk0,Ω≤C(hk+√+1
(1 + 2 C0)n/2) (9)
206
whe e C=C(u, p)is a posi i e cons an jus depending on (u, p). When =O(h2k)and
nla ge enough we ob ain e o bounds O(hk) o eloci y and p essu e
2
X
i=1 |u−un,
i,h |1,Ωi+kp−πn,
hk0,Ω≤C hk
whe e C=C(u, p)is a posi i e cons an jus depending on (u, p).
5 Nume ical expe imen s
We use a known solu ion o he incomp essible S okes equa ions o compu e he e o
be ween he exac solu ion and he nume ical app oxima ion in he case k= 1. In his
es Ω = (0,1) ×(0,1) and he bounda y condi ion is u= 0 on he bounda y ∂Ω o Ω.
The exac solu ion is
u(x, y) = −cos(2πx) sin(2πy) + sin(2πy)
(x, y) = sin(2πx) cos(2πy)−sin(2πx)
p(x, y) = 2π(−cos(2πx) + cos(2πy))
and we ake iscosi y ν= 1. We conside he in e ace Γ as he line y= 0.5 and hen
Ω1= (0,1) ×(0,0.5) and Ω2= (0,1) ×(0.5,1). Nex , we conside a uni o m iangula
mesh o mesh size h=hx=hy, ake =h2and use P1 ini e elemen s wi h he B ezzi-
Pi ka anka s abiliza ion echnique o compu ing he solu ions ui,h and pi,h on each Ωi.
Then we cons uc he app oxima ed eloci y ield uhand p essu e πh∈L2
0(Ω) ia









uh=ui,h,in Ωi
uh= (u1,h +u2,h)/2,on ∂Γ,
πh=p1,h χΩ1+p2,h χΩ2−1
|Ω|(ZΩ1
p1,h +ZΩ2
p2,h),in Ω
whe e |Ω|= 1. Finally we compu e he e o s eu(h) = (P2
i=1 RΩi|∇(uh−un,
ih )|2dx)1/2and
ep(h) = kp−phk0,Ω.The ollowing able shows he alues ob ained o hese measu es.
wesh 16 ×16 (h= 1/16) 32 ×32 (h= 1/32) 64 ×64 (h= 1/64)
eu(h) 0.4600 0.13413 0.0412
ep(h) 0.5773 0.1942 0.066
Indeed, an o de o con e gence sligh ly la ge ha 1 is ob ained on his example.
Re e ences
[1] Chac´
on Rebollo, T., Chac´
on Ve a, E., Dom´
ınguez Delgado, A. Analysis
o a Non-o e lapping Domain Decomposi ion Me hod o Ellip ic Equa ions XVII
CEDYA/VII CMA, Salamanca (Spain) 2001.
207
[2] Chac´
on Rebollo, T., Chac´
on Ve a, E.A non-o e lapping Domain Decomposi-
ion Me hod o he S okes equa ions ia a penal y e m on he in e ace C.R. Acad.
Sci. Pa is, . 334, Se ie I, p. 2002.
[3] J.L. Lions, Pi onneau, O., O e lapping domain decomposi ion o e olu ion ope -
a o s. C.R. Acad. Sci. Pa is, . 330, Se ie I, p. 937-942, 2000.
[4] P.L. Lions On he Schwa z al e na ing me hod III: a a ian o non-o e lapping
subdomains Thi d In e na ional Symposium on Domain Decomposi ion Me hods o
Pa ial Di e en ial Equa ions, T.F. Chan e al. eds., SIAM, Philadelphia, pp. 202-231,
1990.
208