Asymptotic expansion of the solution of the steady Stokes equation with variable viscosity in a two-dimensional tube structure

0.67 MB·English

by  G. Cardone

Asymptotic expansion of the solution of the steady Stokes
equation with variable viscosity in a two-dimensional tube
structure
G.Cardone
University of Sannio, Department of Engineering
Piazza Roma, 21, 84100 Benevento, Italy
email: [email protected]
2
1 R.Fares, G.P.Panasenko
0 Laboratory of Mathematics of the University of Saint Etienne (LaMUSE), EA 3989
2
23, rue P.Michelon, 42023 St. Etienne, France

n
email: [email protected]; [email protected]
a
J
January 11, 2012
0
1

] Abstract
P
A TheStokesequationwiththevaryingviscosityisconsideredinathintubestructure,i.e. inacon-
. nectedunionofthinrectangleswithheightsoforderε<<1andwithbasesoforder1withsmoothened
h
boundary. An asymptotic expansion of the solution is constructed: it contains some Poiseuille type
t
a flows in the channels (rectangles) with some boundary layers correctors in the neighborhoods of the
m bifurcations ofthechannels. Theestimates forthedifferenceof theexact solution and itsasymptotic
[ approximation are proved.

1 Keywords: Stokesequation,thintubestructures,variableviscosity,asymptoticexpansion,bound-
v ary layercorrectors.
1 AMS Subject Classification (2000): 35B27, 35Q30, 76M45, 65N55.
2
1
2 1 Introduction
.
1
0 The blood circulation, the transport of cells and substances in the human body as well as some liquid-
2 cooling systems and oil-recovery/oil-transportprocesses in engineering, are modeled by the equations of
1 fluid motionposedin thin domains. The presentpaper studies the Stokes equationwith varyingviscosity
:
v in a tube structure. In two-dimensionalcase a tube structure1 (or pipe-wise structure) is some connected
i union of thin rectangles with heights of order ε and with bases of order 1 with a smoothened boundary
X
([9],[10]); here ε is a small positive parameter. An asymptotic expansion for the case of a constant
r
a viscosityhavebeenconstructedin[1]. Itis”compiled”ofsomePoiseuilleflowsinsidethe rectanglesglued
by some boundary layersolutions,in the neighborhoodof the junctions. Here we consider a more general
case when the viscosity is not constant but depends on a longitudinal variable for each rectangle. This
situation models a blood flow in a vessel structure where the viscosity depends on the concentration of
some substances diluted in blood or some blood cells. Indeed, the asymptotic analysis of the convection-
diffusion equation set in such domains ([3],[10]) shows that in the case of the Neumann (impermeability)
condition at the lateral boundary and small Reynolds numbers, the concentration is asymptotically close
totheone-dimensionaldescription,thatistheconvection-diffusionequationsetonthegraph. Thesolution
ofthe problemon the graphis the leading termof the asymptotic expansion,andit evidently ondepends
on the longitudinal variable. On the other hand, the viscosity often depends on the concentration of the
dilutedsubstancesordistributedcells,andso,itdependsonthelongitudinalvariable. Ofcourse,thefluid
motion equation is coupled with the diffusion-convection equation in this case. However, if the velocity
is small (in our case, it is of order ε2), then neglecting the convection, in comparison with the diffusion
1seeFig.5
1
term or iterating with respect to the small term 2, we get the steady state diffusion equation; in absence
of the source term in the right hand side, it has a piecewise-linear asymptotic solution on the graph for
the concentration. So, in this simplified situation, the diffusion equation can be solved before the fluid
motionequation,andweobtainfortheflow,theStokesorNavier-Stokesequationwithavariableviscosity
depending (via concentration) on the longitudinal variable. These arguments provide the motivation for
the Stokes equation with variable viscosity; to our knowledge, it has not been studied earlier from the
asymptoticpointofview.Thereare,ofcourse,manyotherpracticalproblemsinvolvingfluidswithvariable
viscosity. For example, the presence of bacteria in suspension (see [6]) may change locally the viscosity.
Inthefirstpart,weconsiderthecaseofaflowinonerectangularchannelwiththeperiodicitycondition
at the end of the channel. An asymptotic expansion of solution is constructed and justified (a regular
ansatz).
In the second part, we consider the case of a flow in one rectangular channel with the inflow/outflow
boundary conditions at the ends. In this case as well we construct an asymptotic expansion of solution,
that contains a regular ansatz and the boundary layer correctors.
In the third part, we construct an asymptotic expansion for solution of the Stokes equation set in a
tube structure. Here as well we associate to each rectangle a regular ansatz and then we glue all the
regular ansatzes with help of the boundary layer correctors exponential decaying from the junctions of
rectangles. This procedure is similar to the procedure described in [10].
In the fourth part we consider some numerical experiments comparing the numerical and asymptotic
solutions.
All constructed expansions are justified by calculation of residual terms and application of a priori
estimates.
2 Flow in one channel
Consider a small parameter ε, ε= 1, q ∈N, and define then a thin domain
q
ε ε
D = (x ,x )∈R2 :0<x <1, − <x < .
ε 1 2 1 2
2 2
n o
Assume that incompressible, viscous fluid fills the domain D . Let f be the exterior force applied to the
ε
fluid.
Figure 1: Thin domain
2Thesimplestcoupleddiffusion-convectionequationis
−div(ν(Cε)Duε)+∇pε=f(x)
divuε=0,
−∆Cε+uε.∇Cε=0
whereCε istheconcentration, uε isthefluidvelocityandpε isthepressure.
Ifuε issmallwithrespecttoCε,wecanwriteanexpansionforCε withrespecttothesmallparameter thatistheratioof
magnitudesof|uε|and|Cε|. Thenforthetermsofthisexpansion,thethirdequationmaybesolvedbeforethefluidmotion
equations. Anotherpossibleapproachisthesuccessiveapproximations(fixedpointiterations):
−div(ν(Cε(n))Du(εn))+∇p(εn)=f(x)
divu(εn)=0,
−∆Cε(n)+u(εn−1).∇Cε(n)=0
wherenisthenumberoftheiteration.
Inbothapproaches, wegetaproblemwiththevariableviscosity.
2
Consider the following steady state Stokes problem:
−div(ν(x )Du )+∇p =f(x) in D ,
1 ε ε ε
divu =0 in D ,
ε ε

u (x , ε)=0 for x ∈(0,1),
(1)  uuεε((x0,11x,−2)2ε=)=ε2ϕ0 (x2) ffoorr xx11 ∈∈((−0,ε1,),ε),
ε 2 0 ε 2 2 2
u (1,x )=ε2ϕ (x2) for x ∈(−ε,ε),
The unknowns of this system aεre th2e velocit1y uεε and the pressure2pε of th2e2fluid.
The non homogeneous boundary conditions for the velocity are given by the functions ϕ ,ϕ whose
0 1
second component is equal to zero, and the first satisfies
1 1
2 2
(2) ϕ (ξ )dξ = ϕ (ξ )dξ ,
01 2 2 11 2 2
Z−12 Z−12
where ϕi1 ∈C02([−12,21]). Here (Duε)ij = 21 ∂∂uxεji + ∂∂uxεij , ν satisfies the following conditions
(cid:16) (cid:17)
(3) ν(x )=ν +ν (x )
1 0 1 1
where ν ∈ C2([0,1]). Moreover there exist ρ > 0, κ > 0 such that ν(x ) ≥ κ for all x ∈ (0,1) and
1 0 1 1
ν (x )=0 for all x ∈(0,ρ)∪(1−ρ,1) (and so, ν >0).
1 1 1 0
2.1 Variational formulation of the problem
In order to obtain the variational formulation of problem (1), let us introduce the following space
(4) H(D )= u∈(H1(D ))2 :div(u)=0
ε 0 ε
and assume that f ∈(L2(D ))2. (cid:8) (cid:9)
ε
Applyingtheextensiontheorem(see[5]),onecanfindafunctionϕε ∈(H1(D ))2 suchthatdivϕε =0,
ε
ϕε = ϕ and ϕε = ϕ and ϕε = 0 . Changing the unknown function u by v = u −ε2ϕε,
|x1=0 0 |x1=1 1 |x2=±ε2 ε ε ε
we give the variational formulation of problem (1):
(5) ν(x )Dv :Dψ = f ·ψ−ε2 ν(x )Dϕ :Dψ, ∀ψ ∈H(D ).
1 ε 1 ε ε
ZDε ZDε ZDε
Definition 2.1. We say that u is a weak solution of problem (1) if v =u −ε2ϕε ∈H(D ) and satisfies
ε ε ε ε
(5).
Proposition 2.1. If u is a weak solution for problem (1), then there exists a distribution p ∈ D′(D )
ε ε ε
such that (u ,p ) satisfies (1) this problem in sense of distributions.
ε ε 1
Proof. If we take v =u −ε2ϕε, then from (5) we have that
ε ε
hdiv(ν(x )Dv +ε2ν(x )Dϕ )+f,ψi=0,∀ψ ∈H(D )
1 ε 1 ε ε
From De Rham lemma, it follows that there exists a distribution p , unique up to an additive constant,
ε
such that
−div(ν(x )Dv )+ε2div(ν(x )Dϕ )−f =−∇p.
1 ε 1 ε
Theorem 2.1. The variational problem (5) admits a unique solution v ∈H(D ).
ε ε
Proof. The Riesz’s theorem gives the existence and the uniqueness of solution because the norms
kvεkI = ν(x1)Dvε :Dvε and kvεk(H01(Dε))2 are equivalent.
sZDε
As a consequence, we have:
Proposition 2.2. Let v ∈ H(D ) be the solution of the variational problem (5). Then the following
ε ε
inequality holds
kvεk(H(Dε))2 ≤C(κ,CPεF) kfk(L2(Dε))2 +kϕεk(H1(Dε))2 ,
where Cε stands for the Poincar´e-Friedrichs ineq(cid:16)uality constant and κ is the lo(cid:17)wer bound of the viscosity
PF
(3).
3
Mention that Cε can be estimated by εC˜, where C˜is independent of ε (see [10]).
PF
Consider the case when ϕε =0. Then we have:
Proposition 2.3. The following inequality holds
k∇pεkH−1(Dε) ≤Ckfk(L2(Dε))2,
where C is a constant independent of ε.
Proof. From De Rham lemma, it follows that there exists a distribution p , unique up to an additive
ε
constant, such that
∇p =div(ν(x )Du )+f.
ε 1 ε
It means that
p divψ = ν(x )Du :Dψ− f ·ψ, ∀ψ ∈H(D ))2,
ε 1 ε ε
ZDε ZDε ZDε
and so,
| p divψ| | ν(x )Du :Dψ− f ·ψ|
sup Dε ε = sup Dε 1 ε Dε ≤
ψ∈(H01(Dε))2 k∇Rψk(L2(Dε))4 ψ∈(H01(Dε))2 R k∇ψk(L2(Dε))4R
≤(C(κ,CPεF)+CPεF)kfk(L2(Dε))2,
where C(κ,Cε ) is the constant of the a priori estimate for u (see above) and Cε is the Poincar´e-
PF ε PF
Friedrichs constant (it is of order ε). This inequality gives the estimate of ∇p in the H−1−norm.
ε
An asymptotic analysis of problem (1) shows that an asymptotic solution is given by a Poiseuille
type flow,with twoboundarylayercorrectorslocalizedinsome neighborhoodsofthe ends ofthe channel.
Didactically,itwouldbebettertoseparatetheconstructionofthePoiseuilletypeflowforvaryingviscosity
and the construction of the boundary layer correctors. That’s why we simplify problem (1), and replace
(1) , (1) by the periodicity condition with respect to x .
5 6 1
Introduce the Sobolev space
H (D )= u∈ H1 (D ) 2 :divu=0 .
per ε per,1,0 ε
n (cid:0) (cid:1) o
HereH1 (D )isthecompletion(withrespecttotheH1(D )-norm)ofthespaceoftheC∞(R×[−ε, ε])-
per,1,0 ε ε 2 2
functions vanishing at the boundary x = ±ε and 1-periodic in x . As in the beginning of the section,
2 2 1
f ∈(L2(D ))2.
ε
Definition 2.2. We say that u ∈H (D ) is a weak solution of the periodic problem
ε per ε
−div(ν(x )Du )+∇p =f(x), in D ,
1 ε ε ε
divu =0, in D ,
ε ε

(6) u (x ,ε)=0, for x ∈(0,1),
 uεε(x11,−2 2ε)=0, for x11 ∈(0,1),
u is 1−periodic in x ,
ε 1
if and only if it satisfies the integral identity
(7) ν(x )Du :Dψ = f ·ψ, ∀ψ ∈H (D ).
1 ε per ε
ZDε ZDε
As in theorem 2.1, we apply the Riesz theorem and prove the existence and the uniqueness of u ∈
ε
H (D ), solution of problem (6). The a priori estimate is given by
per ε
kuεk(H1(Dε))2 ≤C(κ,CPεF)kfk(L2(Dε))2
where C(κ,Cε )=O(ε) and κ is the lower bound of the viscosity (3).
PF
Proposition 2.4. If u is a weak solution for problem (6) then there exists a distribution p ∈ D′(D )
ε ε ε
such that (u ,p ) satisfies (6) in sense of distributions, and
ε ε 1
k∇pεkH−1(Dε) ≤Ckfk(L2(Dε))2,
where C is a constant independent of ε.
4
Figure 2: Infinite layer
Proof. By (7)
ν(x )Du :Dψ = f ·ψ, ∀ψ ∈H (D ).
1 ε per ε
ZDε ZDε
From De Rham lemma, it follows that there exists a distribution p , unique up to an additive constant,
ε
such that
∇p =div(ν(x )Du )+f.
ε 1 ε
It means that
p divψ = ν(x )Du :Dψ− f ·ψ ∀ψ ∈(H1 (D ))2,
ε 1 ε per,1,0 ε
ZDε ZDε ZDε
and so,
| p divψ| | ν(x )Du :Dψ− f.ψ|
sup Dε ε = sup Dε 1 ε Dε ≤
ψ∈(Hp1er,1,0(Dε))2 k∇Rψk(L2(Dε))4 ψ∈(Hp1er,1,0(Dε))2 R k∇ψk(L2(Dε))4R
≤(C(κ,CPεF)+CPεF)kfk(L2(Dε))2,
where C(κ,Cε ) is the constant of the a priori estimate for u (see above) and Cε is the Poincar´e-
PF ε PF
Friedrichs constant (it is of order ε). This inequality gives the estimate of ∇p in the H−1−norm.
ε
2.2 Asymptotic analysis of the problem
Let us first construct the asymptotic expansion for the solution of the periodic problem (6); then we will
study the non periodic one.
Define the infinite layer
ε ε
Ω = (x ,x )∈R2 :x ∈R,− <x < .
ε 1 2 1 2
2 2
n o
Assume that f =f e , f ∈C∞(R), f is 1-periodic in x .
1 1 1 1 1
Denote hψi = 1ψ(x ,x )dx , hψi = 12 ψ(x ,x )dx . An asymptotic solution is written as:
1 0 1 2 1 2 −21 1 2 2
R R
k
x
uk(x ,x ) = εj+2u x , 2 ,
 1 1 2 1,j 1 ε
(8)  uk2(x1,x2) = XXjj==k00εj+3u2,j(cid:16)(cid:16)x1,xε2(cid:17)(cid:17),
k k
x
with u ,u ,p and q 1-peprkio(xd1ic,xfu2)ncti=onsXji=n0xεj+,1spujc(cid:16)hxt1h,atε2h(cid:17)p+i Xj==00ε.jqj(x1),
1,j 2,j j j 1 j 2
Substituting (8), in (6), equating the coefficients of the same powers of ε and denoting ξ = x2 we
2 ε
5
obtain:
− ∂ ν(x )∂u1,j−2 − ν(x1)∂2u1,j − ν(x1)∂2u2,j−2 + ∂pj−1 + ∂qj =f δ ,
∂x1 1 ∂x1 2 ∂ξ22 2 ∂ξ2∂x1 ∂x1 ∂x1 1 j0
 (cid:16) (cid:17)
(9)  ∂21∂u∂x1∂x1,j1 +hν(∂x∂uξ122,)j(cid:16)=∂u0∂1,ξ,j2−1 + ∂u∂2x,j1−3(cid:17)i+ν(x1)∂2u∂2ξ,22j−1 − ∂∂pξ2j =0,
Denote:  uu21,,jj(cid:0)(cid:0)xx11,,±±2211(cid:1)(cid:1)==00,.
1 1
N (ξ )= ξ2− ,
1 2 2 2 4
(cid:18) (cid:19)
(it satisfies N′′ =1, N (±1)=0) and
1 1 2
ξ2
N (ξ )= N (τ)dτ,
2 2 1
Z−12
1
2
(here N (1)= N (τ)dτ =− 1 ). Denote:
2 2 1 12
Z−12
ξ2
D−1 :F −→ F(x ,τ)dτ,
1
 Z−12
 D˜−1 :F −→Z−ξξ1222F(θx1,τ)dτ −Z−2112 Z−θ12 F(x11,τ)dτd21θ, θ
D−2 :F −→ F(x ,τ)dτdθ− ξ + F(x ,τ)dτdθ.
Theorem 2.2.The unknownsZo−f12(Z9−)21u1,j,u12,j,pj,qj ar(cid:18)e g2iven2b(cid:19)yZt−he12 Zfo−ll21owing1relations:
u = −D−2 ∂2u2,j−2 + 2 ∂ ν(x )∂u1,j−2 − ∂pj−1
1,j ∂ξ2∂x1 ν(x1) ∂x1 1 ∂x1 ∂x1
 + 2 nN (ξ ) ∂qj −f δ(cid:16) ,(cid:16) (cid:17) (cid:17)o
(10)  u2,j = D−−ν∂(1xD1)−212n∂2∂3ξu2(cid:16)2∂∂,qj∂xj−x2121−+f∂δ∂1x1j(cid:16)0(cid:17)ν(Nx21)((cid:16)ξ∂)∂x,1 (cid:16)ν(x1)∂u∂1x,j1−2(cid:17)− ∂∂pjx−11(cid:17)(cid:17)o
∂x1 ν(x1) ∂x1 1 j0 2 2
Proof. Integrpajting=twicD˜e−(19)n(cid:16)12a∂n∂xd1 hu(cid:16)νs(inxg1)b(cid:16)o∂uu∂n1ξ,dj2−a(cid:17)1r(cid:17)y+co∂nu∂2dx,ji1−ti3o(cid:17)nis+(9ν)(,x1w)e∂2gu∂e2ξt,22j−(110o).. This relation gives
1 4 1
an expression of the unknown u via q . All other functions contained by this relation are either known
1,j j
fromprevious computations or equalto zero. We integrate next the incompressibility condition (9) with
3
respect to ξ with the boundary condition(9) and get u . Finally, integrating (9) we get p . The
2 5 2,j 2 j
unknownfunction q is determined from the boundary conditionu x ,1 =0. Actually, the boundary
j 2,j 1 2
conditions(9) andu x ,−1 =0aresatisfiedbythedefinitionofD−2 andD−1,whileu x ,1 =0
4 2,j 1 2 (cid:0) (cid:1) 2,j 1 2
gives:
(cid:0) (cid:1) (cid:0) (cid:1)
∂ 1 ∂q
j
(11) − −f δ =
1 j0
∂x 6ν(x ) ∂x
1 (cid:18) 1 (cid:18) 1 (cid:19)(cid:19)
∂3u ∂ 2 ∂ ∂u ∂p
=D−1D−2 2,j−2 + ν(x ) 1,j−2 − j−1
∂ξ ∂x2 ∂x ν(x ) ∂x 1 ∂x ∂x
(cid:26) 2 1 1 (cid:26) 1 (cid:18) 1 (cid:18) 1 (cid:19) 1 (cid:19)(cid:27)(cid:27)(cid:12)(cid:12)ξ2=21.
(cid:12)
(cid:12)
In particular, for j =0, we have the Darcy equation for the leading term of the pressure q :
0
∂ 1 ∂q
0
(12) −f =0,
1
∂x 6ν(x ) ∂x
1 (cid:18) 1 (cid:18) 1 (cid:19)(cid:19)
6
and so q is the 1-periodic function given by the formula
0
x1 hf i x1 hf i
1 1 1 1
q (x )= f (τ)− ν(τ) dτ − f (τ)− ν(τ) dτ .
0 1 1 1
hνi hνi
Z0 (cid:18) 1 (cid:19) (cid:28)Z0 (cid:18) 1 (cid:19) (cid:29)1
Then u (x ,ξ )= 2 ∂q0 −f N (ξ ), u (x ,ξ )=0 and p (x ,ξ )=0.
1,0 1 2 ν(x1) ∂x1 1 1 2 2,0 1 2 0 1 2
For j =1 q1 is the 1-perio(cid:16)dic soluti(cid:17)on of equation
∂ 1 ∂q
1
=0.
∂x 6ν(x )∂x
1 (cid:18) 1 1(cid:19)
It follows: q =0, u (x ,ξ )=0 and u (x ,ξ )=0. For j =1, (10) yields:
1 1,1 1 2 2,1 1 2 3
∂ ∂q
0
p (x ,ξ )= −f (N (ξ )−hN i ).
1 1 2 1 1 2 1 2
∂x ∂x
1 (cid:18) 1 (cid:19)
For j =2, (11) becomes
∂ 1 ∂q ∂ 2 ∂ ∂q
(13) 2 = 0 −f D−1D−2(N (ξ )−hN i ) ,
1 1 2 1 2
∂x1 (cid:18)6ν(x1)∂x1(cid:19) ∂x1 (cid:18)ν(x1)∂x1 (cid:18)∂x1 (cid:19)(cid:19) (cid:12)(cid:12)(cid:12)ξ2=21
(cid:12)
where D−1D−2(N (ξ )−hN i ) =R6=0, q is 1-periodic solution of given by:
1 2 1 2 2
(cid:12)(cid:12)ξ2=12
(cid:12)
(cid:12)
ν(x )
1
q (x )=−12Rhf i −1 .
2 1 1 1
hνi
(cid:18) 1 (cid:19)
It follows:
u (x ,ξ )= 2 ∂2 (∂q0 −f )D−2 N (x2)−hN i + 2 N (x2)∂q2,
1,2 1 2 ν(x1)∂x21 ∂x1 1 1 ε 1 2 ν(x1) 1 ε ∂x1
 (cid:0) (cid:1)
 u2,2(x1,ξ2)=−∂∂x1 ν(x21)∂∂x221 ∂∂xq01 −f1 D−1D−2 N1(xε2)−hN1i2 − ∂∂x1 ν(x21)N2(xε2)∂∂xq21 ,
andwe get the following thheorem. (cid:16) (cid:17)i (cid:0) (cid:1) (cid:16) (cid:17)
Theorem 2.3. The asymptotic solution of problem (6) is given by:
uk(x ,x ) = ε2 2 ∂q0 −f N (x2)
1 1 2 ν(x1) ∂x1 1 1 ε
 +ε4 (cid:16)2 ∂2 (∂q0(cid:17)−f )D−2 N (x2)−hN i + 2 N (x2)∂q2 +O(ε5),
(14)  uk2(x1,x2) = −+ε∂5∂x(cid:16)(cid:16)1ν∂∂(xxν1(1x2h)1∂ν)x(Nx22112)(∂∂xx∂εx21221)(cid:16)∂∂xq∂∂21xq101 −+f1O(cid:0)(cid:17)(iε1D6)−,ε1D−2(cid:0)N112((cid:1)xε2)ν−(xh1N) 1i12(cid:1)ε ∂x1(cid:17)
where q0 apnkd(xq12,xa2r)e pe=riodqi0c+soεlu2(cid:16)t(cid:16)ioq2n(sxo1)f+obt∂a∂xi1n(cid:16)ed(cid:17)∂∂x(cid:17)qa01s−thfe1u(cid:17)n(cid:0)iNqu1e(xsε2o)lu−tiohNn1oif2(cid:1)((cid:17)11+)O(1(3ε)3)r,espectively.
Remark 2.1. Solution given by (14) can be written as follows:
uk(x ,x ) = −ε2 2hf1i1 N (ξ )−ε4 2 hf1i1ν′′(x )D−2 N (x2)−hN i
1 1 2 hν(x1)i1 1 2 ν(x1) hνi1 1 1 ε 1 2
 +24Rhf1i1ν′(x1)N (x2)(cid:16)+O(ε5), (cid:0) (cid:1)
 uk2(x1,x2) = ε+52(cid:16)hh4νν∂R((∂xxxh111f))1hiii111ν((νx2νν(1′x()(x1xhh)1f1ν)1)ii)11′1Nν′2′ε((xxε(cid:19)12))iD+−O1D(ε−62),(cid:0)N1(xε2)−hN1i2(cid:1)
(cid:19)
 pk(x1,x2) = Z+0εx21(cid:16)(cid:18)−f11(2τR)h−f1hihf1ν1(cid:16)ii11νh(ννxi(11τ))−(cid:19)1d(cid:17)τ−−(cid:28)hhfν1Zii101xν1′(cid:18)(xf11)((cid:0)τN)−1(xhεh2fν)1ii−11νhN(τ1)i(cid:19)2(cid:1)d(cid:17)τ(cid:29)+1O(ε3).
7
Introduce the following function:
Fk(x ,x ) = ∂ ν(x )∂u1,k−1 + ν(x1)∂2u2,k−1 − ∂pk e
1 2 ∂x1 1 ∂x1 2 ∂x1∂ξ2 ∂x1 1
(cid:16) (cid:16) (cid:17) (cid:17)
+ 1 ∂ ν(x ) ∂u1,k + ∂u2,k−2 +ν(x )∂2u2,k e
2∂x1 1 ∂ξ2 ∂x1 1 ∂ξ22 2
(15) (cid:16) h (cid:16) (cid:17)i (cid:17)
+ ε ∂ ν(x )∂u1,k + ν(x1) ∂2u2,k e + 1 ∂ ν(x )∂u2,k−1 e
∂x1 1 ∂x1 2 ∂x1∂ξ2 1 2∂x1 1 ∂x1 2
(cid:16) (cid:16) (cid:17) (cid:17) (cid:16) (cid:16) (cid:17)(cid:17)
+ ε2 1 ∂ ν(x )∂u2,k e .
2∂x1 1 ∂x1 2
(cid:16) (cid:16) (cid:17)(cid:17)
Theorem 2.4. Let (uk,pk) be the asymptotic solution given by (8) and (u,p) the solution of (6). Then
the following estimate holds:
kuk−ukH1(Dε)2 =O εk+52 .
(cid:16) (cid:17)
Proof. Denote (Uk,Pk)=(uk−u,pk−p). We obtain the following problems for (Uk,Pk):
−div(ν(x )DUk)+∇Pk =−εk+1Fk(x ,x ) in D ,
1 1 2 ε
divUk =0 in D ,
ε

(16) Uk(x ,ε)=0 for x ∈(0,1),
 Uk(x11,−2 2ε)=0 for x11 ∈(0,1),
Uk is 1−periodic in x .
1
The a priori estimates give us:
kuk−ukH1(Dε)2 ≤C(K,CPF)kεk+1FkkL2(Dε)2 =O εk+52
(cid:16) (cid:17)
and
k∇pk−∇pkH−1(Dε) =O εk+52 .
(cid:16) (cid:17)
Letusconsidernowthenonperiodiccase,i.e. letusconstructanasymptoticexpansionofthesolution
for problem (1). Assume
f =f (x )e , f ∈C∞([0,1]).
1 1 1 1
Define an asymptotic solution by:
uˆk(x ,x ) = uk(x ,x )+uk (x)+uk (x1−1,x2)
(17) 1 2 1 2 BL0 ε BL1 ε ε
pˆk(x ,x ) = pk(x ,x )+pk (x)+pk (x1−1,x2)
(cid:26) 1 2 1 2 BL0 ε BL1 ε ε
The expressions of uk, pk are the same as above: (8) ,(10) ,(11)
Since the functions given by (8) do not satisfy exactly the boundary conditions (1) ,(1) we add the
5 6
boundary layer correctors. They correspond to the left end for i = 0 and to the right end for i = 1 and
their expressions are given by:
k
x −i x
uk (x1−i,x2) = εj+2ui 1 , 2
 BLi ε ε j ε ε
(18)  pk (x1−i,x2) = Xj=k0εj+1pi (cid:18)x1−i,x2(cid:19)
BLi ε ε j ε ε
j=0 (cid:18) (cid:19)
Here i=0,1. To obtain problemsfor the boundary lXayerscorrespondingto the left side, we introduce the
domain Π+ =(0,∞)×(−1,1). The problem
2 2
−ν0∆ u0(ξ)+∇ p0(ξ)=0 if ξ ∈Π+,
2 ξ j ξ j
div u0 =0 if ξ ∈Π+,
 ξ j
(19) u0 =0 if ξ =±1,
 uj0(0,ξ )=ϕ δ − u1,j(0,ξ2) if ξ2 ∈(−21,1),
j 2 01 j0 u (0,ξ ) 2 2 2
2,j−1 2
with the compatibilitycondition (cid:18) (cid:19)
(20) hϕ δ −u (0,ξ )i =0,
01 j0 1,j 2 2
8
1
wherehψi = 2 ψ(x,ξ )dξ ,willgivethe boundarylayercorrectorsforthe velocityandforthe pressure
2 −21 2 2
corresponding to the left end (see [4]). This condition (20) generates a boundary condition for q :
R j
1 ∂q (0) ∂2u ∂2u 1 ∂p
hϕ i δ + j −δ f (0) + D−2 2,j−2 +2D−2 1,j−2 − j−1 =0.
01 2 j0 6ν ∂x j0 1 ∂x ∂ξ ∂x2 ν ∂x
0 (cid:18) 1 (cid:19) *(cid:26) 1 2 (cid:18) 1 0 1 (cid:19)(cid:27)|x1=0+2
Inasimilarwayweintroducetheboundarylayercorrectorscorrespondingtotherightside. Theboundary
layers for the velocity and pressure are defined on Π− = (−∞,0)×(−1,1). The analogous boundary
2 2
conditionforq issatisfiedautomaticallybecauseoftheconservationoftheaverageofu (x ,ξ ). Indeed,
j 1,j 1 2
hu (x ,ξ )i =D−1u (x ,ξ ) ;
1,j 1 2 2 1,j 1 2 |ξ2=12
on the other hand, the condition
∂
u =− D−1u (x ,ξ ) =0
2,j|ξ2=12 ∂x1 1,j 1 2 |ξ2=21
is equivalent to the equation on q , i.e. this equation on q is equivalent to the conservation law for the
j j
averagehu (x ,ξ )i . The problem satisfied by the asymptotic solution of order k is as follows:
1,j 1 2 2
−div(ν(x )Duˆk)+∇pˆk =f −εk+1Fk(x ,x ) in D ,
1 1 2 ε
divuˆk =0 in D ,
 ε
uˆ (x , ε)=0 for x ∈(0,1),
(21)  uuˆˆεεε((0x,11x,−22)2ε=)=ε2ϕ00(xε2)+εk+3u2,k(0,ξ2)e2+ k εj+2u(j1) −1ε,xε2 ffoorr xx121 ∈∈((0−,2ε1,),2ε),
j=0 (cid:18) (cid:19)
X
k
Here Fk isgivuˆeεn(1b,yx2()15=).ε2ϕ1(xε2)+εk+3u2,k(1,ξ2)e2+Xj=0εj+2u(j0)(cid:18)1ε,xε2(cid:19) for x2 ∈(−2ε,2ε),
Mention that the boundary conditions for uk on x = 0,1 are not yet satisfied exactly because the
1
traces of each boundary layer on the opposite side of the rectangle are exponentially small but do not
vanish completely. These traces should be eliminated by a small additional corrector.
Let us construct a new function Uˆk which satisfies the same boundary conditions as uk on x =0,1.
1
Let us describe its construction: let Uk :D −→R2 be a solution of the following problem:
ε
Uk ∈ H1(D ) 2,
ε
divUk =0 in D ,
 (cid:0) (cid:1) ε
(22)  UUUkkk(((xx0,11x,,−22ε))2ε==)=ε0k+03u2,k(0,ξ2)e2+ k εj+2u(j1) −1ε,xε2 fffooorrr xxx112 ∈∈∈(((00−,,2ε11,)),,2ε),
j=0 (cid:18) (cid:19)
X
k
Proposition2.5U.kP(1ro,xbl2e)m=(ε2k2+)3hua2s,ka(1t,lξe2a)set2o+neXj=so0εluj+ti2oun(j0w)i(cid:18)th1εt,hxeε2p(cid:19)roperftoyr x2 ∈(−2ε,2ε).
(23) kUkk(H1(Dε))2 =O εk+32 .
(cid:16) (cid:17)
Proof. Define wk :D −→R2, where D =(0,1)×(−1,1),
ε 2 2
ε Uk x
wk(y ,y )= 1 , with (y ,y )=(x , 2).
ε 1 2 Uk 1 2 1 ε
(cid:18) (cid:0) (cid:1)2 (cid:19)
(cid:0) (cid:1)
9
Obvious computations lead to the following problem for wk:
ε
div wk =0 in D,
y ε
wk(y ,1)=0 for y ∈(0,1),
 ε 1 2 1
wk(y ,−1)=0 for y ∈(0,1),
(24)  wεεk(01,y2)2=εk+3u2,k(0,y2)e2+ XXjj==kkk00εεjj++32uu(1(211,,jj))(cid:18)(cid:18)−−111εε,,yy22(cid:19)(cid:19)  for y12 ∈(−21,12),
εj+3u(0) ,y
Here div is wthεke(1d,iyv2e)rg=enεcke+3inu2y,k−(1v,ayr2ia)eb2le+s.AsXXjj==ki00nεj[+5]2uw(210e,,jj)(cid:18)(cid:18)ca1εεn,yp22r(cid:19)(cid:19)ovethat thfeorreye2x∈ist(s−a12,fu12)n,ction wk ∈
y ε
H1(Dε) 2 so that kwεkk(H1(D))2 ≤ Ckwεkk(cid:16)H12(∂D)(cid:17)2 with the constant C independent of ε. Using the
(cid:0)propertie(cid:1)s of the boundary layer correctors (their exponential decay rate), we get:
kwεkk(H1(D))2 =O εk+3
Direct computations give kUkk(H1(Dε))2 ≤ ε132kwεkk(H1(D))(cid:0)2. Co(cid:1)mbining these two estimates we achieve
the proof.
The function
(25) Uˆk =uˆk−Uk
satisfies the same boundary conditions as u in x = 0,1. The problem for the new functions Uˆk, pˆk is an
obvious consequence of (22) and (21):
−div(ν(x )DUˆk)+∇pˆk =f −εk+1Fk(x ,x )−div(ν(x )DUk) in D ,
1 1 2 1 ε
divUˆk =0 in D ,
 ε
(26)  UUUˆˆˆεε(((x0x,11x,,−2ε))2ε==)=ε020ϕ (x2) fffooorrr xxx11 ∈∈∈(((0−0,,ε11,)),,ε),
ε 2 0 ε 2 2 2
Theorem2U.ˆ5ε.(1L,ext2)(uˆ=k,εpˆ2kϕ)1(bexε2t)he asymptotic solution given by (17) and (u,p)fotrhxe2ex∈a(c−t s2εo,lu2εt)i.on of (1).
Then the following estimates hold:
(27) kuˆk−uεkH(Dε)2 =O εk+23 ,
 k∇pk−∇pεkH−1(Dε)(cid:16)=O ε(cid:17)k+32 .
(cid:16) (cid:17)
Proof. From (25) it follows: 
kuˆk−uεkH(Dε)2 ≤kUˆk−uεkH(Dε)2 +kUkkH(Dε)2 =O εk+25 +O εk+32 .
(cid:16) (cid:17) (cid:16) (cid:17)
From(15),(23)and(26)itfollows: kuˆk−ukH(Dε)2 =O εk+32 . Theestimateforpressureisaconsequence
of (27) and the a priori estimate. (cid:16) (cid:17)
1
3 Flow in tube structures
In this section we are going to construct an asymptotic expansion to the solution of problem (1), stated
in a tube structure containing one bundle. We justify the error estimate. Let us define a tube structure
containing one bundle.
10