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CONDITIONS OF SLIP ON THE BOUNDARY

R. H. W. HOPPE, M. Y. KUZMIN, W. G. LITVINOV, AND V. G. ZVYAGIN Received 26 June 2005; Accepted 1 July 2005

We study a mathematical model describing flows of electrorheological fluids. A theorem of existence of a weak solution is proved. For this purpose the approximating-topological method is used.

Copyright © 2006 R. H. W. Hoppe et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

Electrorheological fluids are smart materials that are concentrated suspensions of polar- izable particles in a nonconducting dielectric liquid. In moderately large electric fields the particles form chains along the field lines and these chains then aggregate into the form of columns. These chainlike and columnar structures yield dramatic changes in the rheolog- ical properties of the suspensions. The fluid becomes anisotropic, the apparent viscosity (the resistance to flow) in the direction, orthogonal to that of the electric field, abruptly increases, while the apparent viscosity in the direction of the electric field changes not so drastically.

LetΩRnbe a bounded domain, in which a fluid flows,n∈ {2, 3}. Let the boundary SofΩbe Lipschitz continuous. As it is well known, the stationary movement of any fluid is described by the equation in Cauchy form:

n j=1

uj∂ui

∂xj n j=1

∂σi j

∂xj =Fi, (1.1)

wherexΩ,u=(u1,. . .,un) is the velocity field of the fluid,{σi j}ni,j=1is the stress tensor, F=(F1,. . .,Fn) is the volume force. We also add the condition of incompressibility to (1.1):

divu= n i=1

∂ui

∂xi =0. (1.2)

Hindawi Publishing Corporation Abstract and Applied Analysis

Volume 2006, Article ID 43560, Pages1–14 DOI10.1155/AAA/2006/43560

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We will consider the following constitutive equation (see [2]):

σi j(p,u)= −i j+ 2ϕI(u),|E|,μ(u,E)εi j(u), (1.3) whereδi jare the components of the unit tensor,εi j(u) are the components of the rate of strain tensor,εi j(u)=(1/2)(∂ui/∂xj+∂uj/∂xi),pis the spherical part of the stress tensor, ϕis the viscosity function,I(u)=n

i,j=1i j(u))2,

μ(u,E)(x)=

αθ+u(x)

αn+u(x), E(x) E(x)

2 Rn

, (1.4)

αis a small positive constant, andRnθ=(1,. . ., 1),E=(E1,. . .,En) is the electric field strength.

We consider the condition of slip onS(see [4,6]). Let f =(f1,. . .,fn) be an external surface force acting on the fluid,

fi= n j=1

i j+ 2ϕI(u),|E|,μ(u,E)εi j(u)ηj|S, (1.5)

whereη=1,. . .,ηn) is the unit outward normal toS. We represent f in the form f(s)=fη(s) +fτ(s) sS, (1.6) where fηand fτare the normal and the tangent vectors:

fη(s)= fη(s)η(s), fη(s)= n i=1

fi(s)ηi(s), fτ(s)=f(s)fη(s)=

n i=1

fτi(s)ei, fτi(s)= fi(s)fη(s)ηi(s),

(1.7)

{e1,. . .,en}is an orthonormal basis inRn.

For the fieldu, a similar decomposition is valid.

The slip conditions on the boundary are the following [4]:

uη(s)=0 sS, (1.8)

fτ(s)= −χfη(s),uτ(s)2uτ(s) sS (1.9)

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(by| · |we denote the norm in Euclidian spaceRn). Instead of (1.9) we will consider the regularized condition

fτ(s)= −χf(s),uτ(s)2uτ(s) sS, (1.10) f(p,u)=

P p+ n i,j=1

I(Pu),|E|,μ(Pu,E)εi j(Pu)ηiηj

S

, Pv(x)=

Rnω|xx´|

v( ´x)dx,´ xΩ,

(1.11)

whereωC(R+), suppω[0,a],aR+,ω(z)0 atzR+,Rnω(|x|)dx=1.

HereR+is the set of nonnegative numbers.

Equation (1.10) means that the model of slip is not local, this is natural from the physical view point (see [1]).

We assume that

Ωp(x)dx=0. (1.12)

Let us describe the concept of a weak solution of (1.1)–(1.3), (1.8), (1.10), (1.12). We introduce some Hilbert spaces (see [4]):

Z=

v:vH1(Ω)n,vη|S=0,

Ω[divv](x)dx=0

, W= {v:vZ, divv=0}.

(1.13)

The expression

(u,v)Z= n i,j=1

ωεi j[u](x)εi j[v](x)dx+ n i=1

Suτi(s)vτi(s)ds (1.14) defines a scalar product onZ(and inW).

Multiplying (1.1) by a function hin L2(Ω)n, and using Green’s formula and (1.3), (1.8), (1.10), we see that

n i,j=1

ωI(u)(x),E(x),μ(u,E)[x]εi j[u](x)εi j[h](x)dx +

n i=1

Sχ frn

p(s),u(s),uτ(s)2uτi(s)hτi(s)ds

+ n i,j=1

ωuj(x)∂ui

∂xj(x)hi(x)dx

ωp(x)[divh](x)dx= n i=1

ωFi(x)hi(x)dx (1.15) (here we suppose thatFL2(Ω)n).

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Definition 1.1. A couple of functions (u,p)W×L2(Ω) is a weak solution of problem (1.1)–(1.3), (1.8), (1.10), (1.12) if it satisfies equality (1.15) for allhZ.

The following conditions are imposed on the functionsϕandχ.

(C1) There are positive constantsa1anda2such that a1ϕy1,y2,y3

a2

y1,y2,y3

R2+×[0, 1]. (1.16) (C2) The functionϕ(y1,·,y3) :R+Ris measurable iny2for all specified (y1,y3)

R+×[0, 1].

(C3) The functionϕ(·,y2,·) :R+×[0, 1]Ris jointly continuous in (y1,y3) for all y2R+.

(C4) The functiony1ϕ(y12,y2,y3)y1is not decreasing at nonnegative values ofy1. (C5) There are positive constantsb1andb2such that

b1χz1,z2

b2 z1,z2

R×R+. (1.17)

(C6) The functionχ:R×R+Ris continuous.

Note that conditions (C1)–(C6) have a physical meaning (see [2,4]).

The main result of this paper is the following theorem.

Theorem 1.2. Suppose that conditions (C1)–(C6) are satisfied. Then there exists a weak solution of problems (1.1)–(1.3), (1.8), (1.10), (1.12).

For the proof ofTheorem 1.2we use the approximating-topological method [8]. For this purpose, in the beginning, we determine an equivalent operational treatment of the problem under consideration. After that for the obtained operational equation, we in- troduce an approximating family of equations depending on a parameterδ and by use of Skrypnik’s version of the topological degree [7], on the basis of a priori estimates, we prove existence of solutions of the approximating equations. As a result, making limiting transition forδ0, we obtain the solvability of problem (1.1)–(1.3), (1.8), (1.10), (1.12).

2. Operational treatment

Let us introduce some notations. ByXwe denote the space, conjugate to some Banach spaceX,g,ydenotes the action of the functionalgXon the element yX,Xm is the topological product ofmcopies of the spaceX.

Determine several mappings as follows:

A:Z−→Z, A(u),h= n i,j=1

ωI(u),|E|,μ(u,E)εi j(u)εi j(h)dx, K:Z×L2(Ω)−→Z,

K(u,p),h= n i=1

Sχ frnp(s),u(s),uτ(s)2uτi(s)hτi(s)ds, M:Z−→Z, M(u),h=

n i,j=1

ωuj(x)∂ui

∂xj(x)hi(x)dx.

(2.1)

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Take

D:Z−→L2(Ω), D(u)=divu. (2.2)

IdentifyingL2(Ω)nand (L2(Ω)n)we obtain D:L2(Ω)n

L2(Ω)n−→Z, D(p),h=

Ωp(x)[divh](x)dx. (2.3) It is obvious that the set of weak solutions of problem (1.1)–(1.3), (1.8), (1.10), (1.12) coincides with the set of couples (u,p)W×L2(Ω) that satisfy the following operational equation:

A(u) +K(u,p) +M(u)D(p)=F. (2.4) 3. Properties of operators

Everywhere below the expressions vk−−−→

k→∞ v0and vk

k→∞ v0 will denote strong and weak convergences, respectively, of the sequence{vk}k=1to an element v0. The case when the sequence{vk}k=1does not converge to v0in strong sense is denoted as vk

k→∞ v0. Lemma 3.1. The following statements hold.

(1) The operatorAis bounded and demicontinuous (the latter means that ifuk−−−→

k→∞ u0

inZ, thenA(uk),h −−−→

k→∞ A(u0),hhZ).

(2) For the operator Ꮽ(u,v),h=

n i,j=1

ωI(u),|E|,μ(v,E)εi j(u)εi j(h)dx, (3.1) the following inequality holds

u1,vᏭ(u2,v),u1u2

0 u1,u2,vZ. (3.2) (3) Specify an elementuZ. Then from any bounded sequence{vk}k=1inZit is possible

to choose a subsequence{vkl}l=1such thatᏭ(u,vkl)−−−→

l→∞ Ꮽ(u,v0) inZ.

Proof. (a) Boundedness of the operatorAis obvious. Let us show that it is demicontinu- ous. So, letuk−−−→

k→∞ u0inZ. Clearly, there is a subsequence{ukl}l=1, for which Iukl[x]−−−→

l→∞ Iu0

[x], ukl[x]−−−→

l→∞ u0[x] for a.e.xΩ. (3.3) Assume that A(uk), ˘h k→∞ A(u0), ˘h for some ˘hZ. Without loss of generality for a sequence{ukl}l=1the following inequality holds with someζ >0:

Aukl, ˘h Au0

, ˘h> ζ. (3.4)

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We estimate AuklAu0

, ˘h

n i,j=1

2

ΩϕIukl,|E|,μukl,Eεi juklu0

εi j( ˘h)dx

+ n i,j=1

2

Ω ϕIukl

,|E|,μukl,EϕIu0

,|E|u0,Eεi j u0

εi j( ˘h)dx 2a2uklu0Zh˘Z

+ 2

Ω ϕIukl,|E|ukl,EϕIu0

,|E|,μu0,E2Iu0

dx 1/2

h˘Z. (3.5) The first component in the last part of inequality (3.5) tends to zero because of the conver- gence ofuk−−−→

k→∞ u0inZ, the second component tends to zero by the Lebesgue theorem.

Hence,A(ukl)A(u0), ˘h −−−→

l→∞ 0, that contradicts inequality (3.4).

(b) Introduce the notationφ(u,v)=ϕ(I(u),|E|,μ(v,E)). With the help of the Cauchy inequality and condition (C4) we obtain that

u1,vu2,v,u1u2

= n i,j=1

2

Ω φu1,vεi j u1

φu2,vεi j

u2 εi j u1u2

dx

=2

Ω φu1,vIu1

+φu2,vIu2

dx

n i,j=1

2

Ω φu1,vεi j u1

εi j u2

φu2,vεi j u1

εi j u2

dx

2

Ω φv,u1

I1/2u1

φu2,vI1/2u2

I1/2u1

I1/2u2

dx0.

(3.6)

(c) Let{vk}k=1be a bounded sequence inZ,uZ. There is a subsequence{vkl}l=1and an elementv0Zsuch that

vkl(x)−−−→

l→∞ v0(x) for a.e.xΩ. (3.7)

We have u,vkl

u,v0Z

= sup

hZ=1

n i,j=1

2

ω ϕI(u),E,μvkl,Eεi j(u)ϕI(u),E,μv0,Eεi j(u)εi j(h)dx

2n

ω ϕI(u),E,μvkl,EϕI(u),E,μv0,E2I(u)dx 1/2

.

(3.8)

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The last expression in inequality (3.8) tends to zero by conditions (C1)–(C3) and by the

Lebesgue theorem.

Lemma 3.2. The operatorKpossesses the following properties.

(1) For any sequence{uk,hk,pk}k=1fromZ×Z×L2(Ω)n, for which (uk,hk,pk)

k→∞

(u0,h0,p0), there is a subsequence{ukl,hkl,pkl}l=1such that K(ukl,pkl),hkl −−−→

l→∞

K(u0,p0),h0 (from this property, in particular, it follows that the operatorK is bounded and demicontinuous).

(2) The operatorK(·,T(·)) is compact for any operatorT:ZL2(Ω).

Proof. (a) So, let the limits from the condition take place. Using the fact that the embed- dingZL2(S)nis compact, we take a subsequence{ukl,hkl}l=1such that

ukl,hkl−−−→

l→∞

u0,h0 inL2(S)n×L2(S)n, (3.9) ukl[s],hkl[s]−−−→

l→∞

u0[s],h0[s] inR2n for a.e.sS. (3.10)

Extend the functions of the sequence{pk}k=1(and the functionp0) onto the entire space Rn. For this purpose we take pk(x)=pk(x) ifxΩ, otherwisepk(x)=0. It is obvious that pkl

l→∞ p0 inL2(Rn). Thus we have

Rnω|ξx´|

pkl( ´x)dx´=Ppkl[ξ]−−−→

l→∞ Pp0[ξ] ξS. (3.11) Similarly

P pkl+ n i,j=1

IPukl,|E|Pukl,Eεi j

Puklηiηj

(s)

=fpkl,ukl[s]−−−→

l→∞ fp0,u0[s]

(3.12)

for allsS.

We estimate

Kukl,pkl,hkl

Ku0,p0,h0

n

i=1

S

χf

pkl,ukl,ukτl2χf

p0,u0,u0τ2u0τih0τids +

n i=1

S

χf

pkl,ukl,ukτl2u0τiukτilh0τids

+ n i=1

S

χf

pkl,ukl,ukτl2ukτilh0τihkτilds.

(3.13)

Using condition (C5) and (3.11), (1.2) we conclude that the first summand in the right- hand side of inequality (3.13) tends to zero by the Lebesgue theorem, the other summands tend to zero by (3.9).

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(b) We will use Gelfand’s criterion, having reformulated it according to the following lemma.

Lemma 3.3. The subsetMof a separable Banach spaceis relatively compact, if from any sequence of functionals{fk}k=1belonging toand such that

fk(y)−−−→

k→∞ 0 yᐄ, (3.14)

it is possible to take a subsequence{fkl}l=1such that (3.14) is valid for it uniformly for ally fromM.

The proof ofLemma 3.3is similar to that of [3, Theorem 3(1.IX), page 274], minor changes are connected with transition to a subsequence in the formulation of the state- ment.

Now, letT:ZL2(Ω) be an operator, letMbe some bounded set fromZ, and let hk

k→∞ 0 in the space (Z)Z. For allufromMwe have hk,K(u,Tu)b2uL2(S)nhk

L2(S)n. (3.15)

However, the embeddingZL2(S)nis compact, therefore for some subsequence{hkl}l=1

we obtain thathkl,K(u,Tu) −−−→

l→∞ 0 uniformly for allufromM. Hence, the setK(M, T(M)) is relatively compact as it was required to show.

Lemma 3.4. LetMδ,δ >0, be an approximation of the operatorM, that is Mδ:Z−→Z, Mδ(u),h= 1

1 +δ1/4uL4(Ω)n

n i,j=1

Ωuj∂ui

∂xjhidx. (3.16) (1) The operatorMδis compact.

(2) Take any sequences{uk}k=1and{hk}k=1in the spaceZ such that uk k→∞ u0 in Z, hk k→∞ h0 inZ. Then there are subsequences{ukl}l=1and{hkl}l=1such that Mδ(ukl),hkl −−−→

l→∞ Mδ(u0),h0.

Proof. (1) The proof of boundedness and continuity of the operatorMδis standard. The property of compactness is shown similarly to item (2) ofLemma 3.2. Here the following inequality is in use:

Mδ(u),huL4(Ω)nuH1(Ω)n

1 +δ1/4uL4(Ω)n hL4(Ω)n. (3.17) (2) Let uk k→∞ u0 and hk k→∞ h0 inZ. It follows from here that there are the sub- sequences{ukl}l=1and{hkl}l=1such that

ukl,hkl−−−→

l→∞

u0,h0 inL4(Ω)n2. (3.18)

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Thus Mδ

u0,h0 Mδ

ukl,hkl

1

1 +δ1/4u0L4(Ω)n

n i,j=1

ωu0j ∂u0i

∂xj

∂ukil

∂xj

h0idx

+

1 1 +δ1/4u0L4(Ω)n

n i,j=1

ω

u0jukjl∂ukil

∂xjh0idx

+ 1

1 +δ1/4u0L4(Ω)n

n i,j=1

ωukjl∂ukil

∂xj

h0ihkildx

+

1 1 +δ1/4u0L4(Ω)n

1

1 +δ1/4uklL4(Ω)n

n

i,j=1

ωukjl∂ukil

∂xjhkildx.

(3.19)

The first summand on the right-hand side of (3.19) tends to zero since ukl l→∞ u0 in

Z, the other summands tend to zero by (3.18).

4. Approximation equation and an a priori estimate

For anyδ >0 we introduce an auxiliary equation in the unknown functionuδ: A+uδ+Auδ+Kδ

uδ+Mδ

uδ+δ1DDuδ=F, (4.1) where

Kδ(u),h= n i=1

Sχfrnδ1Du,u,uτ2uτihτids, A+(u),h=δ

n i,j=1

ωεi j(u)εi j(h)dx.

(4.2)

LetK+(u),h =(b1/2)ni=1Suτihτids.

Lemma 4.1. For the following family of the operational equations, depending on the param- etert[0, 1]:

Λδt

uδ=A+uδ+K+uδ

+tAuδ+KδuδK+uδ+Mδuδ+δ1DDuδF=0, (4.3) the estimate

uδZCFZ,a1,b1,n,Ω (4.4) holds for 0< δc(a1,b1,n,Ω). HerecandCare variables that depend only on the specified parameters.

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Proof. Fort=0 there is only a zero solution, in this case estimation (4.4) is obvious. Let uδbe a solution of (4.3) for somet(0, 1]. Apply both sides of (4.3) touδ. We obtain

A+uδ+K+uδ,uδ0, (4.5) thus

Auδ,uδ+Kδ

uδ,uδ

K+uδ,uδ+Mδ

uδ,uδ+δ1DDuδ,uδf,uδ. (4.6) Notice that

Auδ,uδ+Kδuδ,uδ

K+uδ,uδmin2a1,b1/2uδ2Z. (4.7) At the same time

Mδ

uδ,uδ= 1 1 +δ1/4uδL4(Ω)n

n i,j=1

ωuδj∂uδi

∂xjuδidx

= 1

1 +δ1/4uδL4(Ω)n

1 2

ΩDuδ|uδ|2dx+1 2

n i=1

Suδτiηiuδ2ds

n

2DuδL2(Ω)

uδ2L4(Ω)n

1 +δ1/4uδL4(Ω)n

n

2DuδL2(Ω)

uδ2L4(Ω)n

δ1/4uδL4(Ω)n

DuδL2(Ω) δ1/2 δ1/4n

2ϑuδZDuδ2L2(Ω)

2δ +δ1/2n2

8ϑ2uδ2Z (4.8) (ϑis the norm of the operator of embedding of the spaceZinto the spaceL4(Ω)n).

It is easy to see that

δ1DDuδ,uδ=δ1Duδ,Duδ=δ1Duδ2L2(Ω), FZuδZl2

2F2Z+ 1

2l2uδ2Z. (4.9)

Now for sufficiently largeland smallδwe obtain the required estimate (4.4).

5. Existence of a solution of the approximation equation

For the proof of existence of a solution of the approximation equation we apply the method of topological degree for generalized monotonous maps (see [7]). We show that the familyΛδt carries out homotopy of the maps Λδ0 and Λδ1. For this purpose, we will notice first that from the a priori estimate (4.4) it follows that there is a sphere of nonzero radiusR, with the center at zero, such that on its boundary there are no solutions of the equationΛδt(uδ)=0(t[0, 1]).

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