Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 710214 10 pageshttpdxdoiorg1011552013710214

Research ArticleAnalysis of the Reynolds Equation for Lubrication in Case ofPressure-Dependent Viscosity

Eduard MarušiT-Paloka and Sanja MarušiT

Department of Mathematics University of Zagreb Bijenicka Cesta 30 10000 Zagreb Croatia

Correspondence should be addressed to Eduard Marusic-Paloka emarusicmathhr

Received 30 November 2012 Revised 5 March 2013 Accepted 10 March 2013

Academic Editor Jyh-Hong Chou

Copyright copy 2013 E Marusic-Paloka and S Marusic This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

We study the Reynolds equation describing the ow of a lubricant in case of pressure-dependent viscosity First we prove theexistence and uniqueness of the solution Then we study the asymptotic behavior of the solution in case of periodic roughnessvia homogenization method Some interesting nonlocal effects appear due to the nonlinearity

1 Introduction

The Reynolds equations [1] describe the flow of a thin filmof lubricant separating two rigid surfaces in relative motionControlling the flow of lubricant is an important engineeringissue since inappropriate lubrication would increase the fric-tion and wear finally resulting in degrading the performanceof the device In his classical paper from 1848 [2] Stokespredicted that the viscosity of the fluid can depend on thepressureThose effects for various liquids have beenmeasuredin many engineering papers starting from the beginning ofthe 20th century (see eg [3])That effect is usually neglectedas it becomes important only in case of high pressureMost fluid-lubricated bearings operate with high pressureand in such a flow regime the dependence of the viscosityon the pressure becomes important According to Szeri [4]the idea of pressure-dependent viscosity was introduced inlubrication theory by Gatcombe in 1945 [5] Several modelshave been used to describe that relation since The mostpopular is probably the exponential law

120583 = 1205830exp (120572119901) (1)

usually called the Barus formula [6] Here 1205830and 120572 are the

constants depending on the lubricant The formula seemsto be reasonable for mineral oil unless the pressure is veryhigh (larger then 05MPa) The coefficient 120572 typically rangesbetween 1 and 10minus8The lower end of the range corresponds to

paraffinic and the upper end corresponds to the naphthenicoils (see Jones [7]) That formula is still frequently used byengineersThe simplest viscosity-pressure relation is given bythe the power law

120583 (119901) = 12057210038161003816100381610038161199011003816100381610038161003816120574 (2)

In case of the two above-mentioned laws explicit solutions ofthe equations of motion for some particular situations likeunidirectional and plane-parallel flows were found in [8]Discussion on other possibilities for the viscosity-pressureformula and some historical remarks on the subject can befound in the same paper Several engineering papers can befound discussing other possible laws and their consistencyWe mention for instance [9 10]

We do not make any assumption on the particular formof the function 119901 997891rarr 120583(119901) Some technical assumptions likesmoothness will be needed for the proofs

We study the stationary version of the Reynolds equationUnless the velocity of relative motion is time dependentsteady approximation is reasonable in most cases (see eg[4 Chapter 22])

Our first goal is to prove that the problem is well posedSecondly we investigate the asymptotic behavior of thesolution in case of periodically distributed asperities Usingthe homogenization approach we find the macroscopicReynolds pressure Interesting nonlocal effects appear due tothe nonlinearity caused by the pressure-dependent viscosity

2 Mathematical Problems in Engineering

119909119899+1 119909

119899+1= 120576ℎ(119909)

119909 = (1199091 119909

119899)

119874(120576)

Figure 1

2 Position of the Problem

The fluid domain is bounded by two rigid surfaces The sim-plified mathematical model can be written in the followingform LetO sub R119899 be a bounded domain and let ℎ O rarr R bea bounded strictly positive smooth function such that

0 lt ℎ0le ℎ (119909) le ℎ

1 (3)

Function ℎ describes the shape of the slide By 120576 ≪ 1 wedenote a very small parameter representing the domainthickness Using the shape function ℎ we define the fluiddomain (Figure 1) by

Ω120576= 119883 = (119909 119909

119899+1) isin R119899+1

119909 = (1199091 119909

119899) isin O 0 lt 119909

119899+1lt 120576ℎ (119909)

(4)

We then consider the stationary flow through a domain Ω120576

We want to describe the situation with a lower-dimensionalmodel The velocity of the relative motion of two surfacesis the constant vector denoted by V = V

1e1+ sdot sdot sdot + V

119899e119899

The unknowns in the model are 119906 (the velocity) and 119901 (thepressure) We recall that the stationary motion of the incom-pressible viscous laminar flow is governed by the stationaryNavier-Stokes equationsThus we write the following system

minus div [120583 (119901)Du] + (unabla) u + nabla119901 = 0 div u=0 in Ω120576

u = V for 119909119899+1

= 0 u = 0 for 119909119899+1

= 120576ℎ (119909)

(5)

where Du = (12)(nablau + nablau119905) is the symmetric part of thevelocity gradient It is important to notice that in such systemthe pressure is not defined only up to a constant as in theclassical Navier-Stokes system with constant viscosity Undercertain technical assumptions if the given data are not toolarge the existence of the solution for such system wasdiscussed in [11 12] Neglecting the effects of inertia we getthe Stokes system with pressure-dependent viscosity

div [120583 (119901)Du] + nabla119901 = 0 div u = 0 in Ω120576 (6)

studied in [13]If the thickness of the domain is small the solution can be

fairly approximated by the solution of the Reynolds equations[4 14]

k (119909 119909119899+1

)

=1

12120583 (119901)119909119899+1

(120576ℎ minus 119909119899+1

) nabla119901 (119909) + (1 minus119909119899+1

120576ℎ)V

(7)

div(int120576ℎ

0

k119889119909119899+1

) = 0 (8)

Indeed if we derive a formal asymptotic expansion of thesolution to the system (5) in powers of 120576 then the solution ofthe Reynolds equation (7) makes the first term of the expan-sion (see eg [4]) Here and in the sequel the differentialoperators div andnabla are taken only with respect to 119909 variablethat is

div b =1205971198871

1205971199091

+ sdot sdot sdot +120597119887119899

120597119909119899

nabla120601 =120597120601

1205971199091

e1+ sdot sdot sdot +

120597120601

120597119909119899

e119899

(9)

It leads to an elliptic equation of the form

div( ℎ3

120583 (119901)nabla119901) = 6V sdot nablaℎ in O (10)

119901 = 119902 on 120597O (11)

The goal of this paper is to study that equationWe assume that the function 119901 997891rarr 120583(119901) is of class 1198621(R)

and 120583 gt 0 for any value of 119901 In real life the viscosity increaseswith pressure but such an assumption is not necessary for ourstudy

3 Existence of the Solution

31 Transformed Equation Equation (10) is a quasilinearelliptic PDE but it can be linearized by simple trick To doso we rewrite the equation using the function

119872120590(119901) = int

119901

120590

119889119905

120583 (119905) (12)

We choose 120590 le 119902 Function 119872120590is strictly increasing since

1198721015840

120590(119901) = (1120583(119901)) gt 0 and thus it is bijective Furthermore

119872120590(119901) has the same sign as 119901 minus 120590 that is for 119901 gt 120590 we have

119872120590(119901) gt 0 for 119901 lt 120590 obviously 119872

120590(119901) lt 0 and finally

119872120590(119901) = 0 if and only if 119901 = 120590We introduce the new unknown function

119908 (119909) = 119872120590(119901 (119909)) = int

119901(119909)

120590

119889119905

120583 (119905) (13)

At this point we assume that the integral intminusinfin0

(119889119905120583(119905)) isdivergent that is

int

0

minusinfin

119889119905

120583 (119905)= +infin (14)

As a consequence

lim120590rarrminusinfin

119872120590(119901) = +infin forall119901 isin R (15)

as well as

lim119901rarrminusinfin

119872120590(119901) = minusinfin forall120590 isin R (16)

Mathematical Problems in Engineering 3

Deriving (13) we obtain

1

120583 (119901)nabla119901 = nabla119908 (17)

and the problem can be written as

div(ℎ3nabla119908 ) = 6V sdot nablaℎ in O (18)

119908 = 119872120590(119902) on 120597O (19)

That is a linear elliptic equation for 119908 and it has a uniquesolution To get the existence and uniqueness of the solutionwe quote Theorem 834 from classical book of Gilbarg andTrudinger [15] For simplicity here and in the sequel weassume that 119902 and consequently119872

120590(119902) are defined on whole

O We combine that with the maximum principle from theappendix and it gives the following

Theorem 1 Under the assumption that the boundary 120597O is ofclass 1198621120572 and that ℎ isin 119862

120572(O) 119902 isin 119862

1120572(O) the problem (18)

(19) has a unique solution

119908 isin 1198621120572

(O) (20)

Furthermore

119908 (119909) le 119872(119902) +Z (21)

where

119902 =10038161003816100381610038161199021003816100381610038161003816119871infin(120597120596) (22)

andZ = 0 if V sdot nablah lt 0 Otherwise

Z =

6 |V| [int1

0

119889119905

ℎ(119905)2minusℎ3

0

ℎ5

1

] + (ℎ0

ℎ1

)

3

int

1199021

1199020

119889119904

120583 (119904)

if 119899 = 1

3(85)

(3

2)

(285)(2120587)(14)

ℎ3

0

times119889 (6 |V| ℎ1|O|15 + ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O)) |O|

(75)

if 119899 = 2

(23)

with 119889 = diamO

Proof The existence follows directly from Theorem 834from Gilbarg and Trudinger [15] If V sdot nablaℎ lt 0 then (21)follows directly from the weak maximum principle (see eg[15]) In case 119899 = 1 the problem can be solved by quadraturesand the solution given by (25) can be easily estimated toget (23) In the remaining case 119899 = 2 (21) follows fromthe special variant of the maximum principle proved in theappendix

Remark 2 In case 119899 = 1 (18) is anODE (we takeO =]0 1[ and1199020gt 1199021 without losing generality)

(ℎ31199081015840)1015840

= 6119881ℎ1015840 in ]0 1[

119908 (119894) = 119872120590(119902119894) for 119894 = 0 1

(24)

and it can be solved by quadratures

119908 (119909) = 119872120590(1199020)

+ int

119909

0

1

ℎ(119905)3

[

[

6119881(ℎ (119905)minusint1

0(119889119903ℎ(119903)

2)

int1

0(119889119904ℎ(119904)

3)

)

+119872120590(1199021)minus119872120590(1199020)

int1

0(119889119904ℎ(119904)

3)

]

]

119889119905

(25)

32 Back to the Original Equation Now our goal is not tofind the auxiliary function119908 but to find the pressure 119901 Sincewe have introduced 119908 as

119908 = 119872120590(119901) (26)

we should have 119901 = 119872minus1

120590(119908) In order to do so we have to

make sure that 119908(119909) isin Im119872120590for any 119909 isin O Since 119872

120590is

strictly increasing and we have assumed that (14) holds if wedefine

119872+

120590= lim119904rarr+infin

119872120590(119904) = int

+infin

120590

119889119904

120583 (119904) (27)

due to (16) we obviously have for any 119901 isin R

minusinfin = lim119904rarrminusinfin

119872120590(119904) le 119872

120590(119901) le 119872

+

120590 (28)

Thus

Im119872120590= ]minusinfin119872

+

120590[ (29)

So to fulfill the condition 119908(119909) isin Im119872120590we need to have

119908 (119909) le 119872+

120590 forall119909 isin O (30)

That condition is not necessarily fulfilledIn view of (21) that condition reduces to

Z le int

+infin

119902

119889119905

120583 (119905) (31)

whereZ = Z(V ℎO 120583 119902) is defined by (23)We have proved the following theorem

Theorem 3 Suppose that the conditions of Theorem 1 holdand that in addition (31) is fulfilled Then 119901 = 119872

minus1

120590(119908) is the

unique solution of (10) and (11)

Remark 4 It is important to notice that even though 119908 doesdepend on 120590 the effective pressure 119901 does not For thepurpose of this remark we denote119867

120590(119908) = 119872

minus1

120590(119908) to stress

the dependence on the parameter 120590 which is of interest hereWe start by

119901 = 119867120590(119908) (32)

4 Mathematical Problems in Engineering

It is obvious from the definition of119872120590that

120597119872120590

120597119901(119901) =

1

120583 (119901)

120597119872120590

120597120590(119901) = minus

1

120583 (120590) (33)

As119872120590(119867120590(119908)) = 119908 deriving with respect to 120590 we arrive at

120597119867120590

120597120590(119908) = minus

(120597119872120590120597120590) (119901)

(120597119872120590120597119901) (119901)

=120583 (119901)

120583 (120590)=120583 (119867120590(119908))

120583 (120590)

(34)

Deriving (13) we get (120597119908120597120590) = minus1120583(120590) Using the rulefor deriving the inverse function we have

120597119867120590

120597119908(119908) = 120583 (119901) = 120583 (119867

120590(119908)) (35)

Thus120597119901

120597120590=120597119867120590

120597119908(119908)

120597119908

120597120590(120590) +

120597119867120590

120597120590(119908)

= minus120583 (119867120590(119908))

120583 (120590)+120583 (119867120590(119908))

120583 (120590)= 0

(36)

4 Homogenization

In this section we want to study the effects of rugositiesof surfaces on lubrication process The idea of finding themacroscopic effects of roughness on lubrication process viahomogenization is quite old and well studied Case of con-stant viscosity for incompressible and compressible flows aswell as non-Newtonian deformation dependent viscositieswere investigated The subject was treated by several authorsandwe heremention [16ndash18]The case of pressure-dependentviscosity brings some new interesting nonlocal effects

We assume that the function ℎ describing the form of thefluid domain is periodic with small period 1119898 with119898 isin NTo stress that dependence we denote it by ℎ

119898 More precisely

we denote by 119884 =]0 1[119899 119899 = 1 2 the period We further

assume that ℎ R119899 rarr [1198890 +infin[ 119889

0gt 0 is periodic with

period 119884 and smooth Then we take ℎ119898of the form

ℎ119898(119909) = ℎ (119898119909) (37)

Thus the function ℎ describes the form of periodicallydistributed rugosities

To emphasize that the relative velocity of bearing surfacesV is large we assume that it also depends on 119898 the sameparameter that is taken for description of rugosities In thatcase our equation reads

div(ℎ3

119898

120583 (119901119898)nabla119901119898) = 6V

119898sdot nablaℎ119898

in O (38)

41 One-Dimensional Case If 119899 = 1 the above problem isposed on an intervalO =]0 1[With an appropriate boundarycondition

119901119898(0) = 119902

0 119901

119898(1) = 119902

1 (39)

It forms a boundary value problem for nonlinear ODE

(ℎ3

119898

120583 (119901119898)1199011015840

119898)

1015840

= 6119881119898ℎ1015840

119898in O (40)

To study the asymptotic behavior of the solution with respectto119898we linearize the problem using the transformation119908

119898=

119872120590(119901119898) To simplify in this section we choose 120590 = 119902

0and

dropping the index 120590 in119872120590and119872

+

120590 we denote

119872(119901) = int

119901

1199020

119889119905

120583 (119905) 119872

+= int

+infin

1199020

119889119905

120583 (119905) (41)

Theorem 5 Let

1205940(119910) = 6 (int

1

0

119889119904

ℎ(119904)3)

minus1

times [⟨1

ℎ3⟩int

119910

0

119889V

ℎ(V)2minus⟨

1

ℎ2⟩int

119910

0

119889V

ℎ(V)3]

(42)

⟨sdot⟩ = int

1

0

sdot 119889119910 (43)

and let119867 = 119872minus1 Suppose that there exists a limit

119881 = lim119898rarrinfin

119881119898

119898(44)

and that for 119898 large enough and 119872+ defined in (27) the

following condition holds

119887119898

3(119909)

119887119898

3(1)

119872 (1199021) + 6119881

119898

119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)

119887119898

3(1)

le 119872+

(45)

with

119887119898

120572(119909) = int

119909

0

119889119905

ℎ119898(119905)120572 120572 = 2 3 (46)

Then

119901119898 1199010=int

1

0

119867(119909119872(1199021) + 119881120594

0(119910)) 119889119910

weaklowast in 119871infin(0 1)

(47)

Proof Equation (40) with boundary conditions (39) can besolved by reduction to quadratures after the substitution

119908119898= 119872(119901

119898) (48)

with 119872120590strictly increasing function defined by (12) The

problem for 119908119898now reads

(ℎ3

1198981199081015840

119898)1015840

= 6119881119898ℎ1015840

119898in ]0 1[

119908119898(0) = 119872(119902

0) = 0 119908

119898(1) = 119872(119902

1)

(49)

Mathematical Problems in Engineering 5

It is easy to see that (49) has a unique solution given by (25)Since119872(119902

0) = 0 (25) now reduces to

119908119898=119887119898

3(119909)

119887119898

3(1)

119872120590(1199021) + 6119881

119898

119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)

119887119898

3(1)

(50)

where for 120572 = 2 3

119887119898

120572(119909) = int

119909

0

119889119905

ℎ119898(119905)120572

=1

119898int

119898119909

0

119889119904

ℎ(119904)120572997888rarr 119909int

1

0

119889119904

ℎ(119904)120572

as 119898 997888rarr +infin

(51)

Now 119901119898 the solution to the problem (40) exists if (45) is

fulfilled The second term in (50) thus obviously tends to119909119872(119902

1) as119898 rarr +infin The last term is more interesting The

denominator tends to

⟨1

ℎ3⟩ = int

1

0

119889119904

ℎ(119904)3 (52)

As for its numerator we have

119881119898[119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)]

= 119881119898[int

1

0

int

119909

0

(1

ℎ119898(119905)3ℎ119898(119904)2minus

1

ℎ119898(119905)2ℎ(119904)3

119898

)119889119904 119889119905]

=119881119898

119898(⟨

1

ℎ3⟩int

119898119909

0

119889V

ℎ(V)2minus ⟨

1

ℎ2⟩int

119898119909

0

119889V

ℎ(V)3)

(53)

Suppose that

119881 = lim119898rarr+infin

119881119898

119898(54)

and denote

119866 (119910) = ⟨1

ℎ3⟩int

119910

0

119889V

ℎ(V)2minus ⟨

1

ℎ2⟩int

119910

0

119889V

ℎ(V)3 (55)

Obviously the function119866 is periodicwith period 1 so that dueto the standard periodicity lemma (see eg [19]) as 119898 rarr

+infin

119866 (119898119909) ⟨119866⟩ = int

1

0

119866 (119910) 119889119910 weaklowast in 119871infin(0 1) (56)

By direct computation

⟨119866⟩ = ⟨119910

ℎ3⟩⟨

1

ℎ2⟩ minus ⟨

1

ℎ3⟩⟨

119910

ℎ2⟩

= intint

1

0

119904 minus 119905

ℎ(119904)3ℎ(119905)2119889119904 119889119905

(57)

Thus

119881119898[119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)] 119881 ⟨119866⟩ (58)

Now denoting

1205940(119910) =

119866 (119910)

1198873 (1)

(59)

we have

119908119898asymp 119909119872(119902

1) + 119881120594

0(119898119909) (60)

and thus

119908119898 1199080=119909119872(119902

1) + 119881⟨120594

0⟩ weaklowast in 119871

infin(0 1) (61)

However we are not interested in convergence of theauxiliary function 119908

119898but in the convergence of the pressure

119901119898 Since (45) is assumed to be true we can define 119901

119898=

119867(119908119898) where119867 = 119872

minus1 and we have

119901119898 1199010= int

1

0

119867(119909119872(1199021) + 119881120594

0(119910)) 119889119910

weaklowast in 119871infin(0 1)

(62)

Deriving the expression on the right-hand side we obtainthe effective pressure drop in the form

1199011015840

0(119909) = 119872(119902

1) int

1

0

120583 (119909119872(1199021) + 119881120594

0(119910)) 119889119910 (63)

= int

1199021

1199020

119889120591

120583 (120591)int

1

0

120583 (119909119872(1199021) + 119881120594

0(119910)) 119889119910

= int

1199021

1199020

119889120591

120583 (120591)int

1

0

120583 (119909int

1199021

1199020

119889119904

120583 (119904)+ 1198811205940(119910)) 119889119910

(64)

As we can see the pressure drop is not constant as forthe Newtonian flow The interesting effect appears if 119881 = 0

because in that case the expressions for the pressure and forthe pressure drop are nonlocal due to the integral with respectto 119910 That phenomenon is entirely due to the fact that theviscosity is depending on the pressure

42 Two-Dimensional Case We suppose here that the func-tion ℎ

119898is constructed from positive smooth 119884-periodic

function ℎ R2 rarr [1198890 +infin[ 119889

0gt 0 119884 =]0 1[

2 in the sameway as before that is by taking

ℎ119898(119909) = ℎ (119898119909) (65)

We have seen in the previous section that interesting effectshappen only if we assume that

V119898= 119898V (66)

In that case our equation reads

div(ℎ3

119898

120583 (119901119898)nabla119901119898) = 6V

119898sdot nablaℎ119898

in O (67)

6 Mathematical Problems in Engineering

Aswedid in the existence analysis and in the previous sectionwe linearize the equation by substitution

119908119898= 119872120590(119901119898) (68)

where the function119872120590is defined by (12) Now 119908

119898satisfies

div (ℎ3119872 nabla119908119898) = 6V

119898sdot nablaℎ119898

in O (69)

We postulate the asymptotic expansion in the form

119908119898 (119909) asymp 119908

0(119909 119910) +

1

1198981199081(119909 119910)

+1

11989821199082(119909 119910) + sdot sdot sdot 119910 = 119898119909

(70)

All functions are assumed to be 119884-periodic in 119910 variablePlugging that in (69) and collecting formally terms with

equal powers of119898 we get

1198982 div119910(ℎ3nabla1199101199080) = 6V sdot nabla

119910ℎ

119898 div119910(ℎ3nabla1199101199081) + div

119910(ℎ3nabla1199091199080) + ℎ3div119909nabla1199101199080= 0

1 div119910(ℎ3nabla1199101199082) + div

119910(ℎ3nabla1199091199081) + ℎ3div119909nabla1199101199081

+ ℎ3Δ1199091199080= 0

Denoting

119881 = |V| V0=V119881

(71)

we have

1199080(119909 119910) = V

0(119909) + 119881120594

0(119910)

div119910(ℎ3nabla1199101205940) = 6V

0sdot nabla119910ℎ in 119884

1199081(119909 119910) =

2

sum

119896=1

120594119896(119910)

120597V0

120597119909119896

(119909) + V1(119909)

div119910(ℎ3nabla119910(120594119896+ 119910119896)) = 0

2

sum

119896ℓ=1

a119896ℓ

1205972V0

120597119909119896120597119909ℓ

= 0 in O

a119896ℓ

= int119884

ℎ3 120597

120597119910ℓ

(120594119896+ 119910119896) 119889119910

(72)

Remark 6 Thesame computation can be done in one-dimen-sional case and it gives

(ℎ31205941015840

0)1015840

= 6ℎ1015840997904rArr 1205940= 6int

119910

0

119889119905

ℎ(119905)2+ 1198620int

119910

0

119889119905

ℎ(119905)3+ 1198621

(73)

V0= 119909119872(119902

1) (74)

Constants 1198620 1198621are chosen in a way that boundary condi-

tions 1205940(0) = 120594

0(1) = 0 are met and it follows that

1198621= 0 119862

0= minus 6(int

1

0

119889119905

ℎ(119905)3)

minus1

int

1

0

119889119905

ℎ(119905)2 (75)

Then

119908119898asymp 1199080 (119909119898119909)

= V0(119909) + 120594

0(119898119909)

= 119909119872120590(1199021)

+ 6119881[int

119898119909

0

119889119905

ℎ(119905)2minus (int

1

0

119889119905

ℎ(119905)3)

minus1

times int

1

0

119889119905

ℎ(119905)2

int

119898119909

0

119889119905

ℎ(119905)3]

= 119909119872120590(1199021)

+ 6119881119887119898

2(119909) 1198873(1) minus 119887

2(1) 119887119898

3(119909)

1198873(1)

= 119909119872120590(1199021) + 119881120594

0(119898119909)

(76)

That is a very good approximation of our exact solution (50)It is important to notice that the choice of constants

1198620 1198621was determined from the exterior boundary con-

dition So we should expect the same in two-dimensionalcase However the treatment of boundary conditions intwo-dimensional case is much more complicated and theboundary layer is to be expected

The derived asymptotic expansion should be justified byproving the convergence And we need the strong conver-gence (with corrector of course) for 119908

119898in order to get the

convergence for 119901119898 The form of the approximation

119908119898asymp V0(119909) + 119881120594

0(119898119909) + sdot sdot sdot (77)

suggests that the boundary layer phenomenon should appearon the exterior boundary 120597O since 120594

0term cannot satisfy the

Dirichlet condition on 120597O To get the error estimate and thestrong convergence we need to handle that boundary layerThus at this point we simplify the domain and the boundarycondition in order to be able to avoid it We assume that

O = ]0 1[ times R (78)

119901120576(0 1199092) = 0 119901

120576(1 1199092) = 119902 (119909

2) (79)

1199092997891997888rarr 119901

120576(1199091 1199092) is 1-periodic (80)

ℎ (1199101 1199102) = ℎ (119910

1) (81)

119902 is 1-periodic (82)

Now

119908120576= 119872120590(119901120576) with 119872

120590(119901) =int

119901

120590

119889119904

120583 (119904)997904rArr 119908

120576(0 1199092) = 0

119908120576(1 1199092) = 119872

120590(119902 (1199092))

(83)

and 1199092997891rarr 119908120576(1199091 1199092) is 1-periodic

Mathematical Problems in Engineering 7

In that case we can compute1205940and120594119896 119896 = 1 2 explicitly

and we can impose exterior condition on 1205940 Indeed 120594

0is

exactly the same as in the monodimensional case that is itis given by (73) and (75) Obviously 120594

2= 0 so that

a22

= int

1

0

ℎ(119904)3119889119904 = ⟨ℎ

3⟩ a

12= a21

= 0 (84)

As for the last term

1205941= minus1199101+ (int

1

0

119889119904

ℎ(119904)3)

minus1

int

1199101

0

119889119904

ℎ(119904)3

a11

= (int

1

0

119889119904

ℎ(119904)3)

minus1

=1

⟨1ℎ3⟩

(85)

Finally the function V0satisfies the boundary value problem

1

⟨1ℎ3⟩

1205972V0

12059711990921

+ ⟨ℎ3⟩1205972V0

12059711990922

= 0 in O

V0(0 1199092) = 0

V0(1 1199092) = 119872

120590(119902 (1199092)) V

0is 1-periodic in 119909

2

(86)

It can be solved using the Fourier method and we get

V0(1199091 1199092) =

infin

sum

119896=1

sh (radic⟨ℎ3⟩ ⟨ℎminus3⟩ 1198961205871199091)

times (120572119896sin 119896120587119909

2+ 120573119896cos 119896120587119909

2)

120572119896= 2int

1

0

119872120590(119902 (119905)) sin 119896120587119905119889119905

120573119896= 2int

1

0

119872120590(119902 (119905)) cos 119896120587119905119889119905

(87)

Since the approximation

w119898asymp V0(119909) + 119881120594

0(1198981199091) +

1

1198981205941(1198981199091)120597V0

1205971199091

(119909) (88)

now satisfies the boundary conditions on 120597O it is easy to seethat

1003816100381610038161003816119908119898 minus (V0(119909) + 119881120594

0(119898119909))

1003816100381610038161003816119871infin(O) le 1198621

119898(89)

follows from the maximum principle Assuming that for 119898large enough

V0(119909) + 119881120594

0(119898119909) le 119872

+ (90)

we have1003816100381610038161003816119901119898 minus 119867 (V

0(119909) + 119881120594

0(119898119909))

1003816100381610038161003816119871infin(O) 997888rarr 0 (91)

Finally

119901119898 1199010= int

1

0

119867(V0(119909) + 119881120594

0(1199101)) 1198891199101 (92)

We have proved that

Theorem 7 Let 119901119898be the solution to the problem (67) (79)

and (80) and let V01205940be defined by (87) and (42) respectively

If (90) holds then1003816100381610038161003816119901119898 minus 119867 (V

0 (119909) + 1198811205940 (119898119909))

1003816100381610038161003816119871infin(O) 997888rarr 0

119901119898 1199010= int

1

0

119867(V0(119909) + 119881120594

0(1199101)) 1198891199101

(93)

Remark 8 It is important to notice that 1205870= 119867(119901

0) satisfies

2

sum

119894119896=1

120597

120597119909119894

(a119894119896

120583 (1205870)

1205971205870

120597119909119896

) = 0 in O 1205870= 119902 on 120597O (94)

and thus we would expect it to be the limit of 119901119898in analogy

with the linear case However 1199010

= 1205870

If119881 = lim119898rarrinfin

119898|V119898| is small we can expand119867(V

0(119909)+

1198811205940(119898119909)) in powers of 119881 and we get

119901119898(119909) asymp 119867 (V

0(119909) + 119881120594

0(119898119909))

= 119867 (V0(119909)) + 119881119867

1015840(V0(119909)) 120594

0(119910) + 119874 (119881

2)

= 1205870(119909) + 119881120583 (120587

0(119909)) 120594

0(119910) + 119874 (119881

2)

(95)

Thus

1199010= 1205870+ 120583 (120587

0) ⟨1205940⟩119881 + 119874 (119881

2) (96)

It can be formally written as

119901119898(119909) asymp 120587

0(119909) +

1003816100381610038161003816V1198981003816100381610038161003816

119898120583 (1205870(119909)) 120594

0(119898119909) + 119874(

1003816100381610038161003816V11989810038161003816100381610038162

1198982)

(97)

Appendix

The Maximum Principle

Our goal is to derive maximum principles for the linearReynolds equation with sharp explicit constants in orderto solve the nonlinear Reynolds equation with pressure-dependent viscosity We assume without losing generalitythat V = (16)Vi Indeed we can always choose the coor-dinate system in a way that the first coordinate axis 119909 has adirection of the velocity of relative motion V

The lower bound for 119908(119909 119910) is of no interest justthe upper bound Function 119908(119909 119910) is the solution to theboundary value problem

div (ℎ3nabla119908) = V120597ℎ

120597119909in O sub R2

119908 = 119872120590(119902) on 120597O

(A1)

We assume that if V(120597ℎ120597119909) gt 0 then 119908 cannot have amaximum point in the domain O and thus

119908 (119909 119910) le max(119909119910)isin120597O

119872120590(119902 (119909 119910)) (A2)

8 Mathematical Problems in Engineering

However it is not realistic to assume that (120597ℎ120597119909) does notchange the sign To find the upper bound in the general casewe use the procedure from the DeGiorgi theorem The mainresult of the section is as follows

TheoremA1 Let119908 be the solution to the problem (A1)Then

119908 (119909 119910) le1003816100381610038161003816119872120590 (119902)

1003816100381610038161003816119871infin(120597O) +Z (V ℎO 120583 119902) (A3)

Z=3(85)

times(3

2)

(285)(2120587)(14)

ℎ3

0

diamO (|V| ℎ1|O|15

+ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times |O|(75)

(A4)

Proof The function 119911 = 119908 minus 119866 satisfies

div (ℎ3nabla119911) = V120597ℎ

120597119909+ div (ℎ3nabla119872

120590(119902)) in O sub R2 (A5)

119911 = 0 on 120597O (A6)

Next we introduce the embedding constant for 119882120

(O) sub

119871119903(O) denoted119872

119903 such that

|V|119871119903(O) le 119872119903|nablaV|1198712(O) forallV isin 119867

1

0(O) (A7)

That constant can be estimated as

119872119903le1

2(diamO)

2|O|1119903 minus 12

(119903 + 2

2)

((119903+2)2119903)

radic2120587 (A8)

See for example [20 Lemma 1] Next we define the se-quence

120582119896+1

= 3(120582119896

2+ 1) 120582

1= 2 (A9)

Easy computation yields

120582119896= 8(

3

2)

119896minus1

minus 6 (A10)

Let

119911+(119909 119910) = max 119911 (119909 119910) 0 (A11)

We test (A5) with (119911+)1+120582119896+1 and get

intO

ℎ3nabla119911+nabla(119911+)1+120582119896+1

=1 + 120582119896+1

(1 + 120582119896+1

2)2intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

= intO

[Vℎ120597

120597119909(119911+)1+120582119896+1

+ ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

]

(A12)

For the left-hand side we get the lower bound

ℎ3

0

1 + 120582119896+1

(1 + 120582119896+1

2)2

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

1198712(O)

(A13)

We estimate the terms on the right-hand side using the sameidea

intO

ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

le ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O)

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

intO

ℎV120597

120597119909(119911+)1+120582119896+1

le ℎ1 |V| |O|

15100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

(A14)

Thus it remains to estimate |nabla(119911+)1+120582119896+1 |11987154(O) We have

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

100381610038161003816100381610038161003816

54

11987154(O)

= (1 + 120582119896+1

)54

intO

(119911+)(5120582119896+14)1003816100381610038161003816nabla119911

+1003816100381610038161003816

54

= (1 + 120582119896+1

1 + (120582119896+1

2))

54

intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

541003816100381610038161003816119911+1003816100381610038161003816

(5120582119896+18)

le [Holders inequality with 119901 =8

5 1199011015840=8

3]

le (1 + 120582119896+1

1 + (120582119896+1

2))

54

(intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

)

58

times (intO

(119911+)(53)120582

119896+1

)

38

= [due to (A9) 53120582119896+1

= 5(1 +120582119896

2)

3

8=1

4

120582119896+1

2 + 120582119896

]

= (1 + 120582119896+1

1 + (120582119896+1

2))

54100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

54

1198712(O)

times100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(120582119896+1(2+120582

119896))(54)

1198715(O)

(A15)

Combining with (A14) and (A13) we get

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)1003816100381610038161003816100381610038161198712(O)

le ℎminus3

0(1 +

120582119896+1

2)100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198715(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

Mathematical Problems in Engineering 9

le 119872((120582119896+1)(2+120582

119896))

5ℎminus3

0(1+

120582119896+1

2)

times100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198712(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

(A16)

We recall that

120582119896+1

2 + 120582119896

=3

2(A17)

and define

120572 = 119872(32)

5ℎminus3

0(ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

) (A18)

as well as

120590119896=

1

120572

100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(2(2+120582119896))

1198712(O)

(A19)

Then (A16) implies

120590119896+1

le (1 +120582119896+1

2)

(2(2+120582119896))

120590((120582119896+1)(2+120582

119896+1))

119896 (A20)

Taking the logarithm we arrive at

log120590119896+1

lelog (1 + (120582

119896+12))

1 + (120582119896+1

2)+

120582119896+1

2 + 120582119896+1

log120590119896 (A21)

We first notice that

120582119896+1

2 + 120582119896+1

lt 1 (A22)

and then

1 +120582119896

2= 4(

3

2)

119896minus1

minus 2 gt (3

2)

119896

(A23)

Since the function 119909 997891rarr (log119909119909) is decreasing for 119909 gt 119890 wehave

log (1 + (1205821198962))

1 + (1205821198962)

lelog [(32)119896]

(32)119896

= 119896 log 3

2(2

3)

119896

forall119896 ge 3

(A24)

Then

log120590119896+1

le (119896 + 1) log 3

2(2

3)

119896

+ log120590119896

le log 3

2

119896+1

sum

119895=2

119895(2

3)

119895

+ log1205901le 8 log 3

2+ log120590

1

(A25)

Finally

120590119896+1

le (3

2)

8

1205901 (A26)

Now it remains to estimate 1205901 From the definitionwe see that

1205901=

1

120572

100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

12

1198712(O)

(A27)

To estimate 1205901 we proceed as before and test (A5) with (119911+)3

We get

3

4intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

2

= intO

(Vℎ120597(119911+)3

120597119909+ ℎ3nabla119872120590(119902) nabla(119911

+)3

)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

100381610038161003816100381610038161003816nabla(119911+)310038161003816100381610038161003816100381611987154(O)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times3

2

100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

1003816100381610038161003816119911+1003816100381610038161003816119871103(O)

(A28)

Thus100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

le2

ℎ3

0

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872(103)

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O)

(A29)

Finally testing (A5) with 119911+ we get

intO

ℎ31003816100381610038161003816nabla119911+1003816100381610038161003816

2

= intO

(Vℎ120597119911+

120597119909+ ℎ3nabla119872120590(119902) nabla119911

+) (A30)

so that

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O) le

1

ℎ3

0

(|V| ℎ1|O|12

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198712(O))

le |O|310

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

(A31)

Combining (A29) with (A31) and (A26) gives

120590119896+1

le (3

2)

81

120572

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A32)

Since

120590119896+1

ge1

120572119872(2(2+120582k))2

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) (A33)

10 Mathematical Problems in Engineering

we have arrived to

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) le (

3

2)

8

119872(2(2+120582

119896))

2

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A34)

Since lim119896rarrinfin

120582119896= infin we get

1003816100381610038161003816119911+1003816100381610038161003816119871infin(O) le (

3

2)

8radic2

ℎ3

0

times(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A35)

Finally (A8) implies (A3)

Acknowledgment

This work was supported by MZOS grant 037-0372787-2797

References

[1] O Reynolds ldquoOn the theory and application and its applicationto Mr Beauchamp towerrsquos experiments Including and experi-mental determination of the viscosity of olive oilrdquo PhilosophicalTransactions of the Royal Society vol 117 pp 157ndash234 1886

[2] G G Stokes ldquoNotes on hydrodynamics On the dynamicalequationsrdquoCambridge and DublinMathematical Journal III pp121ndash127 1848

[3] P W Brifgman ldquoThe viscosity of liquids under pressurerdquo Pro-ceedings of the National Academy of Sciences of the United Statesof America vol 11 no 10 pp 603ndash606 1925

[4] A Z Szeri Fluid Film Lubrication Cambridge University PressNew York NY USA 1998

[5] E K Gatcombe ldquoLubrication characteristics of involute spur-gearsmdasha theoretical investigationrdquoTransactions of theAmericanSociety of Mechanical Engineers vol 67 pp 177ndash181 1945

[6] C Barus ldquoIsotherms isopiestics and isometrics relative to vis-cosityrdquo American Journal of Science vol 45 pp 87ndash96 1893

[7] W R Jones ldquoPressure viscosity measurement for several lubri-cantsrdquo ASLE Transactions vol 18 pp 249ndash262 1975

[8] J Hron J Malek and K R Rajagopal ldquoSimple flows of fluidswith pressure-dependent viscositiesrdquo Proceedings of the RoyalSociety A vol 457 no 2011 pp 1603ndash1622 2001

[9] C J A Roelands Correlation aspects of the viscosity-pressurerelationship of lubricating oils [PhD thesis] Delft University ofTechnology Delft The Netherlands 1966

[10] K R Rajagopal and A Z Szeri ldquoOn an inconsistency in thederivation of the equations of elastohydrodynamic lubricationrdquoProceedings of the Royal Society A vol 459 no 2039 pp 2771ndash2786 2003

[11] M Renardy ldquoSome remarks on the Navier-Stokes equationswith a pressure-dependent viscosityrdquo Communications in Par-tial Differential Equations vol 11 no 7 pp 779ndash793 1986

[12] F Gazzola and P Secchi ldquoSome results about stationary Navier-Stokes equations with a pressure-dependent viscosityrdquo in Pro-ceedings of the International Conference on Navier-Stokes Equa-tions vol 388 of Pitman Research Notes in Mathematics Seriespp 174ndash183 Varenna Italy 1998

[13] E Marusic-Paloka ldquoAn analysis of the Stokes system with pres-sure dependent viscosityrdquo In press

[14] S Marusic and E Marusic-Paloka ldquoTwo-scale convergence forthin domains and its applications to some lower-dimensionalmodels in fluid mechanicsrdquo Asymptotic Analysis vol 23 no 1pp 23ndash57 2000

[15] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001

[16] M Kane and B Bou-Said ldquoComparison of homogenization anddirect techniques for the treatment of roughness in incompress-ible lubricationrdquo Journal of Tribology vol 126 no 4 pp 733ndash7372004

[17] M Jai ldquoHomogenization and two-scale convergence of thecompressible Reynolds lubrication equation modelling the fly-ing characteristics of a roughmagnetic head over a rough rigid-disk surfacerdquo RAIRO Modelisation Mathematique et AnalyseNumerique vol 29 no 2 pp 199ndash233 1995

[18] P Wall ldquoHomogenization of Reynolds equation by two-scaleconvergencerdquo Chinese Annals of Mathematics B vol 28 no 3pp 363ndash374 2007

[19] A Bensoussan J-L Lions and G Papanicolaou AsymptoticAnalysis for Periodic Structures North-Holland AmsterdamThe Netherlands 1978

[20] E Marusic-Paloka and A Mikelic ldquoThe derivation of a nonlin-ear filtration law including the inertia effects via homogeniza-tionrdquo Nonlinear Analysis Theory Methods amp Applications vol42 no 1 pp 97ndash137 2000

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

119909119899+1 119909

119899+1= 120576ℎ(119909)

119909 = (1199091 119909

119899)

119874(120576)

Figure 1

2 Position of the Problem

The fluid domain is bounded by two rigid surfaces The sim-plified mathematical model can be written in the followingform LetO sub R119899 be a bounded domain and let ℎ O rarr R bea bounded strictly positive smooth function such that

0 lt ℎ0le ℎ (119909) le ℎ

1 (3)

Function ℎ describes the shape of the slide By 120576 ≪ 1 wedenote a very small parameter representing the domainthickness Using the shape function ℎ we define the fluiddomain (Figure 1) by

Ω120576= 119883 = (119909 119909

119899+1) isin R119899+1

119909 = (1199091 119909

119899) isin O 0 lt 119909

119899+1lt 120576ℎ (119909)

(4)

We then consider the stationary flow through a domain Ω120576

We want to describe the situation with a lower-dimensionalmodel The velocity of the relative motion of two surfacesis the constant vector denoted by V = V

1e1+ sdot sdot sdot + V

119899e119899

The unknowns in the model are 119906 (the velocity) and 119901 (thepressure) We recall that the stationary motion of the incom-pressible viscous laminar flow is governed by the stationaryNavier-Stokes equationsThus we write the following system

minus div [120583 (119901)Du] + (unabla) u + nabla119901 = 0 div u=0 in Ω120576

u = V for 119909119899+1

= 0 u = 0 for 119909119899+1

= 120576ℎ (119909)

(5)

where Du = (12)(nablau + nablau119905) is the symmetric part of thevelocity gradient It is important to notice that in such systemthe pressure is not defined only up to a constant as in theclassical Navier-Stokes system with constant viscosity Undercertain technical assumptions if the given data are not toolarge the existence of the solution for such system wasdiscussed in [11 12] Neglecting the effects of inertia we getthe Stokes system with pressure-dependent viscosity

div [120583 (119901)Du] + nabla119901 = 0 div u = 0 in Ω120576 (6)

studied in [13]If the thickness of the domain is small the solution can be

fairly approximated by the solution of the Reynolds equations[4 14]

k (119909 119909119899+1

)

=1

12120583 (119901)119909119899+1

(120576ℎ minus 119909119899+1

) nabla119901 (119909) + (1 minus119909119899+1

120576ℎ)V

(7)

div(int120576ℎ

0

k119889119909119899+1

) = 0 (8)

Indeed if we derive a formal asymptotic expansion of thesolution to the system (5) in powers of 120576 then the solution ofthe Reynolds equation (7) makes the first term of the expan-sion (see eg [4]) Here and in the sequel the differentialoperators div andnabla are taken only with respect to 119909 variablethat is

div b =1205971198871

1205971199091

+ sdot sdot sdot +120597119887119899

120597119909119899

nabla120601 =120597120601

1205971199091

e1+ sdot sdot sdot +

120597120601

120597119909119899

e119899

(9)

It leads to an elliptic equation of the form

div( ℎ3

120583 (119901)nabla119901) = 6V sdot nablaℎ in O (10)

119901 = 119902 on 120597O (11)

The goal of this paper is to study that equationWe assume that the function 119901 997891rarr 120583(119901) is of class 1198621(R)

and 120583 gt 0 for any value of 119901 In real life the viscosity increaseswith pressure but such an assumption is not necessary for ourstudy

3 Existence of the Solution

31 Transformed Equation Equation (10) is a quasilinearelliptic PDE but it can be linearized by simple trick To doso we rewrite the equation using the function

119872120590(119901) = int

119901

120590

119889119905

120583 (119905) (12)

We choose 120590 le 119902 Function 119872120590is strictly increasing since

1198721015840

120590(119901) = (1120583(119901)) gt 0 and thus it is bijective Furthermore

119872120590(119901) has the same sign as 119901 minus 120590 that is for 119901 gt 120590 we have

119872120590(119901) gt 0 for 119901 lt 120590 obviously 119872

120590(119901) lt 0 and finally

119872120590(119901) = 0 if and only if 119901 = 120590We introduce the new unknown function

119908 (119909) = 119872120590(119901 (119909)) = int

119901(119909)

120590

119889119905

120583 (119905) (13)

At this point we assume that the integral intminusinfin0

(119889119905120583(119905)) isdivergent that is

int

0

minusinfin

119889119905

120583 (119905)= +infin (14)

As a consequence

lim120590rarrminusinfin

119872120590(119901) = +infin forall119901 isin R (15)

as well as

lim119901rarrminusinfin

119872120590(119901) = minusinfin forall120590 isin R (16)

Mathematical Problems in Engineering 3

Deriving (13) we obtain

1

120583 (119901)nabla119901 = nabla119908 (17)

and the problem can be written as

div(ℎ3nabla119908 ) = 6V sdot nablaℎ in O (18)

119908 = 119872120590(119902) on 120597O (19)

That is a linear elliptic equation for 119908 and it has a uniquesolution To get the existence and uniqueness of the solutionwe quote Theorem 834 from classical book of Gilbarg andTrudinger [15] For simplicity here and in the sequel weassume that 119902 and consequently119872

120590(119902) are defined on whole

O We combine that with the maximum principle from theappendix and it gives the following

Theorem 1 Under the assumption that the boundary 120597O is ofclass 1198621120572 and that ℎ isin 119862

120572(O) 119902 isin 119862

1120572(O) the problem (18)

(19) has a unique solution

119908 isin 1198621120572

(O) (20)

Furthermore

119908 (119909) le 119872(119902) +Z (21)

where

119902 =10038161003816100381610038161199021003816100381610038161003816119871infin(120597120596) (22)

andZ = 0 if V sdot nablah lt 0 Otherwise

Z =

6 |V| [int1

0

119889119905

ℎ(119905)2minusℎ3

0

ℎ5

1

] + (ℎ0

ℎ1

)

3

int

1199021

1199020

119889119904

120583 (119904)

if 119899 = 1

3(85)

(3

2)

(285)(2120587)(14)

ℎ3

0

times119889 (6 |V| ℎ1|O|15 + ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O)) |O|

(75)

if 119899 = 2

(23)

with 119889 = diamO

Proof The existence follows directly from Theorem 834from Gilbarg and Trudinger [15] If V sdot nablaℎ lt 0 then (21)follows directly from the weak maximum principle (see eg[15]) In case 119899 = 1 the problem can be solved by quadraturesand the solution given by (25) can be easily estimated toget (23) In the remaining case 119899 = 2 (21) follows fromthe special variant of the maximum principle proved in theappendix

Remark 2 In case 119899 = 1 (18) is anODE (we takeO =]0 1[ and1199020gt 1199021 without losing generality)

(ℎ31199081015840)1015840

= 6119881ℎ1015840 in ]0 1[

119908 (119894) = 119872120590(119902119894) for 119894 = 0 1

(24)

and it can be solved by quadratures

119908 (119909) = 119872120590(1199020)

+ int

119909

0

1

ℎ(119905)3

[

[

6119881(ℎ (119905)minusint1

0(119889119903ℎ(119903)

2)

int1

0(119889119904ℎ(119904)

3)

)

+119872120590(1199021)minus119872120590(1199020)

int1

0(119889119904ℎ(119904)

3)

]

]

119889119905

(25)

32 Back to the Original Equation Now our goal is not tofind the auxiliary function119908 but to find the pressure 119901 Sincewe have introduced 119908 as

119908 = 119872120590(119901) (26)

we should have 119901 = 119872minus1

120590(119908) In order to do so we have to

make sure that 119908(119909) isin Im119872120590for any 119909 isin O Since 119872

120590is

strictly increasing and we have assumed that (14) holds if wedefine

119872+

120590= lim119904rarr+infin

119872120590(119904) = int

+infin

120590

119889119904

120583 (119904) (27)

due to (16) we obviously have for any 119901 isin R

minusinfin = lim119904rarrminusinfin

119872120590(119904) le 119872

120590(119901) le 119872

+

120590 (28)

Thus

Im119872120590= ]minusinfin119872

+

120590[ (29)

So to fulfill the condition 119908(119909) isin Im119872120590we need to have

119908 (119909) le 119872+

120590 forall119909 isin O (30)

That condition is not necessarily fulfilledIn view of (21) that condition reduces to

Z le int

+infin

119902

119889119905

120583 (119905) (31)

whereZ = Z(V ℎO 120583 119902) is defined by (23)We have proved the following theorem

Theorem 3 Suppose that the conditions of Theorem 1 holdand that in addition (31) is fulfilled Then 119901 = 119872

minus1

120590(119908) is the

unique solution of (10) and (11)

Remark 4 It is important to notice that even though 119908 doesdepend on 120590 the effective pressure 119901 does not For thepurpose of this remark we denote119867

120590(119908) = 119872

minus1

120590(119908) to stress

the dependence on the parameter 120590 which is of interest hereWe start by

119901 = 119867120590(119908) (32)

4 Mathematical Problems in Engineering

It is obvious from the definition of119872120590that

120597119872120590

120597119901(119901) =

1

120583 (119901)

120597119872120590

120597120590(119901) = minus

1

120583 (120590) (33)

As119872120590(119867120590(119908)) = 119908 deriving with respect to 120590 we arrive at

120597119867120590

120597120590(119908) = minus

(120597119872120590120597120590) (119901)

(120597119872120590120597119901) (119901)

=120583 (119901)

120583 (120590)=120583 (119867120590(119908))

120583 (120590)

(34)

Deriving (13) we get (120597119908120597120590) = minus1120583(120590) Using the rulefor deriving the inverse function we have

120597119867120590

120597119908(119908) = 120583 (119901) = 120583 (119867

120590(119908)) (35)

Thus120597119901

120597120590=120597119867120590

120597119908(119908)

120597119908

120597120590(120590) +

120597119867120590

120597120590(119908)

= minus120583 (119867120590(119908))

120583 (120590)+120583 (119867120590(119908))

120583 (120590)= 0

(36)

4 Homogenization

In this section we want to study the effects of rugositiesof surfaces on lubrication process The idea of finding themacroscopic effects of roughness on lubrication process viahomogenization is quite old and well studied Case of con-stant viscosity for incompressible and compressible flows aswell as non-Newtonian deformation dependent viscositieswere investigated The subject was treated by several authorsandwe heremention [16ndash18]The case of pressure-dependentviscosity brings some new interesting nonlocal effects

We assume that the function ℎ describing the form of thefluid domain is periodic with small period 1119898 with119898 isin NTo stress that dependence we denote it by ℎ

119898 More precisely

we denote by 119884 =]0 1[119899 119899 = 1 2 the period We further

assume that ℎ R119899 rarr [1198890 +infin[ 119889

0gt 0 is periodic with

period 119884 and smooth Then we take ℎ119898of the form

ℎ119898(119909) = ℎ (119898119909) (37)

Thus the function ℎ describes the form of periodicallydistributed rugosities

To emphasize that the relative velocity of bearing surfacesV is large we assume that it also depends on 119898 the sameparameter that is taken for description of rugosities In thatcase our equation reads

div(ℎ3

119898

120583 (119901119898)nabla119901119898) = 6V

119898sdot nablaℎ119898

in O (38)

41 One-Dimensional Case If 119899 = 1 the above problem isposed on an intervalO =]0 1[With an appropriate boundarycondition

119901119898(0) = 119902

0 119901

119898(1) = 119902

1 (39)

It forms a boundary value problem for nonlinear ODE

(ℎ3

119898

120583 (119901119898)1199011015840

119898)

1015840

= 6119881119898ℎ1015840

119898in O (40)

To study the asymptotic behavior of the solution with respectto119898we linearize the problem using the transformation119908

119898=

119872120590(119901119898) To simplify in this section we choose 120590 = 119902

0and

dropping the index 120590 in119872120590and119872

+

120590 we denote

119872(119901) = int

119901

1199020

119889119905

120583 (119905) 119872

+= int

+infin

1199020

119889119905

120583 (119905) (41)

Theorem 5 Let

1205940(119910) = 6 (int

1

0

119889119904

ℎ(119904)3)

minus1

times [⟨1

ℎ3⟩int

119910

0

119889V

ℎ(V)2minus⟨

1

ℎ2⟩int

119910

0

119889V

ℎ(V)3]

(42)

⟨sdot⟩ = int

1

0

sdot 119889119910 (43)

and let119867 = 119872minus1 Suppose that there exists a limit

119881 = lim119898rarrinfin

119881119898

119898(44)

and that for 119898 large enough and 119872+ defined in (27) the

following condition holds

119887119898

3(119909)

119887119898

3(1)

119872 (1199021) + 6119881

119898

119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)

119887119898

3(1)

le 119872+

(45)

with

119887119898

120572(119909) = int

119909

0

119889119905

ℎ119898(119905)120572 120572 = 2 3 (46)

Then

119901119898 1199010=int

1

0

119867(119909119872(1199021) + 119881120594

0(119910)) 119889119910

weaklowast in 119871infin(0 1)

(47)

Proof Equation (40) with boundary conditions (39) can besolved by reduction to quadratures after the substitution

119908119898= 119872(119901

119898) (48)

with 119872120590strictly increasing function defined by (12) The

problem for 119908119898now reads

(ℎ3

1198981199081015840

119898)1015840

= 6119881119898ℎ1015840

119898in ]0 1[

119908119898(0) = 119872(119902

0) = 0 119908

119898(1) = 119872(119902

1)

(49)

Mathematical Problems in Engineering 5

It is easy to see that (49) has a unique solution given by (25)Since119872(119902

0) = 0 (25) now reduces to

119908119898=119887119898

3(119909)

119887119898

3(1)

119872120590(1199021) + 6119881

119898

119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)

119887119898

3(1)

(50)

where for 120572 = 2 3

119887119898

120572(119909) = int

119909

0

119889119905

ℎ119898(119905)120572

=1

119898int

119898119909

0

119889119904

ℎ(119904)120572997888rarr 119909int

1

0

119889119904

ℎ(119904)120572

as 119898 997888rarr +infin

(51)

Now 119901119898 the solution to the problem (40) exists if (45) is

fulfilled The second term in (50) thus obviously tends to119909119872(119902

1) as119898 rarr +infin The last term is more interesting The

denominator tends to

⟨1

ℎ3⟩ = int

1

0

119889119904

ℎ(119904)3 (52)

As for its numerator we have

119881119898[119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)]

= 119881119898[int

1

0

int

119909

0

(1

ℎ119898(119905)3ℎ119898(119904)2minus

1

ℎ119898(119905)2ℎ(119904)3

119898

)119889119904 119889119905]

=119881119898

119898(⟨

1

ℎ3⟩int

119898119909

0

119889V

ℎ(V)2minus ⟨

1

ℎ2⟩int

119898119909

0

119889V

ℎ(V)3)

(53)

Suppose that

119881 = lim119898rarr+infin

119881119898

119898(54)

and denote

119866 (119910) = ⟨1

ℎ3⟩int

119910

0

119889V

ℎ(V)2minus ⟨

1

ℎ2⟩int

119910

0

119889V

ℎ(V)3 (55)

Obviously the function119866 is periodicwith period 1 so that dueto the standard periodicity lemma (see eg [19]) as 119898 rarr

+infin

119866 (119898119909) ⟨119866⟩ = int

1

0

119866 (119910) 119889119910 weaklowast in 119871infin(0 1) (56)

By direct computation

⟨119866⟩ = ⟨119910

ℎ3⟩⟨

1

ℎ2⟩ minus ⟨

1

ℎ3⟩⟨

119910

ℎ2⟩

= intint

1

0

119904 minus 119905

ℎ(119904)3ℎ(119905)2119889119904 119889119905

(57)

Thus

119881119898[119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)] 119881 ⟨119866⟩ (58)

Now denoting

1205940(119910) =

119866 (119910)

1198873 (1)

(59)

we have

119908119898asymp 119909119872(119902

1) + 119881120594

0(119898119909) (60)

and thus

119908119898 1199080=119909119872(119902

1) + 119881⟨120594

0⟩ weaklowast in 119871

infin(0 1) (61)

However we are not interested in convergence of theauxiliary function 119908

119898but in the convergence of the pressure

119901119898 Since (45) is assumed to be true we can define 119901

119898=

119867(119908119898) where119867 = 119872

minus1 and we have

119901119898 1199010= int

1

0

119867(119909119872(1199021) + 119881120594

0(119910)) 119889119910

weaklowast in 119871infin(0 1)

(62)

Deriving the expression on the right-hand side we obtainthe effective pressure drop in the form

1199011015840

0(119909) = 119872(119902

1) int

1

0

120583 (119909119872(1199021) + 119881120594

0(119910)) 119889119910 (63)

= int

1199021

1199020

119889120591

120583 (120591)int

1

0

120583 (119909119872(1199021) + 119881120594

0(119910)) 119889119910

= int

1199021

1199020

119889120591

120583 (120591)int

1

0

120583 (119909int

1199021

1199020

119889119904

120583 (119904)+ 1198811205940(119910)) 119889119910

(64)

As we can see the pressure drop is not constant as forthe Newtonian flow The interesting effect appears if 119881 = 0

because in that case the expressions for the pressure and forthe pressure drop are nonlocal due to the integral with respectto 119910 That phenomenon is entirely due to the fact that theviscosity is depending on the pressure

42 Two-Dimensional Case We suppose here that the func-tion ℎ

119898is constructed from positive smooth 119884-periodic

function ℎ R2 rarr [1198890 +infin[ 119889

0gt 0 119884 =]0 1[

2 in the sameway as before that is by taking

ℎ119898(119909) = ℎ (119898119909) (65)

We have seen in the previous section that interesting effectshappen only if we assume that

V119898= 119898V (66)

In that case our equation reads

div(ℎ3

119898

120583 (119901119898)nabla119901119898) = 6V

119898sdot nablaℎ119898

in O (67)

6 Mathematical Problems in Engineering

Aswedid in the existence analysis and in the previous sectionwe linearize the equation by substitution

119908119898= 119872120590(119901119898) (68)

where the function119872120590is defined by (12) Now 119908

119898satisfies

div (ℎ3119872 nabla119908119898) = 6V

119898sdot nablaℎ119898

in O (69)

We postulate the asymptotic expansion in the form

119908119898 (119909) asymp 119908

0(119909 119910) +

1

1198981199081(119909 119910)

+1

11989821199082(119909 119910) + sdot sdot sdot 119910 = 119898119909

(70)

All functions are assumed to be 119884-periodic in 119910 variablePlugging that in (69) and collecting formally terms with

equal powers of119898 we get

1198982 div119910(ℎ3nabla1199101199080) = 6V sdot nabla

119910ℎ

119898 div119910(ℎ3nabla1199101199081) + div

119910(ℎ3nabla1199091199080) + ℎ3div119909nabla1199101199080= 0

1 div119910(ℎ3nabla1199101199082) + div

119910(ℎ3nabla1199091199081) + ℎ3div119909nabla1199101199081

+ ℎ3Δ1199091199080= 0

Denoting

119881 = |V| V0=V119881

(71)

we have

1199080(119909 119910) = V

0(119909) + 119881120594

0(119910)

div119910(ℎ3nabla1199101205940) = 6V

0sdot nabla119910ℎ in 119884

1199081(119909 119910) =

2

sum

119896=1

120594119896(119910)

120597V0

120597119909119896

(119909) + V1(119909)

div119910(ℎ3nabla119910(120594119896+ 119910119896)) = 0

2

sum

119896ℓ=1

a119896ℓ

1205972V0

120597119909119896120597119909ℓ

= 0 in O

a119896ℓ

= int119884

ℎ3 120597

120597119910ℓ

(120594119896+ 119910119896) 119889119910

(72)

Remark 6 Thesame computation can be done in one-dimen-sional case and it gives

(ℎ31205941015840

0)1015840

= 6ℎ1015840997904rArr 1205940= 6int

119910

0

119889119905

ℎ(119905)2+ 1198620int

119910

0

119889119905

ℎ(119905)3+ 1198621

(73)

V0= 119909119872(119902

1) (74)

Constants 1198620 1198621are chosen in a way that boundary condi-

tions 1205940(0) = 120594

0(1) = 0 are met and it follows that

1198621= 0 119862

0= minus 6(int

1

0

119889119905

ℎ(119905)3)

minus1

int

1

0

119889119905

ℎ(119905)2 (75)

Then

119908119898asymp 1199080 (119909119898119909)

= V0(119909) + 120594

0(119898119909)

= 119909119872120590(1199021)

+ 6119881[int

119898119909

0

119889119905

ℎ(119905)2minus (int

1

0

119889119905

ℎ(119905)3)

minus1

times int

1

0

119889119905

ℎ(119905)2

int

119898119909

0

119889119905

ℎ(119905)3]

= 119909119872120590(1199021)

+ 6119881119887119898

2(119909) 1198873(1) minus 119887

2(1) 119887119898

3(119909)

1198873(1)

= 119909119872120590(1199021) + 119881120594

0(119898119909)

(76)

That is a very good approximation of our exact solution (50)It is important to notice that the choice of constants

1198620 1198621was determined from the exterior boundary con-

dition So we should expect the same in two-dimensionalcase However the treatment of boundary conditions intwo-dimensional case is much more complicated and theboundary layer is to be expected

The derived asymptotic expansion should be justified byproving the convergence And we need the strong conver-gence (with corrector of course) for 119908

119898in order to get the

convergence for 119901119898 The form of the approximation

119908119898asymp V0(119909) + 119881120594

0(119898119909) + sdot sdot sdot (77)

suggests that the boundary layer phenomenon should appearon the exterior boundary 120597O since 120594

0term cannot satisfy the

Dirichlet condition on 120597O To get the error estimate and thestrong convergence we need to handle that boundary layerThus at this point we simplify the domain and the boundarycondition in order to be able to avoid it We assume that

O = ]0 1[ times R (78)

119901120576(0 1199092) = 0 119901

120576(1 1199092) = 119902 (119909

2) (79)

1199092997891997888rarr 119901

120576(1199091 1199092) is 1-periodic (80)

ℎ (1199101 1199102) = ℎ (119910

1) (81)

119902 is 1-periodic (82)

Now

119908120576= 119872120590(119901120576) with 119872

120590(119901) =int

119901

120590

119889119904

120583 (119904)997904rArr 119908

120576(0 1199092) = 0

119908120576(1 1199092) = 119872

120590(119902 (1199092))

(83)

and 1199092997891rarr 119908120576(1199091 1199092) is 1-periodic

Mathematical Problems in Engineering 7

In that case we can compute1205940and120594119896 119896 = 1 2 explicitly

and we can impose exterior condition on 1205940 Indeed 120594

0is

exactly the same as in the monodimensional case that is itis given by (73) and (75) Obviously 120594

2= 0 so that

a22

= int

1

0

ℎ(119904)3119889119904 = ⟨ℎ

3⟩ a

12= a21

= 0 (84)

As for the last term

1205941= minus1199101+ (int

1

0

119889119904

ℎ(119904)3)

minus1

int

1199101

0

119889119904

ℎ(119904)3

a11

= (int

1

0

119889119904

ℎ(119904)3)

minus1

=1

⟨1ℎ3⟩

(85)

Finally the function V0satisfies the boundary value problem

1

⟨1ℎ3⟩

1205972V0

12059711990921

+ ⟨ℎ3⟩1205972V0

12059711990922

= 0 in O

V0(0 1199092) = 0

V0(1 1199092) = 119872

120590(119902 (1199092)) V

0is 1-periodic in 119909

2

(86)

It can be solved using the Fourier method and we get

V0(1199091 1199092) =

infin

sum

119896=1

sh (radic⟨ℎ3⟩ ⟨ℎminus3⟩ 1198961205871199091)

times (120572119896sin 119896120587119909

2+ 120573119896cos 119896120587119909

2)

120572119896= 2int

1

0

119872120590(119902 (119905)) sin 119896120587119905119889119905

120573119896= 2int

1

0

119872120590(119902 (119905)) cos 119896120587119905119889119905

(87)

Since the approximation

w119898asymp V0(119909) + 119881120594

0(1198981199091) +

1

1198981205941(1198981199091)120597V0

1205971199091

(119909) (88)

now satisfies the boundary conditions on 120597O it is easy to seethat

1003816100381610038161003816119908119898 minus (V0(119909) + 119881120594

0(119898119909))

1003816100381610038161003816119871infin(O) le 1198621

119898(89)

follows from the maximum principle Assuming that for 119898large enough

V0(119909) + 119881120594

0(119898119909) le 119872

+ (90)

we have1003816100381610038161003816119901119898 minus 119867 (V

0(119909) + 119881120594

0(119898119909))

1003816100381610038161003816119871infin(O) 997888rarr 0 (91)

Finally

119901119898 1199010= int

1

0

119867(V0(119909) + 119881120594

0(1199101)) 1198891199101 (92)

We have proved that

Theorem 7 Let 119901119898be the solution to the problem (67) (79)

and (80) and let V01205940be defined by (87) and (42) respectively

If (90) holds then1003816100381610038161003816119901119898 minus 119867 (V

0 (119909) + 1198811205940 (119898119909))

1003816100381610038161003816119871infin(O) 997888rarr 0

119901119898 1199010= int

1

0

119867(V0(119909) + 119881120594

0(1199101)) 1198891199101

(93)

Remark 8 It is important to notice that 1205870= 119867(119901

0) satisfies

2

sum

119894119896=1

120597

120597119909119894

(a119894119896

120583 (1205870)

1205971205870

120597119909119896

) = 0 in O 1205870= 119902 on 120597O (94)

and thus we would expect it to be the limit of 119901119898in analogy

with the linear case However 1199010

= 1205870

If119881 = lim119898rarrinfin

119898|V119898| is small we can expand119867(V

0(119909)+

1198811205940(119898119909)) in powers of 119881 and we get

119901119898(119909) asymp 119867 (V

0(119909) + 119881120594

0(119898119909))

= 119867 (V0(119909)) + 119881119867

1015840(V0(119909)) 120594

0(119910) + 119874 (119881

2)

= 1205870(119909) + 119881120583 (120587

0(119909)) 120594

0(119910) + 119874 (119881

2)

(95)

Thus

1199010= 1205870+ 120583 (120587

0) ⟨1205940⟩119881 + 119874 (119881

2) (96)

It can be formally written as

119901119898(119909) asymp 120587

0(119909) +

1003816100381610038161003816V1198981003816100381610038161003816

119898120583 (1205870(119909)) 120594

0(119898119909) + 119874(

1003816100381610038161003816V11989810038161003816100381610038162

1198982)

(97)

Appendix

The Maximum Principle

Our goal is to derive maximum principles for the linearReynolds equation with sharp explicit constants in orderto solve the nonlinear Reynolds equation with pressure-dependent viscosity We assume without losing generalitythat V = (16)Vi Indeed we can always choose the coor-dinate system in a way that the first coordinate axis 119909 has adirection of the velocity of relative motion V

The lower bound for 119908(119909 119910) is of no interest justthe upper bound Function 119908(119909 119910) is the solution to theboundary value problem

div (ℎ3nabla119908) = V120597ℎ

120597119909in O sub R2

119908 = 119872120590(119902) on 120597O

(A1)

We assume that if V(120597ℎ120597119909) gt 0 then 119908 cannot have amaximum point in the domain O and thus

119908 (119909 119910) le max(119909119910)isin120597O

119872120590(119902 (119909 119910)) (A2)

8 Mathematical Problems in Engineering

However it is not realistic to assume that (120597ℎ120597119909) does notchange the sign To find the upper bound in the general casewe use the procedure from the DeGiorgi theorem The mainresult of the section is as follows

TheoremA1 Let119908 be the solution to the problem (A1)Then

119908 (119909 119910) le1003816100381610038161003816119872120590 (119902)

1003816100381610038161003816119871infin(120597O) +Z (V ℎO 120583 119902) (A3)

Z=3(85)

times(3

2)

(285)(2120587)(14)

ℎ3

0

diamO (|V| ℎ1|O|15

+ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times |O|(75)

(A4)

Proof The function 119911 = 119908 minus 119866 satisfies

div (ℎ3nabla119911) = V120597ℎ

120597119909+ div (ℎ3nabla119872

120590(119902)) in O sub R2 (A5)

119911 = 0 on 120597O (A6)

Next we introduce the embedding constant for 119882120

(O) sub

119871119903(O) denoted119872

119903 such that

|V|119871119903(O) le 119872119903|nablaV|1198712(O) forallV isin 119867

1

0(O) (A7)

That constant can be estimated as

119872119903le1

2(diamO)

2|O|1119903 minus 12

(119903 + 2

2)

((119903+2)2119903)

radic2120587 (A8)

See for example [20 Lemma 1] Next we define the se-quence

120582119896+1

= 3(120582119896

2+ 1) 120582

1= 2 (A9)

Easy computation yields

120582119896= 8(

3

2)

119896minus1

minus 6 (A10)

Let

119911+(119909 119910) = max 119911 (119909 119910) 0 (A11)

We test (A5) with (119911+)1+120582119896+1 and get

intO

ℎ3nabla119911+nabla(119911+)1+120582119896+1

=1 + 120582119896+1

(1 + 120582119896+1

2)2intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

= intO

[Vℎ120597

120597119909(119911+)1+120582119896+1

+ ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

]

(A12)

For the left-hand side we get the lower bound

ℎ3

0

1 + 120582119896+1

(1 + 120582119896+1

2)2

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

1198712(O)

(A13)

We estimate the terms on the right-hand side using the sameidea

intO

ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

le ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O)

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

intO

ℎV120597

120597119909(119911+)1+120582119896+1

le ℎ1 |V| |O|

15100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

(A14)

Thus it remains to estimate |nabla(119911+)1+120582119896+1 |11987154(O) We have

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

100381610038161003816100381610038161003816

54

11987154(O)

= (1 + 120582119896+1

)54

intO

(119911+)(5120582119896+14)1003816100381610038161003816nabla119911

+1003816100381610038161003816

54

= (1 + 120582119896+1

1 + (120582119896+1

2))

54

intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

541003816100381610038161003816119911+1003816100381610038161003816

(5120582119896+18)

le [Holders inequality with 119901 =8

5 1199011015840=8

3]

le (1 + 120582119896+1

1 + (120582119896+1

2))

54

(intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

)

58

times (intO

(119911+)(53)120582

119896+1

)

38

= [due to (A9) 53120582119896+1

= 5(1 +120582119896

2)

3

8=1

4

120582119896+1

2 + 120582119896

]

= (1 + 120582119896+1

1 + (120582119896+1

2))

54100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

54

1198712(O)

times100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(120582119896+1(2+120582

119896))(54)

1198715(O)

(A15)

Combining with (A14) and (A13) we get

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)1003816100381610038161003816100381610038161198712(O)

le ℎminus3

0(1 +

120582119896+1

2)100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198715(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

Mathematical Problems in Engineering 9

le 119872((120582119896+1)(2+120582

119896))

5ℎminus3

0(1+

120582119896+1

2)

times100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198712(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

(A16)

We recall that

120582119896+1

2 + 120582119896

=3

2(A17)

and define

120572 = 119872(32)

5ℎminus3

0(ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

) (A18)

as well as

120590119896=

1

120572

100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(2(2+120582119896))

1198712(O)

(A19)

Then (A16) implies

120590119896+1

le (1 +120582119896+1

2)

(2(2+120582119896))

120590((120582119896+1)(2+120582

119896+1))

119896 (A20)

Taking the logarithm we arrive at

log120590119896+1

lelog (1 + (120582

119896+12))

1 + (120582119896+1

2)+

120582119896+1

2 + 120582119896+1

log120590119896 (A21)

We first notice that

120582119896+1

2 + 120582119896+1

lt 1 (A22)

and then

1 +120582119896

2= 4(

3

2)

119896minus1

minus 2 gt (3

2)

119896

(A23)

Since the function 119909 997891rarr (log119909119909) is decreasing for 119909 gt 119890 wehave

log (1 + (1205821198962))

1 + (1205821198962)

lelog [(32)119896]

(32)119896

= 119896 log 3

2(2

3)

119896

forall119896 ge 3

(A24)

Then

log120590119896+1

le (119896 + 1) log 3

2(2

3)

119896

+ log120590119896

le log 3

2

119896+1

sum

119895=2

119895(2

3)

119895

+ log1205901le 8 log 3

2+ log120590

1

(A25)

Finally

120590119896+1

le (3

2)

8

1205901 (A26)

Now it remains to estimate 1205901 From the definitionwe see that

1205901=

1

120572

100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

12

1198712(O)

(A27)

To estimate 1205901 we proceed as before and test (A5) with (119911+)3

We get

3

4intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

2

= intO

(Vℎ120597(119911+)3

120597119909+ ℎ3nabla119872120590(119902) nabla(119911

+)3

)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

100381610038161003816100381610038161003816nabla(119911+)310038161003816100381610038161003816100381611987154(O)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times3

2

100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

1003816100381610038161003816119911+1003816100381610038161003816119871103(O)

(A28)

Thus100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

le2

ℎ3

0

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872(103)

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O)

(A29)

Finally testing (A5) with 119911+ we get

intO

ℎ31003816100381610038161003816nabla119911+1003816100381610038161003816

2

= intO

(Vℎ120597119911+

120597119909+ ℎ3nabla119872120590(119902) nabla119911

+) (A30)

so that

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O) le

1

ℎ3

0

(|V| ℎ1|O|12

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198712(O))

le |O|310

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

(A31)

Combining (A29) with (A31) and (A26) gives

120590119896+1

le (3

2)

81

120572

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A32)

Since

120590119896+1

ge1

120572119872(2(2+120582k))2

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) (A33)

10 Mathematical Problems in Engineering

we have arrived to

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) le (

3

2)

8

119872(2(2+120582

119896))

2

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A34)

Since lim119896rarrinfin

120582119896= infin we get

1003816100381610038161003816119911+1003816100381610038161003816119871infin(O) le (

3

2)

8radic2

ℎ3

0

times(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A35)

Finally (A8) implies (A3)

Acknowledgment

This work was supported by MZOS grant 037-0372787-2797

References

[1] O Reynolds ldquoOn the theory and application and its applicationto Mr Beauchamp towerrsquos experiments Including and experi-mental determination of the viscosity of olive oilrdquo PhilosophicalTransactions of the Royal Society vol 117 pp 157ndash234 1886

[2] G G Stokes ldquoNotes on hydrodynamics On the dynamicalequationsrdquoCambridge and DublinMathematical Journal III pp121ndash127 1848

[3] P W Brifgman ldquoThe viscosity of liquids under pressurerdquo Pro-ceedings of the National Academy of Sciences of the United Statesof America vol 11 no 10 pp 603ndash606 1925

[4] A Z Szeri Fluid Film Lubrication Cambridge University PressNew York NY USA 1998

[5] E K Gatcombe ldquoLubrication characteristics of involute spur-gearsmdasha theoretical investigationrdquoTransactions of theAmericanSociety of Mechanical Engineers vol 67 pp 177ndash181 1945

[6] C Barus ldquoIsotherms isopiestics and isometrics relative to vis-cosityrdquo American Journal of Science vol 45 pp 87ndash96 1893

[7] W R Jones ldquoPressure viscosity measurement for several lubri-cantsrdquo ASLE Transactions vol 18 pp 249ndash262 1975

[8] J Hron J Malek and K R Rajagopal ldquoSimple flows of fluidswith pressure-dependent viscositiesrdquo Proceedings of the RoyalSociety A vol 457 no 2011 pp 1603ndash1622 2001

[9] C J A Roelands Correlation aspects of the viscosity-pressurerelationship of lubricating oils [PhD thesis] Delft University ofTechnology Delft The Netherlands 1966

[10] K R Rajagopal and A Z Szeri ldquoOn an inconsistency in thederivation of the equations of elastohydrodynamic lubricationrdquoProceedings of the Royal Society A vol 459 no 2039 pp 2771ndash2786 2003

[11] M Renardy ldquoSome remarks on the Navier-Stokes equationswith a pressure-dependent viscosityrdquo Communications in Par-tial Differential Equations vol 11 no 7 pp 779ndash793 1986

[12] F Gazzola and P Secchi ldquoSome results about stationary Navier-Stokes equations with a pressure-dependent viscosityrdquo in Pro-ceedings of the International Conference on Navier-Stokes Equa-tions vol 388 of Pitman Research Notes in Mathematics Seriespp 174ndash183 Varenna Italy 1998

[13] E Marusic-Paloka ldquoAn analysis of the Stokes system with pres-sure dependent viscosityrdquo In press

[14] S Marusic and E Marusic-Paloka ldquoTwo-scale convergence forthin domains and its applications to some lower-dimensionalmodels in fluid mechanicsrdquo Asymptotic Analysis vol 23 no 1pp 23ndash57 2000

[15] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001

[16] M Kane and B Bou-Said ldquoComparison of homogenization anddirect techniques for the treatment of roughness in incompress-ible lubricationrdquo Journal of Tribology vol 126 no 4 pp 733ndash7372004

[17] M Jai ldquoHomogenization and two-scale convergence of thecompressible Reynolds lubrication equation modelling the fly-ing characteristics of a roughmagnetic head over a rough rigid-disk surfacerdquo RAIRO Modelisation Mathematique et AnalyseNumerique vol 29 no 2 pp 199ndash233 1995

[18] P Wall ldquoHomogenization of Reynolds equation by two-scaleconvergencerdquo Chinese Annals of Mathematics B vol 28 no 3pp 363ndash374 2007

[19] A Bensoussan J-L Lions and G Papanicolaou AsymptoticAnalysis for Periodic Structures North-Holland AmsterdamThe Netherlands 1978

[20] E Marusic-Paloka and A Mikelic ldquoThe derivation of a nonlin-ear filtration law including the inertia effects via homogeniza-tionrdquo Nonlinear Analysis Theory Methods amp Applications vol42 no 1 pp 97ndash137 2000

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

Deriving (13) we obtain

1

120583 (119901)nabla119901 = nabla119908 (17)

and the problem can be written as

div(ℎ3nabla119908 ) = 6V sdot nablaℎ in O (18)

119908 = 119872120590(119902) on 120597O (19)

That is a linear elliptic equation for 119908 and it has a uniquesolution To get the existence and uniqueness of the solutionwe quote Theorem 834 from classical book of Gilbarg andTrudinger [15] For simplicity here and in the sequel weassume that 119902 and consequently119872

120590(119902) are defined on whole

O We combine that with the maximum principle from theappendix and it gives the following

Theorem 1 Under the assumption that the boundary 120597O is ofclass 1198621120572 and that ℎ isin 119862

120572(O) 119902 isin 119862

1120572(O) the problem (18)

(19) has a unique solution

119908 isin 1198621120572

(O) (20)

Furthermore

119908 (119909) le 119872(119902) +Z (21)

where

119902 =10038161003816100381610038161199021003816100381610038161003816119871infin(120597120596) (22)

andZ = 0 if V sdot nablah lt 0 Otherwise

Z =

6 |V| [int1

0

119889119905

ℎ(119905)2minusℎ3

0

ℎ5

1

] + (ℎ0

ℎ1

)

3

int

1199021

1199020

119889119904

120583 (119904)

if 119899 = 1

3(85)

(3

2)

(285)(2120587)(14)

ℎ3

0

times119889 (6 |V| ℎ1|O|15 + ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O)) |O|

(75)

if 119899 = 2

(23)

with 119889 = diamO

Proof The existence follows directly from Theorem 834from Gilbarg and Trudinger [15] If V sdot nablaℎ lt 0 then (21)follows directly from the weak maximum principle (see eg[15]) In case 119899 = 1 the problem can be solved by quadraturesand the solution given by (25) can be easily estimated toget (23) In the remaining case 119899 = 2 (21) follows fromthe special variant of the maximum principle proved in theappendix

Remark 2 In case 119899 = 1 (18) is anODE (we takeO =]0 1[ and1199020gt 1199021 without losing generality)

(ℎ31199081015840)1015840

= 6119881ℎ1015840 in ]0 1[

119908 (119894) = 119872120590(119902119894) for 119894 = 0 1

(24)

and it can be solved by quadratures

119908 (119909) = 119872120590(1199020)

+ int

119909

0

1

ℎ(119905)3

[

[

6119881(ℎ (119905)minusint1

0(119889119903ℎ(119903)

2)

int1

0(119889119904ℎ(119904)

3)

)

+119872120590(1199021)minus119872120590(1199020)

int1

0(119889119904ℎ(119904)

3)

]

]

119889119905

(25)

32 Back to the Original Equation Now our goal is not tofind the auxiliary function119908 but to find the pressure 119901 Sincewe have introduced 119908 as

119908 = 119872120590(119901) (26)

we should have 119901 = 119872minus1

120590(119908) In order to do so we have to

make sure that 119908(119909) isin Im119872120590for any 119909 isin O Since 119872

120590is

strictly increasing and we have assumed that (14) holds if wedefine

119872+

120590= lim119904rarr+infin

119872120590(119904) = int

+infin

120590

119889119904

120583 (119904) (27)

due to (16) we obviously have for any 119901 isin R

minusinfin = lim119904rarrminusinfin

119872120590(119904) le 119872

120590(119901) le 119872

+

120590 (28)

Thus

Im119872120590= ]minusinfin119872

+

120590[ (29)

So to fulfill the condition 119908(119909) isin Im119872120590we need to have

119908 (119909) le 119872+

120590 forall119909 isin O (30)

That condition is not necessarily fulfilledIn view of (21) that condition reduces to

Z le int

+infin

119902

119889119905

120583 (119905) (31)

whereZ = Z(V ℎO 120583 119902) is defined by (23)We have proved the following theorem

Theorem 3 Suppose that the conditions of Theorem 1 holdand that in addition (31) is fulfilled Then 119901 = 119872

minus1

120590(119908) is the

unique solution of (10) and (11)

Remark 4 It is important to notice that even though 119908 doesdepend on 120590 the effective pressure 119901 does not For thepurpose of this remark we denote119867

120590(119908) = 119872

minus1

120590(119908) to stress

the dependence on the parameter 120590 which is of interest hereWe start by

119901 = 119867120590(119908) (32)

4 Mathematical Problems in Engineering

It is obvious from the definition of119872120590that

120597119872120590

120597119901(119901) =

1

120583 (119901)

120597119872120590

120597120590(119901) = minus

1

120583 (120590) (33)

As119872120590(119867120590(119908)) = 119908 deriving with respect to 120590 we arrive at

120597119867120590

120597120590(119908) = minus

(120597119872120590120597120590) (119901)

(120597119872120590120597119901) (119901)

=120583 (119901)

120583 (120590)=120583 (119867120590(119908))

120583 (120590)

(34)

Deriving (13) we get (120597119908120597120590) = minus1120583(120590) Using the rulefor deriving the inverse function we have

120597119867120590

120597119908(119908) = 120583 (119901) = 120583 (119867

120590(119908)) (35)

Thus120597119901

120597120590=120597119867120590

120597119908(119908)

120597119908

120597120590(120590) +

120597119867120590

120597120590(119908)

= minus120583 (119867120590(119908))

120583 (120590)+120583 (119867120590(119908))

120583 (120590)= 0

(36)

4 Homogenization

In this section we want to study the effects of rugositiesof surfaces on lubrication process The idea of finding themacroscopic effects of roughness on lubrication process viahomogenization is quite old and well studied Case of con-stant viscosity for incompressible and compressible flows aswell as non-Newtonian deformation dependent viscositieswere investigated The subject was treated by several authorsandwe heremention [16ndash18]The case of pressure-dependentviscosity brings some new interesting nonlocal effects

We assume that the function ℎ describing the form of thefluid domain is periodic with small period 1119898 with119898 isin NTo stress that dependence we denote it by ℎ

119898 More precisely

we denote by 119884 =]0 1[119899 119899 = 1 2 the period We further

assume that ℎ R119899 rarr [1198890 +infin[ 119889

0gt 0 is periodic with

period 119884 and smooth Then we take ℎ119898of the form

ℎ119898(119909) = ℎ (119898119909) (37)

Thus the function ℎ describes the form of periodicallydistributed rugosities

To emphasize that the relative velocity of bearing surfacesV is large we assume that it also depends on 119898 the sameparameter that is taken for description of rugosities In thatcase our equation reads

div(ℎ3

119898

120583 (119901119898)nabla119901119898) = 6V

119898sdot nablaℎ119898

in O (38)

41 One-Dimensional Case If 119899 = 1 the above problem isposed on an intervalO =]0 1[With an appropriate boundarycondition

119901119898(0) = 119902

0 119901

119898(1) = 119902

1 (39)

It forms a boundary value problem for nonlinear ODE

(ℎ3

119898

120583 (119901119898)1199011015840

119898)

1015840

= 6119881119898ℎ1015840

119898in O (40)

To study the asymptotic behavior of the solution with respectto119898we linearize the problem using the transformation119908

119898=

119872120590(119901119898) To simplify in this section we choose 120590 = 119902

0and

dropping the index 120590 in119872120590and119872

+

120590 we denote

119872(119901) = int

119901

1199020

119889119905

120583 (119905) 119872

+= int

+infin

1199020

119889119905

120583 (119905) (41)

Theorem 5 Let

1205940(119910) = 6 (int

1

0

119889119904

ℎ(119904)3)

minus1

times [⟨1

ℎ3⟩int

119910

0

119889V

ℎ(V)2minus⟨

1

ℎ2⟩int

119910

0

119889V

ℎ(V)3]

(42)

⟨sdot⟩ = int

1

0

sdot 119889119910 (43)

and let119867 = 119872minus1 Suppose that there exists a limit

119881 = lim119898rarrinfin

119881119898

119898(44)

and that for 119898 large enough and 119872+ defined in (27) the

following condition holds

119887119898

3(119909)

119887119898

3(1)

119872 (1199021) + 6119881

119898

119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)

119887119898

3(1)

le 119872+

(45)

with

119887119898

120572(119909) = int

119909

0

119889119905

ℎ119898(119905)120572 120572 = 2 3 (46)

Then

119901119898 1199010=int

1

0

119867(119909119872(1199021) + 119881120594

0(119910)) 119889119910

weaklowast in 119871infin(0 1)

(47)

Proof Equation (40) with boundary conditions (39) can besolved by reduction to quadratures after the substitution

119908119898= 119872(119901

119898) (48)

with 119872120590strictly increasing function defined by (12) The

problem for 119908119898now reads

(ℎ3

1198981199081015840

119898)1015840

= 6119881119898ℎ1015840

119898in ]0 1[

119908119898(0) = 119872(119902

0) = 0 119908

119898(1) = 119872(119902

1)

(49)

Mathematical Problems in Engineering 5

It is easy to see that (49) has a unique solution given by (25)Since119872(119902

0) = 0 (25) now reduces to

119908119898=119887119898

3(119909)

119887119898

3(1)

119872120590(1199021) + 6119881

119898

119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)

119887119898

3(1)

(50)

where for 120572 = 2 3

119887119898

120572(119909) = int

119909

0

119889119905

ℎ119898(119905)120572

=1

119898int

119898119909

0

119889119904

ℎ(119904)120572997888rarr 119909int

1

0

119889119904

ℎ(119904)120572

as 119898 997888rarr +infin

(51)

Now 119901119898 the solution to the problem (40) exists if (45) is

fulfilled The second term in (50) thus obviously tends to119909119872(119902

1) as119898 rarr +infin The last term is more interesting The

denominator tends to

⟨1

ℎ3⟩ = int

1

0

119889119904

ℎ(119904)3 (52)

As for its numerator we have

119881119898[119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)]

= 119881119898[int

1

0

int

119909

0

(1

ℎ119898(119905)3ℎ119898(119904)2minus

1

ℎ119898(119905)2ℎ(119904)3

119898

)119889119904 119889119905]

=119881119898

119898(⟨

1

ℎ3⟩int

119898119909

0

119889V

ℎ(V)2minus ⟨

1

ℎ2⟩int

119898119909

0

119889V

ℎ(V)3)

(53)

Suppose that

119881 = lim119898rarr+infin

119881119898

119898(54)

and denote

119866 (119910) = ⟨1

ℎ3⟩int

119910

0

119889V

ℎ(V)2minus ⟨

1

ℎ2⟩int

119910

0

119889V

ℎ(V)3 (55)

Obviously the function119866 is periodicwith period 1 so that dueto the standard periodicity lemma (see eg [19]) as 119898 rarr

+infin

119866 (119898119909) ⟨119866⟩ = int

1

0

119866 (119910) 119889119910 weaklowast in 119871infin(0 1) (56)

By direct computation

⟨119866⟩ = ⟨119910

ℎ3⟩⟨

1

ℎ2⟩ minus ⟨

1

ℎ3⟩⟨

119910

ℎ2⟩

= intint

1

0

119904 minus 119905

ℎ(119904)3ℎ(119905)2119889119904 119889119905

(57)

Thus

119881119898[119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)] 119881 ⟨119866⟩ (58)

Now denoting

1205940(119910) =

119866 (119910)

1198873 (1)

(59)

we have

119908119898asymp 119909119872(119902

1) + 119881120594

0(119898119909) (60)

and thus

119908119898 1199080=119909119872(119902

1) + 119881⟨120594

0⟩ weaklowast in 119871

infin(0 1) (61)

However we are not interested in convergence of theauxiliary function 119908

119898but in the convergence of the pressure

119901119898 Since (45) is assumed to be true we can define 119901

119898=

119867(119908119898) where119867 = 119872

minus1 and we have

119901119898 1199010= int

1

0

119867(119909119872(1199021) + 119881120594

0(119910)) 119889119910

weaklowast in 119871infin(0 1)

(62)

Deriving the expression on the right-hand side we obtainthe effective pressure drop in the form

1199011015840

0(119909) = 119872(119902

1) int

1

0

120583 (119909119872(1199021) + 119881120594

0(119910)) 119889119910 (63)

= int

1199021

1199020

119889120591

120583 (120591)int

1

0

120583 (119909119872(1199021) + 119881120594

0(119910)) 119889119910

= int

1199021

1199020

119889120591

120583 (120591)int

1

0

120583 (119909int

1199021

1199020

119889119904

120583 (119904)+ 1198811205940(119910)) 119889119910

(64)

As we can see the pressure drop is not constant as forthe Newtonian flow The interesting effect appears if 119881 = 0

because in that case the expressions for the pressure and forthe pressure drop are nonlocal due to the integral with respectto 119910 That phenomenon is entirely due to the fact that theviscosity is depending on the pressure

42 Two-Dimensional Case We suppose here that the func-tion ℎ

119898is constructed from positive smooth 119884-periodic

function ℎ R2 rarr [1198890 +infin[ 119889

0gt 0 119884 =]0 1[

2 in the sameway as before that is by taking

ℎ119898(119909) = ℎ (119898119909) (65)

We have seen in the previous section that interesting effectshappen only if we assume that

V119898= 119898V (66)

In that case our equation reads

div(ℎ3

119898

120583 (119901119898)nabla119901119898) = 6V

119898sdot nablaℎ119898

in O (67)

6 Mathematical Problems in Engineering

Aswedid in the existence analysis and in the previous sectionwe linearize the equation by substitution

119908119898= 119872120590(119901119898) (68)

where the function119872120590is defined by (12) Now 119908

119898satisfies

div (ℎ3119872 nabla119908119898) = 6V

119898sdot nablaℎ119898

in O (69)

We postulate the asymptotic expansion in the form

119908119898 (119909) asymp 119908

0(119909 119910) +

1

1198981199081(119909 119910)

+1

11989821199082(119909 119910) + sdot sdot sdot 119910 = 119898119909

(70)

All functions are assumed to be 119884-periodic in 119910 variablePlugging that in (69) and collecting formally terms with

equal powers of119898 we get

1198982 div119910(ℎ3nabla1199101199080) = 6V sdot nabla

119910ℎ

119898 div119910(ℎ3nabla1199101199081) + div

119910(ℎ3nabla1199091199080) + ℎ3div119909nabla1199101199080= 0

1 div119910(ℎ3nabla1199101199082) + div

119910(ℎ3nabla1199091199081) + ℎ3div119909nabla1199101199081

+ ℎ3Δ1199091199080= 0

Denoting

119881 = |V| V0=V119881

(71)

we have

1199080(119909 119910) = V

0(119909) + 119881120594

0(119910)

div119910(ℎ3nabla1199101205940) = 6V

0sdot nabla119910ℎ in 119884

1199081(119909 119910) =

2

sum

119896=1

120594119896(119910)

120597V0

120597119909119896

(119909) + V1(119909)

div119910(ℎ3nabla119910(120594119896+ 119910119896)) = 0

2

sum

119896ℓ=1

a119896ℓ

1205972V0

120597119909119896120597119909ℓ

= 0 in O

a119896ℓ

= int119884

ℎ3 120597

120597119910ℓ

(120594119896+ 119910119896) 119889119910

(72)

Remark 6 Thesame computation can be done in one-dimen-sional case and it gives

(ℎ31205941015840

0)1015840

= 6ℎ1015840997904rArr 1205940= 6int

119910

0

119889119905

ℎ(119905)2+ 1198620int

119910

0

119889119905

ℎ(119905)3+ 1198621

(73)

V0= 119909119872(119902

1) (74)

Constants 1198620 1198621are chosen in a way that boundary condi-

tions 1205940(0) = 120594

0(1) = 0 are met and it follows that

1198621= 0 119862

0= minus 6(int

1

0

119889119905

ℎ(119905)3)

minus1

int

1

0

119889119905

ℎ(119905)2 (75)

Then

119908119898asymp 1199080 (119909119898119909)

= V0(119909) + 120594

0(119898119909)

= 119909119872120590(1199021)

+ 6119881[int

119898119909

0

119889119905

ℎ(119905)2minus (int

1

0

119889119905

ℎ(119905)3)

minus1

times int

1

0

119889119905

ℎ(119905)2

int

119898119909

0

119889119905

ℎ(119905)3]

= 119909119872120590(1199021)

+ 6119881119887119898

2(119909) 1198873(1) minus 119887

2(1) 119887119898

3(119909)

1198873(1)

= 119909119872120590(1199021) + 119881120594

0(119898119909)

(76)

That is a very good approximation of our exact solution (50)It is important to notice that the choice of constants

1198620 1198621was determined from the exterior boundary con-

dition So we should expect the same in two-dimensionalcase However the treatment of boundary conditions intwo-dimensional case is much more complicated and theboundary layer is to be expected

The derived asymptotic expansion should be justified byproving the convergence And we need the strong conver-gence (with corrector of course) for 119908

119898in order to get the

convergence for 119901119898 The form of the approximation

119908119898asymp V0(119909) + 119881120594

0(119898119909) + sdot sdot sdot (77)

suggests that the boundary layer phenomenon should appearon the exterior boundary 120597O since 120594

0term cannot satisfy the

Dirichlet condition on 120597O To get the error estimate and thestrong convergence we need to handle that boundary layerThus at this point we simplify the domain and the boundarycondition in order to be able to avoid it We assume that

O = ]0 1[ times R (78)

119901120576(0 1199092) = 0 119901

120576(1 1199092) = 119902 (119909

2) (79)

1199092997891997888rarr 119901

120576(1199091 1199092) is 1-periodic (80)

ℎ (1199101 1199102) = ℎ (119910

1) (81)

119902 is 1-periodic (82)

Now

119908120576= 119872120590(119901120576) with 119872

120590(119901) =int

119901

120590

119889119904

120583 (119904)997904rArr 119908

120576(0 1199092) = 0

119908120576(1 1199092) = 119872

120590(119902 (1199092))

(83)

and 1199092997891rarr 119908120576(1199091 1199092) is 1-periodic

Mathematical Problems in Engineering 7

In that case we can compute1205940and120594119896 119896 = 1 2 explicitly

and we can impose exterior condition on 1205940 Indeed 120594

0is

exactly the same as in the monodimensional case that is itis given by (73) and (75) Obviously 120594

2= 0 so that

a22

= int

1

0

ℎ(119904)3119889119904 = ⟨ℎ

3⟩ a

12= a21

= 0 (84)

As for the last term

1205941= minus1199101+ (int

1

0

119889119904

ℎ(119904)3)

minus1

int

1199101

0

119889119904

ℎ(119904)3

a11

= (int

1

0

119889119904

ℎ(119904)3)

minus1

=1

⟨1ℎ3⟩

(85)

Finally the function V0satisfies the boundary value problem

1

⟨1ℎ3⟩

1205972V0

12059711990921

+ ⟨ℎ3⟩1205972V0

12059711990922

= 0 in O

V0(0 1199092) = 0

V0(1 1199092) = 119872

120590(119902 (1199092)) V

0is 1-periodic in 119909

2

(86)

It can be solved using the Fourier method and we get

V0(1199091 1199092) =

infin

sum

119896=1

sh (radic⟨ℎ3⟩ ⟨ℎminus3⟩ 1198961205871199091)

times (120572119896sin 119896120587119909

2+ 120573119896cos 119896120587119909

2)

120572119896= 2int

1

0

119872120590(119902 (119905)) sin 119896120587119905119889119905

120573119896= 2int

1

0

119872120590(119902 (119905)) cos 119896120587119905119889119905

(87)

Since the approximation

w119898asymp V0(119909) + 119881120594

0(1198981199091) +

1

1198981205941(1198981199091)120597V0

1205971199091

(119909) (88)

now satisfies the boundary conditions on 120597O it is easy to seethat

1003816100381610038161003816119908119898 minus (V0(119909) + 119881120594

0(119898119909))

1003816100381610038161003816119871infin(O) le 1198621

119898(89)

follows from the maximum principle Assuming that for 119898large enough

V0(119909) + 119881120594

0(119898119909) le 119872

+ (90)

we have1003816100381610038161003816119901119898 minus 119867 (V

0(119909) + 119881120594

0(119898119909))

1003816100381610038161003816119871infin(O) 997888rarr 0 (91)

Finally

119901119898 1199010= int

1

0

119867(V0(119909) + 119881120594

0(1199101)) 1198891199101 (92)

We have proved that

Theorem 7 Let 119901119898be the solution to the problem (67) (79)

and (80) and let V01205940be defined by (87) and (42) respectively

If (90) holds then1003816100381610038161003816119901119898 minus 119867 (V

0 (119909) + 1198811205940 (119898119909))

1003816100381610038161003816119871infin(O) 997888rarr 0

119901119898 1199010= int

1

0

119867(V0(119909) + 119881120594

0(1199101)) 1198891199101

(93)

Remark 8 It is important to notice that 1205870= 119867(119901

0) satisfies

2

sum

119894119896=1

120597

120597119909119894

(a119894119896

120583 (1205870)

1205971205870

120597119909119896

) = 0 in O 1205870= 119902 on 120597O (94)

and thus we would expect it to be the limit of 119901119898in analogy

with the linear case However 1199010

= 1205870

If119881 = lim119898rarrinfin

119898|V119898| is small we can expand119867(V

0(119909)+

1198811205940(119898119909)) in powers of 119881 and we get

119901119898(119909) asymp 119867 (V

0(119909) + 119881120594

0(119898119909))

= 119867 (V0(119909)) + 119881119867

1015840(V0(119909)) 120594

0(119910) + 119874 (119881

2)

= 1205870(119909) + 119881120583 (120587

0(119909)) 120594

0(119910) + 119874 (119881

2)

(95)

Thus

1199010= 1205870+ 120583 (120587

0) ⟨1205940⟩119881 + 119874 (119881

2) (96)

It can be formally written as

119901119898(119909) asymp 120587

0(119909) +

1003816100381610038161003816V1198981003816100381610038161003816

119898120583 (1205870(119909)) 120594

0(119898119909) + 119874(

1003816100381610038161003816V11989810038161003816100381610038162

1198982)

(97)

Appendix

The Maximum Principle

Our goal is to derive maximum principles for the linearReynolds equation with sharp explicit constants in orderto solve the nonlinear Reynolds equation with pressure-dependent viscosity We assume without losing generalitythat V = (16)Vi Indeed we can always choose the coor-dinate system in a way that the first coordinate axis 119909 has adirection of the velocity of relative motion V

The lower bound for 119908(119909 119910) is of no interest justthe upper bound Function 119908(119909 119910) is the solution to theboundary value problem

div (ℎ3nabla119908) = V120597ℎ

120597119909in O sub R2

119908 = 119872120590(119902) on 120597O

(A1)

We assume that if V(120597ℎ120597119909) gt 0 then 119908 cannot have amaximum point in the domain O and thus

119908 (119909 119910) le max(119909119910)isin120597O

119872120590(119902 (119909 119910)) (A2)

8 Mathematical Problems in Engineering

However it is not realistic to assume that (120597ℎ120597119909) does notchange the sign To find the upper bound in the general casewe use the procedure from the DeGiorgi theorem The mainresult of the section is as follows

TheoremA1 Let119908 be the solution to the problem (A1)Then

119908 (119909 119910) le1003816100381610038161003816119872120590 (119902)

1003816100381610038161003816119871infin(120597O) +Z (V ℎO 120583 119902) (A3)

Z=3(85)

times(3

2)

(285)(2120587)(14)

ℎ3

0

diamO (|V| ℎ1|O|15

+ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times |O|(75)

(A4)

Proof The function 119911 = 119908 minus 119866 satisfies

div (ℎ3nabla119911) = V120597ℎ

120597119909+ div (ℎ3nabla119872

120590(119902)) in O sub R2 (A5)

119911 = 0 on 120597O (A6)

Next we introduce the embedding constant for 119882120

(O) sub

119871119903(O) denoted119872

119903 such that

|V|119871119903(O) le 119872119903|nablaV|1198712(O) forallV isin 119867

1

0(O) (A7)

That constant can be estimated as

119872119903le1

2(diamO)

2|O|1119903 minus 12

(119903 + 2

2)

((119903+2)2119903)

radic2120587 (A8)

See for example [20 Lemma 1] Next we define the se-quence

120582119896+1

= 3(120582119896

2+ 1) 120582

1= 2 (A9)

Easy computation yields

120582119896= 8(

3

2)

119896minus1

minus 6 (A10)

Let

119911+(119909 119910) = max 119911 (119909 119910) 0 (A11)

We test (A5) with (119911+)1+120582119896+1 and get

intO

ℎ3nabla119911+nabla(119911+)1+120582119896+1

=1 + 120582119896+1

(1 + 120582119896+1

2)2intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

= intO

[Vℎ120597

120597119909(119911+)1+120582119896+1

+ ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

]

(A12)

For the left-hand side we get the lower bound

ℎ3

0

1 + 120582119896+1

(1 + 120582119896+1

2)2

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

1198712(O)

(A13)

We estimate the terms on the right-hand side using the sameidea

intO

ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

le ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O)

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

intO

ℎV120597

120597119909(119911+)1+120582119896+1

le ℎ1 |V| |O|

15100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

(A14)

Thus it remains to estimate |nabla(119911+)1+120582119896+1 |11987154(O) We have

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

100381610038161003816100381610038161003816

54

11987154(O)

= (1 + 120582119896+1

)54

intO

(119911+)(5120582119896+14)1003816100381610038161003816nabla119911

+1003816100381610038161003816

54

= (1 + 120582119896+1

1 + (120582119896+1

2))

54

intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

541003816100381610038161003816119911+1003816100381610038161003816

(5120582119896+18)

le [Holders inequality with 119901 =8

5 1199011015840=8

3]

le (1 + 120582119896+1

1 + (120582119896+1

2))

54

(intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

)

58

times (intO

(119911+)(53)120582

119896+1

)

38

= [due to (A9) 53120582119896+1

= 5(1 +120582119896

2)

3

8=1

4

120582119896+1

2 + 120582119896

]

= (1 + 120582119896+1

1 + (120582119896+1

2))

54100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

54

1198712(O)

times100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(120582119896+1(2+120582

119896))(54)

1198715(O)

(A15)

Combining with (A14) and (A13) we get

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)1003816100381610038161003816100381610038161198712(O)

le ℎminus3

0(1 +

120582119896+1

2)100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198715(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

Mathematical Problems in Engineering 9

le 119872((120582119896+1)(2+120582

119896))

5ℎminus3

0(1+

120582119896+1

2)

times100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198712(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

(A16)

We recall that

120582119896+1

2 + 120582119896

=3

2(A17)

and define

120572 = 119872(32)

5ℎminus3

0(ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

) (A18)

as well as

120590119896=

1

120572

100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(2(2+120582119896))

1198712(O)

(A19)

Then (A16) implies

120590119896+1

le (1 +120582119896+1

2)

(2(2+120582119896))

120590((120582119896+1)(2+120582

119896+1))

119896 (A20)

Taking the logarithm we arrive at

log120590119896+1

lelog (1 + (120582

119896+12))

1 + (120582119896+1

2)+

120582119896+1

2 + 120582119896+1

log120590119896 (A21)

We first notice that

120582119896+1

2 + 120582119896+1

lt 1 (A22)

and then

1 +120582119896

2= 4(

3

2)

119896minus1

minus 2 gt (3

2)

119896

(A23)

Since the function 119909 997891rarr (log119909119909) is decreasing for 119909 gt 119890 wehave

log (1 + (1205821198962))

1 + (1205821198962)

lelog [(32)119896]

(32)119896

= 119896 log 3

2(2

3)

119896

forall119896 ge 3

(A24)

Then

log120590119896+1

le (119896 + 1) log 3

2(2

3)

119896

+ log120590119896

le log 3

2

119896+1

sum

119895=2

119895(2

3)

119895

+ log1205901le 8 log 3

2+ log120590

1

(A25)

Finally

120590119896+1

le (3

2)

8

1205901 (A26)

Now it remains to estimate 1205901 From the definitionwe see that

1205901=

1

120572

100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

12

1198712(O)

(A27)

To estimate 1205901 we proceed as before and test (A5) with (119911+)3

We get

3

4intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

2

= intO

(Vℎ120597(119911+)3

120597119909+ ℎ3nabla119872120590(119902) nabla(119911

+)3

)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

100381610038161003816100381610038161003816nabla(119911+)310038161003816100381610038161003816100381611987154(O)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times3

2

100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

1003816100381610038161003816119911+1003816100381610038161003816119871103(O)

(A28)

Thus100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

le2

ℎ3

0

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872(103)

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O)

(A29)

Finally testing (A5) with 119911+ we get

intO

ℎ31003816100381610038161003816nabla119911+1003816100381610038161003816

2

= intO

(Vℎ120597119911+

120597119909+ ℎ3nabla119872120590(119902) nabla119911

+) (A30)

so that

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O) le

1

ℎ3

0

(|V| ℎ1|O|12

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198712(O))

le |O|310

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

(A31)

Combining (A29) with (A31) and (A26) gives

120590119896+1

le (3

2)

81

120572

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A32)

Since

120590119896+1

ge1

120572119872(2(2+120582k))2

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) (A33)

10 Mathematical Problems in Engineering

we have arrived to

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) le (

3

2)

8

119872(2(2+120582

119896))

2

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A34)

Since lim119896rarrinfin

120582119896= infin we get

1003816100381610038161003816119911+1003816100381610038161003816119871infin(O) le (

3

2)

8radic2

ℎ3

0

times(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A35)

Finally (A8) implies (A3)

Acknowledgment

This work was supported by MZOS grant 037-0372787-2797

References

[1] O Reynolds ldquoOn the theory and application and its applicationto Mr Beauchamp towerrsquos experiments Including and experi-mental determination of the viscosity of olive oilrdquo PhilosophicalTransactions of the Royal Society vol 117 pp 157ndash234 1886

[2] G G Stokes ldquoNotes on hydrodynamics On the dynamicalequationsrdquoCambridge and DublinMathematical Journal III pp121ndash127 1848

[3] P W Brifgman ldquoThe viscosity of liquids under pressurerdquo Pro-ceedings of the National Academy of Sciences of the United Statesof America vol 11 no 10 pp 603ndash606 1925

[4] A Z Szeri Fluid Film Lubrication Cambridge University PressNew York NY USA 1998

[5] E K Gatcombe ldquoLubrication characteristics of involute spur-gearsmdasha theoretical investigationrdquoTransactions of theAmericanSociety of Mechanical Engineers vol 67 pp 177ndash181 1945

[6] C Barus ldquoIsotherms isopiestics and isometrics relative to vis-cosityrdquo American Journal of Science vol 45 pp 87ndash96 1893

[7] W R Jones ldquoPressure viscosity measurement for several lubri-cantsrdquo ASLE Transactions vol 18 pp 249ndash262 1975

[8] J Hron J Malek and K R Rajagopal ldquoSimple flows of fluidswith pressure-dependent viscositiesrdquo Proceedings of the RoyalSociety A vol 457 no 2011 pp 1603ndash1622 2001

[9] C J A Roelands Correlation aspects of the viscosity-pressurerelationship of lubricating oils [PhD thesis] Delft University ofTechnology Delft The Netherlands 1966

[10] K R Rajagopal and A Z Szeri ldquoOn an inconsistency in thederivation of the equations of elastohydrodynamic lubricationrdquoProceedings of the Royal Society A vol 459 no 2039 pp 2771ndash2786 2003

[11] M Renardy ldquoSome remarks on the Navier-Stokes equationswith a pressure-dependent viscosityrdquo Communications in Par-tial Differential Equations vol 11 no 7 pp 779ndash793 1986

[12] F Gazzola and P Secchi ldquoSome results about stationary Navier-Stokes equations with a pressure-dependent viscosityrdquo in Pro-ceedings of the International Conference on Navier-Stokes Equa-tions vol 388 of Pitman Research Notes in Mathematics Seriespp 174ndash183 Varenna Italy 1998

[13] E Marusic-Paloka ldquoAn analysis of the Stokes system with pres-sure dependent viscosityrdquo In press

[14] S Marusic and E Marusic-Paloka ldquoTwo-scale convergence forthin domains and its applications to some lower-dimensionalmodels in fluid mechanicsrdquo Asymptotic Analysis vol 23 no 1pp 23ndash57 2000

[15] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001

[16] M Kane and B Bou-Said ldquoComparison of homogenization anddirect techniques for the treatment of roughness in incompress-ible lubricationrdquo Journal of Tribology vol 126 no 4 pp 733ndash7372004

[17] M Jai ldquoHomogenization and two-scale convergence of thecompressible Reynolds lubrication equation modelling the fly-ing characteristics of a roughmagnetic head over a rough rigid-disk surfacerdquo RAIRO Modelisation Mathematique et AnalyseNumerique vol 29 no 2 pp 199ndash233 1995

[18] P Wall ldquoHomogenization of Reynolds equation by two-scaleconvergencerdquo Chinese Annals of Mathematics B vol 28 no 3pp 363ndash374 2007

[19] A Bensoussan J-L Lions and G Papanicolaou AsymptoticAnalysis for Periodic Structures North-Holland AmsterdamThe Netherlands 1978

[20] E Marusic-Paloka and A Mikelic ldquoThe derivation of a nonlin-ear filtration law including the inertia effects via homogeniza-tionrdquo Nonlinear Analysis Theory Methods amp Applications vol42 no 1 pp 97ndash137 2000

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Mathematical Problems in Engineering

It is obvious from the definition of119872120590that

120597119872120590

120597119901(119901) =

1

120583 (119901)

120597119872120590

120597120590(119901) = minus

1

120583 (120590) (33)

As119872120590(119867120590(119908)) = 119908 deriving with respect to 120590 we arrive at

120597119867120590

120597120590(119908) = minus

(120597119872120590120597120590) (119901)

(120597119872120590120597119901) (119901)

=120583 (119901)

120583 (120590)=120583 (119867120590(119908))

120583 (120590)

(34)

Deriving (13) we get (120597119908120597120590) = minus1120583(120590) Using the rulefor deriving the inverse function we have

120597119867120590

120597119908(119908) = 120583 (119901) = 120583 (119867

120590(119908)) (35)

Thus120597119901

120597120590=120597119867120590

120597119908(119908)

120597119908

120597120590(120590) +

120597119867120590

120597120590(119908)

= minus120583 (119867120590(119908))

120583 (120590)+120583 (119867120590(119908))

120583 (120590)= 0

(36)

4 Homogenization

In this section we want to study the effects of rugositiesof surfaces on lubrication process The idea of finding themacroscopic effects of roughness on lubrication process viahomogenization is quite old and well studied Case of con-stant viscosity for incompressible and compressible flows aswell as non-Newtonian deformation dependent viscositieswere investigated The subject was treated by several authorsandwe heremention [16ndash18]The case of pressure-dependentviscosity brings some new interesting nonlocal effects

We assume that the function ℎ describing the form of thefluid domain is periodic with small period 1119898 with119898 isin NTo stress that dependence we denote it by ℎ

119898 More precisely

we denote by 119884 =]0 1[119899 119899 = 1 2 the period We further

assume that ℎ R119899 rarr [1198890 +infin[ 119889

0gt 0 is periodic with

period 119884 and smooth Then we take ℎ119898of the form

ℎ119898(119909) = ℎ (119898119909) (37)

Thus the function ℎ describes the form of periodicallydistributed rugosities

To emphasize that the relative velocity of bearing surfacesV is large we assume that it also depends on 119898 the sameparameter that is taken for description of rugosities In thatcase our equation reads

div(ℎ3

119898

120583 (119901119898)nabla119901119898) = 6V

119898sdot nablaℎ119898

in O (38)

41 One-Dimensional Case If 119899 = 1 the above problem isposed on an intervalO =]0 1[With an appropriate boundarycondition

119901119898(0) = 119902

0 119901

119898(1) = 119902

1 (39)

It forms a boundary value problem for nonlinear ODE

(ℎ3

119898

120583 (119901119898)1199011015840

119898)

1015840

= 6119881119898ℎ1015840

119898in O (40)

To study the asymptotic behavior of the solution with respectto119898we linearize the problem using the transformation119908

119898=

119872120590(119901119898) To simplify in this section we choose 120590 = 119902

0and

dropping the index 120590 in119872120590and119872

+

120590 we denote

119872(119901) = int

119901

1199020

119889119905

120583 (119905) 119872

+= int

+infin

1199020

119889119905

120583 (119905) (41)

Theorem 5 Let

1205940(119910) = 6 (int

1

0

119889119904

ℎ(119904)3)

minus1

times [⟨1

ℎ3⟩int

119910

0

119889V

ℎ(V)2minus⟨

1

ℎ2⟩int

119910

0

119889V

ℎ(V)3]

(42)

⟨sdot⟩ = int

1

0

sdot 119889119910 (43)

and let119867 = 119872minus1 Suppose that there exists a limit

119881 = lim119898rarrinfin

119881119898

119898(44)

and that for 119898 large enough and 119872+ defined in (27) the

following condition holds

119887119898

3(119909)

119887119898

3(1)

119872 (1199021) + 6119881

119898

119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)

119887119898

3(1)

le 119872+

(45)

with

119887119898

120572(119909) = int

119909

0

119889119905

ℎ119898(119905)120572 120572 = 2 3 (46)

Then

119901119898 1199010=int

1

0

119867(119909119872(1199021) + 119881120594

0(119910)) 119889119910

weaklowast in 119871infin(0 1)

(47)

Proof Equation (40) with boundary conditions (39) can besolved by reduction to quadratures after the substitution

119908119898= 119872(119901

119898) (48)

with 119872120590strictly increasing function defined by (12) The

problem for 119908119898now reads

(ℎ3

1198981199081015840

119898)1015840

= 6119881119898ℎ1015840

119898in ]0 1[

119908119898(0) = 119872(119902

0) = 0 119908

119898(1) = 119872(119902

1)

(49)

Mathematical Problems in Engineering 5

It is easy to see that (49) has a unique solution given by (25)Since119872(119902

0) = 0 (25) now reduces to

119908119898=119887119898

3(119909)

119887119898

3(1)

119872120590(1199021) + 6119881

119898

119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)

119887119898

3(1)

(50)

where for 120572 = 2 3

119887119898

120572(119909) = int

119909

0

119889119905

ℎ119898(119905)120572

=1

119898int

119898119909

0

119889119904

ℎ(119904)120572997888rarr 119909int

1

0

119889119904

ℎ(119904)120572

as 119898 997888rarr +infin

(51)

Now 119901119898 the solution to the problem (40) exists if (45) is

fulfilled The second term in (50) thus obviously tends to119909119872(119902

1) as119898 rarr +infin The last term is more interesting The

denominator tends to

⟨1

ℎ3⟩ = int

1

0

119889119904

ℎ(119904)3 (52)

As for its numerator we have

119881119898[119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)]

= 119881119898[int

1

0

int

119909

0

(1

ℎ119898(119905)3ℎ119898(119904)2minus

1

ℎ119898(119905)2ℎ(119904)3

119898

)119889119904 119889119905]

=119881119898

119898(⟨

1

ℎ3⟩int

119898119909

0

119889V

ℎ(V)2minus ⟨

1

ℎ2⟩int

119898119909

0

119889V

ℎ(V)3)

(53)

Suppose that

119881 = lim119898rarr+infin

119881119898

119898(54)

and denote

119866 (119910) = ⟨1

ℎ3⟩int

119910

0

119889V

ℎ(V)2minus ⟨

1

ℎ2⟩int

119910

0

119889V

ℎ(V)3 (55)

Obviously the function119866 is periodicwith period 1 so that dueto the standard periodicity lemma (see eg [19]) as 119898 rarr

+infin

119866 (119898119909) ⟨119866⟩ = int

1

0

119866 (119910) 119889119910 weaklowast in 119871infin(0 1) (56)

By direct computation

⟨119866⟩ = ⟨119910

ℎ3⟩⟨

1

ℎ2⟩ minus ⟨

1

ℎ3⟩⟨

119910

ℎ2⟩

= intint

1

0

119904 minus 119905

ℎ(119904)3ℎ(119905)2119889119904 119889119905

(57)

Thus

119881119898[119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)] 119881 ⟨119866⟩ (58)

Now denoting

1205940(119910) =

119866 (119910)

1198873 (1)

(59)

we have

119908119898asymp 119909119872(119902

1) + 119881120594

0(119898119909) (60)

and thus

119908119898 1199080=119909119872(119902

1) + 119881⟨120594

0⟩ weaklowast in 119871

infin(0 1) (61)

However we are not interested in convergence of theauxiliary function 119908

119898but in the convergence of the pressure

119901119898 Since (45) is assumed to be true we can define 119901

119898=

119867(119908119898) where119867 = 119872

minus1 and we have

119901119898 1199010= int

1

0

119867(119909119872(1199021) + 119881120594

0(119910)) 119889119910

weaklowast in 119871infin(0 1)

(62)

Deriving the expression on the right-hand side we obtainthe effective pressure drop in the form

1199011015840

0(119909) = 119872(119902

1) int

1

0

120583 (119909119872(1199021) + 119881120594

0(119910)) 119889119910 (63)

= int

1199021

1199020

119889120591

120583 (120591)int

1

0

120583 (119909119872(1199021) + 119881120594

0(119910)) 119889119910

= int

1199021

1199020

119889120591

120583 (120591)int

1

0

120583 (119909int

1199021

1199020

119889119904

120583 (119904)+ 1198811205940(119910)) 119889119910

(64)

As we can see the pressure drop is not constant as forthe Newtonian flow The interesting effect appears if 119881 = 0

because in that case the expressions for the pressure and forthe pressure drop are nonlocal due to the integral with respectto 119910 That phenomenon is entirely due to the fact that theviscosity is depending on the pressure

42 Two-Dimensional Case We suppose here that the func-tion ℎ

119898is constructed from positive smooth 119884-periodic

function ℎ R2 rarr [1198890 +infin[ 119889

0gt 0 119884 =]0 1[

2 in the sameway as before that is by taking

ℎ119898(119909) = ℎ (119898119909) (65)

We have seen in the previous section that interesting effectshappen only if we assume that

V119898= 119898V (66)

In that case our equation reads

div(ℎ3

119898

120583 (119901119898)nabla119901119898) = 6V

119898sdot nablaℎ119898

in O (67)

6 Mathematical Problems in Engineering

Aswedid in the existence analysis and in the previous sectionwe linearize the equation by substitution

119908119898= 119872120590(119901119898) (68)

where the function119872120590is defined by (12) Now 119908

119898satisfies

div (ℎ3119872 nabla119908119898) = 6V

119898sdot nablaℎ119898

in O (69)

We postulate the asymptotic expansion in the form

119908119898 (119909) asymp 119908

0(119909 119910) +

1

1198981199081(119909 119910)

+1

11989821199082(119909 119910) + sdot sdot sdot 119910 = 119898119909

(70)

All functions are assumed to be 119884-periodic in 119910 variablePlugging that in (69) and collecting formally terms with

equal powers of119898 we get

1198982 div119910(ℎ3nabla1199101199080) = 6V sdot nabla

119910ℎ

119898 div119910(ℎ3nabla1199101199081) + div

119910(ℎ3nabla1199091199080) + ℎ3div119909nabla1199101199080= 0

1 div119910(ℎ3nabla1199101199082) + div

119910(ℎ3nabla1199091199081) + ℎ3div119909nabla1199101199081

+ ℎ3Δ1199091199080= 0

Denoting

119881 = |V| V0=V119881

(71)

we have

1199080(119909 119910) = V

0(119909) + 119881120594

0(119910)

div119910(ℎ3nabla1199101205940) = 6V

0sdot nabla119910ℎ in 119884

1199081(119909 119910) =

2

sum

119896=1

120594119896(119910)

120597V0

120597119909119896

(119909) + V1(119909)

div119910(ℎ3nabla119910(120594119896+ 119910119896)) = 0

2

sum

119896ℓ=1

a119896ℓ

1205972V0

120597119909119896120597119909ℓ

= 0 in O

a119896ℓ

= int119884

ℎ3 120597

120597119910ℓ

(120594119896+ 119910119896) 119889119910

(72)

Remark 6 Thesame computation can be done in one-dimen-sional case and it gives

(ℎ31205941015840

0)1015840

= 6ℎ1015840997904rArr 1205940= 6int

119910

0

119889119905

ℎ(119905)2+ 1198620int

119910

0

119889119905

ℎ(119905)3+ 1198621

(73)

V0= 119909119872(119902

1) (74)

Constants 1198620 1198621are chosen in a way that boundary condi-

tions 1205940(0) = 120594

0(1) = 0 are met and it follows that

1198621= 0 119862

0= minus 6(int

1

0

119889119905

ℎ(119905)3)

minus1

int

1

0

119889119905

ℎ(119905)2 (75)

Then

119908119898asymp 1199080 (119909119898119909)

= V0(119909) + 120594

0(119898119909)

= 119909119872120590(1199021)

+ 6119881[int

119898119909

0

119889119905

ℎ(119905)2minus (int

1

0

119889119905

ℎ(119905)3)

minus1

times int

1

0

119889119905

ℎ(119905)2

int

119898119909

0

119889119905

ℎ(119905)3]

= 119909119872120590(1199021)

+ 6119881119887119898

2(119909) 1198873(1) minus 119887

2(1) 119887119898

3(119909)

1198873(1)

= 119909119872120590(1199021) + 119881120594

0(119898119909)

(76)

That is a very good approximation of our exact solution (50)It is important to notice that the choice of constants

1198620 1198621was determined from the exterior boundary con-

dition So we should expect the same in two-dimensionalcase However the treatment of boundary conditions intwo-dimensional case is much more complicated and theboundary layer is to be expected

The derived asymptotic expansion should be justified byproving the convergence And we need the strong conver-gence (with corrector of course) for 119908

119898in order to get the

convergence for 119901119898 The form of the approximation

119908119898asymp V0(119909) + 119881120594

0(119898119909) + sdot sdot sdot (77)

suggests that the boundary layer phenomenon should appearon the exterior boundary 120597O since 120594

0term cannot satisfy the

Dirichlet condition on 120597O To get the error estimate and thestrong convergence we need to handle that boundary layerThus at this point we simplify the domain and the boundarycondition in order to be able to avoid it We assume that

O = ]0 1[ times R (78)

119901120576(0 1199092) = 0 119901

120576(1 1199092) = 119902 (119909

2) (79)

1199092997891997888rarr 119901

120576(1199091 1199092) is 1-periodic (80)

ℎ (1199101 1199102) = ℎ (119910

1) (81)

119902 is 1-periodic (82)

Now

119908120576= 119872120590(119901120576) with 119872

120590(119901) =int

119901

120590

119889119904

120583 (119904)997904rArr 119908

120576(0 1199092) = 0

119908120576(1 1199092) = 119872

120590(119902 (1199092))

(83)

and 1199092997891rarr 119908120576(1199091 1199092) is 1-periodic

Mathematical Problems in Engineering 7

In that case we can compute1205940and120594119896 119896 = 1 2 explicitly

and we can impose exterior condition on 1205940 Indeed 120594

0is

exactly the same as in the monodimensional case that is itis given by (73) and (75) Obviously 120594

2= 0 so that

a22

= int

1

0

ℎ(119904)3119889119904 = ⟨ℎ

3⟩ a

12= a21

= 0 (84)

As for the last term

1205941= minus1199101+ (int

1

0

119889119904

ℎ(119904)3)

minus1

int

1199101

0

119889119904

ℎ(119904)3

a11

= (int

1

0

119889119904

ℎ(119904)3)

minus1

=1

⟨1ℎ3⟩

(85)

Finally the function V0satisfies the boundary value problem

1

⟨1ℎ3⟩

1205972V0

12059711990921

+ ⟨ℎ3⟩1205972V0

12059711990922

= 0 in O

V0(0 1199092) = 0

V0(1 1199092) = 119872

120590(119902 (1199092)) V

0is 1-periodic in 119909

2

(86)

It can be solved using the Fourier method and we get

V0(1199091 1199092) =

infin

sum

119896=1

sh (radic⟨ℎ3⟩ ⟨ℎminus3⟩ 1198961205871199091)

times (120572119896sin 119896120587119909

2+ 120573119896cos 119896120587119909

2)

120572119896= 2int

1

0

119872120590(119902 (119905)) sin 119896120587119905119889119905

120573119896= 2int

1

0

119872120590(119902 (119905)) cos 119896120587119905119889119905

(87)

Since the approximation

w119898asymp V0(119909) + 119881120594

0(1198981199091) +

1

1198981205941(1198981199091)120597V0

1205971199091

(119909) (88)

now satisfies the boundary conditions on 120597O it is easy to seethat

1003816100381610038161003816119908119898 minus (V0(119909) + 119881120594

0(119898119909))

1003816100381610038161003816119871infin(O) le 1198621

119898(89)

follows from the maximum principle Assuming that for 119898large enough

V0(119909) + 119881120594

0(119898119909) le 119872

+ (90)

we have1003816100381610038161003816119901119898 minus 119867 (V

0(119909) + 119881120594

0(119898119909))

1003816100381610038161003816119871infin(O) 997888rarr 0 (91)

Finally

119901119898 1199010= int

1

0

119867(V0(119909) + 119881120594

0(1199101)) 1198891199101 (92)

We have proved that

Theorem 7 Let 119901119898be the solution to the problem (67) (79)

and (80) and let V01205940be defined by (87) and (42) respectively

If (90) holds then1003816100381610038161003816119901119898 minus 119867 (V

0 (119909) + 1198811205940 (119898119909))

1003816100381610038161003816119871infin(O) 997888rarr 0

119901119898 1199010= int

1

0

119867(V0(119909) + 119881120594

0(1199101)) 1198891199101

(93)

Remark 8 It is important to notice that 1205870= 119867(119901

0) satisfies

2

sum

119894119896=1

120597

120597119909119894

(a119894119896

120583 (1205870)

1205971205870

120597119909119896

) = 0 in O 1205870= 119902 on 120597O (94)

and thus we would expect it to be the limit of 119901119898in analogy

with the linear case However 1199010

= 1205870

If119881 = lim119898rarrinfin

119898|V119898| is small we can expand119867(V

0(119909)+

1198811205940(119898119909)) in powers of 119881 and we get

119901119898(119909) asymp 119867 (V

0(119909) + 119881120594

0(119898119909))

= 119867 (V0(119909)) + 119881119867

1015840(V0(119909)) 120594

0(119910) + 119874 (119881

2)

= 1205870(119909) + 119881120583 (120587

0(119909)) 120594

0(119910) + 119874 (119881

2)

(95)

Thus

1199010= 1205870+ 120583 (120587

0) ⟨1205940⟩119881 + 119874 (119881

2) (96)

It can be formally written as

119901119898(119909) asymp 120587

0(119909) +

1003816100381610038161003816V1198981003816100381610038161003816

119898120583 (1205870(119909)) 120594

0(119898119909) + 119874(

1003816100381610038161003816V11989810038161003816100381610038162

1198982)

(97)

Appendix

The Maximum Principle

Our goal is to derive maximum principles for the linearReynolds equation with sharp explicit constants in orderto solve the nonlinear Reynolds equation with pressure-dependent viscosity We assume without losing generalitythat V = (16)Vi Indeed we can always choose the coor-dinate system in a way that the first coordinate axis 119909 has adirection of the velocity of relative motion V

The lower bound for 119908(119909 119910) is of no interest justthe upper bound Function 119908(119909 119910) is the solution to theboundary value problem

div (ℎ3nabla119908) = V120597ℎ

120597119909in O sub R2

119908 = 119872120590(119902) on 120597O

(A1)

We assume that if V(120597ℎ120597119909) gt 0 then 119908 cannot have amaximum point in the domain O and thus

119908 (119909 119910) le max(119909119910)isin120597O

119872120590(119902 (119909 119910)) (A2)

8 Mathematical Problems in Engineering

However it is not realistic to assume that (120597ℎ120597119909) does notchange the sign To find the upper bound in the general casewe use the procedure from the DeGiorgi theorem The mainresult of the section is as follows

TheoremA1 Let119908 be the solution to the problem (A1)Then

119908 (119909 119910) le1003816100381610038161003816119872120590 (119902)

1003816100381610038161003816119871infin(120597O) +Z (V ℎO 120583 119902) (A3)

Z=3(85)

times(3

2)

(285)(2120587)(14)

ℎ3

0

diamO (|V| ℎ1|O|15

+ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times |O|(75)

(A4)

Proof The function 119911 = 119908 minus 119866 satisfies

div (ℎ3nabla119911) = V120597ℎ

120597119909+ div (ℎ3nabla119872

120590(119902)) in O sub R2 (A5)

119911 = 0 on 120597O (A6)

Next we introduce the embedding constant for 119882120

(O) sub

119871119903(O) denoted119872

119903 such that

|V|119871119903(O) le 119872119903|nablaV|1198712(O) forallV isin 119867

1

0(O) (A7)

That constant can be estimated as

119872119903le1

2(diamO)

2|O|1119903 minus 12

(119903 + 2

2)

((119903+2)2119903)

radic2120587 (A8)

See for example [20 Lemma 1] Next we define the se-quence

120582119896+1

= 3(120582119896

2+ 1) 120582

1= 2 (A9)

Easy computation yields

120582119896= 8(

3

2)

119896minus1

minus 6 (A10)

Let

119911+(119909 119910) = max 119911 (119909 119910) 0 (A11)

We test (A5) with (119911+)1+120582119896+1 and get

intO

ℎ3nabla119911+nabla(119911+)1+120582119896+1

=1 + 120582119896+1

(1 + 120582119896+1

2)2intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

= intO

[Vℎ120597

120597119909(119911+)1+120582119896+1

+ ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

]

(A12)

For the left-hand side we get the lower bound

ℎ3

0

1 + 120582119896+1

(1 + 120582119896+1

2)2

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

1198712(O)

(A13)

We estimate the terms on the right-hand side using the sameidea

intO

ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

le ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O)

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

intO

ℎV120597

120597119909(119911+)1+120582119896+1

le ℎ1 |V| |O|

15100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

(A14)

Thus it remains to estimate |nabla(119911+)1+120582119896+1 |11987154(O) We have

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

100381610038161003816100381610038161003816

54

11987154(O)

= (1 + 120582119896+1

)54

intO

(119911+)(5120582119896+14)1003816100381610038161003816nabla119911

+1003816100381610038161003816

54

= (1 + 120582119896+1

1 + (120582119896+1

2))

54

intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

541003816100381610038161003816119911+1003816100381610038161003816

(5120582119896+18)

le [Holders inequality with 119901 =8

5 1199011015840=8

3]

le (1 + 120582119896+1

1 + (120582119896+1

2))

54

(intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

)

58

times (intO

(119911+)(53)120582

119896+1

)

38

= [due to (A9) 53120582119896+1

= 5(1 +120582119896

2)

3

8=1

4

120582119896+1

2 + 120582119896

]

= (1 + 120582119896+1

1 + (120582119896+1

2))

54100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

54

1198712(O)

times100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(120582119896+1(2+120582

119896))(54)

1198715(O)

(A15)

Combining with (A14) and (A13) we get

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)1003816100381610038161003816100381610038161198712(O)

le ℎminus3

0(1 +

120582119896+1

2)100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198715(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

Mathematical Problems in Engineering 9

le 119872((120582119896+1)(2+120582

119896))

5ℎminus3

0(1+

120582119896+1

2)

times100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198712(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

(A16)

We recall that

120582119896+1

2 + 120582119896

=3

2(A17)

and define

120572 = 119872(32)

5ℎminus3

0(ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

) (A18)

as well as

120590119896=

1

120572

100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(2(2+120582119896))

1198712(O)

(A19)

Then (A16) implies

120590119896+1

le (1 +120582119896+1

2)

(2(2+120582119896))

120590((120582119896+1)(2+120582

119896+1))

119896 (A20)

Taking the logarithm we arrive at

log120590119896+1

lelog (1 + (120582

119896+12))

1 + (120582119896+1

2)+

120582119896+1

2 + 120582119896+1

log120590119896 (A21)

We first notice that

120582119896+1

2 + 120582119896+1

lt 1 (A22)

and then

1 +120582119896

2= 4(

3

2)

119896minus1

minus 2 gt (3

2)

119896

(A23)

Since the function 119909 997891rarr (log119909119909) is decreasing for 119909 gt 119890 wehave

log (1 + (1205821198962))

1 + (1205821198962)

lelog [(32)119896]

(32)119896

= 119896 log 3

2(2

3)

119896

forall119896 ge 3

(A24)

Then

log120590119896+1

le (119896 + 1) log 3

2(2

3)

119896

+ log120590119896

le log 3

2

119896+1

sum

119895=2

119895(2

3)

119895

+ log1205901le 8 log 3

2+ log120590

1

(A25)

Finally

120590119896+1

le (3

2)

8

1205901 (A26)

Now it remains to estimate 1205901 From the definitionwe see that

1205901=

1

120572

100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

12

1198712(O)

(A27)

To estimate 1205901 we proceed as before and test (A5) with (119911+)3

We get

3

4intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

2

= intO

(Vℎ120597(119911+)3

120597119909+ ℎ3nabla119872120590(119902) nabla(119911

+)3

)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

100381610038161003816100381610038161003816nabla(119911+)310038161003816100381610038161003816100381611987154(O)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times3

2

100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

1003816100381610038161003816119911+1003816100381610038161003816119871103(O)

(A28)

Thus100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

le2

ℎ3

0

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872(103)

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O)

(A29)

Finally testing (A5) with 119911+ we get

intO

ℎ31003816100381610038161003816nabla119911+1003816100381610038161003816

2

= intO

(Vℎ120597119911+

120597119909+ ℎ3nabla119872120590(119902) nabla119911

+) (A30)

so that

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O) le

1

ℎ3

0

(|V| ℎ1|O|12

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198712(O))

le |O|310

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

(A31)

Combining (A29) with (A31) and (A26) gives

120590119896+1

le (3

2)

81

120572

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A32)

Since

120590119896+1

ge1

120572119872(2(2+120582k))2

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) (A33)

10 Mathematical Problems in Engineering

we have arrived to

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) le (

3

2)

8

119872(2(2+120582

119896))

2

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A34)

Since lim119896rarrinfin

120582119896= infin we get

1003816100381610038161003816119911+1003816100381610038161003816119871infin(O) le (

3

2)

8radic2

ℎ3

0

times(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A35)

Finally (A8) implies (A3)

Acknowledgment

This work was supported by MZOS grant 037-0372787-2797

References

[1] O Reynolds ldquoOn the theory and application and its applicationto Mr Beauchamp towerrsquos experiments Including and experi-mental determination of the viscosity of olive oilrdquo PhilosophicalTransactions of the Royal Society vol 117 pp 157ndash234 1886

[2] G G Stokes ldquoNotes on hydrodynamics On the dynamicalequationsrdquoCambridge and DublinMathematical Journal III pp121ndash127 1848

[3] P W Brifgman ldquoThe viscosity of liquids under pressurerdquo Pro-ceedings of the National Academy of Sciences of the United Statesof America vol 11 no 10 pp 603ndash606 1925

[4] A Z Szeri Fluid Film Lubrication Cambridge University PressNew York NY USA 1998

[5] E K Gatcombe ldquoLubrication characteristics of involute spur-gearsmdasha theoretical investigationrdquoTransactions of theAmericanSociety of Mechanical Engineers vol 67 pp 177ndash181 1945

[6] C Barus ldquoIsotherms isopiestics and isometrics relative to vis-cosityrdquo American Journal of Science vol 45 pp 87ndash96 1893

[7] W R Jones ldquoPressure viscosity measurement for several lubri-cantsrdquo ASLE Transactions vol 18 pp 249ndash262 1975

[8] J Hron J Malek and K R Rajagopal ldquoSimple flows of fluidswith pressure-dependent viscositiesrdquo Proceedings of the RoyalSociety A vol 457 no 2011 pp 1603ndash1622 2001

[9] C J A Roelands Correlation aspects of the viscosity-pressurerelationship of lubricating oils [PhD thesis] Delft University ofTechnology Delft The Netherlands 1966

[10] K R Rajagopal and A Z Szeri ldquoOn an inconsistency in thederivation of the equations of elastohydrodynamic lubricationrdquoProceedings of the Royal Society A vol 459 no 2039 pp 2771ndash2786 2003

[11] M Renardy ldquoSome remarks on the Navier-Stokes equationswith a pressure-dependent viscosityrdquo Communications in Par-tial Differential Equations vol 11 no 7 pp 779ndash793 1986

[12] F Gazzola and P Secchi ldquoSome results about stationary Navier-Stokes equations with a pressure-dependent viscosityrdquo in Pro-ceedings of the International Conference on Navier-Stokes Equa-tions vol 388 of Pitman Research Notes in Mathematics Seriespp 174ndash183 Varenna Italy 1998

[13] E Marusic-Paloka ldquoAn analysis of the Stokes system with pres-sure dependent viscosityrdquo In press

[14] S Marusic and E Marusic-Paloka ldquoTwo-scale convergence forthin domains and its applications to some lower-dimensionalmodels in fluid mechanicsrdquo Asymptotic Analysis vol 23 no 1pp 23ndash57 2000

[15] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001

[16] M Kane and B Bou-Said ldquoComparison of homogenization anddirect techniques for the treatment of roughness in incompress-ible lubricationrdquo Journal of Tribology vol 126 no 4 pp 733ndash7372004

[17] M Jai ldquoHomogenization and two-scale convergence of thecompressible Reynolds lubrication equation modelling the fly-ing characteristics of a roughmagnetic head over a rough rigid-disk surfacerdquo RAIRO Modelisation Mathematique et AnalyseNumerique vol 29 no 2 pp 199ndash233 1995

[18] P Wall ldquoHomogenization of Reynolds equation by two-scaleconvergencerdquo Chinese Annals of Mathematics B vol 28 no 3pp 363ndash374 2007

[19] A Bensoussan J-L Lions and G Papanicolaou AsymptoticAnalysis for Periodic Structures North-Holland AmsterdamThe Netherlands 1978

[20] E Marusic-Paloka and A Mikelic ldquoThe derivation of a nonlin-ear filtration law including the inertia effects via homogeniza-tionrdquo Nonlinear Analysis Theory Methods amp Applications vol42 no 1 pp 97ndash137 2000

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 5

It is easy to see that (49) has a unique solution given by (25)Since119872(119902

0) = 0 (25) now reduces to

119908119898=119887119898

3(119909)

119887119898

3(1)

119872120590(1199021) + 6119881

119898

119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)

119887119898

3(1)

(50)

where for 120572 = 2 3

119887119898

120572(119909) = int

119909

0

119889119905

ℎ119898(119905)120572

=1

119898int

119898119909

0

119889119904

ℎ(119904)120572997888rarr 119909int

1

0

119889119904

ℎ(119904)120572

as 119898 997888rarr +infin

(51)

Now 119901119898 the solution to the problem (40) exists if (45) is

fulfilled The second term in (50) thus obviously tends to119909119872(119902

1) as119898 rarr +infin The last term is more interesting The

denominator tends to

⟨1

ℎ3⟩ = int

1

0

119889119904

ℎ(119904)3 (52)

As for its numerator we have

119881119898[119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)]

= 119881119898[int

1

0

int

119909

0

(1

ℎ119898(119905)3ℎ119898(119904)2minus

1

ℎ119898(119905)2ℎ(119904)3

119898

)119889119904 119889119905]

=119881119898

119898(⟨

1

ℎ3⟩int

119898119909

0

119889V

ℎ(V)2minus ⟨

1

ℎ2⟩int

119898119909

0

119889V

ℎ(V)3)

(53)

Suppose that

119881 = lim119898rarr+infin

119881119898

119898(54)

and denote

119866 (119910) = ⟨1

ℎ3⟩int

119910

0

119889V

ℎ(V)2minus ⟨

1

ℎ2⟩int

119910

0

119889V

ℎ(V)3 (55)

Obviously the function119866 is periodicwith period 1 so that dueto the standard periodicity lemma (see eg [19]) as 119898 rarr

+infin

119866 (119898119909) ⟨119866⟩ = int

1

0

119866 (119910) 119889119910 weaklowast in 119871infin(0 1) (56)

By direct computation

⟨119866⟩ = ⟨119910

ℎ3⟩⟨

1

ℎ2⟩ minus ⟨

1

ℎ3⟩⟨

119910

ℎ2⟩

= intint

1

0

119904 minus 119905

ℎ(119904)3ℎ(119905)2119889119904 119889119905

(57)

Thus

119881119898[119887119898

3(1) 119887119898

2(119909) minus 119887

119898

3(119909) 119887119898

2(1)] 119881 ⟨119866⟩ (58)

Now denoting

1205940(119910) =

119866 (119910)

1198873 (1)

(59)

we have

119908119898asymp 119909119872(119902

1) + 119881120594

0(119898119909) (60)

and thus

119908119898 1199080=119909119872(119902

1) + 119881⟨120594

0⟩ weaklowast in 119871

infin(0 1) (61)

However we are not interested in convergence of theauxiliary function 119908

119898but in the convergence of the pressure

119901119898 Since (45) is assumed to be true we can define 119901

119898=

119867(119908119898) where119867 = 119872

minus1 and we have

119901119898 1199010= int

1

0

119867(119909119872(1199021) + 119881120594

0(119910)) 119889119910

weaklowast in 119871infin(0 1)

(62)

Deriving the expression on the right-hand side we obtainthe effective pressure drop in the form

1199011015840

0(119909) = 119872(119902

1) int

1

0

120583 (119909119872(1199021) + 119881120594

0(119910)) 119889119910 (63)

= int

1199021

1199020

119889120591

120583 (120591)int

1

0

120583 (119909119872(1199021) + 119881120594

0(119910)) 119889119910

= int

1199021

1199020

119889120591

120583 (120591)int

1

0

120583 (119909int

1199021

1199020

119889119904

120583 (119904)+ 1198811205940(119910)) 119889119910

(64)

As we can see the pressure drop is not constant as forthe Newtonian flow The interesting effect appears if 119881 = 0

because in that case the expressions for the pressure and forthe pressure drop are nonlocal due to the integral with respectto 119910 That phenomenon is entirely due to the fact that theviscosity is depending on the pressure

42 Two-Dimensional Case We suppose here that the func-tion ℎ

119898is constructed from positive smooth 119884-periodic

function ℎ R2 rarr [1198890 +infin[ 119889

0gt 0 119884 =]0 1[

2 in the sameway as before that is by taking

ℎ119898(119909) = ℎ (119898119909) (65)

We have seen in the previous section that interesting effectshappen only if we assume that

V119898= 119898V (66)

In that case our equation reads

div(ℎ3

119898

120583 (119901119898)nabla119901119898) = 6V

119898sdot nablaℎ119898

in O (67)

6 Mathematical Problems in Engineering

Aswedid in the existence analysis and in the previous sectionwe linearize the equation by substitution

119908119898= 119872120590(119901119898) (68)

where the function119872120590is defined by (12) Now 119908

119898satisfies

div (ℎ3119872 nabla119908119898) = 6V

119898sdot nablaℎ119898

in O (69)

We postulate the asymptotic expansion in the form

119908119898 (119909) asymp 119908

0(119909 119910) +

1

1198981199081(119909 119910)

+1

11989821199082(119909 119910) + sdot sdot sdot 119910 = 119898119909

(70)

All functions are assumed to be 119884-periodic in 119910 variablePlugging that in (69) and collecting formally terms with

equal powers of119898 we get

1198982 div119910(ℎ3nabla1199101199080) = 6V sdot nabla

119910ℎ

119898 div119910(ℎ3nabla1199101199081) + div

119910(ℎ3nabla1199091199080) + ℎ3div119909nabla1199101199080= 0

1 div119910(ℎ3nabla1199101199082) + div

119910(ℎ3nabla1199091199081) + ℎ3div119909nabla1199101199081

+ ℎ3Δ1199091199080= 0

Denoting

119881 = |V| V0=V119881

(71)

we have

1199080(119909 119910) = V

0(119909) + 119881120594

0(119910)

div119910(ℎ3nabla1199101205940) = 6V

0sdot nabla119910ℎ in 119884

1199081(119909 119910) =

2

sum

119896=1

120594119896(119910)

120597V0

120597119909119896

(119909) + V1(119909)

div119910(ℎ3nabla119910(120594119896+ 119910119896)) = 0

2

sum

119896ℓ=1

a119896ℓ

1205972V0

120597119909119896120597119909ℓ

= 0 in O

a119896ℓ

= int119884

ℎ3 120597

120597119910ℓ

(120594119896+ 119910119896) 119889119910

(72)

Remark 6 Thesame computation can be done in one-dimen-sional case and it gives

(ℎ31205941015840

0)1015840

= 6ℎ1015840997904rArr 1205940= 6int

119910

0

119889119905

ℎ(119905)2+ 1198620int

119910

0

119889119905

ℎ(119905)3+ 1198621

(73)

V0= 119909119872(119902

1) (74)

Constants 1198620 1198621are chosen in a way that boundary condi-

tions 1205940(0) = 120594

0(1) = 0 are met and it follows that

1198621= 0 119862

0= minus 6(int

1

0

119889119905

ℎ(119905)3)

minus1

int

1

0

119889119905

ℎ(119905)2 (75)

Then

119908119898asymp 1199080 (119909119898119909)

= V0(119909) + 120594

0(119898119909)

= 119909119872120590(1199021)

+ 6119881[int

119898119909

0

119889119905

ℎ(119905)2minus (int

1

0

119889119905

ℎ(119905)3)

minus1

times int

1

0

119889119905

ℎ(119905)2

int

119898119909

0

119889119905

ℎ(119905)3]

= 119909119872120590(1199021)

+ 6119881119887119898

2(119909) 1198873(1) minus 119887

2(1) 119887119898

3(119909)

1198873(1)

= 119909119872120590(1199021) + 119881120594

0(119898119909)

(76)

That is a very good approximation of our exact solution (50)It is important to notice that the choice of constants

1198620 1198621was determined from the exterior boundary con-

dition So we should expect the same in two-dimensionalcase However the treatment of boundary conditions intwo-dimensional case is much more complicated and theboundary layer is to be expected

The derived asymptotic expansion should be justified byproving the convergence And we need the strong conver-gence (with corrector of course) for 119908

119898in order to get the

convergence for 119901119898 The form of the approximation

119908119898asymp V0(119909) + 119881120594

0(119898119909) + sdot sdot sdot (77)

suggests that the boundary layer phenomenon should appearon the exterior boundary 120597O since 120594

0term cannot satisfy the

Dirichlet condition on 120597O To get the error estimate and thestrong convergence we need to handle that boundary layerThus at this point we simplify the domain and the boundarycondition in order to be able to avoid it We assume that

O = ]0 1[ times R (78)

119901120576(0 1199092) = 0 119901

120576(1 1199092) = 119902 (119909

2) (79)

1199092997891997888rarr 119901

120576(1199091 1199092) is 1-periodic (80)

ℎ (1199101 1199102) = ℎ (119910

1) (81)

119902 is 1-periodic (82)

Now

119908120576= 119872120590(119901120576) with 119872

120590(119901) =int

119901

120590

119889119904

120583 (119904)997904rArr 119908

120576(0 1199092) = 0

119908120576(1 1199092) = 119872

120590(119902 (1199092))

(83)

and 1199092997891rarr 119908120576(1199091 1199092) is 1-periodic

Mathematical Problems in Engineering 7

In that case we can compute1205940and120594119896 119896 = 1 2 explicitly

and we can impose exterior condition on 1205940 Indeed 120594

0is

exactly the same as in the monodimensional case that is itis given by (73) and (75) Obviously 120594

2= 0 so that

a22

= int

1

0

ℎ(119904)3119889119904 = ⟨ℎ

3⟩ a

12= a21

= 0 (84)

As for the last term

1205941= minus1199101+ (int

1

0

119889119904

ℎ(119904)3)

minus1

int

1199101

0

119889119904

ℎ(119904)3

a11

= (int

1

0

119889119904

ℎ(119904)3)

minus1

=1

⟨1ℎ3⟩

(85)

Finally the function V0satisfies the boundary value problem

1

⟨1ℎ3⟩

1205972V0

12059711990921

+ ⟨ℎ3⟩1205972V0

12059711990922

= 0 in O

V0(0 1199092) = 0

V0(1 1199092) = 119872

120590(119902 (1199092)) V

0is 1-periodic in 119909

2

(86)

It can be solved using the Fourier method and we get

V0(1199091 1199092) =

infin

sum

119896=1

sh (radic⟨ℎ3⟩ ⟨ℎminus3⟩ 1198961205871199091)

times (120572119896sin 119896120587119909

2+ 120573119896cos 119896120587119909

2)

120572119896= 2int

1

0

119872120590(119902 (119905)) sin 119896120587119905119889119905

120573119896= 2int

1

0

119872120590(119902 (119905)) cos 119896120587119905119889119905

(87)

Since the approximation

w119898asymp V0(119909) + 119881120594

0(1198981199091) +

1

1198981205941(1198981199091)120597V0

1205971199091

(119909) (88)

now satisfies the boundary conditions on 120597O it is easy to seethat

1003816100381610038161003816119908119898 minus (V0(119909) + 119881120594

0(119898119909))

1003816100381610038161003816119871infin(O) le 1198621

119898(89)

follows from the maximum principle Assuming that for 119898large enough

V0(119909) + 119881120594

0(119898119909) le 119872

+ (90)

we have1003816100381610038161003816119901119898 minus 119867 (V

0(119909) + 119881120594

0(119898119909))

1003816100381610038161003816119871infin(O) 997888rarr 0 (91)

Finally

119901119898 1199010= int

1

0

119867(V0(119909) + 119881120594

0(1199101)) 1198891199101 (92)

We have proved that

Theorem 7 Let 119901119898be the solution to the problem (67) (79)

and (80) and let V01205940be defined by (87) and (42) respectively

If (90) holds then1003816100381610038161003816119901119898 minus 119867 (V

0 (119909) + 1198811205940 (119898119909))

1003816100381610038161003816119871infin(O) 997888rarr 0

119901119898 1199010= int

1

0

119867(V0(119909) + 119881120594

0(1199101)) 1198891199101

(93)

Remark 8 It is important to notice that 1205870= 119867(119901

0) satisfies

2

sum

119894119896=1

120597

120597119909119894

(a119894119896

120583 (1205870)

1205971205870

120597119909119896

) = 0 in O 1205870= 119902 on 120597O (94)

and thus we would expect it to be the limit of 119901119898in analogy

with the linear case However 1199010

= 1205870

If119881 = lim119898rarrinfin

119898|V119898| is small we can expand119867(V

0(119909)+

1198811205940(119898119909)) in powers of 119881 and we get

119901119898(119909) asymp 119867 (V

0(119909) + 119881120594

0(119898119909))

= 119867 (V0(119909)) + 119881119867

1015840(V0(119909)) 120594

0(119910) + 119874 (119881

2)

= 1205870(119909) + 119881120583 (120587

0(119909)) 120594

0(119910) + 119874 (119881

2)

(95)

Thus

1199010= 1205870+ 120583 (120587

0) ⟨1205940⟩119881 + 119874 (119881

2) (96)

It can be formally written as

119901119898(119909) asymp 120587

0(119909) +

1003816100381610038161003816V1198981003816100381610038161003816

119898120583 (1205870(119909)) 120594

0(119898119909) + 119874(

1003816100381610038161003816V11989810038161003816100381610038162

1198982)

(97)

Appendix

The Maximum Principle

Our goal is to derive maximum principles for the linearReynolds equation with sharp explicit constants in orderto solve the nonlinear Reynolds equation with pressure-dependent viscosity We assume without losing generalitythat V = (16)Vi Indeed we can always choose the coor-dinate system in a way that the first coordinate axis 119909 has adirection of the velocity of relative motion V

The lower bound for 119908(119909 119910) is of no interest justthe upper bound Function 119908(119909 119910) is the solution to theboundary value problem

div (ℎ3nabla119908) = V120597ℎ

120597119909in O sub R2

119908 = 119872120590(119902) on 120597O

(A1)

We assume that if V(120597ℎ120597119909) gt 0 then 119908 cannot have amaximum point in the domain O and thus

119908 (119909 119910) le max(119909119910)isin120597O

119872120590(119902 (119909 119910)) (A2)

8 Mathematical Problems in Engineering

However it is not realistic to assume that (120597ℎ120597119909) does notchange the sign To find the upper bound in the general casewe use the procedure from the DeGiorgi theorem The mainresult of the section is as follows

TheoremA1 Let119908 be the solution to the problem (A1)Then

119908 (119909 119910) le1003816100381610038161003816119872120590 (119902)

1003816100381610038161003816119871infin(120597O) +Z (V ℎO 120583 119902) (A3)

Z=3(85)

times(3

2)

(285)(2120587)(14)

ℎ3

0

diamO (|V| ℎ1|O|15

+ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times |O|(75)

(A4)

Proof The function 119911 = 119908 minus 119866 satisfies

div (ℎ3nabla119911) = V120597ℎ

120597119909+ div (ℎ3nabla119872

120590(119902)) in O sub R2 (A5)

119911 = 0 on 120597O (A6)

Next we introduce the embedding constant for 119882120

(O) sub

119871119903(O) denoted119872

119903 such that

|V|119871119903(O) le 119872119903|nablaV|1198712(O) forallV isin 119867

1

0(O) (A7)

That constant can be estimated as

119872119903le1

2(diamO)

2|O|1119903 minus 12

(119903 + 2

2)

((119903+2)2119903)

radic2120587 (A8)

See for example [20 Lemma 1] Next we define the se-quence

120582119896+1

= 3(120582119896

2+ 1) 120582

1= 2 (A9)

Easy computation yields

120582119896= 8(

3

2)

119896minus1

minus 6 (A10)

Let

119911+(119909 119910) = max 119911 (119909 119910) 0 (A11)

We test (A5) with (119911+)1+120582119896+1 and get

intO

ℎ3nabla119911+nabla(119911+)1+120582119896+1

=1 + 120582119896+1

(1 + 120582119896+1

2)2intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

= intO

[Vℎ120597

120597119909(119911+)1+120582119896+1

+ ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

]

(A12)

For the left-hand side we get the lower bound

ℎ3

0

1 + 120582119896+1

(1 + 120582119896+1

2)2

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

1198712(O)

(A13)

We estimate the terms on the right-hand side using the sameidea

intO

ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

le ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O)

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

intO

ℎV120597

120597119909(119911+)1+120582119896+1

le ℎ1 |V| |O|

15100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

(A14)

Thus it remains to estimate |nabla(119911+)1+120582119896+1 |11987154(O) We have

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

100381610038161003816100381610038161003816

54

11987154(O)

= (1 + 120582119896+1

)54

intO

(119911+)(5120582119896+14)1003816100381610038161003816nabla119911

+1003816100381610038161003816

54

= (1 + 120582119896+1

1 + (120582119896+1

2))

54

intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

541003816100381610038161003816119911+1003816100381610038161003816

(5120582119896+18)

le [Holders inequality with 119901 =8

5 1199011015840=8

3]

le (1 + 120582119896+1

1 + (120582119896+1

2))

54

(intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

)

58

times (intO

(119911+)(53)120582

119896+1

)

38

= [due to (A9) 53120582119896+1

= 5(1 +120582119896

2)

3

8=1

4

120582119896+1

2 + 120582119896

]

= (1 + 120582119896+1

1 + (120582119896+1

2))

54100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

54

1198712(O)

times100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(120582119896+1(2+120582

119896))(54)

1198715(O)

(A15)

Combining with (A14) and (A13) we get

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)1003816100381610038161003816100381610038161198712(O)

le ℎminus3

0(1 +

120582119896+1

2)100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198715(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

Mathematical Problems in Engineering 9

le 119872((120582119896+1)(2+120582

119896))

5ℎminus3

0(1+

120582119896+1

2)

times100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198712(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

(A16)

We recall that

120582119896+1

2 + 120582119896

=3

2(A17)

and define

120572 = 119872(32)

5ℎminus3

0(ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

) (A18)

as well as

120590119896=

1

120572

100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(2(2+120582119896))

1198712(O)

(A19)

Then (A16) implies

120590119896+1

le (1 +120582119896+1

2)

(2(2+120582119896))

120590((120582119896+1)(2+120582

119896+1))

119896 (A20)

Taking the logarithm we arrive at

log120590119896+1

lelog (1 + (120582

119896+12))

1 + (120582119896+1

2)+

120582119896+1

2 + 120582119896+1

log120590119896 (A21)

We first notice that

120582119896+1

2 + 120582119896+1

lt 1 (A22)

and then

1 +120582119896

2= 4(

3

2)

119896minus1

minus 2 gt (3

2)

119896

(A23)

Since the function 119909 997891rarr (log119909119909) is decreasing for 119909 gt 119890 wehave

log (1 + (1205821198962))

1 + (1205821198962)

lelog [(32)119896]

(32)119896

= 119896 log 3

2(2

3)

119896

forall119896 ge 3

(A24)

Then

log120590119896+1

le (119896 + 1) log 3

2(2

3)

119896

+ log120590119896

le log 3

2

119896+1

sum

119895=2

119895(2

3)

119895

+ log1205901le 8 log 3

2+ log120590

1

(A25)

Finally

120590119896+1

le (3

2)

8

1205901 (A26)

Now it remains to estimate 1205901 From the definitionwe see that

1205901=

1

120572

100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

12

1198712(O)

(A27)

To estimate 1205901 we proceed as before and test (A5) with (119911+)3

We get

3

4intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

2

= intO

(Vℎ120597(119911+)3

120597119909+ ℎ3nabla119872120590(119902) nabla(119911

+)3

)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

100381610038161003816100381610038161003816nabla(119911+)310038161003816100381610038161003816100381611987154(O)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times3

2

100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

1003816100381610038161003816119911+1003816100381610038161003816119871103(O)

(A28)

Thus100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

le2

ℎ3

0

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872(103)

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O)

(A29)

Finally testing (A5) with 119911+ we get

intO

ℎ31003816100381610038161003816nabla119911+1003816100381610038161003816

2

= intO

(Vℎ120597119911+

120597119909+ ℎ3nabla119872120590(119902) nabla119911

+) (A30)

so that

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O) le

1

ℎ3

0

(|V| ℎ1|O|12

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198712(O))

le |O|310

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

(A31)

Combining (A29) with (A31) and (A26) gives

120590119896+1

le (3

2)

81

120572

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A32)

Since

120590119896+1

ge1

120572119872(2(2+120582k))2

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) (A33)

10 Mathematical Problems in Engineering

we have arrived to

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) le (

3

2)

8

119872(2(2+120582

119896))

2

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A34)

Since lim119896rarrinfin

120582119896= infin we get

1003816100381610038161003816119911+1003816100381610038161003816119871infin(O) le (

3

2)

8radic2

ℎ3

0

times(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A35)

Finally (A8) implies (A3)

Acknowledgment

This work was supported by MZOS grant 037-0372787-2797

References

[1] O Reynolds ldquoOn the theory and application and its applicationto Mr Beauchamp towerrsquos experiments Including and experi-mental determination of the viscosity of olive oilrdquo PhilosophicalTransactions of the Royal Society vol 117 pp 157ndash234 1886

[2] G G Stokes ldquoNotes on hydrodynamics On the dynamicalequationsrdquoCambridge and DublinMathematical Journal III pp121ndash127 1848

[3] P W Brifgman ldquoThe viscosity of liquids under pressurerdquo Pro-ceedings of the National Academy of Sciences of the United Statesof America vol 11 no 10 pp 603ndash606 1925

[4] A Z Szeri Fluid Film Lubrication Cambridge University PressNew York NY USA 1998

[5] E K Gatcombe ldquoLubrication characteristics of involute spur-gearsmdasha theoretical investigationrdquoTransactions of theAmericanSociety of Mechanical Engineers vol 67 pp 177ndash181 1945

[6] C Barus ldquoIsotherms isopiestics and isometrics relative to vis-cosityrdquo American Journal of Science vol 45 pp 87ndash96 1893

[7] W R Jones ldquoPressure viscosity measurement for several lubri-cantsrdquo ASLE Transactions vol 18 pp 249ndash262 1975

[8] J Hron J Malek and K R Rajagopal ldquoSimple flows of fluidswith pressure-dependent viscositiesrdquo Proceedings of the RoyalSociety A vol 457 no 2011 pp 1603ndash1622 2001

[9] C J A Roelands Correlation aspects of the viscosity-pressurerelationship of lubricating oils [PhD thesis] Delft University ofTechnology Delft The Netherlands 1966

[10] K R Rajagopal and A Z Szeri ldquoOn an inconsistency in thederivation of the equations of elastohydrodynamic lubricationrdquoProceedings of the Royal Society A vol 459 no 2039 pp 2771ndash2786 2003

[11] M Renardy ldquoSome remarks on the Navier-Stokes equationswith a pressure-dependent viscosityrdquo Communications in Par-tial Differential Equations vol 11 no 7 pp 779ndash793 1986

[12] F Gazzola and P Secchi ldquoSome results about stationary Navier-Stokes equations with a pressure-dependent viscosityrdquo in Pro-ceedings of the International Conference on Navier-Stokes Equa-tions vol 388 of Pitman Research Notes in Mathematics Seriespp 174ndash183 Varenna Italy 1998

[13] E Marusic-Paloka ldquoAn analysis of the Stokes system with pres-sure dependent viscosityrdquo In press

[14] S Marusic and E Marusic-Paloka ldquoTwo-scale convergence forthin domains and its applications to some lower-dimensionalmodels in fluid mechanicsrdquo Asymptotic Analysis vol 23 no 1pp 23ndash57 2000

[15] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001

[16] M Kane and B Bou-Said ldquoComparison of homogenization anddirect techniques for the treatment of roughness in incompress-ible lubricationrdquo Journal of Tribology vol 126 no 4 pp 733ndash7372004

[17] M Jai ldquoHomogenization and two-scale convergence of thecompressible Reynolds lubrication equation modelling the fly-ing characteristics of a roughmagnetic head over a rough rigid-disk surfacerdquo RAIRO Modelisation Mathematique et AnalyseNumerique vol 29 no 2 pp 199ndash233 1995

[18] P Wall ldquoHomogenization of Reynolds equation by two-scaleconvergencerdquo Chinese Annals of Mathematics B vol 28 no 3pp 363ndash374 2007

[19] A Bensoussan J-L Lions and G Papanicolaou AsymptoticAnalysis for Periodic Structures North-Holland AmsterdamThe Netherlands 1978

[20] E Marusic-Paloka and A Mikelic ldquoThe derivation of a nonlin-ear filtration law including the inertia effects via homogeniza-tionrdquo Nonlinear Analysis Theory Methods amp Applications vol42 no 1 pp 97ndash137 2000

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Mathematical Problems in Engineering

Aswedid in the existence analysis and in the previous sectionwe linearize the equation by substitution

119908119898= 119872120590(119901119898) (68)

where the function119872120590is defined by (12) Now 119908

119898satisfies

div (ℎ3119872 nabla119908119898) = 6V

119898sdot nablaℎ119898

in O (69)

We postulate the asymptotic expansion in the form

119908119898 (119909) asymp 119908

0(119909 119910) +

1

1198981199081(119909 119910)

+1

11989821199082(119909 119910) + sdot sdot sdot 119910 = 119898119909

(70)

All functions are assumed to be 119884-periodic in 119910 variablePlugging that in (69) and collecting formally terms with

equal powers of119898 we get

1198982 div119910(ℎ3nabla1199101199080) = 6V sdot nabla

119910ℎ

119898 div119910(ℎ3nabla1199101199081) + div

119910(ℎ3nabla1199091199080) + ℎ3div119909nabla1199101199080= 0

1 div119910(ℎ3nabla1199101199082) + div

119910(ℎ3nabla1199091199081) + ℎ3div119909nabla1199101199081

+ ℎ3Δ1199091199080= 0

Denoting

119881 = |V| V0=V119881

(71)

we have

1199080(119909 119910) = V

0(119909) + 119881120594

0(119910)

div119910(ℎ3nabla1199101205940) = 6V

0sdot nabla119910ℎ in 119884

1199081(119909 119910) =

2

sum

119896=1

120594119896(119910)

120597V0

120597119909119896

(119909) + V1(119909)

div119910(ℎ3nabla119910(120594119896+ 119910119896)) = 0

2

sum

119896ℓ=1

a119896ℓ

1205972V0

120597119909119896120597119909ℓ

= 0 in O

a119896ℓ

= int119884

ℎ3 120597

120597119910ℓ

(120594119896+ 119910119896) 119889119910

(72)

Remark 6 Thesame computation can be done in one-dimen-sional case and it gives

(ℎ31205941015840

0)1015840

= 6ℎ1015840997904rArr 1205940= 6int

119910

0

119889119905

ℎ(119905)2+ 1198620int

119910

0

119889119905

ℎ(119905)3+ 1198621

(73)

V0= 119909119872(119902

1) (74)

Constants 1198620 1198621are chosen in a way that boundary condi-

tions 1205940(0) = 120594

0(1) = 0 are met and it follows that

1198621= 0 119862

0= minus 6(int

1

0

119889119905

ℎ(119905)3)

minus1

int

1

0

119889119905

ℎ(119905)2 (75)

Then

119908119898asymp 1199080 (119909119898119909)

= V0(119909) + 120594

0(119898119909)

= 119909119872120590(1199021)

+ 6119881[int

119898119909

0

119889119905

ℎ(119905)2minus (int

1

0

119889119905

ℎ(119905)3)

minus1

times int

1

0

119889119905

ℎ(119905)2

int

119898119909

0

119889119905

ℎ(119905)3]

= 119909119872120590(1199021)

+ 6119881119887119898

2(119909) 1198873(1) minus 119887

2(1) 119887119898

3(119909)

1198873(1)

= 119909119872120590(1199021) + 119881120594

0(119898119909)

(76)

That is a very good approximation of our exact solution (50)It is important to notice that the choice of constants

1198620 1198621was determined from the exterior boundary con-

dition So we should expect the same in two-dimensionalcase However the treatment of boundary conditions intwo-dimensional case is much more complicated and theboundary layer is to be expected

The derived asymptotic expansion should be justified byproving the convergence And we need the strong conver-gence (with corrector of course) for 119908

119898in order to get the

convergence for 119901119898 The form of the approximation

119908119898asymp V0(119909) + 119881120594

0(119898119909) + sdot sdot sdot (77)

suggests that the boundary layer phenomenon should appearon the exterior boundary 120597O since 120594

0term cannot satisfy the

Dirichlet condition on 120597O To get the error estimate and thestrong convergence we need to handle that boundary layerThus at this point we simplify the domain and the boundarycondition in order to be able to avoid it We assume that

O = ]0 1[ times R (78)

119901120576(0 1199092) = 0 119901

120576(1 1199092) = 119902 (119909

2) (79)

1199092997891997888rarr 119901

120576(1199091 1199092) is 1-periodic (80)

ℎ (1199101 1199102) = ℎ (119910

1) (81)

119902 is 1-periodic (82)

Now

119908120576= 119872120590(119901120576) with 119872

120590(119901) =int

119901

120590

119889119904

120583 (119904)997904rArr 119908

120576(0 1199092) = 0

119908120576(1 1199092) = 119872

120590(119902 (1199092))

(83)

and 1199092997891rarr 119908120576(1199091 1199092) is 1-periodic

Mathematical Problems in Engineering 7

In that case we can compute1205940and120594119896 119896 = 1 2 explicitly

and we can impose exterior condition on 1205940 Indeed 120594

0is

exactly the same as in the monodimensional case that is itis given by (73) and (75) Obviously 120594

2= 0 so that

a22

= int

1

0

ℎ(119904)3119889119904 = ⟨ℎ

3⟩ a

12= a21

= 0 (84)

As for the last term

1205941= minus1199101+ (int

1

0

119889119904

ℎ(119904)3)

minus1

int

1199101

0

119889119904

ℎ(119904)3

a11

= (int

1

0

119889119904

ℎ(119904)3)

minus1

=1

⟨1ℎ3⟩

(85)

Finally the function V0satisfies the boundary value problem

1

⟨1ℎ3⟩

1205972V0

12059711990921

+ ⟨ℎ3⟩1205972V0

12059711990922

= 0 in O

V0(0 1199092) = 0

V0(1 1199092) = 119872

120590(119902 (1199092)) V

0is 1-periodic in 119909

2

(86)

It can be solved using the Fourier method and we get

V0(1199091 1199092) =

infin

sum

119896=1

sh (radic⟨ℎ3⟩ ⟨ℎminus3⟩ 1198961205871199091)

times (120572119896sin 119896120587119909

2+ 120573119896cos 119896120587119909

2)

120572119896= 2int

1

0

119872120590(119902 (119905)) sin 119896120587119905119889119905

120573119896= 2int

1

0

119872120590(119902 (119905)) cos 119896120587119905119889119905

(87)

Since the approximation

w119898asymp V0(119909) + 119881120594

0(1198981199091) +

1

1198981205941(1198981199091)120597V0

1205971199091

(119909) (88)

now satisfies the boundary conditions on 120597O it is easy to seethat

1003816100381610038161003816119908119898 minus (V0(119909) + 119881120594

0(119898119909))

1003816100381610038161003816119871infin(O) le 1198621

119898(89)

follows from the maximum principle Assuming that for 119898large enough

V0(119909) + 119881120594

0(119898119909) le 119872

+ (90)

we have1003816100381610038161003816119901119898 minus 119867 (V

0(119909) + 119881120594

0(119898119909))

1003816100381610038161003816119871infin(O) 997888rarr 0 (91)

Finally

119901119898 1199010= int

1

0

119867(V0(119909) + 119881120594

0(1199101)) 1198891199101 (92)

We have proved that

Theorem 7 Let 119901119898be the solution to the problem (67) (79)

and (80) and let V01205940be defined by (87) and (42) respectively

If (90) holds then1003816100381610038161003816119901119898 minus 119867 (V

0 (119909) + 1198811205940 (119898119909))

1003816100381610038161003816119871infin(O) 997888rarr 0

119901119898 1199010= int

1

0

119867(V0(119909) + 119881120594

0(1199101)) 1198891199101

(93)

Remark 8 It is important to notice that 1205870= 119867(119901

0) satisfies

2

sum

119894119896=1

120597

120597119909119894

(a119894119896

120583 (1205870)

1205971205870

120597119909119896

) = 0 in O 1205870= 119902 on 120597O (94)

and thus we would expect it to be the limit of 119901119898in analogy

with the linear case However 1199010

= 1205870

If119881 = lim119898rarrinfin

119898|V119898| is small we can expand119867(V

0(119909)+

1198811205940(119898119909)) in powers of 119881 and we get

119901119898(119909) asymp 119867 (V

0(119909) + 119881120594

0(119898119909))

= 119867 (V0(119909)) + 119881119867

1015840(V0(119909)) 120594

0(119910) + 119874 (119881

2)

= 1205870(119909) + 119881120583 (120587

0(119909)) 120594

0(119910) + 119874 (119881

2)

(95)

Thus

1199010= 1205870+ 120583 (120587

0) ⟨1205940⟩119881 + 119874 (119881

2) (96)

It can be formally written as

119901119898(119909) asymp 120587

0(119909) +

1003816100381610038161003816V1198981003816100381610038161003816

119898120583 (1205870(119909)) 120594

0(119898119909) + 119874(

1003816100381610038161003816V11989810038161003816100381610038162

1198982)

(97)

Appendix

The Maximum Principle

Our goal is to derive maximum principles for the linearReynolds equation with sharp explicit constants in orderto solve the nonlinear Reynolds equation with pressure-dependent viscosity We assume without losing generalitythat V = (16)Vi Indeed we can always choose the coor-dinate system in a way that the first coordinate axis 119909 has adirection of the velocity of relative motion V

The lower bound for 119908(119909 119910) is of no interest justthe upper bound Function 119908(119909 119910) is the solution to theboundary value problem

div (ℎ3nabla119908) = V120597ℎ

120597119909in O sub R2

119908 = 119872120590(119902) on 120597O

(A1)

We assume that if V(120597ℎ120597119909) gt 0 then 119908 cannot have amaximum point in the domain O and thus

119908 (119909 119910) le max(119909119910)isin120597O

119872120590(119902 (119909 119910)) (A2)

8 Mathematical Problems in Engineering

However it is not realistic to assume that (120597ℎ120597119909) does notchange the sign To find the upper bound in the general casewe use the procedure from the DeGiorgi theorem The mainresult of the section is as follows

TheoremA1 Let119908 be the solution to the problem (A1)Then

119908 (119909 119910) le1003816100381610038161003816119872120590 (119902)

1003816100381610038161003816119871infin(120597O) +Z (V ℎO 120583 119902) (A3)

Z=3(85)

times(3

2)

(285)(2120587)(14)

ℎ3

0

diamO (|V| ℎ1|O|15

+ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times |O|(75)

(A4)

Proof The function 119911 = 119908 minus 119866 satisfies

div (ℎ3nabla119911) = V120597ℎ

120597119909+ div (ℎ3nabla119872

120590(119902)) in O sub R2 (A5)

119911 = 0 on 120597O (A6)

Next we introduce the embedding constant for 119882120

(O) sub

119871119903(O) denoted119872

119903 such that

|V|119871119903(O) le 119872119903|nablaV|1198712(O) forallV isin 119867

1

0(O) (A7)

That constant can be estimated as

119872119903le1

2(diamO)

2|O|1119903 minus 12

(119903 + 2

2)

((119903+2)2119903)

radic2120587 (A8)

See for example [20 Lemma 1] Next we define the se-quence

120582119896+1

= 3(120582119896

2+ 1) 120582

1= 2 (A9)

Easy computation yields

120582119896= 8(

3

2)

119896minus1

minus 6 (A10)

Let

119911+(119909 119910) = max 119911 (119909 119910) 0 (A11)

We test (A5) with (119911+)1+120582119896+1 and get

intO

ℎ3nabla119911+nabla(119911+)1+120582119896+1

=1 + 120582119896+1

(1 + 120582119896+1

2)2intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

= intO

[Vℎ120597

120597119909(119911+)1+120582119896+1

+ ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

]

(A12)

For the left-hand side we get the lower bound

ℎ3

0

1 + 120582119896+1

(1 + 120582119896+1

2)2

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

1198712(O)

(A13)

We estimate the terms on the right-hand side using the sameidea

intO

ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

le ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O)

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

intO

ℎV120597

120597119909(119911+)1+120582119896+1

le ℎ1 |V| |O|

15100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

(A14)

Thus it remains to estimate |nabla(119911+)1+120582119896+1 |11987154(O) We have

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

100381610038161003816100381610038161003816

54

11987154(O)

= (1 + 120582119896+1

)54

intO

(119911+)(5120582119896+14)1003816100381610038161003816nabla119911

+1003816100381610038161003816

54

= (1 + 120582119896+1

1 + (120582119896+1

2))

54

intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

541003816100381610038161003816119911+1003816100381610038161003816

(5120582119896+18)

le [Holders inequality with 119901 =8

5 1199011015840=8

3]

le (1 + 120582119896+1

1 + (120582119896+1

2))

54

(intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

)

58

times (intO

(119911+)(53)120582

119896+1

)

38

= [due to (A9) 53120582119896+1

= 5(1 +120582119896

2)

3

8=1

4

120582119896+1

2 + 120582119896

]

= (1 + 120582119896+1

1 + (120582119896+1

2))

54100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

54

1198712(O)

times100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(120582119896+1(2+120582

119896))(54)

1198715(O)

(A15)

Combining with (A14) and (A13) we get

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)1003816100381610038161003816100381610038161198712(O)

le ℎminus3

0(1 +

120582119896+1

2)100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198715(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

Mathematical Problems in Engineering 9

le 119872((120582119896+1)(2+120582

119896))

5ℎminus3

0(1+

120582119896+1

2)

times100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198712(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

(A16)

We recall that

120582119896+1

2 + 120582119896

=3

2(A17)

and define

120572 = 119872(32)

5ℎminus3

0(ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

) (A18)

as well as

120590119896=

1

120572

100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(2(2+120582119896))

1198712(O)

(A19)

Then (A16) implies

120590119896+1

le (1 +120582119896+1

2)

(2(2+120582119896))

120590((120582119896+1)(2+120582

119896+1))

119896 (A20)

Taking the logarithm we arrive at

log120590119896+1

lelog (1 + (120582

119896+12))

1 + (120582119896+1

2)+

120582119896+1

2 + 120582119896+1

log120590119896 (A21)

We first notice that

120582119896+1

2 + 120582119896+1

lt 1 (A22)

and then

1 +120582119896

2= 4(

3

2)

119896minus1

minus 2 gt (3

2)

119896

(A23)

Since the function 119909 997891rarr (log119909119909) is decreasing for 119909 gt 119890 wehave

log (1 + (1205821198962))

1 + (1205821198962)

lelog [(32)119896]

(32)119896

= 119896 log 3

2(2

3)

119896

forall119896 ge 3

(A24)

Then

log120590119896+1

le (119896 + 1) log 3

2(2

3)

119896

+ log120590119896

le log 3

2

119896+1

sum

119895=2

119895(2

3)

119895

+ log1205901le 8 log 3

2+ log120590

1

(A25)

Finally

120590119896+1

le (3

2)

8

1205901 (A26)

Now it remains to estimate 1205901 From the definitionwe see that

1205901=

1

120572

100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

12

1198712(O)

(A27)

To estimate 1205901 we proceed as before and test (A5) with (119911+)3

We get

3

4intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

2

= intO

(Vℎ120597(119911+)3

120597119909+ ℎ3nabla119872120590(119902) nabla(119911

+)3

)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

100381610038161003816100381610038161003816nabla(119911+)310038161003816100381610038161003816100381611987154(O)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times3

2

100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

1003816100381610038161003816119911+1003816100381610038161003816119871103(O)

(A28)

Thus100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

le2

ℎ3

0

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872(103)

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O)

(A29)

Finally testing (A5) with 119911+ we get

intO

ℎ31003816100381610038161003816nabla119911+1003816100381610038161003816

2

= intO

(Vℎ120597119911+

120597119909+ ℎ3nabla119872120590(119902) nabla119911

+) (A30)

so that

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O) le

1

ℎ3

0

(|V| ℎ1|O|12

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198712(O))

le |O|310

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

(A31)

Combining (A29) with (A31) and (A26) gives

120590119896+1

le (3

2)

81

120572

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A32)

Since

120590119896+1

ge1

120572119872(2(2+120582k))2

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) (A33)

10 Mathematical Problems in Engineering

we have arrived to

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) le (

3

2)

8

119872(2(2+120582

119896))

2

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A34)

Since lim119896rarrinfin

120582119896= infin we get

1003816100381610038161003816119911+1003816100381610038161003816119871infin(O) le (

3

2)

8radic2

ℎ3

0

times(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A35)

Finally (A8) implies (A3)

Acknowledgment

This work was supported by MZOS grant 037-0372787-2797

References

[1] O Reynolds ldquoOn the theory and application and its applicationto Mr Beauchamp towerrsquos experiments Including and experi-mental determination of the viscosity of olive oilrdquo PhilosophicalTransactions of the Royal Society vol 117 pp 157ndash234 1886

[2] G G Stokes ldquoNotes on hydrodynamics On the dynamicalequationsrdquoCambridge and DublinMathematical Journal III pp121ndash127 1848

[3] P W Brifgman ldquoThe viscosity of liquids under pressurerdquo Pro-ceedings of the National Academy of Sciences of the United Statesof America vol 11 no 10 pp 603ndash606 1925

[4] A Z Szeri Fluid Film Lubrication Cambridge University PressNew York NY USA 1998

[5] E K Gatcombe ldquoLubrication characteristics of involute spur-gearsmdasha theoretical investigationrdquoTransactions of theAmericanSociety of Mechanical Engineers vol 67 pp 177ndash181 1945

[6] C Barus ldquoIsotherms isopiestics and isometrics relative to vis-cosityrdquo American Journal of Science vol 45 pp 87ndash96 1893

[7] W R Jones ldquoPressure viscosity measurement for several lubri-cantsrdquo ASLE Transactions vol 18 pp 249ndash262 1975

[8] J Hron J Malek and K R Rajagopal ldquoSimple flows of fluidswith pressure-dependent viscositiesrdquo Proceedings of the RoyalSociety A vol 457 no 2011 pp 1603ndash1622 2001

[9] C J A Roelands Correlation aspects of the viscosity-pressurerelationship of lubricating oils [PhD thesis] Delft University ofTechnology Delft The Netherlands 1966

[10] K R Rajagopal and A Z Szeri ldquoOn an inconsistency in thederivation of the equations of elastohydrodynamic lubricationrdquoProceedings of the Royal Society A vol 459 no 2039 pp 2771ndash2786 2003

[11] M Renardy ldquoSome remarks on the Navier-Stokes equationswith a pressure-dependent viscosityrdquo Communications in Par-tial Differential Equations vol 11 no 7 pp 779ndash793 1986

[12] F Gazzola and P Secchi ldquoSome results about stationary Navier-Stokes equations with a pressure-dependent viscosityrdquo in Pro-ceedings of the International Conference on Navier-Stokes Equa-tions vol 388 of Pitman Research Notes in Mathematics Seriespp 174ndash183 Varenna Italy 1998

[13] E Marusic-Paloka ldquoAn analysis of the Stokes system with pres-sure dependent viscosityrdquo In press

[14] S Marusic and E Marusic-Paloka ldquoTwo-scale convergence forthin domains and its applications to some lower-dimensionalmodels in fluid mechanicsrdquo Asymptotic Analysis vol 23 no 1pp 23ndash57 2000

[15] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001

[16] M Kane and B Bou-Said ldquoComparison of homogenization anddirect techniques for the treatment of roughness in incompress-ible lubricationrdquo Journal of Tribology vol 126 no 4 pp 733ndash7372004

[17] M Jai ldquoHomogenization and two-scale convergence of thecompressible Reynolds lubrication equation modelling the fly-ing characteristics of a roughmagnetic head over a rough rigid-disk surfacerdquo RAIRO Modelisation Mathematique et AnalyseNumerique vol 29 no 2 pp 199ndash233 1995

[18] P Wall ldquoHomogenization of Reynolds equation by two-scaleconvergencerdquo Chinese Annals of Mathematics B vol 28 no 3pp 363ndash374 2007

[19] A Bensoussan J-L Lions and G Papanicolaou AsymptoticAnalysis for Periodic Structures North-Holland AmsterdamThe Netherlands 1978

[20] E Marusic-Paloka and A Mikelic ldquoThe derivation of a nonlin-ear filtration law including the inertia effects via homogeniza-tionrdquo Nonlinear Analysis Theory Methods amp Applications vol42 no 1 pp 97ndash137 2000

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 7

In that case we can compute1205940and120594119896 119896 = 1 2 explicitly

and we can impose exterior condition on 1205940 Indeed 120594

0is

exactly the same as in the monodimensional case that is itis given by (73) and (75) Obviously 120594

2= 0 so that

a22

= int

1

0

ℎ(119904)3119889119904 = ⟨ℎ

3⟩ a

12= a21

= 0 (84)

As for the last term

1205941= minus1199101+ (int

1

0

119889119904

ℎ(119904)3)

minus1

int

1199101

0

119889119904

ℎ(119904)3

a11

= (int

1

0

119889119904

ℎ(119904)3)

minus1

=1

⟨1ℎ3⟩

(85)

Finally the function V0satisfies the boundary value problem

1

⟨1ℎ3⟩

1205972V0

12059711990921

+ ⟨ℎ3⟩1205972V0

12059711990922

= 0 in O

V0(0 1199092) = 0

V0(1 1199092) = 119872

120590(119902 (1199092)) V

0is 1-periodic in 119909

2

(86)

It can be solved using the Fourier method and we get

V0(1199091 1199092) =

infin

sum

119896=1

sh (radic⟨ℎ3⟩ ⟨ℎminus3⟩ 1198961205871199091)

times (120572119896sin 119896120587119909

2+ 120573119896cos 119896120587119909

2)

120572119896= 2int

1

0

119872120590(119902 (119905)) sin 119896120587119905119889119905

120573119896= 2int

1

0

119872120590(119902 (119905)) cos 119896120587119905119889119905

(87)

Since the approximation

w119898asymp V0(119909) + 119881120594

0(1198981199091) +

1

1198981205941(1198981199091)120597V0

1205971199091

(119909) (88)

now satisfies the boundary conditions on 120597O it is easy to seethat

1003816100381610038161003816119908119898 minus (V0(119909) + 119881120594

0(119898119909))

1003816100381610038161003816119871infin(O) le 1198621

119898(89)

follows from the maximum principle Assuming that for 119898large enough

V0(119909) + 119881120594

0(119898119909) le 119872

+ (90)

we have1003816100381610038161003816119901119898 minus 119867 (V

0(119909) + 119881120594

0(119898119909))

1003816100381610038161003816119871infin(O) 997888rarr 0 (91)

Finally

119901119898 1199010= int

1

0

119867(V0(119909) + 119881120594

0(1199101)) 1198891199101 (92)

We have proved that

Theorem 7 Let 119901119898be the solution to the problem (67) (79)

and (80) and let V01205940be defined by (87) and (42) respectively

If (90) holds then1003816100381610038161003816119901119898 minus 119867 (V

0 (119909) + 1198811205940 (119898119909))

1003816100381610038161003816119871infin(O) 997888rarr 0

119901119898 1199010= int

1

0

119867(V0(119909) + 119881120594

0(1199101)) 1198891199101

(93)

Remark 8 It is important to notice that 1205870= 119867(119901

0) satisfies

2

sum

119894119896=1

120597

120597119909119894

(a119894119896

120583 (1205870)

1205971205870

120597119909119896

) = 0 in O 1205870= 119902 on 120597O (94)

and thus we would expect it to be the limit of 119901119898in analogy

with the linear case However 1199010

= 1205870

If119881 = lim119898rarrinfin

119898|V119898| is small we can expand119867(V

0(119909)+

1198811205940(119898119909)) in powers of 119881 and we get

119901119898(119909) asymp 119867 (V

0(119909) + 119881120594

0(119898119909))

= 119867 (V0(119909)) + 119881119867

1015840(V0(119909)) 120594

0(119910) + 119874 (119881

2)

= 1205870(119909) + 119881120583 (120587

0(119909)) 120594

0(119910) + 119874 (119881

2)

(95)

Thus

1199010= 1205870+ 120583 (120587

0) ⟨1205940⟩119881 + 119874 (119881

2) (96)

It can be formally written as

119901119898(119909) asymp 120587

0(119909) +

1003816100381610038161003816V1198981003816100381610038161003816

119898120583 (1205870(119909)) 120594

0(119898119909) + 119874(

1003816100381610038161003816V11989810038161003816100381610038162

1198982)

(97)

Appendix

The Maximum Principle

Our goal is to derive maximum principles for the linearReynolds equation with sharp explicit constants in orderto solve the nonlinear Reynolds equation with pressure-dependent viscosity We assume without losing generalitythat V = (16)Vi Indeed we can always choose the coor-dinate system in a way that the first coordinate axis 119909 has adirection of the velocity of relative motion V

The lower bound for 119908(119909 119910) is of no interest justthe upper bound Function 119908(119909 119910) is the solution to theboundary value problem

div (ℎ3nabla119908) = V120597ℎ

120597119909in O sub R2

119908 = 119872120590(119902) on 120597O

(A1)

We assume that if V(120597ℎ120597119909) gt 0 then 119908 cannot have amaximum point in the domain O and thus

119908 (119909 119910) le max(119909119910)isin120597O

119872120590(119902 (119909 119910)) (A2)

8 Mathematical Problems in Engineering

However it is not realistic to assume that (120597ℎ120597119909) does notchange the sign To find the upper bound in the general casewe use the procedure from the DeGiorgi theorem The mainresult of the section is as follows

TheoremA1 Let119908 be the solution to the problem (A1)Then

119908 (119909 119910) le1003816100381610038161003816119872120590 (119902)

1003816100381610038161003816119871infin(120597O) +Z (V ℎO 120583 119902) (A3)

Z=3(85)

times(3

2)

(285)(2120587)(14)

ℎ3

0

diamO (|V| ℎ1|O|15

+ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times |O|(75)

(A4)

Proof The function 119911 = 119908 minus 119866 satisfies

div (ℎ3nabla119911) = V120597ℎ

120597119909+ div (ℎ3nabla119872

120590(119902)) in O sub R2 (A5)

119911 = 0 on 120597O (A6)

Next we introduce the embedding constant for 119882120

(O) sub

119871119903(O) denoted119872

119903 such that

|V|119871119903(O) le 119872119903|nablaV|1198712(O) forallV isin 119867

1

0(O) (A7)

That constant can be estimated as

119872119903le1

2(diamO)

2|O|1119903 minus 12

(119903 + 2

2)

((119903+2)2119903)

radic2120587 (A8)

See for example [20 Lemma 1] Next we define the se-quence

120582119896+1

= 3(120582119896

2+ 1) 120582

1= 2 (A9)

Easy computation yields

120582119896= 8(

3

2)

119896minus1

minus 6 (A10)

Let

119911+(119909 119910) = max 119911 (119909 119910) 0 (A11)

We test (A5) with (119911+)1+120582119896+1 and get

intO

ℎ3nabla119911+nabla(119911+)1+120582119896+1

=1 + 120582119896+1

(1 + 120582119896+1

2)2intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

= intO

[Vℎ120597

120597119909(119911+)1+120582119896+1

+ ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

]

(A12)

For the left-hand side we get the lower bound

ℎ3

0

1 + 120582119896+1

(1 + 120582119896+1

2)2

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

1198712(O)

(A13)

We estimate the terms on the right-hand side using the sameidea

intO

ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

le ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O)

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

intO

ℎV120597

120597119909(119911+)1+120582119896+1

le ℎ1 |V| |O|

15100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

(A14)

Thus it remains to estimate |nabla(119911+)1+120582119896+1 |11987154(O) We have

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

100381610038161003816100381610038161003816

54

11987154(O)

= (1 + 120582119896+1

)54

intO

(119911+)(5120582119896+14)1003816100381610038161003816nabla119911

+1003816100381610038161003816

54

= (1 + 120582119896+1

1 + (120582119896+1

2))

54

intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

541003816100381610038161003816119911+1003816100381610038161003816

(5120582119896+18)

le [Holders inequality with 119901 =8

5 1199011015840=8

3]

le (1 + 120582119896+1

1 + (120582119896+1

2))

54

(intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

)

58

times (intO

(119911+)(53)120582

119896+1

)

38

= [due to (A9) 53120582119896+1

= 5(1 +120582119896

2)

3

8=1

4

120582119896+1

2 + 120582119896

]

= (1 + 120582119896+1

1 + (120582119896+1

2))

54100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

54

1198712(O)

times100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(120582119896+1(2+120582

119896))(54)

1198715(O)

(A15)

Combining with (A14) and (A13) we get

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)1003816100381610038161003816100381610038161198712(O)

le ℎminus3

0(1 +

120582119896+1

2)100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198715(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

Mathematical Problems in Engineering 9

le 119872((120582119896+1)(2+120582

119896))

5ℎminus3

0(1+

120582119896+1

2)

times100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198712(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

(A16)

We recall that

120582119896+1

2 + 120582119896

=3

2(A17)

and define

120572 = 119872(32)

5ℎminus3

0(ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

) (A18)

as well as

120590119896=

1

120572

100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(2(2+120582119896))

1198712(O)

(A19)

Then (A16) implies

120590119896+1

le (1 +120582119896+1

2)

(2(2+120582119896))

120590((120582119896+1)(2+120582

119896+1))

119896 (A20)

Taking the logarithm we arrive at

log120590119896+1

lelog (1 + (120582

119896+12))

1 + (120582119896+1

2)+

120582119896+1

2 + 120582119896+1

log120590119896 (A21)

We first notice that

120582119896+1

2 + 120582119896+1

lt 1 (A22)

and then

1 +120582119896

2= 4(

3

2)

119896minus1

minus 2 gt (3

2)

119896

(A23)

Since the function 119909 997891rarr (log119909119909) is decreasing for 119909 gt 119890 wehave

log (1 + (1205821198962))

1 + (1205821198962)

lelog [(32)119896]

(32)119896

= 119896 log 3

2(2

3)

119896

forall119896 ge 3

(A24)

Then

log120590119896+1

le (119896 + 1) log 3

2(2

3)

119896

+ log120590119896

le log 3

2

119896+1

sum

119895=2

119895(2

3)

119895

+ log1205901le 8 log 3

2+ log120590

1

(A25)

Finally

120590119896+1

le (3

2)

8

1205901 (A26)

Now it remains to estimate 1205901 From the definitionwe see that

1205901=

1

120572

100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

12

1198712(O)

(A27)

To estimate 1205901 we proceed as before and test (A5) with (119911+)3

We get

3

4intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

2

= intO

(Vℎ120597(119911+)3

120597119909+ ℎ3nabla119872120590(119902) nabla(119911

+)3

)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

100381610038161003816100381610038161003816nabla(119911+)310038161003816100381610038161003816100381611987154(O)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times3

2

100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

1003816100381610038161003816119911+1003816100381610038161003816119871103(O)

(A28)

Thus100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

le2

ℎ3

0

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872(103)

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O)

(A29)

Finally testing (A5) with 119911+ we get

intO

ℎ31003816100381610038161003816nabla119911+1003816100381610038161003816

2

= intO

(Vℎ120597119911+

120597119909+ ℎ3nabla119872120590(119902) nabla119911

+) (A30)

so that

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O) le

1

ℎ3

0

(|V| ℎ1|O|12

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198712(O))

le |O|310

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

(A31)

Combining (A29) with (A31) and (A26) gives

120590119896+1

le (3

2)

81

120572

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A32)

Since

120590119896+1

ge1

120572119872(2(2+120582k))2

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) (A33)

10 Mathematical Problems in Engineering

we have arrived to

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) le (

3

2)

8

119872(2(2+120582

119896))

2

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A34)

Since lim119896rarrinfin

120582119896= infin we get

1003816100381610038161003816119911+1003816100381610038161003816119871infin(O) le (

3

2)

8radic2

ℎ3

0

times(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A35)

Finally (A8) implies (A3)

Acknowledgment

This work was supported by MZOS grant 037-0372787-2797

References

[1] O Reynolds ldquoOn the theory and application and its applicationto Mr Beauchamp towerrsquos experiments Including and experi-mental determination of the viscosity of olive oilrdquo PhilosophicalTransactions of the Royal Society vol 117 pp 157ndash234 1886

[2] G G Stokes ldquoNotes on hydrodynamics On the dynamicalequationsrdquoCambridge and DublinMathematical Journal III pp121ndash127 1848

[3] P W Brifgman ldquoThe viscosity of liquids under pressurerdquo Pro-ceedings of the National Academy of Sciences of the United Statesof America vol 11 no 10 pp 603ndash606 1925

[4] A Z Szeri Fluid Film Lubrication Cambridge University PressNew York NY USA 1998

[5] E K Gatcombe ldquoLubrication characteristics of involute spur-gearsmdasha theoretical investigationrdquoTransactions of theAmericanSociety of Mechanical Engineers vol 67 pp 177ndash181 1945

[6] C Barus ldquoIsotherms isopiestics and isometrics relative to vis-cosityrdquo American Journal of Science vol 45 pp 87ndash96 1893

[7] W R Jones ldquoPressure viscosity measurement for several lubri-cantsrdquo ASLE Transactions vol 18 pp 249ndash262 1975

[8] J Hron J Malek and K R Rajagopal ldquoSimple flows of fluidswith pressure-dependent viscositiesrdquo Proceedings of the RoyalSociety A vol 457 no 2011 pp 1603ndash1622 2001

[9] C J A Roelands Correlation aspects of the viscosity-pressurerelationship of lubricating oils [PhD thesis] Delft University ofTechnology Delft The Netherlands 1966

[10] K R Rajagopal and A Z Szeri ldquoOn an inconsistency in thederivation of the equations of elastohydrodynamic lubricationrdquoProceedings of the Royal Society A vol 459 no 2039 pp 2771ndash2786 2003

[11] M Renardy ldquoSome remarks on the Navier-Stokes equationswith a pressure-dependent viscosityrdquo Communications in Par-tial Differential Equations vol 11 no 7 pp 779ndash793 1986

[12] F Gazzola and P Secchi ldquoSome results about stationary Navier-Stokes equations with a pressure-dependent viscosityrdquo in Pro-ceedings of the International Conference on Navier-Stokes Equa-tions vol 388 of Pitman Research Notes in Mathematics Seriespp 174ndash183 Varenna Italy 1998

[13] E Marusic-Paloka ldquoAn analysis of the Stokes system with pres-sure dependent viscosityrdquo In press

[14] S Marusic and E Marusic-Paloka ldquoTwo-scale convergence forthin domains and its applications to some lower-dimensionalmodels in fluid mechanicsrdquo Asymptotic Analysis vol 23 no 1pp 23ndash57 2000

[15] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001

[16] M Kane and B Bou-Said ldquoComparison of homogenization anddirect techniques for the treatment of roughness in incompress-ible lubricationrdquo Journal of Tribology vol 126 no 4 pp 733ndash7372004

[17] M Jai ldquoHomogenization and two-scale convergence of thecompressible Reynolds lubrication equation modelling the fly-ing characteristics of a roughmagnetic head over a rough rigid-disk surfacerdquo RAIRO Modelisation Mathematique et AnalyseNumerique vol 29 no 2 pp 199ndash233 1995

[18] P Wall ldquoHomogenization of Reynolds equation by two-scaleconvergencerdquo Chinese Annals of Mathematics B vol 28 no 3pp 363ndash374 2007

[19] A Bensoussan J-L Lions and G Papanicolaou AsymptoticAnalysis for Periodic Structures North-Holland AmsterdamThe Netherlands 1978

[20] E Marusic-Paloka and A Mikelic ldquoThe derivation of a nonlin-ear filtration law including the inertia effects via homogeniza-tionrdquo Nonlinear Analysis Theory Methods amp Applications vol42 no 1 pp 97ndash137 2000

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Mathematical Problems in Engineering

However it is not realistic to assume that (120597ℎ120597119909) does notchange the sign To find the upper bound in the general casewe use the procedure from the DeGiorgi theorem The mainresult of the section is as follows

TheoremA1 Let119908 be the solution to the problem (A1)Then

119908 (119909 119910) le1003816100381610038161003816119872120590 (119902)

1003816100381610038161003816119871infin(120597O) +Z (V ℎO 120583 119902) (A3)

Z=3(85)

times(3

2)

(285)(2120587)(14)

ℎ3

0

diamO (|V| ℎ1|O|15

+ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times |O|(75)

(A4)

Proof The function 119911 = 119908 minus 119866 satisfies

div (ℎ3nabla119911) = V120597ℎ

120597119909+ div (ℎ3nabla119872

120590(119902)) in O sub R2 (A5)

119911 = 0 on 120597O (A6)

Next we introduce the embedding constant for 119882120

(O) sub

119871119903(O) denoted119872

119903 such that

|V|119871119903(O) le 119872119903|nablaV|1198712(O) forallV isin 119867

1

0(O) (A7)

That constant can be estimated as

119872119903le1

2(diamO)

2|O|1119903 minus 12

(119903 + 2

2)

((119903+2)2119903)

radic2120587 (A8)

See for example [20 Lemma 1] Next we define the se-quence

120582119896+1

= 3(120582119896

2+ 1) 120582

1= 2 (A9)

Easy computation yields

120582119896= 8(

3

2)

119896minus1

minus 6 (A10)

Let

119911+(119909 119910) = max 119911 (119909 119910) 0 (A11)

We test (A5) with (119911+)1+120582119896+1 and get

intO

ℎ3nabla119911+nabla(119911+)1+120582119896+1

=1 + 120582119896+1

(1 + 120582119896+1

2)2intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

= intO

[Vℎ120597

120597119909(119911+)1+120582119896+1

+ ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

]

(A12)

For the left-hand side we get the lower bound

ℎ3

0

1 + 120582119896+1

(1 + 120582119896+1

2)2

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

1198712(O)

(A13)

We estimate the terms on the right-hand side using the sameidea

intO

ℎ3nabla119872120590(119902) nabla(119911

+)1+120582119896+1

le ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O)

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

intO

ℎV120597

120597119909(119911+)1+120582119896+1

le ℎ1 |V| |O|

15100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

10038161003816100381610038161003816100381611987154(O)

(A14)

Thus it remains to estimate |nabla(119911+)1+120582119896+1 |11987154(O) We have

100381610038161003816100381610038161003816nabla(119911+)1+120582119896+1

100381610038161003816100381610038161003816

54

11987154(O)

= (1 + 120582119896+1

)54

intO

(119911+)(5120582119896+14)1003816100381610038161003816nabla119911

+1003816100381610038161003816

54

= (1 + 120582119896+1

1 + (120582119896+1

2))

54

intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

541003816100381610038161003816119911+1003816100381610038161003816

(5120582119896+18)

le [Holders inequality with 119901 =8

5 1199011015840=8

3]

le (1 + 120582119896+1

1 + (120582119896+1

2))

54

(intO

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

2

)

58

times (intO

(119911+)(53)120582

119896+1

)

38

= [due to (A9) 53120582119896+1

= 5(1 +120582119896

2)

3

8=1

4

120582119896+1

2 + 120582119896

]

= (1 + 120582119896+1

1 + (120582119896+1

2))

54100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)100381610038161003816100381610038161003816

54

1198712(O)

times100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(120582119896+1(2+120582

119896))(54)

1198715(O)

(A15)

Combining with (A14) and (A13) we get

100381610038161003816100381610038161003816nabla(119911+)1+(120582119896+12)1003816100381610038161003816100381610038161198712(O)

le ℎminus3

0(1 +

120582119896+1

2)100381610038161003816100381610038161003816(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198715(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

Mathematical Problems in Engineering 9

le 119872((120582119896+1)(2+120582

119896))

5ℎminus3

0(1+

120582119896+1

2)

times100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198712(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

(A16)

We recall that

120582119896+1

2 + 120582119896

=3

2(A17)

and define

120572 = 119872(32)

5ℎminus3

0(ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

) (A18)

as well as

120590119896=

1

120572

100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(2(2+120582119896))

1198712(O)

(A19)

Then (A16) implies

120590119896+1

le (1 +120582119896+1

2)

(2(2+120582119896))

120590((120582119896+1)(2+120582

119896+1))

119896 (A20)

Taking the logarithm we arrive at

log120590119896+1

lelog (1 + (120582

119896+12))

1 + (120582119896+1

2)+

120582119896+1

2 + 120582119896+1

log120590119896 (A21)

We first notice that

120582119896+1

2 + 120582119896+1

lt 1 (A22)

and then

1 +120582119896

2= 4(

3

2)

119896minus1

minus 2 gt (3

2)

119896

(A23)

Since the function 119909 997891rarr (log119909119909) is decreasing for 119909 gt 119890 wehave

log (1 + (1205821198962))

1 + (1205821198962)

lelog [(32)119896]

(32)119896

= 119896 log 3

2(2

3)

119896

forall119896 ge 3

(A24)

Then

log120590119896+1

le (119896 + 1) log 3

2(2

3)

119896

+ log120590119896

le log 3

2

119896+1

sum

119895=2

119895(2

3)

119895

+ log1205901le 8 log 3

2+ log120590

1

(A25)

Finally

120590119896+1

le (3

2)

8

1205901 (A26)

Now it remains to estimate 1205901 From the definitionwe see that

1205901=

1

120572

100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

12

1198712(O)

(A27)

To estimate 1205901 we proceed as before and test (A5) with (119911+)3

We get

3

4intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

2

= intO

(Vℎ120597(119911+)3

120597119909+ ℎ3nabla119872120590(119902) nabla(119911

+)3

)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

100381610038161003816100381610038161003816nabla(119911+)310038161003816100381610038161003816100381611987154(O)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times3

2

100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

1003816100381610038161003816119911+1003816100381610038161003816119871103(O)

(A28)

Thus100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

le2

ℎ3

0

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872(103)

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O)

(A29)

Finally testing (A5) with 119911+ we get

intO

ℎ31003816100381610038161003816nabla119911+1003816100381610038161003816

2

= intO

(Vℎ120597119911+

120597119909+ ℎ3nabla119872120590(119902) nabla119911

+) (A30)

so that

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O) le

1

ℎ3

0

(|V| ℎ1|O|12

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198712(O))

le |O|310

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

(A31)

Combining (A29) with (A31) and (A26) gives

120590119896+1

le (3

2)

81

120572

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A32)

Since

120590119896+1

ge1

120572119872(2(2+120582k))2

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) (A33)

10 Mathematical Problems in Engineering

we have arrived to

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) le (

3

2)

8

119872(2(2+120582

119896))

2

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A34)

Since lim119896rarrinfin

120582119896= infin we get

1003816100381610038161003816119911+1003816100381610038161003816119871infin(O) le (

3

2)

8radic2

ℎ3

0

times(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A35)

Finally (A8) implies (A3)

Acknowledgment

This work was supported by MZOS grant 037-0372787-2797

References

[1] O Reynolds ldquoOn the theory and application and its applicationto Mr Beauchamp towerrsquos experiments Including and experi-mental determination of the viscosity of olive oilrdquo PhilosophicalTransactions of the Royal Society vol 117 pp 157ndash234 1886

[2] G G Stokes ldquoNotes on hydrodynamics On the dynamicalequationsrdquoCambridge and DublinMathematical Journal III pp121ndash127 1848

[3] P W Brifgman ldquoThe viscosity of liquids under pressurerdquo Pro-ceedings of the National Academy of Sciences of the United Statesof America vol 11 no 10 pp 603ndash606 1925

[4] A Z Szeri Fluid Film Lubrication Cambridge University PressNew York NY USA 1998

[5] E K Gatcombe ldquoLubrication characteristics of involute spur-gearsmdasha theoretical investigationrdquoTransactions of theAmericanSociety of Mechanical Engineers vol 67 pp 177ndash181 1945

[6] C Barus ldquoIsotherms isopiestics and isometrics relative to vis-cosityrdquo American Journal of Science vol 45 pp 87ndash96 1893

[7] W R Jones ldquoPressure viscosity measurement for several lubri-cantsrdquo ASLE Transactions vol 18 pp 249ndash262 1975

[8] J Hron J Malek and K R Rajagopal ldquoSimple flows of fluidswith pressure-dependent viscositiesrdquo Proceedings of the RoyalSociety A vol 457 no 2011 pp 1603ndash1622 2001

[9] C J A Roelands Correlation aspects of the viscosity-pressurerelationship of lubricating oils [PhD thesis] Delft University ofTechnology Delft The Netherlands 1966

[10] K R Rajagopal and A Z Szeri ldquoOn an inconsistency in thederivation of the equations of elastohydrodynamic lubricationrdquoProceedings of the Royal Society A vol 459 no 2039 pp 2771ndash2786 2003

[11] M Renardy ldquoSome remarks on the Navier-Stokes equationswith a pressure-dependent viscosityrdquo Communications in Par-tial Differential Equations vol 11 no 7 pp 779ndash793 1986

[12] F Gazzola and P Secchi ldquoSome results about stationary Navier-Stokes equations with a pressure-dependent viscosityrdquo in Pro-ceedings of the International Conference on Navier-Stokes Equa-tions vol 388 of Pitman Research Notes in Mathematics Seriespp 174ndash183 Varenna Italy 1998

[13] E Marusic-Paloka ldquoAn analysis of the Stokes system with pres-sure dependent viscosityrdquo In press

[14] S Marusic and E Marusic-Paloka ldquoTwo-scale convergence forthin domains and its applications to some lower-dimensionalmodels in fluid mechanicsrdquo Asymptotic Analysis vol 23 no 1pp 23ndash57 2000

[15] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001

[16] M Kane and B Bou-Said ldquoComparison of homogenization anddirect techniques for the treatment of roughness in incompress-ible lubricationrdquo Journal of Tribology vol 126 no 4 pp 733ndash7372004

[17] M Jai ldquoHomogenization and two-scale convergence of thecompressible Reynolds lubrication equation modelling the fly-ing characteristics of a roughmagnetic head over a rough rigid-disk surfacerdquo RAIRO Modelisation Mathematique et AnalyseNumerique vol 29 no 2 pp 199ndash233 1995

[18] P Wall ldquoHomogenization of Reynolds equation by two-scaleconvergencerdquo Chinese Annals of Mathematics B vol 28 no 3pp 363ndash374 2007

[19] A Bensoussan J-L Lions and G Papanicolaou AsymptoticAnalysis for Periodic Structures North-Holland AmsterdamThe Netherlands 1978

[20] E Marusic-Paloka and A Mikelic ldquoThe derivation of a nonlin-ear filtration law including the inertia effects via homogeniza-tionrdquo Nonlinear Analysis Theory Methods amp Applications vol42 no 1 pp 97ndash137 2000

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 9

le 119872((120582119896+1)(2+120582

119896))

5ℎminus3

0(1+

120582119896+1

2)

times100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

((120582119896+1)(2+120582

119896))

1198712(O)

times (ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

)

(A16)

We recall that

120582119896+1

2 + 120582119896

=3

2(A17)

and define

120572 = 119872(32)

5ℎminus3

0(ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O) + ℎ

1 |V| |O|15

) (A18)

as well as

120590119896=

1

120572

100381610038161003816100381610038161003816nabla(119911+)1+(1205821198962)100381610038161003816100381610038161003816

(2(2+120582119896))

1198712(O)

(A19)

Then (A16) implies

120590119896+1

le (1 +120582119896+1

2)

(2(2+120582119896))

120590((120582119896+1)(2+120582

119896+1))

119896 (A20)

Taking the logarithm we arrive at

log120590119896+1

lelog (1 + (120582

119896+12))

1 + (120582119896+1

2)+

120582119896+1

2 + 120582119896+1

log120590119896 (A21)

We first notice that

120582119896+1

2 + 120582119896+1

lt 1 (A22)

and then

1 +120582119896

2= 4(

3

2)

119896minus1

minus 2 gt (3

2)

119896

(A23)

Since the function 119909 997891rarr (log119909119909) is decreasing for 119909 gt 119890 wehave

log (1 + (1205821198962))

1 + (1205821198962)

lelog [(32)119896]

(32)119896

= 119896 log 3

2(2

3)

119896

forall119896 ge 3

(A24)

Then

log120590119896+1

le (119896 + 1) log 3

2(2

3)

119896

+ log120590119896

le log 3

2

119896+1

sum

119895=2

119895(2

3)

119895

+ log1205901le 8 log 3

2+ log120590

1

(A25)

Finally

120590119896+1

le (3

2)

8

1205901 (A26)

Now it remains to estimate 1205901 From the definitionwe see that

1205901=

1

120572

100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

12

1198712(O)

(A27)

To estimate 1205901 we proceed as before and test (A5) with (119911+)3

We get

3

4intO

ℎ3100381610038161003816100381610038161003816nabla(119911+)2100381610038161003816100381610038161003816

2

= intO

(Vℎ120597(119911+)3

120597119909+ ℎ3nabla119872120590(119902) nabla(119911

+)3

)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

100381610038161003816100381610038161003816nabla(119911+)310038161003816100381610038161003816100381611987154(O)

le (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

times3

2

100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

1003816100381610038161003816119911+1003816100381610038161003816119871103(O)

(A28)

Thus100381610038161003816100381610038161003816nabla(119911+)21003816100381610038161003816100381610038161198712(O)

le2

ℎ3

0

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872(103)

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O)

(A29)

Finally testing (A5) with 119911+ we get

intO

ℎ31003816100381610038161003816nabla119911+1003816100381610038161003816

2

= intO

(Vℎ120597119911+

120597119909+ ℎ3nabla119872120590(119902) nabla119911

+) (A30)

so that

1003816100381610038161003816nabla119911+10038161003816100381610038161198712(O) le

1

ℎ3

0

(|V| ℎ1|O|12

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198712(O))

le |O|310

(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))

(A31)

Combining (A29) with (A31) and (A26) gives

120590119896+1

le (3

2)

81

120572

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A32)

Since

120590119896+1

ge1

120572119872(2(2+120582k))2

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) (A33)

10 Mathematical Problems in Engineering

we have arrived to

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) le (

3

2)

8

119872(2(2+120582

119896))

2

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A34)

Since lim119896rarrinfin

120582119896= infin we get

1003816100381610038161003816119911+1003816100381610038161003816119871infin(O) le (

3

2)

8radic2

ℎ3

0

times(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A35)

Finally (A8) implies (A3)

Acknowledgment

This work was supported by MZOS grant 037-0372787-2797

References

[1] O Reynolds ldquoOn the theory and application and its applicationto Mr Beauchamp towerrsquos experiments Including and experi-mental determination of the viscosity of olive oilrdquo PhilosophicalTransactions of the Royal Society vol 117 pp 157ndash234 1886

[2] G G Stokes ldquoNotes on hydrodynamics On the dynamicalequationsrdquoCambridge and DublinMathematical Journal III pp121ndash127 1848

[3] P W Brifgman ldquoThe viscosity of liquids under pressurerdquo Pro-ceedings of the National Academy of Sciences of the United Statesof America vol 11 no 10 pp 603ndash606 1925

[4] A Z Szeri Fluid Film Lubrication Cambridge University PressNew York NY USA 1998

[5] E K Gatcombe ldquoLubrication characteristics of involute spur-gearsmdasha theoretical investigationrdquoTransactions of theAmericanSociety of Mechanical Engineers vol 67 pp 177ndash181 1945

[6] C Barus ldquoIsotherms isopiestics and isometrics relative to vis-cosityrdquo American Journal of Science vol 45 pp 87ndash96 1893

[7] W R Jones ldquoPressure viscosity measurement for several lubri-cantsrdquo ASLE Transactions vol 18 pp 249ndash262 1975

[8] J Hron J Malek and K R Rajagopal ldquoSimple flows of fluidswith pressure-dependent viscositiesrdquo Proceedings of the RoyalSociety A vol 457 no 2011 pp 1603ndash1622 2001

[9] C J A Roelands Correlation aspects of the viscosity-pressurerelationship of lubricating oils [PhD thesis] Delft University ofTechnology Delft The Netherlands 1966

[10] K R Rajagopal and A Z Szeri ldquoOn an inconsistency in thederivation of the equations of elastohydrodynamic lubricationrdquoProceedings of the Royal Society A vol 459 no 2039 pp 2771ndash2786 2003

[11] M Renardy ldquoSome remarks on the Navier-Stokes equationswith a pressure-dependent viscosityrdquo Communications in Par-tial Differential Equations vol 11 no 7 pp 779ndash793 1986

[12] F Gazzola and P Secchi ldquoSome results about stationary Navier-Stokes equations with a pressure-dependent viscosityrdquo in Pro-ceedings of the International Conference on Navier-Stokes Equa-tions vol 388 of Pitman Research Notes in Mathematics Seriespp 174ndash183 Varenna Italy 1998

[13] E Marusic-Paloka ldquoAn analysis of the Stokes system with pres-sure dependent viscosityrdquo In press

[14] S Marusic and E Marusic-Paloka ldquoTwo-scale convergence forthin domains and its applications to some lower-dimensionalmodels in fluid mechanicsrdquo Asymptotic Analysis vol 23 no 1pp 23ndash57 2000

[15] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001

[16] M Kane and B Bou-Said ldquoComparison of homogenization anddirect techniques for the treatment of roughness in incompress-ible lubricationrdquo Journal of Tribology vol 126 no 4 pp 733ndash7372004

[17] M Jai ldquoHomogenization and two-scale convergence of thecompressible Reynolds lubrication equation modelling the fly-ing characteristics of a roughmagnetic head over a rough rigid-disk surfacerdquo RAIRO Modelisation Mathematique et AnalyseNumerique vol 29 no 2 pp 199ndash233 1995

[18] P Wall ldquoHomogenization of Reynolds equation by two-scaleconvergencerdquo Chinese Annals of Mathematics B vol 28 no 3pp 363ndash374 2007

[19] A Bensoussan J-L Lions and G Papanicolaou AsymptoticAnalysis for Periodic Structures North-Holland AmsterdamThe Netherlands 1978

[20] E Marusic-Paloka and A Mikelic ldquoThe derivation of a nonlin-ear filtration law including the inertia effects via homogeniza-tionrdquo Nonlinear Analysis Theory Methods amp Applications vol42 no 1 pp 97ndash137 2000

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Mathematical Problems in Engineering

we have arrived to

1003816100381610038161003816119911+10038161003816100381610038161198712+120582119896 (O) le (

3

2)

8

119872(2(2+120582

119896))

2

radic2

ℎ3

0

times (|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A34)

Since lim119896rarrinfin

120582119896= infin we get

1003816100381610038161003816119911+1003816100381610038161003816119871infin(O) le (

3

2)

8radic2

ℎ3

0

times(|V| ℎ1|O|15

+ ℎ3

1

1003816100381610038161003816nabla119872120590 (119902)10038161003816100381610038161198715(O))119872

12

(103)|O|32

(A35)

Finally (A8) implies (A3)

Acknowledgment

This work was supported by MZOS grant 037-0372787-2797

References

[1] O Reynolds ldquoOn the theory and application and its applicationto Mr Beauchamp towerrsquos experiments Including and experi-mental determination of the viscosity of olive oilrdquo PhilosophicalTransactions of the Royal Society vol 117 pp 157ndash234 1886

[2] G G Stokes ldquoNotes on hydrodynamics On the dynamicalequationsrdquoCambridge and DublinMathematical Journal III pp121ndash127 1848

[3] P W Brifgman ldquoThe viscosity of liquids under pressurerdquo Pro-ceedings of the National Academy of Sciences of the United Statesof America vol 11 no 10 pp 603ndash606 1925

[4] A Z Szeri Fluid Film Lubrication Cambridge University PressNew York NY USA 1998

[5] E K Gatcombe ldquoLubrication characteristics of involute spur-gearsmdasha theoretical investigationrdquoTransactions of theAmericanSociety of Mechanical Engineers vol 67 pp 177ndash181 1945

[6] C Barus ldquoIsotherms isopiestics and isometrics relative to vis-cosityrdquo American Journal of Science vol 45 pp 87ndash96 1893

[7] W R Jones ldquoPressure viscosity measurement for several lubri-cantsrdquo ASLE Transactions vol 18 pp 249ndash262 1975

[8] J Hron J Malek and K R Rajagopal ldquoSimple flows of fluidswith pressure-dependent viscositiesrdquo Proceedings of the RoyalSociety A vol 457 no 2011 pp 1603ndash1622 2001

[9] C J A Roelands Correlation aspects of the viscosity-pressurerelationship of lubricating oils [PhD thesis] Delft University ofTechnology Delft The Netherlands 1966

[10] K R Rajagopal and A Z Szeri ldquoOn an inconsistency in thederivation of the equations of elastohydrodynamic lubricationrdquoProceedings of the Royal Society A vol 459 no 2039 pp 2771ndash2786 2003

[11] M Renardy ldquoSome remarks on the Navier-Stokes equationswith a pressure-dependent viscosityrdquo Communications in Par-tial Differential Equations vol 11 no 7 pp 779ndash793 1986

[12] F Gazzola and P Secchi ldquoSome results about stationary Navier-Stokes equations with a pressure-dependent viscosityrdquo in Pro-ceedings of the International Conference on Navier-Stokes Equa-tions vol 388 of Pitman Research Notes in Mathematics Seriespp 174ndash183 Varenna Italy 1998

[13] E Marusic-Paloka ldquoAn analysis of the Stokes system with pres-sure dependent viscosityrdquo In press

[14] S Marusic and E Marusic-Paloka ldquoTwo-scale convergence forthin domains and its applications to some lower-dimensionalmodels in fluid mechanicsrdquo Asymptotic Analysis vol 23 no 1pp 23ndash57 2000

[15] D Gilbarg and N S Trudinger Elliptic Partial DifferentialEquations of Second Order Springer Berlin Germany 2001

[16] M Kane and B Bou-Said ldquoComparison of homogenization anddirect techniques for the treatment of roughness in incompress-ible lubricationrdquo Journal of Tribology vol 126 no 4 pp 733ndash7372004

[17] M Jai ldquoHomogenization and two-scale convergence of thecompressible Reynolds lubrication equation modelling the fly-ing characteristics of a roughmagnetic head over a rough rigid-disk surfacerdquo RAIRO Modelisation Mathematique et AnalyseNumerique vol 29 no 2 pp 199ndash233 1995

[18] P Wall ldquoHomogenization of Reynolds equation by two-scaleconvergencerdquo Chinese Annals of Mathematics B vol 28 no 3pp 363ndash374 2007

[19] A Bensoussan J-L Lions and G Papanicolaou AsymptoticAnalysis for Periodic Structures North-Holland AmsterdamThe Netherlands 1978

[20] E Marusic-Paloka and A Mikelic ldquoThe derivation of a nonlin-ear filtration law including the inertia effects via homogeniza-tionrdquo Nonlinear Analysis Theory Methods amp Applications vol42 no 1 pp 97ndash137 2000

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of