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Research Article Analysis of Unsteady Axisymmetric Squeezing Fluid Flow with Slip and No-Slip Boundaries Using OHAM Mubashir Qayyum, Hamid Khan, M. Tariq Rahim, and Inayat Ullah Department of Mathematics, National University of Computer & Emerging Sciences, FAST Peshawar Campus, Peshawar 25000, Pakistan Correspondence should be addressed to Hamid Khan; [email protected] Received 27 May 2014; Revised 11 August 2014; Accepted 12 September 2014 Academic Editor: Mar´ ıa Isabel Herreros Copyright © 2015 Mubashir Qayyum et al. is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. In this manuscript, An unsteady axisymmetric flow of nonconducting, Newtonian fluid squeezed between two circular plates is studied with slip and no-slip boundaries. Using similarity transformation, the system of nonlinear partial differential equations is reduced to a single fourth order ordinary differential equation. e resulting boundary value problems are solved by optimal homotopy asymptotic method (OHAM) and fourth order explicit Runge-Kutta method (RK4). It is observed that the results obtained from OHAM are in good agreement with numerical results by means of residuals. Furthermore, the effects of various dimensionless parameters on the velocity profiles are investigated graphically. 1. Introduction e squeezing of an incompressible viscous fluid between two parallel plates is an essential type of flow that is frequently observed in many hydro dynamical tools and machines. In food industries, squeezing flows have several applications particularly in chemical engineering [1, 2]. Compression and injection molding, polymer processing, and modeling of lubrication system are some practical examples of squeezing flows. e modeling and analysis of squeezing flow has started in nineteenth century and continues to receive sig- nificant attention due to its vast applications in biophysical and physical sciences. e initial work in squeezing flows has been done by Stefan [3], who developed an ad hoc asymptotic solution of Newtonian fluids. An explicit solution of the squeeze flow, considering inertial terms, has been established by orpe and Shaw [4]. However, P. S. Gupta and A. S. Gupta [5] proved that the solution set up by [4] fails to satisfy the boundary conditions. Considering fluid inertia effects, Elkouh [6] studied the squeeze film between two plane annuli. Verma [7] and Singh et al. [8] have conducted Numerical solutions of the squeezing flow between parallel plates. Leider and Bird [9] performed theoretical analysis for squeezing flow of power-law fluid between parallel disks. Analytic solution for the squeezing flow of viscous fluid between two parallel disks with suction or blowing effect has been proposed by Domairry and Aziz [10]. Islam et al. [11] studied Newtonian squeezing fluid flow in a porous medium channel. Ullah et al. [12] discussed the Newtonian fluid flow with slip boundary condition keeping MHD effect into account. Siddiqui et al. [13] investigated the unsteady squeezing flow of viscous fluid with magnetic field. Apart from the mentioned researchers, other prominent scholars have conducted various theoretical and experimental studies of squeezing flows [1417]. e difference between fluid and boundary velocity is proportional to the shear stress at the boundary. e dimension of proportionality constant is length, which is known as slip parameter. In fluids with elastic character, slip condition has great importance [18]. It has many applications in medical sciences, for instance, polishing artificial heart valves [19]. ere are various situations in which no-slip boundary condition is inappropriate. Some of these situations include polymeric liquids when the weight of molecule is high, flow on multiple interfaces, fluids containing concerted suspensions, and thin film problems. e general boundary condition which shows the fluid slip at the wall was ini- tially proposed by Navier [20]. Recently, Ebaid [21] studied Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2015, Article ID 860857, 11 pages http://dx.doi.org/10.1155/2015/860857
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Page 1: Research Article Analysis of Unsteady Axisymmetric ...downloads.hindawi.com/journals/mpe/2015/860857.pdfResearch Article Analysis of Unsteady Axisymmetric Squeezing Fluid Flow with

Research ArticleAnalysis of Unsteady Axisymmetric Squeezing Fluid Flow withSlip and No-Slip Boundaries Using OHAM

Mubashir Qayyum Hamid Khan M Tariq Rahim and Inayat Ullah

Department of Mathematics National University of Computer amp Emerging Sciences FAST Peshawar CampusPeshawar 25000 Pakistan

Correspondence should be addressed to Hamid Khan hamidkhannuedupk

Received 27 May 2014 Revised 11 August 2014 Accepted 12 September 2014

Academic Editor Marıa Isabel Herreros

Copyright copy 2015 Mubashir Qayyum et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

In this manuscript An unsteady axisymmetric flow of nonconducting Newtonian fluid squeezed between two circular plates isstudied with slip and no-slip boundaries Using similarity transformation the system of nonlinear partial differential equationsis reduced to a single fourth order ordinary differential equation The resulting boundary value problems are solved by optimalhomotopy asymptotic method (OHAM) and fourth order explicit Runge-Kutta method (RK4) It is observed that the resultsobtained from OHAM are in good agreement with numerical results by means of residuals Furthermore the effects of variousdimensionless parameters on the velocity profiles are investigated graphically

1 Introduction

Thesqueezing of an incompressible viscous fluid between twoparallel plates is an essential type of flow that is frequentlyobserved in many hydro dynamical tools and machines Infood industries squeezing flows have several applicationsparticularly in chemical engineering [1 2] Compression andinjection molding polymer processing and modeling oflubrication system are some practical examples of squeezingflows The modeling and analysis of squeezing flow hasstarted in nineteenth century and continues to receive sig-nificant attention due to its vast applications in biophysicaland physical sciencesThe initial work in squeezing flows hasbeen done by Stefan [3] who developed an ad hoc asymptoticsolution of Newtonian fluids An explicit solution of thesqueeze flow considering inertial terms has been establishedby Thorpe and Shaw [4] However P S Gupta and A SGupta [5] proved that the solution set up by [4] fails tosatisfy the boundary conditions Considering fluid inertiaeffects Elkouh [6] studied the squeeze film between twoplane annuli Verma [7] and Singh et al [8] have conductedNumerical solutions of the squeezing flow between parallelplates Leider and Bird [9] performed theoretical analysisfor squeezing flow of power-law fluid between parallel disks

Analytic solution for the squeezing flow of viscous fluidbetween two parallel disks with suction or blowing effecthas been proposed by Domairry and Aziz [10] Islam etal [11] studied Newtonian squeezing fluid flow in a porousmedium channel Ullah et al [12] discussed the Newtonianfluid flow with slip boundary condition keeping MHD effectinto account Siddiqui et al [13] investigated the unsteadysqueezing flow of viscous fluid with magnetic field Apartfrom the mentioned researchers other prominent scholarshave conducted various theoretical and experimental studiesof squeezing flows [14ndash17]

The difference between fluid and boundary velocityis proportional to the shear stress at the boundary Thedimension of proportionality constant is length which isknown as slip parameter In fluids with elastic character slipcondition has great importance [18] It has many applicationsin medical sciences for instance polishing artificial heartvalves [19] There are various situations in which no-slipboundary condition is inappropriate Some of these situationsinclude polymeric liquids when the weight of molecule ishigh flow on multiple interfaces fluids containing concertedsuspensions and thin film problems The general boundarycondition which shows the fluid slip at the wall was ini-tially proposed by Navier [20] Recently Ebaid [21] studied

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015 Article ID 860857 11 pageshttpdxdoiorg1011552015860857

2 Mathematical Problems in Engineering

the effect of magnetic field in Newtonian fluid in an asym-metric channel with wall slip conditions

Most of scientific incidents are modeled by nonlinearpartial or ordinary differential equations In literature wehave variety of perturbation techniques which can solvenonlinear boundary value problems analytically But thelimitations of these techniques are based on the assumptionof small parameters Detailed review of these methods isgiven by He [22] In recent times the ideas of Homotopy andPerturbation have been combined together Liao [23] and He[24 25] have done the primary work in this regard In seriesof papers Marinca with various scholars used OHAM to findthe approximate solution of nonlinear differential equationsarising in heat transfer steady flow of a fourth-grade fluidand thin film flow [26ndash28]

In this work OHAM is used to analyze an unsteadysqueezing fluid flow between two circular nonrotating diskswith slip and no-slip boundary conditions In additionmovement of the circular plates is considered to be symmetricabout the axial line and the fluid is considered to be Newto-nian incompressible and viscous Sections 2 and 3 include thedescription and mathematical formulation of the problemSections 4 5 and 6 present the basic theory of OHAM andits application in case of no-slip and slip boundaries Resultsand discussions are given in Section 7 while conclusions arementioned in Section 8

2 Description of the Problem

Theunsteady axisymmetric squeezing flow of incompressibleNewtonian fluid with density 120588 viscosity 120583 and kinematicviscosity ] squeezed between two circular plates havingspeed V

119908(119905) is considered It is assumed that at any time 119905 the

distance between two circular plates is 2119904(119905) It is also assumedthat 119903-axis is the central axis of the channel while 119911-axis istaken normal to it Plates move symmetrically with respect tothe central axis 119911 = 0 while the flow is axisymmetric about119903 = 0 The longitudinal and normal velocity componentsin radial and axial directions are 119906

119903(119903 119911 119905) and 119906

119911(119903 119911 119905)

respectivelyThe geometrical interpretation of the problem isgiven in Figure 1

3 Mathematical Formulation

The governing equations of motion are

nabla sdot U = 0

120588 [120597U120597119905

+ (U sdot nabla)U] = 120588119891 + nabla sdot T(1)

where

T = minus119901I + 120583A1

A1= nablaU + (nablaU)119905

(2)

and U is the velocity vector 119901 is the pressure 119891 is the bodyforce T is the Cauchy stress tensor A

1is the Rivlin-Ericksen

2s(t)r = 0

z = 0

r

r

w

w

uz

ur

Figure 1 Geometrical interpretation of the problem

tensor and 120583 is the coefficient of viscosity Now we formulatethe unsteady two-dimensional flow Let us assume that

U = [119906119903(119903 119911 119905) 0 119906

119911(119903 119911 119905)] (3)

and introduce the vorticity functionΩ(119903 119911 119905) and generalizedpressure ℎ(119903 119911 119905) as

Ω (119903 119911 119905) =120597119906119911

120597119903minus120597119906119903

120597119911 (4)

ℎ (119903 119911 119905) =120588

2[1199062

119903+ 1199062

119911] + 119901 (5)

Equations (1) are reduced to120597119906119903

120597119903+119906119903

119903+120597119906119911

120597119911= 0

120597ℎ

120597119903+ 120588(

120597119906119903

120597119905minus 119906119911Ω) = minus120583

120597Ω

120597119911

120597ℎ

120597119911+ 120588(

120597119906119911

120597119905+ 119906119903Ω) =

120583

119903

120597

120597119903(119903Ω)

(6)

The boundary conditions on 119906119903(119903 119911 119905) and 119906

119911(119903 119911 119905) are

119906119903(119903 119911 119905) = 0 119906

119911(119903 119911 119905) = V

119908(119905) at 119911 = 119904

120597

120597119911119906119903(119903 119911 119905) = 0 119906

119911(119903 119911 119905) = 0 at 119911 = 0

(7)

where V119908(119905) = 119889119904119889119905 is the velocity of the plates The

boundary conditions (7) are due to symmetry at 119911 = 0 andno-slip at the upper plate when 119911 = 119904 If we introduce thedimensionless parameter

120578 =119911

119904 (119905) (8)

Equations (4) and (6) transforms to

Ω (119903 119911 119905) =120597119906119911

120597119903minus1

119904

120597119906119903

120597120578 (9)

120597119906119903

120597119903+119906119903

119903+1

119904

120597119906119911

120597120578= 0 (10)

120597ℎ

120597119903+ 120588(

120597119906119903

120597119905minus 119906119911Ω) = minus

120583

119904

120597Ω

120597120578 (11)

1

119904

120597ℎ

120597120578+ 120588(

120597119906119911

120597119905+ 119906119903Ω) =

120583

119903

120597

120597119903(119903Ω) (12)

Mathematical Problems in Engineering 3

The boundary conditions on 119906119903and 119906

119911are

119906119903= 0 119906

119911= V119908(119905) at 120578 = 1

120597119906119903

120597120578= 0 119906

119911= 0 at 120578 = 0

(13)

After eliminating the generalized pressure between (11) and(12) we obtained

120588 [120597Ω

120597119905+ 119906119903

120597Ω

120597119903+119906119911

119904

120597Ω

120597120578minus119906119903

119903Ω] = 120583 [nabla

2

Ω minusΩ

1199032] (14)

where nabla2 is the Laplacian operatorDefining velocity components as [5]

119906119903= minus

119903

2119904 (119905)V119908(119905) 1198921015840

(120578)

119906119911= V119908(119905) 119892 (120578)

(15)

we see that (10) is identically satisfied and (14) becomes

1198894

119892

1198891205784+ 119877[(120578 minus 119892)

1198893

119892

1198891205783+ 2

1198892

119892

1198891205782] minus 119876

1198892

119892

1198891205782= 0 (16)

where

119877 =119904V119908

120592 119876 =

1199042

120592V119908

119889V119908

119889119905 (17)

Here both119877 and119876 are functions of 119905 but we consider119877 and119876constants for similarity solution Since V

119908= 119889119904119889119905 Integrate

first equation of (17) we get

119904 (119905) = (119882119905 + 119878)12

(18)

where119882 and 119878 are constantsTheplatesmove away from eachother symmetrically with respect to 120578 when 119882 gt 0 and 119878 gt0 Also the plates approach to each other and squeezing flowexists with similar velocity profiles when119882 lt 0 119878 gt 0 and119904(119905) gt 0 From (17) and (18) it follows that 119876 = minus119877 Then (16)becomes

1198894

119892

1198891205784+ 119877[(120578 minus 119892)

1198893

119892

1198891205783+ 3

1198892

119892

1198891205782] = 0 (19)

Using (13) and (15) we determine the boundary conditions incase of no-slip and slip at the upper plate as follows

119892 (1) = 1 1198921015840

(1) = 0

119892 (0) = 0 11989210158401015840

(0) = 0(No-slip at the wall) (20)

119892 (1) = 1 1198921015840

(1) = 12057411989210158401015840

(1)

119892 (0) = 0 11989210158401015840

(0) = 0(Slip at the wall) (21)

4 Basic Theory of OHAM [26 29ndash32]Let us apply OHAM to the following differential equation

I [V (119909)] + 119891 (119909) + alefsym [V (119909)] = 0 119861 (V119889V119889119909

) = 0 (22)

where 119909 represents an independent variable V(119909) is unknownfunction and 119891(119909) is known function 119861alefsymI are boundarynonlinear and linear operators respectively

According to OHAM we construct Homotopy Φ(119909 119901)R times [0 1] rarr R which satisfies

(1 minus 119901) [I (Φ (119909 119901)) + 119891 (119909)]

= ℎ (119901) [I (Φ (119909 119901)) + 119891 (119909) + alefsym (Φ (119909 119901))]

119861(Φ (119909 119901) 120597Φ (119909 119901)

120597119909) = 0

(23)

where 119909 isin R and 119901 isin [0 1] is an embedding parameterℎ(119901) is a nonlinear auxiliary function for 119901 = 0 ℎ(0) = 0

and Φ(119909 119901) is an unknown function Clearly when 119901 = 0

and 119901 = 1 it holds that Φ(119909 0) = V0(119909) and Φ(119909 1) = V(119909)

respectivelyThus as 119901 varies from 0 to 1 the solution Φ(119909 119901)

approaches from V0(119909) to V(119909)

We choose the auxiliary function ℎ(119901) in the form of

ℎ (119901) =

119898

sum

119896=0

119901119896

119862119896 (24)

where 119862119896are convergence controlling constants to be deter-

minedTo obtain an approximate solution we expandΦ(119909 119901 119862

119894)

in a Taylor series about 119901 as follows

Φ(119909 119901 119862119894) = V0(119909) +

119898

sum

119896=1

V119896(1198621 1198622 119862

119896) 119901119896

(25)

Substituting (25) into (23) and equating the coefficients of likepowers of 119901 we obtain the following equationsThe zeroth-order problem is

I [V0(119909)] + 119891 (119909) = 0 119861 (V

0119889V0

119889119909) = 0 (26)

First-order problem is

I [V1(119909)] + 119891 (119909) = 119862

1alefsym0[V0(119909)] 119861 (V

1119889V1

119889119909) = 0

(27)

Second-order problem is

I [V2(119909)] +I [V

1(119909)]

= 1198622alefsym0[V0(119909)] + 119862

1I [V1(119909)] + alefsym

1[V0(119909) V

1(119909)]

119861 (V2119889V2

119889119909) = 0

(28)

4 Mathematical Problems in Engineering

The general equations for V119896(119909) are given by

I [V119896(119909)] minusI [V

119896minus1(119909)]

= 1198621alefsym0[V0(119909)]

+

119896minus1

sum

119894=1

119862119894I [V119896minus119894

(119909)]

+alefsym119896minus119894

[V0(119909) V

1(119909) V

119896minus1(119909)]

119861 (V119896119889V119896

119889119909) = 0

119896 = 2 3

(29)

where the coefficient of 119901119898 in the expansion of alefsym(Φ(119909 119901))about 119901 is alefsym

119898[V0(119909) V1(119909) V

119898minus1(119909)]

alefsym(Φ (119909 119901 119862119894))

= alefsym0[V0(119909)] +

119898

sum

119898=1

alefsym119898[V0(119909) V

1(119909) V

119898minus1(119909)] 119901

119898

(30)It is noted that the convergence of the series (25) depends

upon119862119896 For convergence at119901 = 1 the119898th order approxima-

tion V is

V (119909 1198621 1198622 119862

119898) = V0(119909) +

119898

sum

119895=1

V119895(119909 1198621 1198622 119862

119895)

(31)Substituting (31) in (22) the expression for residual is

R (119909 1198621 1198622 119862

119898)

= I [V (119909 1198621 1198622 119862

119898)] + 119891 (119909)

+ alefsym [V (119909 1198621 1198622 119862

119898)]

(32)

If R = 0 then V will be the exact solution but usually thisdoes not happen in nonlinear problems

There are various methods to find the optimal values of119862119894 119894 = 1 2 We apply the method of least square and

Galerkinrsquos method in the following mannerIn method of least square

119869 (119909 1198621 1198622 119862

119898) = int

119887

119886

R2

(119909 1198621 1198622 119862

119898) 119889119909 (33)

Minimizing 119869(119909 1198621 1198622 119862

119898) we have

120597119869

120597119862119894

= 0 119894 = 1 2 119898 (34)

In Galerkinrsquos method we solve the following system for119862119894(119894 = 1 2 119898)

int

119887

119886

R120597V120597119862119894

119889119909 = 0 119894 = 1 2 119898 (35)

To find appropriate 119862119894(119894 = 1 2 119898) we choose 119886 and 119887 in

the domain of the problem Approximate solution of order119898is well-determined with these known constants

5 Application of OHAM in Case ofNo-Slip Boundary

Using (19) and (20) various order problems are as followsZeroth-order problem

V(119894V)0

(120578) = 0

V0(0) = 0 V10158401015840

0(0) = 0 V

0(1) = 1 V1015840

0(1) = 0

(36)

First-order problem

V(119894V)1

(120578) = 31198621119877V101584010158400(120578) + 119862

1119877120578V1015840101584010158400(120578) minus 119862

1119877V0(120578) V1015840101584010158400(120578)

+ V(119894V)0

(120578) + 1198621V(119894V)0

(120578)

V1(0) = 0 V10158401015840

1(0) = 0 V

1(1) = 0 V1015840

1(1) = 0

(37)

Second-order problem

V(119894V)2

(120578) = 31198621119877V101584010158401(120578) minus 119862

1119877V1(120578) V1015840101584010158400(120578) + 119862

1119877120578V1015840101584010158401(120578)

minus 1198621119877V0(120578) V1015840101584010158401(120578) + V(119894V)

1(120578) + 119862

1V(119894V)1

(120578)

V2(0) = 0 V10158401015840

2(0) = 0 V

2(1) = 0 V1015840

2(1) = 0

(38)

Third-order problem

V(119894V)3

(120578) = 31198621119877V101584010158402(120578) minus 119862

1119877V2(120578) V1015840101584010158400(120578)

minus 1198621119877V1(120578) V1015840101584010158401(120578) + 119862

1119877120578V1015840101584010158402(120578)

minus 1198621119877V0(120578) V1015840101584010158402(120578) + V(119894V)

2(120578)

+ 1198621V(119894V)2

(120578)

V3(0) = 0 V10158401015840

3(0) = 0 V

3(1) = 0 V1015840

3(1) = 0

(39)

Fourth-order problem

V(119894V)4

(120578) = 31198621119877V101584010158403(120578) minus 119862

1119877V3(120578) V1015840101584010158400(120578)

minus 1198621119877V2(120578) V1015840101584010158401(120578) minus 119862

1119877V1(120578) V1015840101584010158402(120578)

+ 1198621119877120578V1015840101584010158403(120578) minus 119862

1119877V0(120578) V1015840101584010158403(120578)

+ V(119894V)3

(120578) + 1198621V(119894V)3

(120578)

V4(0) = 0 V10158401015840

4(0) = 0 V

4(1) = 0 V1015840

4(1) = 0

(40)

By considering fourth-order solution we have

V (120578) = V0(120578) + V

1(120578) + V

2(120578) + V

3(120578) + V

4(120578) (41)

The residual of the problem is

R =1198894V (120578)1198891205784

+ 119877[(120578 minus V (120578))1198893V (120578)1198891205783

+ 31198892V (120578)1198891205782

] (42)

Mathematical Problems in Engineering 5

We apply Galerkinrsquos method to find constant 1198621as follows

int

1

0

R120597V1205971198621

119889119909 = 0 (43)

Solving (43) and keeping 119877 = 01 we get

1198621= 101328 (44)

Using above value of 1198621 the approximate solution is

V (120578) =1

2(3120578 minus 120578

3

)

+1

560(374914120578 minus 739694120578

3

+ 3546481205785

+01013281205787

)

+1

776160(956941120578 minus 430222120578

3

+ 514141205785

minus122331205787

minus 5534111205789

minus 19405312057811

)

+1

565045(254414120578 + 143644120578

3

minus 3711641205785

+ 2502321205787

minus 8133381205789

+ 25990112057811

+6961012057813

+ 2296212057815

)

+1

722737(153702120578 minus 187989 times 10

8

1205783

+ 193991 times 108

1205785

+ 828837 times 107

1205787

minus 803587 times 107

1205789

minus 292844 times 107

12057811

+ 167506 times 107

12057813

minus 966217 times 106

12057815

minus16529 times 106

12057817

minus 48038912057819

)

(45)

6 Application of OHAM in Case ofSlip Boundary

Using (19) and (21) different order problems are as followsZeroth-order problem

119906(119894V)0

(120578) = 0

1199060(0) = 0 119906

10158401015840

0(0) = 0 119906

0(1) = 1 119906

1015840

0(1) = 120574119906

10158401015840

0(1)

(46)

First-order problem

119906(119894V)1

(120578)

= 3119862111987711990610158401015840

0(120578) + 119862

1119877120578119906101584010158401015840

0(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

0(120578)

+ 119906(119894V)0

(120578) + 1198621119906(119894V)0

(120578)

1199061(0) = 0 119906

10158401015840

1(0) = 0 119906

1(1) = 0 119906

1015840

1(1) = 120574119906

10158401015840

1(1)

(47)

Second-order problem

119906(119894V)2

(120578) = 3119862111987711990610158401015840

1(120578) minus 119862

11198771199061(120578) 119906101584010158401015840

0(120578)

+ 1198621119877120578119906101584010158401015840

1(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

1(120578)

+ 119906(119894V)1

(120578) + 1198621119906(119894V)1

(120578)

1199062(0) = 0 119906

10158401015840

2(0) = 0 119906

2(1) = 0 119906

1015840

2(1) = 120574119906

10158401015840

2(1)

(48)

Third-order problem

119906(119894V)3

(120578) = 3119862111987711990610158401015840

2(120578) minus 119862

11198771199062(120578) 119906101584010158401015840

0(120578)

minus 11986211198771199061(120578) 119906101584010158401015840

1(120578) + 119862

1119877120578119906101584010158401015840

2(120578)

minus 11986211198771199060(120578) 119906101584010158401015840

2(120578) + 119906

(119894V)2

(120578)

+ 1198621119906(119894V)2

(120578)

1199063(0) = 0 119906

10158401015840

3(0) = 0 119906

3(1) = 0 119906

1015840

3(1) = 120574119906

10158401015840

3(1)

(49)

Fourth-order problem

119906(119894V)4

(120578) = 3119862111987711990610158401015840

3(120578) minus 119862

11198771199063(120578) 119906101584010158401015840

0(120578)

minus 11986211198771199062(120578) 119906101584010158401015840

1(120578) minus 119862

11198771199061(120578) 119906101584010158401015840

2(120578)

+ 1198621119877120578119906101584010158401015840

3(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

3(120578)

+ 119906(119894V)3

(120578) + 1198621119906(119894V)3

(120578)

1199064(0) = 0 119906

10158401015840

4(0) = 0 119906

4(1) = 0 119906

1015840

4(1) = 120574119906

10158401015840

4(1)

(50)

By considering fourth-order solution we have

(120578) =

4

sum

119894=0

119906119894(120578 1198621) (51)

The residual of the problem is

120577 =1198894

(120578)

1198891205784+ 119877[(120578 minus (120578))

1198893

(120578)

1198891205783+ 3

1198892

(120578)

1198891205782] (52)

We apply Galerkinrsquos method to find constant 1198621as follows

int

1

0

120577120597

1205971198621

119889119909 = 0 (53)

Solving (53) and taking 119877 = 02 and 120574 = 1 we get

1198621= minus106872 (54)

6 Mathematical Problems in Engineering

Table 1 OHAM solutions along with residuals for various 119877 in case of no-slip boundary

120578119877 = 01 119877 = 03 119877 = 05

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 0150158 537334 times 10minus11 0151534 minus669283 times 10minus9 0152999 minus623412 times 10minus8

02 0297237 314675 times 10minus10 0299827 166025 times 10minus9 0302582 658388 times 10minus8

03 0438170 851847 times 10minus10 0441661 233504 times 10minus8 0445373 342092 times 10minus7

04 0569900 151431 times 10minus9 0573869 393945 times 10minus8 0578082 493942 times 10minus7

05 0689397 190824 times 10minus9 0693354 281614 times 10minus8 0697548 242092 times 10minus7

06 0793661 133301 times 10minus9 0797122 minus319654 times 10minus8 0800780 minus653046 times 10minus7

07 0879734 minus159255 times 10minus9 0882300 minus221911 times 10minus7 0885004 minus323415 times 10minus6

08 0944705 minus102814 times 10minus8 0946167 minus884707 times 10minus7 0947702 minus12206 times 10minus509 0985722 minus3446 times 10minus8 0986180 minus321268 times 10minus6 0986659 minus441008 times 10minus5

10 1 minus103509 times 10minus7 1 minus111826 times 10minus5 1 minus154022 times 10minus4

Using above value of 1198621 the approximate solution is

(120578) =1

2(3120578 + 120578

3

)

+1

4480(minus113284120578 + 151758120578

3

minus 3890121205785

+04274861205787

)

+1

248372(minus156189120578 + 273825120578

3

minus 1608081205785

+4595261205787

minus 2814271205789

+ 34538712057811

)

+1

723257(minus983844120578 + 1149 times 10

8

1205783

minus 335388 times 107

1205785

+ 404185 times 107

1205787

minus 268964 times 107

1205789

+ 365673 times 106

12057811

minus15753812057813

+ 1724212057815

)

+1

370042(minus197195 times 10

13

120578 + 425372 times 1013

1205783

minus 294213 times 1013

1205785

+ 606765 times 1012

1205787

minus 139445 times 1012

1205789

+ 247846 times 1012

12057811

minus 597188 times 1011

12057813

+ 508263 times 1010

12057815

minus162422 times 109

12057817

+ 152182 times 107

12057819

)

(55)

7 Results and Discussions

In this article we considered the unsteady axisymmetric flowof nonconducting incompressible Newtonian fluid betweentwo circular plates The resulting nonlinear boundary valueproblems are solved with OHAM and fourth-order Runge-Kutta method using Mathematica 70

minus000001

minus000002

minus000003

minus000004

R = 01

R = 03

R = 05

02 04 06 08 10

120578

Res

Figure 2 OHAM residuals at various values of 119877 in case of no-slipboundary

Tables 1 3 and 5 reflect OHAM solutions along withresiduals in case of no-sip and slip boundaries for variousvalues of Reynolds number 119877 and slip parameter 120574 AlsoTables 2 4 and 6 represent RK4 solutions alongwith residualsin case of no-slip and slip boundaries for various values of 119877and 120574 All the tables demonstrate that results obtained usingOHAM are in agreement with RK4 by means of residualsIn addition to above mentioned tables Table 7 shows thecomparison of solutions obtained from OHAM and RK4 forvarious values of Reynolds number 119877

Furthermore Figures 2 3 and 4 indicate the OHAMresiduals in case of no-slip and slip boundaries for variousvalues of 119877 and 120574

The effect of Reynolds number 119877 on velocity profilesin case of no-slip boundary is shown in Figure 5 In theseprofiles we varied 119877 as 119877 = 01 1 2 3 and observed that thenormal velocity is increased with the increase of Reynoldsnumber (Figure 5(a)) It is also noted that the normal velocitymonotonically increases from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time Figure 5(b) describes the impact of119877 on the longitudinal velocity in case of no-slip boundary It

Mathematical Problems in Engineering 7

Table 2 RK4 solutions along with residuals for various 119877 in case of no-slip boundary

120578119877 = 01 119877 = 03 119877 = 05

Solution Residual Solution Residual Solution Residual00 0 288535 times 10minus7 0 966906 times 10minus5 0 530549 times 10minus4

01 0150158 388278 times 10minus8 0151534 571393 times 10minus6 0152999 304600 times 10minus5

02 0297237 minus997789 times 10minus9 0299827 minus141434 times 10minus6 0302582 minus752772 times 10minus6

03 0438170 293767 times 10minus9 0441661 395997 times 10minus7 0445373 210296 times 10minus6

04 0569900 minus751436 times 10minus10 0573869 minus115203 times 10minus7 0578082 minus618081 times 10minus7

05 0689397 minus471705 times 10minus10 0693354 minus96761 times 10minus9 0697548 minus295916 times 10minus8

06 0793661 223004 times 10minus9 0797122 14593 times 10minus7 0800780 714133 times 10minus7

07 0879734 minus718245 times 10minus9 0882300 minus486272 times 10minus7 0885004 minus239245 times 10minus6

08 0944705 268744 times 10minus8 0946167 177796 times 10minus6 0947702 870664 times 10minus6

09 0985722 minus111599 times 10minus7 0986180 minus728952 times 10minus6 0986659 356157 times 10minus5

10 1 minus271683 times 10minus6 1 minus147695 times 10minus4 1 690088 times 10minus4

Table 3 OHAM solutions along with residuals for various values of 119877 in case of slip boundary

120578119877 = 02 119877 = 03 119877 = 04

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 0072692 318345 times 10minus8 0071220 315621 times 10minus7 0069593 174626 times 10minus6

02 0147091 minus172789 times 10minus8 0144268 minus914579 times 10minus8 0141147 minus113148 times 10minus7

03 0224893 minus196995 times 10minus7 0220953 minus166461 times 10minus6 0216599 minus779061 times 10minus6

04 0307773 minus494482 times 10minus7 0303048 minus429173 times 10minus6 0297832 minus207401 times 10minus5

05 0397370 minus799437 times 10minus7 0392277 minus699917 times 10minus6 0386657 minus341643 times 10minus5

06 0495283 minus85784 times 10minus7 0490289 minus75499 times 10minus6 0484784 minus370689 times 10minus5

07 0603056 minus21053 times 10minus7 0598654 minus191141 times 10minus6 0593805 minus96958 times 10minus6

08 0722172 187065 times 10minus6 0718841 162936 times 10minus5 0715176 791082 times 10minus5

09 0854042 640576 times 10minus6 0852211 559717 times 10minus5 0850196 272708 times 10minus4

10 1 146076 times 10minus5 1 127671 times 10minus4 1 622242 times 10minus4

Table 4 RK4 solutions along with residuals for various values of 119877 in case of slip boundary

120578119877 = 02 119877 = 03 119877 = 04

Solution Residual Solution Residual Solution Residual00 0 412179 times 10minus6 0 756765 times 10minus6 0 362897 times 10minus6

01 0072692 174999 times 10minus7 0071220 216125 times 10minus7 0069593 minus349994 times 10minus7

02 0147091 minus421503 times 10minus8 0144268 minus497423 times 10minus8 0141147 953266 times 10minus8

03 0224893 112883 times 10minus8 0220953 122294 times 10minus8 0216598 minus30735 times 10minus8

04 0307773 minus352272 times 10minus9 0303048 minus430187 times 10minus9 0297832 72493 times 10minus9

05 0397370 780684 times 10minus10 0392276 306668 times 10minus9 0386657 846409 times 10minus9

06 0495283 110391 times 10minus9 0490289 minus528586 times 10minus9 0484783 minus336828 times 10minus8

07 0603056 minus435033 times 10minus9 0598654 154159 times 10minus8 0593805 106816 times 10minus7

08 0722172 144198 times 10minus8 0718841 minus605873 times 10minus8 0715175 minus398519 times 10minus7

09 0854042 minus553933 times 10minus8 0852211 259233 times 10minus7 0850196 165518 times 10minus6

10 1 minus197434 times 10minus7 1 796168 times 10minus6 1 389126 times 10minus5

is experienced that this component of velocity deceases nearthe wall but increases near the central axis of the channel

The effect of Reynolds number 119877 on velocity profiles incase of slip boundary is depicted in Figure 6 In these profileswe fixed slip parameter 120574 = 1 and varied Reynolds number119877 as 119877 = 02 06 1 15 It is noted that the normal velocitydecreases as the Reynolds number increases (Figure 6(a))It is also observed that longitudinal velocity decreases near

the central axis of the channel but increases near the wallswhen 119877 increases (Figure 6(b))

Figure 7 demonstrates the effect of slip parameter 120574 on thevelocity profiles After fixing Reynolds number 119877 = 03 wevaried 120574 as 120574 = 06 07 08 1 We find that normal velocityincreases as 120574 increases It is also noted that longitudinalvelocity decreases near the walls but increases near centralaxis of the channel

8 Mathematical Problems in Engineering

Table 5 OHAM solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 minus000880011 minus107933 times 10minus6 00336596 minus754858 times 10minus8 00522402 minus774445 times 10minus9

02 minus0010881 minus22445 times 10minus6 00714091 minus173419 times 10minus7 01074223 minus251485 times 10minus8

03 0000449012 minus345224 times 10minus6 0117324 minus300407 times 10minus7 0168479 minus568899 times 10minus8

04 00318278 minus441058 times 10minus6 0175449 minus432397 times 10minus7 0238322 minus98056 times 10minus8

05 00898133 minus447787 times 10minus6 0249785 minus500627 times 10minus7 0319832 minus129675 times 10minus7

06 018086 minus260114 times 10minus6 0344278 minus377978 times 10minus7 0415852 minus113733 times 10minus7

07 0311298 265136 times 10minus6 0462801 130706 times 10minus7 0529175 120785 times 10minus8

08 0487312 128583 times 10minus5 0609145 128166 times 10minus6 0662538 336724 times 10minus7

09 0714929 291822 times 10minus5 0787012 334515 times 10minus6 0818613 967763 times 10minus7

10 1 514019 times 10minus5 1 649074 times 10minus6 1 200176 times 10minus6

Table 6 RK4 solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 204594 times 10minus5 0 734794 times 10minus6 0 356291 times 10minus6

01 minus000880012 772953 times 10minus7 00336596 321066 times 10minus7 00522402 163598 times 10minus7

02 minus00108811 minus18406 times 10minus7 00714091 minus775305 times 10minus8 01074223 minus396814 times 10minus8

03 0000448983 482747 times 10minus8 0117324 208509 times 10minus8 0168479 10753 times 10minus8

04 00318277 minus154235 times 10minus8 0175449 minus643999 times 10minus9 0238322 minus328063 times 10minus9

05 00898132 514031 times 10minus9 0249785 121029 times 10minus9 0319832 450154 times 10minus10

06 018086 minus651866 times 10minus10 0344278 264494 times 10minus9 0415852 184079 times 10minus9

07 0311298 minus186837 times 10minus9 0462801 minus988382 times 10minus9 0529175 minus651979 times 10minus9

08 0487312 minus133003 times 10minus9 0609145 337336 times 10minus8 0662538 227322 times 10minus8

09 0714929 266400 times 10minus8 0787012 minus132232 times 10minus7 0818613 minus905714 times 10minus8

10 1 555277 times 10minus6 1 minus11671 times 10minus6 1 minus11767 times 10minus6

Table 7 Comparison of OHAM and RK4 solutions for various 119877 in case of slip and no-slip boundary

120578

In case of no-slip boundary In case of slip boundary|RK4 Solution minusOHAM Solution| |RK4 Solution minusOHAM Solution|

119877 = 01 119877 = 03 119877 = 05 119877 = 02 119877 = 03 119877 = 04

00 0 0 0 0 0 001 233147 times 10minus15 162249 times 10minus10 241725 times 10minus9 844379 times 10minus10 790896 times 10minus9 41479 times 10minus8

02 949241 times 10minus14 314228 times 10minus10 467809 times 10minus9 16672 times 10minus9 155832 times 10minus8 815527 times 10minus8

03 382361 times 10minus13 446143 times 10minus10 663065 times 10minus9 244889 times 10minus9 228102 times 10minus8 118952 times 10minus7

04 913492 times 10minus13 547932 times 10minus10 811705 times 10minus9 31663 times 10minus9 293539 times 10minus8 152339 times 10minus7

05 165451 times 10minus12 607476 times 10minus10 895202 times 10minus9 377514 times 10minus9 348026 times 10minus8 179574 times 10minus7

06 242317 times 10minus12 609434 times 10minus10 891445 times 10minus9 418241 times 10minus9 38327 times 10minus8 196532 times 10minus7

07 286218 times 10minus12 536833 times 10minus10 777928 times 10minus9 422088 times 10minus9 384569 times 10minus8 196009 times 10minus7

08 252039 times 10minus12 378563 times 10minus10 54278 times 10minus9 365208 times 10minus9 331085 times 10minus8 167861 times 10minus7

09 120232 times 10minus12 154202 times 10minus10 218716 times 10minus9 224454 times 10minus9 202732 times 10minus8 102385 times 10minus7

10 126281 times 10minus16 253545 times 10minus16 293337 times 10minus17 17436 times 10minus14 492715 times 10minus14 38179 times 10minus14

8 Conclusions

In this article we find the similarity solution for unsteadyaxisymmetric squeezing flow of incompressible Newtonianfluid between two circular plates We observed that thesimilarity solution exists only when distance between theplates varies as (119882119905 + 119878)

12 and squeezing flow occurs when

119882 lt 0 119878 gt 0 and (119882119905+119878) gt 0The key findings of the presentanalysis are as follows

In case of no-slip at boundary

(i) It has been found that increase in Reynolds number 119877increases the normal velocity

Mathematical Problems in Engineering 9

00010

00008

00006

00004

00002

02 04 06 08 10

120578

Res

R = 02

R = 03

R = 04

Figure 3 OHAM residuals at various values of 119877 in case of slip boundary

00001

000008

000006

000004

000002

120574 = 05

120574 = 06

120574 = 07

02 04 06 08 10

120578

Res

Figure 4 OHAM residuals at various values of 120574 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

(120578

120578

)

R = 01

R = 1

R = 2

R = 3

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

R = 01

R = 1

R = 2

R = 3

998400 (120578)

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 5 Velocity profiles for various values of 119877 = 01 1 2 3 in case of no-slip boundary

10 Mathematical Problems in Engineering

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

R = 02

R = 06

R = 1

R = 15

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

R = 02

R = 06

R = 1

R = 15

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 6 Velocity profiles for various values of 119877 = 02 06 1 15 fixing 120574 = 1 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

120574 = 06120574 = 07

120574 = 08

120574 = 1

(a) The effect of 120574 on the Normal velocity profiles

120574 = 06120574 = 07

120574 = 08

120574 = 1

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

(b) The effect of 120574 on the longitudinal velocity profiles

Figure 7 Velocity profiles for various values of 120574 = 06 07 08 1 fixing 119877 = 03 in case of slip boundary

(ii) It has been observed that normal velocity increasesmonotonically from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time

(iii) It has been seen that longitudinal velocity deceasesnear the walls and increases near the central axis ofthe channel

In case of slip at boundary

(i) It has been noted that after fixing slip parameter120574 and varying the Reynolds number 119877 the normalvelocity profile decreases with the increase in 119877 Alsothe longitudinal velocity increases near the walls butdecreases near the central axis of the channel

(ii) It has been examined that for a fixed Reynoldsnumber 119877when we vary slip parameter 120574 the normalvelocity increases with the increase in 120574 Also the

longitudinal velocity decreases near the walls andincreases near the central axis of the channel

(iii) It has been investigated that Reynolds number 119877 andslip parameter 120574 have opposite effects on the normaland longitudinal velocity components

In case of slip versus no-slip boundary

(i) It has been observed that Reynolds number 119877 hasopposite behavior on the normal velocity in case ofslip and no-slip boundaries

(ii) It has been also noticed that Reynolds number 119877 hasopposite effect on the longitudinal velocity near thecentral axis of the channel while near the wall longi-tudinal velocity increases in case of slip boundary anddecease in no-slip boundary This is in conformanceto [33]

Mathematical Problems in Engineering 11

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] Q K Ghori M Ahmed and A M Siddiqui ldquoApplicationof homotopy perturbation method to squeezing flow of anewtonian fluidrdquo International Journal of Nonlinear Sciencesand Numerical Simulation vol 8 no 2 pp 179ndash184 2007

[2] X J Ran Q Y Zhu and Y Li ldquoAn explicit series solution ofthe squeezing flow between two infinite plates by means ofthe homotopy analysis methodrdquo Communications in NonlinearScience and Numerical Simulation vol 14 no 1 pp 119ndash1322009

[3] M J Stefan ldquoVersuch Uber die scheinbare adhasionrdquo Sitzungs-berichteOsterreichische Akademie der Wissenschaften in WienMathematisch-Naturwissenschaftliche Klasse vol 69 pp 713ndash721 1874

[4] J F Thorpe and W A Shaw Eds Developments in Theoreticaland Applied Mechanics vol 3 Pergamon Press Oxford UK1967

[5] P S Gupta and A S Gupta ldquoSqueezing flow between parallelplatesrdquoWear vol 45 no 2 pp 177ndash185 1977

[6] A F Elkouh ldquoFluid inertia effects in squeeze film between twoplane annulirdquo Transactions of the ASME Journal of Tribologyvol 106 no 2 pp 223ndash227 1984

[7] R L Verma ldquoA numerical solution for squeezing flow betweenparallel channelsrdquoWear vol 72 no 1 pp 89ndash95 1981

[8] P Singh V Radhakrishnan and K A Narayan ldquoSqueezing flowbetween parallel platesrdquo Ingenieur-Archiv vol 60 no 4 pp 274ndash281 1990

[9] P J Leider and R B Bird ldquoSqueezing flow between paralleldisks I Theoretical analysisrdquo Industrial and Engineering Chem-istry Fundamentals vol 13 no 4 pp 336ndash341 1974

[10] G Domairry and A Aziz ldquoApproximate analysis of MHDdqueeze flow between two parallel disks with suction orinjection by homotopy perturbation methodrdquo MathematicalProblems in Engineering vol 2009 Article ID 603916 19 pages2009

[11] S Islam H Khan I A Shah and G Zaman ldquoAn axisymmetricsqueezing fluid flow between the two infinite parallel plates in aporous medium channelrdquo Mathematical Problems in Engineer-ing vol 2011 Article ID 349803 10 pages 2011

[12] I Ullah H Khan and M T Rahim ldquoApproximation of firstgrade MHD squeezing fluid flow with slip boundary conditionusing DTM and OHAMrdquo Mathematical Problems in Engineer-ing vol 2013 Article ID 816262 9 pages 2013

[13] A M Siddiqui S Irum and A R Ansari ldquoUnsteady squeezingflowof a viscousMHDfluid between parallel platesrdquoMathemat-ical Modelling and Analysis vol 13 pp 565ndash576 2008

[14] R J Grimm ldquoSqueezing flows of Newtonian liquid films ananalysis including fluid inertiardquo Applied Scientific Research vol32 no 2 pp 149ndash166 1976

[15] M M Rashidi A M Siddiqui and M T Rastegari ldquoAnalyticalsolution of squeezing flow between two circular platesrdquo Interna-tional Journal for ComputationalMethods in Engineering Scienceand Mechanics vol 13 no 5 pp 342ndash349 2012

[16] H M Laun M Rady and O Hassager ldquoAnalytical solutions forsqueeze flow with partial wall sliprdquo Journal of Non-NewtonianFluid Mechanics vol 81 no 1-2 pp 1ndash15 1999

[17] S Ishizawa ldquoThe unsteady flow between two parallel discs witharbitrary varying gap widthrdquo Bulletin of the Japan Society ofMechanical Engineers vol 9 no 35 pp 533ndash550 1966

[18] C le Roux ldquoExistence and uniqueness of the flow of second-grade fluids with slip boundary conditionsrdquoArchive for RationalMechanics and Analysis vol 148 no 4 pp 309ndash356 1999

[19] L J Rhooades R Resnic T OrsquoBradovich and S StegmanldquoAbrasive flow machining of cylinder heads and its positiveeffects on performance and cost characteristicsrdquo Tech RepDearborn Mich USA 1996

[20] C L M H Navier Memoirs de lrsquoCademie Royale des Sciencesde lrsquoInstitut de France vol 1 Royale des Sciences de lrsquoInstitut deFrance 1823

[21] A Ebaid ldquoEffects of magnetic field and wall slip conditions onthe peristaltic transport of a Newtonian fluid in an asymmetricchannelrdquo Physics Letters Section A General Atomic and SolidState Physics vol 372 no 24 pp 4493ndash4499 2008

[22] J-H He ldquoSome asymptotic methods for strongly nonlinearequationsrdquo International Journal of Modern Physics B Con-densed Matter Physics Statistical Physics Applied Physics vol20 no 10 pp 1141ndash1199 2006

[23] S J Liao Proposed homotopy analysis techniques for the solutionof nonlinear problems [PhD thesis] Jiao Tong UniversityShanghai China 1992

[24] J-H He ldquoHomotopy perturbation method for solving bound-ary value problemsrdquo Physics Letters A vol 350 no 1-2 pp 87ndash88 2006

[25] J-H He ldquoApproximate analytical solution for seepage flowwithfractional derivatives in porous mediardquo Computer Methods inApplied Mechanics and Engineering vol 167 no 1-2 pp 57ndash681998

[26] VMarinca andNHerisanu ldquoApplication of optimalHomotopyasymptotic method for solving nonlinear equations arising inheat transferrdquo International Communications in Heat and MassTransfer vol 35 no 6 pp 710ndash715 2008

[27] V Marinca N Herisanu C Bota and B Marinca ldquoAn optimalhomotopy asymptotic method applied to the steady flow of afourth-grade fluid past a porous platerdquo Applied MathematicsLetters vol 22 no 2 pp 245ndash251 2009

[28] V Marinca N Herisanu and I Nemes ldquoOptimal homotopyasymptotic method with application to thin film flowrdquo CentralEuropean Journal of Physics vol 6 no 3 pp 648ndash653 2008

[29] J Ali S Islam and G Zaman ldquoThe solution of multipointboundary value problems by the optimal Homotopy asymptoticMethodrdquoComputers andMathematics with Applications vol 59no 6 pp 2000ndash2006 2010

[30] M Idrees S Islam and S Haq ldquoApplication of the optimalhomotopy asymptotic method to squeezing flowrdquo Computersand Mathematics with Applications vol 59 no 12 pp 3858ndash3866 2010

[31] J Ali S Islam H Khan and S I Shah ldquoThe optimal homotopyasymptotic method for the solution of higher-order boundaryvalue problems in finite domainsrdquo Abstract and Applied Analy-sis vol 2012 Article ID 401217 14 pages 2012

[32] H Khan S Islam J Ali and I Ali Shah ldquoComparison ofdifferent analytic solutions to axisymmetric squeezing fluidflow between two infinite parallel plates with slip boundaryconditionsrdquo Abstract and Applied Analysis vol 2012 Article ID835268 18 pages 2012

[33] T C Papanastasiou G C Georgiou and A N AlexandrouViscous Fluid Flow Library of Congress Cataloging CRC PressNew York NY USA 1999

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Page 2: Research Article Analysis of Unsteady Axisymmetric ...downloads.hindawi.com/journals/mpe/2015/860857.pdfResearch Article Analysis of Unsteady Axisymmetric Squeezing Fluid Flow with

2 Mathematical Problems in Engineering

the effect of magnetic field in Newtonian fluid in an asym-metric channel with wall slip conditions

Most of scientific incidents are modeled by nonlinearpartial or ordinary differential equations In literature wehave variety of perturbation techniques which can solvenonlinear boundary value problems analytically But thelimitations of these techniques are based on the assumptionof small parameters Detailed review of these methods isgiven by He [22] In recent times the ideas of Homotopy andPerturbation have been combined together Liao [23] and He[24 25] have done the primary work in this regard In seriesof papers Marinca with various scholars used OHAM to findthe approximate solution of nonlinear differential equationsarising in heat transfer steady flow of a fourth-grade fluidand thin film flow [26ndash28]

In this work OHAM is used to analyze an unsteadysqueezing fluid flow between two circular nonrotating diskswith slip and no-slip boundary conditions In additionmovement of the circular plates is considered to be symmetricabout the axial line and the fluid is considered to be Newto-nian incompressible and viscous Sections 2 and 3 include thedescription and mathematical formulation of the problemSections 4 5 and 6 present the basic theory of OHAM andits application in case of no-slip and slip boundaries Resultsand discussions are given in Section 7 while conclusions arementioned in Section 8

2 Description of the Problem

Theunsteady axisymmetric squeezing flow of incompressibleNewtonian fluid with density 120588 viscosity 120583 and kinematicviscosity ] squeezed between two circular plates havingspeed V

119908(119905) is considered It is assumed that at any time 119905 the

distance between two circular plates is 2119904(119905) It is also assumedthat 119903-axis is the central axis of the channel while 119911-axis istaken normal to it Plates move symmetrically with respect tothe central axis 119911 = 0 while the flow is axisymmetric about119903 = 0 The longitudinal and normal velocity componentsin radial and axial directions are 119906

119903(119903 119911 119905) and 119906

119911(119903 119911 119905)

respectivelyThe geometrical interpretation of the problem isgiven in Figure 1

3 Mathematical Formulation

The governing equations of motion are

nabla sdot U = 0

120588 [120597U120597119905

+ (U sdot nabla)U] = 120588119891 + nabla sdot T(1)

where

T = minus119901I + 120583A1

A1= nablaU + (nablaU)119905

(2)

and U is the velocity vector 119901 is the pressure 119891 is the bodyforce T is the Cauchy stress tensor A

1is the Rivlin-Ericksen

2s(t)r = 0

z = 0

r

r

w

w

uz

ur

Figure 1 Geometrical interpretation of the problem

tensor and 120583 is the coefficient of viscosity Now we formulatethe unsteady two-dimensional flow Let us assume that

U = [119906119903(119903 119911 119905) 0 119906

119911(119903 119911 119905)] (3)

and introduce the vorticity functionΩ(119903 119911 119905) and generalizedpressure ℎ(119903 119911 119905) as

Ω (119903 119911 119905) =120597119906119911

120597119903minus120597119906119903

120597119911 (4)

ℎ (119903 119911 119905) =120588

2[1199062

119903+ 1199062

119911] + 119901 (5)

Equations (1) are reduced to120597119906119903

120597119903+119906119903

119903+120597119906119911

120597119911= 0

120597ℎ

120597119903+ 120588(

120597119906119903

120597119905minus 119906119911Ω) = minus120583

120597Ω

120597119911

120597ℎ

120597119911+ 120588(

120597119906119911

120597119905+ 119906119903Ω) =

120583

119903

120597

120597119903(119903Ω)

(6)

The boundary conditions on 119906119903(119903 119911 119905) and 119906

119911(119903 119911 119905) are

119906119903(119903 119911 119905) = 0 119906

119911(119903 119911 119905) = V

119908(119905) at 119911 = 119904

120597

120597119911119906119903(119903 119911 119905) = 0 119906

119911(119903 119911 119905) = 0 at 119911 = 0

(7)

where V119908(119905) = 119889119904119889119905 is the velocity of the plates The

boundary conditions (7) are due to symmetry at 119911 = 0 andno-slip at the upper plate when 119911 = 119904 If we introduce thedimensionless parameter

120578 =119911

119904 (119905) (8)

Equations (4) and (6) transforms to

Ω (119903 119911 119905) =120597119906119911

120597119903minus1

119904

120597119906119903

120597120578 (9)

120597119906119903

120597119903+119906119903

119903+1

119904

120597119906119911

120597120578= 0 (10)

120597ℎ

120597119903+ 120588(

120597119906119903

120597119905minus 119906119911Ω) = minus

120583

119904

120597Ω

120597120578 (11)

1

119904

120597ℎ

120597120578+ 120588(

120597119906119911

120597119905+ 119906119903Ω) =

120583

119903

120597

120597119903(119903Ω) (12)

Mathematical Problems in Engineering 3

The boundary conditions on 119906119903and 119906

119911are

119906119903= 0 119906

119911= V119908(119905) at 120578 = 1

120597119906119903

120597120578= 0 119906

119911= 0 at 120578 = 0

(13)

After eliminating the generalized pressure between (11) and(12) we obtained

120588 [120597Ω

120597119905+ 119906119903

120597Ω

120597119903+119906119911

119904

120597Ω

120597120578minus119906119903

119903Ω] = 120583 [nabla

2

Ω minusΩ

1199032] (14)

where nabla2 is the Laplacian operatorDefining velocity components as [5]

119906119903= minus

119903

2119904 (119905)V119908(119905) 1198921015840

(120578)

119906119911= V119908(119905) 119892 (120578)

(15)

we see that (10) is identically satisfied and (14) becomes

1198894

119892

1198891205784+ 119877[(120578 minus 119892)

1198893

119892

1198891205783+ 2

1198892

119892

1198891205782] minus 119876

1198892

119892

1198891205782= 0 (16)

where

119877 =119904V119908

120592 119876 =

1199042

120592V119908

119889V119908

119889119905 (17)

Here both119877 and119876 are functions of 119905 but we consider119877 and119876constants for similarity solution Since V

119908= 119889119904119889119905 Integrate

first equation of (17) we get

119904 (119905) = (119882119905 + 119878)12

(18)

where119882 and 119878 are constantsTheplatesmove away from eachother symmetrically with respect to 120578 when 119882 gt 0 and 119878 gt0 Also the plates approach to each other and squeezing flowexists with similar velocity profiles when119882 lt 0 119878 gt 0 and119904(119905) gt 0 From (17) and (18) it follows that 119876 = minus119877 Then (16)becomes

1198894

119892

1198891205784+ 119877[(120578 minus 119892)

1198893

119892

1198891205783+ 3

1198892

119892

1198891205782] = 0 (19)

Using (13) and (15) we determine the boundary conditions incase of no-slip and slip at the upper plate as follows

119892 (1) = 1 1198921015840

(1) = 0

119892 (0) = 0 11989210158401015840

(0) = 0(No-slip at the wall) (20)

119892 (1) = 1 1198921015840

(1) = 12057411989210158401015840

(1)

119892 (0) = 0 11989210158401015840

(0) = 0(Slip at the wall) (21)

4 Basic Theory of OHAM [26 29ndash32]Let us apply OHAM to the following differential equation

I [V (119909)] + 119891 (119909) + alefsym [V (119909)] = 0 119861 (V119889V119889119909

) = 0 (22)

where 119909 represents an independent variable V(119909) is unknownfunction and 119891(119909) is known function 119861alefsymI are boundarynonlinear and linear operators respectively

According to OHAM we construct Homotopy Φ(119909 119901)R times [0 1] rarr R which satisfies

(1 minus 119901) [I (Φ (119909 119901)) + 119891 (119909)]

= ℎ (119901) [I (Φ (119909 119901)) + 119891 (119909) + alefsym (Φ (119909 119901))]

119861(Φ (119909 119901) 120597Φ (119909 119901)

120597119909) = 0

(23)

where 119909 isin R and 119901 isin [0 1] is an embedding parameterℎ(119901) is a nonlinear auxiliary function for 119901 = 0 ℎ(0) = 0

and Φ(119909 119901) is an unknown function Clearly when 119901 = 0

and 119901 = 1 it holds that Φ(119909 0) = V0(119909) and Φ(119909 1) = V(119909)

respectivelyThus as 119901 varies from 0 to 1 the solution Φ(119909 119901)

approaches from V0(119909) to V(119909)

We choose the auxiliary function ℎ(119901) in the form of

ℎ (119901) =

119898

sum

119896=0

119901119896

119862119896 (24)

where 119862119896are convergence controlling constants to be deter-

minedTo obtain an approximate solution we expandΦ(119909 119901 119862

119894)

in a Taylor series about 119901 as follows

Φ(119909 119901 119862119894) = V0(119909) +

119898

sum

119896=1

V119896(1198621 1198622 119862

119896) 119901119896

(25)

Substituting (25) into (23) and equating the coefficients of likepowers of 119901 we obtain the following equationsThe zeroth-order problem is

I [V0(119909)] + 119891 (119909) = 0 119861 (V

0119889V0

119889119909) = 0 (26)

First-order problem is

I [V1(119909)] + 119891 (119909) = 119862

1alefsym0[V0(119909)] 119861 (V

1119889V1

119889119909) = 0

(27)

Second-order problem is

I [V2(119909)] +I [V

1(119909)]

= 1198622alefsym0[V0(119909)] + 119862

1I [V1(119909)] + alefsym

1[V0(119909) V

1(119909)]

119861 (V2119889V2

119889119909) = 0

(28)

4 Mathematical Problems in Engineering

The general equations for V119896(119909) are given by

I [V119896(119909)] minusI [V

119896minus1(119909)]

= 1198621alefsym0[V0(119909)]

+

119896minus1

sum

119894=1

119862119894I [V119896minus119894

(119909)]

+alefsym119896minus119894

[V0(119909) V

1(119909) V

119896minus1(119909)]

119861 (V119896119889V119896

119889119909) = 0

119896 = 2 3

(29)

where the coefficient of 119901119898 in the expansion of alefsym(Φ(119909 119901))about 119901 is alefsym

119898[V0(119909) V1(119909) V

119898minus1(119909)]

alefsym(Φ (119909 119901 119862119894))

= alefsym0[V0(119909)] +

119898

sum

119898=1

alefsym119898[V0(119909) V

1(119909) V

119898minus1(119909)] 119901

119898

(30)It is noted that the convergence of the series (25) depends

upon119862119896 For convergence at119901 = 1 the119898th order approxima-

tion V is

V (119909 1198621 1198622 119862

119898) = V0(119909) +

119898

sum

119895=1

V119895(119909 1198621 1198622 119862

119895)

(31)Substituting (31) in (22) the expression for residual is

R (119909 1198621 1198622 119862

119898)

= I [V (119909 1198621 1198622 119862

119898)] + 119891 (119909)

+ alefsym [V (119909 1198621 1198622 119862

119898)]

(32)

If R = 0 then V will be the exact solution but usually thisdoes not happen in nonlinear problems

There are various methods to find the optimal values of119862119894 119894 = 1 2 We apply the method of least square and

Galerkinrsquos method in the following mannerIn method of least square

119869 (119909 1198621 1198622 119862

119898) = int

119887

119886

R2

(119909 1198621 1198622 119862

119898) 119889119909 (33)

Minimizing 119869(119909 1198621 1198622 119862

119898) we have

120597119869

120597119862119894

= 0 119894 = 1 2 119898 (34)

In Galerkinrsquos method we solve the following system for119862119894(119894 = 1 2 119898)

int

119887

119886

R120597V120597119862119894

119889119909 = 0 119894 = 1 2 119898 (35)

To find appropriate 119862119894(119894 = 1 2 119898) we choose 119886 and 119887 in

the domain of the problem Approximate solution of order119898is well-determined with these known constants

5 Application of OHAM in Case ofNo-Slip Boundary

Using (19) and (20) various order problems are as followsZeroth-order problem

V(119894V)0

(120578) = 0

V0(0) = 0 V10158401015840

0(0) = 0 V

0(1) = 1 V1015840

0(1) = 0

(36)

First-order problem

V(119894V)1

(120578) = 31198621119877V101584010158400(120578) + 119862

1119877120578V1015840101584010158400(120578) minus 119862

1119877V0(120578) V1015840101584010158400(120578)

+ V(119894V)0

(120578) + 1198621V(119894V)0

(120578)

V1(0) = 0 V10158401015840

1(0) = 0 V

1(1) = 0 V1015840

1(1) = 0

(37)

Second-order problem

V(119894V)2

(120578) = 31198621119877V101584010158401(120578) minus 119862

1119877V1(120578) V1015840101584010158400(120578) + 119862

1119877120578V1015840101584010158401(120578)

minus 1198621119877V0(120578) V1015840101584010158401(120578) + V(119894V)

1(120578) + 119862

1V(119894V)1

(120578)

V2(0) = 0 V10158401015840

2(0) = 0 V

2(1) = 0 V1015840

2(1) = 0

(38)

Third-order problem

V(119894V)3

(120578) = 31198621119877V101584010158402(120578) minus 119862

1119877V2(120578) V1015840101584010158400(120578)

minus 1198621119877V1(120578) V1015840101584010158401(120578) + 119862

1119877120578V1015840101584010158402(120578)

minus 1198621119877V0(120578) V1015840101584010158402(120578) + V(119894V)

2(120578)

+ 1198621V(119894V)2

(120578)

V3(0) = 0 V10158401015840

3(0) = 0 V

3(1) = 0 V1015840

3(1) = 0

(39)

Fourth-order problem

V(119894V)4

(120578) = 31198621119877V101584010158403(120578) minus 119862

1119877V3(120578) V1015840101584010158400(120578)

minus 1198621119877V2(120578) V1015840101584010158401(120578) minus 119862

1119877V1(120578) V1015840101584010158402(120578)

+ 1198621119877120578V1015840101584010158403(120578) minus 119862

1119877V0(120578) V1015840101584010158403(120578)

+ V(119894V)3

(120578) + 1198621V(119894V)3

(120578)

V4(0) = 0 V10158401015840

4(0) = 0 V

4(1) = 0 V1015840

4(1) = 0

(40)

By considering fourth-order solution we have

V (120578) = V0(120578) + V

1(120578) + V

2(120578) + V

3(120578) + V

4(120578) (41)

The residual of the problem is

R =1198894V (120578)1198891205784

+ 119877[(120578 minus V (120578))1198893V (120578)1198891205783

+ 31198892V (120578)1198891205782

] (42)

Mathematical Problems in Engineering 5

We apply Galerkinrsquos method to find constant 1198621as follows

int

1

0

R120597V1205971198621

119889119909 = 0 (43)

Solving (43) and keeping 119877 = 01 we get

1198621= 101328 (44)

Using above value of 1198621 the approximate solution is

V (120578) =1

2(3120578 minus 120578

3

)

+1

560(374914120578 minus 739694120578

3

+ 3546481205785

+01013281205787

)

+1

776160(956941120578 minus 430222120578

3

+ 514141205785

minus122331205787

minus 5534111205789

minus 19405312057811

)

+1

565045(254414120578 + 143644120578

3

minus 3711641205785

+ 2502321205787

minus 8133381205789

+ 25990112057811

+6961012057813

+ 2296212057815

)

+1

722737(153702120578 minus 187989 times 10

8

1205783

+ 193991 times 108

1205785

+ 828837 times 107

1205787

minus 803587 times 107

1205789

minus 292844 times 107

12057811

+ 167506 times 107

12057813

minus 966217 times 106

12057815

minus16529 times 106

12057817

minus 48038912057819

)

(45)

6 Application of OHAM in Case ofSlip Boundary

Using (19) and (21) different order problems are as followsZeroth-order problem

119906(119894V)0

(120578) = 0

1199060(0) = 0 119906

10158401015840

0(0) = 0 119906

0(1) = 1 119906

1015840

0(1) = 120574119906

10158401015840

0(1)

(46)

First-order problem

119906(119894V)1

(120578)

= 3119862111987711990610158401015840

0(120578) + 119862

1119877120578119906101584010158401015840

0(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

0(120578)

+ 119906(119894V)0

(120578) + 1198621119906(119894V)0

(120578)

1199061(0) = 0 119906

10158401015840

1(0) = 0 119906

1(1) = 0 119906

1015840

1(1) = 120574119906

10158401015840

1(1)

(47)

Second-order problem

119906(119894V)2

(120578) = 3119862111987711990610158401015840

1(120578) minus 119862

11198771199061(120578) 119906101584010158401015840

0(120578)

+ 1198621119877120578119906101584010158401015840

1(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

1(120578)

+ 119906(119894V)1

(120578) + 1198621119906(119894V)1

(120578)

1199062(0) = 0 119906

10158401015840

2(0) = 0 119906

2(1) = 0 119906

1015840

2(1) = 120574119906

10158401015840

2(1)

(48)

Third-order problem

119906(119894V)3

(120578) = 3119862111987711990610158401015840

2(120578) minus 119862

11198771199062(120578) 119906101584010158401015840

0(120578)

minus 11986211198771199061(120578) 119906101584010158401015840

1(120578) + 119862

1119877120578119906101584010158401015840

2(120578)

minus 11986211198771199060(120578) 119906101584010158401015840

2(120578) + 119906

(119894V)2

(120578)

+ 1198621119906(119894V)2

(120578)

1199063(0) = 0 119906

10158401015840

3(0) = 0 119906

3(1) = 0 119906

1015840

3(1) = 120574119906

10158401015840

3(1)

(49)

Fourth-order problem

119906(119894V)4

(120578) = 3119862111987711990610158401015840

3(120578) minus 119862

11198771199063(120578) 119906101584010158401015840

0(120578)

minus 11986211198771199062(120578) 119906101584010158401015840

1(120578) minus 119862

11198771199061(120578) 119906101584010158401015840

2(120578)

+ 1198621119877120578119906101584010158401015840

3(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

3(120578)

+ 119906(119894V)3

(120578) + 1198621119906(119894V)3

(120578)

1199064(0) = 0 119906

10158401015840

4(0) = 0 119906

4(1) = 0 119906

1015840

4(1) = 120574119906

10158401015840

4(1)

(50)

By considering fourth-order solution we have

(120578) =

4

sum

119894=0

119906119894(120578 1198621) (51)

The residual of the problem is

120577 =1198894

(120578)

1198891205784+ 119877[(120578 minus (120578))

1198893

(120578)

1198891205783+ 3

1198892

(120578)

1198891205782] (52)

We apply Galerkinrsquos method to find constant 1198621as follows

int

1

0

120577120597

1205971198621

119889119909 = 0 (53)

Solving (53) and taking 119877 = 02 and 120574 = 1 we get

1198621= minus106872 (54)

6 Mathematical Problems in Engineering

Table 1 OHAM solutions along with residuals for various 119877 in case of no-slip boundary

120578119877 = 01 119877 = 03 119877 = 05

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 0150158 537334 times 10minus11 0151534 minus669283 times 10minus9 0152999 minus623412 times 10minus8

02 0297237 314675 times 10minus10 0299827 166025 times 10minus9 0302582 658388 times 10minus8

03 0438170 851847 times 10minus10 0441661 233504 times 10minus8 0445373 342092 times 10minus7

04 0569900 151431 times 10minus9 0573869 393945 times 10minus8 0578082 493942 times 10minus7

05 0689397 190824 times 10minus9 0693354 281614 times 10minus8 0697548 242092 times 10minus7

06 0793661 133301 times 10minus9 0797122 minus319654 times 10minus8 0800780 minus653046 times 10minus7

07 0879734 minus159255 times 10minus9 0882300 minus221911 times 10minus7 0885004 minus323415 times 10minus6

08 0944705 minus102814 times 10minus8 0946167 minus884707 times 10minus7 0947702 minus12206 times 10minus509 0985722 minus3446 times 10minus8 0986180 minus321268 times 10minus6 0986659 minus441008 times 10minus5

10 1 minus103509 times 10minus7 1 minus111826 times 10minus5 1 minus154022 times 10minus4

Using above value of 1198621 the approximate solution is

(120578) =1

2(3120578 + 120578

3

)

+1

4480(minus113284120578 + 151758120578

3

minus 3890121205785

+04274861205787

)

+1

248372(minus156189120578 + 273825120578

3

minus 1608081205785

+4595261205787

minus 2814271205789

+ 34538712057811

)

+1

723257(minus983844120578 + 1149 times 10

8

1205783

minus 335388 times 107

1205785

+ 404185 times 107

1205787

minus 268964 times 107

1205789

+ 365673 times 106

12057811

minus15753812057813

+ 1724212057815

)

+1

370042(minus197195 times 10

13

120578 + 425372 times 1013

1205783

minus 294213 times 1013

1205785

+ 606765 times 1012

1205787

minus 139445 times 1012

1205789

+ 247846 times 1012

12057811

minus 597188 times 1011

12057813

+ 508263 times 1010

12057815

minus162422 times 109

12057817

+ 152182 times 107

12057819

)

(55)

7 Results and Discussions

In this article we considered the unsteady axisymmetric flowof nonconducting incompressible Newtonian fluid betweentwo circular plates The resulting nonlinear boundary valueproblems are solved with OHAM and fourth-order Runge-Kutta method using Mathematica 70

minus000001

minus000002

minus000003

minus000004

R = 01

R = 03

R = 05

02 04 06 08 10

120578

Res

Figure 2 OHAM residuals at various values of 119877 in case of no-slipboundary

Tables 1 3 and 5 reflect OHAM solutions along withresiduals in case of no-sip and slip boundaries for variousvalues of Reynolds number 119877 and slip parameter 120574 AlsoTables 2 4 and 6 represent RK4 solutions alongwith residualsin case of no-slip and slip boundaries for various values of 119877and 120574 All the tables demonstrate that results obtained usingOHAM are in agreement with RK4 by means of residualsIn addition to above mentioned tables Table 7 shows thecomparison of solutions obtained from OHAM and RK4 forvarious values of Reynolds number 119877

Furthermore Figures 2 3 and 4 indicate the OHAMresiduals in case of no-slip and slip boundaries for variousvalues of 119877 and 120574

The effect of Reynolds number 119877 on velocity profilesin case of no-slip boundary is shown in Figure 5 In theseprofiles we varied 119877 as 119877 = 01 1 2 3 and observed that thenormal velocity is increased with the increase of Reynoldsnumber (Figure 5(a)) It is also noted that the normal velocitymonotonically increases from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time Figure 5(b) describes the impact of119877 on the longitudinal velocity in case of no-slip boundary It

Mathematical Problems in Engineering 7

Table 2 RK4 solutions along with residuals for various 119877 in case of no-slip boundary

120578119877 = 01 119877 = 03 119877 = 05

Solution Residual Solution Residual Solution Residual00 0 288535 times 10minus7 0 966906 times 10minus5 0 530549 times 10minus4

01 0150158 388278 times 10minus8 0151534 571393 times 10minus6 0152999 304600 times 10minus5

02 0297237 minus997789 times 10minus9 0299827 minus141434 times 10minus6 0302582 minus752772 times 10minus6

03 0438170 293767 times 10minus9 0441661 395997 times 10minus7 0445373 210296 times 10minus6

04 0569900 minus751436 times 10minus10 0573869 minus115203 times 10minus7 0578082 minus618081 times 10minus7

05 0689397 minus471705 times 10minus10 0693354 minus96761 times 10minus9 0697548 minus295916 times 10minus8

06 0793661 223004 times 10minus9 0797122 14593 times 10minus7 0800780 714133 times 10minus7

07 0879734 minus718245 times 10minus9 0882300 minus486272 times 10minus7 0885004 minus239245 times 10minus6

08 0944705 268744 times 10minus8 0946167 177796 times 10minus6 0947702 870664 times 10minus6

09 0985722 minus111599 times 10minus7 0986180 minus728952 times 10minus6 0986659 356157 times 10minus5

10 1 minus271683 times 10minus6 1 minus147695 times 10minus4 1 690088 times 10minus4

Table 3 OHAM solutions along with residuals for various values of 119877 in case of slip boundary

120578119877 = 02 119877 = 03 119877 = 04

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 0072692 318345 times 10minus8 0071220 315621 times 10minus7 0069593 174626 times 10minus6

02 0147091 minus172789 times 10minus8 0144268 minus914579 times 10minus8 0141147 minus113148 times 10minus7

03 0224893 minus196995 times 10minus7 0220953 minus166461 times 10minus6 0216599 minus779061 times 10minus6

04 0307773 minus494482 times 10minus7 0303048 minus429173 times 10minus6 0297832 minus207401 times 10minus5

05 0397370 minus799437 times 10minus7 0392277 minus699917 times 10minus6 0386657 minus341643 times 10minus5

06 0495283 minus85784 times 10minus7 0490289 minus75499 times 10minus6 0484784 minus370689 times 10minus5

07 0603056 minus21053 times 10minus7 0598654 minus191141 times 10minus6 0593805 minus96958 times 10minus6

08 0722172 187065 times 10minus6 0718841 162936 times 10minus5 0715176 791082 times 10minus5

09 0854042 640576 times 10minus6 0852211 559717 times 10minus5 0850196 272708 times 10minus4

10 1 146076 times 10minus5 1 127671 times 10minus4 1 622242 times 10minus4

Table 4 RK4 solutions along with residuals for various values of 119877 in case of slip boundary

120578119877 = 02 119877 = 03 119877 = 04

Solution Residual Solution Residual Solution Residual00 0 412179 times 10minus6 0 756765 times 10minus6 0 362897 times 10minus6

01 0072692 174999 times 10minus7 0071220 216125 times 10minus7 0069593 minus349994 times 10minus7

02 0147091 minus421503 times 10minus8 0144268 minus497423 times 10minus8 0141147 953266 times 10minus8

03 0224893 112883 times 10minus8 0220953 122294 times 10minus8 0216598 minus30735 times 10minus8

04 0307773 minus352272 times 10minus9 0303048 minus430187 times 10minus9 0297832 72493 times 10minus9

05 0397370 780684 times 10minus10 0392276 306668 times 10minus9 0386657 846409 times 10minus9

06 0495283 110391 times 10minus9 0490289 minus528586 times 10minus9 0484783 minus336828 times 10minus8

07 0603056 minus435033 times 10minus9 0598654 154159 times 10minus8 0593805 106816 times 10minus7

08 0722172 144198 times 10minus8 0718841 minus605873 times 10minus8 0715175 minus398519 times 10minus7

09 0854042 minus553933 times 10minus8 0852211 259233 times 10minus7 0850196 165518 times 10minus6

10 1 minus197434 times 10minus7 1 796168 times 10minus6 1 389126 times 10minus5

is experienced that this component of velocity deceases nearthe wall but increases near the central axis of the channel

The effect of Reynolds number 119877 on velocity profiles incase of slip boundary is depicted in Figure 6 In these profileswe fixed slip parameter 120574 = 1 and varied Reynolds number119877 as 119877 = 02 06 1 15 It is noted that the normal velocitydecreases as the Reynolds number increases (Figure 6(a))It is also observed that longitudinal velocity decreases near

the central axis of the channel but increases near the wallswhen 119877 increases (Figure 6(b))

Figure 7 demonstrates the effect of slip parameter 120574 on thevelocity profiles After fixing Reynolds number 119877 = 03 wevaried 120574 as 120574 = 06 07 08 1 We find that normal velocityincreases as 120574 increases It is also noted that longitudinalvelocity decreases near the walls but increases near centralaxis of the channel

8 Mathematical Problems in Engineering

Table 5 OHAM solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 minus000880011 minus107933 times 10minus6 00336596 minus754858 times 10minus8 00522402 minus774445 times 10minus9

02 minus0010881 minus22445 times 10minus6 00714091 minus173419 times 10minus7 01074223 minus251485 times 10minus8

03 0000449012 minus345224 times 10minus6 0117324 minus300407 times 10minus7 0168479 minus568899 times 10minus8

04 00318278 minus441058 times 10minus6 0175449 minus432397 times 10minus7 0238322 minus98056 times 10minus8

05 00898133 minus447787 times 10minus6 0249785 minus500627 times 10minus7 0319832 minus129675 times 10minus7

06 018086 minus260114 times 10minus6 0344278 minus377978 times 10minus7 0415852 minus113733 times 10minus7

07 0311298 265136 times 10minus6 0462801 130706 times 10minus7 0529175 120785 times 10minus8

08 0487312 128583 times 10minus5 0609145 128166 times 10minus6 0662538 336724 times 10minus7

09 0714929 291822 times 10minus5 0787012 334515 times 10minus6 0818613 967763 times 10minus7

10 1 514019 times 10minus5 1 649074 times 10minus6 1 200176 times 10minus6

Table 6 RK4 solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 204594 times 10minus5 0 734794 times 10minus6 0 356291 times 10minus6

01 minus000880012 772953 times 10minus7 00336596 321066 times 10minus7 00522402 163598 times 10minus7

02 minus00108811 minus18406 times 10minus7 00714091 minus775305 times 10minus8 01074223 minus396814 times 10minus8

03 0000448983 482747 times 10minus8 0117324 208509 times 10minus8 0168479 10753 times 10minus8

04 00318277 minus154235 times 10minus8 0175449 minus643999 times 10minus9 0238322 minus328063 times 10minus9

05 00898132 514031 times 10minus9 0249785 121029 times 10minus9 0319832 450154 times 10minus10

06 018086 minus651866 times 10minus10 0344278 264494 times 10minus9 0415852 184079 times 10minus9

07 0311298 minus186837 times 10minus9 0462801 minus988382 times 10minus9 0529175 minus651979 times 10minus9

08 0487312 minus133003 times 10minus9 0609145 337336 times 10minus8 0662538 227322 times 10minus8

09 0714929 266400 times 10minus8 0787012 minus132232 times 10minus7 0818613 minus905714 times 10minus8

10 1 555277 times 10minus6 1 minus11671 times 10minus6 1 minus11767 times 10minus6

Table 7 Comparison of OHAM and RK4 solutions for various 119877 in case of slip and no-slip boundary

120578

In case of no-slip boundary In case of slip boundary|RK4 Solution minusOHAM Solution| |RK4 Solution minusOHAM Solution|

119877 = 01 119877 = 03 119877 = 05 119877 = 02 119877 = 03 119877 = 04

00 0 0 0 0 0 001 233147 times 10minus15 162249 times 10minus10 241725 times 10minus9 844379 times 10minus10 790896 times 10minus9 41479 times 10minus8

02 949241 times 10minus14 314228 times 10minus10 467809 times 10minus9 16672 times 10minus9 155832 times 10minus8 815527 times 10minus8

03 382361 times 10minus13 446143 times 10minus10 663065 times 10minus9 244889 times 10minus9 228102 times 10minus8 118952 times 10minus7

04 913492 times 10minus13 547932 times 10minus10 811705 times 10minus9 31663 times 10minus9 293539 times 10minus8 152339 times 10minus7

05 165451 times 10minus12 607476 times 10minus10 895202 times 10minus9 377514 times 10minus9 348026 times 10minus8 179574 times 10minus7

06 242317 times 10minus12 609434 times 10minus10 891445 times 10minus9 418241 times 10minus9 38327 times 10minus8 196532 times 10minus7

07 286218 times 10minus12 536833 times 10minus10 777928 times 10minus9 422088 times 10minus9 384569 times 10minus8 196009 times 10minus7

08 252039 times 10minus12 378563 times 10minus10 54278 times 10minus9 365208 times 10minus9 331085 times 10minus8 167861 times 10minus7

09 120232 times 10minus12 154202 times 10minus10 218716 times 10minus9 224454 times 10minus9 202732 times 10minus8 102385 times 10minus7

10 126281 times 10minus16 253545 times 10minus16 293337 times 10minus17 17436 times 10minus14 492715 times 10minus14 38179 times 10minus14

8 Conclusions

In this article we find the similarity solution for unsteadyaxisymmetric squeezing flow of incompressible Newtonianfluid between two circular plates We observed that thesimilarity solution exists only when distance between theplates varies as (119882119905 + 119878)

12 and squeezing flow occurs when

119882 lt 0 119878 gt 0 and (119882119905+119878) gt 0The key findings of the presentanalysis are as follows

In case of no-slip at boundary

(i) It has been found that increase in Reynolds number 119877increases the normal velocity

Mathematical Problems in Engineering 9

00010

00008

00006

00004

00002

02 04 06 08 10

120578

Res

R = 02

R = 03

R = 04

Figure 3 OHAM residuals at various values of 119877 in case of slip boundary

00001

000008

000006

000004

000002

120574 = 05

120574 = 06

120574 = 07

02 04 06 08 10

120578

Res

Figure 4 OHAM residuals at various values of 120574 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

(120578

120578

)

R = 01

R = 1

R = 2

R = 3

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

R = 01

R = 1

R = 2

R = 3

998400 (120578)

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 5 Velocity profiles for various values of 119877 = 01 1 2 3 in case of no-slip boundary

10 Mathematical Problems in Engineering

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

R = 02

R = 06

R = 1

R = 15

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

R = 02

R = 06

R = 1

R = 15

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 6 Velocity profiles for various values of 119877 = 02 06 1 15 fixing 120574 = 1 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

120574 = 06120574 = 07

120574 = 08

120574 = 1

(a) The effect of 120574 on the Normal velocity profiles

120574 = 06120574 = 07

120574 = 08

120574 = 1

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

(b) The effect of 120574 on the longitudinal velocity profiles

Figure 7 Velocity profiles for various values of 120574 = 06 07 08 1 fixing 119877 = 03 in case of slip boundary

(ii) It has been observed that normal velocity increasesmonotonically from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time

(iii) It has been seen that longitudinal velocity deceasesnear the walls and increases near the central axis ofthe channel

In case of slip at boundary

(i) It has been noted that after fixing slip parameter120574 and varying the Reynolds number 119877 the normalvelocity profile decreases with the increase in 119877 Alsothe longitudinal velocity increases near the walls butdecreases near the central axis of the channel

(ii) It has been examined that for a fixed Reynoldsnumber 119877when we vary slip parameter 120574 the normalvelocity increases with the increase in 120574 Also the

longitudinal velocity decreases near the walls andincreases near the central axis of the channel

(iii) It has been investigated that Reynolds number 119877 andslip parameter 120574 have opposite effects on the normaland longitudinal velocity components

In case of slip versus no-slip boundary

(i) It has been observed that Reynolds number 119877 hasopposite behavior on the normal velocity in case ofslip and no-slip boundaries

(ii) It has been also noticed that Reynolds number 119877 hasopposite effect on the longitudinal velocity near thecentral axis of the channel while near the wall longi-tudinal velocity increases in case of slip boundary anddecease in no-slip boundary This is in conformanceto [33]

Mathematical Problems in Engineering 11

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] Q K Ghori M Ahmed and A M Siddiqui ldquoApplicationof homotopy perturbation method to squeezing flow of anewtonian fluidrdquo International Journal of Nonlinear Sciencesand Numerical Simulation vol 8 no 2 pp 179ndash184 2007

[2] X J Ran Q Y Zhu and Y Li ldquoAn explicit series solution ofthe squeezing flow between two infinite plates by means ofthe homotopy analysis methodrdquo Communications in NonlinearScience and Numerical Simulation vol 14 no 1 pp 119ndash1322009

[3] M J Stefan ldquoVersuch Uber die scheinbare adhasionrdquo Sitzungs-berichteOsterreichische Akademie der Wissenschaften in WienMathematisch-Naturwissenschaftliche Klasse vol 69 pp 713ndash721 1874

[4] J F Thorpe and W A Shaw Eds Developments in Theoreticaland Applied Mechanics vol 3 Pergamon Press Oxford UK1967

[5] P S Gupta and A S Gupta ldquoSqueezing flow between parallelplatesrdquoWear vol 45 no 2 pp 177ndash185 1977

[6] A F Elkouh ldquoFluid inertia effects in squeeze film between twoplane annulirdquo Transactions of the ASME Journal of Tribologyvol 106 no 2 pp 223ndash227 1984

[7] R L Verma ldquoA numerical solution for squeezing flow betweenparallel channelsrdquoWear vol 72 no 1 pp 89ndash95 1981

[8] P Singh V Radhakrishnan and K A Narayan ldquoSqueezing flowbetween parallel platesrdquo Ingenieur-Archiv vol 60 no 4 pp 274ndash281 1990

[9] P J Leider and R B Bird ldquoSqueezing flow between paralleldisks I Theoretical analysisrdquo Industrial and Engineering Chem-istry Fundamentals vol 13 no 4 pp 336ndash341 1974

[10] G Domairry and A Aziz ldquoApproximate analysis of MHDdqueeze flow between two parallel disks with suction orinjection by homotopy perturbation methodrdquo MathematicalProblems in Engineering vol 2009 Article ID 603916 19 pages2009

[11] S Islam H Khan I A Shah and G Zaman ldquoAn axisymmetricsqueezing fluid flow between the two infinite parallel plates in aporous medium channelrdquo Mathematical Problems in Engineer-ing vol 2011 Article ID 349803 10 pages 2011

[12] I Ullah H Khan and M T Rahim ldquoApproximation of firstgrade MHD squeezing fluid flow with slip boundary conditionusing DTM and OHAMrdquo Mathematical Problems in Engineer-ing vol 2013 Article ID 816262 9 pages 2013

[13] A M Siddiqui S Irum and A R Ansari ldquoUnsteady squeezingflowof a viscousMHDfluid between parallel platesrdquoMathemat-ical Modelling and Analysis vol 13 pp 565ndash576 2008

[14] R J Grimm ldquoSqueezing flows of Newtonian liquid films ananalysis including fluid inertiardquo Applied Scientific Research vol32 no 2 pp 149ndash166 1976

[15] M M Rashidi A M Siddiqui and M T Rastegari ldquoAnalyticalsolution of squeezing flow between two circular platesrdquo Interna-tional Journal for ComputationalMethods in Engineering Scienceand Mechanics vol 13 no 5 pp 342ndash349 2012

[16] H M Laun M Rady and O Hassager ldquoAnalytical solutions forsqueeze flow with partial wall sliprdquo Journal of Non-NewtonianFluid Mechanics vol 81 no 1-2 pp 1ndash15 1999

[17] S Ishizawa ldquoThe unsteady flow between two parallel discs witharbitrary varying gap widthrdquo Bulletin of the Japan Society ofMechanical Engineers vol 9 no 35 pp 533ndash550 1966

[18] C le Roux ldquoExistence and uniqueness of the flow of second-grade fluids with slip boundary conditionsrdquoArchive for RationalMechanics and Analysis vol 148 no 4 pp 309ndash356 1999

[19] L J Rhooades R Resnic T OrsquoBradovich and S StegmanldquoAbrasive flow machining of cylinder heads and its positiveeffects on performance and cost characteristicsrdquo Tech RepDearborn Mich USA 1996

[20] C L M H Navier Memoirs de lrsquoCademie Royale des Sciencesde lrsquoInstitut de France vol 1 Royale des Sciences de lrsquoInstitut deFrance 1823

[21] A Ebaid ldquoEffects of magnetic field and wall slip conditions onthe peristaltic transport of a Newtonian fluid in an asymmetricchannelrdquo Physics Letters Section A General Atomic and SolidState Physics vol 372 no 24 pp 4493ndash4499 2008

[22] J-H He ldquoSome asymptotic methods for strongly nonlinearequationsrdquo International Journal of Modern Physics B Con-densed Matter Physics Statistical Physics Applied Physics vol20 no 10 pp 1141ndash1199 2006

[23] S J Liao Proposed homotopy analysis techniques for the solutionof nonlinear problems [PhD thesis] Jiao Tong UniversityShanghai China 1992

[24] J-H He ldquoHomotopy perturbation method for solving bound-ary value problemsrdquo Physics Letters A vol 350 no 1-2 pp 87ndash88 2006

[25] J-H He ldquoApproximate analytical solution for seepage flowwithfractional derivatives in porous mediardquo Computer Methods inApplied Mechanics and Engineering vol 167 no 1-2 pp 57ndash681998

[26] VMarinca andNHerisanu ldquoApplication of optimalHomotopyasymptotic method for solving nonlinear equations arising inheat transferrdquo International Communications in Heat and MassTransfer vol 35 no 6 pp 710ndash715 2008

[27] V Marinca N Herisanu C Bota and B Marinca ldquoAn optimalhomotopy asymptotic method applied to the steady flow of afourth-grade fluid past a porous platerdquo Applied MathematicsLetters vol 22 no 2 pp 245ndash251 2009

[28] V Marinca N Herisanu and I Nemes ldquoOptimal homotopyasymptotic method with application to thin film flowrdquo CentralEuropean Journal of Physics vol 6 no 3 pp 648ndash653 2008

[29] J Ali S Islam and G Zaman ldquoThe solution of multipointboundary value problems by the optimal Homotopy asymptoticMethodrdquoComputers andMathematics with Applications vol 59no 6 pp 2000ndash2006 2010

[30] M Idrees S Islam and S Haq ldquoApplication of the optimalhomotopy asymptotic method to squeezing flowrdquo Computersand Mathematics with Applications vol 59 no 12 pp 3858ndash3866 2010

[31] J Ali S Islam H Khan and S I Shah ldquoThe optimal homotopyasymptotic method for the solution of higher-order boundaryvalue problems in finite domainsrdquo Abstract and Applied Analy-sis vol 2012 Article ID 401217 14 pages 2012

[32] H Khan S Islam J Ali and I Ali Shah ldquoComparison ofdifferent analytic solutions to axisymmetric squeezing fluidflow between two infinite parallel plates with slip boundaryconditionsrdquo Abstract and Applied Analysis vol 2012 Article ID835268 18 pages 2012

[33] T C Papanastasiou G C Georgiou and A N AlexandrouViscous Fluid Flow Library of Congress Cataloging CRC PressNew York NY USA 1999

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 3: Research Article Analysis of Unsteady Axisymmetric ...downloads.hindawi.com/journals/mpe/2015/860857.pdfResearch Article Analysis of Unsteady Axisymmetric Squeezing Fluid Flow with

Mathematical Problems in Engineering 3

The boundary conditions on 119906119903and 119906

119911are

119906119903= 0 119906

119911= V119908(119905) at 120578 = 1

120597119906119903

120597120578= 0 119906

119911= 0 at 120578 = 0

(13)

After eliminating the generalized pressure between (11) and(12) we obtained

120588 [120597Ω

120597119905+ 119906119903

120597Ω

120597119903+119906119911

119904

120597Ω

120597120578minus119906119903

119903Ω] = 120583 [nabla

2

Ω minusΩ

1199032] (14)

where nabla2 is the Laplacian operatorDefining velocity components as [5]

119906119903= minus

119903

2119904 (119905)V119908(119905) 1198921015840

(120578)

119906119911= V119908(119905) 119892 (120578)

(15)

we see that (10) is identically satisfied and (14) becomes

1198894

119892

1198891205784+ 119877[(120578 minus 119892)

1198893

119892

1198891205783+ 2

1198892

119892

1198891205782] minus 119876

1198892

119892

1198891205782= 0 (16)

where

119877 =119904V119908

120592 119876 =

1199042

120592V119908

119889V119908

119889119905 (17)

Here both119877 and119876 are functions of 119905 but we consider119877 and119876constants for similarity solution Since V

119908= 119889119904119889119905 Integrate

first equation of (17) we get

119904 (119905) = (119882119905 + 119878)12

(18)

where119882 and 119878 are constantsTheplatesmove away from eachother symmetrically with respect to 120578 when 119882 gt 0 and 119878 gt0 Also the plates approach to each other and squeezing flowexists with similar velocity profiles when119882 lt 0 119878 gt 0 and119904(119905) gt 0 From (17) and (18) it follows that 119876 = minus119877 Then (16)becomes

1198894

119892

1198891205784+ 119877[(120578 minus 119892)

1198893

119892

1198891205783+ 3

1198892

119892

1198891205782] = 0 (19)

Using (13) and (15) we determine the boundary conditions incase of no-slip and slip at the upper plate as follows

119892 (1) = 1 1198921015840

(1) = 0

119892 (0) = 0 11989210158401015840

(0) = 0(No-slip at the wall) (20)

119892 (1) = 1 1198921015840

(1) = 12057411989210158401015840

(1)

119892 (0) = 0 11989210158401015840

(0) = 0(Slip at the wall) (21)

4 Basic Theory of OHAM [26 29ndash32]Let us apply OHAM to the following differential equation

I [V (119909)] + 119891 (119909) + alefsym [V (119909)] = 0 119861 (V119889V119889119909

) = 0 (22)

where 119909 represents an independent variable V(119909) is unknownfunction and 119891(119909) is known function 119861alefsymI are boundarynonlinear and linear operators respectively

According to OHAM we construct Homotopy Φ(119909 119901)R times [0 1] rarr R which satisfies

(1 minus 119901) [I (Φ (119909 119901)) + 119891 (119909)]

= ℎ (119901) [I (Φ (119909 119901)) + 119891 (119909) + alefsym (Φ (119909 119901))]

119861(Φ (119909 119901) 120597Φ (119909 119901)

120597119909) = 0

(23)

where 119909 isin R and 119901 isin [0 1] is an embedding parameterℎ(119901) is a nonlinear auxiliary function for 119901 = 0 ℎ(0) = 0

and Φ(119909 119901) is an unknown function Clearly when 119901 = 0

and 119901 = 1 it holds that Φ(119909 0) = V0(119909) and Φ(119909 1) = V(119909)

respectivelyThus as 119901 varies from 0 to 1 the solution Φ(119909 119901)

approaches from V0(119909) to V(119909)

We choose the auxiliary function ℎ(119901) in the form of

ℎ (119901) =

119898

sum

119896=0

119901119896

119862119896 (24)

where 119862119896are convergence controlling constants to be deter-

minedTo obtain an approximate solution we expandΦ(119909 119901 119862

119894)

in a Taylor series about 119901 as follows

Φ(119909 119901 119862119894) = V0(119909) +

119898

sum

119896=1

V119896(1198621 1198622 119862

119896) 119901119896

(25)

Substituting (25) into (23) and equating the coefficients of likepowers of 119901 we obtain the following equationsThe zeroth-order problem is

I [V0(119909)] + 119891 (119909) = 0 119861 (V

0119889V0

119889119909) = 0 (26)

First-order problem is

I [V1(119909)] + 119891 (119909) = 119862

1alefsym0[V0(119909)] 119861 (V

1119889V1

119889119909) = 0

(27)

Second-order problem is

I [V2(119909)] +I [V

1(119909)]

= 1198622alefsym0[V0(119909)] + 119862

1I [V1(119909)] + alefsym

1[V0(119909) V

1(119909)]

119861 (V2119889V2

119889119909) = 0

(28)

4 Mathematical Problems in Engineering

The general equations for V119896(119909) are given by

I [V119896(119909)] minusI [V

119896minus1(119909)]

= 1198621alefsym0[V0(119909)]

+

119896minus1

sum

119894=1

119862119894I [V119896minus119894

(119909)]

+alefsym119896minus119894

[V0(119909) V

1(119909) V

119896minus1(119909)]

119861 (V119896119889V119896

119889119909) = 0

119896 = 2 3

(29)

where the coefficient of 119901119898 in the expansion of alefsym(Φ(119909 119901))about 119901 is alefsym

119898[V0(119909) V1(119909) V

119898minus1(119909)]

alefsym(Φ (119909 119901 119862119894))

= alefsym0[V0(119909)] +

119898

sum

119898=1

alefsym119898[V0(119909) V

1(119909) V

119898minus1(119909)] 119901

119898

(30)It is noted that the convergence of the series (25) depends

upon119862119896 For convergence at119901 = 1 the119898th order approxima-

tion V is

V (119909 1198621 1198622 119862

119898) = V0(119909) +

119898

sum

119895=1

V119895(119909 1198621 1198622 119862

119895)

(31)Substituting (31) in (22) the expression for residual is

R (119909 1198621 1198622 119862

119898)

= I [V (119909 1198621 1198622 119862

119898)] + 119891 (119909)

+ alefsym [V (119909 1198621 1198622 119862

119898)]

(32)

If R = 0 then V will be the exact solution but usually thisdoes not happen in nonlinear problems

There are various methods to find the optimal values of119862119894 119894 = 1 2 We apply the method of least square and

Galerkinrsquos method in the following mannerIn method of least square

119869 (119909 1198621 1198622 119862

119898) = int

119887

119886

R2

(119909 1198621 1198622 119862

119898) 119889119909 (33)

Minimizing 119869(119909 1198621 1198622 119862

119898) we have

120597119869

120597119862119894

= 0 119894 = 1 2 119898 (34)

In Galerkinrsquos method we solve the following system for119862119894(119894 = 1 2 119898)

int

119887

119886

R120597V120597119862119894

119889119909 = 0 119894 = 1 2 119898 (35)

To find appropriate 119862119894(119894 = 1 2 119898) we choose 119886 and 119887 in

the domain of the problem Approximate solution of order119898is well-determined with these known constants

5 Application of OHAM in Case ofNo-Slip Boundary

Using (19) and (20) various order problems are as followsZeroth-order problem

V(119894V)0

(120578) = 0

V0(0) = 0 V10158401015840

0(0) = 0 V

0(1) = 1 V1015840

0(1) = 0

(36)

First-order problem

V(119894V)1

(120578) = 31198621119877V101584010158400(120578) + 119862

1119877120578V1015840101584010158400(120578) minus 119862

1119877V0(120578) V1015840101584010158400(120578)

+ V(119894V)0

(120578) + 1198621V(119894V)0

(120578)

V1(0) = 0 V10158401015840

1(0) = 0 V

1(1) = 0 V1015840

1(1) = 0

(37)

Second-order problem

V(119894V)2

(120578) = 31198621119877V101584010158401(120578) minus 119862

1119877V1(120578) V1015840101584010158400(120578) + 119862

1119877120578V1015840101584010158401(120578)

minus 1198621119877V0(120578) V1015840101584010158401(120578) + V(119894V)

1(120578) + 119862

1V(119894V)1

(120578)

V2(0) = 0 V10158401015840

2(0) = 0 V

2(1) = 0 V1015840

2(1) = 0

(38)

Third-order problem

V(119894V)3

(120578) = 31198621119877V101584010158402(120578) minus 119862

1119877V2(120578) V1015840101584010158400(120578)

minus 1198621119877V1(120578) V1015840101584010158401(120578) + 119862

1119877120578V1015840101584010158402(120578)

minus 1198621119877V0(120578) V1015840101584010158402(120578) + V(119894V)

2(120578)

+ 1198621V(119894V)2

(120578)

V3(0) = 0 V10158401015840

3(0) = 0 V

3(1) = 0 V1015840

3(1) = 0

(39)

Fourth-order problem

V(119894V)4

(120578) = 31198621119877V101584010158403(120578) minus 119862

1119877V3(120578) V1015840101584010158400(120578)

minus 1198621119877V2(120578) V1015840101584010158401(120578) minus 119862

1119877V1(120578) V1015840101584010158402(120578)

+ 1198621119877120578V1015840101584010158403(120578) minus 119862

1119877V0(120578) V1015840101584010158403(120578)

+ V(119894V)3

(120578) + 1198621V(119894V)3

(120578)

V4(0) = 0 V10158401015840

4(0) = 0 V

4(1) = 0 V1015840

4(1) = 0

(40)

By considering fourth-order solution we have

V (120578) = V0(120578) + V

1(120578) + V

2(120578) + V

3(120578) + V

4(120578) (41)

The residual of the problem is

R =1198894V (120578)1198891205784

+ 119877[(120578 minus V (120578))1198893V (120578)1198891205783

+ 31198892V (120578)1198891205782

] (42)

Mathematical Problems in Engineering 5

We apply Galerkinrsquos method to find constant 1198621as follows

int

1

0

R120597V1205971198621

119889119909 = 0 (43)

Solving (43) and keeping 119877 = 01 we get

1198621= 101328 (44)

Using above value of 1198621 the approximate solution is

V (120578) =1

2(3120578 minus 120578

3

)

+1

560(374914120578 minus 739694120578

3

+ 3546481205785

+01013281205787

)

+1

776160(956941120578 minus 430222120578

3

+ 514141205785

minus122331205787

minus 5534111205789

minus 19405312057811

)

+1

565045(254414120578 + 143644120578

3

minus 3711641205785

+ 2502321205787

minus 8133381205789

+ 25990112057811

+6961012057813

+ 2296212057815

)

+1

722737(153702120578 minus 187989 times 10

8

1205783

+ 193991 times 108

1205785

+ 828837 times 107

1205787

minus 803587 times 107

1205789

minus 292844 times 107

12057811

+ 167506 times 107

12057813

minus 966217 times 106

12057815

minus16529 times 106

12057817

minus 48038912057819

)

(45)

6 Application of OHAM in Case ofSlip Boundary

Using (19) and (21) different order problems are as followsZeroth-order problem

119906(119894V)0

(120578) = 0

1199060(0) = 0 119906

10158401015840

0(0) = 0 119906

0(1) = 1 119906

1015840

0(1) = 120574119906

10158401015840

0(1)

(46)

First-order problem

119906(119894V)1

(120578)

= 3119862111987711990610158401015840

0(120578) + 119862

1119877120578119906101584010158401015840

0(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

0(120578)

+ 119906(119894V)0

(120578) + 1198621119906(119894V)0

(120578)

1199061(0) = 0 119906

10158401015840

1(0) = 0 119906

1(1) = 0 119906

1015840

1(1) = 120574119906

10158401015840

1(1)

(47)

Second-order problem

119906(119894V)2

(120578) = 3119862111987711990610158401015840

1(120578) minus 119862

11198771199061(120578) 119906101584010158401015840

0(120578)

+ 1198621119877120578119906101584010158401015840

1(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

1(120578)

+ 119906(119894V)1

(120578) + 1198621119906(119894V)1

(120578)

1199062(0) = 0 119906

10158401015840

2(0) = 0 119906

2(1) = 0 119906

1015840

2(1) = 120574119906

10158401015840

2(1)

(48)

Third-order problem

119906(119894V)3

(120578) = 3119862111987711990610158401015840

2(120578) minus 119862

11198771199062(120578) 119906101584010158401015840

0(120578)

minus 11986211198771199061(120578) 119906101584010158401015840

1(120578) + 119862

1119877120578119906101584010158401015840

2(120578)

minus 11986211198771199060(120578) 119906101584010158401015840

2(120578) + 119906

(119894V)2

(120578)

+ 1198621119906(119894V)2

(120578)

1199063(0) = 0 119906

10158401015840

3(0) = 0 119906

3(1) = 0 119906

1015840

3(1) = 120574119906

10158401015840

3(1)

(49)

Fourth-order problem

119906(119894V)4

(120578) = 3119862111987711990610158401015840

3(120578) minus 119862

11198771199063(120578) 119906101584010158401015840

0(120578)

minus 11986211198771199062(120578) 119906101584010158401015840

1(120578) minus 119862

11198771199061(120578) 119906101584010158401015840

2(120578)

+ 1198621119877120578119906101584010158401015840

3(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

3(120578)

+ 119906(119894V)3

(120578) + 1198621119906(119894V)3

(120578)

1199064(0) = 0 119906

10158401015840

4(0) = 0 119906

4(1) = 0 119906

1015840

4(1) = 120574119906

10158401015840

4(1)

(50)

By considering fourth-order solution we have

(120578) =

4

sum

119894=0

119906119894(120578 1198621) (51)

The residual of the problem is

120577 =1198894

(120578)

1198891205784+ 119877[(120578 minus (120578))

1198893

(120578)

1198891205783+ 3

1198892

(120578)

1198891205782] (52)

We apply Galerkinrsquos method to find constant 1198621as follows

int

1

0

120577120597

1205971198621

119889119909 = 0 (53)

Solving (53) and taking 119877 = 02 and 120574 = 1 we get

1198621= minus106872 (54)

6 Mathematical Problems in Engineering

Table 1 OHAM solutions along with residuals for various 119877 in case of no-slip boundary

120578119877 = 01 119877 = 03 119877 = 05

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 0150158 537334 times 10minus11 0151534 minus669283 times 10minus9 0152999 minus623412 times 10minus8

02 0297237 314675 times 10minus10 0299827 166025 times 10minus9 0302582 658388 times 10minus8

03 0438170 851847 times 10minus10 0441661 233504 times 10minus8 0445373 342092 times 10minus7

04 0569900 151431 times 10minus9 0573869 393945 times 10minus8 0578082 493942 times 10minus7

05 0689397 190824 times 10minus9 0693354 281614 times 10minus8 0697548 242092 times 10minus7

06 0793661 133301 times 10minus9 0797122 minus319654 times 10minus8 0800780 minus653046 times 10minus7

07 0879734 minus159255 times 10minus9 0882300 minus221911 times 10minus7 0885004 minus323415 times 10minus6

08 0944705 minus102814 times 10minus8 0946167 minus884707 times 10minus7 0947702 minus12206 times 10minus509 0985722 minus3446 times 10minus8 0986180 minus321268 times 10minus6 0986659 minus441008 times 10minus5

10 1 minus103509 times 10minus7 1 minus111826 times 10minus5 1 minus154022 times 10minus4

Using above value of 1198621 the approximate solution is

(120578) =1

2(3120578 + 120578

3

)

+1

4480(minus113284120578 + 151758120578

3

minus 3890121205785

+04274861205787

)

+1

248372(minus156189120578 + 273825120578

3

minus 1608081205785

+4595261205787

minus 2814271205789

+ 34538712057811

)

+1

723257(minus983844120578 + 1149 times 10

8

1205783

minus 335388 times 107

1205785

+ 404185 times 107

1205787

minus 268964 times 107

1205789

+ 365673 times 106

12057811

minus15753812057813

+ 1724212057815

)

+1

370042(minus197195 times 10

13

120578 + 425372 times 1013

1205783

minus 294213 times 1013

1205785

+ 606765 times 1012

1205787

minus 139445 times 1012

1205789

+ 247846 times 1012

12057811

minus 597188 times 1011

12057813

+ 508263 times 1010

12057815

minus162422 times 109

12057817

+ 152182 times 107

12057819

)

(55)

7 Results and Discussions

In this article we considered the unsteady axisymmetric flowof nonconducting incompressible Newtonian fluid betweentwo circular plates The resulting nonlinear boundary valueproblems are solved with OHAM and fourth-order Runge-Kutta method using Mathematica 70

minus000001

minus000002

minus000003

minus000004

R = 01

R = 03

R = 05

02 04 06 08 10

120578

Res

Figure 2 OHAM residuals at various values of 119877 in case of no-slipboundary

Tables 1 3 and 5 reflect OHAM solutions along withresiduals in case of no-sip and slip boundaries for variousvalues of Reynolds number 119877 and slip parameter 120574 AlsoTables 2 4 and 6 represent RK4 solutions alongwith residualsin case of no-slip and slip boundaries for various values of 119877and 120574 All the tables demonstrate that results obtained usingOHAM are in agreement with RK4 by means of residualsIn addition to above mentioned tables Table 7 shows thecomparison of solutions obtained from OHAM and RK4 forvarious values of Reynolds number 119877

Furthermore Figures 2 3 and 4 indicate the OHAMresiduals in case of no-slip and slip boundaries for variousvalues of 119877 and 120574

The effect of Reynolds number 119877 on velocity profilesin case of no-slip boundary is shown in Figure 5 In theseprofiles we varied 119877 as 119877 = 01 1 2 3 and observed that thenormal velocity is increased with the increase of Reynoldsnumber (Figure 5(a)) It is also noted that the normal velocitymonotonically increases from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time Figure 5(b) describes the impact of119877 on the longitudinal velocity in case of no-slip boundary It

Mathematical Problems in Engineering 7

Table 2 RK4 solutions along with residuals for various 119877 in case of no-slip boundary

120578119877 = 01 119877 = 03 119877 = 05

Solution Residual Solution Residual Solution Residual00 0 288535 times 10minus7 0 966906 times 10minus5 0 530549 times 10minus4

01 0150158 388278 times 10minus8 0151534 571393 times 10minus6 0152999 304600 times 10minus5

02 0297237 minus997789 times 10minus9 0299827 minus141434 times 10minus6 0302582 minus752772 times 10minus6

03 0438170 293767 times 10minus9 0441661 395997 times 10minus7 0445373 210296 times 10minus6

04 0569900 minus751436 times 10minus10 0573869 minus115203 times 10minus7 0578082 minus618081 times 10minus7

05 0689397 minus471705 times 10minus10 0693354 minus96761 times 10minus9 0697548 minus295916 times 10minus8

06 0793661 223004 times 10minus9 0797122 14593 times 10minus7 0800780 714133 times 10minus7

07 0879734 minus718245 times 10minus9 0882300 minus486272 times 10minus7 0885004 minus239245 times 10minus6

08 0944705 268744 times 10minus8 0946167 177796 times 10minus6 0947702 870664 times 10minus6

09 0985722 minus111599 times 10minus7 0986180 minus728952 times 10minus6 0986659 356157 times 10minus5

10 1 minus271683 times 10minus6 1 minus147695 times 10minus4 1 690088 times 10minus4

Table 3 OHAM solutions along with residuals for various values of 119877 in case of slip boundary

120578119877 = 02 119877 = 03 119877 = 04

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 0072692 318345 times 10minus8 0071220 315621 times 10minus7 0069593 174626 times 10minus6

02 0147091 minus172789 times 10minus8 0144268 minus914579 times 10minus8 0141147 minus113148 times 10minus7

03 0224893 minus196995 times 10minus7 0220953 minus166461 times 10minus6 0216599 minus779061 times 10minus6

04 0307773 minus494482 times 10minus7 0303048 minus429173 times 10minus6 0297832 minus207401 times 10minus5

05 0397370 minus799437 times 10minus7 0392277 minus699917 times 10minus6 0386657 minus341643 times 10minus5

06 0495283 minus85784 times 10minus7 0490289 minus75499 times 10minus6 0484784 minus370689 times 10minus5

07 0603056 minus21053 times 10minus7 0598654 minus191141 times 10minus6 0593805 minus96958 times 10minus6

08 0722172 187065 times 10minus6 0718841 162936 times 10minus5 0715176 791082 times 10minus5

09 0854042 640576 times 10minus6 0852211 559717 times 10minus5 0850196 272708 times 10minus4

10 1 146076 times 10minus5 1 127671 times 10minus4 1 622242 times 10minus4

Table 4 RK4 solutions along with residuals for various values of 119877 in case of slip boundary

120578119877 = 02 119877 = 03 119877 = 04

Solution Residual Solution Residual Solution Residual00 0 412179 times 10minus6 0 756765 times 10minus6 0 362897 times 10minus6

01 0072692 174999 times 10minus7 0071220 216125 times 10minus7 0069593 minus349994 times 10minus7

02 0147091 minus421503 times 10minus8 0144268 minus497423 times 10minus8 0141147 953266 times 10minus8

03 0224893 112883 times 10minus8 0220953 122294 times 10minus8 0216598 minus30735 times 10minus8

04 0307773 minus352272 times 10minus9 0303048 minus430187 times 10minus9 0297832 72493 times 10minus9

05 0397370 780684 times 10minus10 0392276 306668 times 10minus9 0386657 846409 times 10minus9

06 0495283 110391 times 10minus9 0490289 minus528586 times 10minus9 0484783 minus336828 times 10minus8

07 0603056 minus435033 times 10minus9 0598654 154159 times 10minus8 0593805 106816 times 10minus7

08 0722172 144198 times 10minus8 0718841 minus605873 times 10minus8 0715175 minus398519 times 10minus7

09 0854042 minus553933 times 10minus8 0852211 259233 times 10minus7 0850196 165518 times 10minus6

10 1 minus197434 times 10minus7 1 796168 times 10minus6 1 389126 times 10minus5

is experienced that this component of velocity deceases nearthe wall but increases near the central axis of the channel

The effect of Reynolds number 119877 on velocity profiles incase of slip boundary is depicted in Figure 6 In these profileswe fixed slip parameter 120574 = 1 and varied Reynolds number119877 as 119877 = 02 06 1 15 It is noted that the normal velocitydecreases as the Reynolds number increases (Figure 6(a))It is also observed that longitudinal velocity decreases near

the central axis of the channel but increases near the wallswhen 119877 increases (Figure 6(b))

Figure 7 demonstrates the effect of slip parameter 120574 on thevelocity profiles After fixing Reynolds number 119877 = 03 wevaried 120574 as 120574 = 06 07 08 1 We find that normal velocityincreases as 120574 increases It is also noted that longitudinalvelocity decreases near the walls but increases near centralaxis of the channel

8 Mathematical Problems in Engineering

Table 5 OHAM solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 minus000880011 minus107933 times 10minus6 00336596 minus754858 times 10minus8 00522402 minus774445 times 10minus9

02 minus0010881 minus22445 times 10minus6 00714091 minus173419 times 10minus7 01074223 minus251485 times 10minus8

03 0000449012 minus345224 times 10minus6 0117324 minus300407 times 10minus7 0168479 minus568899 times 10minus8

04 00318278 minus441058 times 10minus6 0175449 minus432397 times 10minus7 0238322 minus98056 times 10minus8

05 00898133 minus447787 times 10minus6 0249785 minus500627 times 10minus7 0319832 minus129675 times 10minus7

06 018086 minus260114 times 10minus6 0344278 minus377978 times 10minus7 0415852 minus113733 times 10minus7

07 0311298 265136 times 10minus6 0462801 130706 times 10minus7 0529175 120785 times 10minus8

08 0487312 128583 times 10minus5 0609145 128166 times 10minus6 0662538 336724 times 10minus7

09 0714929 291822 times 10minus5 0787012 334515 times 10minus6 0818613 967763 times 10minus7

10 1 514019 times 10minus5 1 649074 times 10minus6 1 200176 times 10minus6

Table 6 RK4 solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 204594 times 10minus5 0 734794 times 10minus6 0 356291 times 10minus6

01 minus000880012 772953 times 10minus7 00336596 321066 times 10minus7 00522402 163598 times 10minus7

02 minus00108811 minus18406 times 10minus7 00714091 minus775305 times 10minus8 01074223 minus396814 times 10minus8

03 0000448983 482747 times 10minus8 0117324 208509 times 10minus8 0168479 10753 times 10minus8

04 00318277 minus154235 times 10minus8 0175449 minus643999 times 10minus9 0238322 minus328063 times 10minus9

05 00898132 514031 times 10minus9 0249785 121029 times 10minus9 0319832 450154 times 10minus10

06 018086 minus651866 times 10minus10 0344278 264494 times 10minus9 0415852 184079 times 10minus9

07 0311298 minus186837 times 10minus9 0462801 minus988382 times 10minus9 0529175 minus651979 times 10minus9

08 0487312 minus133003 times 10minus9 0609145 337336 times 10minus8 0662538 227322 times 10minus8

09 0714929 266400 times 10minus8 0787012 minus132232 times 10minus7 0818613 minus905714 times 10minus8

10 1 555277 times 10minus6 1 minus11671 times 10minus6 1 minus11767 times 10minus6

Table 7 Comparison of OHAM and RK4 solutions for various 119877 in case of slip and no-slip boundary

120578

In case of no-slip boundary In case of slip boundary|RK4 Solution minusOHAM Solution| |RK4 Solution minusOHAM Solution|

119877 = 01 119877 = 03 119877 = 05 119877 = 02 119877 = 03 119877 = 04

00 0 0 0 0 0 001 233147 times 10minus15 162249 times 10minus10 241725 times 10minus9 844379 times 10minus10 790896 times 10minus9 41479 times 10minus8

02 949241 times 10minus14 314228 times 10minus10 467809 times 10minus9 16672 times 10minus9 155832 times 10minus8 815527 times 10minus8

03 382361 times 10minus13 446143 times 10minus10 663065 times 10minus9 244889 times 10minus9 228102 times 10minus8 118952 times 10minus7

04 913492 times 10minus13 547932 times 10minus10 811705 times 10minus9 31663 times 10minus9 293539 times 10minus8 152339 times 10minus7

05 165451 times 10minus12 607476 times 10minus10 895202 times 10minus9 377514 times 10minus9 348026 times 10minus8 179574 times 10minus7

06 242317 times 10minus12 609434 times 10minus10 891445 times 10minus9 418241 times 10minus9 38327 times 10minus8 196532 times 10minus7

07 286218 times 10minus12 536833 times 10minus10 777928 times 10minus9 422088 times 10minus9 384569 times 10minus8 196009 times 10minus7

08 252039 times 10minus12 378563 times 10minus10 54278 times 10minus9 365208 times 10minus9 331085 times 10minus8 167861 times 10minus7

09 120232 times 10minus12 154202 times 10minus10 218716 times 10minus9 224454 times 10minus9 202732 times 10minus8 102385 times 10minus7

10 126281 times 10minus16 253545 times 10minus16 293337 times 10minus17 17436 times 10minus14 492715 times 10minus14 38179 times 10minus14

8 Conclusions

In this article we find the similarity solution for unsteadyaxisymmetric squeezing flow of incompressible Newtonianfluid between two circular plates We observed that thesimilarity solution exists only when distance between theplates varies as (119882119905 + 119878)

12 and squeezing flow occurs when

119882 lt 0 119878 gt 0 and (119882119905+119878) gt 0The key findings of the presentanalysis are as follows

In case of no-slip at boundary

(i) It has been found that increase in Reynolds number 119877increases the normal velocity

Mathematical Problems in Engineering 9

00010

00008

00006

00004

00002

02 04 06 08 10

120578

Res

R = 02

R = 03

R = 04

Figure 3 OHAM residuals at various values of 119877 in case of slip boundary

00001

000008

000006

000004

000002

120574 = 05

120574 = 06

120574 = 07

02 04 06 08 10

120578

Res

Figure 4 OHAM residuals at various values of 120574 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

(120578

120578

)

R = 01

R = 1

R = 2

R = 3

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

R = 01

R = 1

R = 2

R = 3

998400 (120578)

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 5 Velocity profiles for various values of 119877 = 01 1 2 3 in case of no-slip boundary

10 Mathematical Problems in Engineering

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

R = 02

R = 06

R = 1

R = 15

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

R = 02

R = 06

R = 1

R = 15

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 6 Velocity profiles for various values of 119877 = 02 06 1 15 fixing 120574 = 1 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

120574 = 06120574 = 07

120574 = 08

120574 = 1

(a) The effect of 120574 on the Normal velocity profiles

120574 = 06120574 = 07

120574 = 08

120574 = 1

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

(b) The effect of 120574 on the longitudinal velocity profiles

Figure 7 Velocity profiles for various values of 120574 = 06 07 08 1 fixing 119877 = 03 in case of slip boundary

(ii) It has been observed that normal velocity increasesmonotonically from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time

(iii) It has been seen that longitudinal velocity deceasesnear the walls and increases near the central axis ofthe channel

In case of slip at boundary

(i) It has been noted that after fixing slip parameter120574 and varying the Reynolds number 119877 the normalvelocity profile decreases with the increase in 119877 Alsothe longitudinal velocity increases near the walls butdecreases near the central axis of the channel

(ii) It has been examined that for a fixed Reynoldsnumber 119877when we vary slip parameter 120574 the normalvelocity increases with the increase in 120574 Also the

longitudinal velocity decreases near the walls andincreases near the central axis of the channel

(iii) It has been investigated that Reynolds number 119877 andslip parameter 120574 have opposite effects on the normaland longitudinal velocity components

In case of slip versus no-slip boundary

(i) It has been observed that Reynolds number 119877 hasopposite behavior on the normal velocity in case ofslip and no-slip boundaries

(ii) It has been also noticed that Reynolds number 119877 hasopposite effect on the longitudinal velocity near thecentral axis of the channel while near the wall longi-tudinal velocity increases in case of slip boundary anddecease in no-slip boundary This is in conformanceto [33]

Mathematical Problems in Engineering 11

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] Q K Ghori M Ahmed and A M Siddiqui ldquoApplicationof homotopy perturbation method to squeezing flow of anewtonian fluidrdquo International Journal of Nonlinear Sciencesand Numerical Simulation vol 8 no 2 pp 179ndash184 2007

[2] X J Ran Q Y Zhu and Y Li ldquoAn explicit series solution ofthe squeezing flow between two infinite plates by means ofthe homotopy analysis methodrdquo Communications in NonlinearScience and Numerical Simulation vol 14 no 1 pp 119ndash1322009

[3] M J Stefan ldquoVersuch Uber die scheinbare adhasionrdquo Sitzungs-berichteOsterreichische Akademie der Wissenschaften in WienMathematisch-Naturwissenschaftliche Klasse vol 69 pp 713ndash721 1874

[4] J F Thorpe and W A Shaw Eds Developments in Theoreticaland Applied Mechanics vol 3 Pergamon Press Oxford UK1967

[5] P S Gupta and A S Gupta ldquoSqueezing flow between parallelplatesrdquoWear vol 45 no 2 pp 177ndash185 1977

[6] A F Elkouh ldquoFluid inertia effects in squeeze film between twoplane annulirdquo Transactions of the ASME Journal of Tribologyvol 106 no 2 pp 223ndash227 1984

[7] R L Verma ldquoA numerical solution for squeezing flow betweenparallel channelsrdquoWear vol 72 no 1 pp 89ndash95 1981

[8] P Singh V Radhakrishnan and K A Narayan ldquoSqueezing flowbetween parallel platesrdquo Ingenieur-Archiv vol 60 no 4 pp 274ndash281 1990

[9] P J Leider and R B Bird ldquoSqueezing flow between paralleldisks I Theoretical analysisrdquo Industrial and Engineering Chem-istry Fundamentals vol 13 no 4 pp 336ndash341 1974

[10] G Domairry and A Aziz ldquoApproximate analysis of MHDdqueeze flow between two parallel disks with suction orinjection by homotopy perturbation methodrdquo MathematicalProblems in Engineering vol 2009 Article ID 603916 19 pages2009

[11] S Islam H Khan I A Shah and G Zaman ldquoAn axisymmetricsqueezing fluid flow between the two infinite parallel plates in aporous medium channelrdquo Mathematical Problems in Engineer-ing vol 2011 Article ID 349803 10 pages 2011

[12] I Ullah H Khan and M T Rahim ldquoApproximation of firstgrade MHD squeezing fluid flow with slip boundary conditionusing DTM and OHAMrdquo Mathematical Problems in Engineer-ing vol 2013 Article ID 816262 9 pages 2013

[13] A M Siddiqui S Irum and A R Ansari ldquoUnsteady squeezingflowof a viscousMHDfluid between parallel platesrdquoMathemat-ical Modelling and Analysis vol 13 pp 565ndash576 2008

[14] R J Grimm ldquoSqueezing flows of Newtonian liquid films ananalysis including fluid inertiardquo Applied Scientific Research vol32 no 2 pp 149ndash166 1976

[15] M M Rashidi A M Siddiqui and M T Rastegari ldquoAnalyticalsolution of squeezing flow between two circular platesrdquo Interna-tional Journal for ComputationalMethods in Engineering Scienceand Mechanics vol 13 no 5 pp 342ndash349 2012

[16] H M Laun M Rady and O Hassager ldquoAnalytical solutions forsqueeze flow with partial wall sliprdquo Journal of Non-NewtonianFluid Mechanics vol 81 no 1-2 pp 1ndash15 1999

[17] S Ishizawa ldquoThe unsteady flow between two parallel discs witharbitrary varying gap widthrdquo Bulletin of the Japan Society ofMechanical Engineers vol 9 no 35 pp 533ndash550 1966

[18] C le Roux ldquoExistence and uniqueness of the flow of second-grade fluids with slip boundary conditionsrdquoArchive for RationalMechanics and Analysis vol 148 no 4 pp 309ndash356 1999

[19] L J Rhooades R Resnic T OrsquoBradovich and S StegmanldquoAbrasive flow machining of cylinder heads and its positiveeffects on performance and cost characteristicsrdquo Tech RepDearborn Mich USA 1996

[20] C L M H Navier Memoirs de lrsquoCademie Royale des Sciencesde lrsquoInstitut de France vol 1 Royale des Sciences de lrsquoInstitut deFrance 1823

[21] A Ebaid ldquoEffects of magnetic field and wall slip conditions onthe peristaltic transport of a Newtonian fluid in an asymmetricchannelrdquo Physics Letters Section A General Atomic and SolidState Physics vol 372 no 24 pp 4493ndash4499 2008

[22] J-H He ldquoSome asymptotic methods for strongly nonlinearequationsrdquo International Journal of Modern Physics B Con-densed Matter Physics Statistical Physics Applied Physics vol20 no 10 pp 1141ndash1199 2006

[23] S J Liao Proposed homotopy analysis techniques for the solutionof nonlinear problems [PhD thesis] Jiao Tong UniversityShanghai China 1992

[24] J-H He ldquoHomotopy perturbation method for solving bound-ary value problemsrdquo Physics Letters A vol 350 no 1-2 pp 87ndash88 2006

[25] J-H He ldquoApproximate analytical solution for seepage flowwithfractional derivatives in porous mediardquo Computer Methods inApplied Mechanics and Engineering vol 167 no 1-2 pp 57ndash681998

[26] VMarinca andNHerisanu ldquoApplication of optimalHomotopyasymptotic method for solving nonlinear equations arising inheat transferrdquo International Communications in Heat and MassTransfer vol 35 no 6 pp 710ndash715 2008

[27] V Marinca N Herisanu C Bota and B Marinca ldquoAn optimalhomotopy asymptotic method applied to the steady flow of afourth-grade fluid past a porous platerdquo Applied MathematicsLetters vol 22 no 2 pp 245ndash251 2009

[28] V Marinca N Herisanu and I Nemes ldquoOptimal homotopyasymptotic method with application to thin film flowrdquo CentralEuropean Journal of Physics vol 6 no 3 pp 648ndash653 2008

[29] J Ali S Islam and G Zaman ldquoThe solution of multipointboundary value problems by the optimal Homotopy asymptoticMethodrdquoComputers andMathematics with Applications vol 59no 6 pp 2000ndash2006 2010

[30] M Idrees S Islam and S Haq ldquoApplication of the optimalhomotopy asymptotic method to squeezing flowrdquo Computersand Mathematics with Applications vol 59 no 12 pp 3858ndash3866 2010

[31] J Ali S Islam H Khan and S I Shah ldquoThe optimal homotopyasymptotic method for the solution of higher-order boundaryvalue problems in finite domainsrdquo Abstract and Applied Analy-sis vol 2012 Article ID 401217 14 pages 2012

[32] H Khan S Islam J Ali and I Ali Shah ldquoComparison ofdifferent analytic solutions to axisymmetric squeezing fluidflow between two infinite parallel plates with slip boundaryconditionsrdquo Abstract and Applied Analysis vol 2012 Article ID835268 18 pages 2012

[33] T C Papanastasiou G C Georgiou and A N AlexandrouViscous Fluid Flow Library of Congress Cataloging CRC PressNew York NY USA 1999

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 4: Research Article Analysis of Unsteady Axisymmetric ...downloads.hindawi.com/journals/mpe/2015/860857.pdfResearch Article Analysis of Unsteady Axisymmetric Squeezing Fluid Flow with

4 Mathematical Problems in Engineering

The general equations for V119896(119909) are given by

I [V119896(119909)] minusI [V

119896minus1(119909)]

= 1198621alefsym0[V0(119909)]

+

119896minus1

sum

119894=1

119862119894I [V119896minus119894

(119909)]

+alefsym119896minus119894

[V0(119909) V

1(119909) V

119896minus1(119909)]

119861 (V119896119889V119896

119889119909) = 0

119896 = 2 3

(29)

where the coefficient of 119901119898 in the expansion of alefsym(Φ(119909 119901))about 119901 is alefsym

119898[V0(119909) V1(119909) V

119898minus1(119909)]

alefsym(Φ (119909 119901 119862119894))

= alefsym0[V0(119909)] +

119898

sum

119898=1

alefsym119898[V0(119909) V

1(119909) V

119898minus1(119909)] 119901

119898

(30)It is noted that the convergence of the series (25) depends

upon119862119896 For convergence at119901 = 1 the119898th order approxima-

tion V is

V (119909 1198621 1198622 119862

119898) = V0(119909) +

119898

sum

119895=1

V119895(119909 1198621 1198622 119862

119895)

(31)Substituting (31) in (22) the expression for residual is

R (119909 1198621 1198622 119862

119898)

= I [V (119909 1198621 1198622 119862

119898)] + 119891 (119909)

+ alefsym [V (119909 1198621 1198622 119862

119898)]

(32)

If R = 0 then V will be the exact solution but usually thisdoes not happen in nonlinear problems

There are various methods to find the optimal values of119862119894 119894 = 1 2 We apply the method of least square and

Galerkinrsquos method in the following mannerIn method of least square

119869 (119909 1198621 1198622 119862

119898) = int

119887

119886

R2

(119909 1198621 1198622 119862

119898) 119889119909 (33)

Minimizing 119869(119909 1198621 1198622 119862

119898) we have

120597119869

120597119862119894

= 0 119894 = 1 2 119898 (34)

In Galerkinrsquos method we solve the following system for119862119894(119894 = 1 2 119898)

int

119887

119886

R120597V120597119862119894

119889119909 = 0 119894 = 1 2 119898 (35)

To find appropriate 119862119894(119894 = 1 2 119898) we choose 119886 and 119887 in

the domain of the problem Approximate solution of order119898is well-determined with these known constants

5 Application of OHAM in Case ofNo-Slip Boundary

Using (19) and (20) various order problems are as followsZeroth-order problem

V(119894V)0

(120578) = 0

V0(0) = 0 V10158401015840

0(0) = 0 V

0(1) = 1 V1015840

0(1) = 0

(36)

First-order problem

V(119894V)1

(120578) = 31198621119877V101584010158400(120578) + 119862

1119877120578V1015840101584010158400(120578) minus 119862

1119877V0(120578) V1015840101584010158400(120578)

+ V(119894V)0

(120578) + 1198621V(119894V)0

(120578)

V1(0) = 0 V10158401015840

1(0) = 0 V

1(1) = 0 V1015840

1(1) = 0

(37)

Second-order problem

V(119894V)2

(120578) = 31198621119877V101584010158401(120578) minus 119862

1119877V1(120578) V1015840101584010158400(120578) + 119862

1119877120578V1015840101584010158401(120578)

minus 1198621119877V0(120578) V1015840101584010158401(120578) + V(119894V)

1(120578) + 119862

1V(119894V)1

(120578)

V2(0) = 0 V10158401015840

2(0) = 0 V

2(1) = 0 V1015840

2(1) = 0

(38)

Third-order problem

V(119894V)3

(120578) = 31198621119877V101584010158402(120578) minus 119862

1119877V2(120578) V1015840101584010158400(120578)

minus 1198621119877V1(120578) V1015840101584010158401(120578) + 119862

1119877120578V1015840101584010158402(120578)

minus 1198621119877V0(120578) V1015840101584010158402(120578) + V(119894V)

2(120578)

+ 1198621V(119894V)2

(120578)

V3(0) = 0 V10158401015840

3(0) = 0 V

3(1) = 0 V1015840

3(1) = 0

(39)

Fourth-order problem

V(119894V)4

(120578) = 31198621119877V101584010158403(120578) minus 119862

1119877V3(120578) V1015840101584010158400(120578)

minus 1198621119877V2(120578) V1015840101584010158401(120578) minus 119862

1119877V1(120578) V1015840101584010158402(120578)

+ 1198621119877120578V1015840101584010158403(120578) minus 119862

1119877V0(120578) V1015840101584010158403(120578)

+ V(119894V)3

(120578) + 1198621V(119894V)3

(120578)

V4(0) = 0 V10158401015840

4(0) = 0 V

4(1) = 0 V1015840

4(1) = 0

(40)

By considering fourth-order solution we have

V (120578) = V0(120578) + V

1(120578) + V

2(120578) + V

3(120578) + V

4(120578) (41)

The residual of the problem is

R =1198894V (120578)1198891205784

+ 119877[(120578 minus V (120578))1198893V (120578)1198891205783

+ 31198892V (120578)1198891205782

] (42)

Mathematical Problems in Engineering 5

We apply Galerkinrsquos method to find constant 1198621as follows

int

1

0

R120597V1205971198621

119889119909 = 0 (43)

Solving (43) and keeping 119877 = 01 we get

1198621= 101328 (44)

Using above value of 1198621 the approximate solution is

V (120578) =1

2(3120578 minus 120578

3

)

+1

560(374914120578 minus 739694120578

3

+ 3546481205785

+01013281205787

)

+1

776160(956941120578 minus 430222120578

3

+ 514141205785

minus122331205787

minus 5534111205789

minus 19405312057811

)

+1

565045(254414120578 + 143644120578

3

minus 3711641205785

+ 2502321205787

minus 8133381205789

+ 25990112057811

+6961012057813

+ 2296212057815

)

+1

722737(153702120578 minus 187989 times 10

8

1205783

+ 193991 times 108

1205785

+ 828837 times 107

1205787

minus 803587 times 107

1205789

minus 292844 times 107

12057811

+ 167506 times 107

12057813

minus 966217 times 106

12057815

minus16529 times 106

12057817

minus 48038912057819

)

(45)

6 Application of OHAM in Case ofSlip Boundary

Using (19) and (21) different order problems are as followsZeroth-order problem

119906(119894V)0

(120578) = 0

1199060(0) = 0 119906

10158401015840

0(0) = 0 119906

0(1) = 1 119906

1015840

0(1) = 120574119906

10158401015840

0(1)

(46)

First-order problem

119906(119894V)1

(120578)

= 3119862111987711990610158401015840

0(120578) + 119862

1119877120578119906101584010158401015840

0(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

0(120578)

+ 119906(119894V)0

(120578) + 1198621119906(119894V)0

(120578)

1199061(0) = 0 119906

10158401015840

1(0) = 0 119906

1(1) = 0 119906

1015840

1(1) = 120574119906

10158401015840

1(1)

(47)

Second-order problem

119906(119894V)2

(120578) = 3119862111987711990610158401015840

1(120578) minus 119862

11198771199061(120578) 119906101584010158401015840

0(120578)

+ 1198621119877120578119906101584010158401015840

1(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

1(120578)

+ 119906(119894V)1

(120578) + 1198621119906(119894V)1

(120578)

1199062(0) = 0 119906

10158401015840

2(0) = 0 119906

2(1) = 0 119906

1015840

2(1) = 120574119906

10158401015840

2(1)

(48)

Third-order problem

119906(119894V)3

(120578) = 3119862111987711990610158401015840

2(120578) minus 119862

11198771199062(120578) 119906101584010158401015840

0(120578)

minus 11986211198771199061(120578) 119906101584010158401015840

1(120578) + 119862

1119877120578119906101584010158401015840

2(120578)

minus 11986211198771199060(120578) 119906101584010158401015840

2(120578) + 119906

(119894V)2

(120578)

+ 1198621119906(119894V)2

(120578)

1199063(0) = 0 119906

10158401015840

3(0) = 0 119906

3(1) = 0 119906

1015840

3(1) = 120574119906

10158401015840

3(1)

(49)

Fourth-order problem

119906(119894V)4

(120578) = 3119862111987711990610158401015840

3(120578) minus 119862

11198771199063(120578) 119906101584010158401015840

0(120578)

minus 11986211198771199062(120578) 119906101584010158401015840

1(120578) minus 119862

11198771199061(120578) 119906101584010158401015840

2(120578)

+ 1198621119877120578119906101584010158401015840

3(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

3(120578)

+ 119906(119894V)3

(120578) + 1198621119906(119894V)3

(120578)

1199064(0) = 0 119906

10158401015840

4(0) = 0 119906

4(1) = 0 119906

1015840

4(1) = 120574119906

10158401015840

4(1)

(50)

By considering fourth-order solution we have

(120578) =

4

sum

119894=0

119906119894(120578 1198621) (51)

The residual of the problem is

120577 =1198894

(120578)

1198891205784+ 119877[(120578 minus (120578))

1198893

(120578)

1198891205783+ 3

1198892

(120578)

1198891205782] (52)

We apply Galerkinrsquos method to find constant 1198621as follows

int

1

0

120577120597

1205971198621

119889119909 = 0 (53)

Solving (53) and taking 119877 = 02 and 120574 = 1 we get

1198621= minus106872 (54)

6 Mathematical Problems in Engineering

Table 1 OHAM solutions along with residuals for various 119877 in case of no-slip boundary

120578119877 = 01 119877 = 03 119877 = 05

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 0150158 537334 times 10minus11 0151534 minus669283 times 10minus9 0152999 minus623412 times 10minus8

02 0297237 314675 times 10minus10 0299827 166025 times 10minus9 0302582 658388 times 10minus8

03 0438170 851847 times 10minus10 0441661 233504 times 10minus8 0445373 342092 times 10minus7

04 0569900 151431 times 10minus9 0573869 393945 times 10minus8 0578082 493942 times 10minus7

05 0689397 190824 times 10minus9 0693354 281614 times 10minus8 0697548 242092 times 10minus7

06 0793661 133301 times 10minus9 0797122 minus319654 times 10minus8 0800780 minus653046 times 10minus7

07 0879734 minus159255 times 10minus9 0882300 minus221911 times 10minus7 0885004 minus323415 times 10minus6

08 0944705 minus102814 times 10minus8 0946167 minus884707 times 10minus7 0947702 minus12206 times 10minus509 0985722 minus3446 times 10minus8 0986180 minus321268 times 10minus6 0986659 minus441008 times 10minus5

10 1 minus103509 times 10minus7 1 minus111826 times 10minus5 1 minus154022 times 10minus4

Using above value of 1198621 the approximate solution is

(120578) =1

2(3120578 + 120578

3

)

+1

4480(minus113284120578 + 151758120578

3

minus 3890121205785

+04274861205787

)

+1

248372(minus156189120578 + 273825120578

3

minus 1608081205785

+4595261205787

minus 2814271205789

+ 34538712057811

)

+1

723257(minus983844120578 + 1149 times 10

8

1205783

minus 335388 times 107

1205785

+ 404185 times 107

1205787

minus 268964 times 107

1205789

+ 365673 times 106

12057811

minus15753812057813

+ 1724212057815

)

+1

370042(minus197195 times 10

13

120578 + 425372 times 1013

1205783

minus 294213 times 1013

1205785

+ 606765 times 1012

1205787

minus 139445 times 1012

1205789

+ 247846 times 1012

12057811

minus 597188 times 1011

12057813

+ 508263 times 1010

12057815

minus162422 times 109

12057817

+ 152182 times 107

12057819

)

(55)

7 Results and Discussions

In this article we considered the unsteady axisymmetric flowof nonconducting incompressible Newtonian fluid betweentwo circular plates The resulting nonlinear boundary valueproblems are solved with OHAM and fourth-order Runge-Kutta method using Mathematica 70

minus000001

minus000002

minus000003

minus000004

R = 01

R = 03

R = 05

02 04 06 08 10

120578

Res

Figure 2 OHAM residuals at various values of 119877 in case of no-slipboundary

Tables 1 3 and 5 reflect OHAM solutions along withresiduals in case of no-sip and slip boundaries for variousvalues of Reynolds number 119877 and slip parameter 120574 AlsoTables 2 4 and 6 represent RK4 solutions alongwith residualsin case of no-slip and slip boundaries for various values of 119877and 120574 All the tables demonstrate that results obtained usingOHAM are in agreement with RK4 by means of residualsIn addition to above mentioned tables Table 7 shows thecomparison of solutions obtained from OHAM and RK4 forvarious values of Reynolds number 119877

Furthermore Figures 2 3 and 4 indicate the OHAMresiduals in case of no-slip and slip boundaries for variousvalues of 119877 and 120574

The effect of Reynolds number 119877 on velocity profilesin case of no-slip boundary is shown in Figure 5 In theseprofiles we varied 119877 as 119877 = 01 1 2 3 and observed that thenormal velocity is increased with the increase of Reynoldsnumber (Figure 5(a)) It is also noted that the normal velocitymonotonically increases from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time Figure 5(b) describes the impact of119877 on the longitudinal velocity in case of no-slip boundary It

Mathematical Problems in Engineering 7

Table 2 RK4 solutions along with residuals for various 119877 in case of no-slip boundary

120578119877 = 01 119877 = 03 119877 = 05

Solution Residual Solution Residual Solution Residual00 0 288535 times 10minus7 0 966906 times 10minus5 0 530549 times 10minus4

01 0150158 388278 times 10minus8 0151534 571393 times 10minus6 0152999 304600 times 10minus5

02 0297237 minus997789 times 10minus9 0299827 minus141434 times 10minus6 0302582 minus752772 times 10minus6

03 0438170 293767 times 10minus9 0441661 395997 times 10minus7 0445373 210296 times 10minus6

04 0569900 minus751436 times 10minus10 0573869 minus115203 times 10minus7 0578082 minus618081 times 10minus7

05 0689397 minus471705 times 10minus10 0693354 minus96761 times 10minus9 0697548 minus295916 times 10minus8

06 0793661 223004 times 10minus9 0797122 14593 times 10minus7 0800780 714133 times 10minus7

07 0879734 minus718245 times 10minus9 0882300 minus486272 times 10minus7 0885004 minus239245 times 10minus6

08 0944705 268744 times 10minus8 0946167 177796 times 10minus6 0947702 870664 times 10minus6

09 0985722 minus111599 times 10minus7 0986180 minus728952 times 10minus6 0986659 356157 times 10minus5

10 1 minus271683 times 10minus6 1 minus147695 times 10minus4 1 690088 times 10minus4

Table 3 OHAM solutions along with residuals for various values of 119877 in case of slip boundary

120578119877 = 02 119877 = 03 119877 = 04

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 0072692 318345 times 10minus8 0071220 315621 times 10minus7 0069593 174626 times 10minus6

02 0147091 minus172789 times 10minus8 0144268 minus914579 times 10minus8 0141147 minus113148 times 10minus7

03 0224893 minus196995 times 10minus7 0220953 minus166461 times 10minus6 0216599 minus779061 times 10minus6

04 0307773 minus494482 times 10minus7 0303048 minus429173 times 10minus6 0297832 minus207401 times 10minus5

05 0397370 minus799437 times 10minus7 0392277 minus699917 times 10minus6 0386657 minus341643 times 10minus5

06 0495283 minus85784 times 10minus7 0490289 minus75499 times 10minus6 0484784 minus370689 times 10minus5

07 0603056 minus21053 times 10minus7 0598654 minus191141 times 10minus6 0593805 minus96958 times 10minus6

08 0722172 187065 times 10minus6 0718841 162936 times 10minus5 0715176 791082 times 10minus5

09 0854042 640576 times 10minus6 0852211 559717 times 10minus5 0850196 272708 times 10minus4

10 1 146076 times 10minus5 1 127671 times 10minus4 1 622242 times 10minus4

Table 4 RK4 solutions along with residuals for various values of 119877 in case of slip boundary

120578119877 = 02 119877 = 03 119877 = 04

Solution Residual Solution Residual Solution Residual00 0 412179 times 10minus6 0 756765 times 10minus6 0 362897 times 10minus6

01 0072692 174999 times 10minus7 0071220 216125 times 10minus7 0069593 minus349994 times 10minus7

02 0147091 minus421503 times 10minus8 0144268 minus497423 times 10minus8 0141147 953266 times 10minus8

03 0224893 112883 times 10minus8 0220953 122294 times 10minus8 0216598 minus30735 times 10minus8

04 0307773 minus352272 times 10minus9 0303048 minus430187 times 10minus9 0297832 72493 times 10minus9

05 0397370 780684 times 10minus10 0392276 306668 times 10minus9 0386657 846409 times 10minus9

06 0495283 110391 times 10minus9 0490289 minus528586 times 10minus9 0484783 minus336828 times 10minus8

07 0603056 minus435033 times 10minus9 0598654 154159 times 10minus8 0593805 106816 times 10minus7

08 0722172 144198 times 10minus8 0718841 minus605873 times 10minus8 0715175 minus398519 times 10minus7

09 0854042 minus553933 times 10minus8 0852211 259233 times 10minus7 0850196 165518 times 10minus6

10 1 minus197434 times 10minus7 1 796168 times 10minus6 1 389126 times 10minus5

is experienced that this component of velocity deceases nearthe wall but increases near the central axis of the channel

The effect of Reynolds number 119877 on velocity profiles incase of slip boundary is depicted in Figure 6 In these profileswe fixed slip parameter 120574 = 1 and varied Reynolds number119877 as 119877 = 02 06 1 15 It is noted that the normal velocitydecreases as the Reynolds number increases (Figure 6(a))It is also observed that longitudinal velocity decreases near

the central axis of the channel but increases near the wallswhen 119877 increases (Figure 6(b))

Figure 7 demonstrates the effect of slip parameter 120574 on thevelocity profiles After fixing Reynolds number 119877 = 03 wevaried 120574 as 120574 = 06 07 08 1 We find that normal velocityincreases as 120574 increases It is also noted that longitudinalvelocity decreases near the walls but increases near centralaxis of the channel

8 Mathematical Problems in Engineering

Table 5 OHAM solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 minus000880011 minus107933 times 10minus6 00336596 minus754858 times 10minus8 00522402 minus774445 times 10minus9

02 minus0010881 minus22445 times 10minus6 00714091 minus173419 times 10minus7 01074223 minus251485 times 10minus8

03 0000449012 minus345224 times 10minus6 0117324 minus300407 times 10minus7 0168479 minus568899 times 10minus8

04 00318278 minus441058 times 10minus6 0175449 minus432397 times 10minus7 0238322 minus98056 times 10minus8

05 00898133 minus447787 times 10minus6 0249785 minus500627 times 10minus7 0319832 minus129675 times 10minus7

06 018086 minus260114 times 10minus6 0344278 minus377978 times 10minus7 0415852 minus113733 times 10minus7

07 0311298 265136 times 10minus6 0462801 130706 times 10minus7 0529175 120785 times 10minus8

08 0487312 128583 times 10minus5 0609145 128166 times 10minus6 0662538 336724 times 10minus7

09 0714929 291822 times 10minus5 0787012 334515 times 10minus6 0818613 967763 times 10minus7

10 1 514019 times 10minus5 1 649074 times 10minus6 1 200176 times 10minus6

Table 6 RK4 solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 204594 times 10minus5 0 734794 times 10minus6 0 356291 times 10minus6

01 minus000880012 772953 times 10minus7 00336596 321066 times 10minus7 00522402 163598 times 10minus7

02 minus00108811 minus18406 times 10minus7 00714091 minus775305 times 10minus8 01074223 minus396814 times 10minus8

03 0000448983 482747 times 10minus8 0117324 208509 times 10minus8 0168479 10753 times 10minus8

04 00318277 minus154235 times 10minus8 0175449 minus643999 times 10minus9 0238322 minus328063 times 10minus9

05 00898132 514031 times 10minus9 0249785 121029 times 10minus9 0319832 450154 times 10minus10

06 018086 minus651866 times 10minus10 0344278 264494 times 10minus9 0415852 184079 times 10minus9

07 0311298 minus186837 times 10minus9 0462801 minus988382 times 10minus9 0529175 minus651979 times 10minus9

08 0487312 minus133003 times 10minus9 0609145 337336 times 10minus8 0662538 227322 times 10minus8

09 0714929 266400 times 10minus8 0787012 minus132232 times 10minus7 0818613 minus905714 times 10minus8

10 1 555277 times 10minus6 1 minus11671 times 10minus6 1 minus11767 times 10minus6

Table 7 Comparison of OHAM and RK4 solutions for various 119877 in case of slip and no-slip boundary

120578

In case of no-slip boundary In case of slip boundary|RK4 Solution minusOHAM Solution| |RK4 Solution minusOHAM Solution|

119877 = 01 119877 = 03 119877 = 05 119877 = 02 119877 = 03 119877 = 04

00 0 0 0 0 0 001 233147 times 10minus15 162249 times 10minus10 241725 times 10minus9 844379 times 10minus10 790896 times 10minus9 41479 times 10minus8

02 949241 times 10minus14 314228 times 10minus10 467809 times 10minus9 16672 times 10minus9 155832 times 10minus8 815527 times 10minus8

03 382361 times 10minus13 446143 times 10minus10 663065 times 10minus9 244889 times 10minus9 228102 times 10minus8 118952 times 10minus7

04 913492 times 10minus13 547932 times 10minus10 811705 times 10minus9 31663 times 10minus9 293539 times 10minus8 152339 times 10minus7

05 165451 times 10minus12 607476 times 10minus10 895202 times 10minus9 377514 times 10minus9 348026 times 10minus8 179574 times 10minus7

06 242317 times 10minus12 609434 times 10minus10 891445 times 10minus9 418241 times 10minus9 38327 times 10minus8 196532 times 10minus7

07 286218 times 10minus12 536833 times 10minus10 777928 times 10minus9 422088 times 10minus9 384569 times 10minus8 196009 times 10minus7

08 252039 times 10minus12 378563 times 10minus10 54278 times 10minus9 365208 times 10minus9 331085 times 10minus8 167861 times 10minus7

09 120232 times 10minus12 154202 times 10minus10 218716 times 10minus9 224454 times 10minus9 202732 times 10minus8 102385 times 10minus7

10 126281 times 10minus16 253545 times 10minus16 293337 times 10minus17 17436 times 10minus14 492715 times 10minus14 38179 times 10minus14

8 Conclusions

In this article we find the similarity solution for unsteadyaxisymmetric squeezing flow of incompressible Newtonianfluid between two circular plates We observed that thesimilarity solution exists only when distance between theplates varies as (119882119905 + 119878)

12 and squeezing flow occurs when

119882 lt 0 119878 gt 0 and (119882119905+119878) gt 0The key findings of the presentanalysis are as follows

In case of no-slip at boundary

(i) It has been found that increase in Reynolds number 119877increases the normal velocity

Mathematical Problems in Engineering 9

00010

00008

00006

00004

00002

02 04 06 08 10

120578

Res

R = 02

R = 03

R = 04

Figure 3 OHAM residuals at various values of 119877 in case of slip boundary

00001

000008

000006

000004

000002

120574 = 05

120574 = 06

120574 = 07

02 04 06 08 10

120578

Res

Figure 4 OHAM residuals at various values of 120574 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

(120578

120578

)

R = 01

R = 1

R = 2

R = 3

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

R = 01

R = 1

R = 2

R = 3

998400 (120578)

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 5 Velocity profiles for various values of 119877 = 01 1 2 3 in case of no-slip boundary

10 Mathematical Problems in Engineering

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

R = 02

R = 06

R = 1

R = 15

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

R = 02

R = 06

R = 1

R = 15

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 6 Velocity profiles for various values of 119877 = 02 06 1 15 fixing 120574 = 1 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

120574 = 06120574 = 07

120574 = 08

120574 = 1

(a) The effect of 120574 on the Normal velocity profiles

120574 = 06120574 = 07

120574 = 08

120574 = 1

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

(b) The effect of 120574 on the longitudinal velocity profiles

Figure 7 Velocity profiles for various values of 120574 = 06 07 08 1 fixing 119877 = 03 in case of slip boundary

(ii) It has been observed that normal velocity increasesmonotonically from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time

(iii) It has been seen that longitudinal velocity deceasesnear the walls and increases near the central axis ofthe channel

In case of slip at boundary

(i) It has been noted that after fixing slip parameter120574 and varying the Reynolds number 119877 the normalvelocity profile decreases with the increase in 119877 Alsothe longitudinal velocity increases near the walls butdecreases near the central axis of the channel

(ii) It has been examined that for a fixed Reynoldsnumber 119877when we vary slip parameter 120574 the normalvelocity increases with the increase in 120574 Also the

longitudinal velocity decreases near the walls andincreases near the central axis of the channel

(iii) It has been investigated that Reynolds number 119877 andslip parameter 120574 have opposite effects on the normaland longitudinal velocity components

In case of slip versus no-slip boundary

(i) It has been observed that Reynolds number 119877 hasopposite behavior on the normal velocity in case ofslip and no-slip boundaries

(ii) It has been also noticed that Reynolds number 119877 hasopposite effect on the longitudinal velocity near thecentral axis of the channel while near the wall longi-tudinal velocity increases in case of slip boundary anddecease in no-slip boundary This is in conformanceto [33]

Mathematical Problems in Engineering 11

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] Q K Ghori M Ahmed and A M Siddiqui ldquoApplicationof homotopy perturbation method to squeezing flow of anewtonian fluidrdquo International Journal of Nonlinear Sciencesand Numerical Simulation vol 8 no 2 pp 179ndash184 2007

[2] X J Ran Q Y Zhu and Y Li ldquoAn explicit series solution ofthe squeezing flow between two infinite plates by means ofthe homotopy analysis methodrdquo Communications in NonlinearScience and Numerical Simulation vol 14 no 1 pp 119ndash1322009

[3] M J Stefan ldquoVersuch Uber die scheinbare adhasionrdquo Sitzungs-berichteOsterreichische Akademie der Wissenschaften in WienMathematisch-Naturwissenschaftliche Klasse vol 69 pp 713ndash721 1874

[4] J F Thorpe and W A Shaw Eds Developments in Theoreticaland Applied Mechanics vol 3 Pergamon Press Oxford UK1967

[5] P S Gupta and A S Gupta ldquoSqueezing flow between parallelplatesrdquoWear vol 45 no 2 pp 177ndash185 1977

[6] A F Elkouh ldquoFluid inertia effects in squeeze film between twoplane annulirdquo Transactions of the ASME Journal of Tribologyvol 106 no 2 pp 223ndash227 1984

[7] R L Verma ldquoA numerical solution for squeezing flow betweenparallel channelsrdquoWear vol 72 no 1 pp 89ndash95 1981

[8] P Singh V Radhakrishnan and K A Narayan ldquoSqueezing flowbetween parallel platesrdquo Ingenieur-Archiv vol 60 no 4 pp 274ndash281 1990

[9] P J Leider and R B Bird ldquoSqueezing flow between paralleldisks I Theoretical analysisrdquo Industrial and Engineering Chem-istry Fundamentals vol 13 no 4 pp 336ndash341 1974

[10] G Domairry and A Aziz ldquoApproximate analysis of MHDdqueeze flow between two parallel disks with suction orinjection by homotopy perturbation methodrdquo MathematicalProblems in Engineering vol 2009 Article ID 603916 19 pages2009

[11] S Islam H Khan I A Shah and G Zaman ldquoAn axisymmetricsqueezing fluid flow between the two infinite parallel plates in aporous medium channelrdquo Mathematical Problems in Engineer-ing vol 2011 Article ID 349803 10 pages 2011

[12] I Ullah H Khan and M T Rahim ldquoApproximation of firstgrade MHD squeezing fluid flow with slip boundary conditionusing DTM and OHAMrdquo Mathematical Problems in Engineer-ing vol 2013 Article ID 816262 9 pages 2013

[13] A M Siddiqui S Irum and A R Ansari ldquoUnsteady squeezingflowof a viscousMHDfluid between parallel platesrdquoMathemat-ical Modelling and Analysis vol 13 pp 565ndash576 2008

[14] R J Grimm ldquoSqueezing flows of Newtonian liquid films ananalysis including fluid inertiardquo Applied Scientific Research vol32 no 2 pp 149ndash166 1976

[15] M M Rashidi A M Siddiqui and M T Rastegari ldquoAnalyticalsolution of squeezing flow between two circular platesrdquo Interna-tional Journal for ComputationalMethods in Engineering Scienceand Mechanics vol 13 no 5 pp 342ndash349 2012

[16] H M Laun M Rady and O Hassager ldquoAnalytical solutions forsqueeze flow with partial wall sliprdquo Journal of Non-NewtonianFluid Mechanics vol 81 no 1-2 pp 1ndash15 1999

[17] S Ishizawa ldquoThe unsteady flow between two parallel discs witharbitrary varying gap widthrdquo Bulletin of the Japan Society ofMechanical Engineers vol 9 no 35 pp 533ndash550 1966

[18] C le Roux ldquoExistence and uniqueness of the flow of second-grade fluids with slip boundary conditionsrdquoArchive for RationalMechanics and Analysis vol 148 no 4 pp 309ndash356 1999

[19] L J Rhooades R Resnic T OrsquoBradovich and S StegmanldquoAbrasive flow machining of cylinder heads and its positiveeffects on performance and cost characteristicsrdquo Tech RepDearborn Mich USA 1996

[20] C L M H Navier Memoirs de lrsquoCademie Royale des Sciencesde lrsquoInstitut de France vol 1 Royale des Sciences de lrsquoInstitut deFrance 1823

[21] A Ebaid ldquoEffects of magnetic field and wall slip conditions onthe peristaltic transport of a Newtonian fluid in an asymmetricchannelrdquo Physics Letters Section A General Atomic and SolidState Physics vol 372 no 24 pp 4493ndash4499 2008

[22] J-H He ldquoSome asymptotic methods for strongly nonlinearequationsrdquo International Journal of Modern Physics B Con-densed Matter Physics Statistical Physics Applied Physics vol20 no 10 pp 1141ndash1199 2006

[23] S J Liao Proposed homotopy analysis techniques for the solutionof nonlinear problems [PhD thesis] Jiao Tong UniversityShanghai China 1992

[24] J-H He ldquoHomotopy perturbation method for solving bound-ary value problemsrdquo Physics Letters A vol 350 no 1-2 pp 87ndash88 2006

[25] J-H He ldquoApproximate analytical solution for seepage flowwithfractional derivatives in porous mediardquo Computer Methods inApplied Mechanics and Engineering vol 167 no 1-2 pp 57ndash681998

[26] VMarinca andNHerisanu ldquoApplication of optimalHomotopyasymptotic method for solving nonlinear equations arising inheat transferrdquo International Communications in Heat and MassTransfer vol 35 no 6 pp 710ndash715 2008

[27] V Marinca N Herisanu C Bota and B Marinca ldquoAn optimalhomotopy asymptotic method applied to the steady flow of afourth-grade fluid past a porous platerdquo Applied MathematicsLetters vol 22 no 2 pp 245ndash251 2009

[28] V Marinca N Herisanu and I Nemes ldquoOptimal homotopyasymptotic method with application to thin film flowrdquo CentralEuropean Journal of Physics vol 6 no 3 pp 648ndash653 2008

[29] J Ali S Islam and G Zaman ldquoThe solution of multipointboundary value problems by the optimal Homotopy asymptoticMethodrdquoComputers andMathematics with Applications vol 59no 6 pp 2000ndash2006 2010

[30] M Idrees S Islam and S Haq ldquoApplication of the optimalhomotopy asymptotic method to squeezing flowrdquo Computersand Mathematics with Applications vol 59 no 12 pp 3858ndash3866 2010

[31] J Ali S Islam H Khan and S I Shah ldquoThe optimal homotopyasymptotic method for the solution of higher-order boundaryvalue problems in finite domainsrdquo Abstract and Applied Analy-sis vol 2012 Article ID 401217 14 pages 2012

[32] H Khan S Islam J Ali and I Ali Shah ldquoComparison ofdifferent analytic solutions to axisymmetric squeezing fluidflow between two infinite parallel plates with slip boundaryconditionsrdquo Abstract and Applied Analysis vol 2012 Article ID835268 18 pages 2012

[33] T C Papanastasiou G C Georgiou and A N AlexandrouViscous Fluid Flow Library of Congress Cataloging CRC PressNew York NY USA 1999

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 5: Research Article Analysis of Unsteady Axisymmetric ...downloads.hindawi.com/journals/mpe/2015/860857.pdfResearch Article Analysis of Unsteady Axisymmetric Squeezing Fluid Flow with

Mathematical Problems in Engineering 5

We apply Galerkinrsquos method to find constant 1198621as follows

int

1

0

R120597V1205971198621

119889119909 = 0 (43)

Solving (43) and keeping 119877 = 01 we get

1198621= 101328 (44)

Using above value of 1198621 the approximate solution is

V (120578) =1

2(3120578 minus 120578

3

)

+1

560(374914120578 minus 739694120578

3

+ 3546481205785

+01013281205787

)

+1

776160(956941120578 minus 430222120578

3

+ 514141205785

minus122331205787

minus 5534111205789

minus 19405312057811

)

+1

565045(254414120578 + 143644120578

3

minus 3711641205785

+ 2502321205787

minus 8133381205789

+ 25990112057811

+6961012057813

+ 2296212057815

)

+1

722737(153702120578 minus 187989 times 10

8

1205783

+ 193991 times 108

1205785

+ 828837 times 107

1205787

minus 803587 times 107

1205789

minus 292844 times 107

12057811

+ 167506 times 107

12057813

minus 966217 times 106

12057815

minus16529 times 106

12057817

minus 48038912057819

)

(45)

6 Application of OHAM in Case ofSlip Boundary

Using (19) and (21) different order problems are as followsZeroth-order problem

119906(119894V)0

(120578) = 0

1199060(0) = 0 119906

10158401015840

0(0) = 0 119906

0(1) = 1 119906

1015840

0(1) = 120574119906

10158401015840

0(1)

(46)

First-order problem

119906(119894V)1

(120578)

= 3119862111987711990610158401015840

0(120578) + 119862

1119877120578119906101584010158401015840

0(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

0(120578)

+ 119906(119894V)0

(120578) + 1198621119906(119894V)0

(120578)

1199061(0) = 0 119906

10158401015840

1(0) = 0 119906

1(1) = 0 119906

1015840

1(1) = 120574119906

10158401015840

1(1)

(47)

Second-order problem

119906(119894V)2

(120578) = 3119862111987711990610158401015840

1(120578) minus 119862

11198771199061(120578) 119906101584010158401015840

0(120578)

+ 1198621119877120578119906101584010158401015840

1(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

1(120578)

+ 119906(119894V)1

(120578) + 1198621119906(119894V)1

(120578)

1199062(0) = 0 119906

10158401015840

2(0) = 0 119906

2(1) = 0 119906

1015840

2(1) = 120574119906

10158401015840

2(1)

(48)

Third-order problem

119906(119894V)3

(120578) = 3119862111987711990610158401015840

2(120578) minus 119862

11198771199062(120578) 119906101584010158401015840

0(120578)

minus 11986211198771199061(120578) 119906101584010158401015840

1(120578) + 119862

1119877120578119906101584010158401015840

2(120578)

minus 11986211198771199060(120578) 119906101584010158401015840

2(120578) + 119906

(119894V)2

(120578)

+ 1198621119906(119894V)2

(120578)

1199063(0) = 0 119906

10158401015840

3(0) = 0 119906

3(1) = 0 119906

1015840

3(1) = 120574119906

10158401015840

3(1)

(49)

Fourth-order problem

119906(119894V)4

(120578) = 3119862111987711990610158401015840

3(120578) minus 119862

11198771199063(120578) 119906101584010158401015840

0(120578)

minus 11986211198771199062(120578) 119906101584010158401015840

1(120578) minus 119862

11198771199061(120578) 119906101584010158401015840

2(120578)

+ 1198621119877120578119906101584010158401015840

3(120578) minus 119862

11198771199060(120578) 119906101584010158401015840

3(120578)

+ 119906(119894V)3

(120578) + 1198621119906(119894V)3

(120578)

1199064(0) = 0 119906

10158401015840

4(0) = 0 119906

4(1) = 0 119906

1015840

4(1) = 120574119906

10158401015840

4(1)

(50)

By considering fourth-order solution we have

(120578) =

4

sum

119894=0

119906119894(120578 1198621) (51)

The residual of the problem is

120577 =1198894

(120578)

1198891205784+ 119877[(120578 minus (120578))

1198893

(120578)

1198891205783+ 3

1198892

(120578)

1198891205782] (52)

We apply Galerkinrsquos method to find constant 1198621as follows

int

1

0

120577120597

1205971198621

119889119909 = 0 (53)

Solving (53) and taking 119877 = 02 and 120574 = 1 we get

1198621= minus106872 (54)

6 Mathematical Problems in Engineering

Table 1 OHAM solutions along with residuals for various 119877 in case of no-slip boundary

120578119877 = 01 119877 = 03 119877 = 05

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 0150158 537334 times 10minus11 0151534 minus669283 times 10minus9 0152999 minus623412 times 10minus8

02 0297237 314675 times 10minus10 0299827 166025 times 10minus9 0302582 658388 times 10minus8

03 0438170 851847 times 10minus10 0441661 233504 times 10minus8 0445373 342092 times 10minus7

04 0569900 151431 times 10minus9 0573869 393945 times 10minus8 0578082 493942 times 10minus7

05 0689397 190824 times 10minus9 0693354 281614 times 10minus8 0697548 242092 times 10minus7

06 0793661 133301 times 10minus9 0797122 minus319654 times 10minus8 0800780 minus653046 times 10minus7

07 0879734 minus159255 times 10minus9 0882300 minus221911 times 10minus7 0885004 minus323415 times 10minus6

08 0944705 minus102814 times 10minus8 0946167 minus884707 times 10minus7 0947702 minus12206 times 10minus509 0985722 minus3446 times 10minus8 0986180 minus321268 times 10minus6 0986659 minus441008 times 10minus5

10 1 minus103509 times 10minus7 1 minus111826 times 10minus5 1 minus154022 times 10minus4

Using above value of 1198621 the approximate solution is

(120578) =1

2(3120578 + 120578

3

)

+1

4480(minus113284120578 + 151758120578

3

minus 3890121205785

+04274861205787

)

+1

248372(minus156189120578 + 273825120578

3

minus 1608081205785

+4595261205787

minus 2814271205789

+ 34538712057811

)

+1

723257(minus983844120578 + 1149 times 10

8

1205783

minus 335388 times 107

1205785

+ 404185 times 107

1205787

minus 268964 times 107

1205789

+ 365673 times 106

12057811

minus15753812057813

+ 1724212057815

)

+1

370042(minus197195 times 10

13

120578 + 425372 times 1013

1205783

minus 294213 times 1013

1205785

+ 606765 times 1012

1205787

minus 139445 times 1012

1205789

+ 247846 times 1012

12057811

minus 597188 times 1011

12057813

+ 508263 times 1010

12057815

minus162422 times 109

12057817

+ 152182 times 107

12057819

)

(55)

7 Results and Discussions

In this article we considered the unsteady axisymmetric flowof nonconducting incompressible Newtonian fluid betweentwo circular plates The resulting nonlinear boundary valueproblems are solved with OHAM and fourth-order Runge-Kutta method using Mathematica 70

minus000001

minus000002

minus000003

minus000004

R = 01

R = 03

R = 05

02 04 06 08 10

120578

Res

Figure 2 OHAM residuals at various values of 119877 in case of no-slipboundary

Tables 1 3 and 5 reflect OHAM solutions along withresiduals in case of no-sip and slip boundaries for variousvalues of Reynolds number 119877 and slip parameter 120574 AlsoTables 2 4 and 6 represent RK4 solutions alongwith residualsin case of no-slip and slip boundaries for various values of 119877and 120574 All the tables demonstrate that results obtained usingOHAM are in agreement with RK4 by means of residualsIn addition to above mentioned tables Table 7 shows thecomparison of solutions obtained from OHAM and RK4 forvarious values of Reynolds number 119877

Furthermore Figures 2 3 and 4 indicate the OHAMresiduals in case of no-slip and slip boundaries for variousvalues of 119877 and 120574

The effect of Reynolds number 119877 on velocity profilesin case of no-slip boundary is shown in Figure 5 In theseprofiles we varied 119877 as 119877 = 01 1 2 3 and observed that thenormal velocity is increased with the increase of Reynoldsnumber (Figure 5(a)) It is also noted that the normal velocitymonotonically increases from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time Figure 5(b) describes the impact of119877 on the longitudinal velocity in case of no-slip boundary It

Mathematical Problems in Engineering 7

Table 2 RK4 solutions along with residuals for various 119877 in case of no-slip boundary

120578119877 = 01 119877 = 03 119877 = 05

Solution Residual Solution Residual Solution Residual00 0 288535 times 10minus7 0 966906 times 10minus5 0 530549 times 10minus4

01 0150158 388278 times 10minus8 0151534 571393 times 10minus6 0152999 304600 times 10minus5

02 0297237 minus997789 times 10minus9 0299827 minus141434 times 10minus6 0302582 minus752772 times 10minus6

03 0438170 293767 times 10minus9 0441661 395997 times 10minus7 0445373 210296 times 10minus6

04 0569900 minus751436 times 10minus10 0573869 minus115203 times 10minus7 0578082 minus618081 times 10minus7

05 0689397 minus471705 times 10minus10 0693354 minus96761 times 10minus9 0697548 minus295916 times 10minus8

06 0793661 223004 times 10minus9 0797122 14593 times 10minus7 0800780 714133 times 10minus7

07 0879734 minus718245 times 10minus9 0882300 minus486272 times 10minus7 0885004 minus239245 times 10minus6

08 0944705 268744 times 10minus8 0946167 177796 times 10minus6 0947702 870664 times 10minus6

09 0985722 minus111599 times 10minus7 0986180 minus728952 times 10minus6 0986659 356157 times 10minus5

10 1 minus271683 times 10minus6 1 minus147695 times 10minus4 1 690088 times 10minus4

Table 3 OHAM solutions along with residuals for various values of 119877 in case of slip boundary

120578119877 = 02 119877 = 03 119877 = 04

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 0072692 318345 times 10minus8 0071220 315621 times 10minus7 0069593 174626 times 10minus6

02 0147091 minus172789 times 10minus8 0144268 minus914579 times 10minus8 0141147 minus113148 times 10minus7

03 0224893 minus196995 times 10minus7 0220953 minus166461 times 10minus6 0216599 minus779061 times 10minus6

04 0307773 minus494482 times 10minus7 0303048 minus429173 times 10minus6 0297832 minus207401 times 10minus5

05 0397370 minus799437 times 10minus7 0392277 minus699917 times 10minus6 0386657 minus341643 times 10minus5

06 0495283 minus85784 times 10minus7 0490289 minus75499 times 10minus6 0484784 minus370689 times 10minus5

07 0603056 minus21053 times 10minus7 0598654 minus191141 times 10minus6 0593805 minus96958 times 10minus6

08 0722172 187065 times 10minus6 0718841 162936 times 10minus5 0715176 791082 times 10minus5

09 0854042 640576 times 10minus6 0852211 559717 times 10minus5 0850196 272708 times 10minus4

10 1 146076 times 10minus5 1 127671 times 10minus4 1 622242 times 10minus4

Table 4 RK4 solutions along with residuals for various values of 119877 in case of slip boundary

120578119877 = 02 119877 = 03 119877 = 04

Solution Residual Solution Residual Solution Residual00 0 412179 times 10minus6 0 756765 times 10minus6 0 362897 times 10minus6

01 0072692 174999 times 10minus7 0071220 216125 times 10minus7 0069593 minus349994 times 10minus7

02 0147091 minus421503 times 10minus8 0144268 minus497423 times 10minus8 0141147 953266 times 10minus8

03 0224893 112883 times 10minus8 0220953 122294 times 10minus8 0216598 minus30735 times 10minus8

04 0307773 minus352272 times 10minus9 0303048 minus430187 times 10minus9 0297832 72493 times 10minus9

05 0397370 780684 times 10minus10 0392276 306668 times 10minus9 0386657 846409 times 10minus9

06 0495283 110391 times 10minus9 0490289 minus528586 times 10minus9 0484783 minus336828 times 10minus8

07 0603056 minus435033 times 10minus9 0598654 154159 times 10minus8 0593805 106816 times 10minus7

08 0722172 144198 times 10minus8 0718841 minus605873 times 10minus8 0715175 minus398519 times 10minus7

09 0854042 minus553933 times 10minus8 0852211 259233 times 10minus7 0850196 165518 times 10minus6

10 1 minus197434 times 10minus7 1 796168 times 10minus6 1 389126 times 10minus5

is experienced that this component of velocity deceases nearthe wall but increases near the central axis of the channel

The effect of Reynolds number 119877 on velocity profiles incase of slip boundary is depicted in Figure 6 In these profileswe fixed slip parameter 120574 = 1 and varied Reynolds number119877 as 119877 = 02 06 1 15 It is noted that the normal velocitydecreases as the Reynolds number increases (Figure 6(a))It is also observed that longitudinal velocity decreases near

the central axis of the channel but increases near the wallswhen 119877 increases (Figure 6(b))

Figure 7 demonstrates the effect of slip parameter 120574 on thevelocity profiles After fixing Reynolds number 119877 = 03 wevaried 120574 as 120574 = 06 07 08 1 We find that normal velocityincreases as 120574 increases It is also noted that longitudinalvelocity decreases near the walls but increases near centralaxis of the channel

8 Mathematical Problems in Engineering

Table 5 OHAM solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 minus000880011 minus107933 times 10minus6 00336596 minus754858 times 10minus8 00522402 minus774445 times 10minus9

02 minus0010881 minus22445 times 10minus6 00714091 minus173419 times 10minus7 01074223 minus251485 times 10minus8

03 0000449012 minus345224 times 10minus6 0117324 minus300407 times 10minus7 0168479 minus568899 times 10minus8

04 00318278 minus441058 times 10minus6 0175449 minus432397 times 10minus7 0238322 minus98056 times 10minus8

05 00898133 minus447787 times 10minus6 0249785 minus500627 times 10minus7 0319832 minus129675 times 10minus7

06 018086 minus260114 times 10minus6 0344278 minus377978 times 10minus7 0415852 minus113733 times 10minus7

07 0311298 265136 times 10minus6 0462801 130706 times 10minus7 0529175 120785 times 10minus8

08 0487312 128583 times 10minus5 0609145 128166 times 10minus6 0662538 336724 times 10minus7

09 0714929 291822 times 10minus5 0787012 334515 times 10minus6 0818613 967763 times 10minus7

10 1 514019 times 10minus5 1 649074 times 10minus6 1 200176 times 10minus6

Table 6 RK4 solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 204594 times 10minus5 0 734794 times 10minus6 0 356291 times 10minus6

01 minus000880012 772953 times 10minus7 00336596 321066 times 10minus7 00522402 163598 times 10minus7

02 minus00108811 minus18406 times 10minus7 00714091 minus775305 times 10minus8 01074223 minus396814 times 10minus8

03 0000448983 482747 times 10minus8 0117324 208509 times 10minus8 0168479 10753 times 10minus8

04 00318277 minus154235 times 10minus8 0175449 minus643999 times 10minus9 0238322 minus328063 times 10minus9

05 00898132 514031 times 10minus9 0249785 121029 times 10minus9 0319832 450154 times 10minus10

06 018086 minus651866 times 10minus10 0344278 264494 times 10minus9 0415852 184079 times 10minus9

07 0311298 minus186837 times 10minus9 0462801 minus988382 times 10minus9 0529175 minus651979 times 10minus9

08 0487312 minus133003 times 10minus9 0609145 337336 times 10minus8 0662538 227322 times 10minus8

09 0714929 266400 times 10minus8 0787012 minus132232 times 10minus7 0818613 minus905714 times 10minus8

10 1 555277 times 10minus6 1 minus11671 times 10minus6 1 minus11767 times 10minus6

Table 7 Comparison of OHAM and RK4 solutions for various 119877 in case of slip and no-slip boundary

120578

In case of no-slip boundary In case of slip boundary|RK4 Solution minusOHAM Solution| |RK4 Solution minusOHAM Solution|

119877 = 01 119877 = 03 119877 = 05 119877 = 02 119877 = 03 119877 = 04

00 0 0 0 0 0 001 233147 times 10minus15 162249 times 10minus10 241725 times 10minus9 844379 times 10minus10 790896 times 10minus9 41479 times 10minus8

02 949241 times 10minus14 314228 times 10minus10 467809 times 10minus9 16672 times 10minus9 155832 times 10minus8 815527 times 10minus8

03 382361 times 10minus13 446143 times 10minus10 663065 times 10minus9 244889 times 10minus9 228102 times 10minus8 118952 times 10minus7

04 913492 times 10minus13 547932 times 10minus10 811705 times 10minus9 31663 times 10minus9 293539 times 10minus8 152339 times 10minus7

05 165451 times 10minus12 607476 times 10minus10 895202 times 10minus9 377514 times 10minus9 348026 times 10minus8 179574 times 10minus7

06 242317 times 10minus12 609434 times 10minus10 891445 times 10minus9 418241 times 10minus9 38327 times 10minus8 196532 times 10minus7

07 286218 times 10minus12 536833 times 10minus10 777928 times 10minus9 422088 times 10minus9 384569 times 10minus8 196009 times 10minus7

08 252039 times 10minus12 378563 times 10minus10 54278 times 10minus9 365208 times 10minus9 331085 times 10minus8 167861 times 10minus7

09 120232 times 10minus12 154202 times 10minus10 218716 times 10minus9 224454 times 10minus9 202732 times 10minus8 102385 times 10minus7

10 126281 times 10minus16 253545 times 10minus16 293337 times 10minus17 17436 times 10minus14 492715 times 10minus14 38179 times 10minus14

8 Conclusions

In this article we find the similarity solution for unsteadyaxisymmetric squeezing flow of incompressible Newtonianfluid between two circular plates We observed that thesimilarity solution exists only when distance between theplates varies as (119882119905 + 119878)

12 and squeezing flow occurs when

119882 lt 0 119878 gt 0 and (119882119905+119878) gt 0The key findings of the presentanalysis are as follows

In case of no-slip at boundary

(i) It has been found that increase in Reynolds number 119877increases the normal velocity

Mathematical Problems in Engineering 9

00010

00008

00006

00004

00002

02 04 06 08 10

120578

Res

R = 02

R = 03

R = 04

Figure 3 OHAM residuals at various values of 119877 in case of slip boundary

00001

000008

000006

000004

000002

120574 = 05

120574 = 06

120574 = 07

02 04 06 08 10

120578

Res

Figure 4 OHAM residuals at various values of 120574 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

(120578

120578

)

R = 01

R = 1

R = 2

R = 3

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

R = 01

R = 1

R = 2

R = 3

998400 (120578)

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 5 Velocity profiles for various values of 119877 = 01 1 2 3 in case of no-slip boundary

10 Mathematical Problems in Engineering

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

R = 02

R = 06

R = 1

R = 15

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

R = 02

R = 06

R = 1

R = 15

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 6 Velocity profiles for various values of 119877 = 02 06 1 15 fixing 120574 = 1 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

120574 = 06120574 = 07

120574 = 08

120574 = 1

(a) The effect of 120574 on the Normal velocity profiles

120574 = 06120574 = 07

120574 = 08

120574 = 1

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

(b) The effect of 120574 on the longitudinal velocity profiles

Figure 7 Velocity profiles for various values of 120574 = 06 07 08 1 fixing 119877 = 03 in case of slip boundary

(ii) It has been observed that normal velocity increasesmonotonically from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time

(iii) It has been seen that longitudinal velocity deceasesnear the walls and increases near the central axis ofthe channel

In case of slip at boundary

(i) It has been noted that after fixing slip parameter120574 and varying the Reynolds number 119877 the normalvelocity profile decreases with the increase in 119877 Alsothe longitudinal velocity increases near the walls butdecreases near the central axis of the channel

(ii) It has been examined that for a fixed Reynoldsnumber 119877when we vary slip parameter 120574 the normalvelocity increases with the increase in 120574 Also the

longitudinal velocity decreases near the walls andincreases near the central axis of the channel

(iii) It has been investigated that Reynolds number 119877 andslip parameter 120574 have opposite effects on the normaland longitudinal velocity components

In case of slip versus no-slip boundary

(i) It has been observed that Reynolds number 119877 hasopposite behavior on the normal velocity in case ofslip and no-slip boundaries

(ii) It has been also noticed that Reynolds number 119877 hasopposite effect on the longitudinal velocity near thecentral axis of the channel while near the wall longi-tudinal velocity increases in case of slip boundary anddecease in no-slip boundary This is in conformanceto [33]

Mathematical Problems in Engineering 11

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] Q K Ghori M Ahmed and A M Siddiqui ldquoApplicationof homotopy perturbation method to squeezing flow of anewtonian fluidrdquo International Journal of Nonlinear Sciencesand Numerical Simulation vol 8 no 2 pp 179ndash184 2007

[2] X J Ran Q Y Zhu and Y Li ldquoAn explicit series solution ofthe squeezing flow between two infinite plates by means ofthe homotopy analysis methodrdquo Communications in NonlinearScience and Numerical Simulation vol 14 no 1 pp 119ndash1322009

[3] M J Stefan ldquoVersuch Uber die scheinbare adhasionrdquo Sitzungs-berichteOsterreichische Akademie der Wissenschaften in WienMathematisch-Naturwissenschaftliche Klasse vol 69 pp 713ndash721 1874

[4] J F Thorpe and W A Shaw Eds Developments in Theoreticaland Applied Mechanics vol 3 Pergamon Press Oxford UK1967

[5] P S Gupta and A S Gupta ldquoSqueezing flow between parallelplatesrdquoWear vol 45 no 2 pp 177ndash185 1977

[6] A F Elkouh ldquoFluid inertia effects in squeeze film between twoplane annulirdquo Transactions of the ASME Journal of Tribologyvol 106 no 2 pp 223ndash227 1984

[7] R L Verma ldquoA numerical solution for squeezing flow betweenparallel channelsrdquoWear vol 72 no 1 pp 89ndash95 1981

[8] P Singh V Radhakrishnan and K A Narayan ldquoSqueezing flowbetween parallel platesrdquo Ingenieur-Archiv vol 60 no 4 pp 274ndash281 1990

[9] P J Leider and R B Bird ldquoSqueezing flow between paralleldisks I Theoretical analysisrdquo Industrial and Engineering Chem-istry Fundamentals vol 13 no 4 pp 336ndash341 1974

[10] G Domairry and A Aziz ldquoApproximate analysis of MHDdqueeze flow between two parallel disks with suction orinjection by homotopy perturbation methodrdquo MathematicalProblems in Engineering vol 2009 Article ID 603916 19 pages2009

[11] S Islam H Khan I A Shah and G Zaman ldquoAn axisymmetricsqueezing fluid flow between the two infinite parallel plates in aporous medium channelrdquo Mathematical Problems in Engineer-ing vol 2011 Article ID 349803 10 pages 2011

[12] I Ullah H Khan and M T Rahim ldquoApproximation of firstgrade MHD squeezing fluid flow with slip boundary conditionusing DTM and OHAMrdquo Mathematical Problems in Engineer-ing vol 2013 Article ID 816262 9 pages 2013

[13] A M Siddiqui S Irum and A R Ansari ldquoUnsteady squeezingflowof a viscousMHDfluid between parallel platesrdquoMathemat-ical Modelling and Analysis vol 13 pp 565ndash576 2008

[14] R J Grimm ldquoSqueezing flows of Newtonian liquid films ananalysis including fluid inertiardquo Applied Scientific Research vol32 no 2 pp 149ndash166 1976

[15] M M Rashidi A M Siddiqui and M T Rastegari ldquoAnalyticalsolution of squeezing flow between two circular platesrdquo Interna-tional Journal for ComputationalMethods in Engineering Scienceand Mechanics vol 13 no 5 pp 342ndash349 2012

[16] H M Laun M Rady and O Hassager ldquoAnalytical solutions forsqueeze flow with partial wall sliprdquo Journal of Non-NewtonianFluid Mechanics vol 81 no 1-2 pp 1ndash15 1999

[17] S Ishizawa ldquoThe unsteady flow between two parallel discs witharbitrary varying gap widthrdquo Bulletin of the Japan Society ofMechanical Engineers vol 9 no 35 pp 533ndash550 1966

[18] C le Roux ldquoExistence and uniqueness of the flow of second-grade fluids with slip boundary conditionsrdquoArchive for RationalMechanics and Analysis vol 148 no 4 pp 309ndash356 1999

[19] L J Rhooades R Resnic T OrsquoBradovich and S StegmanldquoAbrasive flow machining of cylinder heads and its positiveeffects on performance and cost characteristicsrdquo Tech RepDearborn Mich USA 1996

[20] C L M H Navier Memoirs de lrsquoCademie Royale des Sciencesde lrsquoInstitut de France vol 1 Royale des Sciences de lrsquoInstitut deFrance 1823

[21] A Ebaid ldquoEffects of magnetic field and wall slip conditions onthe peristaltic transport of a Newtonian fluid in an asymmetricchannelrdquo Physics Letters Section A General Atomic and SolidState Physics vol 372 no 24 pp 4493ndash4499 2008

[22] J-H He ldquoSome asymptotic methods for strongly nonlinearequationsrdquo International Journal of Modern Physics B Con-densed Matter Physics Statistical Physics Applied Physics vol20 no 10 pp 1141ndash1199 2006

[23] S J Liao Proposed homotopy analysis techniques for the solutionof nonlinear problems [PhD thesis] Jiao Tong UniversityShanghai China 1992

[24] J-H He ldquoHomotopy perturbation method for solving bound-ary value problemsrdquo Physics Letters A vol 350 no 1-2 pp 87ndash88 2006

[25] J-H He ldquoApproximate analytical solution for seepage flowwithfractional derivatives in porous mediardquo Computer Methods inApplied Mechanics and Engineering vol 167 no 1-2 pp 57ndash681998

[26] VMarinca andNHerisanu ldquoApplication of optimalHomotopyasymptotic method for solving nonlinear equations arising inheat transferrdquo International Communications in Heat and MassTransfer vol 35 no 6 pp 710ndash715 2008

[27] V Marinca N Herisanu C Bota and B Marinca ldquoAn optimalhomotopy asymptotic method applied to the steady flow of afourth-grade fluid past a porous platerdquo Applied MathematicsLetters vol 22 no 2 pp 245ndash251 2009

[28] V Marinca N Herisanu and I Nemes ldquoOptimal homotopyasymptotic method with application to thin film flowrdquo CentralEuropean Journal of Physics vol 6 no 3 pp 648ndash653 2008

[29] J Ali S Islam and G Zaman ldquoThe solution of multipointboundary value problems by the optimal Homotopy asymptoticMethodrdquoComputers andMathematics with Applications vol 59no 6 pp 2000ndash2006 2010

[30] M Idrees S Islam and S Haq ldquoApplication of the optimalhomotopy asymptotic method to squeezing flowrdquo Computersand Mathematics with Applications vol 59 no 12 pp 3858ndash3866 2010

[31] J Ali S Islam H Khan and S I Shah ldquoThe optimal homotopyasymptotic method for the solution of higher-order boundaryvalue problems in finite domainsrdquo Abstract and Applied Analy-sis vol 2012 Article ID 401217 14 pages 2012

[32] H Khan S Islam J Ali and I Ali Shah ldquoComparison ofdifferent analytic solutions to axisymmetric squeezing fluidflow between two infinite parallel plates with slip boundaryconditionsrdquo Abstract and Applied Analysis vol 2012 Article ID835268 18 pages 2012

[33] T C Papanastasiou G C Georgiou and A N AlexandrouViscous Fluid Flow Library of Congress Cataloging CRC PressNew York NY USA 1999

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Stochastic AnalysisInternational Journal of

Page 6: Research Article Analysis of Unsteady Axisymmetric ...downloads.hindawi.com/journals/mpe/2015/860857.pdfResearch Article Analysis of Unsteady Axisymmetric Squeezing Fluid Flow with

6 Mathematical Problems in Engineering

Table 1 OHAM solutions along with residuals for various 119877 in case of no-slip boundary

120578119877 = 01 119877 = 03 119877 = 05

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 0150158 537334 times 10minus11 0151534 minus669283 times 10minus9 0152999 minus623412 times 10minus8

02 0297237 314675 times 10minus10 0299827 166025 times 10minus9 0302582 658388 times 10minus8

03 0438170 851847 times 10minus10 0441661 233504 times 10minus8 0445373 342092 times 10minus7

04 0569900 151431 times 10minus9 0573869 393945 times 10minus8 0578082 493942 times 10minus7

05 0689397 190824 times 10minus9 0693354 281614 times 10minus8 0697548 242092 times 10minus7

06 0793661 133301 times 10minus9 0797122 minus319654 times 10minus8 0800780 minus653046 times 10minus7

07 0879734 minus159255 times 10minus9 0882300 minus221911 times 10minus7 0885004 minus323415 times 10minus6

08 0944705 minus102814 times 10minus8 0946167 minus884707 times 10minus7 0947702 minus12206 times 10minus509 0985722 minus3446 times 10minus8 0986180 minus321268 times 10minus6 0986659 minus441008 times 10minus5

10 1 minus103509 times 10minus7 1 minus111826 times 10minus5 1 minus154022 times 10minus4

Using above value of 1198621 the approximate solution is

(120578) =1

2(3120578 + 120578

3

)

+1

4480(minus113284120578 + 151758120578

3

minus 3890121205785

+04274861205787

)

+1

248372(minus156189120578 + 273825120578

3

minus 1608081205785

+4595261205787

minus 2814271205789

+ 34538712057811

)

+1

723257(minus983844120578 + 1149 times 10

8

1205783

minus 335388 times 107

1205785

+ 404185 times 107

1205787

minus 268964 times 107

1205789

+ 365673 times 106

12057811

minus15753812057813

+ 1724212057815

)

+1

370042(minus197195 times 10

13

120578 + 425372 times 1013

1205783

minus 294213 times 1013

1205785

+ 606765 times 1012

1205787

minus 139445 times 1012

1205789

+ 247846 times 1012

12057811

minus 597188 times 1011

12057813

+ 508263 times 1010

12057815

minus162422 times 109

12057817

+ 152182 times 107

12057819

)

(55)

7 Results and Discussions

In this article we considered the unsteady axisymmetric flowof nonconducting incompressible Newtonian fluid betweentwo circular plates The resulting nonlinear boundary valueproblems are solved with OHAM and fourth-order Runge-Kutta method using Mathematica 70

minus000001

minus000002

minus000003

minus000004

R = 01

R = 03

R = 05

02 04 06 08 10

120578

Res

Figure 2 OHAM residuals at various values of 119877 in case of no-slipboundary

Tables 1 3 and 5 reflect OHAM solutions along withresiduals in case of no-sip and slip boundaries for variousvalues of Reynolds number 119877 and slip parameter 120574 AlsoTables 2 4 and 6 represent RK4 solutions alongwith residualsin case of no-slip and slip boundaries for various values of 119877and 120574 All the tables demonstrate that results obtained usingOHAM are in agreement with RK4 by means of residualsIn addition to above mentioned tables Table 7 shows thecomparison of solutions obtained from OHAM and RK4 forvarious values of Reynolds number 119877

Furthermore Figures 2 3 and 4 indicate the OHAMresiduals in case of no-slip and slip boundaries for variousvalues of 119877 and 120574

The effect of Reynolds number 119877 on velocity profilesin case of no-slip boundary is shown in Figure 5 In theseprofiles we varied 119877 as 119877 = 01 1 2 3 and observed that thenormal velocity is increased with the increase of Reynoldsnumber (Figure 5(a)) It is also noted that the normal velocitymonotonically increases from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time Figure 5(b) describes the impact of119877 on the longitudinal velocity in case of no-slip boundary It

Mathematical Problems in Engineering 7

Table 2 RK4 solutions along with residuals for various 119877 in case of no-slip boundary

120578119877 = 01 119877 = 03 119877 = 05

Solution Residual Solution Residual Solution Residual00 0 288535 times 10minus7 0 966906 times 10minus5 0 530549 times 10minus4

01 0150158 388278 times 10minus8 0151534 571393 times 10minus6 0152999 304600 times 10minus5

02 0297237 minus997789 times 10minus9 0299827 minus141434 times 10minus6 0302582 minus752772 times 10minus6

03 0438170 293767 times 10minus9 0441661 395997 times 10minus7 0445373 210296 times 10minus6

04 0569900 minus751436 times 10minus10 0573869 minus115203 times 10minus7 0578082 minus618081 times 10minus7

05 0689397 minus471705 times 10minus10 0693354 minus96761 times 10minus9 0697548 minus295916 times 10minus8

06 0793661 223004 times 10minus9 0797122 14593 times 10minus7 0800780 714133 times 10minus7

07 0879734 minus718245 times 10minus9 0882300 minus486272 times 10minus7 0885004 minus239245 times 10minus6

08 0944705 268744 times 10minus8 0946167 177796 times 10minus6 0947702 870664 times 10minus6

09 0985722 minus111599 times 10minus7 0986180 minus728952 times 10minus6 0986659 356157 times 10minus5

10 1 minus271683 times 10minus6 1 minus147695 times 10minus4 1 690088 times 10minus4

Table 3 OHAM solutions along with residuals for various values of 119877 in case of slip boundary

120578119877 = 02 119877 = 03 119877 = 04

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 0072692 318345 times 10minus8 0071220 315621 times 10minus7 0069593 174626 times 10minus6

02 0147091 minus172789 times 10minus8 0144268 minus914579 times 10minus8 0141147 minus113148 times 10minus7

03 0224893 minus196995 times 10minus7 0220953 minus166461 times 10minus6 0216599 minus779061 times 10minus6

04 0307773 minus494482 times 10minus7 0303048 minus429173 times 10minus6 0297832 minus207401 times 10minus5

05 0397370 minus799437 times 10minus7 0392277 minus699917 times 10minus6 0386657 minus341643 times 10minus5

06 0495283 minus85784 times 10minus7 0490289 minus75499 times 10minus6 0484784 minus370689 times 10minus5

07 0603056 minus21053 times 10minus7 0598654 minus191141 times 10minus6 0593805 minus96958 times 10minus6

08 0722172 187065 times 10minus6 0718841 162936 times 10minus5 0715176 791082 times 10minus5

09 0854042 640576 times 10minus6 0852211 559717 times 10minus5 0850196 272708 times 10minus4

10 1 146076 times 10minus5 1 127671 times 10minus4 1 622242 times 10minus4

Table 4 RK4 solutions along with residuals for various values of 119877 in case of slip boundary

120578119877 = 02 119877 = 03 119877 = 04

Solution Residual Solution Residual Solution Residual00 0 412179 times 10minus6 0 756765 times 10minus6 0 362897 times 10minus6

01 0072692 174999 times 10minus7 0071220 216125 times 10minus7 0069593 minus349994 times 10minus7

02 0147091 minus421503 times 10minus8 0144268 minus497423 times 10minus8 0141147 953266 times 10minus8

03 0224893 112883 times 10minus8 0220953 122294 times 10minus8 0216598 minus30735 times 10minus8

04 0307773 minus352272 times 10minus9 0303048 minus430187 times 10minus9 0297832 72493 times 10minus9

05 0397370 780684 times 10minus10 0392276 306668 times 10minus9 0386657 846409 times 10minus9

06 0495283 110391 times 10minus9 0490289 minus528586 times 10minus9 0484783 minus336828 times 10minus8

07 0603056 minus435033 times 10minus9 0598654 154159 times 10minus8 0593805 106816 times 10minus7

08 0722172 144198 times 10minus8 0718841 minus605873 times 10minus8 0715175 minus398519 times 10minus7

09 0854042 minus553933 times 10minus8 0852211 259233 times 10minus7 0850196 165518 times 10minus6

10 1 minus197434 times 10minus7 1 796168 times 10minus6 1 389126 times 10minus5

is experienced that this component of velocity deceases nearthe wall but increases near the central axis of the channel

The effect of Reynolds number 119877 on velocity profiles incase of slip boundary is depicted in Figure 6 In these profileswe fixed slip parameter 120574 = 1 and varied Reynolds number119877 as 119877 = 02 06 1 15 It is noted that the normal velocitydecreases as the Reynolds number increases (Figure 6(a))It is also observed that longitudinal velocity decreases near

the central axis of the channel but increases near the wallswhen 119877 increases (Figure 6(b))

Figure 7 demonstrates the effect of slip parameter 120574 on thevelocity profiles After fixing Reynolds number 119877 = 03 wevaried 120574 as 120574 = 06 07 08 1 We find that normal velocityincreases as 120574 increases It is also noted that longitudinalvelocity decreases near the walls but increases near centralaxis of the channel

8 Mathematical Problems in Engineering

Table 5 OHAM solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 minus000880011 minus107933 times 10minus6 00336596 minus754858 times 10minus8 00522402 minus774445 times 10minus9

02 minus0010881 minus22445 times 10minus6 00714091 minus173419 times 10minus7 01074223 minus251485 times 10minus8

03 0000449012 minus345224 times 10minus6 0117324 minus300407 times 10minus7 0168479 minus568899 times 10minus8

04 00318278 minus441058 times 10minus6 0175449 minus432397 times 10minus7 0238322 minus98056 times 10minus8

05 00898133 minus447787 times 10minus6 0249785 minus500627 times 10minus7 0319832 minus129675 times 10minus7

06 018086 minus260114 times 10minus6 0344278 minus377978 times 10minus7 0415852 minus113733 times 10minus7

07 0311298 265136 times 10minus6 0462801 130706 times 10minus7 0529175 120785 times 10minus8

08 0487312 128583 times 10minus5 0609145 128166 times 10minus6 0662538 336724 times 10minus7

09 0714929 291822 times 10minus5 0787012 334515 times 10minus6 0818613 967763 times 10minus7

10 1 514019 times 10minus5 1 649074 times 10minus6 1 200176 times 10minus6

Table 6 RK4 solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 204594 times 10minus5 0 734794 times 10minus6 0 356291 times 10minus6

01 minus000880012 772953 times 10minus7 00336596 321066 times 10minus7 00522402 163598 times 10minus7

02 minus00108811 minus18406 times 10minus7 00714091 minus775305 times 10minus8 01074223 minus396814 times 10minus8

03 0000448983 482747 times 10minus8 0117324 208509 times 10minus8 0168479 10753 times 10minus8

04 00318277 minus154235 times 10minus8 0175449 minus643999 times 10minus9 0238322 minus328063 times 10minus9

05 00898132 514031 times 10minus9 0249785 121029 times 10minus9 0319832 450154 times 10minus10

06 018086 minus651866 times 10minus10 0344278 264494 times 10minus9 0415852 184079 times 10minus9

07 0311298 minus186837 times 10minus9 0462801 minus988382 times 10minus9 0529175 minus651979 times 10minus9

08 0487312 minus133003 times 10minus9 0609145 337336 times 10minus8 0662538 227322 times 10minus8

09 0714929 266400 times 10minus8 0787012 minus132232 times 10minus7 0818613 minus905714 times 10minus8

10 1 555277 times 10minus6 1 minus11671 times 10minus6 1 minus11767 times 10minus6

Table 7 Comparison of OHAM and RK4 solutions for various 119877 in case of slip and no-slip boundary

120578

In case of no-slip boundary In case of slip boundary|RK4 Solution minusOHAM Solution| |RK4 Solution minusOHAM Solution|

119877 = 01 119877 = 03 119877 = 05 119877 = 02 119877 = 03 119877 = 04

00 0 0 0 0 0 001 233147 times 10minus15 162249 times 10minus10 241725 times 10minus9 844379 times 10minus10 790896 times 10minus9 41479 times 10minus8

02 949241 times 10minus14 314228 times 10minus10 467809 times 10minus9 16672 times 10minus9 155832 times 10minus8 815527 times 10minus8

03 382361 times 10minus13 446143 times 10minus10 663065 times 10minus9 244889 times 10minus9 228102 times 10minus8 118952 times 10minus7

04 913492 times 10minus13 547932 times 10minus10 811705 times 10minus9 31663 times 10minus9 293539 times 10minus8 152339 times 10minus7

05 165451 times 10minus12 607476 times 10minus10 895202 times 10minus9 377514 times 10minus9 348026 times 10minus8 179574 times 10minus7

06 242317 times 10minus12 609434 times 10minus10 891445 times 10minus9 418241 times 10minus9 38327 times 10minus8 196532 times 10minus7

07 286218 times 10minus12 536833 times 10minus10 777928 times 10minus9 422088 times 10minus9 384569 times 10minus8 196009 times 10minus7

08 252039 times 10minus12 378563 times 10minus10 54278 times 10minus9 365208 times 10minus9 331085 times 10minus8 167861 times 10minus7

09 120232 times 10minus12 154202 times 10minus10 218716 times 10minus9 224454 times 10minus9 202732 times 10minus8 102385 times 10minus7

10 126281 times 10minus16 253545 times 10minus16 293337 times 10minus17 17436 times 10minus14 492715 times 10minus14 38179 times 10minus14

8 Conclusions

In this article we find the similarity solution for unsteadyaxisymmetric squeezing flow of incompressible Newtonianfluid between two circular plates We observed that thesimilarity solution exists only when distance between theplates varies as (119882119905 + 119878)

12 and squeezing flow occurs when

119882 lt 0 119878 gt 0 and (119882119905+119878) gt 0The key findings of the presentanalysis are as follows

In case of no-slip at boundary

(i) It has been found that increase in Reynolds number 119877increases the normal velocity

Mathematical Problems in Engineering 9

00010

00008

00006

00004

00002

02 04 06 08 10

120578

Res

R = 02

R = 03

R = 04

Figure 3 OHAM residuals at various values of 119877 in case of slip boundary

00001

000008

000006

000004

000002

120574 = 05

120574 = 06

120574 = 07

02 04 06 08 10

120578

Res

Figure 4 OHAM residuals at various values of 120574 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

(120578

120578

)

R = 01

R = 1

R = 2

R = 3

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

R = 01

R = 1

R = 2

R = 3

998400 (120578)

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 5 Velocity profiles for various values of 119877 = 01 1 2 3 in case of no-slip boundary

10 Mathematical Problems in Engineering

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

R = 02

R = 06

R = 1

R = 15

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

R = 02

R = 06

R = 1

R = 15

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 6 Velocity profiles for various values of 119877 = 02 06 1 15 fixing 120574 = 1 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

120574 = 06120574 = 07

120574 = 08

120574 = 1

(a) The effect of 120574 on the Normal velocity profiles

120574 = 06120574 = 07

120574 = 08

120574 = 1

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

(b) The effect of 120574 on the longitudinal velocity profiles

Figure 7 Velocity profiles for various values of 120574 = 06 07 08 1 fixing 119877 = 03 in case of slip boundary

(ii) It has been observed that normal velocity increasesmonotonically from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time

(iii) It has been seen that longitudinal velocity deceasesnear the walls and increases near the central axis ofthe channel

In case of slip at boundary

(i) It has been noted that after fixing slip parameter120574 and varying the Reynolds number 119877 the normalvelocity profile decreases with the increase in 119877 Alsothe longitudinal velocity increases near the walls butdecreases near the central axis of the channel

(ii) It has been examined that for a fixed Reynoldsnumber 119877when we vary slip parameter 120574 the normalvelocity increases with the increase in 120574 Also the

longitudinal velocity decreases near the walls andincreases near the central axis of the channel

(iii) It has been investigated that Reynolds number 119877 andslip parameter 120574 have opposite effects on the normaland longitudinal velocity components

In case of slip versus no-slip boundary

(i) It has been observed that Reynolds number 119877 hasopposite behavior on the normal velocity in case ofslip and no-slip boundaries

(ii) It has been also noticed that Reynolds number 119877 hasopposite effect on the longitudinal velocity near thecentral axis of the channel while near the wall longi-tudinal velocity increases in case of slip boundary anddecease in no-slip boundary This is in conformanceto [33]

Mathematical Problems in Engineering 11

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] Q K Ghori M Ahmed and A M Siddiqui ldquoApplicationof homotopy perturbation method to squeezing flow of anewtonian fluidrdquo International Journal of Nonlinear Sciencesand Numerical Simulation vol 8 no 2 pp 179ndash184 2007

[2] X J Ran Q Y Zhu and Y Li ldquoAn explicit series solution ofthe squeezing flow between two infinite plates by means ofthe homotopy analysis methodrdquo Communications in NonlinearScience and Numerical Simulation vol 14 no 1 pp 119ndash1322009

[3] M J Stefan ldquoVersuch Uber die scheinbare adhasionrdquo Sitzungs-berichteOsterreichische Akademie der Wissenschaften in WienMathematisch-Naturwissenschaftliche Klasse vol 69 pp 713ndash721 1874

[4] J F Thorpe and W A Shaw Eds Developments in Theoreticaland Applied Mechanics vol 3 Pergamon Press Oxford UK1967

[5] P S Gupta and A S Gupta ldquoSqueezing flow between parallelplatesrdquoWear vol 45 no 2 pp 177ndash185 1977

[6] A F Elkouh ldquoFluid inertia effects in squeeze film between twoplane annulirdquo Transactions of the ASME Journal of Tribologyvol 106 no 2 pp 223ndash227 1984

[7] R L Verma ldquoA numerical solution for squeezing flow betweenparallel channelsrdquoWear vol 72 no 1 pp 89ndash95 1981

[8] P Singh V Radhakrishnan and K A Narayan ldquoSqueezing flowbetween parallel platesrdquo Ingenieur-Archiv vol 60 no 4 pp 274ndash281 1990

[9] P J Leider and R B Bird ldquoSqueezing flow between paralleldisks I Theoretical analysisrdquo Industrial and Engineering Chem-istry Fundamentals vol 13 no 4 pp 336ndash341 1974

[10] G Domairry and A Aziz ldquoApproximate analysis of MHDdqueeze flow between two parallel disks with suction orinjection by homotopy perturbation methodrdquo MathematicalProblems in Engineering vol 2009 Article ID 603916 19 pages2009

[11] S Islam H Khan I A Shah and G Zaman ldquoAn axisymmetricsqueezing fluid flow between the two infinite parallel plates in aporous medium channelrdquo Mathematical Problems in Engineer-ing vol 2011 Article ID 349803 10 pages 2011

[12] I Ullah H Khan and M T Rahim ldquoApproximation of firstgrade MHD squeezing fluid flow with slip boundary conditionusing DTM and OHAMrdquo Mathematical Problems in Engineer-ing vol 2013 Article ID 816262 9 pages 2013

[13] A M Siddiqui S Irum and A R Ansari ldquoUnsteady squeezingflowof a viscousMHDfluid between parallel platesrdquoMathemat-ical Modelling and Analysis vol 13 pp 565ndash576 2008

[14] R J Grimm ldquoSqueezing flows of Newtonian liquid films ananalysis including fluid inertiardquo Applied Scientific Research vol32 no 2 pp 149ndash166 1976

[15] M M Rashidi A M Siddiqui and M T Rastegari ldquoAnalyticalsolution of squeezing flow between two circular platesrdquo Interna-tional Journal for ComputationalMethods in Engineering Scienceand Mechanics vol 13 no 5 pp 342ndash349 2012

[16] H M Laun M Rady and O Hassager ldquoAnalytical solutions forsqueeze flow with partial wall sliprdquo Journal of Non-NewtonianFluid Mechanics vol 81 no 1-2 pp 1ndash15 1999

[17] S Ishizawa ldquoThe unsteady flow between two parallel discs witharbitrary varying gap widthrdquo Bulletin of the Japan Society ofMechanical Engineers vol 9 no 35 pp 533ndash550 1966

[18] C le Roux ldquoExistence and uniqueness of the flow of second-grade fluids with slip boundary conditionsrdquoArchive for RationalMechanics and Analysis vol 148 no 4 pp 309ndash356 1999

[19] L J Rhooades R Resnic T OrsquoBradovich and S StegmanldquoAbrasive flow machining of cylinder heads and its positiveeffects on performance and cost characteristicsrdquo Tech RepDearborn Mich USA 1996

[20] C L M H Navier Memoirs de lrsquoCademie Royale des Sciencesde lrsquoInstitut de France vol 1 Royale des Sciences de lrsquoInstitut deFrance 1823

[21] A Ebaid ldquoEffects of magnetic field and wall slip conditions onthe peristaltic transport of a Newtonian fluid in an asymmetricchannelrdquo Physics Letters Section A General Atomic and SolidState Physics vol 372 no 24 pp 4493ndash4499 2008

[22] J-H He ldquoSome asymptotic methods for strongly nonlinearequationsrdquo International Journal of Modern Physics B Con-densed Matter Physics Statistical Physics Applied Physics vol20 no 10 pp 1141ndash1199 2006

[23] S J Liao Proposed homotopy analysis techniques for the solutionof nonlinear problems [PhD thesis] Jiao Tong UniversityShanghai China 1992

[24] J-H He ldquoHomotopy perturbation method for solving bound-ary value problemsrdquo Physics Letters A vol 350 no 1-2 pp 87ndash88 2006

[25] J-H He ldquoApproximate analytical solution for seepage flowwithfractional derivatives in porous mediardquo Computer Methods inApplied Mechanics and Engineering vol 167 no 1-2 pp 57ndash681998

[26] VMarinca andNHerisanu ldquoApplication of optimalHomotopyasymptotic method for solving nonlinear equations arising inheat transferrdquo International Communications in Heat and MassTransfer vol 35 no 6 pp 710ndash715 2008

[27] V Marinca N Herisanu C Bota and B Marinca ldquoAn optimalhomotopy asymptotic method applied to the steady flow of afourth-grade fluid past a porous platerdquo Applied MathematicsLetters vol 22 no 2 pp 245ndash251 2009

[28] V Marinca N Herisanu and I Nemes ldquoOptimal homotopyasymptotic method with application to thin film flowrdquo CentralEuropean Journal of Physics vol 6 no 3 pp 648ndash653 2008

[29] J Ali S Islam and G Zaman ldquoThe solution of multipointboundary value problems by the optimal Homotopy asymptoticMethodrdquoComputers andMathematics with Applications vol 59no 6 pp 2000ndash2006 2010

[30] M Idrees S Islam and S Haq ldquoApplication of the optimalhomotopy asymptotic method to squeezing flowrdquo Computersand Mathematics with Applications vol 59 no 12 pp 3858ndash3866 2010

[31] J Ali S Islam H Khan and S I Shah ldquoThe optimal homotopyasymptotic method for the solution of higher-order boundaryvalue problems in finite domainsrdquo Abstract and Applied Analy-sis vol 2012 Article ID 401217 14 pages 2012

[32] H Khan S Islam J Ali and I Ali Shah ldquoComparison ofdifferent analytic solutions to axisymmetric squeezing fluidflow between two infinite parallel plates with slip boundaryconditionsrdquo Abstract and Applied Analysis vol 2012 Article ID835268 18 pages 2012

[33] T C Papanastasiou G C Georgiou and A N AlexandrouViscous Fluid Flow Library of Congress Cataloging CRC PressNew York NY USA 1999

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Stochastic AnalysisInternational Journal of

Page 7: Research Article Analysis of Unsteady Axisymmetric ...downloads.hindawi.com/journals/mpe/2015/860857.pdfResearch Article Analysis of Unsteady Axisymmetric Squeezing Fluid Flow with

Mathematical Problems in Engineering 7

Table 2 RK4 solutions along with residuals for various 119877 in case of no-slip boundary

120578119877 = 01 119877 = 03 119877 = 05

Solution Residual Solution Residual Solution Residual00 0 288535 times 10minus7 0 966906 times 10minus5 0 530549 times 10minus4

01 0150158 388278 times 10minus8 0151534 571393 times 10minus6 0152999 304600 times 10minus5

02 0297237 minus997789 times 10minus9 0299827 minus141434 times 10minus6 0302582 minus752772 times 10minus6

03 0438170 293767 times 10minus9 0441661 395997 times 10minus7 0445373 210296 times 10minus6

04 0569900 minus751436 times 10minus10 0573869 minus115203 times 10minus7 0578082 minus618081 times 10minus7

05 0689397 minus471705 times 10minus10 0693354 minus96761 times 10minus9 0697548 minus295916 times 10minus8

06 0793661 223004 times 10minus9 0797122 14593 times 10minus7 0800780 714133 times 10minus7

07 0879734 minus718245 times 10minus9 0882300 minus486272 times 10minus7 0885004 minus239245 times 10minus6

08 0944705 268744 times 10minus8 0946167 177796 times 10minus6 0947702 870664 times 10minus6

09 0985722 minus111599 times 10minus7 0986180 minus728952 times 10minus6 0986659 356157 times 10minus5

10 1 minus271683 times 10minus6 1 minus147695 times 10minus4 1 690088 times 10minus4

Table 3 OHAM solutions along with residuals for various values of 119877 in case of slip boundary

120578119877 = 02 119877 = 03 119877 = 04

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 0072692 318345 times 10minus8 0071220 315621 times 10minus7 0069593 174626 times 10minus6

02 0147091 minus172789 times 10minus8 0144268 minus914579 times 10minus8 0141147 minus113148 times 10minus7

03 0224893 minus196995 times 10minus7 0220953 minus166461 times 10minus6 0216599 minus779061 times 10minus6

04 0307773 minus494482 times 10minus7 0303048 minus429173 times 10minus6 0297832 minus207401 times 10minus5

05 0397370 minus799437 times 10minus7 0392277 minus699917 times 10minus6 0386657 minus341643 times 10minus5

06 0495283 minus85784 times 10minus7 0490289 minus75499 times 10minus6 0484784 minus370689 times 10minus5

07 0603056 minus21053 times 10minus7 0598654 minus191141 times 10minus6 0593805 minus96958 times 10minus6

08 0722172 187065 times 10minus6 0718841 162936 times 10minus5 0715176 791082 times 10minus5

09 0854042 640576 times 10minus6 0852211 559717 times 10minus5 0850196 272708 times 10minus4

10 1 146076 times 10minus5 1 127671 times 10minus4 1 622242 times 10minus4

Table 4 RK4 solutions along with residuals for various values of 119877 in case of slip boundary

120578119877 = 02 119877 = 03 119877 = 04

Solution Residual Solution Residual Solution Residual00 0 412179 times 10minus6 0 756765 times 10minus6 0 362897 times 10minus6

01 0072692 174999 times 10minus7 0071220 216125 times 10minus7 0069593 minus349994 times 10minus7

02 0147091 minus421503 times 10minus8 0144268 minus497423 times 10minus8 0141147 953266 times 10minus8

03 0224893 112883 times 10minus8 0220953 122294 times 10minus8 0216598 minus30735 times 10minus8

04 0307773 minus352272 times 10minus9 0303048 minus430187 times 10minus9 0297832 72493 times 10minus9

05 0397370 780684 times 10minus10 0392276 306668 times 10minus9 0386657 846409 times 10minus9

06 0495283 110391 times 10minus9 0490289 minus528586 times 10minus9 0484783 minus336828 times 10minus8

07 0603056 minus435033 times 10minus9 0598654 154159 times 10minus8 0593805 106816 times 10minus7

08 0722172 144198 times 10minus8 0718841 minus605873 times 10minus8 0715175 minus398519 times 10minus7

09 0854042 minus553933 times 10minus8 0852211 259233 times 10minus7 0850196 165518 times 10minus6

10 1 minus197434 times 10minus7 1 796168 times 10minus6 1 389126 times 10minus5

is experienced that this component of velocity deceases nearthe wall but increases near the central axis of the channel

The effect of Reynolds number 119877 on velocity profiles incase of slip boundary is depicted in Figure 6 In these profileswe fixed slip parameter 120574 = 1 and varied Reynolds number119877 as 119877 = 02 06 1 15 It is noted that the normal velocitydecreases as the Reynolds number increases (Figure 6(a))It is also observed that longitudinal velocity decreases near

the central axis of the channel but increases near the wallswhen 119877 increases (Figure 6(b))

Figure 7 demonstrates the effect of slip parameter 120574 on thevelocity profiles After fixing Reynolds number 119877 = 03 wevaried 120574 as 120574 = 06 07 08 1 We find that normal velocityincreases as 120574 increases It is also noted that longitudinalvelocity decreases near the walls but increases near centralaxis of the channel

8 Mathematical Problems in Engineering

Table 5 OHAM solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 minus000880011 minus107933 times 10minus6 00336596 minus754858 times 10minus8 00522402 minus774445 times 10minus9

02 minus0010881 minus22445 times 10minus6 00714091 minus173419 times 10minus7 01074223 minus251485 times 10minus8

03 0000449012 minus345224 times 10minus6 0117324 minus300407 times 10minus7 0168479 minus568899 times 10minus8

04 00318278 minus441058 times 10minus6 0175449 minus432397 times 10minus7 0238322 minus98056 times 10minus8

05 00898133 minus447787 times 10minus6 0249785 minus500627 times 10minus7 0319832 minus129675 times 10minus7

06 018086 minus260114 times 10minus6 0344278 minus377978 times 10minus7 0415852 minus113733 times 10minus7

07 0311298 265136 times 10minus6 0462801 130706 times 10minus7 0529175 120785 times 10minus8

08 0487312 128583 times 10minus5 0609145 128166 times 10minus6 0662538 336724 times 10minus7

09 0714929 291822 times 10minus5 0787012 334515 times 10minus6 0818613 967763 times 10minus7

10 1 514019 times 10minus5 1 649074 times 10minus6 1 200176 times 10minus6

Table 6 RK4 solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 204594 times 10minus5 0 734794 times 10minus6 0 356291 times 10minus6

01 minus000880012 772953 times 10minus7 00336596 321066 times 10minus7 00522402 163598 times 10minus7

02 minus00108811 minus18406 times 10minus7 00714091 minus775305 times 10minus8 01074223 minus396814 times 10minus8

03 0000448983 482747 times 10minus8 0117324 208509 times 10minus8 0168479 10753 times 10minus8

04 00318277 minus154235 times 10minus8 0175449 minus643999 times 10minus9 0238322 minus328063 times 10minus9

05 00898132 514031 times 10minus9 0249785 121029 times 10minus9 0319832 450154 times 10minus10

06 018086 minus651866 times 10minus10 0344278 264494 times 10minus9 0415852 184079 times 10minus9

07 0311298 minus186837 times 10minus9 0462801 minus988382 times 10minus9 0529175 minus651979 times 10minus9

08 0487312 minus133003 times 10minus9 0609145 337336 times 10minus8 0662538 227322 times 10minus8

09 0714929 266400 times 10minus8 0787012 minus132232 times 10minus7 0818613 minus905714 times 10minus8

10 1 555277 times 10minus6 1 minus11671 times 10minus6 1 minus11767 times 10minus6

Table 7 Comparison of OHAM and RK4 solutions for various 119877 in case of slip and no-slip boundary

120578

In case of no-slip boundary In case of slip boundary|RK4 Solution minusOHAM Solution| |RK4 Solution minusOHAM Solution|

119877 = 01 119877 = 03 119877 = 05 119877 = 02 119877 = 03 119877 = 04

00 0 0 0 0 0 001 233147 times 10minus15 162249 times 10minus10 241725 times 10minus9 844379 times 10minus10 790896 times 10minus9 41479 times 10minus8

02 949241 times 10minus14 314228 times 10minus10 467809 times 10minus9 16672 times 10minus9 155832 times 10minus8 815527 times 10minus8

03 382361 times 10minus13 446143 times 10minus10 663065 times 10minus9 244889 times 10minus9 228102 times 10minus8 118952 times 10minus7

04 913492 times 10minus13 547932 times 10minus10 811705 times 10minus9 31663 times 10minus9 293539 times 10minus8 152339 times 10minus7

05 165451 times 10minus12 607476 times 10minus10 895202 times 10minus9 377514 times 10minus9 348026 times 10minus8 179574 times 10minus7

06 242317 times 10minus12 609434 times 10minus10 891445 times 10minus9 418241 times 10minus9 38327 times 10minus8 196532 times 10minus7

07 286218 times 10minus12 536833 times 10minus10 777928 times 10minus9 422088 times 10minus9 384569 times 10minus8 196009 times 10minus7

08 252039 times 10minus12 378563 times 10minus10 54278 times 10minus9 365208 times 10minus9 331085 times 10minus8 167861 times 10minus7

09 120232 times 10minus12 154202 times 10minus10 218716 times 10minus9 224454 times 10minus9 202732 times 10minus8 102385 times 10minus7

10 126281 times 10minus16 253545 times 10minus16 293337 times 10minus17 17436 times 10minus14 492715 times 10minus14 38179 times 10minus14

8 Conclusions

In this article we find the similarity solution for unsteadyaxisymmetric squeezing flow of incompressible Newtonianfluid between two circular plates We observed that thesimilarity solution exists only when distance between theplates varies as (119882119905 + 119878)

12 and squeezing flow occurs when

119882 lt 0 119878 gt 0 and (119882119905+119878) gt 0The key findings of the presentanalysis are as follows

In case of no-slip at boundary

(i) It has been found that increase in Reynolds number 119877increases the normal velocity

Mathematical Problems in Engineering 9

00010

00008

00006

00004

00002

02 04 06 08 10

120578

Res

R = 02

R = 03

R = 04

Figure 3 OHAM residuals at various values of 119877 in case of slip boundary

00001

000008

000006

000004

000002

120574 = 05

120574 = 06

120574 = 07

02 04 06 08 10

120578

Res

Figure 4 OHAM residuals at various values of 120574 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

(120578

120578

)

R = 01

R = 1

R = 2

R = 3

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

R = 01

R = 1

R = 2

R = 3

998400 (120578)

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 5 Velocity profiles for various values of 119877 = 01 1 2 3 in case of no-slip boundary

10 Mathematical Problems in Engineering

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

R = 02

R = 06

R = 1

R = 15

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

R = 02

R = 06

R = 1

R = 15

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 6 Velocity profiles for various values of 119877 = 02 06 1 15 fixing 120574 = 1 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

120574 = 06120574 = 07

120574 = 08

120574 = 1

(a) The effect of 120574 on the Normal velocity profiles

120574 = 06120574 = 07

120574 = 08

120574 = 1

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

(b) The effect of 120574 on the longitudinal velocity profiles

Figure 7 Velocity profiles for various values of 120574 = 06 07 08 1 fixing 119877 = 03 in case of slip boundary

(ii) It has been observed that normal velocity increasesmonotonically from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time

(iii) It has been seen that longitudinal velocity deceasesnear the walls and increases near the central axis ofthe channel

In case of slip at boundary

(i) It has been noted that after fixing slip parameter120574 and varying the Reynolds number 119877 the normalvelocity profile decreases with the increase in 119877 Alsothe longitudinal velocity increases near the walls butdecreases near the central axis of the channel

(ii) It has been examined that for a fixed Reynoldsnumber 119877when we vary slip parameter 120574 the normalvelocity increases with the increase in 120574 Also the

longitudinal velocity decreases near the walls andincreases near the central axis of the channel

(iii) It has been investigated that Reynolds number 119877 andslip parameter 120574 have opposite effects on the normaland longitudinal velocity components

In case of slip versus no-slip boundary

(i) It has been observed that Reynolds number 119877 hasopposite behavior on the normal velocity in case ofslip and no-slip boundaries

(ii) It has been also noticed that Reynolds number 119877 hasopposite effect on the longitudinal velocity near thecentral axis of the channel while near the wall longi-tudinal velocity increases in case of slip boundary anddecease in no-slip boundary This is in conformanceto [33]

Mathematical Problems in Engineering 11

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] Q K Ghori M Ahmed and A M Siddiqui ldquoApplicationof homotopy perturbation method to squeezing flow of anewtonian fluidrdquo International Journal of Nonlinear Sciencesand Numerical Simulation vol 8 no 2 pp 179ndash184 2007

[2] X J Ran Q Y Zhu and Y Li ldquoAn explicit series solution ofthe squeezing flow between two infinite plates by means ofthe homotopy analysis methodrdquo Communications in NonlinearScience and Numerical Simulation vol 14 no 1 pp 119ndash1322009

[3] M J Stefan ldquoVersuch Uber die scheinbare adhasionrdquo Sitzungs-berichteOsterreichische Akademie der Wissenschaften in WienMathematisch-Naturwissenschaftliche Klasse vol 69 pp 713ndash721 1874

[4] J F Thorpe and W A Shaw Eds Developments in Theoreticaland Applied Mechanics vol 3 Pergamon Press Oxford UK1967

[5] P S Gupta and A S Gupta ldquoSqueezing flow between parallelplatesrdquoWear vol 45 no 2 pp 177ndash185 1977

[6] A F Elkouh ldquoFluid inertia effects in squeeze film between twoplane annulirdquo Transactions of the ASME Journal of Tribologyvol 106 no 2 pp 223ndash227 1984

[7] R L Verma ldquoA numerical solution for squeezing flow betweenparallel channelsrdquoWear vol 72 no 1 pp 89ndash95 1981

[8] P Singh V Radhakrishnan and K A Narayan ldquoSqueezing flowbetween parallel platesrdquo Ingenieur-Archiv vol 60 no 4 pp 274ndash281 1990

[9] P J Leider and R B Bird ldquoSqueezing flow between paralleldisks I Theoretical analysisrdquo Industrial and Engineering Chem-istry Fundamentals vol 13 no 4 pp 336ndash341 1974

[10] G Domairry and A Aziz ldquoApproximate analysis of MHDdqueeze flow between two parallel disks with suction orinjection by homotopy perturbation methodrdquo MathematicalProblems in Engineering vol 2009 Article ID 603916 19 pages2009

[11] S Islam H Khan I A Shah and G Zaman ldquoAn axisymmetricsqueezing fluid flow between the two infinite parallel plates in aporous medium channelrdquo Mathematical Problems in Engineer-ing vol 2011 Article ID 349803 10 pages 2011

[12] I Ullah H Khan and M T Rahim ldquoApproximation of firstgrade MHD squeezing fluid flow with slip boundary conditionusing DTM and OHAMrdquo Mathematical Problems in Engineer-ing vol 2013 Article ID 816262 9 pages 2013

[13] A M Siddiqui S Irum and A R Ansari ldquoUnsteady squeezingflowof a viscousMHDfluid between parallel platesrdquoMathemat-ical Modelling and Analysis vol 13 pp 565ndash576 2008

[14] R J Grimm ldquoSqueezing flows of Newtonian liquid films ananalysis including fluid inertiardquo Applied Scientific Research vol32 no 2 pp 149ndash166 1976

[15] M M Rashidi A M Siddiqui and M T Rastegari ldquoAnalyticalsolution of squeezing flow between two circular platesrdquo Interna-tional Journal for ComputationalMethods in Engineering Scienceand Mechanics vol 13 no 5 pp 342ndash349 2012

[16] H M Laun M Rady and O Hassager ldquoAnalytical solutions forsqueeze flow with partial wall sliprdquo Journal of Non-NewtonianFluid Mechanics vol 81 no 1-2 pp 1ndash15 1999

[17] S Ishizawa ldquoThe unsteady flow between two parallel discs witharbitrary varying gap widthrdquo Bulletin of the Japan Society ofMechanical Engineers vol 9 no 35 pp 533ndash550 1966

[18] C le Roux ldquoExistence and uniqueness of the flow of second-grade fluids with slip boundary conditionsrdquoArchive for RationalMechanics and Analysis vol 148 no 4 pp 309ndash356 1999

[19] L J Rhooades R Resnic T OrsquoBradovich and S StegmanldquoAbrasive flow machining of cylinder heads and its positiveeffects on performance and cost characteristicsrdquo Tech RepDearborn Mich USA 1996

[20] C L M H Navier Memoirs de lrsquoCademie Royale des Sciencesde lrsquoInstitut de France vol 1 Royale des Sciences de lrsquoInstitut deFrance 1823

[21] A Ebaid ldquoEffects of magnetic field and wall slip conditions onthe peristaltic transport of a Newtonian fluid in an asymmetricchannelrdquo Physics Letters Section A General Atomic and SolidState Physics vol 372 no 24 pp 4493ndash4499 2008

[22] J-H He ldquoSome asymptotic methods for strongly nonlinearequationsrdquo International Journal of Modern Physics B Con-densed Matter Physics Statistical Physics Applied Physics vol20 no 10 pp 1141ndash1199 2006

[23] S J Liao Proposed homotopy analysis techniques for the solutionof nonlinear problems [PhD thesis] Jiao Tong UniversityShanghai China 1992

[24] J-H He ldquoHomotopy perturbation method for solving bound-ary value problemsrdquo Physics Letters A vol 350 no 1-2 pp 87ndash88 2006

[25] J-H He ldquoApproximate analytical solution for seepage flowwithfractional derivatives in porous mediardquo Computer Methods inApplied Mechanics and Engineering vol 167 no 1-2 pp 57ndash681998

[26] VMarinca andNHerisanu ldquoApplication of optimalHomotopyasymptotic method for solving nonlinear equations arising inheat transferrdquo International Communications in Heat and MassTransfer vol 35 no 6 pp 710ndash715 2008

[27] V Marinca N Herisanu C Bota and B Marinca ldquoAn optimalhomotopy asymptotic method applied to the steady flow of afourth-grade fluid past a porous platerdquo Applied MathematicsLetters vol 22 no 2 pp 245ndash251 2009

[28] V Marinca N Herisanu and I Nemes ldquoOptimal homotopyasymptotic method with application to thin film flowrdquo CentralEuropean Journal of Physics vol 6 no 3 pp 648ndash653 2008

[29] J Ali S Islam and G Zaman ldquoThe solution of multipointboundary value problems by the optimal Homotopy asymptoticMethodrdquoComputers andMathematics with Applications vol 59no 6 pp 2000ndash2006 2010

[30] M Idrees S Islam and S Haq ldquoApplication of the optimalhomotopy asymptotic method to squeezing flowrdquo Computersand Mathematics with Applications vol 59 no 12 pp 3858ndash3866 2010

[31] J Ali S Islam H Khan and S I Shah ldquoThe optimal homotopyasymptotic method for the solution of higher-order boundaryvalue problems in finite domainsrdquo Abstract and Applied Analy-sis vol 2012 Article ID 401217 14 pages 2012

[32] H Khan S Islam J Ali and I Ali Shah ldquoComparison ofdifferent analytic solutions to axisymmetric squeezing fluidflow between two infinite parallel plates with slip boundaryconditionsrdquo Abstract and Applied Analysis vol 2012 Article ID835268 18 pages 2012

[33] T C Papanastasiou G C Georgiou and A N AlexandrouViscous Fluid Flow Library of Congress Cataloging CRC PressNew York NY USA 1999

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Page 8: Research Article Analysis of Unsteady Axisymmetric ...downloads.hindawi.com/journals/mpe/2015/860857.pdfResearch Article Analysis of Unsteady Axisymmetric Squeezing Fluid Flow with

8 Mathematical Problems in Engineering

Table 5 OHAM solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 0 0 0 0 001 minus000880011 minus107933 times 10minus6 00336596 minus754858 times 10minus8 00522402 minus774445 times 10minus9

02 minus0010881 minus22445 times 10minus6 00714091 minus173419 times 10minus7 01074223 minus251485 times 10minus8

03 0000449012 minus345224 times 10minus6 0117324 minus300407 times 10minus7 0168479 minus568899 times 10minus8

04 00318278 minus441058 times 10minus6 0175449 minus432397 times 10minus7 0238322 minus98056 times 10minus8

05 00898133 minus447787 times 10minus6 0249785 minus500627 times 10minus7 0319832 minus129675 times 10minus7

06 018086 minus260114 times 10minus6 0344278 minus377978 times 10minus7 0415852 minus113733 times 10minus7

07 0311298 265136 times 10minus6 0462801 130706 times 10minus7 0529175 120785 times 10minus8

08 0487312 128583 times 10minus5 0609145 128166 times 10minus6 0662538 336724 times 10minus7

09 0714929 291822 times 10minus5 0787012 334515 times 10minus6 0818613 967763 times 10minus7

10 1 514019 times 10minus5 1 649074 times 10minus6 1 200176 times 10minus6

Table 6 RK4 solutions along with residuals for various values of 120574 in case of slip boundary

120578120574 = 05 120574 = 06 120574 = 07

Solution Residual Solution Residual Solution Residual00 0 204594 times 10minus5 0 734794 times 10minus6 0 356291 times 10minus6

01 minus000880012 772953 times 10minus7 00336596 321066 times 10minus7 00522402 163598 times 10minus7

02 minus00108811 minus18406 times 10minus7 00714091 minus775305 times 10minus8 01074223 minus396814 times 10minus8

03 0000448983 482747 times 10minus8 0117324 208509 times 10minus8 0168479 10753 times 10minus8

04 00318277 minus154235 times 10minus8 0175449 minus643999 times 10minus9 0238322 minus328063 times 10minus9

05 00898132 514031 times 10minus9 0249785 121029 times 10minus9 0319832 450154 times 10minus10

06 018086 minus651866 times 10minus10 0344278 264494 times 10minus9 0415852 184079 times 10minus9

07 0311298 minus186837 times 10minus9 0462801 minus988382 times 10minus9 0529175 minus651979 times 10minus9

08 0487312 minus133003 times 10minus9 0609145 337336 times 10minus8 0662538 227322 times 10minus8

09 0714929 266400 times 10minus8 0787012 minus132232 times 10minus7 0818613 minus905714 times 10minus8

10 1 555277 times 10minus6 1 minus11671 times 10minus6 1 minus11767 times 10minus6

Table 7 Comparison of OHAM and RK4 solutions for various 119877 in case of slip and no-slip boundary

120578

In case of no-slip boundary In case of slip boundary|RK4 Solution minusOHAM Solution| |RK4 Solution minusOHAM Solution|

119877 = 01 119877 = 03 119877 = 05 119877 = 02 119877 = 03 119877 = 04

00 0 0 0 0 0 001 233147 times 10minus15 162249 times 10minus10 241725 times 10minus9 844379 times 10minus10 790896 times 10minus9 41479 times 10minus8

02 949241 times 10minus14 314228 times 10minus10 467809 times 10minus9 16672 times 10minus9 155832 times 10minus8 815527 times 10minus8

03 382361 times 10minus13 446143 times 10minus10 663065 times 10minus9 244889 times 10minus9 228102 times 10minus8 118952 times 10minus7

04 913492 times 10minus13 547932 times 10minus10 811705 times 10minus9 31663 times 10minus9 293539 times 10minus8 152339 times 10minus7

05 165451 times 10minus12 607476 times 10minus10 895202 times 10minus9 377514 times 10minus9 348026 times 10minus8 179574 times 10minus7

06 242317 times 10minus12 609434 times 10minus10 891445 times 10minus9 418241 times 10minus9 38327 times 10minus8 196532 times 10minus7

07 286218 times 10minus12 536833 times 10minus10 777928 times 10minus9 422088 times 10minus9 384569 times 10minus8 196009 times 10minus7

08 252039 times 10minus12 378563 times 10minus10 54278 times 10minus9 365208 times 10minus9 331085 times 10minus8 167861 times 10minus7

09 120232 times 10minus12 154202 times 10minus10 218716 times 10minus9 224454 times 10minus9 202732 times 10minus8 102385 times 10minus7

10 126281 times 10minus16 253545 times 10minus16 293337 times 10minus17 17436 times 10minus14 492715 times 10minus14 38179 times 10minus14

8 Conclusions

In this article we find the similarity solution for unsteadyaxisymmetric squeezing flow of incompressible Newtonianfluid between two circular plates We observed that thesimilarity solution exists only when distance between theplates varies as (119882119905 + 119878)

12 and squeezing flow occurs when

119882 lt 0 119878 gt 0 and (119882119905+119878) gt 0The key findings of the presentanalysis are as follows

In case of no-slip at boundary

(i) It has been found that increase in Reynolds number 119877increases the normal velocity

Mathematical Problems in Engineering 9

00010

00008

00006

00004

00002

02 04 06 08 10

120578

Res

R = 02

R = 03

R = 04

Figure 3 OHAM residuals at various values of 119877 in case of slip boundary

00001

000008

000006

000004

000002

120574 = 05

120574 = 06

120574 = 07

02 04 06 08 10

120578

Res

Figure 4 OHAM residuals at various values of 120574 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

(120578

120578

)

R = 01

R = 1

R = 2

R = 3

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

R = 01

R = 1

R = 2

R = 3

998400 (120578)

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 5 Velocity profiles for various values of 119877 = 01 1 2 3 in case of no-slip boundary

10 Mathematical Problems in Engineering

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

R = 02

R = 06

R = 1

R = 15

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

R = 02

R = 06

R = 1

R = 15

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 6 Velocity profiles for various values of 119877 = 02 06 1 15 fixing 120574 = 1 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

120574 = 06120574 = 07

120574 = 08

120574 = 1

(a) The effect of 120574 on the Normal velocity profiles

120574 = 06120574 = 07

120574 = 08

120574 = 1

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

(b) The effect of 120574 on the longitudinal velocity profiles

Figure 7 Velocity profiles for various values of 120574 = 06 07 08 1 fixing 119877 = 03 in case of slip boundary

(ii) It has been observed that normal velocity increasesmonotonically from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time

(iii) It has been seen that longitudinal velocity deceasesnear the walls and increases near the central axis ofthe channel

In case of slip at boundary

(i) It has been noted that after fixing slip parameter120574 and varying the Reynolds number 119877 the normalvelocity profile decreases with the increase in 119877 Alsothe longitudinal velocity increases near the walls butdecreases near the central axis of the channel

(ii) It has been examined that for a fixed Reynoldsnumber 119877when we vary slip parameter 120574 the normalvelocity increases with the increase in 120574 Also the

longitudinal velocity decreases near the walls andincreases near the central axis of the channel

(iii) It has been investigated that Reynolds number 119877 andslip parameter 120574 have opposite effects on the normaland longitudinal velocity components

In case of slip versus no-slip boundary

(i) It has been observed that Reynolds number 119877 hasopposite behavior on the normal velocity in case ofslip and no-slip boundaries

(ii) It has been also noticed that Reynolds number 119877 hasopposite effect on the longitudinal velocity near thecentral axis of the channel while near the wall longi-tudinal velocity increases in case of slip boundary anddecease in no-slip boundary This is in conformanceto [33]

Mathematical Problems in Engineering 11

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] Q K Ghori M Ahmed and A M Siddiqui ldquoApplicationof homotopy perturbation method to squeezing flow of anewtonian fluidrdquo International Journal of Nonlinear Sciencesand Numerical Simulation vol 8 no 2 pp 179ndash184 2007

[2] X J Ran Q Y Zhu and Y Li ldquoAn explicit series solution ofthe squeezing flow between two infinite plates by means ofthe homotopy analysis methodrdquo Communications in NonlinearScience and Numerical Simulation vol 14 no 1 pp 119ndash1322009

[3] M J Stefan ldquoVersuch Uber die scheinbare adhasionrdquo Sitzungs-berichteOsterreichische Akademie der Wissenschaften in WienMathematisch-Naturwissenschaftliche Klasse vol 69 pp 713ndash721 1874

[4] J F Thorpe and W A Shaw Eds Developments in Theoreticaland Applied Mechanics vol 3 Pergamon Press Oxford UK1967

[5] P S Gupta and A S Gupta ldquoSqueezing flow between parallelplatesrdquoWear vol 45 no 2 pp 177ndash185 1977

[6] A F Elkouh ldquoFluid inertia effects in squeeze film between twoplane annulirdquo Transactions of the ASME Journal of Tribologyvol 106 no 2 pp 223ndash227 1984

[7] R L Verma ldquoA numerical solution for squeezing flow betweenparallel channelsrdquoWear vol 72 no 1 pp 89ndash95 1981

[8] P Singh V Radhakrishnan and K A Narayan ldquoSqueezing flowbetween parallel platesrdquo Ingenieur-Archiv vol 60 no 4 pp 274ndash281 1990

[9] P J Leider and R B Bird ldquoSqueezing flow between paralleldisks I Theoretical analysisrdquo Industrial and Engineering Chem-istry Fundamentals vol 13 no 4 pp 336ndash341 1974

[10] G Domairry and A Aziz ldquoApproximate analysis of MHDdqueeze flow between two parallel disks with suction orinjection by homotopy perturbation methodrdquo MathematicalProblems in Engineering vol 2009 Article ID 603916 19 pages2009

[11] S Islam H Khan I A Shah and G Zaman ldquoAn axisymmetricsqueezing fluid flow between the two infinite parallel plates in aporous medium channelrdquo Mathematical Problems in Engineer-ing vol 2011 Article ID 349803 10 pages 2011

[12] I Ullah H Khan and M T Rahim ldquoApproximation of firstgrade MHD squeezing fluid flow with slip boundary conditionusing DTM and OHAMrdquo Mathematical Problems in Engineer-ing vol 2013 Article ID 816262 9 pages 2013

[13] A M Siddiqui S Irum and A R Ansari ldquoUnsteady squeezingflowof a viscousMHDfluid between parallel platesrdquoMathemat-ical Modelling and Analysis vol 13 pp 565ndash576 2008

[14] R J Grimm ldquoSqueezing flows of Newtonian liquid films ananalysis including fluid inertiardquo Applied Scientific Research vol32 no 2 pp 149ndash166 1976

[15] M M Rashidi A M Siddiqui and M T Rastegari ldquoAnalyticalsolution of squeezing flow between two circular platesrdquo Interna-tional Journal for ComputationalMethods in Engineering Scienceand Mechanics vol 13 no 5 pp 342ndash349 2012

[16] H M Laun M Rady and O Hassager ldquoAnalytical solutions forsqueeze flow with partial wall sliprdquo Journal of Non-NewtonianFluid Mechanics vol 81 no 1-2 pp 1ndash15 1999

[17] S Ishizawa ldquoThe unsteady flow between two parallel discs witharbitrary varying gap widthrdquo Bulletin of the Japan Society ofMechanical Engineers vol 9 no 35 pp 533ndash550 1966

[18] C le Roux ldquoExistence and uniqueness of the flow of second-grade fluids with slip boundary conditionsrdquoArchive for RationalMechanics and Analysis vol 148 no 4 pp 309ndash356 1999

[19] L J Rhooades R Resnic T OrsquoBradovich and S StegmanldquoAbrasive flow machining of cylinder heads and its positiveeffects on performance and cost characteristicsrdquo Tech RepDearborn Mich USA 1996

[20] C L M H Navier Memoirs de lrsquoCademie Royale des Sciencesde lrsquoInstitut de France vol 1 Royale des Sciences de lrsquoInstitut deFrance 1823

[21] A Ebaid ldquoEffects of magnetic field and wall slip conditions onthe peristaltic transport of a Newtonian fluid in an asymmetricchannelrdquo Physics Letters Section A General Atomic and SolidState Physics vol 372 no 24 pp 4493ndash4499 2008

[22] J-H He ldquoSome asymptotic methods for strongly nonlinearequationsrdquo International Journal of Modern Physics B Con-densed Matter Physics Statistical Physics Applied Physics vol20 no 10 pp 1141ndash1199 2006

[23] S J Liao Proposed homotopy analysis techniques for the solutionof nonlinear problems [PhD thesis] Jiao Tong UniversityShanghai China 1992

[24] J-H He ldquoHomotopy perturbation method for solving bound-ary value problemsrdquo Physics Letters A vol 350 no 1-2 pp 87ndash88 2006

[25] J-H He ldquoApproximate analytical solution for seepage flowwithfractional derivatives in porous mediardquo Computer Methods inApplied Mechanics and Engineering vol 167 no 1-2 pp 57ndash681998

[26] VMarinca andNHerisanu ldquoApplication of optimalHomotopyasymptotic method for solving nonlinear equations arising inheat transferrdquo International Communications in Heat and MassTransfer vol 35 no 6 pp 710ndash715 2008

[27] V Marinca N Herisanu C Bota and B Marinca ldquoAn optimalhomotopy asymptotic method applied to the steady flow of afourth-grade fluid past a porous platerdquo Applied MathematicsLetters vol 22 no 2 pp 245ndash251 2009

[28] V Marinca N Herisanu and I Nemes ldquoOptimal homotopyasymptotic method with application to thin film flowrdquo CentralEuropean Journal of Physics vol 6 no 3 pp 648ndash653 2008

[29] J Ali S Islam and G Zaman ldquoThe solution of multipointboundary value problems by the optimal Homotopy asymptoticMethodrdquoComputers andMathematics with Applications vol 59no 6 pp 2000ndash2006 2010

[30] M Idrees S Islam and S Haq ldquoApplication of the optimalhomotopy asymptotic method to squeezing flowrdquo Computersand Mathematics with Applications vol 59 no 12 pp 3858ndash3866 2010

[31] J Ali S Islam H Khan and S I Shah ldquoThe optimal homotopyasymptotic method for the solution of higher-order boundaryvalue problems in finite domainsrdquo Abstract and Applied Analy-sis vol 2012 Article ID 401217 14 pages 2012

[32] H Khan S Islam J Ali and I Ali Shah ldquoComparison ofdifferent analytic solutions to axisymmetric squeezing fluidflow between two infinite parallel plates with slip boundaryconditionsrdquo Abstract and Applied Analysis vol 2012 Article ID835268 18 pages 2012

[33] T C Papanastasiou G C Georgiou and A N AlexandrouViscous Fluid Flow Library of Congress Cataloging CRC PressNew York NY USA 1999

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

Page 9: Research Article Analysis of Unsteady Axisymmetric ...downloads.hindawi.com/journals/mpe/2015/860857.pdfResearch Article Analysis of Unsteady Axisymmetric Squeezing Fluid Flow with

Mathematical Problems in Engineering 9

00010

00008

00006

00004

00002

02 04 06 08 10

120578

Res

R = 02

R = 03

R = 04

Figure 3 OHAM residuals at various values of 119877 in case of slip boundary

00001

000008

000006

000004

000002

120574 = 05

120574 = 06

120574 = 07

02 04 06 08 10

120578

Res

Figure 4 OHAM residuals at various values of 120574 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

(120578

120578

)

R = 01

R = 1

R = 2

R = 3

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

R = 01

R = 1

R = 2

R = 3

998400 (120578)

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 5 Velocity profiles for various values of 119877 = 01 1 2 3 in case of no-slip boundary

10 Mathematical Problems in Engineering

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

R = 02

R = 06

R = 1

R = 15

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

R = 02

R = 06

R = 1

R = 15

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 6 Velocity profiles for various values of 119877 = 02 06 1 15 fixing 120574 = 1 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

120574 = 06120574 = 07

120574 = 08

120574 = 1

(a) The effect of 120574 on the Normal velocity profiles

120574 = 06120574 = 07

120574 = 08

120574 = 1

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

(b) The effect of 120574 on the longitudinal velocity profiles

Figure 7 Velocity profiles for various values of 120574 = 06 07 08 1 fixing 119877 = 03 in case of slip boundary

(ii) It has been observed that normal velocity increasesmonotonically from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time

(iii) It has been seen that longitudinal velocity deceasesnear the walls and increases near the central axis ofthe channel

In case of slip at boundary

(i) It has been noted that after fixing slip parameter120574 and varying the Reynolds number 119877 the normalvelocity profile decreases with the increase in 119877 Alsothe longitudinal velocity increases near the walls butdecreases near the central axis of the channel

(ii) It has been examined that for a fixed Reynoldsnumber 119877when we vary slip parameter 120574 the normalvelocity increases with the increase in 120574 Also the

longitudinal velocity decreases near the walls andincreases near the central axis of the channel

(iii) It has been investigated that Reynolds number 119877 andslip parameter 120574 have opposite effects on the normaland longitudinal velocity components

In case of slip versus no-slip boundary

(i) It has been observed that Reynolds number 119877 hasopposite behavior on the normal velocity in case ofslip and no-slip boundaries

(ii) It has been also noticed that Reynolds number 119877 hasopposite effect on the longitudinal velocity near thecentral axis of the channel while near the wall longi-tudinal velocity increases in case of slip boundary anddecease in no-slip boundary This is in conformanceto [33]

Mathematical Problems in Engineering 11

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] Q K Ghori M Ahmed and A M Siddiqui ldquoApplicationof homotopy perturbation method to squeezing flow of anewtonian fluidrdquo International Journal of Nonlinear Sciencesand Numerical Simulation vol 8 no 2 pp 179ndash184 2007

[2] X J Ran Q Y Zhu and Y Li ldquoAn explicit series solution ofthe squeezing flow between two infinite plates by means ofthe homotopy analysis methodrdquo Communications in NonlinearScience and Numerical Simulation vol 14 no 1 pp 119ndash1322009

[3] M J Stefan ldquoVersuch Uber die scheinbare adhasionrdquo Sitzungs-berichteOsterreichische Akademie der Wissenschaften in WienMathematisch-Naturwissenschaftliche Klasse vol 69 pp 713ndash721 1874

[4] J F Thorpe and W A Shaw Eds Developments in Theoreticaland Applied Mechanics vol 3 Pergamon Press Oxford UK1967

[5] P S Gupta and A S Gupta ldquoSqueezing flow between parallelplatesrdquoWear vol 45 no 2 pp 177ndash185 1977

[6] A F Elkouh ldquoFluid inertia effects in squeeze film between twoplane annulirdquo Transactions of the ASME Journal of Tribologyvol 106 no 2 pp 223ndash227 1984

[7] R L Verma ldquoA numerical solution for squeezing flow betweenparallel channelsrdquoWear vol 72 no 1 pp 89ndash95 1981

[8] P Singh V Radhakrishnan and K A Narayan ldquoSqueezing flowbetween parallel platesrdquo Ingenieur-Archiv vol 60 no 4 pp 274ndash281 1990

[9] P J Leider and R B Bird ldquoSqueezing flow between paralleldisks I Theoretical analysisrdquo Industrial and Engineering Chem-istry Fundamentals vol 13 no 4 pp 336ndash341 1974

[10] G Domairry and A Aziz ldquoApproximate analysis of MHDdqueeze flow between two parallel disks with suction orinjection by homotopy perturbation methodrdquo MathematicalProblems in Engineering vol 2009 Article ID 603916 19 pages2009

[11] S Islam H Khan I A Shah and G Zaman ldquoAn axisymmetricsqueezing fluid flow between the two infinite parallel plates in aporous medium channelrdquo Mathematical Problems in Engineer-ing vol 2011 Article ID 349803 10 pages 2011

[12] I Ullah H Khan and M T Rahim ldquoApproximation of firstgrade MHD squeezing fluid flow with slip boundary conditionusing DTM and OHAMrdquo Mathematical Problems in Engineer-ing vol 2013 Article ID 816262 9 pages 2013

[13] A M Siddiqui S Irum and A R Ansari ldquoUnsteady squeezingflowof a viscousMHDfluid between parallel platesrdquoMathemat-ical Modelling and Analysis vol 13 pp 565ndash576 2008

[14] R J Grimm ldquoSqueezing flows of Newtonian liquid films ananalysis including fluid inertiardquo Applied Scientific Research vol32 no 2 pp 149ndash166 1976

[15] M M Rashidi A M Siddiqui and M T Rastegari ldquoAnalyticalsolution of squeezing flow between two circular platesrdquo Interna-tional Journal for ComputationalMethods in Engineering Scienceand Mechanics vol 13 no 5 pp 342ndash349 2012

[16] H M Laun M Rady and O Hassager ldquoAnalytical solutions forsqueeze flow with partial wall sliprdquo Journal of Non-NewtonianFluid Mechanics vol 81 no 1-2 pp 1ndash15 1999

[17] S Ishizawa ldquoThe unsteady flow between two parallel discs witharbitrary varying gap widthrdquo Bulletin of the Japan Society ofMechanical Engineers vol 9 no 35 pp 533ndash550 1966

[18] C le Roux ldquoExistence and uniqueness of the flow of second-grade fluids with slip boundary conditionsrdquoArchive for RationalMechanics and Analysis vol 148 no 4 pp 309ndash356 1999

[19] L J Rhooades R Resnic T OrsquoBradovich and S StegmanldquoAbrasive flow machining of cylinder heads and its positiveeffects on performance and cost characteristicsrdquo Tech RepDearborn Mich USA 1996

[20] C L M H Navier Memoirs de lrsquoCademie Royale des Sciencesde lrsquoInstitut de France vol 1 Royale des Sciences de lrsquoInstitut deFrance 1823

[21] A Ebaid ldquoEffects of magnetic field and wall slip conditions onthe peristaltic transport of a Newtonian fluid in an asymmetricchannelrdquo Physics Letters Section A General Atomic and SolidState Physics vol 372 no 24 pp 4493ndash4499 2008

[22] J-H He ldquoSome asymptotic methods for strongly nonlinearequationsrdquo International Journal of Modern Physics B Con-densed Matter Physics Statistical Physics Applied Physics vol20 no 10 pp 1141ndash1199 2006

[23] S J Liao Proposed homotopy analysis techniques for the solutionof nonlinear problems [PhD thesis] Jiao Tong UniversityShanghai China 1992

[24] J-H He ldquoHomotopy perturbation method for solving bound-ary value problemsrdquo Physics Letters A vol 350 no 1-2 pp 87ndash88 2006

[25] J-H He ldquoApproximate analytical solution for seepage flowwithfractional derivatives in porous mediardquo Computer Methods inApplied Mechanics and Engineering vol 167 no 1-2 pp 57ndash681998

[26] VMarinca andNHerisanu ldquoApplication of optimalHomotopyasymptotic method for solving nonlinear equations arising inheat transferrdquo International Communications in Heat and MassTransfer vol 35 no 6 pp 710ndash715 2008

[27] V Marinca N Herisanu C Bota and B Marinca ldquoAn optimalhomotopy asymptotic method applied to the steady flow of afourth-grade fluid past a porous platerdquo Applied MathematicsLetters vol 22 no 2 pp 245ndash251 2009

[28] V Marinca N Herisanu and I Nemes ldquoOptimal homotopyasymptotic method with application to thin film flowrdquo CentralEuropean Journal of Physics vol 6 no 3 pp 648ndash653 2008

[29] J Ali S Islam and G Zaman ldquoThe solution of multipointboundary value problems by the optimal Homotopy asymptoticMethodrdquoComputers andMathematics with Applications vol 59no 6 pp 2000ndash2006 2010

[30] M Idrees S Islam and S Haq ldquoApplication of the optimalhomotopy asymptotic method to squeezing flowrdquo Computersand Mathematics with Applications vol 59 no 12 pp 3858ndash3866 2010

[31] J Ali S Islam H Khan and S I Shah ldquoThe optimal homotopyasymptotic method for the solution of higher-order boundaryvalue problems in finite domainsrdquo Abstract and Applied Analy-sis vol 2012 Article ID 401217 14 pages 2012

[32] H Khan S Islam J Ali and I Ali Shah ldquoComparison ofdifferent analytic solutions to axisymmetric squeezing fluidflow between two infinite parallel plates with slip boundaryconditionsrdquo Abstract and Applied Analysis vol 2012 Article ID835268 18 pages 2012

[33] T C Papanastasiou G C Georgiou and A N AlexandrouViscous Fluid Flow Library of Congress Cataloging CRC PressNew York NY USA 1999

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

Page 10: Research Article Analysis of Unsteady Axisymmetric ...downloads.hindawi.com/journals/mpe/2015/860857.pdfResearch Article Analysis of Unsteady Axisymmetric Squeezing Fluid Flow with

10 Mathematical Problems in Engineering

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

R = 02

R = 06

R = 1

R = 15

(a) The effect of 119877 on the Normal velocity profiles

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

R = 02

R = 06

R = 1

R = 15

(b) The effect of 119877 on the longitudinal velocity profiles

Figure 6 Velocity profiles for various values of 119877 = 02 06 1 15 fixing 120574 = 1 in case of slip boundary

02

04

06

08

10

0200 04 06 08 10

u(120578

120578

)

120574 = 06120574 = 07

120574 = 08

120574 = 1

(a) The effect of 120574 on the Normal velocity profiles

120574 = 06120574 = 07

120574 = 08

120574 = 1

05

10

15

20

0200 04 06 08 10

120578

u998400 (120578)

(b) The effect of 120574 on the longitudinal velocity profiles

Figure 7 Velocity profiles for various values of 120574 = 06 07 08 1 fixing 119877 = 03 in case of slip boundary

(ii) It has been observed that normal velocity increasesmonotonically from 120578 = 0 to 120578 = 1 for fixed positivevalue of 119877 at a given time

(iii) It has been seen that longitudinal velocity deceasesnear the walls and increases near the central axis ofthe channel

In case of slip at boundary

(i) It has been noted that after fixing slip parameter120574 and varying the Reynolds number 119877 the normalvelocity profile decreases with the increase in 119877 Alsothe longitudinal velocity increases near the walls butdecreases near the central axis of the channel

(ii) It has been examined that for a fixed Reynoldsnumber 119877when we vary slip parameter 120574 the normalvelocity increases with the increase in 120574 Also the

longitudinal velocity decreases near the walls andincreases near the central axis of the channel

(iii) It has been investigated that Reynolds number 119877 andslip parameter 120574 have opposite effects on the normaland longitudinal velocity components

In case of slip versus no-slip boundary

(i) It has been observed that Reynolds number 119877 hasopposite behavior on the normal velocity in case ofslip and no-slip boundaries

(ii) It has been also noticed that Reynolds number 119877 hasopposite effect on the longitudinal velocity near thecentral axis of the channel while near the wall longi-tudinal velocity increases in case of slip boundary anddecease in no-slip boundary This is in conformanceto [33]

Mathematical Problems in Engineering 11

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] Q K Ghori M Ahmed and A M Siddiqui ldquoApplicationof homotopy perturbation method to squeezing flow of anewtonian fluidrdquo International Journal of Nonlinear Sciencesand Numerical Simulation vol 8 no 2 pp 179ndash184 2007

[2] X J Ran Q Y Zhu and Y Li ldquoAn explicit series solution ofthe squeezing flow between two infinite plates by means ofthe homotopy analysis methodrdquo Communications in NonlinearScience and Numerical Simulation vol 14 no 1 pp 119ndash1322009

[3] M J Stefan ldquoVersuch Uber die scheinbare adhasionrdquo Sitzungs-berichteOsterreichische Akademie der Wissenschaften in WienMathematisch-Naturwissenschaftliche Klasse vol 69 pp 713ndash721 1874

[4] J F Thorpe and W A Shaw Eds Developments in Theoreticaland Applied Mechanics vol 3 Pergamon Press Oxford UK1967

[5] P S Gupta and A S Gupta ldquoSqueezing flow between parallelplatesrdquoWear vol 45 no 2 pp 177ndash185 1977

[6] A F Elkouh ldquoFluid inertia effects in squeeze film between twoplane annulirdquo Transactions of the ASME Journal of Tribologyvol 106 no 2 pp 223ndash227 1984

[7] R L Verma ldquoA numerical solution for squeezing flow betweenparallel channelsrdquoWear vol 72 no 1 pp 89ndash95 1981

[8] P Singh V Radhakrishnan and K A Narayan ldquoSqueezing flowbetween parallel platesrdquo Ingenieur-Archiv vol 60 no 4 pp 274ndash281 1990

[9] P J Leider and R B Bird ldquoSqueezing flow between paralleldisks I Theoretical analysisrdquo Industrial and Engineering Chem-istry Fundamentals vol 13 no 4 pp 336ndash341 1974

[10] G Domairry and A Aziz ldquoApproximate analysis of MHDdqueeze flow between two parallel disks with suction orinjection by homotopy perturbation methodrdquo MathematicalProblems in Engineering vol 2009 Article ID 603916 19 pages2009

[11] S Islam H Khan I A Shah and G Zaman ldquoAn axisymmetricsqueezing fluid flow between the two infinite parallel plates in aporous medium channelrdquo Mathematical Problems in Engineer-ing vol 2011 Article ID 349803 10 pages 2011

[12] I Ullah H Khan and M T Rahim ldquoApproximation of firstgrade MHD squeezing fluid flow with slip boundary conditionusing DTM and OHAMrdquo Mathematical Problems in Engineer-ing vol 2013 Article ID 816262 9 pages 2013

[13] A M Siddiqui S Irum and A R Ansari ldquoUnsteady squeezingflowof a viscousMHDfluid between parallel platesrdquoMathemat-ical Modelling and Analysis vol 13 pp 565ndash576 2008

[14] R J Grimm ldquoSqueezing flows of Newtonian liquid films ananalysis including fluid inertiardquo Applied Scientific Research vol32 no 2 pp 149ndash166 1976

[15] M M Rashidi A M Siddiqui and M T Rastegari ldquoAnalyticalsolution of squeezing flow between two circular platesrdquo Interna-tional Journal for ComputationalMethods in Engineering Scienceand Mechanics vol 13 no 5 pp 342ndash349 2012

[16] H M Laun M Rady and O Hassager ldquoAnalytical solutions forsqueeze flow with partial wall sliprdquo Journal of Non-NewtonianFluid Mechanics vol 81 no 1-2 pp 1ndash15 1999

[17] S Ishizawa ldquoThe unsteady flow between two parallel discs witharbitrary varying gap widthrdquo Bulletin of the Japan Society ofMechanical Engineers vol 9 no 35 pp 533ndash550 1966

[18] C le Roux ldquoExistence and uniqueness of the flow of second-grade fluids with slip boundary conditionsrdquoArchive for RationalMechanics and Analysis vol 148 no 4 pp 309ndash356 1999

[19] L J Rhooades R Resnic T OrsquoBradovich and S StegmanldquoAbrasive flow machining of cylinder heads and its positiveeffects on performance and cost characteristicsrdquo Tech RepDearborn Mich USA 1996

[20] C L M H Navier Memoirs de lrsquoCademie Royale des Sciencesde lrsquoInstitut de France vol 1 Royale des Sciences de lrsquoInstitut deFrance 1823

[21] A Ebaid ldquoEffects of magnetic field and wall slip conditions onthe peristaltic transport of a Newtonian fluid in an asymmetricchannelrdquo Physics Letters Section A General Atomic and SolidState Physics vol 372 no 24 pp 4493ndash4499 2008

[22] J-H He ldquoSome asymptotic methods for strongly nonlinearequationsrdquo International Journal of Modern Physics B Con-densed Matter Physics Statistical Physics Applied Physics vol20 no 10 pp 1141ndash1199 2006

[23] S J Liao Proposed homotopy analysis techniques for the solutionof nonlinear problems [PhD thesis] Jiao Tong UniversityShanghai China 1992

[24] J-H He ldquoHomotopy perturbation method for solving bound-ary value problemsrdquo Physics Letters A vol 350 no 1-2 pp 87ndash88 2006

[25] J-H He ldquoApproximate analytical solution for seepage flowwithfractional derivatives in porous mediardquo Computer Methods inApplied Mechanics and Engineering vol 167 no 1-2 pp 57ndash681998

[26] VMarinca andNHerisanu ldquoApplication of optimalHomotopyasymptotic method for solving nonlinear equations arising inheat transferrdquo International Communications in Heat and MassTransfer vol 35 no 6 pp 710ndash715 2008

[27] V Marinca N Herisanu C Bota and B Marinca ldquoAn optimalhomotopy asymptotic method applied to the steady flow of afourth-grade fluid past a porous platerdquo Applied MathematicsLetters vol 22 no 2 pp 245ndash251 2009

[28] V Marinca N Herisanu and I Nemes ldquoOptimal homotopyasymptotic method with application to thin film flowrdquo CentralEuropean Journal of Physics vol 6 no 3 pp 648ndash653 2008

[29] J Ali S Islam and G Zaman ldquoThe solution of multipointboundary value problems by the optimal Homotopy asymptoticMethodrdquoComputers andMathematics with Applications vol 59no 6 pp 2000ndash2006 2010

[30] M Idrees S Islam and S Haq ldquoApplication of the optimalhomotopy asymptotic method to squeezing flowrdquo Computersand Mathematics with Applications vol 59 no 12 pp 3858ndash3866 2010

[31] J Ali S Islam H Khan and S I Shah ldquoThe optimal homotopyasymptotic method for the solution of higher-order boundaryvalue problems in finite domainsrdquo Abstract and Applied Analy-sis vol 2012 Article ID 401217 14 pages 2012

[32] H Khan S Islam J Ali and I Ali Shah ldquoComparison ofdifferent analytic solutions to axisymmetric squeezing fluidflow between two infinite parallel plates with slip boundaryconditionsrdquo Abstract and Applied Analysis vol 2012 Article ID835268 18 pages 2012

[33] T C Papanastasiou G C Georgiou and A N AlexandrouViscous Fluid Flow Library of Congress Cataloging CRC PressNew York NY USA 1999

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

Page 11: Research Article Analysis of Unsteady Axisymmetric ...downloads.hindawi.com/journals/mpe/2015/860857.pdfResearch Article Analysis of Unsteady Axisymmetric Squeezing Fluid Flow with

Mathematical Problems in Engineering 11

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] Q K Ghori M Ahmed and A M Siddiqui ldquoApplicationof homotopy perturbation method to squeezing flow of anewtonian fluidrdquo International Journal of Nonlinear Sciencesand Numerical Simulation vol 8 no 2 pp 179ndash184 2007

[2] X J Ran Q Y Zhu and Y Li ldquoAn explicit series solution ofthe squeezing flow between two infinite plates by means ofthe homotopy analysis methodrdquo Communications in NonlinearScience and Numerical Simulation vol 14 no 1 pp 119ndash1322009

[3] M J Stefan ldquoVersuch Uber die scheinbare adhasionrdquo Sitzungs-berichteOsterreichische Akademie der Wissenschaften in WienMathematisch-Naturwissenschaftliche Klasse vol 69 pp 713ndash721 1874

[4] J F Thorpe and W A Shaw Eds Developments in Theoreticaland Applied Mechanics vol 3 Pergamon Press Oxford UK1967

[5] P S Gupta and A S Gupta ldquoSqueezing flow between parallelplatesrdquoWear vol 45 no 2 pp 177ndash185 1977

[6] A F Elkouh ldquoFluid inertia effects in squeeze film between twoplane annulirdquo Transactions of the ASME Journal of Tribologyvol 106 no 2 pp 223ndash227 1984

[7] R L Verma ldquoA numerical solution for squeezing flow betweenparallel channelsrdquoWear vol 72 no 1 pp 89ndash95 1981

[8] P Singh V Radhakrishnan and K A Narayan ldquoSqueezing flowbetween parallel platesrdquo Ingenieur-Archiv vol 60 no 4 pp 274ndash281 1990

[9] P J Leider and R B Bird ldquoSqueezing flow between paralleldisks I Theoretical analysisrdquo Industrial and Engineering Chem-istry Fundamentals vol 13 no 4 pp 336ndash341 1974

[10] G Domairry and A Aziz ldquoApproximate analysis of MHDdqueeze flow between two parallel disks with suction orinjection by homotopy perturbation methodrdquo MathematicalProblems in Engineering vol 2009 Article ID 603916 19 pages2009

[11] S Islam H Khan I A Shah and G Zaman ldquoAn axisymmetricsqueezing fluid flow between the two infinite parallel plates in aporous medium channelrdquo Mathematical Problems in Engineer-ing vol 2011 Article ID 349803 10 pages 2011

[12] I Ullah H Khan and M T Rahim ldquoApproximation of firstgrade MHD squeezing fluid flow with slip boundary conditionusing DTM and OHAMrdquo Mathematical Problems in Engineer-ing vol 2013 Article ID 816262 9 pages 2013

[13] A M Siddiqui S Irum and A R Ansari ldquoUnsteady squeezingflowof a viscousMHDfluid between parallel platesrdquoMathemat-ical Modelling and Analysis vol 13 pp 565ndash576 2008

[14] R J Grimm ldquoSqueezing flows of Newtonian liquid films ananalysis including fluid inertiardquo Applied Scientific Research vol32 no 2 pp 149ndash166 1976

[15] M M Rashidi A M Siddiqui and M T Rastegari ldquoAnalyticalsolution of squeezing flow between two circular platesrdquo Interna-tional Journal for ComputationalMethods in Engineering Scienceand Mechanics vol 13 no 5 pp 342ndash349 2012

[16] H M Laun M Rady and O Hassager ldquoAnalytical solutions forsqueeze flow with partial wall sliprdquo Journal of Non-NewtonianFluid Mechanics vol 81 no 1-2 pp 1ndash15 1999

[17] S Ishizawa ldquoThe unsteady flow between two parallel discs witharbitrary varying gap widthrdquo Bulletin of the Japan Society ofMechanical Engineers vol 9 no 35 pp 533ndash550 1966

[18] C le Roux ldquoExistence and uniqueness of the flow of second-grade fluids with slip boundary conditionsrdquoArchive for RationalMechanics and Analysis vol 148 no 4 pp 309ndash356 1999

[19] L J Rhooades R Resnic T OrsquoBradovich and S StegmanldquoAbrasive flow machining of cylinder heads and its positiveeffects on performance and cost characteristicsrdquo Tech RepDearborn Mich USA 1996

[20] C L M H Navier Memoirs de lrsquoCademie Royale des Sciencesde lrsquoInstitut de France vol 1 Royale des Sciences de lrsquoInstitut deFrance 1823

[21] A Ebaid ldquoEffects of magnetic field and wall slip conditions onthe peristaltic transport of a Newtonian fluid in an asymmetricchannelrdquo Physics Letters Section A General Atomic and SolidState Physics vol 372 no 24 pp 4493ndash4499 2008

[22] J-H He ldquoSome asymptotic methods for strongly nonlinearequationsrdquo International Journal of Modern Physics B Con-densed Matter Physics Statistical Physics Applied Physics vol20 no 10 pp 1141ndash1199 2006

[23] S J Liao Proposed homotopy analysis techniques for the solutionof nonlinear problems [PhD thesis] Jiao Tong UniversityShanghai China 1992

[24] J-H He ldquoHomotopy perturbation method for solving bound-ary value problemsrdquo Physics Letters A vol 350 no 1-2 pp 87ndash88 2006

[25] J-H He ldquoApproximate analytical solution for seepage flowwithfractional derivatives in porous mediardquo Computer Methods inApplied Mechanics and Engineering vol 167 no 1-2 pp 57ndash681998

[26] VMarinca andNHerisanu ldquoApplication of optimalHomotopyasymptotic method for solving nonlinear equations arising inheat transferrdquo International Communications in Heat and MassTransfer vol 35 no 6 pp 710ndash715 2008

[27] V Marinca N Herisanu C Bota and B Marinca ldquoAn optimalhomotopy asymptotic method applied to the steady flow of afourth-grade fluid past a porous platerdquo Applied MathematicsLetters vol 22 no 2 pp 245ndash251 2009

[28] V Marinca N Herisanu and I Nemes ldquoOptimal homotopyasymptotic method with application to thin film flowrdquo CentralEuropean Journal of Physics vol 6 no 3 pp 648ndash653 2008

[29] J Ali S Islam and G Zaman ldquoThe solution of multipointboundary value problems by the optimal Homotopy asymptoticMethodrdquoComputers andMathematics with Applications vol 59no 6 pp 2000ndash2006 2010

[30] M Idrees S Islam and S Haq ldquoApplication of the optimalhomotopy asymptotic method to squeezing flowrdquo Computersand Mathematics with Applications vol 59 no 12 pp 3858ndash3866 2010

[31] J Ali S Islam H Khan and S I Shah ldquoThe optimal homotopyasymptotic method for the solution of higher-order boundaryvalue problems in finite domainsrdquo Abstract and Applied Analy-sis vol 2012 Article ID 401217 14 pages 2012

[32] H Khan S Islam J Ali and I Ali Shah ldquoComparison ofdifferent analytic solutions to axisymmetric squeezing fluidflow between two infinite parallel plates with slip boundaryconditionsrdquo Abstract and Applied Analysis vol 2012 Article ID835268 18 pages 2012

[33] T C Papanastasiou G C Georgiou and A N AlexandrouViscous Fluid Flow Library of Congress Cataloging CRC PressNew York NY USA 1999

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

Page 12: Research Article Analysis of Unsteady Axisymmetric ...downloads.hindawi.com/journals/mpe/2015/860857.pdfResearch Article Analysis of Unsteady Axisymmetric Squeezing Fluid Flow with

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


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