# Solution of the Fractional Form of Unsteady Squeezing Flow through Porous Medium

Authors: A. A. Hemeda; E. E. Eladdad; I. A. Lairje
Journal: Mathematical Problems in Engineering
 (2017)

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## Abstract

We propose two friendly analytical techniques called Adomian decomposition and Picard methods to analyze an unsteady axisymmetric flow of nonconducting, Newtonian fluid. This fluid is assumed to be squeezed between two circular plates passing through porous medium channel with slip and no-slip boundary conditions. A single fractional order nonlinear ordinary differential equation is obtained by means of similarity transformation with the help of the fractional calculus definitions. The resulting fractional boundary value problems are solved by the proposed methods. Convergence of the two methods’ solutions is confirmed by obtaining various approximate solutions and various absolute residuals for different values of the fractional order. Comparison of the results of the two methods for different values of the fractional order confirms that the proposed methods are in a well agreement and therefore they can be used in a simple manner for solving this kind of problems. Finally, graphical study for the longitudinal and normal velocity profiles is obtained for various values of some dimensionless parameters and fractional orders.


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## Body

## 1. Introduction

The importance of the fluid flow through a porous medium goes back a long time in various applications in life such as agriculture, industry, and oil and gas production, where the focus was on estimating and optimizing production. Similarly, another important application is the simulation of ground water pollution, mostly occurring due to leakage of chemicals from tanks and oil pipelines. The objective is to consider ground water as one medium and polluted water as another, so that the spreading in the latter medium and its consequences can be studied.In recent times, after the introduction of the modified Darcy Law [1], analysis through a porous medium has advantage in an important topic for the research community such as reservoir, petroleum, chemical, civil, environmental, and agricultural and biomedical engineering. Some practical applications in these fields include chemical reactors, filtration, geothermal reservoirs, ground water hydrology, and drainage and recovery of crude oil from pores of reservoir rocks [2–7].The importance of squeezing flow is due to its wide applications in many fields such as chemical, biomechanics, food industries, and mechanical and industrial engineering. Practical applications of squeezing flows in these fields are modeling of lubrication systems, polymer processing, compression and injection molding, and so on. These flows are induced by applying normal stresses or vertical velocities by means of a moving boundary, which can be frequently observed in various hydrodynamical tools and machines. Stefan [8] is one of the first researchers who operated on the development of squeezing flows and an ad hoc asymptotic solution to Newtonian fluids. In [9] Thorpe has established an explicit solution to the squeeze flow considering inertial terms. However, in [10], P. S. Gupta and A. S. Gupta proved that the solution given in [9] fails in satisfying boundary conditions. Verma in [11] and Singh et al. in [12] have presented numerical solutions for squeezing flow between parallel plates. Leider and Byron performed in [13] theoretical analysis of power-law fluid between parallel disks. The optimal homotopy asymptotic method (OHAM) has been used by Qayyum et al. [14] to analyze the unsteady axisymmetric flow of nonconducting Newtonian fluid squeezed between two circular plates with slip and no-slip boundaries. Also, the homotopy perturbation method (HPM) has been developed by Qayyum et al. [15], to model and analyze the unsteady axisymmetric flow of nonconducting, Newtonian fluid squeezed between two circular plates passing through a porous medium channel with slip boundary condition. The new iterative and Picard methods had been used by Hemeda and Eladdad in [16] for solving the fractional form of unsteady axisymmetric flow of a nonconducting, Newtonian fluid squeezed between two circular plates with slip and no-slip boundaries. For more studies on squeezing flow through porous medium and also for more theoretical and experimental studies on squeezing flow, you can see [17–25].Fractional calculus becomes one of the important branches of applied mathematics that deals with derivatives and integrals of arbitrary orders. It is the generalized differential and integral calculus of an arbitrary order. Recently, it has the importance of many researchers because of its wide appearance in many applications in fluid mechanics, viscoelasticity, biology, physics, and engineering. Moreover, the ordinary differential operator is a local operator, but the fractional order differential operator is nonlocal. Consequently, attention has been given to the solutions of fractional differential equations for the most important fields of physics, fluid mechanics, and so on. The exact analytical solutions to most nonlinear fractional order problems cannot be found. Therefore, approximation and numerical techniques can be used. Picard method [26–29] and the Adomian decomposition method [30–34] are two powerful approaches of these techniques which can be used in simple manner and short time to obtain analytical approximations to nonlinear problems and they are particularly valuable as tools for researchers, because they provide immediate and visible symbolic terms of analytic solutions, as well as numerical approximate solutions to nonlinear differential equations without linearization or discretization.In the last years, the applications of the proposed methods are extended to the fractional differential equations. Our objective of this article is to prepare and utilize the proposed methods to obtain an analytical solution to the fractional form of an unsteady axisymmetric flow of nonconducting, Newtonian fluid squeezed between two circular plates passing through porous medium with slip and no-slip boundary conditions. Validity of the two methods is confirmed by comparing the obtained results. In addition, the effects of various fractional order, constant containing permeability, Reynolds number, and slip parameter on the solution are studied tabularly and graphically.


## 2. Formulation of the Problem

In this section, the unsteady axisymmetric squeezing flow of incompressible first-grade fluid with densityρ, viscosity μ, and kinematic viscosity ν, squeezed between two circular plates having speed Eωt and passing through porous medium channel, is considered with a fractional order form. At any time t, it is assumed that the distance between the two circular plates is 2ht. Also, it is assumed that r-axis is the central axis of the channel while z-axis is taken as normal to it. Plates move symmetrically with respect to the central axis z=0 while the flow is axisymmetric about r=0. The longitudinal and normal velocity components in radial and axial directions are wrr,z,t and wzr,z,t, respectively. For more physical explanation, see [14–16].The basic system of equations describing the motion of the fluid is(1)∂wr∂r+wrr+∂wz∂z=0,∂p∂r+ρ∂wr∂t-wzΩ=-μ∂Ω∂z+wrk,∂p∂z+ρ∂wr∂t+wrΩ=μ1r∂∂rrΩ-wzk,where Ω=Ωr,z,t is the function of velocity, p=pr,z,t is the function of generalized pressure, and k is the permeability constant.The boundary values onwrr,z,t and wzr,z,t are(2)wrr,z,t=β∂∂zwrr,z,t,wzr,z,t=Eωt,atz=h,∂∂zwrr,z,t=0,wzr,z,t=0,atz=0,where Eωt=dh/dt is the plates velocity. The boundary values in (2) are due to slipping at the upper plate when z=h and symmetry at z=0. If we introduce the dimensionless parameter(3)η=zht, (1) transforms to(4)∂wr∂r+wrr+1h∂wz∂η=0,(5)∂p∂r+ρ∂wr∂t-wzΩ=-μ1h∂Ω∂η+wrk,(6)1h∂p∂η+ρ∂wz∂t+wrΩ=μ1r∂∂rrΩ-wzk.The boundary conditions on wr and wz are(7)wr=β1h∂wr∂η,wz=Eωt,atη=1,∂wr∂η=0,wz=0,atη=0. The elimination of p between (5) and (6) gives(8)ρ∂Ω∂t+wr∂Ω∂r+wzh∂Ω∂η-wrrΩ=μ∇2Ω-1r2+1kΩ,where ∇2 is the Laplacian operator.Defining the components of velocity as follows [10]:(9)wr=-r2htEωtu′η,wz=Eωtuη,we see that (4) is identically satisfied and (8) becomes(10)d4udη4+Rη-ud3udη3+2d2udη2-Qd2udη2-Md2udη2=0,where(11)R=hEωtυ,Q=h2υEωdEωtdt,M=h2k.In the above equations R and Q are functions of t but, for similarity solution, we consider them as constants. Since Eω=dh/dt, integrating first equation of (11), we get(12)ht=ct+d1/2,where c and d are constants. The plates move away from each other symmetrically with respect to η when c>0 and d>0. The squeezing flow exists when the plates approach each other when c<0,d>0 and ht>0. From (11) and (12) it follows that Q=-R. Then (10) becomes(13)d4udη4+Rη-ud3udη3+3d2udη2-Md2udη2=0.Using (7) and (9), the boundary conditions in cases of no-slip and slip boundaries at the upper plate can be defined in the form:(14a)u1=1,u′1=0,u0=0,u′′0=0,No-slip  at  the  wall,(14b)u1=1,u′1=γu′′1,u0=0,u′′0=0,Slipatthewall.


## 3. Fractional Calculus

In the following, we state some definitions of the fractional calculus, which can be used in this work.Definition 1.
The Riemann-Liouville fractional integral operator of orderα>0 and of a function gt∈cμ and μ≥-1 is defined as follows [35]:(15)Itαgt=1Γα∫0tt-τα-1gτdτ,α>0,t>0,It0gt=gt.Also, for Itα, we can define(16)Itαtυ=Γυ+1tυ+αΓυ+1+α.Definition 2.
The fractional order derivativeDtα of gt in the Caputo sense is defined as follows [36]:(17)Dtαgt=Itm-αDtmgt=1Γm-α∫0tt-τm-α-1gmτdτ,form-1<α≤m,t>0.
For the Caputo fractional derivative operator,Dtα, we obtain(18)Dtαtυ=Γυ+1tυ-αΓυ+1-α.
For the Riemann-Liouvill fractional integral and Caputo fractional derivative of orderα, we have the following relations.Lemma 3.
Ifm-1<α≤m, m∈N, and g∈Cμm, μ≥-1, then(19a)DtαItαgt=gt,(19b)ItαDtαgt=gt-∑k=0m-1gk0+tkk!,t>0,m-1<deg·gt,0,m-1≥deg·gt.Remark 4.
According to the above definitions of the fractional calculus, (13) can be rewritten in the fractional order form:(20)dαudηα+Rη-ud3udη3+3d2udη2-Md2udη2=0,3<α≤4.


## 4. Analysis of the Proposed Methods

In this section, we present the analysis of the proposed methods with their suitable algorithms for the fractional order problems.


### 4.1. Picard Method (PM)

To present this method, let us consider the general fractional order problem with an arbitrary orderα>0 [26–29]:(21a)Dtαut=Ft,ukt,m-1<α≤m,m∈N,(21b)dkdtku0=hk,k=0,1,2,…,m-1,where Dtα is the time fractional differential operator of order α>0. According to the fractional integral operators, the fractional order problem (21a) and (21b) takes the equivalent fractional integral form: (22)ut=∑k=0m-1hk·tkk!+ItαFt,utkt=f+Nu,where f=∑k=0m-1hk·tk/k!, Nu=ItαFt,utkt, and Itα is the inverse of Dtα. The required solution ut for (22) which is also the solution for (21a) and (21b) can be obtained as the limit of a sequence of functions ur+1t generated by the recurrence relation:(23)u0=f,ur+1=u0+Nur,r=0,1,2,…,where ut=limr→∞⁡urt.


### 4.2. Adomian Decomposition Method (ADM)

To illustrate the idea of this method, let us consider the fractional order problem [30–34]:(24a)Dtαut=Lut+Nut+gt,m-1<α≤m,m∈N,subject to the initial values:(24b)dkdtku0=hk,k=0,1,2,…,m-1,where L,N are linear and nonlinear operators and gt is a nonhomogeneous term. The method is based on applying the fractional integral operator Itα, the inverse of the fractional differential operator Dtα, to both sides of (24a) and (24b) to obtain(25)ut=∑k=0m-1hk·tkk!+ItαLut+Nut+gt.The ADM suggests that the solutionut be decomposed into the infinite series of components:(26)ut=∑n=0∞unt,and the nonlinear term Nu in (24a) is decomposed as(27)Nu=∑n=0∞An,where An are called Adomian polynomials. Substituting the decomposition series (26) and (27) into both sides of (25) gives(28)∑n=0∞unt=∑k=0m-1hk·tkk!+ItαL∑n=0∞unt+∑n=0∞An+gt.Following the decomposition method, we introduce the recurrence relation as(29)u0t=∑k=0m-1hk·tkk!+Itαgt,uj+1t=ItαLujt+Aj,j≥0.The Adomian polynomialAn can be calculated for all forms of nonlinearity according to specific algorithms constructed by Adomian. The general form of the Adomian polynomials formula for An is (30)An=1n!dndλnN∑k=0nλkukλ=0.This formula is easy to compute. Finally, we approximate the solutionut by the truncated series:(31)ϕNt=∑j=0N-1ujt,limN→∞⁡ϕNt=ut.


## 4.1. Picard Method (PM)

To present this method, let us consider the general fractional order problem with an arbitrary orderα>0 [26–29]:(21a)Dtαut=Ft,ukt,m-1<α≤m,m∈N,(21b)dkdtku0=hk,k=0,1,2,…,m-1,where Dtα is the time fractional differential operator of order α>0. According to the fractional integral operators, the fractional order problem (21a) and (21b) takes the equivalent fractional integral form: (22)ut=∑k=0m-1hk·tkk!+ItαFt,utkt=f+Nu,where f=∑k=0m-1hk·tk/k!, Nu=ItαFt,utkt, and Itα is the inverse of Dtα. The required solution ut for (22) which is also the solution for (21a) and (21b) can be obtained as the limit of a sequence of functions ur+1t generated by the recurrence relation:(23)u0=f,ur+1=u0+Nur,r=0,1,2,…,where ut=limr→∞⁡urt.


## 4.2. Adomian Decomposition Method (ADM)

To illustrate the idea of this method, let us consider the fractional order problem [30–34]:(24a)Dtαut=Lut+Nut+gt,m-1<α≤m,m∈N,subject to the initial values:(24b)dkdtku0=hk,k=0,1,2,…,m-1,where L,N are linear and nonlinear operators and gt is a nonhomogeneous term. The method is based on applying the fractional integral operator Itα, the inverse of the fractional differential operator Dtα, to both sides of (24a) and (24b) to obtain(25)ut=∑k=0m-1hk·tkk!+ItαLut+Nut+gt.The ADM suggests that the solutionut be decomposed into the infinite series of components:(26)ut=∑n=0∞unt,and the nonlinear term Nu in (24a) is decomposed as(27)Nu=∑n=0∞An,where An are called Adomian polynomials. Substituting the decomposition series (26) and (27) into both sides of (25) gives(28)∑n=0∞unt=∑k=0m-1hk·tkk!+ItαL∑n=0∞unt+∑n=0∞An+gt.Following the decomposition method, we introduce the recurrence relation as(29)u0t=∑k=0m-1hk·tkk!+Itαgt,uj+1t=ItαLujt+Aj,j≥0.The Adomian polynomialAn can be calculated for all forms of nonlinearity according to specific algorithms constructed by Adomian. The general form of the Adomian polynomials formula for An is (30)An=1n!dndλnN∑k=0nλkukλ=0.This formula is easy to compute. Finally, we approximate the solutionut by the truncated series:(31)ϕNt=∑j=0N-1ujt,limN→∞⁡ϕNt=ut.


## 5. Applications

In this section, we illustrate the application of the two proposed methods to solve the nonlinear fractional order ordinary differential equation (20) subject to the boundary conditions (14a) and (14b).


### 5.1. PM

Using (20) and (14a) and (14b), the initial value fractional order problem(32)dαdtαuη+Rη-ud3udη3+3d2udη2-Md2udη2=0,3<α≤4,u0=0,u′0=a,u′′0=0,u′′′0=b,according to (22) and (23), becomes(33)un+1η=u0η-IηαRη-und3undη3+3d2undη2-Md2undη2,u0η=aη+bη36,n≥0.Therefore, by Wolframe Mathematica 10 Package, we can obtain the following first few components of the solution:(34)u0η=aη+bη36,u1η=aη+bη36-4bRη1+αΓ2+α+abRη1+αΓ2+α+b2Rη3+αΓ4+α+bMη1+αΓ2+α,u2η=aη+bη36+bMη1+αΓ2+α-4bRη1+αΓ2+α+abRη1+αΓ2+α+b2Rη3+αΓ4+α-bMR-1+αη-1+2αΓ2α+abMR-1+αη-1+2αΓ2α+4bR2-1+αη-1+2αΓ2α-5abR2-1+αη-1+2αΓ2α+⋯,⋮and so on. In the same manner, the rest of the components can be obtained. The4th-order term solution for (32), in series form, is given by(35)u3η=aη+bη36+bMη1+αΓ2+α-4bRη1+αΓ2+α+abRη1+αΓ2+α+b2Rη3+αΓ4+α-bMR-1+αη-1+2αΓ2α+abMR-1+αη-1+2αΓ2α+4bR2-1+αη-1+2αΓ2α-5abR2-1+αη-1+2αΓ2α+⋯.In the special case,α=4, (35) gives(36)u3η=aη+bη36+bMη5120-bRη530+abRη5120+bM2η75040+b2Rη75040-bMRη7504+abMRη71260+bR2η7210-abR2η7280+a2bR2η71680-bM3η9362880+⋯.Using the boundary conditions in (14a) and (14b) with the initial conditions in (32), the unknowns a and b for fixed values of R and M in (36) can be easily determined. In case of no-slip boundary, then a=1.5 and b=-3.0. For R=M=0.5, the solution is(37)uη=1.5η-0.5η3+0.01875η5+0.00078125η7-9.95939×10-5η9-4.67144×10-6η11+8.14242×10-7η13+2.31908×10-8η15-2.27551×10-9η17+7.48163×10-12η19+3.60882×10-12η21-1.75116×10-15η23-4.14156×10-15η25-1.83285×10-17η27+1.31237×10-18η29+1.18008×10-20η31,and in case of slip boundary, then a=0.75 and b=1.5. For R=M=0.5, the solution is(38)uη=0.75η+0.25η3-0.0140625η5+0.000683594η7-5.37933×10-5η9+3.8193×10-6η11-2.57439×10-7η13+1.27696×10-8η15-4.75209×10-10η17+1.56632×10-11η19-4.75822×10-13η21+1.16642×10-14η23-2.03097×10-16η25+2.3071×10-18η27-1.53793×10-20η29+4.6097×10-23η31.


### 5.2. ADM

According to the recurrence relation (29), the initial value fractional order problem (32) gives(39)u0η=aη+bη36,uj+1η=-ItαRηd3ujdη3+3d2ujdη2-Aj-Md2ujdη2,j≥0,where(40)A0=u0d3u0dη3,A1=u0d3u1dη3+u1d3u0dη3,A2=u0d3u2dη3+u1d3u1dη3+u2d3u0dη3,A3=u0d3u3dη3+u1d3u2dη3+u2d3u1dη3+u3d3u0dη3,…,and, therefore, the first few components of the solution are as follows:(41)u0η=aη+bη36,u1η=-4bRη1+αΓ2+α+abRη1+αΓ2+α+b2Rη3+αΓ4+α+bMη1+αΓ2+α,u2η=-bMR-1+αη-1+2αΓ2α+abMR-1+αη-1+2αΓ2α+4bR2α-1η-1+2αΓ2α-5abR2α-1η-1+2αΓ2α+a2bR2-1+αη-1+2αΓ2α+b2MRΓ2+αη1+2α6Γ-1+αΓ2+2α-⋯,⋮and so on. In the same manner the rest of the components can be obtained. The4-term solution is(42)uη=∑i=03uiη=aη+bη36-4bRη1+αΓ2+α+abRη1+αΓ2+α+b2Rη3+αΓ4+α+bMη1+αΓ2+α-bMR-1+αη-1+2αΓ2α+abMR-1+αη-1+2αΓ2α+4bR2-1+αη-1+2αΓ2α-5abR2-1+αη-1+2αΓ2α+a2bR2-1+αη-1+2αΓ2α+b2MRΓ2+αη1+2α6Γ-1+αΓ2+2α+⋯.In the special case,α=4, (42) gives(43)uη=aη+bη36+bMη5120-bRη530+abRη5120+bM2η75040+b2Rη75040-bMRη7504+abMRη71260+bR2η7210-abR2η7280+a2bR2η71680-bM3η9362880+⋯.Similarly, using the boundary conditions in (14a) and (14b) with the initial conditions in (32), the solution in case of no-slip boundary, for R=M=0.5, is(44)uη=1.5η-0.5η3+0.01875η5+0.00078125η7-9.95939×10-5η9-4.67144×10-6η11+8.20035×10-7η13+4.88258×10-8η15,and in case of slip boundary, for R=M=0.5, it is (45)uη=0.75η+0.25η3-0.0140625η5+0.000683594η7-5.37933×10-5η9+3.81932×10-6η11-1.96303×10-7η13+3.05161×10-9η15.It is clear that the number of terms of the solution obtained by ADM in (42), (43), (44), and (45) is less than the number obtained by PM in (35), (36), (37), and (38).The residual error of the problem is(46)Reη=ResidualError=dαu^dηα+Rη-u^d3u^dη3+3d2u^dη2-Md2u^dη2,where u^ is the4-term approximate solution in (35) or (42) for (32).It is important to note that, whenRe=0, u^ will be the exact solution. But, in nonlinear problems, this case does not usually occur.It is clear from the obtained results that the above considered methods are used simply and accurately without linearization or discretization with their difficulties. Therefore, these methods are powerful methods for solving the nonlinear fractional order differential equations.


## 5.1. PM

Using (20) and (14a) and (14b), the initial value fractional order problem(32)dαdtαuη+Rη-ud3udη3+3d2udη2-Md2udη2=0,3<α≤4,u0=0,u′0=a,u′′0=0,u′′′0=b,according to (22) and (23), becomes(33)un+1η=u0η-IηαRη-und3undη3+3d2undη2-Md2undη2,u0η=aη+bη36,n≥0.Therefore, by Wolframe Mathematica 10 Package, we can obtain the following first few components of the solution:(34)u0η=aη+bη36,u1η=aη+bη36-4bRη1+αΓ2+α+abRη1+αΓ2+α+b2Rη3+αΓ4+α+bMη1+αΓ2+α,u2η=aη+bη36+bMη1+αΓ2+α-4bRη1+αΓ2+α+abRη1+αΓ2+α+b2Rη3+αΓ4+α-bMR-1+αη-1+2αΓ2α+abMR-1+αη-1+2αΓ2α+4bR2-1+αη-1+2αΓ2α-5abR2-1+αη-1+2αΓ2α+⋯,⋮and so on. In the same manner, the rest of the components can be obtained. The4th-order term solution for (32), in series form, is given by(35)u3η=aη+bη36+bMη1+αΓ2+α-4bRη1+αΓ2+α+abRη1+αΓ2+α+b2Rη3+αΓ4+α-bMR-1+αη-1+2αΓ2α+abMR-1+αη-1+2αΓ2α+4bR2-1+αη-1+2αΓ2α-5abR2-1+αη-1+2αΓ2α+⋯.In the special case,α=4, (35) gives(36)u3η=aη+bη36+bMη5120-bRη530+abRη5120+bM2η75040+b2Rη75040-bMRη7504+abMRη71260+bR2η7210-abR2η7280+a2bR2η71680-bM3η9362880+⋯.Using the boundary conditions in (14a) and (14b) with the initial conditions in (32), the unknowns a and b for fixed values of R and M in (36) can be easily determined. In case of no-slip boundary, then a=1.5 and b=-3.0. For R=M=0.5, the solution is(37)uη=1.5η-0.5η3+0.01875η5+0.00078125η7-9.95939×10-5η9-4.67144×10-6η11+8.14242×10-7η13+2.31908×10-8η15-2.27551×10-9η17+7.48163×10-12η19+3.60882×10-12η21-1.75116×10-15η23-4.14156×10-15η25-1.83285×10-17η27+1.31237×10-18η29+1.18008×10-20η31,and in case of slip boundary, then a=0.75 and b=1.5. For R=M=0.5, the solution is(38)uη=0.75η+0.25η3-0.0140625η5+0.000683594η7-5.37933×10-5η9+3.8193×10-6η11-2.57439×10-7η13+1.27696×10-8η15-4.75209×10-10η17+1.56632×10-11η19-4.75822×10-13η21+1.16642×10-14η23-2.03097×10-16η25+2.3071×10-18η27-1.53793×10-20η29+4.6097×10-23η31.


## 5.2. ADM

According to the recurrence relation (29), the initial value fractional order problem (32) gives(39)u0η=aη+bη36,uj+1η=-ItαRηd3ujdη3+3d2ujdη2-Aj-Md2ujdη2,j≥0,where(40)A0=u0d3u0dη3,A1=u0d3u1dη3+u1d3u0dη3,A2=u0d3u2dη3+u1d3u1dη3+u2d3u0dη3,A3=u0d3u3dη3+u1d3u2dη3+u2d3u1dη3+u3d3u0dη3,…,and, therefore, the first few components of the solution are as follows:(41)u0η=aη+bη36,u1η=-4bRη1+αΓ2+α+abRη1+αΓ2+α+b2Rη3+αΓ4+α+bMη1+αΓ2+α,u2η=-bMR-1+αη-1+2αΓ2α+abMR-1+αη-1+2αΓ2α+4bR2α-1η-1+2αΓ2α-5abR2α-1η-1+2αΓ2α+a2bR2-1+αη-1+2αΓ2α+b2MRΓ2+αη1+2α6Γ-1+αΓ2+2α-⋯,⋮and so on. In the same manner the rest of the components can be obtained. The4-term solution is(42)uη=∑i=03uiη=aη+bη36-4bRη1+αΓ2+α+abRη1+αΓ2+α+b2Rη3+αΓ4+α+bMη1+αΓ2+α-bMR-1+αη-1+2αΓ2α+abMR-1+αη-1+2αΓ2α+4bR2-1+αη-1+2αΓ2α-5abR2-1+αη-1+2αΓ2α+a2bR2-1+αη-1+2αΓ2α+b2MRΓ2+αη1+2α6Γ-1+αΓ2+2α+⋯.In the special case,α=4, (42) gives(43)uη=aη+bη36+bMη5120-bRη530+abRη5120+bM2η75040+b2Rη75040-bMRη7504+abMRη71260+bR2η7210-abR2η7280+a2bR2η71680-bM3η9362880+⋯.Similarly, using the boundary conditions in (14a) and (14b) with the initial conditions in (32), the solution in case of no-slip boundary, for R=M=0.5, is(44)uη=1.5η-0.5η3+0.01875η5+0.00078125η7-9.95939×10-5η9-4.67144×10-6η11+8.20035×10-7η13+4.88258×10-8η15,and in case of slip boundary, for R=M=0.5, it is (45)uη=0.75η+0.25η3-0.0140625η5+0.000683594η7-5.37933×10-5η9+3.81932×10-6η11-1.96303×10-7η13+3.05161×10-9η15.It is clear that the number of terms of the solution obtained by ADM in (42), (43), (44), and (45) is less than the number obtained by PM in (35), (36), (37), and (38).The residual error of the problem is(46)Reη=ResidualError=dαu^dηα+Rη-u^d3u^dη3+3d2u^dη2-Md2u^dη2,where u^ is the4-term approximate solution in (35) or (42) for (32).It is important to note that, whenRe=0, u^ will be the exact solution. But, in nonlinear problems, this case does not usually occur.It is clear from the obtained results that the above considered methods are used simply and accurately without linearization or discretization with their difficulties. Therefore, these methods are powerful methods for solving the nonlinear fractional order differential equations.


## 6. Numerical Results and Discussion

In this work, an unsteady axisymmetric flow of incompressible, nonconducting Newtonian fluid squeezed between two circular plates passing through porous medium channel with slip and no-slip boundary conditions is considered. The obtained nonlinear fractional order problems are solved analytically through PM and ADM using Wolframe Mathematica 10 Package.In addition to the fractional orderα, there are three parameters: constant containing permeability M, Reynolds number R, and slip parameter γ in the considered problem. Our discussions of the obtained results are based on different values of α and these parameters. First, in case of no-slip boundary, the4th-order approximate solutions obtained by the two methods for different values of α,R, and M are illustrated in Tables 1 and 2. Secondly, Tables 3 and 4 represent the4th-order absolute residual errors Re for different values of α,R, and M. Thirdly, Tables 5, 6, 7, and 8 are in case of slip boundary at γ=1. Fourthly, Tables 9 and 10 and 11 and 12 represent the3rd-order and2nd-order absolute residual errors Re for different values of α,R, and M in case of no-slip and slip boundaries with γ=1 for the two methods. Fifthly, the absolute difference between the ADM and PM solutions is illustrated in Tables 13 and 14 for various values of α,M, and R at γ=1 in case of no-slip and slip boundaries conditions. Finally, the analysis of the absolute residual errors is shown in Tables 15 and 16 for various values of α,while keeping the other parameters fixed for different order solutions obtained by the two methods in case of no-slip and slip boundaries.Table 1
Solutions for different values ofη and α atR=0.5,M=0.3, in case of no-slip boundary, by the two methods.
η
PM
ADM
α
=
3.9
α
=
3.95
α
=
4.0
α
=
4.0
α
=
3.95
α
=
3.9
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1
0.149500
0.149500
0.149500
0.149500
0.149500
0.149500
0.2
0.296011
0.296009
0.296008
0.296008
0.296009
0.296011
0.3
0.436577
0.436567
0.436558
0.436558
0.436567
0.436577
0.4
0.568317
0.568278
0.568244
0.568244
0.568278
0.568316
0.5
0.688448
0.688342
0.688247
0.688247
0.688340
0.688445
0.6
0.794321
0.794080
0.793863
0.793863
0.794074
0.794313
0.7
0.883451
0.882973
0.882539
0.882539
0.882955
0.883428
0.8
0.953549
0.952688
0.951899
0.951899
0.952642
0.953492
0.9
1.002550
1.001110
0.999779
0.999779
1.001000
1.002420
1.0
1.028640
1.026370
1.024260
1.024260
1.026150
1.028380Table 2
Solutions for different values ofη and α at R=0.3, M=0.5, in case of no-slip boundary, by the two methods.
η
PM
ADM
α
=
3.9
α
=
3.95
α
=
4.0
α
=
4.0
α
=
3.95
α
=
3.9
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1
0.149500
0.149500
0.149500
0.149500
0.149500
0.149500
0.2
0.296003
0.296002
0.296002
0.296002
0.296002
0.296003
0.3
0.436520
0.436518
0.436515
0.436515
0.436518
0.436520
0.4
0.568084
0.568074
0.568065
0.568065
0.568074
0.568084
0.5
0.687754
0.687725
0.687700
0.687700
0.687724
0.687753
0.6
0.792625
0.792560
0.792501
0.792501
0.792557
0.792622
0.7
0.879845
0.879714
0.879594
0.879594
0.879706
0.8879835
0.8
0.946620
0.946379
0.946159
0.946159
0.946359
0.946595
0.9
0.990229
0.989819
0.989443
0.989443
0.989773
0.990172
1.0
1.008040
1.007380
1.006770
1.006770
1.007290
1.007920Table 3
Fourth-order absolute residual errors Re for different values of η, α at R=0.3,M=0.5 in case of no-slip boundary, by the two methods.
η
PM
ADM
α
=
3.9
α
=
3.95
α
=
4.0
α
=
4.0
α
=
3.95
α
=
3.9
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1
1.34647
-
9
1.17813
-
9
9.87225
-
10
1.05780
-
9
1.27939
-
9
1.49156
-
9
0.2
2.09141
-
8
2.24087
-
8
2.18121
-
8
3.09468
-
8
3.46406
-
8
3.72693
-
8
0.3
2.76031
-
8
2.33410
-
8
5.34534
-
8
2.12433
-
7
2.27849
-
7
2.35087
-
7
0.4
9.08772
-
7
5.72543
-
7
3.30248
-
7
8.91625
-
7
9.55578
-
7
9.99600
-
7
0.5
4.75214
-
6
3.54731
-
6
2.60346
-
6
3.41554
-
6
3.81990
-
6
4.25255
-
6
0.6
1.40546
-
5
1.12830
-
5
8.97699
-
6
1.34531
-
5
1.57013
-
5
1.83641
-
5
0.7
2.48516
-
5
2.13018
-
5
1.80758
-
5
5.09950
-
5
6.06163
-
5
7.21757
-
5
0.8
6.28431
-
6
9.05690
-
6
1.05177
-
5
1.74732
-
4
2.08025
-
4
2.47783
-
4
0.9
1.46521
-
4
1.12844
-
4
8.62395
-
5
5.33866
-
4
6.32037
-
4
7.48016
-
4
1.0
7.06754
-
4
5.77755
-
4
4.70915
-
4
1.46811
-
3
1.72421
-
3
2.02338
-
3Table 4
Fourth-order absolute residual errors Re for different values of η, α at R=0.3, M=0.7 in case of no-slip boundary, by the two methods.
η
PM
ADM
α
=
3.9
α
=
3.95
α
=
4.0
α
=
4.0
α
=
3.95
α
=
3.9
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1
3.34624
-
9
2.38537
-
9
1.69437
-
9
1.69724
-
9
2.38949
-
9
3.35214
-
9
0.2
8.91047
-
8
6.80440
-
8
5.17797
-
8
5.21676
-
8
6.85643
-
8
8.98014
-
8
0.3
5.71570
-
7
4.54449
-
7
3.60082
-
7
3.67323
-
7
4.63794
-
7
5.83613
-
7
0.4
1.95326
-
6
1.60024
-
6
1.30642
-
6
1.36736
-
6
1.67684
-
6
2.04942
-
6
0.5
4.40399
-
6
3.70732
-
6
3.10861
-
6
3.44222
-
6
4.11856
-
6
4.91033
-
6
0.6
6.66441
-
6
5.80184
-
6
5.02294
-
6
6.41819
-
6
7.49551
-
6
8.71803
-
6
0.7
4.88823
-
6
4.66370
-
6
4.36538
-
6
9.21314
-
6
1.04746
-
5
1.18463
-
5
0.8
6.46530
-
6
4.52650
-
6
3.20404
-
6
1.14928
-
5
1.28039
-
5
1.41866
-
5
0.9
2.64276
-
5
2.20234
-
5
1.82974
-
5
2.18120
-
5
2.51127
-
5
2.89148
-
5
1.0
2.13925
-
5
2.04655
-
5
1.92411
-
5
8.14057
-
5
9.68709
-
5
1.15284
-
4Table 5
Solutions for different values ofη and α at R=0.5, M=0.3, and γ=1, in case of slip boundary, by the two methods.
η
PM
ADM
α
=
3.9
α
=
3.95
α
=
4.0
α
=
4.0
α
=
3.95
α
=
3.9
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1
0.075250
0.075250
0.075250
0.075250
0.075250
0.075250
0.2
0.151993
0.151994
0.151995
0.151995
0.151994
0.151993
0.3
0.231697
0.231704
0.231710
0.231710
0.231704
0.231697
0.4
0.315782
0.315808
0.315832
0.315832
0.315809
0.315782
0.5
0.405603
0.405675
0.405739
0.405739
0.405676
0.405605
0.6
0.502430
0.502590
0.502735
0.502735
0.502594
0.502435
0.7
0.607433
0.607747
0.608033
0.608033
0.607760
0.607449
0.8
0.721674
0.722231
0.722741
0.722741
0.722262
0.721712
0.9
0.846091
0.847006
0.847849
0.847850
0.847075
0.846176
1.0
0.981495
0.982910
0.984220
0.984220
0.983053
0.981668Table 6
Solutions for different values ofη and α at R=0.3, M=0.5, and γ=1, in case of slip boundary, by the two methods.
η
PM
ADM
α
=
3.9
α
=
3.95
α
=
4.0
α
=
4.0
α
=
3.95
α
=
3.9
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1
0.075250
0.075250
0.075250
0.075250
0.075250
0.075250
0.2
0.151997
0.151998
0.151998
0.151998
0.151998
0.151997
0.3
0.231731
0.231733
0.231736
0.231736
0.231733
0.231731
0.4
0.315922
0.315931
0.315940
0.315940
0.315931
0.315922
0.5
0.406017
0.406043
0.406066
0.406066
0.406044
0.406018
0.6
0.503433
0.503492
0.503544
0.503544
0.503494
0.503436
0.7
0.609549
0.609665
0.609770
0.609770
0.609672
0.609559
0.8
0.725704
0.725910
0.726099
0.726099
0.725929
0.725727
0.9
0.853188
0.853530
0.853846
0.853846
0.853572
0.853239
1.0
0.993243
0.993778
0.994273
0.994273
0.993863
0.993346Table 7
Fourth-order absolute residual errors Re for different values of η and α, at R=0.3, M=0.5, and γ=1, in case of slip boundary, by the two methods.
η
PM
ADM
α
=
3.9
α
=
3.95
α
=
4.0
α
=
4.0
α
=
3.95
α
=
3.9
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1
6.14308
-
8
4.19152
-
8
2.85428
-
8
2.84794
-
8
4.18242
-
8
6.13005
-
8
0.2
1.67295
-
6
1.22330
-
6
8.92738
-
7
8.84643
-
7
1.21247
-
6
1.65847
-
6
0.3
1.12755
-
5
8.58463
-
6
6.52302
-
6
6.38533
-
6
8.40763
-
6
1.10483
-
5
0.4
4.24853
-
5
3.32810
-
5
2.60194
-
5
2.49943
-
5
3.20080
-
5
4.08886
-
5
0.5
1.15444
-
4
9.24336
-
5
7.38637
-
5
6.90161
-
5
8.65133
-
5
1.08224
-
4
0.6
2.53158
-
4
2.06288
-
4
1.67767
-
4
1.50576
-
4
1.85676
-
4
2.28478
-
4
0.7
4.75190
-
4
3.92835
-
4
3.24121
-
4
2.74183
-
4
3.33889
-
4
4.05711
-
4
0.8
7.89690
-
4
6.60620
-
4
5.51583
-
4
4.26322
-
4
5.14781
-
4
6.20133
-
4
0.9
1.18531
-
3
1.00111
-
3
8.43922
-
4
5.63283
-
4
6.78348
-
4
8.14641
-
4
1.0
1.62509
-
3
1.38247
-
3
1.17383
-
3
5.99153
-
4
7.28868
-
4
8.82835
-
4Table 8
Fourth-order absolute residual errors Re for different values of η and α, at R=0.3, M=0.7, and γ=1 in case of slip boundary, by the two methods.
η
PM
ADM
α
=
3.9
α
=
3.95
α
=
4.0
α
=
4.0
α
=
3.95
α
=
3.9
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1
1.76581
-
8
1.21259
-
8
8.30938
-
9
8.28814
-
9
1.20954
-
8
1.76144
-
8
0.2
4.75929
-
7
3.50290
-
7
2.57276
-
7
2.54571
-
7
3.46670
-
7
4.71090
-
7
0.3
3.15104
-
6
2.41525
-
6
1.84736
-
6
1.80153
-
6
2.35635
-
6
3.07545
-
6
0.4
1.15658
-
5
9.12382
-
6
7.18218
-
6
6.84284
-
6
8.70018
-
6
1.10377
-
5
0.5
3.03150
-
5
2.44516
-
5
1.96804
-
5
1.80871
-
5
2.25070
-
5
2.79450
-
5
0.6
6.33559
-
5
5.20277
-
5
4.26341
-
5
3.70338
-
5
4.53191
-
5
5.53316
-
5
0.7
1.11599
-
4
9.30102
-
5
7.73532
-
5
6.12585
-
5
7.40371
-
5
8.92657
-
5
0.8
1.70455
-
4
1.43790
-
4
1.21040
-
4
8.11811
-
5
9.74626
-
5
1.16691
-
4
0.9
2.28293
-
4
1.94372
-
4
1.65144
-
4
7.71638
-
5
9.34133
-
5
1.12625
-
4
1.0
2.66966
-
4
2.28522
-
4
1.95203
-
4
1.81316
-
5
2.77087
-
5
3.96049
-
5Table 9
Third-order absolute residual errors Re for different values of η and α, at R=0.3,M=0.5,γ=1 by the two methods.
η
No-slip
Slip
α
=
3.9
α
=
3.95
α
=
4.0
α
=
4.0
α
=
3.95
α
=
3.9
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1
5.48712-6
5.70765-6
5.70833-6
7.38815-5
8.77102-5
1.04048-4
0.2
2.39714-5
2.91487-5
3.25070-5
5.82986-4
6.68560-4
7.66115-4
0.3
1.63919-5
1.19104-5
3.38931-5
1.92251-3
2.16064-3
2.42644-3
0.4
3.52371-4
2.58577-3
1.80174-4
4.40898-3
4.88503-3
5.40837-3
0.5
1.48243-3
1.24374-3
1.03582-3
8.24444-3
9.03521-3
9.89430-3
0.6
4.28883-3
3.77352-3
3.31245-3
1.34847-2
1.46487-2
1.59010-2
0.7
1.01929-2
9.19883-3
8.29226-3
2.00130-2
2.15836-2
2.32596-2
0.8
2.13503-2
1.95839-2
1.79494-2
2.75203-2
2.95023-2
3.16030-2
0.9
4.08930-2
3.79446-2
3.51850-2
3.54931-2
3.78627-2
4.03598-2
1.0
7.32218-2
6.85388-2
6.41147-2
4.32097-2
4.59178-2
4.87584-2Table 10
Third-order absolute residual errors Re for different values of η and α, at R=0.3, M=0.7, and γ=1 by the two methods.
η
No-slip
Slip
α
=
3.9
α
=
3.95
α
=
4.0
α
=
4.0
α
=
3.95
α
=
3.9
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1
8.37934-6
7.24173-6
6.24827-6
2.90165-5
3.43492-5
4.06294-5
0.2
6.24359-5
5.57928-5
4.97782-5
2.27300-4
2.59907-4
2.96952-4
0.3
1.99303-4
1.81455-4
1.64961-4
7.40219-4
8.29420-4
9.28619-4
0.4
4.37350-4
4.03436-4
3.71623-4
1.66653-3
1.84074-3
2.03150-3
0.5
7.43336-4
6.93453-4
6.45982-4
3.03807-3
3.31867-3
3.62221-3
0.6
9.70033-4
9.17089-4
8.65511-4
4.80293-3
5.19979-3
5.62479-3
0.7
7.56286-4
7.39037-4
7.19177-4
6.81333-3
7.32212-3
7.86235-3
0.8
6.18418-4
5.15911-4
4.25325-4
8.81918-3
9.42063-3
1.00546-2
0.9
4.43231-3
4.05396-3
3.70548-3
1.04684-2
1.11297-2
1.18228-2
1.0
1.27795-2
1.18628-2
1.10067-2
1.13149-2
1.19937-2
1.27021-2Table 11
Second-order absolute residual errors Re for different values of η and α, at R=0.3, M=0.5, and γ=1 by the two methods.
η
No-slip
Slip
α
=
3.9
α
=
3.95
α
=
4.0
α
=
4.0
α
=
3.95
α
=
3.9
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1
4.48712-5
4.70765-5
4.70833-5
6.38815-4
7.77102-4
1.02048-3
0.2
1.39714-4
3.91487-4
2.25070-4
4.82986-3
5.68560-3
6.66115-3
0.3
2.63919-4
2.19104-4
2.38931-4
1.72251-2
2.11064-2
2.42644-2
0.4
3.52371-3
3.58577-2
1.80174-3
4.20898-2
4.77503-2
5.50837-2
0.5
2.48243-2
1.24374-2
2.03582-2
7.24444-2
8.03521-2
8.89430-2
0.6
4.28883-2
3.87352-2
3.41245-2
1.14847-1
1.12487-1
1.69010-1
0.7
2.01929-1
8.19883-2
7.29226-2
1.01130-1
2.05836-1
2.42596-1
0.8
2.13503-1
1.85839-1
1.89494-1
2.65203-1
2.75023-1
3.12030-1
0.9
3.08930-1
3.99446-1
3.71850-1
3.44931-1
3.58627-1
4.13598-1
1.0
6.32218-1
5.85388-1
6.71147-1
3.32097-1
4.39178-1
4.77584-1Table 12
Second-order absolute residual errors Re for different values of η and α, at R=0.3, M=0.7, and γ=1 by the two methods.
η
No-slip
Slip
α
=
3.9
α
=
3.95
α
=
4.0
α
=
4.0
α
=
3.95
α
=
3.9
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1
7.37934-5
6.24173-5
5.24827-5
1.90165-4
2.43492-4
3.06294-4
0.2
5.24359-4
5.47928-4
4.77782-4
1.27300-3
1.59907-3
1.96952-3
0.3
1.77303-3
1.61455-3
1.44961-3
6.40219-3
7.29420-3
8.28619-3
0.4
4.32350-3
4.01436-3
3.61623-3
1.55653-2
1.64074-2
1.93150-2
0.5
6.43336-3
6.63453-3
6.45982-3
2.03807-2
3.31867-2
3.63221-2
0.6
8.70033-3
8.17089-3
8.15511-3
3.80293-2
5.29979-2
5.52479-2
0.7
6.56286-3
6.39037-3
6.19177-3
5.81333-2
7.11212-2
7.66235-2
0.8
5.18418-3
5.05911-3
4.35325-3
7.81918-2
8.42063-2
1.00546-1
0.9
4.23231-2
4.01396-2
3.71548-2
1.01684-1
1.11297-1
1.17228-1
1.0
1.17795-1
1.15828-1
1.11067-1
1.12149-1
1.18837-1
1.25021-1Table 13
Absolute difference of ADM and PM solutions;|uADM-uPM| for various values of η and α, at R=0.3, M=0.5, and γ=1.
η
No-slip
Slip
α
=
3.9
α
=
3.95
α
=
4.0
α
=
4.0
α
=
3.95
α
=
3.9
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1
2.77556-17
2.77556-17
0.0
0.0
1.38778-17
1.38778-17
0.2
5.44009-15
3.10862-15
1.83187-15
1.63758-15
2.83107-15
4.77396-15
0.3
4.13447-13
2.57239-13
1.59817-13
1.41387-13
2.27457-13
3.65152-13
0.4
9.09539-12
5.90838-12
3.83005-12
3.33550-12
5.14022-12
7.90479-12
0.5
1.00677-10
6.75984-11
4.52951-11
3.86652-11
5.76179-11
8.56799-11
0.6
7.22511-10
4.98311-10
3.42987-10
2.85796-10
4.14338-10
5.99424-10
0.7
3.84849-9
2.71476-9
1.91119-9
1.54808-9
2.19269-9
3.09911-9
0.8
1.64931-8
1.18619-8
8.51429-9
6.67733-9
9.26833-9
1.28373-8
0.9
5.99027-8
4.38185-8
3.19905-8
2.41963-8
3.29895-8
4.48821-8
1.0
9.91033-8
9.41855-8
9.05135-8
7.64034-8
9.02510-8
9.37241-8Table 14
Absolute difference of ADM and PM solutions;|uADM-uPM| for various values of η and α, at R=0.3, M=0.7, and γ=1.
η
No-slip
Slip
α
=
3.9
α
=
3.95
α
=
4.0
α
=
4.0
α
=
3.95
α
=
3.9
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.1
5.55112-17
2.77556-17
0.0
2.77556-17
1.38778-17
1.38778-17
0.2
2.77556-16
5.55112-17
5.55112-17
5.55112-16
9.15934-16
1.58207-15
0.3
1.77081-14
1.09357-14
6.77236-15
4.72677-14
7.59393-14
1.22014-13
0.4
4.07563-13
2.64122-13
1.70641-13
1.11194-12
1.71324-12
2.63423-12
0.5
4.78895-12
3.20177-12
2.13662-12
1.28488-11
1.91425-11
2.84585-11
0.6
3.67931-11
2.52315-11
1.72713-11
9.46089-11
1.37114-10
1.98292-10
0.7
2.11315-10
1.47993-10
1.03464-10
5.10129-10
7.22201-10
1.02024-9
0.8
9.81739-10
6.99977-10
4.98231-10
2.18861-9
3.03596-9
4.20227-9
0.9
3.88008-9
2.80987-7
2.03147-9
7.88225-9
1.07381-8
1.45969-8
1.0
1.34971-8
9.90959-9
7.26394-9
2.47168-8
3.31290-8
4.43064-8Table 15
Average absolute residual errors forfourth-order solutions for various values of α, at R=0.3, M=0.7, and γ=1 by the two methods.
α
R
e
No-slip
Slip
PM
ADM
PM
ADMα=3.9
α=3.9
Minimum
0.0
0.0
0.0
0.0
Average
8.21978-5
2.83204-4
8.05631-5
4.14605-5
Maximum
2.64276-5
1.15284-4
2.66966-4
1.16691-4
α=3.95
Minimum
0.0
0.0
0.0
0.0
Average
6.69461-5
2.40512-4
6.80068-5
3.38057-5
Maximum
2.20234-5
9.68709-5
2.28522-4
9.34133-5
α=4.0
Minimum
0.0
0.0
0.0
0.0
Average
4.99631-6
1.23248-5
5.73045-5
2.74330-5
Maximum
1.92411-5
8.14057-5
1.95203-4
1.81316-5Table 16
Average absolute residual errors forthird-orderandsecond-order solutions for various values of α, at R=0.3, M=0.7, and γ=1 by the two methods.
α
R
e
Third-order
Second-order
No-slip
Slip
No-slip
Slip
α
=
3.9
Minimum
0.0
0.0
0.0
0.0
Average
1.90976-3
4.99878-3
1.76085-2
4.91696-2
Maximum
1.27795-2
1.27021-2
1.17795-1
1.25021-1
α
=
3.95
Minimum
0.0
0.0
0.0
0.0
Average
1.76729-3
4.66809-3
1.74056-2
4.51991-2
Maximum
1.18628-2
1.19937-2
1.15828-1
1.18837-1
α
=
4.0
Minimum
0.0
0.0
0.0
0.0
Average
1.63280-3
3.79890-3
1.62707-2
3.92726-2
Maximum
1.10067-2
1.13149-2
1.11067-1
1.12149-1For various values ofR and M, the results in Tables 1, 2, 5, and 6 indicate that as α→4, uADM→uPM and the results in Tables 3, 4, 7, and 8 (for4th-order solution), 9 and 10 (for3rd-order solution), and 11 and 12 (for2nd-order solution) indicate that increasing α and converging it to 4 implies decreasing the absolute residual errors Re as M→1. Also, increasing the order of solution implies decreasing Re. Moreover, the results in Tables 3 and 4 indicate that Re obtained by PM are less than Re that obtained by ADM in case of no-slip boundary, while, in case of slip boundary, the results in Tables 7 and 8 indicate the inverse. In addition to the abovementioned, we noted from the results in Tables 13 and 14 that the absolute difference between the ADM and PM solutions are almost equal especially for small values of η for different values of α,M, and R at γ=1 which means that the two methods are in a good agreement in solving the proposed problem and therefore the ADM is a reliable method in solving the fractional order problems Finally, the results in Tables 9 and 10 and 11 and 12 indicate that the3rd-order and2nd-order absolute residual errors Re, respectively, are equal for the two methods in case of no-slip and slip boundaries.Figures1 and 2 show the residual errors Re for various values of α,M, and R by PM in cases of no-slip and slip boundaries with γ=1 (Figure 1), while in Figure 2 it is by ADM.Figure 1
Residuals for various values ofα at R=.5, M=.7, and γ=1 by PM: (a) in case of no-slip boundary and (b) in case of slip boundary.
(a)
(b)Figure 2
Residuals for various values ofα at R=.5, M=.7, and γ=1 by ADM: (a) in case of no-slip boundary and (b) in case of slip boundary.
(a)
(b)The effect of different values of the Reynolds numberR and the constant containing permeability M at the fractional order α=3.7 on velocity profiles of the two methods in case of no-slip boundary is shown in Figures 3 and 4, while the effect in case of slip boundary for different values of R,M, and slip parameter γ at α=3.7 are shown in Figures 5, 6, and 7. In these profiles we varied R as R=1,2,3 for the sake of comparison at M=3 and α=3.7 (Figure 3) and varied M as M=1,2,3 for the sake of comparison at R=3 and α=3.7 (Figure 4) in case of no-slip boundary. In case of slip boundary, we varied R as R=1,2,3 for the sake of comparison at M=3, γ=1, and α=3.7 (Figure 5), varied M as M=1,2,3 for the sake of comparison at R=3, γ=1, and α=3.7 (Figure 6), and varied γ as γ=.75,1,1.25 for the sake of comparison at R=M=3 and α=3.7 (Figure 7).Figure 3
Velocity profiles for various values ofR at α=3.7 and M=3 by PM in case of no-slip boundary: (a) normal velocity and (b) longitudinal velocity.
(a)
(b)Figure 4
Velocity profiles for various values ofM at α=3.7 and R=3 by ADM in case of no-slip boundary: (a) normal velocity and (b) longitudinal velocity.
(a)
(b)Figure 5
Velocity profiles for various values ofR at α=3.7, M=3, and γ=1 by PM in case of slip boundary: (a) normal velocity and (b) longitudinal velocity.
(a)
(b)Figure 6
Velocity profiles for various values ofM at α=3.7, R=3, and γ=1 by ADM in case of slip boundary: (a) normal velocity and (b) longitudinal velocity.
(a)
(b)Figure 7
Velocity profiles for various values ofγ at α=3.7, R=M=3 by ADM in case of slip boundary: (a) normal velocity and (b) longitudinal velocity.
(a)
(b)According to the abovementioned data we have observed that, in case of no-slip boundary, the normal velocity (NV) increases with increasingR (Figure 3(a)) and the longitudinal velocity (LV) increases near the central axis of the channel (Figure 3(b)). It has been analyzed that the NV monotonically increases while the LV monotonically decreases from η=0 to η=1 for fixed value of R at a given time. Almost a similar behavior for the NV and LV is with varying M (Figure 4). In case of slip boundary, we have noted that the NV decreases as R increases (Figure 5(a)) and the LV decreases near the central axis of the channel (Figure 5(b)). Also, almost a similar behavior for the NV and LV is with varying M (Figure 6). Moreover, we have observed that the NV increases with increasing γ (Figure 7(a)) and the LV increases near the wall and decreases near the central axis of the channel (Figure 7(b)).


## 7. Conclusion

In this work, we find the analytical solution and then the similarity solution for the fractional form of an unsteady axisymmetric flow of incompressible, nonconducting Newtonian fluid squeezed between two circular plates passing through porous medium channel using the PM and ADM in cases of no-slip and slip boundary conditions. The convergence of the ADM is determined through various order approximate solutions. Moreover, the validity of ADM is checked by comparing the analytical ADM and the numerical PM solutions. We observed that the similarity solution occurs when the distance between the plates varies asct+d1/2 and squeezing flow exists when c<0,d>0 and ct+d1/2>0. Some key findings related to the present study are as follows.In Case of No-Slip Boundary. (i) The NV increases with the increase in R. (ii) The LV increases near the central axis of the channel with the increase in R. (iii) The NV increases monotonically while the LV decreases monotonically from η=0 to η=1 for fixed value of R at a given time. (iv) For the variation in M, we have noted almost a similar behavior.In Case of Slip Boundary. (i) The NV decreases with the increase in R with fixed M and γ. (ii) The LV decreases near the central axis of the channel with the increase in R with fixed M and γ. (iii) For the variation in M, we have observed almost a similar behavior for fixed R and γ. (iv) The NV increases with the increase in γ with fixed R and M. (v) The LV increases near the wall and decreases near the central axis of the channel with fixed R and M. (vi) R and M have similar effects on the NV and LV, while R,γ have opposite effects on the NV and LV components.In Both Cases of No-Slip and Slip Boundaries. (i) It has been found that the changes in velocity profiles with varying α are less for small values of R and M. (ii) It has been found that the residual errors in the3rd-order solution are equal; also in the2nd-order solution they are equal for the two methods. (iii) It is observed that the Reynolds number R is proportional directly to inertia force and inversely to viscosity force.Therefore, we have concluded that the considered methods can be used simply in various fields of science and engineering.



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*Source: 1421862-2017-08-01.xml*