# Comparison of Different Analytic Solutions to Axisymmetric Squeezing Fluid Flow between Two Infinite Parallel Plates with Slip Boundary Conditions

Authors: Hamid Khan; S. Islam; Javed Ali; Inayat Ali Shah
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We investigate squeezing flow between two large parallel plates by transforming the basic governing equations of the first grade fluid to an ordinary nonlinear differential equation using the stream functionsur(r,z,t)=(1/r)(∂ψ/∂z) and uz(r,z,t)=−(1/r)(∂ψ/∂r) and a transformation ψ(r,z)=r2F(z). The velocity profiles are investigated through various analytical techniques like Adomian decomposition method, new iterative method, homotopy perturbation, optimal homotopy asymptotic method, and differential transform method.


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## Body

## 1. Introduction

The study of squeezing flows has widespread applications in chemical engineering, industrial engineering, mechanical engineering, biomechanics, and food industry. Valves and diarthrodial joints are also the examples for squeeze flows relevant in biology and bioengineering. The first application to squeeze flow problem was made by Stefan in 1874 [1]. The motion of a thin film of lubricant, squeezed flow between two stationary parallel plane surfaces were reported by Tichy and Winner [2] and Wang and Watson [3]. The theoretical and experimental studies of squeezing flows have been conducted by many researchers [4–11]. The mathematical studies of these flows are concerned primarily with the nonlinear partial differential equations which arise from the Navier-Stokes equations. These equations have no general solutions, and only a few exact solutions have been attained by confining some physical aspects of the original problem [12]. To solve these nonlinear differential equations, different perturbation and analytical techniques have been extensively used in fluid mechanics and engineering [13].In the literature only a few papers deal with the comparison of different analytical methods. In this paper we study the squeezed flow between two large parallel plates with slip boundary conditions. The velocity profile is obtained using various analytical techniques like Adomian decomposition method (ADM), new iterative method (NIM), homotopy perturbation (HPM), optimal homotopy asymptotic method (OHAM), and differential transform method (DTM) [14–23]. The residual of each technique is computed and a comparison is made to assess the efficiency of the above techniques. We select DTM for analyzing the velocity profile under different flow parameters.Squeezing flows are produced by vertical movements of boundaries or by applying external normal forces. Commonly two types of boundary conditions are employed. For a viscous fluid at a solid wall, it is generally accepted that the fluid velocity matches the velocity of the solid boundary, and it is known as no slip boundary condition. While the no-slip condition is experimentally proven to be accurate for a number of macroscopic flows. Navier [24] proposed a general boundary condition that incorporates the possibility of fluid slip at a solid boundary. He assumed that the velocityux at a solid surface is proportional to the shear rate at the surface, that is, ux=β∂ux/∂y,where βis the slip length or slip coefficient. Ifβ=0, the generally assumed no-slip boundary condition is obtained, and if β a finite constant, fluid is slip occurs at the wall. Its effect depends upon the length scale of the flow [25–28].


## 2. Basic Equation

We consider a steady axisymmetric flow where the velocity vector,ũ is represented by(2.1)ũ=[ur(r,z),0,uz(r,z)].
In the absence of body forces, the Navier-Stokes equations are obtained for the first grade fluid by using equations of continuity and momentum,(2.2)∇̃⋅ũ=0,-ρ(ũ×w̃)+∇̃(ρ2|ũ|2+p)=-η∇̃×w,̃
where ρis the constant density, p is the pressure, η is the viscosity, and w̃=∇̃×ũ is the vorticity vector.Substituting (2.1) into (2.2), we get the following Continuity equation:(2.3)∂ur∂r+urr+∂uz∂z=0.r-component of N.S equation:(2.4)∂∂r(ρ2(ur2+uz2)+p)-ρuzΩ(r,z)=-η∂Ω(r,z)∂z.z-component of N.S equation:(2.5)∂∂z(ρ2(ur2+uz2)+p)-ρuzΩ(r,z)=η1r∂(rΩ(r,z))∂r,
where Ω(r,z)=∂ur/∂z-∂uz/∂r is the vorticity function.Now we define a function which is known as generalized pressure,(2.6)p̂=ρ2(ur2+uz2)+p.
Using (2.6), (2.4) and (2.5) take the following form:(2.7)∂∂rp̂-ρuzΩ(r,z)=-η∂Ω(r,z)∂z,∂∂zp̂-ρuzΩ(r,z)=η1r∂(rΩ(r,z))∂r.
We now introduce the stream functions, ur(r,z,t)=(1/r)(∂ψ/∂z),uz(r,z,t)=(-1/r)×(∂ψ/∂r),and obtain the following results:
(2.8)Ω(r,z)=-1rE2ψ,
where E2=∂2/∂r2-(1/r)(∂/∂r)+∂2/∂z2,(2.9)∂∂rp̂-ρ1r2∂ψ∂rE2ψ=-ηr∂∂zE2ψ,∂∂zp̂-ρ1r2∂ψ∂zE2ψ=-ηr∂∂rE2ψ.
Eliminating p̂, from (2.9), we obtain(2.10)-ρ∂(ψ,E2ψ/r2)∂(r,z)=ηrE4ψ.
We consider viscous incompressible fluid, squeezed between two large planar and parallel plates, separated by a distance 2d. The plates are moving towards each other with velocity U. The surfaces of both plates are covered by special material with slip length (slip coefficient)β. For small values of U the gape distance 2d between the plates varies slowly with the time t, so the flow can be taken as quasisteady [29–32]. See Figures 1, 2, 3, and 4.Figure 1
Plots of the residuals, blue: ADM, green: NIM, red: HPM, black: OHAM and dashed: DTM. In this figure, a comparison of the residuals is shown graphically, and it reveals that OHAM is more suitable technique for this problem. One can easily observe that the curve of the OHAM residual is approximately normally distributed while the curves of the residuals of other methods areJ-shaped and highly skewed.Figure 2
z-component uz(r,θ,z)=(-1/r)(∂ψ/∂r) of the velocity profile given by OHAM, when R=1,γ=1.Figure 3
r-component ur(r,θ,z)=(1/r)(∂ψ/∂z)of the velocity profile given by OHAM, when R=1, γ=1.Figure 4
Vorticity functionΩ(r,z)=∂ur/∂z-∂uz/∂r where the velocity components have been obtained by OHAM, for R=1, γ=1.The boundary conditions are as follows:(2.11)(i)Atz=d,ur=β∂ur∂z,uz=-U.(ii)Atz=0,uz=0,∂ur∂z=0.
Now we use the transformation
(2.12)ψ(r,z)=r2f(z).By virtue ofE2=∂2/∂r2-(1/r)(∂/∂r)+∂2/∂z2 and (2.12), the compatibility equation (2.10) and the boundary conditions equation (2.11) become(2.13)Fiv(z)+2ρηF(z)F′′′(z)=0,F(0)=0,F′′(0)=0,F(d)=U2,F′(d)=βF′′(d).
Now by introducing dimensionless parameters(2.14)F*=FU/2,z*=zd,γ=βd,R=ρdUη,
and dropping “*” for simplicity, the boundary value problem (2.13) become,(2.15)d4Fdz4+RFd3Fdz3=0,
with boundary conditions(2.16)F(0)=0,F′′(0)=0,F(1)=1,F′(1)=γF′′(1).


## 3. Analytical Techniques

In this section, we give the basic idea of various analytical techniques and evaluate the velocity profile of our problem by consideringR=1,γ=1. The residuals of all the techniques are computed and the results are displayed in Table 2.


### 3.1. Adomian Decomposition Method

According to [13, 14], we consider the differential equation(3.1)L(F(z))+M(F(z))+N(F(z))=g(z),
where Lis the operator of the highest order derivative with respect toz, L=d4/dz4, Ris the reminder of the linear term, and the nonlinear term is represented byN(F(z)). Operating L-1 on both sides of (3.1) we get the following:
(3.2)F(z)=α0+α1z+α2z22!+α3z33!+L-1(g(z))-L-1M(F(z))-L-1N(F(z)),
whereαi:i=0,1,2,3, the constants can be determined by using initial or boundary conditions.The unknown functionF(z)can be expressed by an infinite series of the form(3.3)F(z)=∑n=0∞Fn(z),
whereF0(z)=α0+α1z+α2z2/2!+α3z3/3!+L-1(g(z)) and Fn+1=-L-1M(Fn(z))-L-1(An), n=0,1,2,…. The nonlinear term N(F(z)) is decomposed by an infinite series of polynomial given by(3.4)N(F(z))=∑n=0∞An,
where An are the so-called Adomian polynomials that can be determined by the formula(3.5)An=1n!dndλn[N(∑i=0nλiFi)]λ=0.It has been observed that these polynomials can be constructed for a wide class of nonlinear functions.The solutionF(z)=∑n=0∞Fn(z) is approximated by the truncated series of orderK, that is,(3.6)F̃(z)=∑n=0KFn(z).
In our caseg(z)=0, M(F(z))=0.Now appling ADM on (2.15) and (2.16), for n=0,1,…,5.We obtain(3.7)F0(z)=Az36+BzF1(z)=-(1120ABz5+A2z75040)F2(z)=AB2z71680+A2Bz922680+A3z111108800⋮.Considering the Adomian 5th-order solution,(3.8)F̃(z)=F0(z)+F1(z)+⋅⋅⋅+F5(z)+O(z20).The boundary conditions atz=1 are used to get the following values of Aand B.(3.9)A=1.74911,B=0.71896.By substituting these values our solution is(3.10)F̃(z)=0.718965z+0.291518z3-0.0104795z5-0.0000688452z7+0.0000701133z9-3.49247×10-6z11-4.73949×10-7z13+5.95015×10-8z15-8.90909×10-10z17-1.65991×10-9z19+O(z20).


### 3.2. New Iterative Method

The basic idea of new iterative method [15, 16]. Consider the following nonlinear general differential equation:(3.11)L(F(z))+M(F(z))+N(F(z))=g(z),
where Lis the operator of the highest order derivative with respect toz, L=d4/dz4, M is the reminder of the linear term, and the nonlinear term is represented byN(F(z)). Operating L-1 on both sides of (3.11) we get,(3.12)F(z)=α0+α1z+α2z22!+α3z33!+L-1(g(z))-L-1M(F(z))-L-1N(F(z)),
where αi:i=0,1,2,3, are constants to be determined by using initial or boundary conditions.The unknown functionF(z)can be expressed by an infinite series of the form(3.13)F(z)=∑n=0∞Fn(z),
where F0(z)=α0+α1z+α2z2/2!+α3z3/3!+L-1g(z) and Fn+1=-L-1M(Fn(z))-L-1(Gn),n=0,1,2,… The nonlinear term N(F(z)) is decomposed by an infinite series of polynomials given by(3.14)N(F(z))=∑n=0∞Gn,
where Gn=N(∑k=0nFk)-N(∑k=0n-1Fk) and G0=N(F0(z)).The solutionF(z)=∑n=0∞Fn(z), is approximated by the truncated series of order K, that is,(3.15)F̃(z)=∑n=0KFn(z).In our caseg(z)=0 and M(F(z))=0.Now using NIM on (2.15) and (2.16), for n=0,1,…,5.We obtain the following:(3.16)F0(z)=Az36+BzF1(z)=-(1120ABz5+A2z75040)F2(z)=-AB2z71680+A2Bz922680+A3z111108800-A2B2z111900800-A3Bz1338438400-A4z153962649600⋮.
Considering the NIM 5th-order solution,(3.17)F̃(z)=F0(z)+F1(z)+F3(z)+F4(z)+F5(z)+O(z20),
we have the following:(3.18)F̃(z)=Bz+Az36-1120ABz5+(-A25040+AB21680)z7+(A2B22680-AB324192)z9+(A31108800-241A2B239916800+AB4380160)z11+(-71A3B239500800+131A2B3207567360-AB56589440)z13+(-1051A4217945728000+1357A3B225147584000-4759A2B487178291200)z15+(2243A4B1140023808000-359A3B350523782400+179A2B5130288435200)z17+(21919A5844757641728000-71999A4B2167094918144000+104977A3B4300358272614400-31A2B6649813570560)z19+O(z20).
Using the boundary conditions at z=1, we get the following value of A and B(3.19)A=1.74911,B=0.71897.
The approximate solution is as follows:(3.20)F̃(z)=0.718965z+0.291518z3-0.0104795z5-0.0000688448z7+0.0000701132z9-3.49247×10-6z11-4.73948×10-7z13+5.95015×10-8z15-8.34647×10-11z17-1.18034×10-9z19+O(z20).


### 3.3. HPM

To illustrate the basic idea of homotopy perturbation method [17–21], we consider the following nonlinear differential equation:(3.21)A(F)=f(r).A is a general differential operator, the operator A can usually be divided into two parts Land N, where L is linear, and N is nonlinear:(3.22)A=L+N.
so(3.23)L(F)+N(F)-f(r)=0,r∈Ω.f(r)is a known analytic function.With the boundary conditionB(F,∂F/∂n)=0,r∈Γ.B is a boundary operator, and Γis the boundary of the domain Ω.Now we construct the following homotopy:(3.24)H(v,p)=(1-p)[L(v)-L(F0)]+p[Av-f(r)],
where p∈[0,1]is an embedding parameter and F0 is the first approximation that satisfied the boundary condition. To get an approximate solution, we expand F(r,p)in Taylor’s series aboutp in the following manner:(3.25)F(r)=v0(r)+∑m=1∞vm(r)pm.
Plugging (3.25) into (3.24) and then equating the coefficient of like powers ofp, we get the following problems which are directly integrable.Zeroth-order problem:(3.26)F0(iv)(z)=0,F0(0)=0,F0′′(0)=0,F0(1)=1,F0′(1)=F0′′(1).First-order problem:(3.27)F1iv(z)=-F0(z)F0′′′(z),F1(0)=0,F1′′(0)=0,F1(1)=0,F1′(1)=F1′′(1).Second-order problem:(3.28)F2iv(z)=-F1(z)F0′′′(z)-F0(z)F1′′′(z),F2(0)=0,F2′′(0)=0,F2(1)=0,F2′(1)=F2′′(1)⋮.
We consider the following 5th-order solution,(3.29)F̃(z)=F0(z)+F1(z)+F3(z)+F4(z)+F5(z)+O(z20),F̃(z)=2379640217780939164049z3309814671645081600000+723651501050808628889z32482361003733811200000-43087465806546383z54111571020677120000-1652256496336207z723984164287283200000+121738793951z91735816642560000-31017787z118879270400000-430067207z13909237288960000+74925133z151223973273600000+30859z1715429481267200-16191589z1918688959184896000+O(z20).


### 3.4. OHAM

According to [22–24], we consider the following differential equation:(3.30)A(F(z))=f(r),r∈Ω,
with the boundary conditions(3.31)B(F,∂F(z)∂n)=0,r∈Γ,
where Ais a differential operator, Bis a boundary operator, and f(r)is a known function of r: r∈Ω. The operator A can be written as A=L+N, whereL is linear and Nis a nonlinear operator. In OHAM we first construct a homotopy equation,(3.32)(1-p)L(F(r,p))-h(p)[L(F(r,p))+N(F(r,p))-f(r)]=0,
where p∈[0,1]is an embedding parameter, h(p)is a nonzero auxiliary function for p≠0 andh(0)=0(3.33)L(F(z))=0,forp=0A(F(z))=f(r),forp=1.
The solutionF(r,0)=v0(r) of L(F(z))=0 traces the solution curve v(r) continuously as p approaches to 1, where v0 is the solution of the zeroth-order problem that will come in the next few lines. We next choose the auxiliary function h(p) in the following form:(3.34)h(p)=∑i=1mpiCi,
where C1,C2,…are the convergence controlling constants which are to be determined. To get an approximate solution, we expand F(r,p)in Taylor’s series aboutp in the following manner:(3.35)F(r,p,Ci)=v0(r)+∑m=1∞(r,C1,C1,…,Cm)pm.
Now after substituting the auxiliary function h(p) and F(r,p,Ci)in homotopy equation we compare the coefficient of like powers ofp, to obtain the following linear equations.Zeroth-order problem:(3.36)Lv0=0,B(v0,∂v0∂n)=0.First-order problem:(3.37)Lv1-(1+C1v0)L+C1f(r)=C1N0v0,B(v1,∂v1∂n)=0.Second-order problem:(3.38)Lv2-(1+C1)Lv1-C2Lv0-C2f(r)=C2N0v0+C1N1v1,B(v2,∂v2∂n)=0,
and so on.If the series is convergent atp=1 for suitable auxiliary constants C1,C2,…, then(3.39)v(r)=F(r,Ci)=v0(r)+∑m=1∞vm(r,C1,C1,…,Cm).The result of the mth-order approximations are given by(3.40)v̆(r)=v0(r)+∑i=1mvi(r,C1,C1,…,Ci).Residual of the solution is(3.41)R(r,C1,C2,…,Cm)=Av̆-f(r).
IfR=0, v̆ will be the exact solution, but it does not happen specially in nonlinear problems. To find the optimal values ofCi, many methods can be applied. We follow the method of least squares. According to the method of least squares, we first construct the functional(3.42)J(r,C1,C2,…,Cm)=∫abR2dr,
and then minimizing it, we have(3.43)∂J∂C1=∂J∂C2=⋯=∂J∂Cm=0,
where aand bare in the domain of the problem. With these constants known, the approximate solution (of orderm) is well determined.Now applying OHAM to (2.15) and (2.16), we obtain the following problems which are directly integrable.Zeroth-order problem:(3.44)F0(iv)(z)=0,F0(0)=0,F0′′(0)=0,F0(1)=1,F0′(1)=F0′′(1).First-order problem:(3.45)F1iv(z)=C1F0(z)F0′′′(z)+F0iv(z)+C1F0iv(z),F1(0)=0,F1′′(0)=0,F1(1)=0,F1′(1)=F′′1(1).Second-order problem:(3.46)F2iv(z)=C1F1(z)F0′′′(z)+C1F0(z)F1′′′(z)+F1iv(z)+C1F1iv(z),F2(0)=0,F2′′(0)=0,F2(1)=0,F2′(1)=F2′′(1)⋮.
Considering the OHAM 5th-order solution and using the method of least squares, we obtain C1=-0.93281. Hence the solution is
(3.47)F̃(z)=0.718965z+0.291517z3-0.0104795z5-0.0000688407z7+0.0000700828z9-3.4536×10-6z11-4.87073×10-7z13+5.80616×10-8z15+3.31269×10-9z17-5.61766×10-10z19+O(z20).


### 3.5. NDSolve

NDSolve is a mathematica code, utilized for solution of ordinary and partial differential equations. This code is also used for differential-algebraic equations and system of ordinary differential equation. NDSolve gives solution on discrete points rather than for the functionF itself. List interpolation is used for the construction of approximating polynomial.Apply NDSolve to (2.15) and (2.16), the following approximate solution is obtained:(3.48)F(z)=0.718965z-2.5833×10-6z2+0.291529z3-0.000025575z4-0.0104737z5+0.0000970056z6-0.000294814z7+0.000231011z8-0.0000378459z9+0.0000122062z10.


### 3.6. DTM

According to [33, 34], the basic idea of differential transforms method (DTM) starts from the following definition.IfF(z) is a given function, its differential transform is defined as follows:(3.49)F̅(r)=1r!drF(z)dzr|z=0.
The inverse transform of F̅(r) is defined by(3.50)F(z)=∑r=0∞zrF̅(r).
In actual application, the function F(z) is expressed by a finite series(3.51)F(z)=∑r=0NzrF̅(r).
Equation (3.51) implies that F(z)=∑r=N+1∞zrF̅(r). is negligibly small.The fundamental operations of the DTM are given in Table1.Table 1FunctionTransformed functionK(z)=αK1(z)±βK2(z)K(r)=αK̅1(r)±βK̅2(r)K(z)=F(m)(z)K(r)=(m+r)!r!F(m+r)K(z)=K1(z)K2(z)K̅(r)=∑n=0rK̅1(n)K̅2(r-n)K(z)=zmK̅(r)=δ(r-m),δ(p)={1,p=00,p≠0K(z)=K1(z)K2(z)⋯Km(z)G(r)=∑lm-1=0r∑lm-2=0lm-1⋯∑l2=0l3∑l1=0l2{K̅1(l1)K̅2(l2-l1)⋯K̅m(r-lm-1)}Table 2
Comparison of residuals. In this table residuals of all the techniques at the different domain points are displayed. We observe that OHAM provides more stable solution.
zADMNIMHPMOHAMNDSolveDTM00000-6.14E-400.1-6.66E-16-6.94E-16-3.28E-72.18E-7-3.47E-42.22E-160.2-1.40E-12-1.36E-12-8.73E-74.31E-7-7.35E-5-1.26E-140.3-1.35E-10-1.27E-10-1.79E-65.72E-71.70E-5-1.26E-110.4-3.66E-9-3.30E-9-3.09E-65.64E-7-5.16E-7-1.65E-90.5-5.05E-8-4.34E-8-4.60E-65.17E-7-1.14E-5-7.20E-80.6-4.55E-7-3.72E-7-5.78E-69.07E-71.16E-5-1.56E-60.7-3.06E-6-2.38E-6-5.56E-62.35E-62.11E-6-2.09E-50.8-1.66E-5-1.24E-5-1.74E-64.55E-6-5.40E-5-1.97E-40.9-7.68E-5-5.48E-59.89E-65.08E-62.23E-4-1.41E-31.0-3.11E-4-2.13E-43.60E-51.99E-62.20E-3-8.15E-3


#### 3.6.1. Analysis of the Method

Consider a fourth-order boundary value problem(3.52)F(4)(z)=G(z,F),0<z<H,
with the boundary conditions:(3.53)F(0)=α0,F(H)=α1,F(1)(H)=α2,F(2)(0)=α3,
where αi:i=0,1,2,3 are given values.The differential transform of (3.52) is as follows:(3.54)F̅(r+4)=G̅(r)∏i=14(r+i),
where G̅(r) is the differential transform ofG(z,F).The transformed boundary conditions (3.53) are given by(3.55)F̅(0)=α0,∑r=0NHrF̅(r)=α1,∑r=0NrHrF̅(r)=α2,F̅(2)=α32.
Using (3.54) and (3.55) values of F̅(i):i=4,5,…are obtained which give the following series solution up to O(zN+1),(3.56)F(z)=∑r=0NzrF̅(r)+O(zN+1),
applying DTM on (2.15) and (2.16), the transformed boundary conditions and differential are as follows: (3.57)F̅(0)=0,F̅(1)=a,F̅(2)=0,F̅(3)=b,F̅(r+4)=r!(r+4){-R∑k=0r(k+1)(k+2)(k+3)F̅(k+3)F̅(r-k)}.
Using (3.57), we obtain the following values of F̅(i):i=1,2,3…,19.For these values, the unknownsaand b are determined by the following system:(3.58)∑r=019F̃(r)=1,∑r=019rF̃(r)=∑r=019r(r-1)F̃(r).
We find, a=1.53266andb=-0.570353.Now using the inverse differential transform, the following approximate solution ofO(z20)is obtained:(3.59)F̃(z)=1.53266z-0.570353z3+0.043708z5-0.00710856z7+0.00130068z9-0.000251907z11+0.0000492109z13-9.6216×10-6z15+1.86893×10-6Z17-3.60367×10-7z19+O(z20).


## 3.1. Adomian Decomposition Method

According to [13, 14], we consider the differential equation(3.1)L(F(z))+M(F(z))+N(F(z))=g(z),
where Lis the operator of the highest order derivative with respect toz, L=d4/dz4, Ris the reminder of the linear term, and the nonlinear term is represented byN(F(z)). Operating L-1 on both sides of (3.1) we get the following:
(3.2)F(z)=α0+α1z+α2z22!+α3z33!+L-1(g(z))-L-1M(F(z))-L-1N(F(z)),
whereαi:i=0,1,2,3, the constants can be determined by using initial or boundary conditions.The unknown functionF(z)can be expressed by an infinite series of the form(3.3)F(z)=∑n=0∞Fn(z),
whereF0(z)=α0+α1z+α2z2/2!+α3z3/3!+L-1(g(z)) and Fn+1=-L-1M(Fn(z))-L-1(An), n=0,1,2,…. The nonlinear term N(F(z)) is decomposed by an infinite series of polynomial given by(3.4)N(F(z))=∑n=0∞An,
where An are the so-called Adomian polynomials that can be determined by the formula(3.5)An=1n!dndλn[N(∑i=0nλiFi)]λ=0.It has been observed that these polynomials can be constructed for a wide class of nonlinear functions.The solutionF(z)=∑n=0∞Fn(z) is approximated by the truncated series of orderK, that is,(3.6)F̃(z)=∑n=0KFn(z).
In our caseg(z)=0, M(F(z))=0.Now appling ADM on (2.15) and (2.16), for n=0,1,…,5.We obtain(3.7)F0(z)=Az36+BzF1(z)=-(1120ABz5+A2z75040)F2(z)=AB2z71680+A2Bz922680+A3z111108800⋮.Considering the Adomian 5th-order solution,(3.8)F̃(z)=F0(z)+F1(z)+⋅⋅⋅+F5(z)+O(z20).The boundary conditions atz=1 are used to get the following values of Aand B.(3.9)A=1.74911,B=0.71896.By substituting these values our solution is(3.10)F̃(z)=0.718965z+0.291518z3-0.0104795z5-0.0000688452z7+0.0000701133z9-3.49247×10-6z11-4.73949×10-7z13+5.95015×10-8z15-8.90909×10-10z17-1.65991×10-9z19+O(z20).


## 3.2. New Iterative Method

The basic idea of new iterative method [15, 16]. Consider the following nonlinear general differential equation:(3.11)L(F(z))+M(F(z))+N(F(z))=g(z),
where Lis the operator of the highest order derivative with respect toz, L=d4/dz4, M is the reminder of the linear term, and the nonlinear term is represented byN(F(z)). Operating L-1 on both sides of (3.11) we get,(3.12)F(z)=α0+α1z+α2z22!+α3z33!+L-1(g(z))-L-1M(F(z))-L-1N(F(z)),
where αi:i=0,1,2,3, are constants to be determined by using initial or boundary conditions.The unknown functionF(z)can be expressed by an infinite series of the form(3.13)F(z)=∑n=0∞Fn(z),
where F0(z)=α0+α1z+α2z2/2!+α3z3/3!+L-1g(z) and Fn+1=-L-1M(Fn(z))-L-1(Gn),n=0,1,2,… The nonlinear term N(F(z)) is decomposed by an infinite series of polynomials given by(3.14)N(F(z))=∑n=0∞Gn,
where Gn=N(∑k=0nFk)-N(∑k=0n-1Fk) and G0=N(F0(z)).The solutionF(z)=∑n=0∞Fn(z), is approximated by the truncated series of order K, that is,(3.15)F̃(z)=∑n=0KFn(z).In our caseg(z)=0 and M(F(z))=0.Now using NIM on (2.15) and (2.16), for n=0,1,…,5.We obtain the following:(3.16)F0(z)=Az36+BzF1(z)=-(1120ABz5+A2z75040)F2(z)=-AB2z71680+A2Bz922680+A3z111108800-A2B2z111900800-A3Bz1338438400-A4z153962649600⋮.
Considering the NIM 5th-order solution,(3.17)F̃(z)=F0(z)+F1(z)+F3(z)+F4(z)+F5(z)+O(z20),
we have the following:(3.18)F̃(z)=Bz+Az36-1120ABz5+(-A25040+AB21680)z7+(A2B22680-AB324192)z9+(A31108800-241A2B239916800+AB4380160)z11+(-71A3B239500800+131A2B3207567360-AB56589440)z13+(-1051A4217945728000+1357A3B225147584000-4759A2B487178291200)z15+(2243A4B1140023808000-359A3B350523782400+179A2B5130288435200)z17+(21919A5844757641728000-71999A4B2167094918144000+104977A3B4300358272614400-31A2B6649813570560)z19+O(z20).
Using the boundary conditions at z=1, we get the following value of A and B(3.19)A=1.74911,B=0.71897.
The approximate solution is as follows:(3.20)F̃(z)=0.718965z+0.291518z3-0.0104795z5-0.0000688448z7+0.0000701132z9-3.49247×10-6z11-4.73948×10-7z13+5.95015×10-8z15-8.34647×10-11z17-1.18034×10-9z19+O(z20).


## 3.3. HPM

To illustrate the basic idea of homotopy perturbation method [17–21], we consider the following nonlinear differential equation:(3.21)A(F)=f(r).A is a general differential operator, the operator A can usually be divided into two parts Land N, where L is linear, and N is nonlinear:(3.22)A=L+N.
so(3.23)L(F)+N(F)-f(r)=0,r∈Ω.f(r)is a known analytic function.With the boundary conditionB(F,∂F/∂n)=0,r∈Γ.B is a boundary operator, and Γis the boundary of the domain Ω.Now we construct the following homotopy:(3.24)H(v,p)=(1-p)[L(v)-L(F0)]+p[Av-f(r)],
where p∈[0,1]is an embedding parameter and F0 is the first approximation that satisfied the boundary condition. To get an approximate solution, we expand F(r,p)in Taylor’s series aboutp in the following manner:(3.25)F(r)=v0(r)+∑m=1∞vm(r)pm.
Plugging (3.25) into (3.24) and then equating the coefficient of like powers ofp, we get the following problems which are directly integrable.Zeroth-order problem:(3.26)F0(iv)(z)=0,F0(0)=0,F0′′(0)=0,F0(1)=1,F0′(1)=F0′′(1).First-order problem:(3.27)F1iv(z)=-F0(z)F0′′′(z),F1(0)=0,F1′′(0)=0,F1(1)=0,F1′(1)=F1′′(1).Second-order problem:(3.28)F2iv(z)=-F1(z)F0′′′(z)-F0(z)F1′′′(z),F2(0)=0,F2′′(0)=0,F2(1)=0,F2′(1)=F2′′(1)⋮.
We consider the following 5th-order solution,(3.29)F̃(z)=F0(z)+F1(z)+F3(z)+F4(z)+F5(z)+O(z20),F̃(z)=2379640217780939164049z3309814671645081600000+723651501050808628889z32482361003733811200000-43087465806546383z54111571020677120000-1652256496336207z723984164287283200000+121738793951z91735816642560000-31017787z118879270400000-430067207z13909237288960000+74925133z151223973273600000+30859z1715429481267200-16191589z1918688959184896000+O(z20).


## 3.4. OHAM

According to [22–24], we consider the following differential equation:(3.30)A(F(z))=f(r),r∈Ω,
with the boundary conditions(3.31)B(F,∂F(z)∂n)=0,r∈Γ,
where Ais a differential operator, Bis a boundary operator, and f(r)is a known function of r: r∈Ω. The operator A can be written as A=L+N, whereL is linear and Nis a nonlinear operator. In OHAM we first construct a homotopy equation,(3.32)(1-p)L(F(r,p))-h(p)[L(F(r,p))+N(F(r,p))-f(r)]=0,
where p∈[0,1]is an embedding parameter, h(p)is a nonzero auxiliary function for p≠0 andh(0)=0(3.33)L(F(z))=0,forp=0A(F(z))=f(r),forp=1.
The solutionF(r,0)=v0(r) of L(F(z))=0 traces the solution curve v(r) continuously as p approaches to 1, where v0 is the solution of the zeroth-order problem that will come in the next few lines. We next choose the auxiliary function h(p) in the following form:(3.34)h(p)=∑i=1mpiCi,
where C1,C2,…are the convergence controlling constants which are to be determined. To get an approximate solution, we expand F(r,p)in Taylor’s series aboutp in the following manner:(3.35)F(r,p,Ci)=v0(r)+∑m=1∞(r,C1,C1,…,Cm)pm.
Now after substituting the auxiliary function h(p) and F(r,p,Ci)in homotopy equation we compare the coefficient of like powers ofp, to obtain the following linear equations.Zeroth-order problem:(3.36)Lv0=0,B(v0,∂v0∂n)=0.First-order problem:(3.37)Lv1-(1+C1v0)L+C1f(r)=C1N0v0,B(v1,∂v1∂n)=0.Second-order problem:(3.38)Lv2-(1+C1)Lv1-C2Lv0-C2f(r)=C2N0v0+C1N1v1,B(v2,∂v2∂n)=0,
and so on.If the series is convergent atp=1 for suitable auxiliary constants C1,C2,…, then(3.39)v(r)=F(r,Ci)=v0(r)+∑m=1∞vm(r,C1,C1,…,Cm).The result of the mth-order approximations are given by(3.40)v̆(r)=v0(r)+∑i=1mvi(r,C1,C1,…,Ci).Residual of the solution is(3.41)R(r,C1,C2,…,Cm)=Av̆-f(r).
IfR=0, v̆ will be the exact solution, but it does not happen specially in nonlinear problems. To find the optimal values ofCi, many methods can be applied. We follow the method of least squares. According to the method of least squares, we first construct the functional(3.42)J(r,C1,C2,…,Cm)=∫abR2dr,
and then minimizing it, we have(3.43)∂J∂C1=∂J∂C2=⋯=∂J∂Cm=0,
where aand bare in the domain of the problem. With these constants known, the approximate solution (of orderm) is well determined.Now applying OHAM to (2.15) and (2.16), we obtain the following problems which are directly integrable.Zeroth-order problem:(3.44)F0(iv)(z)=0,F0(0)=0,F0′′(0)=0,F0(1)=1,F0′(1)=F0′′(1).First-order problem:(3.45)F1iv(z)=C1F0(z)F0′′′(z)+F0iv(z)+C1F0iv(z),F1(0)=0,F1′′(0)=0,F1(1)=0,F1′(1)=F′′1(1).Second-order problem:(3.46)F2iv(z)=C1F1(z)F0′′′(z)+C1F0(z)F1′′′(z)+F1iv(z)+C1F1iv(z),F2(0)=0,F2′′(0)=0,F2(1)=0,F2′(1)=F2′′(1)⋮.
Considering the OHAM 5th-order solution and using the method of least squares, we obtain C1=-0.93281. Hence the solution is
(3.47)F̃(z)=0.718965z+0.291517z3-0.0104795z5-0.0000688407z7+0.0000700828z9-3.4536×10-6z11-4.87073×10-7z13+5.80616×10-8z15+3.31269×10-9z17-5.61766×10-10z19+O(z20).


## 3.5. NDSolve

NDSolve is a mathematica code, utilized for solution of ordinary and partial differential equations. This code is also used for differential-algebraic equations and system of ordinary differential equation. NDSolve gives solution on discrete points rather than for the functionF itself. List interpolation is used for the construction of approximating polynomial.Apply NDSolve to (2.15) and (2.16), the following approximate solution is obtained:(3.48)F(z)=0.718965z-2.5833×10-6z2+0.291529z3-0.000025575z4-0.0104737z5+0.0000970056z6-0.000294814z7+0.000231011z8-0.0000378459z9+0.0000122062z10.


## 3.6. DTM

According to [33, 34], the basic idea of differential transforms method (DTM) starts from the following definition.IfF(z) is a given function, its differential transform is defined as follows:(3.49)F̅(r)=1r!drF(z)dzr|z=0.
The inverse transform of F̅(r) is defined by(3.50)F(z)=∑r=0∞zrF̅(r).
In actual application, the function F(z) is expressed by a finite series(3.51)F(z)=∑r=0NzrF̅(r).
Equation (3.51) implies that F(z)=∑r=N+1∞zrF̅(r). is negligibly small.The fundamental operations of the DTM are given in Table1.Table 1FunctionTransformed functionK(z)=αK1(z)±βK2(z)K(r)=αK̅1(r)±βK̅2(r)K(z)=F(m)(z)K(r)=(m+r)!r!F(m+r)K(z)=K1(z)K2(z)K̅(r)=∑n=0rK̅1(n)K̅2(r-n)K(z)=zmK̅(r)=δ(r-m),δ(p)={1,p=00,p≠0K(z)=K1(z)K2(z)⋯Km(z)G(r)=∑lm-1=0r∑lm-2=0lm-1⋯∑l2=0l3∑l1=0l2{K̅1(l1)K̅2(l2-l1)⋯K̅m(r-lm-1)}Table 2
Comparison of residuals. In this table residuals of all the techniques at the different domain points are displayed. We observe that OHAM provides more stable solution.
zADMNIMHPMOHAMNDSolveDTM00000-6.14E-400.1-6.66E-16-6.94E-16-3.28E-72.18E-7-3.47E-42.22E-160.2-1.40E-12-1.36E-12-8.73E-74.31E-7-7.35E-5-1.26E-140.3-1.35E-10-1.27E-10-1.79E-65.72E-71.70E-5-1.26E-110.4-3.66E-9-3.30E-9-3.09E-65.64E-7-5.16E-7-1.65E-90.5-5.05E-8-4.34E-8-4.60E-65.17E-7-1.14E-5-7.20E-80.6-4.55E-7-3.72E-7-5.78E-69.07E-71.16E-5-1.56E-60.7-3.06E-6-2.38E-6-5.56E-62.35E-62.11E-6-2.09E-50.8-1.66E-5-1.24E-5-1.74E-64.55E-6-5.40E-5-1.97E-40.9-7.68E-5-5.48E-59.89E-65.08E-62.23E-4-1.41E-31.0-3.11E-4-2.13E-43.60E-51.99E-62.20E-3-8.15E-3


### 3.6.1. Analysis of the Method

Consider a fourth-order boundary value problem(3.52)F(4)(z)=G(z,F),0<z<H,
with the boundary conditions:(3.53)F(0)=α0,F(H)=α1,F(1)(H)=α2,F(2)(0)=α3,
where αi:i=0,1,2,3 are given values.The differential transform of (3.52) is as follows:(3.54)F̅(r+4)=G̅(r)∏i=14(r+i),
where G̅(r) is the differential transform ofG(z,F).The transformed boundary conditions (3.53) are given by(3.55)F̅(0)=α0,∑r=0NHrF̅(r)=α1,∑r=0NrHrF̅(r)=α2,F̅(2)=α32.
Using (3.54) and (3.55) values of F̅(i):i=4,5,…are obtained which give the following series solution up to O(zN+1),(3.56)F(z)=∑r=0NzrF̅(r)+O(zN+1),
applying DTM on (2.15) and (2.16), the transformed boundary conditions and differential are as follows: (3.57)F̅(0)=0,F̅(1)=a,F̅(2)=0,F̅(3)=b,F̅(r+4)=r!(r+4){-R∑k=0r(k+1)(k+2)(k+3)F̅(k+3)F̅(r-k)}.
Using (3.57), we obtain the following values of F̅(i):i=1,2,3…,19.For these values, the unknownsaand b are determined by the following system:(3.58)∑r=019F̃(r)=1,∑r=019rF̃(r)=∑r=019r(r-1)F̃(r).
We find, a=1.53266andb=-0.570353.Now using the inverse differential transform, the following approximate solution ofO(z20)is obtained:(3.59)F̃(z)=1.53266z-0.570353z3+0.043708z5-0.00710856z7+0.00130068z9-0.000251907z11+0.0000492109z13-9.6216×10-6z15+1.86893×10-6Z17-3.60367×10-7z19+O(z20).


## 3.6.1. Analysis of the Method

Consider a fourth-order boundary value problem(3.52)F(4)(z)=G(z,F),0<z<H,
with the boundary conditions:(3.53)F(0)=α0,F(H)=α1,F(1)(H)=α2,F(2)(0)=α3,
where αi:i=0,1,2,3 are given values.The differential transform of (3.52) is as follows:(3.54)F̅(r+4)=G̅(r)∏i=14(r+i),
where G̅(r) is the differential transform ofG(z,F).The transformed boundary conditions (3.53) are given by(3.55)F̅(0)=α0,∑r=0NHrF̅(r)=α1,∑r=0NrHrF̅(r)=α2,F̅(2)=α32.
Using (3.54) and (3.55) values of F̅(i):i=4,5,…are obtained which give the following series solution up to O(zN+1),(3.56)F(z)=∑r=0NzrF̅(r)+O(zN+1),
applying DTM on (2.15) and (2.16), the transformed boundary conditions and differential are as follows: (3.57)F̅(0)=0,F̅(1)=a,F̅(2)=0,F̅(3)=b,F̅(r+4)=r!(r+4){-R∑k=0r(k+1)(k+2)(k+3)F̅(k+3)F̅(r-k)}.
Using (3.57), we obtain the following values of F̅(i):i=1,2,3…,19.For these values, the unknownsaand b are determined by the following system:(3.58)∑r=019F̃(r)=1,∑r=019rF̃(r)=∑r=019r(r-1)F̃(r).
We find, a=1.53266andb=-0.570353.Now using the inverse differential transform, the following approximate solution ofO(z20)is obtained:(3.59)F̃(z)=1.53266z-0.570353z3+0.043708z5-0.00710856z7+0.00130068z9-0.000251907z11+0.0000492109z13-9.6216×10-6z15+1.86893×10-6Z17-3.60367×10-7z19+O(z20).


## 4. Conclusion

In this paper we have used Adomian’s decomposition method, new iterative method, homotopy perturbation method, optimal homotopy asymptotic method, and differential transform method to an axisymmetric squeezing flow problem. Though all the methods are based on Taylor’s series expansion, they produce different results because each one has its own environment. Homotopy methods combine homotopy from topology and perturbation method. It has been observed that homotopy perturbation method may suffer convergence in many problems while OHAM controls the convergence region by employing the auxiliary function. Differential transform method, Adomian decomposition method, and new iterative method are the straightforward application of Taylor’s series, and they suffer from divergence in general and particularly in initial value problems. Besides all these facts, ADM, HPM, and DTM can lead easily to closed-form solutions of many problems.



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*Source: 835268-2012-02-09.xml*