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BY-NC-ND 3.0 license Open Access Published by De Gruyter Open Access February 20, 2016

Thin film flow of an Oldroyd 6-constant fluid over a moving belt: an analytic approximate solution

  • Remus-Daniel Ene EMAIL logo , Vasile Marinca and Valentin Bogdan Marinca
From the journal Open Physics

Abstract

In this paper the thin film flow of an Oldroyd 6-constant fluid on a vertically moving belt is investigated. The basic equation of a non-Newtonian fluid in a container with a wide moving belt which passes through the container moving vertically upward with constant velocity, is reduced to an ordinary nonlinear differential equation. This equation is solved approximately by means of the Optimal Homotopy Asymptotic Method (OHAM). The solutions take into account the behavior of Newtonian and non-Newtonian fluids. Our procedure intended for solving nonlinear problems does not need small parameters in the equation and provides a convenient way to control the convergence of the approximate solutions.

1 Introduction

It is known that in practical applications the behavior of viscoelastic fluids cannot be represented by that of Newtonian fluids. The inadequacy of the classical Navier-Stokes model to describe the fluids which do not obey the Newtonian postulate that the stress tensor is multiple of the shear strain, has led to the development of several models of non-Newtonian fluids. During the past few decades, various constitutive equations have been proposed for these non-Newtonian fluids. The differential equations that describe non-Newtonian fluids are in general nonlinear and highly complicated, but there is not a single constitutive equation available in the literature by which one can study the flow behavior of viscoelastic fluids. Thin film flows have attracted the attention of numerous researchers, due to their wide applications in engineering and science. Such applications include for example microchip production, the linings of mammalian lungs and so on.

Last few years, many researchers have focused on different problems related to such non-Newtonian fluids. The Oldroyd fluid has acquired a special status amongst the many fluids of the rate type, as it included as special cases the classical Newtonian fluid and the Maxwell fluid [1]. This fluid takes into account the properties of stress relaxation and retardation. The two-dimensional steady flow of an incompressible Maxwell fluid in a corner formed by two planes, one of which is sliding past the other at a certain angle, was first investigated by Strauss [2]. Hancock et al have investigated the effects of inertial forces by constructing regular perturbation series for the stream function in [3]. Rajagopal and Bhatnagar [4] presented two solutions for the flow of an Oldroyd-B fluid: the flow past an infinite porous plate and the longitudinal and torsional oscillations of an infinitely long rod of finite radius. The two-dimensional steady slow flow of an Oldroyd 6-constant fluid between intersecting planes, one of which is fixed and the other moving, has been analyzed by Baris [5]. The effects of the non-Newtonian parameter on the flow pattern are carefully delimited. There is, unlike the case of Newtonian fluid, a secondary flow near the corner. Such flow are of considerable practical interest as they locally appear in a cylinder with a moving piston, or at the edge of a blade used to scrape up liquid from a surface.

Very recently, Hayat et al [6] studied some steady unidimensional flow of an Oldroyd 8-constant MHD fluid in bounded domain. The MHD solutions for a Newtonian fluid as well as those corresponding to the Oldroyd 3 and 6-constant fluids, a Maxwell fluid and a second grade one, appear as limiting cases of the obtained solutions. Hayat et al [7] and Wang et al [8] considered Couette and Poiseuille flows of an Oldroyd 6-constant fluid with magnetic field. Sajid et al [9] investigated the problem of wire coating by withdrawal from a bath of MHD Oldroyd 8-constant fluid, using homotopy analysis method, and Khan et al [10] also discussed an Oldroyd 8-constant MHD fluid but between coaxial cylinders subject to partial slip at the boundaries. The Oldroyd 6-constant fluid and fluid-slip on the three nonlinear boundary value problems are studied by Hayat et al in [11]. Ellahi et al [12] investigated an Oldroyd 8-constant fluid over a suddenly moved plate and Sajid and Hayat [13] obtained an exact solution for thin film flow of an Oldroyd 8-constant fluid. Using Fourier sine and Laplace transforms, Vieru et al [14], established exact solutions for the unsteady flow of an incompressible generalized Oldroyd-B fluid due to an infinite constantly accelerating plate. Hayat et al [15] obtained the exact solution of a thin film flow of an Oldroyd 6-constant fluid and Ellahi et al [16] for flows of an Oldroyd 8-constant fluid with nonlinear slip conditions. The energetic balance for the motion of an Oldroyd-B fluid due to an impulsively moved plate is presented by Fetecău et al in [17]. 3D flow of a generalized Oldroyd-B fluid due to a constant pressure gradient between two side walls perpendicular to a plate is presented by Zheng et al in [18]. Shah et al [19] used the optimal homotopy asymptotic method in the study of the wire coating in a pressure type die with the bath of Oldroyd 8-constant fluid with pressure gradient. The nonlinear differential equations of the steady thin film flow of an Oldroyd 8-constant fluid is solved by Siddiqui et. al.[20] applying the variational iteration method and Adomian decomposition method. In [21], the slip-effect on the rotating flows on an Oldroyd-B fluid in a porous space is examinated by Hayat et al. The fluid is permeated by a transverse magnetic field. The governing equations include the modified Darcy’s law for an Oldroyd-B fluid.

Analytical solutions to nonlinear differential equations play an important role in the study of the flow of an Oldroyd different types fluids, but is difficult to find these solutions in the presence of strong nonlinearity. Many new approaches have been proposed to find and develop approximate solutions of nonlinear differential equations. Perturbation methods have been applied to determine approximate solutions to weakly nonlinear problems [22]. But the use of perturbation theory in many problems is invalid for parameters beyond a certain specified range. Other procedures have been proposed such as: the Adomian decomposition method [23], some linearization methods [24],[25], various modified Lindsedt-Poincaré methods [26], the optimal homotopy perturbation method [27],[28] and so on.

In this paper we consider flows of the Oldroyd 6-constant fluid over a moving belt. A version of the optimal homotopy asymptotic method (OHAM) is applied in this study to derive highly accurate analytical expressions of solutions. Our procedure does not depend upon any small or large parameters, contradistinguishing from other known methods. The main advantage of this approach is the control of the convergence of approximate solutions in a very rigorous way. A very good agreement was found between our approximate solutions and numerical solutions, which proves that our method is very efficient and accurate.

2 Governing equations

The Cauchy stress T is:

(1)T=-pI+S,

where −pI - is the indeterminate part of the stress due to the constraint of incompressibility and the extra stress tensor S is defined by

(2)S+λ1DSDt+λ32(SA1+A1S)+λ52(trS)A1+λ62[tr(SA1)]I=μ[A1+λ2DA1Dt+λ4A12+λ72(tr(A1)2)I],

where λ1, λ2, λ3, λ4, λ5, λ6, λ7, μ are material constants, A1 - is the first Rivlin-Ericksen tensor defined by

(3)A1=L+LT

where L is the spatial velocity gradient, L = gradV, V being the velocity vector. The contravariant derivative D/Dt in terms of the material derivative d/dt is defined by

(4)DSDt+dSdtLSSLT

It should be noted that when λi = 0, i = 1, . . . 7, the model reduces to the classical linearly viscous model of a Newtonian fluid. When λ3 = λ4 = λ5 = λ6 = λ7 = 0 it reduces to a 3-constant model of an Oldroyd-B fluid, while for λ2 = λ3 = λ4 = λ5 = λ6 = λ7 = 0 it reduces to a Maxwell model. For λ1 = λ3 = λ5 = λ6 = λ7 = 0 the model describes a second grade fluid and for λ5 = λ6 = λ7 = 0 it reduces to a Johnson-Segalman model. For λ6 = λ7 = 0 the model reduces to the Oldroyd 6-constant.

Neglecting thermal effects and body forces, the equations of motion of an incompressible fluid are:

(5)divV=0
(6)ρVt+ρ[VgradV]=divT

where ρ is the density and t is the time.

If the extra tensor and velocity are respectively:

(7)S(x)=(SxxSxySxzSyxSyySyzSzxSzySzz),V(x)=(0v(x)0).

From Eq. (7), the continuity equation (5) is satisfied identically and the momentum equation (6) becomes:

(8)dp^dy=ddxSxy

where

(9)Sxy=1M[μdvdx+μα1(dvdx)3]
(10)M=1+α2(dvdx)2
(11)α1=λ1λ4(λ3+λ5)(λ4λ2)
(12)α2=λ1λ3(λ3+λ5)(λ3λ1)

and modified pressure p̂ becomes:

(13)p̂=p11M[μ(λ4λ3)(dvdx)2+μ(λ4α2α1λ3)(dvdx)4].

From Eqs. (9) and (8) we obtain after non-dimensionalization:

(14)d2vdx2+(3α1α2)(dvdx)2d2vdx2+α1α2(dvdx)4d2vdx2m[1+α2(dvdx)2]2=0

where α1 and α2 are given in terms of the nondimensional material constants and m is a nondimensional gravity parameter. The boundary conditions are

(15)v(0)=1,[dvdx+α1(dvdx)3]x=1=0.

3 Basic ideas of optimal homotopy asymptotic method

Eq. (14) with boundary conditions (15) can be written in a more general form

(16)N(Φ(x))=0

where N is a given nonlinear differential operator depending on the unknown function Φ(x), subject to the boundary conditions

(17)B(Φ(x),dΦ(x)dx)=0.

Let Φ0(x) be an initial approximation of Φ(x) and L an arbitrary linear operator such as

(18)L(Φ0(x))f(x)=0,B(Φ0(x),dΦ0(x)dx)=0

where f(x) is an arbitrary continuous function.

We remark that this operator L is not unique.

If p ∈ [0, 1] denotes an embedding parameter and F is a function, then we propose to construct a homotopy [32],[29],[30],[31]:

(19)[L(F(x,p)),H(x,Ci),N(F(x,p))]

with the properties

(20)[L(F(x,0)),H(x,Ci),N(F(x,0))]==L(F(x,0))f(x)=L(Φ0(x))f(x)
(21)[L(F(x,1)),H(x,Ci),N(F(x,1))]=H(x,Ci)N(Φ(x))

where H(x, Ci) ≠ 0 is an arbitrary auxiliary convergence-control function depending on variable x and on a number of auxiliary parameters C1, C2, …, Cm which will be defined later.

Let the function F be in the form

(22)F(x,p)=Φ0(x)+pΦ1(x,Ci)+p2Φ2(x,Ci)+

By substituting Eq. (22) into equation obtained by means of the homotopy (19)

(23)[L(F(x,p)),H(x,Ci),N(F(x,p))]=0

and equating the coefficients of like powers of p, we obtain the governing equation of Φ0(x) given by Eq. (18) and the governing equation of Φ1(x, Ci), Φ2(x, Ci) and so on. If the series (22) is convergent at p = 1, one has

(24)F(x,1)=Φ0(x)+Φ1(x,Ci)+Φ2(x,Ci)+

But in particular we consider only the first-order approximate solution

(25)Φ¯(x,Ci)=Φ0(x)+Φ1(x,Ci),i=1,2,,m

and the homotopy (19) in the form

(26)[L(F(x,p)),H(x,Ci),N(F(x,p))]=L(Φ0(x))f(x)+p[L(Φ1(x,Ci))L(Φ0(x))+f(x)+H(x,Ci)N(Φ0(x))].

Equating only the coefficients of p0 and p1 into Eq. (26), we obtain the governing equation of Φ0(x) given by Eq. (18) and the governing equation of Φ1(x, Ci) i.e.

(27)L(Φ1(x,Ci))=H(x,Ci)N(Φ0(x)),B(Φ1(x,Ci),dΦ1(x,Ci)dx)=0,i=1,2,...,m.

It should be emphasized that Φ0(x) and Φ1(x, Ci) are governed by the linear Eqs. (18) and (27), respectively with boundary conditions that come from the original problem, which can be easily solved. The convergence of the approximate solution (25)depends upon the auxiliary convergence-control function H(x, Ci). There are many possible forms of the function H(x, Ci). Basically the shape of H(x, Ci) must follow the terms appearing in the Eq. (27). Therefore, we try to choose H(x, Ci) so that in Eq. (27), the product H(x, Ci)N(Φ0(x)) be of the same shape with N(Φ0(x)).

Now, substituting Eq. (25) into Eq. (16) results in the following residual

(28)R(x,Ci)=N(Φ¯(x,Ci)).

At this moment, the first-order approximate solution given by Eq. (25) depends on the parameters C1, C2, …, Cm and these parameters can be optimally identified via various methods, such as the least square method, the Galerkin method, the Kantorowich method, the collocation method or by minimizing the square residual error

(29)J(C1,C2,,Cm)=abR2(x,C1,C2,,Cm)dx

where a and b are two values depending on the given problem. The unknown parameters C1, C2, …, Cm can be identified from the condition

(30)JC1=JC2==JCm=0.

With these parameters known (namely convergence-control parameters), the first-order approximate solution (25) is well-determinate.

4 Application of OHAM to thin film flow of an Oldroyd 6-constant fluid

In what follows we apply our procedure to obtain approximate solutions of Eqs. (14) and (15). For this purpose, we choose the linear operator in two cases.

4.1 Case 1.

In the first case we suppose that the linear operator has the form

(31)Lv=v

where prime denotes derivative with respect to x, and Φ(x) = v(x) then the nonlinear operator becomes

(32)Nv=v+(3α1α2)v2v+α1α2v4vm(1+α2v2)2.

The initial conditions (15) become

(33)v(0)=1,v(1)+α1v3(1)=0.

From (33) we obtain two subcases:

– 4.1.a

In the first subcase we consider the following initial conditions

(34)v(0)=1,v(1)=0.

The Eq. (18) can be written as

(35)v0m=0,v0(0)=0,v0(1)=0,f(x)=m.

Its solution is

(36)v0(x)=m(x22x)+1.

The second function v1(x) is obtained from Eq. (27):

(37)v1(x)=H1(x,Ci)N(v0),v1(0)=0,v1(1)=0

where

(38)N(v0)=v0+(3α1α2)v02v0+α1α2v04v0m(1+α2v02)2.

Substituting Eq. (36) into (38), the nonlinear operator N becomes

(39)N(v0)=K1(x1)2+K2(x1)4

where

(40)K1=3(α1α2)m3,K2=α2(α1α2)m5

The expression (39) is a polynomial one and the auxiliary function H1(x, Ci) from Eq. (37) is chosen such that the product H1N be of the same form [32]. Therefore we choose the auxiliary function H1(x, Ci) in the form

(41)H1(x,Ci)=C1(x1)2+C2(x1)1+C3+C4(x1)+C5(x1)2

where C1, C2, …, C5 are unknown parameters.

There are many possible forms of the function H1(x, Ci) which appears into Eq. (37).

For example we can consider the following possibilities:

H1(x,Ci)=C1(x1)1+C2+C3(x1)+C4(x1)3

or

H1(x,Ci)=C1+C2(x1)+C3(x1)2+C4(x1)3+C5(x1)4+C6(x1)5

and so on.

Inserting Eqs. (41) and (39) into Eq. (37) we obtain equation:

(42)v1=K1C1+K1C2(x1)+(K1C3+K2C1)(x1)2+(K1C4+K2C2)(x1)3+(K1C5+K2C3)(x1)4+K2C4(x1)5+K2C5(x1)6,v1(0)=v1(1)=0.

By solving Eq. (42) we obtain

(43)v1(x)=12K1C1[(x1)21]+16K1C2[(x1)3+1]++112(K1C3+K2C1)[(x1)41]+120(K1C4+K2C2)[(x1)5+1]+130(K1C5+K2C3)[(x1)61]+142K2C4[(x1)7+1]+156K2C5[(x1)81].

The first-order approximate solution of Eqs. (14) and (34) is obtained from Eq. (24):

(44)v¯(x)=v0(x)+v1(x)

where v0(x) and v1(x) are given by Eqs. (36) and (43) respectively.

The residual given by Eq. (28) becomes

(45)R1(x,C1,C2,C3,C4,C5)=v¯+(3α1α2)v¯2v¯+α1α2v¯4v¯m(1+α2v¯2)2

with obtained from Eq. (44).

– 4.1.b

In the second subcase, from Eq. (33) we have

(46)v(0)=1,v(1)=±1α1,ifα1<0.

The Eq. (18) becomes

(47)v0m=0,v0(0)=0,v0(1)=±1α1

with the solution

(48)v0(x)=m2(x1)2±1α1(x1)m2±1α1+1.

For the expression (48) of v0(x), the nonlinear operator N becomes

(49)N(v0)=m(α1α2)[K3+K4(x1)+K5(x1)2+K6(x1)3+K7(x1)4]

where

(50)K3=1α1(3α2α1);K4=±2mα1(32α2α1);K5=3m2(12α2α1);K6=±4m3α2α1;K7=m4α2.

Eq. (27) for the function v1(x) is

(51)v1=H1(x,Ci)N(v0),v1(0)=0,v1(1)=0

where the auxiliary function H1(x, Ci) is chosen in the form

(52)H1(x,Ci)=1m(α1α2)[C6+C7(x1)+C8(x1)2+C9(x1)3+C10(x1)4].

Substituting Eqs. (49) and (52) into Eq. (51), and thereafter solving this equation, we obtain the solution

(53)v1(x)=12K3C6[(x1)21]+16(K4C6+K3C7)[(x1)3+1]+112(K5C6+K4C7+K3C8)[(x1)41]+120(K6C6+K5C7+K4C8+K5C9)[(x1)5+1]+130(K7C6+K6C7+K5C8+K4C7+K3C8)[(x1)61]+142(K7C7+K6C8+K5C9+K6C10)[(x1)7+1]+156(K7C8+K6C9+K5C10)[(x1)81]+172(K7C9+K6C8)[(x1)9+1]+190K7C10[(x1)101].

The first-order approximate solution of Eqs. (14) and (46) is

(54)v¯(x)=v0(x)+v1(x)

where v0(x) and v1(x) are given by Eqs. (48) and (53) respectively.

4.2 Case 2.

In the second case, the linear operator can be chosen in the form:

(55)Lv=v+λv

where λ is an unknown parameter, and the nonlinear operator N is given by Eq. (32).

From Eq. (15)2 or Eq. (33) we can consider the following two subcases:

– 4.2.a

In the first subcase, the initial conditions are

(56)v(0)=1,v(1)=0.

The Eq. (18) becomes

(57)v0+λv0λ2eλ=0,v0(0)=1,v0(1)=0,f(x)=λ2eλ.

The solution of Eq. (57) has the form

(58)v0(x)=λeλx+eλx.

Substituting Eq. (58) into Eq. (32), the nonlinear operator N(v0) becomes:

(59)N(v0)=M0+M1eλx+M2e2λx+M3e3λx+M4e4λx+M5e5λx

where

(60)M0=m(1+α2λ2e2λ)2;M1=λ2+(3α1α2)λ4e2λ+α1α2λ6e4λ+4m(α2λ2eλ+α22λ4e3λ);M2=2(3α1α2)λ4eλ4α1α2λ6e3λ2m(α2λ2+3α22λ4e2λ)4mα22λ4e2λ;M3=(3α1α2)λ4+6α1α2λ6e2λ+4mα22λ4eλ;M4=4α1α2λ6eλmα22λ4;M5=α1α2λ6.

The second function v1(x) is obtained from Eq. (27):

(61)v1+λv1=H2(x,Ci)N(v0),v1(0)=0,v1(1)=0.

The expression (59) of the nonlinear operator N(v0) is an exponential one and therefore, the auxiliary function H2(x, Ci) can be chosen in the form

(62)H2(x,Ci)=C1+C2eλx+C3e2λx.

where C1, C2, and C3 are unknown parameters. Also we can use the following expressions for the auxiliary function:

H2(x,Ci)=C1+C2eλx+C3e2λx+C4e3λx

or

H2(x,C1)=C1*+C2*eλx+C3*e4λx+C4*e5λx

and so on.

Now, substituting Eqs. (62) and (59) into Eq. (61), we obtain the equation

(63)v1+λv1=M0C1+(M1C1+M0C2))eλx+(M2C1+M1C2+M0C3)e2λx+(M3C1+M2C2+M1C3)e3λx+(M4C1+M3C2+M2C3)e4λx+(M5C1+M4C2+M3C3)e5λx+(M5C2+M4C3)e6λx+M5C3e7λx,v1(0)=v1(1)=0.

From Eq. (63) we can obtain the solution v1(x), such that the first-order approximate solution of Eqs. (14) and (56) in this subcase, can be written in the form

(64)v¯(x)=v0(x)+v1(x)=A+(M0C1λ+λeλ)x+(B+1M1C1+M0C2λx)eλx+M2C1+M1C2+M0C32λ2e2λx+M3C1+M2C2+M1C36λ2e3λx+M4C1+M3C2+M2C12λ2e4λx+M5C1+M4C2+M3C320λ2e5λx+M5C2+M4C330λ2e6λx+M5C342λ2e7λx

where v0(x) is given by Eq. (58) and

(65)A=(eλλ2M0+1λλ2M1+2eλ12λ2M2+3e2λ16λ2M3+4e3λ112λ2M4+5e4λ120λ2M5)C1+(1λλ2M0+2eλ12λ2M1+3e2λ16λ2M2+4e3λ112λ2M3+5e4λ120λ2M4+6e5λ130λ2M5)C2+(2eλ12λ2M0+3e2λ16λ2M1+4e3λ112λ2M2+5e4λ120λ2M3+6e5λ130λ2M4+7e6λ142λ2M5)C3;B=(eλλ2M0+λ1λ2M1eλλ2M2e2λ2λ2M3e3λ3λ2M4e4λ4λ2M5)C1+(λ1λ2M0eλλ2M1e2λ2λ2M2e3λ3λ2M3e4λ4λ2M4e5λ5λ2M5)C2+(eλλ2M0e2λ2λ2M1e3λ3λ2M2e4λ4λ2M3e5λ5λ2M4e6λ6λ2M5)C3.

In the last subcase, the initial conditions are

(66)v(0)=1,v(1)=±1α1,ifα1<0.

The Eq. (18) has the form

(67)v0+λv0λ2eλ=0,v0(0)=1,v0(1)=±1α1,

and has the solution

(68)v0(x)=(±1α1+λeλ)x+eλx.

By means of Eq. (68), the nonlinear operator N(v0) becomes:

(69)N(v0)=N0+N1eλx+N2e2λx+N3e3λx+N4e4λx+N5e5λx

where

(70)N0=m[1+α2(±1α1+λeλ)2]2N1=λ2+λ2(3α1α2)(±1α1+λeλ)2+α1α2λ2(±1α1+λeλ)4+4mα2λ(±1α1+λeλ)+4mα22λ(±1α1+λeλ)3;N2=2(3α1α2)λ3(±1α1+λeλ)4α1α2λ3(±1α1+λeλ)32mα2λ26mα22λ2(±1α1+λeλ)2;N3=(3α1α2)λ4+6α1α2λ4(±1α1+λeλ)2++4mα22λ3(±1α1+λeλ);N4=4α1α2λ5(±1α1+λeλ)mα22λ4;N5=α1α2λ6.

For the equation in v1(x) we choose the auxiliary function H2(x, Ci) in the form

(71)H2(x,Ci)=C4+C5eλx+C6e2λx.

The Eq. (27) can be written, by means of Eqs. (71) and (69) as

(72)v1+λv1=N0C4+(N1C4+N0C5))eλx+(N2C4+N1C5+N0C6)e2λx+(N3C4+N2C5+N1C6)e3λx+(N4C4+N3C5+N2C6)e4λx+(N5C4+N4C5+N3C6)e5λx+(N5C5+N4C6)e6λx+N5C6e7λx,v1(0)=v1(1)=0.

The first-order approximate solution in the last subcase has the form

(73)v¯(x)=v0(x)+v1(x)=D+(N0C4λ+λeλ±1α1)x+(E+1N1C4+N0C5λx)eλx+N2C4+N1C5+N0C62λ2e2λxN3C4+N2C5+N1C66λ2e3λx+N4C4+N3C5+N2C612λ2e4λxN5C4+N4C5+N3C620λ2e5λx+N5C5+N4C630λ2e6λxN5C642λ2e7λx

where

(74)D=(eλλ2N0+1λλ2N1+2eλ12λ2N2+3e2λ16λ2N3+4e3λ112λ2N4+5e4λ120λN5)C4+(1λλ2N0+2eλ12λ2N1+3e2λ16λ2N2+4e3λ112λ2N3+5e4λ120λ2N4+6e5λ130λ2N5)C5+(2eλ12λ2N0+3e2λ16λ2N1+4e3λ112λ2N2+5e4λ120λ2N3+6e5λ130λ2N4+7e6λ142λ2N5)C6;E=(eλλ2N0+λ1λ2N1eλλ2N2e2λ2λ2N3e3λ3λ2N4e4λ4λ2N5)C4+(λ1λ2N0eλλ2N1e2λ2λ2N2e3λ3λ2N3e4λ4λ2N4e5λ5λ2N5)C5+(eλλ2N0e2λ2λ2N1e3λ3λ2N2e4λ4λ2N3e5λ5λ2N4e6λ6λ2N5)C6.

5 Numerical results and discussions

We illustrate the accuracy of our procedure for different values of the coefficients α1, α2 and m and for every the auxiliary function H1 and H2. Also we represent graphically the variation of velocity profile of the belt problem for an Oldroyd 6-constant fluid. In all cases and subcases, the values of the parameters C1, C2, … and λ are determined using the least square method and the Wolfram Mathematica 6.0 software. In this way the parameters Ci,, i = 1, 2, . . . (namely the convergence-control parameters) are optimally determined and the first-order approximate solutions are known for different values of the coefficients which appear in the Eq. (14).

5.1

If we take into consideration, the solution given by Eq. (44), for α1 = 0.5, α2 = 0.5, m = 1 and v′(1) = 0 (Newtonian case), it is interesting to remark that from Eq. (40) results that K1 = K2 = 0 and therefore v1(x) = 0. It follows that, v¯=v0(x)=m(x22-x)+1 which is the exact solution of Eqs. (14) and (15).

5.2

In this example we consider α1 = 0.5, α2 = 0.75, m = 1 and v′(1) = 0 (non-Newtonian case). The values of the convergence-control parameters for the case 4.1.a are

C1=0.0309007883,C2=0.5673047654,C3=1.7430618401,C4=4.9462074642,C5=3.3490697012.

The first-order approximate solution (44) can be written in the form

(75)v¯(x)=0.4430506264+0.4884122043(x1)20.0709130956(x1)30.1094241898(x1)40.1908012620(x1)50.0946208790(x1)60.0220812833(x1)70.0112134030(x1)8

In Table 1 we present a comparison between the first-order approximate solution given by Eq. (75) and numerical results for some values of variable x and the corresponding relative errors.

5.3

For α1 = 0.5, α2 = 1, m = 1 and v′(1) = 0 we obtain for Eq. (44)

C1=0.0145037003,C2=0.3432811995,C3=1.1572350528,C4=5.1398275696,C5=3.7778914694

and the first-order approximate solution in the form:

(76)v¯(x)=0.3569716548+0.4891222247(1+x)20.0858202998(x1)30.1452587024(x1)40.3940690977(x1)50.2081818243(x1)60.0611884234(x1)70.0337311738(x1)8

In Table 2 a comparison between the first-order approximate solution given by Eq. (76) and numerical results, and the corresponding relative errors, is presented.

5.4

If we consider α1 = 0.5, α2 = 0.75, m = 1.5 and v′(1) = 0, then the convergence-control parameters for the case 4.1.a are

C1=0.0103555184,C2=0.2555006995,C3=2.5463142522,C4=3.1417178283,C5=1.0373234408

and therefore the first-order approximate solution has the form:

(77)v¯(x)=0.0939116440+0.7631062030(x1)2+0.1077893576(x1)3+0.5383418691(x1)4+0.4158131167(x1)5+0.2083746269(x1)6+0.1065063382(x1)7+0.0263744694(x1)8

This approximate solution is compared in Table 3 with the numerical results

5.5

In the case α1 = 0.5, α2 = 1, m = 1.5 and v′(1) = 0, the convergence-control parameters for the case 4.1.a are

C1=0.0430647,C2=0.997138,C3=6.40181,C4=9.14949,C5=3.90268.

The first-order approximate solution becomes:

(78)v¯(x)=0.1330803613+0.8590076019(x1)2+0.8413350971(x1)3+2.7143908882(x1)4+2.5052659024(x1)5+1.4688065504(x1)6+0.8271305377(x1)7+0.2646068580(x1)8

The OHAM solution (78) is compared in Table 4 with the numerical results.

5.6

For α1 = −0.5, α2 = 0.5, m = 1 and v'(1)=105, for the case 4.1.b, the convergence-control parameters are

C1=0.2487584861,C2=0.3794941533,C3=0.4121126222,C4=0.4150647670,C5=0.5193428345

The first-order approximate solution given by Eq. (54) can be written as

(79)v¯(x)=2.4142135623+1.4142135623(x1)0.4950339447[(x1)21]0.0803371704[(x1)3+1]0.0140724640[(x1)41]+0.0102108835[(x1)5+1]0.0603676324[(x1)61]0.1091646985[(x1)7+1]0.0661814531[(x1)81]0.0175193191[(x1)9+1]0.0028852379[(x1)101].

In Table 5 we present a comparison between the approximate solution (79) obtained by means of OHAM, and numerical results, and the corresponding relative errors.

Table 1

Comparison between OHAM results given by Eq. (75) and numerical results for α1 = 0.5, α2 = 0.75, m = 1 and v′(1) = 0 and the auxiliary function (41).

xvnumericOHAM, Eq. (75)relative error = |vnumericv̅OHAM|
01.1.0.
0.10.88660857250.88668217747.3 · 10−5
0.20.78756216710.78758873982.6 · 10−5
0.30.70253389250.70253102492.8 · 10−6
0.40.63089172640.63086674052.4 · 10−5
0.50.57178267050.57179159848.9 · 10−6
0.60.52450861130.52452882492.0 · 10−5
0.70.48843939240.48843480434.5 · 10−6
0.80.46306007520.46303459522.5 · 10−5
0.90.44800062260.44799653464.0 · 10−6
10.44302355080.44305062642.7 · 10−5
Table 2

Comparison between OHAM results given by Eq. (76) and numerical results for α1 = 0.5, α2 = 1, m = 1 and v′(1) = 0 and the auxiliary function (41).

xvnumericv̅OHAM, Eq. (76)relative error = |vnumericv̅OHAM|
01.1.0.
0.10.85717887830.85722299434.4 · 10−5
0.20.73616041380.73617984401.9 · 10−5
0.30.63604154080.63603469796.8 · 10−6
0.40.55485195230.55484352458.4 · 10−6
0.50.49030831440.49030917008.5 · 10−7
0.60.44024881040.44026578701.6 · 10−5
0.70.40294801530.40295019982.1 · 10−6
0.80.37712557230.37710416762.1 · 10−5
0.90.36194704730.36193790989.1 · 10−6
10.35694304370.35697165482.8 · 10−5
5.7

In this case we consider α1 = −0.5, α2 = 0.75, m = 1 and v'(1)=105, such that for Eq. (54), the convergence-control parameters are

C1=0.2484286614,C2=0.4310993122,C3=0.4720538492,C4=0.5090078876,C5=0.6742203389

The first-order approximate solution given by Eq. (54) is

(80)v¯(x)=2.4142135623+1.4142135623(x1)0.6179289764[(x1)21]0.0560133962[(x1)3+1]+0.0071974457[(x1)41]+0.0344624859[(x1)5+1]0.0793096739[(x1)61]0.1669817690[(x1)7+1]0.1122347657[(x1)81]+0.0344266489[(x1)9+1]0.0056185028[(x1)101].

In Table 6 we present a comparison between the present solution (80) and numerical results and corresponding relative errors.

Table 3

Comparison between OHAM results given by Eq. (77) and numerical results for α1 = 0.5, α2 = 0.75, m = 1.5 and v′(1) = 0 and the auxiliary function (41)

xvnumericOHAM, Eq. (77)relative error = |vnumericOHAM|
01.1.0.
0.10.81220304270.81227235286.9 · 10−5
0.20.64805240300.64807571222.3 · 10−5
0.30.50751068180.50749635331.4 · 10−5
0.40.38995870130.38996627477.5 · 10−6
0.50.29435790280.29439353003.5 · 10−5
0.60.21935449320.21933003042.4 · 10−5
0.70.16318895870.16316137552.7 · 10−5
0.80.12430315600.12431390431.0 · 10−5
0.90.10146910980.10148479071.5 · 10−5
1.0.09395989830.09391164404.8 · 10−5
Table 4

Comparison between OHAM results given by Eq. (78) and numerical results for α1 = 0.5, α2 = 1, m = 1.5 and v′(1) = 0 and the auxiliary function (41).

xvnumericv̅OHAM, Eq. (78)relative error = |vnumericOHAM|
01.0.99999999992.2 · 10−16
0.10.74970638200.74983453521.2 · 10−4
0.20.53279631110.53278055331.5 · 10−5
0.30.34989458690.34986056153.4 · 10−5
0.40.20108145640.20122823301.4 · 10−4
0.50.08528859620.08538628859.7 · 10−5
0.6−0.0006296037−0.00081563361.8 · 10−4
0.7−0.0615295648−0.06167972741.5 · 10−4
0.8−0.1020045446−0.10182530401.7 · 10−4
0.9−0.1251782938−0.12508384539.4 · 10−5
1.−0.1327530483−0.1330803613.2 · 10−4
5.8

For α1 = −0.5, α2 = 1, m = 1 and v'(1)=105, the convergence-control parameters are

C1=0.2481244286,C2=0.4823327023,C3=0.5396781473,C4=0.6132014735,C5=0.8540900975

and the first-order approximate solution given by Eq. (54) is

(81)v¯(x)=2.4142135623+1.4142135623(x1)0.7406221430[(x1)21]0.0148810042[(x1)3+1]+0.03592103224[(x1)41]+0.06391596843[(x1)5+1]0.10448164910[(x1)61]0.24482492876[(x1)7+1]0.17646854016[(x1)81]0.0585869683[(x1)9+1]0.0094898899[(x1)101].

In Table 7 we present a comparison between the solution (81) and numerical results, and corresponding relative errors.

In the following Tables, we present only comparisons between the present solutions obtained by means of OHAM and numerical results and corresponding relative errors.

In Figs. 1-23 the velocity v̅ is plotted against the horizontal distance x. Figs 1 and 2 show the variation of velocity field for Newtonian fluid, and Figs 3-6 show the variation of velocity field for non-Newtonian case, for different values of the parameter α2 and α1 = 0.5. It is observed that the speed of belt decreases with the increase of m. Figs 7-12 show the variation of the velocity field for the value α1 = −0.5 and different values of the parameter α2. In these cases the speed of belt increases with the increase of m for different values of the auxiliary function. In Figs 13-16 we present the variation of velocity field for α1 = 0.5 and different values of the parameters m and α2.

Here we observe that for any values of the parameter m, the speed of belt increases with the increase of α2. On the other hand, we remark from Figs 13-16 that the speed of belt decreases for any values of α1 > 0. But, from Figs 17-20 we deduce that the speed of belt increases for any values of α1 > 0. We observe in these last cases like Figs 13-16 that the speed of belt increases with the increase of α2. In Figs 21-23 present graphically the variation of velocity profile with increasing parameter α1. It is observed that the speed of belt decreases with the increase of α1. In all cases, we deduce that the speed of belt decreases for α1 = 0.5 and speed of belt increases for α1 = −0.5 and for any values of m.

Figure 1 Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 0.5, v′(1) = 0 with auxiliary function (41) — numerical solution; ...... OHAM solution.
Figure 1

Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 0.5, v′(1) = 0 with auxiliary function (41) — numerical solution; ...... OHAM solution.

Figure 2 Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 0.5, v′(1) = 0 with auxiliary function (62)—numerical solution; ...... OHAM solution.
Figure 2

Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 0.5, v′(1) = 0 with auxiliary function (62)—numerical solution; ...... OHAM solution.

Figure 3 Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 0.5, v′(1) = 0 with auxiliary function (41) — numerical solution; ...... OHAM solution.
Figure 3

Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 0.5, v′(1) = 0 with auxiliary function (41) — numerical solution; ...... OHAM solution.

Figure 4 Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 0.5, v′(1) = 0 with auxiliary function (62) — numerical solution; ...... OHAM solution.
Figure 4

Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 0.5, v′(1) = 0 with auxiliary function (62) — numerical solution; ...... OHAM solution.

Figure 5 Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 0.5, v′(1) = 0 with auxiliary function (41) — numerical solution; ...... OHAM solution.
Figure 5

Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 0.5, v′(1) = 0 with auxiliary function (41) — numerical solution; ...... OHAM solution.

Figure 6 Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 0.5, v′(1) = 0 with auxiliary function (62) — numerical solution; ...... OHAM solution.
Figure 6

Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 0.5, v′(1) = 0 with auxiliary function (62) — numerical solution; ...... OHAM solution.

Figure 7 Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 1, v'(1)=105$$v'\left( 1 \right) = {1 \over {\sqrt {05} }}$ with auxiliary function (41) — numerical solution; ...... OHAM solution.
Figure 7

Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 1, v'(1)=105 with auxiliary function (41) — numerical solution; ...... OHAM solution.

Figure 8 Variation of velocity profile with increasing parameter m for α1 = -0.5, α2 = 0.5, v'(1)=105$$v'\left( 1 \right) = {1 \over {\sqrt {05} }}$ with auxiliary function (71) — numerical solution; ...... OHAM solution.
Figure 8

Variation of velocity profile with increasing parameter m for α1 = -0.5, α2 = 0.5, v'(1)=105 with auxiliary function (71) — numerical solution; ...... OHAM solution.

Figure 9 Variation of velocity profile with increasing parameter m for α1 = -0.5, α2 = 0.75, v'(1)=105$$v'\left( 1 \right) = {1 \over {\sqrt {05} }}$with auxiliary function (52) — numerical solution; ...... OHAM solution.
Figure 9

Variation of velocity profile with increasing parameter m for α1 = -0.5, α2 = 0.75, v'(1)=105with auxiliary function (52) — numerical solution; ...... OHAM solution.

Figure 10 Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 0.5, v'(1)=105$$v'\left( 1 \right) = {1 \over {\sqrt {05} }}$with auxiliary function (41) — numerical solution; ...... OHAM solution.
Figure 10

Variation of velocity profile with increasing parameter m for α1 = 0.5, α2 = 0.5, v'(1)=105with auxiliary function (41) — numerical solution; ...... OHAM solution.

Figure 11 Variation of velocity profile with increasing parameter m for α1 = -0.5, α2 = 1, v'(1)=105$$v'\left( 1 \right) = {1 \over {\sqrt {05} }}$with auxiliary function (52) — numerical solution; ...... OHAM solution.
Figure 11

Variation of velocity profile with increasing parameter m for α1 = -0.5, α2 = 1, v'(1)=105with auxiliary function (52) — numerical solution; ...... OHAM solution.

Figure 12 Variation of velocity profile with increasing parameter m for α1 = -0.5, α2 = 1, v'(1)=105$$v'\left( 1 \right) = {1 \over {\sqrt {05} }}$ with auxiliary function (41) — numerical solution; ...... OHAM solution.
Figure 12

Variation of velocity profile with increasing parameter m for α1 = -0.5, α2 = 1, v'(1)=105 with auxiliary function (41) — numerical solution; ...... OHAM solution.

Figure 13 Variation of velocity profile with increasing parameter α2 for α1 = 0.5, m = 1, v′(1) = 0 with auxiliary function (41)—numerical solution; ...... OHAM solution.
Figure 13

Variation of velocity profile with increasing parameter α2 for α1 = 0.5, m = 1, v′(1) = 0 with auxiliary function (41)—numerical solution; ...... OHAM solution.

Figure 14 Variation of velocity profile with increasing parameter α2 for α1 = 0.5, m = 1.5, v′(1) = 0 with auxiliary function (41) — numerical solution; ...... OHAM solution.
Figure 14

Variation of velocity profile with increasing parameter α2 for α1 = 0.5, m = 1.5, v′(1) = 0 with auxiliary function (41) — numerical solution; ...... OHAM solution.

Figure 15 Variation of velocity profile with increasing parameter α2 for α1 = 0.5, m = 1, v′(1) = 0 with auxiliary function (62) — numerical solution; ...... OHAM solution.
Figure 15

Variation of velocity profile with increasing parameter α2 for α1 = 0.5, m = 1, v′(1) = 0 with auxiliary function (62) — numerical solution; ...... OHAM solution.

Figure 16 Variation of velocity profile with increasing parameter α2 for α1 = 0.5, m = 1.5, v′ = 0 with auxiliary function (62)—numerical solution; ...... OHAM solution.
Figure 16

Variation of velocity profile with increasing parameter α2 for α1 = 0.5, m = 1.5, v′ = 0 with auxiliary function (62)—numerical solution; ...... OHAM solution.

Figure 17 Variation of velocity profile with increasing parameter α2 for α1 = .0.5, m = 1, v'(1)=105$$v'\left( 1 \right) = {1 \over {\sqrt {05} }}$ with auxiliary function (52) v'(1)=105$$v'\left( 1 \right) = {1 \over {\sqrt {05} }}$numerical solution; ...... OHAM solution..
Figure 17

Variation of velocity profile with increasing parameter α2 for α1 = .0.5, m = 1, v'(1)=105 with auxiliary function (52) v'(1)=105numerical solution; ...... OHAM solution..

Figure 18 Variation of velocity profile with increasing parameter α2 for α1 = .0.5, m = 1.5, v'(1)=105$$v'\left( 1 \right) = {1 \over {\sqrt {05} }}$ with auxiliary function (52) — numerical solution; ...... OHAM solution.
Figure 18

Variation of velocity profile with increasing parameter α2 for α1 = .0.5, m = 1.5, v'(1)=105 with auxiliary function (52) — numerical solution; ...... OHAM solution.

Figure 19 Variation of velocity profile with increasing parameter α2 for α1 = .0.5, m = 1, v'(1)=105$$v'\left( 1 \right) = {1 \over {\sqrt {05} }}$ with auxiliary function (71) — numerical solution; ...... OHAM solution.
Figure 19

Variation of velocity profile with increasing parameter α2 for α1 = .0.5, m = 1, v'(1)=105 with auxiliary function (71) — numerical solution; ...... OHAM solution.

Figure 20 Variation of velocity profile with increasing parameter α2 for α1 = .0.5, m = 1.5, v'(1)=105$$v'\left( 1 \right) = {1 \over {\sqrt {05} }}$ with auxiliary function (71) — numerical solution; ...... OHAM solution.
Figure 20

Variation of velocity profile with increasing parameter α2 for α1 = .0.5, m = 1.5, v'(1)=105 with auxiliary function (71) — numerical solution; ...... OHAM solution.

Figure 21 Variation of velocity profile with increasing parameter α1 for α2 = 0.5, m = 1, Case 4.1, — numerical solution; ...... OHAM solution.
Figure 21

Variation of velocity profile with increasing parameter α1 for α2 = 0.5, m = 1, Case 4.1, — numerical solution; ...... OHAM solution.

Figure 22 Variation of velocity profile with increasing parameter α1 for α2 = 0.75, m = 1, Case 4.1, — numerical solution; ...... OHAM solution.
Figure 22

Variation of velocity profile with increasing parameter α1 for α2 = 0.75, m = 1, Case 4.1, — numerical solution; ...... OHAM solution.

Figure 23 Variation of velocity profile with increasing parameter α1 for α2 = 1, m = 1, Case 4.1, — numerical solution; ...... OHAM solution.
Figure 23

Variation of velocity profile with increasing parameter α1 for α2 = 1, m = 1, Case 4.1, — numerical solution; ...... OHAM solution.

Table 5

Comparison between OHAM results given by Eq. (79) and numerical results for α1 = −0.5, α2 = 0.5, m = 1 and v'(1)=105 and the auxiliary function (52).

xvnumericOHAM, Eq. (79)relative error = |vnumericv̅OHAM|
01.1.0.
0.11.22290628071.22290139714.8 · 10−6
0.21.43779297591.43778074501.2 · 10−5
0.31.64458084071.64458679975.9 · 10−6
0.41.84321596251.84322558659.6 · 10−6
0.52.03360462512.03360473341.0 · 10−7
0.62.21563060532.21562240548.1 · 10−6
0.72.38913559142.38913302252.5 · 10−6
0.82.55390537932.55391420688.8 · 10−6
0.92.70964831352.70965107212.7 · 10−6
1.2.85595396362.85594399019.9 · 10−6
Table 6

Comparison between OHAM results given by Eq. (80) and numerical results for α1 = −0.5, α2 = 0.75, m = 1 and v'(1)=105 and the auxiliary function (52).

xvnumericOHAM, Eq. (80)relative error = |vnumericv̅OHAM|
01.1.0.
0.11.25183516291.25182388881.1 · 10−5
0.21.49206914951.49205315301.5 · 10−5
0.31.72073876431.720743989475.2 · 10−6
0.41.93790318421.937920022031.6 · 10−5
0.52.14362146962.143622911351.4 · 10−6
0.62.33790680702.337894474221.2 · 10−5
0.72.52073476332.520729930824.8 · 10−6
0.82.69202208112.692034222201.2 · 10−5
0.92.85160006672.851604386184.3 · 10−6
1.2.99916368242.999148707371.4 · 10−5
Table 7

Comparison between OHAM results given by Eq. (81) and numerical results for α1 = −0.5, α2 = 1, m = 1 and v'(1)=105 and the auxiliary function (52).

xvnumericOHAM, Eq. (81)relative error = |vnumericv̅OHAM|
01.1.0.
0.11.28396912681.28395410271.5 · 10−5
0.21.55220804171.55218934631.8 · 10−5
0.31.80489976851.80490323663.4 · 10−6
0.42.04224816542.04227314882.4 · 10−5
0.52.26451583312.26451931743.4 · 10−6
0.62.47189892612.47188219171.6 · 10−5
0.72.66456987692.66456208467.7 · 10−6
0.82.84264698752.84266231391.5 · 10−5
0.93.00616188533.00616799396.1 · 10−6
13.15499809063.15497781942.0 · 10−5
Table 8

Comparison between OHAM results given by Eq. (54) and numerical results for α1 = −0.5, α2 = 0.5, m = 1.5 and v'(1)=105 and the auxiliary function (52).

xvnumericOHAMrelative error = |vnumericv̅OHAM|
0.1.1.0.
0.11.26095090911.26075647711.9 · 10−4
0.21.50993159771.50925091826.8 · 10−4
0.31.74690501261.74714334652.3 · 10−4
0.41.97185446891.97261916687.6 · 10−4
0.52.18472293222.18497546012.5 · 10−4
0.62.38539980142.38500035673.9 · 10−4
0.72.57370233272.57350621251.9 · 10−4
0.82.74934029102.74986625975.2 · 10−4
0.92.91185990702.91207165192.1 · 10−4
13.06052431863.05954223159.8 · 10−4
Table 9

Comparison between OHAM results given by Eq. (54) and numerical results for α1 = −0.5, α2 = 0.75, m = 1 and v'(1)=105 and the auxiliary function (52).

xvnumericOHAMrelative error = |vnumericOHAM|
01.1.0.
0.11.30850762861.30862933311.2 · 10−4
0.21.59892006721.59822956946.9 · 10−4
0.31.87142264041.87158317641.6 · 10−4
0.42.12621679292.12702500218.0 · 10−4
0.52.36355214862.36389959343.4 · 10−4
0.62.58359329602.58325353093.3 · 10−4
0.72.78645307582.78632112301.3 · 10−4
0.82.97214249442.97282454806.8 · 10−4
0.93.14050279913.14087321133.7 · 10−4
13.29106061753.29009179099.6 · 10−4
Table 10

Comparison between OHAM results given by Eq. (54) and numerical results for α1 = −0.5, α2 = 1, m = 1 and v'(1)=105 and the auxiliary function (52).

xvnumericOHAMrelative error = |vnumericOHAM|
01.1.0.
0.11.36200215211.36199426037.8 · 10−6
0.21.69891874931.69890136731.7 · 10−5
0.32.01115428182.01115574941.4 · 10−6
0.42.29913880162.29917559013.6 · 10−5
0.52.56346727152.56347335796.0 · 10−6
0.62.80464818092.80462727232.0 · 10−5
0.73.02318811463.02318499103.1 · 10−6
0.83.21952663343.21955978983.3 · 10−5
0.93.39395417623.39396489921.0 · 10−5
13.54643810703.54640818072.9 · 10−5
Table 11

Comparison between OHAM results given by Eq. (64) and numerical results for α1 = 0.5, α2 = 0.5, m = 1 and v′(1) = 0 and the auxiliary function (62).

xvnumericOHAMrelative error = |vnumericOHAM|
01.1.0.
0.10.90534482750.90525799618.6 · 10−5
0.20.82034482750.82036167401.6 · 10−5
0.30.74534482750.74537589093.1 · 10−5
0.40.68034482750.68034965964.8 · 10−6
0.50.62534482750.62531740832.7 · 10−5
0.60.58034482750.58030012314.4 · 10−5
0.70.54534482750.54530638663.8 · 10−5
0.80.52034482750.52033332451.1 · 10−5
0.90.50534482750.50536747042.2 · 10−5
10.50034482750.50038555774.0 · 10−5
Table 12

Comparison between OHAM results given by Eq. (64) and numerical results for α1 = 0.5, α2 = 0.75, m = 1 and v′(1) = 0 and the auxiliary function (62).

xvnumericOHAMrelative error = |vnumericOHAM|
01.1.0.
0.10.88660857250.88633612362.7 · 10−4
0.20.78756216710.78750774875.4 · 10−5
0.30.70253389250.70264388471.0 · 10−4
0.40.63089172640.63095369946.1 · 10−5
0.50.57178267050.57173016045.2 · 10−5
0.60.52450861130.52434826691.6 · 10−4
0.70.48843939240.48825889291.8 · 10−4
0.80.46306007520.46297992688.0 · 10−5
0.90.44800062260.44808619218.5 · 10−5
10.44302355080.44319923371.7 · 10−4
Table 13

Comparison between OHAM results given by Eq. (64) and numerical results for α1 = 0.5, α2 = 1, m = 1 and v′(1) = 0 and the auxiliary function (62).

xvnumericOHAMrelative error = |vnumericOHAM|
01.1.00000000006.6 · 10−16
0.10.85717887830.85744157582.6 · 10−4
0.20.73616041380.73716700171.0 · 10−3
0.30.63604154080.63636533813.2 · 10−4
0.40.55485195230.55409067177.6 · 10−4
0.50.49030831440.48866182481.6 · 10−3
0.60.44024881040.43837232101.8 · 10−3
0.70.40294801530.40169947721.2 · 10−3
0.80.37712557230.37722914641.0 · 10−4
0.90.36194704730.36356992891.6 · 10−3
10.35694304370.35932595502.3 · 10−3
Table 14

Comparison between OHAM results given by Eq. (64) and numerical results for α1 = 0.5, α2 = 0.5, m = 1.5 and v′(1) = 0 and the auxiliary function (62).

xvnumericOHAMrelative error = |vnumericOHAM|
01.1.00000000181.8 · 10−9
0.10.85774896260.85771052353.8 · 10−5
0.20.73024896260.73029059544.1 · 10−5
0.30.61774896260.61779411974.5 · 10−5
0.40.52024896260.52026369981.4 · 10−5
0.50.43774896260.43773059451.8 · 10−5
0.60.37024896260.37021465223.4 · 10−5
0.70.31774896260.31772425862.4 · 10−5
0.80.28024896260.28025627227.3 · 10−6
0.90.25774896260.25779598024.7 · 10−5
10.25024896260.25031701846.8 · 10−5
Table 15

Comparison between OHAM results given by Eq. (64) and numerical results for α1 = 0.5, α2 = 0.75, m = 1.5 and v′(1) = 0 and the auxiliary function (62).

xvnumericOHAMrelative error = |vnumericOHAM|
011.0.
0.10.81220304270.81215186335.1 · 10−5
0.20.64805240300.64797814117.4 · 10−5
0.30.50751068180.50759316248.2 · 10−5
0.40.38995870130.39021555362.5 · 10−4
0.50.29435790280.29459338392.3 · 10−4
0.60.21935449320.21931404894.0 · 10−5
0.70.16318895870.16296775702.2 · 10−4
0.80.12430315600.12421574418.7 · 10−5
0.90.10146910980.10180894313.3 · 10−4
10.09395989830.09458510406.2 · 10−4
Table 16

Comparison between OHAM results given by Eq. (64) and numerical results for α1 = 0.5, α2 = 1, m = 1.5 and v′(1) = 0 and the auxiliary function (62).

xvnumericOHAMrelative error = |vnumericOHAM|
01.0.99999999993.5 · 10−15
0.10.74970638200.74985612241.4 · 10−4
0.20.53279631110.53193775638.5 · 10−4
0.30.34989458690.34975273181.4 · 10−4
0.40.20108145640.20227790501.1 · 10−3
0.50.08528859620.08671490091.4 · 10−3
0.6−0.0006296037−0.00030467423.2 · 10−4
0.7−0.0615295648−0.06224119857.1 · 10−4
0.8−0.1020045446−0.10248546044.8 · 10−4
0.9−0.1251782938−0.12431913948.5 · 10−4
1−0.1327530483−0.13091146421.8 · 10−3
Table 17

Comparison between OHAM results given by Eq. (73) and numerical results for α1 = −0.5, α2 = 0.5, m = 1 and v'(1)=105 and the auxiliary function (71).

xvnumericOHAMrelative error = |vnumericOHAM|
01.1.00000000003.5 · 10−15
0.11.22293843821.22300635326.7 · 10−5
0.21.43782475301.43785255652.7 · 10−5
0.31.64461541751.64459941451.6 · 10−5
0.41.84324802591.84322719852.0 · 10−5
0.52.03363628032.03364538309.1 · 10−6
0.62.21566296472.21570245903.9 · 10−5
0.72.38917004762.38919574592.5 · 10−5
0.82.55394418692.55388106056.3 · 10−5
0.92.70969397332.70948208812.1 · 10−4
12.85601156272.85569930373.1 · 10−4
Table 18

Comparison between OHAM results given by Eq. (73) and numerical results for α1 = −0.5, α2 = 0.75, m = 1 and v'(1)=105 and the auxiliary function (71).

xvnumericOHAMrelative error = |vnumericOHAM|
01.1.0.
0.11.25186560791.25185158881.4 · 10−5
0.21.49208771431.49209106133.3 · 10−6
0.31.72074800541.72075319235.1 · 10−6
0.41.93791338311.93790685946.5 · 10−6
0.52.14362993992.14361633891.3 · 10−5
0.62.33791465932.33791040504.2 · 10−6
0.72.52074364382.52075914461.5 · 10−5
0.82.69203460692.69205795962.3 · 10−5
0.92.85161919352.85161797051.2 · 10−6
12.99919570212.99916191523.3 · 10−5
Table 19

Comparison between OHAM results given by Eq. (73) and numerical results for α1 = −0.5, α2 = 1, m = 1 and v'(1)=105 and the auxiliary function (71).

xvnumericOHAMrelative error = |vnumericOHAM|
01.1.00000000007.1 · 10−15
0.11.28399543621.28398989275.5 · 10−6
0.21.55220905381.55220837866.7 · 10−7
0.31.80487588301.80487794842.0 · 10−6
0.42.04222914692.04222915731.0 · 10−8
0.52.26449315802.26448981173.3 · 10−6
0.62.47187381342.47187056333.2 · 10−6
0.72.66454427662.66454603871.7 · 10−6
0.82.84262400122.84263044026.4 · 10−6
0.93.00614524663.00614632121.0 · 10−6
13.15499572503.15498495651.0 · 10−5
Table 20

Comparison between OHAM results given by Eq. (73) and numerical results for α1 = −0.5, α2 = 0.5, m = 1.5 and v'(1)=105 and the auxiliary function (71).

xvnumericOHAMrelative error = |vnumericOHAM|
01.1.0.
0.11.26098081201.26036643986.1 · 10−4
0.21.50995163241.50936557355.8 · 10−4
0.31.74691986261.74672544001.9 · 10−4
0.41.97186528991.97217417503.0 · 10−4
0.52.18473028672.18544009327.0 · 10−4
0.62.38540578472.38625165828.4 · 10−4
0.72.57370997362.57433740796.2 · 10−4
0.82.74935492522.74942608927.1 · 10−5
0.92.91188875012.91124655306.4 · 10−4
13.06058384973.05952779201.0 · 10−3
Table 21

Comparison between OHAM results given by Eq. (73) and numerical results for α1 = −0.5, α2 = 0.75, m = 1.5 and v'(1)=105 and the auxiliary function (71).

xvnumericOHAMrelative error = |vnumericOHAM|
01.0.99999999992.8 · 10−14
0.11.30852891781.30849345733.5 · 10−5
0.21.59891070091.59891747966.7 · 10−6
0.31.87138422201.87139754061.3 · 10−5
0.42.12617984862.12616546501.4 · 10−5
0.52.36350849742.36347506403.3 · 10−5
0.62.58354516582.58353234381.2 · 10−5
0.72.78640415092.78644038393.6 · 10−5
0.82.97209908282.97215819925.9 · 10−5
0.93.14047389923.14047241011.4 · 10−6
13.29106823983.29098026568.7 · 10−5
Table 22

Comparison between OHAM results given by Eq. (73) and numerical results for α1 = −0.5, α2 = 1, m = 1.5 and v'(1)=105 and the auxiliary function (71).

xvnumericOHAMrelative error = |vnumericOHAM|
01.0.99999999992.2 · 10−13
0.11.36201587571.36197560054.0 · 10−5
0.21.69888327051.69889028317.0 · 10−6
0.32.01106424172.01107939661.5 · 10−5
0.42.29905365732.29903832271.5 · 10−5
0.52.56337049252.56333334873.7 · 10−5
0.62.80454220312.80452697441.5 · 10−5
0.73.02307725903.02311643793.9 · 10−5
0.83.21941854173.21948423246.5 · 10−5
0.93.39386043283.39385939471.0 · 10−6
13.54638646913.54628838409.8 · 10−5

6 Conclusions

In the present work we propose Optimal Homotopy Asymptotic Method to obtain approximate analytical solutions for nonlinear differential equation of thin film flow of an Oldroyd 6-constant fluid. The validity of our procedure was demonstrated choosing the linear operator in two cases and different expressions of the convergence-control functions. Some representative examples are given and very good agreement was found between the approximate analytical results and numerical simulation results. The proposed procedure is valid even if the nonlinear differential equation does not contain any small or large parameter. We examine quantitative effect of the parameters α1, α2 and m and the relative errors of approximate solutions in comparison with numerical results. OHAM is an approach proposed for the first time by Marinca and Herisanu in the study of the thin film flow [29], then is used in many other studies as: the steady flow of a fourth-grade fluid past a porous plate [30], the MHD Jeffery-Hamel flow [31] and so on [32], such that for achieving a very accurate solution, our procedure ensured a very rapid convergence after only one iteration. Instead of infinite series, the OHAM searches for only a few terms and does not need recurrence formula. The parameters which appear in the composition of the auxiliary functions and in the linear operator are optimally identified via various methods by mathematically rigorous point of view. A large number of parameters in the auxiliary functions lead to more accurate results.

In all cases presented in this paper, for different values of the parameters α1, α2 and m, we obtain an excellent agreement of the first-order approximate solutions. Also, the exact solutions of Newtonian fluid is obtained. It is worth mentioning that the proposed method is straightforward, concise and can be applied to other nonlinear problems.

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Received: 2015-5-13
Accepted: 2015-11-10
Published Online: 2016-2-20
Published in Print: 2016-1-1

© 2016 Remus-Daniel Ene, published by De Gruyter Open

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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