# Exact Solutions for the Axial Couette Flow of a Fractional Maxwell Fluid in an Annulus

Authors: M. Imran; A. U. Awan; Mehwish Rana; M. Athar; M. Kamran
Journal: ISRN Mathematical Physics
 (2012)

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## Abstract

The velocity field and the adequate shear stress corresponding to the rotational flow of a fractional Maxwell fluid, between two infinite coaxial circular cylinders, are determined by applying the Laplace and finite Hankel transforms. The solutions that have been obtained are presented in terms of generalizedGa,b,c(·,t) and Ra,b(·,t) functions. Moreover, these solutions satisfy both the governing differential equations and all imposed initial and boundary conditions. The corresponding solutions for ordinary Maxwell and Newtonian fluids are obtained as limiting cases of our general solutions. Finally, the influence of the material parameters on the velocity and shear stress of the fluid is analyzed by graphical illustrations.


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## Body

## 1. Introduction

Due to the several technological applications, the flow analysis of non-Newtonian fluids is very important in the fields of fluid mechanics. Many investigators have not studied the flow behavior of non-Newtonian fluids in various flow fields due to the complex stress-strain relationship [1]. The study of non-Newtonian fluids has got much attention because of their practical applications. Non-Newtonian characteristics are displayed by a number of industrially important fluids including polymers, molten plastic, pulps, microfluids, and food stuff display. Exact analytic solutions for the flows of non-Newtonian fluids are important provided they correspond to physically realistic problems, and they can be used as checks against complicated numerical codes that have been developed for much more complex flows. Many non-Newtonian models such as differential type, rate type, and integral type fluids have been proposed in recent years. Among them, the rate type fluid models have received special attention. The differential type fluids do not predict stress relaxation, and they are not successful for describing the flows of some polymers.The flow between rotating cylinders or through a rotating cylinder has applications in the food industry, it being one of the most important and interesting problems of motion near rotating bodies. As early as 1886, Stokes [2] established an exact solution for the rotational oscillations of an infinite rod immersed in a linearly viscous fluid. However, such motions have been intensively studied since G. I. Taylor (1923) reported the results of his famous investigations [3]. For Newtonian fluids, the velocity distribution for a fluid contained in a circular cylinder can be found in [4]. The first exact solutions corresponding to different motions of non-Newtonian fluids, in cylindrical domains, seem to be those of Ting [5], Srivastava [6], and Waters and King [7]. A lot of interests and studies were also given to the unidirectional start-up pipe flows, which has a significant practical and mathematical meaning. Zhu et al. [8] studied the characteristics of the velocity filed and the shear stress field for an ordinary Maxwell fluid, and Yang and Zhu [9] studied it for a fractional Maxwell fluid. In the last decade, the unidirectional flow of viscoelastic fluid with the fractional Maxwell model was studied by Tan et al. [10, 11] and Hayat et al. [12]. Tong et al. [13, 14] discussed the unsteady flow with a generalized Jeffrey’s model in an annular pipe. In the meantime, a lot of papers regarding such motions have been published. The interested readers can see for instance the papers [15–25] and their related references.The purpose of this paper is to provide exact solutions of the velocity field and the shear stress corresponding to the motion of a fractional Maxwell fluid between two infinite circular cylinders. The Laplace and finite Hankel transforms are used to solve the problem, and the solutions obtained are presented in terms of generalizedGa,b,c(·,t) and Ra,b(·,t) functions. The solutions for ordinary Maxwell and Newtonian fluids are obtained as limiting cases of our general solutions. Furthermore, the solutions for the motion between the cylinders, when one of them is at rest, are also obtained as special cases from our general results. At the end, obtained solutions are discussed graphically for different values of time and material parameters.


## 2. Basic Governing Equations

The constitutive equations of an incompressible Maxwell fluid with fractional calculus are given by [14](2.1)T=-pI+S,S+λDSDt=μA,
where T is the Cauchy stress tensor, -pI denotes the indeterminate spherical stress, S is the extrastress tensor, A=L+LT with L the velocity gradient, μ is the dynamic viscosity of the fluid, λ is the material constant called relaxation time, and DS/Dt is defined by(2.2)DSDt=DtβS+w⋅∇S-LS-SLT.
Here, w is the velocity vector, ∇ is the gradient operator, the superscript T denotes the transpose operation, and the Caputo fractional derivative operator Dtβ is defined as [26](2.3)Dtβf(t)=1Γ(1-β)∫0tf′(τ)(t-τ)βdτ;0≤β<1,
where Γ(·) is the Gamma function which is defined as(2.4)Γ(x)=∫0∞sx-1e-sds;x>0.
This model can be reduced to ordinary Maxwell model when β→1, because in this case Dtβf(t)→df(t)/dt. Furthermore, this model reduces to the classical Newtonian model for β→1 and λ→0.In cylindrical coordinates(r,θ,z), the rotational flow velocity is given by(2.5)w=w(r,t)=w(r,t)eθ,
where eθ is the unit vector in the θ-direction. For such flows, the constraint of incompressibility is automatically satisfied. Since the velocity field w is independent of θ and z, we also assume that S depends only on r and t. Furthermore, if the fluid is assumed to be at rest at the moment t=0, then(2.6)w(r,0)=0,S(r,0)=0.
Equations (2.1), (2.5), and (2.6) imply Srr=Szz=Sθz=0 [18], (2.7)(1+λDtβ)τ(r,t)=μ(∂∂r-1r)w(r,t),
where τ(r,t)=Srθ(r,t) is the nontrivial shear stress. In the absence of body forces and a pressure gradient in the axial direction, the equations of motion lead to the relevant equation(2.8)ρ∂w(r,t)∂t=(∂∂r+2r)τ(r,t),
where ρ is the constant density of the fluid. Eliminating τ between (2.7) and (2.8), we attain to the governing equation(2.9)(1+λDtβ)∂w(r,t)∂t=ν(∂2∂r2+1r∂∂r-1r2)w(r,t),
where ν=μ/ρ is the kinematic viscosity of the fluid. In the following, the fractional partial differential equations (2.9) and (2.7), with appropriate initial and boundary conditions, will be solved by means of Laplace and finite Hankel transforms. In order to avoid lengthy calculations of residues and contours integrals, the discrete inverse Laplace method will be used [13, 14].


## 3. Axial Couette Flow between Two Infinite Circular Cylinders

Let us consider an incompressible fractional Maxwell fluid at rest in an annular region between two coaxial circular cylinders of radiiR1 and R2(>R1). At time t=0+, both cylinders with radii R1 and R2 begin to rotate along their common axis. Owing to the shear, the fluid is gradually moved, its velocity being of the form (2.5). The governing equations are given by (2.9), while the appropriate initial and boundary conditions are(3.1)w(r,0)=∂w(r,0)∂t=0,τ(r,0)=0;r∈[R1,R2],(3.2)w(R1,t)=Ω1R1t,w(R2,t)=Ω2R2tfort≥0,
where Ω1 and Ω2 are constants with dimensions T-2.


### 3.1. Calculation of the Velocity Field

Applying the Laplace transform to (2.9), using the Laplace transform formula for sequential fractional derivatives [26], and having the initial and boundary conditions (3.1) and (3.2) in mind, we find that(3.3)(q+λqβ+1)w¯(r,q)=ν(∂2∂r2+1r∂∂r-1r2)w¯(r,q);r∈[R1,R2],
where w¯(r,q) is the Laplace transform of the function w(r,t) which is defined as(3.4)w¯(r,q)=L{w(r,t)}=∫0∞e-qtw(r,t)dt,
and the image function w¯(r,q) has to satisfy the conditions(3.5)w¯(R1,q)=Ω1R1q2,w¯(R2,q)=Ω2R2q2.In the following, we denote by [27](3.6)w¯H(rn,q)=∫R1R2rw¯(r,q)B(r,rn)dr,
and the Hankel transform of w¯(r,q), where(3.7)B(r,rn)=J1(rrn)Y1(R2rn)-J1(R2rn)Y1(rrn),
and rn are the positive roots of the transcendental equation B(R1,r)=0, while J1(·) and Y1(·) are Bessel functions of the first and second kind of order one.Multiplying both sides of (3.3) by rB(r,rn), integrating with respect to r from R1 to R2, and taking into account the conditions (3.5) and the identity(3.8)∫R1R2r[∂2∂r2+1r∂∂r-1r2]w¯(r,q)B(r,rn)dr=2πq2[Ω2R2J1(R1rn)-Ω1R1J1(R2rn)J1(R1rn)]-rn2w¯H(rn,q),
we find that(3.9)w¯H(rn,q)=2ν[Ω2R2J1(R1rn)-Ω1R1J1(R2rn)]πJ1(R1rn)1q2(λqβ+1+q+νrn2).
Now, for a suitable presentation of the final results, we rewrite (3.9) in the following equivalent form:(3.10)w¯H(rn,q)=2[Ω2R2J1(R1rn)-Ω1R1J1(R2rn)]πrn2J1(R1rn)[1q2-1+λqβq(λqβ+1+q+νrn2)].
Now, applying the inverse Hankel transform formula [27](3.11)w¯(r,q)=π22∑n=1∞rn2J12(R1rn)B(r,rn)J12(R1rn)-J12(R2rn)w¯H(rn,q),
we obtain the Laplace transform of the velocity field w¯(r,q) under the form(3.12)w¯(r,q)=Ω1R12(R22-r2)+Ω2R22(r2-R12)r(R22-R12)1q2-π∑n=1∞J1(R1rn)B(r,rn)J12(R1rn)-J12(R2rn)×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}1+λqβq(λqβ+1+q+νrn2),
writing the last factor of (3.12) in the following equivalent form:(3.13)1+λqβq(λqβ+1+q+νrn2)=1λ∑k=0∞(-νrn2λ)k[q-k-2(qβ+λ-1)k+1+λqβ-k-2(qβ+λ-1)k+1].
Introducing (3.13) into (3.12), applying the discrete inverse Laplace transform, and using the known result [28, equation (97)],(3.14)L-1{qb(qa-d)c}=Ga,b,c(d,t);Re(ac-b),Re(q)>0,|dqa|<1,
where the generalized Ga,b,c(·,·) function is defined by(3.15)Ga,b,c(d,t)=∑j=0∞djΓ(c+j)Γ(c)Γ(j+1)t(c+j)a-b-1Γ[(c+j)a-b],
and we find the velocity field under the form(3.16)w(r,t)=Ω1R12(R22-r2)+Ω2R22(r2-R12)r(R22-R12)t-πλ∑n=1∞J1(R1rn)B(r,rn)J12(R1rn)-J12(R2rn){Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}×∑k=0∞(-νrn2λ)k{Gβ,-k-2,k+1(-λ-1,t)+λGβ,β-k-2,k+1(-λ-1,t)}.


### 3.2. Calculation of the Shear Stress

Applying the Laplace transform to (2.7), we find that(3.17)τ¯(r,q)=μ1+λqβ(∂w¯(r,q)∂r-w¯(r,q)r),
where(3.18)∂w¯(r,q)∂r-w¯(r,q)r=2R12R22(Ω2-Ω1)r2(R22-R12)1q2+π∑n=1∞J1(R1rn)(2/rB(r,rn)-rnB̃(r,rn))J12(R1rn)-J12(R2rn)×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}1+λqβq(λqβ+1+q+νrn2)
is obtained from (3.12) and(3.19)B̃(r,rn)=J0(rrn)Y1(R2rn)-J1(R2rn)Y0(rrn).
Thus, (3.17) becomes(3.20)τ¯(r,q)=2μR12R22(Ω2-Ω1)r2(R22-R12)1q2(1+λqβ)+πμ∑n=1∞J1(R1rn)(2/rB(r,rn)-rnB̃(r,rn))J12(R1rn)-J12(R2rn)×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}1q(λqβ+1+q+νrn2),
applying again the discrete inverse Laplace transform as well as using the known relation [28, equation (21)],(3.21)L-1{qbqa-c}=Ra,b(c,t);Re(a-b)>0,Re(q)>0,
where the generalized Ra,b(c,t) functions are defined by [28](3.22)Ra,b(c,t)=∑n=0∞cnt(n+1)a-b-1Γ[(n+1)a-b]
and the expansion(3.23)1q(λqβ+1+q+νrn2)=1λ∑k=0∞(-νrn2λ)kq-k-2(qβ+λ-1)k+1,
and we obtain the shear stress τ(r,t) under the form(3.24)τ(r,t)=2μR12R22(Ω2-Ω1)λr2(R22-R12)Rβ,-2(-λ-1,t)+πμλ∑n=1∞J1(R1rn)(2/rB(r,rn)-rnB̃(r,rn))J12(R1rn)-J12(R2rn)×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}∑k=0∞(-νrn2λ)kGβ,-k-2,k+1(-λ-1,t).


## 3.1. Calculation of the Velocity Field

Applying the Laplace transform to (2.9), using the Laplace transform formula for sequential fractional derivatives [26], and having the initial and boundary conditions (3.1) and (3.2) in mind, we find that(3.3)(q+λqβ+1)w¯(r,q)=ν(∂2∂r2+1r∂∂r-1r2)w¯(r,q);r∈[R1,R2],
where w¯(r,q) is the Laplace transform of the function w(r,t) which is defined as(3.4)w¯(r,q)=L{w(r,t)}=∫0∞e-qtw(r,t)dt,
and the image function w¯(r,q) has to satisfy the conditions(3.5)w¯(R1,q)=Ω1R1q2,w¯(R2,q)=Ω2R2q2.In the following, we denote by [27](3.6)w¯H(rn,q)=∫R1R2rw¯(r,q)B(r,rn)dr,
and the Hankel transform of w¯(r,q), where(3.7)B(r,rn)=J1(rrn)Y1(R2rn)-J1(R2rn)Y1(rrn),
and rn are the positive roots of the transcendental equation B(R1,r)=0, while J1(·) and Y1(·) are Bessel functions of the first and second kind of order one.Multiplying both sides of (3.3) by rB(r,rn), integrating with respect to r from R1 to R2, and taking into account the conditions (3.5) and the identity(3.8)∫R1R2r[∂2∂r2+1r∂∂r-1r2]w¯(r,q)B(r,rn)dr=2πq2[Ω2R2J1(R1rn)-Ω1R1J1(R2rn)J1(R1rn)]-rn2w¯H(rn,q),
we find that(3.9)w¯H(rn,q)=2ν[Ω2R2J1(R1rn)-Ω1R1J1(R2rn)]πJ1(R1rn)1q2(λqβ+1+q+νrn2).
Now, for a suitable presentation of the final results, we rewrite (3.9) in the following equivalent form:(3.10)w¯H(rn,q)=2[Ω2R2J1(R1rn)-Ω1R1J1(R2rn)]πrn2J1(R1rn)[1q2-1+λqβq(λqβ+1+q+νrn2)].
Now, applying the inverse Hankel transform formula [27](3.11)w¯(r,q)=π22∑n=1∞rn2J12(R1rn)B(r,rn)J12(R1rn)-J12(R2rn)w¯H(rn,q),
we obtain the Laplace transform of the velocity field w¯(r,q) under the form(3.12)w¯(r,q)=Ω1R12(R22-r2)+Ω2R22(r2-R12)r(R22-R12)1q2-π∑n=1∞J1(R1rn)B(r,rn)J12(R1rn)-J12(R2rn)×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}1+λqβq(λqβ+1+q+νrn2),
writing the last factor of (3.12) in the following equivalent form:(3.13)1+λqβq(λqβ+1+q+νrn2)=1λ∑k=0∞(-νrn2λ)k[q-k-2(qβ+λ-1)k+1+λqβ-k-2(qβ+λ-1)k+1].
Introducing (3.13) into (3.12), applying the discrete inverse Laplace transform, and using the known result [28, equation (97)],(3.14)L-1{qb(qa-d)c}=Ga,b,c(d,t);Re(ac-b),Re(q)>0,|dqa|<1,
where the generalized Ga,b,c(·,·) function is defined by(3.15)Ga,b,c(d,t)=∑j=0∞djΓ(c+j)Γ(c)Γ(j+1)t(c+j)a-b-1Γ[(c+j)a-b],
and we find the velocity field under the form(3.16)w(r,t)=Ω1R12(R22-r2)+Ω2R22(r2-R12)r(R22-R12)t-πλ∑n=1∞J1(R1rn)B(r,rn)J12(R1rn)-J12(R2rn){Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}×∑k=0∞(-νrn2λ)k{Gβ,-k-2,k+1(-λ-1,t)+λGβ,β-k-2,k+1(-λ-1,t)}.


## 3.2. Calculation of the Shear Stress

Applying the Laplace transform to (2.7), we find that(3.17)τ¯(r,q)=μ1+λqβ(∂w¯(r,q)∂r-w¯(r,q)r),
where(3.18)∂w¯(r,q)∂r-w¯(r,q)r=2R12R22(Ω2-Ω1)r2(R22-R12)1q2+π∑n=1∞J1(R1rn)(2/rB(r,rn)-rnB̃(r,rn))J12(R1rn)-J12(R2rn)×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}1+λqβq(λqβ+1+q+νrn2)
is obtained from (3.12) and(3.19)B̃(r,rn)=J0(rrn)Y1(R2rn)-J1(R2rn)Y0(rrn).
Thus, (3.17) becomes(3.20)τ¯(r,q)=2μR12R22(Ω2-Ω1)r2(R22-R12)1q2(1+λqβ)+πμ∑n=1∞J1(R1rn)(2/rB(r,rn)-rnB̃(r,rn))J12(R1rn)-J12(R2rn)×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}1q(λqβ+1+q+νrn2),
applying again the discrete inverse Laplace transform as well as using the known relation [28, equation (21)],(3.21)L-1{qbqa-c}=Ra,b(c,t);Re(a-b)>0,Re(q)>0,
where the generalized Ra,b(c,t) functions are defined by [28](3.22)Ra,b(c,t)=∑n=0∞cnt(n+1)a-b-1Γ[(n+1)a-b]
and the expansion(3.23)1q(λqβ+1+q+νrn2)=1λ∑k=0∞(-νrn2λ)kq-k-2(qβ+λ-1)k+1,
and we obtain the shear stress τ(r,t) under the form(3.24)τ(r,t)=2μR12R22(Ω2-Ω1)λr2(R22-R12)Rβ,-2(-λ-1,t)+πμλ∑n=1∞J1(R1rn)(2/rB(r,rn)-rnB̃(r,rn))J12(R1rn)-J12(R2rn)×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}∑k=0∞(-νrn2λ)kGβ,-k-2,k+1(-λ-1,t).


## 4. Limiting Cases

### 4.1. Classical Maxwell Fluid

Makingβ→1 into (3.16) and (3.24), we obtain the velocity field(4.1)wM(r,t)=Ω1R12(R22-r2)+Ω2R22(r2-R12)r(R22-R12)t-πλ∑n=1∞J1(R1rn)B(r,rn)J12(R1rn)-J12(R2rn){Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}×∑k=0∞(-νrn2λ)k{G1,-k-2,k+1(-λ-1,t)+λG1,-k-1,k+1(-λ-1,t)}
and the shear stress(4.2)τM(r,t)=2μR12R22(Ω2-Ω1)λr2(R22-R12)R1,-2(-λ-1,t)+πμλ∑n=1∞J1(R1rn)(2/rB(r,rn)-rnB̃(r,rn))J12(R1rn)-J12(R2rn)×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}∑k=0∞(-νrn2λ)kG1,-k-2,k+1(-λ-1,t),
corresponding to an ordinary Maxwell fluid, performing the same motion. Of course, in view of the identities(4.3)∑k=0∞(-νrn2λ)kG1,-k-1,k+1(-λ-1,t)=eq2nt-eq1ntq2n-q1n,∑k=0∞(-νrn2λ)kG1,-k-2,k+1(-λ-1,t)=λνrn2(1+q1neq2nt-q2neq1ntq2n-q1n),R1,-2(-λ-1,t)=λt-λ2(1-e-t/λ);q1n,q2n=-1±1-4νλrn22λ,
the expressions (4.1) and (4.2) can be written in the simplified form(4.4)wM(r,t)=Ω1R12(R22-r2)+Ω2R22(r2-R12)r(R22-R12)t-πν∑n=1∞J1(R1rn)B(r,rn)rn2[J12(R1rn)-J12(R2rn)]×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}{1-λq1n2eq2nt-q2n2eq1ntq2n-q1n},τM(r,t)=2μR12R22(Ω2-Ω1)r2(R22-R12){t-λ(1-e-t/λ)}+πρ∑n=1∞J1(R1rn)(2/rB(r,rn)-rnB̃(r,rn))rn2[J12(R1rn)-J12(R2rn)]×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}{1+q1neq2nt-q2neq1ntq2n-q1n}.


### 4.2. Newtonian Fluid

By now lettingλ→0 into  (4.4) or β→1 and λ→0 into  (3.16) and (3.24), using(4.5)lim⁡λ→01λkG1,b,k(-1λ,t)=t-b-1Γ(-b),b<0,
we obtain the velocity field(4.6)wN(r,t)=Ω1R12(R22-r2)+Ω2R22(r2-R12)r(R22-R12)t-πν∑n=1∞J1(R1rn)B(r,rn)rn2[J12(R1rn)-J12(R2rn)]×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}{1-e-νrn2t}
and the associated shear stress(4.7)τN(r,t)=2μR12R22(Ω2-Ω1)r2(R22-R12)t+πρ∑n=1∞J1(R1rn)(2/rB(r,rn)-rnB̃(r,rn))rn2[J12(R1rn)-J12(R2rn)]×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}{1-e-νrn2t},
corresponding to a Newtonian fluid, performing the same motion.


## 4.1. Classical Maxwell Fluid

Makingβ→1 into (3.16) and (3.24), we obtain the velocity field(4.1)wM(r,t)=Ω1R12(R22-r2)+Ω2R22(r2-R12)r(R22-R12)t-πλ∑n=1∞J1(R1rn)B(r,rn)J12(R1rn)-J12(R2rn){Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}×∑k=0∞(-νrn2λ)k{G1,-k-2,k+1(-λ-1,t)+λG1,-k-1,k+1(-λ-1,t)}
and the shear stress(4.2)τM(r,t)=2μR12R22(Ω2-Ω1)λr2(R22-R12)R1,-2(-λ-1,t)+πμλ∑n=1∞J1(R1rn)(2/rB(r,rn)-rnB̃(r,rn))J12(R1rn)-J12(R2rn)×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}∑k=0∞(-νrn2λ)kG1,-k-2,k+1(-λ-1,t),
corresponding to an ordinary Maxwell fluid, performing the same motion. Of course, in view of the identities(4.3)∑k=0∞(-νrn2λ)kG1,-k-1,k+1(-λ-1,t)=eq2nt-eq1ntq2n-q1n,∑k=0∞(-νrn2λ)kG1,-k-2,k+1(-λ-1,t)=λνrn2(1+q1neq2nt-q2neq1ntq2n-q1n),R1,-2(-λ-1,t)=λt-λ2(1-e-t/λ);q1n,q2n=-1±1-4νλrn22λ,
the expressions (4.1) and (4.2) can be written in the simplified form(4.4)wM(r,t)=Ω1R12(R22-r2)+Ω2R22(r2-R12)r(R22-R12)t-πν∑n=1∞J1(R1rn)B(r,rn)rn2[J12(R1rn)-J12(R2rn)]×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}{1-λq1n2eq2nt-q2n2eq1ntq2n-q1n},τM(r,t)=2μR12R22(Ω2-Ω1)r2(R22-R12){t-λ(1-e-t/λ)}+πρ∑n=1∞J1(R1rn)(2/rB(r,rn)-rnB̃(r,rn))rn2[J12(R1rn)-J12(R2rn)]×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}{1+q1neq2nt-q2neq1ntq2n-q1n}.


## 4.2. Newtonian Fluid

By now lettingλ→0 into  (4.4) or β→1 and λ→0 into  (3.16) and (3.24), using(4.5)lim⁡λ→01λkG1,b,k(-1λ,t)=t-b-1Γ(-b),b<0,
we obtain the velocity field(4.6)wN(r,t)=Ω1R12(R22-r2)+Ω2R22(r2-R12)r(R22-R12)t-πν∑n=1∞J1(R1rn)B(r,rn)rn2[J12(R1rn)-J12(R2rn)]×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}{1-e-νrn2t}
and the associated shear stress(4.7)τN(r,t)=2μR12R22(Ω2-Ω1)r2(R22-R12)t+πρ∑n=1∞J1(R1rn)(2/rB(r,rn)-rnB̃(r,rn))rn2[J12(R1rn)-J12(R2rn)]×{Ω2R2J1(R1rn)-Ω1R1J1(R2rn)}{1-e-νrn2t},
corresponding to a Newtonian fluid, performing the same motion.


## 5. Special Cases

### 5.1. When the Inner Cylinder Is at Rest

MakingΩ1=0 and Ω2=Ω into  (3.16) and (3.24), for instance, we obtain the velocity field(5.1)w1(r,t)=ΩR22(r2-R12)r(R22-R12)t-πΩR2λ∑n=1∞J12(R1rn)B(r,rn)J12(R1rn)-J12(R2rn)×∑k=0∞(-νrn2λ)k{Gβ,-k-2,k+1(-λ-1,t)+λGβ,β-k-2,k+1(-λ-1,t)}
and the shear stress(5.2)τ1(r,t)=2μΩR12R22λr2(R22-R12)Rβ,-2(-λ-1,t)+πμΩR2λ∑n=1∞J12(R1rn)(2/rB(r,rn)-rnB̃(r,rn))J12(R1rn)-J12(R2rn)∑k=0∞(-νrn2λ)kGβ,-k-2,k+1(-λ-1,t),
corresponding to a fractional Maxwell fluid when the inner cylinder is at rest. Figure 1(a) shows velocity profile corresponding to (5.1) for different values of time, when the inner cylinder is at rest. It shows that velocity is an increasing function with regard to t and r on the whole flow domain.Profiles of the velocityw(r,t) given by (5.1) and (5.3) for R1=0.3, R2=0.5, ν=0.004, μ=2.916, λ=3, β=0.8, and different values of t.
(a)(b)


### 5.2. When the Outer Cylinder Is at Rest

MakingΩ1=Ω and Ω2=0 into (3.16) and (3.24), we obtain the velocity field(5.3)w2(r,t)=ΩR12(R22-r2)r(R22-R12)t+πΩR1λ∑n=1∞J1(R1rn)J1(R2rn)B(r,rn)J12(R1rn)-J12(R2rn)×∑k=0∞(-νrn2λ)k{Gβ,-k-2,k+1(-λ-1,t)+λGβ,β-k-2,k+1(-λ-1,t)}
and the associated shear stress(5.4)τ2(r,t)=2μΩR12R22λr2(R22-R12)Rβ,-2(-λ-1,t)-πμΩR1λ×∑n=1∞J1(R1rn)J1(R2rn)(2/rB(r,rn)-rnB̃(r,rn))J12(R1rn)-J12(R2rn)∑k=0∞(-νrn2λ)kGβ,-k-2,k+1(-λ-1,t),
corresponding to a fractional Maxwell fluid when the outer cylinder is at rest. Figure 1(b) shows the profile of the velocity field corresponding to (5.3) for different values of time, when the outer cylinder is at rest, respectively. It shows that velocity is an increasing function with regard to t like Figure 1(a), but it has opposite effect for r, more exact velocity is decreasing with regard to r on the whole flow domain.


## 5.1. When the Inner Cylinder Is at Rest

MakingΩ1=0 and Ω2=Ω into  (3.16) and (3.24), for instance, we obtain the velocity field(5.1)w1(r,t)=ΩR22(r2-R12)r(R22-R12)t-πΩR2λ∑n=1∞J12(R1rn)B(r,rn)J12(R1rn)-J12(R2rn)×∑k=0∞(-νrn2λ)k{Gβ,-k-2,k+1(-λ-1,t)+λGβ,β-k-2,k+1(-λ-1,t)}
and the shear stress(5.2)τ1(r,t)=2μΩR12R22λr2(R22-R12)Rβ,-2(-λ-1,t)+πμΩR2λ∑n=1∞J12(R1rn)(2/rB(r,rn)-rnB̃(r,rn))J12(R1rn)-J12(R2rn)∑k=0∞(-νrn2λ)kGβ,-k-2,k+1(-λ-1,t),
corresponding to a fractional Maxwell fluid when the inner cylinder is at rest. Figure 1(a) shows velocity profile corresponding to (5.1) for different values of time, when the inner cylinder is at rest. It shows that velocity is an increasing function with regard to t and r on the whole flow domain.Profiles of the velocityw(r,t) given by (5.1) and (5.3) for R1=0.3, R2=0.5, ν=0.004, μ=2.916, λ=3, β=0.8, and different values of t.
(a)(b)


## 5.2. When the Outer Cylinder Is at Rest

MakingΩ1=Ω and Ω2=0 into (3.16) and (3.24), we obtain the velocity field(5.3)w2(r,t)=ΩR12(R22-r2)r(R22-R12)t+πΩR1λ∑n=1∞J1(R1rn)J1(R2rn)B(r,rn)J12(R1rn)-J12(R2rn)×∑k=0∞(-νrn2λ)k{Gβ,-k-2,k+1(-λ-1,t)+λGβ,β-k-2,k+1(-λ-1,t)}
and the associated shear stress(5.4)τ2(r,t)=2μΩR12R22λr2(R22-R12)Rβ,-2(-λ-1,t)-πμΩR1λ×∑n=1∞J1(R1rn)J1(R2rn)(2/rB(r,rn)-rnB̃(r,rn))J12(R1rn)-J12(R2rn)∑k=0∞(-νrn2λ)kGβ,-k-2,k+1(-λ-1,t),
corresponding to a fractional Maxwell fluid when the outer cylinder is at rest. Figure 1(b) shows the profile of the velocity field corresponding to (5.3) for different values of time, when the outer cylinder is at rest, respectively. It shows that velocity is an increasing function with regard to t like Figure 1(a), but it has opposite effect for r, more exact velocity is decreasing with regard to r on the whole flow domain.


## 6. Conclusions

In this paper, the velocityw(r,t) and the shear stress τ(r,t) corresponding to the flow of an incompressible Maxwell fluid with fractional derivatives, in the annular region between two infinite coaxial circular cylinders, have been determined using the Laplace and finite Hankel transforms. The solutions that have been obtained, written under a series form in terms of generalized Ga,b,c(·,t)- and Ra,b(·,t)-functions, satisfy the governing equations and all imposed initial and boundary conditions. In the limiting cases, when β→1 or β→1 and λ→0, the corresponding solutions for the ordinary Maxwell and Newtonian fluids are obtained. These solutions also satisfy the associated initial and boundary conditions (3.1) and (3.2), respectively. Moreover, the solutions for the motion between the cylinders, when one of them is at rest, are also obtained from our general results.In order to reveal some relevant physical aspects of the obtained results, the diagrams of the velocity fieldw(r,t) and the shear stress τ(r,t) given by (3.16) and (3.24) have been drawn against r for different values of the time t and the material parameters. Figures 2 and 3 show the profile of the fluid motion at different values of time when both inner and outer are rotating with the same angular velocity in the same direction and in the opposite direction, respectively. From these figures, one can clearly see that both velocity and shear stress in absolute values are increasing function of t. From Figure 3(a), one can also observe that fluid has zero velocity nearer to inner cylinder.Profiles of the velocityw(r,t) and shear stress τ(r,t) given by (3.16) and (3.24) for R1=0.3, R2=0.5, Ω1=1, Ω2=1, ν=0.004, μ=2.916, λ=4, β=0.5, and different values of t.
(a)(b)Profiles of the velocityw(r,t) and shear stress τ(r,t) given by (3.16) and (3.24) for R1=0.3, R2=0.5, Ω1=-1, Ω2=Ω=1, ν=0.003, μ=2.916, λ=4, β=0.5, and different values of t.
(a)(b)In Figure4, the influence of the relaxation time λ on the fluid motion is shown. As expected, both the velocity and the shear stress (in absolute value) are decreasing functions with respect to λ. Effect of fractional parameter β on the fluid motion is represented in Figure 5, and it is clearly seen that both velocity and shear stress (in absolute value) are increasing with respect to β.Profiles of the velocityw(r,t) and shear stress τ(r,t) given by (3.16) and (3.24) for R1=0.3, R2=0.5, Ω1=1, Ω2=1, t=8, ν=0.003, μ=2.916, β=0.5, and different values of λ.
(a)(b)Profiles of the velocityw(r,t) and shear stress τ(r,t) given by (3.16) and (3.24) for R1=0.3, R2=0.5, Ω1=1, Ω2=1, t=10, ν=0.005, μ=2.916, λ=4, and different values of β.
(a)(b)Finally, for comparison, the diagrams ofw(r,t) and τ(r,t) corresponding to the three models (fractional Maxwell, ordinary Maxwell, and Newtonian) are together depicted in Figure 6 for the same values of the common material constants and time t. The Newtonian fluid is the swiftest, while the fractional Maxwell fluid is the slowest on the whole flow domain. One thing is worth of mentioning that the units of the material constants are SI units in all the figures, and the roots rn have been approximated by nπ/(R2-R1).Figure 6
Profiles of the velocityw(r,t) and shear stress τ(r,t) corresponding to the Newtonian, Maxwell, and fractional Maxwell fluids, for R1=0.3, R2=0.5, Ω1=1, Ω2=1, t=7, ν=0.002, μ=5, λ=2, and β=0.4.



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*Source: 209678-2012-02-13.xml*