# The First Solution for the Helical Flow of a Generalized Maxwell Fluid within Annulus of Cylinders by New Definition of Transcendental FunctionBNrrn

Authors: Fang Wang; Jinling Liu
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

Most articles choose the transcendental functionB1rrn to define the finite Hankel transform, and very few articles choose B0rrn. The derivations of B0rrn and B1rrn are also considered the same. In this paper, we find that the derivative formulas for the transcend function BNrrn are different and prove the derivative formulas for B0rrn and B1rrn. Based on the exact formulas of B0rrn and B1rrn, we keep on studying the helical flow of a generalized Maxwell fluid between two boundless coaxial cylinders. In this case, the inner and outer cylinders start to rotate around their axis of symmetry at different angular frequencies and slide at different linear velocities at time t=0+. We deduced the velocity field and shear stress via Laplace transform and finite Hankel transform and their inverse transforms. According to generalized G and R functions, the solutions we obtained are given in the form of integrals and series. The solution of ordinary Maxwell fluid has been also obtained by solving the limit of the general solution of fractional Maxwell fluid.


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## Body

## 1. Introduction

Most industry and academic workers are interested in the motion of fluids on plates or cylinders. Regarding the flow of Newtonian fluid in cylinders, we can find the transient velocity distribution in [1]. With regard to the flow of non-Newtonian fluid in the cylindrical domain, the first exact solutions are those of Srivastava [2] with Maxwell fluids and Ting [3] with second-grade fluids. Casarella and Laura [4] obtained an analytical expression on a smooth circular cylindrical rod. They studied rods with longitudinal and torsional motion. For flow in the same case, Rajagopal [5] found two solutions. Rajagopal and Bhatnagar [6] obtained these solutions for Oldroyd-B fluid. Khan et al. [7] gave an exact analytical solution for the flow of a Burger’s fluid via Fourier transform. The oscillating pressure gradient they considered was the main driving force for the motion. With regard to second-order fluids, Hayat et al. [8, 9] gave an exact analytical solution for unsteady unidirectional flows. They also considered pressure gradients and the flows induced by the motion of one or two plates or the pressure gradient. In terms of unsteady unidirectional flows, Rajagopal [10] also established two types of exact solutions. Among these two types of flows, one is owing to the rigid plate that oscillates in its own direction and the other is the time-periodic Poiseuille flow on account of the oscillating pressure gradient. Recently, many papers of this type have been obtained by Yang and Wang [11–13].Fractional calculus has made great progress in describing the motion of viscoelastic fluids in recent years. Tong et al. [14, 15] discussed an exact analytic solution for the unsteady transient rotational flow of an Oldroyd-B fluid in terms of the fractional calculus definition. Many researchers have studied the rotational flow of a generalized Maxwell fluid between two boundless coaxial cylinders. For example, Fetecau and Fetecau [16] established analytic solutions for this motion. The longitudinal vibration shear stresses that were imposed by the inner cylinder were the main driving force for the motion. Shaowei and Mingyu [17] considered the unsteady Couette flow with fractional derivative. When the outer cylinder remained immobile, the inner cylinder started to move around their axis of symmetry at a fixed angular velocity. They used the integral transform and the generalized Mittag-Leffler function to get the first solution. Zheng et al. [18] discussed the unsteady rotating flow with a fractional derivative model, where the flow is induced by the rotation of the inner cylinder and the oscillation of the pressure gradient. The first solution is given by integral and series via Laplace transform and Hankel transform. Mahmood et al. [19] also discussed the torsional oscillatory motion. The movement of the fluid was produced when the inner and outer cylinders began to oscillate around their symmetry axis. Chaudry et al. [20] established the first analytic solution for the oscillatory flow, when the outer cylinder oscillated along the axis of symmetry and the inner cylinder was stationary.In the problem of using Hankel transform to solve the unsteady flow in a coaxial cylinder, we need to derive the transcendental functionBNrrn in the Hankel transform. Most articles [17–19] select the transcendental function B1rrn when defining the finite Hankel transform, and very few articles [21] choose B0rrn. However, these documents use the same derivation conclusions when calculating the derivatives of B0rrn and B1rrn; it is proved that the derivative formula is dependent on N; in fact, their derivation conclusions are different. Motivated by this problem, in Section 2 of this paper, we study the derivative of the transcendental function BNrrn, when N takes different values, and use the mathematical induction method to prove the derivative formula of the transcendental function BNrrn. Then, we introduce the basic governing equation in Section 3. Based on the exact formula of B0rrn and B1rrn, we calculate the velocity field and shear stress via Laplace transform and finite Hankel transform and their inverse transforms in Sections 4 and 5. According to generalized G and R functions, the solutions we obtained are given in the form of integrals and series. In Section 6, we obtained the solution of Maxwell fluid by solving the limit of the general solution.


## 2. Derivative of the Transcendental FunctionBNrrn

According to the Bessel functions, Eldabe et al. [22] defined the finite Hankel transformation of fr in a,b as follows:(1)Frn=∫abrBNrrnfrdr,b>a,where the transcendental function is(2)BNrrn=JNrrnYNbrn−JNbrnYNrrn,N=0,1,2,….r
n are the positive roots of the transcendental equation:(3)JNbrnYNarn−JNarnYNbrn=0,and JN is the first-type Bessel function of the N-order and YN is the second-type Bessel function of the N-order [23]. In this part, we will correct the derivative of the transcendental function BNrrn and obtain the following conclusions.Theorem 1.
Derivative formula of transcendental functionBNrrn:(4)∂∂rBNrrn=rnB¯Nrrn−NrBNrrn,N=1,2,…,where(5)BNrrn=JNrrnYNbrn−JNbrnYNrrn,B¯Nrrn=JN−1rrnYNbrn−JNbrnYN−1rrn.Proof.
WhenN=1,(6)B1rrn=J1rrnY1brn−J1brnY1rrn,B¯1rrn=J0rrnY1brn−J1brnY0rrn.
To prove(7)∂∂rB1rrn=rnB¯1rrn−1rB1rrn=rnJ0rrn−1rJ1rrnY1brn−J1brnrnY0rrn−1rY1rrn,we need to obtain(8)∂∂rJ1rrn=rnJ0rrn−1rJ1rrn,∂∂rY1rrn=rnY0rrn−1rY1rrn.
Since(9)∂∂rJ1rrn=∂∂r∑k=0∞−1kk!k+1!rrn22k+1=∑k=0∞2k+1−1⋅rn2⋅−1kk!k+1!rrn22k=rn∑k=0∞−1kk!k!rrn22k−1r∑k=0∞−1kk!k+1!rrn22k+1=rnJ0rrn−1rJ1rrn,we get(10)∂∂rJ1rrn=rnJ0rrn−1rJ1rrn.
Similarly,(11)∂∂rY1rrn=2πrnJ0rrn−1rJ1rrnlnrrn2+1r⋅2πJ1rrn+1π⋅2r2rn−1π∑k=0∞2k+1⋅rn2−1kk!k+1!ψk+1+ψk+2rrn22k=rn⋅2πJ0rrnlnrrn2−1r⋅2πJ1rrnlnrrn2+1r⋅2π∑k=0∞−1kk!k+1!rrn22k+1+1r⋅1π2rrn−1π∑k=0∞2k+1−1⋅rn2⋅−1kk!k+1!ψk+1+ψk+2rrn22k=rn⋅2πJ0rrnlnrrn2−rn⋅1π∑k=0∞−1kk!k!ψk+1+ψk+2−1k+1rrn22k−1rY1rrn.where(12)ψk+1=−γ+1+12+⋯+1k,ψk+2=−γ+1+12+⋯+1k+1k+1,so we establish the following:(13)∂∂rY1rrn=rn2πJ0rrnlnrrn2−1π∑k=0∞−1kk!k!2ψk+1rrn22k−1rY1rrn=rnY0rrn−1rY1rrn,∂∂rB1rrn=rnJ0rrn−1rJ1rrnY1brn−J1brnrnY0rrn−1rY1rrn=rnB¯1rrn−1rB1rrn.
Assume that whenN=m, the conclusion is true. Now, we prove that when N=m+1,(14)Bm+1rrn=Jm+1rrnYm+1brn−Jm+1brnYm+1rrn,B¯m+1rrn=JmrrnYm+1brn−Jm+1brnYmrrn.
To prove(15)∂∂rBm+1rrn=rnB¯m+1rrn−m+1rBm+1rrn=rnJmrrn−m+1rJm+1rrnYm+1brn−Jm+1brnrnYmrrn−m+1rYm+1rrn,we need to prove(16)∂∂rJm+1rrn=rnJmrrn−m+1rJm+1rrn,∂∂rYm+1rrn=rnYmrrn−m+1rYm+1rrn,where(17)Jmrrn=∑k=0∞−1kk!k+m!rrn22k+m,Jm+1rrn=∑k=0∞−1kk!k+m+1!rrn22k+m+1,Ymrrn=2πJmrrnlnrrn2−1π∑k=0m−1m−k−1!k!rrn22k−m−1π∑k=0∞−1kk!k+m!ψk+m+1+ψk+1rrn22k+m,Ym+1rrn=2πJm+1rrnlnrrn2−1π∑k=0mm−k!k!rrn22k−m+1−1π∑k=0∞−1kk!k+m+1!ψk+m+2+ψk+1rrn22k+m+1.
Since(18)∂∂rJm+1rrn=∑k=0∞2k+m+1⋅rn2⋅−1kk!k+m+1!rrn22k+m=∑k=0∞2k+m+1−m+1⋅rn2⋅−1kk!k+m+1!rrn22k+m=rn∑k=0∞−1kk!k+m!rrn22k+m−m+1r∑k=0∞−1kk!k+m+1!rrn22k+m+1=rnJmrrn−m+1rJm+1rrn,then(19)∂∂rJm+1rrn=rnJmrrn−m+1rJm+1rrn.
Similarly,(20)∂∂rYm+1rrn=2πrnJmrrn−m+1rJm+1rrnlnrrn2+1r⋅2π∑k=0∞−1kk!k+m+1!rrn22k+m+1−rn⋅1π∑k=0m−1m−k−1!k!rrn22k−m+m+1r⋅1π∑k=0mm−k!k!rrn22k−m+1−1π∑k=0∞2k+m+1⋅rn2⋅−1kk!k+m+1!ψk+m+2+ψk+1rrn22k+m=rn2πJmrrnlnrrn2−1π∑k=0m−1m−k−1!k!rrn22k−m−1π∑k=0∞−1kk!k+m!ψk+m+2+ψk+1−1k+m+1rrn22k+m−m+1r2πJm+1rrnlnrrn2−1π∑k=0mm−k!k!rrn22k−m+1−1π∑k=0∞−1kk!k+m+1!ψk+m+2+ψk+1rrn22k+m+1=rn2πJmrrnlnrrn2−1π∑k=0m−1m−k−1!k!rrn22k−m−1π∑k=0∞−1kk!k+m!⋅ψk+m+1+ψk+1rrn22k+m−m+1rYm+1rrn=rnYmrrn−m+1rYm+1rrn,∂∂rYm+1rrn=rnYmrrn−m+1rYm+1rrn.
So we obtain(21)∂∂rBm+1rrn=rnJmrrn−m+1rJm+1rrnYm+1brn−Jm+1brnrnYmrrn−m+1rYm+1rrn=rnB¯m+1rrn−m+1rBm+1rrn.Hence, the conclusion is true.Corollary 1.
(1)
Derivative formula of transcendental functionB0rrn is as follows:
(22)
∂
∂
r
B
0
r
r
n
=
−
r
n
J
1
r
r
n
Y
0
b
r
n
−
J
0
b
r
n
Y
1
r
r
n
.
(2)
Derivative formula of transcendental functionB1rrn is as follows:
(23)
∂
∂
r
B
1
r
r
n
=
r
n
J
0
r
r
n
−
1
r
J
1
r
r
n
Y
1
b
r
n
−
J
1
b
r
n
r
n
Y
0
r
r
n
−
1
r
Y
1
r
r
n
.Proof.
(1)
Since
(24)
∂
∂
r
J
0
r
r
n
=
∂
∂
r
∑
k
=
0
∞
−
1
k
k
!
k
!
r
r
n
2
2
k
=
∑
k
=
1
∞
2
k
⋅
r
n
2
⋅
−
1
k
k
!
k
!
r
r
n
2
2
k
−
1
=
−
r
n
∑
k
=
1
∞
−
1
k
−
1
k
−
1
!
k
!
r
r
n
2
2
k
−
1
+
1
=
−
r
n
J
1
r
r
n
,
∂
∂
r
Y
0
r
r
n
=
∂
∂
r
2
π
J
0
r
r
n
ln
r
r
n
2
−
1
π
∑
k
=
0
∞
−
1
k
k
!
k
!
2
ψ
k
+
1
r
r
n
2
2
k
=
−
r
n
⋅
2
π
J
1
r
r
n
ln
r
r
n
2
+
1
r
⋅
2
π
∑
k
=
0
∞
−
1
k
k
!
k
!
r
r
n
2
2
k
−
r
n
⋅
1
π
∑
k
=
1
∞
−
1
k
k
!
k
−
1
!
2
ψ
k
+
1
r
r
n
2
2
k
−
1
=
−
r
n
⋅
2
π
J
1
r
r
n
ln
r
r
n
2
+
1
r
⋅
2
π
+
r
n
⋅
1
π
∑
k
=
1
∞
−
1
k
k
!
k
!
r
r
n
2
2
k
−
1
−
r
n
⋅
1
π
∑
k
=
1
∞
−
1
k
k
!
k
−
1
!
2
ψ
k
+
1
r
r
n
2
2
k
−
1
=
−
r
n
2
π
J
1
r
r
n
ln
r
r
n
2
−
1
π
2
r
r
n
+
1
π
∑
k
=
1
∞
−
1
k
k
!
k
−
1
!
2
ψ
k
+
1
−
1
k
r
r
n
2
2
k
−
1
=
−
r
n
2
π
J
1
r
r
n
ln
r
r
n
2
−
1
π
2
r
r
n
−
1
π
∑
k
=
0
∞
−
1
k
k
!
k
+
1
!
ψ
k
+
2
+
ψ
k
+
1
r
r
n
2
2
k
+
1
=
−
r
n
Y
1
r
r
n
,
we get
(25)
∂
∂
r
B
0
r
r
n
=
∂
∂
r
J
0
r
r
n
Y
0
b
r
n
−
J
0
b
r
n
Y
0
r
r
n
=
−
r
n
J
1
r
r
n
Y
0
b
r
n
−
J
0
b
r
n
−
r
n
Y
1
r
r
n
=
−
r
n
J
1
r
r
n
Y
0
b
r
n
−
J
0
b
r
n
Y
1
r
r
n
.
(2)
It can be directly obtained from Theorem1 conclusion.


## 3. Basic Governing Equations

In this paper, based on the exact formula ofB0rrn and B1rrn, we keep on studying the helical flow of a generalized Maxwell fluid between two boundless coaxial cylinders. We research the velocity of the helical flow of the following form:(26)V=Vr,t=ωr,teθ+vr,tez,where ωr,t is the rotating velocity and vr,t is the sliding velocity. The inner and outer cylinders start to rotate around their axis of symmetry at different angular frequencies ω1 and ω2 and slide along the same axis of symmetry at different linear velocities A1t and A2tA1≠A2 at time t=0+.The initial and boundary conditions we gave are(27)vr,0=ωr,0=∂vr,0∂t=∂ωr,0∂t=0,r∈R1,R2,(28)ωR1,t=ω1sinβ1t,ωR2,t=ω2sinβ2t,t≥0,(29)vR1,t=A1t,vR2,t=A2t,t≥0.By introducing the time derivative with a fractional differential operator to the conservation and constitutive equations of an incompressible Maxwell fluid, the governing equation is given by [15, 24, 25](30)1+λDtαSrθ=μ∂ωr,t∂r−ωr,tr,(31)1+λDtαSrz=μ∂vr,t∂r,where λ is the material constant, μ is the dynamic viscosity of the fluid, and Srθ and Srz are the shear stresses. Without considering the body forces and pressure gradient, the momentum conservation equation is(32)ρ∂ωr,t∂t=∂Srθ∂r+2Srθr,(33)ρ∂vr,t∂t=∂Srz∂r+Srzr,where ρ is the constant density of the fluid.


## 4. Calculation of the Velocity Field

Applying the Laplace transform [26, 27] to (30)–(33), we obtain(34)1+λsαS¯rθ=μ∂ω¯r,s∂r−ω¯r,sr,(35)1+λsαS¯rz=μ∂v¯r,s∂r,(36)ρsω¯r,s=∂S¯rθ∂r+2S¯rθr,(37)ρsv¯r,s=∂S¯rz∂r+S¯rzr.According to (34)–(37), we obtain the following ordinary differential equation:(38)∂2ω¯r,s∂r2+1r∂ω¯r,s∂r−ω¯r,sr2−ρs1+λsαμω¯r,s=0,(39)∂2v¯r,s∂r2+1r∂v¯r,s∂r−ρs1+λsαμv¯r,s=0,where the functions ω¯r,s and v¯r,s are the Laplace transformations of ωr,t and vr,t. Similarly, applying the Laplace transforms (28) and (29), we can obtain(40)ω¯Ri,s=ωiβis2+βi2,v¯Ri,s=Ais2,i=1,2.According to Section2, we know that(41)ω¯Hs=∫R1R2rω¯r,sB1rrndr,n=1,2,3,…,v¯Hs=∫R1R2rv¯r,sB0rrmdr,m=1,2,3,…,are the finite Hankel transforms of ω¯r,s and v¯r,s, where rn and rm are the positive roots of equations B1R1r=0 and B0R1r=0, and(42)B0rrm=J0rrmY0R2rm−J0R2rmY0rrm,B1rrn=J1rrnY1R2rn−J1R2rnY1rrn.Multiplying both sides of (38) and (39) by rB1rrn and rB0rrm, integrating with respect to r from R1 to R2, and considering conditions (40), we can obtain [28](43)∫R1R2r∂2ω¯r,s∂r2+1r∂ω¯r,s∂r−ω¯r,sr2B1rrndr=2πω2β2s2+β22−2πω1β1s2+β12J1R2rnJ1R1rn−rn2ω¯Hs,(44)∫R1R2rρs1+λsαμω¯r,sB1rrndr=ρs1+λsαμω¯Hs.Since(45)∂∂rJ0rrm=−rmJ1rrm,∂∂rY0rrm=−rmY1rrm,Jm+1xYmx−JmxYm+1x=2πx,∫R1R2r∂2v¯r,s∂r2+1r∂v¯r,s∂rB0rrmdr=∫R1R2∂2v¯r,s∂r2rB0rrmdr+∫R1R2∂v¯r,s∂rB0rrmdr=∂v¯r,s∂rrB0rrmR2R1−∫R1R2∂v¯r,s∂r⋅∂∂rrB0rrmdr+∫R1R2∂v¯r,s∂rB0rrmdr=∫R1R2∂v¯r,s∂rrrmJ1rrmY0R2rm−J0R2rmY1rrmdr=v¯r,srrmJ1rrmY0R2rm−J0R2rmY1rrmR2R1−∫R1R2v¯r,srrm2B0rrmdr=2πv¯R2,s−v¯R1,sR1rmJ1R1rmJ0R1rmY0R2rm−J0R2rmJ0R1rmY1R1rmJ0R1rm−rm2v¯Hs=2πv¯R2,s−2πv¯R1,sJ0R2rmJ0R1rm−rm2v¯Hs=2πA2s2−2πA1s2J0R2rmJ0R1rm−rm2v¯Hs,we can get(46)∫R1R2r∂2v¯r,s∂r2+1r∂v¯r,s∂rB0rrmdr=2πA2s2−2πA1s2J0R2rmJ0R1rm−rm2v¯Hs,(47)∫R1R2rρs1+λsαμv¯r,sB0rrmdr=ρs1+λsαμv¯Hs.According to (43)–(47),(48)2πω2β2s2+β22−2πω1β1s2+β12J1R2rnJ1R1rn−rn2ω¯Hs−ρs1+λsαμω¯Hs=0,2πA2s2−2πA1s2J0R2rmJ0R1rm−rm2v¯Hs−ρs1+λsαμv¯Hs=0,or equivalently(49)ω¯Hs=2πω2β2s2+β22μρs+ρλsα+1+μrn2−2πω1β1s2+β12J1R2rnJ1R1rnμρs+ρλsα+1+μrn2,(50)v¯Hs=2πA2s2μρs+ρλsα+1+μrm2−2πA1s2J0R2rmJ0R1rmμρs+ρλsα+1+μrm2.We first write (49) and (50) under the suitable form as follows:(51)ω¯Hs=2πrn2ω2β2s2+β22−2πrn2ω2β2s2+β22ρs1+λsαρs+ρλsα+1+μrn2−2πrn2ω1β1s2+β12J1R2rnJ1R1rn−2πrn2ω1β1s2+β12J1R2rnJ1R1rnρs1+λsαρs+ρλsα+1+μrn2=2πrn2ω2β2s2+β22−2πrn2J1R2rnJ1R1rnω1β1s2+β12−2πrn2ω2β2s2+β22ρs1+λsαρs+ρλsα+1+μrn2+2πrn2ω1β1s2+β12J1R2rnJ1R1rnρs1+λsαρs+ρλsα+1+μrn2,v¯Hs=2πrm2A2s2−2πrm2A2s2ρs1+λsαρs+ρλsα+1+μrm2−2πrm2A1s2J0R2rmJ0R1rm−2πrm2A1s2J0R2rmJ0R1rmρs1+λsαρs+ρλsα+1+μrm2=2πrm2A2s2−2πrm2J0R2rmJ0R1rmA1s2−2πrm2A2s2ρs1+λsαρs+ρλsα+1+μrm2+2πrm2A1s2J0R2rmJ0R1rmρs1+λsαρs+ρλsα+1+μrm2,and we use the inverse Hankel transform formula [29, 30]:(52)ω¯r,s=π22∑n=1∞rn2J12R1rnB1rrnJ12R1rn−J12R2rn×ω¯Hs,v¯r,s=π22∑m=1∞rm2J02R1rmB0rrmJ02R1rm−J02R2rm×v¯Hs.As we know if(53)ar=AR1R22−r2+BR2r2−R12R22−R12r,then(54)an=∫R1R2rarB1rrndr=2πrn2B−2πrn2J1R2rnJ1R1rnA.If(55)br=ClnR2−lnr+Dlnr−lnR1lnR2−lnR1,then(56)∫R1R2rlnR2−lnrlnR2−lnR1B0rrmdr=1lnR2−lnR1R2lnR2rmJ1R2rmY0R2rm−R1lnR2rmJ1R1rmY0R2rm−R2lnR2rmJ0R2rmY1R2rm+R1lnR2rmJ0R2rmY1R1rm−R2lnR2rmJ1R2rmY0R2rm+R1lnR1rmJ1R1rmY0R2rm−J0R2rmY0R2rmrm2+J0R1rmY0R2rmrm2+R2lnR2rmJ0R2rmY1R2rm−R1lnR1rmJ0R2rmY1R1rm+J0R2rmY0R2rmrm2−J0R2rmY0R1rmrm2=1lnR2−lnR1R1lnR2rmJ0R2rmY1R1rm−J1R1rmY0R2rm+R1lnR1rmJ1R1rmY0R2rm−J0R2rmY1R1rm=−2π1rm2J0R2rmJ0R1rm,∫R1R2rlnr−lnR1lnR2−lnR1B0rrmdr=1lnR2−lnR1R2lnR2rmJ1R2rmY0R2rm−R1lnR1rmJ1R1rmY0R2rm+J0R2rmY0R2rmrm2−J0R1rmY0R2rmrm2−R2lnR2rmJ0R2rmY1R2rm+R1lnR1rmJ0R2rmY1R1rm−J0R2rmY0R2rmrm2+J0R2rmY0R1rmrm2−R2lnR1rmJ1R2rmY0R2rm+R1lnR1rmJ1R1rmY0R2rm+R2lnR1rmJ0R2rmY1R2rm−R1lnR1rmJ0R2rmY1R1rm=1lnR2−lnR1R2lnR2rmJ1R2rmY0R2rm−J0R2rmY1R2rm+R1lnR1rmJ0R2rmY1R2rm−J1R2rmY0R2rm=2π1rm2,so (57)bn=∫R1R2rarB0rrmdr=2πrm2D−2πrm2J0R2rmJ0R1rmC.So we can obtain(58)ω¯r,s=ω1β1s2+β12R1R22−r2R22−R12r+ω2β2s2+β22R2r2−R12R22−R12r−π∑n=1∞J1R1rnB1rrnJ12R1rn−J12R2rn×ρs1+λsαρs+ρλsα+1+μrn2J1R1rnω2β2s2+β22−J1R2rnω1β1s2+β12,v¯r,s=A1s2lnR2−lnrlnR2−lnR1+A2s2lnr−lnR1lnR2−lnR1−π∑m=1∞J0R1rmB0rrmJ02R1rm−J02R2rm×ρs1+λsαρs+ρλsα+1+μrm2J0R1rmA2s2−J0R2rmA1s2.To make the calculation of calculus more convenient, we can write [31, 32] the following:(59)ρs1+λsαρs+ρλsα+1+μrn2=11+μrn2/ρs1+λsα=∑k=0∞−μrn2ρk⋅s−k1+λsαk=∑k=0∞−μrn2λρk⋅s−kλ−1+sαk,ρs1+λsαρs+ρλsα+1+μrm2=∑k=0∞−μrm2λρk⋅s−kλ−1+sαk,and we use the Laplace transform of the functions Ga,b,cd,t:(60)L−1sbsa−dc=Ga,b,cd,t,where(61)Ga,b,cd,t=∑j=0∞cjdjtj+ca−b−1j!Γj+ca−b,is the generalized G function and cj is the Pochhammer polynomial [33].Ifft=L−1f¯s and gt=L−1g¯s, then(62)L−1f¯s⋅g¯s=f∗gt=∫0tft−τgτdτ=∫0tftgt−τdτ.We attain the following velocity field:(63)ωr,t=R1R22−r2R22−R12rω1sinβ1t+R2r2−R12R22−R12rω2sinβ2t−π∑n=1∞∑k=0∞J1R1rnB1rrnJ12R1rn−J12R2rn×−μrn2λρk∫0tJ1R1rnω2sinβ2t−τ−J1R2rnω1sinβ1t−τGα,−k,k−λ−1,τdτ.(64)vr,t=lnR2−lnrlnR2−lnR1A1t+lnr−lnR1lnR2−lnR1A2t−π∑m=1∞∑k=0∞J0R1rmB0rrmJ02R1rm−J02R2rm×−μrm2λρk∫0tJ0R1rmA2t−τ−J0R2rmA1t−τGα,−k,k−λ−1,τdτ.We lett=0,r=R1, and r=R2 in (63) and (64) and found that(65)vr,0=ωr,0=0,ωR1,t=ω1sinβ1t,ωR2,t=ω2sinβ2t,vR1,t=A1t,vR2,t=A2t.So the velocity field satisfies our initial and boundary conditions.


## 5. Calculation of the Shear Stress

According to (34) and (35), we can obtain that(66)S¯rθ=μ1+λsα∂ω¯r,s∂r−ω¯r,sr,(67)S¯rz=μ1+λsα∂v¯r,s∂r.By using Corollary 1,(68)∂∂rB1rrn=rnB¯1rrn−1rB1rrn,∂∂rB0rrm=−rmB˜0rrm,where(69)B¯1rrn=J0rrnY1R2rn−J1R2rnY0rrn,B˜0rrm=J1rrmY0R2rm−J0R2rmY1rrm,so(70)∂ω¯r,s∂r−ω¯r,sr=2R12R2R22−R12r2ω2β2s2+β22−2R1R22R22−R12r2ω1β1s2+β12−π∑n=1∞J1R1rnJ12R1rn−J12R2rn×rnB¯1rr1n−2/rB1rrnρs1+λsαρs+ρλsα+1+μrn2J1R1rnω2β2s2+β22−J1R2rnω1β1s2+β12,(71)∂v¯r,s∂r=A2s21rlnR2−lnR1−A1s21rlnR2−lnR1+π∑m=1∞J0R1rmJ02R1rm−J02R2rm×rmB˜0rrmρs1+λsαρs+ρλsα+1+μrm2J0R1rmA2s2−J0R2rmA1s2.Introducing (70) and (71) into (66) and (67), we can obtain(72)S¯rθ=2μR12R2R22−R12r211+λsαω2β2s2+β22−2μR1R22R22−R12r211+λsαω1β1s2+β12−μπ∑n=1∞J1R1rnJ12R1rn−J12R2rn×rnB¯1rr1n−2/rB1rrnρsρs+ρλsα+1+μrn2J1R1rnω2β2s2+β22−J1R2rnω1β1s2+β12,(73)S¯rz=A2s211+λsαμrlnR2−lnR1−A1s211+λsαμrlnR2−lnR1+μπ∑m=1∞J0R1rmJ02R1rm−J02R2rm×rmB˜0rrmρsρs+ρλsα+1+μrm2J0R1rmA2s2−J0R2rmA1s2.To make the calculation of calculus more comfortable, we write(74)ρsρs+ρλsα+1+μrn2=1λ∑k=0∞−μrn2λρk⋅s−kλ−1+sαk+1,ρsρs+ρλsα+1+μrm2=1λ∑k=0∞−μrm2λρk⋅s−kλ−1+sαk+1,11+λsα=1λ1λ−1+sα,and using the Laplace transform of the generalized Ga,b,cd,t function and Ra,bc,d,t function [33], we obtain(75)L−1sbsa−dc=Ga,b,cd,t,L−1e−dssbsa−c=Ra,bc,d,t,where(76)Ga,b,cd,t=∑j=0∞cjdjtj+ca−b−1j!Γj+ca−b,Ra,bc,d,t=∑j=0∞cjt−dj+1a−b−aΓj+1a−b.We apply the inversion Laplace transform to (72) and (73) and obtain the shear stress as follows:(77)Srθ=2μR1R2λR22−R12r2∫0tR1ω2sinβ2t−τ−R2ω1sinβ1t−τRα,0−λ−1,0,τdτ−μπλ∑n=1∞∑k=0∞J1R1rnrnB¯1rrn−2/rB1rrnJ12R1rn−J12R2rn−μrn2λρk×∫0tJ1R1rnω2sinβ2t−τ−J1R2rnω1sinβ1t−τGα,−k,k+1−λ−1,τdτ,(78)Srz=μrlnR2−lnR1∫0tA2t−τ−A1t−τRα,0−λ−1,0,τdτ+μπλ∑m=1∞∑k=0∞rmJ0R1rmB˜0rrmJ02R1rm−J02R2rm−μrm2λρk×∫0tJ0R1rmA2t−τ−J0R2rmA1t−τGα,−k,k+1−λ−1,τdτ.


## 6. Limiting Caseα⟶1

(1)
Whenα⟶1, (63) and (64) can be reduced to
(79)
ω
r
,
t
=
R
1
R
2
2
−
r
2
R
2
2
−
R
1
2
r
ω
1
sin
β
1
t
+
R
2
r
2
−
R
1
2
R
2
2
−
R
1
2
r
ω
2
sin
β
2
t
−
π
∑
n
=
1
∞
∑
k
=
0
∞
J
1
R
1
r
n
B
1
r
r
n
J
1
2
R
1
r
n
−
J
1
2
R
2
r
n
×
−
μ
r
n
2
λ
ρ
k
∫
0
t
J
1
R
1
r
n
ω
2
sin
β
2
t
−
τ
−
J
1
R
2
r
n
ω
1
sin
β
1
t
−
τ
G
1
,
−
k
,
k
−
λ
−
1
,
τ
d
τ
,
(80)
v
r
,
t
=
ln
R
2
−
ln
r
ln
R
2
−
ln
R
1
A
1
t
+
ln
r
−
ln
R
1
ln
R
2
−
ln
R
1
A
2
t
−
π
∑
m
=
1
∞
∑
k
=
0
∞
J
0
R
1
r
m
B
0
r
r
m
J
0
2
R
1
r
m
−
J
0
2
R
2
r
m
×
−
μ
r
m
2
λ
ρ
k
∫
0
t
J
0
R
1
r
m
A
2
t
−
τ
−
J
0
R
2
r
m
A
1
t
−
τ
G
1
,
−
k
,
k
−
λ
−
1
,
τ
d
τ
.
(2)
Whenα⟶1, (77) and (78) can be reduced to
(81)
S
r
θ
=
2
μ
R
1
R
2
λ
R
2
2
−
R
1
2
r
2
∫
0
t
R
1
ω
2
sin
β
2
t
−
τ
−
R
2
ω
1
sin
β
1
t
−
τ
R
1,0
−
λ
−
1
,
0
,
τ
d
τ
−
μ
π
λ
∑
n
=
1
∞
∑
k
=
0
∞
J
1
R
1
r
n
r
n
B
¯
1
r
r
n
−
2
/
r
B
1
r
r
n
J
1
2
R
1
r
n
−
J
1
2
R
2
r
n
−
μ
r
n
2
λ
ρ
k
×
∫
0
t
J
1
R
1
r
n
ω
2
sin
β
2
t
−
τ
−
J
1
R
2
r
n
ω
1
sin
β
1
t
−
τ
G
1
,
−
k
,
k
+
1
−
λ
−
1
,
τ
d
τ
,
(82)
S
r
z
=
μ
r
ln
R
2
−
ln
R
1
∫
0
t
A
2
t
−
τ
−
A
1
t
−
τ
R
1,0
−
λ
−
1
,
0
,
τ
d
τ
+
μ
π
λ
∑
m
=
1
∞
∑
k
=
0
∞
r
m
J
0
R
1
r
m
B
˜
0
r
r
m
J
0
2
R
1
r
m
−
J
0
2
R
2
r
m
−
μ
r
m
2
λ
ρ
k
×
∫
0
t
J
0
R
1
r
m
A
2
t
−
τ
−
J
0
R
2
r
m
A
1
t
−
τ
G
1
,
−
k
,
k
+
1
−
λ
−
1
,
τ
d
τ
.Whenα=1, the fractional Maxwell fluid model is an ordinary Maxwell fluid model. Therefore, (79)–(82) are the analytical solutions for the ordinary Maxwell fluid.


## 7. Conclusion

As we know, most articles choose the transcendental functionB1rrn and few articles choose B0rrn to define the finite Hankel transform; moreover, the derivations of B0rrn and B1rrn are often considered the same. Actually, in this paper, by use of the mathematical induction method, we prove the derivative formula for the transcend function BNrrn. Then, according to the momentum conservation law, the relevant and meaningful equation and constitutive equation are given. Based on the exact formula of B0rrn and B1rrn, we calculate the velocity field and shear stress via Laplace transform and finite Hankel transform and their inverse transforms in Sections 4 and 5. In order to present our solution in a simpler and more comfortable form, we combine the generalized G function and R function. The solutions we obtained are given in the form of integrals and series containing the generalized G function and R function. At the end of Section 4, we explain that these solutions satisfy initial and boundary conditions that we gave. When α=1, the fractional Maxwell fluid model is an ordinary Maxwell fluid model. As before, we obtained the solution of Maxwell fluid by solving the limit of the general solution.



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