RETRACTED ARTICLE: Velocity and shear stress for an Oldroyd-B fluid within two cylinders

Abstract

This paper aims to explore the possible solutions for the movement of an Oldroyd-B fluid placed under certain conditions, i.e. the fluid is present within two cylinders, which are coaxial and oscillating within. Having said that the governing model will be an Oldroyd-B fluid, we wish to achieve our goal of finding the velocity and shear stress by using some common transformations, namely the Laplace transformation and the Hankel transformation. The final results, for the sake of simplicity, will be expressed in the form of generalized G-function and they satisfy all imposed initial and boundary conditions.

1 Introduction

Flow due to an oscillating cylinder is one of the most important and interesting problems of motion near oscillating walls. As early as 1886, Stokes [1] established an exact solution to the rotational oscillations of an infinite rod immersed in a Newtonian fluid. An extension of this problem to the rod undergoing both rotational and longitudinal oscillations has been realized in [2], while the first exact solutions for similar motions of non-Newtonian fluids are those of Rajagopal [3] and Rajagopal and Bhatnagar [4]. However, all these solutions are steady-state solutions to which a transient solution has to be added in order to describe the motion of the fluid for small and large times.

The first closed-form expressions for the starting solutions corresponding to an oscillating motion seem to be those of Erdogan [5] for Newtonian fluids. New exact solutions for the same problem, but presented as a sum of steady-state and transient solutions, have also been established by Corina Fetecau et al. [6]. The extension of these solutions to second grade fluids has been achieved in [7], while the starting solutions for the motion of the same fluids due to longitudinal and torsional oscillations of a circular cylinder have been established in [8]. Recently, starting solutions for oscillating motions of a Maxwell fluid in cylindrical domains have been obtained in [9]. Other interesting results regarding oscillating flows of non-Newtonian fluids have been presented in [10–15].

In this paper, we are interested in the velocity and shear stress for the movement of an Oldroyd-B fluid within two coaxial infinite oscillating cylinders oscillatory motion of a generalized Maxwell fluid between two infinite coaxial circular cylinders, both of them oscillating around their common axis with given constant angular frequencies z. The velocity field and associated tangential stress of the motion are determined by using Laplace and Hankel transforms and are presented by integral and series. It is worthy to point out that the solutions that have been obtained satisfy the governing differential equation and all imposed initial and boundary conditions as well. The solutions correspond to the ordinary Oldroyd-B fluid, performing the same motion.

1.1 Governing equations of problem

The movement of the Oldroyd-B fluid is governed by the following mathematical model:

$$\begin{aligned}& \biggl(1+ \lambda\frac{\partial}{\partial t} \biggr)\tau(r,t)=\mu \biggl(1+ \lambda_{r}\frac{\partial}{\partial t} \biggr) \biggl(\frac{\partial }{\partial r}- \frac{1}{r} \biggr) w(r,t), \end{aligned}$$

(1)

$$\begin{aligned}& \biggl(1+ \lambda\frac{\partial}{\partial t} \biggr)\frac{\partial w(r,t)}{\partial t}=\nu \biggl(1+ \lambda_{r}\frac{\partial}{\partial t} \biggr) \biggl( \frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac {\partial}{\partial r}-\frac{1}{r^{2}} \biggr) w(r,t). \end{aligned}$$

(2)

Here we have labeled the dynamic viscosity as μ, whereas the kinematic viscosity is $\nu=\frac{\mu}{\rho}$, the constant density of the fluid is presented as ρ, the relaxation time is λ, and the retardation time is $\lambda_{r}$. We have labeled the velocity V as $w(r,t)$ and the extra-stress S as $\tau(r,t)$ and the governing model using fractional derivatives eventually becomes

$$\begin{aligned}& \bigl(1+\lambda D^{\xi}_{t}\bigr) \tau(r,t)= \mu \bigl(1+\lambda_{r} D^{\eta}_{t}\bigr) \biggl( \frac{\partial}{\partial r}-\frac{1}{r} \biggr) w(r,t), \end{aligned}$$

(3)

$$\begin{aligned}& \bigl(1+\lambda D^{\xi}_{t}\bigr) \frac{\partial w(r,t)}{\partial t}= \nu \bigl(1+\lambda_{r} D^{\eta}_{t} \bigr) \biggl(\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial}{\partial r}- \frac{1}{r^{2}} \biggr) w(r,t), \end{aligned}$$

(4)

due to the fractional operator defined as follows:

$$\begin{aligned} D^{\xi}_{t} f(t) =& \frac{1}{\Gamma(1-\xi)} \frac {d}{dt} \int\frac{f(\tau)}{(t-\tau)^{\xi}}\,d\tau\quad \mbox{when } 0\leq\xi< 1, \end{aligned}$$

(5)

$$\begin{aligned} =& \frac{d}{dt}f(t) \quad\mbox{when }\xi=1. \end{aligned}$$

(6)

We can notice that for ξ and $\eta \rightarrow1$, our model involving fractional derivatives reduces to the basic model defined earlier due to the fact $D^{1}_{t} f(t)=\frac{d}{dt}f(t)$.

2 Theoretical description of the problem

Suppose a viscoelastic (Oldroyd-B) fluid is at rest in the annulus of coaxial circular cylinders whose lengths are infinite and having $R_{1}$ and $R_{2}$ radii, respectively, where $R_{1} < R_{2}$. Initially at $t = 0$, both the cylinders and the fluid are at rest. At time $t = 0^{+}$, the outer cylinder suddenly begin to oscillate around its axis ($r = 0$) with the velocity $Z \sin(z t)$, where z is the constant angular frequency of the outer cylinder and Z is the constant. Owing to the shear, the fluid between the cylinders is gradually moved, its velocity being of the form

$$\mathbf{V} = \mathbf{V} (r, t)=w (r, t) \mathbf{e}_{\theta}, $$

where $\mathbf{e}_{\theta}$ is the unit vector along θ-direction of the polar coordinate system whose coordinates are $(r, \theta, z)$.

The constraint of incompressibility is automatically satisfied for this kind of flows. The equation for this motion is

$$ \tau(r,t)=\frac{\mu(1+\lambda_{r} D^{\eta}_{t})}{(1+\lambda D^{\xi }_{t})} \biggl(\frac{\partial}{\partial r}- \frac{1}{r} \biggr) w(r,t), $$

(7)

where $\tau (r, t) = S_{r\theta}(r,t)$ is the only non-zero shear stress. When the pressure gradient and the body forces in the axial direction are absent, the following equation is obtained by the balance of the linear momentum:

$$ \rho \frac{\partial w (r, t)}{\partial t}= \biggl(\frac{\partial}{\partial r}+ \frac{2}{r} \biggr) \tau (r, t), $$

(8)

where the constant density of the fluid is ρ.

In this paper, we have determined the velocity and the shear stress when the inner cylinder is fixed and the outer cylinder is moving. The initial and boundary conditions, when the inner cylinder is fixed and the outer cylinder moves gradually become

$$\begin{aligned}& w (r, 0)=0 ;\quad r \in [R_{1}, R_{2}], \end{aligned}$$

(9)

$$\begin{aligned}& w (R_{1}, t)=0 ,\quad\quad w(R_{2}, t)=Z \sin(zt). \end{aligned}$$

(10)

Also

$$ \bar{w} (R_{1}, s)=0 ,\quad\quad \bar{w}(R_{2}, s)=\frac{Zz}{z^{2}+s^{2}}. $$

(11)

Two transformations, namely the Laplace and the Hankel transformations, can be applied to the problem to solve it.

3 Calculation of the velocity field

Let us apply Laplace transformation to equation (4) to obtain the following ODE:

$$ \bigl(s+\lambda s^{\xi+1}\bigr)\bar{w}(r,s)=\nu\bigl(1+ \lambda_{r}s^{\eta}\bigr) \biggl(\frac {\partial^{2}}{\partial r^{2}}+ \frac{1}{r}\frac{\partial}{\partial r}-\frac{1}{r^{2}} \biggr)\bar{w}(r,s), $$

(12)

where ‘s’ is the parameter of the Laplace transform, or

$$ \frac{s+\lambda s^{\xi+1}}{\nu(1+\lambda_{r}s^{\eta})}\bar{w}(r,s)= \biggl(\frac{\partial^{2}}{\partial r^{2}}+ \frac{1}{r}\frac{\partial }{\partial r}-\frac{1}{r^{2}} \biggr)\bar{w}(r,s). $$

(13)

Multiplying both sides of above equation by $r B_{1}(r,r_{n})$ and integrating with respect to ‘r’ from $R_{1}$ to $R_{2}$, where $B_{1}(r,r_{n})= J_{1}(rr_{n})Y_{1}(R_{2}r_{n})-J_{1}(R_{2}r_{n})Y_{1}(rr_{n})$, and $r_{n}$ are the positive roots of the equation $B_{1}(R_{1}r_{n})=0$.

Hence our last equation becomes

$$\begin{aligned}& \frac{s+\lambda s^{\xi+1}}{\nu(1+\lambda_{r}s^{\eta})} \int _{R_{1}}^{R_{2}}r B_{1}(rr_{n}) \bar{w}(r,s)\,dr \\& \quad= \int_{R_{1}}^{R_{2}}r \biggl(\frac{\partial^{2}}{\partial r^{2}}+ \frac {1}{r}\frac{\partial}{\partial r}-\frac{1}{r^{2}} \biggr)B_{1}(rr_{n}) \bar{w}(r,s) \,dr. \end{aligned}$$

(14)

Also we define the Hankel transform of $\bar{w}(r,s)$ as

$$\bar{W}_{H}(r_{n},s)=\int_{R_{1}}^{R_{2}}r\bar{w}(r,s)B_{1}(rr_{n})\,dr. $$

Consider right hand side of the above equation (14), and solving it for simplification purposes, we get

$$ \int_{R_{1}}^{R_{2}}r \biggl(\frac{\partial^{2}}{\partial r^{2}}+ \frac {1}{r}\frac{\partial}{\partial r}-\frac{1}{r^{2}} \biggr)B_{1}(rr_{n}) \bar{w}(r,s)\,dr =\frac{2Zz}{\pi(z^{2}+s^{2})}-r_{n}^{2}\bar{W}_{H}(r_{n},s). $$

(15)

Again, from equation (14), we can deduce that

$$ \frac{s+\lambda s^{\xi+1}}{\nu(1+\lambda_{r}s^{\eta})} \bar {W}_{H}(r_{n},s)= \frac{2Zz}{\pi(z^{2}+s^{2})}-r_{n}^{2}\bar{W}_{H}(r_{n},s). $$

(16)

Again simplifying the above equation for $\bar{W}_{H}(r_{n},s)$, we get

$$ \bar{W}_{H}(r_{n},s)=\frac{2Zz}{\pi(z^{2}+s^{2})} \frac{\nu(1+\lambda _{r}s^{\eta})}{s+\lambda s^{\xi+1}+\nu r_{n}^{2}+\nu r_{n}^{2}\lambda _{r}s^{\eta}}. $$

(17)

More simplification gives us

$$ \bar{W}_{H}(r_{n},s) =\frac{2Zz}{r_{n}^{2}\pi(z^{2}+s^{2})}-\frac{2Zz (s+\lambda s^{\xi +1})}{r_{n}^{2}(\pi(z^{2}+s^{2}))(s+\lambda s^{\xi+1}+\nu r_{n}^{2}+\nu r_{n}^{2}\lambda_{r}s^{\eta})}. $$

(18)

Or equivalently, we write $\bar{W}_{H}(r_{n},s)=\bar {W}_{1H}(r_{n},s)-\bar{W}_{2H}(r_{n},s)$, where

$$ \bar{W}_{1H}(r_{n},s) =\frac{2Zz}{r_{n}^{2}\pi(z^{2}+s^{2})} $$

(19)

and

$$ \bar{W}_{2H}(r_{n},s) = \frac{2Zz (s+\lambda s^{\xi+1})}{r_{n}^{2}(\pi (z^{2}+s^{2}))(s+\lambda s^{\xi+1}+\nu r_{n}^{2}+\nu r_{n}^{2}\lambda _{r}s^{\eta})}. $$

(20)

Before we proceed, let us define the inverse Hankel transform

$$\bar{w}_{1}(r,s)=\frac{Zz}{(s^{2}+z^{2})} \frac{R_{2} (r^{2}-R_{1}^{2})}{(R_{2}^{2}-R_{1}^{2}) r} $$

and

$$\bar{w}_{2}(r,s)=\frac{\pi^{2}}{2}\sum_{n=1}^{\infty} \frac{r_{n}^{2} J_{b1}^{2}(R_{1} r_{n}) B_{1}(r r_{n})}{J_{b1}^{2}(R_{1} r_{n})-J_{b1}^{2}(R_{2} r_{n})} \bar {W}_{2H}(r_{n},s). $$

This leads us to

$$\begin{aligned} \bar{w}(r,s) =&\frac{Zz}{(z^{2}+s^{2})}\frac{R_{2} (r^{2}-R_{1}^{2})}{r (R_{2}^{2}-R_{1}^{2})}-\frac{\pi^{2}}{2}\sum _{n=1}^{\infty}\frac{r_{n}^{2} J_{b1}^{2}(R_{1} r_{n}) B_{1}(r r_{n})}{J_{b1}^{2}(R_{1} r_{n})-J_{b1}^{2}(R_{2} r_{n})} \\ &{}\times \biggl[\frac{2Zz (s+\lambda s^{\xi+1})}{r_{n}^{2}(\pi (z^{2}+s^{2}))(s+\lambda s^{\xi+1}+\nu r_{n}^{2}+\nu r_{n}^{2}\lambda _{r}s^{\eta})} \biggr] \end{aligned}$$

(21)

or equivalently

$$\begin{aligned} \bar{w}(r,s) =&\frac{Zz}{(z^{2}+s^{2})}\frac{R_{2} (r^{2}-R_{1}^{2})}{r (R_{2}^{2}-R_{1}^{2})}-\pi\sum _{n=1}^{\infty}\frac{ J_{b1}^{2}(R_{1} r_{n}) B_{1}(r r_{n})}{J_{b1}^{2}(R_{1} r_{n})-J_{b1}^{2}(R_{2} r_{n})} \\ &{}\times \biggl[\frac{Zz (s+\lambda s^{\xi+1})}{(z^{2}+s^{2})(s+\lambda s^{\xi+1}+\nu r_{n}^{2}+\nu r_{n}^{2}\lambda_{r}s^{\eta})} \biggr]. \end{aligned}$$

(22)

Equivalently,

$$ \frac{1}{s+\lambda s^{\xi+1}+\nu r_{n}^{2}+\nu r_{n}^{2}\lambda _{r}s^{\eta}}=\frac{1}{\lambda}\sum _{k=0}^{\infty}\sum_{m=0}^{k} \lambda _{r}^{m}\biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k}\frac{s^{\eta m- k-1}}{(s^{\xi}+\frac{1}{\lambda})^{k+1}}, $$

(23)

and consequently

$$\begin{aligned} \bar{w}(r,s) =&\frac{Zz}{(z^{2}+s^{2})}\frac{R_{2} (r^{2}-R_{1}^{2})}{r (R_{2}^{2}-R_{1}^{2})}-\pi\sum _{n=1}^{\infty}\frac{ J_{b1}^{2}(R_{1} r_{n}) B_{1}(r r_{n})}{J_{b1}^{2}(R_{1} r_{n})-J_{b1}^{2}(R_{2} r_{n})} \\ &{}\times \Biggl[\frac{Zz (s+\lambda s^{\xi+1})}{(z^{2}+s^{2})} \frac{1}{\lambda}\sum _{k=0}^{\infty}\sum_{m=0}^{k} \lambda_{r}^{m} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k}\frac{s^{\eta m- k-1}}{(s^{\xi}+\frac{1}{\lambda})^{k+1}} \Biggr] \end{aligned}$$

(24)

or

$$\begin{aligned} \bar{w}(r,s) =&\frac{Zz}{(z^{2}+s^{2})}\frac{R_{2} (r^{2}-R_{1}^{2})}{r (R_{2}^{2}-R_{1}^{2})}-\pi\sum _{n=1}^{\infty}\frac{ J_{b1}^{2}(R_{1} r_{n}) B_{1}(r r_{n})}{J_{b1}^{2}(R_{1} r_{n})-J_{b1}^{2}(R_{2} r_{n})} \\ &{}\times\frac{1}{\lambda}\sum_{k=0}^{\infty} \sum_{m=0}^{k}\lambda _{r}^{m} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k} \biggl[ \biggl( \frac{Zz}{(z^{2}+s^{2})} \biggr) \biggl(\frac{s^{\eta m- k}}{ (s^{\xi }+\frac{1}{\lambda} )^{k+1}} \biggr) \\ &{}+\lambda \biggl(\frac{Zz}{(z^{2}+s^{2})} \biggr) \biggl(\frac{s^{\eta m- k+\xi}}{ (s^{\xi}+\frac{1}{\lambda} )^{k+1}} \biggr) \biggr] . \end{aligned}$$

(25)

Taking the Laplace inverse using the convolution theorem and the identity

$$ G_{a, b, c}(d ,t)=L^{-1} \biggl(\frac{s^{b}}{(s^{a}-d)^{c}}, \biggr) $$

(26)

$\operatorname{Re}(ac-b)>0$, $\operatorname{Re}(s)>0$, $|\frac{d}{s^{a}}|>0$, we get the shape of the above equation as

$$\begin{aligned} w(r,t) =&\frac{R_{2} (r^{2}-R_{1}^{2})(Z\sin{zt})}{r (R_{2}^{2}-R_{1}^{2})}-\frac{Z\pi}{\lambda}\sum_{n=1}^{\infty} \frac{ J_{b1}^{2}(R_{1} r_{n}) B_{1}(r r_{n})}{J_{b1}^{2}(R_{1} r_{n})-J_{b1}^{2}(R_{2} r_{n})} \\ &{}\times\sum_{k=0}^{\infty}\sum _{m=0}^{k}\lambda_{r}^{m} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k} \biggl[ \int_{0}^{t}\sin z({t-\tau})G_{\xi,\eta m -k, k+1} \biggl( \frac{-1}{\lambda} ,\tau \biggr)\,d\tau \\ &{}+ \lambda \int_{0}^{t}\sin z({t-\tau})G_{\xi,\eta m -k+\xi, k+1} \biggl( \frac{-1}{\lambda} ,\tau \biggr)\,d\tau \biggr], \end{aligned}$$

(27)

which is the required velocity field.

3.1 Calculation of shear stress

Considering equation (3) and solving it for $\tau(r,t)$, we get

$$ \tau(r,t)=\frac{\mu(1+\lambda_{r} D^{\eta}_{t})}{(1+\lambda D^{\xi }_{t})} \biggl(\frac{\partial}{\partial r}-\frac{1}{r} \biggr) w(r,t), $$

(28)

taking the Laplace transform on both sides, we get

$$ \bar{\tau}(r,s)=\frac{\mu(1+\lambda_{r} s^{\eta})}{(1+\lambda s^{\xi })} \biggl(\frac{\partial}{\partial r}-\frac{1}{r} \biggr)\bar{w}(r,s), $$

(29)

obtaining the value of $\bar{w}(r,s)$ from equation (25) and putting it in the above equation; we need to first calculate $(\frac{\partial}{\partial r}-\frac{1}{r})\bar{w}(r,s)$,

$$\begin{aligned}& \biggl(\frac{\partial}{\partial r}-\frac{1}{r} \biggr)\bar{w}(r,s) = \frac {2Zz}{(z^{2}+s^{2})}\frac{R_{2}R_{1}^{2}}{r^{2}(R_{2}^{2}-R_{1}^{2})}+ \frac{\pi}{\lambda} \sum_{n=0}^{\infty}\frac {J_{1}^{2}(R_{1}r_{n})(2B_{1}(rr_{n})-rr_{n}B_{0}(rr_{n}))}{r(J_{1}^{2}(R_{1}r_{n})-J_{1}^{2}(R_{2}r_{n}))} \\& \hphantom{\biggl(\frac{\partial}{\partial r}-\frac{1}{r} \biggr)\bar{w}(r,s)=} {}\times\sum_{k=0}^{\infty}\sum _{m=0}^{k}\lambda_{r}^{m} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k}\biggl[ \biggl(\frac{Zz}{ (z^{2}+s^{2} )} \biggr) \biggl(\frac {s^{\eta m- k}}{ (s^{\xi}+\frac{1}{\lambda} )^{k+1}} \biggr) \\& \hphantom{\biggl(\frac{\partial}{\partial r}-\frac{1}{r} \biggr)\bar{w}(r,s)=}{} + \lambda \biggl(\frac{Zz}{(z^{2}+s^{2})} \biggr) \biggl(\frac{s^{\eta m- k+\xi}}{ (s^{\xi}+\frac{1}{\lambda} )^{k+1}} \biggr) \biggr], \\& \bar{\tau}(r,s)= \biggl[\frac{\mu(1+\lambda_{r} s^{\eta})}{(1+\lambda s^{\xi})} \biggr] \Biggl[ \frac{2Zz}{(z^{2}+s^{2})} \frac {R_{2}R_{1}^{2}}{r^{2}(R_{2}^{2}-R_{1}^{2})}+\frac{\pi}{\lambda} \sum_{n=0}^{\infty}\frac {J_{1}^{2}(R_{1}r_{n})(2B_{1}(rr_{n})-rr_{n}B_{0}(rr_{n}))}{r(J_{1}^{2}(R_{1}r_{n})-J_{1}^{2}(R_{2}r_{n}))} \\& \hphantom{ \bar{\tau}(r,s)=}{} \times\sum_{k=0}^{\infty}\sum _{m=0}^{k}\lambda_{r}^{m} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k}\biggl[ \frac{Zz}{(z^{2}+s^{2})} \biggl(\frac{s^{\eta m- k}}{ (s^{\xi}+\frac{1}{\lambda} )^{k+1}} \biggr) \\& \hphantom{ \bar{\tau}(r,s)=}{} +\lambda \biggl( \frac {Zz}{(z^{2}+s^{2})} \biggr) \biggl(\frac{s^{\eta m- k+\xi}}{ (s^{\xi }+\frac{1}{\lambda} )^{k+1}} \biggr) \biggr] \Biggr], \end{aligned}$$

or

$$\begin{aligned} \bar{\tau}(r,s) =& \biggl[\frac{\mu}{(1+\lambda s^{\xi})} \biggr] \Biggl[ \frac{2Zz}{(z^{2}+s^{2})} \frac{R_{2}R_{1}^{2}}{r^{2}(R_{2}^{2}-R_{1}^{2})}+\frac{\pi}{\lambda} \sum_{n=0}^{\infty}\frac {J_{1}^{2}(R_{1}r_{n})(2B_{1}(rr_{n})-rr_{n}B_{0}(rr_{n}))}{r(J_{1}^{2}(R_{1}r_{n})-J_{1}^{2}(R_{2}r_{n}))} \\ &{} \times\sum_{k=0}^{\infty}\sum _{m=0}^{k}\lambda_{r}^{m} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k}\biggl[ \frac{Zz}{(z^{2}+s^{2})} \biggl(\frac{s^{\eta m- k}}{ (s^{\xi }+\frac{1}{\lambda} )^{k+1}} \biggr) \\ & {}+\lambda \biggl( \frac {Zz}{(z^{2}+s^{2})} \biggr) \biggl(\frac{s^{\eta m- k+\xi}}{(s^{\xi}+\frac {1}{\lambda})^{k+1}} \biggr) \biggr] \Biggr] \\ & {}+ \biggl[\frac{\mu\lambda_{r} s^{\eta}}{(1+\lambda s^{\xi})} \biggr] \Biggl[ \frac{2Zz}{(z^{2}+s^{2})} \frac{R_{2}R_{1}^{2}}{r^{2}(R_{2}^{2}-R_{1}^{2})}+\frac{\pi}{\lambda} \sum_{n=0}^{\infty}\frac {J_{1}^{2}(R_{1}r_{n})(2B_{1}(rr_{n})-rr_{n}B_{0}(rr_{n}))}{r(J_{1}^{2}(R_{1}r_{n})-J_{1}^{2}(R_{2}r_{n}))} \\ &{}\times\sum_{k=0}^{\infty}\sum _{m=0}^{k}\lambda_{r}^{m} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k}\biggl[ \frac{Zz}{(z^{2}+s^{2})} \biggl(\frac{s^{\eta m- k}}{ (s^{\xi}+\frac{1}{\lambda} )^{k+1}} \biggr) \\ &{} +\lambda \biggl( \frac {Zz}{(z^{2}+s^{2})} \biggr) \biggl(\frac{s^{\eta m- k+\xi}}{ (s^{\xi }+\frac{1}{\lambda} )^{k+1}} \biggr) \biggr] \Biggr]. \end{aligned}$$

Equivalently

$$\begin{aligned} \bar{\tau}(r,s) = &\biggl[\frac{\mu}{\lambda} \biggr] \Biggl[ \frac {2Zz}{(z^{2}+s^{2})} \frac{R_{2}R_{1}^{2}}{r^{2}(R_{2}^{2}-R_{1}^{2})} \frac{s^{0}}{s^{\xi}+\frac{1}{\lambda}}+ \frac{\pi}{\lambda} \sum_{n=0}^{\infty}\frac {J_{1}^{2}(R_{1}r_{n})(2B_{1}(rr_{n})-rr_{n}B_{0}(rr_{n}))}{r(J_{1}^{2}(R_{1}r_{n})-J_{1}^{2}(R_{2}r_{n}))}\\ &{}\times\sum _{k=0}^{\infty}\sum_{m=0}^{k} \lambda_{r}^{m} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k}\biggl[ \frac{Zz}{(z^{2}+s^{2})} \biggl(\frac{s^{\eta m- k}}{(s^{\xi }+\frac{1}{\lambda})^{k+1}} \biggr) \\ &{} +\lambda \biggl( \frac {Zz}{(z^{2}+s^{2})} \biggr) \biggl(\frac{s^{\eta m- k+\xi}}{(s^{\xi}+\frac {1}{\lambda})^{k+1}} \biggr) \biggr] \Biggr] \\ &{}+ \biggl[\frac{\mu\lambda_{r}}{\lambda} \biggr] \Biggl[ \frac {2Zz}{(z^{2}+s^{2})} \frac{R_{2}R_{1}^{2}}{r^{2}(R_{2}^{2}-R_{1}^{2})} + \frac{\pi}{\lambda} \sum_{n=0}^{\infty}\frac {J_{1}^{2}(R_{1}r_{n})(2B_{1}(rr_{n})-rr_{n}B_{0}(rr_{n}))}{r(J_{1}^{2}(R_{1}r_{n})-J_{1}^{2}(R_{2}r_{n}))} \\ &{}\times\sum_{k=0}^{\infty}\sum _{m=0}^{k}\lambda_{r}^{m} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k} \biggl[ \frac{Zz}{(z^{2}+s^{2})} \biggl(\frac{s^{\eta m- k+\eta}}{(s^{\xi }+\frac{1}{\lambda})^{k+2}} \biggr) \\ &{}+\lambda \biggl(\frac{Zz}{(z^{2}+s^{2})} \biggr) \biggl(\frac{s^{\eta m- k+\xi+\eta}}{(s^{\xi}+\frac{1}{\lambda})^{k+2}} \biggr) \biggr] \Biggr] . \end{aligned}$$

Taking the Laplace inverse, using the convolution theorem, and the following identity:

$$ G_{a, b, c}(d ,t)=L^{-1} \biggl(\frac{s^{b}}{(s^{a}-d)^{c}} \biggr), $$

(30)

$\operatorname{Re}(ac-b)>0$, $\operatorname{Re}(s)>0$, $| \frac{d}{s^{a}}|>0$,

$$\begin{aligned} \tau(r,t) =& \biggl[\frac{\mu}{\lambda} \biggr] \Biggl[ \frac{2Z R_{2}R_{1}^{2}}{r^{2}(R_{2}^{2}-R_{1}^{2})} \int_{0}^{t}\sin z({t-\tau})G_{0,\xi,1} \biggl( \frac{-1}{\lambda} ,\tau \biggr)\,d\tau \\ &{}+\frac{Z\pi}{\lambda} \sum_{n=0}^{\infty}\frac {J_{1}^{2}(R_{1}r_{n})(2B_{1}(rr_{n})-rr_{n}B_{0}(rr_{n}))}{r(J_{1}^{2}(R_{1}r_{n})-J_{1}^{2}(R_{2}r_{n}))} \\ &{}\times\sum_{k=0}^{\infty}\sum _{m=0}^{k}\lambda_{r}^{m} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k} \biggl[ \int_{0}^{t}\sin z({t-\tau})G_{\eta m - k,\xi, k+2} \biggl( \frac {-1}{\lambda} ,\tau \biggr)\,d\tau \\ &{} + \lambda \int_{0}^{t}\sin z({t-\tau })G_{\eta m -k+\xi,\xi, k+2} \biggl(\frac{-1}{\lambda} , \tau \biggr)\,d\tau \biggr] \Biggr]\\ &{}+ \biggl[\frac{\mu\lambda_{r}}{\lambda} \biggr] \Biggl[ \frac{2Z R_{2}R_{1}^{2}}{r^{2}(R_{2}^{2}-R_{1}^{2})} \int_{0}^{t}\sin z({t-\tau})G_{\eta,\xi,1} \biggl(\frac{-1}{\lambda} ,\tau \biggr)\,d\tau \\ &{}+ \frac{Z\pi }{\lambda} \sum_{n=0}^{\infty}\frac{J_{1}^{2}(R_{1}r_{n})(2B_{1}(rr_{n})-rr_{n} B_{0}(rr_{n}))}{r(J_{1}^{2}(R_{1}r_{n})-J_{1}^{2}(R_{2}r_{n}))} \\ &{}\times\sum_{k=0}^{\infty}\sum _{m=0}^{k}\lambda_{r}^{m} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k} \biggl[ \int_{0}^{t}\sin z({t-\tau})G_{\eta m - k+\eta,\xi, k+2} \biggl( \frac{-1}{\lambda} ,\tau \biggr)\,d\tau \\ &{} + \lambda \int_{0}^{t}\sin z({t-\tau})G_{\eta m -k+\xi+\eta,\xi , k+2} \biggl( \frac{-1}{\lambda} ,\tau \biggr)\,d\tau \biggr] \Biggr]. \end{aligned}$$

4 Particularization of the above results

The above results are of a general nature and the imposition of certain limits/conditions may bring these to particular fluids.

4.1 Ordinary Oldroyd-B fluid

The velocity field and shear stress of the movement of an ordinary Oldroyd-B fluid can be deduced imposing $\xi, \eta\rightarrow1$ on the obtained results:

$$\begin{aligned} w(r,t) =&\frac{R_{2} (r^{2}-R_{1}^{2})(Z\sin{zt})}{r (R_{2}^{2}-R_{1}^{2})}\\ &{}-\frac{Z\pi}{\lambda}\sum_{n=1}^{\infty} \frac{ J_{b1}^{2}(R_{1} r_{n}) B_{1}(r r_{n})}{J_{b1}^{2}(R_{1} r_{n})-J_{b1}^{2}(R_{2} r_{n})} \\ &{}\times\sum_{k=0}^{\infty}\sum _{m=0}^{k}\lambda_{r}^{m} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k} \biggl[ \int_{0}^{t}\sin z({t-\tau })G_{1,m -k, k+1} \biggl(\frac{-1}{\lambda} ,\tau \biggr)\,d\tau \\ &{}+ \lambda \int_{0}^{t}\sin z({t-\tau})G_{1,m -k+1, k+1} \biggl( \frac {-1}{\lambda} ,\tau \biggr)\,d\tau \biggr], \end{aligned}$$

and the associated shear stress will take the form of

$$\begin{aligned} \tau(r,t) =& \biggl[\frac{\mu}{\lambda} \biggr] \Biggl[ \frac{2Z R_{2}R_{1}^{2}}{r^{2}(R_{2}^{2}-R_{1}^{2})} \int_{0}^{t}\sin z({t-\tau})G_{0,1,1} \biggl( \frac{-1}{\lambda} ,\tau \biggr)\,d\tau \\ &{}+ \frac{Z\pi}{\lambda} \sum_{n=0}^{\infty}\frac {J_{1}^{2}(R_{1}r_{n})(2B_{1}(rr_{n})-rr_{n}B_{0}(rr_{n}))}{r(J_{1}^{2}(R_{1}r_{n})-J_{1}^{2}(R_{2}r_{n}))} \\ &{}\times\sum_{k=0}^{\infty}\sum _{m=0}^{k}\lambda_{r}^{m} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k} \biggl[ \int_{0}^{t}\sin z({t-\tau})G_{m - k,1, k+2} \biggl( \frac {-1}{\lambda} ,\tau \biggr)\,d\tau \\ &{}+ \lambda \int_{0}^{t}\sin z({t-\tau }) G_{m -k+1,1, k+2} \biggl(\frac{-1}{\lambda} ,\tau \biggr)\,d\tau \biggr] \Biggr]\\ &{}+ \biggl[\frac{\mu\lambda_{r}}{\lambda} \biggr] \Biggl[ \frac{2Z R_{2}R_{1}^{2}}{r^{2}(R_{2}^{2}-R_{1}^{2})} \int_{0}^{t}\sin z({t-\tau})G_{1,1,1} \biggl(\frac{-1}{\lambda} ,\tau \biggr)\,d\tau \\ &{}+ \frac{Z\pi }{\lambda} \sum_{n=0}^{\infty}\frac {J_{1}^{2}(R_{1}r_{n})(2B_{1}(rr_{n})-rr_{n}B_{0}(rr_{n}))}{r(J_{1}^{2}(R_{1}r_{n})-J_{1}^{2}(R_{2}r_{n}))} \\ &{}\times\sum_{k=0}^{\infty}\sum _{m=0}^{k}\lambda_{r}^{m} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k} \biggl[ \int_{0}^{t}\sin z({t-\tau})G_{m - k+1,1, k+2} \biggl( \frac{-1}{\lambda} ,\tau \biggr)\,d\tau \\ &{}+\lambda \int_{0}^{t}\sin z({t-\tau})G_{m -k+2,1, k+2} \biggl( \frac {-1}{\lambda} ,\tau \biggr)\,d\tau \biggr] \Biggr] . \end{aligned}$$

4.2 Ordinary Maxwell fluid

If $\xi\rightarrow1$, $\lambda_{r} \rightarrow0$ in the already found results for the velocity and shear stress then the resultants will govern the movement of an ordinary Maxwell fluid under the same circumstances. We have

$$\begin{aligned} w(r,t) =&\frac{R_{2} (r^{2}-R_{1}^{2})(Z\sin{zt})}{r (R_{2}^{2}-R_{1}^{2})}\\ &{}-\frac{Z\pi}{\lambda}\sum_{n=1}^{\infty} \frac{ J_{b1}^{2}(R_{1} r_{n}) B_{1}(r r_{n})}{J_{b1}^{2}(R_{1} r_{n})-J_{b1}^{2}(R_{2} r_{n})} \\ &{}\times\sum_{k=0}^{\infty} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k}\biggl[ \int_{0}^{t}\sin z({t-\tau})G_{1,-k, k+1} \biggl( \frac {-1}{\lambda} ,\tau \biggr)\,d\tau \\ &{}+\lambda \int_{0}^{t}\sin z({t-\tau})G_{1,-k+1, k+1} \biggl( \frac {-1}{\lambda} ,\tau \biggr)\,d\tau \biggr] \end{aligned}$$

and

$$\begin{aligned} \tau(r,t) =& \biggl[\frac{\mu}{\lambda} \biggr] \Biggl[ \frac{2Z R_{2}R_{1}^{2}}{r^{2}(R_{2}^{2}-R_{1}^{2})} \int_{0}^{t}\sin z({t-\tau})G_{0,1,1} \biggl( \frac{-1}{\lambda} ,\tau \biggr)\,d\tau \\ &{} + \frac{Z\pi}{\lambda} \sum_{n=0}^{\infty}\frac {J_{1}^{2}(R_{1}r_{n})(2B_{1}(rr_{n})-rr_{n}B_{0}(rr_{n}))}{r(J_{1}^{2}(R_{1}r_{n})-J_{1}^{2}(R_{2}r_{n}))} \\ &{}\times\sum_{k=0}^{\infty} \biggl( \frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k} \biggl[ \int_{0}^{t}\sin z({t-\tau})G_{- k,1, k+2} \biggl( \frac{-1}{\lambda } ,\tau \biggr)\,d\tau \\ &{}+ \lambda \int_{0}^{t}\sin z({t-\tau })G_{-k+1,1, k+2} \biggl(\frac{-1}{\lambda} ,\tau \biggr)\,d\tau \biggr] \Biggr]\\ &{} + \frac{Z\pi}{\lambda} \sum_{n=0}^{\infty}\frac {J_{1}^{2}(R_{1}r_{n})(2B_{1}(rr_{n})-rr_{n}B_{0}(rr_{n}))}{r(J_{1}^{2}(R_{1}r_{n})-J_{1}^{2}(R_{2}r_{n}))} \\ &{}\times\sum_{k=0}^{\infty} \biggl(\frac{-\nu r_{n}^{2}}{\lambda} \biggr)^{k} \biggl[ \int_{0}^{t}\sin z({t-\tau})G_{- k+\eta,1, k+2} \biggl( \frac {-1}{\lambda} ,\tau \biggr)\,d\tau \\ &{} +\lambda \int_{0}^{t}\sin z({t-\tau})G_{-k+1+\eta,1, k+2} \biggl( \frac {-1}{\lambda} ,\tau \biggr)\,d\tau \biggr] . \end{aligned}$$

5 Conclusion

Our above endeavors were to develop a formula for the calculation of exact solutions for the velocity field and the shear stress of the motion (flow) of an Oldroyd-B fluid present between two rotationally oscillating cylinders of infinite lengths. The use of fractional derivatives and the commonly known transformations, i.e. the Laplace and the Hankel transformations, has made the approach more accessible. The central notion depicts the phenomenon that a viscoelastic (Oldroyd-B) fluid will react under certain conditions and that can we control such flow. At first stage the inner cylinder was supposed to be at rest, i.e. fixed, whereas the movement was produced by the outer cylinder. At the second stage, we analyzed the flow of the fluid produced by the movement of the inner cylinder while considering the outer cylinder at rest or fixed. The obtained solutions satisfy the governing equations and all imposed initial and boundary conditions. The solutions, obtained by means of Laplace and Hankel transforms, are presented in integral and series forms in terms of the generalized G-function. In the end these general solutions have been particularized for ‘ordinary Oldroyd-B fluids’ and for ‘ordinary Maxwell fluids’.

Appendix

The following are some expressions used in the text:

• (A1) The finite Hankel transform of the function

  $$a(r)=\frac{C_{1} R_{1}(R_{2}^{2}-r^{2})+ C_{2} R_{2}(r^{2}-R_{1}^{2})}{(R_{2}^{2}-R_{1}^{2}) r} $$

  satisfying $a(R_{1})=C_{1}$ and $a(R_{2})=C_{2}$ is

  $$a_{n}(r)= \int_{R_{1}}^{R_{2}}r a(r) B_{1}(rr_{n}) \,dr=\frac{2C_{2}}{\pi r_{n}^{2}}-\frac{2C_{1}}{\pi r_{n}^{2}}\frac{J_{1}(R_{2} r_{n})}{J_{1}(R_{1} r_{n})}. $$

• (A2) If $f(t)=L^{-1}\{\overline{f}(q) \}$ and $g(t)=L^{-1}\{\overline{g}(q) \}$, then

  $$\begin{aligned} L^{-1}\bigl\{ \overline{f}(q)\overline{g}(q) \bigr\} =& (f*g) (t) \\ =& \int_{0}^{t} f(t-\tau)g(\tau) \,d\tau \\ =& \int_{0}^{t} f(t)g(t-\tau) \,d\tau. \end{aligned}$$
• (A3)
  $$\sum_{k=0}^{\infty}\bigl(-\nu r_{n}^{2}\bigr)^{k} G_{0,-1-k,k+1}\bigl(-\alpha r_{n}^{2}, t\bigr)=\frac{1}{1+\alpha r_{n}^{2}}\exp \biggl( \frac{-\nu r_{n}^{2} t}{1+\alpha r_{n}^{2}} \biggr). $$