Influence of Magnetic Field and Porous Medium on Taylor–Couette Flows of Second Grade Fluids Due to Time-Dependent Couples on a Circular Cylinder

Abstract

Axially symmetric Taylor–Couette flows of incompressible second grade fluids induced by time-dependent couples inside an infinite circular cylinder are studied under the action of an external magnetic field. The influence of the medium porosity is taken into account in the mathematical modeling. Analytical expressions for the dimensionless non-trivial shear stress and the corresponding fluid velocity were determined using the finite Hankel and Laplace transforms. The solutions obtained are new in the specialized literature and can be customized for various problems of interest in engineering practice. For illustration, the cases of oscillating and constant couples have been considered, and the steady state components of the shear stresses were presented in equivalent forms. Numerical schemes based on finite differences have been formulated for determining the numerical solutions of the proposed problem. It was shown that the numerical results based on analytical solutions and those obtained with the numerical methods have close values with very good accuracy. It was also proved that the fluid flows more slowly and the steady state is reached earlier in the presence of a magnetic field or porous medium.

1. Introduction

There are real fluids and also fluid-like materials used in modern nanotechnologies which are characterized by the Cauchy stress tensor of the form $\tilde{T}=-p\tilde{I}+\tilde{F}({\tilde{A}}_1, {\tilde{A}}_2,\dots ,{\tilde{A}}_N),$ where p is the hydrostatic pressure, $\tilde{I}$ is the identity tensor, $\tilde{F}$ is the response function, and ${\tilde{A}}_k, k=1,2,\dots N$ are the first N Rivlin-Ericksen tensors defined as ${\tilde{A}}_1=\nabla \mathit{V}+{(\nabla \mathit{V})}^T$, ${\tilde{A}}_k={\displaystyle \frac{d{\tilde{A}}_{k-1}}{dt}}+{\tilde{A}}_{k-1}\nabla \mathit{V}+{(\nabla \mathit{V})}^T{\tilde{A}}_{k-1}, k=2,3,\dots ,N$, $\mathit{V}$ being the velocity vector, $\nabla \mathit{V}$ the gradient of velocity vector, ${(\nabla \mathit{V})}^T$ the transpose of the velocity gradient, and ${\displaystyle \frac{d{\tilde{A}}_{k-1}}{dt}}={\displaystyle \frac{\partial {\tilde{A}}_{k-1}}{\partial t}}+(\mathit{V}\cdot \nabla ){\tilde{A}}_{k-1}$ is the material time derivative [1,2].

If the response function $\tilde{F}$ is a polynomial of degree N, then the fluid has been named fluid of grade N. Fluids investigated in the present paper are incompressible fluids of the second grade, with the Cauchy stress tensor $\tilde{T}=-p\tilde{I}+\mu {\tilde{A}}_1+{\alpha }_1 {\tilde{A}}_2+{\alpha }_2{\tilde{A}}_1^2,$ where $\mu >0$ is the dynamic viscosity, and ${\alpha }_1, {\alpha }_2$ are the normal stress module [3,4]. It is known that many dilute polymer solutions belong to the class of second grade fluids [3]. Second grade fluids are used in many industrial and pharmaceutical applications, in polymer processing, and in cosmetic products. For these reasons, many researchers have been concerned with studying the behavior of these fluids in various geometric domains, under different conditions on the boundary of the flow domain. Kanuri et al. [5] carried out an analytical study of the Poiseuille flow of second grade fluids through a cylindrical pipe. Mainly, they have investigated the effects of the viscosity and second-grade fluid parameters on the flow velocity and flow rate. Using similarity transformations and the homotopy analysis method, Nadeem et al. [6] studied some flows of incompressible second grade fluids along a horizontal cylinder. Their results could have applications in the coating of wires and polymer fiber spinning. Relevant excellent contributions to the determination of analytical or semi-analytical solutions of some nonlinear problems in (oceanic) fluid dynamics can be found in the works [7,8,9,10]. Also, the use of modern mathematical tools such as the inverse scattering transform in solving nonlinear dynamics problems is excellently presented in the articles [11,12].

Ozer and Suhubi [13] studied flows of a second grade fluid in a cylindrical tube using the perturbation method in the case when the second order material coefficient is sufficiently small or the characteristic diameter is very large. They investigated flows in a tube of circular cross-section and also applied it to a tube of arbitrary cross-section by using a conformal transformation. The analysis of the effects of a magnetic field on two-dimensional flows of steady third-grade fluids over a stretched circular cylinder was carried out by Ahmed [14]. Numerical solutions of the governing equations have been determined using the Keller–Box method. Uddin et al. [15] studied the effect of a magnetic field on the non-isothermal flows of a second grade fluid in an oscillating vertical cylinder using a mathematical model with a time-fractional Caputo-Fabrizio derivative. The exact solutions of the governing equations were obtained by means of Laplace and finite Hankel transforms. The comparison between the fractional model and the model with derivative of integer order highlights the influence of stress memory on fluid behavior. Hayat et al. [16] and Bano et al. [17] investigated some unsteady flows of second grade fluids, and exact analytic solutions of the governing partial differential equations have been obtained. The authors have recovered the known solutions for a Navier–Stokes fluid in the hydrodynamic case as the limiting cases of their results. Using the generalized method of separation variables and a certain form of the stream function, Siddiqui et al. [18] have determined a class of exact solutions of the equations of transient motion for second grade fluids. Jamil and Zafarullah [19] derived exact solutions for some magnetohydrodynamic (MHD) unsteady flows of such fluids through a porous medium between infinite coaxial circular cylinders. Inertial flows of magneto-hydrodynamic second-grade fluids in a ciliated channel were studied by Maqbool et al. [20] using the homotopy perturbation method. Two-dimensional flows have been considered under the effect of inertial forces, external magnetic field, and Darcy’s resistance.

Petrova et al. [21] focused on the study of the qualitative properties of solutions of governing equations of the flow of second grade fluids. By studying the layered flows and a model problem with a free boundary, the authors have constructed an analog of Karman’s solution, which describes the stationary motion of a second-grade fluid in a half-space induced by the rotation of the plane bounding. Also, the authors proposed a generalization of Steklov’s solution of the problem on unsteady helical flows of a Newtonian fluid to the case of a second-grade fluid. Various aspects of obtaining analytical and semi-analytical approaches in solving equations of hydrodynamics can be found in recent works of Ershkov and Shamin [22] and Baranovskii et al. [23]. Analytic solutions for MHD unsteady motions of incompressible second grade fluids through a porous medium in an infinite circular cylinder were recently established by Fetecau et al. [24]. Fetecau and Vieru [25] have studied the influence of the magnetic field and porous medium on the flow of a second grade fluid in cylindrical domains with time-dependent velocity on the boundary. Exact analytical solutions have been determined using the integral transforms method. Velocity and shear stress fields for some particular flows have been found from the general solutions. However, to the best of our knowledge, exact solutions for MHD motions of such fluids through a porous medium in cylindrical domains are lacking in the existing literature when shear stress is prescribed on the boundary.

This article presents an analytical and numerical study of the axially symmetric Taylor–Couette flows of second grade fluids inside an infinite right circular cylinder on whose surface a time-dependent torque acts. The influence of an external magnetic field and the medium’s porosity is taken into account in the mathematical modeling. Using the finite Hankel and Laplace transforms, exact analytical solutions for the shear stress and the fluid velocity in the case of the torque described by an arbitrary function of time are determined. The solutions obtained are new in the specialized literature and can be customized for various problems of interest in engineering practice. Steady-state solutions have been determined for the case of oscillating torques and constant torque. Also, numerical schemes based on finite differences have been formulated for determining the numerical solutions of the proposed problem. It is shown that the numerical results based on the analytical solutions, and those obtained with the numerical methods have close values with very good accuracy. The influence of the magnetic field and porous medium on the fluid behavior is graphically analyzed and discussed.

2. Problem Presentation

Consider an electrically conducting incompressible second grade fluid (ECISGF) in a stationary state in a porous medium in an infinite circular cylinder of radius R (Figure 1).

After the initial moment $t=0$, the cylinder begins to rotate around its symmetry axis due to a time-dependent torque per unit length $2\pi RSf(t)$. The function $f(\cdot )$ is piecewise continuous and $f(0)=0$ while S is a constant shear stress. In the same moment, a circular magnetic field of constant strength B begins to act in the azimuthally direction [26]. Owing to the shear, the fluid begins to move, and we are looking for a velocity field whose velocity vector [27]

$$\mathit{V}(r,t)=\left(0,u(r,t),0\right) ; 0<r<R, t>0,$$

(1)

in a convenient system of cylindrical coordinate r, $\theta$ and z. Introducing $\mathit{V}(r,t)$ from Equation (1) in the constitutive equation of the incompressible second grade fluids, one finds that the non-trivial shear stress $\eta (r,t)$ is given by the next relation

$$\eta (r,t)=\left(\mu +{\alpha }_1{\displaystyle \frac{\partial }{\partial t}}\right)\left({\displaystyle \frac{\partial }{\partial r}}-{\displaystyle \frac{1}{r}}\right)u(r,t); 0<r<R, t>0.$$

(2)

We assume that the fluid is finitely conducting, its magnetic permeability is constant, the induced magnetic field owing to the applied magnetic field can be neglected, and there is no surplus electric charge distribution inside. In these conditions, the balance of linear momentum for such motions of ECISGFs reduces to the relation [19]

$$\rho {\displaystyle \frac{\partial u(r,t)}{\partial t}}=\left({\displaystyle \frac{\partial \eta (r,t)}{\partial r}}+{\displaystyle \frac{2\eta (r,t)}{r}}\right)-\sigma B^{2}u(r,t)-{\displaystyle \frac{\varphi }{k}}\left(\mu +{\alpha }_1{\displaystyle \frac{\partial }{\partial t}}\right)u(r,t).$$

(3)

Here $\rho$ is the fluid density, $\sigma$ is the electrical conductivity while $\varphi$ and k are the porosity and the permeability of the porous medium, respectively. The two unknown functions $u(r,t)$ and $\eta (r,t)$ have to satisfy the initial and boundary conditions

$$u(r,0)=0, \eta (r,0)=0; 0\le r\le R,$$

(4)

$$\eta (R,t)=\left(\mu +{\alpha }_1{\displaystyle \frac{\partial }{\partial t}}\right)\left({\displaystyle \frac{\partial }{\partial r}}-{\displaystyle \frac{1}{r}}\right){\left.u(r,t)\right|}_{r=R}=Sf(t); t>0.$$

(5)

Introducing the next dimensionless variables, functions, and parameters

$$r^{\ast }={\displaystyle \frac{1}{R}}r, t^{\ast }={\displaystyle \frac{\nu }{R^2}}t, u^{\ast }={\displaystyle \frac{\mu }{RS}}u, {\eta }^{\ast }={\displaystyle \frac{1}{S}}\eta , f^{\ast }(t^{\ast })=f\left({\displaystyle \frac{R^2}{\nu }}{t}^{\ast }\right), \alpha ={\displaystyle \frac{1}{\rho R^2}}{\alpha }_1$$

(6)

and renouncing the star notation, one finds the following non-dimensionless forms

$$\eta (r,t)=\left(1+\alpha {\displaystyle \frac{\partial }{\partial t}}\right)\left({\displaystyle \frac{\partial }{\partial r}}-{\displaystyle \frac{1}{r}}\right)u(r,t); 0<r<1, t>0,$$

(7)

$${\displaystyle \frac{\partial u(r,t)}{\partial t}}=\left({\displaystyle \frac{\partial }{\partial r}}+{\displaystyle \frac{2}{r}}\right)\eta (r,t)-Mu(r,t)-K\left(1+\alpha {\displaystyle \frac{\partial }{\partial t}}\right)u(r,t); 0<r<1, t>0,$$

(8)

for the governing Equations (2) and (3). In relations (6) and (8) $\nu =\mu /\rho$ is the kinematic viscosity of the fluid, while the non-dimensional constants

$$M={\displaystyle \frac{R^2}{\mu }}\sigma B^2 \mathrm{and} K={\displaystyle \frac{\varphi }{k}}{R}^2,$$

(9)

are magnetic and porous parameters, respectively. Let us specify that the magnetic parameter M can be written as the square of the Hartmann number, that is $M=H{a}^2, Ha=BR\sqrt{\sigma /\mu }$ being the Hartmann number. This dimensionless number plays an important role in the study of the motion of fluids driven by the Lorentz force. The corresponding initial and boundary conditions are

$$u(r,0)=0, \eta (r,0)=0; 0\le r\le 1,$$

(10)

$$\eta (1,t)=\left(1+\alpha {\displaystyle \frac{\partial }{\partial t}}\right)\left({\displaystyle \frac{\partial }{\partial r}}-{\displaystyle \frac{1}{r}}\right){\left.u(r,t)\right|}_{r=1}=f(t); t>0.$$

(11)

3. Solutions

In the following, the system of partial differential Equations (7) and (8) with the initial and boundary conditions (10) and (11) will be solved using integral transforms.

3.1. Calculation of the Dimensionless Shear Stress $\eta (r,t)$

The shear stress $\eta (r,t)$ corresponding to this motion can be determined in a simple way using the finite Hankel transform only. To do that, by eliminating the velocity $u(r,t)$ between Equations (7) and (8), one finds the following governing equation

$$\begin{array}{l}{\displaystyle \frac{\partial \eta (r,t)}{\partial t}}=\left(1+\alpha {\displaystyle \frac{\partial }{\partial t}}\right)\left({\displaystyle \frac{{\partial }^2}{\partial r^2}}+{\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial }{\partial r}}-{\displaystyle \frac{4}{r^2}}\right)\eta (r,t)-M\eta (r,t)\\ -K\left(1+\alpha {\displaystyle \frac{\partial }{\partial t}}\right)\eta (r,t); 0<r<1, t>0,\end{array}$$

(12)

for the shear stress $\eta (r,t)$. The corresponding initial and boundary conditions are

$$\eta (r,0)=0, 0\le r\le 1; \eta (1,t)=f(t), t>0.$$

(13)

By multiplying Equation (12) by $r{J}_2(r{r}_n)$, integrating the obtained result with respect to r between zero and one and bearing in mind the conditions (13) and the identity (see for instance the reference [28], Section 14, the equality (59))

$${\displaystyle \underset{0}{\overset{1}{\int }}r{J}_2(r{r}_n)\left(\frac{{\partial }^2}{\partial r^2}+\frac{1}{r}\frac{\partial }{\partial r}-\frac{4}{r^2}\right)\eta (r,t)}dr=-r_n^2{\eta }_H(n,t)-r_n\eta (1,t)J_1(r_n),$$

(14)

one attains the next initial value problem

$${\displaystyle \frac{\partial {\eta }_H(n,t)}{\partial t}}+{\displaystyle \frac{r_n^2+K_{eff}}{1+\alpha (r_n^2+K)}}{\eta }_H(n,t)=-{\displaystyle \frac{r_n{J}_1(r_n)}{1+\alpha (r_n^2+K)}}\left[f(t)+\alpha f^{\prime }(t)\right] ; {\eta }_H(n,0)=0.$$

(15)

In the above relations $J_1(\cdot )$ and $J_2(\cdot )$ are standard Bessel functions of the first kind of one and second order [29], ${\eta }_H(n,t)$ is the finite Hankel transform of $\eta (r,t)$, $r_n$ are the positive roots of the transcendental equation $J_2(r)=0$ and the entity $K_{eff}=M+K$ is called the effective permeability. The roots $r_n, n=1, 2,\dots$ have been determined using the MathCAD15 software. Firstly, using the approximation $J_2(z)\cong \sqrt{{\displaystyle \frac{2}{\pi z}}}\cos \left(z-{\displaystyle \frac{5\pi }{4}}\right)$ we have established the intervals that separate these roots $r_n,$ namely $r_n\in \left({\displaystyle \frac{(4n-1)\pi }{4}},{\displaystyle \frac{(4n+3)\pi }{4}}\right), n=1, 2,\dots$. The subroutine $root\left(J_2(z), z,\left({\displaystyle \frac{(4n-1)\pi }{4}},{\displaystyle \frac{(4n+3)\pi }{4}}\right)\right)$ computes the values of the roots $r_n$, $n\le N_0, N_0>1$, being a fixed natural number.

The solution of the initial value problem (15) can be written in the form

$$\begin{array}{l}{\eta }_H(n,t)=-{\displaystyle \frac{r_n{J}_1(r_n)}{1+\alpha (r_n^2+K)}}\alpha f(t)\\ +(\alpha M-1){\displaystyle \frac{r_n{J}_1(r_n)}{{[1+\alpha (r_n^2+K)]}^2}}{\displaystyle \underset{0}{\overset{t}{\int }}f(t-s)}\exp \left[-{\displaystyle \frac{r_n^2+K_{eff}}{1+\alpha (r_n^2+K)}}s\right]ds.\end{array}$$

(16)

By applying the inverse finite Hankel transform to Equation (16) and using the first entry of Table X from Appendix C of the reference [28], one finds the following suitable form

$$\begin{array}{l}\eta (r,t)=f(t)r^2+2(\alpha K+1)f(t){\displaystyle \sum _{n=1}^{\infty }\frac{J_2(r{r}_n)}{r_n{J}_1(r_n)[1+\alpha (r_n^2+K)]}}\\ +2(\alpha M-1){\displaystyle \sum _{n=1}^{\infty }\frac{r_n{J}_2(r{r}_n)}{J_1(r_n){[1+\alpha (r_n^2+K)]}^2}}{\displaystyle \underset{0}{\overset{t}{\int }}f(t-s)}\exp \left[-{\displaystyle \frac{r_n^2+K_{eff}}{1+\alpha (r_n^2+K)}}s\right]ds,\end{array}$$

(17)

for the dimensionless shear stress $\eta (r,t)$. It satisfies the initial and boundary conditions (10) and (11). Making $\alpha =0$ in Equation (17), one finds the dimensionless shear stress

$${\eta }_N(r,t)=f(t)r^2+2f(t){\displaystyle \sum _{n=1}^{\infty }\frac{J_2(r{r}_n)}{r_n{J}_1(r_n)}}-2{\displaystyle \sum _{n=1}^{\infty }\frac{r_n{J}_2(r{r}_n)}{J_1(r_n)}}{\displaystyle \underset{0}{\overset{t}{\int }}f(t-s)} e^{-(r_n^2+K_{eff})s}ds,$$

(18)

corresponding to electrically conducting incompressible Newtonian fluids performing the same motion. From the relation (18), which is also new in the literature, it results that the shear stress ${\eta }_N(r,t)$ for such motions of the electrical conducting incompressible Newtonian fluids through a porous medium in an infinite circular cylinder does not depend on the parameters M and K independently, but only by means of the effective permeability $K_{eff}=M+K$. Consequently, a two-parameter approach in such situations is superfluous.

3.2. Calculation of the Dimensionless Velocity $u(r,t)$

By applying the Laplace transform to Equations (7) and (8) and bearing in mind the initial conditions (10), one finds the relations

$$\widehat{\eta }(r,s)=(\alpha s+1)\left({\displaystyle \frac{\partial }{\partial r}}-{\displaystyle \frac{1}{r}}\right)\widehat{u}(r,s); 0<r<1,$$

(19)

$$\widehat{u}(r,s)={\displaystyle \frac{1}{(\alpha K+1)s+K_{eff}}}\left({\displaystyle \frac{\partial }{\partial r}}+{\displaystyle \frac{2}{r}}\right)\widehat{\eta }(r,s); 0<r<1,$$

(20)

in which $\widehat{\eta }(r,s)$ and $\widehat{u}(r,s)$ are the Laplace transforms of $\eta (r,t)$ and $u(r,t)$, respectively, while s is the transform parameter. Introducing $\widehat{u}(r,s)$ from Equation (20) in (19), one finds that $\widehat{\eta }(r,s)$ has to satisfy the differential equation

$$\left({\displaystyle \frac{{\partial }^2}{\partial r^2}}+{\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial }{\partial r}}-{\displaystyle \frac{4}{r^2}}\right)\widehat{\eta }(r,s)=a(s)\widehat{\eta }(r,s); 0<r<1,$$

(21)

where

$$a(s)={\displaystyle \frac{(\alpha K+1)s+K_{eff}}{\alpha s+1}}.$$

(22)

The corresponding boundary condition is

$$\widehat{\eta }(1,s)=(\alpha s+1)\left({\displaystyle \frac{\partial }{\partial r}}-{\displaystyle \frac{1}{r}}\right){\left.\widehat{u}(r,s)\right|}_{r=1}=\widehat{f}(s),$$

(23)

where $\widehat{f}(s)$ is the Laplace transform of the function $f(t)$.

Now, applying the finite Hankel transform to Equation (21) and using the boundary condition (23) and the Laplace transform of Equation (14), one finds the finite Hankel transform ${\widehat{\eta }}_H(n,s)$ of $\widehat{\eta }(r,s)$, namely

$${\widehat{\eta }}_H(n,s)=-r_n{J}_1(r_n){\displaystyle \frac{\widehat{f}(s)}{r_n^2+a(s)}},$$

(24)

or equivalently

$${\widehat{\eta }}_H(n,s)=-{\displaystyle \frac{J_1(r_n)}{r_n}}\widehat{f}(s)+{\displaystyle \frac{J_1(r_n)}{r_n}}{\displaystyle \frac{a(s)}{r_n^2+a(s)}}\widehat{f}(s).$$

(25)

By applying the inverse finite Hankel transform to Equation (25), one finds that

$$\widehat{\eta }(r,s)=\widehat{f}(s)r^2+2\widehat{f}(s){\displaystyle \sum _{n=1}^{\infty }\frac{J_2(r{r}_n)}{r_n{J}_1(r_n)}}{\displaystyle \frac{a(s)}{r_n^2+a(s)}}.$$

(26)

By substituting $\widehat{\eta }(r,s)$ from Equation (26) in (20), one obtains for $\widehat{u}(r,s)$ the expression

$$\widehat{u}(r,s)={\displaystyle \frac{\widehat{f}(s)}{(\alpha K+1)s+K_{eff}}}\left[4r+2{\displaystyle \sum _{n=1}^{\infty }\frac{J_1(r{r}_n)}{J_1(r_n)}} {\displaystyle \frac{a(s)}{r_n^2+a(s)}}\right],$$

(27)

from which the fluid velocity $u(r,t)$ can be obtained by applying the inverse Laplace transform. Unfortunately, in this form $\widehat{u}(r,s)$ does not satisfy the boundary condition (23). In order to present it in a convenient form, we rewrite Equation (19) in the following suitable form

$${\displaystyle \frac{1}{r}}\widehat{\eta }(r,s)=(\alpha s+1){\displaystyle \frac{\partial }{\partial r}}\left[{\displaystyle \frac{1}{r}}\widehat{u}(r,s)\right].$$

(28)

Now, by replacing $\widehat{\eta }(r,s)$ from Equation (26) in (28), integrating the result with respect to r and bearing in mind the known identity

$${\displaystyle \int \frac{1}{r}{J}_2(r)}dr=-{\displaystyle \frac{1}{r}}{J}_1(r)+\mathrm{constant},$$

(29)

one finds for $\widehat{u}(r,s)$ the expression

$$\widehat{u}(r,s)={\displaystyle \frac{1}{\alpha s+1}}\left[{\displaystyle \frac{r^3}{2}}\widehat{f}(s)-2\widehat{f}(s){\displaystyle \sum _{n=1}^{\infty }\frac{J_1(r{r}_n)}{r_n^2{J}_1(r_n)}} {\displaystyle \frac{a(s)}{r_n^2+a(s)}}-rC(s)\right].$$

(30)

In order to determine the function $C(\cdot )$, we equalize the expressions of $\widehat{u}(r,s)$ from the relations (27) and (30) and use the identity

$$2{\displaystyle \sum _{n=1}^{\infty }\frac{J_1(r{r}_n)}{r{r}_n^2{J}_1(r_n)}}={\displaystyle \frac{r^2}{2}}+{\displaystyle \sum _{n=1}^{\infty }\frac{1}{r_n{J}_1(r_n)}}.$$

(31)

One finds for $C(s)$ the simple expression

$$C(s)=-\widehat{f}(s)\left[{\displaystyle \frac{4}{a(s)}}+c\right] ; c={\displaystyle \sum _{n=1}^{\infty }\frac{1}{r_n{J}_1(r_n)}}.$$

(32)

Substituting $C(s)$ from the equality (32) in (30), one obtains for $\widehat{u}(r,s)$ the expression

$$\widehat{u}(r,s)={\displaystyle \frac{\widehat{f}(s)}{\alpha s+1}}\left\{{\displaystyle \frac{r^3}{2}}-2{\displaystyle \sum _{n=1}^{\infty }\frac{J_1(r{r}_n)}{r_n^2{J}_1(r_n)}} {\displaystyle \frac{a(s)}{r_n^2+a(s)}}+r\left[{\displaystyle \frac{4}{a(s)}}+c\right]\right\},$$

(33)

which satisfies the boundary condition (23).

Finally, by applying the inverse Laplace transform to the last equality and using the identity

$${\displaystyle \frac{a(s)}{r_n^2+a(s)}}={\displaystyle \frac{\alpha K+1}{1+\alpha (r_n^2+K)}}+{\displaystyle \frac{(\alpha M-1)r_n^2}{{[1+\alpha (r_n^2+K)]}^2}}{\displaystyle \frac{1}{s+\frac{r_n^2+K_{eff}}{1+\alpha (r_n^2+K)}}},$$

(34)

one finds for the dimensionless velocity field $u(r,t)$ the expression

$$\begin{array}{l}u(r,t)={\displaystyle \frac{r^3+2cr}{2\alpha }}f(t)\ast \exp \left(-{\displaystyle \frac{t}{\alpha }}\right)+{\displaystyle \frac{4r}{\alpha K+1}}f(t)\ast \exp \left[-{\displaystyle \frac{K_{eff}}{\alpha K+1}}t\right]\\ -{\displaystyle \frac{2}{\alpha }}f(t)\ast \exp \left(-{\displaystyle \frac{t}{\alpha }}\right)\ast {\displaystyle \sum _{n=1}^{\infty }\frac{J_1(r{r}_n)}{r_n^2{J}_1(r_n)}\left\{\frac{\alpha K+1}{1+\alpha (r_n^2+K)}\delta (t)+\frac{(\alpha M-1)r_n^2}{{[1+\alpha (r_n^2+K)]}^2}\exp \left[-\frac{r_n^2+K_{eff}}{1+\alpha (r_n^2+K)}t\right]\right\}},\end{array}$$

(35)

where $\delta (\cdot )$ is the Dirac delta function and $\ast$ denotes the convolution product.

In the case of Newtonian fluids, when the parameter $\alpha =0$, the relation (33) takes the simpler form

$${\widehat{u}}_N(r,s)=\widehat{f}(s)\left[{\displaystyle \frac{r^3}{2}}-2{\displaystyle \sum _{n=1}^{\infty }\frac{J_1(r{r}_n)}{r_n^2{J}_1(r_n)}\left(1-\frac{r_n^2}{s+r_n^2+K_{eff}}\right)+r\left(\frac{4}{s+K_{eff}}+c\right)}\right] .$$

(36)

Applying the inverse Laplace transform to this last equality, one finds that

$$\begin{array}{l}{u}_N(r,t)=\left({\displaystyle \frac{r^3}{2}}+cr\right)f(t)-2f(t){\displaystyle \sum _{n=1}^{\infty }\frac{J_1(r{r}_n)}{r_n^2{J}_1(r_n)}}+4r{\displaystyle \underset{0}{\overset{t}{\int }}f(t-s){\mathrm{e}}^{-K_{eff}s}ds}\\ +2{\displaystyle \sum _{n=1}^{\infty }\frac{J_1(r{r}_n)}{J_1(r_n)}}{\displaystyle \underset{0}{\overset{t}{\int }}f(t-s){\mathrm{e}}^{-(r_n^2+K_{eff})s}ds}.\end{array}$$

(37)

Direct computations show that ${\eta }_N(r,t)$ and $u_N(r,t)$ given by the relations (18) and (37), respectively, satisfy the governing Equation (7) with $\alpha =0$.

The general expressions of the dimensionless shear stress $\eta (r,t)$ and the velocity $u(r,t)$, given by Equations (17) and (35), can generate exact solutions for any MHD Taylor–Couette flow of ECISGFs through a porous medium in an infinite circular cylinder induced by a time-dependent torque. For illustration, and to bring to light some characteristics of the fluid behavior, different flows with technical relevance are considered, and the corresponding shear stress and velocity fields are provided. Since the respective motions become steady in time, the required time to reach the steady state will be graphically determined. This time is very important for experimental researchers who want to know the transition moment of the fluid motion to the steady or permanent state. Moreover, it is important to know if the steady state is reached earlier or later in the presence of a magnetic field or porous medium.

4. Some Particular Cases and the Corresponding Solutions

4.1. The Cylinder Oscillates Around Its Symmetry Axis

By replacing $f(t)$ by $H(t)\cos (\omega t)$ or $H(t)\sin (\omega t)$ in the relations (17) and (35), one finds the dimensionless starting shear stress and velocity fields ${\eta }_c(r,t)$, ${\eta }_s(r,t)$ and $u_c(r,t)$, $u_s(r,t)$ for MHD motions of ECISGFs through a porous medium in an infinite circular cylinder that oscillates around its symmetry axis due to the time-dependent torques per unit length $2\pi RSH(t)\cos (\omega t)$ or $2\pi RSH(t)\sin (\omega t)$, respectively. Here $H(t)$ is the Heaviside unit step function and $\omega$ is the dimensionless frequency of the oscillations. The steady state (permanent) and transient components ${\eta }_{cs}(r,t)$, ${\eta }_{ct}(r,t)$ and ${\eta }_{ss}(r,t)$, ${\eta }_{st}(r,t)$ of ${\eta }_c(r,t)$ and ${\eta }_s(r,t)$, respectively, are given by the following relations:

$$\begin{array}{l}{\eta }_{cs}(r,t)=r^2\cos (\omega t)+2(\alpha K+1)\cos (\omega t){\displaystyle \sum _{n=1}^{\infty }\frac{J_2(r{r}_n)}{r_n{J}_1(r_n)[1+\alpha (r_n^2+K)]}}\\ +2(\alpha M-1){\displaystyle \sum _{n=1}^{\infty }\frac{(r_n^2+K_{eff})\cos (\omega t)+\omega [1+\alpha (r_n^2+K)]\sin (\omega t)}{1+\alpha (r_n^2+K)}} {\displaystyle \frac{r_n{J}_2(r{r}_n)}{a_n{J}_1(r_n)}},\end{array}$$

(38)

$${\eta }_{ct}(r,t)=2(1-\alpha M){\displaystyle \sum _{n=1}^{\infty }\frac{r_n(r_n^2+K_{eff})J_2(r{r}_n)}{a_n{J}_1(r_n)[1+\alpha (r_n^2+K)]}}\exp \left[-{\displaystyle \frac{r_n^2+K_{eff}}{1+\alpha (r_n^2+K)}}t\right],$$

(39)

$$\begin{array}{l}{\eta }_{ss}(r,t)=r^2\sin (\omega t)+2(\alpha K+1)\sin (\omega t){\displaystyle \sum _{n=1}^{\infty }\frac{J_2(r{r}_n)}{r_n{J}_1(r_n)[1+\alpha (r_n^2+K)]}}\\ +2(\alpha M-1){\displaystyle \sum _{n=1}^{\infty }\frac{(r_n^2+K_{eff})\sin (\omega t)-\omega [1+\alpha (r_n^2+K)]\cos (\omega t)}{1+\alpha (r_n^2+K)}} {\displaystyle \frac{r_n{J}_2(r{r}_n)}{a_n{J}_1(r_n)}},\end{array}$$

(40)

$${\eta }_{st}(r,t)=2\omega (\alpha M-1){\displaystyle \sum _{n=1}^{\infty }\frac{r_n{J}_2(r{r}_n)}{a_n{J}_1(r_n)}}\exp \left[-{\displaystyle \frac{r_n^2+K_{eff}}{1+\alpha (r_n^2+K)}}t\right] .$$

(41)

In above relations, the constants $a_n$ are given by the relation

$$a_n={(r_n^2+K_{eff})}^2+{\omega }^2{[1+\alpha (r_n^2+K)]}^2, n=1, 2, 3\dots$$

(42)

Direct computations show that equivalent forms for dimensionless steady-state components ${\eta }_{cs}(r,t)$ and ${\eta }_{ss}(r,t)$ are given by the next relations

$${\eta }_{cs}(r,t)=\mathrm{Re}\left\{{\displaystyle \frac{I_2(r\sqrt{\gamma })}{I_2(\sqrt{\gamma })}}{e}^{i\omega t}\right\} , {\eta }_{ss}(r,t)=\mathrm{Im}\left\{{\displaystyle \frac{I_2(r\sqrt{\gamma })}{I_2(\sqrt{\gamma })}}{e}^{i\omega t}\right\},$$

(43)

where $I_2(\cdot )$ is the modified Bessel function of the first kind and second order, Re and Im represent the real part and the imaginary part of what follows, and

$$\gamma ={\displaystyle \frac{K_{eff}+i\omega (1+\alpha K)}{1+i\omega \alpha }}.$$

(44)

Figure 2 clearly shows the equivalence of the expressions (38), (43), and (40), (43) of the steady state shear stresses ${\eta }_{cs}(r,t)$ and ${\eta }_{ss}(r,t)$, respectively. Similar solutions corresponding to electrically conducting incompressible Newtonian fluids performing the same motions are immediately obtained by taking $\alpha =0$ in above relations.

Using the relations (8) and (43), it is not difficult to show that the steady state components $u_{cs}(r,t)$ and $u_{ss}(r,t)$ of the dimensionless starting velocity fields $u_c(r,t)$ and $u_s(r,t)$, respectively, are given by the simple relations

$$\begin{array}{l}{u}_{cs}(r,t)=\mathrm{Re}\left\{{\displaystyle \frac{1}{(1+i\omega \alpha )\sqrt{\gamma }}}{\displaystyle \frac{I_1(r\sqrt{\gamma })}{I_2(\sqrt{\gamma })}}{e}^{i\omega t}\right\},\\ u_{ss}(r,t)=\mathrm{Im}\left\{{\displaystyle \frac{1}{(1+i\omega \alpha )\sqrt{\gamma }}}{\displaystyle \frac{I_1(r\sqrt{\gamma })}{I_2(\sqrt{\gamma })}}{e}^{i\omega t}\right\}.\end{array}$$

(45)

Direct computations clearly show that ${\eta }_{cs}(r,t)$, ${\eta }_{ss}(r,t)$ and $u_{cs}(r,t)$, $u_{ss}(r,t)$ given by the relations (43) and (45), respectively, satisfy both governing Equations (7) and (8) and the boundary conditions (11).

4.2. The Cylinder Rotates Around Its Symmetry Axis

By replacing $f(t)$ by $H(t)$ in the relations (17) and (35), one obtains the dimensionless starting shear stress and velocity fields ${\eta }_C(r,t)$ and $u_C(r,t)$ corresponding to the MHD motion of ECISGFs through a porous medium in an infinite circular cylinder that applies a constant torque per unit length $2\pi RSH(t)$ to the fluid [27]. The steady and transient components ${\eta }_{Cs}(r)$ and ${\eta }_{Ct}(r,t)$ of ${\eta }_C(r,t)$, for instance, have the expressions

$${\eta }_{Cs}(r)=r^2+2K_{eff}{\displaystyle \sum _{n=1}^{\infty }\frac{J_2(r{r}_n)}{r_n(r_n^2+K_{eff})J_1(r_n)}},$$

(46)

$${\eta }_{Ct}(r,t)=2(1-\alpha M){\displaystyle \sum _{n=1}^{\infty }\frac{r_n{J}_2(r{r}_n)}{J_1(r_n)(r_n^2+K_{eff})[1+\alpha (r_n^2+K)]}}\exp \left[-{\displaystyle \frac{r_n^2+K_{eff}}{1+\alpha (r_n^2+K)}}t\right] .$$

(47)

Last relations can also be directly obtained by taking $\omega =0$ in Equations (38) and (39). An equivalent form for ${\eta }_{Cs}(r)$, namely

$${\eta }_{Cs}(r)={\displaystyle \frac{I_2(r\sqrt{K_{eff}})}{I_2(\sqrt{K_{eff}})}} ,$$

(48)

is obtained from the first equality (43). It is clear that the steady shear stress ${\eta }_{Cs}(r)$, given by Equations (46) or (48), corresponds to both incompressible Newtonian and second grade fluids. This is not a surprise because the governing equations corresponding to steady motions of these fluids are identical. The steady velocity field corresponding to this motion is

$$u_{Cs}(r)={\displaystyle \frac{I_1(r\sqrt{K_{eff}})}{\sqrt{K_{eff}}{I}_2(\sqrt{K_{eff}})}}.$$

(49)

5. Comparisons Between Analytical and Numerical Solutions

In this section, our purpose is to formulate schemes to determine numerical solutions for the starting shear stress $\eta (r,t)$ and the corresponding velocity field $u(r,t)$.

5.1. Numerical Solutions for the Shear Stress

This paragraph is dedicated to the formulation of the numerical algorithm for solving the partial differential Equation (12) with the initial and boundary conditions (13). Firstly, we write Equation (12) in a convenient equivalent form. By applying the Laplace transform to this form, one obtains

$$\begin{array}{l}r{\displaystyle \frac{\partial }{\partial r}}\left(r{\displaystyle \frac{\partial \widehat{\eta }(r,s)}{\partial r}}\right)={\displaystyle \frac{[(1+\alpha K)r^2+4\alpha ]s+K_{eff}{r}^2+4}{\alpha s+1}}\widehat{\eta }(r,s)\\ =A(r)\widehat{\eta }(r,s)+B(r){\displaystyle \frac{1}{s+1/\alpha }}\widehat{\eta }(r,s),\end{array}$$

(50)

where

$$A(r)={\displaystyle \frac{(1+\alpha K)r^2+4\alpha }{\alpha }}, B(r)={\displaystyle \frac{(K_{eff}-1-\alpha K)r^2}{{\alpha }^2}}.$$

(51)

By applying the inverse Laplace transform to Equation (50), one obtains the following equivalent form of Equation (12)

$$r^2{\displaystyle \frac{{\partial }^2\eta (r,t)}{\partial r^2}}+r{\displaystyle \frac{\partial \eta (r,t)}{\partial r}}=A(r)\eta (r,t)+B(r)\exp (-t/\alpha )\ast \eta (r,t).$$

(52)

To formulate the numerical method, we consider the discrete values of the time variable $t_k=k\mathrm{\Delta }t, k=0, 1,\dots ,N_t,$ $\mathrm{\Delta }t=T_f/N_t, t_k\in [0,T_f],$ respectively, discrete values of the spatial variable $r_j=j\mathrm{\Delta }r, j=0, 1,\dots ,N_r,$ $\mathrm{\Delta }r=1/N_r, r_k\in [0,1]$.

Let us denote by

$$G(r,t)=\exp (-t/\alpha )\ast \eta (r,t)=\exp (-t/\alpha ){\displaystyle \underset{0}{\overset{t}{\int }}\exp (s/\alpha )\eta (r,s)ds,}$$

(53)

a part of the last term in Equation (52). Now, we present a formula for approximating this term at the division points $(r_j,t_k), j=0,1,\dots ,N_r, k=0,1,\dots N_t$. In the following, we use the notations $G(r_j,t_k)=G_j^k, \eta (r_j,t_k)={\eta }_j^k$. Bearing in mind Equation (53), it results that

$$\begin{array}{l}{G}_j^k=G(r_j,t_k)={\mathrm{e}}^{-t_k/\alpha }{\displaystyle \underset{0}{\overset{t_k}{\int }}{\mathrm{e}}^{s/\alpha }\eta (r_j,s)ds={\mathrm{e}}^{-t_k/\alpha }{\displaystyle \sum _{i=0}^{k-1}{\displaystyle \underset{t_i}{\overset{t_{i+1}}{\int }}{\mathrm{e}}^{s/\alpha }\eta (r_j,s)ds}}}={\mathrm{e}}^{-t_k/\alpha }{\displaystyle \sum _{i=0}^{k-1}{\displaystyle \underset{t_i}{\overset{t_{i+1}}{\int }}\frac{{\mathrm{e}}^{s/\alpha }}{2}[\eta (r_j,t_{i+1})+\eta (r_j,t_i)]ds}}\\ ={\mathrm{e}}^{-t_k/\alpha }{\displaystyle \sum _{i=0}^{k-1}\frac{\eta (r_j,t_{i+1})+\eta (r_j,t_i)}{2}}{\displaystyle \underset{t_i}{\overset{t_{i+1}}{\int }}{\mathrm{e}}^{s/\alpha }ds}={\mathrm{e}}^{-t_k/\alpha }{\displaystyle \sum _{i=0}^{k-1}\frac{\alpha [\eta (r_j,t_{i+1})+\eta (r_j,t_i)]}{2}}\left[{\mathrm{e}}^{t_{i+1}/\alpha }-{\mathrm{e}}^{t_i/\alpha }\right] .\end{array}$$

(54)

By introducing the notation $P_i={\mathrm{e}}^{t_{i+1}/\alpha }-{\mathrm{e}}^{t_i/\alpha }$, Equation (54) can be written as

$$G_j^k={\displaystyle \frac{\alpha }{2}}{\mathrm{e}}^{-t_k/\alpha }\left[P_0{\eta }_j^0+{\displaystyle \sum _{i=1}^{k-1}(P_{i-1}+P_i){\eta }_j^i+P_{k-1}{\eta }_j^k}\right].$$

(55)

Now, using the next approximations [30]

$${\left.{\displaystyle \frac{\partial \eta (r,t)}{\partial r}}\right|}_{(r,t)=(r_j,t_k)}\cong {\displaystyle \frac{{\eta }_{j+1}^k-{\eta }_j^k}{\mathrm{\Delta }r}}, {\left.{\displaystyle \frac{{\partial }^2\eta (r,t)}{\partial r^2}}\right|}_{(r,t)=(r_j,t_k)}\cong {\displaystyle \frac{{\eta }_{j+1}^k-2{\eta }_j^k+{\eta }_{j-1}^k}{\mathrm{\Delta }{r}^2}},$$

(56)

of the first and second derivative of the function $\eta (r,t)$ in the point $(r_j,t_k)$, and using (55), Equation (50) can be written as

$$a_j{\eta }_{j-1}^k+b_j{\eta }_j^k+c_j{\eta }_{j+1}^k=q_j^k, j=1,2,\dots ,N_r-1, k=1,2,\dots ,N_t,$$

(57)

where

$$\begin{array}{l}{a}_j={\left({\displaystyle \frac{r_j}{\mathrm{\Delta }r}}\right)}^2=j^2,\\ b_j=-\left[2{\left({\displaystyle \frac{r_j}{\mathrm{\Delta }r}}\right)}^2+{\displaystyle \frac{r_j}{\mathrm{\Delta }r}}+A(r_j)+{\displaystyle \frac{\alpha }{2}}B(r_j){\mathrm{e}}^{-t_k/\alpha }\left({\mathrm{e}}^{t_k/\alpha }-{\mathrm{e}}^{t_{k-1}/\alpha }\right)\right]\\ =-\left[2j^2+j+A(r_j)+{\displaystyle \frac{\alpha }{2}}B(r_j)[1-{\mathrm{e}}^{-\mathrm{\Delta }t/\alpha }]\right],\\ c_j={\left({\displaystyle \frac{r_j}{\mathrm{\Delta }r}}\right)}^2+{\displaystyle \frac{r_j}{\mathrm{\Delta }r}}=j^2+j,\end{array}$$

(58)

and

$$\begin{array}{l}{q}_j^0=0, q_j^1={\displaystyle \frac{\alpha P_0}{2}}B(r_j){\mathrm{e}}^{-t_1/\alpha }{\eta }_j^0, j=0,1,\dots ,N_r,\\ q_j^k={\displaystyle \frac{\alpha }{2}}B(r_j){\mathrm{e}}^{-t_k/\alpha }\left[P_0{\eta }_j^0+{\displaystyle \sum _{i=0}^{k-2}(P_i+P_{i+1}){\eta }_j^{i+1}}\right],\\ k=2,3,\dots ,N_t, j=0,1,\dots ,N_r.\end{array}$$

(59)

The initial and boundary conditions considered along with Equation (57) are

$$\begin{array}{l}\eta (r,0)=0, r\in [0,1],\\ \eta (0,t)=0, \eta (1,t)=f(t), t>0.\end{array}$$

(60)

Last relations can be written in the following form, suitable for the numerical method:

$$\begin{array}{l}{\eta }_j^0=0, j=0,1,\dots ,N_r,\\ {\eta }_0^k=0, {\eta }_{N_r}^k=f(t_k), k=1,2,\dots ,N_t.\end{array}$$

(61)

Using (61), the algebraic system (57) can be written in matrix form

$$M_k{E}_k=Q_k, k=1,2,\dots ,N_t,$$

(62)

where the matrix $M_k$ of order $(N_r-1)\times (N_r-1)$, and matrices $E_k$, $Q_k$ of order $(N_r-1)\times 1$ are defined as

$$M_k=\left(\begin{array}{ccccccc}{b}_1& c_1& 0& 0& 0& \dots & 0\\ a_2& b_2& c_2& 0& 0& \dots & 0\\ 0& a_3& b_3& c_3& 0& \dots & 0\\ 0& 0& a_4& b_4& c_4& \dots & 0\\ \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ 0& 0& 0& \dots & a_{N_r-2}& b_{N_r-2}& c_{N_r-2}\\ 0& 0& 0& 0& \dots & a_{N_r-1}& b_{N_r-1}\end{array}\right), E_k=\left(\begin{array}{c}{\eta }_1^k\\ {\eta }_2^k\\ {\eta }_3^k\\ {\eta }_4^k\\ \dots \\ {\eta }_{N_r-2}^k\\ {\eta }_{N_r-1}^k\end{array}\right), Q_k=\left(\begin{array}{c}{q}_1^k-a_1{\eta }_0^k\\ q_2^k\\ q_3^k\\ q_4^k\\ \dots \\ q_{N_r-2}^k\\ q_{N_r-1}^k-c_{N_r-1}{\eta }_{N_r}^k\end{array}\right).$$

(63)

The elements of the $M_k$ matrix are written, in general, form as follows:

$$\begin{array}{l}{m}_{1n}=b_1{\delta }_{1n}+c_1{\delta }_{1n-1}, n=1,2,\dots ,N_r-1,\\ m_{mn}=a_m{\delta }_{mn+1}+b_m{\delta }_{mn}+c_m{\delta }_{mn-1}, m=2,3,\dots ,N_r-2, n=1,2,\dots ,N_r-1,\\ m_{N_r-1n}=a_{N_r-1}{\delta }_{N_r-1n+1}+b_{N_r-1}{\delta }_{N_r-1n}, n=1,2,\dots ,N_r-1.\end{array}$$

(64)

For each time value $t_k, k=1,2,\dots ,N_t$, the unknown dimensionless shear stress values ${\eta }_j^k, j=1,2,\dots ,N_r-1$ are given by the solution of the matrix equation $M_k{E}_k=Q_k$. Figure 2 shows the graphs of the shear stress $\eta (r,t)$ constructed on the basis of numerical results based on the analytical Formula (17) and the numerical scheme (57). For the numerical simulations presented in this figure, the tension on the cylinder surface was considered as given by the function $\eta (1,t)=1-\exp (-t)$. From Figure 3, one can observe a very good fit of the numerical values obtained by the two methods.

5.2. Numerical Solutions for the Velocity Field

The formulation of the numerical algorithm for determining the velocity is based on Equations (7) and (8). Using the Laplace transform on these equations, eliminating the Laplace transform of the shear stress, it is obtained that the Laplace transform of the velocity field $u(r,t)$ satisfies the differential equation:

$$r^2{\displaystyle \frac{{\partial }^2\widehat{u}(r,s)}{\partial r^2}}+r{\displaystyle \frac{\partial \widehat{u}(r,s)}{\partial r}}=A_1(r)\widehat{u}(r,s)+B_1(r){\displaystyle \frac{1}{s+1/\alpha }}\widehat{u}(r,s),$$

(65)

where

$$A_1(r)={\displaystyle \frac{(1+\alpha K)r^2+\alpha }{\alpha }}, B_1(r)={\displaystyle \frac{(\alpha M-1)r^2}{{\alpha }^2}}.$$

(66)

By applying the inverse Laplace transform to the equality (65), one finds that the velocity field $u(r,t)$ is the solution of the differential equation

$$r^2{\displaystyle \frac{{\partial }^{2}u(r,t)}{\partial r^2}}+r{\displaystyle \frac{\partial u(r,t)}{\partial r}}=A_1(r)u(r,t)+B_1(r)\exp (-t/\alpha )\ast u(r,t).$$

(67)

Let’s introduce the function

$$H(r,t)=\exp (-t/\alpha )\ast u(r,t)=\exp (-t/\alpha ){\displaystyle \underset{0}{\overset{t}{\int }}u(r,\tau )\exp (\tau /\alpha )d\tau }.$$

(68)

Following the same procedure as in the previous section, the function $H(r,t)$ can be approximated by

$$\begin{array}{l}{H}_j^0=0, H_j^1={\displaystyle \frac{\alpha }{2}}{P}_0{u}_j^0\exp (-t_1/\alpha ), j=0,1,\dots ,N_r,\\ H_j^k=H(r_j,t_k)={\displaystyle \frac{\alpha }{2}}\exp (-t_k/\alpha )\left[P_0{u}_j^0+{\displaystyle \sum _{i=1}^{k-1}(P_{i-1}+P_i)u_j^i+P_{k-1}{u}_j^k}\right],\\ j=0,1,\dots ,N_r, k=2,3,\dots ,N_t.\end{array}$$

(69)

By using the approximation formulas

$$\begin{array}{l}{\left.{\displaystyle \frac{\partial u(r,t)}{\partial r}}\right|}_{(r,t)=(r_j,t_k)}={\displaystyle \frac{u_{j+1}^k-u_j^k}{\mathrm{\Delta }r}}, j=1,2,\dots ,N_r-1, k=0,1,\dots ,N_t,\\ {\left.{\displaystyle \frac{{\partial }^{2}u(r,t)}{\partial r^2}}\right|}_{(r,t)=(r_j,t_k)}={\displaystyle \frac{u_{j+1}^k-2u_j^k+u_{j-1}^k}{\mathrm{\Delta }{r}^2}}, j=1,2,\dots ,N_r-1, k=0,1,\dots ,N_t,\end{array}$$

(70)

in Equation (67), and using (69), we attain the following algebraic system for values of the velocity field:

$$S_k{U}_k=R_k, k=1,2,\dots ,N_t,$$

(71)

where the elements of the matrices $S_k, U_k$ and $R_k$ are given by

$$\begin{array}{l}{S}_{1n}={\overline{b}}_1{\delta }_{1n}+{\overline{c}}_1{\delta }_{1n-1}, n=1,2,\dots ,N_r-1,\\ S_{mn}={\overline{a}}_m{\delta }_{mn+1}+{\overline{b}}_m{\delta }_{mn}+{\overline{c}}_m{\delta }_{mn-1}, m=2,3,\dots ,N_r-2, n=1,2,\dots ,N_r-1,\\ S_{N_r-1n}={\overline{a}}_{N_r-1}{\delta }_{N_r-1n+1}+\left({\overline{b}}_{N_r-1}+{\displaystyle \frac{{\overline{c}}_{N_r-1}}{1-\mathrm{\Delta }r}}\right){\delta }_{N_r-1n}, n=1,2,\dots ,N_r-1,\end{array}$$

(72)

$$\begin{array}{l}{\overline{a}}_j={\left({\displaystyle \frac{r_j}{\mathrm{\Delta }r}}\right)}^2=j^2,\\ {\overline{b}}_j=-\left[2{\left({\displaystyle \frac{r_j}{\mathrm{\Delta }r}}\right)}^2+{\displaystyle \frac{r_j}{\mathrm{\Delta }r}}+A_1(r_j)+{\displaystyle \frac{\alpha }{2}}{B}_1(r_j)\exp (-t_k/\alpha )\left[\exp (t_k/\alpha )-\exp (-t_{k-1}/\alpha )\right]\right]\\ =-\left[2j^2+j+A_1(r_j)+{\displaystyle \frac{\alpha }{2}}{B}_1(r_j)\left[1-\exp (-\mathrm{\Delta }t/\alpha )\right]\right], {\overline{c}}_j={\left({\displaystyle \frac{r_j}{\mathrm{\Delta }r}}\right)}^2+{\displaystyle \frac{r_j}{\mathrm{\Delta }r}}=j^2+j,\end{array}$$

(73)

$$U_k={[u_1^k,u_2^k,\dots ,u_{N_r-1}^k]}^T, k=1,2,\dots ,N_t,$$

(74)

$$\begin{array}{l}{R}_1^k=B_1(r_1)H_1^k-u_0^k, k=1,2,\dots ,N_t,\\ R_j^k=B_1(r_j)H_j^k, j=2,3,\dots ,N_r-2, k=1,2,\dots ,N_t,\\ R_{N_r-1}^k=B_1(r_{N_r-1})H_{N_r-1}^k-{\displaystyle \frac{\mathrm{\Delta }r{\overline{c}}_{N_r-1}}{1-\mathrm{\Delta }r}}g(t_k), k=1,2,\dots ,N_t,\end{array}$$

(75)

$$\begin{array}{l}{u}_{N_r}^k={\displaystyle \frac{1}{1-\mathrm{\Delta }r}}\left[u_{N_r-1}^k+\mathrm{\Delta }r g(t_k)\right] , k=0,1,\dots ,N_t,\\ g(t)={\displaystyle \frac{1}{\alpha }}\exp (-t/\alpha )\ast f(t).\end{array}$$

(76)

Figure 4, like Figure 2, for the shear stress $\eta (r,t)$, shows a very good fit of the numerical values of the dimensionless velocity $u(r,t)$ corresponding to the analytic expression from the equality (35) and the numerical scheme (71) when the shear stress is prescribed by the same function $f(t)=1-\exp (-t)$ on the boundary.

Equivalent forms of the analytical solutions of the shear stress and flow velocity obtained from the Bessel equations that are satisfied by the Laplace transforms of these functions have been presented in Appendix A.

6. Some Numerical Results

In order to bring to light the influence of magnetic field and porous medium on the required time to reach the steady state, Figure 5 and Figure 6 have been included here. They underline variations in the dimensionless transient shear stress ${\eta }_{ct}(r,t)$ at distinct values of the parameters M or K and increasing values of the time t. In all cases, it is observed that after small enough values of the dimensionless time t, the absolute values of this shear stress become negligible. The necessary time to reach the steady state, as it clearly results from Figure 5 and

Table 1. Numerical values of the shear stress ${\eta }_{ct}(r,t)$ in different points of Figure 5 corresponding to the value 0.8 of the dimensionless time t.

K = 0.1				K = 0.9
1	0	11	–2.595 × 10−3	1	0	11	–2.417 × 10−3
3	–3.482 × 10−5	12	–2.914 × 10−3	3	–3.204 × 10−5	12	–2.722 × 10−3
3	–1.382 × 10−4	13	–3.175 × 10−3	3	–1.272 × 10−4	13	–2.974 × 10−3
4	–3.068 × 10−4	14	–3.354 × 10−3	4	–2.826 × 10−4	14	–3.151 × 10−3
5	–5.354 × 10−4	15	–3.426 × 10−3	5	–4.936 × 10−4	15	–3.229 × 10−3
6	–8.163 × 10−4	16	–3.364 × 10−3	6	–7.534 × 10−4	16	–3.182 × 10−3
7	–1.140 × 10−3	17	–3.139 × 10−3	7	–1.054 × 10−3	17	–2.981 × 10−3
8	–1.494 × 10−3	18	–2.721 × 10−3	8	–1.383 × 10−3	18	–2.595 × 10−3
9	–1.866 × 10−3	19	–2.080 × 10−3	9	–1.731 × 10−3	19	–1.993 × 10−3
10	–2.239 × 10−3	20	–1.184 × 10−3	10	–2.081 × 10−3	20	–1.140 × 10−3

corresponding to Figure 6, diminishes for increasing values of the parameter M or K. Consequently, the steady state for such motions of ECISGFs is rather obtained in the presence of the magnetic field or porous medium and the transition moment of the motion to the steady state is around the value 0.9 of the dimensionless time t with an error smaller than ${10}^{-3}$.

In Figure 7, for comparison, the surface plots of the dimensionless transient shear stresses ${\eta }_{ct}(r,t)$ and ${\eta }_{st}(r,t)$ are presented together at the same values of the physical parameters. The transition to the steady state is clearly visible in both cases. In addition, as expected, the steady state is reached faster for motions induced by sinusoidal oscillations of the boundary voltage. This is normal because, in this case, at the moment $t=0$ the value of the shear stress $\eta (r,t)$ on cylinder is zero.

Figure 8 and Figure 9 are presented to highlight the variations in the dimensionless velocity $u(r,t)$ with respect to M, K, the time t, and the $\alpha$ parameter of the second grade fluids at different values of the spatial variable r when the function $f(t)=1-\exp (-t)$. In all cases, as expected, the fluid velocity brings up for increasing values of the spatial variable r. It asymptotically tends to constant values for large values of M, K, t, or $\alpha$. In addition, as it clearly results from Figure 8a,b, the fluid velocity declines for increasing values of M or K. Consequently, the fluid flows more slowly in the presence of a magnetic field or porous medium. Figure 9a,b clearly shows that $u(r,t)$ is an increasing function with respect to the time t and diminishes for increasing values of $\alpha$. It means that incompressible second grade fluids flow more slowly in comparison with Newtonian ones. Moreover, from Figure 9a it results that the fluid motion becomes steady in time and the fluid velocity asymptotically tends to constant values at any point of the flow domain for increasing values of the time t. As can be seen from this figure, the steady state is reached later in the vicinity of the cylindrical surface.

Figure 10 shows the contour plot of the fluid velocity $u(r,t)$ in the case where the shear stress on the boundary is given by the function $f(t)=2(1-\cos (t))$. It is observed that for small values of time, the fluid has a slower motion in the central area of the cylinder. As expected, the fluid velocity has higher values in the area near the boundary of the cylinder.

7. Conclusions

In this study the motion problem of ECISGFs in an infinite circular cylinder that applies a time-dependent torque per unit length $2\pi RSf(t)$ to the fluid is completely solved when the magnetic and porous effects are taken into account. The general analytical expressions, that were established for the dimensionless shear stress and velocity fields $\eta (r,t)$ and $u(r,t)$, can generate exact solutions for any motion of this type of the respective fluids. For illustration, some special cases have been considered and the corresponding solutions were provided. Since these motions become steady in time, the corresponding solutions can be written as sum of steady state and transient components. This writing is very important for practical experiments because they allow us to determine the required time to reach the steady state. This is the time after which the numerical values of the transient components of shear stress or velocity are small enough and can be neglected. Consequently, after this time, the fluid behavior will be characterized only by the steady state solutions which are independent of the initial conditions but satisfy the governing equations and boundary conditions.

The main results that have been obtained through this study are:

• General expressions have been established for the velocity field $u(r,t)$ and the shear stress $\eta (r,t)$ corresponding to MHD Taylor–Couette flows of ECISGFs through a porous medium in a right circular infinite cylinder that applies to the fluid the time-dependent torque $2\pi RSf(t)$ per unit length.
• The expressions that have been obtained can generate exact solutions for any such motion of the respective fluids and the problem in question is completely solved. For illustration some particular motions have been highlighted and investigated.
• For results’ validation, the steady shear stresses ${\eta }_{cs}(r,t)$, ${\eta }_{ss}(r,t)$ and ${\eta }_{Cs}(r)$ have been presented in equivalent forms. In addition, the equivalence of analytical and numerical solutions for the fluid velocity $u(r,t)$ and the shear stress $\eta (r,t)$ has been graphically proven when $f(t)=1-\exp (-t)$.
• Graphical representations from Figure 5, Figure 6 and Figure 9b show that:

  -

  ECISGFs flow more slowly and the steady state occurs more quickly in the presence of a magnetic field or porous medium.

  -

  ECISGFs flow more slowly in comparison with incompressible Newtonian fluids.