EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 12, No. 3, 2019, 1149-1175 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global MHD flows of second grade fluid through the moving porous cylindrical domain Muhammad Jamil1,∗, Muhammad Zafarullah1,2 1 Department of Mathematics, NED University of Engineering & Technology, Karachi-75270, Pakistan 2 Department of Mathematics, DJ Sindh Govt. Science College, Karachi, Pakistan Abstract. The flows of Magnetohydrodynamics(MHD) second grade fluid between two infinite porous coaxial circular cylinders are studied. At time t = 0+, the inner cylinder begins to rotate around its axis and to slide along the same axis due to torsional and longitudinal time dependent shear stresses and the outer cylinder is also rotate around its axis and to slide along the same axis with acceleration. The exact solutions obtained with the help of discrete Laplace and finite Hankel transform, satisfy all imposed initial and boundary conditions. The solution presented in convolution product of Laplace transform . The corresponding solutions for second grade and Newtonian fluids are also obtained as limiting cases with and without MHD effect. Finally, the influence of pertinent parameters on the velocity components and shear stresses, as well as a comparison among, second grade and Newtonian fluids with and without MHD is also analyzed by graphical illustrations. 2010 Mathematics Subject Classifications: 76A05, 76A10 Key Words and Phrases: Second grade fluid, MHD flows, velocity field, time-dependent shear stresses, Hankel transform 1. Introduction Now-a-days, the non-Newtonian fluids are more important and significant in science and technological applications than the Newtonian fluids. In several sectors, like polymer and petroleum, food and agriculture, chemical and manufacturing, pharmaceutical and biotechnology industries non-Newtonian fluids are more important, the fluids are the so- lution of either artificial or natural material with other Newtonian fluids as water, oils, red cells and different substances having long chain molecules; the resulting fluids have the non-Newtonian features [5, 6, 10, 16, 18, 20]. There are three famous solution other than exact solutions, can be applicable, those are analytical solutions, numerical solutions ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v12i3.3426 Email addresses: jqrza26@yahoo.com (M. Jamil), muhammad.zafarullah@gmail.com (M. Zafarullah) http://www.ejpam.com 1149 c© 2019 EJPAM All rights reserved. M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1150 and experimental solutions [21]. Analytical solution is the general information and closed- form solution, but it has simple geometry / physics and it can solve only linear problem. Numerical solution is used for complicated physical and nonlinear problems, but its main disadvantages are truncation errors and inappropriate modeling while experimental solu- tion is most realistic and reliable but due to equipment / operation costs, scaling problem, measurement difficulties are not much appreciated [12–14, 17]. The flow issue in porous medium like tube, channel, pipes and space are considered in many research literatures but the only few researcher used the porous medium in the magnetohydrodynamic (MHD) flow of second grade fluid in an infinite cylindrical domain, which is much more realistic and interested when the Reynold’s number is same with any system, such as in blood circulation, digestive and urinary systems in any body, water and other chemical sewerage filtration and purification plants, underground tube wheel, MHD generators for generating electricity and its MHD pumps and other engineering applica- tions [6–8, 10, 12, 16, 18]. The Navier-Stokes equation is used to solve the problem of Newtonian fluid but the governing equations of non-Newtonian fluids are more convoluted and having the higher order than the Nevier-Stokes equation for Newtonian fluids, therefore Navier-Stokes equa- tions are inadequate and insufficient to designate their behavior [1, 4], so we used some mathematical techniques and transforms to solve the flow problem of the non-Newtonian fluid. In cylindrical domain, the Hankel and Laplace transform are the one of the best transform to determine the exact or analytical solution [6, 9, 10, 12, 16]. The objective of this paper is to extend the concepts of boundary conditions and iden- tities of Hankel transformation used in [2, 9] to second grade fluid with more generalized boundary conditions from which we recover many general solutions as special solution to our general solutions. We also point out that the boiundary conditions and indenties used in mentioned paper [9, 15] are very rare in literature. Therefore motivated by these facts, we study, the MHD flow of second grade fluid between two infinite porous coaxial circular cylinders. At time t = 0+, the inner cylinder begins to rotate around its axis and to slide along the same axis due to torsional and longitudinal time dependent shear stresses f1 t p and g1 t p the outer cylinder is also rotate around its axis and to slide along the same axis with velocities f2 t p and g2 t p . The exact solutions obtained with the help of Laplace and finite Hankel transform, presented in convolution product of Laplace transform, which satisfy all imposed initial and boundary conditions. As the limiting and special cases, we find solutions for second grade with and without MHD/porous effects, Newtonian with and without MHD/porous effects and discussed graphically. Furthermore the solutions for the special case for p = 1 is also presented. M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1151 2. Formation of the problem Suppose that an incompressible second grade MHD fluid at rest is situated in the annular region between two infinite coaxial circular cylinders of radii R1 and R2(> R1). At time t = 0+ the inner cylinder begins to rotate around its axis and to slide along the same axis due to the time-dependent shear stresses Figure 1: Geometry of the problem for rotational and translational flows of fluid through an annulus τω(R1, t) = f1 t p and τv(R1, t) = g1 t p, where p ≥ 0, (1) and the outer cylinder also starts to rotate around its axis and slide along same axis with rotational velocity f2 t p and translated velocity g2 t p, where f1, f2, g1 and g2 are constants. Due to the shear, the fluid is gradually moved its velocity and extra stress tensor are considered to be V = V(r, t) = w(r, t)eθ + v(r, t)ez, S = S(r, t), (2) where eθ and ez are unit vectors in the θ and z-directions of the cylindrical coordinate system r, θ and z. The governing equations are given by [11, 19] ∂ ∂t ω(r, t) = ( ϑ+ α ∂ ∂t )( ∂2 ∂r2 + ∂ ∂r − 1 r2 ) ω(r, t)−M ω(r, t)− Φ ( ϑ+ α ∂ ∂t ) ω(r, t); (3) where r ∈ [R1, R2], t > 0, M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1152 ∂ ∂t v(r, t) = ( ϑ+ α ∂ ∂t )( ∂2 ∂r2 + 1 r ∂ ∂r ) v(r, t)−M v(r, t)− Φ ( ϑ+ α ∂ ∂t ) v(r, t); (4) where r ∈ [R1, R2], t > 0, τω = ( µ+ α1 ∂ ∂t )( ∂ ∂r − 1 r ) ω(r, t), (5) τv = ( µ+ α1 ∂ ∂t ) ∂v(r, q) ∂r , (6) where τω = srθ and τv = srz are the shear stresses that are different of zero and ν = µ/ρ is the kinematic viscosity, µ is the dynamic viscosity, ρ is the density of the fluid and Φ = φ k and M = σB2 o ρ are magnetic and porosity constants, where φ is the porosity and κ is the permeability of the porous medium, B0 is the magnitude of applied magnetic field and σ is the electrically conductively of fluid, while the appropriate initial and boundary conditions are w(r, 0) = ∂w(r, 0) ∂t = v(r, 0) = ∂v(r, 0) ∂t = 0 and τ1(r, 0) = τ2(r, 0) = 0; r ∈ [R1, R2], (7) respectively, τω(R1, t) = ( µ+ α1 ∂ ∂t )( ∂ ∂r − 1 r ) ω(r, t) ∣∣∣∣ r=R1 = f1 t p, t > 0, (8) and τv(R1, t) = ( µ+ α1 ∂ ∂t )( ∂ ∂r − 1 r ) v(r, t) ∣∣∣∣ r=R1 = g1 t p, t > 0, (9) ω(R2, t) = f2 t p and v(R2, t) = g2t p; t > 0. (10) In order to solve this problem we shall use Laplace Transforms and the finite Hankel trans- forms. M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1153 2.1. Estimation of the velocity field Applying the Laplace transform to eqs. (3) and (4) and having in mind the initial conditions (7), since the image functions ω(r, q) and v(r, q) be inverse Laplace transform, we find that as q ω(r, q) = ϑ ( ∂2 ∂r2 + 1 r ∂ ∂r − 1 r2 ) ω(r, q) + α ( ∂2 ∂r2 + 1 r ∂ ∂r − 1 r2 ) q ω(r, q) −M ω(r, q)− Φϑ− Φα q ω(r, q); r ∈ (R1, R2), (11) q v(r, q) = ϑ ( ∂2 ∂r2 + 1 r ∂ ∂r − 1 r2 ) v(r, q) + α ( ∂2 ∂r2 + 1 r ∂ ∂r − 1 r2 ) q v(r, q) −M v(r, q)− Φϑ− Φα q v(r, q); r ∈ (R1, R2), (12) Similarly applying Laplace transform on eqs. (8), (9) and (10), we get ( ∂ ∂r − 1 r ) ω(r, t) ∣∣∣∣ r=R1 = f1 p ! qp+1 (µ+ α1 q) ; t > 0, (13) and ∂ ∂r v(r, q) ∣∣∣∣ r=R1 = g1 p ! qp+1 (µ+ α1 q) ; t > 0, (14) also ω(R2, t) = f2 p ! qp+1 and v(R2, t) = g2 p ! qp+1 . (15) The Hankel transformation with respect to r is defined as [2, 3] wH(r, s) = ∫ R2 R1 rw(r, s)Bw(r, rm)dr, m = 1, 2, 3, ... (16) vH(r, s) = ∫ R2 R1 rv(r, s)Bv(r, rn)dr; n = 1, 2, 3, ... (17) M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1154 where Bw(r, rm) = J1(rrm)Y2(R1rm)− J2(R1rm)Y1(rrm), Bv(r, rn) = J0(rrn)Y1(R1rn)− J1(R1rn)Y0(rrn), also ∫ R2 R1 r ( ∂2 ∂r2 + 1 r ∂ ∂r − 1 r2 ) w(r, s)Bw(r, rm)dr = −r2 mwHm + 2 πrm ( ∂ ∂r − 1 r ) w(r, s) ∣∣∣∣ r=R1 +rmR2w(r, s)Dw(R1, R2, rm) (18) ∫ R2 R1 r ( ∂2 ∂r2 + 1 r ∂ ∂r ) v(r, t)Bv(r, rn)dr = −r2 nvHn + 2 πrn ∂ ∂r v(r, s) ∣∣∣∣ r=R1 +rnR2v(r, s)Dv(R1, R2, rn) (19) where Dw(R1, R2, rm) = J2(R2rm)Y2(R1rm)− J2(R1rm)Y2(R2rm), Dv(R1, R2, rn) = J1(R2rn)Y1(R1rn)− J1(R1rn)Y1(R2rn). Multiplying Eqs. (11) and (12) by rBw(rrm) and rBv(rrn), respectively, integrating the results with respect to r from R1 to R2 and using the boundary conditions (13), (14) and (15) and the identities (18), (19), we find that ωH(r, q) = 2f1p ! π rm qp+1ρ (ϑ+ αq) ϑ+ αq q +M + Φ(ϑ+ αq) + r2 m(ϑ+ αq) +f2R2 rm p! Dω(R1, R2, rm) 1 qp+1 ϑ+ αq q +M + Φ(ϑ+ αq) + r2 m(ϑ+ αq) , (20) vH(r, q) = 2g1p ! π rn qp+1ρ (ϑ+ αq) ϑ+ αq q +M + Φ(ϑ+ αq) + r2 n(ϑ+ αq) M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1155 +g2R2 rn p! Dv(R1, R2, rn) 1 qp+1 ϑ+ αq q +M + Φ(ϑ+ αq) + r2 n(ϑ+ αq) . (21) After modifying the aforesaid outcomes (20) and (21) in more appropriate interchangeable pattern, we get ωH(r, q) = 2f1 π µ r3 m p ! qp+1 − 2f1 π µ r3 m p ! qp [1 + α(Φ + r2 m)] + (M + Φϑ)q−1 [1 + α(Φ + r2 m)] + [M + ϑ(Φ + r2 m)] +f2R2 rm Dω(R1, R2, rm) [ p! qp+1 + p ! qp+1 [M + ϑ(Φ + r2 m − 1)] + [1 + α(Φ + r2 m − 1)]q [1 + α(Φ + r2 m)]q + [M + ϑ(Φ + r2 m)] ] , (22) vH(r, q) = 2g1 π µ r3 n p ! qp+1 − 2g1 π µ r3 n p ! qp [1 + α(Φ + r2 n)] + (M + Φϑ)q−1 [1 + α(Φ + r2 n)] + [M + ϑ(Φ + r2 n)] +g2R2 rn Dv(R1, R2, rn) [ p! qp+1 + p ! qp+1 [M + ϑ(Φ + r2 n − 1)] + [1 + α(Φ + r2 n − 1)]q [1 + α(Φ + r2 n)]q + [M + ϑ(Φ + r2 n)] ] . (23) Additionally, we have the relation of discrete inverse Hankel transform, and apply the discrete inverse Hankel transform formulae to (22) and (23), then we get the equations ω(r, q) = f1 2µ ( R1 R2 )2( r − R2 2 r ) p ! qp+1 − f1π µ ∞∑ m=1 J2 1 (R2rm)Bw(rrm) rm[J2 2 (R1rm)− J2 1 (R2rm)] 1 1 + α(Φ + r2 m) × [ p ! qp 1 + α(Φ + r2 m) q + M+ϑ(Φ+r2m) 1+α(Φ+r2m) + p ! qp+1 M + Φ ϑ q + M+ϑ(Φ+r2m) 1+α(Φ+r2m) ] + f2R2π 2 2 ∞∑ m=1 r3 mJ 2 1 (R2rm)Bw(rrm) Dω(R1, R2, rm) J2 2 (R1rm)− J2 1 (R2rm) × [ p! qp+1 − 1 1 + α(Φ + r2 m) { p ! qp+1 M + ϑ(Φ + r2 m − 1) q + M+ϑ(Φ+r2m) 1+α(Φ+r2m) + p ! qp 1 + α(Φ + r2 m − 1) q + M+ϑ(Φ+r2m) 1+α(Φ+r2m) }] , (24) v(r, t) = g1 2µ R1 ln ( r R2 ) p ! qp+1 − g1π µ ∞∑ n=1 J2 0 (R2rn)Bv(rrn) rn[J2 1 (R1rn)− J2 0 (R2rn)] 1 1 + α(Φ + r2 n) M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1156 × [ p ! qp 1 + α(Φ + r2 n) q + M+ϑ(Φ+r2n) 1+α(Φ+r2n) + p ! qp+1 M + Φ ϑ q + M+ϑ(Φ+r2n) 1+α(Φ+r2n) ] + g2R2π 2 2 ∞∑ n=1 r3 nJ 2 0 (R2rn)Bv(rrn) Dv(R1, R2, rn) J2 1 (R1rn)− J2 0 (R2rn) × [ p! qq+1 − 1 1 + α(Φ + r2 n) { p ! qp+1 M + ϑ(Φ + r2 n − 1) q + M+ϑ(Φ+r2n) 1+α(Φ+r2n) + p ! qp 1 + α(Φ + r2 n − 1) q + M+ϑ(Φ+r2n) 1+α(Φ+r2n) }] . (25) In order to get the result of velocity field for rotational and longitude cases as ω(r, t) = L[ω(r, q)] and v(r, t) = L[v(r, q)] respectively, applying the convolution theorem of inverse Laplace transformation on eqs. (24) and (25), we get the velocity field for rotational case ω(r, t) = f1 2µ ( R1 R2 )2( r − R2 2 r ) tp − f1π µ ∞∑ m=1 J2 1 (R2rm)Bw(rrm) rm[J2 2 (R1rm)− J2 1 (R2rm)] 1 1 + α(Φ + r2 m) × ∫ t 0 (t− u)p [ p (1 + α(Φ + r2 m)) t− u +M + Φϑ ] Exp [ − M + ϑ (Φ + r2 m) 1 + α (Φ + r2 m) u ] du + f2R2π 2 2 ∞∑ m=1 r3 mJ 2 1 (R2rm)Bw(rrm) Dω(R1, R2, rm) J2 2 (R1rm)− J2 1 (R2rm) { tp − 1 1 + α(Φ + r2 m) ∫ t 0 (t− u)p × [ [M + ϑ(Φ + r2 m − 1)] + p[1 + α(Φ + r2 m − 1)] t− u ] Exp [ − M + ϑ (Φ + r2 m) 1 + α (Φ + r2 m) u ] du } , (26) and for longitude case, we have v(r, q) = g1 2µ R1 ln ( r R2 ) tp − g1π µ ∞∑ n=1 J2 0 (R2rn)Bv(rrn) rn[J2 1 (R1rn)− J2 0 (R2rn)] 1 1 + α(Φ + r2 n) × ∫ t 0 (t− u)p [ p (1 + α(Φ + r2 n)) t− u +M + Φϑ ] Exp [ − M + ϑ (Φ + r2 n) 1 + α (Φ + r2 n) u ] du + g2R2π 2 2 ∞∑ n=1 r3 nJ 2 0 (R2rn)Bv(rrn) Dv(R1, R2, rn) J2 1 (R1rn)− J2 0 (R2rn) { tp − 1 1 + α(Φ + r2 n) ∫ t 0 (t− u)p × [ [M + ϑ(Φ + r2 n − 1)] + p[1 + α(Φ + r2 n − 1)] t− u ] Exp [ − M + ϑ (Φ + r2 n) 1 + α (Φ + r2 n) u ] du } . (27) M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1157 2.2. Estimation of the shear stresses In order to determine the shear stresses τω(r, t) and τv(r, t), with the initial conditions, Taking eqs. (5) and (6) and applying the Laplace transformation, we get τω(r, q) = (µ+ α1 q) ( ∂ ∂r − 1 r ) ω(r, q), (28) and τv(r, q) = (µ+ α1 q) ∂v(r, q) ∂r . (29) Now taking eqs. (20) and (21), after modifying the outcomes in more appropriate inter- changeable pattern for satisfying the initial and boundary conditions and applying the inverse Hankel transform, then we get the results of ( ∂ ∂r − 1 r ) ω(r, q) = f1 ρ ( R1 r )2 p ! qp+1(ϑ+ αq) + f1π ρ × ∞∑ m=1 J2 1 (R2rm)Bw(rrm) rm[J2 2 (R1rm)− J2 1 (R2rm)] 1 ϑ+ αq p ! qp { 1 + [M + Φ(ϑ+ αq)]q−1 [1 + α(Φ + r2 m)]q + [M + ϑ(Φ + r2 m)] } − f2R2π 2 2 × ∞∑ m=1 r3 mJ 2 1 (R2rm)Bw(rrm) Dω(R1, R2, rm) J2 2 (R1rm)− J2 1 (R2rm) { (ϑ+ αq) [1 + α(Φ + r2 m)]q + [M + ϑ(Φ + r2 m)] } ,(30) ∂v(r, q) ∂r = g1 ρ R1 ln ( r R2 ) p ! qp+1(ϑ+ αq) + g1π ρ × ∞∑ n=1 J2 0 (R2rn)Bv(rrn) rn[J2 1 (R1rn)− J2 0 (R2rn)] 1 ϑ+ αq p ! qp { 1 + [M + Φ(ϑ+ αq)]q−1 [1 + α(Φ + r2 n)]q + [M + ϑ(Φ + r2 n)] } − g2R2π 2 2 × ∞∑ n=1 r3 nJ 2 0 (R2rn)Bv(rrn) Dv(R1, R2, rn) J2 1 (R1rn)− J2 0 (R2rn) { (ϑ+ αq) [1 + α(Φ + r2 n)]q + [M + ϑ(Φ + r2 n)] } . (31) Put (30) and (31) in the eqs. (28) and (29) respectively, then the sample equivalent form, which are M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1158 τω(r, q) = f1 ( R1 r )2 p ! qp+1 + f1π ∞∑ m=1 J2 1 (R2rm)Bw(rrm) J2 2 (R1rm)− J2 1 (R2rm) 1 1 + α(Φ + r2 m) × { (p− 1) ! qp p q + M+ϑ(Φ+r2m) 1+α(Φ+r2m) + p ! qp+1 M + Φ ϑ q + M+ϑ(Φ+r2m) 1+α(Φ+r2m) + (p− 1) ! qp Φαp q + M+ϑ(Φ+r2m) 1+α(Φ+r2m) } + f2R2π 2 2 ∞∑ m=1 r4 mJ 2 1 (R2rm)Bw(rrm) Dω(R1, R2, rm) J2 2 (R1rm)− J2 1 (R2rm) { p ! qp+1 ϑ2 q + M+ϑ(Φ+r2m) 1+α(Φ+r2m) + (p− 1) ! qp 2α ϑp q + M+ϑ(Φ+r2m) 1+α(Φ+r2m) + (p− 2) ! qp−1 α2p (p− 1) q + M+ϑ(Φ+r2m) 1+α(Φ+r2m) } , (32) τv(r, q) = g1 ( R1 r ) p ! qp+1 + g1π ∞∑ n=1 J2 0 (R2rn)Bv(rrn) J2 1 (R1rn)− J2 0 (R2rn) 1 1 + α(Φ + r2 n) × { (p− 1) ! qp p q + M+ϑ(Φ+r2n) 1+α(Φ+r2n) + p ! qp+1 M + Φ ϑ q + M+ϑ(Φ+r2n) 1+α(Φ+r2n) + (p− 1) ! qp Φαp q + M+ϑ(Φ+r2n) 1+α(Φ+r2n) } + g2R2π 2 2 ∞∑ n=1 r4 nJ 2 0 (R2rn)Bv(rrn) Dv(R1, R2, rn) J2 1 (R1rn)− J2 0 (R2rn) { p ! qp+1 ϑ2 q + M+ϑ(Φ+r2n) 1+α(Φ+r2n) + (p− 1) ! qp 2α ϑp q + M+ϑ(Φ+r2n) 1+α(Φ+r2n) + (p− 2) ! qp−1 α2p (p− 1) q + M+ϑ(Φ+r2n) 1+α(Φ+r2n) } . (33) After applying the convolution theorem of inverse Laplace transformation on eqs. (32) and (33), we get the shear stress for rotational case τω(r, t) = f1 ( R1 r ) tp + f1π ∞∑ m=1 J2 1 (R2 rm) Bω(r rm) J2 2 (R1 rm)− J2 1 (R2 rm) 1 1 + α(Φ + r2 m) ∫ t 0 (t− u)p M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1159 × [ p (1 + αΦ) t− u +M + Φϑ ] Exp [ − M + ϑ (Φ + r2 m) 1 + α (Φ + r2 m) u ] du− f2 π2 ρR2 2 × ∞∑ m=1 r4 m J 2 1 (R2 rm) Bω(r rm) Dw(R1, R2, rm) J2 2 (R1 rm)− J2 1 (R2 rm) 1 1 + α(Φ + r2 m) ∫ t 0 (t− u)p × [ ϑ2 + 2αϑP t− u + α2 p(p− 1) (t− u)2 ] Exp [ − M + ϑ (Φ + r2 m) 1 + α (Φ + r2 m) u ] du, (34) and for longitudinal case τv(r, t) = g1 ( R1 r ) tp + g1π ∞∑ n=1 J2 0 (R2 rn) Bv(r rn) J2 1 (R1 rn)− J2 0 (R2 rn) 1 1 + α(Φ + r2 n) ∫ t 0 (t− u)p × [ p (1 + αΦ) t− u +M + Φϑ ] Exp [ − M + ϑ (Φ + r2 n) 1 + α (Φ + r2 n) u ] du− g2 π2 ρR2 2 × ∞∑ n=1 r4 n J 2 0 (R2 rn) Bω(r rn) Dv(R1, R2, rn) J2 1 (R1 rn)− J2 0 (R2 rn) 1 1 + α(Φ + r2 n) ∫ t 0 (t− u)p × [ ϑ2 + 2αϑP t− u + α2 p(p− 1) (t− u)2 ] Exp [ − M + ϑ (Φ + r2 n) 1 + α (Φ + r2 n) u ] du. (35) 3. The limiting and special cases 3.1. Second grade with porous (without MHD effect) Making M→ 0 into (26), (27), (34) and (35), we obtain the velocity field components and shear stresses for Second grade fluid with porous effect and without MHD. 3.2. MHD Second grade (without porous effect) Putting Φ → 0 into resultant eqs. (26), (27), (34) and (35), we obtain the velocity field components and the shear stresses for MHD Second grade without porous effects. M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1160 3.3. Second grade fluid (without MHD and porous effects) Applying M→ 0 & Φ→ 0 into (26), (27), (34) and (35), we get the solutions for velocity fields components respectively the shear stresses for Second grade fluid without MHD and porous effects. 3.4. MHD Newtonian with porous effect Employing α→ 0 into eqs. (26), (27), (34) and (35), we get ω(r, t) = f1 2µ ( R1 R2 )2( r − R2 2 r ) tp − f1π µ ∞∑ m=1 J2 1 (R2rm)Bw(rrm) rm[J2 2 (R1rm)− J2 1 (R2rm)] ∫ t 0 (t− u)p [ p t− u +M + Φϑ ] ×Exp [ − {M + ϑ (Φ + r2 m)}u ] du+ f2R2π 2 2 ∞∑ m=1 r3 mJ 2 1 (R2rm)Bw(rrm) Dω(R1, R2, rm) J2 2 (R1rm)− J2 1 (R2rm) { tp − ∫ t 0 (t− u)p [ M + ϑ(Φ + r2 m − 1) + p t− u ] Exp [ − {M + ϑ (Φ + r2 m)}u ] du } , (36) v(r, q) = g1 2µ R1 ln ( r R2 ) tp − g1π µ ∞∑ n=1 J2 0 (R2rn)Bv(rrn) rn[J2 1 (R1rn)− J2 0 (R2rn)] ∫ t 0 (t− u)p [ p t− u +M + Φϑ ] ×Exp [ − {M + ϑ (Φ + r2 n)}u ] du+ g2R2π 2 2 ∞∑ n=1 r3 nJ 2 0 (R2rn)Bv(rrn) Dv(R1, R2, rn) J2 1 (R1rn)− J2 0 (R2rn) { tp − ∫ t 0 (t− u)p [ M + ϑ(Φ + r2 n − 1) + p t− u ] Exp [ − {M + ϑ (Φ + r2 n)}u ] du } , (37) the solution for the velocity field components, and τω(R1, t) = f1 ( R1 r ) tp + f1π ∞∑ m=1 J2 1 (R2 rm) Bω(r rm) J2 2 (R1 rm)− J2 1 (R2 rm) ∫ t 0 (t− u)p [ p t− u +M + Φϑ ] ×Exp [ − {M + ϑ (Φ + r2 m)}u ] du− f2 π2 ρR2ϑ 2 2 ∞∑ m=1 r4 m J 2 1 (R2 rm)Bω(r rm) Dω(R1, R2, rm) J2 2 (R1 rm)− J2 1 (R2 rm) M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1161 × ∫ t 0 (t− u)pExp [ − {M + ϑ (Φ + r2 m)}u ] du, (38) τv(R1, t) = g1 ( R1 r ) tp + g1π ∞∑ n=1 J2 0 (R2 rn) Bv(r rn) J2 1 (R1 rn)− J2 0 (R2 rn) ∫ t 0 (t− u)p [ p t− u +M + Φϑ ] ×Exp [ − {M + ϑ (Φ + r2 n)}u ] du− g2 π2 ρR2ϑ 2 2 ∞∑ n=1 r4 n J 2 0 (R2 rn) Bv(r rn) Dv(R1, R2, rn) J2 1 (R1 rn)− J2 0 (R2 rn) × ∫ t 0 (t− u)pExp [ − {M + ϑ (Φ + r2 n)}u ] du, (39) the shear stresses for MHD Newtonian with porous effect. 3.5. Newtonian with porous effect Making M → 0 in Eqs. (36), (37), (38) and (39), we get solutions for Newtonian fuid with porous effect and without MHD. 3.6. MHD Newtonian Similarly, putting M → 0 in Eqs. (36), (37), (38) and (39), we get solutions for MHD Newtonian fluid without porous effects. 3.7. Newtonian By vanishing as M→ 0 & Φ→ 0 into the Eqs. (36), (37), (38) and (39), of velocity field components and shear stresses, we get the Newtonian solutions, ω(r, t) = f1 2µ ( R1 R2 )2( r − R2 2 r ) tp − f1π µ ∞∑ m=1 J2 1 (R2rm)Bw(rrm) rm[J2 2 (R1rm)− J2 1 (R2rm)] ∫ t 0 (t− u)p−1 ×Exp(−ϑr2 mu)du+ f2R2π 2 2 ∞∑ m=1 r3 mJ 2 1 (R2rm)Bw(rrm) Dω(R1, R2, rm) J2 2 (R1rm)− J2 1 (R2rm) { tp − ∫ t 0 (t− u)p [ ϑ(r2 m − 1) + p t− u ] Exp(−ϑ r2 m u )du } , (40) M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1162 v(r, q) = g1 2µ R1 ln ( r R2 ) tp − g1p µ ∞∑ n=1 J2 0 (R2rn)Bv(rrn) rn[J2 1 (R1rn)− J2 0 (R2rn)] ∫ t 0 (t− u)p−1 ×Exp(−ϑ r2 n u )du+ g2R2π 2 ∞∑ n=1 r3 nJ 2 0 (R2rn)Bv(rrn) Dv(R1, R2, rn) J2 1 (R1rn)− J2 0 (R2rn) { tp − ∫ t 0 (t− u)p [ ϑ(r2 n − 1) + p t− u ] Exp(−ϑ r2 n u )du } , (41) τω(R1, t) = f1 ( R1 r ) tp + f1π ∞∑ m=1 J2 1 (R2 rm) Bω(r rm) J2 2 (R1 rm)− J2 1 (R2 rm) ∫ t 0 (t− u)p ( p t− u ) ×Exp(−ϑr2 mu)du− f2 π2 ρR2ϑ 2 2 ∞∑ m=1 r4 m J 2 1 (R2 rm) Bω(r rn) Dω(R1, R2, rm) J2 2 (R1 rn)− J2 1 (R2 rm) × ∫ t 0 (t− u)p Exp(−ϑr2 mu)du, (42) τv(R1, t) = g1 ( R1 r ) tp + g1π ∞∑ n=1 J2 0 (R2 rn) Bv(r rn) J2 1 (R1 rn)− J2 0 (R2 rn) ∫ t 0 (t− u)p ( p t− u ) ×Exp(−ϑr2 nu)du− g2 π2 ρR2ϑ 2 2 ∞∑ n=1 r4 n J 2 0 (R2 rn) Bv(r rn)Dv(R1, R2, rn) J2 1 (R1 rn)− J2 0 (R2 rn) × ∫ t 0 (t− u)p Exp(−ϑr2 nu)du. (43) M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1163 3.8. The special solution for p = 1 Finally, switch p = 1 into the resultant equations of velocity field components as well as shear stress (26), (27), (34) and (35), we get longitudinal and rotational cases result by solving the integral from 0 to t ω(r, t) = f1 2µ ( R1 R2 )2( r − R2 2 r ) t− f1[M + ϑ (Φ + r2 m)] µ[1 + α (Φ + r2 m)]3 ∞∑ m=1 J2 1 (R2rm)Bw(rrm) rm[J2 2 (R1rm)− J2 1 (R2rm)] × [ ϑr2 m[1 + α (Φ + r2 m)] ( 1−Exp [ − M + ϑ (Φ + r2 m) 1 + α (Φ + r2 m) ]) + t[M(M + ϑ(2Φ + r2 m)) + Φϑ2(Φ + r2 m)] ] + f2R2π 2[1 + α(Φ + r2 m)] ∞∑ m=1 r3 mJ 2 1 (R2rm)Bw(rrm) Dω(R1, R2, rm) J2 2 (R1rm)− J2 1 (R2rm) { t− 1 [M + ϑ (Φ + r2 m)]2 [ ϑ −α[M + ϑ(Φ + r2 m − 1)]− tϑ[M + ϑ(Φ + r2 m)(Φ + r2 m − 1)] ]( 1−Exp [ − M + ϑ (Φ + r2 m) 1 + α (Φ + r2 m) ]) +t [ M2 + ϑ2(Φ + r2 m)(Φ + r2 m − 1)Exp [ − M + ϑ (Φ + r2 m) 1 + α (Φ + r2 m) ]]} , (44) v(r, q) = g1 2µ R1 ln ( r R2 ) t− g1[M + ϑ (Φ + r2 n)] µ[1 + α (Φ + r2 n)]3 ∞∑ n=1 J2 0 (R2rn)Bv(rrn) rn[J2 1 (R1rn)− J2 0 (R2rn)] × [ ϑr2 n[1 + α (Φ + r2 n)] ( 1−Exp [ − M + ϑ (Φ + r2 n) 1 + α (Φ + r2 n) ]) + t[M(M + ϑ(2Φ + r2 n)) + Φϑ2(Φ + r2 n)] ] + g2R2π 2[1 + α(Φ + r2 n)] ∞∑ n=1 r3 nJ 2 1 (R2rn)Bv(rrn) Dv(R1, R2, rn) J2 2 (R1rn)− J2 1 (R2rn) { t− 1 [M + ϑ (Φ + r2 n)]2 [ ϑ −α[M + ϑ(Φ + r2 n − 1)]− tϑ[M + ϑ(Φ + r2 n)(Φ + r2 n − 1)] ]( 1−Exp [ − M + ϑ (Φ + r2 n) 1 + α (Φ + r2 n) ]) +t [ M2 + ϑ2(Φ + r2 n)(Φ + r2 n − 1)Exp [ − M + ϑ (Φ + r2 n) 1 + α (Φ + r2 n) ]]} , (45) M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1164 τω(R1, t) = f1 ( R1 r ) t+ f1 ∞∑ m=1 J2 1 (R2 rm) Bω(r rm) J2 2 (R1 rm)− J2 1 (R2 rm) 1 1 + α(Φ + r2 m) [ [1 + αΦ] 1 + α (Φ + r2 m) M + ϑ (Φ + r2 m) × { 1−Exp [ − M + ϑ (Φ + r2 m) 1 + α (Φ + r2 m) t ]} + [M + Φϑ] 1 + α (Φ + r2 m) [M + ϑ (Φ + r2 m)]2 { t[M + ϑ (Φ + r2 m)] −[1 + α (Φ + r2 m)] [ 1−Exp [ − M + ϑ (Φ + r2 m) 1 + α (Φ + r2 m) t ]]}] − f2 π ρR2 2[1 + α(Φ + r2 m)] × ∞∑ m=1 r4 m J 2 1 (R2 rm) Bω(r rm) Dw(R1, R2, rm) J2 2 (R1 rm)− J2 1 (R2 rm) [ ϑ2 1 + α (Φ + r2 m) [M + ϑ (Φ + r2 m)]2 { t[M + ϑ (Φ + r2 m)] −[1 + α (Φ + r2 m)] [ 1−Exp [ − M + ϑ (Φ + r2 m) 1 + α (Φ + r2 m) t ]]} + 2αϑ 1 + α (Φ + r2 m) M + ϑ (Φ + r2 m) × { 1−Exp [ − M + ϑ (Φ + r2 m) 1 + α (Φ + r2 m) t ]}] , (46) τv(R1, t) = g1 ( R1 r ) t+ g1 ∞∑ n=1 J2 0 (R2 rn) Bv(r rn) J2 1 (R1 rn)− J2 0 (R2 rn) 1 1 + α(Φ + r2 n) [ [1 + αΦ] 1 + α (Φ + r2 n) M + ϑ (Φ + r2 n) × { 1−Exp [ − M + ϑ (Φ + r2 n) 1 + α (Φ + r2 n) t ]} + [M + Φϑ] 1 + α (Φ + r2 n) [M + ϑ (Φ + r2 n)]2 { t[M + ϑ (Φ + r2 n)] −[1 + α (Φ + r2 n)] [ 1−Exp [ − M + ϑ (Φ + r2 n) 1 + α (Φ + r2 n) t ]]}] − g2 π ρR2 2[1 + α(Φ + r2 n)] × ∞∑ n=1 r4 n J 2 0 (R2 rn) Bv(r rn) Dv(R1, R2, rn) J2 1 (R1 rn)− J2 0 (R2 rn) [ ϑ2 1 + α (Φ + r2 n) [M + ϑ (Φ + r2 n)]2 { t[M + ϑ (Φ + r2 n)] −[1 + α (Φ + r2 n)] [ 1−Exp [ − M + ϑ (Φ + r2 n) 1 + α (Φ + r2 n) t ]]} + 2αϑ 1 + α (Φ + r2 n) M + ϑ (Φ + r2 n) × { 1−Exp [ − M + ϑ (Φ + r2 n) 1 + α (Φ + r2 n) t ]}] . (47) M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1165 4. Numerical results and discussion In this part of the article, the obtained exact solutions are studied numerically in order to determine the effects of several involved parameter such as material parameter α, kinematic viscosity ν, magnetic parameter M , porosity parameter Φ, parameter p, time t and radial distance r. Comparison between two different fluid are also presented for understanding the pre- sented flow problem. Figure 2: Profiles of the velocity field components for MHD Second grade fluid with porous effect w(r, t), u(r, t) and the shear stresses τw(r, t), τv(r, t) given by eqs. (26), (27), (34) and (35), for R1 = 0.1, R2 = 0.3, f1 = g1 = 0.3, f2 = g2 = 0.2, ν = 0.6355, µ = 12.26, α = 0.2, M = 2, Φ = 2, p = 1 and different values of time t. M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1166 Figs. 2 exhibit the influence of time on the fluid motion. The velocity components w(r, t) , v(r, t) and rotational shear stresses τw(r, t) are increasing function of time t. How- ever the amplitude of the longitudinal shear stress τv(r, t) decreases with time. These numerical values demonstrate the influence of above parameter on velocity com- ponents and shear stresses profiles for the flow generated through the inner cylinder that is subject to a rotational and longitudinal time dependent shear stresses and the outer cylinder subject to a rotational and longitudinal velocities along its axes. Figure 3: Profiles of the velocity field components for MHD Second grade fluid with porous effect w(r, t), u(r, t) and the shear stresses τw(r, t), τv(r, t) given by eqs. (26), (27), (34) and (35), for R1 = 0.1, R2 = 0.3, f1 = g1 = 0.3, f2 = g2 = 0.2, ν = 0.6355, µ = 12.26, α = 0.2, M = 2, Φ = 2, p = 1 and different values of radius r. The influence of the annular distance form R1 to R2 is shown in Figs. 3. It is found M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1167 that rotational pair w(r, t), τw(r, t) are increases and longitudinal pair v(r, t), τv(r, t) de- creases along the annular distance. These diagrams of the velocity field components w(r, t) , v(r, t) and the shear stress τw(r, t), τv(r, t) were presented against r for different values of t and of the above men- tioned parameter. All these comments are demonstrated through absolute sense where in the diagrams negative values are appeared. Figure 4: Profiles of the velocity field components for MHD Second grade fluid with porous effect w(r, t), u(r, t) and the shear stresses τw(r, t), τv(r, t) given by eqs. (26), (27), (34) and (35), for R1 = 0.1, R2 = 0.3, f1 = g1 = 0.3, f2 = g2 = 0.2, ν = 0.6355, µ = 12.26, M = 2, Φ = 2, p = 1, t = 1s and different values of material parameter α. Figs. 4 are plotted to show the effects of material parameter on velocities and shear M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1168 stresses profiles. It can be observed that both velocity components and rotational shear stress are decreasing function and longitudinal shear stress is decreasing function of the parameter α when other parameter are fixed. Viscous effects are important for both Newtonian and non-Newtonian fluids. In Figs. 5, the velocities and shear stresses profiles are depicted for different values of the kinematic viscosity ν. Here the values of kinematic viscosity are chosen 0.3, 0.5 and 0.9. It is observed that velocity components and shear stresses are decay with enhancing the viscous effects, which is natural phonomania. Figure 5: Profiles of the velocity field components for MHD Second grade fluid with porous effect w(r, t), u(r, t) and the shear stresses τw(r, t), τv(r, t) given by eqs. (26), (27), (34) and (35), for R1 = 0.1, R2 = 0.3, f1 = g1 = 0.3, f2 = g2 = 0.2, ρ = 19.292, α = 0.2, M = 2, Φ = 2, p = 1, t = 0.2s and different values of kinematic viscosity ν. M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1169 Figs. 6 present the nature of velocities and shear stresses profiles respectively for the variation of magnetic parameter M . The range of the magnetic parameter is taken from 0.1 to 12. It come to the observation that the higher the value of magnetic parameter rise the velocities shear stresses values. Figure 6: Profiles of the velocity field components for MHD Second grade fluid with porous effect w(r, t), u(r, t) and the shear stresses τw(r, t), τv(r, t) given by eqs. (26), (27), (34) and (35), for R1 = 0.1, R2 = 0.3, f1 = g1 = 0.3, f2 = g2 = 0.2, ν = 0.6355, µ = 12.26, α = 0.2, Φ = 2, p = 1, t = 0.5s and different values of magnetic parameter M . Figs. 7 illustrates the effects of the porosity parameter Φ on fluid motion. These fig- ures show the variations of velocities as well as shear stresses profiles for increasing values of porosity parameter Φ when other parameters are kept constant. M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1170 Figs. 7 (a), (b) clearly indicates that porosity parameter have neglige effects on veloc- ity filed components. However share stresses have strong influence on the shear stresses profiles. The rota- tional shear stress τw(r, t) decreases and longitudinal shear stress τv(r, t) decreases with the strength of porosity parameter. Figure 7: Profiles of the velocity field components for MHD Second grade fluid with porous effect w(r, t), u(r, t) and the shear stresses τw(r, t), τv(r, t) given by eqs. (26), (27), (34) and (35), for R1 = 0.1, R2 = 0.3, f1 = g1 = 0.3, f2 = g2 = 0.2, ν = 0.6355, µ = 12.26, α = 0.2, M = 2, p = 1, t = 0.2s and different values of porosity Φ. One of the significant parameter for the flow problem is considered, the power p, that appeared in the boundary conditions (22)-(24). M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1171 We consider only integer values of power p = 1, 2, 3 in Figs 8. It is brought to knowl- edge that increase in the power parameter p causes a increase in rotational pair and decrees the longitudinal pair of quantities for the flow problem under consideration. Figure 8: Profiles of the velocity field components for MHD Second grade fluid with porous effect w(r, t), u(r, t) and the shear stresses τw(r, t), τv(r, t) given by eqs. (26), (27), (34) and (35), for R1 = 0.1, R2 = 0.3, f1 = g1 = 0.3, f2 = g2 = 0.2, ν = 0.6355, µ = 12.26, α = 0.2, M = 2, Φ = 2, t = 1s and different values of the power p. In the end, for comparison the velocities and shear stresses profiles corresponding to two models MHD second grade and MHD Newtonian in porous medium are together de- picted in Figs. 9 for same value of t, α, µ, ν, M and Φ . It is found that MHD Newtonian fluid flows faster than the MHD second grade fluid in porous medium. However the shear stresses are larger of MHD second grade fluid are larger than the MHD Newtonian fluid. M. Jamil, M. Zafarullah / Eur. J. Pure Appl. Math, 12 (3) (2019), 1149-1175 1172 Figure 9: Profiles of the velocity field components for MHD Second grade fluid and MHD Newtonian with porous effect w(r, t), u(r, t) and the shear stresses τw(r, t), τv(r, t) given by eqs. (26), (27), (34), (35), (36), (37), (38) and (39) for R1 = 0.1, R2 = 0.3, f1 = g1 = 0.3, f2 = g2 = 0.2, ν = 0.6355, µ = 12.26, α = 0.2, M = 2, Φ = 2, p = 1 and t = 1s. These comparison is completly agreement the Figs. 4 which also shows that when α→ 0 the MHD second grade fluid flows fast in porous medium. SI units are considered for material parameters in Figs. 2 - 9 and (2m−1)π/2(R2−R1) and (2n−1)π/2(R2−R1) are used as approximated values of the roots rm and rn. REFERENCES 1173 5. Concluding remarks The purpose of this chapter is to present the unsteady flow of second grade fluid be- tween two infinite coaxial circular porous cylinders in the presence of magnetic field is studied by techniques of the discrete Laplace and finite Hankel transforms. The flow of the fluid is generated by the inner cylinder that is subject to a rotational and longitudi- nal time dependent shear stresses and the outer cylinder also subject to a rotational and longitudinal time dependent velocities along its axes when time t = 0+. The exact analyt- ical solutions are obtained for the velocity field components w(r, t) , v(r, t) and the shear stresses τw(r, t), τv(r, t) in series form and presented in term of convolution product of Laplace transforms, satisfy all imposed initial and boundary conditions. The correspond- ing solutions for, second grade, MHD Newtonian and Newtonian fluids in porous medium are also obtained from general solutions for M → 0 and α→ 0. In the special case when Φ→ 0, the general solution reduces to the solutions for second grade and Newtonian fluids in the absence of porous medium. The following conclusions are extracted for this study: The general solutions for velocities and shear stresses are written in simple forms in term of convolution product of Laplace transforms. In absolute sense velocity filed components are increasing functions of time. The rotational velocity increases and longitudinal velocity decreases along radial direction. The effects of material parameter α and kinematic viscosity ν are opposite on ve- locity profiles. For instance rotational velocity is increasing and longitudinal velocity is decreasing functions of α. The increasing values of magnetic values enhance the velocity and shear stress quan- tities. The porosity parameter Φ have ignorable impact on the velocity profiles. When both porous cylindrical helically moved, the MHD Newtonian fluid flows faster than MHD second grade fluid. Acknowledgements The authors, Dr. Muhammad Jamil and Muhammad Zafarullah, are thoroughly obliged to the Department of Mathematics, NED University of Engineering & Technology, Karachi-75270, Pakistan and also would like to express their gratitude to Higher Education Commission of Pakistan for assisting and facilitating this research work. References [1] M. R. Salimpour A. F. Dizaji and F. Jam. 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