# Applications of Magnetohydrodynamic Couple Stress Fluid Flow between Two Parallel Plates with Three Different Kernels

Authors: Imran Siddique; Ali Akgül; Hafte Amsalu Kahsay; Teklay Hailay Tsegay; Kahsay Godifey Wubneh
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we investigate the implementations of newly introduced nonlocal differential operators as convolution of power law, exponential decay law, and the generalized Mittag-Leffler law with fractal derivative in fluid dynamics. The new operators are referred as fractal-fractional differential operators. The governing equations for the problem are constructed with the fractal-fractional differential operators. We present the stability analysis and the error analysis.


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## Body

## 1. Introduction

Magnetohydrodynamics (MHD) deals with the study of the motion of electrically conducting fluids in the presence of the magnetic field. MHD flow has significant importance applications between infinite parallel plates in various areas such as geophysical, astrophysical, and metallurgical processing, MHD generators, pumps, geothermal reservoirs, polymer technology, and mineral industries [1–6]. In last few decades, fractional calculus has taken much interest in many fields [7, 8]. There are many definitions for the fractional derivative operators, and among them are Caputo-Fabrizio (CF) [9] and Atangana and Baleanu (AB) [10] definitions of fractional derivatives with a nonlocal and nonsingular kernels having all the characteristics of the old definitions [7, 11–23]. Farman et al. [24] have analyzed the numerical solution of SEIR Epidemic model of measles with noninteger time fractional derivatives by using the Laplace Adomian decomposition method. Ghanbari and Djilali [25] have taken mathematical analysis of a fractional-order predator-prey model with prey social behavior and infection developed in predator population. Ghanbari and Atangana [26, 27] have given the new edge detecting techniques based on fractional derivatives with nonlocal and nonsingular kernels. Recently, another idea of differentiation has been proposed by Atanagna [28].We organize our manuscript as follows. We present the main definitions in Section2. We construct the problem formulation in Section 3. We present the analysis of the model with the power law kernel in Section 4. We give the analysis of the model with the exponential decay kernel in Section 5. We discuss the analysis of the model with the Mittag-Leffler kernel in Section 6. We present the error analysis in Section 7. We give the conclusion in the last section.


## 2. Preminaries

Definition 1.
Assume thatgς is a continuous function in the c11,d11 and fractal differentiable on c11,d11 with order η, then the fractal-fractional derivative of g of order λ in Riemann-Liouville sense with power law kernel is introduced as [29]
(1)DbFFPtη,λgς=1Γ1−ηddςλ∫bςgxς−x−ηdx,0<η,λ≤1,where
(2)dgxdxλ=limς⟶xgς−gxςλ−xλ.Definition 2.
Assume thatgς is a continuous function in the c11,d11 and fractal differentiable on c11,d11 with order η, then the fractal-fractional derivative of g of order λ in Riemann-Liouville sense with the exponential decay kernel is introduced as [29]
(3)DbFFEtη,λgς=Mη1−ηddςλ∫bςgxexp−η1−ης−xdx,0<η,λ≤1.Definition 3.
Assume thatgς is a continuous function in the c11,d11 and fractal differentiable on c11,d11 with order η, then the fractal-fractional derivative of g of order λ in Riemann-Liouville sense with the generalized Mittag-Leffler kernel is introduced as [29]
(4)DbFFEtη,λgς=ABη1−ηddςλ∫bςgxEη−η1−ης−xηdx,0<η,λ≤1.


## 3. Problem Formulation

We consider(5)ρ∂u11′ξ11′,τ11′∂τ11′=μ∂2u11′ξ11′,τ11′∂ξ11′2−η∂4u11′ξ11′,τ11′∂ξ11′4−σB02u11′ξ11′,τ11′,(6)u11′ξ11′,0=0,forallξ11′,initialcondition(7)u11′0,τ11′=0,u11′d,τ11′=0,0≤ξ11′≤h,(8)∂2u11′ξ11′,τ11′∂ξ11′2=0,atξ11′=0andξ11′=hforanyτ11′>0,(9)v=u11′U0,t=τ11′U0h,y=ξ11′h,into Eqs. (5)-(8), and we obtain
(10)∂vy,t∂t=1Re∂2vy,t∂y2−∂4vy,t∂y4−M11vy,t,(11)vy,0=0,v0,t=0,v1,t=1,(12)∂2vy,t∂y2=0,aty=0andy=1foranyt>0,where M11=σB02d2/μ is the magnetic field parameter, and Re=ρU0d/μ is the Reynold number and h2=η/μ. We demonstrate the geometry of the physical model in Figure 1.Figure 1
Geometry of the physical model.


## 4. Solution of the Problem with the Power Law Kernel

We take into consideration the Eq. (10) with fractal-fractional differential operator using Definition 1 of power law kernel as
(13)D0FFPtα1,β1vy,t=1Re∂2vy,t∂y2−∂4vy,t∂y4−M11vy,t,1Γ1−α1ddt∫0tvy,λt−λ−α1dλ=β1Retβ1−1∂2vy,t∂y2−∂4vy,t∂y4−M11vy,t.The, we get(14)vy,λ=β1ReΓα1∫0tλβ1−1∂2vy,λ∂y2−∂4vy,λ∂y4−M11vy,λt−λα1−1dλ.For simplicity, we take(15)Fy,λ=β1λβ1−1∂2vy,λ∂y2−∂4vy,λ∂y4−M11vy,λ,vy,λ=1ReΓα1∫0tFy,λt−λα1−1dλ.We discretize this equation atyi,t=tn+1 and get
(16)vyi,tn+1=1ReΓα1∫0tn+1Fyi,λtn+1−λα1−1dλ,vyi,tn+1=1ReΓα1∑j=0n∫tjtj+1Fyi,λtn+1−λα1−1dλ.We apply the two-step Lagrange polynomial as(17)pjλ=λ−tj−1tj−tj−1Fyi,tj−λ−tjtj−tj−1Fyi,tj−1.Thus, we will get(18)vyi,tn+1=1ReΓα1∑j=0n∫tjtj+1pjλtn+1−λα1−1dλ=∑j=0nhα1Fyi,tjReΓα1+2n+1−jα1n−j+2+α1−n−jα1n−j+2+2α1−∑j=0nhα1Fyi,tj−1ReΓα1+2n+1−jα1+1−n−jα1n−j+1+α1.We have(19)Fyi,tj=β1tjβ1−1vi+1j−2vij+vi−1jΔy2−vi+2j−4vi+1j+6vij−4vi−1j+vi−2jΔy4−M11vij.Then, we will obtain(20)vyi,tn+1=1ReΓα1∑j=0n∫tjtj+1pjλtn+1−λα1−1dλ=∑j=0nhα1β1tjβ1−1ReΓα1+2vi+1j−2vij+vi−1jΔy2−vi+2j−4vi+1j+6vij−4vi−1j+vi−2jΔy4−M11vij=×n+1−jα1n−j+2+α1−n−jα1n−j+2+2α1−∑j=0nhα1β1tj−1β1−1ReΓα1+2vi+1j−1−2vij−1+vi−1j−1Δy2−vi+2j−1−4vi+1j−1+6vij−1−4vi−1j−1+vi−2j−1Δy4−M11vij−1×n+1−jα1+1−n−jα1n−j+1+α1.We define(21)An,jα1,β1=hα1β1tjβ1−1Kn,jα1ReΓα1+2Δy2,Bn,jα1,β1=hα1β1tjβ1−1Kn,jα1ReΓα1+2Δy4,Cn,jα1,β1=hα1β1tjβ1−1M11Kn,jα1ReΓα1+2,Dn,j−1α1,β1=hα1β1tj−1β1−1γn,jα1ReΓα1+2Δy2,Bn,j−1α1,β1=hα1β1tj−1β1−1γn,jα1ReΓα1+2Δy4,Cn,j−1α1,β1=hα1β1tjβ1−1M11γn,jα1ReΓα1+2,Kn,jα1=n+1−jα1n−j+2+α1−n−jα1n−j+2+2α1,γn,jα1=n+1−jα1+1−n−jα1n−j+1+α1.Then, we get(22)vin+1=∑j=0nAn,jα1,β1vi+1j−2vij+vi−1j−Bn,jα1,β1vi+2j−4vi+1j+6vij−4vi−1j+vi−2j−Cn,jα1,β1vij−∑j=0nDn,j−1α1,β1vi+1j−1−2vij−1+vi−1j−1−En,j−1α1,β1vi+2j−1−4vi+1j−1+6vij−1−4vi−1j−1+vi−2j−1−Fn,j−1α1,β1vij−1.We chooseεin=δnexpikmy. Then, we have
(23)δn+1expikmy=An,nα1,β1δnexpikmy+Δy−2δnexpikmy+δnexpikmy−Δy−Bn,nα1,β1δnexpikmy+2Δy−4δnexpikmy+Δy+6δnexpikmy−4δnexpikmy−Δy+δnexpikmy−2Δy−Cn,nα1,β1δnexpikmy−Dn,n−1α1,β1δn−1expikmy+Δy−2δn−1expikmy+δn−1expikmy−Δy−En,n−1α1,β1δn−1expikmy+2Δy−4δn−1expikmy+Δy+6δn−1expikmy−4δn−1expikmy−Δy+δn−1expikmy−2Δy−Fn,n−1α1,β1δn−1expikmy+∑j=0n−1An,jα1,β1δjexpikmy+Δy−2δjexpikmy+δjexpikmy−Δy−Bn,jα1,β1δjexpikmy+2Δy−4δjexpikmy+Δy+6δjexpikmy−4δjexpikmy−Δy+δjexpikmy−2Δy−Cn,jα1,β1δnexpikmy−∑j=0n−1Dn,j−1α1,β1δj−1expikmy+Δy−2δj−1expikmy+δj−1expikmy−Δy−En,j−1α1,β1δj−1expikmy+2Δy−4δj−1expikmy+Δy+6δj−1expikmy−4δj−1expikmy−Δy+δj−1expikmy−2Δy−Fn,j−1α1,β1δj−1expikmy.After simplification, we obtain(24)δn+1=An,nα1,β1δnexpikmΔy−2δn+δnexp−ikmΔy−Bn,nα1,β1δnexp2ikmΔy−4δnexpikmΔy+6δn−4δnexpikmΔy+δnexp−2ikmΔy−Cn,nα1,β1δn−Dn,n−1α1,β1δn−1expikmΔy−2δn−1+δn−1expikmΔy−En,n−1α1,β1δn−1exp2ikmΔy−4δn−1expikmΔy+6δn−1−4δn−1expikmΔy+δn−1exp−2ikmΔy−Fn,n−1α1,β1δn−1+∑j=0n−1An,jα1,β1δjexpikmΔy−2δj+δjexp−ikmΔy−Bn,jα1,β1δjexp2ikmΔy−4δjexpikmΔy+6δj−4δjexp−ikmΔy+δjexp−2ikmΔy−Cn,jα1,β1δn−∑j=0n−1Dn,j−1α1,β1δj−1expikmΔy−2δj−1+δj−1exp−ikmΔy−En,j−1α1,β1δj−1exp2ikmΔy−4δj−1expikmΔy+6δj−1−4δj−1exp−ikmΔy+δj−1exp−2ikmΔy−Fn,j−1α1,β1δj−1.We prove by induction. Forn=0, we obtain
(25)δ1=A0,0α1,β1δ0expikmΔy−2δ0+δ0exp−ikmΔy−B0,0α1,β1δ0exp2ikmΔy−4δ0expikmΔy+6δn−4δ0expikmΔy+δ0exp−2ikmΔy−C0,0α1,β1δ0=δ0A0,0α1,β1expikmΔy−2+exp−ikmΔy−B0,0α1,β1exp2ikmΔy−4expikmΔy+6δn−4expikmΔy+exp−2ikmΔy−C0,0α1,β1=δ0−4sin2kmΔy2A0,0α1,β1+4B0,0α1,β1sin2kmΔy+4sin2kmΔy2−C0,0α1,β1.We should showδ1/δ0<1. Therefore, we have
(26)−4sin2kmΔy2A0,0α1,β1+4B0,0α1,β1sin2kmΔy+4sin2kmΔy2−C0,0α1,β1<1."Since we have for allm, we obtain
(27)"−4A0,0α1,β1+20B0,0α1,β1−C0,0α1,β1<1."Whenn>1, we have
(28)δn+1=∑j=0nAn,jα1,β1δjexpikmΔy−2δj+δjexp−ikmΔy−Bn,jα1,β1δjexp2ikmΔy−4δjexpikmΔy+6δj−4δjexp−ikmΔy+δjexp−2ikmΔy−Cn,jα1,β1δn−∑j=0n−1Dn,j−1α1,β1δj−1expikmΔy−2δj−1+δj−1exp−ikmΔy−En,j−1α1,β1δj−1exp2ikmΔy−4δj−1expikmΔy+6δj−1−4δj−1exp−ikmΔy+δj−1exp−2ikmΔy−Fn,j−1α1,β1δj−1.Then, we get(29)δn+1=∑j=0nδj−4sin2kmΔy2An,jα1,β1+4Bn,jα1,β1sin2kmΔy+4sin2kmΔy2−Cn,jα1,β1−∑j=0n−1δj−1−4sin2kmΔy2An,j−1α1,β1+4Bn,j−1α1,β1sin2kmΔy+4sin2kmΔy2−Cn,j−1α1,β1.We assume that for alln≥1, δ1/δ0<1. We want to prove that δn+1/δ0<1.. However,
(30)δn+1≤∑j=0nδj−4sin2kmΔy2An,jα1,β1+4Bn,jα1,β1sin2kmΔy+4sin2kmΔy2−Cn,jα1,β1+∑j=0n−1δj−1−4sin2kmΔy2An,j−1α1,β1+4Bn,j−1α1,β1sin2kmΔy+4sin2kmΔy2−Cn,j−1α1,β1."By induction hypothesis for alln≥1,δn<δ0, we have
(31)δn+1<δ0∑j=0n−4sin2kmΔy2An,jα1,β1+4Bn,jα1,β1sin2kmΔy+4sin2kmΔy2−Cn,jα1,β1+∑j=0n−1−4sin2kmΔy2An,j−1α1,β1+4Bn,j−1α1,β1sin2kmΔy+4sin2kmΔy2−Cn,j−1α1,β1.This inequality is true for allm. Thus, we reach
(32)δn+1<δ0∑j=0n−4An,jα1,β1+20Bn,jα1,β1−Cn,jα1,β1+∑j=0n−1−4An,j−1α1,β1+20Bn,j−1α1,β1−Cn,j−1α1,β1.We need to show thatδn+1/δ0<1. Thus, we reach
(33)∑j=0n−4An,jα1,β1+20Bn,jα1,β1−Cn,jα1,β1+∑j=0n−1−4An,j−1α1,β1+20Bn,j−1α1,β1−Cn,j−1α1,β1<1.


## 5. Solution of the Problem with the Exponential Decay Kernel

We consider Eq. (10) with fractal-fractional differential operator using Definition 2 of exponential decay kernel as
(34)D0FFEtα1,β1vy,t=1Re∂2vy,t∂y2−∂4vy,t∂y4−M11vy,t,Mα11−α1ddt∫0tvy,λexp−α11−α1t−λdλ=β1Retβ1−1∂2vy,t∂y2−∂4vy,t∂y4−M11vy,t.For simplicity, we define(35)Fy,t,vy,t=β1Retβ1−1∂2vy,t∂y2−∂4vy,t∂y4−M11vy,t.Then, we reach(36)vy,λ=1−α1Mα1Fy,t,vy,t+α1Mα1∫0tFy,λ,vy,λdλ.We discretize Eq. (36) at yi,tn+1 and yi,tn as
(37)vin+1=1−α1Mα1Fyi,tn,vin+α1Mα1∫0tn+1Fyi,λ,vyi,λdλ,vin=1−α1Mα1Fyi,tn,vin−1+α1Mα1∫0tnFyi,λ,vyi,λdλ.Then, we obtain(38)vin+1=vin+1−α1Mα1Fyi,tn,vin−Fyi,tn−1,vin−1+α1Mα1∫tntn+1Fyi,λ,vyi,λdλ=vin+1−α1Mα1Fyi,tn,vin−Fyi,tn−1,vin−1+α1Mα13h2Fyi,tn,vin−h2Fyi,tn−1,vin−1,where
(39)Fyi,tn,vyi,tn=β1Retnβ1−1vi+1n−2vin+vi−1nΔy2−vi+2n−4vi+1n+6vin−4vi−1n+vi−2nΔy4−M11vin.Thus, we acquire(40)vin+1=vin+β1Retnβ1−1vi+1n−2vin+vi−1nΔy2−vi+2n−4vi+1n+6vin−4vi−1n+vi−2nΔy4−M11vin31−α1+3hα12Mα1−β1Retn−1β1−1vi+1n−1−2vin−1+vi−1n−1Δy2−vi+2n−1−4vi+1n−1+6vin−1−4vi−1n−1+vi−2n−1Δy4−M11vin−121−α1+hα12Mα1.For simplicity, we let(41)An,α1,β1=β1tnβ1−1Nα1ReΔy2,Bn,α1,β1=β1tnβ1−1Nα1ReΔy4,Cn,α1,β1=β1tjβ1−1M11Nα1Re,Dn,α1,β1=β1tn−1β1−1Kα1ReΔy2,En,α1,β1=β1tn−1β1−1Kα1ReΔy4,Fn,α1,β1=β1tn−1β1−1M11Kα1Re,Nα1=21−α1+3hα12Mα1,Kα1=21−α1+hα12Mα1.We chooseεin=δnexpikmy.. Therefore, we reach
(42)δn+1expikmy=δnexpikmy−Cn,nα1,β1δnexpikmy+An,α1,β1δnexpikmy+Δy−2δnexpikmy+δnexpikmy−Δy−Bn,α1,β1δnexpikmy+2Δy−4δnexpikmy+Δy+6δnexpikmy−4δnexpikmy−Δy+δnexpikmy−2Δy−Dn,α1,β1δn−1expikmy+Δy−2δn−1expikmy+δn−1expikmy−Δy−En,α1,β1δn−1expikmy+2Δy−4δn−1expikmy+Δy+6δn−1expikmy−4δn−1expikmy−Δy+δn−1expikmy−2Δy+Fn,n−1α1,β1δn−1expikmy.After simplification, we obtain(43)δn+1=δn−Cn,α1,β1δn+Fn,α1,β1δn−1+An,α1,β1δnexpikmΔy−2δn+δnexp−ikmΔy−Bn,α1,β1δnexp2ikmΔy−4δnexpikmΔy+6δn−4δnexpikmΔy+δnexp−2ikmΔy−Dn,α1,β1δn−1expikmΔy−2δn−1+δn−1expikmΔy+En,α1,β1δn−1exp2ikmΔy−4δn−1expikmΔy+6δn−1−4δn−1expikmΔy+δn−1exp−2ikmΔy.Thus, we have(44)δn+1=δn1−4An,α1,β1sin2kmΔy2+4Bn,jα1,β1sin2kmΔy+4sin2kmΔy2−Cn,α1,β1+δn−14Dn,α1,β1sin2kmΔy2−4En,α1,β1sin2kmΔy+4sin2kmΔy2+Fn,α1,β1.Forn=0, we get
(45)δ1=δ01−4A0,α1,β1sin2kmΔy2+4B0,α1,β1sin2kmΔy+4sin2kmΔy2−C0,α1,β1.Theδ1/δ0<1 implies
(46)1−4A0,α1,β1sin2kmΔy2+4B0,α1,β1sin2kmΔy+4sin2kmΔy2−C0,α1,β1<1.This is true for allm.. Thus, we get
(47)1−4A0,α1,β1+20B0,α1,β1−C0,α1,β1<1.We assume thatδn/δ0<1.. Thus, we need to show δn+1/δ0<1.(48)δn+1≤δn1−4An,α1,β1sin2kmΔy2+4Bn,α1,β1sin2kmΔy+4sin2kmΔy2−Cn,α1,β1+δn−14Dn,α1,β1sin2kmΔy2+4En,α1,β1sin2kmΔy+4sin2kmΔy2+Fn,α1,β1<δ01−4An,α1,β1sin2kmΔy2+4Bn,α1,β1sin2kmΔy+4sin2kmΔy2−Cn,α1,β1+4Dn,α1,β1sin2kmΔy2−4En,α1,β1sin2kmΔy+4sin2kmΔy2+Fn,α1,β1."Thus, we obtain(49)1−4An,α1,β1sin2kmΔy2+4Bn,α1,β1sin2kmΔy+4sin2kmΔy2−Cn,α1,β1+4Dn,α1,β1sin2kmΔy2−4En,α1,β1sin2kmΔy+4sin2kmΔy2+Fn,α1,β1<1.This inequality is true for allm.. Thus, we get
(50)1−4An,α1,β1+20Bn,α1,β1−Cn,α1,β1+4Dn,α1,β1−20En,α1,β1+Fn,α1,β1<1.


## 6. Solution of the Problem with the Generalized Mittag-Leffler Kernel

We take into consideration the Eq. (10) with fractal-fractional differential operator using Definition 3 of Mittag-Leffler kernel as
(51)D0FFMtα1,β1vy,t=1Re∂2vy,t∂y2−∂4vy,t∂y4−M11vy,t,ABα11−α1ddt∫0tvy,λEα1−α11−α1t−λα1dλ=β1Retβ1−1∂2vy,t∂y2−∂4vy,t∂y4−M11vy,t.For simplicity, we define(52)Fy,t,vy,t=β1Retβ1−1∂2vy,t∂y2−∂4vy,t∂y4−M11vy,t.Then, we get(53)vy,λ=1−α1ABα1Fy,t,vy,t+α1Γα1ABα1∫0tFy,λ,vy,λdλ.We discretize above Eq. (53) at yi,tn+1 as
(54)vin+1=1−α1ABα1Fyi,tn+1,vin+α1Γα1ABα1∫0tn+1Fyi,λ,vyi,λtn+1,λα1−1dλ.Then, we obtain(55)vin+1=1−α1ABα1Fyi,tn+1,vin+α1ABα1∑j=0nhα1Fyi,tj,vijΓα1+2n+1−jα1n−j+2+α1−n−jα1n−j+2+2α1−α1ABα1∑j=0nhα1Fyi,tj−1,vij−1ReΓα1+2n+1−jα1+1−n−jα1n−j+1+α1.We have(56)Fyi,tj,vin=β1tnβ1−1vi+1j−2vij+vi−1jΔy2−vi+2j−4vi+1j+6vij−4vi−1j+vi−2jΔy4−M11vij.Then, we will obtain(57)vin+!=β11−α1ReABα1tjβ1−1vi+1j−2vij+vi−1jΔy2−vi+2j−4vi+1j+6vij−4vi−1j+vi−2jΔy4−M11vij+∑j=0nhα1β1tjβ1−1Γα1+2vi+1j−2vij+vi−1jΔy2−vi+2j−4vi+1j+6vij−4vi−1j+vi−2jΔy4−M11vij×n+1−jα1n−j+2+α1−n−jα1n−j+2+2α1α1ReABα1−∑j=0nhα1β1tj−1β1−1Γα1+2vi+1j−1−2vij−1+vi−1j−1Δy2−vi+2j−1−4vi+1j−1+6vij−1−4vi−1j−1+vi−2j−1Δy4−M11vij−1×n+1−jα1+1−n−jα1n−j+1+α1α1ReABα1.For simplicity, we let(58)Mα1=β1tjβ1−11−α1ReΔy2ABα1,Kα1=β1tjβ1−1M111−α1ReABα1,Tα1=β1tjβ1−11−α1ReΔy4ABα1,An,jα1,β1=α1hα1β1tjβ1−1Kn,jα1Γα1+2ReΔy2ABα1,Bn,jα1,β1=α1hα1β1tjβ1−1Kn,jα1Γα1+2ReΔy4ABα1,Cn,jα1,β1=α1hα1β1tjβ1−1M11Kn,jα1Γα1+2ReABα1,An,j−1α1,β1=α1hα1β1tj−1β1−1γn,jα1Γα1+2ReΔy2ABα1,Bn,j−1α1,β1=α1hα1β1tj−1β1−1γn,jα1Γα1+2ReΔy4ABα1,Cn,j−1α1,β1=α1hα1β1tj−1β1−1M11Kn,jα1Γα1+2ReABα1,Kn,jα1=n+1−jα1n−j+2+α1−n−jα1n−j+2+2α1,γn,jα1=n+1−jα1+1−n−jα1n−j+1+α1.Then, we get(59)vin+1=Mα1vi+1n−2vin+vi−1n−Kα1vin−Tα1vi+2n−4vi+1n+6vin−4vi−1n+vi−2n+∑j=0nAn,jα1,β1vi+1j−2vij+vi−1j−Bn,jα1,β1vi+2j−4vi+1j+6vij−4vi−1j+vi−2j−Cn,jα1,β1vij−∑j=0nAn,j−1α1,β1vi+1j−1−2vij−1+vi−1j−1−Bn,j−1α1,β1vi+2j−1−4vi+1j−1+6vij−1−4vi−1j−1+vi−2j−1−Cn,j−1α1,β1vij−1.We chooseεin=δnexpikmy.. Then, we acquire
(60)δn+1expikmy=Mα1δnexpikmy+Δy−2δnexpikmy+δnexpikmy−Δy−Kα1δnexpikmy−Tα1δnexpikmy+2Δy−4δnexpikmy+Δy+6δnexpikmy−4δnexpikmy−Δy+δnexpikmy−2Δy+An,nα1,β1δnexpikmy+Δy−2δnexpikmy+δnexpikmy−Δy−Bn,nα1,β1δnexpikmy+2Δy−4δnexpikmy+Δy+6δnexpikmy−4δnexpikmy−Δy+δnexpikmy−2Δy−Cn,nα1,β1δnexpikmy−An,n−1α1,β1δn−1expikmy+Δy−2δn−1expikmy+δn−1expikmy−Δy+Bn,n−1α1,β1δn−1expikmy+2Δy−4δn−1expikmy+Δy+6δn−1expikmy−4δn−1expikmy−Δy+δn−1expikmy−2Δy+Cn,n−1α1,β1δn−1expikmy+∑j=0n−1An,jα1,β1δjexpikmy+Δy−2δjexpikmy+δjexpikmy−Δy−Bn,jα1,β1δjexpikmy+2Δy−4δjexpikmy+Δy+6δjexpikmy−4δjexpikmy−Δy+δjexpikmy−2Δy−Cn,jα1,β1δnexpikmy−∑j=0n−1An,j−1α1,β1δj−1expikmy+Δy−2δj−1expikmy+δj−1expikmy−Δy−Bn,j−1α1,β1δj−1expikmy+2Δy−4δj−1expikmy+Δy+6δj−1expikmy−4δj−1expikmy−Δy+δj−1expikmy−2Δy−Cn,j−1α1,β1δj−1expikmy.Then, we get(61)δn+1=Mα1δnexpikmΔy−2δn+δnexp−ikmΔy−Kα1δn−Tα1δnexp2ikmΔy−4δnexpikmΔy+6δn−4δnexp−ikmΔy+δnexp−2ikmΔy+An,nα1,β1δnexpikmΔy−2δn+δnexp−ikmΔy−Bn,nα1,β1δnexp2ikmΔy−4δnexpikmΔy+6δn−4δnexpikmΔy+δnexp−2ikmΔy−Cn,nα1,β1δn−An,n−1α1,β1δn−1expikmΔy−2δn−1+δn−1expikmΔy+Bn,n−1α1,β1δn−1exp2ikmΔy−4δn−1expikmΔy+6δn−1−4δn−1expikmΔy+δn−1exp−2ikmΔy+Cn,n−1α1,β1δn−1+∑j=0n−1An,jα1,β1δjexpikmΔy−2δj+δjexp−ikmΔy−Bn,jα1,β1δjexp2ikmΔy−4δjexpikmΔy+6δj−4δjexp−ikmΔy+δjexp−2ikmΔy−Cn,jα1,β1δj−∑j=0n−1An,j−1α1,β1δj−1expikmΔy−2δj−1+δj−1exp−ikmΔy−Bn,j−1α1,β1δj−1exp2ikmΔy−4δj−1expikmΔy+6δj−1−4δj−1exp−ikmΔy+δj−1exp−2ikmΔy−Cn,j−1α1,β1δj−1.We prove by induction. Forn=0, we have
(62)δ1=Mα1δ0expikmΔy−2δ0+δ0exp−ikmΔy−Kα1δ0−Tα1δ0exp2ikmΔy−4δ0expikmΔy+6δ0−4δ0exp−ikmΔy+δ0exp−2ikmΔy+A0,0α1,β1δ0expikmΔy−2δ0+δ0exp−ikmΔy−B0,0α1,β1δ0exp2ikmΔy−4δ0expikmΔy+6δn−4δ0expikmΔy+δ0exp−2ikmΔy−C0,0α1,β1δ0.Then, we get(63)δ1=δ0Mα1expikmΔy−2+exp−ikmΔy−Kα1δ0−δ0Tα1exp2ikmΔy−4expikmΔy+6−4exp−ikmΔy+exp−2ikmΔy+δ0A0,0α1,β1expikmΔy−2+exp−ikmΔy−δ0B0,0α1,β1exp2ikmΔy−4expikmΔy+6−4expikmΔy+exp−2ikmΔy−C0,0α1,β1δ0.Thus, we reach(64)δ1=δ0Mα1−4sin2kmΔy2−Kα1δ0+4δ0Tα1sin2kmΔy2+4sin2kmΔy2+δ0−4sin2kmΔy2A0,0α1,β1+4B0,0α1,β1sin2kmΔy+4sin2kmΔy2−C0,0α1,β1.After simplification, we obtain(65)δ1=δ0−4sin2kmΔy2Mα1+A0,0α1,β1−Kα1−C0,0α1,β1+4Tα1+B0,0α1,β1sin2kmΔy2+4sin2kmΔy2.We should showδ1/δ0<1.. Therefore, we have
(66)−4sin2kmΔy2Mα1+A0,0α1,β1−Kα1−C0,0α1,β1+4Tα1+B0,0α1,β1sin2kmΔy2+4sin2kmΔy2<1."Since we have for allm, we obtain
(67)−4Mα1+A0,0α1,β1−Kα1−C0,0α1,β1+20Tα1+B0,0α1,β1<1."Whenn>1, we have
(68)δn+1=Mα1δnexpikmΔy−2δn+δnexp−ikmΔy−Kα1δn−Tα1δnexp2ikmΔy−4δnexpikmΔy+6δn−4δnexp−ikmΔy+δnexp−2ikmΔy+∑j=0nAn,jα1,β1δjexpikmΔy−2δj+δjexp−ikmΔy−Bn,jα1,β1δjexp2ikmΔy−4δjexpikmΔy+6δj−4δjexp−ikmΔy+δjexp−2ikmΔy−Cn,jα1,β1δn−∑j=0nAn,j−1α1,β1δj−1expikmΔy−2δj−1+δj−1exp−ikmΔy−Bn,j−1α1,β1δj−1exp2ikmΔy−4δj−1expikmΔy+6δj−1−4δj−1exp−ikmΔy+δj−1exp−2ikmΔy−Cn,j−1α1,β1δj−1.We suppose that for alln≥1, δn/δ0<1.. We want to prove that δn+1/δ0<1.. However,
(69)δn+1≤δn−4sin2kmΔy2Mα1+Kα1+∑j=0nδj−4sin2kmΔy2An,jα1,β1+4Bn,jα1,β1sin2kmΔy+4sin2kmΔy2−Cn,jα1,β1+∑j=0n−1δj−1−4sin2kmΔy2An,j−1α1,β1+4Bn,j−1α1,β1sin2kmΔy+4sin2kmΔy2−Cn,j−1α1,β1."By induction hypothesis for alln≥1, δn<δ0, we have
(70)δn+1≤δ0−4sin2kmΔy2Mα1+Kα1+∑j=0nδ0−4sin2kmΔy2An,jα1,β1+4Bn,jα1,β1sin2kmΔy+4sin2kmΔy2−Cn,jα1,β1+∑j=0nδ0−4sin2kmΔy2An,j−1α1,β1+4Bn,j−1α1,β1sin2kmΔy+4sin2kmΔy2−Cn,j−1α1,β1.This inequality is true for allm. Thus, we reach
(71)δn+1≤δ0−4Mα1+Kα1+∑j=0nδ0−4An,jα1,β1+20Bn,jα1,β1−Cn,jα1,β1+∑j=0n−1δ0−4An,j−1α1,β1+20Bn,j−1α1,β1−Cn,j−1α1,β1.We need to show thatδn+1/δ0<1. Thus, we get
(72)−4Mα1+Kα1+∑j=0n−4An,jα1,β1+20Bn,jα1,β1−Cn,jα1,β1+∑j=0n−1−4An,j−1α1,β1+20Bn,j−1α1,β1−Cn,j−1α1,β1<1.


## 7. Error Analysis

In this section, we will consider the error analysis.(73)D0FFPtα1,β1vy,t=1Re∂2vy,t∂y2−∂4vy,t∂y4−M11vy,t,1Γ1−α1ddt∫0tvy,tt−λ−α1dλ=β1Retβ1−1∂2vy,t∂y2−∂4vy,t∂y4−M11vy,t.Then, we get(74)vy,t=β1ReΓα1∫0tλβ1−1∂2vy,λ∂y2−∂4vy,λ∂y4−M11vy,λt−λα1−1dλ.For simplicity, we take(75)Fy,λ=β1λβ1−1∂2vy,λ∂y2−∂4vy,λ∂y4−M11vy,λ.Then, we have(76)vy,t=1ReΓα1∫0tFy,λt−λα1−1dλ.Atyi,t=tn+1, we get
(77)vin+1=1Γα1∫0tn+1Fyi,λtn+1−λα1−1dλ=1Γα1∑j=0n∫tjtj+1Pjλ+λ−tjλ−tj−12!∂2∂λ2Fyi,λλ=ελtn+1−λα1−1dλ=1Γα1∑j=0n∫tjtj+1Pjλtn+1−λα1−1dλ+1Γα1∑j=0n∫tjtj+1λ−tjλ−tj−12!∂2∂λ2Fyi,λλ=ελtn+1−λα1−1dλ=1Γα1∑j=0n∫tjtj+1Pjλtn+1−λα1−1dλ+Rn,jα1,Rn,jα1=1Γα1∑j=0n∫tjtj+1λ−tjλ−tj−12!∂2∂λ2β1λβ1−1∂2vy,λ∂y2−∂4vy,λ∂y4−M11vy,λtn+1−λα1−1dλ≤1Γα1∑j=0n∫tjtj+1λ−tjλ−tj−12!∂2∂λ2β1λβ1−1∂2vy,λ∂y2−∂4vy,λ∂y4−M11vy,λtn+1−λα1−1dλ.Then, we have(78)∂2∂λ2β1λβ1−1∂2vyi,λ∂y2−∂4vyi,λ∂y4−M11vyi,λ=β1β1−1β1−2λβ1−3∂2vyi,λ∂y2−∂4vyi,λ∂y4−M11vyi,λ+2β1−1λβ1−2×∂3vyi,λ∂λ∂y2−∂5vyi,λ∂λ∂y4−M11∂vyi,λ∂λ+λβ1−1∂4vyi,λ∂λ2∂y2−∂6vyi,λ∂λ2∂y4−M11∂2vyi,λ∂λ2,Rn,jα1<1Γα1∑j=0n∫tjtj+1supτ∈τj,τj+1λ−tjsupτ∈τj−1,τj+1λ−tj−1×supτ∈τj,τj+1∂2∂λ2β1λβ1−1∂2vyi,λ∂y2−∂4vyi,λ∂y4−M11vyi,λtn+1−λα1−1dλ<C1C2Γα1∑j=0n∫tjtj+1supτ∈τj,τj+1∂2∂λ2β1λβ1−1∂2vyi,λ∂y2−∂4vyi,λ∂y4−M11vyi,λtn+1−λα1−1dλ.We have(79)supτ∈τj,τj+1∂2∂λ2β1λβ1−1∂2vyi,λ∂y2−∂4vyi,λ∂y4−M11vyi,λ<β1β1−1β1−2λβ1−3∂2v∂y2∞+∂4v∂y4∞+M11v∞+2β1−1λβ1−2×∂3v∂λ∂y2∞+∂5v∂λ∂y4∞−M11∂v∂λ∞+λβ1−1∂4v∂λ2∂y2∞+∂6v∂λ2∂y4∞−M11∂2v∂λ2∞.Therefore, we acquire(80)Rn,jα1<C1C2Γα1β1β1−1β1−2A1∂2v∂y2∞+∂4v∂y4∞+M11v∞+2β1−1A2∂3v∂λ∂y2∞+∂5v∂λ∂y4∞+M11∂v∂λ∞+A3∂4v∂λ2∂y2∞+∂6v∂λ2∂y4∞+M11∂2v∂λ2∞∑j=0n∫tjtj+1tn+1−λα1−1dλ,where
(81)1Γα1∑j=0n∫tjtj+1tn+1−λα1−1dλ=Δtα1n+1α1Γ1+α1.Therefore, we obtain(82)Rn,jα1<C1C2Γα1β1β1−1β1−2A1∂2v∂y2∞+∂4v∂y4∞+M11v∞+2β1−1A2∂3v∂λ∂y2∞+∂5v∂λ∂y4∞+M11∂v∂λ∞+A3∂4v∂λ2∂y2∞+∂6v∂λ2∂y4∞+M11∂2v∂λ2∞∑j=0nΔtα1n+1α1Γ1+α1.Remark 4.
Error analysis with exponential decay kernel and Mittag-Leffler kernel can be obtained likewise. Therefore, we misplaced the error analysis for them.


## 8. Conclusion

In this paper, we investigated the fractional MHD incompressible couple stress fluid flow between two parallel plates. We discussed the discretization and the stability analysis for three different kernels. Additionally, we discussed the error analysis of the model in details.



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