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Analytical solutions of fractional couple stress fluid flow for an engineering problem

  • Rabia Naz , Muhammad Danish Ikram and Muhammad Imran Asjad EMAIL logo
Published/Copyright: May 3, 2023
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Abstract

In this article, analytical solutions of couple stress fluid flow modeled with a power law fractional differential operator are discussed. Stokes’ second problem for an incompressible couple stress fluid is studied for an horizontal plate of infinite length. The governing equations of the flow problem are expressed in terms of a partial differential operator and then converted into a non-dimensional model by using dimensional analysis. Then the integer order problem was formulated in terms of the non-integer order of three types of fractional derivatives and then solved with the help of the Laplace transform method. The obtained solutions are complex and expressed in terms of series. In order to check the memory index of the solutions obtained with three different fractional operators, we have plotted some graphs. It is found that the constant proportional operator provides us a better choice about the memory and maximum enhancement achieved in the comparison of Caputo and Caputo–Fabrizio. Furthermore, in order to check the accuracy of the present results, we have compared the obtained solutions with the existing literature and found a good agreement between them.

1 Introduction

A large portion of the readings underway elaborate on the viscous resources of the classical Naiver–Stokes associations. There are many multifaceted goods, e.g., paints and polymer solutions that cannot be considered over the classical Naiver–Stokes equations. These are famous as non-Newtonian liquids. Numerous models of non-Newtonian liquids are established in the mechanism. The couple stress liquid model is one of them. It has significant configurations because of the existence of body couples, couple stresses, and non-symmetric stress tensor. Inspiring illustrations of the couple stress liquids are blood, oils, and suspension liquids [17].

Umavathi et al. [8] investigated the steady-state heat transfer performance of a couple stress nano-fluid sandwiched between viscous fluids numerically. Gopal et al. [9] discussed the movement of a couple stress liquid over a porous layer bounded by parallel plates. The effects of Newtonian heating on magnetohydrodynamics (MHD) 3-dimensional movement past a stretching region have been studied by refs [10,11]. Turkyilmazoglu [12] analyzed the heat and movement of couple stress MHD liquids over permeable stretching/shrinking surfaces. Hayat et al. [13,14] presented an analysis of MHD extracting movement of couple stress nano-material between two parallel surfaces. Devakar and Iyengar [15] evaluated Stokes’ first and second problems for an incompressible couple stress liquid under isothermal constraints.

Fractional calculus has enlarged significant status and importance due to its valid applications in several branches of science and engineering [1621]. Tahir et al. [22] discussed the natural convection of unsteady movement of Maxwell liquid over a waving upright flat plate with constant temperature at the boundary by Caputo–Fabrizio (CF) fractional operator. Saqib et al. [23] used the CF fractional model to analyze the free convection movement of Jeffrey fluid. Natural convection flow of differential type fluid with Caputo fractional derivatives is determined by Imran et al. [24]. Jarad et al. [25] explained the uniqueness conditions of solutions to a certain class of ODEs with ABC fractional derivative. Abro et al. [26] explored the impacts of carbon nanotubes on the MHD flow of methanol based nanofluids by ABC and CF fractional techniques. Danish Ikram et al. [27] analyzed a fractional model of extraction nanofluids using clay nanoparticles for different based fluids.

Recently, Baleanu et al. [28] introduced a new fractional operator called constant proportional Caputo (CPC) fractional derivative by combining Caputo and conformable fractional derivative. This operator is a power law and is a linear combination of two fractional differential operators. Imran et al. [29] used CPC fractional operator to study the astrophysical properties of clay nanofluids with Brinkman and Maxwell–Garnet terminologies. Ikram et al. [30] presented a comparison of CPC and CF fractional operators using hybrid nanofluids. They concluded that CPC gave better memory effects than CF fractional operator. Chu et al. [31] calculated the MHD free convection movement of hybrid nanofluid in a microchannel through CPC fractional operator. Asadullah et al. [32] explained the heat and mass transfer flow for adhesive fluid between two vertical parallel plates. They found an analytical solution by applying CPC fractional approach.

Hamid et al. [34], introduced an insubstantial method to investigate the nonlinear oscillatory fractional-order differential equations. Hamid et al. [35] proposed a linearized steady-state insubstantial technique to study two-dimensional nonlinear evolutionary and reaction–diffusion models. In the study of Hamid et al. [35], research is related to the growth of a hybrid semi-spectral computational scheme built on multidimensional Chelyshkov polynomials (CPs). In the study of Hamid et al. [36], finite difference method (FDM) and Chelyshkov polynomial method schemes are used. They are used to obtain nonlinear oscillatory difficulties of random orders that do not have particular solutions in the literature. In the study of Hamid et al. [37], an operational matrix-based spectral computational strategy was combined with the Picard system and successfully utilized to find the solution for a family of nonlinear development differential equations. The numerical solutions of non-dimensional system of partial differential equations finite difference method tied with Crank Nicolson technique is established and effectively applied [38]. Crank Nicolson finite difference method is unconditionally stable. Prabhakar fractional model is an alternative, newly proposed analytical approach without the appearance of singularity. This model grants modifications of both Atangana Baleanu (AB) and Caputo Fabrizio models in the literature. In the study of Li and Peng [39], a critical function of utilizing a set of laws or equations is to transform one signal into another. One may wonder why the Laplace transform technique is necessary when other methods are available. The straight forward answer is that this method simplifies ordinary differential equations (ODEs) into algebraic equations, making it easier to solve the ODE problem. There are several methods, each with their own significance. Rubbab et al. [40] used CF time-fractional derivative for the fractional model of the advection–diffusion process. In the study of Guo et al. [41], fractional-order simulations in mass and heat transfer analysis are limited by an elliptic inclined plate through slip effects. MHD fluid flow through a porous medium flowing on a vertical plate for this problem AB time-fractional derivative is used to find the numerical outcomes [42]. In the study of Raza et al. [43], a thermal study of the mixed free convection Casson nanofluid along with heat transfer effects over a vertical plate is discussed. To develop the fractional model, the partial derivative with respect to time is swapped with the recent interpretations of fractional derivatives AB and CF time-fractional derivatives, and the Laplace method is applied to determine the solution of the governing equations. AB fractional derivative is used to find the significances of exponential heating and Darcy’s law for second-grade fluid flow over a wavering plate [44]. Raza et al. [45] discussed the current classification of fractional derivative AB time-fractional derivative on second-grade fluid flowing on an infinite plate under the MHD effect.

By the motivation of the above study, our aim is to report the couple stress fluid by applying CPC fractional operator and comparing its results with other fractional derivatives as well as the classical results obtained by Devakar and Iyengar [15]. There is no such result exists in the literature.

2 Preliminaries

In the literature, there are many basic definitions to deal with fractional derivatives; some of them are provided below.

2.1 Definition 1

The Caputo fractional derivative of order 0 < α < 1 of a function h : ( 0 , ) R n is defined as follows [46]:

(1) D t α C g ( y , t ) = 1 Γ ( 1 α ) 0 t ǵ ( y , s ) ( t s ) α d s = ( h α ǵ ) ( t ) ,

whereas h α ( t ) = t α Γ ( 1 α ) is the singular power-law kernel and ǵ ( y , s ) = g ( y , t ) t t = s = t g ( y , t ) t = s

The Laplace transform of above equation is given as follows:

(2) { D t α C g ( y , t ) } = s α { g ( y , s ) } s α 1 g ( y , o ) ,

2.2 Definition 2

The CPC hybrid operator may be defined as follows [46]:

(3) D t α 0 CPC g ( y , t ) = 1 Γ ( 1 α ) 0 t ( K 1 ( α ) g ( y , s ) + K O ( α ) g ̀ ( τ ) ) ( t τ ) α d τ = K 1 ( α ) 0 R L I t 1 α g ( t ) + K o ( α ) D t α 0 C ,

Here, CPC stands for constant proportional Caputo.

3 Problem formulation

The constitution equation between the force stress tensor t i j and rate of deformation tensor d i j is as follows:

t i j = ( p + λ 1 q ¯ ) δ i j + 2 μ d i j 1 2 ε i j k [ m , k + 4 η ω k , r r + ρ c k ] .

Couple stress tensor m i j that present in the theory have linear constitutive relation

m i j = 1 3 m δ i j + 4 η ́ ω j , i + 4 η ω i , j ,

where ω = 1 2 × q ¯ is spin vector, ω i , j is the spin tensor, d i j is the rate of deformation tensor, m is the trace of couple stress tensor m i j , ρ c k is the body couple vector, and quantities λ 1 , μ and η , η ́ are viscosity and couple stress viscosity coefficients [15].

4 Mathematical formulation

Suppose that the incompressible and unsteady movement of a couple stress liquid that fill up the partial space y > 0 over a smooth (solid) plate covering the x z -plane.

The assumptions are the following:

  • At the start, both the plate and fluid are at rest.

  • On time t = 0 + , the plate is starting to oscillate in its own plane.

  • Beside x -axis, the flow happen in x -direction as presented in the following figure.

  • Consequently, the velocity is V = ( u ( y , t ) , 0 , 0 ) and it generally satisfies the equation of continuity.

The governing equation for couple stress fluid are as follows [15]:

(4) ρ u ( y , t ) t = μ 2 u ( y , t ) y 2 η 4 u ( y , t ) y 4 .

The boundary and initial conditions are

(5) u ( y , t ) 0 if y ,

(6) u ( 0 , t ) = UH ( t ) cos ( ω t ) , when t > 0 .

(7) 2 u ( y , t ) y 2 = 0 , when y = 0 for t > 0 .

(8) u ( y , 0 ) 0 y .

Now the non-dimensional variables are introduced as follows:

(9) Φ = u U , ξ = y L , τ = U L t .

where

L 2 = η μ , Re = ρ U L μ .

As Eq. (4) reduces, we obtain

(10) Re Φ ( ξ , τ ) τ = 2 Φ ( ξ , τ ) ξ 2 4 Φ ( ξ , τ ) ξ 4 .

The dimensionless conditions are

(11) Φ ( ξ , 0 ) = 0 ξ ,

(12) Φ ( 0 , τ ) = H ( τ ) cos ( ω τ ) τ > 0 ,

(13) Φ ( ξ , τ ) 0 as ξ ,

(14) 2 Φ ( ξ , s ) ξ 2 = 0 , at ξ = 0 , τ > 0 .

4.1 Constant proportional model of couple stress fluid

This section will cover the fractional model of a physical problem provided in Eq. (10).

(15) Re CPC D τ α Φ ( ξ , τ ) = 2 Φ ( ξ , τ ) ξ 2 4 Φ ( ξ , τ ) ξ 4 .

With the help of the Laplace transform on Eqs. (10)–(14), we obtain the following:

(16) d 4 Φ ¯ ( ξ , s ) d ξ 4 d 2 Φ ¯ ( ξ , s ) d ξ 2 + Re k 1 ( α ) s + k o ( α ) s α Φ ¯ ( ξ , s ) = 0 ,

satisfying the boundary conditions

(17) Φ ¯ ( 0 , s ) = s s 2 + w 2 ,

(18) Φ ¯ ( ξ , s ) 0 if ξ ,

(19) 2 Φ ¯ ( 0 , s ) ξ 2 = 0 .

After applying the conditions in (11)–(14), the solution of Eq. (17) is

(20) Φ ¯ ( ξ , s ) = s 1 L 1 + k 2 ( α ) s s α 1 2 ( s 2 + w 2 ) 1 L 1 + k 2 ( α ) s s α e ξ 1 + 1 L 1 + k 2 ( α ) s s α 2 + s 1 L 1 + k 2 ( α ) s s α + 1 2 ( s 2 + w 2 ) 1 L 1 + k 2 ( α ) s s α × e ξ 1 1 L 1 + k 2 ( α ) s s α 2 ,

where

(21) L = 4 Re k 0 ( α ) , k 2 ( α ) = k 1 ( α ) k 0 ( α ) .

Eq. (20) can be written as follows:

(22) Φ ¯ ( ξ , s ) = s 2 ( s 2 + w 2 ) e ξ 1 + 1 L 1 + k 2 ( α ) s s α 2 s 2 ( s 2 + w 2 ) 1 L 1 + k 2 ( α ) s s α × e ξ 1 + 1 L 1 + k 2 ( α ) s s α 2 + s 2 ( s 2 + w 2 ) e ξ 1 1 L 1 + k 2 ( α ) s s α 2 + s 2 ( s 2 + w 2 ) 1 L 1 + k 2 ( α ) s s α × e ξ 1 1 L 1 + k 2 ( α ) s s α 2 .

In equivalent form, it can be stated as follows:

(23) Φ ¯ ( ξ , s ) = s s 2 + w 2 + p 1 = 1 p 2 = 0 p 3 = 0 p 4 = 0 r = 0 ( ξ ) p 1 ( L ) p 3 [ k 2 ( α ) ] p 4 ( 1 ) r w 2 r 2 ( 2 ) p 1 p 1 ! p 2 ! p 3 ! p 4 ! s 1 α p 3 + p 4 + 2 r × Γ p 1 2 + 1 Γ p 2 2 + 1 Γ ( p 3 + 1 ) Γ p 1 2 + 1 p 2 Γ p 2 2 + 1 p 3 Γ ( p 3 + 1 p 4 ) p 1 = 1 p 2 = 0 p 3 = 0 p 4 = 0 m = 0 k = 0 r = 0 ( ξ ) p 1 ( 1 ) p 3 + r w 2 r L p 3 + m [ k 2 ( α ) ] p 4 + k 2 ( 2 ) p 1 p 1 ! p 2 ! p 3 ! p 4 ! m ! k ! s 1 + 2 r α p 3 + p 4 α m + k × Γ p 1 2 + 1 Γ p 2 2 + 1 Γ ( p 3 + 1 ) Γ ( m + 1 2 ) Γ ( m + 1 ) Γ p 1 2 + 1 p 2 Γ p 2 2 + 1 p 3 Γ ( p 3 + 1 p 4 ) Γ 1 2 Γ ( m + 1 k ) + q 1 = 1 q 2 = 0 q 3 = 0 q 4 = 0 r = 0 ( ξ ) q 1 ( 1 ) q 2 + q 3 + r w 2 r L q 3 [ k 2 ( α ) ] q 4 2 ( 2 ) q 1 q 1 ! q 2 ! q 3 ! q 4 ! s 1 + 2 r α q 3 + q 4 × Γ q 1 2 + 1 Γ q 2 2 + 1 Γ ( q 3 + 1 ) Γ q 1 2 + 1 q 2 Γ q 2 2 + 1 q 3 Γ ( q 3 + 1 q 4 ) + q 1 = 1 q 2 = 0 q 3 = 0 q 4 = 0 m = 0 k = 0 r = 0 ( ξ ) q 1 ( 1 ) q 2 + q 3 + r w 2 r L q 3 + m [ k 2 ( α ) ] q 4 + k 2 ( 2 ) q 1 q 1 ! q 2 ! q 3 ! q 4 ! m ! k ! s 1 + 2 r α q 3 + q 4 α m + k × Γ q 1 2 + 1 Γ q 2 2 + 1 Γ ( q 3 + 1 ) Γ ( m + 1 2 ) Γ ( m + 1 ) Γ q 1 2 + 1 q 2 Γ q 2 2 + 1 q 3 Γ ( q 3 + 1 q 4 ) Γ 1 2 Γ ( m + 1 k ) .

Taking the Laplace inverse transform to Eq. (23), we obtain

(24) Φ ( ξ , τ ) = cosw τ + p 1 = 1 p 2 = 0 p 3 = 0 p 4 = 0 r = 0 ( ξ ) p 1 ( L ) p 3 [ k 2 ( α ) ] p 4 ( 1 ) r w 2 r τ α p 3 + p 4 + 2 r 2 ( 2 ) p 1 p 1 ! p 2 ! p 3 ! p 4 ! Γ ( 1 α p 3 + p 4 + 2 r ) × Γ p 1 2 + 1 Γ p 2 2 + 1 Γ ( p 3 + 1 ) Γ p 1 2 + 1 p 2 Γ p 2 2 + 1 p 3 Γ ( p 3 + 1 p 4 ) p 1 = 1 p 2 = 0 p 3 = 0 p 4 = 0 m = 0 k = 0 r = 0 ( ξ ) p 1 ( 1 ) p 3 + r w 2 r L p 3 + m [ k 2 ( α ) ] p 4 + k τ 2 r α p 3 + p 4 α m + k 2 ( 2 ) p 1 p 1 ! p 2 ! p 3 ! p 4 ! m ! k ! Γ ( 1 + 2 r α p 3 + p 4 α m + k ) × Γ p 1 2 + 1 Γ p 2 2 + 1 Γ ( p 3 + 1 ) Γ ( m + 1 2 ) Γ ( m + 1 ) Γ p 1 2 + 1 p 2 Γ p 2 2 + 1 p 3 Γ ( p 3 + 1 p 4 ) Γ 1 2 Γ ( m + 1 k ) + q 1 = 1 q 2 = 0 q 3 = 0 q 4 = 0 r = 0 ( ξ ) q 1 ( 1 ) q 2 + q 3 + r w 2 r L q 3 [ k 2 ( α ) ] q 4 τ 2 r α q 3 + q 4 2 ( 2 ) q 1 q 1 ! q 2 ! q 3 ! q 4 ! Γ ( 1 + 2 r α q 3 + q 4 ) × Γ q 1 2 + 1 Γ q 2 2 + 1 Γ ( q 3 + 1 ) Γ q 1 2 + 1 q 2 Γ q 2 2 + 1 q 3 Γ ( q 3 + 1 q 4 ) + q 1 = 1 q 2 = 0 q 3 = 0 q 4 = 0 m = 0 k = 0 r = 0 ( x i ) q 1 ( 1 ) q 2 + q 3 + r w 2 r L q 3 + m [ k 2 ( α ) ] q 4 + k τ 2 r α q 3 + q 4 α m + k 2 ( 2 ) q 1 q 1 ! q 2 ! q 3 ! q 4 ! m ! k ! Γ ( 1 + 2 r α q 3 + q 4 α m + k ) × Γ q 1 2 + 1 Γ q 2 2 + 1 Γ ( q 3 + 1 ) Γ ( m + 1 2 ) Γ ( m + 1 ) Γ q 1 2 + 1 q 2 Γ q 2 2 + 1 q 3 Γ ( q 3 + 1 q 4 ) Γ 1 2 Γ ( m + 1 k ) .

Eq. (24) can be stated in terms of M-function defined in [47],

(25) Φ ( ξ , τ ) = cosw τ + p 1 = 1 p 2 = 0 p 3 = 0 r = 0 ( ξ ) p 1 ( L ) p 3 ( 1 ) r w 2 r 2 ( 2 ) p 1 p 1 ! p 2 ! p 3 ! × M 4 3 ( k 2 ( α ) τ ) ( p 1 2 p 2 + 1 , 0 ) , p 2 2 p 3 + 1 , 0 , ( p 3 + 1 , 1 ) , ( 1 α p 3 + 2 r , 1 ) p 1 2 + 1 , 0 , p 2 2 + 1 , 0 , ( p 3 + 1 , 0 ) p 1 = 1 p 2 = 0 p 3 = 0 p 4 = 0 m = 0 r = 0 ( ξ ) p 1 ( 1 ) p 3 + r w 2 r L p 3 + m [ k 2 ( α ) ] p 4 2 ( 2 ) p 1 p 1 ! p 2 ! p 3 ! p 4 ! m ! × M 6 5 ( [ k 2 ( α ) ] τ ) ( p 1 2 p 2 + 1 , 0 ) , p 2 2 p 3 + 1 , 0 , ( p 3 p 4 + 1 , 0 ) , 1 2 , 0 , ( m + 1 , 1 ) , ( 1 α p 3 + p 4 α m + 2 r , 1 ) p 1 2 + 1 , 0 , p 2 2 + 1 , 0 , ( p 3 + 1 , 0 ) , ( m + 1 2 , 0 ) , ( m + 1 , 0 ) + q 1 = 1 q 2 = 0 q 3 = 0 r = 0 ( ξ ) q 1 ( 1 ) q 2 + q 3 + r w 2 r L q 3 2 ( 2 ) q 1 q 1 ! q 2 ! q 3 ! M 4 3 ( k 2 ( α ) τ ) q 1 2 q 2 + 1 , 0 , q 2 2 q 3 + 1 , 0 , ( q 3 + 1 , 1 ) , ( 1 α q 3 + 2 r , 1 ) q 1 2 + 1 , 0 , q 2 2 + 1 , 0 , ( q 3 + 1 , 0 ) + q 1 = 1 q 2 = 0 q 3 = 0 q 4 = 0 m = 0 r = 0 ( ξ ) q 1 ( 1 ) q 2 + q 3 + r w 2 r L q 3 + m [ k 2 ( α ) ] q 4 2 ( 2 ) q 1 q 1 ! q 2 ! q 3 ! q 4 ! m ! × M 6 5 ( [ k 2 ( α ) ] τ ) q 1 2 q 2 + 1 , 0 , q 2 2 q 3 + 1 , 0 , ( q 3 q 4 + 1 , 0 ) , 1 2 , 0 , ( m + 1 , 1 ) ( 1 α q 3 + q 4 α m + 2 r , 1 ) q 1 2 + 1 , 0 , q 2 2 + 1 , 0 , ( q 3 + 1 , 0 ) , ( m + 1 2 , 0 ) , ( m + 1 , 0 ) .

4.2 Analytical solutions with Caputo derivative

In this section, we will develop a fractional model of the physical problem given in Eq. (10), where

(26) Re C D τ α Φ ( ξ , τ ) = 2 Φ ( ξ , τ ) ξ 2 4 Φ ( ξ , τ ) ξ 4 .

Taking the Laplace transform on Eq. (26), we have

(27) d 4 Φ ¯ ( ξ , s ) d ξ 4 d 2 Φ ¯ ( ξ , s ) d ξ 2 + Re s α Φ ¯ ( ξ , s ) = 0 .

The solution of Eq. (27) becomes, with the conditions given in (11)–(14),

(28) Φ ¯ ( ξ , s ) = s 2 ( s 2 + w 2 ) e ξ 1 + 1 4 Re s α 2 s 2 ( s 2 + w 2 ) 1 4 Re s α e ξ 1 + 1 4 Re s α 2 + s 2 ( s 2 + w 2 ) e ξ 1 1 4 Re s α 2 + s 2 ( s 2 + w 2 ) 1 4 Re s α e ξ 1 1 4 Re s α 2 .

We stated it in its equivalent form

(29) Φ ¯ ( ξ , s ) = s s 2 + w 2 + k 1 = 1 k 2 = 0 k 3 = 0 r = 0 ( ξ ) k 1 ( 4 Re ) k 3 ( 1 ) r w 2 r Γ ( k 1 2 + 1 ) Γ ( k 2 2 + 1 ) 2 ( 2 ) k 1 k 1 ! k 2 ! k 3 ! s 1 + 2 r α k 3 Γ ( k 1 2 + 1 k 2 ) Γ ( k 2 2 + 1 k 3 ) k 1 = 1 k 2 = 0 k 3 = 0 k 4 = 0 r = 0 ( ξ ) k 1 ( 1 ) k 3 + r ( 4 Re ) k 3 + k 4 w 2 r Γ ( k 1 2 + 1 ) Γ ( k 2 2 + 1 ) Γ ( k 4 + 1 2 ) 2 ( 2 ) k 1 k 1 ! k 2 ! k 3 ! k 4 ! s 1 + 2 r α k 3 α k 4 Γ ( k 1 2 + 1 k 2 ) Γ ( k 2 2 + 1 k 3 ) Γ 1 2 + l 1 = 1 l 2 = 0 l 3 = 0 r = 0 ( ξ ) l 1 ( 1 ) l 2 + l 3 + r ( 4 Re ) l 3 w 2 r Γ l 1 2 + 1 Γ l 2 2 + 1 2 ( 2 ) l 1 l 1 ! l 2 ! l 3 ! s 1 + 2 r α l 3 Γ l 1 2 + 1 l 2 Γ l 2 2 + 1 l 3 + l 1 = 1 l 2 = 0 l 3 = 0 l 4 = 0 r = 0 ( ξ ) l 1 ( 1 ) l 2 + l 3 + r ( 4 Re ) l 3 + l 4 w 2 r Γ l 1 2 + 1 Γ l 2 2 + 1 Γ ( l 4 + 1 2 ) 2 ( 2 ) l 1 l 1 ! l 2 ! l 3 ! l 4 ! s 1 + 2 r α l 3 α l 4 Γ l 1 2 + 1 l 2 Γ l 2 2 + 1 l 3 Γ 1 2 .

Taking the Laplace inverse transform to Eq. (29), we have

(30) Φ ( ξ , τ ) = cosw τ + k 1 = 1 k 2 = 0 k 3 = 0 r = 0 ( ξ ) k 1 ( 4 Re ) k 3 ( 1 ) r w 2 r τ 2 r α k 3 Γ ( k 1 2 + 1 ) Γ ( k 2 2 + 1 ) 2 ( 2 ) k 1 k 1 ! k 2 ! k 3 ! Γ ( 1 + 2 r α k 3 ) Γ ( k 1 2 + 1 k 2 ) Γ ( k 2 2 + 1 k 3 ) k 1 = 1 k 2 = 0 k 3 = 0 k 4 = 0 r = 0 ( ξ ) k 1 ( 1 ) k 3 + r ( 4 Re ) k 3 + k 4 w 2 r τ 2 r α k 3 α k 4 Γ ( k 1 2 + 1 ) Γ ( k 2 2 + 1 ) Γ ( k 4 + 1 2 ) 2 ( 2 ) k 1 k 1 ! k 2 ! k 3 ! k 4 ! Γ ( 1 + 2 r α k 3 α k 4 ) Γ ( k 1 2 + 1 k 2 ) Γ ( k 2 2 + 1 k 3 ) Γ 1 2 + l 1 = 1 l 2 = 0 l 3 = 0 r = 0 ( ξ ) l 1 ( 1 ) l 2 + l 3 + r ( 4 Re ) l 3 w 2 r τ 2 r α l 3 Γ l 1 2 + 1 Γ l 2 2 + 1 2 ( 2 ) l 1 l 1 ! l 2 ! l 3 ! Γ ( 1 + 2 r α l 3 ) Γ l 1 2 + 1 l 2 Γ l 2 2 + 1 l 3 + l 1 = 1 l 2 = 0 l 3 = 0 l 4 = 0 r = 0 ( ξ ) l 1 ( 1 ) l 2 + l 3 + r ( 4 Re ) l 3 + l 4 w 2 r τ 2 r α l 3 α l 4 Γ l 1 2 + 1 Γ l 2 2 + 1 Γ ( l 4 + 1 2 ) 2 ( 2 ) l 1 l 1 ! l 2 ! l 3 ! l 4 ! Γ ( 1 + 2 r α l 3 α l 4 ) Γ l 1 2 + 1 l 2 Γ l 2 2 + 1 l 3 Γ 1 2 .

Eq. (30) can be stated in terms of M-function [47] as follows:

(31) Φ ( ξ , τ ) = cosw τ + k 1 = 1 k 2 = 0 r = 0 ( ξ ) k 1 ( 1 ) r w 2 r 2 ( 2 ) k 1 k 1 ! k 2 ! M 3 2 ( 4 Re τ α ) k 1 2 k 2 + 1 , 0 , k 2 2 + 1 , 1 , ( 1 + 2 r , α ) k 1 2 + 1 , 0 , k 2 2 + 1 , 0 k 1 = 1 k 2 = 0 k 3 = 0 r = 0 ( ξ ) k 1 ( 1 ) k 3 + r ( 4 Re ) k 3 w 2 r 2 ( 2 ) k 1 k 1 ! k 2 ! k 3 ! M 4 3 ( 4 Re τ α ) k 1 2 k 2 + 1 , 0 , ( k 2 2 k 3 + 1 , 0 ) , 1 2 , 0 , ( 1 α k 3 + 2 r , α ) k 1 2 + 1 , 0 , k 2 2 + 1 , 0 , 1 2 , 1 + l 1 = 1 l 2 = 0 r = 0 ( ξ ) l 1 ( 1 ) l 2 + r w 2 r 2 ( 2 ) l 1 l 1 ! l 2 ! M 3 2 ( 4 Re τ α ) l 1 2 l 2 + 1 , 0 , l 2 2 + 1 , 1 , ( 1 + 2 r , α ) l 1 2 + 1 , 0 , l 2 2 + 1 , 0 + l 1 = 1 l 2 = 0 l 3 = 0 r = 0 ( ξ ) l 1 ( 1 ) l 2 + l 3 + r ( 4 Re ) l 3 w 2 r 2 ( 2 ) l 1 l 1 ! l 2 ! l 3 ! M 4 3 ( 4 Re τ α ) l 1 2 l 2 + 1 , 0 , l 2 2 l 3 + 1 , 0 , 1 2 , 0 , ( 1 α l 3 + 2 r , α ) l 1 2 + 1 , 0 , l 2 2 + 1 , 0 , 1 2 , 1 .

4.3 Analytical solutions with CF derivative

The solution of the initial value problem Eqs. (11)–(14) with CF fractional derivatives can be obtained via the Laplace transform method. The obtained solution for the velocity field expressed in terms of M-function

(32) Φ ( ξ , τ ) = cosw τ + m 1 = 1 m 2 = 0 m 3 = 0 r = 0 ( ξ ) m 1 ( 4 Re ) m 3 ( 1 ) r w 2 r 2 ( 2 ) m 1 m 1 ! m 2 ! m 3 ! ( 1 α ) m 3 × M 4 3 α 1 α τ m 1 2 m 2 + 1 , 0 , m 2 2 m 3 + 1 , 0 , ( 0 , 1 ) , ( 1 + 2 r , 1 ) m 1 2 + 1 , 0 , m 2 2 + 1 , 0 , ( m 3 , 1 ) m 1 = 1 m 2 = 0 m 3 = 0 m 4 = 0 m 5 = 0 r = 0 ( ξ ) m 1 ( 4 Re ) m 3 + m 5 ( 1 ) r w 2 r ( α ) m 4 2 ( 2 ) m 1 m 1 ! m 2 ! m 3 ! m 4 ! m 5 ! ( 1 α ) m 3 + m 5 × M 6 5 ( α τ ) m 1 2 m 2 + 1 , 0 , m 2 2 m 3 + 1 , 0 , ( m 4 , 0 ) , ( m 5 , 0 ) , ( 1 + m 4 + 2 r , 1 ) , 1 2 , 0 m 1 2 + 1 , 0 , m 2 2 + 1 , 0 , ( m 3 + m 4 , 0 ) , m 5 + 1 2 , 0 , ( m 5 , 1 ) + n 1 = 1 n 2 = 0 n 3 = 0 r = 0 ( ξ ) n 1 ( 1 ) r + n 2 + n 3 w 2 r ( 4 Re ) n 3 2 ( 2 ) n 1 n 1 ! n 2 ! n 3 ! ( 1 α ) n 3 × M 4 3 α 1 α τ n 1 2 n 2 + 1 , 0 , n 2 2 n 3 + 1 , 0 , ( 0 , 1 ) , ( 1 + 2 r , 1 ) n 1 2 + 1 , 0 , n 2 2 + 1 , 0 , ( n 3 , 1 ) + n 1 = 1 n 2 = 0 n 3 = 0 n 4 = 0 n 5 = 0 r = 0 ( ξ ) n 1 ( 1 ) r + n 2 + n 3 w 2 r ( 4 Re ) n 3 + n 5 ( α ) n 4 2 ( 2 ) n 1 n 1 ! n 2 ! n 3 ! n 4 ! n 5 ! ( 1 α ) n 3 + n 5 × M 6 5 ( α τ ) n 1 2 n 2 + 1 , 0 , n 2 2 n 3 + 1 , 0 , ( n 4 , 0 ) , ( n 5 , 0 ) , ( 1 + n 4 + 2 r , 1 ) , 1 2 , 0 n 1 2 + 1 , 0 , n 2 2 + 1 , 0 , ( n 3 + n 4 , 0 ) , n 5 + 1 2 , 0 , ( n 5 , 1 ) .

5 Graphical outcomes and arguments

Figures 1, 2, 3, 4 are designed to show the comparison between CPC, Caputo (C) and CF fractional operators for different values of α . By increasing α , velocity is decreased due to loss in momentum boundary layer. It is concluded that CPC provides better results than all other fractional derivatives. Figures 5 and 6 are designed to show the comparison between CPC fractional operator and classical derivative [15] for different values of α . It is concluded that CPC provides better results than the classical derivative. Figure 7 is designed to show the effect of time on the velocity. With the passage of time, the velocity decreases. The impact of Reynolds number on the velocity is also understood from Figure 8. It is discovered that the velocity drops for larger values of Re.

Figure 1 
               A comparison of CPC with C and CF when 
                     
                        
                        
                           α
                           =
                           0.2
                        
                        \alpha =0.2
                     
                  , 
                     
                        
                        
                           t
                           =
                           5
                        
                        t=5
                     
                  , 
                     
                        
                        
                           w
                           =
                           0.05
                        
                        w=0.05
                     
                  , and 
                     
                        
                        
                           
                              
                              Re
                              
                           
                           =
                           0.09
                        
                        \hspace{0.1em}\text{Re}\hspace{0.1em}=0.09
                     
                  .
Figure 1

A comparison of CPC with C and CF when α = 0.2 , t = 5 , w = 0.05 , and Re = 0.09 .

Figure 2 
               A comparison of CPC with C and CF when 
                     
                        
                        
                           α
                           =
                           0.4
                        
                        \alpha =0.4
                     
                  , 
                     
                        
                        
                           t
                           =
                           5
                        
                        t=5
                     
                  , 
                     
                        
                        
                           w
                           =
                           0.05
                        
                        w=0.05
                     
                  , and 
                     
                        
                        
                           
                              
                              Re
                              
                           
                           =
                           0.09
                        
                        \hspace{0.1em}\text{Re}\hspace{0.1em}=0.09
                     
                  .
Figure 2

A comparison of CPC with C and CF when α = 0.4 , t = 5 , w = 0.05 , and Re = 0.09 .

Figure 3 
               A comparison of CPC with C and CF when 
                     
                        
                        
                           α
                           =
                           0.6
                        
                        \alpha =0.6
                     
                  , 
                     
                        
                        
                           t
                           =
                           5
                        
                        t=5
                     
                  , 
                     
                        
                        
                           w
                           =
                           0.05
                        
                        w=0.05
                     
                  , and 
                     
                        
                        
                           
                              
                              Re
                              
                           
                           =
                           0.09
                        
                        \hspace{0.1em}\text{Re}\hspace{0.1em}=0.09
                     
                  .
Figure 3

A comparison of CPC with C and CF when α = 0.6 , t = 5 , w = 0.05 , and Re = 0.09 .

Figure 4 
               A comparison of CPC with C and CF when 
                     
                        
                        
                           α
                           =
                           0.8
                        
                        \alpha =0.8
                     
                  , 
                     
                        
                        
                           t
                           =
                           5
                        
                        t=5
                     
                  , 
                     
                        
                        
                           w
                           =
                           0.05
                        
                        w=0.05
                     
                  , and 
                     
                        
                        
                           
                              
                              Re
                              
                           
                           =
                           0.09
                        
                        \hspace{0.1em}\text{Re}\hspace{0.1em}=0.09
                     
                  .
Figure 4

A comparison of CPC with C and CF when α = 0.8 , t = 5 , w = 0.05 , and Re = 0.09 .

Figure 5 
               A comparison of CPC with classical derivative [15] when 
                     
                        
                        
                           α
                           =
                           0.2
                        
                        \alpha =0.2
                     
                  , 
                     
                        
                        
                           w
                           =
                           0.05
                        
                        w=0.05
                     
                  , 
                     
                        
                        
                           
                              
                              Re
                              
                           
                           =
                           0.09
                        
                        \hspace{0.1em}\text{Re}\hspace{0.1em}=0.09
                     
                  , and 
                     
                        
                        
                           t
                           =
                           5
                           ,
                           7
                        
                        t=5,7
                     
                  .
Figure 5

A comparison of CPC with classical derivative [15] when α = 0.2 , w = 0.05 , Re = 0.09 , and t = 5 , 7 .

Figure 6 
               A comparison of CPC with classical derivative [15] when 
                     
                        
                        
                           α
                           =
                           0.9
                        
                        \alpha =0.9
                     
                  , 
                     
                        
                        
                           w
                           =
                           0.05
                        
                        w=0.05
                     
                  , 
                     
                        
                        
                           
                              
                              Re
                              
                           
                           =
                           0.09
                        
                        \hspace{0.1em}\text{Re}\hspace{0.1em}=0.09
                     
                  , and 
                     
                        
                        
                           t
                           =
                           5
                           ,
                           10
                        
                        t=5,10
                     
                  .
Figure 6

A comparison of CPC with classical derivative [15] when α = 0.9 , w = 0.05 , Re = 0.09 , and t = 5 , 10 .

Figure 7 
               The effect of 
                     
                        
                        
                           t
                        
                        t
                     
                   on velocity when 
                     
                        
                        
                           α
                           =
                           0.5
                        
                        \alpha =0.5
                     
                  , 
                     
                        
                        
                           w
                           =
                           2
                        
                        w=2
                     
                  , and 
                     
                        
                        
                           
                              
                              Re
                              
                           
                           =
                           0.5
                        
                        \hspace{0.1em}\text{Re}\hspace{0.1em}=0.5
                     
                  .
Figure 7

The effect of t on velocity when α = 0.5 , w = 2 , and Re = 0.5 .

Figure 8 
               The effect of Re on velocity when 
                     
                        
                        
                           t
                           =
                           0.05
                        
                        t=0.05
                     
                  , 
                     
                        
                        
                           w
                           =
                           2
                        
                        w=2
                     
                  , and 
                     
                        
                        
                           α
                           =
                           0.5
                        
                        \alpha =0.5
                     
                  .
Figure 8

The effect of Re on velocity when t = 0.05 , w = 2 , and α = 0.5 .

6 Conclusion

In this article, analytical solutions of couple stress non-Newtonian fluid have been explained. By using the Laplace transform method and fractional derivatives, velocity results are obtained and shown graphically. The key findings of this article are as follows:

  • The fractional approach provides us the choice to control the boundary layers by choosing different values of the fractional parameter.

  • In the limiting case when α 1 , our results are identical to classical obtained by [15].

  • The results obtained with novel power law fractional derivative exhibits the flow property (velocity) decay better in comparison of comparatively (C) and (CF).

  • Velocity falls with the rising values of Re and t .

Acknowledgement

We extend a sense of gratitude to the University of Management and Technology, Lahore, Pakistan, for facilitating and supporting our research.

  1. Funding information: This study did not receive any funding in any form.

  2. Author contributions: All of the authors substantially contributed to this study.

  3. Conflict of interest: The authors declare no conflict of interest in the submission of this article.

  4. Data availability statement: The data used to support the findings of this study are included within the article.

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Received: 2022-04-21
Revised: 2022-12-30
Accepted: 2023-02-08
Published Online: 2023-05-03

© 2023 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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