In this paper, we presents some new exact solutions corresponding to three unsteady flow problems of a generalized Jeffrey fluid produced by a flat plate between two side walls perpendicular to the plate. The fractional calculus approach is used in the governing equations. The exact solutions are established by means of the Fourier sine transform and N-transform. The series solutions of velocity field and associated shear stress in terms of Fox H-function, satisfying all imposed initial and boundary conditions, have been obtained. The similar solutions for ordinary Jeffrey fluid, performing the same motion, appear as limiting case of the solutions are also obtained. Also, the obtained results are analyzed graphically through various pertinent parameters.
Impressive advancement has been made in examining streams of non-Newtonian fluids in the most recent couple of decades. Non-Newtonian fluids have both the properties of elasticity as well as viscosity. The examples of such fluids are very large but we give some of them like honey, toothpaste, ketchup, oils and paints etc. These fluids are widely used in our life and have many interesting applications. It has been proven by many researchers that such kinds of fluids are not only important to academia but also to industry such as polymer processing and making of food and paper.
As we know that Newtonian fluids are modeled by a single equation, the flows of non-Newtonian fluids cannot be explained by a single constitutive model. In general the rheological properties of fluids are specified by their so-called constitutive equations. Exact recent solutions for constitutive equations of viscoelastic fluids are given by Rajagopal and Bhatnagar [
Recently, the fractional derivative [
In 2008, Zafar [
Definition: Let f ( t ) is defined for all t ≥ 0 . The N-transform of f ( t ) is the function f ¯ ( c , s ) defined by
u ¯ s ( c , s ) = N ( u ( t ) ) = ∫ 0 ∞ u ( c t ) e − s t d t , s , c ∈ ( − ∞ , ∞ ) .
The Fox function, also referred as the Fox’s H-function, generalizes the Mellin-Barnes function. The importance of the Fox function lies in the fact that it includes nearly all special functions occurring in applied mathematics and statistics as special cases. In 1961, Fox defined the H-function as the Mellin-Barnes type path integral:
H 1 , p p , q + 1 { − X | ( 1 − a 1 , A 1 ) , ⋯ , ( 1 − a p , A p ) ( 0 , 1 ) ( 1 − b 1 , B 1 ) , ⋯ , ( 1 − b q , B q ) } = ∑ k = 1 ∞ Γ ( a 1 + A 1 K ) ⋯ Γ ( a p + A p K ) k ! Γ ( b 1 + B 1 K ) ⋯ Γ ( b p + B p K ) X k .
Researchers show less attention for the flows of Jeffrey fluids in which the fractional derivatives are appeared. We discuss three different problems related with fractional Jeffrey fluid. In the first problem we assume that the plate is jerked suddenly, in the second problem the plate is moving with uniform acceleration, and in the last problem the plate is moving with non-uniformly acceleration. In this paper we establish exact solutions for the velocity field and the associated shear stress corresponding to the unsteady flow of an incompressible generalized Jeffrey fluid between two side walls perpendicular to the plate. The obtained solutions, expressed under series form in terms of Fox H-functions [
For an incompressible and unsteady generalized Jeffrey fluid the Cauchy stress tensor is defined as [
T = − p I + S , ( 1 + λ ) S = μ ( A + θ β ( D β D t β + ( V ⋅ ∇ ) A ) ) , (1)
where S is the extra stress tensor I is the indeterminate spherical stress, μ is the dynamic viscosity, A = L + L T is the first Rivlin-Ericksen tensor, L is the velocity gradient, λ and θ are relaxation and retardation times, β is the fractional calculus parameter such that 0 ≤ β ≤ 1 , D t β is the fractional differentiation operator of order β based on the Riemann-Liouville definition, defined as [
D t β [ f ( t ) ] = 1 Γ ( 1 − p ) d d t ∫ 0 t f ( τ ) ( t − τ ) p d τ , 0 < p < 1 , (2)
where Γ ( . ) stands for gamma function. Model for ordinary Jeffrey fluid can be obtained by letting β = 1 . For the following problem we consider the velocity field and an extra stress of the form
V = V ( y , z , t ) = u ( y , z , t ) i , S = S ( y , z , t ) , (3)
where u is the velocity and i is the unit vector along the x-direction. The continuity equation for such flows is automatically satisfied. We take the extra stress S independent of x as the velocity field is independent of x. Also, at t = 0 the fluid being at rest is given by
S ( y , z , 0 ) = 0 , (4)
therefore from Equations (1) and (2) it results that S y y = S y z = S z z = 0 and the relevant equations
( 1 + λ ) τ 1 = μ ( 1 + θ β D β D t β ) ∂ y u ( y , z , t ) , (5)
( 1 + λ ) τ 2 = μ ( 1 + θ β D β D t β ) ∂ z u ( y , z , t ) , (5a)
where τ 1 = S x y and τ 2 = S x z are the tangential stresses. In the absence of body forces the balance of linear momentum becomes
∂ y τ 1 + ∂ z τ 2 − ∂ x p = ρ ∂ t u , ∂ y p = ∂ z p = 0 , (6)
here ∂ x p is the pressure gradient and ρ represents the density of the fluid. Eliminating the shear stresses τ 1 and τ 2 between Equations (5) and (6) and neglecting the pressure gradient, the governing equation reduces to the following form
( 1 + λ ) ∂ ∂ t u ( y , z , t ) = v ( 1 + θ β D t β ) ( ∂ 2 ∂ y 2 + ∂ 2 ∂ z 2 ) u ( y , z , t ) , (7)
where υ represents the kinematic viscosity.
We take an unsteady generalized Jeffrey fluid saturating the space above a flat plate which is perpendicular to the y-axis and lies between two side walls perpendicular to the plate. At first the fluid as well as the plane wall is at rest and at time t = 0, the fluid is set into flow by translating the bottom wall in its own plane, with a time dependent velocity Vtm. Its velocity is of the form of Equation (3) and the governing equation is given by Equation (7). The associated initial and boundary conditions are
u ( y , z , 0 ) = ∂ t u ( y , z , 0 ) = 0 ; y > 0 , 0 ≤ z ≤ h
u ( 0 , z , t ) = V t m ; t > 0 , 0 ≤ z ≤ h (8)
u ( y , 0 , t ) = u ( y , h , t ) = 0 ; y , t > 0.
The distance between the two side walls is represented by h. Moreover, the natural conditions
u ( y , z , t ) , ∂ y u ( y , z , t ) → 0 as y → ∞ , − h ≤ z ≤ h , t > 0. . (9)
have to be also satisfied. They are consequences of the fact that the fluid will be at rest at infinity and there is no shear along y-axis.
First we multiply both sides of Equation (7) by sin ( n π z h ) , and then integrate the
obtained result from 0 to h with respect to z, we get the following differential equation
( 1 + λ ) ∂ u n ( y , n , t ) ∂ t = v ( 1 + θ β D t β ) ∂ 2 ∂ y 2 u n ( y , n , t ) − v ( n π h ) 2 ( 1 + θ β D t β ) u n ( y , n , t ) (10)
Applying the N-transform to Equation (10), we find that the image function u ¯ n ( y , n , s ) of u n ( y , n , t ) is given by
∂ 2 ∂ y 2 u ¯ n ( y , n , s ) − [ ξ 2 + s ( 1 + λ ) c v ( 1 + θ β s β c β ) ] u ¯ n ( y , n , s ) = 0 , , (11)
u ¯ n ( 0 , n , s ) = V s ,
u ¯ n ( 0 , n , s ) → 0 as y → ∞ ,
where ξ = n π h . The solution of above differential equation is in the following
form
u ¯ n = c V s exp [ − y ξ 2 + s ( 1 + λ ) c v ( 1 + θ β s β c β ) ] . (12)
We will apply the inverse N-transform technique to obtain analytic solution for the velocity field but to avoid difficult calculations of residues and contour integrals, first we express Equation (12) in series form as
u ¯ n ( y , n , s ) = V ∑ j = 0 ∞ ∑ n = 0 ∞ ∑ q = 0 ∞ ∑ p = 0 ∞ ( − 1 ) j + n + p + q y j ξ j − 2 n v − n j ! n ! q ! p ! Γ ( n ) Γ ( − n ) × λ n − q θ − n − p + β Γ ( p + n ) Γ ( q − n ) Γ ( n − j 2 ) c − n + β ( p + n ) + 1 Γ ( j 2 ) s − n + β ( p + n ) + 1 . (13)
We apply the inverse N-transform to Equation (13), to obtain
u n ( y , n , s ) = V ∑ j = 0 ∞ ∑ n = 0 ∞ ∑ q = 0 ∞ ∑ p = 0 ∞ ( − 1 ) j + n + p + q y j ξ j − 2 n v − n j ! n ! q ! p ! Γ ( n ) Γ ( − n ) Γ ( j 2 ) × λ n − q θ − n − p + β Γ ( p + n ) Γ ( q − n ) Γ ( n − j 2 ) t − n + β ( p + n ) Γ ( − n + β ( p + n ) + 1 ) . (14)
Taking the inverse finite Fourier sine transform to get the analytic solution of the velocity field
u ( y , z , t ) = 2 h ∑ m = 1 ∞ sin ( m π z h ) u n = 2 V h ∑ m = 1 ∞ sin ( m π z h ) u n ∑ j = 0 ∞ ∑ n = 0 ∞ ∑ q = 0 ∞ ∑ p = 0 ∞ ( − 1 ) j + n + p + q j ! n ! q ! p ! × y j ξ j − 2 n v − n Γ ( p + n ) Γ ( q − n ) Γ ( n − j 2 ) t − n + β ( p + n ) Γ ( n ) Γ ( − n ) Γ ( j 2 ) λ q − n θ n + p − β Γ ( − n + β ( p + n ) + 1 ) . (15)
To write Equation (15) in a more compact form, we use the Fox H-function,
u ( y , z , t ) = 2 V h ∑ m = 1 ∞ sin ( m π z h ) ∑ j = 0 ∞ ∑ n = 0 ∞ ∑ q = 0 ∞ ( − 1 ) j + n + p + q y j ξ j − 2 n v − n j ! n ! q ! p ! λ q − n θ n − β t n − β n × H 1 , 3 3 , 5 { t β θ | ( 1 − n , 1 ) , ( 1 − q + n , 0 ) , ( 1 − n + j 2 , 0 ) . ( 0 , 1 ) , ( 1 − n , 0 ) , ( 1 + n , 0 ) , ( 1 − j / 2 , 0 ) , ( n − β n , β ) . } (16)
To obtain (16), the following Fox H-function property is used:
H 1 , p p , q + 1 { − X | ( 1 − a 1 , A 1 ) , ⋯ , ( 1 − a p , A p ) ( 0 , 1 ) ( 1 − b 1 , B 1 ) , ⋯ , ( 1 − b q , B q ) } = ∑ k = 1 ∞ Γ ( a 1 + A 1 K ) ⋯ Γ ( a p + A p K ) k ! Γ ( b 1 + B 1 K ) ⋯ Γ ( b p + B p K ) X k . (17)
To get the shear stress first we apply N-transform to Equations (5) and (5a), to obtain
( 1 + λ ) τ ¯ 1 = μ ( 1 + θ β s β c β ) ∂ y u ¯ ( y , z , s ) , (18)
( 1 + λ ) τ ¯ 2 = μ ( 1 + θ β s β c β ) ∂ z u ¯ ( y , z , s ) . (19)
Taking inverse Fourier transform of Equation (12) to get u ¯ ( y , z , s ) and then putting it into Equation (18), we obtain
τ ¯ 1 = 2 V ξ μ ( 1 + θ β s β c β ) h ( 1 + λ ) ∑ n = 1 ∞ sin ( n π z h ) × exp [ − y ξ 2 + s ( 1 + λ ) v c ( 1 + θ β s β c β ) ] [ 1 + s ( 1 + λ ) ξ 2 v c ( 1 + θ β s β c β ) ] . (20)
We express Equation (20) in series form in order to obtain a more suitable form of τ 1
τ ¯ 1 = 2 V ρ v h ∑ n = 1 ∞ sin ( n π z h ) ∑ j = 0 ∞ ∑ n = 0 ∞ ∑ q = 0 ∞ ∑ p = 0 ∞ ∑ w = 0 ∞ ∑ y = 0 ∞ ∑ z = 0 ∞ ∑ x = 0 ∞ ∑ m = 0 ∞ θ − n − p + w + x + y + z + β × y j ( − 1 ) j + n + p + q + w + y + z + x + m λ n − q s n − β ( p + n − w − y ) − m − 1 / 2 ξ − j + 2 n + 2 m v n − m j ! n ! q ! p ! w ! y ! z ! x ! m ! c n − β ( p + n − w − y ) − m − 1 / 2 × Γ ( p + n ) Γ ( q − n ) Γ ( n − j / 2 ) Γ ( w − 1 2 ) Γ ( x + 1 2 ) Γ ( m − 1 2 ) Γ ( y − m ) Γ ( z + m ) Γ ( n ) Γ ( − n ) Γ ( 1 / 2 ) Γ ( j / 2 ) Γ ( m ) Γ ( − m ) Γ ( 1 / 2 ) Γ ( − 1 / 2 ) . (21)
Taking the inverse N-transform of (21), we obtain
τ 1 = 2 V ρ v h ∑ n = 1 n = 1 sin ( n π z h ) ∑ j = 0 ∞ ∑ n = 0 ∞ ∑ q = 0 ∞ ∑ p = 0 ∞ ∑ w = 0 ∞ ∑ y = 0 ∞ ∑ z = 0 ∞ ∑ x = 0 ∞ ∑ m = 0 ∞ θ − n − p + w + x + y + z + β × y j ( − 1 ) j + n + p + q + w + y + z + x + m λ n − q t − n + β ( p + n − w − y ) + m + 1 / 2 ξ − j + 2 n + 2 m v n − m j ! n ! q ! p ! w ! y ! z ! x ! m ! × Γ ( p + n ) Γ ( q − n ) Γ ( n − j / 2 ) Γ ( w − 1 2 ) Γ ( x + 1 2 ) Γ ( m − 1 2 ) Γ ( y − m ) Γ ( z + m ) Γ ( j / 2 ) Γ ( n ) Γ ( n ) Γ ( − n ) Γ ( 1 / 2 ) Γ ( m ) Γ ( − m ) Γ ( 1 / 2 ) Γ ( − 1 / 2 ) Γ ( − n + β n + 1 / 2 ) (22)
Finally, using the Fox H-function we obtain the stress field as
τ 1 = 2 V v h ∑ n = 1 n = 1 sin ( n π z h ) ∑ j = 0 ∞ ∑ n = 0 ∞ ∑ q = 0 ∞ ∑ p = 0 ∞ ∑ w = 0 ∞ ∑ y = 0 ∞ ∑ z = 0 ∞ ∑ x = 0 ∞ ξ j − 2 n × y j ( − 1 ) j + n + p + q + w + y + z + x λ n − q t − n + β ( p + n − w − y ) + 1 / 2 θ n + p − w − x − y − z − β v n j ! n ! q ! p ! w ! y ! z ! x ! × H 1 , 8 8 , 10 { − ξ 2 t v | ( 1 − p + n , 0 ) , ⋯ , ( 1 − z , 1 ) ( 1 − j / 2 , 0 ) , ⋯ , ( 1 + n − β n , β ) } (23)
In the similar fashion we can find τ 2 ( y , z , t ) from Equations (16) and (19).
Following the procedure of the previous section, the expression for the velocity field is given by
u ( y , z , t ) = 2 V h ∑ m = 1 ∞ sin ( m π z h ) ∑ j = 0 ∞ ∑ n = 0 ∞ ∑ q = 0 ∞ ( − 1 ) j + n + p + q y j ξ j − 2 n v − n j ! n ! q ! p ! λ q − n θ n − β t n − β n − 1 × H 1 , 3 3 , 5 { t β θ | ( 1 − n , 1 ) , ( 1 − q + n , 0 ) , ( 1 − n + j / 2 , 0 ) . ( 0 , 1 ) , ( 1 − n , 0 ) , ( 1 + n , 0 ) , ( 1 − j / 2 , 0 ) , ( 1 + n − β n , β ) . } (24)
Adopting the methodology of the previous section, the resultant expression for the velocity field is given by
u ( y , z , t ) = 2 V h ∑ m = 1 ∞ sin ( m π z h ) ∑ j = 0 ∞ ∑ n = 0 ∞ ∑ q = 0 ∞ ( − 1 ) j + n + p + q y j ξ j − 2 n v − n j ! n ! q ! p ! λ q − n θ n − β t n − β n − 2 × H 1 , 3 3 , 5 { t β θ | ( 1 − n , 1 ) , ( 1 − q + n , 0 ) , ( 1 − n + j / 2 , 0 ) . ( 0 , 1 ) , ( 1 − n , 0 ) , ( 1 + n , 0 ) , ( 1 − j / 2 , 0 ) , ( 2 + n − β n , β ) . } (25)
By letting β = 1 in Equations (16), we get the velocity profile for an ordinary Jeffrey fluid induced by the impulsive motion of the plate
u ( y , z , t ) = 2 V h ∑ m = 1 ∞ sin ( m π z h ) ∑ j = 0 ∞ ∑ n = 0 ∞ ∑ q = 0 ∞ ( − 1 ) j + n + p + q y j ξ j − 2 n v − n j ! n ! q ! p ! λ q − n θ n − 1 × H 1 , 3 3 , 5 { t θ | ( 1 − n , 1 ) , ( 1 − q + n , 0 ) , ( 1 − n + j / 2 , 0 ) . ( 0 , 1 ) , ( 1 − n , 0 ) , ( 1 + n , 0 ) , ( 1 − j / 2 , 0 ) , ( n − n , 1 ) . } (26)
Similarly, we can get velocity field for an ordinary Jeffrey fluid due to impulsive accelerating plate and non-uniformly accelerating plate.
We have presented unsteady flows of a generalized Jeffrey fluid induced by impulsive motion of the plate between two side walls perpendicular to the plate. Exact analytical solutions are established for such flow problem using Fourier sine and N-transforms technique. The obtained solutions are expressed in series form using Fox H-functions. Several graphs are presented here for the analysis of some important physical aspects of the obtained solutions. The corresponding solutions for ordinary Jeffrey fluid is also obtained as limiting case of our general solutions. The numerical results show the profiles of velocity and the adequate shear stress for the flow. We analyze these results by changing different parameters of interest.
The effects of fractional parameters β of the model are important for us to be discussed. In
is noticeable that velocity and shear stress decreases by increasing y. Also, by increasing y the velocity becomes steady, which shows that the boundary condition (9) is satisfied.
In this paper, a new transform is used to obtain some exact solutions regarding Jeffrey fluid model. The N-transform is actually a generalization of Laplace transform. Fractional differential equation is involved in the governing equation, which is solved for the velocity with the help of finite Fourier transform. The flow is set into motion with the help of flat plat which is lying between two perpendicular plates. The series solution of velocity field and the associated shear stress in terms of Fox H-functions, satisfying all imposed initial and boundary
conditions have been obtained. The similar solutions for ordinary Jeffrey fluid, performing the same motion, appear as limiting case of the solutions are obtained here. Also, the obtained results are analyzed graphically through various pertinent parameters. Furthermore, the obtained solutions satisfy the governing equations and all imposed initial and boundary conditions.
Khan, A., Zaman, G., Ahmad, S. and Chohan, M.I. (2017) Some Exact Solutions of Generalized Jeffrey Fluid Using N-Transform. American Journal of Computational Mathematics, 7, 402-412. https://doi.org/10.4236/ajcm.2017.74029