on semi-inverse solutions for the time ...20], hayat et al. [6], and labropulu [8]. the main purpose...

ON SEMI-INVERSE SOLUTIONS FOR THE TIME-DEPENDENTFLOWS OF A SECOND-GRADE FLUID

MUHAMMAD R. MOHYUDDIN, S. ASGHAR, T. HAYAT, AND A. M. SIDDIQUI

Received 7 June 2005; Revised 16 December 2005; Accepted 15 January 2006

This paper deals with analytical solutions for the time-dependent equations arising ina second-grade fluid. The solutions have been developed by assuming certain forms ofthe stream function. Expressions for velocity components are obtained for flows in planepolar, axisymmetric cylindrical, and axisymmetric spherical polar coordinates. The ob-tained solutions are compared with existing results.

Copyright © 2006 Muhammad R. Mohyuddin et al. This is an open access article distrib-uted under the Creative Commons Attribution License, which permits unrestricted use,distribution, and reproduction in any medium, provided the original work is properlycited.

1. Introduction

Under several assumptions, the Navier-Stokes equations can be linearized, and closed-form solutions are available. It is known that in general, the governing equations of non-Newtonian fluids are nonlinear and much more complicated in comparison to Newton-ain fluids. To obtain analytical solutions of such equations is not easy. In order to explainseveral nonstandard features, such as normal-stress effects, rod climbing, shear thinning,and shear thickening, Rivlin-Ericksen fluids [18] of differential type were introduced. Thesecond-grade fluids form a subclass of the differential-type fluid. The Cauchy stress T ina second-grade fluid is [1, 11, 12, 15, 19]

T=−pI +μA1 +α1A2 +α2A21, (1.1)

where

A1 = (gradV) + (gradV)∗,

A2 = dA1dt

+ A1(gradV) + (gradV)∗A1(1.2)

in which V is the velocity vector, (∗) is the matrix transpose, p is the indeterminatepressure constrained by the incompressibility, d/dt is the material derivative, μ, α1, and

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2006, Article ID 54295, Pages 1–22DOI 10.1155/MPE/2006/54295

http://dx.doi.org/10.1155/S1024123X06542954

2 Unsteady semi-inverse solutions

α2 are material constants. They denote, respectively, the viscosity, elasticity, and cross-viscosity. These material constants can be determined from viscometric flows for any realfluid. Second-grade fluids are dilute polymers and the differential-type models can beused to describe the response of dilute polymers. A detailed account of the characteristicsof second-order fluids is well documented by Dunn and Rajagopal [4]. Fosdick and Ra-jagopal [5] have studied that the relation (1.1) is compatible with thermodynamics (theClausius-Duhem inequality and the assumption that the Helmholtz free energy is a min-imum in equilibrium when the fluid is at rest) if the following restrictions are imposedon the material constants:

μ≥ 0, α1 ≥ 0, α1 +α2 = 0. (1.3)

By assuming a certain form of the stream function, solutions for second-grade fluidswere obtained by Mohyuddin et al. [10], Kaloni and Huschilt [7], Siddiqui et al. [2, 19,20], Hayat et al. [6], and Labropulu [8].

The main purpose of the present communication is twofold. Firstly, to present theequations for the unsteady plane and axisymmetric flows of a second-grade fluids in po-lar, cylindrical, and spherical coordinates. Secondly, to obtain some analytical solutions ofthe governing equations in each case. We mention here that no boundary value problemis considered. The governing equations of the second-grade fluids are highly nonlinear.Moreover, these equations are of higher order than the Navier-Stokes equations. Thus,in general, to solve a well-posed problem for such a fluid, one requires additional ini-tial and/or boundary conditions. For a detailed discussion about this issue and for someinteresting examples, we refer the readers to the works by Rajagopal and Gupta [16], Ra-jagopal and Kaloni [17], and Rajagopal [13, 14]. The semi-inverse approach is used todeduce solutions of the governing time-dependent equations.

The outline of the paper is as follows. In Section 2, the basic laws are given. Section 3contains the modelling for unsteady flows for three cases. In Section 4, analytic solutionsare calculated for each case. The analytical solutions obtained are new and comparisonsare made with the solutions available in the literature.

2. Basic equations

The basic laws governing the motion of an incompressible, homogeneous, second-gradefluid are

div V= 0, (2.1)

−grad p+ div(μA1 +α1A2 +α2A21)

+ ρb= ρdVdt

, (2.2)

where div is the divergence and grad is the gradient operator. Substituting (1.2) in (2.2),we obtain

grad[

12ρ|V|2 + p−α1

(V·∇2V+ 1

4

∣∣A1

∣∣2)]

+ ρ[

Vt −V×(∇×V)]

= μ∇2V+α1[∇2Vt +∇2(∇×V)×V

]+(α1 +α2

)div A21 + ρb,

(2.3)

Muhammad R. Mohyuddin et al. 3

where t in the subscript denotes the partial derivative with respect to time and ∇2 is theLaplacian operator.

3. Compatibility equations

In this section, we give the compatibility equations in plane polar, axisymmetric cylin-drical, and axisymmetric spherical polar coordinates. The equations are constructed byeliminating the pressure field from the resulting equations in component form of (2.3)along with the continuity equation (2.1).

3.1. Unsteady plane flows in polar coordinates. Here, the velocity field is of the follow-ing form:

V=[u(r,θ, t),v(r,θ, t),0]. (3.1)

Making use of (3.1) into (2.1) and (2.3), we obtain in the absence of body forces

∂u

∂r+u

r+

1r

∂v

∂θ= 0, (3.2)

∂p̂

∂r+ ρ[∂u

∂t− vω

]=−1

r

(μ+α1

∂

∂t

)∂ω

∂θ−α1v∇2ω, (3.3)

1r

∂p̂

∂θ+ ρ[∂v

∂t+uω

]=(μ+α1

∂

∂t

)∂ω

∂r+α1u∇2ω, (3.4)

in which p̂ is the modified pressure and ω is the vorticity vector. The expressions for p̂and ω are

p̂ = p+ 12ρ(u2 + v2

)+α1

(u

r

∂ω

∂θ− v ∂ω

∂r

)−(3α1 + 2α2

)

4

∣∣A21

∣∣, (3.5a)

∣∣A21

∣∣= 4

(∂u

∂r

)2+ 4(

1r

∂v

∂θ+u

r

)2+ 2(∂v

∂r− vr

+1r

∂u

∂θ

)2, (3.5b)

ω = ∂v∂r

+v

r− 1r

∂u

∂θ, (3.5c)

∇2 = ∂2

∂r2+

1r

∂

∂r+

1r2

∂2

∂θ2. (3.5d)

The existence of continuity equation (2.1) allows us to define a stream function ψ(r,θ, t)as

u= 1r

∂ψ

∂θ, v =−∂ψ

∂r, (3.6)


which satisfies (3.2) identically and (3.3) and (3.4) become

∂p̂

∂r+ ρ[

1r

∂2ψ

∂t∂θ− ∂ψ∂r∇2ψ

]= 1r

(μ+α1

∂

∂t

)∂(∇2ψ)∂θ

−α1 ∂ψ∂r∇4ψ, (3.7)

∂p̂

∂θ− ρ[r∂2ψ

∂t∂r+∂ψ

∂θ∇2ψ

]=−r

(μ+α1

∂

∂t

)∂(∇2ψ)∂r

−α1 ∂ψ∂θ∇4ψ, (3.8)

where ω =−∇2ψ. Differentiating (3.7) with respect to θ and (3.8) with respect to r andusing the integrability condition p̂rθ = p̂θr , we obtain

ρ[r∂

∂t∇2ψ− {ψ,∇2ψ}

]= r(μ+α1

∂

∂t

)∇4ψ−α1

{ψ,∇4ψ}, (3.9)

in which

{ψ,∇2ψ}= ∂ψ

∂r

∂(∇2ψ)∂θ

− ∂ψ∂θ

∂(∇2ψ)∂r

. (3.10)

It is to be noted that the material parameter α2 does not contribute to the compatibilityequation (3.9). However, it does appear in the pressure distribution given in (3.5a).

3.2. Unsteady axisymmetric flows in cylindrical coordinates. The velocity field for thiscase is

V=[u(r,z, t),0,w(r,z, t)]. (3.11)

The continuity and momentum equations are

∂u

∂r+u

r+∂w

∂z= 0, (3.12)

∂p̃

∂r+ ρ[∂u

∂t−wΩ

]

=−(μ+α1

∂

∂t

)∂Ω

∂z−α1w

(∇2Ω− Ω

r2

)+

1r

(α1 +α2

)[

2∂

∂z(uΩ) +Ω2

],

∂p̃

∂z+ ρ[∂w

∂t+uΩ

]=(μ+α1

∂

∂t

)(∂Ω

∂r+Ω

r

)+α1u

(∇2Ω− Ω

r2

)− 2r

(α1 +α2

) ∂

∂r(uΩ),

(3.13)

where

p̃ = p+ 12ρ(u2 +w2

)+α1

[u(∇2u− u

r2

)+w∇2w

]−(3α1 + 2α2

)

4

∣∣A21

∣∣,

∣∣A21

∣∣= 4

(∂u

∂r

)2+ 4(∂w

∂z

)2+ 4(u

r

)2+ 2(∂u

∂z+∂w

∂r

)2,

Ω= ∂w∂r− ∂u∂z

, ∇2 = ∂2

∂r2+

1r

∂

∂r+∂2

∂z2.

(3.14)


Introducing the stream function ψ̃(r,z, t) through

u= 1r

∂ψ̃

∂z, w =−1

r

∂ψ̃

∂r, (3.15)

the continuity equation (3.12) is satisfied identically and (3.13) give

∂p̃

∂r+ ρ[

1r

∂2ψ̃

∂t∂z− ∂ψ̃∂r

E2ψ̃

r2

]

= 1r

(μ+α1

∂

∂t

)∂(E2ψ̃

)

∂z−α1 ∂ψ̃

∂r

E4ψ̃

r2− 1r

(α1 +α2

)[

2∂

∂z

(∂ψ̃

∂z

E2ψ̃

r2

)−(E2ψ̃

r

)2]

,

∂p̃

∂z− ρ[

1r

∂2ψ̃

∂t∂r+∂ψ̃

∂z

E2ψ̃

r2

]

=−1r

(μ+α1

∂

∂t

)∂(E2ψ̃

)

∂r−α1 ∂ψ̃

∂z

E4ψ̃

r2+

2r

(α1 +α2

) ∂

∂r

(∂ψ̃

∂z

E2ψ̃

r2

),

(3.16)

where

E2 = ∂2

∂r2− 1r

∂

∂r+∂2

∂z2, Ω=−1

rE2ψ̃. (3.17)

Elimination of p̃ from the above equations yields

ρ[

1r

∂

∂tE2ψ̃−

{ψ̃,E2ψ̃

r2

}]

= 1r

(μ+α1

∂

∂t

)E4ψ̃−α1

{ψ̃,E4ψ̃

r2

}− 1r

(α1 +α2

)[

2E2(∂ψ̃

∂z

E2ψ̃

r2

)− ∂∂z

(E2ψ̃

r

)2]

,

(3.18)

in which

{ψ̃,E4ψ̃

r2

}= ∂ψ̃∂r

∂

∂z

(E4ψ̃

r2

)− ∂ψ̃∂z

∂

∂r

(E4ψ̃

r2

). (3.19)

3.3. Unsteady axisymmetric flows in spherical coordinates. On using the velocity field

V=[u(R,θ, t),v(R,θ, t),0] (3.20)


into (2.1) and (2.3), we obtain the following equations:

∂u

∂R+

2uR

+1R

∂v

∂θ+v

Rcotθ = 0, (3.21)

∂p

∂R+ ρ[∂u

∂t− vΩ

]=−

(μ+α1

∂

∂t

)1

Rsinθ∂

∂θ(Ωsinθ)−α1v

(∇2Ω− Ω

R2 sin2 θ

)

+(α1 +α2

)[

2Rsinθ

∂

∂θ

(u

R+v

Rcotθ

)Ωsinθ +

Ω2

R

]

,

1R

∂p

∂θ+ ρ[∂v

∂t+uΩ

]=(μ+α1

∂

∂t

)(∂Ω

∂R+Ω

R

)+α1u

(∇2Ω− Ω

R2 sin2 θ

)

+

(α1 +α2

)

R

[− 2 ∂

∂R(u+ v cotθ)Ω+Ω

2cotθ

],

(3.22)

where

p = p+ 12ρ(u2 + v2

)−α1

⎡

⎢⎢⎢⎣

u(∇2u− 2u

R2+

2R2

∂v

∂θ− 2vR2

cotθ)

+v(∇2u+ 2

R2∂u

∂θ− vR2 sin2 θ

)

⎤

⎥⎥⎥⎦−(3α1 + 2α2

)

4

∣∣A21

∣∣,

∣∣A21

∣∣= 4

(∂u

∂R

)2+ 4(

1R

∂v

∂θ+u

R

)2+ 4(u

R+v

Rcotθ

)2+ 2(∂v

∂R− vR

+1R

∂u

∂θ

)2,

Ω= ∂v∂R

+v

R− 1R

∂u

∂θ, ∇2 = 1

R2∂

∂R

(R2

∂

∂R

)+

1R2 sinθ

∂

∂θ

(sinθ

∂

∂θ

).

(3.23)

Define the stream function in spherical polar coordinates as

u= 1R2

∂ψ

∂σ, v = 1

R√

1− σ2∂ψ

∂R, σ = cosθ. (3.24)

We see that (3.21) is satisfied identically and (3.22) lead to

∂p

∂R=−ρ

[1R2

∂2ψ

∂t∂σ− 1R2(1− σ2)

∂ψ

∂RD2ψ

]+

1R2

(μ+α1

∂

∂t

)D2ψ− α1

R2(1− σ2)

∂ψ

∂RD4ψ

(3.25)

+(α1 +α2

)

⎡

⎢⎢⎢⎢⎣

− 2R

∂

∂σ

{(1R4

∂ψ

∂σ+

σ

R3(1− σ2)

∂ψ

∂R

)D2ψ

}

+

(D4ψ

)2

R3(1− σ2)

⎤

⎥⎥⎥⎥⎦

,


∂p

∂σ=− ρ

1− σ2[∂2ψ

∂t∂R+

1R2

∂ψ

∂σD2ψ

]

+1

1− σ2(μ+α1

∂

∂t

)∂(D2ψ

)

∂R+

α1R2(1− σ2)

∂ψ

∂σD4ψ

−(α1 +α2

)

√1− σ2

⎡

⎢⎢⎢⎢⎣

2∂

∂R

(1R3

∂ψ

∂σ+

σ

R2(1− σ2)

∂ψ

∂R

)D2ψ√1− σ2

− σ(D2ψ

)2

R2(1− σ2)3/2

⎤

⎥⎥⎥⎥⎦

,

(3.26)

where

D2 = ∂2

∂R2+

1− σ2R2

∂2

∂σ2. (3.27)

Differentiating (3.25) with respect to σ and (3.26) with respect to R and using integrabil-ity condition pRσ = pσR, we obtain

ρ

⎡

⎢⎢⎢⎢⎣

∂

∂t

(D2ψ

1− σ2)

−{

ψ,D2ψ

R2(1− σ2)

}

⎤

⎥⎥⎥⎥⎦=(μ+α1

∂

∂t

)D4ψ

1− σ2 −α1{

ψ,D4ψ

R2(1− σ2)

}

+2(α1 +α2

)

1− σ2

⎡

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

D2

⎛

⎜⎜⎜⎜⎝

∂ψ

∂σ

D2ψ

R3

+σ

R2(1− σ2)

∂ψ

∂RD2ψ

⎞

⎟⎟⎟⎟⎠

−

⎛

⎜⎜⎜⎜⎝

D2ψ

R3∂

∂σ

+σD2ψ

R2(1− σ2)

∂

∂R

⎞

⎟⎟⎟⎟⎠D2ψ

⎤

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

(3.28)

in which

{

ψ,D2ψ

R2(1− σ2)

}

= ∂ψ∂R

∂

∂σ

(D2ψ

R2(1− σ2)

)

− ∂ψ∂σ

∂

∂R

(D2ψ

R2(1− σ2)

)

. (3.29)

4. Solutions

In this section, we apply the inverse methods to obtain the exact solutions of the nonlinearpartial differential equations (appearing in Section 3) by considering specific forms of thestream function.


4.1. Flow where ψ(r,θ, t)= rnF(θ, t). We choose

ψ(r,θ, t)= rnF(θ, t) (4.1)

in which the arbitrary function F depends upon θ and t, and n is an integer. Upon makinguse of (4.1), (3.9) can be written as

ρ[∂G

∂t−{nF

∂G

∂θ− (n− 2)∂F

∂θG}rn−2

]

=(μ+α1

∂

∂t

)Hr−2−α1

[nF

∂H

∂θ− (n− 4)∂F

∂θH]rn−4.

(4.2)

In the above equation,

G(θ, t)=(n2 +

∂2

∂θ2

)F(θ, t), (4.3a)

H(θ, t)=(

(n− 2)2 + ∂2

∂θ2

)G(θ, t). (4.3b)

Taking n= 0, (4.2), and (4.3) yield

ρ∂G

∂t= 0, (4.4a)

2ρ∂F

∂θG+μH +α1

∂H

∂t= 0, (4.4b)

4α1H∂F

∂θ= 0, (4.4c)

where

G= ∂2F

∂θ2, H = 4G+ ∂

2G

∂θ2. (4.5)

It is worth mentioning that for α1 = 0 and ∂t(·)= 0, we get Jeffery-Hamel flows [21]and for ∂t(·)= 0, we recover the analysis of reference [3].

Equation (4.4a) implies that G �= G(t) which shows that G is steady, and hence from(4.5) H is steady. From (4.4c), we assume ∂F/∂θ �= 0 (since ∂F/∂θ = 0⇒ F �= F(θ) whichcontradicts the assumption (4.1)) which implies that H = 0. Using this information in(4.4b), we get

2ρ∂F

∂θ

∂2F

∂θ2= 0. (4.6)

The solution of the above equation is

F(θ, t)= A(t)θ +B(t), (4.7)

where A(t) and B(t) are arbitrary functions.


Consequently, the expressions for stream function and velocity components are giventhrough (3.2) and (4.1) as

ψ(r,θ, t)= A(t)θ +B(t), u= r−1A(t), v = 0. (4.8)

For n= 1, (4.2) becomes

ρ[∂G

∂t−{F∂G

∂θ+∂F

∂θG}r−1]=(μ+α1

∂

∂t

)Hr−2−α1

[F∂H

∂θ+ 3

∂F

∂θH]r−4, (4.9)

which gives rise to the following equations:

∂G

∂t= 0, (4.10a)

∂

∂θ(FG)= 0, (4.10b)

(μ+α1

∂

∂t

)H = 0, (4.10c)

F∂H

∂θ+ 3

∂F

∂θH = 0, (4.10d)

where

G= F + ∂2F

∂θ2, H =G+ ∂

2G

∂θ2. (4.11)

Equation (4.10a) indicates that G is steady, and hence through (4.11) H is steady andfrom (4.10c), we get

∂2G

∂θ2+G= 0, (4.12)

whose solution is

G(θ, t)=A1(t)cosθ +B1(t)sinθ, (4.13)

where A1(t) and B1(t) are arbitrary functions of t. Substitution of (4.13) into (4.10b)yields

F(θ, t)= C(t)[A1(t)cosθ +B1(t)sinθ]−1

, (4.14)

whereC(t) is a function of integration. The stream function and velocity components are,respectively, given by

ψ(r,θ, t)= rC(t)[A1(t)cosθ +B1(t)sinθ]−1

,

u= C(t)[A1(t)sinθ−B1(t)cosθ][A1(t)cosθ +B1(t)sinθ

]−2,

v =−C(t)[A1(t)cosθ +B1(t)sinθ]−1

.

(4.15)


For n= 2, we have from (4.2) that

ρ[∂G

∂t− 2F ∂G

∂θ

]=(μ+α1

∂

∂t

)Hr−2−α1

[2F

∂H

∂θ+ 2

∂F

∂θH]r−2, (4.16)

which yields

∂G

∂t− 2F ∂G

∂θ= 0,

(μ+α1

∂

∂t

)H − 2α1 ∂

∂θ(FH)= 0,

(4.17)

where

G= 4F + ∂2F

∂θ2, H = ∂

2G

∂θ2. (4.18)

For viscous case (α1 = 0), we get(∂

∂t− 2F ∂

∂θ

)(4F +

∂2F

∂θ2

)= 0,

∂2

∂θ2

(4F +

∂2F

∂θ2

)= 0.

(4.19)

In order to solve (4.19), we introduce the travelling wave solution of the form

F(θ, t)= a2

+Q(s), s= θ + at, (4.20)

which (for Q �= 0) readily givesd3Q

ds3+ 4

dQ

ds= 0. (4.21)

Solving (4.21) and then inserting in (4.20), we obtain

F(θ, t)= a4 + a2 cos2(θ + at) + a3 sin2(θ + at), (4.22)where a, a2, a3, and a4 are arbitrary constants. The stream function and velocity compo-nents are

ψ(r,θ, t)= r2[a4 + a2 cos2(θ + at) + a3 sin2(θ + at)],

u= 2r[− a2 sin2(θ + at) + a3 cos2(θ + at)],

v =−2r[a4 + a2 cos2(θ + at) + a3 sin2(θ + at)].

(4.23)

For α1 �= 0, (4.17) gives

4∂F

∂t+

∂3F

∂t∂θ2− 8F ∂F

∂θ− 2F ∂

3F

∂θ3= 0, (4.24)

(μ+α1

∂

∂t

)(4∂F

∂θ+∂3F

∂θ3

)− 2α1

(4F

∂2F

∂θ2+∂4F

∂θ4

)= a5(t), (4.25)


where a5(t) is an arbitrary function. The possible solution of (4.25) for a5(t)= 0 is givenby

ψ(r,θ, t)= r2[θ0 cos2θ + θ1 sin2θ]eλ2t,

u= 2r[− θ0 sin2θ + θ1 cos2θ]eλ2t,

v =−2r[θ0 cos2θ + θ1 sin2θ]eλ2t,

(4.26)

where λ2, θ0, and θ1 are arbitrary constants.For other values of n, (4.2) requires

∂G

∂t= 0,

(μ+α1

∂

∂t

)H = 0, (4.27a)

nF∂G

∂θ− (n− 2)∂F

∂θG= 0, (4.27b)

nF∂H

∂θ− (n− 4)∂F

∂θH = 0, (4.27c)

where G and H are described in (4.3). From (4.27a), we get G �= G(t), and hence from(4.3b)H �=H(t) and we obtain thatH = 0. Equation (4.27c) is solved to give the followingsolution:

G(θ, t)= C1(t)F(n−2)/2, n �= 0, (4.28)

in which C1(t) is a function of integration.Equation (4.28) together with (4.3) forms a nonlinear partial differential equation for

the determination of F (except when n= 2), which is given as

∂2F

∂θ2+n2F = C1(t)F(n−2)/2. (4.29)

The solution for n= 1 and C1(t)= 0 is given by

ψ = r(A3(t)cosθ +B3(t)sinθ),

u=−A3(t)sinθ +B3(t)cosθ,v =−(A3(t)cosθ +B3(t)sinθ

).

(4.30)

The solution for n= 2 and C1(t) �= 0 is as follows:

ψ = r2[C1(t)

4+A4(t)cos2θ +B4(t)sin2θ

],

u= 2r[−A4(t)sin2θ +B4(t)cos2θ],

v =−2r[C1(t)

4+A4(t)cos2θ +B4(t)sin2θ

],

(4.31)

in which Ai(t) and Bi(t) (i= 3,4) are arbitrary functions.


For ψ = ψ(r, t), (3.9) becomes(μ+α1

∂

∂t

)∇4ψ− ρ∇2ψt = 0. (4.32)

On letting

ψ =Φ(r)eλt, (4.33)

(4.32) reads as

1r

∂

∂r

[r∂

∂r

{1r

∂

∂r

(r∂Φ

∂r

)}]− ξ2 1

r

∂

∂r

(r∂Φ

∂r

)= 0, (4.34)

which on simplification gives

r2d2Φ

dr2+ r

dΦ

dr− r2ξ2Φ= (A4 lnr +B4

)r2, (4.35)

where

ξ2 = ρλμ+α1λ

. (4.36)

For steady case, the solution of (4.32) is

ψ(r)= A5r2 lnr +B5r2 +C3 lnr, (4.37)

where A5, B5, and C3 are arbitrary constants, and the corresponding velocity componentsare

u= 0, v =−(C3r−1 +(A5 + 2B5

)r + 2A5r lnr

). (4.38)

Here we remark that the solution given in (4.37) is in good agreement to that given byMohyuddin et al. in [10].

The solution of (4.35) for A4 = B4 = 0 is

ψ(r)= A4I0(rξ) +B4K0(rξ) (4.39)

and the velocity components are

u= 0, v = ξ[−A4I1(rξ) +B4K1(rξ)], (4.40)

where In(x) and Kn(x) are the modified Bessel functions of the first and second kinds,respectively.

4.2. Flow for ψ̂(r,z, t)= rnF(z, t). Inserting

ψ̂(r,z, t)= rnF(z, t) (4.41)


into (3.18), we get

ρ

⎡

⎢⎢⎢⎣

n(n− 2)rn−3 ∂F∂t

+ rn−1∂3F

∂z2∂t− 4n(n− 2)r2n−5F ∂F

∂z

−{nF

∂3F

∂z3− (n− 2)∂F

∂z

∂2F

∂z2

}r2n−3

⎤

⎥⎥⎥⎦

=(μ+α1

∂

∂t

)

⎡

⎢⎢⎢⎣

n(n− 2)2(n− 4)rn−5F + 2n(n− 2)rn−3 ∂2F

∂z2

+rn−1∂4F

∂z4

⎤

⎥⎥⎥⎦

−α1

⎡

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

6n(n− 2)2(n− 4)r2n−7F ∂F∂z

+2n(n− 2){nF

∂3F

∂z3− (n− 4)∂F

∂z

∂2F

∂z2

}r2n−5

+{nF

∂5F

∂z5− (n− 2)∂F

∂z

∂4F

∂z4

}r2n−3

⎤

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

− 2(α1 +α2)

⎡

⎢⎢⎢⎣

3n(n− 2)2(n− 4)r2n−7F ∂F∂z

+ 2(n− 2)(3n− 2)r2n−5

∂F

∂z

∂2F

∂z2+(∂F

∂z

∂4F

∂z4+ 2

∂2F

∂z2∂3F

∂z3

)r2n−3

⎤

⎥⎥⎥⎦.

(4.42)

For n= 2, (4.42) reduces to

(μ+α1

∂

∂t

)∂4F

∂z4+ ρ(

2F∂3F

∂z3+∂4Ft∂z4

)= 2α1

(F∂5F

∂z5

)+ 2(α1 +α2

)(∂F

∂z

∂4F

∂z4+ 2

∂2F

∂z2∂3F

∂z3

).

(4.43)

The first integral of (4.43) is

μ∂3F

∂z3+ ρ(

2F∂2F

∂z2−(∂F

∂z

)2

− ∂Ft∂z

)= 2α1

⎛

⎜⎜⎜⎝

F∂4F

∂z4− 1

2∂3Ft∂z3

+(∂2F

∂z2

)2

⎞

⎟⎟⎟⎠

+α2

(

2F∂2F

∂z2+(∂2F

∂z2

)2)

,

(4.44)

where we have taken the function of integration in (4.44) equal to zero. In order to solve(4.44), we define

F(z, t)=N +Q(s), s= z+ 2Nt, (4.45)


and obtain the following equation:

μd3Q

ds3+ ρ

(

2Qd2Q

ds2−(dQ

ds

)2)

= 2α1(

Qd4Q

ds4+(d2Q

ds2

)2)

+α2

(

2dQ

ds

d3Q

ds3+(d2Q

ds2

)2)

,

(4.46)

where N is an arbitrary constant.Letting α1 = α2 = 0 in (4.46) and assuming that

Q =Asλ, (4.47)

we get the following relation:

μλ(λ− 1)(λ− 2)sλ−3 +Aρλ(2(λ− 1) + 1)s2(λ−1) = 0. (4.48)

On choosing λ = −1, we readily obtain A = 2ν (ν is the kinematic coefficient of vis-cosity). The expressions for the stream function (4.41) and velocity components (3.15)are

ψ̂(r,z, t)= r2[N + 2ν(z+ 2Nt)−1], (4.49)u=−2νr(z+ 2Nt)−2, (4.50)

v =−2[N + 2ν(z+ 2Nt)−1]. (4.51)

It is noted that the solutions (4.49)–(4.51) reduce to that of Berker solutions [3] whenN = 0.

For α1 �= 0 and α2 �= 0, we assume [9, 21] that

Q =A0(1 +C0eas

), s= z+ 2Nt, (4.52)

in (4.46) and get the following solution:

Q(s)=− μa2(ρ−α1a2

)(1 +C0eas

), (4.53)

where

a=√ρ(4α1 + 3α2

)−1(4.54)

and C0 is a constant. The stream function and velocity components now become

ψ̂(r,z, t)=[N − μa

2(ρ−α1a2

)(1 +C0ea(z+2Nt)

)]r2,

u=− μr2(ρ−α1a2

)C0ea(z+2Nt),

v =[−N + μa

2(ρ−α1a2

)(1 +C0ea(z+2Nt)

)]

2r.

(4.55)


The above expressions reduce to the results of Berker [3] for N = 0.For n= 0, (4.42) gives

ρ∂2Ft∂z3

−(μ+α1

∂

∂t

)∂4F

∂z4= 0, (4.56)

ρ∂F

∂z

∂2F

∂z2= α1 ∂F

∂z

∂4F

∂z4+(α1 +α2

)(∂F

∂z

∂4F

∂z4+ 2

∂2F

∂z2∂3F

∂z3

), (4.57)

(5α1 + 4α2

)∂F

∂z

∂2F

∂z2= 0. (4.58)

Since ∂F/∂z �= 0, (4.58) implies that

F(z, t)= a1(t)z+ a2(t), (4.59)

where a1(t) and a2(t) are arbitrary functions of time and the above expression leads tothe following values of the stream function and velocity components:

ψ̂ = F, u= r−1a1(t), v = 0. (4.60)

Writing

F(z, t)=Φ(z)eλ1t (4.61)

into (4.56) and then solving the resulting equation, we obtain

F(z, t)=(

a5eηz + a6e−ηz −

(a3z+ a4

)

η2

)

eλ1t, (4.62)

in which ai (i= 3 to 6) are arbitrary constants of integration and

η2 = ρλ1μ+α1λ1

, (4.63)

where λ1 is a constant. The stream function and velocity components are of the followingform:

ψ̂ = F, u= r−1 ∂F∂z

, v = 0. (4.64)

4.3. Flow where Ψ(R,σ , t)= RnF(σ , t). On specializing the solution of (3.28) to the form

Ψ(R,σ , t)= RnF(σ , t), (4.65)


we obtain the following equation:

ρ

⎡

⎢⎢⎢⎢⎢⎢⎢⎣

Rn−2∂G1∂t

−

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

nF∂G1∂σ

−(n− 4)∂F∂σG1

⎫⎪⎪⎪⎬

⎪⎪⎪⎭R2n−5

⎤

⎥⎥⎥⎥⎥⎥⎥⎦

= (μ+α1∂t)H1R

n−4−α1(nF

∂H1∂σ

−H1 ∂F∂σ

(n− 6))R2n−7

+2(α1 +α2

)

1− σ2

⎛

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

(1− σ2) ∂

2

∂σ2

(G∂F

∂σ+nσFG1

)

+(2n− 5)(2n− 6)

×(G∂F

∂σ+nσFG1

)

−{G∂G

∂σ+

(n− 2)σ1− σ2 G

2}

⎞

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

R2n−7,

(4.66)

where

G(σ , t)= n(n− 1)F + (1− σ2)∂2F

∂σ2, G1(σ , t)=G

(1− σ2)−1,

H(σ , t)= (n− 2)(n− 3)G+ (1− σ2)∂2G

∂σ2, H1 =H(1− σ2)−1.

(4.67)

For n= 0, we obtain∂G1∂t

= 0, ∂F∂σG1 = 0,

(μ+α1

∂

∂t

)H1 = 0,

−3H1α1 ∂F∂σ=−

(α1 +α2

)

1− σ2

⎡

⎢⎢⎢⎣

(1− σ2) ∂

2

∂σ2

(G∂F

∂σ

)+ 30G

∂F

∂σ

−(G∂G

∂σ− 2 σ

1− σ2G2)

⎤

⎥⎥⎥⎦.

(4.68)

Equations (4.68)1,3 and (4.67)2,4 imply that G and H are not functions of t, and hence(4.68)2 becomes

(1− σ2)∂F

∂σ

∂2F

∂σ2= 0. (4.69)

Since ∂F/∂σ �= 0, the above equation yieldsF(σ , t)= C0(t)σ +C1(t), (4.70)

where C0 and C1 are arbitrary functions and the stream function and velocity compo-nents are

Ψ= C0(t)σ +C1(t), u= C0(t)R2

, v = 0. (4.71)


Now (4.66) for n= 1 leads to the following:∂G1∂t

= 0, −ρ{F∂G1∂σ

+ 3∂F

∂σG1

}=(μ+α1

∂

∂t

)H1,

−α1(F∂H1∂σ

+ 5H1∂F

∂σ

)+

2(α1 +α2

)

1− σ2

×[(1− σ2) ∂

2

∂σ2

(G∂F

∂σ+ σFG1

)12(G∂F

∂σ+ σFG1

)−(G∂G

∂σ+

σ

1− σ2G2)]= 0,

(4.72)

where

G(σ , t)= (1− σ2)∂2F

∂σ2, G1 =

(1− σ2)−1G,

H(σ , t)= 2G+ (1− σ2)∂2G

∂σ2, H1 =

(1− σ2)−1H ,

(4.73)

and (4.3) implies thatG1 is steady, and henceH1 is steady. Using (4.73) in (4.3), we obtain

∂3

∂σ3

[2μ(1− σ2)∂F

∂σ+ 4μσF + ρF2

]= 0. (4.74)

Various steady cases of the above equation have been discussed by Squire [21]. In-tegrating (4.74) and employing a similar procedure as used by Landau and Lifshitz [9](after setting the functions of integrations equal to zero), we obtain a particular solutionof the form

F(σ , t)= 2ν(1− σ2)(σ − a)−1, (4.75)

where a is a constant.Substitution of (4.75) into (4.3)3 yields

96ν2[

1(σ − a)2 −

2σ(σ − a)3 −

1− σ2(σ − a)4

]

×[

α1

{11σ2− 1σ − a +

21σ(1− σ2)

(σ − a)2 +10(1− σ2)2

(σ − a)3}(α1 +α2

)

×{

2σ +1 + σ2

σ − a +7σ(1− σ2)

(σ − a)2 +4σ(1− σ2)2

(σ − a)3}]

= 0.

(4.76)

We note that the solution (4.75) cannot be obtained for all values of a. Siddiqui andKaloni [19] obtained the solution for the steady case when a = −1,0,1. For the sake ofcompleteness, we include it here briefly for the convenience of the reader. Setting a=±1,(4.76) is identically satisfied and (4.75) gives

F1,2 =∓2ν(1± σ) for a=±1. (4.77)


For a= 0 and 7α1 + 2α2 = 0, (4.75) becomes

F3 = 2νσ

(1− σ2). (4.78)

The stream function and velocity fields for the above three cases are

Ψ= RF1(σ , t), u=−2νR

, v =−2νR

(1 + σ)√1− σ2 ,

Ψ= RF2(σ , t), u=−2νR

, v = 2νR

(1− σ)√1− σ2 ,

Ψ= RF3(σ , t), u=−2νR

1 + σ2

σ2, v = 2ν

R

√1− σ2σ

.

(4.79)

For n= 2, (4.66) yields the following form:

∂G1∂t

= 0, ∂∂σ

(G1F

)= 0, (μ+α1∂t)H1 = 0,

−α1(F∂H1∂σ

+ 2H1∂F

∂σ

)+

(α1 +α2

)

1− σ2

×[(

1− σ2)+ ∂2

∂σ2

(G∂F

∂σ+ 2σFG1

)+ 2(G∂F

∂σ+ 2σFG1

)−G∂G

∂σ

]= 0,

(4.80)

in which

G= 2F + (1− σ2)∂2F

∂σ2, G1 =

(1− σ2)−1G,

H = (1− σ2)∂2G

∂σ2, H1 =

(1− σ2)−1H.

(4.81)

Equation (4.80)2 can also be written as

(1− σ2)−1F

[2F + (1− σ2)∂

2F

∂σ2

]= C(t). (4.82)

The solution of the above equation is

F(σ , t)= (σ2− 1)C̃1(t) + 14 C̃2(t)[− 2σ + (1− σ2){ ln(σ − 1)− ln(σ + 1)}] (4.83)

and the expressions for the stream function and velocity components are

Ψ= R2F(σ , t), v = 2(1− σ2)−1/2F(σ , t),

u= σ2

[4C̃1(t) + C̃2(t)

{ln(σ + 1)− ln(σ − 1)− 2

σ

}].

(4.84)


For n= 3, (4.66) gives rise to

∂G1∂t

−(

3F∂G1∂σ

+∂F

∂σG1

)= 0, (4.85)

(μ+α1∂t

)H1−α1

(3∂H1∂σ

F + 3H1∂F

∂σ

)

=−2(α1 +α2)[∂2

∂σ2

(G∂F

∂σ+ 3σFG1

)−(G1∂G

∂σ+ σG21

)],

(4.86)

where

G= 6F + (1− σ2)∂2F

∂σ2, G1 =

(1− σ2)−1G,

H = (1− σ2)∂2G

∂σ2, H1 =

(1− σ2)−1H.

(4.87)

For the steady and viscous cases, (4.86) gives μH1 = 0 which on using (4.87)3 becomes

(1− σ2)∂

2F

∂σ2+ 6F = k1σ + k2, (4.88)

in which k1 and k2 are integration constants. We see that the solution F obtained from(4.88) will only be satisfied by (4.85) when k1 = k2 = 0. The solution of (4.88) is given as

Ψ= R3F(σ), u= RF1(σ), v = 3R(1− σ2)−1/2F1(σ). (4.89)

In (4.89),

F(σ)= σ(σ2− 1)C̃3− 14 C̃4[ −4 + 6σ2

+3σ(σ2− 1)(1− ln(1 + σ))+ ln(σ − 1)

]

,

F1(σ)=(3σ2− 1)C̃3− 92 C̃4σ +

34

(3σ2− 1)C̃4

{ln(1 + σ)− ln(σ − 1)}.

(4.90)

For k1 �= 0 and k2 �= 0, we have the following solutions:

Ψ= R3F(σ), u= RdFdσ

, v = 3R(1− σ2)−1/2F(σ), (4.91)


in which

F(σ)= 124

⎡

⎣c6k2σ2 + 4

{k1σ3 + 6σ

(σ2− 1)C̃5 + 3C̃6

(2− 3σ2)}

+3σ(σ2− 1)(k2− 6C̃6

){ln(σ − 1)− ln(σ + 1)}

⎤

⎦ , (4.92)

where C̃i (i= 3 to 6) are constants.For n= 4, (4.66) reduces to

ρ[R2∂G1∂t

− 4F ∂G1∂σ

R3]= (μ+α1∂t

)H1−α1

[4F

∂H1∂σ

+ 2H1∂F

∂σ

]R

+2(α1 +α2

)

(1− σ2)

[(1− σ2) ∂

2

∂σ2

(G∂F

∂σ+ 4σFG1

)

+ 6(G∂F

∂σ+ 4σFG1

)−(G∂G

∂σ+ 2σ

(1− σ2)−1G2

)]R,

(4.93)

where

G= (1− σ2)∂2F

∂σ2+ 12F, G1 =

(1− σ2)−1G,

H = (1− σ2)∂2G

∂σ2+ 2G, H1 =

(1− σ2)−1H.

(4.94)

The following equations are obtained from (4.93):

∂G1∂t

= 0, F ∂G1∂σ

= 0, (μ+α1∂t)H1 = 0, (4.95)

α1

(2F

∂H1∂σ

+ 2H1∂F

∂σ

)= (α1 +α2

)

⎡

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

(1− σ2) ∂

2

∂σ2

(G∂F

∂σ+ 4σFG1

)

+6(G∂F

∂σ+ 4σFG1

)

−(G∂G

∂σ+ 2σ

(1− σ2)−1G2

)

⎤

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

. (4.96)

The first and third equations in (4.95) imply that G1 is steady, and hence H1 is steady.Since F �= 0, (4.95)2 gives ∂G1/∂σ = 0, which on using (4.94) have the following solution:

G= A6(1− σ2). (4.97)

On substituting (4.97) in (4.95)1, we can write

(1− σ2)∂

2F

∂σ2+ 12F =A6

(1− σ2). (4.98)


The solution of (4.98) for steady case is given by Berker [3]. In order to avoid repe-tition, we directly give the solution with the stream function and velocity componentsas

F = k3σ2(1− σ2),

Ψ= k3σ(1− 2σ2)R2,

u= 2k3σ(1− 2σ2)R2, v = 4k3σ2

√1− σ2R2,

(4.99)

where k3 is a constant.

5. Concluding remarks

In this paper, the governing time-dependent equations for plane polar, axisymmetriccylindrical, and spherical coordinates are constructed. Moreover, the analytical solutionsfor eleven nonlinear equations involving three-coordinate systems are given. The solu-tions obtained are found to be in good agreement to that of the previous steady solutionsfor viscous and second-grade fluids.

Acknowledgments

The authors gratefully acknowledge the support from URF Scheme of Quaid-i-AzamUniversity, Islamabad, Pakistan. Muhammad R. Mohyuddin is thankful to Dr. ArshadM. Khan (Director of Global Change Impact Studies Centre (GCISC)) and Dr. ShoaibRaza for their nice cooperation and encouragement in order to continue his research.

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[10] M. R. Mohyuddin, T. Hayat, F. M. Mahomed, S. Asghar, and A. M. Siddiqui, On solutions ofsome non-linear differential equations arising in Newtonian and non-Newtonian fluids, NonlinearDynamics 35 (2004), no. 3, 229–248.


[11] K. R. Rajagopal, On the decay of vortices in a second grade fluid, Meccanica 15 (1980), no. 3,185–186.

[12] , A note on unsteady unidirectional flows of a non-Newtonian fluid, International Journalof Non-Linear Mechanics 17 (1982), no. 5-6, 369–373.

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[14] , On boundary conditions for fluids of the differential type, Navier-Stokes Equations andRelated Nonlinear Problems (Funchal, 1994) (A. Sequeira, ed.), Plenum Press, New York, 1995,pp. 273–278.

[15] K. R. Rajagopal and A. S. Gupta, On a class of exact solutions to the equations of motion of a secondgrade fluid, International Journal of Engineering Science 19 (1981), no. 7, 1009–1014.

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[17] K. R. Rajagopal and P. N. Kaloni, Some remarks on boundary conditions for flows of fluids of thedifferential type, Continuum Mechanics and Its Applications (G. A. C. Graham and S. K. Malik,eds.), Hemisphere, District of Columbia, 1999, pp. 935–941.

[18] R. S. Rivlin and J. L. Ericksen, Stress-deformation relations for isotropic materials, Journal of Ra-tional and Mechanical Analysis 4 (1955), 323–425.

[19] A. M. Siddiqui and P. N. Kaloni, Certain inverse solutions of a non-Newtonian fluid, InternationalJournal of Non-Linear Mechanics 21 (1986), no. 6, 459–473.

[20] A. M. Siddiqui, M. R. Mohyuddin, T. Hayat, and S. Asghar, Some more inverse solutions for steadyflows of a second-grade fluid, Archives of Mechanics 55 (2003), no. 4, 373–387.

[21] H. B. Squire, The round laminar jet, Quarterly Journal of Mechanics & Applied Mathematics 69(1979), 335–380.

Muhammad R. Mohyuddin: Department of Mathematics, Quaid-i-Azam University,Islamabad-44000, PakistanCurrent address: System and Decision Sciences, Global Change Impact Studies Centre,61/A, 1st Floor, Saudi-Pak Tower, Constitutive Avenue, Islamabad, PakistanE-mail address: m [email protected]

S. Asghar: COMSATS Institute of Information Technology, Plot 30, Sector H-8,Islamabad, PakistanE-mail address: s [email protected]

T. Hayat: Department of Mathematics, Quaid-i-Azam University, Islamabad-44000, PakistanE-mail address: t [email protected]

A. M. Siddiqui: Department of Mathematics, Pennsylvania State University, York Campus,York, PA 17403, USAE-mail address: [email protected]

mailto:[email protected]:[email protected]:[email protected]:[email protected]

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