A theorem for a shear stress-free plane boundary in Stokes flow

Mechanics Research Communications 38 (2011) 529–531

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A theorem for a shear stress-free plane boundary in Stokes flow N. Akhtar a , S.K. Sen b,∗ a b

Department of Mathematics, Shahjalal University of Science and Technology, Sylhet 3114, Bangladesh Research Centre for Mathematical and Physical Sciences, University of Chittagong, Chittagong 4331, Bangladesh

a r t i c l e

i n f o

Article history: Received 24 December 2010 Received in revised form 12 July 2011 Available online 23 July 2011 Keywords: Stokes flow Biharmonic and harmonic functions Fourier transforms

a b s t r a c t A theorem for non-axisymmetrical Stokes flow about a shear stress-free plane boundary is established by using a representation for the velocity and pressure fields for the same flow in terms of biharmonic and harmonic functions. A corollary of the theorem is derived which gives the axisymmetrical Stokes flow in terms of the Stokes function about the same boundary. The formulae for drag and torque on the boundary are also given. A few illustrative examples are presented. © 2011 Elsevier Ltd. All rights reserved.

1. Introduction

2. Basic theory

Akhtar et al. (2004) suggested a representation for the velocity and pressure fields in a general non-axisymmetric Stokes flow incorporating two scalar functions one of which is biharmonic and the other is harmonic. By making use of this representation the authors stated and proved a general theorem for Stokes flow about a no-slip rigid plane boundary, and also recovered the Happel and Brenner solutions (1986) to the Stokes equations. Here the works of Collins (1954, 1958) on the axisymmetrical viscous fluid motion in terms of the Stokes stream function about a plane boundary are notable. Blake (1971), and Blake and Chwang (1974) analyzed, with the help of the Fourier transforms, the velocity and pressure fields in the Stokes flows due to the fundamental singularities, bounded by a no-slip plane boundary. In this paper we examine non-axisymmetric Stokes flow before a shear stress-free plane boundary by using the above mentioned representation. Relevantly, on the other hand Harper (1983) gave a sphere theorem for axisymmetric Stokes flow past a shear stress-free sphere in terms of the Stokes stream function with applications to rising bubbles. Usha and Hemalata (1993) gave a circle theorem for plane Stokes flow around an impermeable and shear stress-free circular cylinder, showing the model for a two phase fluid system. By using another representation due to Palaniappan et al. (1990a) for the velocity and pressure fields in a general threedimensional Stokes flow, a sphere theorem for non-axisymmetric flow past a shear free sphere was stated and proved (Palaniappan et al., 1990b).

The steady and slow three-dimensional motion in an incompressible viscous fluid is governed by the Stokes equations: ∇ 2 qˆ = grad p,

(1)

and div qˆ = 0,

(2)

where qˆ is the fluid velocity, p the fluid pressure,  the constant viscosity coefficient, and where 2 is the Laplacian operator. A general solution (Akhtar et al., 2004) to equations (1) and (2) in cylindrical coordinates (, , z) which appears to be suitable for treating a problem of viscous fluid motion before a plane boundary is, for reference, stated as follows. The fluid velocity qˆ = (q , q , qz ) and the fluid pressure p may represent a solution to the Stokes Eqs. (1) and (2), if q =

1 ∂B ∂2 A + ,  ∂ ∂ ∂z

(3)

q =

1 ∂2 A ∂B − ,  ∂ ∂z ∂

(4)

qz = −

1 ∂  ∂

p = po + 





∂A ∂



∂ (∇ 2 A), ∂z

−

1 ∂2 A , 2 ∂2

(5)

(6)

where po is a constant, and where the scalar functions A and B satisfy respectively the biharmonic equation: ∗ Corresponding author. Tel.: +880 31 2865545. E-mail address: [email protected] (S.K. Sen). 0093-6413/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.mechrescom.2011.07.005

∇ 4 A = 0,

(7)

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N. Akhtar, S.K. Sen / Mechanics Research Communications 38 (2011) 529–531

and the Laplace equation:

∇ 2 B = 0.

(8)

where 0 (, z) is the Stokes stream function for the basic axisymmetric flow in the unbounded fluid, whose singularities are all in the region z ≥ 0.

We now introduce the shear stress-free boundary conditions for a viscous fluid motion about the rigid plane z = 0, and these are, on z = 0:

4. Drag and Torque on the boundary

qz = 0,

Let the fluid exert on the boundary z = 0, the drag Fˆ and torque Tˆ which are obtained from the formulae:

where

z = 0 and



z = 

∂q 1 ∂qz +  ∂ ∂z

z = 0,

(9)



 and

z = 

∂q ∂qz + ∂z ∂



 ,

A = 0,

∂B = 0. ∂z

(11)

3. Stokes flow before a shear free plane boundary Now we state and prove a theorem for non-axisymmetric Stokes flow about a shear free plane boundary. Theorem. Let there be a slow flow, in an unbounded incompressible viscous fluid, characterized by a biharmonic function A0 (, , z) and a harmonic function B0 (, , z), whose singularities are in the region z ≥ 0. If the plane z = 0 is a shear stress-free boundary, then the new flow in the same region is given by the biharmonic function: A(, , z) = A0 (, , z) − A0 (, , −z),

(12)

and the harmonic function: B(, , z) = B0 (, , z) + B0 (, , −z).

(13)

Proof. Since A0 (, , z) is a biharmonic function and B0 (, , z) is a harmonic function, it is simple to prove that A0 (, , −z) and B0 (, , −z) are also respectively biharmonic and harmonic functions. Next by using the expressions (12) and (13) one may easily work out that: on z = 0, A(, , z) = 0,

(14)

∂2 A0 ∂2 A0 ∂2 A (, , z) = (, , z) − (, , −z) = 0, ∂z 2 ∂z 2 ∂z 2

(15)

and ∂B0 ∂B0 ∂B (, , z) = (, , z) + (, , −z) = 0. ∂z ∂z ∂z

(16)

Thus the boundary conditions (11) are satisfied. We then see that if the point (, , z) lies in the region z ≥ 0, the point (, , −z) lies in the region z ≤ 0. Therefore, since the singularities of A0 (, , z) and B0 (, , z) are by hypothesis in the region z ≥ 0, those of A0 (, , −z) and B0 (, , −z) are clearly in the region z ≤ 0; this implies that the expressions (12) and (13) have respectively the same singularities as A0 (, , z)and B0 (, , z) in the region z ≥ 0. Thus the proof of the theorem is complete. Corollary. For the axisymmetric flow about the z-axis, we replace A(, , z) by A(, z), set B(, , z) = 0 and take =  ∂A/∂, where = (, z) is the Stokes stream function. Thus from the results (12) and (13), we obtain the Stokes stream function (, z) for the axisymmetric flow about the z-axis before the shear free plane boundary z = 0 as: (, z) =

0 (, z) −

0 (, −z),

=0

(10)

characterize the shear stresses (Batchelor, 1967). By making use of the velocity components (3)–(5), the shear stresses (10) can be expressed in terms of the biharmonic function A and the harmonic function B; then the boundary conditions (9) clearly result in the simple forms; that is, on the plane z = 0: ∂2 A = 0 and ∂z 2

∞



2

Fˆ =

(17)



∞

(zz )z=0 eˆ z  d d,

(18)

(−zz )z=0 eˆ  2 d d,

(19)

∂qz , ∂z

(20)

=0



2

Tˆ = =0

=0

zz = −p + 2

where  zz denotes the z-component of the stress exerted on a plane element normal to the z-axis, and where eˆ z and eˆ  are respectively the unit vectors in the direction of the positive z-axis and the direction of the increasing azimuthal angle . By making use of the results (6), (12) and (20), the above drag and torque formulae assume the standard forms:



∞



2



Fˆ = =0

and



6



∞

=0



2



Tˆ =

∂qoz ∂z

 6

=0

=0

∂qoz ∂z





− 2 z=0



 − 2 z=0

∂3 Ao ∂z 3

∂3 Ao ∂z 3





ˆ d d, (21) k z=0



 z=0

(ˆi sin  − ˆj cos )2 d d,

(22)

where qoz is the z-component of the basic fluid velocity corresponding to the biharmonic function A0 , while ˆi, ˆj and kˆ are the unit vectors in the directions of the positive x-, y- and z-axes respectively. 5. Illustrative examples 5.1. Stokeslet before a shear free plane boundary ˆ be situated Let a Stokeslet of strength ˛ ˆ = 1/8(1ˆj + 2 k) at the point specified by the Cartesian coordinates (0, 0, c) in an unlimited viscous fluid. In this case the primary fluid velocity in cylindrical coordinates (, , z) may be characterized by its components, which are qo =

1 8

qo =

1 8

qoz =

1 8





1 2 sin  + 2 (z − c) 1 sin  + , R1 R3

 cos 

1 

R1

(23)

1

,

(24)



2 (1  sin  + 2 (z − c))(z − c) + , R1 R3

(25)

1



2 + (z − c)2 . where R1 = We now determine the corresponding biharmonic function A0 and the harmonic function B0 referred to in the above theorem. This is possible by solving the following differential equations formed by substituting (23), (24) and (25) respectively in the left hand sides of the expressions (3), (4) and (5): 1 ∂B0 1 ∂2 A0 + =  ∂ 8 ∂∂z





1 sin  1 2 sin  + 2 (z − c) + , (26) R1 R3 1

N. Akhtar, S.K. Sen / Mechanics Research Communications 38 (2011) 529–531

1 ∂2 A0 ∂B0 1 − =  ∂∂z 8 ∂

−



1 ∂  ∂

 ×





∂A0 ∂

−

 cos 

2 R1

,

(27)

1 ∂2 A0 1 = 8 2 ∂2

(28)

The relevant exact solutions of these equations for A0 and B0 are: 1 8



1



R1 (z − c) sin  − 2 R1 ,

(29)

and B0 = −

1 1 R1 cos . 4 

(30)

Thus the above theorem clearly gives the solution to the problem of the viscous fluid motion before the shear free boundary z = 0 in the forms: A=

+

1 8



1



 

1



R1 (z − c) sin  − 2 R1

R2 (z + c) sin  + 2 R2



 ,

(31)

and B=−

1 1 (R1 + R2 ) cos , 4 

(32)

where R2 = (2 (z + c)2 )1/2 , and where the terms associated with R2 consist of the images of A0 and B0 in the region z ≤ 0. The evaluation of the integrals (21) and (22) then yields the drag Fˆ and torque Tˆ on the shear free boundary, which are ˆ Fˆ = 2 k,

and Tˆ = −1ˆi.

M 16



1 2R1 − R1 2





(z − c) +

1 2R2 − R2 2





(z + c) sin 2, (39)



1

A0 =

By applying the above theorem one now gets the solution to the problem of the viscous flow due to a Stokes-doublet before the shear free boundary z = 0, as: A=

2 (1  sin  + 2 (z − c))(z − c) + , R1 R3

531

(33)

and B=

M 8



× cos 2 −

2R1 1 − 2 R1  1 R2



,

 cos 2 +

1 − R1



2R2 1 − 2 R2 



(40)

where the terms associated with R2 constitute the image system in the region z ≤ 0. Next the drag Fˆ and torque Tˆ on the shear free boundary, on the straightforward evaluation by the formulae (21) and (22) are found to be zero.

6. Conclusions A theorem is derived for the solution of the problem of a nonaxisymmetrical slow viscous fluid motion induced by singularities in a region bounded by an impermeable and shear stress-free plane boundary; and this is done by making use of a representation for the velocity and pressure fields in a slow viscous fluid flow in terms of a biharmonic function A and a harmonic function B. Here one may note the simplicity of the boundary conditions on A and B, which enables us to formulate the theorem. A formula, as a corollary of the same theorem for obtaining the solution to the problem of an axisymmetrical fluid motion before a shear free plane boundary is also given in terms of the Stokes stream function. The application of the theorem is shown in the solution of the problem of a viscous fluid motion due to (i) a Stokeslet before a shear free plane boundary and (ii) a Stokes-doublet before the same boundary; in each case the drag and torque on the plane are obtained.

5.2. Stokes-doublet in the presence of a shear free plane boundary

Acknowledgement

Consider a Stokes-doublet (Palaniappan et al., 1992) of strength M, located at the point with Cartesian coordinates (0, 0, c) in an unbounded viscous fluid. In this case the velocity components of the fluid motion in cylindrical coordinates (, , z) are

The authors are grateful to the referee for his helpful corrections and suggestions.

qo =

3M 3 sin 2, 16 R5 1

(34)

q0 =

3M  , 8 R3 1

(35)

q0z =

3M 2 (z − c) sin 2. 16 R15

(36)

We then determine the corresponding biharmonic function A0 and harmonic function B0 in the usual manner; and their appropriate expressions are Ao =

M 16

and M Bo = 8



2R1 1 − R1 2





1 2R1 − 2 R1 

(z − c) sin 2,



1 cos 2 + R1

(37)

 .

(38)

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