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Stokes Flow Problems

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MSU Is An Affirmative Action/Equal Opportunity Institution
cmmmud

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A MATCHING TECHNIQUE FOR SOLVING
STOKES FLOW PROBLEMS

By

C. Lanette Poteete
A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements

for the Degree of
DOCTOR OF PHILOSOPHY

Department of Mathematics

1993

ABSTRACT

A MATCHING TECHNIQUE FOR SOLVING STOKES FLOW PROBLEMS

By

C. Lanette Poteete

This thesis investigates using a matching technique to analytically solve Stokes
flow problems for those problems whose domains can be divided into contiguous sim-
ple regions. In each region the solution is written in terms of infinite eigenfunction
expansions with unknown coefficients. The expansions arise from a separation of
variables process. Completeness for these expansions is shown. These eigenfunctions
satisfy some (but not all) of the boundary conditions exactly - the other bound-
ary conditions are satisfied using Fourier series methods. Then the solutions and
their higher derivatives (when necessary) along the interfaces of adjacent regions are
equated. The resulting equations are inverted using Fourier series methods and to-
gether with any previously determined equations are truncated and solved for the
remaining unknown coefficients. The technique is applied to seven different Stokes

flow problems with quite satisfactory results.

ACKNOWLEDGEMENTS

I would like to acknowledge many people for their generous and loving support
during the long and tedious process of completing this research. Dr. C. Y. Wang was
invaluable to me and never gave up on me even when I was willing to give up on myself.
I am very grateful to my family who supported me emotionally and prayerfully as
well as to so many friends who contributed much tangible and intangible help. Thank
you to Dr. Hills, Dr. Hill, Dr. Yen, Dr. Lo, and Dr. Kuan for being willing to serve on
my committee and for their patience and support. Thanks also to Peter Springer for
his help on more than one occasion. Above all, I want to acknowledge that without
the help and comfort of the Holy Spirit this would never have been finished. God has

proven Himself faithful to me once again.

iii

Contents

LIST OF TABLES vi
LIST OF FIGURES Vii
1 Introduction 1
1.1 Literature review ............................. 1
1.2 Background ................................ 4
2 Parallel Flows 7
2.1 Formulation ................................ 7
2.2 Completeness Proofs ........................... 10
2.3 Matching Rectangular and Polar Coordinates - Example I ...... 16
2.4 Matching Rectangular Coordinates - Example II ............ 20
3 Two-Dimensional Flows 32
3.1 Forn‘iulation ................................ 32
3.2 Completeness ............................... 35
3.3 Matching Rectangular Coordinates - Example 111 ........... 37
3.4 Matching Rectangular Coordinates - Example IV ........... 40

iv

4 Axisymmetric Flow
4.1 Formulation ................................
4.2 h’latching Spherical Coordinates - Example V .............
4.3 Matching Spherical Coordinates - Example VI .............

4.4 Matching Spherical Coordinates - Example VII ............

5 Discussion

A Example III Revisited

B Definitions of Functions Used in Chapter 3

C Definitions of Gegenbauer Polynomials and Related Identities

LIST OF REFERENCES

54

54

61

70

73

80

82

85

List of Tables

4.1

A.1

Coefficients for Example I ........................ 22
Coefficients for Example 11 ........................ 27
Coefficients for Example III ....................... 4O
Coefficients for Example lV ....................... 49
Coefficients for Example V ........................ 61
Numerical Coefficients .......................... 76

vi

List of Figures

togopiogogogogo
«loam-nearer—

CI)

Cfitthx'Ji-d

3.6
3.7
3.8

4.1
4.2
4.3
4.4

A.1
A.2
A.3

Two Contiguous Rectangular Regions .................. 12
Geometry for Example I ......................... 16
Plots of W1(O,y) and W11(7',7r/6) .................... 21
Geometry for Example II ......................... 23
Plot of W1(a,y) .............................. 28
Plot of W1(a,y) and Wu(a,y) ...................... 29
Plot of W11.(a,y) and Wux(a,y) ..................... 30
Equivelocity in Regions I and II ..................... 31
Geometry for Example III ........................ 37
Plot of \Iil(a.y) .............................. 41
Plot of \III(;r, l) .............................. 42
Streamlines for Example III ....................... 43
Geometry for Example IV ........................ 44
Plot of W1(a,y) .............................. 51
Plot of \P/(a,y) — \Iln(—c,y) ....................... 52
Streamlines for Regions I and II ..................... 53
Geometry for Example V ......................... 55
Plot of w;(1,() and 'wu(1,C) ....................... 60
Geometry for Example VI ........................ 61
Geometry for Example VII ........................ 64
Plot of U’(a,y) .............................. 77
Plot of U'(:1:,1) .............................. 78
Streamlines of U’(:c, y) .......................... 79

vii

Chapter 1

Introduction

1 . 1 Literature review

The Navier-Stokes equations are the basic equations for fluid mechanics, and much
effort has been expended finding analytic solutions when possible. Since the Navier-
Stokes equations are highly nonlinear the number of these solutions is few. The
Navier-Stokes equations are governed by the Reynolds number, a dimensionless pa-

rameter which signifies the relative importance of inertial effects to viscous effects:

fluid density x speed x size

 

Reynolds number = . .
VISCOSlty

Situations in which the Reynolds number is small are called Stokes flows. In these
cases the Navier-Stokes can be simplified by the omission of the inertial terms as com-
pared to the viscous terms (ie., setting the Reynolds number to zero). The resulting
equations are called the creeping equations or Stokes equations. These equations can
be expected to apply to the situation where the Reynolds number is quite small and
they provide, in many cases, physically meaningful solutions of boundary-value prob-
lems involving flow at sufficiently low Reynolds number, such as the flow in narrow

crevices, the microcirculation, or the flow of a highly viscous fluid [28].
The purpose of the research contained in this thesis is to investigate the use of

1

a matching technique to solve Stokes flow problems in various complex geometries.
This method seems to have good success in problems where geometrically the flow
field splits into simple regions. In each of these simple regions the solution may
be written as an infinite series with unknown coefficients. The coefficients are then
determined by matching this representation and its higher derivatives at the common
interfaces between the regions. This idea has been done in various ways by different
authors. Dagan, Weinbaum, and Pfeffer [14] used this method to find an infinite-
series solution for axisymmetric Stokes flow through a pore in a wall. They split the
flow field into two regions: a cylindrical volume bounded by the walls of the pore and
the entrance and exit planes, and an infinite half-space outside the pore. Since their
choice of infinite series gave rise to Fourier-series, Fourier-Bessel series, and Dini-
series representations which were all invertible over the finite matching interface, a
system of linear equations in terms of the unknown coefficients was obtained. They
then solved this infinite system by successive truncations that yielded values for the

unknown coefficents.

The same authors also used, in two separate articles, a boundary collocation
technique to derive the infinite set of linear equations in the case no inversion or
orthogonality relationships were applicable for the common interface. The first of
these papers [15] presented an infinite-series solution to the creeping-flow equations
for the axisymmetric motion of a sphere of arbitrary size towards an orifice whose
diameter is either larger or smaller than the sphere. The flow field was partitioned in
the plane of the opening; and for the flow to either side of the fluid interface separate
solutions were developed that satisfy the viscous-flow boundary conditions in each
region and unknown functions for the axial and radial velocity components on the
plane of the opening. The continuity of the fluid stress tensor at the matching interface

led to a set of dual integral equations which was solved analytically to determine the

unknown functions for the velocity components in the matching plane. A boundary
collocation technique was then used to satisfy the no—slip boundary conditions on the

surface of the sphere.

The second paper [13] examined the general axisymmetric creeping motion of a
spherical particle in a stagnation region of a finite planar surface. This flow was
modeled by studying the axisymmetric motion of a sphere towards a disk of arbitrary
size and by the uniform flow past such a configuration. Again two different stream-
function representations were chosen: one for the region containing the sphere and
one for the remaining half-space. The no-slip boundary conditions for the disk in each
region were satisfied exactly on the disk surface in terms of the unknown velocity at
the plane of the disk. First, the two fields were matched by requiring continuity of
the velocity at the interface. Next continuity of the normal and tangential stress-
tensor components provided a matching condition which gave the solution for the
interfacial velocity in terms of the remaining unknown constant coefficients in the
series solution for the disturbances produced by the sphere. Finally, the no-slip
boundary conditions for the sphere were satisfied at discrete points on the surface
of the sphere. yielding numerical results for the spherical coefficients. In all three of

these papers, convergence and accuracy were tested numerically.

Similarly a matching technique was used by Trogdon and Joseph [64] to study
plane flow of a. fluid over a rectangular slot. They split the region of flow into four
separate subregions. The stream function in each region was then expressed in terms
of the Papkovich—Fadle eigenfunctions which arise in a natural way from separation of
variables. Requiring continuity of velocities and stresses across the common bound-
aries along with satisfying the boundary conditions gave relationships that were then
inverted using certain “biorthogonality” conditions related to the Papkovich-Fadle

eigenfunctions. These biorthogonality conditions led to an infinite system of linear

equations which was then solved by successive truncations.

More recently Phillips [51] used the same eigenfunction expansions to describe
Stokes flow around a two-dimensional constriction. The flow region was decomposed
into a number of simpler rectangular regions, then within each subregion the solution
of the partial differential equation was approximated in terms of the Papkovich-Fadle
eigenfunctions. Biorthogonality conditions were applied and, as before, the resulting

infinite system of linear equations was solved by truncation.

The method described in this thesis avoids several of the difficulties arising in the
previously mentioned papers. One of the major advantages is that all the work here in-
volves r_eal eigenvalues and eigenfunctions whereas the Papovich-Fadle eigenfunctions
are complex-valued. This simplifies the computational work considerably. Another
simplification, due to the choice of eigenfunctions, is that boundary-collocation is not
needed. The boundary-collocation technique can require long computational times

for its numerical evaluations.

1 .2 Background

To begin. a. derivation of the creeping motion or Stokes equations will be made
from the Navier-Stokes equations for constant density fluids. In vector form these
equations are:

21+(v-V)v=f—1Vp+y—V2V (1.1)
0i p p

an d

V-v=0 (1.2)

where v is the \x'elocity vector, p is the constant density, f is external body force vector,

p is the pressure, and )1. is the dynamic viscosity. If the velocites, length scales, and

pressures are normalized and if the flow is steady Equation 1.1 becomes

(v . V)v = fig—v1) + Vzv) (1.3)

where Re is the Reynolds number as defined in Section 1.1. In the limit of very small

Re, we have the Stokes equations:

and

plus any boundary conditions.

The problems examined in this paper fall into three categories; i) parallel flows,
ii) two-dimensional flows, and iii) axisymmetric flow problems. For parallel flow
problems velocity is in one direction only, so Equation 1.5 is automatically satisfied,
and Equation 1.4 becomes

V2u(y,z) = 3:3; (1.6)

where the velocity vector is v = (u(y, z),0, O). This is a Poisson equation which can
be solved by harmonic functions, but only for a very limited number of classes of

simple regions.

For two-dimensional flows the concept of a stream function is introduced which
will automatically satisfy the continuity Equation 1.5. In cartesian coordinates we

define the stream function W as follows:

u(.r,y) 2' wyfxay)

and

Many) = -‘1’r($,y)

where the velocity is v = (11(ar,y),v(a:,y),0). Eliminating the pressure from the
resulting Navier-Stokes equations and using the stream function notation results in

the biharmonic equation:

V411: = 0 (1.7)

For axisyn‘u'netric flows a stream function is again introduced. This time the

resulting equation can be shown to be
E411: = 0 (1.8)

where, in spherical coordinates,

02 1 (92 cot0 8

2——. _————_____—_—
E ‘ arfieaaz r2 09

H"

Analytic solutions to Equations 1.6, 1.7, and 1.8 may be obtained by complex
transforms, integral representations, or separation of variables. These methods, how-
ever, are limited to problems with very specific simple geometries whose boundaries
are coordinate surfaces of a separable orthogonal system. Thus for geometries which
are complex, for example an I.-shape(l region, the above methods fail. The present
thesis introduces a. method which solves problems whose complex domain can be par—
titioned into simple contiguous regions. Each sub-region is solved by separation of
variables using real eigenfunctions. We shall consider in detail the resulting series

solutions for each of the above mentioned three cases.

Chapter 2

Parallel Flows

2. 1 Formulation

In the case of a parallel flow with zero pressure gradient, solving the Stokes equa-

tions reduces to finding solutions for the two-dimensional form of Laplace’s equation,

. 82W 82W
21/ 1’ = — = 2.1
\7 i (9.1.2 + By? 0 ( )

 

where IV(.r. y) is the longitudinal velocity. If we assume a separable solution, where

IV(.r,y) = .\"(.r)l"'(g) then substituting into Equation 2.1 gives

X"Y + XY” = 0

Dividing both sides of this eqation by XY yields the following relationship:

 

 

[Y]! Y'II
_ _T = 0
X + 1'
So that for some real constant 7,
‘xru }/II

If 7 = X2 > 0, we have

X" — A2X = 0 and Y" + A2Y = 0

7

each of which is an ordinary differential equation and easily solved yielding results of
the form

X(.’L') = {416w + Bye—Ax

and

1"(y) = C; cos Ag + D,\ sin Ay
where AA, 31,, CA, and DA are constants of integration.

If '7 = —fi2 < 0, then the roles of x and y are reversed in the above computations

and we get solutions of the form
X(.r) = A3 cos 5:17 + Bg sin ,8]:

and

my) = (Jae-”y + Die-”y
In the degenerate case when 7 = O, we have solutions of the form
13((413): A1417 + A2
and
Vi?!) = A3?! + A4
Thus we have just proven the result:

Theorem 2.1 If H/’(.z:,y) = X(;r)Y(y) and W(:I:,y) satisfies Equation 2.1, then

IV(:1:,y) can be 'l‘fi'])'i‘(i-S‘€iif€(f by a linear combination offunctions of the form

(AM-3"Jr + Bye—”MCI cos Ag + D; sin Ay) (2.3)
(A5 cos fix + BB sin ,BICXCgCfiy + Dfie‘fiy) (2.4)

where A, /3, AA, By, C,\, D3, A5, Be, 03, D3, A1, A2, A3, and A4 are Ed constants.

Completeness of the above representation is discussed the next section.

Next let us consider the solution to Laplace’s equation in polar coordinates (r, 0)

which has form
82W 1 0W 1 WW

2/: __ __
V” as +r8r +r2302

= 0 (2.6)

 

If we assume a separable solution of the form W(r, 0) = R(r)Z(0), then Equation 2.6
becomes

R”Z +11% + 35122" = 0
T T‘

Multiplying both sides of the equation by r2 and dividing by RZ yields the following

relationship:
.2 R” RI Z"
Li + _ = 0
R Z
so that for some real constant 7
7.2 R” + TR’ _ — Z”
R " 7 ‘ z

If 7 = A2 > 0, we have 1'21?” + rR' —— A2}? = 0 and Z” + A2Z = 0 which have solutions
of the form

Rfi) = AVA + BAT—-A

and

Z(0) 2 CA cos A0 + D,\ sin A0

If ’7 = —/32 < 0, we have 7'2R” + rR’ + 62R 2 0 and Z” — 622 = 0 which have

solutions of the form
R(r) = A3 cos(flln r) + Bg sin(flln r)

and

2(0) 2 C3 sinh [30 + Dg cosh 50

10

Finally if 7 = 0, the equations become r212” + rR’ = 0 and Z” = 0. These

equations have solutions of the form

Rf?) 2‘ A] ll'lT' + A2

and

2(6) = A30 + A,

So we have now proven the result:

Theorem 2.2 [fill/"(130) = I?(r)Z(0) and W(r,0) satisfies Equation 2.6, then

I/V(r,0) can be written as a linear combination offunctions of the form

(AVA + B,\r"\)(C,\ cos A0 + 0,», sin A0) (2.7)
(Ag cos(;’3 ln 1') + 8;; sin(,L’31n r))(Cg sinh 00 + Dg cosh 00) (2.8)
(Al 111 r + A2)(A30 + A4) (2.9)

where A, [3, AA, By. (K, ()3, .415, 13,317,), 05, A1, A2, A3, and A4 are real constants.

2.2 Completeness Proofs

Theorem 2.3 The representation of Theorem 2.1 is complete for a rectangle with

Dirichlet 's boundary conditions.

First consider Laplace‘s equation in a rectangle 0 S :1: S a, 0 S y S b defined by
VQIV(:L',y) = 0 (2.10)

l’V(;r,b) = f1(.’r) and lV(a:,0) = f2(a:) (2.11)

Vita-av) = 91(31) and W(0ay) = 92(31) (2-12)

11

This is known as Dirichlet ’s problem. This problem has been shown to have a unique
solution for any bounded domain (for details see [57]). Since Laplace’s equation is

linear we may split the problem into two problems as follows:

V2W1(:r,y) = 0 (2.13)
W1(r,b) = fie), Wicca) = Mm) (214)
W1(a.y) = 0, W1(0,y) = 0 (2.15)
and
VZI’I’f2(.I',y) = O (2.16)
W2(.r, b) = 0, W’2(r,0) = 0 (2.17)
l/V2(a.y) = glib/)9 W2ioay) = 92(9) (2-18)

Then W(.r, y) = I’l»’,(.r,y) + l/I"'42(.r,y) provides a. solution to Equations 2.10, 2.11 and

2.12. Assume W1 has a separable solution in the form

“Hf-17,31) : :anx)Fn(y) (219)

where

Gn(fr)Fn(y) = (An cos that + Bn sin t3n$)(Cn6‘3"y + Due—’3”)

as in Theorem 2.1(2.4) and 73,, are real eigenvalues, A" and 8,, are real constants for

which C3,,(a) = G.,,(0) = 0. (i.e. [3,, = n7r/a and An = 0). Then

2: G,,(.r)F,,(b) = f1(:r)

n=1

and
Z Gn(;1~.)F,,(0) = f2(1’)
n=1
where F,,(0) = ('n + D". anb) = Cne"”b/“ + Dne’""b/“. These are both Fourier sine

series representations of the functions f1(:r) and f2(:r) respectively. If f1(.1:) and f2(:r)

12

are continuous on the interval 0 S :r S a, f1(0) = f1(a), f2(0) = f2(a), and f{(:r) and
f5(.r) both exist on that interval, then the Fourier sine series converge to the value
f,(:r), i=1,2, at each a: in the interval. The Fourier coefficients Fn(b) and Fn(0) are

unique and defined by

Fn(b) = 2/a/Oa f1(:r)sin nnx/a dx

and
1711(0) = 2/a/a f2(a‘.)sinn7r:r/a (la:
0

See [11]. Since [7,, involves two unknown coefficents for each n, these two conditions
uniquely determine W1(;r, 3;). Using a similar argument for W2(:r, y) by expressing it
as a separable function in the form of the representation given in Theorem 2.1(2.3),
with similar continuity requirements for gl(y) and g2(y), we again have determined
the unique solution l/I1’2(.r,y) which in turn gives us a representation for W(:r, y) on the

rectangle. This proves the completeness of the representations given in Theorem 2.1.

Theorem 2.4 The representations in Theorem 2.1 are complete for a region consist-

ing of two contiguous rectangles of arbitrary sizes.

Suppose we have two contiguous rectangular regions as shown in Figure 2.1. The
problem is to find a solution for Dirichlet’s problem for this more complicated geome-

try still using the representations of Theorem 2.1. Since the region is bounded, we are

 

 

 

 

II

 

 

Figure 2.1: Two Contiguous Rectangular Regions

13

guaranteed a unique solution. Cut this region into two simple rectangular regions as
shown and label them Region I and Region II. Let W(O, 3]) along the out be given by
h1(y). Let BW/Bn along the cut be given by h2(y). Then, as shown in Theorem 2.3,
using the original boundary conditions when applicable and using h1(y) when needed,
W1(:r,y) has a unique solution for Region I in the desired series form. Similarly for
Region 11, using the original boundary conditions when applicable and h1(y) when

needed, VV-2(.r, y) has a unique solution in the desired form. Since
Wifoail) = l11(31): W2(07y)
and

0W1 __ 8W2

 

 

along the cut,

nuamunMem=WWaw

uniquely. Thus the representation of Theorem 2.1 is complete for this more compli-

cated rectangular region as well.

Theorem 2.5 The representation of Theorem 2.2 is complete for a polar sector with

Dirichlet ’s boundary conditions.

Consider Laplace’s equation in polar coordinates for a polar sector —a S 0 S a,

a S r S a + 1 with boundary conditions as defined below:

W + 11W. + 162W
0r? r 0r r2 602

W(r, —a) = h1(r)

 

vWV= :0 (an)

”WWH=MU)

WMflL=th

14
we. +1,0) = 114(6)

As before this problem has a unique solution for a bounded domain. To show this
solution can be represented by a linear combination of functions as given in Theo-

rem 2.2, we will split the problem into two separate problems as follows:

V2W1(r,0) = 0 (2.21)
W'1(a,0) = 0 = W1(a +1,0) (2.22)
W1(r, —-a) = hm) and W1(r,a) = 112(7) (2.23)
and
Wit/m, 0) = 0 (2.24)
new, in) = 0 (2.25)
wz(a,0) = 113(9) and W2(a +1,9) = h,(0) (2.26)

Thus W1(r, 0) + l/V-2(r, 0) = l/V(r, 0) is the unique solution to Dirichlet’s problem on a
polar sector. Suppose IV1( r, 0) has a separable solution in the form
I'Vl(r,0) = Z: Gn(r)Fn(0)

71:1

where
Gn(r)F,,(0) = (An cos(fi,, 111 r) + B" sin(fln ln r))(C,, sinh 0,10 + Dn cosh 0,,0)

as in Theorem 2.2 and 0,, are real eigenvalues (i.e., 0,, = W), A, and 8,, are

real constants for which G,,(a) = Gn(a + 1) = 0. Then

(X)

Z G',(7-)F,,(a) = 712(7)

15

where 17,,(01) : Cn sinh tine-+0., cosh 0nd, Fn(—a) = -—C,, sinh fina+Dn cosh find each
of which can be inverted as a Fourier-series with an appropriate change of variables
over the interval a S r S a + 1. If h1(r) is continuous on the interval a S r S a + 1,
h1(a) = 11.1(a+1), 11.2(a) = 112(a+1), and h’1(r) and h’2(r) both exist on that interval,
then the Fourier series converge to the value h,-(r), i = 1,2 at each r in that interval.
The Fourier coefficients F,,((1) and Fn(—a) are unique and since Fn involves two
unknown coefficients for each 11, these two conditions uniquely determine W1(r,0).

Similarly for W2(r, 0), we assume a separable solution of the form

I[V2 ( 7‘ a 0)

||
53
E?
V
:3

Q
V

where

C,,(7')F,,(0) : (A,,1"“" + B,,r_’\")(Cn cos An0 + Dn sin An0)

as in Theorem 2.2 and A" are real eigenvalues, A" and 8,, are real constants for which

Fn(ia) : 0. Then
Gn(a)Fn(6) : [13(6)

71:1

S: Gn(a-+1)F,,(0): [14(0)

11:1

can each be inverted as a Fourier-series in 0 over the interval —a S 0 S 0. Again if
the continuity requirements are met, we have a unique solution for W2(r,0) which in
turn gives a unique solution W(r,0) : W1(r,0) + W2(r,0) in the polar sector. Thus

the representation given in Theorem 2.2 is complete for the polar sector.

Theorem 2.6 The representations of Theorem 2.1 and 2.2 are complete for regions

consisting of contiguous polar sectors and rectangles.

Using arguments as in the proof of Theorem 2.4 is it not difficult to show that any

contiguous regions whose boundaries coincide with orthogonal coordinate axes can

16

be separated, solved independently, and then reunited to give the unique solution to

the one-dimensional flow problem.

2.3 Matching Rectangular and Polar Coordinates
- Example I

For our first example of this technique we will consider shear flow between two
parallel curved plates whose cross section is shown in Figure 2.2. The top plate is
moving with unit velocity normal to the figure and the bottom plate is fixed. Here
we split the region of flow into two simpler regions, Region I and II as indicated in
Figure 2.2. We will use rectangular coordiantes for Region I and polar coordinates

for Region II. The governing equations for Region I are as follows:

V2W1(.r, y) = 0 (2.27)

W,(.r, l.) = 1 (2.28)

W,(.r,0) = 0 (2.29)

Wi(—oo. y) = 1/ (‘3-30)

The boundary condition at .r = 00 arises from the expectation that far downfield

from the bend in the plates the disturbance from the bend will not be felt. This is

      

 
 

Q

E

Figure 2.2: Geometry for Example I

17

a characteristic of Stokes flow. If we subtract off the expected shear flow at infin-
ity, letting W1(;r,y) = y + w[(.z:,y) then the problem reduces to solving Laplace’s
Equation with zero boundary conditions on the walls and at infinity. A flow which
satisfies these boundary conditions can be found from Theorem 2.1 by the following
method. Suppose w,(.r,y) is written as a linear combination of functions as given in

Theorem 2.1. Then forcing the boundary condition at y = 0 gives
0 2 (Ala' + A2)Bg + Dl(Cle’\x + 026'”) + (F1 + F2)(E1 cos 0:1: + E; sin 02:)

for all :r. Hence we must have 82 = 0, D1 = 0, and F1 = —F2. The boundary

condition at y = 1 implies
0 : (A141:+.42)B, + 02 sin A(C1e’\r + 6'26"”) + F2(e’6 — e’3)(El cos 132‘ + E2 sin 0.1:)

for all 1:. Thus we must have Bl : 0, 02 sinA : 0, and F2 2 0. Hence A 2 mr. So

far we have the following
I/I""](.r, y) = 0,, sin 117r(C'1,,e"’”r + Che-n”)

We still need to consider the boundary condition at a: = —oo,wl : 0. That forces
C2,, : 0. So we have the solution for Region I written in the form
00
W1(.r.,y) = y + Z A,,e°’"” sin any (2.31)
11:1

where C1,, = n7r.

For Region II it is more convenient to work with polar coordinates. In this case
we will solve for the velocity after subtracting off a concentric flow which satisfies the

non-homogeneous boundary conditions. Here we have

 

, . _ Ine/a)
W]1(7‘,0) — 111((a+1)/a) + 1011(7', 0) (2.32)

18

Using Theorem 2.2 to describe w11(r, 0) and satisfing the boundary condition at r : a

requires
0 = (A10 + A2)(31 lna + B2) + (01 C03 39 + 02 sin A0)(Dla’\ + DZa-Al'l'

(Elem + Ege’figflFl cos(0ln a) + F2 sin(flln a))

which forces 81 lna = —Bg, 01a" 2 —D2a"‘, and F1 cosfllna = -F2 sinflln a. Next

the boundary condition at r = a + 1 requires
0 = (A10+ A2)Bl(ln(a+1)-—ln a) +(C1 cos A0+C2 sin A0)D1((a+1)A —a2’\(a+ 1)”\)+

—F2(sin(;'31n a) cos(;’3 ln(a + 1)) — cos(,0ln a) sin(flln(a +1)))(Elef39 + Ege‘fig)

for all 0. This forces Bl = 0, D1 = 0, and

cos(/3ln(a + l))sin(flln a) — cos(0ln a) sin(flln(a +1)) = 0

 

or
sin(,01n a +1) : 0
a
So that
an

 

3,, =
/ ln((a+1)/a)
Due to symmetry we require the velocity to be even with respect to 0 = 0. So instead

of using a linear combination of exponential functions, we will use the equivalent form

of cosh 07,0. Thus we now have the solution for Region II in the form

_ ln r/a
I’I/II(T‘, 0) —- W
f: 13,, cosh fln0(sin(fln In a) cos(fl,, ln r) — cos(fl,, In a) sin(fl,, ln r)) (2.33)
11:1
where 0,, = M

l11(a+l)/a‘

19

Next the matching conditions will be applied along the edge :16 = O in Region I and
0 = a in Region II. Along the common interface we have r = y so the first matching

condition gives us:

r—a+ZA,,sina,,(r—a)=

11:1

111 r/a

W + Z 8,, cosh ,‘3na(si11(5,, In a) cos(fi,, ln r) — cos(fl,, In a) sin(fln ln r))

11:1

(2.34)

where on = mr and 13,, = W—a’fi—Tl—a. The second matching condition requires continuity

of the shear stresses along the common interface. For Region I the shear stress is

defined to be 1% while for Region II the shear stress is %. Matching these gives the

following relationship:

1x»
Z Anon sin 01,,(r — a) =

11:1

— l/r 2 13,113,. sinl1/3,,(1(sin(/3,, In a) cos(fl,, ln r) — cos(fl,, In a) sin(fi,, lnr)) (2.35)

11:1

Inverting both of the above equations as Fourier series in r from a to a + 1, results

in the following two linear relationships

A, _ 2 :3 8,, cosh 5,,(1F1(a,n,k) = 201(k) (2.36)
11:1
A), + 2/(kir) Z 31151. sinh finaF2(a,n,k) = 0 (2.37)
11:1

where
n+1

171(a,n,k) = sin(t3,, lna)/ cos(fl,, ln r) sin ak(r — a) dr—

0.

0+1
cos(fi,, In a)/ sin(/3,, ln r) sin 01),.(r — a) dr
n+1
F;(a,n,k) = sin(/3,, In a)/ l/r cos(0,, ln r) sin ak(r — a) dr—
n+1
l/r sin(13,, 111 r) sin ak(r — a) dr

cos(/3,, In a)/

a

20

and

"+1 1n r/a

sin ak(r — a) (W _

01(1) =/ (r — a)) dr

Combining Equations 2.36 and 2.37 and eliminating A), we have a single system of

linear equations to solve using successive truncations:
“(taJlejl = —i7rG'1(i) (2-38)
where the input matrix a(i,j) has form

fljtanh(flja)Fz(a,j.i) +z'7rF1(a,j,i) (2.39)

and

B(j) 2 Bi cosh 0,01

Then
A), = —2/(h7r) Z B(n)t3,, tanh ,0,,aF2(a,n,k)

11:1

The functions F1, F2, and C, were evaluated and the system solved exactly using
MATHEMATICA. The results are shown in Table 2.1 for a = 2.0 and a = 71/6. Both
W1(0, y) and I'I/“(ix 71/6) are shown in Figure 2.3. The convergence is quite rapid, as

is expected, with a relative error of at most approximately 0.05.

2.4 Matching Rectangular Coordinates - Exam-
ple II

In this section we shall look at a simple example of the technique using a matching
of two regions in rectangular coordinates. Let us consider longitudinal parallel flow
between two corrigated plates driven by a pressure gradient as shown in Figure 2.4.
Here we have two regions labeled Region I and Region II. The flow field in each

region will be considered independently of the other region and a series solution for

 

21

 

 

Figure 2.3: Plots of W1(0,y) and Wu(r,1r/6)

[\D
[\D

 

n=2 n=3 n=4 n=5 n=6 n=7
B(1) 0.02610 0.02612 0.02612 0.002612 0.02612 0.02612
B (2) —0.00083 —0.00058 —0.00058 —0.00058 —0.00058 —0.00058
B ( 3) 0.00102 0.00096 0.00098 0.00098 0.00098
B (4) —0.00016 —0.000062 —0.000069 —0.000065
B ( 5) 0.00023 0.00020 0.00021
B (6) —0.000066 —0.000014
B(7) 0.000091
A( 1) 0.02602 0.02602 0.02602 0.02602 0.02602 0.02602
A(2) 0.00123 0.00123 0.00124 0.00123 0.00123 0.00123
A(3) 0.00105 0.00104 0.00104 0.00104 0.00104
A(4) 0.00015 0.00015 0.00015 0.00015
A(5) 0.00022 0.00023 0.00022
A(6) 0.00004 0.00004
A(7) 0.00008

 

 

 

Table 2.1: Coefficients for Example I

each region will be written in terms of the linear combinations of functions as given in

Theorem 2.2. The governing equation for the flow in Region I is the Poisson equation:

VZWI = 1 (2.40)

I«'V1(.r,ib) = 0 (2.41)

ii’,(a,y) = 0 for —b + 2 < y < b (2.42)
I/V1(-a,y) = 0 for —b < y < b — 2 (2.43)

while the governing equations for Region II are:
WW” = 1 (2.44)

W[1(;L",i1) = 0 (2.45)

Equations 2.41, 2.42, 2.43, and 2.45 are the no-slip boundary conditions. The addi-
tional boundary conditions are that W1(:z:,y) and W11,(:r,y) match with Wu(:r’,y’)
and Wn(:r’,y’) along the common interface and the velocities in both regions are

symmetric with respect to their respective polar symmetry points.

23

 

 

 

2b ,_.__.

 

 

 

II

I
I
: l_. .
, X
I 9

1 V I

 

 

 

 

p—- 2a—’|‘— 2c ——b|

Figure 2.4: Geometry for Example II

In order to satisfy Equation 2.40 we consider the flow defined by the parabolic
profile 1/2(y2 — bl). This flow satisfies Equation 2.40 and also satisfies the no slip
boundary conditions at ib. If we define W1(:r, y) = l/2(y2 — b2) + w1(:r,y), then the
problem reduces to solving

V2101 = 0 (2.46)
w,(.r,ib) = 0 (2.47)

A separable flow which satisfies both of these equations and the symmetry condition
can be found from Theorem 2.1 by a. method similar to that used in the previous
example and written in the form:

w1(:z:, y) = 2 sin any/(An sinh ans) + 2 cos 0,,y(Bn cosh 0,,16) (2.48)

11:1 n:0

where 0,, = 1171/ b and 0,, = (n + 1 /2)7r/ b. These choices for an and 0,, are made
for convenience sake and are not necessarily the only choices that could be made.
For example, we choose these deliberately in order to satisfy the no-slip boundary
conditions exactly on the upper and lower walls of the flow. We will force the side-
to-side boundary conditions to be satisfied in series form. Hence we have written

W1(.r, y) in series form using coefficients that will be determined from the remaining

24

boundary conditions. At this point we have

W1(.r, y) = 1/2(y2 — b2) + 2 sin Ont/(An sinh aux) + 2 cos 0,,y(Bn cosh 0,,1) (2.49)

11:] 11:0

where anzmr/b and 0,, (n+1/2)71/b.

Now let us consider Region II. Here we have translated our coordinate system to
the primed system at’ = a: — a — c and y’ : y + b — 1. Again we consider Wn(:r’, y’) =

1/2(y’2 — 1) + wu(.r’, y’) so that the problem reduces to finding solutions for
V21011(:c',y') = 0 (2.50)

1011(13' +1): 0 (2.51)

A separable flow which satisfies both of these equations and the symmetry condition

can be found from Theorem 2.1 and written as:

w”(. :2: si11c1"n(Cn sinh 0112') +2: cos 0:,y'(Dn cosh 022') (2.52)
11:0
where a’ _ -1171 and 0’: (n + 1/2) )71. So we have 1n series form the solution for flow

in Region II:

Wu(:ry’)=1/2(1)+Zsina'y'(Cnsinha/:r')+2cos0ny"(Dncosh011')
11:0
(2.53)
where a; 2 mt and 0:, = (n + 1/2)71. It remains to satisfy the side-to-side boundary

condition:

VV;(a,y) — { 0 If -b+2 < y Sb (2.54)

l/V]1(—C,y + b — 1) If —DS 3] S —b+2
We also need to match the sheer stress along the common interface of the two regions.

This condition gives the relationship:

W1,.(a, y’ — b +1) 2 14/11,,(—c, y’) for — 1 S y’ S 1 (2.55)

25
Thus for Equation 2.54 we have the following relationship which can be inverted as
a Fourier series in y over the interval —b to b.

1/2y2 — b2/2 + Z A], sin any + 2 Bl, cos 0,,y =

n:1 n:0

l/2(y+b—1)2—1/2— :11 C,’,sina£,(y+b—l)+ (2.56)

0 if —b+2 < y Sb
2:6 Di. cos 3:.(1 + b — 1) if —bg y g —b+2

where for computing purposes we put A; = A,,/ sinh ana, B; = Bn/ cosh 0,,a, Cl, :
Cn/ sinh o'gc, and D:, = D,,/ cosh 0;,c. Inverting Equation 2.56 as a Fourier series in

y over the interval -b to b gives the following two relationships:

A11. + Z (mam) — Z D;F,(b,n,k) = G,(b,k) (2.57)
n:1 n:0
31.1) + Z: c; 1215,61) _ 2 D;F,(b,n,k) = Gz(b,k) + G3(b,k) (2.58)

n:l n:0

where

_b+2
F1(l),11,k) : [b sin 0111(1) + b —1)sinakydy

—b+2
F2(b,n,k) = [b cos 0:,(3/ + b — 1) sin aky dy

—b+2
173(b,n,k)=/l sin a:,(y + b — 1) cos 0),}; dy
—)
—b+2
F4(b,n,k) = b cos 0:,(y + b —1)cos 01y dy
—b+2
C1(b,k) = 1/2((y+b-1)2—1)sincrkydy

—b

b
65111.1) = [0 (1.2 — yocos my d1

and
—b+2 1

G'3(b.k) = _b 5((6 + b -1)2 - 1) COS [31y dy

Next let us consider the sheer stress matching condition along the common inter-
face. Here we must have I’ll"),,.(a, y' + b — 1) = Wu,(—c, y’). This equation we invert
as a Fourier series in y’ from —1 to 1. The following two relationships are obtained.

2 A20, coth a,,aF1(b,k,n) + 2 81,0” tanh 0,,aF3(b,k,n) —C,',c1jc coth aficc : 0 (2.59)

n:l n:0

2 211,6, coth (1,,a172(b,k.n)+ Z 8:615:11] 6,,aF,(b,k,n)+D;,6; tanh 6,16 = 0 (2.60)

n:l n:0

where the functions F1(l),11,k), F;(b,n,k), F3(b,n,k), and F4(b,n,k) are defined as
before. Equation 2.77 and Equation 2.59 both hold for k: 1, 2,. . . while Equation 2.58
and Equation 2.60 both hold for k: 0,1,2, . . .. This gives us an infinite set of linear
equations relating the unknown coefficients A1,, Ba. 0;, and 0;. Solving successively
larger truncations of this system for specific values of a, b, and c yields the required
results. The linear system was solved using the Linpack subroutines SGECO and
SGESL. The functions [71(b,n,k), F2(b,n,k), F3(b,n,k), F4(b,n,k), G1(b,k), Gg(b,k),
and C3(b,k) were evaluated exactly using MATHEMATICA. The results for a : 1.3,
b = 4.1, and c = 3.4 are given in Table 2.2. We show the plot of W1(a,y) to show
the convergence of the solution to the side boundary condition in Figure 2.5. Also
plotted in Figures 2.6 and 2.7 are both functions and their derivatives along the
common interface. Some equivelocity lines are shown for both Regions I and II in

Figure 2.8. We have truncated to only the first 6 terms in each set of coefficients.

The accuracy would improve if further terms are kept.

27

 

n=1 n=2 n=3 n=4 n=5 n=6 n=7

A’] 0.1 ‘53 0.1683 0.1590 0.1572 0.1538 0.1531 0.1518
A’2 —0.1899 —0.1723 —0. 1726 —0. 1700 —0. 1698 —0. 1688
A3 0.6225 0.0673 0.0698 0.0704 0.0711
A; 0.0232 0.0153 0.0149 0.0137
[’1’5 —0.0154 —0.0166 —0.0174
2, —0.0164 —0.0125
A5, 0.0131
36 8.569 8.578 8.583 8.584 8.586 8.587 8.5875
B] —0.1204 —0.1342 -0.1355 —0.1392 —0.1398 —0.1411
B; —0.0552 -—0.0578 —0.0574 —0.0576 -0.0575
Bf, —0.0130 —0.0074 —0.0068 —0.0055
Bf, 0.0368 0.0371 0.0366
B; —0.0057 —0.0081
B", —0.0138
C] 0.0877 0.0975 0.1420 0.1122 0.0782 0.0772 0.0725
5 —0.024—2 —0.0308 —0.0301 —0.0359 —0.0362 —0.0458
Cf, 0.0135 0.0132 0.0153 0.0154 0.0174
Cf, —0.0074 -0.0085 —0.0086 —0.0095
C g 0.0054 0.0055 0.0060
C 5, —0.0038 —0.0042
C; 0.0030
D6 —0.3526 —0.24 92 —0.1759 +0.1781 -0.1680 —0.1672 —0.l637
Di 0.043 0.0562 0.0545 0.0709 0.0604 0.0430
D5, —0.0195 —0.0191 —0.0223 —0.0225 —0.0260
D3 0.0097 0.0111 0.0112 0.0125
Df, —0.0067 -0.0067 —0.0075
D’5 0.0045 0.0050
Dg, 0.0054

 

 

Table 2.2: Coefficients for Example II

 

28

O
u
vvrvvvv'VVYV'YVVV

 

Figure 2.5: Plot of W1(a,y)

29

 

 

I-Slsiiifi-lBA T
Wilfaiy)
Wlfany)

Figure 2.6: Plot of W1(a,y) and Wn(a,y)

30

VVI:(&, 3])

v T

      
 

Wllr(a'1 y)

V'Vvivv'vavrfvrvv‘rv

 

 

 

Figure 2.7: Plot of W1,(a,y) and Wux(a,y)

Chapter 3

Two-Dimensional Flows

3. 1 Formulation

For the case of two-dimensional flows in cartesian coordinates the Stokes equations

give rise to the biharmonic equation
WWW = 111,,” + 2111”,, + illyyyy = 0 (3.1)

where ‘I’(1:,y) denotes the stream function. To solve this equation separate Equa-

tion 3.1 into two partial differential equations of second order,
V720 = 11’.1(:r,y), V2111: 0 (3.2)

Following the procedure as outlined in [61] suppose 111(2, y) = X(a:)Y(y). Substituting

into Laplace’s Equation the solutions are as given in Theorem 2.1. Then if \II(:r, y) =

X(:r)—I7(y), Equation 3.2 becomes

II —;II

 

 

 

 

 

: + 11—— : (2410)" :B,\e"\r) (C1 cos Ay : D), sin Ag) (33)
A I A 1’
or
:H + Z _ (.473 cos 01-: BB sin 0:1) (Cgefiy :Dge'fiy) (3.4)
A I A 1
or
—,II 7”

32

33
Each of these equations is of the form
11(1) + K(y) = P($)Q(y)
To solve for Y and '17 differentiate Equation 3.6 with respect to :1: to obtain
1111‘) = P'fl‘)Q(3/)

Qty) = H’(1?)/P'(1‘)

The only way Equation 3.7 can be satisfied is for

(JG/1:1?)

(3.6)

(3.8)

where 1'1 is a nonzero constant. Substituting Equation 3.8 into Equation 3.3 implies

T(y) 2 (CA cos Ag + D), sin Ay)/k1

Then Wen/fig) = —12 so 111.1
71"” — 12.? = Ae’” + 86"”
which has solution of the form
A—(r) = (41+ A2.r)e’\” + (Bl + B2;r)e”\Jr

Similarly, substituting Equation 3.8 into Equation 3.4 gives
To) = (Gee + Bern/k.
so that THO/VFW) : 02 and

A?” + 02T = A cos 0:2: + Bsin 0.1

The solution to this equation is

Th7): (241+ A21‘)COS 0.1' + (81 + ng)sin 0:1:

(3.9)

(3.10)

(3.11)

(3.12)

(3.13)

(3.14)

34

Combining Equations 3.9 with 3.11, and 3.12 with 3.14 the following are solutions

for Equation 3.1:
((A1+ A227)e’\1'+(Bl + Bgr)e"\”)(C1 cos Ag + D1 sin Ay) (3.15)

and

((A1 + 1421') COS 5.1? + (Bl + 82$) SID fl$)(016f3y + D1€_fiy) (3.16)
We may also differentiate Equation 3.6 with respect to y to obtain

K'fy) = Pf$lQ'(y)
13(1’) = I\"(3/)/Q'(3/) (3-17)
Then
13(17): 132 (3-18)

for 112 some non—zero constant. As before, substitute Equation 3.18 into Equations 3.3

—
v
7:

and 3.4, solve for 2771') and then for I (y). The results are solutions in the form:
(Alt/‘1' + Ble‘-“’\J‘)((C'1 + (by) cos Ag + (D1 + Dgy) sin Ay) (3.19)
and

(A. cos (3.1 + B. sin 131)((C.+ 021/193" + (D, + Dine—fly) (3.20)

It remains to consider the degenerate cases for A, 0, In, or 162 = 0. In these cases

we get solutions of the form as given in [61]
A + B1: + C1." + D13 + By + 17y2 + Cy3 + Haty + Kym2 + Lya‘.3 + May2 + Na'y3 (3.21)

These polynomial functions are a subset of complex-valued functions of form <I>(z) +
301(2) and can be described as trigonometric functions as well, so they are already

included in the previous representations.

Thus combining the above results gives

35

Theorem 3.1 If\Il(.r, y) = A’(.1')I"(y), \I’(:r, 3;) satisfies the Biharmonic Equation 3.1,

then \Il(.r,y) can be written as a linear combination offunctions of the form:
A0 '1' C03 ALI/“Al + Azil‘ + A3918” + (Bl + B2113 + Bay)€—’\I)+

+ Sill Ay((Cl '1' C217 + C3y)c\ 1+ (01 + D2113 + D3y)e —A:1:), (3.22)

and

BO + COS ,’3.’l'((/11 + A255 + A3y)6i3y + (81 + BQCB + 83y)6—fiy)+
+ sin 01((01 + 02.1 + Caz/)6” + (D1 + 022? + Dani”) (3.23)

where all the constants A,, 13,-, C,, D,, A, and 0 are real.

3.2 Completeness

Theorem 3.2 The rcprescntations given in Theorem 3.1 are complete for the Bihar-

monic Equation on. a. rectangle.

First we will show that the solution is unique. Suppose \Il(:r,y) is a function
satisfying the Biharmonic Equation 3.1 with \I’(:r,y) and \Iln(:r,y) (where ‘11,, is the

derivative normal to the boundary ) both prescribed on the boundaries. Then
|V2 \I1|2= (V‘IIV2\II) —V\II-V(V2\I') =

= v.(vwv’11)— v . (11111172110) + \PV‘W

So
f/l, [V2312 = [[372 - [Wivzxp — \1V(v211)] =

_//[—v2111—1118—5:}(v2111)] (3.24)

36

Suppose there are two solutions ‘11; and \Ilg. Then the difference function W’ = ‘III —‘112
and W; = ‘11“, — \lI-zn are both identically zero on the boundary. Substituting the
difference function into Equation 3.24 implies that

|V2\Il'|2 = 0
in V. But if VZW’ = 0 and \II’ = O on the boundary S, then \11’ must be identically
zero. Thus the solution to the Biharmonic Equation is unique.

Now suppose the solution to V‘W = 0 is \P = ‘111 + ‘112 where
Val“ = 0
and
We. = m

Also suppose that \Ilz has no homogeneous solutions (they have all been absorbed
into ‘1“). Then ‘1'] is a Fourier series as given in the previous chapter. So \Ilg consists

of linear combinations of functions of the form
("05 /\.’/((-"l2~1' + flat/)6” + (321' + Baylf'—'\T)+

si11,\,y(((’2.17+ C'3y)e'\x + (D21‘ + Dayle—Ar)
and
cos rag-((Agz + Agy)efiy + (ng + Bay)e“’”)+
sin [311(ng + C3y)efiy + (0253 + Dayle-fly)
Note that ‘11; and W; are independent of each other and thus we have two independent

Fourier series which can always satisfy any two boundary conditions \It and ‘11,, on the

boundary. Hence the representation given above is complete for the rectangle.

As before, if there are two contiguous rectangular regions with given boundary

conditions, the regions may be separated and unique solutions found since \II and ‘11,,

37

 

 

 

- i

|¢— 2 a —>|
Figure 3.1: Geometry for Example III

are both unique along the common cut. The resulting piece-wise defined function

constitutes the unique solution for the entire region.

3.3 Matching Rectangular Coordinates - Exam-
ple III

For our first. example of two-dimensional flow, consider the flow through parallel
plates with transverse fins as shown in Figure 3.1. Here each region is identical to
each other region and the flow is periodic with respect to the width of the rectangular

area. The governing equations for the stream function are as follows:
V4\Ill(1t,y) = 0 (3.25)
For given total flow rate we prescribe
‘ll/(r. y) = 1 along upper boundaries (3.26)

‘1’;(.r,y) = —1 along lower boundaries (3.27)

and the no-slip conditions is

Owl/07) = 0 along all solid boundaries (3.28)

38

where 1) is the normal direction. Due to the symmetry of the problem we expect \I']
to be even in :c and odd in y in Region I. A solution which satisfies Equations 3.25
and 3.28 exactly and the symmetry conditions can be found from Theorem 3.1 and
written in the form
\II1(.r,y)= Bo(:y — —— )+ 2: 8n cos EQM )+ Z An sin( __2_2nl)7ryp(x ) (3.29)
n21

where
, mr a mr - mr
ycosh 7‘1 — (:7; + tanh _a’) smh 7,3“

\/(2n7r)(—2n7r/a + sinh 2n1r/a)

 

QWF=

and

 

. .- . (Zn—l)” , . (2n—l)fi& 2 ‘ (2n-1)1rr
1’( I.) _ .1 sinh __2—— — (acoth _2__ + (2n_l)fi)cosh _2_.

— \/(2n — l)((2n-—1)7ra+ sinh(2n —1)7ra)

 

The denominators of P(.z )and Q(y ) are chosen so that the coefficients will be of
order one to ease the numerical computations. It remains to satisfy Equations 3.26

and 3.27. Thus we have the following

\I’](.‘l‘,l) =1
1 CSySI
‘I’I(&.y)= ‘I’Il-aay) -CSZISC
—1 -ISyS-C

Inverting the first relationship as a Fourier series in 1' over the interval —a S :1: S a

gives:
Bo Z l + Z Anan (3.30)
n=1
Bin/3m + 2 1471711171 = 0 a m # 0 (3'31)

The second relationship is already satisfied for the interval —c _<_ y S c since

the stream function is even in 1:. So it suffices to invert the second relationship as

39

a Fourier series in y over the interval c S y S 1 forcing \I‘1(a,y) = 1 there. The

resulting equation is:

 

'2: Antimn + X: BnAmn : c08(m —1/2)7rc

(777 —1/2)7r — g1(m,c) (3.32)

Finally we require \Illnr(a, y) = \Illrm(—a,y) on the interval from -c S y S c. So

00

Z An(n—1/2)l‘5sch(n —1/2,a,1)f1(0,c,(m — l/2)7r/c,(n —1/2)7r) = 0 (3.33)

n 1

where

1)(n+l) fl

T[(n — l/2)7r]"2'ssch(n — l/2,a,1)

, 77W 777.7r a sh(m, 1, a)
a7 a 77171" 2m”-

2 .
7m 2 (—1)"+1———{1721(0,a,yg-,(7i -—1/2)7r,0)-

(2n — 1)7r
sch(77 — 1/2 a l)—(acoth(n—1/2)a7r + ——}-—)
’ ’ (77.-—1/2)7r
h-.’l(0 afl(.2)n—1/0)c.h(77—1/2,a,1)}
a
')
Hum : (—1)"+1-£[(77 — l/2)7r]'2'ssch(n — 1/2,a, 1)gl(m,c)+
+(—asech(n — l/2)7racsch(n ——1/2)na — (—n———11—/2_)7T)
ch(n — 1/2,a, 1)f1(c.1( (777 — 1/2)7r (n —1/2)n)/\/(2n —1)7r

_(-
V .JITF

—('— + tanh—)h1(c, 1,( (.777 —1/2)7r,,n7r/a 0)sh( n,1,a)}

717T

 

)n1{l 31(c, 1( 7(77 — 1/2)7r ,,n7r/a 0) csh(n,1,a)+

The definitions of functions sch, sh, ch, csh, h1, h2, 173, h4, hll, h21, h31, h41, f1,
f2, f3, and g1 are given in Appendix B. The necessary integrations are performed
analytically using symbolic computations by MATHEMATICA. Equations 3.31, 3.32

and 3.33 were truncated and solved using Linpack subroutines SGECO and SGESL.

40

 

 

n=2 n=4 n=6 n=8 n=10
A(l) -26.41 1.35 -12.96 -4.95 -11.2
A(2) 1.87 -3.04 -1.83 -3.07 -2.45
A(3) 2.25 0.66 0.83 0.31
A(4) -0.97 1.41 1.14 1.50
A(5) -2.04 -0.64 -0.76
A(6) 1.19 -1.23 -0.61
A(7) 2.15 0.21
A(8) -1.41 1.59
A(9) -2.58
A(10 1.73
8(0) —9.52 1.68 -4.06 -0.84 -3.35
8(1) -2.23 1.34 -0.72 0.53 —0.45
8(2) 0.50 —0.65 0.43 -0.41 0.37
8(3) 0.32 -0.33 0.33 -0.39
8(4) -0.18 0.23 -0.27 0.36
8(5) -0.16 0.21 -0.31
8(6) 0.12 -0.16 0.25
8(7) 0.13 -0.21
8(8) -0.10 0.17
8(9) -0.14
8(10 0.12

 

 

Table 3.1: Coefficients for Example 111

The results for a = 0.52 and c = 0.5 are shown in Table 3.1. A plot of \II1(a,y) is
shown in Figure 3.2 and \III(.r, 1) is shown in Figure 3.3 both for n=10. Notice the
value of the stream function is very close to 1 on the upper solid boundary, which

is a measure of the error incurred by the truncation. The streamlines are shown in

Figure 3.4.

3.4 Matching Rectangular Coordinates - Exam-

ple IV

The second example of this matching technique involving rectangular coordinates
considers the flow between plates with transverse ridges as shown in Figure 3.5. Here

the flow is split into two regions, Region I and Region II, as shown. The flow field in

41

0.5

Figure 3.2: Plot of \Itl(a, y)

 

 

42

} F ’ ’ .7 r ’ h ’

D F > P . F F F F . F b

h P

P

F D

’

I b F F b I

 

2

1

.1

 

8 6

o o

4.
o

2
o

0.4

0.2

-0.2

.4

 

Figure 3.3: Plot of \II1(:1:, 1)

43

 

 

 

 

 

 

 

 

-1.

 

 

 

-o.? A -c.2 o 0.2 0.4

Figure 3.4: Streamlines for Example 111

44

 

 

 

 

 

 

 

 

 

"’ :2b :12 : :

'—* : :7 II : :
4—-— 2 c——>

flow

 

 

 

 

ll

 

 

G—Za—b

 

Figure 3.5: Geometry for Example IV

each region will be considered independently with a series solution written in terms

of the functions given in Theorem 3.1. The governing equations for Region I are as

follows:
V4\111(.77,y) = 0 (3.34)
\Ill(a;,b) = 1 (3.35)
\P/(ar,—b) = —1 (3.36)
1 if 1 S y S b
‘I’I(a~,y) = WIN—Calf) if — 1 S y S 1 (3-37)
—1 if — b S y S —1
('NI
-,_’(.r,ib) = 0 (3.38)
dy
8‘11 0 ifl S y S b
7%o3r= %%Pfl¢)H—13ysl (3w)
‘1 0 if—bSyS—l

A solution which satisfies Equations 3.35 and 3.36 exactly can be written in the form:

WM. 37) = 1-5y/b - 0.53/3/1)3 + Ao(y/b - 113/173) + Z AnPn(x73/)+

n=l

45

+f(B.Q.( (.y) +)0..,R(a:y))

where

777m: sinh niry/a y cosh n1ry/a

 

 

P,.(;r,y) =

277.7rCSh(n’ b, a) cos a sinh n7rb/a - Ecosh n7rb/a

sin n7ry/b cosh n7rar/b

 

 

:‘ = , , b
Qn( 1.31) \/2n.1r cosh nna/bCh(n a )
sin nary/b sinh n7r:r/b
? = b
Rn(‘17y) 2717f COSh nwa/bCh(n’a, )

The governing equations for Region II are (after translating the coordinate system

to the rimed svstcm :r’ = .7: — a - c and ’ = ' :
P .. y y

 

 

V4q11[($',y’) = 0 (3.40)
‘11”(at', 1) =1 (3.41)
‘Il[[(ft", —1)= —1 (3.42)
‘I’n(-Cay') = ‘I’I(a.y) (3-43)
(NW 7",i1) = 0 (3.44)
()71’
(N), , , __ 8‘1!)
(9.1V (—C.y ) - —8:r_(a’y) (3-45)

A solution which satisfies Equations 3.41 and 3.42 exactly can be found from Theo-

rem 3.1 and written in the form:

‘1’77(:2’y’)=1.5y—0.5y3 +A’(y—y3) +ZA'P'(' 313+

n=l

+:(13,’,Qy)+C’R’(:7:’,)y’)

where

’ ' h ’ coshmr ’ c
csh(n, 1, C) cos mrx {Sln n1ry/c ' y/

l
V2727r c sinhmr/c —y coshmr/c

 

 

Pri($’ay’) : }

sin mry’ cosh mrx’

I I I
l' :
Q"( ’ y ) x/‘Zmr cosh mrc

 

ch(n,c,1)

sin mry’ , sinh mrm’
x _—

x/2n1r cosh mrc

It remains to satisfy Equations 3.37, 3.38, 3.39, 3.43, 3.44, and 3.45. Beginning

 

R:,(:I:',y') = ch(n,c,1)

with Equation 3.38 and inverting it as a Fourier Series on the interval —a _<_ :1: _<_ a

gives the following two relationships:

 

 

°° b sch(n,a,b) 0" (—1)"b b
A = B" —1 " C. l , ,b —— h , ,b 3.46
0 a; ( ) 2a.\/2127r +2 2\/‘2mr(m(n a ) mrasc (n a )) ( )

csh(m, b, a) (m7rb csch m7rb sechmvrb

v2m7r a a

+ Z B,,(—-1)" b 7rch(n, a ,,ab)h4(0 21,121,724:-

0 : Am.

 

 

_ a)+

 

,0)+

\/ 211w m7r mr

sch(n, a, b)h21(0,a ,———,——,0) (3.47)
a b

 

+
[4
Q
EL
J.
U"

Next satisfying Equation 3.44 leads to a Fourier series in .7." on the interval from

—c S .1" S c. Inverting this relationship gives the following two relationships:

 

)n,sch(n,c 1) +ZC,( — 1) ((ch (n c 1)_ sch(n,c,1)

A'2 B;( —— 3.48
0 E 2m mn— ) ‘ )

M20" lwcschm sechm _ C)+
2m7r c c

02A;

"171'

+ Z Bf,( )"‘\/2n7rch(n,c,1)h4(0,c —C—,n7r,+0)

11:1

47

+ Z C';(——1)"\/2nn sch(77, c, 1) h21(0,c, 31m, 0) (3.49)
77:] C
Inverting Equation 3.37 as a Fourier series in y over the interval —b S y S b leads

to the following relationship:
m1r 777r

 

00 (—1)nCSh(Tl, by a') 77171" TM
77 I — a a 7_9 — _ 7 7_7—'_a
"2:14 % (2121( bb 0) h31( bb b a 0))+

 

 

b ch(m,a,b) 0° (—1)"b sch(n,a, b) g2(m,b)
+3". + B); +
2m7r ”22:1 2ax/2n7r

ab sch(777, a, b) 0° (—1)"g‘2(m,b) b2
+Cm m + n; C,, 2% (b ch(n, a,b) — 77—7; sch(n,a,b))+

(_l)n+1CSll(71, 13C)(111(-lw 1, 217: 21,0) — 1131(—1719 %7:’ Ll;,0))+

+ i. A’ .
n; n V2777: b C
77777" (—1) SCh(n’c’1)g3(m,b)}+

—Ch( (77 C 1)
+ B,’,{ ————'——'-——f1 —1,1, ,— —
Z \/‘2777r ( 777r b ) 2C\/2n7r

 

 

°° , csch(77,c, 1) 7777' (—1)" g3(m,b) sch(n,c,1) _
"X231 Cn{ 2777r f1(—1,1,777r, b )+ 2x/277—7r (ch(n,c,1)————mrc )} _
= g4(m,b) (3.50)

Inverting Equation 3.39 as a Fourier series in y over the interval -—b S y S 1) yields

the following equation:

7777rsch(n7,a,b) C bsch(m,a, b) + m7rach(m,a,b)+

Bm ,
\/ 277m 2m7r

 

 

777rsch( (77, C, 1) m7r
+ZB,’, W fl(—1,1,—b—,n7r)+
+ EC” sch( (77, C, 1) +2n7rcch(n, c, ’1)f1(—1,1,-?,n7r) : 0 (3.51)
n7r

In order to ensure Continuity of the stream function and stresses along the common

boundary of the two regions we also require WWI/337:2 to match WWII/03: and

48

WWI/0:03 to match 8311 "/8193 along those boundaries. So along the interval —1 S
y S 1 we have two different Fourier series which when inverted give the next two

equations:

0° (—1)"+1csh(77,b,a) (7777)2

An
1; a2\/2777r

 

(b . h1(_1,1,m1r,%,0) — h31(—1,1,m7r,%7:,0))+

 

 

 

°° ch(77,a.b)(777r)2 mr
+ B,, ’ f1—1,1,m7r,— +
n; b2x/2n7r ( b )
+ f: C1127777‘b ch(n, a, b)2+ (777)2a sch(77,a, b) f1(—1917m7ra n_1r)+
”:1 b 277.77 b
"’0 , (—1)"(777r)2csh(77,1,C) mr 777r
+ A ll —1,1,mw,—,0 ——h31 —1,1,m7r,—,0 +
"2:21 . C, ,—,7r (2 ( C ) ( C ))

 

 

‘ 2 2 . 2
(77m) ch(m,c,1) _ C,’,,(\/‘27—7777Ch(m,c,1)+(m7r) csch(m,C,1)

‘8'" m 2m1r )=° (3'52)

 

f: B (777r)3sch(77,a,b) mr
77:1

f1 —1,1, ,— +
b3\/'2777r ( m7r b)

 

- 3|)(777r)23(‘ll(71.a,b) + (.-...)3.ch(n,.,b) 727
+ C. . f1—1,1,m7r,— +
T; b3v2777r ( b )

+B' (777.7r)3sch(m,c, 1)

m
2m7r

 

3(7777r)2sch(m, C, 1) + (m7r)3c ch(m, C, 1)

+ Cf.
2m7r

 

= 0 (3.53)

Using the six equations, Equations 3.47, 3.49, 3.50, 3.51, 3.52 and 3.53, to solve
for the six sets of unknowns, An, B", C”, A1,, 3;, and C,’,, then using Equations 3.46
and 3.48 to evaluate A0 and A2, yields the values shown in Table 3.2. The numerical
computations were done using a = 2.2, b = 3.1 and C :: 2.3. Once again Linpack
subroutines were used to solve the truncated system for the coefficients. The value of

\Ill(a,y) is plotted in Figure 3.6 for 77 = 10. Also shown is the plot of the difference

49

n=1 n=2 n=3 n=4 n=5

 

 

-0.884 -0.326 -0.509 -0.456 -0.405
0.52 0.12 0.14 0.06 0.02
0.16 -0.035 0.02 0.056

0.037 -0.004 -0.034

-0.001 0.025

-0.019

7.54 4.25 5.52 5.23 4.73
2.78 3.49 3.60 3.32

1.30 1.31 1.09

0.32 0.27
-0.16
2 23 1 14 1.57 -1.48 -1.28

-0.79 --1.13 -1.17 -1.04

-0.39 -0.39 -0.31

-0.10 -0.08

0.04

0.006 0.025 0.007 0.0123 0.0107
-0.07 -0.33 -0.08 -0.17 -0.14
0.23 0.04 0.13 0.11

-0.02 —0.11 -0.08

0.09 0.07

-0.06

-0.36 -0.98 -0.51 -0.50 -0.49
0.41 0.07 0.46 0.02

0.01 -0.19 -0.21

0.11 0.16

—0.12

0.14 0.32 0.20 0.20 0.18
-0.15 -0.03 -0.17 -0.01
-0.003 0.07 0.08

-0.04 -0.06

0.05

 

 

Table 3.2: Coefficients for Example IV

50

of the two functions along their common boundary in Figure 3.7. We see the error is

less than one percent. Figure 3.8 shows some streamlines for both Regions I and II.

51

 

 

 

 

Figure 3.6: Plot of \Ill(a,y)

 

b

52

0.04 -

-0.02

-0.04

 

Figure 3.7: Plot of \111(a, y) — \IIII(—C3y)

~..T-I

 

tau...

53

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

/ \
V”,
d f :2 :1 3
-1 7
.3 )
:2 :1 3 i 3

 

Figure 3.8: Streamlines for Regions I and II

Chapter 4

Axisymmetric Flow

4. 1 Formulation

Stokes equation for axisymmetric flow in spherical coordinates (r, 0, (75) reduces to

the equation:

E4\I7(r,0) = 0 (4.1)
E2!) = 0 (4.2)

where
32 1 02 cot9 a

2 — —_ —_ _. _—
E — 07:2 + 7'2 802 7‘2 80

where the stream function \Il(7', 0) is related to the velocities (u, v) in the (r,0) direc-

tions by
1 ('34!
7‘2 sin 0 BE
—1 391

rsin9—5;

u.(7", =
77(r, 0) =
and Q(7',0) is the swirl related to the azimuthal velocity by

(M730)

rsin0

 

777(7“, 0) =

The general solution to Stokes equations in spherical coordinates is well known, see

54

55

28], and is given by

‘11 = 204.7” + Bnr'”+1 + Cnr'wz + Dar-"+3)I,,(C)+
n=0
+ 2011;" + ELF"+1 + C',’,r"+2 + Dar—”+3)H,,(() (4.3)
n=2
and
(2 = 274117" + Bra-"+1 mm + 37127" + Bz'r-"+‘)H.(C) (4.4)
n=0 n=2

where C = cos 0, and 1,,(C) and Hn(C) are Gegenbauer polynomials of the first and
second kind, respectively. See Appendix C for definitions of the Gegenbauer polyno-

mials.
4.2 Matching Spherical Coordinates - Example V

For an example of using this technique to solve Stokes flow problems in spherical
coordinates consider the flow caused by the rotation of a finite Circular disk of radius
one in an infinite fluid. The flow field is divided into two regions, Region I consists
of a fictitious sphere of radius one circumscribing the disk and Region II is the space

outside Region I as shown in Figure 4.1. Since there is no radial flow, ‘11 is identically

’O'W~\
9 T I’ ‘\
’ \
7 I ‘
—‘
\ I
\ x
\\ I .
\~"’I

 

Figure 4.1: Geometry for Example V

 

56

zero, and the governing equation for this problem is:
E2!) = 0 (4.5)

Solutions to Equation 4.5 may be found by separation of variables as before and
written in the general form:
9(7', 0) = 2(Anr" + Bur—"+1)In(cos 0) (4.6)
77:0

We have eliminated the functions Hn since the velocities must be finite along the axis.

The boundary conditions for Region I are:
77,7,(7‘, 7r/2) : r (4.7)

7771(0, 7)) = 0 (48)
where the angular velocity of the disk is normalized to one. Satisfying Equation 4.7

requires:

1 l —1
w1(r,7r/2) : 7' : ;((A0+Bor)+0+(A27‘2+BzT—1)(§)+0+(A4T‘4+qu_3)(—-8—)+. . .

Matching powers of 7' gives the following necessary relationships:

Ao=0
80:0
A2=2

BQ=0=Bgnfor77_>_1
A4=O=Azn for 7722

Then since 7771 must be finite at r = 0, and in fact w1(0,9) = 0, we have:

 

70,036) = ((417: + Bl)11+ 27312 + (Agr3 + B3r’2)I3 + . ..

rsind

7!

 

 

57

which implies

Bl : 0 : 8277—1

and

A120

Thus the function Q;(7',0) now has form

Q,(r,0):9r212(cos0)+ Z A2n+1r2"+112,,+1(cos 9)

n=1

and
1

sm 7’)

w;(r,0) =

 

[27‘12(cos 0) + Z A2n+17'2"12n+1(cos 0)]

n=1
Let the general form for the swirl in Region ll be given by:
911030) 2 2(Cn7‘" + Dnr‘"+1)1n(cosd)

71:0

The boundary conditions for Region II are:

1011020) —> 0 as r —+ 00

877711
00

 

(7‘,7r/2) : 0

(4.9)

(4.10)

(4.11)

(4.12)

Equation 4.12 is due to the absence of shear stress of the plane of symmetry. Satisfying

Equation 4.11 forces

. C D
0: 11111 (("—O'+DO)IO+(CI+ I

r-"X' sin 0 7‘ r

 

which gives the following:

Do=0
C1=0:Cnforn21

Since 7011 is finite at 0 = 0,

D
)11+(C2r + 33)], + ...)

 

.(u ..‘- -' -l-

 

Satisfying Equation 4.12 requires

D D 1
0:—1(0+—‘(—1)+0+-—33(—)+...)+0
7‘ r 2

So that

D1:O:D2,,+1for7721

The final form before the matching conditions at r = 1 are met is

{211(7‘, 0) =2 D2 Hr‘2"+112 (cos 0)

77:1

and
1

sin 0

D2n

[Zr—-

n12n( cos0 )] (4.13)

 

1011(1‘, 0)

We expect the flow to be symmetric with respect to 0 = 7r/2, ie., even in C = cos 0.
So we will force 777, to be even and then we can determine the matching conditions

by matching 701 with w” and (9701/07: with awn/37' along 7' = 1. So let

 

 

 

 

w (1 0) _ 1 212(cos 0) + 220:, A2n+112n+1(cos 0) 0 S cos0 S 1
I l - sin0 21203030) — Enoozi A2n+112n+1(C036) _1 S C080 S 0
2711(1.0)=1(Z 02.1.7 (cosm)
sin _

(17—11“ 0): 1 21-2 (c030) + :,:_,(277 )A2n+112,,+1(cos 0) 0 S cos0 S1
d7 sin0 212(cos 0) n- °‘,‘(277 77)A2,,+112n+1(cos 0) —l S C030 S 0
a—Lmn 6) —Z( (2)). D I (cos0))

87 :sin0( n1: 2” 2”

Note that since sin 0 is a divisor of each function representation, we may eliminate it

from the matching conditions. Thus letting

00 ~ 212“) + Zoo=1A2n+112n+lfCl 0 SC
: b 771‘ n = 20
if; 2 2 (g) { 212(C) - 271:1 A2n+112n+1(C) ‘1 S

“IA
|/\ 7—2
o

(4.14)

where

 

77(n—1)(2n-1) 1f(C)In(C)
b": 2 2 LITE—dc

59

for n _>_ 2, 77. even (see [28] for details) allows us to match the function f with its

counter part from w, 1. This requires the computation of integrals of the form

, _ 11.40747)
Gm" “/0 7272—“

where 77 2 2 is even and m 2 1 is odd. This was done analytically and the results

are given in Appendix C. The resulting equations are:
02 = 02 = 3(2(1/3) ‘1' 2143632 '1‘ 2A5G52 + . . .)

D4 = ()4 2 42(0 + 2A3G34 + 2A5G54 + . . .)

Similarly setting

._ °°, _ 21(4‘)+>:.°°=(2n)A. I. (C) 03C
9(9) — 2122.12.40 ’ i 21:“) —— 30::(2n)A:.:i1:.:i(<) —1 5

77:1

“IA
|/\ 7—2
O

Matching the function g with its counterpart from Own/07' yields the following rela-

tionships:

4202 2 7’2 2 3(2(1/3) + 4A3G32 + 8.45ng + . . .)

—404 = I); 2 42(0 + 4A3G34 '1‘ 8145054 '1' . . .)

The resulting linear system is truncated and solved using the same Linpack subrou-
tines as before. The results are given in Table 4.1. The velocities w] and w” are
both plotted along the common interface in Figure 4.2. As the truncation size in-
creases, the error decreases. The present example can also be solved using ellipsoidal

coordinates but is very tedious [30]

0.

2

 

wl(17C)

60

    

Figure 4.2: Plot of w1(1,C) and wu(1,C)

61

 

n=2 n=4 n: n=8 n=10 n=12
A(3) -4.00 -3.00 -2.813 -2.734 -2.692 -2.665
A(5) 4.00 2.50 2.185 2.051 1.974
A(7) -4.50 -2.625 -2.215 -2.030
A(9) 5.0 2.813 2.320
A(ll) -5.45 -3.008
A(13) 5.906
D(2) 0.5 0.625 0.684 0.718 0.740 0.756
0(4) 0656 -0.738 -O.790 -0.825 -0.851
D(6) 0.725 0.786 0.828 0.861
0(8) 0764 -0.811 -0.848
D(10) 0.788 0.828
D(l2) —0.806

 

 

 

Table 4.1: Coefficients for Example V

4.3 Matching Spherical Coordinates - Example
VI

Another related example of this technique is given describing the flow generated
by a rotating Circlular disk of radius one in the plane of an infinite solid boundary.
The flow field is separated into two regions as before with Region I consisting of a
fictitious hemisphere of radius one containing the rotating disk and Region II is the
space outside Region I bounded by a solid stationary wall. We will require the flow

to be odd with respect to the stationary wall at 0 = 77/2. See Figure 4.3.

9/

\

-4----

\
I

[.2

I

 

 

Figure 4.3: Geometry for Example VI

62

Boundary conditions for Region I are exactly as those for Region I in Example V
in the previous section. Following the same procedure, the resulting form for the

velocity in Region I is given by

777,(7‘,0):

 

[7712 C080)+ Z A2n+172 12n+1(cos 0)] (4.15)

81110 "_1

As before let the general form for the swirl in Region 11 be given by:

911(7‘,0) = 2(Cnr" + Dnr‘"+1)In(cos 0)

77:0
Boundary conditions for Region II are different. They are now
7011(7',0) —7 0 as 7‘ —> oo (4.16)
77711(7‘,7r/‘2):0 (4.17)

and finite everywhere in the region. Satisfying Equation 4.16 leads to the exact same

result as before in the pi'mfious section:

Since 70” is finite at 0 = 0, D1 = 0. Satisfying Equation 4.17 leads to

0 = (7‘07"—1 + 0 + 027‘-2(1/2) + 0 + D4T—4(—1/8)+

 

So
Co = 0
D2 — 0 : D277
The resulting form for w“ is
1 D_2_77+1
10]](7‘, 0) :81119[:_T2"+1 12,,+1(COS 0)] (4.18)

63

This time we expect the flow to be odd with respect to 0 = 7r/2, ie. odd in C. Thus
we force 7171 to be odd and use a techinque similar to that in the previous section to

perform the match of w and its normal deriviative. Thus

 

_<_
C

701(1,(_‘) _ 1 { 212(C)+£:°=1A2n+112n+1(0 <10

_C
_ sin 0 —212(C) + Z?=1A2n+112n+1(0015

Again the integrals Gnu, are as defined in the previous section. Performing the first

match gives the results below:
D3 215(‘16'23 + A3(1/15))

D5 = 90(46'25 + A5(1/90))

. _ _ 2.. . _
1),, = "l" I?) ” ”402. + A. for n odd (4.19)

H

 

Next matching the normal derivatives in a similar way yields the following:

—3193 =15(4023 + 243(1/15»

:51), = 90(4st + 4A5(1/90))

77(77—1)(277—1)

— 77D,, 2 4G2" + (77 — 1)A,, for 11 odd 77 _>_ 3 (4.20)

 

U

Using Equations 4.19 and 4.20 it is possible to solve for An and Du explicitly.

An = —277(77+1)(77 —1)G2n for n odd

and

Dn = 277(77 —1)(77 — 2)ng for 11 odd

Together with Equations 4.15 and 4.18 the problem is solved analytically, ie. without

any matrix inversion. Note that this example has no previous solution.

 

64

 

fl ’-Q
1” T“
\
‘ \
\ I I
\ ’l
g \
‘\~__,”’
flow

 

Figure 4.4: Geometry for Example VII

4.4 Matching Spherical Coordinates - Example VII

For a final example of matching using spherical coordinates, consider the flow
through a hole in an infinite wall. Divide the flow field into two spherical regions,

Region I and Region II, as shown in Figure 4.4. The governing equations for Region I

are:
E4\P1(r,9) = 0
v = 0 for 0 = 0 (4.21)
u=0 for0= 7r/2 (4.22)
Bu
— = = 4.2
80 0 for 0 0 ( 3)
@-Of0r0—7r/2 (424)
80 _ _ i

We require the velocity to be finite in the entire region and expect the stream function

to be symmetric with respect to 0 = 7r/2. Requiring the velocity to be finite along the

65

axis eliminates H... Requiring velocities to be finite at r = 0 allows us to keep only

2 3 4

those terms containing 7"), r , r , r , and higher powers of r. Requiring the stream

function to be even with respect to 6 = 7r / 2 eliminates all odd subcripted Gegenbauer

polynomials. Thus the stream function is expected to have form:
W10“, C) = (A0 + Co?~2 + 1907‘3)10(C)'i'(Az7‘2 + Czr4)12(C)+

+ (A414 + C4r6)I4(C) + . . . + (A2,,7‘2" + anr2n+2)12n(() + . .. (4.25)

This form identically satisfies Equation 4.22, 4.23 and 4.24. Satisfying Equation 4.21

 

 

forces
_. ‘ _ . —-1 0W1(r,6) _
0_v(7’0)_(191—r.rci(rsin0 37‘ )—
I
: (2C0? + 3D07‘2) lim , O + 0 (4'26)
9—.0 sm0

Thus Co and D0 must both be identically zero. One degree of freedom allows us to

set \II, = 0 at 0 = 0 so the resulting form of the stream function for Region I is:
W1(7‘,C):(Agr2 + Cx'.27'4)12((_f)+(A47‘4 + C4r6)14(C)+
+ (467-6 + C6r8)16(c.‘) + (Agr8 + 08r1°)Is(C) + . .. (4.27)
The governing equations for Region II are:

E4\I’11(7‘, 6) = 0

v = 0 for 0 = 0 (4.28)
811/80 2 0 for 0 = 0 (4.29)
u 2 0 for 0 :2 7r/2 (4.30)
v = 0 for 0 = 7r/2 (4.31)

Equations 4.30 and 4.31 are the no-slip boundary conditions along the wall, whereas

Equations 4.28 and 4.29 require no shear stresses along the axis. We again require

66

the velocities to be finite in the region (r 2 l) and expect the stream function to be
symmetric with respect to 0 = 7r/2. Requiring velocities to be finite along the axis
eliminates Hn. Requiring the velocities to be finite at r = 00 allows us to keep only

1 2 —1

those terms containing 7"), r , r , r , r‘2

, r‘3, and higher negative powers of 1'. At

this point the stream function has form:
‘1’11(7‘~.C)= (146+ 36" + (767‘2)10(C) + (Air + Bi + DiT2)11(C)+

+(A'27‘2+ 337“] +037“ )1-2.(C)+(Bé7"’2 +132013“)+(B.i7"3+195fl"l )14(C)+o -- (43?)

Equation 4.29 is satisfied exactly. Equation 4.28 forces

 

0 = -v(r,0) = lim —1 B‘PHU’O)

0....0 7‘ sin 0 8r

OF

A'l 2 BE, and C’ = 171 (4.33)
Satisfying Equation 4.30 requires

_1 1
0 = 77{(A’11 + B; + D'lw'2)(—l) + 0 + (Bgr‘2 + Dg)(§)+

3
+0 + (Bgr-4 + Dgr‘2)(?) + 0 + . . .}

Matching corresponding powers of r gives the following relationships:

A’, :0
D’, :0
D; = $8;
D. _ g3,

67

Thus using Equation 4.33, A’1 = 0 = B") and D; = 0 = C6. Satisfying Equation 4.31
requires

—1
0 = _r_{o + 0 + (2A’2r — Bgr-2 + Dz)( )+

NIH

—1
+0 + (—3B.,r'4 — arm?) + 0 + ...}

Matching corresponding powers of r gives the following:

Thus far the stream function for Region II is:
' I I 1 I — I — I
‘I’II(T»C) = A010“) + 8111(0'1' E04? 112(C) + (337‘ 2 + 231)13(C)+

3 I — I — ~ I — 4 I — ~
+(606r 3+D4r 1)I4(g)+(Bsr 4+3831: 2)15(g)+... (4.34)

Again using one degree of freedom consistently with Region I, we set ‘11” = 0 at
0 = 0. So that AZ, : B(. If the flow rate through the hole is q = 7r, then at r z 00
the problem reduces to the flow resulting from a point source in a wall with flow rate

q. The solution to this problem is well-known, see [28] and is given by

W = %(1— c0330)
Letting

\II = rlin;W11(r,0)
gives

1/2(1— c0530) = 81+ B{(—cosfi) + 2B{(1/2)(c080 — c0530) + 0

68

This reduces to B; = 1/2. Hence
‘1’][ 2510+511+ZD4T 12+(B3r +1)I3+

3 4 I
+(EDgr'3 + Df,r“)I4 + (Bgr-4 + 31337~--1’)15 + . ..

Since ‘11” as defined above being the solution for the half space is not yet symmetric

with respect to 0 = 7r/2, we force it to be symmetric by redefining \Il” as follows:

(1,, (7. C)={ $10(C)+111(C)+(D’rr“12(C)+(B§,r-2+1)I3(C)+... 05(31
’ ’ 510(C)—%11(C)+ D; r'112(C)- (Bgr-2+1)13(c)+... 43ch
(4.35)

Since ‘1!“ is now symmetric with respect to 0 = 7r/2, it can then be matched with

the symmetric ‘1']. Express \II” as the continuous expansion

\p]1=bo+f)212+b414+...

where

 

 

1),, =

n(n.—1)(2n — 1) 1 ‘I’II(1,C)In(C) -
2 /—1 1— (2 dg

for n 2 2, 72 even. See [28] for details. Then using this representation of ‘11”, we
match the stream functions from Region I and Region II along the common boundary
1‘ = 1. The normal derivatives 8/07', 82/873, and (93/87:3 are also symmetric with
respect to 0 = 7r/2 and are matched similarly. Matching the stream functions at

r = 1 yields the resulting system:

00:0
bz=A2+Cz
()42A4+C4

The normal derivatives at r = 1 have the following form:

04
——’(1 g)— (2 A2+4Cg)12+(4A4+C4)I4+(6A6+8C'6)16+...

69

 

 

8\Im(1 .): filDQIg—2Bglg+(§Dg—4DQ)I4—... ogcgl
ar ’ finglz+2BgI3+(§Dg—4Dg)14+... —1gc_§0
We, .
Frat—(134) = (2A2 +12C2ll2 + (12/14 + 300014 + . ..
82%;“ C) : §0212 + 61331;, + (61);, + 2195,)14 + . .. 0 g c 31
m2 ’ 317212 — 63513 + (60;, + 205,)1. — . .. —1_<_(g 0
we
(79,7104) = 240212 + (24.4. +1200.)14 + . ..
and
mwuu.h: #Durammn—cmmfltpnn—H.ogcgi
aw ’ §mn+m%n—wuyuwuusu.4gcgo

Using the four matching conditions in order to solve for the four sets of unknowns

An, Cu, 8;, and 1):, requires the computation of integrals of the form

Gym" 2/1 Im(C)In(C) dc

4—9

and

 

, _ 11.44) .
AI,_/_11_C2d(

where n 2 2 is even and m 2 1 is odd, need to be evaluated. This was done
analytically and the results are given in Appendix C. The resulting linear system can

be truncated and solved using the the same subroutines as before.

Chapter 5

Discussion

The purpose of this thesis is to introduce a matching technique which can be
used to solve Stokes flow problems in various complex geometries which may not be
solvable in other ways. The previous seven examples were all solved using the same
general method. For each situation, the flow field was divided into simple contigu-
ous regions. The stream function (or similar function) for each simpler region was
described as a linear combination of eigenfunctions with unknown coefficients. Then
these coefficients were determined by using the boundary conditions and by using
the matching conditions along the common interfaces. The method of determination
was by Fourier inversion techniques which led to an infinite system of linear equa-
tions relating the infinite sets of unknown coefficients. By keeping more terms in the

truncations of these infinite systems the results would attain the desired accuracy.

In this paper the Stokes equations were first solved using a separation of variables
technique which gave rise to sets of £631 eigenvalues and gal eigenfunctions. Com-
pleteness and uniqueness was shown. Convergence is guaranteed by Fourier analysis.
This technique compares favorably with the other methods available for solving these

problems.

A common method of solving boundary-value problems involving more complex

70

71

coordinate boundaries is the finite-difference and finite-element methods. The im-
portant advantage to this method over finite-differences is the reduced amount of

computational work required, only order 12 instead of order n2.

Another distinction between the method described here and that used by Trogdon
and Joseph [64] and others is that their work necessitated the evaluation of complex
eigenvalues and eigenfunctions whereas the current work is done completely with gal
eigenvalues and eigenfunctions. The resulting simplicity is highly desirable. Compu-

tation of complex eigenvalues can be time-consuming as well as difficult.

Complex potentials can also be used to solve creeping plane flow (two-dimensional
Stokes flow). But, except for the most degenerate geometries, the selection of the
potentials is an exceedingly difficult task. Conformal mappings are of little help
since the biharmonic equation is not preserved under conformal mapping. Hence the
method in this thesis provides realistic hope for solving creeping plane flow in many

different geometries.

Not only can this technique be used for solving Stokes flow problems, but it
may also be used in elasticity theory where solutions to the biharmonic equation are
needed. It may also be used for other linear partial differential equation boundary-
value problems. If the partial differential equation is linear and separable and gives
rise to Sturm-Louiville type equations, and if the Sturm-Louiville equation has eigen-
values and eigenfunctions which are complete, this technique can be used to find

solutions for such boundary-value problems as well.

Clearly this method is limited to linear problems with geometries that are com-
posed of the union of simple regions whose boundaries constitute orthogonal coordi-
nate axes. The advantage here is that the geometries do not need to be described

in the same coordinate system. One may “mix and match” as it were, to describe

72

the flow differently in each different region. The results of the examples given in this
thesis support the conclusion that this technique is a potentially powerful one for

many problems in both Stokes flow and elasticity.

APPENDICES

Appendix A

Example III Revisited

As a check for the solutions found for Example III, another method of solution
was used similar to that described in the paper by Coffman [12]. He discussed finding
a biharmonic function u. on a rectangle (I which vanishes on 80 and has a preassigned
derivative there. Example III in Chapter 3 of this thesis is a problem needing a bi-
harmonic solution on a rectangle with a preassigned value of u on the boundary while
the derivative vanishes there. Using ideas similar to those developed by Coffman, the

solution to the problem is expressed as an expansion of functions of the form:

(1),,(.r,y) 2 en sin(n. — .5)7ry{:rsinh(n — .5)7r.r — (acoth(n — .5)7ra+

 

 

1
+(n——-3)—7r-)C03h(n — .5)7f$}
and
\Il,,(.z-, y) = (1,, cos nag/(y cosh nay —— (_a_ + ta.11hn47:) sinh m)
mr a a
where
1 1

 

Cn :

\/(2n —1)7r \/(2n —1)7ra + sinh(2n -1)7ra

1 1
— V2717? \/—2n7r/a+sinh2n7r/a

 

 

dn

73

74

Then
A4)" = cn2(n — .5)7rsin(n — .5)7ry cosh(n — .5)7rx

A‘I’n _ -d,, Eflcosflisinh— mry

a a
Write the solution to the problem as
U’(:r,y) = Z An<1>n(;r,y) + Z Bn‘I'n(x,y) + (3/‘2y —1/2y3)

n=1 n=1

Then the formulation becomes:
A2U = 0 in the rectangle

U = U' — (3/2y — 1/2y3) on the boundary

8U

=0 on the boundary
077

where

bit/1nd) <I>.(:1:,)y+ZBnn(\Il.r,y)

n: 1

It can be shown the the functions (1),, and ‘11,, are orthonomal with respect to the
inner product

(4,111) = /_: bi AfbAde/da:
in the following way:
(@n,<bm) = 0 when 71 ;£ m
(111,“ \I1,,,) = 0 when n 75 m
((Dn,<1>n) =

(111., 111,.) :1

(—1)"+m+18m3/2(n — .5)3/2a2 csh(m, 1, a)sch(n — .5, a, 1)
7r[m2 + a2(n — .5)2]2

«pn,th>==

 

As a consequence of Green’s Theorem, the inner product of U with (1)", becomes

A..+§:B,.(<I>,,.,\II,.)=/Q( 11'- (3/2y— 131/23))(-8%A<I>m)ds (A.1)

n=1

75

where 30 is the boundary of the rectangle and 39,—, is the inner normal derivative along
the edges of the rectangle. Performing these integrations along the boundaries leads

to the following linear equations for each m:

Am + Z Bn(<I>m, \Iln) = —2[(2m — 1)7r]3/Zsch(m — .5, a, 1) g(m, c)+

n=1

— E An 27r(771\}-;l_'5):i/2sch(rr1 — .5, a, 1) f1(—c,c, (n — .5)7r, (m — .5)7r)«'

n=1

 

ch(n - .5, a, 1)

{a - SCll(n - ,5,E’t, 1) - (n -— ,5)7l'

_ a . coth(n — .5)7ra - ch(n — .5, a, 1)}+

 

00 r) _ 3/2
2 Bn(_ 1)n+1 -7r(m 5)
n=1 fl

 

sch(m —— .5, a, 1)csh(n,1, a)-

{h31(-—c,c, (m— ..,5)7r 3&3— 0) (———+tanh—)h1(—d(m—.5)7r,%,0)} (A.2)

mr

The functions sch, ch. fl, I131 and hl are all defined in Appendix B. The function g

is defined as:
g(m,c) = /-C(—1 — 3/2y +1/2y3)sin(m — .5)7ry (13/
-1

Similarly, another application of Green’s Theorem, this time to the inner product

of U with ‘11,", gives

B..+V:A.<w..,<1>.)= I... —(3/2y—1/2y>) (5411 >43 (.43)

11:]
Performing these integrations along the boundaries leads to the following linear equa-

tions for each m:

Bm+ZA (\Ilm,<I>) =0 (A.4)

These two infinite sets of linear equations were solved by truncation using Linpack
subroutines for the unknowns An and 8",. The numerical results in the case a = 0.52

and c = 0.5 are given in Table A.l.

76

 

n=1 n=2 n=3 n=4 n=5
A(l) -0.27753 -0.40852 -0.46977 -0.5359 -0.504
A(2) -4.31624 -5.16494 -6.4008 -5.83
A(3) -1.49711 -3.8296 -2.57
A(4) -1.7048 -0.502
A(5) 0.542
8(1) —0.01886 0.98355 0.89739 0.9515 0.953
8(2) -0.34300 -0.23316 —0.2587 —0.275
8(3) 0.08149 0.1111 0.121
8(4) -0.06076 -0.0673
8(5) 0.0420

 

 

 

Table A.1: Numerical Coefficients

The plots of the value of the function along the edge :1: = a is shown in Figure A.1
for n = 5. Figure A.2 shows the boundary values along the edge y = 1. Then some
streamlines are given in Figure A.3. These plots show very good agreement with those
given in Chapter 3 for Example III (Figures 3.2-3.4). This technique of using inner
products is computationally simpler but it is severely limited in its applicability since
the region must be simple, with either the function itself or its normal derivatives
entirely zero on the boundary. Thus the method could not be applicable to the other

examples in this thesis.

 

77

 

 

Figure A.1: Plot of U'(a,y)

 

H)-

78

 

 

Figure A.2: Plot of U'(a:,1)

0.2‘

0.

-0.

_1>

 

 

 

 

 

 

 

Figure A.3: Streamlines of U’(:1:,y)

 

 

Appendix B

Definitions of Functions Used in
Chapter 3

 

 

 

sch(k a b) _ sinh kIra/b
, , \/2k1ra/b + sinh 2k7ra/b

sh(k a b) _ sinh k7ra/b
, , \/—2k7ra/b + sinh 2k7ra/b

ch(lc a b) _ cosh kIra/b
i i \/2k7ra/b + sinh 2k7ra/b

csh(k,a,b) = cosh kna/b

 

\/—2k1ra/b + sinh 2k7ra/b

b .
h1(a,b,d,c,e)=/a %sinc(m+e)dm

b sinh d2:
h2(a, b’ C’ d’ e) :/ sinh db

 

cos C(x + e) da:

b
COSh d2: sin C(a: + e) da:

 

h3(a,b, C1 d’e) = / cosh db

 

b cosh d2:
“(a’b’c’d’el 2] cosh db

cos c(:1: + 6) (1:1:

80

81

sinh dcc ,

sm C(a: + e)d:1:

 

b
h11(a,b,c,d,e)=/ :1: . hdb
a Sln

b sinh da:

‘2 : _—
h..1(a, b, c, d, e) /a xsinh db cos c(:1: + e) d1:

5 cosh d2: ,
h31(a, b, c, d, e) :./a mm sm c(:1: + e)d:1:
b cosh d3:
h41(a,b,c,d,e):/a $WCOSC($+€)d$

b
f1(a,b,c,d)=/ sincxsindxdx

a

b
f2(a,b,c,d) 2/ sincxcosdxda:

a

b
f3(a,b,c,d)=/ coscxcosdrcdx

a

1
g1(m,c) 2/ (1.5g — 0.5y3)sin(m —1/2)7rydy

b
gam, 4) = / Mb — y3/b3) sin % dy

l I

g3(m. b) = f (y — .213) sin "’2” dy

—1

 

m7ry

 

 

b
dy + 2/ sin dy+
1

b

1
+/ (1.53} — 0.5y3) sin fig—y dy
—1

Appendix C

Definitions of Gegenbauer
Polynomials and Related
Identities

The following relationships can be found several places in the literature. For a

complete discussion of Gegenbauer polynomials see [54]. The general formula is given
by

1 d
(n — 1)! dc

 

2—1
C )n—l

171“.) = — )n_2( 2

The explicit forms for the first few functions are given below:
10(0 = 1

11(C) = —C
12(C)=1/2(1—C2)
13(C)=1/‘2(1—C2)C

14(C)=1/8(1—(2)(5C2 — 1)
15(C)=1/8(1—C'")(7C2 — 3)C
16(1) = 1/16(1 — c2)(2144 —14(2 +1)

I7(<)=1/16(1—<2)(33c4 — 3042 + 5)<
82

83
18(() = 1/128(1 — c2)(429(6 — 4956 +13542 — 5)
19(1) _—. 1/1‘28(1 — (2)(715c6 — 100114 + 385C2 — 35):
110(4) 2 1/256(1 —(*2)(2431(‘8 — 4004:6 + 2002c4 — 308(2 + 7)
Note that for n 2 2, In always contains the factor (1 — (2).

Evaluating the integral
1
Mn = / Mm,

—1 1 — $2
is done exactly for n 2 2, 71 even, using MATHEMATICA and the explicit formulas

above. The first few results are as follows:

1
Mg=/_11/2d(j=1

M. =/111/8(5C2 —1)d( = 1/6

I
A162] 1/16(21(“ —14c2 +1)d(=1/15
l

1
M8 = / 1/128(429C6 — 495(4 +135;2 — 5)d( = 1/28
—1

1
M10 = / 1/256(2431(8 — 400446 + 20021‘1 — 308(2 + 7)d( = 1/45
—1

In order to evaluate the integral Gmn as given in Chapter 4, the identity given by

Sampson [54] below is used:

 

In($)Im(‘T) _ l a: ’1. — x I :B
/ I--:L'2 (113— (m—n)(m+n_1){lm( )In( ) In( )Im( )}+C

He used this identity to evaluate integrals with limits of -1 and 1. They also work for

limits of 0 and 1. Thus

. _ ‘Im(C)In(C) ._
Gmn"/O 1_C2 d8—

84

1 I I I I
= (m _ ”Mm + n _1)[Im(1)I,,(1)— 1.,(1)1,,,(1) — I...(0)I..(0) + In(0)Im(0)l

 

In the case that n is greater than 1, 1,,(1) = 0. If m is odd and greater than 1, Im(0)

and Im(1) are both 0. So for our purposes in Chapter 4, the identity reduces to

1 I
Gm” = (m — n)(m + n — 1)In(0)lm(0)

 

when n 2 2, 71 even, m 2 3, m odd. In the special case when m = l,

. _ 1 ,
Gln — 71(71 __ “(171(1) + 111(0))

BIBLIOGRAPHY

Bibliography

[1] T.N.G. Abbott and K. Walters, Rheometrical flow systems: Part 2.
Theory for the orthogonal rheometer, including an exact solution of
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