Title: Stress distribution of a rotating limaçon
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Title: Stress distribution of a rotating limaçon
Physical Description: iv, 71 1 leaves : illus. ; 28 cm.
Language: English
Creator: Tilley, John Leonard, 1928-
Publication Date: 1961
Copyright Date: 1961
 Subjects
Subject: Elasticity   ( lcsh )
Mathematics thesis Ph. D
Dissertations, Academic -- Mathematics -- UF
Genre: bibliography   ( marcgt )
non-fiction   ( marcgt )
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Thesis: Thesis (Ph.D.) - University of Florida.
Bibliography: Bibliography: leaf 69.
General Note: Manuscript copy.
General Note: Vita.
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Bibliographic ID: UF00098418
Volume ID: VID00001
Source Institution: University of Florida
Holding Location: University of Florida
Rights Management: All rights reserved by the source institution and holding location.
Resource Identifier: alephbibnum - 000559468
oclc - 13495724
notis - ACY4925

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STRESS DISTRIBUTION

OF A ROTATING LIMACON











By
JOHN LEONARD TILLEY, JR.










A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY










UNIVERSITY OF FLORIDA
January, 1961














ACKNOWLEDGEMENTS


The author wishes to acknowledge his gratitude to

Dr. C. B. Smith, Chairman of the Supervisory Committee,

without whose encouragement and invaluable aid the suc-

cessful completion of this work would have been immeasur-

ably more difficult. The encouragement received from

Dr. F. W. Kokomoor, now retired as Head of the Department

of Mathematics, throughout the years of the author's grad-

uate study has been most helpful to the successful com-

pletion of this work. The author would like to acknowledge

the tutelage of Drs. L. N. Henderson, C. G. Phipps, J. T.

Moore, and D. E. South throughout his graduate studies.

To Dr. K. P. Kidd, who replaced Dr. L. N. Henderson after

his death, and to Dr. W. A. Gager, who replaced Dr. C. G.

Phipps after his retirement, the author extends his sin-

cere appreciation for their service on the Supervisory

Committee.


















TABLE OF CONTENTS


Page

ACKNOWLEDGEMENTS . . . . . .ii

NOTATION .. ................ .iv

INTRODUCTION ...... ............. 1

PART I ................. ... .2

PART II ............. ..... .18

APPENDIX .................. 67

BIBLIOGRAPHY ................ 69

BIOGRAPHICAL SKETCH . . . . ... .70














NOTATION


x,y rectangular coordinate variables
z,3 complex variables

z,J conjugate complex variables

rp components of a complex variable in polar form

where o ei9

S(z),X(z) complex potentials

ex* y unit elongation in x- and y-directions

Yxy shearing-strain component

oxo y normal components of stress parallel to x- and
y-directions

Txy shearing-stress component

u,v components of displacement in x- and y-directions

X,Y components of body force per unit volume

X,Y components of a distributed surface force per

unit area

E modulus of elasticity in tension and compression

G modulus of elasticity in shear

T) Poisson's ratio.

Cos(x,n) cosine of the angle between the x-axis and the

normal to an arc in the plane

p constant density
w constant angular velocity















INTRODUCTION


A substance for which the elastic properties are

independent of direction is said to be isotropic. For

such material, we may impose a coordinate system at

random in order to designate direction and distance.

This paper will be concerned with a thin plate

of isotropic material in the form of a limaqon. The

object of the work will be to determine the stress dis-

tribution of this plate when it is rotating with con-

stant angular velocity about its centroid.

The first part of the paper will furnish the tools

for the investigation. The second part will then make

use of these tools to solve the problem.












PART I


1.1. Derivation of the Differential Equation Governing
the Stress Distribution in an Isotropic Plate under Gen-
eralized Plane Stress.
Let the plate be referred to an orthogonal space
coordinate system such that the x-, and y-axes are paral-
lel to the plane of the face of the plate. Let the dis-
placements in the directions of x-, y-, and t-axes be
u u(xy,t), v v(x,y,t),and w w(x,y,t), respectively.
Then the components of stress and strain are related as
follows [7, pp. 6,7,9 :

e = u = l -(a + at

v l ( [ +a )
Ey y E y x ty
a j 1 t -AOX + 63)

(1)
au av 1
'Yxy L ay + _Li G txy

au aw 1

yt + y- yt

If the plate is loaded only by forces parallel to
the plane of the plate, the stress components at,yt, and

Txt are zero on the faces of the plate. In the state of










"Generalized Plane Stress," ot = 0 everywhere and this
state is thus specified by ax, Oy, and Txy only, which
are independent of the t-direction and are average values
across the thickness of the plate 5, p.95 .
Under these conditions, equations (1) reduce to
au 1
EX T -f(ax -V y)


(2) Fy =y -(Oy- x
y a"u v + 1
xy = y+ ax Gy

where ex, ey, and Yxy are average values across the thick-
ness of the plate. If these equations are solved for ao,
Oy, and rxy, we have,

x (e + v~) = E /u +
x i----(Tx + T:77 T 5y)


(3) Gy A = (E )y + Yx E) + +

Xxy- -G( )+(.
__u + av) E (u +3v)


Equations (3) are given by Muskhelishvili, (p. 94) in
slightly different notation.
The equations of equilibrium for "Generalized
Plane Stress" are 7, p. 22]:











acx 3tr
x- + + X 0

(4)
+ 1 Y+ Y 0.



These equations may be changed from this form, involving

the stress components, to the form involving the strain

components by the use of equations (3). This yields the

new equilibrium equations of


a + G2u + X 0

(5)
X~i + Gv2v + Y 0


where X T' G 2 e e + y and
2(1+7)Y)x yr

2 2 a2
x2 ay2

If the assumption is made that X = and
ax

and Y = V, where V is a potential function, then
ay
equations (4) can be written as:



x(Ox.-V)+ y" 0

(6)
(o -V)+ x 0
OY y Dx


~









These equations will be satisfied if a new function
4(,x,y) is introduced such that 4j,

Ox-v -
ay2

(7) y-V Z4

xy - ay

By making a substitution from equations (7) into equations
(2) we get:
A.3I'26 2- [@24 1 2 2, 21
X -I +V -2)a ) -TLV-ya +(1-T)V
E y2 ax2 LE y2 ax2 1-'

(8) ax2 7 ayE "lax2 ay2 .

1 -32p 2(1+;) a2O
Gxy axay E axzay

Now the equation of compatibility for "Generalized
Plane Stress" is [7 p. 24]:


2 cE cy a 2,yxy
(9) + 1..- axy"

Substituting equations (8) into (9) yields the differential
equation

(10) + 2- 4 + (l) 2
ax4 ax2ay2 Aay 2 a y
which is the same as
(II) (v2)? (1-) ( + 3).










Thus, if a function can be found to satisfy (11),
it is clear that it will satisfy the equilibrium equations
as well as the compatibility equation.

1.2. A General Solution of (v2)2 = -(1-')V2V and the
Form of the Stress Components for this General Solution.
The differential equation (11) is a nonhomogeneous
partial differential equation whose solution can be given
as the sum of a particular integral and a complementary
function satisfying the homogeneous equation 4, p. 157]

(12) (V2)20 o

Let F(x,y) be the particular integral satisfying (11).
Now (12) can be written as

(13) (Dx2 + Dy2)20 0
whe 2 a2 2
where D2 and D 2 Hence (13) can be written as
sax2 y2
(14) (Dx + iDy)2(Dx iDy)20 = 0.

Thus 0 = fl(x + iy) + f2(x iy) + xf3(x + ly) + xf4(x iy)

i4 P. 159], or

(15) 0 Reyr(z) + Yz(z| i[(z) + ztTT + X(z)+ -?z5J,
where T(z) and X(z) are analytic functions in the region
under consideration and z x + iy. Hence the general
solution of (11) is

(16) 4 F + 2f + + + 7,

where (v2)2F -(l--)V2V.









Now T(z) = u(x,y) + iv(x,y) and thus
Y'(z) ux + ivx Yx and

(17) Yy = Uy + ivy = -vx + lx i(ux + ivx) = iY'(z).

Also Ylz) u(x,y)-iv(x,y) so that Y'(z) ux-ivx-x and

(18) y = Uy ivy -vx lux -i(ux ivx) --iY'(z),

where ux vy and u. -v are the Cauchy-Riemann Equations
3, p. 281. The same forms hold also for ((z) and ~z(T.
Thus from (15), (17),and (18), the following hold:

0x P' (z) + eZP-z + (2)+ fzY+ xI'()+ Z + z)

Oy - r' (z) Z '-- -T () + T () + (Z)- '((z)
(19) 0xx- ILa (z) + zT,"(z)+ 2T'(z)+ 2T'(z) +."(z) +z"(z)
0, z ,"() Z- "()+ "(2) T" "()]
yy-^ [z" "(z) + ei z) 2'I (z) 27(z)T +Z" (z) + 7"7z] .
Hence from (7), (16) and (19), the stress components are
given by
Ox -V+ Fyy "~,,(z) + z -zT-22'(-z) -2~ -z- +X"(z)


(20) Oy -V+ Fxx+ E[ "(z) + zY"(2) 9(2)+ + '(z) +ZX"(z)
+ "(z)]

( y -xy (z) () + X" (Z) "()F .









1.3. Effect on Boundary Forces.
On the boundary of the plate, the following rela-
tions must be satisfied [7, p. 190 :

R %xCos(x,n) + TxyCos(y,n) ao ) Ty
(21)
S= TxyCos(x,n) + OyCos(y,n) TxY(d) .dx

By equation (7), these reduce to

(ds + Ty ds + de y
(22)2
xa ( & 2ds -V5S dsVxP

Hence X + iY = V iV + i0
ds ds ads a
(23) -iv + i + v .i)+ .

By (16) and (19), (23) can be written as
(24) i(X + iY) V( + i) + A[F + iF + (z)
+ 2'(z) + + -2] .
The resultant force on apartof the boundary is thus
(25) i (X + iY)ds = TV(dx + idy) + Fx + iFy + Y(z)
0 Pl
+ zVY(z) + Z'(z).
Now for a part of the boundary of the plate,
if there are no external forces on the boundary [5, p. 134]

(26) (X + iY)ds = 0.
0


L











Thus (25) is reduced to

P
(27) -fV(dx + idy) -(Fx + iFy) Y(Z) + zyr'TW + 7- rz-.

1.4. Method of Attack.

In the solution of the problem of a limagon rotating

about its centroid with constant angular velocity, we shall

consider it fixed in the plane and the forces due to rota-

tion will be considered as body forces. These will be con-

sidered as two separate cases so that the final result will

be the sum of the solutions for each case. Case I will

deal with the body force acting in the y-direction and Case

II with the body force acting in the x-direction. The

boundary will be free from external forces and hence we

shall use equation (27).

As a preliminary example we shall first apply this

method to the problem of a rotating disk which has been

solved by other methods.

Consider a circular disk of radius b having a con-

stant angular velocity u in its plane about its center and

with density p. If we consider the body forces produced

by this as R pw2r in the radial direction, then by fixing

the disk in a certain position, its center at the origin

of the z-plane, we have the body forces

X = pW2x
(28)
Y "p 2p2











Now the equation of the circle is zB = bo, where

a eie. If we construct the relation

(29) z b

where .= ro, 0 r 1 1, we have a mapping function that

transforms the circle from the z-plane into the unit cir-

cle on the S-plane.

We also note that if

F(S) = f(z), and z w(E)

then

F'(S) f'(z)w'(E), or
(30) f(z) =


The parametric equations for the boundary of the disk

in the z-plane are

x bCose
(31)
y = bSinO

1.4.(a) CASE I, Y = p 2y, X = 0. The potential function

V must satisfy


ax
(32)


It is obvious that if

(33) L -" 2

we have the body force components
X1 0
(34)
Y1 o2y.










Now
(35) V2vl -P2.

Thus the particular integral function F may have the form

(36) F1 = ky4

in order that (2)2F1 24k = -(1 -21)2V1 (1 -2)pw2 by
equation (16). Hence

(37) F1- 2( 4a

Now by equation (27), we have

lpw2fy2(dx + idy) 1(l V)y3 Y(z) + z'p(z)

(38) +'(z), or
(38) 1 _2 )y3 + ify2dl

1- u2 1 3+ ify2dx = (
Since this equation holds only on the boundary, we substi-
tute the parametric equations for the boundary of the cir-
cle in the left member of (38) to get

l,2i2 [(b3sin3o) + f(b2Sin2a)(-bSinede)

(39) = u2 b3(Sin30 fSin3de)

= p2 b3(ivSin3e Cos30 + 3Cos). *



All integrals and substitutions for Sinne and Cosne
come from H.B. Dwight, Tables of Integrals and Other Mathe-
matical Data (3rd ed.), New York, 1957.










Now Sine -1(ee e-o) -( o-1) and
21 21
Cos0 (eiB + e-16) (o + o-1). Substituting these
values into expression (39), we have

2 3 2 ( 1 -13 )3o -1 3 1 -13-1


(40) = P2b3 [2(1 + V)o + 3(3 + 2)o + 3(3 )o-1

-(1 )o-3]
Transforming the right member of equation (38) from

the z-plane to the S-plane, we have

(41) y(z) + zy'(z) + X'(z) T(3) + V (5) + X'T3.

Now '(S), ''(S), and I'(S) must be analytic in and
on the unit circle. Hence they have the form


Y(S) ak, *() = kakg 1, and
k-l k-1
(42)
o k
"'(S) = k lbk

On the boundary of the unit circle, S = eia = o and

3 e-io o Thus equation (38) becomes, from (40),

(41), and (42),

8w2b3 [-(1 + +J)o3 + 3(3 + -2)o + 3(3 )o-1

-(1 j)o3]
Sklako + ba 01 + kl 1k'-k
(43) k b k=lkko2_ k l-
kSlakio + k. lkika2-k + .











Equating coefficients in equation (43), we have the
following:


(44)

(45)

(46)

(47)

(48)

(49)


(50)

(51)

(52)



bk-


(53)

By (4

(54)

Equat

(55)

Thus


ak 0, k 4.

(1 + 1VJ2b3
a3 48 2b3

a2 = O.

a+ .-(3 +V)-2b3
al + al 16-3

Za2 0.

3a3+ 1 I 316 l 2b3

4a4 + b2 = 0.

5a5 3 + (.L -V) 2b3.
48

(k + 2)ak+2 +bk O, k t 4.

By (44) ak 0, k t 4, and hence (52) shows that
0, k t 4, and (51) shows that


3 b3 4 -

5), a3 is real and hence a3 = a3. Thus (50) yields

bl1 bl I- )2b3 + 1)62b3 1b J2b3

;ion (47) shows that Im[al is arbitrary and that

(3 + 2b3.
Re sall = 32 b

we shall let a, be real.













(56)


We have now found two finite series for Case I:

l((s) (32))2b3 (1 )Ap2b3j3
32 48


(57) P'i(-) = b3J 4- lp2b33.

1.4.(b) CASE II, X = pt2x, Y 0. The solution to Case II
is done in exactly the same fashion as for Case I. Hence
we shall give only the important equations.


. 2 [fx2(dx


v= -1p x2
V2 2


F2 = kx4, k PW2(1 --)

+ idy) 13 x3] = (z) + zy'(-)

+ *2


2[ + ifx2dy]


= i2 (b3Cos30) + if(b2Cos20) (bCosOde)

= pd2b3 [Cos3 + 31SinO iSin3O]

(62) -= P'b3 (1 + ]/)o3 + 3(3 + )o 3(3 )o-1


-(1 )o-3] .
p2b3 + y)o3 + 3(3 + ) 3(3 -)-1 (1 -3
48 j Y


O(63) k 2 -k 0 k+
6 z aka + kak + k ibka
(63) k-l k-l k-1


(58)

(59)
(60)

(61)










(64) T2 ) (3 + ) 2 3 (1 + )p2b3
= 32 p b + 48 "

(65) 2.) 2b3 (148'b2b

1.4.(c) Final Results for the Rotating Disk. By the ad-
dition of our results from Cases I and II, equations (56)
plus (64) and (57) plus (65), we have the following:

Y() y() + T2(S = (316 + pj2b
(66)
Y'(S) '1() + 2" '(124 2b3 3

Hence
y,(1) = 3 + Y 2b3,b T"(S) 0, and
16
(67)
."(S. ) = (1 7) 2b3 2

From equation (30)
f'(z) F 4j)

and thus
f"(z w '(5)F"(1) w"(S)F'(S)
[w'(5) 3
Hence from (66) and (67)
(3 + 7~ 2 2
'() 16 u b Y"(z) 0, and
(68) X ) - 2b2 2
8

V 1+ V2 = + 5 2(x2 + y2).
(69)
F = F1 + F2 (= -J u2 (x4 y4)










Note that n" + Yn = rn(,n + -n) 2rnCos n and

(70) n n = n = rn( n) = -2irnSin ne.

Now from equations (20) we have

ox I 2(x2 + y2) (1 ).2y2

+ -2 1 22) (1 )p 2bW 2 + j2)]

(71) 2pwb2r2 + 2(1 2)o 2b2r2Sin2 + )38 pf2b2
2 2W 8

+ (1 ) 2b2r2Cos 2 9 I 2b2 [(3 + )
8 8

2(1 + y )r2 (1 Y)r2Cos 209.

S pw 2(x2 + y2) + ) + 2x2

+ [2(3 + 2b2) ()I 2 2b2(b2 + o2

(72) = 2 ( )p2b2r2 Cos2 3 + ) 2pb2
2 2 8
-(1- ) p2b2r2Cos 2 pu 2b2 [(3 + y)

2(1 + -3)r2 + (1 W)r2Cos 2 ].


Txy (1 2b2 (S2 2)
(73)
-( (1- ) pw2r2Sin 29.


The solution for a rotating disk done by other
methods is given as [7, p. 71] :










=(3 + 2) 2 2 2
or 8 b r1

(74) (3 + > 2Ib2 (1 + 3')p 2 2
8 8 r

Tro = 0,

where 0 1 r1A b. The relationships between the stress

distribution in the polar coordinate system and the stress

distribution in the rectangular system is as follows:ss

or + o0 Tr O
ox 2 + 2 Cos 2e TreSin 20

or + 0 or 0
(75) y 2 2 Cos 26 + -rSin 20


Txy = 2 G Sin 20 + TreCos 20


Replacing the rl by br in equation (74) so that 0 L r ~ 1,

then by equations (75) we have


Ox pu2 2 (3 + -) 2(1 + y)r2 (1 )r2Cos 2]

ay P 2b2 [(3 + 7) 2(1 + y)r2 + (1 )r2Cos 2]


Txy -(1 ) 2b2r2Sin 20

which are exactly the forms found in equations (71)through

(73).


ssSee Appendix.














PART II


2.1. Centroid of a Limaqon.


We shall make use of the polar coordinate equa-

tion of a limaqon in the form


(76)


r 2a + 28Cose


where a > p.

Since the equation is symmetric

axis, the centroid lies on the axis.

the centroid lies to the right of the


about the polar

The distance that

pole is given by


SI rCos edA
h A
jf dA
A


The numerator of this
17 2 2BCos a

5 r2drCos Gde
0 0


(78)


s equation is


= (20 + 2ACos 8)3Cos e d&


( f3Cos e + 3 2 Cos2a


+ 3o 2Cos 3 + 3Cos49)de

-8(3- 2 +l 33(4) 2 2


(77)











The denominator in equation (77) is

n 2d 2 3Cos e
Jfrdrde -2 (2- + 20Cos e2de
00 0
17
(79) = 2f(2 + 2 -Cos e + 2Cos2e)de
0

= 2(2r + 1 = (22 + 2.


Hence

(80) h = 8(42 + 2)
2.2 + 2

We wish to consider the limanon rotating in the

plane about this point, (h,0), where h is given by equa-

tion (80). Since we shall fix the limagon in position

and then make use of body forces to solve the problem,

the parametric equations of the limagon will be

x = 2dCos e + 20Cos29 h
(81)
y 2*Sin 6 + 2pSin Cos 8.

2.2. Mapping Function.

In mapping the limaqon from the z-plane to a unit

circle on the S-plane, we shall consider the mapping

given by

(82) Cos 8 I( + a-1) and Sin -ca e-)
2 21
where a = eie.











Thus, from (81) and (82) we have

ZB = 2aCos 6 + 20Cos29 h + 2iaSin e + 2iSin e Cos e


(83) '( + -1) + (0 + a-1)2 h + a(a <-1)

+ (cr o-l)(a + 0-1) p12 + 2g + p h.
2

Now this equation (83) maps the boundary of the limagon

with parametric equations (81) into a unit circle on

the S-plane where Ir 1. Since we shall be dealing

with the interior of the llmaqon, the mapping function

will be

(84) z = 052 + 2o + m


where m - h (4.2 + 2) -2.2p
2,2 + 2 2v2 t p2

If z = w(J) has branch points, they can be found

by solving w'() 0 [8, p. 135]. Since

(85) w'(J) 2p + 2d,

w'(S) = 0 implies that S But a > B, and thus
a
.A
- < -1 so that it .is outside of the unit circle. Also


S) 2* 2 + p h
2 2 -
(86) h = ( )( B) -h
(86) a


=- + 1)(c p) -h.











Since (86) is a point on the negative real axis

in the z-plane, the corresponding point of the limagon is

w(-l) = ( -2a 2a2 24 + p h
2 + 2
(87)
-2(a h) h.


But > B, so that + 1 2. Thus


(88) -( + 1)(a B) h < -2(a ) h

and hence the branch point is also outside of the limagon.

Thus there are no branch cuts inside either the limaqon

or the unit circle.


2.3. CASE I, Y ej2y, X 0.

As was the case for a circle (as done in Part I)

we shall consider the two cases of body forces for the

limaqon. Thus we make use of the results of the first

part of Case I, page 11, for the circle to get the V1

and F1 functions. By equations (33) and (37), we have


(89) V 1 2- 2 2 >- 4.
2 24

Making use of equation (38) is the starting point for

the work with the limagon.


(90) _2[ +-L j y2dx] f(z) + zY'(z) + 1'(z).
2 L3










Since equations (81) give the form of x and y,

a 3 I 3in3e( + OCos 9)3] -T [i8 Sin3e

+ 24a2BCos 8 Sin3e + 24cp2Cos2e Sin3e

+ 8 3Cos3e Sin3e]

-y[- --1)3 _-20(a .-1)3( + -l)
3 ,03_-1p3 3 2 -1 3 -1,

7v(a -- ) (a + a ((o+

S 3 6 -ya2 -5 2_ 4 (42 32) 3
_2- 4 -w--- -r

+ (8c2 + p2) 2 + (2 2 + p2)q
8 2

2(202 + 2)a-1 Y(82 + 2) -2

+ y(42 3P2) -3 +-2 2S-4 + 2 -5




81), x = 2ACos 6 + 20Cos2e h, thus

dx = (-2aSin e 4pCos e Sin e)ds.


(91)


From (

(92)










Hence, from

(y2dx
































(93)


(90)

= 4Sin2e(c +gCos e)2(-2aSin e 49Cos e Sin e)de

= -8 (~ + BCos e)2(a + 20Cos e)Sin39de

-8 (o3Sin3e + 4-21Sin3e Cos e

+ 562Sin3 e Cos2e + 2p3Sin3e Cos39)de

- -8 -Cos3g A3Cos e + a2%Sin4e

+ac2Sin4 8 Cos 6 + 2?Cos3e 2 2Cos e
3
+ Lsin4e Cos2e + -3Sin )
3 6 /
- 8ha(2 + p2)Cos e 8(c2 + t2)Cos3e

(6a2 + 2)Sin4e 8O2Sin4e Cos 8
5-

-83iSin4e Cos2e
3

- 4a(,2 + 22)(a + c-1) 2(.2 + e2)(7 +0-1)3
3

(6c2 + 2)(a 0-1)4 -+(e -1)4@ +&-1)
12 4
.- ( -1)4( + -1)2
24
4 (4 52 )3
24 4 2 12
+ 8(16c2 + 32)a2 + '(6a2 + 5$2)a

(902 + 202) + -(6,2 + 5P2) -1
3 2
+ 2.(16a2 + 322),-2 a(4a2 502),-3
8 12
,A-4 &--4 5 -6
2 4 ~ 24 -










Thus the left hand side of (90) becomes, by (91) and (93)

1 (1 + ) + ] -3 (1 )(2 5 (1 + v)z2 4
2 24 4 2

-24(1 + )a2 (5 + 32)?2]3 + [8(2 + y )a2

+ (3 + )p2]02 + 2(3 + )d2 + (5 + 1)2]c

(94) (9,2 + 22) + a[2(3 +)2 + (5 v) 2] -1

+ 8(2 1)o 2 + (3 T)g 2] -2 [4(1 -),2

(5 3w)2]a-3 _( y).2e _4 (1 4 2-5
22 4
(1 24 ( 3 ,) (

2.3.(a) Determination of thl(). We are working in the
S-plane so that (90) has the form
1e2 2 3 2 + W(S)
e2 3 i ydx + 1() lw'(S (S) + x1'(S), or
(95)
pI r2 1 -v + y2d.) = T ( 11 '
2i ,3 d ( wI(S)

Since we are inside and on the boundary of the unit cir-
cle, the functions Yl(S) and Al(S) must be analytic about
the origin and hence we shall let

(96) ?1(s) = k=la A1(S) = = bkk

Thus

(97) ka -
k-31











On the boundary of the unit circle, S = ei = a, and
T = e-ie --1.
Thus the right hand side of (95) is

(98) 1 a -k (e-2 + 2r-1 + m)'T ka k-l + 1 bak,
k=l k 2pa + 2a k=l 1 k=l

where w(1) is given by (83)


Now 1 1 I 1 I- -
Now2 + 2 1 +

+ .. which is a convergent series since < 1


and Ibl 1. Thus by Cauchy's Method of Multiplication

of Two Series [8, p. 208

1 k-l OD I k-
(99) + 2 kak = l
2ga + 2ak=1 2akl1

where
k k-j
(100) Dk _= ja (- 9-
j=1
2k-l 3
i.e., I Dk k- = a1 + 2af2 + 3a3' + 4a4 +
k-l
8a 26 2 38 3


+ 2) al2 + 2 ) a2 +
3- 3a
a 3 ....










Hence (98) can be written as

S -k 1 -2 -1 c k-1 i k
aka +~0 (a + 2 + + m) Y Dka~ + + bk
k=l k-l k=1

-k k-3 o k-2 J k-1
k-l1 k=1 k=l 2(k-1

Sk
+ bk k
k=l


Now writing the conjugate of (94), we have

12 _i (1 Y)B3 6 (1 y)d25 (1 -2 )o2r4

4(1 ( )a2 -(5 3Y)O2] 3 + 8(2 -)2

+ (3 )p2] .2 + 2(3 2)2

(102) + (5 )02]e (9a2 + 2p2) + .[2(3 + )a2

+ (5 + -)o2] <-1 + [8(2 + )*2 + (3 + )p] -2

2[4(1 + ), (5 + 3y)2] .-3 (1 -) 2 .-4

4 24
(i +4')e -5- (i+4,)e3-


Equating coefficients from (102) and (101) that
do not involve the bk's from (101), we have the follow-
ing set of equations:










(103) D3 + D2 + mD (92 + 22).

(104) -1 + -2 + D1 = [2(3 + v),2 + (5+ 2) 23
1 -- D2 4 2 8(2+ ) 2 2]
(105) a2 + 1 = 8( + + (3 + ) .

(106) (1 + y)2 (5 + 3 ,)p22

(107) a4 = iP,2(1 + ) 2.

(108) 5 = -2(1 + Y)op2.

(109) 6 23
(109) a6 4 2(1 + 2)P3

(110) ak 0, k 7.

We can see from equations (106) through (110)
that, for k 1 3, the ak's are real. Now since D1 = al,
and D2 = 2a2 -al, we have, from (104),

al + a2 _) 281+ a P- [2(3 + ) 2

+ (5 + ) 2], or

16opa2 + 8(2 2 2 )al + 16 1 = 4P2 3 [2(3 + 2)a
(111)
+ (5 + 2)B2].
Also (105) becomes

!2 + a = p23[8(2 + v)2 + (3 + v)2], or
(112 6a2 + 8 8(2 + + (3
(112) 16 ca2 + 8 32 1 = pw2 921[8(2 + 7)) 2 + (3+ )M )2].










Subtracting (112) from (111), we have

8(2o2 2)al + 8(242 _- 2) 1 = eP,2,8(3 + 2).4

+ 4(1 32)a212 (3 + W>)4], or
(113) p2,
al + a P [8(3 + v).4 + 4(1 3Y)-2a2
8(2c2 2)
(3 + )p4].
2
D3 = 3a3 2 a2 +(~2 a and by (106)

a3 -o (1 + V)a2 (5 + 3282].


Thus (103) becomes
.[2 + 2 (5 2 3)32] (t)2 1 R
p16 [4(1 + U)02 (5 + 3 2 ] 2 +a + ) a
0 2fa 2 2
+ 2a2 al + 21al 2- a(9- + 202), or

(114) 2(2a2 2)a2 +[2(m 20) + 83]a

24
224 12(5 -)a2 + (31 + 61)2]

-2o 2
By (84), m = -2 ,2 and thus (114) becomes
2.2 + e24

2n(2a2 2 2)a2 2 2(6 -4 4)a
2- 2a2 + (3
(115)
4 9^ 2(5 -)2 + (31 + 6M .











From (105) we have

2 (6c4 4) + (6,4 p4)a
202 + 2 2 22 + 21
(116) = p,2,(6,4 p4) 8(2 + V)2 + (3 + ,)2]

8(2a2 + p2)

Adding (115) and (116),

2q 20 2 2)a 6+ 4-
S a2 + 2ca2 + e 21
pw2ae [1( )6 9 2
(117) = -- 2 168(1 + i/) 2(34 91)c
24(2N2 + 02)
(79 + 302)a2e4 3(3 + >)6].

Let Im+a23 = A2. By (117), we have

(2 2 )A2 6 -2 -2 0, or 24 = 0.
2 2 + 2m2 + a22

Hence a2 a2 and by (117)

a 218 2 2- [168(1 + >)06 2(34 9-)a462
(118) 96(5.4 p4)
(118)
(79 + 30P)I2 4 3(3 + 7),6].

Since a2 is real, then by (116), al is also real, so

by (113)
a ,, 2 [8(3 + V) .4 + 4(1 3U)c252
( ) 16(2,2 2)
(119) (3 +
(3 + )B4]










Thus the function Y1(S) is known from equations
(119), (118), (106), (107), (108), (109), and (110).

6 k
(120) fI() -= akS,
k-l
where the ak's are given by the above mentioned equations.
Also
6
(121) 1' ka k-1.
k=l

2.3.(b) Determination of 31'(S). From (90) we have

i'() = + 5y2dx) (J) w T or


(122) w'(-) -1'(o-) Pw'(-)( + 2dx)

w'(o) (O-) w(O)pl'(o).

At this point, all parts of (122) are known except for
1-1
1'(a). w'(a) = 2(B-1 + a) by (85). Thus the left
hand side of (122) is

k Ck.k-l -k
(123) 2(po-1 + ,) 1 b_.-k = 2p k b-k-1 + 2 X k b -k
k=1 k-l k-l

By (94), i w' 9 cr)Q 3 + jy2dx)

= p~~(-i1 + )- )36 (1 + >)ae2 5
S )24 4
S 1)26 [4(1 + /)_2 (5 + 31)52 3


+ -i[8(2 + )a.2 + (3 + W)2]c.2 + -[2(3 + -P)a2










+ (5 + 1)o2] (9.2 + 2 2) + [2(3- v)2

+ (5 P) 2] -1 + J[8(2 2)02 + (3 )2 ~2

[4(1 -7)a2 (5 31~~2]o-3 (1 o22 a-4

(1 2)c, 2 -5 (1 )83 -6
4 24

(124) 2--ee 2 4 +g2 + )
24 24

+ +(1 4 )v (2,2 + 2)4 +f2(1 + )2
4

+ (1 + 32) 2] 3 218(5 + 27)a2 + (19 + 9,)B2-2

_- 3 8+ (8,4 + 12a22 + 1[6 M2
82 6

+ (11 + 3P)2] [6(3 1)a4 3(1 + -)aB22

44] -1 8 8(5 2)a2 + (23 5)o j2] '2

+ [8(1 2)4 2(29 15s2 2 3(3 2)4] p 3

+ [l( 2 (5 3)32] .-4 + 3(1 y)22 2-5
+ /12 ( (5 3u)p + e4
S7(1 2)1B3-6 ( 1 ))847 .
24 24

Since Yl() is known to be a finite series, the re-
maining terms on the right hand side of (122) can be writ-
ten as










6 6
1 6 k 2 6 1-k
-2(p-I + C) a.k (pe + 2~ + m) I kak or
k=l k=1

6 6 6
6 k-1 6 k 6 3-k
(125) -20 Y ak. 2a j ak pk lkak'
k=l k=l k=l

6 2-k 6 1-k
2c I kaka m kak -
k=1 k=1

where the ask's are given by (119), (118), (106), (107),

(108), and (109).

Equating the coefficients from (123), (124), and

(125) that involve only the bk's, we have the following

equations:

2bl1 = -[6(3 )4 3(1 + y)c22 44]

(126)
4a4 6aa3 2ma2.

201 + 2ab2 = 8(5 22)a2 + (23 5y)2]

(127)
5pa5 8oa4 3ma3.

2pb2 + 2E3 = --[8(1 2)o4 2(29 15-)22 2

(128)
3(3 -)o4] 6a6 10ca5 4ma4.


2pb3 + 2 b4 = 1Pw2[0( 2)2 (5 37)p2]
(129)
12aa6 5ma5'


(130) 2Bb4 + 2ab5 = 3P2(l )292 6ma6.
4











(131) 2pb5 + 2.b6 7Pw2(1 2),3.
24
Pw2(1 >) 4
(132) 2~bg + 2cb7 = 24

(133) 2pbk-1 + 2dbk = 0, k 2 8.

By (126), (107), (106), and (118), we have

bl = 12 (3 v)C4 3(1 + 8)a28 414]

+ 1P2(l + )ap2 + p2 [4 (1 + 1)c2 (5 + 3 P)82]

p26(3 h) [168(1 + )a6 2(34 9p)482
96a(5o4 -_ 4)

(79 + 302)a284 3(3 + 2)36]

(134) 960,8 12(29 + 197),6p2
963(5o4 B4)
2(142 + 9j)a4p4 + (115 + 42v)c286

+ (41 + 32)08 + ih [168(1 + /)6 2(34 9g '4p2

(79 + 30i))a2p4 3(3 +v)B6] .

Hence b1 b1 as it is real. Since all of the

bk's depend upon bl, all bk's are real. Because of the
length of each of the bk's, only the final results will
be written.








34

b2 962 4 -4 20(7 + ) )8 + 6(173 + 23T)-6p3
2 96; (5 .- 4,;

2(38 + 371)o4 5 (253 + 24)22 07 (41 + 32)IS9

(135) 8 62
-(135) h120(1 + 2)a + 6(3 + 13)a6 02


2(46 + 3l)(4 4 (49 + 12P)O2 6 3(3 + -)B ].


b - w 4 0(1 2,)-0 40(7 + 3),Ia2
3 963(5,4 4)
+ 22(41 + 7))c6 4 + 2(74 + 211)~486

(229 + 24')+2 8 (41 + 3')010
(136)
(136) + h [120(1 + 1) 8 6(3 + 13v) 602


+ 2(22 212)a4 4 + (49 + 12')a2 6

+ 3(3 + V2)88].

b4 2 2 4 '120(l )>10- 30(1 + )- 82
96o (54 4)
6(157 + 194),6) 4 2(43 + 6-)a4 6

+ (229 + 24D))208 + (41 + 32)910
(137)
+ (h[30(1 + V)>8 + 6(3 + 137')a62


2(37 6y)c444 (49 + 12/)A296

3(3 + -) 8] .











b5 _0 4-2 60(1 v),10 + 6(151 + 25),6 4
965(5,4 44)

+ 2(46 + 92)d4 6 (229 + 24)a2.28

(138) (41 + 3-)010 63h[6(3 + 1311)A6

2(34 9,), 4p2 (49 + 12-)a284

3(3 + -)6]}.

= W263 10 6 4
b6= 9-6 283 O10(1 L ),10 4(230 + 31-7/)6~
96a6(54 e4)
2(46 + 9J)o436 + (229 + 24 2 88


(139) + (41 + 3,) 10 + 3h [6(3 + 13,)6
4 2 2 4
2(34 9)-42 (49 + 12)2 64

3(3 + D)56]j.

b7= 7_(5 -7 418(51 + 71)A6' + 2(46 + 9g)04p3
96> (5a 4
(229 + 241)o265 (41 + 3/)7
(140)
h[6(3 + 13V)A-6 2 -(3 9)42


(49 + 12P)o264 3(3 +-)16].

By (133)


bk = bk-l' k 8.
Sk-i'


(141)










Thus
6 7
(142) 1' (S) = bkk+ b7 S (- i-1.
k=l k=l

Now k (- ) -1is an infinite geometric series with
k=l
ratio of But c > and thus < 1 so that

the series converges. Hence

/ a fk-1- 1 _
(143) 2 )/ +
k=l + 0+

Therefore,

6 k b7 f7
(144) 71 Vk + !
k=l + g

where the bk's are given by (134) through (140)
2.4. CASE II, X = pw2x, Y = 0.
As was done in Case I, we make use of the results
for the circle in Part I for Case II. From equations
(58) and (60) we have

1 22 (12- p) 4
(145) V2 - x2, F2 24- 4

Making use of equation (61) is new the starting point for
Case II for work with the limaqon.

9 2 [x2(dx + idy) iL =x3] !2 12[x3
(146)(z) + z(z)
+ i x2dy = T(z) + z''(z) + -'(z).










Since equations (81) give the form for x and y,

yX3 Y (2aCos e + 2pCos 2 h)3

; 3 6 2 5 2 4
a 8P' Cos6 + 2402Cos 9 + 12o(2.2- h)Cos4

+ 80(o(2 3h)Cos3e 6h(2c2 ph)Cos2

+ 6ah2Cos 6 h3]

3 -1 )6 + e242( + 5
3 r 2 4

+ 12 (2a2 h) ()(0 + o-14
2 \ U -1 3
+ 8a(a 3#h) (o)(+ c- )

2 7\ -1 2
6h(2a h) ()(- + -1 )
(147)
+ 6h 2()(c + 0-) h3]

.3 6 3 2 5 3 2 2 4
= 3 -- 4(2+ + C h)o"

1 2 2 3 3 2 3
+ 4(4 + 15 12ph)C + 81 6 P+ 50

2 2 ) 2 3r 2 52
4h(2a + 2? hh) ] + -L 2 +

2h(3 h)] + -[18- + 50 h(12C-

+ 9p2 6Bh + 2h2)] + -[2a 2 + 532

2h(35 h)] r1 + 16 + 53 4h(22

+ 202 fh -2 + -1a(4o2 + 1512 12ph)c-3
S4

+ -3(2a2 + 62 th)a-4 + .oe2-5 + 83C-6
4 4 8










Also
ifx2dy = if(2dCos 9 + 20Cos2, h)2(2aCos a

+ 40Cos2e 2,)de

= 21f [83Cos6E + 20,o32Cos59 + 41(4a2

02 26h)Cos49 + 4a(,2 202

38h)Cos39 2(2028 + 2o2h 2p2h

6h2)Cos2e + okh(40 + h)Cos 9 gh2]d 9

r5.3. 3 383
i 21 8 + 1 Sin 26+ 3 Sin 49

+ -Sin 60 + 2Sin 0 + 2 2Sin 3
24 2 12

+ a2Sin 56 + 3(4a2 p2 2hh)
4 2

+ 0(4c2 82 2Bh)Sin 2e + (4,2 a2
(148)
2ph)Sin 48 + 3cA(2 252 2Sh)Sin e

+ (d2 282 38h)Sin 36 (2a28
3

+ 2.2h 282h gh2)e 1(22 + 2 2h 202h

ph2)Sin 29+ ah(4p + h)Sin G -p h2 e

S2i 3Sin 6 + esin 50 + -(2,2 + 2

12
+h)Sin 4~2 + 4h( 2 + 17+ 22 12Bh)Sin 36

+ L[2, + 763 4h(2c42 + 202 phd Sin 2a










+(14) 2 + 1392 2h(50 h) Sin e
(148) 2 J
Contd. 2 3
+ (4a2 + 0- 2c2h 62h)e .
But h = (4s2 + 82)
But h = and hence the coefficient of e in
22 + 2
expression (148) reduces to

4a20 + 3 2a2p(402 + 22) 53(4a2 + p2)
2o2 + p2 2.2 + p2
(149) 4a2 2) (42 + 2)(2a2 + 52) 0.

2o2 + 2

Also note that Sin N8 = -(o N a- N). Thus expression
21
(148) reduces to the form

(3 6 + .'(a 5 -5) + (2,2 + 02


Oh)(4 -4) + (42 + 17.2 12~h)(3 3)
12
(150)
+ 7 [ + 7 3 4h(22 + 282 h) (o- -2

+ a62 + 13p2 2h(5 h)] ( -1


Thus by adding the coefficients of the c's from
(147) and (150) we have, as the left hand side of (146):

P2 _(1 3 6 1 )25+ (1 + 2,)0(22
+ + (2
2 24 4 4

+ $2 9h)-4 + 2L4(1 + .y)22 + (17 + 157 2
(151)
12(1 + l)h] 3 + [8(3 + 2y)c26 + (7 + 5;),3

4(1 + T)h(2o2 + 202 ph)] 2 + [2(3 + ;)a2










+ (13 + 57)o2 2h((5 + 32)f (1 + /)h) -

+ l[1802 + 503 h(12,2 + 902 6h + 2h2)]

[2(3 -)o2 + (13 5)#p2 2h((5 32)

(1 1)h)] -1 [8(3 2Z )20 + (7 5V)3
(151)
Contd. 4 (1 V)h(2c2 + 202 h)J "2 4(1 7),2

+ (17 1527)2 12(1 >) h] 3

(1- (2c2 + 02 oh)c-4 (1 >)S2 -5
4 4
(1 v)043 c-6
24

2.4.(a) Determination of Y'2(). We shall make use of
part of Section 2.3.(a), namely that the conjugate of
expression (151) is equal to the expression (101). (Note
-ig -1
that c = e = .) Thus we have that

w2 ( (1 )),2 5
2 24 ( 4
(1 -4 (22 + 2 Ph)e4 i(1 o2

+(17 15)h2 12(1 )5h]8-3 [8(3 2)a2

+ (7 5)-53 4(1 2)h(2a2 + 202 sh)] 2
(152)
2(3 7))2 + (13 52/)2 2h (5 37)

(1 /)h)] + '[1802 + 503 h(12o2 + 9 2











69h + 2h2) + 9 2(3 + 2)c2 + (13 + 52)p6

-2h (5 + 32)9 (1 + z2) h)] C7 + 1 [8(3 + 2.)

+ (7 + 52~3 4(1 + /)h(2a + 282- gh)] -2
(152) 2
Contd. + [4(1 + P)o2 + (17 + 15')p2 12(1 + vu)#h

(1 + 2), (2 2 -4 (1 + V)082 -5
+ 4 (22 + p ph)a + 4

(1 + V)33 -6
24 J
-k 00o k-3 k-2
ak + Si Dkr + Y Dk
k-1 k-1 kl=

co k-1 M k
20k-1 D
+ JE Dkr + 1 bkr
k=1 k=1
k k-j
where Dk = Y jaj and m = B h.
j-1

Equating the coefficients from both sides of
(152) that do not involve the bk's, we have the follow-
ing set of equations:

BmD p 2 3 2
D3 + + D2 +~D1 1 B + 50 h(12ot + 9

(153) 2
(153)- 66h + 2h ) .


al + f-D2 + DI = -- [2(3 + 7)a2 + (13 + 5,)?2

(154) ((h)]
2h (5 + 327)8 (1 + )h] .


2




-3


2
6










S2 2 3
+2 + t D1 f [8(3 +2V)a + (7 + 5)3
(155)
(155) 4(1 + )h(22 + 202 ah).

(156) a3 = -~[4(1 + P).2 + (17 + 15Z)2 12(1 + Y),h].

(157) a4- + 2( + (2 2 2 h).

(158) a5 (18 )

(159) a6 = 1 + 48
48

(160) ak = 0, k 2 7.

We can see from equations (156)through (160) that,
for k 3, ak is real. Now since D1 = a1 and D2 2a2
+ -al, we have, from (154),

-2 8w2, [2 )2
822 4
al + 2a2 2~ + al 4 (3 +

+ (13 + 5V) 2 2h((5 + 3V)o (1 + 1)h)], or

16d~a2 + 8(22 2)a + 16c2al = 4 P20 2(3 + )o(
(161)
(161) + (13 + 5,y)2 2h(5 + 32J) (1 + )h) .

Also (155) can be written as

a2 + = i 2-L8(3 + 2V0)20 + (7 + 52P93

4(1 + U)h(202 + 202 h)], or










164a2 + 8 2a-E ew2= [8(3 + 2v)a2p + (7 + 5v)P3
(162) 4(1 + 2/)h(2a2 + 2|2 h)] .

Subtracting (162) from (161), we have

8(2a2 p2)aI + 8(2m2 2) 2 p208(3 + 2J)o4

+ 4(7 + 2)c282 (7 + 5/)84 4h[4(2 + 2)a2

2(1 + 2)83 2(1 + 2)02h + (1 +492h], or

a, + r1 Pw2o2 8(3 + 2)04 + 4(7 + 7j),2 2
1 8(2t2 p2)
(7 + 5)B 4 4h[4(2 + )4o2?- 2(1 + 2)a3
(163)
2(1 + 1/)a2h + (1 + 2,)082h]

Now D3 3a3 VA2 +2 a1 and by (156)

a3 = a (1 + V)2 + (17 + 15)B2 12(1 + /)h] .

Thus (153) becomes
P12p 4(1 + ,)a,2 + (17 + 15202 12(1 + J)0h]
16
-al+ -!tl~8 =s 1__-
2 2+ al + 2a2 1 1 2
Q 2 221 12

+ 533 h(12o2 + 92 60h + 2h2)], or

2o(2o2 ) a2+ [C2(m 20) + 3 ]a

(164) - .3 12(1 52)o(2;f + (51 + 25/)/,63
(24 (

+ 4h [12:Vo2 9p2 6,#h + 2A/h2]].










Since m = h, (164) becomes

28a(2 2 2)a2 2( + h) 3] al

(165) - 2, ,12(1 5:)2p + (51 + 252)83

+ 4h 12VA2 902 62jh + 2Ih2].

From (155), we have
2<42(p + h) A"2+ [,2(6 + h) p] pal

(166)= P2~ F2(0 + h) 3] [8(3 + 27)a2 + (7 + 5;)a3
8
4(1 + 2)h(20 + 22 ph) .

Adding (165) and (166), we have

8(22 82)a2 + [2(p + h) (S~ 2
2 2
= 12(5 + 9)a42 2 2(51+ 29/)'2 4
48
(167) 3(7 + 52)36 + hj 3 [8(2 1)o4 + (19 + 57) 2o2

+ 8(1 + y)94] 12h 2(1 + ( )1 4 + (1 2)a22

+ (1 + y)A4] + 4(3 + -V)a2$h2O }.

But p+ hb = + (442 + 2 2 2 2 ( +2) and
2G2 + p2 2a2 + 4


hence

(168) -2(


+ h) 3 B 2 (64 4)
2h2 + 62










Let Imja2) A2. Thus by (167) and (168), we have

p(22 p2)A2 2 (6a4 4)A2 = 0, or
2a2 + 2
2---.4-2 = 0. Hence a2 = a2 and by (167),
2"2 + 2

a2 p2(2 p2) U12(5 + 92)c 4 02 2(51 + 292/)c2 4
96P(5,4 4)
3(7 + 521)e6 + h 3[8(2 V)o4 + (19 + 52)o2S2

+ 8(1 + y)B4] 12h[2(1 + y)a4 + (1 -_ 2, 2

+ (1 + )(4] + 4(3 + )a2h2

(169) O-2 -- 4 [24(13 + 52/)o6 + 4(33 + 7,)a402
96(5A4 04)
+ (9 + 237/)o2p4 + 3(1 + 32)56]

4(42 + p2)h 3[2(1 + 2>)c4 + (1 2)242

+ (1 + 2)04] (3 + 2)12hjj .

Since a2 is real, then by (164), al is real and (163)
shows that

al 2 is2 8(3 + 2))4 + 4(7 + 2)a2p2
a 16(2,A2 02)
(170) (7 + 52)34 4h [4(2 + 2),2 2(1 + V)S3

2(1 + z)a2h + (1 + 2U),2hJ.











Thus the function Y2(3) is known from equations

(170), (169), (156), (157), (158), (159), and (160). It

is a finite series with the coefficients given by the

above listed equations and it can be written as
6
(171) y2(S) = I ak k, and also
k-l
6 k-1
(172) T2'() = j kak k-
k-l


2.4.(b) Determination of (2'().-

written in terms of 5, we have


From equation (146),


p 32 3
7= L i[x2dy] (5) -w( 7)

or, on the boundary

w'(d) 22(c) S2-w' )^ + iJx2dy


(173)
(173) wTT( 2(-) -


Now w' (-) = 20- + 2o and hence w' (-)

Thus the left hand side of (173) has


w (a)'F2'( .
-1
= 2(- + r ).

the form


o -k-l m -k
(174) 20 Y bk- + 20 I bk"
k-1 k-l

pj2 r3 + ifx2dy is given by (151), so that the
first member of the right hand side of (173) becomes
first member of the right hand side of (173) becomes









2(-1 (1 + 3 6 (1 + /), 62 5
pw2( + ) I( + + 4 25

424 4
+ (1+ 2)8(2c2 + 62 gh)c' + 24(1+ +)a

+ (17 + 15)1 2 12(1 + V)h] a3 + {8(3 + 27-)Io2

+ (7 + 5)263 4(1 + 2)h(2-o2 + 202 ph)] (2

+ [2(3 + ,)o2 + (13 + 5V)p2 2h((5 + 32)

(1 + W)h)l o + k1 82B + 553 h(12o2 + 9 2

6Bh + 2h2)] (3 ) 2 + (13 5V))2

2h (5 31)0? (1 P)h)] <-1 L8(3 22020

+ (7 5-)~-3 -4(1 2)h(2a2 + 202 _h)] c-2

- (1 2)a2 + (17 15))6/ 2 12(1 ~2), h] -3
1 2 -
(1 1)(x2 + e2 h)o-4 (1 ->) 2-5
4 4
(1 z)8e3 -6?
24 )
p ,2 + 2--'g3 (1+ 2 Y)2(60 2 + 2) 5
24 24
(1+ ) 0, 2 2 4 4
(+ y Po6(k2o: + 2p2 8h)c4 + 4(1 + 2)

+ (23 + 21/)c2 02 + 3(1 + 2)S4
(175)
3(1 + P)eh(42 + 2)] 3 + [8(10 + 7z)23


+ 5(11 + 92/)3 12(1 + /)h(2 2 + 402 -h) 7









1 4 2 2 4
+ *8(3 + b)oA + 4(19 + 9 )A + (7 + 52/)0

4ph |(2 + v)2 + (1 + + 2]?+ a6(3 + 4V)2

+ (39 + 204)3 h[12y2 + 3(10 + 9-)p2

6(1 + 21J)fh + 2Ph 2] --16(3 )4

+ 3(13 11)a2 2 5Y4 h[30(1 )2

(175) 9 /3 6(1 )2 h + 6 2 h 2Yh ha2 -1
Contd.
124(2 2))20 + (59 25-) )3 4h [2(1 >)2

+ 4(3 22) 3(1 2)h] j -2 2[8(1- 2)4

+ 2(53 392)d2 2 + 3(7 5y)04



2 -4
+ 2(10 9>2) 15(1 v) h] CF

(1 -7)2(32 2 h)-5 7(1 Q-6
4 24

24

Since (2(S) is known to be a finite series, the
remaining terms of the right hand side of (173) can be
written as
-1 6 k 2 6 1-k
-2(c- + ) Z aka (62 + 2 C + m) E kak" or
k=l k=l









6 6 6
6 k-I 6 k 6 3-k
-20 ak~ 2a 5 akC 0 kakc3
k=l k=l k-l
(176) 6 6
-2a 5 kak- m kaklr
k-l k-l

where the ask's are given by (170), (169), (156), (157), (158),
and (159).
Equating the coefficients from (174), (175), and (176)
that involve only the bk's, we have the following:

2cI = 6( I)(3-)4 + 3(13 1)ma22 5P84

(177) -h [30(1 2)2 983 6(1 v)2h

+ 6282h 2Ydh2] 4a4 6oa3 2ma2.


2pb1 + 2o.2 = P.2O24(2 2)o2, + (59 25.).93

(178) 4h [2(1 )a2 + 4(3 2 ,)2 3(1 />)hh]

50a5 8oa4 3ma3.

2.b2 + 2ab3 "- 2 (1 ,).,4 + 2(53 397)a2)2
24

(179) + 3(7 5V)84 12(1 V)Yh(4a2 + 22 Bh)

6fa6 10a5 4ma4.

2,j3 + 2ab4 = - [10(1 y),~2 + 2(10 9),02
(180)
15(1 3)h] 12Aa6 5ma5.










(181) 2(34

(182) 2p65

(183) 20b6

(184) 2#bk

Noti
and (157),

bl ,


(185)


+ 2,5 2(1 )82(32 + 12 oh) 6ma6.

+ 2ab6 = 7P,2(1 _)aB3
24
2 4
+ 2,b7 = 2(1 -_ )4 .
24

-1 + 2abk = 0, k 8.
ng that m = ( h, then by (177), (169), (156),
we have

- 26 (3 D)o4 + 3(13 117)a 22 5s84

- h[30(1 2)%2 93 6(1 )02h



22
- 4(l + u )a2 + (17 + 15492 12(1 + )Bh]

c-2( h) {f24(13 + 5)a6
960(504 4)

+ 4(33 + 72/)l462 + (9 + 232,)2) 4

+ 3(1 + 32)B6] 4(4a2 + 02)h3 [2(1 + V)o4

+ (1 1)o 22 + (1 + y)84] (3 + Y)am2 h

3- 962(5-4 960-8 + 12(261 5- <6,2
960(504 4)

+ 4(15 13Y).464 (555 59.)a226 (21 25P),8

h [24(97 11V),6 + 4(81 59>) 442

5(63 31k)A25 4 3(3 23~1)6]










+ 4h2 [12(7 3Z)c6 + 2(15 312)d4'62

(185) + 2(3 + 112)l&2v 4 + 3(1 + 5y)+6]
Contd.
4h3 [4(3 4Y')4 + (3 + 2).2t2 + 4V,41.

Hence b, = 51 as it is real. Since all of the
bk's depend upon bl, all bk's are real. Because of the
length of each of the bk's, only the final result for
each one will be given.

b2 -2 -[ 120(9 D)8
9642(54A 94)
6(77 251)a 682 4(117 19),404

+ 7(3 112)o26 + (21 2572~8]

h[120(3 7)A8 + 6(37 + 592/)a662

(186) 4(99 652)d 44 (195 + 1732)62P6

+ 3(3 23Y),8] + 4eh2[12(8 + 32)a6

2(15 31)/)>4 2 + 2(21 + 11/)a284

3(1 + 53)B6] + 42h3 [4(3 4)o44

+ (3 + 2,)222+ 42 4].










b3 -~ 2 80(1 v)o10 + 40(13- 3-)c8 2
96ca3(5o4 84)
+ 2(403 672)o~6p4 + 4(37 7.p)4a6

(93 77P)&2R8 (21 25v)B10

ph i120(3 7)18 + 6(43 5924)6g2

(187) + 4(63 532)o444 + (99 + 173')d206

3(3 23y),8] 442h2[36(1 + .)x6

2(15 311)Q442 + 2(27 + 112)a2#4

3(1 + 5.v),6] 403h3 4(3 4/),4

+ (3 + V)o2e2 + 4,4]j .

b4 P2 !120(1 /)10 + 30(3 1),8 2
96o4(5.4 64)
6(141 292)a6 4 2(135 291)ot4~6

+ (93 77)) 2B8 + (21 25P)$10

fh[30(3 ')a8 6(43 59;/)N6,2

(188) 2(171 121V')4 4 (99 + 173)o\2,6

+ 3(3 23P)O 8] + 442h2[36(1 + y) 6

2(15 31')$,412 + 2(27 + ll')O2 '4

3(1 + 51)/6] + 4h3h3[4(3 427)4

+ (3 + Y)22a + 414 J










b5 9 2& 60(1 .i4,10+ 6(135 232)6 94
96 5(5 4 4)
+ 4(63 132,)4e6 (93 777)26208

(21 25P)O10 f3h 6(43 59z)<6

+ 4(81 592).4B2 + (99 + 1737)c-2 4
(189)
3(3 232)06] 4p2h2 [36(1 +.),6

2(15 31P)o442 + 2(27 + 11)o2e4

3(1 + 57).,6] 4/3h3 4(3 4y)4

+ (3 + 2)A22 + 4 441

b6 2 3 0(1 2,)o10 8(103 -192)o 64
96a\6(504 p4)
4(63 13;/)O4 6 (93 77)d)2.8

(21 25.)f 10 + 31 [6(43 592)a6

+ 4(81 590) 442 + (99 + 173J),r264
(190)
3(3 232))6] + 42h2 [36(1 + V) \6

2(15 312;44p2 + 2(27 + ll)o12 4

3(1 + 5-)36] + 4~3h3[4(3 4A)',4

+ (3 + ;)a22 + 4,4i.
2s2 4~4]1










b7 c2 06 t6(137 25)o642 + 4(63 13;/)404
96o (5o 4 ?4)
(93 772/)2,66 (21 252/)08

Oh[6(43 592)c6 + 4(81 597)a442

(191) + (99 + 1732/)g2fl4 3(3 232) 6]

4h2b6(l + 2)a6 2(15 31/).4P2

+ 2(27 + 11\i2 24 3(1 + 52d),6]

40h3[4(3 42/)4 + (3 + 2)ca2 2+ 4p4]j.

Thus we have the same form as with the case of

j1'(S). Hence by equation (144), we have
6 7k c7
(192) 2' (J) = bkJ
k-i a
where the bk's for this are given by (185) through (191).

2.5. Final Results for the Rotating Limagon.
We shall follow the same procedure as indicated
in Part I for the final results of the Rotating Disk.
2.5.(a) ( Y1() + 2(). To get the final F(S)
we shall add the coefficients of l1(J) and V2(S) as
indicated in (120) and (171). The new coefficients are

al, '2 8- (3 + 2)o(4 + 4(4- 2)2A2
8(2a2 p2) [
(193) (5 + 32)#4 40h [2(2 + Y)a2 (1 + 2)f62]

+ 2(1 + J.)h2(2m2 2)!.
-I










a2 96(54 _4)96(5 + 32)6e + 64(1 + 2)a 3
96(5a 4)
7(10 + Y) 2 5 6(1 y)7

(194)
12(442 + 82)h 2 (1 + )s4 + (1 -7)a2 22

+ (1 + >)?4] + 4(3 + V)(4,2 + i2).2ph2j.

(195) ag' [(11 + 9P)O 6(1 + Y)h] .

(196) a4' P2(1 + 0)12(0 h).
4 8

(197) a5' a6' 0.

Thus f(S) is a finite series and may be written as
4
(198) 4() = ak. k
k=-

where the ak" s are given by (193) through (196).
2.5.(b) X'(S) = '1'(M) + 22'().- Since 71'(f) and

2'a'() both have the same form, we again add the in-
dividual coefficients to get the coefficients for the
final function. The new coefficients, obtained from
(134) through (140) and (185) through (191), are

bi= P 122-4(145 + 72)a6B2
96a(5oa 0 )
+ 2(172 17j)a444 (670 17U)a26

2(31 11),88 -[ h 96(26 2)a6


(199)


+ 2(128 109V)a4P2 -(394 125>).2B4










6(3 11))#6] + 4h2[12(7 32)d6

+ 2(15 312/)o42 + 2(3 + 112)4,2B4
(199)
Contd.
+ 3(1 + 5)l36t 4sh3 [4(3 -4_)a4

+ (3 + v)a2,2 + 4v,4] .

b pw2 240(1 v)a86
96A2(5a 4)
12(175 7).6B3 2(196 41,)495

+ (274 43W)0297 2(31 11)p9
h [240(1 2/)8 + 12(17 + 232)a662

(200) 16(19 2v)A4e4 (146 + 1611)a2,6

+ 6(3 112)98] + 4h2[12(8 + 31c)6

2(15 31,)a44e2 + 2(21 + 117/)l2S4

3(1 + 5)6] + 4,2h3[4(3 41/)04

+ (3 + 7)a2g2 + 4;74] I

=b -2 160(1 )O10
b 96o3 (5o4 4) f )
+ 240(1 /)aa82 + 4(427 + 52)a64
(201)
+ 2(148 + 7)o4 F6 (322 53j,),28

2(31 113),10- h [240(1 21)8










+ 276(1 J)o,662 + 2(148 31),464

+ (50 + 161P)a2#6 6(3 l112) 8

(201) 4p2h2[36(1 + 7),6 2(15 312,)a442
Contd.

+ 2(27 + 11/)cx234 3(1 + 52/)66]

403h3[4(3 4v/)4 + (3 + ,)a202 + 4v4] .

b4 4 2 4 240(1 v),10
96ca4(54 )
+ 60(1 >)A8B2 12(149 5,).6,4

2(178 23V),4,6 +(322 53Y)c2,8

+ 2(31 11)10O h[60(1 -_)o8

(202) 276(1 W)a662 2(134 115.)a464

(50 + 161v)a266 + 6(3 112/)8]

+ 42h2[36(1 + y)a6 2(15 31p)o462

+ 2(27 + 11V)ll 24 3(1 + 5;),6]

+ 403h3 4(3 4y).4 + (3 + y).2,2 + 4),4]8

b5- 2 2 120(1 )10
9645(5a4 4)
(203) + 12(143 +2i)A6 4 2(172 17,)a4,6

(322 53jv)a2,8 2(31 1);10










3 [276(1 )a6 + 2(128 109l2)d42

+ (50 + 1612)a2)4 6(3 11v)/36]

Co203. 42h2 36(1 + l )o6 2(15 314)a42

+ 2(27 + 112)a2 4 3(1 + 51j)P6]

4p3h3 [4(3 4)~44 + (3 + /),A2 + 4+ 4] .

b6 p2- 23 a 0(1 21)610
96o6(5"4 _4) 4
4(436 72)co6 4 2(172 17j)a4,6

+ (322 53P)a208 + 2(31 11),B10

+ 3h [276(1 yi)6 + 2(128 1092)3,42
(204)
(204) + (50 + 1617)o284 6(3 111)f6]

+ 4p2h2 36(1 + y)o6 2(15 31)o4B2

+ 2(27 + lY)o2e4 3(1 + 57)86]

+ 4P3h3[4(3 42)cA4 + (3 + 3),202 + 474]>j'

P P2-6 2(145 2 )o 6 2
967(5a4 4 )
+ 2(172 172,),44
(205)
(322 532/)o26 2(31 11v )t8

h [276(1 2)a6 + 2(128 109),142










+ (50 + 1611)Y2g4 6(3 112)p6]

4h2[36(1 + L/)6 2(15 31-).a4p2
(205)
Contd. + 2(27 + 11,)a24 3(1 + 52),6]


4h3 [4(3 -4,)o4 + (3 + ./)2,2 + 4+ 4]J.

{'(S) has the form
6 k b'7
(205a) '(3) = bk' + -- .
k=-1 + BP

2.5.(c) Derivatives of T(S) and X'(5). Putting equation
(193) into another form, we have
2 4 k
(206) Y(5) T laki
4- k=l

where the ask's can be found from (193) through (196).
Thus the derivatives have the form
PW2 4 k-1
'(~) = kakl-1
kkl
(207)
24 k-2
4"(f) -= k(k-l)akk-2
k-1

From equations (199) through (206), u(') can
be rewritten as
7
p, ,2 r6 k aOb7
(208) '(S) = + bk +*
96o(5 4 4)lk1 +C +BJ











Thus the derivative has the form

(p 2 [k6 kbk k-1
96A(54A4 4) [kl
(209) b(7-S6 + 637)
+ -------.
(a + S3)2

If f(z) = F(S), where z = w(J), then

f'(z) = '(J)
w' (3)
and
f"(z) w' ()F"(3) w"(J)F'(I)

w (J)] 3
Since w(S) = +2 + 2do + 6 h, w'(5) = 285 + 2a, and
w"(S) = 28, we have, from (207)

(65 +a) k(k-1)ak k-2 2 4 kak'-1
=2 k-1 k1-
8( f + o)3


.,2 r 4 k-1
(210) 3 1, I k k(k-2)ak
16(i + a)3 k-l

4 -k-21
+ k(k-l)a -2 .
k=-1

2.5.(d) Form of V and F. From equations(69)

2
t-2 x2 Y2)
V = V1 + V2 - (x2 + 2)
(211) 2(l ) (
F = F1 + F2 -24 x+ y4










Thus
e 1, Pu2(1 2n0
(212) F 2(y2 2 Fxx = 2 2 2, and F = 0.

Since we shall want these functions to hold inside

the limaqon as well as on its boundary, we have, from
equations (82)
z = @S2 + 2af + m
(213) = pr242 + 2ar- + m,

where S = r, 0 A r i 1, o eie, and m = 3 h. Hence
x Or2Cos2e + 2qrCose + m
(214)
y gr2Sin2G+ 2arSine.

Thus
x2 = B2r4Cos229 + 4adr3Cos2eCose + 2#mr2Cos2e

(215) + 4a2r2Cos2g + 4amrCos + m2
y2 ,= 2r4Sin22 + 4d~r3Sin29Sine + 4a2r2Sin2o.

Now rnCosne = (Sn1 + n), rnSinne = !-(Sn 3n), and

2
r2. Thus (215) becomes

x2 = (2 2) S()( + )2 ) + am(j2 + 2)

+ 02(S + -)2 + 2om(S + y) + m2
2= (S4 + -4) + 03(j3 + 3) + (am +2)(S2 + y2)
(216) 2 2 22
+ o(2m + ~r2)(5 + S) + 4 + 22r2 + m2
2
y2 L(2 32)2 8(J2 2)(S ) -2( y)2
4
= -[2 (4 + 34) +0, (3 + T3) + 2(j2 + 32)

(217) rr2(S + 3) -r4 22r2]
2











2.6. Final Stress Components.

Since z 5 2 + 2o(S + m, S C + 2 m + z
or
(218) o + 2 92 + (h + z)].

Because the branch point is outside of the limagon, this
is a single-valued analytic function in and on the
boundary of the limagon. Thus, given any point in or
on the limaqon, equation (218) transforms it to a
3-value. Hence giving the stress components in terms
of S is sufficient for their determination.
By equation (20), we have

Vx V + F [ z" (z) + z9"(z) 2Y' (z)
2+' (z) + X"(z) + I"(z)

(219) "y = V + FXX + [?"'(z) + zY"-Tz) + 2' (z)
+ 2 (z) + )X(z) + x"(z
-xy - Fxy |[zy"(z) z"'(z) + k"(z) X"( .

From (211), (212), (216), and (217), we have

V + Fw(X2 + y2) + L2( a )y2
yy 2 2

= x2 + ;y2)

P W2 )62(4 + 34)
(1 -
(220) 3 _3 2
(20) + 4(1 )o8(13 + 3) + 4 [(1 )

+ ( h)] (2 + 2)







63


(220) + 4o4 L2(0 h) + (+ (1 + 2)r2] ( + )
(220)
Contd.
Contd. + 2(1 + )y2r4 + 8(1 + )42r2 + 4(0 h)2].

V+ Fx ( L(x2 + y2) (1 )x2
Fxx 2 2

= e2(x2 + y2)
2

{(1 )2(4 + 3-4)

+ 4(1 -2)acB(S3 + ) + 4[(1 )l2

(221) vB(6 h)] (S2 + -2) 4L[2Y(S h)

+ (1 + yV)r2](S + y) 2(1 + )42r4

8(1 + J)q2r2 4(O h)23.

Hence from (219), (221), (207), (210), and (209),
we have
=x (1 v)02(S4 + 34) + 4(1 -y)o(3 +3)

+ 4[(1 7)o2 + (6 h)] (S2 + 2)

+ 4a[2(8 h) + (1 + ),)r2](S +
(222)
+ 2(1 + 2)32r4 + 8(1 + V)a2r2 + 4( h)23

T2(~ 2 + 2, ) + h) 4 k-
+. k(k-2)ak 1
32 (J + ,)3 k=1

+ q Z k(k-l)ak~k-
k-1l














4
+(OS2 + 2 + 6 -h) k _ak--1
(6 + a)3 kL-l

+ o I k(k-1)akk-2
k=l J

S4 k-1 4 4 -k-1
ka. Z kak kf
(pW + )k= (-) +" )k=l
(222)
Contd. 6 k-
Z 1 kb k
12a(5a4 64)(8~ + )) Lk=i

+ Ob7(7o$6 + 6B,7)
( S + 3)2

S1 If kb k-1
12(5A4 34)(' + c 1k

+ ob (7t6 + 6g8 7)+
(- + ) J

From (219), (221), (207), (210), and (209),
we have
y 2 (1 )82) 4 +4) + 4(1 )48[ 3 + 3

+ 4 [(1 2/)2 (9 h)] (52 + r2)
(223) 4-[2V(3 h) + (1 + y)r2](S + T)


2(1 + 2)p2r4 8(1 + 2).2r2 42(6 h)2
-j











+ r.1a2 + 2cr + 8 h) k(k-2)ak k-1
32 ( 3f + )3l k

4 k-2
+ i k(l-l)ak'
k-1

(632 + 2 + h) 4 1
+ y k(k-2)ak k'
(Y + ca)3 Lk=1
4
(223) + a 2 k(k-l)akk-
Contd. k=l
4 4
+ ka-1 4 4 kak-1
( + a)k=l (BY + )k-1l



12 (5<4 +4)(1f + a) k1k
+b (70Y 6 + 6p-7)
+ (----- -
(eS + )2
6 -k-1
[4 kbk-1
120(514 84)(8 + 4) "1
c+b (7,36 + 6S7)]"

(0 + a)2
From (219), (212), (210) and (209), we have

iPJ2 T( -2 + 27. + 8 h) 4 k-1
S32 3 k(kS-2)akS
xy 32 ( + C)3 k-1
4 -
(224) + ? k(k-l)aki -
k-l











(OJ2 + 2S + 6 h) [ -k-1
3 _r k(k-2)akS
(i + X)3 Lk l
4 k-2
+a > k(k-l)akS
k=l

1 6 k-1
(224) 41 4 6 kbk Skl
Contd. 12
+b 7(7aS6 + 6cS7)




(+ T6 kb-k
12a (5 4 f34)( 3 + [) 1k- 1

9b (7016 + 6657)
+ +--- -
+ 7 + c)2
It is to be noted that if 6 = 0, then the limaqon
becomes a disk of radius 2a. The final stress components
as given in (222), (223), and (224) reduce to those found
in the first part of this paper, equations (71), (72),
and (73), if 3 = 0 is substituted in equations (222),
(223), and (224).













APPENDIX


If the stress components ox, a"y and rxy are

known for any point of a plate in the state of "Gener-

alized Plane Stress," the stress acting in any plane

through this point perpendicular to the plate and in-

clined to the x- and y-axes can be calculated.

Consider Figure 1. If AB 1, then OA = Cos 6

and OB = Sin G. Thus


Y 0r
f "
S~/t/i


X xCos e + ,xySin 8. A

Y ySin e + xyCos e. O*

or = x Cos e + Y Sin .
tre C = Cos e X Sin e.
Figu



2C
r "xCos20+ ?xySin 2 Cos e + y Si

+ TxySin Cos e
= z(x + + + -( x ~- y)Cos 20 +

're o"ySin Q Cos a + TxyCos2a cxSi

txySin2e
- g(1x -y)Sin 20 + TxyCos 2.


ire 1
2
ne


TxySin 26, and

n e Cos 9


Hence










If instead we consider Figure 2, where B 1,
then OA = Cos e and AB Sin e. Thus f


~ = gSin 8 -TxyCos Q. 6P
xs A
Y =- yCos e TxySin 9.
eY = Y Cos e + X Sin e. O A-- A

-,r = Sin 9 X Cos 8. al
Figure 2
Hence
2 2
o-yCos 9 TxySin e Cos 9 + cxSin8
txySin Cos e
S(C-x + "y) (x -y)Cos 2e txySin 29, and
2
-Tg = ySin 6 Cos G YxySin~ -xSin e Cos 9
+ TxyCos 2
1 (x ry)Sin 29 + zxyCos 20, which is the

same as from Figure 1.














BIBLIOGRAPHY


1. G. B. Airy, Report of the British Association
for the-Advancement of Science, 1862.

2. H. B. Dwight, Tables of Integrals and Other
MathemaTical Data,(3rd ed.), New York, 1957.

3. T. M. MacRobert, Functions of a Complex Variable
(4th ed-), London, 1954.

4. F. H. Miller, Partial Differential Equations,
New YorE, 1941.

5. N. I. Muskhelishvili, Some Basic Problems of
the MatHematical Theory of Elasticity, Gronigen, Holland,
1953.

6. I. S. Sokolnikoff, Mathematical Theory of
ElasticTty (2nd ed.), New York, 1956.

7. S. Timoshenko and J. N. Goodier, Theory of
ElasticTty (2nd ed.), New York, 1951.

8. E. J. Townsend, Functions of a Complex Variable,
New Yore, 1942.














BIOGRAPHICAL SKETCH


John Leonard Tilley was born June 4, 1928, at

Jamaica, New York. In June, 1946, he graduated from

Jenkintown High School, Jenkintown, Pennsylvania. In

June, 1950, he received the degree Bachelor of Science

in Economics from the University of Pennsylvania, Phila-

delphia, Pennsylvania. From January, 1951, through Janu-

ary, 1953, he served as an instructor in the 3rd Army

Supply School, Ft. Jackson, South Carolina. Following

his discharge, he enrolled at the University of Florida

as a Graduate Student in Education. He received the de-

gree Master of Education in August, 1954. From then until

June, 1958, he taught in the public schools of St. Peters-

burg, Florida; first in the Northeast High School, and then

in the St. Petersburg Junior College. At both institutions

he served in the mathematics departments. During this per-

iod he returned each summer to the University of Florida

as a Graduate Student in Mathematics. Having held a 1/3-

time Graduate Assistantship in Mathematics at the Univer-

sity of Florida in the school year 1953-54, he applied fr and

received a 1/2-time Assistantship in the same department

for the school year 1958-59 and again in 1959-60. He was

appointed to the faculty of that department as an Instruc-

tor for the school year 1960-61.

70












John Leonard Tilley is married to the former

Dolores May Cope. He is a member of the Mathematical

Association of America. While at the University of

Pennsylvania, he was an active member of the Alpha Delta

Chapter of Beta Alpha Psi, The National Accounting Frater-

nity and served as student treasurer of that chapter in

his senior year.












This dissertation was prepared under the direction
of the chairman of the candidate's supervisory committee
and has been approved by all members of that committee.
It was submitted to the Dean of the College of Arts and
Sciences and to the Graduate Council, and was approved
as partial fulfillment of the requirements for the degree
of Doctor of Philosophy.



January 28, 1961


Dean, College of Arts aid Sciences



Dean, Graduate School

Supervisory Committee:


Chairman

Th jy7^l






~ i-eZX -> 7-4




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