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 Title Page
 Acknowledgement
 Dedication
 Table of Contents
 Introduction
 Second order linear differential...
 Confluent differential equatio...
 Related differential equations...
 References
 Biographical sketch














Title: Confluent cases of second order linear differential equations with four singular points
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Title: Confluent cases of second order linear differential equations with four singular points
Physical Description: v, 117, 1 leaves : ; 28 cm.
Language: English
Creator: Cundiff, Joyce Coleman, 1927-
Publication Date: 1961
Copyright Date: 1961
 Subjects
Subject: Differential equations, Linear   ( lcsh )
Mathematics thesis Ph. D
Dissertations, Academic -- Mathematics -- UF
Genre: bibliography   ( marcgt )
non-fiction   ( marcgt )
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Thesis: Thesis - University of Florida.
Bibliography: Bibliography: leaf 116.
Additional Physical Form: Also available on World Wide Web
General Note: Manuscript copy.
General Note: Vita.
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Volume ID: VID00001
Source Institution: University of Florida
Holding Location: University of Florida
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Resource Identifier: alephbibnum - 000559149
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Table of Contents
    Title Page
        Page i
        Page i-a
    Acknowledgement
        Page ii
    Dedication
        Page iii
    Table of Contents
        Page iv
        Page v
    Introduction
        Page 1
    Second order linear differential equations
        Page 2
        Page 3
        Page 4
        Page 5
        Page 6
        Page 7
        Page 8
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        Page 32
        Page 33
        Page 34
        Page 35
        Page 36
        Page 37
        Page 38
    Confluent differential equations
        Page 39
        Page 40
        Page 41
        Page 42
        Page 43
        Page 44
        Page 45
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        Page 84
        Page 85
        Page 86
        Page 87
        Page 88
        Page 89
    Related differential equations and classification
        Page 90
        Page 91
        Page 92
        Page 93
        Page 94
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        Page 110
        Page 111
        Page 112
        Page 113
        Page 114
        Page 115
    References
        Page 116
    Biographical sketch
        Page 117
        Page 118
Full Text










CONFLUENT CASES OF SECOND ORDER

LINEAR DIFFERENTIAL EQUATIONS

WITH FOUR SINGULAR POINTS











By

JOYCE COLEMAN CUNDIFF


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
June. 1961

















ACKNOWLEDGMENTS


There are many people who have contributed to the

preparation of this dissertation. First, the writer

wishes to express her deep appreciation to the chairman

of her supervisory committee, Professor Russell W. Cowan,

for suggesting the topic of study and for his guidance

and helpfulness during the research. She is also grateful

to the members of her committee, Professors W.R. Hutcherson,

J. T. Moore, W. P. Morse, C. B. Smith and A. Sobczyk. Warm

thanks are due Professors F. W. Kokomoor and J. E. Maxfield

for their encouragement and interest.

The patience, cooperation and persistence of her

typist, Sandra Fife, is gratefully recognized.

She is humbly grateful to her husband for the less

tangible but real assistance received by his encouragement,

patience and enthusiasm.

































To Mother,

ever uplifting and inspiring













TABLE OF CONTENTS


ACKNOWLEDGMENTS . . . . . . . . . . i

DEDICATION . . . . . . . ... . .ill

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

CHAPTER I Second Order Linear Differential Equations

1.1 Definitions . . . . . . . . . 2

1.2 Formal Solutions of Differential Equations . 3

1.3 Convergence of Formal Solutions . . . .. 10

1.4 Second Solution in Case where the Difference
of the Exponents is an Integer. Example . .24

1.5 Solutions Valid for Large Values of Izi . . 35

1.6 Irregular Singular Points and Confluence . 38

CHAPTER II Confluent Differential Equations

2.1 Differential Equation with Four Singular
Points . . . . . . . . . . 39

2.2 Differential Equation with Singular
Points at z = 0,l, ,-o . . . . . . 44

2.3 Scheffe's Criteria applied to Confluent Case 50

2.4 Solutions of Confluent Case after Normalizing 59

2.5 Factored Solution W ,. Hypergeometric
Solution W .. .. . . .. 67

2.6 Notation and Proofs involving 2F1(a,b;c;z) . 72

2.7 Recurrence Relations, nth Derivative, Sum and
Product Formulas using W .k ......... 83
CL s








CHAPTER III Related Differential Equations and
Classification

3.1 Related Equations of Mathematical Physics
Derived from an Equation having Four
Singularities . . . . . .... ... . 90

(a) Lame's . . . . . . . . . 93

(b) Legendre's . . . . . . . . 94

(c) Jacobi's . . . . . . . ... .95

(d) Gegenbauer's. Derivation of Geganbauer's
Equation . . . . . . . ... 96

(e) Laguerre's . . . . . . . 100

(f) Equation with Solution being Incomplete
Gamma Function . . . . . . 101

(g) Gauss's . . . . . . . . .101

(h) Kummer's . . . . . . ... 102

(i) Whittaker's . . . . ...... .102

3.2 Classification of Differential Equations
having Four Singular Points when the Exponent
Difference is 2 ..........103

LIST OF REFERENCES .................. 116

BIOGRAPHICAL SKETCH . . . . .... . 117












INTRODUCTION

This is a discussion of ordinary second order linear

differential equations with four singular points and the con-

fluent cases which occur by permitting their singular points

to coalesce. Chapter I presents a second order differential

equation, formal solutions with proof of convergence, solu-

tions valid for large values of Izi, irregular singular

points and confluence of singular points. Chapter II de-

velops the general form for a differential equation having

four regular singular points and then considers solutions

of the confluent equation with singularities at z = O,1,o,

o. Using the normalized form of the confluent equation we

find a hypergeometric solution with recurrence relations,

nth derivative,and sum and product formulas. In chapter

III we derive some of the related equations of mathematical

physics (see table of contents) from the general equation

of chapter II and present a classification of this type

equation when the exponent difference is i.

Chapter I elaborates on pertinent material in [8,

chapter 10] Except for the classification in chapter

III, see [2, p. 499], the remaining work is believed to

be entirely original.


The numbers in brackets refer to the list of references.













CHAPTER I

Second Order Linear Differential Equations

1.1 Definitions. Let the standard form of an ordinary

second order linear differential equation be represented as

(1-1) d2u + p(z) du + q(z) u = 0,
dz2 dz

where p(z) and q(z) are assumed to be functions of z

analytic [8, p. 85] in some domain D except at a finite

number of poles. Any point in D at which p(z) and q(z) are

both analytic will be called an ordinary point of the

equation; any other points of D will be called singular

points of the equation. If there exists a point z=c of D

such that, when p(z) and q(z) or both have poles at z=c,

the poles are of such order that (z-c)p(z) and (z-c)2q(z)

are analytic, then z=c is called a regular singular point

for the differential equation. Any poles of p(z) and q(z)

that are not of this nature are called irregular singular

points.

If z=c is a regular singular point, the equation may

be written
(1-2) (z-c) du + (z-c)P(z-c) du + Q(z-c) u = 0 ,
dz2 dz

where P(z-c) and Q(z-c) are analytic at z=c. Expanding

these functions in a Taylor's Series about z=c we have








00

P(z-c) = c) c) + ... + pn(z-c)n + = Zpn(z-c)n
n=o

and

Q(z-c) = q + q1(z-c) + ... + qn(-c)n + ... = q(z-c)n
n=o

where p ,p1,...,q9,q1,...are constants. These series con-

verge in the domain Dc formed by a circle of radius r

(center c) and its interior, where r is chosen sufficiently

small so that c is the only singular point of the equation

which is in DC.
c

Thus

p(z) =P(-c) and q(z) a(z-c)
z-c (z-c)2

in equation (1-1).

1.2 Formal Solutions of Differential Equations. Let

us assume a formal solution of the differential equation to

be



u = (z-c)a 1 + Zan(z-c)n]
n=1

where a,al,a2,... are constants to be determined.

Assuming that the term-by-term differentiation and multi-

plication of the series are valid, we have










U' = c(Z-C)a-'


00oo
[i+ an(z-c)n] +
n=l


CO
(z-c) ann(z-c)n-i
n=l


= (z-c)-1 [a


+ c(a+n)an(z-c)n]
n=1


and


u" = (a-1)(z-c)-2, +I (a+n)a (z-c)n
n=l


+ (z-c)-' Zn(a+n)a (z-c)n-
n=i


co
= (z-c)a-2c (a-1) + (a+n)(a+n-l)a (z-c)
n=l


Now substituting these expressions for u,u' and u"

into equation (1-2) we obtain

cc
(1-5) (z-c)C a(ca-1) + a (a+n)(a+n-1)(z-c)n
n=l



+ (z-c) P(z-c)[a + za (a+n)(z-c)n
n=l


+ (z-c)a Q(z-c)[ 1 + an(z-c)n1
n=l


= 0.


Next substitute the series for P(z-c) and Q(z-c) into


equation (1-3):





5



(z-c) a(a-l) + YIa (u+n) (a+n-l) (z -c)nl



+ (z-c)a[ + p(z-c) + p1(z-cc) +...+ pn(z-c)"

00
+...]-.[ + an(a+n)(z-c)n
n=l


+ (Z-c)aC[q + ql(z-c) + q2(z-c)' ...+ qn(Z-C)n

00
+7 an(Z-C)n = 0.
n=1

Now equate to zero the coefficients of the successive
powers of (z-c):

(z-c)": a(a-1) + poa + qo = 0,


(z-C)Q+l


(1-4)


a 2 + (po-1) a + qo = 0.

: aa(a+l) + alpo(a+l) + apl + alqo + q = 0,

a [2 + a + po(a+l) + q9o + ap, + q = 0,

a (a+1)2-a-1 + p (a+l) + qo] + ap1 + q = O,
1) (-)( ) + + a + 0.
+ (p -1)(a+l) + qo] + op, + q 0.









(z-c)a+2: a2(a+2)(a+l) + poa2(a+2) + plal(a+l) + P2a

+ qa2 + qlal + q2 = 0,


a2[(a+2)(a+l) + po(a+2) + q]


+ a[Pl(a+l) + ql] + p2a


+ q2 = 0,


a2[(a+2)2 a 2


+ p (a+2) +


+ a lP(a+l) + q1


gq]


+ P2a + q = 0,


a2 (a+2)' + (po-l)(a+2) + q


+ a [Pl(a+l) + q]

And in general we can write:


(z-c)+n: a (a+n)(a+n-l) + pa (a-


+ p a + q a
n on n


n-I
+ z q a-
m=l


+ pa + q2 = 0.


n-1
+n) + ma n(a+n-m)
m=1


+ q = 0,
-n


an (a+n)2 a n + po(a+n) + q 0


n-I
+ an-m Pm(a+n-m) + qm] + Pna + q = 0,
m=l









(1-5) a [(c+n)2 + (p-l1)(a+n) + q 0


n-1
+ Z aMp (a+n-m) + q.] + ap + q = 0.
m=l

The first of these equations which is obtained by

equating to zero the coefficients of the lowest power of

(z-c) is called the indicial equation. This equation

determines two values of a (which may or may not be dis-

tinct). If z=c had been an irregular point, the indicial

equation would have been at most of the first degree.

To see this, suppose P(z-c) has a simple pole at z=c

and expand it in a Taylor's series.

P(z-c) = p_l(z-c)-l+ pO + P(z-c)+...+ Pn(z-c)n+.



Pn(z-c)" p_ / 0,
n=-l

and as before
00
Q(z-c) qo + q1(z-c)+...+ qn(z-c)n +... = qn(z-c)n
n=o

Now multiply according to the following scheme [which

is equivalent to substituting the series for P(z-c) and

Q(z-c) into equation (1-3)] and obtain the indicial

equation:








q + ql(z-c) +...



p 1+po(z-c) +...


(z-c)2


u = (z-c)" [ + al(z-c) +...]


S (-c) 1 a + (c+l)a,(z-c) +...


u" = (z-c)-2[a(u-l)+(u+l)ual(z-c) +...]


The indicial equation is


(Z-c)a-1


p_la = 0.


But a 5 0, and p_, 0 or z=c would not be a simple pole

which contradicts our assumption.


Or if we assume that Q(z-c) has a simple pole

and expand it in a Taylor's series:


at z=c


Q(z-c) = q_l(z-c)-~ + qo + ql(z-c)+...+qn(z-c)n +...


03

= q(z-c)n,
n=-l


q_, O,


and

P(z-c) = po + p (z-c) + P2(z-c)2 +...+pn(z-c)n +..



p pn(z-c)n
n=o

Now multiplying similar to the previously mentioned

scheme:








q -(z-c)-'+ q +... u = (z-c)[ 1 + a (z-c) +...


po(z-c)+ pi(z-c)2+... u' = (z-c)a-1 a+(a+l)al(z-c)+...


(z-c)2 u" = (z-c)a-2 a(a) +...]

The indicial equation is

(z-c) -i: q-, = 0,

which is a contradiction of our assumption.

If both P(z-c) and Q(z-c) have simple poles at z=c,
then

q_i(z-c)- + qo+... u = (z-c)al + ai(z-c) +...
pl(Z-C) + u ,a (z -1[ ]
p_-+p.(z-c) +... u (z-c)a + (a+l)ai(z-c)+...]


(z-c)2 u" = (z-c)a-2 a(a-l)+(a+l)aa (z-c)+...

The indicial equation is

(z-c)"-i: pla + q_ = 0,
which is first degree in a.
Hence, if z=c is an irregular singular point, the
indicial equation is at most of first degree.
Now the roots of the indicial equation are called the
exponents of the differential equation at the point z=c.
Returning to our indicial equation which we shall call F(a),
(1-6) F(a) = a2 + (p -l)a + qo = 0,








let a=?l and a=?2 be the roots. Each value for a deter-

mines, in order, a set of coefficients a ,a2,..., thus pro-

ducing two distinct series solutions, provided that 1 2

is not an integer (zero included). It will be shown later

that when the difference is a non-zero integer then our

proof for convergence of the series fails and if the

difference is zero, the two solutions are obviously the

same.

1.3 Convergence of Formal Solutions. We now pre-

sent a proof by induction that these series solutions con-

verge.

Assuming that the exponents ?i and 2 are not equal,

choose them so that Re(-l)aRe(X2), and let A1i-2 = s. Now

since the exponents are a = 1 ,2 we can write (i-o),

(1-7) F(a) = (a-1l)(a-z),

then for a = 1 +n,

F(A +n) = (A?+n-?A)(?l+n-h2)

= n(n+s) .

Let

P(z-c) = p + p1(z-c) + p2(z-c)2 +..+ pn(z-c)n +.

where
p(n)(c)
Pn n

according to Taylor's expansion. By means of repeated

differentiation [4, p. 70]








(n) n_ n P(z) dz
(z) 2= 1 "J Zn+l
C

If we let the curve C be a circle with center at the origin,

z=rei1, then dz=rieie de. Let M1 be the maximum value of

IP(z)l on C. Then

n_ P(z)dz n j IP(z)l ldzl
n+1 n+1
2Ti C zn+1 i C Izn+l


n!2 M1 rie id8
f n+1.l
0O I


27T
M1n! d M1n! 27T
< -- I 0 -_ [9_ ]
21rr n 2rn [0
0
o


M1n!

r


Thus


and


Similarly for qn:


Iql s. 2r


where M2 is the maximum value of IQ(z)I on C.

We next show that
-n
I Pn + qnl M13r-n

INPn + q 1 s Ix1l-IPnl+lqnl
-n -n -n/ n
SIN "M1r-n + Mr- = r (I.11 -M+M, = r M ,

where


MS = I IM1 + M2.


(z) n



pn I Mr-n
^*l)5r


i i * --n







Choose M to be larger than M1,M2, and M3 and also M 1.
Then
(1-8) I|pn < Mr-n; nIq ( Mr-n; IlP + qn < Mr-n

for n = 1,2,3,....
Now with the aid of equation (1-6), we can write
equation (1-4) as
a F(a+l) + ap + ql = 0
and

-(apl + q )
a1 =
SF(a+l)

Now let a = X then

iall l1 pl+ql < Mr M
IF(?h+1)l ll+sl r

since ll+sI > 1. That this is true can be seen by letting
1 = a + pi and X = 7 + 6i. Then s = (a+pi)-(y+6i)
= (a-7)+i(3-6) and

11+s = V(l+a-y) + (p-6) .

If P-6 = 0, or P = 6, then a / y for we have assumed from
the beginning that A1 and Ag are not equal. By our choice
of hi and X2 a > 7 and 1 + a y > 1 and |1+sl > 1. If

a = y, then B 6 andVl + (p-6)2 > 1 or Il+sI > 1. We have
now shown that an has a maximum for n = 1.
Now we assume that

lanl < Mnr- n = 1,2,...,m-l.







From equations (1-5) and (1-6) we can write

n-1
an F(a+n) + a n-pm(+nl-m) + qm] + pa + qn 0,
m=l

and letting a = \,

n-1

,a n-mPm +n-m) + qm] n qn
m=1l
a =
n F( +n)

Now replace n by m and replace m by t. Then

m-1

am-t [Pt(h1n-t) + qt]- m q
(1-9) lam = t=
F(l +m)

m-i m-1

Slam-t p Ip1t +qtl+I lPm+qml+ 7 (m-t)lam-t JiPt
t=1 t=l
m mn+s

m-1 m-1
Z lamt IMr-t + Mr-O + (m-t)ia i '- Mr-
t=1 t=1
\ ------------1,-- -----
m I1+smr

For the first summation:

m-1 m-1
(1-10) a -iM = M a -tIr
t=1 t=1


SM [IaMi 1r-l+jam-2 Ir2+...+Ia r-m+2+ ar-m+l1








< M[Mm-1 -m m-2 -m
M M r + M r


+...+ M2r-m + Mr-m]


< Mr -[M-l + Mm-2 +...+ M2 + M


< Mr- (m-1)M m-1

< (m-l)r-M .

For the second summation:

m-1
(1-11) (M-t)lam-t Mr-t
t=1


= Ml(m-1)


F, r ni-1 -m m-2 -m
< M (m-1)M r +(m-2)Mm-2 r +.


< Mr-m[ (m-)Mm-l+(m-2)Mm-l+...+


< Mmr-m[(m-l)+(m-2)+...+ 2 + 1


< M'r-m M (m-1)


..+ Mr-]


Mm-1I


(m-l)r M + Mr- + Mmr -m (m-l)
|aJ < ------- 2 --
m2 1+sm-1
M-r mm -mrm -m m -m -1
Mr M -r M + Mr + Mr (m-1)
< m 2_2+SM -
m'|+sm-
mn'Il+smn


Thus


lam-lir- 1 +(M-2)la -21r- 2 +...+(l)lallr-nl+ll







min -mm T ,nm- m m -n
mr M r M + Mr + Mr (m-1)
< 2
m' l+sm- |

2m l+sm-

But ll+sm-l ) >1, for letting s = 1-A2 = a+3i (y+6i):

si+ iT m+s m+s|
1+ !1 = m I = m'
m m m
and

Im+s = (a-y+m)2 + (--6)' > m
since 1i and ?2 are distinct.
Thus

Im+sl > 1
m

and

a I ( + rl) < r- M for m > 1.
m 2m

Therefore, by induction, anl < Mnr-n for all values of n.

If the values of the coefficients corresponding to the
exponent y2 be al,a ,... we want to show, by a proof
similar to the previous induction proof, that

la'l MnKr- ,

where K is the upper bound of Il-sl- Ii- sl-1 11- -1
... Let us begin by showing that such a bound exists for
s not a positive integer.







If 0 < Is! < 1, then


1
1ls' '


H1-ls I ,


J = 1,2,3,...,


J 1,2,3,...


Proof: Let s = a+-i, (a,p real); then 0 < Isi = a2V+2 < 1.


1-


= 1- = (1- )2 + () =-- 1 + a '+
5'i -~i--i-i j 2e


j J'


= V l- +a N
),


S1- Qa+f3 2 1 -JaV+02 = 1 IsI = 1-IsI
J
If Is) > 1, then

1 s Is
1-^ 1-

Proof: Let Isl = t + 0, where t is the largest integer in
Isl and 0 g 6 < 1.


J2 j j2
S- 1- (+ ) + J- 1 +

S- -
Jj3


1
1l-Is I


11- s1








We desire I- s to have a minimum value larger than zero,

where j=l,2,3,..., t-n,..., t-l,t,t+l,...t+ ....

(n=l,2,3,..., t-l;=l,2,3, ...). Difficulty could arise,

however, when j=t and if 0 = 0. Then the absolute value

would be zero.

First we show

it+ad t+e 1-0
1- -- > and 1- --
t-n t t+tl -TT
analytically, where n=0,1,2,..., t-l; t=1,2,3,..., 0 < e < 1.

t+8 8
1- t-n n=0,1,2,..., t-l; 0 < 8 < 1.

t-n-t-0 Q
t-n '

n+O e


nt + Ot O 8t n ,

nt -nG, true for n=0.

If n/O, then t > -8.


t+"t~



t+ .-e
t+'t




(t+l)(t- )




d( t+0)


1-8
t- t=1,

1-8
t+l '

1-8
t+l

(1-0)(t+d) ,

_ t+4-8t-8e ,

_ t+ ,

1 t+0 ,

_ 1, which is true.


2,3,...; O < 8 < i.








Therefore,
+0 0 t+ 1-0
(1-12) 1- t+e and 1- -t+
t-n t+' ttT
where n=0,1,2,..., t-1; ,=1,2,3,...' 0 < ( < 1.

Inserting Isl for t+0 and j for t-n and t+L in (1-12)
we see that, for Isl > 1 and 0 < 0 < 1, either

i-- or 1- t-
Jt j I t+1
depending on the values of 0 and t.
Now consider the case where 0=0 (Isl=t) and j=t
= a2+ep P30. We wish to find a 6 such that if s =a+pi,
then

1s 1 I js +6


1- a+3i 6a +P2 + 5
-Va2 1-


IVVa21'32 -a pi | 161

(Va'+P a) + p 62 ,

a'+B 2a Va2+132 + a'+p2 > 52 ,
2t"-2at 62 ,
2t(t-a) a 56

However, since Isl=t > 1 and t =Wa2+8 )> a, if we have
given any value of s (Isl > 1, pBO), then a value of 6
can be determined such that 0 < 6 < 1. Referring to (1-12)
with 0 replaced by 6, we have for Isl 1, 0 < 6 < 1, that







t+6 6 a 1t+6 1-6
1- T and 1-
I F- an t I t+t I '
where n=0,l,2..., t-1; k=1,2,3,...
Summarizing our results we have:
If 0 < Isl < 1, then

1- 1 1-is .

For Isl > 1, Isl / an integer,
(a) -if 0 < 0 < 1, then one of the following


s I Ii = >- i _
6

1 1 + 1 = 1 +6 1 f
(b) if 9 = O, then one of the following




t+1
where J=1,2,3,...; IsI=t+6 (where t is the largest integer
in Isl and 0 0 ( 1); 6 chosen such that 0 < 6 < 1 and
5's 2t(t-2).
We have thus proved the theorem that given any s
(s not an integer) an upper bound x of the sequence
-1
j=1,2,3,...,

is the maximum of

1 t t+l t t+l
1;TT (6/0o); 19/ 0) ;
1- si (eo); 1- ( 6o), 6 1- 6

where 6 is chosen such that 62 2t(t-a) and 0 < 6 < 1.








Now we present an inductive proof that

la'l < Mn K r-n
n

Again using equation (1-6) with a now replaced by 22+ n,

we have

(1-13) F(h2+n) = (h2+n-h?1)(2 +n-2) = n(n-s)

If -i is replaced by Nt, the inequalities of (1-8) will

still be true. However, they will also be found valid if M
1
is chosen larger than as well as M1,M2,M and M 1. Let

this be done. Thus,

IpnI < Mr-n; Iq < Mr-n I2pn+q nI < Mr-n

Also, since M > and M g 1, we have the additional

inequalities which will be utilized later in the proof,

MIc > 1,

(Mc)"- )> 1, where m=1,2,3,...

When a = h2 and n=l in equation (1-4), we can write


a' =
(h2+l)2+(po-l)(h,2+l)+q

and with equation (1-6) this becomes


a' = 2p
1 F(N2+1)








Taking the absolute value of each side and using (1-13) with

n=l, we obtain the inequality

aI p +q < I Mr-
IF(A +1)1 1l-sl

Next, assume


n=1,2,3,... m-l,


and show that

a' M r .

Using (1-9) with A2 and a' replacing 'A and a

respectively,

m-i

m- t[Pt (2- + q t] 2pm
la'll t=l
a'i =-
m F(A2+m)

Tm- m-i

lam- lPt 2+qlt pm+q (m-t) a-tl Pt
t=l t=l
mim-si


m-1

S .Mrm-t t + Mr +
t=l


m-1
i (m-t)'la tl' 'Mr-
t=l


m2 1-sm1 I


lan'i n n -n
a' < M K r








Similar to (1-10) and (1-11) with Jam_t < (MK)lr-m

M > 1,


m-1
am a -tI Mr- ( (m-1)(Mc)M-1 Mr-
t=1



< (m-l)r-mMKm- ,


and


m-1
S(m-t)a I Mr-
t=1


Thus,


(m-l)r-WMmK'- + Mr-m
la'l <
m


m m-1 -m
< (m-1)(MK) mMr



< r (m-l)r-mMmKm-
2



+ M r -m (m-1) KIn-
-1


m ll-sm I


-im m-- -nm r-i -m i -mm iii
mr- MmK -m-_r- M -1 + Mr-m + Mmr-m m (m-l) r-

m I -sm-


Now since (Mc)m- > 1,


Mr-m Mm-m m-1
Mr Mr K


= Mr-m(l, Mm-1 m-I)
= Mr (1-M K )


= Mr 1-(MKc) i]


is negative. Thus,


Ia'l <
m


mr- M K -+ r- M (m-1) m-
m2lsm-1
m |1-sm |








-mm m-l1r+l
Sr M (m+)
2mI 1-sm1


< r-mMmCm (ml) < r-mMmVm
2m


for m > 1.


Therefore,


Ja'l < Mr ,n
n


for all values of n.


The radius of convergence for the power series


[4, p. 80], for the


Mnr-(z-c)n is
n=l


Mn -n
lim M
n -o, Mn+ r-n-l
n ->oo M r


r r
limn -
M M .
nj o


That is, the series is absolutely convergent within the

circle Iz-cl = .. Since laI Mr-", the series



an(z-c)n also must converge within the circle Iz-cl
n=i
r
= and is, therefore, uniformly convergent in the region

Iz-cl ( [L, p. 951.



Similarly an'(z-c)n converges uniformly within the
n=l


region Iz-cl < r
.1-








We have thus obtained two formal solutions


u,(z) = (z-c) 1 + a (z-c)n
n=l

and



u'(z) = (z-c) 1 + an(z-c)n]
n=1

which are uniformly convergent series of analytic functions

when Iz-c( < rM- and Iz-cl < rM1-l respectively, pro-

vided that arg (z-c) is restricted in such a manner that

the series are single-valued. Consequently, the formal

substitution of these series into the differential equation

is justified for -\ 2 not a positive integer. These

solutions are valid in the vicinity of a regular singular

point.

1.4 Second Solution in the Case where the Difference

of the Exponents is an Integer. We now derive a second

solution in the case where the difference of the exponents

is an integer. When l-'2 = s is a positive integer or zero,

the solution u2(z) may break down or coincide with u1(z).

Try the change of variable u = ul(z)-. and substitute
this into

(z-c)2 u" + (z-c)P(z-c) u' + Q(z-c) u = 0

to determine the equation for ).







U = U1 0,
u' = u .)' + u 0,
u" = u. *)" + 2u; 0' + u" 0.

Substituting in we have

(z-c)2 u1()" + 2uj' + u4] + (z-c)P(z-c)(u )' + u )

+ Q(z-c) ul*) = 0,

(z-c)2u 1n" + 2u (z-c)2 + (z-c)P(z-c) -uI 0


+ (z-c)2 u1 + (z-c)P(z-c) u1 + Q(z-c) u1 0 = 0.


Since u (z)

(z-c)2


is a solution

u"1 + (z-c)P(z-c) u! + Q(z-c) u1 = 0,


and
(z-c u 2(-c)2 +2 + (Z-C)P(z-c).uIj = 0.

Dividing by u. we have

(z-c)2 "+ [2 (C) + (z-c)P(Z-c) 0' = 0.


To find a general solution let 0' = Y.

(z-c) Y' + 2 u (z-c) + (z-c)P(z-c) Y = 0,


(z-c)2 d 2 (z-c) + (z-c)P(z-c) Y = 0.








Separating variables and integrating,


-Y +
I dYY


f[


S+
u 1


In Y = In B 21n


-2
Y =Bu-
1


P(z-c dz = In B,

S(z-c)
u (z- )- dz

u/ P z-c)

SP(z-c/ dz
Tz-7 cy


Now since Y = 0',


O [B u-2 e


- dz ] d
FZ --d7


where P(z-c) = po+ pl(z-c) + pp(z-c)'+...+ p(z-c)+..


dz -
P(z-c) dz


P [p(z-c)1 + PI + P2(z-c) +.'. dz


= poln(z-c) + pl(z-c) + p, 2(z +..


Hence,


0= A+B


= A+B


-2 [-p ln(z-c) pl(z-c)-...-pn(z-c)n-...]
uI e


u 2 (z-c)-Po e



( 2
+BC)- dz


00
,P(z-c)n
n n
n=l


= A + B


dz .








But since

oo
u,(z) = (z-c) 1 + an(z-c)
n -1


) = A + B (z-c) -2 g(z)dz,


(z-c)n
ri
SPn n
[ i-2 n=l
where g(z) = 1 + a (z-c)n e
n=1

A and B are arbitrary constants and g(z) is analytic

throughout the interior of any circle C whose center is

z=c, which does not contain any singularities of P(z-c) or
-1
of (z-c) 1u(z), nor any zeros of the latter. g(c) = 1.

Since g(z) is analytic in the interior of C, it can be

expanded in a Taylor's series about its center z=c. Let

00

g(z) = 1 + g(z-c)nl
n=1

We already have a2 + (p0-l) a + q = 0 (1-6) with

roots a = Al,'2 and Al-A2 = s. The sum of the roots l+A 2

= -po, so we can write

-p -2?, = 2 +~2-1-2 = -1 = -s-1.

Then

00
(1-14) = A + B 1f + (zc)n (z-c)-1 dz
n=1










= A + B [


s-1
(z-c)-s-ldz +J gn(-c)n--ldz
n=l


gs(z-c) 1dz


CO

+ n= g ( c)n-+dz
n=s+l


(z-c ) sSjI-
nS
n=1


-n
s-n


(zC)n-s
(z-c)


+ gs In(z-c)


+ in- (z-c)n-s
n=s+l

The general solution, analytic in C except at z=c, is


u = ul(z).* = A ui(z) + B[ u(z).gs.ln(z-c) + (z)1


where


(z-c)-r


s-I

n= c)
s-n
n=l


1
u(z) = u,(z)[-
1 Vs




+
n=s+l


n-
n-s


and


u1(Z) = (Z-C) 11


(ZC)n-]





Co
+Xa(zc)n]
n=1


+


= A + B[









Rewriting, we have



u(z) = (z-c) 1


00 S-1

+ a(z-c) (z-c)f- g-s(z-c)n-
n=l n=l


(z)n-s
(Z-C)


Cn
+ n--s
n=s+l


= (Z-c) 2


uo

la(-c)
n=1l


s-1
n (z-c)n
s s-n
n=l


(z-c)n]


+ n--
n=s+l


= (z-c) -
= (ZC) 2[


+ h(Z-C)nJ
n=l


where h are constants:
n


Sg1 al
s-a +


and


hn(z-c)n = -
n=l


an(z-c) +
n=l


1 + an(z-c)n
n=l


s-1
g-n )n
(z-c)
n=-
n=1


00

Zs gn
Ln-s
n=s+l


(z c)nI
(z-c)








When s = 0, we can write (1-14) in the form

CO
S= A + B 1 + g (z-c)n] (z-c) dz
n=l



A + B [ (z-c)r dz + gn (z-c)ndz
n=1



=A + B ln(z-c) + g(z-c)
n=l

Therefore, when 1 = -2, the second solution is



u = ul(z).* = A ul(z) + B ul(z) ln(z-c) + n (z-c)n+1]
n=l

Thus it is seen that when ,1 -2 is an integer, the

second solution involves a logarithm, except when gs = 0.

A practical way of obtaining this second solution is to

first obtain the solution u1(z), and then determine the

coefficients in a function

00
Z ^.2+n
U1(z) = bn(z-c)
n=o

by substituting

u = ul(z)ln(z-c) + 5 (z)

in the equation and equating to zero the coefficients of

the various powers of z-c in the resulting expression.






31


Example: Find the solutions of the equation

S 1 2
u" +- u' m u = 0
z

regular near z = 0 [8, p. 201].

Equation (1-2) with c = 0 becomes


(1-15) z2u" + zP(z) u' + Q(z) u = 0.

2
Thus, if we multiply our given equation by z we have

2 2 2
z u + zu' m z u = 0,

2 2
where P(z) = 1 and Q(z) = -m z

Assuming

00oo
a n
u = z az ,
n=o

then

2 2 a a+2n a+2n+2
-m z u = az +...+a z +a z +... a /0.
0 2n 2n+2, o

z u' = a aza-l+...+(a+2n+2)a2n+2 a+2n+l+...

z u2 = a a(a-1)z -2+...+(a+2n+2)(a+2n+l)a 2n2+2n+...
0 2n+2 .

Equating the coefficients of powers of z-c to zero,

z : a(a-l) + a = 0,
2
a = 0 and a = 0,0.

a+2n+2 a 2n+(a+2n+2)a2n++((+2n+2)(a+2n+l)a2n+2 = 0.
Z2 -m2+2n









2
m
a2n+2 2
2n+2 (a+2n+2)2

2
m a2n
2n+2 22(n+ )2


If n = 0,


2
m ao
a2 -2
2 212


If n = 1,


2
m a2
a4 2(22
2 (2)2


4
m ao

2 (2!)2


If n = 2,


2
m a4
a6 22()


6
m a

2 6(3!)2


2n
an
n 2"n !2
a n = 2 2 n !n 1)


n=l,2,3, .


Therefore, if we arbitrarily choose a = 1,
o


2n 2n
S2n(n


U1 =
n=o


For the second solution, let


u = ullnz +
n=o


and substitute this into (1-15).


b z
n











u' = u'ln z + ul





u" = u'ln z + 2u ( -


0)

+- + nb z-
n=l


00

+ Zn(n-l)bnzn-2

n=2


00 00

n n
ul'z21n z + 2u 2z u + n(n-l)bnn + u zln z + ul + nbnz
n=2 n=l


00
2 2 2 n+2
- um z in z m b z2
n1 n
n=o


= 0.


Since the coefficient of In z is

2 2 2
z u, + ulz m z u1 = O,


we have left


00 00

2uz + n2bz mbn b zn+2
1 n= n=
n=1 n=o


= 0,


2uz + (n2 m2z2) b z = 0.
n=o

Now, taking the derivative of our first solution ul, we

obtain


2n 2n-1
2n m z
n=A 2n(n!)2
n=1


u z 2










2n 2n
4n m2nz z
22n(n!)2
n=1 n=o


2 =0.
(n -m z )b z = 0.


Equating to zero the coefficients of like powers of z, we

have

z: b 0 = 0, b = arbitrary constant.

1 b 0.
z : b =0.
1


2n
z :


4n m2 + n 2b
2n + 4n b2n
2 n(n!)2


b 2n 1 2 2n-2
2n n 4


- mb =
2n-2


2n
n m
22 (n!)2 ,


n=1,2,jj, .


Choose bo = 0. Then for n = 1,


m2b
b_ o
4


If n =2,


2 2
m -m
22(1!)2 22(1!)2


b -1 bm 2m4
b4 =22 [- b2 4 2
2 4 2 (2!)


2 24(1!)2 24 2 2


4
-m
b4 4 -m
2 (2!)


If n = 3,


1
b6
6 3


(1 + ).


mb4 2 3m6

4 2 (3!)2 j


so that








-1 m6 (1 3m6 1
322 (T(2! 2 26(3!)2

b = -m i + 1+
6 26(5)2 2 3




Finally,
2n
2n ( 1 1 m n
-m 1 1\ mHn
2n 22(n)2 2 n2


1 1 1
H 1 + 1 + 1 +...+ .
n 2 3 n

For n odd,

n 2 2
z : nb -mb = O,
nbn n-2

2
b b n odd,
n 2 n-2
n

since bI = 0.

Therefore, the second solution is

Co 2n 2n
Sm H z
u(z) = u1ln z 2n 2 (n!2
n= 22n(n!)2
n=1


1.5 Solutions Valid for Large Values of Izl. Let us

now consider solutions valid for large values of |z|. To do
1
this we let z = in equation (1-1). If the solution of the
zi

transformed equation is valid for sufficiently small values

of I1zl, then the original equation is said to be valid for








large values of izl. If the point z1 = 0 is an ordinary

(or regular or irregular) point of the transformed equation,

then the point at infinity is said to be an ordinary (or

regular or irregular) point of the original equation.

We begin with the change of variable z in the
zi
equation


d2+ p(z) du +
dz dz


q(z) u = 0.


1
z
z
Z1


1
z = Z
1 z'


dzl 2 dz.
z


du du dz du
dz dzI dz z dzI



du 2 du du dz
2 3 2 2
dz z dz1 z dzI dz


2 2
du 3 du 4 du
2z + z .
dz dz+ dz2
1 1


Substituting in (1-1) we obtain


4 d u 3 2 / 11 du +
z + 2z z p ~q- u = 0.
1 2 1 1 +
dz 1 dz z
1 1
4
Dividing by z4



d u 2 1 du --1 1
dz 2u+ p +l zq u 0.
dz 2 1 z 2dz 1 z
1 1 1 1


(1-1)





37


Now z = 0 (i.e. z = m) will be an ordinary point when

du
the coefficients of and u are both analytic. This re-
dz.
1
quires p and q to have the following forms:

1 2 3
p = 2z + Az + A3a + .


1 4 5 6
q (1) = B4z + B5zl + B6zl +.', B / 0.

Expanding about z = o:

A A2
p(z) + + +...3
w + ~- "


, = 0 wil


z z

B4 B B6
z) + +- +..
z z z

.1 be a regular singular


B4 / 0.


point when


2 1p and q
2l Zi)


1 \ 1
1- are analytic at z = 0.
21 z2 1
1


This means that p and q must be:


p1



z11) =
q 1 =


C11z + C2z2 + C321 +...


2 3 4
D z2 + D3a3 + D4z1 +...


Expanding about z = c:

C C C
C + + +.
P(z) = z' -2 + -3
z z


q( D D3 D
q(z) = z +- + +
z z z


C / 0.



D2 / 0.


C, /0.


D / 0.


.*








Further, it might be helpful to know for what values of

t, m and n z = o will be a regular point in the following

equation having polynomial coefficients (po / 0 and 4, m, n

non-negative integers).


(poz + plzt' +...+ pd) u" + (q zm + qzm +...+ qm) u


+ (r zn + r z +...+ r ) u = 0.


Dividing by the coefficient of u",


q m m-- r)o n-- n-~ -1
u" + z-+ ELz +... u' + z + F z +... = O.
PO O

We must have m-t = -1 or 4 = m+l and n-i = -2, 4 = n+2.

Therefore, if z = o is a regular singular point, then

4 = m+l = n+2.


1.6 Irregular Singular Points and Confluence. We have

already shown in section 1.2 that near an irregular singular

point a second order differential equation cannot have two

solutions because the indicial equation is at most of the

first degree; there may be one or no solutions of this form

near this point.

If a differential equation A is obtained from another

equation B by making two or more singularities of equation B

tend to coincidence, such a limiting process is called con-

fluence. Equation A is called a confluent form of equation

B.













CHAPTER II

Confluent Differential Equations

2.1 Differential Equation with Four Singular Points.
We shall now develop a general form for a second order

linear differential equation which has every point except

a,, a2, a3 and a4 as ordinary points (z = being an ordi-
nary point). These four points are to be regular singular

points and let their exponents be ar, pr at ar (r=1,2,3,4).
Then the form of our equation will have to be


d2u + r + H(z) du+ + + J(z) u = ,
z -i ---a
dz2 z-a)2 z-a) r
dz r=l r=1 -ar

where kr, r', mr (r=1,2,3,4) are constants to be determined,
and H(z), J(z) are polynomials in z.

Let

u = ho(z-a)' + hl(z-al) +l+... h0 / 0,

then
4 M

Z+2 z-] + J(z) u = ho(z-al) + h (z-al) +...

4
Z r- + H(z) u'= ho(z-a )-+...
z-ar0
r=l

1 u" = hoX(?-l)(z-a )-2+...









The indicial equation for z = al is


(z-a )-2:


ho(A-1) + hok 1 + h o = 0,
o o 1 o 1


?2 + (k-1l)? + 1 = 0.


Then



so that


a1 + pi = 1-k ,


k = 1 a1 oi'


and


ali = .1i


Similarly for the other regular points, so that


(2-1) d2u +
dz2
r=l


l-ar-Pr + H( 1 du
z-ar I dz


+ +arr 2 + J(z)
r~l (Zar)2 z-a
L z-a r


u = 0.


1
If z = o (or zi = 0, z = ) is to be an ordinary point,
zi

2z-z p(z) and z4q(z) should be analytic at z = ,
2 1 1 1 1
or - p( ) and -4- q( -) should be analytic at z = 0
1 z 1 z 1
1 1

(section 1.5).

Now


( 21 1
Sz z 2
1 1 21


z l-oai -6(.
r r 1 H 1 Z
1 z 2 z
r=1 r 1
1


2 1
2 zp
1 Z1








41-a r-0r 1
z zl (-arzi) 3 H(
r=l 1


z z 1 -Z1 1
= i -^ (- i H -


21 rr r z 1 2 J/
= r---i--I^ ^T



2 4_, 2 2
I- r 1+a Zl + aZ +.
r=l

1 1

1

4
2 L (1- -r- )
r=1 n 1 (
r=l + h z H
n 1 2 Z
1 n=o 1

where hn are constants. When zI = 0, we need

H z( 0

and
4
2 (1-ar-r) = 0,
r=l

or

4

S(ar+fr) = 2.
r=l







Rewriting q(z) in the form


(z-a1) (z-a2)(z-a3)


(z-a2)2(z-a3)(z-a4)


(z-a4 )2 z-al)(z-a )


(l-a z )2 (-a2 z)(l-a z )


(l-a2z)2(l1-a 3z)(1-a4zO)


(l-a 4z )2(-a zl)(l-a 2z)


(1-a z )2(1-a z )(l-a zl)


1 1
J J(1
z4 z
1


which is analytic at z = O, if J ( -) = 0.

To evaluate nr (r=1,2,3,4), we have


(z-a) 2(z-a2)(z-a3)


(z-a 2 z-a3 )(z-a4)


(z-a3)2(z-a4)(z-a1)


(z-a )2(z-al)(z-a2)


q(z) =


(z-a ) (z-a4)(z-a )


then

1
-. q
zi


+ J(z),


4
S[ar r r+
r= (-r)2 z-ar


1)






Then clearing of fractions,
n (z-az--a3 (z-a4 )2 + n2(z-a )2 (z-a3 )(z-a4)

+ n3(z-al)(z-a2)2(z-a4) + n4(z-al)(z-a )(z-a3)2

SY11z-a2z-a3 )2 )2 + a 22 (z-01)2(z-a3)2 a )2

+a33 (z-a)2 (z-a22(z-a4)2 + aq4 (z-al )2(z-a3 )(z-a )2

+ m (z-a1)(z-a)2 (z-a3)2 (z-a)2

+ m2(z-al)2(z-a2) (z-a3)2(z-a4)

+ m3(z-al)2 (z-a2)2 (z-a3)(-aa4)2

+ m4(z-a )2(z-a2)2(z-a3)2(z-a4).

To determine nl, let z = a :

nl(al-a2)(a -a3)(al-a4)2 = alBl(al-a )(a-a2(a-a )2,

nl "ll(ai-a )(a -a3).

Similarly,

n2 = 22 (a2 -a3)(a2-a4)

n3 = a3 3(a3-a4)(a3-a),

"4 = a44( a4-a)(a4-a2)








Therefore, the general form for a second order linear

differential equation with regular singular points at

z = ar (r= 1,2,3,4) is


2 4
d + l1-ar-Pr du
dzq z-ar dz


a2 2(a2-3 )(a2-a4)_
(z-a )2(z-a )(z-a4)


+ 1(al-a )(al-a 3)
1 (z-a )2(z-a 2)z-a3)


S33 (a3-a1) (a3-a,)
(z-a )2 (z-a )(z-a )


a+ 4134(a4-a,)(a4-a )
(z-a4 )2(z-al)(z-a u =
(z-a ) (z-a )(z-a )


where


4

r=l


(a +p ) = 2,


ar and (r being the exponents at z = ar'


To express the fact that
this type we will write


al

u = al

S1


(2-3)


u satisfies an equation of


a2 a3 a4

a2 "3 "4

P2 33 34


2.2 Differential Equation with Singular Points at
z = 0, 1, . Let us now consider the confluent case

where a2 = 0, a3 = 1, a, = a4 become infinite.


(2-2)


z







First, let a2 = 0 and a3 = 1, then (2-2) becomes


2 [ 1-a -p
(2-4) a u +
dz2 z-aL


l-a ,-P 1-a3 -3 1-a 4-P4 du
+ ----- + +j
z z-1 z-a4


+ Lapiai(ai-l1)
I(z-a) )z2z-l)


a 22a4
+2(- +
z2(z-i)(z-a4)


S3P3(1-a,4)(-al)
(z-1)2 (z-a4)(z-al)


a 44 (a4-al)a4 ] .
----- ----- = 0 .
(z-a4)2 z-al)z

Before letting a4 become infinite, rewrite the equation
in the form


[ -al


+ [a 1al)(al-1)
(z-a1 )2z(z-1)


l-c2-P2 1-a3-3 1-a 4-P4 du
z z-1 z-a4 dz


(2 3 2
z (z-1) a4
(a4


44(1- )

( --- 2(z-al)z-
a(4 1/


Now let a4 become infinite:


(2-5) d2 +
dz2


1-a1-i 1-a -2 +
z- + z
z-a1 z


1-a3-031 du
z-1 dz


+ iOlal(al-1) a 22 + a33(l-al) + a44 ]
(z-a )2z(z) + -2(z- = (ZO
(z-al)2z(z-1) z2(2-1) (z-1)2(z-al) (z-al)z


2
d2u
d+
dz


( P -1 (1-a )
C33(3 a4

(z-1) -a4 1(z-al
a4








which may be represented by


a 0 1 oo


(2-6) u= al a2 a3 z

f1 P 2 z3 4

This equation is actually more general then one might

assume, for by means of the linear transformation

(a4-a3) (z-a2)

(a2-a ) (z-a4)

any three points z = a2, a3, a4 can be carried into

U = 0, 1, 0.

Finally, we want to let aI become infinite to obtain

the confluent case with singular points O 1,, o. Rewrite

the equation (2-5) in the form


d u 1[ -al-P1 1-2- 2 1--3 3 du
--+ -+ + ----
dz L z-a z z-1 dz


[ a a,1) CXP 2 a3 f -l
+ +
S-iz(z-l) z (z-l) (z-1)2 z -)



"4P'4 u 0,
+ (z- = )z,
(z-a1 )z




47

and when al becomes infinite we have the confluent equation


(2-7) du 1-a 2-2 1-a3-P3B du CP a 22
(2-7) +-U + + l + 2(zl)
dz2 z z-1 z(z-l) z(z-)

+ U = O,
C 3P3 =0,
(z-1)

with singular points at z = 0, 1, oo.

Let us proceed to find solutions of this equation.

Multiplying the equation by z 2(z-l)2, we have

2
(2-8) Z2 (-l)2 d + (2---3-3)3
dz


+ (-3+2a +2 +a3+p3)z2 + (1-a 2- )z



+ [(ac + a A3)z2 + (-c,1 a2,B)z + ap2 u = 0.


Assume the solution

N ,+1 A+n-2 A+n-i
u C z+ C +...+ Cn z + Cn z
o 1 n-2 n-1

h+n
+ CnZ +...n C / 0,

then

u' = C Az 1 + Cl(A+l)z +...+ C (A+n-2)z 3
o 1 n-2

+ C (+n-)z+n-2 + C (A+n)z +n +...
nI-1 n







u" = Co0A(-l)z- 2 +...+ Cn2 (+n-2) (+n-)z+n-4

+ Cn_(+n-1)(-+n-2)zh+n-3 + C (+n)(+n-l)zh+n-2+...

If these equations for u and its derivatives were now sub-
stituted in equation (2-8) and the coefficients of the
powers of z equated to zero,we would obtain the following
results.

The indicial equation is

z: 7(N-1) + N(l-a2-P2) + CL22 = 0.

(2-9) -2 (a +2)A + a2 = 0.

Then
= a 2)2'

Finding Cn,

z7+n: Cn-2(A+n-2)(A+n-3)-2 Cn-1 (+n-1)(A+n-2)

+ Cn(+n)(n-) Cn_2(+n-2)(2-a2-2 -3)

+ Cnl(?+n-l)(-3+2a2+2 2 +a3+3) + Cn(?v+n)(l-aC2-P2)

+ Cn 2(0a1 +03~3) + Cn-1(-a1P1- a22) +.CnC2 2 = 0.


Cn = nC._ -(+n-2)(N+n-3)-(-+n-2)(2-a ,-P2-a,3-3 1)- (a L-3)1


+ Cn-1 2(?+n-l)(N+n-2)-(A+n-l) (-3+2+2P +3+33)


+ J (2+n)(7+n-l)+(,+n)(l-a 2_ ) + 2 a







Simplifying the denominator by using the indicial equation
(2-9) we have


C = Cn2[-(N+n-2)(+n-l-cx2-P2-a3-3) p I 033]


+ Cn-1 (?+n-1)(2\+2n-l-22-2a -2 -a3-3)


+ au 1 + a2P1 n(2-+n--1 -)
n(27+n-a -p2)

Let us compute a few terms of our series solution to
see if a general term can be obtained.


Co C (2+l-2a2-2P2-a3-p3) + a 11+ ~ P2]
1 (2A+1-a 2-2)

Using the indicial equation (2-9) we can simplify and write


Co[(1-03-p3) + al 1- a22]
C1 =
22+1-a --


C2 = { [-(+l-a3-3) 3] + C[( )(2


+-2a2-2P2-a3-3) + a P + a P2 -
S2(2)+2-a -6 )







Substituting C1 into this expression:


C2 = C {(2+l-a-2)[-(l-a2-2-a3-~3)-all-a33


+[(1-a3-p3)7 + al"Bl-a2] [ (A+1)(2h+3-2a2-2P2-as3-3)

1
+ a 1P + a P21.
++22 j 2(2X+l-a,-2_)(2X+2-a_-P2)

It seems apparent from the appearance of C1 and C,2 and
from the fact that Cn is a three term recurrence relation,
that a real deal of work might be required to write these
two solutions of (2-7) with a general term, thereby making
available a form suitable for further development. Since
equation (2-7) has five undetermined parameters (aucB a ,2V
L3,33), let us consider the conditions necessary to insure a
solution with a two term recurrence relation. This can be
readily accomplished by applying Scheffe's Criteria [6, p.
240] which Crowson has presented and proved in his dis-
sertation [1, p. 114].
2.3 Scheff4's Criteria Applied to Confluent Case. The
criteria to be applied is: Necessary and sufficient con-
ditions for a solution of a second order linear ordinary
differential equation,


p2(z)u"(z) + )u) + ()( + po(z)u(z) = 0,







to have a two term recurrence equation, relative to the point

z = 0, is that in some neighborhood of the point z = 0,

(2-10) Pj(z) = [Sj Tzh] zJ-m ,

where m is an integer, h is a positive integer, and Sj, Tj

are constants such that Si Tj / 0 for some i = 0,1,2 and
j = 0,1,2.
In assigning values to the exponents, it can be seen

from equation (2-4) that none of the exponents can be a

function of a Since the sum of the exponents must be 2,

if there are any exponents which are functions of a4 there
must be at least two such exponents. ar, r (r = 1,2,3)

could not be functions of a4, otherwise the coefficient of

du will be undefined when a becomes infinite. If a and
dz 4 4
04 are functions of a4, then the term

CL4 (a4-al)a4
(z-a4)2(z-a )z

would be undefined as a4 becomes infinite. Thus the exponents

can not be functions of a4.

Next, suppose that two or more exponents in equation

(2-6) are functions of a Such an equation might be

al 0 1 o0

u = aa^+f a i a -a1L- + t z
a- 2 3 a
a1








where the sum of the exponents is

f + a2 + P2 + a3 + 3 + t + 4 = 2,

f, t and t are arbitrary constants. Using (2-2) we can

write this equation in the form


1
l -aF -f- a 1 i-a -P2 i-a -
u" + + +
z-al z z-1


OP2 3B3 (1-al) ,)
z + + -
z (z-1) (z-12 (z-a )


a1
U" +


a a

z -1
VaI /


f
(,+ L-)al(al-l)
u' + al
(z-az )z(z-l)


(-a --+ t)
(- a'- l a-"]u = 0.


+ -2-2 + u'
z z-1



a2p L3 a1 -1

( al


4 2 1
+ u = 0.
a- -1)z


Letting al become infinite:

u" + + + U-33
z z-1

+ 2 2 + 33 4
z(z-) z 2(z-1) (z-1) Z U = .


+ 1-

z 1)2 1
aW -1 z(z-1)







Clearing of fractions we can write

(2-11) z2(z-1)2u" + [Iz2(z-1)2+(,a 2-B2 )z(z)2


+ (1-a3-P3)Z2(Z-l) U'


+[ z(z-1)-a22(z-l) + a3B3z + Bz(z-l)2] u = O,

where the singular points are at z = 0 1, o, .
Applying the criteria (2-10) we have for J = 2
S[S ,,zh] 2-m 2
P2 = [ T z = z (z-l) ,

2-m h-m+2 4 3 2
S2 T2z z -2z + z

but this is impossible. However, this difficulty can be
avoided by reducing the coefficient of u" by a factor of
at least z-1. To do this a313 must equal zero, so let p3=0.
Thus, with B3=0 and dividing equation (2-11) by z-1, we now
have

z (z-l)u" + Z2 )-( 2)(z-)+ -a3


+ [zz-a 22 + P4'z(z-l) u = 0.


(z 3 )u" + z3 +(2-a-- 2-a 3-)Z2 + (-l+2 +B2)z u'


+ [f4tz2 + (l-P4)tZ 22] U = 0.








Now returning to our criteria (2-10),
S2-m h-m+2 3 2
P2 = S2 Tz = z -Z ,

where h-m+2 = 3 and2-m = 2. Therefore, m = 0, h = 1,

S2 = T2 = -1, S2T2 / 0. We need

P, = [S-Tz ] = tz3 + (2-2- 2-a 3--)z2 + (-l+az +2)z

so that t = O, T1 = a2+ 2 + a3 -2, S1 = a2+82-1 and


p = (2-a2-32-a3)z2 + ( 2+0-l )z.

Finally, we have


PO [S-Toz = -2f2'

where S = -a22 and T = 0.

Thus, if (3 = 0 in equation (2-11) the resulting

equation, which has a two term recurrence relation for a

solution, is

(2-12) z2(z-l)u" + [(2-a2-2-(3)z2 +(U2+2-1)z u'-a282 u = 0.

The requirement that f + a +p0+a3+t+B4 =2 is actually not

restrictive, since f, t and P. do not appear in our final

result.

Consider the alternative of having the exponents in

(2-6) be functions of a and assume that neither al nor a4
appear in the exponents. Apply the criteria (2-10) to

equation








(-) +[i -a3-3 al [(1
(2-7) u" + I-a2-P2 + U' +
z z-1 z(z-l)


z 2
z (z-l)


a3r3 1
+ u_ = 0,
(z-1)
or clearing of fractions

z (z-)2 u" + (1-La -2)z(z-1)2+(l-aa-a)z (z-1) u'


+ [alZz(z-l) a2P2(z-1) + c3P3Z2 u = 0,


where


I
r=l


(ar+pr) = 2.


As before we must remove a factor of z-1, therefore let
f3 = 0 so that a3g3 = 0. Then


z(z-l)u" + (l-a--P2)Z(z-l)+(l-a3)z2]U '+ o1 lZ-a22 u


(z3 zp2)u +[(2-a 2-P23)Z )z 2-+)Z u'+al 1P l- 2 21

Applying the criteria (2-10) we have when j = 2


p2 [S2 -T2zh] 2-m


= 0,


= 0.


3 2
= Z -Z


where h-m+2 = 3 and 2-m = 2. Thence, m = O, h = 1, S2 = T2
= -1, S2T2 / 0. Further,

P = [S-Tiz z = (2-a2-P2-a3)z2+(a2+f2-1)z







and T = a 2+2 + a3 -2, S1 = 2a +2 1. Finally,

Po = S Toz = a 11z a22

with T = -al~l, So = -ar2.

The equation with a solution having a two term re-
currence relation is, therefore,

(2-13) z2 (z-)u" + (2-a 2-B2-a3)z2 + (a 2+0-l)z] u'


+ [ QPz a22] u = ,

where the only restrictions are that in equation (2-7) ar'

Pr (r=1,2,3,4) contain no function of a1 or a4, and P3 = 0.
Since equation (2-12) can be obtained from (2-13) by
letting a181 = O, let us obtain solutions of the latter.

aC 1z-a 2 u = Cz +...+C z +n+C z +n+... C / O,

(2-a2-p2-a3)z2 u' = Coz-1 l+...+Cn_l(A+n-l)zh+n-2
(lzn+.+n- 1 .
+ (a2+B2-1)z + C .(+n)zhn-

z -z u = CoNh(-l)z -2+...

+ Cn-1(+n-l)(h+n-2)z +n-3

+ C (A+n)(A+n-l)zh+n-2
U .
The indicial equation at z = 0 is

z: -A(-1) + (ag+pg-l)N a 0 = 0,

2 (a2+p 2) + a2 2 = 0.






The exponents are
h = C2,P2

z +n: Cn-1(A+n-l)(-+n-2)-Cn (-+n) (h+n-1)+C_n- (X+n-1)(2-a2

-p2-3 ) + Cn (+n)(a2+P2-1) + Cn-la p Cna22 = 0.


=-C n[(h+n-1)(?+n-2)+(\+n-1)(2-a 2- -a ) + alp
C = -- h------------------
n -(2+n)(p+n-1)+(7+n)(2 +( -1l)-a 22


Cn-_l[nl; + (X+n-1)(X+n-a2-p2-a3)]
n (A+n)2 (a2+P2)(?+n) + a2zB


SCn-_1 [a + (?+n-1)(A+n-c2-P2-a3)]
n (X+n-a2)(?+n-~,3)

Let a = a :


Cn=-l1 a + (a2+n-1)(n-P2-a3)]
C = .
n n(a 2-p+n)


Co[ a + a2(1-P2-a3)]
C = +
1 (a2-p2+1)


C [a~l1 + (a2+1)(2-32-a3)]
C 2
2 ((a2-Bg+2)








Co[Cl + a2(l-P2-a3)] i1 [a + (a2+l)(2-P2- 3)]
2 2!(a2-2+1)(a2- 2+2)

Therefore the solutions of (2-13) are:

u = 2 [1 + z1(-2(-2-3)
1 o l!(a2-P2+1)

[a p1+a2 (-132-3 ) ] [a2Pl+(a2+l) (2-p2-a3) ] 2
+ Z +...
2!(a2-P2+1)(C2-P2+2)

+ [all+a2(l-2-a3) ..[a1+(a2+n-1)(n-p2-a3)] n
+ Z +...
nl.(a2-_p2+l) ... (a2-32+n)

(2-14) u1 = Coz2 1

+ [.a 1+a 2(l-p-a 3)] .. .[a1p+(a2+n-1)(n-p2-a 3)] n
n=1 n(a 2-P2+1) .. .(a2-P+n)
and similarly

(2-15) u2 = cz 2 1

+ [ 1p1+2(l-a2-a3 )] ..[al+(P2+n-l)(n-a2-a3)] n
n=1 n!(2p-a+1) ...(p2-a2+n)

If a l = 0, then the solutions of (2-13) will be
hypergeometric:
(2-16) u, = Co z 2F1(a2,-p2-at3; 2-32; Z)








and


(2-17)


U2 = C z 2F1(P2,-C2-2-a3; 2-a2; z)


The hypergeometric form is


(2-18) PF(a,b; c; z) = 1 + z a + a(ab(b+l z2
2 1''' !c 2!c c+l)


+ a(a+l)...(a+n-l)b(b+l)...(b+n-1) ...
n! c(c+l)...(c+n-l)

00
Sa(a+l)...(a+n-l)b(b+l)...(b+n-1) n
n=o n! c(c+l)...(c+n-l)

2.4 Solutions of Confluent Cases after Normalizing.

Another approach to obtaining solutions of equation (2-7),

which we will now use, is to remove the u' term and find

solutions of this transformed equation.

To normalize the general equation

u" + p(z) u' + q(z) u = 0,

let u = vw. Then we have


u = vw.

u' = v'w

u" = v"w


+ vw'.

+ 2v'w' + vw".


Hence,


vw" + (2v' + vp) w' + (v" + v'p + vq) w = 0 .


Let


2v' + vp = 0








and solve for v.


2 d -vp,

dv 1
---p dz.
v 2


Integrating,


In v = - p dz.


v = e


2- p dz


Therefore, to normalize the equation

u" + p(z) u' + q(z) u = 0,


let


v = e


(2-19)

in

(2-20)


- p dz


vw" + (v" + v'p + vq) w = 0.


Now to normalize equation (2-7) we use (2-19) to find


V = Z


Then

S(a -3) (a3+3-1)
v' =. (2+p2-1)z 2 (z-1)


1 1
1 2+ (a(+ +-1) (a3+B-3)
+ (a3+B3-1)z (z-1) 2


1 1
(a +p -1) (a3+p3-1)
(z-1) 2








(-1) 3 (a+2-1)(z-l)


+ (c+1--1)z]


( -1)
f (03 +133-1


S1 -a -(a 2+a-5)z
v = (a2+2-i) (a2+B2-3)z


+ 7(a2+p2- ) (3 3 -1 )


+ -(a3 +3 -1) (a3+P -3)z


2(a2+02-3)


!(a2+ 2-1)


(z-1)


(z-i)


(z-i)


(a +B,-5) (az+ -5)
(Z-1) + [3 (a +3 -1)(a2+-3)(z-1)2


+ (a +p2-1)(ca3+f-1)z(z-1) + (a 3+13-)(a3+z3-3)Z2

Substituting v into equation (2-20) with p and q from (2-7),


1
+ Tz


a2+0-5) -!(a(+3 -5)
(Z-1)(2 (3+3 [ +(a -1)(a+ -3)(Z_1)
12 +f3 -1 2 -3


+ (a + 2-1)(a3+43-1)z(z-l) + -(a3+33-1)(a3+t3-3)z2

1 (a 2+2-3) (a3+3-3)
+ z (z-1) ( z2+(a 2-1)(z-1)


1
V =- Z
2


V, 1
V = ~Z


2(a2+p2-3)


31( C +P 3 -)

1 3- 3
i(a3 +f3-5)


1 1
(:y 2+32-1)) ((c3+f3-1)
(z-1)







+(C +P-) / 1-c22 1-+ -



(+ z +2- Z-1) (3 -)3- 3 12-2
+z (+-1zJ

1 1
(z-l)2 (z-)
z(z-1) z (z-1)

+ )2 w = 0.


S(a2 +02-5) (a3+P3-5)
Dividing by z (z-1) and simplifying,

z(z-l)2 w" + (a2+2-1)(a1+-3)(z-l)2


1 1( + 2
2+ (c2+2-1)(a3+p3-1)z(z-1) + T(Ca3+33-l)(a3+33-3)Z2

1 1-a 33 P -
+ -z(z-1) (a+P,-1l)(z-1)+(a3- l 1)z +2 3 3
z z-1 -

2 21 3 +23 3
+ ( 2 )2 C2 + w 3 0.
Z(z-l) 22(Z-1) (Z-)2

z (z-l) w" + ~-(a,+B-l)(a2,+1-3)(z-1)2


+ (a 2+(2-)(a3+3-1)z(z-1) + 4(a3+0-1)(a3 .3)22


(a2+P2-1)(z-1) + (a3+f33-1)z + a l1z(z-1)-a P3(z-1)


+ a 3z 2 w = 0.







z (z-)2 w" + r


1(a -1)(a--) +2- 33
T(a2++p2-1)(a2+p2-3) + (a 2+p2-1)(a3+P3-1)


+ (a3+P3-1)(a3+33-5) -(a2+p2-l) (a2+2-1)(C3--i)


- (a 3+3-1)2 + a1B


+ a3 12Z2 + [- (a2-2-1)(a2+-2-3)


- -(a2+2-1)(a3+,3-1) + (a2+,2-1)2 + (a +2-l)(a3+P3-)


- a1 al2BZ + [-2+p2-1)(a2+p2-) +2-12


+ aO22]


Sw = 0.


Therefore, the normalized form of equation (2-7) is


(2-21) z2 (-1)2 + 1(az+B)2 T(3-33 )

2(a2+ 2)(a3+33) +) +(a3-3) +
i(a2+B 2)( 0 3- 3) + 2(a2+ B2) +p2()3 -F3) + 4l-1


+a3 ]z2 + 1( a+22)


+ -2(a2+2)(a3+P3)


- 2(a. + ) -( 3) a aL 2Z


+ -(a 2+2 + + a$ 2]


w = 0.


To obtain the solutions of (2-21) let

w = a + + +n-2 + a 1z+n-l + a Z+n+
w = az + az +...+ a z + a z + az +...
0 1 n-i n


ao / 0,








then


'A-2 +n-4
w" = ao(M-l1)z +...+ a n2(+n-2)(A+n-3)z

+n-3 +n-2...
+ a _(+n-l)(A+n-2)z +n an(an)+n)(A+n-1)z .

The indicial equation is

z: X(A-1) (a+P2)2 + 1 + ca = 0,
TIa+p +a0


(2-22)


2 + 1 (a- 2 = .
N -N + = 0.


The exponents are


1 +1


- 4 (a-1 2)


1 (2a 2)
2


Finding an:


z?+n: an-2(A+n-2)(N+n-3) 2an_-l(+n-l)(?+n-2)


+ an(A+n)(h+n-l) + an2 [- -(a+p)2- (a3+3)2


(a2i+2)(a3 3) + -(a2+2) 3+) + 13


+ a 3 + an-1 ()a2+2) + )(a1 3





1
+- 2- 1( S- )- C I1 a2 + an2 (22=02)2


1 p] = 0.
+ + 22









(2-23) a = a 2-(h+n-2)(+n-3) + ("282 + f(a3+ 2

1 1 1
+ (2 .2 ) ( +,) (x2+.., ) -2(3t 3)-~~


a3c3 + a n 2(\A+n-1)(+n-2) (2+e2)


(ac2+ )(al3+ 3) + (C2+,2) + -(a+3 3)


+ + 2
1+ cl1 + 2'2]} (h+n)(h+n-1) (a2-02) + T


In order to simplify further work we can, without loss of
generality, let

(a282) + -(1 (+3)2 + .-(a2+2)(a3+) (a+B2)


2()3+3) a P2 a33 = k

and


- (a2+2) (a2-( )(a+3) + (a2z) + l(a4+33) + 1ap1

+ a2P2 = m.

Making these substitutions in (2-23) and simplifying the
denominator by use of the indicial equation (2-22), we have

an- [-(h+n-2)(7,+n-3)+k] + an- [2+-n-l) (+n-2) + m
a =-
n n(2?+n-l)








a o 2h(-1) + m]

1(2X)


ao(2?2 -2?+m)
1!(2A)


a[-A(\-1) + k] + a [2(A+1)? + mI


2(2,+1)


2?(-_2 ++k) + (22 -2N+m)( 2? +27\+m)
2! 2?(2?+1)

4x4-2A3 + (4m-2)A2 + 2k? + m 2
21 (2?)(2?+1)


a -( \+l)A + k + a2[2(?+2)( \+1) + m]


a3 = a ( 2(-_2--+k)(2X2-2N+m)(2)+l)+(2X2+6+4 +m)[ 4X4-23


+ (4m-2)?2 + 2kA + m2] ,
31 (2\) (2A+1)(2\+2)

a3 ao[8?6 + 12\5 + 4(3m-l)?4 + 6(3m+2k-2)?3



+ 2(3m2+4m-2+4k) 2 + 2(3m2-m+3mk+2k)X


+ m(m +4m+2k)]


a1 =



S=
a


a2 = ao[


a2 = a


3! (2-)(2?+1)(27,+2)








The solutions of equation (2-21) are


(2-24) w = az + (22-2+)
1 1 (2?X)

44 -2X3 +(4m-2)N2 + 2kN + m 2 2
+ Z z +...
2!(2) (2A+l)

I+(a2-82)
where A = +( 2 and k and m have the values:
2

)2
k = -(a +2+2 3+(a2+22+a32.3) a l 33'


m = (a(+2-1)(a2+ 2 3) + i3 + a 2 2

2.5 Factored Solution Wk,m and Hypergeometric
Solution W ,k. Next, we shall consider solutions of the
equation

00 0 1 0O
(2-25) u = I a 2k a4 z

m 1 -a 0 P4


when the u' term has been removed. This can be obtained
from equation (2-21) by letting alB = m, a2 = a, P2 = 1-a,
a3 = 2k, P3=0. Although P3=0 in the normalized equation,
this does not insure a two term recurrence relation as it
would in the original equation (2-7). Our transformed
equation, that is (2-25) in normalized form, is





68


(2-26) z2 (-l)2w" + (m+ k)z2+(a-a-m)z+(l-a) w = 0.

Assume the solution

w = Coz + + Cn-2z+n-2+ Cn- +n-1+ Cnz+n+... C 0,
= 'n+...+C z C + CZ +... C / O,


then


w" = Co(A?-l)z -2+...+ Cn_(+n-2)( +n3)z+n-4


+ Cn _l(+n-l)(A+n-2)zh4n-3+ Cn(N+n)(A+n-l)z+n-2 +...


The indicial equation and exponents are

z: ,(N-1) + a a2 =0,

\2 7 + a(l-a) = 0,

(A-a)(A+a--l) = 0,

X = a, 1-a.

Obtaining C :

z7+n: Cn-2 +n-2)( +n-3)-2Cn-l+n-l)( +n-2)


+ Cn(A+n)(A+n-l) + Cn(m + k2)

+ Cn_(a2 a ) + C (a a2) = 0.
-a-m0+





69


Cn = -C (+-n-2)(h+n-3)+(m+ 1- -k2) +C- 2 (h+n-1)(A+n-2)


(ac a m) 1 2 *------
S(A+n)(?+n-1) + a a

Let ? = a:

S= Cn -(a+n-2)(a+n-3)-m- + k2] + Cn_1a2 + (4n-5)a


+2(n-1)(n-2) + m n- -- 1
S n(2a+n-l)

C a -a2_ +m
C1 1
C 1(2a)


C -a(a-l)-m- + k2] + c 2a+
C2 01 ----------------------
2 2(2a+l)

[2a(-a2+a-m- +k2)+(a2-a+m)(a2+3a+m)
2= o 2!(2a)(2a+l) '


aa4+(2m-l)a2 + (2k2- ')a + m
2 2!(2a)(2a+l)


C -a(a+l)-m- +k' + C[a2+7a+4+m
C =3(2+2)
I5(2a+2)








C = CO 2(a2-a+m)(-a -a-m- 1 +k 2)(2a+l) + a 4+(2m-l)a2


+ (2k2 )a+m2](a 2 +7a+4+m) 1 --
+ (2 1) 3!(2a)(2a+l)(2a+2)



C = Co a6+a +(35m+l4+m+l +(6+6k2- 9)a+(12k+3+3m25)a2


2 2 23 3 3 2 2 1
+(3m2+6k2+6mk2- m- )a+(m 2m2+2k- 1 )

1
3!2a(2a+l)(2a+2)

Thus one solution of (2-26) can be written with a few
terms as:


(2-27 -a+ a +(2m-l)a 2+(2k- Z)a+m2 2
(2-27) w = C z 1+ Z+ z +
L 1! 2a 2. 2a(2a+l) J


A second solution of (2-26) can be obtained by replacing
a by 1-a:


(2-28) w = Coz1-a+ 2-a+
S1(2-2a)

a 4a 3+(5+2m)a 2+(-2k2-4m- 3)a+(m +2m- +2k2) 2
+ +...
21 (2-2a)(3-2a)

Although solution (2-27) has no obvious general term,
it can be shown by a lengthy process of long division (which
is omitted here), that it can be written in the factored form:







2+ (a 2+2ka+m)[ +(2k+2)a+2+2k+m+1
w=C z (l-z) 1+ 2k-z+ z
W=0 (lz 11 2a 2! 2a(2a+l)


(a2+2ka+m )[a2+(2k+2)a+2k+m+l] [a +(2k+4)a+4k+m+4 3
+ z +...
35 2a(2a+l)(2a+2)

+ ((a 2+2ka+m) a +(2k+2)a+2k+m+l] ** a +2(k+n-l)a+m

n
+ (n-l)(2k+n-l1) z+...
n' 2a(2a+l)...(2a+n-1)

Let this solution be indicated W k,m, since a,k,m are the
only parameters; therefore (2-27) can be written


(2-29) Wa,k,m = C0a(1-z)k+ 1 + (a2+2ka+m)
n=

[a2+2(k+l)a+2k+l+m] [a2+2(k+n-l)a+(n-l)(2k+n-l)+m] n

n! 2a(2a+l)***(2a+n-1)

If m = 0, then the solution (2-29) becomes hypergeo-
metric. Let this solution be indicated W ,, a and k being
the only parameters:

Wa,k = Coza(1-z)k+ 1

S(a 2+2ka) a2+2(k+l)a+2k+l [a2+2(k+n-1)a+(n-1)(2k+n-l)

n= nl 2a(2a+l) *.(2a+n-1)







Factoring further we may write

k+ --
Wk Co ZU(-z) I

Sa(a+2k) (a+l)(a+2k+l) ]*[ (a+n-1)(a+2k+n-l) ]

n=1 n! 2a(2a+l)...(2a+n-1) *

Rearranging factors we arrive at the solution

k+
W,k = Coza(l-z) 2


+ a(a+l) '*(a+n-l(+2)( )(a+2k+l) *.(ac+2k+n-l) zn
n= n! 2a(2a+1) **(2a+n-l)

which is hypergeometric in form (2-18) and shall be denoted
by Wa,k Therefore, when m = 0 in (2-26) we have the
solution
1
k+-
(2-30) W,k = (l-z) 2Fl(a,a+2k; 2a; z),

where we have let Co = 1. A second solution can be obtained
by replacing a by 1-a.

2.6 Notation and Proofs Involving 2Fl(a, b; c; z).

Before presenting several interesting properties of Wa,k, we
shall introduce notation and prove seven statements that
will be useful in section 2.7.
Notation:
n,j are integers.

(a)n = a(a+l)(a+2)...(a+n-1).








n(n-1)...(n-J+l)


= 1.


DnW = dnW
Dn,
dzn
W = W a (z).
Wa,k a,kZ).

If F = 2F(a, b; c; z) = F (a, b; c; z), then

F(a+) = F(a+l, b; c; z), [5,p.50] ,
F(b+) = F(a, b+1; c; z),
F(c+) = F(a, b; c+l; z),
F(a-) = F(a-l b; c; z), etc.,


F(+n) = F(a+n, b+n; c+n; z),
F(-1) = F(a-l,,b-1; c-1; z).

We shall prove the following:


n = 1,2,3,...


F = 1 [aF(a+) + (1-c)F(c-)]
a-c+1 L1P

F = 1 F(a-) + (b-c)z F(c+)l
1-z L c J


F(+l) = -- F(b+) F
az I+ Ja


DnF = (a)n(b)n
(c)n


F(+n);


and for F = 2F1(a, a+2k; 2a; z) = F(a, a+2k;


2a; z),


(2-35) F(+n)


(2a)
n
= z
(n,)


-(l- z)
I (1-Z)


n+l n
-k-
2
J=o


(2-31)


(2-32)


(2-33)


(2-34)


(-1) (1-z) W ,k+ n-
J) r~k









(2-36) Dn z(1-z) ]


k-n+
= z"(l-z) 2


(-1) (-a (-k-x) ,(z (-z
j =o


k(2-37) F(-l) zykzl1
(2-37) F(-1) = (1-z)z 2(2a-1)(1-z)-Wak+(-2k)W
(2a-1)(2-z)L a a-j

Finally, we state Leibniz's rule [9, p. 409] for finding
the nth derivative but shall not prove it:


(2-38) Dn(uv) = u D"v + (n)Du D n- +
= V 1


( D2u


Dn- +...


+ >tDu D"n- v +...+ (D u)v

n
(n D= u Dn"-.
t=o


PROOFS

(2-31)


F = 1 [aF(a+) + (l-c)F(c-)]
a-c+ L J


Proof:

aF(a+) + (l-c)F(c-)


(a+l) (b) +
(c) n!
n


co
(1-c) Z
n=o


(a) (b)
(c-1)n!
(c-l) n!


= [a(a+l) (b)n
=o (c) n


(a) (b) z
(1-c) -L I
(c-l)nn!


n=0








c,

S-- [a+n-(c+n-1)
n=o


(a)(b)n
(c) n!


S(a c + 1)F.

Therefore,

F -1 a?(a+) + (1-c)F(c-)]
a-c+1 L


1
F = 1
1-z


I c


(b-c)z F(c+)
C


(a-1) (b)
(C n n
(c),nl


(b-c)z
c


n=o


(a) (b) n

(c+1) nn


(a-1)(b). n + (b-c)

(c) n' c


1 (a-1) (b) (b- )
n=1 (c)nn! c


=1 + [ (a-1) + n(b-c)
S1 L a+n-1 (a+n-l)(b+r


S1 + (a-l)(b+n-l)+n(b-c)
n=l (a+n-1)(b+n-1)


(a)n-1(b)n-1 n

n=l(c+l)n-l(n-l)!

(a)n(b)n z

I-1)J (c) nn


(a) (b)n
(c) n!


(2-32)


Proof:

F(a-) +


00

=L
n=o


00
= 1 +L
n=l


(a)n(b)n n+1
^ ~n ^+Z
S(c+l) n! z
n zo n










=1 [1- n(c+n-1) (a)n(b)n n
n=L (a+n-l)(b+n-l) (c) nn


0o
FZ [ n(c+n-1)
n=l (a+n.1)(b+n-1)


(a) (b)
(cn n
(c) n!


0o
F z
n=l

Cor
F y
n=o


(a) n-1(b) n-1

(c)n-l(n-l)!


(a)n(b)n n+1

(c) n!


= F zF

= (1-z)F.

Thus,


F(a-) + (b-c)z F(c+) = (l-z)F
c


F =1 F(a-) + (b-c)z F(c+
1-z c I


F(+l) = [ F(b+) F]
az L J


F(b+) F =


n=o


(a) (b+1)n
(c) n!
n n=o


(2-33)


Proof:


(a)n(b)n

(c) n!
n









Sb+n
7^


(a)n(b+ln-1 n
(c)n(n-l))!


00
=1
n=l
,00
7-1
*L
n=l

az
c
n


c
n


Hence,


F(b+) F = az F(+1)
c


F(+1) =


-c [F(b+) F]


(a) (b)
Dn = n F(+n).
(c)n


D [F(a, b; c; z)]


=D [ (a)(b)n
n=o (c) n!
n=o ^n


n(a) (b)n n-
(c) n!

(n+)(a) n+(b)n+

(c)n+l(n+l)!


(a) (b+1)1 n
Wn n-1 n
(c)n(n-l)!


S(a+)n-1 (b n-1 n-i
S(c+l) (n-1)! z
=1 n-1

1 (a+l(b+)n n
L (c+1) n:
=0 n


(2-34)


Proof:


zn]


00
oz

n=l



n=o










cn= (c+l)nn!


ab F(+1).
c


Similarly,


D2 [F(a, b; c; z)]


00co
Sa(a+l)b(b+l)
c(c+l) n=l
n=1


(a+2)n-1(b+2)n-1 n-i
Z
(c+2) (n-l)


(a) (b) w

(c) nL
2 n=o


(a+2) (b+2) n
(c+2) n!
n


(a)2(b)2
F(+2).
(c)2


In general,


Dn [F(a, b; c; z)] (a(b)n F(+n),
(c)n


n=1,2,3,...


If F = F(a, a+2k; 2a; z) = z-(l-z)--Wa,k

then

(2-35) F(+n)


(2a) n+1 n
2n z-a-n(l-z)_k-- n-y
Sc (1-z) 2 (-i
()n j=o


j \


J2
1-z) Wk+ n-j
2


Note that when k is replaced by k + m in the premise, then

F(a, a+2k+2m; 2a; z) = z-a(1-z)-k-m-2W
ca,k+m








Proof:
We begin by applying (2-53) to F(+l) and express the
result as a function of W a,k+:

F(+l) = 2a [F(a, a+2k+l; 2a; z) PF
az L I

2a -a( -k-1 -k-1
-- z (1--z) W (1-z)' 2W
az a,k+ az(lz ,k .


(a) ,k+ ,k

Apply the principle of (2-33) to F(+2):

(2-40) F(+2) =2+1) [F(a+l, a+2k+2; 2a+l; z) F(+l)
(a+l)z J

Using the expression (2-39) with k replaced by k + in
(2-40), we have
(2-41) F(+2)
S(2a+l) a 1-a-1l -k-zW kW
(a+l)z a ak+ ak+

2a -a-1 1 w 1
a (1-lz)-- Wa ,k+ -(l-z)2W ,1
a L ak+ akJ/


)2z (1-2 ) a,+-2(1-z)W ak


+ (1-z)Wk]

Let us find F(+3) before generalizing our results to


F(+n). Upon applying (2-33) we have








(2-42) F(+3) = (2a+22) FF(a+2, a+2k+3; 2a+2; z) F(+2)
(a+2)z IJ .

Replace k by k + in (2-41) to write (2-42) in the form


(2+2) (2a)2 -a-2 k-2
F(+3) = z (-z)-k-2 W k+3 -2(-z)
(a+2)z (a)2 2-k+l

(2) 2 a-2(l k-
S(1-z)Wa,k]- (a)--2 ( )- a,k+

1
2(-z) W ,k2 + (l-z) Wak

Simplifying further

(2a)3 -a-3 k12
F(+3) = z (1-z) W,k+ -(1-z)2W,
(a)3 a,k+1

+ 3(1-z)W ,k 1 -(l-z)Wk]

Finally,


2cx) n -a-n -k- n
F(+n) z -a (k (1-) 1) k+ n-2
(a)nk+ ,

2) a,k+ j ak+
n
+...+ (1-z)W ,k .

Therefore,

(2a) an(l-z) n+l1
F(+n) n za-nz (-1) (j\1-z) W k+n-J
(a)nJ=o >a,k+


n=1,2, 11, .








(2-36) Dn [za(l-z)k+

n
= a(1-z)k-n (-1) () -a(-k-)n z(l-z) .
J=0

Using Leibniz's rule where u = za and v = (1-z)k+
find DIu and Dn-v. When u = za

Du = aza-1; D2u = a(a-l)za-2; ...; DDu = (-l)t(-a)za-"

If v = (1-z)k+, then
3
Dv = -(k+)(1-z)k-; D2v = (k+)(k-)(1-z) k- ..

D"'v = (-k-1) (1-z)k+-n+ .

Hence, the nth derivative is found to be


Dn[za(l-z)k+] = ( (-)(-a) --k ) (l-zk+L-n+


n
= za(1z)k-n+ (- (-)(-k-) z -z)
4t=o
If F(-l) = F(a-l, a+2k-l; 2a-l; z) show that

(2-37)
( -kz 1-a
F(-l) = (-z) 2(2a-l)(l-z)2Wa- +(a-2k)Wa k-
(2a- ( (2-z) L

Apply (2-31) to F(-l):


F(a-l, a+2k-1; 2a-l; z)







-1 [(--l)P(a, a+2k-l; 2a-l; z)
-a+1 1

+ (2-2a)F(a-l, a+2k-l; 2a-2; z)]


= -F(a, a+2k-l; 2a-l; z) + 2F(j-1, a+2k-l; 2a-2; z).

Now apply (2-32) to the first term with the result that


F(-l) -1 [(-1) + (2k-a)z F(a, a+2k-l; 2a; z)]
1-z 2a-1

+ 2F(a-1, a+2k-l; 2a-2; z).

Collecting the terms containing F(-l) we have


(2-38) 1 + 1 F(-l) a-2k z F(a, a+2k-l; 2a; z)
1-z 2a-1 1-z

+ 2F(a-l, a+2k-l; 2a-2; z).

Now, since

(2-39) F(a, a+2k; 2a; z) = z-a(l-z)- k2W
l,k
if k is replaced by Ic -, then

F(a, a+2k-l; 2a; z) = z-a(l-z)-kW

If a is replaced by a-1 in (2-39), then
F(a-l, a+2k-l; 2a-2; z) = z-a+1 (-z)-k- -1k

Using these two results and solving equation (2-38)
for F(-l), permits the result








(1-z) a-2k _z z-a(lz)-kW !
(2-z) L2a-1 1-z aK2-

+ 2z (l-z)k 2 k]


Sz- ( j- (a- W 1 + 2(l-z)\
(2-z) -2a-l ak-2 a-1k

or
F(-) (-z) z (a-2k)W' k-1 + 2(2a-l)(l-z)2W-
(2a-1)(2-z) ) a 2 a-1,k .

2.7 Recurrence Relations, nth Derivative, Sum and

Product Formulas for W a By utilizing the statements we

have just verified, it is possible to obtain recurrence

relations for W Several such relations will now be
o,k '
presented.

By replacing a by a-1 in equation (2-30), we may
write

(2-40) W = z-1(l-z)k+ F(a-1, (+2k-1; 2a-2; z).

Since the parameters a and b in F(a, b; c; z) can be inter-

changed without any lose in generality, we can also inter-

change a and b in equation (2-32) to obtain the valid

relationship

F -1 F(b-)+ (a-c)z F(c+)
l-z ce t

Now, if this be applied to (2-40), we may conclude that








(2-41) Wa-1,k = zal(1-z) 2 F(a-l, a+2k-2; 2a-2; z)


+ (1-a)z F(-l)]
2a-2 *

Replacing a by a 1 and k by k - in (2-30), it is seen
that
F(a-1, a+2k-2; 2a-2; z) = z-a(l-z)-k a-,k-"

This, together with (2-37) enables us to write (2-41) in
the form
a-1 k- 1-a -k
Wa = z (1-z) z (-z)- Wa,k


1 (z)kz2-a 2(2a-1)(l1z)2W
2(2a-1) (2-z) a-l,k

+ (a-2k)Wak-


= (1-z) W Z1 W
=a-1,k (2-z) a-1,k


(a-2k) z(l-z)- W ak
2(2a-1)(2-z)
Then
1 + -z W= (I-z1_ Wa-,k- + (a-2k)z w,, i
2-z 2 a-lk2(2a-1)(z-2) ak
Solving for W -1,k, we have the recurrence relation


Wak (2-z) -z- (a-2k)z
a-l,k 2 L2(2a-l)(z-2)








(2-42) (-z 2(2a-1)(z-2) Wa-
(2-2) Wa-,k 4(1-2a) -1,k-2


+ (a-2k)z W kl]

Another recurrence relation is found by first
replacing a by a + 1 in (2-30), giving us

W a+,k = Z+l(1l-z)k+2 F(a+l, a+2k+l; 2a+2; z).

Then apply (2-33) to get
a+1 k+ (2a+)
Wa+ik z 1 (-z) [P(a, a+2k+l; 2a+l; z)


F(a, a+2k; 2a+l; z)]

When (2-31) is applied to each term, we may write

Wa+ (2a+l) (Za(l-z) 1 [aF(a+l, a+2k+l; 2a+l; z)
a+l, a -a


2aF(a, a+2k+l; 2a; z)]


+1 [aF(a+l, a+2k; 2a+l; z) -2aF(a, a+2k; 2a; z)].
aL JJ


(2-43) Wa+ k = (2a+l) za(-z)k+ -F(a+l, a+2k+l; 2a+l; z)


+ 2F(a, a+2k+l; 2a; z)

+ F(a+l, a+2k; 2a+l; z) 2F(a, a+2k; 2a; z)].


From (2-35), letting n = 1, we have








F(+1) = 2z --(l-z)-- [W ,k+ -(i-z)YW ]
k ,k ,

which can be used to rewrite the first and third terms of

(2-43). Hence,

(2a+l) z,(lz)k+-f 2z-a-1 z)-k-1 W
S z -z-2-z) k-
W+,k a k+





1 2
(1-2) Wz- + + 2-a(1-)--1 W ,+


+ 2z'- (1-z) l[wa, (1-z)l W a, j


2z-'(l-z) 2 W
f- WI, -

2(2a+l) [- -1 -z)-- W1 +
a a,2+y



1Z-1








( lz) Wk .]
Thus, we have a second recurrence relation

(2-44 ) axzW C,1j = 2(2a+1) -(1-z)2 W Cx, kj + (2-z) W a~k




To obtain the nth derivative of our solution (2-30)

Wa,k = za(l-z)k+2 F(a, a+2k; 2a; z),







we shall apply Leibniz's rule, where u = z(l-z)k+2 and


v = P(a, a+2k; 2a; z).


Thus, using (2-34) thru (2-36)


we may write


n
D W ak t= D [z(1-z)k] D"-o [F(a, a+2k; 2a; z)]
t,=o
n 4
SZ [()"(l-,z)k-t+ (-, -) z- (1-z)
t-=o J=o


(a) (a+2k)
(2n- ) n-t
n-t


= a(lz-z)k-
t=o


j=o


f+_ _n-lCA 4 3
S*(a+2k) z--2 ( i ( -)2 / Wa,k+ n- .-4
J=o
Therefore,


(2-45) DnW ,k = z-n(-z)-


i(l-z)- zt(a+2k)n-
'c=0


* (-) -a -k-J-j (1-
J-=o
n--4

S (-l)j n- (1z) Wk+ n--3i
o a0k


F(+n-LZ)


(-k-.L j(-)
2 t,_j





88


Now that we have the nth derivative of V (z) we
a, k
can use Taylor's Theorem [8, p. 93] to write formulas for

Wa,k (x+y) and hW (xy). Taylor's expansion can be ex-

pressed in the form
2 n
f(y+a) = f(a) + f'(a)y + f"(a) 1 +...+ f(n) Y +

Now if a be replaced by x, we have
2 n
f(y+x) = f(x) + f'(x)y + f"(x) 2 +...+ f(n)(x) y +..
2! n!

oo
(2-46) f(y+x) = f () (x) n .
n=o

Using (2-46), replace y by (y-l)x to obtain

00
(2-47) f(xy) = f(n)(x) (Y- )nxn.n
n=o

Since Wa,k(z) is a convergent, analytic function for

0
have an addition theorem and a multiplication theorem:


(2-48) Wk (x+y) = W( (x)



n n
= ji x (-x(1-x (a+2k)
n=o t=o



J=o









n-, J n
S (-1 J) n ) ( ) a,k+ n--j
j=-o

and



(2-49) Wa,k(xy) = (x) ( nn
n=O



Sn(y )n (1-x) n x-(a+2k)_
n= n. t=o








J=o
1

J=0


where lyl < 1.













CHAPTER III


Related Differential Equations and Classification

3.1 Related Equations of Mathematical Physics Derived

from an Equation having Four Singularities. The most gener-

al linear differential equation of the second order which

has every point except a ,a2,a3 and o as an ordinary point

with exponents ar'Pr at ar (r=1,2,3) and exponents 41,"2 at

0, is








3 3
2 3 r-ar-B3 du 3rr Az+B



(3-2) 2 + (ar+)-2 + + A = 0.
dz 2 z-a (z-ar) 3(z-a

where A is such that tj and L2 are the roots of




r=1 r=1

To verify this, we can begin with equation (2-1) and let

a4 become infinite, thus obtaining

3 3
u [ 1-ar-r + H(z) u' + r + ---- + J(z) u=0,
+-a + H I z-a ) z-a
r=1 r r=1 +(Z-ar Z-ar

where H(z) and J(z) are polynomials in z. We must now depart

from the previous work of chapter II, for we desire z = o to

be a regular singular point in this case. Referring to

section 1.5, we see that z = ( (z = ) will be a regular
90








singular point when 2 z ) and 21 <1 are analytic

at z1 = O. For the first one,


1l 11-1- a2 3I 3 1
22 p + + + -L1
z1 \Z1 Z1 1 a -- a a
z1 1 z1 2 z1 3

1-a lp1 1-a 2-P2 1-a3 -i 1
l-[l-a11 -a2z l -a3z 11

which is analytic at zi = 0 if H = In the second,


1 c 1l a1P2 a 3 3 1



+ -+ + +
2 -a1 z 11

1 1
a -a a P3
m m
+ + ++
z-- -a z- -a3
1 1

a1 22 3__ _3_

l-2alz +a 1z l-2a 2z+az1 1-2a3z +a1

1 i 2 3 1 -
z 1-a z1 1-a2z 1 -a z z1 z ,

which is analytic at z = O, only if mi + m2 + m3 = 0 and

<1 ) = 0.

Now

m m m,
1 + 2 3+ 1:
+ +[(m1-
z-al z-a z-a

= 2 1
S1 ]2 3 (z-a )(z-a )(z-a 3








where A and B are functions of mr and ar (r = 1,2,3). But

since ml m +m3 = 0, we have equation (3-1). In order to

show that A must satisfy the condition (3-2), we can make

the substitution z =- in (3-1), then the indicial equation
zi
for z = 0 will be this condition. An alternative method

of finding the Indicial equation at z = w (or zi = 0), and

the procedure we shall use here, would be to assume the

solution

u = b0oz + blz--1 +.

and equate the coefficient of the largest power of z (i.e.

z 2) to zero to obtain the indicial equation.

Noting the expansion of the following about z = o(or

- = 0), we have
Z
z
2
a a

1 ( 1 = 1 1a/ 2ar
-+ = 2+ +- + --+...







where r = 1,2,5> and
zar = a r Z +
1- -



(z-a,) z-a,)( z 1 ,
z








1 a a a
(z-a )(z-a2)(z-a3) z -- 1 1
Sz z


1 +a 2+a3
= -T- 1+ + ....
z z"








Then


3 -


/ 2aar Nz-+b
1+ +..) u = bz +b z
j Z2 Z o
1

SAz+BA + al+a2+a3
z \I z +'


3 2
1-a -Br a a
+ -- + r +...)
r= z z z
r=l

1


-A-1 -b-2
u' =-b Az ? +b(--l)z +...



u" = bo (h+l)z-N-2

+ b, (+l)(,+2)z--3..
1. .


3 3
z -: (h41) h (l-ar-r) + arr + A = 0.
r=l r=l

3 3
N2 + 1[i- (1-ar-Pr)] X + Zarfr+ A = 0.
r=l r=1

3 3
S2 + (ar+Pr)-2] N + Carr + A = O,
r=1 r=1

where the exponents at z = c are = 1 and 2

We now proceed to show how many of the related differ-

ential equations of mathematical physics can be obtained

from a differential equation having four singularities, by

assigning values to the exponents ar, Pr, the singularities

ar (r = 1,2,3), and the constant B.

(a) Lame's Equation [8, p. 2051. Beginning with (3-1)

take ar = 1/2 and fr = 0 (r = 1,2,3), 2 = i + n + 2, and


b / 0.


i
r---


- -l+ ...







1
B = - h, where h and n are constants. We can obtain A by

using (3-2) and the fact that the sum of the roots equals

3 3
2 (ar+6r) = and the product of the roots is a arr
r=l r=l

+ A = A, where the roots are 41 and ul + n + 1. Thus,

24 + n + 2 = 1 = n,

and

A = (1 +n--1) = A n(nin ) n(n+l).

Substituting these values into (3-1), the result is

Lame's Equation:

du 2 du n(n+l)z + h
dz2 + z-a dz i3 u -
dz r 4 (z-a )
r=1 1 r
r=l
(b) Legendre's Equation. Let a = a = O, a3 = 1,

a = 3 = = = 0 2 a +1+ i+2+2 = 3/2, B = n(n+l)
3
+ al1 + a2ZB in (3-1). Since the product 1,2 = = ar r
r=l

+ A from (3-2), we have A = -(alP1 + a2p2). Hence,


d2u 1 du 1 al1I + a2 2
++ L)+ + +
2 dz 2
dz z z-1z

-(a lf + a P2 )Z 2 n(n+l) + a 1P + az2 2 u 0.
+ -- -- ^----L 1 2 2 u = 0.
z (z-l)



(3-3) d2 + ] +n(+l u = 0.
dz2 + z z-1 dz 4z2(z-l)
dz kz2(z-1)




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