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
(11) 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 (zc)p(z) and (zc)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
(12) (zc) du + (zc)P(zc) du + Q(zc) u = 0 ,
dz2 dz
where P(zc) and Q(zc) are analytic at z=c. Expanding
these functions in a Taylor's Series about z=c we have
00
P(zc) = c) c) + ... + pn(zc)n + = Zpn(zc)n
n=o
and
Q(zc) = q + q1(zc) + ... + qn(c)n + ... = q(zc)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(zc)
zc (zc)2
in equation (11).
1.2 Formal Solutions of Differential Equations. Let
us assume a formal solution of the differential equation to
be
u = (zc)a 1 + Zan(zc)n]
n=1
where a,al,a2,... are constants to be determined.
Assuming that the termbyterm differentiation and multi
plication of the series are valid, we have
U' = c(ZC)a'
00oo
[i+ an(zc)n] +
n=l
CO
(zc) ann(zc)ni
n=l
= (zc)1 [a
+ c(a+n)an(zc)n]
n=1
and
u" = (a1)(zc)2, +I (a+n)a (zc)n
n=l
+ (zc)' Zn(a+n)a (zc)n
n=i
co
= (zc)a2c (a1) + (a+n)(a+nl)a (zc)
n=l
Now substituting these expressions for u,u' and u"
into equation (12) we obtain
cc
(15) (zc)C a(ca1) + a (a+n)(a+n1)(zc)n
n=l
+ (zc) P(zc)[a + za (a+n)(zc)n
n=l
+ (zc)a Q(zc)[ 1 + an(zc)n1
n=l
= 0.
Next substitute the series for P(zc) and Q(zc) into
equation (13):
5
(zc) a(al) + YIa (u+n) (a+nl) (z c)nl
+ (zc)a[ + p(zc) + p1(zcc) +...+ pn(zc)"
00
+...].[ + an(a+n)(zc)n
n=l
+ (Zc)aC[q + ql(zc) + q2(zc)' ...+ qn(ZC)n
00
+7 an(ZC)n = 0.
n=1
Now equate to zero the coefficients of the successive
powers of (zc):
(zc)": a(a1) + poa + qo = 0,
(zC)Q+l
(14)
a 2 + (po1) 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)2a1 + p (a+l) + qo] + ap1 + q = O,
1) ()( ) + + a + 0.
+ (p 1)(a+l) + qo] + op, + q 0.
(zc)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)' + (pol)(a+2) + q
+ a [Pl(a+l) + q]
And in general we can write:
(zc)+n: a (a+n)(a+nl) + pa (a
+ p a + q a
n on n
nI
+ z q a
m=l
+ pa + q2 = 0.
n1
+n) + ma n(a+nm)
m=1
+ q = 0,
n
an (a+n)2 a n + po(a+n) + q 0
nI
+ anm Pm(a+nm) + qm] + Pna + q = 0,
m=l
(15) a [(c+n)2 + (pl1)(a+n) + q 0
n1
+ Z aMp (a+nm) + 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
(zc) 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(zc) has a simple pole at z=c
and expand it in a Taylor's series.
P(zc) = p_l(zc)l+ pO + P(zc)+...+ Pn(zc)n+.
Pn(zc)" p_ / 0,
n=l
and as before
00
Q(zc) qo + q1(zc)+...+ qn(zc)n +... = qn(zc)n
n=o
Now multiply according to the following scheme [which
is equivalent to substituting the series for P(zc) and
Q(zc) into equation (13)] and obtain the indicial
equation:
q + ql(zc) +...
p 1+po(zc) +...
(zc)2
u = (zc)" [ + al(zc) +...]
S (c) 1 a + (c+l)a,(zc) +...
u" = (zc)2[a(ul)+(u+l)ual(zc) +...]
The indicial equation is
(Zc)a1
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(zc) has a simple pole
and expand it in a Taylor's series:
at z=c
Q(zc) = q_l(zc)~ + qo + ql(zc)+...+qn(zc)n +...
03
= q(zc)n,
n=l
q_, O,
and
P(zc) = po + p (zc) + P2(zc)2 +...+pn(zc)n +..
p pn(zc)n
n=o
Now multiplying similar to the previously mentioned
scheme:
q (zc)'+ q +... u = (zc)[ 1 + a (zc) +...
po(zc)+ pi(zc)2+... u' = (zc)a1 a+(a+l)al(zc)+...
(zc)2 u" = (zc)a2 a(a) +...]
The indicial equation is
(zc) i: q, = 0,
which is a contradiction of our assumption.
If both P(zc) and Q(zc) have simple poles at z=c,
then
q_i(zc) + qo+... u = (zc)al + ai(zc) +...
pl(ZC) + u ,a (z 1[ ]
p_+p.(zc) +... u (zc)a + (a+l)ai(zc)+...]
(zc)2 u" = (zc)a2 a(al)+(a+l)aa (zc)+...
The indicial equation is
(zc)"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),
(16) 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 nonzero 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 A1i2 = s. Now
since the exponents are a = 1 ,2 we can write (io),
(17) F(a) = (a1l)(az),
then for a = 1 +n,
F(A +n) = (A?+n?A)(?l+nh2)
= n(n+s) .
Let
P(zc) = p + p1(zc) + p2(zc)2 +..+ pn(zc)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 M13rn
INPn + q 1 s Ix1lIPnl+lqnl
n n n/ n
SIN "M1rn + Mr = r (I.11 M+M, = r M ,
where
MS = I IM1 + M2.
(z) n
pn I Mrn
^*l)5r
i i * n
Choose M to be larger than M1,M2, and M3 and also M 1.
Then
(18) Ipn < Mrn; nIq ( Mrn; IlP + qn < Mrn
for n = 1,2,3,....
Now with the aid of equation (16), we can write
equation (14) 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)
= (a7)+i(36) and
11+s = V(l+ay) + (p6) .
If P6 = 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 + (p6)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,...,ml.
From equations (15) and (16) we can write
n1
an F(a+n) + a npm(+nlm) + qm] + pa + qn 0,
m=l
and letting a = \,
n1
,a nmPm +nm) + qm] n qn
m=1l
a =
n F( +n)
Now replace n by m and replace m by t. Then
m1
amt [Pt(h1nt) + qt] m q
(19) lam = t=
F(l +m)
mi m1
Slamt p Ip1t +qtl+I lPm+qml+ 7 (mt)lamt JiPt
t=1 t=l
m mn+s
m1 m1
Z lamt IMrt + MrO + (mt)ia i ' Mr
t=1 t=1
\ 1, 
m I1+smr
For the first summation:
m1 m1
(110) a iM = M a tIr
t=1 t=1
SM [IaMi 1rl+jam2 Ir2+...+Ia rm+2+ arm+l1
< M[Mm1 m m2 m
M M r + M r
+...+ M2rm + Mrm]
< Mr [Ml + Mm2 +...+ M2 + M
< Mr (m1)M m1
< (ml)rM .
For the second summation:
m1
(111) (Mt)lamt Mrt
t=1
= Ml(m1)
F, r ni1 m m2 m
< M (m1)M r +(m2)Mm2 r +.
< Mrm[ (m)Mml+(m2)Mml+...+
< Mmrm[(ml)+(m2)+...+ 2 + 1
< M'rm M (m1)
..+ Mr]
Mm1I
(ml)r M + Mr + Mmr m (ml)
aJ <  2 
m2 1+sm1
Mr mm mrm m m m 1
Mr M r M + Mr + Mr (m1)
< m 2_2+SM 
m'+sm
mn'Il+smn
Thus
lamlir 1 +(M2)la 21r 2 +...+(l)lallrnl+ll
min mm T ,nm m m n
mr M r M + Mr + Mr (m1)
< 2
m' l+sm 
2m l+sm
But ll+sml ) >1, for letting s = 1A2 = a+3i (y+6i):
si+ iT m+s m+s
1+ !1 = m I = m'
m m m
and
Im+s = (ay+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 < Mnrn 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 Ilsl Ii sl1 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' '
H1ls 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 ~iii j 2e
j J'
= V l +a N
),
S1 Qa+f3 2 1 JaV+02 = 1 IsI = 1IsI
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
1lIs I
11 s1
We desire I s to have a minimum value larger than zero,
where j=l,2,3,..., tn,..., tl,t,t+l,...t+ ....
(n=l,2,3,..., tl;=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 10
1  > and 1 
tn t t+tl TT
analytically, where n=0,1,2,..., tl; t=1,2,3,..., 0 < e < 1.
t+8 8
1 tn n=0,1,2,..., tl; 0 < 8 < 1.
tnt0 Q
tn '
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)
18
t t=1,
18
t+l '
18
t+l
(10)(t+d) ,
_ t+48t8e ,
_ t+ ,
1 t+0 ,
_ 1, which is true.
2,3,...; O < 8 < i.
Therefore,
+0 0 t+ 10
(112) 1 t+e and 1 t+
tn t+' ttT
where n=0,1,2,..., t1; ,=1,2,3,...' 0 < ( < 1.
Inserting Isl for t+0 and j for tn and t+L in (112)
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(ta) 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 (112)
with 0 replaced by 6, we have for Isl 1, 0 < 6 < 1, that
t+6 6 a 1t+6 16
1 T and 1
I F an t I t+t I '
where n=0,l,2..., t1; k=1,2,3,...
Summarizing our results we have:
If 0 < Isl < 1, then
1 1 1is .
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(t2).
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(ta) and 0 < 6 < 1.
Now we present an inductive proof that
la'l < Mn K rn
n
Again using equation (16) with a now replaced by 22+ n,
we have
(113) F(h2+n) = (h2+nh?1)(2 +n2) = n(ns)
If i is replaced by Nt, the inequalities of (18) 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 < Mrn; Iq < Mrn I2pn+q nI < Mrn
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 (14), we can write
a' =
(h2+l)2+(pol)(h,2+l)+q
and with equation (16) this becomes
a' = 2p
1 F(N2+1)
Taking the absolute value of each side and using (113) with
n=l, we obtain the inequality
aI p +q < I Mr
IF(A +1)1 1lsl
Next, assume
n=1,2,3,... ml,
and show that
a' M r .
Using (19) with A2 and a' replacing 'A and a
respectively,
mi
m t[Pt (2 + q t] 2pm
la'll t=l
a'i =
m F(A2+m)
Tm mi
lam lPt 2+qlt pm+q (mt) atl Pt
t=l t=l
mimsi
m1
S .Mrmt t + Mr +
t=l
m1
i (mt)'la tl' 'Mr
t=l
m2 1sm1 I
lan'i n n n
a' < M K r
Similar to (110) and (111) with Jam_t < (MK)lrm
M > 1,
m1
am a tI Mr ( (m1)(Mc)M1 Mr
t=1
< (ml)rmMKm ,
and
m1
S(mt)a I Mr
t=1
Thus,
(ml)rWMmK' + Mrm
la'l <
m
m m1 m
< (m1)(MK) mMr
< r (ml)rmMmKm
2
+ M r m (m1) KIn
1
m llsm I
im m nm ri m i mm iii
mr MmK m_r M 1 + Mrm + Mmrm m (ml) r
m I sm
Now since (Mc)m > 1,
Mrm Mmm m1
Mr Mr K
= Mrm(l, Mm1 mI)
= Mr (1M K )
= Mr 1(MKc) i]
is negative. Thus,
Ia'l <
m
mr M K + r M (m1) m
m2lsm1
m 1sm 
mm ml1r+l
Sr M (m+)
2mI 1sm1
< rmMmCm (ml) < rmMmVm
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(zc)n is
n=l
Mn n
lim M
n o, Mn+ rnl
n >oo M r
r r
limn 
M M .
nj o
That is, the series is absolutely convergent within the
circle Izcl = .. Since laI Mr", the series
an(zc)n also must converge within the circle Izcl
n=i
r
= and is, therefore, uniformly convergent in the region
Izcl ( [L, p. 951.
Similarly an'(zc)n converges uniformly within the
n=l
region Izcl < r
.1
We have thus obtained two formal solutions
u,(z) = (zc) 1 + a (zc)n
n=l
and
u'(z) = (zc) 1 + an(zc)n]
n=1
which are uniformly convergent series of analytic functions
when Izc( < rM and Izcl < rM1l respectively, pro
vided that arg (zc) is restricted in such a manner that
the series are singlevalued. 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
(zc)2 u" + (zc)P(zc) u' + Q(zc) u = 0
to determine the equation for ).
U = U1 0,
u' = u .)' + u 0,
u" = u. *)" + 2u; 0' + u" 0.
Substituting in we have
(zc)2 u1()" + 2uj' + u4] + (zc)P(zc)(u )' + u )
+ Q(zc) ul*) = 0,
(zc)2u 1n" + 2u (zc)2 + (zc)P(zc) uI 0
+ (zc)2 u1 + (zc)P(zc) u1 + Q(zc) u1 0 = 0.
Since u (z)
(zc)2
is a solution
u"1 + (zc)P(zc) u! + Q(zc) u1 = 0,
and
(zc u 2(c)2 +2 + (ZC)P(zc).uIj = 0.
Dividing by u. we have
(zc)2 "+ [2 (C) + (zc)P(Zc) 0' = 0.
To find a general solution let 0' = Y.
(zc) Y' + 2 u (zc) + (zc)P(zc) Y = 0,
(zc)2 d 2 (zc) + (zc)P(zc) Y = 0.
Separating variables and integrating,
Y +
I dYY
f[
S+
u 1
In Y = In B 21n
2
Y =Bu
1
P(zc dz = In B,
S(zc)
u (z ) dz
u/ P zc)
SP(zc/ dz
Tz7 cy
Now since Y = 0',
O [B u2 e
 dz ] d
FZ d7
where P(zc) = po+ pl(zc) + pp(zc)'+...+ p(zc)+..
dz 
P(zc) dz
P [p(zc)1 + PI + P2(zc) +.'. dz
= poln(zc) + pl(zc) + p, 2(z +..
Hence,
0= A+B
= A+B
2 [p ln(zc) pl(zc)...pn(zc)n...]
uI e
u 2 (zc)Po e
( 2
+BC) dz
00
,P(zc)n
n n
n=l
= A + B
dz .
But since
oo
u,(z) = (zc) 1 + an(zc)
n 1
) = A + B (zc) 2 g(z)dz,
(zc)n
ri
SPn n
[ i2 n=l
where g(z) = 1 + a (zc)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(zc) or
1
of (zc) 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(zc)nl
n=1
We already have a2 + (p0l) a + q = 0 (16) with
roots a = Al,'2 and AlA2 = s. The sum of the roots l+A 2
= po, so we can write
p 2?, = 2 +~212 = 1 = s1.
Then
00
(114) = A + B 1f + (zc)n (zc)1 dz
n=1
= A + B [
s1
(zc)sldz +J gn(c)nldz
n=l
gs(zc) 1dz
CO
+ n= g ( c)n+dz
n=s+l
(zc ) sSjI
nS
n=1
n
sn
(zC)ns
(zc)
+ gs In(zc)
+ in (zc)ns
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(zc) + (z)1
where
(zc)r
sI
n= c)
sn
n=l
1
u(z) = u,(z)[
1 Vs
+
n=s+l
n
ns
and
u1(Z) = (ZC) 11
(ZC)n]
Co
+Xa(zc)n]
n=1
+
= A + B[
Rewriting, we have
u(z) = (zc) 1
00 S1
+ a(zc) (zc)f gs(zc)n
n=l n=l
(z)ns
(ZC)
Cn
+ ns
n=s+l
= (Zc) 2
uo
la(c)
n=1l
s1
n (zc)n
s sn
n=l
(zc)n]
+ n
n=s+l
= (zc) 
= (ZC) 2[
+ h(ZC)nJ
n=l
where h are constants:
n
Sg1 al
sa +
and
hn(zc)n = 
n=l
an(zc) +
n=l
1 + an(zc)n
n=l
s1
gn )n
(zc)
n=
n=1
00
Zs gn
Lns
n=s+l
(z c)nI
(zc)
When s = 0, we can write (114) in the form
CO
S= A + B 1 + g (zc)n] (zc) dz
n=l
A + B [ (zc)r dz + gn (zc)ndz
n=1
=A + B ln(zc) + g(zc)
n=l
Therefore, when 1 = 2, the second solution is
u = ul(z).* = A ul(z) + B ul(z) ln(zc) + n (zc)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(zc)
n=o
by substituting
u = ul(z)ln(zc) + 5 (z)
in the equation and equating to zero the coefficients of
the various powers of zc 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 (12) with c = 0 becomes
(115) 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 azal+...+(a+2n+2)a2n+2 a+2n+l+...
z u2 = a a(a1)z 2+...+(a+2n+2)(a+2n+l)a 2n2+2n+...
0 2n+2 .
Equating the coefficients of powers of zc to zero,
z : a(al) + 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 (115).
b z
n
u' = u'ln z + ul
u" = u'ln z + 2u ( 
0)
+ + nb z
n=l
00
+ Zn(nl)bnzn2
n=2
00 00
n n
ul'z21n z + 2u 2z u + n(nl)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 2n1
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 2n2
2n n 4
 mb =
2n2
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 n2
2
b b n odd,
n 2 n2
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 (11). 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 (11) 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
(11)
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
nonnegative 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 mt = 1 or 4 = m+l and ni = 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 za)2 za) 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(za)' + hl(zal) +l+... h0 / 0,
then
4 M
Z+2 z] + J(z) u = ho(zal) + h (zal) +...
4
Z r + H(z) u'= ho(za )+...
zar0
r=l
1 u" = hoX(?l)(za )2+...
The indicial equation for z = al is
(za )2:
ho(A1) + hok 1 + h o = 0,
o o 1 o 1
?2 + (k1l)? + 1 = 0.
Then
so that
a1 + pi = 1k ,
k = 1 a1 oi'
and
ali = .1i
Similarly for the other regular points, so that
(21) d2u +
dz2
r=l
larPr + H( 1 du
zar I dz
+ +arr 2 + J(z)
r~l (Zar)2 za
L za r
u = 0.
1
If z = o (or zi = 0, z = ) is to be an ordinary point,
zi
2zz 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 loai 6(.
r r 1 H 1 Z
1 z 2 z
r=1 r 1
1
2 1
2 zp
1 Z1
41a r0r 1
z zl (arzi) 3 H(
r=l 1
z z 1 Z1 1
= i ^ ( i H 
21 rr r z 1 2 J/
= riI^ ^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 (1arr) = 0,
r=l
or
4
S(ar+fr) = 2.
r=l
Rewriting q(z) in the form
(za1) (za2)(za3)
(za2)2(za3)(za4)
(za4 )2 zal)(za )
(la z )2 (a2 z)(la z )
(la2z)2(l1a 3z)(1a4zO)
(la 4z )2(a zl)(la 2z)
(1a z )2(1a z )(la 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
(za) 2(za2)(za3)
(za 2 za3 )(za4)
(za3)2(za4)(za1)
(za )2(zal)(za2)
q(z) =
(za ) (za4)(za )
then
1
. q
zi
+ J(z),
4
S[ar r r+
r= (r)2 zar
1)
Then clearing of fractions,
n (zaza3 (za4 )2 + n2(za )2 (za3 )(za4)
+ n3(zal)(za2)2(za4) + n4(zal)(za )(za3)2
SY11za2za3 )2 )2 + a 22 (z01)2(za3)2 a )2
+a33 (za)2 (za22(za4)2 + aq4 (zal )2(za3 )(za )2
+ m (za1)(za)2 (za3)2 (za)2
+ m2(zal)2(za2) (za3)2(za4)
+ m3(zal)2 (za2)2 (za3)(aa4)2
+ m4(za )2(za2)2(za3)2(za4).
To determine nl, let z = a :
nl(ala2)(a a3)(ala4)2 = alBl(ala )(aa2(aa )2,
nl "ll(aia )(a a3).
Similarly,
n2 = 22 (a2 a3)(a2a4)
n3 = a3 3(a3a4)(a3a),
"4 = a44( a4a)(a4a2)
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 + l1arPr du
dzq zar dz
a2 2(a23 )(a2a4)_
(za )2(za )(za4)
+ 1(ala )(ala 3)
1 (za )2(za 2)za3)
S33 (a3a1) (a3a,)
(za )2 (za )(za )
a+ 4134(a4a,)(a4a )
(za4 )2(zal)(za u =
(za ) (za )(za )
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
(23)
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.
(22)
z
First, let a2 = 0 and a3 = 1, then (22) becomes
2 [ 1a p
(24) a u +
dz2 zaL
la ,P 1a3 3 1a 4P4 du
+  + +j
z z1 za4
+ Lapiai(ail1)
I(za) )z2zl)
a 22a4
+2( +
z2(zi)(za4)
S3P3(1a,4)(al)
(z1)2 (za4)(zal)
a 44 (a4al)a4 ] .
  = 0 .
(za4)2 zal)z
Before letting a4 become infinite, rewrite the equation
in the form
[ al
+ [a 1al)(al1)
(za1 )2z(z1)
lc2P2 1a33 1a 4P4 du
z z1 za4 dz
(2 3 2
z (z1) a4
(a4
44(1 )
(  2(zal)z
a(4 1/
Now let a4 become infinite:
(25) d2 +
dz2
1a1i 1a 2 +
z + z
za1 z
1a3031 du
z1 dz
+ iOlal(al1) a 22 + a33(lal) + a44 ]
(za )2z(z) + 2(z = (ZO
(zal)2z(z1) z2(21) (z1)2(zal) (zal)z
2
d2u
d+
dz
( P 1 (1a )
C33(3 a4
(z1) a4 1(zal
a4
which may be represented by
a 0 1 oo
(26) 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
(a4a3) (za2)
(a2a ) (za4)
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 (25) in the form
d u 1[ alP1 12 2 13 3 du
+ + + 
dz L za z z1 dz
[ a a,1) CXP 2 a3 f l
+ +
Siz(zl) z (zl) (z1)2 z )
"4P'4 u 0,
+ (z = )z,
(za1 )z
47
and when al becomes infinite we have the confluent equation
(27) du 1a 22 1a3P3B du CP a 22
(27) +U + + l + 2(zl)
dz2 z z1 z(zl) z(z)
+ U = O,
C 3P3 =0,
(z1)
with singular points at z = 0, 1, oo.
Let us proceed to find solutions of this equation.
Multiplying the equation by z 2(zl)2, we have
2
(28) Z2 (l)2 d + (233)3
dz
+ (3+2a +2 +a3+p3)z2 + (1a 2 )z
+ [(ac + a A3)z2 + (c,1 a2,B)z + ap2 u = 0.
Assume the solution
N ,+1 A+n2 A+ni
u C z+ C +...+ Cn z + Cn z
o 1 n2 n1
h+n
+ CnZ +...n C / 0,
then
u' = C Az 1 + Cl(A+l)z +...+ C (A+n2)z 3
o 1 n2
+ C (+n)z+n2 + C (A+n)z +n +...
nI1 n
u" = Co0A(l)z 2 +...+ Cn2 (+n2) (+n)z+n4
+ Cn_(+n1)(+n2)zh+n3 + C (+n)(+nl)zh+n2+...
If these equations for u and its derivatives were now sub
stituted in equation (28) and the coefficients of the
powers of z equated to zero,we would obtain the following
results.
The indicial equation is
z: 7(N1) + N(la2P2) + CL22 = 0.
(29) 2 (a +2)A + a2 = 0.
Then
= a 2)2'
Finding Cn,
z7+n: Cn2(A+n2)(A+n3)2 Cn1 (+n1)(A+n2)
+ Cn(+n)(n) Cn_2(+n2)(2a22 3)
+ Cnl(?+nl)(3+2a2+2 2 +a3+3) + Cn(?v+n)(laC2P2)
+ Cn 2(0a1 +03~3) + Cn1(a1P1 a22) +.CnC2 2 = 0.
Cn = nC._ (+n2)(N+n3)(+n2)(2a ,P2a,33 1) (a L3)1
+ Cn1 2(?+nl)(N+n2)(A+nl) (3+2+2P +3+33)
+ J (2+n)(7+nl)+(,+n)(la 2_ ) + 2 a
Simplifying the denominator by using the indicial equation
(29) we have
C = Cn2[(N+n2)(+nlcx2P2a33) p I 033]
+ Cn1 (?+n1)(2\+2nl222a 2 a33)
+ au 1 + a2P1 n(2+n1 )
n(27+na p2)
Let us compute a few terms of our series solution to
see if a general term can be obtained.
Co C (2+l2a22P2a3p3) + a 11+ ~ P2]
1 (2A+1a 22)
Using the indicial equation (29) we can simplify and write
Co[(103p3) + al 1 a22]
C1 =
22+1a 
C2 = { [(+la33) 3] + C[( )(2
+2a22P2a33) + a P + a P2 
S2(2)+2a 6 )
Substituting C1 into this expression:
C2 = C {(2+la2)[(la22a3~3)alla33
+[(1a3p3)7 + al"Bla2] [ (A+1)(2h+32a22P2as33)
1
+ a 1P + a P21.
++22 j 2(2X+la,2_)(2X+2a_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 (27) with a general term, thereby making
available a form suitable for further development. Since
equation (27) 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,
(210) Pj(z) = [Sj Tzh] zJm ,
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 (24) 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 (a4al)a4
(za4)2(za )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
(26) 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 (22) we can
write this equation in the form
1
l aF f a 1 ia P2 ia 
u" + + +
zal z z1
OP2 3B3 (1al) ,)
z + + 
z (z1) (z12 (za )
a1
U" +
a a
z 1
VaI /
f
(,+ L)al(all)
u' + al
(zaz )z(zl)
(a + t)
( a' l a"]u = 0.
+ 22 + u'
z z1
a2p L3 a1 1
( al
4 2 1
+ u = 0.
a 1)z
Letting al become infinite:
u" + + + U33
z z1
+ 2 2 + 33 4
z(z) z 2(z1) (z1) Z U = .
+ 1
z 1)2 1
aW 1 z(z1)
Clearing of fractions we can write
(211) z2(z1)2u" + [Iz2(z1)2+(,a 2B2 )z(z)2
+ (1a3P3)Z2(Zl) U'
+[ z(z1)a22(zl) + a3B3z + Bz(zl)2] u = O,
where the singular points are at z = 0 1, o, .
Applying the criteria (210) we have for J = 2
S[S ,,zh] 2m 2
P2 = [ T z = z (zl) ,
2m hm+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 z1. To do this a313 must equal zero, so let p3=0.
Thus, with B3=0 and dividing equation (211) by z1, we now
have
z (zl)u" + Z2 )( 2)(z)+ a3
+ [zza 22 + P4'z(zl) u = 0.
(z 3 )u" + z3 +(2a 2a 3)Z2 + (l+2 +B2)z u'
+ [f4tz2 + (lP4)tZ 22] U = 0.
Now returning to our criteria (210),
S2m hm+2 3 2
P2 = S2 Tz = z Z ,
where hm+2 = 3 and2m = 2. Therefore, m = 0, h = 1,
S2 = T2 = 1, S2T2 / 0. We need
P, = [STz ] = tz3 + (22 2a 3)z2 + (l+az +2)z
so that t = O, T1 = a2+ 2 + a3 2, S1 = a2+821 and
p = (2a232a3)z2 + ( 2+0l )z.
Finally, we have
PO [SToz = 2f2'
where S = a22 and T = 0.
Thus, if (3 = 0 in equation (211) the resulting
equation, which has a two term recurrence relation for a
solution, is
(212) z2(zl)u" + [(2a22(3)z2 +(U2+21)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
(26) be functions of a and assume that neither al nor a4
appear in the exponents. Apply the criteria (210) to
equation
() +[i a33 al [(1
(27) u" + Ia2P2 + U' +
z z1 z(zl)
z 2
z (zl)
a3r3 1
+ u_ = 0,
(z1)
or clearing of fractions
z (z)2 u" + (1La 2)z(z1)2+(laaa)z (z1) u'
+ [alZz(zl) a2P2(z1) + c3P3Z2 u = 0,
where
I
r=l
(ar+pr) = 2.
As before we must remove a factor of z1, therefore let
f3 = 0 so that a3g3 = 0. Then
z(zl)u" + (laP2)Z(zl)+(la3)z2]U '+ o1 lZa22 u
(z3 zp2)u +[(2a 2P23)Z )z 2+)Z u'+al 1P l 2 21
Applying the criteria (210) we have when j = 2
p2 [S2 T2zh] 2m
= 0,
= 0.
3 2
= Z Z
where hm+2 = 3 and 2m = 2. Thence, m = O, h = 1, S2 = T2
= 1, S2T2 / 0. Further,
P = [STiz z = (2a2P2a3)z2+(a2+f21)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,
(213) z2 (z)u" + (2a 2B2a3)z2 + (a 2+0l)z] u'
+ [ QPz a22] u = ,
where the only restrictions are that in equation (27) ar'
Pr (r=1,2,3,4) contain no function of a1 or a4, and P3 = 0.
Since equation (212) can be obtained from (213) by
letting a181 = O, let us obtain solutions of the latter.
aC 1za 2 u = Cz +...+C z +n+C z +n+... C / O,
(2a2p2a3)z2 u' = Coz1 l+...+Cn_l(A+nl)zh+n2
(lzn+.+n 1 .
+ (a2+B21)z + C .(+n)zhn
z z u = CoNh(l)z 2+...
+ Cn1(+nl)(h+n2)z +n3
+ C (A+n)(A+nl)zh+n2
U .
The indicial equation at z = 0 is
z: A(1) + (ag+pgl)N a 0 = 0,
2 (a2+p 2) + a2 2 = 0.
The exponents are
h = C2,P2
z +n: Cn1(A+nl)(+n2)Cn (+n) (h+n1)+C_n (X+n1)(2a2
p23 ) + Cn (+n)(a2+P21) + Cnla p Cna22 = 0.
=C n[(h+n1)(?+n2)+(\+n1)(2a 2 a ) + alp
C =  h
n (2+n)(p+n1)+(7+n)(2 +( 1l)a 22
Cn_l[nl; + (X+n1)(X+na2p2a3)]
n (A+n)2 (a2+P2)(?+n) + a2zB
SCn_1 [a + (?+n1)(A+nc2P2a3)]
n (X+na2)(?+n~,3)
Let a = a :
Cn=l1 a + (a2+n1)(nP2a3)]
C = .
n n(a 2p+n)
Co[ a + a2(1P2a3)]
C = +
1 (a2p2+1)
C [a~l1 + (a2+1)(232a3)]
C 2
2 ((a2Bg+2)
Co[Cl + a2(lP2a3)] i1 [a + (a2+l)(2P2 3)]
2 2!(a22+1)(a2 2+2)
Therefore the solutions of (213) are:
u = 2 [1 + z1(2(23)
1 o l!(a2P2+1)
[a p1+a2 (1323 ) ] [a2Pl+(a2+l) (2p2a3) ] 2
+ Z +...
2!(a2P2+1)(C2P2+2)
+ [all+a2(l2a3) ..[a1+(a2+n1)(np2a3)] n
+ Z +...
nl.(a2_p2+l) ... (a232+n)
(214) u1 = Coz2 1
+ [.a 1+a 2(lpa 3)] .. .[a1p+(a2+n1)(np2a 3)] n
n=1 n(a 2P2+1) .. .(a2P+n)
and similarly
(215) u2 = cz 2 1
+ [ 1p1+2(la2a3 )] ..[al+(P2+nl)(na2a3)] n
n=1 n!(2pa+1) ...(p2a2+n)
If a l = 0, then the solutions of (213) will be
hypergeometric:
(216) u, = Co z 2F1(a2,p2at3; 232; Z)
and
(217)
U2 = C z 2F1(P2,C22a3; 2a2; z)
The hypergeometric form is
(218) PF(a,b; c; z) = 1 + z a + a(ab(b+l z2
2 1''' !c 2!c c+l)
+ a(a+l)...(a+nl)b(b+l)...(b+n1) ...
n! c(c+l)...(c+nl)
00
Sa(a+l)...(a+nl)b(b+l)...(b+n1) n
n=o n! c(c+l)...(c+nl)
2.4 Solutions of Confluent Cases after Normalizing.
Another approach to obtaining solutions of equation (27),
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
(219)
in
(220)
 p dz
vw" + (v" + v'p + vq) w = 0.
Now to normalize equation (27) we use (219) to find
V = Z
Then
S(a 3) (a3+31)
v' =. (2+p21)z 2 (z1)
1 1
1 2+ (a(+ +1) (a3+B3)
+ (a3+B31)z (z1) 2
1 1
(a +p 1) (a3+p31)
(z1) 2
(1) 3 (a+21)(zl)
+ (c+11)z]
( 1)
f (03 +1331
S1 a (a 2+a5)z
v = (a2+2i) (a2+B23)z
+ 7(a2+p2 ) (3 3 1 )
+ (a3 +3 1) (a3+P 3)z
2(a2+023)
!(a2+ 21)
(z1)
(zi)
(zi)
(a +B,5) (az+ 5)
(Z1) + [3 (a +3 1)(a2+3)(z1)2
+ (a +p21)(ca3+f1)z(z1) + (a 3+13)(a3+z33)Z2
Substituting v into equation (220) with p and q from (27),
1
+ Tz
a2+05) !(a(+3 5)
(Z1)(2 (3+3 [ +(a 1)(a+ 3)(Z_1)
12 +f3 1 2 3
+ (a + 21)(a3+431)z(zl) + (a3+331)(a3+t33)z2
1 (a 2+23) (a3+33)
+ z (z1) ( z2+(a 21)(z1)
1
V = Z
2
V, 1
V = ~Z
2(a2+p23)
31( C +P 3 )
1 3 3
i(a3 +f35)
1 1
(:y 2+321)) ((c3+f31)
(z1)
+(C +P) / 1c22 1+ 
(+ z +2 Z1) (3 )3 3 122
+z (+1zJ
1 1
(zl)2 (z)
z(z1) z (z1)
+ )2 w = 0.
S(a2 +025) (a3+P35)
Dividing by z (z1) and simplifying,
z(zl)2 w" + (a2+21)(a1+3)(zl)2
1 1( + 2
2+ (c2+21)(a3+p31)z(z1) + T(Ca3+33l)(a3+333)Z2
1 1a 33 P 
+ z(z1) (a+P,1l)(z1)+(a3 l 1)z +2 3 3
z z1 
2 21 3 +23 3
+ ( 2 )2 C2 + w 3 0.
Z(zl) 22(Z1) (Z)2
z (zl) w" + ~(a,+Bl)(a2,+13)(z1)2
+ (a 2+(2)(a3+31)z(z1) + 4(a3+01)(a3 .3)22
(a2+P21)(z1) + (a3+f331)z + a l1z(z1)a P3(z1)
+ a 3z 2 w = 0.
z (z)2 w" + r
1(a 1)(a) +2 33
T(a2++p21)(a2+p23) + (a 2+p21)(a3+P31)
+ (a3+P31)(a3+335) (a2+p2l) (a2+21)(C3i)
 (a 3+31)2 + a1B
+ a3 12Z2 + [ (a221)(a2+23)
 (a2+21)(a3+,31) + (a2+,21)2 + (a +2l)(a3+P3)
 a1 al2BZ + [2+p21)(a2+p2) +212
+ aO22]
Sw = 0.
Therefore, the normalized form of equation (27) is
(221) z2 (1)2 + 1(az+B)2 T(333 )
2(a2+ 2)(a3+33) +) +(a33) +
i(a2+B 2)( 0 3 3) + 2(a2+ B2) +p2()3 F3) + 4l1
+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 (221) let
w = a + + +n2 + a 1z+nl + a Z+n+
w = az + az +...+ a z + a z + az +...
0 1 ni n
ao / 0,
then
'A2 +n4
w" = ao(Ml1)z +...+ a n2(+n2)(A+n3)z
+n3 +n2...
+ a _(+nl)(A+n2)z +n an(an)+n)(A+n1)z .
The indicial equation is
z: X(A1) (a+P2)2 + 1 + ca = 0,
TIa+p +a0
(222)
2 + 1 (a 2 = .
N N + = 0.
The exponents are
1 +1
 4 (a1 2)
1 (2a 2)
2
Finding an:
z?+n: an2(A+n2)(N+n3) 2an_l(+nl)(?+n2)
+ an(A+n)(h+nl) + an2 [ (a+p)2 (a3+3)2
(a2i+2)(a3 3) + (a2+2) 3+) + 13
+ a 3 + an1 ()a2+2) + )(a1 3
1
+ 2 1( S ) C I1 a2 + an2 (22=02)2
1 p] = 0.
+ + 22
(223) a = a 2(h+n2)(+n3) + ("282 + f(a3+ 2
1 1 1
+ (2 .2 ) ( +,) (x2+.., ) 2(3t 3)~~
a3c3 + a n 2(\A+n1)(+n2) (2+e2)
(ac2+ )(al3+ 3) + (C2+,2) + (a+3 3)
+ + 2
1+ cl1 + 2'2]} (h+n)(h+n1) (a202) + 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 (223) and simplifying the
denominator by use of the indicial equation (222), we have
an [(h+n2)(7,+n3)+k] + an [2+nl) (+n2) + m
a =
n n(2?+nl)
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)
4x42A3 + (4m2)A2 + 2k? + m 2
21 (2?)(2?+1)
a ( \+l)A + k + a2[2(?+2)( \+1) + m]
a3 = a ( 2(_2+k)(2X22N+m)(2)+l)+(2X2+6+4 +m)[ 4X423
+ (4m2)?2 + 2kA + m2] ,
31 (2\) (2A+1)(2\+2)
a3 ao[8?6 + 12\5 + 4(3ml)?4 + 6(3m+2k2)?3
+ 2(3m2+4m2+4k) 2 + 2(3m2m+3mk+2k)X
+ m(m +4m+2k)]
a1 =
S=
a
a2 = ao[
a2 = a
3! (2)(2?+1)(27,+2)
The solutions of equation (221) are
(224) w = az + (222+)
1 1 (2?X)
44 2X3 +(4m2)N2 + 2kN + m 2 2
+ Z z +...
2!(2) (2A+l)
I+(a282)
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(+21)(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
(225) u = I a 2k a4 z
m 1 a 0 P4
when the u' term has been removed. This can be obtained
from equation (221) by letting alB = m, a2 = a, P2 = 1a,
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 (27). Our transformed
equation, that is (225) in normalized form, is
68
(226) z2 (l)2w" + (m+ k)z2+(aam)z+(la) w = 0.
Assume the solution
w = Coz + + Cn2z+n2+ Cn +n1+ Cnz+n+... C 0,
= 'n+...+C z C + CZ +... C / O,
then
w" = Co(A?l)z 2+...+ Cn_(+n2)( +n3)z+n4
+ Cn _l(+nl)(A+n2)zh4n3+ Cn(N+n)(A+nl)z+n2 +...
The indicial equation and exponents are
z: ,(N1) + a a2 =0,
\2 7 + a(la) = 0,
(Aa)(A+al) = 0,
X = a, 1a.
Obtaining C :
z7+n: Cn2 +n2)( +n3)2Cnl+nl)( +n2)
+ Cn(A+n)(A+nl) + Cn(m + k2)
+ Cn_(a2 a ) + C (a a2) = 0.
am0+
69
Cn = C (+n2)(h+n3)+(m+ 1 k2) +C 2 (h+n1)(A+n2)
(ac a m) 1 2 *
S(A+n)(?+n1) + a a
Let ? = a:
S= Cn (a+n2)(a+n3)m + k2] + Cn_1a2 + (4n5)a
+2(n1)(n2) + m n  1
S n(2a+nl)
C a a2_ +m
C1 1
C 1(2a)
C a(al)m + k2] + c 2a+
C2 01 
2 2(2a+l)
[2a(a2+am +k2)+(a2a+m)(a2+3a+m)
2= o 2!(2a)(2a+l) '
aa4+(2ml)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(a2a+m)(a am 1 +k 2)(2a+l) + a 4+(2ml)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 (226) can be written with a few
terms as:
(227 a+ a +(2ml)a 2+(2k Z)a+m2 2
(227) w = C z 1+ Z+ z +
L 1! 2a 2. 2a(2a+l) J
A second solution of (226) can be obtained by replacing
a by 1a:
(228) w = Coz1a+ 2a+
S1(22a)
a 4a 3+(5+2m)a 2+(2k24m 3)a+(m +2m +2k2) 2
+ +...
21 (22a)(32a)
Although solution (227) 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 (lz) 1+ 2kz+ 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+nl)a+m
n
+ (nl)(2k+nl1) z+...
n' 2a(2a+l)...(2a+n1)
Let this solution be indicated W k,m, since a,k,m are the
only parameters; therefore (227) can be written
(229) Wa,k,m = C0a(1z)k+ 1 + (a2+2ka+m)
n=
[a2+2(k+l)a+2k+l+m] [a2+2(k+nl)a+(nl)(2k+nl)+m] n
n! 2a(2a+l)***(2a+n1)
If m = 0, then the solution (229) becomes hypergeo
metric. Let this solution be indicated W ,, a and k being
the only parameters:
Wa,k = Coza(1z)k+ 1
S(a 2+2ka) a2+2(k+l)a+2k+l [a2+2(k+n1)a+(n1)(2k+nl)
n= nl 2a(2a+l) *.(2a+n1)
Factoring further we may write
k+ 
Wk Co ZU(z) I
Sa(a+2k) (a+l)(a+2k+l) ]*[ (a+n1)(a+2k+nl) ]
n=1 n! 2a(2a+l)...(2a+n1) *
Rearranging factors we arrive at the solution
k+
W,k = Coza(lz) 2
+ a(a+l) '*(a+nl(+2)( )(a+2k+l) *.(ac+2k+nl) zn
n= n! 2a(2a+1) **(2a+nl)
which is hypergeometric in form (218) and shall be denoted
by Wa,k Therefore, when m = 0 in (226) we have the
solution
1
k+
(230) W,k = (lz) 2Fl(a,a+2k; 2a; z),
where we have let Co = 1. A second solution can be obtained
by replacing a by 1a.
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+n1).
n(n1)...(nJ+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(al b; c; z), etc.,
F(+n) = F(a+n, b+n; c+n; z),
F(1) = F(al,,b1; c1; z).
We shall prove the following:
n = 1,2,3,...
F = 1 [aF(a+) + (1c)F(c)]
ac+1 L1P
F = 1 F(a) + (bc)z F(c+)l
1z 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),
(235) F(+n)
(2a)
n
= z
(n,)
(l z)
I (1Z)
n+l n
k
2
J=o
(231)
(232)
(233)
(234)
(1) (1z) W ,k+ n
J) r~k
(236) Dn z(1z) ]
kn+
= z"(lz) 2
(1) (a (kx) ,(z (z
j =o
k(237) F(l) zykzl1
(237) F(1) = (1z)z 2(2a1)(1z)Wak+(2k)W
(2a1)(2z)L a aj
Finally, we state Leibniz's rule [9, p. 409] for finding
the nth derivative but shall not prove it:
(238) 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
(231)
F = 1 [aF(a+) + (lc)F(c)]
ac+ L J
Proof:
aF(a+) + (lc)F(c)
(a+l) (b) +
(c) n!
n
co
(1c) Z
n=o
(a) (b)
(c1)n!
(cl) n!
= [a(a+l) (b)n
=o (c) n
(a) (b) z
(1c) L I
(cl)nn!
n=0
c,
S [a+n(c+n1)
n=o
(a)(b)n
(c) n!
S(a c + 1)F.
Therefore,
F 1 a?(a+) + (1c)F(c)]
ac+1 L
1
F = 1
1z
I c
(bc)z F(c+)
C
(a1) (b)
(C n n
(c),nl
(bc)z
c
n=o
(a) (b) n
(c+1) nn
(a1)(b). n + (bc)
(c) n' c
1 (a1) (b) (b )
n=1 (c)nn! c
=1 + [ (a1) + n(bc)
S1 L a+n1 (a+nl)(b+r
S1 + (al)(b+nl)+n(bc)
n=l (a+n1)(b+n1)
(a)n1(b)n1 n
n=l(c+l)nl(nl)!
(a)n(b)n z
I1)J (c) nn
(a) (b)n
(c) n!
(232)
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+n1) (a)n(b)n n
n=L (a+nl)(b+nl) (c) nn
0o
FZ [ n(c+n1)
n=l (a+n.1)(b+n1)
(a) (b)
(cn n
(c) n!
0o
F z
n=l
Cor
F y
n=o
(a) n1(b) n1
(c)nl(nl)!
(a)n(b)n n+1
(c) n!
= F zF
= (1z)F.
Thus,
F(a) + (bc)z F(c+) = (lz)F
c
F =1 F(a) + (bc)z F(c+
1z c I
F(+l) = [ F(b+) F]
az L J
F(b+) F =
n=o
(a) (b+1)n
(c) n!
n n=o
(233)
Proof:
(a)n(b)n
(c) n!
n
Sb+n
7^
(a)n(b+ln1 n
(c)n(nl))!
00
=1
n=l
,00
71
*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 n1 n
(c)n(nl)!
S(a+)n1 (b n1 ni
S(c+l) (n1)! z
=1 n1
1 (a+l(b+)n n
L (c+1) n:
=0 n
(234)
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)n1(b+2)n1 ni
Z
(c+2) (nl)
(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(lz)Wa,k
then
(235) F(+n)
(2a) n+1 n
2n zan(lz)_k ny
Sc (1z) 2 (i
()n j=o
j \
J2
1z) Wk+ nj
2
Note that when k is replaced by k + m in the premise, then
F(a, a+2k+2m; 2a; z) = za(1z)km2W
ca,k+m
Proof:
We begin by applying (253) 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( k1 k1
 z (1z) W (1z)' 2W
az a,k+ az(lz ,k .
(a) ,k+ ,k
Apply the principle of (233) to F(+2):
(240) F(+2) =2+1) [F(a+l, a+2k+2; 2a+l; z) F(+l)
(a+l)z J
Using the expression (239) with k replaced by k + in
(240), we have
(241) F(+2)
S(2a+l) a 1a1l kzW kW
(a+l)z a ak+ ak+
2a a1 1 w 1
a (1lz) Wa ,k+ (lz)2W ,1
a L ak+ akJ/
)2z (12 ) a,+2(1z)W ak
+ (1z)Wk]
Let us find F(+3) before generalizing our results to
F(+n). Upon applying (233) we have
(242) F(+3) = (2a+22) FF(a+2, a+2k+3; 2a+2; z) F(+2)
(a+2)z IJ .
Replace k by k + in (241) to write (242) in the form
(2+2) (2a)2 a2 k2
F(+3) = z (z)k2 W k+3 2(z)
(a+2)z (a)2 2k+l
(2) 2 a2(l k
S(1z)Wa,k] (a)2 ( ) a,k+
1
2(z) W ,k2 + (lz) Wak
Simplifying further
(2a)3 a3 k12
F(+3) = z (1z) W,k+ (1z)2W,
(a)3 a,k+1
+ 3(1z)W ,k 1 (lz)Wk]
Finally,
2cx) n an k n
F(+n) z a (k (1) 1) k+ n2
(a)nk+ ,
2) a,k+ j ak+
n
+...+ (1z)W ,k .
Therefore,
(2a) an(lz) n+l1
F(+n) n zanz (1) (j\1z) W k+nJ
(a)nJ=o >a,k+
n=1,2, 11, .
(236) Dn [za(lz)k+
n
= a(1z)kn (1) () a(k)n z(lz) .
J=0
Using Leibniz's rule where u = za and v = (1z)k+
find DIu and Dnv. When u = za
Du = aza1; D2u = a(al)za2; ...; DDu = (l)t(a)za"
If v = (1z)k+, then
3
Dv = (k+)(1z)k; D2v = (k+)(k)(1z) k ..
D"'v = (k1) (1z)k+n+ .
Hence, the nth derivative is found to be
Dn[za(lz)k+] = ( ()(a) k ) (lzk+Ln+
n
= za(1z)kn+ ( ()(k) z z)
4t=o
If F(l) = F(al, a+2kl; 2al; z) show that
(237)
( kz 1a
F(l) = (z) 2(2al)(lz)2Wa +(a2k)Wa k
(2a ( (2z) L
Apply (231) to F(l):
F(al, a+2k1; 2al; z)
1 [(l)P(a, a+2kl; 2al; z)
a+1 1
+ (22a)F(al, a+2kl; 2a2; z)]
= F(a, a+2kl; 2al; z) + 2F(j1, a+2kl; 2a2; z).
Now apply (232) to the first term with the result that
F(l) 1 [(1) + (2ka)z F(a, a+2kl; 2a; z)]
1z 2a1
+ 2F(a1, a+2kl; 2a2; z).
Collecting the terms containing F(l) we have
(238) 1 + 1 F(l) a2k z F(a, a+2kl; 2a; z)
1z 2a1 1z
+ 2F(al, a+2kl; 2a2; z).
Now, since
(239) F(a, a+2k; 2a; z) = za(lz) k2W
l,k
if k is replaced by Ic , then
F(a, a+2kl; 2a; z) = za(lz)kW
If a is replaced by a1 in (239), then
F(al, a+2kl; 2a2; z) = za+1 (z)k 1k
Using these two results and solving equation (238)
for F(l), permits the result
(1z) a2k _z za(lz)kW !
(2z) L2a1 1z aK2
+ 2z (lz)k 2 k]
Sz ( j (a W 1 + 2(lz)\
(2z) 2al ak2 a1k
or
F() (z) z (a2k)W' k1 + 2(2al)(lz)2W
(2a1)(2z) ) a 2 a1,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 a1 in equation (230), we may
write
(240) W = z1(lz)k+ F(a1, (+2k1; 2a2; 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 (232) to obtain the valid
relationship
F 1 F(b)+ (ac)z F(c+)
lz ce t
Now, if this be applied to (240), we may conclude that
(241) Wa1,k = zal(1z) 2 F(al, a+2k2; 2a2; z)
+ (1a)z F(l)]
2a2 *
Replacing a by a 1 and k by k  in (230), it is seen
that
F(a1, a+2k2; 2a2; z) = za(lz)k a,k"
This, together with (237) enables us to write (241) in
the form
a1 k 1a k
Wa = z (1z) z (z) Wa,k
1 (z)kz2a 2(2a1)(l1z)2W
2(2a1) (2z) al,k
+ (a2k)Wak
= (1z) W Z1 W
=a1,k (2z) a1,k
(a2k) z(lz) W ak
2(2a1)(2z)
Then
1 + z W= (Iz1_ Wa,k + (a2k)z w,, i
2z 2 alk2(2a1)(z2) ak
Solving for W 1,k, we have the recurrence relation
Wak (2z) z (a2k)z
al,k 2 L2(2al)(z2)
(242) (z 2(2a1)(z2) Wa
(22) Wa,k 4(12a) 1,k2
+ (a2k)z W kl]
Another recurrence relation is found by first
replacing a by a + 1 in (230), giving us
W a+,k = Z+l(1lz)k+2 F(a+l, a+2k+l; 2a+2; z).
Then apply (233) 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 (231) is applied to each term, we may write
Wa+ (2a+l) (Za(lz) 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
(243) 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 (235), letting n = 1, we have
F(+1) = 2z (lz) [W ,k+ (iz)YW ]
k ,k ,
which can be used to rewrite the first and third terms of
(243). Hence,
(2a+l) z,(lz)k+f 2za1 z)k1 W
S z z2z) k
W+,k a k+
1 2
(12) Wz + + 2a(1)1 W ,+
+ 2z' (1z) l[wa, (1z)l W a, j
2z'(lz) 2 W
f WI, 
2(2a+l) [ 1 z) W1 +
a a,2+y
1Z1
( lz) Wk .]
Thus, we have a second recurrence relation
(244 ) axzW C,1j = 2(2a+1) (1z)2 W Cx, kj + (2z) W a~k
To obtain the nth derivative of our solution (230)
Wa,k = za(lz)k+2 F(a, a+2k; 2a; z),
we shall apply Leibniz's rule, where u = z(lz)k+2 and
v = P(a, a+2k; 2a; z).
Thus, using (234) thru (236)
we may write
n
D W ak t= D [z(1z)k] D"o [F(a, a+2k; 2a; z)]
t,=o
n 4
SZ [()"(l,z)kt+ (, ) z (1z)
t=o J=o
(a) (a+2k)
(2n ) nt
nt
= a(lzz)k
t=o
j=o
f+_ _nlCA 4 3
S*(a+2k) z2 ( i ( )2 / Wa,k+ n .4
J=o
Therefore,
(245) DnW ,k = zn(z)
i(lz) zt(a+2k)n
'c=0
* () a kJj (1
J=o
n4
S (l)j n (1z) Wk+ n3i
o a0k
F(+nLZ)
(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
(246) f(y+x) = f () (x) n .
n=o
Using (246), replace y by (yl)x to obtain
00
(247) 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:
(248) Wk (x+y) = W( (x)
n n
= ji x (x(1x (a+2k)
n=o t=o
J=o
n, J n
S (1 J) n ) ( ) a,k+ nj
j=o
and
(249) Wa,k(xy) = (x) ( nn
n=O
Sn(y )n (1x) 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 rarB3 du 3rr Az+B
(32) 2 + (ar+)2 + + A = 0.
dz 2 za (zar) 3(za
where A is such that tj and L2 are the roots of
r=1 r=1
To verify this, we can begin with equation (21) and let
a4 become infinite, thus obtaining
3 3
u [ 1arr + H(z) u' + r +  + J(z) u=0,
+a + H I za ) za
r=1 r r=1 +(Zar Zar
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 111 a2 3I 3 1
22 p + + + L1
z1 \Z1 Z1 1 a  a a
z1 1 z1 2 z1 3
1a lp1 1a 2P2 1a3 i 1
l[la11 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_
l2alz +a 1z l2a 2z+az1 12a3z +a1
1 i 2 3 1 
z 1a z1 1a2z 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
zal za za
= 2 1
S1 ]2 3 (za )(za )(za 3
where A and B are functions of mr and ar (r = 1,2,3). But
since ml m +m3 = 0, we have equation (31). In order to
show that A must satisfy the condition (32), we can make
the substitution z = in (31), 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 + blz1 +.
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 
(za,) za,)( z 1 ,
z
1 a a a
(za )(za2)(za3) 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
1a Br a a
+  + r +...)
r= z z z
r=l
1
A1 b2
u' =b Az ? +b(l)z +...
u" = bo (h+l)zN2
+ b, (+l)(,+2)z3..
1. .
3 3
z : (h41) h (larr) + arr + A = 0.
r=l r=l
3 3
N2 + 1[i (1arPr)] 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 (31)
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 (32) 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 +n1) = A n(nin ) n(n+l).
Substituting these values into (31), the result is
Lame's Equation:
du 2 du n(n+l)z + h
dz2 + za dz i3 u 
dz r 4 (za )
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 (31). Since the product 1,2 = = ar r
r=l
+ A from (32), we have A = (alP1 + a2p2). Hence,
d2u 1 du 1 al1I + a2 2
++ L)+ + +
2 dz 2
dz z z1z
(a lf + a P2 )Z 2 n(n+l) + a 1P + az2 2 u 0.
+   ^L 1 2 2 u = 0.
z (zl)
(33) d2 + ] +n(+l u = 0.
dz2 + z z1 dz 4z2(zl)
dz kz2(z1)
