j Study of a Linear Ordinary Second Order
rental Equation With Five Regular
Singular Points
By
HENRY LAWRENCE CROWSON
\I *
M PR
THE
HEGRADUATE COUNCIL OF
OF FLORIDA
OF THE REQUIREMENTS FOR THE
CTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
June, 1959
'* I"
* /I (
This Dissertation is affectionately
dedicated to my wife
Betty
PREFACE
In this dissertation, certain general aspects of
linear ordinary second order differential equations are
initially discussed. Attention is briefly devoted to
establishing criteria under which it is possible to discern
whether or not the point at infinity is an ordinary point
or a regular singular point of a given differential equation.
This is followed by a derivation of the canonical form of a
differential equation with three regular singular points.
Certain constants in this equation are specialized, and the
result is an equation with two regular singular points.
A closed form solution of this latter equation is then given.
Hypergeometric functions are deduced from a study
of a solution of the hypergeometric equation. Criteria for
convergence of hypergeometric functions are also established.
In Chapter II, an equation with five elementary
singular points is introduced. This is followed by a deri
vation and brief discussion of the respective equations of
Lame, Mathieu, Legendre, Bessel, Weber, and Stokes.
In Chapter III, an equation with nregular singular
points is discussed, and certain equations are derived which
are used in Chapters IV and V. A brief discussion of Heun's
equation is given in Chapter IV, which includes its deriva
tion and solution in a general case.
111
A derivation of a general differential equation,
with five nonelementary regular singular points, is given
in Chapter V. Information gained from ChaptersIII and IV
prove useful in this derivation. This general differential
equation is then solved in a general case in Chapter VI.
An exposition of Scheffe's criteria, as applied to a
linear ordinary second order differential equation, is given
in Chapter VII. These criteria are then applied to the equa
tion derived in Chapter V. The result is Chapter VIII. In
this chapter, solutions in a neighborhood of each of the reg
ular singular points z = 0, 1, a, b, and infinity are given
in terms of hypergeometric functions. A solution in a neigh
borhood of z = d, where d is an ordinary point, is also ex
pressed in terms of hypergeometric functions. Chapter VIII
contains certain confluent equations which are classified
in the notation of Ince.
The majority of Chapters V, VI, and VIII is believed
to be new. The other chapters, while not original, are never
theless considered necessary for a logical development of
the study.
Many persons have been helpful in the preparation of
this dissertation. In particular, the writer is most grate
ful to the chairman of his supervisory committee, Professor
Russell W. Cowan, who not only suggested the study, but was
also a constant source of inspiration and guidance. He also
expresses his appreciation to Dr. D. E. South,
Dr. W. R. Hutcherson, Dr. J. T. Moore, and Dr. R. B. Bennett,
who, as members of his committee, were encouraging and help
ful in all phases of his graduate program.
The writer also gratefully acknowledges the inval
uable assistance rendered by Dr. F. W. Kokomoor, Head of the
Mathematics Department.
A note of special thanks is due his typist,
Mrs. Edna Larrick, for her cheerfulness, patience, and
thoroughness while typing this dissertation. He also thanks
Mrs. Jeane Singletary for the adept manner in which she in
serted certain Greek letters into the dissertation.
Finally the writer acknowledges with deep humility
the many times his wife encouraged him in this undertaking.
Henry L. Crowson
Gainesville, Florida
June 8, 1959
TABLE OF CONTENTS
PREFACE . . . . . . . . . . .
Chapter
I. INTRODUCTION . . . . . . . .
Some General Considerations of Linear
Ordinary Differential Equations . .
Differential Equations with Three
Regular Singular Points . . .
Differential Equations with Two
Regular Singular Points . . . .
Hypergeometric Functions . . .
II. DERIVATION OF A DIFFERENTIAL EQUATION WITH
FIVE REGULAR SINGULAR POINTS, HAVING 1/2
AS THE EXPONENT DIFFERENCE AT EACH
SINGULARITY . . . . . .
Lamae's Equation . . . . .
Mathieu's Equation . . . . . .
Legendre's Equation . . . .
Bessel's Equation . . . .
Weber's Equation . . . . .
Stokes' Equation . . . . .
III. SOME CONSIDERATIONS OF A DIFFERENTIAL
EQUATION WITH nREGULAR SINGULAR POINTS
IV. A DISCUSSION OF HEUN'S DIFFERENTIAL
EQUATION . . . . . . . .
V. A DERIVATION OF A GENERAL DIFFERENTIAL
EQUATION WITH FIVE REGULAR SINGULAR
POINTS . . . . . . . .
Page
iii
. 1
. 1
. 9
S 15
S 21
28
S 41
42
S 48
S 52
S 54
S 57
S 61
74
vii
TABLE OF CONTENTS
Page
Chapter
VI. A GENERAL SOLUTION OF A GENERAL DIFFERENTIAL
EQUATION WITH FIVE REGULAR SINGULAR POINTS
VII. ON SCHEFFE'S CRITERIA . . . . . .
VIII. ON THE APPLICATION OF SCHEFFE'S CRITERIA
TO THE EQUATION WITH FIVE REGULAR
SINGULAR POINTS . . . . . . .
Solution in a Neighborhood of z 0 . .
Solution in a Neighborhood of z = 1 . .
Solution in a Neighborhood of z = a,
where a + O,1,b, or oo .. . .....
Solution in a Neighborhood of z = b,
where b + 0,l,a, or oo . . . . .
Solution in a Neighborhood of the
Singular Point at Infinity . . . .
Solution in a Neighborhood of z = d,
where d + 0,1,a,b, or oo . . .
BIBLIOGRAPHY . . . . . . . . . .
BIOGRAPHICAL SKETCH . . . . . . . .
98
104
115
115
137
163
190
218
231
244
245
w
CHAPTER I
INTRODUCTION
Some General Considerations of
Linear Ordinary Differential Equations
A linear ordinary second order differential equation
is frequently written in the form:
2
d2u du
(1) + p(z) + q(z)u = 0.
dz z
We assume u and z to be complex variables with u
designated as the dependent variable and z the independent
variable. We further assume the existence of a closed
region S in the complex plane throughout which both p(z)
and q(z) are analytic [14, p. 83 ] except at a finite
number of poles [7, p. 38 ].1
If p(z) and q(z) are both analytic at some point
z = c in the region S, then z = c is designated as an
ordinary point of S. Other points in S are designated as
singular points. Singular points of p(z) and q(z) are
either regular singular points or irregular singular points.
We define a point z = c, which is in S and not on
the boundary of S, to be a finite regular singular point
of the differential equation (1) if both (zc)p(z) and
(zc)2q(z) are analytic at From this definition it
(zc) q(z) are analytic at c. From this definition it
Numbers in brackets refer to the bibliography.
2
follows that p(z) has a simple pole at c, and q(z) has a
pole of order 2 at c. That is, we can represent p(z) and
q(z) by the following series expansion:
(2) p(z) = 1 + p + pl(zc) + ** + n(zc) +
and
(3) q(z) 2 + = + q+ q(zc) +'**+ ql(zc)n +
(zc)2 zo
where p (n 1,0,1,2,...) and qn, (n = 2,1,0,1,2,...)
are constants, and not all of the quantities p1, q2 and
q1 can be zero.
We now define
(4) P(zc) (zc) p(z), and
(5) Q(zc) = (zc) q(z), so that from equations (2)
and (3) we get:
+ pn(zc) + ** and
2
(7) Q(zc) q2+ q (zo) + qo(zc) + ***
+ qn(zc)n+2 + *
In summation notation, equations (6) and (7) become:
(8) P(zc) = Z (P )(z) and
j0
oo
(9) Q(zo) = E (Q )(zc) where Pj and Q are constants.
From equations (6) (9) we note that P(zo) and Q(zc) are
3
both analytic at z = c, and further that Pj = Pj1, and
Qj = qj2
We may substitute equations (4) and (5) into
equation (1), and rearrange slightly to obtain:
(10) (zc)2 d2 + (zc)P(zc)du + Q(zc)u = 0.
dz2 dz
We will now investigate the conditions under which
infinity is either an ordinary point or a regular singular
point. Initially we make the variable transformation
(11) z = l/w, following which we will investigate the
behavior of p(z) and q(z) in a neighborhood of w = 0.
We note the following equations:
du du dw
(12) dz d= T hence from equation (11)
dz dw dz
(13) dz 2 dw Also from equation (12) we have:
dz z2 dw
2 2 2 2
(14) d2u du d w + (dw) d From equation (11)
dz2 dw dz2 dz dw2
we now obtain:
(15) d2u du +_ d2u
dz2 z3 dw z4 dw2
Substituting equations (13) and (15) into equation
(1) yields the following expression:
(16)
(17)
if both
w 0.
w4 ddwu + 2wq dul u=0
dw2 + 2w p(l/w)[w2 u] + q(l/w)u = 0.
Simplifying equation (16) yields:
du + [2/w p(l/)] + q(l/w)u = 0.
dw2 [ w dw w4
Now w = 0 will be an ordinary point of equation (17)
[2/w p(l/w)] and q(l/w) are analytic at
w2! w4
The condition [2/w 1 p(l/w)] be analytic at
w2
w = 0 will be met if p(l/w) is expressed as follows:
(18) p(l/w) = 2w + p w2 + + p wn +
From equation (18) it follows that
(19) [2/w  p(l/w)] = p2 p3w p w2
pnwn2 .
This is clearly analytic at w = 0 for finite p2,
where'pn, (n = 2,3,'"') are constants.
Further, the condition 1 q(l/w) be analytic at
w = 0 will be met if q(l/w) is expressed as follows:
(20) q(l/w) = qw4 + q5w5 + w. + qnwn + ...; q4 + 0
From equation (20) we obtain:
(21) q(l/w) = q + qw + + q w n++ .** where
w4 4 5 n
5
qn, (n = 4,5...) are constants. By saying an equation is
analytic at a point, we will mean the function defined by
the equation is analytic at that point. Thus, equation (21)
is analytic at w 0.
Substituting equation (11) into equations (18) and
(20), we conclude that infinity will be an ordinary point
of equation (1) provided
(22) p(z) 2/z p2/z2 + .'. + pn/zn + and
(23) q(z) q4/z4 + q5/z5 + *** + qn/z" + .
We will now investigate the conditions under which
infinity is a regular singular point.
From equation (17), if it is true that both
w[2/w  p(l/w)] and W2 1/w ) are analytic at w = 0, and
w2 w4
that either (2/w L p(l/w)3 or (/w) or both fail to be
w2 w4
analytic at w = O, then w = 0 is a regular singular point
of equation (17). We will use this definition consistently.
This implies, from equation (11), that infinity is a regu
lar singular point of equation (1).
Assume an expansion for p(l/w) of the form:
(24) p(l/w) plw + p2w2 + + pnwn + Then
(25) w[2/w  p(l/w)] 2 p p2w
w2 pwn 2 "
Pnwn1
Equation (25) is clearly analytic at w 0, for finite pl
which implies that
(26) zp(z) pl + P2/z + *** + pn/zn1 +  is analytic
at infinity. Again we note that pn, (n = 1,2,...), are
constants.
We next consider L1w ), by first assuming an
expansion of q(l/w) of the form:
(27) q(l/w) q2w2 + q w3 + w n+ .. where not
all of the quantities pl, q2, and q3 are zero.
Then from equation (27) we get:
(28w)!9i 2 + qw + "'" + qawn2 + ".
Equation (28) is evidently analytic at w = 0, so
from equations (11) and (28) we conclude that
(29) z2q(z) q2 + q3/z + + qn/zn2 + *** is analytic
at infinity.
We may summarize the above statements concerning
ordinary points and regular singular points of equation (1)
as follows:
z=c is a finite ordinary point if both p(z) and
q(z) are analytic at z = c. z = o is a finite regular
singular point if both (zc)p(z) and (zc)2q(z) are analytic
at z = c, and if either p(z) or q(z) or both fail to be ana
lytic at z = c.
Next, let z = l/w, then consider the expressions
(2/w p(l/w)] and i. If both of these expres
sions are analytic at w = 0, then w = 0 is an ordinary
point of equation (17). This implies that infinity is an
ordinary point of equation (1). Again we consider the
expressions w[2/w  p(l/w)] and 2qi/W). If both of
these expressions are analytic at w = 0, then w = 0 is
a regular singular point of equation (17). This implies
that infinity is a regular singular point of equation (1).
Points in the extended zplane which do not conform
to the above criteria are designated as irregular singular
points of equation (1).
To further illustrate the conditions under which
infinity is either an ordinary point or a regular singular
point, we give a concrete example. Thus, let p(z) in
equation (1) be a rational function of the form:
k
k k1
E Aizi
(30) p(z) = ; where (A )(B) + 0.
m 0 0
E B zmi
i10
Dividing numerator and denominator of equation (30)
m
by Z B zm gives us:
i0
(31) p(z) (A /Bo)zkm + ...
Again from equation (1), let q(z) be a rational
function of the form:
n
Cizn1
(32) q(z) 10 ; where C + 0.
m
Z Boz z
im0
We may rewrite equation (32) thus:
(33) q(z) = (Co/Bo)znm + .*
Now infinity will be an ordinary point of equation
(1) if both p(z) and q(z) are analytic at infinity. These
conditions are fulfilled, from equations (31) and (33),
if both km < 0 and nm < 0.
Again from equations (31) and (33),
(34) zp(z) (A /B)zkm+l + *** and
(35) z2q(z) (Co/Bo)znm+2 + ..
Thus infinity is a regular singular point of
equation (1) if both zp(z) and z2q(z) are analytic at
infinity. These conditions are met, from equations (34)
and (35) if both km+l < 0 and nm+2 < 0, and if either
p(z) or q(z) or both fail to be analytic at infinity.
The above fundamental considerations, while not
extensive, are nonetheless pertinent to the future develop
ment of this study. Other applicable topics are developed
as the study progresses.
We will now apply the foregoing theory to a dis
cussion of a differential equation with three regular
singular points.
Differential Equations with
Three Regular Singular Points
Let equation (1) have three finite regular singu
lar points z = a, z = b and z = c. We assume that equation
(1) has no other singular points, and that the points a, b,
and c, have respective exponents a, a'; p, p'; and y, '
Then p(z) is a rational function with simple poles at a, b,
and c, and q(z) is a rational function with poles of order
2 at each of a, b, and c.
We may now write p(z) as follows:
E2z2 + E1z + E
(36) p(z) = (za)(zb)(zc) or expanding equation (36)
in partial fractions yields:
(37) p(z) = G1/(za) + G2/(zb) + G3/(zc).
We may write an expression similar to equation (36)
for q(z), thus obtaining:
F z2 + F1z +o 1
(38) q(z) = (za)(zb)(zc) (za)(zb)(zc)
We may expand equation (38) in partial fraction
to get:
(39) q(z) = [H1/(za) + H2/(zb) + H3/(zc)]
*[l/(za)(zb)(zc)].
Substituting equations (37) and (39) into equation
(1) gives us the following expression:
(40) 2 + [Gl/(za) + G2/(zb) + G /(zc)] z
+ [H1/(za) + H2/(zb) + H3/(zc)]
*[l/(za)(zb)(zc)]u = 0.
We now assume a solution of equation (40) in a
neighborhood of z = a to be of the form:
A 00 X+r
(41) u(z) = c0(za) + Z cr(za) ; c0 0.
r=l
Differentiating equation (41) with respect to
(za) yields:
X1 oo X+r1
(42) u'(z) = co,(za) + E cr(X+r)(za)
r=l
du
where it is understood that u'(z) = d(za)
Differentiating equation (42) with respect to
(za) yields the following expression:
X2 00 A+r2
(43) u"(z) = co(l)(za) + ZE r(c+r)(A+rl)(za)
r=l
We now substitute equations (41), (42) and (43)
into equation (40) to get:
X2 0o A+r2
(44) CoMh(l)(za) + Z Cr(x+r)(N+r1)(za)
r=l
Gi+ G2 G Nl 00 X+ r1
+ + c(za) + Ec (A+r)(za)
SLa zb
L r =l
+ H/(za) + H2/(zb) + H3/(zc)
(za)(zb)(zc)
A co 00+r
*[Co(za) + Cr(za) ] = 0.
r=l
Inasmuch as we are expanding in ascending powers
X2
of (za), we equate the coefficient of (za) to zero
thus obtaining the indicial equation:
(45) CoX(hl) + Co0Gl + cok = 0, where k is to be
determined from the coefficient of u(z) in equation (44).
We will now compute k.
We write equation (39) as follows:
Hi I H2 + +
(46) q(z) 1 =2H + 1
(za)2 (zb) (zc) zb zc (za)(zb)(zc)
Let
(47) f(z) = (zb)(z) and expand in a Taylors series
about z a to get:
o0 (za)n f(n)(
(48) f(z) = f(a) + za) f (a)
n=l n.
Substitute equation (48) into equation (46) to get:
H1f(a) oo n2 (n)
(49) q(z) = a) + H1 Z (za) (a)
(za) n=1 n!
+ [H2/(zb) + H3/(zc)][l/(za)(zb)(zc)].
Substituting equation (49) into (44), where f(a) is
computed from equation (47), gives the following expression:
X2 oo X+r2
(50) CoX(X))(za) + E cr(N+r)(X+r1)(za)
r=l
rG G G X1 oo X+r
+ + + coh(za) + E or(+r)(za)
za zb z r=
Soo n2 (n)
+ {H1/(ab)(ac)(za) + H (z fn (a)
n1l n!
+ [H2/(zb) + H3/(zc)][1/(za)(zb)(zc)]}{c0(za)
oo X+r
+ oCr(za) } = 0.
r=l
The indicial equation is now complete. By inspec
tion of equation (50), we note that
(51) k = Hl/(ab)(ac). Substitution into equation (45)
and then division of both members by cogives, upon slight
rearrangement:
(52) X2 + X(G1 1) + H1/(ab)(ac) = 0.
The roots of equation (52) are the exponents of the
solution to equation (40) in a neighborhood of z = a. Hence
from equation (52) we immediately deduce
(53) 1 G1 = + a', and
(54) Hi/(ab)(ac) = as'.
Solving equations (53) and (54) we have:
(55) G1 = 1 a a', and
(56) H1 = a'L (ab)(ac).
Similarly, solving equation (40) in a neighborhood
of z = b and z = c, respectively, yields:
(57) G2 = 1 p p' ,
(58) H2 = Bp' (ba)(bc),
(59) G3 = 1 y and
(60) H3 = YY' (ca)(cb).
Substituting equations (55) (60) into equation (40)
we obtain:
r d d2u  181' 1 qy'du
(6) dz2 za zbzc dz
+ raa'(ab)(ac) + BD '(ba)(bc)
z a zb
+ a'(ca)(cbn. u 0 = o
z c (za)(zb)(zc)J
Equation (61) is the canonical form of a second
order linear ordinary differential equation with three
finite regular singular points.
This canonical form is due to Papperitz [10, p. 213].
We note that equation (61) is determined completely
and uniquely by the singular points and the exponents at the
singular points.
Riemann [12, vol. 7] expressed a solution to
equation (61) as follows:
Sa b c
(62) u = P L p z
a' B' Y'
The singular points are in the first row, with
corresponding exponents, at the respective singular points,
in the second and third rows. The independent variable is
in the fourth column.
Equation (62) is called Riemann's Pequation.
We will now specialize certain of the restrictions
attached to equation (61) and hence derive a differential
equation with two regular singular points. This equation
will then be solved in closed form.
Differential Equations with Two
Regular Singular Points
In equation (61) we let z = c be an ordinary point.
Then both p(z) and q(z) are analytic at c. It then follows
that
(63) 1 y y' = 0 and
(64) yy' = 0, for otherwise z = c is a regular
singular point. Equation (61) now becomes:
(65) du + [(laac)/(za) + (1p p')/(zb)]
dz2 dz
+ [aa'(ab)(ac)/(za) + Mp'(bc)(bay(zb)]
*[u/(za)(zb)(zc)] = 0.
Now in order for z = c to be an ordinary point
z c must divide the coefficient of u/(za)(zb)(zc).
That is,
(66) Q) = aa'(ab)(ac)(zb) + Bp'(bc)(ba)(za)
(z a)(zb)
must be divisible by z c, where Q(z) m (za)(zb)(zc)q(z).
From equation (66) we deduce that
Q(c) aa'(ab)(ac)(cb) + pB'(bc)(ba)(ca) 0
(c a)(c b)
and hence we get:
(67) aLa' = p'.
In Chapter III we will show that for a linear ordin
ary second order differential equation with regular singular
points ap, (r 1,2,...,n), and exponents at these singular
points of ar and r,, (r = 1,2,...,n), then it is true that
n
E (ar + 0 ) = n2.
r=l
Applying this expression to equation (61) we note
that
(68) a + a' + p + P' + y + = but from equation (63)
this simplifies to:
(69) a .+ ca' + p + B' = 0. We will use this relationship
presently.
Now substitute equation (67) into equation (66),
then substitute the result into equation (65) to get:
d2u du
(70) dz + [(laa')/(za) + (lpp')/(zb)] d
dz
+ aa'[(ab)(ac)(zb) + (bc)(ba)(za)]u = 0.
(za)2(zb)2(zc)
Equation (70) simplifies to:
(71) du + [(1aa')/(za) + (1pp')/(zb)] d
dz2
+ [aa'(ab)2/(za)2(zb)2]u = 0.
In order to solve equation (71) we first make the
variable transformation:
w = (za)/(zb) from which it follows that
z = (bw a)/(wl),
za w(ba)/(wl), and
zb = (ba)/(wl).
Now from equations (12),
(14) and (72) we get:
(76) dz (wl) 2/(ab)], and
d.2u du 3 2] d2u[ 4 2
(77) 2 = 2(w1) /(ab) + (wl)/(ab)].
Substituting equations (74) (77) into equation (72)
yields the following equation:
du 3 2 d2u[ 2]
(78) 2(w1) /(ab)2 + [(wl) /(ab)2
aw dw2
+ (w1) 4
S w(ba) (ba)
2 (
w 2(ba)4 u = O.
Sw (ba) a
We now rearrange equation (79) as follows:
4 2
(wl) d u
(ab)2 dw2
+ 2(w)3.du
(ab)2 dw
( wl)3(la')/w + (+1p')1a
(ab)
+ (wl) (a')u = 0.
wiab) 2
(72)
(73)
(74)
(75)
(79)
Dividing both members of equation (79) by
(wb)2 we obtain:
(ab)2
d2u du
(80) [l/(wl)].[(1aa')/w + (p')
+ [2/(wl)]u + (aa'/w2)u = 0.
We simplify equation (80) further as follows:
d2u du
(81) d2u 1 [(1aa') + w(lpp') 2w]dw
dw w(wl)
+ (aa'/w2)u = 0.
Multiplying both members of equation (81) by
w2(w1) to get:
(g2) w2(wl)d + [2(1++p') w(la')]
dw
+ aa'(wl)u = 0.
We now solve equations (67) and (69) simultaneously
to get p = a, and p' = a', hence equation (82) becomes:
2
(83) w2(wl) + [(2w)(1aa')]du + aa'(wl)u = 0.
dw
Since we are assuming a solution in the finite
plane, division by (wl) is permissible, and equation (83)
becomes:
(84) w2 2 + [(laa')(w)] + C'u 0.
dw2 dw
Let D then equation (84) becomes:
w dw
(85) [w2D2 + (laa')wD, + Ma']u = 0.
w
We now let
(86) w i ex, then equation (85) becomes
(87) [Dx(Dx1) + (laa')Dx + aa']u 0, where
D a
x dx
Upon simplification, equation (87) becomes:
(88) [(Dx a)(Dx a')]u = 0.
A general solution of equation (88) is:
(89) u K eax + K2C 'X, where K1 and K2 are arbitrary
constants of integration. [8, p. 18]
(90) Now w= *eax and wC = e 'x, hence combining equa
tion (90) and (72), we see that equation (89) becomes:
(91) u(z) = K 1(za)/(zb)]L + K2 (za)/(zb)]C'.
Equation (91) is the desired solution of equation (71).
Now if we suppose a a', then equation (88) becomes:
(92) (D _ a)2 u 0, and we assume a solution of the form:
(93) u teax, where t and x are complex variables. From
equation (93) we will determine t such that equation (93) is
a general solution of equation (85) with a a'.
Substituting equation (93) into equation (92) we have:
(94) (D a)2(eax.t) eaxD2t.
(95) Now if D2t = 0, then eaxt will satisfy equation (92).
x
Solving equation (95) we get:
(96) t = K + K4x, and hence from equation (93),
(97) u K3eax+K KxeaX.
'Prom equation (86) we have
x In w, and substitution into equation (97) gives
us the following:
u = K e In w + Klan we In w, so that
(98) u wa + K waln w.
Substituting equation (72) into equation (98) gives
us the final result:
(99) u(z) K Z a + K 4 n[& .
3 zb L.bj zb
Equation (99) is a solution in closed form of a linear ordin
ary second order differential equation with two finite regular
singular points.
We will now discuss a particular differential equa
tion from which we will derive the hypergeometric series,
and then define the hypergeometric function.
Hypergeometric Functions
We will now solve a particular differential equation,
thus:
d2u [c (a+b+l)zl du ab
(l00) 2d z(lz) dz z(1z) u = 0.
dz
Multiplying both members of equation (100) by
z(1z) gives the hypergeometric equation:
d2u du
(101) z(1z)2 + [c (a+b+l)z]u abu = 0.
dz2 dz
We now assume a solution of equation (101) in a
neighborhood of z = 0 to be of the form:
oo
u(z) = Z Crz+ c o 0, or
r0O
(102) u(z) = coz + clz+1 + ** + crlX+r1+ crz+r+..
Differentiating equation (102) with respect to z
gives:
(103) u'(z) z cozX1 + cl(X+l)z + + crl(X+rl)z+r2
+ Cp(A+r)z+ + r.
Differentiating equation (103) with respect to z
yields:
(104) u"(z) = CoX(Al)z2 + cl(x+l)(A)zX1 + .
+ c rl(A+r1)(A+r2)z+r
+ 0 (X+r)(X+r)z+r2 + ...
Substituting equations (102) 104) into
equation (101) gives us:
(105) z(lz)[c,(N1)zX2 + cl(1+l)(A)z 1 + *
+ C r1(+r1)(A+r2)zh+r'3
+ c r(+r)(X+r1)zX+r2 + ...
+ [c (a+b+l)z][cohzA1 + cl(A+l)zx + *
+ crl(A+r)z+r2 + Cr(A+r)z+r1 +...
ab[coZ + ciz+ ** + rz z+r1+] = 0.
We now get the indicial equation by equating the
coefficient of z1 to zero:
(106) Coh(Nl) + cohc = 0.
Solving equation (106) yields:
(107) A = 0, 1c.
To get the recurrence equation we equate the coeffi
cient of zN+r1 to zero thus:
(108) Cr(N+r)(N+r1) crl(X+rl)(N+r2) + cr( +r)(c)
(a+b+l)crl(+r1) abcr1 = 0.
Solving equation (108) for or gives the desired
recurrence equation as follows:
(109) cO = [(X+r1)(N+r2) + (a+b+l)(A+r1) + ab]cr1
(A+r)(X+r1) +c(A+r)
Equation (109) simplifies to:
(110) (N+rl+a) (+rl+b) cr,
Or = (X+r)(X+rl+c)
In equation (112) we let X = 0 and set r = 1,thus
obtaining:
(111) cl =' co.
Let r 2 in equation (110) to get
(a+l)(b+l)
2 = 2(c+l)( c1 and substituting cl from
equation (111) yields:
(112) c = (a)(a+l)(b)(b+l)
(2 1"2 (c)(c+l) o
We now generalize by letting r = n in equation (110),
and in conjunction with the formation of equation (112), we
deduce the following equation:
(a)(a+l)*'(a+n1)()(b)(+l)''(b+n1)co
(113) c =
n n! (c)(c+l)***(c+n1)
Substituting equation (113) into equation (102)
gives a general solution of equation (101) in a neighbor
hood of z = 0 whenever X = 0.
This solution is
oo
(114) ul(z) = Kl[1 + + (a)(a+L).''(a+n1)(b)(b+1)...(b+n1)Zn]
n=l n! (o)(c+l)**'(c+n1)
where K1 is an arbitrary constant of integration.
Similar to the above analysis, we note that when
X = 1c, equation (102) becomes:
(115) u2(z) = K2zlc[ + (ac+l)* '(ac+n)(bc+l) *'(bc+n)zn]
n=l n! (2c)(3c)***(n+lc)
w ,a
A general solution of equation (101) in a neighbor
hood of z = 0 is a linear combination of equations (114)
and (115), thus:
(116) u(z) = {K1[1 + Oa(a+l)*'(a+n1)(b)(b+l)* '(b+n1)zn
n=l n! (c)(o+l)*"(c+n1)
+ K2zlC[ + z (ac+l) '(ac+n)(bc+1) (bc+n)zn]bc
2 n=l n! (2c)(3c) ".(n+lc)
We will now investigate the conditions under which
the series given by equation (114) converges.
Let
(117) Sl 1 + E (a)(a+l) **"(a+n1)(b)(b+l)'**(b+nl)zn
n= n! (c)(c+l)'..(c+n1)
and let vn represent the nth term in series (117). Then
by the ratio test:
n+l (a+n)(b+nl) or
v I n(c+n1),
(ll) (1+ 1)(l + b1)
Vn 1+ C1
n
Now from equation (118),
(119) Lim Vn1 z. Hence the series given
n z o Vn
by equation (117) is absolutely and uniformly convergent
if Izl < 1.
We will now establish the conditions under which
series (117) is convergent if Iz = 1.
Let Iz 1, then equation (118) becomes:
(120) ,Vl I [1 + a][1 + bl 1 + 0(1/n2)]
'n n n n
In deriving equation (120), we note that
1 1 + 0(1/n2)
n
Equation (120) oan be written:
n+1 1 + a+b2 1 1 + 0(1/n2)
Tnz n n
1 1 + a+b2 + 0(1/n2
n n
(121) 1 + n+b + 0(1/n2).
vn n
Now writing out the numbers a, b, and c, we get
a x1 + iyl, b = x2 + y2, and c x + iy3, where i2 j 2 .
Substituting into equation (121) we get:
vn+1 x+ + x (y+y2y3) 0
n 1++ + 0(1/n2)
Vn n n
{[1 + (x1+X2x1)/n]2 + [(y1+y2y3)/n]2 + 0(l/n2)}1/2.
(122) + {[1 + (zl+x2x31)/n]2 + 0(1/n2 )1/2
n
Expanding equation (122) in a binomial expansion
gives the following expression:
(123) v 1 + (x+x2x3l)/n + 0(1/n2)
Now S1 as given by equation (117) will be absolutely
convergent for Izl 1 whenever xl+x2x3l < 1, that is
whenever x1+x2x3 < 0.
Expressed in a slightly different way, series (117)
will be absolutely convergent when IzI = 1 if Re (a+bc) < 0.
[14, p. 241.
From the analysis of the convergence of equation (117),
it follows that series (117) defines a function which is ana
lytic when Izl < 1. This function is the hypergeometric
function and is defined as follows:
(124) F(a,b;c;z) =1+ z (a)(a+l)*..(a+nl)((b)(b+)..(b+n1) n
nl n! (c)(c+l) *'(o+n1)
Comparing equation (116) with equation (124) we see
that the solution of equation (101) may be written in terms
of the hypergeometric function. The result is:
(125) u(z) = KlF(a,b;o;z) + K2zl'OF(ac+l, bc+l; 20; z).
This is the solution of equation (101) in a neigh
borhood of z 0, expressed in terms of the hypergeometric
function.
Solutions at the other regular singular points,
unity and infinity, may be derived in a manner similar to the
foregoing.
Some important functions which can be expressed in
terms of hypergeometric functions are as follows:
(126) (1+z)n = F(n,b;b;z),
(127) In(l+z) = zF(l,l;2;z), and
(128) ez = Lim F(l,b;l;z/b).
b  oo
This has been a brief introduction to hypergeometric
functions. However, they will be reviewed and used exten
sively in Chapter VIII when we solve certain differential
equations with nonconfluent regular singular points.
With the introductory material of Chapter I at our
disposal, we are now ready to survey a linear ordinary second
order differential equation with five regular singular points,
and certain special cases. This will be accomplished in
Chapter II.
CHAPTER II
DERIVATION OF A DIFFERENTIAL EQUATION
WITH FIVE REGULAR SINGULAR POINTS,
HAVING 1/2 AS THE EXPONENT DIFFERENCE
AT EACH SINGULARITY
Subject to the restrictions outlined in Chapter I,
we will now derive the most general differential equation
of second order which has every point, except z = ar,
r = 1,2,3,4,and infinity, as an ordinary point. These five
points are regular singular points with respective exponents
ar ,r, r = 1,2,3,4, and the exponents at infinity are yl, '2'
We will show that the above differential equation can
be written thus:
(1) 2u + 1arr du 4 rP3p + Az2+2Bz+C u = 0.
dz2 r=l zar _z r1 (zar)2 4
rdL (zar)
r=l
We begin by writing the differential equation thus:
(2) d u + p(z) du + q(z) u = O, where
dz2 dz
3 1
Z Eizi
i=O
(3) p(z) = 4 and
T(zar)
r=l
6 Ei, 1 0,1,2,3, and
(4) q(z) = j= 3 = 0,1, "**,6
(zar)2 are finite constants.
r=l
3 6
We will now confirm that Z E zi and E F zJ were
i=0 =0
correctly chosen.
From equation (3) we note
(5) p(z) = E3z1 + 0(z2). Now let w = l/z, and
equation (5) becomes
(6) p(l/w) = E3w + 0(w2), and thus
lp(l/w) = E3 + 0(w) which, by analysis in Chapter I,
is analytic at w = 0 for finite E3. This implies by inver
sion that zp(z) is analytic at infinity, hence infinity is
a regular singular point, and Eizi was chosen correctly.
i=0O
6
Similarly, E F zj was chosen correctly.
jo0
We now factor the right member of equation (3)
into partial fractions thus:
3 1
SEiz 4 G
(7) p(z) = iO G
(7) p(z) = 0 and similarly
rl zar
TF1 (za )
r=l
6
z FjZ 4 H 4
(8) q(z) = ~O Fz H4r 4 jr where
S2 r= 2 l z where
T(zar) 2 r=l (zar)2 rl zar
r=l
Gr, Hr, and Jr are constants for r = 1,2,3,4.
On the basis of equations (7) and (8), equation (2)
becomes:
1 a dz l
r4Llrl r j*"
Hr +
(zar)
4 J1
E Z r u 0.
r z I
rul 8r
We note
4 H
that E 2 must be
r1 (zar)
6
E Aiz 4
and Z
2 r=1
T( zar)
r1
of the form
J
 must be of the form
zar
r
2
E Bizi
Q1 in order for infinity to be a regular
TT( zar)
rl
singular point of equation (2).
We now assume a solution of equation (9) in a
neighborhood of z al to be:
u(z) = o (zal) +
00o +n
E cn(zal) ; where c + 0.
n1
Differentiating equation (10) with respect to z,
X1 0o X+n1
u'(z) coX(za)1 + E Cn(+n)(zal)+ and
n1
differentiating equation (11) with respect to z, we get:
u"(z) 0o0(1)(za )2 +
co A+n2
E on(N+n) (+nl)(zal)
nl
(9) dz
dz2
(10)
we get:
(11)
(12)
Substitution of equations (10), (11), and (12)
into equation (9) yields:
(13) [co0(Xl)(zal) +
Gr ]1
7= *[co(zal)
Hr
(zar )2
4
+ E
r=1
00oo +n2
E on(N+n)(N+nl)(zal)
n=l
o00 +n1
+ Z on(X+n)(zal) I
ni n
J [c(zal 00 +n
 *[c (za) + Z c (zal) ]
za 0 n=1
= 0.
We are expanding in ascending powers of (zal);
consequently to obtain the indioial equation, we equate the
coefficient of (za)2
to zero, thus:
(14) co (1) + o Ag(G1,,G32,G,G) + coh(H1,H2,H3,H4) 0,
where g(G^,G2, G3,G) and h(H1,H2,H3,H4) are constants which
we now determine.
From equation (13), let
(15) S4 cox(a)
4
r=l
G,
rza
4( GSL r c
(16) S E _r_ [
5 r=l za" n
(17) S6 Co(za1)
4
r=l
o X+n1
cn(A+n)(zal)
( 2
(zar)2
r4
+ E
r1
[4l
~4
+ E
=1
4 H 00 X+n
(18) S7 = Z ( Cn(zal)
7 r=l (za)2 Ln=1
r4 J
S4 j
(19) S8 = co(zal) z and
r=1 r
(4 oo C+(
(20) S = r c(za)
9 =1 za n=1
L [n" 01 1
Evidently, equation (16), and equations (18) (20)
contribute nothing to equation (14), hence we examine
equations (15) and (17).
We may write equation (15) as follows:
Cl1 4 G X2
S= cX(zal) + c Glh(zal)
r=2 (zal)+(alar)
X1 4
= coh(zal) E
r=2
Xi 4
(21) S4 = coX(zal) E
r=2
G 12
r _1 + coGlx(zal)
alar zaI
1 +
alar
Gr oo j zal
a Z (1) a)
alar J=0 1 ar
X2
+ coGlX(zal)
We observe from equation (21) that the only
is c o X, hence from equation (14),
01
X2
coefficient of (zal)
(22) g(G1,G2,G3,G4) = Gl.
Similarly we may write equation (17) as follows:
4 H oo za
(23) S6 = c(zal) E  (1) (a^r)
r=2 (alar) 1 r
X2
+ coH1(za1)2, and also note from equation (14):
(24) h(H1,H2,H3,H4) = H1. Substitution of equations
(22) and (24) into equation (14) and division of the result
ing expression by co yields:
(25) X(1) + AG1 + H1 = O, or in alternate form:
(26) \2 + (Gl1) + H1 = 0.
Now the roots of equation (26) are the exponents
at z = al, hence we have
al + 0l = 1G1, or
(27) G1 1a1. Similarly,
(28) H1 = al8'.
In general if we expand about z = ar, then
(29) Gr = 1arr; r 1,2,3,4, and
(30) Hr M a rr.
Substituting equations (29) and (30) into equation
(9) yields:
(3) 2u + 4 lar d, u 4 + r_ 4
(31) d2 1 ar E J + u ] 0.
dz2 r=l zar dz 1(1 (zar)2 rl za
We now recall that
2
E Biz
4 Jr izO
(32) r 0 .
r1 zar (
71(za r)
r=l
In equation (32) we arbitrarily let
Bo C, B1 = 2B, and B2 = A, whence
(33) E z
03) ^ Az2+2B + C
r=l Zar 4
TT (za)
r=l r
Substituting equation (33) into equation (31) yields
the desired form of the differential equation. That is,
2 1 r~r Ac2
(34) d2u + 1r du rr + Az + 2Bz + C u = 0.
dz2 = =l zar dz r1 (za r)2
r T (zar)
We pause in our derivations to compute A in equation
(34) in terms of ar after applying the restriction
(35) r ar 1/2. Following this, we will show that
the exponents at infinity may be expressed in terms of ar.
Initially we assume a solution of equation (34) in
a neighborhood of infinity to be:
0o
(36) u(z) = co^ + Z cZnz ; co + 0.
n=l
Differentiating equation (36) with respect to z
yields:
co Xn1
(37) u'(z) c ()z1 + c (Xn)z and
n=l
differentiating equation (37) with respect to z yields:
X2 co Xn2
(38) u"(z) = c (X)(X1)z Z c (Xn)(Xn1)z.
n=l
We now substitute equations (36) (38) into equation
(34) to get:
X2 0o Xn2
(39) c (X)(Nl)z + Z c n(Xn)(Xn1)z
n=l
[41a1 00 Xn
Sr zar co(X)z + E Cn(Nh)z
r1 zar n=l
+ r + Az2 + 2Bz + .C c z + E c z X 0.
r=1 (za,)2 o4 n=l
'T (zar
r=l
Inasmuch as we are expanding in a neighborhood of
infinity, equation (36) is an expansion in descending powers
of z. Thus the indicial equation is found from equation (39)
by equating to zero the coefficient of z 2, where (X2)
is greater than or equal to all other exponents on z.
From equation (39), we let
(40) Sl = co(0)(l)z2
1 4
S2 = c (N)z1
S2 0 rl
1arr
zar
X 4 r
=cz a
0 r1 (zar)2
2
;X (Az2 +2Bz + C)
S4 CoZV 
7T (zar)
r=l
00
S = E
5 nl
c (Xn)(Nnl)z
4 1p 00 n1
6 = . c n(n)z ,
6 r1 zar n=1 = n
4
S = Z
7 r=1
ar r
(zar))2
2
Az + 2Bz + C
S =
4
TT (za)
rl r
00 Nn
Sc nz and
n=l 
oo
00 n
Sc nz
n=l
(41)
(42)
(43)
(44)
(45)
(46)
(47)
U
37
By inspection we note that equations(44) (48)
contribute nothing to the indicial equation. We now examine
equations (40) (43) more closely.
X2
The coefficient of z equated to zero is:
(48) co(X)(x1) + co(X)T2 + coT3 + coT4 = 0,
where Tj are constants which can be determined from the
expression for Sj, (j = 2,3,4).
From equation (41)
Xl 4
S2 = co(X)z E (lar r)/(zar)
r=1
= Oo()z El [(larOr)/z] .
r=l 1 ar/z
4 oo
= o(x)z 1 E [(1arr)/z] E (1)J(ar/z) or
r=1 J3=0
(49) 2 = co(X)zX2 E (1ar) (1) (ar/z)J.
rl j0
Comparing equations (48) and (49), we see that
4
(50) T2 = Z (1arr).
r1
Similar to the above analysis, we may write
equation (42) thus:
X 4 arBr 1
S3 = CoZ E or
r=l z2 2
(1 a/z)2
(51) S3 = CoZ 2 ar Z (l)(ar/z) ]2
r=l r =rJ
Comparing equations (51) and (48) shows that
4
(52) T3 Z ar r.
r=l
We may write equation (43) as follows:
S4 c zX(A/z2 + 2B/z3 + C/z4)
4 o 4
TT(1 ar/z)
r= 1
= coz (A/z2+ 2B/z3 + C/z4) TF ()J(a/z)o
= coz [A/z2 + O(z3)] or finally
(53) S4 = Ac zx2 + c z^ Co(z3)].
Comparing equations (53) and.(48) shows that
(54) T4 = A.
Substituting equations (50), (52) and (54) into
equation (48), and remembering that c + 0 yields the
following expression for the indicial equation:
4 4
(55) x(x+1) x (larPr) + E arPr + A 0.
r"l r=l
We now apply the restriction given by equation (35)
to equation (55), and recall that infinity is an elementary
singularity to get: 15, p. 4953
4 4
(X+1) X Z (1a a 1/2) + ar(a +1/2) + A = 0,
r r r r=l
or upon rearrangement,
2 4 4
(56) \2 + X [1 + 2 E a + E a(ar + 1/2) + A = 0.
r=l r=l
Let the roots of equation (56) be lX and X2, then
4 4 /2 ]
(57) 2 X = {[2 a12 4[ o ar(ar + 1/2) + A]}1/2
r=l r=1
We square both members of equation (57) and simplify
to get:
4 2 4 2 4
(58) A = l E ar 3/2 ar + 3/16.
rl r=1 r1
Equation (58) expresses A, in equation (34), as a
function of ar only. We now substitute equation (58)
into equation (56) and simplify to get:
2 4 4 i2 4
(59) X + N[2 1] + a Z a + 3/16 = 0.
r=l r r= r=l
Solving equation (59) for the exponents at infinity
we get: 4
[1 2 Z a,] + R
r=l
= that is,
2
4
[1 2 E ar] + R
(60) 72 2 and
4
[1 2 E ar] R
(61) 1 a Y, r1= where 1y and 72 are the
exponents at infinity, and
R2 = [2Zr 1]2 4[(Zar) 2 ar + 3/16], or
(62) R2 = 1/4. Substituting the values of R given by
equation (62) into equations (60) and (61) yields:
4
(63) Y1 = 1/4 E ar and
r=F
4
(64) Y, 3/4 E ar.
r=l
Equations (63) and (64) display the exponents at
infinity as functions of oronly. Further, we note that
infinity is an elementary singular point since the expon
ent difference at infinity is 1/2. [5, p. 495 ].
We continue the derivation of certain differential
equations of mathematical physics by first obtaining the
generalized Lame equation, D4, p. 204]. This is accomplished
by applying restriction (35) to equation (34), thus:
2 4 1/2 2arl
du r du
(65) + 
dz2 =1 zar dz
4 OLr(ar + 1/2) Az2 + 2Bz + C =
1 (zar,)2 4
(zar)
rl
where A is given by equation (58), and B and C are constants.
We will now specialize equation (65).
Lame's Equation
Equation (65) may be written as follows:
(2u2 + 3 1/2 2i 1/2 a ,a ( a y
(66) ( Th ar) + l + (zar
r=l dz zar za4 J r1
+(za) 3L a( + 1/2) 4(a4. + 1/2) +Az2+2Bz+C= 0.
r=l 1 (za,)2 za4 z4
In equation (66), we set ar = 0, (r = 1,2,3,4),
let a4o o let SB n(n+l)a4, C = ha4/4, A = 3/16, then
divide both numbers by (zar) to get:
r=l
2 3 1 du
(67) u + + u
7 dz2 r=1 2(za) dz
Lim 3z2 + z(n2+n)a4 + ha
a4 * co
4(za4)
+ 4(zu = 0.
3
TT (zar)
r= 1
Equation (67) immediately simplifies to
d2u 3 du n(n+)z+ h
(68) L + Fdu  u 0.
dz2 r=l 2(zar) dz
r=l
We classify equation (68), in the notation of
2
Ince [ 5, p. 497], as an equation of the [3,1,0] type.2
Equation (68) is Lame's equation.
Mathieu's Equation
In order to derive Mathieu's equation, we consider
equation (65) and let al = a2 = 0, P = p2 = 1/2, al = 0,
a2 = 1, a3 oo, and a4aoo. From this information, and
from equation (38), A = 2a 3c4 3a3/2 3a4/2 + 3/16. We
now choose 2B = Sqa3a4, and C = (a 16q)a3a4/4.
2The expression [a,b,c] characterizes an equation
with the following properties:
a = the number of elementary singularities,
b = the number of nonelementary regular
singularities, and
c = the number of essential singularities.
We substitute the preceding information into
equation (65) to get:
d2u
(69) d2+
dz2
1 1/2 2a 1/2 2a4 du
2z 2(z) (za ) ) dz
1 3 (a4
a (a + 1/2)
+ Za 3
(za )
+ q(a4 + 1/2)
(za4)
(2ac 3a 312 a + ) z2_ (8qa 3a4) z (a16q) a3a y)/ 0.
z(z l)(z2 a3z a4z + a3a4)
We rearrange equation (69) slightly thus:
(70) d +
dz2
1 + 1 1/2 2a3 1/2 2a du
2z 2(zl) (za) (za) dz
a3(a3 + 1/2). a4(a + 1/2)
+ (za3 ) + (za4)
(2a a13 a~+ ) z2
2c 4a2 a
a3aa4
 Sqz (a16q)
4
z(z 1)(z2/a3a4 z/a4 z/a3 + 1)
u = 0.
In equation (70) we let a3 oo, a4oo, hence
a 3a4oo, to get:
(71) [d 1 du al6q+2qzsu
(71) d [2z2(z1) dz z(zl)
dz
We digress at this point to investigate the method
of removing p(z)d from equation (2).
We begin by making the variable transformation:
(72) z = f(w), and recall that
(73)
du du dw
Sd and
dz dw dz '
(74) d2u du d2w 2u d2
(74) d =  u 2 Substitute
dz2 dw dz2 dw2 dz
equations (73) and (74) into equation (2) obtaining:
dw2 dz dz2 dz dw
We now wish to require that
(76) d2w + p(z)w 0. Then we let
dz2 dz
(77) d = v, hence equation (76) becomes:
(78) dv + p(z)v = 0. We separate variables in
dz
equation (78) and integrate to obtain:
(79) v = cle/P(z)dz. Substitute equation (77) into
equation (79) and integrate once more to obtain:
(80) w cl/e!P(z)dz dz + c2, where c1 and c2 are
are arbitrary constants of integration.
Comparing equations (2) and (71) we see that
(81) p(z) = + We now choose c2 O, and
2z 2(z1)
substitute equation (81) into equation (80) to get:
S /efdz/2z /dz/2(z1)dz
In z 1ln(zl)
clfe 2 2 dz.
w = c 1/  This may be written as
[z(z1)]
(82) w = el/ z
[(z1/2)2 1/431/2
In equation (82), let z 1/2 1/2 sec 9,
then dz 1/2 seo Q tan 9 de, and substitution into
equation (82) gives us, after some simplification:
(83) w Cl/ sec 9 de, or finally,
(84) w/co 1 In (sec e + tan 9). But sec 9 2z 1,
hence tan 9 2(z2 z)1/2, hence equation (84) becomes:
w/ol = In [2z1 + 2(z2z)1/2], or
(85) ew/co (2z1) + 2(z2z)1/2.
Solving equation (85) for z, we get
z oosh2 (w/2c1), or
z cos2 (iw/2cl). We now choose cI = 1/2 so that
(86) z cos2 w.
Equation (86) is the desired transformation, which
was outlined in general by equation (72).
For z = cos2 w, equation (73) becomes:
(87) du = csc 2w du
dz dw
From equation (86),
w = cos1 v , whence
dw  1 and
dz 2 vz(lTz)
d2w 2z But z = cos2 w, hence
dz2 4(zz2)3/2
(s8)
d2w
2
dz
1 2 cos2w 3
2 4 3/2
4(cos w cos w)
Substituting equations (87) and
equation (74) yields:
(89) d2u 2 cos 2w .du + (l/sin
dz2 sin3 2 dw
Substituting equations (87) and
equation (71) we get:
2 cos 2w
sin3 2w
(88) into
2 d2u
2w)2 d2
dw2
(s9) into
(90) csc2 2w d2u 2 cos w du
dw2 sin3 2w
11 os 2w d
2 cos2 w 2(cos2 w 1) dw
a 16q + 32q cos2 W u = 0.
[4 cos2 w(cos2 w 1)]
Multiply both members of equation (90) by
sin2 2w to get:
(91) d2u sin 2w sin 2w 2 cos 2w du
912) 2 2 dw
dw2 L2 cos w 2 sin w sin 2w
a 16q + 32q cos2 (sin2 2w)u = 0.
4(oos w)(cos2 w 1)
In equation (91) we notice that
+ sin 2w sin2 w + cos2 w sin 2w cos 2w 2 cot 2w.
2 sin2 w cos2 w sin 2w
2 4
Also
4(cos2 4(cos2 w 1) = 4 sin2w cos2w = sin22w.
Substitute these identities into equation (91)
to get:
2
(92) d2u (a 16q + 32q cos2 w)u = 0.
dw2
Again we note that cos2 w = 1/2 + 1/2 cos 2w,
and substitution into equation (92) yields:
2
u [a 16q + 32q(1/2 + 1/2 cos 2w)]u = 0, or
dw2
(93) d (a + 16q cos 2w)u = 0.
dw2
Equation (93) is Mathieu's equation and is classified
in Ince as being of the [2,0,1] type.
Legendre's Equation
In order to derive Legendre's equation, we refer
to equation (65), and let al = a2 = 1; a3 = a3 = 0,
aI = a2 = a3 = 0, and a4 = 1/4.
Substituting these values into equation (65) yields
the following equation:
(94) d2u + [1/2z + l/(zl)]du + 3_ + Az2 + 2Bz + Cu = 0.
dz2 dz 16z2 z2(zl)2
We can write equation (94) as follows:
2u + [1/2z + l/(2l)]du
dz2 dz
+ n(n+) lj u = O, compare the
IT z z31 z(z1)
coefficients of u, and hence determine values for
A, B, C in terms of m and n.
+ Az2 + 2Bz + C
z2(zl)2
.1 n(n+l) m2
4 z2(zl) z(zl)2
Simplifying the above identity we get:
1(z1) 2 +
16
(Az2 + 2Bz
 3z + 2 + Az2+
g 16
a I[z(n2+nm2) 
4
+ C) a 1[n(n+l)(zl)
4
2Bz + C
n2n],
z2(3/16 + A) + z(2B
 3/8) +
(3/16 + C)
Sz(n2+nm2)
4
n2+n
4
Equating coefficients of like powers of z we
obtain the following set of equations:
(95) A = 3/16
B = 3 + 2(n2+nm2
16
(97) C (4n2+4n+3)
16
Thus
16z2
 m2z],
36 2
16
We now make the variable transformation
z = w2, and deduce from equations(73) and (74) that
du 3 du and
dz 2 dw'
(100) d2u = 6
dz2
d2, + 5 du
dw2
We now substitute equations (95) (100) into
equation (94) to get:
(lol) Hw26 d2u + 45 du w 3 du
dw2w
3w4
+ +
16
S14 + (3+2n2+2n2m2)w2
16 S
4 2 2
w (w 1)
u = 0.
We now simplify equation (101). The coefficient
du
of is:
dw
35 w5 w5 ,55 5 w5
4 2(1w2) 2 2(1w2)
Let T represent the coefficient of u, then
equation (101) becomes:
(98)
(99)
4n2+4in+3
16
(102) 6 d2u
dw2
+ w5 1 ( + Tu 0,
12 4(12 dw
where
(103) T 4 + 3w4 + (6+4n2+4n+4m2)w2 4n24n3
16 16w4(w2 1)2
3w + 3w4 + (6+4n2+4n+4m2)w6 (4n2+4n+3)wg
16 16(lw2)2
= [3w4 6w6 + 3w8 3w4 + 6w6 + 4n2w6 + 4nw6+ 4m2w6
4n2w8 4nw8 3w ]/[16(1w2)2]
n2w6 + nw6 +m2w6
 n2w8 nw
Snw2(n+l)(lw2) +
4(1w2)2
m2w6
S or finally,
4(1w2)2 '
(104) T = Fn+l)
4_lw2 (1w2)2
Substitute equation (104) into equation (102)
to get:
(105) 6 d2u + w5 1
Sdw2 2
1 du
2(1lw2 dw
+ (6 nn+l)+ m2 u = 0.
Slw2 (1w2)2
L A
52
Multiplying both members of equation (105)
4(1w2)
by 6 2,we finally obtain Legendre's equation:
w6
(106) (lw2)d2u 2w d + Cn(n+l) m2/(1w2)]u = 0.
dw2 dw
This equation has been classified by Ince as an
equation of the [1,2,0] class.
Bessel's Equation
From equation (65), we let ar = 0, r = 1,2,3,4;
a1 = a2 = 0; a3 = a and we let a 3 o Substitution
of this information into equation (65) yields the following
equation:
d2u 1 d z2 + 2Bz +C
(107) du + u + Lim z2+ 2Bz + =0.
dz2 z dz a3'O Lz2(za )2 3
From equation (5S), A = 3/16, and we arbitrarily
1 n22
choose 2B = a and C = 
43 4
Substitution of these values into equation (107)
gives us:
(108) du
dz2
1
+ L .d + Lim 16
z dz a c3
z2 + az 
4 3
z2(za )2
We may rearrange equation (10) slightly as follows:
2
z2du + z
dz2
du
dz
Lim
a  oo
_z3 z n2
16a 2
(z/a 1)2
and hence equation (108) finally becomes:
2du
(109) z d +
dz2
du + (zn2)u 0.
dz 4
We now make the variable transformation:
(110) z = w2. Then in accordance with this transfor
mation, equations (73) and (74) yield:
(111) du and
dz 2w dw '
d2u
(112) d
dz2
dw2 w w
4w2
Substitute equation (110) (112) into equation (109)
to obtain:
2a2
4
U = 0.
U = 0,
(113) w ^+ + (w2n2)u 0.
Simplifying equation (113) yields:
2
w2 (w2w)d + (w2n2)u 0, or finally:
dw2 dw
(114) w2 d2u ++ (w2n2)u 0.
dw2 dw
Equation (114) is Bessel's equation and is classified
by Inoe as an equation of the [O,l,1 class.
Weber's Equation
We begin the derivation of Weber's equation by
considering equation (65) where we let arp 0; r = 1,2,3,4;
a1 0; and a2 = a3 a4; and a2 oo. From equation (58)
we note that A 3/16. We choose B a3/32, and
C (n+ 1/2)a /4.
Substitution of the above information into
equation (65) gives us the following equation:
(115) du + ._u. +
\ll.2 d r
dzn, U 0.
dz d 2d L z(za2)3 J
Equation (115) may be rearranged slightly as follows:
(116) du
dz2
which
d2u
(117) d +
(1) z d2u
dz
3  + z/16 (n+ 1)/4
16 a3 2
1. + Lim 2 u = O,
2z dz a2oo zz/a3
L z(z/a 17
becomes upon letting a2 oo,
1 du (n+ /2)/4 z/16 or
F + u = O, or
2z dz z
+ j du + C(n+/2)/4 z/16Ju 0.
We now make the variable transformation:
(119) z w2.
In order to verify that z = w2 is the desired
transformation which will remove the first derivative
term from equation (117), we refer to equation (0O), in
which p(z) We then obtain from equation (SO):
2z
Lim
o p
u = 0.
/dz/2z
w = cl/e dz + c2
(In z)/2
= cle z)/2 dz + C2
1/2
= Cz1/2 dz + c2, or
w = 2cl(z2 ) + c2, where, as before, cl and c2
are arbitrary constants of integration.
Choose c2 0, and cI = 1/2, then w z1/2, or
z = w2 as desired.
Utilizing equations (73) and (74) in accordance with
equation (119) yields:
du 1 du
(120) =  and
dz 2w dw '
(121) d2u " du .u
d 2u 1 d 2 u i u
(dz2 4w2 dw2 4w
We now substitute equations(119), (120), and (121)
into equation (11) to get:
(122) w2 d2 + L
4^2 dw2 4w3 w 2 2w dw
+ [(n+ 1/2)/4 w2/16]u = 0.
Multiplying both members of equation (122) by 4
gives the following equation:
(123) 2u + [(n + 1/2) w2/4)u = 0.
dw
Equation (123) is Weber's equation, and is an
equation of the [1,0,1] class.
Stokes' Equation
To derive Stokes' equation, we refer again to
equation (65), let ar = 0, (r = 1,2,3,4), and let
a *oo, (r = 1,2,3,4). We then get from equation (65):
(124) d u + Lim Az2 + Bz u = 0.
dz2 aloo (zal)
From equation (58), A = 3/16, and we arbitrarily
4 4
choose 2B = Blal and C = Clal .
Substitution of these values into equation (124)
and slight rearrangement gives us the following expression:
2u 3z2/16a 4 + Bz + C1
(125) u + Lim u = 0.
dz2 aloo (z/al1)4
Letting aloo in equation (125) yields:
(d2
(126) + (B + + C)u 0.
dz
We now make the following dependent variable and
independent variable transformations:
1/2
(127) u = (Blz + Cl) v, and
(128) B1z + C1 = (3Bw/2)2 .
Now from equation (127) we get:
du 1/2 dv BlV 1/2
(129) dz = (Blz + C1) d 2 (Bz + Cl) and
d2 (B + 1/2 d2v + B1 + C 1/2 dv
dz (B + dz2 (B C1 
2
Blv 3/2 B 1/2 v
(Blz + C1) + (B1Z + C1)12 d or
d2u 1/2 d2v 1/2 dv
(130) du (B1z + C)l2 d2v + Bl(Blz + C1) 2 dz
dz2 dz2 dz
2
B v 3/2
 (Blz + C1)
We now substitute equatiorn(130) and (127) into
equation (126) to get:
(131) (BIz + C1)/2 d2v + BI(BIz + C1)1/2 d
dz2 1 dz
2
B v 3/2 3/2
(Bz + C)/2 + (Bz + C1)3/2v = 0.
Now from equation (128) we have:
B1 ( 3B1w/2)1/3 (3B1/2) (3Bl/2)1/3, or
(132) dw = (3B1w/2)1/3. Hence it follows that
dz
dv = (3Blw/2)1/3 d
(133) dz /dw
From equation (133) we get:
(134) 2 = (3Bl/2)1/3 d2.dw + Blw/2)2/3B/2)1/3d
dz2 d z 2
Substitution of equations (128), (133), and (134)
into equation (131) gives us the following:
2 B By 1
(2B w) d!2v +2 1 41 + B1 d (3B1w/2)
2 dw2 dw 1 dw 4
+ (3B1w/2) v O, or
(135) (d") a2 + I B + 13Blw/2 B1/6w]v = 0.
(1dw5) 2 2 dw
Multiplying both members of equation (135) by
2w/3B1 gives us the final result:
(136) w2 2 + w + (w2 1/9)v = 0.
dw2 dw
CHAPTER III
SOME CONSIDERATIONS OF A DIFFERENTIAL EQUATION
WITH nREGULAR SINGULAR POINTS
Consider the differential equation
(1) d2u n 1app n ap n J
du E K + E + E u = 0.
dz1 zar dz rl (za) 2 r=1 zar
All points including infinity are ordinary points,
except the points al, a2, ...,an. The points al, a2, ...,an
are regular singular points with exponents al, 10; a2, 02;
...;Ian, On, respectively.
In order to assist in the future development of this
study, we wish to show that
n
(2) E (ar + rp) n 2,
r=l
n
(3) r Jr = 0,
n
(4) E (arJr + aprr) = 0, and
r=1
n 2
(5) rr (arJr + 2arart) = 0.
(6) d_
,d,
Let us write equation (1) as follows:
 Eu + o+E1z + + En1n1 du
I ... _/ N dz
'\ l \a2 \ cn
+
= 0, or alternately,
(7) d4 + p(z) u + q(z) u = 0, where
dz dz
E +E1z + ... + En_ zn1
() p(z) = za z
(zal)(za2)**(zan)
(9) q(z) =
Fo+Flz + ... + F2n2z2n'2
(zal) 2za2) 2(zan) 2
Ei, (i 0,l,2,...,(n1)), and Fi, (1 = 0,1,2,
are constants to be evaluated.
Expanding equations (8) and (9) in partial
...,(2n2))
fractions
yields:
(10) p(z) =
zal za2
H1
(11) q(z) = 2
(za) 2
H
+ +
(za)2
(za 2)
+(z
(zan)2
J + J2
zal za2
. + j
zan
where the constants Gr, Hr and Jp,(r = 1,2,... will be
evaluated presently.
and
z
zan
and
Substituting equations (10) and (11) into equation
(6) we get:
F 01
G
+ `2 +
za2
SH H
+ +
(zal)2 (za2)2
J J
+ 1 + 2+ ** +
zal za 2
Gn_ du
za dz
+ Hn
(zan)2
nJ u 0.
zaz
We now assume a solution of equation (11), in a
neighborhood of z al, of the form:
(13)
00
u E cm(zal) ;
m0
b+ 0.
Differentiating equation (13) with respect to z
yields:
oo
(14) u' = coh(za)1X1 + E cm(A+m)(zal)A+m1, and
m=l
differentiating equation (14) with respect to z yields:
(15) u" C(l)(za)2 00 +m2
(15) u" c(1)(za ) + E o,(+m)(x+ml)(zal)
m1l
We now substitute equations (13), (14), and (15)
into equation (12) to get:
(12)
d2u
dz2
(16) coh(h1)(zal )2 + cm(+m)(+ml)(zal) 2
m1l
n 7%,Gr l)1 o+ E c +nza,)"+m1
+ 1 z ar] L (zal) + cm(+m)(zal )
r1 zap IC01
rn Hr n Jr
+ z r + r
r=1 (zar)2 rl zar
*[ Co(zal) + cm(zal)+ = 0.
M1
We recall that a ,(r = 1,2,...p) are finite distinct
regular singular points. Also we are expanding in ascending
powers of (zal). Therefore, we obtain the indicial equation
x2
by equating to zero the coefficient of (zal) Then
(17) CoN(N1) + CoXg(G1,G2,...~n) + coh(H1,H2,...,Hn) = 0,
where g(Gl,G2,0..,Gn) and h(H1,H2,...1 ) are constants which
we now determine.
From equation (16), let
rn G 7%
(1l) S = y [c coza)1 or
Gn A1
(19) SI zal ] [ coh(za) ]
+L2 zar [ coh(zal) Al .
+ rr=2 zap I 
Now write
(20) S2 [2 co (zal) as follows:
L r 2 1J1 r
[co()za r ..n. .
r.2 alar Za +
alar
(21) S2 = co(zai)] 2 _' E E (1) aza1 *
Substituting equation (21) into equation (19) yields:
(22) Sl = o(za )Xl]{
GI 0 1z al
+ r=2 aar j0 1ar)
+^ (1) 3 f )z
from which we observe that the least exponent on (zal) is
(A2), and the corresponding coefficient of (zal)A2 is
n Gr
coAGl We thus conclude that 2 Z contributes nothing
r=2 Zr
to the indicial equation. The sole contribution
c
comes from hence in equation (17), g(G1iGG2, ,Gn) Gl,
zal
and equation (17) now becomes:
(23) co(Nl1) + CohG1 + cah(H1,H2, ... Hn) = 0, or
since co + O by assumption, equation (23) becomes:
(24) X(X1) + AG1 + h(H1,H2, ... Hn) = 0
We note in passing that from equation (16)
[n Gr oo c+m1
(25) S3 =  *I cm(+m) (za) contributes
S .1 Zar 1
nothing to the indicial equation sinoe the least exponent
on (zal) in the above expression is X1 > X2.
Returning to equation (16), we let
X n Hr
(26) sq 4 co(zal) 2
n H oo00 +m
rl ( ( .m1
(27) S5 = El (za )2 []
(28) S6 = c(zal)A r and
r=l za
n op X+
(29) S7 = (zal)
7 1 Zar =
We observe that the least exponent on (zal) in
equation (27) is X1 > X2. Similarly the least exponent
on (zal) in equation (28) is \1 > X2, and in equation
(29) the least exponent on (zal) is > X2, hence equations
(27) (29) contribute nothing to the indicial equation.
The only contribution to the indicial equation comes from
equation (26).
Let us write equation (26) as follows:
S4 = c (zal1) H1 + o (zal H2 or
(zal)2 r+2 (zaro(
X2 X n Hr
$ oo(zal)2 H + c (zal n HE
S Oo(Z ) H +C(0 r=2 (zal + alar)2
This may be written as follows:
S4 coH1(zal)+2 + co(zal)X 2H
r 2 ( 1ar) za
Sral(ar
We may write the above expression for S4 as follows:
(30) S4 OoH1(zal) 2
r22
+ co(zal) 2 H 0 ( )J .
0 r=2 (a1ar) 2 10 1
Now the least exponent on (zal) in the expression
c (zal)X H5 D( L]zal Zis X,
S 1 r2 (a1ar)2 0 \a1aj
which can be found by actual expansion, so that this expres
sion contributes nothing to the indicial equation. Conse
quently, the only contribution to the indicial equation,
insofar as equations (26) (30) are concerned, comes only
from the first part of equation (30) and is the coefficient
of (zal) 2, that is, (o)(H1).
Returning now to equation (23) we note that
c0h(H1,H2, ...*,H) = coH1, or
h(H1,H2*,...,Hn) H1, thus equation (24) becomes:
(31) X(X1) + XG1 + H1 = O, or in alternate form:
(32) \2 + X(G11) + H1 = 0.
Equation (28) was derived upon expanding in a
neighborhood of z = al. Analogous reasoning shows that
expansion in a neighborhood of any of the regular singular
points z ar, would give an indicial equation of the form:
(33) \2 + A(Gr1) + Hr O;(r = 1,2,...p).
Let the values of A which satisfy equation (33)
be Al and X2. We then have
(34) Xl + N2 = 1G,, and
(35) Xl12 = Hr. But we are given that the exponents at
ar are ar and rp, respectively, hence equations (34) and (35)
become:
Al + 2 = 1Gr = ar + Br, or
(36) Gr = 1arPr. Similarly,
(37) Hr = arBr.
69
Substituting equations (36) and (37) into equation
(12) yields the following differential equation:
n 1arr ]n8 aPp r n Jr
(3o) 2u + du + + E u = 0.
dz2 Lr"l zar Jaz l (zar)2 rl zap
dz2 1
We now make use of the requirement that infinity be
an ordinary point of equation (38).
We make the independent variable transformation:
(39) z = i substitute into equation (39), and require
w 0 to be an ordinary point of the resulting equation.
The equation resulting from the above variable
transformation was derived in Chapter I, and was shown to be:
(40) d2u + ) = 0.
It was further shown that if w = 0 is to be an
ordinary point, then both
E l2 p(1. and 1 q(w) must be analytic at w = 0.
w W
We now expand p(w) in an ascending power series
about w 0 of the form:
(41) p(l) = 2w + A2w2 + ''. + An + and hence
(42) () = +A2 + ** + Awn2 +
= A2A3w ** Anwn2 *, which
is analytic at w = 0, provided A2 is finite.
This implies p(&) is finite at w 0, or
(43) p(z) = + A2 + ... + n + is finite at infinity.
z Z2 Zn
Now from equation (38) we notice that
n (1a rr)
(44) p(z) = z 
r=l zar
n ( 1rOr)
z Ea
Sr 1 ar
z
n
E (1 ^Br'0
Z rl r r
+) + ( )2 + *** + + *], or
n
(45) p(z) E r (1ar r)
z r
n 00 rj)
+ z Z 1arpr) Z .
rl J=l
Now upon comparing equations (43) and (45) and
equating the coefficients of like powers of 1 we get:
z
n
E (1arp1 r 2, or finally,
rl
(46) Z (ar + r) n2.
Similar to the preceding analysis, we expand q(1)
about w = 0 in an ascending power series thus:
(47) q(=) = B1w + B2 2 + *** + Bnwn + '*. which is
finite at w = 0. From equation (47) we have
(48) 1 q()= + B + + B4 + B5w +**+ Bd w4 +*.
w4 w w2 w
In order for q q(w) to be analytic at w = 0,
B1 B2 B3 O,0 in which case equation (47) becomes:
(49) q(1) B44 + B5 5 + .. + Bn + and
since z = L, we have:
B4 Br B
(50) q(z) z + 5 + ... + +
z z
Since q(i), as given by equation (49), is finite
at w = O, then q(z), as given by equation (50), is finite
at infinity. This completes the restrictions under which
infinity will be an ordinary point of equation (38).
From equation (38) we notice that
n n
(51) q(z) W E r 2 + n ; ( r = 1,2***,n).
r1 (zar)2 rl (ar
Equation (51) may be written thus:
n
q(z) = E
z2 rl
ar)r
(z
1 n Jr
z 1
z
Srl [ )+ r + ... + *** ..
Sr=l rE1^ )J
12 2
n ++ (2a a
q(z) + 1.
z2 r fl r~ Z/z 2r
.1
+ rlr l + + +
z rl L \z/ ) I I
n ar r n ararPr n r n arJr
q(z) = + 2 E + Ez +
rl z2 rl z3 r=l z r=l z2
2
+ n arJr +
+ r1
r=l z3
O( 
Finally we may write equation (52) as follows:
n n
(53) q(z) + z Jr E (arr + arJr)
z rl 2 r=l
n 2
+ rZ 1 (2aracrr + arJr) + 0 .
z3 r zl
We now compare equation (53) with equation (50) and
equate coefficients of like powers of i to obtain the
z
following set of equations:
(52)
n
(54) Jr = o,
ril
n
(55) E (arpr + arJr) = 0, and
rl
n 2
(56) E (2ararBr + arJr) = 0.
r1l
Equations (46), and (54) (56) are the relationships
we desired. They will be used in the derivation of Heun's
differential equation C14,p. 576] in Chapter IV, and in the
derivation of a general linear ordinary differential equation
of second order with five regular singular points as shown
in Chapter V.
CHAPTER IV
A DISCUSSION OF
HEUN'S DIFFERENTIAL EQUATION
We will now apply the relationships derived in
Chapter III to the derivation and solution, in a general
case, of Heun's differential equation [14, p. 576].
An intensive study of Heun's differential equation was
recently completed by Neff [9].
Consider the differential equation:
(1) 2+ ~ rar + Z+ n E r u = o.
dz2 _r=l zar dz Lr1 (zar)2 r1 zar
In Chapter III, we derived the following relationships:
n
(2) E (cr +pr) = n2; where n is the number of regular
r=l
singular points.
n
(3) E Jr = 0,
r~l
n
(4) 2 (arJr + ar~r) = 0, and
r=l
n 2
(5) Z (arJr + 2ararr) = 0.
r=l
We arbitrarily let
al = 0; a2 = 1;
a1 = 0; a2 0;
01 = ly; 32 = 16;
a = a;
3 = 0;
3 = Y+6ap;
Substitute the above set of equations into
(1) to get:
a4 = b;
a4 = a;
4 quati on
equation
(6) ad2 + + + 3a+cx8 + 1' du
dz2 z1 z a zb dz
+FOLD + 1 + 52 + 3 + 42 u = 0.
(zb)2 z z1 za zbJ
We also note that the above set of equations when
compared with equations (2) (5) yield the following
expressions:
(7) a + (ly) + (18) + (y+8aP) +p0 2,
(8) Jl + J2 + J3 + J4 = 0,
(9)
(10)
J2 + aJ3 + bJ4 + a = 0, and
J2 + a2J3 + b24 + 2ba = 0.
To continue with the derivation of Heun's differen
tial equation, we multiply both members of equation (6)
by (z)(zl)(za) to get:
(11) z(zl)(za)d2u + [y(zl)(za) + z(za)8
dz
+ z(zl)(a+3y8+l) + z(zl)(za)(laP)/(zb)]Lu
+ [z(zl)(za)(ap)/(zb)2 + (zl)(za)J1 + z(za)J2
+ z(zl)J3 + z(zl)(za)(J4)/(zb)] u = 0.
Let C be the coefficient of u, that is,
2 dz'
C2 = (z2zaz+a'y+ 8(z2az) + (a+pB8+l)(z2z)
+ z(z1)(za)(laP)
z b
= z2(T+S+a+Py8+l) + z(yaya8ap++8+l)
+ z(zl)(za)(lap) or finally we have:
z b
(12) C2 = z2(l+a+p) z(l+a+P8+a8+ay) + ay
+ z(zl)(za)(lap)
z b
In equation (11), we let C3 be the coefficient of u,
that is,
C = 1 2 [z(z1)(za)(ap) + (zb)2(zl)(za)(J1)
(zb)
+ z(za)(zb)2(J 2) + z(zl)(zb)2(J )
+ z(zl)(za)(zb)(J4)].
C 1= {(z3az2z2+az)(ap)
3 (zb)2
+ [z4z3(a+2b+l) + z2(a+2ab+2b+b2)
z(2ab+ab2+b2) + ab2J 1
+ Cz4z3(2b+a) + z2(b2+2ab) z(ab2) J2
+ [z4z3(2b+l) + z2(b2+2b) zb2J33
+ [z4z3(a+b+l) + z2(ab+a+b) z(ab)]J4}.
We now collect the coefficients of like powers of z to get:
(13) C 3  {z[Jl+2+J 3+J41] + z3(ap (a+2b+l)J1
(zb)2
(a+2b)J2 (2b+l)J3 (a+b+l)J4]
+ z2[aapap+(a+2ab+2b+b2)J1
+ (2ab+b2) 2 + (b2+2b)J3 + (a+b+ab)J4]
z[aap+(2ab+ab2+b2)J1+ab2J2+b2J3+abJ4l
+ ab 2 J1
Applying the restriction given by equation (8), we see that
the coefficient of z4 vanishes. We now write the coefficient
of z3 as follows:
z.: a, (a+2b+l)J1 (a+2b+l1)J2 (a+2b+la)J3
(a+2b+lb)J4
= (a+2b+l)(J1+J2+J3+J4) + (aO+J2+aJ3+bJ4)
= 0 by virtue of equations (8) and (9).
We now simplify the coefficient of z2 thus:
z2: (a+2ab+2b+b2)JI + (a+2ab+2b+b2a2b)J2
+ (a+2ab+2b+b2a2ab)J3 + (a+2ab+2b+b2babb2)J4
aa~ ~
= (a+2ab+2b+b2)(J1+J2+J5+J4) (a+2b)J2
(a+2ab)J3 (b+ab+b2)J4 aaP ap
= (a+2ab+2b+b2)(Jl+J2+J3+J4) (a+2b+l1)J2
a(a+2b+la)J3 b(a+2b+lb)J4 (a+2b+l2b)ap
(a+2ab+2b+b2) (J+J2+J3+J4) (a+2b+l)
*(ap+J2+aJ3+bJ4) + (J2+a2J3+b2J4+2baP)
= 0 by virtue of equations (8), (9) and (10).
The expression for C3 has now been reduced to the
following:
C3 = [aap + (2ab+ab2+b2)J1 +ab2J +b2J +abJ ]
S(zb) 2 3
2
ab J
+ a
(zb)2
We now proceed to simplify the coefficient of
z thus:
z: (ab2+b2+ab+ab)J1 + (ab2+b2+abb2ab)J2
+ (ab2+b2+abab2ab)J3 + (ab2+b2+abb2ab2)J
S(ab2+b2+ab)(J1+J2+J3+J4) + abJ1 (ab+b2)J2
(ab+ab2)J3 (ab2+b2)J4 + aac, which upon
applying equation (8) and rearranging somewhat becomes:
(b2+ab+bb)J2 a(b2+ab+bab)J3
b(b+ab+b2b2)J4 + abJ1 +acP
= (b2+ab+bb2abba)aP a(b2+ab+bab)J3
b(b+ab+b2b2)J4 + abJ1 (b2+ab+bb)J2
= (b2+ab+b)(ap+J2+aJ3+bJ4) + abJ1
+ b(J2+a2J3+b2J4) + (b2+ab+a+b)ap.
Utilizing equation (9), the above expression simplifies,
upon slight rearrangement to:
z: b(J2+a2J3+b2J4+2baP) + abJ1 + (a+b+abb2)ac
= abJ1 + (a+b+abb2)ap, by virtue of equation (10).
Equation (13) has now been reduced to the following:
(14) C z[abJ1 + (a+b+abb2)aA] + ab2J
(z b)2
We now substitute equations(12) and (14) into
equation (11) and express the differential equation as
follows:
(15) z(zl)(za)d2u + (z2(l+a+g) z(l+a+8+a+a+ay)
dz2
+ ay + z(zl)(za)(laP)/(zb)] u
{z[abJ1+ (a+b+abb2)aP] ab2J }
u = 0.
(z b)
Since infinity is a regular singular point, we let
boo to get:
(16) z(zl)(za)d2u + [z2(1+a ) z(l+a+8+a86+a y) + ay]!u
dz2
+ (apz + aJi) u = 0.
In equation (16) we let
apz + aJ1 = ap(z q), whence
aJ = apq, or q = aJ
Substitution into equation (16) yields:
(17) z(zl)(za)d2u + [z2(l+a+B) z(l+a+p8+a8+ay)+ay]U
dz2
+ ap(z q) u = 0.
Equation (17) is Heun's differential equation and
has four regular singular points at z 0O,l,a, and oo.
We will now solve equation (17) in a general case,
using the method of Frobenius [3].
Assume a general solution to equation (17) to be
of the form:
(18) u = co + +1+l +* + Crlz +r1 + CzA+r
+ cr+Iz+r+1 + *** where co + 0. Then
(19) u 1 + C (A+l)z + *" + 0 (A+r1)zh+r2
+ o,(A+r)zA+r1 + cr+1(+r+l)z+r + and
(20) d coh(A1)zA2 + cl(X+l)(A)zxl +...
+ cr1(X+r1)(A+r2)z'+r3 + cr(A+r)(A+r1)zh+r2
+ Cor+l(+r+l)(+r)z%+r1 + .
We now substitute equations (18), (19) and (20)
into equation (17), and deduce the indicial equation by
setting the coefficient of zA1 to zero, thus:
(21) aco X(X) + coa0 = 0, where a + 0.
Solving equation (21) yields
X(A1) + TY = O, whence
(22) = 0, 1 Y.
To obtain the recurrence equation connecting respec
tive coefficients in equation (18), we set the coefficient
of zA+r to zero, that is:
(23) crl(X+r1)(X+r2) cr(a+l)(N+r)(X+r1)
+ c r(a)(X+r+l)(X+r) + crl(l+a+p)(N+rl)
cr(l+a+p8+ay+a8)(X+r) + cr+1(ay) (+r+1)
+ c rl(ap) C (aq) = 0 for r 0,1, .
We may rewrite equation (23) as follows;
(24) cr+l(a)(X+r+l)(X+r+y) = cr[(a+l)(X+r)(X+r1) + apq
+ (l+a+p 8+ay+a8)(X+r)]
crl[(X+r1)(x+r2) + (1+a+p)(A+rl) + ap].
Now from equation (22), we let A = O, so that
equation (24) becomes:
(25) cr+l(a)(r+lXr+y) = cr[(a+l)(r)(r1) + apq
+ r(l+a+P8+ay+a8)]
rl[(r1)(r2) + (l+a+p)(rl) + ap].
We would like a general solution of equation (17)
in a neighborhood of z = 0 to be of the following form
when X = 0:
(26) u = co +ap= G(q)
n=l n'(y)(y+l) *** (y+n1)
Setting r = 0 in equation (25), we get:
cl(ay) = c0(aPq) c(l(1a)(l0). But cl = 0,
hence cI apqco Substituting this expression for cl into
ay
equation (18) yields the following equation:
(27) u = coll + z+ a z + 02z2+ + czn+ "'].
Comparing the coefficients of like powers of z in
equations (26) and (27) we note
(2f) G (q) = q.
Next we set r = 1 in equation (25) to get:
(29) c2(a)(2)(l+y) = cl(apq + l+a++8+ay+a8)
co(ap), where cI "
ay
Solving equation (29) for c2 yields:
(30) 2= ap[q(apq + l+a+B8+ay+a8) aT]co
(0) C02
a2.2!(Y)(T+l)
Comparing c2 from equation (30) with G2(q) from
equation (26) we note:
(31) G2(q) = q(apq + l+a+P8+ay+a8) ay.
Letting r = n in equation (25) and comparing cn
so found with G (q) from equation (26) reveals the following:
n
aPG (q) co
(32) Cn c *
an'n( y)(y+1) *** (y+nl)
Substituting equation (32) into equation (25), with appro
priate change of subscripts gives the following equation:
(33) a(n+l)(n+y)Gn+1(q)
a n(n+l)!(y) (+l) *** (y+n1) (+n)
[(a+l)(n)(nl) + apq + n(l+a+p8+ay+aS)]Gn(q)
an (y) (y+l) "' (y+nl)
[(nl)(n2) + (l+a+A)(nl) + a]Gn_l(q)
nI
a (n(n1)!(y)(y+l) (y+n2)
Multiplying both members of equation (33) by
an*n!(7)(y+l) ..* (y+n1), and simplifying the
result, gives us
(34) Gn+l(q) [(a+l)(n)(nl) + aoq + n(l+a+P8+ay+a8)]G (q)
[an(nl+a)(nl+P)(nl+y)]G n_(q).
Equation (34) may be used to compute coefficients
Gn+1(q) which may then be inserted into equation (26) to
give a general solution of equation (17) in a neighborhood
of z =O, when X 0.
CHAPTER V
DERIVATION OF A GENERAL DIFFERENTIAL EQUATION
WITH FIVE REGULAR SINGULAR POINTS
Consider the following differential equation:
(1) d2 +n lar r)diu [ n cror n 2. D
+(1) ( E(r E +Z r lu LJO.
dz rl zar z rl(zar)2 r zar
Analysis in Chapter III revealed that for Jp r Dm ,
n
(2) Dr, 0,
rl
(3) e (arDr + arP4 ) 0,
r=l
n 2
(4) E (arDr + 2ararr) 0, and
rl
(5) 8 (ar +r) n 2.
r=l
Since we are interested in studying a second order
linear ordinary differential equation with five distinct
regular singular points, we choose = 5. Then, following the
scheme of Riemann UL2], we denote a solution of equation (1)
thus:
al a2 a3 a4 a5
(6) u P a a2 a 3 a4 5 z
1 2 3 4 5
As before, we note that the first row displays the regular
singular points, rows two and three display respective ex
ponents at these singular points, and column six contains
the independent variable.
Since a linear transformation will take any three
distinct points into any other three distinct points, there
is no loss of generality if we let al = O, a2 = 1, a3 = a,
a4 = b, and a5 = c. .At the same time, we choose ar = 0 for
r = 1,2,3,4, a5 = a, 9 = 1 7, ^ = 1 6, 3 =1 E ,
P4 = 1 and P5 = p. Substitution of this information
into equation (6) yields:
0 1 a b c
(7) u = P 0 0 0 0 a z
1' 1 1E 1I
where we shall presently let o*oo .
Equation (1) also becomes:
(g) 2du + X + L a .Ll 2 du
7 2 z z1 za zb zc Iz
Sap Di D2 D D4 D
+ [ ++ 2++ +'5]u o,
(zc)2 z z1 za zb zc
where from equations (2) (5) we note that
(9) D1 + D2 + D3 + D4 + D5 = ,
(10) ap + D2 + aD3 + bD4 + oD5 = o,
(11) 2cap + D2 +a2D3 +b2D4 + 2D5 = 0, and
(12) 7y+ 8 + +T a = 1.
Reducing equation (8) to a form which is more amen
able to solution, we initially multiply both members by
z(zl)(za)(zb) to get:
(13) z(zl)(za)(zb)d2u +{(zl)(za)(zb)+8(z)(za)(zb)
dz2
+ E(z)(zL)(zb)+n(z)(z1)(za)+(laP)(z)(1)(za)(zb) ,dz
z 0o
+ {ap(z)(zl)(za)(zb) + D(zl)(za)(zb)+D2(z)(za)(zb)
(zc)
D5(z)(z1)(za)(zb)
+D3(z)(zl)(zb)+D 4(z)(zl)(za)+ } u = 0.
z c
du
Let K2 be the coefficient of d, thus:
(14) K2 y(zl)(za)(zb)+8(z)(za)(zb)+E(z)(zl)(zb)
+ n(z)(zl)(za)+(lap)(z)(zl)(za)(zb)
Z 0
= y[z3z2(a+b+l) +z(ab+a+b) ab]
+ 8[z3(a+b)z2+abz] +E[z3(b+l)z2+bz]
+ n[z3(a+l)z2+az] + (1a3p)(z)(z1)(za)(zb)
z c
(15) K2 = z3[y+8+E+ ]z2[Y(a+b+l)+8(a+b)+E(b+l)+T(a+l)]
+ z[E(a+b+ab) + 8(ab) +Eb +na] aby
+ (1ag)(z)(zl)(za)(zb)
z 0
We now substitute equation (12) into equation (15),
rearrange, and let c oo to obtain:
(16) Lir K z3(l+a+B)z2[(l+t+P)(l+a+b)aEbq8
+ zab( ) +a( ) + b( ab
+ z[ab(7+8) +a(,Y+l) + b('y+E)] abT.
We now let K3 be the coefficient of u, thus:
(aS)(z)(z1)(za)(zb)
(17) K = )(( )(a) + D1(zl)(za)(zb)
(z c)
+ D2(z)(za)(zb) + D3(z)(zl)(zb)
+ D4(z)(zl)(za) + D )(z1)(za)(zb)
5 z c
Expanding, we get:
(18) K1 a [z z3(a+b+l) + z2(a+b+ab) abz]
(zo)2
+ DIz5z4(a+b+l+2c) + z3(a+b+ab+2ac+2bc+2c+c2)
z2(ab+2ac+2bc+2abc+ac2+bc2+2)
+ z(ac2+bc2+abc2+2abc) abc2]
+ D2[z5z (a+b+2c) + z3(ab+2ac+2bc+o2)
z2(2abc+ac2+bc2) + abc2z]
+ D 3z5z4(l+b+2o) + z3(b+2c+2bc+c2)
z2(2bc+c2+bc2) + bc2z]
+ D [z5z4(a+2c+l) + z3(a+2ao+2c+c2)
z2(2ao+ac2+c2) + ac2z]
+ D5[z5z (a+b+c+l) + z3(a+b+ab+ac+bc+o)
z2(ac+bc+abc+ab) + abcz]).
Now from equation (18) we isolate coefficients of
like powers of z to get the following set of equations:
(19) z5: K4 = (D1+D2+D3+D4+D5).
(zc)
(20) z4: K5 
(21) z3: K6 =
(22) z2: K7 =
(23) z1: KS =
1 [a D1(a+b+2c+l) D2(a+b+2c)
(zo)
 D3(b+2c+l) D4(a+2o+l)
 D5(a+b+c+l)].
1 [ap(a+b+l) + D1(a+b+ab+2ac+2bc+2c+o2)
(zc)
+ D2(ab+2ac+2bo+o2)
+ D3(b+2c+2bo+c2) + D4(a+2ac+2c+c2)
+ D5(a+b+ab+ac+bc+c)].
1 [ap(a+b+ab)D (ab+2ac+2bc+2abc+ao2+bc2+c2)
(zc)2 1
 D2(2ab+ao2+bc2) D (2bc+o2+bc2)
 D4(2ac+ac2+c2) D5(ao+bc+abc+ab)].
1 [a ~ab+D (ac2+bc2+abc2+2abc)
(zc)2
+ D2(abc2)+D3(bc2)+D4(ac2)+D5(abc)].
(24) z0: K9 = 1 [D1(abc2)1.
(zo)
We now simplify the above equations.
Substituting equation (9) into equation (19),
(25) we immediately observe that K4 = 0.
Equation (20) can be rearranged as follows;
1
K 5 2 [aPD1(a+b+2c+l) D2(a+b+2c+11)
(zc)
D (a+b+2c+la) D4(a+b+2c+lb)
D5(a+b+2c+lo)]
1
= 2 [(a+b+2c+l) (D1 +D2+D3+D4+D5)
(zc)
+a P+D2+aD3+bD4+cD5 ].
Hence from equations (9) and (10),
K5 = 0.
Simplifying equation (21) slightly, we get:
K6 = 1 [ac(a+b+l)+(a+b+ab+2ac+2bc+2c+c2)
(zc)2
*(D1+D2+D3+D+D5)
D2(a+b+2o) D3(a+ab+2ac)
D4(ab+b+2bc) D5(ac+bc+o+c2)],
by equation (9) reduces to:
1
K6 = 2 [aaapB(b+l) D2(a+b+2o)
(zc)
D (a+ab+2ac+a2a2) D (ab+b+2bc)
D5(ac+bc+c+c2)].
K6 = [a(a3+D2+aD3+bD4+cD5) ap(b+l)
(zc)
D2(b+2c) D3(a+ab+2aca2)
D4(b+2bc) D5(bc+c+c2)],
by equation (10) reduces to:
(26)
which
which
K = 1 [ac(b+l) D2(b+2c) D3(a+ab+2aca2)
(zc)2
D4(b+2bc) D5(bc+c+c2)].
K6 = )1 [ap(2o+b+l2c) D2(b+2c+l1)
(zc)
D3(a+ab+2ac2a2+a2) D4(b+2bc+b2b2)
D5(bc+c+c2)]
S 1 (2cap +D2+a2D3+b2D4c+o2D5)
(zc)
aS(b2c+l) D2(b+2c1)
D3(a+ab+2ac2a2) D4(b+2bcb2)
D5(bo+c)],
which by equation (11) reduces to:
K  CaP(b+2c1) D2(b+2o1)
(zo)2
D (a+ab+2ac2a2) D4(b+2bcb2)
D (bc+c)].
Now letting coo, we get:
(27) Lim K6 = 0.
C 00
Equation (22) may be written as follows:
