Citation
A study of Heun's differential equation

Material Information

Title:
A study of Heun's differential equation
Alternate title:
Heun's differential equation
Creator:
Neff, John David, 1926-
Place of Publication:
[Gainesville]
Publisher:
University of Florida
Publication Date:
Language:
English
Physical Description:
77 leaves : ; 28 cm

Subjects

Subjects / Keywords:
Differential equations, Linear ( fast )
Genre:
bibliography ( marcgt )
theses ( marcgt )
non-fiction ( marcgt )

Notes

Bibliography:
Includes bibliographical references (leaves 75-76).
General Note:
Manuscript copy. Typed on one side of leaf only.
General Note:
Biography.

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Full Text
11 -
= O, a2 1, a, a^ b
(36) 0, a2 53 0, P1 1 7, 02 1 8 ^3 7 + 5 a fit Bk
This choice of exponents at the singularities satisfies
condition (32), which may be verified by addition* Equa
tion (31) may be written, using the values in (36), as
a2u 1 \y +
+ 0 + 1
- 7 -
6 + 1_
- a -
du
'Jit
dz Lz z
- 1
z -
a
z
- b J
dz
+ n
D1
4- 4. _
2 +
D3 ,
% 1
u 0.
L(z b)^
z z
- 1 + z
- a
z b
If we
multiply this
equation
by the expression
z(z -
l)(z a), we
get the
equation
(38)
z(z l)(z -
a)u"(z)
+ U(z -
- l)(z
- a) *
8z(z -
a)
+ (a + /3 +
1 r -
6)z(z -
1)
+ (1 a ft)z{z l)(z a)/(z b)]u(z)
+ [l/(z b)2][a/8z(z l)(z a)
+ D-^iz l)(z a)(z b)2
+ Dgz(z a)(z b)2 + Djz(z l)(z b)2
+ Dj^ziz l)(z a)(z b)]u(z) 0.
We will leave the coefficient of uM(z) unaltered. The co
efficient of u*(z) may be written
(a + ¡3 + l)z2 [a + £- 6 + 1+ (7+ 6)a]z + a7
+ .(A,,"..a A)2j.?..- l)lz.- a),t
The coefficient of u(z) may be written


(68)
_ = 2(27 a + 2r)(27 6 + 2r) .
2r+2 (r + l)(-a £ + 27 + 2r + 2) 2r*
If r = 0,
= 2(27 a) (27 $) .
2 1* (-a -6 + 27 + 2) *
if r = 1,
= 22(27 a) (27 a + 2) (27 fi)(27 P + 2)
^ 21 (-a jg + 27 + 2) (-a ¡5 + 27 + k) 0
and, if r = n 1,
c2n
2n(27- a) (27 a 4- 2)*(27 a + 2n 2)
n!
(27- Q) (27-/S+2) (27-/3+2n-2)
(-a-£f27+2)(-a-0+27+4)*(-a-^+27+2n)
C0 *
Hence, the solution corresponding to the root X = 27 a /3
is
(69) Uj, = cQ (2 2.1! r. /?.
- & \ ?; |)2].
%
The general solution of (63) is then a linear com
bination of (67) and (69) and may be written
(70) u(z) = AjfCf, §; a-* f 37 +-£; 4-(z |)2]
+ b3(z 2^;
- 4(z §)2],
where and Rj are arbitrary constants.


A STUDY OF
HEUNS DIFFERENTIAL EQUATION
By
JOHN DAVID NEFF
A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
August, 1956

ACKNOWLEDGMENTS
The writer wishes to extend his grateful appreci
ation to Dr. R. W. Cowan, who, as chairman of his super
visory ooramittee, not only suggested the problem but also
gave generously of his time and energy. He also wishes
to thank the other members of his supervisory committee
for their assistance during his doctoral program.
ii

TABLE OP CONTENTS
Page
ACKNOWLEDGMENTS il
Chapter
I. DERIVATION OP THE CANONICAL FORM ... 1
II. SOLUTIONS OP HEUN'S EQUATION 33
Solution in a General Case 33
Differential Equations with a Two
Term Recurrence Formula ..... 36
Solutions Relative to the Singular
Point at z = 0 42
Solutions Relative to the Singular
Point at z = l 4S
Solutions Relative to the Singular
Point at z = a, Where a^ 0,1,00. 54
Solutions Relative to the Singular
Point at Infinity 60
Solutions Relative to the Point
z = b, Where b^O, 1, a, 00... 67
BIBLIOGRAPHY ........ 75
BIOGRAPHICAL SKETCH 77
iii

CHAPTER I
DERIVATION OF THE CANONICAL FORM
The standard form of the linear differential
equation of second order is usually taken to be
(1) u"(z) + p(z)u'(z) + q(z)u(z) 0
and it is assumed that there is a domain S in the complex
plane in which both p(z) and q(z) are analytic, except at
a finite number of poles. Any point of S at which both
p(z) and q(z) are analytio is called an ordinary point of
the equation; other points of S are called singular points.
The property that the solution of a linear differential
equation is analytio except at singularities of the coef
ficients of the equation is common to linear differential
equations of all orders.
If a point o of S is such that, although p(z) or
q(z) or both have poles at c but the poles are of such
orders that (z o)p(z) and (z o) q(z) are analytio at o,
the point c is oalled a regular singular point of the equa
tion. Any poles of p(z) or of q(z) which are not of this
nature are called irregular singular points.
It is desired to find the most general ordinary
linear differential equation of second order for which all
points of the complex plane (including infinity), except
- 1 -

- 2 -
for the points z = a-^, a2, a^, are ordinary points,
and these points z a^, a2, ***, a& are regular singular
points with exponents and ¡5^, a2 and $2, , and $n.
This most general form is due to Klein [9* P 4o] and may
also be found in Whittaker and Watson [15, p. 2093.1 It
is desirable that we assume that the points z ** a^, (r 1,
2, n) are distinct, but it is not a necessary re
quirement .
If all points are to be ordinary points, exoept
for ar (r 1, 2, , n), then the coefficient p(z) in
(l) must be of the form
(2) p(z) =* p^/Cz + P2AZ a2^ + *
+ P_/(z a )
n n
= X V(z V*
where the various pr (r 1, 2, n) are non-zero con
stants. It is evident that p(z) has a simple pole at
z ar (r = 1, 2, **, n) and that the function (z ar)p(z)
is analytic at the point z =* ar. Similarly, the coeffi
cient q(z) in (1) must be of the form
(?) q(z) dj/(z ax)2 + d2/(z a2)2 +
+ d^/(z an)2 + Vj/iz ax) + D2/(z a2^
+ * + Dn/(z an),
*Sumerals in square brackets refer to the bibliog
raphy concluding the dissertation.

- 3 -
where the values cL are non-zero oonstants and the values
r
Dp are constants. It is again evident that, although q(z)
has a second order pole at z ap (r 1, 2, **, n), the
p
function (z ap) q(z) Is analytic at z ay for each r.
We may also write equation (3) in the form
(4) q(z) 2 dy/(z ap)2 + 2 Dy/(z ar).
r=l r=l
With the use of equations (2) and (4), equation (1) may he
written
(5)
uw (z)
+ 2 p /(z
r*l *
a^,) u'(z)
First, it is required that z = oo be an ordinary
point of equation (1). We will perform the substitution
z 1/w and require that w 0 be an ordinary point of
the resulting equation. If z = 1/w, then it is found that
o Ij.
u*(z) u(w)w'(z) = -w *u(w) and uM(z) w uMw)
+ 2w^*u'(w), and on substitution of these quantities in
(1), we obtain the equation
w^*u"(w) + [2w^ w2,p(l/w)] u*(w)
+ q(l/w) *u(w) 0,
which may be written in the form (w f 0)

(6)
- 4 -
u"(w) + [(2/w) (1/w2) p(l/w)] u*(w)
+ (l/wVq(l/w)*u(w) 0,
If w a 0 is an ordinary point of (6), then it Is necessary
that the coefficient [(2/w) (l/w2)*p(l/w)] be finite at
w 0. It follows that p(l/w) must begin with the term
2w, so that the first term (2/w) of the coefficient above
will be removed by subtraction, and continue only with
terms of higher degree in w than the first, so that the
entire coefficient will be finite at w = 0. As a conse
quence, p(l/w) must be of the form
(7) p(l/w) = 2w + AgW2 +
where the (i 2, 3, ) are constants. Prom (7), we
may revert to the variable z by inversion, to get
(g) p(z) 2/z + A2/z2 + A^/z^ + .
We may also describe this restriction of the function p(z)
2
by writing p(z) 2/z = 0(z ),
Similarly, in order that w = 0 be an ordinary
point of (6), it is also necessary that the function
q(l/w) be finite at w = 0. By reasoning similar to that
above, we conclude that q(l/w) must be of the form
(9) q(l/w) =* Bjjw^ + Bj-w^ + **,
where the (1 4,5, * *) are constants. With the rela
tion (9), we know that the coefficient (l/w^)*q(l/w) of
(6) is finite at w 0, Again reverting to the variable z

- 5 -
by inversion, we obtain
(10)q(z) + B^/z^ + 0(z^)
Beginning with equation (2), we may perform the
following algebraic manipulations:
p(z) £ Pj/z a ) 88 (1/z) 2 Pj/tl (ap/z)]
r=l r*l
(1/z) £ Pr[l + (ay/z) + (ar2/z2) + ]
(11) = (1/z) £ p_ + (1/z2) £ pa
rl r rl r r
+ (l/z3) E prar2 + .
r=l
On comparison of ooeffioients of (1/z) terms in (3) and
(11), we find that the relation
n
(12) £ p = 2
r=l r
must hold. The remaining sums in (11) are arbitrary,
sinoe the corresponding ooeffioients in (£) are arbitrary.
Similarly, beginning with (3), we may write
q(z) E dj/iz ar)2 + 21 Dj/iz ap)
r*l r*l
- (1/z2) S dp/Cl (V2)]2
r*l
+ (1/z) £ D_/[l (Oj/z)]
r=l

- 6 -
(13)
q(z) = (X/z2) 2 drCl + (ap/z) + (ap2/z2) + ]2
r=l
+ (l/z) 2 Dr[l + (ar/z) + (a-2/z2) + ]
r=l r
(l/z) 2 D + (l/z2) 2 (dn + Drar)
r=l
r=l
+ (l/z5) 2 (2apdr + ap2Dp) +
r=l
On comparison of coefficients of like powers of z in (10)
and (13), we find that the following three relations must
hold:
n
(14)
2
D
peal
n
r
(15)
2
r=l
n
(16)
2
r=l
(2a.
rr
0.
The method of Frobenius [4, p. 214] is used to sat
isfy the requirement that the exponents at z ar be ar
and /9r, Assume a solution to (5) of the form
(17) u(z) f0(z ajJ* + f^z ar)*+1 + f* 3/ 0
(12)
u*(z) f0X(z aj,)^"1 + fj/X + l)(z a )A +
(19) u(z) fftX(X l)(z aJx2
+ fx(X + 1)(X)(z ap)
X-l
+
The lndiolal equation is found by inspection to be

as follows:
the coefficient of the term (z ar)^*2t
f0Cx(x 1) + Xpr + dr3 0.
Sinoe tQ 0, by hypothesis, we may write this as
(20) X2 (1 pr)X+ dp = 0.
That there is no contribution to the lndicial equation
by the factors pj/(z aj) and terms dj/(z a^)2
+ Dj/(z a^) of (5), for i f1 r, may be shown by aotual
substitution of (17)* (13), and (19) into (5)f as follows:
(21)
f0X(X l)(z ar)X~2 + fx(X + 1)(X)(z ap)
X-l
+ +
2 Pj/(z aj) + Py/iz ay)
[f0X(z ap)^1 + fx(X + l)(z a,.)* + ]
r n
2 dj/(z aj)c + dy/(z ap)
U
* A v(z j1 + v ifh
f0(z apV
+ fx(z mj.)^ +
X+1A
* 0
We may write the coefficients 2 p^/(z a^J+Py/iz ap)
as follows:

- g -
n
n
= 2
3=1 Iar aj'TCl (z ar)/Uj a^)!
z a.
r
(22)
+ (z a*)2/^ a*)2 + ] +
It Ib apparent that the lowest exponent of the expression
(z ar) in (22) is -1, from the last term, and so the
last term is the only term used in (20). Similarly, we
may write the other coefficients of (21) as follows:
r
A (apTSjjS
+ 3(2 ar)2/(aj ar)2 + ] + 7 r-
J r (z ar)d
+ Ji sr^sjC1 + (z" ar)/iaJ"ar>
+ (z ar)2/(ai ar)2 + ] + Sc .
Z mm Qip
it is apparent that the lowest power of (z ar)
Again,

- 9 -
2
in (23) is found in the term d^iz ar) and henoe this
is the only term used in the indicial equation.
Returning to the indicial equation (20), the roots
of this equation are to be ar and fipt by hypothesis. Ab a
result, the folio-wing relations hold:
(24) Pr 1 ar ^r
(25) dr = arpr.
The substitution of (24) and (25) into (5) yields
the desired general form which we seek:
(26)
_2 n
+ Z
a*2
1 ar du
rl
!2L
z ar
dz:
afi n
_£_£ + z
2
D,
1 (z ar)
u
r=l z
Substitution of egressions (24) and (25)
w
(l6) and the use of expression (l4) yield
tions on (26):
n n
a 0.
into (12), (15),
the four restrlo-
n
2
r=l
hence
(27)
n
2
r=l
(2*3)
n
2
r=l
(29)
n
2
r=l
p = 2 (1 a § )
r=l
(ar + Pr) n 2,
(ar/3r + ayDj.) 0,
(2ar^par + ar2Dr) <= 0,
* n
2
r=*l
(ar + ^
83
2,

- 10 -
and
(30)
n
2 D o.
r=l
This dissertation is, in general, conoemed with the
differential equation with four regular singular points
and, in particular, concerned with a particular form known
as Heuns equation [6, p. 165]. We let n 4 in expres
sions (26) (30) to get
(3D
uM(z) + 2 ^*u(z)
r=l
z a.
r 4
Vr
+ 2
.r=l (z aj) rl
z a_
r J
u(z) 0,
with the restrictions
(32) 2 (<* + 2,
r=l
(33) 2 (ariSr + SrDr) = 0,
r=l
W 2 + ar2Dr) 0,
r**!
(35) 2 D 0.
r=l r
Heuns equation may he obtained by choosing the
following values for the various unknown constants:

11 -
= O, a2 1, a, a^ b
(36) 0, a2 53 0, P1 1 7, 02 1 8 ^3 7 + 5 a fit Bk
This choice of exponents at the singularities satisfies
condition (32), which may be verified by addition* Equa
tion (31) may be written, using the values in (36), as
a2u 1 \y +
+ 0 + 1
- 7 -
6 + 1_
- a -
du
'Jit
dz Lz z
- 1
z -
a
z
- b J
dz
+ n
D1
4- 4. _
2 +
D3 ,
% 1
u 0.
L(z b)^
z z
- 1 + z
- a
z b
If we
multiply this
equation
by the expression
z(z -
l)(z a), we
get the
equation
(38)
z(z l)(z -
a)u"(z)
+ U(z -
- l)(z
- a) *
8z(z -
a)
+ (a + /3 +
1 r -
6)z(z -
1)
+ (1 a ft)z{z l)(z a)/(z b)]u(z)
+ [l/(z b)2][a/8z(z l)(z a)
+ D-^iz l)(z a)(z b)2
+ Dgz(z a)(z b)2 + Djz(z l)(z b)2
+ Dj^ziz l)(z a)(z b)]u(z) 0.
We will leave the coefficient of uM(z) unaltered. The co
efficient of u*(z) may be written
(a + ¡3 + l)z2 [a + £- 6 + 1+ (7+ 6)a]z + a7
+ .(A,,"..a A)2j.?..- l)lz.- a),t
The coefficient of u(z) may be written

- 12 -
(39) £Di + d2 + D3 + D^z4 [(a + 2b + 1)DX
+ (a + 2b)D2 + (2b + DD^ + (a + b + DD^ + [(b2 + 2ab + 2b + a^ + (2ab + b2)D2
+ (2b + b2)D5 + (ab + a + b)D^ (a + l)a/S]z2
- [(ab2 + b2 + 2ab)D1 + ab2D2 + b2I>3 + abD^
- ao/3]z + ab2D^.
The relations (33) (35) using (36) are
OjS + D2 + aD^ + bDj^ 0
2a/3b + D2 + a2D^ + b2D^ = 0
DX + D2 4- D3 + 0.
The coefficient of z^ in (39) obviously vanishes, as does
the coefficient of -z^ if we write it in the form
(a + 2b + 1)0^ + D2 + Dj + D4)
- (a£ + D2 + aD^ + bD^).
The coefficient of z2 in (39) nay be seen to vanish if we
write it in the form
(b2 + 2ab + 2b + a)(D1 + D2 + + D^)
- (a + 2b + 1)(a/9 + D2 + a+ bD^)
+ (2a/Sb + D2 + b.2J)j + b2D^).
The coefficient of -z in (39) may be written
(ab2 + b2 + abM^ + D2 + + D^)
- (b2 + ab + b) (a/S + D2 + aDj + bD^)
+ b(2ba/9 + D2 + a2D^ + b2D^)
+ c^3(a b)(b 1) + abD^.

- 13 -
The first three terms above vanish, so the coefficient of
z may be written
a/?(a b)(b 1) + abD^.
Using the results of the above simplifications, the dif
ferential equation (3^) may be written
(4o) z(z l)(z a)u"(z) + {(a + ft + l)z2
- [a + £- 5 + 1+ (7 + 6)a]z + aT
It is desired to have the singularity denoted by
a^ b be at infinity. Hence, if we take the limit of
(4o) as b becomes infinite, we have the result known as
Heim's equation:
(4l) z(z l)(z a)u"(z) + -[(a + /5 + l)z2
-[a + /6-8 + l+ (7 + 8)a]z + al] u'(z)
+ ap(z q)u(z) 0.
Since 1^ is arbitrary,
(Jj-2) q aI^/a/3
is an arbitrary constant.
We digress to the case n 3 in (26) to determine
the most general form for the second order differential
equation having three regular singular points in the
finite complex plane. The three singularities are chosen

- 14 -
to be at the points z a,b,c and the exponents at these
singularities are a, a*; fi, ¡3 '; T, y*; respectively.
Prom (26), this general form is
(4-3) u"(z) + C(1 a a')/(z a)
+ (1 P 0')/(z b) + (1 7 7')/(z-o)]u(z)
+ [aa*/(z a)2 + /36*/(z b)2 + 77*/(z o)2
+ Dj/(z a) + Dg/(z b) + D^/(z o)]u(z) =0.
The restrictions (27) (3) in this case are
a + a* + /3 + /S* + r + r' 1,
aa' + /30' + 77* + aP^ + bDg + cDj 0,
2aa'a + 2/3/8b + 277o + a2Dx + b2Dg + o2D^ = 0,
+ Dg + Dj 0.
Factoring the expression [(z a)(z b)(z c)]-^ from
the coefficient of u(z) in (4-3), performing the indicated
divisions and simplifying, we obtain
a!a + ri.-.a- -gl -.A-JJ. + 1.- 2,.- .Ill gu
l1") dz2 Lz-a z-b z-c Jdz
+ + Dg + Dj]z2 + [aa* + fifi' + 77'
- D^(b + c) Dg(a + o) D^(a + b)]z
+ [aa*(a b o) + #5'(b a o)
+ 77(o a b) + Djbo + Dgao + D^ab]
+ oaf(a b)(a o) + /3)f(b a)(b o)
z a z-b
+ T7'(o a)(c JbJI u o.
z c J (z a)(z b)(z o)

- 15 -
Three of the four bracketed terms in the ooeffioient of
(z a) (z bV(z ~J'oT ln vanish with the application
of the above restriotions. The ooeffioient of z2 obviously
vanishes, as does the coefficient of z if we write it in
the form
cea* + fifi* + 77 + aDj. + bD2 + oD^
- (a + b + o)(D1 + D2 + Dj).
The constant term will also vanish if we write it in the
form
- (a + b + c)(aa + /3j3* + 77* + aD-L + bD2 + oD^)
+ (2aaa* + 2bA0* + 2o77* + a2D1 + b2D2 + o2D^)
' + (ab + ao + bo)(D^ + D2 + D^).
The resulting equation is then written
(^5) u"(z) + C(1 a a')/(z a)
+ (1 /3 /3*)/(z b) + (1 7 -7*)/(z o)]u(z)
+|[aa*(a b)(a o)/(z a)
+ /S/S*(b a)(b o)/(z b)
+ 77*(o a)(o b)/(z o)]}
{(z a)(z - o)} *
This form was first given by Papperitz [12, p.213]
and will be referred to as Papperitz1 canonical form. The
interesting feature of this equation is that it is com
pletely and uniquely determined by the three regular sin-

- 16 -
guiar points and the exponents at these singular points.
To express the faot that u satisfies an equation
of this type, Riemann [15, p. 206] used the notation
fa b o
(4-6) u = Mi ft 7 z .
U* 9' 7* ,
The singular points of the equation are plaoed In the first
row with the corresponding exponents direotly beneath them,
and the independent variable is placed in the fourth column.
Returning to Heim's equation (4-1) with its four
singular points, it would seem natural to inquire whether
an analogue to equation (4-5) exists in the case of Heun's
equation (4l). That such an analogue exists may be shown
with extensive algebraic manipulation.
First, let us note the differences between (45)
and (4-1). Equation (4-5) has three singularities at the
arbitrary points z = a, b, c and has arbitrary exponents
a, a; /3, /S'; 7, 7* at these points. Equation (4-1), how
ever, has been specialized with the choice of singularities
at z = 0, 1, a, oo and exponents 0, 1-7; 0, 1 8; 0,
7 + 8 a ¡3; a,/3 at these points respectively. We need
to generalize (4-1) to the extent that its singularities
and exponents are arbitrary also. It may be shown that a
change in the independent variable of (4-1) alters the

- 17 -
location of the singularities, but leaves the exponents of
these translated singularities invariant; also, it may be
shown that a change of the dependent variable in (4-1) al
ters the exponents at the singularities, but leaves the
original location invariant [see 15, p. 207]. In terms of
the notation used in (46),
a b
* { + J /3- 3 k
a*+ J J k
by change of dependent variable
7+ k Zr
7* + k ,
Correspondingly,
fa
b
o n
1
*1
1
r
P
7 z
P
7
zii
|6'
r* j
a'
j8*
r
by change of independent variable
(z-, a1 )(o-, bn) o (z a)(o b)
(z^ bl^l ~ ai) (z b)(o a)
We choose to change the location of the singulari
ties first. As noted, equation (4l) has singularities at
points z 0, 1, a, oo; we desire to shift these singular!-

- 10 -
ties to the points a, b, c, d. The transformation
(40)
z a
a(d b)(z1 a)
(b a)(z^ d)
will cause the singularities z *= 0, 1, a, ao to be moved
to the points b, c, a, d, whioh are distinct non-zero
constants. We may verify that (40) is the desired trans
formation by direct substitution, as follows:
1. Let z = 0 in (40). Then, on division by a, we
get (a bMz^ d) = (d b)^ a), or z^ = b.
2. Let z = 1 in (40). Then (1 a)(b aMz-^ -d)
= a(d bHz^ a), or
abd a2b bd + ad
1 "ad o? b + a "
This expression involving the four singularities is a con
stant and we may designate it by the letter o. Hence,
abd a2b bd + ad
ad ^ -b TV-
3. Let z => a in (40). Then a(d b)(z-^ a) 0,
and since a / 0, b $ d, we find that z^ ** a.
4. The inverse of (40) may be verified to be
(50)
d(b a)(z a) a(d b)
1 (b a)(z a) a(d b)
i
If we let z-> co in (50), we find that z^ d.
From (40), we find that
(51) u*(z) = [(b a)^ d)2/a(d b)(a djju'iz^
and

19
(52) u"(z) C(b a)2(z1 d)Va2(d b)2(a d)2]u"(z1)
+ [2(b a)2(z1 d)5/a2(d b)2(a d)2]u,(z1),
with the use of the relation
(53) z-^iz) (b a)^ d)2/a(d b)(a d).
We substitute (4g), (51), (52) into (4l), to get:
jTa(d aMz-L b)'
a(d b)(zT a)"
\j_ (b a)^ d)
(b a)(zx d)
a(cL a)(z1 b) (b a)(z1 d)
(b a)(z1 d)
v2/_
(b a)2(z1 d)^ a 2(b a)2(z, dp du T
I2(a b)S(a d)5'dzi2 + a2(d b)*(a &)s SjJ
a + p + 1
a2(d a)2(z1 b)2'
. (b aAz^ d)^ .
a + 8 6 + 1 +(Y + 5)a]
(*h a W rl \
+
(b a)(zT d)2
a(d b)(a d)
a/3
a(d a)(z1 b) q(b a)(z1 d)
(b a)(zi d)
u 0.
The coefficient of the second derivative may be simplified
to:
Ca^(d a)2(d b)(b a)2(zi a)^ b)2(z^ d)^
- a2(d a)(b a)2(d b)(z^ a)(z^ b)(z^ d)^]
[a2(b aP(d b)2(a d)2^ -

- 20
Further simplification yields
(5*0
(Zj a)(z1 b)(z1 o)(z1 d)(ad
(d b)(d a)(b a)
b + a)
The coefficient of the first derivative may be
simplified as follows:
[2a^(d a)2(d b)(b a)2^ a)(*1 b)2(z1 dp
- 2a2(d a)(b aP(d bMz^ aHz^ bMz^ dp]
[a2(b aP(d b)2(a d)2(z^ dp]"*3*
4 ta + P + l)a(d a)2(z1 b)2
(b a)(d b)(a d)
[a + /5-8 + l+(r+ 8)a][z1 b]Czx d]
* (d b)
+ ~ a)(z1 d)2
(d b)(a d)
_ ("(z^ aMz^ bHzj^ o)(z1 d)(ad a2 b + a)
"1 f 2a(d a)(z- b) 2(b ajiz^ o)"1
[(ad a2 b + aMz^ cMz^ d) (ad a2 b + a)
a(a + ¡3 + l)(d a)2(z1 b)
| liWWl II11 m I
(ad er b + a)(zj- aMz^ c)(z1 d)
+ Ca + 0 S + 1 + (7 + 6)a][d a][b a]
(ad a2 b + a)^ aHa^ o)
7(b a)2(zn d) 1 #
(ad a2 b + a)(z1 a)(z1 bjiz^ o)J
When expanded in partial fractions, the above expression
beoomes:

21
(55)/(zt a)(z1 b)(z1 o)(z1 d)(ad a2 b + a)"
(d b)(d a)(b a)
a(a + P + 1) [a + J3 6 + 1 + (7 + 6)a] + 7
(ad bt b + a)(o a)
(d a)(b ~ a)
^ a)
7(b a)(d b)
.(ad a2 b + a)(o b) J z^ b
+ C-2a(d a)(c b)^(o a)
-2(b a)(d c)(c a)(o b)
+ a(g + /8 + l)(d a)2(o b)2+ 7(b a)2(d o)2
+ (g + /3 6 + 1 + 7a + 6a)(d a)(b-a)(d-c)(o-b)3
[(ad a2 b + a)(d o)(c a)(o b)]1, 1
J z^ o
r(l a £)a ]
"(d a)(d b)ll
.(ad sid b + a)(d c).
1-* Ji
In order to further simplify the above expression,
we will make use of the following identities, whioh may be
verified by use of expression (49):
(56) (ad a2 b + a)(o a) (a 1)(d a)(b a)
(57) (ad a2 b + a)(o b) = (d b)(b a)
(53) (ad a2 b + a)(d 0) a(d b)(d a).
The use of (56), (57)* a**3- (5^) in (55) yields the
result:

22 -
(59) ('(z1 aHZj ~ 'b)(z1 oH^ d)(ad a2 b + a)
1 (d b)(d a)(b a)
ifa(q + /S + l)-[a+j9-6 + l+(T+ S)a] + T
U (a lMz^ a)
+ y + [(at^-8iltrat6a)
z^l b l a 1
+ (<* + +l)(o b)(d a) 7(b a)(d o
(o a)(d b)
The coefficient of the function u(z^) may also be
simplified to
a/3[a(d a)(z1 b) q(b a)(z1 d)]
(b a)^ d)
_f(z1 a)(z1 b)^ o)(z1 d)(ad a2 b + a)|
I (d b)(d a)(b a) J
f a/3[a(d a)(z1 b) q(b-a)(z1"d)][d~b][d-a]"|
\(z.j-a) (z-j-b) (z^-o) (z-j-d)^(ad - b + a) J
Writing expressions (54), (59) and (6o) together,
we have Heun's equation with four arbitrary singularities
at Zj a a, b, o, d, but with determinate exponents at
these singularities. Careful inspection of these three ex
pressions will reveal that a common factor of the three
terms exists and it may be removed by division. The result
of this division is

- 23 -
(61)
a2u
dz-i
a(g + /3 + l)-[q + /3-& + l + (7 + 5)a] + T
X
Zlrr
(a l)(z1 a)
(a + ¡S 6 + 1 +7a + 8a)
STT-I
(a + /3 + i)(o b)(d a) 7(b a)(d o) ] 1
(o a)(d b)
J Vo
1 a fr) du
zx d J dz^
q/3[a(d-a) (z^-b) q(b-a) (Zl-a)][d-b3Ca-a3
.(z^-a) (z^-b) (z^-c) (Zl-d)(ad a^ b + a)
u 0.
Next, it is desired to change the exponents at the
singularities z^ 53 a, b, c, d, from (0,7 + 6 q /3),
(0, 1 -7), (0, 1 5), and (q,/3), respectively, to
(q1, r + 6- q- /3 + q1), 0^, 1 T + 7^), i/^,1 6 + ^i)
and (q,|6), respectively. Let
1 (zx a)**1^ bf^iz^ oAu
or its inverse
-cu -T-j 'm@\
(62) u (Zl a) (2^ b) (z1 o) i^,
so that we find
, q*i ~Y-t **A> dUi
(63) = {(Zl a) 1(z1 b) 1(z1 o)
-1 -1 -1
- ia1(z1 a) + ^i^zi ** +/^l^zi ) 3t*i}

- 24 -
j2 tti -7 *1 d^Un
(64) {<23. a) 1(z1 b) ^(Z! o)
-1 -1 a -1 du,
- 2[a1(z1-a) + 3Cj^3
-2 ^ -2
+ [ -2 -1 -1
+ + l)(z1 0) + 2a1/31(z1 a) (e^ 0)
-1 -1
+ ScUjT^z^ a) (zx b)
-1 -1
+ 2/5171(z1 b) (z1 c) 3^}.
We substitute (62), (63), and (64) into (6l) direotly and
multiply the result by the factor
(z^ a) ^(z^ b)^1^ of1.
The resulting equation is given an the next page.

- 25
(65)
£> 2
dz-.
A
z-. a
- b
z-, o
du-
dz
*i(ai +11, 7i(Ti+ ;>, y
(zx a)2 (z^ b)2 (z.^ o)2
ab£i.
2ajZi
(z^ aXz-L o) (z1 aJZj^ b)
2/V
ill
Un
(z^ b) (z^ * o)
a(a + & + 1) [a + ff 5 + 1 + (7 + 6)a] + V
z-j_ b
(a 1)(z^ a)
a + fl 6 + l + (7 + S)a
a 1
+ U + Z3 + l)(c b)(d a) 7(b-a)(a-c)
(c a)(d b)
J zi o
+ 1 &1 fO^L
zx d J \dsj_
1 | ^1
_Z1 ~ a Z1 b Zl 0.
"l
AJcx/3[a(d-a)(z1-b) g(b-a)(zi-d)]Cd-b][d-a] } u 0 Q
L (z1-a)(z1-b)(z1-o)(z^-d)2(ad-a2-b+a) j 1
The coefficient of u^iz^) In (65) Is unity. The
coefficient of u^iz^) in (65) may be simplified by using
(56), (57), and (5g). The result is
(66) -2^ + a + /S-7-8 + l_tY-271 8-2/^
zi a Z1 b zx o
+ 1Lz~2L=JL.
Zi d

- 26 -
The coefficient of the function u^(z^) in (65) may
be written, using the simplifying expressions (56), (57)
and (53) wherever possible, as follows:
a. (a. + 1) 7.(7- + 1) 3AL + 1)
ik A + -A... A -f 11
** (zn b)2 (z^ o)2
(z^ a)
2a,7.
2al^l + -1*1
(zi aiz^ o) (z1 a) (z^ b)
(z1 b)(z1 o)
a + /3-y-5 + l
+ T 4. 8 4. A v* /L>
(z1- b J (z1 o) (z1 d)
z^ a
1 ft p
J*L
0i
Zi a zx b zx o
a6[a(d-a)(zn-b) q(b-a)(z^dJjCd-bjCd-a]
+ 1 p p *
(z^-a) (z-^-b) (z^-o) (z^-d) (ad-a-b+a)
As in (43), we remove a factor
**1
[(z^. aMz-j^ b)^ 0)^ d)]
and write the above coefficient in the form given on the
following page.

- 27 -
(67) Uj aHZj^ - oltzj^ 4)j-
{[o^+D-o^a+p-r-S+l] Z ff.t ~ d)
+ [7,(7, + X) 77,3 --Z-^~ a).(z^ ~ ?(Z1 ~ d)
xi 1 (z^ b)
+ viPi. + i) wx] -a).
XX X (zx o)
+ [20^ /^(a+0-7-8+1) a^lzj' b][zx d]
+ [20^ ^(a+zi-r-fi+l) ^ICzj^ o][z1 d]
+ V]Czl a^zi d]
- a^il a /3)(- b)^ o)
- Tx(l a P)(&l a)(Zi o)
- ^^(1 ot <5)(z^ a) (z-^ b)
gj6[a(da)(z1-b) q(b^a)(z1~d)][d-b][d^] 1
(ad-a2-b+a) (z-j-d) i
Expanding in powers of z^, we oolleot coefficients,
beginning with the highest power of z^ (the seoond power).
p
The coefficient of z^ in (67) simplifies to
(63) (o^+^+Tj2) + (20^+211^+2/6^) (^+4+^)
We note from (32) that the sum of the exponents at the
four singularities must be 2. Hence,
T+6-a-/5 + 2a1 + l- r+271 + l- 6 + 2^1
+ a + 0 2,

- 2g -
(69) al +/^1 + 85 *
Thus (62$) may be written as
(<*2 + ^1 + + $2. + Tj.) = *
The coefficient of in (67) is
[a1(a1 + 1) a^U + 0- 7 8 + 1)][a b c d]
+17^! + 1) 771][b a c d]
+ + 1) 5^][c a b d]
- [2a1j81 /3x(a + 6 7 8 +1) 0^8][b + d]
- [2a1)'2 72(a + /3 7 -6 + 1) Yo^Ho + d]
- (2^2 7^2 48)(a + d)
+ a2(l a fi){b + c) + a ^)(a + o)
+ 5^(1 a £)(a + b).
This expression may be written
(a b c dja^2 + (c a b d)/^2
+ (b a c d)")^2 2(b + djc^/^
- 2(c + d)a171 2(a + d^i + ia +01 + 7^
[(d a)a + (d a)/3 + (a b)7 + (a c)8
+ (b + e)].
The term in square brackets vanishes due to (69). The re
maining six terms may be written

- 29 -
(a b c)a+ (c a b)^2 + (b a cJT-j2
- 2ba1^1 200^ 2a^1 d(a1 + Y^2.
The last term again vanishes due to (69). If one writes
(69) as 7^ = a-j. 02.* tlie remaining six terms of the
above expression also vanish.
The constant term in (67) may be written
(a2-ab-ac-ad+bc+bd+cd)[a^2 -a^(a + /3-7-6)]
+ (b2-ab-bo-bd+ac+ad+cd)Cr12 + (l 7)71]
+ (c2-ac-bc-cd+ab+ad+bd) [/5^2 + (1 6)^]
+ [2(^/6 ^(a + /3-7-8 + 1) a-j^S ]bd
+ [2a171 7-j^ia + 3 -7-6 + 1) Ta-j^cd
+ (2/3^ rS T^Jad 0^(1 a /6)bo
- ?i(l a P)ac /^(l a 3)ab
+ a^Ca(d a) q(b a)][d b][d a]
_ b + a
This expression may be written, using (69) again, as
(b a)2^2 + 2(b c)(b aja^^ + (b c)2/^2
+ [(a d) (b a)a + (a d) (b a)# + (b a)2/
+ (c a)(b a)6 + (d b)(b a)]ai
+ C(b c)(a d)a + (b c)(a d)/S
+ (b c)(b a)T + (b c)(c a)8
+ (b c)(a + d b c)]^
+ a^[d a][d c + q(c b)].

- 30 -
This expression may be written
(70) C(b a)a^ + (b c)/^]2 + C(a d)(a + &)
+ (b a)7 + (o a)6 + (d b)]
[(b a)^ + (b c)^] + (b c)(a o)/^
+ a/S[d a][d c + q(o b)].
Collecting coefficients of the various Roman alphabetical
unknowns, we find that the constant term may be written in
the form
(71) K = ax(7 + 8 a /S + a^a2 + 7^1 7 + T^b2
+ /S(1 6 + /51)o2 + o0d2 + Ca1(71 7)
+ 71(7+ 5-a-J6 + a1- l)]ab + [cc^^ 6)
+ ^(0^ a P + y + 8 l)]ao
+ lojia. + (5 l)]ad + C^l(71 7)
+ 8)]bc + 71(a + /6- i)hd
+ i3x(a + 0 l)cd
+ a/3[(ac ad cd) + q(a d)(b 0)].
The remaining terms of (67) may be written
[^(a-^+l) (a+/3-y-6+l)][a b][a c][a d]
Z1 a
+ C-Vi2 + (1 -yiTSHb a]Cb o][b a]
Z1 -b
/5]2 + (X - a][o b][o a]
+ zx o
+ q^Cata a)S(d t>)2] _
(aa-a2-b+a)(zj a)

- 31 -
Combining terms and using (56) we may write the above as
(72) 0^(7 + 5 a 0 + 04 Ha b)(a c)(a d)
- a
+ Tl(1 - Y + r1)(b a)(b o)(b a)
zi -b
+ + a) (0 b)(o d)
z c
4- o^(<3- a) (d b) (d c)
zi -4
Thus, combining (66), (71), and (72), we arrive at
the canonical form for Heun's differential equation, which
vie believe to be new. This canonical form is
(73)
4^7
1 q^ (y + 6 a + a^)
z^ a
x 1-T1- d-r^i) +^L- (i-t+flj)
zx b zx 0
+ l....-.,?. a3
z^ d
f2i +
dzi
ai(T+Sag^ai)(a-b)(a-o)(a-d)
zx a
Y.,(l 7 + Y-,)(b a)(b c)(b d)
zx b
+ ^l(3- ~ 6 +^i)(o a)(c ~ b)(c ~ d)
zi o
+ cu9(d a) (d b) (d c) + K
U1
zi d
(z]_ aTiz! bMz! cMzi d)
= 0,
where K is given by (71).

- 32
For purposes of comparison with (4-5), assume for
the moment that we have the four singularities at the
points z = a, b, o, d, with e^onents a, a; /?'; y, y ;
6, 6*, respectively. The canonical form (73) then becomes
(74) fa + fl 2-,+ ~L.-J r. §.\ + l-.-.X.-.T,'.
dz2 L z a z-b z-c
+ 1-6, r...Ll 1 + rftafr ft), (a rr 1 (fr a)
z-d Jdz L z a
+ P&'Cb a)(b c) (b d)
z-b
+ Tr'(o a) (c b) (c d)
z-c
+ 66 1 (d a) (d b)(d c) + K,1
z-d J
* [iz a) (z b) (z o)(z d)"j *
where K* is the expression
aa'a2 + /3S'b2 + 77'c2 + 66'd2 + [Ha* 1)
+ a(/3 1) ]ab + [a(7' l) + 7(a* l)]ac
+ a(6 + 6 l)ad + [7(p 1) + £(r' l)]bc
+ /3(S + 6* l)bd + 7(6 + 6' l)cd
+ a/3[ao ad cd + q(a d) (b c)].

CHAPTER II
SOLUTIONS OF HEUN'S EQUATION
Solution in a General Case
Using the method of Frobenlus [4], we are able to
find a solution in a general case of Heim's equation* We
will discover on inspection of the solution that it is
essentially in terms of a difference equation which is more
difficult to solve than the original differential equation.
A particular form of the solution is given by Heun
[7, p. 131], and may be derived in the following manner
using (4l) of Chapter I. Assume a formal solution of the
form
u =
c0zX +
LX+1 X+r X+r+1 .
C]Lz + + crz + C^-jZ +
where Cq f 0. Then
= o^Xz*"1 + c,(X + l)zA + + c_(X + r)z?'+r_1
dz u j. i
+ o^fX + r + l)zX+r + "
and
= CqX(X l)zX2 + c,(X + l)XzX"^ +
dz
+ cr(X + r)(X + r l)zX+r2
+ om(X + r + 1)(X + r)zX+lvl + *
The indicial equation is the coefficient of zX-1 and may
be written

a7X + aX(X 1) = 0. (a ^ 0)
The roots of the indiclal equation are X = 0, 1 7. If
we set the coefficient of z* equal to zero, we get the re
lation
- a/6qc0 + a7o^(X + 1)- [a + 0- 6 + 1 + (7+6)a]o0X
+ ao1(X + 1)X (a + 1)oQX(X 1) = 0.
If we choose the root X = 0 of the lndicial equation, the
above relation becomes, when cQ = 1,
(1) (^ Offlq/aY. (y 0)
Similarly, if we set the coefficient of z**-1-
equal to zero, we get the relation
- a^q^ + a/3c0 + a7c2(X + 2) [a + 0 6 + 1
+ (7 + Sjajo^X + 1) + (a + 0 + l)oQX
+ a(X + 2)(X + l)o2 (a + 1)(X + l)Xc1
+ cQX(X 1) = 0.
If we let X = 0, cQ = 1, and use (l) above in this relation,
we find that
(2)
(3)
2 ~ 2 a4fyVI)-{g^5 + C + P 6 + 1
+ (7 + 8)a]q a7 j* .
We accordingly define a series
QD
u 1 + a/fl 2
yq)(^)a
n! 7(7+ 1) * (7 +
TV
and assume it to be a solution to Heun's equation. The
series (3) was denoted by

- 35 -
u = F(a, q; a, fi, 7, 8; 2)
by Heun [7, p. 1S1]. The function F does not refer to the
hypergeometric series, although Heun observed that the
function F(a, q; a, /5, 7, 6; 2) became the hypergeometrio
series F(a, /$; V; z) when a = q = 1.
From (1), we know that
(4) G-^q) = *
and from (2), we know that
(5) Gg(q) = + [a + /S 6 + 1 + (7 + 6)a]q a7.
The derivatives of (3) are
(6) a 2. ? Q-n^^z/a^n1
dz a n=i (n l)i 7(7 + l)***(7 + n l)
and
17) 2u a ^ ? Gn-H(q)(Z//a)n"1
7 dF ?nBl(n-l)!r(7+l)-(7+n)
Substituting (3), (6), and (7) into Heun's equation, the
recurrence relation, after simplification, is
(g) Go+^q) fa[(a + /5-8+n) + (7+6+n l)a]
+ a^q}On(q) (a+n-1) (/f+n-l) (7-Hn-l)anGn<-1(q).
We thus have a solution to Heun's equation in a
general case. The series (3) is the solution, together
with the three term recurrence formula (0) and the initial
values (4) and (5).

- 36 -
Inasmuch as the recurrence formula for the func
tions (^(q) (n = 1, 2, 3 ***) is a difference equation
which is actually more difficult to solve than the origi
nal equation, the solution for particular cases is essen
tially of not much value. On the other hand, if the re
currence formula were to have but two terms, the relation
among the coefficients could be found and a general solu
tion of Heun's equation would be the result. The next
section deals with the two term recurrence formula of a
differential equation in general and the remaining sections
with Heun's equation in particular.
Differential Equations with a Two Term
Recurrence Formula
We propose to find analytic solutions to Heun's
equation relative to the four singular points z = 0, 1, a,ax
It is convenient to use Frobenlus' method [4] and determine
those solutions for which a two term recurrence formula
exists. It will be discovered that some restrictions on
Heun's equation which yield a two term recurrence formula
will also decrease the number of singularities of the
equation. Inasmuch as Heun's equation has four singular
ities at the outset, it is felt that the removal of one or
more of these singularities is not desirable. In such
cases, the resulting equation is given, but further study

- 37 -
and solution was not effeoted.
In order to determine the conditions under which a
solution to Heun*s equation will have a two term recurrence
formula, we propose to use the criteria set forth by H.
Soheff [13] some fourteen years ago. We briefly summarize
this paper, as applied to the second order linear differen
tial equation, in the following paragraphs.
Given a differential equation of the form
(9) Pg(z) + Pj.(z) f§ + P0(z)w = 0,
where the Pj(z) (J =0 1 2) are analytic in some suffi
ciently small neighborhood of z = zQ in the extended z
plane and are not identically zero in this neighborhood, we
assume a series solution to (9) in the form
(10) w = 2 o> (z z )X+r. (c 0)
x=0 A o 0
Multiplying equation (9) by a suitable integral power of
(z zQ), say, (z zq)M, we find that (9) becomes
(11) q2(z)(z-zQ)2 ^5jr + q^zHz-z^ ~ + qQ(z)w = 0,
QZ
where the form of the coefficients qj(z) (j =0, 1, 2) is
(12) q^(z) = The proper selection of the constant M will cause the con
stants qjk (j =0, 1, 2) to obey the restriction

- 3* -
(13) <1202 + The formula obtained by equating to zero the coef
ficient of (z zQ)X+r (X = o, 1, 2, ) when the series
(10) and (12) are substituted into (11) is called the re
currence formula of the differential equation (1) relative
to the point z = zQ. Actual substitution of (10) and (12)
into (11) yields the following expression for the differ
ential equation:
(14) Cq20(z-zo)2 + q21(z-z0P ++ q2*(z-z0)Vf2 + '"3
y. Q y|
[cQr(r l)(z zQ) + c^(r + l)r(z zq) +
+ c^(r + X)(r + X l)(z z0)rfX"2 + ]
+ Uiq(z-z0) + 1ii(z-z0)2 ++ qlx(z-z0)X+1 +]
CoQr(z z0)1^1 + Oj^ir + l)(z zo)r +
+ o^(r + X)(z zo)^1 + ]
+ ^oo + Co (z z )r + 0-(z z)1*1 +
O o 1 o
+ o^(z zQ)X+r + ***] =0.
Prom (l4) we get the recurrence formulas, which are the
coefficients of the various powers of (z zQ). The first
of these recurrence formulas is called the indicial equa
tion.

l* 2
The coefficient of (z z ) is:
o
q200r(r 1) + q10c0r + Wo = 0
or
0 = 0-
pi *1
The coefficient of (z z ) is:
o
i[q20(r + 1)r + qio(r + 11 + W
+ o0[q21r(r 1) + q^r + ^3 = 0
or
2 2
1 qk0[r + 1]k + 0 ,JL kl^k =
k=0
k=0
r+A
The coefficient of (z zQ) is:
(15^ X qk0tr + X^k + X-1 qkltr + x +
k=0
k=0
+ X-m qkmCr + X m]k +
(16)
+ c kS> qkxCr'*k = *
If we define the function Q^ix) by the relation
2
Qj^ix) 2 Q.]anCx]^ ,
k=0
The Pochharamer notation for the factorial poly
nomials is used, where
[x]j x(x l)(x 2) (x j + 1)

equation (15) beoomes
(1?> £, Vr + X m)oX-m = *
m=o
Equation (17) is the recurrence formula which we seek. We
note that, beoause of (13),
(lg) Qq(x) 0.
By definition, the differential equation (9) is
said to have a two term recurrence formula relative to
z = zQ if the recurrence formula is of the type
(19) f(X,r)cx + g(*ir)cx_h = 0,
for X = h, h+1, h+2, where h is some positive integer,
and neither f(X,r) nor g(X,r) is identically zero in both
X and r. Because of (1$), we see that necessary and suf
ficient conditions that (19) be a two term recurrence for
mula are that
(20) QfcU) j 0
and
(21) Q^x) = 0 (m 5* 0, h)
The necessity of (20) is evident, while that of (21) may be
argued as follows. The ^(x) are polynomials of degree <2
and can vanish for x=r+X-m, X-m=h, h+1, h+2
and fixed r, only if Q^x) =0. Hence (20) and (21) are
equivalent to
q,)h 0 for some j
and

-4l-
= 0. (J = 0, 1, 2; m 0, h)
The coefficients p^(z) of (9) are then of the form
Pj(z) = [qJ0 + qJh(z z0^h^z z0^M-
The conditions which we desire then follow readily in the
form of a theorem.
Theorem. A necessary and sufficient condition that
a differential equation have a two term recurrence formula
relative to a point z = zQ is that in some neighborhood of
the point zQ
(22) Pj(z) = [S: Tjtz Z0)h]tz Z0fM,
where M and h > 0 are integers and the constants = q^Q,
Tj = are such that S^Tj 0 for some i,j.
We propose to apply the above criteria to Heun's
equation in the form
(23) z(z l)(z a)- + Hoc + /3 + l)z2 [a + 6 8
dz
+ 1 + (7 + 8)a]z + a7}~ + a^Siz q)u = 0.
dz
Relative to the notation used by Scheff and in the above
summary, we define
(24) p2(z) = z(z l)(z a),
(25) px(z) * (a+0+l)z2 [a+/3-6+l+(7+6)a]z + a7,
(26) P0(z) = afiiz q).
It is apparent that these three functions are analytic at
every finite point in the z plane.

- 42 -
Solutions Relative to the Singular Point at z = 0
The coefficients of equation (23), when expanded
in a finite Taylors series about 2=0, are
(27) P2(z) = z^ (a + l)z2 + az,
(22) px(z) = (a+/3+l)z2 [a+j3-.6+l+(7+6)a]z + a7,
(29) PQ(z) = a^z To begin, we select M = 1 and h = 2 and apply (22) to (27),
(22), and (29). We find that
T p
p2(z) = z-7 (a + l)z + az
= (S2 T2z2)z,
from which S2 = a and Tg = -1. We must accordingly set
(30) a = -1
in (23). Using (30) in (22), we find that
p-^z) = (a + 0 + l)z2 (a + (3 7 26 + l)z 7
from which S^^ = -7 and T^ = -(a + ^ + 1). We then require
that
(31) We next apply (22) to (29), to get
pQ(z) = a/3(z q)
= o ToZ2)*'1,
from which SQ 0 and TQ = -a/5. We require in addition
to (30) and (31) that
(32)q = 0.

Applying the restrictions (30), (31), (32) to (23),
we get the equation
(33) z(z2 l)u"(z) + [(a + /3 + l)z2 7]u'(z)
+ a£zu(z) =0,
In order to solve (33) we assume a series solution
of the form
i \ X X+2 ^ X+2r
u(z) = O0z + CSZ + + e2rz
, X+2r+2
+ Or, +
2r+2
(c0 ¥ 0)
so that
u*(z) = CqXz^ ^ + o2(X + 2)z^+^ +
+ c2r(X + 2r)z
X+2r-l
+ Ogr+2^ + 2r + 2)z
X+2r+l
+ ...
and
X2 X
u"(z) = OqX(X l)z + c2(X + 2)(X + l)z +
+ o2r(X + 2r)(X + 2r l)z^+2r""2
J . . X+2r
C2r+2(X + 2r + + 2r + l)z + ,
and substitute it into (33) The indicial equation, which
is the coefficient of z* \ is
(3^) -o0[x(x 1) + 7X] = 0
and, since cQ $ 0, its roots are
(35) X = 0, 1 7.
The recurrence relation among the coefficients is found by
X+2r+l
setting the coefficient of z
equal to zero:

- 44 -
(X + 2r)(X + 2r l)c2r (X + 2r + 2)(X + 2r + l)c2r+2
+ (X + 2r)(cx + 0 + l)o2r 7(X + 2r + 2)o2l>f2
+ a^c2p 0.
This relation may be written
(36) Oo^p sMV ?: g> + W 0¡>it
J 2r<-2 (X + 2r + 2)(X + 2r + 1 + 7) 2r
where r=0, 1, 2, .
If X = 0, the recurrence relation (36) becomes
(a + 2r)(jS + 2r)
2rf2 2(1 + r)(7 + 1 + 2r) 2r *
For r = 0,
Co
'2 2*1(7 + 1) *
for r = 1,
4 22*2 (7 + 1) (7 + 3)
and, for r = n 1,
c0*
a (a+2) * (a+2n-2) Q(ff+2) (/3+2n-2)
2n*nJ (7+1) (7+3) * (7+2n-l)
This solution may be written as a hypergeometric series:
(37) 00P(|, f; X^-; Z2).
If X = 1 7, the recurrence relation (36) becomes
(a 7 + 2r + l)(ff 7 + 2r + 1)
2r+2 2*(r + l)(-7 + 2r + 3) 2r*
For r = 0,

0 = (g 7 + 1X0 -7+1)
2 2-1! (-7 + 3)
for r = 1,
la .£ + U* r-V.+ 2MLl+. JJLO? ..rJ+3.L .
4 22*2i (-7 + 3)(-7 + 5) *
and, for r = n 1,
>2n
(a 7 + l)(a 7 + 3)-**(a 7 + 2n 1)
2n*ni
(ff 7 + 1)(0 7 + 3)''{ 3 7 + 2n 1)
(-7 + 3) (-7 + 5) *(-7+ 2n + 1)
0'
This solution may also be written as a hypergeometric
series:
(3) 2 o0z1-7(S-_|-tJL,
z2).
A general solution of Heun's equation relative to
the singular point z 0 is then a linear combination of
(37) and (3^) an*1 is written
u(z) f; z2)
+ B1Z
1-7
>
where and are arbitrary constants.
Continuing, we select M = 0 and h = 1 in (22). The
conditions imposed by (22) on (27), (2g), and (29) are then
z p
Pg(z) = vr (a + l)z + az
(S2 T2z)z2,
from which Sg = -(a + 1) and T2 = -1. We must accordingly

- 46 -
set
(39) a = O
In (23). Using (39) in (23), we find that
p-^iz) = (a + p + l)z2 (a + j0 8 + l)z
83 *S1 Tiz)z
from which = -(a + ¡5 6 + l) and = -(a + 3 + 1).
Similarly, applying (22) to (29), we find that
pQ(z) = a(3(z q)
= So V*
from which SQ = -afiq and Tq = -a8. Applying the sole re
striction (39) to (23), we obtain the equation
(40) z2(z l)u"(z) + C(a + /8 + l)z2
- (a + ft 6 + l)z]u'(z) + af3(z q)u(z) = 0.
Equation (4o) has but three regular singular pointy
at z = 0, 1, 00, caused by the coincidence of the singular
points z 0 and z = a. Hence (4o) is a confluent form of
(23). In accordance with the convention suggested in the
second section of this chapter, we will not effect further
study on this equation. We classify it, in the notation
of Inoe [3, p. 497], as an equation of the [0, 3* 0]
7
olass. The solution of (4o), in the notation of the first
^The formula [a, b, o] characterizes the equation
with the property:
a = the number of elementary singularities,
b = the number of non-elementary regular singular
ities,
c = the number of essential singularities

section of this chapter, would be
u = F(0, q; a, ft, 7, 6; z).
We note that in the assignment of values for M and
h in (22), we are restricted to the choice of M an integer
and h a positive integer. Inspection of (27) shows us
that we must assign M and h so that either Sg 1 or
T2 = -1, in order that the term of (27) does not have
to be set equal to zero to satisfy Scheffe's criteria.
This choice of z = 0 is clearly impossible. If we choose
2-M 3
Sg = 1, then z = zy and hence M = -1. Selection of
M = -1 foroes us to choose each of the three values Tg, T^,
Tq as zero, contrary to the hypothesis that S^*T^ 0 for
some i, j. We conclude that we cannot choose Sg = 1. If
h+2-M 3
we choose Tg = -1, then z = z and hence h M = 1.
We have discussed the choice h =* 1, M = 0 and h = 2, M = 1
in this section. The choice of h as any Integer greater
than two and of M as (h 1) may be shown to lead us to
the choice of each of the values Sg, S^, S0 as zero, also
contrary to hypothesis.
Accordingly, we conclude that we have discussed
all possible cases of (23) for which there exists a two
term recurrence formula for the solutions relative to the
singular point at z = 0,

Solutions Relative to the Singular Point at z = 1
In order to find the solutions of Heuns equation
(23) relative to the singular point at z = 1, we transform
the Independent variable by the relation
(4-1) = z 1.
Heun's equation becomes
(4-2) [(1 a)z1 + (2 ajz-j2 + z^W{z^)
+ {6(1 a) + [a + 0 a? + 6(1 a) + l]z^
+ (a + + l)z^}u(z^) + a/6(l q + z^)u(z^) = 0
and the coefficients p^(z^) (J = 0, 1, 2) are, respectively,
(4-3) Ppiz-^ = (1 a)z1 + (2 a)z^ + ZjS,
(44) P1(z1) = 6(1 a) + [a + j3 a7 + 6(1 a) + 1^
+ (a + /5 + l)Zl2,
(4-5) P0(2i) = a/5(l q + zx).
We begin by assuming M = 1 and h = 2 in (22), so
that (4-3) becomes
P2(zi) = (1 a)z1 + (2 a)2]2 +
p
*= (Sg ^2Z1 ^Z1
and hence S2 = 1 a and T2 = -1. We must accordingly re
quire that
(4-6) a = 2
in order to satisfy Scheffes criteria. Applying (4-6) to
(44), we find that

- 49 -
Pl(zl ) = -8 + (a+/3-27-8+l)z1 + (a+jfl+l)^2
= Si v 2>
from which = -8 and T^-(a + jS + l). Accordingly, we
require that
(47) a + /6 27 8 + 1 = 0
in equation (42) above. In the case of (45), we find that
p0(zi) = a/3(l q) + apz1
so that SQ = 0 and TQ = -ajS. We must accordingly set
(4g) q = 1
in equation (42) since, if a = 0 or /3 = 0, the differential
equation (23) could be reduced to a first order differen
tial equation. When we utilize (46), (47), and (4$) in
(42), the latter becomes
(49) z1(z12 liu"^) + [(a + /5 + Dz-j2 6]u,(z1)
+ a^z^uiz-^) = 0.
In order to find a solution to (49) by Probenlus'
method, we assume an infinite series solution of the form
+ " + 2rzl
X+2r
+ o2r+2zl
+
Then

- 50 -
uHz^) = CqXz-l^1 + o2(X + 2)z-jX+'* +
+ c2rU + 2r)z1X+2r~1
+ c2r+2(X + 2r + 2)z1X+2r+1 + *
and
u(Zi) = oQX(X Dz^'2 + o2(X + 2)(X + Dz^ +*
+ o2r(X + 2r)(X + 2r Dz^21*"2
+ o2r+2(X+2r+2)(X+2r+l)z1X+2r +
The indicial equation, which is the coefficient of z^^jis
c0[-5X X(X l)] = 0
and its roots are
(50) X = 0, 1-6.
The recurrence relation for the coefficients of this as-
suraed solution is the coefficient of z-^ and may be
written:
a6c2r 6iX + 2r + 2)2r+2 + (a + 0 + DU + 2r)c2r
- (X+2r+2)(X+2r+l)c2r+2 + (X+2r)(X+2r-l)c2r 0
or
(51)
. (X + 2r + a)(X + 2r + 5)
2r+2~ (X + 2r + 2)(X + 2r + 1 + 6) 2r*
for r = 0, 1, 2, .
Prom (50), if X = 0, then the coefficients are gov
erned by

- 51 -
r = (a + 2r) (3 + 2r)
2rf2 2(r + 1)(6 + 1 + 2r) 2r*
For r = 0,
. gff .
2 2*1! (6 + 1) *
for r = 1,
_ a(a + 2)6(3 +2)
4 22*2' (8 + 1)(6 + 3) 0
and, for r = n 1,
_a(a+2) (a+2n-2)<3(a+2)-**(0+2n-2) ^
Opy. 1 Or\
2nnJ (8+1)(8+3)* *(8+2n-l)
The solution corresponding to X = 0 Is then
(52) ^ = o0P[f, f; (z l)2].
From (50), if X = 1 6, then (51) becomes
0 = U 8 + 1 + 2r)(fl 8 I 2r) 0
2rf2 2(r + 1) (-6 + 2r + 3) 2r
If r = 0,
c2
If P 1,
= La....jLJLL(^ ,5 ,tJL)
2-1 (-8 + 3)
0
(a 8 -> l)(a 5 + 3)(g- 6 + l)(g- ^).f.
22*2¡ (-6 + 3)(-6 + 5)
and, If r = n 1,
2n =
(g 8 + l)(g 8 + 3)*-*(a 8 + 2n 1)
a^n!
. 'S 8 + 1)(6- 8 + 3)-~-(g- 8 + 2n XL
L (-6 + 3)(-6 + 5)* *(-6 + 2n + 1) 1 0

- 52 -
The solution corresponding to X = 1 6 then becomes
(53) = C0(z 1)1-6P[S=|1, Mil; =m.; (z 1)2].
A general solution of Heims equation relative to
the singular point at z = 1 is then a linear combination
of (52) and (53):
u(z) = A2F[|, |; S--L; (z l)2]
+ B2(z l)1_6F[2=|il, =$p.; (z l)2],
where and are arbitrary constants.
We continue by letting M = 0 and h = 1 in (22).
As a result,
Pgz^) = (1 a)zx + (2 a)z^2 +
(Sg ~ ^2Z1^Z1 *
from which S^ = 2 a and T^ = -1. We must accordingly
require that
(54) a = 1
in order to satisfy the hypothesis of a two term recurrence
formula. With the restriction (54) applied to (44), we
find that
Px(zi) = (a + (3 1 + ljz^ + (a + j6 + Dz-j2
= (Si T^z-^)z^,
so that S^=5a + j6-7+l and T^ = -(a + /3 + 1). Simi
larly, applying (54) to (45), we find that

- 53 -
po(zi) = = S0 T0Z1>
so that Sq = a/S(l q) and TQ = -a/3. With the sole re
striction (54), equation (4-2) becomes
(55) zi2(zi + 1)^(2^ + C(a + j0+ l)Zl2
+(a + /S 7 + liz-^Du'iz.^) + <1/6(1 q + z1)u(z1) = 0.
Equation (55) has three regular singular points at
z^ = -1, 0, 00 and because of this decrease in the number
of regular singular points, we do not attempt further study
of this equation. It is a confluent form of equation (42),
caused by the coincidence of the singular points at z 1
and z = a. In the notation of Ince [3, p. 497], we classi
fy (55) as an equation of the [0, 3, 0] class. Its solu
tion may be written
u = F(l, q; a, 3t 7, 6; zx)
using the notation of the first section of this chapter.
It is evident from (43) that we must assign the
coefficient of z^ to either or T^, since we cannot
cause z-jf Itself to vanish in order to satisfy Scheffe's
criteria. Thus, either S2 = 1 or Tg = -1. If we choose
2-M 1
the former, then z^ = zand hence M = -1. If H = -1,
however, we cannot assign T2 as the coefficient of either
of the remaining two terms, for h must be a positive inte-

- 54 -
ger. Failure to assign one of the two remaining terms of
(43) will lead to an impossibility, for we cannot cause
both (1 a) and (2 a) to vanish with a single choice
of a. We conclude that S2 = 1 is an impossibility. If
Tg = -1, then = z^ and hence h M = 1. We have
discussed the cases where h = 1, M = 0 and h = 2, M = 1.
If h is an integer greater than two and M = h 1, then we
are again led to the impossibility that both (1 a) and
(2 a) must vanish simultaneously. We conclude that no
other cases exist where (42) has a two term recurrence
formula relative to the singular point at z = 1.
Solutions Relative to the Singular
Point at z = a, Where a^O, 1, oo
In order to find the solutions of Heun's equation
relative to the regular singular point at z = a (a f 0, 1,
ao ), we transform the independent variable of (23) by the
relation z2 = z a. Heuns equation then becomes
(56) [z2^ + (2a l)z22 + ^ a)z2]u"(z2)
+ ((a + j6 + l)z22 + [(2a + 2/3 7 6 + 2)a
- (a + /3- 6 + 1) ]z2 + (a + $ i 6 + l) (a2-a) }u* (z2)
+ a/3(a q + z2)u(z2) = 0
and the coefficients pj(z2) O = 0, 1, 2) become
(57) P2(z2) z2^ + ^2a l)z22 + ^&2 a)z2,

55 -
(5) p-^Zg) *= - (a + jS-8+l)]z2 + (a + /S-7-8 + l)(a2-a),
(59) Po(z2^ = ^(a We begin by assuming M = 1 and h = 2 in (22), so
that (57) becomes
p2(z2) = (a a)z2 + (2a l)z2 + z^
= (S2 T2Z2 )z2,
p
from which S2 = a a and = -1. We must then require
that
(60) a = 1/2.
Using (6o), equation (520 becomes
Pl(z2) = (ol+/5+1)z22 + g-
= si Tiz2 *
so that =-(a + ft 7 8 + l)/4 and = (a + 0+ 1).
We must accordingly require that (6 7)/2 vanish, so we
set
(61) 8=7.
With (6o), we find that (59) becomes
p0(z2) = a|3(ir = (S0 Tqz2 )z2 1,
so that Sq = 0 and TQ = -a3. We require that
(62) q = |

- 56 -
in order to meet the conditions of the hypothesis.
With restrictions (6o), (6l), (62), equation (56) becomes
(63) z2(z22 j-)uM(z2) + [(a + P + l)z22
- jjKa + $ 27 + lJluizg) + a0z2u(z2) = 0.
In order to find a solution of (63) by Frobenius'
method, we assume a series solution of the form
u(zo) = cnz
0 2
2Z2
X+2
+ +
2rz2
X+2r
+ c z*+2r+2+...
2r+2z2
(cQ ^ 0)
so that
u*(z2) = CqXz^-1 + c2(X + 2)z2*+1 +
+ c2r(X + 2r)z2X+2r"1
+ o2rf2(X + 2r + 2)z2X+2r+1 +
and
u"(z2) = c0X(X-l)z2X"2 + c2(X + 2)(X+ l)z2X +
+ c2p(X + 2r)(X + 2r l)z2X+2r2
+ c2r+2(X + 2r + 2) (X + 2r +l)z2X+2r + .
The indiclal equation, which is the coefficient of z2^\is
o C-J
and its roots are
(64) X = 0, 27 a 0.

- 57
The recurrence relation among the various coefficients of
X121*11
the assumed solution is the coefficient of z2 and may
be written
O|0o2r |(q + 0 27 + 1)(X + 2r + 2)c2r+2
+ (a+3+ l)(X + 2r)o2r j^X + 2r + 2) (X + 2r+l)c2r+2
+ (X + 2r)(X + 2r l)c2r = 0,
or
(65)
_ 4(X + 2r t a)(X + 2r f 3) _
2r+2 (x + 2r + 2)(X + a + £ 2^ + 2r + 2) 2r
From (64), if X = 0, the relation (65) becomes
(66)
= 2(a + 2r)(ff + 2r)
2r+2 (r + l)(a + 0 27 + 2r + 2)
c2r*
If r = 0,
Or
2aj3
11 (a + jB 27 + 2)
o;
if r = 1,
_ 22*q(a + 2)6(6 ^ 2)
4 2* (a + /3 27 + 2) (a + 0 2 + 4) 0
and, if r = n 1,
_ 2n*q(a+2) * (a+2n-2)3(j3+2) * (ft+2n-2)
2n n I (q+0-27+2) (a+$-27+4) " (q+/5-27+2n) 0 *
The solution corresponding to X = 0 Is then
(67) = 00P[|, §; 9L+J-f-gX+2.; 4(Z 1)2].
Corresponding to the selection of X = 2y a ¡9
in (64), equation (65) becomes

(68)
_ = 2(27 a + 2r)(27 6 + 2r) .
2r+2 (r + l)(-a £ + 27 + 2r + 2) 2r*
If r = 0,
= 2(27 a) (27 $) .
2 1* (-a -6 + 27 + 2) *
if r = 1,
= 22(27 a) (27 a + 2) (27 fi)(27 P + 2)
^ 21 (-a jg + 27 + 2) (-a ¡5 + 27 + k) 0
and, if r = n 1,
c2n
2n(27- a) (27 a 4- 2)*(27 a + 2n 2)
n!
(27- Q) (27-/S+2) (27-/3+2n-2)
(-a-£f27+2)(-a-0+27+4)*(-a-^+27+2n)
C0 *
Hence, the solution corresponding to the root X = 27 a /3
is
(69) Uj, = cQ (2 2.1! r. /?.
- & \ ?; |)2].
%
The general solution of (63) is then a linear com
bination of (67) and (69) and may be written
(70) u(z) = AjfCf, §; a-* f 37 +-£; 4-(z |)2]
+ b3(z 2^;
- 4(z §)2],
where and Rj are arbitrary constants.

- 59 -
We next assume that M = 0 and h = 1 In (22), so
that (57) becomes
p2(z2) = (a a)z2 + (2a l)z2 + z2^
== (S2 T2z2)z22
and hence S2 = 2a 1 and T2 = -1. We must require that
the coefficient (a2 a) vanish in order to meet the con
ditions of Scheffe*s criteria. If a2 a is to vanish, we
must choose either a = 0 or a = 1 in (56)* However, these
two values have been excluded by hypothesis. It is inter
esting to note that, were we to continue from the choice
a = 0, we would find that the end result would be precisely
equation (4o), with z replaced by z2. Similarly, contin
uing from the choice a = 1 in (56), we would find that the
end result would be precisely equation (55) with z^ re
placed by z2. Thus, we have already considered the two
cases stemming from the above choice of M and h.
There are no other tenable values of M and h which
we may use in (22). We must assign the coefficient of the
term z2^ In (57) bo either S2 or T2# If S2 =* 1, then
2-M 7
z2 = z2 and M = -1. If M = -1, then we cannot assign
either of the coefficients (2a 1) or (a2 a) to T2, for
h must be positive. We conclude that S2 = 1 is an impossi
bility. If T2 = -1, then z2h+2~M = z25 and h M = 1, We
have discussed the cases h = 1, M = 0 and h = 2, M = 1. If

- 6o -
h is a positive integer greater than two and M = h 1,
then we cannot assign either (2a 1) or (a2 a) and hence
must cause "both to vanish simultaneously. This being
clearly an impossibility, we conclude that no other cases
exist where (56) has a two term recurrence formula relative
to the singular point at z = a, where a ^ 0, 1, 00 .
Solutions Relative to the Singular
Point at Infinity
The determination of solutions of Heuns equation
relative to the singular point at infinity necessitates a
change of variable in the original equation. We will trans
form the Independent variable of (23) by the relation
(71) z = l/w
and find the solutions of the transformed equation relative
to the singular point at w = 0. Prom (71), we determine
that
(72) u'(z) = -w2u(w)
and
(73) u"(z) = w\iM(w) + 2w^u'(w).
Substituting (71), (72), and (73) into (23), we
obtain the transformed equation:
(7*0 [aw^ (a + l)w^ + w2]u"(w) + {a(2 7)w^
+ [a + /3-6-l+(7 + 8- ajajw2
+ (1 a /3)w}u*(w) + a/3(l qw)u(w) 0.

- l -
The three coefficients p.j(w) (3 = 1 2) are determined
by Inspection to be
(75) P2(w) = aw^ (a + Dw? + vr2,
(76) px(w) = a(2 7)w^ + [a + 0- 8-1
+(7 + 8 2)3^ + (1 a 0)w,
(77) P0(w) = a/3(l qw).
We begin by choosing M = 0 and h = 2. Equation
(22) applied to (75) gives
P2(w) = aw* (a + l)w^ + w2
= (S2 T2w2)w2,
so that S2 = 1 and T2 = -a. Accordingly, we require that
the coefficient of w-^ vanish and must set
(7S) a =* -1.
Applying (7S) and (22) to (76) above, we get
p-^(w) = (7- 2)w^ + (a+ 0-7- 25 + l)w2 + (l a 3)w
= (S-j. T.jW2)w,
so that = 1 a 0 and T^ = 2 7. We require that the
second coefficient vanish, so we set
(79) <1 + 0- 7-26 + 1 = 0.
When we apply (22) to (77) above, we find that
p0(w) = a0(l qw)
so that SQ = aft and TQ = 0 and we must set

- 62 -
(go) (1 = 0.
As before, we require that q vanish rather than choose
either a = 0 or j6 = 0, for the latter choices reduce equa
tion (7*0 to essentially a first order differential equa
tion.
When restricted by relations (7^) (79) and- (go),
equation (7*0 becomes
(gl) w2(l w2)u"(w) + C(7 2)w^ + (1 a /3)w]u(w)
+ a/>u(w) = 0.
Using the method of Frobenius [4], we assume a
series solution of (gl) in the form
u(w) = cQwX + c2w*+2 +
* 2y+2r
Then
. X+2r+2 .
+ o2r+2w +
u'(w) = OqXwX^ + c2(X + 2)wX+^ +
(0 0)
, # _ X+2r1
+ o2r(X + 2r)w
+ c2r+2(X + 2r + 2)wX+2r+1 +
and
u"(w) = c0X(X l)wX2 + c2(X + 2)(X + l)w*
+ + c2p(X +.2r)(X + 2r l)wX+2r~2
+ c2r+2(X+2r + 2)(X+ 2r+DwX+2r + *.
The indicial equation, which is the coefficient of wX, is

- 63 -
Oq[X2 (a + j¡3)X + a¡0] = Ot
and its roots are
(62) X = a, (3.
The recurrence relation, the coefficient of is
^c2r+2 + ^ ~ a ~ 6)(X + 2r + 2)o2rf2
+ (7 2)(X + 2r)c2r (X + 2r)(X + 2r l)o2r
+ (X + 2r + 2)(X + 2r + l)c2r+2 = 0
(*?) ~ = (X + 2r)(X 7 + 2r + 1) .
3) 2r+2 (X a + 2r + 2) (X ¡5 + 2r + 2) 2r
or
(23)
c2r+2
for r =
0
H
If we
becomes
2r+2
(a + 2r)(a Of + 1 + 2r) .
+ l)(a + 2 + 2r) 2r*
2(r
If r = 0,
c
- a(a 7+1) 0 .
2 2*1 (a + 2) 0
if r = 1,
O, = L* l^Ha ,,7 + A)!*--.? + 3J. 0 .
4 22*2 (a /3 + 2)(a (5 + 4) *
and, if r = n 1,
2n
g(a + 2)*(a + 2n 2)
2n*nJ
(a I l)(a 7 + 3) * *(tt -7 + 2n 1)
(a jS + 2) (a j3 + 4) (a & + 2n)

- 64 -
This solution may be written In the form of a hypergeomet
ric series:
<*> ! oz"ap(t-
If we choose the root A = & of (22), the recurrence
relation (24) becomes
o = (/3 + 2r)(g 7 + 1 + 2r)
1n rTT ". 1 1 *
'2r+2 2(r + l)(j5 a + 2 + 2r)
2r
If r = 0,
c = &P-:e.
2 2*i: (0 a + 2)
if r = 1,
o, = 121 Q.
4 22'2! (/3- a + 2)(/S a + 4)
and, if r = n 1,
j3(|S + 2) (/6 + 2n 2)
2n "
2n,nJ
[~(/3 r + l)(ff 7 + 3)**(ff Y + 2n 1)1
* [ (0 a + 2)(j8 a + 4)-"(0 a + 2n) J V
This solution may be written
(85) = e^FCf, \ A:1 ~ f* 2; ^1.
A general solution of (21) may be written as a lin
ear combination of (24) and (25):
u(z) = A],z"aF[~, 3-=--X I. A +. ,2.
4^ *2* 2
vM, ^4^ 3.
2 z"
H
2 ~

- 65 -
where and are arbitrary constants.
We continue by choosing M = 0 and h = 1 in (22).
Equation (75) becomes
-2 p
P2(w) = aw (a + ljw-7 + w
= (S2 T2w)w2,
so that Sg = 1 and = a + 1. We accordingly require that
the first term vanish and must set
(36) a = 0.
When we use (36) and (22) in (76), we find that
P;j.(w) = (a + ¡3 6 l)w2 + (1 a $)w
= (S1 Txw)w,
so that = 1 a /5 and T^ = l- a- /3+5. Similarly,
p0(w) = aj3(l qw)
= S T w
0 0
so that Sq = a/3 and TQ = a/Sq.
With the sole restriction (36), equation (74) be
comes
(37) w2(l w)uH(w) + [(a + 0 8 l)w2
+ (1 a £)w]u'(w) + cc/6(1 qw)u(w) = 0.
This equation has but three regular singular points, at
w = 0, 1, and oo By the convention prescribed in the
second section of this chapter, we do not effect further
study of this equation. We classify it in the notation of
Inoe [3, p. 497] as an equation with the formula [0, 3 0]

- 66
and write its solution in the notation of the first section
of this chapter as
u = F(0, q; a, 0, 7, 8; w).
We will again show that no other tenable values of
M and h exist that we may use in (22) applied to (7*0 a-
bove. It is evident that we again must assign the coeffi
cient of w2 in p2(w) to either S2 or for we cannot let
the independent variable Itself vanish. Thus, we must
choose either S2 = 1 or = -1. If we choose the former,
then w^~M = w^ and hence M = 0. If M = 0, then w*1+2*"M
h+2 3 4
= w must be either r or w in (75) for we cannot cause
both of the coefficients a and a + 1 to vanish with a sin
gle choice of a. If w*1*2 = w-^, then h = 1; if w*1*2 = w\
then h = 2 and we have discussed both of these cases. We
conclude that only the two cases discussed above may be
h+?M P
used if S2 = 1. If T2 = -1, then w = w and h = M.
We cannot choose M a non-positive integer, for h cannot
be a non-positive integer, by hypothesis. If M is a posi
tive Integer, we are again led to the impossibility that
both a and a + 1 must vanish simultaneously. We conclude
that no other cases exist where the solution of (7*0 has
a two term recurrence formula relative to the regular
singular point at infinity.

- 67 -
Solutions Relative to the Point z = b,
Where b / 0, 1, a, oo
In order to find the solutions of Heuns equation
relative to the point z = b, where b / 0, 1, a, oo we
transform the independent variable of (23) by the relation
z^ = z b. Heuns equation then becomes
(gg) {b(b l)(b a) + [3b2 2(a + l)b + a]z^
+ (3b a l)z32 + z^Ju"(z^) + {[(a + /3 + l)b2
- (a+/3 5 + 1 +7a + 8a)b + a7]
+ [2(a + /S + l)b (a + /9-6 + l + 7a+ 8a) jz^
+ (a + /3 + 1)z^2}u'(z3) + a£(b q + z^Juiz^) = 0.
The coefficients p^(z^) are then
(S9) p2(z^) = b(b 1)(b a) +[3b2 2(a + l)b + a]z3
+ (3b a l)z^2 + z^,
(90) = {(cc+0+l)b2-[a+(3-8+l + (7 + 8)a]b+a7}
+ {2(a + i6 + i)b-[a + £-6 +1 + (7+ 6)a]}z-.
j
+ (a + 0 + l)z32,
(91) Po^z3^ = apib q + z3).
We begin by choosing M = 2 and h = 3 in (22), so
that equation (&9) may be written

- 6g -
p2(z3) = b(b l)(b a) + [3b2 2(a + l)b + a]z^
+ (3b a l)z32 + z33
= S2 T2z33,
from which S2 = b(b 1)(b a) and Tg = -1. We must then
cause the other two terms to vanish, so we set
(92) 3b2 2(a + l)b + a = 0
and
(93) 3b a 1 = 0.
Solving (92) and (93) simultaneously, we find that there
exists two solutions in a and b:
(94) a = \ U1 = V; b = l*gK = 1,¡¡I
and
(95) a = = _u. b = = a^-B,
p
where to and to are the imaginary cube roots of unity.
Continuing, we may write (90) as
Pl(z3) = {(ct+£+l)b2 [a+,6-S + 1 + (7+8)a]b+a7}
+ {2(a+/3 + l)b-[ + (a + & + l)z32
= (S^ T1z33)z3 1,
and so = 0 and = -(a + j6 + 1). Again, we must set
the other two coefficients of p^(z3) equal to zero, so

- 69 -
(96)
(a + 0 + l)b2- [a 4
P 6 + 1 +
(7 + 6)a]b + a7 = 0
and
(97)
2(a + 0 + l)b
- [a
+ 6 6 + 1 +
(7 + 6)a] = 0.
Similarly, we
find
that (91) may
be written
p0(z^) = a£(b
- q)
+ a^z^
= (Sq *^oz3
3)z -2
! 'z3
and hence SQ = 0 and TQ = -a/3. In order to satisfy
Scheffes criteria, we must set
W) q = b
in (33).
We will apply condition (9*0 now and defer (95)
until later in this section. With restriction (9*0, equa
tions (96) and (97) respectively, may be written
(99) a + 0 + 1 + 7(2id2 + Uf) + 6(6) 1) = 0,
(100) (a + 0 + 1)(26)2 + 1) 37a/2 + 366) = 0.
Eliminating 6 from (99) and (100), we obtain
(101) a + j5 + 1 = 37.
The restrictions (9*0, (9^), and (101), applied to
(33), yield the differential equation whose solution we
propose to find. This differential equation may be written
(102) i$z^ + (6) t62)]uu(z3) + 277z52u(z3)
+ 9a^6z^u(z^) = 0.
Using Frobenius* method, we assume a solution of

-To
dos) in the form
u(z-,) = cnz7^ + c.z *+3 + + o,_zT^+^r
'0*3
+ o
3 3
X+3r+3 + ...
3rt3Z3
3** 3
(o0 f 0)
Then
u(z^) = CqXz^"1 + c^(X + 3)z^X+2 + *'
+ o3r(J> + 3r)zjX+3r_1
+ o3r+J(X + 3r + 3)z3X+3r+2 +
and
u1* (z-j) = c0X(X-1)z5X"2 + c3(X+3)(X+2)z5X+1 +
+ c3r(X + 3r)(X + 3r Dz^^1^2
+ c3r+3(X+ 3r + 3)(^ +3r+ 2)z3X+^r+1 + .
X?
The lndicial equation, which is the coefficient of z3 ,1s
OqCCU >2)X(X 1)] = 0,
and its roots are
(103) x = o, l.
The recurrence relation, the coefficient of z3^+^r+^ is
9a0c3r + 277(X + 3r)c3r + 9U + 3r)(X + 3r l)c3r
+ (d2) (X + 3r + 3) (X + 3r + 2)c3r+3 = 0,
or
9(X + 3r + a)(X + 3r + &)
3^3 (^-STCx V 3r 3)U + '3r ; s) 3r*
for r = 0, 1, 2,


- 71 -
With the use of (9^) and (95) the factor (W2 4)) in the
denominator of the above recurrence relation may be shown
to be equal to -i/3, so that the recurrence relation may
be written
(104)
= a/?+, ?* ++ ?* /, 0,.
u3r+3 (\ + 3r
U + 3r + 3)(x + 3r + 2)
It Is interesting to note that if a = 3 and /8 = 2
or a = 2 and j6 = 3, the recurrence relation reduces to
3r+3 = ^ V
and the solution to (102) is a geometric series. We shall
assume that this is not the case, however, and let (104)
be the recurrence relation for the general case.
If we choose the root X = 0 of (103) the recur
rence formula (104) becomes
- 1/3 (a + 3*0(0 + 3*0
3r+3
If r = 0,
(r + l)(3r + 2)
J3r*
3 V
if r = 1,
0 = (i/3)2(a 4 3)603 + 3) 0 .
6 2* 2-5 V

- 72 -
The solution to (102) may then he written
(105) = c0F[|, i 31/3(2 b)3].
If we choose the root X = 1 in (103), equation
(104) may he written
3r+3
lV3(a + 1 + 3r)(fl + 1 + 3r)
(r + l)(3r + 4) 3r*
= l/3(g + 1)(|8 + 1)
3 l* 4
if r = 1,
_ (i/3)2(a + 1)U + 4)(j8 + 1)(|S + 4) ^ .
6 2! 4-7 *
and, if r = n 1,
(i/3)n(a + 1)(a + 4)**(a t 3a 2)'
3n =
(P + l)(fl + 4)(j3 + 3n 2)
0 *
4*7* **(3n + 1)
The solution to (102) may then he written
(106) Ug = c0(z b)F[2--I, &--1; j; 31/3(2 b)3].
A general solution of (102) is a linear combination
of (105) and (106) and may be written
u(z) = A^Ftj, |-j 3i^(z b)3]
+ Bc(z-b)F[a--lt 31/3(2-b)3],
where A,- and B_ are arbitrary constants,
b 5

- 73
The above solution is the result of the choice of
values (9^)* We now assume the values (95) and prooeed.
With (95)* equations (96) and (97)# respectively, become
(107) a + 0 + 1 + 7(6^ + 2U)) + 8(W2 1) = 0,
(103) (a + 0 + l)(2id + 1) 37W + 3W2 = 0.
Eliminating 6 from (107) and (log), we again obtain equa
tion (101). With restrictions (95) (9^), and (101), equa
tion (gg) becomes
(109) 19Zj3 + (t2 )]u"(z^) + 277z^2u'(z^)
+ 9a/3z^u(z^) =0.
We assume the same series as a solution of (109) as we
assumed for the solution of (102). The indicial equation
is identical to the indicial equation found earlier, except
for sign, and is
cQC(W2 ^)X(X 1)] 0.
Its roots are, as before,
(110) X 0, 1.
The recurrence relation is
9oc/3c^r + 277(X + 3r)c3r + 9U + 3r)(X + r l)o3r
+ (id2 id)(X + 3r + 3) (X + 3r + 2)c3r+3 = 0,
or
9U + 3r + cx)(X + 3rt|g)
3r+3 (id Prom (9^) and (95), the factor ((J (d2) in the denominator

- 74 -
is 1/3 and the recurrence relation may be written
(m) 03rt3 = P, M o
3 (X + 3r + 3)(X + 3r + a) ->r
On comparison of (111) with (104), it is evident that a
general solution of (109) may be written
u(z) = AgP[j, -3l/3(z b)3]
+ Bglz-bW2-^-1, -3i/3(z-b>3],
where Ag and are arbitrary constants.
It is a relatively simple matter to show that we
have chosen the only possible values of M and h in (22) for
the solutions relative to the point z = b, where b / 0, 1,
a, 00. Referring to ($9), It Is readily apparent that we
must assign the terms b(b-l)(b-a) and z^ to S2 and Tg
in that order, since h must be positive. Otherwise, we
would violate the hypothesis that b / 0, 1, a by failure
to assign the first or would require that the independent
variable vanish in the second, this being clearly an impos
sibility. Assignment of the first and fourth terms in
P2(z^) requires that h = 3* If h 3, It is apparent that
we must assign M to be M = 2, in order that we actually
assign the first and fourth terms and not any two terms
whose powers differ by three. We conclude that we have
found solutions of Heun's differential equation in all cases
for which there exists a two term recurrence formula.

BIBLIOGRAPHY
1. A* Erdiyi, Faakalaa flqmtAgn s£ assmsk
order, Duke Hath. J. vol. 9 (1942) pp. 4S-5S.
2. A. Erdlyi, Ifljgg£Bl SQUa&Xsm ISX, £§1& MBfc:
tions. Quart. J. Hath. Oxford Ser. vol. 13 (194-7) pp. 107-
112.
3. A. R. Forsyth, T^gory, gf J&££££gQ&aX Eg.^atipaa,
part IV, Cambridge, I906.
4. G. Frobenius, Ue&er d^e SSi
aren Dlfferentlalglelchungen durch rlehen. J. Reine Angew.
Hath. vol. 76 (IS73) pp. 214-35.
5. L. Fuchs, Zur theorle der linearen Dlfferential-
gleichuncen mit veranderlichen coefflclenten. J. Reine
Angew. Math, vol.' 6 (IS6) pp. 121-60.
1 6. K. Heun, Zur theorle der Riemam^schen funo-
tlonen zweiter ordnung mit vier verzweigungspunkten. Math.
Ann. vol. 33 (1^9) PP l5l-79.
7. K. Heun, Beitrage zur theorle der lame^ohen
functlonen. Hath. Ann. vol. 33 (ISB9) pp. IS0-96.
' S. E. L. Ince, Ordinary Differential Equations.
London, 1927.
9.F. C. Klein, Vpjr.les.un^Qft USbaXL
entlalglelchungen der zweiter ordnung. Goettingen, 1S94-.
10. C. G. Lambe and D. R. Ward, Some differential
efluaUoaS. SZ& ap.spclated integral equation, Quart. J.
Hath. Oxford Ser. vol. 5 (1934) PP. Sl-9/.
11. E. Makai, A t&§. .SSAutjofl q£
differential equation in 3. special casef Publ. Math.
(Debrecen) vol. 3 (19537 pp. l4o-3+3.
712. E. Papperitz. Ueber verwandte s-functlonen.
Hath. Ann. vol. 25 (1SS5) PP* 212-21.
- 75 -

- 76 -
A 13. H. Scheffe, WiepJZ differential eflmlfQnfi wltfo
a tvjo term recurrence formula. J. Math. Physics M. X. T.
vol. 21IT942) pp. 240-4-9.
14. V. w. Wakerllng, T]£ relations between solu
tions. p£ the differential epua.til.Qa 2l second order Him
four* singular noInte. Duke Math. J. vol. 16 (19^9)
PP. 591-99.
15. E. T. Whittaker and G. N, Watson, A Course of
Modern Analysis 4th ed., Cambridge, 1927*

BIOGRAPHICAL SKETCH
John David Neff was born July 30 1926, in Cedar
Rapids, Iowa. As an undergraduate, he attended Iowa State
College and Marquette University as a Naval Reserve Officer
Training Corps student during World War II. He received
the Bachelor of Naval Science degree from Marquette Uni
versity in March, 19^6. On his release from the U, S.
Navy in 19^7 he attended Coe College and graduated with
a Bachelor of Arts degree in mathematics, cum laude, in
June, 19^9.
As a graduate teaching assistant, he attended
Kansas State College and Purdue University and received
the Master of Science degree in mathematics from Kansas
State College in May, 1951* After more than a year as a
Member of the Technical Staff of the Bell Telephone Labo
ratories, New York, New York, he came to the University of
Florida in 1953 where he was an instructor (1953-1955)
and a graduate fellow (1955-1956) in the Department of
Mathematics.
He is a member of Pi Mu Epsilon, honorary mathe
matics fraternity, and Pi Gamma Mu, honorary social studies
fraternity.
- 77 -

This dissertation was prepared under the direction
of the chairman of the candidate's supervisory committee
and has been approved by all members of the committee.
It was submitted to the Dean of the College of Arts and
Sciences and to the Graduate Council and was approved as
partial fulfillment of the requirements for the degree of
Doctor of Philosophy,
August 11, 1956
1/' (2^/
A^'
Dean, College of Arts and Sciences
Dean, Graduate School
SUPERVISORY COMMITTEE;
\ / j / {
V ,-A/ < -



- 5 -
by inversion, we obtain
(10)q(z) + B^/z^ + 0(z^)
Beginning with equation (2), we may perform the
following algebraic manipulations:
p(z) £ Pj/z a ) 88 (1/z) 2 Pj/tl (ap/z)]
r=l r*l
(1/z) £ Pr[l + (ay/z) + (ar2/z2) + ]
(11) = (1/z) £ p_ + (1/z2) £ pa
rl r rl r r
+ (l/z3) E prar2 + .
r=l
On comparison of ooeffioients of (1/z) terms in (3) and
(11), we find that the relation
n
(12) £ p = 2
r=l r
must hold. The remaining sums in (11) are arbitrary,
sinoe the corresponding ooeffioients in (£) are arbitrary.
Similarly, beginning with (3), we may write
q(z) E dj/iz ar)2 + 21 Dj/iz ap)
r*l r*l
- (1/z2) S dp/Cl (V2)]2
r*l
+ (1/z) £ D_/[l (Oj/z)]
r=l


- 27 -
(67) Uj aHZj^ - oltzj^ 4)j-
{[o^+D-o^a+p-r-S+l] Z ff.t ~ d)
+ [7,(7, + X) 77,3 --Z-^~ a).(z^ ~ ?(Z1 ~ d)
xi 1 (z^ b)
+ viPi. + i) wx] -a).
XX X (zx o)
+ [20^ /^(a+0-7-8+1) a^lzj' b][zx d]
+ [20^ ^(a+zi-r-fi+l) ^ICzj^ o][z1 d]
+ V]Czl a^zi d]
- a^il a /3)(- b)^ o)
- Tx(l a P)(&l a)(Zi o)
- ^^(1 ot <5)(z^ a) (z-^ b)
gj6[a(da)(z1-b) q(b^a)(z1~d)][d-b][d^] 1
(ad-a2-b+a) (z-j-d) i
Expanding in powers of z^, we oolleot coefficients,
beginning with the highest power of z^ (the seoond power).
p
The coefficient of z^ in (67) simplifies to
(63) (o^+^+Tj2) + (20^+211^+2/6^) (^+4+^)
We note from (32) that the sum of the exponents at the
four singularities must be 2. Hence,
T+6-a-/5 + 2a1 + l- r+271 + l- 6 + 2^1
+ a + 0 2,


- 9 -
2
in (23) is found in the term d^iz ar) and henoe this
is the only term used in the indicial equation.
Returning to the indicial equation (20), the roots
of this equation are to be ar and fipt by hypothesis. Ab a
result, the folio-wing relations hold:
(24) Pr 1 ar ^r
(25) dr = arpr.
The substitution of (24) and (25) into (5) yields
the desired general form which we seek:
(26)
_2 n
+ Z
a*2
1 ar du
rl
!2L
z ar
dz:
afi n
_£_£ + z
2
D,
1 (z ar)
u
r=l z
Substitution of egressions (24) and (25)
w
(l6) and the use of expression (l4) yield
tions on (26):
n n
a 0.
into (12), (15),
the four restrlo-
n
2
r=l
hence
(27)
n
2
r=l
(2*3)
n
2
r=l
(29)
n
2
r=l
p = 2 (1 a § )
r=l
(ar + Pr) n 2,
(ar/3r + ayDj.) 0,
(2ar^par + ar2Dr) <= 0,
* n
2
r=*l
(ar + ^
83
2,


- 76 -
A 13. H. Scheffe, WiepJZ differential eflmlfQnfi wltfo
a tvjo term recurrence formula. J. Math. Physics M. X. T.
vol. 21IT942) pp. 240-4-9.
14. V. w. Wakerllng, T]£ relations between solu
tions. p£ the differential epua.til.Qa 2l second order Him
four* singular noInte. Duke Math. J. vol. 16 (19^9)
PP. 591-99.
15. E. T. Whittaker and G. N, Watson, A Course of
Modern Analysis 4th ed., Cambridge, 1927*


l* 2
The coefficient of (z z ) is:
o
q200r(r 1) + q10c0r + Wo = 0
or
0 = 0-
pi *1
The coefficient of (z z ) is:
o
i[q20(r + 1)r + qio(r + 11 + W
+ o0[q21r(r 1) + q^r + ^3 = 0
or
2 2
1 qk0[r + 1]k + 0 ,JL kl^k =
k=0
k=0
r+A
The coefficient of (z zQ) is:
(15^ X qk0tr + X^k + X-1 qkltr + x +
k=0
k=0
+ X-m qkmCr + X m]k +
(16)
+ c kS> qkxCr'*k = *
If we define the function Q^ix) by the relation
2
Qj^ix) 2 Q.]anCx]^ ,
k=0
The Pochharamer notation for the factorial poly
nomials is used, where
[x]j x(x l)(x 2) (x j + 1)


CHAPTER I
DERIVATION OF THE CANONICAL FORM
The standard form of the linear differential
equation of second order is usually taken to be
(1) u"(z) + p(z)u'(z) + q(z)u(z) 0
and it is assumed that there is a domain S in the complex
plane in which both p(z) and q(z) are analytic, except at
a finite number of poles. Any point of S at which both
p(z) and q(z) are analytio is called an ordinary point of
the equation; other points of S are called singular points.
The property that the solution of a linear differential
equation is analytio except at singularities of the coef
ficients of the equation is common to linear differential
equations of all orders.
If a point o of S is such that, although p(z) or
q(z) or both have poles at c but the poles are of such
orders that (z o)p(z) and (z o) q(z) are analytio at o,
the point c is oalled a regular singular point of the equa
tion. Any poles of p(z) or of q(z) which are not of this
nature are called irregular singular points.
It is desired to find the most general ordinary
linear differential equation of second order for which all
points of the complex plane (including infinity), except
- 1 -


- 20
Further simplification yields
(5*0
(Zj a)(z1 b)(z1 o)(z1 d)(ad
(d b)(d a)(b a)
b + a)
The coefficient of the first derivative may be
simplified as follows:
[2a^(d a)2(d b)(b a)2^ a)(*1 b)2(z1 dp
- 2a2(d a)(b aP(d bMz^ aHz^ bMz^ dp]
[a2(b aP(d b)2(a d)2(z^ dp]"*3*
4 ta + P + l)a(d a)2(z1 b)2
(b a)(d b)(a d)
[a + /5-8 + l+(r+ 8)a][z1 b]Czx d]
* (d b)
+ ~ a)(z1 d)2
(d b)(a d)
_ ("(z^ aMz^ bHzj^ o)(z1 d)(ad a2 b + a)
"1 f 2a(d a)(z- b) 2(b ajiz^ o)"1
[(ad a2 b + aMz^ cMz^ d) (ad a2 b + a)
a(a + ¡3 + l)(d a)2(z1 b)
| liWWl II11 m I
(ad er b + a)(zj- aMz^ c)(z1 d)
+ Ca + 0 S + 1 + (7 + 6)a][d a][b a]
(ad a2 b + a)^ aHa^ o)
7(b a)2(zn d) 1 #
(ad a2 b + a)(z1 a)(z1 bjiz^ o)J
When expanded in partial fractions, the above expression
beoomes:


- 26 -
The coefficient of the function u^(z^) in (65) may
be written, using the simplifying expressions (56), (57)
and (53) wherever possible, as follows:
a. (a. + 1) 7.(7- + 1) 3AL + 1)
ik A + -A... A -f 11
** (zn b)2 (z^ o)2
(z^ a)
2a,7.
2al^l + -1*1
(zi aiz^ o) (z1 a) (z^ b)
(z1 b)(z1 o)
a + /3-y-5 + l
+ T 4. 8 4. A v* /L>
(z1- b J (z1 o) (z1 d)
z^ a
1 ft p
J*L
0i
Zi a zx b zx o
a6[a(d-a)(zn-b) q(b-a)(z^dJjCd-bjCd-a]
+ 1 p p *
(z^-a) (z-^-b) (z^-o) (z^-d) (ad-a-b+a)
As in (43), we remove a factor
**1
[(z^. aMz-j^ b)^ 0)^ d)]
and write the above coefficient in the form given on the
following page.


- 67 -
Solutions Relative to the Point z = b,
Where b / 0, 1, a, oo
In order to find the solutions of Heuns equation
relative to the point z = b, where b / 0, 1, a, oo we
transform the independent variable of (23) by the relation
z^ = z b. Heuns equation then becomes
(gg) {b(b l)(b a) + [3b2 2(a + l)b + a]z^
+ (3b a l)z32 + z^Ju"(z^) + {[(a + /3 + l)b2
- (a+/3 5 + 1 +7a + 8a)b + a7]
+ [2(a + /S + l)b (a + /9-6 + l + 7a+ 8a) jz^
+ (a + /3 + 1)z^2}u'(z3) + a£(b q + z^Juiz^) = 0.
The coefficients p^(z^) are then
(S9) p2(z^) = b(b 1)(b a) +[3b2 2(a + l)b + a]z3
+ (3b a l)z^2 + z^,
(90) = {(cc+0+l)b2-[a+(3-8+l + (7 + 8)a]b+a7}
+ {2(a + i6 + i)b-[a + £-6 +1 + (7+ 6)a]}z-.
j
+ (a + 0 + l)z32,
(91) Po^z3^ = apib q + z3).
We begin by choosing M = 2 and h = 3 in (22), so
that equation (&9) may be written


19
(52) u"(z) C(b a)2(z1 d)Va2(d b)2(a d)2]u"(z1)
+ [2(b a)2(z1 d)5/a2(d b)2(a d)2]u,(z1),
with the use of the relation
(53) z-^iz) (b a)^ d)2/a(d b)(a d).
We substitute (4g), (51), (52) into (4l), to get:
jTa(d aMz-L b)'
a(d b)(zT a)"
\j_ (b a)^ d)
(b a)(zx d)
a(cL a)(z1 b) (b a)(z1 d)
(b a)(z1 d)
v2/_
(b a)2(z1 d)^ a 2(b a)2(z, dp du T
I2(a b)S(a d)5'dzi2 + a2(d b)*(a &)s SjJ
a + p + 1
a2(d a)2(z1 b)2'
. (b aAz^ d)^ .
a + 8 6 + 1 +(Y + 5)a]
(*h a W rl \
+
(b a)(zT d)2
a(d b)(a d)
a/3
a(d a)(z1 b) q(b a)(z1 d)
(b a)(zi d)
u 0.
The coefficient of the second derivative may be simplified
to:
Ca^(d a)2(d b)(b a)2(zi a)^ b)2(z^ d)^
- a2(d a)(b a)2(d b)(z^ a)(z^ b)(z^ d)^]
[a2(b aP(d b)2(a d)2^ -


a7X + aX(X 1) = 0. (a ^ 0)
The roots of the indiclal equation are X = 0, 1 7. If
we set the coefficient of z* equal to zero, we get the re
lation
- a/6qc0 + a7o^(X + 1)- [a + 0- 6 + 1 + (7+6)a]o0X
+ ao1(X + 1)X (a + 1)oQX(X 1) = 0.
If we choose the root X = 0 of the lndicial equation, the
above relation becomes, when cQ = 1,
(1) (^ Offlq/aY. (y 0)
Similarly, if we set the coefficient of z**-1-
equal to zero, we get the relation
- a^q^ + a/3c0 + a7c2(X + 2) [a + 0 6 + 1
+ (7 + Sjajo^X + 1) + (a + 0 + l)oQX
+ a(X + 2)(X + l)o2 (a + 1)(X + l)Xc1
+ cQX(X 1) = 0.
If we let X = 0, cQ = 1, and use (l) above in this relation,
we find that
(2)
(3)
2 ~ 2 a4fyVI)-{g^5 + C + P 6 + 1
+ (7 + 8)a]q a7 j* .
We accordingly define a series
QD
u 1 + a/fl 2
yq)(^)a
n! 7(7+ 1) * (7 +
TV
and assume it to be a solution to Heun's equation. The
series (3) was denoted by


This dissertation was prepared under the direction
of the chairman of the candidate's supervisory committee
and has been approved by all members of the committee.
It was submitted to the Dean of the College of Arts and
Sciences and to the Graduate Council and was approved as
partial fulfillment of the requirements for the degree of
Doctor of Philosophy,
August 11, 1956
1/' (2^/
A^'
Dean, College of Arts and Sciences
Dean, Graduate School
SUPERVISORY COMMITTEE;
\ / j / {
V ,-A/ < -


as follows:
the coefficient of the term (z ar)^*2t
f0Cx(x 1) + Xpr + dr3 0.
Sinoe tQ 0, by hypothesis, we may write this as
(20) X2 (1 pr)X+ dp = 0.
That there is no contribution to the lndicial equation
by the factors pj/(z aj) and terms dj/(z a^)2
+ Dj/(z a^) of (5), for i f1 r, may be shown by aotual
substitution of (17)* (13), and (19) into (5)f as follows:
(21)
f0X(X l)(z ar)X~2 + fx(X + 1)(X)(z ap)
X-l
+ +
2 Pj/(z aj) + Py/iz ay)
[f0X(z ap)^1 + fx(X + l)(z a,.)* + ]
r n
2 dj/(z aj)c + dy/(z ap)
U
* A v(z j1 + v ifh
f0(z apV
+ fx(z mj.)^ +
X+1A
* 0
We may write the coefficients 2 p^/(z a^J+Py/iz ap)
as follows:


- 59 -
We next assume that M = 0 and h = 1 In (22), so
that (57) becomes
p2(z2) = (a a)z2 + (2a l)z2 + z2^
== (S2 T2z2)z22
and hence S2 = 2a 1 and T2 = -1. We must require that
the coefficient (a2 a) vanish in order to meet the con
ditions of Scheffe*s criteria. If a2 a is to vanish, we
must choose either a = 0 or a = 1 in (56)* However, these
two values have been excluded by hypothesis. It is inter
esting to note that, were we to continue from the choice
a = 0, we would find that the end result would be precisely
equation (4o), with z replaced by z2. Similarly, contin
uing from the choice a = 1 in (56), we would find that the
end result would be precisely equation (55) with z^ re
placed by z2. Thus, we have already considered the two
cases stemming from the above choice of M and h.
There are no other tenable values of M and h which
we may use in (22). We must assign the coefficient of the
term z2^ In (57) bo either S2 or T2# If S2 =* 1, then
2-M 7
z2 = z2 and M = -1. If M = -1, then we cannot assign
either of the coefficients (2a 1) or (a2 a) to T2, for
h must be positive. We conclude that S2 = 1 is an impossi
bility. If T2 = -1, then z2h+2~M = z25 and h M = 1, We
have discussed the cases h = 1, M = 0 and h = 2, M = 1. If


- 15 -
Three of the four bracketed terms in the ooeffioient of
(z a) (z bV(z ~J'oT ln vanish with the application
of the above restriotions. The ooeffioient of z2 obviously
vanishes, as does the coefficient of z if we write it in
the form
cea* + fifi* + 77 + aDj. + bD2 + oD^
- (a + b + o)(D1 + D2 + Dj).
The constant term will also vanish if we write it in the
form
- (a + b + c)(aa + /3j3* + 77* + aD-L + bD2 + oD^)
+ (2aaa* + 2bA0* + 2o77* + a2D1 + b2D2 + o2D^)
' + (ab + ao + bo)(D^ + D2 + D^).
The resulting equation is then written
(^5) u"(z) + C(1 a a')/(z a)
+ (1 /3 /3*)/(z b) + (1 7 -7*)/(z o)]u(z)
+|[aa*(a b)(a o)/(z a)
+ /S/S*(b a)(b o)/(z b)
+ 77*(o a)(o b)/(z o)]}
{(z a)(z - o)} *
This form was first given by Papperitz [12, p.213]
and will be referred to as Papperitz1 canonical form. The
interesting feature of this equation is that it is com
pletely and uniquely determined by the three regular sin-


BIBLIOGRAPHY
1. A* Erdiyi, Faakalaa flqmtAgn s£ assmsk
order, Duke Hath. J. vol. 9 (1942) pp. 4S-5S.
2. A. Erdlyi, Ifljgg£Bl SQUa&Xsm ISX, £§1& MBfc:
tions. Quart. J. Hath. Oxford Ser. vol. 13 (194-7) pp. 107-
112.
3. A. R. Forsyth, T^gory, gf J&££££gQ&aX Eg.^atipaa,
part IV, Cambridge, I906.
4. G. Frobenius, Ue&er d^e SSi
aren Dlfferentlalglelchungen durch rlehen. J. Reine Angew.
Hath. vol. 76 (IS73) pp. 214-35.
5. L. Fuchs, Zur theorle der linearen Dlfferential-
gleichuncen mit veranderlichen coefflclenten. J. Reine
Angew. Math, vol.' 6 (IS6) pp. 121-60.
1 6. K. Heun, Zur theorle der Riemam^schen funo-
tlonen zweiter ordnung mit vier verzweigungspunkten. Math.
Ann. vol. 33 (1^9) PP l5l-79.
7. K. Heun, Beitrage zur theorle der lame^ohen
functlonen. Hath. Ann. vol. 33 (ISB9) pp. IS0-96.
' S. E. L. Ince, Ordinary Differential Equations.
London, 1927.
9.F. C. Klein, Vpjr.les.un^Qft USbaXL
entlalglelchungen der zweiter ordnung. Goettingen, 1S94-.
10. C. G. Lambe and D. R. Ward, Some differential
efluaUoaS. SZ& ap.spclated integral equation, Quart. J.
Hath. Oxford Ser. vol. 5 (1934) PP. Sl-9/.
11. E. Makai, A t&§. .SSAutjofl q£
differential equation in 3. special casef Publ. Math.
(Debrecen) vol. 3 (19537 pp. l4o-3+3.
712. E. Papperitz. Ueber verwandte s-functlonen.
Hath. Ann. vol. 25 (1SS5) PP* 212-21.
- 75 -


- 42 -
Solutions Relative to the Singular Point at z = 0
The coefficients of equation (23), when expanded
in a finite Taylors series about 2=0, are
(27) P2(z) = z^ (a + l)z2 + az,
(22) px(z) = (a+/3+l)z2 [a+j3-.6+l+(7+6)a]z + a7,
(29) PQ(z) = a^z To begin, we select M = 1 and h = 2 and apply (22) to (27),
(22), and (29). We find that
T p
p2(z) = z-7 (a + l)z + az
= (S2 T2z2)z,
from which S2 = a and Tg = -1. We must accordingly set
(30) a = -1
in (23). Using (30) in (22), we find that
p-^z) = (a + 0 + l)z2 (a + (3 7 26 + l)z 7
from which S^^ = -7 and T^ = -(a + ^ + 1). We then require
that
(31) We next apply (22) to (29), to get
pQ(z) = a/3(z q)
= o ToZ2)*'1,
from which SQ 0 and TQ = -a/5. We require in addition
to (30) and (31) that
(32)q = 0.


- 72 -
The solution to (102) may then he written
(105) = c0F[|, i 31/3(2 b)3].
If we choose the root X = 1 in (103), equation
(104) may he written
3r+3
lV3(a + 1 + 3r)(fl + 1 + 3r)
(r + l)(3r + 4) 3r*
= l/3(g + 1)(|8 + 1)
3 l* 4
if r = 1,
_ (i/3)2(a + 1)U + 4)(j8 + 1)(|S + 4) ^ .
6 2! 4-7 *
and, if r = n 1,
(i/3)n(a + 1)(a + 4)**(a t 3a 2)'
3n =
(P + l)(fl + 4)(j3 + 3n 2)
0 *
4*7* **(3n + 1)
The solution to (102) may then he written
(106) Ug = c0(z b)F[2--I, &--1; j; 31/3(2 b)3].
A general solution of (102) is a linear combination
of (105) and (106) and may be written
u(z) = A^Ftj, |-j 3i^(z b)3]
+ Bc(z-b)F[a--lt 31/3(2-b)3],
where A,- and B_ are arbitrary constants,
b 5


- 24 -
j2 tti -7 *1 d^Un
(64) {<23. a) 1(z1 b) ^(Z! o)
-1 -1 a -1 du,
- 2[a1(z1-a) + 3Cj^3
-2 ^ -2
+ [ -2 -1 -1
+ + l)(z1 0) + 2a1/31(z1 a) (e^ 0)
-1 -1
+ ScUjT^z^ a) (zx b)
-1 -1
+ 2/5171(z1 b) (z1 c) 3^}.
We substitute (62), (63), and (64) into (6l) direotly and
multiply the result by the factor
(z^ a) ^(z^ b)^1^ of1.
The resulting equation is given an the next page.


- 65 -
where and are arbitrary constants.
We continue by choosing M = 0 and h = 1 in (22).
Equation (75) becomes
-2 p
P2(w) = aw (a + ljw-7 + w
= (S2 T2w)w2,
so that Sg = 1 and = a + 1. We accordingly require that
the first term vanish and must set
(36) a = 0.
When we use (36) and (22) in (76), we find that
P;j.(w) = (a + ¡3 6 l)w2 + (1 a $)w
= (S1 Txw)w,
so that = 1 a /5 and T^ = l- a- /3+5. Similarly,
p0(w) = aj3(l qw)
= S T w
0 0
so that Sq = a/3 and TQ = a/Sq.
With the sole restriction (36), equation (74) be
comes
(37) w2(l w)uH(w) + [(a + 0 8 l)w2
+ (1 a £)w]u'(w) + cc/6(1 qw)u(w) = 0.
This equation has but three regular singular points, at
w = 0, 1, and oo By the convention prescribed in the
second section of this chapter, we do not effect further
study of this equation. We classify it in the notation of
Inoe [3, p. 497] as an equation with the formula [0, 3 0]


- 12 -
(39) £Di + d2 + D3 + D^z4 [(a + 2b + 1)DX
+ (a + 2b)D2 + (2b + DD^ + (a + b + DD^ + [(b2 + 2ab + 2b + a^ + (2ab + b2)D2
+ (2b + b2)D5 + (ab + a + b)D^ (a + l)a/S]z2
- [(ab2 + b2 + 2ab)D1 + ab2D2 + b2I>3 + abD^
- ao/3]z + ab2D^.
The relations (33) (35) using (36) are
OjS + D2 + aD^ + bDj^ 0
2a/3b + D2 + a2D^ + b2D^ = 0
DX + D2 4- D3 + 0.
The coefficient of z^ in (39) obviously vanishes, as does
the coefficient of -z^ if we write it in the form
(a + 2b + 1)0^ + D2 + Dj + D4)
- (a£ + D2 + aD^ + bD^).
The coefficient of z2 in (39) nay be seen to vanish if we
write it in the form
(b2 + 2ab + 2b + a)(D1 + D2 + + D^)
- (a + 2b + 1)(a/9 + D2 + a+ bD^)
+ (2a/Sb + D2 + b.2J)j + b2D^).
The coefficient of -z in (39) may be written
(ab2 + b2 + abM^ + D2 + + D^)
- (b2 + ab + b) (a/S + D2 + aDj + bD^)
+ b(2ba/9 + D2 + a2D^ + b2D^)
+ c^3(a b)(b 1) + abD^.


A STUDY OF
HEUNS DIFFERENTIAL EQUATION
By
JOHN DAVID NEFF
A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
August, 1956


- 50 -
uHz^) = CqXz-l^1 + o2(X + 2)z-jX+'* +
+ c2rU + 2r)z1X+2r~1
+ c2r+2(X + 2r + 2)z1X+2r+1 + *
and
u(Zi) = oQX(X Dz^'2 + o2(X + 2)(X + Dz^ +*
+ o2r(X + 2r)(X + 2r Dz^21*"2
+ o2r+2(X+2r+2)(X+2r+l)z1X+2r +
The indicial equation, which is the coefficient of z^^jis
c0[-5X X(X l)] = 0
and its roots are
(50) X = 0, 1-6.
The recurrence relation for the coefficients of this as-
suraed solution is the coefficient of z-^ and may be
written:
a6c2r 6iX + 2r + 2)2r+2 + (a + 0 + DU + 2r)c2r
- (X+2r+2)(X+2r+l)c2r+2 + (X+2r)(X+2r-l)c2r 0
or
(51)
. (X + 2r + a)(X + 2r + 5)
2r+2~ (X + 2r + 2)(X + 2r + 1 + 6) 2r*
for r = 0, 1, 2, .
Prom (50), if X = 0, then the coefficients are gov
erned by


- 36 -
Inasmuch as the recurrence formula for the func
tions (^(q) (n = 1, 2, 3 ***) is a difference equation
which is actually more difficult to solve than the origi
nal equation, the solution for particular cases is essen
tially of not much value. On the other hand, if the re
currence formula were to have but two terms, the relation
among the coefficients could be found and a general solu
tion of Heun's equation would be the result. The next
section deals with the two term recurrence formula of a
differential equation in general and the remaining sections
with Heun's equation in particular.
Differential Equations with a Two Term
Recurrence Formula
We propose to find analytic solutions to Heun's
equation relative to the four singular points z = 0, 1, a,ax
It is convenient to use Frobenlus' method [4] and determine
those solutions for which a two term recurrence formula
exists. It will be discovered that some restrictions on
Heun's equation which yield a two term recurrence formula
will also decrease the number of singularities of the
equation. Inasmuch as Heun's equation has four singular
ities at the outset, it is felt that the removal of one or
more of these singularities is not desirable. In such
cases, the resulting equation is given, but further study


- 53 -
po(zi) = = S0 T0Z1>
so that Sq = a/S(l q) and TQ = -a/3. With the sole re
striction (54), equation (4-2) becomes
(55) zi2(zi + 1)^(2^ + C(a + j0+ l)Zl2
+(a + /S 7 + liz-^Du'iz.^) + <1/6(1 q + z1)u(z1) = 0.
Equation (55) has three regular singular points at
z^ = -1, 0, 00 and because of this decrease in the number
of regular singular points, we do not attempt further study
of this equation. It is a confluent form of equation (42),
caused by the coincidence of the singular points at z 1
and z = a. In the notation of Ince [3, p. 497], we classi
fy (55) as an equation of the [0, 3, 0] class. Its solu
tion may be written
u = F(l, q; a, 3t 7, 6; zx)
using the notation of the first section of this chapter.
It is evident from (43) that we must assign the
coefficient of z^ to either or T^, since we cannot
cause z-jf Itself to vanish in order to satisfy Scheffe's
criteria. Thus, either S2 = 1 or Tg = -1. If we choose
2-M 1
the former, then z^ = zand hence M = -1. If H = -1,
however, we cannot assign T2 as the coefficient of either
of the remaining two terms, for h must be a positive inte-


- 74 -
is 1/3 and the recurrence relation may be written
(m) 03rt3 = P, M o
3 (X + 3r + 3)(X + 3r + a) ->r
On comparison of (111) with (104), it is evident that a
general solution of (109) may be written
u(z) = AgP[j, -3l/3(z b)3]
+ Bglz-bW2-^-1, -3i/3(z-b>3],
where Ag and are arbitrary constants.
It is a relatively simple matter to show that we
have chosen the only possible values of M and h in (22) for
the solutions relative to the point z = b, where b / 0, 1,
a, 00. Referring to ($9), It Is readily apparent that we
must assign the terms b(b-l)(b-a) and z^ to S2 and Tg
in that order, since h must be positive. Otherwise, we
would violate the hypothesis that b / 0, 1, a by failure
to assign the first or would require that the independent
variable vanish in the second, this being clearly an impos
sibility. Assignment of the first and fourth terms in
P2(z^) requires that h = 3* If h 3, It is apparent that
we must assign M to be M = 2, in order that we actually
assign the first and fourth terms and not any two terms
whose powers differ by three. We conclude that we have
found solutions of Heun's differential equation in all cases
for which there exists a two term recurrence formula.


- g -
n
n
= 2
3=1 Iar aj'TCl (z ar)/Uj a^)!
z a.
r
(22)
+ (z a*)2/^ a*)2 + ] +
It Ib apparent that the lowest exponent of the expression
(z ar) in (22) is -1, from the last term, and so the
last term is the only term used in (20). Similarly, we
may write the other coefficients of (21) as follows:
r
A (apTSjjS
+ 3(2 ar)2/(aj ar)2 + ] + 7 r-
J r (z ar)d
+ Ji sr^sjC1 + (z" ar)/iaJ"ar>
+ (z ar)2/(ai ar)2 + ] + Sc .
Z mm Qip
it is apparent that the lowest power of (z ar)
Again,


- 56 -
in order to meet the conditions of the hypothesis.
With restrictions (6o), (6l), (62), equation (56) becomes
(63) z2(z22 j-)uM(z2) + [(a + P + l)z22
- jjKa + $ 27 + lJluizg) + a0z2u(z2) = 0.
In order to find a solution of (63) by Frobenius'
method, we assume a series solution of the form
u(zo) = cnz
0 2
2Z2
X+2
+ +
2rz2
X+2r
+ c z*+2r+2+...
2r+2z2
(cQ ^ 0)
so that
u*(z2) = CqXz^-1 + c2(X + 2)z2*+1 +
+ c2r(X + 2r)z2X+2r"1
+ o2rf2(X + 2r + 2)z2X+2r+1 +
and
u"(z2) = c0X(X-l)z2X"2 + c2(X + 2)(X+ l)z2X +
+ c2p(X + 2r)(X + 2r l)z2X+2r2
+ c2r+2(X + 2r + 2) (X + 2r +l)z2X+2r + .
The indiclal equation, which is the coefficient of z2^\is
o C-J
and its roots are
(64) X = 0, 27 a 0.


ACKNOWLEDGMENTS
The writer wishes to extend his grateful appreci
ation to Dr. R. W. Cowan, who, as chairman of his super
visory ooramittee, not only suggested the problem but also
gave generously of his time and energy. He also wishes
to thank the other members of his supervisory committee
for their assistance during his doctoral program.
ii


- 57
The recurrence relation among the various coefficients of
X121*11
the assumed solution is the coefficient of z2 and may
be written
O|0o2r |(q + 0 27 + 1)(X + 2r + 2)c2r+2
+ (a+3+ l)(X + 2r)o2r j^X + 2r + 2) (X + 2r+l)c2r+2
+ (X + 2r)(X + 2r l)c2r = 0,
or
(65)
_ 4(X + 2r t a)(X + 2r f 3) _
2r+2 (x + 2r + 2)(X + a + £ 2^ + 2r + 2) 2r
From (64), if X = 0, the relation (65) becomes
(66)
= 2(a + 2r)(ff + 2r)
2r+2 (r + l)(a + 0 27 + 2r + 2)
c2r*
If r = 0,
Or
2aj3
11 (a + jB 27 + 2)
o;
if r = 1,
_ 22*q(a + 2)6(6 ^ 2)
4 2* (a + /3 27 + 2) (a + 0 2 + 4) 0
and, if r = n 1,
_ 2n*q(a+2) * (a+2n-2)3(j3+2) * (ft+2n-2)
2n n I (q+0-27+2) (a+$-27+4) " (q+/5-27+2n) 0 *
The solution corresponding to X = 0 Is then
(67) = 00P[|, §; 9L+J-f-gX+2.; 4(Z 1)2].
Corresponding to the selection of X = 2y a ¡9
in (64), equation (65) becomes


- 6o -
h is a positive integer greater than two and M = h 1,
then we cannot assign either (2a 1) or (a2 a) and hence
must cause "both to vanish simultaneously. This being
clearly an impossibility, we conclude that no other cases
exist where (56) has a two term recurrence formula relative
to the singular point at z = a, where a ^ 0, 1, 00 .
Solutions Relative to the Singular
Point at Infinity
The determination of solutions of Heuns equation
relative to the singular point at infinity necessitates a
change of variable in the original equation. We will trans
form the Independent variable of (23) by the relation
(71) z = l/w
and find the solutions of the transformed equation relative
to the singular point at w = 0. Prom (71), we determine
that
(72) u'(z) = -w2u(w)
and
(73) u"(z) = w\iM(w) + 2w^u'(w).
Substituting (71), (72), and (73) into (23), we
obtain the transformed equation:
(7*0 [aw^ (a + l)w^ + w2]u"(w) + {a(2 7)w^
+ [a + /3-6-l+(7 + 8- ajajw2
+ (1 a /3)w}u*(w) + a/3(l qw)u(w) 0.


- 62 -
(go) (1 = 0.
As before, we require that q vanish rather than choose
either a = 0 or j6 = 0, for the latter choices reduce equa
tion (7*0 to essentially a first order differential equa
tion.
When restricted by relations (7^) (79) and- (go),
equation (7*0 becomes
(gl) w2(l w2)u"(w) + C(7 2)w^ + (1 a /3)w]u(w)
+ a/>u(w) = 0.
Using the method of Frobenius [4], we assume a
series solution of (gl) in the form
u(w) = cQwX + c2w*+2 +
* 2y+2r
Then
. X+2r+2 .
+ o2r+2w +
u'(w) = OqXwX^ + c2(X + 2)wX+^ +
(0 0)
, # _ X+2r1
+ o2r(X + 2r)w
+ c2r+2(X + 2r + 2)wX+2r+1 +
and
u"(w) = c0X(X l)wX2 + c2(X + 2)(X + l)w*
+ + c2p(X +.2r)(X + 2r l)wX+2r~2
+ c2r+2(X+2r + 2)(X+ 2r+DwX+2r + *.
The indicial equation, which is the coefficient of wX, is


- 73
The above solution is the result of the choice of
values (9^)* We now assume the values (95) and prooeed.
With (95)* equations (96) and (97)# respectively, become
(107) a + 0 + 1 + 7(6^ + 2U)) + 8(W2 1) = 0,
(103) (a + 0 + l)(2id + 1) 37W + 3W2 = 0.
Eliminating 6 from (107) and (log), we again obtain equa
tion (101). With restrictions (95) (9^), and (101), equa
tion (gg) becomes
(109) 19Zj3 + (t2 )]u"(z^) + 277z^2u'(z^)
+ 9a/3z^u(z^) =0.
We assume the same series as a solution of (109) as we
assumed for the solution of (102). The indicial equation
is identical to the indicial equation found earlier, except
for sign, and is
cQC(W2 ^)X(X 1)] 0.
Its roots are, as before,
(110) X 0, 1.
The recurrence relation is
9oc/3c^r + 277(X + 3r)c3r + 9U + 3r)(X + r l)o3r
+ (id2 id)(X + 3r + 3) (X + 3r + 2)c3r+3 = 0,
or
9U + 3r + cx)(X + 3rt|g)
3r+3 (id Prom (9^) and (95), the factor ((J (d2) in the denominator


- 2g -
(69) al +/^1 + 85 *
Thus (62$) may be written as
(<*2 + ^1 + + $2. + Tj.) = *
The coefficient of in (67) is
[a1(a1 + 1) a^U + 0- 7 8 + 1)][a b c d]
+17^! + 1) 771][b a c d]
+ + 1) 5^][c a b d]
- [2a1j81 /3x(a + 6 7 8 +1) 0^8][b + d]
- [2a1)'2 72(a + /3 7 -6 + 1) Yo^Ho + d]
- (2^2 7^2 48)(a + d)
+ a2(l a fi){b + c) + a ^)(a + o)
+ 5^(1 a £)(a + b).
This expression may be written
(a b c dja^2 + (c a b d)/^2
+ (b a c d)")^2 2(b + djc^/^
- 2(c + d)a171 2(a + d^i + ia +01 + 7^
[(d a)a + (d a)/3 + (a b)7 + (a c)8
+ (b + e)].
The term in square brackets vanishes due to (69). The re
maining six terms may be written


BIOGRAPHICAL SKETCH
John David Neff was born July 30 1926, in Cedar
Rapids, Iowa. As an undergraduate, he attended Iowa State
College and Marquette University as a Naval Reserve Officer
Training Corps student during World War II. He received
the Bachelor of Naval Science degree from Marquette Uni
versity in March, 19^6. On his release from the U, S.
Navy in 19^7 he attended Coe College and graduated with
a Bachelor of Arts degree in mathematics, cum laude, in
June, 19^9.
As a graduate teaching assistant, he attended
Kansas State College and Purdue University and received
the Master of Science degree in mathematics from Kansas
State College in May, 1951* After more than a year as a
Member of the Technical Staff of the Bell Telephone Labo
ratories, New York, New York, he came to the University of
Florida in 1953 where he was an instructor (1953-1955)
and a graduate fellow (1955-1956) in the Department of
Mathematics.
He is a member of Pi Mu Epsilon, honorary mathe
matics fraternity, and Pi Gamma Mu, honorary social studies
fraternity.
- 77 -


- 32
For purposes of comparison with (4-5), assume for
the moment that we have the four singularities at the
points z = a, b, o, d, with e^onents a, a; /?'; y, y ;
6, 6*, respectively. The canonical form (73) then becomes
(74) fa + fl 2-,+ ~L.-J r. §.\ + l-.-.X.-.T,'.
dz2 L z a z-b z-c
+ 1-6, r...Ll 1 + rftafr ft), (a rr 1 (fr a)
z-d Jdz L z a
+ P&'Cb a)(b c) (b d)
z-b
+ Tr'(o a) (c b) (c d)
z-c
+ 66 1 (d a) (d b)(d c) + K,1
z-d J
* [iz a) (z b) (z o)(z d)"j *
where K* is the expression
aa'a2 + /3S'b2 + 77'c2 + 66'd2 + [Ha* 1)
+ a(/3 1) ]ab + [a(7' l) + 7(a* l)]ac
+ a(6 + 6 l)ad + [7(p 1) + £(r' l)]bc
+ /3(S + 6* l)bd + 7(6 + 6' l)cd
+ a/3[ao ad cd + q(a d) (b c)].


section of this chapter, would be
u = F(0, q; a, ft, 7, 6; z).
We note that in the assignment of values for M and
h in (22), we are restricted to the choice of M an integer
and h a positive integer. Inspection of (27) shows us
that we must assign M and h so that either Sg 1 or
T2 = -1, in order that the term of (27) does not have
to be set equal to zero to satisfy Scheffe's criteria.
This choice of z = 0 is clearly impossible. If we choose
2-M 3
Sg = 1, then z = zy and hence M = -1. Selection of
M = -1 foroes us to choose each of the three values Tg, T^,
Tq as zero, contrary to the hypothesis that S^*T^ 0 for
some i, j. We conclude that we cannot choose Sg = 1. If
h+2-M 3
we choose Tg = -1, then z = z and hence h M = 1.
We have discussed the choice h =* 1, M = 0 and h = 2, M = 1
in this section. The choice of h as any Integer greater
than two and of M as (h 1) may be shown to lead us to
the choice of each of the values Sg, S^, S0 as zero, also
contrary to hypothesis.
Accordingly, we conclude that we have discussed
all possible cases of (23) for which there exists a two
term recurrence formula for the solutions relative to the
singular point at z = 0,


- 6g -
p2(z3) = b(b l)(b a) + [3b2 2(a + l)b + a]z^
+ (3b a l)z32 + z33
= S2 T2z33,
from which S2 = b(b 1)(b a) and Tg = -1. We must then
cause the other two terms to vanish, so we set
(92) 3b2 2(a + l)b + a = 0
and
(93) 3b a 1 = 0.
Solving (92) and (93) simultaneously, we find that there
exists two solutions in a and b:
(94) a = \ U1 = V; b = l*gK = 1,¡¡I
and
(95) a = = _u. b = = a^-B,
p
where to and to are the imaginary cube roots of unity.
Continuing, we may write (90) as
Pl(z3) = {(ct+£+l)b2 [a+,6-S + 1 + (7+8)a]b+a7}
+ {2(a+/3 + l)b-[ + (a + & + l)z32
= (S^ T1z33)z3 1,
and so = 0 and = -(a + j6 + 1). Again, we must set
the other two coefficients of p^(z3) equal to zero, so


- 3 -
where the values cL are non-zero oonstants and the values
r
Dp are constants. It is again evident that, although q(z)
has a second order pole at z ap (r 1, 2, **, n), the
p
function (z ap) q(z) Is analytic at z ay for each r.
We may also write equation (3) in the form
(4) q(z) 2 dy/(z ap)2 + 2 Dy/(z ar).
r=l r=l
With the use of equations (2) and (4), equation (1) may he
written
(5)
uw (z)
+ 2 p /(z
r*l *
a^,) u'(z)
First, it is required that z = oo be an ordinary
point of equation (1). We will perform the substitution
z 1/w and require that w 0 be an ordinary point of
the resulting equation. If z = 1/w, then it is found that
o Ij.
u*(z) u(w)w'(z) = -w *u(w) and uM(z) w uMw)
+ 2w^*u'(w), and on substitution of these quantities in
(1), we obtain the equation
w^*u"(w) + [2w^ w2,p(l/w)] u*(w)
+ q(l/w) *u(w) 0,
which may be written in the form (w f 0)


- 66
and write its solution in the notation of the first section
of this chapter as
u = F(0, q; a, 0, 7, 8; w).
We will again show that no other tenable values of
M and h exist that we may use in (22) applied to (7*0 a-
bove. It is evident that we again must assign the coeffi
cient of w2 in p2(w) to either S2 or for we cannot let
the independent variable Itself vanish. Thus, we must
choose either S2 = 1 or = -1. If we choose the former,
then w^~M = w^ and hence M = 0. If M = 0, then w*1+2*"M
h+2 3 4
= w must be either r or w in (75) for we cannot cause
both of the coefficients a and a + 1 to vanish with a sin
gle choice of a. If w*1*2 = w-^, then h = 1; if w*1*2 = w\
then h = 2 and we have discussed both of these cases. We
conclude that only the two cases discussed above may be
h+?M P
used if S2 = 1. If T2 = -1, then w = w and h = M.
We cannot choose M a non-positive integer, for h cannot
be a non-positive integer, by hypothesis. If M is a posi
tive Integer, we are again led to the impossibility that
both a and a + 1 must vanish simultaneously. We conclude
that no other cases exist where the solution of (7*0 has
a two term recurrence formula relative to the regular
singular point at infinity.


- 69 -
(96)
(a + 0 + l)b2- [a 4
P 6 + 1 +
(7 + 6)a]b + a7 = 0
and
(97)
2(a + 0 + l)b
- [a
+ 6 6 + 1 +
(7 + 6)a] = 0.
Similarly, we
find
that (91) may
be written
p0(z^) = a£(b
- q)
+ a^z^
= (Sq *^oz3
3)z -2
! 'z3
and hence SQ = 0 and TQ = -a/3. In order to satisfy
Scheffes criteria, we must set
W) q = b
in (33).
We will apply condition (9*0 now and defer (95)
until later in this section. With restriction (9*0, equa
tions (96) and (97) respectively, may be written
(99) a + 0 + 1 + 7(2id2 + Uf) + 6(6) 1) = 0,
(100) (a + 0 + 1)(26)2 + 1) 37a/2 + 366) = 0.
Eliminating 6 from (99) and (100), we obtain
(101) a + j5 + 1 = 37.
The restrictions (9*0, (9^), and (101), applied to
(33), yield the differential equation whose solution we
propose to find. This differential equation may be written
(102) i$z^ + (6) t62)]uu(z3) + 277z52u(z3)
+ 9a^6z^u(z^) = 0.
Using Frobenius* method, we assume a solution of


- l -
The three coefficients p.j(w) (3 = 1 2) are determined
by Inspection to be
(75) P2(w) = aw^ (a + Dw? + vr2,
(76) px(w) = a(2 7)w^ + [a + 0- 8-1
+(7 + 8 2)3^ + (1 a 0)w,
(77) P0(w) = a/3(l qw).
We begin by choosing M = 0 and h = 2. Equation
(22) applied to (75) gives
P2(w) = aw* (a + l)w^ + w2
= (S2 T2w2)w2,
so that S2 = 1 and T2 = -a. Accordingly, we require that
the coefficient of w-^ vanish and must set
(7S) a =* -1.
Applying (7S) and (22) to (76) above, we get
p-^(w) = (7- 2)w^ + (a+ 0-7- 25 + l)w2 + (l a 3)w
= (S-j. T.jW2)w,
so that = 1 a 0 and T^ = 2 7. We require that the
second coefficient vanish, so we set
(79) <1 + 0- 7-26 + 1 = 0.
When we apply (22) to (77) above, we find that
p0(w) = a0(l qw)
so that SQ = aft and TQ = 0 and we must set


- 71 -
With the use of (9^) and (95) the factor (W2 4)) in the
denominator of the above recurrence relation may be shown
to be equal to -i/3, so that the recurrence relation may
be written
(104)
= a/?+, ?* ++ ?* /, 0,.
u3r+3 (\ + 3r
U + 3r + 3)(x + 3r + 2)
It Is interesting to note that if a = 3 and /8 = 2
or a = 2 and j6 = 3, the recurrence relation reduces to
3r+3 = ^ V
and the solution to (102) is a geometric series. We shall
assume that this is not the case, however, and let (104)
be the recurrence relation for the general case.
If we choose the root X = 0 of (103) the recur
rence formula (104) becomes
- 1/3 (a + 3*0(0 + 3*0
3r+3
If r = 0,
(r + l)(3r + 2)
J3r*
3 V
if r = 1,
0 = (i/3)2(a 4 3)603 + 3) 0 .
6 2* 2-5 V


- 23 -
(61)
a2u
dz-i
a(g + /3 + l)-[q + /3-& + l + (7 + 5)a] + T
X
Zlrr
(a l)(z1 a)
(a + ¡S 6 + 1 +7a + 8a)
STT-I
(a + /3 + i)(o b)(d a) 7(b a)(d o) ] 1
(o a)(d b)
J Vo
1 a fr) du
zx d J dz^
q/3[a(d-a) (z^-b) q(b-a) (Zl-a)][d-b3Ca-a3
.(z^-a) (z^-b) (z^-c) (Zl-d)(ad a^ b + a)
u 0.
Next, it is desired to change the exponents at the
singularities z^ 53 a, b, c, d, from (0,7 + 6 q /3),
(0, 1 -7), (0, 1 5), and (q,/3), respectively, to
(q1, r + 6- q- /3 + q1), 0^, 1 T + 7^), i/^,1 6 + ^i)
and (q,|6), respectively. Let
1 (zx a)**1^ bf^iz^ oAu
or its inverse
-cu -T-j 'm@\
(62) u (Zl a) (2^ b) (z1 o) i^,
so that we find
, q*i ~Y-t **A> dUi
(63) = {(Zl a) 1(z1 b) 1(z1 o)
-1 -1 -1
- ia1(z1 a) + ^i^zi ** +/^l^zi ) 3t*i}


- 2 -
for the points z = a-^, a2, a^, are ordinary points,
and these points z a^, a2, ***, a& are regular singular
points with exponents and ¡5^, a2 and $2, , and $n.
This most general form is due to Klein [9* P 4o] and may
also be found in Whittaker and Watson [15, p. 2093.1 It
is desirable that we assume that the points z ** a^, (r 1,
2, n) are distinct, but it is not a necessary re
quirement .
If all points are to be ordinary points, exoept
for ar (r 1, 2, , n), then the coefficient p(z) in
(l) must be of the form
(2) p(z) =* p^/Cz + P2AZ a2^ + *
+ P_/(z a )
n n
= X V(z V*
where the various pr (r 1, 2, n) are non-zero con
stants. It is evident that p(z) has a simple pole at
z ar (r = 1, 2, **, n) and that the function (z ar)p(z)
is analytic at the point z =* ar. Similarly, the coeffi
cient q(z) in (1) must be of the form
(?) q(z) dj/(z ax)2 + d2/(z a2)2 +
+ d^/(z an)2 + Vj/iz ax) + D2/(z a2^
+ * + Dn/(z an),
*Sumerals in square brackets refer to the bibliog
raphy concluding the dissertation.


- 3* -
(13) <1202 + The formula obtained by equating to zero the coef
ficient of (z zQ)X+r (X = o, 1, 2, ) when the series
(10) and (12) are substituted into (11) is called the re
currence formula of the differential equation (1) relative
to the point z = zQ. Actual substitution of (10) and (12)
into (11) yields the following expression for the differ
ential equation:
(14) Cq20(z-zo)2 + q21(z-z0P ++ q2*(z-z0)Vf2 + '"3
y. Q y|
[cQr(r l)(z zQ) + c^(r + l)r(z zq) +
+ c^(r + X)(r + X l)(z z0)rfX"2 + ]
+ Uiq(z-z0) + 1ii(z-z0)2 ++ qlx(z-z0)X+1 +]
CoQr(z z0)1^1 + Oj^ir + l)(z zo)r +
+ o^(r + X)(z zo)^1 + ]
+ ^oo + Co (z z )r + 0-(z z)1*1 +
O o 1 o
+ o^(z zQ)X+r + ***] =0.
Prom (l4) we get the recurrence formulas, which are the
coefficients of the various powers of (z zQ). The first
of these recurrence formulas is called the indicial equa
tion.


- 13 -
The first three terms above vanish, so the coefficient of
z may be written
a/?(a b)(b 1) + abD^.
Using the results of the above simplifications, the dif
ferential equation (3^) may be written
(4o) z(z l)(z a)u"(z) + {(a + ft + l)z2
- [a + £- 5 + 1+ (7 + 6)a]z + aT
It is desired to have the singularity denoted by
a^ b be at infinity. Hence, if we take the limit of
(4o) as b becomes infinite, we have the result known as
Heim's equation:
(4l) z(z l)(z a)u"(z) + -[(a + /5 + l)z2
-[a + /6-8 + l+ (7 + 8)a]z + al] u'(z)
+ ap(z q)u(z) 0.
Since 1^ is arbitrary,
(Jj-2) q aI^/a/3
is an arbitrary constant.
We digress to the case n 3 in (26) to determine
the most general form for the second order differential
equation having three regular singular points in the
finite complex plane. The three singularities are chosen


- 10 -
ties to the points a, b, c, d. The transformation
(40)
z a
a(d b)(z1 a)
(b a)(z^ d)
will cause the singularities z *= 0, 1, a, ao to be moved
to the points b, c, a, d, whioh are distinct non-zero
constants. We may verify that (40) is the desired trans
formation by direct substitution, as follows:
1. Let z = 0 in (40). Then, on division by a, we
get (a bMz^ d) = (d b)^ a), or z^ = b.
2. Let z = 1 in (40). Then (1 a)(b aMz-^ -d)
= a(d bHz^ a), or
abd a2b bd + ad
1 "ad o? b + a "
This expression involving the four singularities is a con
stant and we may designate it by the letter o. Hence,
abd a2b bd + ad
ad ^ -b TV-
3. Let z => a in (40). Then a(d b)(z-^ a) 0,
and since a / 0, b $ d, we find that z^ ** a.
4. The inverse of (40) may be verified to be
(50)
d(b a)(z a) a(d b)
1 (b a)(z a) a(d b)
i
If we let z-> co in (50), we find that z^ d.
From (40), we find that
(51) u*(z) = [(b a)^ d)2/a(d b)(a djju'iz^
and


- 51 -
r = (a + 2r) (3 + 2r)
2rf2 2(r + 1)(6 + 1 + 2r) 2r*
For r = 0,
. gff .
2 2*1! (6 + 1) *
for r = 1,
_ a(a + 2)6(3 +2)
4 22*2' (8 + 1)(6 + 3) 0
and, for r = n 1,
_a(a+2) (a+2n-2)<3(a+2)-**(0+2n-2) ^
Opy. 1 Or\
2nnJ (8+1)(8+3)* *(8+2n-l)
The solution corresponding to X = 0 Is then
(52) ^ = o0P[f, f; (z l)2].
From (50), if X = 1 6, then (51) becomes
0 = U 8 + 1 + 2r)(fl 8 I 2r) 0
2rf2 2(r + 1) (-6 + 2r + 3) 2r
If r = 0,
c2
If P 1,
= La....jLJLL(^ ,5 ,tJL)
2-1 (-8 + 3)
0
(a 8 -> l)(a 5 + 3)(g- 6 + l)(g- ^).f.
22*2¡ (-6 + 3)(-6 + 5)
and, If r = n 1,
2n =
(g 8 + l)(g 8 + 3)*-*(a 8 + 2n 1)
a^n!
. 'S 8 + 1)(6- 8 + 3)-~-(g- 8 + 2n XL
L (-6 + 3)(-6 + 5)* *(-6 + 2n + 1) 1 0


- 25
(65)
£> 2
dz-.
A
z-. a
- b
z-, o
du-
dz
*i(ai +11, 7i(Ti+ ;>, y
(zx a)2 (z^ b)2 (z.^ o)2
ab£i.
2ajZi
(z^ aXz-L o) (z1 aJZj^ b)
2/V
ill
Un
(z^ b) (z^ * o)
a(a + & + 1) [a + ff 5 + 1 + (7 + 6)a] + V
z-j_ b
(a 1)(z^ a)
a + fl 6 + l + (7 + S)a
a 1
+ U + Z3 + l)(c b)(d a) 7(b-a)(a-c)
(c a)(d b)
J zi o
+ 1 &1 fO^L
zx d J \dsj_
1 | ^1
_Z1 ~ a Z1 b Zl 0.
"l
AJcx/3[a(d-a)(z1-b) g(b-a)(zi-d)]Cd-b][d-a] } u 0 Q
L (z1-a)(z1-b)(z1-o)(z^-d)2(ad-a2-b+a) j 1
The coefficient of u^iz^) In (65) Is unity. The
coefficient of u^iz^) in (65) may be simplified by using
(56), (57), and (5g). The result is
(66) -2^ + a + /S-7-8 + l_tY-271 8-2/^
zi a Z1 b zx o
+ 1Lz~2L=JL.
Zi d


Applying the restrictions (30), (31), (32) to (23),
we get the equation
(33) z(z2 l)u"(z) + [(a + /3 + l)z2 7]u'(z)
+ a£zu(z) =0,
In order to solve (33) we assume a series solution
of the form
i \ X X+2 ^ X+2r
u(z) = O0z + CSZ + + e2rz
, X+2r+2
+ Or, +
2r+2
(c0 ¥ 0)
so that
u*(z) = CqXz^ ^ + o2(X + 2)z^+^ +
+ c2r(X + 2r)z
X+2r-l
+ Ogr+2^ + 2r + 2)z
X+2r+l
+ ...
and
X2 X
u"(z) = OqX(X l)z + c2(X + 2)(X + l)z +
+ o2r(X + 2r)(X + 2r l)z^+2r""2
J . . X+2r
C2r+2(X + 2r + + 2r + l)z + ,
and substitute it into (33) The indicial equation, which
is the coefficient of z* \ is
(3^) -o0[x(x 1) + 7X] = 0
and, since cQ $ 0, its roots are
(35) X = 0, 1 7.
The recurrence relation among the coefficients is found by
X+2r+l
setting the coefficient of z
equal to zero:


Solutions Relative to the Singular Point at z = 1
In order to find the solutions of Heuns equation
(23) relative to the singular point at z = 1, we transform
the Independent variable by the relation
(4-1) = z 1.
Heun's equation becomes
(4-2) [(1 a)z1 + (2 ajz-j2 + z^W{z^)
+ {6(1 a) + [a + 0 a? + 6(1 a) + l]z^
+ (a + + l)z^}u(z^) + a/6(l q + z^)u(z^) = 0
and the coefficients p^(z^) (J = 0, 1, 2) are, respectively,
(4-3) Ppiz-^ = (1 a)z1 + (2 a)z^ + ZjS,
(44) P1(z1) = 6(1 a) + [a + j3 a7 + 6(1 a) + 1^
+ (a + /5 + l)Zl2,
(4-5) P0(2i) = a/5(l q + zx).
We begin by assuming M = 1 and h = 2 in (22), so
that (4-3) becomes
P2(zi) = (1 a)z1 + (2 a)2]2 +
p
*= (Sg ^2Z1 ^Z1
and hence S2 = 1 a and T2 = -1. We must accordingly re
quire that
(4-6) a = 2
in order to satisfy Scheffes criteria. Applying (4-6) to
(44), we find that


55 -
(5) p-^Zg) *=
- (a + jS-8+l)]z2 + (a + /S-7-8 + l)(a2-a),
(59) Po(z2^ = ^(a We begin by assuming M = 1 and h = 2 in (22), so
that (57) becomes
p2(z2) = (a a)z2 + (2a l)z2 + z^
= (S2 T2Z2 )z2,
p
from which S2 = a a and = -1. We must then require
that
(60) a = 1/2.
Using (6o), equation (520 becomes
Pl(z2) = (ol+/5+1)z22 + g-
= si Tiz2 *
so that =-(a + ft 7 8 + l)/4 and = (a + 0+ 1).
We must accordingly require that (6 7)/2 vanish, so we
set
(61) 8=7.
With (6o), we find that (59) becomes
p0(z2) = a|3(ir = (S0 Tqz2 )z2 1,
so that Sq = 0 and TQ = -a3. We require that
(62) q = |


-To
dos) in the form
u(z-,) = cnz7^ + c.z *+3 + + o,_zT^+^r
'0*3
+ o
3 3
X+3r+3 + ...
3rt3Z3
3** 3
(o0 f 0)
Then
u(z^) = CqXz^"1 + c^(X + 3)z^X+2 + *'
+ o3r(J> + 3r)zjX+3r_1
+ o3r+J(X + 3r + 3)z3X+3r+2 +
and
u1* (z-j) = c0X(X-1)z5X"2 + c3(X+3)(X+2)z5X+1 +
+ c3r(X + 3r)(X + 3r Dz^^1^2
+ c3r+3(X+ 3r + 3)(^ +3r+ 2)z3X+^r+1 + .
X?
The lndicial equation, which is the coefficient of z3 ,1s
OqCCU >2)X(X 1)] = 0,
and its roots are
(103) x = o, l.
The recurrence relation, the coefficient of z3^+^r+^ is
9a0c3r + 277(X + 3r)c3r + 9U + 3r)(X + 3r l)c3r
+ (d2) (X + 3r + 3) (X + 3r + 2)c3r+3 = 0,
or
9(X + 3r + a)(X + 3r + &)
3^3 (^-STCx V 3r 3)U + '3r ; s) 3r*
for r = 0, 1, 2,



- 16 -
guiar points and the exponents at these singular points.
To express the faot that u satisfies an equation
of this type, Riemann [15, p. 206] used the notation
fa b o
(4-6) u = Mi ft 7 z .
U* 9' 7* ,
The singular points of the equation are plaoed In the first
row with the corresponding exponents direotly beneath them,
and the independent variable is placed in the fourth column.
Returning to Heim's equation (4-1) with its four
singular points, it would seem natural to inquire whether
an analogue to equation (4-5) exists in the case of Heun's
equation (4l). That such an analogue exists may be shown
with extensive algebraic manipulation.
First, let us note the differences between (45)
and (4-1). Equation (4-5) has three singularities at the
arbitrary points z = a, b, c and has arbitrary exponents
a, a; /3, /S'; 7, 7* at these points. Equation (4-1), how
ever, has been specialized with the choice of singularities
at z = 0, 1, a, oo and exponents 0, 1-7; 0, 1 8; 0,
7 + 8 a ¡3; a,/3 at these points respectively. We need
to generalize (4-1) to the extent that its singularities
and exponents are arbitrary also. It may be shown that a
change in the independent variable of (4-1) alters the


- 37 -
and solution was not effeoted.
In order to determine the conditions under which a
solution to Heun*s equation will have a two term recurrence
formula, we propose to use the criteria set forth by H.
Soheff [13] some fourteen years ago. We briefly summarize
this paper, as applied to the second order linear differen
tial equation, in the following paragraphs.
Given a differential equation of the form
(9) Pg(z) + Pj.(z) f§ + P0(z)w = 0,
where the Pj(z) (J =0 1 2) are analytic in some suffi
ciently small neighborhood of z = zQ in the extended z
plane and are not identically zero in this neighborhood, we
assume a series solution to (9) in the form
(10) w = 2 o> (z z )X+r. (c 0)
x=0 A o 0
Multiplying equation (9) by a suitable integral power of
(z zQ), say, (z zq)M, we find that (9) becomes
(11) q2(z)(z-zQ)2 ^5jr + q^zHz-z^ ~ + qQ(z)w = 0,
QZ
where the form of the coefficients qj(z) (j =0, 1, 2) is
(12) q^(z) = The proper selection of the constant M will cause the con
stants qjk (j =0, 1, 2) to obey the restriction


21
(55)/(zt a)(z1 b)(z1 o)(z1 d)(ad a2 b + a)"
(d b)(d a)(b a)
a(a + P + 1) [a + J3 6 + 1 + (7 + 6)a] + 7
(ad bt b + a)(o a)
(d a)(b ~ a)
^ a)
7(b a)(d b)
.(ad a2 b + a)(o b) J z^ b
+ C-2a(d a)(c b)^(o a)
-2(b a)(d c)(c a)(o b)
+ a(g + /8 + l)(d a)2(o b)2+ 7(b a)2(d o)2
+ (g + /3 6 + 1 + 7a + 6a)(d a)(b-a)(d-c)(o-b)3
[(ad a2 b + a)(d o)(c a)(o b)]1, 1
J z^ o
r(l a £)a ]
"(d a)(d b)ll
.(ad sid b + a)(d c).
1-* Ji
In order to further simplify the above expression,
we will make use of the following identities, whioh may be
verified by use of expression (49):
(56) (ad a2 b + a)(o a) (a 1)(d a)(b a)
(57) (ad a2 b + a)(o b) = (d b)(b a)
(53) (ad a2 b + a)(d 0) a(d b)(d a).
The use of (56), (57)* a**3- (5^) in (55) yields the
result:


- 31 -
Combining terms and using (56) we may write the above as
(72) 0^(7 + 5 a 0 + 04 Ha b)(a c)(a d)
- a
+ Tl(1 - Y + r1)(b a)(b o)(b a)
zi -b
+ + a) (0 b)(o d)
z c
4- o^(<3- a) (d b) (d c)
zi -4
Thus, combining (66), (71), and (72), we arrive at
the canonical form for Heun's differential equation, which
vie believe to be new. This canonical form is
(73)
4^7
1 q^ (y + 6 a + a^)
z^ a
x 1-T1- d-r^i) +^L- (i-t+flj)
zx b zx 0
+ l....-.,?. a3
z^ d
f2i +
dzi
ai(T+Sag^ai)(a-b)(a-o)(a-d)
zx a
Y.,(l 7 + Y-,)(b a)(b c)(b d)
zx b
+ ^l(3- ~ 6 +^i)(o a)(c ~ b)(c ~ d)
zi o
+ cu9(d a) (d b) (d c) + K
U1
zi d
(z]_ aTiz! bMz! cMzi d)
= 0,
where K is given by (71).


- 52 -
The solution corresponding to X = 1 6 then becomes
(53) = C0(z 1)1-6P[S=|1, Mil; =m.; (z 1)2].
A general solution of Heims equation relative to
the singular point at z = 1 is then a linear combination
of (52) and (53):
u(z) = A2F[|, |; S--L; (z l)2]
+ B2(z l)1_6F[2=|il, =$p.; (z l)2],
where and are arbitrary constants.
We continue by letting M = 0 and h = 1 in (22).
As a result,
Pgz^) = (1 a)zx + (2 a)z^2 +
(Sg ~ ^2Z1^Z1 *
from which S^ = 2 a and T^ = -1. We must accordingly
require that
(54) a = 1
in order to satisfy the hypothesis of a two term recurrence
formula. With the restriction (54) applied to (44), we
find that
Px(zi) = (a + (3 1 + ljz^ + (a + j6 + Dz-j2
= (Si T^z-^)z^,
so that S^=5a + j6-7+l and T^ = -(a + /3 + 1). Simi
larly, applying (54) to (45), we find that


equation (15) beoomes
(1?> £, Vr + X m)oX-m = *
m=o
Equation (17) is the recurrence formula which we seek. We
note that, beoause of (13),
(lg) Qq(x) 0.
By definition, the differential equation (9) is
said to have a two term recurrence formula relative to
z = zQ if the recurrence formula is of the type
(19) f(X,r)cx + g(*ir)cx_h = 0,
for X = h, h+1, h+2, where h is some positive integer,
and neither f(X,r) nor g(X,r) is identically zero in both
X and r. Because of (1$), we see that necessary and suf
ficient conditions that (19) be a two term recurrence for
mula are that
(20) QfcU) j 0
and
(21) Q^x) = 0 (m 5* 0, h)
The necessity of (20) is evident, while that of (21) may be
argued as follows. The ^(x) are polynomials of degree <2
and can vanish for x=r+X-m, X-m=h, h+1, h+2
and fixed r, only if Q^x) =0. Hence (20) and (21) are
equivalent to
q,)h 0 for some j
and


- 29 -
(a b c)a+ (c a b)^2 + (b a cJT-j2
- 2ba1^1 200^ 2a^1 d(a1 + Y^2.
The last term again vanishes due to (69). If one writes
(69) as 7^ = a-j. 02.* tlie remaining six terms of the
above expression also vanish.
The constant term in (67) may be written
(a2-ab-ac-ad+bc+bd+cd)[a^2 -a^(a + /3-7-6)]
+ (b2-ab-bo-bd+ac+ad+cd)Cr12 + (l 7)71]
+ (c2-ac-bc-cd+ab+ad+bd) [/5^2 + (1 6)^]
+ [2(^/6 ^(a + /3-7-8 + 1) a-j^S ]bd
+ [2a171 7-j^ia + 3 -7-6 + 1) Ta-j^cd
+ (2/3^ rS T^Jad 0^(1 a /6)bo
- ?i(l a P)ac /^(l a 3)ab
+ a^Ca(d a) q(b a)][d b][d a]
_ b + a
This expression may be written, using (69) again, as
(b a)2^2 + 2(b c)(b aja^^ + (b c)2/^2
+ [(a d) (b a)a + (a d) (b a)# + (b a)2/
+ (c a)(b a)6 + (d b)(b a)]ai
+ C(b c)(a d)a + (b c)(a d)/S
+ (b c)(b a)T + (b c)(c a)8
+ (b c)(a + d b c)]^
+ a^[d a][d c + q(c b)].


(6)
- 4 -
u"(w) + [(2/w) (1/w2) p(l/w)] u*(w)
+ (l/wVq(l/w)*u(w) 0,
If w a 0 is an ordinary point of (6), then it Is necessary
that the coefficient [(2/w) (l/w2)*p(l/w)] be finite at
w 0. It follows that p(l/w) must begin with the term
2w, so that the first term (2/w) of the coefficient above
will be removed by subtraction, and continue only with
terms of higher degree in w than the first, so that the
entire coefficient will be finite at w = 0. As a conse
quence, p(l/w) must be of the form
(7) p(l/w) = 2w + AgW2 +
where the (i 2, 3, ) are constants. Prom (7), we
may revert to the variable z by inversion, to get
(g) p(z) 2/z + A2/z2 + A^/z^ + .
We may also describe this restriction of the function p(z)
2
by writing p(z) 2/z = 0(z ),
Similarly, in order that w = 0 be an ordinary
point of (6), it is also necessary that the function
q(l/w) be finite at w = 0. By reasoning similar to that
above, we conclude that q(l/w) must be of the form
(9) q(l/w) =* Bjjw^ + Bj-w^ + **,
where the (1 4,5, * *) are constants. With the rela
tion (9), we know that the coefficient (l/w^)*q(l/w) of
(6) is finite at w 0, Again reverting to the variable z


- 14 -
to be at the points z a,b,c and the exponents at these
singularities are a, a*; fi, ¡3 '; T, y*; respectively.
Prom (26), this general form is
(4-3) u"(z) + C(1 a a')/(z a)
+ (1 P 0')/(z b) + (1 7 7')/(z-o)]u(z)
+ [aa*/(z a)2 + /36*/(z b)2 + 77*/(z o)2
+ Dj/(z a) + Dg/(z b) + D^/(z o)]u(z) =0.
The restrictions (27) (3) in this case are
a + a* + /3 + /S* + r + r' 1,
aa' + /30' + 77* + aP^ + bDg + cDj 0,
2aa'a + 2/3/8b + 277o + a2Dx + b2Dg + o2D^ = 0,
+ Dg + Dj 0.
Factoring the expression [(z a)(z b)(z c)]-^ from
the coefficient of u(z) in (4-3), performing the indicated
divisions and simplifying, we obtain
a!a + ri.-.a- -gl -.A-JJ. + 1.- 2,.- .Ill gu
l1") dz2 Lz-a z-b z-c Jdz
+ + Dg + Dj]z2 + [aa* + fifi' + 77'
- D^(b + c) Dg(a + o) D^(a + b)]z
+ [aa*(a b o) + #5'(b a o)
+ 77(o a b) + Djbo + Dgao + D^ab]
+ oaf(a b)(a o) + /3)f(b a)(b o)
z a z-b
+ T7'(o a)(c JbJI u o.
z c J (z a)(z b)(z o)


CHAPTER II
SOLUTIONS OF HEUN'S EQUATION
Solution in a General Case
Using the method of Frobenlus [4], we are able to
find a solution in a general case of Heim's equation* We
will discover on inspection of the solution that it is
essentially in terms of a difference equation which is more
difficult to solve than the original differential equation.
A particular form of the solution is given by Heun
[7, p. 131], and may be derived in the following manner
using (4l) of Chapter I. Assume a formal solution of the
form
u =
c0zX +
LX+1 X+r X+r+1 .
C]Lz + + crz + C^-jZ +
where Cq f 0. Then
= o^Xz*"1 + c,(X + l)zA + + c_(X + r)z?'+r_1
dz u j. i
+ o^fX + r + l)zX+r + "
and
= CqX(X l)zX2 + c,(X + l)XzX"^ +
dz
+ cr(X + r)(X + r l)zX+r2
+ om(X + r + 1)(X + r)zX+lvl + *
The indicial equation is the coefficient of zX-1 and may
be written


- 10 -
and
(30)
n
2 D o.
r=l
This dissertation is, in general, conoemed with the
differential equation with four regular singular points
and, in particular, concerned with a particular form known
as Heuns equation [6, p. 165]. We let n 4 in expres
sions (26) (30) to get
(3D
uM(z) + 2 ^*u(z)
r=l
z a.
r 4
Vr
+ 2
.r=l (z aj) rl
z a_
r J
u(z) 0,
with the restrictions
(32) 2 (<* + 2,
r=l
(33) 2 (ariSr + SrDr) = 0,
r=l
W 2 + ar2Dr) 0,
r**!
(35) 2 D 0.
r=l r
Heuns equation may he obtained by choosing the
following values for the various unknown constants:


- 44 -
(X + 2r)(X + 2r l)c2r (X + 2r + 2)(X + 2r + l)c2r+2
+ (X + 2r)(cx + 0 + l)o2r 7(X + 2r + 2)o2l>f2
+ a^c2p 0.
This relation may be written
(36) Oo^p sMV ?: g> + W 0¡>it
J 2r<-2 (X + 2r + 2)(X + 2r + 1 + 7) 2r
where r=0, 1, 2, .
If X = 0, the recurrence relation (36) becomes
(a + 2r)(jS + 2r)
2rf2 2(1 + r)(7 + 1 + 2r) 2r *
For r = 0,
Co
'2 2*1(7 + 1) *
for r = 1,
4 22*2 (7 + 1) (7 + 3)
and, for r = n 1,
c0*
a (a+2) * (a+2n-2) Q(ff+2) (/3+2n-2)
2n*nJ (7+1) (7+3) * (7+2n-l)
This solution may be written as a hypergeometric series:
(37) 00P(|, f; X^-; Z2).
If X = 1 7, the recurrence relation (36) becomes
(a 7 + 2r + l)(ff 7 + 2r + 1)
2r+2 2*(r + l)(-7 + 2r + 3) 2r*
For r = 0,


- 49 -
Pl(zl ) = -8 + (a+/3-27-8+l)z1 + (a+jfl+l)^2
= Si v 2>
from which = -8 and T^-(a + jS + l). Accordingly, we
require that
(47) a + /6 27 8 + 1 = 0
in equation (42) above. In the case of (45), we find that
p0(zi) = a/3(l q) + apz1
so that SQ = 0 and TQ = -ajS. We must accordingly set
(4g) q = 1
in equation (42) since, if a = 0 or /3 = 0, the differential
equation (23) could be reduced to a first order differen
tial equation. When we utilize (46), (47), and (4$) in
(42), the latter becomes
(49) z1(z12 liu"^) + [(a + /5 + Dz-j2 6]u,(z1)
+ a^z^uiz-^) = 0.
In order to find a solution to (49) by Probenlus'
method, we assume an infinite series solution of the form
+ " + 2rzl
X+2r
+ o2r+2zl
+
Then


- 6 -
(13)
q(z) = (X/z2) 2 drCl + (ap/z) + (ap2/z2) + ]2
r=l
+ (l/z) 2 Dr[l + (ar/z) + (a-2/z2) + ]
r=l r
(l/z) 2 D + (l/z2) 2 (dn + Drar)
r=l
r=l
+ (l/z5) 2 (2apdr + ap2Dp) +
r=l
On comparison of coefficients of like powers of z in (10)
and (13), we find that the following three relations must
hold:
n
(14)
2
D
peal
n
r
(15)
2
r=l
n
(16)
2
r=l
(2a.
rr
0.
The method of Frobenius [4, p. 214] is used to sat
isfy the requirement that the exponents at z ar be ar
and /9r, Assume a solution to (5) of the form
(17) u(z) f0(z ajJ* + f^z ar)*+1 + f* 3/ 0
(12)
u*(z) f0X(z aj,)^"1 + fj/X + l)(z a )A +
(19) u(z) fftX(X l)(z aJx2
+ fx(X + 1)(X)(z ap)
X-l
+
The lndiolal equation is found by inspection to be


- 46 -
set
(39) a = O
In (23). Using (39) in (23), we find that
p-^iz) = (a + p + l)z2 (a + j0 8 + l)z
83 *S1 Tiz)z
from which = -(a + ¡5 6 + l) and = -(a + 3 + 1).
Similarly, applying (22) to (29), we find that
pQ(z) = a(3(z q)
= So V*
from which SQ = -afiq and Tq = -a8. Applying the sole re
striction (39) to (23), we obtain the equation
(40) z2(z l)u"(z) + C(a + /8 + l)z2
- (a + ft 6 + l)z]u'(z) + af3(z q)u(z) = 0.
Equation (4o) has but three regular singular pointy
at z = 0, 1, 00, caused by the coincidence of the singular
points z 0 and z = a. Hence (4o) is a confluent form of
(23). In accordance with the convention suggested in the
second section of this chapter, we will not effect further
study on this equation. We classify it, in the notation
of Inoe [3, p. 497], as an equation of the [0, 3* 0]
7
olass. The solution of (4o), in the notation of the first
^The formula [a, b, o] characterizes the equation
with the property:
a = the number of elementary singularities,
b = the number of non-elementary regular singular
ities,
c = the number of essential singularities


22 -
(59) ('(z1 aHZj ~ 'b)(z1 oH^ d)(ad a2 b + a)
1 (d b)(d a)(b a)
ifa(q + /S + l)-[a+j9-6 + l+(T+ S)a] + T
U (a lMz^ a)
+ y + [(at^-8iltrat6a)
z^l b l a 1
+ (<* + +l)(o b)(d a) 7(b a)(d o
(o a)(d b)
The coefficient of the function u(z^) may also be
simplified to
a/3[a(d a)(z1 b) q(b a)(z1 d)]
(b a)^ d)
_f(z1 a)(z1 b)^ o)(z1 d)(ad a2 b + a)|
I (d b)(d a)(b a) J
f a/3[a(d a)(z1 b) q(b-a)(z1"d)][d~b][d-a]"|
\(z.j-a) (z-j-b) (z^-o) (z-j-d)^(ad - b + a) J
Writing expressions (54), (59) and (6o) together,
we have Heun's equation with four arbitrary singularities
at Zj a a, b, o, d, but with determinate exponents at
these singularities. Careful inspection of these three ex
pressions will reveal that a common factor of the three
terms exists and it may be removed by division. The result
of this division is


-4l-
= 0. (J = 0, 1, 2; m 0, h)
The coefficients p^(z) of (9) are then of the form
Pj(z) = [qJ0 + qJh(z z0^h^z z0^M-
The conditions which we desire then follow readily in the
form of a theorem.
Theorem. A necessary and sufficient condition that
a differential equation have a two term recurrence formula
relative to a point z = zQ is that in some neighborhood of
the point zQ
(22) Pj(z) = [S: Tjtz Z0)h]tz Z0fM,
where M and h > 0 are integers and the constants = q^Q,
Tj = are such that S^Tj 0 for some i,j.
We propose to apply the above criteria to Heun's
equation in the form
(23) z(z l)(z a)- + Hoc + /3 + l)z2 [a + 6 8
dz
+ 1 + (7 + 8)a]z + a7}~ + a^Siz q)u = 0.
dz
Relative to the notation used by Scheff and in the above
summary, we define
(24) p2(z) = z(z l)(z a),
(25) px(z) * (a+0+l)z2 [a+/3-6+l+(7+6)a]z + a7,
(26) P0(z) = afiiz q).
It is apparent that these three functions are analytic at
every finite point in the z plane.


- 17 -
location of the singularities, but leaves the exponents of
these translated singularities invariant; also, it may be
shown that a change of the dependent variable in (4-1) al
ters the exponents at the singularities, but leaves the
original location invariant [see 15, p. 207]. In terms of
the notation used in (46),
a b
* { + J /3- 3 k
a*+ J J k
by change of dependent variable
7+ k Zr
7* + k ,
Correspondingly,
fa
b
o n
1
*1
1
r
P
7 z
P
7
zii
|6'
r* j
a'
j8*
r
by change of independent variable
(z-, a1 )(o-, bn) o (z a)(o b)
(z^ bl^l ~ ai) (z b)(o a)
We choose to change the location of the singulari
ties first. As noted, equation (4l) has singularities at
points z 0, 1, a, oo; we desire to shift these singular!-


- 35 -
u = F(a, q; a, fi, 7, 8; 2)
by Heun [7, p. 1S1]. The function F does not refer to the
hypergeometric series, although Heun observed that the
function F(a, q; a, /5, 7, 6; 2) became the hypergeometrio
series F(a, /$; V; z) when a = q = 1.
From (1), we know that
(4) G-^q) = *
and from (2), we know that
(5) Gg(q) = + [a + /S 6 + 1 + (7 + 6)a]q a7.
The derivatives of (3) are
(6) a 2. ? Q-n^^z/a^n1
dz a n=i (n l)i 7(7 + l)***(7 + n l)
and
17) 2u a ^ ? Gn-H(q)(Z//a)n"1
7 dF ?nBl(n-l)!r(7+l)-(7+n)
Substituting (3), (6), and (7) into Heun's equation, the
recurrence relation, after simplification, is
(g) Go+^q) fa[(a + /5-8+n) + (7+6+n l)a]
+ a^q}On(q) (a+n-1) (/f+n-l) (7-Hn-l)anGn<-1(q).
We thus have a solution to Heun's equation in a
general case. The series (3) is the solution, together
with the three term recurrence formula (0) and the initial
values (4) and (5).


- 30 -
This expression may be written
(70) C(b a)a^ + (b c)/^]2 + C(a d)(a + &)
+ (b a)7 + (o a)6 + (d b)]
[(b a)^ + (b c)^] + (b c)(a o)/^
+ a/S[d a][d c + q(o b)].
Collecting coefficients of the various Roman alphabetical
unknowns, we find that the constant term may be written in
the form
(71) K = ax(7 + 8 a /S + a^a2 + 7^1 7 + T^b2
+ /S(1 6 + /51)o2 + o0d2 + Ca1(71 7)
+ 71(7+ 5-a-J6 + a1- l)]ab + [cc^^ 6)
+ ^(0^ a P + y + 8 l)]ao
+ lojia. + (5 l)]ad + C^l(71 7)
+ 8)]bc + 71(a + /6- i)hd
+ i3x(a + 0 l)cd
+ a/3[(ac ad cd) + q(a d)(b 0)].
The remaining terms of (67) may be written
[^(a-^+l) (a+/3-y-6+l)][a b][a c][a d]
Z1 a
+ C-Vi2 + (1 -yiTSHb a]Cb o][b a]
Z1 -b
/5]2 + (X - a][o b][o a]
+ zx o
+ q^Cata a)S(d t>)2] _
(aa-a2-b+a)(zj a)


0 = (g 7 + 1X0 -7+1)
2 2-1! (-7 + 3)
for r = 1,
la .£ + U* r-V.+ 2MLl+. JJLO? ..rJ+3.L .
4 22*2i (-7 + 3)(-7 + 5) *
and, for r = n 1,
>2n
(a 7 + l)(a 7 + 3)-**(a 7 + 2n 1)
2n*ni
(ff 7 + 1)(0 7 + 3)''{ 3 7 + 2n 1)
(-7 + 3) (-7 + 5) *(-7+ 2n + 1)
0'
This solution may also be written as a hypergeometric
series:
(3) 2 o0z1-7(S-_|-tJL,
z2).
A general solution of Heun's equation relative to
the singular point z 0 is then a linear combination of
(37) and (3^) an*1 is written
u(z) f; z2)
+ B1Z
1-7
>
where and are arbitrary constants.
Continuing, we select M = 0 and h = 1 in (22). The
conditions imposed by (22) on (27), (2g), and (29) are then
z p
Pg(z) = vr (a + l)z + az
(S2 T2z)z2,
from which Sg = -(a + 1) and T2 = -1. We must accordingly


- 64 -
This solution may be written In the form of a hypergeomet
ric series:
<*> ! oz"ap(t-
If we choose the root A = & of (22), the recurrence
relation (24) becomes
o = (/3 + 2r)(g 7 + 1 + 2r)
1n rTT ". 1 1 *
'2r+2 2(r + l)(j5 a + 2 + 2r)
2r
If r = 0,
c = &P-:e.
2 2*i: (0 a + 2)
if r = 1,
o, = 121 Q.
4 22'2! (/3- a + 2)(/S a + 4)
and, if r = n 1,
j3(|S + 2) (/6 + 2n 2)
2n "
2n,nJ
[~(/3 r + l)(ff 7 + 3)**(ff Y + 2n 1)1
* [ (0 a + 2)(j8 a + 4)-"(0 a + 2n) J V
This solution may be written
(85) = e^FCf, \ A:1 ~ f* 2; ^1.
A general solution of (21) may be written as a lin
ear combination of (24) and (25):
u(z) = A],z"aF[~, 3-=--X I. A +. ,2.
4^ *2* 2
vM, ^4^ 3.
2 z"
H
2 ~


- 63 -
Oq[X2 (a + j¡3)X + a¡0] = Ot
and its roots are
(62) X = a, (3.
The recurrence relation, the coefficient of is
^c2r+2 + ^ ~ a ~ 6)(X + 2r + 2)o2rf2
+ (7 2)(X + 2r)c2r (X + 2r)(X + 2r l)o2r
+ (X + 2r + 2)(X + 2r + l)c2r+2 = 0
(*?) ~ = (X + 2r)(X 7 + 2r + 1) .
3) 2r+2 (X a + 2r + 2) (X ¡5 + 2r + 2) 2r
or
(23)
c2r+2
for r =
0
H
If we
becomes
2r+2
(a + 2r)(a Of + 1 + 2r) .
+ l)(a + 2 + 2r) 2r*
2(r
If r = 0,
c
- a(a 7+1) 0 .
2 2*1 (a + 2) 0
if r = 1,
O, = L* l^Ha ,,7 + A)!*--.? + 3J. 0 .
4 22*2 (a /3 + 2)(a (5 + 4) *
and, if r = n 1,
2n
g(a + 2)*(a + 2n 2)
2n*nJ
(a I l)(a 7 + 3) * *(tt -7 + 2n 1)
(a jS + 2) (a j3 + 4) (a & + 2n)


- 54 -
ger. Failure to assign one of the two remaining terms of
(43) will lead to an impossibility, for we cannot cause
both (1 a) and (2 a) to vanish with a single choice
of a. We conclude that S2 = 1 is an impossibility. If
Tg = -1, then = z^ and hence h M = 1. We have
discussed the cases where h = 1, M = 0 and h = 2, M = 1.
If h is an integer greater than two and M = h 1, then we
are again led to the impossibility that both (1 a) and
(2 a) must vanish simultaneously. We conclude that no
other cases exist where (42) has a two term recurrence
formula relative to the singular point at z = 1.
Solutions Relative to the Singular
Point at z = a, Where a^O, 1, oo
In order to find the solutions of Heun's equation
relative to the regular singular point at z = a (a f 0, 1,
ao ), we transform the independent variable of (23) by the
relation z2 = z a. Heuns equation then becomes
(56) [z2^ + (2a l)z22 + ^ a)z2]u"(z2)
+ ((a + j6 + l)z22 + [(2a + 2/3 7 6 + 2)a
- (a + /3- 6 + 1) ]z2 + (a + $ i 6 + l) (a2-a) }u* (z2)
+ a/3(a q + z2)u(z2) = 0
and the coefficients pj(z2) O = 0, 1, 2) become
(57) P2(z2) z2^ + ^2a l)z22 + ^&2 a)z2,


TABLE OP CONTENTS
Page
ACKNOWLEDGMENTS il
Chapter
I. DERIVATION OP THE CANONICAL FORM ... 1
II. SOLUTIONS OP HEUN'S EQUATION 33
Solution in a General Case 33
Differential Equations with a Two
Term Recurrence Formula ..... 36
Solutions Relative to the Singular
Point at z = 0 42
Solutions Relative to the Singular
Point at z = l 4S
Solutions Relative to the Singular
Point at z = a, Where a^ 0,1,00. 54
Solutions Relative to the Singular
Point at Infinity 60
Solutions Relative to the Point
z = b, Where b^O, 1, a, 00... 67
BIBLIOGRAPHY ........ 75
BIOGRAPHICAL SKETCH 77
iii