Includes bibliographical references (leaves 75-76).

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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 + Â¡3 + l)z2 [a + Â£- 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 Â£ + 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 Â¡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-Â£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, Â§; a-* f 37 +-Â£; 4-(z |)2]
+ b3(z 2^;
- 4(z Â§)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 Â¡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) Â£ Pj/z a ) 88 (1/z) 2 Pj/tl (ap/z)]
r=l r*l
(1/z) Â£ Pr[l + (ay/z) + (ar2/z2) + ]
(11) = (1/z) Â£ p_ + (1/z2) Â£ 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) Â£ p = 2
r=l r
must hold. The remaining sums in (11) are arbitrary,
sinoe the corresponding ooeffioients in (Â£) 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) Â£ 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
_Â£_Â£ + 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 Â§ )
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 + Â¡3 + l)z2 [a + Â£- 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) Â£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)
- (aÂ£ + 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 + Â£- 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, Â¡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 Â¡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 + Â¡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 Â£)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 + Â¡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)
Â£> 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
abÂ£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 Â£)(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. Â§.\ + 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) + Â£(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) fÂ§ + 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?> Â£, 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)
+ aÂ£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 Â¥ 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 0Â¡>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 .Â£ + 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 + Â¡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*2Â¡ (-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,