.1.S.tmbolit'('ontputution (1986) 2, 3 43
An Algorithm for Solving SecondOrder Linear
HomogeneousDifferential Equations
JERALD J. KOVACIC
JYACC Int'..919 Third Ar('tlue
Irlen'York, NY 10022,Lr.S.A.
(Received8 Ma.t' 1985)
I n t h i s p a p e r w r - p r e s e n ta n a l g o n t h m f o r f i n d i n g a " c l o s e d - l o r m " s o l u t i o n o f t h e d i f f e r e n t i a l
e q u a t i o n . 1 " ' + . r I " + b r ' .w h e r e a a n d h a r e r a t i o n a l f u n c t i o n so f a c o m p l e x v a r i a b l e r . p r o v i d e d a
" c l o s e d - [ o r m " s o l u t i o n e x i s t s .T h e a l g o r i t h m i s s o a r r a n g e d t h a t i i n o s o l u t i o n i s f o u n d . t h e n
n t r s o l u t i t r nc a n e x i s t .
l. Introduction
In th i s paper we pr e s e n t a n a l g o ri th m fo r fi n d i ng a " cl osed-form" sol uti on of the
dill-erentialequatiofl -1"'+ql +by, where a and b are rational functions of a complex
va ri a b le. x .pr ov ideda " c l o s e d -fo rm"s o l u ti o ne x i s ts.The al gori thm i s so arrangedthat i f
n o so l u t ic r nis f ound. t h e n n o s o l u ti o n c a n e x i s t.
Th e f r r s t s ec t ionm a k e s p re c i s ew h a t i s m e a n t b y " cl osed-form"and show s that there
a re fo u r pos s ible c as e s .T h e fi rs t th re e c a s e sa r e di scussedi n secti ons 3. 4 and 5
re sp e ct iv elyT. he las t c a s ei s th e c a s ei n w h i c h th e g i ven equati on has no " cl osed-form"
so l u ti o n.I t holds pr ec i s e l yw h e n th e fi rs t th re ec a s e sfai l .
In the s ec onds ec t i o nw e p re s e n tc o n d i ti o n sth a t are necessaryfor each of the three
ca se s.A lt hough t his m a te ri a l c o u l d h a v e b e e n o m i tted. i t seenasdesi rabl eto know i n
a d va n c cwhic h c as esa re p o s s i b l e .
Th e a lgor it hm in c a s e sI a n d 2 i s q u i te s i mp l ea n d can usual l ybe carri edout by hand.
p ro vi d ed t he giv en e q u a ti o n i s re l a ti v e l y s i m p l e . H ow ever. the al gori thm i n case 3
i n vtl l vesquit e ex t ens i v ec o m p u ta ti o n s It
. c a n b e p ro grammedon a computerfor a spepi fi c
d i ffe re nt ialequat ion w i th n o d i fl i c u l ty , In fa c t. th e author has W orked through ;everal
cxa mp lesus ing only a p ro g ra mma b l ec a l c u l a to r.O nl y i n one exampl ew as a coi nputer
n e ce ss ar yand
.
t his wa s b e c a u s ei n te rme d i a ten u m b ers grew .to 20 deci mal di gi ts. more
th a n th e c alc ulat orc o u l d h a n d l e . F o rtu n a te l y ,th e necessarycondi ti ons for case 3 are
quite strong so this casecan often be eliminated from consideration.
Th e a lgor it hm does re q u i reth a t th e p a rti a l fra c ti onexpansi onof the coeffi ci ents
of the
d i ffe rent ialequat ion b e k n o w n . th u s o n e n e e d sto factor a pol ynomi al i n one vari abl e
o ve r the c om plex num b e rs i n to l i n e a r fa c to rs .O n ce the parti al fracti on expansi onsare
kn o u 'n. only linear alg e b rai s re q u i re d .
Usi ng t he M A CS Y M A c o mp u te r a l g e b ra s y s te m .see.for exampl e.P avel l e& W ang
(1 9 8 5 ). B ob Cav ine s s a n d D a v i d S a u n d e rs of R enssel earP ol ytechni c Insti tute
p ro g ra m m edt he ent i re a l g o ri th m (s e eSa u n d e rs(1981)).Meanw hi l e.the al gori thm has
0 1 1 11 l 7 l 8 i ' 0 1 0 0 t ) - 1 - 4S10 . 1 . 000
t
l 9 l t 6 A c a d e m r cP r e s sI n c . ( L o n d o n ) L t d
J. J. Kovacic
been implementedalso in the uaplE computer algebra system,see.for example.Char er
al. ( 1985) .by C a ro l y n Smi th (1 9 8 4 ).
This paper is arranged so that the algorithm may be studied independentlyof the
proofs. In section l, parts I and 2 are necessaryto understandthe algorithm. parts 3 and
4 are devoted to proofs. In the other sections,part I describesthe algorithm, part 2
contains examples,and the remaining parts contain proofs.
Sincethe first appearanceof this paper as a technicalreport. a number of papers have
appearedon the same problem: Baldassarri(1980),Baldassarri& Dwork (lg7g), Sinser
( 1 9 7 9 . 1 9 8 1t 9
, g5).
Special thanks are due to Bob Caviness and David Saunders of RPI for their
encouragementand assistanceduring the preparation of this paper.
l. The Four Cases
In the first part of this section we define preciselywhat we mean by "closed-form"
s olut ion. I n t h e s e c o n dp a rt w e s ta te th e four possi bl ecasesthat can occur. Thesecascs
ar e t r eat ed in d i v i d u a l l y i n th e l a tte r s e cti ons.The thi rd part i s devoted to a bri cf
des c r ipt ionof th e G a l o i s th e o ry o f d i ffe re n ti alequati ons.Thi s theory i s usedi n the prc-rofs
of t he t heor e mso f th e p re s e n tc h a p te r a n d those of secti ons4 and 5. P art 4 contLri nsa
pr oof of t he t h e o re ms ta te di n p a rt 2 .
I.I.
LIOUVILLIANEXTENSIONS
T he goal of th i s p a p e r i s to fi n d " c l o s e d - form"sol uti onsof di fl ' erenti alequati o' s. B ' a
" c los ed- f or m"s o l u ti o n w e me a n . ro u g h l y . o ne that can be w ' ri ttendow n
by a fi rs1-)' cl rr
c alc ulus s t ude n t. S u c h a s o l u ti o n m a y i nvol ve esponenti al s.i ndel i ni te i ntegral s and
s olut ions of p o l y n o m i a l e q u a ti o n s . (As w e are consi deri l g functi ons of a compl cr
v ar iable.we n e e d n o t e x p l i c i tl ym e n ti o n tri gonometri cfuncti ons.they can be w ri ttcn rrr
t er m s of ex p o n e n ti a l s .N o te th a t l o g a ri thms are i ndefi ni te i ntegral s and hence i tre
allowed. )A m o re p re c i s ed e fi n i ti o ni n v o l v e sthe noti on of Li ouvi l l i an fi el d.
DEFtNtrtclN.
Let F be a differentialfield of functionsof a complex variable -x that conrai's
C( x ) . ( T husF i s a fi e l d a n d th e d e ri v a ti o noperaror' (: drd.r) carri esF i nto i tsel f). F i s
said to be Lioutillian if there is a tower of differentialllelds
C('x):FolFt9"
s u c h t h a t . f o r e a c hi :
either
or
Fi:Fi-,(r)
C F , : F
1 .. . . , t 1 .
w h e r ex ' l x e F , _ ,
(F , i s g e n e ra te db y a n e x p o nenti alof an i ntegralover F, )
,
Fi:Fi_,(e) wherer'€F,1
(F , i s g e n e ra te db y a n rn te g ralover F,_,)
or
F , i s f i n i t ea l g e b r a i co v e r F , _ , .
A f unc t ionis s a i d to b e L i o u t' i l l i u n i fi t i s c ontai nedi n some Li ouvi l l i andi fl erenti alfi el d.
S uppos e t ha t ry i s a (n o n -z e ro ) L i o u v i l l i an sol uti on of the di fferenti al equati on
Solving Homogeneous Differential Equations
j ," * a 'y'*bj- , wher eq, b e C (r). It fo l l o w sth a t e v e rys ol uti onof thi s di fferenti alequati oni s
Liouvillian. Indeed. the method of reduction of order producesa secondsolution. namely
ry[k-l 'l rt t ) . T his s ec o n d s o l u ti o n i s e v i d e n tl y L i o uvi l l i an and the tw o sol uti ons are
l i n e a rl y independent .T h u s a n y s o l u ti o n . b e i n g a l i near combi nati on of these tw o. i s
Liouvillian.
We may use a well-known change of variable to eliminate the term involving .),'from
This new equation
t h e d i f f e r e n t i ael q u a t i o n .S e t z : e ! l ' y . T h e n z " * ( b - i o ' - I a ' ) r : 0 .
still has coefficientsin C(.x) and evidently )' is Liouvillian if and only if : is Liouvillian.
Thus no generality is lost by assuming that the term involving y' is missing from the
differentialequation.
L2 rHE FouR cASES
In the remainder of this paper we shall considerthe equation
j"':rj',
re C(-x).
We sh all r ef ert o t his e q u a ti o na s " th e D E " . T o a v o i d tri vi al i ty. w e assumethat rt' C . B y
a so l u tion of t he DE is a l w a v sm e a n t a n o n -z e ros o l uti on.
THgoREl,t.
There are preciselv/our c'd)^es
that c'anoccur.
C a s el .
T he DE ha s a s o l u ti o rto /' th e.fo n n e !' ox ' hereo e C (.x).
Case2.
The DE hasa solutiono/'the./brm et''' v'hereo is algebraicorer C(.x)o/ degree
2. and c'aseI doesnot hold.
Ca se3.
All solutionsof'the DE are algebraic'ot,erC(-x)and cd.res1 untl 2 d o not hol d.
Case4.
The DE httsno Liouttilliansolution.
It is evident that these casesare mutually exclusive.the theorem statesthat they arc
e xh a u st iv e.T he pr oof o f th i s th e o re mw i l l b e p re s e n tedi n part 1.4.
1.3. rHn DTFFERENTTAL
GALOTs
GRoup
Here we presenta brief summary of the Picard-Vessiottheory of differentialequations
(se eKa plans k y ( 1957).o r C h a p te r 6 o f Ko l c h i n (1 973)),w hi ch i s tai l ored speci fi cal l yto
th e DE l" : r j- .
Suppose that ry.( is a fundamental system of solutions of the DE (where q. ( are
functions of a complex variable -x). Form the differential extension field G of C(;)
generatedby 4, (, thus
G:
C ( - x ) Q t , O : C ( x ) ( r yr y, ' , ( . ( ' ) '
Then the Galois group of G over C(x). denoted by G(G/C(,x)), is the group of all
differential automorphisms of G that leave C(r) invariant. (An automorphism o is
differential if o(a') : (oa)' for every a e G.) We refer the reader to the referencescited
above for a proof that the FundamentalTheorem of Galois Theory holds in this context.
There is an isomorphism of G(G/C(,x)) with a subgroup of GL(2), the group of
invertible 2x2 matriceswith coefficientsin C. Let o e G(G/A(-x)).Then
(oq)": o(4"): o(rry)- ron.
J. J. Kovacic
Henc e.oq is a l s o a s o l u ti o n o f th e D E a n d so i s a l i near combi nati ono4:ootl * c' oi .
( ' , , € C .o f r l .( . S i m i l a r l yo. ( : b o r y + d . ( f o r s o m eb o , d o e C .
/
r
\
','\
(u'
(". o ---+
\('
cr.,.
d'/
is im m ediat el ys e e nto b e a n i n j e c ti v eg ro u p homomorphi sm.
This representationr' : G(G/C(-x))-- GL(2) certainly does depend on the choice of
f undam ent als y s te mry .(. l f 4 t,(t i s a n o th er fundamentalsystem.then there i s a matri x
X e G L( 2) s uc h th a t (q ,. (,) : (r/.i )X. T h e refore.
G:
C ( , x ) ( r y . i ): C ( . x ) ( r y(rr. )
a n d c r ( o )- X - 1 c ( o ) X .
T he r epr es en ta ti o n
G(G/C (.x ))--' G L (2 i)s d etermi nedby the D E onl y up to conj ugati on.
By abuseof language.we allow ourselvesto speakof any one of theseconjugategroups as
thc Galois group of the DE. If a fundamental system rl.i is fixed. then we refer tt-r
c ( G ( G iC( . x ) )c) GL (2 ) a s th e G a l o i s g ro u p o f the D E rel ati veto r7.(.
Fix a fundamentalsystemry.( of solutions of the DE and let G c GL(2) be the Galois
gr oup r elat iv eto 1 7 .(.L e t W :t1 ;' -n ' ;
b e the W ronski an of 17.(. A si mpl ecomputati on.
us ing t he DE . s h o w s th a t W ' :0 , s o W rs a (non-zero)constantand i s l eft fi xed by ant'
e l e m e n to f G ( G / C ( , x ) )L. e t o e G ( G l C ( , r ) ) .t h e n . u s i n gt h e n o t a t i o na b o v e .
W : o W : ( a o \ + c , "_ ( ) ( b o 4 ' l d_, ( ' ) - ( a o r y ' * c , i ' ) ( b , r y + d . J
: ( a o d o - h o c ) W : d e Ic ' ( o l. W .
T hus G c S L(2 ),th e g ro u p o f 2 x 2 m a tri c e sw i th determi nant1.
Rec allt hat a s u b g ro u pG o f G L (2 ) i s a n a l gebrai cgroup i f thereexi sta fi ni te number of
poly nom ials
P l . . . . , P , eC [ X , . X , , .X . . X * ]
s u c ht h a t
if and only if
P r ( a . b c, ' ,d ) : ' ' '
(i
").,
: P , ( a h. . r ' . t l ) : 0 .
O ne of t he pr i n c i p a l fa c ts i n th e Pi c a rd V essi ot theory i s that the Gal oi s group of a
differentialequation is an algebraicgroup. For a proof in all generality.seethe references
cited above. Here we sketch a proof in the specialcasethat we are considering.
Let Y . Z , Y r, Z , b e i n d e te rmi n a te s over C (.x) and consi der the substi tuti trn
hom om or phis m
T he k er nel of th i s m a p p i n g i s a p ri me i d e a l p . A n y element
A -
(::)
of S L( 2) induc e sa n a u to m o rp h i s mo f C [.x . y' . Z. Y r.Z,l over C [.r] by the formul a
( Y , Z , Y r ,Z , ) - ( a Y + c Z . b Y + d Z , a Y r + c Z r . b \ + t l Z r ) .
M or eov er ,A e G i f a n d o n l y i f p i s c a rri e di n to i tsel f.The i deal p i s fi ni tel y generated.sav
p : ( q l . . . . . q , ) . w h e r eQ t . . . . , Q " ' d t el i n e a r l yi n d e p e n d e n
o tv e rC . L e t n b e t h e m a x i m u m
of t he degr ee so f q r
1 7 i,n ,x , Y ,Z ,Yr,Z , and l e| V be the vector spaceover C of al l
poly nom ialsin C [,x , Y. Z ,Y r,Z rl o f d e g reerr or l ess.E vi dentl y the acti on of S L(2) on
S o l v i n g H o m o g c n e o u sD i f f e r e n t i a lE q u a t i o n s
q , i s a b a s i so f V . t h e nt h e r ee x i s t
C [ , r . i ' . 1 . . 1 , . / , ] 1 r c s t r i c tlso 1 " .I f 1 1 ,... . , Q , . 4 s + t
p o l y n o r r i i i iP
l s, , € { [ Y , . f . , . \ . , . l * ] s u c ht h a t t h e r e s u l to f t h e a c t i o no f , 4 o n q , i s
b.c'.d)q,.
t P,,(u,
i= L
I t f o l l o w st h a t i e G i f a n d c ' r n l iyf P , t ( t t , b . c ' . d ) : Q
for i:1,....s..i:s*1.....t.
Thgreft,r0Ci is an "rlgetrraicgroup.
1 . 4 .P R o o F
i n th is s ec t ionr v e s h a l l p ro v e th e th e o re mth a t w as statedi n 1.2. W e shal l use several
fa ctsa bout algebr aicg ro u p s .S u i ta b l ere fe re n c easre B orel (1956).K apl ansky (1957).and
. h e f o l l o w i n gr e s u l ti s c o n t a i n e di n K a p l a n s k y( 1 9 5 7 .p . 3 l ) .
C h a p t c r5 o f K o l c h i n ( 1 9 7 3 ) T
L n n a vn.Let G he un ul g e h ru i cs u b g ro u yot/ S L (2 ). T h en one o/ /i tur (' uses(' unoL' (' ur.
Crr.sr'l.
O is triungttlisuhle.
Cuse2.
G is t'ort.jugute
to u xtbgroup of
')
(,, 0\l
(t o ,\l
D - : ] { . . " , ) ,l ' e C . r ' + 0 f \ - {- l, ^ l l, ' e. . r ' * .0l f .
l\0(' 't
) [ \ - . ' - '0 / l
)
urul t'usa I r/oe.s
not holtl.
Ca .se3.
G is linit e u n d c u s e sI u n tl 2 tl o n o t h o l tl .
Case1. G:Sl,(2).
Pro o f. Denot e t he c om p o n e n t o f th e i d e n ti ty o f G by' G . F-i rstw e note that any tw od i me n sionalLie algeb rai s s o l v a b l e "h e n c ee i th e r d i m G : 3 (i n w hi ch caseG : S L(2)) or
e l seG- is s olr able. I n th e l a tte r c a s e .G i s tri a n s u l i sabl ebv the Li e K ol chi n Theorem.
A s s u m et h a t ( i i s t r i a n g u l a r .
':)
I f G i s n o t d i a g o n a l i s a b l et h
. e n G c o n t a i n sa m a t r i r o f t h e f o r n r 0
*itn u * 0
\0 1/
(si n ce an algebr aics ro u p c r-rn ta i n sth e u n i p o te n t and semi -si rnpl e
of al l of i ts
.parts
e l e m e n t s )S. i n c eG i s n o r m a l i n G . a n y m a t r i x i n G c o n j u g a t e rf : i ) l n , o a t r i a n g u l a r
\0 t/
ma tri x. A dir ec t c om p u ta ti o n s h o w s th a t o n l y tri a ngul ar matri ces have thi s property.
T h u s G i t s e l fi s t r i a n g u l a r T
. h i s i s c a s el .
Assu m e nex t t hat 6 rs d i a g o n a l a n d i n fi n i te . s o G contai ns a non-scal ardi agonal
ma tri x , { . B ec aus cG i s n o rm a l i n G. a n y e l e me ntof G conj ugates' 4 i nto a di agonal
m a t r i x .A d i r e c tc o m p u t a t i o ns h o w st h a t a n y m a t r i x w i t h t h i s p r o p e r t vm u s t b e c o n t a i n e d
i n D * . T h e r e f o r ee i t h e rG i s d i a g o n a l .t h i s b e i n gc a s e1 . o r e l s eG i s c o n t a i n c di n D * . t h i s
b e i n g ca s el.
F i n a l l y w e o b s e r i e t h a t i i G i s { r n i t e( a n d t h e r e f o r e6 : l l l ) . t h e n G m u s t a l s o b e
fi n i te . This is c as e3. Th i s p ro v e sth e l e mma .
We shall no\ \ ' pr ov e th e th e o re mo f s e c ti o n2 .
L e t 1 7i" b e a f u n d a m e n t asl y s t e mo f s o i u t i o n so f t h e D E a n d l e t G b e t h e G a l o i s g r o u p
r e l a t i v et o r i .i . S c t G : C ( . x ) ( r 1, )..
C - a s el . G i s t r i a n g u l i s a b l eW
r .h e n . f o r e v e r v
. e m a v a s s u m et h a t G i s t r i a n g r " r l a T
J. J. Kovacic
o € G ( G / C ( r ) ) . o 4 : c o 4 , w h e r e ( ' o € C . c , + 0 . T h e r e f o r eo o ) : r l - l , w h e r er o : q ' 1 4 , w h i c h
implies that rr,re C(.x).
Cas e2. G is c o n j u g a teto b e a s u b g ro u pof D * . W e may assumethat G i s a subgroupof
D r . l f r r ; : r y ' l r \a n d 6 : _ ( l ( . t h e n . f o r e v e r y o e G ( G l C ( , x ) ) ,e i t h e r o @ : o . 0 6 : $
or
o( ) r : Q , oQ : rr;.T h u s a -li s q u a d ra ti co v e r C (.x).
Cnse3. G is finite. In this caseG has only a finite number of differentialautomorphisms
or . . . , o, . S i n c eth e e l e m e n ta rys y mme tri cfuncti on of o, tl . . ., on4 are i nvari ant under
G(G/A(.x)).4 is algebraic over C(.r). Similarly, ( is algebraic over C(.x). Becauseevery
s olut ion of t h e D E i s c o n ta i n e di n G . e v e ry sol uti on of the D E i s al gebrai c.
Cas e4. G : SL (2 ). S u p p o s eth a t th e D E had a Li ouvi l l i an sol uti on. Then, as poi nted
out in l. l. ev e ry s o l u ti o n o f th e D E i s L i o u vi l l i an. Thus G i s contai nedi n a Li ouvi l l i an
f i e l d . I t f o l l o w s t h a t G " i s s o l v a b l e( K o l c h i n , 1 9 7 3 .p . 4 1 5 ) . S i n c e G ' : S L ( 2 ) i s n o t
s olv able,t he D E c a n h a v e n o L i o u v i l l i a n s o l uti on.
This proves the theorem.
2. NecessaryConditions
I n t his s ec t i o nw e d i s c u s ss o m ee a s yc o n di ti onsthat are necessary
for cases1. 2. or 3 to
hold. These conditions give a sufficientcondition for case 4 to hold (namely when the
nec es s ar yc on d i ti o n sfo r c a s e s1 .2 . a n d 3 fai l ). Throughout. w e shal l consi derthe D E
)"':r)'. re C(-x).
2.1. rHE NECESSARy
coNDrrroNs
Sincer is a rational function, we may speak of the poles oi r. by which we shall ali.r'ays
m ean t he pole si n th e fi n i te c o mp l e x p l a n e C. If r:sl t. w i th s. re C [.x]. rel ati vel l pri me.
then the polesof r are the zerosof r and the order of the pole is the multiplicity of the zero
of r. By the order of r at co we shall mean the order of r as a zeroof r. thus the order of r
at r isdegr-degs.
THEonsn. The./ollowingconditionsare necessarl'.fnrthe respectiuec'ases
to hold.
Case l.
Et:erypoleo./r must haueeuenorder or elsehat,eorder 1. The ortler of-r ut r,
must be eL)en
or elsebe greater than 2.
Cttse2.
r nrusthatteat leastone pole that either hasodd order greater thun 2 rtr elsehus
order 2.
Cuse3.
The Ltrdero/'a pole o.f r cannot exceed2 and the order of'r ot :r- nlLtstbe ttt
leust 2. I./ the partial /raction expansiono./'r is
t',+t
,:F
t he , n fr + 4 o ,e Q ,.fn r e a c hi ,
T 4:
0, and i ./'
)':Iat+\Pidi.
I
t h e n1 1 + 4 ; ' e Q .
0t,.
7 x-di
71r-r'1)-
J
S o l v i n g H o m o g e n e o u sD i f f e r e n t i a l E q u a t i o n s
2.2. Exauplss
Ai re y ' s E quat ion ! " : x y h a s n o L i o u v i l l i a n s o l uti on (i .e. case4 hol ds). Thi s i s cl ear
b e ca u s et he nec es s ar yc o n d i ti o n sfo r c a s e s1 ,2 . a n d 3 al l fai l . More general l y,)" ' : P y,
where P e C[x] has odd degree,has no Liouvillian solution.
Fo r B es s el' sE quat i o n
, , ,: 4 ( " 4- x, '' l ) - l , , , n ec
(in self-adjointform), only casesl. 2, and 4 are possible.
For Weber's Equation
y ,, : (i x r _ l _ r)t,. n e C "
only casesI and 4 are possible.
2 .3 . P R o o F
In this sectionwe prove the theorem of Section 1.
Ca se 1. I n t his c as eth e D E h a s a s o l u ti o n o f the form 4:e!" w here rtreC (.r). S i nce
4 ":rq , it f ollows t ha t a ' l (D 2 : r (th e R i c c a tti Eq uati on). B oth co and r have Laurent
seriesexpansionsabout any point c of the complex plane. for easeof notation we take
c:0. Say
a:bx4*
r : a-xl'+
, pez, b+0
,
v€2, a*0.
(The dots representterms involving .x raised to powers higher than that shown.) Using
the Riccatti Equation, we find that
l + . . . + b 2 x 2 +u . . . : 1 . \ ' ' + . . . .
Fbxt'As we need to show that every pole of r either has order 1 or elsehas even order. we may
a s s u m et h a t v < - 3 . S i n c ea * 0 , - 3 > r ' ) m i n ( p - 1 . 2 p ) . I t f o l l o w st h a t p < - 1 a n d
2 p < p - 1 . S i n c eb t + 0 . 2 p : v , w h i c h i m p l i e st h a t y i s e v e n .F o r u s e i n s e c t i o n3 . 3 .w e
re ma rk t hat if r has a p o l e o f o rd e r 2 p )4 a t c . th e n rr-lmust have a pol e of order y at c.
Now considerthe Laurent seriesexpansionsof r and or at r.
at:bxq+
r : C-xr'+
peZ, h+ 0
y € 2 . u * 0.
(Th e d ot s r epr es entt ermsi n v o l v i n gx ra i s e dto a p o w er l ow er than that show n.The order
of r at oc is - r'.) As we need to show that either the order of r at cc is ) 3 or elseis even.
w e ma y as s um et hat r' ).-- l . U s i n g th e R i c c a tti Eq uati on.w e have
1+. ' .*b2x2P+'..: d-xY+.".
FbxuJ u s ta s a b o v e .- I < r ' ( m a x ( p - 1 . 2 p ) . p > - 1 , 2 p > p - 1 . S i n c eh 2+ 0 . 2 p : r , ,s o u i s
even. For use in section3.3, we remark that if r has a pole of order 2p20 at oc,,then rr-r
has a pole of order p at cn.
This verifiesthe necessaryconditions for case l.
Case 2. We analyse this case by consideringthe differential Galois group that must
o b ta i n . B y s ec t ion1. 4 th e g ro u p mu s t b e c o n j u g a teto a subgroup G of D r. w hi ch i s not
triangulisable (otherwise case I would hold). Let q, _( be a fundamental system of
I0
J. J. Kovacic
s o l u t i o n so f t h e D E r e l a t i v et o t h e g r o u p G . F o r e v e r v o e G ( G , C ( . x ) ) .e i t h e r o \ : 1 ' o 1 , .
o ( : r ' , t ( o r o t l : - ( ' ; t i . o i : c o t l .E v i d e n t l v1 1 : . 2i s a n i n v a r i a n to f G ( G , ' C ( x ) a
) nd
- ( . x ) .M o r e o v e r , q - : + C ( . x )f.o r o t h e r w i s eG w o u l d b e a s u b g r o u po f t h c
t h e r e f o r eq 2 ; 2 eC
diagonalgr oup . w h i c h i s c a s e l .
Writing
4 . _ (,: f] I t - (,i ),' ' (e,eZ).
we hav e t hat a t l e a s to n e e x p o n e n tc , i s o d d . W i thout l oss of general i tyw e may assLl l l tc
that
-(2: .x.'fl (.\- c',).',
ry2
and that t, is odd. Let
0 : (q0'i(rlo : )Q\t?)'l\t, _3): \ex 1+ . . ..
where the dots representterms involving -x to non-negativepowers. Since 4" : rn Lrnd
(" : ri.
0"+ 30'0* f/3: 4r0+ 2r'.
L e t r : f l - x r + " ' b e t h e L a u r e n ts e r i e e
s xpansioo
n f r a t 0 . r v h e r et * 0
t hc equat iona b o v e w c o b ta i n
and le Z. From
( n - i r t + | c 3 ) . r ' r +. . ' : 2 x ( c * r ' ) r ' + ' . . .
I f r ' > - 2 . t h e n 0 : 8 e - 6 e 2 * e - r : e ( e - 2 X e - 4 ) . T h i s c o n t r a d i c ttsh e f a c t t h a t e i s o d d .
T l r e r e f o rre' ( - 2 . I f r ' < - 2 . t h e ne * I ' : 0 . s c -l r i s o d d .
T his v er if iesth e n e c e s s a ry
c o n d i ti o n sfo r case2.
C a s e 3 . I n t h i s c a s et h e D E h a s a s o l u t i o n 4 t h a t i s a l g e b r a i co v e r C ( x ) . 1 7 h u s a
P uis eauxs er ie se x p a n s i o na b o u t a n y p o i n t c i n the compl expl ane.for easeof notati t)n\\c
t a k ec ' : 0 . T h e n \ : u x t ' + " ' , w h e r ea e C , a # 0 , / l € Q . S i n c er e C ( _ x )/ . : x . \ ' + . . . .
wher e z * 0 an d l e Z . T h e D E i m p l i e sth a t
'
lt(tt-l)c.xr'-2+
:1rI-\I+rr*
I t f o l l o w s t h a t v ) - 2 . i . e . r h a s n o p o l e o f o r d e r g r e a t e rt h a n 2 . I f r , : - 1 . t h c n
p ( p - l ) : r . B e c a u s1er e 8 . w e m u s t h a v er | + 4 x e Q .
S o f ar we ha v e s h o w n th a t th e p a rti a l fra cti on expansi onof r has the form
, . :I
71r *.x;)- l.r-r/,
wher e P e C[ . x] a n d . I + 4 x ,e Q fo r e a c h i .
Nex t . we c on s i d e rth e s e ri e se x p a n s i o n sa b out r.
tl:uxt'+....
f : l',{'+'...
where the dots representlower powers of -x than those shown. From the DE we c-rbtain
p(p- l)a.rr 2+...:
l'^,d.\\'+rrf
S o l v i n g H o m o g e n e o u sD i f f e r e n t i a l E q u a t i o n s
Just as above. we obtain r' < - 2 and thereforeP : 0. But
'' --T
ai-
-'r - T 0 i
? 1 r _ r , ; ) r? r _ d ,
: q i l r - t + ) ' , t -+2 . . . ,
w h e r ei , : I r , * T
[ ) , d , . T h e r e flo, lr]e, : 0 a n dp ( t t - l ) : l ' . S i n c e
#€Q.y t+aie A.
Thiscompletes
the proofof the theoremstatedin section2.1.
3. The Algorithm for Case I
The first part of this sectionis devotedto a descriptionof the algorithm. It is somewhat
complicatedto describein full generality.yet. as the examplesin part 2 show. it is often
quite easy to apply. The third part is devoted to a proof that the algorithm is correct.
3 .1 . o e s c R rPT rooNF T H EA LGOR TTH M
Th e g oal of t his alg o ri th m i s to h n d a s o l u ti o n o f the D E of the form q: pst' ' ' . w here
P e C [ . x ] a n d r o e C ( . x ) .S i n c e r / m a y b e w r i t t e n a s r y : g ! ( P ' P + ( o )t.h i s i s o f t h c f o r m
d e scri b edin s ec t ion1. 2 .T h e fi rs t s tc p o n th e a l g o ri t hmconsi stsof determi ni ng" parts" of
th c p a r t ial f r ac t ion exp a n s i o no f t,,r.In th e s e c o n dstep w e put these" parts" togetherto
fo rm a c andidat ef or r,.r.T h e ma x i mu m n u mb e r o f candi datespossi bl ei s l p* 1 w here p i s
th e n u m ber of poles o f r. If th e re a re n o c a n d i d a tes.then case I cannot hol d. The thi rd
a n d l a s t s t ep is applie d to e a c h c a n d rd a tefo r ro a n d consi stsof searchi ngfor a sui tabl e
p o l yn o m ial P . I f one i s fo u n d . th e n w e h a v e th e d e si redsol uti on of the D E . If. for each
ca n d i d at ef or o. we f a i l to fi n d a s u i ta b l eP. th e n c a seI cannot hol d.
We a s s um et hat t he n e c e s s a ry
c o n d i ti o n o f s e c ti o n2.1 for case t hol ds. and w e denote
b y I th e s et of poleso f r.
Ste p l. F or eac h c' ef u { ,r} w e d e fi n e a ra ti o n al functi on tJ7].
numbers r,.*.r,. as describedbelow.
and tw o compl cx
(r',) If ce l- and c is a poleof order 1. then
[."7].:o'
(r':)
1.*: x,:l'
I f c ' eI . -and c i s a p o l e o f o rd e r 2 . th e n
: g'
['.'u''i1.'
I - et b be t he c o e fi i c i e n ot f 1 i (.x -r' )2 i n th e p arti al fracti on expansi onfor r. Then
1 . .: l + 1 . , l + 4 h
(c':) I f c e I - and c i s a p _ o l eo f o rd e r 2 v )4
(necessari l even
y
by the condi ti ons of
s e c t i o n2 . 1 ) .t h e n [ . " , ' r ] .i s t h e s u m o f t e r m si n v o l v i n g1 , , ' ( - \ - r ' ) ' f o r :< i < r ' i n t h e
L a u r e n ts e r i e se x p a n s i o no f , i a t r ' . T h e r e a r e t w o p o s s i b i l i t i efso r [ . r ] , . o n e
being t he neg a ti v eo f th e o th e r. e i th e r o n e may be chosen.Thus
[v,i].: n5 + . .,* r.j;,1.
J. J. Kovacic
In practice.one would not form the Laurentseriesfor u6, but rather would
determinel.r/il, by using undeterminedcoefficients.Let b be the coefficientof
ll(x_c)"*1 in r minusthe coefficient
of 1/(.r-c)n*t in (t.r6l.). Then
l( b
*
\
/.,':
t\t;*t').
(ocr) If the orderof r at oc is >2, then
tv6l- : o. dl : o, d,.: 1.
( . c r ) I f t he o rd e r o f r a t o c i s 2 , th e n
f"Fl' : o.
Let b be the coefficientof llx2 in the Laurent seriesexpansion of r at ;c. (If
r : s lt , w h e re s , re C [.x ] a re re l a ti v e l ypri me. then b i s the l eadi ngcoe{l i ci entof s
divided by the leading coefficientof r.) Then
(rr) Ir theorderor r at ' ,, :;' ='r-,;:.:#1,
evenby theconditions
or section
2 . 1 ) . t h e n [ f r ] , i s t h e s u m o f t e r m s i n v o l v i n gr t f o r 0 < t < r , i n t h e L a u r e n t
s er iesfo r .,," ra t o c . (Ei th e r o n e o f the tw o possi bi l i ti esmay be chosen.)Thus
[r '']' : (/'\'+ '+d'
Let b b e th e c o e ffi c i e n o
t f .r' -1 i n r mi nus the cocffi ci entof .r' - 1 i n ([., /.], )'
T hen
, t ( , h \
r;: ,( t ,-r )
.\ "
/
Step 2. For each family s : ( s ( c ) ) . . r - u t x l .w h e r e s ( c ' )i s + o r - . l e t
d :
a')' )-
I
1s(c).
If d is a non-negative
integer,then
ro: I
c e f
/
,
\
( r ( . ) [ . , ,r l . + .":s ('c')- ) * s ( x ) [ , , r, ] ,
\
. \ - L , /
is a candidatefor rr;.If d is not a non-negativeinteger.then the family s may be removed
f r om c ons idera ti o n .
S r ep 3. T his s te p s h o u l d b e a p p l i e d to e a ch of the fami l i esretai nedfrom S tep 2. unti l
s uc c es is
s ac hie v e do r th e s u p p l y o i fa mi l i e sh as beenexhausted.In the l atter event.caseI
c annot hold.
F or eac hf am i l y . s e a rc hfo r a mo n i c p o l y n o mi alP of degreed (as defi nedi n S tep2) that
satisfiesthe differentialequation
P " + 2 t ' . t P ' * ( c o ' * t o 2- r ' ) P : 0 .
This is convenientlydone by using undeterminedcoelicients and is a simple problcm in
linear algebr a .w h i c h ma y o r ma y n o t h a v e a sol uti on. If such a pol ynomi al exi sts.then
Solving Homogeneous Differential Equations
l -l
'
4 : Pe j is a s olut ion o f th e D E. If n o s u c h p o l y n omi al i s found for any fami l y retai ned
from Step 2. then Case 1 cannot hold.
3 .2 . e x n u p l E s
Exantple1. Consider the DE -1"': r'l' where
-!-1l
-20x* 4
, _ o'u + I?L#] !l]-'
: \ 2 - 2 r * 3 + l +L _ : * 1 .
.Y
4.r'
.Y'
\*
Si n c e r has a s ing l ep o l e (a t 0 ) a n d th e o rd e r there i s 4, the necessarycondi ti ons of
se cti on2. 1 f or c as e2 d o n o t h o l d . E v i d e n tl yth e n e cessary
condi ti onsfor case3 al so do
n o t h old. W e apply t h e a l g o ri th m fo r c a s e I to th i s D E .
T h e o r d e r o f r a t t h e p o l e 0 i s 2 r ' : 4 . T h e r e f o r e[ v r ] n : c , ' r : . a n d a 2 : 1 . W e c h o o s e
a : l . s o [ r , r ] o : l / - x 2 b. - - 5 - 0 : - 5 . a n d t h e r e f o r reo t : j ( t ( _ 5 , , , 1 ) + Z ) . w h i c hg i v e s
e J : - 3 2 and y . i : 71 2 .
A t c c , r ' : l . a n d [ . , t 1 r : ( r . \ * r 1 . C o m p a r i n gr a n d t . r / ; ' ] t ,
=ott'*2ctdr+r/2 we see
1
.
(
l
t h a ta 2 : 1 a n d 2 a t l : - 2 . w e c h o o s u
e:1,
T h u s[ / r ] " : , x - r . h - 3 - r : 2 .
a n de t ' :
l 1 2 ,a " : - 3 1 2 .
There are four familiesto consider.
s(0):1.
s ( c c -) + .
s(0):-t-.
s(0): -.
s(:c):-r
s(r'): *.
lt2-(-3t2) - 2
d : -3t2-(-3,i2):0
- -3
tl: | 2-7i2
s(0):-.
.s(:c):--.
d:-32-J,'2
d:
On l y the t r r s t t wo r em a i n fo r c o n s i d e ra ti o n .
We s hall t r eat t he s e c o n dfa m i l y fi rs t. s i n c ed :0
:-5.
i n that case.The candi datefor ro i s
ri -[..
,r: [r r]o*
r ] , : l , - : - . y +l .
-\
-Y- 2r
We now searchfor a monic polynomial P of degree0 such that
P" +2oP' *(@' *ttt2-r)P : 0.
S i n c eP : 1 , t h e e x i s t e n c e
o f P i s a q u e s t i o no f w h e t h e ro r n o t r o ' * t o 2 - r : 0 .
co e ffi c ientof 1/ , xin a' * o )2 -r i s -6 . T h u s n o s u c h pol ynomi al P can exi st.
The only remaining family is the first family. The candidatefor ro is
,r:
,
/ -
r j , F
LVrlo*
- : + L ' f r -' J , : . K|, -
3
2,
But the
*.t- l.
We now searchfor a monic polynomial P of degree2 that satisfiesthe linear differential
e q u a t i o ng i v e n a b o v e .W r i t i n g P : x 2 * a x * b , w e e a s i l yd e t e r m i n et h a t a : 0 , b - - l
p ro vi d e sa s olut ion.
Thereforea solution of the DE is given by
4
:
Pe['u:
-
-X
( . x 2-
1)gltrlx2-3lzx\+x-
t t z 1 r 2- 1 ) e - 1 l x * x 2 t 2 - x .
1)
l.t
J. J. Kovacic
Erample 2. ln this examplewe begin the discussionof Bessel'sEquation
/4n2-l
\
, ) " ': | , .- 4 i - - t
),.
rreC'
The nec es s ar cy o n d i ti o n so f s e c ti o n2 .1 i mp l y that case3 cannot hol d. H ere w e consi der
cas el" c as e2 is w o rk e d o u t i n s e c ti o n3 .2 .
The only pole of r is at c : 0 and the order there is 2. Thus
[v i]o : o.
- l)14. n't : i t +J-l +4b : ! * n.
7,: 14n2
A t ; r , . r has o rd e r 0 a n d [V ' 7 ] , : i . Ev i d e ntl yb:0
There are four familiesto consider.
s(0)
s(0)
I
l
.
I
s(0)
s ( o c ): 1
s (:c ' ): s(r):1
s(r1: -
s(0)
S o a-t - Q.
d - -I12-n
d- -l12-n
d - -lt2+n
tl: -l2*n.
A n e c e s s a rcyo n d i t i o nt h a t c a s eI h o l d si s t h a t - l 2 * r r
b e a n o n - n e g a t i v ien t e g e r .i . e .
t hat n be half a n o d d i n te g e r.We c l a i m th a t thi s condi ti on i s al so suffi ci ent.
I f n is negat iv e a, n d h a l f a n o d d i n te g e r.th e n nr - - I 2-ne N . Thi s correspondsto thc
f ir s t f am ily . in w h i c h c a s e(D : -tn i i X * i . We needto fi nd a pol y' nomi alP of dcgrccrn such
t hat
0 : P " - t 2 o P '* (rr.r'+ t'f / nt \
P "+ 2 1- + i l P ' \ . \ / .
r)P
2int
\
P.
It is straightforward
to verifythat
(2m-jlt_1
I
'o _- ,$? o 1 - 2 i \ ^ - i
'
il@-i)!
is t he des ir edpo l y n o m i a l .A s o l u ti o n to B e s sel ' sE quati on i s gi ven by 17:.y ^P L,' ' .
I f n i s p o s i t i v e ,t h e n m : - l 1 2 * n i s a n o n - n e g a t i v e
i n t e g e r .T h i s c o r r e s p o n d st o t h e
thir d f am ily . I n th i s c a s etrl : -ml x * i . a n d we are back to the caseconsi deredabove.
Erttmple 3. tn this example we treat the general situation where r is a pol,t-nomialof
d e g r e e2 . W e m a y w r i t e 7 " : ( a x + d \ 2 * b f o r s o m eu . h . t l e C ( a a n d t / a r e d e t e r m i n e db 1 ' r '
W e cl ai m that the D E has a
o nly up t o s ign . w e c h o o s ee i th e r o f th e tw o possi bi l i ti es).
if
if
Liouvillian solution and only bla is an odd integer.
T he nec es s a ry
c o n d i ti o n o f s e c ti o n2 .1 i mp l i esthat onl y casesI and 4 are possi bl e.W c
c ons iderc as e l.
E v i d e n t l y[ . u " r ] , ,: d \ * d a n d a i - _j ( * ( b ' ' r r ) - 1 ) .T h e r ea r e n o p o l e s .T ' h u sr / e q u a l sz ,
or r , . . s o one o f th e s etw o n u mb e rsm u s t b e a non-negati vei ntegerfor case I to hol d. It
follows that hlu must be an odd integer.which is the necessitypart of our claim.
F or s uf f ic ien c yw. e m a y a s s u m eth a th l u :2 nl
I i s posi ti ve.si nced may be repl acedbr
- a. Cas e I will h o l d p ro v i d e dth a t th e re i s a moni c pol ynomi al P of degreen such that
-- P" *2oP'
l(ro' *r,t2 - r)P
: P" -l2(ux + d)P' -2naP.
S o l v i n g H o m o g e n e o u sD i f l - e r e n t i aE
l quations
l-5
If w e w r it e
P: i
P,.\i
i"o
a n d s u b s t i t u t ew
. e o b t a i n a s y s t e mo f l i n e a re q u a t i o n si n P , , .
so l u ti o n.nam ely
-i*:
- i * , f+ qt 2xt1]) P
p,: (l" t] xlli ) p,,
r jitr_,.',
tl_t
. . P , , _ ,( P n : l ) t h a t h a s a
(i: rl- l.. . ..0)
w h e r eP u+ | : 0 a n d P , : | .
A spec ialc as eof t h i s e x a mp l ei s W e b e r' sE q u a tion
_ 1 ": ' (i .x . -.! -,t)t' .
H e r eu : - l 2 . h - - 1 , 2 - n , t l : 0 . T h u s h , u : k t + l
a n i n te- uer .
n e C.
i s a n o d d i n t e e e irf a n d o n l v i f r r i s
3 .3 . PR (x )F
l n c a s el . t h c D E h a s a s o l u t i o no f t h e f o r m , / : r , i ( ' . w i t h t ) e C ( . r ) .S i n c e\ " : r q ,
h a ve
( ) '+ 0 2 : r
we
( R i c c a t t iE q u a t i o n ) .
We shall det er m inct h c p a rti a l fra c ti o ne x p a n s i o no f () usi ng the Laurent seri esr-xpansi on
o f r a n d t h i s R i c c a t t iE q u a t i o n .
Fo r c ' eC. u' e deno tc th c " c o m p o n c n t a t r' " o i thc parti al fracti on expansi onof (/ by
[ ( / ] , \+]
(ti.,+-''
:i
t '
\ - t '
i l : ( \ - t ' ) '
I n o r d c r t o s i n t p l i f i t' h e n o t a t i o n .w ' ea s s u m et h a t r ' : 0 a n d d r o p t h e s u b s c r i p t" 0 " . W e
sh a l l als o need t o c on s i d e rth c L a u re n t s c ri c sc -rp a nsi onof (/ about 0
^
t): [r)]+ +{l
'\
ry
" ' . w h e r et h e * d e n o t e sa c o m p l e xn u m b e rw h o s cp a r t i c u l a rv a l u ei s
H e r e0 : x * * r *
i rre l e vantt o our dis c u s s i o n .
We as s um et hat t h e n e c e s s a ryc o n d i ti o n sfo r c ase I (seesecti on2.1) are sati sfi ed.i n
p a rti cu lar we as s um eth a t th e p o l e s o f r e rree i th e r of even order or el seof order l . W e
sp l i t o ur pr oof int o pa rts .d e p e n d i n go n th e n a tu re of r at 0. Thi s pzrral l elthe
s di vi si on of
Ste p 1 of t he algor it h m .
( c t ) S u p p o s et h a t 0 i s a p o l e o f r o f o r d e r l . s o r ' : * , ' . \ + " ' . T h e R i c c a t t ie q u a t i o n
becomes
l ' (J '
4
\r-,
' * u i + " ' : . ,* *
,l
Si n ceui + O . r ' ( I an d [0 ] : 0 .
Su bs t it ut ing0 : r , x * { / i n to th e R i c c a ttiEq u a ti o n.w e have
i.
v.2
2a-
*
_\-
-\-
_\
r
lo
J. J. Kovacic
T h e r e f o r e- c + d . 2 : 0 , S o d : 0 o r a : 1 . W e r e d : 0 , t h e l e f t - h a n ds i d eo f t h i s e q u a t i o n
would have 0 as an ordinary point; however. the right-hand side has a pole at 0. We
concludethat a : I and the component of the partial fraction expansionof 0 at 0 is (in the
not at ion of t he a l g o ri th m )
N
* 'x.
w h e r e1 * : 1 .
x
(cr) Supposethat r has a pole at 0 of order 2, say
b
*
,:?*;*"'.
As in (c,), [(/]:0
of0at0is
and -ala2:b.
T h u s t h e c o m p o n e n to f t h e p a r t i a lf r a c t i o ne x p a n s i o n
c
!
-i,
,
-
c- : +!i.,/ 1+ 4b.
where
( c . ) S uppos eth a t r h a s a p o l e a t 0 o f o r der 2p> - 4. In secti on2.3. w e show ed that
t ' : F . R e c a l l f r o m s e c t i o n3 . 1 t h a t
[r']:tl *
- Ir
*
-.
-\-
where we have dropped the subscript"0".
Ler t -- rG -f"til
Then , : lr,5l2 + 2t[. r f + r - . F r o m t h e R i c c a t t i E q u a t i o n \ \ ' e
- t
obtain the followine formula
(tol- t"6} (tol+ tv6l)
: - [o]'+al, - 0 , - 4 f r l _ 2 e - e )
(&)
_x.\
x.2
)v -
,_:0
-_0'+2i[\ i:1+i..
,X
,X-
An examinationof the right-hand side of this equation determinesthat it is free of terms
( s i n c et , 2 l ) . T h i s i m p l i e st h a t t h e l e f t - h a n ds i d e i s 0 .
i n v o l v i n gl i ' - x ' f o ri : r ' * 2 , . . . , 2
Indeed. since
( t p l - t . ,i l ) + ( t 0 l + t r t l ) : r [ 0 ] .
a t leas tone of t h e fa c to rsi n v o l v e sl /x " . We re the other factor non-zero.i t w oul d i nvol ve
l/.xtfor some i>2. The product would then involve li-x"*tfor some i) 2. which is
a bs ur d.Henc e[ { )] : t t. 7 1 .
T he c oef f ic ie not f l /-r' * 1 i n th e ri g h t-h a n dsi de of (& ) i s * r' a{2aa* b. w herc b i s the
T h e r e f o r ee t : j ( + b i u * r ' ) . W e h a v e
c o e f f i c i e not f l , i , x ' * r j n 2 t l 1 | f + f t - r - t v $ l 2 .
shown t hat if 0 i s a p o l e o f r o f o rd e r )v ).-4 , then the componentof the parti al fracti on
ex pans ionof 0 a t 0 i s
l(
h
\
w h e r er t
tt.i1.t.
(co)Fina'y.we mustconsider
*nr,;;.":-*:
f ; ." lo,nu* pointor r As in
( c ' , ) ,[ 0 ] : 0 a n d - q . l u 2 : 0 .
C o n t r a r y t o t h e s i t u a t i o ni n ( c , ) . h o w e v e r .w e c a n n o t
c onc ludet hat r y1 0 . H e n c e th e c o m p o n e n to f the parti al fracti on expansi onof r at 0 i s
eit her 0 or 1/ . x .
S o l v i n g H o m o g e n e o u sD i f f e r e n t i a lE q u a t i o n s
l7
We collect together what we have proven so far. Let f be the set of poles of r. Then
/
_
d
osrcr \
l
0 : I ( s ( t ' ) [ 1r ] , r , . ' - ] + I . .
-\-(
ce|- \
/
_,+R.
-c{i
i= I -\
wh e re R e C[ . x ] . s ( c ' :) * o r -, a n d tv t].. rf(' r a re as i n the statementof the al gori thm.
Next we considerthe Laurent seriesabout oc. Supposethat
d..f/:R+ -+"'.
(.o,) If r hasorderv > 2 at cr, then
r : * + j - + " ' .
-
-\t'
'
-\t
and -a,
T h e R i c c a t tE
i q u a t i o ni m p l i e st h a t R : 0
(.cr) I f r has or der 2 a t q , th e n
b
'':
*
<
f
,'
Th e Ri c c at t iE quat ion i m p l i e sth a t R :0
.r:
so u.r.:0 or 1.
-F "
"
a n d -c ,, * a2.: b, hence
jtlr
a,:
la1:0.
1aa6
(r:) I n t he ot her c a s e sth
. e o rd e r o f r a t :c m u s t be even.by the necessary
condi ti ons
of se cti on2. F ollor v inga n a rg u m e n ts i mi l a r to th a t u sed i n (r' .) w e fi nd that
t l
: t r - - f( t
R:*[r,'],.
h
\
u-,).
wh e re - 2r ' is t he or de r o f l i tt :r,. (t i s th e l e a d i ng coefhci entof [n' 7],
c o e f f i c i e notf 1 - \ ' - r i n r - [ r r ' ] t , .
We n o w k now t hat th e p a rti a l fra c ti o n e x p a n s i o nof 0 has the form
-
/
-
(t
./sl()\
and h i s the
I
0 : I ( . t ( , ' ) L r ,*' 1.,' . - ) + . ' tv )' L[ \r r f , + I '\- 1,/
r'el
i?t .\-di'
\
Moreover.the coeffrcient
in the Laurentseriesexpansion
of 11.x
of 0 at cc is a1'). Thus
,l : a'f,",,-.I*1i(.)eN.
Let
\
(
"(') + r t x ) [ . r ] , .
c t : *r
]
.
*
L
1s(r')[.
'.\ l- 'L ' ,/ )
cer\
and
"
:
d
,4
(,x- d,).
Th e n 0 : ot * P ' 1' PA
. ga i n . u s i n g th e R i c c a tti E q u a ti on. w e obtai n
P " + 2 t o P 'l ( r D ' * r o 2 - r ) P : 0 .
Th e conv er s e.nam el yth a t i f P i s a s o l u ti o no f th i s equati on.then I sati sfi es
the R i ccatti
E q u a ti on. is a s im ple v e ri fi c a ti o n .It fo l l o w s th a t i f P i s a sol uti on of thi s equati on.then
4 : P e i o ' i sa s o l u t i o no f t h e D E ) " ' : r ! ' .
Th i s p r ov est hat t he a l g o ri th m fo r c a s eI i s c o rre ct.
J. J. Kovacic
llt
4. The Algorithm for Case 2
F ollowing t h e p a tte rn o f s e c ti o n3 . w e s hal l descri bethe al gori thm i n secti on4.l . gi ve
ex am plesin s e c ti o n4 .2 a n d th e p ro o f i n s e cti on4.3. The al gori thm and i ts proof assume
t hat c as c 1 is kn o w n to fa i l .
4 .1 . D ES (-R Ip rIoN
or rH E A LGOR TTH M
Just as for case l. we first cc-rllect
datii for each pole c of r and also for r . The form of
t he dat a is a s e t E ,. (o r rL , ) c o n s i s ti n go f from one to three i ntegers.N cxt w c consi der
f am iliesof ele me n tso f th e s e s e ts .p e rh a p sdi scardi ngsome and retai ni ng others. If no
f am iliesar e r e ta i n e d .c a s e2 c a n n o t h o l d . F or each fami l y retai nedw e searchfor a moni c
poly nom ialt ha t s a ti s fi e a
s c e rta i nl i n e a rd i fferenti alequati on.If no such pol ynomi alexi srs
f or any f am ily. th e n c a s e2 c a n n o t h o l d . If s u cha pol ynomi aldoesexi st.then a sol uti on tcr
t he DE has be e n fo u n d .
Let f be t he s e t o f p o l e so f r.
S t ep l. F or e a c h c e I- w e d e fi n eE ,.a s fo l l ow s.
( ( ' 1 ) t f c i s a p o l eo f r o f o r d e r 1 . t h e n [ , . : l r l ] .
( c ' : ) I f c i s a p o l e o f r o f c ' r r d c2r a n d i f b i s t h e c o e f i l c i e not f | ( . r r ' ) 2i n t h c p a r t i a l
fraction expansionc-rfr. then
E .: l 2 + 4 . I + . l h A : 0 . * 1 1 n r :
I f c ' i s a p o l eo f r o f o r d e rr ' > 2 . t h e n E , : l l l .
( r r ) I f r h a so r d e r > 2 a t : r - .t h e n E , : 1 0 . 2 . 4 1 .
( t , z \ I f r h a s o rd c r 2 a t :f,- a n d b i s the cocfhci entof -x-r i n the Laurcnt scri cs
ex pan s i o no f r a t r,. th e n
(c:)
, + 4 h l f t: 0 .
E, : [2+4.
+2) oZ.
( r : ) I f t h e o r d e ro f r a t r i s r ' < 2 . t h e nE , : [ r ' ) .
S t e p2 . W e c o n s i d e ra l l f a m i l i e s( e . ) . . r u l r r w i t h e . e E . . T h o s e f a m i l i e sa l l o f r l h o s e
c oor dinat esare e v e n ma v b e d i s c a rd e d L
. et
d:rz@,_Ir.).
I f d is a non-n e g a ti v ei n te g e r. th e fa m i l y shoul d be retai ned.otherw i sethe fami l l i s
dis c ar ded.I f n o fa mi l i e sre ma i n u n d e r c o n s i derati on.case2 cannot hol d.
Srep 3. For each family retained from Step 2. we form the rational function
0 : l I - "
crl-.\-('
Next we searchfor a monic polynomial P of degreed (as defined in Step 2) such that
P " ' + 3 0 P "+ ( 3 0 2+ 3 0 ' - 4 r ) P ' + ( 0 " + 3 0 0 '+ 0 3 - 4 r 0 - 2 r ' ) P : 0 .
If no such polynomial is found for any family retained from Step 2. then case 2 cannot
hold.
S uppos et ha t s u c h a p o l y n o m i a l i s fo u n d. Let S :0+ P ' l P
and l et rr-rbe a sol uti on of
t he equat ion
ro2+Qa+G6'++Ot-r):0.
Then ry: el'ois a solution of the DE f' : t'\'.
S o l v i n g H o m o g e n e o u sD i f f e r e n t i a lE q u a t i o n s
l9
4.2. nxnnapr-Es
Era m ple l. Cons ide rth e D E l " ' : rl , w h e re
l
'':.* -
3
16r:'
Th e n e c es s arcyondit io n so f s e c ti o n2 s h o w th a t c a s esI and 3 cannot hol d. (The order of r
a t r i s 1. ) T he only po l e o f r i s a t 0 a n d th e o rd e r therei s 2. The coeffi ci entof 1/,x2i n the
p a rti a l f r ac t ion ex pa n s i o n o f r i s b - - 3 t1 6 . Si nce 2.,1 14h-: 1 i s an i nteger.
E o : 1 1 , 2 . 3 1 . T h e o r d e ro f r a t c i s I a n d E . . : . [ l ] .
We have three familiesto consider.
€o:2.
e:1,
d:-ll2
€o:3.
e:1.
d--l
€o:1,
e:1,
Cl:0.
Only the third family need remain in consideration.For this family. 0 : I /2.xand we need
to fi n d a m onic poly no m i a l P , o f d e g re e0 . s u c h th a t
P " ' + 3 0 P "+ G 0 2 + 3 0 ' - 4 r ) P ' + ( 0 " + 3 0 0 '+ 0 3 - 1 r 0 - 2 r ' ) p : 0 .
E vi d e n tly P m us t be l . s o th e e x i s te n c eo l P i s a questi on of w hether or not
0 " + 3 0 0 ' + ( / 3 - 4 r 0 - 2 r ' i s z e r o .T h a t e x p r e s s i o nd o e s h a p p e n t o b e 0 " s o P : I i s t h e
de si re dp oly nom ial.
Next we form
l
$:olP'iP:
T h e e q u a t i o nf o r l r i s
Th e ro o t s ar e
(!):
t
4.r
It fo l l o w s t hat
t7 :
gi":
--
\ 't
gJ{l l'l'\l+1 . tl :
,,
I'rln2. r
i s a s o l u t i o no f t h e D E . ( A n d - \ t a e - 2 . ' i S a l s oa s o l u t i o n . )
Eru n t plt ' 1. I n t his ex a m p l ew e fi n i s h c o n s i d e ra ti onof B essel ' sE quati on
/4nt-r
\
-'
t ' . r r eC .
. t . ':' (
4.rr
/
th a t wa s s t ar t edir t s c cti o n3 .1 . In th a t s e c ti o nw c o b servedthat case3 cannot hol d and
th a t ca seI holds if and o n l y 'i f rr i s h a l f a n o d d i n te g e r.H ere w e treat casc2 and n-rakethe
a ssu mpt iont hat n is no t h a l f a n o d d i n te g e r.
Th e o nly 'pole of r is a t 0 a n d th e o rd e r th e re i s 2 . S i nce
l\
1 + h h : 2 1 1 + 4 ( 4 , n i2' 1 4 : 4 t r .
c i t h e rE o : l 2 l o r E o : 1 2 .l + L n . 2 - 4 n \ , . d e p c n d i n g
o n w h c t h e r4 n i s a n i n t e g e o
r r not.If
4rr i s n o t an int eger .t he n th c re i s o n l y o n e c a s eto c onsi dcr.
('r:0.
so:f.
tl:- 1.
20
J. J. Kovacic
Thus if 4rr is not an inteser.case2 cannot hold. If 4n is an inteser.there are three casesto
c ons ider .
eo:2.
d, : 0.
eo:2*4n,
d , .: 0 ,
tl:-l
d- -1-2n
€o:2-4n,
9' : 0,
d : -l +2n.
In order that d be a non-negativeinteger,it is necessarythat n be half an integer.Since
n is not half an odd integer,11must be half an even integer,that is n is an integer.But. for
such n, both eo and c. are even. Hence all familiesare discardedand case2 cannot hold.
I n t his ex amp l e .a n d i n Ex a m p l e2 o f s e c t i on3.2,w e have show n that B essel ' sE quati on
has a Liouv illi a n s o l u ti o n i f a n d o n l v i f n i s hal f an odd i nteser.
4.3. PROoF
For the proof of the algorithm for case2 we shall rely heavily on the differentialGalois
gr oup of t he D E . In c a s e2 . th i s g ro u p i s (c o nj ugateto) a subgroupof
, r : { ( ' ^ 9 ) l, e c . , r o,}{ f o ' : ) l, e c , + o }
| . \ 0. - ' l l
I
) |'\-, 0/l
M or eov er . we ma y a s s u meth a t c a s e I d o e s not hol d. so the di fferenti alGal oi s grrl up i s
not t r iangulisa b l eL. e t q , ( b e a fu n d a m e n ta system
l
of sol uti onsof the D E correspondi ng
t o t he s ubgr o u p o f D ' . F o r a n y d i ffe re n tialautomorphi sm o of C (.x)(rl .i ) over C (.x).
eitheroq:cry. o(:t'-r-( or ory:-(' t(. o-(:c4, for some .e C. t'#0. E'idc'tly
o( , lt ?) : nt ?, th e re fo re4 t _ ( e 0 (r). Mo re o ver, q( #C (.r) si ncecase l does not hol d.
We write
q'(t :(/ f] (-x-(')"'
fl t-"-d'1r'.
wher e f is t h e s e t o [ p o l e s o f , n n Al n . . * 0o" ." ,,
det er m inet hes ee x p o n e n ts .
Let
, z ( q . ? ) ' ; ( , / t ;: t: )
O:(tt-J'i(rlO:
f
e,..f' , arei ntegers.Our goal rs to
t"
+l
li
f
,
.\ - rl;
BecauseQ : rl'l,l+ (',,i. it follows that
( *)
,h "+ 3 ,h Q'* Q, : 4rrf + 2r'.
W e f ir s t det e rmi n ee . fo r c e I-. In o rd e r to si mpl i fythe notati on. w e i l ssumethat r:0.
( c , ) S upposeth a t 0 i s a p o l e o f r o f o rd e r l . The Laurent seri esexpansi onsof r and r/
at 0 are of the form
r : rr- I + ' ' ' Q * 0)
q : )ex-, +/'+ . . .
(eeZ. /'eC).
S ubs t it ut ingt h e s es e ri e si n to th e e q u a ti o n 1* ; and retai ni ngal l those terms that i nvol ve
.x I and .r 2. we obtain the following.
e . \ - 3+ ' " - 1 0 t " - t - ) e [ x - 2 + " ' + { e 3 . x - '+ } e t . 1 ' r - +2 ' ' '
:2q.ex-2+ " ' -a.\
2+ ' ' ' .
Solving Homogeneous Differential Equations
2l
T h e r e f o r ee - | e t * f e 3 : 0 ,
s o e : 0 , 2 , 4 . A l s o - | t f + i n ' . f : 2 u e - a . B e c a u s eq . * 0 ,
e + 0 . 2 . H e n c e ,e m u s t b e 4 .
(cr) Supposethat 0 is a pole of r of order 2 and that b is the coefficientof Ilx2 in the
Laurent seriesfor r. That is
r:bx-2+...,
6:)ex-l+....
-3
on the two sidesof equation (*), we obtain
Equating the coeffrcientsof x
e-ie' +*tt :2eb-4b.
O f c o u r s e t, h e l a t t e rt w o r o o t s m a y
T h e r o o t s o f t h i s e q u a t i o na r ee : 2 , e : 2 1 2 \ / 1 + 4 b .
be discardedin the casethat they are non-integral.
(c.) Finally we consider the possibility that 0 is a pole of r of order v > 2. Then
" ' a n d 6 : j e x - r + " ' . E q u a t i n gt h e c o e f f i c i e n tosf y - v - t i n 1 * ; w e o b t a i n
r:x-v+
0 : 2 u e - 2 d v , h e n c e€ : r .
In determiningthe exponentsf we may use the calculationof (c,) above if we replacea
b y 0 (s inc ed, m us t be a n o rd i n a ry p o i n t o f r). W e fi nd that./' ,:0,2, or 4. W e cannot
excludethe possibilitythat .fi:2, but we can, of course,excludethe possibility., : 0.
We have shown so far that
q'e':.U.(r- c)'P'.
w h e ree, eE , ( as def in e di n s e c ti o n4 .1 ) a n d Pe C [r].
S e t 0 : i r' ".- . f
".
-X-C
so 0:g+P'lP.
The next step in our proof is to determinethe degreed of P, which we do by examining
th e L a ur ent s er iesex p a n s i o no f @ a t o c a n d u s i n g e quati on(* ).
t + ",
€,-: \ er+2d.
6: Ie,r
( o o , ) S u p p o s et h a t t h e o r d e ro f r a t c c i s 2 . A s i n , . , ; - .
find that€.:0,2
or 4.
(.or) Supposethat the order of r at cc,is 2 and that b is the coefficientof x-2 in the
L a u re n t s er iesex pan s i o no f r a t o o . T h e n , a s i n ( cr), en:2, 2+ 2f,1 + 4b and e" o i s
integral.
(oo.) Supposethat the order of r at oo is v < 2. As in (cr), it follows that e- : v.
No te t hat at leas to n e o f th e e ,(c e f) i s o d d , s i n c ert(#D (' :r.).
Usi ng equat ion1r ' ;a n d th e e q u a ti o n@ :0 * P' l P , w e obtai n
P " ' + 3 0 P+"Q 0 2+ 3 e ' - 4 r ) P ' + ( 0+"3 e e ' + 0-34 r 0 - 2 r ' ) P: 0 .
This is a linear homogeneousdifferentialequation for P, so there is a polynomial solution
if and only if there is a monic polynomial which is a solution.
Now let ro be a solution of the equation
(**)
a 2- 0 a + I 0 ' + + O t - r : 0 .
To co m plet et he pr oo f w e n e e d to s h o w th a t 1 7 : e l ' i s a sol uti on of the D E y* :ry.
From 1x*) we obtain (by differentiation)
- +0"- 6d' + r'.
(2a- $)rt' : S'co
l1
J. J. Kovacic
(from (**)) so
T h e f a c t o r ( 2 o - q i ) c a n n o tb e z e r o . l n d e e d i.f ( h : 2 t , . r .t h e n t , t l t o 2 - r - 0
:
gl'
<
'
t
w
as
Thi
s
i
s
l
.
w
hi
ch
assumedtt-r
Bu
t
casc
D
E
.
tr-r
C
(.x).
is
o
f
th
e
a
s
ol
u
ti
o
n
\:
)4 ,, e
f ail. us ing ( **) a n d (x ) w e h a v e
2(2o-rf)('t *ro2 -r):
- 4i'-30,b'-4,'+4rcf +2r' :0.
T h u s u ) ' + r o 2 : r ' S of i : g ! ' ' i s A s o l u t i o no f t h e D E .
T his c om ple te sth e p ro o f th a t th e a l g o ri th m for case2 i s correct.
5. The Algorithm for Case3
Following the pattern establishedin the previous two sections. we describe the
algor it hm in s e c ti o n 5 .1 a n d g i v e e x a mp l e si n secti on -5.1.The proof of the i tl gori thrn
r equir esa k no w l e d g e o f th e fi n i te s u b g ro upsof S L(2) and thei r i nvari ants.w hi ch i s
pr ov ided in s e c ti o n5 .3 . T h e p ro o f o f th e a l g ori thm i s presentedi n secti on5.4.
that casesI and I urc
I n c as e3. t h e D E h a s o n l y a l g e b ra i cs o l u ti onsi i nd n,s assLl nrc
k nown t o f ail. ( l t i s p o s s i b l efo r th e D E to h a ve onl y al gebrai csol uti onsand for cascsl or
2 t o a p p l y . F o r e x a m p l ec. a s eI g i v e st h e s o l u t i o r rl i : . x t * t o t h e D E . r " ' : 1 3 l 6 i r ) t ' .
t hen r educ t iono f o rd e r g i v e s( :.x -r * rts a s e condsol uti on.)
5 .1 . D F ,s c R rp rroN
oF TH F.A L(;oR rrH M
Let 4 be a s o l u ti o no f th e D E .)" :' rJ ' a n d Set(D : tl ' l tl .Then" as w e shal l shgu i l l sccti t)l l
5 . 4 , a . its a l g e b r a i co v e r C ( . x )o f d e g r e e4 , 6 o r 1 2 .I t i s t h e m i n i m a l p o l y n o m i a lf o r l r t h u t
we s hall det er mi n e We
a re u n a b l eto d e te rm i nethe mi ni mal cquati onfor 11(w ' hi chri oul d
.
1
be of degr ee24 . 4 8 o r 2 0 ).
T her e ar e t w o p o s s i b l em e th o d sfo r fi n d i ng the mi ni mal cquati on for r' .,.\\' c coul tl l l nd
p
a o l y n o m i a lo f d e g r e el 2 a n d t h e n f a c t o r i t . W e s h a l l p r o r c t h a t i f r , , il s r . r n rs o l u t i o no f
t h e 1 2 t h d e g r e ep o l y n o m i a lf o u n d b y o u r m e t h o d .t h e n r ; : , , r , i s a s o l u t i t - r tn- r ft h r ' [ ) F .
henc e any one o f th e i rre d u c i b l efa c to rsmav be used. Thi s rs thc most ci i rcctnrcth()d:
howev er .t he fa c to ri s n ti o nc a n b e a fo rmi d abl e probl em. eren w i th the assrstuncc
of a
c o m p u t e r .W e i l l u s t r a t et h i s b 1 'c' x a m p l c .i n s e c t i o r5t . 1 .T h e a l t c r n u t i r ci s t o l i r s t a t t e m p t
t o f ind a 4t h d e g re ee q u a ti o nfo r r,r.th e n a 6 th degreecquuti on rrnd fi nal l 1' a l 2th cl cqrcc
equat ion. T he a d v a n ta g ei s th a t i f a n e q uati on i s found. then i t i s guaranteedto bc
ir r educ ible.
I n o u r d e s c r i p t i o no f t h e a l g o r i t h m .r v es h a l lc o m b i n et h e v a r i o u sp o s s i b i l i t i c sd.c n t r t i n g
by n t he degr e eo f th e e q u a ti o nb e i n gs o u g h t.A s before"w e denoteby f the set of pol esof
r . Rec allt hat , b y th e n e c e s s a ry
c o n d i ti o n so f secti on2. r cannot have a pol e of order > l .
Srep l. For each c e I- u I r- | we define a set E. of integersas follows.
( ( ' 1 ) l f t ' i s i r p o l eo f l o f o r d c r l . t h c n E . : l l l l .
( , ' r ) l f c ' i s a p o l e o f r o f o r d e r 2 a n d i f z i s t h e c o e f l l c i e not f I ( \ - ( ' ) r
fraction expansionof r. then
( tzr
rr')
E .: j 6 +
l + 4 xA : 0 .* 1 .+ 1 . . . .! .. , l a z .
\
tt
,
(oc,)If the Laurentseriesfor r at v- is
r:
'
l ' , \ - 2+ . .
/
( ; ' eC . p o s s i b l 0y ) "
in tlrc partial
S o l v i n g H o m o g e n e o u sD i f f e r e n t i a lE q u a t i o n s
then
(
E,:
t2k
lt o * r ? \
l + 4 ^ l, f' t : 0 . +
| +1
'
.
:
-
.
'
.
'
r
+
l
2.1
rrl
',)l n z .
Step2. We consider
all familie s ( e . ) . . r u t r l s u c h t h a t e , e E , . F o r eachsuchfamily.define
.
n
d : 7 ' ^ , ( c , I-. . ) .
lL
c.el'
If ri i s a non- negat iv ei n te g e r.th e fa m i l y i s re ta i n e d.otherw i sethe fami l y i s di scarded.If
no families are retained. then rr.rcannot satisfy a polynomial equation of degree n with
coefficientsin C(.x).
Sterp3. For each family retainedfrom step 2. form the rational function
{): a Y -€'
l2;?r.{-c
Al so d ef ine
S : f] (x-c).
N e xt se ar c hf or a m on i c p o l y n o m i a l Pe C It] o f d e g reed (as defi nedi n step 2) such that
w h e n w e d c f i n e p o l y n o m i a l sP n . P n r . . . . , P - , r e c u r s i v e l yb y t h e f o r m u l a sb e l o w . t h e n
P r :0 (identically).
Pu:-P
Pi I : -sP;+((rr-i)s'-S())PI-(n-i)(i+ l)S2rp,*,
(i:n.n-1. ..0)
Th i s may bc c onv en i e n tl vd o n e b 1 ' u s i n g u n d e te rmi nedcoefl i ci entsfor P . If no such
po l yn o m ial P is f oun d l o r a n 1 ' fa m i l l ' re ta i n e d from step 2. then o cannot sati sfy'a
po l yn o m ialequat iono f d c g re err w i th c o e fi c i e n tsi n C (.x).
Assum et her ta f am i l y a n d i ts a s s o c i a te dp o l y n o m i al P has been found. Let r,r be a
s o l u t i o no f t h c c q u a t i o n
n
S ' Pl ,
T
r,)': 0.
,--, ( n - i ) !
Thenry: ei'"is a solutionof the DE.
5 .2 . rx a u p l n s
Exu n t ple l. O ur f ir s t e x a m p l e i l l u s tra te sth e a l te rnati vetechni quementi oned at the
be g i n n i ngof t he las t s e c ti o n .n a m e l yto b y p a s sth e s earchfor equati onsof degrees4 and 6
[o r ro a n d pr oc eeddir e c tl y to th e s e a rc hfo r a n e q u ati onof degree12.
We co ns idert hc hy p e rg e o m e triec q u a ti o n r.": r._where
1,
r:--
3
2
3
l6F-9a\Jt' * tur("-D'
Th e n ec es s arcyonditi o n so f s e c ti o n2 s h o w th a t a l l four casesare possi bl e.
Ap p l ving t he algor it h m fo r c a s e l . w e fi n d th a t
:',:
;:: Ii ::,
ti :2 3.
r, : 1,,'3.
andd: rl -ri -r,1 can neverbe a non-negative
integer.case I fails.
24
J. J. Kovacic
Applyingthe algorithmfor case2, we find that
E o : { 2 ,3 . 1 }
E, : {2\;
E n': {2\'
c a n n e v e rb e a n o n -n e gati vei nteger.C ase2 fai l s.
and d: €o, - €o-€ t
for case3, searchingfor an equation of degree 12 for (o, thus
the
algorithm
We apply
i
n
a
l
g
o
r
i
t
h
m
.
n:12
the
A t c : 0 , c : - 3 1 1 6a n d
+ + o : I l 2 ( o r - l l 2 ) . H e n c eE o : { 3 , 4 , 5 , 6 , 7 . 8 , 9 } . A t
c : 1 , d . - - 2 l 9 a n d . , / t + 4" /ul : 1 1 3 S. o E , : { 4 , 5 , 6 . 7 , S } .A t o c . i ' : - 2 1 9a n d E . . : { 4 .
5, 6, 7, 8).
for
F ollowing t h e i n s tru c ti o n so f s te p 2 , w e now form the expressi on6l :€.:-eo-e,
w
hi
ch
i
s
E
r.
those
fami
l
i
es
for
d
a
negati
ve
E
-,
E
s
,
e
,
e
We
d
i
scard
e
o
e
€
s
e
c
hoic
e
of
ev er y
possibilities
remain.
four
integer.Only
et:4,
d:0
€o:3,
€t:4,
gt: 5.
d: I
d :0
€o:4,
€t:4.
d:0.
€,r:7,
€ , (: 8 .
d' : 8,
eo:3.
9 o: 3 ,
er:8.
W e now c o n s i d e r th e fi rs t p o s s i b i l i ty ,fo l l ow i ng step 3. W e set (/:3,' .t+ 4 (.r- l ).
S : x 2- - x . and s e a rc hfo r a m o n i c p o l y n o mi al P of degree1 that sati sfi esthe condi ti ons
g i v e ni n s t e p3 . O f c o u r s e .P : l .
T he c om puta ti o n sa re fa r to o c o mp l i c a t edto be accuratel ydone by hand: how ' crer.
t hey ar e eas i l y p ro g ra m m e d i n to a c o mputer. S i nce P i i s al w ays a pol y' nomral
( l: 12, . . . , - 1 ) w h o s ed e g re ei s e a s i l yp re d i cted(i n thi s exampl edeg P ' : 12-i ) arravsof
c oef f ic ient sm a y b e ma n i p u l a te dto c a rry through the computati ons.In order to avoi d
u s i n g 3 3 d i g i t i n t e g e r a r i t h m e t i c .T h c r e s u l t s
roundoff error. we computed l)tz-t,
follow.
Ptz:-7
Ptt:lx-3
P r o : 0 l l 2 ) ( - 5 3 6 1 2* 4 5 9 . r- 9 9 )
- 1485)
ps : ( 3 ! / ( 3 l 2 ' z D (l 8 5 4 4 ,x-2
3 3 7 9 9 1 2+ 1 0 2 60.x
ps : ( 4ll( 16 .1 2 \)(- 1 2 1 4 8 8 1+12 1 7 9 7 2 x 3- 140879,x2
+ 40110.x-4455)
- 138975.x2
- 2673)
p. t : 6 |I Q . l 2 t)X 17 4 0 8 0 -x-s 3 7l l 4 8 ra+ 3 2 0305.x3
+ 30375.x
-227
p o : 6l I 12 t)(- 8 2 5 75 3 6 x u+ 2 1 1 4 5 3
1 6 ,x s
57500,x4
+ I 3 168377.x3
- 4 3 1 8 0 8 3 . 1+27 6 0 3 4 7 x- 5 6 1 3 3 )
ps : Q tlQ t2\)(7929856.x7-2367398416+ 30564708
xs -2210728ix4
+ 377622x 24051)
+ 9668646.x3 2555280.x2
- 135613476816
p+ : ( 8 ! / ( 16 1 2 6 ))(-2 6 4 2 11 5 2 .X+8 9 0 0 9 8 4832x1
+ I 177673400.rs
- 50398362
- 360855)
- 64408232J
xa + 221124972,x3
x2 + 6429780.x
p:: (9!/(3.12t)X114483046.xe-6688997376.x8+11509039440x1-11656902184.x6
- 3316695033.14
- I 9 I 681802,x2
+ 1000183626.x3
+ 1 654170465.xs
+ 2 1 5 5 2 8 8 5 . x1- 0 8 2 5 6 5 )
Solving Homogeneous Differential Equations
25
p z : 00 !lQ. 12o))(
- 228110137
- 1879943809018
6xto+ g7I 3634849,xe
6t 6x1 166931
7476gx6 + 88404I 3683-xs
+ 21766009
- 3217
- 14113941012
319535,x4
10,x3
+ 8387801
5x
+ I 427017
- 649s3e)
- 602r t8t2 0 1 8 8 1 6 , x 1t 03+5 0 7 5 3 1 1 1-61x1e5 9 9 9 2 2 3 g 4 . x 8
Pt: (11!/12t0X134217128
52x1 95 464270
+ 154268489
17,16+ 425263867
2xs- | 3 628| 6657xa
+301684656.x34657653912
+ 4251528,x111t47)
p o : ( 1 2 !I t 2 t , ) ( - g 5 8 g g 3549 2 x 1+24 3 g 3 g g 6 5 4 0 1g-1 1 0 3 61g7 2 0 3 2 0 x t o
- 149I 70390976x8
+ 150145631824.xn
+ 10490nl0g64x1
- 54596424249x6
- 21032969490.xs
- 5948563
455xa
I6.x3- 1652791
+ 12036548
51-x2
+ 13gn466x - 531441\
P r : 0
Therefore
4 : e)"'is a solutionof the DE. wherera is a solutionof the equation
]l (.vr- r)iP, (''r' o'
, L - t 1 2- i ) l
Professors
Caviness
and Saunders
of Rensselaer
Polytechnic
Institutekindly offeredto
attempta factorisationof this polynomialfor ro. They usedthe exceedingly
powerful
systemfor algebraicmanipulationcalledMACSvMA
at MIT. The programtook lessthan 5
minutesto write but took 3 minutesof CPU time to execute.
The resultis that the
polynomialaboveis the cubeof the followingpolynomial.
-(l 3)(.r2
-.x)3(7r- 3)r,r3
--x)2(48
(.rt-.x)aroa
x2 - 41-x
* (1,'24X,x2
+ 9)rr,r2
- (l 432X.*:
-.x)(320.xr
- 409.x2
-2i)o
* 180.x
+ ( I 2 0 7 3 6 ) ( 2 0 4 8- .3x 4 8 4 . x+32 3 1 3 . x- 27 0 2 r +8 l )
Example
2. In thisexamplewe consider
the DE j"' : ry, where
r- -
5 x+ 2 7
36(x-lf
The necessary
conditionsof section2 showthat all four casesare possible.
Note that the partialfractionexpansionof r hasthe form
r--eG+lF+. -rG_ly+ .
and the Laurant seriesfor r about x is
'r : -
5
- L . . .
3612
Applying the algorithm for case I we find that
a!':213'
ui :2 3,
aI : 516,
u-':113
ar : ll3
a- : 116.
For no choiceof signsis d : at-- d! ,,- af a non-negative
integer,thus case1 cannot
hold.
J. J. Kovactc
26
: { 2 ) . a n d c a s e2 d o e s
A p p l y i n g t h e a l g o r i t h mf o r c a s e2 w e f i n d t h a t E - r : E r : E ,
not hold.
W c now app l y th e a l g o ri th mfo r c a s e3 . a t tempti ngto fi nd an equati onof degrec4 over
C(-x)that is satisfiedby ro.
From step I we have that
{ 4 . 5 , 6 . 7 . 8 } a n d E , , : i 2 . 4 , 6 . 8 1. 0 1 .
E , : . t , 4 , 4 . 6 , 7 . 8 1E
. t:
T h e r e a r e f o u r f a m i l i e sw i t h t h e p r o p e r t y t h a t r / : l ( e { - e - r - e r )
int eger ,nam el y
d:0,
€t:4.
€,t:4,
c.-.:8.
€r:10.
€-t:4.
at:6-
r/:0.
8r:10.
c-r:5.
cr:5.
r/:0.
€ t:6,
€t:4,
d:0.
cr:10.
is a non-ltsgrttir,'c
T he f ir s t poss i b i l i tyg i v e s
r{r
4 \
t/ 4
*
.
r
l , l : t ( \ r- t ) '
": , (r+ r
S et t ingS : . x 2 - l . w e h a v e S 0 : 3 , x .S 2 r : - * ( 5 , t t + 2 1 ) . W e t h e n h a v e
P+:-1
P: : (8/3)-x
P2---(l/3)(15,x2+l)
P r : (l /9 X 5 0 ,x 3+ 14.t)
P o : - ( 1 / 5 4 X l 2 5 x a+ 1 3 4 . \ 2 - 3 )
P-r:0'
Let a-rbe a s ol u ti o n o f th e e q u a ti o n
+ +(25.\3 * 7.r)Sru- rllqr,(I l5 ra + I l-lrr - -l)
St,+ : $-xSrrt-*( I 5,x2+ I )Str-r2
I f we m ak e t he s u b s ti tu ti o n6 5 o : i * 4 ,x .the equati onsi mpl i fi esttr
: a : 6 ( , x 2 l ) : t - 8 . r ( . x-2 l ) ; + 3 ( . \ :- l ) r .
T hen
q:
e i , ": ( . x 2_ l ) ' . . * p ( l ( , ( r t - l ) ) d . r )
is a s olut ion o f th e D E .
5 .3 . F rN rrEs u B GR ouPoF
S S L(2)
I n t his s ec t i o nw e d e te rm i n eth e fi n i te s u bgroupsof S L(2). up to conj ugati on"and thei r
inv ar iant s .T h i s w o rk i s c l a s s i c a lb. e i n g fo u nd i n the w ork of K l ei n. Jordan and othcrs.
For the sake of completenesswe sketch the results here in the form needed in the
subsequentsection.
S o l v i n g H o m o g e n e o u sD i f f e r e n t r a lE q u a t i o n s
21
THr-.oRnu
1. Let G he a./rnitesubgroupol' SL(2). Then either
(t) C is coniuguteto a suhgroupo.f the group
(
0 l\
l'D.
\- I 0/
D,: D u l
w'hereD is the diagonulgrlup. 0r
(i l t he or der o/ ' G i s 2 4 (th e " te tra h e d ra l " c a s e ),or
\iii) the ortler o/ G is 48 (the "octahedral" case),or
(it) the ortler of'G is 120 (the "it'osahedral" case).
G containsthe sc'alarmatrix - l.
ln the ltrstthree ('o.se.s
The geometricnameswere usedby Klein; however.our proof will be entirely algebraic.
Pro ol' .W e as s um et h a t G i s n o t c o n j u g a teto a subgroup of D ' . LeI II be the set of
s c a l a rm a t r i c e si n G . t h u s H : l l ]
o r l l . - l ] ' . s o t h e o r d e ro f H i s I o r 2 . F o r a n y
, x eG - H ( i . e . . r eG a n d r d H ) w e d e n o t eb y Z ( x ) t h e c e n t r a l i s eor f . xi n G a n d b y l V ( r ) t h e
n o rma lis erof Z ( . r ) in G .
(The Jordan form of a nonL e t,x e G - H. S inc e-r i s o f fi n i te o rd e r. .x i s d i a g o nal i sabl e.
d i a g o n a l i s a b lm
e a t r i x i n S L ( 2 ) m u s t b e -I (\: 0
lt)1. , t t n . . t h e c e n t r a l i s c irn S L ( 2 ) o f a
d i a g o n a ln o n - s c a l am
r a t r i x i s D ( b y d i r e c tc o m p u t a t i o n )Z ( r ) m u s t b e t h e i n t e r s e c t i o o
nf
G anda conjugate
o f D . H e n c eZ ( x ) : Z ( v ) i f a n d o n l y i f v e / , ( r ) . U s i n gt h i s f a c ta n d t h e
f a c t t h a t Z ( g . r g ' ) : r t Z ( . r ) g t w e m a y ' c o n c l u d et h a t ( f o r a r b i t r a r y . y l. ' . ( / .g ' e G ) e i t h e r
t
u Z ( x ) g a r t ' Z ( t ' ) g ' t- : H o r ( t Z ( x ) t l ' : r l Z ( ) ' ) t t ' I
a n d i n t h e l a t t e rc a s et ' € q ' l q Z ( . r ) t t - t r r ' I. n a d d i t i o nt t Z ( r ) f l 1 : q ' Z ( r ) g ' t i f a n d o n l y i f
g ' t r l e l t ' ( . x )T
. h e r c f o r e\ \ e m A \ u r i t e G a s a d i s l o i n tu n i o n
t :,y ' , ' rJ (tt/tx ,l ,t r
1 1 )' - H
t
(di sj oi nt).
w h e r et h e i n n c r u n i o n i s t a k e no v e r a l l c o s e t sg N ( . r i )i n G ' N ( . x , )..si s s o m en a t u r a ln u m b e r
a n d . r 1 . . . . r .e G - H .
Th e gr oup N( . r i) is e a s y to d e s c ri b es i n c e x , i s d i agonal i sabl e.
Fi rst notc that the onl y
ma tri cesin S L( 2) t hat c o n j u g a tea d i a g o n a ln o n -s cr.rl ar
n" ratri xi nto a di agonalmatri x are
th e ma t r ic esin D- t by d i re c t c rrmp u ta ti o n ).It fo l l o w s that i V (.r,)i s the i ntersecti onof G
a n d i r conjugat eof D* . i n p a rti c u l a rth e i n d e x o f Z (x,) i n N (.x,).p{(,r,) : Z(x,)f. i s ei ther 1
or 2.
L e t M : o r d ( G H ) a n d e ' ,: o r d ( Z ( x , ) H ) . T h e r e p r e s e n t a t i oonf G a s a d i s j o i n tu n i o n
g i ve sthe f ollowing f orm u l a s .
r V ' o r d H : \[ G : N ( . r , ) ] ( c , ' o rH
d-ord H)+ord H.
L,
,
1
or
!)I
,\T: T
lJ
;
or
(#)
I
\1
l
t
,
L
t
t l , r f . , )2, 6 j J ; , ( c i -
l)+ t'
(j, - r) +r
i'ri,.ilz(,,)r
J. J. Kovacic
28
Certainly s + 0 since G + H. If s : 1, then
l l M > 1 / ( [ 1 V (' )- :xZ ( x ' ) ] e ' ) : l l o r d ( N ( x ' ) l H ) , s o G : N ( r , ) .
This contradictsthe fact that G is not conjugateto a subgroup of D'.
S i n c ee , ) - 2 ( i : 1
s) we have
l
"
SO
:
l
l- i
u <
, I , ; r t . *, 1 : z ( x , l f
|
+
' !r U f(." ,): z(x,)l
B ec aus e
p/(r,):Z(x,)f:l
-\ ' 4't
or2,
there are only three solutions of this inequality.
s:2,
s : 2,
s : 3,
[ , V ( x , ) : Z ( x , ) ]: l ,
[l{(xr):Z(xr)): )
U V (,x,): Z (x ,)l : [N(,xr): Z(rr)l : 2,
U { ( . x , ) :Z ( x r ) ) : [ N ( , x 2 ): Z ( r r ) f : p { ( . x 3 ): Z ( r . \ l : ) .
F or all s ol u ti o n s [N (.x r): Z (r-,)f : ). T h us G contai ns a conj ugate of a matri x i n
/ o" _, r \
Dt - D, i. e. t he c o n j u g a teo f a ma tri x o f th e form (
^ ) The squareof such a mi i tri x
\-c-'0,/
i s - 1 . H e n c eo r d H : 2 .
T h e f i r s t s o l u t i o ng i v e sl l M : l i e t * 1 , ( 2 e . ) - 1 , 2 . s o c 1: 3 . a z : 2 a n d i v I : 1 2 . s o
or d G : 24. ( Th e p o i n t b e i n g th a t M > 2 e r,si nce G i s not conj ugateto a subgroupof D .
and t her ef or ee , > 3 .)
T h e s e c o n ds o l u t i o ng i v e sl l M : l l ( 2 e r ) +
I ( 2 e , ) .w h i c h i s i m p o s s i b l es i n c eI I > 2 e , .
T h e t h i r d s o l u t i o ns i v e s
2
l
l
1
M
er'
(':
'
_ lI '
a'-,
A s s u m i n gt h a t e , ( e z ( e 3 w e f i n d t h a t e t 1 3 s o e 1: 2
and
2- I * 1 - 1
M e 2 e 3 2 '
Also e, : 3 sinceM > 2er. The solutions are
€t:2,
€2:3,
€3:3,
M:12.
€3:4,
M:24,
€3: 5, M :60,
ordG:24,
O r dG : 4 8 .
ordG : I20.
This proves the theorem.
In the following sequence of theorems we shall explicitly determine the three
"geometric" groups. To that end we need the following lemma.
L e n u a . L e t G b e a . f i n i t e s u b g r o u p oSfL ( 2 , C ) t h a t i s n o t c o n j u g a t e t o a s u b g r o u p o / ' D r . l , e t
H: 11, - I l. T h e n G IH h a s n o n o rme lc y c l i csubgroup.
Pnoor. If ,xH is a generatorof a normal cyclic subgroup of GIH then the group generated
by r and - r is diagonalisable.Since this group would be normal in G. G would be
c onjugat et o a s u b g ro u po f D t.
Solving Homogeneous Differential Equations
29
THEoREna
2. Let G be a subgroupof SL(2, C) o.forder 24 that is not conjugateto a subgroup
o./ D'. L e t H: { 1, - l} . T h e n G IH i s i s o mo rp h i cto A o, the al ternati nggroup on 4l etters.
Moreouer, G is conjugateto the group generatedby the matrices
(5,e,)
l)
w 'h e re( i s a pr im it iue6th ro o t o f I a n d 3 6 :2 q - l .
Pnrxlp. Since ord G/H is 12, and because of the previous lemma, GIH has 4 Sylow
3-groups, and GIH acts by conjugation on the set of these Sylow 3-groups.This action
ind u ce sa hom om or phi s m G IH --Sn (th e s y mme tri cgroup on 4l etters). The subgroupof
the image consistingof those permutations that leave a particular Sylow 3-group fixed
must have index 4 sinceGIH actstransitively.Thereforethe order of the image is divisible
b y 4 .It f ollows t hat t he o rd e r o f th e k e rn e l i s 1 ,2 o r 3. B y the previ ousl emma, the order
'of the kernel
must be 1, so GIH is isomorphic to a subgroup of S". Now consider the
c o m p o s i t eh o m o m o r p h i s m G I H - , S + - { 1 , - 1 } .
with the last arrow being given by
o--signum (o).By the previous lemma, GIH cannot have a normal subgroup of order 6
(since a subgroup of order 6 contains a unique subgroup of order 3 which would be
normal in GIH). Therefore the composite homomorphism has trivial image and GIH is
iso mo rp hict o A o.
. e may
L e t r G - - + A ob e a h o m o m o r p h i s mw i t h k e r n e lH . L e t A e t 1 ( 1 2 3 )W
coniugate
G so that .4 is a diagonal matrix. Thus
' : (; .9')
S i n c e r . 4 3 : ( l ) , A 3 e H . H o w e v e r ,r A * ( r ) a n d r A 2 + 0 ) , t h u s A # H a n d A r + H .
R e p l a ci ngA by - A , t f n e c e s s a ryw. e ma y a s s u meth at i i s a pri mi ti ve 6th root of 1.
L e t B € r - 1 ( 1 2 ) ( 3 4 ) .S i n c e/ A B ) * { B A ) .
B c a n n o t b e a d i a g o n a lm a t r i x . i . e . n o t b o t h
Brt and 8., are zero.In fact neither is zero becauseif one were zero and the other nonzero, then B would have infinite order.
Wemayconjugare
G by(:
withoutaffecting
,4.l f w e c h o o s ec : B z , a n d d : 2 8 1 2 ,
" \0 i)
dl
then B tras the form
u: (!r !,)
N o w r B 2 : ( l ) s o 8 2 e H . A d i r e c tc o m p u t a t i o ns h o w st h a t
x:6.
N e x t w e o b s e r v et h a t ( B A 2 ) : t ( A B ) 2 s o B A 2 : * ( A B ) 2 . W e p e r f o r mt h e c o m p u t a t i o n
a n d d i s c o v e rt h a t O G ' - 1 ) : * ( a ( u s i n g t h e f a c r t h a t g + 0 ) . R e p l a c i n g B b i _ 8 , i f
ne ce ssar ywe
,
m ay as su me th a t O Gt - I ) : (o . h e nce 36 :2( - | (usi ng the rel ati on
(t: (- l).
Next we use the fact that detB:l
to obtain the formula d2+202:-1.
or
3,1,:+(2(-l).
If necessarw
y .e c o n j u g a t eG b y ( :
:2,:-t:3q5.
- t ?/ ) , " t h a t 3 r l
\u
pro ve sthe t heor em .
The group of this theorem is called the tetrahedralgroup.
This
J. J. Kovacic
to tt suhgroupof
THronEn 3. Let G he a subgroupof'SL(2) of'r>rder48 that is nt'ttcotr.iugcrte
T h e n G t H i s i s o m o r p h i cf o' S r . t h e s r m n t e t r i tg' r o u p o t l 4 l e t t e r s .
Dt. Let H:\1.-l).
Moreoter. G is cortjuguteto the groLtpgenertttedhv the matrices
(; .e') o( l - i )
t+1).
w'here; is a primitire 8th rortt o.l I and 2(b: -4-(
P noor . S inc e o rd G ,' H :2 4 . a n d b e c a u s eof the previ ous l emma. G,' H has 4 S yl ow
3- gr oups .T he a c ti o n o f Gi H o n th e s e t o f S yl ow 3-groups(vi a conj ugati on)i nduccsa
hom om or phi s m G rH -- So . T h e i m a g e c ontai ns a subgroup of i ndex 4. namel y the
s ubgr oup of p e rmu ta ti o n s l e a v i n g a p a rt i cul ar S yl ow 3-group fi xed. si nce G H acts
t r ans it iv elyon th e s e t o f S y l o w 3 -g ro u p s .H e ncethe order of the i magei s di vi si bl eby 4. so
t he or der of t h e k e rn e l i s l . 2 . 3 o r 6 . Were the order of the kernel 6. then the kernel
would c ont a i n a u n i q u e s u b g ro u p o f o r der 3 w hi ch w oul d be normal i n G. Thi s
c ont r adic t st h e l e mma . In d e e d .th e l e m m a impl i esthat ord ker : 1. so G,i H i s i somorphi c
to Sr.
L e t r : G - S + b e a h o m o m o r p h i s mw i t h k c r n e l H a n d l e t A e r l ( t : l + ) . W e m a y
conjugateG so that ,1 is a diagonal matrix
,': (; .9')
S i n c er A a : ( l ) . i o : * l . H c - r w erv. ew e r e i o : l . t h e n : * I a n d 4 2 eI I . B u t r 4 ) + ( 1 ) .
Henc e i is a p ri m i ti v e 8 th ro o t o f l .
. c a n n o tb c a d i a g o n a m
l a t r i r . t h u sn o t b o t h B , ,
L e t B e r t ( t 2 ) . S i n cre( A B l * r ( 8 . , 1 )B
and Brl Are zero. In fact. neither i s z e ro si nceB has fi ni te ordc-r.W c mav con.i ugateG.
n
w i t h o u t d i s t u r b i n el . u r / l
\ rv h e rct' ) : B t, and t l ) : B r , . T l i e n B h a s t h c f o r m
\t) I I
B : ( b -v/ )t
\u
L J s i n gt h e f a c t r B 2 : ( l ) . i . e . 8 2 e H . w e o b t a i n e a s i l l 't h a t 7 : r l t .
B e c a u s te( B A 3 ) : r ( A B ) 2 . 8 . 4 - r : * ( A i l l . M a k i n g t h i s c o m p u t u t i o na. n d u s i n gt h c f a c t
B. if
t h a t , l t+ 0 . w e f i n d t h a t q 5 ( ; t - l ) : * ; . o r 2 r , b : + i ( i t + l ) . R c p l a c i n gB b y
nec es s ar ywe
. ma y a s s u m eth a l 2 tf : i (i r f l ). Then 2Ot - - l . N on' w c usc the fact that
I : d e t B : - 4 r ' - r l r t t o c o n c l u d et h a t 2 t t : _ _ l . C o n i u g a t e( i . r 1 ' n c c e s s a rn\ i' '. ( ' 1 _ :t )/
\0
s o t h a t r L: 6 .
B ec aus er A .rB g e n e ra teSo a n d th e g roup generatedby ,4. B contai ns H . w e can
conclude that A. B generateG. This proves the theorem.
T he gr oup o f th i s th e o re mi s c a l l e dth e o ctahedralgroup.
TnnoRnnt4. Let G he u suhgroupol'SL(2) o./'order120 that is rror t'onjuguteto u suhgroup
o / ' D r . L e t H : [ 1 . - l ] . T h e n G t H i s i s o r n o r p h irco A r . t h e a l t e r n a t i n g r o u p o n 5 l e t t e r s .
to the group generutedbl'the ntatrit'es
G is cotr.jugttte
Nloreot't,r'.
/ ,"(;
.e,)ff -l)
w'herei is a primitit'el}th root of' l, 5 6 : l i t - i t + 4 ( - 2 . u n d5 r !: i 3 + 3 i 2 - 2 i + t .
S o l v r n gH o m o g e n e o u sD i f f e r e n t i a lE q u a t i o n s
rl
PR<-rcrThe
p. pr oof t hat GIH i s i s o mo rp h i cto A . m a y be found i n B urnsi de(1955. 1 2 7 .
p. l6t 2)
L e t r ' . G - - A s b e a h o m o m o r p h i s mw i t h k e r n e l H a n d l e t A e t - r ( t 2 3 + S ) . W e m a y
co n j u g a t eC s o t hat I is a d i a g o n a lma tri x
i:
, : (;
o\
S i n c er A 5 : ( l ) . i s : * l . R e p l a c i n gr { w r t h - / . i f n e c e s s a r yw.e m a y
.,1,).
a s s u m et h a t i s - - l . E v i d c n t l ) ' ( i s a p r i m i t i v c l O t h r o o t o f L
L e t B e r 1 ( t Z ) ( : + ) .A s i n t h e p r o o f o f T h e o r e m3 , w e m a y a s s u m ct h a t B h a s t h e form
u :(\ / ! - ,1h) )
B c c a u s e4 A 1 8 ) : 4 B A ) 2 . A a B : + ( B A ) 2 . M a k i n g t h i s c o m p u t a t i o n w e f i n d t h a t
r b ( l + ( t ) : + ( o . o r 5 $ : * ( 3 ( 3 - i t + 4 . ( - 2 ) .R e p l a c i n g
B by -8. if necessarw
y .e m a y
assu meth at t he plus s ig n o b ta i n s .N o w w e u s e th e fact that 1 : det B to concl udethat
5 { : + ( i r + 3 i : - 2 ; + l ) . C o n j u g aGr eb y f 1 0 \ rl necessary.so that the pl us si gn
_ |
\0
)
obtains.
N o te that t A . t B ge n e ra teAs . (T h i s g ro u p g e n e ratedby rA and rB contai ns an
el e me n to f or der 5. an el e me n to f o rd e r 2 a n d a n e l e m entof order 3. Thus the order of thi s
gro u p i s div is ible by 30 . Si n c e A . i s s i m p l e . th i s g r oup must be A -r.) A l so the group
generatedby A. B contains H. ThereforeA. B generateG. This proves the theorem.
Th e g r oup des c r ibedi n th i s th e o re mi s c a l l e dth e i cosahedralgroup"
Fo r u s e in t he nex t s e c ti o n , w e a l s o n e e d to k now the i nvari ants of the three
. 'ge o me t r ic "
gr oups .
Tunonnv 5 . L e t G b e t h e G u l o i sg r o u p o . / ' t h eD E : " ' : r t ' a n d
s tste mo . fsolutionsrelutit:eto the group G.
0rt
let t7.( he u funtlanrentol
G is the tetruhetlrcrl
group. then (rya+ 8rli3)3e C(.x).
( i i ) r / G is t he oc t ahe d ragl ro u p , th e n (ry s_-( r/i s )2e C (,x).
( i i i ) r / G i s t h e i t ' t t s u h e d rgarl o u p .t h e n 4 t t ( - l l 4 6 i 6 - 4 _ : t 1e C ( . x ) .
Pnrxx.(i) considerthe tetrahedral
group.usingthe notationof Theorem 2. Recall that
a n d3 6 : 2 _ ( - l
i't: -l.ir:<-l
r 7 a + 8 1 7i (s3c a r r i e di n t o ( 1 ( 4 4 + 8 r y ( .b)y r h er r , r i * ( [
\
carries
/t
,9 , ) T h e m a t r i * d ( ,
\-
l)
r / o + 8 r / (:t t t .U + 2 . . ). ( r y+ 2 _ : 2 ) .( r y - 2 i o
lntO
rh1t+2 _O. 3Qr OQ.:- lXrl- 2iO . ,b(t-2 .:)et+ 2i';)
- -3 .
0 1 . ( 2 ( - t ) 2. e 4 + 8 4 i r )
: - 3 ( - 1 3 ) t . ( - 3 ) . ( q ^ i 8 r y i . ) : q+18 , 1 ( t .
Thus(,1'-t8ryit)tis an invariantof G and therefore
is in C(.r).
(ii) considerthe octahedralgroup. using the notationof Theorem 3. R ecal l that
{ * : - I a n d2 6 : i ( i t + t ) .
J. J. Kovacic
12
t t s ; - , / i si s c a r r i e idn t o( 4 1 r 1
i -s. r t f ) b y t h em a t r i x( 5 , ? , ) t n . . r , r ' ^ O ( l - i )
carries
t C ' 0 'Q -t ( 0
' + O '( r y - O (' r +
n 5 ( - 4 ( 5: a ' . ( @
int o
+ 0 ' Q t - 0 ' 2 Q , t2' 0 ( 6 ( + ( ) @ - ( ' O ' O ( - ( ) Q r + ( ' o
6Q1
: 4 . 6 u .( 1- ( o ) .( r y t ( -4 ( )
: 8 ' ( - l l 2 ) ' . ( , t t_ ( - q . ( ) : - ( q t ( - r y ( t ) .
is in O(-x).
Thus(nt(-rt()'is an invariantof G and therefore
groupand usethe notationof Theorem4. Firstwe collect
(iii) Considerthe icosahedral
someeasilyderivableformulas.
(s:_1.
(o:(r_(t+C_1,
5 6 ' : ( t - ( t 3 . 5 r l , :t - ( ' + ( ' - 2 ,
: 2 .( -2 -( - 1 : 5(Q2-rltt).
54r,1,
/ r
o \
.
/' \f
1 1 . .
v 1 1 .
/ , h
T h e m a t r i * ( ; , , ) c a r r i e s r l " ; - l l 4 n ( u - 4 ; ' r i n t o i t s e l f . T h c m a t r i *-1A, / c a r r t e s
\
t
/
\0'
/
q t t . (- l l q u ( u - 4 ( t t : 4 ( ' 0 1 ' - 4 . (- - ( 2 )(' r t ' + ; t r l ( + ( ( t )
Q t ' - . < ' , t-( . ( . ( ) . Q t ' * _ h t ( -( . ( t ) '0 t ' i * , i i + i t ; t )
int o
6 * 0 r '- n ( - _ ( ' )5. 6 t r t (. ( - O Q t-' ( r y- ( -( o { ' ) '( i * X q '+ i r r r i+ i ; : ) '
:5
( - l X r l 2+ ( r y i - ; t ( t ) ' ( - 1X r / 2- ; * r / i + ; t s ' )
( , b { a t ' Q t t t ( -1 1 r / o ;- or y - ; tt ) : r / t t ; - I l r i n i 6 - r / i 1 ' .
1 i s a n i n v a r i a n to f G a n d t h e r e f o r ei s i n C ( . r ) .
Thus n"(-llqu(u-q;t
T his pr ov e sth e th e o re m .
-5.4. PR(x)F oF THE ALGoRITH\I
W e m us t p ro v e th e v a l i d i ty o f fo u r s cparateal gori thms. W e must show that thc
algorithms for finding a 4th. 6th and l2th degree equation for r,r are correct for the
t et r ahedr al. o c ta h e d ra l a n d i c o s a h e d ra lgroups and that the equati on obtai ned i s
ir r educ ible.a n d fi n a l l y th a t th e a l g o ri th m for fi ndi ng a 12th degree equati on i s al l inc lus iv e( alt h o u g hth e e q u a ti o no b ta i n e dneednot be i rreduci bl e).In so far as i s possi bl e.
we c ar r y out th e p ro o fs s i m u l ta n e o u s l y .
We begin by showing that the equationsobtained for a-lin the tetrahcdral.octahcdral
and ic os ahed racl a s e sa re mi n i ma l . T h ro u g hout w e assumethat the Gal oi s group G of the
DE ,r"': r_|is the tetrahedral.octahedralor icosahedralgroup. We also fix a fundamental
s y s t emof s ol u ti o n sry .i o f th e D E re l a ti v eto the group G and set o-r:11' tl .
T s c o n n n tl . L e t q t h e a n t ' s o l u t i t t rot . ft l r c D E a n t l l e t ( r ) 1 : 4 ' r l 4 t .
(i) lf G is the tetrahedralgroup. then
d e g .,..,rtr,
2 4
: 4.
and deg,,,.,,ru
S o l v i n g H o m o g e n e o u sD i f f e r e n t i a l E q u a t i o n s
tt
(ii) I.f G is the octahedralgroup. then
d e g a , . . , r o , 26
a n d d e g . , - " , c: u 6 .
(iii) l/ G is the icosahedralgroup. then
: 12.
d e g .,,,rr-r,
> 1 2 u n d d e g .,.,,r,-;
Pn o o n .S inc ear is lef t f i x e d b y th e g ro u p G, g e n e ra t. dby
' f:
w here < i s a pri mi ti ve
" !,).
\0 i-'l
6th .8 th or 1O t hr oot of I i n th e te tra h e d ra l o. c ta h e d ralor i cosahedralcases.the degreeof
( , o v e r C ( - x )i s < [ G : G l ] :
4 , 6 o r 1 2 .T h e r e v e r s ei n e q u a l i t yi s p r o v e nm o r e g c n c r a l l y a. s
in d i ca tedin t he s t at em e n to f th e th e o re m .
L e t G r be t he s ubgr o u p o f G th a t fi x c s 4 ,. C o m p l ete r7, tc' ra fundamentatlsystemof
-t
s o l u t i o n s4 r , ( r o f t h e D E a n d c o n j u g a t eG t o X G X
so thal XGX-t is the Galois
g r o u p o f t h e D E r e l a t i v et o r i r . ( r . T h e n X G . X 1 c o n s i s t so f m a t r i c e so f t h e f o r m
/,
,/ \
1 i s f i n i t e .t l : 0 a n d c ' : l . w h e r en i s
(:
t h e o r d e ro f G , . E v i d e n t i y
,' l l . S i n c eX G 1 X
\O r'
-t
XG tX
i s a s u b g r o u po f t h e c y c l i cg r o u p
, ,) -l : r }
{[ \(o' ,^g
'-'ll
)
an d th e r ef or eis c y c lic .H e n c e G r' H (w h e reH i s th e centreof G) i s i somorphi cto a cycl i c
su b g ro up of A o in t he tc tra h e d ra lc a s e .o f S o i n th e octahedralcase.and of A , i n the
i c o s a h e d r ac la s e S
. o ord G, H <3.4.5 or c'rrd
G , ( 6 . I t .1 0 .T h u s
d e g .,..,r,-r
r : [G : G ,) 2 4 . 6, 12.
Th i s p ro v est he t heor e m.
Th ro ughout t he r em a i n d e ro f th i s s e c ti o nw c s h a l l be consi deri nga certai ndi fferenti al
e q u a ti o nwr it t en r ec ur s i v e l yn. a m e l v
0u: -1
(#),
oi-1 : -a'i-zoi-(n-i)(i *l)ru,*,
(i :
n,. .
..0)
B y a s o l u t i o n o f ( + ) , i s m e a n t a f u n c t i o n : s u c h t h a t w h e n d , , .. . . . 0 - t
a b o v e . t h e n a - , i s ( i d e n t i c a l l y )z e r o .
a r ed e f i n eads
TneoRnnr2. Let z be a solutiono./'(#),,. ttntl lat o be unt, solutiono.l
di
.,":"ft
,,)'.
i--,, (tt
i)!
Then 4 : et" is ct solutionof' the DE -I"': r-r'.
Proof. Let
A- -,,''.;t
rr- I
:,i 1;;1
11i
,,,-,r'
,r "''
(c,,_.--l).
w h e re n , is an indet er m i n a teW
. e c l a i m th a t
?o*tA, ,
ik*1,4
( t t ' '- t ' ) :
*[(rr
, ^ , r . frt i
r - r r .rA
^,\
l
,.k,4
aft
]A)rr'*t],,,i +k(rr /r+ l ) ^
('lt'n
t7
I
(/.:0.
)
t t
J. J. Kovacic
.r+
F o r A: 0 . w eh a v e
?A
( u ' 2- r )
:
.n,
i t l - - , ,,,)
(t
\i?r (rr i) |
/
r,,,-,.)
ia,
l)rat*,
!'+
u,,
, t t '^Ir r r , * ,* " ) - t
,ll,i+l-'it
i'---o(rr-i)!
i?-o(n-l-i)!
( r r- i ) r 1 i , , . , - ,
( , - i X i + I ) r ' r r i r*, , . i
_'i'
t t * . 4_ " r r
(rr - i) !
;"s (rr i) !
iu-o
nw,
At dnrA-l:, Tr#',' -,4 ffi ,*,'
-i)(i+l)ra'-, i
$ (- n
)
Ir"
(n-i)!
,u-u
: ( r r r*r 'z ) A - I - +
l z a , * o , -, * ( r r- t X i * l ) r c , *1 l u ' i
;;6 (ll-l)i
ni.,,,,''
:
( n i) !
- (rru'*:).4*
;
( 1 1 1 1 ' * : ) . 4- r -
4.1
(" , \
O ur c laim now fo l l o w s b y i n d u c ti o n .
T o s h o w t h a t 4 : e ) " ' i s a s o l u t i o no f t h e D E i s e q u i v a l e ntto s h o w ' i 1 g
thatt,.r'*(,)2:t'.
W e as s um et h a t u )'+ tD 2-r * 0 a n d fo rc e a contradi cti on.
S i n c e. A ( u t \ : 0 . w e h a v e
.A
(t'ltrt *
,.tr.
.' A
(tt') :
o'
,.-,
Therefore
iA
. ( r r ; ) ( r u*' t t t - r ) : ( '11'
t rl
tt.{
^ ( r r r ): ( ' ) .
- (rr;)*(rrtr;*:).{(r,,l)+
("\
( .\
Henc e
iA
- (r,-l): g.
( 11'
A s s u m i n gt h a t
ar-1,,1
- , ,' ( r ' t ):
( Il"
we hav c
0:
Thus
d /t^:1
'\
ikA
^ , ( t r t ): Q .
( ll"
iA*1.,1
aA*11
(rr.r)'
kol :
t (trr)to'n
a'. \..u *
i,,* r
a,i.oi
)
a k + lA
( t o ) ( t r -*rt 't t ' - r )
.
!' + ,
( 'lt'
:
-
?k+tA
. r - (o)*
('lt" (.\
ak+rA
1 r ^ (r'r)*[(lr( ! , 1 "('. \
?kA
iA-r.4
( t r r ) *k ( n - f t + l ) i , , * ' i ( r , . , )
2klto+:1
,\l*
-0.
1.4
a.k*
:
i r'+T (tu)
('11"'
g'
S o l v i n g H o m o g e n e o u sD i f f e r e n t i a lE q u a t i o n s
15
The desiredcontradiction follows from the fact that
?^A - - n l
+0.
. - (,u)
Th i s p ro v est he t heor e m.
THr-.onnu3.
(i l S uppos et ht t t ( #)+ h a s a s o l u ti o nz e C (.x ).T h e n the pol t,nomi al
,r*- i
. ^ " 'i .) l, ,l r ' ' € c t , r ) [ x . ]
t ? o6 -
is irretlucibleot,erC(,r).
(ii) Suppose
t h o t ( # ) , , h u st t s o l u t i o n : eC ( . x ) .T h e n t h e p o l t n o m i u l
.t t ' "- , \
(l'
r r '€' c ( . r) [ nI
t oi r ,
is irretlucihle
orer C(.x).
( i i i ) S u p p o s teh o t ( # ) t z h u s o s o l u t i o n : eC ( . r ) u n d t h u t ( # ) a u n d ( # 1 , . t l o n o t h u t ' a
s olut ior tir
s r C( . r ). T l rc n th e p o l t' n o n ri u l
tt;
l-1
rt''--'L=o
lt''€c(r)[x']
ttl-i )'
i s i r r e t r u c i b,rrtc, r c ( . r ) .
P Rrxx. By T hc or c m s I a n d 2 . a n \, ro o t o f th e p o l l -n omi al
" It
1 ' ., ,r,' (rr,e c(.r))
tr,",
,
i?o0 - i)t
mu st h a v e degr ee4. 6 o r 1 2 o v e r C (.r). Sta te me nt(i ) oi the presenttheorem i s cl car.
Sta te ment( ii) f ollowsf ro m th e fa c t th a t i f a s e x ti ci s reduci bl e.then one of the factorshas
d e g r e e( 3 . T o p r o v e ( i i i ) i t s u l i c e st o s h o w t h a t i f d e g . , . , , t , . rn: . t h e n ( # ) , , h a s a s o l u t i o n
: e C(.r).
L e t , 4 eC ( . x ) [ u ' ]b e t h e m i n i m a l p o l y n o m i a lf o r ( , o v e r C ( . r ) .L e t d e g , , . A : n a n d w , r i t e
,4- -,,,*;t
r#i
,,,':,i,
(rr,,:
1,,1rijr,,, r).
Co n si d ert he poly nom i a l
B:i^4 (,
(It'
l1.r)+,.,*(rr*,*:)..,r.
whgre
The coefhcientof u'"* t in B is
::un,.ir_,
I l t l, , *
lltl,, :
()"
an d th c c oet ir c ientof u ' " i n B i s
-(lt-
l)rt,, 1lu',,+ndr_1*:tt,,:
dr -, -:
:0"
16
J. J. Kovacrc
s i n c ea n : - 1 a n d d , - 1 : : . T h e r e f o rdee g * , 8( n . B u t
i
A -^
?
A
(D')+ . (to)+ (nroI :)Ako)
: = (a,r)(r
B(tr"r)
(.'\
.j'
:
:- (1 4 (ru *)) (rtr,t* :)A (ut)
dr'
:0.
ThereforeB :0. The coefficientof u'' in B is
0:
:
(i+ l)
I
j
a / r) i
ai+t
ir- r-rlJ
t=t" - t '"- - 7 );
+ =!t,,1,
(ir-i)!
(rr*I -i)l
t 1 i ,l ' - . .
"i -.
*+-=
(,,-( r r +l - t l
[(n-t)(i+ l)ra'*1*tt' t'*a'i*:a'f'
w h e r eo - t : 0 . T h e s ea r e p r e c i s e l yt h e e q u a t i o n so f ( # ) , , . T h i s p r o v e st h e t h e o r e m .
F or anv f un c ti o n b w e d e n o teb y l tb : b ' ' b the " l ogari thmi c deri vati ve"of b.
polvnorniul)o/'degreen in solutiortso/ tlrc DE.
THnonnu4. Let F be any,.fornt(hontogeneous
'f
h e n : : l 6 F i s a s o l u t i o no f ( # ) , , .
P nor lr . F ir s t w e p ro v e th a t i f F , a n d F -,a re f uncti onssuch that 1r)f, and /t)l ' , arc sol uti ons
o f ( # ) , , . t h e n / t ) ( c , F l, ( ' - , F - . )i s a s o l u t i o no f ( # ) , , f o r a n \ ' ( ' r . ( ' z € C .L c t , r / . , r i .r r f d c n o t e
d y ( + ) , f o r : : 1 r ) l ' - r . / r ) f1 . / J ( r ' , 1 '*,r ' , l ' . ) r e s p c c t i v e l l .
t h e s e q u c n c edse t e r m i n e b
W e c laim th a t
( c ' , l , * t ' , [ . ) t r f : t ' 1 - F , t*r r/ ' . F - , u i .
T h i s i s c l e a rf o r i : r r . A l s o
( r ' rF r * t' rF .' )c ;t-r : (r'r F r f c r ^ F r)[- Q' :'-/d(c' , F, + t' r Fr)af - (rr- i )(i + 1)rurf*r]
: - [ ( r ' ,F , * c ' r F . ) a l ) ' - ( n - i ) ( i + I ) r ( c 'l,- , + t ' , l ' 2 ) c r ,,i : - - [ c r I - r a !t t ' - , F r a ? ] ' - ( r i - i X i + l ) [ r ' , F , d , l *r * c ' : [ - , u i * r f
: c t F t u ! y * c ' 2 F r t t l,
( i : 1 1. .. . . 0 ) .
Therefore
( r ' , F -+, t ' r F ) a 3 r : c r F r a 1, + c ' , F r a 2 - r: 0 .
whic h v er if ieso u r a s s e rti o n .
T o pr ov e t h e th e o re m.w e m a y a s s u meth at
P: f i , r , .
i= 1
w h e r e4 r . . . . , 4 n a r e s o l u t i o n so f t h e D E .
. hus
L e t r r ; , : 4 ' , 1 4 ia n d d e n o t e b y o . u t h e k t h s y m m e t r i cf u n c t i o n o f t , , r t , . . ( ! ) n t T
6 ^ k : 0 f o r f t< 0 o r A > t n , o m o : 1 a n d
6^k:
1 { i t <
I
e)it'
o)ik
' < i r S m
for 1 ( k ( nr. First we claim that
o ' ^ k : @ t * 1 - k ) r o ^ . k -r - 6 m r 6 ^ k + ( f t + l ) o . . * * , .
Solving Homogeneous Differential Equations
T h i s f o r m u l ai s e a s i l yc h e c k e df . o rm : 1
-17
a n d . f . o rm > 1 ,
o ' ^ u: ( o ^ - r . * * o ^ - 1 . k- t @ ^ ) '
: ( m - k ) r o ^ - r . k - r - o ^ - r . t o ^ - r . r * ( / <+ 1 ) o ^ - r . r ^
+ l ( m + I - k ) r o ^ - r . k - 2 - o ^ - L .r o ^ - r . k - y * k o ^ - r . o ) o ^
* o ^ - r . u- r ( r - @ ' ^ )
: ( m + I - k ) r ( o ^ - 1 . k- t * 6 ^ - 1 . k -z @ ^ ) - ( o ^ - r . r * a ^ ) ( o ^ r . x * 6 ^ - 1 . k -r @ ^ )
+ ( f t+ l ) ( o ^ - r . k + t - l o ^ t . * @ ^ )
: ( m + I - k ) r o ^ . k - r - o m t o m k +( f t+ l ) o ^ . x * r ,
which completes the induction.
Next we use induction on i to prove that
(- 1)'-i+t(n_ i)t on.n-,.
ui:
Evidently
n
an-r : : :
ldF :,I.
ei : onr.
Using (#)^, we have
a i - t : - a ' i - Z a i - ( t t - i ) ( i* l ) r a , * ,
: ( - l ) " - ' ( n - i ) ! o ' n . ni -* o n r ( - l \ n - ' ( n - i ) t o , . n -i
-(n-txi+
l ) r ( - l ) ' - ' ( n - I - i ) t . o n . , - r -,
: ( - l)"(n
- i)!
l o ' n . ni-* o n r o , . n - i - ( i + l ) r o n . n - , _ , ]
: (- l)^-t(n-i)l (n-i + l)on.n_i+t
: (- l)^-t(n-i + l)t.on.n_i+r.
Hence
a_t :
(- 1)'(n+l)! 6n.,*r : 0.
This completes the proof of the theorem.
THEonsu 5.
(i) If G is the tetrahedral group, then (#)o has a solution z : k\u. v'here u3 e C(.x).
(iD I./ G is the octahedral group, then (ff)u has a solution z : I6u. v'here u2 e C(.x).
(iii) I.f G is either the tetrahedral group. the octahedral group or the ic'osahedralgroup,
t h e n ( f t ) r , h a s a s o l u t i o n z : l 6 u . * * h e r eu e C ( , x ) .
Pnoor'. This theorem is a corollary of Theorem 3 of the present section and Theorem 3 of
the previous section. For part (i) we may take u: q4 + 8rl(t. for part (ii) we may take
(rl'(- ry(t)t or
tt:q5(-4f
and for part (iii) we may take u:(,lo+8rl(t)t.
-ry(tt.
1
l
\
u
(
u
ryttiWe shall write
Ltt2'-,LI tt
( ' ) " " eC ( . x ) ,
w h e r en : 4 , 6 o r 12 a n de , e Z . O u . n . * t s t e p i n t h e p r o o f i s t o d e t e r m i n et h e v a r i o u s
p o ssi b ilit iesf or ec . as s ta te d i n s te p 1 o f th c a l g o r i thm. For easeof notati on. w e shal l
assumethat c : 0. To this end we shall use the Laurent seriesfor
: : r d u:
,,;.
i/,)1ur:
-18
J. J. Kovacic
nam ely
tl
( e : e o eZ . p o s s i b 0
l y)
t:T2e.\-'+..'
and for r. namely
r : a - \ 2+ [ ] x - I + ' '
.
( 2 , { l e C . p o s s i b l y0 ) .
( Not e t hat . b y th e n e c e s s a ryc o n d i ti o n s of secti on 2, r can have no pol e of order
ex c eeding2. )
F i r s t w e c o n s i d e rt h e p o s s i b i l i t yt h a t s : 0 a n d l l + 0 . c o r r e s p o n d i n tgo ( r ' , ) o f s t e p I o f
t he algor it hm .
T H E c r n E6u. I / r : 0
u n d f i * 0 . t l r c ne : 1 2 .
Pnoon. We write
,:
.
l)e,\-'+/'+...
and t r eat e an d /' a s i n d e te rm i n a te sT. h e n
0i :
1,ti
u* t+
"+B,-ri-n+1 aC,./-xt
"
where ,,{,. 8 , . C , a r c p o l y n o m i a l s i n c w i t h c o e l i c i e n t s i n C . U s i n g ( # ),,w e fi nd that
An:-1.
8,,:Cr:0.
/
li,:(rr-i
n
\
,' -' . . .,/, , ) , 1 , .
\
/
,
,
B i _ r: ( , - i - l - ; .
t l
\
) B , - ( l - i ) ( i + l ) l lt ,
\
/
c t - t: ( r _ , - r - , ' i . ) c , - { , .
-- ,,
\
fori:n. . ..0.
We leave to the readerthe verification
that the solutior-r
to thesccqu a t i o n si s g i v e nb y
'
A i :- ' I i ( t - , , . )
l-o
I-
\
,/
' - 1 '1( ^ - , ' i
B , : [ t " - i 'U + l X r r - i )
,,).
r-0
'- /
l;1, \
c , : ( n - i ) " - | . . '( , i=o
,. _' i
\
.)
( i : 1 1. . . 0 )
.,
bec aus e
0:rr,r : 1-rr'-t+B-r.\-"+C,/'x-"+
n
/
,.
.
\
0 : A- r : - n ( i- ,".,,
.
r_ )
" r = o\
and
/
0 : B r + C rI
rr r
,_t
i:o
tth\
^__r
tr
: | ] L ( i + l X r l - i ) t I /( A - l,r , ;\ . , } + l i l l * l )/ , I l | A
tJ
/
i-h\
ll
I
S o l v i n g H o m o g e n e o u sD i f f e r e n t i a lE q u a t i o n s
19
Th e f ir s t equat ionim p l i e sth a t
t2,
' - ; '
fo r so m e / : 0.
. . . . t 1.S u p p o s eth a t / I rr.
' fh e n
th e secondequati ongi ves
c : 0(t+lxn- /)
'f]'
(ft- /).
i;?
w h i ch i m pliest hat f : 0 . T h i s c o n tra d i c ti o ns h o w s that / : rr ?rrd thereforee : 12. Thi s
p ro ve st he t heor em .
N e xt we c ons idert h e p o s s i b i l i tyth a t e 1 0 . T h i s correspondsto (c' r) of S tep I of the
a l g o r i t h m .A s a b o v ew e w r i t e u i : A i x ' - " +
in Q[r] x'ltosetlegreeis n - i und v'hose
Lr-.unre.A, is u polrnontiul in e vt'itht'oe./ficients
is ( (n 1 2 \)' -' .
l e u d i n gc oe. f J it ' ient
Pn o o r. Lls ing ( #) , , we h a v e
Au: -l'
/
n \
Ai r : ( rr-i-;
r ,l A , - ( n - i x t + l ) t A , r , .
r\
,/
The lemma is immediatefrom theseformulas.
Th e a ut hor did not s u c c e e di n fi n d i n g a c l o s e d - formsol uti on of theseequati ons.thus
we shall use an indirect argument.
Assu m et hat e + - l r4 . T h e n th e D E )" ' : r!' h a s Pui sseauxseri essol uti onsof the form
-ll4a.
tlr: xlr+.. '.
ltt: j+ j.,
lrz:t- i, tf+,i.
42: xp,+...
of (+), for everyi : 0. ., tl.Since
ByTheorem
4. ln\frry\') is a solution
I 6 ( 4 ' r t t \ -:' ) ( i t r * ( n - i ) p r ) . r I + ' ' '
= (l n; - ( rl -n' ) . -\
th e p o ly nom ial I
- \
t+at)x
'+
'
, m u s t v a n i s hfo r
12
n
ln
\
--g., l++t
t-(a-i)
(i:0."'.rr)'
Tnnonnu 7.
(i) ,4ssunrcthut G is the tetruhedralgroup. Then e is an integer t'hosenjront antong
6 + / < r ,+' i t . A : 0 . + i . + 6 .
(i i ) A s s um et hut G i s th e o c ta h e d ragl ro u p . T h e n e i s un i ntegerc' hosen.fi ' ortt
untong
6 + / . r a l 4 r .k : 0 . t 2 . + 4 . + 6 .
w
-
(iii) Assunte thttt G is either the tetrahetlral group, the oc'ttthetlrul group or the
it'ttstthedrtil .qroup. Then
f t : 0 . + 1 . . . . 1. 6 .
e i,s utt inreger r'/rosen ./i'ttttr ,11y1etr,?
6 + A. | + 4x.
J. J. Kovacrc
10
P Rc x r r .( i) I n t h i s c a s en : 4 . l f u * l /4 . th en w e may use the Lemma and the remarks
f o l l o w i n gi t t o o b t a i n
I
/,,
0 : A _ 1: t l ( ; - 2 + ( 2 - i ) / t + 4 a ) .
i-:^O \ J
T hus
e:6+ft.rzf1*.
/<:0, +3. +6.
I f e : - 114, w e c o m p u ted i re c tl y ,u s i n g the recurrencerel ati onsgi ven above.
A+:'-l
'4, : le
Az: -tnk' -3e+9)
A r : ,? (e i -9 e 2 + T e - 5 4)
A o : - # ( o * - 1 8 e 3- l l 3 5 e 2- 4 5 9 e + r ? 5 )
2 1 6 0 e 2+ 6 4 9 0 e- 7 7 7 6 )
* L s ( e s - 3 0 e 4 + 3 6 0 e- 3
A_t:
:
zlle _ 6)s.
l . / 'a * - 1 1 4 . t h e n w e m a y u s e t h e L e m m a a n d t h e r e m a r k s
( i i ) I n t h i s c a s en : 6 .
l o l l o w i n gi t t o o b t a i n
6
/,,
o : A _ ,: f I ( ; - 3 + r 3 - i ) r | + 4 t ) .
t: o
\z
Thus
e:6*k.., | +4x.
k : 0.+2, +4, +6.
lf z: - | 4. we computedirectll,.
A o : - l
As:
lze
A+: -Ikt
-2e + 6)
At:
l ( . . - 6 e 2+ 2 4 e - 2 4 )
A z : - r ta (n-o l 2 e 3* i 2 e 2 - l 9 2 e+ 2 1 6 )
A t : .tr(.t -20e4* 180er -840e2 * 2040e-2016)
Ar, : -*t(.o
30es+ 390e+-2760e3+ 1l I 60e2-24336e +22320)
A , : J r @ ' , - 4 2 e 6 * 7 5 6 e 5- 7 5 6 0 e a i 4 5 3 6 0 e 3- 1 6 3 2 g 6 e 2+ 3 2 6 5 g 2 e - 2 7 g g 3 6 \
: # 8 ( e_ 6 ) r .
( iii) I n t his ca s e i l : 1 2 . l f t + - 1 ," 4 th
. e n w e may use the Lemma and the remarks
f ollowing it t o o b ta i n
o : . , 1 - , :i i @ _ ' 6 + ( 6 - ri +
t '+ r t .
Thus
i=o
e : 6 * A r l i a x , A: 0 . + 1 . . . . +
. 6.
S o l v i n g H o m o g e n e o u sD i f f e r e n t i a lE q u a t i o n s
11
If r : - | 14, we c om p u te d i re c tl y . U s i n g a p ro g r ammabl ecal cul atorw e obtai ned the
following.
Atz:-l
A t t :
€
Ato:-e2+e-3
A s : e 3- 3 e 2 + ! e - 6
A 8 - - - e * + 6 e 3- 2 7 e 2+ 4 5 e- )
A . ; : e t - l o e a+ 6 0 e 3- 1 8 0 e *2 3 l 5 e - 2 1 6
A o : - e 6 + 1 5 e s- 1 2 0 e ++ 5 4 0 e 3- 1 4 8 5 e 2+ 2 2 4 1 e - 1 4 8 5
A s - - e ' - 2 1 e 6 + + e s - 1 3 6 5 e a- 1 - 5 3 5 5 e 31 -3 0 4 1 e+23 e 1 l ie - 1 1 1 7 8
At:
- e 8 + 2 8 e 1- 3 7 8 e 6* 3 0 6 6 e -s 1 6 1 7 0 e+a 5 6 1 9 6 e-3 1 2 5 1l 8 e 2
+ l623l8e _],Et-aT
At : ee- 36e8* 612e1- 6300e6-t 42903es- 199206e4
+ 628236e3
1293732e2+ 31sQ4e5
e - 862488
A z : - et o *4 5 e e -9 4 5 e 8 -l 1 2 0 6 0 e 1 - 1 0 3 005eu
+ 6l 292l es-2566620e4
2868e7e1
L(]O?;z
llc
12J1
I 17
_
_
e2+
7 453620er
e6
A r : et t -55e10* tTtno -21780e8+ 228195e'- 1690221
+
!E q 31125 e5 -
+ LW *
3461 3 8 65ea +
I u Ll]g s 7 3 5
r,-, -
3 I e 3Q6 16 s
e2
e- 92538045
A o : - e t 2- l 6 6 e l1- 2 0 13 e 1 0+ 3 1 4 5 5 e-eL + - Y t e 8
ea+ 1240169535e3
2817 6 6 8 7 e+6 l 3 1 l 7 9 2 5 l e s -e r-6 -e 2 r58-5
r., {aOTry
42S11+OO2
e _+2bf!2b321
+
1- 6 l l l 6 e l o -+ 9 2 6640ee
- 100011
12e8
A - t : et t - J B e t2* 2 8 0 8 e 1
- 1 205552640ea
+ 8006| 696e1 480370l7 6e6+ 2161665192e5
-28298170368e2
6
+ 28298170368e- 1306069401
+ t7293326336e3
:
(e_6)13.
Th i s pr ov est he t he o re m .
Fi n a lly we c ons ide rw h a t h a p p e n si f r : [] :0 . i .e . at an ordi nary poi nt of r. U si ng the
p re vi o ust heor em .we h a v e th a t (rr/1 2 )ei s a n i n te g e r.
Let f denote the set of poles of r. We have proven the following.
( i ) I n t h e t e t r a h e d r acl a s e .( # ) + h a s a s o l u t i o n: : 1 6 u . w h e r e
Lt3: P3 f] (,t - (')"',
-tiqi,
P e C [ . x ] a n d e " i s a n i n t e g e rc h o s e nf r o m a m o n g 6 + k .
k:0.
(i i ) In t he oc t ahedra lc a s e ,(# )e h a s a s o l u ti o n z : l t\u.w here
L r 2: P 2 f l t t - c ) " ' .
+3. +6.
42
J. J. Kovacic
P e C [ . x ] a n d e ci n a n i n t e g e rc h o s e nf r o m a m o n g6 + / r . l + 4 x . k : 0 . + 2 . + 4 . + 6 .
( iii) I n eit he r th e te tra h e d ra lc a s e .th e o ctahedralcaseor the i cosahedralcase.(#)rz
h a s a s o l u t i o n: : l d u . r v h e r e
u : P,[Jtt - (')"'.
P e C [ . t ] a n d e . i s a n i n t e g e rc h o s e nf r o m a m o n g6 + k r t + + t . A ' : 0 . + 1 . . . . . + 6 .
Let d: dep P. Then the Laurent seriesfor ; at cc has the form
n /1)
\
' : 1 , ) ( ; r +I . . J , * - 1 + ' ' '
and the Laurent seriesfor r at r
has the form
r:1,.\-:+....
(By the necessary''conditions
of section2, the order of r at cc is at least 2.)
I f we let
( ', : t 2 d * f
n
e..
,u.l
the samecondi ti onsas doeseach
t hen. by a t heo re ma n a l o g o u sto T h e o re m7 . e,. sati sfi es
e.. Also
u
/
\
d:i"(e,-Ir,)
lz
cef
\
/
m us t be a non -n e g a ti v ei n te g e r.T h i s i s a re statementof step 2 of the al gori thm.
W e s hall c om p l e teth e p ro o f o f th e a l g o ri thm by show i ngthat the recursi verel ati onsof
s t ep 3 ar e iden ti c a lw i th (# )^ .
Let
u : lr'r;
and s : f] (.x-r').
,1.
-' Pc,.
T hen z - - l6u: P'l P * 0 . A l s o s e t P , : Sn
U si ng (#),. w e have
Pn:-P
Pi-t:
Sn-t+'Po,-,
:
Sn-i*
rP( -a'i-zai-(n-iXi
* 1)ra,*,)
: - S ( S ' - ' r o , ) ' + ( n - i ) S " - ' S ' P a , + S ' - ' |-P* ' o i
- S ( P '+ P 0 X S " o , \ - ( n - i X i + l ) S 2 r ( S ni-- | P t t i l
- - 5 4 + ( ( n- i ) - S 0 ) P-, ( n - l x i + 1 ) S 2 r P , * , .
This is preciselythe equation of step 2 of the algorithm.
F inally . t he e q u a ti o n
ai
, , r n: ' F t
ai
i " : o@ - i ) t
may be rewritten as
o : - snpr.rn
+'t'
Il1
i ' = o( n - i ) l
This completesthe proof of the algorithm.
,r, : i
S'P'
i ' - =(ot t - i ) !
rd'.
1\
Solving Homogeneous Differential Equations
'11
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