DEMONSTRATIO MATHEMATICA
Vol. 49
No 4
2016
Sofiane El-Hadi Miri
PAINLEVÉ EQUATION PII
AND STRONGLY NORMAL EXTENSIONS
Communicated by J. Smith
Abstract. The aim of this paper is to show that if F is a differential field and y is
a PII transcendent such that tr.deg.F xyy “ 2, then every constant in F xyy is in F . We
also show that in this case, F xyy is not contained in any strongly normal extension.
1. Introduction
Today, more than 100 years after their discovery, Painlevé equations still
arouse a lot of interests. Among many other reasons of such an interest we
mention that:
(i) They are irreducible to the classical functions.
(ii) They involve hypergeometric functions and their confluents.
(iii) They have combinatorial features. Particularly they are related with
the combinatorics of Young diagrams in substantial way.
The Painlevé equations were characterized, as those ODE’s of the second
order which have the following property: any local analytic solution extends
to a meromorphic solution on the universal cover P 1 pCqzS, where S is the
finite set of singularities of the equation (including the point at infinity if
necessary).
Using algebraic methods to study the properties of differential equation’s
solutions is nowdays well known, and widely treated especially in the linear
case as presented in the books [4], [16], and [17].
Associating Painlevé equations and Painlevé transcendents to differential
algebra was treated by different authors and in various ways; the most complete work in this direction is due to the Japanese school, namely: Nishioka
2010 Mathematics Subject Classification: 12H05, 34M55, 34M15.
Key words and phrases: Painlevé equation PII, differential algebra, strongly normal
extensions.
DOI: 10.1515/dema-2016-0033
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[14], [15] and Umemura [18], [19], [20] who proved for the first time the
irreducibility of the first Painlevé equation, and by the sequel gave a definition of non linear differential galois theory. We also mention a work due
to Malgrange [12] in which the author propose another way to introduce
nonlinear differential Galois theory, this was used by Casale [2] to show the
(local) irreducibility of the Painlevé equation P1.
We also mention the papers [1] and [13] that exhibit in an elegant way
the relationship between Painlevé equations, differential algebra and normal
forms. And the very pedagogic talk of Weil [21] that focus on the importance
of the constants and the no new constants concept in differential algebra.
I am deeply grateful to the anonymous referee, who have showed me
the very recent and interesting paper of Casale and Weil [3], where the
irreducibility of Painlevé equations is studied.
In this paper we deal with the theory of strongly normal extensions introduced by Kolchin [8] and actualized by Kovacic in [11], which is a third way
to introduce nonlinear differential Galois theory. Strongly normal extension
is a generalization in some sense of the so called Picard–Vessiot extensions,
existence and non existence of strongly normal extensions are closely related
to the ”complexity” of the solutions of a nonlinear differential equation.
In his paper [9] (published after his death), Kolchin showed that the
adjunction of Painlevé transcendent PI to a differential field of characteristic
0 can be done with no new constant, the purpose of the present paper is to
expand this result to Painlevé transcendent PII. We also use a result due to
Kovacic [10] to show that this differential extension is not included in any
strongly normal extension. Finally, the possibility of extending these results
to other Painlevé equations is discussed.
2. Preliminaries
2.1. Painlevé equations. In this section we first recall some well known
definitions concerning Painlevé equations.
Definition 1. The Painlevé equations are the following six special second
order nonlinear differential equations
(1)
y 2 “ 6y 2 ` x,
(2)
y 2 “ 2y 3 ` xy ` α,
1
1
1
δ
y 2 “ py 1 q2 ´ y 1 ` αy 3 ` pβy 2 ` γq ` ,
y
x
x
y
1
3
β
y2 “
py 1 q2 ` y 3 ` 4xy 2 ` 2px2 ´ αqy ` ,
2y
2
y
(3)
(4)
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˙
˙
ˆ
1 1 py ´ 1q2
1
1
β
1 2
py q ´ y `
y “
`
αy `
2y 1 ´ y
x
x2
y
γy δy py ` 1q
`
`
,
x
y´1
ˆ
˙
ˆ
˙
1
1
1
1
1
1 1
`
`
py 1 q2 ´
`
`
y1
y2 “
2 y y´1 y´x
x x´1 y´x
ˆ
˙
ypy ´ 1qpy ´ xq
βx γpx ´ 1q δxpx ´ 1q
`
α` 2 `
`
,
x2 px ´ 1q2
y
py ´ 1q2
py ´ xq2
ˆ
(5)
(6)
2
where α, β, γ, and δ are arbitrary constants. The equations (1),. . . ,(6),
as their solutions are often noted PI, . . . ,PVI, respectively, and where the
d
prime 1 is to be regarded as the usual derivation dx
.
The singularities of an ODE’s solution are called fixed if they do not
depend upon the constants of integration, and they are called movable,
otherwise. The aim of Painlevé was to classify all these equations of the form
`
˘
y 2 “ R y 1 , y, x
where R is rational in all arguments, having solutions with no movable
essential singularities, (nowadays, a nonlinear differential equation having
solutions with no movable essential singularities is called to posses the Painlevé
property). It was shown that the only equations of the form y 2 “ R py 1 , y, xq
with the Painlevé property, which couldn’t be reduced, are the equations
PI,. . . ,PVI.
The general solutions of the Painlevé equations are not expressible with
known functions, but they require the introduction of new transcendental
functions, which are called the Painlevé transcendent.
Painlevé equations are widely treated in the literature, we recommend
the reader to see [5], [6] and [3] and the references therein.
2.2. Differential algebra. In this section, we give some definitions and
theorems related to differential algebra. Throughout this paper, F will design
a field of characteristic 0.
A map D : F Ñ F is called derivation if for all a, b P F we have
D pa ` bq “ Dpaq ` Dpbq and Dpa.bq “ Dpaq.b ` a.Dpbq.
We shall usually denote a derivation by 1 i.e. Dpaq “ a1 . A field equipped
with a derivation is called a differential field.
If F is a differential field, the set C “ tc P F ; c1 “ 0u is a subfield of F
called the field of constants of F and denoted by constpF q.
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Let F and G be two differential fields, we say that G is a differential
extension, or a D-extension of F if F Ă G and the restriction of the Gderivation to F reduces to the F -derivation.
If F Ă G are two differential fields and S is a subset of G, we denote by
F xSy the smallest differential subfield of G containing F and S.
If G is a differential extension of F, an isomorphism σ : G ÝÑ G is called
a D-isomorphism of G over F if
@a P G; σpa1 q “ pσpaqq1
@a P F ; σpaq “ a.
We will say that a function y is of (differential) transcendence degree m
over F, if m is the greatest integer such that, y, y 1 , . . . , y pm´1q don’t verify
any equation of the form, P py, y 1 , . . . , y pm´1q q “ 0, where P is a polynomial
in F rX1 , X2 , . . . , Xm s, we also note in this case tr.degF xyy{F “ m.
Definition 2. A field U is a universal extension of F, if for every finitely
generated extension F1 of F in U and every finitely generated extension
G of F1 not necessarily in U, there is an isomorphism of G into U whose
restriction to F1 is the identity.
Theorem 1. Every differential field has a universal extension.
If we assume that U is a universal extension of F and the field of constants
of U is K, one can introduce the next definitions:
σ is a D-isomorphism of G over F, then we put Cpσq “ CGσG .
Definition 3. Let σ be a D-isomorphism of G over F. Then σ is called
strong if
1/ σ | CG “ id,
2/ σG Ă GCpσq,
3/ G Ă σGCpσq.
Definition 4. We call strongly normal extension of F a finitely Dgenerated extension G with no new constants, such that every D-isomorphism
of G over F is strong.
Remark 1. When F “ Cpxq, roughly speaking, strongly normal extensions
are: extensions by the entries of a fundamental solution of a linear ODE,
extensions by an Abelian function with classical functions as arguments.
Let
B
B
` V2
BX1
Bx2
be a vector field, where, V1 and V2 are polynomials in F rX1 , X2 s and D
is a derivation over F . A non-zero polynomial P is said to be a Darboux
V “ D ` V1
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polynomial, if there exists a polynomial Q such that
BP
BP
` V2
“ QP
BX1
BX2
and Q is called in this case cofactor.
For more details about differential algebra, we refer the reader to [7], [8],
[10], [11], [17] and [2].
P p1q ` V1
3. Painlevé transcendent PII; an algebraic point of view
3.1. No new constants. First of all let us give a theorem to answer the
question: why no new constants?
Theorem 2. Let F Ă G be an extension of differential fields, the following
statements are equivalent:
1/ constpF q “ constpGq.
2/ If f P F admits a primitive in F , then f does not admit a primitive
in GzF.
3/ If g P GzF satisfies g 1 P F , then g 1 has no primitive in F.
Differential fields extensions satisfying any of the condition cited in the last
theorem are called no new constant extensions. They should be regarded as
”economical”; in the sense that they not introduce antiderivtives for elements
of F which can already be integrated in F.
3.2. Painlevé equation PII. In the sequel we will assume that F is a
differential field of characteristic 0, containing an element x such that Dpxq “
x1 “ 1. Let y be an element of a differential extension field of F such that
y is a solution of Painlevé equation PII, i.e. y 2 “ 2y 3 ` xy ` α, α being a
constant in F , we also assume that the transcendence degree of F xyy over
F is 2. Under the assumption that there exist no Darboux polynomial in
F rY, Y 1 s with quadratic cofactor, we will prove the following theorem:
Theorem 3. Let F be a differential field of characteristic zero which
contains an element x such that x1 “ 1. Let y be a PII transcendent i.e.
y 2 “ 2y 3 ` xy ` α if the transcendence degree of G “ F xyy over F is 2, then
the field of constants of G is C.
First we will use some lemmas
Lemma 1. Suppose that F xyy contains a new constant not in F ; then there
is a polynomial P P F ry, y 1 s ´ F and ε P F such that
˘
BP 1 BP ` 3
P p1q `
y ` 1 2y ` xy ` α “ εP
By
By
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where P p1q is the polynomial obtained from P by replacing each coefficient by
its derivative.
P
Proof. Let P, Q P F ry, y 1 s, with Q ‰ 0, be such that Q
is a constant, we
also assume that P R F and that P and
Q
are
relatively
prime.
`P ˘
P
Because Q
is a constant, we have D Q
“ 0, which implies that QDpP q “
P DpQq, in the other hand one have that
˘
BP 1 BP 2
BP 1 BP ` 3
DpP q “ P p1q `
y ` 1 y “ P p1q `
y ` 1 2y ` xy ` α ,
By
By
By
By
˘
BQ 1 BQ 2
BQ 1 BQ ` 3
DpQq “ Qp1q `
y ` 1 y “ Qp1q `
y ` 1 2y ` xy ` α
By
By
By
By
and then
ˆ
˙
˘
BP 1 BP ` 3
Q P p1q `
y ` 1 2y ` xy ` α “ QDpP q,
By
By
ˆ
˙
˘
BQ 1 BQ ` 3
p1q
P Q `
y ` 1 2y ` xy ` α “ P DpQq.
By
By
Because P and Q are relatively prime, there must exists R P F ry, y 1 s such
that
˘
BP 1 BP ` 3
(7)
P p1q `
y ` 1 2y ` xy ` α “ RP,
By
By
comparing the degrees of the two hands of the last equation, we came to
2 ` βy 12 ` γyy 1 ` δy ` y 1 ` ε,
fact that R must be of the form
ř Ri “1 λy
n´i
where λ, β, γ, δ, , ε P F. Let
be the homogeneous part of P
ai y py q
e≤i≤n
of highest degree n, then from formula (7) we obtain
ÿ
ÿ
(8)
2y 3
pn ´ iqai y i py 1 qn´i´1 “ λy 2
ai y i py 1 qn´i `
e≤i≤n
e≤i≤n
βy 12
ÿ
ai y i py 1 qn´i ` γyy 1
e≤i≤n
ÿ
ai y i py 1 qn´i
e≤i≤n
inspecting the coefficients of the terms of the same degree in the two sides of
the last equation, we obtain that: by comparing the coefficients of the monomial y e py 1 qn´e`2 we obtain that β “ 0, doing the same with y e`1 py 1 qn´e`1 ,
and y e`2 py 1 qn´e we obtain, respectively, that γ “ 0 and λ “ 0. Doing
the same with the monomials y e`1 py 1 qn´e and y e py 1 qn´e`1 , we came to the
conclusion δ “ “ 0, which end the proof of the lemma.
Lemma 2. Let t “ y 12 ´ y 4 , P P F ry, y 1 s , A P F ry, y 1 s and m P N.
BP 3
1
m
Suppose that BP
By y ` 2 By 1 y “ A.t . If 2m ą degy 1 P then A “ 0 and
P P F rts. If 2m “ degy1 P then there exists B P F rys such that A “ y 1 BB
By
and P ´ Btm P F rts.
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Proof. One may assume that 2m ≥ degy1 P. To prove the lemma we use
induction.
Suppose that degy1 P “ 0, then we have
`
˘m
BP 1
BP
BP 1
(9)
y ` 2 1 y3 “
y “ A.tm “ A y 12 ´ y 4 .
By
By
By
1
If m ą 0 p2m ą degy1 P q, because the degree of BP
By y is at most 1 then A “ 0
and so BP
By “ 0 thus P P F Ă F rts. On the other hand, if m “ 0 p2m “ 0 “
BB 1
1
degy1 P q then we set B “ P P F rys, by the fact that BP
By y “ By y “ A in
this case, and P ´ Bt0 “ 0 P F rts.
Now suppose that degy1 P “ 1. One must have 2m ą degy1 P, since
`
˘
BP 3
1
degy1 BP
≤ 2, by the equation (9) we must have m “ 1 and
By y ` 2 By 1 y
A P F rys. Let P “ P1 y 1 ` P0 where P1 , P0 P F rys, equation (9) becomes
`
˘
BP 1
BP
y ` 2 1 y 3 “ A y 12 ´ y 4 ,
By
By
`
˘
BP1 12 BP0 1
y `
y ` 2P1 y 3 “ A y 12 ´ y 4 ,
By
By
which gives
BP1
BP0
“ A,
“ 0,
2P1 y 3 “ ´Ay 4 ,
By
By
thus the second equation gives P0 P F, and combining the two others we have
Bp´Ayq
n
1
2 BP
By “ 2A “
By , if A “ An y ` ¨ ¨ ¨ ` A0 we obtain 3An “ ´ pn ` 1q An
so that An “ 0, hence A “ 0 and therefore P1 “ 0, in conclusion P “ P0 P F.
Suppose now that degy1 P ą 1, we make the induction assumption by saying
that the lemma is proved for polynomials of smaller degree in y 1 . Let
Q, R P F ry, y 1 s be such that P “ Q.t ` R where degy1 R ă degy1 t “ 2. We
have
BP
B pQ.t ` Rq 1
B pQ.t ` Rq 3
. BP 1
L pP q “
y ` 2 1 y3 “
y `2
y
By
By
By
By 1
`
˘
ˆ
˙
B y 12 ´ y 4 1
BQ 1
BQ 3
BR 1
BR 3
“
y ` 2 1 y .t `
y ` 2 1y ` Q
y
By
By
By
By
By
`
˘
B y 12 ´ y 4 3
` 2Q
y
By 1
ˆ
˙
BQ 1
BQ
BR 1
BR
“
y ` 2 1 y 3 .t `
y ` 2 1 y 3 ´ 4Qy 3 y 1 ` 4Qy 1 y 3
By
By
By
By
ˆ
˙
BQ 1
BQ
BR 1
BR
“
y ` 2 1 y 3 .t `
y ` 2 1 y3
By
By
By
By
“ L pQq .t ` L pRq ,
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`
˘
thus by equation (9), we have LpRq “ Atm´1 ´ LpQq t, on the other hand,
by the fact that degy1 R ă 2, degy1 LpRq is at most 2.
If degy1 LpRq ≤ 1, then by comparing the degrees of t in the two sides of the
`
˘
equation LpRq “ Atm´1 ´ LpQq t, one must have that Atm´1 ´ LpQq “ 0
BR
and thus BR
By “ By 1 “ 0, in other words R P`F.
˘
If degy1 LpRq “ 2, by equation LpRq “ Atm´1 ´ LpQq t, we must have
Atm´1 ´ LpQq “ ppyq P F rys, thus LpRq “ ppyqt, suppose that ppyq ‰ 0, so
R “ R1 pyqy 1 ` R0 pyq,
BR1 12 BR0 1
LpRq “
y `
y ` 2R1 pyqy 3
By
By
“ ppyqy 12 ´ ppyqy 4
and then
BR1
“ ppyq,
By
BR0
“ 0,
By
2R1 pyqy 3 “ ´ppyqy 4 ,
thus
2R1 pyq “ ´y
BR1
,
By
if
R1 pyq “ an y n ` ¨ ¨ ¨ ` a1 y ` a0 ,
we have
2 pan y n ` ¨ ¨ ¨ ` a1 y ` a0 q “ ´nan y n ´ ¨ ¨ ¨ ´ a1 ,
which gives
p2 ` nq “ 0
Atm´1
and so ppyq “
´ LpQq “ 0; thus we must have in both cases
degy1 LpRq ≤ 1 and degy1 LpRq “ 2 with LpQq “ Atm´1 .
If 2m ą degy1 P then as P “ Q.t ` R, we have that 2 pm ´ 1q ą degy1 Q,
and by the fact that LpQq “ Atm´1 , and using the induction assumption, we
have A “ 0 and Q P F rts, therefore P P F rts.
If 2m “ degy1 P then 2 pm ´ 1q “ degy1 Q and hence by induction asm´1 P F rts
sumption, there exists B P F rys such that A “ y 1 BB
By and Q ´ Bt
m
so P ´ Bt P F rts which completes the proof.
Theorem 4. Let F be a differential field of characteristic zero which
contains an element x such that x1 “ 1. Let y be a PII transcendent i.e.
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y 2 “ 2y 3 ` xy ` α if the transcendence degree of G “ F xyy over F is 2, then
the field of constants of G is C.
Proof. Suppose the contrary, by Lemma 1 there exist P P F ry, y 1 s ´ F and
ε P F such that
˘
BP 1 BP ` 3
(10)
P p1q `
y ` 1 2y ` xy ` α “ εP.
By
By
We consider F ry, y 1 s as a graded ring by assigning to the term by e y 1f the
grade e ` 2f, and we write P “ Pn ` ¨ ¨ ¨ ` P0 , where every term in Pi is
of grade i and Pn ‰ 0. By inspecting the terms in (10) of grade n ` 1, we
BPn 3
12
4
n 1
find that BP
By y ` 2 By 1 y “ 0. By Lemma 2, Pn P F ry ´ y s; because every
term in`Pn is of˘grade n, then we must have n “ 4m for some m P N and
m
Pn “ a y 12 ´ y 4 , where a P F ´ t0u , as m ą 0 we have n ≥ 4.
By examining the terms in (10) of grade n, we find
BPn´1 3
BPn´1 1
y `2
y “ εPn ,
By
By 1
`
˘m BPn´1 1
` 12
˘
BPn´1 3
4 m
a1 y 12 ´ y 4 `
y `2
y
“
εa
y
´
y
,
By
By 1
Pnp1q `
in other words
˘`
˘m
BPn´1 3 `
BPn´1 1
y `2
y “ εa ´ a1 y 12 ´ y 4
1
By
By
“ Atm
as Pn´1 is of grade n ´ 1, so degy1 Pn´1 ≤ 12 pn ´ 1q “ 2m ´ 12 , hence
degy1 Pn´1 ă 2m. By Lemma 1 A “ εa ´ a1 “ 0, thus εa “ a1 .
Now let
`
˘m
Q “ P ´ a y 12 ´ y 4 ´ xy 2 ´ 2α ,
`
˘m
P “ Q ` a y 12 ´ y 4 ´ xy 2 ´ 2α ,
replacing P in the equation (10) we obtain
˘`
`
˘m
`
˘m´1
Qp1q ` a1 y 12 ´ y 4 ´ xy 2 ´ 2α ` ma ´y 2 y 12 ´ y 4 ´ xy 2 ´ 2α
`
˘ `
˘m´1
BQ 1
`
y ` am ´4y 3 ´ 2xy y 1 y 12 ´ y 4 ´ xy 2 ´ 2α
By
˘
BQ `
` 1 2y 3 ` xy ` α
By
`
˘m´1 ` 3
˘
` 2amy 1 y 12 ´ y 4 ´ xy 2 ´ 2α
2y ` xy ` α
`
˘m
“ εQ ` εa y 12 ´ y 4 ´ xy 2 ´ 2α ,
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using the fact that εa “ a1 one have
˘
BQ 1 BQ ` 3
y ` 1 2y ` xy ` α
By
By
`
`
˘
`
˘˘
“ εQ ` ´may 2 ` am ´4y 3 ´ 2xy y 1 ` 2amy 1 2y 3 ` xy ` α ˆ
` 12
˘m´1
y ´ y 4 ´ xy 2 ´ 2α
Qp1q `
thus
Qp1q `
˘
BQ 1 BQ ` 3
y ` 1 2y ` xy ` α
By
By
`
˘`
˘m´1
“ εQ ` am 2αy 1 ´ y 2 y 12 ´ y 4 ´ xy 2 ´ 2α
.
Write Q “ Ql ` ¨ ¨ ¨ ` Q0 where each term in Qi is of grade i, by definition
of Q we must have l ă n “ 4m. By examining the terms of same grade we
obtain
BQl 1
BQl
y ` 2 1 y3 “ 0
if l ą 4m ´ 2,
By
By
`
˘`
˘m´1
BQl 1
BQl
y ` 2 1 y 3 “ am 2αy 1 ´ y 2 y 12 ´ y 4
if l “ 4m ´ 2,
By
By
`
˘`
˘m´1
0 “ am 2αy 1 ´ y 2 y 12 ´ y 4
if l ă 4m ´ 2.
In the first case, using Lemma 1, we come to the conclusion that Ql P
F ry 12 ´ y 4 s, thus l must be divisible by 4 which is absurd.
The second case gives that degy1 Ql ≤ 12 l “ 2m ´ 1, the case degy1 Ql “
2m ´ 1 can not hold (the right and left hand sides of the equation above
wont be homogeneous) so we must have degy1 Ql ≤ 2m ´ 2 “ 2 pm ´ 1q
`
˘
by Lemma 2, since am 2αy 1 ´ y 2 ‰ 0, there exists B P F rts such that
`
˘
am 2αy 1 ´ y 2 “ y 1 BB
By , which is impossible.
The third case is also absurd.
The contradiction complete the proof.
Theorem 5. Let G “ F xyy be an extension of F of transcendence degree 2,
where y is a solution of Painlevé equation PII, i.e. y 2 “ 2y 3 ` xy ` α, then
G is not contained in any strongly normal extension of F.
Proof. The proof is completely similar to the one concerning PI transcendent
given in [10].
4. Conclusion
In this paper, we expanded a work done by Kolchin in the case of Painlevé
transcendent PI to the case of Painlevé transcendent PII, we came to the
conclusions that if F is a differential field of characteristic zero and if y is
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a PII transcendent of differential transcendence degree 2 over F ; then the
subfield of constants of F is equal to the one of F xyy. But even in this case
F xyy is not included in any strongly normal extension of F .
Strongly normal extensions, algebraic dependence over initial conditions,
elementary solutions and transcendence degree are closely related as showed
in [15], [18], [19], and [20], and moreover, in the case of Painlevé equation
PII, it is showed that the transcendence of a solution can be ”measured” by
inspecting its transcendence degree, which is not always true for differential
equations.
To prove Theorem 4 in the case of the PII, we used the essential fact that
F ry, y 1 s can be seen as a graded ring, and y 2 “ p2y 3 ` xy ` αq P F ry, y 1 s, but
in the case of the PIII, we have y 2 “ p y1 py 1 q2 ´ x1 y 1 ` αy 3 ` x1 pβy 2 ` γq ` yδ q R
F ry, y 1 s, so this fact make the proof fails in the case of PIII. By the above
discussion, and because the right hand side in equations (3), (4), (5) and (6)
is rational and not polynomial in y, y 1 , the method used to prove Theorem
4 in the case of PII is inadequate for PIII,. . . , PVI. So one have to find an
adapted method to the rational structure of Painlevé equations PIII, PIV,
PV, and PVI, rather than the algebraic one of PI and PII.
References
[1] L. Brenig, A. Goriely, Painlevé analysis and normal forms, Computer Algebra and
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equation, Journal Reine Angew. Math. (Crelles Journal), 2011.
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Painlevé equation PII
397
[14] K. Nishioka, A note on the transcendency of Painlevé’s first transcendent, Nagoya
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S. E. Miri
LABORATOIRE D’ANALYSE NON LINÉAIRE ET MATHÉMATIQUES APPLIQUÉES
BP 119 TLEMCEN 13000
and
FACULTÉ DE TECHNOLOGIE,
DÉPARTEMENT GEE
BP 230 TLEMCEN 13000,
UNIVERSITÉ DE TLEMCEN,
ALGERIE
E-mail: [email protected]
Received October 21, 2014; revised version July 15, 2015.
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