Download Linear grading function and further reduction of normal forms

Linear grading function and further reduction
of normal forms
Hiroshi Kokubu
Department of Mathematics
Kyoto University
Kyoto 606-01, Japan
Hiroe Okay
Department of Applied Mathematics and Informatics
Faculty of Science and Technology
Ryukoku University
Seta, Otsu 520-21, Japan
[email protected]
and
Duo Wangz
Department of Applied Mathematics
Tsinghua University
Beijing 100084, P. R. China
[email protected]
March 6, 1996
1
Abstract
In this note a kind of new grading functions is introduced, the
denition of N -th order normal form is given and some sucient conditions for the uniqueness of normal forms are derived. A special case
of the unsolved problem in Baider and Sanders paper for the unique
normal form of Bogdanov-Takens singularities is solved.
Keywords: grading function, normal form.
Research was supported in part by Grant-in-Aid for Scientic Research (No.
07740150), Ministry of Education, Science and Culture, Japan.
y Research was supported in part by Science and Technology Fund for Research Grants
of Ryukoku University.
z Research was supported in part by NNSF of China
2
1 Introduction
Normal forms are basic and powerful tools in bifurcation theory of vector
elds. But classical normal form theory, known as Poincare's normal form
(see, e.g., Arnold Ar]), may not give the simplest form since only linear
parts are used for simplifying the nonlinear terms, and hence one can not
apply Poincare normal form theory to vector elds whose linear parts are
identically zero. On the other hand, classical normal forms are not unique
in general. In order to get unique normal forms so that formal classication
could be made, further reduction of the classical normal forms is necessary
and the concept of normal forms should be rened.
Many authors have discussed the further reduction of normal forms and
some of them have discussed uniqueness of normal forms, see, e.g., SM] and
references therein. Ushiki Us] introduced a systematic method by which
nonlinear parts are also used to simplify higher order terms. By classical
method of normal form theory, only one Lie bracket is used to simplify the
higher order terms. Ushiki's method allows more Lie brackets for the simplication CK]. They obtained unique normal forms (simplest normal forms)
up to some degree for some given vector elds. Wang Wa] gave a method
to calculate coecients of normal forms, which needs more parameters in
the transformations due to the non-uniqueness of transformations and hence
may give simplest normal forms (up to some nite order) by suitably choosing
parameters. In fact nonlinear terms play also role in the reduction. Baider
introduced special form Ba], which is in fact unique normal form, in an
abstract sense. Baider and Sanders BS1] introduced new grading functions
to get further reduction of normal forms. They introduced the concept of
n-th order normal form related with the n-th grading function and give the
denition of innite order normal forms (which is unique). They gave unique
normal forms for some nilpotent Hamiltonians. Then they got unique normal forms for some cases of Bogdanov-Takens singularities (BS2]). But some
cases are still unsolved. Results concerning uniqueness of normal forms for
some other cases can be found in BC2] and SM].
In this paper we rst introduce the concept of linear grading function
in section 2, and we give a systematic method to dene some new grading
functions. Then in section 3 we dene n-th order normal forms, in which
we combine Ushiki's method and Baider and Sanders' method. In fact we
need only one grading function. But the n-th order normal forms related
3
to n Lie brackets in the computation. In section 4 we dene innite order
normal form and prove that the innite order normal form must be unique.
In section 5 we give a sucient condition for unique normal forms. Finally
in section 6 we prove the uniqueness of rst order normal form of the special
case = 2 = 1 of Bogdanov-Takens singularities which is a special case of
the unsolved problem in BS2].
2 Linear grading function
Let H be the linear space of all n dimensional real or complex formal vector
elds. We dene a bilinear operator ] : H H H by u v] = Du v Dv u
for any u v H . Then H ] forms a Lie algebra. Now let us dene a
\grading function" such that H ] is a graded Lie algebra.
For the purpose of computing normal forms of formal vector elds, the
\grading function" should satisfy the following properties:
(1) The degree of any monomial is dened to be an integer. The dimension
of the linear space Hk spanned by all monomials of degree k is nite
for any integer k (in the case when there is no monomials of degree k
for some integer k we dene Hk = 0 )
(2) Hm Hn] Hm+n for any integers m n
(3) The grading function should be bounded below.
Let
2
f
!
;
g
f
g
f g
Dn =
(Y
n
i=1
xlii ej
j
li
+
2 Z x
)
i
2
R
(or C ) i j = 1 ::: n where ej is the j -th standard unit vector in Rn (or C n ). Consider the function
: Dn Z dened by
!
Y
n
i=1
xili ej
! X
n
=
i=1
aij li + dj (1)
where aij dj Z i j = 1 ::: n. From the denition of , it is obvious that
condition 1. for a grading function is satised. Now we look for conditions
such that the function dened by (1) satises all other conditions of grading
functions.
2
4
Lemma 2.1 The function de ned by (1) is bounded below if and only if
all faij g are nonnegative integers.
Lemma 2.2 Let the function be de ned by (1) with all coecients aij
f
g
and fdj g nonnegative and Hk be the linear space spanned by all monomials
in ;1 (k). Then dim(Hk ) is nite(or zero) for any integer k if and only if
all faij g are natural numbers.
Lemma 2.3 Let the function be de ned by (1) and Hk be de ned as in
Lemma 2.2. Then Hm Hn] Hm+n if and only if
ai1 = ::: = ain = di for any i = 1 ::: n:
n
n
Let u = Q xli ej v = Q xli ek and (u) = m (v) = n. Then
;
Proof.
i=1
0
i
i=1
i
u v] = Du v Dv u
n
n
n
= xl Q xli xl Q xli Q xli ek
i=1
i=1 n i=1
n l +l
Q
l
= x xi ej xl Q xli +l ek :
;
i
k
k
k
k
i
i=1
0
;
0
j
i
0
i
(2)
i
j
0
;
k
k
i
i=1
0
i
We rst assume that ai1 = ai2 = ::: = ain = di i j = 1
n. Then
;
and
Qn l +l Pn
1
0
x i=1 xi ej = i=1 aij (li + li ) akj + dj
n
n
= P aij li + dj + P aik li0 + dk = m + n
i
0
i
;
k
i=1
i=1
Qn l +l n
x1 xi ek = P aik (li + li0 ) ajk + dk
i=1
i=1n
n
= P aik li + dk + P aij li0 + dj = n + m:
i
0
i
;
j
i=1
i=1
Hence, from(2), we have u v] Hm+n. Note that the operator ] is bilinear. Therefore Hm Hn] Hm+n.
Conversely, we suppose that Hm Hn] Hm+n holds for any integer m n.
For any k N, we x a u Hm with lk > 0. Then from (2) we have
2
2
n
X
i=1
aij (li
2
+ l0 ) ; a
i
kj + dj
=
X
n
i=1
5
! X
n
aij li + dj +
i=1
aik l0 + d
i
k
!
:
Hence
n
X
i=1
i. e.
(aik aij )li0 + dk + akj = 0
;
dk + akj =
(
n
X
i=1
k where l
If we take li0 = l0 ii =
= k
6
(aij aik )lk0 :
2
;
N,
then
akj akk = dk +l akj :
(3)
;
Letting l + , we have akj = akk . Note that j is arbitrary. Therefore
ak1 = ::: = akk = ::: = akn , and hence from (3), dk = ak1 follows.
!
1
;
Denition 2.4 Let
Dn =
(Y
n
i=1
xlii ej
j
li
+
2 Z x
i
2
)
R (or C ) i j = 1 ::: n where ej is the j -th standard unit vector in Rn (or C n ). Then the function
: Dn ! Z de ned by
Y
n
i=1
xlii ej
! X
n
=
i=1
ai li aj ;
(4)
where ai 2 N i = 1 ::: n, is called a linear grading function.
Remark 2.5 1)Suppose that is a linear grading function de ned by a set
of natural numbers fai g. If fai g has a common factor c, then the function 1c is also a linear grading function. So we will always assume that any linear
grading function is de ned by a set of coprime natural numbers fai g.
2)Any linear grading function satises
(xi ei ) = 0 for all i = 1 ::: n:
Hence for any grading function , min
p (p) 0.
3)If the linear grading function is dened by a set of successive natural
numbers a1 ::: an , then for k 1 n, dimHk 1.
f
g
8
;
6
Qn
= P li 1, i.e. a1 = ::: = an = 1, then is
i=1
an linear grading
function.
Note
that
the classical de nition of the degree of
Q xl e is Pn l and hence the grading function de ned above shifts by 1 w.r.t
j
i
i=1 i
i+1
the classical grading.
Example 2.6 If xlii ej
i=1
n
;
i
Example 2.7 In D2 =
Q2
xli ej i=1 i
xl1 xl2 e
1 2
j = 1 2 , de ne
( 2l1 + 3l2 2 j = 1
j = 2l + 3l
1
2 3 j = 2:
;
;
Then (x2 e1) = (x21 e2) = 1. Note that x2e1 is a linear term and x21e2 is a
nonlinear term in the classical sense.
3
N -th
order normal forms
Let be a linear grading function and Hk be the linear space spanned by
all monomials of degree k. Consider a formal vector eld V dened by the
following formal series
X = X + X +1 + ::: + X +k + :::
where Xk Hk k
and denote it as
2
(5)
and X = 0. We call (5) a zeroth order normal form
6
V (0) = V (0) + X +1 + ::: + X +k + ::::
(6)
We may assume that X is already in some simple or satisfactory form(e.g.
X may have been changed to simpler form by classical normal theory).
Let Yk Hk and Y be its time one mapping given by the ow tY
generated from the vector eld corresponding to the equation x_ = Yk (x) x
Rn . Then the transformation y = Y (x), which is a near identity change of
variables, brings (5) to
2
k
k
2
k
Y X = exp(adYk )X
= X + (adYk )X + ::: + n1! (adYk )n X + :::
k
7
where (adYk )X = Yk X ] and (adYk )n = (adYk )n;1 (adYk ) n = 2 3 :::.
For any k N, dene an operator
2
L(1)
k : Hk
!
H +k : Yk
7!
Yk V (0) ]:
(7)
(1)
(0)
It is obvious that L(1)
k is linear. Note that Lk depends on V and can be
(1) (0)
denoted by L(1)
k = Lk V ].
Denition 3.1
V = V + V +1 + ::: + V +k + :::
is called a rst order normal form, if
V +k N (1)+k k = 1 2 :::
2
(1)
(1)
where N (1)+k is a complement subspace to ImL(1)
k in H +k and Lk = Lk V ].
It is easy to see that there is a sequence of near identity formal transformations such that (5) is transformed into a rst order normal form which is
called the rst order normal form of (5) and can be denoted by
V (1) = V (1) + V (1)+1 + ::: + V (1)+k + ::::
(8)
Note that V (1) = V (0) .
In order to make further reduction of a rst order normal form, we dene
a sequence of linear operators L(km) m k = 1 2 3 as follows. Let
V = V + V +1 + V +2 +
+ V +k +
be a formal series, where Vm Hm for each m
. Then we dene
(1) V ] by (7) for any k N if L(m) = L(m) V V V
L(1)
=
L
+1
+m;1]
k
k
k
k
(
m
is dened already for an m 1 and any k N, then we dene Lk +1) =
Lk(m+1)V V +1 V +m] by
2
2
2
L(km+1) : KerL(km) Hm+k H +m+k : ((Yk Yk+1 ::: Yk+m;1) Yk+m)
Yk V +m] + ::: + Yk+m;1 V +1] + Yk+m V ]:
!
8
7!
Remark 3.2 By de nition, it is obvious that
KerL(km) = (Yk Yk+1 ::: Yk+m;1) Hk ::: Hk+m;1
Yk V ] = 0
Yk+1 V ] + Yk V +1] = 0
f
2
j
...
Yk+m;1 V ] + ::: + Yk V +m;1] = 0 :
g
Denition 3.3 A formal vector eld
where Vm 2 Hm for each m , is called an N -th order normal form, if
V +i N (i+) i (1 i N 1)
2
;
and
V +j N (N+j) (j N )
where N (m+k) is a complement to the image of L(km;)m+1V V +1
in H +k for each m 1 and k 1.
2
V +m;1]
Theorem 3.4 For any N
any formal vector eld can be changed by
a sequence of near identity formal transformations to an N -th order normal
form.
Proof.
2
N,
Consider a formal vector eld(a zero-th order normal form)
V (0) = V (0) + X +1 + ::: + X +k + ::::
(9)
(1) (0)
Dene linear operator L(1)
1 = L1 V ] and let
H +1 = ImL(1)
N (1)+1:
1
Then there is a polynomial Y 1 = Y1(1) H1 such that (9) is converted to
2
V (1) = exp(adY 1 )V (0) = V (0) + V (1)+1 + X (1)+2 +
9
(10)
where V (1)
+1
and let
2
(2) (0) (1)
N (1)+1 . Then we dene linear operator L(2)
1 = L1 V V +1]
H +2 = ImL(2)
N (2)+2:
1
Then there is a polynomial Y 2 = Y1(2) + Y2(2) , where Y1(2)
Y2(2) H2 such that (10) is converted to
2
KerL(1)
1 and
2
V (2) = exp(adY 2)V (1) = V (0) + V (1)+1 + V (2)+2 + X (2)+3 +
(11)
(2)
where V (2)
+2 N +2 . Then step by step, for each m = 2 3 N , we dene
a linear operator L(1m) = L(1m) V (0) V (+mm;1)
;1] and then nd a polynomial
(
m
)
(
m
)
(
m
)
m
(
m
)
Y = Y1 + +Ym , where (Y1 Ym;1) KerL(1m;1) and Ym(m) Hm
such that
V (m) = exp(adY m)V (m;1)
= V (0) + + V (+mm) + X (m+)m+1 + 2
2
2
where V (+k)k N (k+)k for k = 1 m, and where N (k+)k is a complement to
ImL(1k) in H +k . Furthermore for V (N ) (denoted also as V (N 1) ) and for each
k = 2 3 , we consider linear operator L(kN ) = L(kN ) V (0) V (+NN;1);1], and
nd Y Nk = Yk(N ) + + Yk(+NN) ;1 where (YkN YkN+N ;2 ) KerL(kN ;1) and
Yk(+NN) ;1 Hk+N ;1 such that
2
2
2
V (Nk) = exp(adY Nk )V (N(k;1))
= V (0) + + V (+NN) + V (+NN) +1 +
+ V (+Nk)+N ;1 + h:o:t:
(12)
where V (+NN) +j N (N+N) +j for each j 1 and where N (N+N) +j is a complement
to ImL(jN+1) in H +N +j . Now the sequence of time one mappings dened by
the sequence of polynomial vector elds Y 1 Y N (= Y N1 ) Y N 2 change
the given vector elds to an N -th order normal form.
2
In what follows, we may always assume that all linear operators L(km) are
dened by the same sequence of homogeneous polynomials V V +1 .
Lemma 3.5
(0 Yk+1 ::: Yk+m;1) KerL(km)
2
()
10
(Yk+1 ::: Yk+m;1) KerL(km+1;1)
2
Lemma 3.6
ImL(km+1) ImL(km+1)
Proof.
8
k m 1:
Note that
ImL(km+1) = X +m+k
(Yk+1 ::: Yk+m;1) KerL(km+1;1)
and Yk+m Hk+m such that
Yk+1 V +m;1] + ::: + Yk+m V ] = X +m+k :
Taking Yk = 0. From Lemma 3.5, if such that
f
j
9
2
2
g
Yk+1 V +m;1] + ::: + Yk+m V ] = X +m+k then (0 Yk+1 ::: Yk+m;1) KerL(km) and
2
0 V +m] + Yk+1 V +m;1] + ::: + Yk+m V ]
= Yk+1 V +m;1] + ::: + Yk+m V ]
= X +m+k :
Hence X +m+k ImL(km+1).
2
Corollary 3.7
dim N (m+k+1)
+m
dim N (m+k) +m
k m
8
2
N:
Remark 3.8 It is reasonable to set
(m)
N (m+k+1)
+m N +k+m 8
k m
2
N:
Remark 3.9 It is obvious that for a given formal vector eld its N -th order
normal form is simpler than its m-th order normal form if m N .
11
4 Unique normal forms
Denition 4.1
V = V + V +1 + ::: + V +m + :::
is called an innite order normal form, if V +m N (m+m) for m N, where
N (m+m) is a complementary subspace to ImL(1m) in H +m and where L(1m) =
L1(m) V V +1 ::: V +m;1] for m N.
2
8
8
2
2
Though in general we have innitely many choices for the complementary
space to the image of L(1m) in H +m, in what follows, we assume that the
choice of the complementary space N (m+m) to ImL(1m) is xed.
Theorem 4.2 Let
V = V + V +1 + ::: + V +m + :::
and
W = V + W +1 + ::: + W +m + :::
be both in nite order normal forms. If there exists a formal series Y =
8
2 N) such that (Y ) V = W , then
2
Y1 + Y2 + ::: + Ym + ::: with Ym Hm ( m
V +m = W +m
m
8
2
N:
Proof. Suppose it would not be the case. Then there exists an m
such that
V +k = W +k (1 k m 1) V +m = W +m :
;
6
Recalling
W = exp(adY )V
1
= V + Y V ] + 2!1 Y Y V ]] + ::: + n! Y ::: Y V ]:::] + :::
we have
Wk = Vk + Y V ]k + 2!1 Y Y V ]]k + ::: + n1! Y ::: Y V ]:::]k + :::
12
2
N
where Y V ]k = Y V ] Hk . Similarly for Y ::: Y V ]:::]k . Notice that
this innite sum has in fact nitely many nontrivial terms, and hence the
summation is well-dened. Therefore we have
1
Y V ]k + Y Y V ]]k + ::: + 1 Y ::: Y V ]:::]k + ::: = 0
(13)
2!
n!
for +1 k
+ m 1. It is easy to see that if Y V ] = 0 and if the lowest
degree of terms in Y V ] is l, then the lowest degree of terms in Y ::: Y V ]:::]
with n-fold bracket operations is l + n 1. Hence from (13), we have
\
;
6
;
Y V ] +1 = 0
1
Y V ] +2 + Y Y V ]] +2 = 0
2!
...
Y V ] +m;1 + 1 Y Y V ]] +m;1 + ::: + 1 Y ::: Y V ]:::] +m;1 = 0:
2!
m!
By induction, we have
Y V ] +1 = Y V ] +2 = ::: = Y V ] +m;1 = 0
and therefore
Y2 V ] + Y1 V +1] = 0
...
Ym;1 V ] + Ym;2 V +1] + ::: + Y1 V +m;2] = 0
namely, from Remark 3.2,
(Y1 Y2 ::: Ym;1) KerL(1m;1) :
2
Thus
W +m = V +m + Y V ] +m :
Note that
Y V ] +m = Y1 V +m;1] + ::: + Ym V1]
This means Y V ] +m ImL(1m) .
2
13
On the other hand, V +m and W +m are both in the same complementary
space N (m+m) to ImL(1m) . Therefore
W +m V +m = Y V ] +m N (m+m) ImL(1m) = 0 ;
2
\
f g
and hence W +m = V +m. This contradiction shows that the conclusion of
the theorem is true.
Corollary 4.3 If
V (N ) = V (0) + V (1)+1 + ::: + V (+NN) + V (+NN+1)+1 + :::
and
W (N ) = V (0) + W (1)+1 + ::: + W (N+N) + W (N+N+1)+1 + :::
are both N -th order normal form of (5), then
V (+k)k = W (k+)k k = 1 ::: N:
5 A special case
In this section we assume that all linear operators L(km) are based one the
same sequence of polynomials V V +1 .
Proposition 5.1 If there exists an N N such that
2
holds, then
KerL(kN +1) = | 0
f g
and
:::
{z
m
0 } KerL(kN+)m f g
ImL(kN +m+1) = ImL(kN++1)
m 14
8
k
2
8
N:
k m
2
N
(14)
(15)
Proof. By assumption, (14) with m = 1 apparently holds. Suppose we
have X +k+N +1 2 ImL(kN +2), namely, there exists (Yk ::: Yk+N +1) satisfying
(Yk ::: Yk+N ) KerL(kN +1) Yk+N +1 Hk+N +1
Yk V +N +1] + ::: + Yk+N +1 V ] = X +k+N +1:
From assumption, Yk = 0 and (Yk+1 ::: Yk+N ) KerL(kN+)N . Hence
Yk+1 V +N ] + ::: + Yk+N +1 V ] = X +k+N +1:
2
2
2
This implies that X +k+N +1 ImL(kN+1+1) and hence ImL(kN +2) ImL(kN+1+1) .
Conversely, if we assume X +k+N +1 ImL(kN+1+1) , namely, there exists
(Yk+1 ::: Yk+N +1 ) satisfying
2
2
(Yk+1 ::: Yk+N ) KerL(kN+1) Yk+N +1 Hk+N +1
Yk+1 V +N ] + ::: + Yk+N +1 V ] = X +k+N +1:
Then taking Yk = 0, it holds that (Yk ::: Yk+N +1) KerL(kN +1) and apparently
Yk V +N +1] + Yk+1 V +N ] + ::: + Yk+N +1 V ] = X +k+N +1:
2
2
2
This implies X +k+N +1
Therefore
2
ImL(kN +1) , and hence ImL(kN+1+1)
ImL(kN +2).
ImL(kN +2) = ImL(kN+1+1)
namely (15) holds for m = 1.
Suppose (14) and (15) hold for a xed m 1. Let (Yk ::: Yk+N +m )
KerLk(N +m+1). Then
Yk V +N +m] + ::: + Yk+N +m V ] = 0:
(16)
2
Note that (Yk ::: Yk+N +m;1) KerL(kN +m) . By induction hypothesis, Yk =
0 ::: Yk+m;1 = 0and(Yk+m Yk+m+N ;1) KerL(kN+)m . Hence from (16),
2
2
L(kN++1)
m (Yk+m ::: Yk+N +m ) = 0
or in other words,
(Yk+m ::: Yk+N +m ) KerL(kN++1)
m :
2
15
By assumption, Yk+m = 0 and (Yk+m+1 ::: Yk+m+N ) KerL(kN+)m+1. Hence
2
KerL(kN +m+1) | 0
f g :::
{z
0 } KerL(kN+)m+1
f g
m+1
8
k
2
N:
Conversely, take Yk = 0 ::: Yk+m = 0 and (Yk+m+1 ::: Yk+m+N )
Then
Lk(N+)m+1.
2
Ker
L(kN +m+1)(Yk ::: Yk+m+N )
= Yk+m+1 V +N ;1] + ::: + Yk+m+N V ]
= 0
namely,
|0
f g :::
{z
0 } KerL(kN+)m+1
f g
m+1
Therefore (14) holds for m + 1 and for any k
be proved as in the case m = 1.
2
KerL(kN +m+1) :
N.
In a similar way, (15) can
Corollary 5.2 If there exists an N such that
KerL(kN +1) = 0
f g KerL(kN+1) k
8
2
N
then an (N +1)-th order normal form must be an in nite order normal form.
Proof.
From Proposition 5.1, we have
ImL(kN +m+1) = ImL(kN++1)
m k m
8
2
N:
Hence we may set
(N +1)
N (N+k++mN+1)
+m = N +k+N +m
as complementary subspaces to ImL(kN +m+1) for k m N. Thus, for any
m > N + 1,
+(m;N ;1)+1) =
N (m+m) = N (N+1+
N +(m;N ;1)
which implies that, if V (+Nm+1) N (N+m+1), then V (+Nm+1) N (m+m) for any m
N + 1. The conclusion thus follows.
8
2
2
2
16
such that ImL(kN +m) = ImL(kN+)m for
any k m 2 N, then the N -th order normal form is an in nite order normal
form.
Example 5.4 If KerL(1)
k = f0g 8k 2 N, then a rst order normal form is
also an in nite order normal form, and hence it is unique normal form of
the original equation.
Corollary 5.3 If there exists an N
2
N
6 The Bogdanov-Takens normal form: the
case = 2 = 1
Baider and Sanders BS2] gave unique normal forms for cases < 2 and
> 2 of Bogdanov-Takens singularities. But the case = 2 is still
unsolved. In this section we consider a special case, i.e., = 2 = 1. By
using our method introduced above we give the unique normal form for this
case.
We consider the following equation:
x_ = y + a11xy + a02y2 + O(3)
(17)
y_ = xy + x3 + b02y2 + O(3)
where = 0.
Dene : D2 Z by
mn!
!
x
y
0
0
= m + 2n 1
xm yn = m + 2n 2:
Then is a linear grading function with
! ! !
2!
0
y
0
0 = xy = x3 = 1 and x0 = 1:
Let
!
y
(0)
V1 = xy + y3 :
6
!
;
;
Then the equation (17) can be written as
V (0) = V1(0) + V2(0) + : : : + Vm(0) + : : :
where Vm(0) Hm m = 1 2 .
2
17
(18)
Lemma 6.1 The following vectors form a basis of the space Hm:
For m = 2k + 1,
! ! 0
0
x2k+3 x2k+1y : : : x3yk
2k+2 ! 2k !
2k
x
x
y
0 : : : x 0y
0
0
For m = 2k + 2,
! !
0
xyk+1 ! k+1 !
y0 ! !
! !
0
0
0
x2k+4 x2k+2y : : : x2yk+1 yk+2 2k+3 ! 2k+1 !
! !
x y x y : : : x3yk xyk+1 :
0
0
0
0
0
In particular, dim Hm = m + 3.
Lemma 6.2
"
!
myn
x
= mxm;1yn+1 + (n 1)xm+1yn + nxm+3yn;1 xmyn
" m n !
#
!
x y V (0) = mxm;1yn+1 + nxm+1yn + nxm+3yn;1 :
1
0
xmyn+1 3xm+2yn
0
!
V1(0)
#
;
;
;
Lemma 6.3 Let
Y2k+1 =
Then
Y2k+1 V1(0) ]
=
kX
+1
i=0
ai
kX
+1
;
! kX
+1 x2k+2;2iyi !
:
x2k+3;2iyi + i=0 bi
0
0
!
0
x2k+3;2iyi+1 (2k + 3 2i)ai + iai+1
+(i + 2)ai+2 bi 3bi+1
k x2k+1;2iyi+1 !
X
+
(2k + 2 2i)bi + (i + 1)bi+1
0
f
i=;1
;
;
;
f
i=;1
+(i + 2)bi+2 ai+1 where ai = 0 bj = 0 if i j < 0 or i j > k + 2.
;
18
g
g
;
Lemma 6.4 Let
Y2k+2 =
kX
+2
ai
i=0
Then
Y2k+2 V1(0) ]
=
kX
+1
! kX
+1 x2k+3;2iyi !
:
0
x2k+4;2iyi + i=0 bi
0
!
0
x2k+3;2iyi+1 (2k + 4 2i)ai + iai+1
+(i + 2)ai+2 bi 3bi+1
kX
+1 x2k+1;2iyi+1 !
+
(2k + 3 2i)bi + (i + 1)bi+1
0
f
i=;1
;
;
;
g
f
i=;1
;
+(i + 2)bi+2 ai+1 ;
g
where ai = 0 bj = 0 if i j < 0 or i j > k + 2.
Using these results, we have a matrix representation for the adjoint opera(1)
Hm+1 : Ym Ym V1(0) ]. Hence
tor ad(V1(0) ),i.e.,L(1)
m . Note that Lm : Hm
the matrix representation L of L(1)
m is given by an (m + 4) (m + 3)-matrix.
Let
!
L
L
1
2
L= L L :
!
7!
3
4
Thenh the
i submatrix L3 is such that L3 is the identity matrix of the size
l = m2 + 2, where q] stands for the integer part of q The other three
submatrices are (almost) tri-diagonal matrices. More specically, they are
given as follows:
For m = 2k + 1,
;
0 B
2
k
+
3
0
B
B
B
2k + 1
B
B
L1 = B
B
B
B
B
B
@
;
0
2
3
... ...
...
0
19
...
...
3
1
0 CCC
CC
CC
CC (k + 1) C
C
A
k C
1
0 3
B
B
B
B
B
B
L2 = B
B
B
B
B
B
@
;
;
3
;
;
;
;
0
0
0
B
B
2k + 2 B
B
2k
B
L4 = B
B
B
B
B
@
0
For m = 2k + 2,
1
0 CCC
CC
3
CC
... ...
CC ... ...
CC
A
3 C
1
0
2
0 CCC
CC
2 3
CC :
... ... ...
C
. . . . . . (k + 1) C
CA
2 (k + 1)
0
BB 2k + 4 0
BB
2k + 2
B
L1 = B
BB
BB
@
;
;
0
2
3
2k 2
...
0
0 3
B
B
B
B
B
B
L2 = B
B
B
B
B
B
@
;
;
3
;
;
;
3
... ...
...
;
1
0 CCC
CC
CC ...
C
. . . (k + 2) C
CA
2 (k + 1)
1
0 CCC
CC
CC
CC ...
CC
A
3 C
;
0
;
;
20
0
0
B
B
2k + 3 B
B
2k + 1
B
L4 = B
B
B
B
B
@
1
0
2
0 CCC
CC
2 3
CC :
... ... ...
C
. . . . . . (k + 1) C
CA
0
1 (k + 1)
Remark 6.5 In order to simplify the expression, we may assume that = 1
since we may make a suitable linear change of variables x y t in the equation
(17) such that the coecient of xy is changed to 1 and the coecient of x3
is changed to =2 accordingly.
Lemma 6.6 For both m = 2k + 1 and m = 2k + 2, the rst k + 3 rows of
matrix L can be reduced to the form
0 M~
by a suitable row transformation. Here the matrix
M~ = (Mij ) (;1 i k + 1 0 j k + 1)
be given, using = 1, as follows:
Case I] For m = 2k + 1,
Mii;1 =
(2k + 3 ; 2i)(2k + 4 ; 2i)
(i = 1 : : : k + 1)
Mii =
(4k + 5 ; 4i)i ; 1
(i = 0 : : : k + 1)
Mii+1 = i(i + 1) + f(4i + 6)k ; (4i + 3)ig
(i = ;1 : : : k)
Mii+2 =
2(i + 1)(i + 2)
(i = ;1 : : : k ; 1)
Mii+3 =
(i + 2)(i + 3) 2
(i = ;1 : : : k ; 2)
and the other entries are all zero.
Case II] For m = 2k + 2,
Mii;1 =
(2k + 4 ; 2i)(2k + 5 ; 2i) (i = 1 : : : k + 1)
Mii =
(4k + 7 ; 4i)i ; 1 (i = 0 : : : k + 1)
Mii+1 = i(i + 1) + f(4i + 2)k ; (4i2 + 3i ; 9)g
(i = ;1 : : : k)
Mii+2 =
2(i + 1)(i + 2) (i = ;1 : : : k ; 1)
Mii+3 =
(i + 2)(i + 3) 2 (i = ;1 : : : k ; 2)
and the other entries are all zero.
21
For convenience, we denote Mii;1 = ai Mii = bi , Mii+1 = ci + di , Mii+2 =
ei mii+3 = fi 2 for both cases.
Lemma 6.7 If is not an algebraic number, then
KerL(1)
m = 0 m
f g
8
2
N:
To show the lemma, it is sucient to show that
det M = 0
6
where M = (Mij )0ij k+1 is a submatrix of M~ .
First we consider the case II.
Lemma 6.8 In the case II, we have
det M =0 = 0
j
Proof.
6
Let Dl be the following subdeterminant:
Dl = det (Mij =0 )1ij l :
j
Then it is easy to see that
det(M =0 ) = ( 1) Dk+1 :
j
;
By induction we can show
k + 1)!! :
Dl = (l(2+k1)!(2
+ 1 2l)!!
;
In fact, it is true for l = 1 and 2. Since M =0 takes a tri-diagonal form, we
have
Dl+1 = bl+1 Dl cl al+1 Dl;1:
Therefore, using
j
;
bl+1 = (4k + 3 4l)(l + 1) 1
al+1 = (2k + 2 2l)(2k + 3 2l)
cl = l(l + 1)
;
;
22
;
;
we have
k + 1)!!
Dl+1 = (4k + 3 4l)(l + 1) 1 (l(2+k1)!(2
+ 1 2l)!!
l(l + 1)(2k + 2 2l)(2k + 3 2l) (2lk!(2+k3+ 1)!!
2l)!!
(l + 1)!(2k + 1)!!
=
(4k + 3 4l)l + (4k + 2 4l) l(2k + 2 2l)
(2k + 1 2l)!!
(l + 1)!(2k + 1)!! (l + 2)(2k + 1 2l)
=
(2k + 1 2l)!!
(l + 1) + 1 !(2k + 1)!!
=
:
2k + 1 2(l + 1) !!
f
;
;
;
g
;
;
f
;
;
;
;
;
;
g
;
;
f
;
g
f
;
g
The conclusion is thus obtained.
It follows that
det(M =0 ) = Dk+1 = (k + 2)!(2k + 1)!! = 0
j
;
;
6
and hence the lemma is proved.
Next we consider the case I. We introduce the same subdeterminant
Dl = det (Mij =0 )1ij l
j
for this case as well. Again, by induction, we can show
Lemma 6.9 For the case I, we have
k!! :
Dl = ((2l +k 1)!2
2l)!!
;
From the lemma, we have
det(M =0 ) = Dk+1 = 0
j
;
and hence we cannot conclude det M = 0 immediately. We therefore di erentiate det M with respect to and will show that
6
@ det M = 0:
=0
@
j
23
6
Let M (l) be the matrix given by di erentiating the l-th column of the
matrix M with respect to . It thus takes the following form:
0
BB a01
BB
B
M (0) =0 = B
BB
BB
B@
j
1
6k 4
0
... ...
...
...
... ...
... ...
b1 c1
0
0 C
C
CC
CC
CC CC
ck C
A
0
ak+1 bk+1
0 1 0
0
B
a1 b1 c1
B
B
... ... ...
B
B
0
B
B
.
.
.
.
.
.
B
. .
.
B
B
al;1 bl;1 cl;1
M (l) =0 = B
B
B
0 0 dl el
B
B
al+1 bl+1 cl+1
B
B
... ... ...
B
B
B
... ... c
B
@
0
k
ak+1 bk+1
0
1
1 0
0
BB a1 b1 c1
CC
BB
CC
... ... ...
B
CC
0
M (k+1) =0 = B
BB
CC :
... ... ...
BB
C
0
ak bk ck C
@
A
;
j
1
CC
CC
CC
CC
CC
CC CC
CC
CC
CC
CA
;
j
0
0 0
The determinants of these matrices are given as follows:
det M (0) =0 = ( a1) 6kk 4a2k;1 det M (l) =0 = Dl;1( al+1) dl k;l al+2el k;l;1 1 l
where
m = det (Mij =0 )k+2;mij k+1 1 m k 1
j
;
j
f
;
;
f
g
;
j
g
24
;
k + 1
and
By induction, we can show:
Lemma 6.10
0 = ;1 = ;2 = 1:
1)!!k!
m = (2(mk m
1 m k 1
)!
;
;
;
From these formulas, we shall compute
@ det M
=0 = ( a1) 6kk 4a2k;1
@
kX
;1
+ ( al+1Dl;1) dl k;l al+2el k;l;1
;
f
l=1
;
g
;
f
;
g
+( ak+1Dk;1)dk :
;
First we compute the second term. Since
k
l;1
Dl;1 = (2k l+!(22k)!!2l)!! = 2k+1;ll(!2k +k!1 l)! = (k2+ 1k!l!l)! k 1 2l)!k! = (2k 2l)!k! k;l = (2k 1l! 2l)!!k! = l!2(2k;1;
l (k 1 l)! 2k;l l!(k l)!
(2k 3 2l)!k!
k;l;1 = (2k (l 3+ 1)!2l)!!k! = (l + 1)!2
k;2;l(k 2 l)!
= k;1;(2l k 2 2l)!k!
2
(l + 1)!(k 1 l)!
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
we have
2l;1 2 (2k 2l)!
dl Dl;1k;l al+2el Dl;1k;l;1 = (2k 5l) (k2 + 1(k!)l)!(
k l)!2k :
;
;
;
;
;
;
Hence
2l;1 2 k + 2 2l)!
al+1Dl;1 (dl k;l al+2el k;l;1) = (2k 5l) 2(k +(k1!) (2l)!(
k l)!2k :
;
;
;
25
;
;
;
We need to compute
kX
;1
2l;1 k + 2 2l)! (k!)2
:
(2k 5l) (2k + (2
1 l)!(k l)! 2k
l=1
;
;
;
;
Let i = k + 1 l and
;
Ap(i) = (2i!ii)!! ip 4k;i p = 1 2 3
for simplicity of notation. Then l = k + 1 i and
;
22l;1(2k + 2 2l)! (k!)2
(3k + 5)(k!)2
5(k!)2
A
(
i
)
+
(2k 5l) (k + 1 l)!(k l)! 2k =
1
2k;1
2k;1 A2(i):
Therefore we need to compute
;
;
;
;
;
k
X
i=2
Ap (i) for p = 1 2:
Lemma 6.11 (1) A1(i + 1) A1(i) = 21 A0(i)
;
(2) A2(i + 1) ; A2(i) = 23 A1(i) + 21 A0(i)
(3) A3(i + 1) ; A3(i) = 52 A2(i) + 2A1(i) + 21 A0(i)
A simple computation shows
Ap(i + 1) Ap(i) = (2i!ii)!! 4k;i
For p = 1 2 3, we get
1
i + 2 (i + 1)1;1
1
i + 2 (i + 1)2;1
1
i + 2 (i + 1)3;1
and hence the lemma follows.
Proof.
;
;
;
;
26
i + 21 (i + 1)p;1 ip :
;
i1 = 12 i0
i2 = 32 i1 + 12 i0 i2 = 52 i2 + 2i1 + 21 i0 Summing them up from i = 2 to k, we have
k
X
A1(i) = 32 A2(k + 1) A2 (2) A1(k + 1) + A1(2)
i=2
(
)
2
(2
k
+
2)!
k
(
k
+
1)
k
;1
=
3 4
3 (k + 1)!(k + 1)!
4
k
X
2
4
1
A2(i) = 5 A3(k + 1) 3 A2(k + 1) + 3 A1(k + 1) A3(2)
i=2
+ 2 4 A2(2) 1 A1(2)
3
(5 3
)
k
(k + 1)(k + 23 )
2
(2k + 2)!
k
;1
5 4
:
= 5 (k + 1)!(k + 1)!
4
f
;
;
g
;
;
;
;
;
From these expressions, we have
;
k
k
(3k + 5)(k!)2 X
5(k!)2 X
(2k + 1)! ( k) + 2k (k!)2 3k:
A
(
i
)
+
A
(
i
)
=
1
2
2k;1
2k;1 i=2
2k;1
i=2
;
Therefore we nally obtain
@ det M = (2k + 1)!k(2k + 3) > 0
=0
@
2k;1
and hence we conclude det M = 0.
6
Theorem 6.12 If =2 is not an algebraic number, then the rst order nor-
mal form of Eqn.(17) with respective to the grading function is unique,
which is
x_ = y
(19)
y_ = xy + x3 + P1m=4 am xm:
Proof. The uniqueness of the rst order normal form follows from Corollary 5.2, Example 5.4 and Lemma 6.7. A simple calculation shows that
spanxm+3@y is a complement to ImL(1)
m for each m 2 N. Then the rst order
normal form (19) is obtained.
27
References
Ar] V.I.Arnold, Geometrical Method in the Theory of Ordinary Di erential
Equations, Springer-Verlag, New York, 1983.
Ba] A.Baider, Unique normal forms for vector elds and Hamiltonians,
J.D.E.,77(1989), 33-52.
BC1] A.Baider and R.C.Churchill, Uniqueness and non-uniqueness of normal forms for vector elds, Proc.Royal Soc.Edinb. 108A(1988),27-33.
BC2] A.Baider and R.C.Churchill, Unique normal forms for planar vector
elds, Math.Z., 199(1988), 303-310.
BS1] A.Baider and J.Sanders, Unique normal forms: the Hamiltonian nilpotent case, J.D.E.
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CK] L.O.Chua and H.Kokubu, Normal forms for nonlinear vector elds,
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36(1989), 51-70.
SM] J.Sanders and J.C.van der Meer, Unique normal form of the Hamiltonian 1:2-resonance, in Geometry and Analysis in Nonlinear Dynamics, H.W.Broer and F.Takens, Eds., Longman Academic, Scientic and
Technical, Harlow(1990), 56-69.
Us] S.Ushiki, Normal forms for singularities of vector elds,
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Wa] D.Wang, A recursive formula and its applications to computations of
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28