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Acta Appl Math
DOI 10.1007/s10440-017-0140-3
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−:
The Inhomogeneous Case
D.I. Barrett1·R. Biggs2·C.C. Remsing1
Received: 16 March 2015 / Accepted: 9 November 2017
© Springer Science+Business Media B.V., part of Springer Nature 2017
Abstract We consider equivalence, stability and integration of quadratic Hamilton–Poisson
systems on the semi-Euclidean Lie–Poisson space se(1,1)∗
−. The inhomogeneous positive
semidefinite systems are classified (up to affine isomorphism); there are 16 normal forms.
For each normal form, we compute the symmetry group and determine the Lyapunov sta-
bility nature of the equilibria. Explicit expressions for the integral curves of a subclass of
the systems are found. Finally, we identify several basic invariants of quadratic Hamilton–
Poisson systems.
Keywords Hamilton–Poisson system ·Lie–Poisson space ·Lyapunov stability
Mathematics Subject Classification (2010) 53D17 ·37J25
1 Introduction
The dual space of a Lie algebra admits a natural Poisson structure, called the Lie–Poisson
structure. Such structures are in a one-to-one correspondence with linear Poisson struc-
tures [23] (i.e., those structures for which the Poisson bracket of two linear functions is
This research was supported in part by the European Union’s Seventh Framework Programme
(FP7/2007-2013, grant no. 317721). The first two authors would also like to acknowledge the financial
support of the National Research Foundation (NRF-DAAD) and Rhodes University towards this
research. Additionally, the second author acknowledges the financial support of the Claude Leon
Foundation.
BC.C. Remsing
c.c.remsing@ru.ac.za
D.I. Barrett
dbarrett6@gmail.com
R. Biggs
rory.biggs@up.ac.za
1Department of Mathematics, Rhodes University, 6140 Grahamstown, South Africa
2Department of Mathematics and Applied Mathematics, University of Pretoria, 0002 Pretoria,
South Africa
D.I. Barrett et al.
linear). Many dynamical systems admit a Hamiltonian formulation in terms of Lie–Poisson
brackets: for instance, the motion of a rigid body (and its generalizations) [20,25]and(on
infinite-dimensional Lie algebras) fluid dynamics in the form of Euler’s equation for an ideal
fluid [25]. Moreover, such systems arise naturally in the study of invariant optimal control
problems [7,21,22].
The study of quadratic Hamilton–Poisson systems (especially on low-dimensional
spaces) has seen a flurry of recent activity. We provide a brief overview here. Systems on
the orthogonal space so(3)∗
−, as well as the Euclidean space se(2)∗
−, have been treated ex-
tensively. In particular, on so(3)∗
−orthogonal equivalence and explicit integration of homo-
geneous systems is considered in [19](seealso[9,29]) whereas in [4,6] affine equiva-
lence, integration and stability of inhomogeneous systems have been studied (in a similar
vein to this paper). Similar questions on se(2)∗
−were considered in [2,3,5]. On the other
hand, spectral and Lyapunov stability, as well as numerical integration, of homogeneous
Hamilton–Poisson systems on se(1,1)∗
−are considered in [13], and spectral stability and
numerical integration on sl(2,R)∗
−in [10]. A number of Hamilton–Poisson systems have
also been studied from the viewpoint of invariant optimal control problems (see, e.g., [2,5,
11,12,15,27,28] and references therein). A thorough treatment of homogeneous systems
on three-dimensional Lie–Poisson spaces has also recently been published [18](seealso
[16]).
This paper serves as a sequel to the earlier work [14], in which the homogeneous systems
on se(1,1)∗
−were treated; here we are chiefly concerned with the inhomogeneous systems.
Together, these papers form an extensive and systematic study of the classification, integra-
tion and stability of positive semidefinite quadratic Hamilton–Poisson systems on se(1,1)∗
−.
(Our restriction to positive semidefinite quadratic forms is motivated by control theoretic
considerations; see, e.g., [14,17].) We expect that this study will complement existing work
on other Lie–Poisson spaces and would be integral to any systematic treatment of (inhomo-
geneous) Hamilton–Poisson systems in three dimensions.
We start by classifying the inhomogeneous and positive semidefinite quadratic Hamilton–
Poisson systems on se(1,1)∗
−. Sixteen normal forms are obtained, including five one-
parameter families of systems, as well as a single two-parameter family of systems. We
distinguish between those systems whose integral curves evolve on lines, on planes, or on
neither lines nor planes. The latter group is further subdivided by separating out those sys-
tems for which the equilibria are the union of lines or planes. Some systems are shown to
be equivalent to systems previously considered on the orthogonal space so(3)∗
−and Eu-
clidean space se(2)∗
−; these systems shall be excluded from our treatment of stability and
integration.
For each normal form, we compute the symmetry group and determine the (Lyapunov)
stability nature of its equilibria. To prove stability, the extended energy-Casimir method [26]
is applied; instability either follows from spectral instability or by a direct approach.
With the exception of a subclass of systems, we find explicit expressions for all (max-
imal) integral curves. (Due to the complexity of the computations required, we exclude
those nonplanar systems whose equilibria are not the union of lines or planes.) We provide
proofs for typical cases. Most integral curves are expressed in terms of elementary func-
tions. However, for one system the Jacobi elliptic functions are used. Also, it turns out that
the Hamiltonian vector fields for two of the normal forms are not complete.
We conclude the paper by identifying some invariants of quadratic Hamilton–Poisson
systems. These invariants may be used to form a “taxonomy” of systems.
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
2 Preliminaries
2.1 Lie–Poisson Spaces
Let gbe an n-dimensional (real) Lie algebra. The dual space g∗admits a natural linear
Poisson structure, called the (minus) Lie–Poisson structure [23,25]. If F,G ∈C∞(g∗),then
the Lie–Poisson bracket is given by
{F,G}(p) =−ad∗
dF(p)p, dG(p)=−p,dF(p),dG(p).
Here [·,·] is the Lie bracket on gand ad∗
dF(p) is the dual of the adjoint map addF(p) =
[dF(p),·].(AsdF(p) and dG(p) are linear functions on g∗, they are identified with el-
ements of g.) The Lie–Poisson space (g∗,{·,·})is denoted g∗
−.Alinear Poisson automor-
phism is a linear isomorphism ψ:g∗→g∗such that {F,G}◦ψ={F◦ψ, G ◦ψ}for every
F,G ∈C∞(g∗). Linear Poisson automorphisms are exactly the dual maps of Lie algebra
automorphisms.
The Hamiltonian vector field
Hcorresponding to a function H∈C∞(g∗)is defined as
H[F]={F,H}for F∈C∞(g∗). Explicitly, we have
H(p)=ad∗
dH(p)p.ACasimir function
is a function C∈C∞(g∗)such that
C=0. (Casimir functions are constants of motion for
any Hamiltonian vector field on g∗.) An integral curve of a Hamiltonian vector field
His an
absolutely continuous curve p(·):(a, b) →g∗such that ˙p(t) =
H(p(t)) for almost every t.
We say that
His complete if the domain of every integral curve of
Hcan be extended to
R(cf. [1]). An integral curve is maximal if it has maximal domain. The following lemma is
easytoprove.
Lemma 1 Let p(·):(a, b ) →g∗be an integral curve of
H.If (i) a=−∞or
limt→ap(t)=∞and (ii) b=∞or limt→bp(t)=∞,then p(·)is maximal.
Aquadratic Hamilton–Poisson system is a pair (g∗
−,H
A,Q),whereg∗
−is a Lie–Poisson
space and HA,Qis a Hamiltonian function of the form
HA,Q(p) =LA(p) +HQ(p) =p,A+Q(p).
Here A∈gand Qis a quadratic form on g∗. (In coordinates, we write Q(p) =1
2pQp,
where Q∈Rn×n.) When the space g∗
−is fixed, (g∗
−,H
A,Q)will be identified with its Hamil-
tonian HA,Q. We shall consider only those systems for which Qis positive semidefinite. If
A=0, then the system is said to be homogeneous; otherwise, it is called inhomogeneous.
Let (g∗
−,H
A,Q)and (h∗
−,H
B,R)be two quadratic Hamilton–Poisson systems. We say
that HA,Qand HB,Rare affinely equivalent (or A-equivalent) if there exists an affine iso-
morphism ψ:g∗→h∗such that ψ∗
HA,Q=
HB,R.Ifψis a linear isomorphism, then HA,Q
and HB,Rare called linearly equivalent (or L-equivalent). It is easy to show that the follow-
ing systems are all L-equivalent to HA,Q:
(E1) HA,Q◦ψ,whereψ:g∗→g∗is a linear Poisson automorphism.
(E2) HA,Q+C,whereCis a Casimir function.
(E3) HA,rQ,wherer=0.
Affine equivalence of two inhomogeneous systems implies linear equivalence of the corre-
sponding homogeneous systems.
D.I. Barrett et al.
Proposition 1 If ψ:p→ψ0(p) +qis an affine isomorphism such that ψ∗
HA,Q=
HB,R,
then (ψ0)∗
HQ=
HR.
Proof We have
HA,Q(p) =ad∗
d(LA+HQ)(p) p=ad∗
Ap+ad∗
dHQ(p) p=
LA(p) +
HQ(p).
Accordingly, ψ0·
HA,Q=ψ0·
LA+ψ0·
HQand
HB,Rψ(p)=
LBψ(p)+
HRψ(p)
=(
LB◦ψ0)(p) +
LB(q) +(
HR◦ψ0)(p) +
HR(q) +F(p)+G(p)
where F(p) =ad∗
dHR(ψ0(p)) qand G(p) =ad∗
dHR(q) ψ0(p). Expanding terms in (ψ0·
HA,Q)(p) −(
HB,R◦ψ)(p) =0, we get
(ψ0·
LA)(p) +(ψ0·
HQ)(p) −(
LB◦ψ0)(p) −(
HR◦ψ0)(p) −F(p)−G(p)
=
LB(q) +
HR(q). (1)
Taking p=0 yields
LB(q) +
HR(q) =0. Interpreting both sides of (1)asmapsg∗→h∗,
we have
T0(ψ0·
LA)+T0(ψ0·
HQ)−T0(
LB◦ψ0)−T0(
HR◦ψ0)−T0F−T0G=0.
(Here T0Fis the tangent map of Fat zero.) Elementary calculations show that T0(ψ0·
HQ)=T0(
HQ◦ψ0)=0. Furthermore, Fand Gcan be shown to be linear; hence we make
the identifications T0F↔Fand T0G↔G.(Likewise,ψ0·
LAand
LB◦ψ0are linear.)
Thus ψ0·
LA−
LB◦ψ0−F−G=0,andso(1) becomes ψ0·
HQ=
HR◦ψ0.Thatis,
(ψ0)∗
HQ=
HR.
2.2 Stability
A point pe∈g∗is called an equilibrium point of a Hamiltonian vector field
Hif
H(p
e)=0.
An equilibrium point peis said to be (Lyapunov) stable if for every neighbourhood Nof
pethere exists a neighbourhood N⊆Nof pesuch that, for every integral curve p(·)of
Hwith p(0)∈N,wehavep(t) ∈Nfor all t>0. The point peis spectrally stable if
all eigenvalues of the linearized dynamical system D
H(p
e)have nonpositive real parts.
Every stable equilibrium point is spectrally stable. The point peis unstable (resp. spectrally
unstable) if it is not stable (resp. spectrally stable).
The (extended) energy-Casimir method and continuous energy-Casimir method [26]pro-
vide sufficient conditions for stability of equilibria. We state simplified versions here.
Proposition 2 Let pebe an equilibrium point and let Cbe a Casimir function.If there
exist λ0,λ
1∈Rsuch that d(λ0H+λ1C)(pe)=0and the quadratic form d2(λ0H+
λ1C)(pe)|W×Wis positive definite,where W=ker dH(p
e)∩ker dC(pe),then peis stable.
Proposition 3 If Cis a Casimir function and H−1(H (pe))∩C−1(C(pe)) ={pe}inaneigh-
bourhood of pe,then peis stable.
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
2.3 The Lie–Poisson Space se(1,1)∗
−
The three-dimensional semi-Euclidean Lie algebra
se(1,1)=⎧
⎨
⎩x1E1+x2E2+x3E3=⎡
⎣
000
x10x3
x2x30⎤
⎦:x1,x
2,x
3∈R⎫
⎬
⎭
is the Lie algebra of the Lie group SE(1,1)of (orientation-preserving) isometries of the
Minkowski plane. The nonzero commutator relations are [E2,E
3]=−E1and [E3,E
1]=E2.
Let (E∗
1,E∗
2,E∗
3)be the dual of the standard basis (E1,E
2,E
3). We write elements
p=p1E∗
1+p2E∗
2+p3E∗
3∈se(1,1)∗as row vectors. For convenience, we take p=
p2
1+p2
2+p2
3. The group of linear Poisson automorphisms of se(1,1)∗
−are
⎧
⎨
⎩p→p⎡
⎣xyv
σy σx w
00σ⎤
⎦:v,w,x, y ∈R,σ∈{−1,1},x
2=y2⎫
⎬
⎭.
Let H:se(1,1)∗→Rbe a Hamiltonian function. The equations of motion take the
following explicit form:
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
˙p1=p2∂H
∂p3
˙p2=p1∂H
∂p3
˙p3=−p1∂H
∂p2−p2∂H
∂p1.
The function C:p→p2
1−p2
2is a Casimir function on se(1,1)∗
−.
Remark 1 The vector field
Hmay be written in the form
H=1
2∇H×∇C.
Lemma 2 The points (0,0,μ),μ∈Rare equilibrium points for any Hamilton–Poisson
system Hon se(1,1)∗
−.If ∂H
∂p3(0,0,μ)= 0, then the state (0,0,μ) is (spectrally)unstable.
Proof The linearization at (0,0,μ) of the vector field
Hhas eigenvalues λ1=0, λ2,3=
±∂H
∂p3(0,0,μ). Hence, if ∂H
∂p3(0,0,μ)= 0, then the state (0,0,μ) is spectrally unstable.
Lemma 3 Let Hbe a Hamilton–Poisson system on se(1,1)∗
−and let p(·)be an absolutely
continuous curve such that ˙p1=p2∂H
∂p3,C(p(t)) =constant and H(p(t))=constant.Then
p(·)is an integral curve of
H.
Proof By assumption, the first equation of motion is satisfied. Differentiating both sides of
C(p(t)) =p1(t)2−p2(t)2yields 0 =2p1˙p1−2p2˙p2, i.e., ˙p2=p1˙p1
p2=p1∂H
∂p3. Hence the
second equation of motion holds. Lastly, differentiate both sides of H(p(t)) =constant,
to get ˙p1∂H
∂p1+˙p2∂H
∂p2+˙p3∂H
∂p3=0. Solving for ˙p3gives ˙p3=−p1∂H
∂p2−p2∂H
∂p1. Thus
˙p(t) =
H(p(t)).
D.I. Barrett et al.
3 Classification
We classify all inhomogeneous quadratic Hamilton–Poisson systems (with positive semidef-
inite quadratic form) on se(1,1)∗
−. This classification is based on a classification of the ho-
mogeneous systems [14]. The following two results essentially comprise that classification;
however, we state a slightly stronger version here. (Nevertheless, the proof is identical, so
we shall not repeat it.)
Proposition 4 (Cf. [14]) Let HQbe a positive semidefinite homogeneous quadratic
Hamilton–Poisson system on se(1,1)∗
−.There exists a linear Poisson automorphism ψand
real numbers r>0, k∈Rsuch that rHQ◦ψ+kC =Hifor exactly one i∈{0,...,5},
where
H0(p) =0H1(p) =1
2p2
1H2(p) =1
2(p1+p2)2
H3(p) =1
2p2
3H4(p) =1
2p2
1+p2
3H5(p) =1
2(p1+p2)2+p2
3.
Corollary 1 Every homogeneous quadratic Hamilton–Poisson system on se(1,1)∗
−is
L-equivalent to exactly one of the systems H0,...,H
5.
For each of the systems H0,...,H
5,letS(Hi)denote the subgroup of linear Poisson
automorphisms ψ:se(1,1)∗→se(1,1)∗satisfying Hi◦ψ=rHi+kC for some r>0and
k∈R.
Lemma 4 The subgroups S(Hi)are given by
S(H0):⎡
⎣xyv
σy σx w
00σ⎤
⎦S(H1):⎡
⎣x0v
0σx w
00σ⎤
⎦,⎡
⎣
0yv
σy 0w
00σ⎤
⎦
S(H2):⎡
⎣xy v
yxw
001
⎤
⎦S(H3):⎡
⎣xy0
σy σx 0
00σ⎤
⎦
S(H4):⎡
⎣σ100
0σ1σ20
00σ2⎤
⎦,⎡
⎣
0σ10
σ2σ200
00σ2⎤
⎦S(H5):⎡
⎣xσ−x0
σ−xx0
001
⎤
⎦.
Here σ, σ1,σ
2∈{−1,1},v,w, x,y ∈Rand the determinant of each matrix is nonzero.
Proof We illustrate by finding S(H1).Wehave
(H1◦ψ)(p) =1
2p⎡
⎣x2σxy 0
σxy y20
000
⎤
⎦p,where ψ:p→p⎡
⎣xyv
σy σx w
00σ⎤
⎦.
If ψ∈S(H1), then either y=0orx=0andsoψis of the given form. If y=0,
then (H1◦ψ)(p) =x2H1(p) and so ψ∈S(H1). Likewise, if x=0, then (H1◦ψ)(p) =
y2H1(p) −y2
2C(p) and so ψ∈S(H1).
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
Theorem 1 Let HA,Q=LA+HQbe an inhomogeneous quadratic Hamilton–Poisson sys-
tem on se(1,1)∗
−.
(i) If HQis L-equivalent to H0(p) =0, then HA,Qis A-equivalent to exactly one of the
following systems:
H(0)
1(p) =p1
H(0)
2,α(p) =αp3.
(ii) If HQis L-equivalent to H1(p) =1
2p2
1,then HA,Qis A-equivalent to exactly one of the
following systems:
H(1)
1(p) =p1+1
2p2
1
H(1)
2(p) =p1+p2+1
2p2
1
H(1)
3,α(p) =αp3+1
2p2
1.
(iii) If HQis L-equivalent to H2(p) =1
2(p1+p2)2,then HA,Qis A-equivalent to exactly
one of the following systems:
H(2)
1(p) =p1+1
2(p1+p2)2
H(2)
2(p) =p1+p2+1
2(p1+p2)2
H(2)
3,δ (p) =δp3+1
2(p1+p2)2.
(iv) If HQis L-equivalent to H3(p) =1
2p2
3,then HA,Qis A-equivalent to exactly one of the
following systems:
H(3)
1(p) =p1+1
2p2
3
H(3)
2(p) =p1+p2+1
2p2
3
H(3)
3(p) =1
2p2
3.
(v) If HQis L-equivalent to H4(p) =1
2(p2
1+p2
3),then HA,Qis A-equivalent to exactly
one of the following systems:
H(4)
1,α(p) =αp1+1
2p2
1+p2
3
H(4)
2,α1,α2(p) =α1p1+α2p2+1
2p2
1+p2
3.
D.I. Barrett et al.
(vi) If HQis L-equivalent to H5(p) =1
2[(p1+p2)2+p2
3],then HA,Qis A-equivalent to
exactly one of the following systems:
H(5)
1,α(p) =αp1+1
2(p1+p2)2+p2
3
H(5)
2(p) =p1−p2+1
2(p1+p2)2+p2
3
H(5)
3,α(p) =α(p1+p2)+1
2(p1+p2)2+p2
3.
Here α>0, α1≥α2>0and δ= 0parametrize families of normal forms,each different
value corresponding to a distinct (non-equivalent)normal form.
Proof Let HA,Qbe an inhomogeneous quadratic Hamilton–Poisson system. By Corollary 1
and (E1), (E2), (E3), we have that HA,Qis A-equivalent to a system H=LB+Hifor some
B∈se(1,1)and i∈{0,...,5}. By Proposition 1,LB+Hiis not A-equivalent to LB+Hj
for any B∈se(1,1)when i= j. Hence there are six cases to consider (corresponding to
each Hi).
(i) Suppose H=LB+H0. There exists ψ∈S(H0)such that LB◦ψ∈{LE1,L
E1+E2,
LαE3:α>0}. Indeed, let B=3
i=1biEi. Suppose b3=0. If b2
1=b2
2,then
ψ:p→p⎡
⎢
⎢
⎣
b1
b2
1−b2
2−b2
b2
1−b2
20
−b2
b2
1−b2
2
b1
b2
1−b2
20
001
⎤
⎥
⎥
⎦
is an element of S(H0)such that LB◦ψ=LE1.Ifb2
1=b2
2, i.e., b1=b= 0andb2=±b,
then ψ:p→pdiag(1
b,±1
b,±1)∈S(H0)and LB◦ψ=LE1+E2. On the other hand, suppose
b3=0. Then
ψ:p→p⎡
⎢
⎣
10 −b1
b3
0sgn(b3)−sgn(b3)b2
b3
00 sgn(b3)
⎤
⎥
⎦
is an element of S(H0)such that LB◦ψ=LαE3,whereα=|b3|>0.
Consequently, His A-equivalent to one of the systems G1(p) =p1,G2(p) =p1+p2or
G3,α (p) =αp3. The systems G1and G2are A-equivalent. Indeed,
ψ:p→p⎡
⎣−1−10
010
001
⎤
⎦
is a linear isomorphism such that ψ∗
G1=
G2.However,G1and G3,α are not A-equivalent.
Suppose there exists an affine isomorphism ψ:p→ pΨ +q,Ψ=[Ψij ]such that
ψ∗
G1=
G3,α . This yields the system of equations
⎧
⎪
⎪
⎨
⎪
⎪
⎩
αΨ12p1+(αΨ22 +Ψ31 )p2+αΨ32p3+αq2=0
αΨ11p1+(αΨ21 +Ψ32 )p2+αΨ31p3+αq1=0
Ψ33p2=0.
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
By inspection, we have Ψ31 =Ψ32 =Ψ33 =0, whence det Ψ=0, a contradiction. Like-
wise, G3,α is A-equivalent to G3,αonly if α=α. Therefore His A-equivalent to either
H(0)
1(p) =p1or H(0)
2,α(p) =αp3.
(ii) Suppose H=LB+H1.Likeincase(i),thereexistsψ∈S(H1)such that LB◦
ψ∈{LE1+βE2,L
αE3:α>0,β ≥0}. Hence His A-equivalent to one of the systems
G1,β (p) =p1+βp2+1
2p2
1or G2,α (p) =αp3+1
2p2
1. The systems G1,β ,β>0and
G1,1are A-equivalent. Indeed, ψ:p→ pdiag(1,1
β,1
β)is a linear isomorphism such
that ψ∗
G1,β =
G1,1. One can now verify that G1,1and G2,α are not A-equivalent, and
G2,α is A-equivalent to G2,αonly if α=α. Hence His A-equivalent to exactly one of
H(1)
1(p) =p1+1
2p2
1,H(1)
2(p) =p1+p2+1
2p2
1or H(1)
3,α(p) =αp3+1
2p2
1.
(iii) Suppose H=LB+H2. There exists ψ∈S(H2)such that LB◦ψ∈{LE1,L
E1+σE
2,
LδE3:δ=0,σ∈{−1,1}},andsoHis A-equivalent to one of the systems G1(p) =p1+
1
2(p1+p2)2,G2,σ (p) =p1+σp2+1
2(p1+p2)2or G3,δ(p) =δp3+1
2(p1+p2)2.Wehave
that G1is A-equivalent to G2,−1. Indeed,
ψ:p→p⎡
⎣
1
21
20
010
001
⎤
⎦
is a linear isomorphism such that ψ∗
G1=
G2,−1.NotwoofthesystemsG1,G2,1and G3,δ ,
δ= 0areA-equivalent. Thus His A-equivalent to exactly one of H(2)
1(p) =p1+1
2(p1+
p2)2,H(2)
2(p) =p1+p2+1
2(p1+p2)2or H(2)
3,δ (p) =δp3+1
2(p1+p2)2.
(iv) Suppose H=LB+H3. There exists ψ∈S(H3)such that LB◦ψ={LE1+βE3,
LE1+E2+γE
3,L
αE3:α>0,β≥0,γ∈R}. Thus His A-equivalent to one of the systems
G1,β (p) =p1+βp3+1
2p2
3,G2,γ (p) =p1+p2+γp3+1
2p2
3or G3,α (p) =αp3+1
2p2
3.
Let ψ:p→ p+βE∗
3,ψ:p→ p+αE∗
3and ψ :p→ p+γE∗
3.Thenψ∗
G1,β =
G1,0,
ψ
∗
G2,α =
G2,0and ψ
∗
G3,γ =
G3,0. No two of the systems G1,0,G2,0and G3,0are A-
equivalent. Therefore His A-equivalent to exactly one of the systems H(3)
1(p) =p1+1
2p2
3,
H(3)
2(p) =p1+p2+1
2p2
3or H(3)
3(p) =1
2p2
3.
(v) Suppose H=LB+H4. There exists ψ∈S(H4)such that LB◦ψ∈{LβE1+αE2,
LγE
1+βE2+αE3:α>0,β≥0,γ∈R}. Thus His A-equivalent to one of the systems
G1,α,β (p) =βp1+αp2+1
2(p2
1+p2
3)or G2,α,β,γ (p) =γp1+βp2+αp3+1
2(p2
1+p2
3).
If ψ:p→ p+αE∗
3,thenψ∗
G2,α,β,γ =
G2,0,β,γ . Likewise, if ψ:p→ pdiag(−1,1,1),
then ψ
∗
G2,0,β,γ =
G2,0,β,−γ. Accordingly, we have a family of potential normal forms
G2,0,β1,β2(p) =β1p1+β2p2+1
2(p2
1+p2
3), with β1,β
2≥0andβ1,β2not both zero. If
β2>0, then G2,0,β1,β2=G1,α,β ,whereα=β2>0andβ=β1≥0. If β1>0, then
ψ :p→p⎡
⎣
010
100
001
⎤
⎦
is a linear isomorphism such that ψ
∗
G2,0,β1,β2=
G1,α,β ,whereα=β1>0andβ=β2≥0.
Let G3,α(p) =αp1+1
2(p2
1+p2
3).Thenψ
∗
G1,α,0=
G3,α . Hence we have the potential
normal forms G1,α1,α2(p) =α1p1+α2p2+1
2(p2
1+p2
3)and G3,α ,whereα, α1,α
2>0. If
α2>α
1,thenψ
∗
G1,α1,α2=
G1,α2,α1, and so we may assume α1≥α2>0. No two of the
systems G1,α1,α2,α1≥α2>0andG3,α ,α>0areA-equivalent. Thus His A-equivalent to
exactly one of H(4)
1,α(p) =αp1+1
2(p2
1+p2
3)or H(4)
1,α1,α2(p) =α1p1+α2p2+1
2(p2
1+p2
3).
D.I. Barrett et al.
(vi) Suppose H=LB+H5. There exists ψ∈S(H5)such that LB◦ψ∈{LβE1+γE
3,
LδE1+αE2+γE
3:α>0,β≥0,γ∈R,δ= 0}. Hence His A-equivalent to one of the sys-
tems G1,β,γ (p) =βp1+γp3+1
2[(p1+p2)2+p2
3]or G2,α,γ,δ (p) =δp1+αp2+γp3+
1
2[(p1+p2)2+p2
3].Ifψ:p→ p+γE∗
3,thenψ∗
G1,β,γ =
G1,β,0.(AsG1,0,0is a homo-
geneous system, we assume β=α>0.) Likewise, ψ∗
G2,α,γ,δ =
G2,α,0,δ . Suppose δ2=α2.
Then
ψ:p→p⎡
⎢
⎣
δ
|δ+α|δ
|δ+α|0
δ
|δ+α|δ
|δ+α|0
001
⎤
⎥
⎦
is a linear isomorphism such that ψ
∗
G2,α,δ,0=
G1,|δ+α|,0. On the other hand, suppose
δ2=α2.LetG3(p) =p1−p2+1
2[(p1+p2)2+p2
3]and G4,α(p) =α(p1+p2)+
1
2[(p1+p2)2+p2
3].Ifα=−δ>0, then
ψ :p→p⎡
⎢
⎣
1+δ
21−δ
20
1−δ
21+δ
20
001
⎤
⎥
⎦
is a linear isomorphism such that ψ
∗
G2,α,δ,0=
G3.Ifα=δ>0, then G2,α,δ,0=G4,α .
No two of the systems G1,α ,α>0, G3and G4,α,α>0areA-equivalent. Thus His
A-equivalent to exactly one of H(5)
1,α(p) =αp1+1
2[(p1+p2)2+p2
3],H(5)
2(p) =p1−p2+
1
2[(p1+p2)2+p2
3]or H(5)
3,α(p) =α(p1+p2)+1
2[(p1+p2)2+p2
3].
We say that a system HA,Qis ruled if the trace of every integral curve of
HA,Qis con-
tained in a line; planar if it is not ruled and if the trace of every integral curve lies in a
plane; and nonplanar, otherwise. A nonplanar system HA,Qis said to be of type I if the set
of all equilibrium points of
HA,Qis the union of lines or planes; otherwise, it is said to be of
type II. The partition of the normal forms into these four classes is given in Table 1.(Asa
concluding remark to the paper, we discuss how these normal forms may be better organized
according to some invariant properties.)
Remark 2 The classes of ruled, planar and nonplanar systems can be characterized in terms
of the curvature and torsion of a system’s integral curves. Indeed, a system is
•ruled, if and only if every integral curve has zero curvature;
•planar, if and only if every integral curve has zero torsion, and there exists an integral
curve with nonzero curvature; and,
•nonplanar, if and only if there exists an integral curve with nonzero curvature and nonzero
torsion.
(Although the curvature and torsion of a curve are not invariant under affine isomorphisms,
whether they vanish or not is invariant.)
3.1 Symmetry Groups
In this section we compute the symmetry group for each normal form. The symmetry group
of a system HA,Q, denoted Sym(HA,Q), is the group of all affine isomorphisms ψ:g∗→g∗
such that ψ∗
HA,Q=
HA,Q. Throughout this section, we shall identify p=[p1p2p3]with
˜p=[1p1p2p3]. An affine isomorphism ψ:p→pΨ +qis then written as ψ:˜p→ ˜p1q
0Ψ.
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
Tab le 1 Normal forms for
inhomogeneous quadratic
Hamilton–Poisson systems Class Systems
Ruled H(0)
1,H(1)
1,H(1)
2,H(2)
1,H(2)
2
Planar H(0)
2,α
Nonplanar, type I H(1)
3,α,H(2)
3,δ ,H(3)
1,H(3)
2,H(4)
1,α,H(5)
3,α
Nonplanar, type II H(4)
2,α1,α2,H(5)
1,α,H(5)
2
Proposition 5 The symmetry groups of the (nontrivial)homogeneous normal forms are
given by
Sym(H1):⎡
⎢
⎢
⎣
100 a
00xv
0y0w
000xy
⎤
⎥
⎥
⎦,⎡
⎢
⎢
⎣
100 a
0x0v
00yw
000xy
⎤
⎥
⎥
⎦Sym(H2):⎡
⎢
⎢
⎣
1a−ab
0xz−xv
0yz−yw
00 0 z2
⎤
⎥
⎥
⎦
Sym(H3):⎡
⎢
⎢
⎣
10 00
0xy0
0σy σx 0
00 0σ
⎤
⎥
⎥
⎦
Sym(H4):⎡
⎢
⎢
⎣
10 0 0
00σ10
0σ200
00 0σ1σ2
⎤
⎥
⎥
⎦,⎡
⎢
⎢
⎣
10 0 0
0σ100
00σ20
00 0σ1σ2
⎤
⎥
⎥
⎦
Sym(H5):⎡
⎢
⎢
⎣
10 00
0xσ−x0
0σ−xx0
00 01
⎤
⎥
⎥
⎦.
Here σ, σ1,σ
2∈{−1,1},a, b,v,w,x,y, z ∈Rand the determinant of each matrix is
nonzero.
Proof As a typical case, we find Sym(H3).Letψ:p→ ψ0(p) +q(where ψ0(p) =p[Ψij ])
be an affine isomorphism such that ψ∗
H3=
H3. By Proposition 1,ψ0is a symmetry of H3.
In particular, we have
ψ0·
H3E∗
3=
H3ψ0·E∗
3⇐⇒ Ψ33Ψ32 Ψ33 Ψ31 0=0.
Suppose Ψ33 =0; then ψ0·
H3(E∗
1+E∗
3)=
H3(ψ0·(E∗
1+E∗
3)) and ψ0·
H3(E∗
2+E∗
3)) =
H3(ψ0·(E∗
2+E∗
3)) imply that Ψ13 =Ψ23 =0, a contradiction. Hence Ψ33 = 0andΨ31 =
Ψ32 =0. Again, as ψ0·
H3(E∗
1+E∗
3)=
H3(ψ0·(E∗
1+E∗
3)) and ψ0·
H3(E∗
2+E∗
3)=
H3(ψ0·(E∗
2+E∗
3)),wegetΨ13 =Ψ23 =0, Ψ33 =σ∈{−1,1}and Ψ21 =σΨ
12,Ψ22 =σΨ
11.
Relabelling Ψ11 and Ψ12 as xand y, respectively, yields
ψ0:p→p⎡
⎣xy0
σy σx 0
00σ⎤
⎦.(2)
D.I. Barrett et al.
It is now easy to show that ψ∗
H3=
H3implies q=0, and so ψis a linear isomorphism of
the form (2). Conversely, every map of this form is a symmetry of H3.
Proposition 6 The symmetry groups of the inhomogeneous normal forms are given below.
(Throughout,we have σ∈{−1,1},a,b,c,v,w,x,y, z ∈Rand the determinant of each
matrix is nonzero.)
(i) For the systems corresponding to H0:
SymH(0)
1:⎡
⎢
⎢
⎣
1a0b
0x0v
0yzw
000z
⎤
⎥
⎥
⎦SymH(0)
2,α:⎡
⎢
⎢
⎣
100a
0xy0
0yx0
000z
⎤
⎥
⎥
⎦.
(ii) For the systems corresponding to H1:
SymH(1)
1:⎡
⎢
⎢
⎣
1−1ab
00av
0x0w
000ax
⎤
⎥
⎥
⎦,⎡
⎢
⎢
⎣
1a0b
01+a0v
00xw
000(1+a)x
⎤
⎥
⎥
⎦
SymH(1)
2:⎡
⎢
⎢
⎣
1abc
01+a0v
001+aw
00 01
⎤
⎥
⎥
⎦,⎡
⎢
⎢
⎣
1abc
001+bv
01+b0w
00 01
⎤
⎥
⎥
⎦
SymH(1)
3,α:⎡
⎢
⎢
⎣
100 a
0x00
00x0
000x2
⎤
⎥
⎥
⎦,⎡
⎢
⎢
⎣
100 a
00x0
0x00
000x2
⎤
⎥
⎥
⎦.
(iii) For the systems corresponding to H2:
SymH(2)
1:⎡
⎢
⎢
⎣
11(σ +a) −a2b
0z(1+2σa) −2σaz v
0z(1−z+2σa) z(z−2σa) w
00 0z2
⎤
⎥
⎥
⎦
SymH(2)
2:⎡
⎢
⎢
⎣
1a−ab
0x1−xv
0y1−yw
00 0 1
⎤
⎥
⎥
⎦,⎡
⎢
⎢
⎣
1a−(1+a) b
0x−(x +1)v
0y−(y +1)w
00 0 1
⎤
⎥
⎥
⎦
SymH(2)
3,δ :⎡
⎢
⎢
⎣
100 a
0xy 0
0yx 0
000(x +y)2
⎤
⎥
⎥
⎦.
(iv) For the systems corresponding to H3:
SymH(3)
1:⎡
⎢
⎢
⎣
100 0
010 0
00σ0
000σ
⎤
⎥
⎥
⎦SymH(3)
2:⎡
⎢
⎢
⎣
10 00
0x1−x0
01−xx0
00 01
⎤
⎥
⎥
⎦.
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
(v) For the systems corresponding to H4:
SymH(4)
1,α:⎡
⎢
⎢
⎣
100 0
010 0
00σ0
000σ
⎤
⎥
⎥
⎦SymH(4)
2,α1,α2:⎡
⎢
⎢
⎣
1000
0100
0010
0001
⎤
⎥
⎥
⎦,⎡
⎢
⎢
⎣
1000
0010
0100
0001
⎤
⎥
⎥
⎦.
(vi) For the systems corresponding to H5:
SymH(5)
1,α:⎡
⎢
⎢
⎣
1000
0100
0010
0001
⎤
⎥
⎥
⎦SymH(5)
2:⎡
⎢
⎢
⎣
1000
0100
0010
0001
⎤
⎥
⎥
⎦,⎡
⎢
⎢
⎣
10 00
00−10
0−100
00 01
⎤
⎥
⎥
⎦
SymH(5)
3,α:⎡
⎢
⎢
⎣
10 00
0x1−x0
01−xx0
00 01
⎤
⎥
⎥
⎦.
Proof The proof is analogous to that for Proposition 5. However, note that by Proposition 1,
if ψ:p→ψ0(p ) +qis an affine isomorphism such that ψ∗
HA,Q=
HA,Q,thenψ0must be
a symmetry of HQ. (This substantially simplifies the calculations.)
3.2 Equivalent Systems on se(2)∗
−and so(3)∗
−
The systems H(4)
1,α and H(4)
2,α1,α2turn out to be affinely equivalent to systems already con-
sidered on the Euclidean space se(2)∗
−and the orthogonal space so(3)∗
−. (Accordingly, we
shall not treat the stability or integration of H(4)
1,α and H(4)
2,α1,α2.) We give explicit isomor-
phisms between the equivalent systems below. The Lie algebras se(2)and so(3)are given
by
se(2)=⎧
⎨
⎩x1
E1+x2
E2+x3
E3=⎡
⎣
00 0
x10−x3
x2x30⎤
⎦:x1,x
2,x
3∈R⎫
⎬
⎭
and
so(3)=⎧
⎨
⎩x1
E1+x2
E2+x3
E3=⎡
⎣
0−x3x2
x30−x1
−x2x10⎤
⎦:x1,x
2,x
3∈R⎫
⎬
⎭
respectively. The non-zero commutator relations are [
E2,
E3]=
E1,[
E3,
E1]=
E2and
[
E2,
E3]=
E1,[
E3,
E1]=
E2,[
E1,
E2]=
E3.
The system (se(1,1)∗
−,H(4)
1,α)is A-equivalent to both the system
se(2)∗
−,H(p) =p1+1
21
c1p2
2+1
c2p2
3,c
1,c
2>0,α=c1
c2
and the system
so(3)∗
−,H(p)=αp1+p2
1+1
2p2
2,α
=√2α
D.I. Barrett et al.
that were treated in [5]and[6], respectively. Indeed,
ψ:se(2)∗→se(1,1)∗,p→ p⎡
⎢
⎣
1
√c1c200
00−1
√c1c2
0−1
c20⎤
⎥
⎦+−c1
c200
and
ψ:so(3)∗→se(1,1)∗,p→ p⎡
⎣−√20 0
00−1
0−√20
⎤
⎦+−√2α00
are affine isomorphisms such that ψ∗
H=
H(4)
1,α and ψ
∗
H=
H(4)
1,α . On the other hand,
(se(1,1)∗
−,H(4)
2,α1,α2)is A-equivalent to the system
so(3)∗
−,H(p) =α
1p1+α
2p3+p2
1+1
2p2
2,α
i=√2αi
considered in [6]. Indeed,
ψ:so(3)∗→se(1,1)∗,p→ p⎡
⎣−√200
001
0√20
⎤
⎦+−√2α
1−√2α
20
is an affine isomorphism such that ψ∗
H=
H(4)
2,α1,α2.
4 Ruled and Planar Systems
For the sake of completeness, we briefly treat the ruled and planar systems. The ruled sys-
tems have the following integral curves and equilibria:
H(0)
1:p(t) =(c1,c
2,c
3−c2t) eη,μ =(η, 0,μ)
H(1)
1:p(t) =c1,c
2,c
3−c2(1+c1)teη,μ
1=(η, 0,μ), eν,μ
2=(−1,ν,μ)
H(1)
2:p(t) =c1,c
2,c
3−(c1+c2+c1c2)teη,μ =−1+eη,−1+e−η,μ
H(2)
1:p(t) =c1−c2,c
2,c
3−c2
1+c2teη,μ =η+η2,−η2,μ
H(2)
2:p(t) =c1,c
2−c1,c
3−c2(c2+1)teη,μ
1=η, −(1+η), μ,eη,μ
2=(η, −η, μ).
(Here c1,c
2,c
3,η,μ,ν ∈Rand ν= 0.) All equilibria for these systems are unstable; we
graph these equilibria in Fig. 1.
The only planar system is H(0)
2,α(p) =αp3,α>0. The equations of motion are
⎧
⎪
⎪
⎨
⎪
⎪
⎩
˙p1=αp2
˙p2=αp1
˙p3=0.
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
Fig. 1 Equilibria of ruled systems (colour figure online)
The equilibrium states of
H(0)
2,α are eμ=(0,0,μ); all equilibria are (spectrally) unstable. The
integral curves are given by
⎧
⎪
⎪
⎨
⎪
⎪
⎩
p1(t) =p1(0)cosh(αt) +p2(0)sinh(αt)
p2(t) =p1(0)sinh(αt) +p2(0)cosh(αt)
p3(t) =p3(0).
5 Nonplanar Systems, Type I
In this section we consider the stability and integration of the nonplanar, type I systems
(see Table 1). The integration for each system is typically subdivided into several cases,
where each case corresponds to a qualitatively different integral curve. Qualitative changes
occur when the level sets corresponding to the Hamiltonian and Casimir functions are
tangent. Since every Hamiltonian vector field
Hon se(1,1)∗
−can be written in the form
H=1
2∇H×∇C, it follows that these level sets are tangent exactly when
H(p)=0, i.e.,
at equilibria. Thus we have a set {(H (pe), C(pe)) :pe∈se(1,1)∗
−,
H(p
e)=0}of critical
energy states corresponding to equilibria. The set of all energy states {(H (p), C(p )) :p∈
se(1,1)∗
−}is subdivided into a number of regions by the critical energy states. As a general
rule, each qualitative case corresponds to a different region of energy states. (The integral
curves corresponding to two different points in the same region can usually be continuously
deformed into each other.)
Remark 3 For some cases, we find it useful to consider critical energy states corresponding
to equilibria “at infinity.” We say that (h0,c
0)is a generalized critical energy state if there
exist two curves fand gin se(1,1)∗such that lims→∞[f(s)−g(s)]=0,
H(g(s)) =0,
H(f(s)) =h0and C(f(s)) =c0. (Every critical energy state is a generalized critical energy
state.)
For each system (except two) we graph the critical energy states. (The analysis of H(1)
3,α
and H(2)
3,δ is straightforward, and hence we omit the graphs for these systems.) Critical states
corresponding to stable equilibria are coloured in blue, whereas those corresponding to un-
stable equilibria are coloured in red. The generalized critical energy states are depicted in
purple. For typical configurations (or rather typical energy states (h0,c
0)) of the system, we
graph the level sets H−1(h0)and C−1(c0)and their intersection. The equilibrium points for
each system are also graphed. (As before, stable equilibria are blue and unstable equilibria
red.)
Throughout this section, we parametrize equilibria by μ, ν ∈Rwith ν=0.
D.I. Barrett et al.
5.1 The Systems H(1)
3,α and H(2)
3,δ
The equations of motion of the system H(1)
3,α(p) =αp3+1
2p2
1,α>0are
⎧
⎪
⎪
⎨
⎪
⎪
⎩
˙p1=αp2
˙p2=αp1
˙p3=−p1p2.
The equilibrium states of
H(1)
3,α are eμ=(0,0,μ); all states are (spectrally) unstable. The
integral curves are given by
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
p1(t) =p1(0)cosh(αt) +p2(0)sinh(αt)
p2(t) =p1(0)sinh(αt) +p2(0)cosh(αt)
p3(t) =1
2αp1(0)2−p1(t)2+p3(0).
The equations of motion of the system H(2)
3,δ (p) =δp3+1
2(p1+p2)2,δ=0are
⎧
⎪
⎪
⎨
⎪
⎪
⎩
˙p1=δp2
˙p2=δp1
˙p3=−(p1+p2)2.
The equilibrium states of
H(2)
3,δ are eμ=(0,0,μ); all states are (spectrally) unstable. The
integral curves are given by
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
p1(t) =p1(0)cosh(δt) +p2(0)sinh(δt)
p2(t) =p1(0)sinh(δt) +p2(0)cosh(δt)
p3(t) =1
2δp1(0)+p2(0)2−p1(t) +p2(t)2+p3(0).
5.2 The System H(3)
1
The equations of motion of the system H(3)
1(p) =p1+1
2p2
3are
⎧
⎪
⎪
⎨
⎪
⎪
⎩
˙p1=p2p3
˙p2=p1p3
˙p3=−p2.
The equilibrium states of
H(3)
1are eμ
1=(μ, 0,0)and eν
2=(0,0,ν).
Proposition 7 The equilibrium states have the following behaviour:
(i) The states eμ
1,μ∈(−∞,0]are unstable.
(ii) The states eμ
1,μ∈(0,∞)are stable.
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
Tab le 2 Index of cases for the
integral curves of H(3)
1Conditions Designation
c0>0h0>√c0Case I-a
h0=√c0Case I-b
−√c0<h
0<√c0Case I-c
h0=−√c0Case I-d
h0<−√c0Case I-e
c0=0h0>0 Case II-a
h0=0 Case II-b
h0<0 Case II-c
c0<0 Case III
(iii) The equilibrium states eν
2are (spectrally)unstable.
Proof (i) Consider the states eμ
1,μ∈(−∞,0). The integral curve
⎧
⎪
⎪
⎨
⎪
⎪
⎩
p1(t) =μ1+2csch2(√−μt)
p2(t) =−2μcoth(√−μt)csch(√−μt)
p3(t) =2√−μcsch(√−μt)
satisfies limt→−∞ p(t) −eμ
1=0. Accordingly, for every neighbourhood Nof eμ
1,there
exists t1<0 such that p(t1)∈N. Furthermore, limt→0p(t) −eμ
1=∞. Hence the states
eμ
1,μ∈(−∞,0)are unstable. Likewise, the integral curve p(t) =(−2
t2,2
t2,2
t)suffices to
show that the state e0
1is unstable.
(ii) Let Hλ=λ0H(3)
1+λ1C,whereλ0=μand λ1=−1
2.ThendHλ(eμ
1)=0and
the restriction of d2Hλ(eμ
1)=diag(−1,1,μ) to W×Wis positive definite, where W=
ker dH(3)
1(eμ
1)∩ker dC(eμ
1)=span{E∗
2,E∗
3}. Hence the states eμ
1,μ∈(0,∞)are stable.
(iii) As ∂H(3)
1
∂p3(eν
2)=ν, it follows from Lemma 2that the states eν
2are spectrally unstable.
Table 2lists the partition of cases used for integration. The critical energy states of the
system H(3)
1aregraphedinFig.2and the typical configurations graphed in Fig. 3and
Fig. 4. The integral curves of
H(3)
1are expressed in terms of the Jacobi elliptic functions
(see, e.g., [8,24]). Given a modulus k∈[0,1](and complementary modulus k=√1−k2),
the basic Jacobi elliptic functions are defined as
sn(x, k ) =sin am(x , k)
cn(x, k ) =cos am(x , k)
dn(x, k ) =1−k2sn2(x , k).
Here am(·,k) =F(·,k)
−1and F(x,k) =x
0dt
√1−k2sin2t
. (For the degenerate cases k=0
and k=1, we recover the circular and hyperbolic functions, respectively.) The functions
sn(·,k) and cn(·,k) have period 4K, whereas dn(·,k) has period 2K,whereK=F(π
2,k).
Furthermore, sn(·,k) is odd, whereas cn(·,k)and dn(·,k)are even.
D.I. Barrett et al.
Fig. 2 Critical energy states
for H(3)
1(colour figure online)
Theorem 2.1 (Case I-a)Let p(·):(−ε, ε) →se(1,1)∗be an integral curve of
H(3)
1such
that H(3)
1(p(0)) =h0,C(p(0)) =c0>0and h0>√c0.
(i) If p1(0)≤−
√c0,then there exist t0∈(0,2K
Ω)and σ∈{−1,1}such that p(t) =
¯p(t +t0)for every t∈(−ε, ε ),where ¯p(·):(0,2K
Ω)→se(1,1)∗is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) =(δ +h0)dn(Ωt , k) +(δ −h0)
dn(Ω t, k) −1
¯p2(t) =2σδ cn(Ωt , k)
dn(Ω t, k) −1
¯p3(t) =σk2Ωsn(Ωt , k)
dn(Ωt , k) −1.
Here δ=h2
0−c0,Ω=√h0+δand k=2δ
h0+δ.
(ii) If p1(0)≥√c0,then there exists t0∈[−2K
Ω,2K
Ω]such that p(t) =¯p(t +t0)for every
t∈(−ε, ε),where ¯p(·):R→se(1,1)∗is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) =√c0kdn(Ωt , k) +1
dn(Ω t, k) +k
¯p2(t) =k2√c0
sn(Ωt , k)
dn(Ωt , k) +k
¯p3(t) =k√2δcn(Ωt , k)
dn(Ω t, k) +k.
Here δ=h2
0−c0,Ω=√h0+δ,k=2δ
h0+δand k=h0−δ
h0+δ.
Proof As p1(0)2≥p1(0)2−p2(0)2=c0,wehavethateitherp1(0)≤−√c0or p1(0)≥√c0.
We describe how the expressions for ¯p(·)were found in the case p1(0)≤−
√c0.(Theex-
pressions for ¯p(·)in (ii) were found in a similar fashion.) Let ¯p(·)be an integral curve of
H(3)
1such that H(3)
1(¯p(0)) =h0,C( ¯p(0)) =c0,h0>√c0and ¯p1(0)≤−√c0.As ˙
¯p1=¯p2¯p3,
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
Fig. 3 Typical configurations of H(3)
1,c0=0 (colour figure online)
D.I. Barrett et al.
Fig. 4 Typical configurations of H(3)
1,c0=0 (colour figure online)
h0=¯p1(t) +1
2¯p3(t) and c0=¯p1(t )2−¯p2(t )2,weget
˙
¯p1=σ1¯p2
1−c0(2h0−2¯p1)
for some σ1∈{−1,1}. Equivalently,
d¯p1
(¯p2
1−c0)(2h0−2¯p1)=σ1dt. (3)
We transform (3) to standard form and apply an integral formula (see, e.g., [8]). First, (3)
may be rewritten as
d¯p1
[A1(¯p1+λ1)2+B1(¯p1+λ2)2][A2(¯p1+λ1)2+B2(¯p1+λ2)2]=σ1dt
where λ1=−(δ +h0),λ2=δ−h0,A1=1
2(1−h0
δ)<0, B1=1
2(1+h0
δ)>0, A2=1
2δ>0,
B2=−1
2δ<0andδ=h2
0−c0. The change of variables u=¯p1+λ1
¯p1+λ2yields
du
−(u2+B1
A1)(u2+B2
A2)=σ1(λ2−λ1)−A1A2dt.
Thus we have du
(h0+δ
h0−δ−u2)(u2−1)=σ1h0−δdt. (4)
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
We now apply the integral formula [8]
x
b
du
(a2−u2)(u2−b2)=1
and−1x
b,√a2−b2
a,b≤x≤a
to the left-hand side of (4), for a=h0+δ
h0−δ,b=1andx=¯p1(t )+λ1
¯p1(t )+λ2.(Herend(x, k) =1
dn(x,k) .)
We get
¯p1(t) +λ1
¯p1(t) +λ2=ndah0−δt,√a2−1
a.
(For convenience, we omit any translation in t.) Let Ω=a√h0−δ=√δ+h0and k=
√a2−1
a=2δ
δ+h0,k=h0−δ
h0+δ.Then ¯p1(t ) takes the form
¯p1(t) =(δ +h0)dn(Ωt , k) +(δ −h0)
dn(Ω t, k) −1.
Using the equation c0=¯p1(t)2−¯p2(t )2and the identity dn2(Ω t, k ) =k2cn2(Ω t, k ) +(k)2,
we obtain
¯p2(t) =2σ2δcn(Ωt , k)
dn(Ω t, k) −1
for some σ2∈{−1,1}. Likewise, as h0=¯p1(t ) +1
2¯p3(t)2,wehave
¯p3(t) =σ3√2δk sn(Ωt , k)
dn(Ω t, k) −1
for some σ3∈{−1,1}. We claim that ¯p(·)is an integral curve of
H(3)
1if and only if σ2=σ3.
By Lemma 3, it suffices to show that ˙
¯p1=¯p2¯p3if and only if σ2=σ3. Indeed,
˙
¯p1(t) −¯p2(t) ¯p3(t ) =2δ√2δk(1−σ2σ3)cn(Ω t, k) sn(Ω t , k)
(dn(Ω t, k) −1)2
and so ˙
¯p1=¯p2¯p3if and only if σ2=σ3.Inthiscasewehave ˙
¯p(t) =
H(3)
1(¯p(t)).Since
dn(Ω t, k) =1fort∈{2nK
Ω:n∈Z}and ¯p(·)has period 2K
Ω, we may take the domain of ¯p(·)
to be (0,2K
Ω). Moreover, by Lemma 1,¯p(·)is maximal.
It remains to be shown that any integral curve takes the form t→ ¯p(t +t0)in each case.
(i) Let σ=−sgn(p3(0)) ∈{−1,1}. (If p3(0)=0, then h0=p1(0)≤−√c0<0, a con-
tradiction.) We have sgn(¯p2|(0,K/Ω )(t )) =σand sgn(¯p2|(K/Ω ,2K/Ω)(t)) =−σ. Moreover,
limt→0¯p2(t) =−σ∞and limt→2K/Ω ¯p2(t) =σ∞. Therefore, since ¯p2(·)is continuous,
there exists t0∈(0,2K
Ω)such that ¯p2(t0)=p2(0).Then
¯p1(t0)2=c0+¯p2(t0)2=c0+p2(0)2=p1(0)2.
We have ¯p1(t0), p1(0)≤−√c0,andso ¯p1(t0)=p1(0).Furthermore,
¯p3(t0)2=2h0−¯p1(t0)=2h0−p1(0)=p3(0)2.
Since sgn(¯p3(t0)) =−σ=sgn(p3(0)), it follows that ¯p3(t0)=p3(0). Therefore, as t→
¯p(t +t0)and t→p(t) are integral curves of
H(3)
1passing through the same point at t=0,
they both solve the same Cauchy problem, and hence are identical.
D.I. Barrett et al.
(ii) Let ω=2h0−2√c0. From the identity h0=p1(t) +1
2p3(t)2we have p3(t )2=
2h0−2p1(t) ≤2h0−2√c0=ω2, i.e., −ω≤p3(t) ≤ω. Likewise, −ω≤¯p3(t) ≤ω.
Moreover, ¯p3(0)=ωand ¯p3(2K
Ω)=−ω. Therefore, since ¯p3(·)is continuous, there exists
t1∈[0,2K
Ω]such that ¯p3(t1)=p3(0).Then ¯p1(t1)=h0−1
2¯p3(t1)2=h0−1
2p3(0)2=p1(0).
Similarly,
¯p2
2(t1)2=¯p1(t1)2−c0=p1(0)2−c0=p2(0)2
and so ¯p2(t1)=±p2(0).Since ¯p1(·)and ¯p3(·)are even and ¯p2(·)is odd, we have ¯p1(−t1)=
¯p1(t1),¯p2(−t1)=−¯p2(t1)and ¯p3(t1)=¯p3(t1). Hence, there exists t0∈[−2K
Ω,2K
Ω](i.e.,
t0=t1or t0=−t1) such that ¯p(t0)=p(0). Therefore, as t→ ¯p(t +t0)and t→ p(t) are
integral curves of
H(3)
1passing through the same point at t=0, they both solve the same
Cauchy problem, and hence are identical.
Theorem 2.2 (Case I-b)Let p(·):(−ε, ε) →se(1,1)∗be an integral curve of
H(3)
1such
that H(3)
1(p(0)) =h0,C(p(0)) =c0>0, h0=√c0and p1(0)≤−
√c0.There exist t0∈
(−π
2Ω,π
2Ω)and σ∈{−1,1}such that p(t) =¯p(t +t0)for every t∈(−ε, ε),where ¯p(·):
(−π
2Ω,π
2Ω)→se(1,1)∗is given by
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
¯p1(t) =−h01+2tan2(h0t)
¯p2(t) =−σ2h0sec(h0t)tan(h0t)
¯p3(t) =2σh0sec(h0t).
Theorem 2.3 (Case I-c)Let p(·):(−ε, ε) →se(1,1)∗be an integral curve of
H(3)
1such
that H(3)
1(p(0)) =h0,C(p(0)) =c0>0and −√c0<h
0<√c0.There exist t0∈(0,2K
Ω)
and σ∈{−1,1}such that p(t) =¯p(t +t0)for every t∈(−ε, ε ),where ¯p(·):(0,2K
Ω)→
se(1,1)∗is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) =(δ +√c0)dn(Ωt , k) +(δ −√c0)
dn(Ωt , k) −1
¯p2(t) =σkδ(δ +2√c0)cn(Ωt , k)√dn(Ω t, k) +1
√dn(Ω t, k) +k[dn(Ω t, k) −1]
¯p3(t) =σ2(δ +√c0−h0)√dn(Ωt , k) +k√1−dn(Ωt , k)
dn(Ω t, k) −1.
Here δ=2(c0−√c0h0),Ω=1
26√c0−2h0+4δ,k=2δ
3√c0−h0+2δand k=
3√c0−h0−2δ
3√c0−h0+2δ.
Theorem 2.4 (Case I-d)Let p(·):(−ε, ε) →se(1,1)∗be an integral curve of
H(3)
1such
that H(3)
1(p(0)) =h0,C(p(0)) =c0>0and h0=−√c0.There exist t0∈Rand σ∈{−1,1}
such that p(t) =¯p(t +t0)for every t∈(−ε, ε ),where ¯p(·)is given by
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
¯p1(t) =h01+2csch2(−h0t)
¯p2(t) =−2σh
0coth(−h0t)csch(−h0t)
¯p3(t) =2σcsch(−h0t).
Furthermore,¯p|(−∞,0)(·)and ¯p|(0,∞)(·)are maximal.
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
Proof The expression for ¯p(·)was obtained by taking the limit h0→−√c0of the expres-
sions in Theorem 2.3.Let ¯p−(·)=¯p|(−∞,0)(·)and ¯p+(·)=¯p|(0,∞)(·). By Lemma 1,we
have that ¯p−(·)and ¯p+(·)are maximal. We prove that any integral curve is of the form
t→ ¯p−(t +t0)or t→ ¯p+(t +t0).Letσ=sgn(p2(0)) ∈{−1,1}and ς=σsgn(p3(0)) ∈
{−1,1}. (If p2(0)=0orp3(0)=0, then p(0)is an equilibrium point of
H(3)
1.) We have
sgn(¯p−,2(t)) =sgn(¯p+,2(t )) =σ. Moreover,
lim
t→0¯p+,2(t) =σ∞,lim
t→∞ ¯p+,2(t) =0,lim
t→−∞ ¯p−,2(t) =0,lim
t→0¯p−,2(t) =σ∞.
Suppose sgn(p3(0)) =1; then ς=σ.Since ¯p−,2(·)and ¯p+,2(·)are continuous and
sgn(p2(0)) =sgn(¯p−,2(t)) =sgn(¯p+,2(t )), there exists
t0∈ (−∞,0)if σ=−1
(0,∞)if σ=+1
such that ¯pσ,2(t0)=p2(0).Then
¯pσ,1(t0)2=¯pσ,2(t0)2+c0=p2(0)2+c0=p1(0)2.
We have p1(0), ¯pσ,1(t0)≤−√c0<0, and so ¯pσ,1(t0)=p1(0). Moreover,
¯pσ,3(t0)2=2h0−2¯pσ,1(t0)=2h0−2p1(0)=p3(0)2.
As sgn(¯pσ,3(t0)) =1=sgn(p3(0)),wehave ¯pσ,3(t0)=p3(0).Thatis, ¯pς(t0)=p(0).There-
fore, as t→ ¯pς(t +t0)and t→ p(t) are integral curves of
H(3)
1passing through the same
point at t=0, they both solve the same Cauchy problem, and hence are identical.
On the other hand, suppose sgn(p3(0)) =−1; then ς=−σ.Since ¯p−,2(·)and ¯p+,2(·)
are continuous and sgn(p2(0)) =sgn(¯p−,2(t)) =sgn(¯p+,2(t)), there exists
t0∈ (−∞,0)if σ=+1
(0,∞)if σ=−1
such that ¯p−σ,2(t0)=p2(0).From ¯pσ,1(t0)2=¯pσ,2(t0)2=p2(0)2=p1(0)2we again get
¯p−σ,1(t0)=p1(0). Similarly, as ¯pσ,3(t0)2=2h0−2¯pσ,1(t0)=2h0−2p1(0)=p3(0)2and
sgn(¯p−σ,3(t0)) =−1=sgn(p3(0)),wehave ¯p−σ,3(t0)=p3(0).Thatis, ¯pς(t0)=p(0).
Therefore t→ ¯pς(t +t0)and t→ p(t) both solve the same Cauchy problem, and so are
identical.
Theorem 2.5 (Case I-e)Let p(·):(−ε, ε) →se(1,1)∗be an integral curve of
H(3)
1such
that H(3)
1(p(0)) =h0,C(p(0)) =c0>0and h0<−√c0.There exist t0∈(−2K
Ω,2K
Ω)and
σ∈{−1,1}such that p(t) =¯p(t +t0)for every t∈(−ε, ε),where ¯p(·):(−2K
Ω,2K
Ω)→
se(1,1)∗is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) =(h0+δ)cn(Ω t , k) +(h0−δ) dn(Ω t, k )
dn(Ω t, k) +cn(Ω t, k)
¯p2(t) =σ2δ
dn(Ω t, k) +cn(Ω t, k)
¯p3(t) =−σ√2δksn(Ω t, k )
dn(Ω t, k) +cn(Ω t, k ) .
D.I. Barrett et al.
Here δ=h2
0−c0,Ω=√δ−h0,k=h0+δ
h0−δand k=2δ
δ−h0.
Theorem 2.6 (Case II-a)Let p(·):(−ε, ε) →se(1,1)∗be an integral curve of
H(3)
1such
that H(3)
1(p(0)) =h0>0and C(p(0)) =0.
(i) If p1(0)<0, then there exist t0∈Rand σ∈{−1,1}such that p(t) =¯p(t +t0)for every
t∈(−ε, ε),where ¯p(·)is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) =−h0csch2h0
2t
¯p2(t) =σh
0csch2h0
2t
¯p3(t) =σ2h0cothh0
2t.
Furthermore,¯p|(−∞,0)(·)and ¯p|(0,∞)(·)are maximal.
(ii) If p1(0)>0, then there exist t0∈Rand σ∈{−1,1}such that p(t ) =¯p(t +t0)for every
t∈(−ε, ε),where ¯p(·):R→se(1,1)∗is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) =h0sech2h0
2t
¯p2(t) =−σh
0sech2h0
2t
¯p3(t) =σ2h0tanhh0
2t.
Theorem 2.7 (Case II-b)Let p(·):(−ε, ε) →se(1,1)∗be an integral curve of
H(3)
1such
that H(3)
1(p(0)) =C(p(0)) =0and p1(0)<0. There exist t0∈Rand σ∈{−1,1}such that
p(t) =¯p(t +t0)for every t∈(−ε, ε),where ¯p(·)is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) =−2
t2
¯p2(t) =2σ
t2
¯p3(t) =2σ
t.
Furthermore,¯p|(−∞,0)(·)and ¯p|(0,∞)(·)are maximal.
Theorem 2.8 (Case II-c)Let p(·):(−ε, ε) →se(1,1)∗be an integral curve of
H(3)
1such
that H(3)
1(p(0)) =h0<0and C(p(0)) =0. There exist t0∈(−π
2Ω,π
2Ω)and σ∈{−1,1}
such that p(t) =¯p(t +t0)for every t∈(−ε, ε),where ¯p(·):(−π
2Ω,π
2Ω)→se(1,1)∗is
given by
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) =h0sec2−h0
2t
¯p2(t) =σh
0sec2−h0
2t
¯p3(t) =σ−2h0tan−h0
2t.
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
Theorem 2.9 (Case III) Let p(·):(−ε,ε ) →se(1,1)∗be an integral curve of
H(3)
1such
that H(3)
1(p(0)) =h0and C(p(0)) =c0<0. There exist t0∈(0,4K
Ω)and σ∈{−1,1}such
that p(t) =¯p(t +t0)for every t∈(−ε, ε),where ¯p(·):(−2K
Ω,2K
Ω)→se(1,1)∗is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) =(h0+δ)cn(Ω t , k) +(h0−δ)
cn(Ω t, k) +1
¯p2(t) =σΩ2dn(Ωt , k)
cn(Ω t, k) +1
¯p3(t) =σΩsn(Ωt , k)
cn(Ω t, k) +1.
Here δ=h2
0−c0,Ω=√2δand k=δ+h0
2δ.
5.3 The System H(3)
2
The equations of motion of the system H(3)
2(p) =p1+p2+1
2p2
3are
⎧
⎪
⎪
⎨
⎪
⎪
⎩
˙p1=p2p3
˙p2=p1p3
˙p3=−(p1+p2).
The equilibrium states of
H(3)
2are eμ
1=(μ, −μ, 0)and eν
2=(0,0,ν).
Proposition 8 The equilibrium states have the following behaviour:
(i) The states eμ
1are unstable.
(ii) The states eν
2are (spectrally)unstable.
Proof (i) Consider the states eμ
1,μ=0. The integral curve p(t) =(μeδt,−μeδt ,−δ),δ>0
satisfies p(0)−eμ
1=δ. Accordingly, for any open neighbourhood Nof eμ
1, there exists
δ>0 such that p(0)∈V. Since limt→∞ p(t)=∞, it follows that the states eμ
1,μ= 0
are unstable. Likewise, the integral curve p(t) =(δeδt ,−δeδt ,−δ) suffices to show that the
state e0
1is unstable.
(ii) As ∂H(3)
2
∂p3(eν
2)=ν, it follows from Lemma 2that the states eν
2are spectrally unsta-
ble.
Remark 4 The states (0,c
0),c0= 0and(h0,0),h0≥0 are generalized critical energy
states of H(3)
2(see Remark 3). Indeed, the points (h0,0),h0≥0 are critical energy
states in the usual sense. On the other hand, the curves f(s)=(−c0s4+4
4s2,c0s4−4
4s2,−2
s)and
g(s) =(−c0s2
4,c0s2
4,0)suffice to show that (0,c
0),c0= 0 are generalized critical energy
states.
The map ψ:(p1,p
2,p
3)→(p2,p
1,p
3)is a symmetry of H(3)
2such that C◦ψ=−C.
Accordingly, we may assume c0≥0. For the cases c0>0andc0=0, there are several
further subcases (see Table 3). The (generalized) critical energy states of H(3)
2are graphed
in Fig. 5; the typical configurations are graphed in Fig. 6.
D.I. Barrett et al.
Tab le 3 Index of cases for the
integral curves of H(3)
2Conditions Designation
c0>0h0>0CaseI-a
h0=0CaseI-b
h0<0CaseI-c
c0=0h0>0 Case II-a
h0=0 Case II-b
h0<0 Case II-c
Fig. 5 Critical energy states
for H(3)
2(colour figure online)
Theorem 3.1 (Case I-a, case II-a)Let p(·):(−ε, ε) →se(1,1)∗be an integral curve of
H(3)
2such that H(3)
2(p(0)) =h0>0and C(p(0)) =c0≥0.
(i) If p1(0)≤−
√c0,then there exists t0∈Rsuch that p(t) =¯p(t +t0)for every t∈
(−ε, ε),where ¯p(·)is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) =− 1
2h0h2
0csch2h0
2t+c0sinh2h0
2t
¯p2(t) =− 1
2h0h2
0csch2h0
2t−c0sinh2h0
2t
¯p3(t) =−2h0cothh0
2t.
Furthermore,¯p|(−∞,0)(·)and ¯p|(0,∞)(·)are maximal.
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
Fig. 6 Typical configurations of H(3)
2(colour figure online)
D.I. Barrett et al.
(ii) If p1(0)≥√c0,then there exists t0∈Rsuch that p(t) =¯p(t +t0)for every t∈(−ε, ε),
where ¯p(·):R→se(1,1)∗is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) =1
2h0h2
0sech2h0
2t+c0cosh2h0
2t
¯p2(t) =1
2h0h2
0sech2h0
2t−c0cosh2h0
2t
¯p3(t) =−2h0tanhh0
2t.
(iii) If c0=0and p1(0)+p2(0)=0(with p1(0)and p2(0)not both zero), then there
exist t0∈Rand σ, ς ∈{−1,1}such that p(t) =¯p(t +t0)for every t∈(−ε, ε),where
¯p(·):R→se(1,1)∗is given by
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
¯p1(t) =ςe−σ√2h0t
¯p2(t) =−ςe−σ√2h0t
¯p3(t) =σ2h0.
Proof Standard computations yield the expressions for ¯p(·)shown. We prove that for
case (i) every integral curve takes the form t→ ¯p−(t +t0)or t→ ¯p+(t +t0),where
¯p−(·)=¯p|(−∞,0)(·)and ¯p+(·)=¯p|(0,∞)(·). (The arguments for (ii) and (iii) are analogous.)
By Lemma 1,wehavethat ¯p−(·)and ¯p+(·)are maximal. Let ς=−sgn(p3(0)) ∈{−1,1}.
(If p3(0)=0, then p1(0)>0, a contradiction.) We have
lim
t→0¯p+,2(t) =−∞,lim
t→∞ ¯p+,2(t) =∞,lim
t→0¯p−,2(t) =−∞,lim
t→−∞ ¯p−,2(t) =∞.
Since ¯p−,2(·)and ¯p+,2(·)are continuous, there exists
t0∈ (−∞,0)if ς=−1
(0,∞)if ς=+1
such that ¯pς,2(t0)=p2(0).Then
¯pς,1(t0)2=¯pς,2(t0)2+c0=p2(0)2+c0=p1(0)2.
We have ¯pς,1(t0), p1(0)≤−√c0≤0, and so ¯pς,1(t0), p1(0)<0. (If c0=0andp1(0)=0,
then p(0)is an equilibrium point.) Hence ¯pς,1(t0)=p1(0).Furthermore,
¯pς,3(t0)2=2h0−¯pς,1(t0)−¯pς,2(t0)=2h0−p1(0)−p2(0)=p3(0)2.
As sgn(¯pς,3(t0)) =−ς=sgn(p3(0)), it follows that ¯pς,3(t0)=p3(0). Therefore, as t→
¯pς(t +t0)and t→p(t) are integral curves of
H(3)
2passing through the same point at t=0,
they both solve the same Cauchy problem, and hence are identical.
Theorem 3.2 (Case I-b, case II-b)Let p(·):(−ε, ε) →se(1,1)∗be an integral curve of
H(3)
2such that H(3)
2(p(0)) =0and C(p(0)) =c0≥0. There exists t0∈Rsuch that p(t) =
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
¯p(t +t0)for every t∈(−ε, ε ),where ¯p(·)is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) =−4+c0t4
4t2
¯p2(t) =−4−c0t4
4t2
¯p3(t) =−2
t.
Furthermore,¯p|(−∞,0)(·)and ¯p|(0,∞)(·)are maximal.
Theorem 3.3 (Case I-c)Let p(·):(−ε, ε) →se(1,1)∗be an integral curve of
H(3)
2such
that H(3)
2(p(0)) =h0<0and C(p(0)) =c0≥0. There exists t0∈(−π
2Ω,π
2Ω)such that
p(t) =¯p(t +t0)for every t∈(−ε, ε),where ¯p(·):(−π
2Ω,π
2Ω)→se(1,1)∗is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) =1
2h0h2
0sec2−h0
2t+c0cos2−h0
2t
¯p2(t) =1
2h0h2
0sec2−h0
2t−c0cos2−h0
2t
¯p3(t) =−2h0tan−h0
2t.
5.4 The System H(5)
3,α
The equations of motion of the system H(5)
3,α(p) =α(p1+p2)+1
2[(p1+p2)2+p2
3],α>0
are
⎧
⎪
⎪
⎨
⎪
⎪
⎩
˙p1=p2p3
˙p2=p1p3
˙p3=−(p1+p2)(p1+p2+α).
The equilibrium states of
H(5)
3,α are eμ
1=(μ, −μ, 0),eμ
2=(μ, −(α +μ), 0)and eν
3=
(0,0,ν).
Proposition 9 The equilibrium states have the following behaviour:
(i) The states eμ
1are unstable.
(ii) The states eμ
2are stable.
(iii) The states eν
3are (spectrally)unstable.
Proof (i) Consider the states eμ
1,μ= 0. We have that p(t) =(μeδt,−μeδt ,−δ) is
an integral curve of
H(5)
3,α (for any δ>0) such that p(0)−eμ
1=δ. Accordingly,
for any neighbourhood Nof eμ
1there exists δ>0 such that p(0)∈N.Furthermore,
limt→∞ p(t)=∞. Therefore the states eμ
1,μ= 0 are unstable. Similarly, the integral
curve p(t) =(δeδt ,−δeδt ,−δ) suffices to show that the state e0
1is unstable.
D.I. Barrett et al.
Tab le 4 Index of cases for
the integral curves of H(5)
3,α Conditions Designation
c0>0h0>0CaseI-a
h0=0CaseI-b
h0<0CaseI-c
c0=0h0>0 Case II-a
h0=0 Case II-b
h0<0 Case II-c
Fig. 7 Critical energy states
for H(5)
3,α (colour figure online)
(ii) Consider the states eμ
2.LetHλ=λ0H(5)
3,α +λ1C,whereλ0=1andλ1=0. Then
dHλ(eμ
2)=0 and the restriction of
d2Hλeμ
2=⎡
⎣
110
110
001
⎤
⎦
to W×Wis positive definite. Here W=ker dH(5)
3,α (eμ
2)∩ker dC(eμ
2)=span{E∗
1−
μ
α+μE∗
2,E∗
3}when α+μ= 0, and W=span{E∗
2,E∗
3}when α+μ=0. Thus the states
eμ
2are stable.
(iii) Consider the states eν
3.Wehave ∂H (5)
3,α
∂p3(eν
3)=ν, hence by Lemma 2the states eν
3are
(spectrally) unstable.
Remark 5 The states (0,c
0),c0= 0and(h0,0),h0≥0 are generalized critical en-
ergy states of H(5)
3,α (see Remark 3). Indeed, (h0,0),h0≥0 are critical energy states in
the usual sense, whereas f(s) =(−c0(1+α2s2)2+4α2
4α(α2s2+1),c0(1+α2s2)2−4α2
4α(α2s2+1),−2α2s
α2s2+1)and g(s) =
(−c0(α2s2+1)
4α,c0(α2s2+1)
4α,0)are sufficient to show that the states (0,c
0),c0= 0 are gener-
alized critical energy states.
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
The cylinders (H (5)
3,α)−1(h0)degenerate to a line exactly when 2h0+α2=0; hence we
assume 2h0+α2>0. The map ψ:(p1,p
2,p
3)→(p2,p
1,p
3)is a symmetry of H(5)
3,α such
that C◦ψ=−C, and so we may assume c0≥0. The cases c0>0andc0=0 are further
subdivided into several subcases (see Table 4). Figure 7illustrates the (generalized) critical
energy states of H(5)
3,α ; the typical configurations of the system are graphed in Fig. 8.
Theorem 4.1 (Case I-a, case II-a)Let p(·):(−ε, ε) →se(1,1)∗be an integral curve of
H(5)
3,α such that C(p(0)) =c0≥0, H(5)
3,α(p(0)) =h0>0and 2h0+α2>0.
(i) If p1(0)≥√c0and p1(0)+p2(0)=0, then there exists t0∈Rsuch that p(t) =¯p(t +
t0)for every t∈(−ε, ε ),where ¯p(·):R→se(1,1)∗is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) +¯p2(t) =2h0
α+2h0+α2cosh(√2h0t)
¯p1(t) −¯p2(t) =c0
2h0α+2h0+α2cosh(2h0t)
¯p3(t) =√2h02h0+α2(α −2h0+α2)tanh(h0
2t)
h0+(h0+α2−α2h0+α2)tanh2(h0
2t)
.
(ii) If p1(0)≤−
√c0and p1(0)+p2(0)= 0, then there exists t0∈Rsuch that p(t) =
¯p(t +t0)for every t∈(−ε, ε ),where ¯p(·):R→se(1,1)∗is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) +¯p2(t) =2h0
α−2h0+α2cosh(√2h0t)
¯p1(t) −¯p2(t) =c0
2h0α−2h0+α2cosh(2h0t)
¯p3(t) =−√2h02h0+α2(α +2h0+α2)tanh(h0
2t)
h0+(h0+α2+α2h0+α2)tanh2(h0
2t)
.
(iii) If c0=0and p1(0)+p2(0)=0(with p1(0)and p2(0)not both zero), then there
exists t0∈Rand σ, ς ∈{−1,1}such that p(t) =¯p(t +t0)for every t∈(−ε, ε ),where
¯p(·):R→se(1,1)∗is given by
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
¯p1(t) +¯p2(t) =0
¯p1(t) −¯p2(t) =2ςe−σ√2h0t
¯p3(t) =σ2h0.
Proof We briefly describe how the expressions for ¯p(·)were found in (i). (A similar ap-
proach may be used for (ii), whereas the integration for case (iii) is straightforward.) Let
ω=2h0+α2and parametrize the cylinder (H (5)
3,α)−1(h0)by θand zas follows:
⎧
⎪
⎪
⎨
⎪
⎪
⎩
¯p1+¯p2=ωcos θ−α
¯p1−¯p2=z
¯p3=ωsin θ.
D.I. Barrett et al.
Fig. 8 Typical configurations of H(5)
3,α (colour figure online)
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
From ˙
¯p3=−(¯p1+¯p2)( ¯p1+¯p2+α),weget ˙
θ=α−ωcos θ. Hence
θ(t)=−2tan−1!(ω −α)
√2h0
tanhh0
2t".
(For convenience, we have omitted any translation in the independent variable.) Using the
identity ¯p1(t)2−¯p2(t)2=c0and solving for ¯p1(t ) and ¯p2(t ) yields the given expressions.
It remains to be shown that every integral curve takes the form t→ ¯p(t +t0).
(i) We have p1(0)+p2(0)>0. (If p1(0)+p2(0)<0, then p1(0)<0, a contradic-
tion.) Thus ω2=(p1(t) +p2(t) +α)2+p3(t)2implies that −ω<p
3(t)<ω. Similarly,
−ω< ¯p3(t) < ω. Furthermore, limt→−∞ ¯p3(t ) =ωand limt→∞ ¯p3(t) =−ω.Since ¯p3(·)is
continuous, there exists t0∈Rsuch that ¯p3(t0)=p3(0).Then
¯p1(t0)+¯p2(t0)+α2=ω2−¯p3(t0)2=ω2−p3(0)2=p1(0)+p2(0)+α2
and so ¯p1(t0)+¯p2(t0)+α=±(p1(0)+p2(0)+α).Butp1(0)+p2(0)+α>0and ¯p1(t0)+
¯p2(t0)+α>0, and so ¯p1(t0)+¯p2(t0)=p1(0)+p2(0). Thus, from
¯p1(t0)−¯p2(t0)¯p1(t0)+¯p2(t0)=c0=p1(0)−p2(0)p1(0)+p2(0)
we get ¯p1(t0)=p1(0)and ¯p2(t0)=p2(0). Therefore, as t→ ¯p(t +t0)and t→ p(t) are
integral curves of
H(5)
3,α passing through the same point at t=0, they both solve the same
Cauchy problem, and hence are identical.
(ii) We have −α−ω≤p1(t ) +p2(t) < 0. (From ω2=(p1(t ) +p2(t) +α)2+p3(t)2we
get p1(t) +p2(t) ≥−α−ω2. Also, if p1(t) +p2(t ) > 0, then p1(t) > 0, a contradiction.)
Furthermore, ¯p1(0)+¯p2(0)=−α−ω2.Sincet→ ¯p1(t ) +¯p2(t) is continuous, there exists
t1∈Rsuch that ¯p1(t1)+¯p2(t1)=p1(0)+p2(0).Then
¯p1(t1)2−¯p2(t1)2=c0=p1(0)2−p2(0)2
implies that ¯p1(t1)=p1(0)and ¯p2(t1)=p2(0). Similarly,
¯p3(t1)2=ω2−¯p1(t1)+¯p2(t1)+α2=ω2−p1(0)+p2(0)+α2=p3(0)2
and so ¯p3(t1)=±p3(0).Since ¯p1(·),¯p2(·)are even and ¯p3(·)is odd, we have ¯p1(−t1)=
¯p1(t1),¯p2(−t1)=¯p2(t1)and ¯p3(−t1)=−¯p3(t1). Hence there exists t0∈R(either t0=−t1
or t0=t1) such that ¯p(t0)=p(0). Therefore t→ ¯p(t +t0)and t→ p(t) both solve the
same Cauchy problem, and hence are identical.
(iii) Let σ=sgn(p3(0)) and ς=sgn(p1(0)). (If p1(0)=0orp3(0)=0, then p(0)is an
equilibrium point.) We have
lim
t→−∞ ¯p1(t) = ς∞if σ=+1
0ifσ=−1and lim
t→∞ ¯p1(t) = 0ifσ=+1
ς∞if σ=−1.
Hence, as sgn(¯p1(t )) =sgn(p1(t)) for every tand ¯p1(·)is continuous, there exists t0∈R
such that ¯p1(t0)=p1(0).Then ¯p2(t0)=−¯p1(t0)=−p1(0)=p2(0)and ¯p3(t0)=σ√2h0=
p3(0). Therefore t→ p(t) and t→ ¯p(t +t0)both solve the same Cauchy problem, and
hence are identical.
D.I. Barrett et al.
Theorem 4.2 (Case I-b, case II-b)Let p(·):(−ε, ε) →se(1,1)∗be an integral curve of
H(5)
3,α such that C(p(0)) =c0≥0and H(5)
3,α(p(0)) =0. There exists t0∈Rsuch that p(t) =
¯p(t +t0)for every t∈(−ε, ε ),where ¯p(·):R→se(1,1)∗is given by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) +¯p2(t) =− 2α
α2t2+1
¯p1(t) −¯p2(t) =−c0
2αα2t2+1
¯p3(t) =− 2α2t2
α2t2+1.
Theorem 4.3 (Case I-c, case II-c)Let p(·):(−ε, ε) →se(1,1)∗be an integral curve of
H(5)
3,α such that C(p(0)) =c0≥0, H(5)
3,α(p(0)) =h0<0and 2h0+α2>0. There exists
t0∈Rsuch that p(t) =¯p(t +t0)for every t∈(−ε, ε ),where ¯p(·):R→se(1,1)∗is given
by
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
¯p1(t) +¯p2(t) =2h0
α−2h0+α2cos(√−2h0t)
¯p1(t) −¯p2(t) =c0
2h0α−2h0+α2cos(−2h0t)
¯p3(t) =√−2h02h0+α2(α +2h0+α2)tan(−h0
2t)
h0−(h0+α2+α2h0+α2)tan2(−h0
2t)
.
6 Nonplanar Systems, Type II
For the nonplanar, type II systems we consider only stability. As before, we graph the critical
energy states for each system, the level sets H−1(h0)and C−1(c0)and their intersection (for
typical configurations), as well as the equilibrium points.
Throughout this section, we again parametrize equilibria by μ, ν ∈R,ν=0.
6.1 The System H(5)
1,α
The equations of motion for H(5)
1,α(p) =αp1+1
2[(p1+p2)2+p2
3],α>0are
⎧
⎪
⎪
⎨
⎪
⎪
⎩
˙p1=p2p3
˙p2=p1p3
˙p3=−αp 2−(p1+p2)2.
The equilibrium states are eμ
1=(1
αμ(μ +α),−1
αμ2,0)and eν
2=(0,0,ν).InFig.9we
graph the critical energy states of this system. The typical configurations are graphed in
Fig. 10 and Fig. 11.
Proposition 10 The equilibrium states have the following behaviour:
(i) The states eμ
1,μ∈(−∞,−α
3)are stable.
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
Fig. 9 Critical energy states
for H(5)
1,α (colour figure online)
(ii) The state eμ
1,μ=−α
3is unstable.
(iii) The states eμ
1,μ∈(−α
3,0)are (spectrally)unstable.
(iv) The state eμ
1,μ=0is unstable.
(v) The states eμ
1,μ∈(0,∞)are stable.
(vi) The states eν
2are (spectrally)unstable.
Proof (i) Let Hλ=λ0H(5)
1,α +λ1C,whereλ0=1andλ1=−α
2μ.ThendHλ(eμ
1)=0and
d2Hλ(eμ
1)|W×Wis positive definite, where W=ker dH(5)
1,α(eμ
1)∩ker dC(eμ
1)=span{E∗
1−
μ+α
μE∗
2,E∗
3}. Hence the states eμ
1,μ∈(−∞,−α
3)are stable.
(ii) The integral curve
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
p1(t) =α6
45 +αt(αt −6)−2
(αt −3)2−2
9
p2(t) =α6
45 +αt(αt −6)−2
(αt −3)2−1
9
p3(t) =− 72α
(αt −3)(α t (αt −6)+45)
satisfies limt→−∞ p(t) −eμ
1=0. Let ε=1
2p(0)−eμ
1>0. Then for every neighbour-
hood Nof eμ
1contained in the ε-ball Bεabout eμ
1, there exists t1<0 such that p(t1)∈N.
However, p(0)/∈Bε, and so the state eμ
1,μ=−α
3is unstable.
(iii) The linearization of the system at has eigenvalues λ1=0andλ2,3=
±√−μ(α +3μ). There exists an eigenvalue with positive real part exactly when μ∈(−α
3,0).
Hence the states eμ
1,μ∈(−α
3,0)are spectrally unstable.
(iv) We have that p(t) =(−2
αt2,2
αt2,2
t)is an integral curve of
H(5)
1,α such that
limt→−∞ p(t)−eμ
1=0. Accordingly, for every neighbourhood Nof eμ
1there exists t1<0
such that p(t1)∈N. Furthermore, limt→0p(t)=∞. Thus the state eμ
1,μ=0 is unstable.
(v) The function Hλof (i) suffices to show that the states eμ
1,μ∈(0,∞)are stable.
(vi) As ∂H(5)
1,α
∂p3(eν
2)=ν, it follows from Lemma 2that the states eν
2are spectrally unsta-
ble.
D.I. Barrett et al.
Fig. 10 Typical configurations of H(5)
1,α , cases athrough f(colour figure online)
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
Fig. 11 Typical configurations of H(5)
1,α , cases gthrough (colour figure online)
D.I. Barrett et al.
Fig. 12 Critical energy states
for H(5)
2(colour figure online)
6.2 The System H(5)
2
The equations of motion for H(5)
2(p) =p1−p2+1
2[(p1+p2)2+p2
3]are
⎧
⎪
⎨
⎪
⎩
˙p1=p2p3
˙p2=p1p3
˙p3=−p1(p1+p2−1)−p2(p1+p2+1).
The equilibrium states are eμ
1=(1
2(μ +μ2), 1
2(μ −μ2), 0)and eν
2=(0,0,ν). The critical
energy states for this system are graphed in Fig. 12; the typical configurations are graphed
in Fig. 13 (by symmetry, we may assume c0≥0).
Proposition 11 The equilibrium states have the following behaviour:
(i) The states eμ
1,μ=0are stable.
(ii) The state eμ
1,μ=0is unstable.
(iii) The states eν
2are (spectrally)unstable.
Proof (i) Suppose μ= 1. Let Hλ=λ0H(5)
2+λ1C,whereλ0=−μand λ1=1. We
have dHλ(eμ
1)=0andd2Hλ(eμ
1)|W×Wpositive definite, where W=ker dH(5)
2(eμ
1)∩
ker dC(eμ
1)=span{E∗
1+μ+1
μ−1E∗
2,E∗
3}. Therefore the states eμ
1,μ/∈{0,1}are stable. Sup-
pose μ=1. We have H(5)
2(e1
1)=3
2and C(e1
1)=1. It is straightforward to show that locally
about e1
1we have (H (5)
2)−1(3
2)∩C−1(1)={e1
1}. Hence, by the continuous energy-Casimir
method, the state e1
1is stable.
(ii) We have that p(t) =(−1
t2,1
t2,2
t)is an integral curve of
H(5)
2such that limt→−∞ p(t)
=0. Accordingly, for every neighbourhood Vof eμ
1there exists t1<0 such that p(t1)∈V.
Furthermore, limt→0p(t)=∞. Thus the state eμ
1,μ=0 is unstable.
(iii) As ∂H(5)
2
p3(eν
2)=ν, it follows from Lemma 2that the states eν
2are spectrally unsta-
ble.
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
Fig. 13 Typical configurations of H(5)
2(colour figure online)
D.I. Barrett et al.
Tab le 5 Taxonomy of homogeneous systems
Type
dim(Sym(HQ))
Eq. index of HQ
Spherical
Hyperboloidal
Cylindrical
Planar
Systems
Ruled 5 (0,2,0,0)••H1
7(0,1,0,0)••H2
Planar 2 (1,1,0,0)••H3
Nonplanar, type I 0 (3,0,0,0)••• H4
1(2,0,0,0)•• H5
7 Concluding Remark
In Sect. 3we partitioned the quadratic (inhomogeneous) systems into four classes, viz., the
ruled, planar, nonplanar type I and nonplanar type II systems. Clearly, if two systems are
affinely equivalent, then they must belong to the same class. We identify several additional
invariants that facilitate the identification of the normal form of a system.
The dimension of the symmetry group is a simple invariant. (Likewise, for an inho-
mogeneous system HA,Q, the dimension of Sym(HQ)is also an invariant.) Regarding the
equilibria, we have that the set of equilibria for each system is the union of (a finite number
of) lines, curves, planes or surfaces. We define the equilibrium index of a system to be a
tuple (i,j,k,),whereiis the number of lines; jthe number of curves that are not lines; k
the number of planes; and the number of surfaces that are not planes. Clearly, equivalent
systems have the same equilibrium index. Moreover, for an inhomogeneous system HA,Q,
the equilibrium index of the corresponding homogeneous system HQis another invariant.
More invariants may be found by identifying the type of quadratic constants of motion
that a system admits. We say that a system has spherical symmetry if it admits a constant
of motion of the form K(p) =Q(p −q),whereQis a positive definite quadratic form.
Likewise, we say that a system has one of the following types of symmetry if it admits a
constant of motion of the form K(p) =Q(p −q),whereQis a quadratic form with the
corresponding signature:
–hyperboloidal symmetry: signature (0,2,1).
–hyp-cylindrical symmetry: signature (1,1,1).
–cylindrical symmetry: signature (1,2,0).
–planar symmetry: signature (2,1,0).
(The signature of Qis the triple (n0,n
+,n
−),wheren0is the number of zero eigenvalues;
n+the number of positive eigenvalues; and n−the number of negative eigenvalues.) Clearly,
every system on se(1,1)∗
−admits the Casimir function as a hyp-cylindrical symmetry.
Equivalent systems must have the same types of symmetry. Accordingly, these invariants
may be useful in a more general classification of inhomogeneous systems. For instance, as
H(5)
3,α does not have a spherical or planar symmetry, it cannot be equivalent to any system
on so(3)∗
−or the Heisenberg Lie–Poisson space (h3)∗
−. (However, as it has cylindrical and
hyperboloidal symmetries, we cannot rule out the possibility of it being equivalent to a
system on se(2)∗
−or the pseudo-orthogonal Lie–Poisson space so(2,1)∗
−.)
Quadratic Hamilton–Poisson Systems on se(1,1)∗
−...
Tab le 6 Taxonomy of inhomogeneous systems
Type
dim(Sym(HA,Q))
dim(Sym(HQ))
Eq. index of HA,Q
Eq. index of HQ
Spherical
Hyperboloidal
Cylindrical
Planar
Systems
Ruled 5 5 (0,0,0,1)(0,2,0,0)••H(1)
2
57(0,0,0,1)(0,1,0,0)••H(2)
1
55(0,2,0,0)(0,2,0,0)••H(1)
1
67(0,2,0,0)(0,1,0,0)••H(2)
2
712(0,1,0,0)–••H(0)
1
Planar 4 12 (1,0,0,0)–••H(0)
2,α
Nonplanar, type I 0 2 (2,0,0,0)(1,1,0,0)•H(3)
1
00(3,0,0,0)(3,0,0,0)••• H(4)
1,α
12(2,0,0,0)(1,1,0,0)•H(3)
2
11(3,0,0,0)(2,0,0,0)•• H(5)
3,α
25(1,0,0,0)(0,2,0,0)H
(1)
3,α
37(1,0,0,0)(0,1,0,0)H
(2)
3,δ
Nonplanar, type II 0 1 (1,0,1,0)(2,0,0,0)•• H(5)
1,α,H(5)
2
00(1,0,2,0)(3,0,0,0)••• H(4)
2,α1,α2
In most cases, these above-mentioned invariants are sufficient to uniquely determine the
equivalence class of a system. In Table 6we list the partition of the inhomogeneous nor-
mal forms according to the invariants discussed above. (For the sake of completeness, the
homogeneous normal forms are likewise partitioned in Table 5.)
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