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NONLINEAR OSCILLATIONS

The aim of this chapter is to investigate the dynamics in the neighbourhood of equilib- ria of Hamiltonian systems, both in the linear and in the nonlinear case. The argument has strong connections with the general problem of equilibria for systems of differen- tial equations, but here emphasis will be given to the aspects that make Hamiltonian system quite peculiar.

Consider for a moment a generic system of differential equations

˙x

j

= X

j

(x

1

, . . . , x

n

) , x ∈ D ⊂ R

n

,

where D is open. Assume that this system has an equilibrium point at (x

1

, . . . , x

n

) . It is well known that in a neighbourhood of the equilibrium the system may often be approximated by a linear one

(5.1) ˙x = Ax

where A is a n×n real matrix with entries A

j,k

=

∂X∂xj

k

(x

1

, . . . , x

n

) . The general method for solving such a system of differential equation was discovered by Lagrange [60]. The complete classification of the equilibria as, e.g., nodes, saddles, foci and centers is due to Poincar´e [85], who used the concept of normal form for the system. The construction of the normal form is based on finding a suitable linear transformation of coordinates which changes the system into a particularly simple one.

The case of Hamiltonian systems may be treated with the same methods, of course. However, it is interesting to investigate to which extent the concept of normal form may be introduced without losing the canonical character of the equations. The question is given a more precise formulation as follows. Let the Hamiltonian H(q, p) have an equilibrium at (q, p). In a neighbourhood of the equilibrium let us introduce local coordinates x = q − q , y = p − p . The transformation is canonical, and we can approximate the Hamiltonian with is quadratic part

(5.2) H(x, y) = X

j,k

A

j,k

x

j

x

k

+ B

j,k

x

j

y

k

+ C

j,k

y

j

y

k

 .

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For, the term H(q, p) can be ignored, being a constant, the linear part vanishes at equilibrium, and we neglect contributions of order higher than two. The resulting canonical equations are linear, which takes us back to the general problem above.

However, we may ask for a normal form of the quadratic Hamiltonian (5.2), i.e., for a linear transformation to normal form that is canonical. This is the subject of the first part of the chapter.

Having settled the linear problem, we may investigate to which extent the linear approximation applies to the dynamics of the complete system. In particular the ques- tion of stability of the equilibrium is raised. As a general fact the theory of Lyapounov states that in most cases the answer can be formulated by analyzing the linear system.

This is indeed the case when the equilibria are nodes, saddle points or foci in the ter- minology of Poincar´e. These points are also collectively called hyperbolic equilibrium points. However, a nontrivial exception to Lyapounov theory is represented by center points, which are the most interesting ones in the Hamiltonian case. These points are also called elliptic equilibrium points, and this is the name that I will adopt from now on. The second part of the chapter is devoted to a discussion of this problem based on the search for first integrals.

5.1 Normal form for linear systems

Let us briefly recall how a general system (5.1) may be given a normal form. Applying the linear transformation

(5.3) x = Mξ , det M 6= 0

the system is changed to

˙ξ = Λξ , Λ = M

−1

AM .

Lemma 5.1: Consider the linear system ˙x = Ax with a real matrix A and assume that the eigenvalues λ

1

, . . . , λ

n

of the matrix A are distinct. Then there exists a com- plex matrix M such that the linear transformation x = Mξ with ξ ∈ C

n

gives the system the normal form

(5.4) ˙ξ = Λξ , Λ = diag(λ

1

, . . . , λ

n

) with Λ = M

−1

AM . The matrix M is written as

M = (w

1

, . . . , w

n

)

where the column vectors w

1

, . . . , w

n

are the complex eigenvectors of A . If the eigen- values are real, then the matrix M is real, and equation (5.4) is restricted to R

n

. The lemma is a standard argument in linear algebra, hence I omit the proof.

Remark. The matrix M is not uniquely determined, due to the fact that the eigen-

values are determined up to a multiplicative factor. This implies that we can multiply

M by a diagonal matrix with non zero diagonal elements. A general statement is: if two

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different matrices M and ˜ M both satisfy the relation M

−1

AM = Λ and ˜ M

−1

A ˜ M = Λ then

(5.5) M ˜ = MD , D = diag(d

1

, . . . , d

n

) with non vanishing d

1

, . . . , d

n

.

If the matrix A has a pair of complex eigenvalues then the following lemma will be useful.

Lemma 5.2: Let λ = µ + iω with ω 6= 0 be a complex eigenvalue of the real matrix A , and let w = u + iv , with u, v real vectors, be the corresponding eigenvector, so that Aw = λw. Then the following statements hold true.

(i) w

= u − iv is the eigenvector corresponding to the eigenvalue λ

, that is Aw

= λ

w

.

(ii) The real vectors u and v are linearly independent.

(iii) We have

Au = µu − ωv , Av = ωu + µv .

Proof. (i) If Aw = λw then by calculating the complex conjugate of both members we have (Aw)

= (λw)

, which in view of A

= A coincides with Aw

= λ

w

. This proves that w

is the eigenvector corresponding to λ

.

(ii) We know that w and w

are linearly independent over the complex numbers, being eigenvectors corresponding to different eigenvalues λ, λ

(this is a known general property). By contradiction, let us assume that u = αv for some real α 6= 0 . Then we have w = u + iv = (α + i)v and w

= (α − i)v, so that w =

α+iα−i

w

, contradicting the linear independence of the eigenvectors over the complex numbers. We conclude that u and v must be linearly independent, as claimed.

(iii) Just calculate

Au = λw + λ

w

2 = µ w + w

2 + iω w − w

2 = µu − ωv , Av = λw − λ

w

2i = µ w − w

2i + iω w + w

2i = ωu + µv ,

and the proof is complete. Q.E.D.

A compact and elegant form of the flow is obtained by introducing a linear evolu- tion operator U(t) as follows. First note that the solution may be written in exponential form as

1

x (t) = e

tA

x

0

, where e

tA

= P

k≥0

t

k

A

k

/k! . The problem is how to explicitly calculate the exponential of the matrix tA . The problem is easily solved in complex

1 It is assumed that the reader knows how to assign a meaning to the exponential of a matrix. If this is not the case, he or she may just check that the expression given here is actually the solution of the equation by formally calculating the time derivative. The question of convergence will be discussed later, in chapter 6, in the more general context of Lie series. For a reference see, e.g., Arnold’s book [6].

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coordinates ξ because we have e

= I + tΛ + t

2

Λ

2

2! + t

3

Λ

3

3! + . . . = diag e

λ1t

, . . . , e

λnt

 ,

where I is the identity matrix. This is easily checked in view of Λ

k

= diag(λ

k1

, . . . , λ

kn

) (prove it by induction). For the real matrix A , using A = MΛM

−1

we have A

0

= I and, by induction,

A

k

= AA

k−1

= MΛM

−1



k−1

M

−1

 = MΛ

k

M

−1

.

Remark that A

k

is a power of real matrix, hence it is real even if the eigenvalues are complex numbers. Thus we conclude that

e

tA

= X

k≥0

t

k

A

k

k! = X

k≥0

t

k

k

M

−1

k! = M

 X

k≥0

t

k

Λ

k

k!



M

−1

= Me

M

−1

. The resulting matrix is real. This leads to the following

Proposition 5.3: Let the eigenvalues of the real matrix A be distinct. Then the flow of the linear system of differential equations ˙x = Ax satisfying the initial condition x(0) = x

0

is

(5.6) φ

t

x

0

= U(t)x

0

, U (t) = Me

M

−1

where e

= diag e

λ1t

, . . . , e

λnt

 is constructed using the eigenvalues λ

1

, . . . , λ

n

of A and M is as in lemma 5.1. The evolution operator U(t) is a real matrix.

A few more consideration can be added in case the real matrix A is symmetric.

Corollary 5.4: If the matrix A is symmetric, then the eigenvalues are real and the matrix M can be chosen to be orthogonal, i.e., M

M = I .

This is a straightforward consequence of known properties of symmetric matrices.

The purpose here is to stress is the analogy with the Hamiltonian case, where the transformation matrix M should be symplectic.

5.2 Normal form of a quadratic Hamiltonian

Let us now enter the discussion concerning a canonical system with a Hamiltonian of the form (5.2). As already pointed out the problem is to look for a normal form of the linear system of Hamilton’s equations keeping the canonical form. This is tantamount to saying that we look for a normal form of the Hamiltonian.

2

2 This problem is a common topic in textbooks when the Lagrangian formalism is used.

One is usually lead to study a Lagrangian L( ˙q, q) = T ( ˙q) − V (q) where both the kinetic energy T and the potential energy V are quadratic forms in the generalized velocities ˙q and in the generalized coordinates q, respectively. In this case the problem reduces to the known algebraic one of simultaneous diagonalization of two quadratic forms. The Hamiltonian case considered here is more general because the quadratic Hamiltonian is allowed to include mixed terms in coordinates and momenta. Moreover, we want all

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It is convenient to simplify the notation by using the compact formalism. Thus, I shall denote z = (x

1

, . . . , x

n

, y

1

, . . . , y

n

)

, a column vector, and the Hamiltonian is the quadratic form

(5.7) H(z) = 1

2 z

C z , C

= C

where C is a 2n × 2n real, symmetric and non degenerate matrix. The corresponding Hamilton’s equations are

(5.8) ˙z = JC z ,

where J is the 2n × 2n antisymmetric matrix J =

 0 I

−I 0



defining the symplectic form. Recall that we have J

= J

−1

= −J .

As in sect. 5.1, let us look for a linear transformation z = Mζ which gives the linear system a diagonal form. We know that if the eigenvalues of the matrix JC are distinct then the matrix M can be determined, and so we have

(5.9) M

−1

JCM = Λ .

The problem is whether the transformation matrix M can be restricted to be symplec- tic, i.e., to satisfy the condition

(5.10) M

JM = J .

5.2.1 The linear canonical transformation

Let us investigate the properties of the eigenvalues of the matrix JC, recalling that C is assumed to be symmetric.

Lemma 5.5: The eigenvalues of the matrix JC satisfy the following properties:

(i) if λ is an eigenvalue, so is its complex conjugate λ

; (ii) if λ is an eigenvalue, so is −λ .

That is: the eigenvalues of JC can be organized either in pairs, when they are real or pure imaginary, or in groups of four, as illustrated in fig. 5.1.

Proof. (i) The characteristic polynomial of a real matrix has real coefficients. Hence its roots are either real or pairs of complex conjugate numbers.

transformations to be canonical, which is a further constraint. In the present notes I limit the discussion to the most common case of distinct eigenvalues, following [94], § 15.

A short exposition of the results for the general case of eigenvalues with multiplicity greater that one is found in [7]. A complete discussion is in [98], where results from [99]

are used.

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Im λ

Re λ

Im λ

Re λ λ

−λ

Im λ

Re λ λ

−λ

λ

−λ

*

*

λ

−λ

a b c

Figure 5.1.

The eigenvalues of the matrix JC with C symmetric. a. A pair of real eigenvalues, namely the complex conjugates are λ = λ and −λ = −λ.

b. A pair of pure imaginary eigenvalues, namely λ , −λ ; we have λ = −λ and

−λ = λ . c. Four distinct complex eigenvalues λ , −λ , λ, −λ with real and imaginary part both non zero.

(ii) Consider the characteristic equation det(JC − λI) = 0 and remark that JC − λI 

= CJ

− λI = − CJ + λI

= −JJ

−1

CJ + λI  = −J J

−1

CJ + J

−1

λI 

= J JCJ + λIJ = J JC + λIJ . This shows that

det JC − λI  = det JC + λI ,

i.e., the characteristic polynomial is symmetric in λ. The claim follows. Q.E.D.

The next lemma may be seen as stating the analogy between symmetric matri- ces, which have orthogonal eigenvectors, and matrices of the form JC , which have eigenvectors possessing a natural symplectic structure.

Lemma 5.6: Let the eigenvalues of the matrix JC be distinct, and let λ, λ

be two eigenvalues corresponding to the eigenvectors w, w

, respectively. Then we have

w

J w

6= 0 if and only if λ

= −λ . Proof. Let us show that we have

(λ + λ

)w

Jw

= 0 .

For, in view of J

2

= I and J

J = I , with a short calculation we get (λ + λ

)w

Jw

= (JCw)

Jw

+ w

J (JCw

)

= w

CJ

Jw

+ w

CJ

2

w

= w

Cw

− w

Cw

= 0 .

If λ+λ

6= 0 then we have w

Jw

= 0 . If λ+λ

= 0 we must prove that w

Jw

6= 0 , so

that w is symplectic orthogonal to all eigenvectors but w

. By contradiction, assume

that also w

Jw

= 0 . This would imply that w

Jw

= 0 for all eigenvectors of the

matrix JC. Since the eigenvectors are a basis, in view of the non degeneration of the

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symplectic form this would imply that w = 0 , contradicting the fact that w is itself an element of the basis. We conclude that w

Jw

6= 0 . Q.E.D.

We now should normalize the eigenvectors so as to be able to construct a symplec- tic matrix. In view of lemma 5.5 we can arrange the eigenvalues and the eigenvectors of the matrix JC in the order

(5.11) λ

1

, . . . , λ

n

, λ

n+1

= −λ

1

, . . . , λ

2n

= −λ

n

, w

1

, . . . , w

n

, w

n+1

, . . . , w

2n

, The n quantities

d

j

= w

j

Jw

j+n

, j = 1, . . . , n are non zero in view of lemma 5.6. Let us construct the matrix (5.12) M = (w

1

/d

1

, . . . , w

n

/d

n

, w

n+1

, . . . , w

2n

) by arranging the eigenvectors in columns in the chosen order.

Lemma 5.7: The matrix M constructed as in (5.12) is symplectic.

Proof. We must prove that M

JM = J . To this end let us denote its elements by a

j,k

. Then in view of lemma 5.6 and of the definition of d

j

we have

a

j,j+n

= 1

d

j

w

j

Jw

j+n

= 1 , a

j+n,j

= 1

d

j

w

j+n

Jw

j+n

= −1 ,

and a

j,k

= 0 in all other cases. Q.E.D.

Since the columns of M are eigenvectors of JC we conclude with the

Proposition 5.8: Let the eigenvalues of the matrix JC be distinct. Then the linear transformation z = Mζ with the matrix M constructed as in (5.12) is canonical, and changes the system of linear differential equations (5.8) into its diagonal form

˙ζ = Λζ , Λ = diag(λ

1

, . . . , λ

n

, −λ

1

, . . . , −λ

n

) .

5.2.2 Non uniqueness of the diagonalizing transformation

The diagonalizing transformation is not unique, for we can multiply it by a symplectic diagonal matrix R = diag(r

1

, . . . , r

2n

) . This is seen as follows. In view of the remark made after lemma 5.1 the matrix ˜ M = MR still gives the system a diagonal form.

Thus we should only check that ˜ M is still symplectic, which is true because the set of symplectic matrices forms a group.

Let us look for the general form of a diagonal symplectic matrix R. To this end let us write it in the convenient form

R =  R

0

0 0 R

1



, R

0

= diag(r

1

, . . . , r

n

) , R

1

= diag(r

1

, . . . , r

n

) .

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Writing the symplecticity condition we have

 R

0

0 0 R

1

  0 I

−I 0

  R

0

0 0 R

1



=  R

0

0 0 R

1

  0 R

1

−R

0

0



=

 0 R

0

R

1

−R

0

R

1

0



=  0 I

−I 0

 .

The last line give us the condition R

0

R

1

= I . We conclude that the symplectic matrix M which takes the linear canonical system in diagonal form may be replaced with the symplectic matrix MR where R is a non degenerate diagonal matrix of the form

(5.13) R =  R

0

0

0 R

−1

0



, R

0

= diag(r

1

, . . . , r

n

) .

5.2.3 Complex normal form of the Hamiltonian

Let us now come to the transformation of the Hamiltonian, using the compact no- tation ζ = (ξ, η)

where ξ = (ξ

1

, . . . , ξ

n

) ∈ C

n

are the coordinates and η = (η

1

, . . . , η

n

) ∈ C

n

are the corresponding conjugate momenta. Furthermore, recall- ing the ordering (5.11) of the eigenvalues, it will be convenient to write the diagonal matrix Λ as

(5.14) Λ =  Λ

0

0

0 −Λ

0



, Λ

0

= diag(λ

1

, . . . , λ

n

) .

Proposition 5.9: Let the eigenvalues of the matrix JC be distinct. Then the lin- ear transformation generated by the matrix M defined by (5.12) gives the quadratic Hamiltonian (5.7) the form

(5.15) H(ξ, η) =

n

X

j=1

λ

j

ξ

j

η

j

.

Proof. By transforming the quadratic form

12

z

Cz we get

12

ζ

M

CMζ. Recalling that M

−1

JCM = Λ we calculate

M

CM = − M

J JCM = − M

JM 

M

−1

JCM  , = −JΛ =  0 Λ

0

Λ

0

0

 , a symmetric matrix. Thus, the transformed Hamiltonian is H(ζ) = −

12

ζ

JΛζ ,

namely (5.15) in complex coordinates (ξ, η) . Q.E.D.

5.2.4 First integrals

The normalized Hamiltonian (5.15) possesses n first integrals which are in involution, namely

(5.16) Ψ

j

j

, η

j

) = ξ

j

η

j

, j = 1, . . . , n .

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This is easily checked by a straightforward calculation of Poisson brackets {Ψ

j

, H} . Thus the system turns out to be integrable in Liouville’s sense.

3

Further first integrals may be found by looking for a solution of the equation {Φ, H} = 0 namely

i

n

X

l=1

λ

l

 η

l

∂Φ

∂η

l

− ξ

l

∂Φ

∂ξ

l



= 0 . Choosing

4

Φ = ξ

j

η

k

we have

hk − j, λiξ

j

η

k

= 0 ,

which can be satisfied only if hk − j, λi = 0 , which is a resonance relation. Therefore further first integrals independent of those in (5.16) do exist only if the eigenvalues λ satisfy a resonance condition. This is an extension of proposition 4.10 to the general case of quadratic Hamiltonians.

The canonical equations for the Hamiltonian (5.16) are (5.17) ˙ξ

j

= λ

j

ξ

j

, ˙η

j

= −λ

j

η

j

,

and the corresponding solutions with initial datum ξ

j,0

, η

j,0

are ξ

j

(t) = ξ

j,0

e

λjt

, η

j

(t) = η

j,0

e

−λjt

.

This is straightforward, but the structure of the orbits in the real phase space remains hidden until we perform all substitutions back to the original variables, or we write in explicit form the evolution operator U(t) = Me

M

−1

as in proposition 5.3.

More generally, the drawback of this section is that the functions so determined have in general complex values, while it seems better to write both the Hamiltonian and the first integrals in real variables. Thus, let us proceed by looking for a normal form in real variables.

5.2.5 The case of real eigenvalues

If the eigenvalues of the matrix JC are real and distinct then the whole normaliza- tion procedure is performed without involving complex objects. Thus, considering for simplicity the case of a system with one degree of freedom, the normal form of the Hamiltonian is

(5.18) H(ξ, η) = λξη , (ξ, η) ∈ R

2

where λ, −λ are the eigenvalues of the matrix JC .

It may be interesting however to remark that one can use as a paradigm Hamil- tonian also the form

(5.19) H(x, y) = λ

2 (x

2

− y

2

) .

3 Integrablity in Arnold–Jost sense is not assured because the invariant surfaces deter- mined by the first integrals need not be compact.

4 I use the multi-index notation, i.e., k, j are integer vectors with non negative entries and ξjηk = ξ1j1· · · ξjnnηk11···ηknn .

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This is a typical form of a Hamiltonian describing the motion of a particle on a one–

dimensional manifold.

Exercise 5.1: Find a canonical transformation that changes the Hamiltonian (5.19) into (5.18).

5.2.6 The case of pure imaginary eigenvalues

Let us consider again the case of a system with one degree of freedom. My aim is to show that the Hamiltonian may be given the normal form

(5.20) H(x, y) = ω

2 (y

2

+ x

2

) ,

which is easily recognized as the Hamiltonian of a harmonic oscillator.

Let the eigenvalues of the matrix JC to be λ = iω and λ

= −iω . Then the corresponding eigenvectors are complex conjugated, i.e., may be written as w = u+iv e w

= u − iv. Recalling that the symplectic form is anticommutative we have

w

Jw

= (u + iv)

J (u − iv)

= i(−u

Jv + v

Ju ) = −2iu

Jv ,

which is a pure imaginary quantity. We may always order the eigenvalues and the eigenvectors as in (5.11) with the further requirement

(5.21) d = u

Jv > 0 .

This may just require exchanging a pair (λ, w) with its conjugate (−λ, w

) . More precisely, we may create the association

iω ↔ u , −iω ↔ v

by choosing the sign of ω so that the condition d > 0 is satisfied.

5

Let us now construct the transformation matrix

(5.22) T = (u/ √

d , v/ √ d) . This is a symplectic matrix, as is checked by calculating

T

JT = 1 d

 u

Ju u

Jv v

Ju v

Jv



=

 0 1

−1 0

 . By property (iii) of lemma 5.2 we have

JCu = −ωv , JCv = ωu ,

and we can write this relation for the columns of the matrix T , thus getting JCT = T

 0 ω

−ω 0

 ,

5 Remark that the sign of ω in the normal form (5.20) depends precisely on this choice.

Since the choice is dictated by the eigenvectors of the matrix JC there is no arbitrariness here.

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that is

T

−1

JCT = J  ω 0 0 ω

 . Thus the equations take the form

˙x = ωy , ˙y = −ωx , and the transformed Hamiltonian is (5.20).

Exercise 5.2: The Hamiltonian of a harmonic oscillator in complex variables is H(ξ, η) = iωξη . Find a canonical transformation that changes the Hamiltonian (5.20) into this one.

5.2.7 The case of complex conjugate eigenvalues

The simplest case is that of a system with two degrees of freedom. My aim is to show that a good paradigm Hamiltonian is

(5.23) H = µx

1

y

1

+ µx

2

y

2

+ ω(x

1

y

2

− x

2

y

1

) where µ and ω are real parameters.

Let me first show how we can construct a real symplectic basis. Write the four eigenvalues and the corresponding eigenvectors of the matrix JC as

λ = µ + iω , w

+

= u

+

+ iv

+

, λ

= µ − iω , w

+

= u

+

− iv

+

,

−λ = −µ + iω , w

= u

+ iv

,

−λ

= −µ − iω , w

= u

− iv

,

the alignment in rows reflecting the correspondence. By lemma 5.6 we know that w

+

Jw

6= 0 and w

+

Jw

6= 0 , and we can always manage that these quantities are equal and pure imaginary,

6

i.e.,

(5.24) w

+

Jw

+ w

+

Jw

= 0 .

Still by lemma 5.6 we know also that, omitting four obvious relations, we have (5.25) w

+

Jw

+

= 0 , w

+

Jw

= 0 , w

Jw

= 0 ,

w

+

Jw

+

= 0 , w

+

Jw

= 0 , w

Jw

= 0 .

Lemma 5.10: With the condition (5.24), the vectors u

+

, u

, v

+

, v

satisfy (5.26) u

+

Jv

= u

Jv

+

6= 0

and

(5.27) u

+

Jv

+

= u

Jv

= u

+

Ju

= v

+

Jv

= 0 .

6 For instance, if w+Jw = ̺e it is enough to replace w+ with ie−iϑw+, with the corresponding change for w+.

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This means that u

+

, u

, v

, v

+

, in this order, form a symplectic basis, still to be normalized.

Proof. From w

+

Jw

+

= 0 we get

(u

+

+ iv

+

)

J (u

+

− iv

+

) = −2iu

+

Jv

+

= 0 .

With a similar calculation, from w

Jw

= 0 we also get u

Jv

= 0 . Thus two of the relations (5.27) are proven.

Using w

+

Jw

= 0 we get

(u

+

+ iv

+

)

J (u

− iv

) = u

+

Ju

+ v

+

Jv

 − i u

+

Jv

− v

+

Ju

 = 0 , and setting separately to zero the real and the imaginary part we get

(5.28) u

+

Ju

+ v

+

Jv

, u

+

Jv

− v

+

Ju

= 0 . Now calculate

(5.29) w

+

Jw

= (u

+

+ iv

+

)

J (u

+ iv

)

= u

+

Ju

− v

+

Jv

 + i u

+

Jv

+ u

Jv

+

 , In view of (5.28) we get

w

+

Jw

= 2u

+

Ju

+ 2iu

+

Jv

In view of the property (5.24) the l.h.s. is a pure imaginary quantity, and since we know that w

+

Jw

6= 0 we get

u

+

Ju

= 0 , u

+

Jv

6= 0 . By (5.28) we also get

v

+

Jv

= 0 , v

+

Ju

6= 0 .

Furthermore by the second of (5.28) we have u

+

Jv

= v

+

Ju

which completes the

proof. Q.E.D.

We are now ready to construct the matrix T that gives the system the normal form (5.23) by suitably adapting the scheme of the previous section. Since u

+

Jv

is a real quantity we can always manage so that

(5.30) u

+

Jv

= v

+

Ju

= d > 0 .

For, if d is negative it is enough to exchange u

+

with u

and v

+

with v

. Then we construct the 4 × 4 matrix

(5.31) T =  u

+

√ d , v

+

√ d , v

√ d , u

√ d



This is a symplectic matrix. For, in view of (5.30) and of lemma 5.10we have

T

JT = 1 d

u

+

Ju

+

u

+

Jv

+

u

+

Jv

u

+

Ju

v

+

Ju

+

v

+

Jv

+

v

+

Jv

v

+

Ju

v

Ju

+

v

Jv

+

v

Jv

v

Ju

u

Ju

+

u

Jv

+

u

Jv

u

Ju

= J

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By property (iii) of lemma 5.2 we have

JCu

+

= µu

+

− ωv

+

, JCv

+

= ωu

+

+ µv

+

, JCu

= −µu

+ ωv

, JCv

= −ωu

− µv

. Using this we also have

JCT =  µu

+

− ωv

+

√ d , ωu

+

+ µv

+

√ d , −µv

− ωu

√ d , ωv

− µu

√ d



= T

µ −ω 0 0

ω µ 0 0

0 0 −µ −ω

0 0 ω −µ

Thus, denoting by (x

1

, x

2

, y

1

, y

2

) the new variables the system (5.8) is transformed into

˙x

1

= µx

1

− ωx

2

, ˙x

2

= ωx

1

+ µx

2

˙y

1

= −µy

1

− ωy

2

, ˙y

2

= ωy

1

− µy

2

which are the canonical equations for the Hamiltonian (5.23).

Exercise 5.3: Find the canonical transformation that changes the Hamilto- nian (5.23) into the Hamiltonian in complex variables H(ξ, η) = λξη .

Hint: this may be used as an example of application of the diagonalizing procedure in sections 5.2.1, 5.2.2 and 5.2.3.

5.3 Nonlinear elliptic equilibrium

The Hamiltonian in a neighbourhood of an elliptic equilibrium can be typically ap- proximated by a quadratic Hamiltonian with the form (see 5.2.6)

(5.32) H

0

(x, y) = 1

2 X

l

ω

l

(x

2l

+ y

l2

) ,

where (x, y) ∈ R

2n

are the canonical variables, and ω = (ω

1

, . . . , ω

n

) ∈ R

n

is the vector of frequencies, that are assumed not to vanish.

By assuming the Hamiltonian to be analytic and expanding it in power series we are led to consider a canonical system with Hamiltonian

(5.33) H(x, y) = H

0

(x, y) + H

1

(x, y) + H

2

(x, y) + . . .

where (x, y) ∈ R

2n

are canonically conjugate variables, H

0

(x, y) has the form (5.32),

and H

s

(x, y), for s ≥ 1, is a homogeneous polynomial of degree s + 2 in the canon-

ical variables. The power series is assumed to be convergent in a neighbourhood of

the origin of R

2n

. This is actually a perturbed system of harmonic oscillators, which

describes many interesting physical models.

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5.3.1 Use of action–angle variables

The canonical transformation to action–angle variables

(5.34) x

l

= p2I

l

cos ϕ

l

, y

l

= p2I

l

sin ϕ

l

, 1 ≤ l ≤ n

gives H

0

the form (4.17) of an isochronous system, which fails to satisfy the condition of non degeneration of Poincar´e’s theorem.

7

The Hamiltonian presents a couple of minor differences with respect to that of the general problem of dynamics as stated by Poincar´e, namely the Hamiltonian (4.1). For, it seems that the Hamiltonian (5.33) is not an expansion in a perturbation parameter, and moreover it is not written in action–angle variables.

The lack of a perturbation parameter is just a trivial matter, because the pa- rameter ε is easily replaced by the distance from the origin. Indeed, if we consider the dynamics inside a sphere of radius ̺ centered at the origin then the homogeneous poly- nomial H

s

(x, y) is of order O(̺

s+2

), so that ̺ plays the role of perturbation parameter.

More formally we may introduce a scaling transformation x

j

= εx

j

, y

j

= εy

j

,

which is not canonical, but preserves the canonical form of the equation if the new Hamiltonian is defined as

H

(x

, y

) = 1

ε

2

H(x, y)

x=εx, y=εy

(see example 2.4). Thus the Hamiltonian is changed to

H

(x

, y

) = H

0

(x

, y

) + εH

1

(x

, y

) + ε

2

H

2

(x

, y

) + . . . ,

which introduces the power expansion in ε . This means that the natural reordering of the power series as homogeneous polynomials corresponds exactly to the use of a parameter.

The transformation to action–angle variables is a bit more delicate. Indeed the canonical transformation (5.34) introduces a singularity at the origin which causes a loss of analyticity for I = 0 . Let us see this point in some more detail.

By transforming a homogeneous polynomial of degree s in x, y we obtain a Fourier series in the angles ϕ with coefficients that depend on the actions I in a particular form. The following rules apply:

8

7 The problem of a degenerate Hamiltonian of the type considered here has been first investigated by Whittaker [96]. As an historical remark, it is curious that in his very exhaustive paper Whittaker did not mention the problem of the consistency of the construction pointed out in sect.4.3.1. A few years later Cherry wrote two papers where a lot of work is devoted to the consistency problem, but without reaching a definite conclusion [20] [21]. An indirect solution was found by Birkhoff in [17], ch. III, § 8, using the method of normal form that usually goes under his name.

8 The reader will easily check that these rules apply by writing the trigonometric functions in complex form.

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(i) The actions I appear only as powers of √

I

1

, . . . , √

I

n

, namely c

j1,...,jn

I

1j1/2

· · · I

njn/2

,

where c

j1,...,jn

∈ R .

(ii) In every term of the Fourier expansion

c

j1,...,jn

I

1j1/2

· · · I

njn/2

e

i(k1ϕ1+...+knϕn)

the exponent k

l

may take only the values −j

l

, −j

l

+ 2, . . . , j

l

− 2, j

l

.

With some patience the reader will be able to check that these properties are preserved by sums, products and Poisson brackets.

9

The polynomial dependence on √

I

1

, . . . , √

I

n

keeps the property of a homoge- neous polynomial of degree s to be of order ̺

s

, so that nothing essential is changed.

However, the lack of analyticity may be somehow annoying when one tries to analyze the convergence. As a matter of fact all these problems actually disappear is we choose to work in cartesian coordinates, as we shall do here. In a first stage the theory will be developed at a purely formal level, in the sense that all calculation will be performed disregarding the problem of convergence of the series that will be constructed. The problem of convergence will be discussed later.

5.3.2 A formally integrable case

In view of the degeneration of the unperturbed system we may expect to be able to construct first integrals in our case by just applying the procedure of sect 4.3.1, namely to solve the system (4.21) of equations.

Let us adapt the scheme to our case. We look for a first integral (5.35) Φ(x, y) = Φ

0

(x, y) + Φ

1

(x, y) + . . .

where Φ

0

(x, y) = I

l

=

12

(x

2l

+ y

2l

) is the action of the l–th oscillator, and Φ

s

(x, y) is a homogeneous polynomial of degree s + 2. By setting l = 1, . . . , n we may construct n independent first integrals. Thus, splitting the equation {H, Φ} in homogeneous polynomials we get the recurrent system

(5.36) {H

0

, Φ

1

} = −{H

1

, Φ

0

}

{H

0

, Φ

s

} = −{H

1

, Φ

s−1

} − . . . − {H

s

, Φ

0

} , s > 0 .

Looking at the algebraic aspect of the equations above makes the problem quite simple.

Denote by P the vector space of formal power series and by P

r

the subspace of

9 The direct verification is straightforward, but we may remark that there is no actual need to do all the calculations. Here is the argument. Recall that the transformation to action–angle variables is canonical and that the functions that we are considering are obtained by transforming polynomials. Since the canonical transformation preserves the Poisson brackets the calculation in action–angle variables can not affect the property that in cartesian variables x, y the result is a polynomial. Thus the properties must be preserved. The same argument, with no use of the canonical structure, applies to the product.

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homogeneous polynomials of degree r. The unperturbed Hamiltonian H

0

acts as a linear operator L

H0

· = {H

0

, ·} from the linear space P

r

into itself. Moreover, using the complex canonical coordinates (ξ, η) ∈ C

2n

defined by

(5.37) x

l

= 1

√ 2 (ξ

l

+ iη

l

) , y

l

= i

√ 2 (ξ

l

− iη

l

) , 1 ≤ l ≤ n we get

(5.38) H

0

= i X

l

ω

l

ξ

l

η

l

,

so that the operator L

H0

above takes a diagonal form. For, by applying it to a mono- mial ξ

j

η

k

≡ ξ

1j1

. . . ξ

njn

η

1k1

. . . η

nkn

we get

L

H0

ξ

j

η

k

= i k − j, ω ξ

j

η

k

.

Define now, as usual, R as the image of P by L

H0

and, correspondingly, by R

r

the image of the subspace P

r

. We see that the eq. (5.36) can be solved if the r.h.s. belongs to R

r

. On the other hand, let us define the null space N as N = L

−1H0

(0) the inverse image of the null element by L

H0

, with the corresponding definition for N

r

. We get that both N

r

and R

r

are linear subspaces of the same space P

r

, which are disjoint, namely satisfy N ∩ R = {0}, and generate P

r

by direct sum, namely satisfy N ⊕ R = P

r

. Thus the system (5.36) can be solved provided the r.h.s. has no component in N . This is actually the consistency problem that has been have pointed out in sect. 4.3.1.

Dealing with the latter problem is actually not easy, in general. However a direct solution may be found in a particular but relevant case which occurs in many physical situations.

10

Let us say that a function f

+

(x, y) is even in the momenta in case f

+

(x, y) = f

+

(x, −y) and that f

(x, y) is odd in the momenta in case f

(x, y) = −f

(x, −y).

10 See [30]. The reader will notice that only the non–resonant case is discussed here, and moreover the proof of the proposition can not be trivially extended to the case of a non–

reversible Hamiltonian. The direct construction of formal first integrals for the resonant case encounters major difficulties due to the nontrivial structure of the null space N . This is indeed the main concern of the second paper of Cherry [21]: he proposes to determine the arbitrary terms in the solution of the linear equation at a given order so as to remove the unwanted terms inN from the known term of the equations for higher orders. However, Cherry fails to prove that the method actually works — although exactly his method has been implemented on computer by Contopoulos in order to perform an explicit calculation (see [23], [24] and [25]). First integrals for both the non–

reversible case and the resonant one may be constructed using the methods of normal form introduced by Poincar´e and widely used by Birkhoff; these methods are indirect, in contrast with the direct and overall simpler approach used here. It may be interesting to remark that an attempt made in [31] to justify the method of Cherry has lead to a connection with the algorithm for giving the Hamiltonian a normal form via the Lie transform methods. These methods will be the subject of chapter 6.

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The formal existence of first integrals is stated by the following

Proposition 5.11: Let H(x, y) be as in (5.33) where H

0

has the form (5.32), and assume:

i. non resonance: for k ∈ Z

n

one has hk, ωi = 0 if and only if k = 0;

ii. reversibility: the Hamiltonian is an even function of the momenta, namely satisfies H(x, −y) = H(x, y).

Then there exist n independent formal integrals Φ

(1)

, . . . , Φ

(n)

of the form (5.35) which are even functions of the momenta. Moreover they form a complete involution system.

The proof is based on the following

Lemma 5.12: The Poisson bracket between even and odd functions of the momenta obeys the rule

{·, ·} +

+ −

+

− +

,

i.e., the Poisson bracket between functions of the same parity is odd; the Poisson bracket between functions of different parity is even.

Proof. The product of even and odd function clearly obeys the opposite rule, i.e., the product of functions of the same parity is even, and the product of functions of different parity is odd. For we have

f

+

· g

+

(x, y) = f

+

(x, y) · g

+

(x, y)

= f

+

(x, −y) · g

+

(x, −y) = f

+

· g

+

(x, −y) , f

· g

(x, y) = f

(x, y) · g

(x, y)

= −f

(x, −y) · −g

(x, −y)  = f

· g

(x, −y) , f

+

· g

(x, y) = f

+

(x, y) · g

(x, y)

= f

+

(x, −y) · −g

(x, −y) = − f

+

· g

(x, −y) .

On the other hand, the derivative with respect of one of the coordinates x does not affect the parity, while the derivative with respect to one of the momenta changes it into the opposite one. Coming to the Poisson bracket, it is enough to take into account that it is defined as products of derivatives, and apply the rules above for derivatives

and products. Q.E.D.

Proof of proposition 5.11. The proof is based on the fact that the non resonance condition implies that every function f ∈ N must be even in the momenta. For, it can depend only on the action variables I

1

= ξ

1

η

1

, . . . , I

n

= ξ

n

η

n

which is real coordinates write I

1

=

x21+y2 12

, . . . , I

n

=

x2n+y2 n2

and are clearly even functions.

Let us now proceed by induction and prove that a first integral can be constructed and

must be an even function. Since the first term Φ

0

must satisfy {H

0

, Φ

0

} = 0 we have

(18)

Φ

0

∈ N , hence it must be an even function. Suppose that Φ

s

has been determined for 0 ≤ s ≤ r as an even function of the momenta, which is true for r = 0. Then the r.h.s. of the equation for Φ

r+1

is an odd function, because it is found as a sum of Poisson brackets between even functions. Therefore it has no component in N , and so Φ

r+1

can also be determined and is an even function, because its Poisson bracket {H

0

, Φ

r+1

} must be an odd function; such a solution is unique up to an arbitrary term ˜ Φ

r+1

∈ N , which adds again an even function. This completes the induction, and shows that Φ may be constructed, and must be an even function. Let Φ

0

= I

l

be any of the actions and construct the corresponding first integrals. Thus we get n first integrals Φ

(1)

, . . . , Φ

(n)

that are independent because the first terms I

1

, . . . , I

n

are independent. Let now Φ and Ψ be two such first integrals; we prove that they are in involution. Let Υ = {Φ, Ψ}. By Poisson’s theorem it is a first integral, hence it must be an even function, as we have proved. On the other hand Υ must be an odd function, because it is the Poisson bracket between even function. Since it must be both odd and even, we conclude that it must be zero. Q.E.D.

We should keep in mind that the proposition just proved is a formal one, in the sense that all the construction is performed by simply using algebra, regardless of the convergence of the series so generated. In the same spirit, one could apply the method of Liouville and Arnold to build the action–angle variables.

11

Hence the system is formally integrable.

Having settled the formal aspect, one should discuss the convergence properties of the series so generated. Indeed the denominators hk, ωi, although non vanishing, are not bounded from below. On the other hand, there are known examples of series involving small denominators which are convergent. An essentially negative answer to the problem of convergence has been given by Siegel in [93]. However, it must be emphasized that the problem of convergence has puzzled the best mathematicians for a couple of centuries. In view of the complexity of the problem, let us make a detour by looking at numerical results.

5.4 Numerical exploration

The present section aims at giving a glimpse on the dynamics around a non linear equilibrium using numerical methods in order to calculate the orbits. Such an approach is motivated by history. The existence of chaotic orbits, that will be illustrated in the present section, had been predicted by Poincar´e [86][87]. However, the phenomenon remained mostly unknown for more that 70 years.

At the dawn of numerical simulations of dynamics with the help of computers, between 1955 and 1960, the method of Poincar´e section has been used in order to visualize the dynamics of systems of two harmonic oscillators with a cubic nonlinearity, a simple model that may describe the dynamics of stars in a Galaxy. Many such studies

11 This is discussed by Whittaker in [97], ch. XVI, §199. Whittaker also includes in §198 a short discussion concerning the problem of the convergence of the formal first integrals.

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Figure 5.2.

The method of Poincar´e section for an orbit in the three dimensional space.

Taking a surface Σ transversal to the flow and an initial point P0

on that surface the orbit is fol- lowed until it crosses again the surface in the same direction at P1, and then again at P2, P3

and so on. An orbit is thus rep- resented by the sequence of the successive intersections.

have been performed by Contopoulos, who also had the idea of calculating the so called third integral (to be added to the two classical ones of the energy and the angular momentum) by a series expansion via the method discussed in sect. 5.3. The series had to be truncated at low order, of course, due to the limited power of computers available at that time.

The existence of chaos has been rediscovered, so to say, thanks to the work of Contopoulos (see, e.g., [23]), followed by the works of H´enon and Heiles [47] in 1964 and of Gustavson [43] in 1966, just to quote the first ones. It should be emphasized that the discovery came as an unexpected surprise, and marked the opening of a wide field of research. After 1970 the interest in numerical simulations of dynamics has raised tumultuously — perhaps chaotically.

12

It goes without saying that a long interval of more than 70 years between the discovery of Poincar´e and the rediscovery and widespread diffusion of knowledge about chaotic phenomena should not be underrated.

It is, I think, a clear symptom of the strong difficulty of imagining such complex phenomena on the mere basis of analytical investigations. This section may help the reader’s imagination.

5.4.1 The Poincar´e section

A first class of works is a numerical implementation of the method known as Poincar´e section, illustrated in figure 5.2. Here we are concerned with a numerical implementa- tion of that method for a class of Hamiltonians that has been widely investigated by

12 A comprehensive exposition of the history from the point of view of one of the protago- nists, namely Contopoulos, can be found in [26].

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