dy i dt = F x i en supposant que la function F peut se développer suivant les puissances d un paramètre très petit µ de la manière suivante: y i
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- Terence Knight
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1 ÁÊËÌ ÁÆÌ Ê ÄË According to Poincaré, the general problem of dynamics is formulated as follows ([87], Vol. I, 13). Nous sommes donc conduit à nous proposer le problème suivant: Étudier les équations canoniques dx i dt = F y i, dy i dt = F x i en supposant que la function F peut se développer suivant les puissances d un paramètre très petit µ de la manière suivante: F = F 0 +µf 1 +µ 2 F , en supposant de plus que F 0 ne dépend que des x et est indépendent des y; et que F 1, F 2,... sont des fonctions périodiques de période 2π par rapport aux y. Let me restate the problem in terms closer to our current language, also introducing some changes of notation coherent with the rest of these notes. To investigate the dynamics of a canonical system of differential equations with Hamiltonian (4.1) H(p,q,ε) = H 0 (p)+εh 1 (p,q)+ε 2 H 2 (p,q)+..., where p (p 1,...,p n ) G R n are action variables in the open set G R n, q (q 1,...,q n ) T n areanglevariables,andεisasmallparameter. TheHamiltonian isassumedtobeananalyticfunctionofallitsarguments;inparticular,weshallassume that it can be expanded in power series of ε in a neighbourhood of the origin. The smallness of the parameter ε (and the implicit assumption that the functions H 1,H 2,...arebounded) meansthatthesystemisasmallperturbationofanintegrable one. This is a typical situation for mechanical systems of interest in Physics and Astronomy. For ε = 0 the unperturbed system is integrable in Arnold Jost sense and all orbits lie on invariant tori parameterized by the actions p 1,...,p n. This leads us to study the character of periodic and quasiperiodic motions on invariant tori. For ε 0 the
2 86 Chapter 4 question is raised whether the dynamics remains similar to that of the unperturbed system. The natural approach consists in trying to prove that the perturbed system is integrable, for instance by constructing a complete set of first integrals in involution. However, it has been recognized by Poincaré that the dynamics of the perturbed system can be very complicated, and by no means it can be reduced to that of the unperturbed system. Inthe first sectionofthischapter Iinclude a detaileddiscussion ofthedynamicsof the unperturbed system, namely the Hamiltonian (4.1) with ε = 0. This includes also the description of resonances and some ergodic properties of the quasiperiodic motions. In the next section I come to the main subject, namely the theorem of Poincaré on non existence of analytic first integrals for Hamiltonians of the form (4.1). 4.1 Periodic and quasi periodic motion on a torus Let us investigate the dynamics of a system which is integrable in Arnold Jost sense. Referring to the general problem of dynamics as stated by Poincaré we consider the (subset of the) phase space F = G T n, where G R n is an open set, q = (q 1,...,q n ) T n are angle variables and p = (p 1,...,p n ) G are actionvariables, and a Hamiltonian function (4.2) H = H(p 1,...,p n ) depending only on the actions p. We know that the canonical equations have the particular form (4.3) q j = ω j (p 1,...,p n ), ṗ j = 0, 1 j n, where (4.4) ω j (p 1,...,p n ) = H p j are the frequencies. We also know that the orbits in phase space are given by (4.5) p j (t) = p j,0, q j (t) = q j,0 +tω(p 1,0,...,p n,0 ), where q 1,0,...,q n,0,p 1,0,...,p n,0 are the initial data. Therefore, the orbits lie on invariant n dimensional tori parameterized by the action variables. The linear flow (4.5) on a torus is named Kronecker flow Motion on a two dimensional torus In the simple case of a two dimensional torus we consider the orbit (4.6) q 1 (t) = q 1,0 +ω 1 t, q 2 (t) = q 2,0 +ω 2 t, where ω 1 and ω 2 are constant frequencies and q 1,0 and q 2,0 are the initial phases. As usual, we represent the torus T 2 as a square of side 2π on the plane; that is, the points (q 1,q 2 ) and (q 1,q 2 ) in R2 are said to represent the same point in case q 1 q 1 and
3 First integrals 87 Figure 4.1. Illustrating the quasi periodic motion on the torus T 2. The torus is represented by a square of side 2π. The orbit is a straight line with angular coefficient ω 2 /ω 1. All segments inside the square are images of that straight line obtained by translation by a multiple of 2π, both in horizontal and vertical direction. q 2 q 2 are integer multiples of 2π. By elimination of t in (4.6) we find that the orbit is the straight line ω 2 (q 1 q 1,0 ) ω 1 (q 2 q 2,0 ) = 0 ; we shall simplify the discussion by putting q 1,0 = q 2,0 = 0, which is tantamount to redefining the origin of the angles, so that the orbit is the straight line ω 2 q 1 ω 1 q 2 = 0. The image of the orbit on the torus is represented in fig All segments representing a slice of the orbit are parallel, so that they never intersect transversally. In order to investigate the dynamics it is convenient to consider the successive intersections of the orbit with the vertical side of the square corresponding to q 1 = 0. It is easily seen that these intersections are represented by the sequence of points with 0, α(mod2π), 2α(mod2π),..., sα(mod2π),..., (4.7) α = 2π ω 2 ω 1. Therefore, we are lead to investigating the behaviour of a sequence of points on a circle defined by a rotation by a fixed angle α (see fig. 5.7). Lemma 4.1: The sequence {sα(mod2π)} s 0 on the circle is periodic if and only if α/(2π) is a rational number. If α/(2π) is irrational, then the sequence is everywhere dense on the circle. Proof. The sequence is periodic in case there is an integer s 0 such that sα(mod2π) = 0, that is, sα = 2rπ for some integer r 0. Hence, α/(2π) = r/s,
4 88 Chapter 4 Figure 4.2. Illustrating the map of a circle into itself defined as a rotation by an angle α. Here, the map generated by the orbit on the torus T 2 of fig. 4.1 is represented. namely a rational number. The condition is clearly necessary and sufficient, and this proves the first claim. Let now α/(2π) be irrational. Then no two points of the sequence can coincide. Since the circle is compact, the sequence has at least one accumulation point. Therefore, for every positive ε there exist integers s, r such that (sα rα)(mod2π) < ε. Setting j = s r, we also have kα (k+j)α (mod2π) < ε for any integer k. The subsequence {njα (mod2π)} n 0 is monotonic, and so it determines an infinite sequence of contiguous intervals of length smaller than ε on the circle. Hence, every interval of length ε contains at least one point of the sequence. Since ε is arbitrary, the claim follows. Q.E.D. By a straightforward application of this lemma we conclude that the orbit (4.6) on the torus T 2 is periodic in case ω 1 /ω 2 is rational, and is everywhere dense on the torus in case ω 1 /ω 2 is irrational The many dimensional case The case n > 2 requires a generalization of the concept of rational ratio of the frequencies. This leads in a natural manner to introducing the concept of resonance module. The resonance module associated to ω R n is the set 1 (4.8) M ω = {k Z n : k,ω = 0}, 1 Geometrically: consider Z n as immersed in R n, and consider also the (n 1) dimensional plane through the origin in R n orthogonal to ω. The resonance module M ω is the intersection of that plane with Z n.
5 First integrals 89 where k,ω = j k jω j. It is an easy matter to check that M ω is a subgroup of Z n. A relation k,ω = 0 is called a resonance, and the number dim M ω is called the multiplicity of the resonance. The latter number actually represents the number of independent resonances to which ω is subjected. The extreme case are dimm ω = 0, which is called the non resonant case, and dimm ω = n 1, the completely resonant case. The main result concerning the dynamics on T n is given by the following Proposition 4.2: Consider the Kronecker flow (4.9) q(t) = q (0) +ωt (mod2π), ω R n, with initial point q 0 on T n, and let M ω be the resonance module associated to the frequency vector ω. Then the orbit (4.9) is dense on a torus T n dim M ω T n. In particular, in the non resonant case the orbit is everywhere dense on T n, and in the completely resonant case dim M ω = n 1 the orbit is periodic. The proof of the proposition above is based on two main results that will be proven in the next sections. Precisely, lemma 4.3 means that the torus T n admits a foliation into (n dimm ω ) dimensional tori which are invariant under the flow, and proposition 4.5 states that the orbits are dense on a noresonant torus. The reader should be able to complete the proof of proposition 4.9 once he or she has understood the proofs of lemma 4.3 and of proposition Changing the frequencies on a torus Recallthatthelinearflowonatorusisgeneratedbythesystemofdifferentialequations (4.10) q = ω, ω R n. A linear transformation (4.11) q = Mq with a unimodular matrix M changes the system (4.10) into (4.12) q = ω, ω = Mω, so that the change of coordinates induces a change of the frequencies. Lemma 4.3: Let ω R n be given, and let dim M ω > 0. Then there is a unimodular matrix M such that ω = Mω has exactly dim M ω vanishing components, while the remaining n dimm ω components form a non resonant vector. Corollary 4.4: The dimension dim M ω of the resonance module is invariant under linear unimodular changes of the angular coordinates. Lemma 4.3 means that we can always introduce coordinates such that the torus T n is actually the Cartesian product T r T n r, where r = dim M ω, with the property that on T r the flow is trivial, all points being fixed, while the frequencies on T n r are non resonant. Thus, the actual frequencies of the motion on a torus are not uniquely defined, depending on the chosen coordinates on the torus, while the dimension of the resonance module is uniquely defined, as stated by corollary 4.4
6 90 Chapter 4 The proof of lemma 4.3 is based on the unimodular transformation (3.46) of lemma Let r = dim M ω and let e 1,...,e r be a basis of the resonance module M ω. The integer vectors e 1,...,e r can be arranged in a rectangular r n matrix with integer entries. The problem is to complete this matrix by adding n r rows so that the resulting n n matrix is a unimodular one. This can be done, indeed, but the proof seem not be elementary. A proof is given in appendix A, lemma A.6. The linear canonical transformation generated by the unimodular matrix thus found transforms the frequencies so that r of them are zero. I leave to the reader the easy completion of the proof of lemma 4.3 and of corollary Time average and phase average Let us consider a function f : T n R. Using the coordinates q 1,...,q n the function f(q 1,...,q n )is2π periodicineachofthevariables.bycompositionwiththekronecker flow q(t) = q 0 +ωt, where q 0 is the initial point, we get a function f(t) of the time t, namely f(t) = f(q 0 +ωt). Let us consider the time average of f(t), defined as (4.13) f 1 (q 0 ) = lim T T T 0 f(q 0 +ωt)dt, assuming that the limit exists. It is an easy matter to check that the result depends only on the orbit, parameterized by the initial point q 0, i.e., by the initial phases. 2 Let us also consider the so called phase average of the function f(q 1,...,q n ), defined as (4.14) f = 1 (2π) n 2π 0 2π... f(q 1,...,q n )dq 1...dq n. 0 Proposition 4.5: Let f:t n R be a Riemann integrable function on T n and let f(t) = f(q 0 +ωt). If dimm ω = 0 then the time average f (q 0 ) defined by (4.13) exists everywhere and is constant on T n, and coincides with the phase average f defined by (4.14). The proof is based on approximation by trigonometric functions. Lemma 4.6: Let f:t n C be a trigonometric polynomial, i.e., f(q) = f k e i k,q, k K where K Z n is a finite set of integer vectors and f k C is a complex coefficient. Then the time average f is constant on T n and coincides with the phase average f. Proof. Let K contain just one vector k. If k = 0 then the function f is constant, namely f = f 0 C, and we trivially have f = f = f 0. If k 0, then we have 2 The reader will immediately realize that taking a different initial point q 0 on the orbit with initial point q 0 the result will not change, i.e., f (q 0) = f (q 0 ). For, the difference is just a quantity O(1/T) which vanishes in the limit T.
7 First integrals 91 f(q) = e i k,q (the actual value of the coefficient is unrelevant) and we may calculate T thus the time average is 0 e i k,(q 0+ωt) dt = e i k,q 0 ei( k,ω )T 1 i k, ω f (q 0 ) = ei k,q 0 i k,ω lim e i( k,ω )T 1 T T for all q 0 T n. On the other hand by direct integration one easily sees that the phase average vanishes, too. Let now K contain a finite set of vectors. Making the averge of every term in the sum we have f 1 (q 0 ) = lim T T k K = 0 T f k e i k,(q0+ωt) dt = f 0, i.e., the coefficient of the term with k = 0, and the same holds true for the phase average. Q.E.D. Lemma 4.7: Let f:t n R be a Riemann integrable function. Then for any ε > 0 there are two trigonometric polynomials P 1 (q 1,...,q n ), P 2 (q 1,...,q n ) such that for every q R n one has P 1 (q) < f(q) < P 2 (q), and moreover 1 (2π) n 0 T n (P 2 P 1 )dq 1...dq n < ε. Proof. Let first f be continuous. Then by Weierstrass theorem there is a trigonometric polynomial P which approximates the function within ε/2, i.e., P(q) f(q) < ε/2 for any q T n. In this case it is enough to set P 1 = P ε/2, P 2 = P +ε/2. If f is not continuous but it is Riemann integrable then there exist two continuous functions f 1,f 2 such that f 1 (q) < f(q) < f 2 (q) on T n, and moreover 1 (2π) n (f 2 f 1 )dq 1...dq n < ε T n 3. By approximating f 1 e f 2 with two polynomials P 1,P 2 such that P 1 < f 1,f 2 < P 2 and 1 (P (2π) n 2 f 2 )dq 1...dq n < ε T n 3, 1 (f (2π) n 1 P 1 )dq 1...dq n < ε T n 3, the claim is proven in view of ε being arbitrary. Q.E.D. Proof of proposition 4.5. For any T > 0 the trigonometric polynomials P 1,P 2 of lemma 4.7 satisfy the relations 1 T T 0 P 1 (q 0 +ωt)dt < 1 T T 0 f(q 0 +ωt)dt < 1 T T 0 ; P 2 (q 0 +ωt)dt. By lemma 4.6, for any ε > 0 there is T 0 (ε) such that for any T > T 0 (ε) the inequality P i 1 T P i (q 0 +ωt)dt T < ε, i = 1, 2 0
8 92 Chapter 4 holds true, where P i is the phase average of the trigonometric polynomial P i. Furthermore one has f P 1 < ε, P 2 f < ε. Thus for T > T 0 (ε) one has also f 1 T f(q 0 +ωt)dt T < 2ε, 0 which proves the claim. Q.E.D Quasiperiodic motions Let us now come to the proof that if the frequencies are non resonant then the orbit is dense on the torus. Actually we can prove more, but we need a further notion. Let D T n be a measurable subset, and consider the set 0 t T q(t) i.e., the part of the orbit corresponding to the time interval [0,T]. Denote by τ D (T) the time spent by the point in the subset D during the latter interval. Letting T we may τ define the average time spent by the orbit in D as lim D (T) T T provided the limit exists. The time τ D is easily computed by introducing the characteristic function χ D of D defined as χ(q) = 1 for q D and χ(q) = 0 elsewhere. For, it is immediately seen that (4.15) τ D (T) = T 0 χ D (q 0 +ωt)dt. Proposition 4.8: Let dimm ω = 0. Then (i) every orbit is dense on the torus T n ; (ii) the average time spent by the orbit in a measurable subset D T n is proportional to the measure of D. Proof. (i) By contradiction, suppose that there is an open subset U T n that is never visited by the orbit, and let f(q) be a continuous function such that f(q) = 0 for q / U and its phase average is f = 1. Then f is zero valued along the whole orbit, and so the time average f is zero, contradicting the statement of proposition 4.5 that f = f. (ii)by(4.15)theaveragetimespent bytheorbitind isthetimeaverageofthecharacteristic function χ D of D, i.e., χ D. By proposition 4.5 we also have χ D = χ D, and the phase average of the characteristic function of D is χ D = (2π) n meas(d). Q.E.D. 4.2 Isochronous and anisochronous systems Let us now go back to considering a Hamiltonian of the form (4.2), i.e., H(p 1,...,p n ) depending only on the action variables, the conjugate variables being angles. This Hamiltonian possesses the complete involution system of first integrals p 1,...,p n. The question that I want to discuss now is whether other first integrals independent of the ones above may exist, although they can not be in involution with all of them.
9 First integrals 93 As we have already remarked, for every p G we have an invariant torus carrying a linear flow with frequencies ω j (p) = H p j (p), j = 1,...,n. It is generally expected that changing the initial value of the actions p causes an actual change of the frequencies. If this happens, then the dynamics on close tori can be very different. For, the dynamics basically depends on dim M ω(p), and this quantity can changeinacrazyway,depending ontheinitialconditions.ontheotherhand, weknow that a system of harmonic oscillators is completely isochronous, since the frequencies of the oscillators do not depend on the initial data, and so are independent of the values of the actions. This leads us to distinguish two extreme cases, namely the cases of anisochronous versus isochronous systems Anisochronous nondegenerate systems A common hypothesis on the Hamiltonian is that ( 2 ) H (4.16) det 0 ; p j p k this is called a nondegeneracy condition, and can be interpreted as the condition that the frequencies ω can be used (at least locally) in place of the actions p as coordinates on G. Under this condition we have the following Proposition 4.9: Let the Hamiltonian H(p) be analytic and nondegenerate, and let Φ(q,p) be an analytic first integral for H. Then Φ must be independent of the angles. Proof. Let us consider the Fourier expansion of Φ(q, p) as Φ(p,q) = k Z n ϕ k (p)e i k,q, where the coefficients ϕ k (p) are analytic functions of the actions p only. Since Φ is a first integral the identity {H,Φ} = i k k,ω(p) ϕ k (p)e i k,q = 0, must be satisfied. This implies that for every k Z n we have either k,ω(p) = 0 or ϕ k (p) = 0. We prove that k,ω(p) = 0 can not happen if k 0. For, by differentiation we must have also n ω l k l = 0, 1 j n, p j l=1 which in view of the nondegeneracy condition gives k = 0. Therefore we have ϕ k (p) = 0 for k 0, and this implies Φ = ϕ 0 (p), namely that Φ does not depend on the angles. Q.E.D.
10 94 Chapter 4 Example 4.1: Uncoupled rotators A paradigm example of anisochronous system that will be used in the following is the system of uncoupled rotators described by the Hamiltonian n p 2 j H = 2, p Rn, q T n. j=1 The system is clarly non degenerate, and the frequencies are ω j = p j, (j = 1,...,n) Isochronous systems The simplest system which violates the nondegeneracy condition is a system of harmonic oscillators, H = 1 n ω j (x 2 j 2 +y2 j ), (x,y) R2n. j=1 Changing to action angle variables via the canonical transformation x j = 2p j cosq j, y j = 2p j sinq j, 1 j n, the Hamiltonian takes the form n (4.17) H = ω j p j, p [0, ) n, q T n, j=1 so that the frequencies ω are actually independent of the initial condition. Proposition 4.10: Let H be as in (4.17), and r = dimm ω. Then the Hamiltonian possesses n + r independent first integrals, that can be taken to be the n actions p 1,...,p n, and r further first integrals of the form Φ k (q) = cos k,q, where k can take r independent values in M ω. The proof is just a verification that Φ k of the form above is a first integral for whatsoever k M. By the way, this is connected with the fact that every torus is foliated into a r parameter family of (n r) dimensional tori with non resonant frequencies A result from diophantine theory Let us now concentrate on anisochronous nondegenerate system. The fact that the frequencies ω 1 (p),...,ω n (p) can be used as coordinates allows us to introduce a kind of geography of resonances on the action domain G. A resonance relation k,ω(p) = 0 with 0 k Z n defines a resonant submanifold of G of dimension n 1; more generally, a resonance module M generates a corresponding resonant manifold of dimension n dimm. These resonant manifolds are clearly dense in G. From the viewpoint of modern perturbation theory it is also interesting to investigate the measure of the nonresonant points with respect to that of the resonant ones. The problem is stated in more precise terms as follows. Let us say that a frequency vector ω R n is strongly nonresonant in case one can find a positive sequence ψ(s), where s is a positive integer, such that the inequality (4.18) k,ω ψ( k ) for 0 k Z n
11 First integrals 95 holds true; here I used the notation k = l k l. Given an open bounded subset D R n, the question is whether one can determine ψ(s) in such a way that the subset of the strongly nonresonant frequencies in D, namely the set Ω = { ω D : k,ω ψ( k ), k > 0 }, in not empty or, better, has positive measure in D. A sequence ψ(s) is determined as follows. Pick a nonzero k Z n, and consider Ω k = { ω D : k,ω < ψ( k ) }, namely the set of the ω s which are close to resonance with k. Such a set is contained in a strip of width smaller than nψ( k )/ k around the plane through the origin orthogonal to k, intersected with D (the factor n is due to the relation k n k, where is the Euclidean norm). Thus, its measure Vol( Ω k ) is bounded by Vol( Ω k ) 2 nc ψ( k ) k where C is a constant depending only on the domain D; an upper bound is C = (diamd) n 1. Then the measure of the complement of Ω in D cannot exceed ( ) Vol Ω k Vol( Ω k ) 2 nc ψ( k ). k k 0 k 0 k 0 Writing now k 0 ψ( k ) k = s>0 k =s, ψ(s) s and using the fact that the number of vectors k Z n satisfying k = s does not exceed 2 n s n 1, one finally gets 3 ( Vol k 0 ) Ω k 2 n+1 nc s>0s n 2 ψ(s). Then it is enough to choose ψ(s) = γs τ with suitable constants γ > 0 and τ > n 1 in order to get that the complement of Ω in D has a measure which is small with γ. Such a result, although obtained with rough estimates, is optimal for what concerns the value of τ. Indeed, in view of an approximation theorem by Dirichlet, for τ < n 1 the set Ω is empty, while for τ = n 1 the set Ω is nonempty, but has zero measure. 3 Let us denote by J n,s the number of vectors k Z n such that k = s. I show that the recursive formula J n,s = J n 1,s + 2J n 1,s J n 1,0 holds true. For, any vector k with k = s may be written as k = (k 1, k) with k Z n 1 such that k s and with k 1 = ±(s k ). For n = 1 we have J 1,0 = 1 and J 1,s = 2 for s > 0, and for n > 1 we have J n,0 = 1. For n = 2 it is immediate to see that J n,s 2 2 s. For n > 2, by a recursive procedure, assume that J n 1,s 2 n 1 s n 2 and calculate J n,s 2 n 1 s n 2 +2[2 n 1 (s 1) n n 1 +1] 2 n 1 s n 2 +2 n (s 1) n n s n 1. This is the wanted relation.,
12 96 Chapter 4 We conclude that the set of frequencies satisfying the diophantine condition (4.19) k,ω γ k τ for 0 k Z n, γ > 0, τ > n 1 has large relative measure. 4.3 The theorem of Poincaré on nonexistence of first integrals A first attempt in investigating the system (4.1) consists in looking for first integrals which are in some sense continuations of the first integrals of the unperturbed system. We first consider the case of a nondegenerate unperturbed Hamiltonian H 0. The nonexistence of first integrals in this case has been proved by Poincaré, generalizing a theorem of Bruns that the problem of three bodies does not admit algebraic first integrals independent of the known ones of the momentum and the angular momentum Equations for a first integral We look for a first integral as a power expansion in the parameter ε, namely (4.20) Φ(p,q,ε) = Φ 0 (p,q)+εφ 1 (p,q)+ε 2 Φ 2 (p,q)+..., where Φ 0,Φ 1,...are analyticfunctions of their arguments. To this end, we try to solve the equation {H,Φ} = 0, replacing the expansions in powers of ε for both the Hamiltonian and the function Φ. Collecting together all coefficients of the same power of ε we get the infinite system of equations (4.21) {H 0,Φ 0 } = 0 {H 0,Φ 1 } = {H 1,Φ 0 }... {H 0,Φ s } = {H 1,Φ s 1 }... {H s,φ 0 }.... Remark that the r.h.s. of the equation for Φ s depends only on H and on Φ 0,...,Φ s 1, which are either known or determined by the equations of the preceding steps. Hence we may attempt a recursive solution of the system. ThefirstequationsimplymeansthatΦ 0 mustbeafirstintegraloftheunperturbed system. This matter has been discussed in sect for anisochronous systems and in sect for isochronous ones. In the rest of this section let us assume (with Poincaré) that the unperturbed Hamiltonian H 0 is non degenerate, i.e., ( 2 ) H 0 det 0, p j p k so that the system is completely anisochronous. Thus, in view of proposition 4.9, Φ 0 must be a function of p 1,...,p n only.
13 First integrals 97 Looking now at the equations for Φ 1, Φ 2,... we see that at every step we are confronted with the problem of solving an equation of the form (4.22) L H0 Φ s = Ψ s, L H0 = {,H 0 } where Ψ s (p,q) is a known function, computed by calculating all Poisson brackets on the right. Since all functions must be periodic in the angles q, we can consider the Fourier expansion of Ψ s in complex form, namely Ψ s (p,q) = ( ) ψ k (p)exp i k,q, k Z n where the coefficients ψ k (p) are known and are analytic functions of the actions p. The fact that in our case Ψ(q,p) is a real function implies that the coefficients satisfy the condition (4.23) ψ k (p) = ψ k (p), as is easily checked. However, this condition will play no role in the rest of the proof. Similarly, we consider the Fourier expansion of Φ s (p,q) with unknown coefficients ϕ k (p). In view of L H0 Φ s = i ( ) k,ω(p) ϕ k (p)exp i k,q, k Z n where ω(p) = H 0, we get the equations p i k,ω(p) ϕ k (p) = ψ k (p). The formal solution of such an equation is easily written as ψ k (p) ϕ k (p) = i k,ω(p), but it is valid only if the denominator k,ω(p) does not vanish. Therefore, we are confronted with two problems. (i) Consistency: the coefficient ψ 0 correspondig to k = 0, namely the average of the known function Ψ(p, q) over the angles, must vanish. Although this happens in the equation for Φ 1, as is easy to see because only derivatives H 1 q j do appear in the r.h.s., it is not evident a priori that the same will occur for all equations (4.21). (ii) Small divisors: for k 0 the divisor k, ω(p) vanishes in case of resonance among the frequencies. Hence, unless the corresponding coefficient ψ k (p) is identically zero on the resonant manifold k, ω(p) = 0, equation (4.22) can be solved only in a subset of the action domain G which excludes the resonant manifolds. Furthermore, even if the divisor is not zero it can assume arbitrarily small values, thus making convergence doubtful.
14 98 Chapter Nonexistence of first integrals The problem of small denominators is actually a major one. This has been discussed in detail by Poincaré, who proved that under some genericness conditions no analytic first integrals exist independent of the Hamiltonian 4. The formal statement is the following Theorem 4.11: Let the Hamiltonian (4.1) satisfy the following hypotheses: (i) nondegeneracy, i.e., ( 2 ) H 0 det 0 ; p j p k (ii) genericness: no coefficient h k (p) of the Fourier expansion of H 1 (p,q) is identically zero on the manifold k,ω(p) = 0. Then there is no analytic first integral independent of H. The proof proceeds in three steps. i. Due to the nondegeneracy condition on the unperturbed Hamiltonian, Φ 0 must be independent of the angles q. ii. The condition that Φ be independent of H may be replaced by the condition that Φ 0 be independent of H 0. iii. Due to the genericness condition, Φ 0 can not be independent of H 0. The first claim is nothing but proposition 4.9; so, we need to prove only the second and the third claims. I split the technical part of the proof into two separate lemmas. Lemma 4.12: If Φ 0 is independent of H 0 then Φ is independent of H. Conversely, if Φ is independent of H then one can construct a first integral Φ = Φ 0 +εφ with Φ 0 independent of H 0. This is equivalent to the claim at point ii. Proof. Let Φ 0 be independent of H 0. This means that the Jacobian matrix (H 0,Φ 0 ) (p 1,...,p n ) has rank 2, that is, it contains a 2 2 submatrix with non zero determinant. Possibly with a rearrangement of the order of the variables we may always assume that ( H0 H 0 ) p det 1 p 2 0. Φ 0 p 1 Φ 0 p 2 Replacing H 0 with H = H 0 +εh and Φ 0 with Φ = Φ 0 +εφ we get the determinant ) det ( (H0 +εh ) p 1 (H 0 +εh ) p 2 (Φ 0 +εφ ) p 1 (Φ 0 +εφ ) p 2 4 This sections is based on the classical treatise of Poincaré [87], Vol. I, chapt. V. The same proof of Poincaré s theorem can be found also in Whittaker s treatise [97], chapt. XIV, 165.
15 First integrals 99 which is a continuos function of ε and coincides with the previous one for ε = 0. Since it is not zero for ε = 0, by continuity it must be non zero for ε 0 small enough, which proves that Φ is independent of H. Coming to the second part of the statement let Φ be independent of H, but Φ 0 be a function of H 0. I show how to construct the first integral Φ = Φ 0 + εφ such that Φ 0 is independent of H 0. This requires some considerations. Let us prove that since Φ 0 = Φ 0 (H 0 ) it must be an analytic function of its argument. For, in view of the nondegeneracy condition one at least of the derivatives H 0 p j, with (j = 1,...,n), must be different from zero, and we may assume that it is H 0 p 1. Thus we can invert the relation H 0 = H 0 (p), getting p 1 = p 1 (H 0,p 2,...,p n ) as an analytic function of its arguments. Replacing this in Φ 0, we construct the function Φ 0 (H 0,p 2,...,p n ) = Φ 0 (p 1,...,p n ) p1 =p 1 (H 0,p 2,...,p n ), which is analytic, being a composition of analytic functions. On the other hand, Φ 0 was assumed to be function of H 0 only, and so it is an analytic function of H 0, as claimed. Let us now consider the function Φ 0 (H) that we obtain by replacing H 0 with H in the expression of Φ 0 (H 0 ). This is clearly a first integral for H, in view of the obvious property { Φ 0 (H),H } = 0. On the other hand it is an analytic function of H, because Φ 0 is an analytic function of its argument, and so we can expand it in power series of ε as Φ 0 (H) = Φ 0 (H 0 )+ε Φ , where Φ 1,... are known functions. By a similar argument also the function Ψ = Φ Φ 0 (H) is a first integral for H, being a difference of first integrals, and is an analytic function of H, so that it can be expanded in power series of ε as Ψ = εψ 1 +ε 2 Ψ , because the first term is Φ 0 Φ 0 (H 0 ) = 0. Dividing by ε and setting Φ 0 = Ψ 1, Φ 1 = Ψ 2,... we get a first integral of the same form of Φ. By possibly iterating this procedure a finite number of times we must end up with a first integral with Φ 0 independent of H 0. For, this process is tantamount to attempting to expand 5 Φ in powers of ε with coefficients that are functions of H, and it must stop at some point because Φ has been assumed to be independent of H. Q.E.D. Lemma 4.13: With the hypotheses of theorem 4.11 the function Φ 0 can not be independent of H 0. 5 Thispointmayrequireashort explanation.afterthefirst stepwewriteφ = F 0 (H)+εΦ 1 where F 0 (H) = Φ 0 (H) and Φ 1 = Ψ 1 + εψ is a first integral with Ψ 1 as its first term. Thus Ψ 1 must be independent of the angles, and we can check if it is independent of H 0. If Ψ 1 is a function of H 0 then, be repeating the same argument, we write Ψ = Φ 1 Ψ 1 (H) = εφ 2 where Φ 2 = Ψ 2 + εψ is again an analytic first integral. Thus we have Φ = F 1 (H) + ε 2 Φ 2 where F 1 (H) = Φ 0 (H) + εψ 1 (H). We repeat this process as many times as we need until we end up with Φ = F s 1 (H) + ε s Φ s where Φ s = Ψ s +εψ s is a first integral with Ψ s the first term of which is independent of H 0, which must happen at some point. For, if this does not occur then letting s we end up with Φ = F (H) as a function of H, contradicting our assumption that Φ is independent of H. The function Φ s is the wanted first integral.
16 100 Chapter 4 Proof. Consider the second equation (4.21), namely L H0 Φ 1 = {H 1,Φ 0 }. This function has zero average, so that the problem of consistency mentioned in sect does not appear. Denoting by h k (p) the Fourier coefficients of H 1 we have {H 1,Φ 0 } = i k Z n k, Φ 0 p h k (p)exp ( i k,q ), so that the equation for the Fourier coefficients ϕ k (p) of Φ 1 reads k,ω(p) ϕ k (p) = k, Φ 0 p (p) h k (p), k Z n. Pick a k 0; on the resonant manifold k,ω(p) = 0 we can solve the equation above only if either condition (4.24) h k (p) = 0 or k, Φ 0 p (p) = 0 is fulfilled. Since h k (p) = 0 has been excluded by the genericness hypothesis, the second condition must apply. Recall now that at every p G we can associate the resonance module M ω(p), and let p be a completely resonant point, i.e., be such that dim M ω(p) = n 1. For that value of p we have k, H 0 p (p) = 0 and k, Φ 0 p (p) = 0 for n 1 independent vector k, so that we conclude that every 2 2 submatrix of the Jacobian matrix (4.25) (H 0,Φ 0 ) (p 1,...,p n ) has zero determinant. Now, completely resonant points are dense in R n, and so, by the nondegeneracy of H 0, are dense in G. Therefore, every 2 2 submatrix of the Jacobian matrix above has zero determinant at a set of points which is dense in G, and so, by continuity, it is zero everywhere. This proves that Φ 0 and H 0 can not be independent, as claimed. Q.E.D. Proof of theorem By lemma 4.12 if a first integral independent of H exists then there is also a first integral Φ(q,p) = Φ 0 (p) + εφ 1 (q,p) +... such that Φ 0 is independent of H 0. Such a first integral must satisfy the system (4.21) of equations, and so it may be constructed by recursively solving that system. But lemma 4.13 claims that the system may be solved only if Φ 0 is a function of H 0. Thus, a first integral independent of H can not exist. Q.E.D.
17 First integrals Some remarks on the theorem of Poincaré The theorem of Poincaré has been sometimes criticized by mathematicians because the hypotheses have been considered either too strong or at least hardly verifiable for a given system. I devote this section to a short discussion concerning the condition of genericness. The condition of degeneration will be removed in the next chapter 5, where a wide discussion will be devoted to the case of isochronous unperturbed Hamiltonians On the genericness condition The genericness condition plays a major role in the proof of Poincaré s theorem, but appears as a very strong limitation on the class of Hamiltonians to which the theorem applies. For, checking this condition requires a thorough knowledge of the coefficients of the Fourier expansion of the Hamiltonian. E.g., does the Hamiltonian of the classical problem of three bodies fulfill the genericness condition? The answer, to the best of my knowledge, is unknown. A long discussion on different ways to replace the genericness condition with a weaker one can be found in Poincaré s work. 6 Following Poincaré, the condition may be weakened, e.g., as follows. Let us consider a partition of Z n \{0} in classes by saying that two vectors k,k belong to the same class if they are parallel, i.e., if k = λk for some rational number λ. Then replace the genericness condition (ii) in theorem 4.11 by the following: For every class there is at least one k such that the coefficient h k (p) is not identically zero on the manifold k,ω(p) = 0. This is clearly enough in order to conclude that the second condition in (4.24) must hold, so that the rest of the argument goes the same way. Still, the condition above may be further weakened. Let p G, and consider the resonance module M ω(p) Z n associated to ω(p), as defined in (4.8). Let us say that p is a completely resonant point in case M ω(p) Z n has multiplicity n 1. Then replace the genericness condition (ii) in theorem 4.11 by the following: There exists in G a dense set of completely resonant points p satisfying the following condition: there are n 1 independent vectors k in M ω(p) such that the coefficient h k (p) is not identically zero on the manifold k,ω(p) = 0. Again, this leads us to conclude that (4.25) must hold true for n 1 independent vectors in a dense subset of G, and the rest of the argument still applies. The reader will immediately realize that the weaker conditions above are not of great help in deciding whether a given Hamiltonian fulfills or not the hypotheses of the theorem of Poincaré. We could keep looking for weaker and weaker conditions, of course. We could also end up with the feeling that if we consider a perturbation containing only a finite number of Fourier harmonics in fact a trigonometric polynomial then we may escape the strong conclusions of the theorem. But let me remark also 6 See [87], chapt. V, The discussion in this section shows that problem is a very delicate one, because it seems that there are many ways to escape the difficulties of the generic case treated by Poincaré.
18 102 Chapter 4 k 2 Φ2 k 2 Φ 3 k 2 Φ 1 k 1 k 1 k 1 order 1 order 2 order 3 Figure 4.3. Illustrating the propagation of Fourier modes through the process of construction of a first integral for the Hamiltonian of example 4.2. that all conditions above are concerned only with H 1 (p,q), the coefficient of the first order in ε, because the proof of the theorem of Poincaré exploits only the first and the second of equations (4.21). Nothing has been said till now about the coefficients of the higher order terms H 2 (p,q), H 3 (p,q),... of the expansion in powers of ε: they could well be zero. Thus, let us see what may happen at higher orders in ε. Example 4.2: Trigonometric polynomials. Consider the Hamiltonian H(p,q,ε) = H 0 (p)+εh 1 (p,q), H 0 (p) = 1 2 (p2 1 +p2 2 ), H 1 (q) = cosq 1 +cos(q 1 q 2 )+cos(q 1 +q 2 )+cosq 2, with p R 2 and q T 2. I emphasize that the perturbation H 1 is a trigonometric polynomial of degree 2, so that the genericness condition does not apply. Let us apply the procedure of sect for a first integral, starting, e.g., with Φ 0 = p 1. Using the exponential form for the trigonometric functions we have {H 1,p 1 } = i 2 [( e iq 1 +e i(q 1 q 2 ) +e i(q 1+q 2 ) ) ] c.c. where c.c. stands for the complex conjugate of the expression in parentheses, so that the resulting function is a real one. In view of ω j (p) = p j, we determine Φ 1 as Φ 1 = 1 [( e iq 1 ) ] + ei(q1 q2) + ei(q1+q2) +c.c.. 2 p 1 p 1 p 2 p 1 +p 2 The solution is determined up an arbitrary function of p 1,p 2, that we choose to be zero. This is consistent provided we exclude from the domain of definition of Φ 1 (and so also of the first integral Φ) the straight lines p 1 = 0, p 1 p 2 = 0 and p 1 +p 2 = 0. Let us go on by constructing Φ 2. Without actually performing the explicit computation, let us focus our attention on the Fourier modes that are generated. The process is illustrated in fig. 4.3, where the Fourier modes that appear in the functions Φ s are represented for orders s = 1,2,3. The Poisson braket makes the exponentials to be multiplied; hence {H 1,Φ 1 } contains the new Fourier modes (among others) e i(2q 1 q 2 ), e i(2q 1+q 2 ), e i(q 1 2q 2 ), e i(q 1+2q 2 ),
19 First integrals 103 (grey squares in the figure for order 2). One will immediately see that {H 1,Φ 1 } is a trigonometric polynomial of degree 4. The corresponding divisors in the solution 7 for Φ 2 will be 2p 1 p 2, 2p 1 +p 2, p 1 2p 2, p 1 +2p 2. Hence, Φ 2 is defined in a domains which excludes also the straight lines 2p 1 p 2 = 0, 2p 1 +p 2 = 0, p 1 +2p 2 = 0, p 1 2p 2 = 0, besides the ones of order 1. This because the procedure itself has generated a number of new Fourier modes, that did not exist in H 1. At order ε 3 we get the new modes represented in the figure as open squares. With a moment s thought we realize that at order ε s the right member {H 1,Φ s 1 } will be a trigonometric polynomial of degree 2s, and will contain practically all Fourier modes e i(k 1q 1 +k 2 q 2 ) with k 1 + k 2 2s; a finite number, but increasing with s. It is quite unlikely that these terms will cancel out each other; so, it is reasonable to expect that all of them will appear. Therefore we must remove more and more resonant straight lines k 1 p 1 +k 2 p 2 = 0 with k 1 + k 2 2s. We conclude that for s we must remove a dense set of resonances, so that a first integral can not be constructed on an open domain, even disregarding the problem of convergence. The example above admits an easy generalization. Suppose that we are given a nondegenerate Hamiltonian of the form (4.1), with H s a trigonometric polynomial of degree sk for some positive integer K. That is, H s is a finite sum of the form H s = k sk h (s) k (p)ei k,q, s 1. Followingthelinesofthediscussioninexample4.2wecaneasilyverifythatther.h.s.of the equation for Φ s is a known trigonometric polynomial of degree sk. Therefore, we are allowed to determine Φ s in an open domain, G (s) say, excluding (generically) all resonances of order not bigger than sk. However, unless there are hidden mechanisms of cancellation, we are not allowed to let s, because we shall be confronted again with the fact that resonances are dense. So, we see that the process of construction of first integrals is able to generate the resonant terms which do not initially appear in the Hamiltonian. With a little extra care, the reader will notice that the process of generation of new Fourier components is fully active if the truncated Hamiltonian contains at least a set of Fourier modes k which is a basis of a n dimensional subgroup of Z n. Indeed the combination of modes fails to produce all possible resonance in case a trivial restriction is imposed on the initial Hamiltonian. This is illustrated by the following Example 4.3: An elementary criterion for the existence of first integrals. Consider an Hamiltonian of the form (4.1), but with the restriction that H 1,...,H s,... contain only Fourier components on some subgroup K Z n, with dimk < n. Then for any α R n satisfying 0 α K the function Φ = α,p is a first integral. For, denoting 7 I forget the problem related to the consistency condition, because this argument does not depend on it.
20 104 Chapter 4 generically by h k (p) the coefficients of the Fourier expansion of H s, we have {H s,φ} = i α,k h k (p)exp ( i k,q ), k K which is zero in view of α,k = 0. In such a case we can perform a linear canonical transformationontheanglesq so thatthehamiltoniandepends onlyondim K angles. If dimk = 1, then the system is actually integrable. If dim K > 1 then we have actually performed a reduction to a system with n dim K degrees of freedom, to which Poincaré s theorem generically applies Truncated first integrals The examples above show that, although first integrals of the form of power series do not exist in general, we can nevertheless construct approximate first integrals on domains excluding resonances. For instance, referring to the case of example 4.2, we are able to construct formal expansions for first integrals up to any finite order ε r provided we restrict our attention to a domain, G (r) say, excluding all straight lines k,p = 0 with k 2r. That is, we construct, e.g., n independent functions (4.26) Φ (j,r) (p,q,ε) = Φ (j) 0 (p)+εφ(j) 1 (p,q)+...+εr Φ r (j) (p,q), Φ (j) 0 (p) = p j, for j = 1,...,n, which are polynomials of degree r in ε, and satisfy (4.27) {H,Φ (j,r) } = O(ε r+1 ). Let us call such functions truncated first integrals. We can argue that truncated integrals can be very useful in view of the following argument. The time derivative of Φ (j,r) is Φ (j,r) = {Φ (j,r),h}; for ε small such a quantity is much smaller that the time derivative of Φ (j) 0 = p j, the latter being of order O(ε). Therefore, during the evolution of the system we have a bound like (4.28) Φ (j,r) (t) Φ (j,r) (0) = O(ε) for t < Tr = O(ε r ). A similar result applies also to the unperturbed actions p 1,...,p n of the system. For, in view of the form (4.26) of the truncated integrals in the domain G (r) T n we have (4.29) pj Φ (j,r) = O(ε). On the other hand we have also pj (t) p j (0) pj (t) Φ (j,r) (t) + Φ (j,r) (t) Φ (j,r) (0) Φ (j,r) (0) p j (0). Using (4.28) and (4.29) we conclude (4.30) p j (t) p j (0) = O(ε) for t < T r = O(ε r ).
21 First integrals 105 This remark lies at the basis of Birkhoff s theory of complete stability. 8 The theory of complete stability of Birkhoff has been later improved in the theory of exponential stability that will the subject of ch. 8. The present section should be considered just as a suggestion that Poincaré s theorem should not considered as the ultimate and hopeless conclusion that nothing can be said about the general problem of dynamics: full integrability is off limits, in general, but many interesting results can nevertheless be found. The present chapter is just the beginning of a long story. 8 I should stress that Birkhoff s theory has been developed for the study of elliptic equilibria, for which the unperturbed Hamiltonian describes a completely isochronous system. I shall discuss in detail this problem later, in chapter 5.
22 106 Chapter 4
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