# Elimination of the angles: first step

Nel documento HAMILTONIAN MECHANICS —————- (pagine 69-73)

### 4.2Elimination of the angles: first step

The canonical transformation eliminating the angles to first order in the perturbation can be looked for in the form of the flow of a certain Hamiltonian. More precisely we set

Cε−1εG11εG1 , (4.7)

where G1 is an unknown Hamiltonian. By defining the Lie derivative

L1:= { ,G1} , (4.8)

and recalling formula (1.21) of Proposition 1.5, one gets H ◦Cε−1 = H ◦ΦεG1= eεL1H =

= (1 + εL1+ ε2. . . )(h + εP1+ ε2. . . ) =

= h + ε(L1h + P1) + ε2. . . .

Requiring that the last row above be equal to h + εZ1+ ε2, one has to solve the so-called homological equation of perturbation theory, namely

L1h + P1= Z1, (4.9)

with unknowns G1and Z1, and the requirement that the latter quantity be independent of the angles. Since h depends only on the actions, equation (4.9) explicitly reads

ω(J) ·∂G1(θ, J)

∂θ = P1(θ, J) − Z1(J), (4.10)

where the frequency vector

ω(J) :=∂h(J)

∂J (4.11)

has been defined. The average of equation (4.10) on the torusTnyields Z1(J) = 〈P1〉 := 1

(2π)n Z

TnP1(θ, J) dnθ . (4.12) Thus, a first result is that if the problem admits a solution, then the angles to first order are removed by averaging the perturbation over the torus. The homological equation is then easily solved by means of the Fourier transform. We recall that any smooth function f :Tn→ R admits the Fourier series expansion

f (θ) = X

k∈Zn

keik·θ , (4.13)

where the Fourier coefficient ˆfkis given by fˆk= 1

(2π)n Z

Tne−ik·θf (θ) dnθ . (4.14)

We recall that the velocity of convergence of the Fourier series, i.e. the rate of decay of the Fourier coefficients ˆfk with |k| is linked to the regularity of the function f . More precisely, one has that if f ∈ Cs(Tn), then there exists a constant As> 0 such that | ˆfk| ≤ As/|k|1s, where

|k|1:=Pn

j=1|kj|; moreover, if there exists ρ > 0 such that f is analytic in the complex extension of the torusTnρ := {z ∈ Cn: Re(zj) ∈ T ; |Im(zj)| ≤ ρ ; j = 1,..., n}, then there exists a constant Bs> 0 such that | ˆfk| ≤ Bse−|k|1ρ. A proof of such statements is reported in Appendix B. Here we make use of the Fourier series at a formal level, in order to solve the homological equation (4.10). Indeed, by inserting the Fourier series expansions

G1(θ, J) = X

k∈Zn

Gb1,k(J)eik·θ (4.15)

and

P1(θ, J) = X

k∈Zn

Pb1,k(J)eik·θ (4.16)

into (4.10), and observing that

Pb1,0(J) = 〈P1〉 (J) = Z1(J), one gets

ik · ω(J) bG1,k(J) = bP1,k(J) − δk,0Pb1,0(J), (4.17) to be solved for any k ∈ Znand any J ∈ B. The formal solution of (4.17), for k 6= 0, reads

Gb1,k(J) = Pb1,k(J)

ik · ω(J) , (4.18)

and immediately points out the major problem of Hamiltonian perturbation theory, i.e. deal-ing with the possible small divisors k · ω(J). Notice that resonant frequencies, namely those frequency vectorsω(J) such that there exists k ∈ Zn\ {0} satisfying k · ω(J) = 0, are just the worst possibility. Indeed, one has to recall that in order to get the homological equation one has expanded the operator eεL1= eε{·,G1}. The latter expansion is meaningful if, roughly speak-ing G1is of order one (with respect to the small parameterε). In particular, any of its Fourier coefficients must be of order one, otherwise its contribution to the expansion would cause prob-lems. Obviously, a denominator k · ω(J) = O(εα), withα ≥ 1, makes the perturbative expansion meaningless. At a deeper level than the one considered here, one finds out that, in general, dangerous small denominators are those satisfying k · ω(J) = O(p

ε).

The problem of small divisors immediately calls for a partition of the integrable systems to be perturbed into the following two classes:

• isochronous systems, with h(J) = ω0· J, i.e. a constant (independent of J) frequency map ω(J) = ∂h/∂J = ω0;

• non-isochronous systems, with h(J) a nonlinear function of J, i.e. a non constant fre-quency mapω(J).

4.2. ELIMINATION OF THE ANGLES: FIRST STEP 71 Of course, small divisors in isochronous systems are easily treated, at least in principle. In-deed, either ω0 is resonant or is not. In the latter case equation (4.17) admits the formal solution (4.18), and one is left with the problem of convergence of the series (4.15). In the for-mer case, one has to characterize the resonance module Mω0:= {k ∈ Zn: k · ω0= 0}. Obviously, for those vectors k ∈ Mω0 the ratio on the right hand side of (4.18) is not defined, unless the numerator Pb1,k(J) = 0 for any J, which, though not generically, may happen in practice. In such a resonant, isochronous case, one cannot simply remove all the angles, but only part of them. More precisely, in this case the transformed/simplified Hamiltonian to first order reads

H(ΦGε1(θ, J)) = h(J) + ε X with arbitrary Fourier coefficients gk(J) on the resonance modulus.

Exercise 4.1. Prove the above statement, showing that (4.20) solves the homological equation (4.10) withZ1(θ, J) as defined in (4.19) in place of Z1(J).

The case of non-isochronous systems is much more complicated, since in that case the small-ness of the denominators on the right hand side of (4.17) varies in the space of the actions, which calls for a major role played by the geometry of the resonances in the problem. Actually, due to the action dependence of the small divisors, non-isochronous systems must be studied by further partitioning them into classes defined by the analytical/geometrical properties of h(J), which in turn establishes precise properties of the frequency map

ω : B → ω(B) : J 7→ ω(J) =∂h(J)

∂J . (4.21)

By far the easiest class of integrable non-isochronous systems to be studied is the following.

Definition 4.2. The integrable system defined by h(J), is said to be non-degenerate in B (open set in the domain ofh) if there exists a constant c > 0 such that

¯

In such a case the frequency map (4.21) is a local diffeomorphism (i.e. a smooth bijection), so that to any frequency vector having a certain property there corresponds one and only one action. As a consequence, any subset with given metric properties in the frequency space is the image of a subset with the same metric properties in the action space.

Example 4.4. A dense, Lebesgue-measure zero set⊂ ω(B) in the frequency space is the image of a dense, Lebesgue-measure zero setA in the action space. Indeed, consider the set A := ω−1(Ω).

The smoothness (actually continuity is enough) of the map ω−1 implies that to close frequency

values there correspond close action values, which implies thatω−1preserves density. For what concerns the measure, observe that

0 = V (Ω) = Z

dnω = Z

A

¯

¯

¯

¯

∂ω

∂J

¯

¯

¯

¯

dnJ ≥ cV (A) , where the non-degeneracy condition (4.22) has been used.

In particular, for non-degenerate systems, one is interested in the sets

r:= {ω ∈ Rn: dim Mω= r} (r = 1,..., n − 1) , (4.23) namely the sets of resonant frequencies whose resonance modulus is generated by r linearly independent integer vectors. The setsΩrare denumerable, dense inRnand have zero relative Lebesgue measure (i.e. for any bounded, measurable set A ⊂ Rn one has V (Ωr∩ A) = 0). The setω(B) \ Sn−1r=1rof non resonant frequencies is dense too and of full measure.

Example 4.5. In the case n = 2 there is only1, the set of frequency vectors ω ∈ R2 such that there exists an integer vectork ∈ Z2satisfyingk1ω1+k2ω2= 0. In other words,Ω1consists of the two coordinate axes and of all the straight lines with rational slope. Ω1is denumerable, dense and has relative measure zero because the rational numbers have the same properties.

Example 4.6. In the case n = 3,1is the set of planes orthogonal to integer vectors, whereasΩ2

is the set of straight lines resulting from the intersection of two planes ofΩ1that are orthogonal to two non parallel integer vectors.

By the non-degeneracy condition (4.22) the sets of resonant actions

Br:= ω−1(Ωr∩ ω(B)) = {J ∈ B : ω(J) ∈r} (4.24) are denumerable, dense and of zero relative measure; the set of non resonant actions B \ Sn−1

r=1Bris dense too and of full measure. The existence of the sets (4.24) in the space of actions implies that the Fourier coefficients (4.18) are not defined on dense sets. In order to get a sharp statement of this kind one needs to control the possibility to have zero numerators on the right hand side of (4.18).

Definition 4.3. The first order perturbation P1(θ, J) is said to be generic in B if for any J ∈ B and anyk ∈ Znthere exists q parallel to k such thatPb1,q(J) 6= 0.

The following theorem is due to Poincaré.

Theorem 4.2 (Unfeasibility of the first step). Suppose that h(J) is non-degenerate and P1(θ, J) is generic in B. Then, the homological equation (4.17) does not admit solutions on any open subset ofB.

C PROOF. For any given J ∈ B \Sn−1

r=1Br and k ∈ Zn\ {0} equation (4.17) can be solved as in (4.18). On the other hand, by the density of the sets Brin B, arbitrarily close to such a J there exists ¯J ∈Sn−1

r=1Br, such that ¯k ·ω( ¯J) = 0 for at least one ¯k ∈ Zn\ {0}. Moreover, since the perturbation P1 is generic in B, there exists q parallel to ¯k such that P1,q( ¯J) 6= 0 and obviously, q · ω( ¯J) = 0. Thus, equation (4.17) cannot be solved in the dense setSn−1

r=1Br. B

Nel documento HAMILTONIAN MECHANICS —————- (pagine 69-73)