An analytic counterexample to the KAM theorem

An Analytic Counterexample to KAM Theorem

Ugo Bessi

Abstract We study the nonexistence of KAM tori for quasi-integrable, analytic lagrangians. Let L: Tm ×Rm → R, ˙ = 1 |Q| ˙ 2 +h(Q) and let ω L(Q, Q) ¯ ∈ Rm be a frequency exponentially close to resonances. We find h analytic 2

of norm arbitrarily small such that L has no invariant torus of frequency ω ¯ projecting diffeomorphically on Tm .

Introduction One of the problems in the theory of Hamiltonian Systems is to understand which are the frequencies for which the KAM theorem holds. To be more precise, let us suppose we are given the lagrangian L: Tm × Rm → R where Tm =

Rm 2πZm

˙ = L(Q, Q)

1 ˙ 2 ˙ |Q| + h(Q, Q) 2

is the m-dimensional torus. The KAM theorem asserts that, if h is sufficiently small in k

a certain norm (C or analytic) then there is a class of ω ∈ Rm for which there is a smooth embedding i: Tm → Tm × Rm satisfying the following three properties. 1) i(Tm ) is invariant for the Euler-Lagrange flow (from now on E-L flow) of L. 2) i conjugates the E-L flow on i(Tm ) with a translation of frequency ω on Tm . n o 3) i(Tm ) is ”close” to the torus Q˙ = ω . If h ∈ C ∞ , then the frequencies ω for which the theorem holds are those satisfying the following condition: ∃C, γ > 0 :

∀k ∈ Zm \ {0}

|hω, ki| ≥

C |k|γ

(∗)

where h·, ·i denotes the standard inner product in Rm . A proof of the sufficiency of (∗) can be found, for instance, in [14]; as a corollary of [9] it follows that, for twist maps, (∗) is also necessary. With the additional condition that i(Tm ) projects diffeomorphically on Tm it is simple to show, using only the most elementary ideas of [9], that (∗) is necessary in any dimension. If h is analytic, the situation is more intricate. It is known (see for instance [6]) that in this case the KAM theorem holds under a weaker condition than (∗), the Bruno condition which we presently explain. 1

Let αr = min |hk, ωi|. 0 0,

∃(pn , qn ) → ∞

:

|qn ω − pn | ≤

C exp(γ(|pn | + |qn |))

do not survive to arbitrarily small analytic perturbations. Our aim is to generalize this part of [5] to higher dimension. We begin introducing some notation. We call Aσ the set of all real analytic functions defined in a complex strip of radius σ around Tm ; for h ∈ Aσ we consider the norm khkσ = sup{|h(z)| :

|Im(z)| ≤ σ}.

We will prove the following theorem. Theorem 1.

Let m ≥ 2 and σ > 0. Let ω ¯ ∈ Rm , |¯ ω | ∈ (1, 2) and let it be rationally independent (i. e.

{¯ ω t}t∈R is dense on Tm .) Moreover, let it satisfy ∃{kn } ⊂ Zm , |kn | → ∞ :

0 < h¯ ω , kn i ≤ e−D|kn | .

(∗ ∗ ∗)

˙ 2 + h(Q) does not have ˙ = 1 |Q| Then if D > σ ∀² ∈ (0, 1] there is h ∈ Aσ with khkσ ≤ ² such that L(Q, Q) 2 any invariant torus T satisfying (i) T projects diffeomorphically on Tm . (ii) There is a C 1 diffeomorphism i: Tm → T which conjugates the motion on T to a translation of frequency ω ¯.

For twist maps, it is a theorem of Birkhoff (see [12] for a proof) that that any homotopically non trivial invariant circle projects diffeomorphically on T1 ; we do not know whether in arbitrary dimension invariant tori of rotation number ω ¯ always project diffeomorphically on Tm . However, if a torus T not satisfying (i) exists, then its Kolmogoroff normal form cannot be positive definite; this is because of the following theorem ˙ Qi ˙ + h.o.t. with A positive definite, then the ˙ = hQ, ˙ ω of [10]. If close to T L behaves like L(Q, Q) ¯ i + 1 hAQ, 2

2

torus supports a minimal measure and this implies that T projects diffeomorphically on Tm . This makes very unlikely that such a T can result from any of the usual proofs of the KAM theorem; indeed these proofs provide positive definite KAM tori whenever the unperturbed torus {Q˙ = ω ¯ } is, like in our case, positive definite. The proof of theorem 1 is contained in the next section and is based on an observation of [13] and [7]: the orbits lying on a KAM torus are minimizers of the action functional. But the projection on Tm of a KAM torus of frequency ω ¯ is surjective: thus through each point of Tm passes a minimal orbit of ”mean frequency” ω ¯ . Thus our counterexample reduces to finding a quasi-integrable Lagrangian whose minimal orbits of mean frequency ω ¯ avoid some points of Tm . This Lagrangian is the same as in [1]; this is not surprising since [1] is practically the Nekhorocheff normal form near a first- order resonance and, by (∗ ∗ ∗), we are very close to a first order resonance.

Section 1 Proof of theorem 1. We begin recalling the definition of [10] of action minimizing orbit; actually, we will ˙ = give a slightly weaker definition, but sufficient for our purposes. Let us consider the lagrangian L(Q, Q) 1 ˙ 2 2 |Q|

+ h(Q). We say that the orbit Q ∈ A.C.(R, Tm ) (i. e. absolutely continuous) is minimal for L if Z t2 Z t2 ˙ ¯ Q)dt ¯˙ ∀t1 < t2 L(Q, Q)dt ≤ L(Q, (1) t1

t1

¯ ∈ A.C. such that for every Q ¯ i ), i = 1, 2. a) Q(ti ) = Q(t ¯ restricted to [t1 , t2 ] are in the same homotopy class or, equivalently, their lifts to Rm connect b) Q and Q the same points. If a) and b) hold for a particular couple of t1 < t2 , we say that Q is minimal for L on [t1 , t2 ]. In the above definition we take minima in the very wide class of A.C. functions; however, since in our case minima are always analytic, it is the same if we restrict ourselves to a smaller class. In particular, in the proof of lemma 2 it will be useful to consider our minima as taken in H 1 ([t1 , t2 ], Tm ), the set of A.C. functions with square-integrable derivative. We now observe that an invariant torus satisfying point (ii) of theorem 1 is lagrangian. Indeed, let Ω be the canonical 2-form on Tm × Rm and let us suppose by contradiction that Ω|T 6= 0. Then there is a point x0 ∈ Tm and two vectors v and w such that i∗ (Ω)(x0 )(v, w) > 0, where i∗ denotes the pullback. By the invariance of i∗ (Ω) under translations of frequency ω ¯ we get that i∗ (Ω)(x)(v, w) > 0 for any x ∈ Tm . Moreover, slightly changing v and w, we can suppose that the plane π which contains them is, quotiented R by 2πZm , a 2- dimensional torus. If we call π ˜ this torus we get that π˜ i∗ (Ω) > 0, which is a contradiction since i∗ (Ω) is exact. Thus we have that T is lagrangian and projects diffeomorphically on Tm ; by the appendix 2 of [10], all ˙ orbits Q with (Q(t), Q(t)) ∈ T ∀t are minimal for L. Moreover, since T is invariant, any point on T is the initial condition of an orbit lying on T . We conclude that, given any Q0 ∈ Tm , we can find Q(t) such that 3

˙ lies on the invariant torus. 1) Q is minimal for L and (Q, Q) 2) Q(0) = Q0 . Thus the proof will go along the following lines: we will define a perturbation h on whose minimal orbits we have much information, then we will suppose by contradiction that an invariant torus exists and show that there is Q0 such that 1) and 2) are violated; this contradiction will prove the theorem. We note that we can always find a vector kn0 which is orthogonal to kn and satisfies ∀n

h¯ ω,

kn0 i ≥ a > 0, |kn0 |

|kn0 | ≤ |kn |

(2)

for some a not depending on n. We choose our perturbation as in [1]: 1 ˙ 2 + hn²,µ (Q) = L˜n²,µ = |Q| 2 0 1 ˙ 2 |Q| + ²e−σ|kn | [1 − coshkn , Qi] + ²µe−σ|kn | [1 − coshkn , Qi]e−σ|kn | coshkn0 , Qi. 2

We note that khn²,µ kσ ≤ 4² if µ ≤ 1. To make L˜n²,µ more wieldy, we make a change of coordinates. We define P and P¯ to be the orthogonal projections on kn and on kn⊥ rispectively and we set w = P¯ Q.

q = |kn |P Q

˙ = Ln²,µ (w, q, w, In the new coordinates we have that L˜n²,µ (Q, Q) ˙ q) ˙ with ˙ q) ˙ = Ln²,µ (w, q, w,

0 1 2 1 1 2 |w| ˙ + |q| ˙ + ²e−σ|kn | [1 − cos(q)] + ²µe−σ|kn | [1 − cos(q)]e−σ|kn | coshkn0 , wi. 2 |kn |2 2

Thus Ln²,µ consists in a rotator w, a pendulum q of lagrangian 1 1 2 |q| ˙ + ²e−σ|kn | [1 − cos(q)] |kn |2 2

˙ = An² (q, q)

and a coupling between the two systems, small with µ. We write below the homoclinic q² of An² such that q˙² > 0, q² (0) = π:

³ √ ´ q² (t) = 4arctg e ²δn t

q˙² (t) =

√ 2 ²δn √ Ch( ²δn t)

where δn = |kn |2 e−σ|kn | and arctg takes values between − π2 and

π 2.

We set

βn = e−σ|kn |

0

αn = e−σ|kn | e−σ|kn |

and define the Melnikoff function as Z M² (ω, w0 , t0 ) = ²αn R

[1 − cos(q² (t − t0 ))] coshkn0 , ω(t − t0 ) + w0 idt. 4

In other words, M² is the integral of the perturbation along a homoclinc of Ln²,0 . It does not depend on t0 and on the projection of w0 on (kn0 )⊥ and is calculated: M² (ω, w0 , t0 ) =

2π 0 | -periodic |kn

in the projection of w0 on kn0 . It can be explicitly

hω, kn0 i 1 −σ|kn0 | e 2π coshw0 , kn0 i. hω,k0 i |kn |2 Sh( π √ n ) 2

²δn

If we specialize the above formula to ω ¯ and recall (2), with a simple calculation we can get the next lemma. Lemma 1.

Let ² > 0, n ∈ N and let hw0 , kn0 i ∈ 2πZ, hw1 , kn0 i ∈ π + 2πZ. Then 0 | −b|kn

M² (¯ ω , w0 , t0 ) − M² (¯ ω , w 1 , t0 ) ≥ e

√

²δn

.

It is easy to see that the transformation : Q → (w, q) preserves the minimality of the orbits; thus, if ˙ (Q, Q) lies on the invariant torus of frequency ω ¯ , we have that (w, q), the image of Q, is minimal for Ln²,µ . We will show that this implies that for all ² ∈ (0, 1) there is n big enough such that q(t) ∈ π + 2πZ

=⇒ hw(t), kn0 i 6∈ 2πZ

(3)

˙ belongs to a torus T enjoying the properties of theorem 1. But this is tantamount to whenever (Q, Q) showing that the properties 1) and 2), listed at the beginning of the proof, do not hold. Thus the proof of theorem 1 reduces to the proof of (3), which contradicts the existence of an invariant torus. We remark that (3) says that minimal orbits do not pass through the maxima of the Melnikoff function; this is the same fact underlying the variational proof of [1] (see for instance [3].) Also as a side remark, the Melnikoff function is closely connected to the ”barrier” of [11]; the lagrangian of [1] is practically the only case where this barrier can be estimated explicitly (see for instance [4].) We now prove a lemma which says that, for µ small enough, M² approximates the integral of the perturbation along the minimal orbits of Ln²,µ . From now on we will consider our orbits as valued in Rm , the universal cover of Tm ; Ci will always denote a constant not depending on ² and n. Lemma 2.

Let M > 16. There are C0 , C1 > 0 and l ∈ N, l ≥ 2 with the following property. Let 0 | C |kn ²δn

0 − √

² ∈ (0, 1]

µ=e

T3 ≥ T1 +

lC1 |kn0 | . ²δn

(4) (5)

Let (w, q) be minimal for Ln²,µ on [T1 , T3 ] and let it satisfy w(T3 ) − w(T1 ) = ω, T3 − T1 q(T1 ) = 0,

|ω| ∈ [1, 2]

q(T3 ) = 2π.

Then for any T2 ∈ (T1 , T3 ) such that q(T2 ) = π we have that 0 | Z T3 ¯ ¯ √ n 1 −b|k ¯ ¯ e ²δn . [1 − cos(q(t))] coshkn0 , w(t)idt − M² (ω, w(T2 ), T2 )¯ ≤ ¯²αn M T1 5

(6) (7)

(8)

Proof.

The proof follows the lines of lemma 1.1 of [3]; here we will only sketch the general argument and

point out those details which are different. First of all we define w² (t) = w(T1 ) + ω(t − T1 ). As in [3], (8) is implied by the following four inequalities: Z

T2 +

²δn T2 −

0 | C1 |kn ²δn

0 | C1 |kn ²δn

Z

0

| coshkn0 , wi

T2 +

²δn T2 −

0 | C1 |kn ²δn

0 | C1 |kn ²δn

−

coshkn0 , w²

−b|kn | √ 1 |kn |2 + w(T2 ) − w² (T2 )i|dt ≤ · −σ|k0 | · e ²δn n 8·M e

0

| cos q(t) − cos q² (t − T2 )|dt ≤

−b|kn | √ |kn |2 1 · −σ|k0 | · e ²δn n 8·M e

Z ²δn

²δn

(10)

0

−b|kn | √ |kn |2 1 ²δn |1 − cos q|dt ≤ · · e 0 0 0 −σ|kn | C1 |kn | C1 |kn | 8 · M e [T1 ,T3 ]\[T2 − ²δ ,T2 + ²δ ] n

Z

(9)

(11)

n

0

−b|kn | √ 1 |kn |2 ²δn · |1 − cos q (t − T )|dt ≤ · e . ² 2 0 0 | 0 | −σ|k | C1 |kn C1 |kn n 8 · M e R\[T2 − ²δ ,T2 + ²δ ] n

(12)

n

In the course of the proof we will also see that T1 ≤ T2 −

C1 |kn0 | ²δn

and

T3 ≥ T2 +

C1 |kn0 | ²δn

so that (11) makes sense. We will always assume that C1 , C0 > 1. First of all, by direct computation it is possible to see that, taking C1 big enough, we have Z

0

²δn R\[T2 −

0 | 0 | C1 |kn C1 |kn ²δn ,T2 + ²δn ]

|1 − cos q² (t − T2 )|dt ≤

−b|kn | √ |kn |2 1 · −σ|k0 | · e ²δn . n 8·M e

This proves (12) and sets the value of C1 . Now we consider φ, a C ∞ cutoff ( φ(t) = and define

   

0

t≤0

1

t≥1

T1 + T3 C1 |kn0 | T1 + T3 )φ(t − ( − )) t ≤ 0 2 2 ²δn q¯(t) =  T + T3 T1 + T3 C1 |kn0 |   (q² (t − 1 ) − 2π)φ(−t + ( + )) + 2π 0 ≤ t. 2 2 ²δn q² (t −

We note that, by (5) and (6), (w² , q¯)(Ti ) = (w, q)(Ti ), i = 1, 3. Since (w, q) is minimal we have that Z

T3

T1

Z Ln²,µ (w² , q¯)dt

T3

≥ T1

Ln²,µ (w, q)dt.

As in [3], from this fact we deduce that, for C0 big enough, Z

T3

{ T1

p 1 1 2 |q| ˙ + ²βn [1 − cos(q)]}dt ≤ C2 ²βn . 2 |kn | 2 6

(13)

By the E-L equation we get that 0

|w| ¨ ≤ µ|kn0 |e−σ|kn | ²βn [1 − cos(q)]. Thus

Z 0

∀t0 < t1 ∈ [T1 , T3 ] |w(t ˙ 1 ) − w(t ˙ 0 )| ≤ µ|kn0 |e−σ|kn |

t1

²βn [1 − cos(q)]dt t0

0

which by (13) and the fact that |kn0 |e−σ|kn | is bounded yields ∀t0 < t1 ∈ [T1 , T3 ]

|w(t ˙ 1 ) − w(t ˙ 0 )| ≤ µC3

implying by (6) ∀t ∈ [T1 , T3 ]

|w(t) ˙ − ω| ≤ µC3

p

²βn

p

²βn .

If we now integrate w˙ from T2 to t and use the above formula we get sup [T2 −

0 | 0 | C1 |kn C1 |kn ²δn ,T2 + ²δn ]

|w(t) − [w² (t) + w(T2 ) − w² (T2 )]| ≤ µC3

p C1 |kn0 | ²βn ≤ ²δn

0

−b|kn | √ 1 |kn |2 1 . · −σ|k0 | · e ²δn · n 8·M e 2C1 |kn0 |

The last inequality follows by (4) taking C0 big enough. Inequality (9) now follows from the above formula and the Lipschitz continuity of the cosine. We are now going to prove (10). By (5) we have that either T2 − T1 ≥

l C1 |kn0 | 2 ²δn

or T3 − T2 ≥

l C1 |kn0 | . 2 ²δn

To fix ideas we suppose that the first case occurs and we consider, on [T1 , T3 ], the solution q¯ of        Since T2 − T1 ≥

0 l C1 |kn | 2 ²δn ,

¨q¯ = ²δn sin(¯ q) q¯(T1 ) = q(T1 ) = 0 q¯(T2 ) = q(T2 ) = π.

it is easy to see that h, the energy of q¯, satisfies 0 < h ≤ ²βn e

We set h(t) =

0 | C l|kn ²δn

1 −C4 √

.

(14)

1 1 2 |q| ˙ − ²βn [1 − cos(q)]. |kn |2 2

As in [3], it is possible to show that there is t0 ∈ [T1 , T2 ] such that h(t0 ) = h; moreover, using the E-L ˙ equation we can estimate h(t); integrating from t0 to t we get sup

p |h(t) − h| ≤ C5 µ ²βn .

t∈[T1 ,T3 ]

7

(15)

We set

p γ = C5 µ ²βn .

As in [3], the minimality of (w, q) implies that q(T ˙ 2 ) > 0; clearly, q¯˙ (T2 ) > 0. If (T¯, T2 ] is the maximal interval to the left of T2 where q˙ > 0 (i. e. where q is invertible) denoting by t(q) the inverse function we have that, by (15), for q < π, Z q Z q 1 ds 1 ds p p T2 + ≤ t(q) ≤ T2 + . |kn | π |kn | π 2(h − γ + ²βn [1 − cos(s)]) 2(h + γ + ²βn [1 − cos(s)]) On the other side, t¯(q), the inverse function of q¯(t), satisfies Z q 1 ds ¯ p t(q) = T2 + . |kn | π 2(h + ²βn [1 − cos(s)]) From the last two formulas we get that |t(q) − t¯(q)| ≤ C6 if γ≤

1 γ |kn | {²βn [1 − cos(q)]} 32

1 ²βn [1 − cos(q)]. 4

(16)

By (14) and (15) it is easy to see that, for l and C0 big enough, kqk ˙ + kq¯˙ k ≤ C7 |kn |

p

²βn = C7

p ²δn .

The above formula and (16) imply that |q(t) − q¯(t)| ≤ C6 C7 as long as γ≤

√ γ ²δn

(17)

3

(²βn [1 − cos(¯ q (t))]) 2

1 ²βn [1 − cos(¯ q (t))]. 4

If l > 1, we have the estimate

(18)

0

q¯(T2 −

C8 |kn | − √ C1 |kn0 | ²δn )≥e . ²δn

If we choose C0 big enough we have by the above formula that the maximal interval on which (18) holds C |k0 |

contains [T2 − 1²δnn , T2 ]. Moreover, it is easily seen by (15) and (14) that on this interval q˙ > 0, i. e. C |k0 | [T¯, T2 ] ⊂ [T2 − 1²δnn , T2 ]. We now re-apply the same estimate to the maximal interval on the right of T2 on which q(t) ˙ > 0; we see that (17) continues to hold and thus |q(t) − q¯(t)| ≤ C6 C7

γ|kn |3 ²δn [1 − cos(¯ q (t))]

3 2

4γ|kn |3

≤ C6 C7

²δn e

3C |k0 | − √8 n ²δn

∀t ∈ [T2 −

C1 |kn0 | C1 |kn0 | , T2 + ]. ²δn ²δn

For l big enough, q¯ and q² (· − T2 ) are close; indeed, an estimate similar to the above one yields |q² (t − T2 ) − q¯(t)| ≤ C9

h|kn |3

∀t ∈ [T2 −

3C |k0 | − √8 n ²δn

²δn e

8

C1 |kn0 | C1 |kn0 | , T2 + ]. ²δn ²δn

If we take l and C0 big enough, from the above two formulas and (14) we get 0

−b|kn | √ 1 |kn |2 1 |q² (t − T2 ) − q(t)| ≤ · −σ|k0 | · e ²δn · . n 8·M e 2C1 |kn0 |

Formula (10) now follows by the Lipschitz continuity of the cosine. The proof of (11) is exactly the same as that of the corresponding inequality in [3]. \\\ We remark that the above lemma continues to hold if we substitute in (4) any constant C bigger that C0 , since this means to reduce further the size of the perturbation. Moreover, with the same proof it is possible to get that Z ¯ ¯ ¯²αn

0 | ¯ √ n 1 −b|k ¯ [1 − cos(q(t))] coshkn0 , w(t)idt − M²− (ω, w(T2 ), T2 )¯ ≤ e ²δn M T1 0 | Z T3 ¯ ¯ √ n 1 −b|k ¯ ¯ 0 + ²δn e [1 − cos(q(t))] coshkn , w(t)idt − M² (ω, w(T2 ), T2 )¯ ≤ ¯²αn M T2

T2

where

Z M²− (ω, w0 , t0 ) = ²αn M²+ (ω, w0 , t0 ) = ²αn

t0

−∞ Z ∞ t0

(19)

[1 − cos(q² (t − t0 ))] coshkn0 , ω(t − t0 ) + w0 idt

[1 − cos(q² (t − t0 ))] coshkn0 , ω(t − t0 ) + w0 idt.

We now note that, if we consider the motion of q(t) as known, the motion of w is given by the non autonomous lagrangian

1 ˙ 2 2 |w|

+ ²µαn [1 − cos(q(t))] coshkn0 , wi whose first derivative in w is bounded by

2²µαn |kn0 |. Moreover, this lagrangian is periodic in w; this is all we need to apply lemma 2 of [2], from which we get the following estimate. Lemma 3.

Let (w, q) be an action minimizing orbit. Then we have that, ∀T1 < T2 , ∀t ∈ [T1 , T2 ]

¯ w(T2 ) − w(T1 ) ¯¯ √ p ¯ ˙ − ¯w(t) ¯ ≤ C10 µ ²δn . T2 − T1

(20)

The next lemma contains an estimate on the action functional of the unperturbed pendulum which we will need in the following. We will use it twice: in lemma 5 we will use point 1) to show that a half-turn around the homoclinic (or whisker, as in [1]) takes a very long time to be done; in the end of the proof of theorem 1 point 2) will tell us that a small translation in time of a homoclinic does not increase its action functional too much. Throughout the rest of the paper, µ and C0 are defined as in (4). Lemma 4.

Let M > 16 be big enough. Then for all C0 big enough there is ∆ = ∆(C0 ) > 0 such that

the following holds. Let T2 , T¯2 ∈ [T1 , T3 ]. Let q be a solution of An² on (T1 , T2 ) and on (T2 , T3 ) with boundary conditions q(T1 ) = 0

q(T2 ) = π 9

q(T3 ) = 2π

and let q¯ be a solution of An² on (T1 , T¯2 ) and on (T¯2 , T3 ) with boundary conditions q¯(T¯2 ) = π

q¯(T1 ) = 0

q¯(T3 ) = 2π.

1) If T2 − T1 ≥ 4∆ T3 − T2 ≤ ∆ T3 − T1 ≥ 5∆ ¯ T1 + T3 ¯¯ 8π ¯¯ ¯T2 − ¯≤ 2 a then we have

Z

T3

T1

Z An² (q, q)dt ˙ −

T¯3

0

An² (¯ q , q¯˙ )dt >

T¯1

| √ n 8µ −b|k e ²δn . M

2) If T3 − T1 ≥ 2∆ T1 + T3 T¯2 = 2 8π |T2 − T¯2 | ≤ a then we have

Z

T3

T1

Moreover, we have

Z An² (q, q)dt ˙ −

T¯3 T¯1

0

An² (¯ q , q¯˙ )dt ≤

| √ n µ −b|k e ²δn . 8

C0 |kn0 | |k 0 | ≤ ∆ ≤ C11 (C0 + M ) n . C11 ²δn ²δn

(21)

Proof.

Let us consider the situation of 1). We define the function # " Z iπ p 2 Z Ti+1 2 X X 1 n 2(h(Ti+1 − Ti ) − V (s))ds − h(Ti+1 − Ti )(Ti+1 − Ti ) L(T2 ) = A² (q, q)dt ˙ = |kn | (i−1)π i=1 Ti i=1

where V (s) = −²βn [1 − cos(s)] and h(T ) denotes the energy of the orbit of the pendulum doing one half turn in time T . A simple calculation yields

dL(T2 ) = (h(T3 − T2 ) − h(T2 − T1 )). dT2

(22)

Since h(T ) is decreasing in T we have that if T2 ∈ [T3 − ∆ − 1, T3 − ∆] then

h(T3 − T2 ) − h(T2 − T1 ) ≥ h(∆ + 1) − h(2∆).

If we now integrate (22) between T¯2 and T2 , we get that 1) holds if 0

h(∆ + 1) − h(2∆) ≥ 10

| √ n 8µ −b|k e ²δn . M

(23)

Let us now consider the situation of 2). Integrating as before we get that 2) holds if 0

| √ n 8π 8π µ −b|k h(∆ − ) ≤ e ²δn . a a 8

It is now easy to see (for instance estimating

d dT

(24)

h(T )) that, for M big but independent on ∆,

h(∆ + 1) − h(2∆) ≥

64 π 8π h(∆ − ). Ma a

A simple calculation now shows ¯

²βn e−C12 T

√

²δn

≤ h(T ) ≤ ²βn e−C12 T

√ ²δn

.

We now choose ∆ such that equality holds in (24); by the last two formulas, also (23) holds, yielding points 1) and 2). Inequality (21) is a consequence of our choice of ∆ and of the last formula. \\\ In the following, we shall fix once and for all M > 16 such that lemma 4 holds. As for C0 , we will choose it so big that both lemmas 2 and 4 hold. Moreover, by (21) we can suppose C0 so big that lC1 |kn0 | . ²δn

(25)

C0 ≥ 2lC12 C1

(26)

∆≥ We also require that

where C12 is the same that appears in the proof of the last lemma; this will come in handy in lemma 5. By the explicit formula for M² we see that, for C0 big enough, 0 | ¯ ¯ √ n 1 −b|k ¯ ¯ ²δn e ω , w0 , t0 )¯ ≤ ¯M² (ω, w0 , t0 ) − M² (¯ M

√ p if |ω − ω ¯ | ≤ C10 µ ²δn

(27)

and the same is true for M²+ and M²− . From now on, C0 is fixed and satisfying the above conditions. The next lemma states that the turns around the homoclinics (or ”whiskers”, as in [1]) are spaced each far enough from the other. It is rather similar to lemma 2 of [2] but the fact that we are close to a homoclinic of the pendulum requires some special care. The constants D and σ appearing below are the same as in the statement of theorem 1. We recall that at the beginning of the proof we have made a change of coordinates on Tm from Q to (w, q). Lemma 5.

Let D > σ. Then for all ² ∈ (0, 1) there is n big enough such that the following holds. Let ˙ lie on an invariant torus of frequency ω (w, q) be minimal for Ln²,µ , let Q be its pre-image and let (Q, Q) ¯. Let us suppose that t0 < t1 and that q(t0 ) = 0, q(t1 ) = π (or that q(t0 ) = π, q(t1 ) = 2π.) Then t1 − t0 ≥ ∆. 11

(28)

Proof.

In the following, given a vector v ∈ Rm , we will denote by v1 its component orthogonal to kn and

kn0 , and by v2 its component along kn0 . Let us consider t0 and t1 as in the hypotheses. We will prove the case q(t0 ) = 0, q(t1 ) = π, the other one being analogous. Since the motion on the invariant torus is conjugate to a rotation of frequency ω ¯ , we can find T¯ > t0 such that, setting Q = (P¯ Q, P Q), ¯µ w(T¯) − w(t ) q(T¯) − q(t ) ¶ ¯ √ p ¯ ¯ 0 0 , − ω ¯ ≤ C µ ²δn ¯ ¯ 10 T¯ − t0 |kn |(T¯ − t0 )

(29)

q(T¯) = N π.

(30)

Moreover, since by (∗ ∗ ∗) the motion in the direction q has mean speed smaller than |kn |e−D|kn | , we can suppose that

1 1 D|kn | T¯ − t0 ≥ N e . 2 |kn |

By (21) and the fact that D > σ we have that, for all n big enough, T¯ − t0 ≥ 16N ∆.

(31)

By (20) and (29) we get that, for C0 big enough, √ p |w˙ 2 (t) − ω ¯ 2 | ≤ 2C10 µ ²δn

∀t ∈ [T0 , T¯].

(32)

We now choose a sequence of times t0 < t1 < . . . < tN = T¯ such that q(tr ) = πr for r ∈ (0, 1, . . . , N ). Let us suppose by contradiction that (28) does not hold. We distinguish two cases: in the first one ∆ > t1 − t0 ≥ lC1

|kn0 | . ²δn

(33)

We will compare (w, q) with (w, ¯ q¯), an orbit which comes out of a surgery in the style of [2]. By (31) there is j ∈ (1, . . . , N − 1) such that tj+1 − tj ≥ 16∆.

(34)

In the new orbit (w, ¯ q¯), q¯ will take a longer time to go from 0 to π and a correspondingly shorter time to go from jπ to (j + 1)π. Thus by lemma 4 the action of An² will decrease; it will decrease enough to compensate the possible increase in the perturbation (actually, everything is choosen in such a way that the Melnikoff function remains the same.) On the other side, (w, ¯ q¯) will be the same as (w, q) (up to a translation) outside the intervals [t0 , t1 ] and [tj , tj+1 ]. We begin defining T =

t1 − t0 + tj+1 − tj . 2

Let s3 be the closest time to tj+1 − T such that w2 (s3 ) − w2 (tj ) ∈ and (2), |tj+1 − T − s3 | ≤

2π a .

2π 0 | Z; |kn

We define w ¯ ∈ A.C. in the following way w ¯1 (t) = w1 (t) 12

we set s2 = s3 − (tj − t1 ). By (32)

and

    w ¯2 (t) =

2 (t0 )+m w2 (t0 ) + w2 (t1 )−w (t − t0 ) s2 −t0 w2 (t − s2 + t1 ) + m w (t )−w2 (tj )−m w ¯2 (s3 ) + 2 j+1 (t − s3 ) tj+1 −s3 w2 (t)

  

where m is given by m = w2 (s3 ) − w2 (tj ) ∈

t 0 ≤ t ≤ s2 s2 ≤ t ≤ s3 s3 ≤ t ≤ tj+1 tj+1 ≤ t ≤ T¯ = tN

2π Z. |kn0 |

(35)

We will see that this choiche of m keeps low the kinetic energy of w ¯ and minimizes the change in the Melnikoff function. We note that w ¯2 (s2 ) − w2 (t1 ) = m ∈ We set

   q¯(t) =

2π Z |kn0 |

w ¯2 (s3 ) − w2 (tj ) = m ∈

q1 (t) q(t − s2 + t1 ) q2 (t) q(t)

 

2π Z. |kn0 |

(36)

t0 ≤ t ≤ s2 s2 ≤ t ≤ s3 s3 ≤ t ≤ tj+1 tj+1 ≤ t ≤ T¯ = tN

where q1 is a solution of An² on (t0 , s1 ) with boundary conditions q1 (t0 ) = q(t0 ) = 0

q1 (s2 ) = q(t0 ) = π

and q2 is a solution of An² on (s3 , tj ) with boundary conditions q1 (s3 ) = q(tj ) = jπ

q1 (tj+1 ) = q(tj+1 ) = (j + 1)π.

In order to deal with shorter formulas, we set 0

pert(w, q) = ²e−σ|kn | e−σ|kn | [1 − cos(q)] coshkn0 , wi. The minimality of (w, q) implies Z

T¯

t0

Z Ln²,µ (w, ¯ q¯)dt −

T¯

t0

Ln²,µ (w, q)dt ≥ 0.

After eliminating all the terms which are the same in the two integrals, we get that Z [t0 ,s2 ]∪[s3 ,tj+1 ]

Z [t0 ,t1 ]∪[tj ,tj+1 ]

We assert that

Z

1 |w ¯˙2 |2 dt + 2

1 |w˙2 |2 dt − 2

Z

[t0 ,s2 ]∪[s3 ,tj+1 ]

An² (¯ q , q¯˙ )dt

+µ

Z

Z

[t0 ,t1 ]∪[tj ,tj+1 ]

Z [t0 ,s2 ]∪[s3 ,tj+1 ]

pert(w ¯2 , q¯)dt− [t0 ,s2 ]∪[s3 ,tj+1 ]

An² (q, q)dt ˙ −µ

1 |w ¯˙2 |2 dt − 2

pert(w2 , q)dt ≥ 0. [t0 ,t1 ]∪[tj ,tj+1 ]

Z [t0 ,t1 ]∪[tj ,tj+1 ]

1 |w˙2 |2 dt ≤ 0. 2

By (34) we have that s3 > tj and, consequently, s2 > t1 . By (35) and (36), on [s3 , tj+1 ] w2 and w ¯2 run the same distance (indeed, w ¯2 (s3 ) = w2 (s3 ) and w ¯2 (tj+1 ) = w2 (tj+1 )) and thus on this interval the action 13

functional of w ¯2 is smaller than that of w2 . On the other side, since w ¯2 (s2 ) − w ¯2 (t0 ) = w2 (t1 ) − w2 (t0 ) + w2 (s3 ) − w2 (tj ), we have that the distance run by w ¯2 on [t0 , s2 ] is less than the distance run by w2 on [t0 , t1 ] ∪ [tj , s3 ]. Moreover, the measure of [t0 , s2 ] is the same as the measure of [t0 , t1 ] ∪ [tj , s3 ]. Using again the minimality of w ¯ for the action functional of kinetic energy, we get the above formula. From the above two formulas we get Z

Z

[t0 ,s2 ]∪[s3 ,tj+1 ]

An² (¯ q , q¯˙ )dt + µ

pert(w ¯2 , q¯)dt− [t0 ,s2 ]∪[s3 ,tj+1 ]

Z

Z [t0 ,t1 ]∪[tj ,tj+1 ]

˙ An² (q, q)dt

−µ

pert(w2 , q)dt ≥ 0.

(37)

[t0 ,t1 ]∪[tj ,tj+1 ]

To fix ideas, let us suppose that q(tj ) is an odd multiple of π and q(tj+1 ) is an even multiple. By (33) we can apply lemma 2 and (19); by (32) we can apply (27); we thus get Z

0

[t0 ,t1 ]∪[tj ,tj+1 ]

pert(w2 , q)dt ≥ M²− (¯ ω , w2 (t1 ), t¯1 ) + M²+ (¯ ω , w2 (tj ), tj ) −

| √ n 4 −b|k e ²δn . M

Analogously, but applying lemma 2 with µ = 0, we get Z

0

pert(w ¯2 , q¯)dt ≤ [t0 ,s2 ]∪[s3 ,tj+1 ]

M²− (¯ ω, w ¯2 (s2 ), s2 )

+

M²+ (¯ ω, w ¯2 (s3 ), s3 )

| √ n 4 −b|k e ²δn . + M

From (36) and the periodicity of M²+ and M²− we get that M²− (¯ ω , w2 (t1 ), t¯1 ) + M²+ (¯ ω , w2 (tj ), tj ) = M²− (¯ ω, w ¯2 (s2 ), s2 ) + M²+ (¯ ω, w ¯2 (s3 ), s3 ). From the last four formulas we get Z

Z [t0 ,s2 ]∪[s3 ,tj+1 ]

q , q¯˙ )dt An² (¯

0

− [t0 ,t1 ]∪[tj ,tj+1 ]

˙ An² (q, q)dt

−b|kn | √ 8 ≥ − µe ²δn . M

(38)

Let now q˜ be a solution of An² on (t0 , t1 ), (tj , tj+1 ) with the same boundary conditions as q in t0 , t1 , tj , tj+1 . Clearly Z

Z [t0 ,s2 ]∪[s3 ,tj+1 ]

An² (¯ q , q¯˙ )dt −

Z [t0 ,t1 ]∪[tj ,tj+1 ]

˙ ≤ An² (q, q)dt

Z

[t0 ,s2 ]∪[s3 ,tj+1 ]

q , q¯˙ )dt − An² (¯

[t0 ,t1 ]∪[tj ,tj+1 ]

An² (˜ q , q˜˙ )dt.

We now note that we are in the hypotheses of point 1) of lemma 4: indeed, t1 − t0 ≤ ∆, tj+1 − tj ≥ 16∆ 2π a

while |s2 − t0 − T | ≤

and |tj+1 − s3 − T | ≤

2π a .

Moreover, the lenghth of [t0 , s2 ] ∪ [s3 , tj+1 ] is the same as

the lenghth of [t0 , t1 ] ∪ [tj , tj+1 ]. Thus we can apply point 1) of lemma 4 and get Z [t0 ,s2 ]∪[s3 ,tj+1 ]

Z An² (¯ q , q¯˙ )dt −

0

[t0 ,t1 ]∪[tj ,tj+1 ]

An² (q, q)dt ˙ −8µ e ²δn M M

14

| √ n 8µ −b|k · e ²δn . M

(39)

a contradiction. It now remains the other case, t1 − t0 < l

C1 |kn0 | . ²δn

(40)

We build (w, ¯ q¯) with exactly the same surgery as before; the previous deductions continue to hold up to formula (37). From (37) and the definition of pert we get that Z

Z

(1 + µ) [t0 ,s2 ]∪[s3 ,tj+1 ]

An² (¯ q , q¯˙ )dt − (1 − µ)

[t0 ,t1 ]∪[tj ,tj+1 ]

An² (q, q)dt ˙ ≥ 0.

With an argument like that of lemma 4 we get that Z

Z [t0 ,s2 ]∪[s3 ,tj+1 ]

An² (¯ q , q¯˙ )dt

− [t0 ,t1 ]∪[tj ,tj+1 ]

An² (q, q)dt ˙

0 | lC |kn ²δn

−2C12 √1

≤ −e

.

Moreover, since [t0 , s2 ] is very big, we have that Z

s2

t0

An² (¯ q , q¯˙ )dt ≤ C2

p

²δn .

By (26), for n big enough the above three formulas are in contradiction. \\\ End of the proof of theorem 1. As we remarked at the beginning of the proof, it suffices to prove (3). Given ² ∈ (0, 1) we fix n such that lemma 5 holds. Let us suppose by contradiction that (3) does not hold, i. e. that there is t¯0 such that 2kπ w2 (t¯0 ) = 0 |kn |

q(t¯0 ) = π + 2lπ

l, k ∈ Z

where w2 denotes, as in lemma 5, the projection of w on kn0 . Simply by changing the lift of (w, q), we an suppose l, k = 0 so that q(t¯0 ) = π

w2 (t¯0 ) = 0.

(41)

We are going to show that (41) contradicts the minimality of (w, q). Indeed, let us consider t0 < t¯0 and t1 > t¯0 such that q(t0 ) = 0

q(t1 ) = 2π.

By lemma 5 we have that t1 − t0 ≥ 2∆.

(42)

We now choose t¯ as the last time to the left of t¯0 or as the first time to the right of t¯0 such that π w2 (t¯) = 0 . |kn | By (32) and (2) we have that |t¯ − t0 | ≤ 15

2π . a

(43)

Thus there are two ways to choose t¯: to the left or to the right of t¯0 . We will explain in the following which is the suitable one. On [t0 , t1 ] we build an orbit in the following way: q¯ is a solution of An² on (t0 , t¯) and on (t¯, t1 ) with boundary conditions q¯(t0 ) = q(t0 ) = 0

q¯(t¯) = q(t¯0 ) = π

q¯(t1 ) = q(t1 ) = 2π.

By the minimality of (w, q) we have that Z

t1

t0

Z Ln²,µ (w, q¯)dt −

t1

t0

Ln²,µ (w, q)dt ≥ 0.

Eliminating all the terms which are the same in the two integrals we get Z

t1

t0

Z An² (¯ q , q¯˙ )dt −

t1 t0

Z An² (q, q)dt ˙ +µ

Z

t1

t1

pert(w, q¯)dt − µ t0

pert(w, q)dt ≥ 0.

(44)

t0

By lemma 2 and the fact that M > 16 we get that Z

Z

t1

µ

pert(w, q¯)dt − µ t0

where ω =

t1

t0

w(t1 )−w(t0 ) . t1 −t0

0

| √ n 2 −b|k π pert(w, q)dt ≤ µ[M² (ω, 0 , t¯) − M² (ω, 0, t¯0 ) + e ²δn ] |kn | 16

√ √ By lemma 3 we have that |ω − ω ¯ | ≤ C10 µ ²δn ; if we recall (27) and lemma 1 we

get

Z

Z

t1

µ

t1

pert(w, q¯)dt − µ t0

t0

0

| √ n 3 −b|k pert(w, q)dt ≤ − µe ²δn . 4

(45)

On the other side, if q˜ is a solution of An² on (t0 , t¯0 ) and on (t¯0 , t1 ) with same boundary conditions as q at t0 , t¯0 , t1 we have that Z

t1

t0

Z An² (¯ q , q¯˙ )dt

t1

− t0

Z An² (q, q)dt ˙

Here is important which t¯ we choose. If |t¯0 −

t1 +t0 2 |

t1

≤ t0

≤

2π a ,

Z An² (¯ q , q¯˙ )dt

t1

− t0

An² (˜ q , q˜˙ )dt.

then we choose the t¯ such that |t¯ −

t1 +t0 2 |

≤

2π a ;

by (42) and (43) part 2) of lemma 4 applies, yielding Z

t1 t0

If |t¯0 −

t1 +t0 2 |

>

2π a ,

Z An² (¯ q , q¯˙ )dt −

t1 t0

we will choose the t¯ closer to

0

An² (q, q)dt ˙ ≤ t1 +t0 2 ,

| √ n µ −b|k e ²δn . 8

thus making the motion of the pendulum more

uniform; clearly, this implies that the above inequality holds with 0 on the right side. From the last formula, (45) and (44) we get that 0

−

| √ n 5µ −b|k e ²δn ≥ 0 8

a contradiction. \\\

References 16

[1] V. I. Arnold, Instability of Dynamical Systems with Several Degrees of Freedom, Soviet Mathematics, vol. 5-1, 581-585, 1964. [2] D. Bernstein, A. Katok, Birkhoff Periodic Orbits for Small Perturbations of Completely Integrable Hamiltonian Systems with Convex Hamiltonians, Invent. Math., vol. 88, 225- 241, 1987. [3] U. Bessi, An approach to Arnold’s Diffusion through the Calculus of Variations, Nonlinear Analysis, T. M. A., vol. 26, 1115-1135, 1996. [4] U. Bessi, Arnold’s Example with Three Rotators, Nonlinearity, vol. 10, 763-781, 1997. [5] G. Forni, Analytic Destruction of Invariant Circles, Ergod. Th. and Dynam. Sys., vol. 14, 267-298, 1994. [6] A. Giorgilli, U. Locatelli, On Classical Series Expansion for Quasi-Periodic Motions, MPEJ, vol. 7, 1997. [7] R. S. MacKay, A Criterion fo Non-Existence of Invariant Tori for Hamiltonia Systems, Physica D, vol. 36, 64-82, 1989. [8] S. Marmi, J. Stark, On the Standard Map Critical Function, Nonlinearity, vol. 5, 743-761, 1992. [9] J. N. Mather, Destruction of Invariant Circles, Ergod. Th. and Dynam. Sys., vol. 8, 199-214, 1988. [10] J. N. Mather, Action Minimizing Invariant Measures for Positive Definite Lagrangian Systems, Math. Z., vol. 207, 169-207, 1991. [11] J. N. Mather, Variational Construction of Connecting Orbits, Annales de l’Institut Fourier, vol. 43, 1349-1368, 1993. [12] J. N. Mather, Variational Construction of Orbits of Twist Diffeomorphisms, J. A. M. S., vol. 4, 207-263, 1991. [13] R. P. A. C. Newman, I. C. Percival, Definite Paths and Upper Bounds on Regular Regions of Velocity Phase Space, Physica D, vol 6, 249-259, 1983. [14] D. Salamon, E. Zehnder, KAM Theory in Configuration Space, Comment. Math. Helvetici, vol. 64, 84-132, 1989. Ugo Bessi, Scuola Normale Superiore, Piazza Cavalieri 7, 56126 Pisa

17