Variational destruction of invariant circles

We construct a sequence of generating functions $(h_n)_{n\in\N}$, arbitrarily close to an integrable system in the $C^r$ topology with $r<4$ for $n$ large enough. With the variational method, we prove that for a given rotation number $\omega$ and $n$ large enough, the exact monotone area-preserving twist maps generated by $(h_n)_{n\in\N}$ admit no invariant circles with rotation number $\omega$.


Introduction
M.Herman constructed an example in [H2]. With geometrical method he proved the example has the property that the invariant circle (i.e. a homotopically nontrivial invariant curve) with a given rotation number can be destructed by an arbitrarily small perturbation in the C 3−ǫ topology for the exact monotone areapreserving twist map. The KAM theory (see [S]) implies the persistence of invariant circle with a Diophantine rotation number under the small perturbation in the C 3+ǫ topology. Hence, Herman's result is optimal. However, the perturbation in the example in [H2] is too artificial. In this paper, following the ideas and techniques developed by J.N.Mather in the series of papers [M1], [M2], [M3] and [M4], we construct an example with a more natural perturbation to achieve the same goal. More precisely, our example has the property as follow: Property: For a given rotation number ω and n large enough , there exists a sequence of generating functions (h n ) n∈N , arbitrarily close to an integrable system in the C r topology with r < 4 such that the exact monotone area-preserving twist maps generated by (h n ) n∈N admit no invariant circles with rotation number ω.
In [M1], [M2], [M3] and [M4], Mather introduced a notion called Peierls's barrier as follows P h ω (ξ) = min where I = Z, if ω is not a rational number, I = {0, ..., q − 1}, if ω = p/q, and (x i ) i∈I ∈ i∈I [x − i , x + i ] satisfying x 0 = ξ. Moreover, he proved that P h ω (ξ) is a nonnegative Lipschitz function with respect to the variable ξ ∈ R with the modulus of continuity with respect to ω and found a criterion of the existence of invariant circles. Namely, the exact area-preserving monotone twist map generated by h admits an invariant circle if and only if P h ω (ξ) ≡ 0 for every ξ ∈ R. For our example, the modulus of continuity can be improved due to the hyperbolicity of the perturbation, which follows from similar ideas of [F]. More precisely, (Lemma 5.1 below) if ω is suitable small, then |P hn ω (ξ) − P hn 0 + (ξ)| ≤ C exp(−n δ ), where δ is a small positive constant independent of n. Based on the improvement, we obtain that there existsξ ∈ R such that P hn ω (ξ) = 0 for n large enough. It follows that h n doesn't admit any invariant circles for n large enough.

Minimal configuration
Let f : T × R → T × R (T = R/Z) be an exact area-preserving monotone twist map and h: R 2 → R 2 be a generating function for the lift F of f to R 2 , namely F is generated by the following equations where F (x, y) = (x ′ , y ′ ). The lift F gives rise to a dynamical system whose orbits are given by the images of points of R 2 under the successive iterates of F . The orbit of the point (x 0 , y 0 ) is the bi-infinite sequence denoted by (x i ) i∈Z is called stationary configuration which stratifies the identity Given a sequence of points (z i , ..., z j ), we can associate its action A configuration (x i ) i∈Z is called minimal if for any i < j ∈ Z, the segment of (x i , ..., x j ) minimizes h(z i , ..., z j ) among all segments (z i , ..., z j ) of the configuration satisfying z i = x i and z j = x j . It is easy to see that every minimal configuration is a stationary configuration. By [B], minimal configurations satisfy a group of remarkable properties as follows: • Two distinct minimal configurations cross at most once, which is so called Aubry's crossing lemma.
• For every minimal configuration x = (x i ) i∈Z , the limit exists and doesn't depend on i ∈ Z. ρ(x) is called the rotation number of x.
• For every ω ∈ R, there exists a minimal configuration with rotation number ω. Following the notations of [B], the set of all minimal configurations with rotation number ω is denoted by M h ω , which can be endowed with the topology induced from the product topology on R Z . If x = (x i ) i∈Z is a minimal configuration, considering the projection pr : M h ω → R defined by pr(x) = x 0 , we set A h ω = pr(M h ω ).
• If ω ∈ Q, say ω = p/q (in lowest terms), then it is convenient to define the rotation symbol to detect the structure of M h p/q . If x is a minimal configuration with rotation number p/q, then the rotation symbol σ(x) of x is defined as follows Moreover, we set M h p/q + = {x a is minimal configuration with rotation symbol p/q or p/q+}, M h p/q − = {x a is minimal configuration with rotation symbol p/q or p/q−}, then both M h p/q + and M h p/q + are totally ordered. Namely, every two configurations in each of them do not cross. We denote pr(M h p/q + ) and pr(M h p/q − ) by A h p/q + and A h p/q − respectively. • If ω ∈ R\Q and x is a minimal configuration with rotation number ω, then σ(x) = ω and M h ω is totally ordered. • A h ω is a closed subset of R for every rotation symbol ω.

Peierls's barrier
In [M3], Mather introduced the notion of Peierls's barrier and gave a criterion of existence of invariant circle. Namely, the exact area-preserving monotone twist map generated by h admits an invariant circle with rotation number ω if and only if the Peierls's barrier P h ω (ξ) vanishes identically for all ξ ∈ R. The Peierls's barrier is defined as follows: where I = Z, if ω is not a rational number, and By [M3], P h ω (ξ) is a non-negative periodic function of the variable ξ ∈ R with the modulus of continuity with respect to ω.
To our example (see Section 3), the modulus of continuity of P h ω (ξ) with respect to ω can be improved significantly. Loosely speaking, the hyperbolicity of the perturbation implies the exponential approximation from P h ω (ξ) to P h 0 + (ξ). The details will be provided in Section 5.

Construction of the generating functions
Consider a completely integrable system with the generating function We construct the perturbation consisting of two parts. The first one is where n ∈ N and a is a positive constant independent of n. The second one is a non negative function v n (x) satisfying where we require s ′ > a. It is enough to take s = (k + 2)a for achieving that. The generating function of the nearly integrable system is constructed as follow: where n ∈ N. Moreover, we have the following theorem.
Theorem 3.1 For ω ∈ R\Q and n large enough, the exact area-preserving monotone twist map generated by h n does not admit any invariant circles with the rotation number satisfying where δ is a small positive constant independent of n.
We will prove Theorem 3.1 in the following sections. First of all, based on the theorem, we verify that our example has the property aforementioned in Section 1.
If ω ∈ Q, then the invariant circles with rotation number ω could be easily destructed even though the perturbation is C ∞ close to 0. Therefore it suffices to consider the irrational ω. The case with a given irrational rotation number can be easily reduced to the one with a small enough rotation number. More precisely, Lemma 3.2 Let h P be a generating function as follow where P is a periodic function of periodic 1. Let Q(x) = q −2 P (qx), q ∈ N, then the exact area-preserving monotone twist map generated by h Q (x, admits an invariant circle with rotation number ω ∈ R\Q if and only if the exact area-preserving monotone twist map generated by h P admits an invariant circle with rotation number qω − p, p ∈ Z.
We omit the proof and for more details, see [H2]. For the sake of simplicity of notations, we denote Q qn by Q n and the same to u qn , v qn and h qn . Let where (q n ) n∈N is a sequence satisfying Dirichlet approximation where p n ∈ Z and q n ∈ N. Since ω ∈ R\Q, we say q n → ∞ as n → ∞. Let 3 For a given rotation number ω ∈ R\Q and every ε, there exists N such that for n > N , the exact area-preserving monotone map generated byh n admits no invariant circle with rotation number ω and where δ ′ is a small positive constant independent of n.
Proof Based on Theorem 3.1 and Dirichlet approximation (3.4), it suffices to take 1 q n ≤ 1 q n a 2 +δ , which implies From (3.1), (3.2) and (3.3), it follows that where C 1 , C 2 are positive constants only depending on r.
To complete the proof, it is enough to make r − a − 2 < 0, which together with (3.5) implies r < a + 2 ≤ 4 − 2δ.
We set δ ′ = 2δ, then the proof of Corollary 3.3 is completed.
The following sections are devoted to prove Theorem 3.1. For simplicity, we don't distinguish the constant C in following different estimate formulas.
4. Estimate of lower bound of P hn 0 + In this section, we will estimate the lower bound of P hn 0 + at the given point. To achieve that, we need to estimate the distances of pairwise adjacent elements of the minimal configuration.

A spacing lemma
Lemma 4.1 Let (x i ) i∈Z be a minimal configuration ofh n with rotation symbol 0 + , then Proof Without loss of generality, we assume x i ∈ [0, 1] for all i ∈ Z. By Aubry's crossing lemma, we have We consider the configuration (ξ i ) i∈Z defined by By the definitions ofh n and (ξ i ) i∈Z , we have Therefore, On the other hand, we consider another configuration (η i ) i∈Z defined by Based on the minimality of (x i ) i∈Z , following the deduction as similar as that above, we have Since (x i ) i∈Z is a stationary configuration, we have Since u ′ n (x) = 2π n a sin(2πx), it follows from (4.1) that Therefore, we have The proof of Lemma 4.1 is completed.

5.
The approximation from P hn 0 + to P hn ω In this section, we will prove the improvement of modulus of continuity of Peierls's barrier based on the hyperbolicity of h n . Namely Lemma 5.1 For every irrational rotation symbol ω satisfying 0 < ω < n − a 2 −δ , we have |P hn ω (ξ) − P hn 0 + (ξ)| ≤ C exp(−n δ ). where ξ ∈ 1 2 − 1 n a , 1 2 + 1 n a and δ is a small positive constant independent of n.

Some counting lemmas
To prove the lemma, we need to do some preliminary work. First of all, we count the number of the elements of a minimal configuration (x i ) i∈Z with arbitrary rotation symbol ω in a given interval. With the method of [F], we can conclude the following lemma.
Lemma 5.2 Let (x i ) i∈Z be a minimal configuration of h n with rotation symbol ω, J n = exp −n δ 2 , 1 2 and Λ n = {i ∈ Z| x i ∈ J n }, then where ♯Λ n denotes the number of elements in Λ n and δ is a small positive constant independent of n.
n } and m k = ♯S k . By the similar deduction as the one in Lemma 4.1, we have For simplicity of notation, we write Cn − a 2 by α n . If there exists k such that i ∈ S k for (x i ) i∈Z , then x i+1 − x i−1 ≥ α n σ k x − , moreover, m k α n σ k x − ≤ 2L(J k n ) = 2(σ − 1)σ k x − , where L(J k n ) denotes the length of the interval of J k n . Hence m k ≤ 2(σ − 1)α −1 n . On the other hand, if i ∈ S k for any k, then there exists l satisfying 1 ≤ l < N such that Hence, ♯{i ∈ Λ n |i ∈ S k for any k} ≤ 2N. Therefore, Since 1 ≤ ln σ ≤ 2 and N = O n δ 2 , then we have The proof of Lemma 5.2 is completed.

Remark 5.3
Let (x i ) i∈Z be a minimal configuration of h n defined by (3.3) with rotation symbol ω, An argument as similar as the one in Lemma 5.2 implies that Second, it is easy to count the number of the elements of a minimal configuration with irrational rotation symbol. More precisely, we have the following lemma.
Lemma 5.4 Let (x i ) i∈Z be a minimal configuration with rotation number ω ∈ R\Q. Then for every interval I k of length k, k ∈ N, Proof For every minimal configuration (x i ) i∈Z with rotation number ω, there exists an orientation-preserving circle homeomorphism φ such that ρ(Φ) = ω, where Φ : R → R denotes the lift of φ. Since ω ∈ R\Q, thanks to [H1], φ has a unique invariant probability measure µ on T such that µ[x, Φ(x)] = ω for every x ∈ R. In particular, which completes the proof of Lemma 5.4.
Based on Lemma 5.2 and Lemma 5.4, if 0 < ω < n − a 2 −δ and ω is irrational, So far, we have proved the following conclusion.
Lemma 5.5 Let (x i ) i∈Z be a minimal configuration of h n defined by (3.3) with rotation symbol 0 < ω < n − a 2 −δ , then there exists j − , j + ∈ Z such that Without loss of generality, we assume that If ξ ∈ A hn ω , then P hn ω (ξ) = 0. Hence, it suffices to consider the case with ξ ∈ A hn ω for destruction of invariant circles. Let (ξ − , ξ + ) be the complementary interval of A hn ω in R and contains ξ. Let ξ ± ξ ± ξ ± = (ξ ± i ) i∈Z be the minimal configurations with rotation symbol ω satisfying ξ ± 0 = ξ ± and let (ξ i ) i∈Z be a minimal configuration of h n with rotation symbol ω satisfying ξ 0 = ξ and ξ − i ≤ ξ i ≤ ξ + i . By the definition of Peierls barrier, we have Since P hn ω (ξ) is 1-periodic with respect to ξ, without loss of generality, we assume that ξ ∈ [0, 1]. We set d(x) = min{|x|, |x − 1|} and write exp(−n δ 2 ) by ǫ(n). By Lemma 5.5, there exist i − , i + such that Thanks to Aubry's crossing lemma, we have

Proof of lemma 5.1
In the following, we will prove Lemma 5.1 with the method similar to the one developed by Mather in [M3]. The proof can be proceeded in the following two steps.

Step 1
We consider the number of the elements in a segment of the configuration as the length of the segment. In the first step, we approximate P hn ω (ξ) for ξ ∈ 1 2 − 1 n a , 1 2 + 1 n a by the difference of the actions of the segments of length i + −i − +1. To achieve that, we define the following configurations It is easy to see that ξ 0 = ξ is contained both of (x i ) i∈Z and (y i ) i∈Z up to the rearrangement of the index i since ξ ∈ 1 2 − 1 n a , 1 2 + 1 n a . Hence, by the minimality of (ξ i ) i∈Z satisfying ξ 0 = ξ, we have Since ω is irrational, then (x i ) i∈Z is asymptotic to (ξ − i ) i∈Z , which together with the minimality of (ξ − i ) i∈Z yields We set By the construction of v n and Lemma 5.5, we have v n (ξ i − ), v n (ξ − i − ) = 0. It follows that It is similar to obtain Hence, Therefore, it follows from (5.3) and (5.4) that
First of all, it is easy to see that K(ξ) and K are bounded. Second, P hn 0 + (ξ) = 0 for ξ = 0 or 1. Hence, we only need to consider the case with ξ ∈ (0, 1). Following the ideas of [M6], let ξ − ξ − ξ − and ξ + ξ + ξ + be minimal configurations of rotation symbol 0 + and let (ξ − 0 , ξ + 0 ) be the complementary interval of A hn 0 + and contains ξ. Based on the definition the proof of (5.7) will be completed when we verify that the configuration (x i ) i∈Z achieving the minimum in the definition of K(ξ) satisfies ξ − i ≤ x i ≤ ξ + i . It can be easily obtained by Aubry's crossing lemma. In fact, since (ξ − i ) i∈Z and (x i ) i∈Z are minimal and both are α-asymptotic to 0 as well as ω-asymptotic to 1, by Aubry's crossing lemma, (ξ − i ) i∈Z and (x i ) i∈Z do not cross. Similarly (ξ + i ) i∈Z and (x i ) i∈Z do not cross. It follows from First, we consider K and h n (ξ − i − , ..., ξ − i + ). Let (z i ) i∈Z be a monotone increasing configuration limiting on 0, 1 such that K = i∈Z h n (z i , z i+1 ). By Lemma 5.2, On the other hand, since (z i ) i∈Z has the rotation number 0 + , then from (5.1), it follows that up to the rearrangement of the index i, there exists a subset of length i + − i − of (z i ) i∈Z , denoted by {z i − , z i − +1 , . . . , z i + −1 , z i + } such that We consider the configuration (x i ) i∈Z defined by Moreover, by the minimality of (z i ) i∈Z , we have By the construction of h n , we have v n ( It is similar to obtain From (5.8) and (5.11) we have To obtain the reverse inequality of (5.12), we consider the configuration as follows From the definition of h n , it follows that v n (z i − +1 ) = 0 and h n (z i , z i+1 ) ≥ 0 for all i ∈ Z. Moreover, we have where η ∈ (0, z i − +1 ). Namely Furthermore, we consider the finite segment of the configuration defined by Then, the minimality of (ξ .., η i + ). Hence, by (5.13), we have By the deduction as similar as (5.9), we have Moreover, Hence, from (5.14) and (5.15), it follows that (5.16) h n (ξ − i − , ..., ξ − i + ) ≤ K + Cǫ(n) 2 , which together with (5.12) and (5.16) implies (5.17) |h n (ξ − i − , ..., ξ − i + ) − K| ≤ Cǫ(n) 2 .
It is an obvious contradiction for n large enough. Therefore, there exists no invariant circle with rotation number 0 < ω < n − a 2 −δ . For −n − a 2 −δ < ω < 0, by comparing P hn ω (ξ) with P hn 0 − (ξ), the proof is similar. We omit the details. Therefore, the proof of Theorem 3.1 is completed.