Universal asymptotics in hyperbolicity breakdown

Transcription

1 IOP PUBLISHING Nonlinearity 2 (2008) NONLINEARITY doi:0.088/ /2/3/00 Universal asymptotics in hyperbolicity breakdown Kristian Bjerklöv and Maria Saprykina Department of Mathematics and Statistics, Queen s University, Kingston, ON, Canada bjerklov@mast.queensu.ca and masha@mast.queensu.ca Received 5 August 2007, in final form 8 January 2008 Published 8 February 2008 Online at stacks.iop.org/non/2/557 Recommended by R de la Llave Abstract We study a scenario for the disappearance of hyperbolicity of invariant tori in a class of quasi-periodic systems. In this scenario, the system loses hyperbolicity because two invariant directions come close to each other, losing their regularity. In a recent paper, based on numerical results, Haro and de la Llave (2006 Chaos ) discovered a quantitative universality in this scenario, namely, that the minimal angle between the two invariant directions has a power law dependence on the parameters and the exponents of the power law are universal. We present an analytic proof of this result. Mathematics Subject Classification: 37C55, 37C60, 37D20, 37D25. Introduction In a recent publication [4], Haro and de la Llave have found spectacular asymptotics when numerically investigating the disappearance of normally hyperbolic invariant tori in quasiperiodically forced systems (see also [5] for a more detailed exposition). The purpose of this paper is to provide analytic proofs of the existence of these asymptotics in a class of systems. We will focus on one concrete model, but our method should be applicable to other systems also. The model we consider in this paper is a quasi-periodically forced Hénon map H : T R 2 T R 2 (T = R/Z) of the form θ θ + ω, (x, y) h(x, y) + ε(x x 0 )V (θ), where h(x, y) = ( + y ax 2, bx) is the Hénon map. Here V is a function V : T R which is at least continuous and ω is an irrational number. The map h(x, y) has a fixed point (x 0, bx 0 ) where x 0 = (b + (b ) 2 + 4a)/(2a). Note that the perturbation in our case is chosen so that the torus (θ, x 0, bx 0 ) is H -invariant for all ε /08/ $ IOP Publishing Ltd and London Mathematical Society Printed in the UK 557

2 558 K Bjerklöv and M Saprykina We shall study the linearization of the above map along the invariant torus (θ, x 0, bx 0 ). This dynamics is given by the cocycle where M ε is the matrix T R 2 (θ, u) (θ + ω, M ε (θ)u) T R 2, (.) M ε (θ) = ( ) 2ax0 + εv (θ). (.2) b 0 The time evolution of this linearized map shall be denoted by M ε (θ + (n )ω) M ε (θ), n > 0, Mε n (θ) = Id, n = 0, M ε (θ + nω) M(θ ω), n < 0. We say that the cocycle (ω, M ε ) is uniformly hyperbolic if there are constants C > 0 and δ < 0 < δ + and a splitting R 2 = Eε (θ) E+ ε (θ) such that Mε n (θ)u Ceδ n u for all u Eε (θ), n > 0, Mε n (θ)u Ce δ+n u for all u E ε + (θ), n > 0. (.3) It is well known that such a splitting, if it exists, is invariant, M ε (θ)e ε ± (θ) = E± ε (θ + ω), and that the subspaces E ε ± (θ), as functions of θ, are as smooth as M (see [7,8]). That the subspaces vary smoothly with θ implies that the angles between them are uniformly bounded away from zero. We define (ε) = min (E ε + (θ), θ T E ε (θ)). (.4) This quantity is the main object of interest in this paper. It is related to the constant C in (.3). Roughly speaking, if the angle between E ε ± (θ) is very small for some θ, then, by continuity, the two vectors Mε n(θ)u± (u ± E ε ± (θ), u± = ) will be close for a long time and hence the constant C must be large. Furthermore, we define the Lyapunov exponents as + (ε) = lim n n (ε) = lim n n T log Mε n (θ) dθ, T log Mε n (θ) dθ. It is a general fact that, for systems of the form (.2), + + = log b = log det(m ε (θ)). In the situation when the cocycle (ω, M ε ) is uniformly hyperbolic, it is well known that lim n ± n log Mn ε (θ)v = + for all θ T, v E ε + (θ) \ {0}, lim n ± n log Mn ε (θ)v = for all θ T, v Eε (θ) \ {0}. Moreover, the convergence is uniform in θ [3]. This implies that ± are the optimal values for δ ± in (.3). We also recall that we can have (ε) < 0 < + (ε) but the cocycle (.) fails to be uniformly hyperbolic. In this case it follows from the Oseledets theorem that the subspaces E ± (θ) exist for a.e. θ and that they vary measurably. They cannot be continuous, since it is a classical fact that non-zero Lyapunov exponents and continuous subspaces E ± (θ) imply uniform hyperbolicity. In [4] it was studied numerically how the uniform hyperbolicity breaks down as ε is increased (it is assumed that M 0 is hyperbolic). The scenario studied there was that (ε) 0,

3 Universal asymptotics in hyperbolicity breakdown 559 Figure. The curves γ ± (θ) when ω = ( 5 )/4, b =, 2ax 0 = 0, ε = and V as in (.5), with λ = 5. but the Lyapunov exponents ± (ε) are uniformly separated from each other. The striking asymptotics observed in [4] are the following. Denoting by ε c > 0 the smallest ε > 0 for which the cocycle (ω, M ε ) fails to be uniformly hyperbolic (note that uniform hyperbolicity is an open condition), it was found that there are constants α, β such that (ε) α(ε c ε) β for all ε < ε c close to ε c. Moreover, + (ε) + (ε c ) + A(ε c ε) B, for some A, B. In both models they had + (ε c ) > 0, and it was found that β =. We will establish the same asymptotic rate in our model (see the main theorem) In figure we have plotted the graphs of the projections of the subspaces E ε ± (θ) in the case when ω = ( 5 )/4, b =, 2ax 0 = 0, ε = and V is as in (.5), with λ = 5. In our model, for any fixed θ, ε, each E ε ± (θ) is a line in R2, passing through the origin. In the figure it is represented by the corresponding angle in ( π/2, π/2]. For a fixed ε < ε c we get two smooth curves γ ε ± corresponding to E ε + and E ε, respectively. When ε approaches ε c from below, the smallest distance between the two curves goes to zero. Still, if the Lyapunov exponent + (ε c ) is positive, then the curves γ ± (θ) for ε must be well separated for most θ, each curve supporting its Lyapunov exponent. This forces the curves to fractalize as ε ε c. The presence of this fractalization process is one of the difficulties in estimating the asymptotics of (ε). Results. In order to make the presentation of our proof as transparent as possible, we have chosen b = and a such that 2ax 0 = 0. The only thing we actually need is that the unperturbed matrix M 0 is of saddle type. What is important, though, is the shape of the forcing

4 560 K Bjerklöv and M Saprykina function V. It has to be flat with a single sharp spike. We have taken it to be V (θ) = + λ sin 2 (πθ). (.5) Here the constant λ should be extremely large. For our method it is enough that V is C 2 ; it is the spike shape that is important. We also need that the frequency ω satisfies the Diophantine condition κ (DC) κ,τ inf qω p > for all q Z \ {0} (.6) p Z q τ for some constants κ > 0, τ. Main theorem. Let M ε (θ) = ( ) εv (θ) 0, where V (θ) = 0 + λ sin 2 (πθ), and assume that ω T satisfies the Diophantine condition (DC) κ,τ for some κ > 0, τ. For all λ > 0 sufficiently large (depending on κ and τ) there is an ε c > 0 (close to 0) such that the cocycle (ω, M ε ) is uniformly hyperbolic for all 0 ε < ε c and the minimal angle (ε), defined in (.4), satisfies lim ε ε c (ε) ε c ε = α for some constant α > 0. ε [0, ε c ]. Remark. Moreover, the Lyapunov exponent + (ε c ) > 0.5 log 5 for all (i) Since b =, we get that (ε) = + (ε). (ii) Note that the asymptotics does not depend on the Diophantine class; we always get β =. (iii) One can use Herman s subharmonic trick [6] to show that for all sufficiently large λ > 0, the following holds: + (ε) > log 0 for all ε and any irrational ω. See the appendix for the details. (iv) The methods of proof of the main theorem permit us to obtain the same result in the multi-frequency case, i.e. in the case when V (θ) is a flat function with a single sharp spike, defined on T d, and ω T d is a Diophantine vector, see []. (v) In this paper we were unable to estimate the asymptotics of the Lyapunov exponent when ε ε c. We believe that + (ε) + (ε c ) + A(ε c ε) /2. The proof of the main theorem is based on a technique developed in []. The general strategy follows the same lines, but the details differ almost everywhere. Therefore we have chosen to present all the details in this paper, without referring to analogous parts in []. The rest of this paper is organized as follows. In section 2 we describe the projective coordinates and the projectivization of the cocycle which we shall work with, and in section 3 we introduce some notation and important definitions. Section 4 includes basic estimates. In section 5 we give a brief explanation of the key ideas behind the proof of the main theorem. To control the geometry of the projective curves γ ε ±, which we shall construct, we will need certain formulae. These formulae are derived in section 6. Section 7 contains abstract help lemmas which are included in order to keep the proof of the main inductive lemma (section 8) to a reasonable size. In section 8 we prove the inductive lemma, which is the heart of the proof of the main theorem. Finally, in the last section, section 9, we put everything together and

5 Universal asymptotics in hyperbolicity breakdown 56 derive the statements of the main theorem. In the appendix we show how to apply Herman s subharmonic trick to our model. We close this section with a discussion of the Schrödinger cocycle: (θ, u) (θ + ω, S(θ)u), S(θ) = ( εv (θ) E 0 This cocycle has been widely studied in the literature (see, e.g., [] and references therein). The same result as in the main theorem also holds in this case when V is as above and E = 0. We believe that the asymptotic of (ε) depends on whether + (ε c ) is positive or not. We have performed computer simulations in the case when E = 2. and V (θ) = cos(2πθ). In this case it is well known that if ε [0, 2] and the cocycle is not uniformly hyperbolic, then + (ε) = 0 (see, e.g., [2]). The numerical results we got are that the cocycle is uniformly hyperbolic for 0 ε < ε c (ε c is close to ), + (ε c ) = 0 and (ε) α(ε c ε) /2. Thus, in this case β = /2. The same asymptotic is also found in the following (highly degenerate) example. Let E = 3 and V (θ) =. Then an easy computation shows that the cocycle is uniformly hyperbolic for 0 ε <, + () = 0 and (ε) = arctan(ψ + ) arctan(ψ ) ε, where ψ ± = (ε 3 ± (ε )(ε 5))/2. ). 2. Projective dynamics The way we are going to investigate cocycle (.) is to study its action on the projective space (the space of lines in R 2 passing through (0, 0)). We will think of the projective space as R { }. Since (εv ) ( ) ( ) (θ) 0 r εv (θ) 0 + /r = r 0 we see that the projective map can be expressed as ε (θ, r) = (θ + ω, εv (θ) 0 + /r), (2.) where θ T and r R { }. We will use the notation (θ k, r k ) = k ε (θ 0, r 0 ). In our estimates we will often use expressions such as r 0 r r k. This can be well defined if r 0. Indeed, if r j = for some j > 0, then we must have r j = 0, and we get r j r j = r j (εv (θ) 0 + /r j ) =. Note that ( ) ( ) Mε n (θ) r0 rn = r 0 r n. Thus the product r 0 r n is directly related to the Lyapunov exponents. 3. Notation and basic definitions. For an interval I T, let I denote its length. We fix ω satisfying the Diophantine condition (DC) κ,τ for some κ > 0, τ (we shall also express this as ω (DC) κ,τ ): inf qω p > p Z κ/ q τ for all q Z \ {0}.

6 562 K Bjerklöv and M Saprykina For our construction to go through we need V (θ) to have a special form: we want it to be close to zero outside a small interval I 0, on this interval V (θ) will have a unique non-degenerate maximum equal to. To be precise, we fix V (θ) = V (θ, λ) = + λ sin 2 (πθ), where λ should be thought of as a sufficiently large constant. We have the bounds Define the first critical interval V C (5π/2) λ; V C 2 = V (0) = 2π 2 λ. (3.) I 0 = [ λ /6, λ /6 ]. (3.2) By this choice V, V and V are small outside I 0 for large λ: max V (θ) λ /2, θ T\I 0 Let max θ T\I 0 V (θ) λ /3, max θ T\I 0 V (θ) λ /4. (3.3) Then I 0 = {θ : V (θ) 0.88}. (3.4) I 0 = c 2λ /2, /20 < c 2 < /8, and I 0 I 0. This interval is introduced because V (θ) is large on it: max V (θ) < I 0 3 V C2. (3.5) We shall consider the values of ε lying in the interval E = [ , 0 2 5]. Inductively we will show that the critical value ε c mentioned in the main theorem lies in this interval. The diffeomorphism ε (θ, r) is defined in (2.). Note that ε (θ, r) = (θ ω, r εv (θ ω) + 0 If I T is an interval centred at c, we denote by ki the interval centred at c of length k I. Define the projections Given θ 0 and r 0, denote Let π θ (θ, r) = θ, π r (θ, r) = r. (θ k, r k ) = k ε (θ 0, r 0 ), k Z. R u = [ 00, 5], R s = [/00, /5]. The notation reflects the fact that the strip T R u is contracted by the forward (and T R s by the backward) iterates of while iterating outside I 0 (respectively, outside I 0 + ω). ).

7 Universal asymptotics in hyperbolicity breakdown 563 Given integers 0 < M 0 <... < M n and intervals I 0 I... I n, define the following sets: and M j + M j + n n F = n (I j + mω), n B = (I j mω) j=0 m= j=0 m=0 n = T \ ( F n B n ). On each scale, F j and B j are the sets of θ, for which the behaviour of the system is irregular ; these sets should be thought of as small. n are the sets of good parameters θ. Note that n. Actually, later in the proof the sets B n and F n will be seen to be disjoint. This will be assured by the Diophantine condition on ω and the choice of M j. The fact that these sets are disjoint will be very important for controlling where the minimum angle can be located. The building blocks of our construction are the following boxes : Mj + à j (ε) = ε (A j ), where A j = {(θ, r) θ I j M j ω, r R u }, (3.6) Mj + B j (ε) = ε (B j ), where B j = {(θ, r) θ I j + M j ω, r R s }, (3.7) for j = 0,... n. We shall see that the sets à j and B j are very thin curvilinear rectangles placed over I j + ω. By construction, they contain pieces of stable and unstable manifolds of, respectively. These sets should be thought of as jth approximations to the stable and unstable manifolds. 4. Preparatory lemmas This section contains the necessary estimates for the mappings V and. The first two lemmas assert that the sets R u T and R s T, defined above, attract forward and backward iterates, respectively. Lemma 4.. Let ε E and assume that λ > 0 is large. Suppose r 0 [ 0, 4] and θ 0 / I 0. Then r R u. Moreover, if r 0 / R s, and θ 0 / I 0, θ 0 + ω / I 0, then r 2 R u. The proof is an easy verification, using estimates (3.3). The corresponding lemma, with R u replaced by R s, is true for backward iterations. Lemma 4.2. Let ε E and assume that λ > 0 is large. Suppose r 0 [/000, /4] and θ 0 ω / I 0. Then r R s. Moreover, if r 0 / R u, θ 0 ω / I 0 and θ 0 2ω / I 0, then r 2 R s. Lemma 4.3. For any θ T and ε E we have and r 0 r k... r 0 ( ) k+ for all k 0 (4.) r 0 r 0... r k k+ for all k 0. (4.2)

8 564 K Bjerklöv and M Saprykina Proof. We shall prove the first of these statements; the second one can be verified in the same way. Suppose that r l < /. Then l < 0 by assumption, and we have ( r l r l+ = r l εv 0 + ) > 0 r l = >, 2 since εv 0 < 0 for ε E. The result follows by induction using the fact that r 0. Since the frequency vector ω T satisfies the Diophantine condition (DC) κ,τ, one can get a lower bound for the return time into a small ball under the rotation by ω. Lemma 4.4. Suppose ω DC(κ, τ), and let I be an interval of length. Then for any N [κ/ ] /τ all the intervals I + jω, j = 0,,..., N, are disjoint. Proof. If x I and x + mω I, then κ inf mω p <. m τ p Z Therefore, m > [ (κ/ ) /τ]. 5. A brief sketch of the proof Since the proof of the main theorem is a quite lengthy inductive argument, we will briefly discuss the idea behind it, at least on the first scale. There will be some overlap here with results of the following sections, but we hope that this discussion will help the reader to better understand the inductive assumptions in section 8 and to have an idea of where we are heading. We stress that the parameter λ in V should always be thought of as being extremely large. What we are going to do is construct the invariant curves Ŵ ε ± (θ), which are the projectivizations of the subspaces E ε ± (θ). In figure the subspaces were represented by angles in ( π/2, π/2]; here they are represented by their slopes, i.e. by the tangent of the angle. Our construction will give us such good estimates that we will know where the minimal angle between the subspaces is located and how the minimum changes with ε. The interval I 0 defined in section 3 is of length 2λ /6. Thus, by lemma 4.4, we know that a point θ starting in I 0 will not return to I 0 (under forward and backward translation by ω) for at least N 0 = const λ /(6τ) steps. We let M 0 = N 0 and define A 0, à 0, B 0, B 0 as in (3.2) and (3.3). The sets à 0 and B 0 will be the first approximations of Ŵ ε + (θ) and Ŵ ε (θ) over I 0 + ω, respectively. We will show that the minimal angle is attained for θ I 0 + ω. Using lemmas (4.) and (4.2) repeatedly we get the following statements, provided that ε E (a) If θ 0 T and r 0 R u, let N 0 be the smallest integer such that θ N I 0. Then r k R u for k = 0,,..., N. (b) If θ 0 T and r 0 R s, let N 0 be the smallest integer such that θ N I 0 + ω. Then r k R u for k = 0,,..., N. From the definition of M 0, we in particular have that (I 0 + mω) I 0 = for 0 < m M 0. Thus the above statements imply B 0 (I 0 + ω) R s

9 Universal asymptotics in hyperbolicity breakdown 565 Figure 2. Approximating the curves Ŵ ±. and M0 (A 0 ) I 0 R u. Given (θ 0, r 0 ) and (θ 0, s 0 ) (that is, r 0 and s 0 are both over θ 0 ), then a trivial computation shows that r s = r 0 s 0 and r s = r s (r 0 s 0 ). r 0 s 0 Thus points in R u (R s ) are contracted by at least a factor of 25 when iterated forwards (backwards). Therefore the sets B 0 and M0 (A 0 ) will be very thin (the thickness is smaller than 25 M0 ). We will also show that they are almost horizontal (using the formulae in section 7) and vary very little with ε. The reason why they are flat is that we have iterated outside I 0, and there V is as flat as we like (by taking λ huge). To get à 0, we have to apply to M0 (A 0 ). Since M0 (A 0 ) lies over I 0, and is almost horizontal, we will get where à 0 = {(θ, r) : θ I 0 + ω, ϕ (θ) r ϕ + (θ)}, ϕ ± (θ) = εv (θ ω) 0 + error ±, so à 0 looks almost like εv 0 over I 0, that is, it looks like the peak of εv, see figure 2. Since V (0) =, we see that the peak of à 0 moves linearly with ε. The set B 0 remains (almost) constant, as we will show. Writing E = [ε, ε+ ], we will see that there is an ε 0 E such that B 0 à 0 = for ε [ε, ε 0 ) and B 0 à 0 for ε [ε 0, ε+ ]. The reason is just that à 0 moves up linearly with ε and B 0 is almost still. If à 0 B 0 =, the cocycle (ω, M ε ) is uniformly hyperbolic. To see this, we proceed as follows. Let 0 F, B 0 and 0 be as in section 3. Take a θ 0 0, and let N > 0 be the smallest integer such that θ N I 0. Then N > M 0 by the definition of 0. By (a) above we get θ k R u for all k [0, N]. In particular we have r N M0 R u, i.e. (θ N M0, r N M0 ) A 0. Thus (θ N+, r N+ ) à 0. From the assumption à 0 B 0 =, we know that r N+M0 / B 0, i.e. r N+M0 / R s. Since θ N I 0,

10 566 K Bjerklöv and M Saprykina we know that θ k / I 0 for k [N +, N + N 0 ]. Recall that N 0 M0 2, so in particular θ N+M0, θ N+M0+ / I 0, and by lemma 4.2 we thus have r N+M0+2 R u, i.e. we are back in the good region. We now let N 0 be the smallest integer such that θ N+M0++N I 0. Then, since θ N I 0, we must have N N 0 M 0. Hence, using condition (a) above, we have r k R u for k [N + M 0 +, N + M N ]. During the passage from k = N + to k = N +M 0 +, we can use lemma 4.4 to get a worst estimate for the product r N+ r N++M0. It is only during this passage that the r k can be outside R u. Note also that θ N++N I 0 M 0 ω and r N++N R u, i.e. the point (θ N++N, r N++N ) A 0. Therefore we can repeat the argument forever. Since N 0 is so much larger than M 0, we can obtain the following. Take θ 0 0 and r 0 R u, and let 0 < T < T 2 < be the times when θ Ti I 0. Then for all i Moreover, for all k 0 r k r Ti > 5 (Ti+ k)/2 for all k [0, T i ]. r k / R u θ k F 0. The first condition shows that the Lyapunov exponent + (ε) 0.5 log 5, since the measure of 0 is positive (recall the discussion in section 2). Note also that if θ 0 0 and r 0, s 0 R u, and if the T i are as above, then we get, using the formulae for contraction, r Ti+ s Ti+ 5 (Ti+) r 0 s 0. (5.) The second condition gives us good control when iterates can be outside the cone R u. This will be important several times, for example, when we control the location of the minimal angle. Analogously, when we consider the backward iterations, we get the following. Take θ 0 0 and r 0 R s, and let 0 < T < T 2 < be the times when θ Ti I 0 + ω. Then for all i Moreover, for all k 0 r Ti r k > 5 (Ti+ k)/2 for all k [0, T i ]. r k / R s θ k B 0. By taking bigger and bigger N such that I 0 Nω 0, and studying the set N+ ((I 0 Nω) R u ) Ã 0, we can use the above estimates to obtain better and better approximations of the curve Ŵ + (θ) over I 0 + ω (recall expression (5.)). In the limit we get a piece of the curve Ŵ + lying over I 0 + ω, which will be continuous by uniform convergence. By iterating this piece under, we get the whole invariant curve Ŵ +. Similarly we obtain Ŵ. This shows that the cocycle (ω, M ε ) is uniformly hyperbolic. Thus, by general results (see, e.g., [7, 8]), the curves Ŵ ± must be as smooth as M ε. The curves will have the following properties. Ŵ + (θ) / R u θ F 0, Ŵ (θ) / R s θ B 0. (5.2) Moreover, Ŵ + (θ) = εv (θ ω) 0 + error over I 0 + ω and Ŵ (θ) will be almost horizontal. We now explain why the minimal distance between the two curves must be attained over I 0 + ω. Note that the sets 0 F and B 0 are disjoint (this is important). This means that if Ŵ + (θ) and Ŵ (θ) are very close for some θ, then either Ŵ + (θ) R u and Ŵ (θ) is in (or very close to) R u or Ŵ (θ) R s and Ŵ + (θ) is in (or very close to) R s. Assume that the minimum distance between Ŵ + and Ŵ was attained for a θ outside I 0 and I 0 + ω. If Ŵ + (θ ), Ŵ (θ ) is close to R u, iterate the points (θ, Ŵ ± (θ )) one step forward. Then they are contracted at

11 Universal asymptotics in hyperbolicity breakdown 567 least by a factor of 20, contradicting the assumption on θ. Similarly, if Ŵ + (θ ) and Ŵ (θ ) are close to R s, iterate one step backward. If the minimal distance was located over θ 0 I 0, then, since Ŵ + (θ) R u for θ I 0, we would have Ŵ ± (θ 0 ) < 4. By iterating the two points (θ 0, Ŵ ± (θ 0 )) one step forward, using the above formula for contraction, we would get an even smaller distance between (θ 0 + ω, Ŵ ± (θ 0 + ω)), which is a contradiction. In the proof later on, we should be able to treat the case à 0 B 0. Then we have to use a multi-scale analysis to zoom in near I 0 + ω. The philosophy is the same as in this first step, but the technicality becomes a bit more involved. 6. Important formulae This section contains important expressions that will be used throughout the proofs. The formulae below give us control of the derivatives, once we have a control on products r 0 r k of the iterates. This is what we will do in this paper: estimate these products. Let r ± 0 = r± 0 (θ, ε) be given and define r ± k = π r( k ε (θ, r± 0 )), k Z. Forward (k > 0). In particular, skipping the ±, r (θ, ε) = εv (θ) 0 + r 0 (θ, ε). We shall write r i instead of r i (θ, ε). Calculating the different derivatives, we get θ r = εv (θ) θr 0 r 2 0 θθ r = εv (θ) θθr 0 r 2 0 For the contraction/expansion we have the formula, ε r = V (θ) εr 0 r0 2, + 2 ( θr 0 ) 2 r0 3, εε r = εεr 0 r ( εr 0 ) 2 r0 3. r + r r+ 0 r 0 r + 0 r 0. (6.) Hence, by induction, we get the expressions k θ r k = εv (θ + (k )ω) + ε ( ) k j V (θ + (j )ω) rk 2... r2 j j= k k j V (θ + (j )ω) ε r k = V (θ + (k )ω) + ( ) rk 2... r2 j j= k θθ r k = εv (θ + (k )ω) + ε ( ) k j V (θ + (j )ω) rk 2... r2 j j= j= + ( ) k θ r 0 r0 2..., (6.2) r2 k + ( ) k ε r 0 r0 2..., (6.3) r2 k + ( ) k θθ r 0 r r2 k k 2 ( ) k j ( θ r j ) 2 rk 2... r2 j r, (6.4) j εε r k = ( ) k εε r 0 r r2 k k 2 j= ( ) k j ( ε r j ) 2 r 2 k... r2 j r j (6.5)

12 568 K Bjerklöv and M Saprykina and r k + r k r 0 + r 0 (r r+ k )(r 0... (6.6) r k ). Backward (k < 0). Similarly, r (θ, ε) = r 0 εv (θ ω) + 0 and θ r = (εv (θ ω) θ r 0 )r 2, εr = (V (θ ω) ε r 0 )r 2, θθ r = (εv (θ ω) θθ r 0 )r 2 + 2( θr ) 2 r, εε r = θθ r 0 r 2 + 2( θr ) 2 r. By induction, we prove k θ r k = ε ( ) k j V (θ jω)r k 2... r2 j + ( )k ( θ r 0 )r k 2... r2, ε r k = j= k ( ) k j V (θ jω)r k 2... r2 j + ( )k ( ε r 0 )r k 2... r2, j= θθ r k = ε k ( ) k j V (θ jω)r k 2... r2 j + ( )k ( θθ r 0 )r k 2... r2 j= k + 2 ( ) k j ( θr j ) 2 r k 2 r... r2 j+ + 2( θr k ) 2, j r k j= k εε r k = ( ) k ( εε r 0 )r k 2... r2 + 2 ( ) k j ( εr j ) 2 r k 2 r... r2 j+ + 2( εr k ) 2, j r k j= r k + r k (r+ 0 r 0 )(r+ k... r+ )(r k... r ). (6.7) 7. Basic lemmas Here we show how to derive geometry control by using the formulae in the previous section, together with certain estimates on products r 0 r r k. The setting in this section is abstract and self-contained, but it is exactly this setting we will have in the inductive construction in section 8. The geometric picture behind the first lemma is that a box A = (I Mω) R u (I I 0 ) is mapped by M+ into a very thin strip M+ (A), which looks like the graph of the function (εv (θ ω) 0) over I 0 + ω. The second lemma shows that a box B = (I + Mω) R s is mapped by M+ into a very thin, almost horizontal strip over I 0 + ω. Recall the picture in figure 2. Lemma 7.. There exists λ 0 such that for λ > λ 0 the following holds. Let ε E and suppose that an interval I I 0 and an integer M > 00 satisfy the following properties: for any point (θ 0, r 0 ) (I Mω) R u we have for all k = 0,... M r k... r M 5 (M k)/2+ and r 2 k r2 p r p r M 5 (M k)/2+ for k p M, if r p /. (7.)

13 Universal asymptotics in hyperbolicity breakdown 569 Denote Then we have where A = {(θ, r) θ I Mω, r R u }. à = M+ ε (A) = {(θ, r) θ I + ω, ϕ (θ, ε) r ϕ + (θ, ε)}, ϕ ± (θ, ε) = εv (θ ω) 0 + φ ± (θ, ε), (7.2) and the functions φ ± satisfy the following estimates: φ ± (θ, ε) 5, (7.3) 0 < φ + (θ, ε) φ (θ, ε), (7.4) 5M θ φ ± (θ, ε) 2λ /3, (7.5) ε φ ± (θ, ε) 20, (7.6) θθ φ ± (θ, ε) V C 2, (7.7) εε φ ± (θ, ε). (7.8) Proof. Here, as before, we use the notation r k (θ, ε) = π r ( k ε (θ, r 0(θ, ε))) for θ (I n Mω), ε E n. Let r 0 (θ, ε) and r+ 0 (θ, ε), defined for θ I n Mω and ε E n, be the horizontal boundaries of the set A. The signs + and are chosen so that the horizontal boundaries of the set à satisfy ϕ (θ, ε) < ϕ + (θ, ε), ϕ ± (θ, ε) = r ± M+ (θ (M + )ω, ε), θ I n + ω, ε E n. One of r 0 + and r 0 equals identically 00, the other one equals 5. Since r M+ = εv 0 + /r M, we can write where ϕ ± (θ, ε) = εv (θ) 0 + φ ± (θ, ε), φ ± (θ, ε) = r ± M (θ (M + )ω, ε), θ I n + ω, ε E n. From (7.) with k = M we get (7.3). Using (7.), (6.3) and the fact that r ± 0 (θ, ε) is a constant, we obtain (7.6): M ε φ(θ, ε) = k= r 2 M r2 k k=0 5 k+2 = 20. (7.9) In order to estimate θ φ(θ, ε), we write (using (6.2) and the fact that r 0 is a constant): M V (θ (k )ω) θ φ(θ, ε) = ε rm 2 r2 k. (7.0) k= Estimating this sum needs a certain care, because V (θ) can become large (of order λ /2 ) when θ I 0.

14 570 K Bjerklöv and M Saprykina Recall the definition of I 0 from (3.2). Let λ be so large that ω / I 0 and take θ I 0. Then, by lemma 4.4, θ jω / I 0 for j =, 2,... N, where N = c λ /(6τ) and c = c (κ, τ). Let N i, i = 0,..., m, be such that 0 = N 0, N i < N i+ and (θ (M N i )ω) I 0, i = 0,..., m. Then N i λ /(6τ) for i =,..., m. (7.) We shall split the sum in (7.0) into sub-sums corresponding to the following periods : let the ith period be k = (M N i+ + ),..., (M N i ) for i = 0,..., m, and the mth period be k =,..., (M N m ). The periods are characterized by the fact that in each period there is at most one value of k such that (θ + (k )ω) I 0. Namely, for the first element of periods with number i = 0,..., m we have (θ + (k )ω) = (θ + (M N i+ )ω) I 0, so the best estimate for V is V C (5π/2) λ. For all the other elements we have V (θ + (k )ω) λ /3. In the mth period all the elements satisfy the latter estimate. Therefore, the part of sum (7.0) over all k from the ith period can be estimated as ε M N i V (θ (k )ω) r 2 k=m N i+ M r2 k + ε V (θ (M N i+ )ω) rm 2 r2 M N i++ 2ελ /3 5 Ni 2 + 5π λ5 N i+ < λ /3 5 i. 2 The last inequality follows from (7.). Therefore, m θ φ(θ, ε) λ /3 5 i < 2λ /3. (7.2) k=0 From formulae (7.) and (6.6) we obtain 0 < φ + φ = ϕ + ϕ r 0 + = r 0 r 0 + r+ M r 0 r M < 00 5 <. (7.3) M+2 5M Now let us estimate the second derivative in θ. The argument in this part of the proof is rather technical. The complications are due to our choice of coordinates: we are working with the tangents of angles, whose range includes infinity. The general form of the second derivative is given by (6.4). Since r 0 is a constant, M θθ φ = ε ( ) M k+ V (θ + (k )ω) M r 2 k= M... 2 ( ) M k+ ( θ r k ) 2 r2 k r 2 k= M... r2 k r := I + II. k First estimate a part of I corresponding to k =,... M 0: M 0 ε k= V (θ + (k )ω) r 2 M r2 k ε k=0 V C 2 5 k+2 ε5 2 V C 2 k=0 5 k 5 0 V C 2. (7.4) Now consider k = M 9,..., M. Since θ + Mω I 0, and the return time to I 0 is large, we know that θ + (M j)ω / I 0 (at least) for j =,..., 20. Hence, (θ + (k )ω) / I 0 for all k = M 9,..., M, and therefore V (θ (k )ω) < λ /4 by (3.3). Hence, M 0 V M V I = ε rm 2 + ε r2 k rm 2 r2 k k= 5 0 V C 2 + ελ /4 k=m 9 M k=m 9 r 2 M r2 k 5 0 V C 2 + 0ελ /4 < 2 V C2. (7.5)

15 Universal asymptotics in hyperbolicity breakdown 57 The latter inequality holds since λ is assumed to be large. The above argument will be used several times during the proof of this lemma. Estimating II requires more work. Denote the kth term of II by A k. Let j i be a subset of indices, j i M such that r ji < /, i =,... I. Note that, by (7.), r M 5, hence in our case j i < M. By definition of ε (θ, r), r ji+. Now we have two cases (a) If k j i and k j i + for all i =,... I, then we estimate A k separately. (b) If k = j i for some i =,... I, then we shall estimate the corresponding pair of terms, i.e. A k + A k+ = ( θ r k ) 2 rm 2 r2 k+ r2 k r ( θr k+ ) 2 k rm 2 r2 k+ r. (7.6) k+ First consider case (a). By using (6.2), and the fact that θ r 0 = 0, we can rewrite A k in the following way: ( A k = ( εv (θ + (k )ω) + ε ) k V 2 (θ + (p )ω) θr k ) 2 p= ( )k p rm 2 r2 k r k = rk 2... r2 p rm 2 r2 k r k 2 = ε 2 V k (θ + (k )ω) r M r k r k + ( ) k p V (θ + (p )ω) /2 r M r k r k /2 rk 2. (7.7) r2 p p= The assumption of this case says that both r k > / and r k > /. This permits us to estimate r k /2 < 4 and apply (7.) for the denominators. Now we use the same argument as in estimating I. For k M 0, A k 6ε 2 V 2 C = 6ε2 V 2 C 5 (M k)+2 r M r k + j=0 k p= 5 j 2ε2 V 2 C 5 (M k). r M r k r 2 k r2 p For each k = M 9,..., M, we split the estimate in the same way as above: for p =,..., k 0, we estimate V V C ; forp = k 9,..., k the point(θ +(p )ω) / I 0, and, hence, V (θ + (p )ω) λ /3. Thus, A k 6ε 2 λ /3 r M r k + k p=k 9 λ /3 r M r k r 2 k r2 p 2 k 0 + p= V C r M r k r 2 k r2 p 2 V 2 C 5 7. Consider case (b). Here we assumed thatk = j i for somei =,..., I, so that r k < /. Recall that in this case k < M. An easy calculation with the definition of ε (θ, r) shows that in this case r k+ and / r k r k+ 2. Using the formula θ r k+ = εv (θ + kω) θrk, we estimate the value in (7.6) by rk 2 A k A k+ (εv (θ + kω)) 2 rm 2 r2 k+ r + 2εV (θ + kω) θ r k k+ rm 2 r2 k+ r k+rk 2 + ( θ r k ) 2 ( rm 2 ) r2 k+2 rk+ 2 r3 k := E k + E2 k + E3 k. r 3 k+ r4 k

16 572 K Bjerklöv and M Saprykina Since r k+, we can write Ek (εv ) 2 rm 2 ε2 V 2 C r2 k+ 5 M k+2 for k =,... M 0 and Ek ε2 λ /3 for k = M 9,..., M. Using r k+ >, (6.2) and the fact that θ r 0 = 0, we estimate Ek 2 2ε2 V (θ + kω) V (θ + (k )ω) k ( ) k p V (θ + (p )ω) rm 2 + r2 k rm 2 r2 p. The latter is estimated in the same way as in (a): E 2 k ε2 V 2 C 5 M k for k =,... M 0, and Ek 2 V 2 C for k = M 9,..., M. 5 7 Note that r k+ r k = r k (εv 0). The reason for distinguishing case (b) is the following cancellation: p= Ek 3 = ( θr k ) 2 εv 0 ( θ r k ) 2 rm 2 r2 k+2 rk+ 3 r3 k 20 rm 2 r2 k+2 r4 k+ r4 k Again, by (6.2) and θ r 0 = 0, we get E 3 k 20ε2 V (θ + (k )ω) r M r k+2 r 2 k+ r2 k k + p= Since r k+, we can use (7.) for the denominators. Then E 3 k 2ε2 V 2 C 5 M k for k =,... M 0 and E 3 k V 2 C 5 7 for k = M 9,..., M. We have obtained that, for k as in (b). A k + A k+ 6ε2 V 2 C 5 M k for k =,... M 0 and A k + A k+ 3 V 2 C for k = M 9,..., M. 5 7 Since we have ε <, 6ε 2 V 2 C II + 30 V 2 C < 5 M k V C 2. k=0 Collecting the above estimates, ( ) k p V (θ + (p )ω) r M r k+2 r 2 k+ r2 k r2 p θθ φ(θ, ε) < V C 2. (7.8) The second derivative in ε is estimated by in a similar way. This finishes the proof of the lemma. 2

17 Universal asymptotics in hyperbolicity breakdown 573 The next lemma gives analogous estimates for the backward iterates. Lemma 7.2. There exists λ 0 such that for λ > λ 0 the following hold. Let ε E and suppose that an interval I I 0 and an integer M > 00 satisfy the following properties: for any point (θ 0, r 0 ) (I + Mω) R s we have for all k = 0,... M r M... r k 5 (M k)/2 and r M r p r p+ 2 r2 k 5 (N k)/2 for M p + k, if r p+. (7.9) Denote B = {(θ, r) θ I + Mω, r R s }. Then we have B = ε M+ (B) = {(θ, r) θ I + ω, ψ (θ, ε) r ψ + (θ, ε)}, where the functions ψ ± satisfy the following estimates: 8. Induction ψ ± (θ, ε) 5, 0 < ψ + (θ, ε) ψ (θ, ε) 5 M, ε ψ ± (θ, ε) 20, (7.20) θ ψ ± (θ, ε) 2λ /3, θθ ψ ± (θ, ε) V C 2, εε ψ ± (θ, ε). This section contains the inductive procedure on which the proof of the main theorem is based. 8.. Conditions (C) 0, (C2) 0 and (C3) 0. In order to keep the formulation of the inductive lemma below (lemma 8.2) more compact, we introduce the following notation. Suppose that integers M 0 <... < M n, non-empty closed intervals I 0 I... I n and E E 0... E n are chosen. Let the sets j, j, Ã j (ε) and B j (ε) be defined as in section 3. Condition (C) F n. Assume that θ 0 n, r 0 R u and ε E n. Let N > 0 be the smallest natural number such that θ N I n. Then for any k = 0,..., N r k... r N 5 (/2+/2n+ )(N k)+, and r 2 k r2 p r p r N 5 (/2+/2n+ )(N k)+ for k p N, if r p /; (8.) n r k / R u θ k n F = M i+ (I i + mω). (8.2) i=0 m= Condition (C) B n. Assume that θ 0 n, r 0 R s and ε E n. Let N > 0 be the smallest natural number such that θ N I n. Then for any k = 0,..., N r N... r k 5 (/2+/2n+ )(N k), and r N r p r 2 p+ r2 k 5 (/2+/2n+ )(N k) for N p + k, if r p+ ; (8.3) n r k / R s θ k n B = M i+ (I i mω). (8.4) i=0 m=0

18 574 K Bjerklöv and M Saprykina Condition (C2) n. For j = 0,, 2, 3, I n ± (M n + j)ω lie in n. Condition (C3) n. Define the functions ϕ ± n (θ, ε), ψ± n (θ, ε) : (I n + ω) E n R: à n (ε) = {(θ, r) θ I n + ω, ϕn (θ, ε) r ϕ+ n (θ, ε)}, B n (ε) = {(θ, r) θ I n + ω, ψn (θ, ε) r ψ n + (θ, ε)}. (8.5) Then these functions are C 2, and for all (θ, ε) they satisfy the following conditions: ϕ n ± (θ, ε) = εv (θ ω) φ± n (θ, ε), (8.6) where φ ± n : (I n + ω) E n R, and Moreover, 0 < φ + n φ n < 5 Mn+, 0 < ψ + n ψ n < 5 Mn+, (8.7) θ φ ± n (θ, ε) 2λ /3, θ ψ ± n (θ, ε) 2λ /3, (8.8) ε φ n ± (θ, ε) 20, εψ n ± (θ, ε) 20, (8.9) θθ φ n ± (θ, ε) V C 2, θθψ n ± (θ, ε) V C2, (8.0) εε φ n ± (θ, ε), εεψ n ± (θ, ε). (8.) ϕ + n (θ, ε) < ψ n (θ, ε) for all θ (I n + ω) \ ( 3 I n + ω ), ε E n. (8.2) Finally, writing E n = [εn, ε+ n ] we have à n B n = for ε [ε n, ε n ), (8.3) there is a unique point θ 3 I n + ω s.t. ϕ n + (θ, εn ) = ψ n (θ, εn ), (8.4) there is a unique point θ 3 I n + ω s.t. ϕ n (θ, ε + n ) = ψ + n (θ, ε + n ). (8.5) 8.2. Basic step Recall that I 0 = 2λ /6. By lemma 4.4, we know that for any θ 0 I 0 we have θ k / I 0 for all k =, 2,... N, where N = [c λ /(6τ) ], c = (κ/2) /τ. We define Then M 0 is of order N. M 0 = [λ /(2τ) ]. (8.6) Lemma 8. (Basic step). Let λ in the definition of V be sufficiently large. Then there exists an interval E 0 = [ε 0, ε+ 0 ] E such that conditions (C) 0, (C2) 0 and (C3) 0 hold. Proof. Assume that λ is sufficiently large depending on κ, τ and V. Condition (C) 0 follows from lemmas 4. and 4.2. Condition (C2) 0 is trivial, since = T. Now we shall choose E 0 = [ε 0, ε+ 0 ] E in such a way that (C3) 0 holds. Define à 0 and B 0 as in (3.6) 0, (3.7) 0, and let ϕ ± 0 (θ, ε) and ψ± 0 (θ, ε) be as in (8.5) 0. Estimates (8.7) 0 (8.) 0 for the functions ϕ ± 0 and ψ± 0 follow from lemmas 7. and 7.2. It can be verified by a calculation that (8.2) 0 holds for all ε E. In fact, it is easy to prove a stronger estimate (recall that V (θ) < 0.88 outside I 0 ): ϕ + 0 (θ, ε) < for all θ (I 0 + ω) \ (I 0 + ω), ε E.

19 Universal asymptotics in hyperbolicity breakdown 575 Let us verify (8.4) 0 and (8.5) 0. Note that B 0 (ε) (I 0 + ω) R s and ε (à 0 (ε)) (I 0 R u ). Let Ã(ε) = ε (I 0 R u ) and denote its upper and lower boundaries by ϕ + (θ, ε) and ϕ (θ, ε), respectively: Ã(ε) = {(θ, r) θ I 0 + ω, ϕ (θ, ε) r ϕ + (θ, ε)}. Then à 0 (ε) Ã(ε) or, in other words, ϕ (θ, ε) ϕ 0 (θ, ε) < ϕ+ 0 (θ, ε) ϕ+ (θ, ε). One can verify by computation the following two statements. For ε < ε = , Ã(ε) (I 0 +ω) R s =, which implies that ϕ + 0 (θ, ε ) ψ 0 (θ, ε ) for θ I 0 +ω. On the other hand, for ε + = 0 2 5, ϕ (0, ε + ) = /5, which implies that ϕ 0 (0, ε+ ) ψ + 0 (0, ε+ ). By (7.2), (7.6) and (7.20), for θ I 0 + ω and all ε E we have ε (ϕ ± 0 (θ, ε) ψ± 0 (θ, ε)) > 0.88 /0 > /2. In particular, this implies that in the interval E = [ε, ε+ ] there is ε 0 such that ϕ 0 + (θ, ε) < ψ 0 (θ, ε) for all θ I 0 + ω and ε < ε 0, and for any ε > ε 0 there is θ (I 0 + ω) such that ϕ 0 + (θ, ε) = ψ 0 (θ, ε). This gives (8.3) 0. To see that there is a unique point θ such that ϕ 0 + (θ, ε 0 ) = ψ 0 (θ 0, ε 0 ), it is enough to show that the function h(θ, ε) = ϕ 0 + (θ, ε) ψ 0 (θ, ε) has a unique non-degenerate maximum. Recall the definition of I 0 from (3.4). For any fixed θ (I 0 + ω) \ (I 0 + ω) and all ε E, we have h(θ, ε) <. Therefore, it is enough to verify that θθ h(θ, ε) is negative for all θ I 0. By (3.5) and (8.0) 0, θθ h(θ) < ε 0 ( 3 V C 2) + 2 V C 2 < 0. This proves (8.4) 0. Similarly we find ε + 0 and θ such that (8.5) 0 holds. Finally, define 8.3. Induction step E 0 = [ε 0, ε+ 0 ]. The following lemma contains the induction step as well as some extra information that is needed for the proof of the main theorem. Lemma 8.2. There exists λ 0 > 0 such that for all λ > λ 0 the following holds. Suppose that M 0, E 0 and I 0 are as above, and we have chosen integers 0 < M 0 <... M n, non-empty closed intervals I 0 I... I n and E 0 E... E n (n 0) with the following properties: M 0 = [λ /2τ ], 5 Mj /4τ M j 2 5 Mj /4τ, j =,..., n, (8.7) I 0 = l 0 = 2λ /6, I j = l 0 5 Mj /2, j =,..., n, (8.8) and (C) n, (C2) n and (C3) n hold. Then there exists an integer M n+ satisfying (8.7) with j = (n + ), a non-empty closed interval I n+ I n (I I 0 if n = 0) satisfying (8.8) and a non-empty closed interval E n+ = [ε n+, ε+ n+ ] E n such that (C) n+, (C2) n+ and (C3) n+ hold. Moreover, if θ (I n \ I n+ ) + ω and (θ, r) à n, (θ, s) B n, then r s > 4 5 Mn+, (8.9) and à n+ à n, B n+ B n (8.20) à n+ B n+ for ε E n+, à n+ B n+ = for ε [εn, ε n+ ). (8.2) Furthermore, suppose that à n B n = for some ε E n. Then the following extension of (C) n+, call it condition (C) n+, holds: assume that θ 0 n, r 0 R u. Let N > 0 be any

20 576 K Bjerklöv and M Saprykina integer such that θ N I n. Then for any k = 0,..., N r k... r N 5 (/2+/2n+2 )(N k)+ and rk 2 r2 p r p r N 5 (/2+/2n+2 )(N k)+ for k p N, if r p /; (8.22) n M i+ r k / R u θ k n F = (I i + mω). (8.23) i=0 m= Let N > 0 be any integer such that θ N I n. Then for any k = 0,..., N r N... r k 5 (/2+/2n+2 )(N k), and r N r p r 2 p+ r2 k 5 (/2+/2n+2 )(N k) for N (p ) k, if r p+ ; n M i+ r k / R s θ k n B = (I i mω). i=0 m=0 The Furthermore... part of the lemma differs from (C) n+ by the fact that in the former we do not assume N > 0 to be the smallest natural number such that θ N I n+ ; we only need that θ N I n. Proof. We assume that λ 0 is sufficiently large, depending on κ, τ and V, to make sure that the statements below hold true. Since the proof of this lemma is quite long, we split it into a number of steps. Basics. It follows from the choice of I n and M n by lemma 4.4 that M n M n (I n M n ω) (I n + mω) = and (I n + M n ω) (I n mω) =, (8.24) m=0 if θ I n M n ω, then k = M n is the smallest positive integer such that θ + kω I n, (8.25) if θ (I n + M n ω), then k = M n is the smallest positive integer such that θ kω (I n + ω). (8.26) Step. Here we define the critical set I n+. The idea is that the projection onto the base T of the intersection à n (ε) B n (ε) should be in (I n+ + ω) for all ε E n. For θ I n + ω consider the functions m=0 ϕ n ± (θ, ε) = εv (θ ω) 0 + φ± n (θ, ε), and ψ± n (θ, ε), defined in (8.5). If n = 0 we note the following. By definition we have V (θ) < 0.88 for θ / I 0. Since φ ± 0 /5 and ψ± 0 /5, it thus follows that ψ 0 (θ, ε) ϕ+ 0 (θ, ε) > for θ I 0 \ I 0, ε E. (8.27) In other words, à 0 (ε) and B n (ε) are far away outside I 0 + ω. For ε E n, we need to estimate the length of the sets: K(ε) = {θ I n + ω ϕ n + (θ, ε) ψ n (θ, ε)}. Recall that, by (8.7), φ + n (θ, ε) φ n (θ, ε) 5 Mn+ and ψ + n (θ, ε) ψ n (θ, ε) 5 Mn+. It is easier to estimate the length of slightly larger sets: K(ε) K (ε) = {θ I n + ω ϕ n (θ, ε) + 5 Mn+ ψ n (θ, ε) 5 Mn+ }.

21 Universal asymptotics in hyperbolicity breakdown 577 By the above argument we must have K (ε) I 0 + ω if n = 0. If n > 0 we have this by assumption, since I n I 0. In order to estimate the length of K (ε), consider h(θ, ε) = ϕ n (θ, ε) + 5 Mn+ (ψ + n (θ, ε) 5 Mn+ ). If n = 0, we think of h as being defined for θ I 0 + ω. By (8.9), for each fixed θ I n + ω we have ε h(θ, ε) min V (θ) φ n C ψ n C = 0.88 /0 > /2, θ I 0 i.e. h(θ, ε) grows with ε for each fixed θ. Therefore, K (ε) K (ε n + ) for all ε E n. It is left to estimate the length of K (ε n + ). Let us fix ε = ε+ n and omit the dependence on ε. By (8.2), (8.5) and by the choice of E n, h(θ) 2 5 Mn+ for all θ I n + ω with the equality only for θ = θ. Therefore, h(θ) has a maximum at the point θ (this is the reason why we estimate K (ε) rather than K(ε)). Recall that θ is in (/3)I n + ω, so it lies well inside I n + ω (when n = 0, it is clear from (8.27) that θ must be in I 0 + ω). In order to show that this maximum is non-degenerate (quadratic), as well as to estimate the length of K, we shall verify that the maximum of θθ h(θ) over θ K is negative and large in absolute value if λ is large. By (8.0), (3.5) and (3.), θθ h(θ) < ε n + ( 3 V C 2) + 2 V C2) < 4λ. Hence, h(θ) has a unique quadratic maximum at θ. The above estimate also permits us to prove that K (ε n + ) is contained in an interval J n+ + ω, centred at θ, of length l0 0 5 Mn/2 (we estimate the set of θ for which h(θ) 0). Recall that l 0 = 2λ /6. Moreover, by estimating the set of θ for which h(θ, ε) 4 5 Mn+, we get ψ n (θ, ε) ϕ+ n (θ, ε) > 4 5 Mn+ for θ (I n \ J n+ ) + ω, ε E n. (8.28) Recall (8.27) in the case when n = 0. In particular we have π θ (Ã n (ε) B n (ε)) J n+ + ω for all ε E n. Let I n+ be an interval of length l 0 5 Mn/2 centred at θ. Since J n+ I n+, we see that (8.28) gives (8.9) n+. For later use we stress that we have and ϕ + n < ψ n for all θ (I n+ + ω) \ (J n+ + ω), ε E n, (8.29) J n+ = (/0)I n+. (8.30) Step 2. Here we verify that condition (C) n+ holds. We shall start by considering forward iterations. Let θ 0 n, r 0 R u, ε E n, and let N > 0 be the smallest positive integer such that θ N I n. For an integer T denote (C)[T ] the condition that for any k = 0,..., T r k... r T 5 (/2+/2n+2 )(T k)+ and r 2 k... r2 p r p r T 5 (/2+/2n+2 )(T k)+, for k p T if r p /; (8.3) r k / R u θ k F n. (8.32) If I n+ is defined, then (C)[N] coincides with (C) F n+. Let 0 < T 0 < T < < T p = N be the times such that θ Tj I n. Since I n = l 0 5 Mn /2 (by assumption) and since (DC) κ,τ holds, it follows from lemma 4.4 that T j+ T j c5 Mn /(2τ), c = c(κ, τ, l 0 ). (8.33) Moreover, since θ 0 n we also have T 0 > M n. (8.34)

22 578 K Bjerklöv and M Saprykina Condition (C)[T 0 ] follows from the assumed (C) F n, which we can apply because n n and E n E n. Assume now that we have proved that (C)[T j ] holds for some j < p. We shall prove (C)[T j+ ]. First we show that (θ Tj +, r Tj +) Ã n. By the definition of T j we have (θ Tj M n, r Tj M n ) I n M n ω. By (C2) n, I n M n ω n. Moreover, from (8.24), Hence, M n (I n M n ω) (I n + mω) =. (I n M n ω) n= M n j (I j + mω) =. j=0 m= By (8.34), T j M n > 0. Hence, we can apply (8.32), which gives us r Tj M n R u. Then (θ Tj M n, r Tj M n ) A n and Since θ Tj (θ Tj +, r Tj +) Ã n. (8.35) / I n+ (by assumption), (θ Tj +, r Tj +) / B n. (8.36) Therefore, (θ Tj +M n, r Tj +M n ) / B n, i.e. r Tj +M n / R s. By (C2) n, neither θ Tj +M n nor θ Tj +M n+ lie in I 0. By lemma 4. we get r Tj +M n+2 R u. Then we can apply lemma 4.3, which gives that for any k [T j +, T j + M n + 2] (note that 2 < 5 3 ) r k... r Tj +M n+2 (/) Tj +Mn k+3 > (/5) 3(Tj +Mn k+3) and rk 2... r2 p r p r Tj +M n+2 = rk 2... r2 p r p r Tj +M n+2 (/) 2(p k) Tj +Mn+3 p (/) (/5) 3(Tj +Mn k+3) for k p T j + M n + 2, if r p /. From (C2) n we have θ Tj +M n+3 I n + (M n + 3)ω n. Now we apply inductive assumption (C) F n to the point (θ T j +M n+3, r Tj +M n+3) and conclude that for each k = T j + M n + 3,..., T j+ we have the following estimate (here N = T j+ (T j +M n + 3) is the smallest positive integer such that θ (Tj +M n+3)+n I n ): r k... r Tj+ 5 (/2+/2n+ )(T j+ k)+ ; and r 2 k r2 p r p r Tj+ 5 (/2+/2n+ )(T j+ k)+ M j + for k p T j+, if r p /; (8.37) n r k / R u θ k n F = (I j + mω). (8.38) j=0 m= Combining the above estimates, we get for any k [T j +, T j + M n + 2] r k r Tj+ = rk r Tj ( +M n+2 rtj +M n+3 r Tj+ > 5 3(T j +M n k+3) 5 and n+ ) ( ) rk 2 r2 p r p r Tj+ 5 3(Tj +Mn k+3) n+ (T j+ T j M n 3))+ (T j+ T j M n 3))+

23 Universal asymptotics in hyperbolicity breakdown 579 for k p T j+, if r p /. One can verify that ( 2 + ) (T 2 n+ j+ T j M n 3)) + 3(T j + M n k + 3) > ( n+2 ) (T j+ k)) +. Indeed, this inequality follows from a stronger one: T j+ T j 5M 2 n+2 n > 0. The latter follows from (8.33) and (8.7) for any n, provided that λ is sufficiently large. Thus, for any k = T j +,... T j + M n + 2 we have r k r Tj+ λ ( 2 + r 2 k r2 p r p r Tj+ λ 2 n+2 )(T j+ k)+, and ( ) n+2 (T j+ k)+ for k p T j+, if r p /. (8.39) Now (8.3) Tj+ follows from (8.3) Tj, (8.37) and (8.39). Thus (C)[T j+ ] holds. By induction we see that (C)[N], i.e., (C) F n+, holds. Note that if à n B n = for some ε E n, then (8.35) implies (8.36) directly. In this case, to make the proof work we do not need to know that θ Tj / I n+. Therefore, we can prove estimates (8.3) and (8.32) for any N > 0 such that θ N I n. This is the content of the Furthermore.. part of lemma 8.2. Note that (8.35) implies the first assertion of (8.20): à n+ à n. The verification of (C) B n+, as well as the second assertion of (8.20), B n+ B n, is very similar. Step 3. Here we chose the number M n+ and verify that (C2) n+ holds. The argument at this step is an exact repetition of the corresponding argument in [], but we decided to include it here for completeness. For each j = 0,,..., n, let N j be the positive integer given by lemma 4.4 when it is applied to I = 3I j. By the inductive estimates (8.8), and the definition of I 0, we get [ ( ) ] [ κ /τ ( ) κ /τ N j = = 5 Mj /(2τ)], j =,..., n. (8.40) 3 I j 3l 0 We thus have (3I j ) 0< m N j ((3I j ) + mω) =, for j = 0,,..., n, Note that M j, given by (8.7), is of the size N j. From this it is easy to deduce that the following is true for each j = 0,..., n: Given any k Z, we can have (I j + pω) 2M j m= 2M j (I j + mω) for at most 4M j + integers p in the interval [k, k + N j ]. Similarly, (I j qω) 2M j m= 2M j (I j + mω)

24 580 K Bjerklöv and M Saprykina for at most 4M j + integers q in [k, k +N j ]. Using this, we see that in the interval [k, k +N n ], there are at most ( ([ ] ) ([ ] )) Nn Nn s := 2 (4M n + ) + (4M n + ) (4M 0 + ) + integers p such that (I n ± pω) n j=0 2M j N n m= 2M j (I j + mω). Since ( ) N n N n s < 00 M n + M n M 0 N n N 0 N 0 ( Mn = 00N n + M n M ) 0, N n N n N 0 it follows from estimates (8.7) and (8.40) that s N n for all large λ, independently of n. Hence we have proved that for any k there is an integer a in the interval [k, k + N n ] such that (I n ± (a + p)ω) n j=0 2M j By definition of n, the latter implies that m= 2M j (I j + mω) = for p = 0,, 2, 3. (I n ± (a + p)ω) n for p = 0,, 2, 3. Take k = 5 Mn/(4τ), and take an integer with the above property in the interval [5 Mn/(4τ), 5 Mn/(4τ) + N n ]. Call this integer M n+. Since I n+ I n, the above expression of course implies the weaker condition I n+ ± (M n+ + p)ω n for j = 0,, 2, 3. (8.4) Hence (C2) n+ holds. Moreover, since N n 5 Mn/(4τ), we have 5 Mn/(4τ) M n+ 2 5 Mn/(4τ), (8.42) as required. From now on M n+ is fixed (this is the M n+ in the statement of lemma 8.2). Step 4. Here we choose the interval E n+ and verify that (C3) n+ and (8.2) hold true. Since (C) n+ holds, it follows from lemmas 7. and 7.2 that the functions ϕ ± n+ and ψ± n+ satisfy ((8.7) (8.)) n+. Recall that we have I n+ (I n+ + mω) =, 0< m M n+ so we can indeed apply lemmas 7. and 7.2. Moreover, (8.20) implies the inequalities and ϕ n (θ, ε) ϕ n+ (θ, ε) < ϕ+ n+ (θ, ε) ϕ+ n (θ, ε), θ I n+ + ω, (8.43) ψ n (θ, ε) ψ n+ (θ, ε) < ψ + n+ (θ, ε) ψ + n (θ, ε), θ I n+ + ω, (8.44) for all ε E n. Recall the definition of the interval J n+ in step. The above two inequalities, combined with (8.29), give ϕ + n+ (θ) < ψ n+ (θ) for all θ (I n+ + ω) \ (J n+ + ω). (8.45) By (8.30), this gives (8.2) n+.

25 Universal asymptotics in hyperbolicity breakdown 58 Since ((8.4) (8.5)) n hold, and since we have the inclusions (8.43) and (8.44) and the derivative estimates on ϕ n+, ψ n+, we can proceed as in the proof of lemma 8. to find a non-degenerate interval E n+ E n such that ((8.4) (8.5)) n+ and (8.2) hold. The induction step is complete. 9. Proof of the main theorem We are now ready to prove the main theorem. As before, assume that λ is sufficiently large. From the inductive construction in the previous section, we get a nested sequence of intervals E n = [εn, ε+ n ]. Since ε n ε n+ for each n, there is an ε such that εn ε as n. From the estimates in section 8 it follows that the cocycle (ω, M ε ) is uniformly hyperbolic on each interval [ε n, ε n ) and therefore on [0, ε ). Indeed, if à n B n = for some n, then the cocycle is uniformly hyperbolic as we shall see below. We shall also see that the cocycle is not uniformly hyperbolic for ε = ε, and thus ε is the value ε c in the statement of the main theorem. Fix an arbitrary n > 0, and take any ε [ε n, ε n ). For this ε, we shall construct the stable and unstable invariant curves, Ŵ + (θ, ε) and Ŵ (θ, ε), and prove that the minimal pointwise distance between these curves is attained for θ in the interval (I n + ω). We shall also estimate the derivative in ε of the minimal distance between Ŵ + (θ, ε) and Ŵ (θ, ε). Finally, we shall derive the statement of the main theorem. We begin by constructing a piece Ŵ 0 + (θ, ε) of the unstable invariant curve for θ I n + ω. 9.. Construction of the invariant curves Ŵ + and Ŵ For an arbitrary n > 0, fix an ε [ε n, ε n ), and let us omit the dependence on ε. From the inductive construction in the previous section we know that hypotheses (C ) n (C 3 ) n hold true. By (8.3), for our fixed ε we have à n (ε) B n (ε) =. The last part of lemma 8.2 implies condition (C ) n+. This condition will be used several times during the proof. Choose a sequence of positive integers T k > T k, k 0, satisfying I n T k ω n for k 0. The possibility of such a choice was proved in step 3 of the inductive procedure above. Denote J k = I n T k ω, C k = J k R u, C k = Tk+ (C k ), k 0. Each curvilinear rectangle C k can be thought of as the kth approximation to Ŵ 0 +. J k, J k n, condition (8.23) implies Tk Tk (C k ) J k R u = C k. Therefore, the curvilinear rectangles C k form a nested sequence. Let Ŵ 0 + = C k. k=0 Since Then Ŵ + 0 is defined for all θ I n + ω. Moreover, the horizontal widths of C k decay very fast with k. Indeed, estimate (8.22), together with the fact that (I n T k ω) n n, permits us to apply lemma 7. with M = T k. Denote by c ± k (θ) the upper and lower boundaries of C k. Then (7.2) and (7.4) imply max θ I n+ω c+ k (θ) c k (θ) 5 Tk+.