2 n intervals in the attractor of the tent map

Transcription

1 2 n intervals in the attractor of the tent map Peter Barendse revised December 2010 Abstract In this paper we give a elementary proof explaining the phenomenon of band-splitting, first explained in [2, 1] and seen most notably in the attractors of certain one-parameter, unimodal maps such as the tent map on the unit interval. We show that there are parameters values, ɛ 0 > ɛ 1 > ɛ 2 such that for any n N, if the parameter is in the interval [1 + ɛ n+1, 1 + ɛ n ), then the attractor consists of exactly 2 n disjoint closed intervals and 2 n isolated points. For the tent map, this parameter is the slope of the tent map. The proof consists of an analytic argument to show that the endpoints of these intervals, which are the iterates of the critical point 1/2, are found in a certain order, then a simple linear mapping argument. 1 Introduction This paper concerns a particular pattern of bifurcations found in the dynamics of many iterated maps, including unimodal maps on the unit interval, perhaps the simplest example being the tent map: { f 1+ɛ (x) = where ɛ << 1. (1 + ɛ)x if x 1/2 (1 + ɛ)(1 x) if x > 1/2 Given any map f : X X, the orbit of a point x X is the sequence is {x, f(x), f 2 (x),...} of successive iterates of x by f. A map is called unimodal if it has only one critical point, c. The orbit of this point we will call the critical sequence, C(f) = c 0 = c, c 1 = f(c), c 2 = f 2 (c),.... The kneading sequence K(f) = k 0, k 1,... is given by k n = sgn(f n (c) 1 2 ) (1 if f n (c) > 1 2, 1 if f n (c) < 1 2, 0 if f n (c) = 1 2 ). These sequences keep track of the successive folds of the unit interval. In a unimodal map they tell nearly all there is to know 1

2 about the dynamics, since the only other action is a varying degree of stretching/compression along the interval. In the case of the tent map f 1+ɛ, each iteration folds the unit interval once at the critical point 1 2, then stretches (by a constant linear factor 1 + ɛ), bringing it back into itself. An attractor is the minimal set Y X toward which every orbit in its basin of attraction (an open set containing Y ) eventually goes. More concretely, In this paper we are interested in the attractor of the entire unit interval., which is that certain maps exhibit what has been called a infinite band-splitting sequence of bifurcations as the parameter approaches a certain value [2, 1]. This following will be the basis for an induction argument to show the main theorem. Theorem 1.1. For any continuous unimodal map of the unit interval into itself such that the critical point c is a maximum and f(x) > x for all x < c, the attractor of the unit interval is contained in [c 2, c 1 ] Proof. Since c is a maximum, the attractor must lie in [0, c 1 ]. Since f is increasing for all x < c, every such point must eventually fall into [c 0, c 1 ]. Since f has only one critical point, which is a maximum, the minimum value f takes on any interval must be an endpoint of that interval. Therefore [c 0, c 1 ] is mapped onto [c 2, c 1 ], and [c 2, c 1 ] is mapped into [c 2, c 1 ], since c 1 is greater than the unique fixed point p of f, and so c 2 < p, so c 2 < c 3. Typically, band-splitting occurs in the attractor when the critical sequence satisfies a particular ordering relationship in which the first 2 n iterates of c are split into pairs by the next 2 n iterates: Let x <> y <> <> z mean that x > y > > z or x < y < < z, and [x <> y] to mean the interval [x, y] or [y, x]. We say a sequence of real numbers satisfies the condition O n if: (1) c 2 < c 1 (2) for every i n, c 2 i 1 <> c 2 i 1 +2 i <> c 2 i+1 <> c 0 <> c 2 i (2) for every i n, c 2i +1 < c 2 i+1 +1 O n is a sufficient condition for band-splitting to occur up to the point of having 2 n bands: Theorem 1.2. If f is a continuous unimodal map on the unit interval, and C(f) satisfies O n, then k 2 n[c k <> c k+2 n] is closed under f. If, 2

3 in addition, sup{(f (x)) n 1 : x [b, a]} < c 1+2 n 1 +2 n c 1+2 n c 1+2 n 1 +2 n c 1+2 n c 1+2 n+1 c 1+2 n 1 +2 n, then the attractor of the unit interval is contained in k 2 n[c k <> c k+2 n]. Proof. The proof proceeds by induction on n. We have already shown the case n = 0 in the previous theorem. Now suppose the theorem holds for all i < n, and f satisfies O n.in other words, assuming that every orbit eventually falls into the 2 n 1 intervals, {[c k <> c k+2 n 1] : 1 k 2 n 1 }, we wish to show using O n that every orbit eventually falls into the 2 n intervals, {[c k <> c k+2 n] : k 2 n }. First note that (2) and (3) of O n imply that c 1 > c 1+2 n+1 > c 1+2 n > c 1+2 n 1 +2 n > c 1+2 n 1. Since f is monotonic on each of the other 2 n 1 1 intervals, we have recursively that for every k < 2 n 1, c k <> c k+2 n <> c k+2 n 1 +2 n <> c k+2 n 1. So the ordering of the iterates is such that each interval [c k <> c k+2 n 1] is therefore broken into three subintervals: [c k <> c k+2 n], [c k+2 n <> c k+2 n 1 +2 n], and [c k+2 n 1 +2 n <> c k+2n 1] so that the leftand right-subintervals (which are disjoint) contain the attractor, after the band-splitting has occurred. We first show that these outer subintervals are closed under f. We then show that orbits in the middle subintervals eventually leave for the outer intervals. First the left- an right-subintervals, {[c k <> c k+2 n] : 1 k 2 n }. Since only [c 2 n <> c 2 n+1] contains the critical point, f is monotonic the other 2 n 1 intervals, so f([c 1 <> c 2n +1]) = [c 1 <> c 2n +1], f([c 2 <> c 2n +2]) = [c 3 <> c 2n +3],. f([c 2 n 1 <> c 2 n+1 1]) = [c 2 n <> c 2 n+1]. Finally, since c 2 n+1 +1 [c 1 <> c 2 n +1] and c 0 [c 2 n O n, <> c 2 n+1] by f([c 2 n <> c 2 n+1]) = [c 1 <> c 2 n +1], completing the cycle. Note that this last map is done with a fold. Without this fold band-splitting could not occur on a continuous map. The middle intervals, [c k+2 n <> c k+2 n 1 +2 n] for k 2n 1, also map into each other, by monotonicity: 3

4 f([c 1+2 n <> c 1+2 n 1 +2 n]) = [c 2+2 n <> c 2+2 n 1 +2 n], f([c 2+2 n <> c 2+2 n 1 +2 n]) = [c 3+2 n <> c 3+2 n 1 +2 n],. f([c 2 n n <> c 2 n n 1 +2 n]) = [c 2 n 1 +2 n <> c 2 n+1] However, the last interval does not: f([c 2 n 1 +2 n <> c 2 n+1]) = [c 1+2 n+1 <> c 1+2 n 1 +2 n] [c 1+2 n <> c 1+2 n 1 +2 n]. Since c 2 n+1 +1 [c 1 <> c 2 n +1], part of the middle interval escapes to the outer pair of intervals with each iteration of f 2n 1. To show that every orbit in the middle interval eventually leaves, so that the attractor is contained by the 2 n+1 outer intervals and our induction is complete, we must take a closer look. Denote c 1+2 n 1 +2 n by a, c 1+2 n by b, and c 2n 1 1+2n+1 by c. Then f maps [b, a] onto [b, c] (or [a, b] onto [c, b], which is essentially the same) by sending a to b and b to c. Let s = a c, and λ = sup{(f (x)) n 1 : x [b, a]}. Then (at least) [b, b + s λ ] escapes after the first application of f 2n 1, [a s λ, a] after the second, etc., since we know a segment is 2 taken off alternating ends of [b, a] by each application of f 2n 1, and if less were taken off, then by the mean value theorem, some point would have to have (f (x)) n 1 > λ (or undefined), contrary to our assumptions. The total length removed is therefore at least n=1 s λ = s n 1 1/λ So, all orbits eventually fall into the outer pair of intervals if the total length s 1 1/λ > a b, i.e., if λ < c a. For the tent map, λ = 1 + ɛ, and the intervals [b, b s s λ ], [a + λ ], 2 etc. are exactly the pieces taken out by successive applications of f 2n 1. Equivalently, since the stretching factor is constant for the tent map, λ =, all orbits leave the middle interval [a, b] iff b c < b c c a, i.e., iff ɛ < c a 1. This is a critical value of ɛ where band-splitting occurs. If, ɛ = c a 1, then the total length removed is equal to a b, but since it is being taken from both sides, orbits all leave that middle interval except one: a s n=1 λ = b+ s 2n n=1 λ. This point is periodic of period 2 n 1, 2n 1 since it must map to a similar point in each of the other 2 n 1 other 4

5 middle intervals. In the next section we see exactly how the tent map satisfies O n, for any given n. 2 A deep analysis of the critical sequence of the tent map In this section we gain a full understanding of critical sequence of the tent map, f 1+ɛ, and similar unimodal maps. In particular we will see how the critical sequence of the tent map satisfies O n for all ɛ < some ɛ n, and the exact ordering of the iterates (resloving every <> into < or >). We do this by looking at the critical sequence in the limit as ɛ 0. What we do, in effect, is create a new dynamical system on a space of polynomials (or infinite sequences if we wish). While this system may not be realized in the actual tent map (for any parameter value), the important characteristics of the behavior of the tent map limits on those of this new system. Definition 2.1. Let {P n, n N} be the limiting sequence of polynomials in ɛ given by the n th iterate of 1 2 : P n = lim ɛ 0 f n 1+ɛ( 1 2 ). Note that P n has degree n, and that the constant term is always 1 2. The first sixteen P n are listed in Appendix A. Definition 2.2. Define a linear ordering by P n P n if lim ɛ 0 sgn(p n P n ) = 1. The coefficient of ɛ i in P n we denote Ci n, and the smallest power of ɛ (other than ɛ 0 ) whose coefficient is not zero we denote min(n). Note that for every n, there is some ν so that f1+ɛ( n 1 2 ) = P n for all ɛ < ν, and, for every n and n, if P n P n, then there is some ν so that f1+ɛ( n 1 2 ) < f 1+ɛ( n 1 2 ) for all ɛ < ν. When ɛ is sufficiently small, analysis shows that P n > 1 2 iff the sign of the least nonzero coefficient, Cmin(n) n, is positive. Thus P n can be computed by P n 1 using the relationship C n+1 i = sgn(c n min(n) )(Cn i + C n i 1). (1) 5

6 We first show that the P n satisfy a certain algebraic property, which we will use to deduce the ordering of the critical sequence. Lemma 2.3. For every i the sequence of coefficients Ci 1, C2 i,... is cyclic of length 2 i. Furthermore, min(2 i ) = i + 1, and Ci+1 2i > 0 (and therefore P 2 i ) iff i even. 1 2 Proof. We proceed by induction on i. It is easy to check that C n 1 = 1 2 if n is odd and C n 1 = 0 if n is even. Therefore C n 1 is cyclic of length 2, and min(2) = 2 and C 2 2 < 0. It is also easy to check that min(4) = 3 and C 4 3 > 0. Assume Ci n is cyclic of length 2i, for all i < N, and min(2 N ) = N+1, and CN+1 2N > 0 iff N is even. Since C2N N = C0 N (both are 0), and because CN n is only affected by the coefficients Cn i, for i < N, which are all cyclic of length 2 N, it follows that C 2N +1 N = CN 1, +2 C2N N = CN 2,.... So CN n is cyclic of length 2N also. Notice also that for all n < 2 N+1, the coefficients C0 n,... CN+1 n are not all 0. This is because for all n < 2 N+1, n = 2 k + 2 k+1 s for some k N and some s N (as can be easily shown by induction on N). So, by the previous paragraph and the inductive assumption, Ck+1 n = C2k k+1 = C2k min(2 k ) 0. We now show that min(2 N+1 ) = N + 2, and CN+2 2N+1 we first show that CN+1 2N+1 = 0. < 0. To do this, Using the basic relationship among coefficients, we have, for any n N: C 0 n+1 = 0, C 1 n+1 = sgn(c 0 min(0) )(C0 n + 0), C 2 n+1 = sgn(c 1 min(1) )(C1 n sgn(c 0 min(0) )(C0 n + 0)),. C 2n n+1 = sgn(c 2n 1 1 min(2 n 1) )(C2n n sgn(c 2n 2 min(2 n 2) ) (C1 n sgn(c 0 min(0) )(C0 n+ 0)) ) now let S = C 2n n+1 > 0. Then: C 2n +1 n+1 = sgn(c2n min(2 n ) )(C2n n + S), C 2n +2 n+1 = sgn(c2n min(2 n +1) )(C2n n sgn(c 2n min(2 n ) )(C2n n + S)),. C 2n +2 n n+1 = sgn(c 2n +2 n 1 +2 n 1 min(2 n +2 n 1) )(C2n n sgn(c 2n +2 n 2 min(2 n +2 n 2) ) +1 (C2n n 6

7 sgn(c 2n min(2 n ) )(C2n n + S)) ) Rearranging and simplifying, C 2n +2 n n+1 = sgn(c 2n +2 n 1 +2 n 1 min(2 n +2 n 1) )(C2n n sgn(c 2n +2 n 2 min(2 n +2 n 2) ) +1 (C2n n sgn(c 2n min(2 n ) )(C2n n )) ) + ( 1) 2n S 2 n+1 1 i=2 n sgn(ci min(i) ) Using the cyclicity of coefficients, and the fact that the coefficients C n 0,... C n n+1 are not all 0, so that sgn(c i min(i) ) = sgn(ci min(i) ) for 1 i < 2 n. C 2n +2 n n+1 = sgn(c 2n +2 n 1 +2 n 1 min(2 n +2 n 1) )(C2n n sgn(c 2n +2 n 2 min(2 n +2 n 2) ) +1 (C2n n sgn(c 2n min(2 n ) )(C2n n )) ) + S 2 n i=1 sgn(ci min(i) ) Since the first term is identical to S, we have, for all n N, C 2n+1 n+1 = S + S 2 n i=1 sgn(c i min(i) ) (2) But for all n < N, we assumed C 2n+1 n+1 = 0, so 2 n i=1 sgn(ci min(i) ) = 1. By repeated use of this fact, we get: 2 N i=1 sgn(ci min(i) ) = sgn(c 2N min(2 N ) )( 2 N 1 i=2 N 1 sgn(ci min(i) )) ( i=2 sgn(ci min(i) ))sgn(c1 min(1) ) = sgn(c 2N min(2 N ) )( 1)N 1 Therefore, CN+1 2N+1 = S+S 2 N i=1 sgn(ci min(i) ) = S+Ssgn(C2N min(2 N ) )( 1)N 1, and since we assumed Cmin(2 2N N ) = C2N N+1 > 0 iff N is even, sgn(c2n min(2 N ) )( 1)N 1 = 1, for either N even or odd. So CN+1 2N+1 = 0. Finally, since C 2N+1 1 N+1 and C 2N+1 1 N+2 have the same sign, CN+2 2N+1 = sgn(c 2N+1 1 )(C2N+1 1 min(2 N+1 1) N+2 + C 2N+1 1 N+1 ) < 0 iff N is even. We now use this lemma to deduce the ordering of the critical sequence: Theorem 2.4. For every n, there is some ɛ n > 0 such that the critical sequence C(f 1+ɛ ) satisfies O n wherever 0 < ɛ < ɛ n. 7

8 Proof. We show that {P n, n N} satisfies O n with the ordering. Since O n involves only finitely many order relations, and each is satisfied whenever ɛ is less than some positive value, we can take the minimum of these values for ɛ n to conclude that {c n, n N} satisfies O n wherever 0 < ɛ < ɛ n. Again we proceed by induction on n. Clearly P P 4 P 3 P 1, so O 0 holds. Now suppose the theorem holds for all n up to some even N. Since we assume O N 1 holds, we have by Theorem 1.2 in the previous section that the attractor is contained in k 2 N 1[P k P k+2 N 1], and these intervals map one into the other, in a cycle. We also know the last of these intervals is [P 2 N 1, P 2 N ], by the lemma. So P 2 N +2 N 1 and P 2 N+1 are in the interval [P 2 N 1, P 2N ], since they are the images of P 2 N by 2 N 1 k applications of f 1+ɛ, for some k N. We must only show that P 2 N +2 N 1 P 2 N By the lemma, Cmin(2 2N 1 N 1 ) = C2N 1 N < 0. Since the coefficients Ci 1, C2 i,... are cyclic of length 2i, for every i N, they are all also cyclic of length 2 N. Therefore, C 2N 1 i = C 2N +2 N 1 i C 2N +2 N 1 min(2 N +2 N 1 ) = C2N +2 N 1 N < 0. So P 2N +2 N for any i N. So Also by the lemma, Cmin(2 2N+1 N+1 ) = C2N+1 N+1 < 0, but since this coefficient is of greater degree, lim ɛ 0 sgn(p 2 N+1 P 2N +2 N 1) = +2 N 1 sgn( C2N N > 0, so P 2N +2 N 1 P 2 N+1. In the case N is odd, replace every by, and vice versa. Question 2.5. Can O n be achieved with a different ordering of intervals? Acknowledgements The author would like to thank James Burgmeier, Jeff Dinitz, and Jianke Yang for their encouragement and guidance, as well as Ivan Zaigrallin for pointing out the possibility of the isolated points. 8

9 3 Appendix: the first sixteen iterates of 1 2, as polynomials in ɛ Figure 1: the first 16 P n References [1] H. Shigematsu, H. Mori, T. Yoshida, and H. Okamoto. Analytic study of power spectra of the tent maps near band-splitting transitions. Journal of Statistical Physics, 30: , /BF [2] T. Yoshida, H. Mori, and H. Shigematsu. Analytic study of chaos of the tent map: Band structures, power spectra, and critical behaviors. Journal of Statistical Physics, 31: , /BF