Superconvergence of period doubling cascade in trapezoid maps Its Rigorous proof and superconvergence of period 0 0 doubling cascade starting from a period p solution 0 2 n Uezu a Tatsuya J 2 1 Graduate School of Human Culture, ] Nara Women’s University, Nara 630-8506 D C . n i l n [ 1 v 7 1 0 1 0 0 0 / n i l n : v i X r a 1 Abstract In thesymmetric andthe asymmetric trapezoidmaps, asa slope a of thetrapezoid is increased, the period doubling cascade occurs and the symbolic sequence of periodic points is the Metropolis-Stein-Stein sequence R m and the convergence of the onset ∗ point a of the period 2m solution to the accumulation point a is exponentially fast. m c In the previous paper, we proved these results. In this paper, we give the detailed description of the proof on the results. Rigor- ously, we show that ba 2mγ ζm ǫ = −c − (1+h ), m m G (a ) ′ c δ = γ( 1∞)m/3(a γ2/3)2m(1+l ), m − c m lim h = 0, m m →∞ lim l = 0, m m →∞ where ǫ ≡ a −a ,δ ≡ ǫm ǫm+1 , b and γ are the smaller size of the trapezoid and m c m m ǫm+−1 ǫm+2 − the ratio of its two slopes, respectively. γ = 1 corresponds to the symmetric trapezoid. Further, we study the period doubling cascade starting from period p(≥ 3) solution. We show ǫ ∝ (γ2ap ) 2mγζm, m p,c − δ ≃ γ( 1)m−1/3(ap γ4/3)2m, m − p,c where a is the accumulation point of the onset of the period p×2m solution. p,c 2 §1. Introduction It is well known that in a class of one-dimensional map, as a parameter is changed, the period doubling bifurcation cascades, and there exist several universal properties1). One of the universal quantities is the so called Feigenbaum constant δ and it depends only on the exponent z characterizing the behavior of the map around the critical point. About a decade ago, there had been controversy on the z → ∞ limit of δ(z). As a typical example, the map x = f(x ) = 1−a|x |z has been extensively studied and two different n+1 n n values were conjectured for this limit, one is finite and the other is infinity2),3). This problem was solved by J. P. Eckmann and H. Epstein and they proved that the limit is finite4). In the previous paper5), instead of considering a map with finite z and taking z → ∞, we investigated a map with z = ∞. As such maps, we treated the symmetric and asymmetric trapezoid maps. 1 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 Fig. 1. Trapezoid map T (x) (a,b) The symmetric trapezoid map x = T (x ) defined in [0,1] is given by n+1 (a,b) n ax for 0 ≤ x ≤ (1−b)/2,  T(a,b)(x) =  a(1−b)/2 for (1−b)/2 ≤ x ≤ (1+b)/2 , (1.1) a(1−x) for (1+b)/2 ≤ x ≤ 1 ,  where 0 < b < 1 and a > 0. See Fig.1. As a is increased from 1 with fixed b, the successive period doubling bifurcations take place. We assign the symbol L,C and R to an orbit x according to the following rule; L for x ∈ i i [0,(1−b)/2] ≡ I , C for x ∈ ((1−b)/2,(1+b)/2) ≡ I , R for x ∈ [(1+b)/2,1] ≡ I . L i C i R We denote this correspondence by H(x ). Further, we define the conjugate of R or L as i R¯ = L or L¯ = R, respectively. Then, in the previous paper5) we proved that as a slope a of the trapezoid map is increased, the period doubling bifurcation cascades and the symbolic sequence is the Metropolis-Stein-Stein sequence R m as usual. And, defining a as the onset ∗ m 3 of a 2m-cycle, we calculated δ ≡ am+1 am and proved that δ scales as m am+2 −am+1 m − δ ≃ a2m m c wherea istheaccumulation pointoftheperioddoublingcascade. Wecalledthisphenomena c the super-convergent period doubling cascade. Further, we investigated the asymmetric trapezoid map x = A (x ) defined in [0,1], n+1 (a,b,γ) n ax for 0 ≤ x ≤ α,  A(a,b,γ)(x) =  αa for α ≤ x ≤ β , (1.2) γa(1−x) for β ≤ x ≤ 1 ,  where γ is the ratio of two slopes of the trapezoid, α = γ (1 − b) and β = α + b = b+γ. 1+γ 1+γ When γ = 1, A (x) is reduced to T (x). In the asymmetric map, when γ and b are (a,b,γ) (a,b) fixed and a is increased, we proved similar results as in the symmetric map, in particular, as the scaling of δ we obtained m δ ≃ γ( 1)m/3(a γ2/3)2m. m − c Finally, we gave approximate expressions for the accumulation point as functions of b which is the length of the smaller side of the trapezoid in both symmetric and asymmetric cases. In this paper, we give the detailed and complete description of the proofs given in the previous paper. New results in this paper are the scaling relations for the period doubling bifurcation starting with a period p(≥ 3) solution which appears by a tangent bifurcation. In the following section, we treat the symmetric trapezoid map, and then in the section 3, we treat the asymmetric map. In sections 4 and 5, we study the period doubling bifurcation for period p(≥ 3) solution in symmetric and asymmetric cases, respectively. We give summary and discussion in the last section. §2. The symmetric case We consider Ta(x) ≡ T (x). For brevity we define a = (a,b). Let x be the maximum (a,b) M value of Ta(x), i.e., xM ≡ a1−2b. Let x0,1 be the non-zero fixed point of Ta(x). Then the following lemma is easily proved. lemma 1 For 1 < a < a1, Ta(x) has the stable fixed point x0,1 = xM, which satisfies 1−2b < xM < 1+b. a is defined by the equation x = 1+b, that is a = 1+b. 2 1 M 2 1 1 b − 4 For a > a1, Ta(x) has the unstable fixed point x0,1 = x∗ ≡ a+a1 and x∗ > 1+2b. At a = a ,x = x . 1 M ∗ See Fig.2. 11 00..88 00..66 00..44 00..22 00 00 00..22 00..44 00..66 00..88 11 Fig. 2. Trapezoid map T (x) for a= a with b = 0.1 (a,b) 1 In the region a > a , by iterating Ta(x) twice and rescaling x such that the domain 1 becomes [0,1], we obtain a trapezoid map with different parameter a(1),a(1) = (a(1),b(1)). 1 11 0.8 00..88 0.6 00..66 0.4 00..44 0.2 00..22 0 00 0 0.2 0.4 0.6 0.8 1 00 00..22 00..44 00..66 00..88 11 Fig. 3. Trapezoid map T2 (x) for a = a Fig. 4. Trapezoid map T2 (x) for a < a < (a,b) 1 (a,b) 1 with b = 0.1. a with b = 0.1. 2 See Fig.3 and 4. In fact, we obtain the following lemma. lemma 2 When Ta(x) has the non-zero unstable fixed point x∗ = a+a1(> 12), that is, for a > a1, by rescaling the coordinate x as x(1) = 2xx∗∗−x1 and restricting x to the interval [1−x∗,x∗], Ta2(x) − is transformed to Ta(1)(x(1)) which is defined for x(1) in [0,1]. Here a(1) ≡ ϕ(a) ≡ (a2,u(a)b) and u(a) = a+1. a 1 − 5 The proof is straightforward. ϕ(a) is defined as long as a 6= 1. In this paper, we restrict ourselves to the case of a > 1. Now, we define the m-th iteration of ϕ. That is, a(m) ≡ (a(m),b(m)) ≡ ϕm(a) = (a2m,u (a)b), (2.1) m m 1 m 1 a2l +1 u (a) ≡ − u(a2l) = − . m Y Y a2l −1 l=0 l=0 These are defined as long as a 6= 1. For a > 1, as functions of a, a(m) is a continuous strictly increasing function andb(m) is a continuous strictly decreasing function. lim b(m) = b < 1 a →∞ for any m > 0. Let us prove the following lemma. lemma 3 For any positive integer m, there exists the unique value of a = a such that m b(m)(a ) = 1. (2.2) m {a } is an increasing sequence, i.e., 1 < a < a < ···. Further, the relation b < b(m) < 1 m ∞m=1 1 2 holds for a < a for m ≥ 1. m Proof Let us consider the case of m = 1. b(1)(a) = 1 has the unique solution a = 1+b > 1, and this is a defined in lemma 1. 1 b 1 − Next, let us assume that b(m)(a ) = 1 and a > 1 for m ≥ 1. Since b(m+1)(a) = m m a(m)+1b(m)(a), b(m+1)(a ) > 1 follows. Thus, there exists the unique value of a = a (> a ) a(m) 1 m m+1 m − such that b(m+1)(a ) = 1. m+1 Therefore, fromthe mathematical induction, b(m)(a) = 1 hasa unique solution a for any m positive integer m and {a } is an increasing sequence. The inequality b < b(m) < 1 for m ∞m=1 a < a is immediately follows from the fact that the function b(m)(a) is strictly decreasing. m Q.E.D. For positive integer m, we define Tϕm(a)(x(m)) for a > am from Tϕm−1(a)(x(m−1)) suc- cessively by the same procedure as in the lemma2. We define xM(m) ≡ a(m)1−b2(m), which is the maximum value of Tϕm(a)(x(m)). Further, we define x(m) ≡ a(m) . x(0) = x and x(0) = x . ∗ a(m)+1 M M ∗ ∗ lemma 4 For any non-negative integer m and for a < a there exists a unique non-zero fixed point m for Tϕm(a)(x(m)) which is defined in [0,1]. For am < a < am+1, 1−b2(m) < xM(m) < 1+b2(m) and 6 xM(m) is the stable fixed point of Tϕm(a)(x(m)). At a = am+1, xM(m) = x(m)∗. For am+1 < a, the non-zero fixed point for Tϕm(a)(x(m)) is x(m)∗ and unstable, and x(m)∗ > 1+b2(m). Here, we define x(0) ≡ x,a ≡ 1,a(0) ≡ a and b(0) ≡ b. 0 Proof First of all, we notice that for m ≥ 1 Tϕm(a)(x(m)) is really a trapezoid map because from lemma 3 b < b(m) < 1 for a < a and for m ≥ 1. m For the case m = 0, the statement follows from lemma 1. Let us assume that the statement holds for m ≥ 0. Since x(m) = a(m) is the unstable ∗ a(m)+1 fixed point of Ta(m)(x(m)) for am+1 < a, from lemma 2, taking Ta2(m)(x(m)) and rescaling x(m) as x(m+1) = x2(mx()m∗−)∗x(m1), we obtain Ta(m+1)(x(m+1)) which is defined in [0,1]. The following − equivalence relations are easily proved for m ≥ 0, 1+b(m) 1+b(m) b(m+1)(a) = 1 ⇐⇒ x(m) = ⇐⇒ a(m) = . (2.3) M 2 1−b(m) Let us consider x(m+1) = a(m+1)1 b(m+1). From the relation (5), since b(m+2)(a ) = 1 it M − 2 m+2 follows that x(m+1) = 1+b(m+1) at a = a . Since x(m+1) is the strictly increasing function M 2 m+2 M w.r.t. a, we obtain 0 < 1 b(m+1) < x(m+1) < 1+b(m+1) for a < a < a . This implies − 2 M 2 m+1 m+2 xM(m+1) is the unique non-zero fixed point for Tϕm+1(a)(x(m+1)) and stable. It is easily shown that for a < a, x(m+1) is no more fixed point but x(m+1) = a(m+1) becomes unstable m+2 M ∗ a(m+1)+1 fixed point and x(m+1) > 1+b(m+1). At a = a ,a(m+1) = x(m+1) holds. Thus, by the ∗ 2 m+2 M ∗ mathematical induction, the proof completes. lemma 5 The unique non-zero fixed point of Tϕm(a)(x(m)) in lemma 4 is a member of a periodic cycle with prime period 2m of Ta(x). Proof The case of m = 0, it is trivial. Now, let us fix m(≥ 0) and assume that for 0 ≤ k ≤ m the unique non-zero fixed point of Tϕk(a)(x(k)) which exists for ak < a is the member of the cycle with prime period 2k. Thus, from lemma 4 for a < a the unique non-zero m+1 fixed point of Tϕm+1(a)(x(m+1)) does not correspond to any periodic points with period less than 2m+1 because it is stable for a < a < a . Thus, the non-zero fixed point of m+1 m+2 Tϕm+1(a)(x(m+1)) is the member of the cycle with prime period 2m+1. This completes the proof. Thus, we obtain the following theorem. 7 Theorem 1 x = 0 is period 1 solution of Ta(x) for any a > 0. Except for this, for any non- negative integer m, for a < a < a there is the unique cycle with prime period 2k for m m+1 k = 0,1,···,m and no other periodic points exist. The periodic solution with prime period 2m is stable for a < a < a and unstable for a < a. At a = a , the period 2m m m+1 m+1 m+1 cycle and the period 2m+1 cycle coincide. Proof For the case of m = 0, the statement is trivial. Let us assume that the statement holds for 0,1,···,m. Then we only have to prove that for a < a < a , except for the period m+1 m+2 2k cycles for k = 1 ∼ m+1, there is no other cycle. Let us consider any initial point x in (0, 1) which does not corresponds to the unstable cycles of period 2k,(k = 0 ∼ m). Then, the orbit Tai (x) finally enters the domain of the map Tϕm(a)(x(m)) if i is appropriately chosen. Thus, since any point except for x(m) = 0 and 1 converges to the fixed point of Tϕm(a)(x(m)) for a < a < a , the nonexistence of other periodic points for a < a < a follows. m+1 m+2 m+1 m+2 The latter part of the theorem follows from lemma 4 immediately. From the above arguments we conclude that as a is increased from 1, the period doubling bifurcation cascades, and at a = a the periodic solution with prime period 2m appears for m m ≥ 0. Now, let us investigate the upper bound of the sequence {a }. x(m)(a) = a(m)1 b(m) is m M −2 (m) defined for a > 1 and the strictly increasing function w.r.t a. For any m > 0,x (a ) = 0 M m (m) and lim x (a) = ∞ follows. Thus, for m > 0 we define a (m) as the unique solution a M M →∞ of the equation x(m)(a) = 1. (2.4) M For m = 0, we define a (0) = 2 . We abbreviate this as a . Then, we obtain the following M 1 b M − lemma. lemma 6 1 < a < a < ··· < a < a < ··· (2.5) 1 2 m m+1 < a (m+1) < a (m)··· < a (2) < a (1) < a . M M M M M Proof (m) (m+1) For m > 0, since x (a ) = 0,a < a (m) follows. For m ≥ 0, x (a) is rewritten as M m m M M 1 1+b(m) x(m+1)(a) = {a(m)}2 (x(m)(a)− ). (2.6) M a(m) −1 M 2 8 Then, x(m+1)(a (m)) = a(m) > 1. Therefore, a (m+ 1) < a (m) for m ≥ 0. Thus, we M M a(m) 1 M M − obtain a < a < a (m+1) < a (m) for m ≥ 0. Q.E.D. m m+1 M M From this, a ≡ lim a ≤ lim a (m) < a (2.7) c m M M m m →∞ →∞ follows. Therefore a is finite. c Now, let us investigate the symbolic sequence. The sequence of R and L for the period 2m solution becomes Metropolis-Stein-Stein(MSS) sequence. Let us prove this. For m ≥ 0, we define the onset of the 2n-cycle for Tϕm(a)(x(m)) as an(m) and the i-th orbit of the period 2n cycle as x(m). a(0) = a . For m = 0, we often omit the superscript n,i n n (0). As x we set the largest x value among 2m members of the periodic cycle. That is, m,1 x = x = a(1−b)/2 when the cycle is stable. Further, form ≥ 0 we divide the coordinate m,1 M space [0,1] of x(m) into the three intervals as I(m) ≡ [0, 1 b(m)],I(m) ≡ (1 b(m), 1+b(m)) and L −2 C −2 2 I(m) ≡ [1+b(m),1]. R 2 lemma 7 For any positive integer m, when a < a, m 1. IC(m) ∋ x(m) is equivalent to IC(m−1) ∋ x(m−1), 2. if IL(m) ∋ x(m) then IR(m−1) ∋ x(m−1), 3. if IR(m) ∋ x(m) then IL(m−1) ∋ x(m−1), wherex(m) andx(m−1) arerelatedbythecoordinationtransformationusedtodefineTϕm(a)(x(m)) from Tϕm−1(a)(x(m−1)). proof Fora < a,x(m) = 0, 1 b(m), 1+b(m) and1correspondtox(m 1) = x(m 1) , 1+b(m−1), 1 b(m−1) m −2 2 − − ∗ 2 − 2 and 1−x(m 1) , respectively. Since this correspondence is linear, the statements hold. − ∗ lemma 8 The fixed point x0(m,1) of Tϕm(a)(x(m)) corresponds to xm,2m. proof Let x be the orbit of 2m solution corresponding to x(m). From lemma 4, for a < a < m,i 0,1 m (m) (m) (m) (m) a ,x = x and I ∋ x . Then, from lemma 7, I ∋ x follows. Therefore, x m+1 0,1 M C 0,1 C m,i m,i is mapped to x by Ta, which is x . Thus, i = 2m. M m,1 lemma 9 9 For any positive integer m, symbols H(x ) for the number of 2m cycle satisfy the m,i followings. 1. H(x ) does not change for a ≤ a and for i = 1,···,2m −1. m,i m 2. For even m, L for a = a ,  m H(xm,2m) =  C for am < a < am+1 , R for a ≤ a .  m+1 and for odd m, R for a = a ,  m H(xm,2m) =  C for am < a < am+1 , L for a ≤ a .  m+1 3. {H(x )} is the MSS sequence R m for a ≤ a. m,i ∗ ∗ m+1 proof Let us consider the case of m = 1. At a = a , x = 1+b ∈ I . Then, at a = a ,H(x ) = 1 0,1 2 R 1 0,1 H(x1,1) = H(x1,2) = R. For a1 < a < a2, x(01,1) = x(M1) = a(1)(12−b(1)) and for a2 ≤ a,x(01,1) = x(1)∗. (1) From lemma 8, x corresponds to x . Thus, from lemma 7, H(x ) is C for a < a < a , 0,1 1,2 1,2 1 2 and is L for a ≤ a. On the other hand, for a ≤ a, since x ≥ x = a then H(x ) = R. 2 1 1,1 0,1 a+1 1,1 Therefore, H(x )H(x ) = RL for a ≤ a. Therefore, for m = 1, the lemma holds. 1,1 1,2 2 Next, we assume that the lemma holds for the case of m(≥ 1). At a = a , x m+1 m+1,i and x emerge from x (i = 1 ∼ 2m). Then, at a = a , H(x ) = H(x ) = m+1,i+2m m,i m+1 m,i m+1,i H(x ) for i = 1,···,2m. These are L or R. Let us assume that at some value of m+1,i+2m a(> a ), H(x ) becomes C. Then, x is mapped to x by Ta. Thus, in this case, m+1 m+1,i m+1,i M a < a and x = x = x follow. So, i should be 2m+1. Thus, x (i = 1 ∼ m+2 m+1,i+1 M m+1,1 m+1,i 2m+1 −1) does not change its symbol for a ≤ a. From lemma 8 x corresponds m+1 m+1,2m+1 (m+1) (m+1) to x = x for a ≤ a ≤ a . Thus, H(x ) = C for a < a < a . 0,1 M m+1 m+2 m+1,2m+1 m+1 m+2 For a ≤ a, x(m+1) = x(m+1) ∈ I(m+1). Thus, from lemma 7, for a ≤ a, m+2 0,1 ∗ R m+2 L for odd m+1,  H(x ) = m+1,2m+1  R for even m+1.  Therefore, the statements 1 and 2 hold for m+1. At a = a ,H(x ) = H(x ). Then, from the assumption, m+1 m+1,2m+1 m,2m R for odd m+1,  H(x ) = m+1,2m+1  L for even m+1.  ∗ Strictly speaking,the lastsymbolofthe sequenceinourdefinitionisRorLandis differentfromthat in the MSS sequence, C 10

Superconvergence of period doubling cascade in trapezoid maps

Similar Superconvergence of period doubling cascade in trapezoid maps