MATH 614 Dynamical Systems and Chaos Lecture 11: Maps of the circle.

Transcription

1 MATH 614 Dynamical Systems and Chaos Lecture 11: Maps of the circle.

2 Circle S 1. S 1 = {(x,y) R 2 : x 2 + y 2 = 1} S 1 = {z C : z = 1} T 1 = R/Z T 1 = R/2πZ α : S 1 [0,2π), angular coordinate α : S 1 R/2πZ R (multi-valued function) α 1

3 φ : R S 1, φ(x) = (cosx,sinx), S 1 R 2. φ(x) = e ix = cosx +i sinx, S 1 C. φ: wrapping map φ(x +2πk) = φ(x), k Z. α R is an angular coordinate of x S 1 if and only if φ(α) = x. For any arc γ S 1 there exists a continuous branch α : γ R of the angular coordinate. If α 1 : γ R and α 2 : γ R are two continuous branches then α 1 α 2 is a constant 2πk, k Z.

4 Examples of continuous branches: α : S 1 \{1} (0,2π), α : S 1 \{ 1} ( π,π). α 1

5 f : S 1 S 1, continuous map Example. D : z z 2 (doubling map) in angular coordinates: α 2α(mod2π). The doubling map: smooth, 2-to-1, no critical points. Theorem The doubling map is chaotic.

6 Orientation-preserving and orientation-reversing The real line R has two orientations. For maps of an interval: orientation-preserving = monotone increasing, orientation-reversing = monotone decreasing. The circle S 1 also has two orientations (clockwise and counterclockwise). Given a map f : S 1 S 1, we say that a map F : R R is a lift of f if f φ = φ F, where φ : R S 1 is the wrapping map. Any continuous map f : S 1 S 1 admits a continuous lift F. The lift satisfies F(x +2π) F(x) = 2πk for some k Z and all x R. If F 0 is another continuous lift of f, then F F 0 is a constant function. A continuous map f : S 1 S 1 is orientation-preserving (resp., orientation-reversing) if so is the continuous lift of f.

7 Maps of the circle f : S 1 S 1, f an orientation-preserving homeomorphism.

8 Rotations of the circle R ω : S 1 S 1, rotation by angle ω R. R ω (z) = e iω z, complex coordinate z; R ω (α) = α+ω (mod2π), angular coordinate α. Each R ω is an orientation-preserving diffeomorphism; each R ω is an isometry; each R ω preserves Lebesgue measure on S 1. R ω is a one-parameter family of maps. R ω is a transformation group. Indeed, R ω1 R ω2 = R ω1 +ω 2, R 1 ω = R ω. It follows that R n ω = R nω, n = 1,2,... Also, R 0 = id and R ω+2πk = R ω, k Z.

9 ω 1

10 An angle ω is called rational if ω = rπ, r Q. Otherwise ω is an irrational angle. If ω is a rational angle then R ω is a periodic map. All points of S 1 are periodic of the same period. If ω = 2πm/n, where m and n are coprime integers, n > 0, then the period of R ω is n. If ω is irrational then R ω has no periodic points. If ω is irrational then R ω is minimal: each orbit is dense in S 1. If ω is irrational then each orbit of R ω is uniformly distributed in S 1.

11 Minimality Theorem Suppose ω is an irrational angle. Then the rotation R ω is minimal: all orbits of R ω are dense in S 1. Proof: Take an arc γ S 1. Then Rω(γ), n n 1, is an arc of the same length as γ. Since S 1 has finite length, the arcs γ,r ω (γ),rω 2 (γ),... cannot all be disjoint. Hence n(γ) Rm ω (γ) for some 0 n < m. But Rω(γ) R n ω m (γ) = Rω(γ n Rω m n (γ)) so γ Rω m n (γ). Thus for any ε > 0 there exists k 1 such that R k ω = R kω is the rotation by an angle ω, ω < ε. Note that ω 0 since ω is an irrational angle. Pick any x S 1. Let n = 2π/ ω. Then points x,r kω (x),r 2 kω (x),...,rn kω (x) divide S 1 into arcs of length < ε.

12 Uniform distribution Let T :S 1 S 1 be a homeomorphism and x S 1. Consider the orbit x,t(x),t 2 (x),...,t n (x),... Let γ S 1 be an arc. By N(x,γ;n) denote the number of integers k {0,1,...,n 1} such that T k (x) γ. The orbit of x is uniformly distributed in S 1 if N(x,γ 1 ;n) lim n N(x,γ 2 ;n) = 1 for any two arcs γ 1 and γ 2 of the same length.

13 An equivalent condition: N(x,γ 1 ;n) lim n N(x,γ 2 ;n) = length(γ 1) length(γ 2 ) for any arcs γ 1 and γ 2. Another equivalent condition: for any arc γ. N(x,γ;n) lim n n = length(γ) 2π Theorem Suppose ω is an irrational angle. Then all orbits of the rotation R ω are uniformly distributed in S 1.

14 Fractional linear transformations of S 1 A fractional linear transformation of the complex plane C is given by f(z) = az +b, a,b,c,d C. cz +d How can we tell if f(s 1 ) = S 1? This happens in the case f(z) = e iψ z z 0 z 0 z 1, where z 0 1 and ψ R. Indeed, if z S 1 then z = e iα, z 0 = re iβ, z z 0 = e iα re iβ = e iα (1 re iβ e iα ), z 0 z 1 = re iβ e iα 1 so that f(z) S 1.

15 Fractional linear transformations of S 1 S 1 = {z C : z = 1}, f : S 1 S 1, f(z) = e iω z z 0 z 0 z 1, where z C, z 0 1 and ω R. Fractional linear transformations of S 1 form a group. Rotations of the circle form a subgroup (z 0 = 0). f is orientation-preserving if z 0 < 1 and orientation-reversing if z 0 > 1.

16 f(g(z)) = f(z) = az +b cz +d, g(z) = a z +b c z +d, z+b aa c z+d +b c a z+b c z+d +d = (aa +bc )z az +b cz +d +ab +bd (ca +dc )z +cb +dd, ( ) a b. c d Composition of fractional linear transformations corresponds to matrix multiplication.

17 f(z) = e iω z z 0 z 0 z 1, ( e e iω/2 iω/2 z 0 e iω/2 z 0 e iω/2 e iω/2 ). det = 1 z 0 2, Tr = e iω/2 +e iω/2 = 2cos(ω/2). Characteristic equation: λ 2 2cos(ω/2)λ+1 z 0 2 = 0. Discriminant: D = cos 2 (ω/2) 1+ z 0 2 = z 0 2 sin 2 (ω/2). If D < 0 then f is elliptic. If D = 0 then f is parabolic. If D > 0 then f is hyperbolic.

18 Theorem (i) If f is elliptic then f has no fixed points and is topologically conjugate to a rotation. (ii) If f is parabolic then f has a unique fixed point, which is neutral. Besides, the fixed point is weakly semi-attracting and semi-repelling. (iii) If f is hyperbolic then f has two fixed points; one is attracting, the other is repelling. Example. Given ω (0, π), the one-parameter family iω z r f r (z) = e 1 rz, 0 r < 1 undergoes a saddle-node bifurcation at r = r 0 = sin(ω/2).

19 Rotation number Suppose T : S 1 S 1 is an orientation-preserving homeomorphism. Is T topologically conjugate to a rotation R ω? Assume this is so, then how can we find ω? For any x S 1 let ω(t,x) denote the length of the shortest arc that goes from x to T(x) in the counterclockwise direction. If T is a rotation then ω(t,x) is a constant.

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21 Consider the average A n (T,x) = 1 n 1 ω(t,t k (x)). n Theorem The limit k=0 lim A n(t,x) n exists for any x S 1 and does not depend on x. The rotation number of T is ρ(t) = 1 2π lim n A n(t,x).

22 For any T, 0 ρ(t) < 1. ρ(r ω ) = ω/(2π)(mod1). Proposition 1 If T 1 and T 2 are topologically conjugate then ρ(t 1 ) = ±ρ(t 2 )(mod1). Proposition 2 ρ(t n ) = nρ(t)(mod1). Proposition 3 If T has a fixed point then ρ(t) = 0. Proposition 4 If T has a periodic point of period n then ρ(t) = k/n, where k Z, 0 k < n.

23 Proposition 5 If ρ(t) = 0 then T has a fixed point. Proof: Suppose T has no fixed points. Then 0 < ω(t,x) < 2π for any x S 1. Since ω(t,x) is a continuous function of x, there exists ε > 0 such that ε ω(t,x) 2π ε for any x S 1. Then ε A n (T,x) 2π ε for all x S 1 and n = 1,2,... It follows that ε/(2π) ρ(t) 1 ε/(2π). Proposition 6 If ρ(t) is rational then T has a periodic point.

24 Theorem (Denjoy) If T is C 2 smooth and the rotation number ρ(t) is irrational, then T is topologically conjugate to a rotation of the circle. Example (Denjoy). There exists C 1 smooth diffeomorphism T of S 1 such that ρ(t) is irrational but T is not minimal.