NOTES AND ERRATA FOR RECURRENCE AND TOPOLOGY. (τ + arg z) + 1

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1 NOTES AND ERRATA FOR RECURRENCE AND TOPOLOGY JOHN M. ALONGI p.11 Replace ρ(t, z) = ρ(t, z) = p.11 Replace sin 2 z e τ ] 2 (τ + arg z) z (e τ dτ. sin 2 (τ + arg z) + 1 z 2 z e τ ] z (e τ dτ. t (t, z) = sin2 (t + arg z) + 1 t (t, z) = sin2 (t + arg z) + 1 z 2 z e t ] z (e t > z e t ] z (e t > p.11 Replace 1 z 2 sin 2 z e τ (τ + arg z) z 2 (e τ 1 z 2 sin 2 (τ + arg z) + 1 z 2 z e τ 1 + z 2 (e τ p.12 Replace Example by the following. (The previous Example was correct, but its purpose is to provide a reparametrization in Example which contains an error.) Example Let φ t be a smooth flow on a manifold M. Let β : M R be a smooth bounded positive function. Define ρ : R M R so that for each x M the function ρ(, x) : R R is the solution of the initial value problem u = β(φ u (x)) u() =. ] 2 ] 2 Date: September 5,

2 2 JOHN M. ALONGI Let x M. Because φ t and β are smooth, the function ρ(, x) : R R is smooth. In particular, t (t, x) = β(φρ(t,x) (x)) >. Thus, ρ(, x) : R R is strictly increasing. Since the function ρ(, x) : R R is increasing, either the limit lim ρ(t, x) t exists, or ρ(t, x) as t. We claim that ρ(t, x) as t. By means of contradiction, assume that there exists a real number S so that lim ρ(t, x) = S. t The continuity of ρ, φ and β implies that lim t (t, x) = lim t t β(φρ(t,x) (x)) = β(φ S (x)) >. However, because β is bounded and ρ(, x) : R R is smooth and increasing, lim (t, x) =. t t This is a contradiction. Therefore, ρ(t, x) as t. Similarly, ρ(t, x) as t. Since the function ρ(, x) : R R is continuous, strictly increasing, and ρ(t, x) ± as t ±, the function ρ(, x) : R R is surjective. Therefore, ρ is a reparametrization. p.13 Replace p.14 Replace ρ(t, z) = ρ(t, z) = sin 2 z e τ ] 2 (τ + arg z) z (e τ dτ sin 2 (τ + arg z) + 1 z 2 z e τ ] z (e τ dτ

3 p.14 Replace p.14 Replace NOTES AND ERRATA FOR RECURRENCE AND TOPOLOGY 3 ρ(t, (r, θ)) = ρ(t, (r, θ)) = sin 2 re τ ] 2 (τ + θ) r(e τ dτ sin 2 (τ + θ) + 1 r 2 re τ ] r(e τ dτ ( r(1 r) 1 ( r(1 r) 1 p.14 Replace ) (sin 2 θ + 1 r 2 ) ) (sin 2 θ + 1 r 4 ) ( r(1 r)(sin 2 θ + 1 r 2 sin 2 θ + 1 r 4 ( r(1 r)(sin 2 θ + 1 r 4 ) sin 2 θ + 1 r 4 pp Replace Example by the following. Example Let φ t be the flow of a smooth vector field f on a manifold M. Let β : M R be a smooth bounded positive function. We will show that scaling the vector field f by the realvalued function β generates a flow which is topologically equivalent to φ t. Define ρ : R M R so that ρ(t, x) is the solution of the initial value problem u = β(φ u (x)) u() = for each x M. By Example the function ρ is a reparametrization. By the Chain Rule, d dt φρ(t,x) (x) = f(φ ρ(t,x) (x)) (t, x) t = f(φ ρ(t,x) (x))β(φ ρ(t,x) (x)) = (βf)(φ ρ(t,x) (x)) for all x M. If ψ t is the flow of the vector field βf on M, then d dt ψt (x) = (βf)(ψ t (x)) ) )

4 4 JOHN M. ALONGI for all x M. Thus, for each x M, the functions ψ t (x) and φ ρ(t,x) (x) are both solutions of the differential equation ẏ = (βf)(y). Since ψ (x) = x = φ (x) = φ ρ(,x) (x) for all x M, the existence and uniqueness theorem for ordinary differential equations implies that ψ t (x) = φ ρ(t,x) (x) for all x M and all t R. Therefore, φ t is topologically equivalent to ψ t. pp In Proposition 2.2.4, the hypothesis that {p} is closed is unnecessary. Proposition If p is a periodic point of a flow φ t period T, then φ T (p) = p. Proof. Let H = {τ R φ τ (p) = p}. If H is discrete, then H is (topologically) closed, so that T = inf{τ > φ τ (p) = p} H. Hence, φ T (p) = p. If H is not discrete, then there is a real number s which is an accumulation point of H. For each ɛ > there exists τ H so that τ s < ɛ. By the group property of flows and Proposition the set H is a subgroup of (R, +). Thus, φ τ s (p) = p. Consequently, T = inf{τ > φ τ (p) = p} =, in which case φ T (p) = φ (p) = p. p.31 The last sentence of Example should read If β, then since sin βt and cos βt are periodic functions, and both have period 2/ β, every point in R 2 is a periodic point of φ t, and the period of each point other than the fixed point at (, ) is 2/ β. p.36 Proposition should read: Proposition Every periodic orbit of a flow is compact, and every periodic orbit of a flow a Hausdorff phase space is compact. The last sentence of the proof should read, Every compact subset of a Hausdorff space is closed. p.36 Replace by. p.36 Replace by. p.39 In Example replace the first two sentences of the second paragraph by: On the other hand, if x O(p), then there exists a positive real number τ such that x = φ τ (p). So, lim n φ±nτ (p) = x by Proposition p.4 In the proof of Proposition (ii) define f : R R by f(t) = φ t (x) x, and replace φ τ (x) x in l.-3 by φ τ (x) x.

5 NOTES AND ERRATA FOR RECURRENCE AND TOPOLOGY 5 pp Proposition (v) should read: The sets ω(x) and α(x) are invariant. The first-countability of X is not necessary. p.48 Replace p.62 Replace p.74 Replace p.75 Replace > > p.99 The last sentence of Example should read: Therefore, Fix(φ t ) Per(φ t ) M(φ t ) for any flow φ t on a Hausdorff phase space. pp There is a stronger version of Proposition 3.2.7: Proposition Let φ t be flow on a topological space X. If A X is topologically transitive respect to φ t, then for each pair of nonempty open sets U and V in A, there exists a nonnegative real number T such that φ T (U) V is nonempty.

6 6 JOHN M. ALONGI Proof. Let U and V be nonempty open sets in A. There exist open sets U and V in X so that U = U A and V = V A. Since A is topologically transitive, there exists x A such that O + (x) and O (x) are dense in A respect to the topology on X. So, there exist nonnegative real numbers r and s such that φ r (x) U and φ s (x) V. By Proposition the set A is invariant respect to φ t. Hence, φ r (x) U and φ s (x) V. Let T = r + s. The number T is nonnegative, and φ s (x) φ s (φ r (U)) V = φ T (U) V. Therefore, φ T (U) V is nonempty. p.16 The word nonempty is missing from each part of the Birkhoff Transitivity Theorem: Theorem The Birkhoff Transitivity Theorem. Let φ t be a flow on a nonempty second-countable Baire space X. (i) If t φt (U) is dense in X for every nonempty open subset U of X, then there exists D X such that D is residual in X, and O (x) is dense in X for all x D. (ii) If t φt (U) is dense in X for every nonempty open subset U of X, then there exists D + X such that D + is residual in X, and O + (x) is dense in X for all x D +. (iii) If t φt (U) and t φt (U) are dense in X for every nonempty open subset U of X, then there exists D X such that D is residual in X, and O + (x) and O (x) are dense in X for all x D. p.17 Proposition is false. Here is a correct proposition: Proposition Let φ t be flow on a topological space X. Let A X be a nonempty second-countable Baire space in the subspace topology and invariant respect to φ t. If for each pair of nonempty open sets U and V in A, there exists a nonnegative real number T such that φ T (U) V is nonempty, then A is topologically transitive respect to φ t. Proof. For each pair of nonempty open subsets U and V in A there exist nonnegative real numbers T and S such that φ T (U) V and φ S (V ) U are nonempty. Consequently, φ S (U) V is nonempty. Since A is invariant respect to φ t, the sets t φt (U) and t φt (U) are dense in A respect to the subspace topology for each nonempty open subset U in A. By the Birkhoff Transitivity Theorem, there exists D A such that D is residual in A, and for each x D the sets O + (x) and O (x) are dense in A respect to the subspace topology. Since A is a nonempty Baire space, the set D is nonempty. Hence, there exists x A such that O + (x) and O (x) are dense in A respect to the subspace topology.

7 NOTES AND ERRATA FOR RECURRENCE AND TOPOLOGY 7 Let W be an open subset of X such that W A is nonempty. The set W A is an open set in A. Since O + (x) and O (x) are dense in A respect to the subspace topology, there exist nonnegative real numbers r and s such that φ r (x) W A and φ s (x) W A. Thus, O + (x) and O (x) are dense in A respect to the topology on X. Therefore, A is topologically transitive respect to φ t. pp There is an easier proof of the forward implication of Proposition Let x, y X be chain equivalent respect to φ t, and let (A, A ) be an attracting-repelling pair for φ t. The collection {A R(φ t ), A R(φ t )} is a partition of R(φ t ) into closed, disjoint sets. If A is empty, then x, y A. If A is empty, then x, y A. Since x and y are chain equivalent, the points x and y are elements of the same connected component of R(φ t ). Therefore, x, y A or x, y A. p.166 Replace θ = sin 2 θ + 1 ( ) φ 2 ( φ = φ 1 φ ) sin 2 θ + 1 ( ) ] φ 2 θ = sin 2 θ + 1 ( ) φ 4 ( φ = φ 1 φ ) sin 2 θ + 1 ( ) ] φ 4 p.167 Replace φ(t) 1 φ(t) ] ( sin 2 θ(t)] + 1 ] ) φ(t) 2 φ(t) 1 φ(t) ] ( sin 2 θ(t)] + 1 ] ) φ(t) 4 p.195 In Exercise 2, replace R(φ t ) = Fix(φ t ) = p(r(φ t )) R(Φ t ) = Fix(Φ t ) = p(r(φ t )).

8 8 JOHN M. ALONGI p.198 Replace Thanks to T.S. Huang, Michael Hurley, Thanate Dhirasakdanon, Keith Burns and Katrin Gelfert.