Synchronization, Chaos, and the Dynamics of Coupled Oscillators. Supplemental 1. Winter Zachary Adams Undergraduate in Mathematics and Biology

Transcription

1 Synchronization, Chaos, and the Dynamics of Coupled Oscillators Supplemental 1 Winter 2017 Zachary Adams Undergraduate in Mathematics and Biology Outline: The shift map is discussed, and a rigorous proof for chaotic dynamics (using Devaney s definition) is given. The Poincaré-Bendixson Theorem is presented and proved, as well as corollaries that exclude the possibility of chaos from two-dimensional ODE systems. The discussion of both of these topics follows that presented by Florin Diacu in Math 442 (Winter semester, 2017). No reference material was given for that course, but there appears to have been significant overlap with the book Differential Equations, Dynamical Systems, & an Introduction to Chaos, by Hirsch, Smale, and Devaney, which has been consulted for further clarification on the proof of the Poincaré-Bendixson Theorem. 1 The Shift Map Let Σ be the set of infinite binary sequences. Elements of Σ are of the form σ = (σ 1, σ 2,...), where σ k {0, 1} for all k N. We equip Σ with a metric d : Σ Σ R + defined for σ, τ Σ as d(σ, τ) = σ k τ k 2 k (1) Checking that d satisfies the properties of a metric is a fairly simple exercise. Lemma 1.1. If σ, τ Σ agree for their first m entries, then d(σ, τ) 2 m. Proof. By hypothesis, σ k = τ k for k {1, 2,..., m}. Thus, d(σ, τ) = σ k τ k 2 k = k=1 k=1 k=m+1 2 m k=1 σ k τ k 2 k 2 k = 2 m. page 1 of 6

2 For any σ = (σ 1, σ 2,...), we define the shift map s : Σ Σ as s(σ) = (σ 2, σ 3,...). The dynamics of the shift map are chaotic on Σ. Rigorously proving that a dynamical system is chaotic is difficult, in no small part because rigorously defining chaos is difficult. There is no universally accepted definition of chaos, but a good working definition has been presented by Devaney. Definition 1.1. A discrete dynamical system on a metric space (X, d) governed by a continuous function f : X X is chaotic if 1. Periodic orbits of f are dense in X: For any x X and ε > 0, there exists a periodic point y of f such that d(x, y) < ε. 2. f is topologically transitive in X: For any nonempty, open sets U, V X, there exists k > 0 such that f k (U) V. 3. Orbits of f depend sensitively on initial conditions: There exists β > 0 such that for any x X and open set U containing x, there is some y U and k > 0 such that d(f k (x), f k (y)) > β. We are now ready to prove that the shift map is chaotic on Σ. 1. Take any σ Σ and ε > 0. Choose k N such that ε > 2 m and set π (mn+k) = σ k for k = 1, 2,..., m, and n = 0, 1, 2,... Clearly, π = (π 1, π 2,...) is periodic under s (with period m) and agrees with σ for its first m entries. Thus by Lemma 1.1, d(σ, π) < 2 m < ε. This proves the density of s-periodic orbits in Σ. 2. Take any open sets U, V Σ. Fix σ = (σ 1, σ 2,...) in U and τ = (τ 1, τ 2,...) in V. Because U is open, there exists ε > 0 such that if α Σ satisfies d(σ, α) < ε, then α U. Choose m such that ε > 2 m, and define α = (σ 1, σ 2,..., σ m, τ 1, τ 2,...). By Lemma 1.1, d(σ, α) < 2 m < ε, and so α U. Moreover, s m (α) = (τ 1, τ 2,...) = τ V. This proves transitivity of s in Σ. 3. Take any σ Σ, and let U be an open set of Σ containing σ. Because U is open, there exists ε > 0 such that if d(x, y) < ε, then y U. Choose m such that 2 m < ε, and define τ = (σ 1, σ 2,..., σ m, 1 σ m+1, 1 σ m+2,...). So τ completely agrees with σ for its first m entries, and completely disagrees with σ for all other entries. By Lemma 1.1, d(σ, τ) < ε, implying that τ U. But, s m (τ) and s m (σ) agree at none of their entries, which means that d(s m (σ), s m (τ)) = 1 =: β. page 2 of 6

3 2 The Poincaré-Bendixson Theorem We begin by stating a few lemmas necessary for the proof of the Poincaré-Bendixson Theorem. We prove the theorem for the case where L is an ω-limit set. The case where L is an α-limit set is nearly identical. For the remainder of Section 2, let U R 2 be open, and let f : U R 2 be a C 1 function. We are concerned with the planar ODE ẋ = f(x). (2) 2.1 Some Definitions Definition 1. The flow of the ODE (2) is a function φ : U R U such that φ(x, t) describes the solution curve of (2) through the point x U. We sometimes write φ(x, t) = φ t (x). Notice that for s, t R, we have φ t+s (x) = φ t (φ s (x)). Definition 2. We say that y is an ω-limit point (α-limit point) of x U if there exists a sequence {t n } diverging to + ( ) such that φ(x, t n ) y as n. The ω-limit set (α-limit set) of x is the set of all ω-limit points (α-limit points) of x. Definition 3. Take x 0 U such that f(x 0 ) 0, and let L be the straight line through x 0 perpendicular to f(x 0 ). Because f is continuous, there is some subinterval S L containing x 0 such that f(x) is not parallel to L for any x S. Such a subinterval S is called a local section at x 0. By the continuity of f, solution curves cross S in a single orientation. Definition 4. Let {x k } be a sequence of points along the solution curve through some point x U i.e. x k = φ(x, t k ) for some t k R. We say that {x k } is monotone along the solution curve if t k < t k+1 for all k. If {x k } is a sequence of points lying on a line I R 2 such that x k lies between x k 1 and x k+1 in the natural order for all k N, then {x k } is monotone along I. 2.2 Lemmas Lemma 2.1. Any limit set L of a point x U is invariant. If K is a limit set of some y in L, then K L. Proof. Consider the solution curve φ(y, t) through y L. There exists a sequence {t n } diverging to + as n such that φ(x, t n ) y as n. Taking the sequence {t + t n } and using continuity page 3 of 6

4 of the flow, lim [φ t+t n (x)] = lim [φ t (φ tn (x))] = φ t ( lim [φ tn (x)]) = φ t (y) n n n So every point along the solution curve through y L is itself in L. The second statement of the Lemma follows immediately. Lemma 2.2. Let S be some local section, and let {x k } be a sequence of points in S that lie along the same solution curve through some x U. If this sequence is monotone along the solution curve, then it is also monotone along S. Proof. It will suffice to prove the statement for x 0, x 1, x 2. Let Γ be a simple closed curve formed by the solution curve between x 0 and x 1, and let Σ be the segment of S between x 0 and x 1. Let D denote the region enclosed by Σ Γ. Recall that flows through the local section S cross S in the same orientation. Without loss of generality, assume that the solution curve through x leaves D at x 1. No solution curve may cross Γ, and hence R 2 \ D is an invariant set for t > 0. Therefore φ(x 1, t) / D for t > 0, and it follows that x 2 S \ Σ. It is fairly clear that S \ Σ = I J, where I and J are disjoint intervals, exactly one of which (say, I) is contained in D. Because x 2 / D, it follows that x 2 J. Lemma 2.3. Let y be a limit point of x U. Then, the solution curve through y can cross any local section at no more than one point. Proof. Without loss of generality, assume y is an ω-limit point of x. Suppose the solution curve through y crosses some local section S at two points y + y (by Lemma 2.1, y ± are ω-limit points of x as well). Let S ± be disjoint segments of S containing y ±, respectively, and let Γ ± be small neighbourhoods of solution curves through S ±. Note that the solution curve through x enters both Γ + and Γ infinitely many times, and therefore must cross both S + and S infinitely many times. We can thus find a sequence x + 1, x 1, x + 2, x 2,... monotone along the solution curve through x, with x ± k S± respectively. But since S ± are disjoint, no such sequence can exist. It is therefore necessary that y + = y. 2.3 The Theorem Theorem 2.1 (Poincaré-Bendixson). Let L be a closed and bounded limit set of some point x U. If L contains no equilibrium points, then L is a limit cycle. page 4 of 6

5 Proof. Take y L. We first show that the solution curve through y is a limit cycle. Following this, we show that if γ L is a limit cycle, then γ = L. Let K be the ω-limit set of y; by Lemma 2.1, K L. Take z K and a local section S through z. Let V be a neighbourhood of solutions curves containing z, and I := S V. Because z is an ω-limit point of y, there is a sequence of positive real numbers {t k } diverging to infinity such that φ(y, t k ) z as k. Thus, there are infinitely many points φ(y, t k ) in V, and we can choose a subsequence {t kn } of {t k } such that φ(x, t kn ) V for all n. Solutions through points in V must cross the local section S (see Definition 3), and thus there exists {r n } such that φ(y, (t kn + r n )) S. For convenience set s n := (t kn + r n ). By Lemma 2.3, the solution curve through y intersects S at exactly one point. So φ(y, s i ) = φ(y, s j ) = z for all i, j N. Taking s j > s i, τ := s j s i > 0, this implies that φ(y, τ) = y. Because there are no equilibria in L, y must lie on a closed orbit. We now show that if γ is a closed orbit in the limit set L, then necessarily γ = L. To prove this, it suffices to show that lim t [d(φ(x, t), γ)] = 0. Take z γ and a local section S through z. We can choose S small enough such that S γ = {z}. Let V be a neighbourhood of solution curves through S containing z. We can find an increasing sequence {t k } such that 1. φ(x, t k ) S (because the curve through x passes arbitrarily close to z, and thus arbitrarily close to S, infinitely many times; thus it must pass through S infinitely many times), 2. lim k [φ(x, t k )] = z, 3. φ(x, t) / S for t (t k 1, t k ). Write x k := φ(x, t k ); by Lemma 2.2, {x k } is a monotone sequence in S converging to z. Claim: The set of differences {t n t n 1 } has an upper bound. Because z lies on a closed orbit, we can choose τ > 0 such that φ(z, τ) = z. For x k sufficiently close to z, φ(x k, τ) V (by continuity of the flow). Then there exists t ( ε, ε) such that φ(x k, τ +t) S (see Definition 3). It follows that (t n t n 1 ) τ + ε, justifying our claim that the set of {t n t n 1 } is bounded. Moreover, since (t n t n 1 ) 2ε, it is necessary that t n as n. Let µ denote the upper bound of {t n t n 1 }. By the continuity of solution curves with respect to initial data, for any α > 0 there exists δ > 0 such that if u lies on another solution curve such that u z < δ, then φ(u.t) φ(z, t) < α for t < µ i.e. the distance from φ(u, t) to γ is less than α for t such that t < µ. We can choose large enough k 0 such that x k z < δ for k k 0. Then, φ(x k, t) φ(z, t) < α whenever t < µ and k k 0. Indeed, if this is the case, then if {t k } is any sequence diverging to + such that φ tk (x) converges, then it must converge to some point on γ. Then, by Lemma 2.1, the ω-limit set of x will be γ. page 5 of 6

6 Take t t k0 and choose k k 0 so that t [t k, t k+1 ]. Since φ t (x) = φ t tk (x k ) and t t k µ, d(φ t (x), γ) φ t (x) φ t tk (z) = φ t tk (x) φ t tk (z) < α. Thus, for any α > 0 we can find T > 0 such that when t > T, the distance from the solution curve through x to γ is less than α, as desired. Putting the pieces together: if the ω-limit set of some x U R 2 is closed, bounded, and contains no equilibria, then it must be a limit cycle. page 6 of 6