Complicated dynamics in nonautonomous ODE’s.

Leszek Pieniążek and Klaudiusz Wójcik ^{∗}

April 22, 1999

IMUJ Preprint 1999/11

Abstract We present a topological method for detecting complicated dynamics in nonautonomous ordinary differential equations (not necesserily periodic with respect to the time variable). Our main result gives a sufficient condition for existence of a class of solutions, whose presence displays some chaotic features of the dynamics. The method is based on the Ważewski Retract Theorem and the Lefschetz Fixed Point Theorem. Some applications to the nonautonomous systems on the plane are considered.

Keywords: dynamical systems, local process, nonautonomous ODE’s, multibump solutions, periodic solutions, Lefschetz number, fixed point index,

Ważewski sets.

1 Introduction.

In this note we study a topological method for detecting complicated dynamics in the local processes generated by the nonautonomous ordinary differential equations (not necesserily periodic with respect to the time variable). We show the existence of a class of solutions, whose presence displays some chaotic features of the dynamics. The results presented here, in the same spirit as in [S1], [SW], [W1] are inspired by a lot of papers on the existence of the multibump orbits ([AM], [BM], [CZN], [CZR], [CZMN], [MN]) starting with the novel minimax method in [Se]. In the context of the Lagrangian systems the multibump orbits mean the orbits which are close to chains of homoclinics of the limit system (see [AM], [BM], [CZR]). The results of [BM] are based on the variational version of the Birkhoff-Smale-Shilnikov theory. In contrast to the above method, we do not need any global information on particular solutions of the considered equation. Our method for detecting chaotic dynamics is based on the existence of some sets, called admissible proper sets and the knowledge of the values of some topological invariants of these sets. The notion of the proper pair is based on the concept of the Ważewski set ([C]) and is an inessential modification of the periodic block in [S1]. In the case when the right-hand side of the considered equation is periodic with respect to the time variable, our main result (up to slightly different notation) improves Th.2 in [SW] (comp. [W1]). In fact, the method presented in this paper, is addapted to the general nonautonomous systems from the periodic case in [SW], [W1], [W4]. In order to apply the topological method introduced in [SW], we have to prove the existence of some sets, called periodic isolating segments, in the extended phase space (see [SW]). Basic property of the segment is that in any point on its boundary the vector field is directed outward or inward with respect to the segment (compare the notion of the isolating block in the Conley index theory). Notice that an admissible proper pair (in our sense) can be easily obtained by gluing translated copies of a periodic isolating segment. It was observed by Roman Srzednicki in [S1], that the fixed point index of the Poincaré map inside the segment is equal to the Lefschetz number of the monodromy homeomorphism given by the segment (see Th.7.1 in [S1]). Our Proposition 1 is a non-periodic version of this result. Theorem 2 in [SW] gives a sufficient condition for the chaotic dynamics in the periodic systems in the sense that the Poincaré map is semiconjugated to the shift on two symbols and the counterimage (by the semiconjugacy) of any periodic point in the shift contains a periodic point of the Poincaré map. It follows by our Theorem 1, that any small perturbation of the T -periodic system for which the results in [SW] show chaos has also complicated dynamics. In non-periodic case we are not able to prove the existence of periodic solutions but the map after time T is still semiconjugated to the Bernoulli shift on some compact set. In particular, the topological entropy is positive.

∗Research partially supported by the Foundation for Polish Science.

2 Proper pairs.

Assume that X is a metric space and φ : D → X is a continuous mapping, D ⊂ ℝ × X × ℝ is an open set. We will denote by φ_{(σ,t)} the function φ(σ, ·, t).

φ is called a local process if the following conditions are satisfied

(1) ∀σ ∈ ℝ, x ∈ X : {t ∈ ℝ : (σ, x, t) ∈ D} is an interval,

(2) ∀σ ∈ ℝ : φ_{(σ,0)} = id_{X}

(3) ∀σ ∈ ℝ : φ_{(σ,s+t)} = φ_{(σ+s,t)} ∘ φ_{(σ,s)}, If D = ℝ × X × ℝ, we call φ a (global) process. For (σ, x) ∈ ℝ × X the set

{(σ + t, φ_{(σ,t)}(x)) ∈ ℝ × X : (σ, x, t) ∈ D}

is called the trajectory of (σ, x) in φ. If T is a positive number such that

(4) ∀σ, t ∈ ℝ : φ_{(σ+T,t)} = φ_{(σ,t)}

we call φ a T -periodic local process. It follows, that the interval in (1) is open and, by (2), it contains 0. Since the domains of the both maps in (3) are equal, (σ, x, s + t) ∈ D if and only if (σ, x, s) ∈ D and (σ + s, φ_{(σ,s)}(x), t) ∈ D.

A local process φ on X determines a local flow Φ on ℝ × X by the formula

Φ_{t}(σ, x) = (σ + t, φ_{(σ,t)}(x)).

Remark 1 The differential equation

(∗) ẋ = f(t, x)

such that f is regular enough to guarantee the uniqueness of solutions of Cauchy problems associated to (∗) generates a local process as follows. For x(t_{0}, x_{0}; ·) the solution of (∗) such that x(t_{0}, x_{0}; t_{0}) = x_{0} we put

φ_{(t_{0},τ)}(x_{0}) = x(t_{0}, x_{0}; t_{0} + τ).

If f is T -periodic with respect to t then φ is a T -periodic local process and in order to determine all T -periodic solutions of the equation (∗) it suffices to look for fixed points of φ_{(0,T)} (called the Poincaré map).

We will use the following notation: by π_{1} : ℝ × X → ℝ and π_{2} : ℝ × X → X we denote the projections and for every Z ⊂ ℝ × X and t ∈ ℝ we put

Z_{t} = {x ∈ X : (t, x) ∈ Z}.

Let (U, U^{−}) be a pair of subsets of ℝ × X (i.e. U^{−} ⊂ U ⊂ ℝ × X). We call (U, U^{−}) a proper pair (for the process φ) and U^{−} the exit set of U if:

(i) U and U^{−} are closed ENR’s, U_{0}, U_{0}^{−} are compact,

(ii) there exists a homeomorphism

h : ℝ × (U_{0}, U_{0}^{−}) ⟶ (U, U^{−})

such that π_{1} = π_{1} ∘ h,

(iii) for every σ ∈ ℝ and x ∈ ∂U_{σ} there exists a t ∈ ℝ such that φ_{(σ,t)}(x) ∉ U_{σ+t},

(iv) U^{−} = {(σ, x) ∈ U : ∃s_{n} > 0 s_{n} → 0 : φ_{(σ,s_{n})}(x) ∉ U_{σ+s_{n}}}.

Define a map

τ_{U} : U_{0} ∋ x ⟶ sup{t ≥ 0 : ∀s ∈ [0, t] : Φ_{s}(0, x) ∈ U} ∈ [0, ∞].

τ_{U} is continuous (by the argument in a proof of the Ważewski Theorem, [C], [S1]).

Let (U, U^{−}) be a proper pair, U_{a} = U_{b}, U_{a}^{−} = U_{b}^{−} for some a, b ∈ ℝ, a < b. Define a homeomorphism

h_{a,b} : (U_{a}, U_{a}^{−}) ⟶ (U_{b}, U_{b}^{−}) = (U_{a}, U_{a}^{−})

by h_{a,b}(x) = π_{2}(h(b, π_{2}h^{−1}(a, x))) for x ∈ U_{a}.

Geometrically, h_{a,b} moves a point x ∈ U_{a} to U_{b} = U_{a} along the arc h([a, b] × {π_{2}h^{−1}(a, x)}). Consider the automorphism

μ_{U_{a,b}} : H(U_{a}, U_{a}^{−}) ⟶ H(U_{a}, U_{a}^{−})

induced by h_{a,b} in the singular homology with rational coefficients. Recall that its Lefschetz number is defined as

Lef(μ_{U_{a,b}}) = ∑_{n=0}^{∞}(−1)^{n}tr H_{n}(h_{a,b}).

In particular, if μ_{U_{a,b}} = id_{H(U_{a},U_{a}^{−})} then Lef(μ_{U_{a,b}}) is equal to the Euler characteristic χ(U_{a}, U_{a}^{−}).

In the sequel we will use the following non-periodic version of the Theorem 7.1 in [S1]

Proposition 1 If (U, U^{−}) is a proper pair, U_{a} = U_{b}, U_{a}^{−} = U_{b}^{−} for some a, b ∈ ℝ, a < b then the set

F_{U_{a,b}} = {x ∈ X : φ_{(a,b−a)}(x) = x, ∀t ∈ [0, b − a] : φ_{(a,t)}(x) ∈ U_{a+t}}

is compact and open in the set of fixed points of φ_{(a,b−a)} and the fixed point index of φ_{(a,b−a)} in F_{U_{a,b}} is given by

ind(φ_{(a,b−a)}, F_{U_{a,b}}) = Lef(μ_{U_{a,b}}).

Proof: Our proof of Proposition 1 is simpler but similar in the spirit to the proof of Theorem 7.1 in [S1]. By Lemma 2.3.1 in [S2] F_{U_{a,b}} is compact and open in the set of fixed points of the φ_{(a,b−a)}. Let τ = τ_{U} : U → [0, ∞]. For s ∈ [a, b] we define a homeomorphism

h_{s,b} : (U_{s}, U_{s}^{−}) → (U_{b}, U_{b}^{−}),

by h_{s,b}(x) = π_{2}h(b, π_{2}h^{−1}(s, x)). Consider a homotopy H : (U_{a}, U_{a}^{−}) × [0, 1] → (U_{b}, U_{b}^{−}) = (U_{a}, U_{a}^{−}) given by

H(x, t) = \{ \begin{matrix}h_{a+τ(x),b}(φ_{(a,τ(x))}(x)), & τ(x) ≤ (1 − t)(b − a) \\ h_{a+(1−t)(b−a),b}(φ_{(a,(1−t)(b−a))})(x)), & τ(x) ≥ (1 − t)(b − a) \end{matrix}

Put H_{t}(x) = H(x, t). It is easy to check that H_{t}(x) = h_{a,b}(x) for x ∈ U_{a}^{−}, t ∈ [0, 1] and H_{1} = h_{a,b}. By the homotopy property of the Lefschetz number

Lef(h_{a,b}) = Lef(H_{0}).

Moreover

Lef(H_{0}) = ind(φ_{(a,b−a)}, F_{U_{a,b}}) + ind(h_{a,b}, Fix(h_{a,b}|_{U_{a}^{−}})),

and by the commutativity property of the fixed point index

ind(h_{a,b}, Fix(h_{a,b}|_{U_{a}^{−}})) = Lef(h_{a,b}|_{U_{a}^{−}}),

so the proof is complete, because

Lef(μ_{U_{a,b}}) = Lef(h_{a,b}) − Lef(h_{a,b}|_{U_{a}^{−}}).
□

3 Main result.

Let (U, U^{−}) be a proper pair for the process φ. We call U admissible iff there is a sequence {t_{n}}_{n∈ℤ} ⊂ ℝ such that t_{n} < t_{n+1}, U_{t_{n}} = U_{t_{0}}, U_{t_{n}}^{−} = U_{t_{0}}^{−} for all n ∈ ℤ.

Suppose that (U, U^{−}), (Z, Z^{−}) are proper pairs, U ⊂ Z are admissible with the same sequence {t_{n}}_{n∈ℤ}, U_{t_{0}} = Z_{t_{0}}, U_{t_{0}}^{−} = Z_{t_{0}}^{−} and the set P = {t ∈ ℝ : U_{t} = Z_{t}, U_{t}^{−} = Z_{t}^{−}} is unbounded both from below and above. Assume that for any n ∈ ℤ there is an s ∈ (t_{n}, t_{n+1}) such that U_{s} ≠ Z_{s}. Consider the following conditions

(a) for all n ∈ ℤ μ_{U_{tn,t_{n+1}}} = id_{H(U_{t_{0}} ,U_{t_{0}}^{−})},

(b) there is an n_{0} ∈ ℕ \ {1} and the authomorphism G : H(U_{t_{0}}, U_{t_{0}}^{−}) → H(U_{t_{0}}, U_{t_{0}}^{−}) such that G^{n_{0}} = id_{H(U_{t_{0}} ,U_{t_{0}}^{−})}, for all n ∈ {1, …, n_{0} − 1} Lef(G^{n}) = Lef(G) and μ_{Z_{tn,t_{n+1}}} = G (n ∈ ℕ),

(c) Lef(G) ≠ χ(U_{t_{0}}, U_{t_{0}}^{−}) ≠ 0.

Remark 2 Let us consider the planar differential equation

ż = e^{it}z̄^{n}.

From the phase portrait one can deduce (comp. [S1]) the existence of the admissible proper pair Z with the sequence t_{k} = 2πk such that Z_{0} is a regular 2(n + 1)-gon, Z_{0}^{−} consist of n + 1 disjoint segments and both the sets Z_{0} and Z_{0}^{−} are invariant with respect to the rotation by the angle \frac{2π}{n+1}. It is easy to see that

Lef(μ_{Z_{t_{k},t_{k+1}}} ) = … = Lef(μ_{Z_{t_{k},t_{k+1}}}^{n}) = 1,
μ_{Z_{t_{k},t_{k+1}}}^{n+1} = id_{H(Z_{0},Z_{0}^{−})},

so the condition (b) holds with n_{0} = n + 1.

We prove the following

Theorem 1 Under the assumptions (a), (b), (c) for any subset S ⊂ ℤ there is x_{0} ∈ U_{t_{0}} such that

(1) for all t ∈ ℝ φ_{(t_{0},t)}(x_{0}) ∈ Z_{t_{0}+t}

(2) if n ∈ S then there exists t ∈ (t_{n} −t_{0}, t_{n+1} −t_{0}) such that φ_{(t_{0},t)}(x_{0}) ∉ U_{t_{0}+t},

(3) if n ∉ S then for all t ∈ [t_{n} − t_{0}, t_{n+1} − t_{0}] φ_{(t_{0},t)}(x_{0}) ∈ U_{t_{0}+t}.

For S ⊂ ℤ, by U^{S} we denote the proper pair such that if n ∈ S then

U^{S} ∩ ([t_{n}, t_{n+1}] × X) = Z ∩ ([t_{n}, t_{n+1}] × X),

and if n ∉ S then

U^{S} ∩ ([t_{n}, t_{n+1}] × X) = U ∩ ([t_{n}, t_{n+1}] × X).

Assume that S is a finite set and card(S) = k for some k ∈ ℕ. Let S = {n_{1}, …, n_{k}}, a, b ∈ {t_{n} : n ∈ ℤ} and a < t_{n_{1}} < … < t_{n_{k}} < b. Recall that the set

F_{U_{a,b}^{S}} = {x ∈ X : φ_{(a,b−a)}(x) = x, ∀t ∈ [0, b − a] : φ_{(a,t)}(x) ∈ U_{a+t}^{S}}

is compact and open in the set of fixed points of φ_{(a,b−a)}. Observe that if n ∈ S, x ∈ F_{U_{a,b}^{S}} then τ_{U}(φ_{(a,t_{n}−a)}(x)) ≠ t_{n+1} − t_{n}. For any subset L ⊂ S by F^{L} we denote the set of points x ∈ F_{U_{a,b}^{S}} such that if n ∈ L then τ_{U}(φ_{(a,t_{n}−a)}(x)) < t_{n+1} − t_{n} and if n ∈ S \ L then τ_{U}(φ_{(a,t_{n}−a)}(x)) > t_{n+1} − t_{n}. The sets F^{L} over all subsets L ⊂ S form a compact and disjoint covering of F_{U_{a,b}^{S}}. The proof of Theorem 1 is based on the following

Lemma 1 If card(S) = k then

ind(φ_{(a,b−a)}, F^{S}) = ∑_{l=0}^{k}(−1)^{k−l}\binom{k}{l}Lef(G^{l}).

Proof: The case k = 0 follows immediately from Proposition 1. For k ≥ 1 we use the induction with respect to k. Let k = 1. Since F^{S} and F^{∅} form a compact and disjoint covering of F_{U_{a,b}^{S}}, by the additivity property of the fixed point index,

ind(φ_{(a,b−a)}, F_{U_{a,b}^{S}}) = ind(φ_{(a,b−a)}, F^{S}) + ind(φ_{(a,b−a)}, F^{∅}),

so by Proposition 1

ind(φ_{(a,b−a)}, F^{S}) = Lef(G) − χ(U_{t_{0}}, U_{t_{0}}^{−}).

Assume now that the lemma holds for p ≤ k. We prove it for k + 1. Again by the additivity of the fixed point index,

ind(φ_{(a,b−a)}, F_{U_{a,b}^{S}}) =

ind(φ_{(a,b−a)}, F^{S}) + ∑_{L⊂S,L≠S} ind(φ_{(a,b−a)}, F^{L}).

By Proposition 1 and the inductive step,

ind(φ_{(a,b−a)}, F^{S}) =

Lef(G^{k+1}) − ∑_{s=0}^{k} \binom{k + 1}{s} ∑_{l=0}^{s}(−1)^{s−l}\binom{s}{l}Lef(G^{l}) =

Lef(G^{k+1}) − ∑_{s=0}^{k} ∑_{l=0}^{s}(−1)^{s−l}\binom{k + 1}{s}\binom{s}{l}Lef(G^{l}) =

Lef(G^{k+1}) − ∑_{s=0}^{k} ∑_{l=0}^{s}(−1)^{s−l}\binom{k + 1}{l}\binom{k + 1 − l}{k + 1 − s}Lef(G^{l}).

Let s_{0} ∈ {0, …, k} be fixed. We show that the coefficient of Lef(G^{s_{0}}) equals (−1)^{k+1−s_{0}} \binom{k+1}{s_{0}}. It is easy to see that this coefficient equals

[− ∑_{r=s_{0}}^{k}(−1)^{r−s_{0}} \binom{k + 1 − s_{0}}{k + 1 − r}] \binom{k + 1}{s_{0}}.

We put m = k + 1 − s_{0}. Then

∑_{r=s_{0}}^{k}(−1)^{r−s_{0}} \binom{k + 1 − s_{0}}{k + 1 − r} =

∑_{w=0}^{m−1}(−1)^{w}\binom{m}{m − w} = ∑_{w=0}^{m−1}(−1)^{w}\binom{m}{w}.

By

∑_{w=0}^{m}(−1)^{w}\binom{m}{w} = 0,

we have that if m = k + 1 − s_{0} is even then ∑_{w=0}^{m−1}(−1)^{w}\binom{m}{w} = −1 and if m = k + 1 − s_{0} is odd then ∑_{w=0}^{m−1} \binom{m}{w} = 1. This finishes the proof.

□

Corollary 1 (1) Under the assumptions of Theorem 1, if card(S) = k then

ind(φ_{(a,b−a)}, F^{S}) = (∑_{n_{0}|s}(−1)^{k−s}\binom{k}{s})(χ(U_{0}, U_{0}^{−}) − Lef(G)).

(2) In particular, if n_{0} is even then ind(φ_{(a,b−a)}, F^{S}) ≠ 0 and if n_{0} is odd then ind(φ_{(a,b−a)}, F^{S}) ≠ 0 iff k is not an odd multiplicity of n_{0}.

Proof: (1) It follows from Lemma 1 because ∑_{s=0}^{k}(−1)^{k−s}\binom{k}{s} = 0 and

Lef(G^{l}) = \{ \begin{matrix}χ(U_{0}, U_{0}^{−}) & , n_{0} | l \\ Lef(G) & , otherwise. \end{matrix}

(2) If n_{0} is even then (−1)^{k−s} = (−1)^{k} for all s such that n_{0} | s, so the conclusion follows by (1). The case of n_{0} odd will be proved in Corollary 2 in the appendix.

□

Proof of the Theorem 1: Assume first that S is a finite set. Let a_{n}, b_{n} ∈ P, a_{n} → −∞, b_{n} → +∞. If a_{n} is sufficiently small and b_{n} sufficiently large then by Lemma 1 there is y_{n} ∈ F^{S} ⊂ F_{U_{an,bn}^{S}}. By the compactness of U_{t_{0}} there is a subsequence of the sequence {φ_{(a_{n},t_{0}−a_{n})}(y_{n})} ⊂ U_{t_{0}} which converges to some x_{0} ∈ U_{t_{0}}. The standard arguments show that the trajectory of x_{0} is defined on the whole real line. The conditions (1) and (3) are easy to verify so we claim only (2). Let t_{n_{0}} ∈ S be fixed. Because

φ_{(a,t_{n_{0}} −a)}(y_{n}) = φ_{(t_{0},t_{n_{0}} −t_{0})}(φ_{(a,t_{0}−a)}(y_{n})),

so by the continuity of τ_{U}

τ_{U}(φ_{(a,t_{n_{0}} −a)}(y_{n})) → τ_{U}(φ_{(t_{0},t_{n_{0}} −t_{0})}(x_{0})).

On the other hand, by the definition of the sequence {y_{n}} we have

0 < τ_{U}(φ_{(a,t_{n_{0}} −a)}(y_{n})) < t_{n_{0}+1} − t_{n_{0}},

so

0 ≤ τ_{U}(φ_{(t_{0},t_{n_{0}} −t_{0})}(x_{0})) ≤ t_{n_{0}+1} − t_{n_{0}}.

Because τ_{Z}(x_{0}) = +∞, Z_{t_{n_{0}}}^{−} = U_{t_{n_{0}}}^{−} and Z_{t_{n_{0}+1}}^{−} = U_{t_{n_{0}+1}}^{−}, we obtain that in fact

0 < τ_{U}(φ_{(t_{0},t_{n_{0}} −t_{0})}(x_{0}) < t_{n_{0}+1} − t_{n_{0}},

so (2) holds for S finite. If S is an infinite set then we use the proved finite case and similar arguments based on the compactness of U_{t_{0}}.

□

4 Periodic case.

Assume that the vector field f (on manifold M) is T -periodic (T > 0) with respect to the time variable. Let U and Z be two proper sets such that the assumptions (a), (b), (c) hold with the sequence t_{n} = nT (n ∈ ℤ). As a corollary we obtain

Theorem 2 (1) There are a compact set I ⊂ M invariant with respect to the Poincaré map φ_{(0,T)} and a continuous surjective map g : I → Σ_{2} such that φ_{(0,T)} is semiconjugated to the shift σ : Σ_{2} → Σ_{2} by the map g in I.

(2) If n_{0} is even then for every n-periodic sequence s ∈ Σ_{2} its counterimage by g contains at least one n-periodic point of the Poincaré map.

(3) If n_{0} is odd and s ∈ Σ_{2} is n-periodic sequence in which the symbol 1 appears k-times, k is not an odd multiplicity of n_{0} then the counterimage of s by g contains n-periodic point of φ_{(0,T)}.

Proof: The set I and map g are defined in the same way like in [SW]. Surjectivity of g follows by density of periodic orbits in Σ_{2} and Corollary 1.

□

Remark 3 Let U and W be two periodic isolating segments over [0, T] for the equation (∗) (see [SW] for the definition). Suppose that U and W fulfil the assumptions of the Theorem 2 in [SW] (see also [W1], [W2], [W3], [W4] for examples of concrete differential equations). We define two admissible proper pairs Ũ, W̃ by the conditions

Ũ_{t} = U_{t modT} ,
W̃_{t} = W_{t modT} ,

for t ∈ ℝ. It follows that for Ũ, W̃ hold all assumptions of our Theorem 1 with t_{n} = nT, so for any sufficiently small perturbation (not necesserily T -periodic) of the system (∗) there is a compact set Λ which is invariant with respect to φ_{(0,T)} and φ_{(0,T)} restricted to Λ is semiconjugated to the Bernoulli shift with two symbols. In particular, the topological entropy of φ_{(0,T)} is positive.

5 Applications.

In the present section we consider the planar nonautonomous equation of the variable z ∈ ℂ

ż = (1 + (cos(t^{2}) + 2)e^{iϕt}|z|^{2})z̄,  (1)

for some ϕ ∈ ℝ.

Theorem 3 The equation (1) fulfils assumptions of theorem 1 for sufficiently small ϕ > 0.

Although a proof is similar to the proof of Th.2 in [SW], we give it in a detailed way for the sake of completeness. Our proof of Theorem 3 consists of the construction of two proper pairs U and Z (admissible with t_{n} = \frac{2π}{ϕ} n) satisfying the conditions (a), (b), and (c) in Theorem 1. Z will be a twisted prism with a square base centered at the origin. Its cross-sections Z_{t} will be obtained by rotating the base with the angle velocity ϕ/2 over the t-interval [0, 2π/ϕ]. The set U will be a regular square-based prism with broadening ends. Its crosssections U_{t} corresponding to t near the centre of the interval will have the small side and they will broaden when t approaches the ends of the interval (because one should get the same cross-sections of U at times 0 and 2π/ϕ as Z).

The reminder of this section will be devoted to presentation of the proof of the above theorem. The equation (1) coincides with the system of two planar equations:

\{ \begin{matrix}ẋ = x + (cos(t^{2}) + 2)(x^{2} + y^{2})(x cos(ϕt) + y sin(ϕt)), \\ ẏ = −y + (cos(t^{2}) + 2)(x^{2} + y^{2})(x sin(ϕt) − y cos(ϕt)). \end{matrix}  (2)

By F we denote the vector-field in the extended phase space ℝ^{3} generated by the right-hand side of the system (2), i.e.

F (t, x, y) = ( \begin{matrix}1 \\ x + (cos(t^{2}) + 2)(x^{2} + y^{2})(x cos(ϕt) + y sin(ϕt)) \\ −y + (cos(t^{2}) + 2)(x^{2} + y^{2})(x sin(ϕt) − y cos(ϕt)) \end{matrix} )

In the sequel we assume that ϕ > 0. In order to construct U and Z we will introduce several auxiliary functions and sets. Let R > 0. Put

Λ_{R}^{1}(t, x, y) = \frac{1}{R^{2}} (x cos(\frac{ϕ}{2}t) + y sin(\frac{ϕ}{2}t))^{2} − 1,

Λ_{R}^{2}(t, x, y) = \frac{1}{R^{2}} (x sin(\frac{ϕ}{2}t) − y cos(\frac{ϕ}{2}t))^{2} − 1,

and

L_{R} = {(t, x, y) ∈ ℝ^{3} : Λ_{R}^{i}(t, x, y) ≤ 0, i = 1, 2},
L_{R}^{−} = {(t, x, y) ∈ ℝ^{3} : Λ_{R}^{1}(t, x, y) = 0, Λ_{R}^{2}(t, x, y) ≤ 0},
L_{R}^{+} = {(t, x, y) ∈ ℝ^{3} : Λ_{R}^{1}(t, x, y) ≤ 0, Λ_{R}^{2}(t, x, y) = 0}.

Lemma 2 If ϕ ≤ 1 and R ≥ 3 then

F (t, x, y) · ∇Λ_{R}^{1}(t, x, y) > 0 ((t, x, y) ∈ L_{R}^{−}),  (3)
F (t, x, y) · ∇Λ_{R}^{2}(t, x, y) < 0 ((t, x, y) ∈ L_{R}^{+}).  (4)

Proof: A direct calculation shows that

F (t, x, y) · ∇Λ_{R}^{1}(t, x, y) =

\frac{2}{R^{2}} ((x cos(\frac{ϕ}{2}t) + y sin(\frac{ϕ}{2}t))(x cos(\frac{ϕ}{2}t) − y sin(\frac{ϕ}{2}t)

−ϕx sin(\frac{ϕ}{2}t) + ϕy cos(\frac{ϕ}{2}t))+

(cos(t^{2}) + 2)(x^{2} + y^{2})(x cos(\frac{ϕ}{2}t) + y sin(\frac{ϕ}{2}t))^{2})

F (t, x, y) · ∇Λ_{R}^{2}(t, x, y) =

\frac{2}{R^{2}} ((x sin(\frac{ϕ}{2}t) − y cos(\frac{ϕ}{2}t))(y cos(\frac{ϕ}{2}t) + x sin(\frac{ϕ}{2}t)

+ϕy sin(\frac{ϕ}{2}t) + ϕx cos(\frac{ϕ}{2}t))−

(cos(t^{2}) + 2)(x^{2} + y^{2})(x sin(\frac{ϕ}{2}t) − y cos(\frac{ϕ}{2}t))^{2})

It is easy to verify that x^{2} + y^{2} ≥ R^{2}, |x| ≤ R\sqrt{2}, and |y| ≤ R\sqrt{2} provided (t, x, y) ∈ ∂L_{R} for some t. If (t, x, y) ∈ L_{R}^{−} then

|x cos(\frac{ϕ}{2}t) + y sin(\frac{ϕ}{2}t)| = R

and

F (t, x, y) · ∇Λ_{R}^{1}(t, x, y) ≥ \frac{2}{R^{2}} (R^{4} − R(2R\sqrt{2} + 2Rϕ\sqrt{2})) = 2(R^{2} − 2\sqrt{2}(1 + ϕ)),

hence (3) follows. In the same manner we see that if (t, x, y) ∈ L_{R}^{+} then

|x sin(\frac{ϕ}{2}t) − y cos(\frac{ϕ}{2}t)| = R

and

F (t, x, y)·∇Λ_{R}^{2}(t, x, y) ≤ \frac{2}{R^{2}} (−R^{4}+R(2R\sqrt{2}+2Rϕ\sqrt{2})) = 2(−R^{2}+2\sqrt{2}(1+ϕ)),

hence (4) is also satisfied and Lemma 2 is proved.

□

Similarly to the previous construction, let r > 0 and put

Ξ_{r}^{1}(t, x, y) = \frac{1}{r^{2}} x^{2} − 1,

Ξ_{r}^{2}(t, x, y) = \frac{1}{r^{2}} y^{2} − 1,

and

K_{r} = {(t, x, y) ∈ ℝ^{3} : Ξ_{r}^{i}(t, x, y) ≤ 0, i = 1, 2},
K_{r}^{−} = {(t, x, y) ∈ ℝ^{3} : Ξ_{r}^{1}(t, x, y) = 0, Ξ_{r}^{2}(t, x, y) ≤ 0},
K_{r}^{+} = {(t, x, y) ∈ ℝ^{3} : Ξ_{r}^{1}(t, x, y) ≤ 0, Ξ_{r}^{2}(t, x, y) = 0}.

Lemma 3 For an arbitrary ϕ and r ≤ \frac{1}{4},

F (t, x, y) · ∇Ξ_{r}^{1}(t, x, y) > 0 ((t, x, y) ∈ K_{r}^{−}),  (5)
F (t, x, y) · ∇Ξ_{r}^{2}(t, x, y) < 0 ((t, x, y) ∈ K_{r}^{+}).  (6)

Proof: We have

F (t, x, y) · ∇Ξ_{r}^{1}(t, x, y) = \frac{2}{r^{2}} (x^{2} + (cos(t^{2}) + 2)(x^{2} + y^{2})(x^{2} cos(ϕt) + xy sin(ϕt)) (7)
F (t, x, y) · ∇Ξ_{r}^{2}(t, x, y) = \frac{2}{r^{2}} (−y^{2} + (cos(t^{2}) + 2)(x^{2} + y^{2})(xy sin(ϕt) − y^{2} cos(ϕt)) (8)

If (t, x, y) ∈ K_{r}^{−} then |x| = r and |y| ≤ r, hence, by (7),

F (t, x, y) · ∇Ξ_{r}^{1}(t, x, y) ≥ \frac{2}{r^{2}} (r^{2} − 6r^{2} · 2r^{2}) = 2 − 24r^{2},

hence (5) is satisfied. By (8), it follows on an analogous way that if (t, x, y) ∈ K_{r}^{+} then

F (t, x, y) · ∇Ξ_{r}^{2}(t, x, y) ≤ −2 + 24r^{2},

hence (6) follows. Lemma 3 is proved.

□

Let ω > 0. For k ∈ ℤ and t ∈ [\frac{2π}{ϕ} k, \frac{2π}{ϕ} k + R/ω] put

Π_{R,ω}^{1}(t, x, y) = \frac{1}{(R − ω(t − \frac{2π}{ϕ} k))^{2}} x^{2} − 1,

Π_{R,ω}^{2}(t, x, y) = \frac{1}{(R − ω(t − \frac{2π}{ϕ} k))^{2}} y^{2} − 1.

and

P_{r,R,ω} = \{(t, x, y) ∈ [\frac{2π}{ϕ} k, \frac{2π}{ϕ} k + \frac{R−r}{ω}] × ℝ^{2} : Π_{R,ω}^{i}(t, x, y) ≤ 0, i = 1, 2\}, P_{r,R,ω}^{−} = \{(t, x, y) ∈ [\frac{2π}{ϕ} k, \frac{2π}{ϕ} k + \frac{R−r}{ω}] × ℝ^{2} : Π_{R,ω}^{1}(t, x, y) = 0, Π_{R,ω}^{2}(t, x, y) ≤ 0\}, P_{r,R,ω}^{+} = \{(t, x, y) ∈ [\frac{2π}{ϕ} , \frac{2π}{ϕ} k + \frac{R−r}{ω}] × ℝ^{2} : Π_{R,ω}^{1}(t, x, y) ≤ 0, Π_{R,ω}^{2}(t, x, y) = 0\}.

Lemma 4 For ϕ > 0 sufficiently small and ω ≤ \frac{r}{2}

F (t, x, y) · ∇Π_{R,ω}^{1}(t, x, y) > 0 ((t, x, y) ∈ P_{r,R,ω}^{−}),  (9)
F (t, x, y) · ∇Π_{R,ω}^{2}(t, x, y) < 0 ((t, x, y) ∈ P_{r,R,ω}^{+}).  (10)

Proof:

F (t, x, y) · ∇Π_{R,ω}^{1}(t, x, y) = \frac{2ωx^{2}}{(R − ω(t − \frac{2π}{ϕ} k))^{3}} +

\frac{2}{(R − ω(t − \frac{2π}{ϕ} k))^{2}} (x^{2} + (cos(t^{2}) + 2)(x^{2} + y^{2})(x^{2} cos(ϕt) + xy sin(ϕt))) ,

F (t, x, y) · ∇Π_{R,ω}^{2}(t, x, y) = \frac{2ωy^{2}}{(R − ω(t − \frac{2π}{ϕ}))^{3}} +

\frac{2}{(R − ω(t − \frac{2π}{ϕ}))^{2}} (−y^{2} + (cos(t^{2}) + 2)(x^{2} + y^{2})(xy sin(ϕt) − y^{2} cos(ϕt))) .

Let ϕ > 0 be small enough that

cos(ϕt) > 0 (t ∈ [\frac{2π}{ϕ} k, \frac{2π}{ϕ} k + \frac{R − r}{ω}]).  (11)

Let (t, x, y) ∈ P_{R,ω}^{−}, hence |x| = R − ω(t − \frac{2π}{ϕ} k) and |y| ≤ R − ω(t − \frac{2π}{ϕ} k). For ϕ > 0 sufficiently small we obtain:

\begin{matrix}F (t, x, y) · ∇Π_{R,ω}^{1}(t, x, y) > \frac{2ω}{r} + 2 − 2(R − ω(t − \frac{2π}{ϕ} k))^{2}ϕ(t − \frac{2π}{ϕ} k) \\ ≥ 2 − 6R^{2}ϕ\frac{R−r}{ω} ≥ 1. \end{matrix}

By a similar argument we conclude that if (t, x, y) ∈ P_{R,ω}^{+} then for ϕ > 0 sufficiently small

F (t, x, y) · ∇Π_{R,ω}^{2}(t, x, y) < −2 + \frac{2ω}{r} + 6R^{2}ϕ\frac{R − r}{ω} ≤ −1 + \frac{2ω}{r} .

Since ω ≤ r/2 is assumed, the proof of Lemma 4 is finished.

□

For t ∈ [\frac{2π}{ϕ} k − \frac{R−r}{ω} , \frac{2π}{ϕ} k] we put

Σ_{R,ω}^{1}(t, x, y) = Π_{R,ω}^{1}(\frac{2π}{ϕ} k − t, x, y),

Σ_{R,ω}^{2}(t, x, y) = Π_{R,ω}^{2}(\frac{2π}{ϕ} k − t, x, y),

S_{r,R,ω} = \{(t, x, y) ∈ [\frac{2π}{ϕ} k − \frac{R−r}{ω} , \frac{2π}{ϕ} k] × ℝ^{2} : Σ_{R,ω}^{i}(t, x, y) ≤ 0, i = 1, 2\}, S_{r,R,ω}^{−} = \{(t, x, y) ∈ [\frac{2π}{ϕ} k − \frac{R−r}{ω} , \frac{2π}{ϕ} k] × ℝ^{2} : Σ_{R,ω}^{1}(t, x, y) = 0, Σ_{R,ω}^{2}(t, x, y) ≤ 0\}, S_{r,R,ω}^{+} = \{(t, x, y) ∈ [\frac{2π}{ϕ} k − \frac{R−r}{ω} , \frac{2π}{ϕ} k] × ℝ^{2} : Σ_{R,ω}^{1}(t, x, y) ≤ 0, Σ_{R,ω}^{2}(t, x, y) = 0\}.

One can check that

Lemma 5 For ϕ > 0 sufficiently small

F (t, x, y) · ∇Σ_{R,ω}^{1}(t, x, y) > 0 ((t, x, y) ∈ S_{r,R,ω}^{−})  (12)
F (t, x, y) · ∇Σ_{R,ω}^{2}(t, x, y) < 0 ((t, x, y) ∈ S_{r,R,ω}^{+})  (13)
□

Lemma 6 If ϕ < \frac{2ω}{R} then P_{r,R,ω} ⊂ L_{R} and S_{r,R,ω} ⊂ L_{R}.

Proof: Essentially it is lemma 6 in [SW].

□

Proof of Theorem 2: All the estimates in the above lemmas are satisfied for R = 3, r = 1/4, ω = 1/8, and ϕ > 0 sufficiently small. Define

U = P_{\frac{1}{4} ,3, \frac{1}{8}} ∪ K_{\frac{1}{4}} ∪ S_{\frac{1}{4} ,3, \frac{1}{8}} ,
Z = L_{3}.

It follows by Lemmas 2, 3, 4, and 5 that U and Z are admissible proper pairs with the sequence t_{n} = \frac{2π}{ϕ} n, and

U_{0} = Z_{0} = {(x, y) ∈ ℝ^{2} : |x| ≤ 3, |y| ≤ 3},

U_{0}^{−} = Z_{0}^{−} = {(x, y) ∈ ℝ^{2} : |x| = 3, |y| ≤ 3}.

The set Z is a twisted prism with a square base, its successive cross-sections Z_{t} are obtained by the rotation of Z_{0} by the angle ϕt/2 (t ∈ [0, 2π]), hence we can take the map (x, y) → (−x, −y) as a homeomorphism h̃ corresponding to Z. Thus

μ_{Z_{tn,tn+1}} ∘ μ_{Z_{tn,tn+1}} = id_{H(U_{0},U_{0}^{−})}, Lef(μ_{W}) = 1.

U is a regular square-based prism, broadening at t_{n}, hence

μ_{U_{tn,tn+1}} = id_{H(U_{0},U_{0}^{−})}.

Moreover, U ⊂ Z because K_{\frac{1}{6}} ⊂ L_{3} and Lemma 6 is valid. Since

χ(U_{0}, U_{0}^{−}) = −1,

all conditions (a), (b), and (c) are satisfied.

□

6 Appendix.

Let n be an odd natural number. For a p ∈ ℕ and an r ∈ ℤ define:

S_{n}(p, r) = ∑_{{0≤s≤p \atop n|s−r}}(−1)^{s−r}\binom{p}{s}.  (14)

It is easy to check that

S_{n}(0, r) = \{ \begin{matrix}(−1)^{r} & if r = k · n; \\ 0 & otherwise. \end{matrix}

Two following lemmas gather some useful properties of S_{n}(p, r)

Lemma 7 S_{n}(p, r) = S_{n}(p − 1, r) + S_{n}(p − 1, r − 1).

Proof:

S_{n}(p, r) = ∑_{{0≤s≤p \atop n|s−r}}(−1)^{s−r}\binom{p}{s}

= ∑_{{0≤s≤p \atop n|s−r}}(−1)^{s−r}\binom{p − 1}{s − 1} + ∑_{{0≤s≤p \atop n|s−r}}(−1)^{s−r}\binom{p − 1}{s}

= ∑_{{1≤s≤p \atop n|s−r}}(−1)^{s−r}\binom{p − 1}{s − 1} + ∑_{{0≤s≤p−1 \atop n|s−r}}(−1)^{s−r}\binom{p − 1}{s}

= ∑_{{0≤s≤p−1 \atop n|s+1−r}}(−1)^{s+1−r}\binom{p − 1}{s} + ∑_{{0≤s≤p−1 \atop n|s−r}}(−1)^{s−r}\binom{p − 1}{s}

= ∑_{{0≤s≤p−1 \atop n|s−(r−1)}}(−1)^{s−(r−1)}\binom{p − 1}{s} + ∑_{{0≤s≤p−1 \atop n|s−r}}(−1)^{s−r}\binom{p − 1}{s}

= S_{n}(p − 1, r − 1) + S_{n}(p − 1, r).
□

Lemma 8 The 2–directional sequence {S_{n}(p, r)}_{r∈ℤ} satisfies following properties:

1. S_{n}(p, r) = −S_{n}(p, r + n);

2. S_{n}(p, p/2 + α) = S_{n}(p, p/2 − α) for such α that p/2 + α ∈ ℤ.

Proof: The lemma is obvious for p = 0.

Now suppose, that it is satisfied for p − 1.

1.

S_{n}(p, r) = S_{n}(p − 1, r) + S_{n}(p − 1, r − 1)
= −S_{n}(p − 1, r + n) − S_{n}(p − 1, r − 1 + n)
= −(S_{n}(p − 1, r + n) + S_{n}(p − 1, r + n − 1))
= −S_{n}(p, r + n);

2.

S_{n}(p, p/2 + α) = S_{n}(p − 1, p/2 + α − 1) + S_{n}(p − 1, p/2 + α)
= S_{n}(p − 1, \frac{(p − 1)}{2} + α − \frac{1}{2}) + S_{n}(p − 1, \frac{(p − 1)}{2} + α + \frac{1}{2})
= S_{n}(p − 1, \frac{(p − 1)}{2} − α + \frac{1}{2}) + S_{n}(p − 1, \frac{(p − 1)}{2} − α − \frac{1}{2})
= S_{n}(p − 1, p/2 − α) + S_{n}(p − 1, p/2 − α − 1)
= S_{n}(p, p/2 − α).
□

Now we are ready to state an important property of numbers S_{n}(p, r):

Theorem 4

S_{n}(p, r) \{ \begin{matrix}= 0 & if \frac{p−2r}{n} ∈ 2ℕ + 1 \\ > 0 & if \frac{p−2r}{n} ∈ (4k − 1, 4k + 1) \\ < 0 & if \frac{p−2r}{n} ∈ (4k + 1, 4k + 3) \end{matrix}

for odd n ∈ ℕ and all p ≥ n − 1.

Proof: Let p = n − 1. Note, that \frac{p−2r}{n} cannot be an odd number.

From the definition,

S_{n}(n − 1, r) = (−1)^{s−r}\binom{n − 1}{s}

for such s ∈ {0, 1, …, n − 1}, that there exists k ∈ ℕ that s − r = kn. But (−1)^{s−r} = (−1)^{kn} = (−1)^{k} = (−1)^{\frac{s−r}{n}} = (−1)^{\frac{r−s}{n}} , so

S_{n}(n − 1, r) > 0 ⇔ \frac{r − s}{n} ∈ 2ℕ ⇔ [\frac{r}{n}] ∈ 2ℕ

⇔ \frac{r}{n} ∈ [2m, 2m + 1) ⇔ \frac{p − 2r}{n} ∈ (−4m − 1 − \frac{1}{n}, −4m + 1 − \frac{1}{n}].

Since p is even number, no number of the form \frac{p−2r}{n} lies in intervals of the form (2l + 1 − \frac{1}{n}, 2l + 1 + \frac{1}{n}). Hence

\frac{p − 2r}{n} ∈ (−4m − 1 − \frac{1}{n}, −4m + 1 − \frac{1}{n}] ⇔ \frac{p − 2r}{n} ∈ (−4m − 1, −4m + 1).

Similarly,

S_{n}(n − 1, r) < 0 ⇔ \frac{p − 2r}{n} ∈ (−4m + 1 − \frac{1}{n}, −4m + 3 − \frac{1}{n}]

⇔ \frac{p − 2r}{n} ∈ (−4m + 1, −4m + 3),

and these equivalences complete the proof for p = n − 1.

Now suppose, that the theorem is fulfilled for p − 1.

1. \frac{p−2r}{n} = 2k + 1 From the Lemma 8 we have S_{n}(p − 1, r) = S_{n}(p − 1, p − 1 − r) so:

S_{n}(p, r) = S_{n}(p−1, r)+S_{n}(p−1, r−1) = S_{n}(p−1, p−1−r)+S_{n}(p−1, r−1)

= S_{n}(p − 1, p − 1 − r) + S_{n}(p − 1, p − 1 − r + (2r − p))
= S_{n}(p − 1, p − 1 − r) + S_{n}(p − 1, p − 1 − r − (2k + 1)n) = 0

because (2k + 1)n is an odd multiplicity of n.

2. \frac{p−2r}{n} ∈ (4k − 1, 4k + 1)

The numbers \frac{(p−1)−2(r−1)}{n} = \frac{p−2r}{n} + \frac{1}{n} and \frac{(p−1)−2r}{n} = \frac{p−2r}{n} − \frac{1}{n} are both of the form \frac{q}{n} and differ from \frac{p−2r}{n} by \frac{1}{n}. Since both ends of the interval [4k − 1, 4k + 1] are of the form \frac{q}{n} and \frac{p−2r}{n} ∈ (4k − 1, 4k + 1), numbers \frac{(p−1)−2(r−1)}{n}, \frac{(p−1)−2r}{n} must lie in [4k − 1, 4k + 1] and at least one of them is in the interior of [4k − 1, 4k + 1]. Thus we have following inequalities:

S_{n}(p − 1, r) ≥ 0, S_{n}(p − 1, r − 1) ≥ 0

and at most one of them is an equality. Hence S_{n}(p, r) = S_{n}(p − 1, r) + S_{n}(p − 1, r − 1) > 0.

3. \frac{p−2r}{n} ∈ (4k + 1, 4k + 3)

Analogically as above we get: S_{n}(p, r) = S_{n}(p−1, r)+S_{n}(p−1, r −1) < 0.

□

As a consequence of the theorem, we obtain:

Corollary 2 ∑_{{0≤s≤p \atop n|s}}(−1)^{p−s}\binom{p}{s} = 0 ⟺ p/n is an odd number.

Proof:

∑_{{0≤s≤p \atop n|s}}(−1)^{p−s}\binom{p}{s} = (−1)^{p} · ∑_{{0≤s≤p \atop n|s}}(−1)^{s}\binom{p}{s} = (−1)^{p} · S_{n}(p, 0).

□

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