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Topological Dynamics and Universality in Chaos I. Proof of Sharkovskis Theorem Ugur G. Abdulla FIT Colloquium May 30, 2014 1 Ugur G. Abdulla Topological Dynamics and Universality in Chaos I. Proof of Sharkovskis Theo Introduction Let f


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Topological Dynamics and Universality in Chaos

  • I. Proof of Sharkovski’s Theorem

Ugur G. Abdulla

FIT Colloquium

May 30, 2014

Ugur G. Abdulla Topological Dynamics and Universality in Chaos I. Proof of Sharkovski’s Theo

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Introduction

Let f : I → I be a continuous map, and I be an interval. Interval is a connected subset of the real line which contains more than one point. < a, b > is a closed interval with endpoints a and b. f 1 = f, f n+1 = f ◦ f n, n ≥ 1

Ugur G. Abdulla Topological Dynamics and Universality in Chaos I. Proof of Sharkovski’s Theo

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Introduction

Let f : I → I be a continuous map, and I be an interval. Interval is a connected subset of the real line which contains more than one point. < a, b > is a closed interval with endpoints a and b. f 1 = f, f n+1 = f ◦ f n, n ≥ 1 x y y = x x0 x1 x2 x3

Ugur G. Abdulla Topological Dynamics and Universality in Chaos I. Proof of Sharkovski’s Theo

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Sharkovski’s Theorem

{f n(x) : n ≥ 0} be an orbit of x.

Ugur G. Abdulla Topological Dynamics and Universality in Chaos I. Proof of Sharkovski’s Theo

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Sharkovski’s Theorem

{f n(x) : n ≥ 0} be an orbit of x. c ∈ I is a fixed point if f(c) = c.

Ugur G. Abdulla Topological Dynamics and Universality in Chaos I. Proof of Sharkovski’s Theo

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Sharkovski’s Theorem

{f n(x) : n ≥ 0} be an orbit of x. c ∈ I is a fixed point if f(c) = c. If I = [a, b] is compact ⇒ ∃ a fixed point c ∈ I, since f(a) − a ≥ 0 ≥ f(b) − b

Ugur G. Abdulla Topological Dynamics and Universality in Chaos I. Proof of Sharkovski’s Theo

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Sharkovski’s Theorem

{f n(x) : n ≥ 0} be an orbit of x. c ∈ I is a fixed point if f(c) = c. If I = [a, b] is compact ⇒ ∃ a fixed point c ∈ I, since f(a) − a ≥ 0 ≥ f(b) − b A point c ∈ I is said to be periodic point of f with period m if f m(c) = c, f k(c) = c for 1 ≤ k < m.

Ugur G. Abdulla Topological Dynamics and Universality in Chaos I. Proof of Sharkovski’s Theo

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Sharkovski’s Theorem

{f n(x) : n ≥ 0} be an orbit of x. c ∈ I is a fixed point if f(c) = c. If I = [a, b] is compact ⇒ ∃ a fixed point c ∈ I, since f(a) − a ≥ 0 ≥ f(b) − b A point c ∈ I is said to be periodic point of f with period m if f m(c) = c, f k(c) = c for 1 ≤ k < m. Periodic m-orbit is c, f(c), f 2(c), · · · , f m−1(c)

Ugur G. Abdulla Topological Dynamics and Universality in Chaos I. Proof of Sharkovski’s Theo

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Sharkovski’s Theorem

{f n(x) : n ≥ 0} be an orbit of x. c ∈ I is a fixed point if f(c) = c. If I = [a, b] is compact ⇒ ∃ a fixed point c ∈ I, since f(a) − a ≥ 0 ≥ f(b) − b A point c ∈ I is said to be periodic point of f with period m if f m(c) = c, f k(c) = c for 1 ≤ k < m. Periodic m-orbit is c, f(c), f 2(c), · · · , f m−1(c)

Theorem 1

(Sharkovsky, 1964) Let the positive integers be totally ordered in the following way: 1 ≻ 2 ≻ 22 ≻ 23 ≻ ... ≻ 22·5 ≻ 22·3 ≻ ... ≻ 2·5 ≻ 2·3 ≻ ... ≻ 7 ≻ 5 ≻ 3 If f has a cycle of period n and m ≻ n, then f also has a periodic orbit

  • f period m.

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Proof of Sharkovski’s Theorem

References

◮ A.N.Sharkovsky, Coexistence of Cycles of a

Continuous Transformation of a Line into itself

  • Ukrain. Mat. Zhurn., 16, 1(1964), 61-71.

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Proof of Sharkovski’s Theorem

References

◮ A.N.Sharkovsky, Coexistence of Cycles of a

Continuous Transformation of a Line into itself

  • Ukrain. Mat. Zhurn., 16, 1(1964), 61-71.

◮ L.S. Block, J. Guckenheimer, M. Misiurewich, L.S.

Young, Periodic Points and Topological Entropy

  • f One-dimensional Maps, Global Theory of

Dynamical Systems, Proc. International Conf., Northwestern Univ., Evanston, Ill.,1979, 18-34.

Ugur G. Abdulla Topological Dynamics and Universality in Chaos I. Proof of Sharkovski’s Theo

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Proof of Sharkovski’s Theorem

References

◮ A.N.Sharkovsky, Coexistence of Cycles of a

Continuous Transformation of a Line into itself

  • Ukrain. Mat. Zhurn., 16, 1(1964), 61-71.

◮ L.S. Block, J. Guckenheimer, M. Misiurewich, L.S.

Young, Periodic Points and Topological Entropy

  • f One-dimensional Maps, Global Theory of

Dynamical Systems, Proc. International Conf., Northwestern Univ., Evanston, Ill.,1979, 18-34.

◮ L.S.Block, W.A.Coppel, Dynamics in One

Dimension, Springer-Verlag, Berlin, 1992.

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Proof of Sharkovski’s Theorem

Lemma 2

If J is a compact subinterval such that J ⊆ f(J), then f has a fixed point in J.

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Proof of Sharkovski’s Theorem

Lemma 2

If J is a compact subinterval such that J ⊆ f(J), then f has a fixed point in J.

  • Proof. If J = [a, b] then for some c, d ∈ J we have f(c) = a, f(d) = b.

Thus f(c) ≤ c, f(d) ≥ d, and by the intermediate value theorem ∃c∗ ∈ [a, b] f(c∗) = c∗.

Ugur G. Abdulla Topological Dynamics and Universality in Chaos I. Proof of Sharkovski’s Theo

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Proof of Sharkovski’s Theorem

Lemma 2

If J is a compact subinterval such that J ⊆ f(J), then f has a fixed point in J.

  • Proof. If J = [a, b] then for some c, d ∈ J we have f(c) = a, f(d) = b.

Thus f(c) ≤ c, f(d) ≥ d, and by the intermediate value theorem ∃c∗ ∈ [a, b] f(c∗) = c∗.

Lemma 3

If J, K are compact subintervals such that K ⊆ f(J), then there is a compact subinterval L ⊆ J such that f(L) = K.

Ugur G. Abdulla Topological Dynamics and Universality in Chaos I. Proof of Sharkovski’s Theo

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Proof of Sharkovski’s Theorem

Lemma 2

If J is a compact subinterval such that J ⊆ f(J), then f has a fixed point in J.

  • Proof. If J = [a, b] then for some c, d ∈ J we have f(c) = a, f(d) = b.

Thus f(c) ≤ c, f(d) ≥ d, and by the intermediate value theorem ∃c∗ ∈ [a, b] f(c∗) = c∗.

Lemma 3

If J, K are compact subintervals such that K ⊆ f(J), then there is a compact subinterval L ⊆ J such that f(L) = K.

  • Proof. Let K = [a, b], c = sup{x ∈ J : f(x) = a}. If f(x) = b for some

x ∈ J with x > c, let d be the least and take L = [c, d].

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Proof of Sharkovski’s Theorem

Lemma 2

If J is a compact subinterval such that J ⊆ f(J), then f has a fixed point in J.

  • Proof. If J = [a, b] then for some c, d ∈ J we have f(c) = a, f(d) = b.

Thus f(c) ≤ c, f(d) ≥ d, and by the intermediate value theorem ∃c∗ ∈ [a, b] f(c∗) = c∗.

Lemma 3

If J, K are compact subintervals such that K ⊆ f(J), then there is a compact subinterval L ⊆ J such that f(L) = K.

  • Proof. Let K = [a, b], c = sup{x ∈ J : f(x) = a}. If f(x) = b for some

x ∈ J with x > c, let d be the least and take L = [c, d]. Otherwise f(x) = b for some x ∈ J with x < c.

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Proof of Sharkovski’s Theorem

Lemma 2

If J is a compact subinterval such that J ⊆ f(J), then f has a fixed point in J.

  • Proof. If J = [a, b] then for some c, d ∈ J we have f(c) = a, f(d) = b.

Thus f(c) ≤ c, f(d) ≥ d, and by the intermediate value theorem ∃c∗ ∈ [a, b] f(c∗) = c∗.

Lemma 3

If J, K are compact subintervals such that K ⊆ f(J), then there is a compact subinterval L ⊆ J such that f(L) = K.

  • Proof. Let K = [a, b], c = sup{x ∈ J : f(x) = a}. If f(x) = b for some

x ∈ J with x > c, let d be the least and take L = [c, d]. Otherwise f(x) = b for some x ∈ J with x < c. Let c′ be the greatest and let d′ ≤ c be the least x ∈ J with x > c′ for which f(x) = a. Then we can take L = [c′, d′].

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Proof of Sharkovski’s Theorem

Lemma 4

If J0, J1, ..., Jm are compact subintervals such that Jk ⊆ f(Jk−1) (1 ≤ k ≤ m), then there is a compact subinterval L ⊆ J0 such that f m(L) = Jm and f k(L) ⊆ Jk (1 ≤ k < m).

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Proof of Sharkovski’s Theorem

Lemma 4

If J0, J1, ..., Jm are compact subintervals such that Jk ⊆ f(Jk−1) (1 ≤ k ≤ m), then there is a compact subinterval L ⊆ J0 such that f m(L) = Jm and f k(L) ⊆ Jk (1 ≤ k < m). If also J0 ⊆ Jm, then there exists a point y such that f m(y) = y and f k(y) ∈ Jk (0 ≤ k < m).

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Proof of Sharkovski’s Theorem

Lemma 4

If J0, J1, ..., Jm are compact subintervals such that Jk ⊆ f(Jk−1) (1 ≤ k ≤ m), then there is a compact subinterval L ⊆ J0 such that f m(L) = Jm and f k(L) ⊆ Jk (1 ≤ k < m). If also J0 ⊆ Jm, then there exists a point y such that f m(y) = y and f k(y) ∈ Jk (0 ≤ k < m).

  • Proof. Note that the first assertion holds for m = 1 due to Lemma 3:

J1 ⊆ f(J0) ⇒ ∃L ⊆ J0, f(L) = J1 Prove by induction: let m > 1 be fixed and assume the assertion is true for all smaller values. We have J1 ⊆ f(J0), J2 ⊆ f(J1), ..., Jm−1 ⊆ f(Jm−2), Jm ⊆ f(Jm−1) Induction assumption applied to last m − 1 relations ⇒ ∃L′ ⊆ J1 : f m−1(L′) = Jm, f k(L′) ⊆ Jk+1, 1 ≤ k < m − 1

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Proof of Sharkovski’s Theorem

Now we have L′ ⊆ J1 ⊆ f(J0) ⇒ ∃L ⊆ J0 such that f(L) = L′ ⊆ J1 ⇒ f m(L) = f m−1(L′) = Jm; f k(L) = f k−1(L′) ⊆ Jk, 2 ≤ k < m Hence first assertion of the lemma is proved.

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Proof of Sharkovski’s Theorem

Now we have L′ ⊆ J1 ⊆ f(J0) ⇒ ∃L ⊆ J0 such that f(L) = L′ ⊆ J1 ⇒ f m(L) = f m−1(L′) = Jm; f k(L) = f k−1(L′) ⊆ Jk, 2 ≤ k < m Hence first assertion of the lemma is proved. If also J0 ⊆ Jm, from the first assertion ⇒ ∃L ⊆ J0 : f m(L) = Jm ⊇ J0 ⊇ L and by Lemma 2 we have ∃c ∈ L ⊆ J0 : f m(c) = c, and f k(c) ⊆ Jk, 1 ≤ k < m Lemma 4 is proved.

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Proof of Sharkovski’s Theorem

Lemma 5

Between any two points of a periodic orbit of period n > 1 there is a point of a periodic orbit of period less than n.

  • Proof. Let a < b are two adjacent points of n-orbit. Consider all m < n

such that f m(b) < b. There is at least one such m. Since there is one more point of the orbit to the left of b than to the left of a, for some m such that 1 ≤ m < n we have f m(a) > a, f m(b) < b. If f m is defined

  • n [a, b], then ∃c ∈ (a, b) f m(c) = c.

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Proof of Sharkovski’s Theorem

Lemma 5

Between any two points of a periodic orbit of period n > 1 there is a point of a periodic orbit of period less than n.

  • Proof. Let a < b are two adjacent points of n-orbit. Consider all m < n

such that f m(b) < b. There is at least one such m. Since there is one more point of the orbit to the left of b than to the left of a, for some m such that 1 ≤ m < n we have f m(a) > a, f m(b) < b. If f m is defined

  • n [a, b], then ∃c ∈ (a, b) f m(c) = c.

Assume f m is not defined throughout [a, b]. Let Jk =< f k(a), f k(b) >, 1 ≤ k ≤ m. We have Jk ⊆ f(Jk−1), 1 ≤ k ≤ m We also have J0 ⊆ Jm, since f m(a) ≥ b, f m(b) ≤ a. hence, the assertion of the Lemma follows from Lemma 4.

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Proof of Sharkovski’s Theorem

Let B = {x1 < x2 < · · · < xn} be n-orbit of f.

Definition 1.1

If f(xi) = xsi, 1 ≤ si ≤ n, i = 1, 2, ..., n, then B is associated with cyclic permutation 1 2 . . . n s1 s2 . . . sn

  • Definition 1.2

Let Ii = [xi, xi+1]. Digraph of a cycle is a directed graph of transitions with vertices I1, I2, · · · , In−1 and oriented edges Ii → Is if Is ⊆ f(Ii).

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Proof of Sharkovski’s Theorem

Properties of Digraphs:

  • 1. ∀ Ij ∃ at least one Ik for which Ij → Ik. Moreover, it is always

possible to choose k = j , unless n = 2.

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Proof of Sharkovski’s Theorem

Properties of Digraphs:

  • 1. ∀ Ij ∃ at least one Ik for which Ij → Ik. Moreover, it is always

possible to choose k = j , unless n = 2.

  • 2. ∀ Ik ∃ at least one Ij for which Ij → Ik. Moreover, it is always

possible to choose j = k, unless n is even and k = n

2 .

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Proof of Sharkovski’s Theorem

Properties of Digraphs:

  • 1. ∀ Ij ∃ at least one Ik for which Ij → Ik. Moreover, it is always

possible to choose k = j , unless n = 2.

  • 2. ∀ Ik ∃ at least one Ij for which Ij → Ik. Moreover, it is always

possible to choose j = k, unless n is even and k = n

2 .

Proof: Suppose there is no j = k such that Ij → Ik. Then if i = k, f(xi) ≤ xk ⇒ f(xi+1) ≤ xk and f(xi) ≥ xk+1 ⇒ f(xi+1) ≥ xk+1.

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Proof of Sharkovski’s Theorem

Properties of Digraphs:

  • 1. ∀ Ij ∃ at least one Ik for which Ij → Ik. Moreover, it is always

possible to choose k = j , unless n = 2.

  • 2. ∀ Ik ∃ at least one Ij for which Ij → Ik. Moreover, it is always

possible to choose j = k, unless n is even and k = n

2 .

Proof: Suppose there is no j = k such that Ij → Ik. Then if i = k, f(xi) ≤ xk ⇒ f(xi+1) ≤ xk and f(xi) ≥ xk+1 ⇒ f(xi+1) ≥ xk+1. If f(xk+1) ≥ xk+1 ⇒ f(xi) ≥ xk+1 for k < i ≤ n ⇒ proper subset {xk+1, ..., xn} is mapped to itself. Hence, f(xk+1) ≤ xk, and similarly f(xk) ≥ xk+1, and accordingly there is a loop Ik → Ik.

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Proof of Sharkovski’s Theorem

Properties of Digraphs:

  • 1. ∀ Ij ∃ at least one Ik for which Ij → Ik. Moreover, it is always

possible to choose k = j , unless n = 2.

  • 2. ∀ Ik ∃ at least one Ij for which Ij → Ik. Moreover, it is always

possible to choose j = k, unless n is even and k = n

2 .

Proof: Suppose there is no j = k such that Ij → Ik. Then if i = k, f(xi) ≤ xk ⇒ f(xi+1) ≤ xk and f(xi) ≥ xk+1 ⇒ f(xi+1) ≥ xk+1. If f(xk+1) ≥ xk+1 ⇒ f(xi) ≥ xk+1 for k < i ≤ n ⇒ proper subset {xk+1, ..., xn} is mapped to itself. Hence, f(xk+1) ≤ xk, and similarly f(xk) ≥ xk+1, and accordingly there is a loop Ik → Ik. f(xi) ≤ xk, k < i ≤ n ⇒ n − k ≤ k ⇒ n ≤ 2k, f(xi) ≥ xk+1, 1 ≤ i ≤ k ⇒ k ≤ n − k ⇒ n ≥ 2k ⇒ n = 2k

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Proof of Sharkovski’s Theorem

Properties of Digraphs:

  • 1. ∀ Ij ∃ at least one Ik for which Ij → Ik. Moreover, it is always

possible to choose k = j , unless n = 2.

  • 2. ∀ Ik ∃ at least one Ij for which Ij → Ik. Moreover, it is always

possible to choose j = k, unless n is even and k = n

2 .

Proof: Suppose there is no j = k such that Ij → Ik. Then if i = k, f(xi) ≤ xk ⇒ f(xi+1) ≤ xk and f(xi) ≥ xk+1 ⇒ f(xi+1) ≥ xk+1. If f(xk+1) ≥ xk+1 ⇒ f(xi) ≥ xk+1 for k < i ≤ n ⇒ proper subset {xk+1, ..., xn} is mapped to itself. Hence, f(xk+1) ≤ xk, and similarly f(xk) ≥ xk+1, and accordingly there is a loop Ik → Ik. f(xi) ≤ xk, k < i ≤ n ⇒ n − k ≤ k ⇒ n ≤ 2k, f(xi) ≥ xk+1, 1 ≤ i ≤ k ⇒ k ≤ n − k ⇒ n ≥ 2k ⇒ n = 2k

  • 3. Digraph always contains a loop

We have f(x1) > x1, f(xn) < xn. Let k = min{1 ≤ j < n : f(xj) ≥ xj+1, f(xj+1) ≤ xj} Then Ik → Ik.

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Proof of Sharkovski’s Theorem

Fundamental Cycle

Definition 1.3

Given n-orbit, a cycle J0 → J1 → · · · → Jn−1 → J0

  • f length n in the digraph is called a Fundamental Cycle (FC) if J0

contains an endpoint c s.t. f k(c) is an endpoint of Jk for 1 ≤ k < n.

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Proof of Sharkovski’s Theorem

Fundamental Cycle

Definition 1.3

Given n-orbit, a cycle J0 → J1 → · · · → Jn−1 → J0

  • f length n in the digraph is called a Fundamental Cycle (FC) if J0

contains an endpoint c s.t. f k(c) is an endpoint of Jk for 1 ≤ k < n. FC always exists and unique.

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Proof of Sharkovski’s Theorem

Fundamental Cycle

Definition 1.3

Given n-orbit, a cycle J0 → J1 → · · · → Jn−1 → J0

  • f length n in the digraph is called a Fundamental Cycle (FC) if J0

contains an endpoint c s.t. f k(c) is an endpoint of Jk for 1 ≤ k < n. FC always exists and unique. In the FC some vertex must occur at least twice among J0, ..., Jn−1, since digraph has only n − 1 vertices.

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Proof of Sharkovski’s Theorem

Fundamental Cycle

Definition 1.3

Given n-orbit, a cycle J0 → J1 → · · · → Jn−1 → J0

  • f length n in the digraph is called a Fundamental Cycle (FC) if J0

contains an endpoint c s.t. f k(c) is an endpoint of Jk for 1 ≤ k < n. FC always exists and unique. In the FC some vertex must occur at least twice among J0, ..., Jn−1, since digraph has only n − 1 vertices. On the

  • ther hand, every vertex occurs at most twice, since interval Ik has two

endpoints.

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Proof of Sharkovski’s Theorem

Fundamental Cycle

Definition 1.3

Given n-orbit, a cycle J0 → J1 → · · · → Jn−1 → J0

  • f length n in the digraph is called a Fundamental Cycle (FC) if J0

contains an endpoint c s.t. f k(c) is an endpoint of Jk for 1 ≤ k < n. FC always exists and unique. In the FC some vertex must occur at least twice among J0, ..., Jn−1, since digraph has only n − 1 vertices. On the

  • ther hand, every vertex occurs at most twice, since interval Ik has two

endpoints.

Definition 1.4

Cycle in a digraph is said to be primitive if it does not consisit entirely of a cycle of smaller length described several times.

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Proof of Sharkovski’s Theorem

Fundamental Cycle

Definition 1.3

Given n-orbit, a cycle J0 → J1 → · · · → Jn−1 → J0

  • f length n in the digraph is called a Fundamental Cycle (FC) if J0

contains an endpoint c s.t. f k(c) is an endpoint of Jk for 1 ≤ k < n. FC always exists and unique. In the FC some vertex must occur at least twice among J0, ..., Jn−1, since digraph has only n − 1 vertices. On the

  • ther hand, every vertex occurs at most twice, since interval Ik has two

endpoints.

Definition 1.4

Cycle in a digraph is said to be primitive if it does not consisit entirely of a cycle of smaller length described several times. If FC contains Ik twice then it can be decomposed into two cycles of smaller length, each of which contains Ik once, and consequently is primitive.

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Proof of Sharkovski’s Theorem

Straffin’s Lemma

Lemma 6

Suppose f has a periodic point of period n > 1. If the associated digraph contains a primitive cycle J0 → J1 → · · · → Jm−1 → J0

  • f length m, then f has a periodic point y of period m such that

f k(y) ∈ Jk(0 ≤ k < m).

  • Proof. J1 ⊆ f(J0), J2 ⊆ f(J1), · · · , Jm−1 ⊆ f(Jm−2), J0 ⊆ f(Jm−1)

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Proof of Sharkovski’s Theorem

Straffin’s Lemma

Lemma 6

Suppose f has a periodic point of period n > 1. If the associated digraph contains a primitive cycle J0 → J1 → · · · → Jm−1 → J0

  • f length m, then f has a periodic point y of period m such that

f k(y) ∈ Jk(0 ≤ k < m).

  • Proof. J1 ⊆ f(J0), J2 ⊆ f(J1), · · · , Jm−1 ⊆ f(Jm−2), J0 ⊆ f(Jm−1)

Lemma 4 (w. Jm = J0) ⇒ ∃ y ∈ J0 f m(y) = y, f k(y) ∈ Jk(0 ≤ k < m)

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Proof of Sharkovski’s Theorem

Straffin’s Lemma

Lemma 6

Suppose f has a periodic point of period n > 1. If the associated digraph contains a primitive cycle J0 → J1 → · · · → Jm−1 → J0

  • f length m, then f has a periodic point y of period m such that

f k(y) ∈ Jk(0 ≤ k < m).

  • Proof. J1 ⊆ f(J0), J2 ⊆ f(J1), · · · , Jm−1 ⊆ f(Jm−2), J0 ⊆ f(Jm−1)

Lemma 4 (w. Jm = J0) ⇒ ∃ y ∈ J0 f m(y) = y, f k(y) ∈ Jk(0 ≤ k < m) Either m is a period of y or period of y is a factor of m.

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Proof of Sharkovski’s Theorem

Straffin’s Lemma

Lemma 6

Suppose f has a periodic point of period n > 1. If the associated digraph contains a primitive cycle J0 → J1 → · · · → Jm−1 → J0

  • f length m, then f has a periodic point y of period m such that

f k(y) ∈ Jk(0 ≤ k < m).

  • Proof. J1 ⊆ f(J0), J2 ⊆ f(J1), · · · , Jm−1 ⊆ f(Jm−2), J0 ⊆ f(Jm−1)

Lemma 4 (w. Jm = J0) ⇒ ∃ y ∈ J0 f m(y) = y, f k(y) ∈ Jk(0 ≤ k < m) Either m is a period of y or period of y is a factor of m. If y is not an endpoint of J0, then m is a period of y since cycle is primitive.

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Proof of Sharkovski’s Theorem

Straffin’s Lemma

Lemma 6

Suppose f has a periodic point of period n > 1. If the associated digraph contains a primitive cycle J0 → J1 → · · · → Jm−1 → J0

  • f length m, then f has a periodic point y of period m such that

f k(y) ∈ Jk(0 ≤ k < m).

  • Proof. J1 ⊆ f(J0), J2 ⊆ f(J1), · · · , Jm−1 ⊆ f(Jm−2), J0 ⊆ f(Jm−1)

Lemma 4 (w. Jm = J0) ⇒ ∃ y ∈ J0 f m(y) = y, f k(y) ∈ Jk(0 ≤ k < m) Either m is a period of y or period of y is a factor of m. If y is not an endpoint of J0, then m is a period of y since cycle is primitive. Assume y is an endpoint of J0.

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Proof of Sharkovski’s Theorem

Straffin’s Lemma

Lemma 6

Suppose f has a periodic point of period n > 1. If the associated digraph contains a primitive cycle J0 → J1 → · · · → Jm−1 → J0

  • f length m, then f has a periodic point y of period m such that

f k(y) ∈ Jk(0 ≤ k < m).

  • Proof. J1 ⊆ f(J0), J2 ⊆ f(J1), · · · , Jm−1 ⊆ f(Jm−2), J0 ⊆ f(Jm−1)

Lemma 4 (w. Jm = J0) ⇒ ∃ y ∈ J0 f m(y) = y, f k(y) ∈ Jk(0 ≤ k < m) Either m is a period of y or period of y is a factor of m. If y is not an endpoint of J0, then m is a period of y since cycle is primitive. Assume y is an endpoint of J0. Since y is an element of n-orbit ⇒ n is a divisor of

  • m. We have Jk ⊆ f(Jk−1) and f k(y) ∈ Jk ⇒ Jk is defined uniquely and

moreover, cycle is a multiple of the FC. Contradiction, since cycle is primitive.

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Proof of Sharkovski’s Theorem

Straffin’s Lemma ⇒ 3 ≺ m ≺ 2 ≺ 1

Suppose f has a 3-orbit: f(c) < c < f 2(c) with corresponding digraph I1 ⇄ I2

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Proof of Sharkovski’s Theorem

Straffin’s Lemma ⇒ 3 ≺ m ≺ 2 ≺ 1

Suppose f has a 3-orbit: f(c) < c < f 2(c) with corresponding digraph I1 ⇄ I2 I1 → I1 ⇒ there is a fixed point;

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Proof of Sharkovski’s Theorem

Straffin’s Lemma ⇒ 3 ≺ m ≺ 2 ≺ 1

Suppose f has a 3-orbit: f(c) < c < f 2(c) with corresponding digraph I1 ⇄ I2 I1 → I1 ⇒ there is a fixed point; I1 → I2 → I1 ⇒ there is a 2-orbit

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Proof of Sharkovski’s Theorem

Straffin’s Lemma ⇒ 3 ≺ m ≺ 2 ≺ 1

Suppose f has a 3-orbit: f(c) < c < f 2(c) with corresponding digraph I1 ⇄ I2 I1 → I1 ⇒ there is a fixed point; I1 → I2 → I1 ⇒ there is a 2-orbit ∀ positive integer m > 2 there is an m-orbit corresponding to primitive cycle of length m: I1 → I2 → I1 → I1 → · · · → I1

Lemma 7

If f has a periodic point of period > 1, then it has a fixed point and a periodic point of period 2.

  • Proof. Digraph has a loop ⇒ there is a fixed point. Let n > 1 be the

least positive integer such that f has a periodic point of period n. If n > 2 decompose FC into two primitive cycles. Since at least one of these has length greater than 1, by Straffin’s lemma we deduce there is a periodic point of period strictly between 1 and n.

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Proof of Sharkovski’s Theorem

Stefan Orbits

Lemma 8

Suppose f has a periodic orbit of odd period n > 1, but no periodic orbit

  • f odd period strictly between 1 and n. If c is the midpoint of the orbit
  • f odd period n, then the points of this orbit have the order

f n−1(c) < f n−3(c) < · · · < f 2(c) < c < f(c) < · · · < f n−2(c)

  • r the inverse order

f n−2(c) < · · · < f(c) < c < f 2(c) < · · · < f n−3(c) < f n−1(c) and associated digraph is given in the figure, where J1 =< c, f(c) > and Jk =< f k−2(c), f k(c) > for 1 < k < n. J1 J2 J3 Jn−3 Jn−2 Jn−1 · · ·

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Proof of Sharkovski’s Theorem

Proof of Lemma 8

  • Proof. Decompose FC into two primitive cycles, one of which has odd

length.

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Proof of Sharkovski’s Theorem

Proof of Lemma 8

  • Proof. Decompose FC into two primitive cycles, one of which has odd
  • length. Its length must be 1!

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Proof of Sharkovski’s Theorem

Proof of Lemma 8

  • Proof. Decompose FC into two primitive cycles, one of which has odd
  • length. Its length must be 1! Thus FC has the form

J1 → J1 → J2 → · · · → Jn−1 → J1 where Ji = J1 for 1 < i < n.

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Proof of Sharkovski’s Theorem

Proof of Lemma 8

  • Proof. Decompose FC into two primitive cycles, one of which has odd
  • length. Its length must be 1! Thus FC has the form

J1 → J1 → J2 → · · · → Jn−1 → J1 where Ji = J1 for 1 < i < n. If Ji = Jk, 1 < i < k < n, then we obtain a smaller primitive cycle, and by excluding the loop at J1 if necessary, we can arrange that its length is odd. Hence, J1, ..., Jn−1 are all distinct and thus a permutation of I1, ..., In−1.

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Proof of Sharkovski’s Theorem

Proof of Lemma 8

  • Proof. Decompose FC into two primitive cycles, one of which has odd
  • length. Its length must be 1! Thus FC has the form

J1 → J1 → J2 → · · · → Jn−1 → J1 where Ji = J1 for 1 < i < n. If Ji = Jk, 1 < i < k < n, then we obtain a smaller primitive cycle, and by excluding the loop at J1 if necessary, we can arrange that its length is odd. Hence, J1, ..., Jn−1 are all distinct and thus a permutation of I1, ..., In−1. Similarly, we cannot have Ji → Jk if k > i + 1 or if k = 1 and i = 1, n + 1.

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Proof of Sharkovski’s Theorem

Proof of Lemma 8

  • Proof. Decompose FC into two primitive cycles, one of which has odd
  • length. Its length must be 1! Thus FC has the form

J1 → J1 → J2 → · · · → Jn−1 → J1 where Ji = J1 for 1 < i < n. If Ji = Jk, 1 < i < k < n, then we obtain a smaller primitive cycle, and by excluding the loop at J1 if necessary, we can arrange that its length is odd. Hence, J1, ..., Jn−1 are all distinct and thus a permutation of I1, ..., In−1. Similarly, we cannot have Ji → Jk if k > i + 1 or if k = 1 and i = 1, n + 1. Suppose J1 = Ik = [a, b]. Since J1 → J2 ⇒ J2 is adjacent to J1 and either xk = a, xk+1 = f(a), xk−1 = f 2(a)

  • r

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Proof of Sharkovski’s Theorem

Proof of Lemma 8

  • Proof. Decompose FC into two primitive cycles, one of which has odd
  • length. Its length must be 1! Thus FC has the form

J1 → J1 → J2 → · · · → Jn−1 → J1 where Ji = J1 for 1 < i < n. If Ji = Jk, 1 < i < k < n, then we obtain a smaller primitive cycle, and by excluding the loop at J1 if necessary, we can arrange that its length is odd. Hence, J1, ..., Jn−1 are all distinct and thus a permutation of I1, ..., In−1. Similarly, we cannot have Ji → Jk if k > i + 1 or if k = 1 and i = 1, n + 1. Suppose J1 = Ik = [a, b]. Since J1 → J2 ⇒ J2 is adjacent to J1 and either xk = a, xk+1 = f(a), xk−1 = f 2(a)

  • r

xk+1 = b, xk = f(b), xk+2 = f 2(b).

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Proof of Sharkovski’s Theorem

Proof of Lemma 8

Consider the first case, the argument in the second being similar. If f 3(a) < f 2(a) then J2 → J1, which is forbidden. Hence f 3(a) > f 2(a).

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Proof of Sharkovski’s Theorem

Proof of Lemma 8

Consider the first case, the argument in the second being similar. If f 3(a) < f 2(a) then J2 → J1, which is forbidden. Hence f 3(a) > f 2(a). Since J2 is not directed to Jk for k > 3 it follows that J3 = [f(a), f 3(a)] is adjacent to J1 on the right.

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Proof of Sharkovski’s Theorem

Proof of Lemma 8

Consider the first case, the argument in the second being similar. If f 3(a) < f 2(a) then J2 → J1, which is forbidden. Hence f 3(a) > f 2(a). Since J2 is not directed to Jk for k > 3 it follows that J3 = [f(a), f 3(a)] is adjacent to J1 on the right. If f 4(a) > f 3(a) then J3 → J1, which is

  • forbidden. Hence f 4(a) < f 2(a) and, since J3 is not directed to Jk for

k > 4, J4 = [f 4(a), f 2(a)] is adjacent to J2 on the left.

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Proof of Sharkovski’s Theorem

Proof of Lemma 8

Consider the first case, the argument in the second being similar. If f 3(a) < f 2(a) then J2 → J1, which is forbidden. Hence f 3(a) > f 2(a). Since J2 is not directed to Jk for k > 3 it follows that J3 = [f(a), f 3(a)] is adjacent to J1 on the right. If f 4(a) > f 3(a) then J3 → J1, which is

  • forbidden. Hence f 4(a) < f 2(a) and, since J3 is not directed to Jk for

k > 4, J4 = [f 4(a), f 2(a)] is adjacent to J2 on the left. Proceeding in this way we see that the order of the Jis on the real line is given by

f n−1(a) f n−3(a) f 4(a) f 2(a) f (a) f 3(a) f n−4(a) f n−2(a) a Jn−1 J4 J2 J1 J3 Jn−2 · · · · · ·

Since the endpoints of Jn−1 = [x1, x2] are mapped into a and f n−2(a) = xn we have Jn−1 → Jk iff k is odd. We found all the arcs in the digraph.

Ugur G. Abdulla Topological Dynamics and Universality in Chaos I. Proof of Sharkovski’s Theo