Equi-kneading of skew tent maps in the square (work in progress) - - PowerPoint PPT Presentation

Equi-kneading of skew tent maps in the square

(work in progress) Zolt´ an Buczolich E¨

• tv¨
• s University, Budapest

www.cs.elte.hu/∼buczo Joint work with: Gabriella Keszthelyi

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Consider a point (α, β) in the unit square [0, 1]2. Denote by Tα,β(x) the skew tent map Tα,β(x) =

• β

αx

if 0 ≤ x < α

β 1−α(1 − x)

if α < x ≤ 1.

2

The topological entropy of Tα,β is denoted by h(α, β).

2

The topological entropy of Tα,β is denoted by h(α, β). The starting point of our paper is the question about the behavior of the function h(α) = h(α, β). The answer to the question about the behavior of the function g(β) = h(α, β) with a fixed α is known. In the case of these β′s where the dynamics of Tα,β(x) is nontrivial the function g(β) is monotone increasing.

2

Skew tent maps and topological entropy were considered by

• M. Misiurewicz and E. Visinescu.

They used different parametrization: Fλ,µ(x) =

• 1 + λx

if x ≤ 0 1 − µx if x ≥ 0

• n R.

2

Skew tent maps and topological entropy were considered by

• M. Misiurewicz and E. Visinescu.

They used different parametrization: Fλ,µ(x) =

• 1 + λx

if x ≤ 0 1 − µx if x ≥ 0

• n R.

It is rather easy to see that if λ = β

α, µ = β 1−α and

h(x) = (β − α)x + α, h−1(x) = x−α

β−α then

Fλ,µ(x) = (h−1 ◦ Tα,β ◦ h)(x), provided that we extend the definition of Tα,β onto R int the obvious way.

2

Results of M. Misiurewicz and E. Visinescu imply that if λ′ ≥ λ, µ′ ≥ µ and at least one of these inequalities is sharp, then h(Fλ′,µ′) > h(Fλ,µ) where h(Fλ,µ) denotes the topological entropy of Fλ,µ. If β is fixed and α increases then λ = β

α decreases, while µ = β 1−α increases.

This implies that the monotonicity result in [MV] is not giving an answer to our question.

2

If β is fixed and α increases then λ = β

α decreases, while µ = β 1−α increases.

This implies that the monotonicity result in [MV] is not giving an answer to our question. In fact, points λ = λ(α, β) and µ = µ(α, β) with fixed β satisfy 1

λ + 1 µ = 1 β, or

λ =

1

1 β − 1 µ and hence λ is a monotone decreasing function of µ along curves in the

(µ, λ) plane corresponding to horizontal line segments in the (α, β) plane.

2

In fact, points λ = λ(α, β) and µ = µ(α, β) with fixed β satisfy 1

λ + 1 µ = 1 β, or

λ =

1

1 β − 1 µ and hence λ is a monotone decreasing function of µ along curves in the

(µ, λ) plane corresponding to horizontal line segments in the (α, β) plane. If α is fixed and β increases then both λ = β

α and µ = β 1−α increase. Therefore,

the monotonicity result in [MV] implies that all the functions g(β) = h(α, β) are monotone increasing in our parameter range. This parameter range 0.5 < β < 1, α ∈ (1 − β, β) corresponds to the region bounded by the curves µ = 1, λ = 1 and 1

µ + 1 λ = 1 in the (µ, λ)-plane.

2

We prove: T.: For any fixed β ∈ (0.5, 1) the function h(α) = h(α, β) is strictly monotone increasing on (1 − β, β).

2

We denote by K(α, β) the kneading sequence of T(α, β).

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We denote by K(α, β) the kneading sequence of T(α, β). If β is fixed we simply write K(α) and T(α). If K(α) = A1A2 . . . then Kn(α) = An ∈ {R, L, C}.

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We denote by K(α, β) the kneading sequence of T(α, β). If β is fixed we simply write K(α) and T(α). If K(α) = A1A2 . . . then Kn(α) = An ∈ {R, L, C}. We denote by M the class of kneading sequences K(0.5, β), β ∈ (0.5, 1], this corresponds to the kneading sequences of functions Fµ,µ with 1 < µ ≤ 2.

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We denote by M the class of kneading sequences K(0.5, β), β ∈ (0.5, 1], this corresponds to the kneading sequences of functions Fµ,µ with 1 < µ ≤ 2. Results of M–V imply: T.: For each M ∈ M there exist two numbers α1(M) < α2(M) and a continuous function ΨM : (α1(M), α2(M)) → U such that for (α, β) ∈ U we have K(α, β) = M if and only if β = ΨM(α). The graphs of the functions ΨM fill up the whole set U. Moreover, limα→α1(M)+ ΨM(α) = 1 if M RLR∞. If M ≺ RLR∞ then the curve (α, ΨM(α)) converges to a point on the line segment {(α, 1 − α) : 0 < α < 1

2}.

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Results of M–V imply: T.: For each M ∈ M there exist two numbers α1(M) < α2(M) and a continuous function ΨM : (α1(M), α2(M)) → U such that for (α, β) ∈ U we have K(α, β) = M if and only if β = ΨM(α). The graphs of the functions ΨM fill up the whole set U. Moreover, limα→α1(M)+ ΨM(α) = 1 if M RLR∞. If M ≺ RLR∞ then the curve (α, ΨM(α)) converges to a point on the line segment {(α, 1 − α) : 0 < α < 1

2}.

If M = RL∞ then α1(M) = 0, α2(M) = 1 and ΨM(α) = 1 for all α ∈ (0, 1).

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The graphs of the functions ΨM fill up the whole set U. Moreover, limα→α1(M)+ ΨM(α) = 1 if M RLR∞. If M ≺ RLR∞ then the curve (α, ΨM(α)) converges to a point on the line segment {(α, 1 − α) : 0 < α < 1

2}.

If M = RL∞ then α1(M) = 0, α2(M) = 1 and ΨM(α) = 1 for all α ∈ (0, 1). We denote by MR∞ the set of those M ∈ M which are of the form M = A1A2 . . . An−1R∞.

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We prove: If M ∈ M \ {RL∞} then ΨM(α) is strictly monotone decreasing. This implies: The function h(α) = h(α, β) is monotone increasing on (1 − β, β). We also show that close to (1, 1) the curves ΨM(α) are almost perpendicular to the y = x line.

3

We prove: If M ∈ M \ {RL∞} then ΨM(α) is strictly monotone decreasing. This implies: The function h(α) = h(α, β) is monotone increasing on (1 − β, β). We also show that close to (1, 1) the curves ΨM(α) are almost perpendicular to the y = x line.

3

We prove: If M ∈ M \ {RL∞} then ΨM(α) is strictly monotone decreasing. This implies: The function h(α) = h(α, β) is monotone increasing on (1 − β, β). We also show that close to (1, 1) the curves ΨM(α) are almost perpendicular to the y = x line.

3

Equi-kneading regions by using the ”square-parametrization” and by the parametrization used in the Misiurewicz–Visinescu paper.

3

L.: Suppose M ∈ M\{RL∞} is given. Then there exists a function ΘM(α, β), ΘM : U → R, such that for (α, β) ∈ U from K(α, β) = M it follows that ΘM(α, β) = 0. (Limitations on reverse implication!)

4

L.: Suppose M ∈ M\{RL∞} is given. Then there exists a function ΘM(α, β), ΘM : U → R, such that for (α, β) ∈ U from K(α, β) = M it follows that ΘM(α, β) = 0. (Limitations on reverse implication!) Moreover, ΘM(α, β) = 1 − β +

∞

• k=1

α − 1 β k α β mk where m1 > 0, mk ≤ mk+1 ≤ mk + m1, k = 0, 1, . . . .

4

L.: Suppose M ∈ M\{RL∞} is given. Then there exists a function ΘM(α, β), ΘM : U → R, such that for (α, β) ∈ U from K(α, β) = M it follows that ΘM(α, β) = 0. (Limitations on reverse implication!) Moreover, ΘM(α, β) = 1 − β +

∞

• k=1

α − 1 β k α β mk where m1 > 0, mk ≤ mk+1 ≤ mk + m1, k = 0, 1, . . . . If K(α, β) ∈ MR,∞ then there exists n s. t. mk+1 = mk for k ≥ n.

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If M is infinite let M′ = M if M is finite, M = A1 . . . An−1C then let M′ = A1 . . . An−1LA1 . . . An−1L . . . . Recall that A1 = R and A2 = L for all M = K(α, β), (α, β) ∈ U. If M / ∈ MR,∞ than there are infinite sequences (λj), (hj) s. t. in M′ the first R is followed by λ1 many L’s, then there are h1 many R’s then λ2 many L’s, then h2 many R’s etc., that is, M′ = R L . . . L R . . . R L . . . L R . . . R . . . . λ1 h1 λ2 h2

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Recall that A1 = R and A2 = L for all M = K(α, β), (α, β) ∈ U. If M / ∈ MR,∞ than there are infinite sequences (λj), (hj) s. t. in M′ the first R is followed by λ1 many L’s, then there are h1 many R’s then λ2 many L’s, then h2 many R’s etc., that is, M′ = R L . . . L R . . . R L . . . L R . . . R . . . . λ1 h1 λ2 h2 ΘM(α, β) can be written as

ΘM(α, β) = (1 − β) + α − 1 β α β λ1 + · · · + α − 1 β h1 α β λ1 + · · · + α − 1 β h1+···+hn−1+1 α β λ1+···+λn + · · · + α − 1 β h1+···+hn α β λ1+···+λn + . . . .

4

Recall: ΘM(α, β) = 1 − β +

∞

• k=1

α − 1 β k α β mk where m1 > 0, mk ≤ mk+1 ≤ mk + m1, k = 0, 1, . . . .

5

Recall: ΘM(α, β) = 1 − β +

∞

• k=1

α − 1 β k α β mk where m1 > 0, mk ≤ mk+1 ≤ mk + m1, k = 0, 1, . . . . L.: Suppose M ∈ M\{RL∞},

1 2 < β < 1 is fixed. Then

Θ(α) = ΘM(α) = ΘM(α, β) is analytic on (1 − β, β).

5

Recall: ΘM(α, β) = 1 − β +

∞

• k=1

α − 1 β k α β mk where m1 > 0, mk ≤ mk+1 ≤ mk + m1, k = 0, 1, . . . . L.: Suppose M ∈ M\{RL∞},

1 2 < β < 1 is fixed. Then

Θ(α) = ΘM(α) = ΘM(α, β) is analytic on (1 − β, β). We need this to verify: L.: For any M ∈ M\{RL∞} the curve (α, ΨM(α)), α ∈ (α1(M), α1(M)) does not contain horizontal intervals.

5

Recall: ΘM(α, β) = 1 − β +

∞

• k=1

α − 1 β k α β mk where m1 > 0, mk ≤ mk+1 ≤ mk + m1, k = 0, 1, . . . . We need this to verify: L.: For any M ∈ M\{RL∞} the curve (α, ΨM(α)), α ∈ (α1(M), α1(M)) does not contain horizontal intervals.

5

We prove that for M ∈ M\{RL∞} the curves (α, ΨM (α)) , α ∈ (α1(M), α2(M)) are strictly monotone decreasing.

6

We prove that for M ∈ M\{RL∞} the curves (α, ΨM (α)) , α ∈ (α1(M), α2(M)) are strictly monotone decreasing. Since 1

2 < ΨM

1

2

• < 1 these curves reach the boundary of U at a point on the

line segment {(β, β) : 1

2 < β < 1}, in fact βM = α2(M)

and limα→α2(M) ΨM(α) = βM.

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We prove that for M ∈ M\{RL∞} the curves (α, ΨM (α)) , α ∈ (α1(M), α2(M)) are strictly monotone decreasing. Since 1

2 < ΨM

1

2

• < 1 these curves reach the boundary of U at a point on the

line segment {(β, β) : 1

2 < β < 1}, in fact βM = α2(M)

and limα→α2(M) ΨM(α) = βM. Reversing our notation for β0 ∈ 1

2, 1

• now we denote by Ψβ0 the curve with the

property limα→β0− Ψβ0(α) = β0.

6

Since 1

2 < ΨM

1

2

• < 1 these curves reach the boundary of U at a point on the

line segment {(β, β) : 1

2 < β < 1}, in fact βM = α2(M)

and limα→α2(M) ΨM(α) = βM. Reversing our notation for β0 ∈ 1

2, 1

• now we denote by Ψβ0 the curve with the

property limα→β0− Ψβ0(α) = β0. In fact, we can extend the definition of Ψβ by setting Ψβ0(β0) = β0.

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Reversing our notation for β0 ∈ 1

2, 1

• now we denote by Ψβ0 the curve with the

property limα→β0− Ψβ0(α) = β0. In fact, we can extend the definition of Ψβ by setting Ψβ0(β0) = β0. Moreover, we show that ΘMβ0(α, Ψβ0(α)) = Θβ0(α, Ψβ0(α)) = 0 for (α, Ψβ0(α)) ∈ U.

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In fact, we can extend the definition of Ψβ by setting Ψβ0(β0) = β0. Moreover, we show that ΘMβ0(α, Ψβ0(α)) = Θβ0(α, Ψβ0(α)) = 0 for (α, Ψβ0(α)) ∈ U. We also show that Θβ0(α, β) is infinitely differentiable in a neighborhood of (β0, β0).

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Moreover, we show that ΘMβ0(α, Ψβ0(α)) = Θβ0(α, Ψβ0(α)) = 0 for (α, Ψβ0(α)) ∈ U. We also show that Θβ0(α, β) is infinitely differentiable in a neighborhood of (β0, β0). Therefore, with the implicit definition one can extend the definition of Ψβ0 a little behind β0, that is one can extend the definition of Ψβ0 onto a small interval (β0, β0 + ε),

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We also show that Θβ0(α, β) is infinitely differentiable in a neighborhood of (β0, β0). Therefore, with the implicit definition one can extend the definition of Ψβ0 a little behind β0, that is one can extend the definition of Ψβ0 onto a small interval (β0, β0 + ε), by setting Θβ0(α, Ψβ0(α)) = 0. A little caution is necessary.

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Therefore, with the implicit definition one can extend the definition of Ψβ0 a little behind β0, that is one can extend the definition of Ψβ0 onto a small interval (β0, β0 + ε), by setting Θβ0(α, Ψβ0(α)) = 0. A little caution is necessary. L.: Suppose β0 ∈ ( 1

2, 1) and we consider Ψβ0, Θβ0 defined as above.

Then Θβ0(β, β) = 0 for all β ∈ ( 1

2, 1) ∂αΘβ0(β0, β0) = 0, ∂βΘβ0(β0, β0) = 0.

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by setting Θβ0(α, Ψβ0(α)) = 0. A little caution is necessary. L.: Suppose β0 ∈ ( 1

2, 1) and we consider Ψβ0, Θβ0 defined as above.

Then Θβ0(β, β) = 0 for all β ∈ ( 1

2, 1) ∂αΘβ0(β0, β0) = 0, ∂βΘβ0(β0, β0) = 0.

Hence we cannot determine the derivative of Ψβ0 at β0 by using implicit differentiation of Θβ0(α, Ψβ0(α)) = 0 at α = β0.

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L.: Suppose β0 ∈ ( 1

2, 1) and we consider Ψβ0, Θβ0 defined as above.

Then Θβ0(β, β) = 0 for all β ∈ ( 1

2, 1) ∂αΘβ0(β0, β0) = 0, ∂βΘβ0(β0, β0) = 0.

Hence we cannot determine the derivative of Ψβ0 at β0 by using implicit differentiation of Θβ0(α, Ψβ0(α)) = 0 at α = β0. To avoid awkward notation we denote by DαΨβ0 the derivative of Ψβ0(α).

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Hence we cannot determine the derivative of Ψβ0 at β0 by using implicit differentiation of Θβ0(α, Ψβ0(α)) = 0 at α = β0. To avoid awkward notation we denote by DαΨβ0 the derivative of Ψβ0(α). T.: With the above notation lim

β0→1− DαΨβ0(β0) = −1.

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To avoid awkward notation we denote by DαΨβ0 the derivative of Ψβ0(α). T.: With the above notation lim

β0→1− DαΨβ0(β0) = −1.

This means that the curves Ψβ0 are almost perpendicular to the diagonal {(β, β) : 1

2 < β < 1} at the point (β0, β0) when β0 is close to 1.

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T.: With the above notation lim

β0→1− DαΨβ0(β0) = −1.

This means that the curves Ψβ0 are almost perpendicular to the diagonal {(β, β) : 1

2 < β < 1} at the point (β0, β0) when β0 is close to 1.

• M. Misiurewicz asked in April in Warwick whether the formula

limβ0→1− DαΨβ0(β0) = −1 can be improved to DαΨβ0(β0) = −1.

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• M. Misiurewicz asked in April in Warwick whether the formula

limβ0→1− DαΨβ0(β0) = −1 can be improved to DαΨβ0(β0) = −1. If α0 = 1/2 and β0 =

√ 5 10 + 1 2, this corresponds to M = RLLRC,

then DαΨβ0(β0) = −

√ 5+3 2 √ 5+2 = −1.

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If α0 = 1/2 and β0 =

√ 5 10 + 1 2, this corresponds to M = RLLRC,

then DαΨβ0(β0) = −

√ 5+3 2 √ 5+2 = −1.

One has to study the implicit curve defined by α5 − 2α4 + α3(β + 1) − α2β − β4(β − 1) = 0 On the other hand, if α0 = 1/2 and β0 = 2/3, this corresponds to M = RLC, then DαΨβ0(β0) = −1. With implicit equation α3 − α2 − β3 + β2 = 0.

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One has to study the implicit curve defined by α5 − 2α4 + α3(β + 1) − α2β − β4(β − 1) = 0 On the other hand, if α0 = 1/2 and β0 = 2/3, this corresponds to M = RLC, then DαΨβ0(β0) = −1. With implicit equation α3 − α2 − β3 + β2 = 0.

6