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¨ otv¨ os University, Budapest www.cs.elte.hu/ ∼ buczo Joint work with: Gabriella Keszthelyi 1

Consider a point ( α, β ) in the unit square [0 , 1] 2 . Denote by T α,β ( x ) the skew tent map � β α x if 0 ≤ x < α T α,β ( x ) = β 1 − α (1 − x ) if α < x ≤ 1 . 2

The topological entropy of T α,β is denoted by h ( α, β ). 2

The topological entropy of T α,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 T α,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. � 1 + λ x if x ≤ 0 They used different parametrization: F λ,µ ( x ) = on R . 1 − µ x if x ≥ 0 2

Skew tent maps and topological entropy were considered by M. Misiurewicz and E. Visinescu. � 1 + λ x if x ≤ 0 They used different parametrization: F λ,µ ( x ) = on R . 1 − µ x if x ≥ 0 It is rather easy to see that if λ = β β α , µ = 1 − α and h ( x ) = ( β − α ) x + α, h − 1 ( x ) = x − α β − α then F λ,µ ( x ) = ( h − 1 ◦ T α,β ◦ h )( x ) , provided that we extend the definition of T α,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 ( F λ ′ ,µ ′ ) > h ( F λ,µ ) where h ( F λ,µ ) denotes the topological entropy of F λ,µ . If β is fixed and α increases then λ = β β α decreases, while µ = 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 β α decreases, while µ = 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 1 λ = µ and hence λ is a monotone decreasing function of µ along curves in the β − 1 1 ( 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 1 λ = µ and hence λ is a monotone decreasing function of µ along curves in the β − 1 1 ( µ, λ ) plane corresponding to horizontal line segments in the ( α, β ) plane. If α is fixed and β increases then both λ = β β α and µ = 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 ). 3

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 ) = A 1 A 2 . . . then K n ( α ) = A n ∈ { R , L , C } . 3

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 ) = A 1 A 2 . . . then K n ( α ) = A n ∈ { 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 F µ,µ with 1 < µ ≤ 2. 3

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 F µ,µ with 1 < µ ≤ 2. Results of M–V imply: T .: For each M ∈ M there exist two numbers α 1 ( M ) < α 2 ( M ) and a continuous function Ψ M : ( α 1 ( M ) , α 2 ( M )) → U such that for ( α, β ) ∈ U we have K ( α, β ) = M if and only if β = Ψ M ( α ). The graphs of the functions Ψ M fill up the whole set U. Moreover, If M ≺ RLR ∞ then the curve lim α → α 1 ( M )+ Ψ M ( α ) = 1 if M � RLR ∞ . ( α, Ψ M ( α )) converges to a point on the line segment { ( α, 1 − α ) : 0 < α < 1 2 } . 3

Results of M–V imply: T .: For each M ∈ M there exist two numbers α 1 ( M ) < α 2 ( M ) and a continuous function Ψ M : ( α 1 ( M ) , α 2 ( M )) → U such that for ( α, β ) ∈ U we have K ( α, β ) = M if and only if β = Ψ M ( α ). The graphs of the functions Ψ M fill up the whole set U. Moreover, If M ≺ RLR ∞ then the curve lim α → α 1 ( M )+ Ψ M ( α ) = 1 if M � RLR ∞ . ( α, Ψ M ( α )) converges to a point on the line segment { ( α, 1 − α ) : 0 < α < 1 2 } . If M = RL ∞ then α 1 ( M ) = 0 , α 2 ( M ) = 1 and Ψ M ( α ) = 1 for all α ∈ (0 , 1) . 3

The graphs of the functions Ψ M fill up the whole set U. Moreover, If M ≺ RLR ∞ then the curve lim α → α 1 ( M )+ Ψ M ( α ) = 1 if M � RLR ∞ . ( α, Ψ M ( α )) converges to a point on the line segment { ( α, 1 − α ) : 0 < α < 1 2 } . If M = RL ∞ then α 1 ( M ) = 0 , α 2 ( M ) = 1 and Ψ M ( α ) = 1 for all α ∈ (0 , 1) . We denote by M R ∞ the set of those M ∈ M which are of the form M = A 1 A 2 . . . A n − 1 R ∞ . 3

We prove: If M ∈ M \ { RL ∞ } then Ψ M ( 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 Ψ M ( α ) are almost perpendicular to the y = x line. 3

We prove: If M ∈ M \ { RL ∞ } then Ψ M ( 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 Ψ M ( α ) are almost perpendicular to the y = x line. 3

We prove: If M ∈ M \ { RL ∞ } then Ψ M ( 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 Ψ M ( 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 Θ M ( α, β ) , Θ M : U → R , such that for ( α, β ) ∈ U from K ( α, β ) = M it follows that Θ M ( α, β ) = 0 . (Limitations on reverse implication!) 4

L .: Suppose M ∈ M \{ RL ∞ } is given. Then there exists a function Θ M ( α, β ) , Θ M : U → R , such that for ( α, β ) ∈ U from K ( α, β ) = M it follows that Θ M ( α, β ) = 0 . (Limitations on reverse implication!) � α − 1 � k � α � m k ∞ � Moreover, Θ M ( α, β ) = 1 − β + β β k =1 where m 1 > 0 , m k ≤ m k +1 ≤ m k + m 1 , k = 0 , 1 , . . . . 4

L .: Suppose M ∈ M \{ RL ∞ } is given. Then there exists a function Θ M ( α, β ) , Θ M : U → R , such that for ( α, β ) ∈ U from K ( α, β ) = M it follows that Θ M ( α, β ) = 0 . (Limitations on reverse implication!) � α − 1 � k � α � m k ∞ � Moreover, Θ M ( α, β ) = 1 − β + β β k =1 where m 1 > 0 , m k ≤ m k +1 ≤ m k + m 1 , k = 0 , 1 , . . . . If K ( α, β ) ∈ M R , ∞ then there exists n s. t. m k +1 = m k for k ≥ n . 4

If M is infinite let M ′ = M if M is finite, M = A 1 . . . A n − 1 C then let M ′ = A 1 . . . A n − 1 LA 1 . . . A n − 1 L . . . . Recall that A 1 = R and A 2 = L for all M = K ( α, β ) , ( α, β ) ∈ U . ∈ M R , ∞ than there are infinite sequences ( λ j ) , ( h j ) s. t. in M ′ the first R is If M / followed by λ 1 many L ’s, then there are h 1 many R ’s then λ 2 many L ’s, then h 2 many R ’s etc., that is, M ′ = R L . . . L � �� � R . . . R � �� � L . . . L � �� � R . . . R � �� � . . . . λ 1 h 1 λ 2 h 2 4

Recall that A 1 = R and A 2 = L for all M = K ( α, β ) , ( α, β ) ∈ U . ∈ M R , ∞ than there are infinite sequences ( λ j ) , ( h j ) s. t. in M ′ the first R is If M / followed by λ 1 many L ’s, then there are h 1 many R ’s then λ 2 many L ’s, then h 2 many R ’s etc., that is, M ′ = R L . . . L � �� � R . . . R � �� � L . . . L � �� � R . . . R � �� � . . . . λ 1 h 1 λ 2 h 2 Θ M ( α, β ) can be written as � α − 1 � � α � λ 1 � α − 1 � h 1 � α � λ 1 Θ M ( α, β ) = (1 − β ) + + · · · + + β β β β � α − 1 � h 1 + ··· + h n − 1 +1 � α � λ 1 + ··· + λ n � α − 1 � h 1 + ··· + h n � α � λ 1 + ··· + λ n · · · + + · · · + + . . . . β β β β 4

Recall: � α − 1 � k � α � m k ∞ � Θ M ( α, β ) = 1 − β + β β k =1 where m 1 > 0 , m k ≤ m k +1 ≤ m k + m 1 , k = 0 , 1 , . . . . 5