Statistical and Deterministic Dynamics of Maps with Memory

We consider a dynamical system to have memory if it remembers the current state as well as the state before that. The dynamics is defined as follows: $x_{n+1}=T_{\alpha}(x_{n-1},x_{n})=\tau (\alpha \cdot x_{n}+(1-\alpha)\cdot x_{n-1}),$ where $\tau$ is a one-dimensional map on $I=[0,1]$ and $0<\alpha<1$ determines how much memory is being used. $T_{\alpha}$ does not define a dynamical system since it maps $U=I\times I$ into $I$. In this note we let $\tau $ to be the symmetric tent map. We shall prove that for $0<\alpha<0.46,$ the orbits of $\{x_{n}\}$ are described statistically by an absolutely continuous invariant measure (acim) in two dimensions. As $\alpha $ approaches $0.5 $ from below, that is, as we approach a balance between the memory state and the present state, the support of the acims become thinner until at $\alpha =0.5$, all points have period 3 or eventually possess period 3. For $0.5<\alpha<0.75$, we have a global attractor: for all starting points in $U$ except $(0,0)$, the orbits are attracted to the fixed point $(2/3,2/3).$ At $\alpha=0.75,$ we have slightly more complicated periodic behavior.


Introduction
In nonlinear discrete chaotic dynamical systems theory we study the statistical long term dynamics of iterated maps which depend only on the present state of the system. In this paper we consider dynamical systems which depend both on the present state as well as on one previous state. Such memory systems find applications in cellular automata and in modeling natural phenomena [1,2].
Let τ be a piecewise, expanding map on I. We refer to it as the base map. At each step, the system remembers the current state x n as well as one previous state x n−1 , which we refer to as the memory. Our dynamical system is defined by x n+1 = T α (x n ) = τ (α · x n + (1 − α) · x n−1 ), where 0 < α < 1 is a fixed number that specifies the ratio between the present state and the memory state. T α does not define a dynamical system, since it is not a map of a space into itself. Rather, it denotes a process. To start a trajectory we need an initial point x 0 and its memory, which we consider to be bundled into the previous state x n−1 . When α is close to 0, the present state x n is weighted down, acting as a perturbation on the memory state x n−1 which is dominant. However, when α is close to 1, the memory state is diminished and the resulting system behaves almost like a regular dynamical system, depending mostly on the present state.
Since we assume that G α is ν α -ergodic, the Birkhoff Ergodic Theorem holds. Thus, for any integrable function f and almost every pair (x, y), we have lim n→∞ 1 n n−1 i=0 f (G i α (x, y)) = f (x, y)dν α (x, y) .
If the function f depends only on the first coordinate, we can rewrite this as that is, Since the limit is independent of the initial condition, the initial history x −1 used by T α is irrelevant. This shows that the marginal measure of the G α -invariant measure determines the behavior of ergodic averages of trajectories of the process T α . Thus, µ α is a good candidate for an "invariant" measure of T α .
In Section 2, we show that for certain α, G α is expanding in both directions and establish the existence of an acim for the memory system defined by any piecewise expanding map τ. In Sections 3 -6 we study the behavior of the memory system defined when the base map is the tent map τ. For 0 < α < 0.46, we prove the orbits of {x n } are described statistically by an acim. As α approaches 0.5 from below, that is, as we approach a balance between the memory state and the present state, the support of the acims become thinner until at α = 0.5, all points have period 3 or eventually possess period 3. In Section 7, we consider 1/2 ≤ α ≤ 3/4. We prove that for α = 1/2 all points (except two fixed points) are eventually periodic with period 3. For α = 3/4 we prove that all points of the line x + y = 4/3 (except the fixed point) are of period 2 and all other points (except (0, 0)) are attracted to this line. For 1/2 < α < 3/4, we prove the existence of a global attractor: for all starting points in the square [0, 1 × [0, 1] except (0, 0), the orbits are attracted to the fixed point (2/3, 2/3).
Additional pictures illustrating the behaviour of the family G α and Maple programs used in this study can be found at http://www.mathstat.concordia.ca/faculty/pgora/G-map/.

Preliminary Results
In this section we show that for certain α, G α is expanding in both directions.(We will ussually suppres the subscript α in the sequel.) Let τ : I → I be a piecewise expanding map is defined on the partition P with endpoints {a 0 = 0, a 1 , a 2 , . . . , a q−1 , a q = 1}. Let Then, the map G α is defined on the partition whose boundaries are the boundaries of the square U = I 2 and the lines Each of the lines L α 0 and L α q intersects [0, 1] 2 at only one point. Let R i denote the region in [0, 1] between the lines L α i−1 and L α i , i = 1, 2, . . . , q. The example for P = {0, 0.25, 0.5, 0.8, 1} is shown in Figure 1 a and the example for P = {0, 0.5, 1} is shown in Figure 1 b.

Figure 1. Examples of partitions for map G
Note that G α is not piecewise expanding. However, we will show that G 2 α is a piecewise expanding map for small values of α. The inverse branches of G 2 which is equal to If α is chosen small enough, since τ is expanding, all the entries of the matrix can be made smaller than one (in absolute value), so the norm is smaller than one. This implies that G 2 α is a piecewise expanding map. By [4] we have the existence of an acim.
One can immediately make the following observation.
, τ (y)), so G α is likely to have an acim because τ has an acim and G α is close to the product τ × τ . On the other hand, if α ≈ 1 (weak memory), then G α (x, y) ≈ (y, τ (y)), which is independent of x and the orbit of any point (x, y) ∈ U is approximately a subset of the graph of τ . In this case it is likely that there is an SRB measure, but that it is singular with respect to the 2D Lebesgue measure.
We now show that in general G α is not piecewise expanding. Suppose τ j : I j = (a j , b j ) → I is a monotonic branch of τ . Then G α is piecewise monotonic on the strips {(x, y) : a j < αy + (1 − α)x < b j }. If G j is the branch of G α corresponding to I j , then the inverse of G α,j is given by Such a matrix has Euclidean norm DG −1 α,j 2 ≥ 1. Indeed, for a square matrix A, this norm is equal to λ max (A T A), where λ max (A T A) denotes the maximum eigenvalue of the symmetric matrix A T A. For us, A is given by (2.4) and is of the form Note that the sum of the eigenvalues of a matrix is equal to the trace of the matrix, which for A T A is 1 + a 2 + b 2 . This means that both eigenvalues cannot be smaller than 1. Therefore, λ max (A T A) ≥ 1, and G is not a piecewise expanding map (see [3], Remark 2.1 item 2) in the sense that all directions are contracted under the branches of the inverse of G.
3. τ is the symmetric tent map.
In the sequel we study the dynamical system where the base map is so G 2 = τ × τ and preserves two-dimensional Lebesgue measure on the square In the sequel we consider only α > 0. Let A 1 denote the part of the square [0, 1]×[0, 1] below the line αy+(1−α)x = 1/2 and A 2 the part above this line. We now collect some simple facts.
Proof. We have For α ∈ (0, 1/2) the coefficient next to y is positive and that next to x is negative so the minimum is reached at (1, 0) and is equal to 2α 2 . This completes the proof.
Proof. We have For α ∈ (0, 1/2) both coefficients next to x and y are negative so the minimum is reached at (1, 1) and is equal to 2α(1 − α) ≥ 2α 2 . This completes the proof. Proof. Proposition 2 implies that every point of A 1 , except (0, 0), enters A 2 after a finite number of steps. Let us consider a point X 0 ∈ A 2 . By Proposition 3 its image X 1 = G(X 0 ) stays above the line αy + (1 − α)x = 2α 2 . Assuming that X 1 ∈ A 1 , by Proposition 4 the point X 2 = G(X 1 ) is also above this line. If X 2 ∈ A 1 the next image X 3 = G(X 2 ) = G 2 (X 1 ) is above the line αy + (1 − α)x = 2α 2 (4α 2 − 2α + 2) (by Proposition 2). Now, if X 3 ∈ A 1 , the next image X 4 = G(X 3 ) = G 2 (X 2 ) is also above this line. We see that further points of the trajectory move up towards A 2 and none of them can go below the line αy + (1 − α)x = 2α 2 .
We define some functions which we will use below. Let G i = G |Ai , i = 1, 2, be the restrictions of G to regions A 1 and A 2 , respectively. Let S(x, y) = αy +(1−α)x.
Proof. We will prove that G 2 satisfies the assumptions of Tsujii ([4]), i.e., it is piecewise analytic and expanding in the sense that for any vector v we have |DG 2 v| > |v|. We will do this by showing that the smaller singular value s 2 (α) of the matrix D i D j , i, j ∈ {1, 2} is above 1 for 0 < α ≤ 0.24760367.
The singular values of the matrices D 2 D 2 and D 1 D 2 are: and They are shown in Figure 3 b). The lower curve intersects level 1 at ∼ 0.3709557543. This shows that at least for 0 < α ≤ α 1 ∼ 0.24760367 the assumptions of [4] are satisfied and thus, G 2 and consequently also G admit an acim.

5.
Proof of the existence of acim for α > α 1 We will prove that high iterates of the map G expand all vectors. We will make estimates of the smaller singular value σ 2 of derivative matrix DG n for large n. The general strategy is as follows: we will consider the admissible products of the derivative matrices ns j=1 D ij , where i j ∈ {1, 2} and the length n s depends on the sequence, for α ∈ (α s , α t ), where (α s , α t ) denotes contiguous intervals. The order of the matrices is natural, e.g., the sequence D 1 D 2 D 2 corresponds to the iteration G 1 G 2 G 2 . We will consider sequences of the form D n 1 D m 2 , n ≤ 3, m ≥ 1, since by Proposition 2 every point (except (0, 0)) visits region A 2 . We will break the long sequence into short "good" sequences for which we can bound σ 2 from below by numbers larger that 1. Since this will allow us to show that the σ 2 of a long product grows to infinity with the length n s . Once we have a good estimate, we proceed as follows: we choose a large number M and find a sequence length n s such that any admissible sequence of length n s starting with D 2 has σ 2 > M . Then, adding at most three matrices D 1 at the beginning of the sequences and a corresponding number of matrices at the end (to keep the length of all sequences equal to n s + 3) we will have derivative matrices of G ns+3 for all non-transient points (we will prove that 3 is enough) and their σ 2 's greater than 1. This proves that G ns+3 on the set of non-transient points expands all vectors and in turn that G admits an acim. Our proofs are based on symbolic calculations using Maple 17, but they are all finite calculations and "in principle" could be done using pen and paper.
Recall G i = G |Ai , i = 1, 2 are the restrictions of G to regions A 1 and A 2 , respectively. The following result holds for all 0 < α < 1/2.
for 0 < α < 1/2. More generally, The singular values of the matrix B are square roots of the eigenvalues of the  Proof. Figure 5 shows the first (green) and second (red) image of A 1 . G −1 (A 2 )∩A 1 is bounded by magenta lines, the blue line is the partition line S(x, y) = 1/2. The important point is v 2 = G(G(v)) = (2α, 2α(1 − 2α)) for v = (0, 1). When v 2 ∈ A 1 , then points can return to A 1 after one visit in A 2 . When v 2 ∈ A 2 , then a point coming from A 1 must stay in A 2 for at least 2 steps. S(v 2 ) = 1/2 for α = ( Proposition 8. The following estimates of σ 2 (D n 1 D m 2 ) for various n and m were obtained using Maple 17: Theorem 2. The map G admits an acim for α 1 ≤ α ≤ α 2 ∼ 0.2797707433.
We define α 2 as a root of the equation 8α 4 − 8α 3 + 8α 2 = 1/2 in [0, 1]. Again, it is explained below in Proposition 10. First, we prove the following: The functions cx, cy and cc are shown in Figure 6 a). We consider the worst case scenario, i.e., y = 1 and x = 0 where cx > 0 and x = 1 where cx < 0. Graphs of cx + cc and cx + cy + cc are shown in Figure 6 b). They both above 1/2 for α > 0.24, and in particular for Proof of Theorem 2: By Proposition 9, Proposition 6 and estimates of Proposition 8 we see that, for α's in the interval [α 1 α 2 ], all admissible "basic" sequences of derivative matrices have σ 2 larger than 1. Note that we have for m > 3 (Proposition 6). This shows that the general strategy described at the beginning of this section will work and proves the theorem.
First, we prove the following: Proposition 10. For α 2 ≤ α ≤ α 3 a point coming from A 2 can stay in A 1 for at most 2 steps. Proof. The proof is similar to that of Proposition 9. It is enough to show that The functions cx, cy and cc are shown in Figure 7 a). Again, we consider the worst case scenario, i.e., y = 1 and x = 0 where cx > 0 and x = 1 where cx < 0. Graphs of cx + cc and cx + cy + cc are shown in Figure 7 b). They are both above 1/2 for α > α 2 .
Proof of Theorem 3: Let us first consider the sequence D 1 D 1 D 2 . By part 3) of Proposition 8 its σ 2 is larger than 1 until α ∼ 0.3149466135. By Proposition 7 the sequence is not admissible after α ∼ 0.3090169942. All other admissible "basic" sequences of derivative matrices have σ 2 larger than 1 for α's in the interval [α 2 , α 3 ]. We used Proposition 10, Proposition 6 and estimates of Proposition 8 as well as inequality (5.3). This shows that the general strategy described at the beginning of this section will work and proves the theorem.
6. Proof of the existence of acim for α > α 3 = 1/3 We will continue the estimates of σ 2 for "basic" admissible sequences.
Proposition 11. For α > α 3 = 1/3 a point coming from A 2 can stay in A 1 for at most 1 step.
Proposition 13. For α > 0.3931078326, the sequence D 1 D 2 D 2 D 2 is followed by With the previous results this extends the interval of the existence of acim up to α ∼ 0.4160029431.
Proof. Figure 9 shows the first four images of B = G(A 2 ) ∩ A 1 (green thick boundary). The blue line is the partition line S(x, y) = 1/2. The images are consecutively G(B) (red), G 2 (B) (blue), G 3 (B) (brown). The set G 3 (B)∩A 2 is bounded by thick brown lines and represents points which stay in A 2 for 3 steps. Its image is bounded by green lines. The set we are interested in is the triangle namely the points which after three steps in A 2 go to A 1 .
Further images of the triangle C are shown in Figure 10 for a) α = 0.391 and b) α = 0.394. The important point is G 7 (w) (the same point w as in the proof of Proposition 11). When G 7 (w) ∈ A 2 , then some points of C stay in A 2 longer than twice. When G 7 (w) ∈ A 1 , all points of C stay in A 2 exactly for two steps. Equation S(G 7 (w)) = 1/2 is equivalent to 192α 7 + 192α 6 − 336α 5 − 144α 4 + 256α 3 − 128α 2 + 53α − 11 = 0 with a root α ∼ 0.3931078326. Since 0.3931078326 < 0.3938896523 for α > 0.3931078326 we replace estimate 1) of Proposition 12 with estimate of Proposition 13 which holds up to α ∼ 0.4160029431. 2 is not admissible. The following estimates hold: Although the group D 1 D 4 2 may not be admissible, this inequality can be used for estimates.  Proof. First, the estimates 1)-5) show that the basic admissible sequences starting with D 1 D 3 2 (followed by D 1 D 2 2 in view of Proposition 13) have σ 2 > 1 up to α ∼ 0.4315221884. Now, we will show that groups D 1 D 4 2 and D 1 D 3 2 are not admissible above some α's. Figure 11 shows further images of Figure 9. The first image G(C 1 ) is bounded in green. These are points which were 3 steps in A 2 , some of them are in A 1 , some stay for the fourth step in A 2 . The region bounded in red is the image G(G(C 1 ) ∩ A 2 ) (thick green), the points which were in A 2 for 4 steps. For α = 0.343 ( a)) some of them land in A 1 , for α = 0.355 ( b)) the whole image is in A 2 . The important point is G 5 (z), where z = (0, 2α) is a vertex of B. Equation S(G 5 (z)) = 1/2 is equivalent to α 6 + 8α 5 − 8α 4 − 40α 3 − 48α 2 − 96α + 320 = 0 with a root α ∼ 0.3510763028. Figure 12 shows the set G 3 (B) ∩ A 2 (thick brown), the set of point which stayed in A 2 for three steps. B = G(A 2 ) ∩ A 1 as in the proof of Proposition 13 and point w is also the same as there. The image G(G 3 (B) ∩ A 2 ) is bounded in green. The important point is G 4 (w). When G 4 (w) ∈ A 1 , then some points can go to A 1 after three steps in A 2 . When G 4 (w) ∈ A 2 , then all points which stayed 3 times in A 2 stay there for at least two more steps (4 times in A 2 were excluded in the previous part of the proof). The equation S(G 4 (w)) = 1/2 is equivalent to 24α 4 + 12α 3 − 36α 2 + 9α + 1 = 0 with a root α ∼ 0.4284630893.
Once the the sequence D 1 D 3 2 is rendered inadmissible, the worst estimate is α ∼ 0.4345268819, estimate 3) of Proposition 12.
To further improve the range of α's for which G has an acim we have to consider sequences starting with sequence D 1 D 6 2 . Proposition 15. Above α ∼ 0.4345268819 the sequence D 1 D 6 2 is followed by the sequence D 1 D 2 2 or D 1 D 5 2 . After α ∼ 0.4397492527 the only possibility is D 1 D 2 2 . Proof. The blue quadrangle in Figure 13 is G 3 (G 3 (B)∩A 2 ), i.e., it is the third image of brown quadrangle of Figure 12. These are images of points which (for our range of α's) were for 5 steps in A 2 . The green triangle O 6 = G(G 3 (G 3 (B)∩A 2 )∩A 2 )∩A 1 are the points which went to A 1 after 6 steps in A 2 . Figure 13 shows the images G(O 6 ) (bigger red), G 2 (O 6 ) (blue) and G 3 (O 6 ) (partially brown, partially red). The points in G 3 (O 6 ) ∩ A 1 (brown part of G 3 (O 6 )) correspond to group D 1 D 2 2 D 1 D 6 2 . Figure 13 shows also three consecutive images of T = G 3 (O 6 ) ∩ A 2 (small red triangles). In particular G 3 (T ) is completely inside A 1 . These points correspond to the group D 1 D 5 2 D 1 D 6 2 . This proves the first claim of the proposition. Figure 14 shows O 6 and its images G(O 6 ), G 2 (O 6 ) and G 3 (O 6 ) for parameters α = 0.434 (part a)) and α = 0.441 (part b)). For larger α's the image G 3 (O 6 ) is completely in A 1 , which means that after group D 1 D 6 2 there must be group D 1 D 2 2 . The group D 1 D 5 2 D 1 D 6 2 is no longer admissible. The important point is G 10 (w), where w is the point used already in Propositions 14 and 13. The equation S(G 10 (w) = 1/2 is equivalent to 1536α 10 + 3840α 9 − 2688α 8 − 7296α 7 + 4128α 6 + 3840α 5 − 3504α 4 + 992α 3 − 160α 2 − 5α + 11 = 0 with a root α ∼ 0.4397492527.
Proposition 16. Above α ∼ 0.4546258153 the sequence D 1 D 6 2 becomes inadmissible. For this range of α the sequence D 1 D 7 2 is also inadmissible. Proof. Figure 15 shows the quadrangle B 1 = G 3 (G 3 (B) ∩ A 2 ) ∩ A 2 (thick blue), the set of points which stay in A 2 for 6 steps. The images G(B 1 ) (brown) and G 2 (B 1 )) (green) are also shown. Part a) is for α = 0.451 and part b) for α = 0.456. For larger α both images are completely inside A 2 . This means that the sequences D 1 D 6 2 and D 1 D 7 2 are inadmissible. The important point is G 7 (w) for the same  point w as before. The equation S(G 7 (w)) = 1/2 is equivalent to 192α 7 + 384α 6 − 432α 5 − 480α 4 + 480α 3 − 69α + 11 = 0 with a root α ∼ 0.4546258153.
Proposition 17. We have proved the existence of acim for alpha's up to α ∼ 0.4345268819 (Proposition 14). We have the following estimates on the σ 2 's of sequences starting with D 1 D 6 2 : Figure 15. Sequence D 1 D 6 2 becomes inadmissible.
Although the group D 1 D 4 2 maybe not admissible, this inequality can be used for useful estimates.
2 ) > 1 at least for α ≤ 0.4718920017. These estimates and previous results extend the range of the existence of acim up to α ∼ 0.4527916100.
Proof. Estimate 1) together with Proposition 15 tell us that all sequences starting with D 1 D 5 2 D 1 D 6 2 have σ 2 > 1 as long as they are admissible. All other sequences starting with D 1 D 6 2 must start with D 1 D 2 2 D 1 D 6 2 . Using inequality (6.1) and estimates 2)-6) we see that they all have σ 2 > 1 at least up to α ∼ 0.4527916100. We want to push α higher to make the sequences starting with D 1 D 6 2 inadmissible. First, we will find out what comes after the sequence D 1 D 2 2 D 1 D 6 2 for α > 0.4527916100.
Proposition 18. After α ∼ 0.4496432201 after the sequence D 1 D 6 2 comes the sequence D 1 D 5 2 D 1 D 2 2 . Proof. Figure 16 a) shows 6 consecutive images of triangle O 6 (introduced in Proposition 15), the set of points which leave A 2 after staying in it for six steps, for α = 0.446. The triangle G 3 (O 6 ) is completely in A 1 . This corresponds to the sequence D 1 D 2 2 D 1 D 6 2 , whose necessity was proved in Proposition 15. The triangle G 6 (O 6 ) intersects the partition line so some points leave A 2 at this moment, some continue staying in A 2 .
Part b) of the same figure show the same 6 images of O 6 and 3 next images, for α = 0.451. Some images have full descriptions, some only numbers. For this α triangle G 6 (O 6 ) is completely inside A 2 so all of its points continue staying in A 2 . The triangle G 9 (O 6 ) is completely in A 1 . This shows that for this range of α's after group D 1 D 6 2 there must be group D 1 D 5 2 D 1 D 2 2 . The important point is G 13 (w) (the same w as before), the left most vertex of G 6 (O 6 ). The equation S(G 13 (w)) = 1/2 implies 12288α 13 + 36864α 12 − 12288α 11 − 86016α 10 +16128α 9 +84480α 8 −43392α 7 −23360α 6 +36288α 5 −19456α 4 +2816α 3 + 1984α 2 − 869α + 91 = 0 with a root α ∼ 0.4496432201.  The exact estimates for α > 0.4600595036 become more and more complicated. We hope to find some more abstract way to prove that G satisfies the expanding conditions of [4]. We performed numerical experiments estimating σ 2 (x 0 , N ) = σ 2 N k=0 DG(G k (x 0 )) for millions of initial points x 0 . Instead of calculating σ 2 directly, we used estimate (see, e.g., [5]) where M F = m 2 1,1 + m 2 1,2 + m 2 2,1 + m 2 2,2 is the Frobenius norm of the matrix M . Since all M k 's are either D 1 or D 2 and det D 1 = det D 2 , the calculations of right hand side of (6.2) are very stable. All trials showed that for α < 1/2 the quantity σ 2 (x 0 , N ) grows to infinity as N increases. This provides numerical evidence for expanding properties of G and the existence of acim.
The Figures 17-18 show the support of acim (or conjectured acim) for α = 0.3, 0.4, 0.43, 0.46, 0.49, 0.495. The pictures were obtained by computer plotting 10 6 iterates long trajectory of G α after skipping the first 1.5 · 10 6 iterations. The experiments show that the obtained support is independent of the typical initial point.
It is clear that the sum of the coordinates grows on each step at least by the value y so eventually it goes above 1, which means that the point goes to the upper triangle. 7.2. α = 3/4. Let α = 3/4 and let us assume that x + y = 4/3 or 3x + 3y = 4. Then, G(x, y) = y, τ 3 4 y + 1 4 x .
The line y = 2/3 − x/3 (or equivalently y + x/3 = 2/3) partitions square [0, 1] × [0, 1] into two parts on which G is defined differently: A 1 below the line and A 2 above it. We partition region A 2 further into three parts: B 1 between the lines y = −x/2 + 5/6 and y = −x/2 + 7/6, B 2 between y = −x/2 + 5/6 and the partition line and B 3 above the line y = −x/2 + 7/6, see Figure 21. Let (x, y) ∈ A 1 \ {(0, 0)}. If (x, y) = (x, 0), then G(x, 0) = (0, x/2), so we can assume that y > 0. It is easy to calculate that G(x, y) = y, with the sum of second coordinate plus one third of the first coordinate equal to 5 2 y + 1 6 x so on each step this sum grows by at least y and eventually every such point will move to the upper half of the square y + x/3 > 2/3.
In Figure 22 a) we see the image G(B 2 ) (green) and both images G(A 1 ) and G(A 2 ) (grey dashed). The points outside G(A 1 ) ∪ G(A 2 ) are transient and unimportant for dynamics because they are eventually mapped into G(A 1 ) ∪ G(A 2 ). Thus, the only part of B 3 we will study is the image G(B 2 ). In Figure 22 b) we see the image G(G(B 2 )) (green). It consists of two parts, upper G 2 (B 2 ) ∩ A 2 and lower G 2 (B 2 ) ∩ A 1 .
In Figure 23 a) we see the image G(G 2 (B 2 ) ∩ A 2 ) (magenta) of the upper part of G 2 (B 2 ). We have G(G 2 (B 2 ) ∩ A 2 ) ⊂ G(B 2 ) ⊂ A 2 so further iterations of these points will be similar to that of whole G(B 2 ). In Figure 23 b) we see the image G(G 2 (B 2 ) ∩ A 1 ) (magenta) of the lower part of G 2 (B 2 ). We see that the points of G(G 2 (B 2 ) ∩ A 1 ) are either in B 1 (and then their future iterates are attracted to the line x + y = 4/3) or they are inside G(B 2 ) above the line y = −x/2 + 7/6 (upper red). The lowest point of G(G 2 (B 2 ) ∩ A 1 ) is (1/6, 3/4) and belongs to the line y = −x/2 + 5/6 (lower red).
Under the action of G every point in A 2 gets closer to the line x+ y = 4/3 (blue). To show that every point of G(B 2 ) is attracted to this line, it is enough to show that for any point (x, y) ∈ G 2 (B 2 ) ∩ A 1 its image (z, w) = (y, 3 2 y + 1 2 x) is either in B 1 or is closer to the line x + y = 4/3 than (x, y). Using the formula for the distance of a point from a line we have to check that Since the point (x, y) is below the partition line we have |x + y − 4/3| = 4/3 − x − y. Since the point (z, w) is above line y = −x/2 + 7/6 (upper red) we have |z + w − 4/3| = z + w − 4/3. Thus, our condition is equivalent to 4/3 − x − y > z + w − 4/3, or (7.2) 4/3 − x − y > y + 3 2 y + 1 2 x − 4/3, or y < − 3 7 x + 16 21 .
We extend Proposition 1 to : Proposition 19. If (x, y) ∈ A 1 and αy + (1 − α)x > a, a < 1/2, then the point (w, z) = G(x, y) satisfies αz + (1 − α)w > 2αa, holds also for the 1/2 < α < 3/4.  Proof. We have Proof. We will construct a trapping region T ⊂ A 2 , containing X 0 , such that G(T ) ⊂ T . Every point whose trajectory stays in A 2 is attracted to X 0 , since G |A2 is an affine map with an attracting point X 0 . We will prove that every point of A 2 eventually enters T . From Proposition 1 we know that every point except (0, 0) eventually enters A 2 .
Construction of T : The trapping region T is shown in Figure 24 a). It is a polygon with vertices p 1 , p 1a , p 2 , p 3 , p 3a , p 4 , p 5 and p 6 (red). Its image G(T ) is bounded by The only thing we have to prove is that any point of W (non-transient points going out of A 2 ) eventually enters the trapping region T . In Figure 24 b) we see that the second image G 2 (W ) is a thin quadrangle (looking like a triangle) adjacent to the upper boundary of the square [0, 1] × [0, 1]. The lowest point of G 2 (W ) is the point (2α(2α − 1), 8α 3 − 8α + 4). Its most to the right point is (α/(α + 1), 1). We will prove in Proposition 23 that for any point (x, y) with x ≤ x w = α/(α + 1) and y ≥ y w = 8α 3 − 8α + 4 and its third image (z, w) = G 3 (x, y) the difference z − x is larger than some positive constant depending on α and w ≥ y w unless (z, w) ∈ T . This shows that any point of G 2 (W ) eventually enters T , and completes the proof of Proposition 22.
Proof. Let (x, y) = (t, 1 − s) satisfy the assumptions. The third iterate G 3 on such point is equal either G 1 • G 2 • G 2 or G 2 • G 2 • G 2 . The first coordinate of (z, w) = G 3 (x, y) does not depend on the whether the last map applied is G 1 or Figure 26. a) T 3 and its images, b) enlargement of T 3 and G 3 (T 3 ).
Since both ct(α) and cs(α) are negative z − t has the least value when both t and s are maximal, i.e., t = x w and s = 1 − y w . Then, The graph of z − t is shown in Figure 25.
To prove the second claim we will consider the images of the rectangle T 3 (see Figures 25 b) and 26 ) with vertices z 1 = (0, y w ), z 2 = (x w , y w ), z 3 = (x w , 1) and z 4 = (0, 1). The second image G 2 2 (T 3 ) has the vertices G 2 (z 1 ), G 2 (z 4 ) ∈ A 1 and G 2 (z 2 ), G 2 (z 3 ) ∈ A 2 . Its sides intersect partition line at points z m between G 2 (z 1 ) and G 2 (z 2 ) and z ′ m between G 2 (z 3 ) and G 2 (z 4 ). The image G 3 (z 4 ) lies on the lower side of the rectangle T 3 and the image G 3 (z 1 ) is higher. The images G(z m ) = ((8α 4 − 12α 3 − 6α 2 + 17α − 6)/(α + 1), 1) and G(z ′ m ) are on the top side of the square. The image The line L(G 3 (z 2 ), G(z m )) intersects right hand side of T 3 at the point z i = (x w , 16α 4 − 32α 3 + 38α − 27 + 6/α) with the second coordinate larger than y w . This shows, that the points of G 3 (T 3 ) lie either in T 3 or in T . Together with the first claim this shows that every point of T 3 eventually enters T .
We continue to prove that the fixed point (2/3, 2/3) is a global attractor for other intervals of parameter α ∈ (1/2, 3/4).  Construction of the trapping region T : T is a pentagon with the vertices: p 3 which is the upper left vertex of W , p 5 = G(p 3 ), p 2 = G(p 5 ), p 4 = G(p 2 ), and p 1 = G −1 2 (p 3 ). Since, for α in the considered interval, G(p 4 ) ∈ T , we G(T ) ⊂ T , i.e., T is a trapping region. Figure 27 a) shows the trapping region T (red) and its image G(T ) (dashed black). The green quadrangle is W = G −1 2 (A 1 ) ∩ G(A 2 ). Below, we will show that fifth or fourth image of W is a subset of T . We consider subintervals of α.
i) (∼ 0.5930703309, ∼ 0.5970091680) α = ( √ 33 − 1)/8 ∼ 0.5930703309 is the largest α for which the sides of W and G 3 (W ) which are on the line x = 1 intersect. For α ∈ (∼ 0.5930703309, ∼ 0.5970091680), W and G 3 (W ) still intersect (the highest vertex of G 3 (W ) is in W ). (∼ 0.5970091680 is a root of 16α 5 − 16α 3 + 10α 2 − 9α + 4 = 0.) This causes a minimal "spill off" of G 4 (W ) outside T . See Figure 28. We also see there that G 5 (W ) ⊂ T .  iii)(∼ 0.6513878188, ∼ 0.7287135539) For α between ∼ 0.6513878188 and 1/4+ (33)/12 =∼ 0.7287135539, the region W is a triangle and G 4 (W ) ⊂ T . See Figure 29. Part a) shows the trapping region T (red) and its image G(T ) (dashed black). Part b) shows region W and its images, G 4 (W ) ⊂ T . For α approaching 0.7287135539 the top vertex of G 4 (W ) approaches boundary of T but stays in T as it is the image of the lowest vertex of G 3 (W ) which is already in T . For α above ∼ 0.7287135539 the image G(p 4 ) goes outside the line L(p 1 , p 2 ) and T is no longer a trapping region.
For the next interval of parameter α we have to make a "micro" adjustment of T adding to its construction two more vertices G(p 4 ) and G 2 (p 4 ).
Proof. Again, we construct a trapping region T and show that fourth image of W = G −1 2 (A 1 ) ∩ G(A 2 ) falls into T . The construction of T is a micro adjustment of the construction from Proposition 24, it is almost not visible on pictures. We add two more vertices , p 1a = G(p 4 ) and p 3a = G 2 (p 4 ), to the the construction and T becomes a heptagon (seven angles figure). Since G(p 3a ) is inside such constructed T , and T is convex, we have G 2 (T ) ⊂ (T ). See Figure 30. Part a) shows the trapping region T (red) and its image G(T ) (dashed black). The green triangle is the region W . G(p 3a ) stays inside T up to α =∼ 0.7464180853 but earlier another problem arises. At α =∼ 0.7360241475 the image G 2 (W ) starts intersecting with W and this needs another approach.   Now, we will consider the last subinterval of α's for which X 0 is an almost global attractor.