By corollary 1.8, if f : I → I is a piecewise-monotone continuous map then lim
n→∞(1
n log V (fn)) = htop(f ) [13].
V (fn) is total variation of fn in an invariant set I. In order to compute the total variation, we first cut the invariant set into m disjoint parts, and these parts will exactly cover another part or union of some parts through the function. Then we count the number of these parts as the total length of the iteration.
Example 3.5. The total variation of n times iterated of tent map 1 for k = 2.
t(x) =
the total length is generated by the length of 2I1, 2I2,
V (t2) = 2V (t(I1)) + 2V (t(I2)) = 4I1+ 4I2, the total length is generated by the length of 4I1, 4I2.
Follow this way, then we distribute a matrix that represent the transition of the quantity of every parts. That is, let T =
1 1 1 1
, so we can compute the total variation by
V (tn) =
From this example, we can see that if we define the graph of the model (as Fig. 13), then we can use this graph to distribute the transition matrix to compute the total variation.
Figure 13: The graph of tent map when coefficient is 2.
Consider the model (7) for κ∈ N, κ ≥ 2, the invariant set is [−κ + 1, 1]. Now we separates this set into κ parts, that is,
I = [−κ + 1, 1] = [−κ + 1, −κ + 2] ∪ [−κ + 2, −κ + 3] ∪ · · · ∪ [−1, 0] ∪ [0, 1] =
Follow this way, Then we can distribute a matrix that represent the transition of the quantity of every parts (also can by the graph as Fig. 14).
Figure 14: The graph of model (7).
Proposition 3.6. If f : I → I is a piecewise-monotone continuous map, and the change of the variation of f can be represent as a matrix T , then htop(f ) = log λ1, λ1 > 0, where λi, i = 1, . . . , n are eigenvalues of T and λ1 >|λm|.
Proof. Because this T is irreducible, so by Perron-Frobenius Theorem [1], there exists λ1 >
0, and λ1 > |λm|, for all m ̸= 1, λ1, . . . , λm are eigenvalues of T. And let v1, . . . , vn are
Definition 3.7 ([32]). The matrix
is called the companion matrix of the characteristic polynomial f (t) = (−1)k(a0+ a1t +· · · + ak−1tk−1+ tk).
By Definition 3.7 the transpose of the matrix Tκ∗κwe obtain before is a companion matrix with a0 = a1 = · · · = ak−1 = −1, and because T and TT has same eigenvalues, so the eigenvalues of Tκ×κ satisfies −λκ+ λκ−1+ λκ−2+· · · + λ + 1 = 0.
Proposition 3.8. The topological entropy of model (7) is bigger than 0 for κ∈ N \ {1}.
Proof. Because −λκ+ λκ−1+ λκ−2+· · · + λ + 1 = −λκ+1λ+2λ−1 κ−1 = 0. Let f (λ) =−λκ+ λκ−1+ λκ−2+· · · + λ + 1 = −λκ+1λ+2λ−1 κ−1 = −λκ + λλκ−1−1, then f (1) = κ− 1 > 0, f(2) = −12 < 0, so by Intermediate Value Theorem, there exists a root in the interval (1, 2).Because f′(λ) =
−κλκ−1+κλκ−1(λ−1)−(λκ−1)
(λ−1)2 =−κλκ−1+ κλλ−1κ−1 − (λλκ−1)−12 =−((1 −λ−11 )κλκ−1+ λκ−1
(λ−1)2) < 0 for λ > 1, so we know f (λ) is strictly decreasing for λ > 1. More precisely, we suppose the root λ = 2− ϵ then f(2 − ϵ) = −(2−ϵ)κ+1(2−ϵ)−1+2(2−ϵ)κ−1 = ϵ(2−ϵ)1−ϵκ−1 = 0 , ϵ(2− ϵ)κ− 1 = 0, (2 − ϵ)κ = 1ϵ, κ is increase as ϵ decrease, so the root is close to 2 when κ is sufficiently large.
Therefore, htop(f ) = log λ1 > log(1) = 0.
By Theorem 1.4 and Corollary 1.8, we can obtain that the model (7) is chaotic in Li and Yorke’s sense for κ∈ N \ {1}.
4 Numerical Analysis
By Definition 1.10 to compute the Lyapunov Exponents of two trees’ model (6), first we need to compute ∥Df(x)∥, then the Lyapunov Exponents
λ(x0) = lim sup
We use MATLAB to compute the Lyapunov Exponents of model (6) as shown in Fig. 15, Fig. 16, Fig. 17, and Fig. 18.
Figure 15: Lyapunov Exponents of model (6) when β = 2.
In Fig. 15, fixed β = 2, and red line is the first Lyapunov exponents and blue line is the second Lyapunov exponents, then we can see that the first Lyapunov exponents are almost positive when κ > 1 except κ∈ N, and second Lyapunov exponents are negative. In Fig. 16 and Fig. 17, these two figures show that the first and the second Lyapunov exponents for a pair of κ and β, we can see that the second Lyapunov exponents are negative. In Fig. 18, the first subfigure is the first Lyapunov exponents for a pair of κ and β, the second subfigure is the phase plane of the model (6) and the period is corresponding to color bar. This way, we can see more careful that the Lyapunov exponents is negative appear in the period part.
Then we compute the Lyapunov exponents of model (7) as shown in Fig. 19, we can also see that the first Lyapunov exponents are negative for κ∈ N, too.
5 Future Work
For κ > 0, when κ ∈ N, there are period solution there, but except here, this model are almost high period. And these solutions are stable, Lyapunov exponents are also negative,
Figure 16: First Lyapunov exponents for a pair of κ and β.
but the topological entropy are bigger than 1. We guess this is computer’s numerical error like tent map when κ = 2.
The line κ|1−Nβ−1| = 1 for n-trees seperates the region to two parts, which is synchronized by the sense of contraction and others. If we make the condition looser, like become dispersed in next m− 1 steps and closer at m step. This way we may can seperates the synchronized part and desynchronized part more accurately.
In original model (5), when a tree is bigger than one, and others are smaller than one.
Next time this tree will become one until another tree is bigger than one. We think this is not reasonable because if one the other tree spent much time to become bigger than one, this model tell us the first tree will stay one and always wait but not withered. So we think this model can be improved. And some plants can flowering itself like papaya, this means it will not wait other trees until they can produce pollen. This way we rewrite model (4) as
Yi(t + 1) =
Yi(t) + 1, if Yi(t)≤ 0,
−κYi(t)Pi(t)β+ 1, if Yi(t) > 0,
(10)
Figure 17: Second Lyapunov exponents for a pair of k and β.
in which Pi(t) is given by
Pi(t) = (
1 N
∑N j=1
[Yj(t)]+ )
, (11)
where [Y ]+= Y if Y > 0; [Y ]+ = 0 if Y ≤ 0. Then we compute the phase plane of this model for two trees as Fig. 20. This is deiiferent as Fig. 4 and the analysis of this model can be an extention of this paper.
Figure 18: Lyapunov exponents and the phase plane of the model (6).
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