\begin{aligned}
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in thus section, we return to the consideration of the quadratic mapping $F_\mu(x)=\mu x(1-x)$. This time we will eoneider the specific parameler value $\mu=3.839$, following some ideas of Smale and Williams. We will drop the subscript $\mu$ for the remainder of this section and let $F(x)=3.839 z(1-z)$. Our goal is to sharpen the results of $\$ 1.10$ regarding the implieatione of the existence of a periodic point of period three. With a calculator, one may check eacily that there is an altracting orbit of period three for $f$ given to six decimals by $$ \begin{aligned} & a_1=.149888 \\ & a_2=.489172 \\ & a_3=.959299 . \end{aligned} $$ Moreover, $\left(F^3\right)^{\prime}\left(a_i\right) \approx-78$ appruxiuately. The existence of such a periodic point can be proved rigorously by hand computation: one simply finds a small interval about $a_1$ which is mapped inside itself by $F^3$ with derivative everywhere lesa than no Wo wo will selegeti tLi titivus deiailu io ine exercises, but it is important to realize that the computations involved ean be done by hand and with complete accuracy. By Sarkovskii's Theorem, $F$ has periodic points of all periods. By the results of $\xi 1.11$, none besides $a_i$ can be altraeting. Hence, for all practical purposes, these points are invisible to the computer. This brings up the question: where are all of these other periodic points and exactly how many of them are there? We auswer these questione below by intmducing a more general concept from symbolic dynamics, the subshift of finite type. We first define the shift on $N$ symbols. Let $\Sigma_N$ denote the set of all possible sequences of natural numbers between 1 and $N$, i.e.,
Our goal is to describe certain subsets of $\Sigma_N$ which arise naiurally und which provide a more general setting for symbolic dynamics. Let $A$ be an $N \times N$ matrix whose entry in the $i^{\text {th }}$ row and $j^{\text {th }}$ column, which we denote by $a_{i j}$, is either 0 or 1 . That is, $A$ is an II $\times$ Ir buaie anim of Q's And 1'e. A ie called the transition matrix for the system. We will use $A$ to describe which sequences in $\Sigma_N$ lie inside a subset which we denote by $\Sigma_A$. A sequence $s=\left(s_0 s_1 s_2 \ldots\right)$ lies in $\Sigma_A$ if it obey* the following rule. Each adjacent pair of entries in the sequence 8 determines a location in the matrix $A$, the $a_{n, \Delta i+1}$ eulry. The sequence lles in $\Sigma_A$ if and only if every such entry is 1 . More concisely, $$ \Sigma_A=\left\{(s)=\left(s_0 s_1 s_2 \ldots\right) \mid a_{e_i s_{i+1}}=1 \text { for all } i\right\} . $$ That is, we use the transition matrix to rule ont certain pairs of entriee in a sequence which lies in $\Sigma_A$. Example 13.2. Let $$ A=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \text {. } $$ Since $a_{12}=a_{21}=0$, it follows that 1 and 2 can never be adjacent in a sequence in $\Sigma_A$. Consequently, there are only two allowable sequences in $\Sigma_A$, the constant sequences $(111 \ldots)$ and $(222, \ldots)$. Example 13.3. Let $$ A=\left(\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right) $$ In this example, 2 may follow 1 , but not vice-versa. Thus, $\mathbf{\Sigma}_{\boldsymbol{A}}$ consists of the constant sequences plus any sequence of the form $(11 \ldots 11222 \ldots)$ where there are arbitrarily many 1's followed by infinitely many 2's. Example 13.4. Let $$ A=\left(\begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array}\right) $$ Any combination of 1 's and 2 's are allowed in a sequence in $\Sigma_A$, exerpt a pair of adjacent 2 's. We denote by $\sigma_A$ the restriction of $\sigma$ to the set $\Sigma_A$. The following propusition guarantees that this makes sense. Proposition 13.5. $\Sigma_A$ is a closed subset of $\Sigma_N$ which is invariant under $\sigma_A$. Proof. Invariance is clear. To prove that $\Sigma_A$ is closed, we suppose that $s_i$ is a sequence of elements of $\Sigma_A$, i.e., n aequence of sequences, which courerge to $t$. If $t \notin \Sigma_{\boldsymbol{A}}$, there is a smallest integer $\alpha$ for which $a_{t_n t_{a+1}}=0$. Fiuce the $a_i$ converge to $t$, there is another integer $K$ such that, if $i>K$, then $d_N\left(s_i, t\right]<1 / N^{a+1}$. By Proposition 13.1, this forces $t_{0,} t_1, \ldots, t_{\alpha+1}$ to agree with the corresponding entries of $s_i$ for $i \geq K$. In particular, we must have $a_{\text {tata+1 }}=1$, sinee $s_i \in \Sigma_A$. This contradictiun establishes the result.
We now return to the quadratic usp $F(z)-3.839 x(1-z)$. Recall that there is an attracting periodic orbit in the vicinity of $a_1=.149888, a_2=$ , 489172 , and $a_{\mathrm{y}}=.959299$. The graphs of $F$ and $F^3$ may be sketched as in Fig. 13.1. There is a second periodic orbit of period 3 for $F$ which we denote by $b_1, b_2, b_3$. These points are given approximately by
Recall from $\$ 1.11$ that there is a maximal open interval about each of Recall from this interval by $W\left(a_i\right)$. From the proof of Theorem 11,4, we see that one of the endpoints of each $W\left(a_i\right)$ is fixed by $F^2$. Hence $b_i$ is une of the eadpoiuts of $W\left(a_i\right)$. Let us denote by $\dot{b}_i$ the point on the opposite side of $a_i$ from $b_i$ which is mapped to $b_i$ by $F^3$. See Fig. 13.2. Let $A_1=\left(\hat{b}_1, b_1\right), A_2-\left(\hat{b}_2, b_2\right)$, and $A_3=\left(b_3, \hat{b}_3\right)$. Note that $F$ maps $A_1$ and $A_3$ monotonically onto $A_2$ and $A_1$ respectively, but $F$ has a critical point at $1 / 2 \in A_2$ so $F$ is not monotonic on this interval. Since, however, the maximum valne of $F$ is 95975 , it follows that $F\left(A_2\right)$ is contained in $A_3$. Also note that $F\left(b_2\right)=F\left(\dot{b}_2\right)=b_3$. As in $\xi 1.5$, it is easy to prove that if $x<0$ or $x>1$ then $F^n(z) \rightarrow-\infty$. Morenver, if $x \in A_i$ for some $i$, then $F^n(z)$ tends to the orbit of $a_1$. Hence all of the other periodic points must lie in the complement of the $A_i$ in $I$. There are iour ciosed intervais in the complement of the $A_i$ in $I_j$ we call these intervals $I_0, I_1, I_2$, and $I_3$, from left to right. Since we know the behavior of the $b_i$ under iteration of $F$, we know how these four intervals are mapped by $F$. Their images are depicted in Fig. 13.3. Since there are no other periodic points for $F$ in the $W\left(a_i\right)$, it follows that all of the infinitely many periodic points must lie in the $I_j$. In fact, we can say more. Proposition 13.6. All periodic points of $F$ lie in $I_1 \cup I_2$ with the exception of the fized point at 0 and the perindie point of period $3: a_1, a_2$, anl $a_3$. Proof. Note that $F$ is monotonic on each of the intervals $I_j$. From Figure 13.3, we see that $F$ maps $I_0$ across both $I_0$ and $I_1, I_1$ onto $I_2, I_2$ across both $I_1$ and $I_2$, and finaliy $I_3$ vato $I_0$. From this it follows that if a periodir point $x \in I_1 \cup I_2$, then $F(x) \in I_1 \cup I_2$. Thus, if $x \in I_1 \cup I_2$ lies on a periodic orbit, then the entire orbit of $x$ lies in $I_1 \cup I_2$. for which $F^{\infty}(x) \notin I_0$. Either $F^n(z) \in A_1$, in which case $x$ is not periodic, or $F^n(z) \in I_1$. In the latter case, the forward orbit of $F^{-\infty}(x)$ can never leave $I_1$ or $I_2$ to return to $x_{\text {, sn, }}$, again, $x$ is not periodic. Finally, if $x \in I_3, F(x) \in I_0$, so $\boldsymbol{x}$ is again not periodic. q.e.d. Consequently, all of the remaining periodic points for $F$ lie in $I_1 \cup I_2$. Let us denote by $\Lambda$ the set of points whose entire orbit is contained in these two intervals. To understand the dynamics of $F$ on $A$, we again invoke symbolic dynamics. We define the sequence associated to $x$ by the rule $$ S(z)=\left(\operatorname{sos}_1 s_2 \ldots\right) $$ where $a_j-1$ if $F^{\prime}(\approx) \subset I_1$ and $s_j-2$ if $F^j(z) \in I_2$. Since $F\left(I_1\right)=I_2$, it follows that a 1 can only be followed by a 2 , i.e., that $S$ takes its values in $\Sigma_{\boldsymbol{A}}$ where $$ A-\left(\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right) $$ We remark that we encountered a simllar phenouruon in the proof of Sarknvakii's Theorem. Indeed, the existence of a point of period 3 forces the existence of a pair of intervals that behave like $I_1$ and $I_2$ under iteration of $F$. In the present case, huwever, we con sey much more. Theorem 13.7. The restriction of $F$ to $A$ is topologically conjugute to the gubghift of finite type giten by $\sigma_A$ on $\Sigma_A$. Proof. The proof of surjectivity and continuity of $S$ proceeds exactly as in $\$ 1.7$ so we omit the details. 'The only difference from $\S 1.7$ arises in the proof that $S$ is one-to-one, an we ennrentrote on thin fet. TLi disacitt luote is that $\left|F^{\prime}(z)\right|$ is not everywhere greater than one on $I_1 \cup I_2$. However, we can say that $\left|F^{\prime}(x)\right|>\nu=F^{\prime}\left(\hat{b}_2\right) \approx, 3$, since $F^{\prime \prime}<0$ and the interval $\left(\vec{b}_2, b_2\right)$ containing the critical point has heen removed. Hence $\left|F^2\right|$ is bounded from below. We claim that there exists $\lambda>1$ such that, if $z \in \Lambda$, then $\left|\left(F^3\right)^{\prime}(x)\right|>\lambda$. To prove this, we note that there are three closed intervals in $I_1 \cup I_2$ in which $\left|\left(F^3\right)^{\prime}(x)\right| \leq 1$. Two of them, $B_1$ and $B_2$, are symmetrically located with reepect to $1 / 2$, while the thiird, $B_3$, lies in $I_2$. We note that the $F^3$-image of $B_3$ is contained in $\left(\vec{b}_1, b_1\right)$ and so $B_3 \cap \Lambda=\square$. See Fig. 13.1 We claim that $B_1$ and $B_2$ are mapped by $F^3$ into $\left(b_3, b_3\right)$. Indeed, one musy check easily that $B_2$ is contained in the interval, $661 We now prove that $A$ is a hyperbolic set. Chnose $K$ such that $\mu^2, \lambda^K>1$. Let $N=3 K+2$. If $n>N$, then we may write $n=3(K+\alpha)+i$ where $\alpha>0$ and $0 \leq i \leq 2$ are integers. Hence, if $x \in A$, using the Chain Fule, we have $$ \left|\left(F^m\right)^{\prime}(x)\right|=\left|\left(F^i\right)^{\prime}\left(F^{3(K+\alpha)}(z)\right)\right| \cdot\left|\left(F^{s(K+\alpha)}\right)^{\prime}(z)\right|>\nu^2 \lambda^{K+\alpha}>1 . $$ This proves that $A$ is a repelling hyperbolic set. With hyperbolicity established, the remainder of the proof is similar to that of Theorem 7.2. This compleles the proof. Thed
Remark. The technigues nked in thic pmot man be wst is pive ihas vise quadratic map $F_\mu(x)=\mu z(1-z)$ admits a hyperbolic set when $4<\mu \leq$ $2+\sqrt{5}$, the cases we omitted in 31.5. See Exercise 7. We now see that there are periodic points for $F$ of all periods, as guaranteed by Sarkovskii's Theorem. Indeed, to produce a periodic point of period $k$ in $\Sigma_A$, we need only list a string of $k-12$ 's followed by a 1 , and then repeat this sequence. These are exactly the orbits produced in the proof of Sarkovskii's Thenrem. Of course there are many olher allowabie repeating sequences in $\Sigma_A$, se this raises the question of exactly how many periodic points of period $k F$ has. To answer this, we first need a definition. Definition 13.8. Let $A=\left(a_{i j}\right)$ be an $N \times N$ matrix. The trace of $A$ is given by $$ \operatorname{Tr}(A)=\sum_{i=1}^N a_i, $$ i.e., by the sum of the diagonal entries of $A$. The trace is an important invariant of the conjugacy class of a matrix which is studied in advanced linear algebra. For us, its purpose is quite different: the trace of the powers of $A$ gives an aceurate count of the periodic points in $\Sigma_A$. Recall that if $A=\left(a_{i j}\right)$ and $B=\left(b_{i j}\right)$ are $N \times N$ matrices, then the product of $A$ and $B$ is the $N \times N$ matrix $A \cdot B=\left(c_{i j}\right)$ where $$ e_{i j}=\sum_{k=1}^N a_{i k} h_{k j} $$ In particular, if $$ A=\left(\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right) $$ then $$ \begin{aligned} & A^2=A \cdot A=\left(\begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right) \\ & A^3=A \cdot A^2=\left(\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right) \\ & A^4=A \cdot A^3=\left(\begin{array}{ll} 2 & 3 \\ 3 & 5 \end{array}\right) \end{aligned} $$ Proposition 13.9. Let $A$ be wn $N \times N$ tranaition matris. Then $$ \text { card } \operatorname{Per}_K \sigma_A=\operatorname{Tr}\left(A^{\hbar}\right) $$ Proof. Recall that a sequence $\mathrm{s}$ in $\Sigma_A$ is fixed by $\sigma^K$ if $s$ is a repeating aequence of the form $\left(i_0 i_1 \ldots i_{K-1}\right.$ iot $\left.\ldots i_{K-1} \ldots\right)$. Such a sequence lies in $\Sigma_A$ iff $a_{i_0 i_1}=a_{i_1 i_2}=\ldots=a_{i_{K-1} i_1}=1$ or, equivalently $a_{i_0 i_1} a_{i_1 i_2} \ldots \ldots a_{i_{K-1} i_0}=$ $i_0 i_1 . . i_{K-1} i_0$ is an allowed piece of a sequence in $\Sigma_A$, and equal to sero otherwise. Consequently, $$ \sum_{i_0, i_1, \ldots, i_{K=1}} a_{i_n i_1} a_{i, i}, \cdots+a_{i_K}, \text { is } $$ gives the cardinality of $\operatorname{Per} K \sigma_A$. On the other hand, it is easily checked that this sum is $\operatorname{Tr}\left(A^K\right)$. q.e.d. Note that we may therefore compute easily that, for $$ A=\left(\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right) $$ we have $$ \begin{aligned} & \operatorname{Tr}(A)=1 \\ & \operatorname{Tr}\left(A^2\right)=3 \\ & \operatorname{Tr}\left(A^3\right)=4 \\ & \operatorname{Tr}\left(A^4\right)=7 \\ & \operatorname{Tr}\left(A^5\right)=11 \end{aligned} $$ In general, for $K>2$, $$ \operatorname{Tr}\left(A^K\right)=\operatorname{Tr}\left(A^{K-1}\right)+\operatorname{Tr}\left(A^{K-2}\right) $$ Tuis recursion is curiousiy the same as the well-known Fibonacci recursion relation $$ p_k=p_{k-1}+p_{k-2} $$ with $k>2$. The usual Fibonacci sequence begins with $p_1=p_2=1$; here we have $p_1=1$ but $p_2-3$. Exercises 1. Let $\mu=3.839$ and $a_1=.149888$. Prove that there is a small interv.: about $a_1$ which is mapped inside itself by $F_{\mu^{+}}^3$ 2. Show that there is a sumaler interval about $a_1$ which is mapped inside itself and on which $\left|\left(F_e^3\right)^{\prime}\right|<1$. This proves that there in a unique attracting periodic point with period 3 near $a_1$. 3. Define a metric on $\Sigma_N$ by $$ d_n[s, t]=\sum_{k=0}^{\infty} \frac{\left|s_k-t_k\right|}{14^2} $$ a. Prove that $d_n$ is a metric. b. Prove the analugue of Proposition 6.2 , i.e. if $s_i-t_i$ for $i=0, \ldots, k$ then $d_n|s, t| \leq 1 / n^k$. Similarly, if $\left.d_n \mid s, t\right]<1 / n^k$, then $s_i=t_i$ for $i \leq k$. 4. $\operatorname{Trt} A=\left(\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right)$. Prove that $\operatorname{Tr}\left(A^K\right)=\operatorname{Tr}\left(A^{K-2}\right)+\operatorname{Tr}\left(A^{K-1}\right)$. 5. Using the intervals $I_0$ and $I_3$ as well as $I_1$ and $I_2$, show that the set of points whose orbits remain for all time in these four intervals also determines a subshift of finite type. What is the transition matrix? 6. Let $$ A=\left(\begin{array}{lll} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) $$ Find a formula for the trace of $A^K$. 7. Prove that $F_\mu(x)=\mu x(1-x)$ admits a hyperbolic set in $[0,1]$ when $4<\mu \leq 2+\sqrt{5}$ Download 34.52 Kb. Do'stlaringiz bilan baham: 
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