Assignment 6

Problem 1

Question

Let $G$ be the logistic map $G(x)=4x(1-x)$ on $[0,1]$ and define the conjugacy to be the change of coordinates $C(x)=ax+b$. Find the map $g$ that is conjugate to $G$.

conjugacy $g=C\circ G\circ C^{-1}$. same setup as HW5 Problem 4 but now general $a,b$.

$C^{-1}(x)=\frac{x-b}{a}$, so:

$$g(x)=C\left(G\left(C^{-1}(x)\right)\right)=a\cdot 4\cdot \frac{x-b}{a}\left(1-\frac{x-b}{a}\right)+b$$ $$=4(x-b)\cdot \frac{a-(x-b)}{a}+b=\frac{4(x-b)(a+b-x)}{a}+b$$

expanding if you want a polynomial form:

$$g(x)=-\frac{4}{a}{x}^2+\frac{4(a+2b)}{a}x-\frac{4b(a+b)}{a}+b$$

sanity check: $C(x)=4x-2$ (i.e. $a=4,b=-2$) should give $g(x)=2-x^2$ like hw5:

$$g(x)=\frac{4}{4}(x+2)(2-x)-2=(x+2)(2-x)-2=-x^2+4-2=2-x^2✓$$

note $C$ maps $[0,1]\to [b,a+b]$, so $g$ acts on $[b,a+b]$.

Answer

$$g(x)=\frac{4(x-b)(a+b-x)}{a}+b$$

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Problem 2

Question

Explain why each infinite itinerary of the tent map $T$ represents the orbit of exactly one point in $[0,1]$.

Tent Map:

$$T(x)=\{\begin{array}{ll}\mu x& for0\le x\le \frac{1}{2}\\ \mu (1-x)& for\frac{1}{2}\le x\le 1\end{array}$$

where $\mu =2$, piecewise linear, slope $\pm 2$.

itinerary of $x$: $S_0{S}_1{S}_2\dots$ where $S_i=L$ if $T^i(x)\in [0,1/2]$, $S_i=R$ if $T^i(x)\in [1/2,1]$.

given infinite itinerary $S_0{S}_1{S}_2\dots$ we define $I_{S_0\dots S_{n-1}}$ as set of all $x\in [0,1]$ whose first $n$ symbols match. claim: this is a closed interval of length $1/2^n$.

induction for int length - base: $I_L=[0,1/2]$, $I_R=[1/2,1]$, both length $1/2$.

induction step: suppose $I_{S_0\dots S_{n-1}}$ has length $1/2^n$. next symbol $S_n$ requires $T^n(x)\in L$ or $R$. since $T$ has slope $\pm 2$ on each half, restrincting $T$ to $L$ or $R$ maps an interval of length $\ell$ onto one of length $2\ell$. equivalently, preimage of any interval under $T{|}_L$ or $T{|}_R$ has half length. so specifying $S_n$ halves $I_{S_0\dots S_{n-1}}$:

$$|I_{S_0\dots S_n}|=\frac{1}{2}|I_{S_0\dots S_{n-1}}|=\frac{1}{2^{n+1}}$$

existence. nested closed intervals: $I_{S_0}\supset I_{S_0{S}_1}\supset I_{S_0{S}_1{S}_2}\supset \cdots$. each is nonempty and compact. by the nested interval theorem, ${\bigcap }_{n=1}^{\infty }{I}_{S_0\dots S_{n-1}}\ne \varnothing$.

uniqueness. diameter $|I_{S_0\dots S_{n-1}}|=1/2^n\to 0$. so intersection contains exactly one point.

Answer

each symbol in itinerary selects one of $T$‘s two branches, halves candidate interval. after $n$ symbols, point confined to closed interval length $1/2^n$. nested interval theorem shows existence; diameter shirnking yeilds uniqueness. thus every infinite itinerary $S_0{S}_1{S}_2\dots$ finds exactly one $x\in [0,1]$.

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Problem 3

Question

(a) Find a scheme to provide for any positive integer $n$, a sequence $S_1\dots S_{n+1}$ of the symbols $L$ and $R$ such that $S_1=S_{n+1}$, and such that the sequence $S_1\dots S_n$ is not the juxtaposition of identical subsequences of shorter length. For example, the sequence $LRLRL$ does NOT guarantee the existence of a period-4 orbit; the sequence $LRRLL$ does.

(b) Prove that the logistic map $G$ has a periodic orbit for each integer period.

A

need: for any $n\ge 1$, a sequence $S_1\dots S_{n+1}$ with $S_1=S_{n+1}$ and $S_1\dots S_n$ not juxtaposition of identical short blocks.

scheme: $S_1\dots S_n=L{R}^{n-1}$, with $S_{n+1}=L=S_1$.

examples:

$n=1$: $LL$ (fixed point in $L$)

$n=2$: $LRL$

$n=3$: $LRRL$

$n=4$: $LRRRL$

proof nonrepeating: suppose $S_1\dots S_n=L{R}^{n-1}$ is a repetition of a block of length $d$, where $d|n$ and $d<n$. then $S_1=S_{1+d}$. but $S_1=L$ and $S_{1+d}=R$ (since $1+d\ge 2$ and all positions $2$ through $n$ are $R$). contradiction.

Answer

$S_1\dots S_n=L{R}^{n-1}$, $S_{n+1}=L$. single $L$ at position 1 breaks any possible repetition, since all other positions are $R$.

B

re-proving HW4 Problem 1, now via symbolic dynamics.

$G$ maps $L=[0,1/2]$ and $R=[1/2,1]$ each onto $[0,1]$ (monotonically). so the transition graph is complete: $L\to L$, $L\to R$, $R\to L$, $R\to R$. every itinerary is realized.

for any itinerary $S_1\dots S_n$, define $I_{S_1\dots S_k}$ as the set of $x\in I_{S_1}$ with $G^{j-1}(x)\in I_{S_j}$ for $j=1,\dots ,k$. since $G{|}_L$ and $G{|}_R$ are monotone bijections onto $[0,1]$, each $I_{S_1\dots S_k}$ is a closed interval.

importantly to our goal, $G^n$ maps $I_{S_1\dots S_n}$ onto $[0,1]$.

proof by induction. $G$ maps $I_{S_1}$ onto $[0,1]$. $G$ maps $I_{S_1\dots S_n}$ onto $I_{S_2\dots S_n}$ (shift the itinerary). so $G^n$ maps $I_{S_1\dots S_n}$ to $G^{n-1}(I_{S_2\dots S_n})$. by inductive hypothesis, $G^{n-1}$ maps $I_{S_2\dots S_n}$ onto $[0,1]$. done.

now take the itinerary $L{R}^{n-1}$ from part (a). $G^n$ maps the closed interval $I_{L{R}^{n-1}}$ continuously onto $[0,1]\supset I_{L{R}^{n-1}}$. by IVT, $G^n$ has a fixed point $x^{\ast }\in I_{L{R}^{n-1}}$.

$x^{\ast }$ has period exactly $n$: its itinerary repeats $L{R}^{n-1}$ cyclically. if the true period were $d|n$ with $d<n$, the itinerary would repeat with period $d$, making $L{R}^{n-1}$ a juxtaposition of length-$d$ blocks. part (a) rules this out.

Answer

for each $n$, the itinerary $L{R}^{n-1}$ from part (a) determines a closed interval $I_{L{R}^{n-1}}$ that $G^n$ maps onto $[0,1]$. IVT gives a fixed point of $G^n$. primitivity of the itinerary guarantees exact period $n$.

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Problem 4

Question

Consider the period-three map of chapter 1, shown in figure 1.14.

(a) Find itineraries that obey the transition graph of the period-three map for any period, which are not periodic for any lower period.

(b) Prove that this period-three map has periodic orbits for each positive integer period.

(c) Why does period-3 imply chaos?

A

period-3 orbit $a<b<c$ with $f(a)=b$, $f(b)=c$, $f(c)=a$. intervals $I=[a,b]$, $J=[b,c]$.

transition graph: $f(I)\supseteq [b,c]=J$ (by IVT, since $f(a)=b$, $f(b)=c$). $f(J)\supseteq [a,c]=I\cup J$ (since $f(b)=c$, $f(c)=a$). so:

graph made with tikz

$"\\usepackage{tikz}\n\\usetikzlibrary{positioning, arrows.meta}\n\\begin{document}\n\\begin{tikzpicture}[\n box/.style={draw, circle, minimum size=1.2cm, font=\\large, thick},\n ->, >=Stealth, semithick, bend angle=25\n]\n \\node[box] (I) {$I$};\n \\node[box, right=3cm of I] (J) {$J$};\n \\draw (I) edge[bend left] (J);\n \\draw (J) edge[bend left] (I);\n \\draw (J) edge[loop above, looseness=6] (J);\n\\end{tikzpicture}\n\\end{document}"$

source code

constraint: from $I$, must go to $J$. from $J$, can go to $I$ or $J$. no $I\to I$ arrow.

scheme: for $n=1$: $J$ (fixed point, $J\to J$). for $n\ge 2$: $I{J}^{n-1}$.

check transitions: $I\to J$ works, $J\to J$ (middle positions) work, wrap $J\to I$ works.

primitivity: if $I{J}^{n-1}$ were a repetition of blocks of length $d|n$, $d<n$, then $S_1=S_{1+d}$. but $S_1=I$ and $S_{1+d}=J$ for $d\ge 1$. contradiction.

Answer

$n=1$: $J$. $n\ge 2$: $I{J}^{n-1}$

B

sim structure to Problem 3(b), adapted to this transition graph.

for itinerary $S_1\dots S_n$ obeying the graph, define $I_{S_1\dots S_k}$ as the set of points in $I_{S_1}$ whose orbit follows $S_1,\dots ,S_k$. since $f$ maps each interval continuously onto a superset of the next, taking preimages gives nested closed intervals.

$f$ shifts the itinerary: $f(I_{S_1\dots S_n})=I_{S_2\dots S_n}$. inductively, $f^{n-1}$ maps $I_{S_1\dots S_n}$ onto $I_{S_n}$.

so $f^n$ maps $I_{S_1\dots S_n}$ onto $f(I_{S_n})$. for our scheme, $S_n=J$ (for $n=1$ it’s $J$; for $n\ge 2$, last symbol is $J$). since $f(J)\supseteq I\cup J\supseteq I_{S_1}\supseteq I_{S_1\dots S_n}$:

$$f^n(I_{S_1\dots S_n})\supseteq I_{S_1\dots S_n}$$

IVT gives a fixed point $x^{\ast }$ of $f^n$ in $I_{S_1\dots S_n}$. primitivity of the itinerary forces exact period $n$.

Answer

for each $n$, the itinerary from part (a) determines a closed interval mapped over itself by $f^n$. IVT gives a period-$n$ point.

C

the transition graph $J\to J$, $J\to I$, $I\to J$ supports:

1. periodic orbits for every period - this alone follows from Sharkovskii’s Theorem: period 3 sits first in the Sharkovskii ordering $3▹5▹7▹\cdots$, so period-3 implies all other periods.
2. uncountably many aperiodic orbits - any non-eventually-periodic sequence over $\{I,J\}$ obeying the transitions gives a distinct orbit
3. Sensitive Dependence on Initial Conditions - two points with different itineraries must eventually land in different intervals ($I$ vs $J$), separated by at least $|b-a|$. nearby points generically have diverging itineraries.

Answer

period-3 implies chaos because the transition graph forces all three Devaney criteria: dense periodic orbits (all periods exist), sensitive dependence (divergent itineraries), and an uncountable scrambled set of aperiodic orbits

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Problem 5

Question

Define the piecewise map

$$f(x)=\{\begin{array}{ll}{f}_1=1-3x& 0\le x\le 1/3\\ f_2=-1/3+x& 1/3\le x\le 1\end{array}$$

Note that $x=0$ is a period-4 point. Let $I=[0,1/3]$, $J=[1/3,2/3]$, $K=[2/3,1]$.

(a) Draw the transition graph.

(b) Fill in the table below

$k$	Periodic Itinerary	Orbit
1	$I$	$\{a_1\}$
2	$IJ$	$\{a_1,a_2\}$
3
4
5

For each periodic orbit of period $p<6$. List the itinerary (in alphabetical order) and label the points of the orbit according to their magnitude $a_1<a_2<\cdots <a_p$.

(c) Show that there is no period-4 orbit with the itinerary $IJIJ$.

A

$f_1(x)=1-3x$ on $I=[0,1/3]$: $f_1(0)=1$, $f_1(1/3)=0$. so $f(I)=[0,1]\supseteq I\cup J\cup K$.

$f_2(x)=-1/3+x$ on $J=[1/3,2/3]$: $f_2(1/3)=0$, $f_2(2/3)=1/3$. so $f(J)=[0,1/3]=I$.

$f_2(x)=-1/3+x$ on $K=[2/3,1]$: $f_2(2/3)=1/3$, $f_2(1)=2/3$. so $f(K)=[1/3,2/3]=J$.

graph made with tikz

$"\\usepackage{tikz}\n\\usetikzlibrary{positioning, arrows.meta}\n\\begin{document}\n\\begin{tikzpicture}[\n box/.style={draw, circle, minimum size=1.2cm, font=\\large, thick},\n ->, >=Stealth, semithick, bend angle=20\n]\n \\node[box] (I) {$I$};\n \\node[box, right=2.5cm of I] (J) {$J$};\n \\node[box, right=2.5cm of J] (K) {$K$};\n \\draw (I) edge[loop above, looseness=6] (I);\n \\draw (I) edge[bend left] (J);\n \\draw (I) edge[bend left=30] (K);\n \\draw (J) edge[bend left] (I);\n \\draw (K) edge[bend left] (J);\n\\end{tikzpicture}\n\\end{document}"$

source code

Answer

$I\to I,J,K$ (full coverage). $J\to I$ only. $K\to J$ only.

B

transition graph: $J$ and $K$ have forced successors. only choice is at $I$: go to $I,J,$ or $K$. forced chain $K\to J\to I$ means $K$ and $J$ always resolve back to $I$.

for enumerating period-$p$ orbits, we list all valid cyclic itineraries of length $p$ and solve linear system for each. Since $f_1$ has slope $-3$ and $f_2$ has slope $1$, all compositions are linear and each valid system has a unique solution.

$k$	Periodic Itinerary	Orbit
1	$I$	$\{1/4\}$
2	$IJ$	$\{1/6,1/2\}$
3	$IIJ$	$\{1/8,7/24,5/8\}$
3	$IKJ$	$\{1/12,5/12,3/4\}$
4	$IIIJ$	$\{1/7,5/21,2/7,4/7\}$
4	$IIKJ$	$\{0,1/3,2/3,1\}$
5	$IIIIJ$	$\{11/80,19/80,61/240,23/80,47/80\}$
5	$IIIKJ$	$\{1/28,19/84,9/28,47/84,25/28\}$
5	$IIJIJ$	$\{11/84,5/28,23/84,13/28,17/28\}$
5	$IJIKJ$	$\{1/24,5/24,3/8,13/24,7/8\}$

C

itinerary $IJIJ$ means $I\to J\to I\to J$ repeating. solve the system:

$$\begin{array}{rl}{x}_2& =1-3x_1\in J\\ x_3& =-1/3+x_2=2/3-3x_1\in I\\ x_4& =1-3x_3=-1+9x_1\in J\\ x_1& =-1/3+x_4=-4/3+9x_1\end{array}$$

closing: $-8x_1=-4/3$, so $x_1=1/6$. then $x_2=1/2$, $x_3=1/6$, $x_4=1/2$.

$x_3=x_1$ and $x_4=x_2$. the orbit collapses to $\{1/6,1/2\}$, which is the period-2 orbit from $IJ$.

this is forced: $IJIJ=(IJ{)}^2$ is period 2 itinerary repeated twice. any fixed point for $f^4$ with this satisfies $f^2(x)=x$, thus has period dividing 2. since $f_1$ and $f_2$ are injective, system has unique solution, and that solution is necessarily the period 2 point.

Answer

unique solution to $IJIJ$ system yields $x_1=x_3=1/6$, $x_2=x_4=1/2$. this is period, not period 4. repeated itinerary structure $(IJ{)}^2$ forces collapse.