Assignment 12

Problem 1

Question

Show that if $x_0$ is in a forward limit set, say $\omega (y)$, then the entire orbit $\{f^n(x_0):n\ge 0\}$ is in $\omega (y)$.

want forward-invariance of $\omega (y)$. recall the definition:

$$\omega (y)=\{z:\exists n_k\to \infty \text{ with }{f}^{n_k}(y)\to z\}$$

assume $f$ continuous (standard).

$x_0\in \omega (y)⟹$ there existxs a subsequence $n_k\to \infty$ with $f^{n_k}(y)\to x_0$. fix $m\ge 0$ and apply $f^m$ to both sides. continuity of $f^m$ yeilds

$$f^{n_k+m}(y)=f^m(f^{n_k}(y))⟶f^m(x_0).$$

let $m_k=n_k+m$. then $m_k\to \infty$ and $f^{m_k}(y)\to f^m(x_0)$, so $f^m(x_0)\in \omega (y)$ by definition.

$m$ arbitrary so $\{f^m(x_0):m\ge 0\}\subset \omega (y)$.

basically just “shift the index by $m$” method: tail of a tail is still a tail, thus continuity passes the limit through $f^m$.

Answer

$f^{n_k}(y)\to x_0$ along $n_k\to \infty$. continuity gives $f^{n_k+m}(y)\to f^m(x_0)$, and $n_k+m\to \infty$, so $f^m(x_0)\in \omega (y)$ for every $m\ge 0$. $\omega$-limit sets are forward-invariant.

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Problem 2

Question

(a) Show that the logistic map $g_a(x)=ax(1-x)$, for $a>4$, has a chaotic set which is not a chaotic attractor.

(b) Show that the Horseshoe Map (described in Chapter 5) contains a chaotic set.

A

peak $g_a(1/2)=a/4>1$, so a middle interval gets ejected from $[0,1]$ on the first iterate. solving $g_a(x)=1$:

$$A_0=(p,q),p,q=\frac{1}{2}\left(1\mp \sqrt{1-4/a}\right)$$

once a point lands in $\mathbb{R}\setminus [0,1]$ it leaves for good. define

$$A_n=\{x\in [0,1]:g_a^n(x)\notin [0,1]\},\Lambda =[0,1]\setminus \bigcup _{n\ge 0}{A}_n.$$

$\Lambda$ is the set of orbits that stay bounded. each $A_n$ is a finite union of open intervals, $\bigcup A_n$ is open dense, so $\Lambda$ is closed nowhere-dense. removing a middle gap and recursing yieldxs the standard Cantor construction, so $\Lambda$ is a Cantor Set, $g_a$-invariant.

for $a>2+\sqrt{5}$, $|g_a^{\prime }|>1$ uniformly on $\Lambda$, so $g_a{|}_{\Lambda }$ is conjugate to the full one-sided 2-shift $\sigma :{\Sigma }_2\to {\Sigma }_2$. that gives:

• dense periodic orbits in $\Lambda$ (periodic sequences are dense in ${\Sigma }_2$)
• a dense orbit in $\Lambda$ (concatenate all finite words)
• sensitive dependence ($\lambda \ge \ln (\inf |g_a^{\prime }|)>0$)

$\Lambda$ is a chaotic set.

not an attractor: every $x\in [0,1]\setminus \Lambda$ is in some $A_n$ and escapes to $-\infty$, so the basin has empty interior. chaotic repeller, not attractor.

B

Smale horseshoe $H:S\to {\mathbb{R}}^2$ on the unit square: contract horizontally by $\lambda <1/2$, stretch vertically by $\mu >2$, fold so $H(S)\cap S$ is two disjoint horizontal strips $H_0,H_1$. backward image $H^{-1}(S)\cap S$ is two vertical strips $V_0,V_1$.

invariant set

$$\Lambda =\bigcap _{n\in \mathbb{Z}}{H}^n(S).$$

at each iterate forward we keep two strips out of one (cantor in vert direction); each iterate backward halves the horizontal direction. so $\Lambda \cong C\times C$, product of two Cantor sets.

itinerary map $\varphi :\Lambda \to {\Sigma }_2^{\mathbb{Z}}$,

$$\varphi (x{)}_n=i⟺H^n(x)\in H_i$$

(equivalently $\in V_i$ via the symmetric setup) is a homeomorphism conjugating $H{|}_{\Lambda }$ to the two-sided shift $\sigma$. semi known conjucgacy

shift $\sigma$ on ${\Sigma }_2$ has dense periodic orbits, a dense orbit, sensitive dependence. conjugacy carries all three accross to $\Lambda$. positive Lyapunov exponent comes for free from the vertical stretch: $\lambda =\ln \mu >0$.

so $\Lambda$ is a chaotic invariant set inside the horseshoe.

Answer

(a) $\Lambda ={\bigcap }_n{g}_a^{-n}([0,1])$ is a Cantor set, $g_a{|}_{\Lambda }$ conjugate to the 2-shift, hence chaotic. all $x\notin \Lambda$ escape to $-\infty$, so the basin has empty interior - not an attractor. (b) $\Lambda ={\bigcap }_{n\in \mathbb{Z}}{H}^n(S)\cong C\times C$, conjugated to the two-sided 2-shift via itineraries. shift dynamics is chaotic (dense periodics, dense orbit, sensitivity), conjugacy transfers it to $\Lambda$.

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Problem 3

Question

For the piecewise linear maps shown in (A) Figure 6.10, (B) 6.13, and (C) 6.14, find all periods for which there are periodic points.

use the transition graphs. for these Markov piecewise-linear maps, a closed path of length $n$ gives a point with $f^n(x)=x$. if the path is not a repeated shorter word, the point has exact period $n$.

A

graph has arrows

$$A\to B,B\to A,B\to B.$$

there is a loop at $B$, so period $1$ occurs. for any $n\ge 2$, take the closed path

$$A\to B\to B\to \cdots \to B\to A$$

with $n$ arrows. the corresponding word contains exactly one $A$, so it cant be a repetition of a shorter word.

so Figure 6.10 has periodic points of every period.

B

the graph splits into two invariant components:

$$\{A_1,A_2\},\{A_3,A_4\}.$$

inside each component both intervals map across both intervals; in particular there are self-loops and arrows both ways. the self-loops give fixed points. for $n\ge 2$, for example,

$$A_1\to A_2\to A_2\to \cdots \to A_2\to A_1$$

is a primitive closed path of length $n$. same story in the $A_3,A_4$ component.

so Figure 6.13 also has periodic points of every period.

C

the graph has arrows

$$A\to C,B\to C,C\to A,C\to B.$$

so away from the boundary fixed point, every orbit alternates between $C$ and $A\cup B$. therefore every closed path in the transition graph has even length.

all even periods occur. for $2m$, use the primitive closed path

$$C\to B\to C\to A\to C\to A\to \cdots \to C$$

of length $2m$ (for $m=1$, just $C\to B\to C$).

there is also a fixed point: the decreasing branch crosses the diagonal at $x=d$, and $f(d)=d$.

so Figure 6.14 has period $1$ and all even periods, but no odd periods greater than $1$.

Answer

(A) all periods $n\ge 1$.
(B) all periods $n\ge 1$.
(C) periods $1,2,4,6,8,\dots$, i.e. $1$ and every even period, with no odd periods larger than $1$.

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Problem 4

Question

Let

$$f(x,y)=\left(\frac{4}{\pi }\arctan x,\frac{y}{2}\right)$$

and find all forward limit sets, attractors, and basins for each attractor. What is the basin boundary, and where do these points go under iteration?

write the map as

$$f(x,y)=(h(x),y/2),h(x)=\frac{4}{\pi }\arctan x.$$

the coordinates decouple, and immediately

$$y_n=\frac{y_0}{2^n}\to 0.$$

so everything is decided by the 1D map $x_{n+1}=h(x_n)$.

fixed points solve

$$x=\frac{4}{\pi }\arctan x.$$

obvious solutions are $x=-1,0,1$. there are no others: for $q(x)=h(x)-x$,

$$q^{\prime }(x)=\frac{4}{\pi (1+x^2)}-1,$$

so on $x>0$, $q$ increases once, then decreases forever; since $q(0)=q(1)=0$, the only positive zero is $1$. oddness gives the negative zero $-1$.

sign picture uields:

$$\begin{array}{ccccc}x& (-\infty ,-1)& (-1,0)& (0,1)& (1,\infty )\\ h(x)-x& +& -& +& -\end{array}$$

so if $x_0>0$, the $x$-orbit is monotone toward $1$. if $x_0<0$, toward $-1$. if $x_0=0$, it stays at $0$.

therefore the forward limit sets are exactly

$$\omega (x_0,y_0)=\{\begin{array}{ll}\{(1,0)\},& x_0>0,\\ \{(0,0)\},& x_0=0,\\ \{(-1,0)\},& x_0<0.\end{array}$$

now check stability:

$$Df(x,y)=\left(\begin{array}{cc}\frac{4}{\pi (1+x^2)}& 0\\ 0& \frac{1}{2}\end{array}\right).$$

at $(\pm 1,0)$ the eigenvalues are $2/\pi$ and $1/2$, both less than $1$ in modulus. so $(1,0)$ and $(-1,0)$ are attracting fixed points (sinks), constitent with the sign picture above. their basins are

$$B(1,0)=\{(x,y):x>0\},B(-1,0)=\{(x,y):x<0\}.$$

at $(0,0)$ the eigenvalues are $4/\pi >1$ and $1/2<1$, so the origin is a saddle, not an attractor. its stable set is only the $y$-axis.

the basin boundary is the common boundary of the two half-plane basins:

$$\partial B(1,0)=\partial B(-1,0)=\{0\}\times \mathbb{R}.$$

points on this boundary stay on it forever:

$$f^n(0,y)=\left(0,\frac{y}{2^n}\right)\to (0,0).$$

Answer

forward limit sets: $\{(1,0)\}$ for $x_0>0$, $\{(-1,0)\}$ for $x_0<0$, and $\{(0,0)\}$ for $x_0=0$. the attractors are the two sinks $(1,0)$ and $(-1,0)$, with basins $x>0$ and $x<0$. their common basin boundary is the $y$-axis; boundary points satisfy $f^n(0,y)=(0,y/2^n)$ and converge to the saddle $(0,0)$.

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Problem 5

Question

Let $g_a(x)=ax(1-x)$. Use matlab to answer the following two questions. Turn in your code with your answers.

(a) Choose a subinterval of length $0.01$ in the parameter $a$ which appears to contain only chaotic attractors (for example, an interval just shy of $a=4$ in Figure 6.3(a)). Magnify the region until a periodic window is located and find the smallest period among the orbits in the window.

(b) Choose a point $p$ on the attractor of $g_a$ for $a=3.9$. Find this point by taking the $10^4$th point of a trajectory in the basin. Choose another point $x$ in the basin and let $m_n$ be the minimum distance between $p$ and the first $n$ iterates of $x$. Plot $m_n$ vs $n$. Does $m_n\to 0$ as $n\to \infty$? Can you quantify the rate at which it goes to zero? In other words, what is the function $f$ such that $f(n)\approx m_n$? Suggest a potential mechanism for this relationship.

Suggestions:

• Only plot the point $(n,m_n)$ if $m_n$ and $m_{n-1}$ are distinct from each other, i.e. if the minimum distance has changed
• Make your plot axes logarithmic and use dots rather than lines, e.g. loglog(n,m,'k.')
• Do this for many different points $x$, e.g. $10^3$ of them, to get a statistical sample of the typical behavior
• As you estimate $m_n$ for $n\to \infty$, do at least $N=10^4$ iterates

used python instead of matlab.

A

picked the length 0.01 interval

$$a\in [3.99,4.00].$$

at normal resolution it looks chaotic basically everywhere. zooming into

$$a\in [3.99024,3.99034]$$

reveals a small stable periodic window. at

$$a=3.99026704$$

orbit repeats with period $5$ after burn-in. the detected cycle is approximately

$$0.00968565\to 0.03827401\to 0.14687816\to 0.50000028\to 0.99756676\to \mathrm{0.00968565.}$$

so the smallest period in this window is $5$.

B

for $a=3.9$, taking the $10^4$th point of the trajectory starting at $x_0=0.2$ gave

$$p\approx \mathrm{0.758063440191.}$$

$1000$ random starting points in $(0,1)$, iterated each for $N=10^5$, tracking

$$m_n=\underset{0\le k\le n}{\min }|x_k-p|.$$

fitting the median accross the $1000$ orbits for $n\ge 1000$ gave

$$m_n\approx 0.594n^{-1.024}\approx \frac{C}{n}.$$

mechanism: after the transient, the chaotic orbit behaves roughly like samples from the invariant density on the attractor. if the density near $p$ is $\rho (p)$, then the chance that one iterate lands within distance $ϵ$ of $p$ is about $2\rho (p)ϵ$. after $n$ roughly independent tries, the nearest hit should satisfy

$$2\rho (p)nϵ\approx 1,$$

so $ϵ\approx C/n$. matchexs the standard recurrence / nearest-neighbor extreme value statistics for a mixing chaotic map.

Answer

(a) On the interval $[3.99,4.00]$, zooming near $a=3.99026704$ reveals a stable period-$5$ window. the smallest period in that window is $5$.
(b) For $a=3.9$, using $p\approx 0.758063440191$ and $1000$ random initial points, the closest-return distance decreases like $m_n\approx 0.594n^{-1.024}$, essentially $C/n$. happens because a chaotic orbit samples invariant measure, so nearest hit to a f.p. among $n$ samples is typically order $1/n$.

Code

Python

import numpy as np import matplotlib.pyplot as plt def g(a, x): return a*x*(1-x) def bifurcation(a_min, a_max, n_a=2600, burn=4000, keep=220): a = np.linspace(a_min, a_max, n_a) x = np.full(n_a, 0.2718281828) for _ in range(burn): x = g(a, x) aa, xx = [], [] for _ in range(keep): x = g(a, x) aa.append(a.copy()) xx.append(x.copy()) return np.concatenate(aa), np.concatenate(xx) def period_estimate(a, burn=100000, keep=2000, max_period=200, tol=1e-10): x = 0.123456789 for _ in range(burn): x = g(a, x) xs = np.empty(keep) for i in range(keep): x = g(a, x) xs[i] = x tail = xs[keep//2:] for p in range(1, max_period+1): if np.max(np.abs(tail[p:] - tail[:-p])) < tol: return p, np.sort(xs[-p:]) return None, xs # part a fig, axes = plt.subplots(1, 2, figsize=(12, 4), sharey=True) for ax, interval, burn, keep in [ (axes[0], (3.99, 4.00), 4000, 220), (axes[1], (3.99024, 3.99034), 8000, 260), ]: aa, xx = bifurcation(*interval, burn=burn, keep=keep) ax.plot(aa, xx, 'k,', markersize=0.4) ax.set_xlim(*interval) ax.set_ylim(0, 1) ax.set_xlabel('a') axes[0].set_ylabel('x') axes[0].set_title('a in [3.99, 4.00]') axes[1].set_title('zoom: period-5 window') axes[1].axvline(3.99026704, color='k', lw=1.0, linestyle='--') period, cycle = period_estimate(3.99026704) print(period, cycle) # part b p = 0.2 for _ in range(10000): p = g(3.9, p) rng = np.random.default_rng(12) M, N = 1000, 100000 x = rng.random(M) mins = np.full(M, np.inf) checkpoints = np.unique(np.logspace(1, np.log10(N), 260).astype(int)) checkpoint_set = set(checkpoints.tolist()) ns, med = [], [] for n in range(1, N+1): x = g(3.9, x) mins = np.minimum(mins, np.abs(x - p)) if n in checkpoint_set: ns.append(n) med.append(np.median(mins)) ns = np.array(ns) med = np.array(med) mask = ns >= 1000 slope, intercept = np.polyfit(np.log(ns[mask]), np.log(med[mask]), 1) C = np.exp(intercept) plt.figure(figsize=(6, 5)) plt.loglog(ns, med, 'k.', markersize=3, label='median $m_n$') plt.loglog(ns[mask], C*ns[mask]**slope, 'k--', lw=1.5, label=f'fit: {C:.3f} n^{{{slope:.3f}}}') plt.xlabel('n') plt.ylabel('$m_n$') print(p, C, slope)

Output