Chaotic Dynamics and Fractals

Home , Orbit (dynamics), Periodic point

Eric Kuennen February 10, 2005

Contents

1 Discrete Dynamics: Iteration and Types of Orbits. 2

2 Graphical Analysis, and Attracting and Repelling Fixed Points 6

3 The Quadratic Family and Bifurcations. 9

4 Transition to Chaos. 9

5 Symbolic Dynamics. 9

6 The Deﬁnition of Chaos. 9

7 Period 3 Theorem and Sarkovskii’s Theorem. 9

8 Fractals: Cantor set, Sierpinski Triangle, Koch Snowﬂake, fractal dimen- sion. 9

9 Create your own fractal: Iterated Function Systems. 9

10 Complex Dynamics, the Julia Set. 9

11 The Mandelbrot Set. 9

12 The Hen´on Map. 9

13 Strange Attractors. The Lorenz Attractor. 9

14 Smale’s Horseshoe and Homoclinic Tangles. 9

15 Fractal Growth Phenomena: Percolation, Diﬀusion-Limited Aggregation, Ballistic Deposition, Eden Clusters. 9

16 Fractal Growth Phenomena: Cellular Automata. 9

1 1 DISCRETE DYNAMICS: ITERATION AND TYPES OF ORBITS. 2

Primary Reference: A First Course in Chaotic Dynamical Systems , Robert Devaney, Addison-Wesley 1992

Secondary References: Fractals, Scaling, and Growth Far From Equilibrium. Peal Meakin, Cambridge University Press, 1998.

Nonlinear Oscillations, Dynamic Systems, and Bifurcations of Vector Fields , J. Guck- enheimer and P. Holmes. Springer-Verlag New York Inc., 1983.

1 Discrete Dynamics: Iteration and Types of Orbits.

We will study how functions behave under iteration. More speciﬁcally, we’ll look at what happens to a point z0 under repeated applications of some function f(z).

Example. Let Pn represent the population of rabbits after n months. Suppose the population grows by 5% each month. Then

Pn+1 = Pn +0.05Pn,

or Pn+1 =1.05Pn. This is called a ﬁrst-order difference equation. The solution to this diﬀerence equation

will be an explicit formula for P which depends only on n and P0, the initial population of

rabbits. To ﬁnd the solution, we can iterate using the diﬀerence equation P +1 = 1.05P , starting with the initial population P0.

P1 = 1.05P0, 2 P2 = 1.05P1 = (1.05) P0,

P = 1.05P ¡1 = (1.05) P0. Since what is going on is simply repeated multiplication by 1.05, let’s deﬁne f(x) =

1.05x. Then we can write the diﬀerence equation as

P = f(P ¡1),

and the solution to the diﬀerence equation as

P = f (P0). 1 DISCRETE DYNAMICS: ITERATION AND TYPES OF ORBITS. 3

To iterate a function means to repeatedly compose the function with itself. Let f(z) represent some function of z. Then the second iterate of f is

f 2(z) = f(f(z)),

and the third iterate of f is

f 3(z) = f(f 2(z)) = f(f(f(z))).

In general, the n-th iterate of f is

f (z) = f(f ¡1(z)) = f(f(f(f(....f(z))))),

meaning f composed with itself n times.

Example 1. Let f(z) = 3z2 1. Then − f 2(z) = 3(3z2 1)2 1 = 27z4 18z2 + 2, − − − and

f 3(z) = 3(3(3z2 1)2 1)2 1 = 2187z8 2916z6 + 1296z4 216z2 + 11. − − − − − A formula for f 4(z) is left to the reader if desired. Note that f 2(z) does not mean [f(z)]2.

The orbit of z0 under f is the following sequence of points:

2 z0, z1 = f(z0), z2 = f (z0), ..., z = f (z0), ... { }

The starting point z0 is referred to as the initial condition or seed.

Types of orbits. A fixed point for f is a point z0 such that f(z0) = z0. Note that 2 if z0 is a ﬁxed point, then f (z0) = f(f(z0)) = f(z0) = z0, so a ﬁxed point will produce an orbit of the form z0, z0, z0,...,z0, ... { }

Fixed points for a function f(z) satisfy the equation f(z) = z, so they can be found

by solving the equation f(z) = z. ¢ Examples. x = 1 is the only fixed point for f(x) = 3x + 2, and 1 3 are − 2 ± 2 the only two fixed points for f(x) = x2 + 1. As an exercise, find all the fixed points for f(x) = x2 + 1 and f(x) = x3 3x.

−

A periodic point for f with period n is a point z0 such that f (z0) = z0, for some n > 0, and i The smallest such n is called the prime period of the orbit, and orbit of z0 is 1 DISCRETE DYNAMICS: ITERATION AND TYPES OF ORBITS. 4

called a periodic orbit, or an n-cycle. If z0 is a periodic point with period n, then the orbit

will look like the following:

¡ ¡1 1 z0, f(z0), ...,f (z0), z0, f(z0), ..., f (z0), z0, ...

A point z0 can also be eventually ﬁxed or eventually periodic if the orbit becomes ﬁxed or periodic after some number of initial iterations.

2 Example. Consider f(z) = z 1. Then the orbit of z0 = 0 is the 2-cycle − 0, -1, 0, -1, 0, -1, 0, -1, 0, ...

Example. Consider f(z) = iz. Then the orbit of z0 = 1 is the 4-cycle

1, i, -1, -i, 1, i, -1, -i, 1, ... ¢ 1 3 Example. Consider f(z)=( + i)z. Then the orbit of z0 = 2i is the 3-cycle − 2 2 2i, -√3 i, √3 i, 2i, -√3 i, √3 i, 2i, ... − − − − 2 Example. Consider f(z) = z . Then the orbit of z0 = i is eventually ﬁxed:

i, 1, 1, 1, 1, 1, ... −

Typically, however, an orbit is not fixed or periodic, but will converge to, or be attracted by an fixed point or periodic orbit. It may be attracted to infinity, in which case the orbit is said to be unbounded. Sometimes, an orbit will never settle down to any pattern and behave quite chaotically. We will make more precise definitions and see examples of these behaviors later.

Example: Finding Square Roots. To ﬁnd the decimal representation of √2, we can

make an initial guess for √2, and call this initial guess x0. If x0 < √2, then √2x0 < 2, 2

√ £ so 2 < 0 , so we have 2 x0 < √2 < . x0

If instead x0 > √2, we will end up with 2 x0 > √2 > . x0

√ £ So either way, 2 is between x0 and 0 . So for our next guess, why not try half-way between 2 x0 and £0 ? That is, let 1 2 x1 = (x0 + ) 2 x0 1 DISCRETE DYNAMICS: ITERATION AND TYPES OF ORBITS. 5

1 2 . As an exercise, iterate f(x) = 2 (x + £) starting with several diﬀerent initial seeds x0 to observe the convergence of the orbit to a ﬁxed point at √2. Which functions iterates will converge to √α?

Exercise. Use a pocket calculator to ﬁnd x50 if you begin with the initial condition £ x0 = 10 under the functions f(x) = √x, f(x) = e , f(x) = sin x and f(x) = cos x.

In each case, what seems to happen to x as n ? → ∞ Let’s close this section with examples of two special functions.

Example. The Doubling Function is deﬁned on [0,1) as D(x) = 2x mod 1. That is,

2x for 0 x < 1/2 D(x) = ≤ 2x 1 for 1/2 x < 1 − ≤ Sketch a graph of the function D(x) on the interval [0,1). Discuss the behavior of D(x) for the following seeds: x0 = 1/3, x0 = 1/5, x0 = 1/7,

Example. The Tent Function is deﬁned on [0,1] as

2x for 0 x 1/2 T (x) = ≤ ≤ 2 2x for 1/2 x 1 − ≤ ≤ Find a formula for T 2(x). Sketch graphs of the functions T (x) and T 2(x) on the interval [0,1]. Find all ﬁxed points for T (x) and T 2(x). 2 GRAPHICAL ANALYSIS, AND ATTRACTING AND REPELLING FIXED POINTS6

2 Graphical Analysis, and Attracting and Repelling Fixed Points

Graphical analysis is a tool to help visualize orbits for functions of a single real variable f(x). First we note that the graphs of y = f(x) and y = x will intersect at the real ﬁxed points for f(x). So we begin our graphical analysis by plotting y = f(x) and the diagonal y = x on the same axes. To sketch an orbit, we pick an initial condition x0, then ﬁnd y0 = f(x0) by moving vertically to the graph of y = f(x). Then, to iterate we wish to let x1 = y0, that is, let the new x-value be the previous y-value. This is switching can be accomplished on the graph by moving horizontally to the diagonal. Then we begin again by moving vertically to the graph of y = f(x), then horizontally to the diagonal y = x, etc.

Examples. On the plots below, use graphical analysis to analyze the orbits of f(x) = x3 and f(x) = x2 1.1. −

Looking at f(x) = x3, from graphical analysis, we can see that f has three fixed points, at x = 0, 1, and 1. Orbits that begin inside the interval (-1,1) tend to the fixed point at − x = 0, while orbits outside of the interval [-1,1] tend to positive or negative infinity. We say that the fixed point at o is attracting, while the fixed points at 1 and -1 are repelling.

Meanwhile, we can see that f(x) = x2 = 1.1 has two fixed points, at x .66 and ≈ − x 1.66. In both cases, orbits that begin the fixed point do not tend towards the fixed ≈ point, and so both are said to be repelling. However, in the case of the fixed point near x = .66, nearby orbits to not seem to be attracted to infinity. − An exellent web-based applet for making these ”cobweb” graphs is found at http://www.emporia.edu/math-cs/yanikjoe/Chaos/CobwebPlot.htm 2 GRAPHICAL ANALYSIS, AND ATTRACTING AND REPELLING FIXED POINTS7

A ﬁxed point z0 is said to be an attracting ﬁxed point for f if there is a neighborhood

D of z0 such that if z D, then f (z) D for all n > 0, and in fact

∈ ∈

f (z) z0 as n . → → ∞

A fixed point z0 is said to be an repelling fixed point for f if there is a deleted neighborhood D of z0 such that if z D, then f (z) D for some n > 0. This means that ∈ 6∈ an orbit with an initial condition starting even very close to z0 will eventually need to move away from z0. Note that the orbit doesn’t have to go to infinity or anywhere in particular, it just has to move away from z0.

Theorem 1. Suppose f(z) has a ﬁxed point at z0. Then z0 is:

1. attracting if f ¤(z0) < 1, | |

2. repelling if f ¤(z0) > 1, and | |

3. neutral if f ¤(z0) = 1. | |

This theorem is a generalization of what we already discovered for linear functions in the ¤ last section, for if f(z) = az, then f ¤(z) = a, and f (z) = a = ρ. | | | | Example. Let f(z) = z2. To ﬁnd the ﬁxed points of f, we solve f(z) = z, or

z2 = z

z2 z = 0 − z(z 1) = 0 −

So the only ﬁxed points are z = 0 and z = 1. Next we compute f ¤(z) = 2z, and

f ¤(z0) = 2z0 . So at the ﬁxed point z0 = 0, | | | |

f ¤(0) = 2 0 = 0 = 0 < 1, | | | · | | | and by the theorem, the origin is an attracting ﬁxed point for f(z) = z2. At the ﬁxed point z0 = 1,

f ¤(0) = 2 1 = 2 = 2 > 1, | | | · | | |

and by the theorem, 1 is a repelling ﬁxed point for f(z) = z2.

If instead z0 is periodic of period n, then since f (z0) = z0, we can view z0 is a ﬁxed point for f (z). So we can apply the theorem above to f to determine the attraction or repulsion of periodic orbits.

Corollary 1. Suppose f(z) has a periodic point z0 of period n. Then the orbit of z0 is:

2 GRAPHICAL ANALYSIS, AND ATTRACTING AND REPELLING FIXED POINTS8

1. attracting if (f ) ¤(z0) < 1,

| |

2. repelling if (f ) ¤(z0) > 1, and

| |

3. neutral if (f ) ¤(z0) = 1. | |

Example. Consider the following function:

f(z) = z2 + i.

2 2 2 2 Let z0 = -1 + i. Since z1 = (z0) + i = (-1 + i) + i = i, and z2 = (z1) + i = i + i = -1 + i = z0, z0 lies on a period 2 cycle:

-1 + i, i, -1 + i, i, -1 + i, ...

To see if this cycle is attracting, repelling, or neutral, we consider that z0 is a ﬁxed point for

f 2(z) = f(z2 + i)=(z2 + i)2 + i = z4 + 2iz2 1 + i. −

2 3 2 3 ¤ We compute (f ) ¤(z) = 4z + 4iz, so (f ) (-1 + i) = 4(-1 + i) + 4i(-1 + i) = 4(2 + 2i) 4i 4=4+4i. So − − 2 (f ) ¤(-1 + i) = 4 + 4i = 4√2 > 1, | | | | which means the 2-cycle is repelling.

Exercise. Find all ﬁxed points for the following functions and determine whether they are attracting, repelling, or neutral.

1. f(z) = z2 + 1

2. f(z) = ¥

3. f(z) = z3 +(i + 1)z

¨ §

2 ¦3 2 Exercise. Show that z0 = e lies on a 2-cycle for f(z) = z . Is this cycle attracting, repelling, or neutral?