# Conjugacy in Dynamical Systems

## Abstract

This paper introduces the concept of topological conjugacy in discrete dynamical systems. It provides teachers with one way to motivate the idea of conjugate functions, contains proofs of basic results, some reasons for studying conjugacies, and brings together a variety of important examples.

Software: The program Mapping, which can be found in Lab2 of PEANUT software for the IBM, may be used to represent conjugacies geometrically in the real plane. To use this, you need to type in H(x) (as defined below) for F(x, y) and H(y) for G(x, y) in real mode. Then input your first function as a parametric curve and the program will sketch the conjugate function.

Prerequisites: Some experience with the dynamics of the two families of functions, Fm(x) = mx(1 - x) and Qc(x) = x2 + c. The final example assumes some knowledge of dynamics in the complex plane, in particular, the notion of a Julia Set as a boundary between basins of attraction.

References:

Devaney, Robert, An Introduction to Chaotic Dynamical Systems (2nd ed.), Addison Wesley, 1989
Devaney, Robert, A First Course in Chaotic Dynamical Systems, Addison Wesley, 1993
Gulick, Denny, Encounters with Chaos, McGraw-Hill, 1992
Peitgen, H-O., H. Jurgens, D. Saupe, Fractals for the Classroom: Part II, Springer-Verlag, 1992

## What are Conjugate Functions? An Example

After investigating the dynamics of the two families of functions, Fm(x) = mx(1 - x) and Qc(x) = x2 + c, one begins to notice that they have much in common. The bifurcation diagrams, in particular, seem to exhibit the same period doubling and chaotic behavior of the two families, although in different directions.

Fortunately, there is a way to describe the relationship between the two families precisely. As a bonus, the technique developed can be put to good use to reveal the unknown or unproved dynamical behavior of one function by showing that it is "conjugate" to a function whose dynamics is known. It is this notion of conjugacy that is described below.

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Notational Conventions.

ƒ: X --> Y is read "the function ƒ maps set X into set Y." X is the domain of the function and Y is a set large enough to contain the range of the function. When Y is the range, we say that "ƒ maps X onto Y." The notation suggests a geometric interpretation of the function, its domain and range. The function ƒ transforms a set of points X into another set of points that lies in some space Y.

For example, we would say that the function defined by ƒ(x) = x2 maps the interval [0, 3] onto the interval [0, 9]. We could also say that ƒ maps [0, 3] into the interval
[0, ), since this set is large enough to contain the range [0, 9].

___________________________________________________________________________

To describe the idea of a conjugacy between two functions, let's look at two familiar quadratic functions,

y = F4(x) = 4x(1 - x) and v = Q-2(u) = u2 - 2.

We are not interested in these functions on the entire real line, but only on restricted domains, and indicate this by writing

F4: [0, 1] --> [0,1] and Q-2: [-2, 2] --> [-2, 2].

The graphs of these two functions will fit snugly into two squares (see figure 1).

Figure 1. F4(x) = 4x(1 - x) and Q-2(u) = u2 - 2.

In this case, one can easily describe geometrically how to transform the first figure into the second. Begin with the graph on the left in Figure 1.

(1) Expand lengths in both horizontal and vertical directions by a scaling factor of 4:

Figure 2.

(2) Reflect this image in the origin:

Figure 3.

(3) Move this image up two units, then two units to the right:

Figure 4.

One sees at a glance that is possible to reverse these steps and recover the first figure uniquely from the last. To do this, invert all the transformations and apply them in the opposite direction: (1) Move the image two units to the left, then down two units. (2) Reflect this image in the origin. (3) Shrink lengths in both directions by a scaling factor of 1/4.

To describe the process algebraically, focus on a single point (x, y) on the graph of F4, so that y = 4x(1 - x). Then apply the transformations:

(1) (x, y) --> (4x, 4y)

(2) (4x, 4y) --> (-4x, -4y)

(3) (-4x, -4y) --> (-4x + 2, -4y + 2).

This last point, (-4x + 2, -4y + 2), should be a point (u, v) on the graph of Q-2, so that v = u2 - 2. To see this algebraically, set u = -4x + 2. We now verify that v = -4y + 2:

v = u2 - 2 = (-4x + 2)2 - 2 = 16x2 - 16x + 2

= -4[4x(1 - x)] + 2 = -4y + 2,

as expected. Similarly, one can show that any point (u, v) on the graph of Q-2 gets transformed back into its preimage (x, y) on the graph of F4 by applying the inverse transformation,

(x, y) = ( - u + , - v + ).

We say that F4: [0, 1] --> [0,1] and Q-2: [-2, 2] --> [-2, 2] are conjugate mappings and call the mapping H: [0,1] --> [-2, 2], with H(x) = -4x + 2, the conjugacy between F4 and Q-2. We want this conjugacy H to preserve the dynamics of F4 as it transforms F4 into Q-2. In particular, fixed points and periodic cycles of F4 should map to fixed points and periodic cycles of Q-2. In fact, for these two functions, more is true: the vertices of the parabolas also correspond. You may easily check the following correspondences between key points on the two graphs.

H

F4 --> Q-2

(0, 0) (2, 2) fixed points

(3/4, 3/4) (-1, -1) fixed points

(1/2, 1) (0, -2) vertices

(1, 0) (-2, 2) endpoints

( , ) ( , ) 2 cycles

Before you continue, it would be a good idea to carry out a similar analysis of the two functions

F2: [0, 1] --> [0, 1], F2(x) = 2x(1 - x) = y,

Q0: [-1, 1] --> [-1, 1], Q0(u) = u2 = v.

You should discover in this case that (u, v) = (-2x + 1, -2y + 1), that is, the conjugacy is the function

H: [0, 1] --> [-1, 1], H(x) = -2x + 1.

Note that in this problem, the functions are not "onto." The image of [0, 1] under F2 is
[0, 1/2], not [0, 1]. But we always want our functions to be defined in the form ƒ: A --> A so that the definition of conjugacy can be applied. That is why we have used [0, 1] for the space containing the range of F2.

How can we describe what the conjugacy does? Let's return to our original example. H changes the coordinate system in such a way that points on the graph of F4 are converted uniquely into points on the graph of Q-2. H-1 converts them back. The diagram in Figure 5, called a commutative diagram, describes this situation pictorially.

The diagram shows that H turns (x, y) into (u, v). If you begin with x [0, 1] in the upper left of the diagram and follow either route to the lower right, you must get the same result: H converts x in [0, 1] into u in [-2, 2]. Q-2 then converts u into v in [-2, 2], where v = u2 - 2. Similarly, F4 converts x in [0, 1] into y in [0, 1], where y =
4x(1 - x). H then converts y into v. In short,

H(F4(x)) = Q-2(H(x)); equivalently, H(y) = Q-2(u).

F4

x [0, 1] --> [0, 1] ' y

H H

u [-2, 2] --> [-2, 2] ' v

Q-2

Figure 5. A commutative diagram showing the conjugacy between F4 and Q-2.

## Formal Definition and Basic Properties of a Conjugacy

Definition. Let ƒ: A --> A and g: B --> B be two functions. Then ƒ and g are conjugate functions if there exists a one-to-one continuous function H from A onto B with a continuous inverse H-1: B --> A such that, for all x A,

H(ƒ(x)) = g(H(x)).

The commutative diagram has the following form:

ƒ

A-->A

H H

B -->B

g

Sometimes it is more convenient to write the conjugacy as

ƒ(x) = H-1 (g ( H(x) ) )

or as

g(u) = H (f ( H-1(u) ) ).

The first of these equations can be interpreted to say that, under the change of coordinates given by H, x is converted into a new domain coordinate, u = H(x), g is then applied to u to obtain a new range coordinate, v = g(u), and finally H-1 converts v back into the old range coordinate, which should, of course, be ƒ(x).

The following properties illustrate how the conjugacy H preserves the basic dynamics of ƒ in converting it into g.

1. (a) Reflexivity. ƒ: A --> A is conjugate to itself.

(b) Symmetry. If ƒ: A --> A is conjugate to g: B --> B, then g: B --> B is
conjugate to ƒ: A --> A.

(c) Transitivity. If ƒ: A --> A is conjugate to g: B --> B and g: B --> B is conjugate
to k: C --> C, then ƒ: A --> A is conjugate to k: C --> C.

Thus, conjugacy is an equivalence relation on the set of functions of the form ƒ: A --> A. The proofs of these properties are straightforward applications of the definition. Only the proof of transitivity is included here.

Proof. Suppose H1: A --> B is the conjugacy between ƒ and g, and H2: B --> C is the conjugacy between g and k. The commutative diagram shows the proof in nugo. H2 H1 must be the conjugacy between ƒ and k. Let's verify this.

ƒ

A-->A

H1 H1

B -->B

g

H2 H2

C -->C

k

k (H2 H1) (x) = k H2 (H1(x))

= H2 g (H1(x))

= H2 (g H1(x))

= H2 (H1 ƒ(x))

= H2 H1 (ƒ(x)),

which shows that ƒ and k are conjugate.

Suppose ƒ and g are conjugate functions, with conjugacy H, so that

H(ƒ(x)) = g(H(x)).

2. If a is a fixed point of ƒ, then H(a) is a fixed point of g.

Proof. a is a fixed point of ƒ

ƒ(a) = a

H(ƒ(a)) = H(a)

g(H(a)) = H(a)

H(a) is a fixed point of g.

3. If {a1, a2} is a two cycle of ƒ, then {H(a1), H(a2)} is a two cycle of g.

Proof. {a1, a2} is a two cycle of ƒ

ƒ(a1) = a2 and ƒ(a2) = a1

H(ƒ(a1)) = H(a2) and H(ƒ(a2)) = H(a1)

g(H(a1) = H(a2) and g(H(a2) = H(a1)

{H(a1), H(a2)} is a two cycle of g.

4. If a has prime period 2 for ƒ, then H(a) has prime period 2 for g.

Note. We say that a has prime period n for ƒ if ƒn(a) = a, but ƒk(a) a, for k = 1, 2, . . . , n - 1. (Here ƒn means the n-fold composite function.)

Proof. Suppose {a, b} is a two cycle for ƒ. By (3),

ƒ(ƒ(a)) = ƒ(b) = a and g(g(H(a)) = g(H(a)) = H(a).

We need to show that if ƒ(a) a, then g(H(a)) H(a). So suppose that g(H(a)) = H(a). Then H(ƒ(a)) = H(a), and since H is one-to-one, by definition, it follows that ƒ(a) = a. But ƒ(a) a. This contradiction establishes the theorem.

One can go on to prove the following results by similar means.

5. For all positive integers n, H(ƒn(x)) = gn(H(x)).

6. If {a1, a2, . . . , an} is an n cycle for ƒ, then {H(a1), H(a2), . . . , H(an)} is an
n cycle for g. Furthermore, if a has prime period n for ƒ, then H(a) has prime period
n for g.

## Examples of Conjugacies

This section discusses some important examples of conjugacies in discrete dynamical systems. Unless the conjugacies are linear (of the form H(x) = ax + b), they are usually difficult or even impossible to write down explicitly. Occasionally, inspired guessing or, in the case of complex variables, knowledge of the geometry of Möbius transformations will yield an explicit formula for H.

1. Fm(x) = mx(1 - x), Qc(x) = x2 + c.

Conjugacy: H(x) = -mx + m/2.

c = m/2 - m2/4, m = 1 + .

Intervals of interest: m [1, 4], c [-2, 1/4].

For this example and the next, we begin by assuming that the conjugacy is linear, set H(x) = ax + b, and substitute this into the equation, H(F4(x)) = Qc(H(x)). We quickly find that a = -m and b = m/2. But also b = (1 + )/ 2, so that m =
1 + .

This conjugacy explains why the bifurcation diagrams (when x is a real variable) and the parameter plane diagrams, i. e., Mandelbrot Sets (when x is a complex variable) look alike for the two functions.

2. Pr(x) = (1 + r)x - rx2, Qc(x) = x2 + c.

Pr is frequently used to model constrained population growth with growth rate r. As the growth rate increases past r = 2, bifurcations occur and the population begins to oscillate, first at regular intervals and subsequently at unpredictable intervals.

Conjugacy: H(x) = -rx + (1 + r).

c = , r = .

Intervals of interest: r [0, 3], c [-2, 1/4].

Because of the transitivity of the conjugacy relation, it follows that Fm and Pr are conjugate.

3. Tent map. T: [0, 1] --> [0, 1], T(x) = ,

Doubling map. D: [0, 1] --> [0, 1], D(x) = .

Figure 6. The Tent Map and The Doubling Map.

These two functions give the key to understanding much chaotic behavior. The tent map stretches and folds the unit interval back on itself; one can view it as a mixing or kneading operation that separates points that are initially close to each other. The doubling function is another example of a kneading operation; in this case, it first stretches the unit interval, then cuts it and pastes it back on itself, as a baker might knead dough. In fact, the doubling function is sometimes called the baker function for this reason. It is also called the sawtooth function, because of the shape of its graph.

If x is written in binary form, then D becomes a fractional shift transformation. For example, x = 1/3 can be written as .010101... in binary form. Each time you apply D, the point shifts one unit to the right and the integral part is deleted:

D(1/3) = 0.10101..., D2(1/3) = 1.010101... - 1 = .010101...,

and so on. Clearly, {1/3, 2/3} is a two cycle for D. Section 10.4 of Fractals for the Classroom, Part II, contains an excellent discussion of the "kneading" properties of these functions and their connection with chaotic behavior.

In any case, one can verify, using binary representations, that the doubling map exhibits chaotic behavior on the unit interval. Remarkably, the doubling map is conjugate to several other well known functions, so that the dynamics of these must illustrate, although in a less obvious manner, the same stretching and folding behavior. Here are a few such functions:

(a) D is conjugate to T, with conjugacy T.

To prove this, consider four cases, with x in each of the four intervals, [0, 1/4], [1/4, 1/2], [1/2, 3/4], and [3/4, 1]. The proof is straightforward, but tedious. The benefit is that one can infer the chaotic dynamics of T from that of D. (See the commutative diagram below.)

D

[0, 1]-->[0, 1]

T T

[0, 1] -->[0, 1]

T

(b) D is conjugate to F4(x) = 4x(1 - x). Since the graphs of these two functions are so much alike on [0,1], this comes as no surprise. What is surprising is that we must use

H: [0, 1] --> [0, 1], H(x) = sin2 (px/2)

as the conjugacy. From the conjugacy, the chaotic dynamics of both F4 and Q-2 follows (in the second case, by the transitivity of the conjugacy relation.) To verify the conjugacy, break the computation into two cases: x [0, 1/2] and x [1/2, 1].

(c) The function ƒ(x) = x2 + 1 has no real zeros. Newton's method applied to this function does not converge; it has no real fixed points. Setting N(x) = x - , we get

N(x) = .

Using, say, x0 = -1, yields a web diagram that looks "chaotic." Amazingly, the doubling function lies behind this behavior.

To see this, add the "point at infinity" to the set of real numbers R to take care of the singularity at x = 0. Let's write R for R » { }. Then N(0) = and N( ) = . The point at infinity is a fixed point of N in this extended domain. With this convention, we can prove that D is conjugate to N, with conjugacy

H: [0, 1] --> R , H(x) = cot (px).

Proof. H(D(x)) = .

But this reduces to H(D(x)) = cot (2px), since p is the period of the cotangent function.

N(H(x)) =

=

=

=

= cot (2px).

Thus, H(D(x)) = N(H(x)).

## Conjugacy in the Complex Plane

4. In this final example, the functions live in the extended complex plane, C = C » { }. Suppose c 0. Let

Nc(z) = and Q0(z) = z2.

Nc is the function one gets by applying Newton's method to the function ƒ(z) = z2 - c. In the extended complex plane, there are always three fixed points of Nc, and ± . Then Nc is conjugate to Q0, with the conjugacy given by

H(z) = .

The verification is straightforward:

H(Nc(z)) =

=

=

= Q0(H(z)).

There are, however, a number of interesting implications of this conjugacy, which it is worthwhile to study in some detail. In figure 7, we can see how the attracting and repelling fixed points of Nc and Q0 correspond.

H

Nc -->Q0

attractor

- 0 attractor

1 repeller

Figure 7.

In Figure 8, the correspondence between the two dynamical systems is shown in far more detail. (The figure actually shows the case c = 3 + 4i, so that the roots are ±(2 + i.) On the one hand, the Julia Set J(Nc) of Nc is the perpendicular bisector of the segment connecting the two fixed points, along with the point at infinity, which is a repelling fixed point. This Set is compact since the ends of the line meet at . (Indeed, on the Riemann sphere, the Set is just a circle.) The two half planes are basins of attraction for the two attracting fixed points, and the Julia Set is the boundary between these basins. Since one would expect and -to attract equally, it is not surprising that the set of points equidistant from each of them should be the Julia Set.

Figure 8. Julia Sets for Nc and Q0.

On the other hand, the Julia Set J(Q0) of Q0 is the unit circle. It contains the repelling fixed point 1 and is the boundary between the basins of attraction of 0 and . We will show that H is a one-to-one mapping of the Julia Set of Nc onto the Julia Set of Q0. Similarly, the half plane containing gets mapped onto the exterior of the unit circle, and the half plane containing -gets mapped onto the interior of the unit circle-again in a one-to-one manner. The conjugacy thus shows a remarkable and intimate relationship between the simple quadratic function and the whole family of Newton functions for the square roots of a complex constant c. The chaotic behavior we saw earlier on the real line in example 3(c) is now seen as one part of a much larger picture. In that example, we let c = -1. The Julia Set in C is thus the real axis, along with . The chaotic behavior on this axis is merely the reflection of the chaotic behavior of Q0 on its Julia Set, the unit circle. And this chaotic behavior is closely related to the fact that squaring z = cis q yields z2 = cis (2q)-the doubling map appears again!

Let's conclude this investigation with a proof that H is a one-to-one mapping of J(Nc) onto J(Q0).

H is clearly one-to-one, with inverse H-1(w) = . Note that H( ) = 1. Otherwise, since J(Nc) is perpendicular to the line connecting to -, any point on it must have the form z = ti, where t is a real number. Now

H(ti) = = = - i.

If we let

u = , v = - ,

then it is easy to verify that u2 + v2 = 1. Thus, H(z) is on J(Q0) if z is on J(Nc). In fact, the formulas for u and v (in terms of t) are a minor variation of a well known parameterization of the unit circle, one than can be used to generate all primitive Pythagorean triples.

If you accept that u + vi as defined above parameterizes the circle in a one-to-one manner as t takes on all values in R , then it is clear that H-1 maps any such point back to J(Nc). (Note that when t = , u = 1, v = 0.) But here is another verification. Let w be any point on J(Q0). Then w = cos q + i sin q. Thus,

H-1(w) = = i = -i.

Since this final expression is of the form ti , the point H-1(w) is certainly on J(Nc) for all q [0, 2p]. Consequently there is a one-to-one correspondence between the two Julia Sets.

Of course, since the conjugacy preserves the dynamics, n-cycles on J(Nc) get mapped to n-cycles on J(Q0), and conversely. An interesting exercise for the reader is to find an algebraic expression for all n-cycles of Q0 (an easy exercise) and then to apply H-1 to them to convert them into algebraic expressions for the n-cycles of Nc. Without knowing the conjugacy, this would be a formidable problem.

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