Relevant section from Encounters with Chaos and Fractals, by D. Gulick: 2.3 "Conjugacy"

Transcription

1 Lecture 8 Topological conjugacy Relevant section from Encounters with Chaos and Fractals, by D. Gulick: 2.3 "Conjugacy" The main goal of this section: Suppose that we have a mapping f : I I and we wish to understand the dynamics of the iteration procedure, x n+ = f(x n ). () For example, is f chaotic on I? Sometimes it may be easier to answer these questions by transforming f into another map g on another interval J. To do this:. We look for a transformation φ : I J which is essentially a change of variable, y = φ(x), so that x I = y J. (2) 2. This change of variable produces a new map g : J J. It is possible that the dynamics of the iteration procedure, y n+ = g(y n ), (3) is better understood. That being said, we still need some conditions on the mapping φ : I J to ensure that the dynamics of g on J is the same as the dynamics of f on I. We start with a basic definition: Definition: Let I and J be intervals. The mapping φ : I J is called a homeomorphism if. φ is one-to-one 2. φ is onto 3. φ and φ are continuous. 5

2 For each x I, there is a unique y J such that y = φ(x). (4) Since φ is onto, then for any y J, there exists an x I such that x = φ (y). (5) Since φ is one-to-one, x is unique. Examples:. φ : [,] [,], where φ(x) = x n, n odd. 2. φ : [,π/2] [,], where φ(x) = sin(x) or cos(x). 3. φ : [,] [,], where φ(x) = sin 2( π 2 x ). Now suppose that we have two mappings on possibly different intervals, i.e., f : I I g : J J, (6) A note of clarification: I and J could be the same interval, but they do not have to be. In order to distinguish the two functions and intervals, we shall let x be the real variable defined on I and y be the real variable on J, i.e., f(x) I, x I g(y) J, y J. (7) Definition: The maps f : I I and g : J J are said to be topologically conjugate if there exists a homeomorphism φ : I J such that φ f = g φ. (8) Mathematical shorthand: f φ g. (9) Let us examine Eq. (8) a little more closely. In terms of variables, we can write φ(f(x)) = g(φ(x)) x I. () 52

3 RHS: The function φ maps x to a point y = φ(x) J. The function g then maps the point y to a point g(y) J. LHS: The function f maps x to a point z = f(x) I. The function φ then maps the point z in I to a point φ(z) J. For these reasons, we cannot write the conjugacy relation between f and g as follows, f φ = φ g, () since it would be meaningless. Many books the book by Gulick included employ Eq. (8) to establish important results involving functions that are topologically conjugate. But it may be more helpful to rewrite Eq. (8) in the following two ways, f = φ g φ (2) and g = φ f φ. (3) These formulas, in which an operator is surrounded by a function and its inverse in some way, might bring back some memories of linear algebra: Let A be an n n matrix, and for a given x R n, let y = Ax. (4) Now suppose that we make a change of basis so that in this new basis, the coordinates of x and y are given by x = Cx, y = Cy, (5) where matrix C is assumed to be invertible. We find the relation between x are y as follows: x = C x, y = C y. (6) Substitution into (4) gives C y = AC x, (7) 53

4 which, after left multiplication by C, y = CAC x = Bx, (8) where is the representation of the matrix A in the new basis. B = CAC (9) Let us now return to Eq. (2), which we now rewrite as a function of x I, f = φ g φ, (2) f(x) = φ (g(φ(x))) (2) schematically as follows, f x f(x) φ φ y g(y) g (22) The arrow x f(x) in the diagram corresponds to the LHS of Eq. (2). Instead of going this way, however, we can take a detour or alternative route, by applying φ to x, to yield y = φ(x), which is represented by the leftmost downward arrow in the diagram. Now apply the map g to y to obtain g(y), shown by the lower horizontal arrow. Then apply φ to g(y) to obtain f(x), shown by the rightmost upper arrow. This entire detour represents the RHS of Eq. (2). This is even more interesting if we consider the iteration of f. For example, from Eq. (2), the map 54

5 f 2 = f f is given by f 2 = f f = (φ g φ) (φ g φ) = φ g φ φ g φ = φ g g φ = φ g 2 φ (23) It should not be difficult to see that if we apply f n times, the cancellation of φ with its inverse φ will take place n times to yield f n = φ g n φ. (24) The diagram presented earlier should now be modified as follows, f n x f n (x) φ φ y g n (y) g n (25) In other words, you can get from x to f n (x) by either:. Applying f n times to x on the interval I. 2. Applying φ to x to produce y = φ(x) J, then applying g n times to y to produce g n (y), and finally applying φ to g n (y) to produce f n (y). This is a kind of indirect proof that if f and g are conjugate, then the dynamics associated with the iteration of these functions over their respective intervals are equivalent. For example, if x is a fixed point of f, then ȳ = φ( x) is a fixed point of g. We ll show this more mathematically below. But first, let s start with fixed points. Theorem : f φ g and x is a fixed point of f, then ȳ = φ( x) is a fixed point of g. 55

6 Proof: Since f φ g, we have φ f = g φ, (26) which means that φ(f(x)) = g(φ(x)) for all x I. (27) Letting x = x, we have, Since f( x) = x and φ( x) = ȳ, the above equation becomes φ(f( x)) = g(φ( x)). (28) φ( x) = g(φ( x)) = ȳ = g(ȳ), (29) which is the desired result. We may now extend this result to period-n points. Theorem 2: Suppose that f φ g. Then φ f n = g n φ. (3) We essentially proved this result earlier, by arguing that f n = φ g n φ. (3) But let s prove it now by induction. Eq. (3) is true for n =, since this is the definition of f φ g. Let us assume that it is true for an n = N > and show that it is true for n = N +. Then, φ f N+ = (φ f N ) f = (g N φ) f (assumed true for n = N) = g N (φ f) = g N g φ (also assumed true for n = ) = g N+ φ. (32) (33) By induction, the formula is true for all n. 56

7 We now use the above result to prove the following result. Theorem: Suppose that f φ g. Then if p is a period-n point of f, the q = φ(p) is a period n-point of g. Proof: Since φ f n = g n φ, (34) we have φ(f n (p)) = g n (φ(p)). (35) By assumption, f n (p) = p so that φ(p) = g n (φ(p)), (36) or q = g n (q). (37) The theorem is proved. Corollary: If f φ g and (p,p 2,...,p n ) is an n-cycle of f, then (q,q 2,,q n ) is an n-cycle of g, where q i = φ(p i ), i n. (38) These two results show that if f and g are topologically conjugate, then they demonstrate the same dynamical behaviour when it comes to periodic points. The next step is to show that if f and g are topologically conjugate, and f is chaotic on I, then g is chaotic on J. Here, we shall simply state the main results, and refer the reader to the appropriate section of Gulick for the proofs (Section 2.3, Theorems 2.3(iii) and 2.4). Theorem: Suppose that f φ g. If f has a dense set of periodic points (on I), then so does g (on J). Remark: All periodic points of f on I are mapped to periodic points of g on J by the mapping φ (and vice versa by φ. To prove that a dense set A I (periodic points of f) is mapped to a dense set B J, we use the fact that φ (and φ ) are continuous. (Technically, continuous maps map open 57

8 sets to open sets. This fact is used in the proof.) Theorem: Suppose that f φ g and that f is transitive on I. Then g is transitive on J. If f is transitive, then it has a dense orbit on I. Once again, the fact that φ (and φ ), assumed to be continuous, mapopensetstoopensets, ensuresthat adenseorbit oni ismappedtoadenseorbitonj. One more result regarding Sensitive Dependence to Initial Conditions is required in order to be able to use topological conjugacy to infer chaotic behaviour. Theorem: Suppose that f φ g and that f has SDIC on I. Then g has SDIC on J. One can easily show, using the continuity of φ and φ, that the SDIC property on I is mapped over to an SDIC property on J. In summary, we have the results necessary to establish the following result. Theorem: Suppose that f φ g and that f is chaotic on I. That is, f:. has a dense set of periodic orbits on I (regularity) 2. has SDIC on I, 3. is transitive on I (i.e., there is an orbit on I that is dense on I). Then g is chaotic on J, i.e., the above properties hold for g on I. 58

9 Using topological conjugacy to show that the logistic map f 4 (x) = 4x( x) is chaotic We now show that the logistic map f 4 (x) is topologically conjugate to the Tent Map T(x). As mentioned in the lectures, one of these may be viewed as a deformed version of the other. The Tent Map T(x) could be viewed as a straightened version of f 4 (x) with a sharp edge at the top. Both of these functions map the interval [, ] to itself, two-to-one and onto. y y x x Logistic map f 4 (x) = 4x( x). Tent Map T(x). We now state the following remarkable result: f 4 (x) and T(x) are topologically transitive with respect to the following mapping φ : [,] [,], To prove this, we need to show that φ(x) = sin 2 π x. (39) 2 φ T = f 4 φ. (4) (Note that we didn t bother to write the tent map T as a function of y, i.e., T(y), in the above figure. We used y earlier in order to emphasize that the two intervals I and J were different. We could have used the same variable, i.e., x, for both cases. Let us now compute both sides of Eq. (4): RHS: (f 4 φ)(x) = 4 (sin 2 π )( sin 2 x 2 π ) 2 x = 4 (sin 2 π ) (cos 2 x 2 π ) 2 x = (2sin π 2 x cos π ) 2 2 x = sin 2 πx. (4) LHS: 59

10 If x 2, (φ T)(x) = φ(2x) = sin 2 πx. (42) If 2 < x, (φ T)(x) = φ(2 2x) = sin 2 π 2 (2 2x) = (sin(π πx)) 2 = sin 2 πx, (43) where the final line comes from the property that sint = sin(π t). Since LHS = RHS for all x [,], we have shown that f 4 (x) and T(x) are topologically conjugate. Before going on, it is interesting to see how we can obtain the tent map T from f 4 using the following modification of Eq. (4): T = φ f 4 φ. (44) We have already computed the composition f 4 φ in (4). We now have to apply the mapping φ to this composition. Let s compute φ as follows: y = φ(x) = x = φ (y), (45) so that Therefore, y = sin 2 π 2 x = x = 2 π Sin y. (46) (φ f 4 φ)(x) = φ (f 4 φ)(x) = φ (sin 2 πx) = 2 π Sin sin 2 πx = 2 π Sin (sinπx). (47) At this point, one would be tempted to simply conclude that the Sin and sin functions cancel each other so that the final result is as follows, 2 (πx) = 2x, (48) π 6

11 This may seem to be half correct, since it yields the first half of the Tent Map. What about the second half, i.e., 2 2x? The problem is that (48) is valid only over the interval x 2. Explanation: The function sinπx in (47) increases in value from to as x increases from to 2. Then, as x increases from 2 to, the function sinπx decreases from to. Over the interval x [ 2,], the Sin function, acting on this linearly decreasing set of values from to, will produce a linearly decreasing set of values from π 2 to. The result is that Sin (sinπx) = π( x), < x. (49) 2 The other half of the result coming from Eq. (47) is therefore 2 π π( x) = 2( x), < x. (5) 2 Putting both results together, we obtain, (φ 2x, x 2 f 4 φ)(x) =, (5) 2 2x, 2 < x, which, of course, is the Tent Map. We have finally arrived at our goal! Since the Tent Map T(x) has been shown to be chaotic, and since T(x) is topologically conjugate to the logistic map f 4 (x), we can conclude that the logistic map f 4 (x) is chaotic on [,]. 6

12 Another interesting set of topologically conjugate functions This section was covered very quickly in the lecture. It is left for the reader to work through in detail. In Problem Set No. 2, you were asked to show that the one-parameter family of quadratic maps, g c (x) = x 2 +c, (52) could be viewed as upside-down versions of the quadratic logistic maps, f a (x) = ax( x). (53) The maps f a are defined on [,] and for a 4, map [,] to itself. If we restrict the domain of definition of the maps g c to the interval I c = [ x (c), x (c)], where x (c) >, is the positive fixed point of g c which is repulsive for c < 4, then the map g c for 2 c 4 maps I c to itself. In that same Problem Set, you showed that as c decreased from 4, the second fixed point x 2(c) < is attractive, then indifferent, giving rise to a two-cycle, which then gives rise to a four-cycle, etc.. In summary, the dynamics for the maps g c (x) behave in a parallel fashion to the dynamics of the logistic map f a (x). As may be suspected, the maps g c and f a are topologically transitive, which we shall show below. But we shall derive a more general result, namely, that the following set of three-parameter quadratic maps, f(x) = ax 2 +bx+c, (54) is topologically transitive to the following one-parameter family of quadratic maps, g(x) = x 2 +d. (55) (We let the parameter c be used in the definition of f(x).) This indicates that you really need only one parameter to vary in order to generate all possible dynamics of quadratic maps the use of three parameters in f(x) is redundant. We need only a linear mapping φ of the form, φ(x) = αx+β, (56) 62

13 to establish the topological transitivity, i.e., that f g φ, i.e., φ g = f φ. (57) We shall have to determine the α and β parameters of φ(x) in terms of the parameters a, b, c and d. First of all, let us compute the LHS of (57): φ(g(x)) = αg(x)+β = α(x 2 +d)+β = αx 2 +αd+β. (58) We now compute the RHS of (57): f(φ(x)) = a(αx+β) 2 +b(αx+β)+c = aα 2 x 2 +[2αβa+αb]x+aβ 2 +βb+c. (59) For (57) to be satisfied for all x, the polynomials in (58) and (59) must be equal for all x, which implies that their respective coefficients must be equal. Equating coefficients of x 2, we have α = aα 2 = α = a or a = α. (6) We now equate coefficients of x, noting that there is no term in x in Eq. (58): 2αβa+αb = α(2aβ +b) =. (6) The above equation is satisfied if α =, but this would imply that the mapping φ(x) = β, a constant, which is not a meaningful mapping from one quadratic to another. In addition, from an earlier result, we would have that a =, which is not valid. We therefore conclude that Finally, we equate coefficients of x : β = b 2a. (62) αd+β = aβ 2 +βb+c. (63) 63

14 After a little algebra, we find that [ b 2 d = a 4a b2 2a +c+ b ] 2a = b2 4 + b +ac. (64) 2 We may summarize these results as follows: The quadratic map, f(x) = ax 2 +bx+c (65) is topologically conjugate to the quadratic map, g(x) = x 2 + [ac ] b2 b2, (66) 4 with respect to the linear homeomorphism, φ(x) = a x b 2a. (67) Let s check these results with some known cases. Case : Logistic map a = 4. We know that the following logistic map, f(x) = 4x( x) = 4x 2 +4x, (68) is atwo-to-one mappingof [,] onto itself. InProblem Set No. 2, you showed that thequadratic map, g(x) = x 2 2, (69) is a two-to-one mapping of [ 2,2] onto itself. It looks like these two maps are topologically transitive. Let us check this conjecture. For the logistic map in (68), the parameters in (54) are a = 4, b = 4, c =. (7) The parameter d for the quadratic map in (66) is ac+ b 2 b2 4 = +2 4 = 2, (7) which is in agreement with the result from Problem Set No

15 Case 2: Logistic map a = 3. We know that the second fixed point, x 2 = 2 3, is attractive for < a < 3. At a = 3 it is indifferent. In Problem Set No. 2, you calculated that the second fixed point of the map g(x) in (55) became indifferent at d = 3 4. For the logistic map with a = 3, the parameters in (54) are a = 3, b = 3, c =. (72) The associated parameter d in the quadratic map in (66) is ac+ b 2 b2 4 = = 3 4, (73) which is in agreement with the result from Problem Set No

16 Lecture 9 Chaotic dynamics (cont d) Spatial distribution of chaotic orbits, visitation frequencies, invariant measures Suppose that we are told that a function f : I I demonstrates chaotic behaviour on I, that is, for almost all seed points x I, the orbit of x, given by the iteration procedure, x n+ = f(x n ), n, (74) exhibits seemingly random behaviour when the x n are plotted vs. n. Moreover, the orbit of x is dense on I: Given any point a I, and any neighbourhood N δ of a, i.e., the interval (a δ,a +δ), there is an element x n to be found in N δ. If this were all the information that could be obtained from the iterates {x n } defined in (74), then there wouldn t seem to be a way to tell whether a chaotic sequence was generated from a function f : I I or another function, say, g : I I. In other words, we wouldn t be able to tell the difference between a chaotic orbit generated by the iteration of the Tent Map and a chaotic orbit generated by iteration of the logistic map f 4 (x). In this section, we show very briefly that there are differences between the two sets of chaotic orbits, even though they are both dense on I. To see these differences, we look at how the iterates {x n } are distributed over the interval I. For example: Do the iterates {x n } tend to be spread out rather evenly over the interval I? Or are they somewhat concentrated at some parts of I and less concentrated at other parts. This question has been at the heart of an enormous amount of research in dynamical systems theory over the past half-century and more. Here we provide a small glimpse into the subject. There is a relatively simple (apart from some possible problems due to finite precision) numerical procedure to visualize how the iterates x n defined in (74) are distributed over the interval I. In what follows, we let I = [a,b]. (Typically, I = [,], but we ll keep the discussion general.) Step No. : For an N relatively large (say -), form a partition of the interval [a,b] in the manner done in first-year Calculus, i.e., let t = b a N, (75) 66

17 and define t k = a+k t, k N, (76) so that t = a and t N = b. Step No. 2: This partition will define a set of N subintervals of I, I k = [t k,t k ], k N. (77) For reasons that will become clear below, define the following set of half-open intervals, J k = [t k,t k ), k N, (78) along with the final interval, J N = [t N,t N ]. (79) In the special case that I[,], for which a = and b =, t = N, (8) and t k = k t = k N, (8) with p = and p =. Step No. 3: Initialize a counting vector, call it c, with N elements, so that The entries of c will be integers. c k =, k N. (82) Step No. 4: Now choose a good seed point x I, i.e., a point that is not preperiodic (or at least hopefully not preperiodic). Start computing the elements {x n } of the forward orbit of x using (74), i.e., x n+ = f(x n ), n. (83) This will involve some kind of loop in your computer program. After you have computed each iterate x n, determine the particular interval J k, in which x n lies. This can done in the following way (or some slight modification of it): k = int [ ] t (x n a) +, (84) 67

18 where, for a y R, int[y] = integer part of y. (85) Rationale: If x n lies in J k, then t k x n < t k = a+(k ) t x n < a+k t = (k ) t x n a < k t = (k ) t (x n a) < k (86) This implies that which yields (85). [ ] k = int t (x n a), (87) After determining k, the index of the interval I k in which x n lies, increase the appropriate entry of c by one, i.e., c k = c k +. (88) Step No. 5: Perform the iteration procedure in (83) for a sufficiently large number M of times, say M = 5,, or, better yet, M =,, or even M = 6. The larger the better: These computations do not take a lot of time. At the end of the computation, you will have produced an N-vector, c. Hopefully, some, if not all, of its entries c k will be nonzero. Question: What is this vector c? Answer: Each element c k of this vector for k N, has recorded the number of times that the interval I k = [t k,t k ) has been visited by an iterate x n over the orbit n M. If you now plot the values of the elements c k vs. k, you will get an idea of how the iterates x n are distributedover theintervali. IntervalsI k withhighernumbersofvisitation willhavehighercountsc k. In order to be able to compare the results of this counting experiment for different choices of M, the total number of iterates computed, it is convenient to normalize the count vector ic by defining 68

19 the following N-vector p, the elements of which will not be integers, but fractions: p k = c k M k N. (89) Technically, we should write p k (M), since our values of p k will depend on M, but we leave the M out for the moment. Note that p k for k N, and N p k =. (9) k= Each element p k, k N, may be interpreted in at least two ways, which are not unrelated:. p k is the fraction of the iterates {x n } M n= that have visited interval J k. If we view the set of iterates {x n } M n= as a huge collection of numbers between and, the p k indicate how they are distributed in the N bins J k, k N. This is, of course, a discrete approximation to how they are distributed over [, ]. 2. p k is the visitation frequency of interval J k or at least an approximation of it by the iterates {x n } M n=. This has a probabilistic interpretation: For each n >, we may view p k as the probability that iterate x n will be found in interval J k. The term visitation frequency sounds statistical and suggests that the results of our computations are approximations to a true visitation frequency that is obtained by letting the number of iterations M. Before going on, let us examine the results of a few computations for two chaotic maps on [,] that we have studied to date: (i) the Tent Map and (ii) the logistic map f 4 (x), the graphs of which are once again plotted below: y y x x Tent Map T(x). Logistic map f 4 (x) = 4x( x). 69

20 In the plots on the next page are shown plots of the elements, p k, k M, of the vector p obtained when the interval [,] is divided into N = subintervals and M = 2 6 iterates are used. The most noticeable difference between the two plots is that one (the tent map) is quite flat compared to the other (logistic-4). Approximations to visitation frequencies for two chaotic maps on [,] x Tent Map T(x) x Logistic map f 4 (x) = 4x( x) In both cases, M = bins and N = 2 6 iterates were used. With regard to the tent map case, note that the value of each p k is roughly., i.e., p k. = N. (9) 7

21 This is consistent with Eq. (9) and indicates that the iterates are visiting the entire interval [,] in a quite uniform manner. We might not be so surprised to see that the distribution associated with the tent tap is flat. After all, the tent map is piecewise linear, i.e., the pieces are straight lines, as oppposed to the logistic function f 4 (x), which is curved. This indeed has something to do with the flatness of the tent map case, and we ll prove that the distribution is flat, i.e., uniform. The question then remains, Why does the distribution for the logistic-4 map curl upwards near the ends of the interval, implying that these outer regions are visited more frequently than the inner region around x = 2? An explanation is now to be provided. In order to understand the shapes of these visitation frequency plots, we shall have to make use of a kind of conservation law, that acknowledges the 2-to- nature of the tent and logistic-4 maps. (The idea extends in a straightforward manner to other many-to-one maps, e.g., 3-to- maps.) Consider an interval [a,b] [,]. Instead of being concerned where points from [a,b] are going under the action of a map f (tent or logistic-4), we are going to be concerned about what points from [,] are coming to [a,b] under the action of f. It is therefore convenient to place the interval [,] on the y-axis for each of the two plots of the graphs of T(x) and f 4 (x), as shown below. y y b b a a a b b 2 a 2 x a b b 2 a 2 x Tent Map T(x). Logistic map f 4 (x) = 4x( x). In each figure are shown the two preimages of the interval [a,b] under the action of the map concerned: Tent map: T([a,b ]) = [a,b] and T([b 2,a 2 ]) = [a,b]. 7

22 Logistic-4 map: f 4 ([a,b ]) = [a,b] and f 4 ([b 2,a 2 ]) = [a,b]. Note that for both maps, the interval [b 2,a 2 ] is flipped, due to the decreasing nature of the map for 2 < x. We now develop our conservation law, which is essentially a law of probabilities. We ll be working in the discrete framework introduced earlier, that is, dividing the interval [, ] into N subintervals J k. Associated with each interval J k is a probability p k that the iterate will visit it during an orbit of length M. We ll also assume that N is sufficiently (enormously?) large so that the discrete approximations that we are employing are sufficiently accurate. We ll also assume that the length M of the orbit is large/enormous, essentially approaching the limit M. Later, we shall actually let N, the number of subintervals, go to infinity in order to arrive at a continuous description of the visitation frequency which will then employ the differential dx instead of the bin width x = t. In what follows, we shall, for the most part, adopt the first of the two interpretations of the quantities p k, k N, defined in Eq. (89), that is, that each p k is the fraction of iterates {x n } M n= found in interval J k. We use this interpretation to adopt the next set of assumptions:. Let K [,] be an interval, and suppose that K is the union of a number of subintervals J k, i.e., This is equivalent to the statement, K = k 2 k=k J k. (92) K = [t n,t n2 ], (93) where the t k are the partition points defined earlier. Then the fraction of iterates {x n } M n= that have visited interval K, to be denoted as F(K) is given by F(K) = n 2 k=n p k. (94) Special case: When K = [,], then n = and n 2 = N so that the sum in (94) is, as it should be: The fraction of all iterates that lie in [,] is, since all iterates x n [,]. Note: Eq. (94) is often written in the more convenient form, F(K) = k p k, (95) 72

23 where the prime on the summation indicates that the summation is over only those k {,2, N} for which J k K. Or, to remove any confusion, we could write, F(K) = p k. (96) {k J k K} 2. This one is very important! With reference to the graphs of the Tent Map and the Logistic-4 Map on the previous page, the fraction of iterates that lie in the interval [a,b] is equal to the sum of the fractions of iterates lying in [a,b ] and [b 2,a 2 ]. Mathematically, F([a,b]) = F([a,b ])+F([b 2,a 2 ]). (97) This is a kind of conservation of iterates (which is essentially a conservation of mass). The iterates in [a,b] come from the two intervals [a,b ] and [b 2,a 2 ]. If we have arrived at a kind of stationary distribution that tells us how the iterates are distributed over the intervals, then the above conservation law has to hold. Later, we shall state this law mathematically. Note: We have relied on the assumption that t, the length of the subintervals J k, is sufficiently small so that the intervals involved above, i.e., [a,b], [a,b ] and [b 2,a ], can be expressed or at least well approximated as unions of the subintervals J k, i.e., no subintervals J k have been split. In the limit N, these approximations will become exact and Eq. (97) is valid without the use of the subintervals J k. We are now in a position to argue not prove, but at least understand why the distribution for the Tent Map is uniform, i.e., all of the probabilites p k are constant and equal. It is, indeed, N because of the fact that the two components of T(x) are straight as well as having slopes of equal magnitude, namely, 2. By simple geometry, the lengths of the intervals [a,b ] and [b 2,a 2 ] are equal and one-half the length of the interval [a,b]. It should be fairly easy to see that if the conservation equation in (97) holds for any interval [a,b] [,], then the p k should all be equal, i.e., p k = N, k N. (98) This is often referred to as a uniform distribution. And what about the distribution associated with the Logistic-4 map? Why does it increase as we approach and? Very loosely speaking, because the magnitude of the tangents to the graph of f 4 (x), i.e., f (x), increases as x + and x and decreases as x 2 73

24 Looking at the graph of f 4 (x) with the interval [a,b] on the y-axis and its preimages, [a,b ] and [b 2,a ], note that the lengths of these intervals are shorter than one-half the length of [a,b]. This is due to the fact that the magnitudes of the tangents to the graph of f 4 (x), i.e., f (x) increase as x + and x, approaching the value of 4 in the limits. As [a,b] is moved downward toward the origin, the lengths of these two preimages gets even smaller. Loosely speaking (or writing), in order for the conservation equation in (97) to hold, the fractions of the iterates on these two intervals has to be greater than the uniform distribution in (98). Admittedly, these are very loose or heuristic descriptions of why we expect the shape of the distributions of the p k fractions to be what they are. Let us now go to a more mathematical description. In order to do so, let us first consider the limit M, where M is the number of iterates. As M increases, the number of iterates falling in the interval J k, which we have called c k, will generally increase with M. In fact, we should denote this number of iterates as c k (M). And we should once again acknowledge that the fractions p k are also functions of M and write We now claim that the following limits exist, p k (M) = c k(m) M. (99) lim p c k (M) k(m) = lim M M M = p k, k N. () Then p k is the limiting fraction of iterates that are found in subinterval J k as we let the number M of iterates go to infinity. Once again, N p k =. () k= We now consider the p k to define a piecewise constant function P(x) on [,]: P N (x) = p k if x J k, k N. (2) The subscript N reminds us that N subintervals J k are used in the construction of this function. Now we do something that will seem quite strange: We define a new function ρ N (x) as follows, ρ N (x) = t P N(x) = p k t if x J k, k N. (3) 74

25 This must seem very strange, indeed, since t is very small, and since we eventually wish to take the limit N, which implies that t. But as N, the number of intervals, increases, and t decreases, each p k decreases there are more intervals in which to find the iterates! The reason we define ρ N (x) in Eq. (3) is that the fraction F(J k ) of iterates found in subinterval J k is now given by F(J k ) = p k = ρ N (x k ) x, x k J k, (4) where x = t (just to keep everything in terms of x). Note that x k can be any point in J k, since F(x) is constant over each subinterval J k. Note: Is this starting to look like something from first-year Calculus, i.e., Riemann integation? Another note: We may view ρ N (x) in Eq. (3) as a density function, i.e., the (normalized) number of iterates per unit length. We write normalized since the total number of normalized iterates over the entire interval [a, b] is, i.e., N p k =. (5) k= In this way, one could think of the iterates as representing electric charges, in which case ρ N (x), x J k, is the lineal charge density (charge per unit length) over the interval J k. We now perform the limiting operation N. In this limit, x, the length of the subintervals J k, will go to zero. The summation over these subintervals of length x will become an integration over the differential dx. We claim that in the limit N, the piecewise constant functions ρ N (x) converge to a function ρ(x), for x [,]. For any subinterval [a,b] [,], the limiting fraction of iterates in [a,b] is no longer a summation over all subintervals J k lying in [a,b] as done in Eq. (94) but rather an integration over the interval [a, b], i.e., F([a,b]) = b a ρ(x) dx. (6) We have arrived at a continuous description of the fractional distribution of iterates over the interval [, ]. Note that the function ρ(x) is a normalized distribution since F([,]) = ρ(x)dx =. (7) Eq. (6) leads to the following continuous version of the conservation equation in (97): For any 75

26 [a,b] K, b b a2 a ρ(x)dx = ρ(x)dx+ ρ(x)dx. (8) a b 2 We are now going to state this conservation result more generally as well as mathematically. In what follows, we let I denote an interval on which a function f : I I is defined. f may or may not be chaotic. For any subset S I, we define the following set, f (S) = {x I,f(x) S}. (9) In other words, f (S) is the set of all points in I that are mapped by f to the set S. In the case of each of the Tent and Logistic-4 maps, S is the interval [a,b] [,] and f ([a,b]) = [a,b ] [b 2,a 2 ], () where the a i and b i depend on the maps. The above relation is true because f([a,b ]) = f([b 2,a 2 ]) = [a,b]. () Definition: Let I be an interval and f : I I. If there exists a function ρ : I R such that for all S I, ρ(x)dx = S f (S) ρ(x) dx, (2) then ρ is said to the be invariant (probability or normalized) density function which defines the invariant measure associated with the mapping f : I I. Note: The notation implies an integration over the entire set f (S) definedearlier. Eq. (8) stated earlier, f (S) b b a2 a ρ(x)dx = ρ(x)dx+ ρ(x)dx, (3) a b 2 is a special case of Eq. (2). Note: The reason for the term invariant measure, is that the density function ρ is considered to define a measure of subsets S I a generalized notion of length, a kind of weighted length. Regions of I that have higher ρ-values, i.e., fractions of iterates, are weighted more heavily that those regions with lower ρ-values. Usually, the invariant measure associated with a dynamical system f : I I is denoted as µ. The invariant measure, or µ-measure of an interval [a,b] is given by µ([a,b]) = b a 76 φ(x) dx, (4)

27 which, as we saw earlier, is the fraction of iterates in the interval [a,b]. The conservation relation in (2) may be expressed as follows, µ(s) = µ(f (S)) for all S I. (5) The invariant density functions for the Tent and Logistic-4 maps Tent Map T(x) Here we simply state that, as expected, the invariant density function ρ(x) for the Tent Map on [,] is a constant function no regions have a higher fraction of iterates than others. In the case that I = [,], ρ(x) =, x [,], (6) Referring to the earlier figure which shows that graph of the Tent Map function T(x) along with the interval [a, b] and its two preimages, the conservation equation in (8) becomes b a dx = b a dx+ Let us finally state explicitly what the a i and b i are: and The integrals in (7) are, of course, simple to evaluate: which is satisfied for all [a,b] [,]. a2 b 2 dx. (7) a = 2 a, b = b, (8) 2 b 2 = 2 b, a 2 = a. (9) 2 b a = (b a )+(a 2 b 2 ) = 2 (b a)+ 2 (b a) = b a. (2) The invariant measure µ defined by the density function ρ is µ([a,b]) = = b a b a φ(x) dx dx = b a. (2) 77

28 In this case, the µ-measure of the interval [a,b] is the length of the interval, the usual notion of the size of an interval. This is somewhat of a coincidence since the interval I on which the Tent Map T(x) is defined is [,]. If we were to consider a tent map on I = [,2], then the constant density function ρ would be so that I ρ(x)dx = ρ(x) = 2, (22) 2 dx =. (23) 2 When the density function ρ(x) is constant, the measure defined by ρ is often called uniform measure. Logistic-4 map f 4 (x) = 4x( x) Referring to the earlier figure which shows the graph of the logistic-4 function f 4 (x) along with the interval [a, b] and its two preimages, let us rewrite the conservation equation in (8) that would have to be solved by the invariant density function ρ(x) associated with the f 4 (x) map. b a ρ(x)dx = Here we also state explicitly what the a i and b i are: b a2 ρ(x)dx+ ρ(x)dx. (24) a b 2 a = 2 2 a, b = 2 2 b, (25) and b 2 = 2 + b, a2 = a, (26) 2 We now state a remarkable result the density function ρ(x) satisfying (24) with the a i and b i defined above, is known analytically: ρ(x) = π x( x). (27) The graph of ρ(x), presented in the figure below, demonstrates a strong similarity to the distribution of iterates of the logistic-4 map obtained numerically and presented earlier. Even though ρ(x) as x + and x, it is integrable: ρ(x)dx = π x( x) dx =. (28) 78

29 Plot of density function ρ(x) = π x( x), x [,], for logistic-4 function f a (x) = 4x( x). The fact that the function ρ(x) satisfies the conservation equation in (24) can be verified after a generous amount of Calculus, starting with the result (left as an exercise) that for a b, b a b ρ(x)dx = dx π a x( x) = [ Sin (2b ) Sin (2a ) ]. (29) π In the special case a = and b =, the above result yields, ρ(x)dx = π = π [ Sin () Sin ( ) ] [( π ( 2) π )] 2 =. (3) This has been a very short introduction to the subject of dynamical systems and invariant measures very little could be done which, of course, means that much has been omitted. But it was intended to be a starting point for anyone who is interested in pursuing the subject further. One final note: The existence of the density function ρ(x) in Eq. (2) is not always guaranteed. But an invariant measure µ generally exists. The complication is that the measure µ is a measure, and measures can be quite irregular. They can include things such as Dirac delta functions, i.e., point masses, which cannot be modelled with normal functions. This will certainly be the case when we study measures on fractal sets at least lightly. 79

30 Lecture 2 Chaotic dynamics (cont d) The Ergodic Theorem It is worthwhile to mention another very important idea arising from dynamical systems theory, the so-called Ergodic Theorem. The Ergodic Theorem has been of special importance in mathematical physics with respect to the idea that Time average = Space average. (3) Here we simply state the basic idea. Let I = [a,b] and f : I I such that associated with f is an invariant density function ρ(x) which satisfies the condition in Eq. (29) from the previous lecture (Lecture 8, Week 6), which we reproduce here: For all S I, ρ(x)dx = S f (S) ρ(x) dx. (32) Given a suitable seed point x I (i.e., an x that is not preperiodic), define In other words, compute the forward orbit O(x ) of x. x n+ = f(x n ), n. (33) Now supposethat g : I R is a continuous function and that we evaluate g at the iterates x n, forming the following average: For N >, S N = N N g(x n ). (34) n= Then (subject to some other technical considerations which we ll ignore here), the limit of the above average exists, i.e., Furthermore, the limit S is related to the density function ρ as follows, Important points: S = lim S N = S. (35) N b a g(x)ρ(x) dx. (36) 8

31 . The quantities S N in (34) are time averages they are the average values of g(x) evaluated at the iterates x n which range over I. We consider the orbit O(x ) = {x n } N n= to define a trajectory over discrete time units, n =,,2,, with g(x) being evaluated at the points in I visited by the iterates. 2. The limit S in (35) is the limiting time average. 3. The integral in (36) is a spatial average. It is a weighted average of the function g(x) the weighting is performed by the invariant density function ρ(x). A region K I that is visited more frequently by the iterates will have higher values of ρ(x). This is also reflected in the time average if the region K I has a higher ρ(x) values, more iterates will be visiting it, which will influence the average S N in (35). 4. Eq. (36) is the mathematical equivalent of the statement in (3). Example : Consider the tent map T(x) on [,] with invariant density function ρ(x) =. In what follows, we consider a few g(x) and compute their time averages in Eq. (34) for two values of N, namely N = 6, N 2 = 8. These results are compared with the corresponding spatial averages in (36). Here, since ρ(x) =, S = g(x)ρ(x)dx = g(x) dx. (37). g(x) = x. We find that S N = S N2 = (38) The time averages are seen to be moving toward the limiting value S = 2. g(x) = x 2. We find that xdx = 2. (39) S N = S N2 = (4) The time averages are seen to be moving toward the limiting value S = x 2 dx = 3. (4) 8

32 3. g(x) = e x. We find that S N = S N2 = (42) The time averages are seen to be moving toward the limiting value (to ten digits) S = e x dx = e = (43) Example 2: Now consider the logistic-4 map f 4 (x) = 4x( x) with invariant density function, ρ(x) = π x( x). (44) Integrals involving the above density function are generally difficult to compute analytically, which provides a good motivation to approximate them using time averages. Here we shall consider only one example: A function that was quite important in an earlier lecture. For the moment, consider a [,4] in general and define the function, g a (x) = ln f a (x) = ln a 2ax. (45) Recall (Lectures 3 and 4, Week 5) that the limiting time average of this function, i.e., S = lim N N n= ln f a (x), (46) is the Lyapunov exponent λ(x ) of f a (x). As was discussed earlier, the Lyapunov exponent is computed as a time average of values of the function ln f a(x) over the iterates x n. What we shall see here, is that the Lyapunov exponent can also be computed as a spatial average over the invariant density function ρ(x). For g a (x) in (45) with a = 4, i.e., g 4 (x) = ln 4 8x, (47) we obtain the following numerical results, once again for N = 6 and N 2 = 8, S N = S N2 = (48) Note that both of these results are positive, in accord with the fact that the map f 4 (x) is chaotic. As such, we expect a positive Lyapunov exponent. 82

33 As N, these time averages should converge to the spatial average of g 4 (x), namely, S = π With a little Calculus, this integral can be computed analytically, S = π = ln2 ln 4 8x x( x) dx. (49) ln 4 8x x( x) dx (5) We see that the estimate S N2 agrees with the theoretical value to part in 8. As a final note, it is interesting that the Lyapunov exponent for the logistic-4 map, λ = ln2, is equal to that of the Tent Map. And one final note: Recall that the Lyapunov exponent for the logistic function f a (x) was defined forall a [,] (thevalueλ = wasalsoacceptable). Formanyotheravalues, especially < a 3, the map f a is not chaotic: Orbits are attracted to fixed points or N-cycles. Clearly, we computed the Lyapunov exponents for these cases as time averages of the function g a (x) over the orbits. The question remains, are these time averages equal to spatial averages over some kind of invariant density functions or measures? The answer is Yes. The problem is that the density functions are no longer functions they are generalized functions, which include the idea of Dirac delta functions. We may discuss this later in the course. 83

34 A return to the dynamics of the Logistic Map: The case a > 4 We now examine the interesting iteration dynamics associated with the logistic map when a > 4. The graph of f a (x) = ax( x) for a = 4.5 is sketched below: y y = x J x The most important qualitative feature of this graph is that f a no longer maps [,] into itself: There is an open interval of points, J, that are mapped outside the interval [,]. For x J, f(x) >. Note what happens to points x J under further iteration of f, as shown in the figure. We note from the graph that: ) f 2 (x) <, 2) f n+ (x) for n 2 and finally 3) f n (x) as n. The natural question that follows is Are there any other points in [,] that eventually leave [,]? The answer is Yes. Note that every point x [,] has two preimages y y 2 such that f(y ) = f(y 2 ) = x. This leads us to consider the set J 2 of all preimages of the set J, i.e. J 2 = {x [,] f(x) J } (5) Since x J 2 implies that f(x) J, this implies in turn that f 2 (x) / [,] and that f n (x) as n. Now, what is the set J 2, the set of preimages of J? We may determinetheset J 2 graphically by copying theset J ontothey-axis andthen applying f backwards, i.e. using the line y = x to determine which points on the x-axis are mapped to J. This is shown below: 84

35 y y = x J J 2 J J 22 x Thus, f(j 2 ) = J and f(j 22 ) = J. The set J 2 = J 2 J 22 is the set of all points x [,] such that f 2 (x) / [,], i.e. the set of points that leave [,] after two applications of f. The reader should see the pattern now. We define the set J 3 of all preimages of J 2, i.e. J 3 = {x [,] f(x) J 2 }. (52) Note that x J 3 implies that f 2 (x) J which in turn implies that f 3 / [,]. Now continue this process, if possible, to define the following sets of points, J n = {x [,] f n (x) / [,]}, n. (53) It is now convenient to define the following sets: C = [,] J (54) C 2 = C J 2 = [,] (J J 2 ) (55). ( n ) C n = C n J n = [,] J k. (56) From the definition of the J k, we see that for n, C n is the subset of points in [,] that remain in [,] after n iterations of f. The natural question is, Is there a set of points J that remain on [,] after any number of iterations of f? In order to answer this, let us examine the sets C,C 2, etc. k= 85

36 From the figures shown earlier, and the definitions in (56), the sets C and C 2 have the following structure: [,] C C 2 Note that C is obtained by means of a dissection procedure a removal of the open set J from [,]. C 2 is obtained from a dissection of C a removal of an open set from each of the subintervals making up C. Note that [,], C and C 2 are closed intervals. The construction of intervals C and C 2 by means of a dissection procedure is reminiscent of the middle-thirds dissection procedure that was used to construct the ternary Cantor set in Section 3.7. At this time, we simply state, without proof, the following result: The sequence of sets C,C 2,... converges, in the limit n, to a Cantor-like set C [,], i.e. lim n C n = C. For any x C, f n (x) J for all n. (The term Cantor-like will be defined shortly.) In other words, the set of points C that remain in [,] after any number of iterations of f a is a Cantor-like set. Note that the structure of this set, i.e. the positions of points x C, x / {, }, is dependent upon the logistic map parameter a. For example, the size/length of the open interval J removed form [,] to produce J is dependent upon a: As a 4 +, this interval is smaller. Of course, at a = 4, no dissection takes place. The reader is encouraged to find endpoints of the intervals that make up the set J, as functions of the parameter a, hence the length of the removed interval. This size will also determine the sizes of the sets J 2 and J 22 removed in the nextdissection procedure, although not in a linear way since themap f a (x) is not a linear function. The reader is invited to consider an alternate definition of the set C constructed above: C = {x [,] x is a periodic point of f a with period n, n }. Graphical analysis should show (as indeed our previous analysis of the bifurcations of f a did) that all periodic orbits of f a are repulsive, as they were for the case a = 4. 86

37 Cantor (or Cantor-like ) sets We now study Cantor-like sets in a little more detail, first by studying the famous ternary Cantor set. We don t really have to do this by looking at the iteration of functions, since Cantor-like sets may be produced by a limiting procedure of dissection, i.e., removal of points from intervals/sets. But it is instructive to consider iteration since, after all, it is a central idea of this course. We shall consider a family of linearized versions of the logistic maps f a (x), namely, the following family of modified Tent maps, ax x 2 T a (x) = (57) a( x) 2 < x. (With apologies: In a previous ( assignment, ) T a (x) was used to denote a shifted Tent Map.) The maximum value of T a (x) is T a = a 2 2. When a = 2, T 2(x) is the Tent Map that we have studied in past lectures. In the case a = 3, the graph of the Tent Map extends out of the box, i.e., [,] [,], as shown below. We see that T 3 maps some points in [,] out of [,]. The fate of these points, as in the case of the logistic map f a for a > 4 is that under further iteration, they go to. y y = T 3 (x) y = x J x We ll proceedaswedidforthelogisticmapf a inthepreviouslecturebymakingafewobservations: 87

38 . Points in the (open) interval J = ( 3, 2 ) 3 (58) are mapped out of [,] by T 3. In other words, these points leave [,] after one application of T 3. ( 2. Points in [,] that are mapped by T 2 to the interval 3, 2 ) are mapped out of [,] after one 3 ( additional application of T 3. These points lie in the preimage of the interval 3, 2 ) which, as 3 in the case of the logistic map, can be determined graphically by backwards iteration to be the set J 2 = In other words, points in J 2 leave [,] after two applications of T 3. ( 9, 2 ) ( 7 9 9, 8 ). (59) 9 This procedure can be continued, but it is perhaps easier to focus on the points in [,] which remain in [,] after a given number of iterations. As we did in the previous lecture for the logistic maps, we determine the sets of points in [,] which remain in [,] after n applications of the map T 3 :. The set of points in [,] which remain in [,] after one application of T 3 comprise the set C = [, ] [ ] 2 3 3, = I I 2. (6) 2. The set of points in [,] which remain in [,] after two applications of T 3 comprise the set C 2 = Graphically, these intervals look as follows, [, ] [ 2 9 9, ] [ 2 3 3, 7 ] [ ] 8 9 9, = I 2 I 22 I 23 I 24. (6) C = [,] C C 2 These intervals look almost identical to the intervals presented earlier for the logistic map, but there is one important difference: All of the line segments that comprise a given set C n have equal length. 88

39 The above results are easily extended to the general case: The set of points in [,] which remain in [,] after n applications of T 3 is the following set of 2 n closed intervals of length 3 n : In other words, C n = 2 n k= I nk. (62) C n = {x [,], T3 n (x) [,]}. (63) From the diagram above, we see that C 2 C C. (64) In general, C n+ C n. (65) In other words, if a point x remains in [,] after n+ iterations, which implies that x = C n+, then it must remain in [,] after only n iterations, i.e., x = C n. But the converse does not necessarily apply: There will be points in C n that leave after the next application of T 3. As such, we have a nested set of closed intervals, C C C 2. (66) We now define the set C to be the infinite intersection of this nested set, i.e., C = C n. (67) n= C is the set of points in [,] that belong to all C n, n, i.e., Eq. (67) is stating that C = {x [,], x C n for all n }. (68) C = lim n C n. (69) The set C described above is commonly known as the ternary Cantor set or simply the Cantor set. Clearly, the set C contains the points,, 3 and 2 3. It contains some multiples of 9 but not all of them, i.e., it contains the points k 9 for k {,,2,3,6,7,8,9} but not for k{4,5}. Later, we ll see that C also contains some points that you may not have guessed, e.g., 4. 89