MAT335H1F Lec0101 Burbulla

Transcription

1 Fall 2011

2 Q 2 (x) = x 2 2 Q 2 has two repelling fixed points, p = 1 and p + = 2. Moreover, if I = [ p +, p + ] = [ 2, 2], it is easy to check that p I and Q 2 : I I. So for any seed x 0 I, the orbit of x 0 for Q 2 will stay in I, but it will not converge to p = 1, or p + = 2, unless x 0 just happens to be on an orbit that is eventually fixed at one of these two fixed points. As an example, take x 0 = 0.2 and consider n = 200 iterations of x 0 under Q 2. Two graphical representations of this orbit are represented on the next slide. A Chaotic Orbit Not only is the orbit repelled from p = 1, it also seems not to be attracted to any periodic cycle at all; it seems to be completely random, or chaotic.

3 Periodic Points of Q 2 Paradoxically, Q 2 has infinitely many periodic points, of all possible periods. We can see why by considering the graphs of Q 2 n, for a few values of n. Figure: Q 2 Figure: Q 2 2 Figure: Q 3 2 Figure: Q 4 2 In general, the function Q 2 has at least 2 n periodic points of period n in the interval I. Thus the proliferation of periodic points for Q c that we have already described, for 5/4 < c < 1, seems to increase without bound. But by c = 2 all the periodic points seem to be repelling. We shall investigate this more fully... later.

4 An Example from the Logistic Family: F 4 (x) = 4x(1 x) F 4 has two repelling fixed points, p = 0 and p = 3 4. Moreover, if J = [0, 1], it is easy to check that F 4 : J J. Again, for any seed x 0 J, the orbit of x 0 for F 4 will stay in J, but it will not converge to p = 0 or p = 3/4 unless x 0 is on an orbit that is eventually fixed at one of these two fixed points. As an example, take x 0 = 0.1 and consider n = 200 iterations of x 0 under F 4. Two graphical representations of this orbit are represented on the next slide. Another Chaotic Orbit Not only is the orbit repelled from both p = 0 and p = 2/3 it also seems not to be attracted to any periodic cycle at all; as with the previous example, it seems to be completely random, or chaotic.

5 Periodic Points of F 4 As with Q 2, F 4 also has infinitely many periodic points, of all possible periods. By considering the graphs of F4 n, for a few values of n you can see behaviour very similar to the quadratic family. Figure: F 4 Figure: F 2 4 Figure: F 3 4 Figure: F 4 4 Q 2 and F 4 are both quadratic functions, one opens upwards and one downwards. But both share one important feature: every subsequent iteration of Q 2 or F 4 doubles the number of valleys or hills in the graph of Q 2 n or F 4 n on the interval I or J, resp.

6 The Range of Q c if c < 2. Consider the graph of Q c on [ p +, p + ] for c < 2. The minimum value of Q c is at x = 0; it is y = c. If c < 2, then c < p + since c < p + = c 2 2c > c 1 2c > 1 4c (1 + 2c) 2 > 1 4c 4c 2 + 4c + 1 > 1 4c c 2 + 2c > 0 c < 2 or c > 0 This means that for c < 2 the image of the interval [ p +, p + ] under Q c is no longer contained in [ p +, p + ]. Let c = 2.5 For the Rest of This Section. Let c = 2.5; then p + = To the right is the graph of Q 2.5 on the interval [ [ p +, p + ] = , 1 + ] Notice that the parabola dips a little below the black square, which is determined by the four points (±p +, ±p + ).

7 Orbits That Escape to Infinity Q 2.5 (x) = p + x = x = ± These two points are eventually fixed points. If < x 0 <, 2 2 then the orbit of x 0 under Q 2.5 will escape to infinity, since x 1 < p +. Some orbits escape after 2 or 3 iterations: Figure: x 0 = 1.5, n = 2 Figure: x 0 = 1, n = 3

8 Here s an orbit that escapes after 17 iterations: take x 0 = 1.8. But not all orbits escape to infinity: the orbit of x 0 could be an eventually fixed point, ending up at p + or p. Let I = [ p +, p + ]; let Λ = {x I Q n 2.5(x) I for all n}. Λ is the set of all points in I for which the orbit under Q 2.5 never leaves I. What kind of set is Λ? One way to describe Λ is to describe all the points in I that are not in Λ; that is, describe the set of all points x 0 I such that the orbit of x 0 under Q 2.5 goes to infinity. If the orbit of x I under Q 2.5 goes to infinity then there is a least value of n for which Q 2.5 n (x) / I. Let A n = {x I n is the least value of k such that Q k 2.5(x) / I }. As we have already calculated, A 1 = ( 2.5 p +, ) 2.5 p +. A 1 is an open interval since Q 2.5 sends both endpoints to p +, which is an eventually fixed point. Figure: A 1

9 We can see the sets A 2 and A 3 by plotting Q and Q3 2.5 on I. Figure: A 2 Figure: A 3 A 2 is the union of two disjoint open intervals and A 3 is the union of four disjoint open intervals. Remarks About Λ and A n 1. A n is the union of 2 n 1 disjoint open intervals. 2. Q 2.5 (A n ) = A n 1 ; equivalently, A n = Q (A n 1). 3. n=1a n is the set of all points in I whose orbit under Q 2.5 escapes to infinity. 4. Λ = I n=1a n is the set of all point in I whose orbits remain in I. 5. Λ must be a closed set, since it is the complement of a union of open intervals. 6. Not so obviously: Λ itself contains no open interval. It is a collection of isolated points. That is, Λ is a closed set with total length zero! Such a set is called totally disconnected, or dust.

10 Λ Contains No Open Interval; Proof by Contradiction Suppose J = (a, b) is an open interval in Λ. J is on the right side of A 1 or on the left side; assume it is on the right side p p+ p + a b p + We have 2.5 p < a < b. For x > 0, Q 2.5 is an increasing function, so Q 2.5 (J) = (Q 2.5 (a), Q 2.5 (b)) = (a 2 2.5, b 2 2.5). Thus Q 2.5 (J) is another interval in Λ (why?) with length b 2 a 2 = (b + a)(b a) > 1.16 (b a). Repeat k times: for each k, Q 2.5 k (J) is another interval in Λ with length greater than 1.16 k (b a). For k large enough, 1.16 k (b a) will be bigger than the length of I, contradicting the fact that Λ I. An Algorithmic Construction of Λ (Not to Scale!) 1. Start with I = [ p +, p + ]: 2. Take out A 1 : 3. Take out A 2 : 4. Take out A 3 : Repeat until every open subset A n has been taken out. What s left is Λ. Just such a set was introduced in 1883 by Georg Cantor; it is called Cantor s middle-thirds set, or simply Cantor s set. NOTE: The arguments in this section are actually valid if c ( )/ , for which c p + 1/2, and for which Q c(x) 1, if x Λ A 1. This last condition is what Devaney uses in his proofs. The results of this section are true for all c < 2, but harder to prove for ( )/4 < c < 2.

11 Definition of the Cantor Middle-Thirds Set: A Recipe 1. Start with the interval [0, 1]. 2. Remove the middle third (1/3, 2/3), leaving two closed intervals left, [0, 1/3] and [2/3, 1], each of length 1/3. 3. From each of these two intervals remove the open middle thirds, (1/9, 2/9) and (7/9, 8/9), leaving the four closed intervals [0, 1/9], [2/9, 3/9] and [6/9, 7/9], [8/9, 9/9] left, each of length 1/9. 4. Repeat this process: remove the open middle third from each of the previous closed intervals. 5. K is the set of points remaining in [0, 1] in the limit as this process is repeated over and over without end. 6. K φ since it contains the endpoints of every closed interval remaining at each step. Picturing the Cantor Set Here are the first few iterations in the construction of the Cantor Set. Figure: The Cantor Middle-Thirds Set Note that the construction of K is very similar to that of Λ, as described in the last section. In fact, K and Λ have many common properties, which are easier to prove for K, but are also true for Λ.

12 Properties of the Cantor Set 1. K is a closed subset of [0, 1], because it is the complement of a union of open sets. 2. K is totally disconnected because, as with Λ, you can show that K contains no open intervals. 3. a K if and only if there is a ternary expansion of a such that a = 0.s 1 s 2 s 3 s 4... with s i = 0 or 2. That is, s i 1. This is base 3 arithmetic! 4. K is uncountable. That is, K cannot be put into a one-to-one correspondence with N, the set of natural numbers. Ternary Expansions Let a [0, 1]. In terms of base 3, a can be written as an infinite series s i a = 3 i = s s s , i=1 with s i {0, 1, 2}. We call 0.s 1 s 2 s 3 s 4... the ternary expansion of a. After all, there is nothing special about decimal expansions; they are base 10. Any base is just as valid as any other. Indeed, in computer science base 2 is very common. In terms of base 2, we could get a binary expansion of a, in which all of its digits would be 0 or 1.

13 Examples of Ternary Expansions Example 1: The ternary expansion of 1/4 is since = 2 9 n=0 n=0 ( ) 1 n = /9 = 1 4. Example 2: The ternary expansion of 5/26 is since = 1 ( ) 1 n + 2 ( ) 1 n = 1 9 n= / /27 = = Numbers With Two Ternary Expansions In the decimal system some numbers have two decimal expansions, eg = Some numbers can also have two ternary expansions. For example, since 1 3 = and 1 3 = , = = 2 9 Similarly, check that 8 9 n=0 = as well as ( ) 1 n = /3 = 1 3.

14 Ternary Expansions and The Cantor Set The ternary digits of a [0, 1] tell you at each step which third a is in: s 0 = 0 s 0 = 1 s 0 = 2 s 0 s 1 = Etc, etc. So a K if all its ternary digits are 2 or 0; a / K if one of its ternary digits is 1. For example 1/4 = K, but 5/26 = / K. But wait: 1/3 = 0.1 K and it has a ternary digit 1. However, as described above, 1/3 can be written as All in all, we can say: a K if and only if there is a ternary expansion of a such that all of its digits are 0 or 2. The Cantor Set is Uncountable Suppose K is countable. Then there is a bijection Φ : N K. Let k n = Φ(n). List all the elements of K as k 1, k 2, k 3,... : k 1 = 0.s 1 1 s1 2 s1 3 s1 4 s with s1 i {0, 2} k 2 = 0.s1 2s2 2 s2 3 s2 4 s with s2 i {0, 2} k 3 = 0.s1 3s3 2 s3 3 s3 4 s with s3 i {0, 2} k 4 = 0.s1 4s4 2 s4 3 s4 4 s with s4 i {0, 2} k 5 = 0.s1 5s5 2 s5 3 s5 4 s with s5 i {0, 2} { 0, if s i Define k = 0.s 1 s 2 s 3 s 4 s 5... such that s i = i = 2 2, if si i = 0. Then k K, because all its ternary digits are 0 or 2; but k is not on the above list because it differs with each ternary expansion in at least one digit. So K is uncountable.

15 What Is An Orbit Diagram? An orbit diagram is an attempt to describe in one picture the dynamics of a family of functions F λ as the parameter λ changes. Values of λ will be on the horizontal axis; values of x n will be on the vertical axis. But not all values of x n : we ignore the first 100 points (say) in the orbit and only plot the values of x n for n > 100. The hope is that if the orbit converges to a point or a periodic cycle, it should have reached it after 100 iterations. So on the orbit diagram we might plot the values of x n, for 100 < n < 300. The choice of x 0 is not random: for x 0 we pick a non-degenerate critical point of F λ, i.e. a point x 0 such that F λ (x 0) = 0 but F λ (x 0) 0. (The reason for this choice of x 0 won t become apparent until Chapter 12.) For the quadratic family, Q c (x) = x 2 + c, the non-denerate critical point is x 0 = 0; for the logistic family, F λ (x) = λx(1 x), the non-degenerate critical point is x 0 = 1/2. The Orbit Diagram for the Quadratic Family Q c So to produce an orbit diagram for the quadratic family we plot, for a given value of c, the values of x n, for 100 < n < 300, in a vertical line. Of course, we always have to pick a finite interval on the vertical axis, so any orbit that escapes to infinity would not show up on the orbit diagram. Based on our previous work we know that for c > 1/4, the orbit of x 0 = 0 goes to infinity, and for c < 2, the orbit of x 0 = 0 also goes to infinity. But for 2 c 1/4, the orbit of x 0 = 0 under Q c is completely contained in the interval [ p +, p + ] [ 2, 2]. Although the production of an orbit diagram is experimental math, our previous work indicates we should expect the orbit to be eventually fixed if 3/4 < c < 1/4, and eventually a 2-cycle if 5/4 < c < 3/4. The orbit diagram is on the next slide....

16 The Orbit Diagram for Q c (x) = x 2 + c, 2 c < 1/4 Remarks Concerning the Orbit Diagram of Q c 1. As c decreases past 3/4, more period doubling occurs: a 2-cycle is followed by a 4-cycle, which is followed by an 8-cycle, etc. 2. In the period-3 window, cycles of order 6, 12,..., appear as c decreases. 3. The orbit diagram seems self-similar: parts of the diagram resemble the whole diagram. 4. The lines in the orbit diagram that connect periodic cycles do so continuously. 5. For many values of c the orbit of x 0 = 0 under Q c is not attracted to any periodic cycle whatsoever.

17 The Orbit Diagram for the Logistic Family, F λ Recall (or check) the following features about the logistic family: 1. F λ (x) = λx(1 x) 2. F λ : [0, 1] [0, 1] for 1 λ p = 1 1/λ is an attracting fixed point if 1 < λ < F λ has an attracting 2-cycle if 3 < λ < λ + 1 ± λ 2 2λ 3 2λ 5. x 0 = 1/2 is a non-degenerate critical point of F λ. The orbit diagram for F λ is on the next slide.... The Orbit Diagram for F λ (x) = λx(1 x), 2.4 λ 4

18 Geometry Behind Some Features of the Orbit Diagram Consider the following six graphs of Q c, for 2 < c < 1/4. Each graph includes the line y = x and the square determined by the vertices (±p +, ±p + ). Figure: c = 1/4 Figure: c = 0 Figure: c = 3/4 Figure: c = 1 Figure: c = 2 Figure: c = 2.2 The first three diagrams illustrate the tangent bifurcation at c = 1/4, the appearance of an attracting fixed point, and the changing of the fixed point to a repelling fixed point at c = 3/4; the second three diagrams illustrate the cases of an attracting 2-cycle, chaotic orbits, and finally, orbits that can escape to infinity.

19 Now Consider the Graphs of Q 2 c for 3/4 < c < 1.65 The following three graphs show that in the marked box, lower left, the graphs of Q 2 c for 5/4 < c < 3/4 are similar to the graphs of Q c for 3/4 < c < 1/4. Figure: c = 3/4 Figure: c = 1 Figure: c = 5/4 Figure: c = 1.3 Figure: c = Figure: c = 1.65 The graphs of Q 2 c in the indicated boxes resemble very much the last three graphs of Q c, in the previous sequence of graphs. Thus in the orbit diagram of Q c we would expect two bits of the diagram for 1.65 < c < 0.75 to resemble very much the whole orbit diagram for 2 c 0.25, one bit on each part of the 2-cycle.

20 Period Doubling What the previous graphs illustrate is that just as the fixed point for Q c becomes repelling an attracting 2-cycle appears, namely the fixed points for Q 2 c. Similarly, just as the attracting 2-cycle for Q c becomes repelling, an attracting 4-cycle appears, namely the fixed points of Q 4 c. But the graphs of Q 2 c, Q 4 c, Q 8 c,... are getting steeper and steeper, so the bifurcations occur for values of c closer and closer together. This behaviour is not peculiar to the period-1 window; the same pattern of period doubling occurs in the other windows in the orbit diagram, though some of these windows are very thin compared to the period-1 window. Some Graphs of Q 3 c for 1.8 < c < 1.7 For example, the graphs of Q 3 c pass the line y = x in three different places for 1.8 < c < Thus we expect the orbit diagram in the 3-window to resemble three copies of the whole diagram. Figure: c = 1.7 Figure: c = 1.78 Figure: c = 1.8