Chaos and Dynamical Systems

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1 Chaos and Dynamical Systems y Megan Richards Astract: In this paper, we will discuss the notion of chaos. We will start y introducing certain mathematical concepts needed in the understanding of chaos, such as iterates of functions and stale and unstale fixed points. We will discuss the graphical analysis of an orit and show a couple examples of an orit diagram. We will riefly show how the Lyapunov exponent can e found and used to determine whether a system is chaotic and if so where it is chaotic. Then we will egin discussing the aker map, which this paper focuses on the most. In particular, we will discuss the values at which the aker map exhiits chaotic ehavior and why it is chaotic at those values. This has een focused on ecause, while it can e easy for someone to pick up a ook and read aout chaos, it may not e so easy to research the aker map and find the values at which it is chaotic and why. Finally, we will look at an example of a known chaotic system the doule-pendulum. We will use a computer program to model the ehavior of the pendulum and oserve how it is sensitive to initial conditions. In this paper, we will assume that the reader has a mathematical ackground up to the calculus level and can understand certain proof techniques, such as a proof y induction or y contradiction. Introduction Chaos is a word we all know usually meaning a lack of order or predictaility. Most people might associate the utterfly effect with the notion of chaos. This utterfly effect descries how a utterfly flapping its wings in some part of the world might e largely responsile for a huge storm in another part of the world several weeks later, ecause of the weather s extreme sensitivity to initial conditions. The weather is indeed one example of a chaotic system. A weather reporter might make nearly identical forecasts for a city in Europe as for a city in the United States, meaning that the initial conditions of the system are quite similar, while y the next week the weather patterns are completely different. Because of this sensitive dependence on initial conditions, weather forecasters have a very difficult time predicting the weather far in advance. Another example of chaos is the doule pendulum, and later we will use a computer program to model the motion of a doule pendulum. We will place the doule pendulum at a certain initial point and again 1

2 at another initial point just slightly different than the first to oserve how the resulting motion will e very different. What makes certain systems chaotic? Chaos descries the ehavior of a system that is highly sensitive to initial conditions. Chaotic systems are not predictale over a long period of time and are typically associated with fractal structures. Understanding chaos will help us understand why some systems exhiit seemingly erratic and random ehavior yet are still deterministic systems (that is, systems determined y their initial conditions). It will show us why chaos is not complete disorder, ut rather is associated with a geometrical structure. Understanding chaos is somewhat complicated, so we first need to e ale to understand certain mathematical concepts and the instaility and staility of non-chaotic systems. Hence, this paper will egin y explaining iterates of functions, unstale and stale fixed points, and some examples of maps and orit diagrams. We will then move on to discuss the aker map and at which points the aker map exhiits chaotic ehavior. Finally we will show the motion of a doule-pendulum using a computer program. Maps First we will look at the iterates of functions. Suppose that f is a function and x 0 is an element of the domain of f. The iterates of x 0 will consist of x 0, f(x 0 ), f(f(x 0 )), f(f(f(x 0 ))),... These iterates together are called the orit of x 0. Note we can write x n+1 = f(x n ), which we often refer to as a map, so that the orit of x ecomes x 0, x 1, x, x,... As an example, consider the map x n+1 = cos(x n ), and let x 0 = 0.5. To find the iterates of x 0, we enter 0.5 into a calculator. Then we enter cos(ans) repeatedly. The first iterate of x 0 for f is x 1 = f(x 0 ) = The second iterate is x = f(f(x 0 )) = , and so on. The orit of x 0 will start looking like 0.5, , , , ,... If we press cos(ans) enough times, we end up repeatedly getting the calculator to spit out the numer This numer descries the value of x for which x = cos(x). This value of x is called a fixed point of f. In general, if we have a function f and a numer c in the domain of f, c is a fixed point of f if

3 f(c) = c. We can visualize this graphically y plotting the line y = x with the graph of f and finding where they intersect. Fixed points can e either stale or unstale. An unstale fixed point is a fixed point for which iterates that start neary the fixed point will move away from the point. A stale fixed point is a fixed point for which iterates neary the fixed point approach the point. In our previous example, since our iterates of 0.5 approached the fixed point , we have reason to assume that this fixed point is stale (of course we would need to make sure that our fixed point is approached y iterates on oth sides of it). We can check this assumption using the following procedure. Assuming that f is a differentiale function at a fixed point c, we can use the following to determine whether or not c is stale or unstale: If df dx x=c < 1, then c is stale, whereas if df dx x=c > 1, then c is unstale. If df dx x=c = 1, then c can e either stale, unstale, or neither. Note, in our example aove, df = sin(x) x= = < 1, which proves dx x=c our earlier assumption that x = is a stale fixed point of f(x) = cos(x). Figure 1: Graphical analysis of the orit of.5 for f(x) = cos(x), in the neighorhood of a stale fixed point Figure : Graphical analysis of the orit of.9 for f(x) = x, in the neighorhood of an unstale fixed point We can easily visualize the orit of a point x 0 for a function f graphically. First we plot the function f as well as the line given y y = x. We start at the point (x 0, x 0 ), which is along the line y = x, and draw a line vertically to the function f, stopping at the point (x 0, f(x 0 )). Then we draw a horizontal line to the line y = x, where we stop at the point (x 1, x 1 ), where x 1 = f(x 0 ) is the first iterate of x 0. This process is

4 then repeated. The graphical interpretation for the iterates of x 0 =.5 for f(x) = cos(x) is given in Figure 1. (Note the Matla codes used to produce each figure in this paper are given in the Appendix.) Rememer that these iterates converged to the stale fixed point x = We can also see this graphically in the figure. Note that for unstale fixed points, iterates will diverge from the fixed point, which can also e seen graphically. Consider the graphical analysis of the orit of x 0 =.9 for the function f(x) = x shown in Figure. First, note that 0 and 1 are fixed points for f since f(1) = 1 = 1 and f(0) = 0 = 0. Note 1 is unstale since df dx x=1 = d dx (x ) x=1 = x x=1 = > 1, while 0 is stale since df dx x=0 = x x=0 = 0 < 1. Thus, iterates of x 0 =.9 will move away from the unstale fixed point at 1 and converge to the stale fixed point at 0. Now consider iterates of x 0 = 1 for the function f(x) = 4x 4x. The orit of 1 will egin looking like {0., , , ,...}. We might e expecting these iterates to converge to a fixed point, as they did with f(x) = cos(x). However, this turns out not to e the case. It turns out that the iterates are numers in the interval (0, 1) that have no predictale pattern. This is an example of a function that exhiits chaos. Suppose f is a function and x 0 is in the domain of f. Then if x n = x 0 and if x 0, x 1, x,..., x n 1 are distinct, we say that x 0 has period n. Note if x 0 has period n, then the orit of x 0 is given y {x 0, x 1, x,..., x n 1 }. This orit is called a periodic orit or an n-cycle. Note that if a point has period 1, it is a fixed point. A periodic point is a value x for which some iterate is again x. For instance, if we consider the function f(x) = x 1, the orit of x 0 = 1 is given y 1, 0, 1, 0, 1,.... Note f(f(x 0 )) = 1, and so 1 is a periodic point. In fact, { 1, 0} forms a -cycle since f( 1) = 0 and f(f( 1)) = f(0) = 1. Now consider the map x n+1 = r cos(x n ), where r is a real numer. To see how this map ehaves for all values of r at once, we plot what is known as an orit diagram, shown in Figure. The darkened patches represent the areas where the iterates do not converge to a fixed point, ut spread out from (r, r). It is important to note that this spreading out does not indicate chaos rather it is the sensitivity to initial conditions that determines whether or not a system is chaotic. However, in this case sensitivity to initial conditions does occur in these darkened areas, and so they show the values of r for which the map 4

5 Figure : Orit diagram for x n+1 = r cos x n x n+1 = r cos(x n ) is exhiiting chaotic ehavior. A function that is parametrized, such as f(x) = r cos(x), is said to have a ifurcation at a point r 0 if the type (stale or unstale) or numer of periodic points change at that point. This point r 0 is then known as a ifurcation point for the function f. Thus, as can e seen from the plot in Figure, a ifurcation exists around r 0 = 1. since at that point our 1-cycle changes to a -cycle. This is known as a period-douling ifurcation. In general, a period douling ifurcation is a ifurcation where an n-cycle gives rise to a n-cycle. Period-douling is a common route to chaos. The Lyapunov Exponent The Lyapunov exponent determines whether or not a system is chaotic. The Lyapunov exponent for the orit of a function f starting at x 0 is given y { n 1 1 } λ = lim ln f (x i ). n n i=0 If λ > 0 we have chaotic dynamics. Otherwise, we have a non-chaotic situation. Calculating the Lyapunov exponent numerically is fairly simple. In general, the Lyapunov exponent is hard to find analytically, ut we will calculate it for a simple example. Consider the tent map given y f(x) = { rx 0 x 0.5 r rx 0.5 x 1 5

6 for 0 r and 0 x 1. The derivative of f is then f (x) = { r 0 x 0.5 r 0.5 x 1 The Lyapunov exponent is given y { n 1 1 λ = lim n n i=0 } { n 1 1 ln f (x i ) = lim n n i=0 } { } 1 ln ±r = lim ln r n = ln r. n n Thus, since λ > 0 when r > 1, and λ 0 when 0 < r 1, we get chaos when r > 1 and order when 0 < r 1. Figure 4 shows a plot of the Lyapunov exponent as a function of r for the tent map, while Figure 5 shows the orit diagram for the tent map. Figure 4: Liapunov exponent for the tent map Figure 5: Orit diagram for the tent map Note that from the plot of the Lyapunov exponent, we can see that λ > 0 when r > 1, as we found aove. As can e seen in Figure 5, chaotic ehavior starts occurring at r = 1, which agrees with the plot in Figure 4. We also note from the diagram that as r increases the chaotic ehavior for the iterates of the corresponding tent functions is also increasing. 6

7 The Baker Map One Dimensional Baker Function: Let us first introduce a version of the one-dimensional aker function B 1, given y { x 0 x 0.5 B 1 = x < x 1 First we will show that B 1 is extremely sensitive to initial conditions. Let us find some iterates of = and to see if they are much different. iterates Note that the tenth iterates of and are, respectively, and 0.008, which differ y over 0.6. Thus, even though we started with two numers fairly close together, the tenth iterates of those numers are already quite far apart. Hence, we cannot relate the higher iterates of to the corresponding iterates of. If we were to experiment further, we would find that choosing a different pair of numers close together will exhiit similar ehavior. Two Dimensional Baker Map: Now, consider the two-dimensional aker map, given y { (cx n, y n ) 0 y n 0.5 B(x n, y n ) = (1 + c(x n 1), y n 1) 0.5 < y n 1 where 0 < c < 1 First, note that, setting B(x 0, y 0 ) = (x 0, y 0 ), we otain two fixed points, namely (0, 0) and (1, 1). Thus, the points (0, 0) and (1, 1) remain unchanged under the mapping of B. Other points move differently depending on if they are aove or elow the line y n = 1. 7

8 If we map a fine grid of the unit square through B, we otain the plot shown in Figure 6. In other words, Figure 6 shows the first iteration of all points taken from the unit square (shown in lue). Consequently, putting these first-iteration points through B again gives the second iteration of these points, which are plotted in Figure 7. Similarly, Figure 8 shows the third iteration and Figure 9 shows the fourth iteration. Figure 6: First iteration of points y the aker map Figure 7: Second iteration of points y the aker map Figure 8: Third iteration of points y the aker map Figure 9: Fourth iteration of points y the aker map Now we know how the aker map acts on the unit square. So next we are interested in finding out whether or not the aker map exhiits chaotic ehavior and, if so, the points at which chaos occurs. In the next couple of pages, we will look solely at the y-component of the aker map and ignore the x-component, reducing the aker map to the one dimensional aker map. We will show and go into detail 8

9 aout what happens if y 0 is rational and conclude that in this case the iterations of y 0 repeat. Then we will show that if y 0 is irrational, then the aker map is chaotic. Finally, we will show that chaos in the aker map actually does not depend on the initial x-value ut rather only on the initial y-value. So, now we will prove that if y 0 is a rational numer, the iterations of y 0 will eventually repeat, which will result in a non-chaotic situation. More specifically, if y 0 = a, where a, Z (so y 0 is rational), then y 0 will have at most iterations efore an iteration is repeated, or, in other words, y 0 or an iteration of y 0 will have at most period. Choose y 0 = a, where a, Z and 0 < y 0 1. Then a. If y 0 0.5, then y 1 = y 0 = a. If y 0 > 0.5, then y 1 = y 0 1 = a 1 = a. Note in oth cases the numerator is an integer and the denominator is. I can repeat this process and the numerator will still e an integer, while the denominator will still e the integer. Since the numerator is an integer and is less than or equal to, we only have choices for the numerator. Thus, we will have at most iterations of y 0 efore the iterations egin to repeat. The first iteration that will e repeated will e either y 0 (if the denominator of y 0 is not divisile y ) or an iteration of y 0 (if the denominator is divisile y ). The previous statement was a it of a claim to make. Let us prove this. Our method of proof will e to introduce a function γ which mimics the aker function when y 0 is a rational numer etween 0 and 1 whose denominator is a fixed numer that is not divisile y. We will show that γ is injective and hence show that if is not divisile y, then y 0 will e repeated in a later iterate. Let γ (a) = B( a ), where 0 < a < and is an integer not divisile y. Note we can use our γ function to represent the numerator values in our iterations of y 0 = a when a and are integers and is not divisile y. Since in each iteration the denominator does not change, we can let y 0 = a0, y 1 = a1, and so on. Then we can represent our iterations {y 0, y 1, y,...} as { a0, a1, a,..., }. Thus, y 0 = a 0, y 1 = a 1, and so on. So, for instance, γ (a 0 ) = B( a0 ) = B(y 0) = y 1 = a 1. Similarly, we have that γ (a 1 ) = B( a1 ) = y = a, and so on. 9

10 Claim 1: The function γ is injective. Proof. Let a 1, a Z. Suppose γ (a 1 ) = γ (a ). We consider the three possile cases and show in each case a 1 = a. Case 1: a 1, a a1. Note, a Simplifying, we get a 1 = a. Thus, γ is injective. Case : a 1, a > a1. Note, a 1. Since γ (a 1 ) = γ (a ), we get B( a1 a. Simplifying, we get a 1 = a. Thus, γ is injective. > 1. Since γ (a 1 ) = γ (a ), we get B( a1 a a1 ) = B( ). So = a. a a1 ) = B( ). So = Case : a 1 and a > a1. Note 1 and a > 1. Since γ (a 1 ) = γ (a ), we get B( a1 a ) = B( ). So a1 = a. So (a a 1 ) =, a contradiction since is not divisile y. Hence, we have proven that γ is injective. Claim : If y 0 = a when a and are integers and is not divisile y, then y 0 will e repeated in a later iterate. Proof. Suppose y 0 = a0 where is not divisile y, and suppose, for the sake of contradiction, that y 0 does not ecome repeated later, ut that another iterate y i is the first iterate repeated later, where i 1. Let y k = y i, where k > i. Then γ (a i 1 ) = a i = a k = γ(a k 1 ). Since γ is injective, a i 1 = a k 1, a contradiction since a i was the first iterate repeated. Claim : Let y 0 e in lowest terms. If y 0 = a, where a, Z and is divisile y, then y 0 will not e repeated ut rather an iterate of y 0 will e repeated. Proof. Suppose y 0 = a, where a, Z and is divisile y. Then we can write = m for some m Z. Thus, y 0 = a m. If y 0 0.5, then y 1 = y 0 = a m. If y 0 > 0.5, then y 1 = y 0 1 = a m 1 = a m m. In oth cases, the denominator reduces to m. If m is again divisile y, we repeat the process until we have an iterate where the denominator is no longer divisile y. Let us call this iterate y i. Then, since the denominator of y i is no longer divisile y and we still have an integer in the numerator, y our previous proof, y i is the first iterate that will e repeated later. Hence, our inital y 0 will not e repeated. Now let us look at some examples. 10

11 Suppose y 0 = 1 9. We expect, since 9 is not divisile y, that y 0 should e repeated later in at most nine iterations. As can e seen in the iterates of y 0 shown elow, we are correct. iterates y 1 y y y 4 y 5 y Note that the 6 th iteration returns y 0 and so y 0 = 1 9 has period 6. So here we have a non-chaotic situation. See Figure 10 for a plot of the iterations of (x 0, y 0 ) = ( 1, 1 9 ). Figure 10: Iterations of (x 0, y 0) = ( 1, 1 ): a non-chaotic orit 9 Now suppose y 0 = 10. In this case the denominator is divisile y, and this causes the fraction to e reduced in the second iterate. The initial value y 0 is therefore never repeated, ut the first iterate of y 0, namely y 1, is repeated. The iterates of y 0 are shown elow. iterates y 1 y y y 4 y Note y 5 = y 1, and so y 1 has period 4. The aker map thus repeatedly spits out the values of y 1 through 11

12 y 4 for the y-values and so in this case exhiits non-chaotic ehavior. We have shown that we will get a non-chaotic situation if y 0 is rational and have shown a couple examples of this. So what happens if y 0 is irrational? Let us look at when y 0 =. The tale elow shows the iterations of this value of y 0 up to 4 iterations. iterates y 1 y y y 4 y 5 y 6 y 7 y iterates y 9 y 10 y 11 y 1 y 1 y 14 y 15 y iterates y 17 y 18 y 19 y 0 y 1 y y y To see how the aker function is sensitive to initial conditions, we will also plot a tale of the iterations of y 0 = and oserve how they differ from the iterations of y 0 =. iterates y 1 y y y 4 y 5 y 6 y 7 y iterates y 9 y 10 y 11 y 1 y 1 y 14 y 15 y iterates y 17 y 18 y 19 y 0 y 1 y y y Comparing the iterations of the two initial points, we see that the iterations stay somewhat close together at the eginning, ut then diverge. By the 17 th iteration they differ y aout Note we also see no repetitions so far with the iterations and the ehavior of the iterations seem somewhat random, giving us reason to elieve that we have found a point at which the aker function exhiits chaotic ehavior. However, it is very hard to make this claim from just 4 iterates. Thus, we use a computer program to make a plot showing many more iterates. See Figure 11 for a plot of iterations of (x 0, y 0 ) = ( 1, ) up to n = 00. 1

13 Figure 11: Iterations of (x 0, y 0) = ( 1, ) The plot shows very irregular ehavior. Along with the data earlier showing how y 0 was sensitive to initial conditions, the plot certainly makes it seem like we have found a point at which the aker map is chaotic. However, we have not proven this we would need to calculate the Lyapunov exponent to do so. Although we will not do this here, numerically calcuating the Lyapunov exponent would prove that the aker map is chaotic at this point. Will all irrational numers y 0 cause the aker map to e chaotic? To answer this question, first we show that we can write any iterate of an irrational y 0 in a particular form and from then show that iterates of y 0 can not repeat. Claim 4: Suppose y 0 is an irrational numer etween 0 and 1, so that y 0 = r, where r is an irrational numer and is an integer and 0 < r < 1. Then the kth iterate of our irrational numer can e written as y k = k r m, where m Z. Proof. We will proceed y induction. For the ase case, let k = 1. Note 1

14 y 1 = B(y 0 ) = B( r ) = { r 0 < r 1 r r, where oth 1 < r < 1 m = 0 in the first case and m = 1 in the second case). Now suppose y k = k r m. Note that r and are in the form y 1 = 1 r m (note y k+1 = B(y k ) = B( k r m ) = { k+1 r m y k 1 k+1 r (m+1) y k > 1 In either case, y k+1 can e written in the form k+1 r m for some m Z. Thus, we have shown that the k th iterate of our irrational numer r can e written in the form y k = k r m, where m Z.. Claim 5: The iterates of an irrational numer y 0 do not repeat. Proof. Suppose, for the sake of contradiction, that two iterates of our irrational numer y 0 = r are equal. That is, suppose y n and y k are iterates of y 0, where n < k, such that y n = y k. By our previous proof, y k = k r m 1 and y n = n r m, where m 1, m Z. Since y n = y k, we have k r m 1 = n r m. Simplifying, we get r( k n ) = (m 1 m ). Thus, r = (m1 m) k n. Note that k n 0 (since n < k). Also, note that since, m 1, m Z and the integers are closed with respect to multiplication and addition, (m 1 m ) Z. And note that since k, n Z, we have that k n Z. Thus, r is rational, a contradiction since r is irrational. Hence, we have shown that if y 0 is irrational, our iterations will not repeat. Rather, as shown y an earlier example, they will jump ack and forth in a seemingly irregular fashion, indicating that we have chaotic ehavior on our hands. Next we want to show that the values for x 0 do not affect whether or not the aker map exhiits chaotic ehavior, nor do the values of c. It is solely the values we choose for y 0 that determine whether or not we get a chaotic situation. 14

15 Suppose y 0 is a rational numer, and either y 0 or an iteration of y 0 has a certain period. Let y k e a repeated iteration. Claim 6: If y 0 1, we can write the kth iteration of x 0 as x k = c k x 0 + k 1 i=0 α ic i. Proof. We will proceed y induction. If k = 1, then x k = x 1 = cx 0, which is in the desired form. Now, suppose x k = c k x 0 + k 1 i=0 α ic i. Note { cx k y k 1 x k+1 = B(x k ) = 1 + c(x k 1) y k > 1 So x k+1 can e written in the desired form. = { c k+1 x 0 + k 1 i=0 α ic i+1 y k 1 1 c + c k+1 x 0 + k 1 i=0 α ic i+1 y k > 1 Claim 7: When k is large, x k x k. Proof. Case 1: y 0 1 Since y k is a repeated value for y 0 or an iteration of y 0, the aker map will do the same operation on x k+k as it will on x k. Note k 1 k 1 x k = c k x k + α i c i = c k (c k x 0 + α i c i ) + i=0 k 1 = c k x 0 + i=0 k 1 = c k x 0 + i=k k 1 α i c i+k + i=0 k 1 α i c i + i=0 i=0 k α i c i i=0 k 1 α i c i = c k x 0 + j=k k 1 α i c i = c k x 0 + i=k k 1 α j k c j + i=0 α i c i + x k c k x 0 Thus, x k = c k x 0 + k 1 i=k α ic i + x k c k x 0. Note when k is very large, all three terms except x k will e very small, and so x k x k. α i c i Case : y 0 > 1 Now we simply proceed until some i th iteration at which y i 1. We then treat y i as y 0 and x i and x 0 and proceed as we have in Case 1 to get the same result. Let s look at a couple of examples. First suppose y 0 = 1. Let us choose any c such that 0 < c < 1 and any x 0 such that 0 < x 0 < 1. Note y 1 =, y = 1, and so on. Note if i is even, then y i < 1, while if i is 15

16 odd, then y i > 1. Thus, we get x 1 = cx 0 x = 1 c + cx 1 = 1 c + c x 0 x = cx = c c + c x 0 x 4 = 1 c + cx = 1 c + c c + c 4 x 0 x 5 = c c + c c 4 + c 5 x 0 x 6 = 1 c + c c + c 4 c 4 + c 6 x 0 x 7 = c c + c c 4 + c 5 c 6 + c 7 x 0 x 8 = 1 c + c c + c 4 c 5 + c 6 c 7 + c 8 x 0 Note that, since 0 < c < 1, large powers of c will e very small. Thus, for large k, x k x k and x k+1 x k 1. To see this clearer, suppose c = 1 and x 0 =. Then x 1 = x = x = x 4 = x 5 = x 6 = x 7 = x 8 = x 9 = x 10 = x 0 = 0.75 x 1 = 0.5 x = 0.75 x = 0.5 Note we will still get the same convergence if we change x 0, ut leave y 0 and c the same. For instance, if instead we let x 0 = 1, we get the following iterations: 16

17 x 1 = x = x = x 4 = x 5 = x 6 = x 7 = x 8 = x 9 = x 10 = x = 0.75 x = 0.5 If we change c, ut leave y 0 = 1, then the even and odd iterations will converge to two different numers (not 0.5 and 0.75 ut something else). Meanwhile, if we change y 0 this will change how many distinct numers in our (approximately) repeating iterations we have. For instance, if y 0 = 1 5, say, then we will have five distinct numers that keep getting repeated in our iterations. So for large k, y k y k+5. So we will have, for instance, y 0 y 5 y 40 and y 1 y 6 y 41, and so on. Our final conclusion is that x 0 and c do not have any influence on whether or not we get chaotic ehavior. It is solely our choice for y 0 that determines this. Now suppose we create a plot showing the locations of many iterations of a single point (x 0, y 0 ), where y 0 is irrational, in the unit square. What will this look like? Consider the iterations of the point (x 0, y 0 ) = ( 1, ). Figures 1-17 show plots of 10, 50, 100, 500, 1000, and iterations, respectively. The initial four points are also shown and laeled in each plot. 17

18 Figure 1: 10 iterations of (x 0, y 0) = ( 1, ) Figure 1: 50 iterations of (x0, y0) = ( 1, ) Figure 14: 100 iterations of (x 0, y 0) = ( 1, ) Figure 15: 500 iterations of (x0, y0) = ( 1, ) Figure 16: 1000 iterations of (x 0, y 0) = ( 1, ) Figure 17: iterations of (x 0, y 0) = ( 1, ) 18

19 These plots emphasize the fact that the iterations of our point (x 0, y 0 ) = ( 1, ) really do spread out over many areas of the unit square and do not settle down. But, we do see that there are wide gaps where the later iterations of the points do not touch. In fact, note that our plot of iterations in Figure 17 is strikingly similar to the plot in Figure 9 of the fourth iteration of a fine grid of points taken from the unit square. Both are ecause of the way the aker map acts on a point. The plot in Figure 17 resemles what is known as a fractal pattern. This is an important aspect of chaotic systems. We can actually find geometrical structures associated with chaos, which tells us that chaos is not total disorder. If it were, the iterations of the point (x 0, y 0 ) = ( 1, ) would fill up the whole unit square instead of creating a pattern. We will not go into detail aout fractals, ut they are important in understanding chaos, and whole ooks can e devoted just to the study of fractals or the relationship etween fractals and chaos. The Doule Pendulum Now we will show how a doule pendulum exhiits chaos. A doule pendulum consists of two pendulums, one hanging from a fixed point and another pendulum hanging from the first. We will consider a doule pendulum in which the masses and lengths of the two pendulums are equal. We also do not consider friction in the system. To talk aout the position of the doule pendulum, we will say that the first pendulum is at an angle θ 1 from the vertical and the second pendulum is at an angle θ from the vertical, as shown in the drawing in Figure 17. Note that the red circle in the drawing indicates the fixed point at which the first pendulum is attached. 19

20 Figure 18: Doule Pendulum We will position two doule pendulums at nearly identical locations and oserve how their motion differs over just a short period of time. We position our first doule pendulum with its first pendulum at θ 1 = π 4 radians, and its second pendulum at θ = π radians. Our second doule pendulum is placed with its first pendulum at θ 1 =.14 4 radians and its second pendulum at θ =.14 radians; that is, it has een placed in nearly the same initial position as that of the first doule pendulum except that we have approximated π as.14. Figure 19 shows how the motion of each doule pendulum varies over a period of five seconds. Our first doule pendulum is shown in the right side of each figure, while the second doule pendulum (with π.14) is shown on the left side. As can e seen in the figure, the pendulums are still in nearly the same positions after three seconds. By four seconds, they have started to change slightly, and y five seconds they are in completely different locations, and their locations thereafter will not e related to one another. 0

21 Figure 19: Motion of two doule pendulums over a period of 5 seconds. Conclusion Chaos is everywhere in the world. This paper is meant to e a rief introduction to chaos and dynamical systems in hopes that the reader can egin to understand what chaos is and how it can e studied. We started y looking at the iterates of functions, stale and unstale fixed points, the graphical analysis of an orit, and some examples of maps and orit diagrams. Then we discussed the Lyapunov exponent and its importance in determining whether or not a system is chaotic. Next we went into a discussion of the aker map, including how it acts on the unit square and at what points it exhiits chaotic ehavior. Finally, we riefly discussed an example of a more well-known ut chaotic system that we can all visualize the doule pendulum and showed an example of how it is sensitive to initial conditions. 1

22 Resources Gulick, Denny. Encounters with Chaos. New York: McGraw-Hill, 199. Tél, Tamás and Márton Gruiz. Chaotic Dynamics. New York: Camridge University Press, 006.

23 Appendix This appendix contains the Matla codes that were written to produce the figures in this paper.

24 4

25 5

26 The following function and code were used to produce Figures 6-9: 6

27 Note Figure 11 was made simply y changing the first line x0 = sym(1/); y0 = 1/9; in the previous code to x0 = sym(1/); y0 = sqrt()/; 7

28 8

29 9

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