Synchronizing Chaotic Systems

Summer 2006

Student: Karl J. Dutson Advisor: Dr. Robert Indik

An appropriate definition of Chaos Theory (and one of many) is “the quali-tative study of unstable, aperiodic behavior in deterministic, non-linear, dynam-ical systems” [1]. Chaos implies that a system exhibits sensitive dependence oninitial conditions. For the purposes of this research project, we are interestedin two types of dynamical systems which can be chaotic.

The first is a map, of the general form:

xn+1 = F (xn), where F : Rm −→ Rm.

A second system of equal interest is an autonomous system of ordinarydifferential equations (ODE’s), of the form:

dy

dt= y′ = F (y), where F : Rm −→ Rm.

The specific examples we will use include the logistic map:

xn+1 = axn(1− xn) , (1)

and the well known Lorenz system [2]:

x = −σx + σyy = −xz + rx− y

z = xy − bz ,

where a, σ, r, and b are positive parameters.

Synchronization is a phenomenon that can occur when two or more copiesof some dynamical system couple together. If two systems are coupled, theyinteract with each other. If they synchronize, their behavior is nearly identical,even though it may still be unpredictable.

A question that arises concerning two (or more) such systems is “what kindof coupling leads to synchronization?” “Can we predict which values of thesystem parameters will cause synchronization to occur?”

It turns out that the answer is yes, as verified by previous research [3], andthis is possible through determining what are known as Lyapunov Exponents(LE’s) of a system. Calculating the LE’s of a chaotic system provides a greatdeal of information about its dynamics. In particular, the largest LE measuresthe rate at which initially close solutions to the system diverge from each other.Also, previous research [3] revealed that the critical coupling strength for whichtwo coupled dynamical systems will synchronize is given in terms of the firstLE. Although the LE is a crucial piece of information regarding a complexsystem, in practice it can be difficult to calculate. Fortunately, the results of

1

synchronization offer an adequate method of measuring the largest LE of thesystem. However, there are other LE’s of complex systems, which are alsoof interest, but are much harder to compute. Our objective is to investigatethe coupling of more than two copies of a dynamical system, and see if thesynchronization of these systems can be described in terms of higher order LE.LE will be explained in further detail later.

The outline of this report is as follows: We will first examine the logisticmap. This will be done by generating a bifurcation diagram to show how itsbehavior depends on its single parameter a. In addition, we will analyticallyfind some bifurcation points and expressions for fixed points. LE’s will then beintroduced within this context.

Next we will consider two coupled copies of the logistic map, which form a 2-dimensional system. We will find what parameter value(s) for coupling strengthsynchronize the system, for different values of a, and their dependence on theLE. Following this we will look at higher-dimensional coupled systems, and wehope to find relations between the synchronization of these systems and theirhigher order LE’s.

Recall that a map, in general, is of the form

F : Rm −→ Rm.

For any initial condition x0 ∈ Rm, the map defines a sequence recursively by

xn+1 = F (xn). (2)

Thus the recursive sequence produced by evaluating (or iterating) the map is afunction of the initial condition(s) and the number of iterations, n. For example,if we set x0 = 0.5 and a = 1, and iterate the logistic map (Equation 1) once weobtain x1 = 0.25. Then, to iterate the map again, we start with x1 = 0.25 asour initial input, and the result is x2 = 0.1875, and so on. In this way, for anyinteger n ≥ 0, xn is defined.

The iteration of a map can be thought of as the evolution of the state ofsome system. A common example is how the population of a species growsor decreases from one year to the next, based on an initial population. Forexample, a caterpillar population changing from generation to generation wouldbe a suitable application of a map. This means that the number of iterations inthe sequence, n, must be a positive integer; it does not make sense to evaluatesomething like x1.75.

Anytime we have a function such as F , we can call it a map and can defineiterates of that function (or map) by

~xn = ~xn(x0). (3)

Also, the same function F can be used to define an autonomous ODE:

d~x

dt= F (~x), (4)

2

and for each initial condition x0 there exists a solution

~x = ~x(t, ~x0). (5)

Note that while both solutions (Equations 3 and 5) depend on an initial con-dition, the evolution of the map depends on n, the number of iterations, andthe ODE solution depends on time, t. Thus with a continuous flow, such asEquation 4, it makes sense to evaluate non-integer points or intervals in time,as time is continuous and non-discrete.

A fixed point of the iterated map satisfies x = F (x), or xn+1 = xn. Forthe logistic map, a nice trivial example of a fixed point is x = 0, from whichxn+1 = 0 = xn.

We say that a list x0, x1, . . . , xP−1 is a period P orbit for the map F ifxP = F (xP−1) = x0. If the list x0, . . . , xP−1 is indeed a period P orbit, thenevery xj in the list is a fixed point of the map G defined by iterating F P times:

G(x) = F (F (· · · (F (x)) · · · )), (6)

where the dots above mean F is composed with itself P times.A good way to tell if a system is stable is to determine whether its long term

behavior is highly dependent on initial conditions. If a system is stable, initialconditions close to each other will eventually converge into the same orbit, or thesame fixed point. The system will not display the level of sensitivity to initialconditions that is the trademark of chaos. Conversely, if the system is not stable,initially close conditions will eventually diverge from each other exponentially,and perhaps even indefinitely. So, from these two possible outcomes, we cannarrow down whether or not the system is stable by taking two initial conditionsseparated by a very small difference, and observing how that difference changesin the long run.

Suppose that x is a fixed point x ∈ Rm of a map F such that x = F (x). Todetermine whether x is stable we slightly perturb x to x + δ0 (where |δ0| < 1 isvery small) and apply the map F to x0 = x + δ0. Then δ0 = x0 − x and thefixed point is stable if δn = xn − x −→ 0.

Because we are assuming |δn| = |xn − x| is small, a linearization in theneighborhood of x suffices to check stability:

F (x + δ0) ≈ F (x) + F ′(x)δ0, (7)

by the Taylor Theorem. Since x = F (x), F (x + δn) = x + δn, and

δn+1 ≈ F ′(x)δn ≈ [F ′(x)]n+1δ0. (8)

Therefore

δn ≈ [F ′(x)]nδ0. (9)

Thus the fixed point is stable if [F ′(x)]n → 0. This is true if and only if theeigenvalues λ of F ′(x) all have the property |λ| < 1 [4]. Note that because F isin Rm, F ′(x) is an m×m matrix - namely, the Jacobian Matrix:

3

J(x1, x2, . . . , xm) =

∂F1∂x1

∂F1∂x2

. . . ∂F1∂xm

∂F2∂x1

∂F2∂x2

......

. . .∂Fm

∂x1. . . ∂Fm

∂xm

.

We will explore the fixed points and stability of the logistic map, which, asstated above, is our first dynamical system of interest. It is written:

xn+1 = axn(1− xn) ,

where 0 < a < 4. While this may seem like a very simple function, its behaviorcan quickly become quite complex - chaotic even. The difference between chaosand stability depends only on the value of the parameter “a”.

We started exploring the map numerically, by choosing an initial condition,an appropriate number of iterations of the map, and varying the value of a. Foreach calculation in the following table, the initial condition was x0 = 0.7, andthe number of iterations, n, was 5000. Initial conditions other than x0 = 0.7(where 0 < x0 < 1) yielded results that varied only slightly from those listed.The quantity denoted x in the table is the value that the iteration converged toafter the map was applied an appropriate number of times (this was differentdepending on a, but was always < 5000), and is an approximate fixed point.

a x f ′(x)0.10 0 0.10000.25 0 0.25000.50 0 0.50000.75 0 0.75000.991 0.0001 0.98991.00 0.0002 0.99991.10 0.0909 0.90001.25 0.2000 0.75001.50 0.3333 0.50001.75 0.3333 0.24991.90 0.3333 0.09992.00 0.5000 02.10 0.5238 -0.10002.25 0.5560 -0.25202.50 0.6000 -0.50002.75 0.6364 -0.75022.90 0.6552 -0.9002

a x1 x2 f ′(x1) f ′(x2)3.01 0.6700 0.6330 -1.0200 -0.97983.1 0.5580 0.7646 -0.3596 -1.64003.2 0.5130 0.7995 -0.0832 -1.91683.3 0.4794 0.8236 0.1360 -2.13583.4 0.4520 0.8422 0.3264 -2.3270

1For these a values, the iteration seems to converge very slowly. The number of iterations

selected was sufficient for the other values of a, but for these many more applications of the

map would be necessary to obtain the same accuracy. Theoretically, at a = 0.99, x should =

0, and at a = 3, it should be true that x1 = x2.

4

This data suggests that there are about three major intervals containingthree different types of stable behavior for this map. When 0 < a < 1, thefixed point, x, seems to always = 0. But for 1 ≤ a ≤ 3, the eventual fixedpoint is different for each a, and is > 0. Finally, when 3 < a < 3.45, thereseems to be a convergence to a period 2 solution, where those two values are x1

and x2 in the second data table above. The behavior is even more complicatedwhen a ≥ 3.45, because the map converges to a period 4 solution! Furthermore,based on several extensive numerical measurements, it seems that the system’sbehavior becomes chaotic for any a > 3.544, as very slight changes in a causeadditional (even multiple) period doublings. It is for this reason that the datafor a ≥ 3.45 has not been included in the table. We will elaborate more on thislater, but for now, let us only concern ourselves with these first three intervals,as anything beyond them gets much more complicated.

For these intervals, we want to analyze the behavior of the system, whichmeans finding fixed points and analyzing their stability. Let us begin with astarting value of x0, and define

f(x) = xn+1 = axn(1− xn) = f(xn). (10)

Now, suppose we perturb this starting value by a very small amount δ0, where|δ0| ¿ 1, and call it

y0 = x0 + δ0 ; (11)

Then

δ0 = y0 − x0, and δn = yn − xn. (12)

So the question then becomes “what happens to δn in the long run?” Doesδn → 0, such that x0 ≈ y0 and the system (or at least the interval) is stable?Or does δn diverge and the system is unstable?

To proceed, let us also define yn by

f(yn) = yn+1 = ayn(1− yn) . (13)

Thus

yn+1 = xn+1 + δn+1 = f(yn) , (14)

and since yn = xn + δn,

f(yn) = f(xn + δn) ≈ f(xn) + δnf ′(xn), (15)

by the Taylor Theorem. So we have

xn+1 + δn+1 ≈ f(xn) + δnf ′(xn), (16)

and therefore, by subtracting f(xn) = xn+1 from both sides,

δn+1 ≈ δnf ′(xn). (17)

5

And from this, we obtain an important result:

δn ≈ δ0

n−1∏

k=0

f ′(xk). (18)

Note that if xn = x is a fixed point, then this Equation is a one-dimensionalcase of the general result found in Equation 9. If

limn→∞

n−1∏

k=0

f ′(xk) → 0, (19)

then δn → 0. A sufficient condition for this to be true is for |f ′(xn)| ≤ α, forsome α < 1, since then

n−1∏

k=0

|f ′(xk)| ≤ αn → 0. (20)

For our first two intervals, 0 < a < 1 and 1 ≤ a ≤ 3, the long term behaviorseems to converge to a fixed point, x. Thus

n−1∏

k=0

f ′(xk) = |f ′(x)|n , (21)

meaning that we can substitute and simplify things greatly:

limn→∞

n−1∏

k=0

f ′(xk) = limn→∞

|f ′(x)|n . (22)

Now f(x) = ax(1− x), so

f ′(x) = a(1− x) + ax(−1) (23)= a− ax− ax (24)= a(1− 2x). (25)

From the data it seems that for all 0 < a < 1, the system converges to x = 0,so we will start by evaluating the stability of this point:

f ′(0) = a(1− 0) = a. (26)

Hence we find that

limn→∞

|f ′(x)|n = limn→∞

|a|n . (27)

And if 0 < a < 1, then

limn→∞

an → 0 . (28)

6

So for the interval 0 < a < 1, we know that the fixed point x = 0 is stable,and that δn → 0. Perturbed initial conditions do not diverge from each other,so there is no sensitive dependence on initial conditions and the system is notchaotic.

This same argument (Equation 27) also shows that for a > 1, x = 0 is anunstable fixed point. This fixed point still exists, it is just not stable for anyinterval other than 0 < a < 1. So what are stable point(s) for a > 1? Accordingto our data, unlike for 0 < a < 1, it seems the fate of x is not always the sameregardless of a, and that x 6= 0. Instead, x converges to some positive non-zerovalue. This value, a fixed point, seems to depend on the initial condition x0

and the magnitude of a. We can check all this by analytically obtaining anexpression for a fixed point 6= 0 as a function of a on the interval 1 < a < 3 .We begin with the definition of a fixed point, f(x) = x, which for the logisticmap becomes

ax(1− x) = x. (29)

Solving for x, we obtain

x = 1− 1/a . (30)

This equation agrees with the data, and we can see that the stable fixed pointwhich the system eventually settles onto does indeed increase with a, and isalways positive. Also, for a < 1 or a > 3, the fixed points given by Equation30 are unstable, though they still exist. Using this result (Equation 30), we canshow that

limn→∞

n−1∏

k=0

|f ′(xk)| → 0 . (31)

This can be done by evaluating f ′(x), and substituting in Equation 30, ourstable fixed point solution for the interval:

f ′(x) = a(1− 2x)f ′((1− 1/a)) = a[1− 2(1− 1/a)] (32)

= a− 2a + 2 = 2− a.

So f ′(x) = 2− a, and

limn→∞

n−1∏

k=0

|f ′(xk)| = limn→∞

n−1∏

k=0

|2− a| → 0, (33)

because 1 < a < 3, so |2− a| < 1.For a ≥ 3, things quickly become more complicated. Recall from the result

in Equation 6 that because we seem to have a period 2 orbit at a > 3, we havetwo stable period 2 solutions. These are different from our two stable solutions

7

(x = 0 and x = 1− 1/a) of the previous intervals. Thus we experience our firstperiod doubling at a > 3. This first period doubling also coincides with a lossof fixed point stability. To find period 2 solutions, we look for fixed points off ◦ f :

f(f(x)) = x = f(ax(1− x)) . (34)

After a fair helping of algebra, we deduce from this the following 4th degreepolynomial:

x(a3x3 − 2a3x2 + a3x + a2x− a2 + 1) = 0 , (35)

whose roots will provide expressions for the four fixed points, which includethe period 2 solutions. Finding these roots might have been difficult, but wealready know two of them: x = 0 and x = 1 − 1/a, so we do not have to solvefor all four roots! These two were the stable fixed points for 0 < a < 1 and1 < a < 3 respectively, so if we divide them out of our quartic polynomial, weare left with a quadratic equation whose roots are the expressions for the period2 fixed points.

We start by dividing out x, since it is clear that x = 0 satisfies the statement,and obtain:

a3x3 − 2a3x2 + a3x + a2x− a2 + 1 = 0 . (36)

Next, since x = 1 − 1/a was a solution (and has been double-checked as asolution of Equation 35 ), we divide the remaining cubic function by the factor

x− (1− 1/a) = x− 1 + 1/a = x + (1/a− 1) = 0 . (37)

The quotient of Equation 36 and Equation 37 is

ax2 − ax− x + 1/a + 1 = ax2 − (a + 1)x + (1/a + 1) = 0 , (38)

and thus our quartic equation is reduced to a more manageable quadratic, whoseroots were found to be

x =a + 12a

±√

(a + 1)(a− 3)2a

. (39)

There are 2 fixed points here, (the map oscillates between the 2 as long as3 < a), so by letting xp1 represent the first and xp2 represent the second, wemay state

xp1 =a + 12a

+

√(a + 1)(a− 3)

2a. (40)

xp2 =a + 12a

−√

(a + 1)(a− 3)2a

. (41)

8

Now we have an expression for each of the 2 fixed points entirely as a function ofa. Note that these are only valid for a ≥ 3 (otherwise the results are imaginary),and that for a = 3, xp1 = xp2, which gives the same result as x = 1− 1/a.

Previously, for the intervals 0 < a < 1 and 1 < a ≤ 3, we obtained anequation for the fixed point(s), and then for the derivative evaluated at thispoint. The same was done here; but because we have 2 fixed points we havetwo linearizations, given by

f ′(xp1) = −1 +√

(a + 1)(a− 3) , (42)

f ′(xp2) = −1−√

(a + 1)(a− 3) . (43)

Our expressions for f ′(xp1) and f ′(xp2) offer a method of determining thestability of the period 2 solution. Upon looking at the data table once again,you will notice that within the interval 1 ≤ a < 3, f ′(x) −→ −1 as a in-creases. Indeed, the next period doubling occurs precisely when the productf ′(xp1)f ′(xp2) = −1. Hence, by setting f ′(xp1)f ′(xp2) = −1, we can solve for ato find the “maximum a value” for this interval, as well as locate exactly wherethe next period doubling occurs.

f ′(xp1)f ′(xp2) = −1 (44)1− (a + 1)(a− 3) = −1 (45)

a2 − 2a− 5 = 0 . (46)

The roots of this final quadratic turn out to be a = 1±√6. Since we are onlylooking at 0 < a < 4, a = 1+

√6 ≈ 3.4495 is the correct upper limit of stability

for to this period 2 interval. Also for this interval, it’s still true that

limn→∞

δn = limn→∞

n−1∏

k=0

|f ′(xk)| δ0 → 0 . (47)

because

−1 < f ′(xp1)f ′(xp2) < 1. (48)

At a ≈ 3.4495 there will be another period doubling. This 2nd perioddoubling also coincides with an additional loss of fixed point stability, as withthe first. Here there will be a period 4 orbit for the map, and so it will oscillatebetween four fixed points. Similar to the previous case, where the map wouldreturn to a given fixed point every other iteration, the map will now only returnto a fixed point every four iterations.

If we wanted to find expressions for each of these four fixed points, theprocedure is the same, starting with

x = f(f(f(f(x)))) , (49)

9

which represents a fixed point in this situation. We could then substitute f(x) =ax(1 − x) into this and solve for x, which would yield a degree 8 equation.Its roots would provide expressions for each fixed point. This equation canbe factored, because we already have four fixed point equations, and thus wealready have four of its roots. But dividing out all four roots still leaves uswith a 4th degree polynomial. Its roots would be much more unpleasant thanthose previously procured, and would require some careful computing to collect.However, if we were to do so, and found the roots, next we could linearize ateach fixed point expression. Finally, by evaluating

f ′(xp1)f ′(xp2)f ′(xp3)f ′(xp4) = −1 (50)

(where xp1, xp2, etc. denote fixed points) and solving for a, we could find wherethe next period doubling takes place, along with the upper boundary of theperiod 4 fixed point interval.

We did not attempt this, but we were able, however, to determine that theinterval where these four fixed points exist was something near 1 +

√6 < a <

3.544. Notice that these intervals between period doublings are decreasing. Thisis no coincidence; the decrease in interval length with each successive perioddoubling is geometric, as proven by Feigenbaum’s Theorem [5]. Also, upon firstanalyzing the map the value of a = 3.544 seemed to introduce the onset of chaos.

The findings from all the data and analytical calculations are summarizedby the following bifurcation diagram:

0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

a (Parameter Value)

x (F

ixed

Poi

nt)

Bifurcation Diagram: Behavior of Logistic Map as a Function of Parameter "a"

10

Hopefully you can see from the trend and from the diagram that followingthis last interval another period doubling occurs, and then another and an-other. It appears this will continue indefinitely as a increases. But the questionremains, “where exactly does the system go chaotic?” True, the behavior be-comes seemingly more and more complicated with each period doubling, butwhere (and how) do you draw the line between an enormous number of fixedpoints and the chaotic regime? From the data it seems that this occurs ataround a = 3.544, but how do we know for sure?

This was where the Lyapunov Exponent came in. Recall that the LyapunovExponent (LE) of a system is a measurement or gauge of sorts as to how chaotica system is. More accurately, the LE of a system measures the rate at whichsolutions to the system slightly perturbed from each other will diverge. If werefer back to Equation 18

δn = δ0

n−1∏

k=0

f ′(xk) ,

we have a nice way of looking at this mathematically. Another way to viewthis expression is that we are linearizing at each point in the trajectory (or timeseries), and between each linearization and the next, getting a sample of thestability and dynamics near that point, and then combining all of these pieces(or stability gauges) to get an overall idea of the entire system dynamics.

Hence this can be done for the long term by evaluating

limn→∞

δn

δ0= lim

n→∞

n−1∏

k=0

|f ′(xk)| , (51)

so to observe the average level of stability, we look at the nth root ofEquation 51

limn→∞

(δn

δ0

) 1n

= limn→∞

(n−1∏

k=0

|f ′(xk)|) 1

n

. (52)

To get the Lyapunov Exponent we take the log of Equation 52:

limn→∞

1n

logn−1∏

k=0

|f ′(xk)| . (53)

By determining this quantity (the LE) we have a means of measuring the averagesystem dynamics in the long run, and the amount of diverging or stretching thattakes place. Equation 53 was used to find the first LE for the logistic map for3 ≤ a ≤ 4, and the results are in the table on the next page:

11

Parameter a Lyapunov Exponent3.00 -0.00013.10 -0.26383.20 -0.91633.25 -1.38633.30 -0.61893.40 -0.13723.50 -0.87253.56 -0.07713.57 0.01063.60 0.18183.63 -0.01913.70 0.35243.74 -0.11193.75 0.36393.80 0.43493.83 -0.36953.90 0.49684.00 0.6933

A negative LE corresponds to the system’s behavior being non-chaotic, whilea positive LE corresponds to when it is chaotic. The magnitude of the LE playsa key role as well. If it is positive, then the larger the LE, the more chaotic thesystem’s overall behavior. Likewise, if the LE is negative, the opposite is true.Note that at values such as a = 3.63, 3.74, and 3.83 (and others) the behaviorbriefly exits the chaotic regime, as seen in the figure below.

3.5 3.6 3.7 3.8 3.9 4

−1.2

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

Parameter a

Lyap

unov

Exp

onen

t / F

ixed

Poi

nt

Comparison of Lyapunov Exponents and Bifurcation Diagram, both as a function of "a"’

12

In fact, an interesting discovery at this stage of the research was that of aperiod 3 orbit and three fixed points (well after the 2nd period doubling), whichwas unintentionally detected by arbitrarily setting a = 3.83. This is depicted inthe figure below:

0 50 100 150 200 250 300 350 400 4500.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

n (# of Iterations of Logistic Map)

x (F

ixed

Poi

nts)

Period 3 Fixed Points, Found in Map at a = 3.83, x0 = 0.7

Even though this is not a new result, the independent discovery of a period 3orbit almost immediately following the chaotic regime was still a fascinating oneregardless.

The data and figures provided show that the LE is useful for more than justdetermining when we crossed the line into chaos. Being able to calculate theLE for various a’s and plot it alongside the bifurcation diagram showed thatthe system does not become increasingly chaotic with increasing a, somethingwhich was not obvious, and perhaps even counterintuitive.

Now that we have thoroughly investigated the behavior of the logistic map,we wish to explore the concept of synchronization. A 2-dimensional coupled sys-tem can be created from the logistic map using two copies of it and introducinga new parameter ε, where ε < 1. The logistic map

f(xn) = xn+1 = axn(1− xn) (54)

then becomes the coupled system

13

xn+1 = f( (1− ε)xn + εyn) , (55)

yn+1 = f( (1− ε)yn + εxn) , (56)

which is given by

xn+1 = a[(1− ε)xn + εyn][1− [(1− ε)xn + εyn]] , (57)

yn+1 = a[(1− ε)yn + εxn][1− [(1− ε)yn + εxn]] . (58)

Note that the coupled system is derived by replacing xn (or yn) with its weightedaverage yn (or xn).

If ε = 0.5, then regardless of the initial values x0 and y0, xn will = yn,for n ≥ 1. On the other hand, if ε = 0, the systems decouple. If the initialconditions in this case are nearby but unequal they will diverge if the logisticmap is chaotic. But what are the ε values that synchronize the coupled system?And how do they relate to the Lyapunov Exponent?

There are indeed precise ε values that synchronize the coupled system, andthey are in fact different for different values of a. Moreover, there is not justone value of ε, but an entire interval where synchronization occurs. Thus, for agiven value of a, there exists a range of different ε values that synchronize thesystem.

These facts were found numerically, and can be seen from the table below.For all values in this table, the initial perturbation was δ0 = 10−4, the initialcondition was x0 = 0.7, and the total number of iterations was n = 4000.

SynchronizingParameter a LE (Measured) ε Value (Measured)

3.6 0.1818 0.0863.7 0.3524 0.1583.8 0.4349 0.1803.9 0.4968 0.200

Note that for each different a, the interval of synchronization is [εsynch., 0.5];between the ε value which first synchronized the system and 0.5. This is il-lustrated by the following pair of figures, which were obtained by first running5000 iterates of xn, and then introducing and iterating the coupled system,with yn = xn + δn. The first shows the system in an unsynchronized state, witha = 3.9 and ε = 0.197904:

14

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y

Coupled System, a = 3.9, x0 = 0.7, d = 0.0001, e = 0.197904, n = 4000

By slightly increasing the value of ε to 0.197905, synchronization is achieved,and the graph becomes:

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y

Coupled System, a = 3.9, x0 = 0.7, d = 0.0001, e = 0.197905, n = 4000

We can tell that the two systems are synchronized because their behavior isvirtually the same.

15

There were many other interesting phenomena surrounding the behavior ofthe system and how it developed in approaching synchrony or moving out of it.A few of these are depicted below. They demonstrate how the coupled systems’dynamics changed drastically with minor variations of a (specifically changes of+0.1). Here the value of ε was held constant (at 0.09), along with everythingelse except a.

0.4 0.5 0.6 0.7 0.8 0.9 1

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y

Coupled System, a = 3.6, x0 = 0.7, d = 0.0001, e = 0.09, n = 5000

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y

Coupled System, a = 3.7, x0 = 0.7, d = 0.0001, e = 0.09, n = 5000

16

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y

Coupled System, a = 3.8, x0 = 0.7, d = 0.0001, e = 0.09, n = 5000

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y

Coupled System, a = 3.9, x0 = 0.7, d = 0.0001, e = 0.09, n = 5000

17

At this point it is necessary to introduce the Singular Value Decompositionof a matrix. Let A ∈ Mm×n be any matrix, where

A : Rn −→ Rm.

Then ∃ orthogonal matrices U ∈ Mm×m and V ∈ Mn×n, along with a diagonalmatrix Σ ∈ Mm×n, such that A may be factored as:

A = UΣV T (59)

The elements of Σ, denoted a11 = σ1, a22 = σ2, . . . , amn = σmin(m,n), areuniquely determined by A, and appear in descending order σ1 ≥ σ2 ≥ . . . ≥σmin(m,n) ≥ 0. These diagonal entries of Σ are known as the Singular Valuesof A, and Equation 59 is the Singular Value Decomposition of A. Recall thatbecause U is orthogonal, its columns form an orthonormal set and thereforeUT U = I = UUT . The same is obviously true for V as well. Another way ofsaying this is that there are orthonormal bases vj for Rn and ui for Rm such thatAvj = σjuj and σj ≥ 0. Here vj and uji are columns of V and U respectively.

The implications of the Singular Value Decomposition are far greater thanthe fact that any matrix can be factored into A = UΣV T . Let S represent theunit sphere (which is the n-dimensional equivalent of the unit circle in R2, andas such, is centered at the origin), such that

S = { s ∈ Rn | s = Σ cjvj , | Σ c2j = 1},

since the columns vj of V form an orthonormal basis for Rn, and since ‖ s ‖2= 1.Then the image of S under the application of any linear map A is always a hy-perellipsoid (like the unit sphere, a hyperellipsoid is an n-dimensional equivalentof an ellipse in R2) whose principle axes are parallel to the columns of U :

AS = { w ∈ Rn | w = A

n∑

j=1

cjvj =n∑

j=1

cjσjuj , |n∑

j=1

c2j = 1 }.

Thus a geometrical interpretation of the Singular Value Decomposition is thatany linear mapping can be written as a reflection or rotation by U , a stretching(or shrinking) along the principal axis of each dimension by Σ, followed by areflection or rotation by V .

That Avj = σjuj may come as a surprise, but this fact and the relevancethereof can easily be shown. We have A = UΣV T , and because V is orthogonal,we can write

AV = ΣU. (60)

Now V has columns v1, v2, . . . , vn, so the jth column of AV (Avj), is equal tothe jth column of UΣ. But Σ only has diagonal entries, so the jth column ofthe m× n matrix UΣ is just σjuj . Therefore

Avj = σjuj . (61)

18

If each σj is distinct, then the vectors uj and vj are unique up to a sign.Because σ1 is the First Singular Value of A, since σ1 ≥ σ2 ≥ . . . ≥ σmin(m,n) ≥

0, then σ1u1 is the First Singular Vector of A, and corresponds to the largestprincipal axis of the resulting hyperellipsoid. Likewise σ2u2 is the Second Sin-gular Vector and the second largest principal axis, and so on.

The fact that A = UΣV T allows us to write

AT A = (UΣV T )T (UΣV T )= (V ΣUT )(UΣV T ) (62)= V Σ2V T

because ΣT = Σ, and UT U = Im, since U is orthogonal. In the same mannerAAT = UΣ2UT . Now let W = AT A, and let x ∈ Rn be an eigenvector of W ,with λ an eigenvalue of W , such that

Wx = λx. (63)

Substituting AT A for W , we have

V Σ2V T x = λx. (64)

Now we introduce another quantity, R = V T x, so that x = V R. By replacingthis expression for x into Equation 64, we obtain

V Σ2V T x = V Σ2V T (V R) = V Σ2R = λV R (65)

and then, applying the transpose of V to both sides,

Σ2R = λR. (66)

This implies a very useful result; that σ2j = |λ| for some j.

Suppose that instead of some matrix A ∈ Mm×n, we have a symmetricmatrix B ∈ Mm×m, such that BT = B, with eigenvalues λi. B will haveSingular Value Decomposition B = UΣV T , as before. However, the quantityW will be

W = BT B = B2 = V Σ2V T . (67)

Now recall that in general two matrices, say C and D, are similar if ∃ someinvertible matrix X such that C = X−1DX and D = XCX−1. Furthermore,if C and D are similar, then their eigenvalues are the same. We can see inEquation 67 that the matrix W = B2 is similar to Σ2. This implies anotheruseful result - namely, that the eigenvalues of B2 and Σ2 are the same. SinceΣ2 is diagonal with entries σ2

1 , σ22 , . . . , σ2

min(m,n), these are the eigenvalues ofboth matrices, and therefore

(σj)2 = (λi)2. (68)

19

Taking the square root of both quantities gives

|σj | = |λi|, (69)

but we have specified that each σj be non-negative, so then

σj = |λi|. (70)

Two important results have been demonstrated: that the singular values ofa matrix A are equal to the square root of the absolute value of the eigenvaluesof A, and that for a symmetric matrix, the singular values are equal to theabsolute value of its eigenvalues. These facts can be used to gain further insightinto synchronization and to verify the numerical findings in the last table.

Our coupled system, based on the logistic map, can be written

(xn+1

yn+1

)=

(f( (1− ε)xn + εyn)f( (1− ε)yn + εxn)

)= ~F

((xn

yn

); a, ε

). (71)

Let

~Vn =(

xn

yn

)(72)

so that Equation 71 may be more concisely written as ~Vn+1 = ~F (~Vn). Also, letx0 and y0 be initial conditions, and let

~V0 =(

x0

y0

)(73)

be the initial state of the system. Similar to the uncoupled system, if we perturb~V0 by ~δ0, the eventual fate of the system will be determined by:

~δn ≈ ~F ′(~Vn−1) ~F ′(~Vn−2) · · · ~F ′(~V1) ~F ′(~V0) ~δ0. (74)

But previously F ′ was just a scalar, so what is it in this case? Since ~F (~Vn) is

~F

(xn

yn

)=

(f( (1− ε)xn + εyn)f( εxn + (1− ε)yn)

)(75)

and more importantly since the coupled system is 2-dimensional, ~F ′(~Vn) here isa 2× 2 Jacobian, given by

20

~F ′(

xn

yn

)=

∂ ~F1∂x1

∂ ~F1∂x2

∂ ~F2∂x1

∂ ~F2∂x2

(76)

=

∂∂xf( (1− ε)xn + εyn) ∂

∂y f( (1− ε)xn + εyn)

∂∂xf( εxn + (1− ε)yn) ∂

∂y f( εxn + (1− ε)yn)

.

Or, after being evaluated

~F ′(~Vn) =

0@ f ′( (1− ε)xn + εyn) (1− ε) f ′( (1− ε)xn + εyn) ε

f ′( εxn + (1− ε)yn) ε f ′( εxn + (1− ε)yn) (1− ε)

1A . (77)

We are interested in the stability of the synchronized solution; that is, if we startwith V0 where x0 = y0, such that xn = yn, we want to know what happens to δn. Doits entries become equal in the limit? To determine this, we substitute xn = yn = vn

into our last expression, and obtain

~F ′(~Vn) =

0@ f ′(vn) (1− ε) f ′(vn) ε

f ′(vn) ε f ′(vn) (1− ε)

1A (78)

= f ′(vn)

0@ (1− ε) ε

ε (1− ε)

1Awhere f ′(vn) = f ′(yn) = f ′(xn). Now that we have determined ~F ′(~Vn), we can rewriteEquation 74, the expression for how the initial perturbation changes:

~δn ≈n−1Yk=0

f ′(vn)

0@ (1− ε) ε

ε (1− ε)

1An

~δ0. (79)

Now let

E =

0@ (1− ε) ε

ε (1− ε)

1An

. (80)

This is actually why the Singular Values are relevant to our purpose. They correspondto how far a system diverges, or stretches, from its initial state, and therefore canmeasure if the eventual behavior is chaotic. Let σk(n) ≥ 0 ∈ Rm denote the evolution

21

of the kth Singular Value of some matrix as a function of n applications of the map. Ifσ1 = σ1(n) → ∞ as n → ∞, then the stretching persists indefinitely, and indeed thebehavior is chaotic. Moreover, if σ1(n) and σ2(n) → ∞ as n → ∞, then even wilderstretching is occurring, and so forth.

If we can find the Singular Values of E we can determine if it will synchronize thesystem (that is, if δn → 0) and therefore if the synchronization is entirely determinedby ε. Since E is symmetric, we need only find its eigenvalues to obtain the SingularValues (as previously shown). The eigenvalues of E were found to be λ1 = 1, andλ2 = (1− 2ε), and the corresponding eigenvectors are h1 = (1, 1)T and h2 = (1,−1)T

respectively.The initial perturbation, ~δ0, may be written as a linear combination of these eigen-

vectors, since they form a basis:

~δ0 = α

�11

�+ β

�1−1

�, (81)

where

α =x0 + y0

2, and β =

x0 − y0

2. (82)

This can then be placed into Equation 79 to give

~δn ≈n−1Yk=0

f ′(vn)

0@ (1− ε) ε

ε (1− ε)

1An �α

�11

�+ β

�1−1

��. (83)

However, a further generalization can be induced if we observe a simple trend. Wealready know ~δ0, so let us continue on and evaluate ~δ1 and then ~δ2:

~δ1 = f ′(v0)

0@ (1− ε) ε

ε (1− ε)

1A1 �α

�11

�+ β

�1−1

��= f ′(v0)

�α

�11

�+ (1− 2ε)β

�1−1

��,

~δ2 = f ′(v1)f′(v0)

0@ (1− ε) ε

ε (1− ε)

1A2 �α

�11

�+ β

�1−1

��= f ′(v1)f

′(v0)

�α

�11

�+ (1− 2ε)2β

�1−1

��.

The matrix multiplication is not entirely obvious, but the trend continues, and itfollows that Equation 83 becomes

22

~δn =

n−1Yk=0

f ′(vn)

�α

�11

�+ (1− 2ε)nβ

�1−1

��. (84)

We know that if the original system (the logistic map) is chaotic, then the coupledsystem is chaotic as well. However, we are interested in how close to synchronizationthe system is, which is determined by xn − yn = (1,−1)~δn, or whether |xn − yn| → 0.

Our last equation for ~δn, Equation 84, can be rearranged to reflect this:

(1,−1)~δn = yn − xn

= (1,−1)

n−1Yk=0

f ′(vn)

�α

�11

�+ (1− 2ε)nβ

�1−1

��(85)

=

n−1Yk=0

f ′(vn)(1− 2ε)n 2β (86)

=

n−1Yk=0

f ′(vn)(1− 2ε)n (x0 − y0) (87)

because 2β = x0 − y0. Take note that for n = 0, the expression collapses down to theinitial conditions, as it should.

This last quantity should be a familiar one: it was found in Equation 17, and theLE can be obtained from it by taking the limit and the log, as in Equation 53. Let usdenote L as the Lyapunov Exponent, so that

L = limn→∞

1

nlog

n−1Yk=0

f ′(vn). (88)

Multiplying both sides by n, and then raising them as exponents of e, we may roughlysay that

enL =

n−1Yk=0

f ′(vn). (89)

Though in these operations we are side-stepping the limit present in this expressionand neglecting some deeper mathematics necessary to handle it, this is sufficient forour purposes.

Using this last result, we can determine precisely which values of ε will synchronizethe system. If enL(1 − 2ε)n < 1, then |xn − yn| → 0 and the system synchronizes; ifenL(1−2ε)n > 1 it does not. Thus, synchronization occurs exactly when enL(1−2ε)n =1. From this it follows that

23

enL(1− 2ε)n = 1 (90)heL(1− 2ε)

in

= 1

eL(1− 2ε) = 1

(1− 2ε) = e−L

ε =1− e−L

2. (91)

We have just shown that whether or not the system synchronizes is entirely dependenton ε, which is a function of the LE of the original system, and is therefore a functionof a. These analytical findings were used to verify and test the accuracy of the thosefound numerically, in the previous table:

Synchronizing

Parameter a LE (Measured) ε Value (Measured)1−e−L

2

3.6 0.1818 0.086 0.0833.7 0.3524 0.158 0.1493.8 0.4349 0.180 0.1763.9 0.4968 0.200 0.196

Apparently the accuracy of the numerical results was right on the mark - these ana-lytical values are quite close to those found in the previous data set!

We have determined the values of ε for which the coupled system synchronizes,and related this to the LE for our one dimensional dynamical system. But we wish toextend the analysis to more complex coupling schemes, and attempt to relate higherorder LE’s to the synchronization of more than two copies of higher dimensional maps.

This can be approached by building on the generalization already obtained inEquation 9:

δn ≈ [F ′(x)]nδ0,

where F ′(x) was an m×m Jacobian Matrix. It governs the fate of some perturbation δ0

and thus the stability of an m-dimensional map. This generalization has been appliedin two cases: for the logistic map, F ′(x) was simply 1 × 1; for the coupled system,F ′(x) was 2× 2. Now we will explore the case where F ′(x) is m×m.

To first approach this, we attempted to make not just 2, but m copies some functionF : Rm −→ Rm, not necessarily the logistic map. We also begin by letting

Z =

0BBB@z1

z2

...zm

1CCCA ∈ Rmn, where zj ∈ Rm. (92)

24

As with the coupled one dimensional systems, we need to create a coupled systemby replacing each vector zj ∈ Z with a weighted average. Our first approach is to startwith a simple case. One possible weighted average, which is similar to that chosen forthe 2-dimensional coupled system, is

z1 −→ (1− (m− 1)ε)z1 + εz2 + · · · + εzm

z2 −→ εz1 + (1− (m− 1)ε)z2 + · · · + εzm

...

zm −→ εz1 + εz2 + · · · + (1− (m− 1)ε)zm

for some ε ≤ 0.5.One way to create this weighted average is to apply to Z the following block matrix:

Q =

0BBBB@(1− (m− 1)ε)In ε In . . . ε In

ε In (1− (m− 1)ε)In

......

. . . ε In

ε In ε In . . . (1− (m− 1)ε)In

1CCCCA .

Here Q is mn ×mn, consisting of m rows by m columns of blocks. Each block of Qconsists of an n×n matrix; specifically, an n×n identity matrix multiplied by ε or someconstant based on ε. For each block element Qij of Q, the entry is (1 − (m − 1)ε)In

for i = j, and ε In for i 6= j.For our purposes, a more useful way to write Q is to separate the diagonal and off

diagonal entries. Expanding the quantity (1− (m− 1)ε), we obtain 1−mε + ε, so wemay write

Q = (1−mε)Imn + ε

0B@ In . . . In

.... . .

...In . . . In

1CA .

Furthermore, we will let

K =

0B@ In . . . In

.... . .

...In . . . In

1CAso that Q may more concisely be written as Q = (1−mε)Imn + εK.

Thus by applying Q to Z we obtain the desired weighted average, which will bedenoted T = QZ. Next we apply F to each block of the new matrix T , and we willcall this result

N(Z) = F (QZ) = F (T ). (93)

You will notice the procedure is essentially the same as with the coupling of the 2one-dimensional systems. But here, instead of replacing x with its weighted average,and applying the logistic map (thereby creating a 2-dimensional coupled system), we

25

are replacing each vector z ∈ Z with a weighted average, and applying some generalfunction F .

Having created an m-dimensional coupled system with an appropriate weightedaverage, we now evaluate whether the synchronization of this system is given entirelyin terms of ε. As with the coupled one-dimensional system, the long term fate of thegeneralized system will be given by

δn ≈ [F ′(QZ)]nδ0

≈ N ′(Zn−1) N ′(Zn−2) · · · N ′(Z1) N ′(Z0) δ0. (94)

As with each of the previous cases, this means we need to find F ′. Since N(Z) =F (T ) = F (QZ), N ′(Z) = F ′(T ) = F ′(QZ) = F ′(QZ)Q, by the chain rule, we have

δn ≈ F ′(Tn−1)Q F ′(Tn−2)Q · · · F ′(T1)Q F ′(T0)Q δ0. (95)

For the logistic map, F ′ was 1× 1, and was a scalar; for the coupled system, F ′ was a2× 2 Jacobian Matrix; for this generalized system, F ′ will have dimension mn×mn.Furthermore, F ′(T ) will be given by

F ′(T ) =

0BBBB@F ′(T ) 0 . . . 0

0 F ′(T )...

.... . . 0

0 0 . . . F ′(T )

1CCCCAwhere each of the blocks F ′(T ) along the principle diagonal are n × n Jacobiansthemselves.

Now we introduce some useful properties needed to complete our analysis. SayC and D are any two matrices that ∈ Mm×n, with Singular Values σ1, . . . , σk andρ1, . . . , ρk respectively. In general, the Singular Values of the matrix CD are notsimply σ1ρ1, . . . , σkρk. However, if CD = DC, and CDT = DT C, then the SingularValues of CD and DC are in fact σ1ρ1, . . . , σkρk. Also, the eigenvalues of the matrixC + D are equal to the sum of the eigenvalues of C with the eigenvalues of D.

Why are these facts important? Because if they apply, that means by simplyfinding the eigenvalues of F ′(T ) and Q (which are both symmetric matrices) separately,and then multiplying them together (followed by taking the absolute value of theproduct), we will have obtained the Singular Values of F ′(T )Q. From there, we willbe able to determine if the synchronization of our general system is based on ε alone.

So the next step was to see if the products F ′(T )Q and F ′(T )QT were commutative.Again, knowing that F ′(T ) and Q are symmetric, we know F ′(T )Q = F ′(T )QT . Somecalculations also showed that F ′(T )Q and F ′(T )QT were indeed commutative, andwe computationally confirmed that F ′(T )Q = QF ′(T ), which implies F ′(T )QT =QT F ′(T ).

However, this can easily seen, given that F ′(T ) and Q both have the property thatall of their diagonal blocks are the same. So the product of F ′(T ) and Q will simplybe the only elements of F ′(T ) (which are the blocks along the diagonal) times thediagonal blocks of Q, regardless of multiplicative order. Either way the product is:

26

0BBBB@F ′(T )(1− (m− 1)ε)In ε In . . . ε In

ε In F ′(T )(1− (m− 1)ε)In

......

. . . ε In

ε In ε In . . . F ′(T )(1− (m− 1)ε)In

1CCCCAThis useful fact allows us to rearrange Equation 95 from

δn ≈ F ′(Tn−1)Q F ′(Tn−2)Q · · · F ′(T1)Q F ′(T0)Q δ0

into the more simplified

δn ≈

0B@ F ′(Tn−1)F′(Tn−2) · · ·F ′(T0) 0 0

0. . . 0

0 0 F ′(Tn−1)F′(Tn−2) · · ·F ′(T0)

1CAQnδ0.

We will refer to the matrix product on the right hand side of the above equation as Ψ,and the matrix with diagonal block entries F ′(Tn−1)F

′(Tn−2) · · ·F ′(T0) we will callΥ.

Whether δn → 0 as n → 0 can be determined from the Singular Values of Ψ. Infinding these, the aforementioned facts come into play. We begin with the fact thatthe eigenvalues of C + D are the same as the eigenvalues of C plus the eigenvalues ofD, for any two commutative matrices C and D. Thus the eigenvalues of Q are simplythe sum of the eigenvalues of (1−mε)Imn and εK.

The rank of εK is n, so the null space has dimension nm−n = n(m−1), and there-fore one eigenvalue of εK is 0, with multiplicity n(m− 1). The only other eigenvaluewas found to be mε, with multiplicity n. So, since all of the eigenvalues of (1−mε)Imn

are simply (1−mε), Q has eigenvalues (1−mε)+ εm = 1 and (1−mε)+0 = (1−mε),each with multiplicities n and n(m − 1) respectively. As we established above, sinceQ is symmetric its Singular Values are equal to the absolute value of its eigenvalues.So then

27

|λ1| = · · · = |λm| = σ1 = · · · = σm = 1 (96)

and|λm+1| = · · · = |λmn| = σm+1 = · · · = σmn = (1−mε). (97)

Similarly, the Singular Values of Qn will be 1n = 1, with multiplicity n, and |(1−mε)n|,with multiplicity n(m− 1).

Unfortunately, this means that synchronization is not entirely controlled by ε, aswe would have hoped. To illustrate, let the Singular Values of Ψ, Q, and Υ be ψj(n),qj(n), and σj(n) respectively. Then

ψj(n) = qj(n)σj(n). (98)

But q1(n) = · · · = qm(n) = 1, and qm+1(n) = · · · = qmn(n) = (1−mε)n. Therefore

ψ1(n) = · · · = ψm(n) = σ1(n) = · · · = σm(n), (99)

and

ψm+1(n) = · · · = ψmn(n) = σ1(n)(1−mε)n = · · · = σmn(n)(1−mε)n. (100)

From this it follows that Qn does not completely control synchronization, which showsthat the values of ε which synchronize the system depend only on the first LyapunovExponent. This is, of course, disappointing, as we wanted to describe the synchroniza-tion of a generalized coupled system in terms of any LE but the first! So it appearsthat our basic first approach and weighted average were not sufficient to relate thehigher order LE’s of an m-dimensional system to its synchronization.

Though it is disappointing our initial, simplistic attempt did not yield furtherinsight into this relation, this possibility was anticipated, and at the least, we havemade another small step toward our stated objective by ruling out the weighted averageused. Based upon this result, we know that further investigation into generalizing theresults of synchronization would require a very different and more complex weightedaverage. In pursuing the research topic beyond this point, the next step would beto find a weighted average as stated, and complete the same analytical steps with it.However, the challenge would be to properly select a weighted average that wouldisolate the synchronization of a generalized coupled system purely as a function of itsparameter. Moreover, finding exactly which type of weighted average would provide ameans to the end we seek would be a task in and of itself. In addition, other couplingschemes should probably be investigated as well.

28

References

[1] Kellert, S.H. (1993). In the Wake of Chaos: Unpredictable Order in DynamicalSystems, (Chicago and London, The University of Chicago Press).

[2] Lorenz, E.N. (1963). Deterministic Nonperiodic Flow, (Journal of the AtmosphericSciences: Vol. 20, No. 2, pp. 130141)

[3] Bergevin, C., and Steinke, S. (1999). Undergraduate Research Projects - Univer-sity of Arizona, University of Arizona Department of Mathematics. 22 Apr. 2006.http://math.arizona.edu/ ura/971/bergevin/.

[4] Olver, P.J., and Chehrzad, S. (2006). Applied Linear Algebra, (Pearson PrenticeHall: Theorem 10.12).

[5] Glendinning, P. (1994). Stability, Instability, and Chaos, (Cambridge UniversityPress: Ch. 11).

[6] Trefethen, L.N., and Bau, D. (1997) Numerical Linear Algebra, (SIAM - Societyfor Industrial and Applied Mathematics)

[7] Gleick, J. (1987). Chaos: Making a New Science, (Viking).

29