Untangling Complex Systems, page 73
Ichimaru and Moody 1999). For its analysis, you can freely download the TISEAN soft-
ware (Hegger et al. 1999). As a first attempt, you may try to determine the phase space of
the time series “b1.txt” (https://physionet.org/physiobank/database/santa-fe/) by calculating the Mutual Information and/or the Autocorrelation, and the False Nearest Neighbors.
After estimating the time delay ( τ) and the embedding dimension ( m), try to calculate the
largest Lyapunov exponent by using the Kantz’s method.
10.14 SOLUTIONS TO THE EXERCISES
10.1. A linear equation is an expression of the type: y = a + b * x, where x and y are the independent and dependent variables, respectively. The equation ( g) is an example. When we plot
a linear equation, it looks like a straight line. An equation is nonlinear when the indepen-
dent variable does not appear simply at the first power. Nonlinear equations contain power
functions, product functions and/or transcendental functions. Examples of nonlinear equa-
tions are (a), (f), and (i). A differential equation is linear when the dependent variables and
their derivatives appear only to the first power, and there are not products of dependent
variables. Examples are the equations (c) and (d). In equation (d) there is a second deriva-
tive of the dependent variable, but it is to the first power. On the other hand, the equation
(e) is nonlinear because the second derivative is elevated to the power of 2. Other examples
of nonlinear differential equations are (b) and (h).
10.2. The trends of the angles ϑ and over the first 20 seconds are shown in Figure 10.24.
1
ϑ 2
The dynamics are periodic for both angles, in agreement with what we found in the
first and second run plotted in Figure 10.3. A difference is that now the oscillations are
anti-phase, whereas they were in phase condition in Figure 10.3. The trends for the spatial
coordinates ( x , ) and ( , ) are plotted as follows. They have been calculated by using
1 y 1
x 2 y 2
trigonometry and the estimated values of θ and (Figure 10.25).
1
θ 2
x =
1
L 1 sinθ1
y 1 = − L 1 cos 1
θ
The Emergence of Chaos in Time
361
θ 1
θ 2
0.4
0.0
θ (radians)
−0.4
0
5
10
15
20
Time (s)
FIGURE 10.24 Trends of θ and over time.
1
θ 2
−1.983
−2.85
−1.986
−2.88
−1.989
−2.91
−1.992
y 1
y 2 −2.94
−1.995
−2.97
−1.998
−3.00
−2.001
−0.3 −0.2 −0.1 0.0 0.1 0.2 0.3
−0.3−0.2−0.1 0.0 0.1 0.2 0.3
x 1
x 2
FIGURE 10.25 Profiles of the Cartesian coordinates for the masses of the double pendulum.
x 2 = x 1 + L 2sinθ2
y 2 = y 1 − L 2 cosθ2
10.3. Since the initial values of θ and are large, the motion of the double pendulum is highly
1
θ 2
nonlinear. In fact, it is aperiodic. The trends for the spatial coordinates ( x , ) and ( , )
1 y 1
x 2 y 2
are plotted in Figure 10.26.
It is interesting to compare the motion for θ (
(
1 t = 0) = 1.57 (90°), θ 2 t = 0) = 1.57 (90°),
with that for θ (
(
1 t = 0) = 1.57 (90°), θ 2 t = 0) = 4.71 (270°). In the two conditions, the double
pendulum has the same potential energy. Although the total energy of the Hamiltonian
system is equal, the motion is significantly different, especially for mass 2 of the double
pendulum (see Figure 10.27).
362
Untangling Complex Systems
2
0.5
0.0
0
−0.5
y 1
y 2
−1.0
−2
−1.5
−2.0
−4
−2 −1
0
1
2
−4
−2
0
2
4
x 1
x 2
FIGURE 10.26 Profiles of the Cartesian coordinates for the two masses of the double pendulum.
θ 1( t = 0)= 1,57; θ 2( t = 0) = 1,57.
θ 1( t = 0) = 1,57; θ 2( t = 0) = 4,71.
2.0
1.5
1.0
0.5
adians) 0.0
(R −0.5
θ 1 −1.0
−1.5
−2.0
0
20
40
60
80
100
Time (S)
20
0
−20
adians)(R −40
θ 2 −60
−80
0
20
40
60
80
100
Time (S)
FIGURE 10.27 Trends of θ and over time for two initial conditions that correspond to the identical total 1
θ 2
energy of the double pendulum.
10.4. For the integration of the differential equation [10.10], ( dN/ dt) = rN (1− ( N/ K)), first of all, we separate the variables:
KdN
rdt
N ( K − N) =
Then, we set
KdN
A
B
A
( K − N) + BN dN
dN
dN
rdt
N ( K − N) =
+
N
( K − N) =
N ( K − N)
=
The Emergence of Chaos in Time
363
When
N = 0, AK = K, i.e., A = 1. When N = K, BK = K, i.e., B = 1. Therefore, Nt
Nt
t
1
1
dN +
dN = rdt
∫
∫ − ∫
N
K N
N 0
N 0
0
Finally, we achieve the logistic function [10.11].
10.5. From the data reported into Table 10.1 and depicted in Figure 10.28, it is evident that when r = 3, the evolution of the population is not sensitive to the initial conditions. The
two sequences oscillate between two values, and the discrepancy between the two series
maintains equal to the initial value of 0.00001. An entirely different behavior emerges in
the case of r = 4. The evolution is extremely sensitive to the initial conditions. The two
sequences stay reasonably close together for the first ten iterations. Then, they diverge,
and there is no way to tell that the sequences ever started off a mere 10−5 away from each
other. When r = 3, the population evolves towards an ordered state, whereas when r = 4,
the evolution is chaotic because it is aperiodic and really sensitive to the initial conditions.
TABLE 10.1
Values of the Populations According to the Logistic Map
Calculated for 50 Iterations and Fixing the Parameter r
Equal to 3 and 4, Respectively
r = 3
r = 4
0
0.30000
0.30001
0.30000
0.30001
1
0.63000
0.63001
0.84000
0.84002
2
0.6993
0.69929
0.5376
0.53756
3
0.63084
0.63085
0.99434
0.99436
4
0.69864
0.69864
0.02249
0.02244
5
0.63162
0.63163
0.08794
0.08775
6
0.69803
0.69802
0.32084
0.32019
7
0.63236
0.63237
0.87161
0.87068
8
0.69745
0.69744
0.44762
0.4504
9
0.63305
0.63305
0.98902
0.99016
10
0.6969
0.69689
0.04342
0.03898
11
0.6337
0.6337
0.16615
0.14985
12
0.69638
0.69637
0.55416
0.50958
13
0.63431
0.63432
0.98826
0.99963
14
0.69588
0.69588
0.04639
0.00147
15
0.63489
0.6349
0.17695
0.00586
16
0.69542
0.69541
0.58256
0.02331
17
0.63544
0.63545
0.97273
0.09108
18
0.69497
0.69496
0.1061
0.33115
19
0.63596
0.63597
0.37936
0.88595
20
0.69454
0.69454
0.94178
0.40416
21
0.63646
0.63647
0.21931
0.96326
22
0.69414
0.69413
0.68484
0.14158
23
0.63693
0.63694
0.86333
0.48614
24
0.69375
0.69374
0.47196
0.99923
25
0.63738
0.63739
0.99685
0.00307
26
0.69338
0.69337
0.01254
0.01225
27
0.63782
0.63782
0.04955
0.04842
( Continued)
364
Untangling Complex Systems
TABLE 10.1 ( Continued)
Values of the Populations According to the Logistic Map
Calculated for 50 Iterations and Fixing the Parameter r
Equal to 3 and 4, Respectively
r = 3
r = 4
28
0.69302
0.69302
0.18837
0.18428
29
0.63823
0.63824
0.61155
0.60129
30
0.69268
0.69267
0.95023
0.95896
31
0.63863
0.63863
0.18918
0.15742
32
0.69235
0.69234
0.61356
0.53055
33
0.63901
0.63901
0.94841
0.99627
34
0.69203
0.69203
0.1957
0.01488
35
0.63937
0.63938
0.62961
0.05863
36
0.69173
0.69172
0.93281
0.22078
37
0.63972
0.63973
0.25071
0.68815
38
0.69143
0.69143
0.75142
0.8584
39
0.64006
0.64006
0.74715
0.48621
40
0.69115
0.69115
0.75566
0.99924
41
0.64039
0.64039
0.73855
0.00304
42
0.69088
0.69087
0.77237
0.01213
43
0.6407
0.6407
0.70325
0.04793
44
0.69061
0.69061
0.83475
0.18254
45
0.641
0.64101
0.55176
0.59688
46
0.69035
0.69035
0.98928
0.96246
47
0.6413
0.6413
0.04241
0.14454
48
0.69011
0.6901
0.16243
0.49459
49
0.64158
0.64158
0.5442
0.99988
50
0.68987
0.68986
0.99219
4.69E-4
0.6
r = 3
xn
x′0 = 0.30000
0.4
x″0 = 0.30001
10
20
30
40
50
Generation
r = 4
0.8
xn
0.4
0.0
10
20
30
40
50
Generation
FIGURE 10.28 Evolutions of two populations for two different initial conditions: x′0 = 0 30000
.
(dark points)
and x′′0 = 0.30001 (gray points), and for r = 3 (upper graph) and r = 4 (lower graph).
The Emergence of Chaos in Time
365
If you want to calculate the iterations by using MATLAB, you can use the following script:
Logistic map.m
numsteps=50;
x(1)=0.30000;
r=3.0;
for i=1:numsteps
x(i+1)=r*x(i)*(1-x(i));
end;
plot(1:numsteps, x, ‘bo-’);
10.6. The unimodal map x
π
1
( )
n+ = rsin
xn has a shape that is pretty similar to that of the logistic
map (see Figure 10.29). It is smooth, concave down, and its maximum is equal to xn+ 1 = r at x
y
rπ cos π x becomes null when x
n = 0.5 (in fact, its first derivative ′ =
( n)
n = 1/2).
The bifurcation diagram can be built by using the following script of MATLAB:
bifurcation.m
numr=1000;
startr=0.00;
endr=1.00;
r=linspace(startr, endr, numr);
skipnum=500;
num=500;
for j=1:length(r)
x=0.1;
for i=1:skipnum
x=r(j) * sin(pi*x);
end;
r
xn+1
0.0
0.2
0.4
0.6
0.8
1.0
xn
FIGURE 10.29 Profile of the unimodal map xn+1 = rsin(π xn).
366
Untangling Complex Systems
for i=1:num
x=r(j) * sin(pi*x);
results(j, i)=x;
end;
end;
plot(r, results, “b.”, “MarkerSize”, 0.5);
xlabel(“r (Growth Rate)”)
ylabel(“Final x Values (Population)”)
title (“Bifurcation Diagram for Unimodal Map”)
return;
The unimodal map x
π
1
( )
n+ = rsin
xn exhibits a bifurcation diagram (built by discarding
the first 500 iterations) that is pretty similar to that obtained in the case of the logistic map
(compare Figure 10.30 with Figure 10.9). Both unimodal maps undergo period-doubling evolutions up to reach a chaotic regime. The chaotic regimes are interspersed with periodic
windows.
10.7. We rewrite equation [10.17] in the following simplified form:
rk r
δ ≈ − k−1
∞
r − rk
After rearranging it, we achieve
rk r 1
r ≈ r
k−
∞
k +
−
δ
Introducing the numerical values of rk, rk−1 and δ, the result is r∞ = 3 569680
.
. A better esti-
mate can be obtained using values of r for larger k.
10.8. By rearranging equation [10.22] in the following form
( Ra )µα
∆ T
c
=
β gh 3
and introducing the values of the parameters listed in the text of the exercise, we find that
when h = 0.1 cm, Δ T = 0.003 K, whereas when h = 1 cm, Δ T = 3 × 10−6 K. Libchaber used a cell with h = 0.1 cm because its microbolometers could detect thermal gradients of the
order of one thousandth of degree K. He did not use a cell with h = 1 cm, because he could
not measure the thermal gradients with an accuracy of one millionth of degree K.
10.9. From the list of the critical ratio values, which are Ra /
/
, and
/
, we can
4 Ra , Ra Ra
Ra Ra
c
3
c
2
c
estimate the parameter δ, through the equation as follows:
( Ra 3 / Rac)−( Ra 2 / Rac) 3 6183
.
3 4850
.
δ = (
= 4 4
.
Ra
−
4 / Rac ) − ( Ra 3 / Rac ) =
