MATH 345: Assignment 5 Solutions
Q1
1
0
-1
-2
L-exp
-3
-4
-5
-6
-1
-0.5
0
0.5
1
1.5
2
2.5
3
a
Q2
ẋ = −y − z,
ẏ = x + ay,
ż = bx − cz + xz
(1)
(a)
See p. (4) - plot of single trajectory in R3
See p. (5) - plot of y vs. t for two trajectories that start close together. For small t, the trajectories
are indistinguishable on the plot, but as t increases the trajectories can be seen to separate. This
gives evidence of sensitive dependence on initial conditions.
Using three different initial conditions, XPP computed the maximal Lyapunov exponent to be
0.076689, 0.0936659, and 0.0858617. These are all reasonably close, about 0.08, so we can have
reasonable confidence that the maximal Lyapunov exponent is positive, which implies sensitive
dependence on initial conditions (and chaos, assuming there is an attractor like the one suggested
by the plot on p. (4)).
(b)
See p. (6) - plot of iterate of the Poincaré map on the yz-plane.
See p. (7) - Same iterates, zoomed in. Appear to lie on a curve.
See p. (8) - trajectory, section, initial value, three iterates.
(c)
See p. (9) - Ruelle plot, looks like parabola.
(d)
i) With the default initial condition and Total = 150, the trajectory looks chaotic and XPP
computes the Lyapunov exponent λ = 0.356721 > 0. Thus at first sight, it looks like a chaotic
1
attractor. However, increasing Total or choosing another initial condition reveals that there is
not a chaotic attractor, but there is a stable period-5 cycle with negative Lyapunov exponent
λ ≈ −0.3. (This shows that what appears to be a choatic attractor in a simulation might in fact
be a transient - some care needs to be taken.)
ii) Substitute y = αx + β into yn+1 = ã + b̃(yn − c̃)2 , where (ã, b̃, c̃) = (−6, 0.65, −3.5). This
gives:
αxn+1 + β = ã + b̃(αxn + β − c̃)2
Solving for xn+1 gives:
i
1h 2
b̃(β − 2c̃β + c̃2 ) + ã − β
α
If we compare this to the logistic map (??), we get
xn+1 = b̃αx2n + 2b̃(β − c̃)xn +
b̃α = −r,
2b̃(β − c̃) = r,
2
2
b̃(β − 2c̃β + c̃ ) + ã − β = 0
(2)
(3)
(4)
Solving (4) for β, with ã = −6, b̃ = 0.65, c̃ = −3.5 gives us
q
2
2
β = 12b̃ 2b̃c̃ + 1 ± (2b̃c̃ + 1) − 4b̃(b̃c̃ + ã)
= −0.624144096, −4.837394452
and then using (3), we get r = 3.738612788, −1.738612788. Rejecting the negative value of r,
we have β = −0.624144096. Then, from (2), we have α = −5.751711982 and the quadratic map is
transformed into the logistic map with r = 3.738612788. For this value of r, the logistic map does
not have a chaotic attractor, it has a stable period-5 cycle. (However a sufficiently accurate value
of r must be used, and a large enough number of iterations must be used. Otherwise it will look
like a choatic attractor.)
(e)
Numerical approximations are used to solve the Rössler system (1) and construct the Poincaré
map considered in (b).
Visually, the Poincaré mape iterates appear to lie on a one-dimensional curve. Approximating
the two-dimensional Poincareé map by a one-dimensional map, we make a Ruelle plot to find the
one dimensional map.
From the Ruelle plot, we approximate the one-dimensional map as yn+1 = −6 + 0.65(yn + 3.5).
This last map can be transformed into the logistic map exactly with a certain value of r, but
round-off error gives an approximate value of r.
It is not really clear whether the attractor seen in part (a), even if it truly is an attractor
as it seems to be, is chaotic or not. The appearance of computed trajectories and Lyapunov
exponent calculations (approximations) seem to include a chaotic attractor. The one dimensional
map approximations include a stable period-5 cycle. Whether or not there is chaos is sensitive
2
to parameter values and approximations, and better approximations may show different results.
For example, what if a different step size in the ODE solver was used, or a different numerical
method?. What if the values of ã, b̃ c̃ were slightly different.
3
8
6
0
-6
6
-6
2
0
-2
-4
-6
0
50
100
150
200
z
8
6
4
2
0
-7
-6
-5
-4
-3
y
-2
-1
0
z
-0.02
-0.03
-0.04
-0.05
-0.06
-6
-5
-4
-3
y
-2
-1
y(n+1)
0
-1
-2
-3
-4
-5
-6
-7
-7
-6
-5
-4
-3
y(n)
-2
-1
0