# Simple chaotic behavior in non-linear systems¶

In the last lecture, we discussed how floating point arithmetic can be subtle: its finite precision leads to the fact that even simple properties of elementary arithmetic, like the associativity of addition, don't hold:

In [1]:
a, b, c = 1, 1e-16, 1e-16
print(f"(a + b) + c = {(a + b) + c}")
print(f"a + (b + c) = {a + (b + c)}")

(a + b) + c = 1.0
a + (b + c) = 1.0000000000000002


This behavior can have serious implications in a variety of numerical work scenarios.

Consider the seemingly trivial problem of evaluating with a computer the expression

$$f(x) = r x (1-x)$$

where $r$ and $x$ are real numbers with $r \in [0,4]$ and $x \in (0,1)$. This expression can also be written in an algebraically equivalent form:

$$f_2(x) = rx - rx^2.$$

We will see, however, that when using a computer these two forms don't necessarily produce the same answer. Computers can not represent the real numbers (a mathematical abstraction with infinite precision) but instead must use finite-precision numbers that can fit in finite memory. The two expressions above can, therefore, lead to slightly different answers as the various (algebraically equivalent) operations are carried by the computer.

First a look at a few simple tests:

In [2]:
def f1(x): return r*x*(1-x)
def f2(x): return r*x - r*x**2

r = 1.9
x = 0.8
print('f1:', f1(x))
print('f2:', f2(x))

f1: 0.30399999999999994
f2: 0.3039999999999998

In [3]:
r = 3.9
x = 0.8
print('f1:', f1(x))
print('f2:', f2(x))

f1: 0.6239999999999999
f2: 0.6239999999999997


The difference is small but not zero:

In [4]:
print('difference:', (f1(x)-f2(x)))

difference: 2.220446049250313e-16


More importantly, this difference begins to accumulate as we perform the same operations over and over. Let's illustrate this behavior by using the formulas above iteratively, that is, by feeding the result of the evaluation back into the same formula:

$$x_{n+1} = f(x_n), n=0,1, \ldots$$

We can experiment with different values of $r$ and different starting points $x_0$ to observe the different results. We will simply build a python list that contains the results of three different (algebraically equivalent) forms of evaluating the above expression.

Exercise

Build a little script that computes the iteration of $f(x)$ using three different ways of writing the expression. Store your results and plot them using the plt.plot() function (the solution follows).

For completeness, we define three algebraically equivalent formulations:

In [5]:
def f1(x): return r*x*(1-x)
def f2(x): return r*x - r*x**2
def f3(x): return r*(x-x**2)


In order to see the difference between the initial behavior and the later evolution, let's declare two variables to control our plotting:

In [6]:
num_points = 100  # total number of points to compute
drop_points = 0  # don't display the first drop_points


Using these, we can get the following figure:

In [7]:
%matplotlib notebook
import matplotlib.pyplot as plt
import numpy as np

x0 = 0.55 # Any starting value
r  = 3.9  # Change this to see changes in behavior
fp = (r-1.0)/r
x1 = x2 = x3 = x0
data = []
data.append([x1,x2,x3])
for i in range(num_points):
x1 = f1(x1)
x2 = f2(x2)
x3 = f3(x3)
data.append([x1,x2,x3])

# Display the results
plt.figure()
plt.title('r=%1.1f' % r)
plt.axhline(fp, color='black')
plt.plot(data[drop_points:], '-o', markersize=4)
plt.show()


Exercise

Now, experiment with different values of $r$ as well as different starting points $x_0$. What do you see? What happens when $r$ is small (close to 0)? Experiment with these values of $r$: $1.9, 2.9, 3.1, 3.5, 3.9$ and think about the behavior of the system as you change $r$.

Once we've understood the basic pattern, let's try to think of the entire evolution of the system as a function of $r$. First, observe that a sequence generated by an iterative process of the form

$$x_{n+1} = f(x_n), n=0,1, \ldots$$

will stop producing new values if there is a certain $x^*$ such that

$$x^* = f(x^*).$$

This special $x^*$ is called a fixed point of the iterative process. It is easy to show that for our $f(x)$, the fixed point is

$$x^* = \frac{r-1}{r}$$

(in fact, that's the value plotted as a thin black line in the earlier script).

Exercise

Study whether the iteration converges to the fixed point or not by letting it run for each value of r for a few hundred points and discarding those, and then plotting the rest. Make a diagram with these plots as a function of r.

The following code is a simple solution:

In [8]:
def f(x, r):
return r*x*(1-x)

num_points = 500
drop_points = 50
rmin, rmax = 3.4, 4
xmin, xmax = 0, 1
x0 = 0.65
fig, ax = plt.subplots()
ax.set_xlim(rmin, rmax)
ax.set_ylim(xmin, xmax)
for r in np.linspace(rmin, rmax, 2000):
x = np.empty(num_points)
x[0] = x0
for n in range(1, num_points):
x[n] = f(x[n-1], r)
x = x[drop_points:]
rplot = r*np.ones_like(x)
ax.plot(rplot, x, 'b,')

plt.show()


Exercise

Can you relate the features of this figure to the behavior you saw in your earlier plots? Zoom in the region past $r=3$, what finer features do you see? Where is the fixed point we discussed earlier?