Random experiments

The logistic map

Consider a petri dish with a few bacteria. They clone themselves at a constant rate over time, and in the same amount of time a fraction of them die. So, if we start with \(X_0\) bacteria, after a fixed time (a day, a week or any reasonable period of time), the population is \[ X_1 = a\,X_0, \] where \(a\) accounts for the death, newborn and surviving individuals. Of course, after another time step the population is \(X_2 = a\,X_1=a^2\,X_0\). Continuing in this way we see that the population at time step \(n\) is \(X_n = a^n\,X_0\), which are just iterations of \(X\mapsto a\,X\).

The evolution heavily depends on \(a\). Forget about \(a=0\), because everything dies in one time step. If \(0 < a < 1\) all the bacteria will eventually die, since in each step we have less individuals than before. In fact, the population exponentially goes down to \(0\). Mathematically, the population will never reach \(0\) unless it starts at \(0\), but the model says that the population is doomed. If \(a=1\), the population is constant over time. Finally, if \(a>1\) the population grows exponentially without limit, it will quickly reach the maximal amount of bacteria the petri dish can sustain. Since the environment cannot nourish all of them some would die from starvation.

To add this behavior to out model we replace the variable \(X\) by a the relative variable \(x\in[0,1]\) where \(1\) is the maximal population the petri dish can support. We consider a simple model (the logistic model) for the evolution of the population \[ f(x) = a\,x\,(1-x), \] where \(a\) is a parameter related with both birth and dead rates. Hence, starting from a population of \(x_0\), the population evolves as \(x_n = f(x_{n-1})\). The graph of the function \(f\) is drawn in blue below. The identity map \(x\mapsto x\) is plotted in red. The intersection of both are the fixed points of the system. You can also see the orbit (up to 100 iterations) of a initial condition \(x_0\).

\(a\):

\(x_0\):

The dynamics are quite wild! Why are the orbits wander around and not going to the fixed points? Because there are other attractors, the periodic points of \(f\), i.e., the fixed points of \(f^n\). Take a look to them... Specially interesting is the case $n=3$.

\(a\):

\(n\):

To visualize this we can plot a bifurcation diagram: the diagram below shows the attractors of the system (stable points where the iteration converges, drawn in black) for \(a\in[0,4]\).

Drag to zoom in, double click to zoom out.