Logistic Map

Logistic Map

We define the logistic map $f:[0,1]\to [0,1]$ by $f(x)=rx(1-x)r\in [0,4]$

Stable Fixed Point

To determine if a fixed point $x_0$ is stable (meaning nearby points are attracted to it over time), we evaluate its derivative $f^{\prime }(x)$.

• Stable: $|f^{\prime }(x_0)|<1$
• Unstable: $|f^{\prime }(x_0)|>1$

Fixed Points of Logistic Map

Parameter Range	Fixed Points (x=f(x))	Stability & Long-Term Behavior
$0\le r\le 1$	$x=0$	$x=0$ is the only stable fixed point. All populations eventually decrease to 0 (die off). The approach is fast when $r<1$ and slow when $r=1$.
$1<r\le 2$	$x=0$ (unstable) $x=1-\frac{1}{r}$ (stable)	Populations eventually move monotonically to the non-zero equilibrium $x=1-\frac{1}{r}$.
$2<r\le 3$	$x=0$ (unstable) $x=1-\frac{1}{r}$ (stable)	Populations still approach the non-zero equilibrium, but not monotonically. The values oscillate from one side of the equilibrium to the other as they settle.

Period Doubling and Chaos

When $r>3$, the previously stable fixed point $x=1-\frac{1}{r}$ becomes unstable. The system enters a phase of period doubling:

• $r>3$: The system develops a stable 2-cycle. The population oscillates back and forth between two distinct values ($f^2(x)=x$).
• As $r$ gets larger: The 2-cycle becomes unstable, giving way to a stable 4-cycle, then an 8-cycle, and so on.
• The “Chaotic” Regime: Past a certain threshold ($r_{\infty }$), the map becomes completely chaotic. It exhibits extreme sensitivity to initial conditions, commonly known as the “butterfly effect.”
• At $r=4$: The logistic map is considered “fully chaotic” and is dynamically equivalent to the simple period-doubling map.

Sharkovsky's Theorem

If the logistic map $f(x)=rx(1-x)$ has a 3-cycle, then it has cycles of all finite sizes. This relies on the Sharkovsky Order of positive integers, which organizes numbers as follows: $3▹5▹7▹9\cdots ▹2\cdot 3▹2\cdot 5\cdots ▹2^2\cdot 3\cdots ▹2^3▹2^2▹2▹1$ If a continuous function has a cycle of length $k$, and $k▹n$ in the Sharkovsky order, then the function is guaranteed to also have an orbit of length $n$. Because $3$ is at the very top of this hierarchy, the existence of a 3-cycle guarantees the existence of every other cycle length.