Logistic Map Bifurcation diagram

Question

So in the bifurcation diagram of the logistic map, there is period doubling from about $r=3$ to about $r=3.54409$. There are two fluctuation points between $r=3$ and $r=1+\sqrt{6}$. My question is, how would one obtain the $r$ value of $1+\sqrt{6}$?

Answer 1

The Logistic Map is given by:

$$f(x) = r~ x~ (1-x),~~ x \in [0,1]~,~ r > 0$$

Taking the 2-cycle, we solve:

$$f^2(x) = x$$

A plot of $f^2(x)$ and the line $x$ is:

This gives us the four roots:

$$x = 0, 1 - \dfrac{1}{r}, \dfrac{r+1 ~\pm ~ \sqrt{(r-3)(r+1)}}{2r}$$

The first two are repelling points and the second two are periodic points.

We now want to find the range of stability for those two points. For stability:

$$(f^2)'(x_0) = f'(x_1)f'(x_0)$$

where $x_1 = f(x_0)$.

When $x_0 = \dfrac{r+1 ~\pm ~ \sqrt{(r-3)(r+1)}}{2r} \implies (f^2)(x_0) = 4 + 2 r - r^2$.

Therefore, the 2-cycle is attracting when:

$$ |4 + 2 r - r^2| \lt 1$$

• For the positive absolute value we have $(4 + 2 r - r^2) \lt 1 \implies r \gt 3$
• For the negative absolute value we have $-(4 + 2 r - r^2) \lt 1 \implies r \lt 1 + \sqrt{6}$

This gives stability when:

$$ 3 \lt r \lt 1 + \sqrt{6}$$

If we plot a time series of $x_n$ versus $n$, for a value of $r = 3.4$, we can see this 2-cycle as:

However, if we do the same plot for $r = 3.5$, we now see a 4-cycle as:

If we plot the cobweb of each, we can see the stable 2-cycle, versus the 4-cycle as:

We can also plot the bifurcation diagram of the logistic map as: