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Show that the logistic map $x_{n+1}=Ax_n(1-x_n)$ has stable $2$-cycle for all $A>3$.

I am a newbie in non-linear dynamics and chaos theory. All I have been able to find out about the map is that it has equilibrium points at $x^*=0$ and $x^*=1-\frac{1}{A}$. At $x^*=0$, the system attains asymptotic stability and for $1<A<3$, also the system is asymptotically stable. But how to geometrically/graphically analyze about the stable $2$-cycle for values of $A$ above $3$?

am_11235...
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I will show you here a numerical example of how to determine a 2-cycle in iterative maps. Showing that for $A>3$ 2-cycles are obtained should follow smoothly then.

Start from the basics: a 1-cycle is a fixed point. Meaning, that after a single iteration, the value obtained from the map is the same. Here, to find a fixed point, we solve:

(1): $x = Ax(1-x)$

and this is how you obtain the first two answers in your problem definition.

Next, a 2-cycle means that every two iterations you will repeat the same value. Hence, what you want is:

Define $f(x) = Ax(1-x)$, and the root solving problem is:

(2): $f(f(x)) = x$

you can solve that analytically (I believe) or show graphically that for $ A> 3 $ a point satisfying (2) without satisfying (1) exists. Actually, these will be pairs of points, I will let you deduce the reasoning for that.

I have written a small MATLAB script to illustrate the manner of thinking: Figure 1 In figure 1, intersection of yellow with blue solves (2). Intersection of yellow with orange solves (1).

Here, I have tried A = 5. And you see a pair of points solving equation (2) specifically.

The value of one of the points is 0.95 (approx). I have tried this value in the mapping and, voila, a two cycle is found: Figure 2

Snifkes
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  • I solved the equation $f(f(x))=x$ and got two ranges for its real roots. One is $A<-1$ and another $A>3$. So does that imply another stable $2$-cycle for all $A<-1$? – am_11235... Aug 21 '18 at 18:16
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    From numerical checks it seems like you are right. Although for stability of cycles you must check the derivative of $f^2(x)$ which will involve both roots found for (2). It is a linear test though so it may not be too conclusive... – Snifkes Aug 22 '18 at 14:18
  • Your plotted "2-cycle test" should not be stable. You used a more exact solution than 0.95 to keep the instability out of the first 20 cycles, right? – bobuhito Feb 04 '22 at 17:24
  • @bobuhito I did not use 0.95, I used a more accurate representation of the fixed point that is in the proximity of 0.95. Indeed, it is an unstable solution – Snifkes Feb 10 '22 at 22:50