Consider the following map in the interval $u\in[-1,1]$ ($U\in[-1,1]$ also)


It has 3 fixed points at $u=0,\pm 1$. If we compute the derivative of the map

$$\frac{dU}{du}=1-64u^{2}+5\times 64u^{4}-7\times 64 u^{6}+3\times 64u^{8}$$

we see that it is $1$ at the fixed points. So we cant determine whether the fixed points are stable/unstable linearly.

Numerically we can easily see that $0$ is a stable point and $\pm 1$ are unstable globally.

Is there a way to show/prove this analytically?

If I compute the Lyapunov exponent for several initial values of $u$ in the $-1,1]$ interval and plot it, together with the iteration map I get the following graph

enter image description here

The exponent is positive in certain ranges where the map seems to bounce around an infinite number of times. In other ranges the Lyapounov exponent is negative and orbits in this range seem to end in the fixed point 0. The fact the exponent is positive would seem to indicate the map is chaotic in those ranges but looking at the maps it seems it just bounces around in a square.

For example for $u_{0}=0.39$ the orbit goes to $u=0$ enter image description here but for $u_{0}=0.4$ it seems to move around in a square enter image description here

Is the map chaotic in any range? Or all the orbits for positive Lyapunov exponent are just periodic (2,4,6, etc cycles)? Is there a way to show that all orbits with negative Lyapunov exponent approach the fixed point $u=0$?


We have $f(u) < u$ for $0 < u < 1$ and $f(u) > u$ for $u > 1$, so the fixed point $1$ is unstable. Similarly $f(u) < u$ for $u < -1$ and $f(u) > u$ for $-1 < u < 0$ implies $-1$ is unstable. Since $|f(u)| < |u|$ for $u$ close to $0$, $0$ is stable.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.