!Dynamical Systems! Show the existence of a non-fixed periodic point of $q_\mu$ of period 3 for $\mu>4$ and $q_\mu=\mu x(1-x)$.

Question

I'm taking a course in dynamical systems and this question popped up in the course-litterature. In the litterature they show a similar example where we do the same thing exept for us instead of showing the existence of a non-fixed periodic point of period 3, we show the existence of a non-fixed periodic point of period 2. What they do in this example is the following: $$q_\mu([1/\mu,1/2])\supset[1-1/\mu,1]$$ $$q_\mu([1-1/\mu,1])\supset[0,1-1/\mu]\supset[1/\mu,1/2]$$ And so at this point we can clearly see that $q_\mu^2([1/\mu,1/2])\supset[1/\mu,1/2]$ and then we can see by the intermediate value theorem $q_\mu^2$ has a fixed point $p_2\in[1/\mu,1/2]$. Thus $p_2$ and $q_\mu(p_2)$ are non-fixed periodic points of period 2.

What i'm not getting here is how am I supposed to come up with an interval to start with? And if i'm not supposed to come up with an interval to start with, to show the existence of a non-fixed periodic point. Then how am I supposed to show the existence of such a point?

Repost of the question:

Show the existence of a non-fixed periodic point of $q_\mu$ of period 3 for $\mu>4$ and $q_\mu=\mu x(1-x)$.

Answer 1

It is not a simple problem. This is a sketch of the solution, which maybe you will study at the end of the course.

For convenience, fix $\mu>4$ and drop the subscript $\mu$.

Define $\Lambda=\{x\in[0,1]:f^n(x)\in[0,1]\ \forall n\}$. If $x\notin\Lambda$, then $f^n(x)\to-\infty$. The dynamics of $f$ are concentrated on $\Lambda$. It can be shown that $\Lambda$ is a Cantor set. The system $f\colon\Lambda\to\Lambda$ is equivalent to $\sigma\colon\Sigma\to\Sigma$, where $\Sigma$ is the set of sequences of $0$ 'sand $1$'s and $\sigma$ is the shift operator: $\sigma(a_1a_2a_3\dots)=a_2a_3\dots$ Finally, it is clear that $$ 100100100100\dots $$ is a periodic point of $\sigma$ of period $3$. By equivalence, $f$ has a periodic point of period $3$ in $\Lambda$.

Answer 2

By letting $p_{\mu}(x)=\mu x(1-x)$ we have that $$ q_{\mu}(x) = \frac{p_{\mu}(p_{\mu}(p_{\mu}(x))))-x}{x} $$ is a seventh-degree polynomial fulfilling $q_{\mu}(0)=\mu^3-1$, $q_{\mu}(1/\mu) = \mu-2$ and $$q_\mu(1/(2\mu))=\frac{1}{128} \left(-129-16 \mu +8 \mu ^2+64 \mu ^3-16 \mu ^4\right)<0,$$ hence there is a root of $q_\mu$ between $\frac{1}{2\mu}$ and $\frac{1}{\mu}$. It is not difficult to show that this root cannot be a root of $\frac{p_\mu(x)-x}{x}$ or $\frac{p_\mu(p_\mu(x))-x}{x}$, hence it is a periodic point of period $3$.

With a little refinement, Newton's method locates the first positive 3-periodic point around $\frac{\mu^2-3\mu-1}{\mu^2(\mu-3)}$.