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$f(x) = \begin{cases} \mu x &\text{ for } 0\leq x\leq 0.5 \\ \mu(1 - x) &\text{ for } 0.5 < x \leq 1 \end{cases}$

I am to find the orbits of period 2.

By using google I found an example I tried to follow, which is located here.

First, I find the fixed points of $f(x)$, which I denote by $x^*$: $$x^* = \begin{cases} 0 &\text{ for } 0\leq x\leq 0.5 \\ \frac{\mu}{\mu + 1} &\text{ for } 0.5 < x \leq 1 \end{cases}$$

Second, I find $f^2(x)=f(f(x))$: $$f^2(x) = \begin{cases} \mu ^2 x &\text{for $0 \leq x \leq 0.5$ } \\ \mu - \mu^2 + \mu^2x &\text{for $0.5 < x \leq 1$} \end{cases}$$

Third, I find fixed points $f^2(x)$: $$x^* = \begin{cases} 0 &\text{for $0 \leq x \leq 0.5$ } \\ \frac{\mu}{\mu + 1} &\text{for $0.5 < x \leq 1$} \end{cases}$$ Which are the same as in $f(x)$

I think that the fixed point $x^* = 0$ has an orbit of period $1 $ (correct me if I am wrong). So does that leave $x^* = \frac{\mu}{\mu + 1}$ to have an orbit of period $2$?

SOULed_Outt
  • 963

1 Answers1

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Notice,

$$ u - u^2 + u^2x = x $$ $$ u - u^2 = (1 - u^2)x $$ $$ x = \frac{u - u^2}{1 - u^2} = \frac{(\sqrt{u}-u)(\sqrt{u} + u)}{(1-u)(1 + u)} $$

This $x$ does not seem to look like your $x^*$.

In addition, you have to worry about whether the function moves outside the inequality it was originally in. I don't know what $u$ stands for, but say $x = 0.6$, and we get $u(1 - 0.6) = 0.4u$ and $u < 1$, then it would no longer be true that $0.5 < x < 1$.

Make sure you do some thinking about how $u$ affects the system. There is a variation of behaviour that is kind of interesting. Namely, think about

$$ u(1 - x) \leq\frac{1}{2} $$