0
$\begingroup$

I have been given a difference equation: $$x_{t+1} = a\cdot x_t\cdot(1-x_t)$$ and I want to find out the equilibrium points of the system. Could you recommend something to read in order to solve this problem?

The only thing I would know is to set $x_{t+1}= x_t$.

How do I solve this ? :(

  • 0
    You can find a lot of fascinating properties of this iteration by searching for "Feigenbaum map". It's one of the classic examples of how chaotic behavior emerges from regular behavior under continuous parameter changes.2011-10-03
  • 0
    @HenningMakholm yeah i already saw that. and i know that it has one (stable) solution for a<1 and so on ... but i want to proof it mathematically. How do i proof it ?2011-10-03
  • 0
    As to "how do I solve this," what's mysterious about a (barely) quadratic equation?2011-10-03
  • 0
    Maybe a further question : http://en.wikipedia.org/wiki/Logistic_map#Behavior_dependent_on_r .How do we know, these different behaivors for a ? There is also a proof, which i don't understand : http://mathworld.wolfram.com/LogisticMap.html2011-10-03
  • 0
    @AndréNicolas can you write down the quadratic equation, i have to solve ? ... maybe i'm just tooo blind to see it :(2011-10-03
  • 1
    If you have stability, $x_{t+1}=x_t$; use that to get your quadratic equation.2011-10-03
  • 1
    @fragant1996: taking your idea of setting $x_{t+1}=x_t$ and using $x$ for the variable, you have $x=ax(1-x)$ or $ax^2+(1-a)x=0$ with solutions $x=0,\frac{a-1}{a}$2011-10-03
  • 1
    $x=ax(1-x)$, or equivalently $ax^2 -(a-1)x=0$, with the solutions $x=0$ and $x=(a-1)/a$ (if $a \ne 0$).2011-10-03
  • 0
    @RossMillikan thanks .. now i understand2011-10-03
  • 1
    May I suggest that someone promote a comment to an answer? Brian, Ross, Andre?2011-10-04

1 Answers 1

1

$x=ax(1−x)$, or equivalently $ax^2−(a−1)x=0$, with the solutions $x=0$ and $x=(a−1)/a\ $ (if $a≠0$)

Thanks @André Nicolas and @Ross Millikan