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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 ? :(

Did
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fragant1996
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  • 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. – hmakholm left over Monica Oct 03 '11 at 22:49
  • @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 ? – fragant1996 Oct 03 '11 at 22:52
  • As to "how do I solve this," what's mysterious about a (barely) quadratic equation? – André Nicolas Oct 03 '11 at 23:02
  • 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.html – fragant1996 Oct 03 '11 at 23:06
  • @AndréNicolas can you write down the quadratic equation, i have to solve ? ... maybe i'm just tooo blind to see it :( – fragant1996 Oct 03 '11 at 23:08
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    If you have stability, $x_{t+1}=x_t$; use that to get your quadratic equation. – Brian M. Scott Oct 03 '11 at 23:11
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    @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 $x=0,\frac{a-1}{a}$ – Ross Millikan Oct 03 '11 at 23:11
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    $x=ax(1-x)$, or equivalently $ax^2 -(a-1)x=0$, with the solutions $x=0$ and $x=(a-1)/a$ (if $a \ne 0$). – André Nicolas Oct 03 '11 at 23:12
  • @RossMillikan thanks .. now i understand – fragant1996 Oct 03 '11 at 23:13
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    May I suggest that someone promote a comment to an answer? Brian, Ross, Andre? – Gerry Myerson Oct 04 '11 at 01:04

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$x=ax(1−x)$, 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

Andrew
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fragant1996
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