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after getting fixed points of this system :

$$x_{t+1} = a\cdot x_t\cdot(1-x_t)$$

i want to analyze the stability of the system for a = 0.9 , a = 2.1 and a = 3,58.

given :

An enlargement of the previous diagram around is illustrated above, with value of at which a -cycle first appears indicated by blue lines.

from the bifurcation diagram, i can deduce for a = 2.1 we have 1 stable state and 3.58 we have chaotic behaivor. And from wikipedia i know that a < 1 ...

the population will eventually die, independent of the initial population.

I don't understand the criteria for stability in this equation .
Can someone explain this ?

edit : for the record. i have the solutions, but i want to understand !

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The "why" for $a=0.9$ is I hope easy to understand, at least for the "reasonable" assumption $0 \le x_0 \le 1$. Note that $x_{t+1}=(0.9)x_t(1-x_t)<(0.9)x_t$, so $x_t$ approaches $0$ at least as fast as a geometric series with common ratio $0.9$. –  André Nicolas Oct 4 '11 at 0:34

1 Answer 1

up vote 2 down vote accepted

So, you know how to find fixed points. Stability is the question of whether, if you start close to a fixed point, you approach the fixed point; if so, the fixed point is stable.

Let's look at $a=2.1$, so $$x_{t+1}=2.1x_t(1-x_t)$$ and the fixed point is $x=1-(1/2.1)=11/21$. If you take $x_t$ close to 11/21, say, $x_t=(11/21)+\epsilon$, where it is understood that $\epsilon$ is small, then is $x_{t+1}$ closer to 11/21, or farther away? Well, $$x_{t+1}=(2.1)(11/21+\epsilon)(10/21-\epsilon)=(2.1)(110/441-\epsilon/21-\epsilon^2)=11/21-\epsilon/10-2.1\epsilon^2$$ For small $\epsilon$, we can ignore that term with $\epsilon^2$, and we see that $|x_{t+1}-11/21|$ is about one-tenth as big as $|x_t-11/21|$. That looks pretty stable to me.

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