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We have the equation

$$x_{n+1} = ax_n(1-x_n) - v_n$$

  • Why are there only fixed points for $(a-1)^2 - 4av_0 \geq 0$?

  • Show that if $ 1<a<4$, there are 2 fixed points with $0<p_1 < p_2 <1$

For the first problem, I was able to calculate that the fixed points would be given by:

$$ p_{1,2} = \dfrac{a-1}{2a} \pm \sqrt{(\dfrac{a-1}{2a})^2 - \dfrac{v_o}{a}}$$

But after that I'm stuck. The second problem I have no idea how to prove but I think that is because I don't understand the first problem.

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1 Answer 1

up vote 2 down vote accepted

Taking from where you stopped, consider the expression under the root and factor out $(2a)^2 = 4a^2$:

$$ \begin{split} p_\pm = \frac{a-1}{2a} \pm \sqrt{\left( \frac{a-1}{2a}\right)^2 - \frac{v_0}{a}} = \frac{a-1}{2a} \pm \frac{1}{2a} \sqrt{(a-1)^2 - 4av_0}, \end{split} $$ which is a real number provided $(a-1)^2 \geq 4av_0$, other wise the $\sqrt{\cdot}$ is complex.

2nd Problem For the last problem it suffices to prove that $$ 0 \leq \frac{a-1}{2a} \pm \frac{1}{2a} \sqrt{(a-1)^2 - 4av_0} \leq 1, $$

or equivalently that $$ 0 \leq a-1 \pm \sqrt{(a-1)^2 - 4av_0} \leq 2a. $$

Right inequality:

(assuming $1\leq a \leq 4$), certainly $a - 1 \leq a$ and also $$ \sqrt{(a-1)^2 - 4av_0} \leq \sqrt{(a-1)^2} = (a-1) < a, $$ so the sum (and surely the difference) should be no more than $a+a = 2a$.

Left inequality

Since $$ \sqrt{(a-1)^2 - 4av_0} \leq \sqrt{(a-1)^2} = a-1 $$ and so subtracting both sides from $a-1$ yields the result.

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Thank you very much! The first crucial step made everything easy, wouldn't have figured it out by myself though. –  iAm90offFrom90 Jun 19 '13 at 22:44
    
@iAm90offFrom90 No problem, glad to help you. –  gt6989b Jun 19 '13 at 22:45
    
Wait, how did $\dfrac{v_0}{a}$ change into $4av_0$? –  iAm90offFrom90 Jun 19 '13 at 23:00
    
@iAm90offFrom90 You are factoring out $1/(4a^2)$ from $v_0/a$ to get $$ \frac{v_0}{a} = \frac{v_0}{a} \frac{4a^2}{4a^2} = \frac{4av_0}{4a^2} $$ –  gt6989b Jun 19 '13 at 23:16

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