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How do I control the growth rate of this dynamical system?

$$ \left\{ \begin{aligned} &a_{n+1} = a_n + ka_n (2000 - a_n)\\ &a_0 = 4 \end{aligned} \right. $$

I experimented with multiple values of $k$ and i still can't find a way to control the growth.

(For additional context, is there a way such that this converges to a finite number and without this- ever exceeding $2000$?)

Egor Ivanov
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Lowell0803
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1 Answers1

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$$a_{n+1} = ka_n\left(\left(2000+\frac1 k \right) -a_n \right)$$ Define $x_n =\frac{k}{2000k+1} \cdot a_n$ then $$x_{n+1} = (2000k+1)x_n(1-x_n) \tag{1}$$

$(1)$ is the logistic map and converges to a finite value if and only if $(2000k+1) \in [0,3]$, that is $$-\frac{1}{2000} \le k \le \frac{1}{1000}$$

In this case, the sequence converges to $l \in [0,1)$ such that: $$l = \cases{0 &if $k \in \left[-\frac1 {2000}, 0 \right]$\\ 1-\frac{1}{2000k+1} &if $k \in \left[ 0,\frac{1}{1000} \right]$\\} $$

NN2
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  • why was this converted to xₙ and xₙ₊₁? and also, does this mean that if k falls within that range then the model will never reach 2000? (sorry if obvious questions, i'm barely introductory -level on dynamical systems) – Lowell0803 Feb 18 '24 at 04:25
  • @Lowell0803 It suffices to replace $a_n = x_n\cdot (2000k+1)/k$ and $a_{n+1} = x_{n+1}\cdot (2000k+1)/k$ in your initial equation, you will obtain the equation $(1)$. And as you asked in the question the condition of the convergence of the sequence (to 1 single finite number), we need to have $(2000k+1) \in [0,3]$ according to the wiki page. – NN2 Feb 18 '24 at 07:52
  • @Lowell0803 I added the limit of the sequence in the answer. – NN2 Feb 18 '24 at 07:59