Logistic map (discrete dynamical system) vs logistic differential equation

Question

I have to roughly illustrate the logistic discrete dynamical system (as a model for population growth) to some non mathematics students. I'm not an analyst or an expert of dynamical systems.

Looking things up in the internet, I find the logistic map

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

with initial condition $x_0\in [0,1]$, and where $r\in[0,4]$ is a parameter (the condition $0\leq r \leq 4$ guarantees that $x_n$ doesn't escape the unit interval $[0,1]$ throughout the evolution of the system). Here $x_n$ represents the ratio between the population at time $n$ and the total population the environment is able to support.

I find also different behaviors according to the value of the parameter $r\in[0,4]$. For example,

1. For $0<r\leq 1$ there's extinction of the population.
2. For $1<r\leq 3$ the sequence tends to a stable equilibrium $x_\infty:=1-1/r$.
3. For $3<r\leq 1+\sqrt{6}$ there's convergence to a period-$2$ cycle.
4. For $1+\sqrt{6}<r\leq r^*$ (where $r^*$ is a certain constant) several bifurcations occur with limit a cycle of period that doubles as $r$ traverses that range.
5. For $r>r^*$ there's chaotic behavior.

I would expect that a similar span of different behaviors also happens for the logistic differential equation

$$\dot{x}(t)=rx(t)(1-x(t))$$

upon varying the parameter $r$. But on the internet I found no reference to anything like this. On the contrary, many pages care to solve the differential equation explicitely and illustrate the solution, which is the famous logistic function: the S-shaped increasing curve (depending on $r$) with a horizontal asymptote and an inflection point. It seems this solution is obtainable no matter what $r$ is. This looks only like case number 2. of the discrete dynamical system above.

So where are the analogous to cases 1.,3.,4. and 5. where the time is continuous??

Or am I misunderstanding some aspects of how a continuous-time dynamical system gets discretized?

Also,

Which correspondence is there between the $r$ of the discrete version and the $r$ of the continuous version?

Answer 1

Only some elements

A differential equation and it associated discretized problem may have different stability behaviors depending on the way you perform the discretization. See for example Wikipedia - discretization, paragraph Approximations.

Nevertheless, approximating $\dot{x}(t) = rx(t)(1-x(t))$ by $x_{n+1}=r x_n(1-x_n)$ is rather strange. It should be better discretized by $x_{n+1} -x_n =r x_n(1-x_n)$. Because usually $$\dot{x}(t) = \frac{x_{n+1}-x_n}{t_{n+1}-t_n} =x_{n+1}-x_n$$ if you consider equal discretization of the time.

Answer 2

In the logistic equation you get transforming it as Bernoulli equation $$ \frac{d}{dt}x(t)^{-1}=r(1-x(t)^{-1})\implies x(t)^{-1}=1+ce^{-rt}. $$ Now compare the values for $t$ and $t+h$, $$ x(t+h)^{-1}-1 = ce^{-rt-rh}=e^{-rh}(x(t)^{-1}-1). $$ For the sequence $x_k=x(kh)$ you get thus the recursion formula $$ x_{k+1}=\frac{x_k}{1-e^{-rh}+e^{-rh}x_k} $$ which looks rather different than the logistic map.

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Any sane discretization using the Euler method or similar would use a sufficiently small $h$, so that in the Euler forward discretization $$x_{k+1}=x_k(1+hr)-hrx_k^2.$$ To get to the normal form one would have to rescale $y_k=ax_k$, so that then $$y_{k+1}=y_k(1+hr)-\frac{hr}ay_k^2=(1+hr)y_k\left(1-\frac{hr}{a(1+hr)}y_k\right)$$ giving $a=\frac{hr}{1+hr}$. This is indeed case $2$ in the list, with equilibrium at $y_\infty=1-\frac1{1+hr}=\frac{hr}{1+hr}=a$ or translated back at $x_\infty=1$.