Population Dynamics - part 4

Fourth in our series of problems on population dynamics for advanced students.

Age

16 to 18

Challenge level

Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving

Being curious Being resourceful Being resilient Being collaborative

The Logistic Map

The logistic map is the discrete case of the logistic equation, given by: $\frac {\mathrm{d}y}{\mathrm{d}t}=ry(1-\frac{y}{Y})$

We then approximate to deduce the discrete case:$$ \begin{align*} \frac{y_{n+1}-y_n}{\Delta t} &\approx ry_n\left(1-\frac{y_n}{Y}\right) \\ y_{n+1} &\approx r \Delta t y_n \left(1-\frac{y_n}{Y}\right)+y_n \\ y_{n+1}&=(1+r \Delta t)y_n-\frac {r\Delta t}{Y}{(y_n)}^2 \\ y_{n+1}&=(1+r \Delta t)y_n\Bigg( 1-\bigg(\frac{r\Delta t}{1+r\Delta t}\bigg)\frac{y_n}{Y}\Bigg) \end{align*} $$

Let $\lambda=1+r\Delta t$ and $x_n=\frac {r\Delta t}{1+r \Delta t} \frac {y_n}{Y}$ . Then our equation becomes: $$x_{n+1}=\lambda x_n (1-x_n) $$ This is the logistic map. We can also think of it as a function $x_{n+1}=f(x_n)$.

Finding Equilibrium Points

Question: A fixed point implies $x_{n+1}=x_n$ . Find the fixed points by solving $$ \lambda x_n (1-x_n) = x_n $$ To determine the stability of these points, we are going to find the stability, by investigating the function for values nearby the equilibrium points.

Start by supposing that $x_n=X$ is a fixed point. This means that $f(X)=X$.

To find a value near the equilibrium point, let $x_n=X+\epsilon{_n}$ where $\epsilon_n < < 1$. Then using the Taylor expansion: $$ \begin{align*} x_{n+1}&=f(x_n) \\ X+\epsilon_{n+1} &= f(X+\epsilon_n) \\ &=f(X)+\epsilon_n f'(X)+... \end{align*}$$

We neglect the higher-order terms to get: $$X+\epsilon_{n+1}=f(X)+\epsilon_n f'(X)$$ Now from above we saw that $f(X)=X$ , so we can simplify to get: $$\epsilon_{n+1} \approx f'(X) \epsilon_n$$ A fixed point, X, is then stable if: $\Bigg|\frac{\epsilon_{n+1}}{\epsilon_n}\Bigg | =\Bigg |f'(X)\Bigg | < 1$

Question: Given that $f'(x)=\lambda-2\lambda x$ , find the stability of the fixed points $x_n=0$ and $x_n=1-\frac{1}{\lambda}$