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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)
Question: A fixed point implies x_{n+1}=x_n . Find the fixed points by solving \lambda x_n (1-x_n) = x_n
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)
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}
Below are some graphs of the logistic map for different values of \lambda .
Case 1: \lambda< 1
Only fixed point is 0, which is stable:
Case 2: 1< \lambda < 2
Unstable fixed point at 0 and stable fixed point at 1-\frac{1}{\lambda}
Question: Can you find the stability for the case 2< \lambda < 3 ?
Below is a picture of some fantastic fractal behaviour which occurs for 3< \lambda< 4.
Question: Can you relate these values of \lambda to what would actually be occuring in a population of organisms?
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