Population Dynamics

7253 /www/nrich/html/content/id/7253/ 1 2011-11-28T15:03:44 <mdoxml version="1.0"> <h3>The Logistic Map</h3> The logistic map is the discrete case of the logistic <a href="http://nrich.maths.org/public/viewer.php?time=1311004466&obj_id=7048&part=index">equation</a>, 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 $x_{n+1}=f(x_n)$. <h3>Finding Equilibrium Points</h3> 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 <a href="http://mathworld.wolfram.com/TaylorSeries.html">Taylor expansion</a>: $$ \begin{align*} x_{n+1}&=f(x_n) \\ X+\epsilon_{n+1} &= f(X+\epsilon_n) \\ &=f(X)+\epsilon_n f'(X)+... \end{align*}$$ 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 $x_n=1-\frac{1}{\lambda}$ <h3>Different Cases of Stability</h3> 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: <mdo:image style="width: 442px; height: 193px;" src="Logistic%201.5.jpg" alt=""></mdo:image> Case 2: $1< \lambda < 2$ Unstable fixed point at 0 and stable fixed point at $1-\frac{1}{\lambda}$ <mdo:image style="width: 453px; height: 281px;" src="Logistic%202.5.jpg" alt=""></mdo:image> Question: Can you find the stability for the case $2< \lambda < 3$ ? Below is a picture of some fantastic fractal behaviour which occurs for $3< \lambda< 4$. <mdo:image style="width: 271px; height: 271px;" src="logisticfractal.png" alt=""></mdo:image> Question: Can you relate these values of $\lambda$ to what would actually be occuring in a population of organisms? </mdoxml> <mdoxml version="1.0"> So $x=0$ is stable for $-1< \lambda < 1$ and $x=1-\frac{1}{\lambda}$ for $1< \lambda < 3$ Oscillatory convergence to the stable fixed point. <mdo:image alt="" src="Logistic%203.5.jpg" style="width: 373px; height: 263px;"></mdo:image> </mdoxml> 2 4 0 0 0 0 1 Population Dynamics - part 4 Fourth in our series of problems on population dynamics for advanced students. Using, Applying and Reasoning about Mathematics Mathematical modelling Applications biology Pre-Calculus and Calculus Calculus generally Collections Population Dynamics