Discrete Logistic Growth (cont'd)

A unique solution to a discrete map is an infinite sequence of points:

\[ P_{0}, P_{1}, P_{2}, P_{3}, \ldots \]

Let's simulate the equation using Desmos:

\[ P_{t} \ll K \ \ \textrm{(much smaller than)} \]

\[ \frac{P_{t}}{K} \ll 1 \]

\[ \left(1-\frac{P_{t}}{K}\right) \approx 1 \]

\[ P_{t+1} \approx (1+r)P_{t} \ \ (\textrm{Exponential growth}) \]

\[ \begin{align} P_{t+1} & = (1+r)P_{t} \\ & = (1+r)(1+r)P_{t-1}, \ \textrm{since} \ P_{t} = (1+r)P_{t-1} \\ & = (1+r)^2P_{t-1} \\ & = (1+r)^3P_{t-2}, \ \textrm{since} \ P_{t-1} = (1+r)P_{t-2} \\ & = (1+r)^?P_{0} \\ & = (1+r)^{t+1}P_{0} \end{align} \]

We have an explicit solution (exponential function)!!!

\[ P_{t} = F(t,P_{0}) = (1+r)^tP_{0} \]

What happens when \( t \rightarrow \infty \)? Depends on \( r \)!

Exponential decay when \( r < 0 \) and exponential growth when \( r > 0 \). What about \( r = 0 \)?

\[ P_{t} \approx \frac{K}{2} \Longrightarrow \left(1-\frac{P_{t}}{K}\right) \approx \frac{1}{2} \]

\[ P_{t+1} \approx P_{t} + \frac{r}{2}P_{t} = \left(1+\frac{r}{2}\right)P_{t} \]

\[ P_{t} \approx K \Longrightarrow \left(1-\frac{P_{t}}{K}\right) \approx 0 \]

\[ P_{t+1} \approx P_{t} + rP_{t}\cdot 0 = P_{t} = K \]

\( K \) is called a fixed point.

There are a few ways to study a discrete map like this.

• Simulate in time using a computer
  • Pro: Easy to do; Con: Not rigorous
• Explicitly solve the equation, i.e. convert to a form \( P_{t} = F(t, P_{0}) \)
  • Pro: If exists, know everything; Con: Very few models can be explicitly solved.
• Study the behavior of solutions (without solving them!)

For the discrete logistic model, even though it is relatively simple, there does not exist an explicit solution!

We've simulated using a computer - what about exploring behavior of solutions?? Dynamical Systems Theory!!

In dynamical systems theory, one of the most important questions is:

Question: What happens to solutions as \( t \rightarrow \infty \)?

We saw this in the simulations - there is a transient period (\( t < \infty \)) where the solution seems to converge to an equilibrium (\( t \rightarrow \infty \)).

The simplest example of equilibrium solutions are known as fixed points. Let's explore this idea through the simulations.

Definition: Fixed points of a discrete map are denoted by \( P_{\infty} \) and satisfy the relation

\[ P_{t} = P_{\infty}, \ \textrm{for all $t$} \]

This also means \( P_{t+1} = P_{t} = P_{\infty} \). So how do we find fixed points for a discrete map?

If the discrete map is given by

\[ P_{t+1} = F(P_{t}) \]

then fixed points satisfy the equation

\[ P_{t+1} = {\bf F(P_{t}) = P_{t}} \Rightarrow F(P_{t})-P_{t} = 0 \]

In other words, fixed points simultaneously solve (algebraically) the two equations:

\[ \begin{align} P_{t+1} & = F(P_{t}) \\ P_{t+1} & = P_{t} \end{align} \]

We can visual the dynamics AND these two equations in a cobweb plot.

To do this, we will (graphically) assign \( y = P_{t+1} \) and \( x = P_{t} \) to give

\[ \begin{align} y & = F(x) \\ y & = x \end{align} \]

Let's plot these two equations.

\[ \begin{align} P_{t+1} & = P_{t} + rP_{t}\left(1 - \frac{P_{t}}{K}\right) \\ P_{t+1} & = P_{t} \end{align} \]

\( \Longrightarrow \)

\[ \begin{align} P_{t} & = P_{t} + rP_{t}\left(1 - \frac{P_{t}}{K}\right) \\ 0 & = rP_{t}\left(1 - \frac{P_{t}}{K}\right) \\ \end{align} \]

\( \Longrightarrow \ P_{t} = 0 \ \) (or) \( \ 1 - \frac{P_{t}}{K} = 0 \ \Longrightarrow \ P_{t} = 0 \ \) (or) \( \ P_{t} = K \).

Question: How do we determine the stability of fixed points?