Section3.3The Logistic Discrete-Time Dynamical System
Motivating Questions
What is the logistic dynamical system and what can it be used for?
How can we use calculus tools we’ve developed so far to understand how the logistic dynamical system behaves?
In Section 1.9, we analyzed two discrete-time dynamical systems, the lung model and the competing species model, using tools from Chapter 1. In this section, we will again analyze an important discrete-time dynamical system, the logistic DTDS, using our newly developed tools involving the derivative from Section 3.2.
In Subsection 1.9.2, we modeled population growth with the updating function \(p_{t+1} =rp_t\text{,}\) where \(r\) represents the per capita reproduction rate. We know from Section 1.7 that a solution function associated with this updating function is exponential. While exponential growth can be a good model for population growth at first, over time it becomes unrealistic since exponential growth does not reflect limited resources or a carrying capacity that we observe with real populations over time. One use case for the logistic dynamical system is as a population model that takes into account the carrying capacity of a population.
Warm-Up3.3.1.
Consider the updating function \(x_{t+1} = rx_t(1-x_t)\text{.}\)
Write the updating function rule. Identify which letters are parameters and which letters are variables.
Compute the derivative of the updating function rule.
Note: you can use the product rule, or do some algebraic simplification first and use the power and sum rules.
\(x^*=0\) is an equilibrium point (why?). Compute \(f'(0)\text{.}\)
According to the Stability Theorem, for what values of \(r\) is \(x^*=0\) stable?
Subsection3.3.1The logistic map
Let \(p_t\) be the population size in year \(t\text{,}\) and \(M\) the maximum possible population size. We would like a model that reflects our observation that population growth slows as it gets nearer to \(M\text{.}\) We can achieve this by letting the per capita reproduction rate change as the population changes, as opposed to remaining constant as in the exponential model. To that end, if \(r \gt 0\) is the maximum possible per capita reproduction rate, we set the per capita reproduction rate at year \(t\) to be
Note that this has the desired effect of decreasing the per capita reproduction rate when the population is large (close to \(M\)). The updating function for the population then becomes
\begin{align*}
p_{t+1} \amp=\text{per capita reproduction at time } t \cdot \text{population at time } t \\
\amp= r_t \cdot p_t\\
\amp=r\left(1- \dfrac{p_t}{M}\right) \cdot p_t\text{.}
\end{align*}
Thus, \(p_{t+1}=r\left(1- \dfrac{p_t}{M}\right) \cdot p_t\) is an updating function which models population size at year \(t\text{,}\) taking into account a maximum population size. Since we typically will not care about the actual value of \(M\text{,}\) but just the size of the population relative to \(M\text{,}\) we define the variable \(x_t=\dfrac{p_t}{M}\text{,}\) which gives the decimal percentage of the maximum population at year \(t\text{.}\) We can then re-write our updating function as
The logistic map is the recursive function \(x_{t+1}=rx_t(1-x_t)\text{.}\)
In the context of population growth, \(r\) is a positive constant representing the maximum possible per capita reproduction rate, and \(x_t\) is the percentage of the maximum population in year \(t\text{.}\)
Subsection3.3.2Equilibrium points and stability of the logistic map
We now have a new model for population growth in the logistic map, but we don’t know many details of how this system behaves outside of the general decrease in per capita reproduction with increase in population size that we built into the model. What are the equilibrium points of this system? How does the maximum per capita reproduction rate \(r\) impact their stability? What do the equilibrium points and their stability represent in the context of the model?
We know from Section 1.8 that we can find equilibrium points graphically or algebraically. Since our model contains a parameter and thus has no single graph to analyze, we will compute equilibrium points algebraically. Letting \(x_{t+1}=x_t = x^*\) in the updating function \(x_{t+1}=rx_t(1-x_t)\) gives \(x^*=rx^*(1-x^*)\text{,}\) or equivalently \(x^*= rx^*-r(x^*)^2\text{.}\) We now solve for \(x^*\text{:}\)
Thus we have two factors, \(x^*\) and \(rx^*+ (1-r)\) that multiply to \(0\text{,}\) meaning the zeros of both factors can give us an equilibrium point. The first equilibrium point is \(x^*=0\text{,}\) and the second equilibrium point is the solution to the equation \(rx^*+ (1-r)=0\text{,}\) which we solve as
Division by \(r\) is valid since \(r \gt 0\text{,}\) and the second equilibrium point is only valid for \(r \gt 1\) since \(x^*\) represents a percentage and cannot be negative.
We are now interested in classifying the stability of the equilibrium points \(x^*=0\) and \(x^*=\dfrac{r-1}{r}\text{.}\) Cob-webbing is not an ideal method, since our model contains a parameter and thus these equilibrium points represent many different systems (and graphs). Therefore, we will use the first derivative of the updating function rule to analyze stability using the Stability Theorem.
The updating function rule of \(x_{t+1} = rx_t(1-x_t)\) is \(f(x)=rx(1-x)\text{,}\) or equivalently \(f(x)=rx - rx^2\text{.}\) The first derivative is then \(f'(x)= r-2rx\text{.}\) For the equilibrium point \(x^*=0\text{,}\) we compute
By the Stability Theorem, \(x^*=0\) is stable when \(|r| \lt 1\text{,}\) which is equivalent to \(-1 \lt r \lt 1\text{.}\) Since \(r \gt 0\) in our model, \(x^*=0\) is stable when \(0 \lt r \lt 1\text{.}\)
For the equilibrium point \(x^*=\dfrac{r-1}{r}\text{,}\) we compute
Thus, \(x^* = \dfrac{r-1}{r}\) is stable when \(1 \lt r \lt 3\text{.}\)
We can summarize our analysis of the equilibrium points in the following table:
Table3.3.1.Stability of Equilibrium Points for the logistic map
\(x^*=0\)
\(x^*=\dfrac{r-1}{r}\)
Stable
\(0 \lt r \lt 1\)
\(1 \lt r \lt 3\)
Unstable
\(r \gt 1\)
\(0 \lt r \lt 1\) and \(r \gt 3\)
Note that our analysis does not say what happens when \(r=1\) or \(r=3\text{,}\) since these are cases in which the Stability Theorem is inconclusive. Cob-webbing or numerical methods may be used to determine the stability of each equilibrium point for these values of \(r\text{.}\)
Take a moment with the interactive below to visually verify our results regarding equilibrium points and stability in the logistic dynamical system.
Activity3.3.2.
The following questions are in reference to the equilibrium points of the logistic map discussed above.
Will the solution function ever oscillate around either equilibrium point? If so, for which value(s) of \(r\) does this occur?
In the context of modeling population growth, write a carefully worded sentence describing what it means for \(x^*=0\) to be a stable equilibrium point.
In the context of modeling population growth, write a carefully worded sentence describing what it means for \(x^*=0.5\) to be a stable equilibrium point.
There are many similar models to the logistic dynamical system with different equilibrium points and stability behaviors. In the final activity, you will perform a similar analysis as we have in this section on a DTDS whose updating function is a modified logistic map.
Activity3.3.3.
Consider a modified logistic dynamical system whose updating function is \(y_{t+1} = ry_t(1-(y_t)^2)\text{,}\) where \(r \gt 0\text{.}\)
Determine all of the positive equilibrium points of this system.
For each positive equilibrium point, determine the values of \(r\) which make that point stable.
What is the logistic dynamical system and what can it be used for?Answer.
The logistic dynamical system is the DTDS with updating function \(x_{t+1} = rx_t(1-x_t)\text{.}\) It can be used as a model for population growth which takes into account a maximum possible population, among other things.
Question3.3.3.
How can we use calculus tools we’ve developed so far to understand how the logistic dynamical system behaves?Answer.
Since the logistic DTDS contains a parameter \(r\text{,}\) we cannot look at a single graph to analyze its behavior. Instead, we can compute equilibrium points algebraically and compute the derivative of the updating function, which gives us answers in terms of the parameter \(r\text{.}\) We can then analyze these expressions to determine how the parameter \(r\) effects the system’s behavior. For example, we can use the Stability Theorem to determine which values of \(r\) make a given equilibrium point stable.
Exercises3.3.4Exercises
1.
Consider the logistic map with \(r=2\text{:}\)\(x_{t+1}=2x_t(1-x_t)\text{.}\) We will explore what the solution function looks like when \(x_0 = 0.01\text{.}\)
Based on our analysis in this section, what are the equilibrium points of this system? Classify each as stable or unstable.
Iterate the updating function to determine the first \(10\) points of the solution function (up to \(x_{9}\)). Use Desmos^{ 2 } to plot the points on a graph.
Looking at your graph, how is the solution function similar to an exponential solution function? How is it different?
How are the stable equilibrium point(s) of this system represented in the graph of the solution function?