Math Insight

We will explore how to use the program R to simulate the dynamics of the elk population. We assume that you already have R and the R package deSolve installed on your computer.

1. Download the R script run_logistic.R and execute it in an R command shell. (One way to execute the file is to enter source("run_logistic.R") when you are in the directory where you saved the file.) Once you have run the script file, it will have defined the function run_logistic, which will simulate the above logistic model.

   To check if it worked, simulate the logistic model with parameters $r=0.1$ and $K=5000$. Simulate it starting with many different initial conditions $p_0 \ge 0$, including some smaller and larger than $5000$. For example, to simulate it with initial condition $p_0=100$, run the commands

   results = run_logistic(r=0.1, K=5000, p0=100)
   plot(results)

   which save the results to the variable results and then plot a graph of the population size versus time. (See comments at beginning of script for more details on how to run it as well as how to look at the output results.)

   You should find that for most initial conditions $p_0 \ge 0$, the population size approaches the same value for large $t$. (Depending on the initial condition you choose, you may need to increase tmax beyond 100 to see the value, for example, by adding tmax=400 to your call of run_logistic.)

   Summarize your results.
   As long as $p_0 >$
   , the population size approaches
   .
   If we start with $p_0=$
   , then the population size stays at
   .
   The latter result makes sense because elk are being constantly beamed in from Mars, so if we start with no elk, the population size should still grow. elk can spontaneously appear anywhere, so if we start with no elk, the population size should still grow. the elk population size will always grow regardless if the current population size is positive. if we don't start with any elk, then no elk will be born and the population size won't grow.
   .

   The former result, though, might make you wonder. If you start with a population that contains one-tenth of one elk, i.e., with $p_0=0.1$, what does the logistic model predict should happen to the elk population? The one-tenth elk successfully reproduces and the population eventually recovers to 5000 elk. This is stupid; it doesn't even make sense to have a population of one-tenth elk. The one-tenth elk cannot find a mate, dies without reproducing, and the population goes extinct. A male one-twentieth elk and a female one-twentieth elk mate and reproduce so that the population recovers to 5000 elk. The population will go extinct.
   

   (Although we might be uncomfortable with the results when starting with $p_0=0.1$, in general, don't worry about having fractions of elks, such as 102.3 elk. Allowing fractional population sizes just makes the math easier.)

2. If $r=0.1$ and $K=5000$, and the initial population size is $p_0=100$, how long does it take for the population to reach $p(t)=4000$ elk?
   years. (Your answer should be within 1 year of the correct answer; examine the output of the simulation to determine correct time. If needed, you can specify a value of nt in your call of run_logistic to increase the number of points where it displays the results.)

   If you increase $r$ to 0.2, how long does it take for the population to reach 4000 elk?
   years.

   If you leave $r$ at $0.1$ but increase $K$ to 10000 and start with $p_0=200$, how long does it take for the population to reach 8000 elk?
   years.

We'll find equilibria for the logistic model, only we'll use $p$ rather than $x$.

1. To find the equilibria of the logistic growth model with $r=0.1$ and $K=5000$, i.e., $$\diff{p}{t} = 0.1 p \left(1 - \frac{p}{5000}\right),$$ we set $\diff{p}{t}=0$, i.e., we set
   $= 0$.

   The above equation for the equilibria should be a product of three factors set equal to zero. One of them should be 0.1. Divide both sides of the equation by 0.1 so that you have the product of just two factors equal to zero.
   $= 0$.

   Since a product of two factors is zero, you know one of those two factors must be zero, i.e.,
   
   $=0$ or
   $=0$.

   We therefore have two equilibria $p=$
   . (Enter in increasing order, separated by commas.)

   We often denote these equilibria by $p_{e}$. We could rewrite your above answer to state that the equilibria are $$p_{e} = ＿.$$

2. Simulate the logistic equation $\diff{p}{t} = 0.1 p \left(1 - \frac{p}{5000}\right)$ in R using the program run_logistic with initial condition $p_0$ set to each of the equilibrium values you calculated above. For either of the initial conditions, the plot of $p(t)$ versus time $t$ looks like a constantly decreasing curve a vertical line a constantly increasing curve a horizontal line a curve that changes directions
   , which means that the solution was decreasing a constant increasing both increasing and decreasing
   .
3. The two equilibria, however, have quite different properties. Let's start with the second equilibrium, the larger one, which is $p_e = $
   . Simulate the logistic model with initial conditions slightly smaller and slightly larger than this second equilibrium. What does the solution do? It moves toward the equilibrium. It is constant. It moves away from the equilibrium.
   

   If whenever you start at nearby initial conditions, the solution moves toward an equilibrium, we say the equilibrium is stable. If when starting at some nearby initial conditions, the solution moves away from the equilibrium, we say the equilibrium is unstable. The larger equilibrium of the logistic model is unstable neither stable nor unstable stable
   .

4. The first equilbrium, $p_e=$
   , though, has different properties. Simulate the logistic model with initial conditions slightly larger than the smaller equilibrium. What does the solution do? It moves toward the equilibrium. It is constant. It moves away from the equilibrium.
   (You can also simulate it with initial conditions smaller than the smaller equilibrium, but you'll probably get an error message. If you set tmax to something small like tmax=20 it should work. The problem is that the solution blows up all the way to $-\infty$, i.e., has a vertical asymptote, and the computer can't compute the solution.)

   The smaller equilibrium of the logistic model is stable neither stable nor unstable unstable
   .

5. We can repeat the procedure to find the equilibria of the logistic growth model with general parameters $r$ and $K$, i.e., $$\diff{p}{t} =r p \left(1 - \frac{p}{K}\right).$$ To find the equilibria, we set $\diff{p}{t}=0$, i.e., we set
   $= 0$.

   The above equation for the equilibria should be a product of three factors set equal to zero. Since we assume $r \ne 0$, you can divide both sides of the equation by $r$ so that you have the product of just two factors equal to zero.
   $= 0$.

   Since a product of two factors is zero, you know one of those two factors must be zero, i.e.,
   
   $=0$ or
   $=0$.

   We therefore have two equilibria $p_{eq}=$
   . (Enter in increasing order, separated by commas. The carrying capacity $K$ is a positive number.)

   Since we can't simulate the equation in R with general parameters $r$ and $K$, we can't use simulations to determine stability (though you could get a pretty good guess by simulating for many different values of $r$ and $K$). Instead, we need to determine stability analytically just from the equations.

We'll calculate the stability of the equilibria of the logistic model, where the state variable is $p$.

1. As with the previous problem, we start with the logistic growth model with $r=0.1$ and $K=5000$, i.e., $$\diff{p}{t} = 0.1 p \left(1 - \frac{p}{5000}\right).$$ The equilibria are $p_e= $
   (enter in increasing order, separated by commas).

   To apply the stability theorem, we must calculate the derivative of what function $f(p)$?
   $f(p) = $
   

   Calculate the derivative of that function.
   $f'(p) = $
   

   Evaluate the function at the equilibria to determine their stability.
   For the first equilibrium $f'(p_e) = f'($
   $)= $
   
   Since $f'(p_e)$ > < =
   
   , the equilibrium $p_e=$
   is neither stable nor unstable stable unstable
   .

   For the second equilibrium $f'(p_e) = f'($
   $)= $
   
   Since $f'(p_e)$ > < =
   
   , the equilibrium $p_e=$
   is neither stable nor unstable unstable stable
   .

2. Now we can calculate the stability of the equilibria of the logistic equation even with general parameters $r$ and $K$. Both $r$ and $K$ are positive numbers. $$\diff{p}{t} =r p \left(1 - \frac{p}{K}\right).$$ The equilibria are $p_e=$
   . (Enter in increasing order, separated by commas.)

   To apply the stability theorem, we must calculate the derivative of what function $f(p)$?
   $f(p) = $
   

   Calculate the derivative of that function.
   $f'(p) = $
   

   Evaluate the function at the equilibria to determine their stability.
   For the first equilibrium $f'(p_e) = f'($
   $)= $
   
   Since $f'(p_e)$ > = <
   
   , the equilibrium $p_e=$
   is stable neither stable nor unstable unstable
   .

   For the second equilibrium $f'(p_e) = f'($
   $)= $
   
   Since $f'(p_e)$ = > <
   
   , the equilibrium $p_e=$
   is neither stable nor unstable unstable stable
   .