4.4 The logistic equation  (Page 2/12)

We use the variable $K$ to denote the carrying capacity. The growth rate is represented by the variable $r.$ Using these variables, we can define the logistic differential equation.

The logistic equation was first published by Pierre Verhulst in $1845.$ This differential equation can be coupled with the initial condition $P\left(0\right)=P_0$ to form an initial-value problem for $P\left(t\right).$

Suppose that the initial population is small relative to the carrying capacity. Then $\frac{P}{K}$ is small, possibly close to zero. Thus, the quantity in parentheses on the right-hand side of [link] is close to $1,$ and the right-hand side of this equation is close to $rP.$ If $r>0,$ then the population grows rapidly, resembling exponential growth.

However, as the population grows, the ratio $\frac{P}{K}$ also grows, because $K$ is constant. If the population remains below the carrying capacity, then $\frac{P}{K}$ is less than $1,$ so $1-\frac{P}{K}>0.$ Therefore the right-hand side of [link] is still positive, but the quantity in parentheses gets smaller, and the growth rate decreases as a result. If $P=K$ then the right-hand side is equal to zero, and the population does not change.

Now suppose that the population starts at a value higher than the carrying capacity. Then $\frac{P}{K}>1,$ and $1-\frac{P}{K}<0.$ Then the right-hand side of [link] is negative, and the population decreases. As long as $P>K,$ the population decreases. It never actually reaches $K$ because $\frac{dP}{dt}$ will get smaller and smaller, but the population approaches the carrying capacity as $t$ approaches infinity. This analysis can be represented visually by way of a phase line. A phase line    describes the general behavior of a solution to an autonomous differential equation, depending on the initial condition. For the case of a carrying capacity in the logistic equation, the phase line is as shown in [link] .

This phase line shows that when $P$ is less than zero or greater than $K,$ the population decreases over time. When $P$ is between $0$ and $K,$ the population increases over time.

Chapter opener: examining the carrying capacity of a deer population

Let’s consider the population of white-tailed deer ( Odocoileus virginianus ) in the state of Kentucky. The Kentucky Department of Fish and Wildlife Resources (KDFWR) sets guidelines for hunting and fishing in the state. Before the hunting season of $2004,$ it estimated a population of $\mathrm{900,000}$ deer. Johnson notes: “A deer population that has plenty to eat and is not hunted by humans or other predators will double every three years.” (George Johnson, “The Problem of Exploding Deer Populations Has No Attractive Solutions,” January $12,2001,$ accessed April 9, 2015, http://www.txtwriter.com/onscience/Articles/deerpops.html.) This observation corresponds to a rate of increase $r=\frac{\text{ln}\left(2\right)}{3}=0.2311,$ so the approximate growth rate is $23.11\text{\%}$ per year . (This assumes that the population grows exponentially, which is reasonable––at least in the short term––with plentiful food supply and no predators.) The KDFWR also reports deer population densities for $32$ counties in Kentucky, the average of which is approximately $27$ deer per square mile. Suppose this is the deer density for the whole state $(\mathrm{39,732}$ square miles). The carrying capacity $K$ is $\mathrm{39,732}$ square miles times $27$ deer per square mile, or $\mathrm{1,072,764}$ deer .

1. For this application, we have $P_0=\mathrm{900,000},K=\mathrm{1,072,764},$ and $r=0.2311.$ Substitute these values into [link] and form the initial-value problem.
2. Solve the initial-value problem from part a.
3. According to this model, what will be the population in $3$ years? Recall that the doubling time predicted by Johnson for the deer population was $3$ years. How do these values compare?
4. Suppose the population managed to reach $\mathrm{1,200,000}$ deer. What does the logistic equation predict will happen to the population in this scenario?

1. The initial value problem is
   $\frac{dP}{dt}=0.2311P\left(1-\frac{P}{\mathrm{1,072,764}}\right),P\left(0\right)=\mathrm{900,000.}$
2. The logistic equation is an autonomous differential equation, so we can use the method of separation of variables.
   Step 1: Setting the right-hand side equal to zero gives $P=0$ and $P=\mathrm{1,072,764}.$ This means that if the population starts at zero it will never change, and if it starts at the carrying capacity, it will never change.
   Step 2: Rewrite the differential equation and multiply both sides by:
   
   $\begin{array}{ccc}\hfill \frac{dP}{dt}& =\hfill & 0.2311P\left(\frac{\mathrm{1,072,764}-P}{\mathrm{1,072,764}}\right)\hfill \\ \hfill dP& =\hfill & 0.2311P\left(\frac{\mathrm{1,072,764}-P}{\mathrm{1,072,764}}\right)dt.\hfill \end{array}$
   
   Divide both sides by $P\left(\mathrm{1,072,764}-P\right)\text{:}$
   
   $\frac{dP}{P\left(\mathrm{1,072,764}-P\right)}=\frac{0.2311}{\mathrm{1,072,764}}dt.$
   
   Step 3: Integrate both sides of the equation using partial fraction decomposition:
   
   $\begin{array}{ccc}\hfill {\displaystyle \int \frac{dP}{P\left(\mathrm{1,072,764}-P\right)}}& =\hfill & {\displaystyle \int \frac{0.2311}{\mathrm{1,072,764}}dt}\hfill \\ \hfill \frac{1}{\mathrm{1,072,764}}{\displaystyle \int \left(\frac{1}{P}+\frac{1}{\mathrm{1,072,764}-P}\right)dP}& =\hfill & \frac{0.2311t}{\mathrm{1,072,764}}+C\hfill \\ \hfill \frac{1}{\mathrm{1,072,764}}\left(\text{ln}\left|P\right|-\text{ln}\left|\mathrm{1,072,764}-P\right|\right)& =\hfill & \frac{0.2311t}{\mathrm{1,072,764}}+C.\hfill \end{array}$
   
   Step 4: Multiply both sides by $\mathrm{1,072,764}$ and use the quotient rule for logarithms:
   
   $\text{ln}\left|\frac{P}{\mathrm{1,072,764}-P}\right|=0.2311t+C_1.$
   
   Here $C_1=\mathrm{1,072,764}C.$ Next exponentiate both sides and eliminate the absolute value:
   
   $\begin{array}{ccc}\hfill e^{\text{ln}\left|\frac{P}{\mathrm{1,072,764}-P}\right|}& =\hfill & e^{0.2311t+C_1}\hfill \\ \hfill \left|\frac{P}{\mathrm{1,072,764}-P}\right|& =\hfill & C_2{e}^{0.2311t}\hfill \\ \hfill \frac{P}{\mathrm{1,072,764}-P}& =\hfill & C_2{e}^{0.2311t}.\hfill \end{array}$
   
   Here $C_2=e^{C_1}$ but after eliminating the absolute value, it can be negative as well. Now solve for:
   
   $\begin{array}{ccc}\hfill P& =\hfill & C_2{e}^{0.2311t}\left(\mathrm{1,072,764}-P\right).\hfill \\ \hfill P& =\hfill & \mathrm{1,072,764}{C}_2{e}^{0.2311t}-C_{2}P{e}^{0.2311t}\hfill \\ \hfill P+C_{2}P{e}^{0.2311t}& =\hfill & \mathrm{1,072,764}{C}_2{e}^{0.2311t}\hfill \\ \hfill P\left(1+C_2{e}^{0.2311t}\right)& =\hfill & \mathrm{1,072,764}{C}_2{e}^{0.2311t}\hfill \\ \hfill P\left(t\right)& =\hfill & \frac{\mathrm{1,072,764}{C}_2{e}^{0.2311t}}{1+C_2{e}^{0.2311t}}.\hfill \end{array}$
   
   Step 5: To determine the value of $C_2,$ it is actually easier to go back a couple of steps to where $C_2$ was defined. In particular, use the equation
   
   $\frac{P}{\mathrm{1,072,764}-P}=C_2{e}^{0.2311t}.$
   
   The initial condition is $P\left(0\right)=\mathrm{900,000}.$ Replace $P$ with $\mathrm{900,000}$ and $t$ with zero:
   
   $\begin{array}{ccc}\hfill \frac{P}{\mathrm{1,072,764}-P}& =\hfill & C_2{e}^{0.2311t}\hfill \\ \hfill \frac{\mathrm{900,000}}{\mathrm{1,072,764}-\mathrm{900,000}}& =\hfill & C_2{e}^{0.2311\left(0\right)}\hfill \\ \hfill \frac{\mathrm{900,000}}{\mathrm{172,764}}& =\hfill & C_2\hfill \\ \hfill C_2& =\hfill & \frac{\mathrm{25,000}}{\mathrm{4,799}}\approx 5.209.\hfill \end{array}$
   
   Therefore
   
   $\begin{array}{cc}\hfill P\left(t\right)& =\frac{\mathrm{1,072,764}\left(\frac{25000}{4799}\right)e^{0.2311t}}{1+\left(\frac{25000}{4799}\right)e^{0.2311t}}\hfill \\ & =\frac{\mathrm{1,072,764}\left(25000\right)e^{0.2311t}}{4799+25000e^{0.2311t}}.\hfill \end{array}$
   
   Dividing the numerator and denominator by $\mathrm{25,000}$ gives
   
   $P\left(t\right)=\frac{\mathrm{1,072,764}{e}^{0.2311t}}{0.19196+e^{0.2311t}}.$
   
   [link] is a graph of this equation.
   

   

3. Using this model we can predict the population in $3$ years.
   
   $P\left(3\right)=\frac{\mathrm{1,072,764}{e}^{0.2311\left(3\right)}}{0.19196+e^{0.2311(3)}}\approx \mathrm{978,830}\text{deer}$
   
   This is far short of twice the initial population of $\mathrm{900,000}.$ Remember that the doubling time is based on the assumption that the growth rate never changes, but the logistic model takes this possibility into account.
4. If the population reached $\mathrm{1,200,000}$ deer, then the new initial-value problem would be
   
   $\frac{dP}{dt}=0.2311P\left(1-\frac{P}{\mathrm{1,072,764}}\right),P\left(0\right)=\mathrm{1,200,000.}$
   
   The general solution to the differential equation would remain the same.
   
   $P\left(t\right)=\frac{\mathrm{1,072,764}{C}_2{e}^{0.2311t}}{1+C_2{e}^{0.2311t}}$
   
   To determine the value of the constant, return to the equation
   
   $\frac{P}{\mathrm{1,072,764}-P}=C_2{e}^{0.2311t}.$
   
   Substituting the values $t=0$ and $P=\mathrm{1,200,000},$ you get
   
   $\begin{array}{ccc}\hfill C_2{e}^{0.2311\left(0\right)}& =\hfill & \frac{\mathrm{1,200,000}}{\mathrm{1,072,764}-\mathrm{1,200,000}}\hfill \\ \hfill C_2& =\hfill & -\frac{\mathrm{100,000}}{\mathrm{10,603}}\approx -9.431.\hfill \end{array}$
   
   Therefore
   
   $\begin{array}{cc}\hfill P\left(t\right)& =\frac{\mathrm{1,072,764}{C}_2{e}^{0.2311t}}{1+C_2{e}^{0.2311t}}\hfill \\ & =\frac{\mathrm{1,072,764}\left(-\frac{\mathrm{100,000}}{\mathrm{10,603}}\right)e^{0.2311t}}{1+\left(-\frac{\mathrm{100,000}}{\mathrm{10,603}}\right)e^{0.2311t}}\hfill \\ & =-\frac{\mathrm{107,276,400,000}{e}^{0.2311t}}{\mathrm{100,000}{e}^{0.2311t}-\mathrm{10,603}}\hfill \\ & \approx \frac{\mathrm{10,117,551}{e}^{0.2311t}}{9.43129e^{0.2311t}-1}.\hfill \end{array}$
   
   This equation is graphed in [link] .
   

   

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