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The Logistic Equation

Describe the concept of environmental carrying capacity in the logistic model of population growth.

Population Growth and Carrying Capacity

To model population growth using a differential equation, we first need to introduce some variables and relevant terms. The variable \(t.\) will represent time. The units of time can be hours, days, weeks, months, or even years. Any given problem must specify the units used in that particular problem. The variable \(P\) will represent population. Since the population varies over time, it is understood to be a function of time. Therefore we use the notation \(P(t)\) for the population as a function of time. If \(P(t)\) is a differentiable function, then the first derivative \(\frac{dP}{dt}\) represents the instantaneous rate of change of the population as a function of time.

In Exponential Growth and Decay, we studied the exponential growth and decay of populations and radioactive substances. An example of an exponential growth function is \(P(t)={P}_{0}{e}^{rt}.\) In this function, \(P(t)\) represents the population at time \(t,{P}_{0}\) represents the initial population (population at time \(t=0),\) and the constant \(r>0\) is called the growth rate. shows a graph of \(P(t)=100{e}^{0.03t}.\) Here \({P}_{0}=100\) and \(r=0.03.\)

We can verify that the function \(P(t)={P}_{0}{e}^{rt}\) satisfies the initial-value problem

\[\frac{dP}{dt}=rP,\ P(0)={P}_{0}.\]

This differential equation has an interesting interpretation. The left-hand side represents the rate at which the population increases (or decreases). The right-hand side is equal to a positive constant multiplied by the current population. Therefore the differential equation states that the rate at which the population increases is proportional to the population at that point in time. Furthermore, it states that the constant of proportionality never changes.

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.

Condensed: the full section is in OpenStax Calculus Volume 2.

Solving the Logistic Differential Equation

The logistic differential equation is an autonomous differential equation, so we can use separation of variables to find the general solution, as we just did in .

Step 1: Setting the right-hand side equal to zero leads to \(P=0\) and \(P=K\) as constant solutions. The first solution indicates that when there are no organisms present, the population will never grow. The second solution indicates that when the population starts at the carrying capacity, it will never change.

Step 2: Rewrite the differential equation in the form

\[\frac{dP}{dt}=\frac{rP(K-P)}{K}.\]

Then multiply both sides by \(dt\) and divide both sides by \(P(K-P).\) This leads to

\[\frac{dP}{P(K-P)}=\frac{r}{K}dt.\]

Multiply both sides of the equation by \(K\) and integrate:

\[\int \frac{K}{P(K-P)}dP=\int rdt.\]

The left-hand side of this equation can be integrated using partial fraction decomposition. We leave it to you to verify that

\[\frac{K}{P(K-P)}=\frac{1}{P}+\frac{1}{K-P}.\]

Then the equation becomes

\[\begin{array}{lll}\int \frac{1}{P}+\frac{1}{K-P}dP & = & \int rdt \\ \text{ln}|P|-\text{ln}|K-P| & = & rt+C \\ \text{ln}|\frac{P}{K-P}| & = & rt+C.\end{array}\]\[\begin{array}{lll}{e}^{\text{ln}|\frac{P}{K-P}|} & = & {e}^{rt+C} \\ |\frac{P}{K-P}| & = & {e}^{C}{e}^{rt}.\end{array}\]\[\frac{P}{K-P}={C}_{1}{e}^{rt}.\]\[\begin{array}{lll}P & = & {C}_{1}{e}^{rt}(K-P) \\ P & = & {C}_{1}K{e}^{rt}-{C}_{1}P{e}^{rt} \\ P+{C}_{1}P{e}^{rt} & = & {C}_{1}K{e}^{rt}.\end{array}\]\[\begin{array}{lll}P(1+{C}_{1}{e}^{rt}) & = & {C}_{1}K{e}^{rt} \\ P(t) & = & \frac{{C}_{1}K{e}^{rt}}{1+{C}_{1}{e}^{rt}}.\end{array}\]\[\begin{array}{lll}\frac{P}{K-P} & = & {C}_{1}{e}^{rt} \\ \frac{{P}_{0}}{K-{P}_{0}} & = & {C}_{1}{e}^{r(0)} \\ {C}_{1} & = & \frac{{P}_{0}}{K-{P}_{0}}.\end{array}\]\[P(t)=\frac{{C}_{1}K{e}^{rt}}{1+{C}_{1}{e}^{rt}}=\frac{\frac{{P}_{0}}{K-{P}_{0}}K{e}^{rt}}{1+\frac{{P}_{0}}{K-{P}_{0}}{e}^{rt}}\]\[\begin{array}{ll}P(t) & =\frac{\frac{{P}_{0}}{K-{P}_{0}}K{e}^{rt}}{1+\frac{{P}_{0}}{K-{P}_{0}}{e}^{rt}} \\ & =\frac{\frac{{P}_{0}}{K-{P}_{0}}K{e}^{rt}}{1+\frac{{P}_{0}}{K-{P}_{0}}{e}^{rt}}\cdot \frac{K-{P}_{0}}{K-{P}_{0}} \\ & =\frac{{P}_{0}K{e}^{rt}}{(K-{P}_{0})+{P}_{0}{e}^{rt}}.\end{array}\]\[P(t)=\frac{1,072,764{e}^{0.2311t}}{0.19196+{e}^{0.2311t}}.\]\[{r}^{2}{P}_{0}K(K-{P}_{0}){e}^{rt}((K-{P}_{0})-{P}_{0}{e}^{rt})=0.\]\[(K-{P}_{0})-{P}_{0}{e}^{rt}=0.\]

Condensed: the full section is in OpenStax Calculus Volume 2.

Key Concepts

  • When studying population functions, different assumptions, such as exponential growth, logistic growth, or threshold population, lead to different rates of growth.
  • The logistic differential equation incorporates the concept of a carrying capacity. This value is a limiting value on the population for any given environment.
  • The logistic differential equation can be solved for any positive growth rate, initial population, and carrying capacity.

Key Equations

Logistic differential equation and initial-value problem\(\frac{dP}{dt}=rP(1-\frac{P}{K}),\ P(0)={P}_{0}\)
Solution to the logistic differential equation/initial-value problem\(P(t)=\frac{{P}_{0}K{e}^{rt}}{(K-{P}_{0})+{P}_{0}{e}^{rt}}\)
Threshold population model\(\frac{dP}{dt}=\text{-}rP(1-\frac{P}{K})(1-\frac{P}{T})\)

The Logistic Equation

For the following problems, consider the logistic equation in the form \(P'=CP-{P}^{2}.\) Draw the directional field and find the stability of the equilibria.

The following problems consider the logistic equation with an added term for depletion, either through death or emigration.

It is more likely that the amount of fishing is governed by the current number of fish present, so instead of a constant number of fish being caught, the rate is proportional to the current number of fish present, with proportionality constant \(k,\) as

\(P'=0.4P(1-\frac{P}{10000})-kP.\)

The following problems add in a minimal threshold value for the species to survive, \(T,\) which changes the differential equation to \(P'(t)=rP(1-\frac{P}{K})(1-\frac{T}{P}).\)

The following questions consider the Gompertz equation, a modification for logistic growth, which is often used for modeling cancer growth, specifically the number of tumor cells.

Below is a table of the populations of whooping cranes in the wild from \(1940\ \text{to}\ 2000.\) The population rebounded from near extinction after conservation efforts began. The following problems consider applying population models to fit the data. Assume a carrying capacity of \(10,000\) cranes. Fit the data assuming years since \(1940\) (so your initial population at time \(0\) would be \(22\) cranes).

Year (years since conservation began)Whooping Crane Population
\(1940(0)\)\(22\)
\(1950(10)\)\(31\)
\(1960(20)\)\(36\)
\(1970(30)\)\(57\)
\(1980(40)\)\(91\)
\(1990(50)\)\(159\)
\(2000(60)\)\(256\)

さあ 計算機では解けませんが、計算可能です。下の一つを試してみましょう。もしくは自分で入力してください。

実践 (38)

まず紙で試してみてください。チェックするには答えを明らかにしてください。確認した答えは各ステップの解法で開きます。

  1. A population of rabbits in a meadow is observed to be \(200\) rabbits at time \(t=0.\) After a month, the rabbit population is observed to have increased by \(4\text{\%}.\) Using an initial population of \(200\) and a growth rate of \(0.04,\) with a carrying capacity of \(750\) rabbits,

    1. Write the logistic differential equation and initial condition for this model.
    2. Draw a slope field for this logistic differential equation, and sketch the solution corresponding to an initial population of \(200\) rabbits.
    3. Solve the initial-value problem for \(P(t).\)
    4. Use the solution to predict the population after \(1\) year.
    答えを明らかにしろ

    1. \(\frac{dP}{dt}=0.04(1-\frac{P}{750}),\ P(0)=200\)

    3. \(P(t)=\frac{3000{e}^{.04t}}{11+4{e}^{.04t}}\)

    4. After \(12\) months, the population will be \(P(12)\approx 278\) rabbits.

  2. Solve the logistic equation for \(C=10\) and an initial condition of \(P(0)=2.\)

    答えを明らかにしろ

    \(P=\frac{10{e}^{10x}}{{e}^{10x}+4}\)

  3. Solve the logistic equation for \(C=-10\) and an initial condition of \(P(0)=2.\)

  4. A population of deer inside a park has a carrying capacity of \(200\) and a growth rate of \(2\text{\%}.\) If the initial population is \(50\) deer, what is the population of deer at any given time?

    答えを明らかにしろ

    \(P(t)=\frac{10000{e}^{0.02t}}{150+50{e}^{0.02t}}\)

  5. A population of frogs in a pond has a growth rate of \(5\text{\%}.\) If the initial population is \(1000\) frogs and the carrying capacity is \(6000,\) what is the population of frogs at any given time?

  6. [T] Bacteria grow at a rate of \(20\text{\%}\) per hour in a petri dish. If there is initially one bacterium and a carrying capacity of \(1\) million cells, how long does it take to reach \(500,000\) cells?

    答えを明らかにしろ

    \(69\) hours \(5\) minutes

  7. [T] Rabbits in a park have an initial population of \(10\) and grow at a rate of \(4\text{\%}\) per year. If the carrying capacity is \(500,\) at what time does the population reach \(100\) rabbits?

  8. [T] Two monkeys are placed on an island. After \(5\) years, there are \(8\) monkeys, and the estimated carrying capacity is \(25\) monkeys. When does the population of monkeys reach \(16\) monkeys?

    答えを明らかにしろ

    \(8\) years \(11\) months

  9. [T] A butterfly sanctuary is built that can hold \(2000\) butterflies, and \(400\) butterflies are initially moved in. If after \(2\) months there are now \(800\) butterflies, when does the population get to \(1500\) butterflies?

  10. [T] The population of trout in a pond is given by \(P'=0.4P(1-\frac{P}{10000})-400,\) where \(400\) trout are caught per year. Use your calculator or computer software to draw a directional field and draw a few sample solutions. What do you expect for the behavior?

  11. In the preceding problem, what are the stabilities of the equilibria \(0<{P}_{1}<{P}_{2}?\)

  12. [T] For the preceding problem, use software to generate a directional field for the value \(f=400.\) What are the stabilities of the equilibria?

    答えを明らかにしろ

    \({P}_{1}\) semi-stable

  13. [T] For the preceding problems, use software to generate a directional field for the value \(f=600.\) What are the stabilities of the equilibria?

  14. [T] For the preceding problems, consider the case where a certain number of fish are added to the pond, or \(f=-200.\) What are the nonnegative equilibria and their stabilities?

    答えを明らかにしろ

    \({P}_{2}>0\) stable

  15. [T] For the previous fishing problem, draw a directional field assuming \(k=0.1.\) Draw some solutions that exhibit this behavior. What are the equilibria and what are their stabilities?

  16. [T] Use software or a calculator to draw directional fields for \(k=0.4.\) What are the nonnegative equilibria and their stabilities?

    答えを明らかにしろ

    \({P}_{1}=0\) is semi-stable

  17. [T] Use software or a calculator to draw directional fields for \(k=0.6.\) What are the equilibria and their stabilities?

  18. Solve this equation, assuming a value of \(k=0.05\) and an initial condition of \(2000\) fish.

    答えを明らかにしろ

    \(P\left(t\right)=\frac{3500}{\left(4+3{e}^{-035t}\right)}\)

  19. Solve this equation, assuming a value of \(k=0.05\) and an initial condition of \(5000\) fish.

  20. Draw the directional field of the threshold logistic equation, assuming \(K=10,r=0.1,T=2.\) When does the population survive? When does it go extinct?

  21. For the preceding problem, solve the logistic threshold equation, assuming the initial condition \(P(0)={P}_{0}.\)

  22. Bengal tigers in a conservation park have a carrying capacity of \(100\) and need a minimum of \(10\) to survive. If they grow in population at a rate of \(1\text{\%}\) per year, with an initial population of \(15\) tigers, solve for the number of tigers present.

    答えを明らかにしろ

    \(P(t)=\frac{850+500{e}^{0.009t}}{85+5{e}^{0.009t}}\)

  23. A forest containing ring-tailed lemurs in Madagascar has the potential to support \(5000\) individuals, and the lemur population grows at a rate of \(5\text{\%}\) per year. A minimum of \(500\) individuals is needed for the lemurs to survive. Given an initial population of \(600\) lemurs, solve for the population of lemurs.

  24. The population of Florida panthers in the southwestern tip of Florida has an estimated carrying capacity of \(250\) and grows at a rate of \(0.25\text{\%}\) per year and there must be \(25\) for the population to survive. With an initial population of \(30\) Florida panthers, how many years will it take to get the Florida panthers off the endangered species list (at least \(100)?\)

    答えを明らかにしろ

    \(13\) years months

  25. The Gompertz equation is given by \(P(t)'=\alpha \ \text{ln}\ (\frac{K}{P(t)})P(t).\) Draw the directional fields for this equation assuming all parameters are positive, and given that \(K=1.\)

  26. Assume that for a population, \(K=1000\) and \(\alpha =0.05.\) Draw the directional field associated with this differential equation and draw a few solutions. What is the behavior of the population?

  27. Solve the Gompertz equation for generic \(\alpha\) and \(K\) and \(P(0)={P}_{0}.\)

  28. [T] The Gompertz equation has been used to model tumor growth in the human body. Starting from one tumor cell on day \(1\) and assuming \(\alpha =0.1\) and a carrying capacity of \(10\) million cells, how long does it take to reach “detection” stage at \(5\) million cells?

    答えを明らかにしろ

    \(31.465\) days

  29. [T] It is estimated that the world human population reached \(3\) billion people in \(1959\) and \(6\) billion in \(1999.\) Assuming a carrying capacity of \(16\) billion humans, write and solve the differential equation for logistic growth, and determine what year the population reached \(7\) billion.

  30. [T] It is estimated that the world human population reached \(3\) billion people in \(1959\) and \(6\) billion in \(1999.\) Assuming a carrying capacity of \(16\) billion humans, write and solve the differential equation for Gompertz growth, and determine what year the population reached \(7\) billion. Was logistic growth or Gompertz growth more accurate, considering world population reached \(7\) billion on October \(31,2011?\)

    答えを明らかにしろ

    September \(2008\)

  31. Show that the population grows fastest when it reaches half the carrying capacity for the logistic equation \(P'=rP(1-\frac{P}{K}).\)

  32. When does population increase the fastest in the threshold logistic equation \(P'(t)=rP(1-\frac{P}{K})(1-\frac{T}{P})?\)

    答えを明らかにしろ

    \(\frac{K+T}{2}\)

  33. When does population increase the fastest for the Gompertz equation \(P(t)'=\alpha \ \text{ln}\ (\frac{K}{P(t)})P(t)?\)

  34. Find the equation and parameter \(r\) that best fit the data for the logistic equation.

    答えを明らかにしろ

    \(r=0.0405\)

  35. Find the equation and parameters \(r\) and \(T\) that best fit the data for the threshold logistic equation.

  36. Find the equation and parameter \(\alpha\) that best fit the data for the Gompertz equation.

    答えを明らかにしろ

    \(\alpha =0.0081\)

  37. Graph all three solutions and the data on the same graph. Which model appears to be most accurate?

  38. Using the three equations found in the previous problems, estimate the population in \(2010\) (year \(70\) after conservation). The real population measured at that time was \(437.\) Which model is most accurate?

    答えを明らかにしろ

    Logistic: \(361,\) Threshold: \(436,\) Gompertz: \(309.\)

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どうやって: The Logistic Equation

  1. Describe the concept of environmental carrying capacity in the logistic model of population growth.
  2. Draw a direction field for a logistic equation and interpret the solution curves.
  3. Solve a logistic equation and interpret the results.
  4. For this application, we have
  5. Solve the initial-value problem from part a.
  6. According to this model, what will be the population in
  7. Suppose the population managed to reach
  8. The initial value problem is

質問

What is a differential equation?

An equation whose unknown is a function, relating it to its own derivatives. "The rate of growth is proportional to the population" is y′ = ky, and solving it means finding y as a function of time.

Why does the solution have arbitrary constants?

Integrating loses information: many functions share the same derivative. An n-th order equation has n constants, fixed by n initial or boundary conditions.

このページの一部は、 OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). ここで簡略化して再説明する 誤りは我々の責任だ

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