# Lesson Video: The Logistic Model Mathematics • Higher Education

In this video, we will learn how to use the logistic differential equation to model situations where the growth of a quantity is limited by a carrying capacity.

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### Video Transcript

In this video, we will introduce the logistic model for representing population growth. Weβll first recap the model used for simple population growth and discuss why this model may not be appropriate in certain circumstances. Weβll then look at how we can develop this model further to take account of some of the practical constraints of population growth, such as the maximum population that can be supported. And weβll then consider a series of examples in which we work with this new model.

You should already be familiar with a model for simple population growth, in which the population increases at a rate proportional to the population itself. Such population growth can be modelled by the differential equation dπ by dπ‘ equals ππ, where π is the constant of proportionality or the rate at which the population is growing. By separating the variables in this differential equation and then integrating, we know that the general solution to this differential equation is of the form π equals π΄π to the power of ππ‘. And if we know the initial value of the population at time π‘ equals zero, π nought, we can find the particular solution to be π equals π nought π to the power of ππ‘. So we find that the population grows exponentially. And so the graph of the population over time would look a little like this.

However, we need to consider the practicality of this model because it implies that the population will continue to grow at an ever-increasing rate. In reality, this is unlikely to be possible as the environment in which the population live will not have unlimited resources. There will perhaps come a point where the population is too large to be supported by the available resources. And at this point, our model will cease to be accurate in predicting how this population will change over time. We therefore want to change our model to include two new assumptions.

Firstly, the relative growth of the population will decrease as the population itself increases. So whilst exponential growth might be appropriate initially, the rate at which the population grows will eventually slow down. The second assumption is that there is a maximum population that can be supported in a given environment. This is known as the carrying capacity. And we denote it using the letter πΏ.

Weβll look at the model that incorporates these assumptions in a moment. But the graph of the solution to this model would look a little something like this. The population initially grows in an exponential way. But this rate of growth then decreases. And finally, the population levels off at a level that is equal to the carrying capacity, πΏ, of the environment.

The simplest model for population growth, which incorporates both of the assumptions we just discussed, is as follows. One over π dπ by dπ‘ equals π multiplied by one minus π over πΏ. And proving this is beyond the scope of this video. Multiplying both sides of this equation by π gives dπ by dπ‘ equals ππ multiplied by one minus π over πΏ. And this is known as the logistic differential equation for population growth.

In this model, π represents the growth rate of the population and πΏ represents the carrying capacity. Notice that if the population is small compared to the carrying capacity, then π over πΏ will be close to zero. One minus π over πΏ will be close to one. And hence, dπ by dπ‘ will be approximately ππ. And we have this simple model for population growth.

So initially, when π is small relative to πΏ, the exponential model for population growth is appropriate. However, as the population approaches its carrying capacity, π over πΏ will tend to one. Therefore, one minus π over πΏ will tend to zero. And hence, the rate of change of the population, dπ by dπ‘, will also tend to zero. If the population exceeds the carrying capacity, then π over πΏ will be greater than one, which means that one minus π over πΏ will be negative. And so the rate of change of the population will also be negative, meaning that the population is decreasing, because it has exceeded the maximum population that the environment can support.

Now letβs consider how we can solve this logistic equation. Although itβs more complicated than the model for simple population growth, it is still a separable differential equation. We can rewrite the right-hand side as ππ multiplied by πΏ minus π over πΏ and then separate the variables, to give πΏ over π multiplied by πΏ minus π dπ is equal to π dπ‘. We then solve by integrating both sides of this equation.

Now the right-hand side will be straightforward. But to integrate the left-hand side, we need to use partial fractions. We can express πΏ over π multiplied by πΏ minus π as π΄ over π plus π΅ over πΏ minus π. We can then multiply both sides of this equation by π πΏ minus π to give πΏ equals π΄ multiplied by πΏ minus π plus π΅π. We can then substitute values of our population π to determine the values of the constants π΄ and π΅.

When π is equal to zero, we find that πΏ is equal to πΏπ΄. And therefore, π΄ is equal to one. When the population π is equal to πΏ, we find that πΏ is equal to π΅πΏ. And therefore, π΅ is also equal to one. So the fraction πΏ over π multiplied by πΏ minus π can be expressed as the partial fractions one over π plus one over πΏ minus π.

Substituting back into our integral, weβre now able to perform this integration. We recall that the integral of one over π₯ with respect to π₯ is equal to the natural logarithm of the absolute value of π₯ plus some constant of integration π. And if we were to use the method of substitution, weβd see that the integral of one over π minus π₯ with respect to π₯ for some constant π is equal to negative the natural logarithm of the absolute value of π minus π₯ plus some constant π.

So we have that the natural logarithm of the absolute value of π minus the natural logarithm of the absolute value of πΏ minus π is equal to ππ‘. And weβve just included one constant of integration π on the right-hand side.

Now the population π is greater than zero. And as we need the population to be less than the carrying capacity πΏ in order for this model to be meaningful, πΏ minus π is also greater than zero. So the absolute value of π is just π and the absolute value of πΏ minus π is πΏ minus π. We therefore have the natural logarithm of π minus the natural logarithm of πΏ minus π equals ππ‘ plus π.

Multiplying through by negative one and then applying laws of logarithms gives that the natural logarithm of πΏ minus π over π is equal to negative ππ‘ plus some constant π two. We can then raise π to the power of each side, knowing that this will cancel out the natural logarithm on the left-hand side, giving πΏ minus π over π equals π to the power of negative ππ‘ plus π two. And using laws of exponents, we can express this as π΄π to the power of negative ππ‘.

We can then rewrite the fraction on the left-hand side as πΏ over π minus π over π, which simplifies to πΏ over π minus one. And then add one to each side of the equation to give πΏ over π is equal to π΄π to the negative ππ‘ plus one. Finally, we can multiply both sides of the equation by π and then divide by π΄π to the negative ππ‘ plus one to give π equals πΏ over one plus π΄π to the negative ππ‘. And this is the general solution to the logistic model for population growth.

Now we can also determine the value of the constant π΄ if weβre told the population at time zero. If π equals some value π nought at π‘ equals zero, then we have the equation π nought equals πΏ over one plus π΄π to the power of zero. But π to the power of zero is just one. Rearranging this equation gives an expression for π΄ in terms of the carrying capacity πΏ and the initial population π nought. π΄ is equal to πΏ minus π nought over π nought.

And so we see the solution to the logistic model for population growth. We can quote this as a general result when answering questions on this topic. Notice that this time as π‘ approaches infinity, π to the power of negative ππ‘ will tend to zero. And so the population will tend to πΏ over one, which is just πΏ, the carrying capacity of the population. Letβs now consider some examples of this logistic model. In our first example, weβll see how to write a logistic differential equation from a physical description.

Suppose a population grows according to a logistic model with a carrying capacity of 7500 and π equals 0.006. Write the logistic differential equation for this information.

Weβre told which type of population growth model to use in this question. So we can quote the standard logistic differential equation. Itβs dπ by dπ‘ equals ππ multiplied by one minus π over πΏ, where π is the growth rate of the population and πΏ is the carrying capacity. Weβve been given both of these values in the question. So we just need to substitute them into the logistic differential equation.

We have then that dπ by dπ‘ is equal to 0.006 β thatβs ππ β multiplied by one minus π over 7500 β thatβs πΏ. Weβre only asked to write the logistic differential equation. So thereβs no need for us to attempt to solve this and weβre done.

In our next example, weβll see how to find the particular solution for a given logistic differential equation with an initial condition.

Suppose a populationβs growth is governed by the logistic equation dπ by dπ‘ equals 0.07π multiplied by one minus π over 900, where π of zero is equal to 50. Write the formula for π of π‘.

Writing π of π‘ means that we need to find the solution to this logistic equation. We can begin by writing down its general form. We know that for the logistic equation dπ by dπ‘ equals ππ multiplied by one minus π over πΏ that its solution is given by π equals πΏ over one plus π΄π to the negative ππ‘, where π΄ is equal to πΏ minus π nought over π nought. Here π represents the growth rate of the population. πΏ is the carrying capacity. And π nought is the initial population. We can identify each of these values from the information given in the question.

First, we see that π is equal to 0.07 and πΏ is equal to 900. Weβre also told that π zero is equal to 50. So we can fill in each of the values in the general solution. Letβs work out π΄ first of all. π΄ is equal to πΏ minus π nought over π nought. Thatβs 900 minus 50 over 50 or 850 over 50, which is equal to 17.

Now we can substitute into the general form of the solution. π is equal to πΏ β thatβs 900 β over one plus π΄ β thatβs 17 β π to the power of negative π β thatβs negative 0.7 β π‘. So we have our solution for π or π of π‘. Itβs equal to 900 over one plus 17π to the power of negative 0.07π‘.

In our final example, weβll see how to calculate the growth rate of the population and then calculate the population at a given time π‘.

Suppose a population grows according to a logistic model with an initial population of 1000 and a carrying capacity of 10000. If the population grows to 2500 after one year, what will the population be after another three years?

We know that the general solution to the logistic model is given by π of π‘ is equal to πΏ over one plus π΄π to the negative ππ‘, where π΄ is equal to πΏ minus π nought over π nought. Here πΏ is the carrying capacity of the population, π nought is the initial population, and π is the growth rate of the population.

Weβve been given some of this information in the question. Weβre told that the initial population, π nought, is 1000. And weβre told that the carrying capacity, πΏ, is 10000. But we havenβt been given the growth rate of the population. Instead, weβve been given another pair of values for π and π‘. Weβre told that the population after one year is 2500. Weβll be able to combine this information with our values of πΏ and π nought in order to determine the growth rate of the population.

First, we can work out the value of the constant π΄. Itβs πΏ minus π nought over π nought, 10000 minus 1000 over 1000, which is 9000 over 1000, which is nine. Substituting πΏ and π΄ into our model then, we have that π of π‘ is equal to 10000 over one plus nine π to the power of negative ππ‘.

Now we can use the population after one year in order to determine the value of π. Substituting 2500 for π and one for π‘, we have 2500 equals 10000 over one plus nine π to the negative π. To solve for π, we first multiply by one plus nine π to the negative π and then divide by 2500, giving one plus nine π to the negative π is equal to four. We can then subtract one and divide by nine, giving π to the negative π equals four minus one over nine, which is three-ninths or one-third.

We then take natural logarithms of each side, knowing that this will cancel out with the exponential on the left-hand side, to give negative π equals the natural logarithm of one-third. We can then multiply by negative one to give π equals negative the natural logarithm of one-third. And using laws of logarithms, this is equal to the natural logarithm of three. So we found the value of π, the growth rate of the population. Our model therefore becomes π of π‘ equals 10000 over one plus nine π to the negative π‘ multiplied by the natural logarithm of three.

Now weβre asked for the population after another three years, which means weβre looking for the population when π‘ is equal to four. So the final step is to substitute π‘ equals four into our model. We have then that π of four is equal to 10000 over one plus nine π to the negative four ln three. This actually works out very nicely, which you can see if you apply laws of logarithms in the denominator. We get 10000 over 10 over nine, which is 10000 times nine over 10, which is equal to 9000. So we find that the population after another three years, that is, the population four years after weβve started, is 9000.

Letβs now summarise what weβve seen in this video. Weβve introduced the logistic model, which can be used as a more realistic model of population growth to reflect the fact that populations usually cannot grow without limitation due to finite resources. The logistic differential equation for population growth is dπ by dπ‘ equals ππ multiplied by one minus π over πΏ, where π is the growth rate and πΏ is the carrying capacity.

By separating the variables and integrating, weβve found that the solution to the logistic differential equation with an initial population of π nought is π of π‘ equals πΏ over one plus π΄π to the negative ππ‘, where π΄ is equal to πΏ minus π nought over π nought. This produces a much more realistic model of population growth, where the population initially grows at an exponential rate. But then the rate of the growth slows down. And finally, the population levels off at the carrying capacity of the population.