### Video Transcript

In this video, we will introduce
the logistic model for representing population growth. Weβll first recap the model used
for simple population growth and discuss why this model may not be appropriate in
certain circumstances. Weβll 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 weβll 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 ππ, 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 ππ‘. 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 ππ‘. 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
πΏ.

Weβll 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, πΏ, 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 πΏ. And proving this is beyond the
scope of this video. Multiplying both sides of this
equation by π gives dπ by dπ‘ equals ππ multiplied by one minus π over πΏ. And this is known as the logistic
differential equation for population growth.

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

So initially, when π is small
relative to πΏ, the exponential model for population growth is appropriate. However, as the population
approaches its carrying capacity, π over πΏ will tend to one. Therefore, one minus π over πΏ
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 πΏ will be greater than one, which means that one
minus π over πΏ 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 letβs consider how we can solve
this logistic equation. Although itβs more complicated than
the model for simple population growth, it is still a separable differential
equation. We can rewrite the right-hand side
as ππ multiplied by πΏ minus π over πΏ and then separate the variables, to give
πΏ over π multiplied by πΏ 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 πΏ over π
multiplied by πΏ minus π as π΄ over π plus π΅ over πΏ minus π. We can then multiply both sides of
this equation by π πΏ minus π to give πΏ equals π΄ multiplied by πΏ 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 πΏ is equal to πΏπ΄. And therefore, π΄ is equal to
one. When the population π is equal to
πΏ, we find that πΏ is equal to π΅πΏ. And therefore, π΅ is also equal to
one. So the fraction πΏ over π
multiplied by πΏ minus π can be expressed as the partial fractions one over π plus
one over πΏ minus π.

Substituting back into our
integral, weβre 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, weβd 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 πΏ minus π is equal to ππ‘. And weβve 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 πΏ in order for this model to be meaningful, πΏ
minus π is also greater than zero. So the absolute value of π is just
π and the absolute value of πΏ minus π is πΏ minus π. We therefore have the natural
logarithm of π minus the natural logarithm of πΏ minus π equals ππ‘ plus π.

Multiplying through by negative one
and then applying laws of logarithms gives that the natural logarithm of πΏ minus π
over π is equal to negative ππ‘ 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 πΏ minus π over π equals π to the power of negative ππ‘
plus π two. And using laws of exponents, we can
express this as π΄π to the power of negative ππ‘.

We can then rewrite the fraction on
the left-hand side as πΏ over π minus π over π, which simplifies to πΏ over π
minus one. And then add one to each side of
the equation to give πΏ over π is equal to π΄π to the negative ππ‘ plus one. Finally, we can multiply both sides
of the equation by π and then divide by π΄π to the negative ππ‘ plus one to give
π equals πΏ over one plus π΄π to the negative ππ‘. 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 weβre 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 πΏ 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 πΏ and the initial population π
nought. π΄ is equal to πΏ 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 ππ‘ will tend to zero. And so the population will tend to
πΏ over one, which is just πΏ, the carrying capacity of the population. Letβs now consider some examples of
this logistic model. In our first example, weβll 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.

Weβre told which type of population
growth model to use in this question. So we can quote the standard
logistic differential equation. Itβs dπ by dπ‘ equals ππ
multiplied by one minus π over πΏ, where π is the growth rate of the population
and πΏ is the carrying capacity. Weβve 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 thatβs ππ β multiplied by one minus π over 7500 β thatβs πΏ. Weβre only asked to write the
logistic differential equation. So thereβs no need for us to
attempt to solve this and weβre done.

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

Suppose a populationβs 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 ππ multiplied by one minus π over πΏ that its solution
is given by π equals πΏ over one plus π΄π to the negative ππ‘, where π΄ is equal
to πΏ minus π nought over π nought. Here π represents the growth rate
of the population. πΏ 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 πΏ is equal to 900. Weβre also told that π zero is
equal to 50. So we can fill in each of the
values in the general solution. Letβs work out π΄ first of all. π΄ is equal to πΏ minus π nought
over π nought. Thatβs 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 πΏ β thatβs 900 β
over one plus π΄ β thatβs 17 β π to the power of negative π β thatβs negative 0.7
β π‘. So we have our solution for π or
π of π‘. Itβs equal to 900 over one plus
17π to the power of negative 0.07π‘.

In our final example, weβll 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 πΏ over one plus π΄π to the
negative ππ‘, where π΄ is equal to πΏ minus π nought over π nought. Here πΏ is the carrying capacity of
the population, π nought is the initial population, and π is the growth rate of
the population.

Weβve been given some of this
information in the question. Weβre told that the initial
population, π nought, is 1000. And weβre told that the carrying
capacity, πΏ, is 10000. But we havenβt been given the
growth rate of the population. Instead, weβve been given another
pair of values for π and π‘. Weβre told that the population
after one year is 2500. Weβll be able to combine this
information with our values of πΏ and π nought in order to determine the growth
rate of the population.

First, we can work out the value of
the constant π΄. Itβs πΏ minus π nought over π
nought, 10000 minus 1000 over 1000, which is 9000 over 1000, which is nine. Substituting πΏ 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 ππ‘.

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 weβre asked for the population
after another three years, which means weβre 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 weβve started,
is 9000.

Letβs now summarise what weβve seen
in this video. Weβve 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 ππ multiplied by one minus π over πΏ,
where π is the growth rate and πΏ is the carrying capacity.

By separating the variables and
integrating, weβve found that the solution to the logistic differential equation
with an initial population of π nought is π of π‘ equals πΏ over one plus π΄π to
the negative ππ‘, where π΄ is equal to πΏ 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.