Modelling Population Growth and Decay

# Modelling Population Growth and Decay

One important application of differential equations arises in modelling population growth or decay within a species. Let $y = y(t)$ denote the population size of a species at time $t$. If the rate of change of this population is proportional by some constant $r$ (where $r > 0$ implies that the population is growing and $r < 0$ implies that the population is decaying) then we can model the change in population with the following differential equation:

(1)
\begin{align} \quad \frac{dy}{dt} = ry \end{align}

Suppose that the initial population at time $t = 0$ is $y_0$, that is $y(0) = y_0$ . We can solve the differential equation above by either using the method of integrating factors or as a separable equation. Let's use integrating factors. We first rewrite the differential equation above in the form:

(2)
\begin{align} \quad y' - ry = 0 \end{align}

Then $p(t) = -r$. Let $\mu (t) = e^{\int p(t) \: dt} = e^{\int -r \: dt} = e^{-rt}$ be our integrating factor. Multiplying both sides of this differential equation by $\mu (t)$ and we get that:

(3)
\begin{align} \quad \mu (t) y' - \mu (t) ry = 0 \\ \quad e^{-rt} y' - r e^{-rt} y = 0 \\ \quad \frac{d}{dt} \left ( e^{-rt} y \right ) = 0 \\ \quad e^{-rt} y = \int 0 \: dt \\ \quad e^{-rt} y = C \: dt \\ \quad y = \frac{C}{e^{-rt}} \\ \quad y = Ce^{rt} \end{align}

With the initial condition $y(0) = y_0$ we see that $C = y_0$ and so the solution to our differential equation is:

(4)
\begin{align} \quad y = y_0e^{rt} \end{align}

As we see, if $r > 0$ then the growth of the population exponentially increases and if $r < 0$ then the decay of the population exponentially decreases. Of course, if $r > 0$ then $\lim_{t \to \infty} y_0e^{rt} = \infty$. Realistically, as time increases, the population $y$ will not increase indefinitely as population dynamics are based off of many other variables that are bounded by limits. For example, a finite amount of space or food may impede on a population from growing indefinitely.

Our original assumption that the population growth/decay rate $r$ was constant is not realistic. Instead, the growth/decay of a population is also dependent on the population itself. We replace the constant $r$ with the function $h(y)$ which depends on the population $y$. Therefore, we have the new differential equation:

(5)
\begin{align} \quad \frac{dy}{dt} = h(y) y \end{align}

The choice of function $h(y)$ can be determined based on various conditions that affect the rate of growth/decay in a population. A generic choice of this function is $h(y) = (r - ay)$, $a > 0$, and where $r$ is the growth/decay rate from earlier. Note that as $y$ gets very large, $h(y)$ does not get too large. Substituting this into the differential equation and letting $K = \frac{r}{a}$ gives us:

(6)
\begin{align} \quad \frac{dy}{dt} = (r - ay)y \\ \quad \frac{dy}{dt} = r(1 - \frac{a}{r} y) y \\ \quad \frac{dy}{dt} = r \left (1 - \frac{y}{K} \right )y \end{align}

The constant $K$ is known as the Carrying Capacity for the population. Now, this differential equation is separable since it can be rewritten as:

(7)
\begin{align} \quad \frac{1}{\left (1 - \frac{y}{K} \right )y} \frac{dy}{dt}= r \\ \end{align}

We we will now solve this separable differential equation by integrating both sides:

(8)
\begin{align} \quad \frac{1}{\left (1 - \frac{y}{K} \right )y} \: dy = r \: dt \\ \quad \int \frac{1}{\left (1 - \frac{y}{K} \right )y} \: dy = \int r \: dt \\ \quad \int \left [ \frac{1}{y} + \frac{K^{-1}}{1 - yK^{-1}} \right ] \: dy = rt + C \\ \quad \ln \mid y \mid - \ln \biggr \rvert 1 - \frac{y}{K} \biggr \rvert = rt + C \\ \quad e^{\ln y - \ln \left ( 1 - \frac{y}{K} \right ) } = e^{rt + C} \\ \quad \frac{y}{1 - \frac{y}{K}} = De^{rt} \end{align}

If we plug in our initial condition $y(0) = y_0$ we get that:

(9)
\begin{align} \quad D = \frac{y_0}{1 - \frac{y_0}{K}} \end{align}

When we plug this into our differential equation and isolated for $y$ we get that:

(10)
\begin{align} \quad y = \frac{y_0 K}{y_0 + (K - y_0)e^{-rt}} \end{align}

The solution to this initial value problem models population growth in constraining the population growth as it approaches the carrying capacity.