An exponential growth model

During the COVID-19 pandemic, there has been a lot of talk about infection rates and mathematical models.

The problem I chose for this article is:
Find a simple mathematical model to describe the growth of the number of people infected with a virus.

Introduction #

The model I describe here is the least sophisticated of all possible growth models. For all its simplicity, it is very useful because it predicts the behaviour of many growth patterns found in nature.

The resulting equation can follow directly from the definition of Euler's number, $e$[1]. But this article takes a slightly longer path. The purpose of this less trodden path is to illustrate a method often employed to discover mathematical models of all kinds. This method builds on the knowledge needed to solve ordinary differential equations[2].

The method itself is described here as a process that consists of three steps.

• Step 1 starts with explaining the problem using some equations that describe different kinds of changes (i.e. the differences). Also, stipulate some initial values for the functions we seek.
• Step 2 is to solve the system of equations (i.e. find the antiderivatives).
• Step 3 is there to find concrete values for those integration constants that emerged from step 2.

Notational conveniences #

For modelling problems like these, it is convenient to use the symbol $t$ as the dependant variable. This makes sense because the changes typically happen over time. For example, the COVID-19 infections started on some particular day (where $t = 0$), and $t$ increases by $1$ for each day that goes by. Keep in mind that our data comes in discrete increments, but the growth of the infected population happens continuously throughout every day.

It is also typical to use the name $y$ for the function we want to solve. Note that this is a real-valued function, so we have $y : \mathbb R \mapsto \mathbb R$. We use $y'(t)$ (pronounced y-prime) to denote derivative of $y(t)$. And we usually omit the $(t)$ from the expressions to make them easier to read.

There is also an alternative way to write the derivative of a function, $f$, which happens to be quite handy for this particular problem.

$$\begin{array}{r}{f}^{\prime }=\frac{\text{d}f}{\text{d}t}\end{array}\text{\hspace{1em}(1)}$$

A final convenience is to use $c_1, c_2, c_3, \ldots$ for the constants of integration.

Step 1: Defining the problem #

For our growth function, $y$ assume that the growth rate is some constant $\rho$. Under this simplifying assumption, the number of new people infected is a constant fraction of the number of people that are already infected. In other words, the change in infections can be expressed as this difference function:
$$\begin{array}{r}{y}^{\prime }=\rho y\end{array}\text{\hspace{1em}(2)}$$

We also assume that we started with $1$ infected person. So we have this initial value:
$$\begin{array}{r}y(0)=1\end{array}\text{\hspace{1em}(3)}$$

We have two unknowns ($y$ and $y'$) and two equations; so we have enough information to proceed.

Step 2: Finding the antiderivatives #

We start by applying the form from Eq 1 to Eq. 2 and moving all the $y$ parts to the left and taking the other parts across to the right-hand side.

$$\begin{array}{r}\frac{\text{d}y}{\text{d}t}=\rho y\\ \frac{1}{y}\text{d}y=\rho \text{d}t\end{array}\text{\hspace{1em}(4)}$$

We note that

$$\begin{array}{r}\int \frac{1}{x}\text{d}x=\ln x+c_1\end{array}\text{\hspace{1em}(5)}$$

Using Eq 5 we can now proceed to integrate Eq 4

$$\begin{array}{r}\int \frac{1}{y}\text{d}y=\int \rho \text{d}t\\ \ln y+c_1=\rho t+c_2\\ \ln y=\rho t+c_2-c_1\\ \ln y=\rho t+c_3\end{array}$$

And solve for $y$
$$\begin{array}{r}{e}^{\ln y}=e^{\rho t+c_3}\\ y=e^{\rho t}{e}^{c_3}\\ y=c_4{e}^{\rho t}\end{array}\text{\hspace{1em}(6)}$$

Step 3: Resolving the constants #

We get the value for $c_4$ by substituting Eq 3 into Eq 6.

$$\begin{array}{r}y(t)=c_4{e}^{\rho t}\\ y(0)=c_4{e}^{\rho 0}\\ 1=c_4\end{array}$$

And finally replacing the $c_4$ in Eq 6, we end up with our answer:

$$\begin{array}{r}y(t)=e^{\rho t}\end{array}\text{\hspace{1em}(7)}$$

Conclusion #

We have derived the simplest model for the growth of the infected population. It is described by Eq 7.

In our model, the particulars of the growth from one infection event to the next is captured by the value of $\rho$.

This one parameter approach is a crude one, but because of the exponential nature of the equation, it covers a vast range of possibilities. In Fig 1 below you can see the impact that changes in $\rho$ has on the growth rate, even over this very small period of $5$ days.

Fig 1: Showing $y(x)$ for different values of $\rho$

References #

1. Euler's number - Wikipedia
2. Ordinary differential equations - Wikipedia
3. Antiderivatives - Wikipedia
4. Constants of integration

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