Modelling the spread of disease

Thursday 7th May 2020

Despite all the danger and disruption, the Covid-19 outbreak is proving fertile ground for exploring and demonstrating the power of maths. We have heard a lot in the news about modelling and, amazingly, correct use of the word 'exponential' by journalists! I thought I would give a very brief outline of a possible activity that should be accessible to most A level Mathematics students, and incorporates the topics and themes of Differential Equations, Exponential Functions, Modelling and Use of Technology.

Let’s start with a very basic question: how could we model the spread of a disease amongst a population? What initial assumptions might we make that would allow us to start tackling this using A level Mathematics? Some possible assumptions could be:

  • We have a very large (effectively infinite) population.
  • Anyone who is infected remains so from then on and is equally infectious.

Putting this in more formal terms, we could call the number of infected people \(y\) and the number of people each infected person themselves infects per day \(r\). This leads to the differential equation $$\frac{dy}{dt}=ry$$ where \(t\) is measured in days (note there is a slight approximation here, as we are treating small finite changes as infinitesimals). By separating variables or otherwise, we find the solution $$y=Ae^{rt}$$ where \(A\) is the number of people infected at the point where \(t=0\). This is 'exponential growth', the 'steepness' of which depends on the parameter \(r\). Now, evaluating this model, we might note the following:

  • Exponential growth describes well the growth of the infected population in the early stages of an outbreak.
  • As time goes on, the model breaks down, most obviously because \(y\) will always eventually exceed the actual population.

So, refining the model slightly, let’s change our assumptions to these:

  • We have a fixed total population, \(L\).
  • People are either infected or uninfected.
  • Anyone who is uninfected is equally susceptible to infection.
  • Anyone who is infected remains so from then on and is equally infectious.

Again, formalising this a little, the daily rate of infection will be proportional both to the number of infected people, \(y\), and to the number of people who remain susceptible, \(L-y\), so we will be dealing with a differential equation of this form: $$\frac{dy}{dt}=ry(L-y)$$

Can you separate variables here and show that this leads to a solution of this form?

$$ y=\frac{L}{1+e^{-k(t-a)}} $$

(This is not trivial – you will need to use partial fractions to do the integration and a bit of work to get the solution into this form).

How does \(k\) relate to \(r\) and \(L\)?

This is called a 'logistic function'. It may not be obvious what the graph of this curve will look like, so this is a good opportunity to use dynamic graphing software such as GeoGebra to plot the curve, using sliders to vary the parameters. The red curve below shows its basic shape.

This model is by no means perfect, but it is at least getting close to what happens in real life. The shape may not look like the 'bell-shaped' curve that we have seen so much of in the news, however. This is because that is showing the number of new cases each day so, to see something like that, we need to differentiate the logistic function to obtain $$\frac{dy}{dt}=\frac{kLe^{-k(t-a)}}{(1+e^{-k(t-a)})^2}$$

The blue curve below shows the basic shape of this differentiated function. It is worth pointing out that, despite appearances, this is not a normal distribution curve!

ne-3-May20

Can you vary \(k\) and 'flatten the curve'?

This is probably as far as we can get with A level Mathematics. However, there are very obvious reasons why we would need to go much further to model the actual situation effectively. In particular, we have taken no account of the spatial distribution of populations, or the fact that most people become better and are both non-infectious and likely immune from then on. There is plenty of further reading available online if you search for it!

By Michael Gibson

Share this news article