Logistic proportions and the first whole week

A flu virus is spreading in a community which has a fixed population of NN people. Scientists discover that at time tt weeks from the beginning, the rate of change of the number of people who are infected is proportional to the product of the number of people infected, II, and the number of people who are not infected.

(i) Write down a differential equation relating I,tI,t. mark

It is noted that at the beginning, 1% of the people in the community is infected with the virus. 12 weeks later, the proportion of people infected in the community has risen to 25%.

(ii) By solving the differential equation obtained in part (i), show that I=N99(33−t/12)+1I=\frac N{99(33^{-t/12})+1}. marks

(iii) Find the least number of weeks needed for the proportion of people infected to exceed 50%, giving your answer to the nearest integer. marks

(iv) Sketch the graph of II against tt. Explain, with justification, what happens to the number of people infected if this situation continues indefinitely. marks

(v) State a possible limitation of the model in part (i). mark

[2021/HCI/Prelim/I/10]

Total marks: 12

Solution

(i)

Forming the equation

Write down a differential equation for the infected number II in a population NN.

Hint

  1. Rate proportional to I×(N−I)I\times(N-I).

Answer · (i)

I′=kI(N−I)I\prime=kI(N-I), k>0k>0.

(ii)

Logistic solution

With 1% infected at the start and 25% after 12 weeks, show I=N99(33−t/12)+1I=\dfrac N{99(33^{-t/12})+1}.

Explore this idea

Hint

  1. Partial fractions: 1I(N−I)=1N(1I+1N−I)\frac1{I(N-I)}=\frac1N\left(\frac1I+\frac1{N-I}\right).

Answer · (ii)

I=N/[1+99(33)−t/12]I=N/[1+99(33)^{-t/12}].

Explore the idea

The proportion infected (P = I/N)

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of P at t = 00.01
Read the solution at t12
Try an example
P at t = 12
0.25
Rate dP/dt there
0.055
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

The proportion rises in an S-shape from 1% towards 100%.

Ideas to try
  1. Read when PP reaches 0.5.
  2. Move the starting fraction to 1 to see the fully infected equilibrium.
(iii)

First whole week

Find the least number of weeks for the proportion infected to exceed 50%.

Hint

  1. 99(33−t/12)+1<299(33^{-t/12})+1<2.

Answer · (iii)

1616 weeks.

(iv)

Long-term behaviour

Sketch II against tt and describe what happens if the situation continues indefinitely.

Hint

  1. 33−t/12→033^{-t/12}\to0.

Logistic infection proportion I/N versus time in weeks
Logistic infection proportion I/N versus time in weeksScroll across to see the full diagram.

Answer · (iv)

Rises sigmoidally from 0.01N0.01N toward NN, with inflection at I=N/2I=N/2.

(v)

Limitation

State a possible limitation of the model.

Hint

  1. Think about the assumptions: fixed population, no recovery or immunity.

Answer · (v)

Fixed population and no recovery/immunity limit the model.