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Like a lot of people around the globe, I'm a little worried about the Coronavirus spreading and I thought rather than reading statistics from the news and media, I would take matters into my own hands and work things out (approximately).

I am trying to express an exponential virus growth in a population. So every day, the number of people infected doubles and once people have recovered they cannot re catch it. However, when I try and formulate the problem, I feel that my functions start to get recursive and that I am doing something wrong.

This is what I have tried so far and I am feeling confused by it.

p is the population of the country and k is the duration of viral infection in days.

$$ p = 6.6\times10^{7}\, $$ $$ k = 14 $$ $$ Y(d) = (1-(Y(d-1)/p)) 2^{d} - Y(d-k)\, $$

I thought I would see something in this kind of shape:

enter image description here

Please help.

schlebe
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apollowebdesigns
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    It is never bad to try by oneself, but in this case, you should now take a look at the numerous models already existing : keywords "epidemiology, models, SIR, compartments..." – Jean Marie Mar 13 '20 at 17:23
  • Thanks, I was not aware of these fields existing. I will research epidemiology some more. – apollowebdesigns Mar 13 '20 at 20:18
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    You might be interested in this YouTube video on "Exponential Growth and Epidemics" by 3blue1brown: https://www.youtube.com/watch?v=Kas0tIxDvrg – awkward Mar 14 '20 at 13:15

1 Answers1

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According to your assumptions your model should rather look like $$ Y(d) = 2Y(d-1) -Y(d-k)$$ What you describe doubles the healthy portion of the population "minus the recovered". Although the above does not take into account the population cap, namely that $Y\le p$. A more suitable model would be e.g. $$ Y(d)= 2\bigg(1-\frac{Y(d-1)}p\bigg)Y(d-1) - Y(d-k) $$ while $Y$ is small compared to $p$ this is basically doubling the infected, but as soon as Y assumes values comparable to p the rate of growth slows. The other option to the first equation would be $$ Y(d)=\begin{cases} 2Y(d-1) - Y(d-k) & \mbox{if }Y(d)< p \\ p &\mbox{if } 2Y(d-1)-Y(d-k)\ge p \\ Y(d-1)-Y(d-k) & \mbox{if }Y(d-1)=p, \mbox{and after reaching p} \end{cases} $$

Mick
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  • Thanks for the answer, I was wondering if there was a way of having this function re written just so that Y(d) on the left hand side of the equation was the only mention of Y(d) in the equation. I thought the right hand side could just be written with exponentials? – apollowebdesigns Mar 13 '20 at 20:16
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    If it is just $Y(d)=2Y(d-1)$ with $Y(0)=1$, then yes, $Y(d)=2^d$. – Mick Mar 16 '20 at 06:00