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How to approximate the peak of an epidemic (depending on epidemiological parameters)?

Mathematics Asked on December 8, 2021

[I posted a follow-up question at MathOverflow.]


Numerical solutions of the SEIR equations (describing the spreading of an epidemic disease)

  • $dot{S} = – N$

  • $dot{E} = + N – E/lambda$

  • $dot{I} = + E/lambda – I/delta$

  • $dot{R} = + I/delta$

with

  • $N = beta I S / M$ = number of newly infected individuals

  • $beta = $ infection rate

  • $lambda = $ latence period

  • $delta = $ duration of infectiosity

  • $M = S + E + I + R = $ size of the population

yield characteristic peaks of the number $I$ of infectious individuals which can very coarsely be approximated by a Gauss curve

$$widetilde{I}(t) = I_0 e^{-(t-t_0)^2/sigma_1}$$

or slightly less coarsely by a skewed Gauss curve:

$$
widetilde{I}(t) =
begin{cases}
I_0 e^{-(t-t_0)^2/sigma_1} text{ if } t leq t_0\
I_0 e^{-(t-t_0)^2/sigma_2} text{ if } t > t_0
end{cases}
$$

with $I_0$ the maximal value of $I(t)$, $I(t_0) = I_0$, and $sigma_1$ such that $widetilde{I}(0) = 1$, i.e. $sigma_1 = t_0^2 / text{ln} I_0$.

For each combination $(beta,lambda,delta,M)$ the numbers $I_0(beta,lambda,delta,M)$ and $t_0(beta,lambda,delta,M)$ can be determined numerically (and thus $sigma_1(beta,lambda,delta,M) = t_0^2 / text{ln} I_0$). So the numbers $I_0$ and $t_0$ depend somehow on $(beta,lambda,delta,M)$ and by carefully plotting them one might see the dependency.

But how would one approximate $I_0(beta,lambda,delta,M)$ and $t_0(beta,lambda,delta,M)$ by some friendly explicit functions telling (for example and only guessed) that

for fixed $(lambda,delta,M)$ the function $I_0^{lambdadelta M}(beta)$ depends on $beta$ roughly like a shifted sigmoid $1 / (1 + e^{-gamma(beta – beta_0)})$ with parameters $gamma$ and $beta_0$ depending so-and-so on
$(lambda,delta,M)$.

By "how" I mean "by which (explicit functions)".

One Answer

You can always make some reasonable approximations which simplify the problem for certain special cases. For example, you can assume that the infection spreads very quickly. Then, at the beginning, nearly nobody is infected, i.e.$S/Mapprox 1$, while towards the end, nearly everybody is infected (or recovered), i.e. $S/Mapprox 0$. If everything happens quickly enough, you can ignore the time in between, i.e. get a reasonable approximation by assuming that the switch between nearly nobody being infected and nearly everybody being infected is nearly instantaneous.

The approximation $S/Mapprox 1$ for the beginning results in an ODE system which is linear and for which you can thus simply write down the solution. You then determine for which time $T$, $S(T)=0$. At this time, you then assume $S/M=0$, which again results in a linear ODE system which you initialize with the state of the first one at $t=T$. You then set $I(t)$ to the corresponding solution of the first ODE system for $t<T$ and to the solution of the second system for $t>T$.

PS: surprisingly, the approximations seem to become even better if you assume a fixed time delay between $I$ and $E$, i.e. if you set $E(t)=I(t-tau)$ with $tau$ the incubation time. Depending on the case, this might be even more precise than the exponential distribution you currently assume. Alternatively, you might also assume a mixture of both to e.g. say that there incubation period is at least 3 days and from there on follows an exponential distri. The point is that an explicit time delay somewhat decouples the ODEs. Essentially, this means that $S$ and $E$ are "living" at time $t$ and everything else at time $t-tau$. If you now manage to quickly deplete most of the susceptible population in a big final wave of infections lasting for $tau$ or less, the concentration of $I$ will just rise exponentially troughout this final wave even though $S$ will already quickly decrease (essentially, $I$ only "realizes" that a significant part of the population is already infected and thus that $S$ is already quickly decreasing with a time delay $tau$). Once you are through the final wave, $I$ will start to stop growing exponentially and do complicated things, but this doesn't matter anymore because everybody is already infected.

In short: what seems to make the problem more complicated (adding explicit time delays) actually makes it easier to find good analytic approximations. This is thus one of the few occasions where being more precise and realistic is simplifying the calculations.

Answered by NeitherNor on December 8, 2021

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