[math-fun] 99.9% solution of SIR equations
I asked the question: to what extent is the logistic iteration, x_{n+1} = x_n + x_n*(1-x_n)*dt, a solution of the SIR equations, S' = -S*I, I' = S*I - r*I? The answer, apparently, is 99.9% when dt in [1/10,1], as is shown in the following plots: https://0x0.st/iQv6.png , where the red points are from the discrete iteration and the blue curve is from the continuous ODE solution. The particular function from dt -> r can be found by minimizing err = abs[I' - (S*I - r*I)], after eliminating S by S = Exp[-Int(I*dt)], and after accounting for the non-zero asymptote of S. One shape parameter is one shape parameter, whether it is "dt" or "r", doesn't seem to matter very much at all. --Brad
See also: https://0x0.st/iQEw.png r = 0.7083 - 0.2443*dt - 1.371*dt^2 + 1.263*dt^3; r in [0.28,0.71] suggests natural cutoff at dt=0.8; however, 99.9% accuracy persists to dt=1. --Brad On Thu, Apr 16, 2020 at 12:00 AM Brad Klee <bradklee@gmail.com> wrote:
I asked the question: to what extent is the logistic iteration, x_{n+1} = x_n + x_n*(1-x_n)*dt, a solution of the SIR equations, S' = -S*I, I' = S*I - r*I?
The answer, apparently, is 99.9% when dt in [1/10,1], as is shown in the following plots: https://0x0.st/iQv6.png , where the red points are from the discrete iteration and the blue curve is from the continuous ODE solution.
The particular function from dt -> r can be found by minimizing err = abs[I' - (S*I - r*I)], after eliminating S by S = Exp[-Int(I*dt)], and after accounting for the non-zero asymptote of S.
One shape parameter is one shape parameter, whether it is "dt" or "r", doesn't seem to matter very much at all.
--Brad
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Brad Klee