Derivation of an epidemic equations with calculus. What I do in coffee shops. 28th, 2009 at 9:26 AM. Jorma Jyrkkanen

Derivation of an epidemic equations with calculus. Apr. 28th, 2009 at 9:26 AM Jorma’s Coffee Math jjyrkkanen76@outlook.com

An Epidemic Equation Derivation To model an epidemic assume N is susceptible and this is known, y is the number infected at time t, while susceptible and not infected is N-y. Assume that the rate of change of y is proportional to the number infected y and that the fewer uninfected that remain, the slower the infection rate should be, so it should diminish as the number increases. Assume then that rate of change of infected individuals is proportional to the number infected and also to number susceptible but not yet infected. Since y denotes no. infected, rate of change of y is dy/dt. The number of susceptible persons is N. As a result of these assumptions, for some constant k, dy/dt = ky(N-y) (1) or: dy/(y(N-y)) = k dt (2) This rate equation can then be integrated to find the Dynamic ∫[1/(y(N-y)]dy = ∫k dt (3) However, integration requires partial fractions since it is not so straightforward. Here's how. 1/(y(N-y)) = [A(N-y) + By]/(y(N-y)) = [AN –Ay +By] /(y(N-y)) = [AN – y(A-B)] /(y(N-y)) But A-B=0, so A=B--> AN/(y(N-y)) But AN = 1 So A=1/N, and (1/N)-B=0, so that 1/(y(N-y)) = A/y + B/(N-y) (4) This leads to the happy solution that is also easier to integrate: ∫[1/(y(N-y)]dy = ∫ k dt (5) ∫[A/y + B/(N-y)] dy = ∫k dt A ln y – (B) ln (N-y) = kt + C The dynamic: (1/N) ln (y/(N-y)) = kt + C (6) This is the basic equation of the epidemic but we must now find C. Let number N0 known, at time t0 be denoted by y0 so that in the initial condition; (1/N0) ln (y/(N0- y0)) = C (7) and substituting into (6) (1/N) ln (y/(N-y)) = kt + (1/N0) ln (y/(N0 -y0)) kt = (1/N) ln (y/(N-y)) - (1/N0) ln (y/(N0-y0)) kt = (1/N) [ln (y/(N-y))/ (1/N0) ln (y/(N0-y0))] kt = (1/N) ln [(y/(N-y))x ((N0-y0)/ y0)] (8) The constant k can be determined by making one other observation and substituting into this last equation. If we want an explicit solution we can express y as a function of t, so that; ln [(y(N-y0)/(N-y) y0] = Nkt (9) Taking exponents yields [y(N- y0)/(N-y) y0] = e Nkt (10) Solving for y y(N- y0) = (N-y) y0 e Nkt So y(N - y0 + y0 e Nkt) = N y0e Nkt And y = [N y0 e Nkt ]/[ N – y0 + y0e Nkt ] and finally Wheeyouuu Y(t) = [N y0]/[(N- y0) e -Nkt + y0] (11) Pant! Pant! Man, that was tough! Test with some iterations. There may be a typo in copying out the syntax. See if you can find one