1Physiology 472/572 - 2009 - Quantitative modeling of biological systems Lecture 23: Infectious diseases Models of epidemics • SIR model: Susceptible, Infected and Removed (Kermack and MacKendrick, 1927) IdtdRI-ISdtdIISdtdSννββ==−= • SIRS model: allows for loss of immunity RIdtdR)IS(IISdtdIRISdtdSγννβνβγβ−=−=−=+−= • S + I + R = N, constant total population • two steady states: γνβννγνβνγβν+−=+−=====/N R,/NI,S;0 R,0I,NS222111 • the second solution can only exist if N > ν/β • the disease can only become established in the population if R0 = Nβ/ν > 1 • Nβ = rate of growth of infective population (when S = N), ν = rate of loss of infectiveness • eliminate R using R = N - S - I I)G(S,I-ISdtdII)F(S,)ISN(ISdtdS≡=≡−−+−=νβγβ2Phase-plane analysis of a second order nonlinear system • consider S-I plane, define nullclines as locations where dS/dt = 0 or dI/dt = 0 • nullclines cross at the stationary points (steady-state solutions) • to determine if stationary points are stable, examine behavior of system near them • two-dimensional Taylor series: IIFSSF)I,SF(I)IS,SF( Δ∂∂+Δ∂∂+≈Δ+Δ+ and similarly for G, where the partial derivatives are evaluated at )I,S( • at a stationary point 0)I,SG()I,SF( == , so ⎥⎦⎤⎢⎣⎡ΔΔ=⎥⎦⎤⎢⎣⎡ΔΔ⎥⎦⎤⎢⎣⎡∂∂∂∂∂∂∂∂=⎥⎦⎤⎢⎣⎡ΔΔISISIG/SG/IF/SF/ISdtdJ where J is the Jacobian matrix at )I,S( • if we look for solutions proportional to eλt then λ must be an eigenvalue of J • the eigenvalues of ⎥⎦⎤⎢⎣⎡dcba satisfy λ2 - (a + d) λ + ad - bc = (λ - λ1)( λ - λ2) = 0 • so λ1 + λ2 = a + d = tr(J), the trace, and λ1λ2 = ad - bc = det(J) the determinant • if λ1 and λ2 are real, the solution is stable if both are negative • if λ1 and λ2 are complex, they must be complex conjugates, and the solution is stable if the real part is negative • stability is ensured if the trace is negative and the determinant is positive Phase-plane for this case • nullclines: dI/dt = 0 if I = 0 or if S = ν/β • dS/dt = 0 if βIS = γ(N - S - I), i.e. if γβγ+−=SS)(NI3• ⎥⎦⎤⎢⎣⎡−−−−−=νββγβγβSISIJ • at )I,S(11 λ1λ2 = det(J) = -γ(βN - ν) < 0 if βN/ν > 1 - must be one negative root: unstable • at )I,S(22 λ1 + λ2 = tr(J) = -γ(βN + γ)/(ν + γ) < 0 and λ1λ2 = det(J) = γ(βN - ν) > 0 if βN/ν > 1 - equilibrium is stable • example: β = 0.01, γ = 0.05, ν = 0.2, N = 100 0 20 40 60 80
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