Introduction to the Modeling and Analysis of Complex Systems

418CHAPTER 18. DYNAMICAL NETWORKS III: ANALYSIS OF NETWORK DYNAMICSside for simplicity):q t+1 = (1 − q) ( 1 − (1 − p e qp i ) n−1) + q(1 − p r ) (18.19)= 1 − q − (1 − q)(1 − p e qp i ) n−1 + q − qp r (18.20)≈ 1 − (1 − q)(1 − (n − 1)p e qp i ) − qp r (because p e qp i is small) (18.21)= q ((1 − q)(n − 1)p e p i + 1 − p r ) (18.22)= q ((1 − q)s + 1 − p r ) = f(q) (with s = (n − 1)p e p i ) (18.23)Now this is a simple iterative map about q t , which we already know how to study. Bysolving f(q eq ) = q eq , we can easily find that there are the following two equilibrium points:q eq = 0,1 − p rs(18.24)And the stability of each of these points can be studied by calculating the derivative off(q):df(q)= 1 − p r + (1 − 2q)s (18.25)dqdf(q)dq ∣ = 1 − p r + s (18.26)q=0df(q)dq ∣ = 1 + p r − s (18.27)q=1−pr/sSo, it looks like p r − s is playing an important role here. Note that 0 ≤ p r ≤ 1 because itis a probability, and also 0 ≤ s ≤ 1 because (1 − s) is an approximation of (1 − p e qp i ) n−1 ,which is also a probability. Therefore, the valid range of p r − s is between -1 and 1. Bycomparing the absolute values of Eqs. (18.26) and (18.27) to 1 within this range, we findthe stabilities of those two equilibrium points as summarized in Table 18.2.Table 18.2: Stabilities of the two equilibrium points in the network SIS model.Equilibrium point −1 ≤ p r − s < 0 0 < p r − s ≤ 1q = 0 unstable stableq = 1 − prsstable unstableNow we know that there is a critical epidemic threshold between the two regimes. Ifp r > s = (n − 1)p e p i , the equilibrium point q eq = 0 becomes stable, so the disease shouldgo away quickly. But otherwise, the other equilibrium point becomes stable instead, which