Networks-p5

Problem 5

Background: SIR model

\(S(t)\): number of Susceptible people at time \(t\)
\(I(t)\): number of Infected people at time \(t\)
\(R(t)\): number of people who are Removed/Recovered because they've gained immunity
Total population: \(N = S(t) + I(t) + R(t)\)

\(I(t)\)

\(S(t)\)

\(R(t)\)

a.

Explain the differential equation \[\frac{dS}{dt} = -bs(t) I(t)\] for the number of susceptibles

A: \(S(t+1) - S(t) \approx \frac{dS}{dt} = -bs(t) I(t) \)

Divide the above equation by \(N\) on both sides:

\[\frac{ds}{dt} = -bs(t) i(t) \]

since \(s(t) = S(t) / N\) and \(i(t) = I(t) / N\)

\(b\): number of contacts that each infected individual meets per day

a.

Explain the differential equation \[\frac{dr}{dt} = ki(t)\] for the fraction of recovered

A: \(R(t+1) - R(t) \approx \frac{dR}{dt} = kI(t) \)

Divide the above equation by \(N\) on both sides:

\[\frac{dr}{dt} = k i(t) \]

since \(r(t) = R(t) / N\)

\(k\): recovery rate

b.

Derive \(\displaystyle \frac{di}{dt}\) and explain

A: Given \(S(t) + I(t) + R(t) = N\) we have

\[s(t) + i(t) + r(t) = N / N = 1\]

\[\implies \frac{ds}{dt} + \frac{di}{dt} + \frac{dr}{dt} = 0\]

Substitute and obtain

\[\frac{di}{dt}=(bs(t)-k)i(t) \]

From a:

\(\frac{ds}{dt} = -bs(t) i(t)\)

\(\frac{dr}{dt} = ki(t) \)

c.

Is fixed population a good assumption for modeling COVID-19?

A: It depends.

Deterministic view vs Probabilistic view
Travel restrictions were lax during the early days of the spread
Captures local spread but not between communities

d.

Under what condition does the \(i(t)\) curve attain its maximum?

A: By the chain rule, \[\frac{di}{dt} = \frac{di}{ds}\frac{ds}{dt} \implies \frac{di}{ds} = \frac{di}{dt}/\frac{ds}{dt}\]

From a & b:

\(\frac{ds}{dt} = -bs(t) i(t)\)

\(\frac{di}{dt} = (bs(t) -k)i(t) \)

\frac{di}{ds}=\frac{bs(t)-k}{-bs(t)}=-1+\frac{k}{bs(t)}

\implies i(s)= -s + \frac{k}{b}\ln(s) + C

\(i''(s) \leq 0 \implies \) \(i\) is a concave function in \(s\)
FOC: \(i'(s) = -1 + \frac{k}{c}\frac{1}{s} = 0\) tells us that maximum is reached when \(s = \frac{k}{b}\)

integrate

e.

Explain why \(\frac{b}{k} = \frac{\ln s_\infty}{s_\infty - 1}\). Plot \(s_\infty\) as a function of \(b/k\).

A: Consider \(t=0\) and \(t = \infty\) respectively.

\begin{cases} i(s_\infty) = -s_\infty+ \frac{k}{b} \ln s_\infty + C\\ i(s_0) = -s_0+ \frac{k}{b} \ln s_0+ C \end{cases}

i(s)= -s + \frac{k}{b}\ln(s) + C

Since \(s_0 \approx 1, i(0) \approx 0, i(s_\infty) = 0\), we get the equation

\[\frac{b}{k} = \frac{\ln s_\infty}{s_\infty - 1}\]

e.

Explain why \(\frac{b}{k} = \frac{\ln s_\infty}{s_\infty - 1}\). Plot \(s_\infty\) as a function of \(b/k\).

\frac{b}{k} = \frac{\ln s_\infty}{s_\infty - 1}

\(s_\infty\)

\(b/k\)

Recall that

\(s_\infty\) is upperbounded by 1