Problem Statement
This task investigates the nonlinear Kermack-McKendrick SIR compartmental differential equation model and the transcendental final size equation.
A fixed population of size $N$ is divided into Susceptible $S(t)$, Infectious $I(t)$, and Recovered $R(t)$ individuals:
$$\frac{dS}{dt} = -\beta S I, \quad \frac{dI}{dt} = \beta S I - \gamma I, \quad \frac{dR}{dt} = \gamma I$$
where $\beta > 0$ is the transmission rate and $\gamma > 0$ is the recovery rate. Initial conditions: $S(0) = S_0 \approx N, I(0) = I_0 > 0, R(0) = 0$.
**(a)** [3 marks] Show that the total population $S(t) + I(t) + R(t) = N$ is strictly constant for all $t \ge 0$.
**(b)** [4 marks] The basic reproduction number is $R_0 = \frac{\beta S_0}{\gamma}$. Show that an initial epidemic outbreak occurs if and only if $R_0 > 1$.
**(c)** [6 marks] Show that the phase plane trajectory is governed by:
$$\frac{dI}{dS} = -1 + \frac{\gamma}{\beta S}$$
Solve this separable differential equation with initial conditions to show that:
$$I(S) = I_0 + S_0 - S + \frac{\gamma}{\beta} \ln\left( \frac{S}{S_0} \right)$$
**(d)** [6 marks] Determine the maximum infection level $I_{\max}$ in terms of $S_0, I_0,$ and $R_0$. Show that when $I_0 \ll S_0$:
$$\frac{I_{\max}}{S_0} \approx 1 - \frac{1}{R_0} - \frac{\ln R_0}{R_0}$$
**(e)** [7 marks] As $t \to \infty$, $I(t) \to 0$ and $S(t) \to S_\infty$. Show that the proportion $s_\infty = \frac{S_\infty}{S_0}$ satisfies:
$$\ln(s_\infty) = R_0(s_\infty - 1)$$
For $R_0 = 2.5$, use numerical methods to determine $s_\infty$ correct to 3 significant figures, and calculate the critical vaccination coverage $p_c = 1 - \frac{1}{R_0}$.
Verified Solution & Marking Scheme
Part (a): Conservation of Population [M1, A1]
$$\frac{d}{dt}(S + I + R) = \frac{dS}{dt} + \frac{dI}{dt} + \frac{dR}{dt} = -\beta SI + (\beta SI - \gamma I) + \gamma I = 0$$
Since the derivative is identically zero for all $t \ge 0$, $S(t) + I(t) + R(t) = S_0 + I_0 + R_0 = N$ is conserved.
Part (b): Outbreak Condition R0 > 1 [M1, A1, R1]
At $t = 0$:
$$\left.\frac{dI}{dt}\right|_{t=0} = (\beta S_0 - \gamma) I_0$$
For an outbreak, $\frac{dI}{dt} > 0$ initially. Since $I_0 > 0$:
$$\beta S_0 - \gamma > 0 \implies \beta S_0 > \gamma \implies \frac{\beta S_0}{\gamma} > 1 \implies R_0 > 1$$
Part (c): Phase-Plane Trajectory Integration [M1, M1, A1]
$$\frac{dI}{dS} = \frac{dI/dt}{dS/dt} = \frac{(\beta S - \gamma)I}{-\beta S I} = -\frac{\beta S - \gamma}{\beta S} = -1 + \frac{\gamma}{\beta S}$$
Integrating with respect to $S$:
$$\int_{I_0}^{I} dI = \int_{S_0}^{S} \left( -1 + \frac{\gamma}{\beta u} \right) du$$
$$I - I_0 = [ -u + \frac{\gamma}{\beta} \ln u ]_{S_0}^{S} = -(S - S_0) + \frac{\gamma}{\beta} \ln\left( \frac{S}{S_0} \right)$$
$$I(S) = I_0 + S_0 - S + \frac{\gamma}{\beta} \ln\left( \frac{S}{S_0} \right)$$
Part (d): Peak Infection Level [M1, A1, A1]
Peak infection occurs when $\frac{dI}{dt} = 0 \implies S^* = \frac{\gamma}{\beta} = \frac{S_0}{R_0}$.
Substitute $S^* = \frac{S_0}{R_0}$ into $I(S)$:
$$I_{\max} = I_0 + S_0 - \frac{S_0}{R_0} + \frac{S_0}{R_0} \ln\left( \frac{1}{R_0} \right) = I_0 + S_0\left( 1 - \frac{1}{R_0} - \frac{\ln R_0}{R_0} \right)$$
Dividing by $S_0$ with $I_0 \ll S_0$:
$$\frac{I_{\max}}{S_0} \approx 1 - \frac{1}{R_0} - \frac{\ln R_0}{R_0}$$
Part (e): Final Size Equation & Numerical Solution [M1, A1, A1]
As $t \to \infty, I \to 0, I_0 \to 0$:
$$0 = S_0 - S_\infty + \frac{S_0}{R_0} \ln\left( \frac{S_\infty}{S_0} \right) \implies \frac{S_0}{R_0} \ln(s_\infty) = S_\infty - S_0 = S_0(s_\infty - 1)$$
$$\ln(s_\infty) = R_0(s_\infty - 1)$$
For $R_0 = 2.5$:
$$\ln(s_\infty) = 2.5(s_\infty - 1)$$
Using GDC / Newton-Raphson iteration with initial guess $s = 0.1$:
$$s_\infty \approx 0.107 \; (10.7\% \text{ escape infection})$$
Vaccination threshold for herd immunity:
$$p_c = 1 - \frac{1}{R_0} = 1 - \frac{1}{2.5} = 1 - 0.40 = 0.60 \; (60\%)$$