Modeling — Question 6

Question 6 – Solution

Strategy. Conservation confines the dynamics; eliminating time gives an exact curve and identifies both the peak and final-size constraint.

Step 1: Balance the three populations. The equations are $\boxed{s'=-2si,\ i'=2si-i,\ r'=i}$, with $s+i+r=1$. Boundary rates cannot create negative fractions. More explicitly, $s=0.99e^{-2\int_0^t i}$ and $i=0.01e^{\int_0^t(2s-1)}$ are positive at finite times; $r'=i\ge 0$. Use $r=1-s-i$ to reduce the system.

Step 2: Eliminate time and locate the peak. Since $s,i>0$, $di/ds=-1+1/(2s)$. Integrating through the initial point gives $$\boxed{i=1-s+\tfrac 12\log(s/0.99).}$$ Here $i'=i(2s-1)$ changes sign at $s=1/2$. This value must be reached: if $s\ge 1/2$ forever, then $i\ge 0.01$ and $s'\le-0.02s$, a contradiction. Strict decrease of $s$ permits only one crossing. Hence $\boxed{i_{\max}=\tfrac 12+\tfrac 12\log(0.5/0.99)\approx 0.15845}$.

Step 3: Identify the physical final size. Because $\int_0^\infty i=r_\infty\le 1$, $s\ge 0.99e^{-2}>0$. Thus decreasing $s$ has a positive limit. Also $i$ is integrable and has a bounded derivative on the invariant simplex, so $i\to 0$: nonvanishing peaks would have widths bounded below and contradict integrability. The peak crossing implies $s_\infty<1/2$. Substitution gives $$\boxed{1-s_\infty+\tfrac 12\log(s_\infty/0.99)=0,\quad s_\infty\approx 0.200.}$$ On $(0,1/2)$ the left side is strictly increasing, tends to $-\infty$ at zero and is positive at $1/2$, so this root is unique.

Step 4: Interpret the physical trajectory. Time moves toward smaller $s$, up to the peak and then down toward zero infectious fraction. The limiting endpoint is open because $i(t)>0$ at finite times. At the peak, incidence and removal balance; both are positive. The peak marks zero net infectious growth, not the end of infection.