Modeling viral disease

Consider the spreading of a highly communicable disease on an isolated island with population size \(N\). A portion of the population travels abroad and returns to the island infected with the disease. You would like to predict the number of people \(X\) who will have been infected by some time \(t\). Consider the following model, where \(k > 0\) is constant:

\[\begin{equation*} \frac{dX}{dt}=k\textcolor{red}{X}(N-X) \end{equation*}\]

1. List two major assumptions implicit in the preceding model. How reasonable are your assumptions?
   Answer: Here are two major assumptions:
   

1. Fixed population \(\implies\) all infected. We assume the population size remain unchanged that is no one gets in the island or no one gets out of the island. This will lead everyone affected by the disease eventually.

2. No immediate cure or vaccination. We also assume that there is no immediate hard immunity build up among the population or invention of vaccination.

2. Graph \(\frac{dX}{dt}\) versus \(X\)

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3. Graph \(X\) versus \(t\) if the initial number of infections is \(X_1 < \frac{N}{2}\). Graph \(X\) versus \(t\) if the initial number of infections is \(X_2 >\frac{N}{2}\).
   Answer: For equilibrium of the model

\[\begin{align*} f(X)&=kX(N-X)=0\\ \implies kX&=0 & N-X=0\\ \implies X=&0 & X=N \end{align*}\] For the stability analysis:

\[\begin{align*} f(X)&=(kN)X-kX^2 & \implies f'(X)=kN-2kX \end{align*}\]

Now, \(f'(0)=kN>0\) therefore, \(X=0\) is an unstable equilibrium. And \(f'(N)=kN-2kN=-kN<0\) since \(k, N>0\). So, \(X=N\) is a stable equilibrium.

equilibrium

If the initial infection \(X_1<\frac{N}{2}\) it might decrease and reach to 0 but that is not a stable equilibrium. So eventually it will hit \(N\).

4. Solve the model given earlier for \(X\) as a function of \(t\).
   Answer: Solving the ODE we have

\[\begin{align*} \frac{dX}{dt}&=kX(N-X)\\ \text{Since}\hspace{2mm} X&>0\\ \frac{dX}{X(N-X)}&=kdt\\ \implies \int \frac{dX}{X(N-X)}&=\int kdt\\ \implies \frac{1}{N}\int \left(\frac{1}{X}+\frac{1}{N-X}\right)dX&= \int kdt\\ \implies \frac{1}{N} \ln\left(\frac{X}{N-X}\right)&=kt+c\\ \implies \ln\left(\frac{X}{N-X}\right)&=Nkt+Nc\\ \implies \frac{X}{N-X}&=e^{Nkt+Nc}\\ \implies X&=Ne^{Nkt+Nc}-Xe^{Nkt+Nc}\\ \implies X\left(1+e^{Nkt+Nc}\right)&=Ne^{Nkt+Nc}\\ \implies X(t)&=\frac{Ne^{Nkt+Nc}}{1+e^{Nkt+Nc}}\\ \implies X(t)&=\frac{N}{1+e^{-(Nkt+Nc)}} \end{align*}\]

5. From part (d), find the limit of \(X\) as \(t\) approaches infinity.
   Answer:

\[\begin{align*} \lim_{t\longrightarrow \infty} X(t)&=\lim_{t\longrightarrow \infty} \frac{N}{1+e^{-(Nkt+Nc)}}=N \end{align*}\]

6. Consider an island with a population of \(5000\). At various times during the epidemic the number of people infected was recorded as follows:

\(t\) (days)	2	6
\(X\) (People infected)	\(1887\)	\(4087\)
\(\ln{\left(\frac{X}{N-X}\right)}\)	\(-0.5\)	\(1.5\)

Do the collected data support the given model?
Answer: If we look at part (d) we have

\[\begin{align*} \ln\left(\frac{X}{N-X}\right)&=Nkt+Nc & \text(And),\\ X(t)&=\frac{N}{1+e^{-(Nkt+Nc)}} \end{align*}\]

So we get if \(2Nk+Nc=-0.5\) then \(X(2)=\frac{5000}{1+e^{0.5}}=1887.703\), if \(6Nk+Nc=1.5\) then \(X(6)=\frac{5000}{e^{-1.5}}=4087.87\), and if \(10Nk+Nc=3.5\) then \(X(10)=\frac{5000}{e^{-3.5}}=4853.44\)
Therefore, the collected data supports the model.

7. Use the results in part (f) to estimate the constants in the model, and predict the number of people who will be infected by \(t = 12\) days.
   Answer: We have

\[\begin{align*} 2Nk+Nc&=-0.5\\ 6Nk+Nc&=1.5 \end{align*}\]

Solving the above system we have \(k=\frac{1}{2N}\) and \(c=\frac{-1.5}{N}\). If we substitute these values in the solution we got in part (d) we have

\[\begin{align*} X(t)&=\frac{N}{1+e^{-\left(\frac{t}{2}-1.5\right)}} \end{align*}\]
So, \(X(12)\approx 4945\)

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