Question

Prove that the infected population I(t)in the SIR model does not increase if S(0) is less than or equal to \(\frac{{\bf{k}}}{{\bf{a}}}\).

Short answer

Thus, it is proved that the infected population I(t)in the SIR model does not increase if S(0) is less than or equal to \(\frac{{\bf{k}}}{{\bf{a}}}\).

Step-by-step solution

Apply the SIR model.

If S(t) is the number of susceptible individuals, I(t) the number of the currently infected individuals, and R(t) the number of individuals who have recovered from the infection.

Then N = I+S+R.

The SIR model assumes that N does not change with time then the possibilities are;

1. The number of infected individuals increases, and the number of S(t) decreases. And eventually, increases and eventually, the number of R(t) increases.

1. The number of S(t) n does not change, the number of I(t) decreases and they all eventually recover.

In both cases\({\bf{S(0}}) \ge {\bf{S(t)}}\).

The change in the numbers is:

\(\frac{{{\bf{di}}}}{{{\bf{dt}}}}{\bf{ = a}}\left( {{\bf{s - }}\frac{{\bf{k}}}{{\bf{a}}}} \right){\bf{i}}\)

And

\({\bf{i = }}\frac{{\bf{I}}}{{\bf{N}}}{\bf{,s = }}\frac{{\bf{S}}}{{\bf{N}}}\)

The solution of the equation is\({\bf{i(t) = }}{{\bf{e}}^{{\bf{(as - k)t}}}}\)or\({\bf{i(t) = C}}\).

Get the result. 

The solution is\({\bf{a}}\left( {{\bf{s - }}\frac{{\bf{k}}}{{\bf{a}}}} \right){\bf{ = 0}}\)and the first form.

Now, the infected population will not decrease if and only if\({\bf{a}}\left( {{\bf{s - }}\frac{{\bf{k}}}{{\bf{a}}}} \right) \le 0\).