Q5E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Suppose for a certain disease described by the SIR model it is determined that a = 0.003 and b = 0.5.

1. In the SI-phase plane, sketch the trajectory corresponding to the initial condition that one person is infected and 700 persons are susceptible.
2. From your graph in part (a), estimate the peak number of infected persons. Compare this with the theoretical prediction\({\bf{S = }}\frac{{\bf{k}}}{{\bf{a}}}\)=167 persons when the epidemic is at its peak.

Step-by-Step Solution

Answer

1. The equation is \({\bf{I =  - S + }}\frac{{{\bf{500}}}}{{\bf{3}}}{\bf{lnS - 390}}{\bf{.8467}}\)
2. The peak of infected persons is 295.

Step 1: Use SIR Model

While using the SIR model\({\bf{i =  - s + }}\frac{{\bf{k}}}{{\bf{a}}}{\bf{lns + K}}\).

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

 

\({\bf{I =  - S + }}\frac{{\bf{k}}}{{\bf{a}}}{\bf{lnS + K}}\)

 

Where K is some constant.

 

Put a = 0.003 and b = 0.5 then 

 

\(\begin{aligned}{c}{\bf{I =  - S + }}\frac{{{\bf{0}}{\bf{.5}}}}{{{\bf{0}}{\bf{.003}}}}{\bf{lnS + K}}\\{\bf{I =  - S + }}\frac{{{\bf{500}}}}{{\bf{3}}}{\bf{lnS + K}}\end{aligned}\)

 

Now, put I=1, S=700 then;

 

\(\begin{aligned}{c}{\bf{1 =  - 700 + }}\frac{{{\bf{500}}}}{{\bf{3}}}{\bf{ln700 + K}}\\{\bf{K =  - 390}}{\bf{.8467}}\end{aligned}\)

 

The equation is\({\bf{I =  - S + }}\frac{{{\bf{500}}}}{{\bf{3}}}{\bf{lnS - 390}}{\bf{.8467}}\).

Step 2: Trajectory

Step 3: Find the result

While seeing the graph it shows that the peak of infected persons is 295. This is kind of far off from the theoretical prediction.

This is the required result.