SIRS Model Analysis

Question

Consider the SIRS model given by \begin{align} \frac{dS}{dt}&=-aSI+cR, \\ \frac{dI}{dt}&=aSI-bI, \\ \frac{dR}{dt}&=bI-cR. \\ \end{align} Here $a,b,$ and $c$ are constants and $N=S+I+R$ such that $N'(t)=0$. Using the substitution $R=N-S-I$, the system of equations reduces to \begin{align} \frac{dS}{dt}&=-aSI+c\left(N-S-I\right), \\ \frac{dI}{dt}&=aSI-bI. \\ \end{align} My question is, what happened to the equation for $R'(t)$? How has the system of three equations reduced to a system of two equations?

Answer 1

$$\begin{cases} \frac{dS}{dt}=-aSI+cR \\ \frac{dI}{dt}=aSI-bI \\ \frac{dR}{dt}=bI-cR \\ \end{cases} \quad\implies\quad \frac{dS}{dt}+\frac{dI}{dt}+\frac{dR}{dt}=0$$ $$\frac{d(S+I+R)}{dt}=0\quad\implies\quad S+I+R=C_1$$ $C_1=$constant. $\quad R=C_1-S-I$ . Puting it into the original system of three equations leads to :

$$\frac{dS}{dt}=-aSI+c(C_1-S-I) \tag A$$ $$\frac{dI}{dt}=aSI-bI \tag B$$ $$-\frac{dS}{dt}-\frac{dI}{dt}=bI-c(C_1-S-I) \tag C$$

We observe that equation $(C)$ is equivalent to the sum of the equations $(A)$ and $(B)$. Thus equation $(C)$ can be forgotten. Only two equations are remaining. The system is reduced to two equations : $$\begin{cases} \frac{dS}{dt}=-aSI+c(C_1-S-I) \\ \frac{dI}{dt}=aSI-bI \end{cases} \tag 2$$ Your question is, what happened to the equation for R′(t)?

Answer : This third equation didn't vanished as if by magic. The third equation is still here because it is the above equation $\frac{d(S+I+R)}{dt}=0$ which already led to : $$R=C_1-S-I$$ This third equation will be used after solving the above system $(2)$ for $S(t)$ and $I(t)$ in order to find $R(t)$.