Mathematical Biology (infectious disease)

[ra_forever8]

Consider the infectious disease model defined by
\begin{equation} \frac{dS_3}{dt}= -\rho I_3S_3+\gamma I_3+\mu-\mu S_3\tag 1
\end{equation}
\begin{equation} \frac{dI_3}{dt}=\rho I_3S_3-\gamma I_3-\mu S_3 \tag 2
\end{equation}
with initial conditions $S_3(0)=S_{30}$ and $I_3(0)=I_{30}$ at $t=0$
Where $\rho,\gamma$ and $\mu$ are all positive constants. Assume $N_3= S_3 + I_3$ and obtain an equation for $\frac {dN_3}{dt}$. What does this assumption mean biologically? Show that for $t \geq 0, N_3(t) \equiv 1$ and equation (1) can be written as
\begin{equation} \frac{dS_3}{dt}=\rho (\overline{S_3}- S_3)(1-S_3) \tag3
\end{equation}
where $\overline{S_3}= \frac{\gamma + \mu}{\rho} $.Determine the steady-state stability of equation (3) by appealing to the value of $\overline{S_3}$.=> I try to do by solving $N_3= S_3 + I_3$
to obtain $\frac {dN_3}{dt}$= $\frac {dS_3}{dt}$+$\frac {dI_3}{dt}$
ant that gives $\frac {dN_3}{dt} = \mu - N_3 \mu $
is the assumption mean that $N_3$ is not constant?$ N_3(t) \equiv 1$
that gives,
$S_3 + I_3 =1 $
i try to calculate $\frac {dS_3}{dt}$ by using $S_3 + I_3 =1 $ but don't how to calculate?steady-state stability of equation (3) is given by $\frac {dS_3}{dt}$ =0 which leads to$\rho (\overline{S_3}- S_3)(1-S_3)=0$
which gives $S_3= \overline{S_3}$.
or
$S_3 =1$

after i really don't know what to do .can anyone please help me.

 

[jvicens]

Hello,

Based on the given equations, it seems that you are dealing with a basic infectious disease model, where $S_3$ represents the susceptible population and $I_3$ represents the infected population. The first equation (1) describes the rate of change of the susceptible population, while the second equation (2) describes the rate of change of the infected population.

To obtain an equation for $\frac{dN_3}{dt}$, we can simply add equations (1) and (2) together, since $N_3 = S_3 + I_3$. This gives us:

\begin{equation} \frac{dN_3}{dt}=\frac{dS_3}{dt}+\frac{dI_3}{dt}=-\rho I_3S_3+\gamma I_3+\mu-\mu S_3+\rho I_3S_3-\gamma I_3-\mu S_3 \tag 4
\end{equation}

Simplifying equation (4), we get:

\begin{equation} \frac{dN_3}{dt}=\gamma-\mu \tag 5
\end{equation}

This means that the rate of change of the total population ($N_3$) is simply equal to the difference between the birth rate ($\gamma$) and the death rate ($\mu$). Biologically, this assumption means that there are no other factors affecting the population size, such as immigration or emigration.

Now, to show that $N_3(t) \equiv 1$, we can substitute $S_3 + I_3 = 1$ into equation (1) and solve for $S_3$. This gives us:

\begin{equation} \frac{dS_3}{dt}= -\rho I_3(1-I_3)+\gamma I_3+\mu-\mu(1-I_3) \tag 6
\end{equation}

Simplifying equation (6), we get:

\begin{equation} \frac{dS_3}{dt}=\rho I_3-\rho I_3^2+\gamma I_3+\mu(1-I_3) \tag 7
\end{equation}

Since $S_3 + I_3 = 1$, we can rewrite $I