mathematical biology1

Question

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?

=> 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?

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} $.

$ 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?

Determine the steady-state stability of equation (3) by appealing to the value of $\overline{S_3}$.

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.

Answer 1

The notations are horrible, the phrasing of the questions themselves is odd and the OP did not copy the text faithfully, but here is the (corrected) setting:

Assume that two populations $S$ and $I$ evolve according to the equations $$ S'(t)= -\rho I(t)S(t)+\gamma I(t)+\mu-\mu S(t), $$ and $$ I'(t)=\rho I(t)S(t)-\gamma I(t)-\mu I(t), $$ where $\rho$, $\gamma$ and $\mu$ are positive constants. Define $N(t)=S(t)+I(t)$.

• Determine the evolution of $N(t)$.
• Assuming that $N(0)=1$, deduce from the system above an equation describing the evolution of $S(t)$ with no $I(t)$ term.
• Finally, determine the limit of $S(t)$ when $t\to\infty$.

To solve this, note that $N'(t)=S'(t)+I'(t)$ hence $$ N'(t)=\mu-\mu S(t)-\mu I(t)=\mu(1-N(t)). $$ Thus, for every starting value $N(0)$, $$ \lim_{t\to\infty}N(t)=1. $$ From now on, one assumes that $N(0)=1$. Then $N(t)=1$ for every $t$, hence $I(t)=1-S(t)$ and the equation giving $S'(t)$ becomes $$ S'(t)= -\rho (1-S(t))S(t)+\gamma (1-S(t))+\mu-\mu S(t), $$ that is, $$ S'(t)=\rho(\nu-S(t))(1-S(t)),\qquad\nu=\frac{\mu+\gamma}\rho. $$ Let $S(0)=S_0$ with $S_0$ in $[0,1]$. If $S_0=1$, then $S(t)=1$ for every $t$ hence $S(t)\to1$. If $S_0$ is in $[0,1)$, then two cases arise:

• If $\nu\geqslant1$, then $S'(t)\gt0$ for every $S(t)$ in $[0,1)$ hence $S(t)\to1$ for every $S_0$ in $[0,1)$.
• If $\nu\lt1$, then $S'(t)\gt0$ for every $S(t)$ in $[0,\nu)$ and $S'(t)\lt0$ for every $S(t)$ in $(\nu,1)$ hence $S(t)\to\nu$ for every $S_0$ in $[0,1)$.

Finally, for every $S_0$ in $[0,1)$, $$ \lim_{t\to\infty}S(t)=\min\{1,\nu\}. $$

Answer 2

'What does this assumption mean biologically?' Imposible to say if you don't tell us what $S_3$ and $I_3$ mean biologically. Besides, it's not clear what assumption are you referring to. In fact, the whole statement of the problem is quite unclear, which may stem from your lack of understanding of it and makes it hard to help you. I'll do my best guess.

If N3(t)≡1 then $S_3+I_3=1$, and differentiating with respect to $t$ gives $\frac {dS_3}{dt}+\frac{dI_3}{dt}=0$. But adding the first two equations you provide, you get $\frac {dS_3}{dt}+\frac{dI_3}{dt}=\mu(1-2 S_3)$ so $\mu(1-2 S_3)=0$ and $S_3=1/2$, and since $S_3+I_3=1$, then $I_3=1/2$.

Something must be wrong with the statement of the problem, since it doesn't make much sense to provide differential equations for $S_3$ and $I_3$ if you get they are just constants (forced by the definition of N3 and the condition N3(t)≡1)