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Can someone here help me establish that $\lim\limits_{t \rightarrow \infty} S(t) = \Lambda/\mu$, given that:

$\frac{dS}{dt}=\Lambda-\mu S-\beta \frac{S}{N}(H+C+C_1+C_2)-\tau \frac{S}{N}(T+C)$

$N(t)=S(t)+H(t)+C(t)+C_1(t)+C_2(t)$

$\Lambda,\beta,\tau,\mu > 0$

$\text{As} \ \ t \rightarrow \infty, \ T,H,C,C_1,C_2 \rightarrow 0?$


My attempt:

$\frac{dS}{dt}=\Lambda-\mu S-\beta \frac{S}{N}(H+C+C_1+C_2)-\tau \frac{S}{N}(T+C) \leq \Lambda-\mu S.$ Clearly if $S > \Lambda/\mu,$ then $dS/dt<0.$ Since $dS/dt$ is bounded by $\Lambda-\mu S,$ a standard comparison theorem can be used to show that $S(t) \leq S(0)e^{-\mu t} + \frac{\Lambda}{\mu}(1-e^{-\mu t}).$ In particular, $S(0) \leq \frac{\Lambda}{\mu} \Rightarrow S(t) \leq \frac{\Lambda}{\mu}.$

Also, since $T,H,C,C_1,C_2 \rightarrow 0 \ \ \text{as} \ \ t \rightarrow \infty,$ $$(\forall \delta>0)(\exists T>0)\left(t \geq T \implies \left|-\beta\frac{S}{N}(H+C+C_1+C_2)-\tau\frac{S}{N}(T+C)\right|<\frac{\delta}{2}\right).$$ If we take $S \geq \frac{\Lambda-\delta}{\mu},$ then $\Lambda-\mu S \leq \delta.$

Hence $\frac{dS}{dt}=\Lambda-\mu S-\beta \frac{S}{N}(H+C+C_1+C_2)-\tau \frac{S}{N}(T+C) \leq \frac{3\delta}{2}.$

If we take $S \leq \frac{\Lambda+\delta}{\mu},$ then $\Lambda-\mu S \geq -\delta.$

Hence $\frac{dS}{dt}=\Lambda-\mu S-\beta \frac{S}{N}(H+C+C_1+C_2)-\tau \frac{S}{N}(T+C) \geq -\frac{3\delta}{2}.$


Question: How do I conclude?

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One can synthetize the relevant assumptions as the fact that $S'(t)=\Lambda-\mu(t)S(t)$ where $\mu(t)\to\mu$ when $t\to\infty$ and $\mu\gt0$.

For every $\nu\gt\mu$, there exists some $t_\nu$ such that $\mu(t)\leqslant\nu$ for every $t\geqslant t_\nu$. Then $S'(t)\geqslant\Lambda-\nu S(t)$ for every $t\geqslant t_\nu$. The RHS is positive for every $S(t)\lt S_\nu$, where $$S_\nu=\Lambda/\nu,$$ hence, if $S(t)\lt S_\nu$ for some $t\geqslant t_\nu$, then the function $t\mapsto S(t)$ is increasing until $S(t)\geqslant S_\nu$ hence $\liminf\limits_{t\to\infty}S(t)\geqslant S_\nu$. Likewise, for every positive $\nu\lt\mu$, $\limsup\limits_{t\to\infty}S(t)\leqslant S_\nu$.

Thus, $S_{\mu+\varepsilon}\leqslant\liminf\limits_{t\to\infty}S(t)\leqslant\limsup\limits_{t\to\infty}S(t)\leqslant S_{\mu-\varepsilon}$ for every positive $\varepsilon\lt\mu$. When $\varepsilon\to0^+$, $S_{\mu+\varepsilon}\to S_\mu$ and $S_{\mu-\varepsilon}\to S_\mu$, hence $\liminf\limits_{t\to\infty}S(t)=\limsup\limits_{t\to\infty}S(t)=S_\mu$, that is, $\lim\limits_{t\to\infty}S(t)=S_\mu$.

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  • $\begingroup$ Finish what? First, my answer is complete. Second, the proof in your comment is absurd. $\endgroup$ – Did Jan 4 '15 at 16:40
  • $\begingroup$ ((Comment by the OP deleted.)) $\endgroup$ – Did Jan 4 '15 at 17:18
  • $\begingroup$ Before "thus", one establishes a lower bound on the liminf and an upper bound on the limsup. After "thus", one puts them together. Thus, "thus". (And imbecility is off-topic here.) $\endgroup$ – Did Jan 4 '15 at 18:05
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    $\begingroup$ Sure, since the lower bound on the liminf holds for every $\nu\gt\mu$ and the upper bound on the limsup holds for every $\nu\lt\mu$. $\endgroup$ – Did Jan 4 '15 at 18:11

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