Establish the following limit?

[kalish1]

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

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

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

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

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

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**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 $\frac{dS}{dt}=\Lambda-\mu S-\beta \frac{S}{N}(H+C+C_1+C_2)-\tau \frac{S}{N}(T+C) \geq -\frac{3\delta}{2}.$
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**Question:** How do I conclude?

I have crossposted this question here: differential equations - How to show that $\lim\limits_{t \rightarrow \infty} S(t) = \Lambda/\mu$? - Mathematics Stack Exchange

 

[Prove It]

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

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

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

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

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

_________________________________________________________________________________________
**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?

I have crossposted this question here: differential equations - How to show that $\lim\limits_{t \rightarrow \infty} S(t) = \Lambda/\mu$? - Mathematics Stack Exchange

Are all the parameters constants? If so, it's first order linear...

 

[kalish1]

Are all the parameters constants? If so, it's first order linear...

$\Lambda,\tau,\beta,\mu$ are all constants. I understand that it's first order linear, but it's a differential inequality. How do I establish equality in the limit?

 

[Prove It]

$\Lambda,\tau,\beta,\mu$ are all constants. I understand that it's first order linear, but it's a differential inequality. How do I establish equality in the limit?

What I am saying is that since the DE is first order linear, you can use the integrating factor method to solve it. Then once you have S(t), you can see what happens to S as t goes to infinity...

 

[kalish1]

What I am saying is that since the DE is first order linear, you can use the integrating factor method to solve it. Then once you have S(t), you can see what happens to S as t goes to infinity...

Ok. I obtain $S(t) \leq \Lambda/\mu + Ce^{-\mu t}$, which shows that $S \leq \Lambda/\mu.$ But how do get equality? Don't I need to use $liminf$ and $limsup$ somewhere? I don't know how.

 

[Prove It]

Ok. I obtain $S(t) \leq \Lambda/\mu + Ce^{-\mu t}$, which shows that $S \leq \Lambda/\mu.$ But how do get equality? Don't I need to use $liminf$ and $limsup$ somewhere? I don't know how.

You aren't listening to me. SOLVE THE DE!

 

[I like Serena]

It's not a linear first order ODE since N(t) contains S(t).

And I see that you already have an answer on MSE.

 

[kalish1]

It's not a linear first order ODE since N(t) contains S(t).

And I see that you already have an answer on MSE.

The answer on MSE is incomplete.

Can you help me finish this? This is what I have:

Note that $\lim\limits_{t \rightarrow \infty}\frac{dS}{dt}=\lim\limits_{t \rightarrow \infty}\bigg[\Lambda-\mu S-\beta \frac{S}{N}(H+C+C_1+C_2)-\tau \frac{S}{N}(T+C) \bigg] = \lim\limits_{t \rightarrow \infty}(\Lambda-\mu S).$ Solving the differential equation within the limit, we obtain $S(t)=\frac{\Lambda}{\mu}+e^{-\mu t}.$ Applying the limit, we see that $\lim\limits_{t \rightarrow \infty}S = \frac{\Lambda}{\mu}$.

 

[I like Serena]

The answer on MSE is incomplete.

Can you help me finish this? This is what I have:

Note that $\lim\limits_{t \rightarrow \infty}\frac{dS}{dt}=\lim\limits_{t \rightarrow \infty}\bigg[\Lambda-\mu S-\beta \frac{S}{N}(H+C+C_1+C_2)-\tau \frac{S}{N}(T+C) \bigg] = \lim\limits_{t \rightarrow \infty}(\Lambda-\mu S).$

Yep. We can split the limit and make the small stuff go away.

This should be written out though and your original attempt is not mathematically solid.

Solving the differential equation within the limit, we obtain $S(t)=\frac{\Lambda}{\mu}+e^{-\mu t}.$ Applying the limit, we see that $\lim\limits_{t \rightarrow \infty}S = \frac{\Lambda}{\mu}$.

This looks a bit tricky.

Let's make it a bit more formal.

We have:

\(\displaystyle \forall \epsilon >0 \quad \exists T\) such that \(\displaystyle \forall t>T: \left|\frac{dS}{dt} - (\Lambda-\mu S) \right| < \epsilon\).

From this we can conclude:
$$
\left|\frac{dS}{dt} e^{\mu t} - (\Lambda e^{\mu t}-\mu S e^{\mu t}) \right| < \epsilon e^{\mu t}$$

$$\left|\frac{d}{dt}(S e^{\mu t}) - \Lambda e^{\mu t} \right| < \epsilon e^{\mu t}$$

$$(\Lambda - \epsilon) e^{\mu t} < \frac{d}{dt}(S e^{\mu t}) < (\Lambda + \epsilon) e^{\mu t}$$

$$\left(\frac \Lambda \mu - \frac \epsilon {|\mu|}\right) e^{\mu t} + C_1< S e^{\mu t} < \left(\frac \Lambda \mu + \frac \epsilon {|\mu|}\right) e^{\mu t} + C_2$$

$$\frac \Lambda \mu - \frac \epsilon {|\mu|} + C_1 e^{-\mu t} < S < \frac \Lambda \mu + \frac \epsilon {|\mu|} + C_2 e^{-\mu t}
$$

In other words, $S$ is squeezed to \(\displaystyle \frac \Lambda \mu\), but only if $\mu > 0$.

So if $\mu > 0$, then:
$$\lim_{t\to \infty} S(t) = \frac \Lambda \mu$$