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I am working with the SIR model with vital dynamics:

\begin{align} \frac{dS}{dt} &= \mu (K - S)- \beta SI \label{eq3}\\ \frac{dI}{dt} &= I(\beta S - \gamma -\mu) \label{eq4} \end{align}

where I discarded the last ODE since it does not appear in the first two. The region for this problem is $D={(S,I) : S>0, I \geq 0, S+I \leq K}$

I found that one equilibrium point is $(K,0)$ which is a local asymptotically stable point if $\frac{\beta K}{\gamma + \mu} < 1$. How can I show that is indeed a globally stable point? I thought about using the lyapunov function $V(S,I) = S+I-K$ from which get $\dot{V} = \mu(K-S-I) - \gamma I$. But since $S+I\leq K $ then I cannot conclude that $\dot{V}<0$. What am I doing wrong? Do I need to choose another lyapunv function?

Thank you

Fras
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    Well, $V(S,I) = S+I$ doesn't even have a strict local minimum at $(K,0)$ to begin with, so it's not a Lyapunov function no matter what $\dot V$ is... – Hans Lundmark Feb 17 '24 at 10:43
  • Also, it's not true that $(K,0)$ is stable for all values of the parameters! It makes a difference whether $(\gamma+\mu)/\beta$ is greater or less than $K$. – Hans Lundmark Feb 17 '24 at 10:47
  • Yes sorry I forgot to add that condition.. I am quite new to this area of maths so I am not sure how to choose a Lyapunov function for this problem. – Fras Feb 17 '24 at 10:56
  • Even if I try to use $V(S,I) = S + I - K$ which has a local minimum at $(K,0)$ I still get the same answer as above – Fras Feb 17 '24 at 11:00
  • $V(S,I)=S+I-K$ also doesn't have a local minimum there! Lyapunov functions are typically chosen as a sum of squares or something similiar. – Hans Lundmark Feb 17 '24 at 11:02
  • What could I choose then for my problem? – Fras Feb 17 '24 at 11:04

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I suggest you try another approach, namely that you rule out periodic orbits using the Dulac–Bendixson theorem with $1/I$ as your Dulac function.

Hans Lundmark
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