SECOND BOUNTY!
i) SI MODEL
Consider
\begin{align} \frac{dS}{dt} &= \mu N -\frac{\beta S I}{N} - \nu S\\[2ex] \frac{dI}{dt} &= \frac{\beta S I}{N} -\nu I \end{align} Where $N=S+I$ is the total population.
If we assume $\mu =\nu$, the above reduces to:
\begin{align} \frac{dS}{dt} &= -\beta S I + \nu I\\[2ex] \frac{dI}{dt} &= \beta S I -\nu I \end{align}
The equilibrium points:
\begin{align*} e_1 : \left( S_1^*, I_1^*\right)&= \left(1, 0\right), \\[2ex] e_2 : \left( S_2^*, I_2^*\right)&= \left(\frac{\nu }{\beta}, \frac{\nu}{\beta}\left(\frac{\beta}{\nu}-1 \right)\right) \end{align*}
where $\mathcal{R}_0 = \beta/\nu$.
The set $\Omega = \left\lbrace \left(S,I\right)\in \mathbb{R}_+^2 : S\geq 0, I \geq 0, S+I \leq 1 \right\rbrace$ is our domain of definition. This set is a positively invariant set for our system.
Now to prove the global (asymptotically) stability of $e_1$ is straightforward, we use the function $V = I$ and the result follows.
Now my question is, how would I construct a Lyapunov function for $e_2$? I know it should be in the form of a Volterra function but, how can I choose a parricular one? any ideas?
ii) SIS MODEL
Consider
\begin{align} \frac{dS}{dt} &= \mu N -\frac{\beta S I}{N}+ \gamma I - \nu S\\[2ex] \frac{dI}{dt} &= \frac{\beta S I}{N} -(\gamma+\nu) I \end{align} Where $N=S+I$ is the total population.
If we assume $\mu =\nu$, the above reduces to:
\begin{align} \frac{dS}{dt} &= -\beta S I + (\gamma +\nu) I\\[2ex] \frac{dI}{dt} &= \beta S I -(\gamma+\nu) I \end{align}
The equilibrium points:
\begin{align*} e_1 : \left( S_1^*, I_1^*\right)&= \left(1, 0\right), \\[2ex] e_2 : \left( S_2^*, I_2^*\right)&= \left(\frac{\left(\gamma+\nu\right)}{\beta}, \frac{\left(\gamma+\nu\right)}{\beta}\left(\frac{\beta}{\gamma+\nu} -1\right)\right), \end{align*}
where $\mathcal{R}_0 = \beta/(\gamma+\nu)$.
Again, as in case (i) we have the set $\Omega = \left\lbrace \left(S,I\right)\in \mathbb{R}_+^2 : S\geq 0, I \geq 0, S+I \leq 1 \right\rbrace$ as our domain of definition. This set is a positively invariant set for our system.
Analogously as earlier, to prove the global (asymptotically) stability of $e_1$ is straightforward, we use the function $V = I$ and the result follows.
How would I chose a suitable Lyapunov function for this model to analyse the endemic equilibrium $e_2$? Well, if we find a suitable Lyapunov function for case (i) then we can just replace $\nu$ with $\gamma+\nu$ for this system.
iii) SIR MODEL
Consider
\begin{align} \frac{dS}{dt} &= \mu N -\frac{\beta S I}{N} - \nu S\\[2ex] \frac{dI}{dt} &= \frac{\beta S I}{N} -(\gamma+\nu) I\\[2ex] \frac{dR}{dt} &= \gamma I -\nu R \end{align} Where $N=S+I+R$ is the total population.
If we assume $\mu =\nu$, the above reduces to:
\begin{align} \frac{dS}{dt} &= -\beta S I + \nu I +\nu R \\[2ex] \frac{dI}{dt} &= \beta S I -(\gamma +\nu) I\\[2ex] \frac{dR}{dt} &= \gamma I -\nu R -\xi R \end{align}
Using $R=N-S-I$ to kill degeneracy, this further reduces to
\begin{align} \frac{dS}{dt} &= -\beta S I + \nu -\nu S \\[2ex] \frac{dI}{dt} &= \beta S I -(\gamma +\nu) I\\[2ex] \end{align}
The equilibrium points read
\begin{align*} e_1 : \left( S_1^*, I_1^*, R_1^*\right)&= \left(1, 0, 0\right), \\[2ex] e_2 : \left( S_2^*, I_2^*, R_2^*\right)&= \left(\frac{\left(\gamma+\nu\right)}{\beta}, \frac{\nu}{\beta}\left(\frac{\beta}{\gamma +\nu}-1\right), \frac{\gamma}{\beta}\left(\frac{\beta}{\gamma +\nu}-1\right)\right)\\[1ex] \end{align*}
where $\mathcal{R}_0 = \beta/(\gamma+\nu)$.
Again, as in case (i) and (ii) we have the set $\Omega = \left\lbrace \left(S,I\right)\in \mathbb{R}_+^2 : S\geq 0, I \geq 0, S+I \leq 1 \right\rbrace$ as our domain of definition. This set is a positively invariant set for our system.
Theorem
If $\mathcal{R}_0 \leq 1$, then the disease-free equilibrium $e_1$ is globally asymptotically stable in $\Omega$.
Proof
We use the function $V = I$ and the result follows.
Theorem
If $\mathcal{R}_0 > 1$, then the endemic equilibrium $e_2$ is globally asymptotically stable in the interior of $\Omega$.
Proof
To prove the global asymptotically stability for $e_2$, consider the Lyapunov function (in the form of a Volterra function):
$$V(S,I) = \left(S-S_2^*\right)+ \left( I-I_2^*\right) -S_2^* \ln \frac{S}{S_2^*} - I_2^* \ln \frac{I}{I_2^*}. $$
working out the time derivatives along our reduced system, we arrive to:
$$\dot V = -\beta (S_2^* -S)^2 [\frac{1}{S}] \leq 0 $$
We see $\dot V$ is negative except when $\dot V$ takes on the equilibrium values so that $\dot V =0$, so we have semi-definiteness. The largest compact invariant set in $\Omega$ so that $\dot V$ is $0$ is the singleton $\lbrace{ e_2\rbrace}$. Hence, concluding from LaSalle's invariance principle, $e_2$ is globally asymptotically stable in $\Omega$.
iv) SIRS MODEL
Consider
\begin{align} \frac{dS}{dt} &= \mu N -\frac{\beta S I}{N} +\xi R - \nu S\\[2ex] \frac{dI}{dt} &= \frac{\beta S I}{N} -(\gamma+\nu) I\\[2ex] \frac{dR}{dt} &= \gamma I-\xi R -\nu R \end{align} Where $N=S+I+R$ is the total population.
If we assume $\mu =\nu$, the above reduces to:
\begin{align} \frac{dS}{dt} &= -\beta S I + \nu I +(\xi+\nu) R \\[2ex] \frac{dI}{dt} &= \beta S I -(\gamma +\nu) I \\[2ex] \frac{dR}{dt} &= \gamma I -(\xi +\nu) R \end{align}
with equilibrium points:
\begin{align*} e_1 : \left( S_1^*, I_1^*, R_1^*\right)&= \left(1, 0, 0\right), \\[2ex] e_2 : \left( S_2^*, I_2^*, R_2^*\right)&= \left(\frac{\left(\gamma + \nu\right)}{\beta}, \frac{\left(\gamma+\nu \right) \left( \xi + \nu \right) \left( \frac{\beta}{\gamma+\nu} -1 \right) }{\beta\left(\gamma + \xi+\nu \right)}, \frac{\gamma \left(\gamma+\nu \right)\left( \frac{\beta}{\gamma+\nu} -1 \right) }{\beta\left(\gamma + \xi+\nu \right)}\right) \end{align*}
So my question(s) are; how do I find suitable Lyapunov functions for systems (i), (ii) and (iv) similar to my example in (iii)? Can you also show the full solution to systems (i), (ii) and (iv) like how I presented in (iii)? This will be appreciated!
EDIT
As per Hans comments for (i):
By substituting $S=N-I$ into (2.4) we have
\begin{align} \frac{dI}{dt} &= (\beta - \nu)I - \beta I^2 \end{align}
Solving (2.5) with initial condition $I(0)= I_0$ analytically, we have the solution to the system:
\begin{align} I(t) &= \frac{I_0 (\beta - \nu)}{\beta I_0 - e^{-\left(\beta - \nu\right)t} \left[ \beta I_0 - \left(\beta - \nu\right)\right]}.\\[2ex] S(t) &= 1-I(t). \end{align}
We can make inferences of the long term behaviour of this model by examining the possible values of $\left(\beta -\nu\right)$, that is, of course when (2.6) is feasible. We have two cases \begin{align*} \beta - \nu & < 0 \\[2ex] \beta - \nu & > 0 \end{align*}
If $\beta - \nu < 0$, then $e^{-\left(\beta - \nu\right)t} \rightarrow \infty \text{ as } t \rightarrow \infty$ hence
\begin{align} \lim_{t \to \infty} I(t) = 0. \end{align}
If $\beta - \nu > 0$, then $e^{-\left(\beta - \nu\right)t} \rightarrow 0 \text{ as } t \rightarrow \infty$ hence we have the limit
\begin{align} \lim_{t \to \infty} I(t) = \frac{I_0\left(\beta-\nu\right)}{\beta I_0} = 1-\frac{\nu}{\beta}. \end{align}
$R_0 \leq 1$:
$R_0 >1$:
For (ii):
Analogously as in section 2.1, we obtain the complete solution to our system:
\begin{align} I(t) &= \frac{I_0 (\beta - \gamma-\nu)}{\beta I_0 - e^{-\left(\beta - \gamma -\nu\right)t} \left[ \beta I_0 - \left(\beta -\gamma - \nu\right)\right]}.\\[2ex] S(t) &= 1-I(t). \end{align}
As before, we make inferences of the long term behaviour of this model by examining the possible values of $\left(\beta -\gamma -\nu\right)$, that is, of course when (2.14) is feasible. We have two cases \begin{align*} \beta - \gamma-\nu & < 0 \\[2ex] \beta - \gamma -\nu & > 0 \end{align*}
If $\beta - \gamma-\nu < 0$, then $e^{-\left(\beta - \gamma -\nu\right)t} \rightarrow \infty \text{ as } t \rightarrow \infty$ hence
\begin{align} \lim_{t \to \infty} I(t) = 0. \end{align}
If $\beta - \gamma -\nu > 0$, then $e^{-\left(\beta - \gamma -\nu\right)t} \rightarrow 0 \text{ as } t \rightarrow \infty$ hence we have the limit
\begin{align} \lim_{t \to \infty} I(t) = \frac{I_0\left(\beta-\gamma -\nu\right)}{\beta I_0} = 1-\frac{\gamma+\nu}{\beta}. \end{align}
$R_0 \leq 1$:
$R_0 >1$:
Ignore the wrong equation tags..
But I'm not sure whether this edit is proving global stability for models (i) and (ii)..