1
$\begingroup$

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 I -\nu I \end{align} where $N=S+I$ is the total population.

If $\mu=\nu$, the above reduces to

\begin{align} \frac{dS}{dt} &= \nu -\beta S I +\gamma I- \nu S\\[2ex] \frac{dI}{dt} &= \beta S I -\nu I -\gamma I \end{align}

with 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{\gamma+\nu}{\beta}, \frac{\gamma+\nu}{\beta}\left(\frac{\beta}{\gamma+\nu} -1\right)\right) \end{align*}

I have analysed the global stability of $e_1$ however I haven't managed to find a global Lyapunov function to check stability for $e_2$. I know the Lyapunov function should composites of the Volterra function, any ideas?

EDIT

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.

Which reduces to

\begin{align} \frac{dS}{dt} &= \nu -\beta S I - \nu S\\[2ex] \frac{dI}{dt} &= \beta S I -\nu I \end{align}

The equilibrium points now read

\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)= \left(\frac{1}{\mathcal{R}_0}, \frac{1}{\mathcal{R}_0}\left(\mathcal{R}_0-1\right)\right). \end{align*}

The biological feasible region for this system is the simplex in the first quadrant, the set being $\Omega = \left\lbrace \left(S,I\right)\in \mathbb{R}_+^2 : S\geq 0, I \geq 0, S+I \leq 1 \right\rbrace$. This set is a positively invariant set for our system.

Theorem: If $\mathcal{R}_0 > 1$, then the endemic equilibrium $e_2$ is globally asymptotically stable in the interior of $\Omega$.

Proof: Consider the Lyapunov function

\begin{equation} 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^*}. \end{equation}

This function is positive definite since $V\left(S,I\right) \geq 0$ and $V\left(S,I\right)=0$ at the equilibrium point $e_2$. The derivative of $V$ along solutions of (2.3) and (2.4) we have

\begin{align*} \dot V &= \dot S +\dot I - \frac{S_2^*}{S}\dot S-\frac{I_2^*}{I}\dot I\\ &= \nu(1-S) -\beta S I + \beta S I -\nu I -\frac{\nu}{\beta S}\left( \nu -\nu S -\beta S I \right) - I_2^*\left(\beta S - \nu\right)\\ &= \nu(1-S) -\frac{\nu}{\beta S}\left( \nu -\nu S \right) - I_2^*\left(\beta S - \nu\right)\\ &= \nu(1-S) -\frac{\nu}{\beta S}\left( \nu -\nu S \right) - \left(1-S_2^*\right)\left(\beta S - \beta S_2^*\right)\\ &= \nu(1-S)\left[1-\frac{\nu}{\beta S}\right] - \left(1-S_2^*\right)\left(\beta S - \beta S_2^*\right)\\[1ex] &= \nu(1-S)\left[\frac{\beta S -\beta S_2^*}{\beta S}\right] - \left(1-S_2^*\right)\left(\beta S - \beta S_2^*\right)\\[1ex] &= \nu\beta\left(S-S_2^*\right)\left[\frac{1-S}{\beta S} -\frac{1-S_2^*}{\beta S_2^*} \right]\\[1ex] &= -\nu\beta\left(S_2^*-S\right)\left[\frac{\beta S_2^*\left(1-S\right) - \beta S\left(1-S_2^*\right)}{\beta S\left(\beta S_2^*\right)}\right]\\[1ex] &= -\beta\left(S_2^*-S\right)\left[\frac{\beta S_2^* -\beta S}{\beta S}\right]\\[1ex] &= -\beta\left(S_2^*-S\right)^2\left[\frac{1}{S}\right]\\[1ex] & \leq 0 \end{align*}

We notice $\dot V$ is always negative with the exception of the special case where $S$ and $I$ take on the the endemic equilibrium values, thus we can say $\dot V$ is semi positive definite. We see, when $S\rightarrow 0$ or $S \rightarrow \infty$, $\dot V \rightarrow \infty$. Similarly, when $I\rightarrow 0$ or $I \rightarrow \infty$, $\dot V \rightarrow \infty$. We can now conclude that $V$ is a Lyapunov function for our system and according to Lyapunov stability theorems, the endemic equilibrium $e_2$ is globally asymptotically stable in $\Omega$.

$\endgroup$
6
  • $\begingroup$ Shouldn't it be impossible to have a global Lyapunov function, since there are multiple equilibria? Or are you ok with a negative semi-definite derivative of the Lyapunov function? $\endgroup$ Commented Aug 5, 2021 at 20:01
  • $\begingroup$ @Math I don't think your reduction in the case $\mu=\nu$ is right. For example, $N$ is no longer nessecairly preserved by it, despite being constant in the original equation. $\endgroup$
    – Pax
    Commented Aug 5, 2021 at 22:35
  • $\begingroup$ @KwinvanderVeen I am okay with negative semi-definite then I can apply LaSalle's invariance principle to prove GAS on endemic equilibrium. $\endgroup$
    – user644376
    Commented Aug 10, 2021 at 11:27
  • $\begingroup$ @pax if we take $S=S/N$ and $I=I/N$(I know this isn't good notation...) this reduces to the above and since $\mu = \nu$ surely we can substitute? $\endgroup$
    – user644376
    Commented Aug 10, 2021 at 11:33
  • $\begingroup$ @KwinvanderVeen Actually we don't need to apply LaSalle's Invariance principle here since this system is simple, but we will in more complicated models later. $\endgroup$
    – user644376
    Commented Aug 10, 2021 at 11:49

1 Answer 1

1
$\begingroup$

This is just in reference to the $\mu=\nu$ case. In this case one can check that $\dot{N}=0$ so we may consider it as a fixed variable. We now compute that $$\dot{I}=\frac{\beta}{N}(N-I)I-(\gamma+\nu)I=-\frac{\beta}{N}I^2+bI,$$

where we notate $b=\beta-\gamma-\nu$. We assume that $b\neq 0$, though this case can be dealt with similarly. This admits two stable points, $I=0$ and $I=bN/\beta$. This equation is solved by $I=0$ and $$I(t)=\frac{b\exp(bt)}{c+N^{-1}\beta\exp(bt)},$$

The behavior around $c\sim 0$ corrosponds to $I=bN/\beta$, and the behavior around $c^{-1}\sim 0$ corrosponds to $I=0$.

For $b<0$, the point $I=0$ is stable, and $I=bN/\beta$ is unstable. When $b>0$ the point $I=0$ is unstable, and $I=bN/\beta$ is stable. In both cases, the Lyponauv exponent away from the unstable point is $|b|$.

$\endgroup$
4
  • $\begingroup$ Maybe I'll edit another model I did before to see what I need, please have a look in due time! $\endgroup$
    – user644376
    Commented Aug 10, 2021 at 11:36
  • $\begingroup$ I did find the solution you posted however I am looking for proving the global stability of the endemic equilibrium when $\mathcal{R}_0>1$. If you have a look at the edited question, you'll see what I mean :) $\endgroup$
    – user644376
    Commented Aug 10, 2021 at 12:22
  • $\begingroup$ why isn't my reduction right in the original post? we have $\mu = \nu$ so this implies $\dot S = \nu N - \frac{\beta S I}{N} -\nu S$. and taking the fractions $S/N$ and $I/N$ as $S$ and $I$, we have have $\dot S = \nu -\beta S I -\nu S$. $\endgroup$
    – user644376
    Commented Aug 19, 2021 at 12:35
  • $\begingroup$ And is your solution analysing local stability or global stability? $\endgroup$
    – user644376
    Commented Aug 19, 2021 at 13:10

You must log in to answer this question.