# Global Lyapunov function to examine global stability?

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}

\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

$$$$V(S,I) = \left(S-S_2^*\right)+ \left( I-I_2^*\right) -S_2^* \ln \frac{S}{S_2^*} - I_2^* \ln \frac{I}{I_2^*}.$$$$

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 $$\Omega$$.

• 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? Commented Aug 5, 2021 at 20:01
• @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.
– Pax
Commented Aug 5, 2021 at 22:35
• @KwinvanderVeen I am okay with negative semi-definite then I can apply LaSalle's invariance principle to prove GAS on endemic equilibrium.
– user644376
Commented Aug 10, 2021 at 11:27
• @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?
– user644376
Commented Aug 10, 2021 at 11:33
• @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.
– user644376
Commented Aug 10, 2021 at 11:49

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 $$|b|$$.

• Maybe I'll edit another model I did before to see what I need, please have a look in due time!
– user644376
Commented Aug 10, 2021 at 11:36
• 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 :)
– user644376
Commented Aug 10, 2021 at 12:22
• 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 $\dot S = \nu -\beta S I -\nu S$.
– user644376
Commented Aug 19, 2021 at 12:35
• And is your solution analysing local stability or global stability?
– user644376
Commented Aug 19, 2021 at 13:10