This question seems interesting but I cant figure out how to apply Micahel E2's solution to my system:

How to write that Lyapunov function?

My system is very simple, I already have a working Lyapunov function for the following system however I would like to see what Mathematica has to offer:

\begin{align} \frac{dS}{dt} &= \nu N -\frac{\beta S I}{N} + \xi R - \nu S\\[2ex] \frac{dI}{dt} &= \frac{\beta S I}{N} -\gamma I -\nu I \\[2ex] \frac{dR}{dt} &= \gamma I -\xi R - \nu R \end{align} Let $N=1$; other parameters are arbitrary.

Working Lyapunov function:

\begin{equation} V(S,I) = S_2^*\left(\frac{S}{S_2^*}-\ln \frac{S}{S_2^*} \right)+ \frac{\gamma + \xi + \nu }{\gamma + \nu}I_2^*\left(\frac{I}{I_2^*}-\ln \frac{I}{I_2^*} \right) . \end{equation}


\begin{align*} S_2^* &= \frac{\gamma + \nu}{\beta}\\ I_2^*&= \frac{\left(\gamma+\nu \right) \left( \xi + \nu \right) \left( \frac{\beta}{\gamma+\nu} -1 \right) }{\beta\left(\gamma + \xi+\nu \right)} \end{align*}

A more complex system which I need a Lyapunov function would be:

\begin{align} \frac{dS}{dt} &= \Lambda-\beta S I + \xi R - \mu S\\[2ex] \frac{dL}{dt} &= \beta S I+(1-\eta) \delta T -(\mu+\epsilon) L \\[2ex] \frac{dI}{dt} &= \epsilon L +\eta \delta T - (\mu+\gamma+\sigma_1) I\\[2ex] \frac{dT}{dt} &= \gamma I -(\mu+\delta+\sigma_2+\alpha) T \\[2ex] \frac{dR}{dt} &= \alpha T -(\mu+\xi)R \end{align}

Another interesting system is:

\begin{align} \frac{dS}{dt} &= \Lambda-\beta S I - (\mu+\sigma) S + \theta V\\[2ex] \frac{dV}{dt} &= \sigma S -(\mu +\theta)V \\[2ex] \frac{dE}{dt} &= \beta S I -(\mu +\omega)E\\[2ex] \frac{dI}{dt} &= \omega E -(\mu+\kappa) I \\[2ex] \frac{dR}{dt} &= \kappa I -\mu R \end{align}

  • 1
    $\begingroup$ I don't know if I will have time to work on this, but I will offer this piece of standard advice: "People here generally like users to post code as Mathematica code instead of just images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find the meta Q&A, How to copy code from Mathematica so it looks good on this site, helpful." $\endgroup$
    – Michael E2
    Mar 21, 2022 at 17:58
  • $\begingroup$ I am aware of the previous solution by @MichaelE2. A couple of questions for clarity. Do you want a way of generalizing this to more differential equations? The previous example had 2 and now there are 3 in the simplest case. Or is this a question about leaving some parameters arbitrary? If they are left arbitrary and we pass a Manipulate or Animate or something, could you please give some ranges so we know what we are after? Many thanks! $\endgroup$
    – bmf
    Mar 21, 2022 at 18:50
  • $\begingroup$ @bmf Yes, the goal is to generalise to more differential equations. All parameters are arbitrary as so the lyapunov function theoretically should have some parameters as constants(see my the first example of the Lyapunov function). I don’t know what you mean by ranges.. if you mean what values the parameters take, they are all strictly positive. $\endgroup$
    – Math
    Mar 21, 2022 at 19:03


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