I was trying to prove the asymptotic stability of the trivial equilibrium $(0,0)$ of the two-dimensional non linear ODE system:

\begin{align} \frac{dH}{dt}=\mu\frac{(H+F)^2}{K^2+(H+F)^2}-d_1 H-H\left(\sigma_1-\sigma_2\frac{F}{H+F}\right)\\ \frac{dF}{dt}= H\left(\sigma_1-\sigma_2\frac{F}{H+F}\right)-(p+d_2)F \end{align} where $H$ and $F$ are dependent variable and positive.

All other parameters are non negative with $\sigma_2>\sigma_1$.

I tried using Jacobian matrix but it blows up when I plug in $(0,0)$ into the Jacobian matrix. So I thought of defining a positively invariant region around the equilibrium point and then try to prove that it is positively invariant. That is, the solutions that enter into the region must converge to the equilibrium and hence it is stable. But I don't know how to prove it using differential inequalities!

Any kind of help/suggestion/guidance will be appreciated! Thanks!

  • $\begingroup$ No, I haven't tried yet because I couldn't come up with the right Lyapunov function. How do you know if the function you chose, is right? Sorry if that's a too basic question. $\endgroup$ – math Jul 3 '14 at 21:02

You need $K$ to be strictly positive. Because if $K=0$ and $\mu>0$, then in a neighborhood of the origin, $dH/dt \approx \mu >0$. So, suppose $K>0$.

In the first equation, the terms $\mu\frac{(H+F)^2}{K^2+(H+F)^2}$ and $H\sigma_2\frac{F}{H+F}$ are of second order of smallness near the origin. Thus, the sign of $dH/dt$ is determined by $-(d_1+\sigma_1)H$ which of course suggests stability.

In the second equation, $-H\sigma_2\frac{F}{H+F}$ is of second order. Neglecting it, we are left with $H \sigma_1 - (p+d_2)F$. This looks troublesome, but if $H$ goes to zero, $F$ will be forced to follow.

Let's summarize. For every $\epsilon>0$ there is a neighborhood of the origin in which $$\begin{split}\frac{dH}{dt}&<-(d_1+\sigma_1 )H +\epsilon (H+F) \\ \frac{dF}{dt} & < H \sigma_1 - (p+d_2)F +\epsilon (H+F) \end{split}$$ (I work in the positive quadrant $H,F>0$, which is what you are interested in). Hence, $$\frac{d(2H+F)}{dt} <-(2d_1+\sigma_1 )H - (p+d_2)F +3\epsilon (H+F) $$ which is negative, provided $2d_1+\sigma_1>0$, $p+d_2>0$, and $\epsilon$ is chosen sufficiently small.

I took $2H+F$ instead of $H+F$ so that the coefficient of $H$ on the right would have $\sigma_1$, increasing the chance of the coefficient being negative.

| cite | improve this answer | |
  • $\begingroup$ Thank you very much for the wonderful explanation and too step by step. Thanks a lot! In the meanwhile, I tried something similar. Just wanted to make sure if this is right too: \begin{align} \frac{dH}{dt}+\frac{dF}{dt}\leq \mu\frac{(H+F)^2}{K^2+(H+F)^2}-\delta (H+F) \end{align} where $\delta=min\{d_1,d_2,p\}$. Suppose $H+F=y$. Thus for small enough $\tilde{y}$, the set $$Z_{\tilde{y}}=\{y:y<\tilde{y}, H>0,F>0\}$$ is positively invariant with inward pointing flux. So every solution starting from a point in the set enters a positively invariant region. Thus converges to origin. $\endgroup$ – math Jul 3 '14 at 22:54
  • 1
    $\begingroup$ @math Yes, if $d_1,d_2,p>0$ (you said "nonnegative" in the post), this works. $\endgroup$ – user147263 Jul 3 '14 at 22:57
  • $\begingroup$ Oh yes, you are right! Thanks for pointing it out! $\endgroup$ – math Jul 3 '14 at 22:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.