Why is the largest invariant set the following?

Question

Consider this paper:

Hai-Feng Huo and Li-Xiang Feng, "Global stability for an HIV/AIDS epidemic model with different latent stages and treatment", Applied Mathematical Modelling, Volume 37, Issue 3, February 2013, Pages 1480-1489, MR3002236, Zbl 1351.34044.

Everything is understood apart from page 7 (page 1486 in the journal). Why/how did the author come to the conclusion that the largest invariant set is the singleton $\{(1,1,1,1)\}$? Is it because this singleton kills the variable terms in the Lyapunov function?

Addendum:

The authors then go on to say by LaSalle's invariance principle, the equilibrium point is globally stable, how/why can they conclude in this manner?

EDIT:

As per Martin M. W. comment:

Here is the equivalent system(please double check):

\begin{align} \dot x&= x[\beta_1 I_2^*(\frac{1}{x}-z)+\beta_2J^*(\frac{1}{x}-u)+\mu(\frac{1}{x}-1)]\\[1ex] \dot y &= y[\frac{p \beta_1 S^* I_2^*}{I_1^*}(\frac{xz}{y}-1)+\frac{q \beta_2 S^* J^*}{I_1^*}(\frac{xu}{y}-1)+\frac{\xi_1 J^*}{I_1^*}(\frac{u}{y}-1)]\\[1ex] \dot z&= z[(1-p)\beta_1 S^*(x-1)+(1-q)\frac{\beta_2 S^* J^*}{I_2^*}(\frac{xu}{z}-1)+\frac{\epsilon I_1^*}{I_2^*}(\frac{y}{z}-1)+\frac{\xi_2 J^*}{I_2^*}(\frac{u}{z}-1)]\\[1ex] \dot u&= u[\frac{p_1 I_2^*}{J^*}(\frac{z}{u}-1)] \end{align}

I omitted the fifth equation as we can decouple this from the original system.

Answer 1

Here's how I interpret that paper.

Why (1,1,1,1) is special

First, to make things simpler, I'll rewrite the equations by collapsing some constants as follows.

$x' = x[a(\frac{1}{x} - z) + b(\frac{1}{x} - u) + \mu (\frac{1}{x} - 1)]$

$y' = y[c(\frac{xz}{y} - 1) + d(\frac{xu}{y} - 1) + e(\frac{u}{y} - 1)]$

$z' = z[m(x-1) + n(\frac{xu}{z} - 1) + p(\frac{y}{z} - 1) + q(\frac{u}{z} - 1)]$

$u' = ru(\frac{z}{u} - 1)$

The paper defines a Lyapunov function $V$ such that $V' \leq 0$ everywhere, and $V' = 0$ precisely on the set $S = \{(x,y,z,u) : x=1, y=z=u\}$.

On $S$, since $x = 1$ and $u = z$,

$x' = (a + b)(1 - u)$

If I understand correctly, the constant $a + b \neq 0$, so $x' = 0$ only when $u = y = z = 1$.

That means that the trajectory of every point in $S$ will leave $S$, with the exception of the point $P = (1, 1, 1, 1)$, which we can see is actually a fixed point. That means the maximal invariant set in $S$ is just $P$.

Global stability

The authors then invoke the LaSalle invariance principle to say that the fixed point $P$ is globally asymptotically stable. There are some steps left implicit, which I will try to flesh out. So far, what they've shown is enough to say that the only possible accumulation point of a trajectory is the point $P$.

However, the standard LaSalle statement for global asymptotic stability also requires $V \geq 0$ everywhere, and that $V$ is "radially unbounded." These are close enough to true here that everything works; here's what is going on.

First of all, the authors are only interested in solutions where the variables are non-negative, and Lemma 2 says that any solution that begins with non-negative values has positive values for the rest of time. So we can restrict attention to the positive quadrant.

Second, the Lyapunov function $V$ is a sum of terms that look like $ax - b \ln x$, for each variable. However, the function $f(x) = ax - b \ln x$ has a global minimum for $x > 0$, and goes to infinity as $x \to 0$ and $x \to \infty$. It follows that any trajectory in the positive quadrant has to stay in a bounded region, away from the edges of the positive quadrant, for all time. Given that the only possible accumulation point is $P$, we can see $P$ is globally asymptotically stable.