# Why is the largest invariant set the following?

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?

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.

• Consider adding a bit more context: the question seems to be a nice one, but not all the potential answerers may have the time to read the first seven pages of the paper in order to realize what's the problem. Jan 12 at 10:55
• @DanieleTampieri I don't think reading all pages are required. We need to look at the end form of the Lyapunov function(equation 3.16) and the paragraph directly below eq. (3.16) to draw the conclusion if I am not mistaken.
– Math
Jan 12 at 11:17
• I agree with you that this is not required: however, a bit more contexts customarily helps the potential to decide whether he/she can answer. It's just a matter of (hig level) marketing ;). As you can see, a just few edits helped to perceive the question in a better way (even if an answer is still not guaranteed). Jan 12 at 11:40
• @DanieleTampieri I agree in the general sense with you! However, in this case, I don't know whether extra details will help with my question(s) :)
– Math
Jan 12 at 11:44
• Is your question about LaSalle's principle, or how it applies to the equation? If the latter, here's an idea. Can you rewrite Eq. 2.1 (the differential equation) in the variables (x,y,z,u) used in the paragraph you have questions about, and then post it here? That would make this question self-contained and more accessible to MathOverflow readers (and in the process, the answer might become clear anyway). Jan 12 at 13:26

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.

• Good answer, can you expand a little on the last paragraph? I mean why is it applicable here?
– Math
Jan 14 at 10:34
• Added some detail about how LaSalle applies here, let me know if this is useful. Jan 14 at 12:44
• Thank you, this was useful!
– Math
Jan 14 at 14:05