# Can we combine this two Lyapunov functions (which imply local stability by separate) to conclude global stability?

Let $$x(t)\in\mathbb{R}^n$$ constrained to a dynamical system $$\dot{x}(t) = f(x(t))$$ for some vector field $$f:\mathbb{R}^n\to\mathbb{R}^n$$. Moreover, the dynamical system has a unique equilibrium point at the origin.

My goal is to conclude that for this system (with a particular $$f$$ I have) the origin is globally asymptotically stable. So far, I have found two Lyapunov function candidates $$V_1(x),V_2(x)$$ which both are positive definite and radially unbounded. Moreover, I managed to show that they comply

1. $$\dot{V}_1(x)<0$$ $$\ \forall x : \|x\|
2. $$\dot{V}_2(x)<-c$$ $$\ \forall x : \|x\|>r_2$$ and some constant $$c>0$$
3. $$0

If it was the case that $$r_1>r_2>0$$ instead of item 3), we would be done: global stability follows since item 2) ensures that $$x$$ reaches $$\|x\|\leq r_2$$ in finite time from any initial condition, and then one can use 1) to conclude $$x$$ reaches the origin asymptotically from here.(as pointed out in the comments, this argument requires nuance, nevertheless, the important part is the following). However, from item 3) this is not the case. Thus, I have this disk $$r_1<\|x\| in which my Lyapunov stability test is inconclusive. My hope is to show that trajectories $$x(t)$$ cannot stay in the disk $$r_1<\|x\| forever but ultimately reach $$\|x\|.

My question is: from the information given here, do you think is possible to conclude global asymptotic stability either by constructing a new Lyapunov function from $$V_1,V_2$$ or by checking some condition on $$f$$ to conclude trajectories cannot stay in $$r_1<\|x\|?

Indeed, there are many counterexamples in which this is not possible. But my question here is if there is a way to check if it is possible: what properties do you suggest me to check for $$f$$. I'm looking for suggestions, ideas or relevant bibliography that you think can help me.

EDIT: In case this problem is easier, or adds useful information, I can also show the following instead of the original items:

1. $$\dot{V}_1(x)<0$$ $$\ \forall x \in M_1 = \{x : V_1(x)
2. $$\dot{V}_2(x)<-c$$ $$\ \forall x \in M_2=\{ V_2(x)>r_2'\}$$ and some constant $$c>0$$
3. $$D:=\mathbb{R}^n\setminus(M_1\cup M_2)\neq \emptyset$$

We are left to show that there are no invariants in $$D$$.

EDIT 2: After some discussion in the comments, I think the main essence of my question, is more on what happens when the Lyapunov arguments are inconclusive on this non empty set $$D$$ (bounded, but not containing the origin). A concrete diagram for the state space of what $$r_1,r_2,M_1,M_2,D$$ looks in the case I am interested is the following: • I assume you are trying to show this for a particular system. Did you simulate it for multiple initial conditions with $r_1<\|x(0)\|<r_2$ and did all converge to the origin? Jan 12 at 18:39
• Yes, I am interested in a particular system. I have made many simulations with those initial conditions and Im very confident it is globally asymthotically stable (it always converge to the origin) Jan 12 at 18:45
• I am not sure if it is sufficient for your case but you could try the Bendixon criterion to rule out a limit cycle exists in your disk. Jan 12 at 19:00
• @SampleTime Does that criterion also rule out limit manifolds? Jan 12 at 19:15
• I would say that even $r_1>r_2$ is not sufficient. Generally, it is not enough to say that $\dot{V}_1<0$ to prove that $\|x\|$ remains below $r_1$ as we do not know if the level set of $V_1$ is a circle. Jan 12 at 22:28

## Bounds on $$V$$

We start considering the case when we have bounds on the levels of $$V$$, and we will use these results to study the original question later.

Let us say that there exist $$C_1$$ such that $$\dot{V}_1(x)<0$$ for all $$x$$ such that $$V_1(x)>C_1$$ and $$C_2$$ such that $$\dot{V}_2(x)<0$$ for all $$x$$ such that $$V_2(x). I.e., the $$C_1$$ constant defines the level set for $$V_1$$ such that $$x$$ is converging for large initial conditions, and $$C_2$$ defines the level set for $$V_2$$ such that $$x$$ is converging for small initial conditions. When the system is converging? The intuition is that we need a safe switch from $$V_1$$ to $$V_2$$.

### One way...

Choose an arbitrary small constant $$\delta>0$$ and define $$\Omega = \{x: V_1(x)\le C_1+\delta\}$$. From the properties of $$V_1$$ we know that $$x$$ reaches $$\Omega$$ in finite time, and since $$V_1$$ is radially unbounded, the set $$\Omega$$ is bounded and contains the origin. Then to prove the convergence we need that $$V_2(x) for all $$x\in\Omega$$. Then for any $$x_0\in\Omega$$ we have $$V_2(x_0) and due to the properties of $$V_2$$ we conclude that $$V_2\to 0$$. Note that the trajectory $$x$$ can leave the set $$\Omega$$ as it is not invariant.

We need the constant $$\delta$$ because otherwise, it is possible that $$x\to\Omega$$ asymptotically and we never switch to $$V_2$$.

### ...or another

Another way to see the same idea is to define the set $$\mathcal{M} = \{x:V_2(x). From the properties of $$V_2$$ we know that the set $$\mathcal{M}$$ is invariant ($$V_2$$ is monotonic decaying for all $$x\in\mathcal{M}$$ and thus $$V_2$$ will always remain below $$C_2$$), contains the origin, and the origin is attractive for all initial conditions in $$\mathcal{M}$$ since $$V_2\to 0$$.

Then to prove the convergence we need to show that the set $$\Omega = \{x: V_1(x)\le C_1+\delta\}$$ lies inside $$\mathcal{M}$$, $$\Omega\subset\mathcal{M}$$. Then any initial condition outside $$\mathcal{M}$$ converges to $$\Omega$$ (and thus to $$\mathcal{M}$$) due to the $$V_1$$ analysis, and for all trajectories starting in $$\mathcal{M}$$ the origin is attractive due to the $$V_2$$ analysis.

### NB

We do not say that $$C_1>C_2$$ or $$C_2>C_1$$. What is important is the relation between the sets $$\Omega$$ and $$\mathcal{M}$$.

## UPD: Bounds on $$x$$

Ok, let us see the original question. Suppose there exists $$r_1$$ such that $$\dot{V}_1<0$$ for all $$\|x\|>r_1$$ and $$r_2$$ such that $$\dot{V}_2<0$$ for all $$\|x\|, and $$r_2. Let us replace $$r_1$$ with $$R_1 = r_1+\delta$$, then we know for sure that $$x$$ enters the ball $$\|x\| in finite time. Choose $$C_1$$ as the largest value of $$V_1$$ on the boundary $$\|x\|=R_1$$. Then the set $$\Omega=\{x:V_1(x)\le C_1\}$$ is attractive and invariant, and the ball $$\|x\| lies inside $$\Omega$$. Note also that the ball $$\|x\|\le R_1$$ is not necessarily invariant.

Next, choose $$C_2$$ as the smallest value of $$V_2$$ on the boundary $$\|x\|=r_2$$. Then the set $$\mathcal{M} = \{x:V_2(x) lies inside the ball $$\|x\|\le r_2$$, this set is invariant and the origin is attractive for all trajectories starting in $$\mathcal{M}$$.

So what we know at the moment is that the set $$\Omega$$ is invariant and attractive for all trajectories starting outside of $$\Omega$$ and that the set $$\mathcal{M}\subset \Omega$$ is invariant and all trajectories starting in $$\mathcal{M}$$ converge to the origin.

What is left is the set $$\mathcal{D} = \Omega-\mathcal{M}$$. As far as I see, it is sufficient to show that no trajectory staring in $$\mathcal{D}$$ can stay in $$\mathcal{D}$$, i.e., there are no invariants in $$\mathcal{D}$$.

• Thanks! I agree with your arguments about the relation between $r_1,r_2$, they are not important by themselves. Indeed, the only thing that matters is the relation between $\Omega, \mathcal{M}$. For the first part of your explanation, I agree that if these two sets overlap as you exposed, stability would follow. However, my case is the opposite. I have this non empty set $\mathcal{D}$ for which I want to show that there are no invariants. Unfortunately I failed to expose this correctly in my question, by incorrectly introducing this $r_1,r_2$ values, sorry for that. Jan 13 at 10:32
• I modified my question at the end, using some of the ideas from here. I think the main point is how to approach the problem of invariants in $\mathcal{D}$. That was the main essence of what I wanted to ask: I have a nonempty set in which my lyapunov arguments are inconclusive. Hopefully, I understood your answer correctly. Thanks! Jan 13 at 10:39