Hi, Abbas. The Lyapunov theorem says that a point of equilibrium of a system is stable if there exists a positive definite function V such that its time derivative Vdot is non-positive for all trajectories of a system. In your case, stability is thus guaranteed if B < |A|. More in p.108... of my book Continuous-Time Systems, Springer, 2007. It is available here.

Dear Yuriy and P. Subramanyam,

Thank you for your reply. The system is indeed stable. If A was function of the tracking error e, it would be possible to say that the system is ISS (input-to-state-stable) under some conditions. However, in this case A is not function of e. Do you know if there is any workaround to find the bounds of the tracking error?

Dear Abbas, I think it does not matter in your case. If A

Consider my answer as an intuitive description of what happens. You can find more detailed info in the book "Nonlinear analysis" of H. Khalil. Well, your system is practically stable. This means that the trajectories converge in a region around the equilibrium point. Such a region is contained in a ball which radius depends on B (as B is large, as the ball which contains such a region is large). Moreover, if your inequality holds for any time interval and for any arbitrarily large B, then you have Input-to-state stability (ISS). Moreover, if your inequality holds for any initial condition, then you have a global result.

This means that your tracking is not pefect. However, in stedy-state (i.e. after a sufficient large time) your trajectories go close to the reference signal, but without never perfectly track it. A certain steady-state bound such a tracking error is given by a function of the magnitude of B (which in turns provides the ball that I cited above). If B=0, then you have perfect tracking.

Finally, it can happen that \dot V does not depend on e, but is in general a function of time. You can see by setting d/dt V(e(t)) := d/dt W(t) := F(t), where I have considered the composition W(t):=V(e(t)).

You can find an application of those concepts in one of my paper also.

Something doesn't sound right in the formulation of the problem. Positive or Negative Definite is function of the argument. So, V is Positive definite. However, if, as it is claimed, -A is always negative with no relation with the error AND also A>B, the Lyapunov derivative would ALWAYS be Negative, not Negative Definite. However, this would bring the POSITIVE Definite V to -infinity, which is a contradiction.

So, if you say that your system is stable, there must be another mechanism there that ultimately does bring Vdot to zero. If, for example, B can become large enough to affect the sign of Vdot and yet Vdot remains Negative Semi-definite, all may end at Vdot=0 and from there you get what the ultimate errorr is as a function of B.

If B is such that could make Vdot sometimes positive sometimes negative, you may get or not get stability, yet this is beyond Lyapunov (and Barbalat and LaSalle) theory.

Dear Yuriy, Bilal, Ubaldo and Itzhak,

I thank you for your interesting and helpful discussions. I will take some time to analyze your comments along with your research papers. I will allow myself, if you don't mind, to contact you for more precision.

Itzhak, yes, A is not related to the error and it is always positive. However, A is not always larger than B. So, as you pointed out, Vdot is sometimes positive and other times negative. You are saying that this is beyond Lyapunov theory. Did you encounter or come across such situation before?

Even before I am trying to respond: You may ask (me, at least) any question any time. Besides, no one is under any test here, so "Don't know" is a perfectly legitimate answer.

Now, for your specific problem, the first general answer indeed is "Don't know." Only in LTI systems can one say "If... then... if not... not." In nonlinear systems, if you are lucky, or after some experience, you find the right Lyapunov function and prove stability. Otherwise, you don't know and must keep looking for a better function that may end with the desired result, UNLESS you found a way to prove instability.

So, before Lyapunov, one would have had to solve any specific equation for all possible situations. Lyapunov's genius was to avoid having to do it. So, if you choose a Positive Definite function V(x) (energy-like, for intuition) of the state x, i.e, that is positive everywhere except for x=0 AND its derivative Vdot(x,t) is Negative Definite, i.e. negative everywhere except for x=0, it is easy to show that the system must end at x=0. (Otherwise it would imply that V may reach -oo, which is a contradiction).

In general

V(x(t))=V(0)+integral(Vdot(x(tau), tau) dtau) from zero to t, so as t -->oo, you reach Vfinal .

For your problem, when Vdot is positive on some segments and negative on others, you would have to prove that the integral of Vdot must be finite and in the lucky case that it is zero. However, this is some mathematics and is not part of the present stability theory. I am not sure it can have any better theory.

A more common case is when Vdot is Negative Semidefinite. Even here, there are many problems. The most commonly used tool is called Barbalat Lemma, which says that if it can be shown that V(t) has a final value and its derivative Vdot(t) is UNIFORMLY Continuous, then Vdot tends to zero. If you have the guarantee of continuity, then this alows you to find some response by solving Vdot=0.

This was good, yet better results were obtained by LaSalle with his Invariance Principle, only these results are not too commonly known.

Sorry if I am advertising a paper of my own, yet in response to some respected Colleague, who threatened to "kill all your proofs with even the tiniest discontinuity," I wrote a paper that makes order in what Barbalat and LaSalle did (in the 70's) and what has been done afterwards.

I. Barkana: "Defending the Beauty of the Invariance Principle," International Journal of Control (Published On-Line 06 Sep 2013), DOI:10.1080/00207179.2013.826385.

Then, although I said that there is no general theory, if you KNOW that your system is stable, a converse theorem says that a proper Lyapunov function exists. How to find it, this is art. Maybe you must incorporate some term in V, maybe function of A or its integral, and then it may give you some effect in the derivative.

Last but not least, I would not even dare to say linear and nonlinear in same sentence. If so much has been written about the "trivial" polynomial of order one, each nonlinear system is a new world and the miracle is that there are some general rules and theories. Knowing your specific system MIGHT help you reach conclusions that cannot be found within the general theory.

Actually, there's always good to leave it for a while and then to have a second thought.

In spite of my previous "Don't know," your case could be a good test for Barbalat Lemma BECAUSE your derivative is not negative definite or even semidefinite. If you can show, or have reson to assume, that V has a finite limit and that Vdot is continuous, then the system ends with Vdot =0, i.e., at B=A.

Dear Barkana,

Thank you for the interesting discussion. I will take a look into your paper. As you suggested, I may have to look for a better Lyapunov function or go for Barbalat Lemma.

Hi Abbas,

(...and let's stay with Itzhak)

Most publications use Barbalat Lemma that requires uniform continuity of the Lyapunov derivative, pretty strong requirement for practical applications. My paper mostly shows that you do not need this IF Vdot is negative semidefinite.

About using Barbalat Lemma when Vdot is not even semidefinite, the basic assumption that "V has a finite limit" above is a very strong requirement. In many cases, as my paper shows, the (bounded) trajectory may end oscillating around (the line, surface, hypersurface) Vdot=0, this takes us back to selecting the "right" V.

We are instinctively led to choose some simple V, yet it may not necessarily represent the system. The natural selection V=e^2 means that V=const is a simple circle (sphere, hypersphere) and that the trajectory that takes you from V=k1 to a lower V=k2 would go along the radius. This does not necessarily fit the system.

Take the simplest, linear case xdot = Ax. If you choose V=x^2, it leads you nowhere. So, we choose V=x'Px and compute the derivative Vdot=x'(PA+A'P)x=-x'Qx. So far, all was just playing with notations, yet it leads you to the algebraic Lyapunov equation

PA+A'P=-Q. But now look what curiosity. IF the system is asymptotically stable, then for any Q you choose, you will find a Positive Definite Symmetric (PDS) solution for P. Not viceversa. It seems to say that you may not decide what closed surfaces V=c may represent, yet you may chose how Vdot should behave and then the corresponding V will show.

Of course, in nonlinear systems, the differential equation somehow leads us to selection and maybe adjustments of V, yet (as far as I know) it does not give us any good idea about how we should possibly select a Vdot.

This is a lot of thought with no much help. However, along with reading some of the books and thinking about your specific problem may finally lead you towards suddenly "seeing the point" and towards the desired solution.

Because some luck can never hurt, I wish you Good Luck.

Dear Itzhak, Thank you for the advice. I certainly need to read more in the hope to find a better Lyapunov function.

Dear Abbas,

I was just made aware of a paper that you can get for free at this link:

http://cas.ensmp.fr/~praly/Telechargement/Conferences/2011_CDC_%20Astolfi-Praly.pdf

I don't know your system, so I don't know if it helps, yet it is worth a look, as it deals with the case when the Lyapunov derivative is not necessarily negative definite or semidefinite.

Dear Itzhak, thank you for your additional suggestion. I will check this paper.

Well if magnitude of A is greater than that of B then the system will be stable but nothing can be concluded about exponential stability