Hi Jad,

No question is too naive.

Moreover, people say "The only stupid question is the question which is not asked." :-)

Actually, before I try to answer, I must make sure that I understand your question alright.

Are you asking about stability of Linear Time-Invariant (LTI) systems?

In such a case, the eigenvalues of the system matrix, or the poles of the transfer function, are sufficient to define the stability or instability of the system.

If you take the simple function of time x(t)=exp(-t), it is clear that it vanishes (goes to x=zero) as time tends to infinity

If instead you take the simple function of time x(t)=exp(t), it is clear that it diverges (x tends to infinity) as time tends to infinity.

Now, the transfer function of first function is 1/(s+1), which implies that its “pole” is negative (or in the Left Half Plane – LHP- of the s plane)

Instead, the transfer function of second function is 1/(s-1), which implies that its “pole” is positive (or in the Right Half Plane- LHP- of the s plane).

If you now use the state space representation of the systems, the poles become eigenvalues of the system.

The transfer function of any complex system can be decomposed in its simple components. So, if all poles of the transfer function are in LHF, then all components converge to zero, yet if any one of the poles is in the RHP, then the result is divergence.

Second order terms in transfer function may correspond to complex poles (and corresponding eigenvalues) which in the time domain correspond to sinusoidal time-response. If the pole is in LHP, the sinusoidal response converges and if in RHP diverges, while on the imaginary axis may result in steady oscillations..

Stability of nonlinear systems is more complex and requires Lyapunov stability theory, etc.

 In an attempt to simplify the analysis, people calculate the location of fixed points of the nonlinear system and try to use a linear approximation of the nonlinear system around those fixed points. If the linearized system comes out to be asymptotically stable, then the system is asymptotically stable in a small neighborhood around the fixed point. Here, your “naïve” question actually is not naïve at all. All is based on the simple fact that second or third power of a small entity is smaller than first power (which came out negative) and can be neglected.  However, there is no way to know how large the region of stability really is. If the linearized system comes out to be simply stable (zero real-part eigenvalues) and not asymptotically stable, no conclusion at all can be drawn.

I will not make it a dissertation, so I will stop here. If you need more explanations, or if your question is different, please feel free to come back.

Best regards,

Itzhak  

 Thanks a lot Itzhak! That is helpful.

Do you have any comment about the last part of my question? "Maybe more specifically, can one find a nonlinear problem..."

Let's stay in the continuous-time domain for an instant.

If you system is x'=f(x)=-x+x^3, then f(x)=0 gives you x^3-x=(x^2-1)x=0, with 3 fixed points x=0, x=1, x=-1.

Now, around x=0, you can ignore x^3 and approximate the system as the linearized xdot=-x and... everybody is happy: around x=0, we can say that the system is asymptotically stable. How far away can you go from x=0 and be sure that the trajectory would be attracted back to x=0 is YOGIAGAM (i.e., your guess is as good as mine :-)

The other points are more difficult to treat this way. One way is to first write z=x-1 (or x=z+1), zdot=xdot, then zdot=(x-1)(x+1)x=z(z+2)(z+1) and again you brought a fixed point to the origin. Make the computations and leave the first order term in z and get conclusions.

Then, same for z=x+1 ( or x=z-1)...

Now, Lyapunov analysis is more rigorous, yet finding the appropriate Lyapunov function for a really complex system is not  an easy issue. Power systems are  highly nonlinear and hard to analyze and they do use linearized analysis around some equilibrium points. Everything had been working alright for decades until ...  one day, the entire US network came down.

In Japan, they said: "Oh, this cannot happen here" and... the Japanese net came down a couple of weeks later.

Never happened before, never happened since, and yet... it happened.

Let's hope for the best. :-)

Itzhak

Hi Jad,

You may want to read about the Kalman conjecture:

https://en.wikipedia.org/wiki/Kalman%27s_conjecture

In discrete-time, I have a paper about it:

http://personalpages.manchester.ac.uk/staff/joaquin.carrascogomez/Kalman.pdf

We did an analysis of the "attraction" of the orbit. I have found cases where the orbit is unstable. So the linearisation is stable at any point, but it is possible to find an orbit that is unstable in the sense that an small variation makes the system reach a different stability point. You can see a simulation here:

http://maplecloud.maplesoft.com/application.jsp?appId=5742954790518784

In the case of an orbit with just one point, I think that you will always find a region about it where it is locally stable if the Jacobian is "stable".  The main problem with linearisation is that you are not really sure what will happen around the point. From a practical perspective, we want to have global stability or at least local stability within a known set. Global and input/output stability of nonlinear system have been study for the last 50-60 years within control engineering, 

BR,

J

Thanks a lot Joaquin.

Would you and Itzhak then agree with me that blind faith application of linearization in order to study stability using eigenvalues comes with a bit of hand-waving and oversimplification in non-specialized texts?

Can I draw a vague analogy with the case of pointwise differentiability (analogy hope that linear stability applies) versus continuity over an interval (versus analogy study of stability over a set)?

"Blind faith application" of even well established concepts and formulas may lead to great mistakes and false conclusions, as i was surprised to find out when dealing with stability of nonlinear systems. 

However, I would be more careful here. Simple problems are simple, yet tough problems are just that: tough. If this is a theoretical example fro some paper, you may decided to throw it away, yet if this is your power plant or some other nonlinear/flexible/noisy creature that you have to live with, you do your best to get the best of it. Linearization of a very complex nonlinear system when you have nothing else in hand is the best you can do. The problems only tell you that you must be careful. In the simple scalar (i.e., first-order) example that I gave you, there is a small neighborhood around x=0 . How large it is and when and how the higher order terms affect stability is another issue, so you must be careful. 

BTW, I would not use "linear stability" here (or maybe anywhere), because could be confusing. I would rather use "stability of the linearized system" if this is what you mean.

I am working in adaptive control, where I cannot assume any initial situation even close to stable, so I have to deal with stability of the nonlinear system.

Sorry, yet I don't really understand the second paragraph with the analogies. 

Thank you for your infinite patience.

I have an author saying that "any smooth nonlinear system is approximable by a linear one at the fixed points, points of steady state", he goes on to do just that for an example, studies the eigenvalues of the approximating linear system, and then assumes the conclusion are valid for the nonlinear one. Is there a theorem establishing the validity of this approach, in the sense of a theorem that the eigenvalue stability conclusions of the linear approximating system imply the existence of an open set around the fixed point of the original system where the stability claims are the same? It can very well be that this is trivially true and I am too stupid, or that I am reading a book that is far too high-level. 

In other words, the fact that the nonlinear system is approximable by the linear one around a fixed point, does not imply (for me, intuitively) that the stability analysis of latter is transferrable to an open set around the fixed point of the former, and not only exactly at the fixed point.

About my sloppy analogy. A function can be differentiable at exactly a single point. When it is so, the linear approximation does not tell us anything reliable about the the function, more specifically I quote this: 

"Example 2 If f : R −→ R is a continuously differentiable function which has a strict local minimum at some point x0, then most students tend to expect that there is a small neighbourhood (x0 − δ, x0 + δ) such that f is decreasing on (x0 − δ, x0] and increasing on [x0,x0 +δ).  The function ...shows that this is wrong..." (https://arxiv.org/pdf/1302.1065.pdf)

I thought the situation in the quote can be analogous to the pitfalls of applying linearisation+eigenvalue analysis to nonlinear functions without additional conditions and restrictions (I don't know what those would be, and I would like to know if they exist).

Very clear answer thank you!

"Take Taylor expansion (the most popular one). If your function is analytic at the equilibrium point then you get a nonzero sized neighborhood around this equilibrium where the approximation converges and stability properties will be "transferred"."

May I ask where I would find a proof of this?

The Jacobean stability is based on the approximate linearization of the system. But in Lyapunov stability it is possible to investigate the stability of a system without approximately liberalization technique.

Thanks! Also for the book recommendations!

I hope it's not too easy :) I also hope you can help with other related questions as they come by.

This is related:

"If we have a n-dimensional linear system of differential equations ... with a single fixed point at the origin we can observe several types of behaviors, such as saddle points, spirals, cycles, stars and nodes, which are well-understood. We classify these cases based on the eigenvalues of the matrix A used to classify the system. With a nonlinear system the behavior of the system is more difficult to analyze. Fortunately, we are not left completely in the dark. We can find the Jacobian matrix, or “total derivative”, J, corresponding to the system and evaluate it at a fixed point to obtain a linear system with a characteristic coefficient matrix. The Hartman- Grobman theorem tells us that, at least in a neighborhood of the fixed point, if J’s eigenvalues all have nonzero real part then we can get a qualitative idea of the behavior of solutions in the nonlinear system. Such qualitative characteristics we can glean include whether solution trajectories approach or move away from the equilibrium point over time, and whether the solutions spiral or if the equilibrium point acts as a node. " (An Undergraduate’s Guide to the Hartman-Grobman and Poincar ́e-Bendixon Theorems) (https://www.math.hmc.edu/~levy/181_web/Zimmerman_web.pdf)

Very good reference.

I happen to deal with real systems and so, first of all, I must know what they are doing. I mean, are they stable, at least? If you start from finite points, will all trajectories remain bounded? Some of them? Can we define some region or regions where they are guaranteed to be bounded? 

Then, we may go and see where those bounded trajectories are going.

For the first part, which in some cases may also be the last, we have the Lyapunov approach.  If for a second order system you select a positive definite Lyapunov function, such as V=x1^2+x2^2, and the derivative "along the trajectories" comes out to be negative definite, such as Vdot=-(x1^2+x2^4), then everything is bounded and all trajectories must end at Vdot=0, namely, global asymptotic stability.

If the result is only negative semidefinite, such as Vdot=-x2^2, you still know that V cannot increase, so boundedness is guaranteed. You actually get stability in the sense of Lyapunov, which is less than  asymptotic stability, yet more than simple boundedness. To put is simply, no matter how close you start, you don't go away. 

If, however, Vdot=-(x1^2+x2^2)+100, then it is not even negative semidefinite. However, it is easy to see that, if the trajectories go out of the  region (x1^2+x2^2)

Before I even properly read your answer I wonder: what did I do to deserve such welcoming and detailed answers?! Thanks!

Take it easy. First of all, I see humans and I am only trying to be human. 

Besides, I passed same path you are passing now, with lots of questions, confusions and misunderstandings.

If you also reach the stage when you actually read some of the recent publications, I was also shocked to discover that correct mathematical concepts, which had been automatically applied to become well-established stability concepts, actually had been applied where they do not hold and ended in wrong, though also "well-established," conclusions.

I can only hope my attempts at explaining things help and do not add confusion.

Makes a lot of sense, and I am not as surprised as I would have been some years ago.

I thought to maybe add another point, because, while simple cases are simple, tough cases are just that: tough.

You analyse stability of a system that could be unstable. If you managed to find a positive definite Lyapunov function with positive definite derivative, then it tells you that the system is unstable. However, the derivative may not necessarily be positive definite and then, maybe you try to linearize your system around its equilibrium point and at least see what happens there.

If, in the opposite case, you have a positive definite Lyapunov function and you managed to find a positive definite Lyapunov function with negative definite derivative, then it tells you that the system is globally asymptotically stable.

However, in other than class-room examples, the Lyapunov derivative ends being at most negative semidefinite. A lot of effort has been dedicated to reach the conclusion that the bounded trajectories tend to Vdot=0, which I managed to replace by "trajectories end at Vdot identically equal zero." This new concusion implies that the trajectories not only tend to Vdot=0, but also to remain there for ever. 

Still, only if Vdot identically equal zero gives you just one solution, say x=0, you can still end with the asymptotic stability conclusion.

However, in more complex (realistic) cases, you may end with more than one  solution, say x=0, x=-1, x=+1. Vdot negative semidefinite tells you that the system does not diverge. However, are all points stable equilibrium points? Not necessarily. Here, you linearize around each one and, as it happened in one of my examples, you may find out that x=0 is unstable while the others are stable. Then why does x=0 come out as equilibrium point? What we call "unstable equilibrium point" actually is an "accident" of pure Math. It implies that, if the trajectory starts there and no perturbation, even infinitesimally small, is present, it will stay there. Otherwise, it goes away. Anyway, in such cases, I think linearization after the Lyapunov analysis is a must. 

Can I have references to the related papers you wrote please? This is well summarized!

During the last few years, I had a few works on stability. The Conference paper with an example of the kind is the 2016 ICNPAA paper  

Can Stability Analysis be really simplified? (Revisiting Lyapunov, Barbalat, LaSalle and all that)

The paper itself is only going to be published in the Conference Proceedings. However, I uploaded the Technical Report with same title and content on Research Gate and you can download it.

Example 3 shows how you can get the desired results with one Lyapunov function, while others would use more; also, it shows how to decide about each equilibrium point.

(But, please, don't forget that you asked for it! :-)     

Let me return the threat back in this form: please don't forget that *you* offered advise, I might come back with questions :)

Any question (or comment), any time and as many times as you feel you need!

Nevertheless, please don't forget that you asked for the paper on nonlinear stability. :-)

I found a good book about this: "Geometric Control of Mechanical Systems" by Bullo and Lewis.

Answering myself, this is a related theorem: https://en.wikipedia.org/wiki/Hartman–Grobman_theorem