Hi Masahiro,

Yep, you are right! For linear systems it is enough to

look at the eigenvalues of the matrices. 

Certainly it is not necessary to use Lyapunov method in the linear case : the exponentials with negative real part let the solutions converge to zero (the only stable point since A is invertible by hypothesis on its eigenvalues).

Dear Marc,Gabriel

Thanks for your replies.

I would like to ask a related question.

If Lyapunov method is redundant in the linear case, what is the usefulness of Lyapunov equation which can be used  only in the linear case?

Is  Lyapunov equation totally useless?

Best regards

Dear Masahiro!

Looking at eigenvalues is nice from a theoretical point of view, but you will find it difficult to determine the eigenvalues of a large matrix in practice (even numerically). Solving a Lyapunov equation might be easier. And without being an expert in this field, I know that Lyapunov equation play an important role in control theory.

Regards Werner

Although stability or instability of linear systems could be extracted from eigenvalues, but  a good lyapunov function (among many possible lyapunov functions) can give us further informations about the robustness of stability with respect to additive drift terms or perturbations, often in terms of lipschitz-like deductions. while the effect of this kind of perturbations on the eigenvalues could not be determined easily.

In addition, in case of time varying linear systems, where matrix A=A(t) is time dependent, eigenvalues having negative real parts does not yield stability. But existence of a  lyapunov function (perhaps time varying) always results in "global stablity".

 Dear Mohammad

Could you suggest materials on the robustness of stability with respect to additive drift terms or perturbations and time-varying linear systems?

Best regards

Dear Masahiro!

For robustness of stability, for example you can take a look at this note:

http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-241j-dynamic-systems-and-control-spring-2011/readings/MIT6_241JS11_chap14.pdf

where it says:

"Example 14.2 Set of Bounded Norm

In this problem, we are interested in studying the stability of linear time-invariant systems of the form x_dot=(A +DELTA)x  where DELTA is a real matrix perturbation with bounded norm. In particular, we are interested in calculating a good bound on the size of the smallest perturbation that will destabilize a stable matrix A.

and for the linear time varying systems, take a look at this one:

http://www.egr.msu.edu/classes/me851/jchoi/lecture/Lect_24.pdf

Best regards

Dear Mohammad

Thanks for valuable materials.

I will read them and I will ask if any questions.

Best regards

Dear Mohammad

I have read 'Lectures on Dynamic Systems and Control'.

This is a very nice material for me!!

Now I could understand the usefulness of the Lyapunov equation for a linear system.

I would like to ask some implementation questions.

1. What Q should we give?  Can we give -E as Q (where E is an identity matrix and -E is negative definite)?

2. Given Q, we can get a solution P by solving a linear equation with n(n+1)/2 unknown variables. After we get P, we have to confirm whether P is positive definite or not.

In order to confirm it, we have to get eigenvalues of P whose size is same as A.

Is it easier to get eigenvalues of P and solve P than to get eigenvalues of A directly?

Is it due to P's symmetricity?

As for the example 14.2:

3. How do we give Q and get P under a negative definite condition of  A'P+PA'+Q ? 

4. What is the meaning of σmax(P)?

I think the eigenvalue is denoted by λ(P), so I dont't see what σmeans.

Best regards

Dear Mohammad

About question 3, I got an idea of  a solution  as follows.

Given Q>0 and e>0 where e is a scalar,  solve the equation A'P+PA'+Q +eE=0 as a usual linear equation. Such solution satisfies the original negative definite condition because A'P+PA'+Q =-eE

Dear Mohammad

I have read 'Linear Control Systems Lecture No 24 Time-Varying Systems Transition Matrix and Stability and Control'.

This is very nice material too.

I would like to ask some questions.

I have understood the solution  p.3/24 and the example at p.16/24 which shows the negative real part of eigenvalues don't guarantee the stability.

In the general case of dot(x(t))=A(t)x(t) , how can we see the stability of the system?

Do we use Lyapunov equation of form such as A'(t)P+PA'(t)=-Q where P and Q are time-invariant?

Best regards

Dear Masahiro!

i have some partial answers to your questions:

1- according to  theorem 14.1, given a Q>0 , there exists a P>0 for lyapunov equation A*P+PA=-Q  if and only if all the eigenvalues of A are in the open left half plane. So you can chose any Q>0 that you like. but  a good choice for Q , which affects P, could yield a bigger upper bound for minimum perturbation that destabilizes system, which is more desirable. (see answer 3)

2- Yes. you are right. solving lyapunov equation and computing the eigenvalues of P is computationally less expensive than computing eigen-decomposition of A. specially when dimension of A is large. there exists iterative methods for solving lyapunov equation. specially when Q is structured. 

3- if all eigenvalues of A  are strictly in the open left half plane, then for a Q>0, there exists a P>0 such that  A*P+PA=-Q. now for any Q1>0 such that ||Q1||

for stability analysis of  time-varying linear systems, the situation is much more difficult. see this similar question and answers:

https://www.researchgate.net/post/Is_anyone_familiar_with_the_stability_analysis_of_time_varying_systems

Best regards

Dear Masahiro,

The Lyapunov idea is to find a (Lyapunov) function which clearly increases as you go away from the origin and decreases as you come closer to the origin. 

So, to introduce the idea on a simple example, if a system has two variable, x and y, you may try something like V=x2+y2. If you can show that the derivative "along the trajectories of the systems" is negative definite, then it is clear that V can only decrease. As V cannot be less than zero, the derivative must end while it stops being negative and so, both Vdot and V end at zero.

Before moving from LTI to time-varying and nonlinear systems, for  the LTI system xdot=Ax you choose some positive definite V(x)=x'Px and, if you differentiate "along the trajectories of the systems, you get 

Vdot=x'(PA+A'P)x. Because you you want the result to be negative definite, you end with

 Vdot=x'(PA+A'P)x=-x'Qx. From here, you get the algebraic equation PA+A'P=-Q which must have solutions for positive definite symmetric matrices Q and P. It is interesting that, if the system is asymptotically stable, then for any Q you get a P, but not necessarily vice-versa.

For time-varying systems xdot=A(t)x(t), constant P and Q do not wok (in general), so you select the function V(x)=x'P(t)x and now the derivative is Vdot=x'(Pdot(t)+P(t)A(t)+A'(t)P(t))x=-x'Q(t)x which gives the differential equation 

Pdot(t)+P(t)A(t)+A'(t)P(t)=-Q(t) that must have solutions for uniformly positive definite symmetric matrices P(t) and Q(t). More difficult, yet there is no much choice, as in time varying systems eigenvalues are not relevant for stability any more.

So, in nonlinear systems, or even in LTI systems with some nonstationary or nonlinear controller (adaptive, etc) there is no other way but Lyapunov.