If we multiply a positive definite matrix with a negative scalar, the resulting matrix would be negative definite. Proof of this is available in the internet. Please do take a look :)
Krasovskii's Theorem deals with choosing Lyapunov function. If F^(x) is not negative definite, it does not mean the system is unstable. It simply means system stability can't be ascertained by the chosen Lyapunov function. So, it is not a necessary condition :)
Thank you for the good explanations. My doubt is since V(X) is our choice as you said in the previous videos, can we always take the positive definite function x1^2+x2^2 as V(X) ? then why this method is used?
But that V(x) does not always necessarily represent the energy function of our system satisfactorily. According to Lyapunov Stability, we need our dV/dt to be negative definite for system to be stable. But V = x1^2 + x2^2 doesnot always guarantee a negative definite dV/dt. In such cases, randomly choosing a V(x) that satisfies lyapunov stability will be a painful exercise. Here, Krasovskii' Method or Variable Gradient Method makes our search for V(x) easier :)
Thank you for a great video! Do we have a necessary version of this result? Could you also add some references (e.g., books, papers) to the description? Thanks.
Sorry, I didn't exactly understand what you meant by necessary version. Could you please elaborate? As for references, 1. Applied Nonlinear Control by Slotine and Li - amzn.to/2Ed8Rw6 2. Nonlinear Control Systems by Alberto Isidori - amzn.to/3l5VeQv 3. Nonlinear Systems by Hassan K Khalil - amzn.to/3aG0zsA You can also find original papers, but I'm afraid most of them are behind paywalls.
@@Topperly This method is a sufficient condition for stability. By necessary, I meant that are these conditions (must be) necessary if the system is asymptotically stable. Thank you for the references, I got them all.
Krasovskii's Theorem deals with choosing Lyapunov function. If F^(x) is not negative definite, it does not mean the system is unstable. It simply means system stability can't be ascertained by the chosen Lyapunov function. So, it is not a necessary condition :)
Hey Topperly I am very early viewer of your channel and Its help me a lot. I am really thankful. I have one doubt that if in -F^(x) the first minor have (-2 +6x1^2) and second minor have |{(-2 +6x1^2)(-2 +6x2^2)}-0| then how we will say that -F^(x) is +ve definite or not. Or does F^(x) is semi -ve definite? then system will be stable or unstable.....or what is your opinion. Thanking you for your great work.
I'm a little confused here. What's your F^(x) again? Can you please write all it's element in a row and mention the dimensions of the matrix so that I can take a look?
Here since the principal determinants are sign indefinite, (using sylvester's theorem)we can't conclude sign definiteness of F^(x). And by definition, Krasovskii's theorem only state that if F^(x) is negative definite, then eqbm state at origin is asymptotically stable. Here, we can't conclude sign definiteness of F^(x), hence stability cannot be ascertained using Krasovskii's theorem.
Thank you for video. But, I have a question for you. if in a mathematical model there are two equilibrium points. how to find the lyapunov function using the krasovskii method?
Hi Intan, Lyapunov Theorems only comment about the stability of equilibrium point at origin. And Krasovskii's Theorem also thereby deal with the equilibrium point at origin only :)
@@Topperly For example, in the sir model, there are two equilibrium points and will be analyzed for global stability. could it be the krasovskii method?
Thank you madam for your good explanation..
You are most welcome
Poli.. Sanam👌
Thankseeey! 😌😂
very good visualization of the content. Could you provide subtitles or a script for this video? Would appreciate it very much
Will add subtitles at the earliest :)
Subtitles done.
Thank you
you took negetive common in last step from the matrix is that step nessassary
@8.49 why did u add f*(X)+f(x) to get f^(x)
Please watch the video again from 02:40 :)
Can you please tell me what are the similarities between lyapunov stability criteria for linear systems and krovaskki method??????????????
Hi Chirag,
Krasovskii's method is an aid to check lyapunov stability. It's explained in this and previous video of the playlist :)
At 9:35 we can't take -ve common since matrix determinant doesn't work like this, please have a look
If we multiply a positive definite matrix with a negative scalar, the resulting matrix would be negative definite. Proof of this is available in the internet. Please do take a look :)
mam thank you for video but i want to ask is it necessary condition for stability?
Krasovskii's Theorem deals with choosing Lyapunov function. If F^(x) is not negative definite, it does not mean the system is unstable. It simply means system stability can't be ascertained by the chosen Lyapunov function. So, it is not a necessary condition :)
Thank you for the good explanations. My doubt is since V(X) is our choice as you said in the previous videos, can we always take the positive definite function x1^2+x2^2 as V(X) ? then why this method is used?
But that V(x) does not always necessarily represent the energy function of our system satisfactorily. According to Lyapunov Stability, we need our dV/dt to be negative definite for system to be stable. But V = x1^2 + x2^2 doesnot always guarantee a negative definite dV/dt. In such cases, randomly choosing a V(x) that satisfies lyapunov stability will be a painful exercise. Here, Krasovskii' Method or Variable Gradient Method makes our search for V(x) easier :)
@@Topperly understood, thank you for the quick response and your concern in doubt clearing
@@Topperly Thanks very much for these videos
Thank you for a great video! Do we have a necessary version of this result? Could you also add some references (e.g., books, papers) to the description? Thanks.
Sorry, I didn't exactly understand what you meant by necessary version. Could you please elaborate?
As for references,
1. Applied Nonlinear Control by Slotine and Li - amzn.to/2Ed8Rw6
2. Nonlinear Control Systems by Alberto Isidori - amzn.to/3l5VeQv
3. Nonlinear Systems by Hassan K Khalil - amzn.to/3aG0zsA
You can also find original papers, but I'm afraid most of them are behind paywalls.
@@Topperly This method is a sufficient condition for stability. By necessary, I meant that are these conditions (must be) necessary if the system is asymptotically stable. Thank you for the references, I got them all.
Krasovskii's Theorem deals with choosing Lyapunov function. If F^(x) is not negative definite, it does not mean the system is unstable. It simply means system stability can't be ascertained by the chosen Lyapunov function. So, it is not a necessary condition :)
Madam if Fcap(X) is not negative definite(or positive definate) then we can say the system is unstable?
thank you
your content is very good
keep going
If Fcap(X) is not negative definite, then dV/dt is not negative definte either. Means system is unstable
@@Topperly Can't there be any other V(x) that may satisfy V(dot) is negative definite?
Hey Topperly I am very early viewer of your channel and Its help me a lot. I am really thankful. I have one doubt that if in -F^(x) the first minor have (-2 +6x1^2) and second minor have |{(-2 +6x1^2)(-2 +6x2^2)}-0| then how we will say that -F^(x) is +ve definite or not. Or does F^(x) is semi -ve definite? then system will be stable or unstable.....or what is your opinion. Thanking you for your great work.
I'm a little confused here. What's your F^(x) again? Can you please write all it's element in a row and mention the dimensions of the matrix so that I can take a look?
@@Topperly the ; F^(x)= [R1;R2]= (2-6x1^2) 0 ; 0 (2-6x2^2)]
Here since the principal determinants are sign indefinite, (using sylvester's theorem)we can't conclude sign definiteness of F^(x).
And by definition, Krasovskii's theorem only state that if F^(x) is negative definite, then eqbm state at origin is asymptotically stable.
Here, we can't conclude sign definiteness of F^(x), hence stability cannot be ascertained using Krasovskii's theorem.
@@Topperly Interesting what I had thought. Thank you.
Thank you for video. But, I have a question for you. if in a mathematical model there are two equilibrium points. how to find the lyapunov function using the krasovskii method?
Hi Intan,
Lyapunov Theorems only comment about the stability of equilibrium point at origin. And Krasovskii's Theorem also thereby deal with the equilibrium point at origin only :)
@@Topperly For example, in the sir model, there are two equilibrium points and will be analyzed for global stability. could it be the krasovskii method?
I don't think so. Krasovskii can only determine the stability of equilibrium point at origin.
@@Topperly I'm dizzy now because I think about college assignments
Haha...I can understand.
By the way, if you only need to check stability of equilibrium points, you can use phase plane analysis for that :)