Linear Stability Analysis | Dynamical Systems 3

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Faculty of Khan

Күн бұрын

Пікірлер: 41

@estershaw7944 5 жыл бұрын

OMG this is amazing. you are literally my hero for posting this derivation. THANK YOU.

@umarbashir7196 Жыл бұрын

The video was really educating. I really enjoyed watching it.

@md.khurshedulislam1762 7 жыл бұрын

very good explanation for beginners. Keep it up sir. Want more video

@funtofun321 6 жыл бұрын

Very nice video. It helped me to understand the concept of linearization around a critical point..., lot of thanks. Keen to watch your more videos..

@ytjoemoore94 3 жыл бұрын

Imagine the expanding brain meme that starts with Kahn Academy and ends with this guy!!

@man9mj 2 жыл бұрын

Very excellent demonstration. Thank you

@jnxmaster 6 жыл бұрын

Great videos. You have a new subscriber.

@biaschatterjee9836 4 жыл бұрын

Very helpful video, thank you sir😊👍

@marrytesfu3163 6 жыл бұрын

Simply WOW!!

@loganthrashercollins 7 жыл бұрын

What exactly is meant by "very small perturbations"? Is this related to limits or is does "very small" simply denote that it is within a small enough interval that the behavior doesn't change? Also: great job explaining. I really enjoy these nonlinear dynamics videos in particular. I look forward to the rest of the series!

@FacultyofKhan 7 жыл бұрын

Thank you! It's not related to limits, even though I've used epsilon (the same greek letter you see in the epsilon-delta limit proofs). Small epsilon just means that the amount by which we disturb the particle from its fixed point/steady state is so small that epsilon^2, epsilon^3 etc are negligible. For example, if epsilon were 0.01, then 0.01^2 = 0.001, which is very small (1/100th) of the original epsilon. When epsilon^2/^3 etc are negligible, we can safely apply the linear approximation as I did in this video. The main caveat is that you're restricted to looking at disturbances in the local area of x_f. You don't know about global behavior via linear stability analysis. Hope that helps, and feel free to ask more questions!

@tjmitchell21 5 жыл бұрын

Wonderful video, thanks! It's really helped. :)

@shivrajahirwar4334 5 жыл бұрын

nice explanation

@farzanehazargoshasbi284 2 жыл бұрын

Your videos are very useful.Thank you very much🙏🏻

@pierreretief Жыл бұрын

Stunning. Thanks

@薛启谭 2 жыл бұрын

thanks for teaching, it's helpful

@TheCerebralOne 2 жыл бұрын

@6:00 to clarify, if the derivative f' at the fixed point is zero then the linear stability analysis is inconclusive?

@dantecalanza1330 3 жыл бұрын

Thank you po Sir.☝️

@VictorHugo-xn9jz Ай бұрын

But when would we actually make use of a large perturbation to analyze a stable point? Why would global fixed points be of interest?

@FilipeSilva1 5 жыл бұрын

Thank you.

@harshitjuneja9462 3 жыл бұрын

so it is basically the second derivative test from single-variable calculus, right?

@FacultyofKhan 3 жыл бұрын

It's similar, in that it allows you to determine the nature of the point where the first derivative is 0 (i.e. the fixed point). The difference here is that you're differentiating dx/dt = f(x) with respect to x as opposed to taking the straight second derivative (i.e. d^2 x/dt^2).

@اممحمد-ق2ه 3 жыл бұрын

Can you help please How classification of critical points of system in three equation in 3d

@aissamaissam5618 4 жыл бұрын

Thanks alot for this nice video. I want know the condition that makes the following discrete system stable: x(k+1)=Ax(k)+B Where A and B are a known square matrices an x' in R^n. Thanks in advance sir.

@rajatmishra4189 2 жыл бұрын

Thanks a lot buddy 😁

@hayleycoyle9726 2 жыл бұрын

You totally lost me at f'(x_f) = -1 . Where did this come from? How did you figure this out?

@Noumankhan-nx9nz 4 жыл бұрын

thanks alot

@maurocruz1824 3 жыл бұрын

You make that exponential solution for epsilon. That reminds me of the generator of a Lie group. Is there any connection?

@zhiyongli7526 4 жыл бұрын

4:34, The fixed pint is not the stationary point, why should it be zero.

@FacultyofKhan 4 жыл бұрын

For this video, they mean the same thing. A fixed point is where the function x(t) is momentarily fixed (i.e. dx/dt = f(x) = 0), and a stationary point is where the function x(t) is momentarily stationary (i.e. dx/dt = f(x) = 0).

@zhiyongli7526 4 жыл бұрын

@@FacultyofKhan Ok, thanks for the explanation. That is a stationary point other than a fixed point. Fixed point x(t)-t=0, stationary point: dx(t)/dt=f(x)=0.