3.1 Linearization PROOF | Nonlinear Dynamics

[@virtually_passed]

Hi guys, minor correction at 4:33. In the first line, I should have said that linearization cannot be trusted when: the trace = 0 AND when the determinant > 0. Thanks to those that have corrected me.

[@gustruan]

Great explanations:) loved the series and it’s graphics

[@virtually_passed]

Thanks!

[@amritawasthi7030]

I'm so sad and happy at the same time. Sad cuz I'm really very late to discover your channel. Happy because atleast finally i found your channel. Thank you so much for all your work.

[@virtually_passed]

Thanks for your kind words

[@notnilc2107]

The edge cases at 4:24 sound wrong. Gorbman Hartmann theorem only says it can't be trusted if each eigenvalue has a purely imaginary eigenvalue. I haven't checked, but I'm pretty sure a matrix with zero determinant can still have a non-zero imaginary part.

[@virtually_passed]

Hi! Thanks for the question. It turns out that all centers, degenerate nodes, stars and non-isolated fixed points are cases where the linearization theory cannot be trusted. Non isolated fixed points occur when the determinant is = 0. Centers occur when the trace = 0 and determinant > 0 stars and degenerate noes occur when trace - 4*determinant = 0 If you want to see more information about this I'd recommend 'Nonlinear Dynamics and Chaos' by Steven Strogatz Chapter 5.2. In answer to your other question, it turns out that the eigenvalues of the A matrix can be written as: λ1,2 = 0.5*(trace +- sqrt(trace^2-4*determinant)) So that means if the determinant = 0 then its impossible to have a complex eigenvalue because tau^2 >= 0 and the stuff inside the sqrt() won't be negative. Hope that helps :)

[@snah70]

The Hartman-Grobman theorem requires that there are no eigenvalues on the imaginary axis. If the determinant is zero, then at least one eigenvalue is zero, and the number zero lies on the imaginary axis, so the HG thm doesn't apply in that case. But there is another inaccuracy in the statement in the video: if the determinant is negative, then it doesn't matter whether the trace is zero or not; it's always a saddle in that case. (Centers correspond to the trace being zero *and* the determinant positive.)

[@virtually_passed]

@@snah70 Thanks so much for your comment. Indeed, I think you're right! What I was trying to communicate is that linearization can't be trust for the following fixed points: 1) centers, (det>0, tau=0) 2) stars & degenerate nodes (tau^2-4*det = 0) 3) non isolated fixed points (det = 0) But you are correct in saying that saddles can be trusted when tau = 0 and det < 0. So in my video I have made a mistake when I claimed all fixed points with tau = 0 cannot be trusted. Instead I should have said where tau = 0 and det > 0.

[@nazishahmad1337]

Can you please tell me how you create your videos , which tools you're using and all that stuff.

[@virtually_passed]

Sure, this set of videos was made using "manim" which is a python library that was created by Grant Sanderson (3b1b). I create the hand written stuff using microsoft onenote and do the editing using filmora pro. I use a blue yetti microphone.

[@nazishahmad1337]

Thanks alot .If in future it's possible for you to make a video on how you make your video that would be really helpful.

[@berry4862]

Can you recommend books to go deeper?

[@virtually_passed]

It depends how deep you want to go. Here are three books I'd recommend based on difficulty :) 1) Nonlinear Dynamics and Chaos by Steven Strogatz (easy) 2) Nonlinear Systems by Khalil (medium) 3) Nonlinear Differential Equations and Dynamical Systems (second edition) by Ferdinand Verhulst (hard)

[@angelespinosa906]

Is the A matrix a jacobian?

[@virtually_passed]

Yes :)

[@jjing_jji4282]

The fixed points are equilibrium point right?

[@virtually_passed]

Yes, same thing.