1.1 Stability of Fixed Points PROOF | Nonlinear Dynamics

[@XahhaTheCrimson]

What a clear explanation!

[@virtually_passed]

Thanks!

[@zaynbashtash]

Love the mathematical proof videos

[@r2k314]

You do such a good job. It seems so straightforward to teach this way. Why is it so rare. It's a disgrace!

[@jackzweifel4493]

This is a really great video, thanks!

[@virtually_passed]

glad you liked it :)

[@vitalysarmaev]

Perfect explanation!

[@Patowallac3]

Easy to follow, straight to the point & (background isn't white) loved it

[@GoalWalker]

5:17 I don't understand how the second fixed point is stable. In my mind, it seems like a ball on slope that is slightly to the right of the second fixed point would fall to the right by "gravity," rather than get "sucked/pulled" upwards (positive y-axis) (4:25). I guess I am thinking of an imaginary ball on the function's slope with -y axis as gravity. But is this wrong?

[@GoalWalker]

After review, so if I am understanding this right, for a parabola of f(x) = x^2 - 25. The fixed points would be x = -5 and x = 5. Due to the negative slope at x = -5, this point would be stable, thus a ball on the left of x=-5.1 would roll towards x=-5, while a ball at x = -4.9 would get magically sucked up to x = -5 (this still makes no sense to me with gravity). Then, fixed point at x = 5 would be unstable because of the positive slope at that point (this I can agree with ease). What am I missing here?

[@virtually_passed]

@@GoalWalker Hey, let's consider your parabola dx/dt = f(x) = x^2-25. You are right there will be two fixed points x_1* = -5 and x_2* = +5. Note that the derivative at each of these fixed points are df/dx(x_1*) = -10 < 0 and df/dx(x_2*) = +10 > 0. This means that the first fixed point x_1* is stable and the second fixed point x_2* is unstable. But now let's try and understand this intuitively. Consider the first fixed point again x_1* = -5. A little bit to the left of this fixed point the flow will be to the right (since f(-5.1) > 0), and a little bit to the right of this fixed point the flow will be to the left (since f(-4.9) < 0). This means if I start really close to -5, i'm going to be sucked back towards -5. That's what makes it stable. Now let's intuitively understand the second fixed point x_2* = +5. Note that f(4.9) < 0 and so the flow is to the left, and note that f(5.1) > 0 so the flow is to the right. This means the flow is AWAY from the fixed point. This means that if you start near +5 you will diverge away from that point. That's what makes this fixed point unstable. Hope this helps

[@GoalWalker]

@@virtually_passed Ah ok! Thank you so much! To me, if I treat dx/dt = f(x) as velocity, then it really makes sense instead of viewing f(x) as position.

[@zuhair95]

Incredible explanation! Thanks !

[@klevisimeri607]

I inspire to make videos like you!

[@goldeer7129]

Amazing video, very clear ! What is the software that you use to draw your visuals ?

[@virtually_passed]

I use manim CE. I python library :)

[@malawigw]

f-dash? why not f prime?

[@virtually_passed]

Arbitrary preference :)

[@malawigw]

@@virtually_passed ok I thought it was standard lingo in non linear diff eqs

[@XuanJr.]

This is not totaly right, because you can only throw the higher order term when (x-x*) is small.

[@virtually_passed]

Linearization can only ever be trusted in a very small region near the fixed point (where x-x* is close to 0)

[@thomasanderson9539]

Hartman-Grossman theory be like: am I a joke to you Edit: H-G is just a sufficient condition for justifying the linearization. Even if H-G does not hold, one can still get away with a fairly good linear approximation for common cases.

[@willie333b]

Topic