as before, thank you for explaining so thoroughly. As you said, even if the math itself is rather confusing (to me, as i'm not too familiar with the symbology or the more complex grammar), the graphics are pretty intuitive and make sense... So far at least!
Dear sir, at the time stamp around 29:39, you showed how the signs of f'(x*) determines the character of the stability of the point x*. But f'(x*) is evidently the second-order derivative of x(t); hence when its sign is positive, it should indicate local minima, an intuitively stable equilibrium point (opposite for -ve sign of f'(x)). But your qualitative result is yielding the opposite outcome, i.e., unstable point. Don't they contradict each other? Can you please shed some light on it? Thank you
I think your confusion results from conflating a derivative with respect to TIME t ( d/dt, which is represented here by an over dot) and a derivative with respect to the STATE x ( d/dx, which is represented here by a prime). So while dx/dt = f(x), f'(x) means df/dx, so it does not denote a second derivative of x(t) with respect to time. It is the instantaneous derivative of the state change (dx/dt) with respect to the state.
Dear Prof. Ross, thank you for posting this series of lectures. May I ask why is it the case that when x dot > 0, it moves to the right, and when x dot < 0, it moves to the left?
Maria, this is because we usually depict the x-axis such that x increases to the right. So x dot > 0 means x is increasing; thus, motion to the right. And vice-versa.
The example equation here at 1:23, dx/dt = sin(x), is just a mathematical example. It is not from a population dynamics model, so that's why x can be any real number, positive or negative. In this example, x = -π is just a state of this abstract system that is stable in the sense that any small deviation of the state x from -π from shrink, taking the state back to -π. Hopefully that explanation helps.
Rupa, yes, just equilibrium of motion. The treatment done here is general for any system that can be described mathematically by ordinary differential equations, and is it necessarily tied to physics or physical forces.
I want to say thank you from my heart 😻 Very well-explained🙏🙏
Always welcome
Such a fantastic series, thank you so much!
as before, thank you for explaining so thoroughly. As you said, even if the math itself is rather confusing (to me, as i'm not too familiar with the symbology or the more complex grammar), the graphics are pretty intuitive and make sense... So far at least!
Dear sir, at the time stamp around 29:39, you showed how the signs of f'(x*) determines the character of the stability of the point x*. But f'(x*) is evidently the second-order derivative of x(t); hence when its sign is positive, it should indicate local minima, an intuitively stable equilibrium point (opposite for -ve sign of f'(x)). But your qualitative result is yielding the opposite outcome, i.e., unstable point. Don't they contradict each other? Can you please shed some light on it?
Thank you
And one more, at 30:51, can I call this point x* a "semi equilibrium" point?
I think your confusion results from conflating a derivative with respect to TIME t ( d/dt, which is represented here by an over dot) and a derivative with respect to the STATE x ( d/dx, which is represented here by a prime). So while dx/dt = f(x), f'(x) means df/dx, so it does not denote a second derivative of x(t) with respect to time. It is the instantaneous derivative of the state change (dx/dt) with respect to the state.
@@sayanjitb As long as you define a semi equilibrium point this way, yes. It is not a terminology I use in these lectures.
@@ProfessorRoss yes sir, it's my bad. Now I understand. Thank you sir.
Dear Prof. Ross, thank you for posting this series of lectures. May I ask why is it the case that when x dot > 0, it moves to the right, and when x dot < 0, it moves to the left?
Maria, this is because we usually depict the x-axis such that x increases to the right. So x dot > 0 means x is increasing; thus, motion to the right. And vice-versa.
Another engaging video. I am going slower than I expected to but hopefully the end of this course is the stable x* for my journey.
x is stable at -π
What does it mean for the population?
The example equation here at 1:23, dx/dt = sin(x), is just a mathematical example. It is not from a population dynamics model, so that's why x can be any real number, positive or negative. In this example, x = -π is just a state of this abstract system that is stable in the sense that any small deviation of the state x from -π from shrink, taking the state back to -π. Hopefully that explanation helps.
sir are you following the book Strogatz
Indeed. All the details regarding the section and other helpful info are in the video description
Great Lecture. Thanks
I'm glad it was helpful.
Sir, just to clarify, this treatment is for equilibrium of motion, not equilibrium of force right?
Rupa, yes, just equilibrium of motion. The treatment done here is general for any system that can be described mathematically by ordinary differential equations, and is it necessarily tied to physics or physical forces.