1:04:45 - Actually there is a "universal trick" for getting conserved quantities. Noether's theorem tells us that a conserved quantity exists for every continuous symmetry of the system, and it tells us how to identify that quantity. So you first look for your symmetries, and as soon as you know them you can turn the crank and grind out the conservation equations. It is still a system dependent thing, though, because to do the above-mentioned procedure you need to know the "Lagrangian" of the system, and sometimes it might not be obvious what the Lagrangian is. In purely mechanical systems it's generally kinetic energy minus potential energy. But in more exotic systems the Lagrangian is, well, more exotic.
In case anyone is confused in the future, Dr. Strogatz makes a mistake interpreting the homoclinic orbit in the double-well example. The homoclinic orbit doesn't start from the right side and approach the middle. It starts from the middle, taking forever to move significantly far from the saddle point, eventually rolls down, past the trough and up the far side until it reaches a stop, THEN comes back and approaches the middle point again, taking forever to do so.
Hi I would appreciate the help of someone who took the course or has the material to provide me with the assignments or problem sets in this course which are typically chosen from the textbook just problem numbers from the textbook for each assignment. Kind regards
All trajectories "inside" the homoclinic orbit take a finite time before getting back at their "starting" point and all the trajectories "outside" the homoclinic orbit, also, take a finite time before getting back at their starting point. Why do you elect to say that on the homoclinic orbit, it would take an infinite amount of time to reach the saddle point ? rather than it would take an infinite amount of time to reach the saddle point and then, stay there infinitely (because v=0, there)? While the other orbit may have a period which is not constant, I fail to see why their period would become closer and closer to infinity as they approach the monoclinic orbit (mainly for those "inside" the homoclinic orbit, at least). I tried to solve analytically the problem, hoping to get a finite expression for x(t) invertible for t=t(x) given a starting x, but it seems to imply a division by zero for the particular solution of the ODE (I may have made an error though) for all possible solutions. I am even at the point to ask myself if both interpretations (asymptotical approach, versus finite time + infinitely idle) are different, or not, to start with, ... are they? Another way to twist my mind against that asymptotical approach is to perturb V(x) with eta(x) while hopefully keeping the conservative E. In the perturbed case, the trajectory will take a finite amount of time to reach x=0 ( with eta(x) negative ) since it is now on a "peanut" orbit rather than the original homoclinic orbit..
Here is my explanation, why it takes infinite time for a solution starting on the homoclinic trajectory to reach (0, 0). The system (x', y')=(y, x - x^3) has the same local behavior as its linearized version (x', y') = (y, x). Indeed, close to 0 the cubic term x^3 becomes negligibly small compared to x. (In fact, it has the exact same topological behavior around (0,0) see Hartman Grobman Theorem) Now solving the linearized system we get the two eigenspace solutions given by: (1, 1)*e^(1*t) ("unstable solution") and (1, -1)*e^(-1*t) ("stable solution"). Now let t go to infinity for the stable solution. You will approach (0, 0), but you will never reach it.
This is such a fascinating topic. It's just kind of a shame that it inevitably seems to become a slog through a mountain of problem-specific algebra. The concepts are simple, and it doesn't take much time to cover them, I guess.
1:04:45 - Actually there is a "universal trick" for getting conserved quantities. Noether's theorem tells us that a conserved quantity exists for every continuous symmetry of the system, and it tells us how to identify that quantity. So you first look for your symmetries, and as soon as you know them you can turn the crank and grind out the conservation equations.
It is still a system dependent thing, though, because to do the above-mentioned procedure you need to know the "Lagrangian" of the system, and sometimes it might not be obvious what the Lagrangian is. In purely mechanical systems it's generally kinetic energy minus potential energy. But in more exotic systems the Lagrangian is, well, more exotic.
Not all systems of the form dx/dt = f(x) are Lagrangian systems, so the point about there not being a universal trick stands.
In case anyone is confused in the future, Dr. Strogatz makes a mistake interpreting the homoclinic orbit in the double-well example. The homoclinic orbit doesn't start from the right side and approach the middle. It starts from the middle, taking forever to move significantly far from the saddle point, eventually rolls down, past the trough and up the far side until it reaches a stop, THEN comes back and approaches the middle point again, taking forever to do so.
39:06-Thanks Dr. Strogatz for such informative and cheerful lecture.
DR. Strogatz, thank for a fantastic lecture on Conservative Systems. Mechanical and Nonlinear systems are good examples of Conservative Systems.
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(sorry just going through this course and thought of how youtube normally works lol. I promise not to s***post on these again)
just a once-upon-a-time math major learning some new stuff in her free time.
@ 42:37 face on the board looking at the point.
Hi
I would appreciate the help of someone who took the course
or has the material to provide me with the assignments or problem sets in this
course which are typically chosen from the textbook just problem numbers from
the textbook for each assignment.
Kind regards
All trajectories "inside" the homoclinic orbit take a finite time before getting back at their "starting" point and all the trajectories "outside" the homoclinic orbit, also, take a finite time before getting back at their starting point. Why do you elect to say that on the homoclinic orbit, it would take an infinite amount of time to reach the saddle point ? rather than it would take an infinite amount of time to reach the saddle point and then, stay there infinitely (because v=0, there)? While the other orbit may have a period which is not constant, I fail to see why their period would become closer and closer to infinity as they approach the monoclinic orbit (mainly for those "inside" the homoclinic orbit, at least). I tried to solve analytically the problem, hoping to get a finite expression for x(t) invertible for t=t(x) given a starting x, but it seems to imply a division by zero for the particular solution of the ODE (I may have made an error though) for all possible solutions. I am even at the point to ask myself if both interpretations (asymptotical approach, versus finite time + infinitely idle) are different, or not, to start with, ... are they?
Another way to twist my mind against that asymptotical approach is to perturb V(x) with eta(x) while hopefully keeping the conservative E. In the perturbed case, the trajectory will take a finite amount of time to reach x=0 ( with eta(x) negative ) since it is now on a "peanut" orbit rather than the original homoclinic orbit..
infinite 💫stil not accurate enough, by means of 0 dot would be more accountable for the 0 being not that at all. t= 0 dot or 0. and 0.. or even 0 ...
Here is my explanation, why it takes infinite time for a solution starting on the homoclinic trajectory to reach (0, 0).
The system (x', y')=(y, x - x^3) has the same local behavior as its linearized version (x', y') = (y, x). Indeed, close to 0 the cubic term x^3 becomes negligibly small compared to x. (In fact, it has the exact same topological behavior around (0,0) see Hartman Grobman Theorem) Now solving the linearized system we get the two eigenspace solutions given by: (1, 1)*e^(1*t) ("unstable solution") and (1, -1)*e^(-1*t) ("stable solution"). Now let t go to infinity for the stable solution. You will approach (0, 0), but you will never reach it.
Nothing is being "elected." He's just telling us how an object would behave on the various paths.
This is such a fascinating topic. It's just kind of a shame that it inevitably seems to become a slog through a mountain of problem-specific algebra. The concepts are simple, and it doesn't take much time to cover them, I guess.
Was the camera person sleeping? Professor:: excellent, CameraMan:: worst...