- #1

ShayanJ

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1) Stands still at a point.

2) Moves periodically.

3) Moves from one equilibrium point to another with no periodicity.

Thanks

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- Thread starter ShayanJ
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- #1

ShayanJ

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1) Stands still at a point.

2) Moves periodically.

3) Moves from one equilibrium point to another with no periodicity.

Thanks

- #2

Pythagorean

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4) Unphysical systems can also run off to infinity

5) a chaotic attractor is a non periodic attractor

- #3

ShayanJ

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4) Unphysical systems can also run off to infinity

5) a chaotic attractor is a non periodic attractor

1) What is the definition of unphysical?

2) It seems to me that that #5 is just the same as #3!

- #4

Pythagorean

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2) the chaotic attractor is itself an attractor, no equilibrium point is reached or the system wouldn't be chaotic (or would only be transiently chaotic).

- #5

ShayanJ

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2) the chaotic attractor is itself an attractor, no equilibrium point is reached or the system wouldn't be chaotic (or would only be transiently chaotic).

At first I should ask does the fact that you're discussing these issues mean that I had a right understanding of the theorem? If yes, then why it seems that what you're saying is contradicting the theorem, by introducing additional behaviours?(Maybe you're talking about 3D dynamical systems while the theorem is only about 2D dynamical systems!) If no, then why aren't you answering my initial question? Because you don't know enough about the theorem too?

- #6

Pythagorean

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I thought you were speaking of dynamical systems in general as a background to PB. You haven't actually seemed to touch on PB itself yet.

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- #7

ShayanJ

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Actually my main reason in posting this thread is clarifying PB theorem for myself. Of course further discussions are interesting and I'll appreciate them but now I'm just lost more because your previous posts seem to be incompatible with the theorem.I thought you were speaking of dynamical systems in general as a background to PB. You haven't actually seemed to touch on PB itself yet (which pertains only to the existence of a limit cycle in a region).

So can you state the theorem in a self-contained manner which doesn't need much pre-study?

- #8

Pythagorean

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- #9

Pythagorean

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- #10

ShayanJ

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So I have problem with the definition of a fixed point. Is it simply where a spatial derivative becomes zero? Or there is more to it?

- #11

Pythagorean

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So I have problem with the definition of a fixed point. Is it simply where a spatial derivative becomes zero? Or there is more to it?

It is. However, for an unstable point (or a saddle node) this doesn't mean trajectories will ever come to the node. An unstable fixed point repels all trajectories unless it starts right on the unstable node. Think of it like the tip of a mountain. A nearby ball will quickly leave the vicinity of the mountain tip; but you can carefully balance the ball at the tip and (assuming no wind or earthquakes) it will stay there indefinitely. However, all balls don't start on the mountain tip.

- #12

ShayanJ

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But saddle points are minimums of a curve passing from that point and maximums of another(necessarily orthogonal to the former curve?) and a regular point on other curves. So its possible that the particle starts moving in the curve where the saddle point is a minimum and moves toward the saddle point. Then, when the particle reaches the saddle point, its in a stable equilibrium w.r.t. on curve and in an unstable one on another curve and a regular one on any other curves. So if there'll be no push, the particle should stand still there, right?It is. However, for an unstable point (or a saddle node) this doesn't mean trajectories will ever come to the node. An unstable fixed point repels all trajectories unless it starts right on the unstable node. Think of it like the tip of a mountain. A nearby ball will quickly leave the vicinity of the mountain tip; but you can carefully balance the ball at the tip and (assuming no wind or earthquakes) it will stay there indefinitely. However, all balls don't start on the mountain tip.

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