examstudy
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My text book states right after defining a strict Lyapunov function, "the only equilibrium is the origin x=y=0".
I'm a bit confused as to what this is referring to?
Can you not have strict Lyapunov functions for equilibrium points that are not the origin?
So you could not use Liapunov's stability theorem to show an equilibrium point that is not the origin is aymptotically stable?
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DFranklin
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So, all I've never studied this, and all I've done is skim a couple of wiki pages, but in the discussions it seems standard to assume you translate the equilibrium point to the origin,in which case the statement you quote is more saying "the equilibrium point is unique" than "the equilibrium point is always at the origin".

Is there any chance you can post the page in question (or link to the book pages on Google books or something)?
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examstudy
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(Original post by DFranklin)
So, all I've never studied this, and all I've done is skim a couple of wiki pages, but in the discussions it seems standard to assume you translate the equilibrium point to the origin,in which case the statement you quote is more saying "the equilibrium point is unique" than "the equilibrium point is always at the origin".

Is there any chance you can post the page in question (or link to the book pages on Google books or something)?
Thank you for helping.
Here is the link to the book, its pg 193
http://sgpwe.izt.uam.mx/files/users/...rsch-Smale.pdf
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DFranklin
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(Original post by examstudy)
Thank you for helping.
Here is the link to the book, its pg 193
http://sgpwe.izt.uam.mx/files/users/...rsch-Smale.pdf
The statement in question doesn't seem strictly correct, which is causing the confusion, but I do think it's what I said - that it's so standard to translate the equilibrium point to the origin, (or equivalently, reframe so you are considering the perturbation from the equilibrium point) that he's switched to (x, y) being the distance from the equilibrium point without actually ever explicitly saying so.

(The other paragraphs around this seem to explicitly contradict "the only equilibrium point is at the origin", he has a named point \bar{x} for the equilibrium point rather than just saying "the origin" in the paragraph above, and in the example immediately after he states "the z-axis consists entirely of equilibrium points").

But as I said, I'm not familiar with this area, so take my thoughts at your own risk... (Although to be honest I think this is the only reasonable interpretation).
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examstudy
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(Original post by DFranklin)
The statement in question doesn't seem strictly correct, which is causing the confusion, but I do think it's what I said - that it's so standard to translate the equilibrium point to the origin, (or equivalently, reframe so you are considering the perturbation from the equilibrium point) that he's switched to (x, y) being the distance from the equilibrium point without actually ever explicitly saying so.

(The other paragraphs around this seem to explicitly contradict "the only equilibrium point is at the origin", he has a named point \bar{x} for the equilibrium point rather than just saying "the origin" in the paragraph above, and in the example immediately after he states "the z-axis consists entirely of equilibrium points").

But as I said, I'm not familiar with this area, so take my thoughts at your own risk... (Although to be honest I think this is the only reasonable interpretation).
Yes I agree. Thank you
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