Dynamical Systems: Stability of Fixed Points

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RDKGames
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Just to make sure, am I able to knock parts (a) and (b) down with one function; i.e. f(x) = 2-e^{1-x}, where the fixed point is x=1.

This point is both an attractor and repeller since starting with x_0 > 1 will always take you to the fixed point no matter how far off you start, while with x_0 < 1 you will always repel away from the fixed point no matter how close or far you start as well.

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Gregorius
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I would have thought not; don't you require an attractor to have a basin of attraction that contains an open neighbourhood?
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mqb2766
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(Original post by Gregorius)
I would have thought not; don't you require an attractor to have a basin of attraction that contains an open neighbourhood?
Would be my gut reaction as well, you'd need it to attract / diverge on both sides. When the function has a unity gradient, its generally classified as neither.
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RDKGames
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(Original post by Gregorius)
I would have thought not; don't you require an attractor to have a basin of attraction that contains an open neighbourhood?
Hm, the definition I was given makes no mention of that.

Def (attractor): A fixed point x^{*} is called an attractor if it attracts orbit with initial x_0 close to x^{*}. In other words, \forall \epsilon > 0, \exists \delta >0 such that |x_0-x^{*}| < \delta \implies |f^{n}(x_0)-x^{*}| < \epsilon.

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mqb2766
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(Original post by RDKGames)
Hm, the definition I was given makes no mention of that.

Def (attractor): A fixed point x^{*} is called an attractor if it attracts orbit with initial x_0 close to x^{*}. In other words, \forall \epsilon > 0, \exists \delta >0 such that |x_0-x^{*}| < \delta \implies |f^{n}(x_0)-x^{*}| < \epsilon.

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It would only attract on one side though (as you point out in the OP), not all initial points in a ball around the "attractor".
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RDKGames
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(Original post by mqb2766)
It would only attract on one side though (as you point out in the OP), not all initial points in a ball around the "attractor".
So you're saying that the given definition is incorrect?
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mqb2766
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(Original post by RDKGames)
So you're saying that the given definition is incorrect?
The definition sounds fine to me. It would have to attract on both sides to satisfy the defn?
If all the curves have to have a gradient of 1 at the fixed point, your example sounds like c)
Can you not create two antisymmetric curves (about 1) by cutting your example into two, where one will attract and the other diverge?
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RDKGames
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(Original post by mqb2766)
The definition sounds fine to me. It would have to attract on both sides to satisfy the defn?
But I don't see where in the definition it requires that it must attract on both sides of the fixed point.
It just says 'close to x^{*}' but then being as close as we want from only one side is sufficient to satisfy this.

If all the curves have to have a gradient of 1 at the fixed point, your example sounds like c)
Can you not create two antisymmetric curves (about 1) by cutting your example into two, where one will attract and the other diverge?
But ok, in the case that we need a neighbourhood around x^{*} to attract then indeed I can just split the current curve so that:

a) [attractor] can have

f(x) = \begin{cases} 2-e^{\left(1-x\right)}, & x \geq 1 \\ 1-\ln\left(2-x\right), & x < 1

and looks like
Spoiler:
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Image



b) [repeller] can have

f(x) = \begin{cases} 1-\ln\left(2-x\right), & x \geq 1 \\ 2-e^{\left(1-x\right)}, & x < 1

and looks like
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Image



and so c) is obviously fit by my original example in OP.
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mqb2766
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(Original post by RDKGames)
But I don't see where in the definition it requires that it must attract on both sides of the fixed point.
It just says 'close to x^{*}' but then being as close as we want from only one side is sufficient to satisfy this.



But ok, in the case that we need a neighbourhood around x^{*} to attract then indeed I can just split the current curve so that:

a) [attractor] can have

f(x) = \begin{cases} 2-e^{\left(1-x\right)}, & x \geq 1 \\ 1-\ln\left(2-x\right), & x < 1

and looks like
Spoiler:
Show


Image


b) [repeller] can have

f(x) = \begin{cases} 1-\ln\left(2-x\right), & x \geq 1 \\ 2-e^{\left(1-x\right)}, & x < 1

and looks like
Spoiler:
Show


Image


and so c) is obviously fit by my original example in OP.
Looks good to me.

Edit: For c) in your original example (didn't read your reply properly)


A fixed point x^{*} is called an attractor if it attracts orbit with initial x_0 close to x^{*}. In other words, \forall \epsilon > 0, \exists \delta >0 such that |x_0-x^{*}| < \delta \implies |f^{n}(x_0)-x^{*}| < \epsilon.

You can pick an x_0 close to x* and make it diverge. The sign of x_0-x^{*} is irrelevant in the definition, but in your example it is relevant?
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Gregorius
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(Original post by RDKGames)
But I don't see where in the definition it requires that it must attract on both sides of the fixed point.
 |x_0-x^{*}| < \delta \implies |f^{n}(x_0)-x^{*}| < \epsilon

Note carefully the modulus signs.
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RDKGames
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(Original post by mqb2766)
Looks good to me.

Edit: For c) in your original example (didn't read your reply properly)


A fixed point x^{*} is called an attractor if it attracts orbit with initial x_0 close to x^{*}. In other words, \forall \epsilon > 0, \exists \delta >0 such that |x_0-x^{*}| < \delta \implies |f^{n}(x_0)-x^{*}| < \epsilon.

You can pick an x_0 close to x* and make it diverge. The sign of x_0-x^{*} is irrelevant in the definition, but in your example it is relevant?
(Original post by Gregorius)
 |x_0-x^{*}| < \delta \implies |f^{n}(x_0)-x^{*}| < \epsilon

Note carefully the modulus signs.
Ok thanks, I was overlooking and misunderstanding that a bit there for a moment.
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