Mixing
Why Lorenz attractor can be embedded by a 3-step time delay map?
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I'm investigating attractor reconstruction of Lorenz system. I saw a bunch of work claiming that the time delay map $[x(t), x(t -tau), x(t - 2tau)]$ is sufficient to reconstruct the attracotr, e.g. http://www.scholarpedia.org/article/Attractor_reconstruction,
https://www.youtube.com/watch?v=6i57udsPKms.
If I'm understanding this correctly, this means that the state space of Lorenz system can be embedded into $mathbb{R}^3$. However as far as I have known by Takens' theorem, the time delay step $n$ to embed a strange attractor of dimension $d$ should be $n geq 2d+1$. In this sense, since the fractal dimension of Lorenz attractor is slightly greater than $2$, there should be at least $5$ delay steps in order to achieve the embedding.
Is there any specific theorem/paper claiming that the Lorenz attractor can be embedded by a 3-step time delay embedding?
dynamical-systems mathematical-physics chaos-theory
asked Jan 16 at 21:10
mw19930312mw19930312
163
$endgroup$
add a comment |
$begingroup$
I'm investigating attractor reconstruction of Lorenz system. I saw a bunch of work claiming that the time delay map $[x(t), x(t -tau), x(t - 2tau)]$ is sufficient to reconstruct the attracotr, e.g. http://www.scholarpedia.org/article/Attractor_reconstruction,
https://www.youtube.com/watch?v=6i57udsPKms.
If I'm understanding this correctly, this means that the state space of Lorenz system can be embedded into $mathbb{R}^3$. However as far as I have known by Takens' theorem, the time delay step $n$ to embed a strange attractor of dimension $d$ should be $n geq 2d+1$. In this sense, since the fractal dimension of Lorenz attractor is slightly greater than $2$, there should be at least $5$ delay steps in order to achieve the embedding.
Is there any specific theorem/paper claiming that the Lorenz attractor can be embedded by a 3-step time delay embedding?
dynamical-systems mathematical-physics chaos-theory
asked Jan 16 at 21:10
mw19930312mw19930312
163
$endgroup$
$begingroup$
Welcome to Mathematics Stack Exchange! Great post! A quick tour will hep you understand how best to form questions and answers.
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– dantopa
Jan 16 at 21:30
$begingroup$
The theorem gives a guarantee for $nge 2d+1$, but does not claim that smaller $n$ do not work. // The similar image is for $τ=0.1$, just doubling it to $τ=0.2$ gives a much curvier surface.
$endgroup$
– LutzL
Jan 16 at 22:26
$begingroup$
@LutzL Thanks for the reply! I understand that Takens' theorem only provides an upper limit. What I'm concerned is that $[x(t), x(t-tau), x(t-2tau)]$ does not seem to be an embedding since I could not find any rigorous theorem says so. Do you happen to now any reference?
$endgroup$
– mw19930312
Jan 17 at 15:06
$begingroup$
No, only the argument put forward in the answer that the system is sufficiently coupled that the map from $[x(t),y(t),z(t)]$ to the states at $t−τ$ and $t-2τ$ and then projecting down to $[x(t),x(t−τ),x(t−2τ)]$ is bijective, so that the dynamics is faithfully captured in the 3-step delay map. // When plotting, it helps much for smaller $τ$ to apply a linear transformation and plot the curve [(x(t)+x(t−τ)+x(t−2τ)),(x(t)-x(t−2τ)),(x(t)-2x(t−τ)+x(t−2τ))]$.
$endgroup$
– LutzL
Jan 17 at 15:13
add a comment |
$begingroup$
I'm investigating attractor reconstruction of Lorenz system. I saw a bunch of work claiming that the time delay map $[x(t), x(t -tau), x(t - 2tau)]$ is sufficient to reconstruct the attracotr, e.g. http://www.scholarpedia.org/article/Attractor_reconstruction,
https://www.youtube.com/watch?v=6i57udsPKms.
If I'm understanding this correctly, this means that the state space of Lorenz system can be embedded into $mathbb{R}^3$. However as far as I have known by Takens' theorem, the time delay step $n$ to embed a strange attractor of dimension $d$ should be $n geq 2d+1$. In this sense, since the fractal dimension of Lorenz attractor is slightly greater than $2$, there should be at least $5$ delay steps in order to achieve the embedding.
Is there any specific theorem/paper claiming that the Lorenz attractor can be embedded by a 3-step time delay embedding?
dynamical-systems mathematical-physics chaos-theory
asked Jan 16 at 21:10
mw19930312mw19930312
163
$endgroup$
I'm investigating attractor reconstruction of Lorenz system. I saw a bunch of work claiming that the time delay map $[x(t), x(t -tau), x(t - 2tau)]$ is sufficient to reconstruct the attracotr, e.g. http://www.scholarpedia.org/article/Attractor_reconstruction,
https://www.youtube.com/watch?v=6i57udsPKms.
If I'm understanding this correctly, this means that the state space of Lorenz system can be embedded into $mathbb{R}^3$. However as far as I have known by Takens' theorem, the time delay step $n$ to embed a strange attractor of dimension $d$ should be $n geq 2d+1$. In this sense, since the fractal dimension of Lorenz attractor is slightly greater than $2$, there should be at least $5$ delay steps in order to achieve the embedding.
Is there any specific theorem/paper claiming that the Lorenz attractor can be embedded by a 3-step time delay embedding?
dynamical-systems mathematical-physics chaos-theory
dynamical-systems mathematical-physics chaos-theory
asked Jan 16 at 21:10
mw19930312mw19930312
163
asked Jan 16 at 21:10
mw19930312mw19930312
163
asked Jan 16 at 21:10
mw19930312mw19930312
163
asked Jan 16 at 21:10
mw19930312mw19930312
163
asked Jan 16 at 21:10
mw19930312mw19930312
163
163
$begingroup$
Welcome to Mathematics Stack Exchange! Great post! A quick tour will hep you understand how best to form questions and answers.
$endgroup$
– dantopa
Jan 16 at 21:30
$begingroup$
The theorem gives a guarantee for $nge 2d+1$, but does not claim that smaller $n$ do not work. // The similar image is for $τ=0.1$, just doubling it to $τ=0.2$ gives a much curvier surface.
$endgroup$
– LutzL
Jan 16 at 22:26
$begingroup$
@LutzL Thanks for the reply! I understand that Takens' theorem only provides an upper limit. What I'm concerned is that $[x(t), x(t-tau), x(t-2tau)]$ does not seem to be an embedding since I could not find any rigorous theorem says so. Do you happen to now any reference?
$endgroup$
– mw19930312
Jan 17 at 15:06
$begingroup$
No, only the argument put forward in the answer that the system is sufficiently coupled that the map from $[x(t),y(t),z(t)]$ to the states at $t−τ$ and $t-2τ$ and then projecting down to $[x(t),x(t−τ),x(t−2τ)]$ is bijective, so that the dynamics is faithfully captured in the 3-step delay map. // When plotting, it helps much for smaller $τ$ to apply a linear transformation and plot the curve [(x(t)+x(t−τ)+x(t−2τ)),(x(t)-x(t−2τ)),(x(t)-2x(t−τ)+x(t−2τ))]$.
$endgroup$
– LutzL
Jan 17 at 15:13
add a comment |
$begingroup$
Welcome to Mathematics Stack Exchange! Great post! A quick tour will hep you understand how best to form questions and answers.
$endgroup$
– dantopa
Jan 16 at 21:30
$begingroup$
The theorem gives a guarantee for $nge 2d+1$, but does not claim that smaller $n$ do not work. // The similar image is for $τ=0.1$, just doubling it to $τ=0.2$ gives a much curvier surface.
$endgroup$
– LutzL
Jan 16 at 22:26
$begingroup$
@LutzL Thanks for the reply! I understand that Takens' theorem only provides an upper limit. What I'm concerned is that $[x(t), x(t-tau), x(t-2tau)]$ does not seem to be an embedding since I could not find any rigorous theorem says so. Do you happen to now any reference?
$endgroup$
– mw19930312
Jan 17 at 15:06
$begingroup$
No, only the argument put forward in the answer that the system is sufficiently coupled that the map from $[x(t),y(t),z(t)]$ to the states at $t−τ$ and $t-2τ$ and then projecting down to $[x(t),x(t−τ),x(t−2τ)]$ is bijective, so that the dynamics is faithfully captured in the 3-step delay map. // When plotting, it helps much for smaller $τ$ to apply a linear transformation and plot the curve [(x(t)+x(t−τ)+x(t−2τ)),(x(t)-x(t−2τ)),(x(t)-2x(t−τ)+x(t−2τ))]$.
$endgroup$
– LutzL
Jan 17 at 15:13
$begingroup$
Welcome to Mathematics Stack Exchange! Great post! A quick tour will hep you understand how best to form questions and answers.
$endgroup$
– dantopa
Jan 16 at 21:30
$begingroup$
Welcome to Mathematics Stack Exchange! Great post! A quick tour will hep you understand how best to form questions and answers.
$endgroup$
– dantopa
Jan 16 at 21:30
$begingroup$
The theorem gives a guarantee for $nge 2d+1$, but does not claim that smaller $n$ do not work. // The similar image is for $τ=0.1$, just doubling it to $τ=0.2$ gives a much curvier surface.
$endgroup$
– LutzL
Jan 16 at 22:26
$begingroup$
The theorem gives a guarantee for $nge 2d+1$, but does not claim that smaller $n$ do not work. // The similar image is for $τ=0.1$, just doubling it to $τ=0.2$ gives a much curvier surface.
$endgroup$
– LutzL
Jan 16 at 22:26
$begingroup$
@LutzL Thanks for the reply! I understand that Takens' theorem only provides an upper limit. What I'm concerned is that $[x(t), x(t-tau), x(t-2tau)]$ does not seem to be an embedding since I could not find any rigorous theorem says so. Do you happen to now any reference?
$endgroup$
– mw19930312
Jan 17 at 15:06
$begingroup$
@LutzL Thanks for the reply! I understand that Takens' theorem only provides an upper limit. What I'm concerned is that $[x(t), x(t-tau), x(t-2tau)]$ does not seem to be an embedding since I could not find any rigorous theorem says so. Do you happen to now any reference?
$endgroup$
– mw19930312
Jan 17 at 15:06
$begingroup$
No, only the argument put forward in the answer that the system is sufficiently coupled that the map from $[x(t),y(t),z(t)]$ to the states at $t−τ$ and $t-2τ$ and then projecting down to $[x(t),x(t−τ),x(t−2τ)]$ is bijective, so that the dynamics is faithfully captured in the 3-step delay map. // When plotting, it helps much for smaller $τ$ to apply a linear transformation and plot the curve [(x(t)+x(t−τ)+x(t−2τ)),(x(t)-x(t−2τ)),(x(t)-2x(t−τ)+x(t−2τ))]$.
$endgroup$
– LutzL
Jan 17 at 15:13
$begingroup$
No, only the argument put forward in the answer that the system is sufficiently coupled that the map from $[x(t),y(t),z(t)]$ to the states at $t−τ$ and $t-2τ$ and then projecting down to $[x(t),x(t−τ),x(t−2τ)]$ is bijective, so that the dynamics is faithfully captured in the 3-step delay map. // When plotting, it helps much for smaller $τ$ to apply a linear transformation and plot the curve [(x(t)+x(t−τ)+x(t−2τ)),(x(t)-x(t−2τ)),(x(t)-2x(t−τ)+x(t−2τ))]$.
$endgroup$
– LutzL
Jan 17 at 15:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As to why 3 delay steps are sufficient for the Lorenz system:
We know that by Taylor
$$
frac{x(t+τ)-x(t-τ)}{2τ}=dot x(t)+frac{τ^2}6dddot x(t)+...
$$
and
$$
frac{x(t+τ)-2x(t)+x(t-τ)}{τ^2}=ddot x(t)+frac{τ^2}{12}x^{(4)}(t)+...
$$
Now insert the Lorenz differential equations
$$
left.begin{aligned}
dot x&=σ(y-x)\
ddot x&=σ(x(rho-z)-y-dot x)
end{aligned}right}
implies
left.begin{aligned}
y&=x+frac{dot x}σ\
z&=rho-frac{y+dot x+frac{ddot x}σ}{x}
end{aligned}right}
$$
to see that up to order $τ^2$ the values of $y(t)$ and $z(t)$ are easy to extract from the difference quotients and the first derivative terms on the right. Involving higher order derivative terms gives a system of higher degree that will provide a more exact relation between the two sets of data.
But even this first approximation shows that it is possible to invoke the inverse function theorem as long as $x>0$ to get a bijection.
answered Jan 17 at 15:33
LutzLLutzL
58.9k42056
$endgroup$
$begingroup$
Got it. Thanks a lot!
$endgroup$
– mw19930312
Jan 18 at 16:04
add a comment |
$begingroup$
this means that the state space of Lorenz system can be embedded into $mathbb{R}^3$.
Without a restriction to delay embedding, this is trivial since the Lorenz system consists of three differential equations.
However as far as I have known by Takens' theorem, the time-delay step $n$ to embed a strange attractor of dimension $d$ should be $n geq 2d+1$.
The dimension given by Takens’s theorem is only an upper limit. A lower embedding dimension may suffice. Also see this question and answer.
Also note that Takens’s theorem doesn’t use fractal dimensions at all; it’s the Sauer–Yorke–Casdagli theorem that does.
Is there any specific theorem/paper claiming that the Lorenz attractor can be embedded by a 3-step time delay embedding?
Given that the Lorenz attractor can be embedded in three dimensions (see above), it would be intuitively surprising if a three-dimensional delay embedding fails here (in particular for all delays).
Moreover, and maybe most importantly, three-dimensional delay embeddings of the Lorenz attractor have been used extensively investigated for benchmarking, proofs of principle, or similar – which, to my knowledge, hasn’t turned up any inconsistencies as to be expected for a failed embedding.
I am not aware of rigorous investigations of this, but I wouldn’t be surprised if none exist, due to a lack of relevance: The entire point of a Takens embedding is to reconstruct attractors of unknown dynamics. Applying it to something like the Lorenz system is only for benchmarking, proofs of principle, etc.
answered Jan 17 at 6:57
WrzlprmftWrzlprmft
3,08611233
$endgroup$
$begingroup$
Thanks for the reply! By saying "Given that the Lorenz attractor can be embedded in three dimensions (see above), it would be surprising if a delay embedding fails here (in particular for all delays).", do you mean that since Lorenz attractor can be embedded into $mathbb{R}^3$, there must exist time delay embedding with $ngeq 7$? Actually I don't know whether $[x(t), x(t - tau), x(t - 2tau)]$ is an embedding since there is no theorem substantiate it. All existing work I've found assume this by default since trajectory of original system looks similar to time-delayed trajectory....
$endgroup$
– mw19930312
Jan 17 at 15:11
$begingroup$
@mw19930312: Please see my edit.
$endgroup$
– Wrzlprmft
Jan 17 at 15:36
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
As to why 3 delay steps are sufficient for the Lorenz system:
We know that by Taylor
$$
frac{x(t+τ)-x(t-τ)}{2τ}=dot x(t)+frac{τ^2}6dddot x(t)+...
$$
and
$$
frac{x(t+τ)-2x(t)+x(t-τ)}{τ^2}=ddot x(t)+frac{τ^2}{12}x^{(4)}(t)+...
$$
Now insert the Lorenz differential equations
$$
left.begin{aligned}
dot x&=σ(y-x)\
ddot x&=σ(x(rho-z)-y-dot x)
end{aligned}right}
implies
left.begin{aligned}
y&=x+frac{dot x}σ\
z&=rho-frac{y+dot x+frac{ddot x}σ}{x}
end{aligned}right}
$$
to see that up to order $τ^2$ the values of $y(t)$ and $z(t)$ are easy to extract from the difference quotients and the first derivative terms on the right. Involving higher order derivative terms gives a system of higher degree that will provide a more exact relation between the two sets of data.
But even this first approximation shows that it is possible to invoke the inverse function theorem as long as $x>0$ to get a bijection.
answered Jan 17 at 15:33
LutzLLutzL
58.9k42056
$endgroup$
$begingroup$
Got it. Thanks a lot!
$endgroup$
– mw19930312
Jan 18 at 16:04
add a comment |
$begingroup$
As to why 3 delay steps are sufficient for the Lorenz system:
We know that by Taylor
$$
frac{x(t+τ)-x(t-τ)}{2τ}=dot x(t)+frac{τ^2}6dddot x(t)+...
$$
and
$$
frac{x(t+τ)-2x(t)+x(t-τ)}{τ^2}=ddot x(t)+frac{τ^2}{12}x^{(4)}(t)+...
$$
Now insert the Lorenz differential equations
$$
left.begin{aligned}
dot x&=σ(y-x)\
ddot x&=σ(x(rho-z)-y-dot x)
end{aligned}right}
implies
left.begin{aligned}
y&=x+frac{dot x}σ\
z&=rho-frac{y+dot x+frac{ddot x}σ}{x}
end{aligned}right}
$$
to see that up to order $τ^2$ the values of $y(t)$ and $z(t)$ are easy to extract from the difference quotients and the first derivative terms on the right. Involving higher order derivative terms gives a system of higher degree that will provide a more exact relation between the two sets of data.
But even this first approximation shows that it is possible to invoke the inverse function theorem as long as $x>0$ to get a bijection.
answered Jan 17 at 15:33
LutzLLutzL
58.9k42056
$endgroup$
$begingroup$
Got it. Thanks a lot!
$endgroup$
– mw19930312
Jan 18 at 16:04
add a comment |
$begingroup$
As to why 3 delay steps are sufficient for the Lorenz system:
We know that by Taylor
$$
frac{x(t+τ)-x(t-τ)}{2τ}=dot x(t)+frac{τ^2}6dddot x(t)+...
$$
and
$$
frac{x(t+τ)-2x(t)+x(t-τ)}{τ^2}=ddot x(t)+frac{τ^2}{12}x^{(4)}(t)+...
$$
Now insert the Lorenz differential equations
$$
left.begin{aligned}
dot x&=σ(y-x)\
ddot x&=σ(x(rho-z)-y-dot x)
end{aligned}right}
implies
left.begin{aligned}
y&=x+frac{dot x}σ\
z&=rho-frac{y+dot x+frac{ddot x}σ}{x}
end{aligned}right}
$$
to see that up to order $τ^2$ the values of $y(t)$ and $z(t)$ are easy to extract from the difference quotients and the first derivative terms on the right. Involving higher order derivative terms gives a system of higher degree that will provide a more exact relation between the two sets of data.
But even this first approximation shows that it is possible to invoke the inverse function theorem as long as $x>0$ to get a bijection.
answered Jan 17 at 15:33
LutzLLutzL
58.9k42056
$endgroup$
As to why 3 delay steps are sufficient for the Lorenz system:
We know that by Taylor
$$
frac{x(t+τ)-x(t-τ)}{2τ}=dot x(t)+frac{τ^2}6dddot x(t)+...
$$
and
$$
frac{x(t+τ)-2x(t)+x(t-τ)}{τ^2}=ddot x(t)+frac{τ^2}{12}x^{(4)}(t)+...
$$
Now insert the Lorenz differential equations
$$
left.begin{aligned}
dot x&=σ(y-x)\
ddot x&=σ(x(rho-z)-y-dot x)
end{aligned}right}
implies
left.begin{aligned}
y&=x+frac{dot x}σ\
z&=rho-frac{y+dot x+frac{ddot x}σ}{x}
end{aligned}right}
$$
to see that up to order $τ^2$ the values of $y(t)$ and $z(t)$ are easy to extract from the difference quotients and the first derivative terms on the right. Involving higher order derivative terms gives a system of higher degree that will provide a more exact relation between the two sets of data.
But even this first approximation shows that it is possible to invoke the inverse function theorem as long as $x>0$ to get a bijection.
answered Jan 17 at 15:33
LutzLLutzL
58.9k42056
answered Jan 17 at 15:33
LutzLLutzL
58.9k42056
answered Jan 17 at 15:33
LutzLLutzL
58.9k42056
answered Jan 17 at 15:33
LutzLLutzL
58.9k42056
58.9k42056
$begingroup$
Got it. Thanks a lot!
$endgroup$
– mw19930312
Jan 18 at 16:04
add a comment |
$begingroup$
Got it. Thanks a lot!
$endgroup$
– mw19930312
Jan 18 at 16:04
$begingroup$
Got it. Thanks a lot!
$endgroup$
– mw19930312
Jan 18 at 16:04
$begingroup$
Got it. Thanks a lot!
$endgroup$
– mw19930312
Jan 18 at 16:04
add a comment |
$begingroup$
this means that the state space of Lorenz system can be embedded into $mathbb{R}^3$.
Without a restriction to delay embedding, this is trivial since the Lorenz system consists of three differential equations.
However as far as I have known by Takens' theorem, the time-delay step $n$ to embed a strange attractor of dimension $d$ should be $n geq 2d+1$.
The dimension given by Takens’s theorem is only an upper limit. A lower embedding dimension may suffice. Also see this question and answer.
Also note that Takens’s theorem doesn’t use fractal dimensions at all; it’s the Sauer–Yorke–Casdagli theorem that does.
Is there any specific theorem/paper claiming that the Lorenz attractor can be embedded by a 3-step time delay embedding?
Given that the Lorenz attractor can be embedded in three dimensions (see above), it would be intuitively surprising if a three-dimensional delay embedding fails here (in particular for all delays).
Moreover, and maybe most importantly, three-dimensional delay embeddings of the Lorenz attractor have been used extensively investigated for benchmarking, proofs of principle, or similar – which, to my knowledge, hasn’t turned up any inconsistencies as to be expected for a failed embedding.
I am not aware of rigorous investigations of this, but I wouldn’t be surprised if none exist, due to a lack of relevance: The entire point of a Takens embedding is to reconstruct attractors of unknown dynamics. Applying it to something like the Lorenz system is only for benchmarking, proofs of principle, etc.
answered Jan 17 at 6:57
WrzlprmftWrzlprmft
3,08611233
$endgroup$
$begingroup$
Thanks for the reply! By saying "Given that the Lorenz attractor can be embedded in three dimensions (see above), it would be surprising if a delay embedding fails here (in particular for all delays).", do you mean that since Lorenz attractor can be embedded into $mathbb{R}^3$, there must exist time delay embedding with $ngeq 7$? Actually I don't know whether $[x(t), x(t - tau), x(t - 2tau)]$ is an embedding since there is no theorem substantiate it. All existing work I've found assume this by default since trajectory of original system looks similar to time-delayed trajectory....
$endgroup$
– mw19930312
Jan 17 at 15:11
$begingroup$
@mw19930312: Please see my edit.
$endgroup$
– Wrzlprmft
Jan 17 at 15:36
add a comment |
$begingroup$
this means that the state space of Lorenz system can be embedded into $mathbb{R}^3$.
Without a restriction to delay embedding, this is trivial since the Lorenz system consists of three differential equations.
However as far as I have known by Takens' theorem, the time-delay step $n$ to embed a strange attractor of dimension $d$ should be $n geq 2d+1$.
The dimension given by Takens’s theorem is only an upper limit. A lower embedding dimension may suffice. Also see this question and answer.
Also note that Takens’s theorem doesn’t use fractal dimensions at all; it’s the Sauer–Yorke–Casdagli theorem that does.
Is there any specific theorem/paper claiming that the Lorenz attractor can be embedded by a 3-step time delay embedding?
Given that the Lorenz attractor can be embedded in three dimensions (see above), it would be intuitively surprising if a three-dimensional delay embedding fails here (in particular for all delays).
Moreover, and maybe most importantly, three-dimensional delay embeddings of the Lorenz attractor have been used extensively investigated for benchmarking, proofs of principle, or similar – which, to my knowledge, hasn’t turned up any inconsistencies as to be expected for a failed embedding.
I am not aware of rigorous investigations of this, but I wouldn’t be surprised if none exist, due to a lack of relevance: The entire point of a Takens embedding is to reconstruct attractors of unknown dynamics. Applying it to something like the Lorenz system is only for benchmarking, proofs of principle, etc.
answered Jan 17 at 6:57
WrzlprmftWrzlprmft
3,08611233
$endgroup$
$begingroup$
Thanks for the reply! By saying "Given that the Lorenz attractor can be embedded in three dimensions (see above), it would be surprising if a delay embedding fails here (in particular for all delays).", do you mean that since Lorenz attractor can be embedded into $mathbb{R}^3$, there must exist time delay embedding with $ngeq 7$? Actually I don't know whether $[x(t), x(t - tau), x(t - 2tau)]$ is an embedding since there is no theorem substantiate it. All existing work I've found assume this by default since trajectory of original system looks similar to time-delayed trajectory....
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– mw19930312
Jan 17 at 15:11
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@mw19930312: Please see my edit.
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– Wrzlprmft
Jan 17 at 15:36
add a comment |
$begingroup$
this means that the state space of Lorenz system can be embedded into $mathbb{R}^3$.
Without a restriction to delay embedding, this is trivial since the Lorenz system consists of three differential equations.
However as far as I have known by Takens' theorem, the time-delay step $n$ to embed a strange attractor of dimension $d$ should be $n geq 2d+1$.
The dimension given by Takens’s theorem is only an upper limit. A lower embedding dimension may suffice. Also see this question and answer.
Also note that Takens’s theorem doesn’t use fractal dimensions at all; it’s the Sauer–Yorke–Casdagli theorem that does.
Is there any specific theorem/paper claiming that the Lorenz attractor can be embedded by a 3-step time delay embedding?
Given that the Lorenz attractor can be embedded in three dimensions (see above), it would be intuitively surprising if a three-dimensional delay embedding fails here (in particular for all delays).
Moreover, and maybe most importantly, three-dimensional delay embeddings of the Lorenz attractor have been used extensively investigated for benchmarking, proofs of principle, or similar – which, to my knowledge, hasn’t turned up any inconsistencies as to be expected for a failed embedding.
I am not aware of rigorous investigations of this, but I wouldn’t be surprised if none exist, due to a lack of relevance: The entire point of a Takens embedding is to reconstruct attractors of unknown dynamics. Applying it to something like the Lorenz system is only for benchmarking, proofs of principle, etc.
answered Jan 17 at 6:57
WrzlprmftWrzlprmft
3,08611233
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this means that the state space of Lorenz system can be embedded into $mathbb{R}^3$.
Without a restriction to delay embedding, this is trivial since the Lorenz system consists of three differential equations.
However as far as I have known by Takens' theorem, the time-delay step $n$ to embed a strange attractor of dimension $d$ should be $n geq 2d+1$.
The dimension given by Takens’s theorem is only an upper limit. A lower embedding dimension may suffice. Also see this question and answer.
Also note that Takens’s theorem doesn’t use fractal dimensions at all; it’s the Sauer–Yorke–Casdagli theorem that does.
Is there any specific theorem/paper claiming that the Lorenz attractor can be embedded by a 3-step time delay embedding?
Given that the Lorenz attractor can be embedded in three dimensions (see above), it would be intuitively surprising if a three-dimensional delay embedding fails here (in particular for all delays).
Moreover, and maybe most importantly, three-dimensional delay embeddings of the Lorenz attractor have been used extensively investigated for benchmarking, proofs of principle, or similar – which, to my knowledge, hasn’t turned up any inconsistencies as to be expected for a failed embedding.
I am not aware of rigorous investigations of this, but I wouldn’t be surprised if none exist, due to a lack of relevance: The entire point of a Takens embedding is to reconstruct attractors of unknown dynamics. Applying it to something like the Lorenz system is only for benchmarking, proofs of principle, etc.
answered Jan 17 at 6:57
WrzlprmftWrzlprmft
3,08611233
edited Jan 17 at 15:36
answered Jan 17 at 6:57
WrzlprmftWrzlprmft
3,08611233
answered Jan 17 at 6:57
WrzlprmftWrzlprmft
3,08611233
answered Jan 17 at 6:57
WrzlprmftWrzlprmft
3,08611233
3,08611233
$begingroup$
Thanks for the reply! By saying "Given that the Lorenz attractor can be embedded in three dimensions (see above), it would be surprising if a delay embedding fails here (in particular for all delays).", do you mean that since Lorenz attractor can be embedded into $mathbb{R}^3$, there must exist time delay embedding with $ngeq 7$? Actually I don't know whether $[x(t), x(t - tau), x(t - 2tau)]$ is an embedding since there is no theorem substantiate it. All existing work I've found assume this by default since trajectory of original system looks similar to time-delayed trajectory....
$endgroup$
– mw19930312
Jan 17 at 15:11
$begingroup$
@mw19930312: Please see my edit.
$endgroup$
– Wrzlprmft
Jan 17 at 15:36
add a comment |
$begingroup$
Thanks for the reply! By saying "Given that the Lorenz attractor can be embedded in three dimensions (see above), it would be surprising if a delay embedding fails here (in particular for all delays).", do you mean that since Lorenz attractor can be embedded into $mathbb{R}^3$, there must exist time delay embedding with $ngeq 7$? Actually I don't know whether $[x(t), x(t - tau), x(t - 2tau)]$ is an embedding since there is no theorem substantiate it. All existing work I've found assume this by default since trajectory of original system looks similar to time-delayed trajectory....
$endgroup$
– mw19930312
Jan 17 at 15:11
$begingroup$
@mw19930312: Please see my edit.
$endgroup$
– Wrzlprmft
Jan 17 at 15:36
$begingroup$
Thanks for the reply! By saying "Given that the Lorenz attractor can be embedded in three dimensions (see above), it would be surprising if a delay embedding fails here (in particular for all delays).", do you mean that since Lorenz attractor can be embedded into $mathbb{R}^3$, there must exist time delay embedding with $ngeq 7$? Actually I don't know whether $[x(t), x(t - tau), x(t - 2tau)]$ is an embedding since there is no theorem substantiate it. All existing work I've found assume this by default since trajectory of original system looks similar to time-delayed trajectory....
$endgroup$
– mw19930312
Jan 17 at 15:11
$begingroup$
Thanks for the reply! By saying "Given that the Lorenz attractor can be embedded in three dimensions (see above), it would be surprising if a delay embedding fails here (in particular for all delays).", do you mean that since Lorenz attractor can be embedded into $mathbb{R}^3$, there must exist time delay embedding with $ngeq 7$? Actually I don't know whether $[x(t), x(t - tau), x(t - 2tau)]$ is an embedding since there is no theorem substantiate it. All existing work I've found assume this by default since trajectory of original system looks similar to time-delayed trajectory....
$endgroup$
– mw19930312
Jan 17 at 15:11
$begingroup$
@mw19930312: Please see my edit.
$endgroup$
– Wrzlprmft
Jan 17 at 15:36
$begingroup$
@mw19930312: Please see my edit.
$endgroup$
– Wrzlprmft
Jan 17 at 15:36
add a comment |
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– dantopa
Jan 16 at 21:30
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The theorem gives a guarantee for $nge 2d+1$, but does not claim that smaller $n$ do not work. // The similar image is for $τ=0.1$, just doubling it to $τ=0.2$ gives a much curvier surface.
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– LutzL
Jan 16 at 22:26
$begingroup$
@LutzL Thanks for the reply! I understand that Takens' theorem only provides an upper limit. What I'm concerned is that $[x(t), x(t-tau), x(t-2tau)]$ does not seem to be an embedding since I could not find any rigorous theorem says so. Do you happen to now any reference?
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– mw19930312
Jan 17 at 15:06
$begingroup$
No, only the argument put forward in the answer that the system is sufficiently coupled that the map from $[x(t),y(t),z(t)]$ to the states at $t−τ$ and $t-2τ$ and then projecting down to $[x(t),x(t−τ),x(t−2τ)]$ is bijective, so that the dynamics is faithfully captured in the 3-step delay map. // When plotting, it helps much for smaller $τ$ to apply a linear transformation and plot the curve [(x(t)+x(t−τ)+x(t−2τ)),(x(t)-x(t−2τ)),(x(t)-2x(t−τ)+x(t−2τ))]$.
$endgroup$
– LutzL
Jan 17 at 15:13