Dimensionality of the Poincare section?

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Consider the Poincare map $P$, for which we place the Poincare section transverse to the trajectories and it intersects at various points which forms a set of discrete points.

For $dotx = f(x)$ where $x$ is an $n$ dimensional vector, why the section $S$ is an $n-1$ dimensional surface? How can I visualize this?

dynamical-systems dimensional-analysis

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asked Jul 31 at 14:34

BAYMAX

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Consider the Poincare map $P$, for which we place the Poincare section transverse to the trajectories and it intersects at various points which forms a set of discrete points.

For $dotx = f(x)$ where $x$ is an $n$ dimensional vector, why the section $S$ is an $n-1$ dimensional surface? How can I visualize this?

dynamical-systems dimensional-analysis

share|cite|improve this question

asked Jul 31 at 14:34

BAYMAX

2,43121021

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Consider the Poincare map $P$, for which we place the Poincare section transverse to the trajectories and it intersects at various points which forms a set of discrete points.

For $dotx = f(x)$ where $x$ is an $n$ dimensional vector, why the section $S$ is an $n-1$ dimensional surface? How can I visualize this?

dynamical-systems dimensional-analysis

share|cite|improve this question

asked Jul 31 at 14:34

BAYMAX

2,43121021

Consider the Poincare map $P$, for which we place the Poincare section transverse to the trajectories and it intersects at various points which forms a set of discrete points.

For $dotx = f(x)$ where $x$ is an $n$ dimensional vector, why the section $S$ is an $n-1$ dimensional surface? How can I visualize this?

dynamical-systems dimensional-analysis

share|cite|improve this question

asked Jul 31 at 14:34

BAYMAX

2,43121021

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 14:34

BAYMAX

2,43121021

asked Jul 31 at 14:34

BAYMAX

2,43121021

asked Jul 31 at 14:34

BAYMAX

2,43121021

2,43121021

add a comment | 

add a comment | 

1 Answer
1

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up vote
2
down vote



accepted

Two angles for explanation:

• The Poincaré map is supposed to remove exactly one dimension from the dynamics: The one corresponding to time. All other dimensions must still be present in the Poincaré map. Hence it’s state space must have dimension $n-1$. As the state space consists of points of the Poincaré section, the section also must have dimension $n-1$.




• For most applications, you want the Poincaré map to have one iteration per oscillation¹. Hence, the Poincaré section has to be frequently intersected. Now, very loosely speaking, a trajectory segment (a one-dimensional object) has a probability of zero to hit any object (the Poincaré section) with dimension $n-2$ or lower, while it cannot miss an $n-1$-dimensional object. Of course, for some dynamics you may carefully place an $n-2$-dimensional section such that it is hit for sure, but this would mean that your dynamics has a spurious dimension.

---

^{¹ otherwise it’s every other oscillation, every high-amplitude oscillation, or similar, but the argument stays essentially the same}

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edited Aug 1 at 9:21

answered Jul 31 at 19:24

Wrzlprmft

2,62111032

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How the Poincare map removes exactly one dimension(the time) from the dynamics? is it because the map is time independent? like $x_n+1 = f(x_n)$? where as its continous version still has time like $fracdxdt=f(x)$ ?
  – BAYMAX
  Aug 1 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @BAYMAX: I am not sure I get your question here, but the question in your comment seems to be asked the wrong way around. The Poincaré map removes one dimension because $S$ is $n-1$-dimensional. $S$ is chosen this way because otherwise we wouldn’t get a useful map. (Also see my edit.)
  – Wrzlprmft
  Aug 1 at 9:25
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Two angles for explanation:

• The Poincaré map is supposed to remove exactly one dimension from the dynamics: The one corresponding to time. All other dimensions must still be present in the Poincaré map. Hence it’s state space must have dimension $n-1$. As the state space consists of points of the Poincaré section, the section also must have dimension $n-1$.




• For most applications, you want the Poincaré map to have one iteration per oscillation¹. Hence, the Poincaré section has to be frequently intersected. Now, very loosely speaking, a trajectory segment (a one-dimensional object) has a probability of zero to hit any object (the Poincaré section) with dimension $n-2$ or lower, while it cannot miss an $n-1$-dimensional object. Of course, for some dynamics you may carefully place an $n-2$-dimensional section such that it is hit for sure, but this would mean that your dynamics has a spurious dimension.

---

^{¹ otherwise it’s every other oscillation, every high-amplitude oscillation, or similar, but the argument stays essentially the same}

share|cite|improve this answer

edited Aug 1 at 9:21

answered Jul 31 at 19:24

Wrzlprmft

2,62111032

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How the Poincare map removes exactly one dimension(the time) from the dynamics? is it because the map is time independent? like $x_n+1 = f(x_n)$? where as its continous version still has time like $fracdxdt=f(x)$ ?
  – BAYMAX
  Aug 1 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @BAYMAX: I am not sure I get your question here, but the question in your comment seems to be asked the wrong way around. The Poincaré map removes one dimension because $S$ is $n-1$-dimensional. $S$ is chosen this way because otherwise we wouldn’t get a useful map. (Also see my edit.)
  – Wrzlprmft
  Aug 1 at 9:25
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Two angles for explanation:

• The Poincaré map is supposed to remove exactly one dimension from the dynamics: The one corresponding to time. All other dimensions must still be present in the Poincaré map. Hence it’s state space must have dimension $n-1$. As the state space consists of points of the Poincaré section, the section also must have dimension $n-1$.




• For most applications, you want the Poincaré map to have one iteration per oscillation¹. Hence, the Poincaré section has to be frequently intersected. Now, very loosely speaking, a trajectory segment (a one-dimensional object) has a probability of zero to hit any object (the Poincaré section) with dimension $n-2$ or lower, while it cannot miss an $n-1$-dimensional object. Of course, for some dynamics you may carefully place an $n-2$-dimensional section such that it is hit for sure, but this would mean that your dynamics has a spurious dimension.

---

^{¹ otherwise it’s every other oscillation, every high-amplitude oscillation, or similar, but the argument stays essentially the same}

share|cite|improve this answer

edited Aug 1 at 9:21

answered Jul 31 at 19:24

Wrzlprmft

2,62111032

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How the Poincare map removes exactly one dimension(the time) from the dynamics? is it because the map is time independent? like $x_n+1 = f(x_n)$? where as its continous version still has time like $fracdxdt=f(x)$ ?
  – BAYMAX
  Aug 1 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @BAYMAX: I am not sure I get your question here, but the question in your comment seems to be asked the wrong way around. The Poincaré map removes one dimension because $S$ is $n-1$-dimensional. $S$ is chosen this way because otherwise we wouldn’t get a useful map. (Also see my edit.)
  – Wrzlprmft
  Aug 1 at 9:25
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Two angles for explanation:

• The Poincaré map is supposed to remove exactly one dimension from the dynamics: The one corresponding to time. All other dimensions must still be present in the Poincaré map. Hence it’s state space must have dimension $n-1$. As the state space consists of points of the Poincaré section, the section also must have dimension $n-1$.




• For most applications, you want the Poincaré map to have one iteration per oscillation¹. Hence, the Poincaré section has to be frequently intersected. Now, very loosely speaking, a trajectory segment (a one-dimensional object) has a probability of zero to hit any object (the Poincaré section) with dimension $n-2$ or lower, while it cannot miss an $n-1$-dimensional object. Of course, for some dynamics you may carefully place an $n-2$-dimensional section such that it is hit for sure, but this would mean that your dynamics has a spurious dimension.

---

^{¹ otherwise it’s every other oscillation, every high-amplitude oscillation, or similar, but the argument stays essentially the same}

share|cite|improve this answer

edited Aug 1 at 9:21

answered Jul 31 at 19:24

Wrzlprmft

2,62111032

Two angles for explanation:

• The Poincaré map is supposed to remove exactly one dimension from the dynamics: The one corresponding to time. All other dimensions must still be present in the Poincaré map. Hence it’s state space must have dimension $n-1$. As the state space consists of points of the Poincaré section, the section also must have dimension $n-1$.




• For most applications, you want the Poincaré map to have one iteration per oscillation¹. Hence, the Poincaré section has to be frequently intersected. Now, very loosely speaking, a trajectory segment (a one-dimensional object) has a probability of zero to hit any object (the Poincaré section) with dimension $n-2$ or lower, while it cannot miss an $n-1$-dimensional object. Of course, for some dynamics you may carefully place an $n-2$-dimensional section such that it is hit for sure, but this would mean that your dynamics has a spurious dimension.

---

^{¹ otherwise it’s every other oscillation, every high-amplitude oscillation, or similar, but the argument stays essentially the same}

share|cite|improve this answer

edited Aug 1 at 9:21

answered Jul 31 at 19:24

Wrzlprmft

2,62111032

share|cite|improve this answer

share|cite|improve this answer

edited Aug 1 at 9:21

edited Aug 1 at 9:21

edited Aug 1 at 9:21

answered Jul 31 at 19:24

Wrzlprmft

2,62111032

answered Jul 31 at 19:24

Wrzlprmft

2,62111032

answered Jul 31 at 19:24

Wrzlprmft

2,62111032

2,62111032

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How the Poincare map removes exactly one dimension(the time) from the dynamics? is it because the map is time independent? like $x_n+1 = f(x_n)$? where as its continous version still has time like $fracdxdt=f(x)$ ?
  – BAYMAX
  Aug 1 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @BAYMAX: I am not sure I get your question here, but the question in your comment seems to be asked the wrong way around. The Poincaré map removes one dimension because $S$ is $n-1$-dimensional. $S$ is chosen this way because otherwise we wouldn’t get a useful map. (Also see my edit.)
  – Wrzlprmft
  Aug 1 at 9:25
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How the Poincare map removes exactly one dimension(the time) from the dynamics? is it because the map is time independent? like $x_n+1 = f(x_n)$? where as its continous version still has time like $fracdxdt=f(x)$ ?
  – BAYMAX
  Aug 1 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @BAYMAX: I am not sure I get your question here, but the question in your comment seems to be asked the wrong way around. The Poincaré map removes one dimension because $S$ is $n-1$-dimensional. $S$ is chosen this way because otherwise we wouldn’t get a useful map. (Also see my edit.)
  – Wrzlprmft
  Aug 1 at 9:25
  

  
  
  
  

How the Poincare map removes exactly one dimension(the time) from the dynamics? is it because the map is time independent? like $x_n+1 = f(x_n)$? where as its continous version still has time like $fracdxdt=f(x)$ ?
– BAYMAX
Aug 1 at 9:09

How the Poincare map removes exactly one dimension(the time) from the dynamics? is it because the map is time independent? like $x_n+1 = f(x_n)$? where as its continous version still has time like $fracdxdt=f(x)$ ?
– BAYMAX
Aug 1 at 9:09

@BAYMAX: I am not sure I get your question here, but the question in your comment seems to be asked the wrong way around. The Poincaré map removes one dimension because $S$ is $n-1$-dimensional. $S$ is chosen this way because otherwise we wouldn’t get a useful map. (Also see my edit.)
– Wrzlprmft
Aug 1 at 9:25

@BAYMAX: I am not sure I get your question here, but the question in your comment seems to be asked the wrong way around. The Poincaré map removes one dimension because $S$ is $n-1$-dimensional. $S$ is chosen this way because otherwise we wouldn’t get a useful map. (Also see my edit.)
– Wrzlprmft
Aug 1 at 9:25

add a comment | 

 

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