Constructing a Poincare map for dynamical system
$begingroup$
I'm trying to construct a Poincare map for the system:
$$dot{x} = y$$
$$dot{y} = -a^2x + bcos(theta)$$
$$dot{theta} = a$$
I have always thought of the Poincare map as more of a theoretical tool and have never had to calculate one before.
My first difficulty here is deciding a surface on which to find the intersections. My initial thought was to use $theta = 0$ and then take snapshots of the system every time $theta$ is a multiple of $2pi$ but I'm having trouble finding the actual map.
Thanks for any help
ordinary-differential-equations dynamical-systems
edited Dec 17 '18 at 20:30
Adam
1,1951919
asked May 12 '15 at 15:01
WoosterWooster
1,400935
$endgroup$
add a comment |
$begingroup$
I'm trying to construct a Poincare map for the system:
$$dot{x} = y$$
$$dot{y} = -a^2x + bcos(theta)$$
$$dot{theta} = a$$
I have always thought of the Poincare map as more of a theoretical tool and have never had to calculate one before.
My first difficulty here is deciding a surface on which to find the intersections. My initial thought was to use $theta = 0$ and then take snapshots of the system every time $theta$ is a multiple of $2pi$ but I'm having trouble finding the actual map.
Thanks for any help
ordinary-differential-equations dynamical-systems
edited Dec 17 '18 at 20:30
Adam
1,1951919
asked May 12 '15 at 15:01
WoosterWooster
1,400935
$endgroup$
add a comment |
$begingroup$
I'm trying to construct a Poincare map for the system:
$$dot{x} = y$$
$$dot{y} = -a^2x + bcos(theta)$$
$$dot{theta} = a$$
I have always thought of the Poincare map as more of a theoretical tool and have never had to calculate one before.
My first difficulty here is deciding a surface on which to find the intersections. My initial thought was to use $theta = 0$ and then take snapshots of the system every time $theta$ is a multiple of $2pi$ but I'm having trouble finding the actual map.
Thanks for any help
ordinary-differential-equations dynamical-systems
edited Dec 17 '18 at 20:30
Adam
1,1951919
asked May 12 '15 at 15:01
WoosterWooster
1,400935
$endgroup$
I'm trying to construct a Poincare map for the system:
$$dot{x} = y$$
$$dot{y} = -a^2x + bcos(theta)$$
$$dot{theta} = a$$
I have always thought of the Poincare map as more of a theoretical tool and have never had to calculate one before.
My first difficulty here is deciding a surface on which to find the intersections. My initial thought was to use $theta = 0$ and then take snapshots of the system every time $theta$ is a multiple of $2pi$ but I'm having trouble finding the actual map.
Thanks for any help
ordinary-differential-equations dynamical-systems
ordinary-differential-equations dynamical-systems
edited Dec 17 '18 at 20:30
Adam
1,1951919
asked May 12 '15 at 15:01
WoosterWooster
1,400935
edited Dec 17 '18 at 20:30
Adam
1,1951919
asked May 12 '15 at 15:01
WoosterWooster
1,400935
edited Dec 17 '18 at 20:30
Adam
1,1951919
edited Dec 17 '18 at 20:30
Adam
1,1951919
edited Dec 17 '18 at 20:30
Adam
1,1951919
1,1951919
asked May 12 '15 at 15:01
WoosterWooster
1,400935
asked May 12 '15 at 15:01
WoosterWooster
1,400935
asked May 12 '15 at 15:01
WoosterWooster
1,400935
1,400935
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, $theta = 0$ is a good place to start. Thus take $theta = at$, and then solve the second-order linear DE
$ddot{x} + a^2 x = b cos(at)$ with $x(0) = x_0$, $y(0) = dot{x}(0) = y_0$.
Your Poincare map takes $(x_0, y_0)$ to $(x(2pi), y(2pi))$.
answered May 12 '15 at 15:17
Robert IsraelRobert Israel
327k23216470
$endgroup$
$begingroup$
Okay great - thanks for your help. My difficulty was first believing I had the right choice with $theta = 0$ and then secondly realising that the $x$ and $y$ equations could be made into a 2nd order ODE.
$endgroup$
– Wooster
May 12 '15 at 15:45
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
$begingroup$
Yes, $theta = 0$ is a good place to start. Thus take $theta = at$, and then solve the second-order linear DE
$ddot{x} + a^2 x = b cos(at)$ with $x(0) = x_0$, $y(0) = dot{x}(0) = y_0$.
Your Poincare map takes $(x_0, y_0)$ to $(x(2pi), y(2pi))$.
answered May 12 '15 at 15:17
Robert IsraelRobert Israel
327k23216470
$endgroup$
$begingroup$
Okay great - thanks for your help. My difficulty was first believing I had the right choice with $theta = 0$ and then secondly realising that the $x$ and $y$ equations could be made into a 2nd order ODE.
$endgroup$
– Wooster
May 12 '15 at 15:45
add a comment |
$begingroup$
Yes, $theta = 0$ is a good place to start. Thus take $theta = at$, and then solve the second-order linear DE
$ddot{x} + a^2 x = b cos(at)$ with $x(0) = x_0$, $y(0) = dot{x}(0) = y_0$.
Your Poincare map takes $(x_0, y_0)$ to $(x(2pi), y(2pi))$.
answered May 12 '15 at 15:17
Robert IsraelRobert Israel
327k23216470
$endgroup$
$begingroup$
Okay great - thanks for your help. My difficulty was first believing I had the right choice with $theta = 0$ and then secondly realising that the $x$ and $y$ equations could be made into a 2nd order ODE.
$endgroup$
– Wooster
May 12 '15 at 15:45
add a comment |
$begingroup$
Yes, $theta = 0$ is a good place to start. Thus take $theta = at$, and then solve the second-order linear DE
$ddot{x} + a^2 x = b cos(at)$ with $x(0) = x_0$, $y(0) = dot{x}(0) = y_0$.
Your Poincare map takes $(x_0, y_0)$ to $(x(2pi), y(2pi))$.
answered May 12 '15 at 15:17
Robert IsraelRobert Israel
327k23216470
$endgroup$
Yes, $theta = 0$ is a good place to start. Thus take $theta = at$, and then solve the second-order linear DE
$ddot{x} + a^2 x = b cos(at)$ with $x(0) = x_0$, $y(0) = dot{x}(0) = y_0$.
Your Poincare map takes $(x_0, y_0)$ to $(x(2pi), y(2pi))$.
answered May 12 '15 at 15:17
Robert IsraelRobert Israel
327k23216470
answered May 12 '15 at 15:17
Robert IsraelRobert Israel
327k23216470
answered May 12 '15 at 15:17
Robert IsraelRobert Israel
327k23216470
answered May 12 '15 at 15:17
Robert IsraelRobert Israel
327k23216470
327k23216470
$begingroup$
Okay great - thanks for your help. My difficulty was first believing I had the right choice with $theta = 0$ and then secondly realising that the $x$ and $y$ equations could be made into a 2nd order ODE.
$endgroup$
– Wooster
May 12 '15 at 15:45
add a comment |
$begingroup$
Okay great - thanks for your help. My difficulty was first believing I had the right choice with $theta = 0$ and then secondly realising that the $x$ and $y$ equations could be made into a 2nd order ODE.
$endgroup$
– Wooster
May 12 '15 at 15:45
$begingroup$
Okay great - thanks for your help. My difficulty was first believing I had the right choice with $theta = 0$ and then secondly realising that the $x$ and $y$ equations could be made into a 2nd order ODE.
$endgroup$
– Wooster
May 12 '15 at 15:45
$begingroup$
Okay great - thanks for your help. My difficulty was first believing I had the right choice with $theta = 0$ and then secondly realising that the $x$ and $y$ equations could be made into a 2nd order ODE.
$endgroup$
– Wooster
May 12 '15 at 15:45
add a comment |
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