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How to determine the existence of limit cycle?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
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I'm student learning about dynamical system. I understood well how to find fixed points and determine stability thanks to eigenvalues of the jacobian matrix, but not how to find limit cycle...
I heard about Poincaré–Bendixson theorem, but it remain totally unclear how to applied it in "pratical" way...
Thanks
dynamical-systems bifurcation
asked 3 hours ago
Dadep
1135
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Dadep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
2
down vote
favorite
I'm student learning about dynamical system. I understood well how to find fixed points and determine stability thanks to eigenvalues of the jacobian matrix, but not how to find limit cycle...
I heard about Poincaré–Bendixson theorem, but it remain totally unclear how to applied it in "pratical" way...
Thanks
dynamical-systems bifurcation
asked 3 hours ago
Dadep
1135
New contributor
Dadep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Hey, that's actually a good question. I'm working through an extensive answer below.
– Rebellos
3 hours ago
2
Note that the Poincaré-Bendixson theorem can be used to establish the existence (or absence) of a limit cycle for certain two-dimensional dynamical systems. I believe the problem of finding or even proving the existence of limit cycles in higher dimensional systems is in general much harder because of the possibility of strange atrractors.
– gandalf61
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm student learning about dynamical system. I understood well how to find fixed points and determine stability thanks to eigenvalues of the jacobian matrix, but not how to find limit cycle...
I heard about Poincaré–Bendixson theorem, but it remain totally unclear how to applied it in "pratical" way...
Thanks
dynamical-systems bifurcation
asked 3 hours ago
Dadep
1135
New contributor
Dadep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm student learning about dynamical system. I understood well how to find fixed points and determine stability thanks to eigenvalues of the jacobian matrix, but not how to find limit cycle...
I heard about Poincaré–Bendixson theorem, but it remain totally unclear how to applied it in "pratical" way...
Thanks
dynamical-systems bifurcation
dynamical-systems bifurcation
asked 3 hours ago
Dadep
1135
New contributor
Dadep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Dadep
1135
New contributor
Dadep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Dadep
1135
New contributor
Dadep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Dadep
1135
asked 3 hours ago
Dadep
1135
1135
New contributor
Dadep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dadep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dadep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Hey, that's actually a good question. I'm working through an extensive answer below.
– Rebellos
3 hours ago
2
Note that the Poincaré-Bendixson theorem can be used to establish the existence (or absence) of a limit cycle for certain two-dimensional dynamical systems. I believe the problem of finding or even proving the existence of limit cycles in higher dimensional systems is in general much harder because of the possibility of strange atrractors.
– gandalf61
2 hours ago
add a comment |Â
1
Hey, that's actually a good question. I'm working through an extensive answer below.
– Rebellos
3 hours ago
2
Note that the Poincaré-Bendixson theorem can be used to establish the existence (or absence) of a limit cycle for certain two-dimensional dynamical systems. I believe the problem of finding or even proving the existence of limit cycles in higher dimensional systems is in general much harder because of the possibility of strange atrractors.
– gandalf61
2 hours ago
1
1
Hey, that's actually a good question. I'm working through an extensive answer below.
– Rebellos
3 hours ago
Hey, that's actually a good question. I'm working through an extensive answer below.
– Rebellos
3 hours ago
2
2
Note that the Poincaré-Bendixson theorem can be used to establish the existence (or absence) of a limit cycle for certain two-dimensional dynamical systems. I believe the problem of finding or even proving the existence of limit cycles in higher dimensional systems is in general much harder because of the possibility of strange atrractors.
– gandalf61
2 hours ago
Note that the Poincaré-Bendixson theorem can be used to establish the existence (or absence) of a limit cycle for certain two-dimensional dynamical systems. I believe the problem of finding or even proving the existence of limit cycles in higher dimensional systems is in general much harder because of the possibility of strange atrractors.
– gandalf61
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
The Poincare-Bendixson theorem, states that :
Theorem : Given a differentiable real dynamical system defined on an open subset of the plane, then every non-empty compact É-limit set of an orbit, which contains only finitely many fixed points, is either :
a fixed point, a periodic orbit, or a connected set composed of a finite number of fixed points together with homoclinic and heteroclinic orbits connecting these.
Moreover, there is at most one orbit connecting different fixed points in the same direction. However, there could be countably many homoclinic orbits connecting one fixed point.
I think the best way to get an understanding over the existence of $omega$ limit sets and thus limit cycles on real dynamical systems, the best way is to elaborate some examples. Read carefully through the following :
Exercise :
Consider the following differential system, given in polar coordinates:
$$r' = r(μ-r^2)$$
$$θ' = ÃÂ(r^2)$$
where $μ>0$ is a constant and $p: mathbb R_+ to mathbb R$ smooth function with $ÃÂ(μ) > 0$. Show that there exists a unique limit cycle.
Solution :
The interesting part of the given system, is the expression $r(μ-r^2)$. Recall that since we are working over polar coordinates, it is $r>0$. Thus, for the given expression, it is :
$$r(μ-r^2) = 0 Rightarrow r=sqrtμ$$
Now, to derive a conclusion about the existence of an $omega-$limit set, all you have to do is draw an one-dimensional phase portrait, where an $longrightarrow$ arrow deonotes the area where $r'>0$ and on the other hand, an $longleftarrow$ deones the area where $r'<0$. Thus, the following portrait is yielded :
$$longrightarrow quad sqrtmu quad longleftarrow$$
From this one-dimension portrait, we can see that $r'$ gets "bounded" around $sqrtmu$. Thus we have proved the existence of an $omega$ limit set, which proves to be a clockwise limit-cycle, since $θ'(sqrtmu) = ÃÂ(μ) > 0$.
Exercise :
Given the dynamical system :
$$x_1' = x_2+2μx_1(5-x_1^2-x_2^2)$$
$$x_2' = -x_1+2μx_2(5-x_1^2-x_2^2)$$
where $(x_1,x_2) in mathbb R^2$ and $μ>0$ a constant. Applying polar coordinates, determine the omega limit set $É(x_0)$ for any given vector $(x_1,x_2)$.
Discussion :
I used the polar coordinates substitution :
$$x_1 = rcosθ$$
$$x_2 = rsinθ$$
and via the expressions :
$$rr' = x_1x_1' + x_2x_2'$$
$$θ' = fracx_1x_2' - x_2x_1'r^2$$
Using a polar coordinate substitution correctly for $x_1$ and $x_2$, one yields :
$$r' = 2μr(5-r^2)$$
$$θ' = -1$$
Now, what we can see is that the sign of $r'$ entirely depends on the factor $5-r^2$, since by it's definition $r>0$ and $μ>0$. This means that until $r=sqrt5$ $r'>0$ and after that $r'<0$. Also, noting that $θ'=-1$ means that the direction of the phase portrait flow follows an anticlockwise flow.
Other than that, defining an one dimension phase portrait for $r'$ and noting the arrows just like the previous exercise, we get :
$$ longrightarrow sqrt5 longleftarrow$$
Again, we see that $r'$ is bounded around $sqrt5$, thus this means that there exists a limit cycle defined with $r=sqrt5$.
This implies that the omega limit set for any given values, will be :
$$textFor space x_0 neq 0 space rightarrow space É(x_0) = S_sqrt5$$
$$textFor space x_0 = 0 space rightarrow space É(x_0) = 0$$
answered 3 hours ago
Rebellos
10.9k21039
1
This is very clear, thanks a lot.
– Dadep
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The Poincare-Bendixson theorem, states that :
Theorem : Given a differentiable real dynamical system defined on an open subset of the plane, then every non-empty compact É-limit set of an orbit, which contains only finitely many fixed points, is either :
a fixed point, a periodic orbit, or a connected set composed of a finite number of fixed points together with homoclinic and heteroclinic orbits connecting these.
Moreover, there is at most one orbit connecting different fixed points in the same direction. However, there could be countably many homoclinic orbits connecting one fixed point.
I think the best way to get an understanding over the existence of $omega$ limit sets and thus limit cycles on real dynamical systems, the best way is to elaborate some examples. Read carefully through the following :
Exercise :
Consider the following differential system, given in polar coordinates:
$$r' = r(μ-r^2)$$
$$θ' = ÃÂ(r^2)$$
where $μ>0$ is a constant and $p: mathbb R_+ to mathbb R$ smooth function with $ÃÂ(μ) > 0$. Show that there exists a unique limit cycle.
Solution :
The interesting part of the given system, is the expression $r(μ-r^2)$. Recall that since we are working over polar coordinates, it is $r>0$. Thus, for the given expression, it is :
$$r(μ-r^2) = 0 Rightarrow r=sqrtμ$$
Now, to derive a conclusion about the existence of an $omega-$limit set, all you have to do is draw an one-dimensional phase portrait, where an $longrightarrow$ arrow deonotes the area where $r'>0$ and on the other hand, an $longleftarrow$ deones the area where $r'<0$. Thus, the following portrait is yielded :
$$longrightarrow quad sqrtmu quad longleftarrow$$
From this one-dimension portrait, we can see that $r'$ gets "bounded" around $sqrtmu$. Thus we have proved the existence of an $omega$ limit set, which proves to be a clockwise limit-cycle, since $θ'(sqrtmu) = ÃÂ(μ) > 0$.
Exercise :
Given the dynamical system :
$$x_1' = x_2+2μx_1(5-x_1^2-x_2^2)$$
$$x_2' = -x_1+2μx_2(5-x_1^2-x_2^2)$$
where $(x_1,x_2) in mathbb R^2$ and $μ>0$ a constant. Applying polar coordinates, determine the omega limit set $É(x_0)$ for any given vector $(x_1,x_2)$.
Discussion :
I used the polar coordinates substitution :
$$x_1 = rcosθ$$
$$x_2 = rsinθ$$
and via the expressions :
$$rr' = x_1x_1' + x_2x_2'$$
$$θ' = fracx_1x_2' - x_2x_1'r^2$$
Using a polar coordinate substitution correctly for $x_1$ and $x_2$, one yields :
$$r' = 2μr(5-r^2)$$
$$θ' = -1$$
Now, what we can see is that the sign of $r'$ entirely depends on the factor $5-r^2$, since by it's definition $r>0$ and $μ>0$. This means that until $r=sqrt5$ $r'>0$ and after that $r'<0$. Also, noting that $θ'=-1$ means that the direction of the phase portrait flow follows an anticlockwise flow.
Other than that, defining an one dimension phase portrait for $r'$ and noting the arrows just like the previous exercise, we get :
$$ longrightarrow sqrt5 longleftarrow$$
Again, we see that $r'$ is bounded around $sqrt5$, thus this means that there exists a limit cycle defined with $r=sqrt5$.
This implies that the omega limit set for any given values, will be :
$$textFor space x_0 neq 0 space rightarrow space É(x_0) = S_sqrt5$$
$$textFor space x_0 = 0 space rightarrow space É(x_0) = 0$$
answered 3 hours ago
Rebellos
10.9k21039
1
This is very clear, thanks a lot.
– Dadep
2 hours ago
add a comment |Â
up vote
4
down vote
accepted
The Poincare-Bendixson theorem, states that :
Theorem : Given a differentiable real dynamical system defined on an open subset of the plane, then every non-empty compact É-limit set of an orbit, which contains only finitely many fixed points, is either :
a fixed point, a periodic orbit, or a connected set composed of a finite number of fixed points together with homoclinic and heteroclinic orbits connecting these.
Moreover, there is at most one orbit connecting different fixed points in the same direction. However, there could be countably many homoclinic orbits connecting one fixed point.
I think the best way to get an understanding over the existence of $omega$ limit sets and thus limit cycles on real dynamical systems, the best way is to elaborate some examples. Read carefully through the following :
Exercise :
Consider the following differential system, given in polar coordinates:
$$r' = r(μ-r^2)$$
$$θ' = ÃÂ(r^2)$$
where $μ>0$ is a constant and $p: mathbb R_+ to mathbb R$ smooth function with $ÃÂ(μ) > 0$. Show that there exists a unique limit cycle.
Solution :
The interesting part of the given system, is the expression $r(μ-r^2)$. Recall that since we are working over polar coordinates, it is $r>0$. Thus, for the given expression, it is :
$$r(μ-r^2) = 0 Rightarrow r=sqrtμ$$
Now, to derive a conclusion about the existence of an $omega-$limit set, all you have to do is draw an one-dimensional phase portrait, where an $longrightarrow$ arrow deonotes the area where $r'>0$ and on the other hand, an $longleftarrow$ deones the area where $r'<0$. Thus, the following portrait is yielded :
$$longrightarrow quad sqrtmu quad longleftarrow$$
From this one-dimension portrait, we can see that $r'$ gets "bounded" around $sqrtmu$. Thus we have proved the existence of an $omega$ limit set, which proves to be a clockwise limit-cycle, since $θ'(sqrtmu) = ÃÂ(μ) > 0$.
Exercise :
Given the dynamical system :
$$x_1' = x_2+2μx_1(5-x_1^2-x_2^2)$$
$$x_2' = -x_1+2μx_2(5-x_1^2-x_2^2)$$
where $(x_1,x_2) in mathbb R^2$ and $μ>0$ a constant. Applying polar coordinates, determine the omega limit set $É(x_0)$ for any given vector $(x_1,x_2)$.
Discussion :
I used the polar coordinates substitution :
$$x_1 = rcosθ$$
$$x_2 = rsinθ$$
and via the expressions :
$$rr' = x_1x_1' + x_2x_2'$$
$$θ' = fracx_1x_2' - x_2x_1'r^2$$
Using a polar coordinate substitution correctly for $x_1$ and $x_2$, one yields :
$$r' = 2μr(5-r^2)$$
$$θ' = -1$$
Now, what we can see is that the sign of $r'$ entirely depends on the factor $5-r^2$, since by it's definition $r>0$ and $μ>0$. This means that until $r=sqrt5$ $r'>0$ and after that $r'<0$. Also, noting that $θ'=-1$ means that the direction of the phase portrait flow follows an anticlockwise flow.
Other than that, defining an one dimension phase portrait for $r'$ and noting the arrows just like the previous exercise, we get :
$$ longrightarrow sqrt5 longleftarrow$$
Again, we see that $r'$ is bounded around $sqrt5$, thus this means that there exists a limit cycle defined with $r=sqrt5$.
This implies that the omega limit set for any given values, will be :
$$textFor space x_0 neq 0 space rightarrow space É(x_0) = S_sqrt5$$
$$textFor space x_0 = 0 space rightarrow space É(x_0) = 0$$
answered 3 hours ago
Rebellos
10.9k21039
1
This is very clear, thanks a lot.
– Dadep
2 hours ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The Poincare-Bendixson theorem, states that :
Theorem : Given a differentiable real dynamical system defined on an open subset of the plane, then every non-empty compact É-limit set of an orbit, which contains only finitely many fixed points, is either :
a fixed point, a periodic orbit, or a connected set composed of a finite number of fixed points together with homoclinic and heteroclinic orbits connecting these.
Moreover, there is at most one orbit connecting different fixed points in the same direction. However, there could be countably many homoclinic orbits connecting one fixed point.
I think the best way to get an understanding over the existence of $omega$ limit sets and thus limit cycles on real dynamical systems, the best way is to elaborate some examples. Read carefully through the following :
Exercise :
Consider the following differential system, given in polar coordinates:
$$r' = r(μ-r^2)$$
$$θ' = ÃÂ(r^2)$$
where $μ>0$ is a constant and $p: mathbb R_+ to mathbb R$ smooth function with $ÃÂ(μ) > 0$. Show that there exists a unique limit cycle.
Solution :
The interesting part of the given system, is the expression $r(μ-r^2)$. Recall that since we are working over polar coordinates, it is $r>0$. Thus, for the given expression, it is :
$$r(μ-r^2) = 0 Rightarrow r=sqrtμ$$
Now, to derive a conclusion about the existence of an $omega-$limit set, all you have to do is draw an one-dimensional phase portrait, where an $longrightarrow$ arrow deonotes the area where $r'>0$ and on the other hand, an $longleftarrow$ deones the area where $r'<0$. Thus, the following portrait is yielded :
$$longrightarrow quad sqrtmu quad longleftarrow$$
From this one-dimension portrait, we can see that $r'$ gets "bounded" around $sqrtmu$. Thus we have proved the existence of an $omega$ limit set, which proves to be a clockwise limit-cycle, since $θ'(sqrtmu) = ÃÂ(μ) > 0$.
Exercise :
Given the dynamical system :
$$x_1' = x_2+2μx_1(5-x_1^2-x_2^2)$$
$$x_2' = -x_1+2μx_2(5-x_1^2-x_2^2)$$
where $(x_1,x_2) in mathbb R^2$ and $μ>0$ a constant. Applying polar coordinates, determine the omega limit set $É(x_0)$ for any given vector $(x_1,x_2)$.
Discussion :
I used the polar coordinates substitution :
$$x_1 = rcosθ$$
$$x_2 = rsinθ$$
and via the expressions :
$$rr' = x_1x_1' + x_2x_2'$$
$$θ' = fracx_1x_2' - x_2x_1'r^2$$
Using a polar coordinate substitution correctly for $x_1$ and $x_2$, one yields :
$$r' = 2μr(5-r^2)$$
$$θ' = -1$$
Now, what we can see is that the sign of $r'$ entirely depends on the factor $5-r^2$, since by it's definition $r>0$ and $μ>0$. This means that until $r=sqrt5$ $r'>0$ and after that $r'<0$. Also, noting that $θ'=-1$ means that the direction of the phase portrait flow follows an anticlockwise flow.
Other than that, defining an one dimension phase portrait for $r'$ and noting the arrows just like the previous exercise, we get :
$$ longrightarrow sqrt5 longleftarrow$$
Again, we see that $r'$ is bounded around $sqrt5$, thus this means that there exists a limit cycle defined with $r=sqrt5$.
This implies that the omega limit set for any given values, will be :
$$textFor space x_0 neq 0 space rightarrow space É(x_0) = S_sqrt5$$
$$textFor space x_0 = 0 space rightarrow space É(x_0) = 0$$
answered 3 hours ago
Rebellos
10.9k21039
The Poincare-Bendixson theorem, states that :
Theorem : Given a differentiable real dynamical system defined on an open subset of the plane, then every non-empty compact É-limit set of an orbit, which contains only finitely many fixed points, is either :
a fixed point, a periodic orbit, or a connected set composed of a finite number of fixed points together with homoclinic and heteroclinic orbits connecting these.
Moreover, there is at most one orbit connecting different fixed points in the same direction. However, there could be countably many homoclinic orbits connecting one fixed point.
I think the best way to get an understanding over the existence of $omega$ limit sets and thus limit cycles on real dynamical systems, the best way is to elaborate some examples. Read carefully through the following :
Exercise :
Consider the following differential system, given in polar coordinates:
$$r' = r(μ-r^2)$$
$$θ' = ÃÂ(r^2)$$
where $μ>0$ is a constant and $p: mathbb R_+ to mathbb R$ smooth function with $ÃÂ(μ) > 0$. Show that there exists a unique limit cycle.
Solution :
The interesting part of the given system, is the expression $r(μ-r^2)$. Recall that since we are working over polar coordinates, it is $r>0$. Thus, for the given expression, it is :
$$r(μ-r^2) = 0 Rightarrow r=sqrtμ$$
Now, to derive a conclusion about the existence of an $omega-$limit set, all you have to do is draw an one-dimensional phase portrait, where an $longrightarrow$ arrow deonotes the area where $r'>0$ and on the other hand, an $longleftarrow$ deones the area where $r'<0$. Thus, the following portrait is yielded :
$$longrightarrow quad sqrtmu quad longleftarrow$$
From this one-dimension portrait, we can see that $r'$ gets "bounded" around $sqrtmu$. Thus we have proved the existence of an $omega$ limit set, which proves to be a clockwise limit-cycle, since $θ'(sqrtmu) = ÃÂ(μ) > 0$.
Exercise :
Given the dynamical system :
$$x_1' = x_2+2μx_1(5-x_1^2-x_2^2)$$
$$x_2' = -x_1+2μx_2(5-x_1^2-x_2^2)$$
where $(x_1,x_2) in mathbb R^2$ and $μ>0$ a constant. Applying polar coordinates, determine the omega limit set $É(x_0)$ for any given vector $(x_1,x_2)$.
Discussion :
I used the polar coordinates substitution :
$$x_1 = rcosθ$$
$$x_2 = rsinθ$$
and via the expressions :
$$rr' = x_1x_1' + x_2x_2'$$
$$θ' = fracx_1x_2' - x_2x_1'r^2$$
Using a polar coordinate substitution correctly for $x_1$ and $x_2$, one yields :
$$r' = 2μr(5-r^2)$$
$$θ' = -1$$
Now, what we can see is that the sign of $r'$ entirely depends on the factor $5-r^2$, since by it's definition $r>0$ and $μ>0$. This means that until $r=sqrt5$ $r'>0$ and after that $r'<0$. Also, noting that $θ'=-1$ means that the direction of the phase portrait flow follows an anticlockwise flow.
Other than that, defining an one dimension phase portrait for $r'$ and noting the arrows just like the previous exercise, we get :
$$ longrightarrow sqrt5 longleftarrow$$
Again, we see that $r'$ is bounded around $sqrt5$, thus this means that there exists a limit cycle defined with $r=sqrt5$.
This implies that the omega limit set for any given values, will be :
$$textFor space x_0 neq 0 space rightarrow space É(x_0) = S_sqrt5$$
$$textFor space x_0 = 0 space rightarrow space É(x_0) = 0$$
answered 3 hours ago
Rebellos
10.9k21039
answered 3 hours ago
Rebellos
10.9k21039
answered 3 hours ago
Rebellos
10.9k21039
answered 3 hours ago
Rebellos
10.9k21039
10.9k21039
1
This is very clear, thanks a lot.
– Dadep
2 hours ago
add a comment |Â
1
This is very clear, thanks a lot.
– Dadep
2 hours ago
1
1
This is very clear, thanks a lot.
– Dadep
2 hours ago
This is very clear, thanks a lot.
– Dadep
2 hours ago
add a comment |Â
Dadep is a new contributor. Be nice, and check out our Code of Conduct.
Dadep is a new contributor. Be nice, and check out our Code of Conduct.
Dadep is a new contributor. Be nice, and check out our Code of Conduct.
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1
Hey, that's actually a good question. I'm working through an extensive answer below.
– Rebellos
3 hours ago
2
Note that the Poincaré-Bendixson theorem can be used to establish the existence (or absence) of a limit cycle for certain two-dimensional dynamical systems. I believe the problem of finding or even proving the existence of limit cycles in higher dimensional systems is in general much harder because of the possibility of strange atrractors.
– gandalf61
2 hours ago