Global Attractor Existence
up vote
1
down vote
favorite
How do I know if a system of differential equations has a global attractor?
I am an undergraduate physics student. I am trying to understand what is the domain of applicability of Convergent Cross Mapping. I tested it on nonlinear and linear systems of equations and gives counter-intuitive results. My hypothesis is that it applies only to dynamical systems of differential equations which have a global attractor. But I am new to this area. How do I know if system of diff eqs has a global attractor? Are there criteria for it?
I understood the definition of the global attractor from here and maybe here. But I am interested only to know if it has or not. Not what the global attractor is.
Thanks in advance!
PS: I know this is a soft question and I am not sure if I posted it in the right place.
differential-equations dynamical-systems
asked Nov 18 at 10:57
asd11
1937
add a comment |
up vote
1
down vote
favorite
How do I know if a system of differential equations has a global attractor?
I am an undergraduate physics student. I am trying to understand what is the domain of applicability of Convergent Cross Mapping. I tested it on nonlinear and linear systems of equations and gives counter-intuitive results. My hypothesis is that it applies only to dynamical systems of differential equations which have a global attractor. But I am new to this area. How do I know if system of diff eqs has a global attractor? Are there criteria for it?
I understood the definition of the global attractor from here and maybe here. But I am interested only to know if it has or not. Not what the global attractor is.
Thanks in advance!
PS: I know this is a soft question and I am not sure if I posted it in the right place.
differential-equations dynamical-systems
asked Nov 18 at 10:57
asd11
1937
1
You might be interested in "Lyapunov stability theory", for a formal treatment the classic paper by Kalman and Bertram is one of the best: fluidsengineering.asmedigitalcollection.asme.org/…
– WalterJ
Nov 18 at 15:40
Thank you! I'll study your paper!
– asd11
Nov 18 at 16:34
Please see the edit to my answer.
– Wrzlprmft
Nov 18 at 21:36
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How do I know if a system of differential equations has a global attractor?
I am an undergraduate physics student. I am trying to understand what is the domain of applicability of Convergent Cross Mapping. I tested it on nonlinear and linear systems of equations and gives counter-intuitive results. My hypothesis is that it applies only to dynamical systems of differential equations which have a global attractor. But I am new to this area. How do I know if system of diff eqs has a global attractor? Are there criteria for it?
I understood the definition of the global attractor from here and maybe here. But I am interested only to know if it has or not. Not what the global attractor is.
Thanks in advance!
PS: I know this is a soft question and I am not sure if I posted it in the right place.
differential-equations dynamical-systems
asked Nov 18 at 10:57
asd11
1937
How do I know if a system of differential equations has a global attractor?
I am an undergraduate physics student. I am trying to understand what is the domain of applicability of Convergent Cross Mapping. I tested it on nonlinear and linear systems of equations and gives counter-intuitive results. My hypothesis is that it applies only to dynamical systems of differential equations which have a global attractor. But I am new to this area. How do I know if system of diff eqs has a global attractor? Are there criteria for it?
I understood the definition of the global attractor from here and maybe here. But I am interested only to know if it has or not. Not what the global attractor is.
Thanks in advance!
PS: I know this is a soft question and I am not sure if I posted it in the right place.
differential-equations dynamical-systems
differential-equations dynamical-systems
asked Nov 18 at 10:57
asd11
1937
asked Nov 18 at 10:57
asd11
1937
edited Nov 18 at 11:09
asked Nov 18 at 10:57
asd11
1937
asked Nov 18 at 10:57
asd11
1937
asked Nov 18 at 10:57
asd11
1937
1937
1
You might be interested in "Lyapunov stability theory", for a formal treatment the classic paper by Kalman and Bertram is one of the best: fluidsengineering.asmedigitalcollection.asme.org/…
– WalterJ
Nov 18 at 15:40
Thank you! I'll study your paper!
– asd11
Nov 18 at 16:34
Please see the edit to my answer.
– Wrzlprmft
Nov 18 at 21:36
add a comment |
1
You might be interested in "Lyapunov stability theory", for a formal treatment the classic paper by Kalman and Bertram is one of the best: fluidsengineering.asmedigitalcollection.asme.org/…
– WalterJ
Nov 18 at 15:40
Thank you! I'll study your paper!
– asd11
Nov 18 at 16:34
Please see the edit to my answer.
– Wrzlprmft
Nov 18 at 21:36
1
1
You might be interested in "Lyapunov stability theory", for a formal treatment the classic paper by Kalman and Bertram is one of the best: fluidsengineering.asmedigitalcollection.asme.org/…
– WalterJ
Nov 18 at 15:40
You might be interested in "Lyapunov stability theory", for a formal treatment the classic paper by Kalman and Bertram is one of the best: fluidsengineering.asmedigitalcollection.asme.org/…
– WalterJ
Nov 18 at 15:40
Thank you! I'll study your paper!
– asd11
Nov 18 at 16:34
Thank you! I'll study your paper!
– asd11
Nov 18 at 16:34
Please see the edit to my answer.
– Wrzlprmft
Nov 18 at 21:36
Please see the edit to my answer.
– Wrzlprmft
Nov 18 at 21:36
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
There seem to be multiple definitions of global attractor, so I cover them separately.
Case 1: Global attractors are attractors
Here, I assume that a global attractor is an attractor whose basin of attraction is the entire phase space. In particular, a system with a global attractor is never multistable.
I am not aware of any straightforward criteria that allow you determine the existence of global attractors and are applicable to a relevant number of cases.
For most complex dynamical systems the existence of attractors and their number is usually determined by simulations or thinking very hard™.
That being said, there are a few classes of systems which are known to have global attractors and, in certain cases it is easy to show that a system does not have global attractors.
Some examples:
Linear differential-equation systems have a known solution, which has at most one attractor.
Hamiltonian systems do not have attractors at all (only dissipative systems have).
If your differential equation is $dot{x} = f(x)$, a point $x$ is a stable fixed point (a certain type of attractor) if and only if $f(x)=0$ and the real parts of all eigenvalues $nabla f(x)$ are negative. If you can find two such stable fixed points, your system does not have a global attractor. Mind that the inverse does not hold as not all attractors are stable fixed points.
Ecological or metabolic systems are known to need a positive feedback loop to exhibit multistability.
Case 2: Global attractors contain all attractors
Here, I assume that a global attractor is a set whose basin of attraction is the entire phase space. Such a global attractor would be the union of all attractors. Most confusingly, such a global attractor would not need to be an attractor itself, since it is not minimal. Finally note that the union of all attractors of a system needs not be a global attractor.
In this case, all you need to show is that your dynamics is bounded, i.e., there are no trajectories going towards infinity.
Again there are some cases where it has been shown that such global attractors exist and in some cases it can be easy to show that the system has such a global attractor.
For example, if you know that your system’s dynamics is usually evolving around the origin, it may be easy to show that for points that have a certain distance to the origin, this distance is decreasing over time.
answered Nov 18 at 14:02
Wrzlprmft
3,04111233
Take $dot{x}=x(1-x^2)$ and $A=[-1,1]$. Then $A$ satisfies the continuous-time analogs of the definitions of global attractor given in the OP's references, but it contains two asymptotically stable equilibria, namely $-1$ and $1$.
– user539887
Nov 18 at 16:10
@user539887: How is $A$ an attractor? It is not invariant under time-evolution. With the exception of $0$ all points in $A$ converge towards the stable equlibria ($-1$ and $1$) under time-evolution.
– Wrzlprmft
Nov 18 at 16:23
Yes, it is invariant.
– user539887
Nov 18 at 16:25
@user539887: And how do you reconcile this with all points converging towards $-1$ and $1$? Just so we are on the same page, I consider a set $A$ invariant under time evolution if $ϕ(A,t) = A$ for all $t≥0$, where $ϕ$ is the time-evolution operator.
– Wrzlprmft
Nov 18 at 16:40
Take $x=0$ or $pm1$. As they are equilibria, $phi(x,t)=x$ for all $tinmathbb{R}$. Take $xin(-1,0)$, Then we have $phi(x,t)in(-1,0)$ for all $tinmathbb{R}$. Take $xin(0,1)$, Then we have $phi(x,t)in(0,1)$ for all $tinmathbb{R}$, too. Consequently, $phi(A,t)subset A$ for all $tinmathbb{R}$, from which it follows, by "algebraic" properties of the flow, that $phi(A,t){color{red}=}A$ for all $tinmathbb{R}$, in particular for all $tge0$.
– user539887
Nov 18 at 16:50
|
show 2 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
There seem to be multiple definitions of global attractor, so I cover them separately.
Case 1: Global attractors are attractors
Here, I assume that a global attractor is an attractor whose basin of attraction is the entire phase space. In particular, a system with a global attractor is never multistable.
I am not aware of any straightforward criteria that allow you determine the existence of global attractors and are applicable to a relevant number of cases.
For most complex dynamical systems the existence of attractors and their number is usually determined by simulations or thinking very hard™.
That being said, there are a few classes of systems which are known to have global attractors and, in certain cases it is easy to show that a system does not have global attractors.
Some examples:
Linear differential-equation systems have a known solution, which has at most one attractor.
Hamiltonian systems do not have attractors at all (only dissipative systems have).
If your differential equation is $dot{x} = f(x)$, a point $x$ is a stable fixed point (a certain type of attractor) if and only if $f(x)=0$ and the real parts of all eigenvalues $nabla f(x)$ are negative. If you can find two such stable fixed points, your system does not have a global attractor. Mind that the inverse does not hold as not all attractors are stable fixed points.
Ecological or metabolic systems are known to need a positive feedback loop to exhibit multistability.
Case 2: Global attractors contain all attractors
Here, I assume that a global attractor is a set whose basin of attraction is the entire phase space. Such a global attractor would be the union of all attractors. Most confusingly, such a global attractor would not need to be an attractor itself, since it is not minimal. Finally note that the union of all attractors of a system needs not be a global attractor.
In this case, all you need to show is that your dynamics is bounded, i.e., there are no trajectories going towards infinity.
Again there are some cases where it has been shown that such global attractors exist and in some cases it can be easy to show that the system has such a global attractor.
For example, if you know that your system’s dynamics is usually evolving around the origin, it may be easy to show that for points that have a certain distance to the origin, this distance is decreasing over time.
answered Nov 18 at 14:02
Wrzlprmft
3,04111233
Take $dot{x}=x(1-x^2)$ and $A=[-1,1]$. Then $A$ satisfies the continuous-time analogs of the definitions of global attractor given in the OP's references, but it contains two asymptotically stable equilibria, namely $-1$ and $1$.
– user539887
Nov 18 at 16:10
@user539887: How is $A$ an attractor? It is not invariant under time-evolution. With the exception of $0$ all points in $A$ converge towards the stable equlibria ($-1$ and $1$) under time-evolution.
– Wrzlprmft
Nov 18 at 16:23
Yes, it is invariant.
– user539887
Nov 18 at 16:25
@user539887: And how do you reconcile this with all points converging towards $-1$ and $1$? Just so we are on the same page, I consider a set $A$ invariant under time evolution if $ϕ(A,t) = A$ for all $t≥0$, where $ϕ$ is the time-evolution operator.
– Wrzlprmft
Nov 18 at 16:40
Take $x=0$ or $pm1$. As they are equilibria, $phi(x,t)=x$ for all $tinmathbb{R}$. Take $xin(-1,0)$, Then we have $phi(x,t)in(-1,0)$ for all $tinmathbb{R}$. Take $xin(0,1)$, Then we have $phi(x,t)in(0,1)$ for all $tinmathbb{R}$, too. Consequently, $phi(A,t)subset A$ for all $tinmathbb{R}$, from which it follows, by "algebraic" properties of the flow, that $phi(A,t){color{red}=}A$ for all $tinmathbb{R}$, in particular for all $tge0$.
– user539887
Nov 18 at 16:50
|
show 2 more comments
up vote
2
down vote
accepted
There seem to be multiple definitions of global attractor, so I cover them separately.
Case 1: Global attractors are attractors
Here, I assume that a global attractor is an attractor whose basin of attraction is the entire phase space. In particular, a system with a global attractor is never multistable.
I am not aware of any straightforward criteria that allow you determine the existence of global attractors and are applicable to a relevant number of cases.
For most complex dynamical systems the existence of attractors and their number is usually determined by simulations or thinking very hard™.
That being said, there are a few classes of systems which are known to have global attractors and, in certain cases it is easy to show that a system does not have global attractors.
Some examples:
Linear differential-equation systems have a known solution, which has at most one attractor.
Hamiltonian systems do not have attractors at all (only dissipative systems have).
If your differential equation is $dot{x} = f(x)$, a point $x$ is a stable fixed point (a certain type of attractor) if and only if $f(x)=0$ and the real parts of all eigenvalues $nabla f(x)$ are negative. If you can find two such stable fixed points, your system does not have a global attractor. Mind that the inverse does not hold as not all attractors are stable fixed points.
Ecological or metabolic systems are known to need a positive feedback loop to exhibit multistability.
Case 2: Global attractors contain all attractors
Here, I assume that a global attractor is a set whose basin of attraction is the entire phase space. Such a global attractor would be the union of all attractors. Most confusingly, such a global attractor would not need to be an attractor itself, since it is not minimal. Finally note that the union of all attractors of a system needs not be a global attractor.
In this case, all you need to show is that your dynamics is bounded, i.e., there are no trajectories going towards infinity.
Again there are some cases where it has been shown that such global attractors exist and in some cases it can be easy to show that the system has such a global attractor.
For example, if you know that your system’s dynamics is usually evolving around the origin, it may be easy to show that for points that have a certain distance to the origin, this distance is decreasing over time.
answered Nov 18 at 14:02
Wrzlprmft
3,04111233
Take $dot{x}=x(1-x^2)$ and $A=[-1,1]$. Then $A$ satisfies the continuous-time analogs of the definitions of global attractor given in the OP's references, but it contains two asymptotically stable equilibria, namely $-1$ and $1$.
– user539887
Nov 18 at 16:10
@user539887: How is $A$ an attractor? It is not invariant under time-evolution. With the exception of $0$ all points in $A$ converge towards the stable equlibria ($-1$ and $1$) under time-evolution.
– Wrzlprmft
Nov 18 at 16:23
Yes, it is invariant.
– user539887
Nov 18 at 16:25
@user539887: And how do you reconcile this with all points converging towards $-1$ and $1$? Just so we are on the same page, I consider a set $A$ invariant under time evolution if $ϕ(A,t) = A$ for all $t≥0$, where $ϕ$ is the time-evolution operator.
– Wrzlprmft
Nov 18 at 16:40
Take $x=0$ or $pm1$. As they are equilibria, $phi(x,t)=x$ for all $tinmathbb{R}$. Take $xin(-1,0)$, Then we have $phi(x,t)in(-1,0)$ for all $tinmathbb{R}$. Take $xin(0,1)$, Then we have $phi(x,t)in(0,1)$ for all $tinmathbb{R}$, too. Consequently, $phi(A,t)subset A$ for all $tinmathbb{R}$, from which it follows, by "algebraic" properties of the flow, that $phi(A,t){color{red}=}A$ for all $tinmathbb{R}$, in particular for all $tge0$.
– user539887
Nov 18 at 16:50
|
show 2 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
There seem to be multiple definitions of global attractor, so I cover them separately.
Case 1: Global attractors are attractors
Here, I assume that a global attractor is an attractor whose basin of attraction is the entire phase space. In particular, a system with a global attractor is never multistable.
I am not aware of any straightforward criteria that allow you determine the existence of global attractors and are applicable to a relevant number of cases.
For most complex dynamical systems the existence of attractors and their number is usually determined by simulations or thinking very hard™.
That being said, there are a few classes of systems which are known to have global attractors and, in certain cases it is easy to show that a system does not have global attractors.
Some examples:
Linear differential-equation systems have a known solution, which has at most one attractor.
Hamiltonian systems do not have attractors at all (only dissipative systems have).
If your differential equation is $dot{x} = f(x)$, a point $x$ is a stable fixed point (a certain type of attractor) if and only if $f(x)=0$ and the real parts of all eigenvalues $nabla f(x)$ are negative. If you can find two such stable fixed points, your system does not have a global attractor. Mind that the inverse does not hold as not all attractors are stable fixed points.
Ecological or metabolic systems are known to need a positive feedback loop to exhibit multistability.
Case 2: Global attractors contain all attractors
Here, I assume that a global attractor is a set whose basin of attraction is the entire phase space. Such a global attractor would be the union of all attractors. Most confusingly, such a global attractor would not need to be an attractor itself, since it is not minimal. Finally note that the union of all attractors of a system needs not be a global attractor.
In this case, all you need to show is that your dynamics is bounded, i.e., there are no trajectories going towards infinity.
Again there are some cases where it has been shown that such global attractors exist and in some cases it can be easy to show that the system has such a global attractor.
For example, if you know that your system’s dynamics is usually evolving around the origin, it may be easy to show that for points that have a certain distance to the origin, this distance is decreasing over time.
answered Nov 18 at 14:02
Wrzlprmft
3,04111233
There seem to be multiple definitions of global attractor, so I cover them separately.
Case 1: Global attractors are attractors
Here, I assume that a global attractor is an attractor whose basin of attraction is the entire phase space. In particular, a system with a global attractor is never multistable.
I am not aware of any straightforward criteria that allow you determine the existence of global attractors and are applicable to a relevant number of cases.
For most complex dynamical systems the existence of attractors and their number is usually determined by simulations or thinking very hard™.
That being said, there are a few classes of systems which are known to have global attractors and, in certain cases it is easy to show that a system does not have global attractors.
Some examples:
Linear differential-equation systems have a known solution, which has at most one attractor.
Hamiltonian systems do not have attractors at all (only dissipative systems have).
If your differential equation is $dot{x} = f(x)$, a point $x$ is a stable fixed point (a certain type of attractor) if and only if $f(x)=0$ and the real parts of all eigenvalues $nabla f(x)$ are negative. If you can find two such stable fixed points, your system does not have a global attractor. Mind that the inverse does not hold as not all attractors are stable fixed points.
Ecological or metabolic systems are known to need a positive feedback loop to exhibit multistability.
Case 2: Global attractors contain all attractors
Here, I assume that a global attractor is a set whose basin of attraction is the entire phase space. Such a global attractor would be the union of all attractors. Most confusingly, such a global attractor would not need to be an attractor itself, since it is not minimal. Finally note that the union of all attractors of a system needs not be a global attractor.
In this case, all you need to show is that your dynamics is bounded, i.e., there are no trajectories going towards infinity.
Again there are some cases where it has been shown that such global attractors exist and in some cases it can be easy to show that the system has such a global attractor.
For example, if you know that your system’s dynamics is usually evolving around the origin, it may be easy to show that for points that have a certain distance to the origin, this distance is decreasing over time.
answered Nov 18 at 14:02
Wrzlprmft
3,04111233
edited Nov 18 at 21:36
answered Nov 18 at 14:02
Wrzlprmft
3,04111233
answered Nov 18 at 14:02
Wrzlprmft
3,04111233
answered Nov 18 at 14:02
Wrzlprmft
3,04111233
3,04111233
Take $dot{x}=x(1-x^2)$ and $A=[-1,1]$. Then $A$ satisfies the continuous-time analogs of the definitions of global attractor given in the OP's references, but it contains two asymptotically stable equilibria, namely $-1$ and $1$.
– user539887
Nov 18 at 16:10
@user539887: How is $A$ an attractor? It is not invariant under time-evolution. With the exception of $0$ all points in $A$ converge towards the stable equlibria ($-1$ and $1$) under time-evolution.
– Wrzlprmft
Nov 18 at 16:23
Yes, it is invariant.
– user539887
Nov 18 at 16:25
@user539887: And how do you reconcile this with all points converging towards $-1$ and $1$? Just so we are on the same page, I consider a set $A$ invariant under time evolution if $ϕ(A,t) = A$ for all $t≥0$, where $ϕ$ is the time-evolution operator.
– Wrzlprmft
Nov 18 at 16:40
Take $x=0$ or $pm1$. As they are equilibria, $phi(x,t)=x$ for all $tinmathbb{R}$. Take $xin(-1,0)$, Then we have $phi(x,t)in(-1,0)$ for all $tinmathbb{R}$. Take $xin(0,1)$, Then we have $phi(x,t)in(0,1)$ for all $tinmathbb{R}$, too. Consequently, $phi(A,t)subset A$ for all $tinmathbb{R}$, from which it follows, by "algebraic" properties of the flow, that $phi(A,t){color{red}=}A$ for all $tinmathbb{R}$, in particular for all $tge0$.
– user539887
Nov 18 at 16:50
|
show 2 more comments
Take $dot{x}=x(1-x^2)$ and $A=[-1,1]$. Then $A$ satisfies the continuous-time analogs of the definitions of global attractor given in the OP's references, but it contains two asymptotically stable equilibria, namely $-1$ and $1$.
– user539887
Nov 18 at 16:10
@user539887: How is $A$ an attractor? It is not invariant under time-evolution. With the exception of $0$ all points in $A$ converge towards the stable equlibria ($-1$ and $1$) under time-evolution.
– Wrzlprmft
Nov 18 at 16:23
Yes, it is invariant.
– user539887
Nov 18 at 16:25
@user539887: And how do you reconcile this with all points converging towards $-1$ and $1$? Just so we are on the same page, I consider a set $A$ invariant under time evolution if $ϕ(A,t) = A$ for all $t≥0$, where $ϕ$ is the time-evolution operator.
– Wrzlprmft
Nov 18 at 16:40
Take $x=0$ or $pm1$. As they are equilibria, $phi(x,t)=x$ for all $tinmathbb{R}$. Take $xin(-1,0)$, Then we have $phi(x,t)in(-1,0)$ for all $tinmathbb{R}$. Take $xin(0,1)$, Then we have $phi(x,t)in(0,1)$ for all $tinmathbb{R}$, too. Consequently, $phi(A,t)subset A$ for all $tinmathbb{R}$, from which it follows, by "algebraic" properties of the flow, that $phi(A,t){color{red}=}A$ for all $tinmathbb{R}$, in particular for all $tge0$.
– user539887
Nov 18 at 16:50
Take $dot{x}=x(1-x^2)$ and $A=[-1,1]$. Then $A$ satisfies the continuous-time analogs of the definitions of global attractor given in the OP's references, but it contains two asymptotically stable equilibria, namely $-1$ and $1$.
– user539887
Nov 18 at 16:10
Take $dot{x}=x(1-x^2)$ and $A=[-1,1]$. Then $A$ satisfies the continuous-time analogs of the definitions of global attractor given in the OP's references, but it contains two asymptotically stable equilibria, namely $-1$ and $1$.
– user539887
Nov 18 at 16:10
@user539887: How is $A$ an attractor? It is not invariant under time-evolution. With the exception of $0$ all points in $A$ converge towards the stable equlibria ($-1$ and $1$) under time-evolution.
– Wrzlprmft
Nov 18 at 16:23
@user539887: How is $A$ an attractor? It is not invariant under time-evolution. With the exception of $0$ all points in $A$ converge towards the stable equlibria ($-1$ and $1$) under time-evolution.
– Wrzlprmft
Nov 18 at 16:23
Yes, it is invariant.
– user539887
Nov 18 at 16:25
Yes, it is invariant.
– user539887
Nov 18 at 16:25
@user539887: And how do you reconcile this with all points converging towards $-1$ and $1$? Just so we are on the same page, I consider a set $A$ invariant under time evolution if $ϕ(A,t) = A$ for all $t≥0$, where $ϕ$ is the time-evolution operator.
– Wrzlprmft
Nov 18 at 16:40
@user539887: And how do you reconcile this with all points converging towards $-1$ and $1$? Just so we are on the same page, I consider a set $A$ invariant under time evolution if $ϕ(A,t) = A$ for all $t≥0$, where $ϕ$ is the time-evolution operator.
– Wrzlprmft
Nov 18 at 16:40
Take $x=0$ or $pm1$. As they are equilibria, $phi(x,t)=x$ for all $tinmathbb{R}$. Take $xin(-1,0)$, Then we have $phi(x,t)in(-1,0)$ for all $tinmathbb{R}$. Take $xin(0,1)$, Then we have $phi(x,t)in(0,1)$ for all $tinmathbb{R}$, too. Consequently, $phi(A,t)subset A$ for all $tinmathbb{R}$, from which it follows, by "algebraic" properties of the flow, that $phi(A,t){color{red}=}A$ for all $tinmathbb{R}$, in particular for all $tge0$.
– user539887
Nov 18 at 16:50
Take $x=0$ or $pm1$. As they are equilibria, $phi(x,t)=x$ for all $tinmathbb{R}$. Take $xin(-1,0)$, Then we have $phi(x,t)in(-1,0)$ for all $tinmathbb{R}$. Take $xin(0,1)$, Then we have $phi(x,t)in(0,1)$ for all $tinmathbb{R}$, too. Consequently, $phi(A,t)subset A$ for all $tinmathbb{R}$, from which it follows, by "algebraic" properties of the flow, that $phi(A,t){color{red}=}A$ for all $tinmathbb{R}$, in particular for all $tge0$.
– user539887
Nov 18 at 16:50
|
show 2 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3003395%2fglobal-attractor-existence%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
You might be interested in "Lyapunov stability theory", for a formal treatment the classic paper by Kalman and Bertram is one of the best: fluidsengineering.asmedigitalcollection.asme.org/…
– WalterJ
Nov 18 at 15:40
Thank you! I'll study your paper!
– asd11
Nov 18 at 16:34
Please see the edit to my answer.
– Wrzlprmft
Nov 18 at 21:36