Stability VS orbital stability in ODEs
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Let $dot x=f(x)$ and ODE.
We say that $x$ is a stable solution if $forall varepsilon >0$, $exists delta >0$ s.t for all solution $y(t)$ s.t. $|x(t_0)-y(t_0)|<delta implies |x(t)-y(t)|<varepsilon .$
Let $Phi_t$ denote the flow of the ODE, i.e. $x(t)=Phi^{t,t_0}(x_0)$ for $x(t)$ is a solution.
We say that $x$ is orbitaly stable if $forall varepsilon >0$, $exists delta >0$ s.t. for all $xi$ s.t. $|x(t_0)-Phi^{t_0,0}(xi)|<delta implies text{dist}(mathcal O^+(x_0,t_0),Phi^{t,t_0}(xi))<varepsilon $ where $mathcal O^+(x_0,t_0)={Phi^{t,t_0}(x_0)mid tgeq t_0}$.
I see what is an stable solution, but I can't see what is an orbitaly stable solution. I don't really understand the example. Could someone explain ?
real-analysis ordinary-differential-equations
$endgroup$
add a comment |
$begingroup$
Let $dot x=f(x)$ and ODE.
We say that $x$ is a stable solution if $forall varepsilon >0$, $exists delta >0$ s.t for all solution $y(t)$ s.t. $|x(t_0)-y(t_0)|<delta implies |x(t)-y(t)|<varepsilon .$
Let $Phi_t$ denote the flow of the ODE, i.e. $x(t)=Phi^{t,t_0}(x_0)$ for $x(t)$ is a solution.
We say that $x$ is orbitaly stable if $forall varepsilon >0$, $exists delta >0$ s.t. for all $xi$ s.t. $|x(t_0)-Phi^{t_0,0}(xi)|<delta implies text{dist}(mathcal O^+(x_0,t_0),Phi^{t,t_0}(xi))<varepsilon $ where $mathcal O^+(x_0,t_0)={Phi^{t,t_0}(x_0)mid tgeq t_0}$.
I see what is an stable solution, but I can't see what is an orbitaly stable solution. I don't really understand the example. Could someone explain ?
real-analysis ordinary-differential-equations
$endgroup$
add a comment |
$begingroup$
Let $dot x=f(x)$ and ODE.
We say that $x$ is a stable solution if $forall varepsilon >0$, $exists delta >0$ s.t for all solution $y(t)$ s.t. $|x(t_0)-y(t_0)|<delta implies |x(t)-y(t)|<varepsilon .$
Let $Phi_t$ denote the flow of the ODE, i.e. $x(t)=Phi^{t,t_0}(x_0)$ for $x(t)$ is a solution.
We say that $x$ is orbitaly stable if $forall varepsilon >0$, $exists delta >0$ s.t. for all $xi$ s.t. $|x(t_0)-Phi^{t_0,0}(xi)|<delta implies text{dist}(mathcal O^+(x_0,t_0),Phi^{t,t_0}(xi))<varepsilon $ where $mathcal O^+(x_0,t_0)={Phi^{t,t_0}(x_0)mid tgeq t_0}$.
I see what is an stable solution, but I can't see what is an orbitaly stable solution. I don't really understand the example. Could someone explain ?
real-analysis ordinary-differential-equations
$endgroup$
Let $dot x=f(x)$ and ODE.
We say that $x$ is a stable solution if $forall varepsilon >0$, $exists delta >0$ s.t for all solution $y(t)$ s.t. $|x(t_0)-y(t_0)|<delta implies |x(t)-y(t)|<varepsilon .$
Let $Phi_t$ denote the flow of the ODE, i.e. $x(t)=Phi^{t,t_0}(x_0)$ for $x(t)$ is a solution.
We say that $x$ is orbitaly stable if $forall varepsilon >0$, $exists delta >0$ s.t. for all $xi$ s.t. $|x(t_0)-Phi^{t_0,0}(xi)|<delta implies text{dist}(mathcal O^+(x_0,t_0),Phi^{t,t_0}(xi))<varepsilon $ where $mathcal O^+(x_0,t_0)={Phi^{t,t_0}(x_0)mid tgeq t_0}$.
I see what is an stable solution, but I can't see what is an orbitaly stable solution. I don't really understand the example. Could someone explain ?
real-analysis ordinary-differential-equations
real-analysis ordinary-differential-equations
asked Dec 16 '18 at 13:44
NewMathNewMath
4059
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2 Answers
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$begingroup$
To understand the difference you need first carefully understand the difference between the integral curves, which are the graph of solutions in $mathbb Rtimes mathbb R^n$ and orbits, which are the images of the solutions in $mathbb R^n$ parametrized by the time $t$.
For instance, take
$$
dot x=y,\
dot y=-x.
$$
Clearly your solutions will be
$$
x(t)=C cos t,quad y(t)= Csin t
$$
and you should sketch the corresponding graphs in $mathbb Rtimes mathbb R^2=mathbb R^3$ and convince yourself that all the solutions are stable.
Now, the orbits are given by
$$
x^2+y^2=C^2,
$$
which are circles, and it should be clear that each orbit starting close to another one stays close for any $t$, hence they are also orbitally stable.
Now take
$$
dot x=y,\
dot y=-sin x.
$$
The orbits will look almost the same (and since you are discussing orbital stability you should be able to sketch a phase portrait of this seen-in-any-texbook system), and therefore the corresponding solutions are orbitally stable. It is also well known (but not elementary to show) that each solution has its own period (depending on the initial conditions, this is a general property of nonlinear oscillators). This means that if you take two sufficiently close initial conditions the corresponding integral curves, given enough time, will part sufficiently far from each other and hence are not stable.
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Thank you for your answer. At the end, can we say that stability $implies $ orbit stability, but the converse is not true ?
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– NewMath
Dec 18 '18 at 9:52
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Yes, and this is the reason to introduce new stability concept since the (Lyapunov) stability is quite demanding on the solutions.
$endgroup$
– Artem
Dec 18 '18 at 13:43
add a comment |
$begingroup$
Removing the orbit from the definition for orbitally stable gives
We say that x is orbitaly stable if $∀ε>0$, $∃δ>0$ s.t for any solution $y(t)$ with $∥x(t_0)−y(t_0)∥<δ$ it holds that for all $t>t_0$ there is some $s>t_0$ so that $∥x(s)−y(t)∥<ε$.
The distance in the second definition is to the curve, the set of points on the trajectory of $x$, not to a specific point on the path as in the first definition.
Take as example $dot r = r(1-r^2)$, $dot theta=r$ in polar coordinates, that is, with $r=sqrt{x_1^2+x_2^2}$
$$dot x_1 = -rx_2+(1-r^2)x_1,\ dot x_2=~~~rx_1+(1-r^2)x_2,$$ with the closed loop solution $x(t)=(x_1(t),x_2(t))=(cos t,sin t)$. The general solution is given by
begin{align}
%frac{d}{dt}r(t)^{-2}&=2(1-r(t)^{-2})implies (r(t)^{-2}-1)=e^{-2t}(r_0^{-2}-1)\
r(t)&=frac{r_0}{sqrt{r_0^2+e^{-2t}(1-r_0^2)}}
=frac{r_0e^t}{sqrt{r_0^2(e^{2t}-1)+1}}
end{align}
and
$$
θ(t)
%=int_0^tfrac{d(r_0e^t)}{sqrt{(r_0^2e^t)^2-r_0^2+1}}
=t+lnleft(r_0+sqrt{e^{-2t}+r_0^2(1-e^{-2t})}right)-ln(r_0+1)
$$
While the orbit starting at $r(0)=r_0=x_{1,0}$, $θ(0)=0$ (with $x_{2,0}=0$) converges visually to the limit cycle, the angle evolution is initially very different from the unit speed on the limit cycle. Asymptotically $θ(t)sim t+ln(2r_0)-ln(r_0+1)$, so that the angle difference stabilizes, but does not go to zero.
$endgroup$
$begingroup$
What does it mean graphically ? (because as written, I don't really see how it look). Do you know where I can find an example ?
$endgroup$
– NewMath
Dec 16 '18 at 14:23
1
$begingroup$
Suppose $x$ is a closed loop, a stable limit cycle, like in $dot r = r(1-r^2)$, $dot theta=r$ the solution $(x,y)=(cos t,sin t)$. Then any other solution converging to it will have a slightly different angular speed, thus while it comes closer to the orbit, the points of same times run away from each other.
$endgroup$
– LutzL
Dec 16 '18 at 14:43
$begingroup$
You mix to much thing, sorry I'm not confortable at all with all this. You say "$x$ a closed loop", then you introduce $(x,y)=(cos t,sin t)$, is it the same $x$ as before ? And to finish, what is the relation between $dot r=r(1-r^2)$, $dot theta =r$ and $x$, and $(cos t,sin t)$ ?
$endgroup$
– NewMath
Dec 16 '18 at 14:56
$begingroup$
Yes, should have been $x(t)=(x_1(t),x_2(t))$. $(r,θ)$ are the polar coordinates of $x$, or just consider $θ$ modulo $2pi$.
$endgroup$
– LutzL
Dec 16 '18 at 15:23
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
To understand the difference you need first carefully understand the difference between the integral curves, which are the graph of solutions in $mathbb Rtimes mathbb R^n$ and orbits, which are the images of the solutions in $mathbb R^n$ parametrized by the time $t$.
For instance, take
$$
dot x=y,\
dot y=-x.
$$
Clearly your solutions will be
$$
x(t)=C cos t,quad y(t)= Csin t
$$
and you should sketch the corresponding graphs in $mathbb Rtimes mathbb R^2=mathbb R^3$ and convince yourself that all the solutions are stable.
Now, the orbits are given by
$$
x^2+y^2=C^2,
$$
which are circles, and it should be clear that each orbit starting close to another one stays close for any $t$, hence they are also orbitally stable.
Now take
$$
dot x=y,\
dot y=-sin x.
$$
The orbits will look almost the same (and since you are discussing orbital stability you should be able to sketch a phase portrait of this seen-in-any-texbook system), and therefore the corresponding solutions are orbitally stable. It is also well known (but not elementary to show) that each solution has its own period (depending on the initial conditions, this is a general property of nonlinear oscillators). This means that if you take two sufficiently close initial conditions the corresponding integral curves, given enough time, will part sufficiently far from each other and hence are not stable.
$endgroup$
$begingroup$
Thank you for your answer. At the end, can we say that stability $implies $ orbit stability, but the converse is not true ?
$endgroup$
– NewMath
Dec 18 '18 at 9:52
$begingroup$
Yes, and this is the reason to introduce new stability concept since the (Lyapunov) stability is quite demanding on the solutions.
$endgroup$
– Artem
Dec 18 '18 at 13:43
add a comment |
$begingroup$
To understand the difference you need first carefully understand the difference between the integral curves, which are the graph of solutions in $mathbb Rtimes mathbb R^n$ and orbits, which are the images of the solutions in $mathbb R^n$ parametrized by the time $t$.
For instance, take
$$
dot x=y,\
dot y=-x.
$$
Clearly your solutions will be
$$
x(t)=C cos t,quad y(t)= Csin t
$$
and you should sketch the corresponding graphs in $mathbb Rtimes mathbb R^2=mathbb R^3$ and convince yourself that all the solutions are stable.
Now, the orbits are given by
$$
x^2+y^2=C^2,
$$
which are circles, and it should be clear that each orbit starting close to another one stays close for any $t$, hence they are also orbitally stable.
Now take
$$
dot x=y,\
dot y=-sin x.
$$
The orbits will look almost the same (and since you are discussing orbital stability you should be able to sketch a phase portrait of this seen-in-any-texbook system), and therefore the corresponding solutions are orbitally stable. It is also well known (but not elementary to show) that each solution has its own period (depending on the initial conditions, this is a general property of nonlinear oscillators). This means that if you take two sufficiently close initial conditions the corresponding integral curves, given enough time, will part sufficiently far from each other and hence are not stable.
$endgroup$
$begingroup$
Thank you for your answer. At the end, can we say that stability $implies $ orbit stability, but the converse is not true ?
$endgroup$
– NewMath
Dec 18 '18 at 9:52
$begingroup$
Yes, and this is the reason to introduce new stability concept since the (Lyapunov) stability is quite demanding on the solutions.
$endgroup$
– Artem
Dec 18 '18 at 13:43
add a comment |
$begingroup$
To understand the difference you need first carefully understand the difference between the integral curves, which are the graph of solutions in $mathbb Rtimes mathbb R^n$ and orbits, which are the images of the solutions in $mathbb R^n$ parametrized by the time $t$.
For instance, take
$$
dot x=y,\
dot y=-x.
$$
Clearly your solutions will be
$$
x(t)=C cos t,quad y(t)= Csin t
$$
and you should sketch the corresponding graphs in $mathbb Rtimes mathbb R^2=mathbb R^3$ and convince yourself that all the solutions are stable.
Now, the orbits are given by
$$
x^2+y^2=C^2,
$$
which are circles, and it should be clear that each orbit starting close to another one stays close for any $t$, hence they are also orbitally stable.
Now take
$$
dot x=y,\
dot y=-sin x.
$$
The orbits will look almost the same (and since you are discussing orbital stability you should be able to sketch a phase portrait of this seen-in-any-texbook system), and therefore the corresponding solutions are orbitally stable. It is also well known (but not elementary to show) that each solution has its own period (depending on the initial conditions, this is a general property of nonlinear oscillators). This means that if you take two sufficiently close initial conditions the corresponding integral curves, given enough time, will part sufficiently far from each other and hence are not stable.
$endgroup$
To understand the difference you need first carefully understand the difference between the integral curves, which are the graph of solutions in $mathbb Rtimes mathbb R^n$ and orbits, which are the images of the solutions in $mathbb R^n$ parametrized by the time $t$.
For instance, take
$$
dot x=y,\
dot y=-x.
$$
Clearly your solutions will be
$$
x(t)=C cos t,quad y(t)= Csin t
$$
and you should sketch the corresponding graphs in $mathbb Rtimes mathbb R^2=mathbb R^3$ and convince yourself that all the solutions are stable.
Now, the orbits are given by
$$
x^2+y^2=C^2,
$$
which are circles, and it should be clear that each orbit starting close to another one stays close for any $t$, hence they are also orbitally stable.
Now take
$$
dot x=y,\
dot y=-sin x.
$$
The orbits will look almost the same (and since you are discussing orbital stability you should be able to sketch a phase portrait of this seen-in-any-texbook system), and therefore the corresponding solutions are orbitally stable. It is also well known (but not elementary to show) that each solution has its own period (depending on the initial conditions, this is a general property of nonlinear oscillators). This means that if you take two sufficiently close initial conditions the corresponding integral curves, given enough time, will part sufficiently far from each other and hence are not stable.
answered Dec 18 '18 at 1:35
ArtemArtem
11.5k32245
11.5k32245
$begingroup$
Thank you for your answer. At the end, can we say that stability $implies $ orbit stability, but the converse is not true ?
$endgroup$
– NewMath
Dec 18 '18 at 9:52
$begingroup$
Yes, and this is the reason to introduce new stability concept since the (Lyapunov) stability is quite demanding on the solutions.
$endgroup$
– Artem
Dec 18 '18 at 13:43
add a comment |
$begingroup$
Thank you for your answer. At the end, can we say that stability $implies $ orbit stability, but the converse is not true ?
$endgroup$
– NewMath
Dec 18 '18 at 9:52
$begingroup$
Yes, and this is the reason to introduce new stability concept since the (Lyapunov) stability is quite demanding on the solutions.
$endgroup$
– Artem
Dec 18 '18 at 13:43
$begingroup$
Thank you for your answer. At the end, can we say that stability $implies $ orbit stability, but the converse is not true ?
$endgroup$
– NewMath
Dec 18 '18 at 9:52
$begingroup$
Thank you for your answer. At the end, can we say that stability $implies $ orbit stability, but the converse is not true ?
$endgroup$
– NewMath
Dec 18 '18 at 9:52
$begingroup$
Yes, and this is the reason to introduce new stability concept since the (Lyapunov) stability is quite demanding on the solutions.
$endgroup$
– Artem
Dec 18 '18 at 13:43
$begingroup$
Yes, and this is the reason to introduce new stability concept since the (Lyapunov) stability is quite demanding on the solutions.
$endgroup$
– Artem
Dec 18 '18 at 13:43
add a comment |
$begingroup$
Removing the orbit from the definition for orbitally stable gives
We say that x is orbitaly stable if $∀ε>0$, $∃δ>0$ s.t for any solution $y(t)$ with $∥x(t_0)−y(t_0)∥<δ$ it holds that for all $t>t_0$ there is some $s>t_0$ so that $∥x(s)−y(t)∥<ε$.
The distance in the second definition is to the curve, the set of points on the trajectory of $x$, not to a specific point on the path as in the first definition.
Take as example $dot r = r(1-r^2)$, $dot theta=r$ in polar coordinates, that is, with $r=sqrt{x_1^2+x_2^2}$
$$dot x_1 = -rx_2+(1-r^2)x_1,\ dot x_2=~~~rx_1+(1-r^2)x_2,$$ with the closed loop solution $x(t)=(x_1(t),x_2(t))=(cos t,sin t)$. The general solution is given by
begin{align}
%frac{d}{dt}r(t)^{-2}&=2(1-r(t)^{-2})implies (r(t)^{-2}-1)=e^{-2t}(r_0^{-2}-1)\
r(t)&=frac{r_0}{sqrt{r_0^2+e^{-2t}(1-r_0^2)}}
=frac{r_0e^t}{sqrt{r_0^2(e^{2t}-1)+1}}
end{align}
and
$$
θ(t)
%=int_0^tfrac{d(r_0e^t)}{sqrt{(r_0^2e^t)^2-r_0^2+1}}
=t+lnleft(r_0+sqrt{e^{-2t}+r_0^2(1-e^{-2t})}right)-ln(r_0+1)
$$
While the orbit starting at $r(0)=r_0=x_{1,0}$, $θ(0)=0$ (with $x_{2,0}=0$) converges visually to the limit cycle, the angle evolution is initially very different from the unit speed on the limit cycle. Asymptotically $θ(t)sim t+ln(2r_0)-ln(r_0+1)$, so that the angle difference stabilizes, but does not go to zero.
$endgroup$
$begingroup$
What does it mean graphically ? (because as written, I don't really see how it look). Do you know where I can find an example ?
$endgroup$
– NewMath
Dec 16 '18 at 14:23
1
$begingroup$
Suppose $x$ is a closed loop, a stable limit cycle, like in $dot r = r(1-r^2)$, $dot theta=r$ the solution $(x,y)=(cos t,sin t)$. Then any other solution converging to it will have a slightly different angular speed, thus while it comes closer to the orbit, the points of same times run away from each other.
$endgroup$
– LutzL
Dec 16 '18 at 14:43
$begingroup$
You mix to much thing, sorry I'm not confortable at all with all this. You say "$x$ a closed loop", then you introduce $(x,y)=(cos t,sin t)$, is it the same $x$ as before ? And to finish, what is the relation between $dot r=r(1-r^2)$, $dot theta =r$ and $x$, and $(cos t,sin t)$ ?
$endgroup$
– NewMath
Dec 16 '18 at 14:56
$begingroup$
Yes, should have been $x(t)=(x_1(t),x_2(t))$. $(r,θ)$ are the polar coordinates of $x$, or just consider $θ$ modulo $2pi$.
$endgroup$
– LutzL
Dec 16 '18 at 15:23
add a comment |
$begingroup$
Removing the orbit from the definition for orbitally stable gives
We say that x is orbitaly stable if $∀ε>0$, $∃δ>0$ s.t for any solution $y(t)$ with $∥x(t_0)−y(t_0)∥<δ$ it holds that for all $t>t_0$ there is some $s>t_0$ so that $∥x(s)−y(t)∥<ε$.
The distance in the second definition is to the curve, the set of points on the trajectory of $x$, not to a specific point on the path as in the first definition.
Take as example $dot r = r(1-r^2)$, $dot theta=r$ in polar coordinates, that is, with $r=sqrt{x_1^2+x_2^2}$
$$dot x_1 = -rx_2+(1-r^2)x_1,\ dot x_2=~~~rx_1+(1-r^2)x_2,$$ with the closed loop solution $x(t)=(x_1(t),x_2(t))=(cos t,sin t)$. The general solution is given by
begin{align}
%frac{d}{dt}r(t)^{-2}&=2(1-r(t)^{-2})implies (r(t)^{-2}-1)=e^{-2t}(r_0^{-2}-1)\
r(t)&=frac{r_0}{sqrt{r_0^2+e^{-2t}(1-r_0^2)}}
=frac{r_0e^t}{sqrt{r_0^2(e^{2t}-1)+1}}
end{align}
and
$$
θ(t)
%=int_0^tfrac{d(r_0e^t)}{sqrt{(r_0^2e^t)^2-r_0^2+1}}
=t+lnleft(r_0+sqrt{e^{-2t}+r_0^2(1-e^{-2t})}right)-ln(r_0+1)
$$
While the orbit starting at $r(0)=r_0=x_{1,0}$, $θ(0)=0$ (with $x_{2,0}=0$) converges visually to the limit cycle, the angle evolution is initially very different from the unit speed on the limit cycle. Asymptotically $θ(t)sim t+ln(2r_0)-ln(r_0+1)$, so that the angle difference stabilizes, but does not go to zero.
$endgroup$
$begingroup$
What does it mean graphically ? (because as written, I don't really see how it look). Do you know where I can find an example ?
$endgroup$
– NewMath
Dec 16 '18 at 14:23
1
$begingroup$
Suppose $x$ is a closed loop, a stable limit cycle, like in $dot r = r(1-r^2)$, $dot theta=r$ the solution $(x,y)=(cos t,sin t)$. Then any other solution converging to it will have a slightly different angular speed, thus while it comes closer to the orbit, the points of same times run away from each other.
$endgroup$
– LutzL
Dec 16 '18 at 14:43
$begingroup$
You mix to much thing, sorry I'm not confortable at all with all this. You say "$x$ a closed loop", then you introduce $(x,y)=(cos t,sin t)$, is it the same $x$ as before ? And to finish, what is the relation between $dot r=r(1-r^2)$, $dot theta =r$ and $x$, and $(cos t,sin t)$ ?
$endgroup$
– NewMath
Dec 16 '18 at 14:56
$begingroup$
Yes, should have been $x(t)=(x_1(t),x_2(t))$. $(r,θ)$ are the polar coordinates of $x$, or just consider $θ$ modulo $2pi$.
$endgroup$
– LutzL
Dec 16 '18 at 15:23
add a comment |
$begingroup$
Removing the orbit from the definition for orbitally stable gives
We say that x is orbitaly stable if $∀ε>0$, $∃δ>0$ s.t for any solution $y(t)$ with $∥x(t_0)−y(t_0)∥<δ$ it holds that for all $t>t_0$ there is some $s>t_0$ so that $∥x(s)−y(t)∥<ε$.
The distance in the second definition is to the curve, the set of points on the trajectory of $x$, not to a specific point on the path as in the first definition.
Take as example $dot r = r(1-r^2)$, $dot theta=r$ in polar coordinates, that is, with $r=sqrt{x_1^2+x_2^2}$
$$dot x_1 = -rx_2+(1-r^2)x_1,\ dot x_2=~~~rx_1+(1-r^2)x_2,$$ with the closed loop solution $x(t)=(x_1(t),x_2(t))=(cos t,sin t)$. The general solution is given by
begin{align}
%frac{d}{dt}r(t)^{-2}&=2(1-r(t)^{-2})implies (r(t)^{-2}-1)=e^{-2t}(r_0^{-2}-1)\
r(t)&=frac{r_0}{sqrt{r_0^2+e^{-2t}(1-r_0^2)}}
=frac{r_0e^t}{sqrt{r_0^2(e^{2t}-1)+1}}
end{align}
and
$$
θ(t)
%=int_0^tfrac{d(r_0e^t)}{sqrt{(r_0^2e^t)^2-r_0^2+1}}
=t+lnleft(r_0+sqrt{e^{-2t}+r_0^2(1-e^{-2t})}right)-ln(r_0+1)
$$
While the orbit starting at $r(0)=r_0=x_{1,0}$, $θ(0)=0$ (with $x_{2,0}=0$) converges visually to the limit cycle, the angle evolution is initially very different from the unit speed on the limit cycle. Asymptotically $θ(t)sim t+ln(2r_0)-ln(r_0+1)$, so that the angle difference stabilizes, but does not go to zero.
$endgroup$
Removing the orbit from the definition for orbitally stable gives
We say that x is orbitaly stable if $∀ε>0$, $∃δ>0$ s.t for any solution $y(t)$ with $∥x(t_0)−y(t_0)∥<δ$ it holds that for all $t>t_0$ there is some $s>t_0$ so that $∥x(s)−y(t)∥<ε$.
The distance in the second definition is to the curve, the set of points on the trajectory of $x$, not to a specific point on the path as in the first definition.
Take as example $dot r = r(1-r^2)$, $dot theta=r$ in polar coordinates, that is, with $r=sqrt{x_1^2+x_2^2}$
$$dot x_1 = -rx_2+(1-r^2)x_1,\ dot x_2=~~~rx_1+(1-r^2)x_2,$$ with the closed loop solution $x(t)=(x_1(t),x_2(t))=(cos t,sin t)$. The general solution is given by
begin{align}
%frac{d}{dt}r(t)^{-2}&=2(1-r(t)^{-2})implies (r(t)^{-2}-1)=e^{-2t}(r_0^{-2}-1)\
r(t)&=frac{r_0}{sqrt{r_0^2+e^{-2t}(1-r_0^2)}}
=frac{r_0e^t}{sqrt{r_0^2(e^{2t}-1)+1}}
end{align}
and
$$
θ(t)
%=int_0^tfrac{d(r_0e^t)}{sqrt{(r_0^2e^t)^2-r_0^2+1}}
=t+lnleft(r_0+sqrt{e^{-2t}+r_0^2(1-e^{-2t})}right)-ln(r_0+1)
$$
While the orbit starting at $r(0)=r_0=x_{1,0}$, $θ(0)=0$ (with $x_{2,0}=0$) converges visually to the limit cycle, the angle evolution is initially very different from the unit speed on the limit cycle. Asymptotically $θ(t)sim t+ln(2r_0)-ln(r_0+1)$, so that the angle difference stabilizes, but does not go to zero.
edited Dec 16 '18 at 15:52
answered Dec 16 '18 at 14:18
LutzLLutzL
58.8k42056
58.8k42056
$begingroup$
What does it mean graphically ? (because as written, I don't really see how it look). Do you know where I can find an example ?
$endgroup$
– NewMath
Dec 16 '18 at 14:23
1
$begingroup$
Suppose $x$ is a closed loop, a stable limit cycle, like in $dot r = r(1-r^2)$, $dot theta=r$ the solution $(x,y)=(cos t,sin t)$. Then any other solution converging to it will have a slightly different angular speed, thus while it comes closer to the orbit, the points of same times run away from each other.
$endgroup$
– LutzL
Dec 16 '18 at 14:43
$begingroup$
You mix to much thing, sorry I'm not confortable at all with all this. You say "$x$ a closed loop", then you introduce $(x,y)=(cos t,sin t)$, is it the same $x$ as before ? And to finish, what is the relation between $dot r=r(1-r^2)$, $dot theta =r$ and $x$, and $(cos t,sin t)$ ?
$endgroup$
– NewMath
Dec 16 '18 at 14:56
$begingroup$
Yes, should have been $x(t)=(x_1(t),x_2(t))$. $(r,θ)$ are the polar coordinates of $x$, or just consider $θ$ modulo $2pi$.
$endgroup$
– LutzL
Dec 16 '18 at 15:23
add a comment |
$begingroup$
What does it mean graphically ? (because as written, I don't really see how it look). Do you know where I can find an example ?
$endgroup$
– NewMath
Dec 16 '18 at 14:23
1
$begingroup$
Suppose $x$ is a closed loop, a stable limit cycle, like in $dot r = r(1-r^2)$, $dot theta=r$ the solution $(x,y)=(cos t,sin t)$. Then any other solution converging to it will have a slightly different angular speed, thus while it comes closer to the orbit, the points of same times run away from each other.
$endgroup$
– LutzL
Dec 16 '18 at 14:43
$begingroup$
You mix to much thing, sorry I'm not confortable at all with all this. You say "$x$ a closed loop", then you introduce $(x,y)=(cos t,sin t)$, is it the same $x$ as before ? And to finish, what is the relation between $dot r=r(1-r^2)$, $dot theta =r$ and $x$, and $(cos t,sin t)$ ?
$endgroup$
– NewMath
Dec 16 '18 at 14:56
$begingroup$
Yes, should have been $x(t)=(x_1(t),x_2(t))$. $(r,θ)$ are the polar coordinates of $x$, or just consider $θ$ modulo $2pi$.
$endgroup$
– LutzL
Dec 16 '18 at 15:23
$begingroup$
What does it mean graphically ? (because as written, I don't really see how it look). Do you know where I can find an example ?
$endgroup$
– NewMath
Dec 16 '18 at 14:23
$begingroup$
What does it mean graphically ? (because as written, I don't really see how it look). Do you know where I can find an example ?
$endgroup$
– NewMath
Dec 16 '18 at 14:23
1
1
$begingroup$
Suppose $x$ is a closed loop, a stable limit cycle, like in $dot r = r(1-r^2)$, $dot theta=r$ the solution $(x,y)=(cos t,sin t)$. Then any other solution converging to it will have a slightly different angular speed, thus while it comes closer to the orbit, the points of same times run away from each other.
$endgroup$
– LutzL
Dec 16 '18 at 14:43
$begingroup$
Suppose $x$ is a closed loop, a stable limit cycle, like in $dot r = r(1-r^2)$, $dot theta=r$ the solution $(x,y)=(cos t,sin t)$. Then any other solution converging to it will have a slightly different angular speed, thus while it comes closer to the orbit, the points of same times run away from each other.
$endgroup$
– LutzL
Dec 16 '18 at 14:43
$begingroup$
You mix to much thing, sorry I'm not confortable at all with all this. You say "$x$ a closed loop", then you introduce $(x,y)=(cos t,sin t)$, is it the same $x$ as before ? And to finish, what is the relation between $dot r=r(1-r^2)$, $dot theta =r$ and $x$, and $(cos t,sin t)$ ?
$endgroup$
– NewMath
Dec 16 '18 at 14:56
$begingroup$
You mix to much thing, sorry I'm not confortable at all with all this. You say "$x$ a closed loop", then you introduce $(x,y)=(cos t,sin t)$, is it the same $x$ as before ? And to finish, what is the relation between $dot r=r(1-r^2)$, $dot theta =r$ and $x$, and $(cos t,sin t)$ ?
$endgroup$
– NewMath
Dec 16 '18 at 14:56
$begingroup$
Yes, should have been $x(t)=(x_1(t),x_2(t))$. $(r,θ)$ are the polar coordinates of $x$, or just consider $θ$ modulo $2pi$.
$endgroup$
– LutzL
Dec 16 '18 at 15:23
$begingroup$
Yes, should have been $x(t)=(x_1(t),x_2(t))$. $(r,θ)$ are the polar coordinates of $x$, or just consider $θ$ modulo $2pi$.
$endgroup$
– LutzL
Dec 16 '18 at 15:23
add a comment |
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