Does the existence of a superstable fixed point imply a $-infty$ Lyapunov exponent?
$begingroup$
This is the Lyapunov exponent as a function of $r$ for the logistic map ($x_{n+1}=f(x_n)=r(x_{n}-x_{n}^2)$)
The big dips are centered around points where $f'(x)=0$ for some $x$ in the trajectory used to calculate the exponent $lambda$, and therefore for those values $lambda=-infty$. These are precisely the $r$ values where there exists a superstable fixed point/cycle, which makes sense as $f'(x^*)=0$ is a necessary condition for superstability (where $x^*$ is some fixed point).
This naturally led me to think that perhaps the existence of a superstable fixed point/cycle is a sufficient condition for any 1D map to have $lambda = -infty$
However, this is now a Lyapunov graph for $x_{n+1}=g(x_n)=r(x_{n}^2-x_{n}^3)$
For which $0$ is a superstable fixed point for all values of $r$. For $r<4$ we indeed see that $lambda = -infty$ but for some values of $r$ there is a real $lambda$, for some even positive.
So what's the difference? Are these superstable fixed points different somehow? Also, the bluriness of the line in the second image results from most of the $r$ values in between still being $-infty$ - why is this so?
Notes:
I used the standard algorithm to generate the $lambda$ values. Namely, for each value of $r$ I draw uniformly some $x_0in [0, 1]$, and calculate $lambda = left(frac{1}{n}sum_{i=1}^{n}lnleft|f'left(x_{i}right)right|right)$ for $n=10^4$
The maximal $r$ value, $6.75$, was chosen to ensure the interval $[0,1]$ maps onto itself, as in the logistic map.
dynamical-systems chaos-theory
asked Dec 21 '18 at 0:06
H.RappeportH.Rappeport
6841510
$endgroup$
add a comment |
$begingroup$
This is the Lyapunov exponent as a function of $r$ for the logistic map ($x_{n+1}=f(x_n)=r(x_{n}-x_{n}^2)$)
The big dips are centered around points where $f'(x)=0$ for some $x$ in the trajectory used to calculate the exponent $lambda$, and therefore for those values $lambda=-infty$. These are precisely the $r$ values where there exists a superstable fixed point/cycle, which makes sense as $f'(x^*)=0$ is a necessary condition for superstability (where $x^*$ is some fixed point).
This naturally led me to think that perhaps the existence of a superstable fixed point/cycle is a sufficient condition for any 1D map to have $lambda = -infty$
However, this is now a Lyapunov graph for $x_{n+1}=g(x_n)=r(x_{n}^2-x_{n}^3)$
For which $0$ is a superstable fixed point for all values of $r$. For $r<4$ we indeed see that $lambda = -infty$ but for some values of $r$ there is a real $lambda$, for some even positive.
So what's the difference? Are these superstable fixed points different somehow? Also, the bluriness of the line in the second image results from most of the $r$ values in between still being $-infty$ - why is this so?
Notes:
I used the standard algorithm to generate the $lambda$ values. Namely, for each value of $r$ I draw uniformly some $x_0in [0, 1]$, and calculate $lambda = left(frac{1}{n}sum_{i=1}^{n}lnleft|f'left(x_{i}right)right|right)$ for $n=10^4$
The maximal $r$ value, $6.75$, was chosen to ensure the interval $[0,1]$ maps onto itself, as in the logistic map.
dynamical-systems chaos-theory
asked Dec 21 '18 at 0:06
H.RappeportH.Rappeport
6841510
$endgroup$
$begingroup$
Can you please check your system for bistability? Looking at this graph I can see the second stable fixed point. It is possible that when $r$ is not big enough, the only attractors of this system are these two fixed points ...
$endgroup$
– Evgeny
Dec 22 '18 at 6:48
$begingroup$
... However, when you sample initial point randomly, because of bistability you have a chance to fall to a different attractor. You are right that the fixed point at $x = 0$ is always stable, but you can simply miss its basin of attraction by choosing starting point in another basin of attraction. Usually to avoid problem with multistability you have to have some form of attractor continuation: when plotting Lyapunov exponent at the next point in parameter space just choose the initial point in phase space close to the attractor on the previous step.
$endgroup$
– Evgeny
Dec 22 '18 at 7:24
$begingroup$
Aha, reading Strogatz, I got the impression that the Lyapunov exponent was uniquely defined for a given 1D system, but it seems if there is more than one stable attractor this is not true. However, I still don't see how there can be a positive exponent in the presence of a [super]stable attractor.
$endgroup$
– H.Rappeport
Dec 22 '18 at 9:34
1
$begingroup$
Can you please clarify what exactly you don't understand? If there are two attractors in the system, they shouldn't influence dynamics inside of the basins of other attractor. So, if you pick a point in a basin of superstable fixed point, you'll see $-infty$ as a Lyapunov exponent. However, if another attractor is strange, like in logistic map, then its largest Lyapunov exponent for sure can be positive.
$endgroup$
– Evgeny
Dec 22 '18 at 11:00
add a comment |
$begingroup$
This is the Lyapunov exponent as a function of $r$ for the logistic map ($x_{n+1}=f(x_n)=r(x_{n}-x_{n}^2)$)
The big dips are centered around points where $f'(x)=0$ for some $x$ in the trajectory used to calculate the exponent $lambda$, and therefore for those values $lambda=-infty$. These are precisely the $r$ values where there exists a superstable fixed point/cycle, which makes sense as $f'(x^*)=0$ is a necessary condition for superstability (where $x^*$ is some fixed point).
This naturally led me to think that perhaps the existence of a superstable fixed point/cycle is a sufficient condition for any 1D map to have $lambda = -infty$
However, this is now a Lyapunov graph for $x_{n+1}=g(x_n)=r(x_{n}^2-x_{n}^3)$
For which $0$ is a superstable fixed point for all values of $r$. For $r<4$ we indeed see that $lambda = -infty$ but for some values of $r$ there is a real $lambda$, for some even positive.
So what's the difference? Are these superstable fixed points different somehow? Also, the bluriness of the line in the second image results from most of the $r$ values in between still being $-infty$ - why is this so?
Notes:
I used the standard algorithm to generate the $lambda$ values. Namely, for each value of $r$ I draw uniformly some $x_0in [0, 1]$, and calculate $lambda = left(frac{1}{n}sum_{i=1}^{n}lnleft|f'left(x_{i}right)right|right)$ for $n=10^4$
The maximal $r$ value, $6.75$, was chosen to ensure the interval $[0,1]$ maps onto itself, as in the logistic map.
dynamical-systems chaos-theory
asked Dec 21 '18 at 0:06
H.RappeportH.Rappeport
6841510
$endgroup$
This is the Lyapunov exponent as a function of $r$ for the logistic map ($x_{n+1}=f(x_n)=r(x_{n}-x_{n}^2)$)
The big dips are centered around points where $f'(x)=0$ for some $x$ in the trajectory used to calculate the exponent $lambda$, and therefore for those values $lambda=-infty$. These are precisely the $r$ values where there exists a superstable fixed point/cycle, which makes sense as $f'(x^*)=0$ is a necessary condition for superstability (where $x^*$ is some fixed point).
This naturally led me to think that perhaps the existence of a superstable fixed point/cycle is a sufficient condition for any 1D map to have $lambda = -infty$
However, this is now a Lyapunov graph for $x_{n+1}=g(x_n)=r(x_{n}^2-x_{n}^3)$
For which $0$ is a superstable fixed point for all values of $r$. For $r<4$ we indeed see that $lambda = -infty$ but for some values of $r$ there is a real $lambda$, for some even positive.
So what's the difference? Are these superstable fixed points different somehow? Also, the bluriness of the line in the second image results from most of the $r$ values in between still being $-infty$ - why is this so?
Notes:
I used the standard algorithm to generate the $lambda$ values. Namely, for each value of $r$ I draw uniformly some $x_0in [0, 1]$, and calculate $lambda = left(frac{1}{n}sum_{i=1}^{n}lnleft|f'left(x_{i}right)right|right)$ for $n=10^4$
The maximal $r$ value, $6.75$, was chosen to ensure the interval $[0,1]$ maps onto itself, as in the logistic map.
dynamical-systems chaos-theory
dynamical-systems chaos-theory
asked Dec 21 '18 at 0:06
H.RappeportH.Rappeport
6841510
asked Dec 21 '18 at 0:06
H.RappeportH.Rappeport
6841510
edited Dec 21 '18 at 11:57
H.Rappeport
asked Dec 21 '18 at 0:06
H.RappeportH.Rappeport
6841510
asked Dec 21 '18 at 0:06
H.RappeportH.Rappeport
6841510
asked Dec 21 '18 at 0:06
H.RappeportH.Rappeport
6841510
6841510
$begingroup$
Can you please check your system for bistability? Looking at this graph I can see the second stable fixed point. It is possible that when $r$ is not big enough, the only attractors of this system are these two fixed points ...
$endgroup$
– Evgeny
Dec 22 '18 at 6:48
$begingroup$
... However, when you sample initial point randomly, because of bistability you have a chance to fall to a different attractor. You are right that the fixed point at $x = 0$ is always stable, but you can simply miss its basin of attraction by choosing starting point in another basin of attraction. Usually to avoid problem with multistability you have to have some form of attractor continuation: when plotting Lyapunov exponent at the next point in parameter space just choose the initial point in phase space close to the attractor on the previous step.
$endgroup$
– Evgeny
Dec 22 '18 at 7:24
$begingroup$
Aha, reading Strogatz, I got the impression that the Lyapunov exponent was uniquely defined for a given 1D system, but it seems if there is more than one stable attractor this is not true. However, I still don't see how there can be a positive exponent in the presence of a [super]stable attractor.
$endgroup$
– H.Rappeport
Dec 22 '18 at 9:34
1
$begingroup$
Can you please clarify what exactly you don't understand? If there are two attractors in the system, they shouldn't influence dynamics inside of the basins of other attractor. So, if you pick a point in a basin of superstable fixed point, you'll see $-infty$ as a Lyapunov exponent. However, if another attractor is strange, like in logistic map, then its largest Lyapunov exponent for sure can be positive.
$endgroup$
– Evgeny
Dec 22 '18 at 11:00
add a comment |
$begingroup$
Can you please check your system for bistability? Looking at this graph I can see the second stable fixed point. It is possible that when $r$ is not big enough, the only attractors of this system are these two fixed points ...
$endgroup$
– Evgeny
Dec 22 '18 at 6:48
$begingroup$
... However, when you sample initial point randomly, because of bistability you have a chance to fall to a different attractor. You are right that the fixed point at $x = 0$ is always stable, but you can simply miss its basin of attraction by choosing starting point in another basin of attraction. Usually to avoid problem with multistability you have to have some form of attractor continuation: when plotting Lyapunov exponent at the next point in parameter space just choose the initial point in phase space close to the attractor on the previous step.
$endgroup$
– Evgeny
Dec 22 '18 at 7:24
$begingroup$
Aha, reading Strogatz, I got the impression that the Lyapunov exponent was uniquely defined for a given 1D system, but it seems if there is more than one stable attractor this is not true. However, I still don't see how there can be a positive exponent in the presence of a [super]stable attractor.
$endgroup$
– H.Rappeport
Dec 22 '18 at 9:34
1
$begingroup$
Can you please clarify what exactly you don't understand? If there are two attractors in the system, they shouldn't influence dynamics inside of the basins of other attractor. So, if you pick a point in a basin of superstable fixed point, you'll see $-infty$ as a Lyapunov exponent. However, if another attractor is strange, like in logistic map, then its largest Lyapunov exponent for sure can be positive.
$endgroup$
– Evgeny
Dec 22 '18 at 11:00
$begingroup$
Can you please check your system for bistability? Looking at this graph I can see the second stable fixed point. It is possible that when $r$ is not big enough, the only attractors of this system are these two fixed points ...
$endgroup$
– Evgeny
Dec 22 '18 at 6:48
$begingroup$
Can you please check your system for bistability? Looking at this graph I can see the second stable fixed point. It is possible that when $r$ is not big enough, the only attractors of this system are these two fixed points ...
$endgroup$
– Evgeny
Dec 22 '18 at 6:48
$begingroup$
... However, when you sample initial point randomly, because of bistability you have a chance to fall to a different attractor. You are right that the fixed point at $x = 0$ is always stable, but you can simply miss its basin of attraction by choosing starting point in another basin of attraction. Usually to avoid problem with multistability you have to have some form of attractor continuation: when plotting Lyapunov exponent at the next point in parameter space just choose the initial point in phase space close to the attractor on the previous step.
$endgroup$
– Evgeny
Dec 22 '18 at 7:24
$begingroup$
... However, when you sample initial point randomly, because of bistability you have a chance to fall to a different attractor. You are right that the fixed point at $x = 0$ is always stable, but you can simply miss its basin of attraction by choosing starting point in another basin of attraction. Usually to avoid problem with multistability you have to have some form of attractor continuation: when plotting Lyapunov exponent at the next point in parameter space just choose the initial point in phase space close to the attractor on the previous step.
$endgroup$
– Evgeny
Dec 22 '18 at 7:24
$begingroup$
Aha, reading Strogatz, I got the impression that the Lyapunov exponent was uniquely defined for a given 1D system, but it seems if there is more than one stable attractor this is not true. However, I still don't see how there can be a positive exponent in the presence of a [super]stable attractor.
$endgroup$
– H.Rappeport
Dec 22 '18 at 9:34
$begingroup$
Aha, reading Strogatz, I got the impression that the Lyapunov exponent was uniquely defined for a given 1D system, but it seems if there is more than one stable attractor this is not true. However, I still don't see how there can be a positive exponent in the presence of a [super]stable attractor.
$endgroup$
– H.Rappeport
Dec 22 '18 at 9:34
1
1
$begingroup$
Can you please clarify what exactly you don't understand? If there are two attractors in the system, they shouldn't influence dynamics inside of the basins of other attractor. So, if you pick a point in a basin of superstable fixed point, you'll see $-infty$ as a Lyapunov exponent. However, if another attractor is strange, like in logistic map, then its largest Lyapunov exponent for sure can be positive.
$endgroup$
– Evgeny
Dec 22 '18 at 11:00
$begingroup$
Can you please clarify what exactly you don't understand? If there are two attractors in the system, they shouldn't influence dynamics inside of the basins of other attractor. So, if you pick a point in a basin of superstable fixed point, you'll see $-infty$ as a Lyapunov exponent. However, if another attractor is strange, like in logistic map, then its largest Lyapunov exponent for sure can be positive.
$endgroup$
– Evgeny
Dec 22 '18 at 11:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As Evgeny helpfully points out, the second system in the question exhibits bistability for $r>4$.
If we plot the attractor of each $x_0in [0,1]$ (by iterating $g$ for 1000 iterations to get rid of the transients and then plotting the next 100 points), for $r<4$ this looks like
(Note: The equation in the title of these figures is wrong, didn't notice at the time. Should be $r(x^2-x^3)$)
So 0 is a [super]stable fixed point, and its basin is all of $[0,1]$
However, for $r$ values greater than 4 and less than ~6.2 (where the Lyapunov exponents are still negative), a typical attraction map looks like
So some (a little less than half) of the starting points are attracted to 0, but if we have $x_0$ in the middle of the range (approx. 0.25 to 0.95), the trajectory gets attracted to a 2-cycle.
Finally, for $r$ values greater than ~5.8, there appears a strange attractor, and hence the Lyapunov exponent is positive for some starting points $x_0$
Bonus: The full orbit diagram for the map
answered Dec 22 '18 at 14:09
H.RappeportH.Rappeport
6841510
$endgroup$
$begingroup$
Seems good to me, nice way to visualize basins of attraction! If you also can verify analytically the basin of stability for superstable fixed point, that would be a good addition. Also, it feels like that somewhere in $4 < r < 5.5$ a period doubling happens: before that an attractor is just a fixed point, like I've mentioned before. Also also, some people prefer 10000 iterations and plotting last few hundred points as a numerical attractor.
$endgroup$
– Evgeny
Dec 22 '18 at 18:18
$begingroup$
@Evgeny Not a period doubling, but period doublings - plural(: I've added the bifurcation diagram. Thank you for your helpful comments
$endgroup$
– H.Rappeport
Dec 22 '18 at 19:22
$begingroup$
Yeah, sure there are many of them, I've meant the period doubling that happens before $r = 5.5$, there is only one such :) Glad that my advice helped!
$endgroup$
– Evgeny
Dec 22 '18 at 22:31
add a comment |
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$begingroup$
As Evgeny helpfully points out, the second system in the question exhibits bistability for $r>4$.
If we plot the attractor of each $x_0in [0,1]$ (by iterating $g$ for 1000 iterations to get rid of the transients and then plotting the next 100 points), for $r<4$ this looks like
(Note: The equation in the title of these figures is wrong, didn't notice at the time. Should be $r(x^2-x^3)$)
So 0 is a [super]stable fixed point, and its basin is all of $[0,1]$
However, for $r$ values greater than 4 and less than ~6.2 (where the Lyapunov exponents are still negative), a typical attraction map looks like
So some (a little less than half) of the starting points are attracted to 0, but if we have $x_0$ in the middle of the range (approx. 0.25 to 0.95), the trajectory gets attracted to a 2-cycle.
Finally, for $r$ values greater than ~5.8, there appears a strange attractor, and hence the Lyapunov exponent is positive for some starting points $x_0$
Bonus: The full orbit diagram for the map
answered Dec 22 '18 at 14:09
H.RappeportH.Rappeport
6841510
$endgroup$
$begingroup$
Seems good to me, nice way to visualize basins of attraction! If you also can verify analytically the basin of stability for superstable fixed point, that would be a good addition. Also, it feels like that somewhere in $4 < r < 5.5$ a period doubling happens: before that an attractor is just a fixed point, like I've mentioned before. Also also, some people prefer 10000 iterations and plotting last few hundred points as a numerical attractor.
$endgroup$
– Evgeny
Dec 22 '18 at 18:18
$begingroup$
@Evgeny Not a period doubling, but period doublings - plural(: I've added the bifurcation diagram. Thank you for your helpful comments
$endgroup$
– H.Rappeport
Dec 22 '18 at 19:22
$begingroup$
Yeah, sure there are many of them, I've meant the period doubling that happens before $r = 5.5$, there is only one such :) Glad that my advice helped!
$endgroup$
– Evgeny
Dec 22 '18 at 22:31
add a comment |
$begingroup$
As Evgeny helpfully points out, the second system in the question exhibits bistability for $r>4$.
If we plot the attractor of each $x_0in [0,1]$ (by iterating $g$ for 1000 iterations to get rid of the transients and then plotting the next 100 points), for $r<4$ this looks like
(Note: The equation in the title of these figures is wrong, didn't notice at the time. Should be $r(x^2-x^3)$)
So 0 is a [super]stable fixed point, and its basin is all of $[0,1]$
However, for $r$ values greater than 4 and less than ~6.2 (where the Lyapunov exponents are still negative), a typical attraction map looks like
So some (a little less than half) of the starting points are attracted to 0, but if we have $x_0$ in the middle of the range (approx. 0.25 to 0.95), the trajectory gets attracted to a 2-cycle.
Finally, for $r$ values greater than ~5.8, there appears a strange attractor, and hence the Lyapunov exponent is positive for some starting points $x_0$
Bonus: The full orbit diagram for the map
answered Dec 22 '18 at 14:09
H.RappeportH.Rappeport
6841510
$endgroup$
$begingroup$
Seems good to me, nice way to visualize basins of attraction! If you also can verify analytically the basin of stability for superstable fixed point, that would be a good addition. Also, it feels like that somewhere in $4 < r < 5.5$ a period doubling happens: before that an attractor is just a fixed point, like I've mentioned before. Also also, some people prefer 10000 iterations and plotting last few hundred points as a numerical attractor.
$endgroup$
– Evgeny
Dec 22 '18 at 18:18
$begingroup$
@Evgeny Not a period doubling, but period doublings - plural(: I've added the bifurcation diagram. Thank you for your helpful comments
$endgroup$
– H.Rappeport
Dec 22 '18 at 19:22
$begingroup$
Yeah, sure there are many of them, I've meant the period doubling that happens before $r = 5.5$, there is only one such :) Glad that my advice helped!
$endgroup$
– Evgeny
Dec 22 '18 at 22:31
add a comment |
$begingroup$
As Evgeny helpfully points out, the second system in the question exhibits bistability for $r>4$.
If we plot the attractor of each $x_0in [0,1]$ (by iterating $g$ for 1000 iterations to get rid of the transients and then plotting the next 100 points), for $r<4$ this looks like
(Note: The equation in the title of these figures is wrong, didn't notice at the time. Should be $r(x^2-x^3)$)
So 0 is a [super]stable fixed point, and its basin is all of $[0,1]$
However, for $r$ values greater than 4 and less than ~6.2 (where the Lyapunov exponents are still negative), a typical attraction map looks like
So some (a little less than half) of the starting points are attracted to 0, but if we have $x_0$ in the middle of the range (approx. 0.25 to 0.95), the trajectory gets attracted to a 2-cycle.
Finally, for $r$ values greater than ~5.8, there appears a strange attractor, and hence the Lyapunov exponent is positive for some starting points $x_0$
Bonus: The full orbit diagram for the map
answered Dec 22 '18 at 14:09
H.RappeportH.Rappeport
6841510
$endgroup$
As Evgeny helpfully points out, the second system in the question exhibits bistability for $r>4$.
If we plot the attractor of each $x_0in [0,1]$ (by iterating $g$ for 1000 iterations to get rid of the transients and then plotting the next 100 points), for $r<4$ this looks like
(Note: The equation in the title of these figures is wrong, didn't notice at the time. Should be $r(x^2-x^3)$)
So 0 is a [super]stable fixed point, and its basin is all of $[0,1]$
However, for $r$ values greater than 4 and less than ~6.2 (where the Lyapunov exponents are still negative), a typical attraction map looks like
So some (a little less than half) of the starting points are attracted to 0, but if we have $x_0$ in the middle of the range (approx. 0.25 to 0.95), the trajectory gets attracted to a 2-cycle.
Finally, for $r$ values greater than ~5.8, there appears a strange attractor, and hence the Lyapunov exponent is positive for some starting points $x_0$
Bonus: The full orbit diagram for the map
answered Dec 22 '18 at 14:09
H.RappeportH.Rappeport
6841510
edited Dec 22 '18 at 19:18
answered Dec 22 '18 at 14:09
H.RappeportH.Rappeport
6841510
answered Dec 22 '18 at 14:09
H.RappeportH.Rappeport
6841510
answered Dec 22 '18 at 14:09
H.RappeportH.Rappeport
6841510
6841510
$begingroup$
Seems good to me, nice way to visualize basins of attraction! If you also can verify analytically the basin of stability for superstable fixed point, that would be a good addition. Also, it feels like that somewhere in $4 < r < 5.5$ a period doubling happens: before that an attractor is just a fixed point, like I've mentioned before. Also also, some people prefer 10000 iterations and plotting last few hundred points as a numerical attractor.
$endgroup$
– Evgeny
Dec 22 '18 at 18:18
$begingroup$
@Evgeny Not a period doubling, but period doublings - plural(: I've added the bifurcation diagram. Thank you for your helpful comments
$endgroup$
– H.Rappeport
Dec 22 '18 at 19:22
$begingroup$
Yeah, sure there are many of them, I've meant the period doubling that happens before $r = 5.5$, there is only one such :) Glad that my advice helped!
$endgroup$
– Evgeny
Dec 22 '18 at 22:31
add a comment |
$begingroup$
Seems good to me, nice way to visualize basins of attraction! If you also can verify analytically the basin of stability for superstable fixed point, that would be a good addition. Also, it feels like that somewhere in $4 < r < 5.5$ a period doubling happens: before that an attractor is just a fixed point, like I've mentioned before. Also also, some people prefer 10000 iterations and plotting last few hundred points as a numerical attractor.
$endgroup$
– Evgeny
Dec 22 '18 at 18:18
$begingroup$
@Evgeny Not a period doubling, but period doublings - plural(: I've added the bifurcation diagram. Thank you for your helpful comments
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– H.Rappeport
Dec 22 '18 at 19:22
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Yeah, sure there are many of them, I've meant the period doubling that happens before $r = 5.5$, there is only one such :) Glad that my advice helped!
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– Evgeny
Dec 22 '18 at 22:31
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Seems good to me, nice way to visualize basins of attraction! If you also can verify analytically the basin of stability for superstable fixed point, that would be a good addition. Also, it feels like that somewhere in $4 < r < 5.5$ a period doubling happens: before that an attractor is just a fixed point, like I've mentioned before. Also also, some people prefer 10000 iterations and plotting last few hundred points as a numerical attractor.
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– Evgeny
Dec 22 '18 at 18:18
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Seems good to me, nice way to visualize basins of attraction! If you also can verify analytically the basin of stability for superstable fixed point, that would be a good addition. Also, it feels like that somewhere in $4 < r < 5.5$ a period doubling happens: before that an attractor is just a fixed point, like I've mentioned before. Also also, some people prefer 10000 iterations and plotting last few hundred points as a numerical attractor.
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– Evgeny
Dec 22 '18 at 18:18
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@Evgeny Not a period doubling, but period doublings - plural(: I've added the bifurcation diagram. Thank you for your helpful comments
$endgroup$
– H.Rappeport
Dec 22 '18 at 19:22
$begingroup$
@Evgeny Not a period doubling, but period doublings - plural(: I've added the bifurcation diagram. Thank you for your helpful comments
$endgroup$
– H.Rappeport
Dec 22 '18 at 19:22
$begingroup$
Yeah, sure there are many of them, I've meant the period doubling that happens before $r = 5.5$, there is only one such :) Glad that my advice helped!
$endgroup$
– Evgeny
Dec 22 '18 at 22:31
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Yeah, sure there are many of them, I've meant the period doubling that happens before $r = 5.5$, there is only one such :) Glad that my advice helped!
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– Evgeny
Dec 22 '18 at 22:31
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Can you please check your system for bistability? Looking at this graph I can see the second stable fixed point. It is possible that when $r$ is not big enough, the only attractors of this system are these two fixed points ...
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– Evgeny
Dec 22 '18 at 6:48
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... However, when you sample initial point randomly, because of bistability you have a chance to fall to a different attractor. You are right that the fixed point at $x = 0$ is always stable, but you can simply miss its basin of attraction by choosing starting point in another basin of attraction. Usually to avoid problem with multistability you have to have some form of attractor continuation: when plotting Lyapunov exponent at the next point in parameter space just choose the initial point in phase space close to the attractor on the previous step.
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– Evgeny
Dec 22 '18 at 7:24
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Aha, reading Strogatz, I got the impression that the Lyapunov exponent was uniquely defined for a given 1D system, but it seems if there is more than one stable attractor this is not true. However, I still don't see how there can be a positive exponent in the presence of a [super]stable attractor.
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– H.Rappeport
Dec 22 '18 at 9:34
1
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Can you please clarify what exactly you don't understand? If there are two attractors in the system, they shouldn't influence dynamics inside of the basins of other attractor. So, if you pick a point in a basin of superstable fixed point, you'll see $-infty$ as a Lyapunov exponent. However, if another attractor is strange, like in logistic map, then its largest Lyapunov exponent for sure can be positive.
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– Evgeny
Dec 22 '18 at 11:00