Does $x_{t} = 1-x_{t-1}$ have a stable steady state solution?
$begingroup$
At steady state, $x = x_{t} = x_{t-1}$. So I can solve for the steady state value of $x=0.5$.
The general rule of determining the stability of the steady state is that the $|text{slope}|<1$. But here slope = 1, so can I say that the steady state is unstable?
Graphically the picture looks like this
I have a picture of an oscillation motion around the steady state. But will it converge to the steady state and will it be stable?
stability-theory steady-state
asked Jan 18 at 3:59
OGCOGC
1,43821229
$endgroup$
add a comment |
$begingroup$
At steady state, $x = x_{t} = x_{t-1}$. So I can solve for the steady state value of $x=0.5$.
The general rule of determining the stability of the steady state is that the $|text{slope}|<1$. But here slope = 1, so can I say that the steady state is unstable?
Graphically the picture looks like this
I have a picture of an oscillation motion around the steady state. But will it converge to the steady state and will it be stable?
stability-theory steady-state
asked Jan 18 at 3:59
OGCOGC
1,43821229
$endgroup$
$begingroup$
Perhaps you can draw it as a vertical segment between the two lines.
$endgroup$
– Michael Burr
Jan 18 at 4:05
$begingroup$
You need to actually try to write out a few sample paths, that is, write out the first few values of the sequence. The answer becomes obvious when you do.
$endgroup$
– Michael
Jan 18 at 4:33
add a comment |
$begingroup$
At steady state, $x = x_{t} = x_{t-1}$. So I can solve for the steady state value of $x=0.5$.
The general rule of determining the stability of the steady state is that the $|text{slope}|<1$. But here slope = 1, so can I say that the steady state is unstable?
Graphically the picture looks like this
I have a picture of an oscillation motion around the steady state. But will it converge to the steady state and will it be stable?
stability-theory steady-state
asked Jan 18 at 3:59
OGCOGC
1,43821229
$endgroup$
At steady state, $x = x_{t} = x_{t-1}$. So I can solve for the steady state value of $x=0.5$.
The general rule of determining the stability of the steady state is that the $|text{slope}|<1$. But here slope = 1, so can I say that the steady state is unstable?
Graphically the picture looks like this
I have a picture of an oscillation motion around the steady state. But will it converge to the steady state and will it be stable?
stability-theory steady-state
stability-theory steady-state
asked Jan 18 at 3:59
OGCOGC
1,43821229
asked Jan 18 at 3:59
OGCOGC
1,43821229
edited Jan 18 at 7:51
OGC
asked Jan 18 at 3:59
OGCOGC
1,43821229
asked Jan 18 at 3:59
OGCOGC
1,43821229
asked Jan 18 at 3:59
OGCOGC
1,43821229
1,43821229
$begingroup$
Perhaps you can draw it as a vertical segment between the two lines.
$endgroup$
– Michael Burr
Jan 18 at 4:05
$begingroup$
You need to actually try to write out a few sample paths, that is, write out the first few values of the sequence. The answer becomes obvious when you do.
$endgroup$
– Michael
Jan 18 at 4:33
add a comment |
$begingroup$
Perhaps you can draw it as a vertical segment between the two lines.
$endgroup$
– Michael Burr
Jan 18 at 4:05
$begingroup$
You need to actually try to write out a few sample paths, that is, write out the first few values of the sequence. The answer becomes obvious when you do.
$endgroup$
– Michael
Jan 18 at 4:33
$begingroup$
Perhaps you can draw it as a vertical segment between the two lines.
$endgroup$
– Michael Burr
Jan 18 at 4:05
$begingroup$
Perhaps you can draw it as a vertical segment between the two lines.
$endgroup$
– Michael Burr
Jan 18 at 4:05
$begingroup$
You need to actually try to write out a few sample paths, that is, write out the first few values of the sequence. The answer becomes obvious when you do.
$endgroup$
– Michael
Jan 18 at 4:33
$begingroup$
You need to actually try to write out a few sample paths, that is, write out the first few values of the sequence. The answer becomes obvious when you do.
$endgroup$
– Michael
Jan 18 at 4:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The solutions of
$x_t = 1 - x_{t - 1} tag 1$
are of two kinds, depending on whether or not there exists an $x_t$ such that
$x_t = dfrac{1}{2} tag 2$
or not. If (1) holds for some $t$, then it is easy to see that
$forall t, ; x_t = dfrac{1}{2}; tag 2$
if, on the other hand,
$exists t, ; x_t ne dfrac{1}{2} , tag 3$
then since
$x_{t + 1} = 1 - x_t, tag 4$
we have
$x_{t + 2} = 1 - (1 - x_t) = x_t, tag 5$
and every solution is of period $2$.
In this case, the oscillating solutions are in fact stable, alternating as they do 'twixt two values, one greater and the other less than $1/2$.
It is worth observing in this context that the somewhat more general dynamic
$x_t = 1 - kx_{t - 1}, ; k in Bbb R, tag 6$
satisfies
$x_t - x_{t - 1} = (1 - kx_{t - 1}) - (1 - kx_{t - 2}) = k(x_{t - 2} - x_{t - 1}), tag 7$
and has a fixed point satisfying
$x = 1 - kx, tag 8$
that is,
$x = dfrac{1}{k + 1}, ; k ne -1; tag 9$
also, (7) yields
$vert x_t - x_{t - 1} vert = vert k vert vert x_{t - 1} - x_{t - 2} vert, tag{10}$
so the distance 'twixt successive iterates grows or shrinks according to whether $vert k vert$ is larger or smaller than $1$; furthermore it is easy to see that
$x_t - x = k(x - x_{t - 1}), tag{11}$
$vert x_t - x vert = vert k vert vert x - x_{t - 1} vert, tag{12}$
which show the $x_t$ move farther or closer to the fixed point, again as $vert k vert > < 1$; from (11);we see that the $x_t$ either oscillate around $x$ or converge/diverge directly to/from $x$, depending on the sign of $k$. When $k = -1$ we find
$x_t = 1 + x_{t - 1}; tag{13}$
the iterates march off to $+infty$; there is no equilibrium state. The stability here is thus determined by the magnitude of $vert k vert$ relative to $1$.
A rather cute example, this, illustrating stable, decaying, or growing oscillation as well as monotonic decay to or growth from equilibrium, or no equilibrium at all.
answered Jan 18 at 6:39
Robert LewisRobert Lewis
46.3k23066
$endgroup$
1
$begingroup$
Thanks for the very detailed explanation and the introduction of a general dynamic equation.
$endgroup$
– OGC
Jan 18 at 8:01
1
$begingroup$
Thank you my friend! A great example this, a very simple dynamical system with very rich behaviour. By the way, if you really found my answer useful, you might consider "accepting" it. Thanks again!
$endgroup$
– Robert Lewis
Jan 18 at 8:04
$begingroup$
And thanks for the "acceptance"! Cheers!
$endgroup$
– Robert Lewis
Jan 18 at 8:06
add a comment |
Your Answer
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$begingroup$
The solutions of
$x_t = 1 - x_{t - 1} tag 1$
are of two kinds, depending on whether or not there exists an $x_t$ such that
$x_t = dfrac{1}{2} tag 2$
or not. If (1) holds for some $t$, then it is easy to see that
$forall t, ; x_t = dfrac{1}{2}; tag 2$
if, on the other hand,
$exists t, ; x_t ne dfrac{1}{2} , tag 3$
then since
$x_{t + 1} = 1 - x_t, tag 4$
we have
$x_{t + 2} = 1 - (1 - x_t) = x_t, tag 5$
and every solution is of period $2$.
In this case, the oscillating solutions are in fact stable, alternating as they do 'twixt two values, one greater and the other less than $1/2$.
It is worth observing in this context that the somewhat more general dynamic
$x_t = 1 - kx_{t - 1}, ; k in Bbb R, tag 6$
satisfies
$x_t - x_{t - 1} = (1 - kx_{t - 1}) - (1 - kx_{t - 2}) = k(x_{t - 2} - x_{t - 1}), tag 7$
and has a fixed point satisfying
$x = 1 - kx, tag 8$
that is,
$x = dfrac{1}{k + 1}, ; k ne -1; tag 9$
also, (7) yields
$vert x_t - x_{t - 1} vert = vert k vert vert x_{t - 1} - x_{t - 2} vert, tag{10}$
so the distance 'twixt successive iterates grows or shrinks according to whether $vert k vert$ is larger or smaller than $1$; furthermore it is easy to see that
$x_t - x = k(x - x_{t - 1}), tag{11}$
$vert x_t - x vert = vert k vert vert x - x_{t - 1} vert, tag{12}$
which show the $x_t$ move farther or closer to the fixed point, again as $vert k vert > < 1$; from (11);we see that the $x_t$ either oscillate around $x$ or converge/diverge directly to/from $x$, depending on the sign of $k$. When $k = -1$ we find
$x_t = 1 + x_{t - 1}; tag{13}$
the iterates march off to $+infty$; there is no equilibrium state. The stability here is thus determined by the magnitude of $vert k vert$ relative to $1$.
A rather cute example, this, illustrating stable, decaying, or growing oscillation as well as monotonic decay to or growth from equilibrium, or no equilibrium at all.
answered Jan 18 at 6:39
Robert LewisRobert Lewis
46.3k23066
$endgroup$
1
$begingroup$
Thanks for the very detailed explanation and the introduction of a general dynamic equation.
$endgroup$
– OGC
Jan 18 at 8:01
1
$begingroup$
Thank you my friend! A great example this, a very simple dynamical system with very rich behaviour. By the way, if you really found my answer useful, you might consider "accepting" it. Thanks again!
$endgroup$
– Robert Lewis
Jan 18 at 8:04
$begingroup$
And thanks for the "acceptance"! Cheers!
$endgroup$
– Robert Lewis
Jan 18 at 8:06
add a comment |
$begingroup$
The solutions of
$x_t = 1 - x_{t - 1} tag 1$
are of two kinds, depending on whether or not there exists an $x_t$ such that
$x_t = dfrac{1}{2} tag 2$
or not. If (1) holds for some $t$, then it is easy to see that
$forall t, ; x_t = dfrac{1}{2}; tag 2$
if, on the other hand,
$exists t, ; x_t ne dfrac{1}{2} , tag 3$
then since
$x_{t + 1} = 1 - x_t, tag 4$
we have
$x_{t + 2} = 1 - (1 - x_t) = x_t, tag 5$
and every solution is of period $2$.
In this case, the oscillating solutions are in fact stable, alternating as they do 'twixt two values, one greater and the other less than $1/2$.
It is worth observing in this context that the somewhat more general dynamic
$x_t = 1 - kx_{t - 1}, ; k in Bbb R, tag 6$
satisfies
$x_t - x_{t - 1} = (1 - kx_{t - 1}) - (1 - kx_{t - 2}) = k(x_{t - 2} - x_{t - 1}), tag 7$
and has a fixed point satisfying
$x = 1 - kx, tag 8$
that is,
$x = dfrac{1}{k + 1}, ; k ne -1; tag 9$
also, (7) yields
$vert x_t - x_{t - 1} vert = vert k vert vert x_{t - 1} - x_{t - 2} vert, tag{10}$
so the distance 'twixt successive iterates grows or shrinks according to whether $vert k vert$ is larger or smaller than $1$; furthermore it is easy to see that
$x_t - x = k(x - x_{t - 1}), tag{11}$
$vert x_t - x vert = vert k vert vert x - x_{t - 1} vert, tag{12}$
which show the $x_t$ move farther or closer to the fixed point, again as $vert k vert > < 1$; from (11);we see that the $x_t$ either oscillate around $x$ or converge/diverge directly to/from $x$, depending on the sign of $k$. When $k = -1$ we find
$x_t = 1 + x_{t - 1}; tag{13}$
the iterates march off to $+infty$; there is no equilibrium state. The stability here is thus determined by the magnitude of $vert k vert$ relative to $1$.
A rather cute example, this, illustrating stable, decaying, or growing oscillation as well as monotonic decay to or growth from equilibrium, or no equilibrium at all.
answered Jan 18 at 6:39
Robert LewisRobert Lewis
46.3k23066
$endgroup$
1
$begingroup$
Thanks for the very detailed explanation and the introduction of a general dynamic equation.
$endgroup$
– OGC
Jan 18 at 8:01
1
$begingroup$
Thank you my friend! A great example this, a very simple dynamical system with very rich behaviour. By the way, if you really found my answer useful, you might consider "accepting" it. Thanks again!
$endgroup$
– Robert Lewis
Jan 18 at 8:04
$begingroup$
And thanks for the "acceptance"! Cheers!
$endgroup$
– Robert Lewis
Jan 18 at 8:06
add a comment |
$begingroup$
The solutions of
$x_t = 1 - x_{t - 1} tag 1$
are of two kinds, depending on whether or not there exists an $x_t$ such that
$x_t = dfrac{1}{2} tag 2$
or not. If (1) holds for some $t$, then it is easy to see that
$forall t, ; x_t = dfrac{1}{2}; tag 2$
if, on the other hand,
$exists t, ; x_t ne dfrac{1}{2} , tag 3$
then since
$x_{t + 1} = 1 - x_t, tag 4$
we have
$x_{t + 2} = 1 - (1 - x_t) = x_t, tag 5$
and every solution is of period $2$.
In this case, the oscillating solutions are in fact stable, alternating as they do 'twixt two values, one greater and the other less than $1/2$.
It is worth observing in this context that the somewhat more general dynamic
$x_t = 1 - kx_{t - 1}, ; k in Bbb R, tag 6$
satisfies
$x_t - x_{t - 1} = (1 - kx_{t - 1}) - (1 - kx_{t - 2}) = k(x_{t - 2} - x_{t - 1}), tag 7$
and has a fixed point satisfying
$x = 1 - kx, tag 8$
that is,
$x = dfrac{1}{k + 1}, ; k ne -1; tag 9$
also, (7) yields
$vert x_t - x_{t - 1} vert = vert k vert vert x_{t - 1} - x_{t - 2} vert, tag{10}$
so the distance 'twixt successive iterates grows or shrinks according to whether $vert k vert$ is larger or smaller than $1$; furthermore it is easy to see that
$x_t - x = k(x - x_{t - 1}), tag{11}$
$vert x_t - x vert = vert k vert vert x - x_{t - 1} vert, tag{12}$
which show the $x_t$ move farther or closer to the fixed point, again as $vert k vert > < 1$; from (11);we see that the $x_t$ either oscillate around $x$ or converge/diverge directly to/from $x$, depending on the sign of $k$. When $k = -1$ we find
$x_t = 1 + x_{t - 1}; tag{13}$
the iterates march off to $+infty$; there is no equilibrium state. The stability here is thus determined by the magnitude of $vert k vert$ relative to $1$.
A rather cute example, this, illustrating stable, decaying, or growing oscillation as well as monotonic decay to or growth from equilibrium, or no equilibrium at all.
answered Jan 18 at 6:39
Robert LewisRobert Lewis
46.3k23066
$endgroup$
The solutions of
$x_t = 1 - x_{t - 1} tag 1$
are of two kinds, depending on whether or not there exists an $x_t$ such that
$x_t = dfrac{1}{2} tag 2$
or not. If (1) holds for some $t$, then it is easy to see that
$forall t, ; x_t = dfrac{1}{2}; tag 2$
if, on the other hand,
$exists t, ; x_t ne dfrac{1}{2} , tag 3$
then since
$x_{t + 1} = 1 - x_t, tag 4$
we have
$x_{t + 2} = 1 - (1 - x_t) = x_t, tag 5$
and every solution is of period $2$.
In this case, the oscillating solutions are in fact stable, alternating as they do 'twixt two values, one greater and the other less than $1/2$.
It is worth observing in this context that the somewhat more general dynamic
$x_t = 1 - kx_{t - 1}, ; k in Bbb R, tag 6$
satisfies
$x_t - x_{t - 1} = (1 - kx_{t - 1}) - (1 - kx_{t - 2}) = k(x_{t - 2} - x_{t - 1}), tag 7$
and has a fixed point satisfying
$x = 1 - kx, tag 8$
that is,
$x = dfrac{1}{k + 1}, ; k ne -1; tag 9$
also, (7) yields
$vert x_t - x_{t - 1} vert = vert k vert vert x_{t - 1} - x_{t - 2} vert, tag{10}$
so the distance 'twixt successive iterates grows or shrinks according to whether $vert k vert$ is larger or smaller than $1$; furthermore it is easy to see that
$x_t - x = k(x - x_{t - 1}), tag{11}$
$vert x_t - x vert = vert k vert vert x - x_{t - 1} vert, tag{12}$
which show the $x_t$ move farther or closer to the fixed point, again as $vert k vert > < 1$; from (11);we see that the $x_t$ either oscillate around $x$ or converge/diverge directly to/from $x$, depending on the sign of $k$. When $k = -1$ we find
$x_t = 1 + x_{t - 1}; tag{13}$
the iterates march off to $+infty$; there is no equilibrium state. The stability here is thus determined by the magnitude of $vert k vert$ relative to $1$.
A rather cute example, this, illustrating stable, decaying, or growing oscillation as well as monotonic decay to or growth from equilibrium, or no equilibrium at all.
answered Jan 18 at 6:39
Robert LewisRobert Lewis
46.3k23066
answered Jan 18 at 6:39
Robert LewisRobert Lewis
46.3k23066
answered Jan 18 at 6:39
Robert LewisRobert Lewis
46.3k23066
answered Jan 18 at 6:39
Robert LewisRobert Lewis
46.3k23066
46.3k23066
1
$begingroup$
Thanks for the very detailed explanation and the introduction of a general dynamic equation.
$endgroup$
– OGC
Jan 18 at 8:01
1
$begingroup$
Thank you my friend! A great example this, a very simple dynamical system with very rich behaviour. By the way, if you really found my answer useful, you might consider "accepting" it. Thanks again!
$endgroup$
– Robert Lewis
Jan 18 at 8:04
$begingroup$
And thanks for the "acceptance"! Cheers!
$endgroup$
– Robert Lewis
Jan 18 at 8:06
add a comment |
1
$begingroup$
Thanks for the very detailed explanation and the introduction of a general dynamic equation.
$endgroup$
– OGC
Jan 18 at 8:01
1
$begingroup$
Thank you my friend! A great example this, a very simple dynamical system with very rich behaviour. By the way, if you really found my answer useful, you might consider "accepting" it. Thanks again!
$endgroup$
– Robert Lewis
Jan 18 at 8:04
$begingroup$
And thanks for the "acceptance"! Cheers!
$endgroup$
– Robert Lewis
Jan 18 at 8:06
1
1
$begingroup$
Thanks for the very detailed explanation and the introduction of a general dynamic equation.
$endgroup$
– OGC
Jan 18 at 8:01
$begingroup$
Thanks for the very detailed explanation and the introduction of a general dynamic equation.
$endgroup$
– OGC
Jan 18 at 8:01
1
1
$begingroup$
Thank you my friend! A great example this, a very simple dynamical system with very rich behaviour. By the way, if you really found my answer useful, you might consider "accepting" it. Thanks again!
$endgroup$
– Robert Lewis
Jan 18 at 8:04
$begingroup$
Thank you my friend! A great example this, a very simple dynamical system with very rich behaviour. By the way, if you really found my answer useful, you might consider "accepting" it. Thanks again!
$endgroup$
– Robert Lewis
Jan 18 at 8:04
$begingroup$
And thanks for the "acceptance"! Cheers!
$endgroup$
– Robert Lewis
Jan 18 at 8:06
$begingroup$
And thanks for the "acceptance"! Cheers!
$endgroup$
– Robert Lewis
Jan 18 at 8:06
add a comment |
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$begingroup$
Perhaps you can draw it as a vertical segment between the two lines.
$endgroup$
– Michael Burr
Jan 18 at 4:05
$begingroup$
You need to actually try to write out a few sample paths, that is, write out the first few values of the sequence. The answer becomes obvious when you do.
$endgroup$
– Michael
Jan 18 at 4:33