Iterated map $x_{n+1} = rx_n(1-x_n^2)$
$begingroup$
In my differential equations homework, which is mostly extremely easy, there is one question which I cannot grasp.
For easier reference, here is the full question as I have been given it:
Consider the iterated map $x_{n+1} = rx_n(1-x_n^2)$
Show that for $0 < r < 3sqrt3/2$ if $0 ≤ x_n ≤ 1$ then $0 ≤ x_{n+1} ≤ 1$. Show that if
$r < 1$ then the only fixed point in $(0, 1)$ is zero, and that this is stable. When
$r > 1$ there is another fixed point in $(0, 1)$. Find the value of this fixed point
(as a function of $r$).
For which values of $r$ is it stable, and for which values is
it unstable? What would you expect to happen when $r > 2$?
For the first part, I have managed to get a cubic and have found the minimum and maximum values of $r$ as $x_n$ goes from $0$ to $1$, which I think is the correct way to show it, although I am not sure why $0 < r < 3sqrt 3/2$ is used instead of $0 ≤ r ≤ 3sqrt 3/2$ as I think it's still true for the case when r is equal to those two things.
For the $r<1$ fixed point, I know that $1>x_{n+1}≥0$ as $1≥x_n(1-x_n^2)≥0$, but I don't know what to do from there.
I am quite unsure about how to go about the rest of the question.
Thank you for any help.
ordinary-differential-equations recurrence-relations
edited Dec 6 '18 at 5:21
Tianlalu
3,08121038
asked Dec 6 '18 at 3:54
Plz help mePlz help me
766
$endgroup$
add a comment |
$begingroup$
In my differential equations homework, which is mostly extremely easy, there is one question which I cannot grasp.
For easier reference, here is the full question as I have been given it:
Consider the iterated map $x_{n+1} = rx_n(1-x_n^2)$
Show that for $0 < r < 3sqrt3/2$ if $0 ≤ x_n ≤ 1$ then $0 ≤ x_{n+1} ≤ 1$. Show that if
$r < 1$ then the only fixed point in $(0, 1)$ is zero, and that this is stable. When
$r > 1$ there is another fixed point in $(0, 1)$. Find the value of this fixed point
(as a function of $r$).
For which values of $r$ is it stable, and for which values is
it unstable? What would you expect to happen when $r > 2$?
For the first part, I have managed to get a cubic and have found the minimum and maximum values of $r$ as $x_n$ goes from $0$ to $1$, which I think is the correct way to show it, although I am not sure why $0 < r < 3sqrt 3/2$ is used instead of $0 ≤ r ≤ 3sqrt 3/2$ as I think it's still true for the case when r is equal to those two things.
For the $r<1$ fixed point, I know that $1>x_{n+1}≥0$ as $1≥x_n(1-x_n^2)≥0$, but I don't know what to do from there.
I am quite unsure about how to go about the rest of the question.
Thank you for any help.
ordinary-differential-equations recurrence-relations
edited Dec 6 '18 at 5:21
Tianlalu
3,08121038
asked Dec 6 '18 at 3:54
Plz help mePlz help me
766
$endgroup$
$begingroup$
If $rin(0,1)$, and $x_nin(0,1)$, so $(1-x_n^2)in(0,1)$. This means that $x_{n+1} = delta_nx_n$ for $delta_nin(0,1)$, so $x_n$ is decreasing as $ntoinfty$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 4:46
add a comment |
$begingroup$
In my differential equations homework, which is mostly extremely easy, there is one question which I cannot grasp.
For easier reference, here is the full question as I have been given it:
Consider the iterated map $x_{n+1} = rx_n(1-x_n^2)$
Show that for $0 < r < 3sqrt3/2$ if $0 ≤ x_n ≤ 1$ then $0 ≤ x_{n+1} ≤ 1$. Show that if
$r < 1$ then the only fixed point in $(0, 1)$ is zero, and that this is stable. When
$r > 1$ there is another fixed point in $(0, 1)$. Find the value of this fixed point
(as a function of $r$).
For which values of $r$ is it stable, and for which values is
it unstable? What would you expect to happen when $r > 2$?
For the first part, I have managed to get a cubic and have found the minimum and maximum values of $r$ as $x_n$ goes from $0$ to $1$, which I think is the correct way to show it, although I am not sure why $0 < r < 3sqrt 3/2$ is used instead of $0 ≤ r ≤ 3sqrt 3/2$ as I think it's still true for the case when r is equal to those two things.
For the $r<1$ fixed point, I know that $1>x_{n+1}≥0$ as $1≥x_n(1-x_n^2)≥0$, but I don't know what to do from there.
I am quite unsure about how to go about the rest of the question.
Thank you for any help.
ordinary-differential-equations recurrence-relations
edited Dec 6 '18 at 5:21
Tianlalu
3,08121038
asked Dec 6 '18 at 3:54
Plz help mePlz help me
766
$endgroup$
In my differential equations homework, which is mostly extremely easy, there is one question which I cannot grasp.
For easier reference, here is the full question as I have been given it:
Consider the iterated map $x_{n+1} = rx_n(1-x_n^2)$
Show that for $0 < r < 3sqrt3/2$ if $0 ≤ x_n ≤ 1$ then $0 ≤ x_{n+1} ≤ 1$. Show that if
$r < 1$ then the only fixed point in $(0, 1)$ is zero, and that this is stable. When
$r > 1$ there is another fixed point in $(0, 1)$. Find the value of this fixed point
(as a function of $r$).
For which values of $r$ is it stable, and for which values is
it unstable? What would you expect to happen when $r > 2$?
For the first part, I have managed to get a cubic and have found the minimum and maximum values of $r$ as $x_n$ goes from $0$ to $1$, which I think is the correct way to show it, although I am not sure why $0 < r < 3sqrt 3/2$ is used instead of $0 ≤ r ≤ 3sqrt 3/2$ as I think it's still true for the case when r is equal to those two things.
For the $r<1$ fixed point, I know that $1>x_{n+1}≥0$ as $1≥x_n(1-x_n^2)≥0$, but I don't know what to do from there.
I am quite unsure about how to go about the rest of the question.
Thank you for any help.
ordinary-differential-equations recurrence-relations
ordinary-differential-equations recurrence-relations
edited Dec 6 '18 at 5:21
Tianlalu
3,08121038
asked Dec 6 '18 at 3:54
Plz help mePlz help me
766
edited Dec 6 '18 at 5:21
Tianlalu
3,08121038
asked Dec 6 '18 at 3:54
Plz help mePlz help me
766
edited Dec 6 '18 at 5:21
Tianlalu
3,08121038
edited Dec 6 '18 at 5:21
Tianlalu
3,08121038
edited Dec 6 '18 at 5:21
Tianlalu
3,08121038
3,08121038
asked Dec 6 '18 at 3:54
Plz help mePlz help me
766
asked Dec 6 '18 at 3:54
Plz help mePlz help me
766
asked Dec 6 '18 at 3:54
Plz help mePlz help me
766
766
$begingroup$
If $rin(0,1)$, and $x_nin(0,1)$, so $(1-x_n^2)in(0,1)$. This means that $x_{n+1} = delta_nx_n$ for $delta_nin(0,1)$, so $x_n$ is decreasing as $ntoinfty$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 4:46
add a comment |
$begingroup$
If $rin(0,1)$, and $x_nin(0,1)$, so $(1-x_n^2)in(0,1)$. This means that $x_{n+1} = delta_nx_n$ for $delta_nin(0,1)$, so $x_n$ is decreasing as $ntoinfty$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 4:46
$begingroup$
If $rin(0,1)$, and $x_nin(0,1)$, so $(1-x_n^2)in(0,1)$. This means that $x_{n+1} = delta_nx_n$ for $delta_nin(0,1)$, so $x_n$ is decreasing as $ntoinfty$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 4:46
$begingroup$
If $rin(0,1)$, and $x_nin(0,1)$, so $(1-x_n^2)in(0,1)$. This means that $x_{n+1} = delta_nx_n$ for $delta_nin(0,1)$, so $x_n$ is decreasing as $ntoinfty$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 4:46
add a comment |
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$begingroup$
If $rin(0,1)$, and $x_nin(0,1)$, so $(1-x_n^2)in(0,1)$. This means that $x_{n+1} = delta_nx_n$ for $delta_nin(0,1)$, so $x_n$ is decreasing as $ntoinfty$.
$endgroup$
– AlexanderJ93
Dec 6 '18 at 4:46