Stability of $y = cos(2t)$ for the ode $y'' + 4y = 0$
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I want to examine the stability of the solution $phi = cos(2t)$ of the ode $y'' + 4y = 0$.
I know the general solution of this ode is $y = c_1 cos(2t) + c_2 sin(2t)$. So to examine the stability of my solution, I need to see if other solutions stay close to it starting at $t = 0$. It'll be asymptotically stable if other solutions converge to it.
So, when $t = 0$, $phi(0) = cos(0) = 1$. and $phi'(0) = -sin(0) =0$. But.. how do I pick $y(t)$ so that $y(0)$ starts out close to $1$ and $y'(0)$ is close to $0$? Then what?
Well, $y(t) = 1$ when $t = 0$ and when $t = 0$, $y'(0) = 1$. I'm not sure how to proceed.
ordinary-differential-equations
asked Dec 14 '18 at 17:48
NaltNalt
756
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add a comment |
$begingroup$
I want to examine the stability of the solution $phi = cos(2t)$ of the ode $y'' + 4y = 0$.
I know the general solution of this ode is $y = c_1 cos(2t) + c_2 sin(2t)$. So to examine the stability of my solution, I need to see if other solutions stay close to it starting at $t = 0$. It'll be asymptotically stable if other solutions converge to it.
So, when $t = 0$, $phi(0) = cos(0) = 1$. and $phi'(0) = -sin(0) =0$. But.. how do I pick $y(t)$ so that $y(0)$ starts out close to $1$ and $y'(0)$ is close to $0$? Then what?
Well, $y(t) = 1$ when $t = 0$ and when $t = 0$, $y'(0) = 1$. I'm not sure how to proceed.
ordinary-differential-equations
asked Dec 14 '18 at 17:48
NaltNalt
756
$endgroup$
add a comment |
$begingroup$
I want to examine the stability of the solution $phi = cos(2t)$ of the ode $y'' + 4y = 0$.
I know the general solution of this ode is $y = c_1 cos(2t) + c_2 sin(2t)$. So to examine the stability of my solution, I need to see if other solutions stay close to it starting at $t = 0$. It'll be asymptotically stable if other solutions converge to it.
So, when $t = 0$, $phi(0) = cos(0) = 1$. and $phi'(0) = -sin(0) =0$. But.. how do I pick $y(t)$ so that $y(0)$ starts out close to $1$ and $y'(0)$ is close to $0$? Then what?
Well, $y(t) = 1$ when $t = 0$ and when $t = 0$, $y'(0) = 1$. I'm not sure how to proceed.
ordinary-differential-equations
asked Dec 14 '18 at 17:48
NaltNalt
756
$endgroup$
I want to examine the stability of the solution $phi = cos(2t)$ of the ode $y'' + 4y = 0$.
I know the general solution of this ode is $y = c_1 cos(2t) + c_2 sin(2t)$. So to examine the stability of my solution, I need to see if other solutions stay close to it starting at $t = 0$. It'll be asymptotically stable if other solutions converge to it.
So, when $t = 0$, $phi(0) = cos(0) = 1$. and $phi'(0) = -sin(0) =0$. But.. how do I pick $y(t)$ so that $y(0)$ starts out close to $1$ and $y'(0)$ is close to $0$? Then what?
Well, $y(t) = 1$ when $t = 0$ and when $t = 0$, $y'(0) = 1$. I'm not sure how to proceed.
ordinary-differential-equations
ordinary-differential-equations
asked Dec 14 '18 at 17:48
NaltNalt
756
asked Dec 14 '18 at 17:48
NaltNalt
756
asked Dec 14 '18 at 17:48
NaltNalt
756
asked Dec 14 '18 at 17:48
NaltNalt
756
asked Dec 14 '18 at 17:48
NaltNalt
756
756
add a comment |
add a comment |
2 Answers
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Since all solutions are periodic, this is stable but not asymptotically stable: solutions that start close stay close, but do not converge to your solution.
answered Dec 14 '18 at 18:28
Robert IsraelRobert Israel
324k23213467
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$begingroup$
Hi Robert, thanks for your answer. I am little new at learning about stability so: This is stable because it is periodic. The reasoning being that all solutions stay within a range of the solution $cos(2t)$, between $-1$ and $1$. Is this it?
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– Nalt
Dec 14 '18 at 18:31
add a comment |
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Note that :
$$-1 leq sin(2t) leq 1 quad text{and} quad -1 leq cos(2t) leq 1$$
That means that your solution will be bounded, by let's say, some $m in mathbb R$ and $M in mathbb R$, such that :
$$m leq y(t) leq M$$
This automatically tells us that the system is stable, since its solutions do not diverge for any values of $c_1,c_2 in mathbb R$ arbitrary constants.
But, note that the limits $lim_{tto infty} sin(2t)$ and $lim_{tto infty} cos(2t)$ do not exists, which means that your solutions will not get arbitrary close to each other.
Thus, the given ODE solution is simply stable.
For a more formal definition (which proves our conclusions above) recall that :
Definition : A solution of an ODE is stable if for every $varepsilon >0$ there exists $delta >0 $ when if $hat{y}(t)$ satisfies the ODE, it is :
$$|hat{y}(t) - y(t_0) | leq delta implies |hat{y}(t) - y(t)| leq varepsilon ; forall t geq t_0$$
answered Dec 14 '18 at 18:35
RebellosRebellos
14.8k31248
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Thanks Rebellos! Your answer and Prof. Israel's really helped me to understand!
$endgroup$
– Nalt
Dec 14 '18 at 18:37
1
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@Nalt Glad I could help ! You may also use the "vote up" arrows (buttons) to reward users for answering something that helps you !
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– Rebellos
Dec 14 '18 at 18:38
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since all solutions are periodic, this is stable but not asymptotically stable: solutions that start close stay close, but do not converge to your solution.
answered Dec 14 '18 at 18:28
Robert IsraelRobert Israel
324k23213467
$endgroup$
$begingroup$
Hi Robert, thanks for your answer. I am little new at learning about stability so: This is stable because it is periodic. The reasoning being that all solutions stay within a range of the solution $cos(2t)$, between $-1$ and $1$. Is this it?
$endgroup$
– Nalt
Dec 14 '18 at 18:31
add a comment |
$begingroup$
Since all solutions are periodic, this is stable but not asymptotically stable: solutions that start close stay close, but do not converge to your solution.
answered Dec 14 '18 at 18:28
Robert IsraelRobert Israel
324k23213467
$endgroup$
$begingroup$
Hi Robert, thanks for your answer. I am little new at learning about stability so: This is stable because it is periodic. The reasoning being that all solutions stay within a range of the solution $cos(2t)$, between $-1$ and $1$. Is this it?
$endgroup$
– Nalt
Dec 14 '18 at 18:31
add a comment |
$begingroup$
Since all solutions are periodic, this is stable but not asymptotically stable: solutions that start close stay close, but do not converge to your solution.
answered Dec 14 '18 at 18:28
Robert IsraelRobert Israel
324k23213467
$endgroup$
Since all solutions are periodic, this is stable but not asymptotically stable: solutions that start close stay close, but do not converge to your solution.
answered Dec 14 '18 at 18:28
Robert IsraelRobert Israel
324k23213467
answered Dec 14 '18 at 18:28
Robert IsraelRobert Israel
324k23213467
answered Dec 14 '18 at 18:28
Robert IsraelRobert Israel
324k23213467
answered Dec 14 '18 at 18:28
Robert IsraelRobert Israel
324k23213467
324k23213467
$begingroup$
Hi Robert, thanks for your answer. I am little new at learning about stability so: This is stable because it is periodic. The reasoning being that all solutions stay within a range of the solution $cos(2t)$, between $-1$ and $1$. Is this it?
$endgroup$
– Nalt
Dec 14 '18 at 18:31
add a comment |
$begingroup$
Hi Robert, thanks for your answer. I am little new at learning about stability so: This is stable because it is periodic. The reasoning being that all solutions stay within a range of the solution $cos(2t)$, between $-1$ and $1$. Is this it?
$endgroup$
– Nalt
Dec 14 '18 at 18:31
$begingroup$
Hi Robert, thanks for your answer. I am little new at learning about stability so: This is stable because it is periodic. The reasoning being that all solutions stay within a range of the solution $cos(2t)$, between $-1$ and $1$. Is this it?
$endgroup$
– Nalt
Dec 14 '18 at 18:31
$begingroup$
Hi Robert, thanks for your answer. I am little new at learning about stability so: This is stable because it is periodic. The reasoning being that all solutions stay within a range of the solution $cos(2t)$, between $-1$ and $1$. Is this it?
$endgroup$
– Nalt
Dec 14 '18 at 18:31
add a comment |
$begingroup$
Note that :
$$-1 leq sin(2t) leq 1 quad text{and} quad -1 leq cos(2t) leq 1$$
That means that your solution will be bounded, by let's say, some $m in mathbb R$ and $M in mathbb R$, such that :
$$m leq y(t) leq M$$
This automatically tells us that the system is stable, since its solutions do not diverge for any values of $c_1,c_2 in mathbb R$ arbitrary constants.
But, note that the limits $lim_{tto infty} sin(2t)$ and $lim_{tto infty} cos(2t)$ do not exists, which means that your solutions will not get arbitrary close to each other.
Thus, the given ODE solution is simply stable.
For a more formal definition (which proves our conclusions above) recall that :
Definition : A solution of an ODE is stable if for every $varepsilon >0$ there exists $delta >0 $ when if $hat{y}(t)$ satisfies the ODE, it is :
$$|hat{y}(t) - y(t_0) | leq delta implies |hat{y}(t) - y(t)| leq varepsilon ; forall t geq t_0$$
answered Dec 14 '18 at 18:35
RebellosRebellos
14.8k31248
$endgroup$
$begingroup$
Thanks Rebellos! Your answer and Prof. Israel's really helped me to understand!
$endgroup$
– Nalt
Dec 14 '18 at 18:37
1
$begingroup$
@Nalt Glad I could help ! You may also use the "vote up" arrows (buttons) to reward users for answering something that helps you !
$endgroup$
– Rebellos
Dec 14 '18 at 18:38
add a comment |
$begingroup$
Note that :
$$-1 leq sin(2t) leq 1 quad text{and} quad -1 leq cos(2t) leq 1$$
That means that your solution will be bounded, by let's say, some $m in mathbb R$ and $M in mathbb R$, such that :
$$m leq y(t) leq M$$
This automatically tells us that the system is stable, since its solutions do not diverge for any values of $c_1,c_2 in mathbb R$ arbitrary constants.
But, note that the limits $lim_{tto infty} sin(2t)$ and $lim_{tto infty} cos(2t)$ do not exists, which means that your solutions will not get arbitrary close to each other.
Thus, the given ODE solution is simply stable.
For a more formal definition (which proves our conclusions above) recall that :
Definition : A solution of an ODE is stable if for every $varepsilon >0$ there exists $delta >0 $ when if $hat{y}(t)$ satisfies the ODE, it is :
$$|hat{y}(t) - y(t_0) | leq delta implies |hat{y}(t) - y(t)| leq varepsilon ; forall t geq t_0$$
answered Dec 14 '18 at 18:35
RebellosRebellos
14.8k31248
$endgroup$
$begingroup$
Thanks Rebellos! Your answer and Prof. Israel's really helped me to understand!
$endgroup$
– Nalt
Dec 14 '18 at 18:37
1
$begingroup$
@Nalt Glad I could help ! You may also use the "vote up" arrows (buttons) to reward users for answering something that helps you !
$endgroup$
– Rebellos
Dec 14 '18 at 18:38
add a comment |
$begingroup$
Note that :
$$-1 leq sin(2t) leq 1 quad text{and} quad -1 leq cos(2t) leq 1$$
That means that your solution will be bounded, by let's say, some $m in mathbb R$ and $M in mathbb R$, such that :
$$m leq y(t) leq M$$
This automatically tells us that the system is stable, since its solutions do not diverge for any values of $c_1,c_2 in mathbb R$ arbitrary constants.
But, note that the limits $lim_{tto infty} sin(2t)$ and $lim_{tto infty} cos(2t)$ do not exists, which means that your solutions will not get arbitrary close to each other.
Thus, the given ODE solution is simply stable.
For a more formal definition (which proves our conclusions above) recall that :
Definition : A solution of an ODE is stable if for every $varepsilon >0$ there exists $delta >0 $ when if $hat{y}(t)$ satisfies the ODE, it is :
$$|hat{y}(t) - y(t_0) | leq delta implies |hat{y}(t) - y(t)| leq varepsilon ; forall t geq t_0$$
answered Dec 14 '18 at 18:35
RebellosRebellos
14.8k31248
$endgroup$
Note that :
$$-1 leq sin(2t) leq 1 quad text{and} quad -1 leq cos(2t) leq 1$$
That means that your solution will be bounded, by let's say, some $m in mathbb R$ and $M in mathbb R$, such that :
$$m leq y(t) leq M$$
This automatically tells us that the system is stable, since its solutions do not diverge for any values of $c_1,c_2 in mathbb R$ arbitrary constants.
But, note that the limits $lim_{tto infty} sin(2t)$ and $lim_{tto infty} cos(2t)$ do not exists, which means that your solutions will not get arbitrary close to each other.
Thus, the given ODE solution is simply stable.
For a more formal definition (which proves our conclusions above) recall that :
Definition : A solution of an ODE is stable if for every $varepsilon >0$ there exists $delta >0 $ when if $hat{y}(t)$ satisfies the ODE, it is :
$$|hat{y}(t) - y(t_0) | leq delta implies |hat{y}(t) - y(t)| leq varepsilon ; forall t geq t_0$$
answered Dec 14 '18 at 18:35
RebellosRebellos
14.8k31248
edited Dec 14 '18 at 18:37
answered Dec 14 '18 at 18:35
RebellosRebellos
14.8k31248
answered Dec 14 '18 at 18:35
RebellosRebellos
14.8k31248
answered Dec 14 '18 at 18:35
RebellosRebellos
14.8k31248
14.8k31248
$begingroup$
Thanks Rebellos! Your answer and Prof. Israel's really helped me to understand!
$endgroup$
– Nalt
Dec 14 '18 at 18:37
1
$begingroup$
@Nalt Glad I could help ! You may also use the "vote up" arrows (buttons) to reward users for answering something that helps you !
$endgroup$
– Rebellos
Dec 14 '18 at 18:38
add a comment |
$begingroup$
Thanks Rebellos! Your answer and Prof. Israel's really helped me to understand!
$endgroup$
– Nalt
Dec 14 '18 at 18:37
1
$begingroup$
@Nalt Glad I could help ! You may also use the "vote up" arrows (buttons) to reward users for answering something that helps you !
$endgroup$
– Rebellos
Dec 14 '18 at 18:38
$begingroup$
Thanks Rebellos! Your answer and Prof. Israel's really helped me to understand!
$endgroup$
– Nalt
Dec 14 '18 at 18:37
$begingroup$
Thanks Rebellos! Your answer and Prof. Israel's really helped me to understand!
$endgroup$
– Nalt
Dec 14 '18 at 18:37
1
1
$begingroup$
@Nalt Glad I could help ! You may also use the "vote up" arrows (buttons) to reward users for answering something that helps you !
$endgroup$
– Rebellos
Dec 14 '18 at 18:38
$begingroup$
@Nalt Glad I could help ! You may also use the "vote up" arrows (buttons) to reward users for answering something that helps you !
$endgroup$
– Rebellos
Dec 14 '18 at 18:38
add a comment |
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