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







share|cite|improve this question









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    5












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      5












      5








      5


      1



      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 14 '18 at 17:48









      NaltNalt

      756




      756






















          2 Answers
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          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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



















          3












          $begingroup$

          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 :
          $$|hat{y}(t) - y(t_0) | leq delta implies |hat{y}(t) - y(t)| leq varepsilon ; forall t geq t_0$$







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          3












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          3












          3








          3





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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











          3












          $begingroup$

          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 :
          $$|hat{y}(t) - y(t_0) | leq delta implies |hat{y}(t) - y(t)| leq varepsilon ; forall t geq t_0$$







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          3












          $begingroup$

          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 :
          $$|hat{y}(t) - y(t_0) | leq delta implies |hat{y}(t) - y(t)| leq varepsilon ; forall t geq t_0$$







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          3












          3








          3





          $begingroup$

          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 :
          $$|hat{y}(t) - y(t_0) | leq delta implies |hat{y}(t) - y(t)| leq varepsilon ; forall t geq t_0$$







          share|cite|improve this answer











          $endgroup$



          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 :
          $$|hat{y}(t) - y(t_0) | leq delta implies |hat{y}(t) - y(t)| leq varepsilon ; forall t geq t_0$$








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 18:37

























          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


















          • $begingroup$
            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


















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