Damped & Negatively damped harmonic oscillator [not physics question]












1












$begingroup$


$$frac{d^2Q(t)}{dt^2} + frac{omega(t)}{2pi} frac{dQ(t)}{dt} + omega^{2}(t)Q(t) = 0, $$



where $ omega(t)= frac{2pi}{v(t)} frac{dv}{dt} $ and $v>0$ is a known parameter which is given from a data feed.



I have already built some visualization by using mathematica.



Can I transform this equation by using the variable $T(t)$ instead of $t$?:



$$frac{1}{T(t)}= frac{1}{v(t)} frac{dv}{dt} .$$






ordinary-differential-equations







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  • $begingroup$
    So you want to compute $Q(T)$?
    $endgroup$
    – Andrei
    Jan 18 at 21:37










  • $begingroup$
    I need a more explicit equation because the relaxation time $tau = 2T $
    $endgroup$
    – Bouarfa Mahi
    Jan 18 at 21:42












  • $begingroup$
    Great first post and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
    $endgroup$
    – dantopa
    Jan 18 at 22:23
















1












$begingroup$


$$frac{d^2Q(t)}{dt^2} + frac{omega(t)}{2pi} frac{dQ(t)}{dt} + omega^{2}(t)Q(t) = 0, $$



where $ omega(t)= frac{2pi}{v(t)} frac{dv}{dt} $ and $v>0$ is a known parameter which is given from a data feed.



I have already built some visualization by using mathematica.



Can I transform this equation by using the variable $T(t)$ instead of $t$?:



$$frac{1}{T(t)}= frac{1}{v(t)} frac{dv}{dt} .$$






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    So you want to compute $Q(T)$?
    $endgroup$
    – Andrei
    Jan 18 at 21:37










  • $begingroup$
    I need a more explicit equation because the relaxation time $tau = 2T $
    $endgroup$
    – Bouarfa Mahi
    Jan 18 at 21:42












  • $begingroup$
    Great first post and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
    $endgroup$
    – dantopa
    Jan 18 at 22:23














1












1








1





$begingroup$


$$frac{d^2Q(t)}{dt^2} + frac{omega(t)}{2pi} frac{dQ(t)}{dt} + omega^{2}(t)Q(t) = 0, $$



where $ omega(t)= frac{2pi}{v(t)} frac{dv}{dt} $ and $v>0$ is a known parameter which is given from a data feed.



I have already built some visualization by using mathematica.



Can I transform this equation by using the variable $T(t)$ instead of $t$?:



$$frac{1}{T(t)}= frac{1}{v(t)} frac{dv}{dt} .$$






ordinary-differential-equations







share|cite|improve this question











$endgroup$




$$frac{d^2Q(t)}{dt^2} + frac{omega(t)}{2pi} frac{dQ(t)}{dt} + omega^{2}(t)Q(t) = 0, $$



where $ omega(t)= frac{2pi}{v(t)} frac{dv}{dt} $ and $v>0$ is a known parameter which is given from a data feed.



I have already built some visualization by using mathematica.



Can I transform this equation by using the variable $T(t)$ instead of $t$?:



$$frac{1}{T(t)}= frac{1}{v(t)} frac{dv}{dt} .$$






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








edited Jan 18 at 22:20









Dog_69

6361523




6361523










asked Jan 18 at 21:31









Bouarfa MahiBouarfa Mahi

84




84












  • $begingroup$
    So you want to compute $Q(T)$?
    $endgroup$
    – Andrei
    Jan 18 at 21:37










  • $begingroup$
    I need a more explicit equation because the relaxation time $tau = 2T $
    $endgroup$
    – Bouarfa Mahi
    Jan 18 at 21:42












  • $begingroup$
    Great first post and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
    $endgroup$
    – dantopa
    Jan 18 at 22:23


















  • $begingroup$
    So you want to compute $Q(T)$?
    $endgroup$
    – Andrei
    Jan 18 at 21:37










  • $begingroup$
    I need a more explicit equation because the relaxation time $tau = 2T $
    $endgroup$
    – Bouarfa Mahi
    Jan 18 at 21:42












  • $begingroup$
    Great first post and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
    $endgroup$
    – dantopa
    Jan 18 at 22:23
















$begingroup$
So you want to compute $Q(T)$?
$endgroup$
– Andrei
Jan 18 at 21:37




$begingroup$
So you want to compute $Q(T)$?
$endgroup$
– Andrei
Jan 18 at 21:37












$begingroup$
I need a more explicit equation because the relaxation time $tau = 2T $
$endgroup$
– Bouarfa Mahi
Jan 18 at 21:42






$begingroup$
I need a more explicit equation because the relaxation time $tau = 2T $
$endgroup$
– Bouarfa Mahi
Jan 18 at 21:42














$begingroup$
Great first post and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 18 at 22:23




$begingroup$
Great first post and welcome to Mathematics Stack Exchange! A quick tour will enhance your experience.
$endgroup$
– dantopa
Jan 18 at 22:23










1 Answer
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$begingroup$

This transformation is possible iff $T(t)$ is a bijection. You can for example guarantee this by demanding that
$$0< frac{d}{dt} T(t)^{-1} = frac{d^2}{dt^2} ln nu(t),.$$



Assuming that the above holds, we can invert the relation and obtain $t(T)$. You are asking what differential equation $f(T)=Q(t(T))$ obeys. The relevant chain rule reads
$$frac{dQ}{dt}= frac{df(T(t))}{dt} = frac{df}{dT} frac{dT}{dt}
= -frac{df}{dT} frac{2pi omega'}{omega^2}$$
where we have used that
$$frac{dT}{dt} = frac{d}{dt} left(frac{2pi}{omega} right)= -frac{2pi omega'}{omega^2} ,.$$



Similarly, we obtain
$$ frac{d^2 Q}{dt^2} = frac{d}{dt}left( -frac{df}{dT} frac{2pi omega'}{omega^2}right) = frac{d^2 f}{dT^2}left(frac{2pi omega'}{omega^2}right)^2 -frac{df}{dT} underbrace{frac{d}{dt} left(frac{2pi omega'}{omega^2}right)}_{=2pi(omega omega''-2 omega')/omega^3},. $$



As a result, your differential equation transforms into
$$left(frac{2pi omega'}{omega^2}right)^2f''-frac{omega^2 omega'+2 pi (omega omega
''-2 omega'^2)}{omega^3} f' + omega^2 f =0,.$$






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    1 Answer
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    1 Answer
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    active

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    0












    $begingroup$

    This transformation is possible iff $T(t)$ is a bijection. You can for example guarantee this by demanding that
    $$0< frac{d}{dt} T(t)^{-1} = frac{d^2}{dt^2} ln nu(t),.$$



    Assuming that the above holds, we can invert the relation and obtain $t(T)$. You are asking what differential equation $f(T)=Q(t(T))$ obeys. The relevant chain rule reads
    $$frac{dQ}{dt}= frac{df(T(t))}{dt} = frac{df}{dT} frac{dT}{dt}
    = -frac{df}{dT} frac{2pi omega'}{omega^2}$$
    where we have used that
    $$frac{dT}{dt} = frac{d}{dt} left(frac{2pi}{omega} right)= -frac{2pi omega'}{omega^2} ,.$$



    Similarly, we obtain
    $$ frac{d^2 Q}{dt^2} = frac{d}{dt}left( -frac{df}{dT} frac{2pi omega'}{omega^2}right) = frac{d^2 f}{dT^2}left(frac{2pi omega'}{omega^2}right)^2 -frac{df}{dT} underbrace{frac{d}{dt} left(frac{2pi omega'}{omega^2}right)}_{=2pi(omega omega''-2 omega')/omega^3},. $$



    As a result, your differential equation transforms into
    $$left(frac{2pi omega'}{omega^2}right)^2f''-frac{omega^2 omega'+2 pi (omega omega
    ''-2 omega'^2)}{omega^3} f' + omega^2 f =0,.$$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      This transformation is possible iff $T(t)$ is a bijection. You can for example guarantee this by demanding that
      $$0< frac{d}{dt} T(t)^{-1} = frac{d^2}{dt^2} ln nu(t),.$$



      Assuming that the above holds, we can invert the relation and obtain $t(T)$. You are asking what differential equation $f(T)=Q(t(T))$ obeys. The relevant chain rule reads
      $$frac{dQ}{dt}= frac{df(T(t))}{dt} = frac{df}{dT} frac{dT}{dt}
      = -frac{df}{dT} frac{2pi omega'}{omega^2}$$
      where we have used that
      $$frac{dT}{dt} = frac{d}{dt} left(frac{2pi}{omega} right)= -frac{2pi omega'}{omega^2} ,.$$



      Similarly, we obtain
      $$ frac{d^2 Q}{dt^2} = frac{d}{dt}left( -frac{df}{dT} frac{2pi omega'}{omega^2}right) = frac{d^2 f}{dT^2}left(frac{2pi omega'}{omega^2}right)^2 -frac{df}{dT} underbrace{frac{d}{dt} left(frac{2pi omega'}{omega^2}right)}_{=2pi(omega omega''-2 omega')/omega^3},. $$



      As a result, your differential equation transforms into
      $$left(frac{2pi omega'}{omega^2}right)^2f''-frac{omega^2 omega'+2 pi (omega omega
      ''-2 omega'^2)}{omega^3} f' + omega^2 f =0,.$$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        This transformation is possible iff $T(t)$ is a bijection. You can for example guarantee this by demanding that
        $$0< frac{d}{dt} T(t)^{-1} = frac{d^2}{dt^2} ln nu(t),.$$



        Assuming that the above holds, we can invert the relation and obtain $t(T)$. You are asking what differential equation $f(T)=Q(t(T))$ obeys. The relevant chain rule reads
        $$frac{dQ}{dt}= frac{df(T(t))}{dt} = frac{df}{dT} frac{dT}{dt}
        = -frac{df}{dT} frac{2pi omega'}{omega^2}$$
        where we have used that
        $$frac{dT}{dt} = frac{d}{dt} left(frac{2pi}{omega} right)= -frac{2pi omega'}{omega^2} ,.$$



        Similarly, we obtain
        $$ frac{d^2 Q}{dt^2} = frac{d}{dt}left( -frac{df}{dT} frac{2pi omega'}{omega^2}right) = frac{d^2 f}{dT^2}left(frac{2pi omega'}{omega^2}right)^2 -frac{df}{dT} underbrace{frac{d}{dt} left(frac{2pi omega'}{omega^2}right)}_{=2pi(omega omega''-2 omega')/omega^3},. $$



        As a result, your differential equation transforms into
        $$left(frac{2pi omega'}{omega^2}right)^2f''-frac{omega^2 omega'+2 pi (omega omega
        ''-2 omega'^2)}{omega^3} f' + omega^2 f =0,.$$






        share|cite|improve this answer









        $endgroup$



        This transformation is possible iff $T(t)$ is a bijection. You can for example guarantee this by demanding that
        $$0< frac{d}{dt} T(t)^{-1} = frac{d^2}{dt^2} ln nu(t),.$$



        Assuming that the above holds, we can invert the relation and obtain $t(T)$. You are asking what differential equation $f(T)=Q(t(T))$ obeys. The relevant chain rule reads
        $$frac{dQ}{dt}= frac{df(T(t))}{dt} = frac{df}{dT} frac{dT}{dt}
        = -frac{df}{dT} frac{2pi omega'}{omega^2}$$
        where we have used that
        $$frac{dT}{dt} = frac{d}{dt} left(frac{2pi}{omega} right)= -frac{2pi omega'}{omega^2} ,.$$



        Similarly, we obtain
        $$ frac{d^2 Q}{dt^2} = frac{d}{dt}left( -frac{df}{dT} frac{2pi omega'}{omega^2}right) = frac{d^2 f}{dT^2}left(frac{2pi omega'}{omega^2}right)^2 -frac{df}{dT} underbrace{frac{d}{dt} left(frac{2pi omega'}{omega^2}right)}_{=2pi(omega omega''-2 omega')/omega^3},. $$



        As a result, your differential equation transforms into
        $$left(frac{2pi omega'}{omega^2}right)^2f''-frac{omega^2 omega'+2 pi (omega omega
        ''-2 omega'^2)}{omega^3} f' + omega^2 f =0,.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 19 at 1:54









        FabianFabian

        20.1k3774




        20.1k3774






























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