Lagrangian corresponding to these equations of motion












0












$begingroup$



I have the following equations of motion for a system with two degrees of freedom:
$$ddot{q_1}+q_1^2-q_2^2=0$$
and
$$ddot{q_2}+2q_1q_2=0.$$




I have tried to deduce the Lagragian corresponding to this system, but I could not figure out how to obtain the second term in each equation.



$$L=frac{1}{2}left(dot{q_1}+dot{q_2}right)-frac{q_1^3}{3}+q_1q_2^2$$



for example works for the first equation, but not for the second one. Is it possible that no Lagrangian exists for such a system?






homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators







share|cite|improve this question











$endgroup$












  • $begingroup$
    Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
    $endgroup$
    – Aaron Stevens
    4 hours ago










  • $begingroup$
    It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
    $endgroup$
    – TheAverageHijano
    4 hours ago










  • $begingroup$
    You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
    $endgroup$
    – Ballanzor
    4 hours ago
















0












$begingroup$



I have the following equations of motion for a system with two degrees of freedom:
$$ddot{q_1}+q_1^2-q_2^2=0$$
and
$$ddot{q_2}+2q_1q_2=0.$$




I have tried to deduce the Lagragian corresponding to this system, but I could not figure out how to obtain the second term in each equation.



$$L=frac{1}{2}left(dot{q_1}+dot{q_2}right)-frac{q_1^3}{3}+q_1q_2^2$$



for example works for the first equation, but not for the second one. Is it possible that no Lagrangian exists for such a system?






homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators







share|cite|improve this question











$endgroup$












  • $begingroup$
    Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
    $endgroup$
    – Aaron Stevens
    4 hours ago










  • $begingroup$
    It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
    $endgroup$
    – TheAverageHijano
    4 hours ago










  • $begingroup$
    You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
    $endgroup$
    – Ballanzor
    4 hours ago














0












0








0


1



$begingroup$



I have the following equations of motion for a system with two degrees of freedom:
$$ddot{q_1}+q_1^2-q_2^2=0$$
and
$$ddot{q_2}+2q_1q_2=0.$$




I have tried to deduce the Lagragian corresponding to this system, but I could not figure out how to obtain the second term in each equation.



$$L=frac{1}{2}left(dot{q_1}+dot{q_2}right)-frac{q_1^3}{3}+q_1q_2^2$$



for example works for the first equation, but not for the second one. Is it possible that no Lagrangian exists for such a system?






homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators







share|cite|improve this question











$endgroup$





I have the following equations of motion for a system with two degrees of freedom:
$$ddot{q_1}+q_1^2-q_2^2=0$$
and
$$ddot{q_2}+2q_1q_2=0.$$




I have tried to deduce the Lagragian corresponding to this system, but I could not figure out how to obtain the second term in each equation.



$$L=frac{1}{2}left(dot{q_1}+dot{q_2}right)-frac{q_1^3}{3}+q_1q_2^2$$



for example works for the first equation, but not for the second one. Is it possible that no Lagrangian exists for such a system?






homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators




homework-and-exercises classical-mechanics lagrangian-formalism coupled-oscillators






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Qmechanic♦

105k121911206




105k121911206










asked 5 hours ago









TheAverageHijanoTheAverageHijano

4739




4739












  • $begingroup$
    Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
    $endgroup$
    – Aaron Stevens
    4 hours ago










  • $begingroup$
    It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
    $endgroup$
    – TheAverageHijano
    4 hours ago










  • $begingroup$
    You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
    $endgroup$
    – Ballanzor
    4 hours ago


















  • $begingroup$
    Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
    $endgroup$
    – Aaron Stevens
    4 hours ago










  • $begingroup$
    It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
    $endgroup$
    – TheAverageHijano
    4 hours ago










  • $begingroup$
    You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
    $endgroup$
    – Ballanzor
    4 hours ago
















$begingroup$
Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
$endgroup$
– Aaron Stevens
4 hours ago




$begingroup$
Does this describe something physical, or are you just taking two coupled differential equations and seeing if you can pull a Lagrangian from them?
$endgroup$
– Aaron Stevens
4 hours ago












$begingroup$
It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
$endgroup$
– TheAverageHijano
4 hours ago




$begingroup$
It does not represent a physical system in the sense that $q_1$ and $q_2$ are spatial coordinates, but I am interested in calculating a possible Lagrangian for the equations in case some properties (such a conservation of a magnitude) can be exploited.
$endgroup$
– TheAverageHijano
4 hours ago












$begingroup$
You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
$endgroup$
– Ballanzor
4 hours ago




$begingroup$
You can try and deduce the lagrangian from the Lagrange equations that give such equation of motion.
$endgroup$
– Ballanzor
4 hours ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

I suppose the Lagrangian you found is actually
$$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
$$ddot{q}_2 - 2 q_1 q_2 = 0$$
There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
$$-ddot{q}_2 - 2 q_1 q_2 = 0$$
This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
$$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$






share|cite|improve this answer









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    1 Answer
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    1 Answer
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    $begingroup$

    I suppose the Lagrangian you found is actually
    $$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
    Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
    $$ddot{q}_2 - 2 q_1 q_2 = 0$$
    There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
    $$-ddot{q}_2 - 2 q_1 q_2 = 0$$
    This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
    $$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      I suppose the Lagrangian you found is actually
      $$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
      Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
      $$ddot{q}_2 - 2 q_1 q_2 = 0$$
      There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
      $$-ddot{q}_2 - 2 q_1 q_2 = 0$$
      This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
      $$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        I suppose the Lagrangian you found is actually
        $$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
        Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
        $$ddot{q}_2 - 2 q_1 q_2 = 0$$
        There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
        $$-ddot{q}_2 - 2 q_1 q_2 = 0$$
        This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
        $$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$






        share|cite|improve this answer









        $endgroup$



        I suppose the Lagrangian you found is actually
        $$L = frac{1}{2} left( dot{q}_1^2 + dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$
        Well, your solution is already almost correct. If you calculate the equation for $q_2$, you can find
        $$ddot{q}_2 - 2 q_1 q_2 = 0$$
        There's no way that I can see to change the sign of the second term without affecting the first equation too. But you can act on the first term, and if that becomes negative too, then the second equation matches the one you posted.
        $$-ddot{q}_2 - 2 q_1 q_2 = 0$$
        This can be achieved simply by changing the sign of $dot{q}_2^2$ in the Lagrangian:
        $$L = frac{1}{2} left( dot{q}_1^2 - dot{q}_2^2 right) - frac{q_1^3}{3} + q_1 q_2^2$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 4 hours ago









        GRBGRB

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        9551722






























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