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Lagrangian corresponding to these equations of motion

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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

edited 1 hour ago

Qmechanic♦

105k121911206

asked 5 hours ago

TheAverageHijanoTheAverageHijano

4739

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

edited 1 hour ago

Qmechanic♦

105k121911206

asked 5 hours ago

TheAverageHijanoTheAverageHijano

4739

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 1 hour ago

Qmechanic♦

105k121911206

asked 5 hours ago

TheAverageHijanoTheAverageHijano

4739

$endgroup$

share|cite|improve this question

edited 1 hour ago

Qmechanic♦

105k121911206

asked 5 hours ago

TheAverageHijanoTheAverageHijano

4739

share|cite|improve this question

edited 1 hour ago

Qmechanic♦

105k121911206

asked 5 hours ago

TheAverageHijanoTheAverageHijano

4739

edited 1 hour ago

Qmechanic♦

105k121911206

edited 1 hour ago

Qmechanic♦

105k121911206

asked 5 hours ago

TheAverageHijanoTheAverageHijano

4739

asked 5 hours ago

TheAverageHijanoTheAverageHijano

4739

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

answered 4 hours ago

GRBGRB

9551722

$endgroup$

add a comment | 

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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

answered 4 hours ago

GRBGRB

9551722

$endgroup$

add a comment | 

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

answered 4 hours ago

GRBGRB

9551722

$endgroup$

add a comment | 

$begingroup$

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

answered 4 hours ago

GRBGRB

9551722

$endgroup$

share|cite|improve this answer

answered 4 hours ago

GRBGRB

9551722

answered 4 hours ago

GRBGRB

9551722

answered 4 hours ago

GRBGRB

9551722