Solving 2nd Order ODE: r\ddot\theta-g\sin\theta=0

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SUMMARY

The discussion focuses on solving the second-order ordinary differential equation (ODE) given by r\ddot\theta-g\sin\theta=0, where r and g are constants. Participants suggest using numerical methods such as the Runge-Kutta method and highlight the importance of elliptic integrals for analytical solutions. The conversation emphasizes the reduction of the equation to quadratures and the approximation for small angles, where \sin\theta can be approximated by \theta. A recommended resource for further understanding is Derek F. Lawden's "Elliptic Functions and Applications".

PREREQUISITES
  • Understanding of second-order ordinary differential equations
  • Familiarity with numerical methods, specifically the Runge-Kutta method
  • Knowledge of elliptic integrals and their applications
  • Basic concepts of Lagrangian dynamics and small angle approximations
NEXT STEPS
  • Study the Runge-Kutta method for numerical solutions of ODEs
  • Explore elliptic integrals and their role in solving differential equations
  • Learn about Lagrangian dynamics and how to apply small angle approximations
  • Read Derek F. Lawden's "Elliptic Functions and Applications" for deeper insights
USEFUL FOR

Mathematicians, physicists, engineers, and students studying dynamics or differential equations who seek to understand the analytical and numerical solutions of second-order ODEs.

NeutronStar
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How would I go about finding a solution to this differential equation?

r\ddot\theta-g\sin\theta=0

Where r and g are constants.
 
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Oh pooh,...

That's what I didn't want to hear! :cry:
 
Did you intend for the second term to have a plus sign?

IIRC, you should be able to reduce the solution to quadratures expressed in terms of elliptic integrals.
 
Tide said:
Did you intend for the second term to have a plus sign?

IIRC, you should be able to reduce the solution to quadratures expressed in terms of elliptic integrals.
Yes, it should be plus. Sorry about that.

Hey, thanks for the tip about the elliptic integrals! That may be just what I'm looking for! :approve:
 
NeutronStar said:
How would I go about finding a solution to this differential equation?

r\ddot\theta-g\sin\theta=0

Where r and g are constants.

If you want to solve the equations of your Lagrange Dynamics problems, you could also post it ¡n that thread you wrote. I didn't mention it to you, but the next step after writing the equations is solving them analytically. The usual assumptions made here by phsicists and engineers are to consider small displacements (i.e \theta\rightarrow 0). Then you could remove sen\theta and cos\theta of your equations and made it quasi-linears. Try to go about that, because it is the usual estrategy in Lagrange Dynamic courses.
 
Such an equation usually appears for oscillating motions.
You can get a reasonably good approximation for small angles, where:

\sin \theta \approx \theta

and

\cos \theta \approx 1-\frac{1}{2}\theta^2
 
NeutronStar said:
How would I go about finding a solution to this differential equation?

r\ddot\theta-g\sin\theta=0

Where r and g are constants.

The solution should be sinus amplitudinis,the Jacobi elliptic function.I'm sure of it.
 
I was curious as well to learn the solution of the simple pendulum ODE.The best approach i came across is the one in
Derek F.Lawden:"Elliptic Functions and Applications",Springer Verlag,1989,p.114 pp.117.
But the chapters 1 pp.3 (p.1 pp.94) are essential to understanding properly what he's doing when speaking of the simple pendulum.
 

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