Solving non-linear ODE with cosine term [closed]

I want to find an equation of motion for non-uniform circular motion under the inlfuence of gravity (like, say, a pendulum) of the form r(t)=R cos⁡(θ(t)) i + R sin⁡(θ(t)) j where $$\theta (t)$$ is only influenced by gravity, and is measured like a trigonometric angle in the unit circle (counter-clockwise from x axis).

The Lagrangian of the system would be:

$$L = \frac{1}{2}m{R^2}{\dot \theta ^2} - mgR\sin \theta$$

I'd then apply the E-L equation and get:

$$\frac{{\partial L}}{{\partial \theta }} = - mgR\cos \theta$$ $$\frac{{\partial L}}{{\partial \dot \theta }} = m{R^2}\dot \theta$$

$$\frac{d}{{dt}}\left( {\frac{{\partial L}}{{\partial \dot \theta }}} \right) = m{R^2}\ddot \theta$$

And the differential equation I'd need to solve would be:

$$\ddot \theta = - \frac{g}{R}\cos \theta$$

The initial conditions I had in mind were that $$\dot \theta (t=0) = \dot \theta _0$$ while $$\theta (t=0) = 0$$.

I, however, have no clue how to deal with the cosine term. I know that this non-linear ODE has no analytical solution, but I have no clue how to approximate it either. I cannot use small angles because I want $$\theta$$ to be able to do a full circle.

Any help would be appreciated, cheers.

• If you don't want to make an approximation you could just solve it numerically using Eulers method or RK4 Mar 29, 2020 at 21:22
• since there is no analytical solution you have to do it numerically. but you can find numerical solutions as graphs in the net Mar 29, 2020 at 21:22
• I know that this non-linear ODE has no analytical solution It does have an analytic solution in terms of the Jacobi elliptic sine function sn. See pgccphy.net/ref/nonlin-pendulum.pdf Mar 29, 2020 at 21:28
• Your initial conditions are strange. You want to call the initial angular velocity $\theta_0$? Mar 29, 2020 at 21:36
• Exact solution for thé non-linear pendulum. Mar 29, 2020 at 22:48

You can use $$E=\frac 12 \dot \theta^2+ V(\theta)$$ being constant and separate variables to get something like $$\int_0^t dt +const. = \int_0^{\theta(t)} \frac{ d\theta}{\sqrt{2(V(\theta)-E)}},$$ but, except for the case that the pendulum starts at rest at the topmost point, you are looking at an elliptic integral on the RHS. The start-at-top case is easy though. The answer (measuring $$\theta$$ from the topmost point) is $$\theta(t)= 4 \tan^{-1} (e^{At})$$ for some constant $$A$$ depending on $$m$$, $$g$$, $$R$$ etc. The general solution requires Jacobi functions.
• Pretty sure the constant $A$ won't depend on $m$. Also, the kinetic energy should be $\frac{1}{2} m R^2 \dot{\theta}^2$. Mar 29, 2020 at 22:57
• In that case, the constant $A$ should be a function of 1, 1, 1, etc. :-) Mar 29, 2020 at 23:00
• @G.Smith At top plus $\epsilon$... It takes infinitely long for the $\theta(t)=4\tan^{-1} (e^t)$ to make the round trip. This is the space part of the Sine-Gordon Soliton solution. Mar 30, 2020 at 11:58