I tried to find the equation of this pendulum, but I think I did something wrong. I know I have to get the Bessel's equation but I can't see it. It's a simple 2-D pendulum, without any dissipation. The length is $l$.

\begin{gather} x(t)= l(t) \sin\theta(t), \\ y(t)=l(t) \cos\theta(t), \\ \dot x(t)= \dot l(t) \sin\theta(t) + l(t) \dot \theta(t) \cos\theta(t) \\ \dot y(t)= \dot l(t) \cos\theta(t) - l(t) \dot \theta(t) \sin\theta \end{gather}

and: \begin{equation} \dot x^2 + \dot y^2 = \dot l^2 + l^2 \dot \theta^2 \end{equation}

The potential and kinetic energy: \begin{gather} V=-mgy= -mgl(t) \cos\theta(t), \\ T=\frac{m}{2} (\dot l^2(t)+ l^2(t) \dot \theta^2(t)) \end{gather}

So the Langrangian is: \begin{equation} L=T-V= \frac{m}{2}(\dot l^2 + l^2 \dot \theta^2) + mgl \cos\theta \end{equation}

After this, we get: \begin{gather} \frac{ \partial L}{\partial \theta} = -mgl \sin\theta, \\ \frac{ d}{dt} \frac{\partial L}{\partial \dot \theta}= m2l \dot l \dot \theta +ml^2 \ddot \theta. \end{gather}

From thism we get the equation: \begin{equation} \ddot \theta + 2 \dot l \dot \theta = g \sin\theta. \end{equation}

And for the radial part:

\begin{gather} \frac{\partial L}{\partial l} = ml \dot \theta^2 + mg \cos\theta , \\ \frac{d}{dt} \frac{\partial L}{\partial \dot l}= m \ddot l. \end{gather}

The equation is: \begin{equation} \dot l = l \dot \theta^2 + g \cos\theta \end{equation}

My question is how can I get to the Bessel's equation from here?


2 Answers 2


The correct equation of motion for $\theta$ is \begin{equation} l\ddot \theta + 2 \dot l \dot \theta = -g \sin\theta. \end{equation} Now if you assume $l(t)=l_{0}+vt$ then you will get a Bessel Differential Equations. \begin{equation} \ddot \theta+\frac{2}{l}\dot \theta+\frac{g}{v^{2}l}\theta=0 \end{equation} The solution for which is the Bessel Function. ($\sin\theta$ approximated by $\theta$ for small angle)


First of all, two tiny mistakes: your $\theta$ equation is missing an $\ell$, and the sign of $g \sin \theta$ is flipped. That is, it should be

$$ 2 \dot{\ell} \dot{\theta} + \ell \ddot{\theta} = -g \sin \theta $$

Now, I have not seen Bessel equation emerge for pendulum dynamics for arbitrary length variability. It only arises, as far as I have seen, for the linearly-changing length, that is, for $\ell=\ell_0+v t$, and even in that case, under the small-$\theta$ approximation, to the first non-zero order or $\theta$.

In this special case, the equation reduces to $(\ell_0+v t) \ddot{\theta}+2 v \dot{\theta}+g \theta = 0$. There is a change-of-variables that transforms this into Bessel's ode, which is $\psi= \theta \sqrt{\ell}$ and $\phi=\frac{2}{v}\sqrt{\ell g}$. The calculations can be googled, and for example can be found here. This eventuates in the Bessel equation for the $\psi$ and $\phi$ pair.

Another pendulum-related case that I know where Bessel's equation emerges is when the deformation of the rope from a straight line is taken into account, and the first corrections invoke Bessel equations.

Other than these two special cases, I don't think one can evoke Bessel's equation in pendulum dynamics.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.