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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?

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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.

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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)

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