Figure 1 shows a sketch of a double pendulum: two identical rods of length $2L$ and mass $m$ are freely hinged together at one end, and the linked pair is freely pivoted to a fixed support at one end. We can describe the position of the two pendulums at any instant with the two angles $\theta$ and $\phi$ as shown.

We can derive the governing equations for free motion by using Lagrange’s equations. First, we need expressions for the potential and kinetic energies. In terms of Cartesian x-y axes with an origin at the fixed pivot, the centre of mass of the top pendulum is at the point

$$[L \sin \theta, -L \cos \theta] \tag{1}$$

while that of the lower pendulum is at the point

$$[2L \sin \theta + L \sin \phi, -2L \cos \theta – L \cos \phi]. \tag{2}$$

We can immediately write down an expression for the potential energy, which is entirely due to gravity:

$$V=mgL(1 – \cos \theta) +mgL[2(1- \cos \theta) + (1-\cos \phi)] \tag{3}$$

where the reference configuration with $V=0$ has been taken to be the equilibrium position $\theta=\phi=0$.

For the kinetic energy, we first need the velocities of the two centres of mass. For the top pendulum this has components

$$L \dot{\theta} [\cos \theta, \sin \theta] \tag{4}$$

while for the lower pendulum the components are

$$2L \dot{\theta} [\cos \theta, \sin \theta] + L \dot{\phi} [\cos \phi, \sin \phi]. \tag{5}$$

We can now write down an expression for the kinetic energy: for each pendulum separately, we can use the general rigid-body expression $(1/2)m v_G^2 + (1/2) I_G \Omega^2$ where $m$ is the mass, $I_G$ is the moment of inertia about the centre of mass, $v_G$ is the magnitude of the velocity of the centre of mass, and $\Omega$ is the angular velocity.

The result is

$$T=\dfrac{1}{2}mL^2 \dot{\theta}^2 (\cos^2 \theta + \sin^2 \theta) +\dfrac{1}{2}\left(\dfrac{1}{3}mL^2\right) \dot{\theta}^2+\dfrac{1}{2}\left(\dfrac{1}{3}mL^2\right) \dot{\phi}^2$$

$$+ \dfrac{1}{2}mL^2 \left[ \left(2 \dot{\theta}\cos \theta + \dot{\phi}\cos \phi\right)^2 + \left(2 \dot{\theta}\sin \theta + \dot{\phi}\sin \phi\right)^2 \right] \tag{6}$$

which, after multiplying out and simplifying, comes down to

$$T=\dfrac{1}{2}mL^2 \left[ \dfrac{16}{3} \dot{\theta}^2 + \dfrac{4}{3} \dot{\phi}^2 + 4 \dot{\theta} \dot{\phi} \cos (\phi-\theta) \right] . \tag{7}$$

Now we use the Lagrange equations. Applied to the angle $\theta$, the equation is

$$\dfrac{d}{dt}\left[ \dfrac{\partial T}{\partial \dot{\theta}} \right] – \dfrac{\partial T}{\partial \theta} + \dfrac{\partial V}{\partial \theta} = 0 \tag{8}$$

leading to

$$\dfrac{d}{dt}\left[\dfrac{16}{3}mL^2 \dot{\theta} + 2mL^2 \dot{\phi} \cos (\phi – \theta) \right] $$

$$-2mL^2 \dot{\theta} \dot{\phi} \sin (\phi – \theta) + 3mgL \sin \theta = 0 \tag{9}$$

which expands and simplifies to the final result

$$\dfrac{16}{3}\ddot{\theta}+ 2\ddot{\phi}\cos (\phi – \theta) – 2 \dot{\phi}^2 \sin (\phi – \theta) + \dfrac{3g}{L} \sin \theta = 0. \tag{10}$$

The corresponding Lagrange equation for the angle $\phi$ gives

$$\dfrac{d}{dt}\left[\dfrac{4}{3}mL^2 \dot{\phi} + 2mL^2 \dot{\theta} \cos (\phi – \theta) \right] $$

$$+2mL^2 \dot{\theta} \dot{\phi} \sin (\phi – \theta) + mgL \sin \phi = 0 \tag{11}$$

which expands and simplifies to give

$$\dfrac{4}{3}\ddot{\phi}+ 2\ddot{\theta}\cos (\phi – \theta) + 2 \dot{\theta}^2 \sin (\phi – \theta) + \dfrac{g}{L} \sin \phi = 0. \tag{12}$$

Equations (10) and (12) determine the free motion of the pendulum system. Because we have two second-order differential equations, we need four initial values: the most natural things to specify are the values at time $t=0$ of $\theta$, $\dot{\theta}$, $\phi$ and $\dot{\phi}$.