Recall from Part 2 that for the double pendulum,
L(\theta,\dot \theta) = T - V = \frac 1 2 (m_1+m_2) \ell_1^2 \dot{\theta}_1^2 +\frac 1 2 m_2\left( 2 \dot{\theta}_1 \dot{\theta}_2\ell_1 \ell_2 \cos(\theta_2 - \theta_1) + \ell_2^2 \dot{\theta}_2^2\right) + (m_1+m_2) g \ell_1 \cos \theta_1 + m_2 g \ell_2 \cos \theta_2,
where, \theta_1, \theta_2 are the angles from vertical of the two pendulums, \dot \theta_1, \dot \theta_2 are their angular speeds, \ell_1, \ell_2 are the length of the rods, and m_1, m_2 are their masses. This looks very imposing, so let's remark a little on the structure of that equation. In simpler situations, one often sees different squared speed terms, but here we have the interaction of the two parts of the system, in terms with products like \dot \theta_1 \dot \theta_2, as well as the \cos(\theta_2 - \theta_1). This distinguishes the system from two completely isolated (uncoupled) pendulums; the Lagrangian already captures the aspect that energy gets transferred back and forth between the two pendulums.
The equations of motion are derived from considering an optimization problem, namely, what paths \gamma through state space connecting two states optimizes the action integral
\int_a^b L(\gamma(t),\dot\gamma(t)) dt.
In a process quite similar to what's done in first-year calculus, we "take a derivative and set it to zero", which derives the equations
\frac{\partial L}{\partial \theta_i} - \frac d {dt} \frac{\partial L}{\partial \dot \theta_i} = 0
where \partial L/\partial \dot \theta_i means the partial derivative with respect to \dot \theta_i as if it were an independent variable. In fact, one way is to start off with a formula involving two independent variables \theta and \omega; only after solving for the extremal path do we actually find \omega = \dot \theta (this is a generalization called the Hamilton-Pontryagin principle).
\partial L/\partial \dot {\boldsymbol \theta}, angular momentum, and the moment of inertia
Let's now do the computation \frac{\partial L}{\partial \dot \theta_1} (this quantity actually is the angular momentum of the first pendulum, p_1)—treating everything except \dot \theta_1 as a constant:p_1 = \frac{\partial L}{\partial \dot \theta_1} = (m_1 + m_2) \ell_1^2 \dot \theta_1 + m_2 \ell_1\ell_2 \cos(\theta_2-\theta_1)\dot \theta_2 .
Note that unlike the more basic examples of angular momentum, this isn't simply dependent only on the (angular) velocity \dot \theta_1, but also on \dot \theta_2 and nonlinearly on \theta_1 and \theta_2. Similarly,
p_2 = \frac{\partial L}{\partial \dot \theta_2} = m_2 \ell_2^2 \dot \theta_2 + m_2\ell_1\ell_2 \cos(\theta_2 - \theta_1) \dot \theta_1.
This is more compactly expressed as
\frac{\partial L}{\partial \dot{\boldsymbol{\theta}}}=\begin{pmatrix}p_1\\p_2\end{pmatrix} = \begin{pmatrix} (m_1 +m_2) \ell_1^2 & m_2\ell_1\ell_2\cos(\theta_2-\theta_1) \\ m_2 \ell_1 \ell_2 \cos(\theta_2 -\theta_1) & m_2 \ell_2^2\end{pmatrix} \begin{pmatrix}\dot \theta_1 \\ \dot\theta_2 \end{pmatrix}.
This helps keep the unruliness under one umbrella. The square matrix in the above will be important, so let's call it I (it is something like a moment of inertia for our system—but it's more complicated, principally because there is an interaction between its two component parts... It's not even a rigid body!). It should also be noted that the kinetic energy is T = \frac 1 2 \dot{\boldsymbol{\theta}}^t I \dot{\boldsymbol \theta}, so it still is quadratic in \dot {\boldsymbol \theta}, just like good ol' \frac{1}{2} mv^2.
The forces, \partial L/\partial \boldsymbol{\theta}
Due to the nonlinear dependence of the Lagrangian on the angular variables, the derivatives are trickier. But we forge on, because sticking to just the simplest examples won't help us gain as much insight into how things are used in practice. We have
\frac{\partial L}{\partial \theta_1} = m_2 \ell_1 \ell_2 \dot \theta_1 \dot \theta_2 \sin(\theta_2-\theta_1) - (m_1 + m_2) g \ell_1 \sin \theta_1
\frac{\partial L}{\partial \theta_1} = m_2 \ell_1 \ell_2 \dot \theta_1 \dot \theta_2 \sin(\theta_2-\theta_1) - (m_1 + m_2) g \ell_1 \sin \theta_1
and
\frac{\partial L}{\partial \theta_2} = -m_2 \ell_1 \ell_2 \dot \theta_1 \dot \theta_2 \sin(\theta_2-\theta_1) - m_2 g \ell_2 \sin \theta_2.
\frac{\partial L}{\partial \theta_2} = -m_2 \ell_1 \ell_2 \dot \theta_1 \dot \theta_2 \sin(\theta_2-\theta_1) - m_2 g \ell_2 \sin \theta_2.
Putting it together
Now we take the time derivatives of the momenta we derived above and subtract and set to zero to derive the Euler-Lagrange equations. That's straightfoward calculus; but different from simpler examples in that, of course, the chain rule must be applied to the \cos(\theta_2 - \theta_1), so there will be an extra \dot \theta_1 or \dot \theta_2 term:
\frac{d}{dt} \frac{\partial L}{\partial \dot{\boldsymbol{\theta}}} = \begin{pmatrix} 0 & -m_2\ell_1\ell_2\sin(\theta_2-\theta_1)(\dot \theta_2 - \dot \theta_1) \\ -m_2 \ell_1 \ell_2 \sin(\theta_2 -\theta_1)(\dot \theta_2 - \dot \theta_1) & 0\end{pmatrix} \begin{pmatrix}\dot \theta_1 \\ \dot\theta_2 \end{pmatrix}
+\begin{pmatrix} (m_1 +m_2) \ell_1^2 & m_2\ell_1\ell_2\cos(\theta_2-\theta_1) \\ m_2 \ell_1 \ell_2 \cos(\theta_2 -\theta_1) & m_2 \ell_2^2\end{pmatrix} \begin{pmatrix}\ddot \theta_1 \\ \ddot\theta_2 \end{pmatrix}.
The nice thing about it is that when it is subtracted from \partial L/\partial \boldsymbol \theta, the cross terms with the \dot \theta_1 \dot\theta_2 cancel. Also, the matrix multiplying the \ddot{\theta}_i's is none other than our moment-of-inertia matrix I. This finally gives us
0=\frac{\partial L}{\partial \boldsymbol\theta} - \frac{d}{dt} \frac{\partial L}{\partial \dot{\boldsymbol\theta}} =\begin{pmatrix}-(m_1+m_2) g \ell_1 \sin \theta_1 - m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_2^2 \\ -m_2 g \ell_2 \sin \theta_2 +m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_1^2 \end{pmatrix} -I \begin{pmatrix} \ddot \theta_1 \\ \ddot \theta_2 \end{pmatrix}
as the Euler-Lagrange equations. As the final equation of motion, we bring the \ddot {\boldsymbol \theta} term to the other side:
I \begin{pmatrix} \ddot \theta_1 \\ \ddot \theta_2 \end{pmatrix} = \begin{pmatrix}-(m_1+m_2) g \ell_1 \sin \theta_1 - m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_2^2 \\ -m_2 g \ell_2 \sin \theta_2 +m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_1^2 \end{pmatrix}
(keep in mind that I varies with \theta_1 and \theta_2, however). We discretize and solve this next time. For now, simply take note of the interesting nonlinear interactions: in the positions as the sine of the difference, and quadratic nonlinearity in the derivatives (note the reversal: the \ddot\theta_1 gets a \dot \theta_2^2 and vice versa).
+\begin{pmatrix} (m_1 +m_2) \ell_1^2 & m_2\ell_1\ell_2\cos(\theta_2-\theta_1) \\ m_2 \ell_1 \ell_2 \cos(\theta_2 -\theta_1) & m_2 \ell_2^2\end{pmatrix} \begin{pmatrix}\ddot \theta_1 \\ \ddot\theta_2 \end{pmatrix}.
The nice thing about it is that when it is subtracted from \partial L/\partial \boldsymbol \theta, the cross terms with the \dot \theta_1 \dot\theta_2 cancel. Also, the matrix multiplying the \ddot{\theta}_i's is none other than our moment-of-inertia matrix I. This finally gives us
0=\frac{\partial L}{\partial \boldsymbol\theta} - \frac{d}{dt} \frac{\partial L}{\partial \dot{\boldsymbol\theta}} =\begin{pmatrix}-(m_1+m_2) g \ell_1 \sin \theta_1 - m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_2^2 \\ -m_2 g \ell_2 \sin \theta_2 +m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_1^2 \end{pmatrix} -I \begin{pmatrix} \ddot \theta_1 \\ \ddot \theta_2 \end{pmatrix}
as the Euler-Lagrange equations. As the final equation of motion, we bring the \ddot {\boldsymbol \theta} term to the other side:
I \begin{pmatrix} \ddot \theta_1 \\ \ddot \theta_2 \end{pmatrix} = \begin{pmatrix}-(m_1+m_2) g \ell_1 \sin \theta_1 - m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_2^2 \\ -m_2 g \ell_2 \sin \theta_2 +m_2 \ell_1\ell_2 \sin(\theta_2-\theta_1)\dot{\theta}_1^2 \end{pmatrix}
(keep in mind that I varies with \theta_1 and \theta_2, however). We discretize and solve this next time. For now, simply take note of the interesting nonlinear interactions: in the positions as the sine of the difference, and quadratic nonlinearity in the derivatives (note the reversal: the \ddot\theta_1 gets a \dot \theta_2^2 and vice versa).
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