Editing Double pendulum (section)

===Lagrangian===
The [[Lagrangian mechanics|Lagrangian]] is given by
<math display="block">\begin{align}
L &= \text{kinetic energy} - \text{potential energy} \\
&= \tfrac{1}{2} m \left ( v_1^2 + v_2^2 \right ) + \tfrac{1}{2} I \left ( \dot\theta_1^2 + \dot\theta_2^2 \right ) - m g \left ( y_1 + y_2 \right ) \\
&= \tfrac{1}{2} m \left ( \dot x_1^2 + \dot y_1^2 + \dot x_2^2 + \dot y_2^2 \right ) + \tfrac{1}{2} I \left ( \dot\theta_1^2 + \dot\theta_2^2 \right ) - m g \left ( y_1 + y_2 \right )
\end{align}</math>
The first term is the ''linear'' [[kinetic energy]] of the [[center of mass]] of the bodies and the second term is the ''rotational'' kinetic energy around the center of mass of each rod. The last term is the [[potential energy]] of the bodies in a uniform gravitational field. The [[Newton's notation|dot-notation]] indicates the [[time derivative]] of the variable in question.

Using the values of <math>x_1</math> and <math>y_1</math> defined above, we have
<math display="block">
\begin{align}
\dot x_1 &= \dot \theta_1 \left(\tfrac{1}{2}\ell \cos \theta_1 \right)  \\[1ex]
\dot y_1 &= \dot \theta_1 \left(\tfrac{1}{2} \ell \sin \theta_1 \right)
\end{align} 
</math>
which leads to
<math display="block">
v_1^2 = \dot x_1^2 + \dot y_1^2
= \tfrac{1}{4} \dot \theta_1^2 \ell^2 \left(\cos^2 \theta_1 + \sin^2 \theta_1 \right)
= \tfrac{1}{4} \ell^2 \dot \theta_1^2 .
</math>

Similarly, for <math>x_2</math> and <math>y_2</math> we have
<math display="block">
\begin{align}
\dot x_2 &= \ell \left(\dot \theta_1 \cos \theta_1 + \tfrac{1}{2} \dot \theta_2 \cos \theta_2 \right) \\
\dot y_2 &= \ell \left(\dot \theta_1 \sin \theta_1 + \tfrac{1}{2} \dot \theta_2 \sin \theta_2 \right)
\end{align} 
</math>

and therefore

<math display="block">
\begin{align}
v_2^2
&= \dot x_2^2 + \dot y_2^2 \\[1ex]
&= \ell^2 \left(
    \dot \theta_1^2 \cos^2 \theta_1 + \dot \theta_1^2 \sin^2 \theta_1
    + \tfrac{1}{4} \dot \theta_2^2 \cos^2 \theta_2 + \tfrac{1}{4} \dot \theta_2^2 \sin^2 \theta_2
    + \dot \theta_1 \dot \theta_2 \cos \theta_1 \cos \theta_2 + \dot \theta_1 \dot \theta_2 \sin \theta_1 \sin \theta_2
\right) \\[1ex]
&= \ell^2 \left(
\dot \theta_1^2 + \tfrac{1}{4} \dot \theta_2^2 
+ \dot \theta_1 \dot \theta_2 \cos \left(\theta_1 - \theta_2 \right)
 \right).
\end{align} 
</math>

Substituting the coordinates above into the definition of the Lagrangian, and rearranging the equation, gives
<math display="block">
\begin{align}
L &=
\tfrac{1}{2} m \ell^2 \left(
    \dot \theta_1^2
    + \tfrac{1}{4} \dot \theta_1^2 + \tfrac{1}{4} \dot \theta_2^2 
    + \dot \theta_1 \dot \theta_2 \cos \left(\theta_1 - \theta_2 \right)
\right)
+ \tfrac{1}{24} m \ell^2 \left( \dot \theta_1^2 + \dot \theta_2^2 \right)
- m g \left(y_1 + y_2 \right)
\\[1ex]
&= \tfrac{1}{6} m \ell^2 \left (
    \dot \theta_2^2 + 4 \dot \theta_1^2
    + 3 {\dot \theta_1} {\dot \theta_2} \cos (\theta_1-\theta_2)
\right)
+ \tfrac{1}{2} m g \ell \left ( 3 \cos \theta_1 + \cos \theta_2 \right ).
\end{align}
</math>

The equations of motion can now be derived using the [[Euler–Lagrange equation]]s, which are given by
<math display="block">
\frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}_i} - \frac{\partial L}{\partial \theta_i} = 0, \quad i = 1,2.
</math>
We begin with the equation of motion for <math>\theta_1</math>. The derivatives of the Lagrangian are given by
<math display="block">
\frac{\partial L}{\partial \theta_1} = -\tfrac{1}{2} m \ell^2 \dot{\theta}_1 \dot{\theta}_2 \sin(\theta_1 - \theta_2) - \tfrac{3}{2} mg\ell \sin\theta_1
</math>
and
<math display="block">
\frac{\partial L}{\partial \dot{\theta}_1} = \tfrac{4}{3} m\ell^2 \dot{\theta}_1 + \tfrac{1}{2} m\ell^2 \dot{\theta}_2 \cos(\theta_1-\theta_2).
</math>
Thus
<math display="block">
\frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}_1} 
= \tfrac{4}{3} m\ell^2 \ddot{\theta}_1 
+ \tfrac{1}{2} m\ell^2 \ddot{\theta}_2 \cos(\theta_1-\theta_2)
- \tfrac{1}{2} m\ell^2 \dot{\theta}_2(\dot{\theta}_1 - \dot{\theta}_2) \sin(\theta_1 - \theta_2).
</math>
Combining these results and simplifying yields the first equation of motion,
<math display="block">
\tfrac{4}{3} \ell \ddot{\theta}_1 + \tfrac{1}{2} \ell \ddot{\theta}_2 \cos(\theta_1 - \theta_2) + \tfrac{1}{2} \ell \dot{\theta}_2^2 \sin(\theta_1-\theta_2) + \tfrac{3}{2} g \sin\theta_1 = 0.
</math>

Similarly, the derivatives of the Lagrangian with respect to <math>\theta_2</math> and <math>\dot{\theta}_2</math> are given by
<math display="block">
\frac{\partial L}{\partial \theta_2} = \tfrac{1}{2} m \ell^2 \dot{\theta}_1 \dot{\theta}_2 \sin(\theta_1 - \theta_2) - \tfrac{1}{2} mg\ell \sin\theta_2
</math>
and
<math display="block">
\frac{\partial L}{\partial \dot{\theta}_2} = \tfrac{1}{3} m\ell^2 \dot{\theta}_2 + \tfrac{1}{2} m\ell^2 \dot{\theta}_1 \cos(\theta_1-\theta_2).
</math>
Thus
<math display="block">
\frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}_2} 
= \tfrac{1}{3} m\ell^2 \ddot{\theta}_2 
+ \tfrac{1}{2} m\ell^2 \ddot{\theta}_1 \cos(\theta_1-\theta_2)
- \tfrac{1}{2} m\ell^2 \dot{\theta}_1(\dot{\theta}_1 - \dot{\theta}_2) \sin(\theta_1 - \theta_2).
</math>
Plugging these results into the Euler-Lagrange equation and simplifying yields the second equation of motion,
<math display="block">
\tfrac{1}{3} \ell \ddot{\theta}_2 + \tfrac{1}{2} \ell \ddot{\theta}_1 \cos(\theta_1 - \theta_2) - \tfrac{1}{2} \ell \dot{\theta}_1^2 \sin(\theta_1-\theta_2) + \tfrac{1}{2} g \sin\theta_2 = 0.
</math>

No [[closed form expression|closed form]] solutions for <math>\theta_1</math> and <math>\theta_2</math> as functions of time are known, therefore the system can only be solved [[numerical integration|numerically]], using the [[Runge–Kutta methods|Runge Kutta method]] or [[numerical methods for ordinary differential equations|similar techniques]].

[[File:Double-pendulum.png|thumb|Parametric plot for the time evolution of the angles of a double pendulum. Note that the graph resembles [[Brownian motion]].]]