Editing Double pendulum (section)

==Analysis and interpretation==
Several variants of the double pendulum may be considered; the two limbs may be of equal or unequal lengths and masses, they may be [[simple pendulum]]s or [[compound pendulum]]s (also called complex pendulums) and the motion may be in three dimensions or restricted to one vertical plane. In the following analysis, the limbs are taken to be identical compound pendulums of length {{mvar|ℓ}} and mass {{mvar|m}}, and the motion is restricted to two dimensions.

[[Image:Double-compound-pendulum-dimensioned.svg|right|thumb|Double compound pendulum]]
[[Image:Double pendulum simulation.gif|right|thumb|Motion of the double compound pendulum (from [[numerical integration]] of the equations of motion)]]

In a compound pendulum, the mass is distributed along its length. If the double pendulum mass is evenly distributed, then the [[center of mass]] of each limb is at its midpoint, and the limb has a [[moment of inertia]] of {{math|1=''I'' = {{sfrac|1|12}}''mℓ''<sup>2</sup>}} about that point.<!-- The moment of inertia of a rod rotating around an axis attached to one of its ends equals {{math|1=''I'' = {{sfrac|1|3}}''mℓ''<sup>2</sup>}}. -->

It is convenient to use the angles between each limb and the vertical as the [[generalized coordinates]] defining the [[Configuration space (physics)|configuration]] of the system. These angles are denoted {{math|''θ''<sub>1</sub>}} and {{math|''θ''<sub>2</sub>}}. The position of the center of mass of each rod may be written in terms of these two coordinates. If the origin of the [[Cartesian coordinate system]] is taken to be at the point of suspension of the first pendulum, then the center of mass of this pendulum is at:<math display="block">\begin{align}
x_1 &= \tfrac{1}{2} \ell \sin \theta_1 \\
y_1 &= -\tfrac{1}{2} \ell \cos \theta_1 
\end{align}</math>

and the center of mass of the second pendulum is at
<math display="block">\begin{align}
x_2 &= \ell \left ( \sin \theta_1 + \tfrac{1}{2} \sin \theta_2 \right ) \\
y_2 &= -\ell \left ( \cos \theta_1 + \tfrac{1}{2} \cos \theta_2 \right )
\end{align}</math>
This is enough information to write out the Lagrangian.

===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]].]]