I am wondering how I can convert the function $\theta'' + k \sin(\theta) = 0$ into a system of first order ODEs? I understand that I should set for example:
$y_2' = \theta''$
$y_2 = y_1' = \theta'$
$y_1 = \theta $
And then generally it is easy to set up the $(y_2', y_1')$ matrix. However I am wondering how I can do it as the variable is a function of sine?
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Converting a $2^{\rm nd}$ order ODE $\theta''= F(t,\theta,\theta')$ for a scalar function $t\mapsto \theta(t)$ into a system of $2$ first order ODEs is easy: Introduce new unknown functions $$\theta_0(t):=\theta(t),\qquad \theta_1(t):=\theta'(t)\ .$$ Then you have to deal with the system $$\eqalign{\theta_0'&=\theta_1\cr \theta_1'&=F(t,\theta_0,\theta_1)\ .\cr}$$ In the case at hand this amounts to $$\eqalign{\theta_0'&=\theta_1\cr \theta_1'&=-k\>\sin\theta_0\ .\cr}$$ This rewriting has not converted the given ODE into a linear system. At this point one either restricts to "small $|\theta|$", which allows to replace $\sin\theta$ with $\theta$, or one has to go into a full-fledged global discussion of the problem. This would then allow also for pendula making full turns, or the interesting case when the pendulum comes to a stop in a labile position at $t=\infty$.
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You need to bring in three suitable auxiliary parameters $ p, \phi, \theta_1 : $
$$ \theta_1^{\prime} = \sin \phi / k \tag{1}$$
$$ \phi^{\prime} = \sin \theta_1/ p \tag{2}$$
with constraint function
$$ \cos \phi = - p \cos \theta_1 \tag{3}$$
where $$\theta_1 = \theta/2 \tag {4} $$
the half angle.
I like to know the significance of these parameters in the pendulum problem.
Set $y_1 = \theta$, and then we have:
$y_1' = \theta' = y_2$
$y_2' = \theta'' = -k \sin y_1$
As a system, we would write $Y = A y + f$, that is:
$$Y = \begin{pmatrix} y_1' \\ y_2' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} y_1 \\ y_2 \end{pmatrix} + \begin{pmatrix} 0 \\ -k \sin y_1 \end{pmatrix}$$