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The classical (undamped) pendulum ODE is

$$\ddot \theta = -\frac{g}{\ell} \sin(\theta)$$

Defining state ${\bf x} := (\theta, \dot \theta)$, we have a system of $2$ ODEs, $\dot {\bf x} = {\bf f} ({\bf x})$, where ${\bf f} : \Bbb R^2 \to \Bbb R^2$.

Consider the change of coordinates ${\bf y} = {\bf \Phi} ({\bf x})$, where vector field ${\bf \Phi} : \Bbb R^2 \to \Bbb R^2$ is a diffeomorphism. Differentiating with respect to time,

$$\dot {\bf y} = \left(\left({\bf D \, \Phi} \cdot {\bf f} \right) \circ {\bf \Phi}^{-1} \right) ({\bf y})$$

where ${\bf D \, \Phi}$ denotes the Jacobian of vector field ${\bf \Phi}$, $\cdot$ denotes multiplication, $\circ$ denotes composition and ${\bf \Phi}^{-1}$ denotes the inverse of ${\bf \Phi}$. Are there "nice" choices of vector field ${\bf \Phi}$ that make vector field $$\color{blue}{\left( {\bf D \, \Phi} \cdot {\bf f} \right) \circ {\bf \Phi}^{-1}}$$ "nice" in some sense?

What is "nice"? For example, polynomial and rational functions are nice. All elementary functions are nice. Anything of relatively "low" descriptive complexity would be nice. In fact, any alternative to the classical pendulum ODE would be nice, even if highly contrived.

My work

The only idea I had was to rewrite the pendulum ODE in Cartesian coordinates, which led to

$$\begin{aligned} (\ddot x - g) y - x \ddot y &= 0\\ x^2 + y^2 &= \ell^2 \end{aligned}$$

which isn't really a system of ODEs. Rather, it is a system of differential-algebraic equations (DAEs). Nothing else occurred to me. I am not familiarized with diffeomorphisms. Ideas are most welcome.

Rodrigo de Azevedo
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    Write the energy as $\dot θ^2+4\omega^2\sin^2(θ/2)=R^2$, so that one could use polar coordinates $\dotθ=R\cos\phi$, $2\omega\sin(θ/2)=R\sin\phi$, with $\dot R=0$, $\dot\phi=\omega\cos(θ/2)$. – Lutz Lehmann Apr 21 '21 at 08:24
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    While it may not be the type of "nice" you are looking for, the action-angle coordinates are often considered the "nicest" coordinates for this type of dynamical system. – Kajelad Apr 22 '21 at 23:32
  • Still, why $\Phi=\mathrm{id}$ is not okay for you? if sine function is not sufficiently elementary, then what do you mean by elementary? – Start wearing purple May 04 '21 at 20:06
  • @Startwearingpurple Because of this. – Rodrigo de Azevedo May 04 '21 at 20:15
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    But if even trigonometric functions are not okay and you insist on the word "diffeomorphism" (defined globally), the problem has no solution. One way to see it is this: the initial system has an infinite number of fixed points in $\mathbb R^2$, and the system you want (where trigonometry is forbidden) would only have finitely many. – Start wearing purple May 04 '21 at 20:30
  • @Startwearingpurple I have nothing against "local" diffeomorphisms. This question was inspired by chemistry — can one "simulate" a pendulum using a chemical computer, even if only on a "small" region of the non-negative quadrant? — but is now a separate matter. At first, I was curious if my claim on Chemistry SE was correct. Now, I am curious about other things, too. If you believe the question should be refined, suggestions on how to refine it are most welcome. – Rodrigo de Azevedo May 04 '21 at 20:50
  • If you just want to map the initial system to a polynomial/rational one, one way is to rewrite the equation for $\theta$ as an equation for $u=\tan \frac{\theta}{2}$. Obviously, $\dot \theta=\frac{2\dot u}{1+u^2}$, $\ddot \theta=\frac{2\ddot u}{1+u^2}-\frac{4u{\dot u}^2}{(1+u^2)^2}$, $\sin\theta=\frac{2 u}{1+u^2}$, so that the differential equation $\ddot \theta+\sin\theta=0$ becomes $\ddot u-\frac{2u{\dot u}^2}{1+u^2}+u=0$ (which of course can be written as a 1st order system). But it might be that I misunderstand completely your aim. – Start wearing purple May 04 '21 at 21:00
  • @Startwearingpurple Thank you for the input. Very roughly speaking, in state space, a pendulum's trajectory is a curve, though not an ellipse. Is there a "coordinate system" in which the trajectory is indeed an ellipse? This "deformation" of the state space could be local rather than global. I am interested in behavior close to the origin, not in all the other equilibria. – Rodrigo de Azevedo May 04 '21 at 22:05

2 Answers2

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It is possible to find another way for the solution of $$y''+k\sin(y)=0 \qquad \text{with} \qquad y=y(x)\qquad \text{and} \qquad k>0$$ Switch variables and then write $$\frac {x''}{[x']^3}=k\sin(y)$$ The usual reduction of order gives $$x'=\pm\frac{1}{\sqrt{2 k \cos (y)+c_1}}$$ So $$x+c_2=\pm\frac{2}{\sqrt{c_1+2 k}}F\left(\frac{y}{2}|\frac{4 k}{c_1+2 k}\right)$$ where appears the elliptic integral of the first kind.

This can be inversed easily leading to the amplitude for the Jacobi elliptic functions.

Claude Leibovici
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To solve the ODE in the form of $$2y''+ p^2\sin y = 0,\quad p=\sqrt{\dfrac l{2g}},$$ can be used the integrating factor $\;y'\not=0,\;$ and that leads to the equation $$y'^2 - p^2\cos y = y'^2\big|_{y=0} -p^2 = (k^2-1)p^2,$$ $$\dfrac{\text dy}{\text dx} = \pm p\sqrt{k^2 - 1-\cos y}\,,$$ with the solution $$\pm p(x\bigg|_{y=0}+x) = \dfrac2{\sqrt{k^2-2}}\operatorname{F}\left(\dfrac y2\,\bigg|-\dfrac2{k^2-2}\right),$$ where $\;F\;$is the elliptic integral of the first kind.

Yuri Negometyanov
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