A simple pendulum is modeled by the (nondimensionalized) differential equation $$\frac{d^{2}\theta}{dt^2} = -\sin(\theta)$$
With the auxiliary variable $w$, this is equivalent to the first-order system $$\begin{align*}\frac{d\theta}{dt}&= w\\ \frac{dw}{dt}&= -\sin(\theta)\end{align*}\qquad (3)$$
We know that this system has equilibrium at $(\theta, w) = (2n\pi,0)$ and $(\theta,w) = ((2n+1)\pi,0)$. The second set of equilibria were saddle points, and therefore unstable, but the first set were centers whose stability we were able to determine. In this problem, we'll use a conserved quantity of this system to prove that these fixed points are in fact stable.
The quantity $E = (1/2)w^2 – \cos(\theta)$ is proportional to the total energy (kinetic plus potential) of the pendulum. We expect this to be constant on physical grounds, Use $(3)$ to prove that $dE/dt = 0$
I think I need to solve the first differential because I tried to solve the problem without it and got nowhere. I'm not really sure how to solve it though. Any ideas on how to proceed?
Best Answer
You only need to differentiate using common derivative rules and then susbtitute the expressions from the given system into the result. See:
$$ \frac{dE}{dt} = \frac{d}{dt}\left(\frac{1}{2}\omega^2-\cos\theta\right)=\omega\frac{d\omega}{dt}+(\sin\theta)\frac{d\theta}{dt} = \omega(-\sin\theta)+(\sin\theta)\omega=0.$$