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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?

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2 Answers 2

You only need to differentiate using common derivative rules and then susbtitute the expressions from the given system into the result. See:

$$ \frac{dE}{dt} = \frac{d}{dt}\left(\frac{1}{2}\omega^2-\cos\theta\right)=\omega\frac{d\omega}{dt}+(\sin\theta)\frac{d\theta}{dt} = \omega(-\sin\theta)+(\sin\theta)\omega=0.$$

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Thanks for the help – Math Student Aug 5 '11 at 21:09

Since this looks like homework, I'll give you a hint: have you tried evaluating $\frac{dE}{dt}$ directly and seeing where it leads you? Also, are the fixed points relevant to the problem?

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