# 8.3.1 Stability of equilibrium for the pendulum

There is a standard procedure for examining the stability of an equilibrium position, and the pendulum gives a good illustration of it. We already know the governing equation, from section 8.2.1:

$$\ddot{\theta} = -\dfrac{g}{L} \sin \theta . \tag{1}$$

For equilibrium, $\ddot{\theta}=0$ and so we need $\sin \theta =0$. The two relevant solutions are $\theta=0,\pi$. Now we take those in turn, and make use of a Taylor series expansion to linearise the governing equation very close to the equilibrium. Near $\theta =0$ we simply use the familiar result $\sin \theta \approx \theta$ to obtain

$$\ddot{\theta} \approx -\dfrac{g}{L} \theta . \tag{2}$$

This is the simple harmonic equation, and its general solution takes the form

$$\theta = A \cos \Omega t + B \sin \Omega t \tag{3}$$

where $\Omega^2=g/L$ and $A$ and $B$ are arbitrary constants. Whatever small perturbation we make around the equilibrium position, and thus whatever the values of $A$ and $B$, $\theta$ remains small because sine and cosine never have a magnitude greater than unity. So the equilibrium is stable: any small perturbation results in the pendulum remaining close to the equilibrium for all later times.

For the case $\theta = \pi$, we first change variable so that we have a small parameter available for the purposes of expansion: so let $\theta = \pi + \alpha$. Now we need to remember, from the properties of the sine function, that $\sin (\pi + \alpha) = -\sin \alpha$. So the approximate governing equation this time is

$$\ddot{\alpha} = \dfrac{g}{L} \sin \alpha \approx \dfrac{g}{L} \alpha = \Omega^2 \alpha. \tag{4}$$

This time the general solution takes the form

$$\alpha = A e^{\Omega t} + B e^{-\Omega t} \tag{5}$$

where $A$ and $B$ are again arbitrary constants. This time, virtually any small perturbation away from the equilibrium position will produce a non-zero value of $A$. However small that value may be, eventually the growing exponential will make that term large. We conclude that the equilibrium is unstable: a small perturbation does not remain small. This analysis does not mean that the actual motion of the pendulum grows exponentially for all subsequent time: remember that we have obtained this solution by assuming that $\alpha$ is small. All we can deduce is that it does not remain small, so that eventually the approximation must break down.

For the undamped pendulum case, there is an alternative approach based on an energy argument. The potential energy, relative to the position $\theta =0$, is

$$V=mgL(1- \cos \theta) \tag{6}$$

and the kinetic energy is

$$T=\dfrac{1}{2} m L^2 \dot{\theta}^2 \tag{7}$$.

A position of stable equilibrium corresponds to a minimum of potential energy, because any perturbation away from that position requires energy, and thus reduces the kinetic energy. But near a maximum of potential energy, any perturbation releases energy and increases kinetic energy, leading to an unstable runaway. It is obvious from visualising the graph of $1-\cos \theta$ that $\theta = 0$ is a minimum of energy, while $\theta = \pi$ is a maximum, leading to the same result obtained above.

As a footnote, energy also provides an easy way to graph the phase portrait of an undamped system. Any possible motion must conserve total energy $E=T+V$, or in other words trajectories must be contour lines of constant energy. So all we need to do is compute $E$ at a grid of points in the phase plane and then draw a contour map: this is how Fig. 3 of section 8.3 was in fact computed.