The governing equation for free motion of a pendulum with damping is

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

A class of equations which includes this case would be

$$\ddot{x}=g(x,\dot{x}) \tag{2}$$

where $g$ is any function. In the spirit of the phase-plane view, we can recast this as a pair of first-order equations by writing

$$\dot{x} = y,\mathrm{~~~~~}\dot{y}=g(x,y). \tag{3}$$

Again we could generalise this: the approach to be followed here would apply to any equations of the form

$$\dot{x} = f(x,y),\mathrm{~~~~~}\dot{y}=g(x,y). \tag{4}$$

where $f$ is another function. Singular points (or equilibrium points) occur where $\dot{x}=\dot{y}=0$, so they occur at positions $(x_0,y_0)$ which are roots of

$$f(x_0,y_0)=0, \mathrm{~~~~~}g(x_0,y_0)=0. \tag{5}$$

Provided $f$ and $g$ are both smooth, we can expand both functions by Taylor series in the vicinity of each singular point, and keep only the linear terms:

$$f \approx a_{11} (x-x_0) + a_{12} (y-y_0)$$

$$g \approx a_{21} (x-x_0) + a_{22} (y-y_0) \tag{6}$$

so that

$$\begin{bmatrix}\dot{x}\\ \dot{y}\end{bmatrix} \approx A \begin{bmatrix}x-x_0\\ y-y_0\end{bmatrix} \tag{7}$$

where

$$A=\begin{bmatrix} a_{11} & a_{12}\\a_{21} & a_{22} \end{bmatrix} . \tag{8}$$

To see the behaviour determined by these linearised equations, it is useful to look first for solutions satisfying

$$\begin{bmatrix} \dot{x} \\ \dot{y} \end{bmatrix} = \lambda \begin{bmatrix} x-x_0 \\ y-t_0 \end{bmatrix} \tag{9}$$

where $\lambda$ is a constant. In other words, we seek the eigenvalues and eigenvectors of the matrix $A$ such that

$$A \mathbf{u} = \lambda \mathbf{u} . \tag{10}$$

The associated solution of eq. (9) is simply

$$\mathbf{u} = \mathbf{u_0} e^{\lambda t} . \tag{11}$$

If the eigenvalues are real, this describes motion radially inwards or outwards along the eigenvector direction. If the eigenvalues are complex, they must form a complex conjugate pair, and the corresponding motion involves circulation around the singular point.

From the usual determinant, the eigenvalues are the roots of

$$(a_{11}-\lambda)(a_{22}-\lambda)-a_{12} a_{21} =0 \tag{12}$$

which leads to

$$\lambda^2 – T \lambda +D =0 \tag{13}$$

where $T=a_{11}+a_{22}$ is the trace of the matrix $A$ and $D$ is its determinant.

We can recognise this as the same as the characteristic equation for the damped harmonic oscillator (see section 2.2.7), where $D$ plays the role of the squared frequency, and $T$ is the negative of the damping coefficient. So we already know what behaviour to expect: if $D$ is negative, the system is unstable regardless of the value of $T$. On the other hand if $D$ is positive, oscillatory behaviour can occur. If $T=0$ the system is undamped and will exhibit steady oscillation. If $T<0$ we have positive damping and the motion decays, while if $T>0$ the motion grows in an unstable manner. If $T^2 < 4D$, damping is below the critical value so oscillation will be seen. But if $T^2 > 4D$, damping is greater than the critical value and decay without oscillation will occur.

These conditions map directly onto six categories of behaviour, each with its own phase portrait:

$D<0$ gives a *saddle point*

$D>0$ with $T=0$ gives a *centre*

$D>0$ with $0 < |T| < 2\sqrt{D}$ gives a *spiral*, either *stable* or *unstable* depending on whether $T<0$ or $T>0$

$D>0$ with $|T| > 2\sqrt{D}$ gives a *node*, either *stable *or* unstable* depending on whether $T<0$ or $T>0$

An undamped system always has $T=0$, so the only options are saddle points and centres.

Typical examples of these cases are illustrated in Fig. 1. We have seen the phase portraits for most of these in section 8.3, but they are repeated here for completeness. The top row shows the saddle point and the centre. The second row shows shows a spiral and a node: for stable versions of these, the arrows will point inwards, whereas for unstable versions they point outwards.