# 8.3.3 Singular points of Duffing’s equation

Back in section 8.2.2 we met Duffing’s equation. For undamped, free motion the equation can be written:

$$\ddot{x} +\alpha x +\mu x^3 = 0 \tag{1}$$

where $\alpha$ represents the stiffness of the linear spring and $\mu$ is the nonlinear coefficient. If $\mu > 0$ it is a hardening spring, if $\mu < 0$ it is a softening spring. To put this into a suitable form for looking at the phase space representation, we write it as the pair of equations

$$\dot{x}=y \mathrm{,~~~~}\dot{y}=-\alpha x -\mu x^3 . \tag{2}$$

To find singular points we need to solve $\dot{x}=0$, $\dot{y}=0$ so that

$$\alpha x +\mu x^3 = 0 \mathrm{~~and~~} y=0 \tag{3}$$

The solutions of the first of these equations are $x=0$, together with

$$x^2=-\alpha / \mu \tag{4}$$

if the latter yields a real solution for $x$, in other words if $\alpha / \mu \le 0$. So $(0,0)$ is always a singular point, and there are also singular points at

$$x=\pm \sqrt{\dfrac{-\alpha}{\mu}} \mathrm{,~~~} y=0 \tag{5}$$

if one of $\alpha$ or $\mu$ is positive while the other is negative.

To find out what kind of singular point these are, we need to perform the linearised analysis described in section 8.3.2. For the point $(0,0)$, the linearised matrix $A$ can be written down immediately by ignoring the cubic term in eq. (2): it is

$$A=\begin{bmatrix}0 & 1\\ -\alpha & 0\end{bmatrix} \tag{6}$$

so the trace $T=0$ as expected for an undamped oscillator, while the determinant $D=\alpha$. The results from section 8.3.2 then tell us that if $\alpha > 0$ the singular point is a centre, while if $\alpha < 0$ it is a saddle point.

For the other singular point, we are only interested in the case $\alpha < 0$ so write $\beta = -\alpha$ with $\beta > 0$. Now write

$$x=\pm \sqrt{\dfrac{\beta}{\mu}} + \epsilon \tag{7}$$

so that $\epsilon$ is small close to one of the pair of singular points. Equations (2) then read

$$\dot{\epsilon}=y \mathrm{,~~~~} \dot{y}=\beta \left[\pm \sqrt{\dfrac{\beta}{\mu}} + \epsilon \right] -\mu \left[\pm \sqrt{\dfrac{\beta}{\mu}} + \epsilon\right]^3 \tag{8}$$

with the linearised approximation

$$\dot{\epsilon}=y \mathrm{,~~~~} \dot{y} \approx \beta \left[\pm \sqrt{\dfrac{\beta}{\mu}} + \epsilon \right] -\mu \left[\pm \left(\dfrac{\beta}{\mu}\right)^{3/2} + 3\epsilon\dfrac{\beta}{\mu} \right] \tag{9}$$

and so

$$\dot{\epsilon}=y \mathrm{,~~~~} \dot{y} \approx -2 \beta \epsilon . \tag{10}$$

The matrix $A$ is thus

$$A=\begin{bmatrix}0 & 1\\ -2\beta & 0\end{bmatrix}, \tag{11}$$

which has trace $T=0$ and determinant $D=2\beta$, so since $\beta > 0$, both these singular points are centres.

These results can be summarised in the bifurcation diagram shown in Fig. 1: this shows the positions $x$ of the singular points as a function of $\alpha$, for the case when $\mu > 0$. Stable points (centres) are indicated by solid lines, unstable points (saddles) by a dashed line. At the value $\alpha=0$ a bifurcation occurs. If you think of $\alpha$ decreasing, then at this point the single stable solution turns into three solutions, the outer pair stable and the middle one unstable. This pattern occurs in many systems: it is called a pitchfork bifurcation.

The behaviour of Duffing’s equation can also be viewed from the energy perspective. Equation (1) can be obtained by using the kinetic energy

$$T=\dfrac{1}{2} \dot{x}^2 \tag{12}$$

and the potential energy

$$V=\dfrac{\alpha}{2} x^2 + \dfrac{\mu}{4} x^4, \tag{13}$$

then using energy conservation in the form $\frac{d}{dt}[T+V]=0$.

Figure 2 shows a plot of $V(x)$ for various values of $\alpha$. When $\alpha > 0$, the curves show a single minimum at $x=0$, indicating the stable centre. But when $\alpha < 0$ the curve changes shape so that it has a maximum at $x=0$, indicating the unstable saddle point, flanked symmetrically by two minima indicating the pair of centres. Figure 2. Potential energy $V(x)$ for the Duffing equation, for the case $\mu =1$. The upper curve (yellow) corresponds to $\alpha = 1$, the middle curve (red) to $\alpha = 0.1$, the lower curve (blue) to $\alpha = -1$.