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.

Figure 1. Bifurcation diagram for Duffing’s equation

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 $\alpha = -1$.