8.2.2 Duffing’s equation and harmonic balance

Duffing’s equation describes a single-degree-of-freedom oscillator with a nonlinear spring. For weak nonlinearity it is reasonable to expect that the spring force-displacement relation could be expanded in a Taylor series, and that this might then be approximated by discarding higher powers of displacement. For a spring force that behaves in a symmetrical manner under tension or compression, we expect only odd powers of displacement in the Taylor expansion, so the simplest approximation is to include just a cubic term alongside the linear variation.

We can write the resulting equation in this form:

$$\ddot{x}+p \eta x +p^2 x +\mu x^3 = F(t) \tag{1}$$

where $p$ is the angular frequency of the linear oscillator resonance, $\eta$ is the linear damping factor, $\mu$ is the nonlinear coefficient, and $F(t)$ is the externally applied excitation force. If $\mu > 0$ it is a hardening spring, if $\mu < 0$ it is a softening spring. We will consider the case of harmonic excitation with amplitude $a$ and frequency $\omega$:

$$F(t)=a \cos \omega t . \tag{2}$$

We will first look at the undamped case, with $\eta =0$. The cubic spring law means that the nonlinearity generates only odd harmonics. So a simple 2-term harmonic expression to try would be

$$x \approx \alpha \cos \omega t + \beta \cos 3\omega t . \tag{3}$$

Substituting in eq. (1) gives

$$-\omega^2 \left(\alpha \cos \omega t + 9\beta \cos 3\omega t\right) + p^2 \left(\alpha \cos \omega t + \beta \cos 3\omega t\right)$$

$$+\mu \left(\alpha \cos \omega t + \beta \cos 3\omega t\right)^3 = a \cos \omega t. \tag{4}$$

We now expand out the cubed term, and make use of three standard trigonometrical formulae:

$$\cos \theta \cos \phi = \frac{1}{2} \left[\cos (\theta – \phi) + \cos (\theta + \phi) \right], \tag{5}$$

$$\cos^2 \theta = \dfrac{1}{2}\left(1 + \cos 2 \theta \right) \tag{6}$$

and

$$\cos^3 \theta =\frac{3}{4}\cos \theta + \frac{1}{4} \cos 3 \theta . \tag{7}$$

The result is a linear combination of terms in $\cos n\omega t$ for various odd-number values of $n$. We equate the coefficients of all the terms in $\cos \omega t$ to give one relation between the coefficients $\alpha$ and $\beta$, and we do the same with the terms in $\cos 3 \omega t$ to obtain another. We simply ignore the terms with $n$ values greater than 3; but if we wanted information about higher values we could extend the approach to include more terms in the initial guess (3).

The two equations that result are:

$$-\alpha \omega^2 + \alpha p^2 + \mu \left[\frac{3}{4}\alpha^3 + \frac{3}{4}\alpha^2 \beta +\frac{3}{2}\alpha \beta^2 \right] = a \tag{8}$$

and

$$-9\beta \omega^2+ \beta p^2 + \mu \left[\frac{1}{4}\alpha^3 + \frac{3}{2}\alpha^2 \beta +\frac{3}{4} \beta^3 \right] = 0 . \tag{9}$$

We can make a further approximation: we expect the third harmonic generated by the nonlinearity to be small in magnitude, so that $ |\beta | \ll |\alpha |$. The terms in square brackets, which are also multiplied by the small parameter $\mu$, can thus be simplified to give

$$\alpha (p^2 – \omega^2) +\dfrac{3}{4} \mu \alpha^3 \approx a \tag{10}$$

and

$$\beta (p^2-9 \omega^2) + \dfrac{1}{4} \mu \alpha^3 \approx 0 . \tag{11}$$

Equation (10) is a cubic equation for $\alpha$ for given excitation level $a$. Once $\alpha$ is found, eq. (11) gives the associated value of $\beta$.

In principle there is a formula for the roots of a cubic equation, but for the purposes of this discussion it is simpler to compute a numerical example. Figure 1 shows a plot for a hardening spring using the parameter values $a=1$, $p=1$, $\eta =0$ and $\mu = 10^{-7}$. Damping is zero here so that both curves, linear and nonlinear, head off to infinity. The plot shows the fundamental amplitude $\alpha$ only: the third harmonic governed by $\beta$ affects the frequency spectrum of the motion, but makes little difference to the amplitude. Figure 3 of section 8.2 shows a matching case except with $\eta = 0.002$, while Fig. 5 of that section shows the corresponding case for a softening spring, with $\mu = -10^{-7}$.

Figure 1. Amplitude of response to sinusoidal driving of a linear oscillator (black) and a Duffing oscillator with a hardening spring (red). Both cases are undamped.

The final step is to add the damping term in. For that, we need to replace eq. (3) by something involving both $\sin \omega t$ and $\cos \omega t$, because damping introduces phase shifts. Having seen how the previous calculation worked out, we can save effort by looking only at the fundamental component of the Fourier series: so this time we try

$$x=\alpha \cos \omega t + \gamma \sin \omega t . \tag{12}$$

We substitute into eq. (1), and make similar use of trigonometrical identities. Ignoring all higher harmonic components, we can obtain two equations for $\alpha$ and $\gamma$ by equating coefficients of all terms in $\cos \omega t$, and all terms in $\sin \omega t$. The result is

$$(p^2 – \omega^2) \alpha +p \eta \gamma \omega +\frac{3}{4} \mu \alpha (\alpha^2 + \gamma^2) = a \tag{13}$$

and

$$(p^2 – \omega^2) \gamma -p \eta \alpha \omega +\frac{3}{4} \mu \gamma (\alpha^2 + \gamma^2) = 0 . \tag{14}$$

With a little manipulation, these two equations can be combined to yield a single equation for the amplitude $z$, defined by $z^2=\alpha^2 + \gamma^2$:

$$\left[ \left(\omega^2 – p^2 – \frac{3}{4} \mu z^2 \right)^2 + (p \eta \omega)^2 \right] z^2 = a^2. \tag{15}$$

This cubic equation for $z^2$ was solved numerically to generate the damped response shown in Figs. 3, 4 and 5 of section 8.2.

Finally, to add the blue circles indicating the “backbone curve” in Figs. 3 and 5 of section 8.2 we need to find the condition that defines the “nose” of the leaning curves. We take our cue from the linear harmonic oscillator: the peak response occurs at the point where the motion is exactly $90^\circ$ out of phase with the excitation. In other words, we want the point where $\alpha = 0$. Substituting this into eqs. (13) and (14) and doing minor manipulations, we obtain the equation

$$\gamma^4 +\dfrac{4p^2}{3 \mu} \gamma^2 -\dfrac{4a^2}{3 \mu p^2 \eta^2}=0 \tag{16}$$

which is a quadratic equation for $\gamma^2$. Once $\gamma$ is known, the corresponding frequency satisfies

$$p \eta \gamma \omega = a. \tag{17}$$

These two equations, evaluated with different values of $a$ representing a slow decay of the oscillation, give the blue circles shown in Figs. 3 and 5 of section 8.2.