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Duffing Differential Equation

The most general forced form of the Duffing equation is

\ddot x+\delta\dot x+(\beta x^3\pm{\omega_0}^2 x)=A\sin(\omega t+\phi).
\end{displaymath} (1)

If there is no forcing, the right side vanishes, leaving
\ddot x+\delta\dot x+(\beta x^3\pm{\omega_0}^2 x)=0.
\end{displaymath} (2)

If $\delta=0$ and we take the plus sign,
\ddot x+{\omega_0}^2x+\beta x^3=0.
\end{displaymath} (3)

This equation can display chaotic behavior. For $\beta>0$, the equation represents a ``hard spring,'' and for $\beta
<0$, it represents a ``soft spring.'' If $\beta
<0$, the phase portrait curves are closed. Returning to (1), take $\beta =1$, $\omega_0=1$, $A=0$, and use the minus sign. Then the equation is
\ddot x+\delta\dot x+(x^3-x) = 0
\end{displaymath} (4)

(Ott 1993, p. 3). This can be written as a system of first-order ordinary differential equations by writing
$\displaystyle \dot x$ $\textstyle =$ $\displaystyle y,$ (5)
$\displaystyle \dot y$ $\textstyle =$ $\displaystyle x-x^3-\delta y.$ (6)

The fixed points of these differential equations
$\displaystyle \dot x$ $\textstyle =$ $\displaystyle y = 0,$ (7)

so $y=0$, and
$\displaystyle \dot y$ $\textstyle =$ $\displaystyle x-x^3-\delta y=x(1-x^2)-0$ (8)

giving $x=0, \pm 1$. Differentiating,
$\displaystyle \ddot x$ $\textstyle =$ $\displaystyle \dot y = x-x^3-\delta y$ (9)
$\displaystyle \ddot y$ $\textstyle =$ $\displaystyle (1-3x^2)\dot x-\delta \dot y$ (10)

\left[{\matrix{\ddot x\cr \ddot y\cr}}\right] = \left[{\matr...\cr}}\right] \left[{\matrix{\dot x\cr \dot y\cr}}\right].
\end{displaymath} (11)

Examine the stability of the point (0,0):
\left\vert\matrix{0-\lambda & 1\cr 1 & -\delta-\lambda \cr}\...
...rt = \lambda(\lambda+\delta)-1 = \lambda^2+\lambda\delta-1 = 0
\end{displaymath} (12)

\lambda^{(0,0)}_\pm = {\textstyle{1\over 2}}(-\delta\pm\sqrt{\delta^2+4}\,).
\end{displaymath} (13)

But $\delta^2 \geq 0$, so $\lambda^{(0,0)}_\pm$ is real. Since $\sqrt{\delta^2+4} > \vert\delta\vert$, there will always be one Positive Root, so this fixed point is unstable. Now look at ($\pm 1$, 0).
\left\vert\matrix{0-\lambda & 1\cr -2 & -\delta-\lambda\cr}\...
...rt = \lambda(\lambda+\delta)+2 = \lambda^2+\lambda\delta+2 = 0
\end{displaymath} (14)

\lambda^{(\pm 1,0)}_\pm = {\textstyle{1\over 2}}(-\delta\pm\sqrt{\delta^2-8}\,).
\end{displaymath} (15)

For $\delta>0$, $\Re[\lambda^{(\pm 1,0)}_\pm] < 0$, so the point is asymptotically stable. If $\delta=0$, $\lambda^{(\pm
1,0)}_\pm = \pm i\sqrt{2}$, so the point is linearly stable. If $\delta \in (-2\sqrt{2}, 0)$, the radical gives an Imaginary Part and the Real Part is $> 0$, so the point is unstable. If $\delta=-2\sqrt{2}$, $\lambda^{(\pm
1,0)}_\pm = \sqrt{2}$, which has a Positive Real Root, so the point is unstable. If $\delta
<-2\sqrt{2}$, then $\vert\delta\vert < \sqrt{\delta^2-8}$, so both Roots are Positive and the point is unstable. The following table summarizes these results.

$\delta>0$ asymptotically stable
$\delta=0$ linearly stable (superstable)
$\delta<0$ unstable

Now specialize to the case $\delta=0$, which can be integrated by quadratures. In this case, the equations become

$\displaystyle \dot x$ $\textstyle =$ $\displaystyle y$ (16)
$\displaystyle \dot y$ $\textstyle =$ $\displaystyle x-x^3.$ (17)

Differentiating (16) and plugging in (17) gives
\ddot x=\dot y=x-x^3.
\end{displaymath} (18)

Multiplying both sides by $\dot x$ gives
\ddot x \dot x -\dot x x+\dot x x^3=0
\end{displaymath} (19)

{d\over dt} ({\textstyle{1\over 2}}\dot x^2-{\textstyle{1\over 2}}x^2+{\textstyle{1\over 4}}x^4)=0,
\end{displaymath} (20)

so we have an invariant of motion $h$,
h\equiv {\textstyle{1\over 2}}\dot x^2-{\textstyle{1\over 2}}x^2+{\textstyle{1\over 4}}x^4.
\end{displaymath} (21)

Solving for $\dot x^2$ gives
\dot x^2=\left({dx\over dt}\right)^2=2h+x^2-{\textstyle{1\over 2}}x^4
\end{displaymath} (22)

{dx\over dt}=\sqrt{2h+x^2+{\textstyle{1\over 2}}x^2},
\end{displaymath} (23)

t=\int dt=\int{dx\over\sqrt{2h+x^2+{\textstyle{1\over 2}}x^2}}.
\end{displaymath} (24)

Note that the invariant of motion $h$ satisfies

\dot x={\partial h\over\partial\dot x} ={\partial h\over\partial y}
\end{displaymath} (25)

{\partial h\over \partial x}=-x+x^3=-\dot y,
\end{displaymath} (26)

so the equations of the Duffing oscillator are given by the Hamiltonian System
\dot x = {\partial h\over\partial y}\cr
\dot y =-{\partial h\over\partial x}.\cr}
\end{displaymath} (27)


Ott, E. Chaos in Dynamical Systems. New York: Cambridge University Press, 1993.

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© 1996-9 Eric W. Weisstein