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Floquet Analysis

Given a system of periodic Ordinary Differential Equations of the form

{d\over dt} \left[{\matrix{x\cr y\cr v_x\cr v_y\cr}}\right] ...
... 0\cr}}\right]\left[{\matrix{x\cr y\cr v_x\cr v_y\cr}}\right],
\end{displaymath} (1)

the solution can be written as a linear combination of functions of the form
\left[{\matrix{x(t)\cr y(t)\cr v_x(t)\cr v_y(t)\cr}}\right] ...
...x{x_0\cr y_0\cr v_{x0}\cr v_{y0}\cr}}\right]e^{\mu t}P_\mu(t),
\end{displaymath} (2)

where $P_\mu(t)$ is a function periodic with the same period $T$ as the equations themselves. Given an Ordinary Differential Equation of the form
\ddot x+g(t)x=0,
\end{displaymath} (3)

where $g(t)$ is periodic with period $T$, the ODE has a pair of independent solutions given by the Real and Imaginary Parts of
$\displaystyle x(t)$ $\textstyle =$ $\displaystyle w(t)e^{i\psi(t)}$ (4)
$\displaystyle \dot x$ $\textstyle =$ $\displaystyle (\dot w+iw\dot\psi)e^{i\psi}$ (5)
$\displaystyle \ddot x$ $\textstyle =$ $\displaystyle [\ddot w+i\dot w\dot \psi+i(\dot w\dot \psi+w\ddot \psi+iw\dot\psi ^2)]e^{i\psi}$  
  $\textstyle =$ $\displaystyle [(\ddot w-w\dot\psi^2)+i(2\dot w\dot \psi+w\ddot \psi)]e^{i\psi}.$ (6)

Plugging these into (3) gives
\ddot w+2i\dot w\dot\psi+w(g+i\ddot\psi-\dot\psi^2)=0,
\end{displaymath} (7)

so the Real and Imaginary Parts are
\ddot w+w(g-\dot\psi^2)=0
\end{displaymath} (8)

2\dot w\dot\psi+w\ddot\psi=0.
\end{displaymath} (9)

From (9),
$\displaystyle {2\dot w\over w}+{\ddot\psi\over \dot\psi}$ $\textstyle =$ $\displaystyle 2 {d\over dt} (\ln w)+{d\over dt} [\ln(\dot\psi)]$  
  $\textstyle =$ $\displaystyle {d\over dt} \ln(\dot\psi w^2) =0.$ (10)

Integrating gives
\dot\psi={C\over w^2},
\end{displaymath} (11)

where $C$ is a constant which must equal 1, so $\psi$ is given by
\psi=\int_{t_0}^t {dt\over w^2}.
\end{displaymath} (12)

The Real solution is then
\end{displaymath} (13)

$\displaystyle \dot x$ $\textstyle =$ $\displaystyle \dot w\cos \psi-w\dot\psi\sin \psi = \dot w {x\over w} -w\dot\psi \sin \psi$  
  $\textstyle =$ $\displaystyle \dot w{x\over w}-w {1\over w^2}\sin \psi = \dot w {x\over w}-{1\over w}\sin \psi$ (14)

$\displaystyle 1$ $\textstyle =$ $\displaystyle \cos ^2\psi+\sin ^2\psi=x^2w^{-2}+\left[{w\left({\dot w {x\over w}-\dot x}\right)}\right]^2$  
  $\textstyle =$ $\displaystyle x^2w^{-2}+(\dot w x-w\dot x)^2 \equiv I(x,\dot x,t),$ (15)

which is an integral of motion. Therefore, although $w(t)$ is not explicitly known, an integral $I$ always exists. Plugging (10) into (8) gives
\ddot w+g(t)w-{1\over w^3}=0,
\end{displaymath} (16)

which, however, is not any easier to solve than (3).


Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, p. 727, 1972.

Binney, J. and Tremaine, S. Galactic Dynamics. Princeton, NJ: Princeton University Press, p. 175, 1987.

Lichtenberg, A. and Lieberman, M. Regular and Stochastic Motion. New York: Springer-Verlag, p. 32, 1983.

Margenau, H. and Murphy, G. M. The Mathematics of Physics and Chemistry, 2 vols. Princeton, NJ: Van Nostrand, 1956-64.

Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 556-557, 1953.

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© 1996-9 Eric W. Weisstein