Two oscillators—solving the differential equations
The original problem
Suppose the evolution of some system is described by a vector $X(t)\in\R^4$ and
suppose that $X(t)$ satisfies
\begin{equation}
\frac{dX} {dt} = AX
\label{eq:X-system}
\end{equation}
where the real $4\times 4$ matrix $A$ has two purely imaginary eigenvalues
$i\omega_1$ and $i\omega_2$. Since $A$ is real, $-i\omega_1$ and $-i\omega_2$
are also eigenvalues.
If $W_1=U_1+iV_1$ and $W_2=U_2+iV_2$ are the eigenvectors corresponding to
$i\omega_1$ and $i\omega_2$, then the eigenvectors for $-i\omega_1$ and
$-i\omega_2$ are $\bar W_1 = U_1-iV_1$ and $\bar W_2 = U_2-iV_2$.
These assumptions are very often satisfied by differential equations that
describe mechanical systems. For a very typical example do
the three springs problem.
Standard form of the equations
To put the system of equations $X'=AX$
in standard form we choose $\{U_1, -V_1, U_2, -V_2\}$ as basis for $\R^4$ and
introduce the new vector $Y$ which is related to $X$ by
$X = y_1U_1 - y_2V_1 + y_3U_2 - y_4 V_2$, i.e. $X=TY$
where
\[
T = \bigl[ U_1,\, -V_1,\, U_2,\, -V_2\bigr].
\]
(why the minus signs? We can of course choose any basis we like; putting
the minus signs here reduces the number of minus signs we get in the final result.)
Then $Y(t)$ satisfies
$Y'= BY$, where
$B= T^{-1}AT
=
\begin{pmatrix}
0 & -\omega_1 & & \\
\omega_1 & 0 & & \\
& & 0 & -\omega_2 \\
& & \omega_2 & 0
\end{pmatrix}
$.
Written out in components, these equation are
\begin{equation}
y_1' = -\omega_1 y_2, \; y_2'=\omega_1y_1 \qquad
y_3' = -\omega_2 y_4, \; y_4'=\omega_2y_3.
\label{eq:y-system}
\end{equation}
Double Polar Coordinates
Introduce polar coordinates for both $(y_1,y_2)$ and $(y_3, y_4)$:
\[
\begin{pmatrix}
y_1 \\ y_2
\end{pmatrix}
=
\begin{pmatrix}
r_1 \cos\phi_1 \\ r_1 \sin\phi_1
\end{pmatrix}\, ,
\qquad
\begin{pmatrix}
y_3 \\ y_4
\end{pmatrix}
=
\begin{pmatrix}
r_2 \cos\phi_2 \\ r_2 \sin\phi_2
\end{pmatrix}\, .
\]
The system $Y'=BY$, (i.e. \eqref{eq:y-system}) in these polar coordinates is
\begin{equation}
r_1'=0, \; r_2'=0,\qquad \phi_1'=\omega_1,\; \phi_2'=\omega_2.
\label{eq:r-theta-system}
\end{equation}
Thus the solution to the original system can be written as
\begin{equation}
X(t) = r_1\cos\phi_1(t)\; U_1 - r_1\sin\phi_1(t)\; V_1 +
r_2\cos\phi_2(t)\; U_2 - r_2\sin\phi_2(t)\; V_2.
\label{eq:X-solution}
\end{equation}
where the “phases” $\phi_1(t)$ and $\phi_2(t)$ evolve according to
\[
\phi_1(t) = \phi_1(0) + \omega_1t,\qquad
\phi_2(t) = \phi_2(0) + \omega_2t.
\]
Interpretation
In view of the solution \eqref{eq:X-solution} we can
think of the system as consisting of two independent oscillators: one rotates
with frequency $\omega_1$, the other with frequency $\omega_2$. The phases
$(\phi_1, \phi_2)$ move along a straight line in the direction of the vector
$\begin{pmatrix} \omega_1\\ \omega_2\end{pmatrix}$.
