Two oscillators-visualizing the solutions

In twooscillators-solution.html we began with a linear system $X'=AX$ which we could put in standard form $Y'=BY$ by means of a transformation $X=TY$ with $B=T^{-1}AT$. Our assumptions implied that the standard form of the system is \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} The solution is easiest decribed by introducing polar coordinates for $y_1y_2$ and for $y_3y_4$: \[ \begin{pmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \end{pmatrix} = \begin{pmatrix} r_1 \cos\phi_1 \\ r_1 \sin\phi_1 \\ r_2 \cos\phi_2 \\ r_2 \sin\phi_2 \end{pmatrix} \] In these double polar coordinates the differential equations are \[ r_1'=0, \; r_2'=0,\qquad \phi_1'=\omega_1,\; \phi_2'=\omega_2. \]

The “energy sphere”

The solutions of the four dimensional system $Y'=BY$ are curves in $\R^4$, which we cannot visualize. To draw pictures of orbits we will have to settle for less.

For solutions of the system $Y'=BY$ the quantity \[ E(y_1, y_2, y_3, y_4) = y_1^2 + y_2^2 + y_3^2 + y_4^2 \] is constant. In the example of the three springs problem this quantity turns out to be precisely the total energy of the system (i.e. the kinetic energy of the two moving masses plus the potential energy that is stored in the three springs).

Instead of trying to visualize all solutions we pick a value for the energy $E$ and try to draw all solutions with that given energy. For definiteness we pick $E=1$. The set of all $Y\in\R^4$ for which \[ y_1^2 + y_2^2 + y_3^2 + y_4^2 = 1 \] is the unit sphere in $\R^4$. This unit sphere $S^3$ is a three dimensional object. Using stereographic projection from the northpole $(1,0,0,0)$ in $S^3$ onto the $y_2y_3y_4$-space we can visualize all of $S^3$ except for the northpole.

The “total energy” is the sum of two partial energies, \[ E_{1} = \tfrac12 r_1^2= \tfrac12\bigl(y_1^2+y_2^2\bigr), \qquad E_{2} = \tfrac12 r_2^2 = \tfrac12\bigl(y_3^2+y_4^2\bigr). \] However, along solutions of the differential equation the quantity \[ E = y_1^2+y_2^2+y_3^2+y_4^2 \] is constant.

Let $n=4$. Within the three dimensional sphere \[ S^3 = \bigl\{ Y\in\R^4 : y_1^2+\cdots +y_4^2=1\bigr\} \] we have the subsets \[ T_{a,b} = \bigl\{ Y : y_1^2+y_2^2 = a^2, \; y_3^2 +y_4^2=b^2\bigr\} \] where $a^2+b^2=1$. Points on $T_{a,b}$ are determined by two angles, $\phi_1$ and $\phi_2$, via \[ Y_{a,b}(\phi_1, \phi_2) = \begin{pmatrix} a \cos\phi_1 \\ a \sin\phi_1 \\ b \cos\phi_2 \\ b \sin\phi_2 \end{pmatrix} \] The image under stereographic projection of $T_{a,b}$ is \[ Z_{a,b}(\phi_1, \phi_2) = \Bigl( \frac{a\cos\phi_1} {1-b\sin\phi_2}, \frac{a\sin\phi_1} {1-b\sin\phi_2}, \frac{b\cos\phi_2} {1-b\sin\phi_2} \Bigr) \] In cylindrical coordinates \[ (z_1, z_2, z_3) = (r\cos\theta, r\sin\theta, z) \] the image $Z_{a,b}$ is given by \[ \theta = \phi_1, \quad r=\frac{a} {1-b\sin\phi_2}, \quad z= \frac{b\cos\phi_2} {1-b\sin\phi_2}. \] As $\phi_1$ increases from $0$ to $2\pi$ the point $Z_{a,b}(\phi_1, \phi_2)$ traces out a circle around the $z_3$ axis with radius $a/(1-b\sin\phi_2)$.

As $\phi_2$ runs from $0$ to $2\pi$, we have \begin{align*} z^2+r^2 &= \frac{a^2 + b^2\cos^2\phi_2} {(1-b\sin\phi_2)^2} \\ &= \frac{a^2+b^2 - b^2\sin^2\phi_2} {1-2b\sin\phi_2+b^2\sin^2\phi_2} \\ &= \frac{1 - b^2\sin^2\phi_2} {1-2b\sin\phi_2+b^2\sin^2\phi_2} \\ &= \frac{-1+2b\sin\phi_2 - b^2\sin^2\phi_2} {1-2b\sin\phi_2+b^2\sin^2\phi_2} +\frac{2-2b\sin\phi_2} {1-2b\sin\phi_2+b^2\sin^2\phi_2} \\ &= -1 + \frac{2(1-b\sin\phi_2)} {(1-b\sin\phi_2)^2} \\ &= -1 + \frac{2} {1-b\sin\phi_2} \\ &= -1 + \frac2a r. \end{align*} Hence \[ z^2+\bigl(r-\frac1a\bigr)^2 = \frac1{a^2} - 1 = \frac{1-a^2} {a^2} = \left(\frac ba\right)^2. \] Therefore, as $\phi_2$ runs from $0$ to $2\pi$, the point $Z_{a,b}(\phi_1, \phi_2)$ traces out a circle with center at $r=1/a$ and with radius $b/a$.