When do the solutions of a linear ODE system lie on ellipses?

Question

Let $x(t) \in \mathbb{R}^n$ and $A(t)$ be a $n\times n$ matrix. When $A(t)$ is a skew-symmmetric matrix, the solutions of the linear system $\dot{x} = A(t) x$ lie on spheres centered at the origin since $$ \frac{1}{2} \frac{d}{dt} \lvert x \rvert^2 = \langle \dot{x}, x \rangle = \langle Ax, x \rangle = 0 \implies \lvert x \rvert^2 = \lvert x(0) \rvert^2 $$ Is there a specific class of matrices $A(t)$ for which the solutions lie on ellipses instead?

Answer 1

An $n$-dimensional ellipsoid is defined by an equation of the form

$\displaystyle \sum_1^n a_i x_i^2 = c > 0, \; c \; \text{a constant} \tag 1$

where

$a_i > 0, \; 1 \le i \le n; \tag 2$

if we define the matrix $D$ via

$D = [\delta_{ij} a_i] = \text{diag} (a_1, a_2, \ldots, a_n), \tag 3$

then (1) may be written

$\langle x, Dx \rangle = c, \tag 4$

where $\langle \cdot, \cdot \rangle$ denotes the ordinary Euclidean inner product on $\Bbb R^n$

$\langle x, y \rangle = \displaystyle \sum_1^n x_i y_i. \tag 5$

Now if $x(t)$ follows a path such that

$\dot x(t) = A(t) x(t) \tag 6$

for some $t$-dependent matrix $A(t)$, then since differentiation of (4) yields

$\langle \dot x(t), Dx \rangle + \langle x(t), D \dot x(t) \rangle = 0, \tag 7$

in accord with (6) we have

$\langle A(t) x(t), Dx(t) \rangle + \langle x(t), D A x(t) \rangle = 0; \tag 8$

we re-arrenge this, thusly, using the symmetry of $\langle \cdot, \cdot \rangle$:

$\langle Dx(t), A(t) x(t) \rangle + \langle x(t), D A x(t) \rangle = 0, \tag 9$

and since $D$, being diagonal, is a symmetric matrix,

$D^T = D, \tag{10}$

$2 \langle x(t), D A(t) x(t) \rangle = \langle x(t), D A(t) x(t) \rangle + \langle x(t), D A x(t) \rangle$ $= \langle x(t), D^T A(t) x(t) \rangle + \langle x(t), DA(t) x(t) \rangle$ $= \langle D x(t), A(t) x(t) \rangle + \langle x(t), DA(t) x(t) \rangle = 0, \tag{11}$

i.e.,

$\langle x(t), D A(t) x(t) \rangle = 0; \tag{12}$

by virtue of the fact that we may arbitrarily choose

$x(t) \in \Bbb R^n, \tag{13}$

by if neccessary adjusting $x(0)$ accordingly, we conclude that

$\forall y \in \Bbb R^n, \; \langle y, D A(t) y \rangle = 0, \tag{14}$

which as is well-known forces $DA(t)$ to be skew-symmetric:

$(DA(t))^T = -DA(t); \tag{15}$

this may also be expressed in the form

$A^T(t)D = A^T(t) D^T = (DA(t))^T = -DA(t), \tag{16}$

or

$A^T(t) = -DA(t)D^{-1}; \tag{17}$

we note that for all $t \in \Bbb R$, $A^T(t)$ is similar to $-A(t)$ via conjugation by the matrix $D$. This condition is seen by the above to be necessary for the motion of $x(t)$ to lie in the ellipsoid (1), (4); however, careful scrutiny of our argument shows that it may be reversed, and thus (17) is also sufficient.

Note Added in Edit, Tuesday 24 December 2019 1:39 PM PST: Returning to (14)-(15), we show why

$\langle y, Cy \rangle = 0, \; \forall y \in \Bbb R^n \tag{18}$

forces $C$ to be skew-symmetric, that is,

$C^T = -C; \tag{19}$

for (18) implies

$\langle y + z, C(y +z) \rangle = 0; \tag{20}$

expanding:

$\langle y, Cy \rangle + \langle z, Cy \rangle + \langle y, Cz \rangle + \langle z, Cz \rangle = 0; \tag{21}$

again by virtue of (18),

$\langle z, Cy \rangle + \langle y, Cz \rangle = 0, \tag{22}$

whence

$\langle y, Cz \rangle = -\langle z, Cy \rangle = -\langle C^T z, y \rangle = -\langle y, C^T z \rangle; \tag{23}$

this binds for every $y$, and so also

$Cz = -C^T z \tag{24}$

for every $z$; hence (19) We may also run this argument in reverse if we observe that (24) yields

$\langle z, Cz \rangle = -\langle z, C^T z \rangle = -\langle Cz, z \rangle = -\langle z, Cz \rangle \Longrightarrow \langle z, C z \rangle = 0. \tag{25}$

End of Note.