For which $p,q\in \mathbb R$ is the following system stable?$$\frac{\mathrm dx}{\mathrm dt} = \begin{bmatrix} p & -q \\ q & p \end{bmatrix}x(t)$$
If I'm correct about this, isn't it just when the eigenvalues are $< 1?$ Or is there something more fancy to it? Any and all help appreciated.
On
Assuming that $p, q \in \Bbb R$ are constants, the system
$\dfrac{d \vec x(t)}{dt} = \begin{bmatrix} p & -q \\ q & p \end{bmatrix} \vec x(t) \tag 1$
has a well-known explicit solution. Suppose we set
$A = \begin{bmatrix} p & -q \\ q & p \end{bmatrix}, \tag 2$
and let
$\vec x_0 = \vec x(t_0) \tag 3$
be an initialization of $\vec x(t)$; then we have
$\vec x(t) = e^{A(t - t_0)} \vec x_0 = \exp(A(t - t_0)) \vec x_0; \tag 4$
equation (4) is easily validated by direct differentiation; indeed,
$\dot{\vec x}(t) = \dfrac{d \exp(A(t - t_0))}{dt} \vec x_0 = A \exp(A(t - t_0)) \vec x_0 = A \vec x(t); \tag 5$
we also see from (4) that
$\vec x(t_0) = \exp(A(t_0 - t_0)) \vec x_0 = \exp(A(0)) \vec x_0 = \exp(0) \vec x_0 = I \vec x_0 = \vec x_0; \tag 6$
thus (4) is consistent with our choice of initial condition. We compute the matrix $\exp(A(t - t_0))$ as follows:
set
$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag 7$
then we have
$A = \begin{bmatrix} p & -q \\ q & p \end{bmatrix} = pI + qJ; \tag 8$
it follows that
$\exp(A(t - t_0)) = \exp((pI + qJ)(t - t_0)) = \exp(p(t - t_0)I + q(t - t_0)J); \tag 9$
now since $I$ and $J$ commute,
$IJ = JI, \tag{10}$
so do $p(t - t_0)I$ and $q(t - t_0)J$
$(p(t - t_0)I)(q(t - t_0)J) = (q(t - t_0)J)(p(t - t_0)I); \tag{11}$
this allows us to conclude that, just as if $I$ and $J$ were ordinary real or complex numbers,
$\exp((p(t - t_0)I + q(t - t_0)J) = \exp(p(t - t_0)I) \exp(q(t - t_0)J); \tag{12}$
indeed, a demonstration of (12) given (11) is virtually identical the the proof that
$\exp(a + b) = \exp(a) \exp (b), \; a, b \in \Bbb C; \tag{13}$
for more information, see this wikipedia page.
So far we have
$\exp(A(t - t_0)) = \exp(p(t - t_0)I + q(t - t_0)J) = \exp(p(t - t_0)I) \exp(q(t - t_0)J); \tag{14}$
the next step is the evaluation of $\exp(p(t - t_0)I)$ and $\exp(q(t - t_0)J)$; the first of these two is easy, using the power series:
$\exp(p(t - t_0)I) = \displaystyle \sum_0^\infty \dfrac{(p(t - t_0)I)^n}{n!} = \sum_0^\infty \dfrac{(p(t - t_0))^n}{n!}I^n$ $= \left (\displaystyle \sum_0^\infty \dfrac{(p(t - t_0))}{n!} \right ) I = \exp(p(t - t_0))I; \tag {15}$
in order to find $\exp(q(t - t_0)J)$ we note that
$J^2 = -I; \; J^3 = -J; \; J^4 = -J^2 = I; \; J^5 = J^4J = J; \; J^6 = J^4J^2 = -I, \tag{16}$
and so forth, just as $i = \sqrt{-1}$ satisfies
$i^2 = -1; \; i^3 = -i; \; i^4 = 1; \; i^5 = i; \; i^6 = -1, \tag{17}$
etc.; it is easy to see that $i$ and $J$ follow essentially the same pattern, so it follows that the power series for $\exp(q(t - t_0)J$ will, just as
$e^{iq(t - t_0))} = \cos q(t - t_0) + i \sin q(t - t_0), \tag{18}$
yield
$\exp(q(t - t_0)J) = \cos q(t - t_0) I + \sin q(t - t_0) J$ $= \begin{bmatrix} \cos q(t - t_0) & -\sin q(t - t_0) \\ \sin q(t - t_0) & \cos q(t - t_0) \end{bmatrix} = R(t); \tag{19}$
we see the matrix above, which we have chosen to name $R(t)$, is in fact an ordinary $2$-dimensional rotation matrix, and as such is orthogonal; it is easy to see that
$R(t)R^T(t) = R^T(t)R(t) = I, \; \forall t \in \Bbb R. \tag{20}$
If we now assemble (4), (14), (15) and (19) together we may write
$\vec x(t) = \exp(p(t - t_0)) R(t) \vec x_0 \tag{21}$
as the solution to (1) with $x(t_0) = x_0$; this equation allows us to address stability questions concerning (1) directly in terms of $p$: first of all, suppose that
$p = 0; \tag{22}$
then (21) becomes
$\vec x(t) = R(t) \vec x_0, \tag{23}$
which has several relvant consequences: first of all, it implies
$\Vert x(t) \Vert = \Vert x_0 \Vert, \tag{24}$
since $R(t)$ is orthogonal; thus the solutions may be seen to be circles surrounding the origin; as such the system trajectories maintain a fixed distance from $(0, 0)$ and hence the system is stable; furthermore, (1) manifests another form of stability in the event that $p = 0$: suppose that we have two solutions $\vec x(t)$ and $\vec y(t)$ of (1), with
$\vec x(t_0) = \vec x_0, \; \vec y(t_0) = \vec y_0; \tag{25}$
then
$\dfrac{d}{dt}(\vec x(t) - \vec y(t)) = \dot{\vec x}(t) - \dot{\vec y}(t) = A \vec x(t) - A \vec y(t) = A(\vec x(t) - \vec y(t)); \tag{26}$
also,
$\vec x(t_0) - \vec y(t_0) = \vec x_0 - \vec y_0; \tag{27}$
based upon (25)-(27) we see that $\vec x(t) - \vec y(t)$ is a solution initialized at $\vec x_0 - \vec y_0$; thus, in accord with (23), (24) we may write
$(\vec x(t) - \vec y(t)) = R(t) (x_0 - y_0), \tag{28}$
$\Vert (\vec x(t) - \vec y(t) \Vert = \Vert x_0 - y_0 \Vert, \tag{29}$
which shows that the system (1) with $p = 0$ enjoys an even more comprehensive stability property: not only is $(0, 0)$ a stable equilibrium, but any two system points remain the same distance apart for all time; indeed, it is almost as if the flow of (1) is in fact a rigid motion of the plane; in any event, it is clear from (29) that any two solutions remain as close as they started out forever, an enhancement upon the simple stability of the origin.
We now turn to the case $p < 0$; we then see from (21) that
$\Vert \vec x(t) \Vert = \exp(p(t - t_0)) \Vert \vec x_0 \Vert \to 0 \; \text{as} \; t \to \infty \tag{30}$
for every solution $\vec x(t)$; thus the origin is globally asymptotically stable; also, it is easily seen that (25)-(27) in this case imply, via (21), that
$\Vert \vec x(t) - \vec y(t) \Vert = \exp(p(t - t_0)) \Vert \vec x_0 - \vec y_0 \Vert \to 0 \; \text{as} \; t \to \infty, \tag{31}$
i.e., all trajectories converge to one another with the passage of time. In this case, the origin is a spiral stable point if $q \ne 0$, and a stable node when $q$ vanishes.
When $p > 0$ the origin is of course unstable and all trajectories diverge from one another; the reader may easily work out the details along lines analogous to the above.
Finally, we have presented the above discussion in terms of $p$ even though the problem statement spoke in terms of eigenvalues of the matrix $A$; but the characteristic polynomial of the matrix $A$ is
$\det(A - \lambda I) = \lambda^2 - 2p \lambda + (p^2 + q^2); \tag{32}$
that the zeroes of (32) are $p \pm iq$ is an easy calculation, so in dealing with $p$ we are actually addressing the real part of the eigenvalues of $A$, as per request.
No, the condition you think of is for discrete-time dynamical systems.
Here in the continuous-time case you need that the eigenvalues $p\pm i\,q$ have non-positive real part. You need an extra test for the purely imaginary case, as that can not in general be decided from the eigenvalues alone.