stability (Lyapunov) কী?

Example 19.9 (The plane, classified)

For $x' = Ax$ with $A \in M_2(\R)$ invertible, the phase portrait near $0$ is decided by $\tau = \operatorname{tr}A$ and $\delta = \det A$, through the eigenvalues $\lambda_\pm = \frac{\tau \pm \sqrt{\tau^2 - 4\delta}}2$:

• $\delta < 0$: real eigenvalues of opposite signs — a saddle; two trajectories enter, two leave, all others fly by. Always unstable.
• $\delta > 0$, $\tau^2 \geq 4\delta$: real eigenvalues of the same sign ($= \operatorname{sign}\tau$) — a node, stable iff $\tau < 0$; trajectories are tangent to the slow eigendirection.
• $\delta > 0$, $\tau^2 < 4\delta$, $\tau \neq 0$: complex conjugate eigenvalues $\frac\tau2 \pm \iu\omega$ — a spiral (focus), stable iff $\tau < 0$; the solutions are $\eu^{\tau t/2}\times$ rotations of period $\frac{2\pi}\omega$.
• $\tau = 0$, $\delta > 0$: purely imaginary eigenvalues — a center: closed orbits (ellipses), stability without asymptotic stability, exactly the borderline that Theorem 19.12 cannot decide for nonlinear systems (the pendulum’s bottom equilibrium, Problem 19.1, sits here).

The boundary parabola $\tau^2 = 4\delta$ carries the degenerate nodes (Jordan blocks: trajectories with a single tangent direction). Everything is read off two numbers — which is why the first reflex before any planar phase portrait is to compute $\operatorname{tr}$ and $\det$; e.g. $A = \bigl(\begin{smallmatrix}0 & 1\\ -1 & -c\end{smallmatrix}\bigr)$ (damped oscillator): $\delta = 1 > 0$, $\tau = -c$: stable spiral for $0 < c < 2$, stable node for $c \geq 2$ — underdamping versus overdamping, in one glance.