Asymptotic stability

An equilibrium is asymptotically stable when it is Lyapunov stable and every trajectory starting sufficiently nearby converges to it.

An equilibrium has exponential stability when sufficiently small displacements satisfy an estimate $\Vert x(t)-x_{\ast }\Vert \le C{e}^{-\alpha t}\Vert x(0)-x_{\ast }\Vert$ for some $C,\alpha >0$.

An equilibrium point is quasi-asymptotically stable when some neighborhood of initial conditions has forward solutions converging to it. This attraction condition alone does not include Lyapunov stability. Asymptotic stability requires both attraction and Lyapunov stability. A positive Lyapunov function with nonpositive derivative and no nonstationary invariant trajectory in its zero-derivative set proves both when its sufficiently small sublevel sets are compact and forward invariant.

For $\dot{e}=Ae$, differentiation gives $d\Vert e{\Vert }^2/dt=e^T(A+A^T)e$. A uniform negative-definiteness bound with $\kappa >0$ yields exponential contraction by direct integration of this differential inequality. This is a sufficient criterion for asymptotic stability, stronger than requiring all eigenvalues of $A$ to have negative real parts. A merely semidefinite symmetric part needs additional analysis.

A Hurwitz stable matrix has all eigenvalues strictly in the open left half-plane. For $\dot{x}=Ax$, this is equivalent to $e^{At}x\to 0$ for every initial vector. Jordan blocks contribute polynomial factors times $e^{\lambda t}$, which decay exactly when their eigenvalues have negative real part. Translating an affine equilibrium reduces its attractivity to this same condition.