Lyapunov stability does not imply attraction: the harmonic model is stable at its equilibrium while every nonzero datum keeps its distance and never converges

The equilibrium is Lyapunov stable: $U(q)=\frac{1}{2}{q}^2$ has the strict local minimum $q=0$ ($U(q)>U(0)$ for $q\mathrm{\ne }0$), so [L1] makes $(0,0)$ Lyapunov stable in the chart model.

The nonzero datum does not converge: by [L2] the datum $(\delta ,0)$ at time $t_0=0$ has the maximal motion $x(t)=\delta \cos t$, with $A=\delta$, $B=0$, $R=\delta >0$; the least positive period is $2\pi$ and $x(2\pi n)=\delta \cos (2\pi n)=x(0)=\delta$ for every integer $n$. Hence the sequence $t_n=2\pi n$ tends to infinity while $\parallel (x(2\pi n),x^{\mathrm{\prime }}(2\pi n))-(0,0){\parallel }_2\ge \delta$; the state does not converge to the equilibrium.

Conclusion: the equilibrium is Lyapunov stable (step 1.1) but for every $\delta >0$ the datum $(\delta ,0)$, arbitrarily close to the equilibrium, fails to converge (step 1.2); the equilibrium is therefore not asymptotically stable, and the implication of the refuted statement is false for the declared model. The stronger statement of Energy stability and instability of the mechanical model at an equilibrium: the strict minimum is Lyapunov stable, a negative Hessian eigenvalue gives linear instability, a negative definite potential Hessian gives nonlinear instability, and exact energy conservation rules out attraction clause 4 explains the failure: the energy $\frac{1}{2}(v^2+q^2)$ is conserved, equals ${\delta }^2\mathrm{/}2$ on this datum, and no state of different energy can converge to the equilibrium.