A bounded sinusoidal drive with unbounded motions: the undamped driven oscillator at $\mathrm{\Omega }={\omega }_0$ refutes the unqualified boundedness claim

Take the model data $m=1\mathrm{k}\mathrm{g}$, $k=1\mathrm{N}\mathrm{/}\mathrm{m}$, $c=0$, $f_0=1\mathrm{N}$, $\mathrm{\Omega }=1{\mathrm{s}}^{-1}$, so that ${\omega }_0=1$ and the drive is $t\mapsto \cos t$; it is bounded: $\mid \cos t\mid \le 1$ for all $t$ by [L4]. The hypothesis of the refuted statement therefore holds for this model.

By [L2] the function $x_p(t)=t\sin t\mathrm{/}2$ is a motion of this model on $\mathbb{R}$. At the times $t_k=\pi \mathrm{/}2+2\pi k$, $k\in \mathbb{N}$, one has $\sin t_k=1$ and hence $x_p(t_k)=t_k\mathrm{/}2\to +\mathrm{\infty }$ as $k\to \mathrm{\infty }$ by [L4]; therefore $x_p$ is not bounded on $[0,\mathrm{\infty })$, and the conclusion of the refuted statement fails.

In fact every maximal motion of this model is unbounded on $[0,\mathrm{\infty })$: by [L2] it equals $x_p+y$ on $\mathbb{R}$ with $y$ harmonic, and a bounded $y$ cannot compensate the unbounded $x_p$. Local motions are restrictions of these maximal motions; no unboundedness is claimed for a restriction to a bounded time interval. Hence the refuted statement fails at the level of a declared model of the framework, not merely through one exceptional solution.

The failure exhibited here is specific to the nonzero resonant drive. For the same $m,k,f_0,\mathrm{\Omega }$ but with $c=1\mathrm{/}5>0$, [L3] gives that every maximal motion is bounded; and for $c=0$ but with $\mathrm{\Omega }=2\mathrm{\ne }{\omega }_0$, [L3] again gives that every maximal motion is bounded. Clause 6 also records the zero-drive resonant case: when $c=0$, $\mathrm{\Omega }={\omega }_0$ and $f_0=0$, the equation is homogeneous and every maximal motion is bounded. Thus this example refutes the implication for nonzero undamped resonant data, while the stated bounded cases remain bounded.