Averaging for a weakly damped oscillator: the rotating-frame amplitude, the averaged exponential energy decay and the order-$\epsilon$ bound on the $O(1\mathrm{/}\epsilon )$ time scale

Identification of the slow system. Let $x$ be a motion and put $u(t):={\omega }_{0}x(t)+i{x}^{\mathrm{\prime }}(t)$, so that $\zeta =e^{i{\omega }_{0}t}u\mathrm{/}\sqrt{2{\omega }_0}$ and $\mathrm{Im}u=\sqrt{2{\omega }_0}(\cos ({\omega }_{0}t)\mathrm{Im}\zeta -\sin ({\omega }_{0}t)\mathrm{Re}\zeta )$ by [F2]. From $x^{\mathrm{\prime }\mathrm{\prime }}=-{\omega }_0^{2}x-(\epsilon \gamma \mathrm{/}m)x^{\mathrm{\prime }}$ of the model, $u^{\mathrm{\prime }}={\omega }_0{x}^{\mathrm{\prime }}+i{x}^{\mathrm{\prime }\mathrm{\prime }}={\omega }_0{x}^{\mathrm{\prime }}-i{\omega }_0^{2}x-i\frac{\epsilon \gamma }{m}{x}^{\mathrm{\prime }}=-i{\omega }_{0}u-i\frac{\epsilon \gamma }{m}\mathrm{Im}u,$ so differentiating $\zeta$ with the product and chain rules of [F2] gives ${\zeta }^{\mathrm{\prime }}=-\frac{i\epsilon \gamma }{m}\frac{e^{i{\omega }_{0}t}}{\sqrt{2{\omega }_0}}\mathrm{Im}u=\epsilon F(t,\zeta )$. Conversely, if $\zeta$ solves ${\zeta }^{\mathrm{\prime }}=\epsilon F(t,\zeta )$ and $u:=\sqrt{2{\omega }_0}{e}^{-i{\omega }_{0}t}\zeta$, then $u^{\mathrm{\prime }}=-i{\omega }_{0}u-i(\epsilon \gamma \mathrm{/}m)\mathrm{Im}u$, so $x:=\mathrm{Re}(u)\mathrm{/}{\omega }_0$ satisfies $x^{\mathrm{\prime }}=\mathrm{Re}(u^{\mathrm{\prime }})\mathrm{/}{\omega }_0=\mathrm{Im}u$ and $x^{\mathrm{\prime }\mathrm{\prime }}=\mathrm{Im}(u^{\mathrm{\prime }})=-{\omega }_0\mathrm{Re}u-(\epsilon \gamma \mathrm{/}m)\mathrm{Im}u=-{\omega }_0^{2}x-(\epsilon \gamma \mathrm{/}m)x^{\mathrm{\prime }}$, that is, $x$ is a motion with the stated inverse formula. The two constructions are inverse to each other because $x$ and $x^{\mathrm{\prime }}$ are recovered from $\zeta$ and $\zeta$ is recovered from $x,x^{\mathrm{\prime }}$.

Smoothness, periodicity and average. $F$ is smooth on $\mathbb{R}\times \mathbb{C}$ because $\zeta \mapsto F(t,\zeta )$ is real-linear and the coefficients $t\mapsto e^{i{\omega }_{0}t}\cos ({\omega }_{0}t)$ and $t\mapsto e^{i{\omega }_{0}t}\sin ({\omega }_{0}t)$ are smooth; it is $P$-periodic with $P=2\pi \mathrm{/}{\omega }_0$ because $e^{i{\omega }_0(t+P)}=e^{i{\omega }_{0}t}$ and $\cos ({\omega }_0(t+P))=\cos ({\omega }_{0}t)$, $\sin ({\omega }_0(t+P))=\sin ({\omega }_{0}t)$. Expanding the product, $F(t,\zeta )=\frac{\gamma }{m}(\sin ({\omega }_{0}t)\cos ({\omega }_{0}t)\mathrm{Im}\zeta -{\sin }^2({\omega }_{0}t)\mathrm{Re}\zeta )-\frac{i\gamma }{m}({\cos }^2({\omega }_{0}t)\mathrm{Im}\zeta -\sin ({\omega }_{0}t)\cos ({\omega }_{0}t)\mathrm{Re}\zeta ),$ so by [F3] the one-period average of the first bracket is $-\gamma \mathrm{Re}\zeta \mathrm{/}(2m)$ and that of the second is $-\gamma \mathrm{Im}\zeta \mathrm{/}(2m)$, giving $\bar{F}(\zeta )=-\frac{\gamma }{2m}(\mathrm{Re}\zeta +i\mathrm{Im}\zeta )=-\frac{\gamma }{2m}\zeta$. The averaged system is linear with constant coefficient and has the displayed global exponential solution. This is clause 2.

Bounds on the ball. For $\mid \zeta -{\zeta }_0\mid \le R$ one has $\mid \zeta \mid \le \mid {\zeta }_0\mid +R$, so $\parallel F(t,\zeta ){\parallel }_2=\frac{\gamma }{m}\mid \cos ({\omega }_{0}t)\mathrm{Im}\zeta -\sin ({\omega }_{0}t)\mathrm{Re}\zeta \mid \le \frac{\gamma }{m}(\mid \mathrm{Im}\zeta \mid +\mid \mathrm{Re}\zeta \mid )\le \frac{\gamma }{m}\sqrt{2}(\mid {\zeta }_0\mid +R)=M\mathrm{.}$ The state derivative is $D_{\zeta }F(t,\zeta )h=-(i\gamma \mathrm{/}m)e^{i{\omega }_{0}t}(\cos ({\omega }_{0}t)\mathrm{Im}h-\sin ({\omega }_{0}t)\mathrm{Re}h)$, whose modulus is at most $(\gamma \mathrm{/}m)\sqrt{2}\mid h\mid$ by the same inequality, so $\parallel D_{\zeta }F(t,\zeta )\parallel \le L=(\gamma \mathrm{/}m)\sqrt{2}$ by [F5].

Existence and uniqueness. By [F1] every datum has exactly one maximal motion and it is defined on all of $\mathbb{R}$; by step 1.1 its amplitude solves the slow system on $\mathbb{R}$ with the datum $\zeta (0)$, and distinct motions have distinct amplitudes. Conversely a solution of the slow system on $\mathbb{R}$ produces a motion by step 1.1; this proves the bijection of clause 1.

Averaged trajectory and the estimate. The averaged solution is $\bar{\zeta }(t)={\zeta }_0{e}^{-\epsilon \gamma t\mathrm{/}(2m)}$, and its distance from ${\zeta }_0$ is $\mathrm{\mid }{\zeta }_0\mathrm{\mid }(1-e^{-\epsilon \gamma t\mathrm{/}(2m)})$. Thus the trajectory stays in the required ball exactly when the endpoint displacement is at most $R-\rho$; [F6] gives the equivalent logarithmic sufficient condition in clause 3. Apply the numerical rescaling specified in the Statement; write its numerical variables and constants without hats in this argument. The exact amplitude is already global by step 2.1. For $g(t,w):={\int }_0^t(F(s,w)-\bar{F}(w))ds$, periodicity and the vanishing one-period integral give $g(0,w)=0$, $\mathrm{\mid }g(t,w)\mathrm{\mid }\le 2MP=B$ and $\mathrm{\parallel }{D}_{w}g(t,w)\mathrm{\parallel }\le 2LP=B_1$ on $\mathrm{\mid }w-{\zeta }_0\mathrm{\mid }\le R$: reduce $t$ modulo $P$ and integrate the bounds $2M$, $2L$ on the remaining interval. Smoothness of $F$ justifies differentiating this primitive in $w$. Before the first exit of the global exact amplitude from this ball, put $d=\zeta -\epsilon g(t,\zeta )$ and $w=d-\bar{\zeta }$. The chain rule gives $w^{\mathrm{\prime }}=\epsilon (\bar{F}(\zeta )-\bar{F}(\bar{\zeta }))-{\epsilon }^2{D}_{w}g(t,\zeta )F(t,\zeta )$, with $w(0)=0$, hence $\mathrm{\mid }w(t)\mathrm{\mid }\le {\int }_0^t(\epsilon L\mathrm{\mid }w(s)\mathrm{\mid }+{\epsilon }^2(LB+B_{1}M))ds$. Put $a=\epsilon L$ and $b={\epsilon }^2(LB+B_{1}M)$, and $r(t)={\int }_0^t(a\mathrm{\mid }w(s)\mathrm{\mid }+b)ds$. Then $\mathrm{\mid }w\mathrm{\mid }\le r$, $r(0)=0$, $r^{\mathrm{\prime }}\le ar+b$ and $(e^{-at}r(t){)}^{\mathrm{\prime }}\le b{e}^{-at}$; integration gives $\mathrm{\mid }w(t)\mathrm{\mid }\le r(t)\le bt{e}^{at}={\epsilon }^2(LB+B_{1}M)t{e}^{\epsilon Lt}$. Thus on $0\le t\le A\mathrm{/}\epsilon$ before any exit, $\mathrm{\mid }\zeta -\bar{\zeta }\mathrm{\mid }\le \epsilon [B+(LB+B_{1}M)A{e}^{LA}]\le \epsilon K$. At a first exit time in this horizon, continuity would imply $R=\mathrm{\mid }\zeta -{\zeta }_0\mathrm{\mid }\le \epsilon K+(R-\rho )<R$, a contradiction. There is no finite-existence-endpoint issue because the exact curve is global. The estimate therefore holds over the whole horizon; undoing the explicitly stated time/amplitude rescaling gives clause 3 with its physical units.

Energy identity and estimate. With $\zeta =(1\mathrm{/}\sqrt{2{\omega }_0})({\omega }_{0}x+i{x}^{\mathrm{\prime }})e^{i{\omega }_{0}t}$ one has $2{\omega }_0\mid \zeta {\mid }^2={\omega }_0^2{x}^2+x^{\mathrm{\prime }2}$ by [F2], hence $m{\omega }_0\mid \zeta {\mid }^2=\frac{1}{2}m{x}^{\mathrm{\prime }2}+\frac{1}{2}m{\omega }_0^2{x}^2=\frac{1}{2}m{x}^{\mathrm{\prime }2}+\frac{1}{2}k{x}^2=E\mathrm{.}$ For $0\le t\le A\mathrm{/}\epsilon$, $\mid \mid \zeta (t){\mid }^2-\mid \bar{\zeta }(t){\mid }^2\mid =\mid \mid \zeta (t)\mid -\mid \bar{\zeta }(t)\mid \mid (\mid \zeta (t)\mid +\mid \bar{\zeta }(t)\mid )\le \epsilon K(2\mid {\zeta }_0\mid +\epsilon K)$, because $\mid \zeta (t)\mid \le \mid \bar{\zeta }(t)\mid +\epsilon K\le \mid {\zeta }_0\mid +\epsilon K$; multiplying by $m{\omega }_0$ and using $m{\omega }_0\mid \bar{\zeta }(t){\mid }^2=E_0{e}^{-\epsilon \gamma t\mathrm{/}m}$ gives the displayed energy estimate; this is clause 4.

Clauses 1-4 are proved. The averaged equation defines a different model from the exact one; the estimate is finite-time, of order $\epsilon$ in the amplitude, and holds only on the stated horizon under the smallness and trapping hypotheses; no exact decay law and no empirical qualification is asserted.