Averaging for fast periodic oscillations: order-$\epsilon$ agreement on the $O(1\mathrm{/}\epsilon )$ time scale

Given: The data $n,P,O,f,x_0,\rho ,R,M,L$, the derived $\bar{f},g,B,B_1$, and the parameters $\epsilon >0$, $A>0$, with a solution $y:[0,A\mathrm{/}\epsilon ]\to \bar{B}(x_0,R-\rho )$ of the averaged system.

Derivative-to-Lipschitz bound and periodic-deviation bounds: for each fixed $t$, the map $x\mapsto f(t,x)$ is $L$-Lipschitz on the convex ball $\bar{B}(x_0,R)$. Indeed, for $x,y$ in the ball the segment $x+s(y-x)$ lies in it, the composite $s\mapsto f(t,x+s(y-x))$ is differentiable with derivative $D_{x}f(t,x+s(y-x))(y-x)$, and the vector-valued fundamental theorem estimate gives $\mathrm{\parallel }f(t,x)-f(t,y){\mathrm{\parallel }}_2\le L\mathrm{\parallel }x-y{\mathrm{\parallel }}_2$ (If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative, The chain rule for total derivatives: $D(g\circ f)(a)=Dg(f(a))\circ Df(a)$, If $f:[a,b]\to {\mathbb{R}}^m$ is differentiable with integrable $f^{\mathrm{\prime }}$ then ${\int }_a^b{f}^{\mathrm{\prime }}=f(b)-f(a)$; and a bounded derivative makes $f$ Lipschitz). Thus Elementary bounds for the periodic average, the deviation and its time derivative applies and gives $\mathrm{\parallel }\bar{f}(x){\mathrm{\parallel }}_2\le M$ and $\mathrm{\parallel }\bar{f}(x)-\bar{f}(y){\mathrm{\parallel }}_2\le L\mathrm{\parallel }x-y{\mathrm{\parallel }}_2$ on the ball; $\mathrm{\parallel }g(t,x){\mathrm{\parallel }}_2\le B=2MP$ for $t\in \mathbb{R}$, $x\in \bar{B}(x_0,R)$; $\mathrm{\parallel }{D}_{x}g(t,x)\mathrm{\parallel }\le B_1=2LP$ there; ${\mathrm{\partial }}_{t}g(t,x)=f(t,x)-\bar{f}(x)$; and $g$ is totally differentiable on $\mathbb{R}\times O$. [given]

The field $F_{\epsilon }(t,x):=\epsilon f(t,x)$ is continuous. Near any $(t_{\ast },x_{\ast })\in \mathbb{R}\times O$, choose a time interval and a convex state ball contained in $\mathbb{R}\times O$ on which continuity of $D_{x}f$ bounds its operator norm by some finite $C$. For each fixed time in that interval, the line-segment argument of [F1], with $C$ in place of $L$, gives $\mathrm{\parallel }{F}_{\epsilon }(t,x)-F_{\epsilon }(t,y){\mathrm{\parallel }}_2\le \epsilon C\mathrm{\parallel }x-y{\mathrm{\parallel }}_2$ for states in that ball. This is local Lipschitz continuity in the state variable as defined in Local Lipschitz continuity in the state variable, locally uniform in time and parameters, so the Picard-Lindelof theorem gives a unique maximal solution of the initial-value problem (First-order systems, initial value problems, and solutions on intervals, Every Picard–Lindelöf initial value problem has one maximal solution on an open interval, Extensions, maximal solutions, maximal intervals, and global solutions). Every such solution satisfies the Volterra equation $x(t)=x_0+{\int }_0^t\epsilon f(s,x(s))ds$ (A first-order initial value problem is equivalent to its Volterra integral equation), and compact continuation extends a maximal solution whose graph stays in a compact subset of $[0,\mathrm{\infty })\times O$ near a finite endpoint (Compact continuation for locally Lipschitz ODEs: a solution whose graph stays in a compact subset of the domain near a finite endpoint extends past it). [given]

Chain rule: for a totally differentiable curve $G$ on $\mathbb{R}\times O$ and a differentiable curve $t\mapsto x(t)$, the composite $t\mapsto G(t,x(t))$ is differentiable with derivative ${\mathrm{\partial }}_{t}G+D_{x}G\cdot x^{\mathrm{\prime }}$ (The chain rule for total derivatives: $D(g\circ f)(a)=Dg(f(a))\circ Df(a)$). [given]

Integral inequalities: $\mathrm{\parallel }{\int }_a^b\varphi {\mathrm{\parallel }}_2\le \mathrm{\mid }{\int }_a^b\mathrm{\parallel }\varphi {\mathrm{\parallel }}_2\mathrm{\mid }$ and monotonicity for real integrands (For $a\le b$ and $f:[a,b]\to {\mathbb{R}}^m$ integrable when $a<b$, $\parallel {\int }_a^{b}f{\parallel }_2\le {\int }_a^b\parallel f{\parallel }_2$; for $a<b$, $\parallel f{\parallel }_2$ is integrable, If $f\le g$ on $[a,b]$ and both are integrable then ${\int }_a^{b}f\le {\int }_a^{b}g$; and $m(b-a)\le {\int }_a^{b}f\le M(b-a)$, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral); Gronwall: if continuous $u\ge 0$ satisfies $u(t)\le L_0+M_{0}t+K_0{\int }_0^{t}u$ on $[0,T]$ with constants $\ge 0$, then $u(t)\le (L_0+M_{0}t)e^{K_{0}t}$ (apply the nondecreasing-$a$ clause of Gronwall's integral inequality with variable and constant coefficients with $a(t)=L_0+M_{0}t$ and $b(t)=K_0$). [given]