Periodic averaging data: the average field, the oscillatory part and the periodic deviation

Let $n\ge 1$, let $P>0$, let $O\subseteq {\mathbb{R}}^n$ be open, and let $f:\mathbb{R}\times O\to {\mathbb{R}}^n$ be continuous (Vector-valued functions $f:A\to {\mathbb{R}}^m$, their limits and continuity, with the dictionary to the metric notions). Assume that for each standard coordinate vector $e_j$ of ${\mathbb{R}}^n$, $0\le j<n$, and every $(t,x)\in \mathbb{R}\times O$, the vector-valued derivative at $r=0$ of $r\mapsto f(t,x+r{e}_j)$ exists, and that each resulting partial derivative $(t,x)\mapsto {\mathrm{\partial }}_{x_j}f(t,x)$ is continuous on $\mathbb{R}\times O$ (Directional derivatives and partial derivatives of a map $U\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^n$). This is the meaning of continuously partially differentiable in the state variable here. Suppose also that $f$ is $P$-periodic in time, that is

$f(t+P,x)=f(t,x)\text{for all }(t,x)\in \mathbb{R}\times O\mathrm{.}$

The average of such an $f$ is the map

$\bar{f}:O⟶{\mathbb{R}}^n,\bar{f}(x):=\frac{1}{P}{\int }_0^{P}f(s,x)ds,$

where the integral is the componentwise Riemann integral of the continuous curve $s\mapsto f(s,x)$ over the compact interval $[0,P]$ (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length); it is well defined and finite for every $x\in O$ by componentwise application of A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion. The oscillatory part of $f$ is $f-\bar{f}$, and the deviation of $f$ is

$g:[0,P]\times O⟶{\mathbb{R}}^n,g(t,x):={\int }_0^t(f(s,x)-\bar{f}(x))ds\mathrm{.}$

Because ${\int }_0^{P}f(s,x)-\bar{f}(x)ds=P\bar{f}(x)-P\bar{f}(x)=0$ by the linearity clause of The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral and the componentwise constant-integral formula of If $m\le f\le M$ on $[a,b]$ then $m(b-a)\le L(f,P)\le \underset{‾}{{\int }_a^b}f\le \overline{{\int }_a^b}f\le U(f,P)\le M(b-a)$ for every partition $P$; in particular every constant function is integrable, with ${\int }_a^{b}c=c(b-a)$, the formula $g(t+kP,x):=g(t,x)$ for $k\in \mathbb{Z}$ and $t\in [0,P)$ defines a unique extension, again written $g$, to all of $\mathbb{R}\times O$; it satisfies $g(t+P,x)=g(t,x)$ and $g(0,x)=0$ for all $(t,x)$, and on $[0,P]$ it is the integral displayed above.

The averaged (or reduced) system of $f$ for a parameter $\epsilon >0$ is the autonomous first-order system on $O$

$y^{\mathrm{\prime }}(t)=\epsilon \bar{f}(y(t))\mathrm{.}$

• Scope. The objects $\bar{f}$, $g$ and the averaged system are pure mathematics: $f$ is an arbitrary continuous, $P$-periodic, continuously differentiable-in-state map on $\mathbb{R}\times O$. No physical reading of $t$, $x$, $\epsilon$ or $f$ is asserted here; a physical model that identifies them with time, state, a small parameter and a force field is the consumer's responsibility.

• The deviation is bounded on one period. For nonempty compact $K\subseteq O$, $[0,P]\times K$ is compact by Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, so $f$ is uniformly continuous there by Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous. The integral norm inequality and monotonicity (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 $m(b-a)\le {\int }_a^{b}f\le M(b-a)$) give $\mathrm{\parallel }\bar{f}(x)-\bar{f}(y){\mathrm{\parallel }}_2\le {\sup }_{s\in [0,P]}\mathrm{\parallel }f(s,x)-f(s,y){\mathrm{\parallel }}_2$, hence continuity of $\bar{f}$ on $K$. Put $u(s,x)=f(s,x)-\bar{f}(x)$. Then $u$ is continuous on the compact product and $C_K:={\max }_{[0,P]\times K}\mathrm{\parallel }u{\mathrm{\parallel }}_2$ is finite by A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value. For $t,r\in [0,P]$ and $x,y\in K$, splitting the integral at $r$ and using its orientation gives $\mathrm{\parallel }g(t,x)-g(r,y){\mathrm{\parallel }}_2\le P{\sup }_{s\in [0,P]}\mathrm{\parallel }u(s,x)-u(s,y){\mathrm{\parallel }}_2+C_K\mathrm{\mid }t-r\mathrm{\mid }$. Uniform continuity of $u$ proves joint continuity of $g$ on that product, so its norm has a finite maximum there. The periodic extension preserves this bound; for $K=\mathrm{\varnothing }$ boundedness is vacuous. Quantitative bounds with explicit constants are supplied separately.