Krylov–Bogolyubov averaging

Remarks

The first averaging theorem justifies the method. Let $x$ and $y$ solve the two equations with $x(0) = y(0)$, and let $L > 0$ be such that $y(t)$ stays in a compact subset of $D$ for $0 \le t \le L/\varepsilon$. Then there is a constant $C$, independent of $\varepsilon$, with $\norm{x(t) - y(t)} \le C \varepsilon$ for $0 \le t \le L/\varepsilon$ and all small $\varepsilon$.

Example: $f(x, t) = x \sin^2 t$ on $\R$ with $T = \pi$. Then $\bar f(y) = \frac{1}{\pi} \int_0^\pi y \sin^2 t\, dt = \frac{1}{2} y$. The solutions are $x(t) = x_0 e^{\varepsilon (t/2 - \sin(2t)/4)}$ and $y(t) = x_0 e^{\varepsilon t/2}$. Their ratio is $e^{-\varepsilon \sin(2t)/4} = 1 + O(\varepsilon)$, so the difference is $O(\varepsilon)$ on intervals of length of order $1/\varepsilon$.

Non-example: $x' = \varepsilon x t$ is not periodic in $t$, so it is not in the standard form above, and there is no period over which to average.

Weakly nonlinear oscillators $u'' + u = \varepsilon h(u, u')$ are put in standard form by writing $u = r \cos(t + \phi)$, $u' = -r \sin(t + \phi)$ and using $(r, \phi)$ as the new unknowns.