I have a query regarding two equalities in the lemma in the book.
But first I'll provide two definitions that one needs for this lemma.
Definition 4.2.4: Consider the vector field $f(x,t)$ with $f:\mathbb{R}^n\times \mathbb{R}\to \mathbb{R}^n$, Lipschitz continuous in $x$ on $D\subset \mathbb{R}^n$, $t\ge 0$;$f$ continuous in $t$ and $x$ on $\mathbb{R}^+\times D$. If the average $$\bar{f}(x)=\lim_{T\to\infty} \frac{1}{T}\int_{0}^T f(x,s)ds$$ exists and the limit is uniform in $x$ on compact sets $K\subset D$, then $f$ is called a KBM-vector field.
Notation 4.3.3: In the general setup, we have to define inductively a number of order function $\delta(\epsilon)$. Let $\kappa$ be a counter, starting at $0$. Let $\delta(\epsilon)=\epsilon$ and increase $\kappa$ by one. The general induction step runs as follows. Let $I_{\kappa}$ be a multi-index, written as $I_{\kappa}=\iota_0|\ldots | \iota_m$, $m<\kappa$, where we don't write trailing zeros. Each $\iota_j$ stands for multiplicity of the order function $\delta_{I_j}(\epsilon)$ in the expression: $$\delta_{I_j}(\epsilon)=\sup_{x\in D}\sup_{t\in [0,L/\epsilon)}\pi_{I_j}(\epsilon)\|\int_0^t [f^{I_j}(x,s)-\bar{f}^{I_j}(x)]ds\|,$$ with $$\pi_{I_j}(\epsilon) = \prod_{k=0}^{j-1}\delta_{I_k}^{\iota_k}(\epsilon).$$ By putting $j=\kappa$ in these formulae, we obtain the definition of $\delta_{I_\kappa}(\epsilon)$. If the RHS doesn't exist for $j=\kappa$, the theory stops here. If it does exist, we proceed to define: $$\delta_{I_{\kappa}}(\epsilon)u^{I_{\kappa}}(w,t)=\pi_{I_{\kappa}}(\epsilon)\int_0^t[f^{I_{\kappa}}(w,s)-\bar{f}^{I_\kappa}(w)]ds$$ This definition implies that $u^{I_\kappa}$ is bounded by a constant, independent of epsilon. We then increase $\kappa$ by $1$ and repeat our induction step, as far as necessary for the estimates we want to obtain.
Now for the lemma and its proof plus the parts that I don't understand from the proof.
Lemma 4.5.1 Suppose $f^1$ is a KBM-vector field which has a Lipschitz continuous first derivative in $x$; $x\in D\subset \mathbb{R}^n$, $t$ on the time scale $1/\epsilon$; $x$ is the solution of: $$\dot{x}=\epsilon f^1(x,t), \ \ \ x(0)=a$$ We define $y$ by $x(t)=y(t)+\delta_1(\epsilon)u^1(y(t),t)$. Then $$y(t)=a+\epsilon\int_0^t f^1(y(s))ds+\epsilon\delta_1(\epsilon)\int_0^t(Df^1(y(s),s)\cdot u^1(y(s),s)-Du^1(y(s),s)\cdot\bar{f}^1(y(s)))ds+\mathcal{O}(\delta_1^2)$$ on the time scale $1/\epsilon$.
Proof This is the standard computation: $$y(t)=x(t)-\delta_1(\epsilon)u^1(y(t),t)=$$ $$= a+\epsilon \int_0^t f^1(x(s),s)ds-\epsilon \int_0^t (f^1(y(s),s)-\bar{f}^1(y(s)))ds-\delta_1(\epsilon)\int_0^t Du^1(y(s),s)\cdot \frac{dy}{ds}ds=$$ $$= a+\epsilon \int_0^t \bar{f}^1(y(s))ds+\epsilon\delta_1(\epsilon)\int_0^t(Df^1(y(s),s)\cdot u^1(y(s),s)-Du^1(y(s),s)\cdot \bar{f}^1(y(s),s))ds+\mathcal{O}(\delta_1^2)$$
Now, for my question how did they get the second equality? I.e, why is: $\delta_1(\epsilon)u^1(y(t),t)=\epsilon \int_0^t (f^1(y(s),s)-\bar{f}^1(y(s),s))ds+\delta_1(\epsilon)\int_0^t Du^1(y(s),s)\cdot \frac{dy}{ds}ds$? I understand that it follows from the Notation 4.3.3, but I don't see how exactly?
My second question is how to infer the third equality, it seems to me I should show that: $$\epsilon\delta_1(\epsilon)\int_0^t(Df^1(y(s),s)\cdot u^1(y(s),s)-Du^1(y(s),s)\cdot \bar{f}^1(y(s),s))ds+\mathcal{O}(\delta_1^2)=\epsilon \int_0^t (f^1(x(s),s)-f^1(y(s),s))ds+\delta_1(\epsilon)\int_0^t Du^1(y(s),s)\cdot \frac{dy}{ds}ds$$ but how exactly?
I appreciate your help!
P.S the book was written by Jan A. Sanders, Ferdinand Verhulst and James Murdock, and it's the second edition.
