we consider the equation $$ -u''(x)+ q(\xi)=0, $$ where $\xi=\dfrac{x}{\epsilon}$. We get $$ u(x)= \sum_{l+p=0}^k \epsilon^{l+p} N_l^p(\xi) D^l v(x) $$ where $N_l^p$ are $1-$ periodic functions, and $D^l= \dfrac{d^l}{dx^l}$. We plug $u$ in the equation, then the autor find that $$ -D^2 u+q(\xi) u= -\sum_{l+p}^{k+1} \epsilon^{l+p-2} H_l^p D^l v(x)+ B^{\epsilon,k}_l $$ where $$ H_l^p= \dfrac{d^2 N_l^p}{d \xi^2}(\xi)+2 \dfrac{d N^p_{l-1}}{d \xi}(\xi) + N_{l-2}^p(\xi) - q(\xi) N_l^{p-2}(\xi) $$ and $B_l^{\epsilon,k}$ is given by $$ B_l^{\epsilon,k}= \epsilon^k \left\{ -\sum_{p+l=k+2} (2 \dfrac{d N^p_{l-1}}{d\xi}(\xi)+ N^p_{l-2}(\xi)- q(\xi) N_l^{p-2}(\xi)) D^l v(x) -\sum_{p+l=k+3} \epsilon (N_{l-2}^p(\xi) q(\xi) N_l^{p-2}(\xi)) D^l v(x) \right\} $$
I try to find the same result, so first we have that
\begin{align} -u''+q(\dfrac{x}{\epsilon})u &= -D^2\left(\sum_{l,p=0}^k \epsilon^{l+p} N_l^p(\xi) D^l v(x)\right) + q(\xi) \sum_{l+p=0}^k \epsilon^{l+p} N_l^p(\xi) D^l v(x)\\ &= -\sum_{l+p=0}^k \epsilon^{l+p} D^2\left(N_l^p(\xi) D^l v(x)\right) + q(\xi)\sum_{l+p=0}^k \epsilon^{l+p} N_l^p(\xi) D^lv(x) \end{align} and we have that $$ D^2\left(N_l^p(\xi) D^lv(x)\right)= \epsilon^{-2} D^2 N_l^p(\xi) D^l v(x) + 2 \epsilon^{-1}DN_l^p(\xi) D^{l+1} v + N_l^p(\xi) D^{l+2} v(x) $$
But i don't understand and i don't find how we find the same result of the author. Thank's to the help.