Proving simple homogenization in a 1-dimensional case If i have an interval I=]-1,1[, a smooth 1-periodic function $a:I \rightarrow \mathbb{R}$ and a function $u_\epsilon \in W^{1,2}_{0}(I)$ that satisfies following conditions: $$(1) \:-\frac{d}{dx}(a(\frac{x}{\epsilon})u'_\epsilon(x))=1$$
$$(2) \quad u_\epsilon(-1)=u_\epsilon(1)=0$$ how could i prove that $u_\epsilon \rightarrow \frac{(1-x^2)}{2}\int_{0}^{1}\frac{ds}{a(s)}$ uniformly, as $\epsilon \rightarrow 0$ in $L^{2}(I)$ (or in $W^{1,2}(I)$, im not exactly sure which is the proper space to inspect the convergence).
I have already concluded that $u'_{\epsilon}(x)=-\frac{x-C}{a(\frac{x}{\epsilon})}$, but if i integrate that it becomes quite a mess and i have no idea how to continue. For uniform convergence i dont know any other strategy than the definition. Anyone could give me a hint?
 A: Note that $a(x/ϵ)$ is $ϵ$-periodic, thus oscillating increasingly fast around a middle position.
Thus using $A=\int_0^1\frac{ds}{a(s)}$
$$
\int_{x}^{x+ϵ}\frac{t-C}{a(t/ϵ)}dt=\int_0^1\frac{x+ϵs-C}{a(x/ϵ+s)}d(x+ϵs)
=ϵA(x-C)+O(ϵ^2)=\frac A2[(x+ϵ-C)^2-(x-C)^2]+O(ϵ^2)
$$
In total this gives
$$
\int_{-1}^x\frac{C-t}{a(t/ϵ)}dt=\frac A2[-(x-C)^2+(-1-C)^2]+O(ϵ)=\frac A2[1-x^2-2C(1-x)]+O(ϵ)
$$
Applying the boundary conditions finds $C=O(ϵ)$ and thus the result.
$$
u_ϵ(x)=\frac{1-x^2}2\int_0^1\frac{ds}{a(s)}+O(ϵ).
$$
A: $\newcommand{\eps}{\varepsilon}$
Let me denote $a_\eps (x) := a(x/\eps)$ and $\bar{a}^{-1} := \int_{0}^{1} a^{-1}$, so that the homogenization becomes 
$$
-(a_\eps u_\eps)' = 1 
\quad \leadsto \quad 
-(\bar{a} \bar{u})' = 1.
$$
Note that 
$$
\int_{t_1}^{t_2} f a_\eps^{-1} \to \bar{a}^{-1} \int_{t_1}^{t_2} f
$$
for any reasonable function $f$ and any interval $[t_1,t_2]$. 

Once you know this, it's easy to finish your solution. You already know that 
\begin{equation}
\label{eq}
\tag{$\star$}
u'_\eps = - \frac{x-C_\eps}{a_\eps}.
\end{equation}
First, $C_\eps$ cannot be an arbitrary constant (we expect uniqueness of solutions, right?). Indeed, since $u_\eps$ is zero at the boundary, $C_\eps$ needs to be chosen so that $\int_{-1}^1 u'_\eps = 0$. There are two more things to do: 


*

*Use \eqref{eq} to determine $C_\eps$ and compute its limit. 

*Integrate \eqref{eq} to obtain the formula for $u_\eps$ and then take the limit.


In both steps, don't be afraid of the mess, because you just need to find the limit (the observation from the top should help). 

Edit. Let me be a little more explicit. 
Ad 1. Plugging in \eqref{eq} into $\int_0^1 u'_\eps = 0$ gives us 
$$
C_\eps \int_{-1}^1 a_\eps^{-1} = \int_{-1}^1 a_\eps^{-1} x. 
$$
Since the first integral tends to $\bar{a}^{-1}$ and the second tends to $\bar{a}^{-1} \int_{-1}^1 x = 0$, so in the limit we have $C_\eps \to 0$. 
Ad 2. The formula for $u_\eps$ is simply $u_\eps(t) = \int_0^t u'_\eps$. Looking at the two integrals 
\begin{align*}
\int_{-1}^t \frac{C_\eps}{a_\eps}
& \to 0, \\
\int_{-1}^t \frac{x}{a_\eps} 
& \to \bar{a}^{-1} \int_{-1}^t x = \bar{a}^{-1} \cdot \tfrac{x^2-1}{2},
\end{align*}
we end up with 
$$
u_\eps(t) \to \bar{a}^{-1} \cdot \tfrac{1-x^2}{2},
$$
as required. 
This only shows pointwise convergence, but it's not difficult to show more. Actually, $u'_\eps \to -\bar{a}^{-1} x$ uniformly, so we obtain strong $C^1$-convergence. 
