Proving simple homogenization in a 1-dimensional case

Question

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?

Answer 1

$\newcommand{\eps}{\varepsilon}$ 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 $[t_1,t_2]$.

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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:

1. Use \eqref{eq} to determine $C_\eps$ and compute its limit.
2. 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).

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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.

Answer 2

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. $$ u_ϵ(x)=\frac{1-x^2}2\int_0^1\frac{ds}{a(s)}+O(ϵ). $$