= Solution
Use the standard bounded <uniform ellipticity> assumption: $\xi\cdot a(y)\xi\ge\alpha|\xi|^2$ for some $\alpha>0$, with bounded periodic coefficients. It holds for the usual finite set of positive-conductivity phases. Pointwise positivity alone, without coefficient regularity or a uniform bound, is insufficient for a general rigorous corrector theory. The calculation below is the requested formal bulk expansion; for nonsymmetric <uniformly elliptic> coefficients the flux derivation still works.
Set $y=x/\epsilon$. In a <two-scale expansion for periodic conductivity>, differentiation becomes
$$
D_i=\partial_{x_i}+\epsilon^{-1}\partial_{y_i}.
$$
The leading, order $\epsilon^{-2}$, equation is
$$
-\partial_{y_i}(a_{ij}\partial_{y_j}u_0)=0.
$$
For fixed $x$, multiply by $u_0$ minus its cell mean and integrate over the periodic cell. Opposite-face terms cancel, giving
$$
0=\int_Y(\nabla_yu_0)\cdot a(y)\nabla_yu_0\,dy
\ge\alpha\int_Y|\nabla_yu_0|^2\,dy.
$$
Hence \b[$u_0$ is independent of $y$].
The order $\epsilon^{-1}$ equation is now
$$
\partial_{y_i}\left[a_{ij}\bigl(\partial_{x_j}u_0+\partial_{y_j}u_1\bigr)\right]=0.
$$
For each coordinate direction, choose a periodic zero-mean corrector $w_j$ satisfying the <periodic conductivity cell problem>
$$
\boxed{\partial_{y_i}\left(a_{ik}\partial_{y_k}w_j\right)=-\partial_{y_i}a_{ij}.}
$$
The right side has zero cell mean. In weak form,
$$
\int_Y\nabla\phi\cdot a\nabla w_j\,dy
=-\int_Y\nabla\phi\cdot ae_j\,dy
$$
for every periodic test function $\phi$. On the space of zero-mean <periodic functions>, the <Poincaré inequality> and <uniform ellipticity> give coercivity, so the corrector exists and is unique by the <Lax-Milgram theorem>. Discontinuous phase coefficients are handled by this <weak formulation>.
The order $\epsilon^{-1}$ solution is
$$
u_1(x,y)=w_j(y)\partial_{x_j}u_0(x)+\bar u_1(x),
$$
where the last term is independent of $y$ and does not affect leading flux. Thus the leading microscopic flux is
$$
h_i^{(0)}=a_{ik}(y)\left(\delta_{kj}+\partial_{y_k}w_j\right)\partial_{x_j}u_0.
$$
At order one the equation reads
$$
-\partial_{x_i}h_i^{(0)}
-\partial_{y_i}\left[a_{ij}\bigl(\partial_{x_j}u_1+\partial_{y_j}u_2\bigr)\right]=f.
$$
Average over $Y$. The second term integrates to zero by periodicity, and $|Y|=1$. Therefore
$$
\boxed{
a_{ij}^*=\int_Y\left(a_{ij}+a_{ik}\partial_{y_k}w_j\right)\,dy,\qquad
-\partial_{x_i}(a_{ij}^*\partial_{x_j}u_0)=f.
}
$$
The effective <tensor> is constant, so the last equation is exactly $-a_{ij}^*\partial_{x_i}\partial_{x_j}u_0=f$. The macroscopic <boundary condition> is $u_0=0$ on $\partial\Omega$. Higher bulk correctors usually do not individually vanish on the physical boundary; a <boundary corrector in periodic homogenization> handles this without changing the leading homogenized equation.
For symmetric $a$, a useful check on the <effective conductivity> is
$$
a_{ij}^*=\int_Y(e_i+\nabla w_i)\cdot a(e_j+\nabla w_j)\,dy.
$$
The extra term involving $\nabla w_i$ vanishes by the weak cell equation. This formula proves <symmetry>. With $w_\xi=\xi_jw_j$, it also gives $\xi\cdot a^*\xi\ge\alpha\int_Y|\xi+\nabla w_\xi|^2\ge\alpha|\xi|^2$, confirming that the effective <tensor> is <positive-definite>.
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