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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 $y=x/ϵ$. In a two-scale expansion for periodic conductivity, differentiation becomesThe leading, order ${ϵ}^{-2}$, equation isFor fixed $x$, multiply by $u_0$ minus its cell mean and integrate over the periodic cell. Opposite-face terms cancel, givingHence $u_0$ is independent of $y$.

The order ${ϵ}^{-1}$ equation is nowFor each coordinate direction, choose a periodic zero-mean corrector $w_j$ satisfying the periodic conductivity cell problemThe right side has zero cell mean. In weak form,for every periodic test function $\varphi$. 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 ${ϵ}^{-1}$ solution iswhere the last term is independent of $y$ and does not affect leading flux. Thus the leading microscopic flux isAt order one the equation readsAverage over $Y$. The second term integrates to zero by periodicity, and $|Y|=1$. ThereforeThe effective tensor is constant, so the last equation is exactly $-a_{ij}^{\ast }{\partial }_{x_i}{\partial }_{x_j}{u}_0=f$. 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 isThe extra term involving $\nabla w_i$ vanishes by the weak cell equation. This formula proves symmetry. With $w_{\xi }={\xi }_j{w}_j$, it also gives $\xi \cdot a^{\ast }\xi \ge \alpha {\int }_Y|\xi +\nabla w_{\xi }{|}^2\ge \alpha |\xi {|}^2$, confirming that the effective tensor is positive-definite.