Boundary-cutoff corrector estimate for regular periodic data

Let $\theta (s)=0$ for $s\le 1$, $\theta (s)=1$ for $s\ge 2$, and linear on $[1,2]$, and set ${\eta }_{ϵ}(x)=\theta (\mathrm{dist}(x,\mathrm{\partial }Q)\mathrm{/}ϵ)$. Distance to a closed set is $1$-Lipschitz, so this cutoff has the required gradient bound. Set $U_{ϵ}=u_0+ϵ{\sum }_j{\chi }_j(x\mathrm{/}ϵ)g_j(x)$ with $g_j={\mathrm{\partial }}_j{u}_0$. For each corrector, its defining smooth periodic approximants converge in $H^1$ on any bounded union of period cells, since that union meets only finitely many cells. Applying the affine pullback rule [F8] with $y=x\mathrm{/}ϵ$ shows ${\chi }_j(x\mathrm{/}ϵ)\in H^1(Q)$ with weak gradient ${ϵ}^{-1}{\mathrm{\nabla }}_y{\chi }_j(x\mathrm{/}ϵ)$. Multiplication by $g_j\in C^1(\bar{Q})$ preserves $H^1$: apply the classical product rule to the smooth periodic approximants and pass in $H^1$, using the boundedness of $g_j$ and $\mathrm{\nabla }{g}_j$. The cutoff correction is an $H^1(Q)$ function supported in the compact set $\{x\in \bar{Q}:\mathrm{dist}(x,\mathrm{\partial }Q)\ge ϵ\}\subset Q$; [F9] puts it in $H_0^1(Q)$, so $w_{ϵ}\in H_0^1(Q)$. Existence and uniqueness of $u_{ϵ}$ follow from [F7]. Differentiating the uncut expression by the weak chain and product rules gives $a(x\mathrm{/}ϵ)\mathrm{\nabla }{U}_{ϵ}=A^{\ast }\mathrm{\nabla }{u}_0+{\sum }_j{q}^j(x\mathrm{/}ϵ)g_j+ϵa(x\mathrm{/}ϵ){\sum }_j{\chi }_j(x\mathrm{/}ϵ)\mathrm{\nabla }{g}_j\mathrm{.}$ Subtract the homogenized weak equation for $u_0$.

Extend each periodic potential ${\sigma }^j$ to the bounded region ${ϵ}^{-1}Q$ as the $H^1$ limit of its smooth periodic approximants; the finite-cell bound gives convergence there. For $v\in C_c^{\mathrm{\infty }}(Q)$, [F8] gives $q_i^j(x\mathrm{/}ϵ)=ϵ{\mathrm{\partial }}_{x_k}{\sigma }_{ik}^j(x\mathrm{/}ϵ)$ in distributions. The product $g_j{\mathrm{\partial }}_{i}v$ is $H^1(Q)$ with compact support, so [F9] places it in $H_0^1(Q)$ and the distributional identity extends to this test by density. Integration by parts and antisymmetry of ${\sigma }_{ik}^j$ cancel the term containing ${\mathrm{\partial }}_{ki}v$, since the Hessian is symmetric; the remaining term is ${\int }_Q{q}_i^j(x\mathrm{/}ϵ)g_j{\mathrm{\partial }}_{i}v=-ϵ{\int }_Q{\sigma }_{ik}^j(x\mathrm{/}ϵ)({\mathrm{\partial }}_k{g}_j)({\mathrm{\partial }}_{i}v)\mathrm{.}$ Approximate the periodic potential by smooth periodic functions and pass in $L^2$. By [F5] and boundedness of $D^2{u}_0$, this has magnitude at most $Cϵ\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2$. The last flux term in step 1.1 has the same bound by [F1], [F5], and boundedness of $D^2{u}_0$. Density in [F6] extends the residual estimate to every $v\in H_0^1(Q)$: $\mid {\int }_{Q}a(x\mathrm{/}ϵ)\mathrm{\nabla }{U}_{ϵ}\cdot \mathrm{\nabla }v-{\int }_{Q}fv\mid \le Cϵ\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2\mathrm{.}$.

Let $S_{2ϵ}=\{x\in Q:\mathrm{dist}(x,\mathrm{\partial }Q)<2ϵ\}$. It is covered by at most $C_d{ϵ}^{1-d}$ cubes of the lattice $ϵ(k+Y)$: along each of the $2d$ faces there are at most a fixed number of normal layers and at most $C_d{ϵ}^{-(d-1)}$ tangential cells. Each cell contributes at most ${ϵ}^d\mathrm{\parallel }g{\mathrm{\parallel }}_{L^2(Y)}^2$ to $\int \mathrm{\mid }g(x\mathrm{/}ϵ){\mathrm{\mid }}^2$. Thus $\mathrm{\parallel }g(\cdot \mathrm{/}ϵ){\mathrm{\parallel }}_{L^2(S_{2ϵ})}\le C_d\sqrt{ϵ}\mathrm{\parallel }g{\mathrm{\parallel }}_{L^2(Y)}$ for every periodic $g\in L^2(Y)$. Applying this to ${\chi }_j$ and $\mathrm{\nabla }{\chi }_j$, and differentiating $w_{ϵ}-U_{ϵ}=ϵ({\eta }_{ϵ}-1){\sum }_j{\chi }_j(x\mathrm{/}ϵ)g_j$, gives $\mathrm{\parallel }\mathrm{\nabla }(w_{ϵ}-U_{ϵ}){\mathrm{\parallel }}_2\le C\sqrt{ϵ}$: the three product-rule terms are bounded respectively by $C\mathrm{\parallel }\mathrm{\nabla }{\chi }_j(\cdot \mathrm{/}ϵ){\mathrm{\parallel }}_{L^2(S)}$, $Cϵ\mathrm{\parallel }{\chi }_j(\cdot \mathrm{/}ϵ){\mathrm{\parallel }}_{L^2(S)}$, and $Cϵ\mathrm{\parallel }\mathrm{\nabla }{\eta }_{ϵ}{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{\chi }_j(\cdot \mathrm{/}ϵ){\mathrm{\parallel }}_{L^2(S)}$.

The residual for $w_{ϵ}$ is the residual for $U_{ϵ}$ plus ${\int }_{Q}a(x\mathrm{/}ϵ)\mathrm{\nabla }(w_{ϵ}-U_{ϵ})\cdot \mathrm{\nabla }v$. By [F1], steps 2.1 and 2.2, its magnitude is at most $C\sqrt{ϵ}\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2$. Put $e_{ϵ}=u_{ϵ}-w_{ϵ}\in H_0^1(Q)$ and test the difference equation with $e_{ϵ}$. Coercivity gives $\lambda \mathrm{\parallel }\mathrm{\nabla }{e}_{ϵ}{\mathrm{\parallel }}_2^2\le C\sqrt{ϵ}\mathrm{\parallel }\mathrm{\nabla }{e}_{ϵ}{\mathrm{\parallel }}_2,$ so $\mathrm{\parallel }\mathrm{\nabla }{e}_{ϵ}{\mathrm{\parallel }}_2\le C\sqrt{ϵ}$ (trivial if it vanishes). Finally [F6] gives $\mathrm{\parallel }{e}_{ϵ}{\mathrm{\parallel }}_2\le C\sqrt{ϵ}$, while $w_{ϵ}-u_0=ϵ{\eta }_{ϵ}{\sum }_j{\chi }_j(x\mathrm{/}ϵ)g_j$ has $L^2$ norm at most $Cϵ$ by [F5]. The triangle inequality proves both claimed bounds.