Homogenization, multiscale FEM and heterogeneous media

Let $v\in C_c^{\mathrm{\infty }}(Q)$ and write $x=(t,x^{\mathrm{\prime }})$ with first library coordinate $t=x_0$ and $x^{\mathrm{\prime }}=(x_1,\dots ,x_{d-1})$ (the empty tuple when $d=1$). For fixed $x^{\mathrm{\prime }}$ and $0<t<1$, the one-dimensional fundamental theorem gives $v(t,x^{\mathrm{\prime }})={\int }_0^t{\mathrm{\partial }}_{0}v(s,x^{\mathrm{\prime }})ds$. Cauchy--Schwarz and $t\le 1$ imply $\mathrm{\mid }v(t,x^{\mathrm{\prime }}){\mathrm{\mid }}^2\le {\int }_0^1\mathrm{\mid }{\mathrm{\partial }}_{0}v(s,x^{\mathrm{\prime }}){\mathrm{\mid }}^{2}ds$. Integrating first in $t$ and then in $x^{\mathrm{\prime }}$ gives $\mathrm{\parallel }v{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }{\mathrm{\partial }}_{0}v{\mathrm{\parallel }}_2^2$. By [F1], approximating any $v\in H_0^1(Q)$ in $H^1$ and passing to the limit proves the displayed Poincaré bound.

The map $J:H^1(Q)\to L^2(Q{)}^{d+1}$, $J(v)=(v,{\mathrm{\partial }}_{0}v,\dots ,{\mathrm{\partial }}_{d-1}v)$, is an isometry for the Sobolev norm by [F1]. The target is separable: extend each coordinate by zero to ${\mathbb{R}}^d$ and restrict the countable dense rational-box family [F3]. The image $J(H^1(Q))$ and its subset $J(H_0^1(Q))$ are separable subspaces of a second-countable metric space: intersect a countable base with the subspace and choose one point from each nonempty basic set, a countable selection allowed by Countable Choice. Thus $H_0^1(Q)$ is separable.

For $v\in C_c^{\mathrm{\infty }}(Q)$, extend it by zero to $D=(-1,2{)}^d$ and identify opposite faces of $D$ to obtain a smooth function on the torus of side length $3$. The extension preserves the $H^1$ norms. By [F1], for every $v\in H_0^1(Q)$ choose test functions converging to $v$ in $H^1$; their periodic zero extensions are Cauchy in $H^1$ on this torus, so the extension is in periodic $H^1(D)$ and the same norm bound holds.

Rescale $D$ affinely to the unit torus using [F7]; the characters on the side-length-$3$ torus are $e^{2\pi ik\cdot x\mathrm{/}3}$. For a periodic $H^1(D)$ function $V$, integration by parts on smooth periodic approximants and passage in $H^1$ give $\widehat{{\mathrm{\partial }}_{j}V}(k)=(2\pi i{k}_j\mathrm{/}3)\widehat{V}(k)$ for $0\le j<d$. The zero extension in step 1.3 is supported away from the identified faces, so its weak derivatives on the torus are exactly its zero-extended weak derivatives on $Q$. Parseval [F4] therefore yields ${\sum }_{k\in {\mathbb{Z}}^d}(1+\mathrm{\mid }k{\mathrm{\mid }}^2)\mathrm{\mid }\widehat{V}(k){\mathrm{\mid }}^2\le C_d\mathrm{\parallel }V{\mathrm{\parallel }}_{H^1(D)}^2$, with a fixed constant accounting for the side length $3$. Hence for every integer $N\ge 1$, ${\sum }_{\mathrm{\mid }k\mathrm{\mid }>N}\mathrm{\mid }\widehat{V}(k){\mathrm{\mid }}^2\le C_d{N}^{-2}\mathrm{\parallel }V{\mathrm{\parallel }}_{H^1(D)}^2$.

Let $(v_n)$ be bounded in $H_0^1(Q)$, and write $V_n$ for the periodic zero extensions from step 1.3. For each fixed Fourier mode $k$, the complex coefficients are bounded, so their real and imaginary parts are bounded real sequences. Use the following fixed extraction rule for any bounded real sequence $(b_n)$: let $m$ be the least positive integer bounding all $\mathrm{\mid }{b}_n\mathrm{\mid }$, start with $[-m,m]$, and bisect repeatedly, keeping the left closed half if it contains infinitely many terms and otherwise the right closed half. At stage $r$ take the least index larger than the previous index whose term lies in the retained interval. Each such index exists, and the selected terms are Cauchy because the intervals are nested with lengths tending to zero; [F5] then implies they converge. This defines a unique subsequence from the input sequence, without choice. Enumerate the real and imaginary parts of the Fourier modes, recursively apply this fixed rule to the current subsequence for the next coordinate, and let $n_r$ be the $r$th index of the $r$th nested subsequence. These diagonal indices increase strictly and form a tail subsequence of every fixed earlier extraction, so every coefficient converges; this recursion uses a specified function, not Dependent Choice. If its terms are $V_{n_r},V_{n_s}$, then for fixed $N$ the sum of squared coefficient differences over $\mathrm{\mid }k\mathrm{\mid }\le N$ tends to zero as $r,s\to \mathrm{\infty }$, while by step 2.1 the sum over $\mathrm{\mid }k\mathrm{\mid }>N$ is at most $C{N}^{-2}$ uniformly in $r,s$. First choose $N$ large, then $r,s$ large; Parseval shows $(V_{n_r})$ is Cauchy in $L^2(D)$. Completeness gives an $L^2(D)$ limit, and restriction to $Q$ gives the required $L^2(Q)$ convergence.

For smooth periodic $v$ and each coordinate $0\le i<d$, opposite-face values agree, so integrating that coordinate derivative over $Y$ and applying the one-dimensional FTC [F2] gives ${\int }_Y{\mathrm{\partial }}_{i}v=0$. Approximate an arbitrary $v\in H_{\mathrm{\#}}^1(Y)$ in $H^1$ by smooth periodic functions using [F1]; convergence of the derivatives in $L^2(Y)$ implies convergence of their integrals by [F3] and $\mathrm{\mid }Y\mathrm{\mid }=1$. Thus every component of ${\int }_Y\mathrm{\nabla }v$ is zero.

For smooth periodic $v$, write $m={\int }_{Y}v(y)dy$. For $x,y\in Y$, telescope from $y$ to $x$ one coordinate at a time, in order $0,\dots ,d-1$. The $i$th difference is the integral of ${\mathrm{\partial }}_{i}v$ along a segment of length at most one, with the earlier coordinates fixed at $x$ and later coordinates fixed at $y$. Cauchy--Schwarz [F3] bounds its square by the integral of $\mathrm{\mid }{\mathrm{\partial }}_{i}v{\mathrm{\mid }}^2$ along the full unit coordinate interval. The inequality $\mathrm{\mid }{\sum }_{i<d}{b}_i{\mathrm{\mid }}^2\le d{\sum }_{i<d}\mathrm{\mid }{b}_i{\mathrm{\mid }}^2$ and Tonelli [F4], integrating in $x$ and $y$, then give ${\int }_Y{\int }_Y\mathrm{\mid }v(x)-v(y){\mathrm{\mid }}^{2}dydx\le d{\sum }_{i<d}\mathrm{\parallel }{\mathrm{\partial }}_{i}v{\mathrm{\parallel }}_2^2\mathrm{.}$ For each $x$, Jensen's inequality gives $\mathrm{\mid }v(x)-m{\mathrm{\mid }}^2=\mathrm{\mid }{\int }_Y(v(x)-v(y))dy{\mathrm{\mid }}^2\le {\int }_Y\mathrm{\mid }v(x)-v(y){\mathrm{\mid }}^{2}dy$. Integrating in $x$ proves the stated inequality for smooth periodic mean-zero $v$.

Approximate any $v\in V_{\mathrm{\#}}$ in $H^1$ by smooth periodic $v_n$ using Countable Choice [F5]. Their means converge to zero by Cauchy--Schwarz, so $v_n-{\int }_Y{v}_n$ are smooth periodic mean-zero functions converging to $v$ in $H^1$. Passing to the limit in step 1.2 proves the inequality on $V_{\mathrm{\#}}$. By [F7], $H^1(Y)$ is complete. Its closure-defined subspace $H_{\mathrm{\#}}^1(Y)$ is closed, and the mean is continuous by $\mathrm{\mid }{\int }_{Y}v\mathrm{\mid }\le \mathrm{\parallel }v{\mathrm{\parallel }}_2\le \mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}$, so $V_{\mathrm{\#}}$ is closed in $H^1(Y)$. Every $H^1$-Cauchy sequence in $V_{\mathrm{\#}}$ therefore has an ambient limit in $V_{\mathrm{\#}}$, proving completeness of its inherited norm. The inequality shows the gradient norm and $H^1$ norm are equivalent, hence the gradient norm is complete too.

Tile ${\mathbb{R}}^d$ by cells $C_k^{ϵ}=ϵ(k+Y)$. Only $O_d({ϵ}^{-d})$ cells meet the fixed compact support of $\phi$. On each such cell choose its lower corner $x_k=ϵk$. The mean-value bound gives $\mathrm{\mid }\phi (x)-\phi (x_k)\mathrm{\mid }\le \sqrt{d}ϵ\mathrm{\parallel }\mathrm{\nabla }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}$. By [F3], ${\int }_{C_k^{ϵ}}\mathrm{\mid }g(x\mathrm{/}ϵ)\mathrm{\mid }dx={ϵ}^d\mathrm{\parallel }g{\mathrm{\parallel }}_{L^1(Y)}$. Therefore replacing $\phi$ by $\phi (x_k)$ on every intersecting cell changes the integral by at most $Cϵ\mathrm{\parallel }g{\mathrm{\parallel }}_1\mathrm{\parallel }\mathrm{\nabla }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}$; for sufficiently small $ϵ$, every such cell is contained in $Q$ because the support is compact in $Q$. Extend $\phi$ by zero to a compactly supported $C^1$ function on ${\mathbb{R}}^d$. For all sufficiently small $ϵ$, the grid sum has only cells in a fixed bounded box, and $\mid {ϵ}^d{\sum }_k\phi (ϵk)-{\int }_{{\mathbb{R}}^d}\phi (x)dx\mid \le {\sum }_k{\int }_{C_k^{ϵ}}\mathrm{\mid }\phi (ϵk)-\phi (x)\mathrm{\mid }dx\le Cϵ\mathrm{\parallel }\mathrm{\nabla }\phi {\mathrm{\parallel }}_{\mathrm{\infty }},$ since the total volume of those cells is uniformly bounded. Thus the grid Riemann sums converge. By [F3], the change-of-variables identity first holds for $g^{+}$ and $g^{-}$ separately; subtracting their finite integrals gives ${\int }_{C_k^{ϵ}}g(x\mathrm{/}ϵ)dx={ϵ}^d{\int }_{Y}g$ for signed $g$. The cell-corner sum is therefore ${ϵ}^d{\sum }_k\phi (ϵk){\int }_{Y}g$, which converges to the stated mean times integral.

The same tiling bounds the $L^2$ rescaling: each cell has ${\int }_{C_k^{ϵ}}\mathrm{\mid }g(x\mathrm{/}ϵ){\mathrm{\mid }}^{2}dx={ϵ}^d\mathrm{\parallel }g{\mathrm{\parallel }}_2^2$ by [F3], and at most $C_d{ϵ}^{-d}$ cells meet $Q$ when $ϵ\le 1$. Summing gives the asserted uniform bound.

The averaging result applied to every $C_c^{\mathrm{\infty }}(Q)$ test gives convergence of pairings against that dense class. The uniform $L^2$ bound in step 1.2, together with $\mathrm{\mid }{\int }_{Y}g\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_2$, lets one approximate an arbitrary $L^2(Q)$ test by [F2] and pass the pairing limit. This proves weak $L^2$ convergence.

On $V_{\mathrm{\#}}$ with gradient inner product, define $b(u,v)={\int }_{Y}a\mathrm{\nabla }u\cdot \mathrm{\nabla }v$ and $F_p(v)=-{\int }_{Y}ap\cdot \mathrm{\nabla }v$. By [F5] and Cauchy--Schwarz, $\mathrm{\mid }b(u,v)\mathrm{\mid }\le \mathrm{\Lambda }\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2$, $b(v,v)\ge \lambda \mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2^2$, and $\mathrm{\mid }{F}_p(v)\mathrm{\mid }\le \mathrm{\Lambda }\mathrm{\mid }p\mathrm{\mid }\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2$. Thus [F2]–[F4] apply and give a unique ${\chi }_p$ with $b({\chi }_p,v)=F_p(v)$ for every $v$. Rearranging is exactly [F1].

Test the equation with $v={\chi }_p$. Then $\lambda \mathrm{\parallel }\mathrm{\nabla }{\chi }_p{\mathrm{\parallel }}_2^2\le {\int }_{Y}a\mathrm{\mid }\mathrm{\nabla }{\chi }_p{\mathrm{\mid }}^2=-{\int }_{Y}ap\cdot \mathrm{\nabla }{\chi }_p\le \mathrm{\Lambda }\mathrm{\mid }p\mathrm{\mid }\mathrm{\parallel }\mathrm{\nabla }{\chi }_p{\mathrm{\parallel }}_2$. If the gradient norm is zero the estimate is immediate; otherwise divide by it. The $L^2$ bound follows from [F2].

For $p,q\in {\mathbb{R}}^d$ and scalars $s,t$, the function $s{\chi }_p+t{\chi }_q$ satisfies the cell equation with datum $sp+tq$, by linearity of the integral in both $p$ and the corrector. Uniqueness in step 1.1 therefore gives ${\chi }_{sp+tq}=s{\chi }_p+t{\chi }_q$.

Let $M={\sup }_n\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_2$. By [F7], extend functions by zero to see that $L^2(Q\times Y)$ is separable; [F3] says the allowed test subspace is dense there. Intersect a countable base of its metric topology with that subspace and, using [F4], choose one point from every nonempty intersection. The rational span of these points is a countable dense family $D$. For each fixed $\mathrm{\Phi }\in D$, Cauchy--Schwarz gives $\mid {\int }_Q{u}_n(x)\mathrm{\Phi }(x,x\mathrm{/}{ϵ}_n)dx\mid \le M\mathrm{\parallel }\mathrm{\Phi }(\cdot ,\cdot \mathrm{/}{ϵ}_n){\mathrm{\parallel }}_2,$ which is bounded by [F2]. For a bounded real sequence $(b_n)$ fix this extraction rule: let $m$ be the least positive integer bounding every $\mathrm{\mid }{b}_n\mathrm{\mid }$, start with $[-m,m]$, bisect repeatedly, and retain the left closed half when it contains infinitely many terms, otherwise the right closed half. From the $r$th retained interval take the term with least index larger than the previously selected index. Such indices exist; nestedness and vanishing interval lengths make the selected sequence Cauchy, and [F5] makes it convergent. This uniquely specified extraction map requires no choice. Enumerate $D$, recursively apply this map to each scalar pairing on the current subsequence, and take as $n_r$ the $r$th index of the $r$th nested subsequence. The indices increase strictly and their tail lies in every fixed earlier subsequence, so all pairings with $D$ converge. Recursion of the specified map does not use Dependent Choice.

Denote the limit on $D$ by $L_0$. For each $\mathrm{\Phi }\in D$, step 1.1 and [F2] give $\mathrm{\mid }{L}_0(\mathrm{\Phi })\mathrm{\mid }\le M\mathrm{\parallel }\mathrm{\Phi }{\mathrm{\parallel }}_{L^2(Q\times Y)}$. Limits preserve rational linearity; the bound makes $L_0$ uniformly continuous, so it has a unique continuous extension $L$ to all of the real Hilbert space $L^2(Q\times Y)$, with $\mathrm{\parallel }L\mathrm{\parallel }\le M$. To verify real homogeneity, first fix $\mathrm{\Phi }\in D$ and $t\in \mathbb{R}$, and choose rationals $q_m\to t$. Since $q_m\mathrm{\Phi }\in D$, $L(q_m\mathrm{\Phi })=q_m{L}_0(\mathrm{\Phi })$; continuity and $q_m\mathrm{\Phi }\to t\mathrm{\Phi }$ give $L(t\mathrm{\Phi })=tL(\mathrm{\Phi })$. For arbitrary $h\in L^2(Q\times Y)$, choose ${\mathrm{\Phi }}_j\in D$ with ${\mathrm{\Phi }}_j\to h$; then $t{\mathrm{\Phi }}_j\to th$, and continuity gives $L(th)=tL(h)$. Additivity also passes from $D$ by continuity, so $L$ is bounded and real-linear.

By [F6], choose $u\in L^2(Q\times Y)$ such that $L(\mathrm{\Phi })={\int }_{Q\times Y}u\mathrm{\Phi }$. Fix any allowed smooth product test $\mathrm{\Psi }$ and choose ${\mathrm{\Phi }}_j\in D$ with ${\mathrm{\Phi }}_j\to \mathrm{\Psi }$ in product $L^2$. Along the extracted subsequence, [F2] applied to $\mathrm{\Psi }-{\mathrm{\Phi }}_j$ gives ${\mathrm{lim\hspace{0.17em}sup}}_n\mid {\int }_Q{u}_n(\mathrm{\Psi }-{\mathrm{\Phi }}_j)(x,x\mathrm{/}{ϵ}_n)dx\mid \le M\mathrm{\parallel }\mathrm{\Psi }-{\mathrm{\Phi }}_j{\mathrm{\parallel }}_2\mathrm{.}$ Also $\mathrm{\mid }L(\mathrm{\Psi }-{\mathrm{\Phi }}_j)\mathrm{\mid }\le M\mathrm{\parallel }\mathrm{\Psi }-{\mathrm{\Phi }}_j{\mathrm{\parallel }}_2$. Since the pairings for ${\mathrm{\Phi }}_j$ converge to $L({\mathrm{\Phi }}_j)$, letting $j\to \mathrm{\infty }$ proves convergence for $\mathrm{\Psi }$ to $L(\mathrm{\Psi })=\int u\mathrm{\Psi }$. This is precisely the two-scale convergence in [F1].

By [F1]–[F2], first extract a subsequence converging weakly in $H_0^1(Q)$ to some $u\in H_0^1(Q)$, and then a further subsequence converging strongly in $L^2(Q)$ to the same $u$. By [F3], extract successive further subsequences on which each of the $d$ components of $\mathrm{\nabla }{u}_n$ two-scale converges; since there are finitely many components, one subsequence works for all. Write its vector limit as $G(x,y)$.

For every allowed product test $\varphi (x)\psi (y)$, strong $L^2$ convergence and the uniform sampled-test bound in [F4] show that replacing $u_n$ by $u$ in its pairing changes the limit by zero. Weak $L^2$ averaging in [F4], applied to $\psi (x\mathrm{/}{ϵ}_n)$ against $u\varphi \in L^2(Q)$, then gives limit ${\int }_{Q}u\varphi {\int }_Y\psi$. Thus the two-scale limit of $u_n$ is the $y$-independent function $u(x)$.

Set $H(x,y):=G(x,y)-{\mathrm{\nabla }}_{x}u(x)$. For a smooth compactly supported $\varphi (x)$ and constant vector $c$, test gradient convergence componentwise with $\varphi c$. Ordinary weak $H^1$ convergence gives ${\int }_{Q\times Y}G\cdot \varphi c={\int }_Q\mathrm{\nabla }u\cdot \varphi c$. By Cauchy--Schwarz and Tonelli [F8], $x\mapsto {\int }_{Y}H(x,y)dy$ lies in $L^2(Q)$ and pairs to zero with every $\varphi \in C_c^{\mathrm{\infty }}(Q)$; [F9] makes it zero almost everywhere.

Let $\mathrm{\Phi }(x,y)={\sum }_r{\varphi }_r(x){\mathrm{\Psi }}_r(y)$ be a smooth periodic vector field, compactly supported in $x$, with ${\mathrm{div}}_y\mathrm{\Phi }=0$, and put ${\mathrm{\Phi }}_n(x)=\mathrm{\Phi }(x,x\mathrm{/}{ϵ}_n)$. By [F10], $\mathrm{div}{\mathrm{\Phi }}_n=({\mathrm{div}}_x\mathrm{\Phi })(x,x\mathrm{/}{ϵ}_n)+{ϵ}_n^{-1}({\mathrm{div}}_y\mathrm{\Phi })(x,x\mathrm{/}{ϵ}_n)=({\mathrm{div}}_x\mathrm{\Phi })(x,x\mathrm{/}{ϵ}_n)$. The field ${\mathrm{\Phi }}_n$ is a valid compactly supported smooth test in the weak-derivative identity [F7], so ${\int }_Q\mathrm{\nabla }{u}_n\cdot {\mathrm{\Phi }}_n=-{\int }_Q{u}_n\mathrm{div}{\mathrm{\Phi }}_n$. Taking limits by componentwise two-scale convergence for the two allowed tests $\mathrm{\Phi }$ and ${\mathrm{div}}_x\mathrm{\Phi }$ gives ${\int }_{Q\times Y}G\cdot \mathrm{\Phi }=-{\int }_{Q\times Y}u{\mathrm{div}}_x\mathrm{\Phi }={\int }_{Q\times Y}{\mathrm{\nabla }}_{x}u\cdot \mathrm{\Phi }$, where the last equality is weak integration by parts. Thus $H$ is orthogonal to every such divergence-free field.

For every nonzero $k\in {\mathbb{Z}}^d$ and every real $c\perp k$, use slow tests $\varphi (x)c\cos (2\pi k\cdot y)$ and $\varphi (x)c\sin (2\pi k\cdot y)$ in step 3.1. They are divergence-free in $y$. The resulting pairings show that each real and imaginary component of $\widehat{H}(x,k)\cdot c$ pairs to zero with every $\varphi \in C_c^{\mathrm{\infty }}(Q)$, so [F9] makes it zero almost everywhere. There are countably many $k$ and a finite basis of each $k^{\perp }$, so [A1] permits one null set outside which $\widehat{H}(x,k)$ is parallel to $k$ for all $k$; the zero coefficient is zero by step 2.2.

For $k\mathrm{\ne }0$ define ${\widehat{u}}_1(x,k)=(\widehat{H}(x,k)\cdot k)\mathrm{/}(2\pi i\mathrm{\mid }k{\mathrm{\mid }}^2)$ and set ${\widehat{u}}_1(x,0)=0$. The parallel property from step 4.1 gives $2\pi ik{\widehat{u}}_1(x,k)=\widehat{H}(x,k)$. Moreover $\mathrm{\mid }{\widehat{u}}_1(x,k)\mathrm{\mid }\le \mathrm{\mid }\widehat{H}(x,k)\mathrm{\mid }\mathrm{/}(2\pi \mathrm{\mid }k\mathrm{\mid })$, so Parseval [F6] and Tonelli [F8] give ${\int }_Q{\sum }_k(1+\mathrm{\mid }k{\mathrm{\mid }}^2)\mathrm{\mid }{\widehat{u}}_1(x,k){\mathrm{\mid }}^{2}dx\le C_d\mathrm{\parallel }H{\mathrm{\parallel }}_{L^2(Q\times Y)}^2$. For almost every $x$, the sum is finite; the partial Fourier sums over $\mathrm{\mid }k\mathrm{\mid }\le N$ are then Cauchy in $H^1(Y)$ by Parseval, and their limit belongs to $H_{\mathrm{\#}}^1(Y)$ because each partial sum is smooth periodic. The zero coefficient makes this limit mean-zero. Define $u_1$ as the product-space $L^2(Q\times Y)$ limit of these partial sums. Their $y$-gradients converge there, and the coefficient identity identifies the gradient limit with $H$. Conjugate symmetry makes the limit real, and the integrated estimate gives the stated square-integrable $H^1(Y)$ slice norm. Hence $u_1\in L^2(Q;V_{\mathrm{\#}})$ with ${\mathrm{\nabla }}_y{u}_1=H$, completing the decomposition.