Geometry of the positive definite cone, versus homogenization of elliptic PDEs

Question

Homogenization is a process that assigns to a positive definite-valued map $x\mapsto S(x)$ a non-trivial but physically meaningful average $\bar S$. There are various settings, for instance stochastic homogenization, but I shall restrict to the deterministic one, in which $x\mapsto S(x)$ is ${\mathbb Z}^d$-periodic.

To define $\bar S$, we consider the solutions of the elliptic PDE $${\rm div}(S(x)\nabla\phi)=0,$$ where $\nabla\phi$ is periodic too (but $\phi$ is not !). For every vector $q\in{\mathbb R}^d$, there exists a unique solution (up to an irrelevant additive constant) $\phi_q$ whose gradient averages to $q$ : $\langle\phi_q\rangle=q$. Then $\bar S$ is defined by $$\bar Sq=\langle S\nabla\phi_q\rangle.$$

I am interested in a homogenization process that provides the geometrical mean $A\sharp B$ of two positive definite symmetric matrices, as $\bar S$, when $S(x)$ takes the value $A$ on the half of a fundamental domain, and $B$ otherwise. I guess that such a process must exist (but I should be grateful to receive a reference). Is there a known explicit process ?

My motivation comes from an observation made in the book Homogenization of differential operators and integral functionals by V. V. Jikov, S. M. Kozlov, O. A. Oleinik: Let $d=2$ and $\sigma$ be the rotation by $\frac\pi2$. If $S(x)\sigma S(\sigma x)\sigma\equiv-k^2I_2$ where $k>0$ is a constant, then $\bar S=kI_2$. When applied to a chess-board with $S(x)=a_\pm I_2$, this provides $\bar S=\sqrt{a_-a_+}I_2$, which answers a particular case of my question.

For the sake of completeness, I recall that $A\sharp B$ is maximal among those positive definite $X$ such that $$\begin{pmatrix} A & X \\ X & B \end{pmatrix}\ge0_n.$$ It is given by the (amazingly non-symmetric) formula $$A\sharp B=A^{1/2}(A^{-1/2}BA^{-1/2})^{1/2}A^{1/2}.$$

One argument in favour of the existence of such a process is the fact that $A\sharp B$ satisfies the same necessary conditions (see again the same book) as $\bar S$ : $$\left(\frac{A^{-1}+B^{-1}}2\right)^{-1}\le A\sharp B\le\frac{A+B}2\,.$$

Edit. Let me be much more ambitious.

If $d=2$ and $\chi$ is the characteristic function of the black cells in a checkerboard. Suppose $S(x)=\chi(x)A+(1-\chi(x))B$. Do we have $\bar S=A\sharp B$ ?

This formula turns out to be true in the case where $B=-\sigma A\sigma$ (that is $B=(\det A)A^{-1}$). It is also consistent with the well-known observation that if $x\mapsto \det S$ is constant, then $\det \bar S$ equals this constant ; actually $\det A\sharp B=\sqrt{\det A\det B}$.

Answer 1

I discussed this question at lunch with my colleague Jean-Claude Sikorav, and we came up with a solution when $d=2$.

We first consider a matrix $P$ such that $A\sharp B=PP^T$, then form $A'=P^{-1}AP^{-T}$ and $B'=P^{-1}BP^{-T}$, both symmetric positive definite. One has $A'\sharp B'=P^{-1}(A\sharp B)P^{-T}$ (use the characterization as the maximal $X$ such that ...), hence $A'\sharp B'=I_2$, which means $B'=(A')^{-1}$.

Let us make the change of variables $x=Py$. The operator ${\rm div}_xS\nabla_x$ becomes ${\rm div}_yS'\nabla_y$, where $S'(y)=P^{-1}S(x)P^{-T}$.

Here is the homogenization process: we choose $S'(y)$ with values $A'$ on the black cells of a checkerboard, and $B'$ on the white cells. Because $B'=(A')^{-1}$, we know that $\overline{S'}=I_2$ (this is where we use the dimension $d=2$). Going back to the original variables, we have $S(x)=PS'(y)P^T$, which is $\Lambda$-periodic for some lattice $\Lambda$. Then $\bar S=P\overline{S'}P^T=PP^T=A\sharp B$.

Remarks

1. This leaves open the higher dimensional case ($d\ge3$).
2. Our diffusion matrix is not ${\mathbb Z}^2$-periodic in general, but only periodic according to some lattice.
3. The process is non-unique, because there are infinitely many matrices $P$ such that $A\sharp B=PP^T$.
4. Our solution is not entirely satisfactory from a practical point of view, because it does use $A\sharp B$ in the construction of $S$.