Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 80 / 3

Use symmetric positive conductivity, for example $\alpha I⪯a(x)⪯MI$ with $\alpha >0$. Let $v$ have the same Dirichlet boundary data as $u$, and write $v=u+\varphi$ with $\varphi \in H_0^1(\Omega )$. Symmetry givesThe middle integral vanishes by the weak field equation; equivalently integrate by parts, use $\varphi =0$ on the boundary and $\nabla \cdot (a\nabla u)=0$. HenceThereforeEquality forces $\nabla \varphi =0$, and its zero boundary trace gives $\varphi =0$, proving uniqueness. This is the Dirichlet principle for the conductivity equation.

The positivity is essential to the physical minimum-energy interpretation; symmetry by itself would give only stationarity. For example, take $a=-I$, $\lambda =0$ and $u=0$. The field equation holds, but every nonzero smooth compactly supported $v$ has $J(v)=-{\int }_{\Omega }|\nabla v{|}^2<0$, and scaling $v$ makes the energy unbounded below. Thus if the printed word “conductivity” were not understood to include positivity, the stated minimum claim would require that additional hypothesis.

Choose the affine trial function $v(x)=\lambda \cdot x$. It has the required trace and constant gradient, so the Dirichlet principle givesFor the macroscopic isotropic effective medium, the corresponding affine-boundary energy is $|\Omega |a^{\ast }|\lambda {|}^2$. Dividing by $|\Omega ||\lambda {|}^2$ for a nonzero imposed gradient yields the Voigt boundMore generally the same calculation is the tensor inequality in the Loewner order $a^{\ast }⪯⟨a⟩$. For a finite unresolved specimen the calculation bounds its apparent affine-boundary response; identification with the bulk effective conductivity uses the macroscopic scale separation.

Write $\Delta =a_1-a_2<0$, and let $q=t\Delta \lambda {\chi }_1$. The inverse contrast in the Hashin-Shtrikman conductivity variational principle is interpreted only in phase 1: the admissible $q$ vanishes in phase 2, where $a-a_{2}I=0$. Thus

For the nonlocal term define $K(x,x^{\prime })={\nabla }_x{\nabla }_{x^{\prime }}g(x,x^{\prime })$. The given Green representation impliesThe two-point phase correlation function therefore givesBecause the unweighted double integral of $K$ is zero, subtract $p_1^2$ inside the bracket. Set $h(r)=p_{11}(r)-p_1^2$; it vanishes for $r>\rho$, so only the microscopic near-diagonal kernel contributes. Its contact part isAt coincident points ${\chi }_1^2={\chi }_1$, hence $h(0)=p_1-p_1^2=p_1{p}_2$. The ordinary dipole part has zero angular integral by the stipulated spherical-shell identity. Since $h$ is radial, its weighting preserves this cancellation:Consequently, in the macroscopic bulk regime implicit in the supplied local Green approximation,andThis is the dipole Green tensor angular cancellation: the constant background correlation disappears and the connected contact value supplies $p_1{p}_2$.

Insert all three averaged terms in the variational upper inequality and normalize by $|\Omega ||\lambda {|}^2$. For the bulk effective conductivity one obtainsThe kernel in the question has an approximation sign. At fixed finite size, spheres around points near $\partial \Omega$ are truncated and the radial cancellation is not an exact finite-body identity. Their contribution and the regular boundary part disappear in the stated microscopic-to-macroscopic limit. The calculation derives the intended homogenized bound, rather than silently treating that Green approximation as exact for arbitrary finite $\rho$.

Let $C=1+\Delta p_2/(3a_2)$. Since $a_1>0$, $\Delta >-a_2$, so $C>0$. Also $\Delta p_1<0$ when both phases are present. Completing the square in the trial upper bound givesThus the tightest upper bound is obtained by minimizing at $t=1/C$, not maximizing. It isPure-phase cases follow directly or by continuity. These Hashin-Shtrikman bounds for conductivity are volume-fraction bounds on the statistically isotropic homogenized material; the full variational framework is discussed in Chapter 23 of www.math.utah.edu/~milton/TheoryCompositesNOPRINT.pdf .

For the lower variational inequality choose reference conductivity $a_1$ and a trial polarization $q$ that vanishes in phase 1. Let $v_q\in H_0^1(\Omega )$ solveThe Hashin-Shtrikman conductivity variational principle isIt holds for every admissible trial $q$; the exact principle takes the supremum of the right side. Writing the last integral only over phase 2 avoids applying an inverse to the zero contrast in the reference phase.

With $D=a_2-a_1>0$, the isotropic phase trial $q=tD\lambda {\chi }_2$ gives the lower expressionHere the quadratic is concave, so maximize it at $t=[1+D{p}_1/(3a_1){]}^{-1}$. The lower member of the Hashin-Shtrikman conductivity bounds isTogether with part (iii), this gives the required upper and lower interval. All denominators are positive for positive phase conductivities.