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Write $a=\min (C_1,C_2)>0$, $b=\max (C_1,C_2)$ and $c=C^0>0$. At crystal orientation $\theta$, the conductivity tensor is $Q_{\theta }\mathrm{diag}(a,b)Q_{\theta }^T$. Because the comparison tensor is scalar, its resolvent rotates in the same way:Uniform angular averaging uses $⟨{\cos }^2\theta ⟩=⟨{\sin }^2\theta ⟩=1/2$ and $⟨\sin \theta \cos \theta ⟩=0$. It therefore sends any such rotated diagonal tensor to half its trace times $I$. In particular,Dividing these scalar coefficients in the stated Hashin-Shtrikman conductivity variational principle givesThis averaging calculation uses the uniform orientation model in the question; a nonuniform distribution need not give isotropic effective conductivity.

The lower Hashin-Shtrikman conductivity bounds use a comparison no larger than every local principal conductivity, so $0<c\le a$. The upper variational bound uses $c\ge b$. These opposite directions come from the sign of the contrast $C-cI$ in the polarization functional: positive contrast gives a maximizing lower principle, and negative contrast a minimizing upper principle. The endpoint values are understood as limits when a contrast eigenvalue vanishes. Sincethe strongest lower and upper values in their admissible comparison ranges are, respectively,Thus $C_{l}I⪯C^{\mathrm{eff}}⪯C_{u}I$ in the statistically isotropic polycrystal setting of these Hashin-Shtrikman conductivity bounds.

The self-consistent conductivity approximation chooses the comparison conductivity equal to the predicted conductivity, $c=C_{\mathrm{SC}}=H(c)$. Multiplication by the positive denominator cancels the terms $c(a+b)$ and leaves $2c^2=2ab$. Only the positive root is an admissible conductivity:For a direct proof of its ordering relative to the variational bounds, set $t=\sqrt{b/a}\ge 1$. ThenBoth quantities are nonnegative, proving $C_l\le C_{\mathrm{SC}}\le C_u$, with equality throughout when $a=b$. The self-consistent conductivity approximation is a closure calculation; no unrequested claim of exactness for every spatial arrangement follows just from this fixed-point computation.

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.

For uniformly distributed rotations of a two-dimensional conductivity tensor with positive principal values $a\le b$, the Hashin-Shtrikman conductivity variational principle with isotropic comparison $cI$ yields $H(c)I$. Averaging a rotated diagonal tensor replaces it by half its trace times the identity. Since $H^{\prime }(c)=(a-b{)}^2/(a+b+2c{)}^2$, the admissible comparison endpoints give $a(a+3b)/(3a+b)\le C_{\mathrm{eff}}\le b(3a+b)/(a+3b)$ within the statistically isotropic polycrystal model.