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For a homogeneous comparison elastic stiffness tensor $C^0$, the elastic strain Green operator maps an elastic polarization to minus the induced infinitesimal strain tensor. In a bounded domain the induced displacement field has zero boundary data; in an infinite medium it is the decaying perturbation. The defining static elastic equilibrium equation is $\nabla \cdot (C^0\mathrm{sym}\nabla u+\tau )=0$. The bounded-domain operator is self-adjoint, annihilates constant elastic polarizations, has zero mean output, and satisfies $\Gamma C^0\Gamma =\Gamma$.

For a three-dimensional isotropic comparison material, the contraction of its elastic strain Green operator is the displayed distributional derivative. Its double trace is ${\Gamma }_{iikk}=\delta /L$, because $\Delta |x{|}^{-1}=-4\pi \delta$. Away from zero the contraction is $({\delta }_{ij}|x{|}^2-3x_i{x}_j)/(4\pi L|x{|}^5)$; the full distribution also includes ${\delta }_{ij}\delta /(3L)$. This contact term must not be discarded when computing hydrostatic strain.

Put $\mu ={\mu }^0$, $\lambda ={\kappa }^0-2\mu /3$, and $L=\lambda +2\mu ={\kappa }^0+4\mu /3$. Assume positive bulk modulus and shear modulus. For a unit vector $\xi$, direct contraction of the isotropic elastic stiffness tensor gives the acoustic tensor and its inverse:These are its transverse and longitudinal orthogonal projections. Symmetrizing the definition of the directional elastic strain Green operator consequently givesSubstitution into the spherical representation is already the isotropic specialization. We can also evaluate its angular integrals as distributions. For $r=|x|>0$, direct integration with the polar axis parallel to $x$ givesThe first identity holds as a locally integrable distribution. Since the second derivative of $|s|$ is $2\delta (s)$, taking two derivatives in the first identity and four in the second givesThus an explicit complete isotropic elastic strain Green kernel isAll derivatives here are distributional derivatives, which retain the contribution at the origin.

For the requested contraction, ${\tilde{\Gamma }}_{ijkk}={\xi }_i{\xi }_j/L$, or equivalently trace the complete kernel and use $\Delta r=2/r$. ThereforeMore explicitly, its distribution form iswhere the Cauchy principal value uses spherical exclusions about the origin. The coefficient of the Dirac delta can be checked by integrating by parts over a punctured ball: ${\partial }_i{\partial }_j(r^{-1})$ has contact term $-4\pi {\delta }_{ij}\delta /3$. Taking the trace, the principal-value part has zero trace and $\Delta (r^{-1})=-4\pi \delta$, soKeeping only the ordinary kernel away from the origin would incorrectly give zero for this double trace.

Now extend the elastic polarization by zero outside $V$. Since the shear modulus is unchanged, the stiffness contrast acts only on the spherical strain:Hence $\tau =(\kappa -{\kappa }^0)\mathrm{tr}\epsilon$ is its scalar coefficient. Relative to the same external loading, the elastic strain Green operator gives $\epsilon ={\epsilon }^0-\Gamma \ast (\tau I)$. Taking the trace and using the contact identity reduces this nonlocal equation toSolving the scalar equation gives the constant-shear bulk-modulus inclusion formulaUsing the single contraction instead of the double trace givesFinally eliminate the actual trace using the scalar formula:The reference trace ${\epsilon }_{kk}^0$ in this last formula is essential and is what the original PDF prints; the converted TeX loses its superscript. The second derivatives are interpreted distributionally, or with their principal-value and contact terms explicitly included. With ordinary regularity they give the corresponding pointwise formula at regular interior points.

Let $u$ be the actual displacement field, let $u^{\prime }$ be an admissible trial displacement, and set $v=u^{\prime }-u$. On the displacement-prescribed part of the boundary, $v=0$. With actual stress $\sigma =D_{\epsilon }W(\epsilon )$, the supporting-plane inequality for the convex strain energy density givesBoth $\sigma$ and ${\sigma }_0$ satisfy static elastic equilibrium with the same body force, so $\nabla \cdot (\sigma -{\sigma }_0)=0$. On the traction-prescribed boundary their tractions agree. Since ${\epsilon }^{\prime }-\epsilon =\mathrm{sym}\nabla v$, integration by parts givesSubtracting the prescribed external-work representative proves the minimum potential energy principle in elasticity:Strict convexity gives uniqueness of the minimizing strain, although rigid displacements can remain when the boundary conditions allow them.

For the comparison bound, put $F(e,x)=W(e,x)-W_0(e)$ and use the lower conjugate, with its infimum sign convention,Its defining inequality is $F_{\ast }(\tau ,x)\le \tau :e-F(e,x)$, hence $W(e,x)\le W_0(e)+\tau :e-F_{\ast }(\tau ,x)$ for every $e$. Integrate this pointwise majorant for each compatible strain, then take the infimum over the same admissible set. This provesNo convexity of $F$ is used for this inequality. If $F_{\ast }(\tau ,x)=-\infty$, it yields only the extended-real upper bound $+\infty$. For example, $W_0=0$ and $W(e)=|e{|}^2/2$ give $F_{\ast }(\tau )=-\infty$ for every finite $\tau$. A useful finite bound requires a suitable comparison energy; choosing a sufficiently stiff quadratic comparison makes the usual excess energies concave. The derivative used below additionally presupposes finite differentiable duals, or an appropriate supergradient formulation.

Now normalize $|\Omega |=1$, take positive constant $C^0$, and write $F_r(e)=W_r(e)-e:C^{0}e/2$. For phasewise constant elastic polarization $\tau ={\sum }_r{\chi }_r{\tau }^r$, its lower conjugate contribution is constant within each phase. The remaining minimization isWrite the minimizing strain as $\hat{e}=E+\eta$, where $\eta$ comes from a zero-boundary displacement field. The static elastic equilibrium equation is $\nabla \cdot (C^0\eta +\tau )=0$, so the supplied elastic strain Green operator gives $\eta =-\Gamma \tau$. Its zero mean removes the cross term with the constant $E$. Using its reciprocity and projection property $\Gamma C^0\Gamma =\Gamma$, we obtainHere the brackets in the last two terms denote the spatial integral and tensor contraction, not just a phase average.

Let $c_r={\int }_{\Omega }{\chi }_r$. The zero output mean and reciprocity of $\Gamma$ imply that it annihilates constant elastic polarizations on either side. Thus subtracting $c_r$ and $c_s$ from the phase indicators does not change their bilinear integral. Define the fourth-order block tensorsConsequently $A_{rs}=A_{sr}^T$, ${\sum }_s{A}_{rs}=0$, andThese identities follow from reciprocity and ${\sum }_s{\chi }_s={\sum }_s{c}_s=1$; no phase independence assumption is needed. The phase polarization comparison bound, valid for every polarization with finite dual terms, is thereforeIn particular, it can be optimized over the phase elastic polarizations without changing the upper-bound direction.

To derive the final displayed form, impose the stationarity equations in the question. For $c_r>0$, put $e^r=D{F}_{r\ast }({\tau }^r)$; these are dual strains, initially distinct from the comparison solution's phase averages. Reciprocity of $A$ givesThus stationarity is exactlyThe comparison solution has phase meanso the dual strains become its phase means at stationarity. Multiplying the stationarity equations by ${\tau }^r$ and summing gives ${\sum }_{r,s}{\tau }^r:A_{rs}{\tau }^s={\sum }_r{c}_r{\tau }^r:(E-e^r)$.

For clarity, the concave biconjugate here uses another infimum: $F_{r\ast \ast }(e)={\inf }_s\{s:e-F_{r\ast }(s)\}$. The lower conjugate $F_{r\ast }$ is concave, since it is an infimum of affine functions. At a differentiable finite point, its supporting-plane inequality shows that $s={\tau }^r$ minimizes this last expression when $e=e^r$. HenceSubstitute both identities into $\mathcal{B}$ to obtainThis is the requested final bound with the accompanying phase equations. It remains valid at any well-defined stationary choice; asserting existence of such a differentiable choice for arbitrary $W_0$ would require extra hypotheses. Without stationarity, the earlier $\mathcal{B}(E,\tau )$ is the correct general bound. In the concave closed excess-energy case $F_{r\ast \ast }=F_r$; for general excess energies replacing the concave biconjugate by $F_r$ without proof would reverse the needed majorant logic.