= Solution
At a <perfectly bonded elastic interface> without an interfacial force, <displacement> and <traction> are continuous:
$$
[u]=0,\qquad[\sigma]n=0.
$$
Tangential derivatives of the common <displacement> trace are also continuous. Thus the jump of a piecewise constant <displacement gradient tensor> annihilates every tangential vector, and has the form $\alpha\otimes n$. Taking its symmetric part shows that the <strain> jump is $B\alpha$, where
$$
(B\alpha)_{ij}=\tfrac12(\alpha_i n_j+\alpha_j n_i),
\qquad B^T\tau=\tau n.
$$
The adjoint formula follows from $\tau:B\alpha=(\tau n)\cdot\alpha$ for symmetric <stress tensors>. Tangential-tangential <strain> components are therefore common to all layers. Conversely, such gradient jumps can be realized by continuous piecewise affine <displacement> in the layered geometry.
Take positive <elastic stiffness tensors> with their usual minor and major symmetries, and a positive comparison <elastic stiffness tensor> $C^0$. A fictitious layer can have the same tangential displacement derivatives and the same <traction> as the physical layers. Hence $\varepsilon^r=\varepsilon^0+B\alpha^r$. The common <traction> condition gives
$$
B^TC^r(\varepsilon^0+B\alpha^r)=B^TC^0\varepsilon^0,
\qquad K^r\alpha^r=-B^T(C^r-C^0)\varepsilon^0,
$$
where $K^r=B^TC^rB$ is the positive <acoustic tensor>, with components $K^r_{ik}=C^r_{ijkl}n_j n_l$. Solving for $\alpha^r$ gives
$$
\boxed{\varepsilon^r=\bigl[I-\widetilde\Gamma^r(C^r-C^0)\bigr]\varepsilon^0,
\qquad \widetilde\Gamma^r=B(K^r)^{-1}B^T.}
$$
Expanding the two occurrences of $B$ gives exactly the symmetrization over $(ij)$ and $(kl)$ in the <directional elastic strain Green operator> specified in the paper.
For any two layers, the inverse identity for their <acoustic tensors> proves the required resolvent relation without interchanging noncommuting <tensors>:
$$
\begin{aligned}
\widetilde\Gamma^r(C^r-C^s)\widetilde\Gamma^s
&=B(K^r)^{-1}(K^r-K^s)(K^s)^{-1}B^T\\
&=B\bigl[(K^s)^{-1}-(K^r)^{-1}\bigr]B^T\\
&=\boxed{\widetilde\Gamma^s-\widetilde\Gamma^r}.
\end{aligned}
$$
With $D_r=C^r-C^0$, this identity also gives $\widetilde\Gamma^0D_r\widetilde\Gamma^r=\widetilde\Gamma^0-\widetilde\Gamma^r$. Therefore
$$
\bigl[I+\widetilde\Gamma^0D_r\bigr]
\bigl[I-\widetilde\Gamma^rD_r\bigr]=I.
$$
These finite-dimensional operators are inverses. Define $A_r=[I+\widetilde\Gamma^0(C^r-C^0)]^{-1}$. Then $\varepsilon^r=A_r\varepsilon^0$, and averaging <strains> and <stresses> gives
$$
E=\langle A\rangle\varepsilon^0,
\qquad S=\langle CA\rangle\varepsilon^0.
$$
Eliminating the fictitious <strain> yields
$$
\boxed{C^{\mathrm{eff}}=\langle CA\rangle\langle A\rangle^{-1}.}
$$
All products have the displayed order; the <elastic stiffness tensors> need not commute.
It remains to justify the inverse and comparison independence. Consider two compatible equilibrated layer solutions with the same mean <strain>. Their difference has zero common tangential-tangential <strain>, so $\delta\varepsilon^r=B\beta^r$, with $\sum_r c_r\beta^r=0$ because $B$ is injective and the mean difference is zero. Their <traction> difference is one common vector $g$. Therefore
$$
\sum_r c_r\delta\varepsilon^r:C^r\delta\varepsilon^r
=\sum_r c_r\beta^r\cdot g=0.
$$
Positive <elastic energy> forces every $\delta\varepsilon^r$ to vanish. In particular, if $\langle A\rangle\varepsilon^0=0$, all physical layer <strains> vanish. Their common tangential components then force $\varepsilon^0=B\beta^0$, and their zero <traction> gives $K^0\beta^0=0$. Hence $\varepsilon^0=0$, proving that $\langle A\rangle$ is invertible. Finally, every positive choice of $C^0$ constructs the unique layer solution for the same $E$. Its averaged <stress> is unique, so \b[the displayed effective stiffness is independent of the fictitious comparison material]. This is the exact layered constitutive response, rather than a claim that piecewise constant fields meet arbitrary finite-body boundary data.
Back to article page
