## 2 General Approach to Homogenization
In this way, we can see that Equation (23) takes the following form:  
\[D^{-1}(\mathbf{k};\mathbf{G},0)+\overset{=-1}{T}^{-1}(\mathbf{G})\frac{\Omega}{i }\sum\
olimits_{\mathbf{G}^{{}^{\prime}}}\Delta A(-\mathbf{G}^{{}^{\prime}})D ^{-1}(\mathbf{k};\mathbf{G}^{{}^{\prime}},0)=\overset{=-1}{T}^{-1}(\mathbf{G})I \delta_{\mathbf{G},0}, \tag{24}\]  
or what is the same:  
\[D^{-1}(\mathbf{k};\mathbf{GMultiplying Equation (25) by \(\Delta A(-\mathbf{G}^{{}^{\prime}})\) and adding the overall reciprocal vectors, \(\mathbf{G}\), the equation needed to determine \(\overset{=}{\widetilde{C}}\) is found:  
\[\overset{=}{\widetilde{C}}+\sum_{\mathbf{G}}\Delta A(-\mathbf{G})\overset{=-1} {T}(\mathbf{G})\frac{\Omega}{i}\overset{=}{\widetilde{C}}=\sum_{\mathbf{G}} \Delta A(-\mathbf{G})\overset{=-1}{T}(\mathbf{G})\delta_{\mathbf{G},0}, \tag{27}\]  
from which, we have:  
\[\overset{=}{\widetilde{C}}+\sum_{\mathbf{G}}\Delta A(-\mathbf{G})\overset{=- 1}{T}(\mathbf{G})\frac{\Omega}{i}\overset{=}{\widetilde{C}}=\Delta A(0) \overset{=-1}{T}(0), \tag{28}\]  
from here:  
\[\overset{=}{\widetilde{C}}=\bigg{[}I+\sum_{\mathbf{G}}\Delta A(-\mathbf{G}) \overset{=-1}{T}(\mathbf{G})\frac{\Omega}{i}\bigg{]}^{-1}\cdot\Delta A(0) \overset{=-1}{T}(0), \tag{29}\]  
Substituting this expression for \(\overset{=}{\widetilde{C}}\) in Equation (25), we finally obtain the values for the \(D^{-1}(\mathbf{k};0,0)\) matrix block:  
\[D^{-1}(\mathbf{k};0,0)=\overset{=-1}{T}(0)-\overset{=-1}{T}(0)\frac{\Omega}{i} \cdot\bigg{[}I+\sum_{\mathbf{G}}\Delta A(-\mathbf{G})\overset{=-1}{T}(\mathbf{ G})\frac{\Omega}{i}\bigg{]}^{-1}\cdot\Delta A(0)\overset{=-1}{T}(0). \tag{30}\]  
Thus, substituting Equation (30) in Equation (6), the value of \(A_{eff}(\mathbf{k})\) is obtained:  
\[ \begin{split} A_{eff}(\mathbf{k})=i\Omega^{-1}\Bigg{[}\Bigg{\{} \overset{=-1}{T}(0)-\overset{=-1}{T}(0)\frac{\Omega}{i}\cdot\bigg{[}I+\sum_ {\mathbf{G}}\Delta A(-\mathbf{G})\overset{=-1}{T}(\mathbf{G})\frac{\Omega}{i} \bigg{]}^{-1}\cdot\Delta A(0)\overset{=-1}{T}(0)\Bigg{\}}^{-1}+\\ \bigg{[}\begin{array}{cc}0_{3}&K_{3\times 6}(0,0)\\ (K_{3\times 6}(0,0))^{T}&0_{6}\end{array}\bigg{]}\Bigg{]}.\end{split}  \tag{31}\]  
By factoring, we reach the following expression:  
\[ \begin{split} A_{eff}(\mathbf{k})=i\Omega^{-1}\Bigg{\{}\bigg{[}I+\sum _{\mathbf{G}}^{\prime}\Delta A(-\mathbf{G})\overset{=-1}{T}(\mathbf{G})\frac {\Omega}{i}\bigg{]}^{-1}\cdot\bigg{[}I+\sum_{\mathbf{G}}\Delta A(-\mathbf{G}) \overset{=-1}{T}(\mathbf{G})\frac{\Omega}{i}\bigg{]}^{-1}\overset{=}{T}(0)+\\ \bigg{[}\begin{array}{cc}0_{3}&K_{3\times 6}(0,0)\\ (K_{3\times 6}(0,0))^{T}&0_{6}\end{array}\bigg{]}\Bigg{\}}.\end{split}  \tag{32}\]  
The apostrophe in the sum indicates that the term \(\mathbf{G}=0\) is excluded. This expression is valid for small-filling fractions; the inversion of the matrix is not too large. Finally, let us apply Equation (32) to calculate the effective tensors for a phononic crystal in the quasi-static limit (\(\mathbf{k}\to 0,\omega\to 0\)). In this case, from Equation (9), we have:  
\[A_{eff}(\mathbf{k}\to 0)=i\Omega^{-1}\bigg{[}I+\sum_{\mathbf{G}}^{\prime} \Delta A(-\mathbf{G})\overset{=-1}{T}(\mathbf{G})\frac{\Omega}{i}\bigg{]}^{- 1}\cdot\bigg{[}I+\sum_{\mathbf{G}}\Delta A(-\mathbf{G})\overset{=-1}{T}( \mathbf{G})\frac{\Omega}{i}\bigg{]}^{-1}\overset{=}{T}(0). \tag{33}\]