[TOC]
1 The different ways to construct effective medium
1.1 The problem and the direct solution
Consider two mass-in-mass elements, $j=1,2$, interacting through a soft spring of constant $h$. Each element has an outer mass $M_{j}$ and an inner mass $m_{j}$, and a coupling spring of constant $k_{j}$. Calling the displacement of the outer mass $U_{j}$ and $u_{j}$ that of the inner mass, it comes that the system obeys the motion equation
\[\left[\begin{array}{cccc} k_{1}+h & -k_{1} & -h & 0 \\ -k_{1} & k_{1} & 0 & 0 \\ -h & 0 & k_{2}+h & -k_{2} \\ 0 & 0 & -k_{2} & k_{2} \end{array}\right]\left[\begin{array}{l} U_{1} \\ u_{1} \\ U_{2} \\ u_{2} \end{array}\right]=\omega^{2}\left[\begin{array}{cccc} M_{1} & & & \\ & m_{1} & & \\ & & M_{2} & \\ & & & m_{2} \end{array}\right]\left[\begin{array}{l} U_{1} \\ u_{1} \\ U_{2} \\ u_{2} \end{array}\right]\]For the unperturbated system, the equations are
\[\left[\begin{array}{cccc} k_{1} & -k_{1} & 0 & 0 \\ -k_{1} & k_{1} & 0 & 0 \\ 0 & 0 & k_{2} & -k_{2} \\ 0 & 0 & -k_{2} & k_{2} \end{array}\right]\left[\begin{array}{l} U_{1} \\ u_{1} \\ U_{2} \\ u_{2} \end{array}\right]=\omega^{2}\left[\begin{array}{cccc} M_{1} & & & \\ & m_{1} & & \\ & & M_{2} & \\ & & & m_{2} \end{array}\right]\left[\begin{array}{l} U_{1} \\ u_{1} \\ U_{2} \\ u_{2} \end{array}\right]\]The eigenvalues are
\[0, \ 0, \ \omega_1 = \frac{M_{1} k_{1}+k_{1} m_{1}}{M_{1} m_{1}}, \ \omega_2 = \frac{M_{2} k_{2}+k_{2} m_{2}}{M_{2} m_{2}}\]and the corresponding eigenvectors are
\[e_1 = [1,1,0,0]^T \\ e_2 = [0,0,1,1]^T \\ e_3 = [k_1,-k_1 M_1/m_1,0,0]^T \\ e_4 = [0,0,k_2,-k_2 M/m_2]^T\]1.2 First order degenerate pertubation theory
Assume the solution as the superposition of vector $e_3$ and $e_4$
\[u = x e_1 + y e_2\]Subsituting that into Eq. (1), we have
\[K(xe_1+ye_2) = \omega M (x e_1 + y e_2)\]Let product of $e_1$ and $e_2$ gives the equations of the coefficients $x,y$. The equations are
\[\left[\begin{array}{cc} k_{1}^{2} h + M_1^2 k_1 \omega_1^4 & -k_{1} k_{2} h \\ -k_{1} k_{2} h & k_{2}^{2} h + M_2^2 k_2 \omega_2^4 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\omega^{2}\left[\begin{array}{cc} M_{1} k_{1}^{2}\left(1+\frac{M_{1}}{m_{1}}\right) & 0 \\ 0 & M_{2} k_{2}^{2}\left(1+\frac{M_{2}}{m_{2}}\right) \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]\]If $k_1=k_2$, $m_1=m_2$, $M_1=M_2$, then $\omega_1=\omega_2$. And Eq. (7) can be reduced to
\[\left[\begin{array}{cc} k_{1}^{2} h & -k_{1} k_{2} h \\ -k_{1} k_{2} h & k_{2}^{2} h \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left(\omega^{2}-\omega_0^2\right)\left[\begin{array}{cc} M_{1} k_{1}^{2}\left(1+\frac{M_{1}}{m_{1}}\right) & 0 \\ 0 & M_{2} k_{2}^{2}\left(1+\frac{M_{2}}{m_{2}}\right) \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]\]Since this method is the first order degenerate perturbation theory, so the error exhibit $\mathcal{O}(h^2)$.
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Fig. 1. Frequency spectrum for $k_1 = k_2 = 10, h=1, m_1 = m_2 = M_1 = M_2 = 1$. (a) Black lines are from exact solution Eq. (3), and red dash lines are from first degerate perturbation theory. (b) The error is propotional to $h^2$.
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Fig. 2. Frequency spectrum for $k_1 =10, k_2 = 15, h=1, m_1 = m_2 = M_1 = M_2 = 1$. (a) Black lines are from exact solution Eq. (3), and red dash lines are from first degerate perturbation theory. (b) The error is propotional to $h^2$.
1.3 The exact way
We transform the displacement vector to the new vector in the basis of eigenstates, or
\[\left[\begin{array}{l} U_{1} \\ u_{1} \\ U_{2} \\ u_{2} \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & k1 & 0 \\ 1 & 0 & -k_1 M_1/m_1 & 0 \\ 0 & 1 & 0 & -k_2 \\ 0 & 1 & 0 & -k_2 M_2/m_2 \end{array}\right] \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right] \\ \text{or} \ u = Uv.\]Substituting Eq. (9) into Eq. (1), we have
\[U^T K U v = \omega^2 U^T M U v\]or
\[\left[\begin{array}{cccc} h & -h & h k_{1} & -h k_{2} \\ -h & h & -h k_{1} & h k_{2} \\ h k_{1} & -h k_{1} & \frac{M_{1}^{2} k_{1}^{3}+2 m_{1} M_{1} k_{1}^{3}}{m_{1}^{2}}+k_{1}^{2}\left(h+k_{1}\right) & -h k_{1} k_{2} \\ -h k_{2} & h k_{2} & -h k_{1} k_{2} & \frac{M_{2}^{2} k_{2}^{3}+2 m_{2} M_{2} k_{2}^{3}}{m_{2}^{2}}+k_{2}^{2}\left(h+k_{2}\right) \end{array}\right] \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right] = \omega^2 \left(\begin{array}{cccc} M_{1}+m_{1} & 0 & 0 & 0 \\ 0 & M_{2}+m_{2} & 0 & 0 \\ 0 & 0 & \frac{M_{1} k_{1}^{2}\left(M_{1}+m_{1}\right)}{m_{1}} & 0 \\ 0 & 0 & 0 & \frac{M_{2} k_{2}^{2}\left(M_{2}+m_{2}\right)}{m_{2}} \end{array}\right) \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right]\]Define $H = K-\omega^2 M$, and rewirte $H=\left[\begin{array}{cc} H_{11} & H_{12} \ H_{21} & H_{22} \end{array}\right]$ by 2 by 2 block matrices $H_{ij}$. Then the effective matrix of 2 $v_3,v_4$ is
\[H_{\text{eff}} = H_{22} - H_{21}H_{11}^{-1}H_{12}.\]The spectrum can be obtained from the vanishing of determinant of Eq. (12), or
\[|H_{\text{eff}}|=0.\]| 
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Fig. 3. Frequency spectrum for (a) $k_1 = k_2 = 10, h=1, m_1 = m_2 = M_1 = M_2 = 1$. and (b) $k_1 =10, k_2 = 15, h=1, m_1 = m_2 = M_1 = M_2 = 1$. Black lines are from Eq. (3) and red dash lines are from Eq. (13).
1.4 Schrieffer-Wolff transform
For simplicity, we set $m_1 = m_2 = M_1 = M_2=1$, so Eq. (11) can be transformed into
\[\frac{1}{2}\left[\begin{array}{cccc} h & -h & h & -h \\ -h & h & -h & h \\ h & -h & h+4 k_{1} & -h \\ -h & h & -h & h+4 k_{2} \end{array}\right] \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right] = \omega^2 \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right]\]We separate Eq. (14) into the leading term and perturbative term, so we have
\[\frac{1}{2}\left[\begin{array}{cccc} h & -h & 0 & 0 \\ -h & h & 0 & 0 \\ 0 & 0 & h+4 k_{1} & -h \\ 0 & 0 & -h & h+4 k_{2} \end{array}\right] \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right] + \left[\begin{array}{cccc} 0 & 0 & h & -h \\ 0 & 0 & -h & h \\ h & -h & 0 & 0 \\ -h & h & 0 & 0 \end{array}\right] \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right] = \omega^2 \left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \end{array}\right] \\ \text{or} \quad K v = \omega^2 v, \ \text{where} \ K = K_0+K_1.\]Now we want to do the Schrieffer-Wolff transform to separate the low frequency subspace to the high frequency subspace. We define the unitary transform $v=\mathcal{U}w$ where $\mathcal{U}=e^{S}$ and $S$ is an anti-Hermitian matrix. After the unitary transform the stiffness matrix is transformed into
\[K' = e^{S} K e^{-S}\]By using the Baker-Campbell-Haussdorf formula , we have
\[K^{\prime}=K+[S, K]+\frac{1}{2}[S,[S, K]]+\ldots\]or
\[K^{\prime}=K_{0}+K_1+\left[S, K_{0}\right]+ [S, K_1]-\frac{1}{2}\left[S,\left[S, K_{0}\right]\right]-\frac{1}{2}[S,[S, K_1]]+\ldots\]The Hamiltonian can be made diagonal to first order in $K_1$ by choosing the generator $S$ such that
\[\left[K_{0}, S\right]= K_1\]This equation always has a definite solution under the assumption that $h K_1$ is off-diagonal in the eigenbasis of $K_{0}$. Substituting this choice in the previous transformation yields:
\[K^{\prime}=K_{0}+\frac{1}{2}[S, K_1]+\mathcal{O}\left(h^{3}\right)\]where $K’$ is block diagonalized so we finish the separation of low frequency subspace and high frequency subspace. The error after SW transform are $O\left(h^{3}\right)$ while it is $O\left(h^{2}\right)$ for first degenerate perturbation theory. Actually, the SW transform is the second degenerate pertubation theory which is not commonly used.
To get $S$, let’s diagonalize $K_0$ by producting a unitary transform matrix $\mathcal{U}_1$
\[\left[\begin{array}{cccc} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 & 0 \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & E_{11} & E_{12} \\ 0 & 0 & E_{21} & E_{22} \end{array}\right]\]where $K_1 E_i = d_i E_i$. And $K_0$ is diagonalized as
\[K_0' = \left[\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & h & 0 & 0 \\ 0 & 0 & d_1 & 0 \\ 0 & 0 & 0 & d_2 \end{array}\right]\]where
\[d_1 = \frac{h}{2}+k_{1}+k_{2}-\frac{\sqrt{h^{2}+4 k_{1}^{2}-8 k_{1} k_{2}+4 k_{2}^{2}}}{2}\\ d_2 = \frac{h}{2}+k_{1}+k_{2}+\frac{\sqrt{h^{2}+4 k_{1}^{2}-8 k_{1} k_{2}+4 k_{2}^{2}}}{2}.\]And $K_1$ are transformed into
\[K_1' = \mathcal{U}_1^T K_1 \mathcal{U}_1\]and it is
\[\left[\begin{array}{cccc} 0 & 0 & K_{11}' & K_{12}' \\ 0 & 0 & K_{21}' & K_{22}' \\ K_{11}' & K_{12}' & 0 & 0 \\ K_{21}' & K_{22}' & 0 & 0 \end{array}\right]\]And the solution of $S$ is
\[S_{i j}= \left\{\begin{array}{cc} \frac{K_{i j}^{\prime}}{d_{j}-d_{i}} & i \neq j \\ 0 & i=j \end{array}\right.\]and then from Eq. (20), we have
\[K'_d =K_{0}'+\frac{1}{2}[S, K_1']+\mathcal{O}\left(h^{3}\right)\]Rotating back give the stiffness matrix whose low frequency subspace and high frequency subspace are separated
\[K' = \mathcal{U}_1 K_d' \mathcal{U}_1^T\]| 
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Fig. 4. Frequency spectrum for $k_1 = k_2 = 10, h=1, m_1 = m_2 = M_1 = M_2 = 1$. (a) Black lines are from exact solution Eq. (3), and red dash lines are from SW transformation. (b) The error is propotional to $h^3$.
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Fig. 5. Frequency spectrum for $k_1 =10, k_2 = 15, h=1, m_1 = m_2 = M_1 = M_2 = 1$. (a) Black lines are from exact solution Eq. (3), and red dash lines are from SW transformation. (b) The error is propotional to $h^3$.