The direct approach is used to compute the wave solutions of the unit cell. The dynamic of the unit cell is described with the dynamic stiffness matrix \mathbf{D} (we assume no damping, \mathbf{C}=0)

\underbrace{\left(\mathbf{K}-\omega^2 \mathbf{M}\right)}_{\mathbf{D}}\mathbf{q}=\mathbf{f}

The matrices \mathbf{K} and \mathbf{M} are the stiffness and mass matrix, \mathbf{f} is the load vector (see article Equation of Motion). The entries in the dynamic stiffness matrix of the unit cell can be ordered with respect to the left, interior and right degrees of freedom

(1) \begin{equation} \begin{bmatrix} \mathbf{D}_{LL} & \mathbf{D}_{LI} & \mathbf{D}_{LR} \\ \mathbf{D}_{IL} & \mathbf{D}_{II} & \mathbf{D}_{IR} \\ \mathbf{D}_{RL} & \mathbf{D}_{RI} & \mathbf{D}_{RR} \end{bmatrix} \begin{bmatrix} \mathbf{q}_{L}\\ \mathbf{q}_{I}\\ \mathbf{q}_{R} \end{bmatrix} = \begin{bmatrix} \mathbf{f}_{L}\\ \mathbf{f}_{I}\\ \mathbf{f}_{R} \end{bmatrix} \label{eq:orderDynamicStiffness} \end{equation}

Theory

The inverse approach formulates a different eigenvalue problem. In this case, we reduce the vector of displacements \mathbf{q} and forces \mathbf{f} according to (Periodic Structures:2), (Periodic Structures:3) and (Periodic Structures:4)

(2) \begin{align} \begin{bmatrix} \mathbf{q}_{L}\\ \mathbf{q}_{I}\\ \mathbf{q}_{R} \end{bmatrix} = \underbrace{ \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{0} & \mathbf{1} \\ \mathbf{1}e^{-ikL} & \mathbf{0} \end{bmatrix} }_{\mathbf{R}} \begin{bmatrix} \mathbf{q}_{L}\\ \mathbf{q}_{I} \end{bmatrix} \end{align}
(3) \begin{align} \underbrace{ \begin{bmatrix} \mathbf{1} & \mathbf{0} & \mathbf{1}e^{ikL}\\ \mathbf{0} & \mathbf{1} & \mathbf{0} \end{bmatrix} }_{\mathbf{R}^{*T}} \begin{bmatrix} \mathbf{f}_{L}\\ \mathbf{f}_{I}\\ \mathbf{f}_{R} \end{bmatrix} =\begin{bmatrix} \mathbf{0}\\ \mathbf{0} \end{bmatrix} \label{eq:f_red} \end{align}

We assume that there is no external loading (\mathbf{f}_I=0). Inserting equation (2), we modify the discretized equation of motion

\begin{align} \underbrace{\left(\mathbf{K}-\omega^2 \mathbf{M}\right)}_{\mathbf{D}}\mathbf{q}=\mathbf{f} \quad\Longleftrightarrow\quad \left( \mathbf{K}-\omega^2 \mathbf{M} \right) \mathbf{R} \begin{bmatrix} \mathbf{q}_{L}\\ \mathbf{q}_{I} \end{bmatrix} = \begin{bmatrix} \mathbf{f}_{L}\\ \mathbf{f}_{I}\\ \mathbf{f}_{R} \end{bmatrix} \end{align}

In the next step, we multiply both sides from the left with the conjugate transpose \mathbf{R}^{*T}

\begin{align} \mathbf{R}^{*T}\left( \mathbf{K}-\omega^2 \mathbf{M} \right) \mathbf{R} \begin{bmatrix} \mathbf{q}_{L}\\ \mathbf{q}_{I} \end{bmatrix} = \mathbf{R}^{*T} \begin{bmatrix} \mathbf{f}_{L}\\ \mathbf{f}_{I}\\ \mathbf{f}_{R} \end{bmatrix} \end{align}

According to (3) the right side vanishes and one obtains an eigenvalue problem in \omega^2

\begin{align} \mathbf{R}^{*T}\left( \mathbf{K}-\omega^2 \mathbf{M} \right) \mathbf{R} \begin{bmatrix} \mathbf{q}_{L}\\ \mathbf{q}_{I} \end{bmatrix} = \mathbf{0} \end{align}

Prescribing the wave type by defining a specific exponential term e^{ikL}, we can compute the frequencies at which the prescribed wave solution occurs. Due to the prescribed wave type, the inverse approach is mostly used to identify stop bands. Stop bands are frequency ranges where propagating wave do not occur.

Example

An example for the inverse approach can be found here

Outlook

The eigenvectors \Phi_i of the underlying eigenvalue problem describe the wave shapes. The eigenpairs \left(\mu_i,\Phi_i\right) can be used to transform the problem into the wave domain. This transformation is comparable to the transformation of the problem into modal coordinates. In the wave domain the problem is solved in terms of wave amplitudes. The benefit of the WFEM is that the wave solutions can be computed based on a single unit cell. However, it is possible to compute the frequency response of a structure consisting of multiple unit cells.