The Problem
I have the following system of linear second order differential equations which I would like to solve:
$$\mathbf{J}\ddot{\mathbf{\Theta}} + \mathbf{C}\dot{\mathbf{\Theta}} + \mathbf{K\Theta} = \mathbf{ T}_t(t) \tag{1}$$
I would like to solve it using the technique involving Fourier series.
My Attempt
These are the steps I follow:
Assume that $\mathbf{\Theta}$ is a Fourier series with unknown coefficients $\mathbf{c}_n$, and finds its derivatives: \begin{align} \mathbf{\Theta} & = \sum^{+\infty}_{n=-\infty} \mathbf{c}_n e^{in\omega t} \label{eq: fourier disp}\\ \dot{\mathbf{\Theta}} & = \sum^{+\infty}_{n=-\infty} (in\omega) \mathbf{c}_n e^{in\omega t} \label{eq: fourier speed} \\ \ddot{\mathbf{\Theta}} & = \sum^{+\infty}_{n=-\infty} -(n\omega)^2 \mathbf{c}_n e^{in\omega t} \label{eq: fourier accel} \end{align}
Expand the right hand side of Eq.(1) into the following Fourier series with known coefficients $\mathbf{f}_n$:
\begin{equation} \mathbf{T}_t = \sum^{+\infty}_{n=-\infty} \mathbf{f}_n e^{in\omega t} \end{equation}
- Substitue all of the above equations from Steps 1 and 2 into Eq.(1), also knowning that $\mathbf{c}_{n}=\mathbf{c}_{n,\text{Re}} + i\mathbf{c}_{n,\text{Im}}$ and $\mathbf{f}_{n}=\mathbf{f}_{n,\text{Re}} + i\mathbf{f}_{n,\text{Im}}$, to get the following:
\begin{gather} -(n\omega)^2 \mathbf{J}\mathbf{c}_n e^{in\omega t} + (in\omega)\mathbf{C} \mathbf{c}_n e^{in\omega t} + \mathbf{K}\mathbf{c}_n e^{in\omega t} = \mathbf{f}_n e^{in\omega t} \nonumber \\ % -(n\omega)^2 \mathbf{J}\mathbf{c}_n + (in\omega)\mathbf{C} \mathbf{c}_n + \mathbf{K}\mathbf{c}_n = \mathbf{f}_n \nonumber \\ % -(n\omega)^2 \mathbf{J}(\mathbf{c}_{n,\text{Re}} + i\mathbf{c}_{n,\text{Im}}) + (in\omega)\mathbf{C} (\mathbf{c}_{n,\text{Re}} + i\mathbf{c}_{n,\text{Im}}) + \mathbf{K}(\mathbf{c}_{n,\text{Re}} + i\mathbf{c}_{n,\text{Im}}) = \mathbf{f}_{n,\text{Re}} + i\mathbf{f}_{n,\text{Im}} \tag{2} \end{gather}
- Equate the real and imaginary parts of Eq.(2) and create a linear system of equations:
\begin{equation} \begin{bmatrix} -(n\omega)^2 \mathbf{J} + \mathbf{K} & - (n\omega)\mathbf{C} \\ -(n\omega)^2 \mathbf{J} + \mathbf{K} & (n\omega)\mathbf{C} \end{bmatrix} % \begin{bmatrix} \mathbf{c}_{n,\text{Re}} \\ \mathbf{c}_{n,\text{Im}} \end{bmatrix} = \begin{bmatrix} \mathbf{f}_{n,\text{Re}} \\ \mathbf{f}_{n,\text{Im}} \end{bmatrix} \tag{3} \end{equation}
You can probably see this by eye as well, but when I substitute all the numbers to solve Eq.(3) above, the square block matrix is singular. Therefore I am making a mistake somewhere along the way. My hunch tells me that something might be wrong in step 3 but I can't tell what.
Equation $(2)$ is correct, but equation $(3)$ should be : $$\begin{pmatrix} \mathbf K - (n\omega)^2 \mathbf J & -n\omega \mathbf C \\ n \omega \mathbf C & \mathbf K - (n\omega)^2 \mathbf J \end{pmatrix} \begin{pmatrix} \mathbf c_{n,\rm Re} \\ \mathbf c_{n,\rm Im}\end{pmatrix} = \begin{pmatrix} \mathbf f_{n,\rm Re} \\ \mathbf f_{n,\rm Im}\end{pmatrix} \tag{3}$$