Introduction:
I have a mass-spring-damper system with $q$ degrees of freedom. We model it by using a 2nd-order vectorial ODE \eqref{1} on variable $\left\{X\right\}$:
$$ \left[M\right] \cdot \left\{\ddot{X}\right\} + \left[C\right] \cdot \left\{\dot{X}\right\} + \left[K\right] \cdot \left\{X\right\} = \left\{0\right\} \label{1}\tag{1} $$ $$\sum_{j=1}^{q} M_{ij} \cdot \ddot{x}_{j} + C_{ij} \cdot \dot{x}_{j} + K_{ij} \cdot x_{j} = 0 \ \ \ \ \ \ \forall \ i = 1, \ 2, \ \cdots, \ q$$ $$\left\{X(t)\right\} = \left(x_1(t), \ x_2(t), \ \cdots, \ x_q(t)\right)$$ With $2q$ initial conditions: $$\left\{X(t=0)\right\} = \left\{X_{0}\right\}$$ $$\left\{\dot{X}(t=0)\right\} = \left\{\dot{X}_{0}\right\}$$
- The objective is to find $\left\{X(t)\right\} = \left(x_1(t), \ \cdots, \ x_q(t)\right)$
This is a complicated task. A particular case is when $[C]=[0]$:
\eqref{1} turns to \eqref{2}: $$ \left[M\right] \cdot \left\{\ddot{X}\right\} + \left[K\right] \cdot \left\{X\right\} = \left\{0\right\}\label{2}\tag{2} $$ With matrix $[M]$ and $[K]$, it's possible to find $[B]$ and $[\omega_n^2]$, such we get a diagonal system \eqref{3} (a new coordinate system):
$$ \underbrace{\left(\left[B\right]^{T} \left[M\right] \left[B\right]\right)}_{\left[I\right]} \cdot \underbrace{\left(\left[B\right]^{-1} \left\{\ddot{X}\right\}\right)}_{\left\{\ddot{Y}\right\}} + \underbrace{\left(\left[B\right]^{T} \left[K\right] \left[B\right]\right)}_{\left[\omega_n^2\right]} \cdot \underbrace{\left(\left[B\right]^{-1}\left\{X\right\}\right)}_{\left\{Y\right\}} = \left\{0\right\} $$
$$ \left\{\ddot{Y}\right\} + \underbrace{\left[\omega_n^2\right]}_{diagonal} \cdot \left\{Y\right\} = \left\{0\right\}\label{3}\tag{3} $$ $$ \ddot{y}_i + \omega_{n,i}^2 \cdot y_i = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \forall i = 1, \ 2, \ \cdots, \ q$$ Then, solving for each $y_i(t)$, we come back to the original variable: $$\left\{X(t)\right\} = \left[B\right] \cdot \left\{Y(t)\right\}$$
Question:
What's if $\left[C\right] \ne [0]$? Is it possible to write \eqref{1} as an unbound system \eqref{4}?
$$ \left\{\ddot{Y}\right\} + \underbrace{\left[2\xi \omega_n\right]}_{diagonal} \cdot \left\{\dot{Y}\right\} + \underbrace{\left[\omega_n^2\right]}_{diagonal} \cdot \left\{Y\right\} = \left\{0\right\}\label{4}\tag{4} $$ $$ \ddot{y}_i + 2\xi_{i} \omega_{n,i}\cdot \dot{y}_i +\omega_i^2 \cdot y_i = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \forall i = 1, \ 2, \ \cdots, \ q $$
- If yes:
How can I do it? That means, find the matrix $\left[B\right]$ (vectorial base) and values $\xi_i$, $\omega_{n,i}$ such that
$$\left[B\right]^{T} \cdot \left[M\right] \cdot \left[B\right] = \left[I\right]$$ $$\left[B\right]^{T} \cdot \left[C\right] \cdot \left[B\right] = \underbrace{\left[2\xi \omega_n\right]}_{diagonal}$$ $$\left[B\right]^{T} \cdot \left[K\right] \cdot \left[B\right] = \underbrace{\left[\omega_n^2\right]}_{diagonal}$$
- If not: Is there a method to solve $(1)$?
Additional information:
- About the matrix
- $\left[M\right]$, $\left[C\right]$ and $\left[K\right]$ are $q\times q$ matrix;
- All coefficients of $\left[M\right]$, $\left[C\right]$ and $\left[K\right]$ are constants;
- $\left[M\right]$ and $\left[K\right]$ are symmetric and definite-positive
- One DOF problem:
A $1$ DOF(degree of freedom) mass-spring-damper system can be described by the equation
$$m\cdot \dot{x} + c\cdot \dot{x} + k \cdot x = 0$$
With $m$, $c$ and $k$ positive constants, initial conditions $x(0) = x_0$, $\dot{x}(0)=v_0$. We rewrite it as
$$\ddot{x} + 2\xi \omega_n \cdot \dot{x} + \omega_n^2 \cdot x = 0$$ $$\omega_n = \sqrt{\dfrac{k}{m}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \xi = \dfrac{c}{2\sqrt{km}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mu = \sqrt{1-\xi^2}$$ For a overdamped system we have $0 \le \xi < 1$ and the solution becomes
$$x(t) = x_0 \cdot x_{pos}(t) + v_0 \cdot x_{vel}(t)$$ $$ x_{pos}(t) = \exp\left(-\xi \omega t\right) \left[\cos \left(\mu \omega t\right) + \dfrac{\xi}{\mu} \cdot \sin \left(\mu \omega t\right)\right]$$ $$ x_{vel}(t) = \dfrac{1}{\mu \omega} \exp\left(-\xi \omega t\right) \sin \left(\mu \omega t\right)$$
Numerical solution:
To verify if the values of $\xi$ and $\omega$ are correct, I use a numerical solution.
For $1$ DOF system I can rewrite the system as
$$\begin{bmatrix} \dot{x} \\ \ddot{x} \end{bmatrix} = \underbrace{\begin{bmatrix} 0 & 1 \\ \frac{-k}{m} & \frac{-c}{m} \end{bmatrix}}_{\left[A\right]}\begin{bmatrix} x \\ \dot{x} \end{bmatrix} $$ And solve using Runge-Kutta for example. It's nice to see that the eigenvalues of $A$ are
- $\lambda_1 = -\xi \omega + i\mu \omega$
- $\lambda_2 = -\xi \omega - i\mu \omega$
For a $n$ degree system, I can do the same
$$ \begin{bmatrix} \left\{\dot{X}\right\} \\ \left\{\ddot{X}\right\} \end{bmatrix} = \underbrace{\begin{bmatrix} \left[0\right] & \left[I\right] \\ -\left[M\right]^{-1} \left[K\right] & -\left[M\right]^{-1} \left[C\right] \end{bmatrix}}_{\left[A\right]} \begin{bmatrix} \left\{X\right\} \\ \left\{\dot{X}\right\} \end{bmatrix}$$ Getting the eigenvalues of $A$, let's call $\lambda_A$, I get $2q$ complex values conjugated in pairs. I noticed that $|\lambda_{A}|$ are the same as $\omega$ (gotten from the generalized eigenvalue problem).
Then I tried to get the values of $\xi_i$ from $\lambda_A$, but I did not succeed.
This is just a hint / pointer
Special Case
If you can find the matrix $\mathbf{B}$ such that $\mathbf{B}^{T}\mathbf{M}\mathbf{B}$ and $\mathbf{B}^{T}\mathbf{K}\mathbf{B}$ are diagonal, will $\mathbf{B}^{T}\mathbf{C}\mathbf{B}$ be also diagonal if $\mathbf{C}=c_{1}\mathbf{M}+c_{2}\mathbf{K}$ ?
General Case
Define $\mathbf{D}=0.5\cdot\mathbf{M}^{-1}\mathbf{C}$ and $\mathbf{W}=\left(\mathbf{M}^{-1}\mathbf{K}\right)^{0.5}$. Then the solution is
$$ \mathbf{x}= c_{1}\cdot e^{\left[-\mathbf{D}+\left(\mathbf{D}^{2}-\mathbf{W}^{2}\right)^{0.5}\right]t}+c_{2}\cdot e^{\left[-\mathbf{D}-\left(\mathbf{D}^{2}-\mathbf{W}^{2}\right)^{0.5}\right]t}\phantom{xx}\forall\phantom{xx}c_{1},c_{2}\in\mathbb{C} $$
Compare this solution with solution for single DOF system, also take a look at definition of exponential of a matrix.