Example

We continue to use the example introduced in §2.1 and Figure 2.1. We now consider the fully general case of nonzero masses $m_i$ and damping constants $b_i$. This leads to the equations of motion $M \ddot{x}(t) = -B \dot{x} (t) -K x(t)$. We solve them by changing variables to

$$y(t) = \bmat{c} \dot{x}(t) \\ x(t) \emat,$$

yielding

$$\bmat{cc} M & 0 \\ 0 & I \emat \dot{y}(t) = \bmat{c} M \dd... ...ot{x}(t) \emat = \bmat{cc} -B & -K \\ I & 0 \emat \cdot y(t)$$

or $A \dot{y}(t) = C y(t)$. We solve this by substituting $y(t) = e^{\lambda t} y$, where $y$ is a constant vector and $\lambda$ is a constant scalar to be determined. This yields

$$Cy = \lambda Ay.$$

Thus $y$ is an eigenvector and $\lambda$ is an eigenvalue of the generalized nonsymmetric eigenvalue problem.