Damped quadratic eigenvalue problem ($({\lambda }^{2}M+\lambda C+K)r=0$)

The damped quadratic eigenvalue problem finds exponential motions of a linear mechanical model with mass matrix $M$, damping matrix $C$ and stiffness matrix $K$. Substituting $u(t)=r{e}^{\lambda t}$ into $M\ddot{u}+C\dot{u}+Ku=0$ givesThe dependence on $\lambda$ is quadratic, so solving the undamped problem $Kr={\omega }^{2}Mr$ alone does not find the damped decay rates and mode shapes.^[ref]

A useful linearization sets $z=(r,\lambda r{)}^T$ and rewrites the problem asThis generalized eigenvalue problem has twice the displacement dimension. With invertible $M$, it becomes an ordinary first-order state eigenproblem by multiplication by the block mass inverse. Singular $M$ requires more care and cannot be handled by silently using $M^{-1}$.

For an original one-coordinate example in consistent units, take $M=1$, $C=2$ and $K=5$. Then ${\lambda }^2+2\lambda +5=0$, giving $\lambda =-1\pm 2i$. Real free motions have the formThe decay rate is $1$ and damped oscillation frequency is $2$, while the undamped natural frequency is $\sqrt{5}$. Increasing $C$ to6 instead gives real roots $-1,-5$, showing that the quadratic problem also describes nonoscillatory decay.

For real symmetric positive-definite $M,K$ and positive-semidefinite $C$, an oscillatory eigenpair satisfies a useful energy check. Multiply by the conjugate transpose $r^{\ast }$ and write $m=r^{\ast }Mr>0$, $c=r^{\ast }Cr\ge 0$, $k=r^{\ast }Kr>0$. The imaginary part of $m{\lambda }^2+c\lambda +k=0$ gives $\mathrm{Re}\lambda =-c/(2m)\le 0$ when $\mathrm{Im}\lambda \ne 0$. In a multi-coordinate system, complex eigenvectors can encode relative phase differences between coordinates; nonclassical damping generally prevents replacing them by the undamped real mode shapes.