ALBEpack can easily be used to solve the quadratic eigenproblem. The problem has the following form

( lambda^2 M + lambda C + K ) x = 0where M, C, and K are matrices, lambda is an eigenvalue, and x a corresponding right eigenvector.

To solve the quadratic eigenproblem, first give it the following generalized eigenvalue problem form, by linearizing.

A z = lambda B zVector z is a right eigenvector. The matrices A and B are constructed in the following manner.

-- -- -- -- | 0 I | | I 0 | A = | | B = | | |-K -C | | 0 M | -- -- -- --This linearization is detailed in

Please note that A could easily be singular. If it is, do not use a shift value

Once the linear problem is solved by

-- -- -- -- | x | | ( lambda M + C )' y | z = | | w = | | | lambda x | | y | -- -- -- --The example script

Briefly, the script does the following:

- load matrices M, C, K
- specify shift value
`alpha`, to be used in linear problem - specify number of eigenvalues to find,
`neig` - construct matrices A and B
- initialize
`ppp`input parameters - call
`ppp` - convert linear problem eigenvectors to quadratic problem eigenvectors
- display eigenvalues and residuals

problem
| Modal analysis of a mass / damper / spring system. |

matrix size | 100 x 100 |

shift value | -1.0 |

eigenvalues desired | 4 |

eigenvalues found | 5 |

Lanczos steps | 22 |

In this example, M is a diagonal mass matrix, C is a diagonal damping matrix, and K is the stiffness matrix. The i'th row of matrix K has three nonzero elements. K(i,i-1) = -ki, K(i,i) = ki + ki+1, K(i,i+1) = -ki+1, where ki is the stiffness of the i'th spring. Masses, damping coefficients, and spring coefficients were random values between 0 and 1.

matrix M - mass

matrix C - damping

matrix K - spring constants

Four eigenvalues were requested, five converged. All have small residual norms, even the two largest on the order of 10e-7.

* shift value alpha + approximate eigenvalue, Ritz value o exact eigenvalue x right residual norm x left residual norm |

approximate left residual right residual eigenvalues norms norms -0.95 0.00 0.00 -0.91 -0.0042i 0.0004 e-8 0.00 -0.91 +0.0042i 0.0004 e-8 0.00 -1.15 -0.0419i 0.3358 e-8 0.1112 e-6 -1.15 +0.0419i 0.3358 e-8 0.1112 e-6

problem
| modal analysis of a vibrating mass electrical gap closing device |

matrix size | 30 x 30 |

shift value | -1000.0 + 190000i |

eigenvalues desired | 4 |

eigenvalues found | 11 |

Lanczos steps | 12 |

Observe that the right residual norms are small, but the left residual norms are rather large. Only the third eigenvalue has good residual norms. Matrix A for this problem was very ill conditioned, with a condition number of 2.2342 e22. Finding a small number of approximate eigenvalues using

* shift value alpha + approximate eigenvalue, Ritz value o exact eigenvalue x right residual norm x left residual norm |

approximate left residual right residual eigenvalues norms norms -0.01 +1.89i e5 0.0084 e-12 0.0041 -0.01 +1.91i e5 0.0075 e-12 0.0038 -0.05 +2.02i e5 0.1127 e-12 0.0001 the "good" eigenvalue 0.23 +1.91i e5 0.1799 e-12 0.0014

problem
| model of a bearing support shaft |

matrix size | 80 x 80 |

shift value | -0.2 + 2500i |

eigenvalues desired | 4 |

eigenvalues found | 4 |

Lanczos steps | 15 |

All the approximate eigenvalues which are close to the actual eigenvalues, but that is due to a well chosen shift value

* shift value alpha + approximate eigenvalue, Ritz value o exact eigenvalue x right residual norm x left residual norm |

approximate left residual right residual eigenvalues norms norms -0.0000 +2.37i e3 0.1528 e-3 0.00 -0.0001 +3.14i e3 0.1757 e-3 0.00 -0.0006 +4.68i e3 0.3133 e-3 0.3362 e-5 -0.0001 +1.07i e3 0.1602 e-3 0.0081 e-5