Quadratic eigenvalue problem: Difference between revisions

Revision as of 22:09, 9 March 2023

Latest revision as of 16:18, 21 March 2025

(8 intermediate revisions by 4 users not shown)

Line 4:

:<math> Q(\lambda)x = 0 ~ \text{ and } ~ y^\ast Q(\lambda) = 0,</math>

where <math>Q(\lambda)=\lambda^2 A_2M + \lambda A_1C + A_0K</math>, with matrix coefficients <math>A_2M, \, A_1C, A_0K \in \mathbb{C}^{n \times n}</math> and we require that <math>A_2M\,\neq 0</math>, (so that we have a nonzero leading coefficient). There are <math>2n</math> eigenvalues that may be ''infinite'' or finite, and possibly zero. This is a special case of a [[nonlinear eigenproblem]]. <math>Q(\lambda)</math> is also known as a quadratic [[polynomial matrix]].

==Spectral theory==

A QEP is said to be <em>regular</em> if <math>\text{det} (Q(\lambda)) \not \equiv 0</math> identically. The coefficient of the <math>\lambda^{2n}</math> term in <math>\text{det}(Q(\lambda))</math> is <math>\text{det}(M)</math>, implying that the QEP is regular if <math>M</math> is nonsingular.

Eigenvalues at infinity and eigenvalues at 0 may be exchanged by considering the reversed polynomial, <math> \lambda^2 Q(\lambda^{-1}) = \lambda^2 K + \lambda C + M </math>. As there are <math> 2n</math> eigenvectors in a <math>n</math> dimensional space, the eigenvectors cannot be orthogonal. It is possible to have the same eigenvector attached to different eigenvalues.

==Applications==

=== Systems of differential equations ===

Quadratic eigenvalue problems arise naturally in the solution of systems of second order [[Linear differential equation|linear differential equations]] without forcing:

:<math> M q''(t) +C q'(t) + K q(t) = 0 </math>

Line 17 ⟶ 23:

q(t) = \sum_{j=1}^{2n} \alpha_j x_j e^{\lambda_j t} = X e^{\Lambda t} \alpha

</math>

Where <math>\Lambda = \text{Diag}([\lambda_1, \ldots, \lambda_{2n}]) \in \mathbb{R}^{2n, \times 2n} </math> are the quadratic eigenvalues, <math> X = [x_1, \ldots, x_{2n}] \in \mathbb{R}^{n, \times 2n} </math> are the <math> 2n</math> right quadratic eigenvectors, and <math> \alpha = [\alpha_1, \cdots, \alpha_{2n}]^\top \in \mathbb{R}^{2n}</math> is a parameter vector determined from the initial conditions on <math> q</math> and <math> q'</math>.

[[Stability theory]] for linear systems can now be applied, as the behavior of a solution depends explicitly on the (quadratic) eigenvalues.

Line 23 ⟶ 29:

A QEP can result in part of the dynamic analysis of structures [[Discretization|discretized]] by the [[finite element method]]. In this case the quadratic, <math>Q(\lambda)</math> has the form <math>Q(\lambda)=\lambda^2 M + \lambda C + K</math>, where <math>M</math> is the [[mass matrix]], <math>C</math> is the [[damping matrix]] and <math>K</math> is the [[stiffness matrix]].

Other applications include vibro-acoustics and [[fluid dynamics]].

==Methods of solution==

Line 29 ⟶ 35:

Direct methods for solving the standard or [[Generalized eigenvalue problem|generalized eigenvalue problems]] <math> Ax = \lambda x</math> and <math> Ax = \lambda B x </math>

are based on transforming the problem to [[Schur form|Schur]] or [[Schur decomposition#Generalized Schur decomposition|Generalized Schur]] form. However, there is no analogous form for quadratic matrix polynomials.

One approach is to transform the quadratic [[matrix polynomial]] to a linear [[matrix pencil]] (<math> A-\lambda B</math>), and solve a generalized

eigenvalue problem. Once eigenvalues and eigenvectors of the linear problem have been determined, eigenvectors and eigenvalues of the quadratic can be determined.

The most common linearization is the first [[Companion matrix|companion]] linearization

:<math>

LL1(\lambda) =

\begin{bmatrix}

0 & I_nN \\

-K & -C

\end{bmatrix}

-

\lambda\begin{bmatrix}

I_nN & 0 \\

0 & M

\end{bmatrix},

</math>

where <math>I_n</math> is the <math>n</math>-by-<math>n</math> identity matrix, with corresponding eigenvector

:<math>

z =

Line 53 ⟶ 59:

\end{bmatrix}.

</math>

For convenience, one often takes <math>N</math> to be the <math>n\times n</math> [[identity matrix]]. We solve <math> L(\lambda) z = 0 </math> for <math> \lambda </math> and <math>z</math>, for example by computing the Generalized Schur form. We can then

take the first <math>n</math> components of <math>z</math> as the eigenvector <math>x</math> of the original quadratic <math>Q(\lambda)</math>.

Another common linearization is given by

:<math>

L2(\lambda)= \begin{bmatrix}

-K & 0 \\

0 & N

\end{bmatrix}

-

\lambda\begin{bmatrix}

C & M \\

N & 0

\end{bmatrix}.

</math>

In the case when either <math>A</math> or <math>B</math> is a [[Hamiltonian matrix]] and the other is a [[skew-Hamiltonian matrix]], the following linearizations can be used.

:<math>

L3(\lambda)= \begin{bmatrix}

K & 0 \\

C & K

\end{bmatrix}

-

\lambda\begin{bmatrix}

0 & K \\

-M & 0

\end{bmatrix}.

</math>

:<math>

L4(\lambda)= \begin{bmatrix}

0 & -K \\

M & 0

\end{bmatrix}

-

\lambda\begin{bmatrix}

M & C \\

0 & M

\end{bmatrix}.

</math>

{{mathapplied-stub}}