**3.3. Physical Background**

We look now at the application of eigenvalues of quadratic problems in engineering. The largest review of applications QEP is in the [10].We have already mentioned in the introduction to eigenvalue problem arises in connection with differential equations or systems of differen‐ tial equations. In structural mechanics the most commonly are used differential equations and therefore the problem of eigenvalues. Note that the ultimate goal is to determine the effect of vibrations on the performance and reliability of the system, and to control these effects.

We will now demonstrate the linearization of QEP on a concrete example from the engineering. Low vibration system on *n* unknowns are described by the following system of differential equations

$$\mathbf{M}\ddot{\mathbf{y}} + \mathbf{C}\dot{\mathbf{y}} + \mathbf{K}\mathbf{y} = \mathbf{0},\tag{23}$$

where *M* is mass matrix, *C* is viscous damping matrix, and *K* is the stiffness matrix thus Because of the conditions in physics *M* and *K* are related to the kinetic and strain energy, respectively, by a quadratic form which makes them symmetric. For most structures *M* and *K* are positive definite and space.

The introduction of shift **y** = **x**e<sup>λ</sup> after rearrangement, we get

$$\left(\lambda^2 \mathbf{M} + \lambda \mathbf{C} + \mathbf{K}\right) \mathbf{x} \mathbf{e}^{\lambda} = \mathbf{0}$$

Respectively

$$\left(\lambda^2 \mathbf{M} + \lambda \mathbf{C} + \mathbf{K}\right) \mathbf{x} = \mathbf{0} \tag{24}$$

Therefore, the system (23) has a nontrivial solution **y** is selected, such that λ that QEP (24) has a nontrivial solution **x**.

Now we're going to QED (24) apply linearization method presented in section 3.2. Thus we have the appropriate GEP

$$
\begin{pmatrix}
I & \mathbf{O}
\end{pmatrix}
\begin{pmatrix}
\mathbf{z} \\
\mathbf{x}
\end{pmatrix} = \lambda \begin{pmatrix}
M & \mathbf{O} \\
\mathbf{O} & I
\end{pmatrix}
\begin{pmatrix}
\mathbf{z} \\
\mathbf{x}
\end{pmatrix}.
$$

When the system is undamped (*C*=*O)* we get

$$\boldsymbol{\alpha} \mathbf{M} \mathbf{x} := \boldsymbol{\lambda}^{\top} \boldsymbol{M} \mathbf{x} = \mathbf{K} \mathbf{x} = \mathbf{0} \dots$$

Because the most common matrix *M* and *K* are symmetric, positive definite and space obtained GEP is easy to solve.
