3. Eigenmode norm of the quadratic problem

3.1. Euclidean norms and « largest component at 1 »

For the quadratic problem, we have the same standards as for the generalized problem. Since natural modes are complex, we work with the Hermitian product. The various « classical » standards become:

• Hermitian norm: \({\mathrm{\parallel }\Phi \mathrm{\parallel }}_{2}\mathrm{=}{(\mathrm{\sum }_{k\mathrm{=}1}^{m}{\mathrm{\mid }{\Phi }_{{\mathrm{l}}_{\mathrm{k}}}\mathrm{\mid }}^{2})}^{1\mathrm{/}2}\mathrm{=}{(\mathrm{\sum }_{k\mathrm{=}1}^{m}({\stackrel{ˉ}{\Phi }}_{{\mathrm{l}}_{\mathrm{k}}}{\Phi }_{{\mathrm{l}}_{\mathrm{k}}}))}^{1\mathrm{/}2}\) where \(\stackrel{ˉ}{{\Phi }_{{l}_{k}}}\) is the conjugate of \({\Phi }_{{l}_{k}}\) (the absolute value in the real domain becomes the module in the complex domain),

• « largest component to 1 » standard: \({\mathrm{\parallel }\Phi \mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{=}\underset{k\mathrm{=}\mathrm{1,}m}{\text{max}}\mathrm{\mid }{\Phi }_{{\mathrm{l}}_{\mathrm{k}}}\mathrm{\mid }\mathrm{=}\underset{k\mathrm{=}\mathrm{1,}m}{\text{max}}({({\stackrel{ˉ}{\Phi }}_{{l}_{k}}{\Phi }_{{\mathrm{l}}_{\mathrm{k}}})}^{1\mathrm{/}2})\).

3.2. Standard mass or generalized unit stiffness

With regard to the norm « mass or generalized rigidity », a term by analogy with the generalized problem, the matrix associated with the norm is used the one that intervenes in the writing of the quadratic problem put in the reduced form [éq 1.3-1].

We then have:

• generalized mass standard:

\({\parallel \Phi \parallel }_{\stackrel{ˆ}{B}}=(\lambda {\Phi }^{T},{\Phi }^{T})\stackrel{ˆ}{B}(\begin{array}{c}\lambda \Phi \\ \Phi \end{array})=(\lambda {\Phi }^{T},{\Phi }^{T})\left[\begin{array}{cc}0& B\\ B& C\end{array}\right](\begin{array}{c}\lambda \Phi \\ \Phi \end{array})=2\lambda {\Phi }^{T}B\Phi +{\Phi }^{T}C\Phi ,\)

\(\stackrel{ˆ}{\Phi }=\frac{1}{{\parallel \Phi \parallel }_{\stackrel{ˆ}{B}}}\Phi\),

• generalized stiffness standard:

\({\parallel \Phi \parallel }_{\stackrel{ˆ}{B}}=(\lambda {\Phi }^{T},{\Phi }^{T})\stackrel{ˆ}{A}(\begin{array}{c}\lambda \Phi \\ \Phi \end{array})=(\lambda {\Phi }^{T},{\Phi }^{T})\left[\begin{array}{cc}-B& 0\\ 0& A\end{array}\right](\begin{array}{c}\lambda \Phi \\ \Phi \end{array})=-{\lambda }^{2}{\Phi }^{T}B\Phi +{\Phi }^{T}A\Phi ,\)

\(\stackrel{ˆ}{\Phi }=\frac{1}{{\parallel \Phi \parallel }_{\stackrel{ˆ}{A}}}\Phi\).