# Boundaries of Oscillatory Motion in Structures with Nonviscous Dampers
## 2 New Results on Critical Damping of Structures with Viscoelastic Dampers
However, the inverse statement is not true in general; that is, eigensolutions holding \(\mathbf{D}^{\prime}(s)\mathbf{u}\
eq\mathbf{0}\) might lead to \(\operatorname{rank}\bigl{[}\mathbf{\mathcal{J}}_{\mathbf{F},\mathbf{y}}\bigr{]}<n+1\). In the current paper, that result will be improved by finding the necessary and sufficient conditions for which Jacobian matrix is non-invertible, enabling a much more general characterization of the critical damping. These conditions are presented in the following theorem:  
Theorem 1: _Assume that \(s\in\mathbb{R}^{-}\) and \(\mathbf{u}\in\mathbb{R}^{n}\) are a real eigensolution for a certain value of the damping parameters \(\mathbf{\theta}\in\mathbb{R}^{p}\). In addition, it will be assumed that \(\operatorname{rank}[\mathbf{D}(s,\mathbf{\theta})]=n-1\). Under these hypotheses, the Jacobian is non-invertible, i.e., \(\det\bigl{[}\mathbf{\mathcal{J}}_{\mathbf{F},\mathbf{y}}\bigr{]}=0\), if and only if \(\mathbf{u}^{\mathsf{T}}\mathbf{D}^{\prime}(s,\mathbf{\theta})\mathbf{u}=0\)._  
Proof: First, let us consider that \(\det\bigl{[}\mathbf{\mathcal{J}}_{\mathbf{F},\mathbf{y}}\bigr{]}=0\). Since we are free to choose \(f(\mathbf{u})\), it can be assumed that the last row of \(\mathbf{\mathcal{J}}_{\mathbf{F},\mathbf{y}}\), i.e., the vector \(\{0,\mathbf{\
abla}^{\mathsf{T}}f\}\), is linearly independent of the previous \(n\) rows; therefore, \(\operatorname{rank}[\mathbf{D}^{\prime}(s,\mathbf{\theta})\mathbf{u},\mathbf{D}(s, \mathbf{\theta})]\leq n-1\). Then, there exist \(n\) real coefficients, \(\alpha_{1},\dots,\alpha_{n}\), not all of which are zero, such that:  
\[\mathbf{D}^{\prime}(s,\mathbf{\theta})\mathbf{u}=\alpha_{1}\,\mathbf{d}_{1}+\dots+ \alpha_{n}\,\mathbf{d}_{n}, \tag{15}\]where \(\mathbf{d}_{j}=\{D_{j1}(s,\boldsymbol{\theta}),\ldots,D_{jn}(s,\boldsymbol{\theta}) \}^{T}\) denotes the \(j\)th column/row of the matrix \(\mathbf{D}(s,\boldsymbol{\theta})\), due to the symmetry. Writting Equation (15) in matrix form yields:  
\[\mathbf{D}^{\prime}(s,\boldsymbol{\theta})\mathbf{u}=\mathbf{D}(s,\boldsymbol{ \theta})\,\boldsymbol{\alpha}, \tag{16}\]  
where \(\boldsymbol{\alpha}=\{\alpha_{1},\cdots\alpha_{n}\}^{T}\
eq\mathbf{0}^{T}\). Since \(\mathbf{u}\) is an eigenvector, \(\mathbf{D}(s,\boldsymbol{\theta})\mathbf{u}=\mathbf{0}\), and consequently:  
\[\mathbf{u}^{T}\mathbf{D}^{\prime}(s,\boldsymbol{\theta})\mathbf{u}=\mathbf{u} ^{T}(\mathbf{D}(s,\boldsymbol{\theta})\,\boldsymbol{\alpha})=\left(\mathbf{u} ^{T}\mathbf{D}(s,\boldsymbol{\theta})\right)\boldsymbol{\alpha}=\mathbf{0}^{T }\,\boldsymbol{\alpha}=0. \tag{17}\]  
Inversely, assume that \(\mathbf{u}^{T}\mathbf{D}^{\prime}(s,\boldsymbol{\theta})\mathbf{u}=0\). Then, the vector \(\mathbf{D}^{\prime}(s,\boldsymbol{\theta})\mathbf{u}\) belongs to the vector subspace:  
\[\mathcal{S}=\{\mathbf{x}\in\mathbb{R}^{n}:\mathbf{u}^{T}\,\mathbf{x}=0\}.\]  
It is straightforward that the \(\dim(\mathcal{S})=n-1\) because \(\mathbf{u}\in\mathbb{R}^{n}\) is a non-zero vector. Moreover, since \(\mathbf{u}\) is an eigenvector associated to \(s\), the \(n\) following relationships hold:  
\[\mathbf{D}(s,\boldsymbol{\theta})\mathbf{u}=\{\mathbf{u}^{T}\,\mathbf{d}_{1}, \ldots,\mathbf{u}^{T}\,\mathbf{d}_{n}\}^{T}=\mathbf{0}, \tag{18}\]  
where \(\mathbf{d}_{j}\) has already been defined above. It is clear that \(\mathbf{d}_{j}\in\mathcal{S}\), for \(1\leq j\leq n\). Since \(\dim(\mathcal{S})=n-1\) and \(\operatorname{rank}[\mathbf{D}(s,\boldsymbol{\theta})]=n-1\), then (without loss of generality) the set of vectors \(\mathbf{d}_{1},\ldots,\mathbf{d}_{n-1}\) is a basis of \(\mathcal{S}\).