## 2 Non-Autonomous Dynamic Systems
### Eigenvalue Problem Formulation  
The eigenvalue problem is first expressed in terms of the aforementioned generalized coordinates \(q_{i}(t)\), for \(i=3\) eigenvalues, which are deemed sufficient in the analysis, as follows:  
\[ \begin{array}{c}[M(t)]\big{\{}\tilde{q}(t)\big{\}}+[C]\big{\{}\dot{q}(t)\big{\}} +[K]\big{\{}q(t)\big{\}}=\{0\}\\ [M(t)]=\left[\begin{array}{ccc}1+2(R+kt)&-2(R+kt)&2(R+kt)\\ -2(R+kt)&1+2(R+kt)&-2(R+kt)\\ 2(R+kt)&-2(R+kt)&1+2(R+kt)\end{array}\right],\\ [C]=\left[\begin{array}{ccc}2Q&-2Q&2Q\\ -2Q&2Q&-2Q\end{array}\right],\ K=\left[\begin{array}{ccc}\omega_{1}^{2}&0&0 \\ 0&\omega_{2}^{2}&0\\ 0&0&\omega_{3}^{2}\end{array}\right]\end{array}  \tag{6}\]  
In the above, \([M(t)]\), \([C]\), and \([K]\) are the mass matrix, a damping-type matrix and the stiffness matrix, respectively, written in a normalized form. Two mass ratios are defined here as \(R=m_{0}/m_{P}\) (dimensionless) and \(Q=\mu/m_{P}\) (a normalized mass rate). Furthermore, \(\omega_{i}\) [rad/s] are the eigenfrequencies of the stand-alone pylon. In order to recover a closed-form solution, we proceed to recast the above second-order, matrix differential equation as a first-order differential equation of double size as  
\[ \begin{array}{c}[A(t)]\big{\{}\dot{y}(t)\big{\}}+[B(t)]\big{\{}y(t)\big{\}}= \{0\},\ \big{\{}y(t)\big{\}}^{T}=\big{\lfloor}\big{\{}\dot{q}(t)\big{\}},\ \big{\{}q(t)\big{\}}\big{\rfloor},\\ [A]=\left[\begin{array}{ccc}[0]&[M(t)]\\ [M(t)]&[C]\end{array}\right],[B]=\left[\begin{array}{ccc}-[M(t)]&[0]\\ [0]&[K]\end{array}\right]\end{array}  \tag{7}\]  
The more compact form for the above matrix equation is  
\[\big{\{}\dot{y}(t)\big{\}}=-\Big{[}\Lambda^{-1}(t)\Big{]}[B(t)]\big{\{}y(t) \big{\}}=-[D(t)]\big{\{}y(t)\big{\}} \tag{8}\]  
At this point, we opted for a time-stepping solution of Equation (8) by defining a total time interval of interest as \(T=N\Delta t\) [s], with \(N\) the total number of time steps \(\Delta t\). Thus, \(T\) is the time it takes for the mass ratio \(R\) to commence from an initial value and reach a final value. During a given time instant \(t_{n}=t_{n-1}+\Delta t\), the system matrix \([D(t_{n})]\) is assumed to remain constant. Once the number of eigenvalue/eigenfunction pairs has been decided upon (in our case, \(i=3\)), the eigenvalue problem is solved starting at time \(t=0\) with \([D(0)]=-\big{[}A^{-1}(0)\big{]}[B(0)]\). This is continued for all time steps \(t_{n}\) with system matrix \([D(t_{n})]\), and initial conditions equal to the final conditions for the displacement and velocity of the immediately previous time step \(t_{n-1}\). In more detail, since system matrices \([A(t)]\), \([B(t)]\) are symmetric, a solution in the form \(\{y(t)\}=\{\dot{y}\}\)\(\exp(\lambda t)\), where the carat indicates the amplitude of the kinematic variables, namely of the displacement and velocity. Upon substitution in Equation (8), the eigenvalue problem assumes its standard form as follows:  
\[([A(t)]\lambda+[B(t)])\{\dot{y}\}=\{0\} \tag{9}\]  
Solution yields the complex-valued eigenvalues \(\lambda_{i},\ i=1,2,3\). Specifically, the condition \(|[A]\lambda+[B]|=0\) yields  
\[ \begin{array}{c}\left[\begin{matrix}-[M]&\lambda[M]\\ \lambda[M]&\lambda[C]+[K]\end{matrix}\right]\begin{pmatrix}\lambda\{\dot{y} \}\\ \{\dot{y}\}\end{pmatrix}=\left[\begin{matrix}0\\ 0\end{matrix}\right]\Rightarrow\begin{matrix}-[M]\ \lambda\{\dot{y}\}+\lambda[M]\{\dot{y}\}=0\\ \big{(}\lambda^{2}[M]+\lambda[C]+[K]\big{)}\{\dot{y}\}=0\end{array}  \tag{10}\]  
Since the first of the above equations is merely an identity, we impose on the second one the determinant equal to zero as \(\big{|}\lambda^{2}[M(t)]+\lambda[C]+[K]\big{|}=0\). This is the equation that can now be solved for every time increment from the onset of motion at \(t=0\). The time-dependent eigenvalue code is depicted in Table 2 and all computations were carried out using the Python [23] software.