Abstract
A mechanical system with n degrees of freedom subjected to the action of positional forces and a small smooth control is considered. It is assumed that in the absence of control, the system may have a family of single-frequency oscillations. A universal control—a nonlinear force that implements and simultaneously stabilizes a cycle in the system—is found. An illustrative example is given. In the previous paper [5], the universal control was designed for a two-dimensional manifold of the system.
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Appendices
Appendix
1.1 Lyapunov Transformation for System of Variational Equations
A cycle of the controlled mechanical system satisfies variational equations of the form
For μ = 0 the system (A.1) is a reversible linear periodic system described by
with two stationary sets of the form
Here the plus (minus) sign indicates the matrices and vectors containing the even (odd, respectively) 2π-periodic functions.
The adjoint system for (A.2),
where T means transposition, is also reversible and admits of two stationary sets:
The solution (ξ(t), η(t)) of the system (A.3) is used to write the first integral
of the system (A.2). By the way, this result shows that the function ψ(t) in the amplitude Eq. (2) is odd.
The system (A.2) is reduced to a system with constant coefficients using an appropriate Lyapunov transformation. Recall that the system (A.2) contains n solutions symmetrical with respect to Mξ, as well as the same number of solutions symmetrical with respect to Mη. Choose these 2n solutions to obtain 2n integrals f for reducing the reversible system (A.2) to a system with constant coefficients. Following [8], select the transformation
with the 2π-periodic vector functions \({p}_{s}^{\pm }(t)\) and \({q}_{s}^{\pm }(t)\); the transformed system contains the stationary sets
The system (A.1) is transformed simultaneously.
The reference SPM has characteristics with zero real parts. Therefore, consider the following cases: (a) a pair of zero characteristics in a Jordan cell; (b) a pair ± iω, ω > 0, of pure imaginary characteristics; (c) a pair of simple zero characteristics. Write the integrals (A.4) and the corresponding reduced equations in the three cases mentioned.
(a) A pair of zero characteristics in a Jordan cell. This pair corresponds to the first integrals
Find the derivatives \({\dot{f}}_{1}\) and \({\dot{f}}_{2}\) along the trajectories of the system (A.1):
Denoting \({x}_{* }={g}_{* }^{+}\) and \({y}_{* }={g}_{* }^{-}\), for μ = 0 obtain \({\dot{y}}_{* }={\dot{f}}_{1}=0\), \({\dot{x}}_{* }={y}_{* }-{\dot{f}}_{2}\), and \({\dot{f}}_{2}=0\). For μ ≠ 0, from (A.1) it follows that
(b) A pair ± iω of pure imaginary characteristics. In this case, the first integrals have the complex representation
Calculate the total derivatives of the functions f± along the trajectories of the system (A.1):
Next, find
Then the variables \({x}_{\omega }={g}_{\omega }^{+}\) and \({y}_{\omega }=i{g}_{\omega }^{-}\) satisfy the equations
Finally, transition to the real variables yields two equations for the system (7).
(c) A pair of simple zero characteristics. Here the first integrals are given by
The corresponding equations of the system (A.1) take the form
Thus, the groups of variables (x*, y*), (xω, yω) and (x+, y−) can be adopted for reducing the system (A.1) to a convenient form for further calculation of characteristics. For this purpose, apply the transformation (A.5) in which u = δq, \(v=\delta \dot{q}\), the vector x(y) consists of the vectors x*, xω, x+ (y*, yω, y+, respectively), and the functions \(\delta \,R(\delta \,q,\delta \dot{q})\) and the vectors δq and \(\delta \dot{q}\) are replaced by the vectors x and y using the inverse of (A.5).
Funding
This work was supported by the Russian Foundation for Basic Research, project no. 19-01-00146.
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Tkhai, V. Stabilizing the Oscillations of a Controlled Mechanical System with n Degrees of Freedom. Autom Remote Control 81, 1637–1646 (2020). https://doi.org/10.1134/S0005117920090040
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DOI: https://doi.org/10.1134/S0005117920090040