An Attracting Cycle in a Coupled Mechanical System with Phase Shifts in Subsystem Oscillations

Abstract

This paper considers the set of reversible mechanical systems with single-period oscillations and individual phase shifts in them. The problem of aggregating a coupled system with an attracting cycle is solved. The approach developed below is to choose a leader (control) system that acts on the other (follower) systems through one-way coupling control: in an aggregated system, there are no links between follower systems. Universal coupling controls are used. Particular attention is paid to conservative systems. Possible scenarios for the operation of the aggregated system are presented.

APPENDIX

The adjoint solution can be calculated using Lemma 1.

Consider a smooth reversible mechanical system of the second order:

$$\dot {u} = U(u,\,\,{v}),\quad {\dot {v}} = V(u,\,\,{v}),\quad U(u,\,\, - {v}) = - U(u,\,\,{v}),\quad V(u,\,\, - {v}) = V(u,\,\,{v}).$$

Let this system admit an SPM described by the functions

$$u = \varphi (t),\quad {v} = \theta (t),\quad \varphi ( - t) = \varphi (t),\quad \theta ( - t) = - \theta (t).$$

The variational equations for the SPM have the form

$$\begin{gathered} \dot {x} = {{a}_{ - }}(t)x + {{a}_{ + }}(t)y, \hfill \\ \dot {y} = {{b}_{ + }}(t)x + {{b}_{ - }}(t)y, \hfill \\ \end{gathered} $$

(A.1)

where a_±(t), b_±(t) denote even (+) and odd (–) periodic functions. They have the solution x = $\dot {\varphi }(t)$, y = $\dot {\theta }(t)$.

Lemma 1. For a given SPM, the solution of the system adjoint to (A.1) is calculated by constructive formulas.

Proof. Let us apply the transformation

$$x = {{\xi }_{ + }}(t)\tilde {x},\quad y = {{\eta }_{ + }}(t)\tilde {y}$$

with even periodic functions ξ₊(t) and η₊(t) with nonzero means. As a result,

$${{\xi }_{ + }}(t)\dot {\tilde {x}} + {{\dot {\xi }}_{ + }}(t)\tilde {x} = {{a}_{ - }}(t){{\xi }_{ + }}(t)\tilde {x} + {{a}_{ + }}(t){{\eta }_{ + }}(t)\tilde {y},$$

$${{\eta }_{ + }}(t)\dot {\tilde {y}} + {{\dot {\eta }}_{ + }}(t)\tilde {y} = {{b}_{ + }}(t){{\xi }_{ + }}(t)\tilde {x} + {{b}_{ - }}(t){{\eta }_{ + }}(t)\tilde {y}.$$

The functions ξ₊(t) and η₊(t) are appropriately chosen to satisfy the equalities

$${{\dot {\xi }}_{ + }} = {{a}_{ - }}(t){{\xi }_{ + }},\quad {{\dot {\eta }}_{ + }} = {{b}_{ - }}(t){{\eta }_{ + }}.$$

Then the transformed system

$$\dot {\tilde {x}} = {{\tilde {a}}_{ + }}(t)\tilde {y},\quad \dot {\tilde {y}} = {{\tilde {b}}_{ + }}(t)\tilde {x}$$

(A.2)

contains no odd functions of t.

The adjoint system of

$${{x}_{1}} = {{\xi }_{{1 + }}}(t){{\tilde {x}}_{1}},\quad {{y}_{1}} = {{\eta }_{{1 + }}}(t){{\tilde {y}}_{1}}$$

is transformed by analogy. We obtain

$${{\dot {\tilde {x}}}_{1}} = - {{\tilde {b}}_{ + }}(t){{\tilde {y}}_{1}},\quad {{\dot {\tilde {y}}}_{1}} = - {{\tilde {a}}_{ + }}(t){{\tilde {x}}_{1}}.$$

(A.3)

In the variables ${{\tilde {x}}_{1}}$ = $- \tilde {y}$ and ${{\tilde {y}}_{1}} = \tilde {x}$, the resulting system (A.3) coincides with (A.2). Hence, its solution is given by ${{\tilde {x}}_{1}}$ = –ξ₊(t)^–1$\dot {\theta }$(t), ${{\tilde {y}}_{1}}$ = η₊(t)^–1$\dot {\varphi }$(t). Therefore, the solution of the adjoint system can be written as

$${{x}_{1}} = - {{\xi }_{{1 + }}}(t){{\xi }_{ + }}{{(t)}^{{ - 1}}}\dot {\theta }(t),\quad {{y}_{1}} = {{\eta }_{{1 + }}}(t){{\eta }_{ + }}{{(t)}^{{ - 1}}}\dot {\varphi }(t).$$

The proof of Lemma 1 is complete.