Huygens synchronization of three clocks equidistant from each other

Abstract

This paper investigates the synchronization of three identical oscillators, or clocks, suspended from a common rigid support. We consider scenarios where each clock interacts with the other two, achieving synchronization through small impacts exchanged between oscillator pairs. The fundamental outcome of our study reveals that the ultimate synchronized state maintains a phase difference of $\frac{2\pi }{3}$ between successive clocks, either clockwise or counter-clockwise. Furthermore, these locked states exhibit an attracting set, the closure of which encompasses the entire initial conditions space. Our analytical approach involves constructing a nonlinear discrete dynamical system in dimension two. These findings hold significance for sets of three weakly coupled periodic oscillators engaged in mutual symmetric impact periodic interaction, irrespective of the specific oscillator models employed. Lastly, we explore the amplitude of oscillations at the final locked state in the context of two and three interacting Andronov pendulum clocks. Our analysis reveals a precise small change in the amplitude of the locked-state oscillations, as quantified in this paper.

Appendix

1.1 The steps in the construction of the model

Step 1 first impact. Interactions of $O_{1}$ on $O_{2}$ and of $O_{1}$ on $O_{3}$, at $t=0$.

When the system in position A attains phase 0 $({\text {mod}}2\pi )$ it receives a sudden supply of energy, for short “a kick”, from its escape mechanism, this kick propagates in the common support of the three clocks and reaches the other two clocks.

Now, the phase difference between $O_{3}$ and $O_{1}$ is corrected by the perturbative value P:

$$\begin{aligned} \left( CA\right) _{I}=\left( CA\right) _{0}+P\left( \left( CA\right) _{0}\right) ={\psi }_{3}^{0}+P\left( {\psi }_{3}^{0}\right) . \end{aligned}$$

The phase difference between $O_{1}$ and $O_{3}$ is

$$\begin{aligned} \left( AC\right) _{I}= & {} \left( AC\right) _{0}+P\left( \left( AC\right) _{0}\right) \\= & {} -{\psi }_{3}^{0}+P\left( -{\psi }_{3}^{0}\right) =-(CA)_{I}, \end{aligned}$$

since P must be an odd function of the mutual phase difference.

The phase difference between $O_{2}$ and $O_{1}$ is

$$\begin{aligned} \left( BA\right) _{I}=\left( BA\right) _{0}+P\left( \left( BA\right) _{0}\right) ={\psi }_{2}^{0}+P\left( {\psi }_{2}^{0}\right) , \end{aligned}$$

and the symmetric phase difference between $O_{1}$ and $O_{2}$ is

$$\begin{aligned} \left( AB\right) _{I}= & {} \left( AB\right) _{0}+P\left( \left( AB\right) _{0}\right) \\= & {} -{\psi }_{2}^{0}+P\left( -{\psi }_{2}^{0}\right) )=-\left( BA\right) _{I}. \end{aligned}$$

The phase difference between $O_{3}$ and $O_{2}$ depends on $\left( CA\right) _{I}$ and $\left( BA\right) _{I}$, and it is

$$\begin{aligned} \left( CB\right) _{I}= & {} \left( CA\right) _{I}-\left( BA\right) _{I}\\= & {} {\psi }_{3}^{0}-{\psi }_{2}^{0}+P({\psi }_{3}^{0})-P({\psi }_{2}^{0})=-(CA)_{I}, \end{aligned}$$

Step 2 first natural time shift. The next clock to arrive at $2\pi ^{-}$, from working hypothesis 3.2 (2), is the clock $O_{3}$ at vertex C. The situation right before $O_{3}$ receives its kick of energy is when the phase of this clock is $2\pi ^{-}$.

At this point we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \psi _{3}^{2} &{} \!\!={2\pi }^{-}\\ \psi _{1}^{2} &{} \!\!=2\pi -(CA)_{I}=2\pi +(AC)_{I}=2\pi -\left( {\psi }_{3}^{0}+P({\psi }_{3}^{0})\right) \\ \psi _{2}^{2} &{} \!\!{=}2\pi {-}\left( CB\right) _{I}{=}2\pi {+}\left( BC\right) _{I}{=}2\pi {+}{\psi }_{2}^{0}{-}{\psi }_{3}^{0}{+}P({\psi }_{2}^{0}){-}P({\psi }_{3}^{0}). \end{array}\right. } \end{aligned}$$

Step 3 second impact. Clock $O_{3}$ receives its internal kick, at the position $2\pi$.

Now, we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \psi _{3}^{3} &{} ={2\pi }\\ \psi _{1}^{3} &{} =\psi _{1}^{2}+P(\psi _{1}^{2})\\ &{} =2\pi -\left( {\psi }_{3}^{0}+P({\psi }_{3}^{0})\right) +P\left( 2\pi -\left( {\psi }_{3}^{0}+P({\psi }_{3}^{0})\right) \right) \\ &{} =2\pi -\left( {\psi }_{3}^{0}+P({\psi }_{3}^{0})\right) -P\left( {\psi }_{3}^{0}+P({\psi }_{3}^{0})\right) \\ &{} \simeq 2\pi -{\psi }_{3}^{0}-2P\left( {\psi }_{3}^{0}\right) \\ \psi _{2}^{3} &{} =\psi _{2}^{2}+P(\psi _{2}^{2})\\ &{} =2\pi +{\psi }_{2}^{0}-{\psi }_{3}^{0}+P({\psi }_{2}^{0})-P({\psi }_{3}^{0})\\ &{} +P(2\pi +{\psi }_{2}^{0}-{\psi }_{3}^{0}+P({\psi }_{2}^{0})-P({\psi }_{3}^{0}))\\ &{} =2\pi +{\psi }_{2}^{0}-{\psi }_{3}^{0}+P({\psi }_{2}^{0})-P({\psi }_{3}^{0})\\ &{} +P({\psi }_{2}^{0}-{\psi }_{3}^{0}+P({\psi }_{2}^{0})-P({\psi }_{3}^{0}))\\ &{} \simeq 2\pi +{\psi }_{2}^{0}-{\psi }_{3}^{0}+P\left( {\psi }_{2}^{0}\right) -P\left( {\psi }_{3}^{0}\right) +P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) \\ &{} \end{array}\right. } \end{aligned}$$

Step 4 second natural time shift. The next clock to arrive at $2\pi ^{-}$, from working hypothesis 3.2 (2), is the clock $O_{2}$ at vertex B. The situation right before $O_{2}$ receives its kick of energy is when the phase of this clock is $2\pi ^{-}$.

Then we have

$$\begin{aligned} {\left\{ \begin{array}{ll} \psi _{2}^{4} &{} =2\pi ^{-}\\ \psi _{1}^{4} &{} =\psi _{1}^{3}+2\pi -\psi _{2}^{3}\\ &{} \simeq 2\pi -{\psi }_{3}^{0}-2P\left( {\psi }_{3}^{0}\right) +2\pi \\ &{} -\left( 2\pi +{\psi }_{2}^{0}-{\psi }_{3}^{0}+P\left( {\psi }_{2}^{0}\right) -P\left( {\psi }_{3}^{0}\right) +P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) \right) \\ &{} =2\pi -{\psi }_{2}^{0}-P\left( {\psi }_{2}^{0}\right) -P\left( {\psi }_{3}^{0}\right) -P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) \\ \psi _{3}^{4} &{} =\psi _{3}^{3}+2\pi -\psi _{2}^{3}\\ &{} \simeq 2\pi +{2\pi }-\left( 2\pi +{\psi }_{2}^{0}-{\psi }_{3}^{0}+P\left( {\psi }_{2}^{0}\right) \right. \\ &{}\left. -P\left( {\psi }_{3}^{0}\right) +P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) \right) \\ &{} \simeq 2\pi -{\psi }_{2}^{0}+{\psi }_{3}^{0}-P\left( {\psi }_{2}^{0}\right) +P\left( {\psi }_{3}^{0}\right) -P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) . \end{array}\right. } \end{aligned}$$

Step 5 third impact. Clock $O_{2}$ receives its internal energy kick. It reaches the position $2\pi$.

Then we have

$$\begin{aligned}{\left\{ \begin{array}{ll} \psi _{2}^{5} &{} ={2\pi }\\ \psi _{3}^{5} &{} =\psi _{3}^{4}+P(\psi _{3}^{4})\\ &{} \simeq 2\pi -{\psi }_{2}^{0}+{\psi }_{3}^{0}-P\left( {\psi }_{2}^{0}\right) +P\left( {\psi }_{3}^{0}\right) -P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) \\ &{} +P(2\pi -{\psi }_{2}^{0}+{\psi }_{3}^{0}-P\left( {\psi }_{2}^{0}\right) +P\left( {\psi }_{3}^{0}\right) -P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) )\\ &{} \simeq 2\pi -{\psi }_{2}^{0}+{\psi }_{3}^{0}-P\left( {\psi }_{2}^{0}\right) +P\left( {\psi }_{3}^{0}\right) \\ &{}-P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) -P({\psi }_{2}^{0}-{\psi }_{3}^{0})\\ &{} =2\pi -{\psi }_{2}^{0}+{\psi }_{3}^{0}-P\left( {\psi }_{2}^{0}\right) +P\left( {\psi }_{3}^{0}\right) -2P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) \\ \psi _{1}^{5} &{} =\psi _{1}^{4}+P\left( \psi _{1}^{4}\right) \\ &{} \simeq 2\pi -{\psi }_{2}^{0}-P\left( {\psi }_{2}^{0}\right) -P\left( {\psi }_{3}^{0}\right) -P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) +\\ &{} P\left( 2\pi -{\psi }_{2}^{0}-P\left( {\psi }_{2}^{0}\right) -P\left( {\psi }_{3}^{0}\right) -P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) \right) \\ &{} \simeq 2\pi -{\psi }_{2}^{0}{-}P\left( {\psi }_{2}^{0}\right) {-}P\left( {\psi }_{3}^{0}\right) {-}P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) {-}P({\psi }_{2}^{0})\\ &{} =2\pi -{\psi }_{2}^{0}-2P\left( {\psi }_{2}^{0}\right) -P\left( {\psi }_{3}^{0}\right) -P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) . \end{array}\right. } \end{aligned}$$

Step 6 (the final) third natural time shift. The next clock to arrive at $2\pi ^{-}$, from working hypothesis 3.2 (2), is the clock $O_{1}$ at vertex A. The situation before $O_{1}$ receives its kick of energy is when the phase of this clock is $2\pi ^{-}$, i.e., the cycles are complete.

At this point we are able to describe what happens to the phases after a complete cycle of the reference clock.

We have

$$\begin{aligned} {\left\{ \begin{array}{ll} \psi _{1}^{6} &{} ={2\pi }^{-}\\ \psi _{2}^{6} &{} =\psi _{2}^{5}+2\pi -\psi _{1}^{5}\\ &{} \simeq 2\pi +2\pi -\left( 2\pi -{\psi }_{2}^{0}-2P\left( {\psi }_{2}^{0}\right) -P\left( {\psi }_{3}^{0}\right) \right. \\ &{}\left. -P\left( {\psi }_{2}^{0}-{\psi }_{3} ^{0}\right) \right) \\ &{} =2\pi +{\psi }_{2}^{0}+2P\left( {\psi }_{2}^{0}\right) +P\left( {\psi } _{3}^{0}\right) +P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) ;\\ \psi _{3}^{6} &{} =\psi _{3}^{5}+2\pi -\psi _{1}^{5}\\ &{} \simeq 2\pi -{\psi }_{2}^{0}+{\psi }_{3}^{0}-P\left( {\psi }_{2}^{0}\right) +P\left( {\psi }_{3}^{0}\right) \\ &{}-2P\left( {\psi }_{2}^{0}-{\psi }_{3} ^{0}\right) +2\pi \\ &{} -\left( 2\pi -{\psi }_{2}^{0}-2P\left( {\psi }_{2}^{0}\right) -P\left( {\psi }_{3}^{0}\right) -P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) \right) \\ &{} =2\pi +{\psi }_{3}^{0}+P({\psi }_{2}^{0})+2P({\psi }_{3}^{0})-P({\psi }_{2} ^{0}-{\psi }_{3}^{0}). \end{array}\right. } \end{aligned}$$

Now, we compute the phase differences after the first cycle of $O_{1}$.

We have

$$\begin{aligned} (BA)_{I}&=-(AB)_{I}=\psi _{2}^{6}-\psi _{1}^{6}\\&\simeq 2\pi +{\psi }_{2}^{0}+2P\left( {\psi }_{2}^{0}\right) \\&+P\left( {\psi }_{3}^{0}\right) +P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) -2\pi \\&={\psi }_{2}^{0}+2P\left( {\psi }_{2}^{0}\right) +P\left( {\psi }_{3}^{0}\right) +P\left( {\psi }_{2}^{0}-{\psi }_{3}^{0}\right) \\&=(BA)_{0}+2P((BA)_{0})\\&+P((CA)_{0})+P((BA)_{0}-(CA)_{0})\\ \end{aligned}$$

and

$$\begin{aligned}&(CA)_{I}\\&=-(AC)_{I}=\psi _{3}^{6}-\psi _{1}^{6}\\&=2\pi +{\psi }_{3}^{0}+P({\psi }_{2}^{0})+2P({\psi }_{3}^{0})-P({\psi }_{2}^{0}-{\psi }_{3}^{0})-2\pi \\&={\psi }_{3}^{0}+P({\psi }_{2}^{0})+2P({\psi }_{3}^{0})-P({\psi }_{2}^{0}-{\psi }_{3}^{0})\\&=({(CA)}_{0})+P({(BA)}_{0})+2P({(CA)}_{0})\\&-P((BA)_{0}-(CA)_{0})\\ \end{aligned}$$

Hence, if we set $x=BA$ and $y=CA$, we obtain the system

$$\begin{aligned} \left\{ \begin{array}{c}x_{1}=x_{0}+2P(x_{0})+P(y_{0})+P(x_{0}-y_{0})\\ y_{1}=x_{0}+P(x_{0})+2P({y}_{0})-P(x_{0}-y_{0}). \end{array} \right. \end{aligned}$$