Unidirectional energy transport in the symmetric system of non-linearly coupled oscillators and oscillatory chains

Abstract

In the present paper, we study the fundamental mechanism of unidirectional energy transport in the symmetric, weakly dissipative system of two coupled nonlinear oscillators and weakly coupled oscillatory chains. We demonstrate that under particular choice of system parameters the model under consideration allows the irreversible transfer of energy from the initially excited oscillator to the initially resting one. In the second part of the paper, we implement the mechanism for control of spatially localized nonlinear waves (discrete breathers) in weakly dissipative, coupled oscillatory chains. This mechanism is implemented on the symmetric system of two weakly dissipative, non-linearly coupled oscillatory chains. Using the regular multi-scale asymptotic analysis, we derive the slow flow system. Further applying the method of collective coordinates on the slow flow system, we were able to describe analytically the mechanism of unidirectional inter-chain transport of static breathers. In general, the analytical method developed in the paper allows one to predict the special regions in the parametric space corresponding to the formation of the aforementioned regimes of irreversible energy transfer in the system of non-linearly coupled oscillators and oscillatory chains.

Appendix

Projection of the resonant nonlinear terms for the resonant conditions brought in (8).

Complexification and averaging are obtained via the following integral expression:

$$\begin{aligned} \left\langle f \right\rangle= & {} \frac{1}{2\pi }\int _0^{2\pi } {f\left( {x\left( {\varphi e^{it},\varphi ^{*}e^{-it}} \right) } \right) e^{-it}\mathrm{d}t} \nonumber \\\equiv & {} \frac{i\gamma }{2}\varphi \left| \varphi \right| ^{a},\quad \gamma ,a>0. \end{aligned}$$

(50)

The pure imaginary constant $\frac{i\gamma }{2}$ was added due to the Hamiltonian structure of the conservative force. 1 / 2 was added due to the averaging of the linear term, and $\,\gamma >0$ because the force is of the restoring type. We need to calculate the parameter $\gamma$. The complex representation of the displacement is $x=\frac{-i}{2}\left( {\varphi e^{it}-\varphi ^{*}e^{-it}} \right) \,$ where $\varphi =\varphi \left( {\tau _{1} } \right) =\left| \varphi \right| e^{i\lambda }$ is the zero order of the asymptotic approximation with the amplitude $A\left( {\tau _{1} } \right) =\left| \varphi \right|$ and the phase $angle\left( \varphi \right) =\lambda \left( {\tau _{1} } \right)$. Both amplitude and phase vary on the slow timescale $\tau _{1} =\varepsilon t$. Substituting the complex form in (50) leads to the following expression of the force,

$$\begin{aligned}&f=-x\left| x \right| ^{a}=-\left( {\frac{-i}{2}} \right) \frac{1}{2^{a}}\left( {\varphi e^{it}-\varphi ^{*}e^{-it}} \right) \nonumber \\&\qquad \left| {\varphi e^{it}-\varphi ^{*}e^{-it}} \right| ^{a}\nonumber \\&\quad =\frac{i}{2}\left( {\varphi e^{it}-\varphi ^{*}e^{-it}} \right) \left| \varphi \right| ^{a}\left| {\sin \left( {\lambda +t} \right) } \right| ^{a} \end{aligned}$$

(51)

or

$$\begin{aligned} f=-\left| \varphi \right| ^{a+1}\left| {\sin \left( {\lambda +t} \right) } \right| ^{a}\sin \left( {\lambda +t} \right) , \end{aligned}$$

(52)

where the non-smooth expression was simplified as follows:

$$\begin{aligned} \frac{1}{2^{a}}\left| {\varphi e^{it}-\varphi ^{*}e^{-it}} \right| ^{a}= & {} \frac{1}{2^{a}}\left| \varphi \right| ^{a}\left| {e^{it}e^{i\lambda }-e^{-it}e^{-i\lambda }} \right| ^{a}\nonumber \\= & {} \left| \varphi \right| ^{a}\left| {\sin \left( {\lambda +t} \right) } \right| ^{a} \end{aligned}$$

(53)

and the smooth term can be transformed in the similar way,

$$\begin{aligned} \frac{i}{2}\left( {\varphi e^{it}-\varphi ^{*}e^{-it}} \right) =-\left| \varphi \right| \sin \left( {\lambda +t} \right) . \end{aligned}$$

(54)

By direct substitution of (52) or (54) into the integral yields the following expression,

$$\begin{aligned} \left\langle f \right\rangle= & {} \frac{1}{2\pi }\frac{i}{2}\left| \varphi \right| ^{a}\int _0^{2\pi } {\left( {\varphi e^{it}-\varphi ^{*}e^{-it}} \right) } e^{-it}\nonumber \\&\left| {\sin \left( {\lambda +t} \right) } \right| ^{a}\mathrm{d}t; \nonumber \\ \left\langle f \right\rangle= & {} \frac{-1}{2\pi }\left| \varphi \right| ^{a+1}\int _0^{2\pi } {\sin \left( {\lambda +t} \right) } e^{-it}\nonumber \\&\left| {\sin \left( {\lambda +t} \right) } \right| ^{a}\mathrm{d}t. \end{aligned}$$

(55)

Thus, we need to evaluate the integral $I=\int _0^{2\pi } {\sin \left( {\lambda + t} \right) }$$\left| {\sin \left( {\lambda +t} \right) } \right| ^{a}e^{-it}\mathrm{d}t$. Transforming

$$\begin{aligned}&u=\lambda +t;\nonumber \\&-t=-u+\lambda ; \nonumber \\&\mathrm{d}t=\mathrm{d}u. \end{aligned}$$

(56)

The integral yields

$$\begin{aligned} I=e^{i\lambda } \int _0^{2\pi }\sin (u)|\sin (u)|^{a} e^{-iu} \mathrm{d}u. \end{aligned}$$

(57)

For better understanding, we can make the symmetric boundaries as well by transforming again $u=w+\pi , w=u-\pi , \mathrm{d}u=\mathrm{d}w$,

$$\begin{aligned} I=e^{i\lambda } \int _{-\pi }^{\pi }\sin (w)|\sin (w)|^{a} e^{-iw} \mathrm{d}w. \end{aligned}$$

(58)

The last integral is simply

$$\begin{aligned}&I=-ie^{i\lambda } \int _{-\pi }^{\pi }\sin ^2 (w)|\sin (w)|^{a} \nonumber \\&\mathrm{d}w=-2ie^{i\lambda }\int _0^{\pi } |\sin (w)|^{a+2}\nonumber \\&\mathrm{d}w=-4ie^{i\lambda } W(a+2)\nonumber \\&W(x)=\int _0^{\frac{\pi }{2}} \cos ^x (\xi ) \mathrm{d}\xi \,{\hbox {or}}\, \nonumber \\&W(x)=\int _0^{\frac{\pi }{2}}\sin ^{x} (\xi )\mathrm{d}\xi , \end{aligned}$$

(59)

where W stands for the Wallis integral [51]. Thus, the averaged nonlinear term reads,

$$\begin{aligned} \langle f\rangle= & {} \frac{-1}{2\pi } |\varphi | \times (-4ie^{i\lambda } W (a+2))\nonumber \\= & {} \frac{i}{2}\varphi |\varphi |^a \times \left( \frac{4}{\pi }W(a+2)\right) =\frac{i\gamma }{2}\varphi |\varphi |^a\nonumber \\= & {} \frac{4}{\pi }W(a+2)=\frac{4}{\pi }\times \frac{\sqrt{\pi }}{2} \frac{\varGamma \left( \frac{(a+2)}{2}+1\right) }{\varGamma \left( \frac{a+2}{2}+1\right) }\nonumber \\= & {} \frac{2}{\pi }\frac{\frac{a+1}{2}}{\frac{a}{2}+1} \frac{\varGamma \left( \frac{a+1}{2}\right) }{\varGamma \left( \frac{a}{2}+1\right) }. \end{aligned}$$

(60)