Energy supply into a semi-infinite $\beta$-Fermi–Pasta–Ulam–Tsingou chain by periodic force loading

Abstract

We deal with dynamics of the $\beta$-Fermi–Pasta–Ulam–Tsingou chain with one free end, subjected to the sudden sinusoidal force. We examine the evolution of the total energy supplied into the chain at large times. In the harmonic case ($\beta =0$), the energy grows in time linearly at the driving frequencies, corresponding to nonzero group velocities. The rate of energy supply is shown to decrease with increasing excitation frequency. Loading with the cut-off frequency, corresponding to zero group velocity, results in the energy growth proportionally to $\sqrt{t}$. Explanation of behavior in time of the energy is proposed by analysis of the obtained approximate closed-form expressions for the particle velocity. In the weak anharmonic case, large-time asymptotic approximation for the total energy is obtained by using the renormalized dispersion relation. The approximation allows one to estimate rate of the energy supply at the driving frequencies, which belong both in the pass-band and in the stop-band of the harmonic chain. Consistency of the asymptotic estimates with the results of numerical simulations is discussed.

Appendices

Appendix A Derivation of Eq. (16)

Here, we derive Eq. for the contribution $v_{n}^\Omega$ in the closed form. Using the following approximations due to $\eta /\omega _e$ is infinitesimal:

$$\begin{aligned} \begin{array}{l} \displaystyle \phi _2\approx \left[ \begin{aligned} \pi , \, \omega < \Omega ; \\ 0 ,\, \omega > \Omega , \\ \end{aligned} \right. , \quad \sqrt{\Omega ^2-\omega ^2\pm 2\text{ i }\eta \Omega }\approx \sqrt{\Omega ^2-\omega ^2}\pm \frac{\text{ i }\eta \Omega }{\sqrt{\Omega ^2-\omega ^2}}, \end{array} \end{aligned}$$

(A1)

rewrite the expression for $v_{n}^\Omega$ in Eq. (15) as follows:

$$\begin{aligned} v_{n}^\Omega \approx -\frac{F_0\Omega \cos (\Omega t)H(t)}{m}\left( \text {I}_{n}+\text {I}_{n+1}+\text {c.c.}\right) ,\quad \text {I}_{n}=\frac{1}{4\pi }\int _{-\pi }^{\pi }\frac{e^{\text {i}n\theta }\text {d}\theta }{\Omega ^2-\omega ^2-\text {i}\eta \Omega }, \end{aligned}$$

(A2)

where $\text {c.c.}$ stands for the complex conjugate terms. We transform the integral in Eq. (A2) to the unit circle integral. Put $z=e^{\text {i}\theta }$. Then

$$\begin{aligned} \text {d}\theta =-\text {i}\frac{\text {d}z}{z}, \qquad 2\cos \theta =z+\frac{1}{z}, \end{aligned}$$

(A3)

and, therefore,

$$\begin{aligned} \text {I}_n=-\frac{\text {i}}{4 \pi }\oint _{\vert z\vert =1}\frac{z^{n}\text {d}z}{\omega _e^2z^2+(\Omega ^2-2\omega _e^2-\text {i}\eta \Omega )z+\omega _e^2 }. \end{aligned}$$

(A4)

Rewrite Eq. (A4) as

$$\begin{aligned} \text {I}_n= & {} -\frac{\text {i}}{4\pi \omega _e^2(z_1-z_2)}\left( \oint _{\vert z\vert =1}\frac{z^n\text {d}z}{z-z_1}-\oint _{\vert z\vert =1}\frac{z^n\text {d}z}{z-z_2}\right) , \end{aligned}$$

(A5)

$$\begin{aligned} z_{1,2}= & {} \frac{2\omega _e^2-\Omega ^2+\text {i}\eta \Omega \pm \text {i}\sqrt{\Omega }\sqrt{\Omega -\text {i}\eta }\sqrt{4\omega _e^2-\Omega ^2+\text {i}\eta \Omega }}{2\omega _e^2}. \end{aligned}$$

(A6)

where $z_1$ and $z_2$ are roots of the denominator of the integrand in Eq. (A4), i.e., poles. Rewrite the latter in the limit $\eta \rightarrow 0+$:

$$\begin{aligned} z_{1,2}=1-\frac{\Omega ^2}{2\omega _e^2}\pm \text{ i }\left( \frac{\Omega \sqrt{4\omega _e^2-\Omega ^2}}{2\omega _e^2}\pm 0\right) . \end{aligned}$$

(A7)

Consider $0<\Omega <2\omega _e$. In this case, the pole $z_2$ lies inside the unit circle $\vert z\vert <1$ only. Therefore, we take into account the residue at $z_2$ only and the first term in (A5) equals zero. Therefore, we have

$$\begin{aligned} \text {I}_{n}=\frac{\text {i}z_2^n}{2\Omega \sqrt{4\omega _e^2-\Omega ^2}}=\frac{(-1)^ne^{2\text {i}\varphi n}}{-2\text {i}\Omega \sqrt{4\omega _e^2-\Omega ^2}},\quad \varphi =\arccos {\frac{\Omega }{2\omega _e}}. \end{aligned}$$

(A8)

Consider $\Omega >2\omega _e$. In this case, Eqs. for $z_{1,2}$ have the form

$$\begin{aligned} z_{1,2}=1-\frac{\Omega ^2\pm \Omega \sqrt{\Omega ^2-4\omega _e^2}}{2\omega _e^2}. \end{aligned}$$

(A9)

Substituting $z_{1,2}$ to Eq. (A5) with taking into account $\vert z_2 \vert <1$ yields

$$\begin{aligned} \begin{array}{l} \displaystyle \text {I}_n=\frac{z_2^n}{2\Omega \sqrt{\Omega ^2-4\omega _e^2}}=\frac{(-1)^ne^{-\gamma n}}{2\Omega \sqrt{\Omega ^2-4\omega _e^2}},\\ \displaystyle \gamma =-\ln \left( \frac{\Omega ^2-\Omega \sqrt{\Omega ^2-4\omega _e^2}}{2\omega _e^2}-1\right) =2\text {arccosh}\frac{\Omega }{2\omega _e}. \end{array} \end{aligned}$$

(A10)

Consider $\Omega =2\omega _e$. The integral for $\text {I}_n$ and its complex conjugation, $\bar{\text {I}}_n$, determined in (A2), can be analogously transformed into the unit circle integral and then to the form

$$\begin{aligned}{} & {} \begin{array}{l} \displaystyle \text{ I}_n=-\frac{\text{ i }}{4\pi \omega _e^2}\oint _{\vert z\vert =1}\frac{z^n\text{ d }z}{(z+1-0\text{ i})^2}=0, \end{array} \end{aligned}$$

(A11)

$$\begin{aligned}{} & {} \bar{\text{ I }}_n=-\frac{\text{ i }}{4\pi \omega _e^2}\oint _{\vert z\vert =1}\frac{z^{-n}\text{ d }z}{(z+1+0\text{ i})^2}=\frac{(-1)^{(n+1)}n}{2\omega _e^2}. \end{aligned}$$

(A12)

Here, the following identities [75]

$$\begin{aligned} \frac{1}{(z\pm 0\text{ i})^2}=\pm \text{ i }\pi \delta '(z)+\text{ p.v. }\frac{1}{z^2}, \end{aligned}$$

(A13)

where $\text {p.v.}$ stands for the Cauchy principle value, are used. Substitution of Eqs. (A8), (A10), (A11), (A12) to Eq. (A2) with respect to the corresponding values of the driving frequency and with further simplifications yields Eq. (16). Note that the expressions (A8), (A10), (A12) were previously obtained in [54]. However, Eqs. (A8)+$\text {c.c.}$ and (A11) were obtained in the sense of the Cauchy principal value only, while, here, we do this exactly by using the limiting absorption principle.

Appendix B Derivation of the expression (18)

Rewrite Eq. for $v_{n}^\omega$ in Eq. (14) as follows:

$$\begin{aligned} v_{n}^\omega =\frac{2F_0\Omega e^{-\eta t} H(t)}{m\pi }\int _0^{\pi }\frac{(\Omega ^2-\omega (\theta )^2)\cos (\omega (\theta ) t)}{(\Omega ^2-\omega (\theta )^2)^2+4\eta ^2\Omega ^2}\cos \frac{(2n+1)\theta }{2} \cos \frac{\theta }{2} \text {d}\theta . \end{aligned}$$

(B14)

Here, we use the following approximations due to $\eta /\omega _e$ is infinitesimal:

$$\begin{aligned}{} & {} \begin{array}{l} \displaystyle \phi _1\approx \left[ \begin{aligned} 0, \, \omega < \Omega ; \\ \pi ,\, \omega > \Omega , \\ \end{aligned} \right. \\ \end{array} \end{aligned}$$

(B15)

$$\begin{aligned}{} & {} \sqrt{\omega ^2-\eta ^2}\approx \omega ,\quad \sqrt{4\eta ^2(\omega ^2-\eta ^2)+(\Omega ^2-\omega ^2+2\eta ^2)^2}\approx \vert \Omega ^2-\omega ^2\vert . \end{aligned}$$

(B16)

Rewrite the wavenumber integral (B14), changing to the frequency integral:

$$\begin{aligned} \begin{array}{l} \displaystyle v_{n}^\omega =\frac{2F_0\Omega e^{-\eta t} H(t)}{m\omega _e\pi }\int _0^{2\omega _e}\frac{(\Omega ^2-\omega ^2)\cos (\omega (\theta ) t)}{(\Omega ^2-\omega ^2)^2+4\eta ^2\Omega ^2}\cos \left( (2n+1)\arcsin \frac{\omega }{2\omega _e} \right) \text {d}\omega . \end{array}{} \end{aligned}$$

(B17)

Introduce a function, $\text {I}_n^\text {cr}$, which denotes the contribution to Eq. (B14), coming from the vicinity of the point $\Omega =\omega$. We introduce a variable $\epsilon =\Omega -\omega$, which is infinitesimal for infinitesimal value $\eta /\omega _e$. Consider two cases: $0<\Omega <2\omega _e$ and $\Omega =2\omega _e$. In the first case, we transform (B14) using the following asymptotic expansions with respect to $\epsilon /\omega _e$:

$$\begin{aligned}{} & {} \begin{array}{l} \displaystyle \Omega ^2-\omega ^2=2\Omega \epsilon +O\left( \left( \frac{\epsilon }{\omega _e}\right) ^2\right) ,\\ \end{array} \end{aligned}$$

(B18)

$$\begin{aligned}{} & {} \cos \left( (2n+1)\arcsin \frac{\omega }{2\omega _e}\right) =\cos \left( (2n+1)\arcsin \frac{\Omega }{2\omega _e}\right) +O\left( \frac{\epsilon }{\omega _e}\right) . \end{aligned}$$

(B19)

Based on aforesaid, we write equation for $\text {I}^\text {cr}_n$ as the integral along the vicinity of $\epsilon =0$:

$$\begin{aligned} \begin{array}{ll} \text{ I}^\text{ cr}_n&{}{}=\displaystyle \frac{F_0e^{-\eta t}H(t){\mathcal {T}}_{2n+1}\left( \sqrt{1-\frac{\Omega ^2}{4\omega _e^2}}\right) }{m\omega _e \pi }\Bigg [\cos (\Omega t)\int _{\Omega -2\omega _e}^{\Omega }\frac{\epsilon \cos (\epsilon t)\text{ d }\epsilon }{\epsilon ^2+\eta ^2}\\ {} &{}{}\quad +\displaystyle \sin (\Omega t)\int _{\Omega -2\omega _e}^{\Omega }\frac{\epsilon \sin (\epsilon t)\text{ d }\epsilon }{\epsilon ^2+\eta ^2}\Bigg ]. \end{array} \end{aligned}$$

(B20)

Introduce a function $v^\omega _{n,\text {cr}}$ such as

$$\begin{aligned} v^\omega _{n,\text{ cr }}\backsim \text{ I}^\text{ cr}_n \vert _{\Omega t\gg 1,\, \eta \rightarrow 0+}. \end{aligned}$$

(B21)

Then, tending $\epsilon t\rightarrow \infty$ in the integrals in Eq. (B20) and using the following identities, which are true for any $\tilde{\eta }>0$

$$\begin{aligned} \int _{-\infty }^{\infty }\frac{x\sin x}{x^2+\tilde{\eta }^2}\text {d}x=e^{-\tilde{\eta }}\pi ,\qquad \int _{-\infty }^{\infty }\frac{x\cos x}{x^2+\tilde{\eta }^2}\text {d}x=0, \end{aligned}$$

(B22)

we write the expression for $v^\omega _{n,\text {cr}}$ in the form

$$\begin{aligned} v^\omega _{n,\text {cr}}\approx \displaystyle \frac{F_0{\mathcal {T}}_{2n+1}\left( \sqrt{1-\frac{\Omega ^2}{4\omega _e^2}}\right) }{m\omega _e}\sin (\Omega t). \end{aligned}$$

(B23)

Consider the case when $\Omega =2\omega _e$. Putting $\eta =0$ and  $\epsilon =2\omega _e-\omega$, we rewrite the asymptotic expansions (B19) as

$$\begin{aligned}{} & {} \displaystyle \Omega ^2-\omega ^2=4\omega _e\epsilon +O\left( \left( \frac{\epsilon }{\omega _e}\right) ^2\right) , \end{aligned}$$

(B24)

$$\begin{aligned}{} & {} \displaystyle \cos \left( (2n+1)\arcsin \frac{\omega }{2\omega _e}\right) =(-1)^n(2n+1)\sqrt{\frac{\epsilon }{\omega _e}}+O\left( \left( \frac{\epsilon }{\omega _e}\right) ^\frac{3}{2}\right) . \end{aligned}$$

(B25)

Therefore,

$$\begin{aligned} \text{ I}^\text{ cr}_n=\frac{(-1)^nF_0H(t)(2n+1)}{m\pi \sqrt{\omega _e^3}}\Bigg [\cos (2\omega _e t)\int _{0}^{2\omega _e}\frac{\cos (\epsilon t)\text{ d }\epsilon }{\sqrt{\epsilon }}+\sin (2\omega _e t)\int _{0}^{2\omega _e}\frac{\sin (\epsilon t)\text{ d }\epsilon }{\sqrt{\epsilon }}\Bigg ]. \end{aligned}$$

(B26)

Tending $\epsilon t\rightarrow \infty$ in (B26) and using Eq. (B21) and the identities

$$\begin{aligned} \int _0^\infty \frac{\cos x\,\text {d}x}{\sqrt{x}}=\sqrt{\frac{\pi }{2}},\qquad \int _0^\infty \frac{\sin x\text {d}x}{\sqrt{x}}=\sqrt{\frac{\pi }{2}}, \end{aligned}$$

(B27)

we write the expression for $v^\omega _{n,\text {cr}}$ as

$$\begin{aligned} v^\omega _{n,\text{ cr }}\approx \displaystyle \frac{(-1)^nF_0(2n+1)}{m\sqrt{2\pi \omega _e^3 t}} \left( \cos (2\omega _e t)+\sin (2\omega _e t)\right) , \end{aligned}$$

(B28)

which can be transformed to the form shown in Eq. (18).

Appendix C Derivation of the expression (19)

Put $\eta =0$ in Eq. (14) and rewrite Eq. for $v_{n}^\omega$ in as follows:

$$\begin{aligned} \displaystyle v_{n}^\omega= & {} \frac{F_0\Omega H(t)}{m}\left( \text {I}_1-\text {I}_2+\text {I}_3+\text {I}_4\right) , \end{aligned}$$

(C29)

$$\begin{aligned} \text {I}_{1,3}= & {} \frac{1}{\pi }\int _0^{\pi }\frac{\cos (\omega (\theta )t\pm n\theta )}{\Omega ^2-\omega (\theta )^2}\cos ^2\frac{\theta }{2}\text {d}\theta ,\quad \text {I}_{2,4}=\frac{1}{\pi }\int _0^{\pi }\frac{\sin (\omega (\theta )t\pm n\theta )}{\Omega ^2-\omega (\theta )^2}\cos \frac{\theta }{2}\sin \frac{\theta }{2}\text {d}\theta . \end{aligned}$$

(C30)

In order to estimate the large-time asymptotics of the integrals in Eq. (C29) at the moving front, firstly, we transform them to the form with the structure of the Fourier integral:

$$\begin{aligned} \tilde{\text {I}}=\int f(\theta )e^{\text {i}\varphi (\theta ) t}\text {d}\theta . \end{aligned}$$

(C31)

Following [76,77,78], put $n=w\omega _e t$, where w is the dimensionless constant in time speed of the observation point, $0<w<1$. Then, we consider the following integrals:

$$\begin{aligned} \displaystyle \tilde{\text{ I }}_{1\pm }= & {} {} \frac{1}{\pi }\int _0^{\pi } f_{1}(\theta )e^{\text{ i }\varphi _\pm (\theta )\omega _e t}\text{ d }\theta ,\qquad \tilde{\text{ I }}_{2\pm }=\frac{1}{\pi }\int _0^{\pi } f_{2}(\theta )e^{\text{ i }\varphi _\pm (\theta )\omega _e t}\text{ d }\theta , \end{aligned}$$

(C32)

$$\begin{aligned} f_1(\theta )= & {} \frac{\displaystyle \cos ^2\frac{\theta }{2}}{\Omega ^2-\omega (\theta )^2},\qquad f_2(\theta )=\frac{\displaystyle \cos \frac{\theta }{2}\sin \frac{\theta }{2}}{\Omega ^2-\omega (\theta )^2},\qquad \varphi _\pm (\theta )=2\sin \frac{\theta }{2}\pm w\theta . \end{aligned}$$

(C33)

For estimation of large-time asymptotics, we use the stationary phase method [79]. The stationary points, $\theta _s$, corresponding to the integrals in Eq. (C32) satisfy the condition $\displaystyle \frac{\text {d}\varphi _{\pm }}{\text {d}\theta }\Big \vert _{\theta =\theta _s}=0$. As well as the negative stationary points, the stationary point $\theta _s=2(\pi -\arccos {w})$ does not belong to the interval $(0; \pi )$. Therefore,

$$\begin{aligned} \tilde{\text {I}}_{1+}=\tilde{\text {I}}_{2+}=O((\omega _et)^{-\infty }) \end{aligned}$$

(C34)

and thus $\text {I}_1=\text {I}_2=O((\omega _et)^{-\infty })$. Consider $\theta _s=2\arccos w$. Then

$$\begin{aligned} \begin{array}{l} \displaystyle \varphi _{-}(\theta _s)=2\sqrt{1-w^2}-2w\arccos w,\quad \displaystyle \frac{\text {d}^2\varphi _{-}}{\text {d}\theta ^2}\Big \vert _{\theta =\theta _s}=-\frac{1}{2}\sqrt{1-w^2}<0. \end{array} \end{aligned}$$

(C35)

Since the stationary point is not degenerate, we can therefore write the following expression for the principal term of the asymptotics of the integral $\tilde{\text {I}}_{1-}$ [79]:

$$\begin{aligned} \begin{array}{ll} \displaystyle \tilde{\text{ I }}_{1-}&{}{}\backsim \frac{1}{\pi }\sqrt{\frac{2\pi }{\omega _e t\Big \vert \frac{\text{ d}^2\varphi _{-}}{\text{ d }\theta ^2}\big \vert _{\theta =\theta _s}\Big \vert }}f_1(\theta _s)e^{\text{ i }\left( \varphi _{-}(\theta _s)\omega _e t+\frac{\pi }{4}\text{ sgn } \frac{\text{ d}^2\varphi _{-}}{\text{ d }\theta ^2}\Big \vert _{\theta =\theta _s}\right) }\\ {} &{}{}\displaystyle =\frac{2}{\sqrt{\pi \omega _e t\sqrt{1-w^2}}}\frac{w^2}{\Omega ^2-4\omega _e^2(1-w^2)}e^{\text{ i }\left( (2\sqrt{1-w^2}-2w\arccos w)\omega _et-\frac{\pi }{4}\right) }. \end{array} \end{aligned}$$

(C36)

Analogously,

$$\begin{aligned} \displaystyle \tilde{\text{ I }}_{2-}\backsim \frac{2}{\sqrt{\pi \omega _e t\sqrt{1-w^2}}}\frac{w\sqrt{1-w^2}}{\Omega ^2-4\omega _e^2(1-w^2)}e^{\text{ i }\left( (2\sqrt{1-w^2}-2w\arccos w)\omega _et-\frac{\pi }{4}\right) }. \end{aligned}$$

(C37)

The integrals $\text {I}_3$ and $\text {I}_4$ are calculated as

$$\begin{aligned} \text {I}_3=\Re (\tilde{\text {I}}_{1-})H(1-w),\qquad \text {I}_4=\Im (\tilde{\text {I}}_{2-})H(1-w). \end{aligned}$$

(C38)

Substitution of $w=n/(\omega _et)$ to Eqs. (C36), (C37) and (C38) and the final result to Eq. (C29) with further simplifications yields the expressions (19) and (20).

Appendix D Solution of Eq. (51)

We seek solution of Eq. (51) with accuracy up to order of $\beta F_0^2/c^3$. Knowing that $\Omega _\text {cr}>2\omega _e$ but not much more, than $2\omega _e$ we can write expression for $\mu (\Omega _\text {cr})$ as

$$\begin{aligned} \mu (\Omega _\text {cr})=\mu (2\omega _e)+O \left( \frac{\beta F_0^2}{c^3}\right) . \end{aligned}$$

(D39)

Substituting (D39) to (51) with preserving terms of order of $\beta F_0^2/c^3$ yields

$$\begin{aligned} \Omega _\text {cr}\left( 1-\frac{\beta F_0^2}{c^3} \mu (2\omega _e)\right) =2\omega _e, \end{aligned}$$

(D40)

whereas

$$\begin{aligned} \Omega _\text {cr}=\frac{2\omega _e}{1-\frac{\beta F_0^2}{c^3} \mu (2\omega _e)}= 2\omega _e\left( 1+\frac{\beta F_0^2}{c^3}\mu (2\omega _e)+O \left( \frac{\beta ^2 F_0^4}{c^6}\right) \right) . \end{aligned}$$

(D41)