Homoclinic, subharmonic, and superharmonic bifurcations for a pendulum with periodically varying length

Abstract

Dynamic behavior of a weightless rod with a point mass sliding along the rod axis according to periodic law is studied. This is the simplest model of child’s swing. Melnikov’s analysis is carried out to find bifurcations of homoclinic, subharmonic oscillatory, and subharmonic rotational orbits. For the analysis of superharmonic rotational orbits, the averaging method is used and stability of obtained approximate solution is checked. The analytical results are compared with numerical simulation results.

Appendices

Appendices

Melnikov function for homoclinic orbit

$$\begin{aligned} M^{\pm }&= 8\varepsilon \omega ^2\int _{-\infty }^{\infty }\frac{\sin (\tau )\,\mathrm{\,d}\tau }{\cosh ^2\!\left( \omega \left( \tau -\tau _0\right) \right) }\nonumber \\&-\,4\beta \omega ^3\int _{-\infty }^{\infty }\frac{\mathrm{\,d}\tau }{\cosh ^2\!\left( \omega \left( \tau -\tau _0\right) \right) }\nonumber \\&+\, 4\varepsilon \omega ^3\int _{-\infty }^{\infty }\frac{\cos \!\left( \tau \right) \sinh \!\left( \omega \left( \tau -\tau _0\right) \right) }{\cosh ^3\! \left( \omega \left( \tau -\tau _0\right) \right) }\mathrm{\,d}\tau , \end{aligned}$$

(51)

we denote $M^{\pm } = 8\varepsilon \omega ^2 I_1 - 4\beta \omega ^3 I_2 + 4\varepsilon \omega ^3 I_3$, where

$$\begin{aligned} I_{1}&= \int _{-\infty }^{\infty }\frac{\sin (\tau ) \mathrm{\,d}\tau }{\cosh ^2\!\left( \omega \left( \tau -\tau _0\right) \right) } = \frac{1}{\omega }\!\int _{-\infty }^{\infty }\frac{\sin \!\left( \tau _0+\eta /\omega \right) \mathrm{\,d}\eta }{\cosh ^2\!\left( \eta \right) } \nonumber \\&= \frac{\sin \!\left( \tau _0\right) }{\omega }\!\int _{-\infty }^{\infty }\!\frac{\cos \!\left( \eta /\omega \right) \!\mathrm{\,d}\eta }{\cosh ^2\!\left( \eta \right) }\nonumber \\&+ \frac{\cos \!\left( \tau _0\right) }{\omega }\!\int _{-\infty }^{\infty }\!\frac{\sin \!\left( \eta /\omega \right) \!\mathrm{\,d}\eta }{\cosh ^2\!\left( \eta \right) }. \end{aligned}$$

(52)

The integral $\int _{-\infty }^{\infty }\frac{\sin \!\left( \eta /\omega \right) \mathrm{\,d}\eta }{\cosh ^2\!\left( \eta \right) }$ is zero because its integrand is an odd function, while the other integral has even integrand and can be calculated as follows: $\int _{-\infty }^{\infty }\frac{\cos (t/\omega )}{\cosh ^2 t}\mathrm{\,d}t = \frac{\pi }{\omega \sinh (\pi /2\omega )}$; hence, the first term has the expression

$$\begin{aligned} I_{1}&= \frac{\pi \sin \left( \tau _0\right) }{\omega ^2\sinh (\pi /2\omega )}. \end{aligned}$$

(53)

The second integral can be calculated as follows:

$$\begin{aligned} I_2&= \int _{-\infty }^{\infty }\frac{\mathrm{\,d}\tau }{\cosh ^2\!\left( \omega \left( \tau -\tau _0\right) \right) } = \frac{1}{\omega }\int _{-\infty }^{\infty }\frac{\mathrm{\,d}s}{\cosh ^2\!\left( s\right) } = \frac{2}{\omega }, \nonumber \\ \end{aligned}$$

(54)

while the integral $I_3$ can be converted to $I_1$ via integration by parts using the relation $\frac{\sinh (s)\mathrm{\,d}s}{\cosh ^3(s)} = -\frac{1}{2}\mathrm{\,d}\frac{1}{\cosh ^2(s)}$ as

$$\begin{aligned} I_3&= \int _{-\infty }^{\infty }\frac{\cos (\tau )\,\sinh \left( \omega \left( \tau -\tau _0\right) \right) }{\cosh ^3\!\left( \omega \left( \tau -\tau _0\right) \right) }\mathrm{\,d}\tau ,\nonumber \\&= \frac{1}{\omega }\int _{-\infty }^{\infty }\frac{\cos (\tau _0 + \eta /\omega )\,\sinh \left( \eta \right) }{\cosh ^3\!\left( \eta \right) }\mathrm{\,d}\eta \nonumber \\&= -\frac{1}{2\omega ^2}\left. \frac{\cos (\tau _0 + \eta /\omega )}{\cosh ^2\!\left( \eta \right) }\right| _{-\infty }^{\infty }\nonumber \\&- \frac{1}{2\omega ^2}\int _{-\infty }^{\infty }\frac{\sin (\tau _0 + \eta /\omega )}{\cosh ^2\!\left( \eta \right) }\mathrm{\,d}\eta \nonumber \\&= - \frac{I_1}{2\omega }. \end{aligned}$$

(55)

Thus, $M^{\pm } = 8\varepsilon \omega ^2 I_1 - 4\beta \omega ^3 I_2 + 4\varepsilon \omega ^3 I_3 = \frac{6\pi \varepsilon \sin \left( \tau _0\right) }{\sinh (\pi /2\omega )} - 8\beta \omega ^2$.

Melnikov function for subharmonic oscillations

$$\begin{aligned} M^{p/q}&= 4\omega ^2 k^2 \int _{0}^{2\pi q}\left( 2\varepsilon \sin (\tau )-\beta \omega \right) \mathrm{\,cn}^2\nonumber \\&\times \left( \omega \left( \tau -\tau _0\right) ,k\right) \mathrm{\,d}\tau \nonumber \\&+\,4\varepsilon \omega ^3 k^2 \int _{0}^{2\pi q}\cos (\tau )\mathrm{\,sn}\!\left( \omega \left( \tau -\tau _0\right) ,k\right) \nonumber \\&\times \mathrm{\,dn}\!\left( \omega \left( \tau -\tau _0\right) ,k\right) \mathrm{\,cn}\!\left( \omega \left( \tau -\tau _0\right) ,k\right) \mathrm{\,d}\tau ,\nonumber \\ \end{aligned}$$

(56)

so we denote $M^{p/q} = 8\varepsilon \omega ^2 k^2 I_1 - 4\beta \omega ^3 k^2 I_2 + 4\varepsilon \omega ^3 k^2 I_3$,

$$\begin{aligned} I_1&= \int _{0}^{2\pi q}\sin (\tau )\mathrm{\,cn}^2\!\left( \omega \left( \tau -\tau _0\right) ,k\right) \mathrm{\,d}\tau \nonumber \\&= -\left. \cos (\tau )\mathrm{\,cn}^2\!\left( \omega \left( \tau -\tau _0\right) ,k\right) \right| _{0}^{2\pi q} - 2\omega I_3\nonumber \\&= - 2\omega I_3, \end{aligned}$$

(57)

$$\begin{aligned} I_2&= \int _{0}^{2\pi q}\mathrm{\,cn}^2\!\left( \omega \left( \tau -\tau _0\right) ,k\right) \mathrm{\,d}\tau , \end{aligned}$$

(58)

$$\begin{aligned} I_3&= \int _{0}^{2\pi q}\cos (\tau )\mathrm{\,cn}\!\left( \omega \left( \tau -\tau _0\right) ,k\right) \mathrm{\,sn}\!\left( \omega \left( \tau -\tau _0\right) ,k\right) \nonumber \\&\times \mathrm{\,dn}\!\left( \omega \left( \tau -\tau _0\right) ,k\right) \mathrm{\,d}\tau , \end{aligned}$$

(59)

where we use the formula $\frac{\mathrm{\,d}\mathrm{\,cn}(u)}{\mathrm{\,d}u} = - \mathrm{\,sn}(u)\,\mathrm{\,dn}(u)$. Thus, we have $M^{q/p} = -12\varepsilon \omega ^{3} k^2 I_3 - 4\beta \omega ^3 k^2 I_2$.

$$\begin{aligned} I_1&= \int _{0}^{2\pi q}\sin (\tau )\mathrm{\,cn}^2\!\left( \omega \left( \tau -\tau _0\right) ,k\right) \mathrm{\,d}\tau \nonumber \\&= \frac{1}{\omega }\int _{0}^{2\pi q \omega }\sin (\tau _0 + s/\omega )\mathrm{\,cn}^2\!\left( s,k\right) \mathrm{\,d}s\nonumber \\&= \frac{\sin (\tau _0)}{\omega }\int _{0}^{2\pi q \omega }\cos (s/\omega )\mathrm{\,cn}^2\!\left( s,k\right) \mathrm{\,d}s. \end{aligned}$$

(60)

The integral $I_3$ vanishes except for $p=1$ and even $q$. In this case we have from (312.02) in [22]

$$\begin{aligned}&I_2 = \frac{4\omega }{k^2}\left( E(k)-(k'^2)K(k)\right) , \\&I_3 = -\frac{\pi }{k^2\omega }\frac{\sin \tau _0}{\sinh (K(k'))}, \end{aligned}$$

where $k'^2=1-k^2$. So we have (24).

Melnikov function for subharmonic rotations

$$\begin{aligned} M^{q/r}&= 4\omega ^2 k^2 \int _{0}^{2\pi q}\left( 2\varepsilon \sin (\tau )-\beta \omega \right) \mathrm{\,dn}^2\nonumber \\&\times \left( \omega k \left( \tau -\tau _0\right) ,\frac{1}{k}\right) \mathrm{\,d}\tau \nonumber \\&+\,4\varepsilon \omega ^3 k \int _{0}^{2\pi q}\cos (\tau )\mathrm{\,sn}\!\left( \omega k\left( \tau -\tau _0\right) ,\frac{1}{k}\right) \nonumber \\&\times \mathrm{\,dn}\!\left( \omega k\left( \tau -\tau _0\right) ,\frac{1}{k}\right) \mathrm{\,cn}\!\left( \omega k\left( \tau -\tau _0\right) ,\frac{1}{k}\right) \mathrm{\,d}\tau ,\nonumber \\ \end{aligned}$$

(61)

so we denote $M^{q/r} = 8\varepsilon \omega ^2 k^2 I_1 - 4\beta \omega ^3 k^2 I_2 + 4\varepsilon \omega ^3 k I_3$,

$$\begin{aligned} I_1&= \int _{0}^{2\pi q}\sin (\tau )\mathrm{\,dn}^2\!\left( \omega k\left( \tau -\tau _0\right) ,\frac{1}{k}\right) \mathrm{\,d}\tau = - \frac{2\omega }{k} I_3\nonumber \\&-\left. \cos (\tau )\,\mathrm{\,dn}^2\left( \omega k\left( \tau -\tau _0\right) ,\frac{1}{k}\right) \right| _{0}^{2\pi q} = - \frac{2\omega }{k} I_3, \end{aligned}$$

(62)

$$\begin{aligned} I_2&= \int _{0}^{2\pi q}\mathrm{\,dn}^2\!\left( \omega k\left( \tau -\tau _0\right) ,\frac{1}{k}\right) \mathrm{\,d}\tau , \end{aligned}$$

(63)

$$\begin{aligned} I_3&= \int _{0}^{2\pi q}\cos \!\left( \tau \right) \mathrm{\,sn}\!\left( \omega k\left( \tau -\tau _0\right) , \frac{1}{k}\right) \nonumber \\&\times \mathrm{\,dn}\!\left( \omega k\left( \tau -\tau _0\right) ,\frac{1}{k}\right) \mathrm{\,cn}\! \left( \omega k\left( \tau -\tau _0\right) \!,\frac{1}{k}\right) \!\mathrm{\,d}\tau ,\nonumber \\ \end{aligned}$$

(64)

where we use the formula $\frac{\mathrm{\,d}\mathrm{\,dn}(u,\frac{1}{k})}{\mathrm{\,d}u} = - \frac{\mathrm{\,sn}(u,\frac{1}{k})\,\mathrm{\,cn}(u,\frac{1}{k})}{k^2}$. Thus, we have $M^{q/r} = -12\varepsilon \omega ^3 k^2 I_3 - 4\beta \omega ^3 k^2 I_2$.

The integral $I_3$ vanishes except for $r=1$. In this case we have

$$\begin{aligned} I_2 = \frac{2}{\omega k}\, E\!\left( \frac{1}{k}\right) , \quad I_3 = -\frac{\pi }{k^2\omega }\frac{\sin \tau _0}{\sinh (K')}, \end{aligned}$$

where $K' = K\!\left( \sqrt{1-1/k^2}\right)$. So we have (30)