Nonintegrability of forced nonlinear oscillators

Abstract

In recent papers by the authors (Motonaga and Yagasaki, Arch. Ration. Mech. Anal. 247:44 (2023), and Yagasaki, J. Nonlinear Sci. 32:43 (2022)), two different techniques which allow us to prove the real-analytic or complex-meromorphic nonintegrability of forced nonlinear oscillators having the form of time-periodic perturbations of single-degree-of-freedom Hamiltonian systems were provided. Here the concept of nonintegrability in the Bogoyavlenskij sense is adopted and the first integrals and commutative vector fields are also required to depend real-analytically or complex-meromorphically on the small parameter. In this paper we review the theories and continue to demonstrate their usefulness. In particular, we consider the periodically forced damped pendulum, which provides an especially important differential equation not only in dynamical systems and mechanics but also in other fields such as mechanical and electrical engineering and robotics, and prove its nonintegrability in the above meaning.

Appendix A: Derivation of (4.9) and (4.10)

In this appendix we use the method of residues and compute the integrals (4.7) and (4.8) to obtain (4.9) and (4.10). A similar calculation is found in [34].

We begin with the first term in (4.7). Letting $s=1/{{\,\textrm{sn}\,}}t$, we have

$$\begin{aligned} \int {{\,\textrm{cn}\,}}^2 t\, {\textrm{d}} t=-\int \frac{1}{s^2} \sqrt{\frac{1-s^2}{k^2-s^2}}{\textrm{d}} s \end{aligned}$$

(A.1)

from the basic properties of the Jacobi elliptic functions

$$\begin{aligned} \frac{{\textrm{d}}}{{\textrm{d}} t} {{\,\textrm{sn}\,}}t={{\,\textrm{cn}\,}}t {{\,\textrm{dn}\,}}t, \quad {{\,\textrm{cn}\,}}^2 t=1-{{\,\textrm{sn}\,}}^2 t, \quad {{\,\textrm{dn}\,}}^2 t=1-k^2{{\,\textrm{sn}\,}}^2 t. \end{aligned}$$

Obviously, the integrand in the right hand side of (A.1) has a pole of order 2 and

$$\begin{aligned} \frac{{\textrm{d}}}{{\textrm{d}} s}\sqrt{\frac{1-s^2}{k^2-s^2}}=0 \end{aligned}$$

at $s=0$. Noting that $s=0$ when $t=iK(k')$, we obtain

$$\begin{aligned} \int _{\bar{\gamma }_\theta }{{\,\textrm{cn}\,}}^2 t\, {\textrm{d}} t =\int _{|s|=\rho } \frac{1}{s^2} \sqrt{\frac{1-s^2}{k^2-s^2}}{\textrm{d}} s=0 \end{aligned}$$

(A.2)

by the method of residues, where $\bar{\gamma }_\theta =\{t\in \mathbb {C}\mid t+\tfrac{1}{2}\hat{T}^k\in \gamma _\theta \}$ and $\rho >0$ is sufficiently small.

We turn to the second term in (4.7). Since

$$\begin{aligned} \cos \omega t=\cosh (\omega K(k')) +O(t-iK(k')) \end{aligned}$$

(A.3)

and

$$\begin{aligned} {{\,\textrm{cn}\,}}t=-\frac{i}{k(t-iK(k'))}+O(1) \end{aligned}$$

near $t=iK(k')$, we have

$$\begin{aligned} \int _{\bar{\gamma }_\theta } {{\,\textrm{cn}\,}}t\cos \omega t\,{\textrm{d}} t=\frac{2\pi }{k} \cosh \omega K(k'). \end{aligned}$$

Similarly, since

$$\begin{aligned} \sin \omega t&=i\sinh (\omega K(k'))+O(t-iK(k')) \end{aligned}$$

(A.4)

near $t=iK(k')$, we have

$$\begin{aligned} \int _{\bar{\gamma }_\theta } {{\,\textrm{cn}\,}}t\sin \omega t\,{\textrm{d}} t=\frac{2\pi i}{k} \sinh \omega K(k'). \end{aligned}$$

Thus, we obtain (4.9).

We next compute (4.8). We easily see that the first term vanishes by (A.2) since

$$\begin{aligned} {{\,\textrm{dn}\,}}^2 t =k'^2-k^2{{\,\textrm{cn}\,}}^2 t. \end{aligned}$$

On the other hand, since

$$\begin{aligned} {{\,\textrm{dn}\,}}t=-\frac{i}{t-iK(k')}+O(1), \end{aligned}$$

we have

$$\begin{aligned} \int _{\bar{\gamma }_\theta } {{\,\textrm{dn}\,}}\left( \frac{t}{k}\right) \cos \omega t\,{\textrm{d}} t =2\pi k \cosh (\omega kK(k')) \end{aligned}$$

by (A.3), where $\bar{\gamma }_\theta =\{t\in \mathbb {C}\mid t+\tfrac{1}{2}\tilde{T}^k\in \tilde{\gamma }_\theta \}$. Similarly, by (A.4) we have

$$\begin{aligned} \int _{\bar{\gamma }_\theta } {{\,\textrm{dn}\,}}\left( \frac{t}{k}\right) \sin \omega t\,{\textrm{d}} t =2\pi ik \sinh (\omega kK(k')). \end{aligned}$$

Thus, we obtain (4.10).