Obstructions to Integrability of Nearly Integrable Dynamical Systems Near Regular Level Sets

Abstract

We study the existence of real-analytic first integrals and real-analytic integrability for perturbations of integrable systems in the sense of Bogoyavlenskij, including non-Hamiltonian ones. In particular, we assume that there exists a family of periodic orbits on a regular level set of the first integrals having a connected and compact component and give sufficient conditions for nonexistence of the same number of real-analytic first integrals in the perturbed systems as the unperturbed ones and for their real-analytic nonintegrability near the level set such that the first integrals and commutative vector fields depend analytically on the small parameter. We compare our results with the classical results of Poincaré and Kozlov for systems written in action and angle coordinates and discuss their relationships with the subharmonic and homoclinic Melnikov methods for periodic perturbations of single-degree-of-freedom Hamiltonian systems. In particular, the latter discussion reveals that the perturbed systems can be real-analytically nonintgrable even if there exists no transverse homoclinic orbit to a periodic orbit. We illustrate our theory with three examples containing the periodically forced Duffing oscillator.

Appendix A. Proof of Proposition 3.6

In this Appendix, we prove Proposition 3.6, which is an extension of Ziglin’s lemma [4, 10, 33]. Our approach is to reduce the functional independence of first integrals near the level set $F^{-1}(c)$ to that near a point on it using action and angle variables. We begin with the following lemma:

Lemma A.1

Let $\Omega$ be an open subset of ${\mathbb {R}}^k$ and let $\chi _j:\Omega \rightarrow {\mathbb {R}}$, $j=1,\ldots ,m$, be analytic, where $k,m\in {\mathbb {N}}$. Let $\chi (x)=(\chi _1(x),\ldots ,\chi _m(x))$. If ${{\,\textrm{rank}\,}}d\chi$ is constant on $\Omega$ and less than m, then for any $x\in \Omega$ there exists a neighborhood V of x on which $\chi _1, \ldots ,\chi _m$ are analytically dependent, i.e., there exist an open set $\Omega ' \subset {\mathbb {R}}^m$ and a non-constant analytic map $\zeta :\Omega '\rightarrow {\mathbb {R}}$ such that $\chi (V)\subset \Omega '$ and $\zeta (\chi (y))=0$ for any $y\in V$.

Proof

Using Theorem 1.3.14 of [22] and an argument in the proof of Theorem 1.4.15 of [22], we can immediately obtain the desired result as follows. The theorem says that there exist a neighborhood V (resp. $V'$) of x (resp. of $\chi (x)$), a cube Q (resp. $Q'$) in ${\mathbb {R}}^k$ (resp. in ${\mathbb {R}}^m$) and analytic isomorphisms $u:Q\rightarrow V$ and $u':V'\rightarrow Q'$ such that the composite map $u'\circ \chi \circ u$ has the form $(x_1, \ldots , x_k)\rightarrow (x_1, \ldots , x_{m'}, 0, \ldots , 0)$, where $x_j$ is the jth element of x for $j=1,\ldots ,k$ and $m'={{\,\textrm{rank}\,}}d\chi <m$. Here a cube in ${\mathbb {R}}^k$ is an open set of the form

$$\begin{aligned} \{x\mid |x_j-a_j|<r_j,j=1,\ldots ,k\} \end{aligned}$$

for some $a_j\in {\mathbb {R}}$ and $r_j>0$, $j=1,\ldots ,k$. Letting $u'=(u'_1, \ldots , u'_m)$ and $\zeta =u'_m$, we have $\zeta (\chi (y))=0$ for every $y\in V$. $\square$

Let $f_\varepsilon : {\mathscr {M}}\rightarrow {\mathbb {R}}$ be an analytic function such that it depends on $\varepsilon$ analytically. We expand it near $\varepsilon =0$ as $f_\varepsilon (x)=\sum _{j=0}^\infty f^j(x)\varepsilon ^j$, where $f^j(x)$, $j\in {\mathbb {Z}}_0:={\mathbb {N}}\cup \{0\}$, are analytic functions on ${\mathscr {M}}$. Define the order function $\sigma (f_\varepsilon )$ by

$$\begin{aligned} \sigma (f_\varepsilon ):=\min \{j\in {\mathbb {Z}}_0 \mid f^j(x)\not \equiv 0\} \end{aligned}$$

if $f_\varepsilon \not \equiv 0$ and $\sigma (0):=+\infty$, as in [6].

Lemma A.2

Suppose that $f_\varepsilon (x)$ is a nonconstant analytic first integral of (1.4) depending analytically on $\varepsilon$ near $\varepsilon =0$. Then there exists an analytic first integral $\tilde{f}_\varepsilon (x)=\tilde{f}^0(x)+O(\varepsilon )$ depending analytically on $\varepsilon$ near $\varepsilon =0$ such that $\tilde{f}^0(x)$ is not constant.

Proof

Since $f_\varepsilon$ is not constant, $\sigma (df_\varepsilon )$ takes a finite value. Let $k=\sigma (df_\varepsilon )$ and $f_\varepsilon (x)=\sum _{j=0}^\infty \varepsilon ^j f^j(x)$. Define

$$\begin{aligned} \tilde{f}_\varepsilon (x):=\frac{1}{\varepsilon ^k}\left( f_\varepsilon (x) -\sum _{j=0}^{k-1}\varepsilon ^j f^j(x)\right) . \end{aligned}$$

Then $\tilde{f}_\varepsilon (x)=f^k(x)+O(\varepsilon )$ and $\tilde{f}^0(x)=f^k(x)$ is not constant. Moreover, $\tilde{f}_\varepsilon (x)$ is a first integral of (1.4) since $X_\varepsilon (f_\varepsilon )=0$ and $\sum _{j=0}^{k-1}\varepsilon ^j f^j$ is constant. $\square$

We are now in a position to give a proof of Proposition 3.6.

Proof of Proposition 3.6

For $k=1$ the statement of the proposition holds by Lemma A.2. Let ${k}>1$ and suppose that it is true up to $k-1$. Let $G_1^\varepsilon (x),\ldots ,G_k^\varepsilon (x)$ be analytic first integrals of (1.4) in a neighborhood of $F^{-1}(c)$ near $\varepsilon =0$ such that they are functionally independent for $\varepsilon \ne 0$ and depend analytically on $\varepsilon$. Without loss of generality, we assume that $G_1^0(x),\ldots , G_{k-1}^0(x)$ are functionally independent near $F^{-1}(c)$. Letting $G^\varepsilon (x)=(G_1^\varepsilon (x),\ldots ,G_k^\varepsilon (x))$, we see that $G^0(\varphi (I,\theta ))$ depends only on I, where $\varphi$ denotes the analytic diffeomorphism in Proposition 2.1(ii). Let $\tilde{G}_j(I)=G_j^0(\varphi (I,\theta ))$ for $j=1,\ldots ,k$. Note that, if $d\tilde{G}_1(I), \ldots , d\tilde{G}_k(I)$ are linearly independent at $I=I_0\in U$, then so are $dG_1(x), \ldots , dG_k(x)$ on $\varphi (\{I_0\}\times {\mathbb {T}}^q)\subset {\mathcal {U}}$.

Assume that $\tilde{G}_1(I),\ldots ,\tilde{G}_{k-1}(I), \tilde{G}_k(I)$ are functionally dependent in an open set $U'\subset U$. So $\Omega :=\{p\in U'\mid {{\,\textrm{rank}\,}}d_p\tilde{G}=k-1\}$ contains a dense open set in $U'$ since $d\tilde{G}_1(I), \ldots , d\tilde{G}_{k-1}(I)$ are functionally independent on U. By Lemma A.1, there exist an open set $\Omega '\subset {\mathbb {R}}^{k}$ and a nonzero analytic function $\zeta :\Omega '\rightarrow {\mathbb {R}}$ such that $\tilde{G}(V) =(\tilde{G}_1(V),\ldots , \tilde{G}_k(V))\subset \Omega '$ and

$$\begin{aligned} \zeta (\tilde{G}_1(I),\ldots ,\tilde{G}_k(I))=0 \end{aligned}$$

in a neighborhood V of $p\in \Omega$. Moreover, there is a positive integer s such that $({\partial ^s \zeta }/{\partial y_k^s})(\tilde{G}(I))\ne 0$, since if not, then $\zeta (\tilde{G}_1(I),\ldots ,\tilde{G}_{k-1}(I),y_k)$ depends on $y_k$ near $\tilde{G}(V)$ and consequently $\tilde{G}_1(I),\ldots ,\tilde{G}_{k-1}(I)$ are functionally dependent. Let s be the smallest one of such integers and let $\tilde{\zeta }(y)=({\partial ^{s-1} \zeta }/{\partial y_k^{s-1}})(y)$. Then $\tilde{\zeta }$ satisfies

$$\begin{aligned} \tilde{\zeta }(\tilde{G}_1(I),\ldots ,\tilde{G}_k(I))=0, \end{aligned}$$

and $(\partial \tilde{\zeta }/\partial y_k)(\tilde{G}(I))\ne 0$ on V. Hence,

$$\begin{aligned} \tilde{\zeta }(G^0(x))=0, \end{aligned}$$

(A1)

and $(\partial \tilde{\zeta }/\partial y_k)(G^0(x))\ne 0$ on $\varphi (V\times {\mathbb {T}}^q)$.

Let $\hat{G}_k^\varepsilon (x)=\tilde{\zeta }(G^\varepsilon (x))/\varepsilon$. By (A1) $\hat{G}_k^\varepsilon$ is an analytic first integral depending analytically on $\varepsilon$. We have

$$\begin{aligned} d\hat{G}_k^\varepsilon =\varepsilon ^{-1}d(\tilde{\zeta }(G^\varepsilon (x))) =\varepsilon ^{-1}\sum _{j=1}^k \frac{\partial \tilde{\zeta }}{\partial y_j} (G^\varepsilon (x)) dG_j^\varepsilon \end{aligned}$$

and

$$\begin{aligned} N(\hat{G}^\varepsilon ) :=dG_1^\varepsilon \wedge \ldots \wedge dG_{k-1}^\varepsilon \wedge d\hat{G}_k^\varepsilon =\varepsilon ^{-1} \frac{\partial \tilde{\zeta }}{\partial y_k}(G^\varepsilon (x)) dG_1^\varepsilon \wedge \ldots \wedge d{G}_k^\varepsilon . \end{aligned}$$

Since $(\partial \tilde{\zeta }/\partial y_k)(G^0(x))\ne 0$ on $\varphi (V\times {\mathbb {T}}^q)$, we have $\sigma ((\partial \tilde{\zeta }/\partial y_k)(G^0(x)))=0$, so that

$$\begin{aligned} \sigma (N(\hat{G}^\varepsilon ))=\sigma (\varepsilon ^{-1}N({G}^\varepsilon )) +\sigma \left( \frac{\partial \tilde{\zeta }}{\partial y_k} (G^\varepsilon (x))\right) =\sigma (N({G}^\varepsilon ))-1. \end{aligned}$$

Repeating this procedure till $\sigma (N(\hat{G}^\varepsilon ))=0$, we obtain

$$\begin{aligned} dG_1^0\wedge \ldots \wedge dG_{k-1}^0\wedge d\hat{G}_k^0\ne 0, \end{aligned}$$

which gives the desired result. $\square$