2 Determination of non-integrability for equation (1.5)

In this section, we give a technique for determining whether the system (1.5) is not meromorphically Bogoyavlenskij-integrable such that the first integrals and commutative vector fields also depend meromorphically on $\varepsilon $ near $\varepsilon =0$ .

When $\varepsilon =0$ , equation (1.5) becomes

(2.1) $$ \begin{align} \dot{I}=0,\quad \dot{\theta}=\omega(I). \end{align} $$

We assume the following on the unperturbed system (2.1).

1. (A1) For some $I^\ast \in \mathbb {R}^\ell $ , a resonance of multiplicity $m-1$ ,

   $$ \begin{align*} \dim_{\mathbb{Q}}\langle\omega_1(I^\ast),\ldots,\omega_m(I^\ast)\rangle=1, \end{align*} $$
   occurs with $\omega (I^\ast )\neq 0$ , that is, there exists a constant $\omega ^\ast>0$ such that
   $$ \begin{align*} \frac{\omega(I^\ast)}{\omega^\ast}\in\mathbb{Z}^m\setminus\{0\}, \end{align*} $$
   where $\omega _j(I)$ is the jth element of $\omega (I)$ for $j=1,\ldots ,m$ .

Note that we can replace $\omega ^\ast $ with $\omega ^\ast /n$ for any $n\in \mathbb {N}$ in assumption (A1). We refer to the m-dimensional torus $\mathscr {T}^\ast =\{(I^\ast ,\theta )\mid \theta \in \mathbb {T}^m\}$ as the resonant torus and to periodic orbits $(I,\theta )=(I^\ast ,\omega (I^\ast )t+\theta _0)$ , $\theta _0\in \mathbb {T}^m$ , on $\mathscr {T}^\ast $ as the resonant periodic orbits. Let $T^\ast =2\pi /\omega ^\ast $ . We also make the following assumption.

1. (A2) For some $k\in \mathbb {Z}_{\ge 0}:=\mathbb {N}\cup \{0\}$ and $\theta \in \mathbb {T}^m$ , there exists a closed loop $\gamma _\theta $ in a region including $(0,T^\ast )\subset \mathbb {R}$ in $\mathbb {C}$ such that $\gamma _\theta \cap (i\mathbb {R}\cup (T^\ast +i\mathbb {R}))=\emptyset $ and

   (2.2) $$ \begin{align} \mathscr{I}^k(\theta):=\frac{1}{k!}\mathrm{D}\omega(I^\ast)\int_{\gamma_\theta} \mathrm{D}_\varepsilon^k h(I^\ast,\omega(I^\ast)\tau+\theta;0)\,d\tau \end{align} $$
   is not zero (see Figure 2).
   

   Figure 2 Assumption (A2).

Note that the condition $\gamma _\theta \cap (i\mathbb {R}\cup (T^\ast +i\mathbb {R}))=\emptyset $ is not essential in assumption (A2), since it always holds by replacing $\omega ^\ast $ with $\omega ^\ast /n$ for sufficiently large $n\in \mathbb {N}$ if necessary. We prove the following theorem which guarantees that conditions (A1) and (A2) are sufficient for non-integrability of equation (1.5) in a certain meaning.

Our basic idea of the proof of Theorem 2.1 is similar to that of Morales-Ruiz [Reference Morales-Ruiz, Crespo and Hajto21], who studied time-periodic Hamiltonian perturbations of single-degree-of-freedom Hamiltonian systems and showed a relationship of their non-integrability with a version due to Ziglin [Reference Ziglin44] of the Melnikov method [Reference Melnikov18] when the small parameter $\varepsilon $ is regarded as a state variable. Here the Melnikov method enables us to detect transversal self-intersection of complex separatrices of periodic orbits unlike the standard version [Reference Guckenheimer and Holmes13, Reference Melnikov18, Reference Wiggins39]. More concretely, under some restrictive conditions, he essentially proved that they are meromorphically non-integrable when the small parameter is taken as one of the state variables if the Melnikov functions are not identically zero, based on a generalized version due to Ayoul and Zung [Reference Ayoul and Zung5] of the Morales–Ramis theory [Reference Morales-Ruiz20, Reference Morales-Ruiz and Ramis22]. We also use their generalized versions of the Morales–Ramis theory and its extension, the Morales–Ramis–Simó theory [Reference Morales-Ruiz, Ramis and Simó23], to prove Theorem 2.1. These generalized theories enable us to show the non-integrability of general differential equations in the Bogayavlenskij sense by using differential Galois groups [Reference Crespo and Hajto11, Reference van der Put and Singer37] of their variational or higher-order variational equations along non-constant particular solutions. We extend the idea of Morales-Ruiz [Reference Morales-Ruiz, Crespo and Hajto21] to higher-dimensional non-Hamiltonian systems near periodic orbits.

For the proof of Theorem 2.1, we first consider systems of the general form

(2.3) $$ \begin{align} \dot{x}=f(x;\varepsilon),\quad x\in\mathbb{C}^n, \end{align} $$

where $f:\mathbb {C}^n\times \mathbb {C}\to \mathbb {C}^n$ is meromorphic, and describe direct consequences of the generalized versions due to Ayoul and Zung [Reference Ayoul and Zung5] of the Morales–Ramis theory [Reference Morales-Ruiz20, Reference Morales-Ruiz and Ramis22] when the parameter $\varepsilon $ is regarded as a state variable in equation (2.3) near $\varepsilon =0$ . Let $x=\bar {x}(t)$ be a periodic orbit in the unperturbed system

$$ \begin{align*} \dot{x}=f(x;0). \end{align*} $$

Taking $\varepsilon $ as another state variable, we extend equation (2.3) as

(2.4) $$ \begin{align} \dot{x}=f(x;\varepsilon),\quad \dot{\varepsilon}=0, \end{align} $$

in which $(x,\varepsilon )=(\bar {x}(t),0)$ is a periodic orbit. The variational equation (VE) of equation (2.4) along the periodic solution $(\bar {x}(t),0)$ is given by

(2.5) $$ \begin{align} \dot{y}=\mathrm{D}_x f(\bar{x}(t);0)y+\mathrm{D}_\varepsilon f(\bar{x}(t);0)\unicode{x3bb},\quad \dot{\unicode{x3bb}}=0. \end{align} $$

We regard equation (2.5) as a linear differential equation on a Riemann surface. Applying the version due to Ayoul and Zung [Reference Ayoul and Zung5] of the Morales–Ramis theory [Reference Morales-Ruiz20, Reference Morales-Ruiz and Ramis22] to equation (2.4), we obtain the following result.

See Appendix A for necessary information on the differential Galois theory.

We can obtain a more general result for equation (2.4) as follows. For simplicity, we assume that $n=1$ . The general case can be treated similarly. Letting

$$ \begin{align*} \bar{f}^{(j,l)}=\mathrm{D}_x^j\mathrm{D}_\varepsilon^l f(\bar{x}(t);0), \end{align*} $$

we express the Taylor series of $f(x;\varepsilon )$ about $(x,\varepsilon )=(\bar {x}(t),0)$ as

$$ \begin{align*} f(x;\varepsilon)=\sum_{l=0}^\infty\sum_{j=0}^\infty \frac{1}{j! l!}\bar{f}^{(j,l)}(x-\bar{x}(t))^j\varepsilon^l. \end{align*} $$

Let

$$ \begin{align*} x=\bar{x}(t)+\sum_{j=1}\varepsilon^j y^{(j)}. \end{align*} $$

Using these expressions, we write the kth-order VE of equation (2.4) along the periodic orbit $(x,\varepsilon )=(\bar {x}(t),0)$ as

(2.6) $$ \begin{align} \dot{y}^{(1)}&=\bar{f}^{(1,0)}y^{(1)}+\bar{f}^{(0,1)}\unicode{x3bb}^{(1)},\quad \dot{\unicode{x3bb}}^{(1)}=0, \nonumber\\ \dot{y}^{(2)}&=\bar{f}^{(1,0)}y^{(2)}+\bar{f}^{(0,1)}\unicode{x3bb}^{(2)} \nonumber\\ &\quad +\bar{f}^{(2,0)}(y^{(1)})^2+2\bar{f}^{(1,1)}(y^{(1)},\unicode{x3bb}^{(1)}) +\bar{f}^{(0,2)}(\unicode{x3bb}^{(1)})^2,\quad \dot{\unicode{x3bb}}^{(2)}=0, \nonumber\\ \dot{y}^{(3)}&=\bar{f}^{(1,0)}(y^{(3)})+\bar{f}^{(0,1)}(\unicode{x3bb}^{(3)}) +2\bar{f}^{(2,0)}(y^{(1)},y^{(2)})\nonumber\\ &\quad+2\bar{f}^{(1,1)}(y^{(1)},\unicode{x3bb}^{(2)})+2\bar{f}^{(1,1)}(y^{(1)},\unicode{x3bb}^{(2)}) +2\bar{f}^{(0,2)}(\unicode{x3bb}^{(1)},\unicode{x3bb}^{(2)}) \\ &\quad +\bar{f}^{(3,0)}(y^{(1)})^3+3\bar{f}^{(2,1)}((y^{(1)})^2,\unicode{x3bb}^{(1)})+3\bar{f}^{(1,2)}(y^{(1)},(\unicode{x3bb}^{(1)})^2) \nonumber\\ &\quad +\bar{f}^{(0,3)}(\unicode{x3bb}^{(1)})^3,\quad \dot{\unicode{x3bb}}^{(3)}=0, \nonumber\\ & \vdots \nonumber\\ \dot{y}^{(k)}&=\sum\frac{(j+l)!}{j_1!\cdots j_r! l_1!\cdots l_s!} \nonumber\\ & \quad\times\bar{f}^{(j,l)} ((y^{(i_1)})^{j_1},\ldots,(y^{(i_r)})^{j_r},(\unicode{x3bb}^{(i_1')})^{l_1},\ldots,(\unicode{x3bb}^{(i_s')})^{l_s}),\quad \dot{\unicode{x3bb}}^{(k)}=0, \nonumber \end{align} $$

where such terms as $(y^{(1)})^0,(\unicode{x3bb} ^{(1)})^0=1$ have been substituted and the summation in the last equation has been taken over all integers

$$ \begin{align*} j,l,r,s,i_1,\ldots,i_r,i_1',\ldots,i_s',j_1,\ldots,j_r,l_1,\ldots,l_s \end{align*} $$

such that

$$ \begin{align*} & 1\le j+l\le k,\quad i_1<i_2<\cdots<i_r,\quad i_1'<i_2'<\cdots<i_s',\\ & \sum_{r'=1}^r j_{r'}=j,\quad \sum_{s'=1}^s l_{s'}=l,\quad \sum_{r'=1}^r j_{r'}i_{r'}+\sum_{s'=1}^s l_{s'}i_{s'}=k. \end{align*} $$

See [Reference Morales-Ruiz, Ramis and Simó23] for the details on derivation of higher-order VEs in a general setting. Substituting $y^{(j)}=0$ , $j=1,\ldots ,k-1$ , and $\unicode{x3bb} ^{(l)}=0$ , $l=2,\ldots ,k$ , into equation (2.6), we obtain

$$ \begin{align*} \dot{\unicode{x3bb}}^{(1)}=0,\quad \dot{y}^{(k)}=\bar{f}^{(1,0)}(y^{(k)})+\bar{f}^{(0,k)}(\unicode{x3bb}^{(1)})^k, \end{align*} $$

which is equivalent to

(2.7) $$ \begin{align} \dot{y}=\mathrm{D}_x f(\bar{x}(t);0)y+\frac{1}{k!}\mathrm{D}_\varepsilon^k f(\bar{x}(t);0)\unicode{x3bb},\quad \dot{\unicode{x3bb}}=0 \end{align} $$

with $y=y^{(k)}$ and $\unicode{x3bb} =k!(\unicode{x3bb} ^{(1)})^k$ . Equation (2.7) is derived for $n>1$ similarly. We regard equation (2.7) as a linear differential equation on a Riemann surface, again. Such a reduction of higher-order VEs was used for planar systems in [Reference Acosta-Humánez, Lázaro, Morales-Ruiz and Pantazi1, Reference Acosta-Humánez and Yagasaki2]. We call equation (2.7) the kth-order reduced variational equation (RVE) of equation (2.4) around the periodic orbit $(x,\varepsilon )=(\bar {x}(t),0)$ . Using the version due to Ayoul and Zung [Reference Ayoul and Zung5] of the Morales–Ramis–Simó theory [Reference Morales-Ruiz, Ramis and Simó23], we obtain the following result.

We return to the system (1.5) and regard $\varepsilon $ as a state variable to rewrite it as

(2.8) $$ \begin{align} \dot{I}=\varepsilon h(I,\theta;\varepsilon),\quad \dot{\theta}=\omega(I)+\varepsilon g(I,\theta;\varepsilon),\quad \dot{\varepsilon}=0, \end{align} $$

like equation (2.4). We extend the domain of the independent variable t to a region including $\mathbb {R}$ in $\mathbb {C}$ , as stated in Theorem 2.1. The $(k+1)$ th-order RVE of equation (2.8) along the periodic orbit $(I,\theta ,\varepsilon )=(I^\ast ,\omega (I^\ast )t+\theta _0,0)$ is given by

(2.9) $$ \begin{align} & \dot{\xi}=h^k(I^\ast,\omega^\ast t+\theta_0;0)\unicode{x3bb}, \nonumber\\& \dot{\eta}=\mathrm{D}\omega(I^\ast)\xi+g^k(I^\ast,\omega^\ast t+\theta_0;0)\unicode{x3bb}, \quad (\xi,\eta,\chi)\in\mathbb{C}^\ell\times\mathbb{C}^m\times\mathbb{C},\\& \dot{\unicode{x3bb}}=0,\nonumber \end{align} $$

where

$$ \begin{align*} h^k(I,\theta)=\frac{1}{k!}\mathrm{D}_\varepsilon^k h(I,\theta;0),\quad g^k(I,\theta)=\frac{1}{k!}\mathrm{D}_\varepsilon^k g(I,\theta;0). \end{align*} $$

As a Riemann surface, we take any region $\Gamma $ in $\mathbb {C}/T^\ast \mathbb {Z}$ such that the closed loop $\gamma _\theta $ in assumption (A2), as well as $\mathbb {R}/T^\ast \mathbb {Z}$ , is contained in $\Gamma $ , as in Theorem 2.1 (see Figure 3). Let $\mathbb {K}_\theta \neq \mathbb {C}$ be a differential field that consists of $T^\ast $ -periodic functions and contains the elements of $h^k(I^\ast ,\omega (I^\ast )t+\theta )$ and $g^k(I^\ast ,\omega (I^\ast )t+\theta )$ with $t\in \Gamma $ . We regard the $(k+1)$ th-order RVE (2.9) as a linear differential equation over $\mathbb {K}_\theta $ on the Riemann surface $\Gamma $ . We obtain a fundamental matrix of equation (2.9) as

$$ \begin{align*} \Phi^k(t;\theta_0)= \begin{pmatrix} \mathrm{id}_\ell & 0 & \Xi^k(t;\theta_0)\\ \mathrm{D}\omega(I^\ast)t & \mathrm{id}_m & \Psi^k(t;\theta_0)\\ 0 & 0 & 1 \end{pmatrix}, \end{align*} $$

where $\mathrm{id}_\ell$ is the $\ell\times\ell$ identity matrix and

$$ \begin{align*} \Xi^k(t;\theta) &=\int_0^t h^k(I^\ast,\omega(I^\ast)\tau+\theta)\,d\tau,\\ \Psi^k(t;\theta) &=\int_0^t(\mathrm{D}\omega(I^\ast)\Xi(\tau;\theta) +g^k(I^\ast,\omega(I^\ast)\tau+\theta))\,d\tau. \end{align*} $$

Figure 3 Riemann surface $\Gamma $ . The monodromy matrix $M_\gamma $ is computed along the loop $\gamma $ .

Let $\mathscr {G}_\theta $ be the differential Galois group of equation (2.9) and let $\sigma \in \mathscr {G}_\theta $ . Then

$$ \begin{align*} \frac{\mathrm{d}}{\mathrm{d} t}\sigma(\Xi^k(t;\theta))=\sigma\bigg(\frac{\mathrm{d}}{\mathrm{d} t}\Xi^k(t;\theta)\bigg) =h^k(I^\ast,\omega(I^\ast)t+\theta), \end{align*} $$

so that

(2.10) $$ \begin{align} \sigma(\Xi^k(t;\theta))=\int_0^t h^k(I^\ast,\omega(I^\ast)\tau+\theta;0)\,d\tau+C =\Xi^k(t;\theta)+C, \end{align} $$

where C is a constant $\ell $ -dimensional vector depending on $\sigma $ . If $\Xi ^k(t;\theta )\in \mathbb {K}_\theta $ , then $C=0$ for any $\sigma \in \mathscr {G}_\theta $ . Similarly, we have

$$ \begin{align*} \sigma(\mathrm{D}\omega(I^\ast)t)=\mathrm{D}\omega(I^\ast)t+C', \end{align*} $$

where $C'$ is a constant $m\times \ell $ matrix depending on $\sigma $ . If $\mathrm {D}\omega (I^\ast )\neq 0$ , then $C'\neq 0$ for some $\sigma \in \mathscr {G}_\theta $ since $\mathrm {D}\omega (I^\ast )t\notin \mathbb {K}_\theta $ . However,

$$ \begin{align*} \frac{\mathrm{d}}{\mathrm{d} t}\sigma(\Psi^k(t;\theta)) &=\sigma\bigg(\frac{\mathrm{d}}{\mathrm{d} t}\Psi^k(t;\theta)\bigg) =\sigma(\mathrm{D}\omega(I^\ast)\Xi^k(t;\theta) +g^k(I^\ast,\omega(I^\ast)t+\theta))\\ &=\mathrm{D}\omega(I^\ast)\Xi^k(t;\theta)+g^k(I^\ast,\omega(I^\ast)t+\theta) +\mathrm{D}\omega(I^\ast)C. \end{align*} $$

Hence,

$$ \begin{align*} \sigma(\Psi^k(t;\theta))=\Psi^k(t;\theta)+\mathrm{D}\omega(I^\ast)Ct+C", \end{align*} $$

where $C"$ is a constant m-dimensional vector depending on $\sigma $ . If $\Xi ^k(t;\theta ),\Psi ^k(t;\theta )\in \mathbb {K}_\theta $ , then $C"=0$ for any $\sigma \in \mathscr {G}_\theta $ . Thus, we see that

$$ \begin{align*} \mathscr{G}_\theta\subset\tilde{\mathscr{G}}:=\{ M(C_1,C_2,C_3)\mid C_1\in\mathbb{C}^\ell,C_2\in\mathbb{C}^m,C_3\in\mathbb{C}^{m\times\ell}\}, \end{align*} $$

where

$$ \begin{align*} M(C_1,C_2,C_3)= \begin{pmatrix} \mathrm{id}_\ell & 0 & C_1\\ C_3 & \mathrm{id}_m & C_2\\ 0 & 0 & 1 \end{pmatrix}. \end{align*} $$

Proof of Theorem 2.1

Assume that the hypotheses of the theorem hold. We fix $\theta \in \mathbb {T}^m$ such that the integral (2.2) is not zero. We continue the fundamental matrix $\Phi ^k(t;\theta )$ analytically along the loop $\gamma =\gamma _\theta $ to obtain the monodromy matrix as

(2.11) $$ \begin{align} M_{\gamma}= \begin{pmatrix} \mathrm{id}_\ell & 0 & \hat{C}_1\\ 0 & \mathrm{id}_m & \hat{C}_2\\ 0 & 0 & 1 \end{pmatrix}, \end{align} $$

where

$$ \begin{align*} & \hat{C}_1=\int_{\gamma_\theta}h^k(I^\ast,\omega(I^\ast)t+\theta;0)\,d\tau,\\ & \hat{C}_2=\int_{\gamma_\theta}(\mathrm{D}\omega(I^\ast)\Xi^k(\tau;\theta) +g^k(I^\ast,\omega(I^\ast)\tau+\theta))\,d\tau. \end{align*} $$

See Appendix B for basic information on monodromy matrices. In particular, we have $M_\gamma \in \mathscr {G}_\theta $ . Note that $\mathrm {D}\omega (I^\ast )\hat {C}_1\neq 0$ by assumption (A2).

Let $\bar {\gamma }=\{T^\ast s\mid s\in [0,1]\}$ , which is also a closed loop on the Riemann surface $\Gamma $ (see Figure 3). We continue $\Phi ^k(t;\theta )$ analytically along the loop $\bar {\gamma }$ to obtain the monodromy matrix as

$$ \begin{align*} M_{\bar{\gamma}}= \begin{pmatrix} \mathrm{id}_\ell & 0 & \Xi^k(T^\ast;\theta)\\ \mathrm{D}\omega(I^\ast)T^\ast & \mathrm{id}_m & \Psi^k(T^\ast;\theta)\\ 0 & 0 & 1 \end{pmatrix}. \end{align*} $$

Let $\bar {C}_1=\Xi ^k(T^\ast ;\theta )$ , $\bar {C}_2=\Psi ^k(T^\ast ;\theta )$ and $\bar {C}_3=\mathrm {D}\omega (I^\ast )T^\ast $ . We see that $M_{\bar {\gamma }}=M(\bar {C}_1,\bar {C}_2, \bar {C}_3)\in \mathscr {G}_\theta $ and $\bar {C}_3\hat {C}_1\neq 0$ by $\mathrm {D}\omega (I^\ast )\hat {C}_1\neq 0$ .

Lemma 2.5. Suppose that $M(C_1,C_2,C_3),M(C_1',C_2',C_3')\in \mathscr {G}_\theta $ for some $C_j,C_j'$ , $j=1,2,3$ , with $C_3C_1'\neq C_3'C_1$ . Then the identity component $\mathscr {G}_\theta ^0$ of $\mathscr {G}_\theta $ is not commutative.

Proof. Assume that the hypothesis holds. We easily see that $M(C_1,C_2,C_3)$ and $M(C_1',C_2',C_3')$ is not commutative since

$$ \begin{align*} M(C_1,C_2,C_3)M(C_1',C_2',C_3) =\begin{pmatrix} \mathrm{id}_\ell & 0 & C_1+C_1'\\ C_3+C_3' & \mathrm{id}_m & C_3C_1'+C_2+C_2'\\ 0 & 0 & 1 \end{pmatrix} \end{align*} $$

while

$$ \begin{align*} M(C_1',C_2',C_3)M(C_1',C_2',C_3) =\begin{pmatrix} \mathrm{id}_\ell & 0 & C_1+C_1'\\ C_3+C_3' & \mathrm{id}_m & C_3'C_1+C_2+C_2'\\ 0 & 0 & 1 \end{pmatrix}\!. \end{align*} $$

However, we compute

$$ \begin{align*} M(C_1,C_2,C_3)^2 &=\begin{pmatrix} \mathrm{id}_\ell & 0 & C_1\\ C_3 & \mathrm{id}_m & C_2\\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} \mathrm{id}_\ell & 0 & C_1\\ C_3 & \mathrm{id}_m & C_2\\ 0 & 0 & 1 \end{pmatrix}\\ &=\begin{pmatrix} \mathrm{id}_\ell & 0 & 2C_1\\ 2C_3 & \mathrm{id}_m & C_3C_1+2C_2\\ 0 & 0 & 1 \end{pmatrix} \end{align*} $$

and easily show by induction that

$$ \begin{align*} M(C_1,C_2,C_3)^k =\begin{pmatrix} \mathrm{id}_\ell & 0 & kC_1\\ kC_3 & \mathrm{id}_m & (k-1)C_3C_1+kC_2\\ 0 & 0 & 1 \end{pmatrix} \end{align*} $$

for any $k\in \mathbb {N}$ . Since $\mathscr {G}_\theta ^0$ is a subgroup of finite index in $\mathscr {G}_\theta $ (see Appendix A), we show that

$$ \begin{align*} \mathscr{G}_\theta^0\supset\{M(cC_1,(c-1)C_3C_1+cC_2,cC_3)\mid c\in\mathbb{C}\} \end{align*} $$

if $C_3C_1+C_2\neq 0$ and

$$ \begin{align*} \mathscr{G}_\theta^0\supset\{M(cC_1,C_2,cC_3)\mid c\in\mathbb{C}\} \end{align*} $$

if $C_3C_1+C_2=0$ . Thus, we show that $\mathscr {G}_\theta ^0$ is not commutative in both cases.

By Lemma 2.5, the identity component $\mathscr {G}_\theta ^0$ is not commutative. Applying Theorem 2.3, we see that the system (1.5) is meromorphically non-integrable near the resonant periodic orbit $(I^\ast ,\omega (I^\ast )t+\theta )$ in the meaning of Theorem 2.1. If this statement holds for $\theta $ on a dense set $\Delta \subset \mathbb {T}^m$ , then it does so on $\mathbb {T}^m$ . Thus, we complete the proof.

3 Planar case

We prove Theorem 1.1 for the planar case (1.1). We only consider a neighborhood of $(x,y)=(-\mu ,0)$ since we only have to replace x and $\mu $ with $-x$ and $1-\mu $ to obtain the result for a neighborhood of $(1-\mu ,0)$ . We introduce a small parameter $\varepsilon $ such that $0<\varepsilon \ll 1$ . Letting

$$ \begin{align*} \varepsilon^2\xi=x+\mu,\quad \varepsilon^2\eta=y,\quad \varepsilon^{-1}p_\xi=p_x,\quad \varepsilon^{-1}p_\eta=p_y+\mu \end{align*} $$

and scaling the time variable $t\to \varepsilon ^3 t$ , we rewrite equation (1.1) as

$$ \begin{align*} & \dot{\xi}=p_\xi+\varepsilon^3\eta,\quad \dot{p}_\xi=-\frac{(1-\mu)\xi}{(\xi^2+\eta^2)^{3/2}} +\varepsilon^3 p_\eta-\varepsilon^4\mu -\varepsilon^4\frac{\mu(\varepsilon^2\xi-1)}{((\varepsilon^2\xi-1)^2+\varepsilon^4\eta^2)^{3/2}},\\ & \dot{\eta}=p_\eta-\varepsilon^3\xi,\quad \dot{p}_\eta=-\frac{(1-\mu)\eta}{(\xi^2+\eta^2)^{3/2}} -\varepsilon^3 p_\xi-\varepsilon^6\frac{\mu\eta}{((\varepsilon^2\xi-1)^2+\varepsilon^4\eta^2)^{3/2}}, \end{align*} $$

or up to the order of $\varepsilon ^6$ ,

(3.1) $$ \begin{align} & \dot{\xi}=p_\xi+\varepsilon^3\eta,\quad \dot{p}_\xi=-\frac{(1-\mu)\xi}{(\xi^2+\eta^2)^{3/2}} +\varepsilon^3 p_\eta+2\varepsilon^6\mu\xi, \nonumber\\ & \dot{\eta}=p_\eta-\varepsilon^3\xi,\quad \dot{p}_\eta=-\frac{(1-\mu)\eta}{(\xi^2+\eta^2)^{3/2}}-\varepsilon^3 p_\xi -\varepsilon^6\mu\eta, \end{align} $$

where the $O(\varepsilon ^8)$ terms have been eliminated. Equation (3.1) is a Hamiltonian system with the Hamiltonian

(3.2) $$ \begin{align} H=\frac{1}{2}(p_\xi^2+p_\eta^2)-\frac{1-\mu}{\sqrt{\xi^2+\eta^2}} +\varepsilon^3(\eta p_\xi-\xi p_\eta)-\frac{1}{2}\varepsilon^6\mu(2\xi^2-\eta^2). \end{align} $$

Non-integrability of a system which is similar to equation (3.1) but does not contain a small parameter was proven by using the Morales–Ramis theory [Reference Morales-Ruiz20, Reference Morales-Ruiz and Ramis22] in [Reference Morales-Ruiz, Simó and Simon24]. See also Remark 3.1(ii).

We next rewrite equation (3.2) in the polar coordinates. Let

$$ \begin{align*} \xi=r\cos\phi,\quad \eta=r\sin\phi. \end{align*} $$

The momenta $(p_r,p_\phi )$ corresponding to $(r,\phi )$ satisfy

$$ \begin{align*} p_\xi=p_r\cos\phi-\frac{p_\phi}{r}\sin\phi,\quad p_\eta=p_r\sin\phi+\frac{p_\phi}{r}\cos\phi. \end{align*} $$

See, e.g., [Reference Meyer and Offin19, §8.6.1]. The Hamiltonian becomes

$$ \begin{align*} H=\frac{1}{2}\bigg(p_r^2+\frac{p_\phi^2}{r^2}\bigg)-\frac{1-\mu}{r}-\varepsilon^3 p_\phi -\frac{1}{4}\varepsilon^6\mu r^2(3\cos 2\phi+1). \end{align*} $$

Up to $O(1)$ , the corresponding Hamiltonian system becomes

(3.3) $$ \begin{align} \dot{r}=p_r,\quad \dot{p}_r=\frac{p_\phi^2}{r^3}-\frac{1-\mu}{r^2},\quad \dot{\phi}=\frac{p_\phi}{r^2},\quad \dot{p}_\phi=0, \end{align} $$

which is easily solved since $p_\phi $ is a constant. Let $u=1/r$ . From equation (3.3), we have

$$ \begin{align*} \frac{\mathrm{d}^2u}{\mathrm{d}\phi^2}+u=\frac{1-\mu}{p_\phi^2}, \end{align*} $$

from which we obtain the relation

(3.4) $$ \begin{align} r=\frac{p_\phi^2}{(1-\mu)(1+e\cos\phi)}, \end{align} $$

where the position $\phi =0$ is appropriately chosen and e is a constant. We choose $e\in (0,1)$ , so that equation (3.4) represents an elliptic orbit with the eccentricity e. Moreover, its period is given by

(3.5) $$ \begin{align} T=\frac{p_\phi^3}{(1-\mu)^2}\int_0^{2\pi}\frac{\mathrm{d}\phi}{(1+e\cos\phi)^2} =\frac{2\pi p_\phi^3}{(1-\mu)^2(1-e^2)^{3/2}}. \end{align} $$

Now we introduce the Delaunay elements obtained from the generating function

(3.6) $$ \begin{align} W(r,\phi,I_1,I_2)=I_2\phi+\chi(r,I_1,I_2), \end{align} $$

where

(3.7) $$ \begin{align} \chi(r,I_1,I_2) &=\int_{r_-}^r\bigg(\frac{2(1-\mu)}{\rho}-\frac{(1-\mu)}{I_1^2} -\frac{I_2^2}{\rho^2}\bigg)^{1/2}\mathrm{d}\rho\notag\\ &= -2I_1^2\arcsin\sqrt{\frac{r_+-r}{r_+-r_-}} +\sqrt{(r_+-r)(r-r_-)}\notag\\ &\quad +\frac{2I_1I_2}{\sqrt{1-\mu}}\arctan\sqrt{\frac{r_-(r_+-r)}{r_+(r-r_-)}} \end{align} $$

with

$$ \begin{align*} r_\pm=I_1\bigg(I_1\pm\sqrt{I_1^2-\frac{I_2^2}{1-\mu}}\bigg) \end{align*} $$

(see, e.g., [Reference Meyer and Offin19, §8.9.1]). We have

$$ \begin{align*} & p_r=\frac{\partial W}{\partial r} =\frac{\partial\chi}{\partial r}(r,I_1,I_2),\quad p_\phi=\frac{\partial W}{\partial\phi} =I_2,\\ & \theta_1=\frac{\partial W}{\partial I_1}=\chi_1(r,I_1,I_2),\quad \theta_2=\frac{\partial W}{\partial I_2}=\phi+\chi_2(r,I_1,I_2), \end{align*} $$

where

$$ \begin{align*} \chi_1(r,I_1,I_2)=\frac{\partial\chi}{\partial I_1}(r,I_1,I_2),\quad \chi_2(r,I_1,I_2)=\frac{\partial\chi}{\partial I_2}(r,I_1,I_2). \end{align*} $$

Since the transformation from $(r,\phi ,p_r,p_\phi )$ to $(\theta _1,\theta _2,I_1,I_2)$ is symplectic, the transformed system is also Hamiltonian and its Hamiltonian is given by

$$ \begin{align*} H&=-\frac{(1-\mu)}{2I_1^2}-\varepsilon^3 I_2\notag\\ &\quad -\frac{1}{4}\varepsilon^6\mu R(\theta_1,I_1,I_2)^2 (3\cos 2(\theta_2-\chi_2(R(\theta_1,I_1,I_2),I_1,I_2))+1), \end{align*} $$

where $r=R(\theta _1,I_1,I_2)$ is the r-component of the symplectic transformation satisfying

(3.8) $$ \begin{align} \theta_1=\chi_1(R(\theta_1,I_1,I_2),I_1,I_2). \end{align} $$

Thus, we obtain the Hamiltonian system

(3.9) $$ \begin{align} \dot{I}_1&= \frac{1}{2}\varepsilon^6\mu\frac{\partial R}{\partial\theta_1}(\theta_1,I_1,I_2)R(\theta_1,I_1,I_2) \bigg(3\cos 2(\theta_2-\chi_2(R(\theta_1,I_1,I_2),I_1,I_2)) \nonumber\\ &\quad +1+3R(\theta_1,I_1,I_2)\frac{\partial\chi_2}{\partial r}(R(\theta_1,I_1,I_2),I_1,I_2) \nonumber\\ &\quad \times\sin 2(\theta_2-\chi_2(R(\theta_1,I_1,I_2),I_1,I_2))\bigg), \\ \dot{I}_2&= -\frac{3}{2}\varepsilon^6\mu R(\theta_1,I_1,I_2)^2 \sin 2(\theta_2-\chi_2(R(\theta_1,I_1,I_2),I_1,I_2)),\nonumber\\ \dot{\theta}_1&= \frac{1-\mu}{I_1^3}+O(\varepsilon^6),\quad \dot{\theta}_2=-\varepsilon^3+O(\varepsilon^6).\nonumber \end{align} $$

Similarly to the treatment for equation (1.1) stated just above Theorem 1.1, the new variables $(v_1,v_2,v_3)\in \mathbb {C}\times (\mathbb {C}/2\pi \mathbb {Z})^2$ given by

$$ \begin{align*} V_1(v_1,r,I_1,I_2) :&=v_1^2+r^2-2I_1^2r+\frac{I_1^2I_2^2}{1-\mu}=0,\\ V_2(v_2,r,I_1,I_2) :&=I_1^2\bigg(I_1^2-\frac{I_2^2}{1-\mu}\bigg)(2\sin^2v_2-1)^2-(r-I_1^2)^2=0,\\ V_3(v_3,r,I_1,I_2) :&= I_1^2\bigg(r-\frac{I_2^2}{1-\mu}\bigg)^2(\tan^2v_3+1)^2\\ &\quad -r^2\bigg(I_1^2-\frac{I_2^2}{1-\mu}\bigg)^2(\tan^2v_3-1)^2=0 \end{align*} $$

are introduced, so that the generating function (3.6) is regarded as an analytic one on the four-dimensional complex manifold

$$ \begin{align*} \bar{\mathscr{S}}_2 =\{&(r,\phi,I_1,I_2,v_1,v_2,v_3) \in\mathbb{C}\times(\mathbb{C}/2\pi\mathbb{Z})\times\mathbb{C}^3\times(\mathbb{C}/2\pi\mathbb{Z})^2\\ & \mid V_j(v_j,r,I_1,I_2)=0,\ j=1,2,3\} \end{align*} $$

since equation (3.7) is represented by

$$ \begin{align*} \chi(I_1,I_2,v_1,v_2,v_3)=v_1-2I_1^2v_2+\frac{2I_1I_2v_3}{\sqrt{1-\mu}}. \end{align*} $$

Hence, we can regard equation (3.9) as a meromorphic two-degree-of-freedom Hamiltonian system on the four-dimensional complex manifold

$$ \begin{align*} \hat{\mathscr{S}}_2&=\{(I_1,I_2,\theta_1,\theta_2,r,v_1,v_2,v_3) \in\mathbb{C}^2\times(\mathbb{C}/2\pi\mathbb{Z})^2\times\mathbb{C}^2\times(\mathbb{C}/2\pi\mathbb{Z})^2\\ & \quad \ \ \mid\theta_1-\chi_1(r,I_1,I_2)=V_j(v_j,r,I_1,I_2)=0,\ j=1,2,3\}, \end{align*} $$

like equation (1.3) on $\mathscr {S}_2$ for equation (1.1). Actually, we have

$$ \begin{align*} \frac{\partial V_j}{\partial v_j}\frac{\partial v_j}{\partial r} +\frac{\partial V_j}{\partial r}=0,\quad \frac{\partial V_j}{\partial v_j}\frac{\partial v_j}{\partial I_l} +\frac{\partial V_j}{\partial I_l}=0,\quad j=1,2,3,\ l=1,2, \end{align*} $$

to express

$$ \begin{align*} \frac{\partial\chi}{\partial r} =\sum_{j=1}^3\frac{\partial\chi}{\partial v_j}\frac{\partial V_j}{\partial r},\quad \chi_l=\frac{\partial\chi}{\partial I_l} +\sum_{j=1}^3\frac{\partial\chi}{\partial v_j}\frac{\partial V_j}{\partial I_l},\quad l=1,2 \end{align*} $$

as meromorophic functions of $(r,I_1,I_2,v_1,v_2,v_3)$ on $\bar {\mathscr {S}}_2$ . In particular, the Hamiltonian system has an additional first integral that is meromorphic in $(I_1,I_2, \theta _1,\theta _2, v_1,v_2,v_3,\varepsilon )$ on $\hat {\mathscr {S}}_2\setminus \Sigma (\hat {\mathscr {S}}_2)$ near $\varepsilon =0$ if the system (1.1) has an additional first integral that is meromorphic in $(x,y,p_x,p_y,u_1,u_2)$ on $\mathscr {S}_2\setminus \Sigma (\mathscr {S}_2)$ near $(x,y)=(-\mu ,0)$ , as in [Reference Combot10, Theorem 2], since the corresponding Hamiltonian system has the same expression as equation (3.9) on $\hat {\mathscr {S}}_2\setminus \Sigma (\hat {\mathscr {S}}_2)$ , where $\Sigma (\hat {\mathscr {S}}_2)$ is the critical set of $\hat {\mathscr {S}}_2$ on which the projection $\hat {\pi }_2:\hat {\mathscr {S}}_2\to \mathbb {C}^2\times (\mathbb {C}/2\pi \mathbb {Z})^2$ given by

$$ \begin{align*} \hat{\pi}_2(I_1,I_2,\theta_1,\theta_2,r,v_1,v_2,v_3)=(I_1,I_2,\theta_1,\theta_2) \end{align*} $$

is singular.

We next estimate the $O(\varepsilon ^6)$ -term in the first equation of equation (3.9) for the unperturbed solutions. When $\varepsilon =0$ , we see that $I_1,I_2,\theta _2$ are constants and can write $\theta _1=\omega _1 t+\theta _{10}$ for any solution to equation (3.9), where

(3.10) $$ \begin{align} \omega_1=\frac{1-\mu}{I_1^3} \end{align} $$

and $\theta _{10}\in \mathbb {S}^1$ is a constant. Since $r=R(\omega _1 t+\theta _{10},I_1,I_2)$ and

$$ \begin{align*} \phi =-\chi_2(R(\omega_1 t+\theta_{10},I_1,I_2),I_1,I_2) \end{align*} $$

respectively become the r- and $\phi $ -components of a solution to equation (3.3), we have

(3.11) $$ \begin{equation} \begin{aligned} &R(\omega_1 t+\theta_{10},I_1,I_2) =\frac{I_2^2}{(1-\mu)(1+e\cos(\phi(t)+\bar{\phi}(\theta_{10})))},\\&\quad -\chi_2(R(\omega_1 t+\theta_{10},I_1,I_2),I_1,I_2) =\phi(t)+\bar{\phi}(\theta_{10}) \end{aligned}\end{equation} $$

by equation (3.4), where $\phi (t)$ is the $\phi $ -component of a solution to equation (3.3) and $\bar {\phi }(\theta _{10})$ is a constant depending on $\theta _{10}$ . Differentiating both equations in equation (3.11) with respect to t yields

(3.12) $$ \begin{equation} \begin{aligned} &\omega_1\frac{\partial R}{\partial\theta_1}(\omega_1t+\theta_{10},I_1,I_2) =\frac{eI_2^2\sin(\phi(t)+\bar{\phi}(\theta_{10}))\dot{\phi}(t)} {(1-\mu)(1+e\cos(\phi(t)+\bar{\phi}(\theta_{10})))^2},\\ &\quad-\omega_1\frac{\partial\chi_2}{\partial r} (R(\omega_1 t+\theta_{10},I_1,I_2),I_1,I_2) \frac{\partial R}{\partial\theta_1}(\omega_1t+\theta_{10},I_1,I_2) =\dot{\phi}(t). \end{aligned}\end{equation} $$

Using equations (3.11) and (3.12), we can obtain the necessary expression of the $O(\varepsilon ^6)$ -term.

We are ready to check the hypotheses of Theorem 2.1 for the system (3.9). Assumption (A1) holds for any $I_1>0$ . Fix the values of $I_1,I_2$ at some $I_1^\ast ,I_2^\ast>0$ , and let $\omega ^\ast =\omega _1/3$ . Since by the second equation of equation (3.11) $\phi (t)$ is $2\pi /\omega _1$ -periodic, we have

$$ \begin{align*} \frac{2\pi I_2^{\ast 3}}{(1-\mu)^2(1-e^2)^{3/2}}=\frac{2\pi I_1^3}{1-\mu} \end{align*} $$

by equations (3.5) and (3.10), so that

$$ \begin{align*} I_2^\ast=I_1^\ast(1-\mu)^{1/3}\sqrt{1-e^2}. \end{align*} $$

From equation (3.10), we also have

$$ \begin{align*} \mathrm{D}\omega(I^\ast) &=\begin{pmatrix} -3(1-\mu)/I_1^{\ast 4} & 0\\ 0 & 0 \end{pmatrix}\!{,} \end{align*} $$

where $I^\ast {\kern-1.2pt}={\kern-1.2pt}(I_1^\ast ,{\kern-1pt}I_2^\ast )$ . We write the first component of equation (2.2) with $k{\kern-1.2pt}={\kern-1.2pt}5$ for $I{\kern-1.2pt}={\kern-1.2pt}I^\ast $ as

$$ \begin{align*} \mathscr{I}_1^5(\theta) &=-\frac{3\mu(1-\mu)}{2I_1^{\ast 4}}\int_{\gamma_\theta}\bigg( \frac{\partial R}{\partial\theta_1}(\omega_1 t+\theta_1,I_1^\ast,I_2^\ast) R(\omega_1 t+\theta_1,I_1^\ast,I_2^\ast)\notag\\ &\quad \times \bigg(3\cos 2(\theta_2 -\chi_2(R(\omega_1 t+\theta_1,I_1^\ast,I_2^\ast),I_1^\ast,I_2^\ast))+1\notag\\ &\quad +3R(\omega_1 t+\theta_1,I_1^\ast,I_2^\ast) \frac{\partial\chi_2}{\partial r} (R(\omega_1 t+\theta_1,I_1^\ast,I_2^\ast),I_1^\ast,I_2^\ast)\notag\\ &\quad \times\sin 2(\theta_2 -\chi_2(R(\omega_1 t+\theta_1,I_1^\ast,I_2^\ast),I_1^\ast,I_2^\ast))\bigg)\,d t, \end{align*} $$

where the closed loop $\gamma _\theta $ is specified below. Using equations (3.11) and (3.12), we compute

(3.13) $$ \begin{align} \mathscr{I}_1^5(\theta)&=\frac{3\mu I_2^{\ast 4}}{2(1-\mu)^2I_1^\ast} \int_{\gamma_\theta}\dot{\phi}(t)\bigg( \frac{3\sin 2(\phi(t)+\bar{\phi}(\theta_1)+\theta_2)} {(1+e\cos(\phi(t)+\bar{\phi}(\theta_1)))^2}\notag\\ &\quad -\frac{e\sin(\phi(t)+\bar{\phi}(\theta_1)) (3\cos 2(\phi(t)+\bar{\phi}(\theta_1)+\theta_2)+1)} {(1+e\cos(\phi(t)+\bar{\phi}(\theta_1)))^3}\bigg)\,d t. \end{align} $$

By equations (3.3) and (3.4), we have

(3.14) $$ \begin{align} \frac{\dot{\phi}(t)}{(1+e\cos\phi(t))^2}=\frac{(1-\mu)^2}{I_2^{\ast 3}} =\frac{\omega_1}{(1-e^2)^{3/2}}. \end{align} $$

Using integration by substitution and the relation (3.14), we rewrite the above integral as

(3.15) $$ \begin{align} \mathscr{I}_1^5(\theta) &=\frac{9\mu I_2^\ast}{2I_1^\ast}\int_{\gamma_\theta}\bigg(\sin 2(\phi(t)+\theta_2) -\frac{e\sin\phi(t)(\cos 2(\phi(t)+\theta_2)+{1}/{3})}{1+e\cos\phi(t)}\bigg)\,d t, \end{align} $$

where the path of integration might change but the same notation $\gamma _\theta $ has still been used for it.

Here we integrate equation (3.14) to obtain

(3.16) $$ \begin{align} \omega_1 t=2\arctan\bigg(\frac{(1-e)\tan{\phi}/{2}}{\sqrt{1-e^2}}\bigg) -\frac{e\sqrt{1-e^2}\sin\phi}{1+e\cos\phi}\quad \mbox{for}\ \phi\in(-\pi,\pi), \end{align} $$

which is rewritten as

(3.17) $$ \begin{align} \omega_1 t=2\operatorname{\mathrm{arccot}}\bigg(\frac{(1-e)\cot({\phi+\pi})/{2}}{\sqrt{1-e^2}}\bigg) -\frac{e\sqrt{1-e^2}\sin\phi}{1+e\cos\phi}+\pi\quad \mbox{for}\ \phi\in(0,2\pi), \end{align} $$

when $\phi (0)=0$ or $\lim _{t\to 0}\phi (t)=0$ . From equation (3.16), we see that as $\operatorname {\mathrm {Im}}\phi \to +\infty $ , $\omega _1 t\to iK_1$ , where

$$ \begin{align*} K_1=2\operatorname{\mathrm{arctanh}}\bigg(\frac{1-e}{\sqrt{1-e^2}}\bigg)-\sqrt{1-e^2}>0. \end{align*} $$

See Figure 4. So the integrand in equation (3.15) is singular at $t=iK_1$ . Let $K_2=\operatorname {\mathrm {arccosh}}(1/e)$ . Then $1+e\cos \phi =0$ at $\phi =\pi +iK_2$ , and by equation (3.17),

$$ \begin{align*} \frac{1}{\omega_1 t}=\frac{1}{\sqrt{1-e^2}}\Delta\phi+o(\Delta\phi) \end{align*} $$

near $\phi =\pi +iK_2$ , where $\Delta \phi =\phi -(\pi +iK_2)$ . Moreover, near $\phi =\pi +iK_2$ ,

$$ \begin{align*} & \sin\phi=-\frac{i\sqrt{1-e^2}}{e} +O(\Delta\phi),\quad \cos\phi=-\frac{1}{e}+\frac{i\sqrt{1-e^2}}{e}\Delta\phi +O(\Delta\phi^2),\\ & \sin 2\phi=\frac{2i\sqrt{1-e^2}}{e^2} +O(\Delta\phi),\quad \cos 2\phi=\frac{2-e^2}{e^2} +O(\Delta\phi). \end{align*} $$

Figure 4 Dependence of $K_1$ on e.

Figure 5 Closed path $\gamma _\theta $ .

We take a closed path starting and ending at $t=\tfrac {1}{3}T^\ast $ and passing through $t=\tfrac {2}{3}T^\ast $ , $\tfrac {2}{3}T^\ast +i(K_1\mp \delta )$ , $\tfrac {2}{3}T^\ast +iM$ , $\tfrac {1}{3}T^\ast +iM$ , and $\tfrac {1}{3}T^\ast +i(K_1\pm \delta )$ as $\gamma _\theta $ in $\mathbb {C}/T^\ast \mathbb {Z}$ , where $\delta $ and M are respectively sufficiently small and large positive constants (see Figure 5). Here $\gamma _\theta $ passes along the left circular arc centered at $\tfrac {2}{3}T^\ast +iK_1$ (respectively at $\tfrac {1}{3}T^\ast +iK_1$ ) with radius $\delta $ between $\tfrac {2}{3}T^\ast +i(K_1-\delta )$ and $\tfrac {2}{3}T^\ast +i(K_1+\delta )$ (respectively between $\tfrac {1}{3}T^\ast +i(K_1+\delta )$ and $\tfrac {1}{3}T^\ast +i(K_1-\delta )$ ). We compute

$$ \begin{align*} & \int_{2T^\ast\!/3+iM}^{T^\ast\!/3+iM} \frac{\sin\phi(t)(\cos 2(\phi(t)+\theta_2)+{1}/{3})}{1+e\cos\phi(t)}\,d t\\ &\quad =-\int_{2T^\ast\!/3}^{T^\ast\!/3}\bigg(\frac{2-e^2}{e^3\sqrt{1-e^2}}\cos 2\theta_2 -\frac{2i}{e^3}\sin 2\theta_2+\frac{1}{3e\sqrt{1-e^2}}\bigg)i\omega_1 M\,d t+O(1)\\ &\quad =\frac{2\pi}{e^3}\bigg(\frac{2-e^2}{\sqrt{1-e^2}}i\cos2\theta_2+2\sin2\theta_2 +\frac{ie^2}{3\sqrt{1-e^2}}\bigg)M+O(1), \end{align*} $$

while

$$ \begin{align*} \int_{2T^\ast\!/3+iM}^{T^\ast\!/3+iM}\sin 2(\phi(t)+\theta_2)\,d t=O(1). \end{align*} $$

Moreover, the integral on $[\tfrac {1}{3}T^\ast ,\tfrac {2}{3}T^\ast ]$ in equation (3.15) is $O(1)$ , and the integrals from $\tfrac {2}{3}T^\ast $ to $\tfrac {2}{3}T^\ast +iM$ and from $\tfrac {1}{3}T^\ast +iM$ to $\tfrac {1}{3}T^\ast $ cancel since the integrand is $\tfrac {1}{3}T^\ast $ -periodic. Thus, we see that the integral (3.15) is not zero for $M>0$ sufficiently large, so that assumption (A2) holds.

Finally, we apply Theorem 2.1 to show that the meromorphic Hamiltonian system corresponding to equation (3.9) is not meromorphically integrable such that the first integral depends meromorphically on $\varepsilon $ near $\varepsilon =0$ even if any higher-order terms are included. Thus, we obtain the conclusion of Theorem 1.1 for the planar case.

4 Spatial case

We prove Theorem 1.1 for the spatial case (1.2). As in the planar case, we only consider a neighborhood of $(x,y,z)=(-\mu ,0,0)$ and introduce a small parameter $\varepsilon $ such that $0<\varepsilon \ll 1$ . Letting

$$ \begin{align*}& \varepsilon^2\xi=x+\mu,\quad \varepsilon^2\eta=y,\quad \varepsilon^2\zeta=z,\\ & \varepsilon^{-1}p_\xi=p_x,\quad \varepsilon^{-1}p_\eta=p_y+\mu,\quad \varepsilon^{-1}p_\zeta=p_z \end{align*} $$

and scaling the time variable $t\to \varepsilon ^3 t$ , we rewrite equation (1.2) as

$$ \begin{align*} & \dot{\xi}=p_\xi+\varepsilon^3\eta,\quad \dot{\eta}=p_\eta-\varepsilon^3\xi,\quad \dot{\zeta}=p_\zeta,\\ & \dot{p}_\xi=-\frac{(1-\mu)\xi}{(\xi^2+\eta^2+\zeta^2)^{3/2}} +\varepsilon^3p_\eta-\varepsilon^4\mu-\varepsilon^4\frac{\mu(\varepsilon^2\xi-1)} {((\varepsilon^2\xi-1)^2+\varepsilon^4(\eta^2+\zeta^2))^{3/2}},\\ & \dot{p}_\eta=-\frac{(1-\mu)\eta}{(\xi^2+\eta^2+\zeta^2)^{3/2}} -\varepsilon^3 p_\xi-\varepsilon^6\frac{\mu\eta} {((\varepsilon^2\xi-1)^2+\varepsilon^4(\eta^2+\zeta^2))^{3/2}},\\ & \dot{p}_\zeta=-\frac{(1-\mu)\zeta}{(\xi^2+\eta^2+\zeta^2)^{3/2}} -\varepsilon^6\frac{\mu\zeta} {((\varepsilon^2\xi-1)^2+\varepsilon^4(\eta^2+\zeta^2)))^{3/2}}, \end{align*} $$

or up to the order of $\varepsilon ^6$ ,

(4.1) $$ \begin{align} & \dot{\xi}=p_\xi+\varepsilon^3\eta,\quad \dot{p}_\xi=-\frac{(1-\mu)\xi}{(\xi^2+\eta^2+\zeta^2)^{3/2}} +\varepsilon^3p_\eta+2\varepsilon^6\mu\xi,\nonumber\\ & \dot{\eta}=p_\eta-\varepsilon^3\xi,\quad \dot{p}_\eta=-\frac{(1-\mu)\eta}{(\xi^2+\eta^2+\zeta^2)^{3/2}} -\varepsilon^3p_\xi-\varepsilon^6\mu\eta,\\ & \dot{\zeta}=p_\zeta,\quad \dot{p}_\zeta=-\frac{(1-\mu)\zeta}{(\xi^2+\eta^2+\zeta^2)^{3/2}}-\varepsilon^6\mu\zeta,\nonumber \end{align} $$

like equation (3.1), where the $O(\varepsilon ^8)$ terms have been eliminated. Equation (4.1) is a Hamiltonian system with the Hamiltonian

(4.2) $$ \begin{equation} \begin{aligned} H&=\tfrac{1}{2}(p_\xi^2+p_\eta^2+p_\zeta^2)-\frac{1-\mu}{\sqrt{\xi^2+\eta^2+\zeta^2}}\\ &\quad +\varepsilon^3(\eta p_\xi-\xi p_\eta)-\tfrac{1}{2}\varepsilon^6(2\xi^2-\eta^2-\zeta^2). \end{aligned}\end{equation} $$

We next rewrite equation (4.2) in the spherical coordinates (see Figure 6). Let

$$ \begin{align*} \xi=r\sin\psi\cos\phi,\quad \eta=r\sin\psi\sin\phi,\quad \zeta=r\cos\psi. \end{align*} $$

The momenta $(p_r,p_\phi ,p_\psi )$ corresponding to $(r,\phi ,\psi )$ satisfy

$$ \begin{align*} & p_\xi=p_r\cos\phi\sin\psi-\frac{p_\phi\sin\phi}{r\sin\psi}+\frac{p_\psi}{r}\cos\phi\cos\psi,\\ & p_\eta=p_r\sin\phi\sin\psi+\frac{p_\phi\cos\phi}{r\sin\psi}+\frac{p_\psi}{r}\sin\phi\cos\psi,\\ & p_\zeta=p_r\cos\psi-\frac{p_\psi}{r}\sin\psi \end{align*} $$

(see, e.g., [Reference Meyer and Offin19, §8.7]). The Hamiltonian becomes

$$ \begin{align*} H&=\frac{1}{2}\bigg(p_r^2+\frac{p_\psi^2}{r^2}+\frac{p_\phi^2}{r^2\sin^2\psi}\bigg) -\frac{1-\mu}{r}\notag\\ &\quad -\varepsilon^3 p_\phi -\varepsilon^6\mu r^2\bigg(\frac{1}{4}\sin^2\psi(3\cos 2\phi+1)-\frac{1}{2}\cos^2\psi \bigg). \end{align*} $$

Up to $O(1)$ , the corresponding Hamiltonian system becomes

(4.3) $$ \begin{equation} \begin{aligned}& \dot{r}=p_r,\quad \dot{p}_r=\frac{p_\psi^2}{r^3}+\frac{p_\phi^2}{r^3\sin^2\psi}-\frac{1-\mu}{r^2}, \\ & \dot{\phi}=\frac{p_\phi}{r^2\sin^2\psi},\quad \dot{p}_\phi=0,\quad \dot{\psi}=\frac{p_\psi}{r^2},\quad \dot{p}_\psi=\frac{p_\phi^2\cos\psi}{r^2\sin^3\psi}. \end{aligned}\end{equation} $$

We have the relation (3.4) for periodic orbits on the $(\xi ,\eta )$ -plane since equation (4.3) reduces to equation (3.3) when $\psi =\tfrac {1}{2}\pi $ and $p_\psi =0$ .

Figure 6 Spherical coordinates.

As in the planar case, we introduce the Delaunay elements obtained from the generating function

(4.4) $$ \begin{align} \hat{W}(r,\phi,\psi,I_1,I_2.I_3)=I_3\phi+\chi(r,I_1,I_2)+\hat{\chi}(\psi,I_2,I_3), \end{align} $$

where

(4.5) $$ \begin{align} \hat{\chi}(\psi,I_2,I_3) &=\int_{\psi_0}^\psi\bigg(I_2^2-\frac{I_3^2}{\sin^2s}\bigg)^{1/2}\,d s\notag\\ &= I_2\arctan\frac{\sqrt{I_2^2\sin^2\psi-I_3^2}}{I_2\cos\psi} -I_3\arctan\frac{\sqrt{I_2^2\sin^2\psi-I_3^2}}{I_3\cos\psi} \end{align} $$

with $\psi _0=\arcsin (I_3/I_2)$ . See, e.g., [Reference Meyer and Offin19, §8.9.3], although a slightly modified generating function is used here. We have

(4.6) $$ \begin{align} p_r&=\frac{\partial \hat{W}}{\partial r} =\frac{\partial\chi}{\partial r}(r,I_1,I_2),\quad p_\phi=\frac{\partial \hat{W}}{\partial\phi} =I_3, \nonumber\\ p_\psi&=\frac{\partial\hat{W}}{\partial\psi} =\frac{\partial\hat{\chi}}{\partial\psi}(\psi,I_2,I_3),\quad \theta_1=\frac{\partial \hat{W}}{\partial I_1} =\chi_1(r,I_1,I_2), \\ \theta_2&=\frac{\partial\hat{W}}{\partial I_2} =\chi_2(r,I_1,I_2)+\hat{\chi}_2(\psi,I_2,I_3),\quad \theta_3=\frac{\partial \hat{W}}{\partial I_3} =\phi+\hat{\chi}_3(\psi,I_2,I_3), \nonumber \end{align} $$

where

$$ \begin{align*} \hat{\chi}_2(\psi,I_2,I_3) =\frac{\partial\hat{\chi}}{\partial I_2}(\psi,I_2,I_3),\quad \hat{\chi}_3(\psi,I_2,I_3) =\frac{\partial\hat{\chi}}{\partial I_3}(\psi,I_2,I_3). \end{align*} $$

Since the transformation from $(r,\phi ,\psi ,p_r,p_\phi ,p_\psi )$ to $(\theta _1,\theta _2,\theta _3,I_1,I_2,I_3)$ is symplectic, the transformed system is also Hamiltonian and its Hamiltonian is given by

$$ \begin{align*} H&=-\frac{(1-\mu)}{2I_1^2}-\varepsilon^3 I_3 -\varepsilon^6\mu R(\theta_1,I_1,I_2)^2\bigg(\frac{1}{4}\sin^2\Psi(\theta_1,\theta_2,I_1,I_2,I_3)\\&\quad \times (3\cos 2(\theta_3-\hat{\chi}_3(\Psi(\theta_1,\theta_2,I_1,I_2,I_3),I_1,I_2))+1)\\&\quad -\frac{1}{2}\cos^2\Psi(\theta_1,\theta_2,I_1,I_2,I_3)\bigg), \end{align*} $$

where $r=R(\theta _1,I_1,I_2)$ and $\psi =\Psi (\theta _1,\theta _2,I_1,I_2,I_3)$ are the r- and $\psi $ -components of the symplectic transformation satisfying equation (3.8) and

$$ \begin{align*} \hat{\chi}_2(\Psi(\theta_1,\theta_2,I_1,I_2,I_3),I_2,I_3)+\chi_2(R(\theta_1,I_1,I_2),I_1,I_2) =\theta_2, \end{align*} $$

respectively. Thus, we obtain the Hamiltonian system as

(4.7) $$ \begin{align} \dot{I}_1&= \frac{1}{2}\varepsilon^6\mu\hat{h}_1(I,\theta),\quad \dot{I}_2=O(\varepsilon^6),\quad \dot{I}_3=O(\varepsilon^6), \nonumber\\ \dot{\theta}_1&= \frac{1-\mu}{I_1^3}+O(\varepsilon^6),\quad \dot{\theta}_2=O(\varepsilon^6),\quad \dot{\theta}_3=-\varepsilon^3+O(\varepsilon^6), \end{align} $$

where

$$ \begin{align*} \hat{h}_1(I,\theta) &= \frac{\partial R}{\partial\theta_1}(\theta_1,I_1,I_2)R(\theta_1,I_1,I_2) (\sin^2\Psi(\theta_1,\theta_2,I_1,I_2,I_3)\\ &\qquad \times(3\cos 2(\theta_3-\hat{\chi}_3(\Psi(\theta_1,\theta_2,I_1,I_2,I_3),I_2,I_3))+1)\\ &\qquad -2\cos^2\Psi(\theta_1,\theta_2,I_1,I_2,I_3))\\ &\quad +R(\theta_1,I_1,I_2)^2\frac{\partial\Psi}{\partial\theta_1}(\theta_1,\theta_2,I_1,I_2,I_3)\\ &\qquad \times 3\sin\Psi(\theta_1,\theta_2,I_1,I_2,I_3)\cos\Psi(\theta_1,\theta_2,I_1,I_2,I_3)\\ &\qquad \times(\cos 2(\theta_3-\hat{\chi}_3(\Psi(\theta_1,\theta_2,I_1,I_2,I_3),I_2,I_3))+1)\\ & \quad+3R(\theta_1,I_1,I_2)^2\sin^2\Psi(\theta_1,\theta_2,I_1,I_2,I_3)\\ &\qquad \times\frac{\partial\Psi}{\partial\theta_1}(\theta_1,\theta_2,I_1,I_2,I_3) \frac{\partial\hat{\chi}_3}{\partial\psi}(\Psi(\theta_1,\theta_2,I_1,I_2,I_3),I_2,I_3)\\ & \qquad \times\sin2(\theta_3-\hat{\chi}_3(\Psi(\theta_1,\theta_2,I_1,I_2,I_3),I_2,I_3)). \end{align*} $$

As in the planar case, the new variables $w_1,w_2\in (\mathbb {C}/2\pi )$ given by

$$ \begin{align*} & W_1(w_1,\psi,I_2,I_3):=I_2^2\cos^2\psi\tan^2w_1-I_2^2\sin^2\psi+I_3^2=0,\\ & W_2(w_2,\psi,I_2,I_3):=I_3^2\cos^2\psi\tan^2w_2-I_2^2\sin^2\psi+I_3^2=0 \end{align*} $$

are introduced, so that the generating function (4.4) is regarded as an analytic one on the six-dimensional complex manifold

$$ \begin{align*} \bar{\mathscr{S}}_3&=\{(r,\phi,\psi,I_1,I_2,I_3,v_1,v_2,v_3,w_1,w_2) \in\mathbb{C}\times(\mathbb{C}/2\pi\mathbb{Z})^2\times\mathbb{C}^4\times(\mathbb{C}/2\pi\mathbb{Z})^4\\ &\quad \ \ \mid V_j(v_j,r,I_1,I_2)=W_l(w_l,\psi,I_2,I_3)=0,\ j=1,2,3,\ l=1,2\} \end{align*} $$

since equation (4.5) is represented by

$$ \begin{align*} \hat{\chi}(I_2,I_3;v_1,v_2)=I_2w_1-I_3w_2. \end{align*} $$

Moreover, we can regard equation (4.7) as a meromorphic three-degree-of-freedom Hamiltonian systems on the six-dimensional complex manifold

$$ \begin{align*} \hat{\mathscr{S}}_3 &=\{(I_1,I_2,I_3,\theta_1,\theta_2,\theta_3,r,v_1,v_2,v_3,w_1,w_2) \in\mathbb{C}^3\times(\mathbb{C}/2\pi\mathbb{Z})^3\times\mathbb{C}^2\times(\mathbb{C}/2\pi\mathbb{Z})^4\\ &\quad \ \ \mid \theta_1-\chi_1(r,I_1,I_2)=V_j(v_j,r,I_1,I_2)=W_l(w_l,\psi,I_2,I_3)=0,\\ &\qquad\ j=1,2,3,\ l=1,2\}, \end{align*} $$

like equation (1.4) on $\mathscr {S}_3$ for equation (1.2). Actually, we have

$$ \begin{align*} \frac{\partial W_j}{\partial w_j}\frac{\partial w_j}{\partial\psi} +\frac{\partial W_j}{\partial\psi}=0,\quad \frac{\partial W_j}{\partial w_j}\frac{\partial w_j}{\partial I_l} +\frac{\partial W_j}{\partial I_l}=0,\quad j=1,2,\ l=2,3 \end{align*} $$

to express

$$ \begin{align*} \frac{\partial\hat{\chi}}{\partial\psi} =\sum_{j=1}^2\frac{\partial\hat{\chi}}{\partial w_j}\frac{\partial W_j}{\partial\psi},\quad \hat{\chi}_l=\frac{\partial\hat{\chi}}{\partial I_l} +\sum_{j=1}^2\frac{\partial\hat\chi}{\partial w_j}\frac{\partial W_j}{\partial I_l},\quad l=2,3 \end{align*} $$

as meromorophic functions of $(\psi ,I_2,I_3,w_1,w_2)$ on $\bar {\mathscr {S}}_3$ . In particular, the Hamiltonian system has two additional meromorphic integrals that are meromorphic in $(I_1,I_2,I_3,\theta _1,\theta _2,\theta _3,r,v_1,v_2,v_3,w_1,w_ 2,\varepsilon )$ on $\hat {\mathscr {S}}_3\setminus \Sigma (\hat {\mathscr {S}}_3)$ near $\varepsilon =0$ , if the system (1.2) has two additional meromorphic integrals that are meromorphic in $(x,y,z,p_x,p_y,p_z,u_1,u_2)$ on $\mathscr {S}_3\setminus \Sigma (\mathscr {S}_3)$ near $(x,y,z)=(-\mu ,0,0)$ , as in the planar case. Here $\Sigma (\hat {\mathscr {S}}_3)$ is the critical set of $\hat {\mathscr {S}}_3$ on which the projection ${\hat {\pi }_3:\hat {\mathscr {S}}_3\to \mathbb {C}^3\times (\mathbb {C}/2\pi \mathbb {Z})^3}$ given by

$$ \begin{align*} \hat{\pi}_3(I_1,I_2,I_3,\theta_1,\theta_2,\theta_3,r,v_1,v_2,v_3,w_1,w_2) =(I_1,I_2,I_3,\theta_1,\theta_2,\theta_3) \end{align*} $$

is singular.

We next estimate the function $\hat {h}_1(I,\theta )$ for solutions to equation (4.7) with $\varepsilon =0$ on the plane of $\psi =\tfrac {1}{2}\pi $ . When $\varepsilon =0$ , we see that $I_1,I_2,I_3,\theta _2,\theta _3$ are constants and can write $\theta _1=\omega _1 t+\theta _{10}$ for any solution to equation (4.7) with equation (3.10), where $\theta _{10}\in \mathbb {S}^1$ is a constant. Note that if $\psi =\tfrac {1}{2}\pi $ and $p_\psi =0$ , then $I_2=I_3$ by equation (4.6). Since $r=R(\omega _1 t+\theta _{10},I_1,I_2)$ and

$$ \begin{align*} \phi=-\hat{\chi}_3(\Psi(\omega_1 t+\theta_{10},\theta_2,I_1,I_3,I_3),I_3,I_3),\quad \Psi(\omega_1 t+\theta_{10},\theta_2,I_1,I_3,I_3)=\tfrac{1}{2}\pi \end{align*} $$

respectively become the r- and $\phi $ -components of a solution to equation (4.3) with $\psi =\tfrac {1}{2}\pi $ and $p_\psi =0$ , we have the first equation of equation (3.11) with

(4.8) $$ \begin{align} -\hat{\chi}_3(\Psi(\omega_1 t+\theta_{10}, \theta_2,I_1,I_3,I_3),I_3,I_3) =\phi(t)+\bar{\phi}(\theta_{10}), \end{align} $$

where $\phi (t)$ is the $\phi $ -component of a solution to equation (3.3) and $\bar {\phi }(\theta _1)$ is a constant depending only on $\theta _1$ as in the planar case. Differentiating equation (4.8) with respect to t yields

(4.9) $$ \begin{align} & -\omega_1\frac{\partial \hat{\chi}_3}{\partial\psi} (\Psi(\omega_1 t+\theta_{10}, \theta_2,I_1,I_3,I_3),I_3,I_3)\notag\\ &\qquad \times\frac{\partial\Psi}{\partial\theta_1}(\omega_1 t+\theta_{10}, \theta_2,I_1,I_3,I_3)=\dot{\phi}(t). \end{align} $$

Using equations (3.11), (3.12), (4.8), and (4.9), we can obtain the necessary expression of $\hat {h}_1(I,\theta )$ .

We are ready to check the hypotheses of Theorem 2.1 for the system (4.7). Assumption (A1) holds for any $I_1>0$ . Fix the value of $I_1$ at some $I_1^\ast>0$ , and let $\omega ^\ast =\omega _1/3$ . By the first equation of equation (4.8), $\phi (t)$ is $2\pi /\omega _1$ -periodic, so that by equations (3.5) and (3.10),

$$ \begin{align*} I_2=I_3=I_1^\ast(1-\mu)^{1/3}\sqrt{1-e^2}\ (=I_2^\ast). \end{align*} $$

From equation (3.10), we have

$$ \begin{align*} \mathrm{D}\omega(I^\ast)= \begin{pmatrix} -3(1-\mu)/I_1^{\ast 4} & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}\!, \end{align*} $$

where $I^\ast =(I_1^\ast ,I_2^\ast ,I_2^\ast )$ . Using the first equations of equations (3.11) and (3.12), equations (4.8) and (4.9), we compute the first component of equation (2.2) with $k=5$ for $I=I^\ast $ as

$$ \begin{align*} \mathscr{I}_1^5(\theta) &=-\frac{3(1-\mu)}{I_1^{\ast 4}}\int_{\gamma_\theta} h_1(I^\ast,\omega^\ast t+\theta_1,\theta_2,\theta_3)\,d\omega^\ast t\\ &=\frac{3\mu I_2^{\ast 4}}{2(1-\mu)^2I_1^\ast} \int_{\gamma_\theta}\dot{\phi}(t)\bigg( \frac{3\sin 2(\phi(t)+\bar{\phi}(\theta_1)+\theta_3)} {(1+e\cos(\phi(t)+\bar{\phi}(\theta_1)))^2}\\ &\quad +\frac{e\sin(\phi(t)+\bar{\phi}(\theta_1)) (3\cos 2(\phi(t)+\bar{\phi}(\theta_1)+\theta_3)+1)} {(1+e\cos(\phi(t)+\bar{\phi}(\theta_1)))^3}\bigg)\,d t, \end{align*} $$

which has the same expression as equation (3.13) with $\theta _2=\theta _3$ . Repeating the arguments given in §3, we can show that assumption (A2) holds as in the planar case. Finally, we apply Theorem 2.1 to show that the meromorphic Hamiltonian system corresponding to equation (4.7) is not meromorphically integrable such that the first integrals depend meromorphically on $\varepsilon $ near $\varepsilon =0$ . Thus, we complete the proof of Theorem 1.1 for the spatial case.