Summability of Transseries Solution of Non-integrable Hamiltonian System

Abstract

This paper studies the summability of the transseries solution of a nonintegrable Hamiltonian system. Since our system has a resonance and is not integrable a general transseries theory does not work well as far as the author knows. In order to construct a formal transseries solution and prove its summability our main idea is to use the superintegrability of a Hamiltonian system in a class of transseries. More precisely we first show the superintegrability of a Hamiltonian system in the category of transseries via the key Lemmas 1 and 4 which follow. By virtue of the superintegrability we show the existence of a formal transseries solution. Then its summability is proved via the superintegrability. We note that the argument based on the superintegrability is elementary.

1 Introduction

Let $(q_1, q_2, q_3, \cdots ,q_n)$, $(p_1, p_2, p_3, \cdots , p_n)$ be the variables in some neighborhood of the origin of ${\mathbb R}^{2n}$ or ${\mathbb C}^{2n}$ $(n \ge 1)$ with respect to a standard symplectic form. Let ${\mathbb Z}_+$ be the set of nonnegative integers. In this paper we study the summability of the transseries solution of the $C^\infty$-integrable and $C^\omega$-nonintegrable Hamiltonian system studied by Bolsinov, Taimanov, Gorni and Zampieri. (cf. [3] and [6]). Consider the Hamitonian system with n degrees of freedom

$$\begin{aligned} \dot{q}_j = \nabla _{p_j} H, \quad \dot{p}_j = - \nabla _{q_j} H, \quad j=1,2, \ldots , n, \end{aligned}$$

(1)

for the Hamiltonian H whose precise definition is given in the next section.

Starting from the pioneering work of Ecalle, [5] a transseries attracts more attention in mathematical physics and dynamical systems. (See also [7, 8] and the references therein). It is also noted in [4] that a transseries solution gives a general solution. In order to construct a transseries solution we first construct a formal transseries solution. Then the summability theory gives the solution of the equation via the sum of a formal transseries.

In this paper we give the rigorous proof of the above assertion by using a superintegrability. More precisely we first show a superintegrability in the category of transseries. By virtue of the superintegrability one can construct a formal transseries solution and show its summability. We consider a transserial first integral since the nonintegrability in the analytic category implies that Eq. 1 does not have n $C^\omega$-first integrals. On the other hand, a superintegrability holds in the category of transseries. Namely, we can construct $2n-1$ functionally independent first integrals of Eq. 1 containing transseries. The idea of a transserial superintegrability for a nonintegrable Hamiltonian system seems natural although no rigorous proof seems to be known.

This paper is organized as follows. In Section 2 we formulate the existence of a formal transseries solution as Theorem 1. In Section 3 we prove a superintegrability in the category of formal transseries. In Section 4 we prove Theorem 1 by virtue of the superintegrability. In Section 5 we state the summability of the formal transseries solution constructed in Theorem 1 as Theorem 5. In Section 6 we prove the summability of formal first integrals and show the superintegrability. We prove Theorem 5 in Section 7.

2 Formal Transseries Solution

We set $q=(q_2, \ldots , q_n)$ and write $(q_1, q_2, \ldots , q_n) = (q_1,q)$. Similarly, we write $(p_1, p_2, \ldots , p_n) = (p_1, p)$, where $p=(p_2, \ldots , p_n)$. Let $H=H(q_1, p_1,q,p)$ be a smooth function. Define the Hamiltonian vector field $\chi _H$ by

$$\begin{aligned} \chi _H := \{H, \cdot \} = \sum _{j=1}^n \left( \frac{\partial H}{\partial p_j}\frac{\partial }{\partial q_j} - \frac{\partial H}{\partial q_j}\frac{\partial }{\partial p_j} \right) , \end{aligned}$$

(2)

where $\{ \cdot , \cdot \}$ denotes the Poisson bracket. We say that $\phi$ is the first integral of $\chi _H$ if $\chi _H \phi =0$. Let $\varOmega \subset {\mathbb R}^{2n}$ or $\varOmega \subset {\mathbb C}^{2n}$ be an open set. Let $C^\omega \equiv C^\omega (\varOmega )$ be the set of functions analytic in $\varOmega$. We say that $\chi _H$ is $C^\omega$-Liouville integrable in $\varOmega$ if there exist first integrals $\phi _j \in C^\omega (\varOmega )$ $(j=1,2, \ldots ,n)$ which are Poisson commuting, i.e., $\{ \phi _j, \phi _k \} =0$, $\{ H, \phi _k \} =0$, and that are functionally independent on an open set $\varOmega ' \subset \varOmega$ with $\varOmega '$ being dense in $\varOmega$. If $\phi _j \in C^\infty$ $(j=1,2, \ldots ,n)$, then we say $C^\infty$-Liouville integrable. We say that $\chi _H$ is superintegrable if $\chi _H$ has $2n-1$ functionally independent first integrals on an open set. We say that v is the formal first integral of $\chi _H$ if $\chi _H v=0$ as a formal power series.

Let $H_0$ and $H_1$ be given, respectively, by

$$\begin{aligned} H_0= & {} q_1^{2\sigma }p_1 + \sum _{j=2}^n \lambda _j q_j p_j ,\end{aligned}$$

(3)

$$\begin{aligned} H_1= & {} \sum _{j=2}^n q_j^2 B_j(q_1,q_1^{2\sigma }p_1, q), \ \end{aligned}$$

(4)

where $B_j(q_1,s,q)$’s are holomorphic in some neighborhood of the origin and $\sigma \ge 1$ is an integer. Define $H := H_0+H_1$. Assume

$$\begin{aligned} B_\nu \equiv B_\nu (q_1,q_1^{2\sigma } p_1 ,q) = B_{\nu ,0}(q_1,q) + q_1^{2\sigma } p_1 B_{\nu ,1}(q_1,q), \ \ \nu =2,\ldots , n, \end{aligned}$$

(5)

where $B_{\nu ,0}$ and $B_{\nu ,1}$ are analytic at $(q_1,q)= (0,0)$. Suppose that the Poincaré condition holds

$$\begin{aligned} Re \,\lambda _j >0, \quad j=2,3, \ldots ,n. \end{aligned}$$

(6)

Assume the nonresonance condition

$$\begin{aligned} \sum _{\nu =2}^n \lambda _\nu k_\nu - \lambda _j \ne 0, \quad \forall \ k_\nu \in {\mathbb Z}_+, \ \nu = 2, \ldots ,n, \ j = 2, \ldots ,n . \end{aligned}$$

(7)

Set $\lambda =(\lambda _2, \ldots , \lambda _n)$. Consider the formal power series solution

$(q_1(t),\ldots , q_n(t), p_1(t), \ldots , p_n(t))$ whose component has the following form

$$\begin{aligned} \sum _{k \ge k_0, \ \ell \ge \ell _0} c_{k, \ell } t^{-\frac{\ell }{2\sigma -1}} e^{ \lambda k t}, \end{aligned}$$

(8)

where $k=(k_2, \ldots , k_n)$, $\lambda k = \lambda _2 k_2 + \cdots + \lambda _n k_n$, and where $c_{k, \ell }$’s are complex constants and $k_0$ is a multiinteger and $\ell _0 \ge 0$ is an integer. The series Eq. 8 is called a transseries. As for the general definition of a transseries and its property we refer to Ecalle [5], Costin, [4] and Kuik, [7] and the references therein.

Our first aim is to construct the formal transseries solution of Eq. 1. In [1] a formal transseries solution similar to Eq. 8 is constructed for the first order system of ordinary differential equations under the nonresonance condition. See also the related works [4] and [7]. Now, we have

Theorem 1

Suppose that Eqs. 5, 6 and 7 are satisfied. Then there exists a neighborhood of $t= \infty$, $\varOmega _1$ such that there exists a formal transseries solution $(q_1(t), \ldots , q_n(t), p_1(t), \ldots , p_n(t))$ of Eq. 1 at every point of $\{ t| Re (\lambda _j t ) <0, j=2, \ldots , n \} \cap \varOmega _1$.

Example 1

The following Hamiltonian is the local counterpart of the Hamiltonian studied by Taimanov related with the nonintegrability of a geodesic flow (cf. [3, 6])

$$\begin{aligned} H_1 := c q_1^{4\sigma } p_1^2 + \sum _{j=2}^n B_j(q_1) p_j^2, \end{aligned}$$

(9)

where c is a constant and $B_j(q_1)$ is an analytic function in some neighborhood of $q_1=0$. For $H_0$ in Eq. 3, we define $H:= H_0 + H_1$. $\chi _H$ is $C^\infty$-Liouville integrable at the origin. (cf. [10]). We know that it is not $C^\omega$-Liouville integrable under a certain condition. (cf. [9]).

In this paper we study the superintegrability and the solvability of a Hamiltonian system including H in the category of transseries. In the next section we show that the system is superintegrable in the class of transseries. By virtue of the superintegrability we prove the solvability of a Hamiltonian system in the class of transseries. As for the property of the solution we see from Theorem 1 that the solution has a formal transseries expansion. Moreover, the formal transseries is a true solution in a certain sector since it is summable by Theorem 5 which follows. The study of the global behavior of the summed transseries solution is left as a future problem.

3 Superintegrability in the Category of Transseries

3.1 Formal First Integral and Superintegrability

In order to prove Theorem 1 we first show the superintegrability of Eq. 1 in the category of formal transseries. For this purpose, we first construct $2n-1$ functionally independent first integrals. We recall that the formal power series v is said to be the formal first integral of the Hamiltonian vector field $\chi _H$ if $\chi _H v=0$ as a formal power series. As for the precise definition of functionally independent first integrals in a formal series we need to introduce some notation. For $c\in {\mathbb C}$ and $\alpha = (\alpha _2, \ldots , \alpha _n)\in {\mathbb Z}^{n-1}$, define

$$\begin{aligned} E_c \equiv E_c ( q_1) = \exp \left( \frac{ c q_1^{-2\sigma +1}}{ 2\sigma - 1} \right) , \ E^\alpha = E_{\lambda _2}^{\alpha _2} \cdots E_{\lambda _n}^{\alpha _n}. \end{aligned}$$

(10)

We denote by $e_j$ the j-th unit vector, $e_j =(0, \cdots ,1, \cdots 0)$, $j=2,3,\ldots ,n$.

We construct the first integral v of $\chi _H$ given by

$$\begin{aligned} v = \phi ^{(\alpha )} (q_1,p_1,q,p) E^\alpha , \end{aligned}$$

(11)

where $\phi ^{(\alpha )} (q_1,p_1,q, p)$ is a formal power series in $q_1$, q, $p_1$ and p of the following form.

• i) Case $\alpha = 0$. We have

  $$\begin{aligned} \phi ^{(0)} \equiv \phi ^{(0)}_j = p_j q_j + U_{0,j} + q_1^{2\sigma }p_1 U_{1,j}, \quad j=2, \ldots , n, \end{aligned}$$

  (12)

  where

  $$\begin{aligned} U_{0,j}&= U_{0,j}(q_1,q) = \sum _{\nu =0}^\infty U_{0,j,\nu }(q) q_1^\nu , \end{aligned}$$

  (13)
  $$\begin{aligned} U_{1,j}&= U_{1,j}(q_1,q) = \sum _{\nu =0}^\infty U_{1,j,\nu }(q) q_1^\nu , \end{aligned}$$

  (14)

  are formal power series in $q_1$ with coefficients analytic in q.

• ii) Case $\alpha = e_j$, $(2 \le j \le n)$. We have

  $$\begin{aligned} \phi ^{(e_j)} = p_j q_j^2 ( 1 + U_{2, j}) + U_{0,j} + q_1^{2\sigma }p_1 U_{1,j}, \quad j=2, \ldots , n, \end{aligned}$$

  (15)

  where $U_{0,j}$, $U_{1,j}$ and $U_{2,j}$ are formal power series in $q_1$ with coefficients analytic in q. Here the formal power series $U_{0,j}$ and $U_{1,j}$ may not coincide with the power series in i), respectively.

• iii) Case $\alpha = - e_j$, $(2 \le j \le n)$. We have

  $$\begin{aligned} \phi ^{(-e_j)} = p_j ( 1 + U_{2, j}) + U_{0,j} + q_1^{2\sigma }p_1 U_{1,j}, \quad j=2, \ldots , n, \end{aligned}$$

  (16)

  where $U_{0,j}$, $U_{1,j}$ and $U_{2,j}$ are formal power series in $q_1$ with coefficients analytic in q. Here $U_{0,j}$, $U_{1,j}$ and $U_{2,j}$ may not coincide with the power series in ii), respectively. And $U_{0,j}$ and $U_{1,j}$ may not coincide with the ones in i), respectively.

Definition 1

We say that v in Eq. 11 is the formal first integral of $\chi _H$ if the following conditions are satisfied.

• (i) $\chi _H v=0$ as a formal power series.

• (ii) If $\alpha =0$, then $\phi ^{(0)} \equiv \phi ^{(0)}_j$, $(j=2, \ldots , n)$ satisfies Eqs. 12, 13 and 14 with $U_{0,j,\nu }(q)$’s and $U_{1,j,\nu }(q)$’s analytic in some neighborhood of the origin $q=0$ independent of $\nu$ and j. If $\alpha = e_j$ (resp. $\alpha = -e_j$), $(j=2, \ldots , n)$, then $\phi ^{(\alpha )}$ has the form Eq. 15 (resp. Eq. 16), with $U_{0,j}$’s, $U_{1,j}$’s and $U_{2,j}$’s satisfying the same conditions as the case $\alpha =0$.

We say that the formal series $U_{0,j}$ in Eq. 13 is Gevrey of order s (in short, s- Gevrey), for some $s \ge 0$, if there exist a neighborhood of the origin $q=0$, $\varOmega _0$ and constants $C>0$, $K>0$ for which

$$\begin{aligned} \sup _{q\in \varOmega _0}| U_{0,j,\nu }(q)| \le C K^\nu \Gamma (1 + s\nu ), \end{aligned}$$

hold for all $\nu \ge 0$, where $\Gamma$ denotes the Gamma function. If both $U_{0,j}$ and $U_{1,j}$ are s-Gevrey, then we say that $\phi ^{(0)}_j$ is s-Gevrey. We say that $\phi ^{(e_j)}$ ( resp. $\phi ^{(-e_j)}$) is s-Gevrey if $U_{0,j}$, $U_{1,j}$ and $U_{2,j}$ are s-Gevrey.

Then we have

Theorem 2

Assume Eqs. 5, 6 and 7. Then $\chi _H$ has the formal first integrals, $\phi ^{(0)}_j$, $\phi ^{(e_j)} E^{e_j}$ and $\phi ^{(-e_j)} E^{-e_j}$, $(j=2, \ldots ,n)$, which are $(2\sigma -1)$- Gevrey. If $\alpha =0$, then we have $U_{0,j} = O(|q|^2)$ and $U_{1,j} = O(|q|^2)$ as $q\rightarrow 0$ . If $\alpha =e_j$, then we have $U_{0,j} = O(|q|^3)$, $U_{1,j} = O(|q|^3)$ and $U_{2,j} = O(|q|^2)$ as $q\rightarrow 0$. If $\alpha = -e_j$, then we have $U_{0,j} = O(|q|^2)$, $U_{1,j} = O(|q|^2)$ and $U_{2,j} = O(|q|)$ as $q\rightarrow 0$.

Consider the system of formal first integrals $v_j^{(\alpha )} = v_{j,0}^{(\alpha )} E^\alpha + \tilde{v_j^{(\alpha )}} E^\alpha$, H $( \alpha =0, - e_j; j=2, \ldots , n-1 )$ with $v_{j,0}^{(\alpha )}$ being analytic and independent of $q_1$ and $\tilde{v_j^{(\alpha )}} =O(q_1)$. We say that the system of functions H, $v_j^{(\alpha )}$ $( \alpha =0, - e_j; j=2, \ldots , n-1 )$ is functionally independent if the Jacobian of the functions $v_{j,0}^{(\alpha )} E^\alpha$ $(\alpha =0, - e_j; j=2, \ldots , n-1)$, $H|_{q_1=0}$ does not vanish.

In order to prove Theorem 2 we introduce some notation. By definition we have, for $\mathcal {L}:= \{ H_0, \cdot \}$ and $R:= \{ H_1, \cdot \}$,

$$\begin{aligned} \mathcal {L} = q_1^{2\sigma } \frac{\partial }{\partial q_1} - 2\sigma q_1^{2\sigma -1}p_1\frac{\partial }{\partial p_1} + \sum _{j=2}^n \lambda _j \left( q_j \frac{\partial }{\partial q_j} - p_j \frac{\partial }{\partial p_j} \right) , \end{aligned}$$

(17)

$$\begin{aligned} R = \sum _{j=2}^n \left( - 2 q_j B_j \frac{\partial }{\partial p_j} + q_j^2 ( \partial _{p_1} B_j) \frac{\partial }{\partial q_1} - q_j^2 ( \partial _{q_1} B_j) \frac{\partial }{\partial p_1} - q_j^2 \nabla _{q} B_j \cdot \frac{\partial }{\partial p} \right) . \end{aligned}$$

(18)

By using the formula

$$\begin{aligned} \partial _{p_1} B_j = B_{j,1} q_1^{2\sigma }, \quad q_1^{2\sigma } (\partial / \partial q_1) E^\alpha = - (\sum _{j=2}^n \lambda _j \alpha _j ) E^\alpha = -\langle \lambda ,\alpha \rangle E^\alpha , \end{aligned}$$

we have

$$\begin{aligned} \mathcal {L} (\phi ^{(\alpha )} E^\alpha )= & {} E^\alpha \left( q_1^{2\sigma } \frac{\partial }{ \partial q_1} -2\sigma q_1^{2\sigma -1} p_1 \frac{\partial }{ \partial p_1}\right. \\+ & {} \left. \sum _{j=2}^n \lambda _j \left( q_j \frac{\partial }{ \partial q_j} - p_j \frac{\partial }{ \partial p_j} -\alpha _j \right) \right) \phi ^{(\alpha )} , \nonumber \end{aligned}$$

(19)

and

$$\begin{aligned} R ( \phi ^{(\alpha )} E^\alpha ) = E^\alpha \left( - \langle \lambda ,\alpha \rangle \sum _{j=2}^n q_j^2 B_{j,1} + R \right) \phi ^{(\alpha )} . \end{aligned}$$

(20)

It follows that if $v=E^\alpha \phi ^{(\alpha )}$ is the first integral of $\chi _H$, then $\phi ^{(\alpha )}$ satisfies

$$\begin{aligned}{} & {} \left( q_1^{2\sigma } \frac{\partial }{ \partial q_1} -2\sigma q_1^{2\sigma -1} p_1 \frac{\partial }{ \partial p_1} + \sum _{j=2}^n \lambda _j \left( q_j \frac{\partial }{ \partial q_j} - p_j \frac{\partial }{ \partial p_j} -\alpha _j \right) \right) \phi ^{(\alpha )} \nonumber \\+ & {} \left( - \sum _{j=2}^n \langle \lambda ,\alpha \rangle q_j^2 B_{j,1} + R \right) \phi ^{(\alpha )} =0. \end{aligned}$$

(21)

Let $\alpha =0$. We look for the equations which $U_{0,j}$ and $U_{1,j}$ in Eq. 12 satisfy. Substitute Eqs. 12 and 5 into Eq. 21. For the sake of simplicity, we set $U_0 := U_{0,j}$ and $U_1 := U_{1,j}$. We consider the terms which appear from $p_j q_j$ in $\phi ^{(\alpha )}$. The nonvanishing terms appearing from $R (p_j q_j)$ are given by

$$\begin{aligned} -2 q_j^2 B_j - \sum _{\nu =2}^n q_\nu ^2 (\partial _{q_j} B_\nu ) q_j = -2 q_j^2 ( B_{j,0} + q_1^{2\sigma } p_1 B_{j,1}) - \sum _{\nu =2}^n q_\nu ^2 q_j \partial _{q_j}(B_{\nu ,0} + q_1^{2\sigma } p_1 B_{\nu ,1} ). \end{aligned}$$

(22)

Consider the terms which contain $p_1$. The terms containing $p_1$ which appear from the differentiation $\mathcal {L}$ in Eq. 21 are given by

$$\begin{aligned} q_1^{2\sigma } p_1 \left( q_1^{2\sigma }\frac{\partial U_1}{\partial q_1} + 2\sigma q_1^{2\sigma -1}U_1 - 2\sigma q_1^{2\sigma -1}U_1 + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial U_1}{\partial q_\nu } \right) . \end{aligned}$$

(23)

Next, the terms containing $p_1$ that appear from R are given by the second and the third terms of R:

$$\begin{aligned}&- q_1^{2\sigma } p_1 \left( \sum _{\nu =2}^n q_\nu ^2 q_1^{2\sigma } ( \partial _{q_1} B_{\nu ,1})U_1 + \sum _{\nu =2}^n q_\nu ^2 2\sigma q_1^{2\sigma -1}B_{\nu ,1}U_1 \right) \nonumber \\&+ q_1^{2\sigma } p_1 \left( \sum _{\nu =2}^n q_\nu ^2 2\sigma q_1^{2\sigma -1}B_{\nu ,1}U_1 + \sum _\nu q_\nu ^2 B_{\nu ,1} q_1^{2\sigma } \frac{\partial U_1}{\partial q_1} \right) \nonumber \\&= q_1^{2\sigma } p_1 \left( - \sum _{\nu =2}^n q_\nu ^2 q_1^{2\sigma } ( \partial _{q_1} B_{\nu ,1})U_1 + \sum _\nu q_\nu ^2 B_{\nu ,1} q_1^{2\sigma } \frac{\partial U_1}{\partial q_1} \right) . \end{aligned}$$

(24)

These expressions give the terms which contain $p_1$.

Next, we calculate the terms which do not contain $p_1$ in Eq. 21 in a similar way. By substituting Eq. 12 and 5 into Eq. 21 we have

$$\begin{aligned}&q_1^{2\sigma } p_1 \left( q_1^{2\sigma } \frac{\partial U_1}{\partial q_1} + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial U_1}{\partial q_\nu } - \sum _{\nu =2}^n q_\nu ^2 q_1^{2\sigma } ( \partial _{q_1} B_{\nu ,1})U_1 \right. \nonumber \\&\left. + \sum _\nu q_\nu ^2 B_{\nu ,1} q_1^{2\sigma } \frac{\partial U_1}{\partial q_1} - 2 q_j^2 B_{j,1} - \sum _\nu q_\nu ^2 q_j \partial _{q_j} B_{\nu ,1} \right) \nonumber \\&+ q_1^{2\sigma }\frac{\partial U_0}{\partial q_1} + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial U_0}{\partial q_\nu } - \sum _\nu q_\nu ^2 (\partial _{q_1} B_{\nu ,0}) q_1^{2\sigma } U_1 \nonumber \\&+ \sum _\nu q_\nu ^2 q_1^{2\sigma } B_{\nu ,1} \frac{\partial U_0}{\partial q_1} - 2 q_j^2 B_{j,0} - \sum _\nu q_\nu ^2 (q_j \partial _{q_j}B_{\nu ,0} ) =0. \end{aligned}$$

(25)

Hence we have

$$\begin{aligned}&q_1^{2\sigma }\frac{\partial U_1}{\partial q_1} + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial U_1}{\partial q_\nu } - \sum _{\nu =2}^n q_\nu ^2 q_1^{2\sigma } ( \partial _{q_1} B_{\nu ,1})U_1 \nonumber \\&+ \sum _\nu q_\nu ^2 B_{\nu ,1} q_1^{2\sigma } \frac{\partial U_1}{\partial q_1} - 2 q_j^2 B_{j,1} - \sum _\nu q_\nu ^2 q_j \partial _{q_j} B_{\nu ,1} =0. \end{aligned}$$

(26)

$$\begin{aligned}&q_1^{2\sigma }\frac{\partial U_0}{\partial q_1} + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial U_0}{\partial q_\nu } + \sum _\nu q_\nu ^2 q_1^{2\sigma } B_{\nu ,1} \frac{\partial U_0}{\partial q_1} \nonumber \\&- 2 q_j^2 B_{j,0} - \sum _\nu q_\nu ^2 (q_j \partial _{q_j}B_{\nu ,0} ) - \sum _\nu q_\nu ^2 (\partial _{q_1} B_{\nu ,0}) q_1^{2\sigma } U_1 = 0. \end{aligned}$$

(27)

We next consider the case $\alpha = - e_j$. The calculation is the same as for the case $\alpha =0$. Set $U_0 := U_{0,j}$, $U_1 := U_{1,j}$ and $U_2 := U_{2,j}$. Then we have

$$\begin{aligned}&q_1^{2\sigma }\frac{\partial U_2}{\partial q_1} + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial U_2}{\partial q_\nu } + \lambda _j q_j^2 B_{j,1} U_2 \nonumber \\&+ \sum _\nu q_\nu ^2 B_{\nu ,1} q_1^{2\sigma } \frac{\partial U_2}{\partial q_1} + \lambda _j q_j^2 B_{j,1} = 0 . \end{aligned}$$

(28)

$$\begin{aligned}&q_1^{2\sigma }\frac{\partial U_1}{\partial q_1} + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial U_1}{\partial q_\nu } + \lambda _j U_1 - \sum _{\nu =2}^n q_\nu ^2 q_1^{2\sigma } (\partial _{q_1} B_{\nu ,1}) U_1 + \lambda _j q_j^2 B_{j,1} U_1 \nonumber \\&- 2 q_j B_{j,1} U_2 - q_j^2 (\partial _{q_j} B_{j,1}) U_2 - \sum _\nu q_\nu ^2 q_j^2 (\partial _{q_j} B_{\nu ,1}) - 2 q_j^2 B_{j,1} = 0. \end{aligned}$$

(29)

$$\begin{aligned}&q_1^{2\sigma }\frac{\partial U_0}{\partial q_1} + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial U_0}{\partial q_\nu } + \lambda _j U_0 + \sum _{\nu =2}^n q_\nu ^2 q_1^{2\sigma } B_{\nu ,1}\frac{\partial U_0}{\partial q_1} + \lambda _j q_j^2 B_{j,1} U_0 \nonumber \\&- \sum _{\nu =2}^n q_\nu ^2 q_1^{2\sigma } (\partial _{q_1} B_{\nu ,0})U_1 - 2 q_j B_{j,0} U_2 \nonumber \\&- \sum _{\nu =2}^n q_\nu ^2 (\partial _{q_j} B_{\nu ,0})U_2 - \sum _{\nu =2}^n q_\nu ^2 q_j (\partial _{q_j} B_{\nu ,0}) - 2 q_j^2 B_{j,0} = 0. \end{aligned}$$

(30)

Next we consider the case $\alpha =e_j$. Set $U_0 := U_{0,j}$, $U_1 := U_{1,j}$ and $U_2 := U_{2,j}$. Then we have

$$\begin{aligned} q_1^{2\sigma } \frac{\partial U_2}{\partial q_1} + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial U_2}{\partial q_\nu } - \lambda _j q_j^2 B_{j,1} U_2 - \lambda _j q_j^2 B_{j,1} + \sum _{\nu =2}^n q_\nu ^2 q_1^{2\sigma } B_{\nu ,1}\frac{\partial U_2}{\partial q_1} = 0. \end{aligned}$$

(31)

$$\begin{aligned}&q_1^{2\sigma }\frac{\partial U_1}{\partial q_1} + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial U_1}{\partial q_\nu } - \lambda _j U_1 + \sum _{\nu =2}^n q_\nu ^2 q_1^{2\sigma } B_{\nu ,1}\frac{\partial U_1}{\partial q_1} - \lambda _j q_j^2 B_{j,1} U_1 \nonumber \\&- \sum _{\nu =2}^n q_\nu ^2 q_1^{2\sigma } (\partial _{q_1} B_{\nu ,1})U_1 - 2 q_j^3 B_{j,1}U_2 \nonumber \\&- \sum _{\nu =2}^n q_\nu ^2 q_j^2 (\partial _{q_j} B_{\nu ,1})U_2 - \sum _{\nu =2}^n q_\nu ^2 q_j^2 (\partial _{q_j} B_{\nu ,1}) - 2 q_j^3 B_{j,1} = 0. \end{aligned}$$

(32)

$$\begin{aligned}&q_1^{2\sigma }\frac{\partial U_0}{\partial q_1} + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial U_0}{\partial q_\nu } - \lambda _j U_0 + \sum _{\nu =2}^n q_\nu ^2 q_1^{2\sigma } B_{\nu ,1}\frac{\partial U_0}{\partial q_1} - \lambda _j q_j^2 B_{j,1} U_0 \nonumber \\&- 2 q_j^3 B_{j,0} U_2 - \sum _{\nu =2}^n q_\nu ^2 q_1^{2\sigma } (\partial _{q_1} B_{\nu ,0}) U_1 \nonumber \\&- \sum _{\nu =2}^n q_\nu ^2 q_j^2 (\partial _{q_j} B_{\nu ,0})U_2 - \sum _{\nu =2}^n q_\nu ^2 q_j^2 (\partial _{q_j} B_{\nu ,0}) - 2 q_j^3 B_{j,0} = 0. \end{aligned}$$

(33)

In order to prove Theorem 2 we prepare a lemma. Let $R_j>0$ $(j=2,\ldots ,n)$ be given. Define $V_0 := \prod _{j=2}^n \{ z_j \ | \ |z_j|<R_j \}$. Let $\mathcal{O}(V_0)$ be the set of holomorphic functions in $V_0$ continuous up to the boundary. Define $M_0(q) := \prod _{j=2}^n (R_j - |q_j|)$. For $f\in \mathcal{O}(V_0)$ we define the norm $\Vert f \Vert$ and the weighted norm $\Vert | f |\Vert$ by $\Vert f \Vert := \sup _{q\in V_0} |f(q)|$ and $\Vert |f|\Vert := \sup _{q\in V_0} |f(q) M_0(q)|$, respectively. Equipped with the norm $\Vert |\cdot | \Vert$, the space $\mathcal{O}(V_0)$ is a Banach space.

Let $\lambda :=(\lambda _2, \ldots , \lambda _n)$ and $\alpha = (\alpha _2, \ldots , \alpha _n)$. We consider the equation

$$\begin{aligned} Lu \equiv \left( \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial }{\partial q_\nu } - \lambda \cdot \alpha \right) u = f \in \mathcal{O}(V_0), \quad f = O(|q|). \end{aligned}$$

(34)

Then we have

Lemma 1

Let $\alpha = 0, \pm e_j, j=2, \ldots , n$. Assume Eqs. 6 and 7. Then there exists a constant $K>0$ such that, for every $f \in \mathcal{O}(V_0)$ with $f = O(|q|)$ there exist a unique holomorphic solution u of Eq. 34 in $\mathcal{O}(V_0)$ such that $\Vert |u|\Vert \le K \Vert f \Vert$.

Proof

Expand $f(q) =\sum _{\eta \ne 0}f_\eta q^\eta$. Then the unique analytic solution of Eq. 34 is given by $u(q) = \sum _{\eta \ne 0}f_\eta q^\eta (\lambda \cdot (\eta - \alpha ))^{-1}$. Take $0< R_j'<R_j$ for $j=2,\ldots ,n$. Let $|q_j|<R_j'<R_j$. By Cauchy’s theorem we have

$$\begin{aligned} u(q) = \sum _{\eta \ne 0} q^\eta (\lambda \cdot (\eta - \alpha ))^{-1} (2\pi i)^{-n+1} \int _{|z_j|= R_j'} \frac{ f(z)}{ z^{\eta +e}} dz, \end{aligned}$$

(35)

where $e=(1,\ldots ,1)$. By Eqs. 6 and 7 there exists $K_1>0$ independent of $\eta$ such that $|\lambda \cdot (\eta - \alpha )|^{-1} \le K_1$ for all $\eta$. Hence |u(q)| is estimated by

$$\begin{aligned}&K_1 (2\pi )^{-n+1} \Vert f\Vert \sum _{\eta \ne 0} \prod _j \left( (|q_j|/R_j')^{\eta _j}2\pi \right) \nonumber \\&\le K_1 \Vert f\Vert \prod _j \frac{1}{1 - |q_j|/R_j'}\nonumber \\&\le K_1 \Vert f\Vert \prod _j \frac{R_j'}{R_j' - |q_j|} \le K_1 \Vert f\Vert (\prod _j R_j) \prod _j \frac{1}{R_j' - |q_j|}. \end{aligned}$$

(36)

We let $R_j'$ with $R_j' < R_j$ tend to $R_j$. Then $|u(q)| M_0(q)$ is estimated by $K_1 \Vert f\Vert (\prod _j R_j)$.

3.2 Proof of Theorem 2

Proof

(Proof of Theorem 2) Let $\kappa =2\sigma -1$. We show that $\phi ^{(\alpha )}$ with $\alpha =0$ is a formal first integral having the $\kappa ^{-1}$-Gevrey estimate. Consider $\sum _\nu q_\nu \partial _{q_\nu } u =g$. By Lemma 1 we have

$$\begin{aligned} \Vert u \Vert \le \Vert M_0^{-1} M_0 u\Vert \le C_2 \Vert |u|\Vert \le C_2 K \Vert g\Vert , \end{aligned}$$

where $C_2 = \sup _q |M_0(q)^{-1}|$.

Consider Eq. 12. For simplicity we denote $U_{0,j}$ and $U_{1,j}$ by $U_0$ and $U_1$, respectively. Determine $U_1$ by Eq. 26 and define

$$\begin{aligned} f(q_1,q):= - 2 q_j^2 B_{j,1} - \sum _\nu q_\nu ^2 q_j \partial _{q_j} B_{\nu ,1}. \end{aligned}$$

Expand the functions $f(q_1,q)$, $U_1$, $q_1^{2\sigma }\sum q_\nu ^2 (\partial _{q_1} B_{\nu ,1})$ and $q_1^{2\sigma -1} \sum q_\nu ^2 B_{\nu ,1}$ in the power series of $q_1$

$$\begin{aligned}&f(q_1,q) = \sum _{\ell =0}^\infty f_\ell (q) q_1^\ell , \ U_1 = \sum _\ell u_\ell (q) q_1^\ell , \end{aligned}$$

(37)

$$\begin{aligned}&q_1^{2\sigma }\sum _{\nu =2}^n q_\nu ^2 (\partial _{q_1} B_{\nu ,1}) = \sum _{\ell \ge 2\sigma } a_\ell (q)q_1^\ell , \end{aligned}$$

(38)

$$\begin{aligned}&q_1^{2\sigma -1} \sum _{\nu =2}^n q_\nu ^2 B_{\nu ,1} = \sum _{\ell \ge 2\sigma -1} b_\ell (q)q_1^\ell . \end{aligned}$$

(39)

By substituting the expansions Eqs. 37, 38, 39 into 26 and by comparing the power of $q_1^\ell$ we have

$$\begin{aligned} (\ell -\kappa ) u_{\ell -\kappa } + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{ \partial u_\ell }{\partial q_\nu } + \sum _{m+k=\ell , m \ge 2\sigma } a_m u_k + \sum _{m+k=\ell , m \ge \kappa } k b_m u_k = f_\ell . \end{aligned}$$

(40)

Hence, we can easily show that $U_1$ has the $\kappa ^{-1}$-Gevrey estimate. We can similarly show that $U_0$ has the $\kappa ^{-1}$-Gevrey estimate. Hence $\phi ^{(\alpha )}$ with $\alpha =0$ is a formal first integral with the $\kappa ^{-1}$-Gevrey estimate. We can similarly show that $\phi ^{(\alpha )}$’s with $\alpha = \pm e_j$ $(j=2, \ldots , n)$ have the same property.

4 Proof of Theorem 1

4.1 Preparatory Lemma

Define

$$\begin{aligned} \mathcal {C}:= \left\{ z\in {\mathbb C} | \, Re \, (\overline{\lambda _j}z^{2\sigma -1})>0 \quad j=2,\ldots ,n \right\} . \end{aligned}$$

(41)

Let $\phi _j^{(0)}$ and $\phi ^{(-e_j)}E^{-e_j}$ $(j=2,\ldots ,n)$ be the formal first integrals given by Theorem 2 with $q_1=z$.

Let $C_j$, $\tilde{C_j}$ and $C_0$ be constants. For $z \in \mathcal {C}$, we solve the system of equations for q, p, $p_1$

$$\begin{aligned} \phi _j^{(0)} = C_j, \ \phi ^{(-e_j)}E^{-e_j} = \tilde{C_j}, \ H= C_0, \quad j=2,\ldots ,n , \end{aligned}$$

(42)

where $H=H_0+H_1$ is given by

$$\begin{aligned} H = z^{2\sigma }p_1 + \sum _{j=2}^n \lambda _j q_j p_j + \sum _{j=2}^n q_j^2 B_j(z,z^{2\sigma }p_1, q). \end{aligned}$$

(43)

Here the unknown quantities are

$$\begin{aligned} q=q(z,T), \ p = p(z,T), \ p_1 = p_1(z,T), \end{aligned}$$

(44)

where

$$\begin{aligned} q = \sum _{n=0}^\infty c_n z^n, c_n=c_n(T^{-1}), \quad T = (T_j)_j, \quad T_j = \tilde{C_j}E^{e_j}, \end{aligned}$$

(45)

is a formal series in z with $c_n(T^{-1})$ convergent in T in some neighborhood of $T=\infty$. The Taylor series of p has the same form as q. (cf. Eq. 54 which follows). Concerning $p_1$ we have

$$\begin{aligned} p_1 z^{2\sigma } = \sum _{n=0}^\infty \rho _n z^n, \rho _n= \rho _n(T^{-1}), \end{aligned}$$

(46)

with $\rho _n(T^{-1})$ convergent in T in some neighborhood of $T=\infty$. By Eqs. 12 and 16 we have

$$\begin{aligned} p_j q_j + \tilde{A_j}(z,z^{2\sigma }p_1, q )&= C_j, \quad j=2,\ldots ,n, \end{aligned}$$

(47)

$$\begin{aligned} p_j (1 + D_j(z,q)) + \tilde{D_j}(z,z^{2\sigma }p_1, q )&= T_j,\quad j=2,\ldots ,n, \end{aligned}$$

(48)

$$\begin{aligned} H&= C_0. \end{aligned}$$

(49)

Then we have

Lemma 2

Assume Eq. 6. Then Eqs. 47, 48 and 49 has the formal solution $(q,p,p_1)$ for $z \in \mathcal {C}$ given by Eqs. 44, 45 and 46.

Proof

We prove the proposition in two steps. By Theorem 2 we have $\tilde{A}_j = \tilde{A}_j(z,z^{2\sigma }p_1, q ) = O(|q|^2)$, $\tilde{D}_j = \tilde{D}_j(z,z^{2\sigma }p_1, q ) = O(|q|)$ and $D_j(z,q)= O(|q|^2)$. Moreover, $\tilde{A}_j$ and $\tilde{D}_j$ are the polynomials of $p_1 z^{2\sigma }$ of degree 1 whose coefficients are independent of p.

Step 1. Since $E_\lambda (z) = \exp ( \frac{\lambda z^{-2\sigma +1}}{2\sigma -1})$ and

$$\begin{aligned} Re\,(\lambda z^{-2\sigma +1})= Re\,(\overline{\lambda }\overline{z}^{-2\sigma +1} ) = Re\, (\overline{\lambda } z^{2\sigma -1}/|z|^{2(2\sigma -1)}), \end{aligned}$$

we see that $z\in \mathcal {C}$ if and only if $Re\,(\lambda _j z^{-2\sigma +1})>0$ for $j=2,\ldots ,n$. Hence, we have $T \rightarrow \infty$ when $z\rightarrow 0$ in $z\in \mathcal {C}$. Set

$$\begin{aligned} D_j = \sum _k q_k D_{j,0,k}(q) + z \tilde{D}_{j,1}(z,q), \end{aligned}$$

(50)

where $D_{j,0,k}(q)$ is analytic at $q=0$ and where $\tilde{D}_{j,1}(z,q)$ is a formal power series in z with coefficients analytic in q. Then $( 1 + z \tilde{D}_{j,1})^{-1}$ exists as a formal power series.

By Eq. 48 we have

$$\begin{aligned} p_j \sum _k q_k D_{j,0,k}(q) + p_j (1 + z \tilde{D}_{j,1}) + \tilde{D}_j = T_j . \end{aligned}$$

(51)

Since $(1 + z \tilde{D}_{j,1})^{-1}$ exists as a formal power series, we have

$$\begin{aligned} p_j \sum _k q_k D_{j,0,k}(q) (1 + z \tilde{D}_{j,1})^{-1} + p_j = T_j (1 + z \tilde{D}_{j,1})^{-1} - \tilde{D}_j(1 + z \tilde{D}_{j,1})^{-1}. \end{aligned}$$

(52)

We decompose

$$\begin{aligned} p_j \sum _k q_k D_{j,0,k}(q)(1 + z \tilde{D}_{j,1})^{-1} = E_{j,0} + E_{j,1}, \end{aligned}$$

(53)

where $E_{j,0}$ either vanishes identically or is unbounded when $T_j \rightarrow \infty$ and $E_{j,1}=O(1)$ when $T_j \rightarrow \infty$. Indeed, by Eqs. 45 and 46 we have $q_j =O(1)$ and $z^{2\sigma }p_1=O(1)$. Hence $p_j q_j = O(1)$ by Eq. 47. By Eq. 48 we have $p_j\rightarrow \infty$. It follows that $q_j \rightarrow 0$. Hence we have Eq. 53. Define

$$\begin{aligned} p_j = \tilde{p_j} + \tilde{T}_j, \quad \tilde{T}_j := T_j (1 + z \tilde{D}_{j,1})^{-1} - E_{j,0}. \end{aligned}$$

(54)

Then Eq. 52 is written in

$$\begin{aligned} \tilde{p_j} = - E_{j,1} - \tilde{D}_j(1 + z \tilde{D}_{j,1})^{-1} =:F_{j,0}. \end{aligned}$$

(55)

Next, let $\tilde{A}_j(z,z^{2\sigma }p_1, q) = \tilde{A}_{j,0}(z, q) + z^{2\sigma }p_1 \tilde{A}_{j,1}(z, q)$. Substitute Eqs. 47 into 49. Then we have

$$\begin{aligned}&C_0 - \sum \lambda _\nu C_\nu + \sum \lambda _\nu \tilde{A}_{\nu ,0}(z, q) + z^{2\sigma }p_1 \sum \lambda _\nu \tilde{A}_{\nu ,1}(z, q)\nonumber \\&= z^{2\sigma }p_1 + \sum _\nu q_\nu ^2 ( B_{\nu ,0} + z^{2\sigma }p_1 B_{\nu ,1} ). \end{aligned}$$

(56)

Hence we have

$$\begin{aligned} z^{2\sigma }p_1 (1 + \sum q_\nu ^2 B_{\nu ,1} - \sum \lambda _\nu \tilde{A}_{\nu ,1}) = \sum \lambda _\nu \tilde{A}_{\nu ,0} - \sum q_\nu ^2 B_{\nu ,0} + C_0 - \sum \lambda _\nu C_\nu . \end{aligned}$$

(57)

If $c_0=q(0)$ is sufficiently small, then $(1 + \sum q_\nu ^2 B_1 - \sum \lambda _\nu \tilde{A}_{\nu ,1})^{-1}$ exists as a formal series. Hence we have

$$\begin{aligned}&\qquad z^{2\sigma }p_1 = \nonumber \\&\left( \sum \lambda _\nu \tilde{A}_{\nu ,0} - \sum q_\nu ^2 B_{\nu ,0} + C_0 - \sum \lambda _\nu C_\nu \right) \left( 1 + \sum q_\nu ^2 B_1 - \sum \lambda _\nu \tilde{A}_{\nu ,1}\right) ^{-1} . \end{aligned}$$

(58)

The right-hand side of Eq. 58 is a formal power series in z whose coefficients are analytic in q at $q=0$. Note that, if q is a formal power series in z with a sufficiently small constant term $c_0$, then the right-hand side of Eq. 58 is a formal power series in z as well. We also note that, by substituting $z^{2\sigma }p_1$ in $F_{j,0}$, Eq. 55, $F_{j,0}$ is a formal power series in z with coefficients analytic in q. If q is a formal power series in z whose constant term $c_0$ is sufficiently small, then $F_{j,0}$ is also a formal power series in z.

We look for the equation of q. By Eq. 55 we have $p_j = \tilde{T}_j + \tilde{p}_j = \tilde{T}_j + F_{j,0}(z, q)$. $(\tilde{T}_j + F_{j,0})^{-1}$ exists as a formal series in z with coefficients analytic in q at $q=0$. By Eq. 47 we have

$$\begin{aligned} q_j = (\tilde{T}_j + F_{j,0})^{-1} C_j - (\tilde{T}_j + F_{j,0})^{-1} \tilde{A}_j(z,z^{2}p_1, q) =:G_j. \end{aligned}$$

(59)

Substitute $z^{2}p_1$ in Eq. 58 into $G_j$. Then $G_j$ is a formal power series in z whose coefficients are analytic in q and $\xi =(\xi _j)_j$, $\xi _j:=T_j^{-1}$ at $q = 0$ and $\xi = 0$, respectively. Indeed, by the form of $G_j$ in Eq. 59 we consider $(\tilde{T}_j + F_{j,0})^{-1}$. Since $z \tilde{D}_{j,1} = z O(|q|^2)$ is small, we have $1 + z \tilde{D}_{j,1} \ne 0$. Hence we have

$$\begin{aligned} (\tilde{T}_j + F_{j,0})^{-1}&= T_j^{-1} \left( (1 + z \tilde{D}_{j,1})^{-1} - T_j^{-1} E_{j,0} + T_j^{-1} F_{j,0} \right) ^{-1}\\&= \xi _j \left( (1 + z \tilde{D}_{j,1})^{-1} - \xi _j E_{j,0} + \xi _j F_{j,0} \right) ^{-1}. \nonumber \end{aligned}$$

(60)

The right-hand side of Eq. 60 is holomorphic in $\xi _j$ in some neighborhood of $\xi _j = 0$. On the other hand, the second term of $G_j$, $- (\tilde{T}_j + F_{j,0})^{-1} \tilde{A}_j(z, z^{2\sigma } p_1, q)$ is holomorphic in $\xi _j$. Hence $G_j$ is holomorphic in $\xi _j$ at $\xi _j =0$. Similarly, we see that $G_j$ is holomorphic in q at $q = 0$.

Note that, if we substitute q with a formal power series in z whose constant term $c_0$ is sufficiently small, then $G_j$ is a formal power series in z. Therefore we consider

$$\begin{aligned} q_j = G_j(z,q, \xi ), \quad j=2,\ldots ,n. \end{aligned}$$

(61)

Step 2. Let $q_j = \sum _{n=0}^\infty c_{j,n} z^n$, $c_{j,n} = c_{j,n}(\xi )$ be the Taylor series of $q_j$. Compare the constant term of both sides of Eq. 61. We have

$$\begin{aligned} c_{j,0} = G_j(0,c_0, \xi ). \end{aligned}$$

(62)

In order to estimate $G_j(0,c_0, \xi )$ by Eq. 59 we consider $F_{j,0}$. By Eqs. 55, 43 and 58 and $D_{j,0}(q_2)=O(1)$ we see that $F_{j,0} = O(1)$. On the other hand, the constant part of $\tilde{A}_j$ is $O(c_0^2)$. The second term of $G_j(0,c_0, \xi )$ in the expression Eq. 59 is $O(c_0^2 \xi )$. Hence, by Eq. 59 we have $G_j(0,c_0, \xi ) = O(\xi ) + O(c_0^2 \xi )$. By the implicit function theorem there exists a solution $c_0$ being arbitrarily small if $T_j$ is sufficiently large.

Suppose that the coefficient $c_\nu (\xi )$’s are determined for $\nu =0, \ldots , n-1$. Then the coefficient of $c_n(\xi )$ of the Taylor series of q is determined by the relation

$$\begin{aligned} c_n = (G_j)_{q_2}(0,c_0, \xi )c_n + \tilde{R}_j, \end{aligned}$$

(63)

where $\tilde{R}_j$ is the function of $c_\nu$ with $\nu \le n-1$. Since $(G_j)_{q_2}(0,c_0, \xi )$ is arbitrarily small if $c_0$ and $\xi$ are sufficiently small, we determine $c_n$ if $c_0$ is sufficiently small. The $\kappa$- Gevrey estimate of q is shown by differentiating Eq. 61 and by recalling that $G_j$ is $\kappa$- Gevrey series with respect to z. This ends the proof.

Remark 1

In Eq. 62 the value $c_{0,j}$ depends on T. We can verify that $c_{0,j} T_j$ does not vanish for some j as $T_j$ tends to infinity.

4.2 Construction of Formal Transseries Solution

Let z satisfy $\dot{z} = z^{2\sigma }$. Namely

$$\begin{aligned} t = - \frac{z^{1- 2\sigma }}{2\sigma -1} . \end{aligned}$$

(64)

Let $q \equiv q(z)$, $p_1 \equiv p_1(z)$ and $p \equiv p(z)$ be the formal series given by Lemma 2. From Eq. 64 one can deduce the transseries representations of $q = q(z)$, $p_1 = p_1(z)$ and $p = p(z)$ in the t-variable, by changing the variable in a transseries given by Lemma 2. The exponential part is given by $e^{\lambda k t}$ for $k\ge -1$. For the sake of simplicity, we write the transseries with the same letter $q \equiv q(t)$, $p_1 \equiv p_1(t)$ and $p \equiv p(t)$. Then we have

Theorem 3

Suppose that Eqs. 5, 6 and 7 are satisfied. Then there exists a neighborhood of $t=\infty$, $\varOmega _1$ such that there exists a formal transseries solution $q_1(t)$ of $\dot{q}_1 = H_{p_1}$ at every point of $\{ t | Re (\lambda _j t ) <0, \ j=2,\ldots , n \} \cap \varOmega _1$ for which $(q_1(t), q(t), p_1(t), p(t))$ is the formal transseries solution of Eq. 1.

Theorem 1 follows from Theorem 3. Set $\lambda =(\lambda _2, \ldots , \lambda _n)$ and

$Re \, \lambda =(Re \, \lambda _2, \ldots , Re \, \lambda _n)$. Then we have

Lemma 3

Assume Eq. 6. Then there exists a real number $\theta$ such that

$$\begin{aligned} Re\, (e^{i\theta } \lambda _j) >0, \quad j=2,\ldots ,n, \end{aligned}$$

(65)

and, for every pair $\alpha , \beta \in {\mathbb Z}_+^{n-1}\setminus 0$ we have $\alpha \ne \beta$ if and only if

$$\begin{aligned} \alpha \cdot Re\,(e^{i\theta } \lambda ) \ne \beta \cdot Re\,(e^{i\theta } \lambda ). \end{aligned}$$

(66)

Moreover, $\alpha \cdot Re\,(e^{i\theta } \lambda )$ does not accumulate on a finite value when $|\alpha |\rightarrow \infty$, $\alpha \in {\mathbb Z}_+^{n-1}\setminus 0$.

Proof

The last statement follows from Eq. 65 easily. We prove Eq. 66. Set $m =\alpha - \beta$ and consider $m\cdot Re\,(e^{i\theta } \lambda ) \ne 0$, namely $Re\, (e^{i\theta } m\cdot \lambda ) \ne 0$. The last condition is equivalent to $\theta \ne - \arg (m\cdot \lambda ) \pm \pi /2$ for $m \in {\mathbb Z}^{n-1}\setminus 0$ and $m\ne 0$. The condition holds for some sufficiently small real number $\theta$.

Multiplying the equation with $e^{i\theta }$, if necessary, we assume that Eq. 66 with $\theta = 0$ holds in the following.

Proof

(Proof of Theorem 3) The proof is done in four steps.

Step 1. By definition we have

$$\begin{aligned} H_{p_1} = q_1^{2\sigma } + q_1^{2\sigma } \sum _{\nu =2}^n q_\nu ^2 B_{\nu ,1}(q_1,q). \end{aligned}$$

(67)

We construct the solution $q_1$ of $\dot{q}_1 = H_{p_1}$ in the form

$$\begin{aligned} q_1 = \sum _{m\ge 0, n \ge 0} a_{m,n} t^{- n/(2\sigma -1)} e^{\lambda m t} = \sum _{m \ge 0} A_m(t) e^{\lambda m t}. \end{aligned}$$

(68)

We note

$$\begin{aligned} \frac{d}{dt}\left( A_m(t) e^{\lambda m t} \right) = e^{\lambda m t} \left( \frac{d}{dt} A_m + \lambda m A_m \right) . \end{aligned}$$

(69)

Insert Eq. 68 into $\dot{q}_1 = H_{p_1}$ and compare the terms with $m=0$. Note also that the expansion of q does not contain the terms with $m=0$. Hence we have

$$\begin{aligned} \frac{d}{dt} A_0 = A_0^{2\sigma } . \end{aligned}$$

(70)

By Eq. 70 we have

$$\begin{aligned} A_0 = (1 -2\sigma )^{-1/(2\sigma -1)} t^{-1/(2\sigma -1)}. \end{aligned}$$

(71)

By Lemma 3 we line up all multiindices $m \in {\mathbb Z}_+^{n-1}\setminus 0$ in the ascending series $m_1, m_2, \ldots$ such that $0< m_j \cdot Re \lambda < m_{j+1} \cdot Re \lambda$ $(j=1,2, \ldots )$. Next we determine $A_1$ for $m=m_1$. By considering the terms of the order $O(e^{\lambda m_1 t})$ in $\dot{q}_1 = H_{p_1}$ we obtain

$$\begin{aligned} \frac{d}{dt} A_1 + \lambda \cdot m_1 A_1 = 2\sigma A_0^{2\sigma -1}A_1 = \frac{2\sigma }{(1 - 2\sigma )t} A_1. \end{aligned}$$

(72)

First we see that the constant term of $A_1$ vanishes. Then, by the inductive arguments we see that $A_1$ vanishes.

Next we consider $A_2$ for $m=m_2$. We compare the terms of order $e^{\lambda m_2 t}$ of both sides of $\dot{q}_1 = H_{p_1}$. The term of order $e^{ \lambda m_2 t}$ in $\dot{q}_1 - q_1^{2\sigma }$ is given by

$$\begin{aligned} \frac{d}{dt} A_2 + \lambda \cdot m_2 A_2 - 2\sigma A_0^{2\sigma -1}A_2 = \frac{d}{dt} A_2 + \lambda \cdot m_2 A_2 - \frac{2\sigma }{(1 - 2\sigma )t} A_2. \end{aligned}$$

(73)

We write $q_j = \sum _{m=1}^\infty \alpha _{j,m}(t) e^{\lambda m t}$. Since $q_j = O(e^{ \lambda _j t})$ and

$$A_0^{2\sigma } = (1-2\sigma )^{-1 - \frac{1}{2\sigma -1}}t^{-1 - \frac{1}{2\sigma -1}},$$

the terms of order $e^{\lambda m_2 t}$ appearing from $q_1^{2\sigma } \sum _\nu q_\nu ^2 B_{\nu ,1}(q_1,q)$ are written as

$$\begin{aligned} A_0^{2\sigma }F_2(t) = (1-2\sigma )^{-1-1/(2\sigma -1)} t^{-1-1/(2\sigma -1)} F_2(t), \end{aligned}$$

(74)

with $F_2(t)$ being determined by $A_0(t)$ and has the nonpositive power of t. Note that $A_2$ does not appear in Eq. 74. By Eqs. 73 and 74$A_2$ satisfies

$$\begin{aligned} \frac{d}{dt} A_2 + \lambda \cdot m_2 A_2 - \frac{2\sigma }{(1 - 2\sigma )t} A_2 = (1-2\sigma )^{-1-1/(2\sigma -1)} t^{-1-1/(2\sigma -1)} F_2(t). \end{aligned}$$

(75)

One can determine $A_2$ from Eq. 75 by the same argument as the case of $A_1$. One also sees that $A_2 = O(t^{-1-1/(2\sigma -1)})$.

The general case $m_\nu$ can be shown by the same argument. The term containing $A_\nu$ is given by

$$\begin{aligned} \frac{d A_\nu }{dt} + m_\nu \cdot \lambda A_\nu + \frac{2\sigma }{2\sigma -1}\frac{A_\nu }{t}. \end{aligned}$$

(76)

On the other hand, by substituting the expansion of $q_1$ into $q_1^{2\sigma }$ there appears the polynomial of $A_0, \ldots , A_{\nu -1}$ in the inhomogeneous term. These terms are $O(t^{-1-1/(2\sigma -1)})$. Similarly, the polynomial of $A_0, \ldots , A_{\nu -1}$ appears from $H_{p_1} - q_1^{2\sigma }$ since the term $q_\nu ^2$ exists. By the definition of H these terms are $O(t^{-1-1/(2\sigma -1)})$. By constructing the formal series of the negative power of t inductively like in the case $A_2$, we can determine the formal expansion of $A_\nu$ inductively. We easily see that $A_\nu = O(t^{-1-1/(2\sigma -1)})$ for $\nu \ge 0$.

Step 2. Let $q = q(t)$, $p_1 = p_1(t)$ and $p = p(t)$ be the formal solution of Eqs. 47, 48 and 49, where we use Eq. 64. Let $q_1=q_1(t)$ be the solution of $\dot{q}_1 = H_{p_1}$ given in Step 1. Let $\phi _j^{(0)}$ and $\phi ^{(-e_j)} E^{(-e_j)}$ $( j=2,\ldots ,n)$ be the formal first integral given by Theorem 2.

We shall verify that the substitution of the transseries $q_1(t)$, q(t), $p_1(t)$ and p(t) into the formal first integral is well defined. Since the first integral is the linear function of $p_1$ and p, the assertion is trivial. Next, consider the formal expansions of $q_1$ and $q= \sum _{n=0}^\infty q_1^n a_n(q)$. We recall that $a_n(q)$’s are analytic in some neighborhood of the origin $q =0$ independent of n. Let

$$\begin{aligned} q = q^{(0)} + \sum _{\nu =0}^\infty b_{m_\nu }(t) E_\lambda ^{- m_\nu } \equiv q^{(0)} + \tilde{q}, \end{aligned}$$

(77)

where $m_0=0$ and $b_0(t)$ does not contain the constant term and consists of negative powers of t. On the other hand, $b_{m_\nu }(t)$ consists of nonnegative powers of t. Note also that the constant term $q^{(0)}$ can be taken sufficiently small. By Step 1, $q_1(t)$ also has an expansion like Eq. 77 without a constant term.

Since $a_n(q)$ is analytic in some neighborhood of the origin independent of n and $\tilde{q}(t)$ does not have a constant term, $a_n(q^{(0)} + \tilde{q})$ is well defined and is analytic at $\tilde{q} =0$. Similarly, by noting that $q_1(t)$ does not have a constant term, the substitution $q_1 =q_1(t)$, $q =q(t)$ in $\sum _{n=0}^\infty q_1^n a_n(q)$ is also well defined.

Step 3. Substitute $z=q_1(t)$ in Eq. 49 and differentiate it with respect to t. Then we have

$$\begin{aligned} \dot{q_1}H_{q_1} + \dot{q} H_{q} + \dot{p_1} H_{p_1} + \dot{p} H_{p} = 0 . \end{aligned}$$

(78)

Substitute the formal series $q_1=q_1(t)$, $q = q(t)$, $p_1 = p_1(t)$ and $p = p(t)$ into Eq. 78. Since H is the first integral of the Hamiltonian vector field we get the trivial relation

$$\begin{aligned} H_{p_1}H_{q_1}+ H_{p} H_{q} - H_{q_1}H_{p_1} - H_{q} H_{p} = 0. \end{aligned}$$

(79)

By subtracting Eqs. 79 from 78 and by using $\dot{q_1}=H_{p_1}$ we obtain

$$\begin{aligned} (\dot{q} - H_{p} )H_{q} + (\dot{p_1} + H_{q_1})H_{p_1} + (\dot{p}+ H_{q})H_{p} = 0. \end{aligned}$$

(80)

Let $G^{(j)}$ and $\tilde{G}^{(j)}E^{- e_{j}}$ $(j= 2, \ldots ,n)$ be the formal first integrals given by Theorem 2; $G^{(j)} = \phi _j^{(0)}$, $\tilde{G}^{(j)} = \phi ^{(- e_{j})}$ $(j= 2, \ldots ,n)$, respectively. By applying the same argument to $G^{(j)}$ we have

$$\begin{aligned} \dot{q_1}(G^{(j)})_{q_1} + \dot{q}(G^{(j)})_{q} + \dot{p_1}(G^{(j)})_{p_1} + \dot{p}(G^{(j)})_{p} = 0, \ j = 2, \ldots ,n. \end{aligned}$$

(81)

On the other hand, since $G^{(j)}$ is the first integral of the Hamiltonian vector field of H we have

$$\begin{aligned} H_{p_1}(G^{(j)})_{q_1} + H_{p} (G^{(j)})_{q} - H_{q_1}(G^{(j)})_{p_1} - H_{q} (G^{(j)})_{p} = 0, \ j= 2, \ldots ,n. \end{aligned}$$

(82)

By subtracting Eqs. 82 from 54 and by using $\dot{q_1}=H_{p_1}$ we have

$$\begin{aligned} (\dot{q} - H_{p} )(G^{(j)})_{q} + (\dot{p_1} + H_{q_1})(G^{(j)})_{p_1} + (\dot{p}+ H_{q})(G^{(j)})_{p} = 0, \end{aligned}$$

(83)

where $j= 2, \ldots ,n$. Similarly, we have

$$\begin{aligned} (\dot{q} - H_{p} )(\tilde{G}^{(j)})_{q} + (\dot{p_1} + H_{q_1})(\tilde{G}^{(j)})_{p_1} + (\dot{p}+ H_{q})(\tilde{G}^{(j)})_{p} = 0 , \end{aligned}$$

(84)

where $j= 2, \ldots ,n$.

Step 4. We prove

$$\begin{aligned} \dot{p}_1 + H_{q_1}=0, \quad \dot{p} + H_{q}=0, \quad \dot{q} - H_{p}=0 , \end{aligned}$$

(85)

where the equalities are understood in the sense of transseries. By definition we expand

$$\begin{aligned} \dot{p}_1 + H_{q_1}&= \sum _{k=0}^\infty (\dot{p}_1 + H_{q_1})_k(t) e^{\lambda m_k t}, \end{aligned}$$

(86)

$$\begin{aligned} \dot{q} - H_{p}&= \sum _{k= 1}^\infty (\dot{q} - H_{p})_k(t) e^{\lambda m_k t}, \end{aligned}$$

(87)

$$\begin{aligned} \dot{p}_j + H_{q_j}&= \sum _{k= 0}^\infty (\dot{p}_j + H_{q_j})_k(t) e^{ (-\lambda _j +\lambda m_k )t}, \quad j=2, \ldots ,n, \end{aligned}$$

(88)

where $m_0 = 0$ and $(\dot{p}_1 + H_{q_1})_k(t)$ denotes the coefficients of $e^{\lambda m_k t}$ in the expansion of $\dot{p}_1 + H_{q_1}$. The expansion Eqs. 87 follows from 59. The relation Eqs. 88 follows from 53 and 54.

We show

$$\begin{aligned}&(\dot{p}_1 + H_{q_1})_k (t) =0, \quad k=0,1,2,\ldots , \end{aligned}$$

(89)

$$\begin{aligned}&(\dot{p}_j + H_{q_j})_k(t) = 0, \quad k= 0,1,2,\ldots , \ j=2,\ldots ,n, \end{aligned}$$

(90)

$$\begin{aligned}&(\dot{q} - H_{p})_k(t) = 0, \quad k= 1,2,\ldots \end{aligned}$$

(91)

Clearly, Eqs. 89, 90 and 91 imply 85.

Let $G^{(j)}$ and $\tilde{G}^{(j)}E^{- e_{j}}$ be the formal first integrals given in Step 3. Then, by the result of Step 3 we have Eqs. 80, 83 and 84. On the other hand, we have, by definition,

$$\begin{aligned}&G^{(j)} = q_j p_j + U_{0,j} + q_1^{2\sigma } p_1 U_{1,j} , \quad j=2,\ldots ,n, \end{aligned}$$

(92)

$$\begin{aligned}&\tilde{G}^{(j)} = p_{j} (1 + \tilde{U}_{2,j}) + \tilde{U}_{0,j} + q_1^{2\sigma } p_1 \tilde{U}_{1,j}, \quad j=2, \ldots , n. \end{aligned}$$

(93)

By Eqs. 3, 92 and 93 we have

$$\begin{aligned} (H_{q},H_{p_1},H_{p} )&= \left( \lambda _2 p_2 + \partial _{q_2}(\sum _\nu q_\nu ^2 B\nu ), \ldots , \lambda _n p_n + \partial _{q_n}(\sum _\nu q_\nu ^2 B\nu ), \right. \\&\left. q_1^{2\sigma }+ q_1^{2\sigma }\sum _\nu B_{\nu ,1} q_\nu ^2, \lambda _2 q_2, \ldots , \lambda _n q_n \right) ,\nonumber \end{aligned}$$

(94)

$$\begin{aligned} ( (G^{(j)})_{q},(G^{(j)})_{p_1},(G^{(j)})_{p} ) = \left( \partial _{q}(q_jp_j +\tilde{U}_{0,j}) + q_1^{2\sigma } p_1 \partial _{q}\tilde{U}_{1,j} , q_1^{2\sigma }\tilde{U}_{1,j} , \partial _{p}(q_jp_j )\right) , \end{aligned}$$

(95)

$$\begin{aligned}&( (\tilde{G}^{(j)})_{q},(\tilde{G}^{(j)})_{p_1},(\tilde{G}^{(j)})_{p} ) \\&= \left( p_j \partial _{q} \tilde{U}_{2,j} + \partial _{q} \tilde{U}_{0,j} + q_1^{2\sigma } p_1 \partial _{q} \tilde{U}_{1,j}, q_1^{2\sigma }\tilde{U}_{1,j}, \partial _p (p_j(1+ \tilde{U}_{2,j})) \right) , \nonumber \end{aligned}$$

(96)

where $j=2, \ldots , n$. By Theorem 2 we have

$$\begin{aligned} U_{0,j}&= O(|q|^2), \ U_{1,j} = O(|q|^2), \ \tilde{U}_{0,j} = O(|q|^2), \nonumber \\ \tilde{U}_{1,j}&= O(|q|^2), \ \tilde{U}_{2,j} = O(|q|). \end{aligned}$$

(97)

We compare the terms with the order $O(e^{-\lambda _j t})$ of both sides of Eq. 84. By the proof of Theorem 2 we have

$$\begin{aligned} q_j = O(e^{\lambda _j t}), \ p_j = O(e^{-\lambda _j t}), \ p_1 = O(1), \quad Re\, (\lambda _j t) \rightarrow - \infty . \end{aligned}$$

(98)

We have

$$\begin{aligned} (\tilde{G}^{(j)})_{q}(\dot{q} - H_{p})&= O(1), \ (\tilde{G}^{(j)})_{p_1} (\dot{p}_1 + H_{q_1}) = O( 1), \nonumber \\&(\tilde{G}^{(j)})_{p_j} (\dot{p}_j + H_{q_j}) = O(e^{-\lambda _j t}). \end{aligned}$$

Thus the term with the order $e^{-\lambda _j t}$ of both sides of Eq. 84 is given by $(\dot{p}_j + H_{q_j})_{0}$. Therefore we have $(\dot{p}_j + H_{q_j})_{0}=0$ for $j=2, \ldots ,n$.

Next, compare terms with the order O(1) of both sides of Eq. 83. We have $(G^{(j)})_{q}(\dot{q} - H_{p}) = O(1)$ and the term $(\dot{q}_j - H_{p_j})_{\ell } (p_j)_{0}$ appears for some $\ell \ge 1$ with $\lambda m_\ell = \lambda _j$. The second term has the order $(G^{(j)})_{p_1} (\dot{p}_1 + H_{q_1}) = o(1)$. The third term has the order $(G^{(j)})_{p} (\dot{p} + H_{q}) = O(1)$ and the term $(\dot{p}_j + H_{q_j})_{0}(q_j)_\nu$ appears for some $\nu$, which vanishes by induction. Hence, we obtain $(\dot{q}_j - H_{p_j})_{\ell } (p_j)_{0} =0$. Since $(p_j)_{0}\ne 0$ by Theorem 2 we have $(\dot{q}_j - H_{p_j})_{\ell } =0$ for $j=2,\ldots ,n$. We note that $\ell$ satisfies $\lambda m_\ell =\lambda _j$ and there appears no lower order term than $e^{\lambda _j t}$ in $\dot{q}_j - H_{p_j}$ by definition.

Next we compare terms with the order O(1) of both sides of Eq. 80. The term $(\dot{q}_j - H_{p_j})_{\ell } (p_j)_{0}$ appears from $(\dot{q} - H_{p}) H_{q}$. It vanishes by induction. Next the term $(\dot{p}_1 + H_{q_1})_0 (q_1^{2\sigma })_0$ appears from $(\dot{p}_1 + H_{q_1}) H_{p_1}$.On the other hand, the term $(\dot{p}_j + H_{q_j})_{0} (q_j)_\nu$ appears from $(\dot{p} + H_{q}) H_{p}$, which vanishes by induction. Hence we obtain $(\dot{p}_1 + H_{q_1})_0 (q_1^{2\sigma })_0 = 0$. Since $(q_1^{2\sigma })_0 \ne 0$ by Step 1 we have $(\dot{p}_1 + H_{q_1})_0 =0$.

We proceed in the same way. Namely, we compare the terms with the order O(1) of both sides of Eq. 84. Then we obtain $(\dot{p}_j + H_{q_j})_{1}=0$ for $j=2,\ldots ,n$. Next we compare terms with the order $O(e^{\lambda m_1 t})$ of both sides of Eq. 83. Then we obtain $(\dot{q} - H_{p})_{2} =0$ and so on. Next we consider terms with the order $O(e^{\lambda m_1 t})$ of both sides of Eq. 80. We have $(\dot{p}_1 + H_{q_1})_1 =0$. By induction we have Eqs. 89, 90 and 91.

4.3 Analytic Solution of Eq. 1

By the method of the proof of Theorem 3 we have the analogue of Theorem 3 in the analytic or in the class of formal series. Given analytic first integrals of Eq. 1, $G^{(j)}$, $\tilde{G}^{(j)}$ $(j= 2,\ldots ,n)$. We say that H, $G^{(j)}$’s and $\tilde{G}^{(j)}$’s are functionally independent if the vectors $(H_{q_1}, H_{q}, H_{p_1}, H_{p})$, $( G^{(j)}_{q_1}, G^{(j)}_{q}, G^{(j)}_{p_1}, G^{(j)}_{p})$, $( \tilde{G}^{(j)}_{q_1}, \tilde{G}^{(j)}_{q}, \tilde{G}^{(j)}_{p_1}, \tilde{G}^{(j)}_{p})$ $(j= 2,\ldots ,n)$ are linearly independent in some domain, where $H_{q_1} = \partial H/ \partial q_1$, $H_q = \nabla _q H$ and so on. Here $\nabla _q$ denotes the nabla with respect to q. Then we have

Theorem 4

Suppose that H, $G^{(j)}$ and $\tilde{G}^{(j)}$ $(j= 2,\ldots ,n)$ are functionally independent in some neighborhood of $(q_1^{(0)}, p_1^{(0)}, q^{(0)}, p^{(0)})$ with $q_1^{(0)}\ne 0$. Let q(z), p(z) and $p_1(z)$ be the formal transseries solution given by Lemma 2. Suppose that q(z), p(z) and $p_1(z)$ are analytic at $z= q_1^{(0)}\in \mathcal {C}$ such that $q(q_1^{(0)}) = q^{(0)}$, $p(q_1^{(0)})= p^{(0)}$, $p_1(q_1^{(0)})= p_1^{(0)}$. Assume that there exists an analytic solution $q_1 = q_1(t)$ of $\dot{q}_1 =H_{p_1}$ for $q = q(q_1), p= p(q_1)$ and $p_1=p_1(q_1)$. Then $(q_1, q,p, p_1)$ ( $q = q(q_1(t))$, $p_1= p_1(q_1(t))$, $p = p(q_1(t))$) is an analytic solution of Eq. 1 in some neighborhood of $t_0$ with $q_1^{(0)} = q_1(t_0)$.

Proof

For simplicity we assume that $q_1^{(0)} =0$. First we show that

$$\begin{aligned} \dot{q}(0) = H_{p}(0), \quad \dot{p}(0) = - H_{q}(0), \quad \dot{p}_1(0) = - H_{q_1}(0), \end{aligned}$$

(99)

where the dot denotes the derivative with respect to t and $H_{p_2}(0)= H_{p_2}|_{t=0}$.

Substitute $z=q_1$ in Eq. 49 and differentiate the equation with respect to t. We obtain Eq. 78. Since H is the first integral of the Hamiltonian vector field we have the trivial relation Eq. 79. By subtracting Eqs. 79 from 78 and by using $\dot{q_1}=H_{p_1}$ we obtain Eq. 80.

Next, by applying the same argument to $G^{(j)}$ we have Eq. 81. On the other hand, since $G^{(j)}$ is the first integral of the Hamiltonian vector field of H we have Eq. 82. By subtracting Eqs. 82 from 81 and by using $\dot{q_1}=H_{p_1}$ we have Eq. 83. Similarly, we have Eq. 84. By Eqs. 80, 83 and 84 and the assumption on the functional independency of H, $G^{(j)}$’s and $\tilde{G}^{(j)}$’s we obtain Eq. 99.

Next, we differentiate Eqs. 80, 83 and 84 with respect to t. Then the functional independency of H, $G^{(j)}$’s and $\tilde{G}^{(j)}$’s and the inductive argument we see that $\dot{q}_2 - H_{p_2}$ vanishes up to first derivative. The same assertion holds for $\dot{p}_2 + H_{q_2}$ and $\dot{p}_1 + H_{q_1}$. By induction we see that all derivatives of $\dot{q}_2 - H_{p_2}$, $\dot{p}_2 + H_{q_2}$ and $\dot{p}_1 + H_{q_1}$ vanish.

5 Summability of the Formal Transseries Solution

5.1 Borel Summability

In order to state the result we begin with the definitions of the Borel summability of a formal first integral and a transseries.

Summability of Formal First Integral   Consider the first integrals constructed in Theorem 2. We consider the formal series

$$\begin{aligned} \psi (q_1,q) = q_1^\kappa \sum _{n=0}^\infty v_n (q)q_1^n, \end{aligned}$$

(100)

where $\kappa \ge 1$ is an integer and $v_n(q)$’s are holomorphic in $q\in V_0$ for some open sets $V_0$ independent of n. The formal $\kappa$- Borel transform $\hat{\mathcal {B}}_\kappa$ is defined by

$$\begin{aligned} \hat{\mathcal {B}}_\kappa (\psi )(\zeta ,q) := \sum _{n=0}^\infty v_n(q)\frac{\zeta ^n}{\Gamma (\frac{n+\kappa }{\kappa })}, \end{aligned}$$

(101)

where $\zeta$ is the dual variable of $q_1$ and $\Gamma (z)$ is the Gamma function. For $\phi$ in Eq. 100 we have

$$\begin{aligned} \hat{\mathcal {B}}_\kappa (q_1^{\kappa +1}\frac{d}{d q_1} \psi )(\zeta ,q) = \kappa \zeta ^\kappa \hat{\mathcal {B}}_\kappa (\psi )(\zeta , q). \end{aligned}$$

(102)

For the bisecting direction $d\in {\mathbb R}$ and the opening $\eta >0$, define $S(d,\eta ): =\{ z\in {\mathbb C}; |\arg \, z -d |<\eta /2 \}$. For the neighborhood $\varOmega _0 \subset {\mathbb C}$ of the origin, define

$$\begin{aligned} \Sigma _0 := \varOmega _0 \cup S(d,\eta ) . \end{aligned}$$

(103)

We say that the formal power series $\psi (q_1,q)$ is $\kappa$-summable with respect to $q_1$ in the direction d if there exist $\theta >0$ and a neighborhood $\varOmega _1$ of $\zeta =0$ such that $\hat{\mathcal {B}}_\kappa (\psi )(\zeta ,q)$ converges when $(\zeta ,q)\in \varOmega _1 \times V_0$ and $\hat{\mathcal {B}}_\kappa (\psi )(\zeta ,q)$ can be analytically continued to $(\zeta ,q)\in S(d,\eta ) \times V_0$ and is of exponential type of order $\kappa$ in $\zeta \in S(d,\eta )$. Namely, there exist $K_0>0$ and $K_2>0$ such that

$$|\hat{\mathcal {B}}_\kappa (\psi )(\zeta ,q) | \le K_0 e^{K_2 |\zeta |^\kappa }, \quad \zeta \in S(d,\eta ), \ q\in V_0.$$

For simplicity, we denote the analytic continuation with the same notation. Then the $\kappa$- sum of the formal series $\psi (q_1,q)$, $\Psi (q_1,q)$ is defined by the Laplace transform

$$\begin{aligned} \Psi (q_1,q):= \int _0^{\infty e^{id}} e^{-(\zeta / q_1)^\kappa }\hat{\mathcal {B}}_\kappa (\psi )(\zeta ,q) d\zeta ^\kappa . \end{aligned}$$

(104)

Summability of Transseries We give the definition of the Borel summability of the transseries Eq. 8. As for a general definition and properties we refer to [4] and the references therein.

Consider the transseries u given by Eq. 8. We write

$$\begin{aligned} u = \sum _{k \ge k_0, \ell \ge 0} c_{k,\ell } t^{-\ell /(2\sigma -1)} e^{\lambda k t} = \sum _{k\ge k_0} e^{\lambda k t} u_k(t), \end{aligned}$$

(105)

where

$$\begin{aligned} u_k(t) = \sum _{j=0}^{2\sigma -2} t^{- j/(2\sigma -1)} u_{k,j}(t), \ \ u_{k,j}(t) = \sum _{m=0}^\infty c_{k, m(2\sigma -1)+j} t^{-m}. \end{aligned}$$

(106)

We say that u is $\kappa$- Borel summable in the direction d if there exist $\Sigma _0$ in Eq. 103 and the constant $K_0$ such that, for every j, $j=0, \ldots , 2\sigma -1$ and every integer $k \ge 0$ the formal $\kappa$- Borel transform of $f_{k,j}(t):= e^{\lambda k t} u_{k,j}(t)$, $\mathcal {B}_\kappa (f_{k,j})(\tau )$ is extended to a holomorphic function on $\Sigma _0$ of order $\kappa$ uniformly in k, namely there exist $\exists C_k>0$ satisfying $\sum _k C_k <\infty$ such that

$$\begin{aligned} |\mathcal {B}(f_{k,j})(\tau )| \le C_k e^{K_0 |\tau |^\kappa }, \quad \forall \tau \in \Sigma _0, \end{aligned}$$

(107)

where $\tau$ is the dual variable of t.

5.2 Statement of the Result

Let $(q_1(t), p_1(t), q(t), p(t))$ be the formal transseries solution given by Theorem 2. Then we have

Theorem 5

Suppose that Eqs. 5, 6 and 7 are satisfied. Then the formal transseries solution $(q_1(t), p_1(t), q(t), p(t))$ is $(2\sigma -1)$-Borel summable in every direction in $\{ t| Re (\lambda _j t) <0, j=2,\ldots ,n \}$. There exists a neighborhood of $t = \infty$, $\varOmega _1$ such that the Borel sum is the analytic transseries solution of Eq. 1 in the set $\{ t| Re (\lambda _j t) <0, j=2,\ldots ,n \} \cap \varOmega _1$.

In order to prove Theorem 5 we show a superintegrability by virtue of the Borel sum of formal first integrals.

5.3 Function Space

Let $c>0$. Let $A_c^{(\kappa )}(\Sigma _0\times V_0)$ be the set of all f holomorphic and of exponential growth of order $\kappa$ in $\Sigma _0$ such that

$$\begin{aligned} \Vert f\Vert _c := \sup _{z\in \Sigma _0, q\in V_0} |f(z,q) e^{-c Re\, (z/d)^\kappa } (1+ |z^\kappa |)^2|<\infty . \end{aligned}$$

(108)

Equipped with the norm Eq. 108 the space $A^{(\kappa )}_c(\Sigma _0\times V_0)$ is a Banach space.

For $f, g \in A^{(\kappa )}_c(\Sigma _0\times V_0)$ we define the $\kappa$-convolution by

$$\begin{aligned} (f*_\kappa g)(\zeta , q) := \int _0^{\zeta ^\kappa } f(\tau , q) g((\zeta ^\kappa - \tau ^\kappa )^{1/\kappa },q) d\tau ^\kappa . \end{aligned}$$

(109)

We can easily verify that there exists a constant $K_0>0$ independent of f and g such that $\Vert f*_\kappa g \Vert _c \le K_0 \Vert f \Vert _c \Vert g\Vert _c$ for every $f, g \in A^{(\kappa )}_c(\Sigma _0 \times V_0)$.

6 Summability of First Integral

6.1 Statement of Result

Set $\kappa =2\sigma -1$ and $\lambda := (\lambda _2, \ldots , \lambda _n)$. Let $\alpha = (\alpha _2, \ldots , \alpha _n)\in {\mathbb Z}_+^{n-1}$ and $k = (k_2, \ldots , k_n) \in {\mathbb Z}_+^{n-1}$. Define

$$\begin{aligned} S_0(\alpha ):= \left\{ z\in {\mathbb C}\, | \ \kappa z^\kappa + \lambda \cdot (k - \alpha ) =0, \forall k\in {\mathbb Z}_+^{n-1} \setminus \{0\} \right\} . \end{aligned}$$

(110)

Let $B_{\nu , 0}$ and $B_{\nu , 1}$ be given by Eq. 5. Assume

$$\begin{aligned} B_{\nu , 0}(q_1,q) = O(q_1^\kappa ), \quad B_{\nu ,1}(q_1,q) = O(q_1^\kappa ), \quad \nu = 2, \ldots , n. \end{aligned}$$

(111)

Then we have

Theorem 6

Assume Eqs. 5, 6, 7 and 111. Let $v=E^\alpha \phi ^{(\alpha )}$ $(\alpha = 0, \pm e_j, j=2, \ldots , n)$ be the formal first integrals given by Theorem 2. Then $\phi ^{(\alpha )}$ is $\kappa$-summable with respect to $q_1$ in every direction d such that $d \not \in S_0(\alpha )$.

Remark

We can drop Eq. 111 if we use Lemma 2 of [2].

We give the corollary of Theorem 6. Assume Eq. 6. Let $0 \le \theta _0 \le \theta _1<2\pi$ be such that the closure of $\{ - \lambda \cdot k \, | \, k \in {\mathbb Z}_+^{n-1}\setminus \{0\} \}$ is contained in the sector $\{ z \, | \, \theta _0 \le \arg \, z \le \theta _1\}$. Then we have

Corollary 1

Assume Eqs. 5, 6, 7 and 111. Suppose $\sigma =1$. Then there exist neighborhoods $\varOmega _1$, $\tilde{\varOmega }_1$, $\varOmega _0$, $\tilde{\varOmega }_0$ of $q_1=0$, $p_1 = 0$, $q =0$ and $p =0$, respectively such that the Hamiltonian system for H is Liouville-integrable in the domain $(q_1, p_1, q, p)\in \varOmega _1 \times \tilde{\varOmega }_1 \times \varOmega _0 \times \tilde{\varOmega }_0$ with $q_1 \in \{ z \, | \, \theta _1 -\pi /2< \arg \, z < \theta _0 + 5\pi /2 \}$.

Proof

By Eq. 6 we have $\theta _1 -\theta _0 <\pi$. We show that the first integrals $\phi ^{(-e_j)} E^{-e_j}$ $(j=2, \ldots ,n)$ and H are functionally independent first integrals. By the definition of $S_0( -e_j )$ and by Theorem 6 together with $\kappa = 2\sigma -1 =1$, the summability of the formal first integrals holds for the directions $\theta _1 \le \arg z \le \theta _0 +2\pi$. Hence the Borel sums of formal first integrals exist in the sector given in the theorem. Elementary calculations show that $\nabla \phi ^{(-e_j)}$ $(j=2,\ldots ,n)$ and $\nabla H$ are linearly independent when |q| is sufficiently small.

6.2 Estimate of a Linearized Operator

In order to prove Theorem 6 we prepare a lemma. Let $R_j>0$ $(j=2,\ldots ,n)$ be given. Define $V_0 := \prod _{j=2}^n \{ z_j | |z_j| < R_j \}$. Let $\hat{f} \in A_c^{(\kappa )}(\Sigma _0 \times V_0)$ be given. Consider the equation of $\hat{u} \in A_c^{(\kappa )}(\Sigma _0 \times V_0)$

$$\begin{aligned} L \hat{u} := \left( \kappa \zeta ^\kappa + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial }{\partial q_\nu } \right) \hat{u} = \hat{f}, \quad \zeta \in \Sigma _0. \end{aligned}$$

(112)

For $\hat{f} \in A_c^{(\kappa )}(\Sigma _0 \times V_0)$ we define the weighted norm $|\!\Vert \hat{f} |\!\Vert _c$ by $|\!\Vert \hat{f} |\!\Vert _c := \Vert M_0(q) \hat{f} \Vert _c$, where $M_0(q)$ is given in Lemma 1. Then we have

Lemma 4

Let $\alpha = 0, \pm e_j, j=2, \ldots , n$. Suppose that Eqs. 6 and 7 and $\Sigma _0 \cap S_0(\alpha ) = \emptyset$ are satisfied. Assume that $\hat{f} =O(|q|)$ when $q\rightarrow 0$. Then Eq. 112 has a unique solution $\hat{u}$ in $A_c^{(\kappa )}(\Sigma _0 \times V_0)$. Moreover, there exists $K>0$ independent of $\zeta \in \Sigma _0$ such that $|\!\Vert \hat{u}|\!\Vert _c \le K \Vert \hat{f}\Vert _c$ and $|\!\Vert \zeta ^\kappa \hat{u}|\!\Vert _c \le K \Vert \hat{f}\Vert _c$ hold for every $\hat{f}\in A_c^{(\kappa )}(\Sigma _0 \times V_0)$.

The proof of Lemma 4 is almost identical with that of Lemma 1.

6.3 Proof of Theorem 6

Proof

The proof is done in three steps.

Step 1. Let d be the direction not contained in $S_0(\alpha )$. Define $\Sigma _0$ by Eq. 103. By Eqs. 6 and 7 we have $\Sigma _0 \cap S_0(\alpha ) = \emptyset$ if we take $\eta >0$ and $\varOmega \equiv \varOmega _0$ in Eq. 103 sufficiently small.

We prove the case $\alpha =0$. The proof of other cases $\alpha = \pm e_j$ $(j=2,\ldots ,n)$ is similar to the case $\alpha =0$. By Eq. 12 it is sufficient to prove the $\kappa$-summability of $U_0$ and $U_1$. Since the proof is similar we consider $U_1$. Let $\sum _{\nu =0}^\infty U_{1,j,\nu }q_1^\nu$ be the formal series solution of Eq. 26. Define

$$\begin{aligned} U_1 = V + u_0, \quad u_0 := \sum _{\nu =0}^{\kappa -1} U_{1,j,\nu }q_1^\nu . \end{aligned}$$

(113)

Then we have $V = O(q_1^\kappa )$. Substitute Eqs. 113 into 26. Then, by denoting the inhomogeneous term by $f=O(q_1^\kappa )$ we have

$$\begin{aligned}&q_1^{2\sigma } \frac{\partial V}{\partial q_1} + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial V}{\partial q_\nu } - \sum _{\nu =2}^n q_\nu ^2 q_1^{2\sigma } ( \partial _{q_1} B_{\nu ,1}) V \\&+ \sum _\nu q_\nu ^2 B_{\nu ,1}q_1^{2\sigma } \frac{\partial V}{\partial q_1} + f(q_1,q) = 0 . \nonumber \end{aligned}$$

(114)

Step 2. Recalling Eq. 111 we apply the $\kappa$-Borel transform to Eq. 114. Let $\zeta$ be the dual variable of $q_1$. Let $\hat{V}$ and $\hat{f}$ be the $\kappa$-Borel transform of V and f, respectively. Then we have

$$\begin{aligned} L \hat{V} - \sum _{\nu =2}^n q_\nu ^2 \hat{B_2} * \hat{V} + \sum _\nu q_\nu ^2 \kappa \hat{B_1} * \zeta ^\kappa \hat{V} + \hat{f}(\zeta ,q) = 0 , \end{aligned}$$

(115)

where

$$\begin{aligned} L \hat{V} := \left( \kappa \zeta ^\kappa + \sum _{\nu =2}^n \lambda _\nu q_\nu \frac{\partial }{\partial q_\nu }\right) \hat{V}, \end{aligned}$$

(116)

and $\hat{B_1}:= \widehat{B_{\nu ,1}}$ and $\hat{B_2}:= \widehat{q_1^{2\sigma }\partial _{q_1} B_{\nu ,1}}$ are the $\kappa$-Borel transform of $B_{\nu ,1}$ and $q_1^{2\sigma }\partial _{q_1} B_{\nu ,1}$, respectively.

Define $L \hat{V} =: \hat{W}$. Since $L^{-1}$ exists, by Lemma 4 we have

$$\begin{aligned} \hat{W} - \sum _{\nu } q_\nu ^2 \hat{B_2} * L^{-1} \hat{W} + \sum _\nu q_\nu ^2 \kappa \hat{B_1} * \zeta ^\kappa L^{-1} \hat{W} + \hat{f}(\zeta ,q) = 0 . \end{aligned}$$

(117)

Define

$$\begin{aligned} L_1 \hat{W} := \kappa \hat{B_1} * \zeta ^\kappa L^{-1} \hat{W}, \quad L_2 \hat{W} := \hat{B_2} * L^{-1} \hat{W}. \end{aligned}$$

(118)

Then, by Eq. 117 we obtain

$$\begin{aligned} \hat{W} + \sum _{\nu } q_\nu ^2 (L_1 - L_2) \hat{W} + \hat{f}(\zeta ,q) = 0 . \end{aligned}$$

(119)

Step 3. We construct the approximate sequence

$$\begin{aligned} \hat{W_0}:= - \hat{f}(\zeta ,q), \quad \hat{W_j}:= -\sum _\nu q_\nu ^2 (L_1 - L_2) \hat{W}_{j-1}, \ j=1,2, \ldots \end{aligned}$$

(120)

If the limit $\hat{W}:= \sum _{j=0}^\infty \hat{W_j}$ exists in $\mathcal{A}_c^{(\kappa )}$ with $R_j$ replaced by $R_j'<R_j$, $r_j-R_j'$ small, then $\hat{W}$ is the desired solution.

We show the convergence. By the estimate of a convolution and Eq. 118 we have

$$\begin{aligned}&\Vert L_1 \hat{W} \Vert _c = \Vert M_0^{-1} M_0 L_1 \hat{W} \Vert _c \le K_1 \Vert M_0 L_1 \hat{W} \Vert _c = K_1 \Vert |L_1 \hat{W}|\Vert _c \\ \le&\, K_1 K_2 \Vert | \zeta ^\kappa L^{-1} \hat{W} |\Vert _c \le K_1 K_2 K \Vert \hat{W} \Vert _c , \nonumber \end{aligned}$$

(121)

where $K_1 = \sup _q |M_0(q)^{-1}|$, $K_2 = \Vert \kappa \hat{B_1} \Vert _c$. Similarly, we have $\Vert L_2 \hat{W} \Vert _c \le K_1 K_2 K \Vert \hat{W} \Vert _c$. Let $\epsilon >0$ be given. By taking $R_j>0$ sufficiently small we have

$$\begin{aligned} \Vert \hat{W_j} \Vert _c \le K_1 K_2 K \epsilon \Vert \hat{W}_{j-1} \Vert _c , \quad j=1,2, \ldots \end{aligned}$$

(122)

The estimate Eq. 122 implies the desired convergence of the sequence.

7 Proof of Theorem 5

Proof

(Proof of Theorem 5) We prove the theorem by five steps.

Step 1. Consider Eqs. 47, 48 and 49. These equations are equivalent to Eqs. 55, 58 and 61. Hence, if we show the summability of q we have the summability of p and $p_1$ as well.

Set $\kappa = 2\sigma -1$. Let t and z satisfy $t = - \kappa ^{-1} z^{ -\kappa }$. Let $z_0$ be such that $Re (\lambda _j z_0^{-\kappa }) >0$. Define $\Sigma _0$ by Eq. 103 with $d = \arg z_0$. We show that there exist constants $C_0>0$, $C_1>0$ and $\eta >0$ such that, for $\Sigma _0$ given by Eq. 103 with $d = \arg z_0$ we have

$$\begin{aligned} \left| \kappa z^{\kappa } + \lambda _j k \right| \ge C_0 |z|^{\kappa },&\quad \forall z, \ |z|>C_1, z\in \Sigma _0, \end{aligned}$$

(123)

$$\begin{aligned} \left| \kappa z^{\kappa } + \lambda _j k \right| > C_0,&\quad \forall z, \ |z| \le C_1, \ z\in \Sigma _0, \end{aligned}$$

(124)

for $j=2,\ldots ,n$ and $k=1,2, \ldots$. Indeed, we have

$$\begin{aligned} \left| \kappa z^{\kappa }\lambda _j^{-1} + k \right| = \left| \frac{\kappa }{\lambda _j z^{-\kappa }} + k \right| = \left| -\frac{1}{\lambda _j t} + k \right| > C_0/|\lambda _j|, \end{aligned}$$

(125)

for some $C_0>0$ and all t, $Re (\lambda _j t)<0$, $j=2,\ldots ,n$, $k=1,2, \ldots$ Eqs. 124 follows from 125.

We show Eq. 123. We have

$$\begin{aligned}&\left| \kappa z^{\kappa }\lambda _j^{-1} + k \right| \ge \left| Re \left( \kappa z^{\kappa }\lambda _j^{-1} \right) +k \right| \nonumber \\&\ge Re \left( \kappa z^{\kappa }\lambda _j^{-1} \right) \ge \epsilon _1 |z|^{\kappa }, \end{aligned}$$

(126)

for some $\epsilon _1 >0$ and all $z \in \Sigma _0$ and all $k=1,2, \ldots$, $j=2,\ldots ,n$.

Step 2. We solve Eq. 61. Let $\zeta$ be the dual variable of z. Let

$$\begin{aligned} q_{0}= \sum _{n=0}^\infty c_{n}(\xi ) z^n \end{aligned}$$

(127)

be the formal series solution of Eq. 61. Define $\tilde{q}$ by

$$\begin{aligned} q = \tilde{q} + \rho , \quad \rho = \sum _{n=0}^{\kappa -1} c_{n}(\xi ) z^n . \end{aligned}$$

(128)

Clearly we have $\tilde{q} = O(z^\kappa )$. Let $G=(G_j)$ be given by Eq. 59. Next, consider

$$G(z,q, \xi ) = \sum _{n=0}^\infty z^n s_n(q, \xi ) = \sum _{n, \ell } s_{n, \ell }(\xi ) z^n q^\ell , $$

where the Taylor expansions of $s_n(q, \xi )$’s with respect to q converge in some neighborhood of the origin $q = 0$ and $\xi =0$ independent of n. Then we consider

$$\begin{aligned} \sum _n z^n s_n(\tilde{q} + \rho , \xi ) - \sum _{n=0}^{\kappa -1} c_{n}(\xi ) z^n = \sum _{n, \ell } s_{n, \ell }(\xi ) z^n (\tilde{q} + \rho )^\ell - \sum _{n=0}^{\kappa -1} c_{n}(\xi ) z^n. \end{aligned}$$

(129)

We note that Eq. 129 is $O(z^\kappa )$ by the definition of formal series. By the formula $(\tilde{q} + \rho )^\ell = \sum _\beta \tilde{q}^\beta \rho ^{\ell -\beta } \ell !/(\beta ! (\ell -\beta )!)$ the formal $\kappa$-Borel transform of the left-hand side of Eq. 129 converges. Denoting the right-hand side of Eq. 129 by $G(z,\tilde{q},\xi )$ for the sake of simplicity and rewriting $\tilde{q}$ as q we consider

$$\begin{aligned} q = G(z,q, \xi ). \end{aligned}$$

(130)

We note that the $\kappa$- Borel transform of $\frac{\partial G}{\partial q}(z,0, \xi )$ in $A_c^{(\kappa )}$ is sufficiently small for sufficiently small $\xi$ by the definition of G.

Step 3. $G(z,q,\xi )$ is a formal power series in z with coefficients being holomorphic in $\xi$ and q in some neighborhood of the origin $\xi =0$, $q = 0$ which is uniform among the coefficients. By expanding the coefficients in the power series of $\xi$ and q and rearranging them we obtain the series of $\xi$ and q whose coefficients are formal series in z. We show that the coefficients of the series of G with respect to $\xi$ and q are summable in z which are uniform among the coefficients. We denote the uniform summability property by (P).

By the definition of $G= (G_j)$ it is sufficient to show that (P) holds for the first integrals constructed in Theorem 2. Indeed, since $G_j$ is given by Eq. 59, it is sufficient to show that $(\tilde{T}_j + F_{j,0})^{-1}$ and $\tilde{A}_j(z,z^{2}p_1, q)$ satisfy (P). The latter term satisfies (P) if the first integral satisfies (P). As for the former one, consider $F_{j,0}$ in Eq. 55. The terms $\tilde{D}_j$ and $\tilde{D}_{j,1}$ satisfy (P) by the assumption on the first integral. By Eq. 53 we see that $E_{j,1}$ satisfies (P). Next, by Eq. 54$\tilde{T}_j$ satisfies (P). It follows that $(\tilde{T}_j + F_{j,0})^{-1}$ satisfies (P). Hence $G_j$ satisfies (P).

Let $C(z,q, \xi )$ be any formal first integral constructed in Theorem 2. For every pair of multiintegers $m \ge 0, n \ge 0$ we consider the coefficient of $q^m \xi ^n$ of the Taylor series of $C(z,q, \xi )$

$$\begin{aligned} C_{m,n}(z) = \frac{1}{ (2 \pi i)^2} \int \int _{| w_j | = \epsilon _1, |s_\nu |= \epsilon _2} \frac{C(z,w, s)}{w^{m+1}s^{n+1}} dwds , \end{aligned}$$

(131)

where $\epsilon _1 >0$ and $\epsilon _2 >0$ are small constants. Let $\hat{C}(\zeta , w, \xi )$ be the formal Borel transform of $C(z,w, \xi )$ with respect to z, where $\zeta$ is the dual variable of z. By the formal Borel transform of Eq. 131 we have

$$\begin{aligned} \hat{C}_{m,n}(\zeta ) = \frac{1}{ (2 \pi i)^2} \int \int _{| w_j | = \epsilon _1, |s_\nu |= \epsilon _2} \frac{\hat{C}(\zeta ,w,s )}{w^{m+1}s^{n+1}} dw ds . \end{aligned}$$

(132)

Since C(z, w, s) is Borel summable, there exist $\Sigma _0$ in Eq. 103 and the neighborhoods $V_0$ and $V_1$ of $q = 0$ and $\xi =0$, respectively, such that $\hat{C}(\zeta ,w, s)$ is holomorphic in $(\zeta , w, s) \in \Sigma _0 \times V_0 \times V_1$. Moreover, $\hat{C}(\zeta ,w, s)$ is of exponential order of one in $\zeta \in \Sigma _0$ for every $(w, s)\in V_0 \times V_1$. By the scale change of variables $q \mapsto \epsilon q$ and $\xi \mapsto \epsilon \xi$ we may assume that $V_0$ and $V_1$ contain a disk with sufficiently large radius. Therefore, by Eq. 132 we have the summability of $C_{m,n}(z)$ uniformly in m and n. In the following we assume the condition.

Step 4. We prove the summability of q as a transseries. It is sufficient to show the summability with respect to the variable z instead of t. Expand $c_n(\xi )$ in Eq. 127 in the power series of $\xi$ and consider

$$\begin{aligned} q(z) = \sum _{j \ge 0} \xi ^j q_{j}(z). \end{aligned}$$

(133)

By Eq. 128 it is sufficient to show the summability of q in Eq. 130. Note that, by the definition of the summability of a transseries it is sufficient to show the uniform summability of $q_{j}$’s and the convergence of the sum Eq. 133 with $q_{j}$ replaced by its Borel sum.

If $j=0$, then the summability of $q_{0}\equiv 0$ is trivial. Suppose that the uniform summability of $q_{j}$ for $j=0,\ldots , k-1$ holds. Namely, the formal Borel transform of $q_{j}$, $\hat{q}_{j}$ is holomorphic in $\Sigma _0$ and has the same exponential order for $j=0,\ldots , k-1$. Consider $q_{k}$. Substitute Eqs. 133 into 130.

Since G is analytic at $q =0$ we consider the term

$$\begin{aligned} C_\ell (\xi , z) ( \sum _{j, |j|>0} q_{j}\xi ^j )^\ell , \end{aligned}$$

(134)

where $\ell \ge 0$ is a multiinteger and $C_\ell (\xi , z)$ is analytic in $\xi$ and is a formal power series in z. Expand $C_\ell (\xi , z)$ in a power series in $\xi$, $C_\ell (\xi , z) = \sum _{|\nu |\ge 1} K_{\ell ,\nu }(z) \xi ^\nu$. We introduce the weight $\epsilon _0^j$ in front of $q_{j}$ by the scale change $\xi \mapsto \epsilon _0 \xi$ (cf. step 4), where $\epsilon _0>0$ is a sufficiently small number. Then the coefficient of $\xi ^k$ appearing from $G(z, q, \xi )$ is given by

$$\begin{aligned} \sum \frac{K_{\ell ,\nu }(z) \ell ! \epsilon _0^{|k|} }{ m_1 ! \cdots m_{\mu }!} q_{j_1}^{m_1} q_{j_2}^{m_2} \cdots q_{j_\mu }^{m_{\mu }}, \end{aligned}$$

(135)

where the summation is taken over the pair of multiintegers, $m_1, \ldots , m_{\mu }$ satisfying

$$\begin{aligned} m_1 + \cdots + m_{\mu }= \ell , \quad j_1|m_1| + j_2 |m_2| + \cdots + j_\mu |m_{\mu }| = k - \nu , \end{aligned}$$

(136)

where $\mu$ is an integer and $j_1, \ldots , j_\mu \ge 0$ are multiintegers. By the result of Step 3 $K_{\ell ,\nu }(z)$ is uniformly summable in $\ell$ and $\nu$ and $\sum _{\ell ,\nu }\Vert K_{\ell ,\nu }\Vert <\infty$. By Eqs. 135 and 130 we see that the formal Borel transform of $q_{k}(z)$, $\hat{q}_{k}(\zeta )$ is holomorphic in $\Sigma _0$ and has the same exponential order as $\hat{q}_{j}$’s.

It remains to estimate $\Vert q_{k} \Vert$, where $\Vert q_{k} \Vert$ is a certain maximal norm. Suppose that

$$\begin{aligned} \Vert q_{j}\Vert \le K_1 \epsilon _2^{|j|}, \quad |j|<|k|, \end{aligned}$$

(137)

for some positive constants $K_1$ and $\epsilon _2$, where $\epsilon _2$ is chosen sufficiently small. Take $\epsilon _0 \le 1$ and $2\epsilon _0 < \epsilon _2$. We have

$$\begin{aligned}&\sum \frac{ \ell ! }{ m_1 ! \cdots m_{\mu }!} (\Vert q_{j_1}\Vert )^{m_1}(\Vert q_{j_2}\Vert )^{m_2} \cdots (\Vert q_{j_\mu }\Vert )^{m_{\mu }} \\&\le \sum \frac{ \ell ! }{ m_1 ! \cdots m_{\mu }!} ( K_1 \epsilon _2^{|j_1|})^{|m_1|}(K_1 \epsilon _2^{|j_2|})^{|m_2|} \cdots (K_1 \epsilon _2^{|j_\mu |})^{|m_{\mu }|}\nonumber \\&\le ( K_1 \sum _{j, |j|>0} \epsilon _2^{|j|} )^{|\ell |} \le (CK_1 \epsilon _2)^{|\ell |},\nonumber \end{aligned}$$

(138)

where the summation is taken over all combinations satisfying Eq. 136 and where C satisfies $\sum _{j, |j|>0} \epsilon _2^{|j|} \le C\epsilon _2$.

Then the term Eq. 135 is estimated by

$$\begin{aligned} \epsilon _2^k \sum _{\ell , \nu } (C K_1 \epsilon _2)^{|\ell |} \Vert K_{\ell ,\nu }\Vert . \end{aligned}$$

(139)

By taking $\epsilon _2$ sufficiently small we have

$$\begin{aligned} \sum _{\ell \ne 0, \nu } (C K_1 \epsilon _2)^{|\ell |} \Vert K_{\ell ,\nu }\Vert \le \frac{K_1}{2}. \end{aligned}$$

(140)

On the other hand, we may assume $\Vert K_{0,\nu }\Vert \le K_1/2$ since $\nu \ge 1$. Hence Eq. 139 is estimated by $K_1 \epsilon _2^n$, which proves the convergence of the sum.

Step 5. We prove the summability of $q_1$. If we prove the summability of $q_1$ and q, then we have the summability of $p_1$ and p as well by Eqs. 58 and 55, respectively. Let q(z) be given by Eq. 133. Consider

$$\begin{aligned} \frac{d q_1}{dt} = H_{p_1} = q_1^{2\sigma } \left( 1 + \sum q_\nu ^2 B_{1,\nu }(q_1,q) \right) . \end{aligned}$$

(141)

Introduce z by Eq. 64.

Let q be given by Eq. 133. Set

$$\begin{aligned} q = q^{(0)} + \tilde{q}, \quad q^{(0)} = \sum _{|k| >0} q_{k}^{(0)}(z) \xi ^k, \end{aligned}$$

(142)

where $q^{(0)}$ is the polynomial of z with degree $\kappa -1$ and analytic in $\xi$ at $\xi =0$. Expand

$$\begin{aligned} \tilde{q} = \sum _{|k| > 0} \tilde{q}_{k}(z) \xi ^k. \end{aligned}$$

(143)

By what we have proved in the above, the $\kappa$- Borel transform of $\tilde{q}_{k}(z)$ exists for every k. Substitute Eq. 142 into the right-hand side of Eq. 141. Since $B_1$ is the polynomial in q by assumption we see that the right-hand side of Eq. 141 is the polynomial of $\tilde{q}$ whose coefficients are the polynomials of z and analytic in $q_1$ and $\xi$.

Substitute the Taylor series of $q_1$

$$\begin{aligned} q_1(z) = \sum _{|j| \ge 0} \xi ^j q_{1,j}(z), \end{aligned}$$

(144)

into Eq. 141 and compare the coefficients of the power $\xi ^k$ with $k=0$ of both sides of Eq. 141. We have $q_{1,0}=A_0$, where $A_0$ is given by Eq. 71. One can easily verify that $A_0=z$ by the definition of z. By Eq. 72 we have $q_{1,1}= A_1 = 0$. Let $k \ge 2$. By Eq. 76 and $A_k = q_{1,k}$ we see that the left-hand side of Eq. 141 yields $\frac{q_{1,k}}{dt} + k\lambda q_{1,k}$. It is equal to

$$\begin{aligned} z^{\kappa +1}\frac{q_{1,k}}{dz} + k\lambda q_{1,k}. \end{aligned}$$

Therefore $q_{1,k}$’s satisfy the recurrence relation

$$\begin{aligned} z^{\kappa +1}\frac{d}{dz}q_{1,k} + k\lambda q_{1,k} = c_0 z^{\kappa }q_{1,k} + f \left( z, \xi , q_{1,j}, \tilde{q}_{\ell }; j< k, \ell <k \right) , \end{aligned}$$

(145)

where f is the polynomial of $q_{1,j}$’s, $\tilde{q}_{\ell }$’s and z and analytic in $\xi$ at $\xi =0$, and where $c_0$ is a constant. By the scale change of z we may assume that $c_0$ is sufficiently small.

In order to see the form of f, note that the partial Taylor expansion of the right-hand side of Eq. 141 has the form

$$\begin{aligned} C_{\ell , \nu }(z, \xi ) ( \sum _{|j|>0} q_{1,j}\xi ^j )^{2\sigma + \ell } ( \sum _{|k|>0} \tilde{q}_{k}\xi ^k )^{\nu }, \end{aligned}$$

(146)

for some non-negative integer $\ell$ and a multiinteger $\nu$ with $|\nu |\ge 2$. Expand

$$\begin{aligned} C_{\ell , \nu }(z, \xi ) = \sum _{s \ge 0} C_{\ell , \nu , s}(z)\xi ^s. \end{aligned}$$

(147)

Then, by comparing the coefficients of the power $\xi ^k$ in Eq. 146 we have

$$\begin{aligned} \sum C_{\ell , \nu , s}(z) \frac{\nu !}{\alpha _1! \cdots \alpha _i!}\frac{(2\sigma + \ell )!}{\mu _1! \cdots \mu _m!} (q_{1,j_1})^{\mu _1} \cdots (q_{1,j_m})^{\mu _m} (\tilde{q}_{k_1})^{\alpha _1} \cdots (\tilde{q}_{k_i})^{\alpha _i}, \end{aligned}$$

(148)

where the summation is taken over the combinations

$$\begin{aligned}&\mu _1 + \mu _2 + \cdots + \mu _m = 2\sigma +\ell , \quad \alpha _1 + \alpha _2 + \cdots + \alpha _i = \nu \\&j_1\mu _1 + j_2\mu _2 + \cdots + j_m\mu _m + k_1 |\alpha _1| + k_2|\alpha _2| + \cdots + k_i|\alpha _i| = k - s, \ 0 \le s \le k. \nonumber \end{aligned}$$

(149)

By Step 1 of the proof of Theorem 3 we have $q_{1,j}= O(t^{-1-1/\kappa })=O(z^{\kappa +1})$. Hence the formal $\kappa$-Borel transform of $q_{1,j}$ is well defined. By the formal $\kappa$-Borel transform of Eq. 145 we have

$$\begin{aligned} (\kappa \zeta ^{\kappa } + k\lambda ) \widehat{q_{1,k}} = c_0 \left( 1* \widehat{q_{1,k}}\right) + f^* \left( \zeta , \xi ,\widehat{q_{1,j}}, \widehat{\tilde{q}_{\ell }}; j<k, \ell <k \right) , \end{aligned}$$

(150)

where $\widehat{q_{1,k}}$ is the formal $\kappa$- Borel transform of $q_{1,k}$ and $*$ denotes the convolution. $f^*$ denotes the formal $\kappa$- Borel transform of f, where the product is understood as the convolution product. Define $v:= \widehat{q_{1,k}}$ and denote the second term of the right-hand side of Eq. 150 by $g_k$, for simplicity. Then Eq. 150 is written as

$$\begin{aligned} (\kappa \zeta ^{\kappa } + k\lambda ) v = c_0 \left( 1* v\right) + g_k . \end{aligned}$$

(151)

We define the sequence $v_0, v_1,\cdots$ by

$$\begin{aligned}&(\kappa \zeta ^{\kappa } + k\lambda ) v_0 = g_k\end{aligned}$$

(152)

$$\begin{aligned}&(\kappa \zeta ^{\kappa } + k\lambda ) v_j = c_0 (1 * v_{j-1}), \quad j=1,2, \ldots \end{aligned}$$

(153)

If the limit $v:= v_0 + v_1 + \cdots$ exists, then v gives the solution of Eq. 150. By definition we have the summability of $q_{1,k}$. It is easy to see the convergence of v.

In order to see the summability of $q_1$ as a transseries we need to show the uniform summmability of $q_{1,k}$’s in k. Since we have the uniform summability of $q_{k}$’s, we can inductively show that $g_k$ is holomorphic in the domain uniform in k with exponential growth order being uniform in k. In view of the definition of the approximate sequence of v in Eqs. 152 and 153 the same property holds for $v= \widehat{q_{1,k}}$.

Next we show the convergence of $q_1 = \sum _k q_{1,k} \xi ^k$. In order to have the estimate of $\Vert q_{1,k}\Vert$ it is sufficient to estimate $\Vert g_k \Vert$ in view of the definition of the approximate sequence of $q_{1,k}$. The estimate of $g_k$ follows from the expression Eq. 148. Since the expression Eq. 148 has the same form as Eq. 135, we can estimate Eq. 148 by the same argument as the estimate of Eq. 135 in the step 4. Hence we can show the estimate of $q_{1,k}$ by the same calculation in proving Eq. 137. Therefore we have the summability of $q_1$ as a transseries. This ends the proof.