Nonintegrability of Hamiltonian system perturbed from integrable system with two singular points

Abstract

We give a Hamiltonian system which is nonintegrable in a domain containing two singular points and that is integrable in some neighborhood of a singular point. The system is an arbitrarily small nontrivial perturbation of an integrable Hamiltonian system given by confluence of regular singular points of a generalized hypergeometric system.

1 Introduction

Let $n\ge 2$ be an integer, and consider the Hamiltonian system

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle z^{2}\frac{dq}{dz} =\ \nabla _{p}\mathcal {H}(z,q,p), &{}\\ \displaystyle z^{2}\frac{dp}{dz} =- \nabla _{q}\mathcal {H}(z,q,p), &{} \end{array}\right. } \end{aligned}$$

(1)

where $q=(q_2,\ldots ,q_n)$, $p=(p_2,\ldots ,p_n)$. Here

$$\begin{aligned} \nabla _{q}:=\left( \frac{\partial }{\partial q_2},\ldots , \frac{\partial }{\partial q_n} \right) , \quad \nabla _{p}:=\left( \frac{\partial }{\partial p_2},\ldots , \frac{\partial }{\partial p_n} \right) . \end{aligned}$$

The system (1) is equivalent to an autonomous one

$$\begin{aligned} {\left\{ \begin{array}{ll} \dot{q}_1=\ H_{p_1}, &{}\dot{q}=\ \nabla _{p}H, \\ \dot{p}_1= -H_{q_1}, &{}\dot{p}= -\nabla _{q}H, \end{array}\right. } \end{aligned}$$

(2)

where $q_1=z$ and $H(q_1,q,p_1,p) :=q_1^2 p_1+\mathcal {H}(q_1,q,p)$ or $H(q_1,q,p_1,p) :=$ $p_1+ q_1^{-2}\mathcal {H}(q_1,q,p)$. We say that the Hamiltonian system (2) is $C^\omega$-Liouville integrable if there exist first integrals $\phi _j \in C^\omega$ $(j=1,\ldots ,n)$ which are functionally independent on an open dense set and Poisson commuting, i.e., $\{ \phi _j, \phi _k \} =0$, $\{ H, \phi _k \} =0$, where $\{ \cdot , \cdot \}$ denotes the Poisson bracket. The Hamiltonian H is a first integral of this autonomous system. We abbreviate $C^\omega$-Liouville integrable to $C^\omega$-integrable or integrable if there is no fear of confusion.

In [2] Bolsinov and Taimanov showed a non $C^\omega$-integrability of some Hamiltonian system related with geodesic flow on a Riemannian manifold. Then Gorni and Zampieri showed similar results in the local setting, namely for a Hamiltonian system being singular at the origin they showed the non $C^\omega$-integrability (cf. [3, 5, 6]). In this paper we study the nonintegrability from a semi-global point of view. Namely we consider Hamiltonian system which is singular at the origin $q_1=0$ as well as $q_1=1$. We shall show that the system is integrable near the origin, while it is not integrable in the domain containing both $q_1=0$ and $q_1=1$. The Hamiltonian function is given by the arbitrary small non zero perturbation of an integrable Hamiltonian of the confluent generalized hypergeometric system (cf. Sect. 2).

More precisely, we consider

$$\begin{aligned} H =\sum _{j\in J'}\mu _{j}q_jp_j +\frac{q_1^2}{(q_1-1)^2} \sum _{j\in J }\mu _{j}q_jp_j +q_1^2p_1, \end{aligned}$$

(3)

where $\mu _j$ are complex constants and J and $J'$ are the sets of multi-indices such that

$$\begin{aligned} J \ne \emptyset , J'\ne \emptyset , J\cap J'=\emptyset , J\cup J'=\{2,\ldots ,n\}. \end{aligned}$$

(4)

The Hamiltonian is derived from the generalized hypergeometric system by confluence of singularities (cf. Sect. 2). The Hamiltonian system (2)–(3) determines the Hamiltonian vector field

$$\begin{aligned} \chi _{H}= & {} \ q_1^2\frac{\partial }{\partial q_1} - 2q_1p_1\frac{\partial }{\partial p_1} + \frac{2q_1}{(q_1-1)^3}\left( \sum _{j\in J }\mu _{j}q_jp_j \right) \frac{\partial }{\partial p_1} \nonumber \\&+ \sum _{j\in J'}\mu _{j}\left( q_j\frac{\partial }{\partial q_j} -p_j\frac{\partial }{\partial p_j} \right) + \frac{q_1^2}{(q_1-1)^2} \sum _{j\in J }\mu _{j}\left( q_j\frac{\partial }{\partial q_j} -p_j\frac{\partial }{\partial p_j} \right) . \end{aligned}$$

(5)

Let

$$\begin{aligned} H_1:=\sum _{j=2}^{n} p_j^2B_j(q_1,p). \end{aligned}$$

(6)

Note that $H_1$ does not depend on q. Suppose that the nonresonance condition (NRC) holds:

$$\begin{aligned} \forall \gamma =(\gamma _2,\ldots ,\gamma _n) \in \mathbb {Z}^{n-1}{\setminus }\{0\},\ \sum _{j=2}^{n}\mu _{j}\gamma _j\ne 0, \end{aligned}$$

(7)

i.e. $\mu _{j}$’s are linearly independent over $\mathbb {Z}^{n-1}$. Moreover, assume

(TC): For $k\in J'$, the equation

$$\begin{aligned} q_1^2\frac{d}{dq_1}v - 2\mu _{k}v = B_k(q_1,0) \end{aligned}$$

(8)

has no solution v holomorphic at $q_1=0$, and for $k\in J$, the equation

$$\begin{aligned} q_1^2\frac{d}{dq_1}w - 2\mu _{k}\frac{q_1^2 w}{(q_1-1)^2} = B_k(q_1,0) +\mu _{k} \frac{q_1B_k(0,0)}{(q_1-1)^2} + B_k(0,0) \end{aligned}$$

(9)

has no solution w holomorphic at $q_1=1$.

Let $\varOmega _1\subset {\mathbb C}$ be a domain containing $\{q_1 = 0, 1\}$, and $\varOmega _2\subset {\mathbb C}^{2n-1}$ be a neighborhood of $(p_1,q,p)=(0,0,0)$ and define $\varOmega :=\varOmega _1\times \varOmega _2$. Then we have

Theorem 1

Assume that (NRC) and (TC) are satisfied. Then, there exists $\varOmega$ such that the Hamiltonian system (2) is not $C^\omega$-integrable in $\varOmega$. More precisely, for every first integral $\phi$ satisfying $\chi _{H+H_1}\phi =0$ and holomorphic in $\varOmega$, there exists a holomorphic function $\psi$ defined in some neighborhood of the origin $t=0$ such that $\phi (q_1,q,p_1,p)=\psi (H+H_1)$ in some neighborhood of the origin.

In spite of the non integrability shown in Theorem 1 we have the integrability about a singular point of $\chi _{H+H_1}$. We recall that the Hamiltonian system corresponding to $H+H_1$ has irregular singularities at $q_1=0$ and $q_1=1$. We have

Proposition 1

Suppose that $H_1(q_1,p)$ be independent of $p_\nu$ for every $\nu \in J'$. Then, $\chi _{H+H_1}$ is analytically Liouville-integrable in some neighborhood of the origin.

Remark

(i) In Sect. 5 we show that (TC) holds on an open dense set in the set of analytic functions. (TC) also implies that $H_1$ could be replaced by $\varepsilon H_1$ with an arbitrary small $\varepsilon \ne 0$. On the other hand, it is necessary in Theorem 1 that $H_1$ does not vanish identically because H is integrable in view of Lemma 1 (cf. Sect. 3). Hence the non-integrability occurs by an arbitrary small non-zero generic perturbation.

By Proposition 1 we see that our class of Hamiltonians contains subclass for each of which the integrability at the origin holds. Hence the (non-) integrability in Theorem 1 is caused by the interference of singular points.

(ii) Of course, a globally integrable system is locally integrable. So, it is sufficient for the proof of Theorem 1 to prove the local non-integrability.

(iii) In these days, monodromy is usually treated from the point of view of the differential Galois theory (for example, see [7]) because of enrichment of the theory however, we treat it from another point of view.

2 Confluence of singularities

In this section we deduce (3) from the genelarized hypergeometric system

$$\begin{aligned} (z-C)\frac{dv}{dz}=Av, \end{aligned}$$

(10)

where $C=\mathrm{diag}(\varLambda _1,{}^t\varLambda _1)$, $\varLambda _1$ being $(n-1)\times (n-1)$ matrix with eigenvalues $\lambda _2,\ldots ,\lambda _n$ such that $\lambda _j\ne 0$ for all j (cf. [1, 4] ). For the sake of simplicity, we assume $\varLambda _1=\mathrm{diag}(\lambda _2,\ldots ,\lambda _n)$. We assume $A=\mathrm{diag}(A_1,A_1)$, where $A_1$ is an $(n-1)\times (n-1)$ constant matrix satisfying $\varLambda _1A_1=A_1\varLambda _1$. For simplicity, we further assume $A_1=\mathrm{diag}(\tau _2,\ldots ,\tau _n)$.

Let $v={}^t(q,p)\in \mathbb {C}^{2(n-1)}$. Define

$$\begin{aligned} H=\langle (z-\varLambda _1)^{-1}p,A_1q\rangle , \end{aligned}$$

(11)

where $\langle (x_2,\ldots ,x_n),{}^t(y_2,\ldots ,y_n)\rangle :=\sum _{2\le k\le n}x_ky_k$. Then, (10) is written in the Hamiltonian system

$$\begin{aligned} \frac{dq}{dz}= H_p(z,q,p),\ \frac{dp}{dz}=-H_q(z,q,p). \end{aligned}$$

(12)

Now we operate the confluence of regular singularities. Let $v_{\nu }$ and $(Av)_{\nu }$ denote the $\nu$th entry of v and Av, respectively. Then we can write (12) in the form

$$\begin{aligned} (z-\lambda _{\nu })\frac{dv_{\nu }}{dz}=(Av)_{\nu }. \end{aligned}$$

Substituting $z=1/\zeta$, we have

$$\begin{aligned} -\zeta ^2\frac{dv_{\nu }}{d\zeta } =(\zeta ^{-1}-\lambda _{\nu })^{-1}(Av)_{\nu } . \end{aligned}$$

(13)

In the following, $a\mapsto b$ denotes the replacement of a by b.

Let $\zeta \mapsto \epsilon ^{-1}\eta$; and $\lambda _{\nu }\mapsto \epsilon \lambda _{\nu }$ for $\nu \in J$, $\lambda _{\nu }\mapsto \lambda _{\nu }$ for $\nu \in J'$. Multiply the $\nu$th row of A in (13) by $\epsilon ^{-1}$ if $\nu \in J'$ and take the limit $\epsilon \rightarrow 0$. Then (12) is reduced to the Hamiltonian system

$$\begin{aligned} -\eta ^2\frac{dq}{d\eta }=\mathfrak {A}A_1q,\ -\eta ^2\frac{dp}{d\eta }=-{}^tA_1\mathfrak {A}p, \end{aligned}$$

(14)

where $\mathfrak {A}=\mathrm{diag}(\mathfrak {A}_2,\ldots ,\mathfrak {A}_{n})$ and

$$\begin{aligned} \mathfrak {A}_{\nu }:= {\left\{ \begin{array}{ll} -\lambda _{\nu }^{-1} &{}(\nu \in J'), \\ (\eta ^{-1}-\lambda _{\nu })^{-1} &{}(\nu \in J). \end{array}\right. } \end{aligned}$$

(15)

Note that (14) is irregular singular at $\eta =0$.

In order to introduce another singular point, choose any $a\ne 0$ such that $a\ne \lambda _j^{-1}$ for all j and put $\zeta =\eta -a$. Let $\zeta \mapsto \epsilon ^{-1}\zeta$ and $(A)_{\nu }\mapsto \epsilon ^{-1}(A)_{\nu }$. Make substitution $a\mapsto \epsilon ^{-1}a$ for $j\in J'$ and $a\mapsto a$ for $j\in J$ and take the limit $\epsilon \rightarrow 0$. Then (12) is reduced to a Hamiltonian system with irregular points at 0 and $-a$. Set $a=-1$. Finally, by transforming to the autonomous system and putting $\mu _j:=\mu _{j}$, we obtain (3).

3 Proof of Proposition 1

Let H and $H_1$ be given by (3) and (6), respectively. First we show

Lemma 1

If $k\in J$, then $\chi _{H}$ has first integrals

$$\begin{aligned} q_k \exp \left( \frac{\mu _{k}}{q_1-1} \right) , \quad p_k \exp \left( -\frac{\mu _{k}}{q_1-1} \right) , \end{aligned}$$

(16)

while, for $k\in J'$ it has

$$\begin{aligned} q_k \exp \left( \frac{\mu _{k}}{q_1} \right) , \quad p_k \exp \left( -\frac{\mu _{k}}{q_1} \right) . \end{aligned}$$

(17)

Note that $\chi _{H}$ is analytically integrable at $q_1=0$ or $q_1=1$, because $q_kp_k$ is an analytic first integral about the singular point $q_1=0$ or $q_1=1$.

Proof of Lemma 1

The assertion is easily verified in view of the definition of first integrals.

Remark

Lemma 1 says that in the $C^\infty$ class the Hamiltonian is superintegrable. The perturbation in Proposition 1 breaks some first integrals, but not all of them. The remaining ones are not either sufficiently regular for integrability near both points.

Proof of Proposition 1

We have $H_1$ not depending on $p_k$, $k\in J'$, $q_1, q_k$, $k=2,\ldots ,n$ by hypothesis and (6). So the dynamical equations give that $q_k$, $k \in J'$, $q_1, p_k$, $k=2,\ldots ,n$ are first integrals of $H_1$. Thus in particular

$$\begin{aligned} p_k q_k, \ (k\in J'), \quad p_k \exp \left( -\frac{\mu _{k}}{q_1-1} \right) , \quad (k\in J) \end{aligned}$$

(18)

are first integrals of $H_1$, and are analytic at 0. As these are also first integrals of H, they are in involution and first integrals of $H+ H_1$. This ends the proof.

4 Proof of Theorem 1

Let $\phi =:u$ be a holomorphic first integral in $\varOmega$ and expand u at $p=0$

$$\begin{aligned} u=\sum _{\alpha }u_{\alpha }(q_1,q,p_1)p^{\alpha }. \end{aligned}$$

(19)

Substitute (19) into $\chi _{H+H_1}u=0$ and compare the powers like $p^0=1$ of both sides. Then we have the equation of $u_0=u_0(q_1,q,p_1)$

$$\begin{aligned} \left\{ q_1^2p_1,u_0\right\} + \sum _{j\in J'} \mu _{j}q_j \frac{\partial }{\partial q_j}u_0 + \frac{q_1^2}{(q_1-1)^2} \sum _{j\in J} \mu _{j}q_j \frac{\partial }{\partial q_j}u_0 = 0. \end{aligned}$$

(20)

Indeed, no constant term in p appears from $\chi _{H_1}u$ in view of the definition of $\chi _{H_1}$.

Substituting the expansion $u_0=\sum _{\beta }u_{0,\beta }(q_1,p_1)q^{\beta }$ into (20), we see that $U_0:=u_{0,0}$ satisfies $\{q_1^2p_1, U_0\}=0$, namely

$$\begin{aligned} \left( q_1\frac{\partial }{\partial q_1} -2p_1\frac{\partial }{\partial p_1} \right) U_0=0. \end{aligned}$$

(21)

Substitute the expansion $U_0=\sum _{\nu ,\mu } c_{\nu \mu }q_1^{\mu }p_1^{\nu }$ into (21). Then we have

$\sum _{\nu ,\mu }c_{\nu ,\mu }(\mu -2\nu )q_1^{\mu }p_1^{\nu } =0$. It follows that $c_{\nu ,\mu }=0$ for $\mu \ne 2\nu$. Hence we obtain

$$\begin{aligned} U_0 = \sum _{\nu }c_{\nu ,2\nu }q_1^{2\nu }p_1^{\nu } = \sum _{\nu }c_{\nu ,2\nu }(q_1^{2}p_1)^{\nu }. \end{aligned}$$

(22)

It follows that there exists a function of one variable t, $\phi _0(t)$ holomorphic in some neighborhood of $t=0$ such that $U_0=\phi _0(q_1^2p_1)$.

Next, we focus on the equation of $u_{0,\beta }$ with $\beta \ne 0$

$$\begin{aligned} \{q_1^2p_1,u_{0,\beta }\} + \sum _{j\in J'}\mu _{j} \beta _ju_{0,\beta } +\frac{q_1^2}{(q_1-1)^2} \sum _{j\in J }\mu _{j} \beta _ju_{0,\beta } = 0. \end{aligned}$$

Expand

$$\begin{aligned} u_{0,\beta } = \sum _{\nu }\omega _{\beta ,\nu }(q_1)p_1^{\nu }, \end{aligned}$$

(23)

and consider the equation of $\omega _{\beta ,\nu }$. If $\nu =0$, then, by comparing the coefficients of $p_1^0=1$, we have

$$\begin{aligned} q_1^2\frac{d}{dq_1}\omega _{\beta ,0} + \left( \sum _{j\in J'}\mu _{j}\beta _j +\frac{q_1^2}{(q_1-1)^2} \sum _{j\in J }\mu _{j}\beta _j \right) \omega _{\beta ,0} = 0. \end{aligned}$$

(24)

Since $\beta \ne 0$, it follows from (NRC), (7), that either $A':= \sum _{j\in J'}\mu _{j}\beta _j\ne 0$ or $A := \sum _{j\in J }\mu _{j}\beta _j\ne 0$ is valid. If $A'\ne 0$, then we have $\omega _{\beta ,0}=0$ in some neighborhood of $q_1=0$. Indeed, by subsituting the expansion $\omega _{\beta ,0} =\sum _{l=0}^{\infty }C_lq_1^{l}$ into (24) and by using the relations

$$\begin{aligned} q_1^2\frac{d}{dq_1}\omega _{\beta ,0} =\sum _{l=0}^{\infty }C_l{l}q_1^{l+1} \end{aligned}$$

and

$$\begin{aligned} \frac{q_1^2}{(q_1-1)^2} \sum _{j\in J }\mu _{j} \beta _j\omega _{\beta ,0} =\sum _{l=0}^{\infty } C_l'\ {q_1}^{l+2} \end{aligned}$$

for some $C_l'$, we obtain

$$\begin{aligned} C_0A'&=0\text {\quad i.e. }C_0=0, \\ C_1A'&+C_0\cdot 0=0\text {\quad i.e. }C_1=0, \\ C_2A'&+C_0' +C_1=0\text {\quad i.e. }C_2=0, \\&\cdots \end{aligned}$$

Note that $C_0'=0$ since $C_0=0$. Hence we have $\omega _{\beta ,0}=0$.

In the case where $A'=0$ and $A\ne 0$, (24) is written in

$$\begin{aligned} (q_1-1)^2\frac{d}{dq_1}\omega _{\beta ,0} + A\omega _{\beta ,0} = 0. \end{aligned}$$

(25)

Similarly to the case $A'\ne 0$, we obtain $\omega _{\beta ,0}=0$ in some neighborhood of $q_1=1$. Therefore, we have $\omega _{\beta ,0}=0$ in $\varOmega _1$.

Next, by comparing the coefficients of $p_1^1=p_1$, we have the equation of $\omega _{\beta ,1}(q_1)$

$$\begin{aligned} \left( q_1^2\frac{d}{dq_1}-2q_1 \right) \omega _{\beta ,1} + \left( A' +\frac{q_1^2}{(q_1-1)^2} A \right) \omega _{\beta ,1} =0. \end{aligned}$$

(26)

Similarly to the above, $A'\ne 0$ implies $\omega _{\beta ,1}=0$ near $q_1=0$, while $A'=0$ and $A\ne 0$ imply $\omega _{\beta ,1}=0$ near $q_1=1$. Hence we have $\omega _{\beta ,1}=0$ in $\varOmega _1$. By the same argument we obtain $\omega _{\beta ,\nu }=0$ in $\varOmega _1$ for all $\nu \in \mathbb {N}\cup \{0\}$. It follows that $u_{0,\beta }=0$ for all $\beta \ne 0$.

Therefore, we have

$$\begin{aligned} u_0 = u_{0,0}(q_1^2p_1) + \sum _{\beta \ne 0} u_{0,\beta }(q_1^2p_1)q^{\beta } = \phi _0(q_1^2p_1) \end{aligned}$$

(27)

for some $\phi _0(t)$ of one variable being analytic at $t=0$. Note that

$$\begin{aligned} u|_{p=0}-\phi _0(H+H_1)|_{p=0}= & {} u_0(q_1,p_1)-\phi _0(H|_{p=0}) \\= & {} \phi _0(q_1^2p_1)-\phi _0(q_1^2p_1) \equiv 0. \end{aligned}$$

Hence, without loss of generality, we may assume $u|_{p=0}=0$.

Next we consider $u_{\alpha }=u_{\alpha }(q_1,p_1,q)$ for $|\alpha |=1$. Write $\alpha =e_k$ $(2\le k\le n)$ where $e_k:=(0,\ldots ,0,1,0,\ldots ,0)$ is the kth unit vector. Then, $u_{\alpha }$ satisfies

$$\begin{aligned}&\left\{ q_1^2p_1,u_{\alpha }\right\} + \sum _{j\in J'} \mu _{j} \left( q_j\frac{\partial }{\partial q_j}-\delta _{k,j} \right) u_{\alpha } \nonumber \\&\quad + \frac{q_1^2}{(q_1-1)^2} \sum _{j\in J } \mu _{j} \left( q_j\frac{\partial }{\partial q_j}-\delta _{k,j} \right) u_{\alpha } = 0, \end{aligned}$$

(28)

where $\delta _{k,j}$ is the Kronecker’s delta, $\delta _{k,j}=1$ if $k=j$, and =0 if otherwise. Note that, because $u_0=0$, $\chi _{H_1}$ gives no term.

Substitute the expansion $u_{\alpha } =\sum _{\beta }u_{\alpha ,\beta }(q_1,p_1)q^{\beta }$ into (28), and compare the powers like $q^0=1$. Then we have the equation of $u_{\alpha ,0}$

$$\begin{aligned} \left\{ q_1^2p_1,u_{\alpha ,0}\right\} - \mu _{k}\left( \sum _{j\in J'}\delta _{k,j} \right) u_{\alpha ,0} - \frac{q_1^2}{(q_1-1)^2}\left( \sum _{j\in J } \mu _{j}\delta _{k,j} \right) u_{\alpha ,0} = 0. \end{aligned}$$

(29)

If $k\in J'$, then

$$\begin{aligned} \left\{ q_1^2p_1,u_{\alpha ,0}\right\} - \mu _{k}u_{\alpha ,0} = 0. \end{aligned}$$

Because $\mu _{k}\ne 0$ by (NRC) condition, we have $u_{\alpha ,0}=0$.

On the other hand, if $k\in J$, then

$$\begin{aligned} \left\{ q_1^2p_1,u_{\alpha ,0}\right\} - \frac{q_1^2}{(q_1-1)^2} \mu _{k}u_{\alpha ,0} = 0. \end{aligned}$$

By considering the equation around $q_1=1$ together with (NRC) condition we obtain $u_{\alpha ,0}=0$.

Next we consider $u_{\alpha ,\beta }$($\beta \ne 0$) ($\alpha =(\alpha _2,\ldots ,\alpha _n)$, $\alpha _j=\delta _{j,k}$).

$$\begin{aligned}&\left\{ q_1^2p_1,u_{\alpha ,\beta }\right\} + \sum _{j\in J'} \mu _{j} (\beta _j-\alpha _j)u_{\alpha ,\beta } \nonumber \\&\quad +\, \frac{q_1^2}{(q_1-1)^2} \sum _{j\in J } \mu _{j} (\beta _j-\alpha _j)u_{\alpha ,\beta } = 0. \end{aligned}$$

(30)

If $\beta \ne \alpha$, then (NRC) condition yields $u_{\alpha ,\beta }=0$, by the similar argument as in the above. If $\beta =\alpha$, then we have $\{q_1^2p_1,u_{\alpha ,\alpha }\}=0$. Hence, there exists $\phi _\alpha (t)$ of one variable t such that $u_{\alpha ,\alpha }= \phi _{\alpha }(q_1^2p_1)$. Therefore we obtain

$$\begin{aligned} u= \sum _{|\alpha |=1} \phi _{\alpha }(q_1^2p_1) q^{\alpha }p^{\alpha }+O(|p|^2). \end{aligned}$$

(31)

Now we consider the equation for $u_{\alpha }$ when $|\alpha |=2$. We substitute (19) and (31) into the equation $\chi _{H+H_1}u=0$ and compare the powers like $p^\alpha$ $(|\alpha |=2)$. In order to get the expressions of the powers like $p^\alpha$, we note that the following terms appear from $\chi _{H}u$:

$$\begin{aligned}&\left\{ q_1^2p_1,u_{\alpha }\right\} + \sum _{j\in J'} \mu _{j}\left( q_j\frac{\partial }{\partial q_j}-\alpha _j \right) u_{\alpha } + \frac{q_1^2}{(q_1-1)^2}\sum _{j\in J } \mu _{j}\left( q_j\frac{\partial }{\partial q_j}-\alpha _j \right) u_{\alpha } \nonumber \\&\quad +\, \frac{2q_1}{(q_1-1)^3}\sum _{j\in J } \mu _{j}q^{\alpha } \frac{\partial }{\partial p_1}\phi _{\alpha -e_j}. \end{aligned}$$

(32)

On the other hand, the following terms appear from $\chi _{H_1}u$.

$$\begin{aligned}&\sum _{\nu }\frac{\partial }{\partial p_\nu } \left( \sum _{j}p_j^2B_j(q_1,p)\right) \frac{\partial }{\partial q_\nu } (\phi _{e_\nu } q_\nu p_\nu )\nonumber \\&\quad - \frac{\partial }{\partial q_1} \left( \sum _{j}p_j^2B_j(q_1,p)\right) \frac{\partial }{\partial p_1} (\sum _{|\alpha |=1} \phi _\alpha q^\alpha p^\alpha ). \end{aligned}$$

(33)

Note that the second term in (33) is $O(|p|^3)$. Hence it does not appear in the recurrence formula because $|\alpha |=2$. Moreover, since we consider terms of $O(|p|^2)$, the first term yields

$$\begin{aligned} 2 \sum _\nu \phi _{e_\nu } B_\nu (q_1,0) \delta _{\alpha ,2e_\nu }. \end{aligned}$$

(34)

Therefore, by comparing the powers like $p^\alpha$ in $\chi _{H+H_1}u=0$ we have

$$\begin{aligned}&\left\{ q_1^2p_1,u_{\alpha }\right\} + \sum _{j\in J'} \mu _{j}\left( q_j\frac{\partial }{\partial q_j}-\alpha _j \right) u_{\alpha } \nonumber \\&\quad + \frac{q_1^2}{(q_1-1)^2}\sum _{j\in J } \mu _{j}\left( q_j\frac{\partial }{\partial q_j}-\alpha _j \right) u_{\alpha } \nonumber \\&\quad + \frac{2q_1}{(q_1-1)^3} q^{\alpha } \sum _{j\in J } \mu _{j} \frac{\partial }{\partial p_1}\phi _{\alpha -e_j} + 2 \sum _\nu \phi _{e_\nu } B_\nu (q_1,0) \delta _{\alpha ,2e_\nu } =0. \end{aligned}$$

(35)

Expand $u_{\alpha }$ with respect to q, $u_{\alpha } =\sum _{\beta } u_{\alpha ,\beta }(q_1,p_1)q^{\beta }$ and insert the expansion into (35). By comparing the power of $q^{\beta }$ we obtain the recurrence relation for $u_{\alpha ,\beta }(q_1,p_1)$. We consider 4 cases:

• (i) $\alpha \ne 2e_\nu$ for every $\nu$ and $\beta \ne \alpha$.

• (ii) $\alpha =2e_k$ for some k and $\beta \ne \alpha ,0$.

• (iii) $\alpha =2e_k$ for some k and $\beta =0$.

• (iv) $\beta =\alpha$.

Case (i): We note that the fourth and the fifth terms of the left-hand side of (35) yield no term in the recurrence relation for $u_{\alpha ,\beta }$. Indeed, the fourth term is a monomial of $q^\alpha$. Hence, $u_{\alpha ,\beta }$ satisfies

$$\begin{aligned} \left\{ q_1^2p_1,u_{\alpha ,\beta }\right\} + \sum _{j\in J'} \mu _{j}(\beta _j-\alpha _j ) u_{\alpha ,\beta } +\frac{q_1^2}{(q_1-1)^2}\sum _{j\in J } \mu _{j}( \beta _j-\alpha _j ) u_{\alpha ,\beta } = 0. \end{aligned}$$

(36)

By virtue of (NRC) and $\beta \ne \alpha$, either $\sum _{j\in J'}\mu _j (\beta _j-\alpha _j )\ne 0$ or

$\sum _{j\in J } \mu _j ( \beta _j-\alpha _j )\ne 0$ holds. One can easily show that $u_{\alpha ,\beta } = 0$ by the holomorphy of $u_{\alpha ,\beta }$.

Case (ii): Because the fourth and fifth terms of the left-hand side of (35) do not yield terms by the assumption $\beta \ne \alpha ,0$, we see that $u_{\alpha ,\beta }$ satisfies (36). Therefore, we have $u_{\alpha ,\beta }=0$.

Case (iii): Let $k\in J'$. Because the fourth term of the left-hand side of (35) is a monomial $q^\alpha$, $u_{\alpha ,0}$ satisfies

$$\begin{aligned} \left\{ q_1^2p_1,u_{\alpha ,0}\right\} -2\mu _{k}u_{\alpha ,0} + 2\phi _{e_k}(q_1^2p_1)B_k(q_1,0) = 0. \end{aligned}$$

(37)

Expand $u_{\alpha ,0}(q_1,p_1) =\sum _{\nu }u_{\alpha ,0,\nu }(q_1)p_1^{\nu }$ and compare the constant terms in $p_1$ of both sides of (37). Then we have

$$\begin{aligned} q_1^2\frac{d}{dq_1}u_{\alpha ,0,0} - 2\mu _{k}u_{\alpha ,0,0} + 2\phi _{e_k}(0)B_k(q_1,0) = 0. \end{aligned}$$

(38)

If $\phi _{e_k}(0)\ne 0$, then $v:= u_{\alpha ,0,0}/(-2\phi _{e_k}(0))$ satisfies

$$\begin{aligned} q_1^2\frac{d}{dq_1}v - 2\mu _{k}v = B_k(q_1,0), \end{aligned}$$

which contradicts (TC). Hence, $\phi _{e_k}(0)=0$ and (38) reduces to

$$\begin{aligned} q_1^2\frac{d}{dq_1}u_{\alpha ,0,0} - 2\mu _{k}u_{\alpha ,0,0} = 0. \end{aligned}$$

(NRC) condition implies $2\mu _{k}\ne 0$, and the holomorphcity of $u_{\alpha ,0,0}$ at $q_1=0$ tells us $u_{\alpha ,0,0}=0$.

Next, $u_{\alpha ,0,1}$ satisfies

$$\begin{aligned} \left( q_1^2\frac{d}{dq_1}-2q_1\right) u_{\alpha ,0,1} - 2\mu _{k}u_{\alpha ,0,1} + 2B_k(q_1,0)\phi _{e_k}'(0)q_1^2 =0. \end{aligned}$$

(39)

Since $u_{\alpha ,0,1}(q_1)=O(q_1^2)$, we put $u_{\alpha ,0,1}(q_1)= q_1^2\tilde{u}_{\alpha ,0,1}(q_1)$ with $\tilde{u}:=\tilde{u}_{\alpha ,0,1}(q_1)$ satisfying

$$\begin{aligned} q_1^2\frac{d}{dq_1} \tilde{u} -2\mu _{k}\tilde{u} =-2B_k(q_1,0)\phi _{e_k}'(0). \end{aligned}$$

If $\phi _{e_k}'(0)\ne 0$, then, by putting $v=\tilde{u}/(-2\phi _{e_k}'(0))$, we have a contradiction to (TC). Therefore, $\phi _{e_k}'(0)=0$ and $\tilde{u}=0$.

Similarly we can show $u_{\alpha ,0,\nu }=0$ and $\phi _{e_k}^{(\nu )}(0)=0$ for $\nu \in \mathbb {N}\cup \{0\}$, which implies $u_{\alpha ,0}=0$ and $\phi _{e_k}=0$ for every $k\in J'$.

Let $k\in J$. Then $u_{\alpha ,0}$ satisfies

$$\begin{aligned} \{q_1^2p_1,u_{\alpha ,0}\} -2\mu _{k}\frac{q_1^2}{(q_1-1)^2}u_{\alpha ,0} +2\phi _{e_k}(q_1^2p_1)B_k(q_1,0) =0. \end{aligned}$$

Expand $u_{\alpha ,0}(q_1,p_1)=\sum _{\nu }u_{\alpha ,0,\nu }(q_1)p_1^{\nu }$. Then $u_{\alpha ,0,0}$ satisfies

$$\begin{aligned} q_1^2\frac{d}{dq_1}u_{\alpha ,0,0} - 2\mu _{k}\frac{q_1^2}{(q_1-1)^2}u_{\alpha ,0,0} + 2\phi _{e_k}(0)B_k(q_1,0) =0. \end{aligned}$$

(40)

If $\phi _{e_k}(0)\ne 0$, then, by (40) we have $B_k(0,0) =0$. On the other hand, $v:= u_{\alpha ,0,0}/(-2\phi _{e_k}(0))$ satisfies

$$\begin{aligned} q_1^2\frac{d}{dq_1}v - 2\mu _{k}\frac{q_1^2}{(q_1-1)^2}v = B_k(q_1,0), \end{aligned}$$

which contradicts (TC). So, $\phi _{e_k}(0)=0$ and (40) reduces to

$$\begin{aligned} (q_1-1)^2\frac{d}{dq_1}u_{\alpha ,0,0} - 2\mu _{k}u_{\alpha ,0,0} =0. \end{aligned}$$

Again we have $u_{\alpha ,0,0}=0$.

Next, consider the equation of $u_{\alpha ,0,1}$

$$\begin{aligned} \left( q_1^2\frac{d}{dq_1}-2q_1\right) u_{\alpha ,0,1} - 2\mu _{k} \frac{q_1^2}{(q_1-1)^2}u_{\alpha ,0,1} = -2\phi _{e_k}'(0)q_1^2B_k(q_1,0). \end{aligned}$$

(41)

Observing $u_{\alpha ,0,1}(0)=0$, we put $u_{\alpha ,0,1}(q_1)=cq_1+q_1^2 v$. Substituting it into (41), we have $c=-2\phi _{e_k}'(0)B_k(0,0)$ and v satisfies

$$\begin{aligned}&-2\phi _{e_k}'(0)\left\{ B_k(q_1,0) + B_k(0,0) + 2B_k(0,0)\mu _{k} \frac{q_1}{(q_1-1)^2} \right\} \\&\quad =\left( q_1^2\frac{d}{dq_1} -2\mu _{k} \frac{q_1}{(q_1-1)^2} \right) v. \end{aligned}$$

By use of (TC), we obtain $\phi _{e_k}'(0)=0$ and $u_{\alpha ,0,1}=0$.

In general, $u_{\alpha ,0,\nu }$ ($\nu \ge 2$) satisfies

$$\begin{aligned} \left( q_1^2\frac{d}{dq_1}-2\nu q_1\right) u_{\alpha ,0,\nu } - 2\mu _{k} \frac{q_1^2}{(q_1-1)^2}u_{\alpha ,0,\nu } = -2\frac{\phi _{e_k}^{(\nu )}(0)}{\nu !}q_1^{2\nu } B_k(q_1,0). \end{aligned}$$

(42)

Since we easily see $u_{\alpha ,0,\nu }=O(q^{2\nu -1})$, we put $u_{\alpha ,0,\nu }=cq_1^{2\nu -1}+q_1^{2\nu }w$. Then we have $c=-2\phi _{e_k}^{(\nu )}(0)B_k(0,0)/\nu !$ and w satisfies

$$\begin{aligned}&-\frac{2\phi _{e_k}'(0)}{\nu !}\left\{ B_k(q_1,0) + B_k(0,0) + 2B_k(0,0)\mu _{k} \frac{q_1}{(q_1-1)^2} \right\} \\&\quad = \left( q_1^2\frac{d}{dq_1} -2\mu _{k} \frac{q_1^2}{(q_1-1)^2} \right) w. \end{aligned}$$

By virtue of (TC), we obtain $\phi _{e_k}^{(\nu )}(0)=0$ and $w=0$. Therefore, $u_{\alpha ,0,\nu }=0$ for all $\nu \in \mathbb {N}\cup \{0\}$. Because of analyticity, we have $u_{\alpha ,0}=0$ and $\phi _{e_k}=0$ for every $k\in J$. Consequently, $\phi _{e_k}=0$ holds for all $k\in J'\cup J$.

Case (iv): Because $\phi _{e_k}=0$ for every k by what we have proved in the above, the fourth and fifth terms of the left-hand side of (35) do not yield terms in the recurrence relation. Hence, $u_{\alpha ,\alpha }$ satisfies $\{q_1^2p_1,u_{\alpha ,\alpha }\}=0$. It follows that there exists a function of one variable $\phi _{\alpha }(t)$ such that $u_{\alpha ,\alpha }=\phi _{\alpha }(q_1^2p_1)$.

Therefore we have proved

$$\begin{aligned} u= \sum _{|\alpha |=2} \phi _{\alpha }(q_1^2p_1)\ q^{\alpha }p^{\alpha } + O(|p|^3). \end{aligned}$$

Finally we shall prove

Lemma 2

Suppose

$$\begin{aligned} u= \sum _{|\alpha |=\nu } \phi _{\alpha }(q_1^2p_1) q^{\alpha }p^{\alpha } + O(|p|^{\nu +1}) \end{aligned}$$

(43)

for some $\nu \ge 1$. Then we have

• (i) $\phi _{\alpha }=0$ for all $\alpha$ satisfying $|\alpha |=\nu$.

• (ii) For every $\alpha$ satisfying $|\alpha |=\nu +1$, there exists a holomorphic function $\phi _{\alpha }$ of one variable such that

  $$\begin{aligned} u= \sum _{|\alpha |=\nu +1} \phi _{\alpha }(q_1^2p_1) q^{\alpha }p^{\alpha } + O(|p|^{\nu +2}). \end{aligned}$$

  (44)

We have already proved (43) for $\nu =1,2$. Note that the lemma ends the proof of Theorem 1 because we have $u=0$ as an analytic function of q and p.

Proof of Lemma 2

By comparing the coefficients of $p^\alpha$ in $\chi _{H+H_1}u=0$ we have

$$\begin{aligned}&\left\{ q_1^2p_1,u_{\alpha }\right\} + \sum _{J'}\mu _{j} \left( q_j\frac{\partial }{\partial q_j} -\alpha _j \right) u_{\alpha } \nonumber \\&\quad + \frac{q_1^2}{(q_1-1)^2} \sum _{J }\mu _{j} \left( q_j\frac{\partial }{\partial q_j} -\alpha _j \right) u_{\alpha } \nonumber \\&\quad + \frac{2q_1}{(q_1-1)^3} \left( \sum _{J }\mu _{j}q_jp_j \right) \frac{\partial }{\partial p_1} u_{\gamma } + \sum _{j,\gamma } \frac{\partial H_1}{\partial p_j} \frac{\partial }{\partial q_j} u_{\gamma } = 0, \end{aligned}$$

(45)

where $|\gamma |<|\alpha |$ and $\alpha =\gamma +e_j$.

Let $|\alpha |=\nu +1$. Substituting the expansion $u_{\alpha }=\sum _{\beta } u_{\alpha ,\beta }(q_1,p_1)q^{\beta }$ into (45) and by using (43), we obtain the relation for $u_{\alpha ,\beta }$

$$\begin{aligned}&\left\{ q_1^2p_1,u_{\alpha ,\beta }\right\} + \sum _{J'}\mu _{j} ( \beta _j-\alpha _j )u_{\alpha ,\beta } + \frac{q_1^2}{(q_1-1)^2} \sum _{J }\mu _{j} ( \beta _j-\alpha _j )u_{\alpha ,\beta }\nonumber \\&\quad + 2\frac{q_1}{(q_1-1)^3} \sum _{J }\mu _{j} \frac{\partial }{\partial p_1} \phi _{\alpha -e_j}(q_1^2p_1)\delta _{\alpha ,\beta } \nonumber \\&\quad + 2\sum _{j\in J'\cup J} \delta _{\alpha -2e_j,\beta } B_j(q_1,0) \phi _{\alpha -e_j} (\alpha _j-1) = 0. \end{aligned}$$

(46)

Indeed, because it is easy to show the expressions up to the fourth term in the left-hand side of (46), we consider the fifth term, which corresponds to the fifth term in the left-hand side of (45). In view of (43) we may consider $2\sum _{j} p_j B_j(q_1,0)$ in $\frac{\partial H_1}{\partial p_j}$ because other terms have no effect to (45). Hence we may consider terms containing $p^{\alpha -e_j}$ in $\frac{\partial }{\partial q_j} u_{\gamma }$. By (43) the coefficient of the term containing $p^{\alpha -e_j}$ is $(\alpha _j-1)q^{\alpha -2e_j}B_j(q_1,0)\phi _{\alpha -e_j}$. Hence we have the desired expression.

Set $B':=\sum _{\in J'}\mu _{j} (\beta _j-\alpha _j)$ and $B :=\sum _{j\in J }\mu _{j} (\beta _j-\alpha _j)$. We consider 4 cases.

Case (1) The case where $\alpha -2e_j\ne \beta$ for $j=2,\ldots ,n$ and $B'\ne 0$. Clearly we have $\beta \ne \alpha$. It follows that the fourth and the fifth terms in the left-hand side of (46) vanish. Hence we have $u_{\alpha ,\beta }=0$ by considering (46) at $q_1=0$.

Case (2) The case where $\alpha -2e_j\ne \beta$ for $j=2,\ldots ,n$, $\beta \ne \alpha$ and $B' =0$. By (NRC) we have $B\ne 0$. Hence the fourth and the fifth terms in the left-hand side of (46) vanish. We have $u_{\alpha ,\beta }=0$ by considering (46) at $q_1=1$.

Case (3) The case where $\alpha -2e_k=\beta$ for some k. Clearly, we have $\beta \ne \alpha$. Assume $k\in J$. Then, for every $j\in J'$ we have $j\ne k$, and hence $\alpha _j=\beta _j$, which implies $B'=0$. Equation (46) is reduced to

$$\begin{aligned} \left\{ q_1^2p_1, u_{\alpha ,\beta } \right\} - 2\mu _{k}\frac{q_1^2}{(q_1-1)^2} u_{\alpha ,\beta } + 2(\alpha _k-1)\phi _{\alpha -e_k}B_k(q_1,0) = 0. \end{aligned}$$

Expand $u_{\alpha ,\beta } =\sum _{\nu =0}^{\infty } u_{\alpha ,\beta ,\nu }(q_1)p_1^{\nu }$. We will show that $\phi _{\alpha -e_k}$ vanishes.

Indeed, $v:= u_{\alpha ,\beta ,0}$ satisfies

$$\begin{aligned} q_1^2\frac{dv}{dq_1} - 2\mu _{k} \frac{q_1^2}{(q_1-1)^2}v = -2(\alpha _k-1)\phi _{\alpha -e_k}(0) B_k(q_1,0). \end{aligned}$$

Note that $\alpha _k =2+\beta _k \ge 2$. If $\phi _{\alpha -e_k}(0)\ne 0$, then $w :=v/(-2(\alpha _k-1)\phi _{\alpha -e_k}(0))$ is a holomorphic solution at $q_1=0$ of the equation

$$\begin{aligned} q_1^2\frac{dw}{dq_1} - 2\mu _{k} \frac{q_1^2}{(q_1-1)^2}w = B_k(q_1,0). \end{aligned}$$

Because one can verify $B_k(0,0)=0$, we have a contradiction to (TC). Hence we have $\phi _{\alpha -e_k}(0)=0$ and $u_{\alpha ,\beta ,0}=0$.

Next, $v=u_{\alpha ,\beta ,1}$ satisfies

$$\begin{aligned} q_1^2\frac{dv}{dq_1} - 2\mu _{k} \frac{q_1^2}{(q_1-1)^2}v - 2q_1 v = -2(\alpha _k-1) \phi _{\alpha -e_k}'(0) q_1^{2}B_k(q_1,0). \end{aligned}$$

By comparing the coefficients of $q_1^2$ of both sides we see that $v=O(q_1^2)$. Similarly to the above, $w:= v q_1^{-2}$ leads to a contradiction to (TC). Hence, we have $\phi _{\alpha -e_k}'(0)=0$ and $u_{\alpha ,\beta ,1}=0$.

In general, $v=u_{\alpha ,\beta ,\nu }$($\nu \ge 2$) satisfies

$$\begin{aligned} q_1^2\frac{dv}{dq_1} - 2\mu _{k} \frac{q_1^2}{(q_1-1)^2}v - 2q_1\nu v = -\frac{2(\alpha _k-1)}{\nu !} \phi _{\alpha -e_k}^{(\nu )}(0) q_1^{2\nu }B_k(q_1,0). \end{aligned}$$

Similarly to the above, we have $\phi _{\alpha -e_k}^{(\nu )}(0)=0$ and $u_{\alpha ,\beta ,\nu }=0$. Therefore, $\phi _{\alpha -e_k}=0$ and $u_{\alpha ,\beta }=0$ for $k\in J$.

Let $k\in J'$. Equation (46) is reduced to

$$\begin{aligned} \left\{ q_1^2p_1,u_{\alpha ,\beta }\right\} -2\mu _{k} u_{\alpha ,\beta } + 2(\alpha _k-1)\phi _{\alpha -e_k}B_k(q_1,0) =0. \end{aligned}$$

The holomorphicity of $u_{\alpha ,\beta }$ at $q_1=0$ and (TC) implies $\phi _{\alpha -e_k}(0)=0$ and $u_{\alpha ,\beta }=0$ for $k\in J'$. Therefore, $\phi _{\alpha }=0$ for $k\in J'$. Because $\phi _{\alpha }=0$ for $k\in J$, we have $\phi _{\alpha }=0$ for all $\alpha$ with $|\alpha |=\nu$.

Case (4) The case $\beta =\alpha$. We have $\{q_1^2p_1,u_{\alpha ,\alpha }\}=0$, since we have proved $\phi _\gamma =0$ for $|\gamma |=\nu$. Hence, there exists $\phi _\alpha$ such that $u_{\alpha ,\alpha }= \phi _{\alpha }(q_1^2p_1)$.

Consequently, we have proved the lemma. $\square$

5 Properties of (TC)

We will show that (TC) holds for almost all $B_k(q_1,0)$. Set $q_1=t$, $B_k(t,0)=: a(t)$ and $c:= \mu _{k}$, and write (8) in the form

$$\begin{aligned} t^2\frac{d}{d t}v - 2c v = a(t). \end{aligned}$$

(47)

Clearly, if a(t) is a constant function, then (TC) does not hold since (47) has a constant solution $v=-a(0)/(2c)$. We first prove

Proposition 2

Suppose that a(t) is a polynomial of degree $\ell \ge 1$. Then (47) has an analytic solution at $t=0$ if and only if (47) has a polynomial solution v of degree $\ell -1$. The set of a(t) for which (47) has a polynomial solution is contained in the set of codimension one of the set of polynomials of degree $\ell$.

Remark

For a given polynomial v of degree $\ell -1$, define a(t) by (47). Clearly the set of a’s such that (47) has a polynomial solution is an infinite set.

Proof of Proposition 2

Let $a(t)=\sum _{j=0}^\ell a_j t^j$ $(a_\ell \ne 0)$ and let $v(t)= \sum _{j=0}^\infty v_j t^j$ be the analytic solution of (47). By inserting the expansions into (47) and by comparing the powers of t we obtain

$$\begin{aligned} v_0= -a_0/(2c), \quad v_n=(n-1)v_{n-1}/(2c) - a_n/(2c), \ n=1,2,\ldots \end{aligned}$$

(48)

If $n>\ell$, then we have $v_n=(n-1)v_{n-1}/(2c)$. Therefore, if $v_\ell =0$, then $v_n=0$ for $n>\ell$. Hence v is a polynomial. On the other hand, if $v_\ell \ne 0$, then $v_n = (2c)^{\ell -n} (n-1)(n-2)\cdots \ell v_\ell$. It follows that v(t) is not analytic in any neighborhood of the origin, which contradicts to the assumption. Hence v is a polynomial of degree $\ell -1$. The converse statement is trivial.

We will show the latter half. By the recurrence formula (48), one easily sees that $v_\ell$ is a nontrivial linear function of $a_0, \ldots , a_\ell$. Hence the condition $v_\ell =0$ is satisfied for a polynomial a(t) on the set of codimension 1. This completes the proof. $\square$

Example

We give an example of $B_k(q_1,0)$’s satisfying the condition (TC) in Theorem 1. We use the notation in Proposition 2. If $k\in J'$, then we look for $a(t) \equiv B_k(t,0)$ such that $a(t) = \alpha t + \beta t^2$ for some complex constants $\alpha$ and $\beta$. In order to verify that (47) has no solution v being analytic at $t=0$, we expand $v(t)= \sum _{j=0}^\infty v_j t^j$ and consider the recurrence relation (48). We assume that $c= \mu _{k}\ne 0$. Clearly, we have $v_1= -\alpha /(2c)$ and $v_1 - 2c v_2 =\beta$. It follows that $v_2 = -(\alpha /(2c) +\beta )/(2c)$. For $n \ge 3$, we have $v_n=(n-1)v_{n-1}/(2c)$, which implies $v_n = (n-1)! (2c)^{2-n} v_2$. Therefore, if $v_2\ne 0$, then v does not converge. Hence (47) has no analytic solution. We observe that $v_2\ne 0$ holds if $\alpha /(2c) +\beta \ne 0$.

Next we assume $k\in J$, and we consider (9) in (TC). (9) is rewritten in (53) which follows. We look for b(t) such that $b(t)= \gamma t^2 + \delta t^3$ for some complex constants $\gamma$ and $\delta$. We set $q_1=t+1$. Since $b(0)=0$, we have $a(0)=0$. Hence, by (53) we have the relation

$$\begin{aligned} a(t+1) =a(q_1)= (\gamma + \delta t )(t+1)^2= q_1^2 (\gamma -\delta + \delta q_1). \end{aligned}$$

In order to verify (TC) we argue as in the above. We expand w(t) in the series $w(t) = w_2 t^2 + w_3 t^3 + \cdots$ and we subsitutute it into (53). By comparing the powers of $t^2$ of both sides we have $w_2= -\gamma /(2c)$. Similarly, we have $w_3= -(\gamma /c + \delta )/(2c)$. If $\gamma + c\delta \ne 0$, then we have $w_3\ne 0$ and we see that the formal power series expansion of $w(t) = w_2 t^2 + w_3 t^3 + \cdots$ diverges. Hence we have the desired property. Consequently, we choose $B_k(q_1,0)= \alpha q_1 + \beta q_1^2$ with $\alpha /(2c) +\beta \ne 0$ for $k\in J'$, and $B_k(q_1,0)= q_1^2(\gamma -\delta + \delta q_1)$ with $\gamma + c\delta \ne 0$ for $k\in J$. Then we see that (TC) is satisfied.

Next we study (TC) when a(t) is an analytic function. By replacing v(t) and a(t) with $v(t)-v(0)$ and $a(t)-a(0)$, $(2c v(0)= -a(0))$, respectively, we may assume that $v(0)=0$ and $a(0)=0$ in (47). Then we have

Proposition 3

The set of analytic functions a(t)’s at the origin such that (47) has an analytic solution v is contained in the set of codimension 1 of the set of germs of analytic functions at $t=0$.

Proof

Let v be the analytic solution of (47) at $t=0$. Set $v(t)=t\tilde{v}(t)$ and $a(t)=t\tilde{a}(t)$. Then

$$\begin{aligned} t^2\frac{d}{d t}\tilde{v} + t\tilde{v} -2c \tilde{v} = \tilde{a}(t). \end{aligned}$$

(49)

We make the (formal) Borel transform ${\mathcal B}(\tilde{v})$ to (49)

$$\begin{aligned} {\mathcal {B}}(\tilde{v})(z)\equiv \widehat{\tilde{v}}(z):= \sum _{n=1}^\infty v_n \frac{z^{n-1}}{(n-1)!}. \end{aligned}$$

(50)

Because $\tilde{v}(t)$ and $\tilde{a}(t)$ are analytic at $t=0$, it follows that ${\mathcal B}(\tilde{v})(z)$ and ${\mathcal B}(\tilde{a})(z)$ are entire functions of exponential type of order 1. Recalling that ${\mathcal {B}} \left( (t^2\frac{d}{d t} + t)\tilde{v}\right) (z) =z{\mathcal {B}}(\tilde{v})(z)$ we have

$$\begin{aligned} (z -2c){\mathcal {B}}(\tilde{v}) = {\mathcal {B}}(\tilde{a})(z). \end{aligned}$$

(51)

It follows that

$$\begin{aligned} {\mathcal {B}}(\tilde{a})(2c)=0. \end{aligned}$$

(52)

This shows that the germ $\{ a_n \}_{n=1}^\infty$ of a(t) at $t=0$ is contained in the hyperplane. This ends the proof. $\square$

Next we consider (9) in (TC). We set $t=q_1-1$, $a(t+1):= B_k(t+1,0)$, $c=\mu _{k}$ and $a(0)=B_k(0,0)$. Then (9) can be written in

$$\begin{aligned} \left( t^2\frac{d}{dt} - 2c \right) w = \frac{t^2}{(t+1)^2} a(t+1) + \frac{a(0)}{(t+1)^2} (t^2+c(t+1))=:b(t). \end{aligned}$$

(53)

This equation has the same form as (47). We determine w(0) by $-2c w(0)=b(0)$. If we make the appropriate change of unknown functions w and b as before, one may assume that $w(0)=0$ and $b(0)=0$. In view of the definition of b(t) we have $c a(0)=0$. Hence we have $a(0)=0$. It follows that $b(t)=t^2 a(t+1)/(t+1)^2$. In the following we assume $w(0)=0$ and $a(0)=0$. Then we have

Proposition 4

Suppose that a(t) is holomorphic in a connected domain containing $t=0$ and $t=1$. Then the set of a(t) for which (53) has an analytic solution is contained in the set of codimension one of the set of germs of analytic functions at $t=0$.

Proof

Let w(t) be an analytic solution of (53) at $t=0$. We set $\alpha :=a'(0)$ and $a(z) = \alpha z +A(z)z^2$ for some analytic function A(z). Then, by the general formula w is given by

$$\begin{aligned} w= \exp \left( - \frac{2c}{t} \right) \left( K + \int _\tau ^t \exp \left( \frac{2c}{s}\right) \left( \frac{\alpha }{s+1} +A(s+1)\right) ds \right) , \end{aligned}$$

(54)

where K and $\tau \ne 0$ are some constants. We take a smooth curve $\gamma$ which connects $\tau$ and the origin such that it stays in the half space, $\mathfrak {R}\, (c/t)<0$ near the origin. Then the limit

$$\begin{aligned}&\int _\tau ^0 \exp \left( \frac{2c}{s}\right) \left( \frac{\alpha }{s+1} +A(s+1)\right) ds \qquad \qquad \nonumber \\&\quad := \lim _{t\in \gamma , t\rightarrow 0} \int _\tau ^t \exp \left( \frac{2c}{s}\right) \left( \frac{\alpha }{s+1} +A(s+1)\right) ds \end{aligned}$$

(55)

exists and it is a non-constant analytic function of $\tau$. If the condition

$$\begin{aligned} K + \int _\tau ^0 \exp \left( \frac{2c}{s}\right) \left( \frac{\alpha }{s+1} +A(s+1)\right) ds \ne 0 \end{aligned}$$

(56)

holds, then, by taking the limit $t\rightarrow 0$, $\mathfrak {R}\, (c/t)<0$ in (54) we see that w(t) tends to infinity, which contradicts to the analyticity of w at the origin. Hence we have

$$\begin{aligned} K= \int ^\tau _0 \exp \left( \frac{2c}{s}\right) \left( \frac{\alpha }{s+1} +A(s+1)\right) ds. \end{aligned}$$

(57)

By substituting (57) to (54) we have

$$\begin{aligned} w(t) = \exp \left( - \frac{2c}{t} \right) \int _0^t \left( \frac{2c}{s}\right) \left( \frac{\alpha }{s+1} +A(s+1)\right) ds. \end{aligned}$$

(58)

We take t sufficiently close to the origin such that the Taylor expansion $A(s+1)= \sum _{n=0}^\infty a_n s^n$ converges for $|s|\le |t|$. Because $w(te^{2\pi i})=w(t)$ holds by the analyticity of w, it follows that

$$\begin{aligned} \int _t^{te^{2\pi i}} \exp \left( \frac{2c}{s}\right) \left( \frac{\alpha }{s+1} +A(s+1)\right) ds =0. \end{aligned}$$

(59)

By calculating the residue we have $\int _t^{te^{2\pi i}} \exp \left( \frac{2c}{s}\right) \frac{\alpha }{s+1}ds = 2\pi i \alpha (1-e^{-2c})$. The non-resonance condition implies $c=\mu _{k}\ne 0$, and hence $1-e^{-2c}\ne 0$. Hence, by (59) the germ of $A(z)/\alpha$ at $z=1$ (in case $\alpha \ne 0$) or that of A(z) at $z=1$( in case $\alpha =0$) is contained in some hyperplane of the set of germs of analytic functions.

We recall that A(z) is analytic in some domain containing $z=0$ and $z=1$. We will show that by the analytic continuation from $z=1$ to $z=0$ the germ of A(z) at $z=1$ is transformed to that of A(z) at $z=0$ by an infinite matrix. If we can prove this, then the germ of A(z) or $A(z)/\alpha$ at $z=0$ is contained in some hyperplane. In view of $a(z) = \alpha z +A(z)z^2$, the germ of a(z) at $z=0$ is contained in some hyperplane.

We take a rectifiable curve which connects $z=1$ and $z=0$. First we consider the analytic continuation from $z=1$ to $z=z_0$, where $z_0$ is contained in the disk centered at $z=1$ in which A(z) is analytic. Let $A(z) = \sum _{n=0}^\infty a_n (z-1)^n$ be the expansion at $z=1$. Then the Taylor expansion of A(z) at $z=z_0$ is given by

$$\begin{aligned} \sum _{k=0}^\infty \frac{(z-z_0)^k}{k!}\sum _{n=k}^\infty a_n (z_0-1)^{n-k}\frac{n!}{(n-k)!}. \end{aligned}$$

(60)

It follows that the germ at $z=z_0$ is given by

$$\begin{aligned} \left( \sum _{n=k}^\infty a_n {n\atopwithdelims ()k} (z_0-1)^{n-k} \right) _{k=0}^\infty . \end{aligned}$$

(61)

Hence the germ at $z=1$ is transformed to the one in (60) by the infinite matrix

$$\begin{aligned} {\mathcal A}:= \left( (z_0-1)^{n-k} {n\atopwithdelims ()k} \right) _{k\downarrow 0,1,\ldots ; n\rightarrow 0,1,\ldots }, \end{aligned}$$

(62)

where we set the (k, n)-component $(k>n)$ to be zero. Note that if $|z_0-1|$ is sufficiently small, then ${\mathcal A}$ defines a continuous linear operator on the space of sequences with an appropriate norm. Therefore, if the germ of A(z) at $z=1$ is contained in the hyperplane, then the germ of A(z) at $z=z_0$ is contained in some hyperplane. By finite times of analytic continuation we see that the germ of A(z) at $z=0$ is contained in some hyperplane. This completes the proof. $\square$