These authors contributed equally to this work.

[1]Boris A. Percino-Figueroa These authors contributed equally to this work.

These authors contributed equally to this work.

[1]Facultad de Ciencias en Física y Matemáticas, Universidad Autónoma de
Chiapas, Carr. Emiliano Zapata km 8, Tuxtla Gutiérrez, 29050, Chiapas, México

2] CONAHCyT, Av. de los Insurgentes Sur 1528, Benito Juarez, 03940, Ciudad de
México, México

# On the integrability of generalized $N$-center problems

Eddaly Guerra-Velasco  edaly.velasco<EMAIL_ADDRESS><EMAIL_ADDRESS>Russell-Aarón Quiñones-Estrella
<EMAIL_ADDRESS>*  [

###### Abstract

In this paper, we study the rational integrability of the $N$-center problem
with rational weak and moderate forces. We show that the problem is not
rationally integrable for all but a finite number of values $\alpha\in]0,2[$,
where $\alpha$ is the order of the singularities. We identify the remaining
cases and give the necessary conditions for integrability.

###### keywords:

Non-integrability, Celestial Mechanics, $N$-center problem, Hamiltonian
Systems

###### pacs:

[

MSC Classification]37J30,70F15

##  1 Introduction

This paper deals with the rational integrability of the $N$\- center problem
in $\mathbb{R}^{d}$. To be more specific, let $m_{1},\dots,m_{N}$ positive
masses located at fixed position vectors $c_{1},\dots,c_{N}\in\mathbb{R}^{d}$,
assume that $c_{i}\neq c_{j}$ if $i\neq j$. The potential gives the law
governing the motion of a particle $x\in\mathbb{R}^{d}$,

$U(q)=-\sum_{i=1}^{N}\frac{m_{i}}{\|q-c_{i}\|^{\alpha}},$ (1)

where $\alpha\in]0,2[$ and for $q=(q_{1},\dots,q_{d})\in\mathbb{R}^{d}$,
$\|q\|$ denotes the Euclidian norm

$\|q\|^{2}=\sum_{i=1}^{d}q_{i}^{2}.$

In this case, the Hamiltonian of the system is defined on the set
$\Omega\times\mathbb{R}^{d}$, where $\Omega=\\{q\in\mathbb{R}^{d}|q\neq
c_{k},\;k=1,\dots,N\\}$, and is given by

$H(q,p)=\frac{1}{2}\left\|\,p\,\right\|^{2}+U(q).$ (2)

The equations of motion are

$\begin{split}\frac{q_{j}}{dt}&=\frac{\partial H}{\partial p_{j}}(q,p),\\\
\frac{p_{j}}{dt}&=-\frac{\partial H}{\partial q_{j}}(q,p).\end{split}$ (3)

where $(q,p)=(q_{1},\dots,q_{d},p_{1},\dots,p_{d})$.

A special case of this class of Hamiltonians, and also the most studied, is
the $N$-center problem of celestial mechanics, in which $d=2$ or $3$,
$\alpha=1$ and the potential is given by

$U(q)=-\;\sum_{j=1}^{N}\;\frac{m_{j}}{\left\|\,q-c_{j}\,\right\|}.$

This is a simplified model for a Newtonian $(N+1)$-body problem, assuming that
one of masses moves faster than the other $N$. For $N=1$, this is the Kepler
problem, which describes the evolution of the $2$-body problem and whose
dynamics is well understood actually [1].

For a general Hamiltonian system (3) defined by a function
$H:M\subset\mathbb{R}^{2d}\to\mathbb{R}$, $\;M\neq\emptyset$ open, a _first
integral_ is a $C^{1}$ function $F:M\to\mathbb{R}$ such that is constant along
each solution of (3). This is equivalent to the fact that the _Poisson
bracket_ of $F$ and $H$ is zero, that is to say:

$\\{F,H\\}=\sum_{i=1}^{d}(\frac{\partial F}{\partial q_{i}}\frac{\partial
H}{\partial p_{i}}-\frac{\partial F}{\partial p_{i}}\frac{\partial H}{\partial
q_{i}})=0.$

Two functions $F,G$ such that $\\{F,G\\}=0$ are said to be in _involution_.

A Hamiltonian system defined by $H:M\subset\mathbb{R}^{2d}\to\mathbb{R}$ is
called _(rationally) integrable_ if there are $d$ (rational) first integrals
$F_{1},\dots F_{d}$ that are in involution and such that $\nabla
F_{1},\dots,\nabla F_{d}$ are linearly independent in an dense subset of $M.$
It is well known that orbits in an integrable system are conjugated to orbits
of linear flows on $d-$dimensional tori contained in $M$; on the other hand,
orbits of a nonintegrable system tend to be chaotic [1]. Therefore, it is
crucial to determine whether or not a Hamiltonian system is integrable.

Different methods have been studied for a long time to prove the integrability
of some Hamiltonian systems. Noether’s Theorem, for instance, establishes that
if the system possesses symmetry, then it possesses some first integrals [1,
2] Kepler’s problem features rotational symmetry; thus, the angular momentum
is a first integral. Another approach is the Hamilton-Jacobi method, which
consists of finding smooth global solutions to the associated Hamilton-Jacobi
equation. If these solutions exist, they can be used to define new coordinates
so that the system in these coordinates is decoupled [3].

Differential Galois theory has recently provided a powerful tool to determine
the integrability of certain differential equations. In the case of the
$N$-body problem, using the Morales–Ramis theory, Maciejewski and Przybylska
proved in [4] the non-integrability of the three-body problem for any fixed
masses.

As we mentioned before, the Kepler problem is integrable. On the other hand,
Euler found a first integral of motion for the planar 2-center problem and,
therefore, remains in the class of integrable systems.

As already discovered by Poincaré, in the case $\alpha\geq 2$ (usually known
as a strong force system), it is easy to find periodic solutions using
variational methods [5]. By using differential Galois theory, Shibayama [6]
showed, in the Newtonian spatial case (_i.e._ $\alpha=1,d=3$), the non-
integrability of the $N-$center problem. In contrast, Bolotin [5] and Koslov
[7] proved, by geometric methods, that the planar $N-$center problem, $N\geq
3$, is not integrable for moderate forces. By moderate forces, we mean that
$1<\alpha<2$; recall that by weak forces, it is meant $0<\alpha<1$. This paper
follows Shibayama’s ideas to extend his results to non-planar weak and
moderate forces $\alpha\in]0,2[$, with $\alpha$ a rational number.

The method is the following. Since the potential (1) is neither meromorphic
nor homogeneous, it is necessary to find a new related meromorphic and
homogeneous Hamiltonian $H_{0}$ such that if $H$ is rationally integrable,
then $H_{0}$ is also rationally integrable. Thus, we focus our attention on
the integrability of $H_{0}$. Theorem 2 in [8] in particular implies that the
variational equation along a non-stationary orbit of the system associated
with $H_{0}$ is solvable; see [8], [9] for more details in the Differential
Galois Theory in Hamiltonian systems. Therefore, we compute a non-stationary
solution of the Hamiltonian system associated with $H_{0}$ and the variational
equation along this solution. Finally, we use a theorem by Kimura (Theorem 3)
to give conditions for solvability, resulting in conditions for the
integrability of the original Hamiltonian system. An example of this procedure
arises from the N-center problem, where some centers are in a regular convex
polygon (see Example 5)

##  2 Homogenization procedure

As we said before, we aim to apply Combot’s Theorem [8] on meromorphic
homogeneous potentials to have necessary conditions for the integrability of
system (3) in the general $N$-center problem setting. The main difficulty is
that once extended to $\mathbb{C}^{d}$, the potential (1) is neither
meromorphic nor homogeneous. Therefore, we must switch the problem to a
convenient meromorphic homogeneous potential. To this end, let
$c_{k}=(c_{k}^{1},\dots c_{k}^{d})\in\mathbb{R}^{d}$ be the centers in the
canonical coordinates and extend the domain and Hamiltonian function from
$\mathbb{R}^{d}\times\mathbb{R}^{d}$ to $\mathbb{C}^{d}\times\mathbb{C}^{d}$.
We will denote the extension $H$ with the same symbol. Now classify the set of
singularities through

$\mathcal{S}_{k}=\\{(x_{1},\dots,x_{d})\in\mathbb{C}^{d}|(x_{1}-c_{k}^{1})^{2}+\cdots+(x_{d}-c_{k}^{d})^{2}=0\\},\;k=1,\dots,N.$

Notice that the $\mathcal{S}_{k}$’s are complex analytic varieties of co-
dimension 1; some can intersect.

We can assume, without loss of generality, that
$\mathcal{S}_{1}\cap\dots\cap\mathcal{S}_{l}\neq\emptyset$, Consider a
singularity $s\in\mathcal{S}_{1}\cap\dots\cap\mathcal{S}_{l}$ but
$s\notin\mathcal{S}_{l+1}\cap\dots\cap\mathcal{S}_{N}$. Let us notice that,
even when the kinetic term $\frac{1}{2}\left\|\,p\,\right\|^{2}$ is
holomorphic and homogeneous, the potential terms

$U(q)=-\sum_{i=1}^{N}\frac{m_{i}}{\|q-c_{i}\|^{\alpha}},$

are not. From now on, $\alpha$ will denote a rational number, say
$\alpha=\frac{r}{s}$ with $r,s\in\mathbb{Z}$ and $r,s>0$.

In order to homogenize the problem consider $x\cdot
y=x_{1}y_{1}+\dots+x_{d}y_{d},\,x_{i},y_{i}\in\mathbb{C}$ and define

$H_{\lambda}(q,p)=\lambda^{2r}H(\lambda^{4s}q^{+}e,\lambda^{-r}p)\\\
=\frac{1}{2}\left\|\,p\,\right\|^{2}-\sum_{i=1}^{l}\frac{m_{i}}{[\lambda^{4s}q\cdot
q-2q\cdot(e-c_{i})]^{\frac{r}{2s}}}\\\
-\lambda^{2r}\sum_{i=l+1}^{N}\frac{m_{i}}{\lambda^{8s}(q\cdot
q+\lambda^{4s}2q\cdot(e-c_{i}))+(e-c_{i})\cdot(e-c_{i})}$ (4)

Using the principal branch of logarithm and expanding in the Taylor series
near $\lambda=0$, we get

$H_{\lambda}(q,p)=H_{0}(q,p)+\lambda H_{1}(q,p)+\dots,$

where

$\begin{split}H_{0}(q,p)&=\frac{1}{2}\left\|\,p\,\right\|^{2}-\sum_{i=1}^{l}\frac{m_{i}}{[2q(e-c_{i})]^{\frac{r}{2s}}}\\\
&=\frac{1}{2}\left\|\,p\,\right\|^{2}-\sum_{i=1}^{l}\frac{m_{i}}{[2q(e-c_{i})]^{\frac{\alpha}{2}}}.\end{split}$
(5)

Let us notice that the new Hamiltonian $H_{0}$ is meromorphic and homogeneous,
and the corresponding Hamilton’s equations are:

$\displaystyle\dot{q}$ $\displaystyle=p$ $\displaystyle\dot{p}$
$\displaystyle=\sum_{i=1}^{l}\frac{m_{i}}{\frac{\alpha}{2}[2q\cdot(e-c_{i})]^{\frac{\alpha+2}{2}}}(e-c_{i}).$

Now let $F:\mathbb{C}^{d}\to\mathbb{C}$ be a rational function and consider

$F_{\lambda}(q,p)=F(\lambda^{4s}q+e,\lambda^{-r}p);$

the singularity $\lambda=0$ is a pole, since $F$ is rational. Hence we can
write the Laurent series of $F(\lambda^{4s}q+e,\lambda^{-r}p)$ with respect to
$\lambda$ as follows

$F_{\lambda}(q,p)=F(\lambda^{4s}q+e,\lambda^{-r}p)=\lambda^{k}F_{0}(q,p)+\lambda^{k+1}F_{1}(q,p)+\dots,\hskip
14.22636ptk<0.$ (6)

######  Remark 1.

If $\nabla F$ and $\nabla H$ are linearly independent, then $\nabla
F_{\lambda}$ and $\nabla H_{\lambda}$ are also linearly independent.

For the next part, we introduce the following terminology about the Laurent
series. Given a Laurent series

$F(\lambda)=a_{k}\lambda^{k}+a_{k+1}\lambda^{k+1}+\dots$

with $a_{k}\neq 0$, we call the coefficient $a_{k}$ the _lowest part of_
$F(\lambda)$.

######  Lemma 1.

If $H$ has a rational first integral $F$ such that $\nabla F$ is linearly
independent to $\nabla H$ then there is a rational first integral $\tilde{F}$
for $H$ such that $\nabla\tilde{F},\nabla H$ are linearly independent and
$\nabla\tilde{F}_{0},\nabla H_{0}$ are linearly independent.

###### Proof.

If $\nabla F_{0}$ and $\nabla H_{0}$ are linearly independent, it is enough to
take $\tilde{F}=F$.

Assume that $\nabla F_{0},\nabla H_{0}$ are linearly dependent. Since $\nabla
H_{0}$ is different from zero, there is a constant $C_{1}$ such that

$\nabla F_{0}=C_{1}\nabla H_{0}.$ (7)

Let

$F^{(1)}=F-\lambda^{k+2r}C_{1}H$

and

$\displaystyle F_{\lambda}^{(1)}$
$\displaystyle=F_{\lambda}-\lambda^{k}C_{1}H_{\lambda}$
$\displaystyle=(F_{0}-C_{1}H_{0})\lambda^{k}+(F_{1}-C_{1}H_{1})\lambda^{k+1}+\dots$

then by (7)

$\nabla F_{\lambda}^{(1)}=(\nabla F_{1}-C_{1}H_{1})\lambda^{k+1}+\dots$

and the lowest part of $\nabla F_{\lambda}^{(1)}$ is different from $\nabla
F_{0}$. Notice that

$\nabla F_{\lambda}-\lambda^{k}C_{1}\nabla H_{\lambda}\neq 0,$

due to Remark 1. Therefore there is $j_{1}\geq 1$ such that lowest part of
$\nabla F^{(1)}$ is $(\nabla F_{j_{1}}-C_{1}\nabla H_{j_{1}})$. Now if
$(\nabla F_{j_{1}}-C_{1}\nabla H_{j_{1}})$ and $\nabla H_{0}$ are linearly
independent, we take $\tilde{F}=F^{(1)}$.

If $(\nabla F_{j_{1}}-C_{1}\nabla H_{j_{1}})$ and $\nabla H_{0}$ are linearly
dependent, there exists a constant $C_{2}$ such that

$(\nabla F_{j_{1}}-C_{1}\nabla H_{j_{1}})=C_{2}\nabla H_{0},$

and we define

$\displaystyle F^{(2)}$ $\displaystyle=F^{(1)}-\lambda^{k+j_{1}+2r}C_{2}H$
$\displaystyle=F-\lambda^{k+2r}C_{1}H-\lambda^{k+j_{1}+2r}C_{2}H$
$\displaystyle=F-\lambda^{k+2r}(C_{1}+\lambda^{j_{1}}C_{2})H.$

Proceeding recursively, we construct functions

$F^{(m)}=F-\lambda^{k+2r}(C_{1}+\lambda^{j_{1}}C_{2}+\lambda^{j_{2}}C_{2}+\dots+\lambda^{j_{m}-1}C_{m})H.$
(8)

We claim that this process can not go further indefinitely. Otherwise, we
would have the following series

$\phi(\lambda)=C_{1}+\lambda^{j_{1}}C_{2}+\lambda^{j_{2}}C_{2}+\dots+\lambda^{j_{m}-1}C_{m}\dots,$

and it is not difficult to see that $\phi(\lambda)$ converges if and only if
$\nabla F^{(m)}$ converges to zero.

The last is true due to the convergence of the series of $F_{\lambda}$ and
$H_{\lambda}$. This, in turn, implies that

$\nabla F-\lambda^{k+2r}\phi(\lambda)\nabla H=0$

contradicting the linear independence of $\nabla F$ and $\nabla H$.

Finally, by the properties of the Poisson bracket, each $F^{(m)}$ is a first
integral of $H$. In particular, the same is true for $\tilde{F}$. ∎

We can generalize a proposition by Shibayama [6].

######  Proposition 2.

If the Hamiltonian $H$ given in (2) is integrable, then the Hamiltonian
$H_{0}$ is integrable.

###### Proof.

Suppose that $H$ is rationally integrable and let
$F:\mathbb{C}^{d}\to\mathbb{C}$ be a rational first integral for $H$ such that
$\nabla F$ and $\nabla H$ are linearly independent. As in the previous
discussion, consider the function
$F_{\lambda}(p,q)=F(\lambda^{4s}q^{+}e,\lambda^{-r}p)$, and expand in Laurent
series

$F(\lambda^{4s}q^{+}e,\lambda^{-r}p)=\lambda^{K}F_{0}(q,p)+\lambda^{K+1}F_{1}(q,p)+\lambda^{K+2}F_{2}(q,p)+...,\,K<0.$

We can assume that $\nabla F_{0}$ and $\nabla H_{0}$ are linearly independent;
if this is not the case, we switch $F$ by the correspondent first integral
$\tilde{F}$ of $H$ given by Lemma 1.

On the other hand, by using the properties of the Poisson bracket, we obtain

$\begin{split}\\{H(\lambda^{4s}q^{+}e,\lambda^{-r}p),F(\lambda^{4s}q^{+}e,\lambda^{-r}p)\\}&=\\\
\lambda^{K-2r}\\{H_{0},F_{0}\\}(q,p)&+\lambda^{K+1-2r}(\\{H_{1},F_{0}\\}(q,p)+\\{H_{0},F_{1}\\}(q,p))\\\
&+o(\lambda^{K+1-2r})\,\text{ as }\,\lambda\to 0.\end{split}$

From $\\{H,F\\}=0$ and the uniqueness of the Laurent series expansion, we have
that

$\\{H_{0},F_{0}\\}=0,$

this means, $H_{0}$ is a rational first integral of $H$ such that $\nabla
F_{0}$ and $\nabla H_{0}$ are linearly independent. ∎

##  3 Variational equation

This section aims to compute the variational equation along a particular
solution and reduce it conveniently to a second-order differential equation.
The equations of motion for $H_{0}$ are given as

$\begin{split}\frac{dq}{dt}&=p\\\
\frac{dp}{dt}&=\sum_{i}\frac{m_{i}}{\frac{\alpha}{2}[2q\cdot(e-c_{i})]^{\frac{\alpha+2}{2}}}(e-c_{i}).\end{split}$
(9)

As customary, we look for homographic type solutions for (9), this means,
solutions of the form

$q(t)=g(t)(v_{1},\dots,v_{d})$ (10)

where $V=(v_{1},\dots,v_{d})\in\mathbb{C}^{d}$ is a fixed vector to be
determined. Let us notice that this type of solution is inspired by the
homographic solutions associated with the N-body problem, giving rise to the
so-called central configurations. We will not have any difficulties finding
such a vector $V$ since we are taking advantage of the fact that $\mathbb{C}$
is algebraically closed.

Plugin (10) in (9), we get that there is a constant $C$ such that

$\sum_{i=1}^{l}\frac{m_{i}}{\alpha[v_{i}(e-c_{i}]^{\frac{\alpha+2}{2}}}(e-c_{i})=C$
(11)

and

$\frac{d^{2}g(t)}{dt^{2}}=\frac{C}{(g(t))^{\frac{\alpha+2}{2}}}.$ (12)

Equation (11) has solutions for any constant $C\in\mathbb{C}$, while equation
(12) is easily solvable by separation of variables. Once a particular solution
$q(t)$ of the form (10) has been chosen, we have that the variational equation
for (9) around $q(t)$ is given by

$\begin{bmatrix}\dot{Y}_{1}&\dot{Y}_{2}\\\
\dot{Y}_{3}&\dot{Y}_{4}\end{bmatrix}=\begin{bmatrix}Y_{3}&Y_{4}\\\
B(q(t))Y_{1}&B(q(t))Y_{2}\end{bmatrix},$ (13)

where each block is a $d\times d$ matrix and $B(q)$ has entries

$B_{ij}=\sum_{k}\frac{(\alpha+2)m_{k}(e_{i}-c_{i}^{k})(e_{j}-c_{j}^{k})}{[2q\cdot(e-c_{k})]^{\frac{\alpha+2}{2}}}.$
(14)

Analyzing the blocks of (13), we see that we have to deal with two equations
of the form

$\ddot{X}=B(q(t))X.$ (15)

From now on, we focus on (15).

Recalling that $q(t)=g(t)V$, substituting in (15) and writing

$A=B(g(t)V),$ (16)

we have an explicit form of the variational equation, namely

$\frac{d^{2}X}{dt^{2}}=\frac{AX}{g^{\frac{\alpha+4}{2}}}.$ (17)

On the other hand, system (12) is Hamiltonian with Hamiltonian function

$\bar{H}(x,v)=\frac{1}{2}v^{2}+\frac{2}{\alpha}Cx^{-\frac{\alpha}{2}},$

thus, we can fix the energy of $g(t)$, that is to say,

$\frac{1}{2}g^{\prime 2}+\frac{2}{\alpha}Cg^{-\frac{\alpha}{2}}=h$ (18)

for some constant $h$. Let us introduce the new variable

$w=\frac{K}{g(t)^{\frac{\alpha}{2}}},\text{ with }K=2Ch,$ (19)

then

$\begin{split}\frac{dw}{dt}&=-\frac{\alpha}{2}Kg(t)^{-\frac{\alpha+2}{2}}g^{\prime}(t),\\\
\frac{d^{2}w}{dt^{2}}&=\frac{\alpha(\alpha+2)}{4}Kg^{-\frac{\alpha+4}{2}}(g^{\prime})^{2}-\frac{\alpha}{2}Kg^{-\frac{\alpha+2}{2}}g^{\prime\prime}.\end{split}$
(20)

By using (12),(17),(18), (19) and (20) we obtain

$\begin{split}\frac{d^{2}X}{dt^{2}}&=\frac{d^{2}X}{dw^{2}}\left(\frac{dw}{dt}\right)^{2}+\frac{dX}{dw}\frac{d^{2}w}{dt^{2}}\\\
&=\frac{d^{2}X}{dw^{2}}\left(\frac{\alpha^{2}K^{2}h}{2}g^{-(\alpha+2)}-\alpha
CK^{2}g^{\frac{-3\alpha+4}{2}}\right)\\\
&+\frac{dX}{dw}\left(\frac{Kh\alpha(\alpha+2)}{2}g^{-\frac{\alpha+4}{2}}-(\frac{3\alpha}{2}+2)KCg^{-(\alpha+2)}\right).\end{split}$
(21)

Therefore

$\begin{split}AX&=g^{\frac{\alpha+4}{2}}\frac{d^{2}X}{dt^{2}}\\\
&=\frac{d^{2}X}{dw^{2}}\left(\frac{\alpha^{2}K^{2}h}{2}g^{-\frac{\alpha}{2}}-\alpha
CK^{2}g^{-\alpha}\right)\\\
&+\frac{dx}{dw}\left(\frac{Kh\alpha(\alpha+2)}{2}-\frac{3}{2}(\alpha+2)KCg^{-\frac{\alpha}{2}}\right),\end{split}$
(22)

then

$C^{-1}AX=\frac{d^{2}X}{dw^{2}}\left(\frac{\alpha^{2}Kh}{2C}w-\alpha
w^{2}\right)+\frac{dX}{dw}\left(\frac{Kh\alpha(\alpha+2)}{2C}-(\frac{3}{2}\alpha+2)w\right).$
(23)

Defining $\tilde{w}=\sqrt{\alpha}\,w$, we have

$\frac{dX}{d\tilde{w}}=\frac{1}{\sqrt{\alpha}}\frac{dX}{dw},$

and we can rewrite (23) as

$C^{-1}AX=\frac{1}{\alpha}\frac{d^{2}X}{d\tilde{w}}(1-\tilde{w})\tilde{w}+((\alpha+2)-(\frac{3}{2}\alpha+2)\tilde{w})\frac{dX}{d\tilde{w}}.$
(24)

If we assume that $A$ is diagonalizable with eigenvalues $a_{1},\dots,a_{d}$,
after diagonalizing $A$ each component of (24) is represented by

$\alpha
C^{-1}a_{k}\xi=\frac{d^{2}\xi}{d\tilde{w}^{2}}(1-\tilde{w})+((\alpha+2)-(\frac{3}{2}\alpha+2)\tilde{w})\frac{d\xi}{d\tilde{w}},$
(25)

or equivalently

$\tilde{w}(1-\tilde{w})\frac{d^{2}\xi}{d\tilde{w}^{2}}+(\gamma-(\delta+\beta+1)\tilde{w})\frac{d\xi}{d\tilde{w}}-\delta\beta\xi=0,$
(26)

where

$\gamma=\alpha+2,\,\delta+\beta=\frac{3}{2}\alpha+1\text{ and
}\delta\beta=\frac{\alpha a_{k}}{C}.$ (27)

Equation (26) is the announced reduced form of (17).

##  4 Integrability criteria

To have criteria of integrability for (3), we need to analyze the solvability
of equation (26); we will make use of the following result by T. Kimura:

######  Theorem 3 ([10]).

Define the constants $\lambda=1-\gamma,\mu=\gamma-\delta-\beta$ and
$\nu=\beta-\delta$. Then equation (26) has only rational solutions if and only
if one of the following statements holds.

  1. 1.

Exactly two or four of the constants $\lambda+\mu+\nu$,
$\lambda-\mu-\nu,\,\lambda-\mu+\nu,\,\lambda+\mu-\nu$ are odd integers.

  2. 2.

The quantities $\pm\lambda,\pm\mu,\pm\nu$ take, in arbitrary order, values
given in the Schwarz table, see Table 1, with $l,m,q\in\mathbb{Z}.$

1 | $\frac{1}{2}+l$ | $\frac{1}{2}+m$ | rational number |
---|---|---|---|---
2 | $\frac{1}{2}+l$ | $\frac{1}{3}+m$ | $\frac{1}{3}+q$ |
3 | $\frac{2}{3}+l$ | $\frac{1}{3}+m$ | $\frac{1}{3}+q$ |  $l+m+q$ even integer
4 | $\frac{1}{2}+l$ | $\frac{1}{3}+m$ | $\frac{1}{4}+q$ |
5 | $\frac{2}{3}+l$ | $\frac{1}{4}+m$ | $\frac{1}{4}+q$ |  $l+m+q$ even integer
6 | $\frac{1}{2}+l$ | $\frac{1}{3}+m$ | $\frac{1}{5}+q$ |
7 | $\frac{2}{5}+l$ | $\frac{1}{3}+m$ | $\frac{1}{3}+q$ |  $l+m+q$ even integer
8 | $\frac{2}{3}+l$ | $\frac{1}{5}+m$ | $\frac{1}{5}+q$ |  $l+m+q$ even integer
9 | $\frac{1}{2}+l$ | $\frac{2}{5}+m$ | $\frac{1}{5}+q$ |  $l+m+q$ even integer
10 | $\frac{3}{5}+l$ | $\frac{1}{3}+m$ | $\frac{1}{5}+q$ |  $l+m+q$ even integer
11 | $\frac{2}{5}+l$ | $\frac{2}{5}+m$ | $\frac{2}{5}+q$ |  $l+m+q$ even integer
12 | $\frac{2}{3}+l$ | $\frac{1}{3}+m$ | $\frac{1}{5}+q$ |  $l+m+q$ even integer
13 | $\frac{4}{5}+l$ | $\frac{1}{5}+m$ | $\frac{1}{5}+q$ |  $l+m+q$ even integer
14 | $\frac{1}{2}+l$ | $\frac{2}{5}+m$ | $\frac{1}{3}+q$ |  $l+m+q$ even integer
15 | $\frac{3}{5}+l$ | $\frac{2}{5}+m$ | $\frac{1}{3}+q$ |  $l+m+q$ even integer
Table 1: Schwarz table

Substituting (27) in the definitions of $\lambda,\mu$, and $\nu$, we have

$\begin{split}\lambda&=-\alpha-1\\\ \mu&=1-\frac{1}{2}\alpha\\\
\nu_{k}&=\pm\sqrt{\Big{(}\frac{3}{2}\alpha+1\Big{)}^{2}-4\frac{\alpha
a_{k}}{C}}\end{split}$ (28)

Notice that $\lambda$ and $\mu$ are completely determined by $\alpha$. We can
now compare (28) with Kimura’s Theorem 3 to prove the main result of this
paper, giving necessary conditions for the integrability of the $N$-center
problem system (3).

######  Main Theorem 4.

If the $N$-center problem (3) is rationally integrable, then for every
$k\in\\{1,\dots,d\\}$, one of the cases in the following table holds for
$\alpha$ and $a_{k}/C$.

| $\alpha$ | $\nu_{k}$
---|---|---
1 | $1$ | $\sqrt{\frac{25}{4}-\frac{4a_{k}}{C}}\in\frac{1}{2}+\mathbb{Z}$
2 | $\frac{4}{3}$ | $\sqrt{9-\frac{16a_{k}}{3C}}\in(2\mathbb{Z}+1)\cup\left(\frac{4}{3}+2\mathbb{Z}\right)\cup\left(\frac{6}{5}+2\mathbb{Z}\right)$
3 | $\frac{2}{3}$ | $\sqrt{1-\frac{2a_{k}}{3C}}\in\frac{1}{2}\mathbb{Z}\cup\left(\frac{1}{3}+2\mathbb{Z}\right)\cup\left(\frac{1}{5}+2\mathbb{Z}\right)$
4 | $\frac{1}{2}$ | $\sqrt{\frac{49}{16}-\frac{2a_{k}}{C}}\in\frac{1}{2}+\mathbb{Z}$
5 | $\frac{3}{2}$ | $\sqrt{\frac{169}{16}-\frac{6a_{k}}{C}}\in\left(\frac{1}{3}+\mathbb{Z}\right)\cup\left(\frac{1}{2}+\mathbb{Z}\right)$
6 | $\frac{4}{5}$ | $\frac{1}{5}\sqrt{121-\frac{80a_{k}}{C}}\in\frac{1}{3}+2\mathbb{Z}$
7 | $\frac{8}{5}$ | $\sqrt{\frac{289}{25}-\frac{32a_{k}}{5C}}\in\frac{3}{2}+2\mathbb{Z}$
Table 2: Integrability Criteria

In particular, if $\alpha$ is not in the previous table, the N-center problem
(3) is not integrable.

###### Proof.

To verify the first condition in Theorem 3, assume that $\lambda+\mu+\nu_{k}$,
$\lambda+\mu-\nu_{k}$ are odd integers, adding these up, we obtain that
$\lambda+\mu\in\mathbb{Z}$. On the other hand,
$\lambda+\mu=-\frac{3}{2}\alpha$, since $\alpha\in(0,2)\cap\mathbb{Q}$ we have
that $\alpha=\frac{2}{3}$ or $\alpha=\frac{4}{3}$.

If $\alpha=\frac{2}{3}$ then $\lambda+\mu=-1$, thus $\nu_{k}\in 2\mathbb{Z}$.
But $\nu_{k}=2\sqrt{1-\frac{2a_{k}}{3C}}$, we conclude that
$\sqrt{1-\frac{2a_{k}}{3C}}\in\mathbb{Z}$.

If $\alpha=\frac{4}{3}$, we have that $\lambda+\mu=-2$, so $\nu_{k}\in
2\mathbb{Z}+1$, or $\sqrt{9-\frac{16a_{k}}{3C}}\in 2\mathbb{Z}+1.$

Assume now that $\lambda+\mu+\nu_{k},\lambda-\mu-\nu_{k}\in 2\mathbb{Z}+1$,
adding up again, we deduce that $\lambda\in\mathbb{Z}$, therefore
$\lambda=-1-af\in\mathbb{Z}$ we obtain that $\alpha\in\mathbb{Z}$, and so
$\alpha=1$. Thus
$\sqrt{\frac{25}{4}-\frac{4a_{k}}{C}}\in\frac{1}{2}+\mathbb{Z}$

If $\lambda+\mu-\nu_{k},\lambda-\mu-\nu_{k}\in 2\mathbb{Z}+1$ then
$\nu_{k}=1-\frac{1}{2}\alpha\in\mathbb{Z}$ and therefore $\alpha\in
2\mathbb{Z}$, since $\alpha\in(0,2)$ we conclude that this case is not
possible. The other cases can be discarded similarly.

Based on the previous computations, the four constants cannot be odd integers
simultaneously.

Suppose now that $\lambda,\mu,\nu_{k}$ appear in the Schwarz Table 1 in
theorem 3, in particular, $\lambda$ has to be a value in the same table, using
that $\alpha\in(0,2)$ and recalling that $\lambda=-1-\alpha$ we can determine
the possible values of $\alpha$ obtaining that

$\alpha\in\\\
\Big{\\{}1,\frac{1}{2},\frac{3}{2},\frac{2}{3},\frac{5}{3},\frac{1}{3},\frac{4}{3},\frac{3}{4},\frac{7}{4},\frac{4}{5},\frac{9}{5},\frac{3}{5},\frac{8}{5},\frac{2}{5},\frac{7}{5},\frac{1}{5},\frac{6}{5}\Big{\\}},$
(29)

substituting in $\mu$ and $\nu_{k}$ and contrasting again with the Schwarz
Table 1 we can locate the values of $\alpha$ such that the triple
$\lambda,\mu,\nu_{k}$ apears in the table. For instance, if $\alpha=1$, then
$\lambda=-2,\mu=\frac{1}{2}$ and
$\nu_{k}=\sqrt{\frac{25}{4}-\frac{4a_{k}}{C}}$ this triple apears in the
table, whenever
$\sqrt{\frac{25}{4}-\frac{4a_{k}}{C}}\in\frac{1}{2}+\mathbb{Z}.$

If $\alpha=\frac{1}{4}$, then $\lambda=-\frac{4}{3},\mu=\frac{5}{6}$, notice
that $\frac{5}{6}$ is not a value in the Schwarz Table, therefore we can
discard $\alpha=\frac{1}{4}.$

Exhausting every case in (29), putting it all together, we obtain Table 2.

∎

##  5 Generalized N-center problem with polygonal configuration

In this section, we aim to prove that, in the 3-dimensional case, if some
centers feature certain rotational symmetry, the $N$ center problem with
unitary masses is not rationally integrable. We have the following result.

######  Theorem 5.

In the $3$ dimensional space $d=3$, assume $N\geq 3$ and that there is an
integer $3\leq l\leq N$ such that $l$ centers are located at the vertices of a
plane regular convex $l$-gon. The $N$-center problem with masses
$m_{i}=1,\,i=1\dots,N$, $\alpha\in(0,2)\cap\mathbb{Q}$ is not rationally
integrable.

###### Proof.

We can assume, without loss of generality, that the plane containing the
$l$-gone is the plane $z=0$; we also assume that the center of the $l$-gone is
the origin of coordinates, then we can write, relabeling the centers if
necessary, $c_{k}=(c_{k}^{1},c_{k}^{2},0),k=1,\dots,l$, and

$c_{k}^{1}=\cos\Big{(}\frac{2k\pi}{l}\Big{)},\,c_{k}^{2}=\sin\Big{(}\frac{2k\pi}{l}\Big{)},\,k=1,\dots,l.$
(30)

Suppose that $V=(0,0,v_{3})$ and $(e_{1},e_{2},e_{3})=(0,0,\pm i)$ satisfy

$\displaystyle C(0,0,v_{3})$
$\displaystyle=\sum_{k}^{l}\frac{1}{[2(-c_{k}^{1},-c_{k}^{2},\pm
i)\cdot(0,0,v_{3})]^{\frac{\alpha+2}{2}}}(-c_{k}^{1},-c_{k}^{2},\pm i)$ (31)
$\displaystyle=\sum_{k}^{l}\frac{(-\cos(\frac{2\pi k}{l}),-\sin(\frac{2\pi
k}{l}),\pm i)}{[2(\pm iv_{3})]^{\frac{\alpha+2}{2}}},$ (32)

It follows that

$v_{3}^{\alpha+4}=\frac{2^{2}l^{2}}{2^{\alpha+4}C^{2}i^{\alpha+4}}.$ (33)

Now we can compute the matrix $A$ given in (16). By using (14) and (16) and
assuming that $l\geq 3$ we have, we have that

$A_{ij}=\sum_{k}\frac{(\alpha+2)(e_{i}-c_{i}^{k})(e_{j}-c_{j}^{k})}{[2V\cdot(e-c_{k})]^{\frac{\alpha+4}{2}}},$

by using this and with (30), we obtain that

$\displaystyle A_{11}$
$\displaystyle=\frac{\alpha+2}{(2iv_{3})^{\frac{\alpha+4}{2}}}\sum_{k=1}^{l}\cos^{2}\Big{(}\frac{2\pi
k}{l}\Big{)}=\frac{(\alpha+2)l}{2(2iv_{3})^{\frac{\alpha+4}{2}}};$ (34)
$\displaystyle A_{12}$
$\displaystyle=A_{21}=\frac{(\alpha+2)}{(2iv_{3})^{\frac{\alpha+4}{2}}}\sum_{k=1}^{l}\cos(\frac{2\pi
k}{l})\sin(\frac{2\pi k}{l})=0;$ (35) $\displaystyle A_{22}$
$\displaystyle=\frac{(\alpha+2)}{(2iv_{3})^{\frac{\alpha+4}{2}}}\sum_{k=1}^{l}\sin^{2}\Big{(}\frac{2\pi
k}{l}\Big{)}=\frac{(\alpha+2)l}{2(2iv_{3})^{\frac{\alpha+4}{2}}};$ (36)
$\displaystyle A_{13}$
$\displaystyle=A_{31}=\frac{(\alpha+2)}{(2iv_{3})^{\frac{\alpha+4}{2}}}\sum_{k=1}^{l}-\cos\Big{(}\frac{2\pi
k}{l}\Big{)}i=0;$ (37) $\displaystyle A_{23}$
$\displaystyle=A_{32}=\frac{(\alpha+2)}{(2iv_{3})^{\frac{\alpha+4}{2}}}\sum_{k=1}^{l}-\sin\Big{(}\frac{2\pi
k}{l}\Big{)}l=0;$ (38) $\displaystyle A_{33}$
$\displaystyle=\frac{(\alpha+2)}{(2iv_{3})^{\frac{\alpha+4}{2}}}\sum_{k=1}^{l}i^{2}=-\frac{(\alpha+2)l}{(2iv_{3})^{\frac{\alpha+4}{2}}}.$
(39)

therefore, the matrix $A$ is diagonal, with eigenvalues
$A_{11},A_{22},A_{33}$. From (33) we have

$\displaystyle\frac{a_{1}}{C}$
$\displaystyle=\frac{A_{11}}{C}=\pm\frac{\alpha+2}{4}=\frac{a_{2}}{C},$ (40)
$\displaystyle\frac{a_{3}}{C}$
$\displaystyle=\frac{A_{22}}{C}=\mp\frac{\alpha+2}{2}$ (41)

we conclude that

$\displaystyle\nu_{1}=\nu_{2}$
$\displaystyle=\sqrt{\frac{13}{4}\alpha^{2}+5\alpha+1};$ (42)
$\displaystyle\nu_{3}$ $\displaystyle=1-\frac{\alpha}{2}.$ (43)

Comparing these formulas with the Table 2 of Theorem 4, we have the following
values for $\nu_{1}$ and $\nu_{3}$.

$\alpha$ | $\nu_{1}=\nu_{2}$ | $\nu_{3}$
---|---|---
$1$ | $\frac{\sqrt{37}}{2}$ | $\frac{1}{2}$
$\frac{4}{3}$ | $\frac{11}{3}$ | $\frac{1}{3}$
$\frac{2}{3}$ | $\frac{2\sqrt{13}}{3}$ | $\frac{2}{3}$
$\frac{1}{2}$ | $\frac{\sqrt{64}}{4}$ | $\frac{3}{4}$
$\frac{3}{2}$ | $\frac{\sqrt{253}}{4}$ | $\frac{1}{4}$
$\frac{4}{5}$ | $\frac{\sqrt{177}}{5}$ | $\frac{3}{5}$
$\frac{8}{5}$ | $\frac{\sqrt{433}}{5}$ | $\frac{1}{5}$

We conclude that for any $\alpha$ on the Table 2, $\nu_{1}$ does not satisfy
the intregrability condition given in Theorem 4. ∎

######  Remark 2.

Notice that $\alpha=1$ corresponds to Shibayama’s result [6]; actually, Case 1
in Table 2 recovers Shibayama’s criterion. It is remarkable that, in Theorem
5, we only need a subset of centers forming a regular polygon to obtain non-
integrability; there isn’t any restriction on the configuration of the
remaining centers.

We finish this work with the following open problems.

Question Is there an irrational number $\alpha\in(0,2)$ such that the
associated $N$-center problem is integrable?

Problem It is still an open problem if whether or not non rational first
integrals exist for the $N$-center problem.

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