2 Existence of the N-gon periodic solutions with quasi-homogeneous potential

In accordance with (8), the equations of motion for the planar $ N $ -body problem with quasi-homogeneous potential assuming equal masses $ m_k = m $ for each $ k=1,\ldots , N $ are as follows:

(10) $$ \begin{align} \ddot{\boldsymbol{r}}_k=m \sum_{j=1, k\neq j}^{N} \left[\frac{\boldsymbol{r}_j-\boldsymbol{r}_k}{r_{jk}^3}+\frac{pC_{kj}(\boldsymbol{r}_j-\boldsymbol{r}_k)}{r_{jk}^{p+2}} \right], \quad k=1,\ldots, N. \end{align} $$

Next, our aim will be to characterize the existence of periodic solutions where all the particles have the same mass m, are situated on a circle of radius $\rho $ , and rotate with a constant angular velocity $\omega $ , thereby forming a regular polygon (see Figure 1). We propose that such circular solutions are parametrized by

(11) $$ \begin{align} \boldsymbol{r}_k(t)= \rho e^{i(\omega t + \theta_k)}, \quad t\in\mathbb{R}, \end{align} $$

where

$$ \begin{align*} \theta_k=\theta_0+ 2\pi\left( \frac{k-1}{N}\right), \quad k=1,\ldots, N. \end{align*} $$

Figure 1: Representation of the periodic solution of the quasi-homogeneous problem (10) forming a regular N-gon.

For the sake of simplicity, we will assume that $ \theta _0=0 $ , then

(12) $$ \begin{align} \theta_k= 2\pi\left( \frac{k-1}{N}\right). \end{align} $$

From (11), we obtain the first and second derivative of $ \boldsymbol{r}_{k} $

(13) $$ \begin{align} \dot{\boldsymbol{r}}_k = i\omega \boldsymbol{r}_k=i\omega \rho e^{i(\omega t+\theta_k)}, \quad \ddot{\boldsymbol{r}}_k = i\omega \dot{\boldsymbol{r}}_k = -\omega^2 \rho e^{i\theta_k} e^{i\omega t}. \end{align} $$

Next, we calculate the mutual distances

(14) $$ \begin{align} \boldsymbol{r}_j-\boldsymbol{r}_k &= \rho e^{i(\omega t+\theta_{j})}-\rho e^{i(\omega t+\theta_{k})} \nonumber\\ &= \rho e^{i\omega t}\left(e^{i\theta_{j}}-e^{i\theta_{k}}\right) \nonumber\\ &= \rho e^{i\omega t}e^{i\theta_k}\left( e^{i(\theta_j-\theta_k)}-1\right). \end{align} $$

So,

(15) $$ \begin{align} r_{jk}=\lVert\boldsymbol{r}_j-\boldsymbol{r}_k\rVert=\rho\underbrace{\lVert e^{i(\theta_j-\theta_k)}-1\rVert}_{d_{jk}}= \rho d_{jk}, \end{align} $$

where

(16) $$ \begin{align} d_{jk}&= \lVert e^{i(\theta_j-\theta_k)}-1\rVert= \sqrt{2}\sqrt{1-\cos(\theta_j-\theta_k)}. \end{align} $$

Substituting in (10), we have that

$$ \begin{align*} -\frac{\omega^2}{m} e^{i\theta_k} &=\sum_{j=1, j\neq k}^{N} \left( \frac{(e^{i\theta_j}-e^{i\theta_k})}{\rho^{3}d_{jk}^{3}} +\frac{pC_{kj}(e^{i\theta_j}-e^{i\theta_k})}{\rho^{p+2}d_{jk}^{p+2}}\right)\\ &=\sum_{j=1, j\neq k}^{N}\frac{(e^{i\theta_j}-e^{i\theta_k})}{\rho^3}\left( \frac{1}{d_{jk}^{3}} + \frac{pC_{kj}}{\rho^{p-1}d_{jk}^{p+2}}\right)\\ &=e^{i\theta_k}\sum_{j=1, j\neq k}^{N}\frac{(e^{i(\theta_j-\theta_k)}-1)}{\rho^3}\left( \frac{1}{d_{jk}^{3}} + \frac{pC_{kj}}{\rho^{p-1}d_{jk}^{p+2}}\right). \end{align*} $$

Therefore,

(17) $$ \begin{align} -\frac{\omega^2}{m} = \sum_{j=1, j\neq k}^{N}\frac{(e^{i(\theta_j-\theta_k)}-1)}{\rho^3}\left( \frac{1}{d_{jk}^{3}} + \frac{pC_{kj}}{\rho^{p-1}d_{jk}^{p+2}}\right). \end{align} $$

From (12), it follows that

(18) $$ \begin{align} \theta_j-\theta_k = 2\left(\frac{\pi(j-k)}{N}\right), \end{align} $$

and

(19) $$ \begin{align} 1-\cos(\theta_j-\theta_k)=2\sin^2\left( \frac{(\theta_j-\theta_k)}{2}\right)= 2\sin^2\left( \frac{\pi(j-k)}{N}\right). \end{align} $$

Thus,

(20) $$ \begin{align} d_{jk}= 2\left| \sin\left( \frac{\pi(j-k)}{N}\right) \right|. \end{align} $$

Therefore, taking the real and imaginary parts in the system (17), it is equivalent to

(21) $$ \begin{align} \frac{\omega^2}{m}\rho^3 = \displaystyle\sum_{j=1, j\neq k}^{N}\left(\frac{1}{4}\left|\csc\left( \frac{\pi(j-k)}{N}\right) \right| + \frac{p C_{kj}}{2^{p+1}} \frac{1}{\rho^{p-1}}\left|\csc\left( \frac{\pi(j-k)}{N}\right)\right|{}^{p} \right), \end{align} $$

and

(22) $$ \begin{align} \displaystyle\sum_{j=1, j\neq k}^{N} \left(\frac{1}{8} \frac{\sin\left(\frac{2\pi(j-k)}{N}\right)}{\left|\sin^3\left( \frac{\pi(j-k)}{N}\right) \right|} + \frac{p C_{kj}}{2^{p+2}}\frac{1}{\rho^{p-1}}\frac{\sin\left(\frac{2\pi(j-k)}{N}\right)}{\left|\sin\left( \frac{\pi(j-k)}{N}\right) \right|{}^{p+2}}\right) =0, \end{align} $$

for every $k=1,\ldots ,N$ .

In order to study system (21) and (22), from now on, we will assume that

$$ \begin{align*}C_{kj}= C, \end{align*} $$

for all $k,j=1, \ldots , N$ .

Next, define

$$ \begin{align*}f_k= \displaystyle\sum_{j=1, j\neq k}^{N} \frac{\sin^2\left( \frac{(\theta_j-\theta_k)}{2}\right) }{4\left|\sin\left( \frac{\pi(j-k)}{N}\right) \right|{}^3} + \frac{p}{\rho^{p-1}}\frac{C \sin^2\left( \frac{(\theta_j-\theta_k)}{2}\right) }{\left|\sin\left( \frac{\pi(j-k)}{N}\right) \right|{}^3}, \end{align*} $$

for $k=1, 2, \ldots , N$ .

Now, we point out that rearranging the sum in $f_k$ , we observe that the relations $f_1= f_2= \cdots = f_N$ are valid. On the other hand, since

$$ \begin{align*}\displaystyle\sum_{j=2}^{N} \frac{\sin\left(\frac{2\pi(j-1)}{N}\right)}{\left|\sin\left( \frac{\pi(j-1)}{N}\right) \right|{}^{\alpha}}=0, \end{align*} $$

for all $\alpha>0$ . So $f_1=0$ and then (22) is always satisfied for $k=1, \cdots , N$ .

Subsequently, upon rearranging the system of equations in (21), for $k=1, 2, 3, \ldots , N$ , it becomes evident that all of the equations in the system are equal. So, we can conclude that the following result has been proven.

Lemma 1 Assume $C_{kj}= C$ and maintain the previous notations. Then, $\boldsymbol{r}_k(t)= \rho e^{i(\omega t + \theta _k)}, \quad t\in \mathbb {R}$ , where $\theta _k= 2\pi \left ( \frac {k-1}{N}\right )$ with equal masses $m_k = m$ , $k=1,\ldots , N$ is a N-gon solution of the system (10), if and only if, the following condition is satisfied:

(23) $$ \begin{align} \begin{array}{rl} \frac{\omega^2}{m} \rho^3 =& \displaystyle\sum_{j= 2}^{N}\left(\frac{1}{4}\left|\csc\left( \frac{\pi(j-1)}{N}\right) \right| + \frac{p C}{2^{p+1}} \frac{1}{\rho^{p-1}}\left|\csc\left( \frac{\pi(j-1)}{N}\right)\right|{}^{p} \right). \end{array} \end{align} $$

Now, we see that the equation (23) can be rewritten as

(24) $$ \begin{align} \frac{\omega^2}{m} \rho^3 = \frac{1}{4} A + \frac{p B C}{2^{p+1}} \frac{1}{\rho^{p-1}}, \end{align} $$

with

(25) $$ \begin{align} A= \displaystyle\sum_{j= 2}^{N}\left|\csc\left( \frac{\pi(j-1)}{N}\right) \right|,\quad B= \sum_{j=2}^{N} \left|\csc\left( \frac{\pi(j-1)}{N}\right)\right|{}^{p}. \end{align} $$

From all this, we have proved the following theorem.

Theorem 2.1 Let $ N\geq 2 $ , $ C_{kj}=C\neq 0 $ for each $k,j=1,\ldots ,N$ , $m_l=m$ , $ \omega>0 $ for $l= 1,\ldots ,N$ , then the functions $ \boldsymbol{r}_{k}(t) $ given in (11) define a $ 2\pi /\omega $ -periodic solution of the N-body problem with quasi-homogeneous potential (10), which forms a regular polygon on a circle of radius $\rho $ , provided that the following equation is satisfied:

(26) $$ \begin{align} \frac{\omega^2}{m}\rho^{p+2}-\frac{A}{4}\rho^{p-1}-\frac{p B C}{2^{p+1}}=0. \end{align} $$

Next, we will study the admissible values of $ \rho $ that satisfy (26). The subsequent analysis will be divided into two distinct cases: one where $ C>0 $ and another where $ C <0 $ . It should be noted that, by employing (26) we can determine the mass m as a function of $ \rho $ , assuming that $ C, \omega>0 $ and $ p>1 $ are fixed. In fact, we have

(27) $$ \begin{align} m=f(\rho):=\omega^2 \rho^{p+2} \left(\frac{A}{4} \rho^{p-1} + \frac{p C B}{2^{p+1}} \right)^{-1}, \end{align} $$

where $A>0$ , $B>0$ are as in (25).

For the case $ C>0$ , the following result is obtained.

Proof We will show that the function $ f(\rho ) $ is a monotone increasing. Indeed, differentiating the function $ f $ with respect to $ \rho $ yields

(28) $$ \begin{align} f'(\rho)= \frac{\omega^2 \rho^{p+1}}{\left(\frac{A}{4} \rho^{p-1} + \frac{p B C}{2^{p+1}} \right)^2} \left(\frac{3 A}{4} \rho^{p-1} + \frac{p(p+2)}{2^{p+1}} B C \right). \end{align} $$

Since $ C>0 $ clearly $ f'(\rho )>0 $ , then $ f $ is a monotonic increasing function, thus $ f(\rho )>f(0)=0 $ . Also, $ \lim _{\rho \rightarrow +\infty } f(\rho )= +\infty $ . Consequently, for any value of $ m>0 $ , there exists a unique $ \rho>0 $ such that $ f(\rho )=m $ as in (27) (see Figure (2)). Therefore, we have that $f(\rho (0))=0$ and that $f(\rho (m)) \rightarrow +\infty $ as $m \rightarrow +\infty $ . This concludes the proof of the theorem.

Remark 1 Figure 3 illustrates the existence of a unique regular N-gon solution $ \rho (m) $ for the case $C>0$ . This solution represents the only possible configuration of the $ N $ bodies that can be arranged in a regular polygon at any given instant. These results agree with those obtained for the Newtonian case in [Reference Perko and Walter9].

Figure 2: Study of the function $ f $ given in (27) for the case $ C>0 $ and $ p>1$ .

Figure 3: Existence of a regular polygon solution given $ C>0 $ and $ p>1 $ .

The following theorem presents the analysis of the case $C<0$ .

Theorem 2.3 Assume case $ C<0 $ , $ p>1 $ , $ N>2 $ and $ \omega>0 $ . Then there exists a bifurcation value of the mass, denoted by

(29) $$ \begin{align} m_{\sharp}= \frac{\omega^2 (p+2) \left(-B C p (p+2) A^{-1} \right)^{3/(p-1)}}{2 A (p-1)}, \end{align} $$

such that the following conditions are satisfied:

1. (1) If $ m= m_{\sharp }$ , then there exists a unique of $\rho $ , which is given by the formula $ \rho _{\sharp }= \frac {1}{2}\left (-\frac {B C}{3 A}p(p+2) \right )^{1/(p-1)}$ which gives rise to a N-gon solution of the N-body problem with quasi-homogeneous potential.

2. (2) If $ m> m_{\sharp }$ , then there exist two admissible radii, $ \rho _1 , \rho _2 $ such that $ \rho _*= \left (-\frac {p B C}{2^{p-1} A} \right )^{1/(p-1)} < \rho _1 < \rho _{\sharp } < \rho _2 $ which give rise to two N-gon solutions of the N-body problem with quasi-homogeneous potential.

3. (3) If $ 0<m< m_{\sharp }$ , then there are no admissible radii.

Proof In accordance with the definition of $ f(\rho ) $ in (27), it can be established that, in the case $ C<0 $ , in order to have $ m>0 $ , it is necessary to impose the following condition:

$$ \begin{align*}\left(\frac{A}{4} \rho^{p-1} + \frac{p B C}{2^{p+1}} \right)>0 \Leftrightarrow \rho^{p-1}> -\frac{4p}{2^{p+1}} \frac{B C}{A} \Leftrightarrow 0<\rho< \rho_*,\end{align*} $$

where

(30) $$ \begin{align} \rho_* = \left(-\frac{p B C}{2^{p-1} A} \right)^{1/(p-1)}= \frac{1}{2}\left(- \frac{p B C}{A} \right)^{1/p-1}. \end{align} $$

On the other hand, according to (28), the sign of $ f'(\rho ) $ is determined by the sign of the expression

$$ \begin{align*}\frac{3 A}{4} \rho^{p-1} + \frac{p (p+2)}{2^{p+1}} B C.\end{align*} $$

It should be noted that $ f'(\rho )=0 \Leftrightarrow \rho = \rho _\sharp $ where

(31) $$ \begin{align} \rho_\sharp = \left( -\frac{p (p+2)}{2^{p-1}} \frac{B C}{3 A} \right)^{1/(p-1)}= \frac{1}{2}\left(-\frac{B C}{3 A}p(p+2) \right)^{1/(p-1)}. \end{align} $$

Let us determine which domain, $ f(\rho ) $ and $ f'(\rho ) $ , are positive. It can be verified that

$$ \begin{align*}f(\rho)>0 \,\ \wedge \,\ f'(\rho)>0 \Leftrightarrow \rho> \frac{1}{2}\left(-\frac{ B C}{3 A}p(p+2) \right)^{1/(p-1)}\Leftrightarrow \rho > \rho_\sharp. \end{align*} $$

Let us now examine which domain $f(\rho )>0 $ and $f'(\rho )<0 $ . In this case,

$$ \begin{align*}f(\rho)>0\, \wedge\, f'(\rho)<0 \quad \Leftrightarrow \rho_* < \rho < \rho_\sharp. \end{align*} $$

Besides, $ \lim _{\rho \rightarrow \rho _{*}^{+}} f(\rho )= +\infty $ and $\lim _{\rho \rightarrow +\infty } f(\rho ) = +\infty $ .

In this context, we will denote the number obtained by substituting $ \rho _\sharp $ in (27) by $ m_\sharp $ , which is, $ m_\sharp = f(\rho _\sharp ) $ (see Figure 4). In consequence, we can state that:

1. (1) If $ m = m_\sharp $ , then there is a unique admissible radius $ \rho>0 $ , denoted by $ \rho _\sharp $ , which yields a N-gon solution.

2. (2) If $ m> m_\sharp $ , then there exist two distinct admissible radii, namely $ \rho _1 $ and $ \rho _2 $ such that $ \rho _1< \rho _\sharp < \rho _2 $ and each gives rise to a N-gon solution.

3. (3) If $ 0<m< m_\sharp $ , there are no admissible radii.

Figure 4: Graphic of the function $ f $ given in (27) for the case $ C<0 $ and $ p>1$ .

Thus, we have concluded the proof of the theorem.

Figure 5 illustrates the existence of two N-gon solutions $ \rho _1 (m) $ and $ \rho _2 (m) $ , that is, there are at least two possible of arrangements of the $ N $ bodies such that they form a regular polygon at each instant.

Figure 5: Existence of two regular polygon solutions given $ C<0 $ and $ p>1$ .

3 Relative equilibrium for the quasi-homogeneous N-body problem

The equations of motion of the particles are given by (10), where the quasi-homogeneous potential is given by

(32) $$ \begin{align} W(\boldsymbol{r})=\sum_{1\leq k<j\leq N }\left[\frac{m^2}{\|\boldsymbol{r}_k-\boldsymbol{r}_j\|}+\frac{m^2 C}{\|\boldsymbol{r}_k-\boldsymbol{r}_j\|^p}\right]. \end{align} $$

The linear momentum of the $ k $ th particle is given by the equation $ \boldsymbol{p}_k= m \dot {\boldsymbol{r}_k} $ with $ k=1,\dots ,N $ . Then, the equations of motion are written in their Hamiltonian form

(33) $$ \begin{align} \dot{\boldsymbol{r}}&=\boldsymbol{M}^{-1}\boldsymbol{p}=H_{\boldsymbol{p}} \nonumber\\ \dot{\boldsymbol{p}}&=\nabla W(\boldsymbol{r})=-H_{\boldsymbol{r}}, \end{align} $$

whose Hamiltonian function is

$$ \begin{align*} H(\boldsymbol{r},\boldsymbol{p})=\frac{1}{2}\boldsymbol{p}^T \boldsymbol{M}^{-1}\boldsymbol{p}-W(\boldsymbol{r}), \end{align*} $$

where $ \boldsymbol{r} =(\boldsymbol{r}_1, \boldsymbol{r}_2,\dots ,\boldsymbol{r}_N),\,\ \boldsymbol{p}=(\boldsymbol{p}_1,\boldsymbol{p}_2,\cdots ,\boldsymbol{p}_N) $ and $ \boldsymbol{M} = diag(M_1,M_2,\dots ,M_N) $ with

$$ \begin{align*} M_k=\left( \begin{array}{ccccc} m_k & 0 & 0 & \cdots & 0 \\ 0 & m_k & 0 & \cdots & 0 \\ 0 & 0 & m_k & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m_k \end{array} \right)_{3N\times 3N}, \end{align*} $$

for each $ k=1,\dots ,N $ , where $ \boldsymbol{M}=mI $ .

The system described in (33) can be expressed as follows:

(34) $$ \begin{align} \dot{\boldsymbol{r}}_k&=m^{-1}\boldsymbol{p}_k=H_{\boldsymbol{p}_k}, \nonumber\\ \dot{\boldsymbol{p}}_k&=\nabla W(\boldsymbol{r}_k)=-H_{\boldsymbol{r}_k}, \, k=1, \ldots, N. \end{align} $$

Let us consider a systems in which the center of mass is fixed at the origin of the coordinates. We assume the existence of equilibrium solutions as described in (11) and then we introduce a time-dependent symplectic change of coordinates (rotating coordinates) by

(35) $$ \begin{align} \boldsymbol{r}_k = e^{\omega \boldsymbol{L}_k t}\boldsymbol{x}_k \quad \text{and}\quad \boldsymbol{p}_k = e^{\omega \boldsymbol{L}_k t}\boldsymbol{y}_k, \end{align} $$

whose system rotates with constant angular velocity $ \omega $ and where $ \displaystyle \boldsymbol{L}_k=\left (\begin {matrix} 0 & -1 \\ 1 & 0 \end {matrix}\right ) $ .

Differentiating the equations in (35), we get

(36) $$ \begin{align} \dot{\boldsymbol{r}}_k&= \omega \boldsymbol{L}_k e^{\omega \boldsymbol{L}_k t} \boldsymbol{x}_k + e^{\omega \boldsymbol{L}_k t} \dot{\boldsymbol{x}}_k, \nonumber\\ \dot{\boldsymbol{p}}_k&= \omega \boldsymbol{L}_k e^{\omega \boldsymbol{L}_k t} \boldsymbol{y}_k + e^{\omega \boldsymbol{L}_k t} \dot{\boldsymbol{y}}_k, \, k=1, \ldots, N. \end{align} $$

Then, substituting the equations of the system (34) in (36), we obtain the new equations of motion

(37) $$ \begin{align} \dot{\boldsymbol{x}}_k & = m^{-1}\boldsymbol{y}_k-\omega \boldsymbol{L}_k\boldsymbol{x}_k, \nonumber\\ \displaystyle\dot{\boldsymbol{y}}_k &= \sum_{j=1, j\neq k}^{N} \left[ \frac{m^2(\boldsymbol{x}_j-\boldsymbol{x}_k)}{\lVert\boldsymbol{x}_j-\boldsymbol{x}_k\rVert^3} + \frac{pm^2 C (\boldsymbol{x}_j-\boldsymbol{x}_k)}{\lVert\boldsymbol{x}_j-\boldsymbol{x}_k\rVert^{p+2}} - \omega \boldsymbol{L}_k\boldsymbol{y}_k\right], \, k=1, \ldots, N. \end{align} $$

These equations have as their Hamiltonian function

(38) $$ \begin{align} H(\boldsymbol{x},\boldsymbol{y})=\frac{1}{2}\boldsymbol{y}^T m^{-1}\boldsymbol{y} + \omega \boldsymbol{x}^T \boldsymbol{L}\boldsymbol{y} - W(\boldsymbol{x}), \end{align} $$

with $ \boldsymbol{L}=diag(\boldsymbol{L}_1,\boldsymbol{L}_2,\dots , \boldsymbol{L}_N) $ and $ W(x) $ is the quasi-homogeneous potential (32) at the new coordinates.

An equilibrium point of (33) now corresponds to a relative equilibrium of the solutions (37), that is,

(39) $$ \begin{align} m^{-1}\boldsymbol{y}-\omega \boldsymbol{L}\boldsymbol{x} = 0, & \quad \nabla W(\boldsymbol{x})-\omega \boldsymbol{L}\boldsymbol{y} = 0, \end{align} $$

hence

(40) $$ \begin{align} \nabla W(\boldsymbol{x})=-\omega^2 m\boldsymbol{x}. \end{align} $$

By construction, it is clear that $x= \boldsymbol{x}^*=(\boldsymbol{x}_{1}^{*}, \boldsymbol{x}_{2}^{*},\dots , \boldsymbol{x}_{N}^{*})=\rho \boldsymbol{r}^*$ with $\boldsymbol{x}_{k}^{*}=\rho \boldsymbol{r}_{k}^{*}$ and $ \boldsymbol{r}_{k}^{*}= e^{i\theta _k} $ is an equilibrium point of the previous system (or, a relative equilibrium point, since the system is rotating uniformly).

A significant outcome of this section is to demonstrate that the relative equilibrium point, represented by the vector $ \boldsymbol{x}^* $ , is a simultaneous relative equilibrium for the functions U and V. This implies that both U and V evaluated at the relative equilibrium are multiples of the equilibrium position.

Proposition 1 The relative equilibrium solution $(\boldsymbol{x}^*, \boldsymbol{y}^*)$ of (39), which is associated with the N-gon solution for the quasi-homogeneous N-body problem i.e., $\boldsymbol{x}^* $ , satisfies (40), more precisely,

(41) $$ \begin{align} \boldsymbol{x}^*=(\boldsymbol{x}_{1}^{*}, \boldsymbol{x}_{2}^{*},\dots, \boldsymbol{x}_{N}^{*})=\rho \boldsymbol{r}^*, \end{align} $$

where $\boldsymbol{x}_{k}^{*}=\rho \boldsymbol{r}_{k}^{*}$ , and such that $ \boldsymbol{r}_{k}^{*}= e^{i \frac {2 \pi (k-1)}{N}}$ , is a simultaneous relative equilibrium for $ U $ and $ V $ , that is, there exist $\mu , \nu \in \mathbb {R}$ such that

(42) $$ \begin{align} \nabla U(\boldsymbol{x}^*)=-\mu m \boldsymbol{x}^* \quad \text{and} \quad \nabla V(\boldsymbol{x}^*)=-\nu m \boldsymbol{x}^*. \end{align} $$

Proof By calculating the gradient of $ U(\boldsymbol{x}^*) $ and given that the imaginary part of the equation (21) is null, we conclude that

(43) $$ \begin{align} \frac{\partial U (\boldsymbol{x}^*)}{\partial \boldsymbol{r}_{k}}&=\sum_{j=1, j\neq k}^{N}\frac{m^2(\boldsymbol{r}_j-\boldsymbol{r}_k)}{\lVert\boldsymbol{r}_j-\boldsymbol{r}_k\rVert^3} \nonumber\\ &= e^{i\theta_k} m^2 \frac{\rho}{\rho^3}\sum_{j=1, j\neq k}^{N} \frac{1}{d_{jk}^{3}}[(\cos(\theta_j-\theta_k)-1),-\sin(\theta_j-\theta_k)] \nonumber\\ &= e^{i\theta_k} m^2 \frac{\rho}{\rho^3}\sum_{j=1, j\neq k}^{N} \frac{[(\cos(\theta_j-\theta_k)-1)]}{d_{jk}^{3}}\\ &= -\left( \frac{Am^2}{4 \rho^3}\right) \rho e^{i\theta_k}= -\left( \frac{Am^2}{4 \rho^3}\right) \rho \boldsymbol{r}_{k}^{*} = -\left( \frac{Am^2}{4 \rho^3}\right) \boldsymbol{x}_{k}^{*}.\nonumber \end{align} $$

In particular, we have that $ \nabla U(\boldsymbol{x}^*)=-\mu m\boldsymbol{x}^{*} $ , with

(44) $$ \begin{align} \mu= \frac{Am}{4 \rho^3}. \end{align} $$

Similarly, by computing $ \nabla V(\boldsymbol{x}^*) $ and noting that the imaginary part of the equation (21) is zero, we obtain

(45) $$ \begin{align} \frac{\partial V (\boldsymbol{x}^*)}{\partial \boldsymbol{r}_{k}}&=\sum_{j=1, j\neq k}^{N}\frac{m^2 pC (\boldsymbol{r}_j-\boldsymbol{r}_k)}{\lVert\boldsymbol{r}_j-\boldsymbol{r}_k\rVert^{p+2}} \nonumber\\ &= e^{i\theta_k} m^2 p C \frac{\rho}{\rho^{p+2}}\sum_{j=1, j\neq k}^{N} \frac{1}{d_{jk}^{p+2}}[(\cos(\theta_j-\theta_k)-1),-\sin(\theta_j-\theta_k)] \nonumber\\ &= e^{i\theta_k} m^2 \frac{pBC}{2^{p+1}} \frac{\rho}{\rho^{p+2}}\sum_{j=1, j\neq k}^{N} \frac{[(\cos(\theta_j-\theta_k)-1)]}{d_{jk}^{p+2}}\\ &= -\left( \frac{m^2 pBC}{2^{p+1} \rho^{p+2}}\right) \rho e^{i\theta_k}= - \left( \frac{m^2 p BC}{2^{p+1} \rho^{p+2}}\right) \boldsymbol{x}_{k}^{*}.\nonumber \end{align} $$

Thus, $ \nabla V(\boldsymbol{x}^*)=-\nu m\boldsymbol{x}^{*} $ , with

(46) $$ \begin{align} \nu= \frac{m p B C}{2^{p+1} \rho^{p+2}}. \end{align} $$

In particular, we have proved that the gradients of U and V evaluated at the relative equilibrium are multiples of the equilibrium position. It can be concluded that the point $x^*$ represents a simultaneous relative equilibrium for both potentials.

4 Stability of the relative equilibrium associated with the N-gon for the planar N-body problem with quasi-homogeneous potential

In this section, we will study the linear stability of the relative equilibrium associated with the N-gon periodic solution. We begin rewriting the system (37) by setting $ \boldsymbol{z}=(\boldsymbol{x},\boldsymbol{y})$ , which allows us to express the equations in the following form:

(47) $$ \begin{align} \dot{\boldsymbol{z}}=\boldsymbol{J}\nabla H(\boldsymbol{z}), \end{align} $$

where

$$ \begin{align*} \boldsymbol{J}=\left( \begin{matrix} \boldsymbol{0} & \boldsymbol{I} \\ -\boldsymbol{I} & \boldsymbol{0} \end{matrix}\right), \end{align*} $$

where $ \boldsymbol {I} $ is the identity matrix of order $2n$ .

Let $\boldsymbol{z}^*$ be an equilibrium point of (47), then $ \nabla H(\boldsymbol{z}^*)=0 $ and by developing in Taylor series the function $ H $ around $ \boldsymbol{z}^* $ , we arrive to

$$ \begin{align*} \dot{\boldsymbol{z}}=\boldsymbol{J}D\nabla H(\boldsymbol{z}^*)(\boldsymbol{z}-\boldsymbol{z}^*). \end{align*} $$

Using this expression, the linearized system of the equations (47) at the equilibrium $ \boldsymbol{z}^* $ can be expressed as follows:

(48) $$ \begin{align} \dot{\boldsymbol{z}}= \boldsymbol{A} \boldsymbol{z}, \end{align} $$

where $ \boldsymbol {S} $ is the symmetric matrix $ \boldsymbol {S}=D\nabla H(\boldsymbol{z}^*)= Hess H(\boldsymbol{z}^*)$ , $\boldsymbol {A}=\boldsymbol {J}\boldsymbol {S}$ . In this equation, we have used the same notation $ \boldsymbol{z} $ for $ \boldsymbol{z}-\boldsymbol{z}^* $ . We point out that

(49) $$ \begin{align} \boldsymbol{A}=\left[\begin{matrix} -\omega \boldsymbol{L} & m^{-1}I \\ D\nabla W(\boldsymbol{x}^*) & -\omega \boldsymbol{L} \end{matrix} \right]. \end{align} $$

Proof As in Moeckel’s work [Reference Moeckel8], the proof focus on the subspace $ Q_1(\boldsymbol{x}^*) $ spanned by the vectors of $ \mathbb {R}^{4n} $

(50) $$ \begin{align} (\boldsymbol{x}^*,\boldsymbol{0}), (\boldsymbol{0},\boldsymbol{x}^*), (\boldsymbol{L}\boldsymbol{x}^*,\boldsymbol{0}), (\boldsymbol{0},\boldsymbol{L}\boldsymbol{x}^*), \end{align} $$

where $ \boldsymbol{x}^* $ is the relative equilibrium associated with the N-gon solution defined in (41) for the problem of $ N $ -bodies with quasi-homogeneous potential. Since $ \boldsymbol{x}^* $ and $ \boldsymbol{L}\boldsymbol{x}^* $ are orthogonal, it follows that the dimension of $ Q_1(\boldsymbol{x}^*) $ is 4.

Given that the potential functions $ U $ and $ V $ are homogeneous of degrees $ -1 $ and $ -p $ , respectively, we have that their gradients are homogeneous of degrees $ -2 $ and $ -(p+1) $ . Consequently, by Euler’s theorem for homogeneous functions and since $ \boldsymbol{x}^* $ is a simultaneous relative equilibrium, we have that

(51) $$ \begin{align} D\nabla W(\boldsymbol{x}^{*}) \boldsymbol{x}&= -2\nabla U(\boldsymbol{x}^{*})-(p+1)\nabla V(\boldsymbol{x}^{*}) \nonumber\\ &= -2(-\mu mI\boldsymbol{x}^*)-(p+1)(-\nu mI\boldsymbol{x}^*) \nonumber\\ & = 2 \mu m\boldsymbol{x}^* + (p+1) \nu m\boldsymbol{x}^*. \end{align} $$

As a consequence of the fact that $ W $ is invariant under rotations, it follows that $ \nabla W $ is also invariant under rotations, in particular,

$$ \begin{align*} \nabla W(e^{\boldsymbol{L}t}\boldsymbol{x})=e^{\boldsymbol{L}t}\nabla W(\boldsymbol{x}) \end{align*} $$

for all $ t $ , with $ e^{\boldsymbol{L}t}=diag(e^{\boldsymbol{L}_1t},e^{\boldsymbol{L}_2t}, \ldots , e^{\boldsymbol{L}_nt}) $ . Therefore, it follows that

$$ \begin{align*} \nabla W(\boldsymbol{x}) \boldsymbol{L}\boldsymbol{x}=\boldsymbol{L}\nabla W(\boldsymbol{x}). \end{align*} $$

Now, since $ \boldsymbol{x}^* $ is a relative equilibrium, then from (40) we have

(52) $$ \begin{align} D\nabla W(\boldsymbol{x}^*) \boldsymbol{L}\boldsymbol{x}^*=-\omega^2\boldsymbol{L}mI\boldsymbol{x}^*. \end{align} $$

This equality will be employed to prove that $ Q_1(\boldsymbol{x}^*) $ is invariant by $ \boldsymbol {A} $ . In fact, note that

1. (1) $ \boldsymbol {A} (\boldsymbol{x}^*, \boldsymbol {0})= -\omega (\boldsymbol{L}\boldsymbol{x}^*,0)+(0, D\nabla W(\boldsymbol{x}^*) \boldsymbol{x}^*)= -\omega (\boldsymbol{L}\boldsymbol{x}^*,0)+ m ( 2\mu + (p+1)\nu ) (0,\boldsymbol{x}^*)$ , because $ D\nabla W(\boldsymbol{x}^*) \boldsymbol{x}^*= m (2 \mu m+(p+1) \nu ) \boldsymbol{x}^*$ .

2. (2) $ \boldsymbol {A} (\boldsymbol {0}, \boldsymbol{x}^*)= m^{-1}(\boldsymbol{x}^*,0)-\omega (0,\boldsymbol{L}\boldsymbol{x}^*) $ .

3. (3) $ \boldsymbol {A} (\boldsymbol{L}\boldsymbol{x}^*, \boldsymbol {0})= -\omega (\boldsymbol{L}^2\boldsymbol{x}^*,0)+(0,D\nabla W(\boldsymbol{x}^*) \boldsymbol{L}\boldsymbol{x}^*) = \omega (\boldsymbol{x}^*,0)-\omega ^2 m(0,\boldsymbol{L}\boldsymbol{x}^*)$ , because by (52) $ D\nabla W(\boldsymbol{x}^*) \boldsymbol{L}\boldsymbol{x}^*=-\omega ^2\boldsymbol{L}m\boldsymbol{x}^* $ and as $ \boldsymbol{L}^2=-I $ .

4. (4) $ \boldsymbol {A} (\boldsymbol {0}, \boldsymbol{L}\boldsymbol{x}^*) = m^{-1} (\boldsymbol{L}\boldsymbol{x}^*,0)+ \omega (0, \boldsymbol{x}^*) $ .

Thus, we have proved that $ Q_1(\boldsymbol{x}^*) $ is invariant by $ \boldsymbol {A} $ . Moreover,

(53) $$ \begin{align} \beta=\{(\boldsymbol{x}^*,\boldsymbol{0}), (\boldsymbol{0}, \boldsymbol{x}^*), (\boldsymbol{L}\boldsymbol{x}^*,\boldsymbol{0}) (\boldsymbol{0},\boldsymbol{L}\boldsymbol{x}^*)\}, \end{align} $$

is a basis for $ Q_1(\boldsymbol{x}^*) $ , and the matrix of $ \boldsymbol {A} $ restricted to this invariant subspace has the form

(54) $$ \begin{align} [\boldsymbol{A}]_{\beta}=\left[ \begin{array}{cccc} 0 & m^{-1} & \omega & 0 \\ \left( 2\mu +(p+1)\nu \right) m & 0 & 0 & \omega \\ -\omega & 0 & 0 & m^{-1} \\ 0 & -\omega & -\omega^2 m & 0 \end{array}\right], \end{align} $$

whose factor of the characteristic polynomial is

$$ \begin{align*} \lambda^2 (\lambda^2-\alpha+3 \omega^2), \end{align*} $$

with $ \alpha =2\mu +(p+1)\nu $ . Therefore, it follows that the characteristic polynomial is written as the product of two polynomials: one of which is the factor $ \lambda ^2 (\lambda ^2-\alpha +3\omega ^2)$ , which is associated with the last block to be found, and the other is a polynomial, which we shall refer to as $ \hat {p}(\lambda ) $ , the form of which is unknown.

The following proposition will serve to analyze the eigenvalues $\lambda = \pm \sqrt {\alpha -3 \omega ^2}$ .

Proof In fact,

$$ \begin{align*} \gamma = \alpha - 3 \omega^2 = 2 \mu + (p+1) \nu - 3 \omega^2= -\mu + \nu (p-2). \end{align*} $$

Using the equality in (44) and (46), we arrive at

$$ \begin{align*}\gamma= -\frac{m}{4 \rho^3} \left( 2 A - 2^{1-p} \rho^{1-p} B C (p+1) p\right) - 3\omega^2. \end{align*} $$

Then, if $ p\geq 2 $ with $ C<0 $ , it is verified that $ \gamma <0 $ , which consequently establishes that $ \lambda $ is a pure imaginary number.

We proceed to define another subspace, denoted $ Q_2(\boldsymbol{x}^*) $ , of dimension $ 4 $ . This subspace is spanned by the vectors of $ \mathbb {R}^{4n} $

(55) $$ \begin{align} (\xi,\boldsymbol{0}), (\boldsymbol{0},\xi), (\eta,\boldsymbol{0}), (\boldsymbol{0},\eta), \end{align} $$

where $ \xi =(1,0,1,0,\dots ) $ and $\eta =(0,1,0,1,\dots )$ are points of $ \mathbb {R}^{2n} $ . In this case, we have the following result.

Lemma 4 The characteristic polynomial of the matrix $ \boldsymbol {A} $ , restricted to $ Q_2(\boldsymbol{x}^*) $ , can be expressed as a factor of the form

$$ \begin{align*} (\lambda^2+\omega^2)^2. \end{align*} $$

The eigenvalues of this polynomial are $ \pm i \omega $ , with multiplicity $ 2 $ .

Proof It is evident that these vectors are linearly independent in $ \mathbb {R}^{4n} $ . On the other hand, it is known that the function $ W $ is invariant under translations, so $ \nabla W $ is also invariant under translations. Then,

$$ \begin{align*} \nabla W(\boldsymbol{x}+t\xi)=\nabla W(\boldsymbol{x}),\quad\text{for all } t. \end{align*} $$

By differentiating this equality with respect to $ t $ and employing the chain rule, we obtain

(56) $$ \begin{align} D\nabla W(\boldsymbol{x}) \xi=0, \end{align} $$

and analogously,

(57) $$ \begin{align} D\nabla W(\boldsymbol{x}) \eta=0. \end{align} $$

Considering (56) and (57), it follows that:

1. (1) $ \boldsymbol {A}(\xi , \boldsymbol {0})= (-\omega \boldsymbol{L} \xi , D\nabla W(\boldsymbol{x}) \xi )= -\omega (\eta , 0)$ , because $\boldsymbol{L} \xi = \eta $ .

2. (2) $ \boldsymbol {A} (\boldsymbol {0}, \xi ) = m^{-1}(\xi , -\omega \boldsymbol{L} \xi )= m^{-1}(\xi ,\boldsymbol {0})-\omega (\boldsymbol {0},\boldsymbol{L} \xi ) = m^{-1}(\xi ,\boldsymbol {0})-\omega (\boldsymbol {0},\eta )$ .

3. (3) $ \boldsymbol {A}(\eta , \boldsymbol {0}) = (-\omega \boldsymbol{L} \eta , D\nabla W(\boldsymbol{x}) \eta )= \omega (\xi ,\boldsymbol {0}) $ , because $\boldsymbol{L} \eta = -\xi $ .

4. (4) $ \boldsymbol {A}(\eta , \boldsymbol {0}) = m^{-1}(\eta , -\omega \boldsymbol{L} \eta )=m^{-1}(\eta ,\boldsymbol {0})-\omega (\boldsymbol {0},\boldsymbol{L} \eta )=m^{-1}(\eta ,\boldsymbol {0})+\omega (\boldsymbol {0},\xi ) $ .

Thus, we have proved that $ Q_2(\boldsymbol{x}^*) $ is invariant by $ \boldsymbol {A} $ . Furthermore,

(58) $$ \begin{align} \beta'=\{(\xi,\boldsymbol{0}), (\boldsymbol{0},\xi), (\eta,\boldsymbol{0}), (\boldsymbol{0},\eta)\}, \end{align} $$

is a basis for $ Q_2(\boldsymbol{x}^*)$ , and the matrix of A associated with this basis has the form

$$ \begin{align*} [A]_{\beta'}=\left[ \begin{array}{cccc} 0 & m^{-1} & \omega & 0 \\ 0 & 0 & 0 & \omega \\ -\omega & 0 & 0 & m^{-1} \\ 0 & -\omega & 0 & 0 \end{array}\right]. \end{align*} $$

Therefore, it follows that the polynomial $ (\lambda ^2+\omega ^2)^2 $ is a factor of the characteristic polynomial of A. This finishes the proof.

5 Study of the stability of the $3$ -gon solution for the planar $3$ -body problem with quasi-homogeneous potential and $p>1$ arbitrary

In this section, we will study the stability of the N-gon solutions previously obtained, focusing on the case where $ N=3 $ and $ p>1 $ is arbitrary. First, it is observed that for $N=3$ the relative equilibrium solution is parametrized by

(59) $$ \begin{align} x_1&=\rho, y_1=0, x_2=-\frac{1}{2} \rho, y_2=\frac{\sqrt{3}}{2} \rho, x_3=-\frac{1}{2} \rho, y_3=-\frac{\sqrt{3}}{2} \rho. \nonumber\\ X_1&=0, Y_1=m \rho \omega, X_2=-\frac{\sqrt{3}}{2} m \rho \omega, Y_2=-\frac{1}{2} m \rho \omega,\nonumber\\ X_3&=\frac{\sqrt{3}}{2} m \rho \omega, Y_3=-\frac{1}{2} m \rho \omega, \end{align} $$

where $\rho $ is an admissible radius satisfying the equation (27). In particular, the parameterization of the particles forming the equilateral triangle is given by

$$ \begin{align*} q_1=\rho(1,0),\,\ q_2=\rho\left( -\frac{1}{2},\frac{\sqrt{3}}{2}\right),\,\ q_3=\rho\left( -\frac{1}{2},-\frac{\sqrt{3}}{2}\right). \end{align*} $$

The velocities of the particles are

$$ \begin{align*} p_1= m \rho \omega (1,0),\,\ p_2= m \rho \omega \left( \frac{\sqrt{3}}{2},-\frac{1}{2}\right),\,\ p_3= m \rho \omega \left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right). \end{align*} $$

We check that for the case $ N=3 $ , we must have $ \displaystyle \theta _1=0 $ , $ \theta _2= \displaystyle \frac {2\pi }{3} $ , $ \displaystyle \theta _3=\frac {4\pi }{3} $ . Moreover, it follows by equations in (25) that the values of A and B are as follows:

(60) $$ \begin{align} A = 2 \csc\left( \frac{\pi}{3}\right) = \frac{4}{\sqrt{3}}, \quad B = 2\csc\left( \frac{\pi}{3}\right)^{p} = \frac{2^{p+1}}{3^{p/2}}. \end{align} $$

Then, substituting the values of A and B into (60), we get that $ \rho , C, m $ , and $ \omega $ must satisfy

(61) $$ \begin{align} \frac{\omega^2}{m}\rho^{p+2} - \frac{\rho^{p-1}}{\sqrt{3}}- 3^{-p/2}Cp=0. \end{align} $$

In general, it is not straightforward to ascertain the admissible values of $\rho $ from the above equation. Nevertheless, the explicit relationship between m and the remaining parameters, or between C and the remaining parameters, can be determined for a given solution $\rho $ , namely:

(62) $$ \begin{align} m =\frac{3^{1+p/2} \rho^{p+3} \omega^2}{3 C p + 3^{(p+1)/2} \rho^p}, \quad C =\frac{3^{p/2-1}}{mp} \rho^{p-1}\left( -\sqrt{3}m + 3\rho^3 \omega^2\right). \end{align} $$

For $N=3$ , the matrix $ \boldsymbol {A} $ is of dimension $ 12\times 12 $ and in this situation, we have

$$ \begin{align*} Hess\ W(\boldsymbol{x}^*) =\left[ \begin{array}{cccccc} a_{11} & a_{12} & a_{13} & 0 & a_{15} & a_{16} \\ a_{21} & a_{22} & a_{23} & a_{24} & a_{25} & 0 \\ a_{31} & a_{32} & a_{33} & a_{34} & 0 & a_{36} \\ 0 & a_{42} & a_{43} & a_{44} & a_{45} & a_{46} \\ a_{51} & a_{52} & 0 & a_{54} & a_{55} & a_{56} \\ a_{61} & 0 & a_{63} & a_{64} & a_{65} & a_{66} \end{array} \right], \end{align*} $$

where the entries of this matrix are

$$ \begin{align*} a_{11}&= \frac{m \left(-\sqrt{3} m (p-1) + (3p+2) \rho^3 \omega^2 \right) }{6 \rho^3}= -2 a_{12}= -2 a_{13}= -2 a_{21}=-2 a_{31},\\ a_{15}&= \frac{m \left(m- mp +\sqrt{3} (p+2) \rho^3 \omega^2 \right) }{12 \rho^3}= a_{24}= a_{36}= a_{42}= a_{51}= a_{63},\\ a_{16}&= \frac{m \left(m (p-1) -\sqrt{3} (p+2) \rho^3 \omega^2 \right) }{12 \rho^3}= a_{25}= a_{34}= a_{43}= a_{52}= a_{61},\\ a_{22}&= -\frac{m \left(\sqrt{3} m (p-1) + (2-3p) \rho^3 \omega^2 \right) }{12 \rho^3}= a_{33},\\ a_{23}&= \frac{m \omega^2}{3}= a_{32},\\ a_{44}&= \frac{m \left(-\sqrt{3} m (p-1) + 3(p-2) \rho^3 \omega^2 \right) }{18 \rho^3}=-2 a_{45}= -2 a_{46}=-2 a_{54}=-2 a_{64},\\ a_{55}&= \frac{m \left(-5 \sqrt{3} m (p-1) + 3 (5p+2) \rho^3 \omega^2 \right) }{36 \rho^3}= a_{66},\\ a_{56}&= \frac{m \left(\sqrt{3} m (p-1) - 3 (p+1) \rho^3 \omega^2 \right) }{9 \rho^3}= a_{65}. \end{align*} $$

We verify that the characteristic polynomial associated with the matrix $ \boldsymbol {A} $ has the form

(63) $$ \begin{align} C_{\boldsymbol{A}}(\lambda)= \frac{1}{36 \rho^9} \lambda^2(\lambda^2+\omega^2)^2 (\lambda^2- \alpha+ 3 \omega^2)\ \hat{p}(\lambda), \end{align} $$

with $ \bar {p}(\lambda )$ a polynomial of degree four, given by

(64) $$ \begin{align} \hat{p}(x) &= x^2 + a_1 x + a_0,\quad x= \lambda^2 \end{align} $$

where,

(65) $$ \begin{align} a_1= \frac{1}{12 \rho^6} \left(8 \sqrt{3} p m \rho^3 + 4\,\ 3^{(2-p)/2} p m C (p+1) \rho^{4-p} + 12 (1-2 p) \rho^6 \omega^2 \right), \end{align} $$

and

(66) $$ \begin{align} a_0 &= \frac{1}{12 \rho^6} \left(m^2 (5 p^2 + 6 p + 5) + 4\,\ 3^{1-p} p^2 (p+1)^2 m^2 C^2 \rho^{2(1-p)} \right) \nonumber\\ &\quad + \frac{1}{12 \rho^6} \left( 8\,\ 3^{(1-p)/2} p (p+1)^2 m^2 C \rho^{1-p} - 4 \sqrt{3} m \rho^3 \omega^2\right) \\ & \quad+ \frac{1}{12 \rho^6} \left(- 18 \sqrt{3} p m \rho^3 \omega^2 - 10 \sqrt{3} p^2 m \rho^3 \omega^2 \right) \nonumber\\ & \quad- \frac{1}{12 \rho^6} \left( 8 \,\ 3^{(2-p)/2} p (p+1)^2 m C \rho^{4-p} \omega^2 + 3 (5 p^2 + 12 p + 8) \rho^6 \omega^4 \right)\hspace{-1pt}.\nonumber \end{align} $$

At this stage, it is important to highlight that the values of $ \mu $ and $ \nu $ , as presented in (44) and (46), respectively, for $ N=3 $ and in accordance with (62), assume the following form:

$$ \begin{align*}\mu = \frac{m}{\sqrt{3} \rho^3} \quad\text{and}\quad \nu= -\frac{m}{\sqrt{3} \rho^3} + \omega^2, \end{align*} $$

as a function of m, or equivalently,

$$ \begin{align*}\mu = \frac{p \omega^2}{C p + 3^{(p-1)/2} \rho^{p-1}} \quad\text{and}\quad \nu = \frac{C p \omega^2}{C p + 3^{(p-1)/2} \rho^{p-1}}, \end{align*} $$

as function of C.

On the other hand, as previously established in Lemmas 2 and 4 the characteristic polynomial exhibits eight known eigenvalues: two null eigenvalues, four pure imaginary represented by $ \pm i\omega $ , and the remaining two eigenvalues are given by $ \pm \sqrt {\alpha - 3 \omega ^2}= \pm \sqrt {-\frac {m (p-1) }{\sqrt {3} \rho ^3} + (p-2) \omega ^2} $ . We must therefore characterize the two roots of the polynomial $\hat {p}(x)$ .

At the outset, we note that the discriminant of $ \hat {p}(x) $ in (64) is given by

(67) $$ \begin{align} D &= -9 p^2 (p+1)^2 m^2 C^2 \rho^2\nonumber\\ &\quad - 2\,\ 3^{p/2} p (p+1) m C \rho^{p+1} \left(2 \sqrt{3} m (p+2) - 3 (2p+5) \rho^3 \omega^2 \right) \nonumber \\ &- 3^p \rho^{2p} (m^2 (p+1) (p+5) - 2 \sqrt{3} m (p (p+11) + 2) \rho^3 \omega^2\nonumber\\ &+ 3(p (p + 16) + 7) \rho^6 \omega^4 ). \end{align} $$

To carry out the stability study, we evaluate $a_1$ and the discriminant D with respect to the mass specified in (62). Then (65) and (67) are equivalent to

(68) $$ \begin{align} a_1= \frac{\omega^2 \left(-3 p (p-2) C + v \right) }{3 p C + v}, \end{align} $$

and

(69) $$ \begin{align} D= -\frac{8\,\ \omega^4}{(3 p C + v)^2}\left(9 p^3 C^2 + 3 p (p+1) C v + v^2 \right), \end{align} $$

where $ v= \kappa u$ , $\kappa = 3^{(p+1)/2} $ and $ u= \rho ^{p-1} $ .

The roots of the quadratic polynomial in $ x $ are given by

(70) $$ \begin{align} x_- = -\frac{a_1 + \sqrt{D}}{2}, \end{align} $$

and

(71) $$ \begin{align} x_+ = \frac{-a_1 + \sqrt{D}}{2}. \end{align} $$

As demonstrated in Theorem 2.2, for a given $ C>0 $ and $ p>1 $ , there exists a unique admissible radius $ \rho>0 $ , for the fixed mass $ m $ . The following theorem will be employed for the purpose of verifying the type of stability of the associated solution.

Next we will analyze the case $C<0$ . We note that the $ sgn(D)=-sgn(v^2 + 3 p (p+1) C v + 9 p^3 C^2) $ . Therefore, the objective is now to examine the behavior of the quadratic polynomial

(72) $$ \begin{align} h(v)= v^2 + 3 p(p+1) C v + 9 p^3 C^2. \end{align} $$

Lemma 5 Let $C<0$ and $p>1$ . So we have that:

1. (1) $D=0 \Leftrightarrow v=v_+= -3 p^2 C $ or $ v= v_-= -3 p C $ .

2. (2) $ D>0 \Leftrightarrow v_- < v <v_+ $ .

3. (3) $D <0 \Leftrightarrow v>v_+$ or $0 < v < v_-$ .

The graph of D is shown in Figure 6.

Figure 6: Graphic of D with $ C<0 $ and $ p>1 $ .

Proof In fact,

$$ \begin{align*} h(v)=0 & \Leftrightarrow v= \frac{1}{2} \left(-3 p(p+1) C \pm \sqrt{9 p^2 (p+1)^2 C^2 - 36 p^3 C^2} \right) \\ & \Leftrightarrow v= \frac{1}{2} \left(-3 p(p+1) C \pm \sqrt{9 p^2 C^2 ((p+1)^2 - 4p)} \right)\\ & \Leftrightarrow v= \frac{1}{2} \left(-3 p(p+1) C \pm 3 p |C| \sqrt{p^2 -2p + 1} \right)\\ & \Leftrightarrow v= \frac{1}{2} \left(-3 p(p+1) C \pm 3 p (p-1) C \right)\hspace{-1pt}, \end{align*} $$

because $|C|= -C$ . Thus,

(73) $$ \begin{align} v_+ &= \displaystyle \frac{-3 p C}{2} ((p+1) + (p-1))= -3 p^2 C>0, \end{align} $$

(74) $$ \begin{align} v_- &= \displaystyle \frac{-3 p C}{2} ((p+1) - (p-1))= -3 p C>0, \end{align} $$

clearly $ v_- < v_+ $ . This proves part 1 of the lemma.

The proofs of items 2 and 3 of the lemma are immediate, as it can be observed that $ h(v)<0 $ if $ v_- < v < v_+ $ and $ h(v)>0 $ if $ v>v_+ $ . As $ sgn(D)=-sgn(h) $ , the proof of the lemma follows.

Next, we will analyze the root $ x_+ $ of the quadratic polynomial $ \hat {p}(x) $ given in (71). Substituting the values of $a_1$ given in (68) and D given in (69), we obtain that

$$ \begin{align*}x_+ = \frac{\omega^2}{3 p C + v}\left[(3 p (p-2) C - v) + 2 \sqrt{6}\,\ \sqrt{-3 p^3 C^2 + p (p+1) C v + \frac{v^2}{3}} \right]. \end{align*} $$

Now, we know by Lemma 6 that $ 3 p C + v>0 $ . Let us then define the auxiliary function

(75) $$ \begin{align} g(v)= 3 p (p-2) C - v + 2 \sqrt{6}\,\ \sqrt{-3 p^3 C^2 + p (p+1) C v + \frac{v^2}{3}}. \end{align} $$

Of course, it is clear that $sgn(x_+)= sgn(g(v))$ for $v \in (v_-, v_+)$ .

Proof Let $v \in (v_-, v_+)$ . Differentiating the auxiliary function $ g $ given in (75) with respect to v, we have that

(76) $$ \begin{align} g'(v)=-1-\frac{\sqrt{2}\,\ (3 p (p+1) C + 2v)}{\sqrt{-9 p^3 C^2 - 3 p (p+1) C v - v^2}}, \end{align} $$

and differentiating again, we get

(77) $$ \begin{align} g"(v)= -\frac{9 p^2 (p-1)^2 C^2}{\sqrt{2}\,\ \left[-9 p^3 C^2 - 3 p (p+1) C v - v^2 \right]^{3/2}}. \end{align} $$

Note that the second derivative is always negative, so $ g' $ is decreasing for all $ v $ . In particular, for $v \in (v_-, v_\sharp )$ it is verified that

(78) $$ \begin{align} g'(v_\sharp)< g'(v) < g'(v_-). \end{align} $$

We assert that $ g'(v_\sharp )=0 $ . In fact,

$$ \begin{align*} g'(v_\sharp) &= -1-\frac{\sqrt{2}\,\ [3 p (p+1) C - 2 p (p+2) C]}{\left[-9 p^3 C^2 + 3 p^2 (p+1)(p+2) C^2 - p^2 (p+2)^2 C^2 - p^2 (p+2)^2 C^2\right]^{1/2}}\\ & =-1+\frac{\sqrt{2}\,\ p C (p-1)}{p C \left[-9 p + 3 p (p+1)(p+2) - (p+2)^2\right]^{1/2}}= -1 + \frac{\sqrt{2}\,\ (p-1)}{\sqrt{2 (p-1)^2}}= 0. \end{align*} $$

Thus, $ 0 = g'(v_\sharp )< g'(v) < g'(v_-) $ , therefore $ g $ is increasing by $v \in (v_*, v_\sharp ) $ . Then,

(79) $$ \begin{align} g(v_-)< g(v) < g(v_\sharp). \end{align} $$

In an analogous way, we affirm that $g(v_\sharp )=0$ . In fact,

$$ \begin{align*} g(v_\sharp)&= 3 p (p-2) C + p (p+2) C + 2 \sqrt{6}\,\\ &\quad\sqrt{\left(-3 p^3 C^2 - p^2 (p+1)(p+2) C^2 + \frac{1}{3} (p+2)^2 p^2 C^2\right)}\\ &= 3 p (p-2) C + p (p+2) C + 2 \sqrt{6}\,\\ &\quad\sqrt{p^2 C^2 \left(-3 p - (p+1)(p+2) + \frac{1}{3} (p+2)^2\right)}\\ &= p C \left(4 (p-1) - 2\sqrt{2}\,\ \sqrt{-2 p^2 - 4p + 2} \right)\\ &= p C (4 (p-1) - 4\,\ \sqrt{p^2 - 2p +1})= p C (4 (p-1)-4 (p-1))=0. \end{align*} $$

Then, $g(v) <0 $ for $v \in (v_-, v_\sharp ) $ .

Similarly, since $ g" $ is negative, then $ g' $ is decreasing in $ (v_\sharp , v_+) $ , so $ g'(v_+) < g'(v) < g'(v_\sharp ) $ and as $ g'(v_\sharp )=0 $ , we have that $ g $ is decreasing in $ (v_\sharp , v_+) $ . Then $ g(v_+) < g(v) < g(v_\sharp ) $ and again since $ g(v_\sharp )=0 $ , we conclude that $ g < 0 $ in $ (v_\sharp , v_+ ) $ . This finishes the proof of the lemma.

Now, we know from Theorem 2.3 of Section 2, that for $ C<0 $ , $ p>1 $ , and $ m> m_\sharp $ defined in (29) there exist two radii $ \rho _1, \rho _2>0 $ admissible for fixed mass $ m $ . Next, we will check the type of stability of these solutions.

6 Concluding remarks

In this work, we consider the planar N-body problem with quasi-homogeneous potential given by

$$ \begin{align*}W = \sum_{1\leq k<j\leq N} \left[ \frac{ m_k m_j}{\|\boldsymbol{r}_k-\boldsymbol{r}_j\|} +\frac{m_k m_j C_{jk}}{\|\boldsymbol{r}_k-\boldsymbol{r}_j\|^p} \right], \end{align*} $$

where $ m_k>0 $ are the masses and $C_{jk}= C_{kj}$ is a nonzero real constant, with the exponent $ p> $ 1. We characterize the periodic solutions that form a regular polygon with equal masses ( $m_k= m$ , $k=1, \cdots , N$ ) and equal constants $C_{jk}$ ( $C_{jk}= C$ ( $j, k=1, \cdots , N$ ) (for short, N-gon solutions).

Initially, we study the existence of periodic solutions with angular velocity $ \omega $ and equal masses that form a regular polygon on a circle of radius $ \rho $ (as we said before), assuming that $ C_{kj}=C $ for all $ k,j=1,\ldots , N $ . We characterize the possible radii as a function of $C, m$ , and p. We verify that the existence of an N-gon solution with quasi-homogeneous potential is given by the fulfillment of the following condition:

$$ \begin{align*} \frac{\omega^2}{m}\rho^{p+2}-\frac{A}{4}\rho^{p-1}-\frac{pBC}{2^{p+1}}=0, \end{align*} $$

with

$$ \begin{align*} A= \displaystyle\sum_{j= 2}^{N}\left|\csc\left( \frac{\pi(j-1)}{N}\right) \right|,\quad B= \sum_{j=2}^{N} \left|\csc\left( \frac{\pi(j-1)}{N}\right)\right|{}^{p}, \end{align*} $$

positive constants. We have shown that for $ C_{kj}=C>0 $ there is a unique admissible radius which leads to a N-gon solution for the planar $N $ -body problem with quasi-homogeneous potential. On the other hand, for $ C_{kj}=C< 0 $ there are 2, 1, or no admissible radii. In fact, there is a critical value of $ m $ denoted by $ m_\sharp = \frac {\omega ^2 (p+2) \left (-B C p (p+2) A^{-1} \right )^{3 /(p-1)}}{2 A (p-1) }$ such that if $ m< m_\sharp $ there are no admissible radii. If $ m = m_\sharp $ there is a unique admissible radius which gives rise to a unique N-gon solution. If $ m> m_\sharp $ , then there exist two admissible radii denoted by $ \rho _1 $ and $ \rho _2 $ such that $ \rho _* = 1/2 \left (- p B C / A \right )^{1/ p-1} < \rho _1 < \rho _\sharp = 1/2 \left (-(B C /3 A) p(p+2) \right )^{1/p-1} $ and $ \rho _2> \rho _\sharp $ each of them gives rise to a N-gon solution.

Furthermore, we guarantee that one of the factors of the characteristic polynomial of the Hamiltonian matrix associated with a relative equilibrium is given by the expression $ \lambda ^2(\lambda ^2-\alpha +3\omega ^2)$ , where $ \displaystyle \alpha = 2 \mu +(p+1)\nu $ and the values of the parameters are as follows: $ \mu = A m /4 \rho ^3 $ and $ \nu = m p B C / 2^{p+1} \rho ^{p+2}$ . The remaining factor of the characteristic polynomial is of the form $ (\lambda ^2+\omega ^2)^2 $ .

We conclude this work by studying the stability for the case $ N=3 $ . We find that for $ C>0 $ the $3$ -gon solution is always unstable in the Lyapunov sense. For $ C<0 $ , defining the auxiliary values of $\rho $ , namely, $\rho _*=\left (-3 p C/3^{(p+1)/2}\right )^{1/(p-1)}$ , $\rho _\sharp = \left (-p(p+1)C/3^{(p+1)/2}\right )^{1/(p-1)}$ and $\rho _+ = \left (-3 p^2 C/3^{(p+1)/2}\right )^{1/(p-1)}$ , we have that the $3$ -gon solution with $\rho = \rho _1 $ is spectrally stable throughout its entire interval of definition $\left (\rho _*, \rho _\sharp \right ]$ . The solution for $\rho =\rho _2$ is spectrally stable in the interval $(\rho _\sharp , \rho _+]$ . It is unstable in the Lyapunov sense in the interval $ \rho _2> \rho _+ $ .