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$ .

Figure 4: Long description

The vertical axis is labeled m and the horizontal axis is labeled rho. A blue curve labeled f opens upward. Two horizontal dashed lines are drawn: one at m equals m sub H and one above it at m greater than m sub H. The curve intersects the lower line at three points, marked as rho star, rho sub 1, and rho sub H, with rho star at the far left, rho sub 1 to its right, and rho sub H at the curve's minimum. The curve then rises to intersect the upper line at rho sub 2, further right. All intersection points are marked with black dots. Vertical dashed lines connect each intersection to the rho axis.

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$ .

Figure 5: Long description

At the center is the origin marked O, with two perpendicular axes forming a cross. Two arrows labeled rho sub 1 and rho sub 2 extend from the origin to the vertices of two concentric dashed polygons. The inner polygon has vertices labeled P sub 1, P sub 2, and so on, up to P sub N, arranged counterclockwise. The outer polygon mirrors this structure, with corresponding vertices labeled similarly. The angle theta sub 0 is marked at the origin along the positive horizontal axis. Each vertex is represented by a large black dot. The polygons are connected by dashed lines, and the overall structure is symmetric about the origin.

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 $ .

Figure 6: Long description

The vertical axis is labeled D and the horizontal axis is labeled v. A blue curve opens downward, intersecting the horizontal axis at two points marked with black dots labeled v minus on the left and v plus on the right. Between these points, the region above the axis is labeled D greater than zero. To the right of v plus, the region below the axis is labeled D less than zero. At v plus, the curve is labeled D equals zero. The left region below the axis is also D less than zero.

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 _+ $ .