On Strictly Convex Central Configurations of the 2n-Body Problem


We consider planar central configurations of the Newtonian 2n-body problem consisting in two twisted regular n-gons of equal masses. We prove the conjecture that for \(n\ge 5\) all convex central configurations of two twisted regular n-gons are strictly convex.

This is a preview of subscription content, access via your institution.

Fig. 1
Fig. 2
Fig. 3
Fig. 4


  1. 1.

    Barrabés, E., Cors, J.: On central configurations of twisted crowns. arXiv:1612.07135 (2016)

  2. 2.

    Chen, K.C., Hsiao, J.S.: Convex central configurations of the \(n\)-body problem which are not strictly convex. J. Dyn. Differ. Equ. 24, 119–128 (2012)

    MathSciNet  Article  Google Scholar 

  3. 3.

    Fernandes, A., Garcia, B., Mello, L.: Convex but not strictly convex central configurations. J. Dyn. Differ. Equ. (2017). https://doi.org/10.1007/s10884-017-9596-0

    Article  MATH  Google Scholar 

  4. 4.

    Roberts, G.: Existence and stability of relative equilibria in the n-body problem. PhD thesis, Boston University (1999)

  5. 5.

    Saari, D.G.: Collisions, rings, and other Newtonian \(N\)-body problems, CBMS Regional Conference Series in Mathematics, vol 104. Published for the Conference Board of the Mathematical Sciences, Washington, DC; by the American Mathematical Society, Providence, RI (2005)

  6. 6.

    Yu, X., Zhang, S.: Twisted angles for central configurations formed by two twisted regular polygons. J. Differ. Equ. 253, 2106–2122 (2012)

    MathSciNet  Article  Google Scholar 

Download references

Author information



Corresponding author

Correspondence to E. Barrabés.

Additional information

This work has been realized thanks to the MINECO Grants MTM2016-80117-P and MTM2016-77278-P (FEDER) and Catalan (AGAUR) Grants 2017 SGR 1374 and SGR 1617.



Proof of Lemma 1

We want to prove that, for any fixed value \(a\in [\cos (\pi /7),1)\), the function

$$\begin{aligned} T(a,u)= \dfrac{a}{2}-\dfrac{4}{b^3} \dfrac{bu-a}{(1+u^2)^{3/2}}, \end{aligned}$$

is positive in the closed interval \(u\in {{\mathscr {K}}}=\left[ u_m(a),\dfrac{3a}{b}\right] \), where \(b=\sqrt{1-a^2}\) and

$$\begin{aligned} u_m(a)=\dfrac{4a^3-2a^2+a-1}{b(4a^2-2a-1)}. \end{aligned}$$

The function T is clearly continuous and differentiable in the domain, so T must attain an absolute minimum on \({\mathscr {K}}\). On one hand, if we look for critical points in the interior of \({\mathscr {K}}\),

$$\begin{aligned} \dfrac{\partial T}{\partial u} = \dfrac{-4}{b^3}\dfrac{b-2bu^2+3au}{(1+u^2)^{5/2}}=0 \end{aligned}$$

occurs at \(u_c=(3a+\sqrt{a^2+8})/(4b)\). It is not difficult to see that \(u_c<u_m\) for \(a\in [\cos (\pi /7),1)\), and \(\partial T/\partial u (a,u) >0\) for \(u\in [u_m,3a/b]\). Therefore, the minima of T occurs at \(u=u_m(a)\).

To conclude the proof it is enough to see that \(T(a,u_m(a))>0\). Simplifying, we want to see that

$$\begin{aligned} T(a,u_m(a))= \dfrac{a}{2}-\sqrt{2} \dfrac{(2a-1)(4a^2-2a-1)^2}{(16a^4-16a^3+a-1)^{3/2}}>0, \end{aligned}$$

which is equivalent to prove that

$$\begin{aligned} a (16a^4-16a^3+a-1)^{3/2} > 2\sqrt{2} (2a-1) (4a^2-2a-1)^2. \end{aligned}$$

Squaring both sides, the expression factors into

$$\begin{aligned} (a-1)p(x) >0, \end{aligned}$$

where p(x) is a polynomial of degree 13 with integer coefficients, which using Sturm’s Theorem contains no real zeros in the interval [9 / 10, 1].

Proof of Lemma 2

We want to prove that, for any fixed value of \(a\in [\cos (\pi /6),1)\), the function

$$\begin{aligned} T(a,u)= \dfrac{1}{2}-2\dfrac{a^2}{b^3} \dfrac{1+(bu-a)^2}{(1+u^2)^{3/2}} \end{aligned}$$

is negative for any \(u\in {{\mathscr {K}}}=\left[ \dfrac{a}{b},u_m(a)\right] \), where \(b=\sqrt{1-a^2}\) and

$$\begin{aligned} u_m(a)=\dfrac{2a(3a^2-2)}{b(2a^2-1)}. \end{aligned}$$

The function is clearly continuous and differentiable in \({\mathscr {K}}\). On one hand, the function T(au) is strictly increasing with respect the variable u because

$$\begin{aligned} \dfrac{\partial T}{\partial u}= & {} \dfrac{2a^3}{b^3}\dfrac{b^2u^3-4abu^2+(5a^2+1)u+2ab}{(1+u^2)^{5/2}} \\\ge & {} \dfrac{2a^3}{b^3} \dfrac{u(b^2u^2-4abu+5a^2+1}{(1+u^2)^{5/2}} >0. \end{aligned}$$

Therefore, \(T(a,u) \le T(a,u_m(a))\) and

$$\begin{aligned} p(a)=T(a,u_m)=\dfrac{1}{2} - 2a^3 \dfrac{(2a^2-1)(16a^6-20a^4+5a^2+1)}{(32a^6-40a^4+11a^2+1)^{3/2}}. \end{aligned}$$

Finally, it is not difficult to see that \(p'(a)\) has no roots in \(a\in [\cos (\pi /6),1]\) (for example, using Sturm’s theory) and \(p'(a)>0\). Therefore, \(T(a,u_m(a)) < p(1) =0\), which concludes the proof.

Rights and permissions

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Barrabés, E., Cors, J.M. On Strictly Convex Central Configurations of the 2n-Body Problem. J Dyn Diff Equat 31, 2293–2304 (2019). https://doi.org/10.1007/s10884-018-9708-5

Download citation


  • Central configuration
  • Convex central configuration
  • n-Body problem
  • Twisted central configuration

Mathematics Subject Classification

  • 70F10
  • 70F15