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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.

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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.

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Barrabés, E., Cors, J.M. On Strictly Convex Central Configurations of the 2n-Body Problem. J Dyn Diff Equat 31, 2293–2304 (2019).

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

Mathematics Subject Classification

  • 70F10
  • 70F15