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

Abstract

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.

Appendix

1.1 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].

1.2 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(a, u) 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.