This is a preprint of: “On Strictly Convex Central Configurations of the 2n-Body Problem”, Esther Barrabés, Josep Maria Cors, J. Dyn. Diff. Equat., vol. 31, 2293–2304, 2019. DOI: [10.1007/s10884-018-9708-5] On strictly convex central configurations of the 2n-body problem E. Barrabés · J.M. Cors Abstract We consider planar central configurations of the Newtonian 2n-body prob- lem consisting in two twisted regular n-gons of equal masses. We prove the conjec- ture that for n ≥ 5 all convex central configurations of two twisted regular n-gons are strictly convex. Keywords central configuration · convex central configuration · n-body problem · twisted central configuration Mathematics Subject Classification (2010) 70F10 · 70F15 1 Introduction and Main Result Consider the planar Newtonian 2n-body problem of two groups of n bodies in the same plane (x, y) at positions q ji ∈ R2 , i = 1, . . . , n, j = 1, 2, such that all the bod- ies in the same group have equal mass, m1 and m2 , respectively. Without lost of generality, we can consider the center of mass at the origin. It is well known (see Saari (2005)) that a central configuration of the planar 2n-body problem is a solution q = (q11 , q12 , . . . , q2n ) ∈ R4n of the equation ∇U(q) + w2 Mq = 0, 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. E. Barrabés Universitat de Girona, Escola Politècnica Superior, 17071 Girona, Spain E-mail: esther.barrabes@udg.edu J.M. Cors Universitat Politècnica de Catalunya, Escola Politècnica Superior d’Enginyeria de Manresa, 08242 Man- resa, Spain E-mail: cors@epsem.upc.edu 2 E. Barrabés, J.M. Cors for some value of w, where U is the Newtonian potential 2 n−1 n m2j n n m1 m2 U(q) = ∑∑ ∑ +∑ ∑ , j=1 i=1 l=i+1 ||q ji − q jl || i=1 m=1 ||q1i − q2m || and M is the diagonal mass matrix with diagonal m1 , . . . , m1 , m2 , . . . , m2 (each mass repeated n times). We are interested in central configurations such that the bodies are located at the vertices of two regular n-gons. A configuration of n equal masses at the vertices of a regular n-gon is also called a ring. Yu and Zhang (2012) show that in any central configuration of two regular n-gons of equal masses within each gon, the two rings are nested, i.e., the vertices of the two n-gons are aligned, or they are twisted, that is, the vertices of the second n-gon are rotated an angle π/n with respect the vertices of the first one. Definition 1 Let q ji ∈ R2 , i = 1, . . . , n, j = 1, 2 be a central configuration of two rings. The configuration is convex if the interior of the polygon defined by the sequence of vertex q11 , q21 , q12 , q22 , . . . , q1n , q2n is a convex set. It is strictly convex if every internal angle is strictly less than π. Clearly, from the definition, in a convex central configuration of two rings each vertex qji is on the boundary of the convex hull defined by all the vertices. Moreover, a configuration is convex but not strictly convex if one or more interior angles are equal to π. Since no convex configurations are possible when the two rings are nested, we focuss on central configurations of two twisted rings. Without lost of generality, we can consider that the bodies of one ring have mass m1 = 1 and are located on a circle of radius 1. The second ring is formed by bodies of mass m2 = m and are located at a circle of radius r. It is not difficult to see that the configuration is convex if and only if π  1 1 an = cos ≤r≤ π  = . (1) n cos an n The values of an , 1/an correspond to convex configurations which are not strictly convex. See Figure 1. In Roberts (1999), Yu and Zhang (2012) and Barrabés and Cors (2016), it is shown that two twisted rings are in central configuration if and only if m and r satisfy the following equation:   S (C(r) − rS) + − rD(r) m = 0, (2) r2 where S is the constant 1 n−1 1 S= ∑ sin(kπ/n) , 4 k=1 (3) On strictly convex central configurations of the 2n-body problem 3 Fig. 1 The points X and Y give the limits for the strictly convex configurations of two rings. and C and D are functions of r that depend on mutual distances: n r − cos((2k − 1)π/n) C(r) = ∑ (1 + r2 − 2r cos((2k − 1)π/n))3/2 , (4) k=1 n 1 − r cos((2k − 1)π/n) D(r) = ∑ (1 + r2 − 2r cos((2k − 1)π/n))3/2 . (5) k=1 Notice that equation (2) is linear with respect to the mass m. Then, solving for m as a function of the radius r > 0 we have that r2 (rS −C(r)) F(r) m = H(r) = = r2 . (6) S − r3 D(r) G(r) Therefore, not all values of r are admissible, only those for which H(r) > 0. In Barrabés and Cors (2016), we characterize the admissible values of r and the number of central configurations for each value of n. In Figure 2 we show two central configurations of two√twisted rings for n = 4 and n = 5. For n = 4, a solution of equation (6) is given by r = 2 and m = 16.05679942... that corresponds a convex central configuration of the 8-body problem which is not strictly convex. For n = 5, a solution of equation (6) is given by r = 1.1 and m = 2.365332699..., which is a strictly convex central configuration of the 10-body problem. In the general N-body problem, it is known that when N = 4 any convex central configuration is strictly convex. In the 5-body problem an example of a con- vex central configuration which is not strictly convex is given by Chen and Hsiao (2012). In Fernandes et al. (2017) the authors prove that any convex central configuration of two twisted rings when n = 3, 5, 6 is strictly convex. For n = 4 they show that there exists a convex central configuration which is not strictly convex (represented in Figure 2, left). They also conjecture that n = 4 is the only value of n for which there exist non-strictly convex central configurations of two twisted rings. Our aim is to give proof of that conjecture. 4 E. Barrabés, J.M. Cors Fig. 2 Left, a convex central configuration √ of the 8-body problem which is not strictly convex formed by two twisted rings with r = 2. Right, a strictly convex central configuration of the 10-body problem formed by two twisted rings with with r = 1.1 Theorem 1 Let us consider 2n bodies in a central configuration of two twisted rings. Then, for n ≥ 5 all convex central configurations are strictly convex. All the numerical computations carried out throughout the paper use integer arith- metics. For instance, in the computation of the Sturm sequence of a univariate poly- nomial with integer coefficients to count the number of real roots in a given interval. Since no floating-point arithmetic have been used nowhere, all the results are analytic. 2 Proof of Theorem 1 First, we prove that the set of strictly convex central configurations of two twisted rings is not empty. In fact, r = 1 and m = 1 is a solution of (6) that correspond to the regular 2n-gon with equal masses, which is obviously a strictly convex central configuration. Second, recall from (1) that the values r = an , 1/an correspond to convex config- urations which are not strictly convex. Therefore, to proof Theorem 1 it is enough to prove that both H(an ) < 0 and H(1/an ) < 0, where m = H(r) is the function defined in (6). Next result comes straightforward from the definitions of the functions F and G in (6). Proposition 1 Let F(r) and G(r) be the functions defined in (6) for r ∈ (0, ∞). Then, F(r) and G(r) are analytic functions and F(r) = r G(1/r). From Proposition 1, for all positive values of r, H(1/r) = 1/H(r). Thus, the proof of Theorem 1 reduces to see that H(an ) < 0. This is equivalently to see that F(an ) · G(an ) < 0, which follows from the next two Propositions. π  Proposition 2 Let F(r) be defined in (6) for r ∈ (0, ∞) and an = cos . Then, n F(an ) > 0 for n ≥ 5. On strictly convex central configurations of the 2n-body problem 5 π  Proposition 3 Let G(r) be defined in (6) for r ∈ (0, ∞) and an = cos . Then, n G(an ) < 0 for n ≥ 5. Next subsections are devoted to prove Propositions 2 and 3. 2.1 Proof of Proposition 2 From equation (6), F(r) = rS −C(r), where S and C(r) are given in (3) and (4). First, notice that due to the symmetry of the angles involved in the sums we can write    1 (n/2)−1 1 1  2 ∑ sin(kπ/n) + 4 n even,   S= k=1 (7)  (n−1)/2  1 1  2 ∑ sin(kπ/n)   n odd, k=1 and   n/2  r − cos((2k − 1)π/n)  2 ∑ (1 + r2 − 2r cos((2k − 1)π/n))3/2   n even, C(r) = k=1  (n−1)/2  r − cos((2k − 1)π/n) 1  2 ∑ (1 + r2 − 2r cos((2k − 1)π/n))3/2 + (1 + r)2 n odd.   k=1 Let [·] be the floor function. For convenience,   we split S in the following way: S = S1 + S2 , where S1 keeps the first n−1 2 − 2 terms and S2 the remaining ones. Explicitly,  0  n = 5, 6, [ n−1 2 ]−2 S1 = 1 1   2 ∑ sin(kπ/n) n ≥ 7, k=1 and      1 1 1 1   n π + n π + 6 ≤ n even,   2 sin(( 2 − 2) n ) sin(( 2 − 1) n ) 4 S2 = !     1 1 1   π + π 5 ≤ n odd. 2 sin(( n−3 2 )n) sin(( n−1 2 )n)  n−1We  also split the function C(r) = C1 (r) + C2 (r), such that C1 keeps the first 2 − 1 terms and C2 the remaining ones:  0  n = 5, 6, [ n−1 2 ]−1 C1 (r) = r − cos((2k − 1)π/n) 2  ∑ (1 + r2 − 2r cos((2k − 1)π/n))3/2 n ≥ 7, k=1 6 E. Barrabés, J.M. Cors and   2(r − cos ((n − 3)π/n))    2 3/2  (1 + r − 2r cos ((n − 3)π/n))    2(r − cos ((n − 1)π/n)) C2 (r) = + 6 ≤ n even,   (1 + r2 − 2r cos ((n − 1)π/n))3/2     2(r − cos ((n − 2)π/n)) 1   + 5 ≤ n odd. 2 (1 + r − 2r cos ((n − 2)π/n)) 3/2 (1 + r)2 Therefore, F = F1 + F2 where Fi (r) = rSi − Ci (r), i = 1, 2. Next, we will show that F1 (an ) ≥ 0 and F2 (an ) > 0 for n ≥ 5, which concludes the proof of Proposition 2. 1. First we consider the function F1 (r) for n ≥ 7 (F1 = 0 for n = 5, 6). Recall that an = cos(π/n). Notice that the first sumand in the expression of C1 (an ) vanish. Therefore [ n−1 2 ]−2   an 2(an − cos((2k + 1)π/n)) F1 (an ) = ∑ − . k=1 2 sin(kπ/n) (1 + a2n − 2an cos((2k + 1)π/n))3/2 Expanding cos((2k + 1)π/n) in terms of xnk = cos(kπ/n) we get [ n−1 2 ]−2 1 F1 (an ) = ∑ sin(kπ/n) T (an , xnk ), k=1 where a ay + bx T (a, x) = − 4y2 2 2 , (8) 2 (b y + (bx + 2ay)2 )3/2 √ √ and b = 1 − a2 , y = 1 − x2 . For any fixed value n, we have that     π  n−1 π n−1 cos −2 ≤ xnk ≤ cos , k = 1, . . . , [ ] − 2. 2 n n 2 Thus, we consider the function T (an , x) defined in x ∈ [xmin (an ), an ], where p   1 − a2n (4a2n − 1) for n even,  xmin (an ) = r   1 − an  (4a2n + 2an − 1) for n odd. 2   Clearly, F1 (an ) > 0 if T (an , xnk ) > 0 for all k = 1, . . . , n−1 2 − 2 and n ≥ 7. To show that we use the following Lemma. Its proof can be found in the Appendix. √ Lemma 1 For any fixed value a ∈ [cos(π/7), 1), let b = 1 − a2 and a 4 bu − a T (a, u) =− 3 . 2 b (1 + u2 )3/2  3  4a − 2a2 + a − 1 3a Then, T (a, u) > 0 for u ∈ , . b(4a2 − 2a − 1) b On strictly convex central configurations of the 2n-body problem 7 Introducing the new variable u = x/y + 2a/b in the expression (8), we get that T (a, x) = T (a, u) where T is the function defined in Lemma 1. Using that u is an increasing function of x, the interval x ∈ [xmin (an ), an ] transforms into u∈ [umin (an ), 3an /bn ] where   4a4 − a2 − 1  uem (a) =  for n even, umin (a) = ab(4a2 − 3) (9) 3 2 4a − 2a + a − 1    uom (a) = for n odd. b(4a2 − 2a − 1) It is important to notice that to derive the expressions of umin (a) we have used that 4a2 −3 > 0 (n even) and 4a2 −2a−1 > 0 (n odd), which are true for a > cos(π/8) (n even) or a > cos(π/7) (n odd). It is a straightforward computation to verify that uom (a) < uem (a) when a < 1 (see Figure 3). Thus, using the result of Lemma 1, we finish the proof that F1 (an ) > 0. 30 20 u 10 0 0,92 0,94 0,96 0,98 1 a Fig. 3 Domain of T (a, u) given in Lemma 1: uem (a), red solid curve; uom (a), blue dashed curve, are given by (9) and u = 3a/b, black solid curve. 2. Second, we want to show that F2 (an ) > 0 for n ≥ 6 even. As before, recall that an = cos(π/n). Therefore, F2 (an ) = q1 (an ) − q2 (an ), where 1 a a q1 (a) = + + , 2 4 2(2a2 − 1) 4a(2a2 − 1) 4a q2 (a) = 3/2 + , 2 4 (1 − 5a + 8a ) (1 + 3a2 )3/2 are continuous decreasing functions in [cos(π/8), 1]. Then, F2 (an ) ≥ q1 (1) − q2 (cos(π/8)) > 0, for n ≥ 8. The case n = 6 can be checked directly. 8 E. Barrabés, J.M. Cors 3. Finally, we want to show that F2 (an ) > 0 for n ≥ 7 odd. In this case,   an 1 2(2a2n + an − 1) 1 F2 (an ) =  π  + 1 − − . 2 cos 2a n − 1 (1 − 2an + a2 + 4a3 )3/2 n n (1 + an )2 2n It ismore suitable to write the above expression in terms of the half angle pn = π cos through the identity an = 2p2n − 1. Thus, F2 (an ) = q1 (pn ) − q2 (pn ) 2n where (2p2 − 1)2 1 q1 (p) = − , p(4p2 − 3) 4p4 4p2 (4p2 − 3) q2 (p) = , (40p6 − 60p4 + 26p2 − 2)3/2 for p ∈ [cos(π/14), 1]. Both functions are continuous and decreasing in that in- terval, so F2 (an ) ≥ q1 (1) − q2 (cos(π/14)) > 0, for n ≥ 7. 2.2 Proof of Proposition 3 From equation (6), G(r) = S − r3 D(r), where S and D(r) are given in (3) and (5). We will follow the same ideas than in the previous Section in order to prove that G(an ) < 0 . First, we use the symmetry of the angles to write the constant S as in (7) and   n/2  1 − r cos((2k − 1)π/n)  ∑ (1 + r2 − 2r cos((2k − 1)π/n))3/2   2 n even, D(r) = k=1  (n−1)/2  1 − r cos((2k − 1)π/n) 1  2 ∑ (1 + r2 − 2r cos((2k − 1)π/n))3/2 + (1 + r)2 n odd.   k=1 n Next, we split S = S1 + S2 , where S1 keeps the first 2 − 2 terms and S2 the remaining ones. Explicitly, for n ≥ 5  0 n  n = 5, [ 2 ]−2 S1 = 1 1  2 ∑ sin(kπ/n) n ≥ 6,  k=1 and   1 1 1   n π + 6 ≤ n even,   2 sin(( − 1) ) 4  2 n S2 = !   1 1 1     π + π 5 ≤ n odd. 2 sin(( n−3 2 )n) sin(( n−1 2 )n) On strictly convex central configurations of the 2n-body problem 9  n  We also split the function D(r) = D1 (r) + D2 (r), such that D1 keeps the first 2 − 2 terms and D2 the remaining ones:  0 n  n = 5, [ 2 ]−2 D1 (r) = 1 − r cos((2k − 1)π/n) 2 ∑  2 3/2 n ≥ 6, k=1 (1 + r − 2r cos((2k − 1)π/n)) and    2(1 − r cos ((n − 3)π/n))   2 3/2  (1 + r − 2r cos ((n − 3)π/n))       2(1 − r cos ((n − 1)π/n))   + 6 ≤ n even, (1 + r2 − 2r cos ((n − 1)π/n))3/2 D2 (r) =   2(1 − r cos ((n − 4)π/n))     (1 + r2 − 2r cos ((n − 4)π/n))3/2     2(1 − r cos ((n − 2)π/n)) 1    + + 5 ≤ n odd. 2 (1 + r − 2r cos ((n − 2)π/n)) 3/2 (1 + r)2 Therefore, G = G1 + G2 where Gi (r) = Si − r3 Di (r), i = 1, 2. Next, we will show that G1 (an ) ≤ 0 and G2 (an ) < 0 for n ≥ 5. Recall that an = cos(π/n). 1. First we consider the function G1 (r) for n ≥ 6 (G1 = 0 for n = 5), which can be written as [ 2n ]−2   1 2a3n (1 − an cos((2k − 1)π/n)) G1 (an ) = ∑ − . k=1 2 sin(kπ/n) (1 + a2n − 2an cos((2k − 1)π/n))3/2 Expanding cos((2k − 1)π/n) in terms of xnk = cos(kπ/n) we get [ 2n ]−2 1 G1 (an ) = ∑ sin(kπ/n) T (an , xnk ), k=1 where 1 y2 + (ay − bx)2 T (a, x) = − 2a3 y 2 2 , (10) 2 (b y + (2ay − bx)2 )3/2 √ √ and b = 1 − a2 , y = 1 − x2 . For any fixed value n, we have that h n i π π  n−1 cos −2 ≤ xnk ≤ cos , k = 1, . . . , [ ] − 2. 2 n n 2 Therefore, it is enough to see that T (an , x) < 0 for x ∈ [xmin (an ), an ] where   2an bn   for n even, xmin (an ) = r   1 − an  (4a2n + 2an − 1) for n odd. 2 In order to do that we use the following Lemma. Its proof can be found in the Appendix. 10 E. Barrabés, J.M. Cors √ Lemma 2 For any fixed value a ∈ [cos(π/6), 1), let b = 1 − a2 and 1 a2 1 + (bu − a)2 T (a, u) = −2 3 . 2 b (1 + u2 )3/2   a 2a(3a2 − 2) Then, T (a, u) < 0 for u ∈ , . b b(2a2 − 1) Introducing the new variable u = −x/y + 2a/b in the expression (10), we get that T (a, x) = T (a, u) where T is the function defined in Lemma 2. Using that u is a decreasing function of x, the interval x ∈ [xmin (an ), an ] transforms into u ∈ [an /bn , umax (an )] where   2a(3a2 − 2)  uem (a) =  for n even, b(2a2 − 1) umax (a) = (11)   12a3 − 6a2 − 5a + 1  uom (a) = for n odd. b(4a2 − 2a − 1) It is important to notice that to derive the expressions of umax (a) we have used that 2a2 −1 > 0 (n even) and 4a2 −2a−1 > 0 (n odd), which are true for a > cos(π/6) (n even) or a > cos(π/7) (n odd). It is a straightforward computation to verify that uom (a) < uem (a) when a < 1 (see Figure 4). So, using the result of Lemma 2, we finish the proof that G1 (an ) > 0 for n ≥ 6. 30 20 u 10 0 0,88 0,9 0,92 0,94 0,96 0,98 a Fig. 4 Domain of T (a, u) given in Lemma 2: uem (a), red solid curve; uom (a), blue dashed curve, are given by (11). u = a/b, black solid curve. On strictly convex central configurations of the 2n-body problem 11 2. Second, we want to show that G2 (an ) < 0 for n ≥ 6 even. Using that an = cos(π/n), G2 (an ) = q1 (an ) − q2 (an ), where 1 1 2a3 (1 + a2 ) q1 (a) = + − , 4 2a (1 + 3a2 )3/2 2a3 (4a4 − 3a2 + 1) q2 (a) = , (8a4 − 5a2 + 1)3/2 are continuous decreasing functions in [cos(π/6), 1]. Then, √ √ 3 1 21 39 1 G2 (an ) ≤ q1 (cos(π/6)) − q2 (1) = + − − < 0. 3 4 338 2 3. Finally, we want to show that G2 (an ) < 0 for n ≥ 5 odd. In this case, a a3n G2 (an ) =  π n − 2 cos (2an − 1) (1 + an ) 2n   3 2a3n − an + 1 8a5n − 8a3n + an + 1 − 2an − . (4a3n + a2n − 2an + 1)3/2 (16a5n − 16a3n + a2n + 2an + 1)3/2 It ismore suitable to write the above expression in terms of the half angle pn = π cos through the identity an = 2p2n − 1. Then, 2n (2p2n − 1) G2 (an ) = (q1 (pn ) − q2 (pn )), 2pn where   1 8p4 − 12p2 + 5 q1 (p) = −(2p2 − 1)2 3 + , 2p (8p4 − 11p2 + 4)3/2 2 2 2 128p8 − 320p6 + 288p4 − 112p2 + 17 q2 (p) = − + (2p − 1) . 4p2 − 3 (128p8 − 320p6 + 288p4 − 111p2 + 16)3/2 It h is not difficult to see that q1 decreases and that q2 (p) > −11/10 for p ∈ π i cos , 1 . Therefore, 10   π  11 q1 (pn ) − q2 (pn ) 0 for u ∈ [um , 3a/b]. Therefore, the minima of T occurs at u = um (a). To conclude the proof it is enough to see that T (a, um (a)) > 0. Simplifying, we want to see that a √ (2a − 1)(4a2 − 2a − 1)2 T (a, um (a)) = − 2 > 0, 2 (16a4 − 16a3 + a − 1)3/2 which is equivalent to prove that √ a(16a4 − 16a3 + a − 1)3/2 > 2 2(2a − 1)(4a2 − 2a − 1)2 . Squaring both sides, the expression factors into (a − 1)p(x) > 0, 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]. 3.2 Proof of Lemma 2 We want to prove that, for any fixed value of a ∈ [cos(π/6), 1), the function 1 a2 1 + (bu − a)2 T (a, u) = −2 3 2 b (1 + u2 )3/2 On strictly convex central configurations of the 2n-body problem 13 ha i √ is negative for any u ∈ K = , um (a) , where b = 1 − a2 and b 2a(3a2 − 2) um (a) = . b(2a2 − 1) The function is clearly continuous and differentiable in K . On one hand, the function T (a, u) is strictly increasing with respect the variable u because ∂T 2a3 b2 u3 − 4abu2 + (5a2 + 1)u + 2ab = 3 ∂u b (1 + u2 )5/2 2a3 u(b2 u2 − 4abu + 5a2 + 1 ≥ > 0. b3 (1 + u2 )5/2 Therefore, T (a, u) ≤ T (a, um (a)) and 1 (2a2 − 1)(16a6 − 20a4 + 5a2 + 1) p(a) = T (a, um ) = − 2a3 . 2 (32a6 − 40a4 + 11a2 + 1)3/2 Finally, it is not difficult to see that p0 (a) has no roots in a ∈ [cos(π/6), 1] (for exam- ple, using Sturm’s theory) and p0 (a) > 0. Therefore, T (a, um (a)) < p(1) = 0, which concludes the proof. References Barrabés E, Cors J (2016) On central configurations of twisted crowns, arXiv:1612.07135 190 Chen KC, Hsiao JS (2012) Convex central configurations of the n-body problem which are not strictly convex. Journal of Dynamics and Differential Equations 24:119–128 Fernandes A, Garcia B, Mello L (2017) Convex but not strictly convex central configurations. Journal of Dynamics and Differential Equations Roberts G (1999) Existence and stability of relative equilibria in the n-body problem. PhD thesis, Boston University Saari DG (2005) 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 Yu X, Zhang S (2012) Twisted angles for central configurations formed by two twisted regular polygons. J Differential Equations 253:2106–2122