This is a preprint of: “On the convex central configurations of the symmetric (l+2)-body problem”, Montserrat Corbera, Jaume Llibre, Pengfei Yuan, Regul. Chaotic Dyn., vol. 25(3), 250–272, 2020. Archive for Rational Mechanics and Analysis manuscript No. (will be inserted by the editor) On the convex central configurations of the symmetric (` + 2)–body problem Montserrat Corbera · Jaume Llibre · Pengfei Yuan Received: date / Accepted: date Abstract For the 4–body problem there is the following conjecture: Given arbitrary positive masses the planar 4–body problem has a unique convex central configuration for each ordering of the masses on its convex hull. Until now this conjecture remains open. Our aim is to prove that this conjecture cannot be extended to the (` + 2)–body problem with ` ≥ 3. In particular, we prove that the symmetric (2n + 1)–body problem with masses m1 = · · · = m2n−1 = 1 and m2n = m2n+1 = m sufficiently small has at least two classes of convex central configuration when n = 2, five when n = 3, and four when n = 4. We conjecture that the (2n + 1)-body problem has at least n classes of convex central configurations for n > 4 and we give some numerical evidences that the conjecture can be true. We also prove that the symmetric (2n + 2)– body problem with masses m1 = · · · = m2n = 1 and m2n+1 = m2n+2 = m sufficiently small has at least three classes of convex central configuration when n = 3, two when n = 4, and three when n = 5. We also conjecture The first two authors are partially supported by the Ministerio de Economı́a, Industria y Competitividad, Agencia Estatal de Investigación grants MTM2016-77278-P (FEDER). The second author is also partially supported by the Agència de Gestió d’Ajuts Universitaris i de Recerca grant 2017SGR1617, and the H2020 European Research Council grant MSCA-RISE- 2017-777911. The third author is partially supported by Fundamental Research Funds for the Central Universities (NO.XDJK2015C139), China Scholarship Council(No.201708505030), the National Natural Science Foundation of China(grant No. 11626193) Montserrat Corbera Departament d’Enginyeries, Universitat de Vic–Universitat Central de Catalunya, 08500 Vic, Barcelona, Spain E-mail: montserrat.corbera@uvic.cat Jaume Llibre Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Catalonia, Spain E-mail: jllibre@mat.uab.cat Pengfei Yuan School of Mathematics and Statistics, Southwest University, 400715, Chongqing, China E-mail: yuanpengfei@swu.edu.cn 2 M. Corbera, J. Llibre and P. Yuan that the (2n + 2)-body problem has at least [(n + 1)/2] classes of convex central configurations for n > 5 and we give some numerical evidences that the conjecture can be true. Keywords Convex central configuration · (` + 2)-body problem Mathematics Subject Classification (2010) 70F10 · 70F15 1 Introduction and statement of the main result The equations for the classical Newtonian N -body problem are given by N X qj − qi mi q̈i = G mi mj , |qj − qi |3 j=1 j 6= i i = 1, . . . , N , where qi = (xi , yi ) ∈ R2 is the position vector of the punctual mass mi in an inertial coordinate system, |qj − qi | is the Euclidean distance between the masses mj and mi , and G is the gravitational constant which can be taken equal to one by choosing conveniently the unit of time. The configuration space of the planar N –body problem is E = {(q1 , . . . , qN ) ∈ R2N : qi 6= qj , for i 6= j}. Given m1 , . . . , mN a configuration (q1 , . . . , qN ) ∈ E is central if there exists a positive constant λ such that q̈i = −λ (qi − cm) for i = 1, . . . , N , where PN PN ! i=1 mi x i mi y i cm = (cmx , cmy ) = PN , Pi=1 N , i=1 m i k=1 mi is the center of mass of the system. Thus a central configuration (q1 , . . . , qN ) ∈ E of the N –body problem with positive masses m1 , . . . , mN is a solution of the system of 2N equations N X xj − xi ei = mj + λ (xi − cmx ) = 0, |qj − qi |3 j=1 j 6= i (1) N X yj − yi eN +i = mj + λ (yi − cmy ) = 0, |qj − qi |3 j=1 j 6= i with i = 1, . . . , N , for some λ. On the convex central configurations of the symmetric (` + 2)–body problem 3 A configuration is called convex if none of the masses is contained in the interior of the convex hull of the other remaining masses. McMillan and Bartky in 1932 (see [8]) proved that the 4–body problem for any set of positive masses has a convex central configuration. This result was reproved in a simpler way by Xia in 2004 (see [11]). In the paper of MacMillan and Bartky is implicit the following conjecture: For the planar 4-body problem with positive masses there is a unique convex central configuration for each ordering of the masses on its convex hull. This conjecture also appear explicitly in the papers of Pérez–Chavela and Santo- prete [10], and of Albouy and Fu [1]. Until now this conjecture remains open and the opinion of the people that have thought in it is that its proof can be hard. Numerically it is known that the five body problem with equal masses has only one convex central configuration, see [7] and [9]. Chen and Hsiao [5] provided necessary conditions in order that a central configuration of the five body problem to be convex and also show some numerical convex central configurations for different values of the masses. The objective of this paper is to show analytically that this conjecture cannot be extended to the (` + 2)–body problem with ` > 2 except perhaps for ` = 4. Here we consider symmetric (` + 2)–body problem with masses m1 = m2 = · · · = m` = 1 and m`+1 = m`+2 = m having the following symmetries. When ` = 2n − 1 is odd we consider the S1 –symmetric (2n + 1)– body problem having the S1 –symmetry: y1 = 0, x2n+1−i = xi , y2n+1−i = −yi , yi > 0, 2 ≤ i ≤ n, x2n+1 = x2n , y2n+1 = −y2n , y2n > 0. When ` = 2n is even we consider the S2 –symmetric (2n + 2)–body problem having the S2 – symmetry: y1 = 0, yn+1 = 0, x2n−i = xi+2 , y2n−i = −yi+2 , yi+2 > 0, 0 ≤ i ≤ n − 2, x2n+2 = x2n+1 , y2n+2 = −y2n+1 , y2n+1 > 0, and the S3 –symmetric (2n + 2)–body problem having the S3 –symmetry: x2n−i = xi+1 , y2n−i = −yi+1 , yi+1 > 0, 0 ≤ i ≤ n − 1, x2n+2 = x2n+1 , y2n+2 = −y2n+1 , y2n+1 > 0. Our results are the following. Theorem 1 For each m > 0 sufficiently small the S1 –symmetric ` + 2–body problem with masses mi = 1, i = 1, . . . , ` = 2n − 1 and m`+1 = m`+2 = m has at least two, five, four convex central configurations for ` = 3, 5, 7, respectively. Theorem 1 is proved in Section 3. 4 M. Corbera, J. Llibre and P. Yuan Conjecture 1 For each m > 0 sufficiently small and n ≥ 5, the S1 –symmetric ` + 2–body problem with masses mi = 1, i = 1, . . . , ` = 2n − 1 and m`+1 = m`+2 = m has at least n convex central configurations. In Subsection 3.5 we give numerical evidences that Conjecture 1 holds. Theorem 2 For each m > 0 sufficiently small the S2 –symmetric ` + 2–body problem with masses mi = 1, i = 1, . . . , ` = 2n and m`+1 = m`+2 = m has at least three, two, three convex central configurations for ` = 6, 8, 10, respectively. For ` = 4 there are no convex S2 –symmetric central configurations that can be obtained from continuation of central configurations of the restricted S2 – symmetric ` + 2–body problem. Theorem 2 is proved in Section 4. Conjecture 2 For each m > 0 sufficiently small and for n ≥ 6, the S2 – symmetric ` + 2–body problem with masses mi = 1, i = 1, . . . , ` = 2n and m`+1 = m`+2 = m has at least [ n+1 2 ] convex central configurations. In Subsection 4.6 we give numerical evidences that Conjecture 2 holds. Theorem 3 For each m > 0 sufficiently small the S3 –symmetric ` + 2–body problem with masses mi = 1, i = 1, . . . , ` = 2n and m`+1 = m`+2 = m has no convex central configurations for ` = 4, 6, 8, 10 that can be obtained from continuation of central configurations of the restricted S2 –symmetric `+2–body problem. Theorem 3 is proved in Section 5. Conjecture 3 For each m > 0 sufficiently small and for n ≥ 6, the S3 – symmetric ` + 2–body problem with masses mi = 1, i = 1, . . . , ` = 2n and m`+1 = m`+2 = m has no convex central configurations that can be obtained from continuation of central configurations of the restricted S3 –symmetric ` + 2–body problem. 2 Symmetric restricted ` + 2 body problem with ` equal masses at the vertices of a regular `–gon Now for k = 1, 2, 3 we consider the Sk –symmetric restricted (` + 2)–body problem with ` masses equal to one at the vertices of a regular `–gon with positions xi = cos αi , yi = sin αi , 1 ≤ i ≤ ` where αi = 2π(i − 1)/` + α and α = 0 when k = 1, 2 and α = π/` when k = 3, and two infinitesimal masses with m = 0 located at the points (x`+1 , y`+1 ) and (x`+1 , −y`+1 ). Here (x`+1 , y`+1 ) corresponds to the position of the infinitesimal mass in a central configuration of the restricted ` + 1–body problem with ` equal masses at the vertices of a regular `-gon. The restricted (` + 1)–body problem with ` equal masses at the vertices of a regular `-gon have been studied by several authors. Arenstorf in [2] proved On the convex central configurations of the symmetric (` + 2)–body problem 5 that for the central configurations of the restricted (3 + 1)–body problem with 3 equal masses at the vertices of an equilateral triangle, the infinitesimal mass must be on one of the three straight lines passing through barycenter and a vertex of the triangle. He also proved that on each straight line there are exactly four positions for the infinitesimal mass in a central configuration. Bang and Elmabsout in [3] and [4] and later on Fernandes et al. in [6] studied the problem for ` ≥ 3. The results of [3, 4, 6] are summarized in the following lemma. Lemma 1 For ` ≥ 3 consider the central configurations of the restricted (` + 1)–body problem with equal masses at the vertices of a regular `–gon of radius 1. (a) The infinitesimal mass is located on an axis of symmetry of the `-gon (see Theorem 2 in [3]). (b) Assuming that the bodies are arranged so that the positive x–axis is a semi- axis of symmetry containing one of the primaries, the possible positions for the infinitesimal mass are (0, 0) and (ρ1 , 0) with ρ1 > 1. (c) Assuming that the bodies are arranged so that the positive x–axis is a semi- axis of symmetry that does not contain any of the primaries, the possi- ble positions for the infinitesimal mass are (0, 0), (ρ2 , 0) and (ρ3 , 0) with 0 < ρ2 < 1 < ρ3 . Neither [2], [4] nor [6] provide the explicit values of ρ1 , ρ2 and ρ3 . In this paper these values are necessary and they would be given when we need them. 3 S1 –symmetric (2n + 1)–body problem 3.1 The equations Since the set of central configurations is invariant under rotations and dilations, and we have the first integral of the center of mass, without loss of generality we can assume that the center of masses is at the origin of coordinates and that q1 = (1, 0). Since the center of masses is at the origin we have (cmx , cmy ) = P2n+1 Pn (0, 0), so i=1 mi xi = 0, and x2 = −(1/2 + i=3 xi + mx2n ). Notice that P2n+1 due to the symmetry i=1 mi yi is identically zero. Taking into account these conditions together with the S1 –symmetry the 4n + 2 equations e1 = 0, . . . , e4n+2 = 0 given by (1) with N = 2n + 1 for the central configurations of the S1 –symmetric (2n + 1)–body problem reduce to the following 2n equations ei = 0, e2n+1+i = 0, i = 2, . . . , n, e2n = 0, e4n+1 = 0. P2n−1 Indeed, the equations e1 and e2n+2 are omitted because i=1 ei + me2n + P2n−1 me2n+1 = 0 and i=1 ei+2n+1 + me4n+1 + me4n+2 = 0. And from the S1 – symmetry, we have ei = e2n+1−i , ei+2n+1 = −e4n+2−i , for 2 ≤ i ≤ n, e2n+1 = e2n and e4n+2 = −e4n+1 . 6 M. Corbera, J. Llibre and P. Yuan From the first equation e2 = 0 we isolate λ and we substitute it into the other 2n − 1 equations that we denote by E3 = 0, · · · , En = 0, E2n = 0, E2n+3 , · · · , E3n+1 = 0, E4n+1 = 0. (2) In short, we have 2n − 1 equations and 2n − 1 unknowns y2 , xi , yi , 3 ≤ i ≤ n, x2n , y2n . We are interested in the central configurations of the S1 –symmetric re- stricted 2n + 1–body problem, see Section 2. Since the number of primaries is odd, without loss of generality we can assume that the line of symmetry is the x–axis and that one of the primaries is in the positive x–axis. We assume also that the infinitesimal mass m2n is in the x–axis, i.e. y2n = 0. Straight- forward computations show that when m = 0, xi = cos αi , yi = sin αi , with αi = 2π(i − 1)/(2n − 1), 1 ≤ i ≤ 2n − 1 and y2n = 0 all the equations in (2) are identically zero except the equation E2n = 0 which becomes fn (x2n ) = 0 with 2n−1 X cos αj − x2n fn (x2n ) = + λx2n , 2 3/2 j=1 (1 − 2 cos αj x2n + x2n ) where 2n−1 X cos αj / cos α2 − 1 λ=− 3/2 . j=1 (2 − 2 cos αj−1 ) j 6= 2 3.2 Proof of Theorem 1 for ` = 3 For the central configurations of the S1 –symmetric 5–body problem (that is n = 2) with masses m1 = m2 = m3 = 1 and m4 = m5 = m equations (2) taking into account that x2 = −mx4 − 1/2 (by the fist integral of the center of masses) reduce to E4 = 0, E7 = 0 and E9 = 0 where (2mx4 + 1)/2 + x4 (2mx4 + 1)/2 + x4 1 − x4 E4 = λx4 − 3 − 3 + 3 , r24 r34 r14 m(y4 − y2 ) m(y2 + y4 ) y2 1 E7 = λy2 + 3 − 3 − 3 − 2, r24 r34 r12 4y2 m y4 y2 − y4 y2 + y4 E9 = λy4 − 2 − 3 + 3 − 3 , 4y4 r14 r24 r34 with q r12 = (2mx4 + 3)2 /4 + y22 , q r14 = (x4 − 1)2 + y42 , p r24 = (2(m + 1)x4 + 1)2 /4 + (y2 − y4 )2 , p r34 = (2(m + 1)x4 + 1)2 /4 + (y2 + y4 )2 , On the convex central configurations of the symmetric (` + 2)–body problem 7 and 3 3 3 3 3 3 3 3 3 3 3 mr12 r24 + mr12 r34 + 3r24 r34 + 2m(r24 r34 + (1 + m)r12 (r24 + r34 ))x4 λ= 3 3 3 . r12 r24 r34 (1 + 2mx4 ) In short, we have three equations and three unknowns y2 , x4 and y4 . In the next result we provide all the convex central configurations of the S1 –symmetric restricted (3+2)–body problem having the three primaries with masses equal to one located√ at the vertices of the equilateral √ triangle (x1 , y1 ) = (1, 0), (x2 , y2 ) = (−1/2, 3/2) and (x3 , y3 ) = (−1/2, − 3/2), and the two infinitesimal masses m = 0 m located at (x4 , y4 ) and (x5 , y5 ) = (x4 , −y4 ). Proposition 1 The S1 –symmetric restricted (3+2)–body problem has exactly two classes of convex central configurations which are given by √ (a) (x4 , y4 ) = (−a/2, − 3a/2), see Figure 1(b), (b) (x4 , y4 ) = (a, 0), see Figure 1(a). Note that in this configuration the two infinitesimal masses are colliding. Here a = −1.6197896088 . . . is a root of the polynomial √  8 √  7 √ 6 √  5 P (x)√= x11 4 − x  4 10 4 + 2 3√− 3  3x4 + 3 + 2 √ 3 x  2 4 + 2 3x 4√− 2  3 + 2 √3 x4 − 4 3 − 42 x4 − 9 + 4 3 x4 + 8 + 2 3 x4 + 37 + 2 3 x4 + 2 3 + 9. Proof First we look for the central configurations of the S1 –symmetric re- stricted (3 + 2)–body problem with m4 on the axis of symmetry containing m1 , i.e. with y4 = 0. In the previous subsection we have seen that under this condition x4 must satisfy the equation f2 (x4 ) = 0. Since we are only interested in convex central configurations, we can restrict to x4 < −1/2. Under this hypothesis equation f2 (x4 ) = 0 becomes 3(2x4 + 1) √ 3 = 3x4 + . (3) (x24 + x4 + 1) 3/2 (x4 − 1)2 Squaring both sides of this equation and dropping the denominators we get the polynomial equation 3x4 P (x) = 0. Solving numerically this polynomial equation we get exactly two solutions with x4 < −1/2, but only x4 = a is a solution for the initial equation (3). This proves statement (b). Taking the axis of symmetry through the mass m3 , or equivalently rotating the solution of statement (b) an angle of −2π/3 we get statement (a). Clearly the configuration √ with (x4 , y4 ) = (x5 , y5 ) √ = (a, 0) and the one with (x4 , y4 ) = (−a/2, − 3a/2) and (x5 , y5 ) = (−a/2, 3a/2) are both convex. Moreover, if we take the axis of symmetry through the mass m2 , by symmetry we get again the configuration given in statement (a). Thus these are the unique convex central configurations of the restricted (3 + 2)–body problem. Proposition 2 The two convex central configurations of the symmetric re- stricted (3 + 2)–body problem given in Proposition 1 can be continued to two families of central configurations of the symmetric 5–body problem with masses mi = 1, i = 1, 2, 3 and m4 = m5 = m > 0 sufficiently small. 8 M. Corbera, J. Llibre and P. Yuan m4 = 0 m2 m2 m4 = 0 m1 m1 m5 = 0 (1, 0) (1, 0) m3 m3 (a) (b) m5 = 0 Fig. 1 Convex central configurations of the symmetric 5–body problem with m4 = m5 = 0. Proof Using the Implicit Function Theorem we will see that the solutions of system E4 = 0, E7 = 0 and E9 = 0 with m = 0 given in Proposition 1 can be continued to solutions of this system with m > 0 sufficiently √ small. We start with the solution (x4 , y4 ) = (−a/2, − 3a/2). The determinant of the Jacobian of system E4 = 0, E7 = 0 and E9 √ = 0 (with respect to the variables √ y2 , x 4 , y 4 ), evaluated at m = 0, y 2 = 3/2, and (x4 , y4 ) = (−a/2, − 3a/2) takes the value √  √  3 −16a2 − 16a + 11 3 32a4 + 64a3 − 12a2 − 44a + 5 5/2 + 5 2 4(a − 1)3 (a2 + a + 1) √ 8 (a + a + 1) 1 1 3 1 − 3/2 + 3 + 6 − √ = −0.6885602058 · · · 6= 0. (a2 + a + 1) 2(a − 1) (a − 1) 2 3 Since this determinant is different from zero from the Implicit Function The- orem there exist an analytic family y21 (m), x14 (m) and 1 √ y4 (m) of solutions of 1 system E4√= 0, E7 = 0 and E9 = 0 with y2 (0) = 3/2, x14 (0) = −a/2 and y41 (0) = − 3a/2, defined for m > 0 sufficiently small. Notice that system E4 = 0, E7 = 0 and E9 = 0 is√not analytic with respect to all its variables in a neighborhood of m = 0, y2 = 3/2 and (x4 , y4 ) = (a, 0) because equation E9 = 0 contains the term m/y42 . After doing the change of variables y4 = µY4 /2 with µ = m1/3 we obtain a new system of equations which√is analytic with respect to all its variables in a neighborhood of µ = 0, y2 = 3/2, x4 = a and Y4 6= 0 and it is given by e4 = − 8(2x4 + 1) 24x4 1 − x4 E 3/2 + 3/2 + 3/2 + O(µ2 ), ((2x4 + 1)2 + 4y22 ) (4y22 + 9) ((x4 − 1)2 ) e7 = 16y2 y2 E 3/2 − 3/2 + O(µ3 ), (4y22 + 9) 4 (y22 )  e9 = 1 Y4 16 (2x4 + 1)2 − 8y22 1 24 E − 5/2 − 3/2 + 3/2 2 ((2x4 + 1)2 + 4y22 ) ((x4 − 1)2 ) (4y22 + 9) ! 2 − 3/2 µ + O(µ3 ). (Y42 ) On the convex central configurations of the symmetric (` + 2)–body problem 9 Consider now the system of equations e4 , Ē4 = E e7 , Ē7 = E e9 /µ, Ē9 = E which√is analytic with respect to all its variables in a neighborhood √ of µ = 0, y2 = 3/2, x4 = a and Y4 6= 0. Substituting µ = 0, y2 = 3/2, x4 = a into system Ē4 = 0, Ē7 = 0 and Ē9 = 0 we get that Ē4 and Ē7 are identically zero and that equation Ē9 = 0 is equivalent to ! 1 −4a2 − 4a + 5 1 2 1 Y4 5/2 + 3 − 3/2 +√ = 0. 2 2 (a2 + a + 1) (a − 1) 2 (Y4 ) 3 The solutions of this equation are Y4 = 0 and Y4 = ±b = ±1.4869700925 . . . . Clearly, system Ē4 = 0, Ē7 = 0 and Ē9 = 0 is analytic with √ respect to all its variables in a neighborhood of the solution µ = 0, y2 = 3/2, x4 = a and Y4 = b. Moreover the determinant of the Jacobian of the system (with respect to the variables y2 , x4 , Y4 ) evaluated at this solution is ! 1 3(2a + 1)2 2 2 1 − √ 5/2 − 3/2 − 3 +√ 8 3 2 (a2 + a + 1) (a2 + a + 1) (a − 1) 3  ! 3 4a2 + 4a − 5 6 24 √ − 5/2 + 3 + 3 +2 3 = −1.0328403087 · · · 6= 0. (a2 + a + 1) (a − 1) b Therefore, applying the Implicit Function Theorem, there exist a unique ana- lytic family of solutions √ y2 (µ), x4 (µ) and Y4 (µ) of system Ē4 = 0, Ē7 = 0 and Ē9 = 0 with y2 (0) = 3/2, x4 (0) = a and Y4 (0) = b defined for µ > 0 suffi- ciently small. This family provides the√family of solutions of system E4 = 0, E7 = 0 and E9 = 0 given by y22 (m) = 3/2 + O(m1/3 ), x24 (m) = a + O(m1/3 ) and y42 (m) = b/2 m1/3 + O(m2/3 ) defined for m > 0 sufficiently small. This completes the proof of the proposition. Proof (Proof of Theorem 1 for ` = 3) Let x12 (m) = −mx14 (m) − 1/2 and x22 (m) = −mx24 (m) − 1/2. Given m > 0 sufficiently small, Proposition 2 pro- vides two central configurations: the one given by (x1 , y1 ) = (1, 0), (x2 , y2 ) = (x12 (m), y21 (m)), (x3 , y3 ) = (x12 (m), −y21 (m)), (x4 , y4 ) = (x14 (m), y41 (m)), and (x5 , y5 ) = (x14 (m), −y41 (m)), and the one given by (x1 , y1 ) = (1, 0), (x2 , y2 ) = (x22 (m), y22 (m)), (x3 , y3 ) = (x22 (m), −y22 (m)), (x4 , y4 ) = (x24 (m), y42 (m)), and (x5 , y5 ) = (x24 (m), −y42 (m)). Since for m = 0 they are two different convex central configurations, it follows that for m > 0 sufficiently small they continue being two different convex central configurations. This completes the proof of the theorem. 10 M. Corbera, J. Llibre and P. Yuan 3.3 Proof of Theorem 1 for ` = 5 For the symmetric 7-body problem (that is n = 3), the equations of central configuration (2) are reduced to 5 equations: E3 = 0, E6 = 0, E9 = 0, E10 = 0, E13 = 0 where x2 = −mx6 − 1/2 − x3 (by the fist integral of the center of masses), and the ones for the symmetric restricted (5 + 2)–body problem are reduced to the equation f3 (x6 ) = 0, see Subsection 3.1. In order to prove the convexity of our central configurations we will use the following result. Lemma 2 Let qi for i = 1, . . . , N be the vertices of a n-gon ordered sequen- tially counterclockwise. If the signed areas of all the triangles formed by three consecutive vertices of the n-gon are positive, then the n-gon is convex. That is, if − q− −−→ −−−−−−→ i qi+1 × qi+1 qi+2 > 0, for all i = 1, . . . , N , then the n-gon is convex. Here qN +1 = q1 and qN +2 = q2 . Recall that the signed area of a triangle formed by the consecutive vertices in counterclockwise of a n-gon not necessarily regular p1 = (x1 , y1 ), p2 = (x2 , y2 ) and p3 = (x3 , y3 ) is given by 1 −−−→ −−−→ 1 x2 − x1 x3 − x2 A= p1 p2 × p2 p3 = . 2 2 y2 − y1 y3 − y2 Proposition 3 Let a1 = −1.5979217289 . . . and a2 = −0.8221828699 . . . be the two roots of f3 (x6 ) = 0 with x6 < cos(4π/5). The symmetric restricted (5 + 2)–body problem has exactly five classes of convex central configurations which are given by (a) (x6 , y6 ) = (cos (−2π/5)a1 , sin (−2π/5)a1 ), see Figure 2 (b), (b) (x6 , y6 ) = (cos (−2π/5)a2 , sin (−2π/5)a2 ), see Figure 2 (d), (c) (x6 , y6 ) = (cos (−4π/5)a2 , sin (−4π/5)a2 ), see Figure 2 (e), (d) (x6 , y6 ) = (a1 , 0), see Figure 2 (a). Note that in this configuration the two infinitesimal masses are colliding, (e) (x6 , y6 ) = (a2 , 0), see Figure 2 (c). In this configuration the two infinites- imal masses are colliding. Proof Taking the mass m6 on the axis of symmetry that passes through m1 , i.e. y6 = 0, the equations for the central configurations of the S1 –symmetric restricted (5 + 2)–body problem are reduced to the equation f3 (x6 ) = 0. Since we are only interested in convex central configurations we can assume that x6 < cos(4π/5), under this assumption equation f3 (x6 ) = 0 becomes s √ √ 1 2 2( 5 − 1 − 4x6 ) 2 + 1 + √ x6 + √ (x6 − 1) 5 (2 + x6 − 5x6 + 2x26 )3/2 √ √ (4) 2( 5 + 1 + 4x6 ) − √ =0 (2 + x6 + 5x6 + 2x26 )3/2 On the convex central configurations of the symmetric (` + 2)–body problem 11 m2 m2 m3 m3 m6 = 0 m1 m6 = 0 m1 m7 = 0 (1, 0) m7 = 0 (1, 0) m4 m4 m5 m5 (a) (c) m6 = 0 m2 m6 = 0 m2 m3 m3 m1 m1 (1, 0) (1, 0) m4 m4 m5 m7 = 0 m5 m7 = 0 (b) (d) m2 m3 m6 = 0 m1 (1, 0) m4 m7 = 0 m5 (e) Fig. 2 Convex central configurations of the symmetric 7–body problem with m6 = m7 = 0 . Squaring both sides of this equation and dropping the denominators we get the polynomial equation −8192x6 P (x6 ) = 0 where P (x6 ) is a polynomial of degree 35 in the variable x6 . This polynomial equation has exactly four real roots with x6 < cos(4π/5) of those only x6 = a1 and x6 = a2 are solutions of the initial equation (4). It is easy to see that the configurations with (x6 , y6 ) = (a1 , 0) and the one with (x6 , y6 ) = (a2 , 0) are both convex because they satisfy the conditions of Lemma 2. This proves statements (d) and (e). 12 M. Corbera, J. Llibre and P. Yuan We also can take the mass m6 on an axis of symmetry passing through m4 and m5 . This corresponds to rotate the solutions (x6 , y6 ) = (a1 , 0) and (x6 , y6 ) = (a2 , 0) and angle of −4π/5 and −2π/5, respectively. Note that by symmetry if we take the mass m6 on the axis of symmetry through m2 (respectively, m3 ) we obtain the same configuration than taking the axis of symmetry through m5 (respectively, m4 ). By rotating the solution (x6 , y6 ) = (a1 , 0) with an angle of −2π/5 and rotating the solution (x6 , y6 ) = (a2 , 0) with an angle of −2π/5 and −4π/5 we get the configurations of statements (a), (b) and (c) respectively. Applying Lemma 2 again we see that all these configurations are convex. Note that the configuration given by rotating the solution (x6 , y6 ) = (a1 , 0) with an angle of −4π/5 is not convex because a1 cos (−4π/5) > 1. So these are the unique convex central configurations of the S1 –symmetric restricted (5 + 2)–body problem. Proposition 4 Each of the five convex central configurations of the symmetric restricted (5 + 2)–body problem can be continued to a family of central configu- rations of the symmetric 7–body problem with masses mi = 1, i = 1, . . . , 5 and m6 = m7 = m > 0 sufficiently small. Proof Using the Implicit Function Theorem we will see that the solutions of system E3 = 0, E6 = 0, E9 = 0, E10 = 0, E13 = 0 with m = 0 given in Proposition 3 can be continued to solutions of this system with m > 0 sufficiently small. We use the following notation: U = (y2 , x3 , y3 ), V = (x6 , y6 ), and Ũ = (sin(2π/5), cos(4π/5), sin(4π/5)), Ṽ1 = (cos (−2π/5)a1 , sin (−2π/5)a1 ), Ṽ2 = (cos (−2π/5)a2 , sin (−2π/5)a2 ), and Ṽ3 = (cos (−4π/5)a2 , sin (−4π/5)a2 ). System E3 = 0, E6 = 0, E9 = 0, E10 = 0, E13 = 0 is analytic with respect to all its variables in a neighborhood of the solutions m = 0, U = Ũ and V = Ṽi for i = 1, 2, 3. Let J (U, V, m) denote the Jacobian of the system E3 = 0, E6 = 0, E9 = 0, E10 = 0, E13 = 0 with respect to the variables U , V . Straightforward computations show that det (J (U, V, m)) |U =Ũ ,V =Ṽ1 ,m=0 = 61.9657422025 . . . , det (J (U, V, m)) |U =Ũ ,V =Ṽ2 ,m=0 = −1928.5558212278 . . . , det (J (U, V, m)) |U =Ũ ,V =Ṽ3 ,m=0 = −1928.5558212278 . . . . Since these determinants are different from zero from the Implicit Function Theorem there exist three analytic families of solutions of system E3 = 0, E6 = 0, E9 = 0, E10 = 0, E13 = 0 defined for m > 0 sufficiently small. They are given by (U i (m), V i (m)) for i = 1, 2, 3 with U i (m) = (y2i (m), xi3 (m), y3i (m)) and V i (m) = (xi6 (m), y6i (m)) satisfying U i (0) = Ũ , V i (0) = Ṽi . System E3 = 0, E6 = 0, E9 = 0, E10 = 0, E13 = 0 is not analytic with respect to all its variables in a neighborhood of the solution m = 0, X = X̃ and Y = (x6 , y6 ) = (a, 0) with either a = a1 or a = a2 because equation E13 = 0 contains the term m/y62 . After doing the change of variables y6 = µY6 /2 with µ = m1/3 we obtain a new system of equations which is analytic with respect On the convex central configurations of the symmetric (` + 2)–body problem 13 to all its variables in a neighborhood of µ = 0, U = Ũ , x6 = a and Y6 6= 0 and it can be written as e3 = E E e3 (y2 , x3 , y3 ) + O(µ3 ), e6 = E E e6 (y2 , x3 , y3 , x6 ) + O(µ2 ), e9 = E E e9 (y2 , x3 , y3 ) + O(µ3 ), e10 = E E e10 (y2 , x3 , y3 ) + O(µ3 ), e13 = µE E e13 (y2 , x3 , y3 , x6 , Y6 ) + O(µ3 ). Now we consider the system of equations e3 , Ē3 = E e6 , Ē6 = E e9 , Ē9 = E e10 , Ē10 = E e13 /µ, Ē13 = E which is also analytic with respect to all its variables in a neighborhood of µ = 0, U = Ũ , x6 = a, and Y6 6= 0. Substituting µ = 0, U = Ũ and x6 = a into system Ē3 = 0, Ē6 = 0, Ē9 = 0, Ē10 = 0, Ē13 = 0 we get that Ē3 , Ē6 , Ē9 , Ē10 are identically zero and that equation Ē13 = 0 is equivalent to p 1 Y62 Ē13 (Y6 ) = AY6 + , 4 Y63 where √ √  √  2 2 −8a2 + 4 5 − 1 a + 3 5 + 7 A=− √ 5/2 + 2a2 − 5a + a + 2 √ √  √  s 2 2 8a2 + 4 1 + 5 a + 3 5 − 7 2 2 √ 5/2 − 3 −2 1+ √ . 2 2a + 5a + a + 2 (a − 1) 5 The solutions of this equation are r 3 4 Y6 = ± − . A If a = a1 then Y6 = ±b1 = ±1.1302790764 . . . and if a = a2 then Y6 = ±b2 = ±0.4564622776 . . . . By symmetry, we are only interested in the positive values of Y6 . Clearly, system Ē3 = 0, Ē6 = 0, Ē9 = 0, Ē10 = 0, Ē13 = 0 is analytic with respect to all its variables in a neighborhood of the solutions µ = 0, U = Ũ , x6 = ai , Y6 = bi with i = 1, 2. Moreover the Jacobian of the system (with respect to the variables U , x6 , Y6 ) evaluated at these solutions take the values 92.9486133 . . . when i = 1 and −2892.8337318417 . . . when i = 2. Therefore, applying the Implicit Function Theorem, there exist two analytic families of solutions of system Ē3 = 0, Ē6 = 0, Ē9 = 0, Ē10 = 0, Ē13 = 0 defined for µ > 0 sufficiently small. They are given by (U i+3 (µ), xi+3 6 (µ), Y6 i+3 (µ)) for i+3 i+3 i+3 i = 1, 2 with U (µ) = (y2 (µ), x3 (µ), y3 (µ)) and satisfying U i+3 (0) = i+3 Ũ , xi+3 6 (0) = ai and Y6 i+3 (0) = bi . Let y6i+3 (µ) = µ Y6i+3 (µ)/2, then these 14 M. Corbera, J. Llibre and P. Yuan two families provide the two families of solutions of system E3 = 0, E6 = 0, E9 = 0, E10 = 0, E13 = 0 given by (U i+3 (m1/3 ), V i+3 (m1/3 )) for i = 1, 2 where V i+3 (µ) = (xi+3 i+3 6 (µ), y6 (µ)), defined for m > 0 sufficiently small. In particular, y2 (m) = sin(2π/5) + O(m1/3 ), xi+3 i+3 3 (m) = cos(4π/5) + O(m 1/3 ), i+3 1/3 i+3 1/3 i+3 y3 (m) = sin(4π/5) + O(m ), x6 (m) = ai + O(m ) and y6 (m) = bi /2 m1/3 + O(m1/3 ). Proof (Proof of Theorem 1 for ` = 5) Let xi2 (m) = −mxi6 (m) − 1/2 − xi3 (m) for i = 1, . . . , 5. Given m > 0 sufficiently small, Proposition 4 provides the five central configurations given by (x1 , y1 ) = (1, 0), (x2 , y2 ) = (xi2 (m), y2i (m)), (x3 , y3 ) = (xi3 (m), y3i (m)), (x4 , y4 ) = (xi3 (m), −y3i (m)), (x5 , y5 ) = (xi2 (m), −y2i (m)), (x6 , y6 ) = (xi6 (m), y6i (m)), and (x7 , y7 ) = (xi6 (m), −y6i (m)) for i = 1, . . . , 5. Since for m = 0 we have five different central configurations that are con- vex, see Figure 2, it follows that for m > 0 sufficiently small they continue being five different convex central configurations. This completes the proof of the theorem. 3.4 Proof of Theorem 1 for ` = 7 For the symmetric 9–body problem (that is n = 4), the equations of central configuration are reduced to 7 equations: E3 = 0, E4 = 0, E8 = 0, E11 = 0, E12 = 0, E13 = 0, E17 = 0 where x2 = −m x8 − 1/2 − x3 − x4 and the ones for the symmetric restricted (7 + 2)–body problem are reduced to the equation f4 (x8 ) = 0, see Subsection 3.1. Proposition 5 Let a3 = −0.9189903637 . . . be the biggest root of f4 (x8 ) = 0 with x8 < cos(6π/7). The symmetric restricted (7+2)–body problem has exactly four convex central configurations which are given by (a) (x8 , y8 ) = (cos (−2π/7)a3 , sin (−2π/7)a3 ), see Figure 3 (b), (b) (x8 , y8 ) = (cos (−4π/7)a3 , sin (−4π/7)a3 ), see Figure 3 (c), (c) (x8 , y8 ) = (cos (−6π/7)a3 , sin (−6π/7)a3 ), see Figure 3 (d), (d) (x8 , y8 ) = (a3 , 0), see Figure 3 (a). Note that in this configuration the two infinitesimal masses are colliding. Proof Assuming that the mass m8 is on the axis of symmetry that passes through m1 , i.e. y8 = 0, the equations for the central configurations of the S1 –symmetric restricted (7 + 2)–body problem are reduced to the equation f4 (x8 ) = 0 with 7 X cos (2(j − 1)π/7) − x8 f4 (x8 ) = 3/2 j=1 (1 − 2 cos (2(j − 1)π/7)x8 + x28 ) 7 X cos (2(j − 1)π/7)/ cos (2π/7) − 1 − x8 3/2 , j=1 (2 − 2 cos (2(j − 2)π/7)) j 6= 2 On the convex central configurations of the symmetric (` + 2)–body problem 15 m3 m3 m2 m8 = 0 m2 m4 m4 m8 = 0 m1 m1 m9 = 0 (1, 0) (1, 0) m5 m5 m7 m9 = 0 m7 m6 m6 (a) (b) m3 m8 = 0 m3 m2 m2 m4 m4 m8 = 0 m1 m1 (1, 0) (1, 0) m5 m5 m9 = 0 m7 m7 m6 m9 = 0 m6 (c) (d) Fig. 3 Convex central configurations of the symmetric 9–body problem with m8 = m9 = 0. see Subsection 3.1. Since we are only interested in convex central configura- tions we can assume that x8 < cos(6π/7). Rearranging conveniently equation f4 (x8 ) = 0, squaring both sides of the equation as many times as we need and dropping the denominators we get the polynomial equation kx8 P (x8 ) = 0, where k is a constant and P is a polynomial of degree 95. This polynomial equa- tion has exactly eight real roots with x8 < cos(6π/7) of those only x8 = a3 and x8 = a4 = −1.5841209012 . . . are solutions of the initial equation f4 (x8 ) = 0. We also can take the mass m8 on the axis of symmetry passing through any other of the masses. By symmetry it is sufficient to take m8 on the axes of symmetry passing through m5 , m6 , and m7 , this corresponds to rotate the solutions (x8 , y8 ) = (a3 , 0) and (x8 , y8 ) = (a4 , 0) an angle of −6π/7, −4π/7 and −2π/7, respectively. By taking the mass m8 on the axes of symmetry passing through m4 , m3 and m2 respectively we do not obtain new configurations. It is easy to see that the configuration with (x8 , y8 ) = (a3 , 0) is con- vex because it satisfies the conditions of Lemma 2. This proves statements (d). By rotating the solution (x8 , y8 ) = (a3 , 0) with an angle of −2π/7, −4π/7 and −6π/7 we get the configurations of statements (a), (b) and (c) respectively. Applying Lemma 2 again we see that all these configurations are convex. Note that the configurations with (x8 , y8 ) = (a4 , 0), (x8 , y8 ) = (cos (−2π/7)a4 , sin (−2π/7)a4 ), (x8 , y8 ) = (cos (−4π/7)a4 , sin (−4π/7)a4 ), and (x8 , y8 ) = (cos (−6π/7)a4 , sin (−6π/7)a4 ) are not convex because they do not satisfy Lemma 2. So these are the unique convex central configurations of the S1 –symmetric restricted (7 + 2)–body problem. 16 M. Corbera, J. Llibre and P. Yuan Proposition 6 Each of the four convex central configurations of the symmet- ric restricted (7 + 2)–body problem can be continued to a family of central con- figurations of the symmetric 9–body problem with masses mi = 1, i = 1, . . . , 7 and m8 = m9 = m > 0 sufficiently small. Proof The proof is similar to the proof for ` = 3 and ` = 5. Using the Implicit Function Theorem we will see that the solutions of system E3 = 0, E4 = 0, E8 = 0, E11 = 0, E12 = 0, E13 = 0, E17 = 0 with m = 0 given in Proposition 5 can be continued to solutions of this system with m > 0 sufficiently small. Let U = (y2 , x3 , y3 , x4 , y4 ), V = (x8 , y8 ) and Ũ = (sin (2π/7), cos (4π/7), sin (4π/7), cos (6π/7), sin (6π/7)), Ṽi = (cos (−2πi/7)a3 , sin (−2πi/7)a3 ), i = 1, 2, 3. Then E3 , E4 , E8 , E11 , E12 , E13 , E17 are functions with variables U , V . System E3 = 0, E4 = 0, E8 = 0, E11 = 0, E12 = 0, E13 = 0, E17 = 0 is analytic with respect to U , V in a neighborhood of the solutions m = 0, U = Ũ and V = Ṽi for i = 1, 2, 3. Let J (U, V, m) denote the Jacobian of the system with respect to the variables U , V . Straightforward computations show that det (J (X, Y, m)) |U =Ũ ,V =Ṽi ,m=0 = −203091.2451410419 . . . , i = 1, 2, 3. Since these determinants are different from zero from the Implicit Function Theorem there exist three analytic families of solutions of system E3 = 0, E4 = 0, E8 = 0, E11 = 0, E12 = 0, E13 = 0, E17 = 0 defined for m > 0 sufficiently small. They are given by by (U i (m), V i (m)) for i = 1, 2, 3 with U i (m) = (y2i (m), xi3 (m), y3i (m) xi4 (m), y4i (m)) and V i (m) = (xi8 (m), y8i (m)) satisfying U i (0) = Ũ , V i (0) = Ṽi . System E3 = 0, E4 = 0, E8 = 0, E11 = 0, E12 = 0, E13 = 0, E17 = 0 is not analytic with respect to all its variables in a neighborhood of the solution m = 0, U = Ũ and (x8 , y8 ) = (a3 , 0) because equation E17 = 0 contains the term m/y82 . After doing the change of variables y8 = µY8 /2 with µ = m1/3 we obtain a new equivalent system of equations which is analytic with respect to all its variables in a neighborhood of µ = 0, U = Ũ , x8 = a3 and Y8 6= 0 and it can be written as Eei = E ei (y2 , x3 , y3 , x4 , y4 ) + O(µ3 ), i = 3, 4, 11, 12, 13, Ee8 = E e8 (y2 , x3 , y3 , x4 , y4 , x8 ) + O(µ2 ), e17 = µE E e17 (y2 , x3 , y3 , x4 , y4 , x8 , Y8 ) + O(µ3 ). Now we consider the system of equations ei , Ēi = E i = 3, 4, 8, 11, 12, 13, e17 /µ, Ē17 = E which is also analytic with respect to all its variables in a neighborhood of µ = 0, U = Ũ , x8 = a3 , and Y8 6= 0. Substituting µ = 0, U = Ũ , x8 = a3 and Y8 into system Ē3 = 0, Ē4 = 0, Ē8 = 0, Ē11 = 0, Ē12 = 0, Ē13 = 0, Ē17 = 0 On the convex central configurations of the symmetric (` + 2)–body problem 17 we get that all these equations are identically zero except the equation Ē17 = 0 which is equivalent to p Y82 Ē17 (Y8 ) = ĀY8 − = 0, Y83 where Ā = 25.9395013227 . . . . The solutions of this equation are Y8 = ±d = ± 0.3378154118 . . . . By symmetry, we are only interested in the positive values of Y8 . Clearly, system Ē3 = 0, Ē4 = 0, Ē8 = 0, Ē11 = 0, Ē12 = 0, Ē13 = 0, Ē17 = 0 is analytic with respect to all its variables in a neighborhood of the solutions µ = 0, U = Ũ , x8 = a3 , and Y8 = d. Moreover the Jacobian of the system (with respect to the variables U , x8 , Y8 ) evaluated at this solution takes the value −304636.8677115628 . . . . Therefore, applying the Implicit Function Theorem, this solution can be continued to an analytical family of solutions of system Ē3 = 0, Ē4 = 0, Ē8 = 0, Ē11 = 0, Ē12 = 0, Ē13 = 0, Ē17 = 0 defined for µ > 0 sufficiently small. This family provides the family of solutions of system E3 = 0, E4 = 0, E8 = 0, E11 = 0, E12 = 0, E13 = 0, E17 = 0 given by y24 = sin (2π/7) + O(m1/3 ), x43 = cos (4π/7) + O(m1/3 ), y34 = sin (4π/7) + O(m1/3 ), x44 = cos (6π/7) + O(m1/3 ), y44 = sin (6π/7) + O(m1/3 ), x48 = a3 + O(m1/3 ), and y84 = d/2 m1/3 + O(m2/3 ). Proof (Proof of Theorem 1 for ` = 7) Let xi2 (m) = −mxi8 (m) − 1/2 − xi3 (m) − xi4 (m) for i = 1, . . . , 5. Given m > 0 sufficiently small, from Proposition 6, we have the four central configurations given by (x1 , y1 ) = (1, 0), (x2 , y2 ) = (xi2 (m), y2i (m)), (x3 , y3 ) = (xi3 (m), y3i (m)), (x4 , y4 ) = (xi4 (m), y4i (m)), (x5 , y5 ) = (xi4 (m), −y4i (m)), (x6 , y6 ) = (xi3 (m), −y3i (m)), (x7 , y7 ) = (xi2 (m), −y2i (m)), (x8 , y8 ) = (xi8 (m), y8i (m)), (x9 , y9 ) = (xi8 (m), −y8i (m)) for i = 1, . . . , 4. Since for m = 0 we have four different central configurations that are convex, see Figure 3, it follows that for m > 0 sufficiently small they continue being four different convex central configurations. This completes the proof of the theorem. 3.5 Numerical evidences than Conjecture 1 is true From Lemma 1 we know that for all n ≥ 3 equation fn (x2n ) = 0 has two solutions with x2n < 0 that we denote by a1 (n) and a2 (n), where a1 (n) < a2 (n). We have computed numerically these two solutions for n = 4, . . . , 150 (see Figure 3.5 for the plot of these solutions as a function of n). Each solution provides n central configurations of the S1 -symmetric restricted (2n + 1)–body problem that are given by       −2π(i − 1) −2π(i − 1) (x2n , y2n )i,j = cos aj (n), sin aj (n) , 2n − 1 2n − 1 18 M. Corbera, J. Llibre and P. Yuan a1 HnL a2 HnL -1.30 -1.35 -0.9980 -1.40 -0.9985 -1.45 -0.9990 -1.50 -0.9995 -1.55 n n 20 40 60 80 100 120 140 20 40 60 80 100 120 140 Fig. 4 The plot of a1 (n) and a2 (n) for n = 4, . . . , 150. with j = 1, 2 and i = 1, . . . , n − 1. Applying Lemma 2 we see that the central configurations (x2n , y2n )i,1 are concave for all i = 1, . . . , n − 1 whereas the central configurations (x2n , y2n )i,2 are convex for all i = 1, . . . , n − 1. In short for n = 4, . . . , 150 the restricted (2n + 1)–body problem with 2n − 1 masses at the vertices of a regular n–gon has exactly n classes of convex central configurations. We have applied the Implicit function Theorem as in the proof of Theorem 1 for ` = 3, 5, 7 to continue these central configurations to the symmetric (2n + 1)–body problem with m > 0 sufficiently small for n = 4, . . . , 20 and in all cases we have seen that the corresponding determinants are different form zero. In particular, we see that the absolute value of these determinants increases as n increases. This gives numerical evidence that Conjecture 1 is true. 4 S2 –symmetric (2n + 2)–body problem 4.1 S2 –symmetric (2n + 2)–body problem We assume that q1 = (1, 0). Since the center of masses is at the origin we have P2n+2 Pn (cmx , cmy ) = (0, 0), so i=1 mi xi = 0, and x2 = −( i=3 xi + xn+1 /2 + P2n+2 1/2 + mx2n+1 ). Notice that due to the symmetry i=1 mi yi is identically zero. Taking into account all these conditions the 4n + 4 equations e1 = 0, . . . , e4n+4 = 0 given by (1) with N = 2n + 2 for the central configurations of the planar S2 –symmetric (2n + 2)–body problem reduce to the following 2n + 1 equations ei = 0, i = 1, . . . , n, e2n+1 = 0, e2n+2+i = 0 i = 2, . . . , n, e4n+3 = 0. P2n Indeed, the equation en+1 are omitted because i=1 ei +me2n+1 +me2n+2 = 0. And from the S2 –symmetry, we have ei = e2n+2−i , ei+2n+2 = −e4n+4−i , for 2 ≤ i ≤ n, e2n+1 = e2n+2 and e4n+3 = −e4n+4 , e2n+3 ≡ 0, e3n+3 ≡ 0. From the first equation e1 = 0 we isolate λ and we substitute it into the other 2n equations that we denote by Ei = 0, 2 ≤ i ≤ n, E2n+1 = 0, E2n+2+i = 0, 2 ≤ i ≤ n, E4n+3 = 0. On the convex central configurations of the symmetric (` + 2)–body problem 19 In short, we have 2n equations and 2n unknowns y2 , xi , yi , 3 ≤ i ≤ n, xn+1 , x2n+1 , y2n+1 . Now we consider the symmetric restricted (2n + 2)–body problem. From the results in Section 2 we know that the infinitesimal masses must be on one of the axis of symmetry of the 2n–gon. Clearly when the infinitesimal masses are on an axis of symmetry containing one of the primaries the resulting central configuration cannot be convex, so we only consider the case where the primaries are on an axis of symmetry that does not contain any of the primaries. Without loss of generality we can assume that the 2n primaries with masses equal to one are at the vertices of the regular 2n–gon xi = cos αi , yi = sin αi , 1 ≤ i ≤ 2n with αi = 2π(i − 1)/2n and the two infinitesimal masses with m = 0 are at the points (x2n+1 , y2n+1 ) = (r cos (π/2n), r sin (π/2n)) and (x2n+2 , y2n+2 ) = (r cos (π/2n), −r sin (π/2n)) for some r > 0. Straightforward computations show that if m = 0, xi = cos αi , yi = sin αi , 1 ≤ i ≤ 2n, (x2n+1 , y2n+1 ) = (r cos (π/2n), r sin (π/2n)) and (x2n+2 , y2n+2 ) = (r cos (π/2n), −r sin (π/2n)), then all the equations in (2) are identically zero except the equations E2n+1 = 0 and E4n+3 = 0 which satisfy E4n+3 = E2n+1 tan (π/2n). So we only need to solve E2n+1 = 0, which becomes fn (r) = 0 with 2n X cos αj − r cos (π/2n) fn (r) = 3/2 + λr cos (π/2n), j=1 (1 − 2r cos (αj − π/2n) + r2 ) where 2n X 1 λ= . j=2 [23 (1 − cos αj )]1/2 4.2 Proof of Theorem 2 for ` = 4 Next we see that there are no convex central configurations of the S2 –symmetric restricted (4 + 2)–body problem. Indeed, in Subsection 4.1 we have seen that the equations for the cen- tral configurations of the symmetric restricted (4 + 2)–body problem with (x1 , y1 ) = (1, √0), (x√ 2 , y2 ) = (0, 1), (x3 , y3 ) √ √ (x4 , y4 ) = (0, −1), = (−1, 0), (x5 , y5 ) = (r 2/2, r 2/2) and (x6 , y6 ) = (r 2/2, −r 2/2) are reduced to the equation f2 (r) = 0 with √ √ 1 − 2r − 2r − 1 1 √  f2 (r) = √ 3/2 + √ 3/2 + 8 4 + 2 r, r2 − 2r + 1 r2 + 2r + 1 see Subsection 3.1. Dropping the square roots and the denominators of equa- tion f2 (r) = 0 as above we get a polynomial equation kr2 P (r) = 0, where k is a constant and P is a polynomial of degree 26. This polynomial equation has exactly four positive real roots of those only r = r1 = 0.6973805098 . . . 20 M. Corbera, J. Llibre and P. Yuan and r = r2 = 1.6024084862 . . . are solutions of the initial equation f2 (r) = 0. Applying Lemma 2 we see that neither r = r1 nor r = r2 provides convex central configurations of the S2 –symmetric restricted (4 + 2)–body problem. Due to the symmetry if we take the position on m5 on any other axis of symmetry that does not contain any primary we obtain the same central con- figurations. So there are no convex central configurations of the S2 –symmetric restricted (4 + 2)–body problem and consequently there are no convex central configurations of the S2 –symmetric 6–body problem with m > 0 sufficiently small coming from continuation of convex central configurations of the S2 – symmetric restricted (4 + 2)–body problem. 4.3 Proof of Theorem 2 for ` = 6 For the S2 –symmetric 8–body problem (that is n = 3), the equations of central configuration are reduced to E2 = 0, E3 = 0, E7 = 0, E10 = 0, E11 = 0, E15 = 0 where x2 = −m x7 − 1/2 − x3 − x4 /2 and the ones for the symmetric restricted (6 + 2)–body problem are reduced to the equation f3 (r) = 0, see Subsection 4.1. Proposition 7 Let a1 = 1.5922353553 . . . and a2 = 0.8843211381 . . . be the two roots of f3 (r) = 0 with r > 0. The symmetric restricted (6 + 2)–body problem has exactly three classes of convex central configurations which are given by (a) (x7 , y7 ) = (0, a1 ), (b) (x7 , y7 ) = (a2 cos (π/6), a2 sin (π/6)), (c) (x7 , y7 ) = (0, a2 ). Proof Assuming that m7 is on the axis of symmetry between m1 and m2 , i.e. (x7 , y7 ) = (r cos (π/6), r sin (π/6)), the equations of the S2 –symmetric re- stricted (6 + 2)–body reduce to the equation f3 (r) = 0 which becomes √ √ √ 3r 3 − 2 3r −2 3r − 3 − 3/2 + √ 3/2 + √ 3/2 (r2 + 1) 2 r2 − 3r + 1 2 r2 + 3r + 1 1 √  + 4 + 5 3 r = 0. 8 By squaring both sides of this equation as many times as we need and dropping the denominators we get the polynomial equation r4 P (r) = 0, where P (r) is a polynomial of degree 76. Solving numerically this polynomial equation we get exactly ten real solutions with r > 0 of which only the solutions r = a1 , r = a2 are solutions of the initial equation. We also can take the mass m7 on the axis of symmetry passing between of any other pair of consecutive masses. By symmetry the only axis of symmetry that provides different central configurations is the axis passing between m2 and m3 which correspond to a rotation with an angle of 2π/6 of the solutions On the convex central configurations of the symmetric (` + 2)–body problem 21 (x7 , y7 ) = (a1 cos (π/6), a1 sin (π/6)) and (x7 , y7 ) = (a2 cos (π/6), a2 sin (π/6)). However, when (x7 , y7 ) = (a1 cos (π/6), a1 sin (π/6)), the corresponding cen- tral configuration is not convex. So we have only the three convex central configurations given in statements (a), (b) and (c). Proposition 8 The three convex central configurations of the symmetric re- stricted (6 + 2)–body problem can be continued to three families of central con- figurations of the symmetric 8–body problem with masses mi = 1, 1 ≤ i ≤ 6 and m7 = m8 = m > 0 sufficiently small. Proof Using the Implicit Function Theorem we will see that the solutions of system E2 = 0, E3 = 0, E7 = 0, E10 = 0, E11 = 0, E15 = 0 with m = 0 given in Proposition 7 can be continued to solutions of this system with m > 0 sufficiently small. We use the notation: U = (y2 , x3 , y3 , x4 ), V = (x7 , y7 ), Ũ = (sin (2π/8), cos (4π/6), sin (4π/6), −1), Ṽ1 = (0, a1 ), Ṽ2 = (a2 cos (π/6), a2 sin (π/6)), and Ṽ3 = (0, a2 ). The determinant of the Jacobian of system E2 = 0, E3 = 0, E7 = 0, E10 = 0, E11 = 0, E15 = 0 (with respect to the variables U , V ), evaluated at m = 0, U = Ũ and V = Ṽ1 takes the value 1.0896102708 . . . Since this determinant is different from zero and the system is analytic with respect to the variables (U, V ), from the Implicit Function Theorem there exist analytic families U 1 (m) = (y21 (m), x13 (m), y31 (m), x14 (m)) and V 1 (m) = (x17 (m), y71 (m)) of solutions of system E2 = 0, E3 = 0, E7 = 0, E10 = 0, E11 = 0, E15 = 0 satisfying U 1 (0) = Ũ and V 1 (0) = Ṽ1 defined for m > 0 sufficiently small. The determinant of the Jacobian of system E2 = 0, E3 = 0, E7 = 0, E10 = 0, E11 = 0, E15 = 0 (with respect to the variables U , V ), evaluated at m = 0, U = Ũ and V = Ṽi for i = 2, 3 takes the value −57.55271347 . . . which is also different from zero. So these two solutions can be continued to the two families U 1 (m) = (y2i (m), x13 (m), y3i (m), x14 (m)) and V 1 (m) = (xi7 (m), y7i (m)) for i = 2, 3 of solutions of system E2 = 0, E3 = 0, E7 = 0, E10 = 0, E11 = 0, E15 = 0 satisfying U 1 (0) = Ũ and V i (0) = Ṽi defined for m > 0 sufficiently small. Since for m = 0 the three central configurations of the S2 –symmetric re- stricted (8 + 2)–body problem are different, the continued central configura- tions for m > 0 small continue being different. So if xi2 (m) = −mxi7 (m) − 1/2 − xi3 (m) − xi4 (m)/2 for i = 1, 2, 3, then for m > 0 sufficiently small Propo- sition 8 provides the three different families of central configurations given by (x1 , y1 ) = (1, 0), (x2 , y2 ) = (xi2 (m), y2i (m)), (x3 , y3 ) = (xi3 (m), y3i (m)), (x4 , y4 ) = (xi4 (m), 0), (x5 , y5 ) = (xi3 (m), −y3i (m)), (x6 , y6 ) = (xi2 (m), −y2i (m)), (x7 , y7 ) = (xi7 (m), y7i (m)), (x8 , y8 ) = (xi7 (m), −y7i (m)) for i = 1, 2, 3. This proves Theorem 2 for ` = 6. 22 M. Corbera, J. Llibre and P. Yuan 4.4 Proof of Theorem 2 for ` = 8 For the central configurations of the S2 –symmetric 10–body problem (that is n = 4) the equations of central configurations are reduced to E2 = 0, E3 = 0 ,E4 = 0,E9 = 0, E12 = 0, E13 = 0, E14 = 0, E19 = 0 where x2 = −mx9 − 1/2 − x3 − x4 − x5 /2 and the ones for the restricted S2 –symmetric (8 + 2)–body problem are reduce to f4 (r) = 0, see Subsection 4.1. Proposition 9 Let a3 = 0.9401381791 · · · be the smallest positive root of f4 (r) = 0. The symmetric restricted (8 + 2)–body problem has exactly two classes of convex central configurations which are given by (a) (x9 , y9 ) = (a3 cos (π/8), a3 sin (π/8)), (b) (x9 , y9 ) = (a3 cos (3π/8), a3 sin (3π/8)). Proof Let (x9 , y9 ) = (r cos (π/8), r sin (π/8)) be the position of an infinitesi- mal mass in a central configuration of the restricted (8 + 2)–body problem. Under this hypothesis r satisfies equation f4 (r) = 0. Solving this equation we get exactly two solutions with r > 0, they are r = a4 = 1.5745151766 · · · and r = a3 = 0.9401381791 · · · . By rotating the positions (x9 , y9 ) = (ai cos (π/8), ai sin (π/8)) for i = 3, 4 with an angle of 2π/8 we get another two solutions. However, when (x9 , y9 ) = (a4 cos (π/8), a4 sin (π/8)), (x9 , y9 ) = (a4 cos (3π/8), a4 sin (3π/8)) the corresponding central configurations are not convex. Any additional rotation of an angle of 2π/8 does not provide any new central con- figuration. So we have only the two convex central configurations given in statements (a) and (b). Proceeding as in the case ` = 6, the proof of Theorem 2 for ` = 8 follows from the following result. Proposition 10 The two convex central configurations of the symmetric re- stricted (8 + 2)–body problem can be continued to two families of central con- figurations of the symmetric 10–body problem with masses mi = 1, 1 ≤ i ≤ 8 and m9 = m10 = m > 0 sufficiently small. Proof Using the Implicit Function Theorem we will see that the solutions of system E2 = 0, E3 = 0 ,E4 = 0,E9 = 0, E12 = 0, E13 = 0, E14 = 0, E19 = 0 with m = 0 given in Proposition 9 can be continued to solutions of this system with m > 0 sufficiently small. Let U = (y2 , x3 , y3 , x4 , y4 , x5 ), V = (x9 , y9 ) and Ũ = (sin (2π/8), cos (4π/8), sin (4π/8), cos (6π/8), sin (6π/8), −1), Ṽ1 = (a3 cos (π/8), a3 sin (π/8)), Ṽ2 = (a3 cos (3π/8), a3 sin (3π/8)). The determinant of the Jacobian of system E2 = 0, E3 = 0 ,E4 = 0, E9 = 0, E12 = 0, E13 = 0, E14 = 0, E19 = 0 (with respect to the variables U , V ), evaluated at m = 0, U = Ũ and V = Vi for i = 1, 2 takes the value On the convex central configurations of the symmetric (` + 2)–body problem 23 −2.9072711711 × 106 . Since this determinant is different from zero from the Implicit Function Theorem the two solutions can be continued to solutions of E2 = 0, E3 = 0 ,E4 = 0, E9 = 0, E12 = 0, E13 = 0, E14 = 0, E19 = 0 for m > 0 sufficiently small. 4.5 Proof of Theorem 2 for ` = 10 For the central configurations of the S2 –symmetric 12–body problem (that is n = 5) the equations of central configurations reduced to E2 = 0, E3 = 0, E4 = 0, E5 = 0, E11 = 0, E14 = 0, E15 = 0, E16 = 0, E17 = 0, E23 = 0 where x2 = −mx11 − 1/2 − x3 − x4 − x5 − x6 /2 and the ones for the restricted S2 –symmetric (10 + 2)–body problem reduce to f5 (r) = 0, see Subsection 4.1. Proposition 11 Let a4 = 0.9634598812 . . . be the smallest positive root of f5 (r) = 0. The S2 –symmetric restricted (10 + 2)–body problem has exactly three classes of convex central configurations which are given by (a) (x11 , y11 ) = (a4 cos (π/10), a4 sin (π/10)), (b) (x11 , y11 ) = (a4 cos (3π/10), a4 sin (3π/10)), (c) (x11 , y11 ) = (a4 cos (5π/10), a4 sin (5π/10)). Proof Let (x11 , y11 ) = (r cos (π/10), r sin (π/10)) be the position of an in- finitesimal mass in a central configuration of the S2 –symmetric restricted (10 + 2)–body problem. Under this hypothesis r satisfies equation f5 (r) = 0. Solving this equation we get exactly two solutions r = a5 = 1.5541234676 · · · , r = a4 = 0.9634598812 · · · with r > 0. By rotating the positions (x9 , y9 ) = (ai cos (π/8), ai sin (π/8)) for i = 4, 5 with angles of 2π/10 and of 4π/10 we get another four solutions. However if (x11 , y11 ) = (a5 cos (π/10), a5 sin (π/10)), (x11 , y11 ) = (a5 cos (3π/10), a5 sin (3π/10)), and (x11 , y11 ) = (a5 cos (5π/10), a5 sin (5π/10)), then the corresponding central configurations are not convex. Any additional rotation of an angle of 2π/10 does not provide any new central configuration. So we have only the three convex central configurations given in statements (a), (b) and (c). The proof of Theorem 2 for ` = 10 follows from the following result. Proposition 12 The three convex central configurations of the S2 –symmetric restricted (10 + 2)–body problem can be continued to three families of central configurations of the S2 –symmetric 12–body problem with masses mi = 1, 1 ≤ i ≤ 10 and m11 = m12 = m > 0 sufficiently small. Proof Using the Implicit Function Theorem we will see that the solutions of system E2 = 0, E3 = 0 ,E4 = 0, E5 = 0, E11 = 0, E14 = 0, E15 = 0, E16 = 0, E17 = 0, E23 = 0 with m = 0 given in Proposition 11 can be continued to solutions of this system with m > 0 sufficiently small. Let U = (y2 , x3 , y3 , x4 , y4 , x5 , y5 , x6 ), V = (x11 , y11 ) and Ũ =(sin (2π/10), cos (4π/10), sin (4π/10), cos (6π/10), sin (6π/10), 24 M. Corbera, J. Llibre and P. Yuan cos (8π/10), sin (8π/10), −1), Ṽi =(a4 cos (π/10 + 2πi/10), a4 sin (π/10 + 2πi/10)), i = 1, 2, 3. The determinant of the Jacobian of system E2 = 0, E3 = 0 ,E4 = 0, E5 = 0, E11 = 0, E14 = 0, E15 = 0, E16 = 0, E17 = 0, E23 = 0 (with respect to the variables U , V ), evaluated at m = 0, U = Ũ and V = Ṽi for i = 1, 2, 3 is −5.3833737767 × 108 Since this determinant is different from zero from the Implicit Function Theorem the three central configurations of the S2 –symmetric restricted (10 + 2)–body problem given by Proposition 11 can be continued for m > 0 sufficiently small. 4.6 Some numerical evidences than Conjecture 2 is true Proceeding as in Subsection 3.5 for the S2 –symmetric (2n + 2)–body problem we have computed the two solutions of fn (r) for n = 6, . . . , 151. We denote these solutions by r = a1 (n) and r = a2 (n) with a1 (n) < a2 (n). Then applying Lemma 2 we have seen that the solution a2 (n) does not lead to convex central configurations of the S2 –symmetric rectricted (2n + 2)–body problem whereas the solution a1 (n) provides exactly [ n+12 ] convex central configurations. We have also applied the Implicit Function Theorem as in the proof of Theorem 2 for ` = 6, 8, 10 to continue these central configurations to the symmetric (2n + 2)–body problem with m > 0 sufficiently small for n = 6, . . . , 20 and in all cases we have seen that the corresponding determinants are different form zero. Moreover the absolute value of these determinants increases as n increases. This gives numerical evidence that Conjecture 2 is true. 5 S3 –symmetric (2n + 2)–body problem 5.1 Equations of the S3 –symmetric (2n + 2)–body problem Without loss of generality we can assume that the position of m1 is fixed at the point q1 = (cos(π/(2n)), sin(π/(2n))) and that the center of masses is at the P2n+2 origin of coordinates, so Pn i=1 mi xi = 0, and using the S3 –symmetry we have x2 = −(cos(π/(2n)) + i=3 xi + mx2n+1 ). Notice that using the S2 –symmetry P2n+2 again i=1 mi yi is identically zero. Taking into account all these conditions the 4n + 4 equations e1 = 0, . . . , e4n+4 = 0 given by (1) with N = 2n + 2 for the central configurations of the planar symmetric (2n + 2)–body problem reduce to the following 2n + 1 equations ei = 0, e2n+2+i = 0, i = 2, . . . , n, e2n+1 = 0, e2n+3 = 0, e4n+3 = 0. Indeed, from the S3 –symmetry we have ei+1 = e2n−i , ei+2n+3 = −e4n+2−i , for 0 ≤ i ≤ n − 1, e2n+2 = e2n+1 and e4n+4 = −e4n+3 . So equations en+i , On the convex central configurations of the symmetric (` + 2)–body problem 25 e2n+2 , e3n+i+2 P2n and e4n+4 can be omitted. P Moreover equation e1 can be omitted n because i=1 ei + me2n+1 + me2n+2 = 2( i=1 ei + me2n+1 ) = 0. Nevertheless in this case equation e2n+3 cannot be omitted because e2n+3 = −e4n+2 and we have already omitted e4n+2 . From equation e2n+3 = 0 we isolate λ and we substitute it into the other 2n equations that we denote by E2 = 0, · · · , En = 0; E2n+1 = 0; E2n+4 = 0, · · · , E3n+2 = 0; E4n+3 = 0, (5) where Ek is ek after the substitution. In short, we have 2n equations and 2n−1 unknowns y2 , xi , yi , 3 ≤ i ≤ n, x2n+1 , y2n+1 . Since we have more equations than unknowns this system could have no solution. Now we consider the S3 –symmetric restricted (2n+2)–body problem where the 2n primaries with masses equal to one are at the vertices of the regu- lar 2n-gon xi = cos(αi + π/(2n)), yi = sin(αi + π/(2n)), 1 ≤ i ≤ 2n with αi = π(i − 1)/n and the two infinitesimal masses with m = 0 are at the points (x2n+1 , y2n+1 ) and (x2n+1 , −y2n+1 ). Here (x2n+1 , y2n+1 ) is the position of the infinitesimal mass in a central configuration of the restricted (2n + 1)– body problem with 2n equal masses at the vertices of a regular 2n-gon. From Lemma 1, the position of the infinitesimal mass is on an axis of symmetry of the 2n–gon. Without loss of generality we can assume that the line of symmetry is either y2n+1 = 0 (in this case none of the primaries are on the axis of symmetry) or y2n+1 = tan(π/(2n))x2n+1 (in this case two primaries are on the axis of sym- metry). We start with the case y2n+1 = 0. Straightforward computations show that when m = 0, xi = cos(αi + π/(2n)), yi = sin(αi + π/(2n)), 1 ≤ i ≤ 2n and y2n+1 = 0, all equations (5) are identically zero except the equation E2n+1 = 0. Using Lemma 1 again the solutions of E2n+1 = 0 are (0, 0), (ρ2 , 0) with 0 < ρ2 < 1 and (ρ3 , 0) with ρ3 > 1. Taking into account the symme- tries of the problem, it is sufficient to consider the following positions for (x2n+1 , y2n+1 ): (0, 0), ρ2 (cos(πi/n), sin(πi/n)), and ρ3 (cos(πi/n), sin(πi/n)) for i = 0, . . . , [n/2] where [x] denotes the integer part function of x. When m = 0, xi = cos(αi + π/(2n)), yi = sin(αi + π/(2n)), 1 ≤ i ≤ 2n and y2n+1 = tan(π/(2n))x2n+1 , all equations (5) are identically zero except the equations E2n+1 = 0 and E4n+3 = 0, moreover E4n+3 = tan π/(2n)E2n+1 . Us- ing Lemma 1 the solutions of equation E2n+1 = 0 are (0, 0), and ρ1 (cos(π/(2n)), sin(π/(2n))) with ρ1 > 1. Taking into account the symmetries of the problem, it is sufficient to consider the following positions for (x2n+1 , y2n ): (0, 0) and ρ1 (cos(πi/n + π/(2n)), sin(πi/n + π/(2n)) for i = 0, . . . , [(n + 1)/2] − 1. Since we are interested only in convex central configurations we do not consider (x2n+1 , y2n+1 ) = (0, 0) and (x2n+1 , y2n+1 ) = ρ1 (cos(πi/n + π/(2n)), sin(πi/n + π/(2n)) for i = 0, . . . , [(n + 1)/2] − 1. Clearly (x2n+1 , y2n+1 ) = (0, 0) does not provide a convex central configuration. Consider now the three con- secutive vertices of the configuration p1 = (cos(π(i − 1)/n + π/(2n)), sin(π(i − 1)/n + π/(2n)), p2 = (cos(πi/n + π/(2n)), sin(πi/n + π/(2n)) and p3 = 26 M. Corbera, J. Llibre and P. Yuan ρ1 (cos(πi/n + π/(2n)), sin(πi/n + π/(2n)), it is easy to see that   − p−1− →×− p → = (1 − ρ ) sin π < 0, p−2− p 2 3 1 n because ρ1 > 1. Therefore the configuration is not convex because it does not satisfy the conditions of Lemma 2. 5.2 Proof of Theorem 3 for ` = 4 For the central configurations of the S3 –symmetric 6–body problem (that is n = 2) equations (2) reduce to the four equations E2 = 0, E5 = 0, E8 = 0, E11 = 0, (6) and the√three unknowns y2 , x5 and y5 , plus the parameter m. Here x2 = mx5 − 2/2. The equation E5 = 0 for the S2 –symmetric restricted (4+2)–body problem when y5 = 0 becomes     2 x5 − √12   2 x + √1 √ 1 1 5 2 √ 3/2 + 2 − 2 − √ x5 +   2 3/2 = 0. (7) 2 x5 − 2x5 + 1 4 2 1 1 x 5 + √2 + 2 We note that by the symmetry of the problem, we only are interested in solutions with x5 > 0. Dropping off the denominators and the square roots in (7), the equation can be transformed to a polynomial equation of the form x25 P (x5 ) = 0, where P (x5 ) is a polynomial of degree 26 in the variable x5 whose set of roots contains all solutions of (7). Solving numerically equation P (x5 ) = 0 we get four positive real root of which only x5 = ρ2 = 0.6973805098.. and x5 = ρ3 = 1.6024084862.. are solutions of the initial equation (7). Since cos(π/4) = sin(π/4) > ρ2 the solution x5 = ρ2 cannot give convex central configurations. This proves the following result. Proposition 13 Up to symmetry the unique positions for the infinitesimal mass m5 that can provide convex central configurations of the S3 –symmetric restricted (4 + 2)–body problem are (x5 , y5 ) = (ρ3 , 0) and (x5 , y5 ) = (0, ρ3 ). Now we see that the central configuration of the S3 –symmetric restricted (4 + 2)–body problem with (x5 , y5 ) = (0, ρ3 ) cannot be continued to a family of central configurations of the S3 –symmetric 6–body problem with m > 0 small. To do that we consider system (6) as a system of the four unknowns y2 , x5 , y5 and m and we apply the Inverse Function Theorem. The Jacobian of the system with respect to the variables y2 , x5 , y5 and m evaluated at the solution y2 = sin(3π/4), x5 = 0, y5 = ρ3 and m = 0 becomes 3.9841276914.. 6= 0. Therefore the solution is isolated and it cannot be continued to a family of solutions of system (6) for m > 0 sufficiently small. On the convex central configurations of the symmetric (` + 2)–body problem 27 We consider now the central configuration of the S3 –symmetric restricted (4 + 2)–body problem with (x5 , y5 ) = (ρ3 , 0). We also will see that this central configuration cannot be continued to a family of central configurations of the S3 –symmetric restricted 6–body problem with µ > 0 sufficiently small. Notice that in this case m5 and m6 collide, so system (6) is not analytic with respect to all its variables in a neighborhood of y2 = sin(3π/4), x5 = ρ3 , y5 = 0 and m = 0. Proceeding as in the proofs of Theorems 1 and 2, after doing the change of variables y5 = µY5 /2 with µ = m1/3 we obtain a new system of equations Ẽ2 = Ẽ20 + O(µ3 ) = 0, Ẽ5 = Ẽ50 + O(µ2 ) = 0, Ẽ8 = Ẽ80 + O(µ3 ) = 0, Ẽ11 = Ẽ110 µ + O(µ3 ) = 0, which is analytic with respect to its variables in a neighborhood of y2 = sin(3π/4), x5 = ρ3 , Y5 6= 0, µ = 0. (8) Here Ẽ20 , Ẽ50 , Ẽ80 and Ẽ110 are functions that does not depend on µ. Finally we consider the system of equations Ē2 = Ẽ2 = 0, Ē5 = Ẽ5 = 0, Ē8 = Ẽ8 = 0, Ē11 = Ẽ11 /µ = 0, (9) which is also analytic with respect to its variables in a neighborhood of (8). Substituting (8) into system (9) we get that Ē2 , Ē5 and Ē8 are identically zero, and that equation Ē11 = 0 is equivalent to equation p Y52 −A Y5 + = 0, Y53 with √  √  √  2ρ23 + 2 2ρ3 − 1 −2ρ23 + 2 2ρ3 + 1 4+ 2 A=− √ 5/2 + √ 5/2 + √ 2 ρ23 + 2ρ3 + 1 2 ρ23 − 2ρ3 + 1 8 2 = 0.5285045250... The solutions of this equation are Y5 = 0 and Y5 = ±b = ±A−1/3 . Clearly the derivatives of Ē2 , Ē5 , Ē8 , and Ē11 with respect to µ when µ = 0 are zero, so the Jacobian of system (9) with respect to the variables y2 , x5 , Y5 and µ evaluated at the solution (8) with Y5 = b is zero. In this case the Inverse Function Theorem is not sufficient to prove that the solution (8) with Y5 = b cannot be continued analytically to a solution with µ > 0 sufficiently small. We assume that it can be continued; that is, that y2 = sin(3π/4) + y21 µ + y22 µ2 + y23 µ3 + O(µ4 ), x5 = ρ3 + x51 µ + x52 µ2 + x53 µ3 + O(µ4 ), Y5 = b + Y51 µ + Y52 µ2 + Y53 µ3 + O(µ4 ), 28 M. Corbera, J. Llibre and P. Yuan is a solution of (9) and we shall arrive to a contradiction. Indeed, we substitute this solutions into (9), then expanding in power series of µ we get Ē2 =Ē21 (y21 )µ + Ē22 (y21 , y22 )µ2 + Ē23 (y21 , y22 , y23 )µ3 + O(µ4 ) = 0, Ē5 =Ē51 (y21 , x51 )µ + Ē52 (y21 , x51 , Y51 , y22 , x52 )µ2 + Ē53 (y21 , x51 , Y51 , y22 , x52 , Y52 , y23 , x53 )µ3 + O(µ4 ) = 0, (10) Ē8 =Ē81 (y21 )µ + Ē82 (y21 , y22 )µ2 + Ē83 (y21 , y22 , y23 )µ3 + O(µ4 ) = 0, Ē11 =Ē111 (y21 , x51 , Y51 )µ + Ē112 (y21 , x51 , Y51 , y22 , x52 , Y52 )µ2 + Ē113 (y21 , x51 , Y51 , y22 , x52 , Y52 , y23 , x53 , Y53 )µ3 + O(µ4 ) = 0. Solving system (10) at first order in µ, i.e. Ē21 (y21 ) = 0, Ē51 (y21 , x51 ) = 0, Ē81 (y21 ) = 0, Ē111 (y21 , x51 , Y51 ) = 0, we get y21 = 0, x51 = 0, and Y51 = 0. Solving system (10) at second order in µ, i.e. Ē22 (0, y22 ) = 0, Ē52 (0, 0, 0, y22 , x52 ) = 0, Ē82 (0, y22 ) = 0, Ē112 (0, 0, 0, y22 , x52 , Y52 ) = 0, we get y22 = 0, x52 = x52 (ρ3 , b) = 0.2007484218.., and Y52 = Y52 (ρ3 , b) = 0.4509017133... Finally solving equation Ē2 at third order in µ, i.e. Ē23 (0, 0, y23 ) = 0, we get  √ √ ∗ 8 √   2 −2ρ3 − 2 y23 = y23 =− 1+2 2 − √ 3/2 − √  3/2 7 ρ23 − 2ρ3 + 1 ρ3 ρ3 + 2 + 1  1 √  − 1 + 8 2 ρ3 = 13.5243491694... 8 ∗ We can see that Ē83 (0, 0, y23 ) = −23.3594334957.. 6= 0, Therefore the solution (8) with Y5 = b cannot be continued analytically to a solution of system (6) with µ > 0 sufficiently small. 5.3 Proof of Theorem 3 for ` > 4 Proceeding in a similar way than in the proof of Theorem 3 for ` = 4 we get that when n = 3 the solutions of equation E7 = 0 for the restricted (6 + 2)– body problem when m = 0 are ρ2 = 0.8843211381.., ρ3 = 1.5922353553.. Applying Lemma 2 we see that, up to symmetry, the unique positions for the infinitesimal mass m7 that can provide convex central configurations of the S3 – symmetric restricted (6 + 2)–body problem are (x7 , y7 ) = (ρ2 , 0), (x7 , y7 ) = On the convex central configurations of the symmetric (` + 2)–body problem 29 ρ2 (cos(π/3), sin(π/3)), (x7 , y7 ) = (ρ3 , 0), (x7 , y7 ) = ρ3 (cos(π/3), sin(π/3)). Us- ing the Inverse Function Theorem as in the case n = 2 we can prove that the central configurations of the S3 –symmetric restricted (6 + 2)–body problem with (x7 , y7 ) = ρ2 (cos(π/3), sin(π/3)) and (x7 , y7 ) = ρ3 (cos(π/3), sin(π/3)) cannot be continued to convex central configurations of the S3 –symmetric 8– body problem with m > 0 sufficiently small. Moreover doing the change of variables y7 = µY7 /2 with µ = m1/3 and proceeding as in the case ` = 4 we prove also that the central configurations of the S3 –symmetric restricted (6 + 2)–body problem with (x7 , y7 ) = (ρ2 , 0) and (x7 , y7 ) = (ρ3 , 0) cannot be continued to convex central configurations of the S3 –symmetric 8–body problem with m > 0 sufficiently small. When n = 4 the solutions of equation E9 = 0 for the restricted (8+2)–body problem when m = 0 are ρ2 = 0.9401381791.., ρ3 = 1.574515766.. Applying Lemma 2 we can see that, up to symmetry, the unique positions for the infinitesimal mass m7 that can provide convex central configurations of the S3 –symmetric restricted (8+2)–body problem are (x7 , y7 ) = (ρ2 , 0), (x7 , y7 ) = ρ2 (cos(π/4), sin(π/4)), and (x7 , y7 ) = ρ2 (cos(π/2), sin(π/2)). 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