This is a preprint of: “Central Configurations in the Five-Body Problem: Rhombus Plus One”, José Lino Cornelio Soberano, Martha Álvarez-Ramı́rez, Josep Maria Cors, Qual. Theory Dyn. Syst., vol. 20, 51:1–13, 2021. DOI: [10.1007/s12346-021-00486-6] Central configurations in the five-body problem: Rhombus plus one J. Lino Cornelio, Martha Alvarez-Ramı́rez and Josep M. Cors Abstract. We show the existence of central configurations in the planar five-body problem where four bodies are located at the vertices of a rhombus, called rhombus plus one central configurations. Concretely we prove analytically their existence when one diagonal is nearly equal to the sides of the rhombus and when the two diagonals are either equal or nearly equal. In addition, we prove that given a rhombus plus one configuration, the corresponding vector of positive masses that makes the configuration central, if exists, is unique. Mathematics Subject Classification (2010). 70F10; 37N05; 70F15. Keywords. five-body problem, central configuration, rhombus configu- ration, square configuration. 1. Introduction In the n-body problem a configuration is central if the acceleration vector of every body is proportional, same common scalar, to its position vector with respect to the center of mass. There are many reasons why such configurations are of special importance, and regardless of everything many fundamental questions about them are still open. For instance, they allow to obtain explicit solutions of the n-body problem, concretely, solutions where the shape of the configuration is preserved along the orbit up to rescaling and rotations. In [15, 16, 18] and references therein, the reader can found a detailed introduction about central configurations, together with new ideas and techniques which have been developed recently in their study. The problem of the finiteness of central configurations for every choice of the positive masses, also known as the Chazy-Wintner conjecture, was in- cluded in Smale’s list as a challenge question for the 21st century [20]. It is known that for n = 3 there are exactly five classes of central configurations M. Alvarez-Ramı́rez is partially supported by Programa Especial de Apoyo a la Inves- tigación (UAM) grant number I5-2019. Josep M. Cors has been supported by grant MTM2016-77278-P (FEDER), and Catalan (AGAUR) grant SGR 1617. 2 J. Lino Cornelio, M. Alvarez-Ramı́rez and Josep M. Cors for each choice of positive masses, counting up rotations and translations in the plane. Namely, three collinear, when the three bodies lie on the same line, and two equilateral, when the three bodies are located at the vertices of an equilateral triangle. When n = 4, Hampton and Moeckel [10] proved by computer-assisted that for each choice of masses there is a finite number of central configurations. Recently, for the planar five-body problem the conjec- ture has been solved by Albouy and Kaloshin [1], except for masses in some codimension two variety of the mass space. Hampton and Jensen [12] showed that in the five-body problem the number of spatial central configurations is finite, except for some special cases. So, the finiteness conjecture is wide open for n ≥ 6. The determination of possible shapes for central configurations, at least, the ones possessing some type of symmetry or defining a geometric property, is a relevant issue analogous to the finiteness conjecture. In spite of the fact that the finiteness conjecture is now settled for n = 5, in the planar case, there are still many unsolved problems regarding shapes of five-body central configurations. Williams ([23]) and Chen and Hsiao ([2]) studied the existence of convex central configurations in the five-body problem and gave some geo- metric properties. Apart from these fundamental results, little more is known, in contrast with the planar four-body problem, where recently, Corbera, Cors and Roberts [3] have classified the full set of convex central configurations. In this paper we will investigate symmetric central configurations in the five-body problem where three bodies lie on the axis of symmetry of the configuration. Similar symmetric configurations were previously studied in [4, 5, 7, 13]. Works when only one of the five bodies lie on the axis of sym- metry are in [9, 14]. Notice that in all the previous cited works the central configurations studied are staked, that is, a proper subset of the configuration is also a central configuration. See [9], where the concept of stacked central configurations was introduced. Since in the present work we consider config- urations that, in addition to the axial symmetry, four of the five bodies lies at the vertices of a rhombus, such a central configurations will not be stacked anymore due to a result of Fernandes and Mello [6]. They proved that the only stacked central configuration of the five-body problem when one body is removed is the square with equal positive masses at its vertices and the removed body, with any positive mass, located at its center. Among previous cited works on the five-body problem considering three bodies on the axis of symmetry, we distinguish the article of Shoaib et al. [21] for its similitude with the present paper. The authors, as in our set up, in addition to the axial symmetry, they impose that four of the five bodies have to be located at the vertices of a rhomboid, although a careful reading suggest that the use of term kite instead a rhomboid should be more accurate, see Fig. 1 in [21]. Rhombus plus one 3 Besides of we consider four bodies at the vertices of a rhombus is worthy to mention that in our work any constraint on the masses is considered. We present in Figure 2 a complete classification of all central configurations of the five-body problem where in addition to place three bodies on the axis of symmetry of the configuration, four of the five bodies are at the vertices of a rhombus, that we call rhombus plus one configuration. Concretely, in Section 2, we introduce the equations for central configurations and state the two main theorems that insure analytically the existence of rhombus plus one central configurations in the five-body, when one diagonal is nearly equal to the sides of the rhombus and when the two diagonal are equal. Section 3 is devoted to the unicity question of what is called the inverse problem. We prove that, for a given rhombus plus one configuration, the corresponding vector of positive masses that makes the configuration central, if exists, is unique. Finally, the two existence theorems are proved in sections 4 and 5. 2. Statement of the problem and main results Let qi ∈ R2 and mi > 0 denote the position and mass, respectively, of the i-th body. Let rij = kqi − qj k be the Euclidean distance between the i-th P5 and j-th bodies. If M = i=1 mi denotes the sum of the masses, then the 1 P5 center of mass is given by c = M i=1 mi qi . The planar five-body problem is governed by the equations ∂U mi q̈i = , i = 1, . . . , 5 (2.1) ∂qi where U denotes the Newtonian potential given by X mi mj U (q) = . rij 1≤i 0 when m1 lies outside the rhombus and a < 0 when m1 lies inside the rhombus, |a| = r12 , and let b be the half of the distance of the diagonal of the rhombus joining bodies with masses m4 and m5 , that is 2b = r45 . Then any point (a, b)-plane belonging to the open shaded region M in Figure 2 represents a rhombus plus one central configuration. Boundary of M is given by curves where the values of the masses are zero, m1 = 0 (black), m2 = 0 (blue) and m3 = 0 (red), and √ curves where the denominators in (2.6) are zero, the straight lines b = 3/2 (R24 = 1) and a = −1/2 (R12 = R13 ), both in brown. The particular case where the rhombus becomes a square is represented in Figure 2 by the green straight line b = 1/2 (R45 = 1) and has nonempty intersection with M. In Figure 3, we show the values of the masses m1 (black), m2 (blue) and m3 (red) along the set of square plus one central configurations. When the fifth mass is outside the square the admissible values of a go from 0.4402277051 . . . to 0.59267461067 . . . . Specifically, at a = 0.4402277051 . . . the value of the masses are m1 = 0 and m2 = m3 = 1, and at the other end of the range a = 0.5926746107 . . . we have m1 = 0.7298273415 . . . , m2 = 0.7909651413 . . . and m3 = 0. On the other hand, when the fifth mass is inside the rhombus the admissible values of a go from −1/2 to −0.2495724426 . . . . Rhombus plus one 7 Figure 3. The value of the masses m1 (black), m2 (blue) and m3 (red) when b = 1/2, that is, a square plus one config- uration. The admissible values of a go from 0.4402277051 . . . to 0.5926746107 . . . when the fifth mass is outside the square, and from −1/2 to −0.2495724426 . . . when the fifth mass is inside the square. At a = −0.2495724426 . . . the masses are m1 = 0.7268534679 . . . , m2 = 0 and m3 = 6.814939450 . . . . Recall that when the fifth mass is located at the intersection of the two diagonals of the square (a = −1/2) the configuration is central for any value of m1 and m2 = m3 = 1. See [7]. From Figure 3 the question about the number of central configurations for a given positive masses is also answered numerically when the rhombus becomes a square. Clearly, given a normalized mass vector (m1 , m2 , m3 , 1, 1) of the five-body problem, if a square plus one central configuration exists, such a configuration is unique. In the next section we settle the inverse question in the rhombus plus one central configurations. 3. The uniqueness of rhombus plus one central configurations In this section we prove that given a rhombus plus one configuration the nor- malized mass vector (m1 , m2 , m3 , 1, 1) of the five-body problem that makes the configuration central is unique. Theorem 3.1. Given a configuration of the planar five-body problem, where four bodies m2 , m3 , m4 and m5 are located at the vertices of a rhombus and the fifth body m1 is placed on the diagonal containing m2 and m3 , then the positive normalized mass vector (m1 , m2 , m3 , 1, 1) that makes the configura- tion central is unique, as long as the particular case where the fifth body is at the intersection of the two diagonals of the rhombus is excluded. 8 J. Lino Cornelio, M. Alvarez-Ramı́rez and Josep M. Cors Proof. Equations (2.5) can be written as a non homogeneous system Bm = b where   0 (R24 − R12 )∆124 (R24 − R13 )∆134 B =  (R12 − R14 )∆124 0 (R24 − 1)∆243  , (R13 − R14 )∆134 (1 − R24 )∆234 0     (−R14 + R45 )∆145 m1 b =  (−R24 + R45 )∆245  , m =  m2  . (R24 − R45 )∆245 m3 The existence and uniqueness of m1 , m2 and m3 (positive or not) depends on whether the determinant of B, det(B) is zero or not, where det(B) = ∆124 ∆134 ∆234 (R24 − 1)(R14 − R24 )(R12 − R13 ). (3.1) Clearly ∆234 6= 0, as well as the other two areas ∆124 and ∆134 in (3.1), otherwise there would be collision between m1 and m2 , which is excluded. Again R14 = R24 implies collision between m1 and m2 , along with R12 = R13 implies that m1 is located at the intersection of the two diagonals of the rhombus. Last case is excluded and was studied by Gidea and Llibre in [7]. It corresponds to case (i) of the proof of their Theorem 1 (a). Then the normalized mass vector that solve equations (2.5) will be unique, as long as R24 6= 1, that is, the sides of the rhombus are equal to the diagonal containing the fifth body, that is, when the rhombus becomes a diamond. Claim: if r24 = 1 the equations (2.5) has no solution. √ When r24 = 1, we have r45 = 3. Indeed, by adding equations f24 and f34 , from (2.5), the resulting equation is (R12 − R14 )∆124 + (R13 − R14 )∆134 = 0, (3.2) that can be written as a function of a. At this point we distinguish two cases. The case where m1 is outside the rhombus (a > 0), and where is inside (a < 0). Equation (3.2) when a > 0 becomes 3 (1/2 + a)(1 + a)2 a2 − (a2 + a + 1/2)(a2 + a + 1) 2 3 = 0. (3.3) 2a2 (1 + a)2 (a2 + a + 1) 2 Trivially the denominator in this last expression do not vanish when a > 0. So we have to prove that the numerator have the same sign for all positive values of a. In order to eliminate the square root of the numerator in (3.3), we equals the numerator to zero and then we square it conveniently to obtain a polynomial equation of degree 8, 15a8 + 60a7 + 115a6 + 135a5 + 109a4 + 63a3 + 26a2 + 7a + 1 = 0. Rhombus plus one 9 Since all coefficients are positive, from Descartes’ rule of signs, see [22], there are no positive roots, and consequently equation (3.3) is not satisfied for any positive value of a. We now compute (3.2) for the case where m1 is inside the rhombus (a < 0), and is found to be 3  (1/2 + a) (a2 + a + 1) 2 + a2 (1 + a)2 3 = 0. (3.4) 2a2 (1 + a)2 (a2 + a + 1) 2 To obtain the zeros of this equation, if any, we proceed as in the previous case, arriving at the polynomial of degree 8 given by −a8 − 4a7 − 5a6 − a5 + 5a4 + 7a3 + 6a2 + 3a + 1. Now using the Sturm’s theorem in the interval (−1/2, 0) we check that has no root on it. Remember that from the symmetry of the configuration is enough to study when the fifth body, m5 , is located between the body m2 and the intersection of the two diagonals of the rhombus. Also notice that the denominator of equation (3.2) when a < 0 is equal to one obtained in the case a > 0, and as before is different from zero in the interval (−1/2, 0).  4. Proof of Theorem 2.1 The sides of the rhombus are nearly equal to the diagonal of the rhombus, r23 = 1, containing the fifth mass, m1 , when r24 ≈ 1. From the proof of Theorem 3.1 we know that the determinant (3.1), that appears at the de- nominator of the expressions of the masses (2.6), is different from zero unless r24 = 1. Also is easy to check that such a determinant is negative when r24 is sufficiently close to one and r24 − 1 < 0. We are going to prove that when r24 = 1 there exists an open interval of values of r12 such that the numerators N1 , N2 and N3 of m1 , m2 and m3 , respectively, in (2.6) have constant sign. Therefore there will exist values of r12 such that m1 > 0 and the other two masses, m2 and m3 , are unbounded with positive sign. Finally, using the continuity of the values of the masses with respect to the mutual distances, for values of r24 sufficiently close to one and less than one, there will exist an open set of dimension two in a neighborhood of the r24 = 1 where the three masses are positive. Setting r24 = 1 and r12 = |a|, all other mutual distances and signed areas that appear in N1 , N2 and N3 in equations (2.6) can be only written as a function of real variable a. From now on, we will distinguish when the fifth mass is outside (a > 0) or inside (−1/2 < a < 0) the rhombus, as usual. • Case a > 0 10 J. Lino Cornelio, M. Alvarez-Ramı́rez and Josep M. Cors Numerator, N1 , of m1 in equations (2.6) is equal to √ (9 − 3)(2a5 + 5a4 + 4a3 − a2 − 2a − 1) N1 = − . (4.1) 48a2 (1 + a)2 Let P (a) be the above polynomial of degree 5. Applying the Descartes’s rule of signs, P (a) has only one positive root. Moreover, P5 (6/10) < 0 and P5 (7/10) > 0. Therefore P5 (a) > 0, and consequently N1 < 0, for all a > 7/10. Numerators N2 and N3 in equations (2.6) are equal to √ (9 − 3)(a2 + 3a + 3)F (a) N2 = 3 , 24a(1 + a)5 (a2 + a + 1) 2 √ (9 − 3)(1 − a)F (a) N3 = 1 , 24a5 (1 + a)2 (a2 + a + 1) 2 3 where F (a) = (1/2 + a)(1 + a)2 a2 − (a2 + a + 12 )(a2 + a + 1) 2 , which is equal to the numerator of the left hand side equation (3.3), and we know that has no positive roots. Since F (0) < −1/2, then F (a) < 0 and N2 < 0 for all a > 0 and N3 < 0 only for all 0 < a < 1, due to the factor (1 − a) present in the expression of N3 . • Case −1/2 < a < 0 Numerator, N1 , of m1 in equations (2.6) is equal to √ (9 − 3)(2a + 1)(a4 + 2a3 + a2 + 1) N1 = − < 0. (4.2) 48a2 (1 + a)2 On the other hand N2 and N3 have similar expressions √ ( 3 − 9)(1/2 + a)(a2 + 3a + 3)G(a) N2 = − 3 , 24a(1 + a)5 (a2 + a + 1) 2 √ ( 3 − 9)(a2 − a + 1)G(a) N3 = 3 , 24a5 (1 + a)(a2 + a + 1) 2 3 where G(a) = a2 (1 + a)2 + (a2 + a + 1) 2 , and again coincides with the numerator of the left hand side equation (3.2) when a < 0, and we know that has no roots in the interval (−1/2, 0). Since G(0) = 1 > 0 then G(a) > 0 and N2 < 0 in the interval (−1/2, 0). Notice that N3 > 0 when −1/2 < a < 0 and apparently the mass m3 is unbounded, but negative. However, m3 really has positive sign, since ∆124 , that has negative sign when the fifth mass is inside the rhombus and is the only negative factor in the determinant (3.1), do not appear in the denominator of m3 .  In summary we have proved that m1 , m2 and m3 are positives for all values of r12 between 7/10 and 1, when the fifth body, m1 , is outside the rhombus, and for all values of r12 between 0 and 1/2, when the fifth body, Rhombus plus one 11 m1 , is inside the rhombus, if r24 is sufficiently close to one but smaller. See Figure 2. 5. Proof of Theorem 2.2 Following the ideas in the previous section we are going to prove that when r45 = 1, that is, when the diagonals of the rhombus are equal, the masses m1 , m2 and m3 in (2.6) are positive for some values of r12 . In what follows we will show that in (2.6) when r45 = 1 the numerators and denominators have the same sign, respectively. Once again, we will distinguish when the fifth mass is outside (a > 0) or inside (−1/2 < a < 0) the square. • Case a > 0 Numerator, N1 , of m1 in equations (2.6) is equal to P1 (a) N1 = − √ , 16a2 (1 + a)2 (a2 + a + 12 ) 4a2 + 4a + 2 where √ P1 (a) = (14a5 + 35a4 + 28a3 + 9a2 + 2a + 1 − (4a2 + 4a + 2) 2 ) 1 √ (a2 + a + 1/2)(4a2 + 4a + 2) 2 − 8(1 + a)2 (1/2 + a)( 2 − 1/2)a2 . In order to study the zeros of N1 we use the same techniques as in Section 3. P1 (a) can be written as a polynomial of degree 32, P32 (a), with two number of sign changes. So from Descartes’ rule of signs the number of zeros of P32 (a) is zero or two. √ Since P (3/10) √ √ √ 32 < 0, P32 (2/5) > 0, P32 (1/2) < 0 and N1 (1/2) = 5 9 2 + 2 25 2 5 − 83 5 72 − 25 < 0, follows that N1 (a) < 0, and consequently m1 > 0, for all a > 1/2. Numerator, N2 , of m2 in equations (2.6) is equal to P2 (a) N2 = − √ , 16a2 (1 + a)5 (2a2 + 2a + 1)3 4a2 + 4a + 2 where P2 (a) = ((−48a10 − 288a9 − 808a8 − 1352a7 − 1516a6 − 1248a5 − 818a4 √ − 430a3 − 164a2 − 38a − 4) 2 + 112a11 + 808a10 + 2720a9 + 5668a8 + 8124a7 + 8430a6 + 6504a5 + 3769a4 + 1622a3 1 + 495a2 + 96a + 9)(4a2 + 4a + 2) 2 − 224a2 (1 + a)5 (1/2 + a)(a2 + a + 1/2)2 . Again, P2 (a) can be written as a polynomial √ of degree √ 42 with no zeros 15737 2 7 5 66931 in the interval ∈ (0, 1). Since N2 (1/2) = 30375 + 25 − 30375 < 0, then N2 (a) < 0, so that m2 > 0 for all 0 < a < 1. 12 J. Lino Cornelio, M. Alvarez-Ramı́rez and Josep M. Cors Numerator, N3 , of m3 in equations (2.6) is equal to P3 (a) N3 = − √ , 48a5 (1 + a)2 (2a2 + 2a + 1)3 4a2 + 4a + 2 where P3 (a) = ((−144a10 − 576a9 − 1128a8 − 1224a7 − 540a6 + 288a5 + 582a4 √ + 426a3 + 186a2 + 48a + 6) 2 − 336a11 − 1272a10 − 2400a9 − 2796a8 − 2244a7 − 1410a6 − 816a5 − 459a4 − 234a3 1 − 93a2 − 24a − 3)(4a2 + 4a + 2) 2 + 672a5 (1 + a)2 (1/2 + a)(a2 + a + 1/2)2 . By using the same method as before, we proceed to find the zeros of P3 (a). Thus we transform its expression to a polynomial of degree 42. Applying Sturm’s theorem in the interval (0, 1), we obtain two zeros, one between 1/2 and 51/100 and another between 59/100 and 3/5. √ On 59238755508940350 1247292558547000 2 the other hand, N3 (55/100) = 24506348699914631 − 790527377416601 − √ 7350 541 51 59 292681 < 0. Then N3 (a) < 0 and m3 > 0 for all 100 < a < 100 . In summary we have proved that masses m1 , m2 and m3 are posi- tives as long as the values of a (the distance between m1 and m2 ) belong 51 59 to the interval ( 100 , 100 ). So there exist square plus one central configu- rations in the five-body problem when the fifth mass is located outside 51 59 the square and r12 ∈ ( 100 , 100 ). • Case −1/2 < a < 0 Using similar arguments to the previous case, (a > 0), it can be proved that m1 and m3 are always positive while m2 is only positive for values less than −1/4. So there exist square plus one central configura- tions in the five-body problem when the fifth mass is located inside the square and r12 ∈ (1/4, 1/2).  We would like to remark, first, that the bounds obtained along the proof of the theorem for the admissible values of r12 are not sharp. 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ORCID: 0000-0001-7571-3716 e-mail: kornelio 85@hotmail.com Martha Alvarez-Ramı́rez Departamento de Matemáticas UAM-Iztapalapa, México City, México. ORCID: 0000-0001-9187-1757 e-mail: mar@xanum.uam.mx Josep M. Cors Departament de Matemàtiques, Universitat Politècnica de Catalunya, 08242 Manresa, Catalonia. ORCID: 0000-0002-9803-8490 e-mail: cors@epsem.upc.edu