Convex Central Configurations of the 4-Body Problem with Two Pairs of Equal Adjacent Masses

We study the convex central configurations of the 4-body problem assuming that they have two pairs of equal masses located at two adjacent vertices of a convex quadrilateral. Under these assumptions we prove that the isosceles trapezoid is the unique central configuration.


Introduction and Statement of the Main Results
The classical Newtonian n-body problem studies a system formed by n punctual bodies with positives masses m 1 , . . . , m n and position vectors r 1 , . . . , r n in R d , d = 2, 3, interacting under the Newton's gravitational law [20]. The equations of motion of this problem arë for i = 1, . . . , n, where r i j = |r i −r j | is the Euclidean distance between the bodies at r i and r j , and t is the independent variable called time. Taking the unit of mass conveniently we can assume that the gravitational constant G = 1 in (1). An interesting class of particular solutions of the n-body problem (1) are the homographic solutions in which the shape of the configuration is preserved as time varies. The first homographic solutions were found by Euler [10] and Lagrange [13] in the 3-body problem.
We say that at a given instant t = t 0 the n bodies are in a central configuration if for all i = 1, . . . , n there exists a constant λ = 0 such thatr i = λ(r i − c) where c is the center of mass of the n bodies, that is c = 1 m 1 + · · · + m n n j=1 m j r j . Such configurations are closely related with homographic solutions. In fact, the configuration of bodies at any time in a homographic solution is a central configuration. For more details see, for instance, [19,22,23] and [25].
To find a central configuration is reduced to find a solution of a nonlinear system of equations, because from Equation (1) and the definition of a central configuration, we must solve the system of equations for i = 1, . . . , n. Equation (2) are called the equations of the central configurations. Two central configurations (r 1 , . . . , r n ) and (r 1 , . . . ,r n ) of the n bodies are related if we can pass from one to the other through a dilation and a rotation (centered at the center of mass). Thus we can study the classes of central configurations defined by the above equivalence relation.
Taking into account this equivalence relation we have exactly five classes of central configurations in the 3-body problem. The finiteness of the number of central configurations performed by n bodies with positive masses is a question posed by Chazy [6], Wintner [25] and reformulated to the planar case by Smale [24]. For n = 4 this problem has an affirmative answer given by Hampton and Moeckel [12]. Recently, another proof of this finiteness for n = 4 has been given by Albouy and Kaloshin, see [4], where some results on the finiteness for n = 5 are also given. However, the problem on the finiteness of the classes of central configurations remains open for n 5.
In the planar 4-body problem a configuration is convex if there is not a body located in the interior of the convex hull of the other three, otherwise the configuration is concave, see Figure 1.
In [16], a landmark for the study of convex central configurations in the planar 4-body problem, MacMillan and Bartky proved the following existence theorem: MacMillan and Bartky provided information on the admissible shapes of the 4-body convex central configurations.

Theorem 2. In a convex 4-body central configuration
(i) the diagonals are greater than all exterior sides, and (ii) the biggest side is opposite to the smallest one.
MacMillan and Bartky also provided information on the isosceles trapezoid central configuration in the 4-body problem assuming the isosceles trapezoid symmetry in the hypotheses. In [16] the authors showed that there exists a curve of central configurations connecting the equilateral triangle central configuration and the square central configuration in which the mass ratio m/μ is strictly increasing.
In Llibre [15], assuming that the planar central configurations of the 4-body problem with equal masses have some symmetry, showed numerically that the 4body problem with equal masses have 50 classes of central configurations. Later on Albouy, in [1,2], proved that such symmetries always exist and provide an analytical proof of the 50 classes.
Albouy et al. [3] studied some symmetric central configurations in the 4-body problem. In particular they showed that in a convex planar central configuration of 4 bodies if two opposite masses are equal then there exists an axis of symmetry passing through the other two masses. The converse of this statement is also true. This kind of central configurations are called kite central configurations. Several papers were written studying kite central configurations and their properties, see [5,14,17,18] and references therein. In [21] Perez-Chavela and Santoprete proved that the unique convex planar central configuration with two opposite equal masses is the kite central configuration or the rhombus central configuration when the other two masses are also equal.
Albouy et al. [3] stated the following conjecture: There is a unique convex planar central configuration having two pairs of equal masses located at the adjacent vertices of the configuration and it is an isosceles trapezoid.
Recently Corbera and Llibre [7] proved this conjecture assuming that two equal masses are sufficiently small.
In this paper we prove Conjecture 1 for all values of the masses. We consider the 4-body problem in the plane with masses m 1 = m 2 and m 3 = m 4 located at This article is organized as follows. We prove Theorem 5 in Section 3. In Section 2 we prove some preliminary results used in the proof of Theorem 5.

Preliminary Results
In this section we present a set of equations equivalent to the central configuration equations. The following result is well known (see for instance [11]): Note that i jk is twice the oriented area of the triangle formed by the bodies at r i , r j and r k (see [11]). The n(n − 1)/2 Equations (3) are called the Dziobek-Laura-Andoyer equations or simply the Andoyer equations.
Using the notation of Lemma 1 we can state the main theorem of [3]. Of course an analogous result to Theorem 6 is true having a symmetry with respect to the other diagonal.
Without loss of generality we can assume m μ. Moreover since we consider convex configurations, by the Perpendicular Bisector Theorem (see [19]), we also can assume that x 4 < 0, x 3 > 0, y 3 > 0 and y 4 > 0. See Figure 2.
The six Andoyer equations for our problem are Since m > 0 and μ > 0, if we define  (8) we have m = μ. Then, from Theorem 6 the configuration must be a square with four equal masses at the vertices, which is a type of isosceles trapezoid. Hence in what follows we can assume that R 14 = R 34 and R 12 = R 23 .
Again from Equation (5), if R 34 < R 14 then R 23 < R 12 , or equivalently, if r 34 > r 14 , then r 23 > r 12 , which implies that 124 < 234 . So, from Theorem 6 it follows that m > μ, in contradiction with our hypothesis. Thus we must have R 34 > R 14 which implies that R 23 > R 12 . A similar argument can be used to show that we must have R 34 > R 23 which implies that R 14 > R 12 . In order to have a central configuration, taking out the case of the square and using Theorem 2 the following inequalities must hold: Since r 12 = 2, inequalities (12) imply that Without loss of generality we can assume that y 4 y 3 . Then from Theorem 6 the following inequalities must hold: The explicit expressions for these areas are the following: In the rest of this section we consider the hypotheses of Theorem 5. Thus the configuration is like described in Figure 2 satisfying (12). So we have the first lemma.

Lemma 2.
Under the assumptions of Theorem 5, if y 4 = y 3 then the configuration must be an isosceles trapezoid.
Proof. If y 3 = y 4 , using (13), Equations (4) and (9) can be written as Since the areas are positive, these equations are satisfied if and only if R 13 = R 24 and R 14 = R 23 . But in this case the configuration is an isosceles trapezoid.
Thus henceforth consider y 3 > y 4 . Proof. First consider the inequalities (12), in which we must have r 14 > r 34 , or equivalently In order that this inequality be satisfied for x 3 > 0 it is necessary that (x 4 , y 4 ) belongs to the region (open and connected set) A 1 which is determined by the parabola y 2 4 + 2x 4 + 1 = 0 and the circles r 14 = 2 and r 24 = 2, see Figure 3. Now consider (x 4 , y 4 ) fixed. Computing the partial derivative of G, defined in (10), with respect to x 3 we get Computing the partial derivative of G with respect to y 3 we get or equivalently, The last part of the above inequality arises from the fact that the points (x 3 , y 3 ) must belong to the region A 2 symmetric to A 1 determined by the parabola y 2 3 − 2x 3 + 1 = 0 and the circles r 23 = 2 and r 13 = 2, where we have Q 23 > Q 13 and 2y 2 3   Since y 3 > y 4 we must have Consider the point (x 3 , y 3 ) = (2 + x 4 , y 4 ) in the circle r 23 = r 14 (remember r 4 is fixed). Thus However, at this point G is positive since the point (x 3 , y 3 ) = (2 + x 4 , y 4 ) belongs to the interior of the circle r 13 = r 24 (remember r 4 is fixed). Notice that the gradient of G remains pointing northwest as in Lemma 3, because x 3 ∈ (0, 2 + x 4 ) ⊂ (0, 1). So G > 0 for all points in B 3 , which is the subset of A 2 with x 3 ∈ (0, 2 + x 4 ) and y 3 > y 4 . See Figure 5. Thus f 12 > 0. From the above calculations we only need to study the case where x 4 ∈ [−1, 0) and x 3 ∈ (−x 4 , 2 + x 4 ). In order to satisfy (12) with x 3 > −x 4 we need that Thus it is necessary that (x 4 , y 4 ) belongs to the region A 3 determined by the hyperbola y 2 4 − 3x 2 4 + 2x 4 + 1 = 0 and the circles r 14 = 2 and r 24 = 2, see Figure 6. The intersection of the hyperbola y 2 4 − 3x 2 4 + 2x 4 + 1 = 0 with the circle r 24 = 2 is the point Thus we must have y 4 > y c . Since we are considering values of x 3 ∈ (−x 4 , 2 + x 4 ) and y 3 > y 4 , for a fixed pair (x 4 , y 4 ) the region of interest for (x 3 , y 3 ) is the region B 4 defined by the points of A 2 where x 3 ∈ (−x 4 , 2 + x 4 ) and y 3 > y 4 , see Figure 7. Now define the region A 4 bounded by the hyperbola y 2 3 − 3x 2 3 − 2x 3 + 1 = 0 and the circles r 23 = 2 and r 13 = 2, see again Figure 7. Note that the points on the straight line x 3 = −x 4 between the line y 3 = y 4 and the circle r 23 = 2 always belong to A 4 . Note also that in region B 4 we have r 13 > r 24 .
For a fixed pair (x 4 , y 4 ), consider the function The zero level set of this function, denoted by T 0 , is the straight line passing through  In the region B 4 the coefficient of m is always negative, so in order to satisfy this equation the coefficient of μ must be positive. We define the following function: In Lemmas 5 and 7 we use the sets defined below: Thus, T 0 in B 41 and T 0 in B 42 ; See Figure 8. Proof. Consider the function L restricted to T 0 . Note that L(−x 4 , y 4 ) = 0. Using Equation (13) and the definition of R i j the expression (15) restricted to T 0 can be written as Solving T = 0 for x 3 and replacing the result in the last equation we have Note that the expression between the brackets is a function P of y 3 whose graph is a parabola concave downward. We will compare the position of the roots of P with y 3 = y 4 in order to study the sign of L restricted to T 0 . Evaluating P at y 3 = y 4 we get 2x 4 )).
From the last equation, define The zero level set of L 1 is given by (1 + x 4 ) 2 .
. Evaluating the derivative of P with respect to y 3 at y 3 = y 4 we get From the last equation, define The zero level set of K 1 is given by Thus the zero level set of K 1 for y 4 > 0 is a function of x 4 passing through the point (0, √ 3) and going to +∞ when x 4 goes to −(1/2) + . See Figure 10. In conclusion, K 1 is negative in the region A 3 and this implies that L 1 is negative in the region A 3 . So the function L restricted to T 0 is always negative when y 3 > y 4 .
To see that the function L is negative in B 41 we compute the partial derivative of L with respect to x 3 Denote the first two terms in the above expression by the following function: This function vanishes at the point (x 3 , y 3 ) = (−x 4 , y 4 ) and its gradient points southwest in B 4 . In fact, Thus the partial derivative of L with respect to x 3 is always negative when y 3 > y 4 . See Figure 9. So the function L is always negative in B 41 . In short, the equations f 14 = 0 and f 23 = 0 are not satisfied simultaneously in B 41 .
For a fixed pair (x 4 , y 4 ) ∈ A 3 , define the following two functions: In the next lemmas we prove some properties of the above functions.

Lemma 6. Under the assumptions of Theorem 5, if (x 3 , y 3 ) in B 4 then the function H 2 is negative.
Proof. Note that H 2 (−x 4 , y 4 ) = 0. The derivative of H 2 with respect to x 3 is given by while the derivative of H 2 with respect to y 3 is given by The zero level set of this last derivative is formed by the two straight lines  Fig. 11. The numerator of a 0 − a 1 is negative in the cone defined by the straight lines (16) containing the region B 4 The zero level set of this derivative is formed by the two straight lines In the region B 4 the derivative of H 1 with respect to y 3 is negative. Thus the zero level set of H 1 is a function of x 3 which has implicit derivative given by The slope of the straight line T 0 is given by

Now we study the function
The denominator of the above expression is negative in B 4 according to the previous analysis. The numerator of the above expression vanishes on the straight lines See Figure 11. Thus in the set B 4 the difference a 0 − a 1 is positive.
Since y 3 > y 4 we will prove that H (x 3 , y 3 ) < 0 in B 42 by proving that h(x 3 , y 3 ) < 0 in this set. Note that the zero level set of h is the straight line given by This straight line always pass through (x 3 , y 3 ) = (−x 4 , −y 4 ). Thus in order to complete the proof we need to analyze the slope of this straight line which is The numerator of this last expression is positive so the sign of the slope is given by the denominator The zero level set of this expression for y 4 > 0 is a function of x 4 given by whose graph is depicted in Figure 12. Thus for all points in the region A 3 the sign of the slope is negative. Therefore the function H is always negative in B 42 and this implies that the equation f 34 = 0 is not satisfied in B 42 .

Proof of Theorem 5
In this section we give the proof of Theorem 5. We will prove that the symmetry in the masses implies the symmetry in the positions in order to satisfy all the Andoyer equations. Thus we will be under the hypotheses of MacMillan and Bartky Theorem, that is of Theorem 3. In other words, if we have symmetry in the masses and the positions then the uniqueness follows from that theorem. From the previous analyses a necessary condition to satisfy all the Andoyer equations is, the symmetry (x 3 , y 3 ) = (−x 4 , y 4 ), that is, the quadrilateral must be an isosceles trapezoid, see Theorem 3. This completes the proof of Theorem 5. For a modern and very well written work about the isosceles trapezoid central configuration, see Cors and Roberts [8].