# Central Configurations and Action Minimizing Orbits in Kite Four-Body Problem

Authors: B. Benhammouda; A. Mansur; M. Shoaib; I. Szücs-Csillik; D. Offin
Journal: Advances in Astronomy
 (2020)

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## Abstract

In the current article, we study the kite four-body problems with the goal of identifying global regions in the mass parameter space which admits a corresponding central configuration of the four masses. We consider two different types of symmetrical configurations. In each of the two cases, the existence of a continuous family of central configurations for positive masses is shown. We address the dynamical aspect of periodic solutions in the settings considered and show that the minimizers of the classical action functional restricted to the homographic solutions are the Keplerian elliptical solutions. Finally, we provide numerical explorations via Poincaré cross-sections, to show the existence of periodic and quasiperiodic solutions within the broader dynamical context of the four-body problem.


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## Body

## 1. Introduction

To understand the dynamics presented by a total collision of the masses or the equilibrium state of a rotating system, we are led to the concept of central configurations. A configuration ofn bodies is central if the acceleration of each body is a scalar multiple of its position [1–4]. Let ri∈ℝ2 and mi,i=1,…,n, denote the position and the mass of the ith body, respectively. Also, let rij=ri−rj represent the distance between the ith and jth bodies. An n-body system forms a planar noncollinear central configuration [5, 6] if(1)fij=∑k=0,k≠i,jn−1mkRik−RjkΔijk=0,where Rij=rij−3 and Δijk=ri−rj∧ri−rk represent the area of the triangle determined by the sides ri−rj and ri−rk.The four-body problem can be considered from two different perspectives. The perturbative approach where we study the dynamical aspects as a perturbation of the three-body dynamics and assume that one of the masses is vanishingly small, or the global approach where we allow the masses to vary independently and stay positive. In this paper, we take the global approach and will study analytically the problem of central configurations and their dynamical aspects.The computation of central configurations is a difficult problem forn≥4. To overcome this difficulty, symmetries or other restriction methods are used to reduce the number of variables and obtain partial answers; see, for example, Cors and Roberts [7]; Albouy et al. [8]; Shoaib et al. [9]; Érdi and Czirják [10]. In this paper, we consider a four-body problem with one axis of symmetry so that the four different masses make a convex or concave kite.Since the classification of central configurations as one of the problems for the 21st century by Smale [11], it has attracted a lot of attention in recent years and has helped in the understanding of the n-body problem [12–23]. Ji et al. [24] and Waldvogel [25] study a rhomboidal four-body problem with two pairs of masses and use Poincaré sections to find regions of stability for the rhomboidal four-body problem. In addition, Waldvogel [25] also takes advantage of the simplicity of the equations and study its collisions and escape manifolds. Yan [26] considers the same problem for four equal masses and studies the linear stability of its periodic orbits. One of our results will consider the same model but with only one pair of masses. We will analytically derive regions of central configurations and will also investigate the existence of periodic orbits. Mello and Fernandez [27] prove the existence of kite central configurations for four- and five-body problems on a circle. In one special case, our rhomboidal model is similar to their model for which we also identify a number of periodic orbits and discuss its action minimizing orbits. Gordon [28] has proved that the elliptic Keplerian orbit minimizes the Lagrangian action of the two-body problem with periodic boundary conditions. It is also known that the Eulerian and Lagrangian elliptical solutions for the planar three-body problem are the variational minimizers of the Lagrangian action functional [29, 30]. In the study by Mansur and Offin [31]; Mansur et al. [32]; Mansur et al. [33], the authors have extended these ideas to prove that the homographic solutions to the constrained parallelogram four-body problem are the variational minimizers of the Lagrangian action functional. In this paper, we prove that the minimizers for the action functional restricted to the homographic solutions are the Keplerian elliptical solutions for the four-body problem with three equal and unequal masses. Perez-Chavela and Santoprete [13] show the existence of kite central configurations for a pair of symmetric masses and show that such a configuration must always possess a symmetry. Similarly, Celli [34] proves the existence of planar diamond and trapezoidal central configurations for two pairs of equal masses. Corbera and Llibre [35] give a complete classification of the same problem and show that this setup has exactly 34 different classes of central configurations.More recently, Deng et al. [36] and Corbera et al. [37] prove that any four-body setup with perpendicular diagonals must be a kite [35, 38]. Santoprete [39] studies a four-body problem with a pair of equal masses and a pair of parallel opposite sides and show that if the opposite masses are equal, then the four-body arrangement must have a line of symmetry and will be a kite.The paper is organized as follows: Section2 discusses the equations of motion for the four-body problem. Section 3 discusses the existence of central configurations and the action minimizing orbits for the four-body problem where three masses are equal and arranged at vertices of an isosceles triangle and the fourth mass is on the axis of symmetry. In Section 4, we discuss the variational techniques where the action functional corresponding to these family of solutions is shown to be a minimizer. Section 5 discusses the existence of central configurations and the action minimizing orbits for the four-body problem where two symmetric masses are equal on the horizontal axis and two nonequal masses are on the vertical axis.


## 2. Equations of Motion

Consider four positive point massesm0, m1, m2, and m3 having position vectors ri and interbody distances rij. For a general four-body setup, equation (1) gives the following six central configuration equations when n=4:(2)f01=m2R02−R12Δ012+m3R03−R13Δ013=0,f02=m1R01−R21Δ021+m3R03−R23Δ023=0,f03=m1R01−R31Δ031+m2R02−R32Δ032=0,f12=m0R10−R20Δ120+m3R13−R23Δ123=0,f13=m0R10−R30Δ130+m2R12−R32Δ132=0,f23=m0R20−R30Δ230+m1R21−R31Δ231=0.Lemma 1.
Consider a four-body problem with massesm0, m1, m2, and m3 and position vectors r0=0,c, r1=−1,−a, r2=0,b, and r3=1,−a, where a>0 and b>c, then(a)
The symmetric massesm1 and m3 are equal.(b)
The central configuration equations are(3)f01=m1R03−R13Δ013+m2R02−R12Δ012,f12=m0R10−R20Δ120+m1R13−R23Δ123.Proof.
Consider four positive massesm0, m1, m2, and m3 with position vectors r0=0,c, r1=−1,−a, r2=0,b, and r3=1,−a, where a>0 and b>c. Using the definitions of Rij, Δijk, and rii=0,1,2,3,4, we obtain(4)R01=R03=R30=1a+c2+13/2,R13=18,R02=1b−c3,R12=R21=R32=1a+b2+13/2,with(5)Δijk=−Δjik=−Δikj=−Δkji,Δijk=Δjki=Δkij,Δijk=0,ifi=jori=korj=k,Δ012=Δ023=c−b,Δ013=2a+c,Δ123=−2a+b.
Using the symmetry of the problem and the relations (4) and (5), it is trivial to see that(6)f02=m1−m3R01−R21Δ021=0.
SinceΔ021≠0 and R21≠R01, therefore m3=m1. This completes the proof of Lemma 1 (a).
From the geometry of the problem,Δm1m0m3 and Δm1m2m3 are both isosceles, and therefore R01=R03, R12=R32, and Δ032=Δ012 and hence f03=f01. This also implies that f13≡0. By a similar argument, it can be shown that f23=f12. This leaves two independent equations f01 and f12 from the set of equations given in (2). This completes the proof of Lemma 1.


## 3. Three Equal Masses at the Vertices of a Triangle and a Fourth Mass on the Axis of Symmetry

In this section, we consider a four-body problem where three equal massesm1=m2=m3 are arranged at the vertices of an isosceles triangle and a fourth mass m0 is on the axis of symmetry as shown in Figure 1. We start by showing the existence of central configuration for a concave kite four-body problem and then explicitly find regions where such a configuration exists for positive masses. We also discuss the action minimizing orbits for this particular problem.Figure 1
Concave kite four-body configurations with three equal massesm1=m2=m3.


### 3.1. Central Configurations

Theorem 1.
Consider four point massesm0 and m1=m2=m3 having position vectors r0=0,0, r1=−1,−a, r2=0,b, and r3=1,−a, where a>0 and b>0. Then, there exists a unique mass ratio μ0a,b:(7)μ0=m0m1=a+b2+13/2−8a+bb2a2+13/24a2+13/2−b3a+b2+13/2,such that r=r0,r1,r2,r3 is a central configuration for μ0>0 in Rμ0a,b subject to the constraint ga,b=0. The region Rμ0a,b and the constraint ga,b are given below:(8)Rμ0a,b=a,b23<b<3∧0<a<3−b∨a>b2−1∨b≥3∧a>b2−1,ga,b=−b1b3−1α+2a−18+1β=0,where α=a+b2+13/2 and β=a2+13/2.
Before we attempt to prove Theorem1, we will need help from the following lemmas.Lemma 2.
The functiong defined by (8) is negative for all 0<a<3 and 0<b≤B=2/33/7+23/71/4≃1.13.Proof.
Let0<a<3 and 0<b≤B, then we have(9)ga,bb=1α+2ab1β−4+ab24b3.
Consider the equation of the straight line segment that lies in the first quadrant:(10)x+2aby−4+ab24b3=0.
For positivex and y,(11)x+2aby−4+ab24b3<0,is equivalent to(12)x<4+ab24b3,y<4+ab28ab2.
To prove thatga,bb−1<0, we need to show that(13)1α<4+ab24b3,1β<4+ab28ab2.
For1/α<4+ab2/4b3, it is trivial to see that(14)1α=11+a+b23/2<1b3<4+ab24b3.
Similarly, to show that1/β<4+ab2/8ab2, we need(15)β>8ab24+ab2,(16)or1+a23>8ab24+ab22.
Inequality (16) is equivalent to(17)1+a234+ab22−64a2b4>0.
Expanding the left-hand side of (17), we obtain the following:(18)pa,b=b4a8+8b2a7+16+3b4a6+24b2a5+48+3b4a4+24b2a3+48−63b4a2+8b2a+16.
For positivea and b,(19)pa,b>48a4+48−63b4a2+16,≥48a4+48−63B4a2+16,=48a2−132≥0.
Consequently,ga,b<0 for 0<a<3 and 0<b≤B.
This completes the proof of Lemma2.Lemma 3.
The partial derivative ofg defined by (3) satisfies gba,b>3/α2, for all a>0 and b>0.Proof.
For positivea and b, we have(20)gba,b=2b3+α−bαbα2,=2b3+α−3ba+bγα2,where γ=1+a+b21/2 and αb is the derivative of α w.r.t. b.
Sinceb3<α for all a>0 and b>0, then(21)gba,b>2α+α−3ba+bγα2.
Using the fact thatα=γ3 and γ2=1+a+b2, we get(22)gba,b>31+a+b21+a2+abγ6.
Sincea and b are positive, therefore(23)gba,b>3α2.
This completes the proof of Lemma3.Remark 1.
Numerically, it is easy to show thatga,2=0 when a=0.14 and a=1.32. For a∈I=0.14,1.32, ga,1<0 and ga,2>0, when a∈I, see Figure 2. Therefore, by intermediate value theorem ga,b has at least one root b∈1,2 when a∈I.Figure 2
Graph ofga,b for values of b=0.5,1,and1.5.Lemma 4.
Consider the functiong defined by (3). Then, for any a0∈I=0.14,1.32 there exists an interval U⊂I containing a0 and an interval V⊂1,2 containing b0 such that there is a unique continuously differentiable function b=ψa defined on U with b∈V that satisfies ga,b=0.Proof.
Leta0 be any number in the interval I=0.14,1.32; then, using Lemma 3, we have ga0,1<0. Then, numerically, one can check that ga0,2>0 for a0∈I. Thus, by the mean value theorem, there exists at least one b0∈1,2, solution of ga0,b=0. By Lemma 3, gba0,b>0 for all b>0, hence the solution b0 is unique. Since g has continuous partial derivatives and ga0,b0=0, with gba0,b0≠0, then by the implicit function theorem, there exists an open interval U⊂I containing a0 and an interval V⊂1,2 containing b0 such that there is a unique continuously differentiable function b=ψa defined on U with b∈V that satisfies ga,b=0. This completes the proof of Lemma 4.Proof of Theorem1.
Letm1=m2=m3 and c=0, and then from Lemma 1, we obtain the following central configuration equations:(24)f01=R01−R13Δ013+R02−R12Δ012=0,f12=m0R01−R02Δ012+m1R13−R12Δ123=0.
Solving the above equations, we obtain(25)μ0a,b=m0m1=ϕ1a,ba+bb2a2+13/24ϕ2a,ba+b2+13/2,such that constraint (3) holds, where ϕ1a,b=a+b2+13/2−8 and ϕ2a,b=a2+13/2−b3. It is proved in Lemmas 2, 3, and 4 that constraint (8) is satisfied by showing the existence of a smooth curve:(26)ψa=a,bga,b=0,0<a<3,1<b<2.
To find the region whereμ0>0, we solve the following inequality for a and b:(27)ϕ1a,bϕ2a,b>0.
The functionsϕ1a,b and ϕ2a,b are positive in Rϕ1 and Rϕ2, respectively, where(28)Rϕ1=a,b0<b≤3∧a>3−b∨b>3∧a>0,Rϕ2=a,b0<b≤1∧a>0∨b>1∧a>b2−1.
Therefore, the configurationr=r0,r1,r2,r3 shown in Figure 1 forms a central configuration in Rμ0=Rϕ1∩Rϕ2∪Rϕ1c∩Rϕ2c, where(29)Rμ0=Rϕ1∩Rϕ2∪Rϕ1c∩Rϕ2c=a,b0<b≤1∧a>3−b∨1<b<23∧0<a<b2−1∨a>3−b∨b=23∧0<a<13∨a>13∨23<b<3∧0<a<3−b∨a>b2−1∨b≥3∧a>b2−1,such that (8) holds. Since ga,b=0 has an absolute minimum at a0,b0=0.53,1.15 and is monotonically decreasing for 0<a<0.53 and increasing for a>0.53. Therefore, ga,b≠0 for b<1.15. Hence, the region Rμ0 simplifies to(30)Rμ0=a,b23<b<3∧0<a<3−b∨a>b2−1∨b≥3∧a>b2−1.
The regionRμ0 is shown in Figure 3. This completes the proof of Theorem 1.
To be able to comment on the values ofμ0 in the central configuration region, we use interpolation techniques and write the solution of ga,b=0 as b=ψa, where(31)ψa=−99.4142a11+713.882a10−2296.59a9+4371.32a8−5475.35a7+4748.21a6−2919.84a5+1282.71a4−400.891a3+88.7673a2−13.8585a+2.4259.
The functionψa accurately approximates the solution of ga,b=0 in the central configuration region where μ0 is positive. The approximation error is between 10−10 and 10−6. This gives μ0 as a function of a as follows:(32)μ0a=a+ψa2+13/2−8a+ψaψ2aa2+13/24a2+13/2−ψa3a+ψa2+13/2.
The functionμ0a is a bounded, well-defined continuous function of a except when qa=a2+1−ψa2=0. To identify the values of a where qa=0, we write it as(33)qa=−9883.18a22+141940.a21−966255.a20+4.14813×106a19−1.26042×107a18+2.88398×107a17−5.16176×107a16+7.41023×107a15−8.68138×107a14+8.40051×107a13−6.77049×107a12+4.57074×107a11−2.59398×107a10+1.2402×107a9−5.00236×106a8+1.705×106a7−492405.a6+120891.a5−25214.6a4+4405.41a3−621.741a2+67.2389a−4.88501=0.
The numerical solution ofqa=0 shows that it has three real roots a=1.14605, a=1.2471, and a=1.44556. However, a careful observation of the region of existence of central configuration for the four-body problem in Figure 1 shows that a=1.14605 defines a boundary between the region of existence and nonexistence and a=1.2471 and a=1.44556 are outside the domain of interest. Hence, μ0a uniquely defines the positive values of the mass ratio m0/m3 for the four-body problem as described in Theorem 1. The region of existence of central configuration for the four-body problem with three equal masses is given in Figure 3. Taking advantage of the interpolated expression b=ψa, μ0a is given for b∈0.1,1.14 in Figure 4 which shows μ0a to be an increasing function of a with the minimum and maximum at the end points of the domain.Figure 3
Region of existence of central configuration for the concave kite four-body problem with (m1=m2=m3, m0.Figure 4
Values ofμ0a in the region of central configuration.


### 3.2. Action Minimizing Orbits in the Triangular Four-Body Problem

In this section, we introduce the analytical description of a family of periodic solutions in the four-body problem using variational techniques.Theorem 2.
For the four-body problem considered in Theorem1, the minimizers for Aq restricted to the homographic solutions qit=ϕtqi,0 are precisely the Keplerian elliptical solutions and the minimum of the action is equal to 3/22π2/3T1/3ξa,b/ηa,b2/3.
Define the Lagrangian action as(34)Aq=∫0TLqt,q˙tdt,where the Lagrangian L is defined by(35)Lq,q˙=∑i=14mi2q˙2−Uq,Uq=∑i<jmimjqi−qj.
Let us callΩcm the y-coordinate of the center of mass in the configuration described earlier in Section 2, then(36)Ωcm=b−2am13m1+m0,and the center of mass is(37)COM=0,Ωcm.
In this case, we have the following Cartesian coordinates for the pointsq0,0,q1,0,q2,0,q3,0:(38)q0,0=0,Ωcm,q1,0=−1,−a−Ωcm,q2,0=0,b−Ωcm,q3,0=1,−a−Ωcm.
Observe that(39)r0=q0,0=Ωcm,r1=r3=1+a+Ωcm2,r2=b−Ωcm.
We focus on solutions of the formqit=ϕtqi,0. These solutions are called homographic solutions. We will restrict the action functional to solutions of this type.Proof of Theorem 2.
The kinetic energy termK is equal to(40)K=12∑i=03miq˙it2=ϕ˙t2r12m1+12m0r0r12+m1r2r12=q˙it2m1+12m0r0r12+m1r2r12,where we have used q˙1t2=ϕ˙t2r12. The potential energy is given by(41)U=∑0≤i<j≤3mimjqi−qj,and using qi−qj=ϕtqi,0−qj,0, we get(42)U=m0m1ϕt2a2+1+1b+m12ϕt2a+b2+1+12.
Multiplying and dividing byr1, we obtain(43)U=1q1m0m1r12a2+1+1b+m12r12a+b2+1+12.
As defined previously, we usem0/m1=μ0, and by letting m1=1, we have(44)U=1q1μ0r12a2+1+1b+r12a+b2+1+12.
Now, we are ready to compute the action restricted to this class of homographic solutions. We have(45)A=∫0T1+12m0r0r12+r2r12q1.22dt+∫0Tμ0r12a2+1+1b+r12a+b2+1+121q1dt=2+m0r0r12+r2r12∫0Tq1.22dt+μ0r12a2+1+1b+r12a+b2+1+12∫0T1q1dt.
Let(46)ηa,b=2+m0r0r12+r2r12,ξa,b=μ0r12a2+1+1b+r12a+b2+1+12.
Then,(47)Aq=η∫0Tq1.22dt+ξ∫0T1q1dt.
The infimum ofAq is(48)infqAq=infa>0,b>0infq1η∫0Tq1.22dt+ξ∫0T1q1dt=infa>0,b>0ηinfq1∫0Tq1.22dt+ξη∫0T1q1dt.
By Gordon’s result, we have(49)infq1∫0Tq1.22dt+ξη∫0T1q1dt=322π2/3T1/3ξa,bηa,b2/3.
Then,(50)infqAq=infa>0,b>0ηa,b322π2/3T1/3ξb,cηb,c2/3=infa>0,b>0322π2/3T1/3ηa,b1/3ξa,b2/3.
Letϕa,b=3/22π2/3T1/3ηa,b1/3ξa,b2/3, then ϕa,b attains its infimum at a0,b0 if and only if ηa,bξa,b2 attains its infimum at a0,b0. It is challenging to see that for positive values of a0 and b0, the function ϕa,b is convex and coercive. However, we have proved that b can be written as b=ψa and used interpolation to find ψa as given in equation (31). Hence, we can now write ηa,b,ξa,b, and ϕa,b as functions of one variable. For convexity, we use the second derivative test and show that ∂2ϕa/∂a2>0 when a∈0.1,1.14 Hence, ϕa is convex when a∈0.1,1.14. For coercivity, we see that ϕa is continuous for all positive values of a, ϕa⟶∞ as a⟶0, and when a⟶∞, ϕa⟶∞, which implies ϕa is coercive. Hence, ϕa attains infa>0ϕa at unique a0>0 and satisfies ϕa0=0.


## 3.1. Central Configurations

Theorem 1.
Consider four point massesm0 and m1=m2=m3 having position vectors r0=0,0, r1=−1,−a, r2=0,b, and r3=1,−a, where a>0 and b>0. Then, there exists a unique mass ratio μ0a,b:(7)μ0=m0m1=a+b2+13/2−8a+bb2a2+13/24a2+13/2−b3a+b2+13/2,such that r=r0,r1,r2,r3 is a central configuration for μ0>0 in Rμ0a,b subject to the constraint ga,b=0. The region Rμ0a,b and the constraint ga,b are given below:(8)Rμ0a,b=a,b23<b<3∧0<a<3−b∨a>b2−1∨b≥3∧a>b2−1,ga,b=−b1b3−1α+2a−18+1β=0,where α=a+b2+13/2 and β=a2+13/2.
Before we attempt to prove Theorem1, we will need help from the following lemmas.Lemma 2.
The functiong defined by (8) is negative for all 0<a<3 and 0<b≤B=2/33/7+23/71/4≃1.13.Proof.
Let0<a<3 and 0<b≤B, then we have(9)ga,bb=1α+2ab1β−4+ab24b3.
Consider the equation of the straight line segment that lies in the first quadrant:(10)x+2aby−4+ab24b3=0.
For positivex and y,(11)x+2aby−4+ab24b3<0,is equivalent to(12)x<4+ab24b3,y<4+ab28ab2.
To prove thatga,bb−1<0, we need to show that(13)1α<4+ab24b3,1β<4+ab28ab2.
For1/α<4+ab2/4b3, it is trivial to see that(14)1α=11+a+b23/2<1b3<4+ab24b3.
Similarly, to show that1/β<4+ab2/8ab2, we need(15)β>8ab24+ab2,(16)or1+a23>8ab24+ab22.
Inequality (16) is equivalent to(17)1+a234+ab22−64a2b4>0.
Expanding the left-hand side of (17), we obtain the following:(18)pa,b=b4a8+8b2a7+16+3b4a6+24b2a5+48+3b4a4+24b2a3+48−63b4a2+8b2a+16.
For positivea and b,(19)pa,b>48a4+48−63b4a2+16,≥48a4+48−63B4a2+16,=48a2−132≥0.
Consequently,ga,b<0 for 0<a<3 and 0<b≤B.
This completes the proof of Lemma2.Lemma 3.
The partial derivative ofg defined by (3) satisfies gba,b>3/α2, for all a>0 and b>0.Proof.
For positivea and b, we have(20)gba,b=2b3+α−bαbα2,=2b3+α−3ba+bγα2,where γ=1+a+b21/2 and αb is the derivative of α w.r.t. b.
Sinceb3<α for all a>0 and b>0, then(21)gba,b>2α+α−3ba+bγα2.
Using the fact thatα=γ3 and γ2=1+a+b2, we get(22)gba,b>31+a+b21+a2+abγ6.
Sincea and b are positive, therefore(23)gba,b>3α2.
This completes the proof of Lemma3.Remark 1.
Numerically, it is easy to show thatga,2=0 when a=0.14 and a=1.32. For a∈I=0.14,1.32, ga,1<0 and ga,2>0, when a∈I, see Figure 2. Therefore, by intermediate value theorem ga,b has at least one root b∈1,2 when a∈I.Figure 2
Graph ofga,b for values of b=0.5,1,and1.5.Lemma 4.
Consider the functiong defined by (3). Then, for any a0∈I=0.14,1.32 there exists an interval U⊂I containing a0 and an interval V⊂1,2 containing b0 such that there is a unique continuously differentiable function b=ψa defined on U with b∈V that satisfies ga,b=0.Proof.
Leta0 be any number in the interval I=0.14,1.32; then, using Lemma 3, we have ga0,1<0. Then, numerically, one can check that ga0,2>0 for a0∈I. Thus, by the mean value theorem, there exists at least one b0∈1,2, solution of ga0,b=0. By Lemma 3, gba0,b>0 for all b>0, hence the solution b0 is unique. Since g has continuous partial derivatives and ga0,b0=0, with gba0,b0≠0, then by the implicit function theorem, there exists an open interval U⊂I containing a0 and an interval V⊂1,2 containing b0 such that there is a unique continuously differentiable function b=ψa defined on U with b∈V that satisfies ga,b=0. This completes the proof of Lemma 4.Proof of Theorem1.
Letm1=m2=m3 and c=0, and then from Lemma 1, we obtain the following central configuration equations:(24)f01=R01−R13Δ013+R02−R12Δ012=0,f12=m0R01−R02Δ012+m1R13−R12Δ123=0.
Solving the above equations, we obtain(25)μ0a,b=m0m1=ϕ1a,ba+bb2a2+13/24ϕ2a,ba+b2+13/2,such that constraint (3) holds, where ϕ1a,b=a+b2+13/2−8 and ϕ2a,b=a2+13/2−b3. It is proved in Lemmas 2, 3, and 4 that constraint (8) is satisfied by showing the existence of a smooth curve:(26)ψa=a,bga,b=0,0<a<3,1<b<2.
To find the region whereμ0>0, we solve the following inequality for a and b:(27)ϕ1a,bϕ2a,b>0.
The functionsϕ1a,b and ϕ2a,b are positive in Rϕ1 and Rϕ2, respectively, where(28)Rϕ1=a,b0<b≤3∧a>3−b∨b>3∧a>0,Rϕ2=a,b0<b≤1∧a>0∨b>1∧a>b2−1.
Therefore, the configurationr=r0,r1,r2,r3 shown in Figure 1 forms a central configuration in Rμ0=Rϕ1∩Rϕ2∪Rϕ1c∩Rϕ2c, where(29)Rμ0=Rϕ1∩Rϕ2∪Rϕ1c∩Rϕ2c=a,b0<b≤1∧a>3−b∨1<b<23∧0<a<b2−1∨a>3−b∨b=23∧0<a<13∨a>13∨23<b<3∧0<a<3−b∨a>b2−1∨b≥3∧a>b2−1,such that (8) holds. Since ga,b=0 has an absolute minimum at a0,b0=0.53,1.15 and is monotonically decreasing for 0<a<0.53 and increasing for a>0.53. Therefore, ga,b≠0 for b<1.15. Hence, the region Rμ0 simplifies to(30)Rμ0=a,b23<b<3∧0<a<3−b∨a>b2−1∨b≥3∧a>b2−1.
The regionRμ0 is shown in Figure 3. This completes the proof of Theorem 1.
To be able to comment on the values ofμ0 in the central configuration region, we use interpolation techniques and write the solution of ga,b=0 as b=ψa, where(31)ψa=−99.4142a11+713.882a10−2296.59a9+4371.32a8−5475.35a7+4748.21a6−2919.84a5+1282.71a4−400.891a3+88.7673a2−13.8585a+2.4259.
The functionψa accurately approximates the solution of ga,b=0 in the central configuration region where μ0 is positive. The approximation error is between 10−10 and 10−6. This gives μ0 as a function of a as follows:(32)μ0a=a+ψa2+13/2−8a+ψaψ2aa2+13/24a2+13/2−ψa3a+ψa2+13/2.
The functionμ0a is a bounded, well-defined continuous function of a except when qa=a2+1−ψa2=0. To identify the values of a where qa=0, we write it as(33)qa=−9883.18a22+141940.a21−966255.a20+4.14813×106a19−1.26042×107a18+2.88398×107a17−5.16176×107a16+7.41023×107a15−8.68138×107a14+8.40051×107a13−6.77049×107a12+4.57074×107a11−2.59398×107a10+1.2402×107a9−5.00236×106a8+1.705×106a7−492405.a6+120891.a5−25214.6a4+4405.41a3−621.741a2+67.2389a−4.88501=0.
The numerical solution ofqa=0 shows that it has three real roots a=1.14605, a=1.2471, and a=1.44556. However, a careful observation of the region of existence of central configuration for the four-body problem in Figure 1 shows that a=1.14605 defines a boundary between the region of existence and nonexistence and a=1.2471 and a=1.44556 are outside the domain of interest. Hence, μ0a uniquely defines the positive values of the mass ratio m0/m3 for the four-body problem as described in Theorem 1. The region of existence of central configuration for the four-body problem with three equal masses is given in Figure 3. Taking advantage of the interpolated expression b=ψa, μ0a is given for b∈0.1,1.14 in Figure 4 which shows μ0a to be an increasing function of a with the minimum and maximum at the end points of the domain.Figure 3
Region of existence of central configuration for the concave kite four-body problem with (m1=m2=m3, m0.Figure 4
Values ofμ0a in the region of central configuration.


## 3.2. Action Minimizing Orbits in the Triangular Four-Body Problem

In this section, we introduce the analytical description of a family of periodic solutions in the four-body problem using variational techniques.Theorem 2.
For the four-body problem considered in Theorem1, the minimizers for Aq restricted to the homographic solutions qit=ϕtqi,0 are precisely the Keplerian elliptical solutions and the minimum of the action is equal to 3/22π2/3T1/3ξa,b/ηa,b2/3.
Define the Lagrangian action as(34)Aq=∫0TLqt,q˙tdt,where the Lagrangian L is defined by(35)Lq,q˙=∑i=14mi2q˙2−Uq,Uq=∑i<jmimjqi−qj.
Let us callΩcm the y-coordinate of the center of mass in the configuration described earlier in Section 2, then(36)Ωcm=b−2am13m1+m0,and the center of mass is(37)COM=0,Ωcm.
In this case, we have the following Cartesian coordinates for the pointsq0,0,q1,0,q2,0,q3,0:(38)q0,0=0,Ωcm,q1,0=−1,−a−Ωcm,q2,0=0,b−Ωcm,q3,0=1,−a−Ωcm.
Observe that(39)r0=q0,0=Ωcm,r1=r3=1+a+Ωcm2,r2=b−Ωcm.
We focus on solutions of the formqit=ϕtqi,0. These solutions are called homographic solutions. We will restrict the action functional to solutions of this type.Proof of Theorem 2.
The kinetic energy termK is equal to(40)K=12∑i=03miq˙it2=ϕ˙t2r12m1+12m0r0r12+m1r2r12=q˙it2m1+12m0r0r12+m1r2r12,where we have used q˙1t2=ϕ˙t2r12. The potential energy is given by(41)U=∑0≤i<j≤3mimjqi−qj,and using qi−qj=ϕtqi,0−qj,0, we get(42)U=m0m1ϕt2a2+1+1b+m12ϕt2a+b2+1+12.
Multiplying and dividing byr1, we obtain(43)U=1q1m0m1r12a2+1+1b+m12r12a+b2+1+12.
As defined previously, we usem0/m1=μ0, and by letting m1=1, we have(44)U=1q1μ0r12a2+1+1b+r12a+b2+1+12.
Now, we are ready to compute the action restricted to this class of homographic solutions. We have(45)A=∫0T1+12m0r0r12+r2r12q1.22dt+∫0Tμ0r12a2+1+1b+r12a+b2+1+121q1dt=2+m0r0r12+r2r12∫0Tq1.22dt+μ0r12a2+1+1b+r12a+b2+1+12∫0T1q1dt.
Let(46)ηa,b=2+m0r0r12+r2r12,ξa,b=μ0r12a2+1+1b+r12a+b2+1+12.
Then,(47)Aq=η∫0Tq1.22dt+ξ∫0T1q1dt.
The infimum ofAq is(48)infqAq=infa>0,b>0infq1η∫0Tq1.22dt+ξ∫0T1q1dt=infa>0,b>0ηinfq1∫0Tq1.22dt+ξη∫0T1q1dt.
By Gordon’s result, we have(49)infq1∫0Tq1.22dt+ξη∫0T1q1dt=322π2/3T1/3ξa,bηa,b2/3.
Then,(50)infqAq=infa>0,b>0ηa,b322π2/3T1/3ξb,cηb,c2/3=infa>0,b>0322π2/3T1/3ηa,b1/3ξa,b2/3.
Letϕa,b=3/22π2/3T1/3ηa,b1/3ξa,b2/3, then ϕa,b attains its infimum at a0,b0 if and only if ηa,bξa,b2 attains its infimum at a0,b0. It is challenging to see that for positive values of a0 and b0, the function ϕa,b is convex and coercive. However, we have proved that b can be written as b=ψa and used interpolation to find ψa as given in equation (31). Hence, we can now write ηa,b,ξa,b, and ϕa,b as functions of one variable. For convexity, we use the second derivative test and show that ∂2ϕa/∂a2>0 when a∈0.1,1.14 Hence, ϕa is convex when a∈0.1,1.14. For coercivity, we see that ϕa is continuous for all positive values of a, ϕa⟶∞ as a⟶0, and when a⟶∞, ϕa⟶∞, which implies ϕa is coercive. Hence, ϕa attains infa>0ϕa at unique a0>0 and satisfies ϕa0=0.


## 4. Three Unequal Massesm3=m1≠m2≠m0

In this section, we discuss a four-body problem which has two symmetric equal massesm3=m1 on the horizontal axis and two nonequal masses m0 and m2 on the vertical axis. The position vectors of the four masses m0, m1, m2, and m3 are r0=0,c, r1=−1,0, r2=0,b, and r3=1,0, respectively. For c>0, the four masses make an isosceles triangle with m0 inside the triangle (Figure 5), and for c<0, the convex kite configuration is obtained (Figure 6).Figure 5
Concave four-body configurations with a pair of equal massesm1=m3.Figure 6
Convex four-body kite configurations with two equal massesm1=m3.


### 4.1. Central Configurations

Theorem 3.
Consider four point massesm0, m1=m3≠m2 having position vectors r0=0,c, r1=−1,0, r2=0,b, and r3=1,0, where b>0 and b>c such that r=r0,r1,r2,r3 is a central configuration.(a)
Then, there exist unique mass ratios:(51)μ0=m0m1=b8−b2+13/2b−c2c2+13/24b−c3−c2+13/2b2+13/2,μ2=m2m1=c8−c2+13/2c−b2b2+13/24b2+13/2−b−c3c2+13/2.(b)
The region of existence of central configuration where the four positive masses are arranged in a concave kite configuration is given by(52)Rt=b,c0<c<13∧c+c2+1<b<3∨13<c<3∧3<b<c+c2+1.(c)
The region of existence of central configuration where the four positive masses are arranged in a convex kite configuration is given by(53)Rr=b,c−3<c≤−13∧c+c2+1<b<3∨−13<c<−2+3∧c2−12c<b<3.Proof of Theorem3.
Consider four point masses with position vectorsr0,r1,r2,r3 and masses m0,m1, m2, and m3, where m3=m1. The solution of f01=0 and f12=0 gives(54)μ0=m0m1=R12−R13Δ123R01−R02Δ012=βγbα−84αb−cβ−γ,μ2=m2m1=R01−R13Δ013R12−R02Δ012=αγc8−β4βb−cα−γ,where α=1+b23/2, β=1+c23/2, and γ=b−c3. The mass ratios μ0 and μ2 are well-defined functions of b and c except at b=c and b=c±1+c2.
To find central configuration regions whereμ0>0, it is sufficient to show that α−8 and β−γ have the same sign. It is trivial to see that Rα−8=0 when b=3. Similarly, β−γ=0 when b=c+c2+1. Hence, β−γ is positive in(55)Ra=a,bb>0∧c<b∧b<c+c2+1.
The complement ofRa, where β−γ<0, is given by(56)Rac=a,bb>0∧c<b2−12b−1.
It is to be noted that the sign ofc will be determined according to whether the four-body configuration is concave c>0 or convexc<0. After some simplifications, the central configuration region for μ0>0,c>0 is given by(57)Rt1=b,c0<c<13∧c2+1+c<b<3∨13<c≤3∧3<b<c2+1+c∨c>3∧c<b<c2+1+c.
Similarly, the central configuration region forμ0>0,c<0 is given by(58)Rr1=b,cc<0,c+c2+1<b<3.
Consider the mass ratioμ2. Let c>0. Since b>c, therefore for μ2>0, β−8α−γ−1 must have the same sign. It is trivial to see that 8−β>0 when c∈0,3 and α−γ>0 for all b>0 and c>0. Therefore, μ2>0 in(59)Rt2=b,c0<b≤3∧0<c<b∨b>3∧0<c<3.
Similarly, whenc<0 (rhomboidal configuration), the central configuration region where μ2>0 is given by(60)Rr2=b,c0<b≤3∧0<c<b∨b>3∧0<c<3.
Hence, the region of existence of central configuration for the concave kite four-body problem where all the massesm0, m1=m3, and m2 are positive is given by Rt=Rt1∩Rt2 and the corresponding convex kite central configuration region is given by Rr=Rr1∩Rr2, where(61)Rt=b,c0<c<3−1∧c+c2+1<b<3∨3−1<c<3∧3<b<c+c2+1,Rr=b,c−3<c≤−3−1∧c+c2+1<b<3∨−3−1<c<3−2∧c2−12c−1<b<3.
The regionsRt and Rr are shown in Figures 7 and 8, respectively.Figure 7
Region of existence of central configuration for the concave kite four-body configuration wherem1=m3=m≠m0≠m2.Figure 8
Region of existence of central configuration for the convex kite four-body configuration wherem1=m3=m≠m0≠m2.


### 4.2. Action Minimizing Orbits in the Convex Kite Four-Body Problem

In this section, we discuss the minimization property of a four-body problem which has two equal massesm1=m3 on the horizontal axis and two positive masses m2 and m0 on the vertical axis, which is also the axis of symmetry.Theorem 4.
For the four-body problem considered in Theorem1, the minimizers for Aq restricted to the homographic solutions qit=ϕtqi,0 are precisely the Keplerian elliptical solutions, and the minimum of the action is equal to 3/22π2/3T1/3ξb,c/ηb,c2/3.
Let us callΩcm the y-coordinate of the center of mass in the configuration described earlier in Section 3.1, then(62)Ωcm=m0c+m2bm0+2m1+m2,and the center of mass is COM=0,Ωcm. Observe that(63)r0=q0,0=c−Ωcm,r1=r3=1+Ωcm2,r2=b−Ωcm.Proof of Theorem4.
The kinetic energy termK is equal to(64)K=12∑i=03miq˙it2=ϕ˙t2m1r12+m22r22+m02r02=ϕ˙t2r12m1+12m2r22+m0r02r12=q˙1t2m1+12m2r22+m0r02r12,where we have used the fact that q˙1t2=ϕ˙t2r12.
On the other hand, the potential energy is given by(65)U=∑0≤i<j≤3mimjqi−qj,and using qi−qj=ϕtqi,0−qj,0, we get(66)U=1ϕtm122+2m1m21+b2+2m1m01+c2+m2m0c−b.
Multiplying and dividing byr1, we obtain(67)U=1q1m12r12+2m1m2r11+b2+2m1m0r11+c2+m2m0r1c−b.
Definingμ0=m0/m1, μ2=m2/m1, and m1=1,(68)U=1q1r12+2μ2r11+b2+2μ0r11+c2+μ2μ0r1c−b.
Now, we are ready to compute the action restricted to this class of homographic solutions:(69)A=∫0T1+12μ2r22+μ0r02r12q1.22dt+∫0Tr12+2μ2r11+b2+2μ0r11+c2+μ2μ0r1c−b1q1dt=2+μ2r22+μ0r02r12∫0Tq1.22dt+r12+2μ2r11+b2+2μ0r11+c2+μ2μ0r1c−b∫0T1q1dt.
Let(70)ηb,c=1+12μ2r22+μ0r02r12,ξb,c=r12+2μ2r11+b2+2μ0r11+c2+μ2μ0r1c−b.
Therefore,(71)Aq=η∫0Tq1.22dt+ξ∫0T1q1dt.
The infimum ofAq is(72)infqAq=infb>0,c>0infq1η∫0Tq1.22dt+ξ∫0T1q1dt=infb>0,c>0ηinfq1∫0Tq1.22dt+ξη∫0T1q1dt.
By Gordon’s result, we have(73)infq1∫0Tq1.22dt+ξη∫0T1q1dt=322π2/3T1/3ξb,cηb,c2/3.
Then,(74)infqAq=infb>0,c>0ηb,c322π2/3T1/3ξb,cηb,c2/3=infb>0,c>0322π2/3T1/3ηb,c1/3ξb,c2/3.
Letϕb,c=3/22π2/3T1/3ηb,c1/3ξb,c2/3, and then ϕb,c attains its infimum at b0,c0 if and only if ηb,cξb,c2 attains its infimum at b0,c0. Similar to the concave case, we need to show that for positive values of b0 and c0, the function ϕb,c is convex and coercive. For convexity, we compute the Hessian matrix for ϕb,c and numerically show that Hb,c is positive semidefinite in the region Rr.
That concludes that the functionϕb,c is convex. For coercivity, we see that ϕb,c is continuous for all positive values of b and c, ϕb,c⟶∞ as b,c⟶0,0, and when b⟶∞ and c⟶∞, ϕb,c tends to ∞, which implies ϕb,c is coercive. Hence, ϕb,c attains infb>0,c>0ϕb,c at unique b0,c0, b0>0, c0>0, and satisfiesϕb0,c0=0.


## 4.1. Central Configurations

Theorem 3.
Consider four point massesm0, m1=m3≠m2 having position vectors r0=0,c, r1=−1,0, r2=0,b, and r3=1,0, where b>0 and b>c such that r=r0,r1,r2,r3 is a central configuration.(a)
Then, there exist unique mass ratios:(51)μ0=m0m1=b8−b2+13/2b−c2c2+13/24b−c3−c2+13/2b2+13/2,μ2=m2m1=c8−c2+13/2c−b2b2+13/24b2+13/2−b−c3c2+13/2.(b)
The region of existence of central configuration where the four positive masses are arranged in a concave kite configuration is given by(52)Rt=b,c0<c<13∧c+c2+1<b<3∨13<c<3∧3<b<c+c2+1.(c)
The region of existence of central configuration where the four positive masses are arranged in a convex kite configuration is given by(53)Rr=b,c−3<c≤−13∧c+c2+1<b<3∨−13<c<−2+3∧c2−12c<b<3.Proof of Theorem3.
Consider four point masses with position vectorsr0,r1,r2,r3 and masses m0,m1, m2, and m3, where m3=m1. The solution of f01=0 and f12=0 gives(54)μ0=m0m1=R12−R13Δ123R01−R02Δ012=βγbα−84αb−cβ−γ,μ2=m2m1=R01−R13Δ013R12−R02Δ012=αγc8−β4βb−cα−γ,where α=1+b23/2, β=1+c23/2, and γ=b−c3. The mass ratios μ0 and μ2 are well-defined functions of b and c except at b=c and b=c±1+c2.
To find central configuration regions whereμ0>0, it is sufficient to show that α−8 and β−γ have the same sign. It is trivial to see that Rα−8=0 when b=3. Similarly, β−γ=0 when b=c+c2+1. Hence, β−γ is positive in(55)Ra=a,bb>0∧c<b∧b<c+c2+1.
The complement ofRa, where β−γ<0, is given by(56)Rac=a,bb>0∧c<b2−12b−1.
It is to be noted that the sign ofc will be determined according to whether the four-body configuration is concave c>0 or convexc<0. After some simplifications, the central configuration region for μ0>0,c>0 is given by(57)Rt1=b,c0<c<13∧c2+1+c<b<3∨13<c≤3∧3<b<c2+1+c∨c>3∧c<b<c2+1+c.
Similarly, the central configuration region forμ0>0,c<0 is given by(58)Rr1=b,cc<0,c+c2+1<b<3.
Consider the mass ratioμ2. Let c>0. Since b>c, therefore for μ2>0, β−8α−γ−1 must have the same sign. It is trivial to see that 8−β>0 when c∈0,3 and α−γ>0 for all b>0 and c>0. Therefore, μ2>0 in(59)Rt2=b,c0<b≤3∧0<c<b∨b>3∧0<c<3.
Similarly, whenc<0 (rhomboidal configuration), the central configuration region where μ2>0 is given by(60)Rr2=b,c0<b≤3∧0<c<b∨b>3∧0<c<3.
Hence, the region of existence of central configuration for the concave kite four-body problem where all the massesm0, m1=m3, and m2 are positive is given by Rt=Rt1∩Rt2 and the corresponding convex kite central configuration region is given by Rr=Rr1∩Rr2, where(61)Rt=b,c0<c<3−1∧c+c2+1<b<3∨3−1<c<3∧3<b<c+c2+1,Rr=b,c−3<c≤−3−1∧c+c2+1<b<3∨−3−1<c<3−2∧c2−12c−1<b<3.
The regionsRt and Rr are shown in Figures 7 and 8, respectively.Figure 7
Region of existence of central configuration for the concave kite four-body configuration wherem1=m3=m≠m0≠m2.Figure 8
Region of existence of central configuration for the convex kite four-body configuration wherem1=m3=m≠m0≠m2.


## 4.2. Action Minimizing Orbits in the Convex Kite Four-Body Problem

In this section, we discuss the minimization property of a four-body problem which has two equal massesm1=m3 on the horizontal axis and two positive masses m2 and m0 on the vertical axis, which is also the axis of symmetry.Theorem 4.
For the four-body problem considered in Theorem1, the minimizers for Aq restricted to the homographic solutions qit=ϕtqi,0 are precisely the Keplerian elliptical solutions, and the minimum of the action is equal to 3/22π2/3T1/3ξb,c/ηb,c2/3.
Let us callΩcm the y-coordinate of the center of mass in the configuration described earlier in Section 3.1, then(62)Ωcm=m0c+m2bm0+2m1+m2,and the center of mass is COM=0,Ωcm. Observe that(63)r0=q0,0=c−Ωcm,r1=r3=1+Ωcm2,r2=b−Ωcm.Proof of Theorem4.
The kinetic energy termK is equal to(64)K=12∑i=03miq˙it2=ϕ˙t2m1r12+m22r22+m02r02=ϕ˙t2r12m1+12m2r22+m0r02r12=q˙1t2m1+12m2r22+m0r02r12,where we have used the fact that q˙1t2=ϕ˙t2r12.
On the other hand, the potential energy is given by(65)U=∑0≤i<j≤3mimjqi−qj,and using qi−qj=ϕtqi,0−qj,0, we get(66)U=1ϕtm122+2m1m21+b2+2m1m01+c2+m2m0c−b.
Multiplying and dividing byr1, we obtain(67)U=1q1m12r12+2m1m2r11+b2+2m1m0r11+c2+m2m0r1c−b.
Definingμ0=m0/m1, μ2=m2/m1, and m1=1,(68)U=1q1r12+2μ2r11+b2+2μ0r11+c2+μ2μ0r1c−b.
Now, we are ready to compute the action restricted to this class of homographic solutions:(69)A=∫0T1+12μ2r22+μ0r02r12q1.22dt+∫0Tr12+2μ2r11+b2+2μ0r11+c2+μ2μ0r1c−b1q1dt=2+μ2r22+μ0r02r12∫0Tq1.22dt+r12+2μ2r11+b2+2μ0r11+c2+μ2μ0r1c−b∫0T1q1dt.
Let(70)ηb,c=1+12μ2r22+μ0r02r12,ξb,c=r12+2μ2r11+b2+2μ0r11+c2+μ2μ0r1c−b.
Therefore,(71)Aq=η∫0Tq1.22dt+ξ∫0T1q1dt.
The infimum ofAq is(72)infqAq=infb>0,c>0infq1η∫0Tq1.22dt+ξ∫0T1q1dt=infb>0,c>0ηinfq1∫0Tq1.22dt+ξη∫0T1q1dt.
By Gordon’s result, we have(73)infq1∫0Tq1.22dt+ξη∫0T1q1dt=322π2/3T1/3ξb,cηb,c2/3.
Then,(74)infqAq=infb>0,c>0ηb,c322π2/3T1/3ξb,cηb,c2/3=infb>0,c>0322π2/3T1/3ηb,c1/3ξb,c2/3.
Letϕb,c=3/22π2/3T1/3ηb,c1/3ξb,c2/3, and then ϕb,c attains its infimum at b0,c0 if and only if ηb,cξb,c2 attains its infimum at b0,c0. Similar to the concave case, we need to show that for positive values of b0 and c0, the function ϕb,c is convex and coercive. For convexity, we compute the Hessian matrix for ϕb,c and numerically show that Hb,c is positive semidefinite in the region Rr.
That concludes that the functionϕb,c is convex. For coercivity, we see that ϕb,c is continuous for all positive values of b and c, ϕb,c⟶∞ as b,c⟶0,0, and when b⟶∞ and c⟶∞, ϕb,c tends to ∞, which implies ϕb,c is coercive. Hence, ϕb,c attains infb>0,c>0ϕb,c at unique b0,c0, b0>0, c0>0, and satisfiesϕb0,c0=0.


## 5. Hamiltonian Formulation of the Problem: Some Numerical Examples

It is well known that the study of the trajectories of celestial bodies under their mutual gravitational attractions is important for understanding their movement and navigation. In that sense, a special type of the four-body problem with analytical and numerical investigation can contribute to the understanding of the dynamical behavior of quadruple stellar systems (e.g., the HD 98800 quadruple system with two pairs of stars orbiting each other). Several authors studied the stability and dynamical evolution of symmetric quadruple systems for stars and exoplanetary systems of two planets [40, 41]. In this section, we will study the periodic behavior of the kite four-body problem.Consider the four-body problem introduced in Lemma1, which has a pair of equal masses and one axis of symmetry. Using the symmetries and position coordinates from Lemma 1, we obtain the following reduced Hamiltonian in the case of unequal masses:(75)H=∑i=03pi22mi−m0m1r01−m0m2r02−m0m3r03−m1m2r12−m1m3r13−m2m3r23,where(76)r012=1+q1−c2,r022=b−c2,r032=1+q3−c2,r122=1+q1−b2,r132=4+q1−q32,r232=1+b−q32,b>c,and qi and pi, i=0,3¯, are the generalized coordinates and momenta (we assume for simplicity that the gravitational constant is equal to 1).For the investigation of the reduced Hamiltonian equations of motion (75), we have selected examples from concave and convex kite four-body problems introduced in Section 2. We have used classical numerical methods and found periodic orbits for a given vector field. The stability of a periodic orbit for an autonomous vector field can be calculated by Poincaré maps, which replaces the flow of the n-dimensional continuous vector field with an n−1-dimensional map [42]. In this article, the analysis of periodic and quasiperiodic orbits is performed using the Poincaré surface of the section technique by picking the phase element p1=0. In the n-dimensional case, the Poincaré surface of the section has dimensionality 2n−2. The intersection points of the solution curves with the corresponding (pi, qi), i=0,n−1¯, plane still lie on a smooth curve [43].In order to construct the Poincaré surface of the section and to find the corresponding periodic orbits, we plot the motion from the4D phase space q1,q3,p1,p3 in a “cut plane” p1t=0,q1t>0. Since H is conserved, any point on this surface of the section will uniquely define the orbit. Our Poincaré surface of sections (Figures 9–11) describe the concave four-body problem in the unequal mass case with energy values E∈−3.52,…,−2, plotting q3t versus p3t, and m0=1.28155, m2=0.31, m3=m1=1, b=1.1, and c=0.1. The initial conditions considered here satisfy the central configuration equations introduced in Section 2. We wish to note that in case of the energy level −3.52, we can see many little islands of quasiperiodic orbits. It should be emphasized that increasing the energy levels increase the Poincaré surface of sections in size, as a blowing up effect (see Figures 9–11).Figure 9
The progressive evolution of a Poincaré surface of sections for different energy levels in the unequal masses case (m0=1.28155, m2=0.31, m3=m1=1, b=1.1, and c=0.1).Figure 10
The progressive evolution of a Poincaré surface of sections for different energy levels in the unequal masses case (m0=1.28155, m2=0.31, m3=m1=1, b=1.1, and c=0.1).Figure 11
The progressive evolution of a Poincaré surface of sections for different energy levels in the unequal masses case (m0=1.28155, m2=0.31, m3=m1=1, b=1.1, and c=0.1).Another interesting unequal mass case is for increased massesm0=2.658, m2=1.03, and m3=m1=1 compared to the previous case. Similarly, using different energy levels, in this case for energy values E∈−8.34,…,−7.3, the variation of the “horseshoe with open side to the left” as the letter D shape to the “horseshoe with open side to the right” as the letter C shape can be observed (Figures 12–14). As earlier, in this particular case, we observe quasiperiodic orbits and the increasing size of the Poincaré surface of sections.Figure 12
The progressive changing of a Poincaré surface of sections for different energy levels in the unequal masses case (m0=2.658, m2=1.03, m3=m1=1, b=1.9, and c=0.7).Figure 13
The progressive changing of a Poincaré surface of sections for different energy levels in the unequal masses case (m0=2.658, m2=1.03, m3=m1=1, b=1.9, and c=0.7).Figure 14
The progressive changing of a Poincaré surface of sections for different energy levels in the unequal masses case (m0=2.658, m2=1.03, m3=m1=1, b=1.9, and c=0.7).In the equal mass case, we have chosen the following particular case:m4=5.17662, m=1, b=1.9, and c=0.7. In this situation, we detected an interesting transformation of the Poincaré surface of sections through the energy levels E∈−13.6,…,−11. In the case of lower energy levels, the outside part of the orbit indicates chaotic behavior. Moreover, by increasing the energy levels, the inside part of the orbit disintegrates, but the size of the Poincaré surface of sections has just a minor variation (Figure 15).Figure 15
The progressive transformation of a Poincaré surface of sections for different energy levels in the equal masses case (m0=5.17662, m=1, b=1.9, and c=0.7).Consequently, we developed the surface of sections with around 300 points, which presents the progressive disintegration of the surfaces of solution curves of regular motion (KAM surface) generated with the increase inH from −3.615 to −2 in our first example (Figures 9–11), from −8.34 to −7 in our second example (Figures 12–14), and from −13.6 to −11 in our third example (Figure 15). It seems that there exist invariant curves (for example, in the plots for H equal to −3.5) near resonances. These invariant curves could form “island sequences” (small groups in a row) and “islands cycles” (small islands inside of islands). One can observe that transitions between levels fuse at progressive integration, and the KAM surfaces progressively disappear.In both cases (nonequal and equal masses), the surfaces show unique types of orbits, including quasiperiodic and island orbits. Let us mention that comparing both cases, when the values ofm0 are increased, then there is a noticeable effect influence on the stability and the existence of quasiperiodic orbits in concave and convex kite four-body problems. In other words, we conjecture that the increasing central mass plays a stabilizing role.The described Poincaré surface of sections allows to study the local stability of the kite four-body problem, the transition from ordered to stochastic motion. They contribute significantly in numerical studies and in verification of the concordance between analytical and numerical results.


## 6. Conclusions

In this paper, we investigated the central configurations of convex and concave kite four-body problems, deriving regions of central configuration. Three four-body arrangements are discussed which include two concave four-body configurations and one convex four-body configuration. In one of the concave configurations, three of the masses on the vertices of the triangle are equal and the mass on the axis of symmetry can take various positive values. In the second case of concave configuration and the case of the convex configuration, there is only a pair of equal masses and the two masses on the axis of symmetry are nonequal. In each of the three cases, regions of central configuration are derived for positive masses. In the first concave case, we can write the mass ratios as a function of one variable and show its optimum values give values of the parameter. In the other two cases, the mass ratios are written as functions of two variables. The action minimizing orbits for both the concave and convex configurations is analyzed, and it is shown that the minimizers of the action functional restricted to the homographic solutions are the Keplerian elliptical solutions. Using the Hamiltonian formalism, we have identified regions with periodic and quasiperiodic orbits. Moreover, we studied the chaotic behavior in the phase space utilizing the Poincaré surface of sections. It was shown that increasing the value of the central massm0 plays a stabilizing role in the case of both convex and concave four-body problems.



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