# On the Symmetric Central Configurations for the Planar 1 + 4-Body Problem

Authors: Chunhua Deng; Fengying Li; Shiqing Zhang
Journal: Complexity
 (2019)

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## Abstract

We study the planar symmetric central configurations of the 1 + 4-body problem where the symmetry axis does not contain any infinitesimal masses. Under certain assumptions, we find analytically some central configurations for suitable positive masses and also get some numerical results of symmetric central configurations where infinitesimal masses are not necessarily equal.


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## Body

## 1. Introduction

A very old problem in celestial mechanics is the study of central configurations for then-body problem. One of the reasons why central configurations are interesting is that they allow us to construct exact solutions of the n-body problem. Central configurations also have other interesting properties in the study of the n-body problem (see [1–14] for details).In this paper, we consider a restricted version of the problem of planar central configurations; that is, we study the limit case of one large mass andN small masses as the small masses tend to zero (planar 1+N-body problem). This problem was first considered by Maxwell [15] trying to construct a model for Saturn’s rings. Hall [16] studied the planar central configuration of the 1+N-body problem where the N small masses are equal. He found that when N is sufficiently large, the only possible relative equilibrium is Maxwell’s ring, that is, a regular N-gon with a central mass, and that other configurations are possible for small N. Moeckel [17] found a necessary and sufficient condition for the linear stability of relative equilibria of the 1+N-body problem with N small but not necessarily equal masses. Recently, these configurations have attracted the attention of astronomers. Renner and Sicardy [18] suggested that the presence of coorbital satellites might explain, at least partly, the confinement of Neptune’s ring arcs. Cors et al. [19] proved that there are only three symmetric central configurations of the 1+4-body problem with four separate identical satellites. Albouy and Fu [2] proved that all central configurations of the 1+4-body problem are symmetric which settles the question in this case. Oliveira and Cabral [20] showed that, for the planar 1+4-body problem where the satellites have different infinitesimal masses and two of them are diametrically opposite in a circle, the configurations are necessarily symmetric and the other satellites have the same mass. Moreover, they prove that the number of central configurations in this case is in general one, two, or three, and in the special case, where the satellites diametrically opposite have the same mass, they prove that the number of central configurations is one or two and give the exact value of the ratio of the masses that provides this bifurcation. Many other results can be found in [9, 21, 22, 23, 24, 25]. Here, we study the planar symmetric central configurations of the 1+4-body problem where the symmetry axis does not contain any infinitesimal masses, and also the satellites may have different masses.


## 2. Preliminaries

Considern particles of masses m1,…,mn in ℝ2 subject to their mutual Newtonian gravitational interaction. In an inertial reference frame and choosing appropriate units, the equations of motion are(1)miq¨i=∂Uq∂qi,i=1,2,…,n,where(2)Uq=Uq1,q2,…,qn=∑1≤k<j≤nmkmjqk−qj,is the Newtonian potential of system (1). The position vector q=q1,q2,…,qn∈ℝ2n is often referred to as the configuration of the system.A configurationq=q1,…,qn∈ℝ2n with ∑i=1nmiqi=0 is called a central configuration if there exists some positive constant λ, called the Lagrangian multiplier, such that(3)−λqi=∑j=1,j≠inmjqj−qiqj−qi3,i=1,2,…,n.We are interested in the planarn=1+N-body problem, where the big mass is equal to 1 with position q0=0. The remaining N bodies with positions qi, called satellites, have masses mi=μiϵ,i=1,2,…,N, where μi∈ℝ+ and ϵ>0 is a small parameter that tends to zero. In all central configurations of the planar 1+N-body problem, the satellites lie on a circle centered at the big mass ([26]); that is, the satellites are coorbital.We exclude collisions in the definition of central configuration and take the anglesθi between two consecutive particles as coordinates; we refer to [2, 19, 26] for details. In these coordinates, the space of configuration is the simplex:(4)Δ=θ=θ1,θ2,…,θN:∑i=1Nθi=2π,θi>0,i=1,2,…,N,and the equations characterizing the central configurations of the planar 1+N-body problem are(5)μ2fθ1+μ3fθ1+θ2+⋯+μNfθ1+⋯+θN−1=0,μ3fθ2+μ4fθ2+θ3+⋯+μ1fθ2+⋯+θN=0,μ4fθ3+μ5fθ3+θ4+⋯+μ2fθ3+⋯+θN+θ1=0,…μNfθN−1+μ1fθN−1+θN+⋯+μN−2fθN−1+θN+θ1+⋯+θN−3=0,μ1fθN+μ2fθN+θ1+⋯+μN−1fθN+θ1+⋯+θN−2=0,θ1+⋯+θN=2π.where fθ=sinθ1−1/8sin3θ/2.In the case of the four-satellite system, Equation (5) is(6)μ2fθ1+μ3fθ1+θ2+μ4fθ1+θ2+θ3=0,μ3fθ2+μ4fθ2+θ3+μ1fθ2+θ3+θ4=0,μ4fθ3+μ1fθ3+θ4+μ2fθ3+θ4+θ1=0,μ1fθ4+μ2fθ4+θ1+μ3fθ4+θ1+θ2=0,θ1+θ2+θ3+θ4=2π.The functionfθ defined above plays a key role in this problem (Figure 1). The following two lemmas state some properties of f and its derivatives which will be used to prove our results. Lemma 1 can be found in [2, 20], and Lemma 2 can be proved straightforwardly.Figure 1
Graph off.Lemma 1.
The functionfθ satisfies the following:(1)
fπ/3=fπ=f5π/3=0.(2)
fπ−θ=−fπ+θ,f2π−θ=−fθ,∀θ∈0,π.(3)
f′θ=cosθ+3+cosθ/16sin3θ/2≥f′π=−7/8,forallθ∈0,2π.(4)
In0,π, there is a unique critical point θc of f, where 3π/5<θc<2π/3, such that f′θ>0 in 0,θc and f′θ<0 in θc,π. In π,2π, there is a unique critical point θl=2π−θc of f, where 4π/3<θl<7π/5, such that f′θ>0 in θl,2π and f′θ<0 in π,θl.Lemma 2.
f
″
θ
=
−
sin
θ
−
11
+
cos
θ
cos
θ
/
2
/
32
sin
4
θ
/
2, and(1)
f″θ<0 in 0,π and f″θ>0 in π,2π(2)
f′θ<0 in θc,θl and f′θ>0 in 0,θc∪θl,2π(3)
fθ<0 in 0,π/3∪π,5π/3 and fθ>0 in π/3,π∪5π/3,2πA central configurationθ1,θ2,…,θN of the planar 1+N-body problem is symmetric with respect to a straight line L containing the central body; since central configuration is invariant under rotation transformation, we have the following:(1)
WhenN is even, either(7)θ1=θN,θ2=θN−1,…,θN/2=θN+2/2,where the symmetry axisL contains two satellites (Figure 2(a)), or(8)θ1=θN−1,θ2=θN−2,…,θN−2/2=θN+2/2,where the symmetry axisL contains no satellites (Figure 2(b)).(2)
WhenN is odd,(9)θ1=θN,θ2=θN−1,…,θN−1/2=θN+3/2,where the symmetry axisL contains one satellite. In the case of the 1+4-body problem, the symmetric central configuration contains θ1=θ3 or θ1=θ4. Oliveira and Cabral [20] have completed the first case θ1=θ4, where the symmetry axis contains q1 and q3, and θ2=θ3. We consider the symmetric central configuration of the 1+4-body problem with θ1=θ3, where the symmetry axis does not contain any satellites, and θ2 and θ4 are not necessarily equal.Figure 2
Symmetric configuration for the 1 + 4 body problem. (a) Symmetric configuration withθ1=θ4. (b) Symmetric configuration with θ1=θ3.
(a)
(b)


## 3. Main Results

For the symmetric central configuration of the1+4-body problem, we consider the case θ1=θ3. Using the property of f that f2π−x=−fx, system (6) becomes(10)μ2fθ1+μ3fθ1+θ2−μ4fθ4=0,μ3fθ2+μ4fθ1+θ2−μ1fθ1=0,μ4fθ1−μ1fθ1+θ2−μ2fθ2=0,μ1fθ4−μ2fθ1+θ2−μ3fθ1=0,θ1+θ2+θ3+θ4=2π,where 0<θ2,θ4<2π,and0<θ1=θ3<π.Let(11)A=μ20−μ4μ3−μ1μ30μ4μ4−μ20−μ1−μ30μ1−μ2,X=fθ1fθ2fθ4fθ1+θ2,then system (10) can be rewritten as(12)AX=0.With simple computation, we havedetA=0 for any μ1,μ2,μ3,andμ4.Lemma 3.
Letθ1,θ2,θ3,θ4 be a symmetric central configuration solution of system (6) with θ1=θ3; then for fθi,i=1,2,4, and fθ1+θ2, there can be at most one to be zero.Proof.
First, we prove thatfθi,i=1,2,4, and fθ1+θ2 cannot be zero simultaneously. Since fθ1=0, and 0<θ1=θ3<π, θ1=θ3=π/3. By fθ1+θ2=0, we have θ1+θ2=π or 5π/3.Ifθ1+θ2=π, then θ2=θ4=2π/3, which is a contradiction with the fact fθ4=0.Ifθ1+θ2=5π/3, then θ2=4π/3 and θ4=0 which is a contradiction with θ4>0.
Moreover, when some two offθ1+θ2 and fθi,i=1,2,4, equal zero, with simple calculation, we obtain that they are all equal to zero. Thus, this proves Lemma 3.
By Lemma 3, we consider the following cases that one offθ1+θ2 and fθi,i=1,2,4, equals zero while the others are not zero, and also the case that all of fθi, i=1,2,4, and fθ1+θ2 are not zero.
f
θ
1
+
θ
2
=
0 implies θ1+θ2=π/3,π, or 5π/3.
Whenθ1+θ2=π, we get the following proposition which has already been obtained in [20], and we omit the proof.Proposition 1 (see [20]).
Consider a symmetric configurationθ1,θ2,θ3,θ4 of the 1 + 4-body problem with θ1=θ3 and θ1+θ2=π. There exists exactly one central configuration π/2,π/2,π/2,π/2 (Figure 3(a)) if and only if μ1=μ3andμ2=μ4.
By renumbering these satellites from the third one counterclockwisely, we can find that the case withθ1+θ2=5π/3 is equivalent to the case with θ1+θ2=π/3. Thus, we just consider the case that θ1+θ2=π/3 and get the following theorem.Figure 3
Central configurations for the 1 + 4 body problem withθ1+θ2=π and θ1+θ2=π/3. (a) Square central configuration for Proposition 1. (b) The only central configuration according to Theorem 1.
(a)
(b)Theorem 1.
Consider a symmetric configurationθ1,θ2,θ3,θ4 of the 1 + 4-body problem with θ1=θ3 and θ1+θ2=π/3. There exists exactly one central configuration θ0,π/3−θ0,θ0,5π/3−θ0 (Figure 3(b)) with θ0≈0.6281 if and only if μ1,μ2,μ3,andμ4 with μi∈ℝ+ satisfy μ1/μ3=μ4/μ2=fπ/3−θ0/fθ0, where θ0 is the unique root of fθ2−fπ/3−θf5π/3−θ=0 in 0,π/3.Whenfθ1=0, we get the following theorem.Theorem 2.
Consider a symmetric configurationθ1,θ2,θ3,θ4 of the 1 + 4-body problem with θ1=θ3 and fθ1=0. There exists no central configuration for any μ1,μ2,μ3,andμ4 with μi∈ℝ+.Also by renumbering these satellites from the third one counterclockwisely, we can find that the case withfθ4=0 is equivalent to the case with fθ2=0. Thus, we just consider the case that fθ2=0 and get the following theorem.Theorem 3.
Consider a symmetric configurationθ1,θ2,θ3,θ4 of the 1 + 4-body problem with θ1=θ3 and fθ2=0. There exists exactly one central configuration θ0,π/3,θ0,5π/3−2θ0 (Figure 4) with θ0≈1.4127 if and only if μ1,μ2,μ3,andμ4 with μi∈ℝ+ satisfy μ1=μ4 and μ2+μ3/μ1=f5π/3−2θ0/fθ0, where θ0 is the unique root of fπ/3+θ1−fθ1=0 in π/3,2π/3.Whenfθi≠0,i=1,2,4, and fθ1+θ2≠0, with numerical evidence, we get the following conjecture.Figure 4
The only central configuration according to Theorem 3.Conjecture 1.
Consider a symmetric configurationθ1,θ2,θ3,θ4 of the 1 + 4-body problem with θ1=θ3, fθi≠0,i=1,2,4, and fθ1+θ2≠0 in order that the four satellites are in a central configuration; μ1,μ2,μ3,andμ4 follow μ1=μ4, μ2=μ3 or μ1≠μ4, and μ2≠μ3. Furthermore, for suitable μ1,μ2,μ3,andμ4 with μi∈ℝ+, there exist two central configurations for any θ2∈0,π/3 (in Table 1) and one symmetric central configuration for any θ2∈π/3,π∪π,5π/3 (in Tables 2 and 3).Table 1
Some numerical results ofθ1, θ2, μ1+μ4/μ2+μ3, and μ1−μ4/μ3−μ2 along F−10∩D1 and related conditions of μ1,μ2,μ3,andμ4.
θ
1
θ
2
μ
1
+
μ
4
/
μ
2
+
μ
3
μ
1
−
μ
4
/
μ
3
−
μ
2
0.4922
0.1
73.2972
17.0322
0.5199
0.2
12.8052
5.6441
0.5692
0.3
5.4154
3.5674
0.6190
0.4
3.0292
2.8349
0.6647
0.5
1.8989
2.5298
0.7053
0.6
1.2437
2.4327
0.7404
0.7
0.8125
2.4765
0.7695
0.8
0.5023
2.6403
0.7927
0.9
0.2653
2.9459
0.8102
1.0
0.0766
3.4274
1.5212
0.1
1.8820e + 03
−77.5064
1.4739
0.2
234.2838
−19.3970
1.4312
0.3
69.1611
−8.5529
1.3951
0.4
29.1372
−4.6917
1.3675
0.5
14.9637
−2.8510
1.3498
0.6
8.7330
−1.8086
1.3432
0.7
5.5832
−1.1469
1.3486
0.8
3.8296
−0.6917
1.3662
0.9
2.7810
−0.3594
1.3954
1.0
2.1227
−0.1033Table 2
Some numerical results ofθ1, θ2, μ1+μ4/μ2+μ3, and μ1−μ4/μ3−μ2 along F−10∩D2 and related conditions of μ1,μ2,μ3,andμ4.
θ
1
θ
2
μ
1
+
μ
4
/
μ
2
+
μ
3
μ
1
−
μ
4
/
μ
3
−
μ
2
1.4341
1.1
1.6921
0.1065
1.4779
1.2
1.4123
0.2922
1.5196
1.3
1.2310
0.4710
1.5511
1.4
1.1158
0.6551
1.5677
1.5
1.0406
0.8512
1.5703
1.6
0.9848
1.0645
1.5626
1.7
0.9355
1.3035
1.5479
1.8
0.8868
1.5839
1.5289
1.9
0.8357
1.9341
1.5073
2.0
0.7816
2.4105
1.4843
2.1
0.7242
3.1410
1.4608
2.2
0.6637
4.4956
1.4375
2.3
0.6003
8.1788
1.4150
2.4
0.5343
79.7431
1.3938
2.5
0.4661
−8.7964
1.3748
2.6
0.3959
−3.7527
1.3588
2.7
0.3241
−2.1595
1.3473
2.8
0.2509
−1.3575
1.3432
2.9
0.1768
−0.8588
1.3528
3.0
0.1024
−0.5000
1.4020
3.1
0.0292
−0.1902Table 3
Some numerical results ofθ1, θ2, μ1+μ4/μ2+μ3, and μ1−μ4/μ3−μ2 along F−10∩D3 and related conditions of μ1,μ2,μ3,andμ4.
θ
1
θ
2
μ
1
+
μ
4
/
μ
2
+
μ
3
μ
1
−
μ
4
/
μ
3
−
μ
2
0.8058
3.7
7.9282
0.3052
0.7924
3.8
3.7257
0.3402
0.7769
3.9
2.3431
0.3684
0.7597
4.0
1.6559
0.3897
0.7408
4.1
1.2385
0.4036
0.7207
4.2
0.9579
0.4105
0.6994
4.3
0.7535
0.4101
0.6771
4.4
0.5969
0.4024
0.6539
4.5
0.4723
0.3874
0.6300
4.6
0.3704
0.3652
0.6055
4.7
0.2855
0.3359
0.5805
4.8
0.2135
0.2992
0.5553
4.9
0.1519
0.2552
0.5303
5.0
0.0988
0.2028
0.5071
5.1
0.0530
0.1404
0.4922
5.2
0.0132
0.0573


## 4. The Proof of Theorems 1–3

### 4.1. The Proof of Theorem 1

Whenfθ1+θ2=0, θ1=θ3, and fθi≠0,i=1,2,4, system (10) reduces to(13)μ2fθ1=μ4fθ4,μ3fθ2=μ1fθ1,μ4fθ1=μ2fθ2,μ1fθ4=μ3fθ1,2θ1+θ2+θ4=2π.By the first and the fourth equation of (13), we have μ1μ2=μ3μ4, and by the second and the third equation of (13), we get the same conclusion. This means μ1μ2=μ3μ4 is a necessary condition for the existence of the central configuration under these assumptions. Thus, Equation (13) is equivalent to(14)μ2fθ1=μ4fθ4,μ3fθ2=μ1fθ1,μ1μ2=μ3μ4,2θ1+θ2+θ4=2π.System (14) yields(15)fθ12=fθ2fθ4,and the sign of fθi must be the same for all i=1,2,4.Whileθ1+θ2=π/3, let θ1=θ3=θ, then θ2=π/3−θ and θ4=5π/3−θ, where 0<θ<π/3. Substituting these assumptions into (15), we have(16)F1θ=fθ2−fπ3−θf5π3−θ=0.For0<θ<π/3, f′θ>0 and fθ<0. So(17)fθ2′=2fθf′θ<0.We are going to prove that−fπ/3−θf5π/3−θ monotonically decreases in 0<θ<π/3. With simple computation,(18)−fπ3−θf5π3−θ′=f′π3−θf5π3−θ+fπ3−θf′5π3−θ.We divide the proof into two cases:Case 1.
(0<θ≤5π/3−θl=θc−π/3). When 0<θ≤5π/3−θl=θc−π/3<π/3, fπ/3−θ<0, f5π/3−θ<0, f′π/3−θ>0, and f′5π/3−θ≥0. So(19)f′π3−θf5π3−θ+fπ3−θf′5π3−θ<0.Case 2.
(5π/3−θl=θc−π/3<θ<π/3). When 0<5π/3−θl=θc−π/3<θ<π/3, (20)3+cosπ3−θ−2sinπ3−θ>0,sinπ3−θ>0,cosπ3−θ>0.So(21)fπ3−θ+f′π3−θ=sinπ3−θ+cosπ3−θ+3+cosπ/3−θ−2sinπ/3−θ16sin3π/3−θ/2>0,that is,(22)f′π3−θ>−fπ3−θ>0.Considering the functionG1θ=f5π/3−θ−f′5π/3−θ, it is easily computed that(23)G1′θ=f″5π3−θ−f′5π3−θ>0,for f″5π/3−θ>0, −f′5π/3−θ>0. Then, G1θ<G1π/3=−23/363+2/3<0. So(24)−f5π3−θ>−f′5π3−θ>0.Hence, by (22) and (24), we again get(25)f′π3−θf5π3−θ+fπ3−θf′5π3−θ<0.Cases 1 and 2, and (17) give that F1θ=fθ2−fπ/3−θf5π/3−θ monotonically decreases in 0<θ<π/3. It is easy to see that F1θ⟶+∞ as θ⟶0+ and F1θ⟶−∞ as θ⟶π/3−. So there is exactly one root θ0 of F1θ=fθ2−fπ/3−θf5π/3−θ=0 in 0,π/3. By (13), we get μ1/μ3=μ4/μ2=fπ/3−θ0/fθ0>0. Also by choosing suitable μi∈ℝ+ satisfying μ1/μ3=μ4/μ2=fπ/3−θ0/fθ0, we can get exactly one central configuration θ0,π/3−θ0,θ0,5π/3−θ0 satisfying (13). Thus, Theorem 1 is proved.


### 4.2. The Proof of Theorem 2

Whenfθ1=0, θ1=θ3, system (10) becomes(26)μ3fθ1+θ2−μ4fθ4=0,μ3fθ2+μ4fθ1+θ2=0,μ1fθ1+θ2+μ2fθ2=0,μ1fθ4−μ2fθ1+θ2=0,2θ1+θ2+θ4=2π.With simple calculation, system (26) is equivalent to(27)μ1μ3=μ2μ4,fθ2fθ4+fθ1+θ22=0,μ1fθ1+θ2+μ2fθ2=0,2θ1+θ2+θ4=2π.Forfθ1=0 and θ1=θ3, we get θ1=θ3=π/3 and θ4=4π/3−θ2 where 0<θ2<4π/3. The second equation of (27) becomes(28)fθ2f4π3−θ2+fπ3+θ22=0.We consider four subcases according toθ2 in 0,π/3, π/3,π, or π,4π/3, respectively, or according to θ2 equal to π/3 or π. Assuming that θ2∈0,π/3, we have 4π/3−θ2∈π,4π/3, and then from the plot of f, we have fθ2<0 and f4π/3−θ2<0. Now suppose that θ2∈π/3,π, then 4π/3−θ2∈π/3,π; similarly, we have fθ2>0 and f4π/3−θ2>0. When θ2∈π,4π/3, we have 4π/3−θ2∈0,π/3, and this implies fθ2<0 and f4π/3−θ2>0. Finally, when θ2=π/3 or π, fθ2f4π/3−θ2+fπ/3+θ22=fπ/3+θ22>0. The above analysis shows that (28) does not hold. Theorem 2 is proved.


### 4.3. The Proof of Theorem 3

Whenfθ2=0 and θ1=θ3, system (10) becomes(29)μ2fθ1+μ3fθ1+θ2−μ4fθ4=0,μ4fθ1+θ2−μ1fθ1=0,μ4fθ1−μ1fθ1+θ2=0,μ1fθ4−μ2fθ1+θ2−μ3fθ1=0,2θ1+θ2+θ4=2π.From the second and the third equation, we obtainμ1=μ4 and fθ1+θ2=fθ1. This means μ1=μ4 is a necessary condition for the existence of the central configuration under these assumptions. Then, system (29) is equivalent to(30)μ1=μ4,fθ1+θ2=fθ1,μ2+μ3fθ1−μ1fθ4=0,2θ1+θ2+θ4=2π.f
θ
2
=
0 implies that θ2=π/3,π, or 5π/3.Whenθ2=π/3, θ4=5π/3−2θ1 where 0<θ1<5π/6 and θ1≠π/3 for fθ1≠0. The third equation of (30) means that fθ1 and fθ4 have the same sign. When fθ1>0 and fθ4>0, we have π/3<θ1<2π/3. When fθ1<0 and fθ4<0, we have 0<θ1<π/3. Then, we consider two subcases according to θ1 in 0,π/3 or π/3,2π/3. Hence, we must solve(31)fπ3+θ1−fθ1=0,forθ∈0,π3∪π3,2π3.Clearly,f′π/3+θ1−f′θ1<0 in 0,2π/3 because f″x<0 in 0,π.It is easy to see thatfπ/3+θ1−fθ1⟶+∞ as θ1⟶0+, fπ/3+θ1−fθ1⟶f2π/3>0 as θ1⟶π/3, and fπ/3+θ1−fθ1⟶−f2π/3<0 as θ1⟶2π/3−. So fπ/3+θ1−fθ1=0 must have exactly one root θ0≈1.4127 in π/3,2π/3. Then, by the third equation of system (30), we get μ2+μ3/μ1=f5π/3−2θ0/fθ0>0.Whenθ2=π, θ4=π−2θ1 where 0<θ1<π/2. If θ1∈0,π/3, then θ4=π−2θ1∈π/3,π, which is a contradiction with the fact that fθ1 and fθ4 must have the same sign. If θ1∈π/3,π/2, then θ4=π−2θ1∈0,π/3, and again, we have the same contradiction.Now considerθ2=5π/3, then θ4=π/3−2θ1 where 0<θ1<π/6. From the plot of f, we have fθ1<0 and fθ1+θ2=fθ1+5π/3>0. This means the second equation of (30) does not hold.The above analysis shows thatμ1=μ4 and μ2+μ3/μ1=f5π/3−2θ0/fθ0 must be satisfied for the existence of the central configuration. Also by choosing suitable μi∈ℝ+ satisfying μ1=μ4 and μ2+μ3/μ1=f5π/3−2θ0/fθ0, we can get exactly one central configuration θ0,π/3,θ0,5π/3−2θ0 satisfying (29). Thus, Theorem 3 is proved.


## 4.1. The Proof of Theorem 1

Whenfθ1+θ2=0, θ1=θ3, and fθi≠0,i=1,2,4, system (10) reduces to(13)μ2fθ1=μ4fθ4,μ3fθ2=μ1fθ1,μ4fθ1=μ2fθ2,μ1fθ4=μ3fθ1,2θ1+θ2+θ4=2π.By the first and the fourth equation of (13), we have μ1μ2=μ3μ4, and by the second and the third equation of (13), we get the same conclusion. This means μ1μ2=μ3μ4 is a necessary condition for the existence of the central configuration under these assumptions. Thus, Equation (13) is equivalent to(14)μ2fθ1=μ4fθ4,μ3fθ2=μ1fθ1,μ1μ2=μ3μ4,2θ1+θ2+θ4=2π.System (14) yields(15)fθ12=fθ2fθ4,and the sign of fθi must be the same for all i=1,2,4.Whileθ1+θ2=π/3, let θ1=θ3=θ, then θ2=π/3−θ and θ4=5π/3−θ, where 0<θ<π/3. Substituting these assumptions into (15), we have(16)F1θ=fθ2−fπ3−θf5π3−θ=0.For0<θ<π/3, f′θ>0 and fθ<0. So(17)fθ2′=2fθf′θ<0.We are going to prove that−fπ/3−θf5π/3−θ monotonically decreases in 0<θ<π/3. With simple computation,(18)−fπ3−θf5π3−θ′=f′π3−θf5π3−θ+fπ3−θf′5π3−θ.We divide the proof into two cases:Case 1.
(0<θ≤5π/3−θl=θc−π/3). When 0<θ≤5π/3−θl=θc−π/3<π/3, fπ/3−θ<0, f5π/3−θ<0, f′π/3−θ>0, and f′5π/3−θ≥0. So(19)f′π3−θf5π3−θ+fπ3−θf′5π3−θ<0.Case 2.
(5π/3−θl=θc−π/3<θ<π/3). When 0<5π/3−θl=θc−π/3<θ<π/3, (20)3+cosπ3−θ−2sinπ3−θ>0,sinπ3−θ>0,cosπ3−θ>0.So(21)fπ3−θ+f′π3−θ=sinπ3−θ+cosπ3−θ+3+cosπ/3−θ−2sinπ/3−θ16sin3π/3−θ/2>0,that is,(22)f′π3−θ>−fπ3−θ>0.Considering the functionG1θ=f5π/3−θ−f′5π/3−θ, it is easily computed that(23)G1′θ=f″5π3−θ−f′5π3−θ>0,for f″5π/3−θ>0, −f′5π/3−θ>0. Then, G1θ<G1π/3=−23/363+2/3<0. So(24)−f5π3−θ>−f′5π3−θ>0.Hence, by (22) and (24), we again get(25)f′π3−θf5π3−θ+fπ3−θf′5π3−θ<0.Cases 1 and 2, and (17) give that F1θ=fθ2−fπ/3−θf5π/3−θ monotonically decreases in 0<θ<π/3. It is easy to see that F1θ⟶+∞ as θ⟶0+ and F1θ⟶−∞ as θ⟶π/3−. So there is exactly one root θ0 of F1θ=fθ2−fπ/3−θf5π/3−θ=0 in 0,π/3. By (13), we get μ1/μ3=μ4/μ2=fπ/3−θ0/fθ0>0. Also by choosing suitable μi∈ℝ+ satisfying μ1/μ3=μ4/μ2=fπ/3−θ0/fθ0, we can get exactly one central configuration θ0,π/3−θ0,θ0,5π/3−θ0 satisfying (13). Thus, Theorem 1 is proved.


## 4.2. The Proof of Theorem 2

Whenfθ1=0, θ1=θ3, system (10) becomes(26)μ3fθ1+θ2−μ4fθ4=0,μ3fθ2+μ4fθ1+θ2=0,μ1fθ1+θ2+μ2fθ2=0,μ1fθ4−μ2fθ1+θ2=0,2θ1+θ2+θ4=2π.With simple calculation, system (26) is equivalent to(27)μ1μ3=μ2μ4,fθ2fθ4+fθ1+θ22=0,μ1fθ1+θ2+μ2fθ2=0,2θ1+θ2+θ4=2π.Forfθ1=0 and θ1=θ3, we get θ1=θ3=π/3 and θ4=4π/3−θ2 where 0<θ2<4π/3. The second equation of (27) becomes(28)fθ2f4π3−θ2+fπ3+θ22=0.We consider four subcases according toθ2 in 0,π/3, π/3,π, or π,4π/3, respectively, or according to θ2 equal to π/3 or π. Assuming that θ2∈0,π/3, we have 4π/3−θ2∈π,4π/3, and then from the plot of f, we have fθ2<0 and f4π/3−θ2<0. Now suppose that θ2∈π/3,π, then 4π/3−θ2∈π/3,π; similarly, we have fθ2>0 and f4π/3−θ2>0. When θ2∈π,4π/3, we have 4π/3−θ2∈0,π/3, and this implies fθ2<0 and f4π/3−θ2>0. Finally, when θ2=π/3 or π, fθ2f4π/3−θ2+fπ/3+θ22=fπ/3+θ22>0. The above analysis shows that (28) does not hold. Theorem 2 is proved.


## 4.3. The Proof of Theorem 3

Whenfθ2=0 and θ1=θ3, system (10) becomes(29)μ2fθ1+μ3fθ1+θ2−μ4fθ4=0,μ4fθ1+θ2−μ1fθ1=0,μ4fθ1−μ1fθ1+θ2=0,μ1fθ4−μ2fθ1+θ2−μ3fθ1=0,2θ1+θ2+θ4=2π.From the second and the third equation, we obtainμ1=μ4 and fθ1+θ2=fθ1. This means μ1=μ4 is a necessary condition for the existence of the central configuration under these assumptions. Then, system (29) is equivalent to(30)μ1=μ4,fθ1+θ2=fθ1,μ2+μ3fθ1−μ1fθ4=0,2θ1+θ2+θ4=2π.f
θ
2
=
0 implies that θ2=π/3,π, or 5π/3.Whenθ2=π/3, θ4=5π/3−2θ1 where 0<θ1<5π/6 and θ1≠π/3 for fθ1≠0. The third equation of (30) means that fθ1 and fθ4 have the same sign. When fθ1>0 and fθ4>0, we have π/3<θ1<2π/3. When fθ1<0 and fθ4<0, we have 0<θ1<π/3. Then, we consider two subcases according to θ1 in 0,π/3 or π/3,2π/3. Hence, we must solve(31)fπ3+θ1−fθ1=0,forθ∈0,π3∪π3,2π3.Clearly,f′π/3+θ1−f′θ1<0 in 0,2π/3 because f″x<0 in 0,π.It is easy to see thatfπ/3+θ1−fθ1⟶+∞ as θ1⟶0+, fπ/3+θ1−fθ1⟶f2π/3>0 as θ1⟶π/3, and fπ/3+θ1−fθ1⟶−f2π/3<0 as θ1⟶2π/3−. So fπ/3+θ1−fθ1=0 must have exactly one root θ0≈1.4127 in π/3,2π/3. Then, by the third equation of system (30), we get μ2+μ3/μ1=f5π/3−2θ0/fθ0>0.Whenθ2=π, θ4=π−2θ1 where 0<θ1<π/2. If θ1∈0,π/3, then θ4=π−2θ1∈π/3,π, which is a contradiction with the fact that fθ1 and fθ4 must have the same sign. If θ1∈π/3,π/2, then θ4=π−2θ1∈0,π/3, and again, we have the same contradiction.Now considerθ2=5π/3, then θ4=π/3−2θ1 where 0<θ1<π/6. From the plot of f, we have fθ1<0 and fθ1+θ2=fθ1+5π/3>0. This means the second equation of (30) does not hold.The above analysis shows thatμ1=μ4 and μ2+μ3/μ1=f5π/3−2θ0/fθ0 must be satisfied for the existence of the central configuration. Also by choosing suitable μi∈ℝ+ satisfying μ1=μ4 and μ2+μ3/μ1=f5π/3−2θ0/fθ0, we can get exactly one central configuration θ0,π/3,θ0,5π/3−2θ0 satisfying (29). Thus, Theorem 3 is proved.


## 5. The Numerical Evidence for Conjecture 1

Now considerfθi≠0,i=1,2,4, and fθ1+θ2≠0. System (10) can be written as the following homogeneous linear system of equations:(32)0fθ1fθ1+θ2−fθ4−fθ10fθ2fθ1+θ2−fθ1+θ2−fθ20fθ1fθ4−fθ1+θ2−fθ10μ1μ2μ3μ4=Aμ1μ2μ3μ4=0,where θ4=2π−2θ1−θ2, 0<θ1<π.In order to get the nonzero mass vector, the determinant of the matrixA vanishes, that is,(33)fθ1+θ22−fθ12+fθ2fθ4=0,and the determinant of the matrix A is the square of the left-hand side of the equation above.Since the matrixA is antisymmetric, from a classical theorem of linear algebra, the rank of A is even. Under the above constraint, it is easy to get that the rank of matrix A is 2. This means we can perform row operations on the matrix A to create a new matrix that produces an equivalent system. The following are some notations of three elementary matrix row operations. Ri⟺Rj means interchanging the i-th and j-th rows. aRi⟶Ri means replacing the i-th row with a times itself. Ri+Rj⟶Rj means replacing the j-th row with the sum of i-th and j-th rows. For fθi≠0,i=1,2,4, and fθ1+θ2≠0, performing the row operations fθ1+θ2/fθ4R4+R3⟶R3 and fθ1/fθ4R4+R2⟶R2 on matrix A, we get the new matrix and denote it by A1. And then performing the row operations −fθ4/fθ1+θ2R2⟶R2 and −fθ4/fθ1R3⟶R3 on matrix A1, we get matrix A2. Finally, performing the row operations −R1+R2⟶R2 and −R1+R3⟶R3 on matrix A2, we get the following matrix:(34)A3=0fθ1fθ1+θ2−fθ400000000fθ4−fθ1+θ2−fθ10.Thus, under these assumptions and constraint (33), system (32) reduces to(35)fθ1+fθ1+θ2μ2+μ3=fθ4μ4+μ1,fθ1−fθ1+θ2μ2−μ3=fθ4μ4−μ1.By (33), fθ1+fθ1+θ2 and fθ1−fθ1+θ2 cannot be zero; otherwise, fθ2=0 or fθ4=0 which is a contradiction with the above assumptions in this section. Then, Equation (35) implies that μ1,μ2,μ3,andμ4 follow μ1=μ4, μ2=μ3 or μ1≠μ4, and μ2≠μ3.Whenμ1≠μ4 and μ2≠μ3, because of constraint (33), Equation (35) becomes(36)μ4+μ1μ2+μ3=fθ2fθ1−fθ1+θ2,μ4−μ1μ2−μ3=fθ2fθ1+fθ1+θ2.Whenμ1=μ4 and μ2=μ3, because of constraint (33), Equation (35) reduces to(37)μ1μ2=fθ2fθ1−fθ1+θ2.We only consider the definite positive values ofμ1,μ2,μ3,andμ4, so the signs of fθ2 and fθ1−fθ1+θ2 must be the same. When 5π/3<θ2<2π, we have 5π/3<θ1+θ2<2π and 0<θ1<π/3; then, fθ2>0 and fθ1−fθ1+θ2<0. Thus, μ1/μ2<0. In the following, we just consider θ2∈0,π/3∪π/3,π∪π,5π/3.Forθ4=2π−2θ1−θ2, we get 0<2θ1+θ2<2π. Let(38)Fθ1,θ2=fθ12−fθ1+θ22−fθ2f2π−2θ1−θ2,Hθ1,θ2=fθ1−fθ1+θ2,D1=θ1,θ2:0<θ2<π3,Hθ1,θ2<0,0<θ1<π,0<2θ1+θ2<2π,D2=θ1,θ2:π3<θ2<π,Hθ1,θ2>0,0<θ1<π,0<2θ1+θ2<2π,D3=θ1,θ2:π<θ2<5π3,Hθ1,θ2<0,0<θ1<π,0<2θ1+θ2<2π,F−10=θ1,θ2:Fθ1,θ2=0.Obviously, the signs offθ2 and fθ1−fθ1+θ2 in D1, D2, and D3 are the same; thus, the mass ratio μ1+μ4/μ2+μ3 is guaranteed to be positive.For0<2θ1+θ2<2π, we just consider the region below the straight line 2θ1+θ2=2π (Figure 5(a)). The sign of fθ1−fθ1+θ2 is determined in Figure 5(b). The regions D1, D2, and D3 can be seen in Figure 5(c), and the curve Fθ1,θ2=0 is plotted in Figure 5(d).Figure 5
Graphs for Conjecture 1. (a) The straight linesθ2=π/3, θ2=π, θ2=5π/3 and 2θ1+θ2=2π⋅2θ1+θ2<2π must be satisfied for θ4=2π−2θ1−θ2>0, fθ2>0, for θ2∈0,π/3∪π,5π/3. (b) The sign of fθ1−fθ1+θ2 and the curve Hθ1+θ2=0. (c) The regions D1, D2, and D3 defined in (38). (d) The curve Fθ1,θ2=0 in D1, D2, and D3: the curve segments AC, DE, AB, and GH without the end points. Points A, K, and J correspond to the central configurations in Theorems 1 and 3 and Proposition 1, respectively. Points L and M correspond to the same central configurations in Theorems 1 and 3, respectively.
(a)
(b)
(c)
(d)The end points ofFθ1,θ2=0 in D1, D2, and D3 are the following, respectively: A, E, D, and C; F, B, A, and N; and H, G, and F.D and C are the intersection points of the curve Fθ1,θ2=0 and the horizontal line θ2=0. Along the curve Fθ1,θ2=0, as θ2 tends to zero, we get f2π−2θ1−θ2⟶0 which is equivalent to θ1⟶π/6,π/2, or 5π/6. Thus, we get the coordinates of D and C: π/6,0 and π/2,0. Both of them do not correspond to any central configuration.A, E, and N are the intersection points of the curve Fθ1,θ2=0 and the horizontal line θ2=π/3. Notice that point A is also on the curve Hθ1,θ2=0, and then A satisfies θ2=π/3 and fθ1−fθ1+θ2=0 which corresponds to the central configuration θ0,π/3,θ0,5π/3−2θ0 with θ0≈1.4127 of Theorem 3. Point E is not on the curve Hθ1,θ2=0, and we have fθ1−fθ1+θ2≠0 which is a contradiction with the second equation of (30). Furthermore, we get fθ1+fθ1+θ2=0 from Fθ1,θ2=0 and θ2=π/3. Thus, point E satisfying θ2=π/3 and fθ1+fθ1+θ2=0 does not correspond to any central configuration. Point N is also on the line 2θ1+θ2=2π, and then the coordinates of N are 5π/6,π/3. Furthermore, we get θ4=2π−2θ1−θ2=0 which does not correspond to any central configuration.B and F are the intersection points of the curve Fθ1,θ2=0 and the horizontal line θ2=π. Point B is also on the line 2θ1+θ2=2π; then, the coordinates of B are π/2,π, and we get θ4=2π−2θ1−θ2=0 which does not correspond to any central configuration. Point F is also on the curve Hθ1,θ2=0, and we can get the coordinates of F by θ2=π and fθ1−fθ1+θ2=0 where θ1∈0,π/2. Theorem 3 implies that point F does not correspond to any central configuration.H is the intersection point of the curve Fθ1,θ2=0 and the horizontal line θ2=5π/3. Point H is also on the line 2θ1+θ2=2π, and then the coordinates of H are π/6,5π/3. Furthermore, we get θ4=2π−2θ1−θ2=0 which does not correspond to any central configuration.G is the intersection point of the curves Fθ1,θ2=0 and Hθ1,θ2=0. It is easy to find that fθ4=0 for fθ2≠0. Unfortunately, substituting fθ4=0 into (10), we have fθ1+fθ1+θ2=0 which is a contradiction with Hθ1,θ2=0. Thus, G does not correspond to any central configuration.PointJ satisfying θ1+θ2=π/3 and Fθ1,θ2=0 in D1 corresponds to the central configuration θ0,π/3−θ0,θ0,5π/3−θ0 in Theorem 1 where θ0≈0.6281 is the unique root of fθ2−fπ/3−θf5π/3−θ=0 in 0,π/3. Point L satisfying θ1+θ2=5π/3 and Fθ1,θ2=0 in D3 corresponds to the central configuration θ0,5π/3−θ0,θ0,π/3−θ0 which is the same with central configuration in Theorem 1. Point M satisfying fθ4=0 and Fθ1,θ2=0 in D2 corresponds to the central configuration θ0,5π/3−2θ0,θ0,π/3 which is the same with central configuration in Theorem 3, where θ0≈1.4127 is the unique root of fπ/3+θ1−fθ1=0 in π/3,2π/3. Point Kπ/2,π/2 corresponds to the central configuration π/2,π/2,π/2,π/2 in Proposition 1.In Tables1–3, we give some numerical results for θ1, θ2, μ1+μ4/μ2+μ3, and μ1−μ4/μ3−μ2 along F−10∩D1, F−10∩D2, and F−10∩D3, respectively, when μ2≠μ3andμ1≠μ4. If μ2=μ3andμ1=μ4, we get the same values for θ1, θ2, and μ1/μ2 in the first three columns of tables. In Table 1, for any θ2∈0,π/3, there exist two values of θ1 which correspond to two central configurations for suitable μ1,μ2,μ3,andμ4 with μi∈ℝ+. In Tables 2 and 3, for any θ2 in π/3,π and π,5π/3, there exists exactly one value of θ1 which corresponds to one central configuration for suitable μ1,μ2,μ3,andμ4 with μi∈ℝ+.



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