Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 173639, 8 pages
http://dx.doi.org/10.1155/2013/173639
Research Article
Periodic Solutions for Circular Restricted 𝑁 + 1-Body Problems
Xiaoxiao Zhao,1 Wei Li,2 and Peng Wang1
1
2
College of Mathematics, Sichuan University, Chengdu 610064, China
School of Economic Mathematics, Southwestern University of Finance and Economics, Chengdu 610074, China
Correspondence should be addressed to Xiaoxiao Zhao; zxgg2007@163.com
Received 18 February 2013; Accepted 3 April 2013
Academic Editor: Baodong Zheng
Copyright © 2013 Xiaoxiao Zhao et al. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
For circular restricted 𝑁 + 1-body problems, we study the motion of a sufficiently small mass point (called the zero mass point) in
the plane of N equal masses located at the vertices of a regular polygon. By using variational minimizing methods, for some N, we
prove the existence of the noncollision periodic solution for the zero mass point with some fixed wingding number.
1. Introduction and Main Results
In this paper, we study the planar circular restricted 𝑁 + 1body problems. Suppose that points of positive masses 𝑚1 =
⋅ ⋅ ⋅ = 𝑚𝑁 = 1 move on a circular orbit around the center
of masses, and that the sufficiently small mass point, called
the zero mass point, moves on the moving plane of the
given 𝑁 equal masses and does not influence the motion of
𝑚1 , . . . , 𝑚𝑁, but the motion of the zero mass point is affected
√
by the given 𝑁 equal mass points. Let 𝜌𝑖 = 𝑒 −1(2𝜋𝑖/𝑁) and
we denote the position vectors of the given 𝑁 bodies by 𝑞𝑖 (𝑡),
𝑖 = 1, . . . , 𝑁, then
√−12𝜋𝑡
𝑞1 (𝑡) = 𝑟𝑒
√−12𝜋𝑡
𝑞𝑖 (𝑡) = 𝑟𝑒
𝜌1 , . . . ,
√−12𝜋𝑡
𝑞𝑁 (𝑡) = 𝑟𝑒
𝜌𝑖 , . . . ,
(1)
,
where the radius 𝑟 > 0. It is known that 𝑞1 (𝑡), . . . , 𝑞𝑁(𝑡) satisfy
the following Newtonian equations:
𝑚𝑖 𝑞𝑖̈ =
where
𝑈=
𝜕𝑈
,
𝜕𝑞𝑖
𝑖 = 1, . . . , 𝑁,
𝑚𝑖 𝑚𝑗
∑ 󵄨󵄨
󵄨󵄨 .
1≤𝑖<𝑗≤𝑁 󵄨󵄨󵄨𝑞𝑖 − 𝑞𝑗 󵄨󵄨󵄨
(2)
(3)
We also assume that
𝑁
𝑁
𝑖=1
𝑖=1
∑𝑚𝑖 𝑞𝑖 = ∑𝑞𝑖 = 0.
(4)
The orbit 𝑞(𝑡) ∈ 𝑅2 for the zero mass point is governed by the
gravitational forces of 𝑚1 , . . . , 𝑚𝑁 and therefore it satisfies the
following equation:
𝑁
𝑚 (𝑞 − 𝑞)
𝑞 ̈ = ∑ 󵄨𝑖 𝑖 󵄨3 .
󵄨
󵄨
𝑖=1 󵄨󵄨𝑞𝑖 − 𝑞󵄨󵄨
(5)
For 𝑁-body problems, there are many papers concerned
with the periodic solutions by using variational methods;
see [1–15] and the references therein. In [1], Chenciner and
Montgomery proved the existence of the remarkable figure“8” type periodic solution for planar Newtonian 3-body
problems with equal masses. Marchal [4] studied the fixed
end problem for Newtonian 𝑛-body problems and proved
that the minimizer for the Lagrangian action has no interior
collision. In [6], Simó used computer to discover many new
periodic solutions for Newtonian 𝑛-body problems. Zhang
and Zhou [10–12] decomposed the Lagrangian action for 𝑛body problems into some sum for two body problems and
[11, 12] avoided collisions by comparing the lower bound for
the Lagrangian action on the symmetry collision orbits and
the upper bound for the Lagrangian action on test orbits in
some cases.
Motivated by the above works, we use variational methods to study the circular restricted 𝑁 + 1-body problem with
some fixed wingding numbers and 𝑁 equal masses.
For the readers’ conveniences, we recall the definition of
the winding number, which can be found in many books on
the classical differential geometry.
2
Abstract and Applied Analysis
Definition 1. Let Γ : 𝑥(𝑡), 𝑡 ∈ [𝑎, 𝑏] be a given oriented
continuous closed curve, and 𝑝 be a point on the plane not
on the curve. Then, the mapping 𝜑 : Γ → 𝑆1 given by
𝑥 (𝑡) − 𝑝
𝜑 (𝑥 (𝑡)) = 󵄨󵄨
󵄨,
󵄨󵄨𝑥 (𝑡) − 𝑝󵄨󵄨󵄨
𝑡 ∈ [𝑎, 𝑏]
(6)
is defined to be the position mapping of the curve Γ relative
to 𝑝, and when the point on Γ goes around the curve once, its
image point 𝜑(𝑥(𝑡)) will go around 𝑆1 a number of times; this
number is called the winding number of the curve Γ relative
to 𝑝, and we denote it by deg(Γ, 𝑝). If 𝑝 is the origin, we write
deg Γ.
Let
2𝜋
2𝜋
− sin
cos
𝑛
𝑛
) ∈ 𝑆𝑂 (2)
𝑅 (𝑛) = (
2𝜋
2𝜋
sin
cos
𝑛
𝑛
Define
𝑅
𝑊1,2 ( , 𝑅2 )
𝑍
= {𝑥 (𝑡) | 𝑥 (𝑡) , 𝑥̇ (𝑡) ∈ 𝐿2 (𝑅, 𝑅2 ) , 𝑥 (𝑡 + 1) = 𝑥 (𝑡)} .
(8)
The norm of 𝑊1,2 (𝑅/𝑍, 𝑅2 ) is
1
‖𝑥‖ = [∫ |𝑥|2 𝑑𝑡]
0
1/2
1
+ [∫ |𝑥|̇ 2 𝑑𝑡]
1/2
0
.
We consider the Lagrangian functional of (5) as
(7)
be a counter-clockwise rotation of angle 2𝜋/𝑛 in 𝑅2 .
1
1 󵄨 󵄨2 𝑁 1
𝑓 (𝑞) = ∫ [ 󵄨󵄨󵄨𝑞󵄨󵄨󵄨̇ + ∑ 󵄨󵄨
󵄨󵄨 ] 𝑑𝑡
2
0
𝑖=1 󵄨󵄨𝑞 − 𝑞𝑖 󵄨󵄨
Our main results are the following.
Theorem 2. For 𝑁 = 2, 3, the minimizer of 𝑓(𝑞) on the closure
Λ− of Λ − is a noncollision 1-periodic solution of (5).
Remark 3. In proving Theorem 2, we need to use test
functions. We find that when 𝑁 = 3, if the test functions
are circular orbits, we cannot get the desired results on Λ− .
Therefore, we select elliptic orbits as test functions.
Theorem 4. For 2 ≤ 𝑁 ≤ 32402, the minimizer of 𝑓(𝑞) on the
closure Λ+ of Λ + is a noncollision 1-periodic solution of (5).
2. Preliminaries
In this section, we will list some basic lemmas and inequality
to prove our theorems.
Lemma 5 (see [8]). The radius 𝑟 for the moving orbit of 𝑁
equal mass points is
1/3
.
(10)
on Λ ± , where
󵄨󵄨
1
󵄨
{
1,2 𝑅
2 󵄨󵄨󵄨 𝑞 (𝑡 + ) = 𝑅 (2) 𝑞 (𝑡) , 𝑞 (𝑡) ≠ 𝑞𝑖 (𝑡) , ∀𝑡 ∈ [0, 1] , 𝑖 = 1, . . . , 𝑁, }
Λ − = {𝑞 ∈ 𝑊 ( , 𝑅 ) 󵄨󵄨
2
},
󵄨󵄨
𝑍
)
=
−1,
deg
(𝑞
−
𝑞
)
=
1,
𝑗
=
1,
.
.
.
,
𝑁
−
1
deg
(𝑞
−
𝑞
󵄨
𝑁
𝑗
𝑁
󵄨
{
}
󵄨󵄨
󵄨󵄨 𝑞 (𝑡 + 1 ) = 𝑅 (𝑁) 𝑞 (𝑡) , 𝑞 (𝑡) ≠ 𝑞 (𝑡) , ∀𝑡 ∈ [0, 1] , 𝑖 = 1, . . . , 𝑁, }
{
𝑅
󵄨
𝑖
Λ + = {𝑞 ∈ 𝑊1,2 ( , 𝑅2 ) 󵄨󵄨󵄨
𝑁
}.
󵄨󵄨
𝑍
deg
(𝑞
−
𝑞
)
=
1,
deg
(𝑞
−
𝑞
)
=
1,
𝑗
=
1,
.
.
.
,
𝑁
−
1
󵄨
𝑁
𝑗
𝑁
󵄨
}
{
𝜋
1 2/3
𝑟 = ( ) [ ∑ csc ( 𝑗)]
4𝜋
𝑁
[1≤𝑗≤𝑁−1
]
(9)
(12)
Lemma 6 (Zhang and Zhou [9]). Suppose that 𝑔 ∈ 𝑆𝑂(𝑘)
is an element of finite order 𝑠 and it has no fixed point
(11)
other than the origin (i.e., 1 is not an eigenvalue of 𝑔). Then
𝑇
∫0 𝑋(𝑡)𝑑𝑡 = 0 for all 𝑋(𝑡) = (𝑥1 (𝑡), . . . , 𝑥𝑁(𝑡)) ∈ 𝐸𝑔𝑁, where
𝐸𝑔𝑁 = {(𝑥1 , . . . , 𝑥𝑁) | 𝑥𝑖 ∈ 𝐻1 (𝑅/𝑇𝑍, 𝑅𝐾 ), 𝑥𝑖 (𝑡 + 𝑇/𝑠) =
𝑔(𝑥𝑖 (𝑡)), ∀𝑡 ∈ [0, 𝑇], 𝑖 = 1, . . . , 𝑁}.
Lemma 7 (Poincare-Wirtinger inequality [16]). Let 𝑞
𝑇
𝑊1,2 (𝑅/𝑇𝑍, 𝑅𝐾 ) and ∫0 𝑞(𝑡)𝑑𝑡 = 0, and then
𝑇
𝑇2 𝑇 󵄨
󵄨2
󵄨
󵄨2
∫ 󵄨󵄨󵄨𝑞 (𝑡)󵄨󵄨󵄨 𝑑𝑡 ≤ 2 ∫ 󵄨󵄨󵄨𝑞 ̇ (𝑡)󵄨󵄨󵄨 𝑑𝑡.
4𝜋 0
0
∈
(13)
Lemma 8 (see [17]). Let 𝑋 be a reflexive Banach space, 𝑆 be a
weakly closed subset of 𝑋, and 𝑓 : 𝑆 → 𝑅∪{+∞}. If 𝑓 ≢ + ∞
is weakly lower semicontinuous and coercive (𝑓(𝑥) → +∞ as
‖ 𝑥 ‖ → +∞), then 𝑓 attains its infimum on 𝑆.
Lemma 9 (Palais’s symmetry principle [18]). Let 𝜎 be an
orthogonal representation of a finite or compact group 𝐺, let
𝐻 be a real Hilbert space, and let 𝑓 : 𝐻 → 𝑅 satisfy 𝑓(𝜎 ⋅ 𝑥) =
𝑓(𝑥), ∀𝜎 ∈ 𝐺, ∀𝑥 ∈ 𝐻.
Set 𝐹 = {𝑥 ∈ 𝐻 | 𝜎 ⋅ 𝑥 = 𝑥, ∀𝜎 ∈ 𝐺}. Then the critical
point of 𝑓 in 𝐹 is also a critical point of 𝑓 in 𝐻.
Remark 10. By Palais’s symmetry principle and the perturbation invariance for wingding numbers, we know that
Abstract and Applied Analysis
3
the critical point of 𝑓(𝑞) in Λ ± is a periodic solution of
Newtonian equation (5).
Lemma 11. (1) (Gordon’s theorem [19]) Let 𝑥
∈
𝑊1,2 ([𝑡1 , 𝑡2 ], 𝑅𝐾 ) and 𝑥(𝑡1 ) = 𝑥(𝑡2 ) = 0. Then for any
𝑎 > 0, one has
𝑡2
1
𝑎
3
1/3
) 𝑑𝑡 ≥ (2𝜋)2/3 𝑎2/3 (𝑡2 − 𝑡1 ) .
∫ ( |𝑥|̇ 2 +
2
2
|𝑥|
𝑡1
1,2
(14)
Lemma 13. For 𝜕Λ − = {𝑞 ∈ 𝑊1,2 (𝑅/𝑍, 𝑅2 ) | 𝑞(𝑡 + 1/2) =
𝑅(2)𝑞(𝑡), ∃1 ≤ 𝑖0 ≤ 𝑁, 𝑡0 ∈ [0, 1] 𝑠.𝑡. 𝑞𝑖0 (𝑡0 ) = 𝑞(𝑡0 )}, we
have
inf 𝑓 (𝑞)
𝑞∈𝜕Λ −
3
𝑁−1 3
1
≥ (4𝜋)2/3 𝑁−1/3 + (
) (2𝜋)2/3 𝑁2/3 − (2𝜋)2 𝑟2
2
𝑁
2
2
≜ 𝑑0 .
(19)
𝐾
(2) (Long and Zhang [20]) Let 𝑥 ∈ 𝑊 (𝑅/𝑇𝑍, 𝑅 ), and
𝑇
∫0 𝑥𝑑𝑡 = 0, then for any 𝑎 > 0, we have
Proof. It follows from (4) that
𝑁
∑𝑞𝑖̇ = 0,
𝑇
1
𝑎
3
) 𝑑𝑡 ≥ (2𝜋)2/3 𝑎2/3 𝑇1/3 .
∫ ( |𝑥|̇ 2 +
2
2
|𝑥|
0
(15)
(20)
𝑖=1
which implies
𝑁
󵄨
󵄨2
∑󵄨󵄨󵄨𝑞 ̇ − 𝑞𝑖̇ 󵄨󵄨󵄨
3. Proof of Theorems
𝑖=1
In order to get our theorems, we need two steps to complete
the proof.
𝑁
󵄨 󵄨2 󵄨 󵄨2
= ∑ (󵄨󵄨󵄨𝑞󵄨󵄨󵄨̇ + 󵄨󵄨󵄨𝑞𝑖̇ 󵄨󵄨󵄨 − 2 ⟨𝑞,̇ 𝑞𝑖̇ ⟩)
𝑖=1
Step 1. We will establish the existence of variational minimizers of 𝑓(𝑞) in (10) on Λ± .
𝑖=1
𝑁
𝑖=1
2𝜋
𝜆 = cos ,
𝑛
2𝜋
sin
= 0,
𝑛
Therefore,
(16)
which implies 𝜆 ≠ 1 for 𝑛 ≥ 2. Then by Lemma 6, we have
1
∫ 𝑞 (𝑡) = 0 ∀𝑞 ∈ Λ± .
0
(17)
𝑁
󵄨󵄨 󵄨󵄨̇ 2 1
󵄨
󵄨2 󵄨 󵄨2
󵄨󵄨𝑞󵄨󵄨 = ∑ (󵄨󵄨󵄨𝑞 ̇ − 𝑞𝑖̇ 󵄨󵄨󵄨 − 󵄨󵄨󵄨𝑞𝑖̇ 󵄨󵄨󵄨 ) .
𝑁 𝑖=1
1
1 󵄨 󵄨2 𝑁 1
𝑓 (𝑞) = ∫ [ 󵄨󵄨󵄨𝑞󵄨󵄨󵄨̇ + ∑ 󵄨󵄨
󵄨󵄨 ] 𝑑𝑡
2
0
𝑖=1 󵄨󵄨𝑞 − 𝑞𝑖 󵄨󵄨
=
(18)
Hence, by the definition of 𝑓(𝑞), 𝑓 is coercive on Λ± .
Similar to the proof of Lemma 2.4 in [13], we can get the
conclusion that 𝑓 is weakly lower semi-continuous on Λ± .
Therefore, by using Lemma 8, it can be concluded that
𝑓(𝑞) in (10) attains its infimum on Λ± .
Step 2. We will prove that the variational minimizer in
Lemma 12 is the noncollision 1-period solution of (5).
For any collision generalized solution 𝑞, we can estimate
the lower bound for the value of Lagrangian action functional.
1 1 𝑁 1 󵄨󵄨
𝑁
󵄨2
∫ ∑ [ 󵄨󵄨𝑞 ̇ − 𝑞𝑖̇ 󵄨󵄨󵄨 + 󵄨󵄨
󵄨 ] 𝑑𝑡
𝑁 0 𝑖=1 2
󵄨󵄨𝑞 − 𝑞𝑖 󵄨󵄨󵄨
−
1/2
.
(22)
Hence,
So by using Lemma 7, for all 𝑞 ∈ Λ± , we see that an equivalent
norm of (9) on Λ± is
󵄩󵄩 󵄩󵄩
󵄨 󵄨2
󵄩󵄩𝑞󵄩󵄩 ≅ [∫ 󵄨󵄨󵄨𝑞󵄨󵄨󵄨̇ 𝑑𝑡]
0
𝑖=1
(21)
󵄨 󵄨2
󵄨 󵄨2
= 𝑁󵄨󵄨󵄨𝑞󵄨󵄨󵄨̇ + ∑󵄨󵄨󵄨𝑞𝑖̇ 󵄨󵄨󵄨 .
Proof. It is easy to check that the eigenvalue 𝜆 of 𝑅(𝑛) satisfies
𝑇
𝑁
󵄨 󵄨2
󵄨 󵄨2
= 𝑁󵄨󵄨󵄨𝑞󵄨󵄨󵄨̇ + ∑󵄨󵄨󵄨𝑞𝑖̇ 󵄨󵄨󵄨 − 2 ⟨𝑞,̇ ∑𝑞𝑖̇ ⟩
Lemma 12. 𝑓(𝑞) in (10) attains its infimum on Λ± .
{
{
{
{
{
𝑁
(23)
1 1 𝑁 󵄨󵄨 󵄨󵄨2
∫ ∑󵄨𝑞 ̇ 󵄨 𝑑𝑡.
2𝑁 0 𝑖=1󵄨 𝑖 󵄨
If 𝑞 ∈ Λ− is a collision solution, then there exists 𝑡0 ∈ [0, 1]
and 1 ≤ 𝑖0 ≤ 𝑁 s.t 𝑞(𝑡0 ) = 𝑞𝑖0 (𝑡0 ). Since 𝑞𝑖 (𝑡+1/2) = 𝑅(2)𝑞𝑖 (𝑡),
one has 𝑞(𝑡0 + 𝑘/2) = 𝑞𝑖0 (𝑡0 + 𝑘/2), ∀0 ≤ 𝑘 ≤ 2. So, by (1) of
Lemma 11, we get
1 1 1 󵄨󵄨
𝑁
󵄨2
∫ [ 󵄨󵄨𝑞 ̇ − 𝑞𝑖̇ 0 󵄨󵄨󵄨󵄨 + 󵄨󵄨
󵄨 ] 𝑑𝑡
𝑁 0 2󵄨
󵄨󵄨𝑞 − 𝑞𝑖0 󵄨󵄨󵄨
󵄨
󵄨
=
2 1/2 1 󵄨󵄨
𝑁
󵄨2
∫ [ 󵄨󵄨󵄨𝑞 ̇ − 𝑞𝑖̇ 0 󵄨󵄨󵄨󵄨 + 󵄨󵄨
󵄨 ] 𝑑𝑡
𝑁 0
2
󵄨󵄨𝑞 − 𝑞𝑖0 󵄨󵄨󵄨
󵄨
󵄨
3
≥ (4𝜋)2/3 𝑁−1/3 .
2
(24)
4
Abstract and Applied Analysis
1
For noncollision pair 𝑞, 𝑞𝑖 (𝑖 ≠ 𝑖0 ), we claim ∫0 (𝑞(𝑡) −
𝑞𝑖 (𝑡))𝑑𝑡 = 0. In fact, it follows from (1) that 𝑞𝑖+1 (𝑡) = 𝑞𝑖 (𝑡 +
1/𝑁), which implies
𝑞𝑖 (𝑡) = 𝑞1 (𝑡 +
𝑖−1
).
𝑁
(25)
Lemma 14. For 𝜕Λ + = {𝑞 ∈ 𝑊1,2 (𝑅/𝑍, 𝑅2 ) | 𝑞(𝑡 + 1/𝑁) =
𝑅(𝑁)𝑞(𝑡), ∃1 ≤ ̃𝑖0 ≤ 𝑁, ̃𝑡0 ∈ [0, 1] 𝑠.𝑡. 𝑞̃𝑖0 (̃𝑡0 ) = 𝑞(̃𝑡0 )}, we
have
inf 𝑓 (𝑞)
𝑞∈𝜕Λ +
3
𝑁−1 3
1
≥ (2𝜋)2/3 𝑁1/3 + (
) (2𝜋)2/3 𝑁2/3 − (2𝜋)2 𝑟2
2
𝑁
2
2
Therefore, by (4), we have
1
≜ 𝑑1 .
∫ 𝑞1 (𝑡) 𝑑𝑡
(31)
0
1/𝑁
=∫
0
(𝑁−1)/𝑁
1/𝑁
=∫
0
1
𝑞1 (𝑡) 𝑑𝑡 + ⋅ ⋅ ⋅ + ∫
1/𝑁
𝑞1 (𝑡) 𝑑𝑡 + ⋅ ⋅ ⋅ + ∫
0
𝑞1 (𝑡) 𝑑𝑡
𝑞1 (𝑡 +
𝑁−1
) 𝑑𝑡
𝑁
(26)
Proof of Theorem 2. In order to get Theorem 2, we are going
𝑞) = 𝑑2 . Then the
to find a test loop 𝑞̃ ∈ Λ − such that 𝑓(̃
minimizer of 𝑓 on Λ− must be a noncollision solution if 𝑑2 <
𝑑0 .
Let 𝑎 > 0, 𝑏 > 0, 𝜃 ∈ [0, 2𝜋) and
1/𝑁 𝑁
=∫
0
∑𝑞𝑖 (𝑡) 𝑑𝑡
𝑖=1
= 0.
Since 𝑞1 (𝑡 + 1) = 𝑞1 (𝑡), by (25) and (26), we obtain
1
1
0
0
∫ 𝑞𝑖 (𝑡) 𝑑𝑡 = ∫ 𝑞1 (𝑡 +
Proof. Similar to the proof of Lemma 13, Lemma 14
holds.
𝑞̃ − 𝑞𝑁 = (𝑎 cos (−2𝜋𝑡 + 𝜃) , 𝑏 sin (−2𝜋𝑡 + 𝜃))𝑇 .
(32)
1
𝑖−1
) 𝑑𝑡 = ∫ 𝑞1 (𝑡) 𝑑𝑡 = 0. (27)
𝑁
0
1
Combined with (17) and (27), we have ∫0 (𝑞(𝑡) − 𝑞𝑖 (𝑡))𝑑𝑡 = 0.
Hence, by (2) of Lemma 11, we can get
1󵄨
𝑁
1 1
󵄨2
∫ ∑ [ 󵄨󵄨𝑞 ̇ − 𝑞𝑖̇ 󵄨󵄨󵄨 + 󵄨󵄨
󵄨 ] 𝑑𝑡
𝑁 0 𝑖 ≠ 𝑖0 2 󵄨
󵄨󵄨𝑞 − 𝑞𝑖 󵄨󵄨󵄨
Hence,
𝑞̃ − 𝑞1 = (𝑟 cos 2𝜋𝑡 − 𝑟 cos (2𝜋𝑡 +
+ 𝑎 cos (−2𝜋𝑡 + 𝜃) , 𝑟 sin 2𝜋𝑡
(28)
𝑁−1 3
≥(
) (2𝜋)2/3 𝑁2/3 .
𝑁
2
− 𝑟 sin (2𝜋𝑡 +
For the other term of 𝑓, using the expression for the orbits
𝑞1 , . . . , 𝑞𝑁 as in (1), one has
−
1
1 1 𝑁 󵄨󵄨 󵄨󵄨2
∫ ∑󵄨𝑞 ̇ 󵄨 𝑑𝑡 = − (2𝜋)2 𝑟2 .
2𝑁 0 𝑖=1󵄨 𝑖 󵄨
2
2𝜋
)
𝑁
(29)
𝑇
2𝜋
) + 𝑏 sin(−2𝜋𝑡 + 𝜃)) ,
𝑁
..
.
𝑞̃ − 𝑞𝑖 = (𝑟 cos 2𝜋𝑡 − 𝑟 cos (2𝜋𝑡 +
𝑖
2𝜋)
𝑁
+ 𝑎 cos (−2𝜋𝑡 + 𝜃) , 𝑟 sin 2𝜋𝑡
𝑇
𝑖
− 𝑟 sin (2𝜋𝑡 + 2𝜋) + 𝑏 sin(−2𝜋𝑡 + 𝜃)) ,
𝑁
Therefore, it follows from (24), (28), and (29) that
(33)
(34)
..
.
inf 𝑓 (𝑞)
𝑞∈𝜕Λ −
3
𝑁−1 3
1
≥ (4𝜋)2/3 𝑁−1/3 + (
) (2𝜋)2/3 𝑁2/3 − (2𝜋)2 𝑟2
2
𝑁
2
2
≜ 𝑑0 .
(30)
𝑞̃ − 𝑞𝑁−1 = (𝑟 cos 2𝜋𝑡 − 𝑟 cos (2𝜋𝑡 +
𝑁−1
2𝜋)
𝑁
+ 𝑎 cos (−2𝜋𝑡 + 𝜃) , 𝑟 sin 2𝜋𝑡
− 𝑟 sin(2𝜋𝑡 +
𝑇
𝑁−1
2𝜋)+ 𝑏 sin(−2𝜋𝑡 + 𝜃)) .
𝑁
(35)
Abstract and Applied Analysis
5
It is easy to see that 𝑞̃ ∈ Λ − and
2
2
2
𝑎2 + 𝑏2 𝑎2 − 𝑏2
󵄨
󵄨󵄨 ̃
+
cos (4𝜋𝑡 − 2𝜃),
󵄨󵄨𝑞 − 𝑞𝑁󵄨󵄨󵄨 = √
2
2
2
𝑎 +𝑏
𝑎 −𝑏
󵄨󵄨 ̇
󵄨2
󵄨󵄨𝑞̃ − 𝑞1̇ 󵄨󵄨󵄨 = (2𝜋)2 {
−
cos (4𝜋𝑡 − 2𝜃)
󵄨
󵄨
2
2
+ 𝑟2 (2 − 2 cos
󵄨 ̇ 󵄨󵄨2
󵄨󵄨 ̇ 󵄨󵄨2
2 2
󵄨󵄨 = (2𝜋) 𝑟 .
󵄨󵄨𝑞1 󵄨󵄨 = ⋅ ⋅ ⋅ = 󵄨󵄨󵄨𝑞𝑁
2𝜋
) − (𝑎 + 𝑏) 𝑟
𝑁
× [cos (4𝜋𝑡 − 𝜃) − cos (4𝜋𝑡 +
+ (𝑎 − 𝑏) 𝑟 [cos 𝜃 − cos (
2𝜋
− 𝜃)]
𝑁
2𝜋
+ 𝜃)] } ,
𝑁
(36)
𝑓 (̃
𝑞)
=
1 1 𝑁 󵄨󵄨 󵄨󵄨2
1 1 𝑁 1 󵄨󵄨 ̇
𝑁
󵄨2
]
𝑑𝑡
−
∫ ∑ [ 󵄨󵄨󵄨𝑞̃ − 𝑞𝑖̇ 󵄨󵄨󵄨󵄨 + 󵄨󵄨
∫ ∑󵄨𝑞 ̇ 󵄨 𝑑𝑡
󵄨
𝑁 0 𝑖=1 2
2𝑁 0 𝑖=1󵄨 𝑖 󵄨
󵄨󵄨𝑞̃ − 𝑞𝑖 󵄨󵄨󵄨
= (𝑎2 + 𝑏2 ) 𝜋2 + (2𝜋)2 𝑟2
+ 2𝜋2 (𝑎 − 𝑏) 𝑟 [cos 𝜃 −
2𝜋
) + (𝑎 + 𝑏) 𝑟
𝑁
× [cos (4𝜋𝑡 − 𝜃) − cos (4𝜋𝑡 +
(42)
Therefore by (36)–(42), we get
𝑎2 + 𝑏2 𝑎2 − 𝑏2
󵄨
󵄨󵄨 ̃
+
cos (4𝜋𝑡 − 2𝜃)
󵄨󵄨𝑞 − 𝑞1 󵄨󵄨󵄨 = {
2
2
+ 𝑟2 (2 − 2 cos
(41)
𝑁
1
+ ∑∫ {
2𝜋
− 𝜃)]
𝑁
𝑖=1 0
𝑎2 + 𝑏2 𝑎2 − 𝑏2
+
cos (4𝜋𝑡 − 2𝜃)
2
2
+ 𝑟2 (2 − 2 cos
1/2
2𝜋
+ 𝜃)] }
+ (𝑎 − 𝑏) 𝑟 [cos 𝜃 − cos (
𝑁
𝑖
1 𝑁
∑ cos ( 2𝜋 + 𝜃)]
𝑁 𝑖=1
𝑁
,
𝑖
2𝜋) + (𝑎 + 𝑏) 𝑟
𝑁
× [cos (4𝜋𝑡 − 𝜃) − cos (4𝜋𝑡 +
..
.
𝑖
2𝜋 − 𝜃)]
𝑁
−1/2
+ (𝑎 − 𝑏) 𝑟 [cos 𝜃 − cos (
(37)
𝑎2 + 𝑏2 𝑎2 − 𝑏2
󵄨󵄨 ̇
󵄨2
󵄨󵄨𝑞̃ − 𝑞𝑖̇ 󵄨󵄨󵄨 = (2𝜋)2 {
−
cos (4𝜋𝑡 − 2𝜃)
󵄨
󵄨
2
2
+ 𝑟2 (2 − 2 cos
+ (𝑎 − 𝑏) 𝑟 [cos 𝜃 − cos (
= 2𝜋2 (
𝑖
2𝜋 − 𝜃)]
𝑁
𝑖
2𝜋 + 𝜃)] } ,
𝑁
(38)
𝑎2 + 𝑏2 𝑎2 − 𝑏2
󵄨
󵄨󵄨 ̃
+
cos (4𝜋𝑡 − 2𝜃)
󵄨󵄨𝑞 − 𝑞𝑖 󵄨󵄨󵄨 = {
2
2
+ 𝑟2 (2 − 2 cos
𝑎2 + 𝑏2
+ 𝑟2 )
2
+ 2𝜋2 (𝑎 − 𝑏) 𝑟 [cos 𝜃 −
𝑁
1
+ ∑∫ {
𝑖=1 0
1 𝑁
𝑖
∑ cos ( 2𝜋 + 𝜃)]
𝑁 𝑖=1
𝑁
𝑎2 + 𝑏2 𝑎2 − 𝑏2
+
cos (4𝜋𝑡 − 2𝜃)
2
2
+ 𝑟2 (2 − 2 cos
𝑖
2𝜋) + (𝑎 + 𝑏) 𝑟
𝑁
𝑖
2𝜋) + (𝑎 + 𝑏) 𝑟
𝑁
× [cos (4𝜋𝑡 − 𝜃) − cos (4𝜋𝑡 +
𝑖
× [cos (4𝜋𝑡 − 𝜃) − cos (4𝜋𝑡 + 2𝜋 − 𝜃)]
𝑁
1/2
+ (𝑎 − 𝑏) 𝑟 [cos 𝜃 − cos (
𝑑𝑡
1
− (2𝜋)2 𝑟2
2
𝑖
2𝜋) − (𝑎 + 𝑏) 𝑟
𝑁
× [cos (4𝜋𝑡 − 𝜃) − cos (4𝜋𝑡+
𝑖
2𝜋 + 𝜃)] }
𝑁
𝑖
2𝜋 + 𝜃)] }
𝑁
,
..
.
(39)
𝑎2 + 𝑏2 𝑎2 − 𝑏2
󵄨󵄨 ̇
󵄨󵄨2
󵄨󵄨𝑞̃ − 𝑞𝑁
󵄨󵄨 = (2𝜋)2 (
̇
−
cos (4𝜋𝑡 − 2𝜃)) ,
󵄨
󵄨
2
2
(40)
𝑖
2𝜋 − 𝜃)]
𝑁
−1/2
+ (𝑎 − 𝑏) 𝑟 [cos 𝜃 − cos (
𝑖
2𝜋 + 𝜃)] }
𝑁
𝑑𝑡
= 𝑑2 (𝑎, 𝑏, 𝜃) .
(43)
In order to estimate 𝑑2 , we have computed the numerical
values of 𝑑2 = 𝑓(𝑞) over some selected test loops. The
computation of the integral that appears in (43) has been done
using the function {𝑞𝑢𝑎𝑑} of Mathematica 7.1 with an error
less than 10−6 . The results of the numerical explorations are
given in Table 1.
6
Abstract and Applied Analysis
Hence,
Table 1: Parameters of test loops for Theoerm 2.
𝑁
2
2
2
2
2
2
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
𝑎
0.11
0.13
0.13
0.15
0.15
0.17
0.17
0.19
0.19
0.21
0.23
0.23
0.25
0.25
0.15
0.15
0.17
0.17
0.17
0.19
0.19
0.19
0.61
0.63
0.63
0.65
0.65
0.65
0.67
0.67
0.67
0.69
0.69
0.69
𝑏
0.29
0.25
0.27
0.21
0.23
0.19
0.21
0.17
0.19
0.21
0.15
0.23
0.13
0.25
0.67
0.69
0.65
0.67
0.69
0.63
0.65
0.67
0.23
0.19
0.21
0.17
0.19
0.21
0.15
0.17
0.19
0.13
0.15
0.17
𝜃
𝜋/2
𝜋
𝜋
𝜋
𝜋/5
𝜋
𝜋/5
𝜋
𝜋
𝜋
𝜋
𝜋/5
𝜋/2
𝜋/5
𝜋/30
𝜋/30
𝜋/30
𝜋/20
𝜋/20
𝜋/30
𝜋/20
𝜋/20
𝜋
𝜋
𝜋
𝜋
𝜋
𝜋
𝜋
𝜋
𝜋
𝜋
𝜋
𝜋
𝑑0
9.813298
9.813298
9.813298
9.813298
9.813298
9.813298
9.813298
9.813298
9.813298
9.813298
9.813298
9.813298
9.813298
9.813298
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
11.523843
𝑑2
9.753899
9.753243
9.643849
9.790068
9.746803
9.751237
9.625193
9.784859
9.564603
9.279445
9.793205
9.109854
9.730925
9.038706
11.505860
11.444212
11.493238
11.452135
11.400124
11.519350
11.455969
11.386608
11.516685
11.489791
11.436105
11.461786
11.392115
11.349366
11.472422
11.383978
11.324970
11.522980
11.412094
11.334189
For the parameters 𝑎, 𝑏, and 𝜃 given in Table 1, we all
have 𝑑2 < 𝑑0 . For 4 ≤ 𝑁 ≤ 1000, we do not find
suitable parameters 𝑎 ∈ [0.01, 2𝑟), 𝑏 ∈ [0.01, 2𝑟), and 𝜃 =
𝜋, 𝜋/2, 𝜋/3, . . . , 𝜋/29, 𝜋/30 such that 𝑑2 < 𝑑0 , where 𝑟 is the
radius of the first 𝑁 bodies with equal masses. Hence we
only consider the case 𝑁 = 2, 3. This completes the proof of
Theorem 2.
𝑞 − 𝑞1
√−12𝜋𝑡 √−1𝜃
= 𝑞𝑁 + 𝑎𝑒
𝑒
√−12𝜋𝑡
(1 − 𝑒
= 𝑟𝑒
− 𝑞1
√−1(2𝜋/𝑁)
√−12𝜋𝑡 √−1𝜃
) + 𝑎𝑒
√−12𝜋𝑡 √−1𝜃
𝑞 − 𝑞𝑁 = 𝑎𝑒
𝑒
.
(44)
,
(45)
..
.
𝑞 − 𝑞𝑖
√−12𝜋𝑡 √−1𝜃
= 𝑞𝑁 + 𝑎𝑒
𝑒
√−12𝜋𝑡
(1 − 𝑒
= 𝑟𝑒
− 𝑞𝑖
√−1(𝑖/𝑁)2𝜋
√−12𝜋𝑡 √−1𝜃
) + 𝑎𝑒
𝑒
,
(46)
..
.
𝑞 − 𝑞𝑁−1
√−12𝜋𝑡 √−1𝜃
= 𝑞𝑁 + 𝑎𝑒
𝑒
√−12𝜋𝑡
(1 − 𝑒
= 𝑟𝑒
− 𝑞𝑁−1
√−1((𝑁−1)/𝑁)2𝜋
√−12𝜋𝑡 √−1𝜃
) + 𝑎𝑒
𝑒
.
(47)
It is easy to see that 𝑞 ∈ Λ + and
2𝜋
󵄨󵄨 ̇
󵄨2
󵄨󵄨𝑞 − 𝑞1̇ 󵄨󵄨󵄨 = (2𝜋)2 {𝑟2 (2 − 2 cos ) + 𝑎2
󵄨
󵄨
𝑁
2𝜋
+2𝑎𝑟 [cos 𝜃 − cos (𝜃 −
)]} ,
𝑁
(48)
2𝜋
󵄨
󵄨󵄨
2
2
󵄨󵄨𝑞 − 𝑞1 󵄨󵄨󵄨 = {𝑟 (2 − 2 cos ) + 𝑎
𝑁
+2𝑎𝑟 [cos 𝜃 − cos (𝜃 −
1/2
2𝜋
)]} ,
𝑁
(49)
..
.
𝑖
󵄨󵄨 ̇
󵄨2
󵄨󵄨𝑞 − 𝑞𝑖̇ 󵄨󵄨󵄨 = (2𝜋)2 {𝑟2 (2 − 2 cos 2𝜋) + 𝑎2
󵄨
󵄨
𝑁
+2𝑎𝑟 [cos 𝜃 − cos (𝜃 −
𝑖
2𝜋)]} ,
𝑁
(50)
𝑖
󵄨
󵄨󵄨
2
2
󵄨󵄨𝑞 − 𝑞𝑖 󵄨󵄨󵄨 = {𝑟 (2 − 2 cos 2𝜋) + 𝑎
𝑁
+2𝑎𝑟 [cos 𝜃 − cos (𝜃 −
Proof of Theorem 4. To get Theorem 4, we are going to find a
test loop 𝑞 ∈ Λ + such that 𝑓(𝑞) = 𝑑3 . Then the minimizer of
𝑓 on Λ+ must be a noncollision solution if 𝑑3 < 𝑑1 .
Let 𝑎 > 0, 𝜃 ∈ [0, 2𝜋), and
𝑒
1/2
𝑖
2𝜋)]} ,
𝑁
(51)
..
.
󵄨󵄨 ̇
󵄨2
󵄨
󵄨󵄨
󵄨󵄨𝑞 − 𝑞𝑁
̇ 󵄨󵄨󵄨󵄨 = (2𝜋)2 𝑎2 ,
󵄨󵄨𝑞 − 𝑞𝑁󵄨󵄨󵄨 = 𝑎,
󵄨
󵄨󵄨2
2 2
󵄨󵄨󵄨𝑞1̇ 󵄨󵄨󵄨2 = ⋅ ⋅ ⋅ = 󵄨󵄨󵄨𝑞𝑁
󵄨 󵄨
󵄨 ̇ 󵄨󵄨 = (2𝜋) 𝑟 .
(52)
(53)
Abstract and Applied Analysis
7
2970
Therefore by (48)–(53), we get
2960
𝑓 (𝑞)
𝑑-val
2950
1 1 𝑁 1 󵄨󵄨 ̇
𝑁
󵄨2
=
∫ ∑ [ 󵄨󵄨󵄨𝑞 − 𝑞𝑖̇ 󵄨󵄨󵄨󵄨 + 󵄨󵄨
󵄨 ] 𝑑𝑡
𝑁 0 𝑖=1 2
󵄨󵄨𝑞 − 𝑞𝑖 󵄨󵄨󵄨
2930
2920
1 1 𝑁 󵄨󵄨 󵄨󵄨2
−
∫ ∑󵄨𝑞 ̇ 󵄨 𝑑𝑡
2𝑁 0 𝑖=1󵄨 𝑖 󵄨
2910
4
0
𝑎𝑟 𝑁
𝑖
∑ cos (𝜃 − 2𝜋)]
𝑁 𝑖=1
𝑁
+ 𝑎−1 + ∑ {𝑟2 (2 − 2 cos
𝑖=1
𝑖
2𝜋)]}
𝑁
12.5
11.5
12
12.5
𝑎-val
2960
2950
𝑑-val
+ (2𝜋)2 [arcos𝜃 −
+𝑎
12
2970
2
𝑁−1
11.5
Figure 1: 𝑁 = 32403.
= 2𝜋 (𝑎 + 𝑟 )
−1
11
−1/2
1
− (2𝜋)2 𝑟2
2
2
10.5
10
𝑑1
𝑑3
𝑖
2𝜋) + 𝑎2
𝑁
+2𝑎𝑟 [cos 𝜃 − cos (𝜃 −
2
2
al
𝑁−1
𝜃 -v
= 2𝜋2 𝑎2 + (2𝜋)2 𝑟2
+ (2𝜋)2 [arcos𝜃 −
2940
2930
𝑎𝑟 𝑁
𝑖
∑ cos (𝜃 − 2𝜋)]
𝑁 𝑖=1
𝑁
2920
2910
4
𝑖
+ ∑ {𝑟 (2 − 2 cos 2𝜋) + 𝑎2
𝑁
𝑖=1
2
𝜃 -v
−1/2
𝑖
2𝜋)]}
𝑁
2
al
+2𝑎𝑟 [cos 𝜃 − cos (𝜃 −
2940
0
= 𝑑3 (𝑎, 𝜃) .
11
𝑎-val
𝑑1
𝑑3
(54)
In order to estimate 𝑑3 , we have computed the numerical
values of 𝑑3 = 𝑓(𝑞) over some selected test loops. The
results of the numerical explorations show that by choosing
some 𝑎 ∈ [0.11, 12.50] and 𝜃 = 𝜋, one has 𝑑3 < 𝑑1
for 2 ≤ 𝑁 ≤ 32402. From the computational data, let
𝜃 = 𝜋, 𝜋/2, 𝜋/3, . . . , 𝜋/19, 𝜋/20, and we find that 𝑑3 < 𝑑1
holds for 2 ≤ 𝑁 ≤ 19 with some 𝑎 ∈ [0.11, 0.90].
But for 20 ≤ 𝑁 ≤ 32402, no matter how we choose
the value of 𝑎 ∈ [0.11, 12.50], we obtain 𝑑3 < 𝑑1 only
if 𝜃 = 𝜋. For 𝑁 = 32403, the computational data
shows that 𝑑3 ≫ 𝑑1 for all 𝑎 ∈ [0.11, 10.00] and 𝜃 =
𝜋, 𝜋/2, 𝜋/3, . . . , 𝜋/99, 𝜋/100. Then combined with Figure 1, it
is easy to see that 𝑎.𝑒. 𝑑3 > 𝑑1 for all 𝑎 ∈ [0.11, 12.50] and 𝜃 =
𝜋, 𝜋/2, 𝜋/3, . . . , 𝜋/99, 𝜋/100. The same holds for 𝑁 = 32404
from Figure 2 and the computational data. Moreover, we can
get the same conclusions for 32405 ≤ 𝑁 ≤ 100000. Hence,
we only consider the case 2 ≤ 𝑁 ≤ 32402. This completes the
proof of Theorem 4.
10.5
10
Figure 2: 𝑁 = 32404.
Acknowledgments
This study is supported by the national natural Science
Foundation of China. The authors would like to thank
Professor Shiqing Zhang, Taisong Xiong, Donglun Wu and
Yiyang Deng for some helpful discussions, and the referee
for his/her many valuable comments and suggestions on this
paper.
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