New Periodic Solutions for Newtonian $n$-Body Problems with Dihedral Group Symmetry and Topological Constraints

0.23 MB·English

by  Zhiqiang Wang

NEW PERIODIC SOLUTIONS FOR NEWTONIAN n-BODY
PROBLEMS WITH DIHEDRAL GROUP SYMMETRY AND
5 TOPOLOGICAL CONSTRAINTS
1
0
2 ZHIQIANGWANGANDSHIQINGZHANG

r
a
M Abstract. Inthispaper,weprovetheexistenceofafamilyofnewnon-
collisionperiodicsolutionsfortheclassicalNewtoniann-bodyproblems.

2 In our assumption, the n=2l ≥4 particles are invariant under the di-
1
hedralrotationgroupDl inR3suchthat,ateachinstant,thenparticles

] form two twisted l-regular polygons. Our approach is variational min-
h
p imizing method and we show that the minimizers are collision-free by
-
level estimates and local deformations.
h
at Keywords: Periodic Solutions, Newtonian n-body Problems,
m
Variational Method, DihedralGroup Symmetry,Topological
[ Constraints.

2
v
1
4
1. Introduction and Main Result
7
7
0 Many authors (for example[6][8][9][18][22]) used the variational method
.
1 to discover many new periodic solutions of the classical Newtonian n-body
0
5 problemsin thelast fifteen years. In particular, Chencinerand Montgomery
1
[6] proved the existence of the remarkable figure-8 type periodic solution for
:
v
planar Newtonian 3-body problems with equal masses. Ferrario and Ter-
i
X
racini [8] simplified and developed Marchal’s [12] important works and in-
r
a troduced the rotating circle property, proved that if the motion has certain
symmetry under some group action having the rotation circle property, the
solution exists and has no collision. Also Fusco et al.[9] proved the exis-
tence and collisionless of a number of new and interesting motions with the
invariance of certain platonic polyhedra group action and some topological
constraints.
In this paper, we consider a system of n = 2l positive masses with
their positions x(t) = (x (t),x (t),...,x (t))T moving in the space under
1 2 n
2010 Mathematics Subject Classification. 70F10,70F16,37C80,70G75.
1
2 ZHIQIANGWANGANDSHIQINGZHANG
Newton’s law of gravitation:
∂U(x(t))
(1.1) x¨(t) = ,
M ∂x
wherethepotentialfunctionU(x)= mimj and = diag m ,m ,...,m .
i<j |xi−xj| M { 1 2 n}
Itiswellknownthatlookingforperiodicsolutionsfor(1.1)isequivalent
P
to seeking the critical points of the Lagrange functional : Λ R +
A → ∪{ ∞}
T T
(x(t)) = (x)dt = K +U dt
A L
Z0 Z0
T n 1 (cid:0) m(cid:1) m
= ( m x˙ 2+ i j )dt
i i
2 | | x x
Z0 i=1 i<j | i − j|
X X
on the set
Λ = x(t) H1(R/TZ,R3n)x (t) = x (t), i = j, t R ,
i j
{ ∈ | 6 ∀ 6 ∀ ∈ }
and our approach is based on the following basic lemma:
Lemma 1.1. ([17]) Let X be a reflexive Banach space, M X is a weakly
⊂
closed subset, f : M R is weakly lower semi-continuous; if f is coercive,
→
that is, f(x) + as x + , then f attains its infimum on M.
→ ∞ | |→ ∞
There are two difficulties in this approach: one is the lack of coercivity
oftheaction functional onthewholesetΛ; andtheotheristhatincritical
A
points, there might be trajectories with collisions. To obtain the coercivity,
one can consider the functional on some symmetric subspace Λ Λ
G
A ⊂
such that is coercive. And the following famous lemma proves that
A|ΛG
the critical point on Λ is also a critical point on the whole space Λ.
G
Lemma 1.2. (Palais principle of symmetric criticality [15])
Let G be an orthogonal group on a Hilbert space Λ. Define the fixed
point space: Λ = x Λg x = x, g G ; if f C1(Λ,R) and satisfies
G
{ ∈ | · ∀ ∈ } ∈
f(g x)= f(x)for anyg Gand x Λ, thenthe critical point of f restricted
· ∈ ∈
on Λ is also a critical point of f on Λ.
G
Here is a traditional way to define the group action on loop space Λ
suchthat (g x) = (x). LetGbeafinitegroupwiththreerepresentations
A · A
ρ:G O(d), τ :G O(2) and σ :G such that
→ → → n
g x(t) = (ρ(g)xσ(g−1)(1)(τ(g−1)t),..P.,ρ(g)xσ(g−1)(n)(τ(g−1)t)).
·
RUNNING HEAD 3
Where we only consider homomorphisms σ with property that g G :
∀ ∈
(σ(g)(i) = j m = m ) and for more detail, we refer the readers to [8].
i j
⇒
Also we can add some topological constraints on Λ to get an open cone
G
Λ with the property
G
K ⊂
∂ = u Λ : t R,i = j : u (t ) = u (t ) ,
G G c i c j c
K ⊂ △ { ∈ ∃ ∈ 6 }
and find critical points inside ([9][14][18]). By the above arguments, we
K
can distinguish geometrically different solutions, and get the coercivity even
if isnotcoercive. Indeed,ifweareabletoprove thataminimizeru of
A|ΛG ∗
exists and for any collision trajectories u ∂ : (u )< (u ),
c G c
A|K ∈ K ⊂ △ A ∗ A
thenwemusthaveu Λ andiscollisionfree. ThusbyLemma1.2,u
G
∗ ∈ ∗ ∈ K
is a critical point of and therefore a solution of the n-body problem.
Λ
A|
Suppose the motions are in the space O ξ ξ ξ and let e be the unit
1 2 3 j
−
vectors of the coordinate axes ξ for j = 1,2,3. Denote the rotation of
j
cos 2π sin2π 0
l − l
angle 2π around ξ -axis R = sin2π cos 2π 0 , the rotation of angle π
l 3  l l 
0 0 1
 
 
1 0 0
around ξ -axis S = 0 1 0 . Then D = R,S is the dihedral group
1   l
− h i
0 0 1
 − 
([11]) of order n= 2l and it is a group of rotations with their rotation axes
kπ
ξ = ξ tan ,
2 1
Γ = ξ L l 1, where L is the line l
3 { k}j−=0 k 
ξ = 0.
 3
S
For simplicity, we denote D = R n 1, where R = Rk, R = RkS
l { j}j=−0 k l+k
for k = 0,1,...,l 1, i.e.
−
cos 2kπ sin 2kπ 0 cos 2kπ sin 2kπ 0
l − l l l
R = sin 2kπ cos 2kπ 0 ,R = sin 2kπ cos 2kπ 0 .
k  l l  l+k  l − l 
0 0 1 0 0 1
   − 
   
Nowweconsidern= 2l 4pointparticles u(t) = u (t),u (t),...,u (t)
0 1 n 1
≥ −
with equal masses in space with the following symmetry:
(cid:0) (cid:1)
(1.2) u (t) = R u (t), j = 0,1,...,n 1.
j j 0
−
Underthissymmetrythetrajectorieshavethepropertythat,ateachinstant,
thenpointparticlesformatwonestedregularl-polygonswiththesamesize.
4 ZHIQIANGWANGANDSHIQINGZHANG
Bylettingthethemassesm = 1andundertheassumptionofthesymmetry
i
(1.2), the action functional can be written as
n T n−1 1
u(t) = u (t) = u˙ (t)2 + dt.
0 0
A A 2 | | (R R )u (t)
(cid:0) (cid:1) (cid:0) (cid:1) Z0 (cid:16) Xj=1 | j − 0 0 |(cid:17)
In [7], Ferrario and Portaluri studied central configurations with this dihe-
dral symmetry. In Section 2 of [9], Fusco et al. got a new periodic solution
in the case of n = 4 by applying some topological constraint, here our result
is a generalization of theirs. Obviously, the trajectories u(t) are uniquely
determined by the trajectory u (t), and in [9], u (t) is called the generating
0 0
particle of the motion. Also we need some other symmetric constraints on
the loop of u :
0
u (t)= Rˆ u ( t),
0 0 0
−
(1.3)
T
u (t)= Rˆ u ( t).
 0 s 0 h −

cos 2kπ sin2kπ 0
l l
Wheres,h laresomepositiveintegersandRˆ = sin 2kπ cos 2kπ 0
≤ k  l − l 
0 0 1
 
are reflections along the plane P : ξ = ξ tan kπ, k= 0,1,...,l 1. 
k 2 1 l −
We notice that (1.3) implies u (t) = Rˆ Rˆ u (t T) = R u (t T).
0 s 0 0 − h s 0 − h
Since u (t) is a T-periodic solution i.e. u (t) = u (t + T), we must have
0 0 0
Rh = R = id which implies
s 0
(1.4) sh l mod l.
≡
Moreover if T is the minimal positive period, then h is the minimal positive
integer satisfying (1.4) and of course h l. Also applying (1.2) we see that
|
u (t) = u (t T) and
0 s − h
T T
u (t) = Su (t) = SR u (t ) = R Su (t )
l 0 s 0 l s 0
− h − − h
T T
= R u (t ) = u (t ).
2l s 0 2l s
− − h − − h
RUNNING HEAD 5
If we denote ¯j to be the least nonnegative residue of j modulo l, i.e. ¯j j
≡
mod l and 0 ¯j < l, we have
≤
T 2T (h 1)T
u (t) = u (t ) = u (t ) = = u (t − ),
0 s¯ − h 2s − h ··· (h 1)s − h
−

T 2T (h 1)T
 u1(t) = us+1(t− h) = u2s+1(t− h )= ··· = u(h 1)s+1(t− −h ),
 −

 ...



 T 2T (h 1)T

 ul/h−1(t) = us+l/h−1(t− h) = u2s+l/h−1(t− h ) = ··· = u(h−1)s+l/h−1(t− −h ),

 T 2T (h 1)T

 ul(t) = u2l−s¯(t− h) = u2l−2s(t− h ) = ··· = u2l−(h−1)s(t− −h ),

T 2T (h 1)T
u (t) = u (t ) = u (t )= = u (t − ),
 2l−1 2l−(s+1) − h 2l−(2s+1) − h ··· 2l−((h−1)s+1) − h

 ...



 T 2T
u (t) = u (t ) = u (t ) = ...
 2l−l/h+1 2l−(s+l/h−1) − h 2l−(2s+l/h−1) − h

 (h 1)T

 = u (t − ).
 2l−((h−1)s+l/h−1) − h



That is to say the n = 2l particles’ motion are composed of 2l/h choreog-
raphy trajectories. For example when l = 12,s = 9, then we have h = 4
and
T T 3T
u (t) = u (t ) = u (t ) = u (t ),
0 9 6 3
− 4 − 2 − 4
 T T 3T
 u (t) = u (t ) = u (t ) = u (t ),
 1 10 − 4 7 − 2 4 − 4

 T T 3T

 u (t) = u (t ) = u (t ) = u (t ),
 2 11 − 4 8 − 2 5 − 4

 T T 3T

u12(t) = u15(t ) = u18(t )= u21(t ),
− 4 − 2 − 4
T T 3T

u23(t) = u14(t− 4) = u17(t− 2)= u20(t− 4 ),


 T T 3T
u22(t) = u13(t− 4) = u16(t− 2)= u19(t− 4 ).




Remark 1.3. We notice that the angle between the plane P and P is the
s 0
same with the angle between P and P , and hs l mod l if and only if
l s 0
− ≡
(l s)h l mod l which means l s and s implies the same h. So in (1.3),
− ≡ −
the two symmetric constraints are the same in geometry, and we can only
consider the integer 1 s l = n/4.
≤ ≤ 2
6 ZHIQIANGWANGANDSHIQINGZHANG
Remark 1.4. By the condition (1.3), we see that I = [0, T ] is a fundamen-
2h
tal domain of dihedral type for the trajectories (see [8]) for more details),
which implies that the motion of the particles on the whole period [0,T] is
determinedbytheir motiononI = [0, T ]throughthesymmetricconditions.
2h
And from (1.2) we see that u is the generating particle, so in Section 3 we
0
can only consider the motion of the generating particle u in the interval
0
I= [0, T ].
2h
Let G = R,S,Rˆ ,Rˆ with the following representations:
s s 0
h i
ρ(R)= R,τ(R)t = t,σ(R) = (0,1,...,l 1)(l,l+1,...,n 1),
− −
l 1
−
ρ(S) = S,τ(S)t = t,σ(S) = (0,l) (l k,l+k),
−
k=1
Y
T
ρ(Rˆ )= Rˆ ,τ(Rˆ )t = t,σ(Rˆ ) = id,
s s s s
h −
ρ(Rˆ )= Rˆ ,τ(Rˆ )t = t,σ(Rˆ ) = id,
0 0 0 0
−
and set
Λ = u(t) Λ :u(t) satisfy (1.2)(1.3) .
s
{ ∈ }
ItiseasytocheckthatΛ = Λ ,i.e. lookingfortrajectories withproperties
s Gs
(1.2)(1.3) is equivalent to seeking for critical point of on Λ .
A Gs
In [8], Ferrario and Terracini proved that is coercive if and only
A|ΛG
if , x g x = x, g G = 0 where is the configuration space
G
X { ∈ X| · ∀ ∈ } X
of the particles. Obviously, in our assumption, is not coercive since
A|Λs
(e ,...,e , e ,..., e ) where e = (0,0,1). So motivated by [9], we
3 3 3 3 G 3
− − ∈ X
add some topological condition on Λ , to get the open cone described
Gs Ks
in the previous. From (1.3) we see that u (0) P and u (T ) P , so we
0 ∈ 0 0 2h ∈ s
let
T
(1.5) = u(t) Λ : u (0) P ,u ( ) P+
Ks { ∈ s 0 ∈ 0− 0 2h ∈ s }
where we have set P = p P : p e < 0 and P+ = p P :p e > 0 .
0− { ∈ 0 · 3 } s { ∈ s · 3 }
Now we state our main theorem:
Theorem1.5. Forn = 2l 4andeveryintegers max 1,(n 1)3/2 π n
≥ ≤ { −n 23/2logn+γ−
1 where γ 0.57721566490153286 is the Euler-Mascheroni constant, there
} ≈
exists a T-periodic solution u of the classical n-body problem.
s
∗ ∈ K
RUNNING HEAD 7
Corollary 1.6. For n = 2l 4 and s = 1, then h = l and there exists a
≥
T-periodic solution u of the classical Newtonian n-body problem.
1
∗ ∈ K
The case for n = 4 was discussed in Section 2 of [9], here we generalize
theirresult. ByRemark1.3,scanbechosenlargerthan1whenn 8. Ifwe
≥
let f(n) = (n 1)3/2 π n 1, we see that f(n) is monotone increasing
−n 23/2logn+γ −
and we have f(4) 0.4697, f(6) 1.1400, f(8) 1.7376, f(10) 2.2931,
≈ ≈ ≈ ≈
f(14) 3.3262, f(26) 6.0995, etc. That is to say, for n 10, we can
≈ ≈ ≥
choose s = 2 such that there exists a T-periodic solution u of the
2
∗ ∈ K
classical n-body problem.
Remark 1.7. Actually, the value of integer s depends on the estimate of
excluding total collisions. By doing more explicit computation we can prove
that, for n = 8, there exists a T-periodic solution u of the classical
2
∗ ∈ K
8-body problem.
2. Coercivity
Proposition 2.1. is coercive and ∂ .
A|Ks Ks ⊂ △Gs
Proof. From (1.2) we see that there is a collision if and only if there exists
some t R such that u (t ) Γ, where Γ are the rotating axes of the
c 0 c
∈ ∈
dihedral group D . So it is obvious that ∂ and in the following we
l Ks ⊂ △Gs
will prove the coercivity.
Applied Newton-Leibniz Formula and Ho¨lder Inequality, we have
T T
2h 2h
u (T/2h) u (0) = u˙ (t)dt u˙ (t)dt
0 0 0 0
| − | | | ≤ | |
Z0 Z0 .
T
( T )12( 2h u˙0(t)2dt)12
≤ 2h | |
Z0
Since u (0) P ,u (T ) P+, we must have
0 ∈ 0− 0 2h ∈ s
T T sπ
u (0) u ( ) u (0) u ( ) cos ,
0 0 0 0
· 2h ≤| || 2h | l
8 ZHIQIANGWANGANDSHIQINGZHANG
which induces that
T
u˙ (t) 2 C u ( ) u (0)2
|| 0 ||L2 ≥ 1| 0 2h − 0 |
T T sπ
C u ( )2+ u (0)2 2u (0) u ( ) cos
1 0 0 0 0
≥ | 2h | | | − | || 2h | l
h sπ T 2 siπ
= C u (0) cos u ( ) +C u (0)2sin2
1 0 0 1 0
| | l −| 2h | | | l
h i
C u (0)2,
0
≥ | |
which implies that u˙ is an equivalent norm of H1 = W1,2 and the
0 L2
k k
coercivity for the functional follows. (cid:3)
A
3. Estimate on Collisions and the Proof of Theorem 1.5
In this section, we show that the minimizer u is free of collisions.
s
∗ ∈ K
Firstly we exclude total collisions and in Section 3.2 we discuss the partial
collisions.
3.1. Total Collision. Our way to show that there is no total collision is
based on level estimate. Firstly assuming that a total collision happens
in u , we show that there is a lower bound of the action functional
s
∗ ∈ K
(u ), and we construct a test loop u˜ without collisions such that
s
B ≤ A ∗ ∈ K
(u˜)< . Then we have (u˜) < (u ) which contradicts with that u is a
A B A A ∗ ∗
minimizer. Thus the minimizers u is free of total collisions.
s
∗ ∈ K
Lemma3.1. (Gordon’s Theorem [10]) Letx W1,2([t ,t ],Rd)and x(t ) =
1 2 1
∈
x(t ) = 0. Then for any a > 0, we have
2
t2(1 x˙ 2+ a )dt 3(2π)23a32(t2 t1)13.
2| | x ≥ 2 −
Zt1 | |
Proposition 3.2. (Lower bound estimates for with total collisions)
A
Assume that u has a total collision. Then
s
∈ K
(u) 3(n−1)(2h)23π23n32T13 , .
A ≥ 4 B
Proof. Since the center of mass is at origin, the functional can be written
A
as
1 T 1 M
(3.1) (u) = m m [ u˙ u˙ 2+ ]dt,
i j i j
A M 2| − | u u
i<j Z0 | i − j|
X
RUNNING HEAD 9
where the total mass M = N m . This formulation came from ([19][20])
i=1 i
and has been widely used to obtain the lower bound estimate of collision
P
paths ([3],[5],[22],[21]etc). In our assumption, we have m = = m = 1
1 n
···
and M = n. Moreover, if u has a total collision, then they collide at
s
∈ K
least h times in the interval [0,T). Applying Lemma 3.1, we have
1 T 1 n
(u)= [ u˙ u˙ 2+ ]dt
i j
A n 2| − | u u
i<j Z0 | i− j|
X
1n(n 1)3 2 2 T 1
− (2π)3n3( )3 h
≥ n 2 2 h ×
= 3(n−1)(2h)23π32n23T13 , .
4 B
(cid:3)
Nextweconstructtestloopsu˜ suchthat (u˜) < intwodifferent
s
∈ K A B
ways. The idea of the first one is from Fusco et al.[9], but it holds only for
s = 1. The second one holds for s < (n 1)3/2 π n 1 and it needs
−n 23/2logn+γ −
some more explicit analysis on the potential U.
Proposition 3.3. (Upper bound estimate for s = 1)
When s = 1, there exists u˜ such that
1
∈ K
(u˜) 3(2h2)13n(n 1)23π32T13.
A ≤ 4 −
Proof. We construct the test loop u˜ similar to that in Proposition 5.3 in
[9]. Assume the generating particle u˜ moves with constant speed on a
0
curve which is the union of two quarters of circumferences C , C of radius
1 2
rtan π. C has the center (r,0,0) and lies on the plane ξ = r. C has
n 1 1 2
the center (rcos π,rsin π,0) and lies on the plane ξ cos π +ξ sinπ = r, see
l l 1 l 2 l
figure 1. So the constant speed of generating particle u˜˙ = πrtannπ, and the
0 T/2h
kinetic energy
n 2hπrtan π
K(u˜˙) = ( n)2.
2 T
From the definition of u˜, we see that u˜ u˜ 2rtan π for all i = j,
| i − j| ≥ n 6
which implies
n(n 1) 1
V(u˜) − ,
≤ 2 2rtan π
n
therefore we have
T 2h2nπ2r2tan2 π n(n 1)T
(u˜) = K(u˜˙)+V(u˜)dt n + − ,
A ≤ T 4rtan π
Z0 n
10 ZHIQIANGWANGANDSHIQINGZHANG
ξ
3
u (T )
0 2h
ξ
2
u (T )
0 4h
ξ
1
u (0)
0
Figure 1.
and the conclusion follows if we choose r = (n−11)1/3T2/3 . (cid:3)
(16h2)3 tanππ2/3
n
From Proposition 3.2 and 3.3, we see that A(u˜) ( n )1/3 < 1, i.e.
≤ 2(n 1)
B −
(u˜) < . Next we consider the situation s 2, since s n, in the
A B ≥ ≤ 4
following we suppose n = 2l 8.
≥
Proposition 3.4. (Upper bound estimate for the general s)
When n = 2l 8, there is a test loop u˜ such that (u˜) < for
s
≥ ∈ K A B
s+1 < (n 1)3/2 π n .
−n 23/2logn+γ
Proof. Assumethattheparticle u˜ moves withconstant speedonthesphere
0
u˜ = a, with the radius a to be determined later. More precisely, suppose
0
| |
u˜ (ϕ(t),θ(t)) = a(cosϕeθ(t)√ 1,sinϕ(t)) where
0 −
π T
(ϕ(t),θ(t)) = ( ,ωt), t [0, ]
−n ∈ 4h(s+1)

 2π π T 3T
(ϕ(t),θ(t)) = (ωt , ), t ( , ]

 − n n ∈ 4h(s+1) 4h(s+1)



π 2π 3T T
(ϕ(t),θ(t)) = ( ,ωt ), t ( , ]
 n − n ∈ 4h(s+1) 2h




where ω =(s+1)π/l = 2h(s+1)π, see Figure 2.
T/2h lT
First we claim that, for every t [0, T ], the potential
∈ 2h
n2
(3.2) U(u˜(t)) < (logn+γ).
2aπ