A Noncollision Periodic Solution for N-Body Problems

A Noncollision Periodic Solution for N-Body Problems

Journal of Mathematical Analysis and Applications 257, 332–342 (2001) doi:10.1006/jmaa.2000.7349, available online at http://www.idealibrary.com on A...

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Journal of Mathematical Analysis and Applications 257, 332–342 (2001) doi:10.1006/jmaa.2000.7349, available online at http://www.idealibrary.com on

A Noncollision Periodic Solution for N-Body Problems1 Shiqing Zhang Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200030, People’s Republic of China; and Department of Mathematics, Chongqing University, Chongqing 400044, People’s Republic of China

and Qing Zhou Department of Mathematics, East China Normal University, Shanghai 200062, People’s Republic of China Submitted by F. E. Udwadia Received April 11, 2000 Using variational minimization methods, we prove the existence of one noncollision periodic solution for N-body type problems whose potentials are pinched between two homogeneous potentials in R k (k ≥ 2). © 2001 Academic Press Key Words: N-body problems; noncollision periodic solution; variational methods.

1. INTRODUCTION AND MAIN RESULTS The motion of N-body type problems [1, 2, 9, 12, 19, 20] is related with solving the following second order differential equations, ∂U mi q¨ i =  (1.1) ∂qi where mi > 0 is the mass of the ith body and qi ∈ R k k ≥ 2 is the position of the ith body, and the potential  Uq = Uq1   qN  = Uij qi − qj  (1.2) 1≤i
where Uij x ∈ C 1 R k \0 R. 1

Partially supported by grants of MOST, NSFC, and QSSTF. 332

0022-247X/01 $35.00 Copyright © 2001 by Academic Press All rights of reproduction in any form reserved.

noncollision periodic solutions

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In the last 20 years, some researchers applied variational methods to study the periodic solutions of N-body type problems [3–8, 10, 13, 14, 18, 21, 22], but they didn’t get the existence of one noncollision periodic solution for any given masses of N bodies. Observing the symmetry and choosing a suitable domain of the Lagrangian action integral for (1.1), we prove that the minimizer of the Lagrangian action integral is one noncollision periodic solution of (1.1)–(1.2) assuming the potential Uq is pinched between two homogeneous potentials. Let Okk ≥ 2 denote the rotational group in R k and   Bθ 0 ∈ Ok Aθ = (1.3) 0 −Ik−2 θ ∈ 0 2π where Ik−2 is a unit matrix with order k − 2 and  cos θ − sin θ  Bθ = sin θ cos θ Let

(1.4)

  (1.5) H = W 12 R/TZ  R k       2π T H# = x ∈ H x t + =A xt r ≥ 2 an integer (1.6) r r  E = q = q1   qN  qi − qj ∈ H#  i j = 1  N   N 

= q = q1   qN  ∈ E mi qi t ≡ 0 E (1.7) 

i=1

i t = qj t  = q = q1   qN  ∈ E q



∀ t ∈ R 1 ≤ i = j ≤ N f q =

(1.8)

T N 1 T mi q˙i 2 dt + Uqdt 2 0 i=1 0

(1.9)

Theorem 1.1. Assume Uq satisfies (1) mi mj mi mj b  a  ≤ Uq ≤  α 2 1≤i
2 1≤i
α>0

(1.10)

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zhang and zhou (2) UAq = Uq

(1.11)

for some A = A2π/r ∈ Ok where r will be defined later. Then there is an integer r depending on α and masses m1   mN such that the minimizer of f q on   is one noncollision T-periodic solution for (1.1)–(1.2). Theorem 1.2. If α = 1 N = 3, m1 = m2 = m3 = 1, and a = b = 1 then the minimizer of f q on   with r = 2 is one noncollision T-periodic solution. Remark. The domain  for f q is different from the one defined by Bessi and Coti Zelati [4]. 2. THE PROOF OF THEOREM 1.1 In order to prove Theorem 1.1, first we give some lemmas: Lemma 2.1. The critical points of f q in  are noncollision T-periodic solutions of (1.1)–(1.2).

for f q restricted on E

is First, we prove that the critical point q ∈ E also a critical point for f q on E. In fact, the condition that the center of masses is fixed at the origin is equivalent to N 2  gq = mi qi t ≡ 0 (2.1) i=1

Hence by the Lagrangian multiplier rule, for any critical point q of f q on

we have E, f  q + λg q = 0

(2.2)

that is, for any ϕ = ϕ1   ϕN , ϕi ∈ W 12 R/TZ  R k  we have f  q ϕ + λg q ϕ = 0 N N   f  q ϕ + 2λ mi qi mi q˙i  ϕi  = 0 i=1

i=1

(2.3) (2.4)

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335

f  q ϕ = 0

(2.5)

that is,

Now assume q ∈ E is a critical point of f q on E. Then f  q y = 0

∀ y ∈ E

(2.6)

Hence we have p = p1   pN  ∈ E ⊥ , where pi ≡ q¨ i −

∂U  ∂qi

i = 1  N

(2.7)

On the other hand, we notice that pi − pj = q¨ i − q¨ j −

∂U ∂U + ∂qi ∂qj

= qi − qj  − +

N  aαmi mj qj − qi  j =ij=1

qi − qj α+2

N  aαmi mj qi − qj  i =ji=1

qj − qi α+2



(2.8)

Hence pi − pj ∈ H#  p ∈ E. Hence p ∈ E ⊥ ∩ E = 0; that is, q is a solution of (1.1). Lemma 2.2. The functional f is coercive on ; that is, for any qn    qn H → +∞ f qn  → +∞. Proof.

For any q = q1   qN  ∈  we have       T T T ≡ qi t + − qj t + qi − qj  t + r r r    2π  =A qi t − qj t r   2π ≡A qi − qj t r 2   qi − qj  t + T − qi − qj t t  2  2π = A qi − qj t − qi − qj t r 2 π 2 = 2 sin qi − qj t r

(2.9)

(2.10)

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On the other hand, 2   qi − qj  t + T − qi − qj t r T 2 t+ r = q˙i − q˙j dt t

T ≤ r =



t+ Tr

t



2

q˙i − q˙j t dt

2 T T q˙i − q˙j t dt 2 r 0

(2.11)

Hence we have 2 T 2 q˙i − q˙j t 2 dt ≥ r 2 sin π qi − qj t 2 T r 0 T  2 mi mj q˙i − q˙j dt

(2.12)

0 1≤i
≥ M



π 2  r 2 m m q − qj 2 2 sin T r 1≤i
N 2 π 2  r 2 mi q˙i dt ≥ 2 sin M mi qi 2  T r i=1 i=1

N T 

0

(2.13) (2.14)

where M=

N  i=1

mi

(2.15)

Hence the standard norm for H is equivalent to q ˙ 2=



N 1

0 i=1

mi q˙i 2 dt

1/2



(2.16)

Hence the definition of f q implies f is coercive. Lemma 2.3. The system (1.1)–(1.2) has a weak T-periodic solution q = q1   qN  ∈   in the sense of Bari and Rabinowitz [3]: (1◦ )

qi ∈ W 12 R/TZ  R k .

(2◦ ) The collision set C = t ∈ 0 T  qi t = qj t for some 1 ≤ i = j ≤ N has Lebesgue measure 0.

noncollision periodic solutions (3◦ )

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qi is C 2 on 0 T \C and satisfies (1.1) and energy conservation, N 1 m q˙ 2 − Uq1   qN  = h 2 i=1 i i

(2.17)

Proof. It’s easy to prove f q has positive lower bound and is weakly lower semi-continuous, so Lemma 2.2 implies Lemma 2.3. In order to get a good lower bound estimate of f q on the collision solutions, we need another lemma:

 Lemma 2.4. Let index sets A and B satisfy A B =  and A B = 1 2  N. Then 

m i mj

qi − qj α ij∈A×B 





ij∈A×B

mi m j

1+ α2 

 ij∈A×B

mi mj qi − qj

2

− α2



(2.18)

For the proof of Lemma 2.4 refer to Long and Zhang [13]. In order to facilitate the reader, we repeat it. By H¨ older’s inequality, we have  2       mi mj α (2.19) mi mj ≤ m m

q − q

i j i j

qi − qj α i∈A j∈B i∈A j∈B i∈A j∈B By H¨ older’s inequality, we have  mi mj qi − qj α i∈A j∈B

 ≤

 i∈A j∈B

mi m j

 2−α  2

 i∈A j∈B

mi mj qi − qj 2

α/2



(2.20)

By (2.19) and (2.20) we get (2.18). Now we estimate the lower bound of f q on the collision solutions. Let SN denote the group of all the permutations of 1  N. For l = 2  N, we set  ∂l = q ∈ E ∃ s ∈ SN  ∃ t¯ ∈ 0 T  s.t. qs1 t¯ = · · · = qsl t¯ (2.21) First, we assume l = 2, s is the identity, and t¯ = 0. Then by the Lagrangian identity and the symmetry property qi − qj t + Tr  = A 2π qi − qj t r we have f q ≥ g1 q + g2 q + g3 q

(2.22)

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where

 g1 q = r

×  g2 q = r

g3 q = r



1≤i =j≤2 T/r



  1 1 2 dt

q˙ − q˙j + Ma 2 i

qi − qj α

3≤i =j≤N T/r



0



m i mj



  1 1 dt

q˙i − q˙j 2 + Ma 2

qi − qj α



0

(2.24)

mi m j

1≤i≤2 3≤j≤N T/r

(2.23)

mi mj



2 2M ×



0

1 2M ×





1 2M

  1 1 dt

q˙i − q˙j 2 + Ma 2

qi − qj α

(2.25)

Using the estimates of the Lagrangian action integral on collision solutions of two body [7] problems we have Lemma 2.5. 2α

2−α

g1 q ≥ C1 r 2+α T 2+α  where −α

C1 = AM22+α M2 =

2  i=1

 1≤i =j≤2

mi m j

mi

(2.26)

(2.27) (2.28)

  2α 1 1 1 A= + αa2/α+2 2π 2+α 2 2 α Let

 1≤i =j≤2

B2 = mins∈SN 

2  i=1

(2.29)

msi msj

msi

α/2+α

 = AB C 1 2

(2.30)

(2.31)

Then 2α 2−α r 2+α T 2+α infg1 q q ∈ ∂2  ≥ C 1

(2.32)

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By the arguments of Degiovanni and Giannoni [7], we can get the lower bound estimate on g2 q, 2−α

g2 q ≥ C2 T 2+α 

(2.33)

where −α

2+α C2 = AMN−2

MN−2 = Let

N  i=3

N  i=3

mi mj

(2.34)

mi

(2.35)



3≤i =j≤N msi msj BN−2 = min N α/2+α s∈SN  i=3 msi 

(2.36)

 = AB C 2 N−2

(2.37)

Then T 2+α infg2 q q ∈ ∂2  ≥ C 2 2−α

(2.38)

We use inequality (2.18) of Lemma 2.4, Sundman’s inequality [19], and the arguments of Degiovanni and Giannoni [7] to estimate the lower bound for g3 q: 1+ α2    1 T 2 g3 q ≥ m m q˙ − q˙j dt + a m i mj 2M 0 1≤i≤2 3≤j≤N i j i 1≤i≤2 3≤j≤N 



×

1≤i≤2 3≤j≤N



1 T 2M 0  +a

mi mj qi − qj 2

m i mj

1≤i≤2 3≤j≤N

×



T 0



dt

(2.39)

 1/2  2 d  2 dt m m

q − q

i j i j dt 1≤i≤2 3≤j≤N 



−α/2

 1≤i≤2 3≤j≤N

1+ α2

mi mj qi − qj 2

−α/2

dt

1+ α2   1 T 2 ≥ inf

r

˙ dt + a mi m j 2M 0 1≤i≤2 3≤j≤N

(2.40)

×



T

0



r −α dt r ∈ W 12 0 T  R+ 

(2.41)

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zhang and zhou 

1 = T inf 2M



2π T

2

 R +a 2

 1≤i≤2 3≤j≤N

×

mi mj

1+ α2

1 R > 0 Rα

 (2.42)

2  2   α+2 1+ α2  α+2     α   α 2 1 2π 2 α+2 = 1+ a m i mj 2 α 2M T 1≤i≤2 3≤j≤N 2      α+2  α  α 2 2−α 2  2  α+2 α − α+2 1+ = M a α+2 mi mj T 2+α 2π 2 α 1≤i≤2 3≤j≤N      2 2α α 2−α 2 1 1+ αa α+2 2π α+2 M − α+2 = mi mj T 2+α 2 α 1≤i≤2 3≤j≤N    α 2−α 2−α = 2AM − α+2 mi mj T 2+α ≡ C3 T 2+α (2.43)

1≤i≤2 3≤j≤N

Let B=

mins∈SN



1≤i≤2 3≤j≤N M α/α+2

msi msj

(2.44)

 = 2AB C 3

(2.45)

Then T 2−α 2+α infg3 q q ∈ ∂2  ≥ C 3 r inff q q ∈ ∂2  ≥ C 1

2α 2+α

(2.46) +C T +C 2 3

2−α 2+α



(2.47)

r 2+α + C +C T 2+α inff q q ∈ ∂l  ≥ C 1 2 3

(2.48)

It’s easy to see that for l > 2 we also have 2α

2−α

Remark. The corresponding lower bound estimate in Bessi and Coti Zelati [4] is not correct since the symmetry breaks down when they move the binary collision to the origin and they work on their domain. Lemma 2.6 [6]. Let ρ=



m i mj

sinπi − j/N α 2  πi − j σ= mi mj sin N 1≤i =j≤N

(2.49)

1≤i =j≤N

(2.50)

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Then the minimizing value for Lagrangian action f q has upper bound estimate   2 2α 2 α 2−α 1 1 1 f q ≤ + bα 2+α 2π α+2 ρ α+2 σ α+2 M −α/α+2 T 2+α 2 2 α

2−α 2+α ≡ CT

(2.51)

Now we can prove Theorem 1.1. Assume the minimizer q for f q on   has a collision time t¯ ∈ 0 T . + r 2α/2+α + C Then by Lemma 2.5, we can get a lower bound estimate C 1 2 2−α/2+α T for f q, which depends on the integer r ≥ 2. By Lemma 2.6 C 3 we can choose r large enough so that 2α +C  > C

r 2+α +C C 1 2 3

(2.52)

This is a contradication. Now we prove Theorem 1.2. If α = 1 m1 = m2 = m3 = 1 a = 1, and r = 2 then we have A=

3 2π2/3 4

B2 = 22/3  = 3 · 2−4/3 2π2/3  C 1

2α r 2+α C = 3 · 2−2/3 2π2/3 1

=0 C 2 B = 2 · 3−1/3 2α r 2+α C 1

 = 2AB = 32/3 2π2/3 C 3     + C2 + C3 = 3 · 2−2/3 + 32/3 2π2/3

On the other hand, we compute 

1 2 = √ 6 = 4 · 31/2

sinπi − j/3

3 1≤i =j≤3  π 2 9 πi − j 2 σ= = 6 sin = sin 3 3 2 1≤i =j≤3 ρ=

 1/3 13 2/3 1/2 2/3 9

C= 2π 4 · 3  · 3−1/3 22 2

1 5/3 3 2π2/3 < 3 · 2−2/3 + 32/3 2π2/3 2 2α r 2+α +C  =C +C =

1

2

3

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zhang and zhou REFERENCES

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