Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 845795, 5 pages http://dx.doi.org/10.1155/2013/845795 Research Article The Characterization of the Variational Minimizers for Spatial Restricted π + 1-Body Problems Fengying Li, Shiqing Zhang, and Xiaoxiao Zhao Yangtze Center of Mathematics and College of Mathematics, Sichuan University, Chengdu 610064, China Correspondence should be addressed to Fengying Li; lify0308@163.com Received 8 February 2013; Accepted 27 April 2013 Academic Editor: Maoan Han Copyright © 2013 Fengying Li 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. We use Jacobi’s necessary condition for the variational minimizer to study the periodic solution for spatial restricted π + 1-body problems with a zero mass on the vertical axis of the plane for π equal masses. We prove that the minimizer of the Lagrangian action on the anti-T/2 or odd symmetric loop space must be a nonconstant periodic solution for any 2 ≤ π ≤ 472; hence the zero mass must oscillate, so that it cannot be always in the same plane with the other bodies. This result contradicts with our intuition that the small mass should always be at the origin. 1. Introduction and Main Result where The Newtonian n-body problem [1] is a classical problem. Spatial restricted 3-body model was studied by Sitnikov [2]. Mathlouthi [3] et al. studied the periodic solutions for the spatial circular restricted 3-body problems by mini-max variational methods. In this paper, we study spatial circular restricted π + 1body problems with a zero mass on the vertical axis of the plane for π equal masses. Suppose point masses π1 = ⋅ ⋅ ⋅ = ππ = 1 move on a circular orbit around the center of masses. The motion for the zero mass is governed by the gravitational √ forces of π1 , . . . , ππ. Let ππ = π −1(2ππ/π) and √−12ππ‘ π1 (π‘) = ππ π1 , . . . , ππ (π‘) √−12ππ‘ (1) π= π π (π − π) π Μ = ∑ σ΅¨π π σ΅¨3 . σ΅¨ σ΅¨ π=1 σ΅¨σ΅¨ππ − πσ΅¨σ΅¨ ππ , πππ π = 1, . . . , π, (4) Obviously, πΜ(π‘) = (0, 0, 0) satisfies (4); it seems that πΜ(π‘) ≡ 0 is a variational minimizer, but we will prove it is not this is the goal of this paper. Define 1 1 σ΅¨ σ΅¨2 π 1 π (π) = ∫ [ σ΅¨σ΅¨σ΅¨πσ΅¨σ΅¨σ΅¨Μ + ∑ σ΅¨σ΅¨ σ΅¨σ΅¨ ] ππ‘, 2 0 π=1 σ΅¨σ΅¨π − ππ σ΅¨σ΅¨ satisfy the Newtonian equations: (3) The orbit π(π‘) = (0, 0, π§(π‘)) ∈ π 3 for zero mass satisfies the following equation: = ππ π1 (π‘) , . . . , ππ (π‘) = ππ ππ ππΜ = ππ ππ ∑ σ΅¨σ΅¨ σ΅¨σ΅¨ . 1≤π<π≤π σ΅¨σ΅¨σ΅¨ππ − ππ σ΅¨σ΅¨σ΅¨ π ∈ Λ π, (5) then (2) 1 π 1 σ΅¨ σ΅¨2 ] ππ‘ β π (π§) , π (π) = ∫ [ σ΅¨σ΅¨σ΅¨σ΅¨π§σΈ σ΅¨σ΅¨σ΅¨σ΅¨ + 2 √ 2 0 π + π§2 π ∈ Λ π , (6) 2 Abstract and Applied Analysis Lemma 5. π(π) in (6) attains its infimum on Λ1 = Λ 1 or Λ2 = Λ 2. where {π (π‘) = (0, 0, π§ (π‘)) | π§ (π‘) ∈ π1,2 ( π , π )} Λ1 = { π }, π (−π‘) = −π (π‘) } { Proof. By Lemmas 3 and 4, it is easy to prove Lemma 5. 1,2 π { {π (π‘) = (0, 0, π§ (π‘)) | π§ (π‘) ∈ π ( , π )} } (7) π Λ2 = { , } 1 { } π§ (π‘ + ) = −π§ (π‘) { } 2 σ΅¨ 2 σ΅¨ π σ΅¨ π₯ (π‘) , π₯Μ (π‘) ∈ πΏ ((0, 1) , π ) }. π1,2 ( , π ) = {π₯ (π‘) σ΅¨σ΅¨σ΅¨ π₯ (π‘ + 1) = π₯ (π‘) σ΅¨σ΅¨ π Notice that the symmetry in Λ 2 is related to the Italian symmetry [4]. In this paper, our main result is the following. Theorem 1. The minimizer of π(π) on the closure Λπ of Λ π (π = 1, 2) is a nonconstant periodic solution for 2 ≤ π ≤ 472; hence the zero mass must oscillate, so that it can not be always in the same plane with the other bodies. 2. Proof of Theorem 1 We define the inner product and equivalent norm of π1,2 (π /π, π ): 1 β¨π’, Vβ© = ∫ (π’V + π’σΈ ⋅ VσΈ ) ππ‘, 0 1 βπ’β = [∫ |π’|2 ππ‘] 1/2 0 1 σ΅¨ σ΅¨2 + [∫ σ΅¨σ΅¨σ΅¨σ΅¨π’σΈ σ΅¨σ΅¨σ΅¨σ΅¨ ππ‘] 0 (8) 1/2 , which is equivalent to 1 σ΅¨ σ΅¨2 [∫ σ΅¨σ΅¨σ΅¨σ΅¨π’σΈ σ΅¨σ΅¨σ΅¨σ΅¨ ππ‘] 0 1/2 + |π’ (0)| . (9) Lemma 2 (Palais’ Symmetry Principle [5]). By Palais’ Symmetry Principle, we know that the critical point of π(π) in Λπ is a noncollission periodic solution of Newtonian equation (4). Let π be an orthogonal representation of a finite or compact group πΊ in the real Hilbert space π» such that for all π ∈ πΊ, π(π⋅ π₯) = π(π₯), where π : π» → π . Let π = {π₯ ∈ π» | π ⋅ π₯ = π₯, for all π ∈ πΊ}. Then the critical point of π in π is also a critical point of π in π». Lemma 6 (Jacobi’s Necessary Condition [7]). If the critical point π’ = π’Μ(π‘) corresponds to a minimum of the functional π ∫π πΉ(π‘, π’(π‘), π’σΈ (π‘))ππ‘ and if πΉπ’σΈ π’σΈ > 0 along this critical point, then the open interval (π, π) contains no points conjugate to π; that is, for all π ∈ (π, π), the following boundary value problem − β (π) = 0, Lemma 4 (Poincare-Wirtinger Inequality). Let π ∈ π1,2 π × (π /π, π π) and ∫0 π(π‘)ππ‘ = 0; then π 2 π 2π σ΅¨2 σ΅¨ Μ σ΅¨σ΅¨2 σ΅¨ ∫ σ΅¨σ΅¨σ΅¨π(π‘) σ΅¨σ΅¨ ππ‘ ≥ ( ) ∫ σ΅¨σ΅¨σ΅¨π(π‘)σ΅¨σ΅¨σ΅¨ ππ‘. π 0 0 (10) (11) β (π) = 0, has only the trivial solution β(π‘) ≡ 0, for all π‘ ∈ (π, π), where 1 σ΅¨σ΅¨ π = πΉπ’σΈ π’σΈ σ΅¨σ΅¨σ΅¨σ΅¨ 2 σ΅¨π’=Μπ’ (12) σ΅¨σ΅¨ 1 π σ΅¨σ΅¨ π = (πΉπ’π’ − πΉπ’π’σΈ )σ΅¨σ΅¨ . σ΅¨σ΅¨π’=Μπ’ 2 ππ‘ Remark 7. It is easy to see that Lemma 6 is suitable for the fixed end problem. In this paper, we consider the periodic solutions of (2) on Λπ = Λ π (π = 1, 2); hence we need to establish a similar conclusion as Lemma 6 for the periodic boundary problem. Lemma 8. Let πΉ ∈ πΆ3 (π × π × π , π ). Assume that π’ = π’Μ(π‘) π is a critical point of the functional ∫0 πΉ(π‘, π’(π‘), π’σΈ (π‘))ππ‘ on {π’ ∈ π1,2 (π /ππ, π ), π’σΈ (0) = 0} and πΉπ’σΈ π’σΈ |π’=Μπ’ > 0. If the open interval (0, π) contains a point π conjugate to 0, then π’ = π’Μ(π‘) π is not a minimum of the functional ∫0 πΉ(π‘, π’(π‘), π’σΈ (π‘))ππ‘. Proof. Suppose π’ = π’Μ(π‘) is a minimum of the π functional ∫0 πΉ(π‘, π’(π‘), π’σΈ (π‘))ππ‘. The second variation of π ∫0 πΉ(π‘, π’(π‘), π’σΈ (π‘))ππ‘ is π ∫ (πβσΈ 2 + πβ2 ) ππ‘, 0 where σ΅¨σ΅¨ 1 πΉπ’σΈ π’σΈ σ΅¨σ΅¨σ΅¨σ΅¨ , 2 σ΅¨π’=Μπ’ π= σ΅¨σ΅¨ π 1 σ΅¨ π = (πΉπ’π’ − πΉπ’π’σΈ )σ΅¨σ΅¨σ΅¨ . σ΅¨σ΅¨π’=Μπ’ 2 ππ‘ In order to prove Theorem 1, we need the following lemmas: Lemma 3 (see [6]). Let π be a reflexive Banach space, π be a weakly closed subset of π, π : π → π ∪ {+∞}, and π ≡ΜΈ +∞ is weakly lower semicontinuous and coercive (π(π₯) → +∞ as βπ₯β → +∞); then π attains its infimum on π. π (πβσΈ ) + πβ = 0, ππ‘ (13) (14) Set π ππ’Μ (β) = ∫ (πβσΈ 2 + πβ2 ) ππ‘. 0 (15) For all β ∈ {π’ ∈ π1,2 (π /ππ, π ), π’σΈ (0) = 0}, it is easy to see that ππ’Μ (β) ≥ 0. Then by ππ’Μ (π) = 0, π is a minimum of ππ’Μ (β). The Euler-Lagrange equation which is called the Jacobi equation of (15) is − π (πβσΈ ) + πβ = 0. ππ‘ (16) Abstract and Applied Analysis 3 Since the interval (0, π) contains a point π conjugate to 0, there exists a nonzero Jacobi field β0 ∈ πΆ2 ([0, π], π ) satisfying π − (πβσΈ 0 ) + πβ0 = 0, ππ‘ β0 (0) = 0, β0 (π) = 0, (17) β0σΈ Then π3 = (0) = 0. π‘ ∈ [0, π] , π‘ ∈ (π, π] , 1 2/3 π π = ( ) [ ∑ csc ( π)] 4π π [1≤π≤π−1 ] (18) π ππ’Μ (Μβ) = ∫ (πΜβσΈ 2 + πΜβ2 ) ππ‘ = ∫ π 0 (19) (πβ0σΈ 2 + πβ02 ) ππ‘ 1 σ΅¨ σ΅¨2 π . πΉ (π§, π§σΈ ) = σ΅¨σ΅¨σ΅¨σ΅¨π§σΈ σ΅¨σ΅¨σ΅¨σ΅¨ + 2 √π + π§2 2 (20) Combining with Μβ(0) = Μβ(π) = 0 and ΜβσΈ (0) = 0, by the uniqueness of initial value problems for secondorder differential equation, we have Μβ(π‘) ≡ 0 on [0, π], which contradicts the definition of Μβ. Therefore, Lemma 8 holds. Lemma 9. The radius π for the moving orbit of π equal masses is π=( 1 ) 4π 2/3 [ ∑ csc ( π π)] π [1≤π≤π−1 ] 1 where σ΅¨σ΅¨ 1 1 πΉπ§σΈ π§σΈ σ΅¨σ΅¨σ΅¨σ΅¨ = , 2 σ΅¨π§=0 2 σ΅¨σ΅¨σ΅¨ π π 1 π = (πΉπ§π§ − πΉπ§π§σΈ )σ΅¨σ΅¨σ΅¨ = − 3 . σ΅¨σ΅¨π§=0 2 ππ‘ 2π (28) π= The Euler equation of (27) is called the Jacobi equation of the original functional (6), which is − π (πβσΈ ) + πβ = 0, ππ‘ Proof. By (1)–(3), we have ππ − ππ Μ = ∑σ΅¨ ππ σ΅¨σ΅¨3 . σ΅¨ π =ΜΈ π σ΅¨σ΅¨ππ − ππ σ΅¨σ΅¨ σ΅¨ σ΅¨ π (30) β = 0. π3 Next, we study the solution of (30) with initial values β(0) = 0, βσΈ (0) = 1. It is easy to get βσΈ σΈ + √ (31) π π = ππ. π3 (32) We have π=[ −4π2 = ∑ 1 − ππ 1 π 4π2 π3 = ∑ σ΅¨ = ∑ csc ( π) . 3 σ΅¨ σ΅¨ σ΅¨ 4 1≤π≤π−1 π π =ΜΈ π σ΅¨σ΅¨1 − ππ σ΅¨σ΅¨ σ΅¨ σ΅¨ π3 π ⋅ sin √ 3 π‘, π π which is not identically zero on [0, 1], but we will prove β(π) = 0 for some π ∈ (0, 1). Suppose there exists π ∈ (0, 1) such that β(π) = 0. Hence, for some π ∈ π+ (22) Substituting (1) into (22), we have ππ − ππ σ΅¨σ΅¨ σ΅¨σ΅¨3 , π =ΜΈ π π3 σ΅¨σ΅¨ππ − ππ σ΅¨σ΅¨ σ΅¨ σ΅¨ (29) that is, β (π‘) = √ (21) (26) (27) 1/3 . (25) ∫ (πβσΈ 2 + πβ2 ) ππ‘, 0 π (πΜβσΈ ) + πΜβ = 0. ππ‘ . Then the second variation of (6) in the neighborhood of π§ = 0 is given by = 0. Notice that we can extend Μβ periodically when we take π as the period, so Μβ ∈ π01,2 (π /ππ, π ). For all β ∈ πΆ01 ([0, π], π ), it is easy to check that ππ’Μ (β) ≥ 0. Then by (19), one has Μβ ∈ πΆ2 ([0, π] \ {π}, π ) ∩ π01,2 (π /ππ, π ) is a minimum of ππ’Μ (β). Hence we get − 1/3 Proof of Theorem 1. Clearly, π(π‘) = (0, 0, 0) is a critical point of π(π) on Λπ = Λ π (π = 1, 2). For the functional (6), let we have Μβ ∈ πΆ2 ([0, π] \ {π}, π ), Μβ(0) = Μβ(π) = Μβ(π) = 0 and 0 (24) Therefore Letting Μβ (π‘) = {β0 (π‘) 0 1 π ∑ csc ( π) . 2 16π 1≤π≤π−1 π =[ [ (23) by using (24). π2 π2 π3 ] π 1/2 π2 (∑π−1 π=1 csc ((π/π) π)) 16π 1/2 ] ] (33) 4 Abstract and Applied Analysis Case 1 (Minimizing π(π) on Λ1 = Λ 1 ). Letting 0 < π < 1/2, β (π‘) { { β (π‘) = {0 { {−β (1 − π‘) π‘ ∈ [0, π] , π‘ ∈ (π, 1 − π] , π‘ ∈ (1 − π, 1] . π−1 It is easy to check that β(π‘) ∈ πΆ ([0, 1] \ {π, 1 − π}, π ) ∩ π1,2 (π , π ), β(−π‘) = −β(π‘), β(0) = β(0) = 0, β(π) = β(π) = 0, and β is a nonzero solution of (30). If we take π = 1, π=[ [ 16π 1/2 ] ] 1 < . 2 (35) It is equivalent to π−1 ∑ csc ( π=1 π π) < 4π. π ∑ csc ( (34) 2 ∑π−1 π=1 csc ((π/π) π) It implies that π=1 { { {π‘ π§ (π‘) = { 1 { { {4 Let (37) It is not hard to check that π(π₯) is nonmonotone. But for 2 ≤ π ≤ 472, (36) holds by writing program to calculate it. Therefore, for 2 ≤ π ≤ 472, we have π ∈ (0, 1) such that sin √ π π = sin π = 0. π3 Case 2 (Minimizing π(π) on Λ2 = Λ 2 ). Let β (π‘) { { { { { { { 0 { { { { Μβ (π‘) = 1 { { −β (π‘ − ) { { { 2 { { { { { {0 { π‘ ∈ [0, π] , 1 π‘ ∈ (π, ] , 2 1 1 π‘ ∈ ( , + π] , 2 2 1 π‘ ∈ ( + π, 1] . 2 [ ∑π−1 π=1 csc ((π/π) π) 16π 1/2 ] ] 1 < . 2 (42) π (π§) = 1/2 π 1 ππ‘ + 2∫ 2 √ 4 0 π + π§2 = 1/4 1/2 1 π π ππ‘ ππ‘ + 2 ∫ + 2∫ 2 2 2 √π + π‘ 4 0 1/4 √π + (1/16) = 1 1 + √1 + 16π2 2π . + 2π ln + √1 + 16π2 4 4π (43) Writing program to calculate, we find an π0 = 9 ∗ 1010 such that π(π§) < π(0). Hence π(π‘) = (0, 0, 0) is not a local minimum for π(π) on Λπ = Λ π (π = 1, 2). So the minimizers of π(π) on Λ π are not always at the center of masses; they must oscillate periodically on the vertical axis; that is, the minimizers are not always coplanar with the other bodies; hence, we get the nonplanar periodic solutions. Acknowledgments (39) It is easy to check that Μβ(π‘) ∈ πΆ2 ([0, 1]\{π, 1/2, (1/2)+π}, π )∩ π1,2 (π , π ), Μβ(π‘ + (1/2)) = −Μβ(π‘), Μβ(0) = β(0) = 0, Μβ(π) = β(π) = 0, and Μβ is a nonzero solution of (30). We hope π ∈ (0, 1/2); that is, π=[ 1 π‘ ∈ [0, ] , 4 1 1 π‘ ∈ [ , ], 4 2 and we extend π§(π‘) by π§(π‘ + (1/2)) = −π§(π‘). We have (38) Notice that we can extend β periodically when we take 1 as the period, so β ∈ Λ 1 . Then by Lemma 8, π(π‘) = (0, 0, 0) is not a local minimum for π(π) on Λ 1 . Hence the minimizers of π(π) on Λ 1 are not always at the center of masses; they must oscillate periodically on the vertical axis; that is, the minimizers are not always coplanar with the other bodies; therefore, we get the nonplanar periodic solutions. (41) Calculated by program, for 2 ≤ π ≤ π0 = 472, we have π ∈ (0, 1/2) such that β(π) = 0. Notice that we can extend Μβ periodically when we take 1 as the period, so Μβ ∈ Λ 2 . Then by Lemma 8, π(π‘) = (0, 0, 0) is not a local minimum for π(π) on Λ 2 . Hence the minimizers of π(π) on Λ 2 are not always at the center of masses; they must oscillate periodically on the vertical axis; that is, the minimizers are not always coplanar with the other bodies; therefore, we get the nonplanar periodic solutions. We can use another argument to get much larger π0 . We construct a test function π§(π‘) such that π(π§) < π(0) = π/π for π ≤ π0 , where π0 is a very large number. Let (36) π π (π₯) = ∑ csc ( π) . π₯ π =ΜΈ π₯ π π) < 4π. π (40) The authors would like to thank the referee for his/her many valuable comments and suggestions. This paper is Supported by the national Natural Science Foundation of China (11071175) and the Ph.D Programs Foundation of Ministry of Education of China. References [1] A. Wintner, The Analytical Foundations of Celestial Mechanics, vol. 5 of Princeton Mathematical Series, Princeton University Press, Princeton, NJ, USA, 1941. [2] K. Sitnikov, “Existence of oscillating motions for the three-body problem,” Doklady Akademii Nauk USSR, vol. 133, pp. 303–306, 1960. Abstract and Applied Analysis [3] S. Mathlouthi, “Periodic orbits of the restricted three-body problem,” Transactions of the American Mathematical Society, vol. 350, no. 6, pp. 2265–2276, 1998. [4] U. Bessi and V. Coti Zelati, “Symmetries and noncollision closed orbits for planar π-body-type problems,” Nonlinear Analysis. Theory, Methods & Applications, vol. 16, no. 6, pp. 587–598, 1991. [5] R. S. Palais, “The principle of symmetric criticality,” Communications in Mathematical Physics, vol. 69, no. 1, pp. 19–30, 1979. [6] M. Struwe, Variational Methods, vol. 34 of Results in Mathematics and Related Areas, Springer, Berlin, Germany, 3rd edition, 2000. [7] I. Gelfand and S. 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