The Characterization of the Variational Minimizers for Spatial Restricted N + 1-Body Problems

Theorem 1. The minimizer of f(q) on the closure $$ of Λ_i  (i = 1,  2) is a nonconstant periodic solution for 2 ≤ N ≤ 472; hence the zero mass must oscillate, so that it can not be always in the same plane with the other bodies.

Lemma 2 (Palais′ Symmetry Principle [5]). By Palais′ Symmetry Principle, we know that the critical point of f(q) in $$ is a noncollission periodic solution of Newtonian equation (4).

Let σ be an orthogonal representation of a finite or compact group G in the real Hilbert space H such that for all σ ∈ G, f(σ · x) = f(x), where f : H → R.

Let S = {x ∈ H∣σ · x = x,  for all σ ∈ G}. Then the critical point of f in S is also a critical point of f in H.

Lemma 3 (see [6].)Let X be a reflexive Banach space, S be a weakly closed subset of X, f : S → R ∪ {+∞}, and f≢+∞ is weakly lower semicontinuous and coercive (f(x)→+∞ as ∥x∥ → +∞); then f attains its infimum on S.

Lemma 4 (Poincare-Wirtinger Inequality). Let q    ∈ W^1,2(R/Z, R^N) and $$; then

Lemma 5. f(q) in (6) attains its infimum on $$ or $$.

Proof. By Lemmas 3 and 4, it is easy to prove Lemma 5.

Lemma 6 (Jacobi′s Necessary Condition [7]). If the critical point $$ corresponds to a minimum of the functional $$ and if $$ along this critical point, then the open interval (a, b) contains no points conjugate to a; that is, for all c ∈ (a, b), the following boundary value problem

has only the trivial solution h(t) ≡ 0, for all t ∈ (a, c), where

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 $$; hence we need to establish a similar conclusion as Lemma 6 for the periodic boundary problem.

Lemma 8. Let F ∈ C³(R × R × R, R). Assume that $$ is a critical point of the functional $$ on {u ∈ W^1,2(R/TZ, R), u^′(0) = 0} and $$. If the open interval (0, T) contains a point c conjugate to 0, then $$ is not a minimum of the functional $$.

Proof. Suppose $$ is a minimum of the functional $$. The second variation of $$ is

where Set For all h ∈ {u ∈ W^1,2(R/TZ, R), u^′(0) = 0}, it is easy to see that $$. Then by $$, θ is a minimum of $$. The Euler-Lagrange equation which is called the Jacobi equation of (15) is Since the interval (0, T) contains a point c conjugate to 0, there exists a nonzero Jacobi field h₀ ∈ C²([0, T], R) satisfying Letting we have $$ and Notice that we can extend $$ periodically when we take T as the period, so $$. For all $$, it is easy to check that $$. Then by (19), one has $$ is a minimum of $$. Hence we get Combining with $$ and $$, by the uniqueness of initial value problems for second-order differential equation, we have $$ on [0, c], which contradicts the definition of $$. Therefore, Lemma 8 holds.

Lemma 9. The radius r for the moving orbit of N equal masses is

Proof. By (1)–(3), we have

Substituting (1) into (22), we have Then Therefore

Proof of Theorem 1. Clearly, q(t) = (0,0, 0) is a critical point of f(q) on $$. For the functional (6), let

Then the second variation of (6) in the neighborhood of z = 0 is given by where The Euler equation of (27) is called the Jacobi equation of the original functional (6), which is that is, Next, we study the solution of (30) with initial values h(0) = 0,  h^′(0) = 1. It is easy to get which is not identically zero on [0,1], but we will prove h(c) = 0 for some c ∈ (0,1).

Suppose there exists c ∈ (0,1) such that h(c) = 0.

Hence, for some k ∈ Z⁺

We have by using (24).

Case 1 (Minimizing f(q) on <!--${ifMathjaxEnabled: 10.1155%2F2013%2F845795}-->Λ-1=Λ1<!--${/ifMathjaxEnabled:}--><!--${ifMathjaxDisabled: 10.1155%2F2013%2F845795}--><!--${/ifMathjaxDisabled:}-->). Letting 0 < c < 1/2,

It is easy to check that $$, $$, $$, and $$ is a nonzero solution of (30).

If we take k = 1,

It is equivalent to

Let

It is not hard to check that f(x) is nonmonotone. But for 2 ≤ N ≤ 472, (36) holds by writing program to calculate it.

Therefore, for 2 ≤ N ≤ 472, we have c ∈ (0,1) such that

Notice that we can extend $$ periodically when we take 1 as the period, so $$. Then by Lemma 8, q(t) = (0,0, 0) is not a local minimum for f(q) on Λ₁. Hence the minimizers of f(q) on Λ₁ 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.

Case 2 (Minimizing f(q) on <!--${ifMathjaxEnabled: 10.1155%2F2013%2F845795}-->Λ-2=Λ2<!--${/ifMathjaxEnabled:}--><!--${ifMathjaxDisabled: 10.1155%2F2013%2F845795}--><!--${/ifMathjaxDisabled:}-->). Let

It is easy to check that $$, $$,$$, and $$ is a nonzero solution of (30).

We hope c ∈ (0,1/2); that is,

It implies that

Calculated by program, for 2 ≤ N ≤ N₀ = 472, we have c ∈ (0,1/2) such that h(c) = 0.

Notice that we can extend $$ periodically when we take 1 as the period, so $$. Then by Lemma 8, q(t) = (0,0, 0) is not a local minimum for f(q) on Λ₂. Hence the minimizers of f(q) on Λ₂ 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 N₀. We construct a test function z(t) such that f(z) < f(0) = N/r for N ≤ N₀, where N₀ is a very large number. Let

and we extend z(t) by z(t + (1/2)) = −z(t). We have

Writing program to calculate, we find an N₀ = 9*10¹⁰ such that f(z) < f(0).

Hence q(t) = (0,0, 0) is not a local minimum for f(q) on $$. So the minimizers of f(q) on Λ_i 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.