# The Characterization of the Variational Minimizers for Spatial RestrictedN+1-Body Problems

Authors: Fengying Li; Shiqing Zhang; Xiaoxiao Zhao
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We use Jacobi's necessary condition for the variational minimizer to study the periodic solution for spatial restrictedN+1-body problems with a zero mass on the vertical axis of the plane for N 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≤N≤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.


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## Body

## 1. Introduction and Main Result

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 restrictedN+1-body problems with a zero mass on the vertical axis of the plane for N equal masses. Suppose point masses m1=⋯=mN=1 move on a circular orbit around the center of masses. The motion for the zero mass is governed by the gravitational forces of m1,…,mN. Let ρj=e-1(2πj/N) and
(1)q1(t)=re-12πtρ1,…,qj(t)=ρjq1(t),…,qN(t)=re-12πt
satisfy the Newtonian equations:
(2)miq¨i=∂U∂qi,i=1,…,N,
where
(3)U=∑1≤i<j≤N‍mimj|qi-qj|.
The orbit q(t)=(0,0,z(t))∈R3 for zero mass satisfies the following equation:
(4)q¨=∑i=1N‍mi(qi-q)|qi-q|3.Obviously,q~(t)=(0,0,0) satisfies (4); it seems that q~(t)≡0 is a variational minimizer, but we will prove it is not this is the goal of this paper.Define(5)f(q)=∫01‍[12|q˙|2+∑i=1N‍1|q-qi|]dt,q∈Λi,
then
(6)f(q)=∫01‍[12|z′|2+Nr2+z2]dt≜f(z),q∈Λi,
where
(7)Λ1={q(t)=(0,0,z(t))∣z(t)∈W1,2(RZ,R)q(-t)=-q(t)},Λ2={q(t)=(0,0,z(t))∣z(t)∈W1,2(RZ,R)z(t+12)=-z(t)},W1,2(RZ,R)={x(t)|x(t),x˙(t)∈L2((0,1),R)x(t+1)=x(t)}.
Notice that the symmetry in Λ2 is related to the Italian symmetry [4].In this paper, our main result is the following.Theorem 1.
The minimizer off(q) on the closure Λ¯i of Λi(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.


## 2. Proof of Theorem1

We define the inner product and equivalent norm ofW1,2(R/Z,R):
(8)〈u,v〉=∫01‍(uv+u′·v′)dt,∥u∥=[∫01‍|u|2dt]1/2+[∫01‍|u′|2dt]1/2,
which is equivalent to
(9)[∫01‍|u′|2dt]1/2+|u(0)|.Lemma 2 (Palais' Symmetry Principle [5]).
By Palais' Symmetry Principle, we know that the critical point off(q) in Λ¯i is a noncollission periodic solution of Newtonian equation (4).
Letσ 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.
LetS={x∈H∣σ·x=x,forallσ∈G}. Then the critical point of f in S is also a critical point of f in H.In order to prove Theorem1, we need the following lemmas:Lemma 3 (see [6]).
LetX 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).
Letq∈W1,2(R/Z,RN) and ∫0Tq(t)dt=0; then
(10)∫0T‍|q˙(t)|2dt≥(2πT)2∫0T‍|q(t)|2dt.Lemma 5.
f
(
q
) in (6) attains its infimum on Λ-1=Λ1 or Λ-2=Λ2.Proof.
By Lemmas3 and 4, it is easy to prove Lemma 5.Lemma 6 (Jacobi's Necessary Condition [7]).
If the critical pointu=u~(t) corresponds to a minimum of the functional ∫ab‍F(t,u(t),u′(t))dt and if Fu′u′>0 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
(11)-ddt(Ph′)+Qh=0,h(a)=0,h(c)=0,
has only the trivial solution h(t)≡0, for all t∈(a,c), where
(12)P=12Fu′u′|u=u~Q=12(Fuu-ddtFuu′)|u=u~.Remark 7.
It is easy to see that Lemma6 is suitable for the fixed end problem. In this paper, we consider the periodic solutions of (2) on Λ¯i=Λi(i=1,2); hence we need to establish a similar conclusion as Lemma 6 for the periodic boundary problem.Lemma 8.
LetF∈C3(R×R×R,R). Assume that u=u~(t) is a critical point of the functional ∫0T‍F(t,u(t),u′(t))dt on {u∈W1,2(R/TZ,R),u′(0)=0} and Fu′u′|u=u~>0. If the open interval (0,T) contains a point c conjugate to 0, then u=u~(t) is not a minimum of the functional ∫0TF(t,u(t),u′(t))dt.Proof.
Supposeu=u~(t) is a minimum of the functional ∫0TF(t,u(t),u′(t))dt. The second variation of ∫0TF(t,u(t),u′(t))dt is
(13)∫0T‍(Ph′2+Qh2)dt,
where
(14)P=12Fu′u′|u=u~,Q=12(Fuu-ddtFuu′)|u=u~.
Set
(15)Qu~(h)=∫0T‍(Ph′2+Qh2)dt.
For all h∈{u∈W1,2(R/TZ,R),u′(0)=0}, it is easy to see that Qu~(h)≥0. Then by Qu~(θ)=0, θ is a minimum of Qu~(h). The Euler-Lagrange equation which is called the Jacobi equation of (15) is
(16)-ddt(Ph′)+Qh=0.
Since the interval (0,T) contains a point c conjugate to 0, there exists a nonzero Jacobi field h0∈C2([0,T],R) satisfying
(17)-ddt(Ph0′)+Qh0=0,h0(0)=0,h0(c)=0,h0′(0)=0.
Letting
(18)h^(t)={h0(t)t∈[0,c],0t∈(c,T],
we have h^∈C2([0,T]∖{c},R),h^(0)=h^(c)=h^(T)=0 and
(19)Qu~(h^)=∫0T‍(Ph^′2+Qh^2)dt=∫0c‍(Ph0′2+Qh02)dt=0.
Notice that we can extend h^ periodically when we take T as the period, so h^∈W01,2(R/TZ,R). For all h∈C01([0,T],R), it is easy to check that Qu~(h)≥0. Then by (19), one has h^∈C2([0,T]∖{c},R)∩W01,2(R/TZ,R) is a minimum of Qu~(h). Hence we get
(20)-ddt(Ph^′)+Qh^=0.
Combining with h^(0)=h^(c)=0 and h^′(0)=0, by the uniqueness of initial value problems for second-order differential equation, we have h^(t)≡0 on [0,c], which contradicts the definition of h^. Therefore, Lemma 8 holds.Lemma 9.
The radiusr for the moving orbit of N equal masses is
(21)r=(14π)2/3[∑1≤j≤N-1‍csc(πNj)]1/3.Proof.
By (1)–(3), we have
(22)q¨N=∑j≠Nqj-qN|qj-qN|3.
Substituting (1) into (22), we have
(23)-4π2=∑j≠Nρj-ρNr3|ρj-ρN|3,4π2r3=∑j≠N‍1-ρj|1-ρj|3=14∑1≤j≤N-1‍csc(πNj).
Then
(24)r3=116π2∑1≤j≤N-1‍csc(πNj).
Therefore
(25)r=(14π)2/3[∑1≤j≤N-1‍csc(πNj)]1/3.Proof of Theorem1.
Clearly,q(t)=(0,0,0) is a critical point of f(q) on Λ-i=Λi(i=1,2). For the functional (6), let
(26)F(z,z′)=12|z′|2+Nr2+z2.
Then the second variation of (6) in the neighborhood of z=0 is given by
(27)∫01‍(Ph′2+Qh2)dt,
where
(28)P=12Fz′z′|z=0=12,Q=12(Fzz-ddtFzz′)|z=0=-N2r3.
The Euler equation of (27) is called the Jacobi equation of the original functional (6), which is
(29)-ddt(Ph′)+Qh=0,
that is,
(30)h′′+Nr3h=0.
Next, we study the solution of (30) with initial values h(0)=0,h′(0)=1. It is easy to get
(31)h(t)=r3N·sinNr3t,
which is not identically zero on [0,1], but we will prove h(c)=0 for some c∈(0,1).
Suppose there existsc∈(0,1) such that h(c)=0.
Hence, for somek∈Z+(32)Nr3c=kπ.
We have
(33)c=[π2k2r3N]1/2=[k2(∑j=1N-1csc((π/N)j))16N]1/2
by using (24).Case 1 (Minimizingf(q) on Λ-1=Λ1).
Letting0<c<1/2,
(34)h-(t)={h(t)t∈[0,c],0t∈(c,1-c],-h(1-t)t∈(1-c,1].
It is easy to check that h-(t)∈C2([0,1]∖{c,1-c},R)∩W1,2(R,R), h-(-t)=-h-(t), h-(0)=h(0)=0,h-(c)=h(c)=0, and h- is a nonzero solution of (30).
If we takek=1,
(35)c=[∑j=1N-1csc((π/N)j)16N]1/2<12.
It is equivalent to
(36)∑j=1N-1‍csc(πNj)<4N.
Let(37)f(x)=∑j≠x‍csc(πxj);
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, for2≤N≤472, we have c∈(0,1) such that
(38)sinNr3c=sinπ=0.
Notice that we can extend h- periodically when we take 1 as the period, so h-∈Λ1. 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 (Minimizingf(q) on Λ-2=Λ2).
Let(39)h~(t)={h(t)t∈[0,c],0t∈(c,12],-h(t-12)t∈(12,12+c],0t∈(12+c,1].
It is easy to check that h~(t)∈C2([0,1]∖{c,1/2,(1/2)+c},R)∩W1,2(R,R), h~(t+(1/2))=-h~(t),h~(0)=h(0)=0,h~(c)=h(c)=0, and h~ is a nonzero solution of (30).
We hopec∈(0,1/2); that is,
(40)c=[∑j=1N-1csc((π/N)j)16N]1/2<12.
It implies that
(41)∑j=1N-1‍csc(πNj)<4N.
Calculated by program, for2≤N≤N0=472, we have c∈(0,1/2) such that h(c)=0.
Notice that we can extendh~ periodically when we take 1 as the period, so h~∈Λ2. 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 largerN0. 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
(42)z(t)={tt∈[0,14],14t∈[14,12],
and we extend z(t) by z(t+(1/2))=-z(t). We have
(43)f(z)=14+2∫01/2‍Nr2+z2dt=14+2∫01/4‍Nr2+t2dt+2∫1/41/2‍Nr2+(1/16)dt=14+2Nln1+1+16r24r+2N1+16r2.
Writing program to calculate, we find anN0=9*1010 such that f(z)<f(0).
Henceq(t)=(0,0,0) is not a local minimum for f(q) on Λ-i=Λi(i=1,2). So the minimizers of f(q) on Λi 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.



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