An existence proof of a symmetric periodic orbit in the octahedral six-body problem

We present a proof of the existence of a periodic orbit for the Newtonian six-body problem with equal masses. This orbit has three double collisions each period and no multiple collisions. Our proof is based on the minimization of the Lagrangian action functional on a well chosen class of symmetric loops.


Introduction
The Variational Methods applied to the n-body Newtonian problem allows to prove the existence of periodic orbits, in most cases with some symmetry. It was exploited by the Italian school in the 90's [4], [5], [10]. They gave new periodic solutions for a mechanical systems with potentials satisfying a hypothesis called strong force, which excludes the Newtonian potential. The strong force hypothesis was introduced by Poincaré, see [9]. This method was only exploited for the Newtonian potential much later. We can cite three works in this area [3], [13] and [11]. In the first one it is proved the existence of a new periodic solution in the planar three-body problem, where all bodies are on the same curve, and this curve has a figure-eight shape; in the second one (our most prominent reference for the present paper) it is proved the existence of the collinear solution for the three-body problem already discovery numerically in 1956 by J. Schubart [12]); in the third one, the author studies the existence of periodic solutions for subsytems where the dimension of the configuration space can be reduced to d = 2. In this paper we proved a variational existence proof of a periodic orbit in the octahedral six-body problem with equal masses. Next we explain the main ideas on the prove.
Let us consider six bodies with equal masses in R 3 . We assume that every coordinate axis contains a couple of bodies, and they are symmetric with respect to the origin, that is the center of mass of the system. This is the octahedral six-body problem. Our orbit starts with a collision in the x−axis and the other 4 bodies form a square on the ortogonal plane. Let us denote by T the period of the solution. During the first sixth of the period, the bodies on the x−axis move away, the two bodies on the z−axis also move away, while the bodies on y−axis approach. At time t = T /6 the bodies on the z−axis are point and they approach each other as t ∈ [T /6, T /3]. At time t = T /3 the bodies on y−axis are at a double collision, and the other bodies form a square on the orthogonal plane. In the second third of the period, the motion is the same as above, after having exchanged x by y, y by z and z by x. That is to say the solution satisfies the symmetry condition x(t − T /3) = y(t), y(t − T /3) = z(t) and z(t − T /3) = x(t). See figure 1 In section 2 we will introduce the formal aspects of the problem, we will write the equations of motion and explain the variational setting. In particular, we will prove that the lagrangian action functional is coercive. In section 3 we show that a minimizer has no other collisions besides those imposed by the choice of set of loops where we minimize the action funcional, and that all collisions are double (i.e. there are no quadruple collision). In section 4 we regularize all possible collisions. Our conclusions are summarized in the Main Theorem, which is proved in section 4.

Equations of Motion and Variational Setting
Notice that if at a certain instant the configuration of the bodies is octahedral (and the velocities too), the configuration is octahedral for all time. To describe the octahedral six-body problem with equal masses we will identify the configuration space with R 3 . Let the masses m 1 , m 2 be on x-axis, at positions q 1 = (x, 0, 0) and q 2 = (−x, 0, 0), respectively . Analogously, m 3 , m 4 are on q 3 = (0, y, 0), q 4 = (0, −y, 0) and m 5 , m 6 on q 5 = (0, 0, z), q 6 = (0, 0, −z).
We can identify the configuration space X with R 3 , and every configuration with a vector X = (x, y, z) ∈ R 3 = X.
Without loss of generality we can normalize the masses, m i = 1 2 , i = 1, ..., 6. So the kinetic energy and potential function assume the form: U (X) = 1 A configuration is said to be a collision if at least two bodies occupy the same position in space. Due to the symmetry of the configuration, when this happens, one of the three Cartesian coordinate is equal to zero. We denote by ∆ the set of collision configurations, and byX = X \ ∆ the open set of collision-free configurations. A configuration q = (x, y, z) is said to be a double collision if only one of the three Cartesian coordinates is equal to zero. It is said that to be a quadruple collision if only two of the three Cartesian coordinates are equal to zero, and it is said to be a total collision if x = y = z = 0. One can see that the potential becomes singular at collisions. Therefore the Hamiltonian (H = K − U ) and the Lagrangian (L = K + U ) beccome sigular at the collisions. We will prove in section 3 that there is no quadruple and total collision. In section 4 we will show that for an action minimizer the double collisions are regularize. The equations of motion are Euler-Lagrange equations for the lagrangean L = K + U . They can be written: Throughout the paper, X · Y will denote the standard scalar product in R 3 .
Recall the Sobolev space H 1 (R/T Z, X) is the vector space of absolutely continuous T -periodic loops X : H 1 (R/T Z, X) is endowed with the following Hilbert product: and · H 1 denote the associated norm. The Lagrangian action functional is defined by where L(X(t),Ẋ(t)) = +∞ if X is a collision configuration. The Lagrangian action functional is differentiable on the open set of collision-free orbits. A classical computation shows that collision-free critical points of A are T -periodic solutions of the problem (see [14]). The action functional is also semicontinuous with respect to the weak topology of the space H 1 (R/T Z, X) (see [14]). Let G be a group with two generators which satirfy the relations: G is in fact the diedral group D 3 . We consider the action of G on H 1 (R/T Z, X) defined by h(x, y, z)(t) = (x, z, y)(−t).
So, it defines an action of G on H 1 (R/T Z, X). Since we assume that all masses are equal, it is easy to see that G acts by isometries on H 1 (R/T Z, X) and leaves invariant the action functional. Notice that, since we identify X with R 3 , the invariance of a loop X ∈ H 1 (R/T Z, X) by g can be interpreted by saving that X(t+T /3) is obtained by a rotaion X(t) of an angle 2π/3 around the axis generated by the vector (1, 1, 1); the invariance by h is equivalent to says that X(−t) is symmetric to X(t) with respect to the plan y = z. Let Λ G = {X ∈ H 1 (R/T Z, X) : gX = X and hX = X}, the subspace of invariant loops. Acoording to Palais principle (see [8]), a critical point of A| Λ G is still a critical point of A, thus it is a T -periodic solution of the problem. Remark that since invariant loops X ∈ Λ G , have X(T /3 − T ) and X(t) symmetric with respect to the plane x = y, they satisfy statement (iii) of the Main Theorem. In order to use the Direct Method in the Calculus of Variations, see [14], we need to restrict the action functional to a closed subset Ω of Λ G , such that A| Ω is coercive, i.e., the set of loops verifying A| Ω (X) < C is bounded for every C > 0. Notice that A| Λ G is not coercive. Indeed, we can construct a sequence of constant loops, for instance X (n) (t) = (n, n, n), such that X (n) ∈ Λ G and A(X (n) ) → 0 as n → +∞. Therefore, since A| Λ G (X) is everywhere strictly positive, its infimum is not achieved. Let we will define Ω as Ω = {X = (x(t), y(t), z(t)) ∈ Λ G : x(0) = 0 and X(t) ∈ S ∀ t ∈ R}.
Proposition 2.1. A| Ω is coercive and has a minimizer.
Proof. Consider X ∈ Ω, by symmetry we will study only the first third of the orbits. Let X(t) the maximum of X(t) with t ∈ [0, T /3]. In order to prove the coercivity we will check that L(X) ≥ a X(t) , where L is the length of the curve X| [0,T /3] and a is a positive real number independent of X. If X(t) = 0 the inequality, L(X) ≥ a X(t) , is trivially satisfied, so we can suppose X(t) = 0. Let us write l = y 2 (0) + z 2 (0) = x 2 (T /3) + z 2 (T /3) and d = X(t) . Let α be the angle between X(t) and the bissectrix of the first quadrant of the plane x = 0; let β be the angle between X(t) and the bissectrix of the first quadrant of the plane y = 0; and let θ be the angle between X(0) and X(T /3), so θ = π/3. Since X ∈ Ω, by definition of Ω and Λ G , we know that X(0) is in the bissectrix of the first quadrant of the plane x = 0, and X(T /3) is in the bissectrix of the first quadrant of the plane y = 0. The distance between X(0) and X(t) is √ l 2 + d 2 − 2ld cos α, and the distance between X(t) and X(T /3) is l 2 + d 2 − 2ld cos β. Hence: Since the two terms on the right-hand side of the inequality above are decreasing with respect to α and β, and α + β ≥ θ = π/3, we can write Notice that √ l 2 + d 2 − 2ld cos α is greater than or equal to the distance from X(t) to the bissectrix of first quadrant of the plane x = 0, which is d sin α. In a similar way we have that Let us prove that this set is bounded, this will imply that A| Ω is coercive. By the Cauchy-Schwartz inequality, (2.2) and the invariance of A by the action of G, we have: In order to prove that A| Ω has a minimizer, let (c n ) n∈N be a decreasing sequence of real positive numbers converging to infA| Ω . By the last inequality and the weak lower semicontinuity of A Ω every set is bounded in the norm of H 1 and closed in the weak topology. Since H 1 (R/T Z, X) is a separable Hilbert space, Ω n is compact in the weak topology [1]. Since the sequence (c n ) n∈N is decreasing we have Ω n+1 ⊂ Ω n , therefore Notice that any element of ∩ n∈N Ω n is a minimizer.

Elimination of Extra Collisions
In this section we will study collisions. We will prove that for a minimizer of A| Ω , each segment of orbit without collisions are solutions of the octahedral problem, hence it is indeed a solution of the Newtonian six-body problem. We shall prove that there are no other collisions, like quadruple or total collision, and no extra double collision. In the next section we will show that the double collisions (in Proof. Denote by T c (X) ⊂ R the set of collision times of X. Since X is a continuous path with finite action, T c (X) is closed and has zero measure. R\T c (X) is the union of open intervals; let (a, b) be one of those intervals. Since in our orbit t = 0, T /3 are collision times, b − a ≤ T /3. By invariance of X under the action of G, we can assume that (a, b) ⊂ (0, T /3). Let , and consider the unique T -periodic constructed from ξ, still denoted ξ. In order to symmetrize ξ we will define ξ G : where |G| = 6 is the cardinality of the group G. It is clear that ξ G and supp(ξ G ) are containd in the complement of T C (X), hence s → A(X + sξ G ) is differentiable for s small enough. Moreover: Since X is a minimizer of A, in every direction where A is differentiable at X, the derivate is equal to zero, hence [15]) shows that (3.1) implies: so, X| (a,b) is a solution of the six-body problem. This finishes the proof.
Proof. To simplify the proof, we will ommit dependence on t. It is easy to check On the other hand, suppose that X = (x, y, z) is a solution of (3.2) and, by contradiction, that the condition x = y = z is not satisfied. Without loss of generality, we can assume that x > y. We have and we obtain a contradiction with the two first equations of (3.2).
Proof. The invariance of A by the action of G implies that we just need to prove that if X is a minimizer, X| [0,T /6] is free of total collision. Following [13], we introduce the notation ; is a minimizer of A 1 6 . By contradiction, suppose that there is a total collision at the instant t ∈ [0, T /6]. First of all we will prove that X| [0,T /6] is one half of the homothetic ejection-collision orbit with central configuration the regular octahedron. Afterwards we explicit a deformation of X, X ε , which decreases the action, hence producing a contradiction.
We denote by X c + the normalized central configuration in our problem: X c + 2 = 1. Let us term r(t) = X(t) andŨ (X) = X U (X). Since X c + is the unique central configuration for the octahedral six-body problem, G =Ũ (X c + ) is a minimum ofŨ on S. Then Let us consider now the function The minimum of a is achived by the path t → v(t) obtained by joining one-half of a collision-ejection solution of period 2t, with one-half of an ejection-collision solution of period 2(T /6 − t), for the one-dimensional Kepler problem: We remark that the path t → v(t)X c + is not necessarly an element of Ω 1 6 , but we still have , By the Virial Theorem, the action of a homothetic collision-ejection motion of period τ , denoted here a(τ ), is equal to the action of a circular motion of same period, τ , thus Using an argument introduced by Gordon [6], essentially the concavity of τ → α 0 τ 1 3 , we have: is one-half of the homothetic ejection-collision solution of period 2T /6. Notice that X is now an element of Ω 1 6 . Now we will construct a deformation of X = (x, y , z) decreasing the action. Let Let us define X ε = X + εf ε (t)Z, where Z = (z 1 , z 2 , z 3 ) is a fixed configuration satisfying z 2 = z 3 > 0 and z 1 = 0. We can say, for instance, Z = (0, 1, 1). We remark that X ε is still in Ω 1 6 , we can write the diference between the action of X ε and the action of X as the sum of three terms: Using the classical Sundman's estimation we have We can write X c + = 1 √ 3 (1, 1, 1). Notice that each A i , i = 1, 2 is non-positive. Let us prove that A 1 (ε) < 0. Let A 1 (ε) be defined as introducing the variable τ define by t 2 3 = ετ , we get We know that 3 Now, we will prove that A 3 (ε) = O(ε), and with (3.5) we have so ∆A 1 6 < 0 for ε > 0 sufficiently small. And we have a contradiction, which proves our proposition.
Let us study A 3 (ε), where the last inequality holds because X c+ · Z is postive. This ends our proof.
In order to prove that there are no quadruple collisions, we need to prove that there is not an extra double collisions. A double collision is called extra when it is a double collision with masses on the same axis at an instant different from t = kT /3, where k ∈ Z.
Without loss of generality, we just have to prove that there is no extra double collisions for the cluster {1, 2} (masses m 1 , m 2 ) in the interval (0, T /6), and the result could be extended for all real time. Extra double collisions for the cluster {1, 2} means that x = 0, y > 0 and z > 0. The other cases (y = 0, x > 0, z > 0 and z = 0, x > 0, y > 0) are analogous. Let us define the lagrangean and energy of that cluster in the following way: and h x (X,Ẋ) = 1 2ẋ 2 − 1 8x , following [13], we introduce this notation for the cluster. Then, the kinetic and potential energies are define as K x (X,Ẋ) = 1 2ẋ 2 and U x (X,Ẋ) = 1 8x .
Definition 3.1. Let f (t, ε) be a function of two real values and g(ε) a function of ε. We say that  is monotonic and bijective if ε is sufficiently small. Let us denote by t → τ ε (t) the inverse of the map τ → t ε (t), hence The path X ε (τ ) = (x(t ε (τ )), y(τ ), z(τ )) is a variation of X| [0,T /6] , and it is an element of Ω 1 6 . We define: and U 0 = 1 When we make the change of variables τ = τ ε (t) in A x 1 6 (X ε ), we have Since X| [0,T /6] is a minimizer of A 1 6 Ω and ε → A 1 6 (X ε ) is a differentiable function, using (3.8) we have Like in proposition 3.1, we will use the Fundamental Lemma of the Calculus of Variations (see [15]), to conclude that t → h x (t) is absolutely continuous in [a, b] andḣ for almost every t ∈ [a, b]. Proof. Let X a minimizer of A| Ω , t ∈ (0, T /6] an extra double collision and T c (X) = {t * : x(t * )y(t * )z(t * ) = 0} the set of collision times. By contradiction, suppose that t is not isolated in T c (X). Without loss of generality, we can suppose that x(t) = 0, y(t) > 0 and z(t) > 0. By continuity of X, only collisions of m 1 , m 2 can accumulate in t. We can choose an interval [a, b] such that t ∈ (a, b) and for all t ∈ [a, b] there is no double collision between m 1 or m 2 and any other masses. T c is closed with zero measure, so there exists a sequence of compact intervals ([a n , b n ]) n∈N such that a n , b n → t and x(a n ) = x(b n ) = 0 and x(t) > 0 ∀t ∈ (a n , b n ).
The moment of inertia of the cluster could be written as I x (t) = I x (X)(t) = x(t) 2 . Therefore, I x (a n ) = I x (b n ) = 0 and I x (t) = 0 ∀t ∈ (a n , b n ). By proposition 3.1, X is a solution of octahedral symmetric six-body problem in (a n , b n ) and I x (X)(t) is C 2 in (a n , b n ). If s n ∈ (a n , b n ) denotes the maximum of I x in [a n , b n ], I x (s n ) ≤ 0. The equation of motion tells us and we have the generalized Lagrange-Jacobi relation Using proposition 3.3 and this corollary we will prove the following theorem. Proof. We will make a small deformation of the hypothetical minimizer with a collision decreasing the action, similar to that we did in proposition 3.2. Suppose that there is a minimizer of A| Ω with an extra double collision in the instant t ∈ (0, T /6) between the masses on the x-axis. By Sundman's estimates we have Set X ε (t) = (x ε (t), y ε (t), z ε (t)) where: Notice that X ε ∈ Ω 1 6 . We will prove that A 1 6 (X ε ) < A 1 6 (X| [0,T /6] ), which is a contradiction. This difference beteween the action of X ε and that that of X| [0,T /6] can be written as As in proposition 3.2, we decompose ∆A + (ε) as the sum of three terms: To estimate A + 1 , observe that A + 1 (ε) ≤ A + 1 (ε), where we set . Changing the variables t = t + (ετ ) 3 2 , and making the same computation as in proposition 3.2, we get the existence of a constant k + > 0, independent of ε, such that A so A + 1 (ε) < 0 for ε sufficiently small. Since x(t) and f ε (t) are positive, we have (U 0 (X ε (t)) − U 0 (X(t))) dt. We conclude that ∆A + and we have a contradiction. Notice that if t = T /6 we can define f ε (t) as (ε). This finishes our prove.
In order to conclude our study of the collisions in the octahedral symmetric sixbody problem we have to prove that there is no quadruple collision, and that the double collisions are regularized. To prove the there is no quadruple collision we follow the method used to prove that there is no extra double collisions.
We argue by contradiction. First of all we remark that at t = 0 we have x(0) = 0, y(0) = z(0), therefore we can have only a double collision or a total collision. We have already prove that a minimizer is free of total collisions. Hence at t = 0 we have necessarily a double collision. The same argument holds for t = T /3 and t = 2T /3. Assume now, by contradiction, that we have a quadruple collision at time t. By symmetry we can assume that t ∈ (0, T /6]. Without loss of generality, we will say that this quadruple collision is of the cluster {1, 2, 3, 4}, that is to say: x(t) = y(t) and z(t) > 0. The lagrangian and energy of the cluster are L x,y = 1 2 (ẋ 2 +ẏ 2 ) + 1 8 With this notations, we will prove the following proposition.  Proof. Since X is a minimizer of A| Ω , X| [0,T /6] is a minimizer of A| Ω 1 6 = A 1 6 . So the function h x,y (t) = h x,y (X(t),Ẋ(t)) is well defined on a subset of [0, T /6] with full measure as in (3.9), and such that only quadruple collisions of the masses on the x, y-axis occur at t ∈ [a, b]. Let δ be a function of class C 1 with support contained in [a, b]. Let As in proposition 3.3, t ε is monotonic and bijective if ε > 0 is small, and the equations (3.7), (3.8) hold. Let X : [0, T /6] → S, X ε (τ ) = (x(t ε ), y(t ε ), z(τ )) a variation of X| [0, T 6 ] . Define where we set After making the change of variables τ = τ ε (t) in A x,y 1 6 (X ε ), we have Since X| [0,T /6] is a minimizer of A 1 6 Ω and ε → A 1 6 (X ε ) is a differentiable function, using (3.8) we obtain As in proposition 3.3, we use the Fundamental Lemma of the Calculus of Variations, in order to conclude that t → h x,y (t) is absolutely continuous and alsȯ where we set ∇ x,y U 1 (X(t)) = ∂U 1 ∂x , ∂U 1 ∂y .
Corollary 3.4. Let X = (x, y, z) ∈ Ω be a minimizer and t ∈ [0, T /6] an instant of quadruple collision, say x(t) = y(t) = 0. Then t is isolated in the set of collision times, T c (X).
Remark. As we already proved in theorem 3.3, double collisions can not accumulate in a quadruple colllision and there is no extra double collision! We have proved also that there is no total colission. So, we can choose a neighborhood of t such that there is no double collision such that x = 0 or y = 0.
Proof. Let X be a minimizer of A| Ω , t ∈ (0, T /6) a quadruple collision time at which x = y = 0 and T c (X) = {t * : x(t * )y(t * )z(t * ) = 0}. By contradiction, suppose that t is not isolated in T c (X). Notice, by the remark above, that there is no possible acumulations of extra double collisions or total collisions to a quadruple collision. So, only collisions of m 1 , m 2 , m 3 , m 4 can accumulate in t, we can choose an interval [a, b] such that t ∈ (a, b) and for all t ∈ [a, b] there is no collision of m 1 , m 2 , m 3 or m 4 with the others masses, m 5 , m 6 . We know that T c is closed and has zero measure. So there exists a sequence of compact intervals ([a n , b n ]) n∈N such that a n , b n → t and x(a n ) = x(b n ) = y(a n ) = y(b n ) = 0 and I x,y (t) = 0 ∀t ∈ (a n , b n ), where I x,y (t) is the moment of inertia of the cluster, which can be written as I x,y (t) = I x,y (X(t)) = x(t) 2 + y(t) 2 . Therefore, I x,y (a n ) = I x,y (b n ) = 0 and I x,y (t) = 0 ∀t ∈ (a n , b n ). By proposition 3.1, X is a solution of the octahedral symmetric six-body problem in (a n , b n ) and I x,y (X(t)) is C 2 in (a n , b n ). If s n ∈ (a n , b n ) denotes the maximum of I x,y in [a n , b n ], we haveÏ x,y (s n ) ≤ 0. The equations of motion tells us that (ẍ,ÿ) = ∇ x,y U x,y + ∇ x,y U 1 .
And with this we have the generalized Lagrange-Jacobi relation I x,y (t) = 4K x,y + 2(x, y) · (∇ x,y U x,y + ∇ x,y U 1 ) = 4h x,y (t) + 2U x,y + 2(x, y) · ∇ x,y U 1 .  Proof. We will make a small deformation of the hypothetical minimizer with a collision which decreases the action, similar to what we did in propositions 3.2 and 3.3. Notice that by the symmetry of the problem a quadruple collision in {t ∈ R : t ∈ T 3 Z} means a total collision. From proposition 3.2 it can not happen. Moreover, by theorem 3.3 there is no collision in (0, T /3). Suppose that there is a minimizer of A| Ω with an extra quadruple collision at the instant t ∈ (0, T /6) among the masses on the x, y-axis. By Sundman's estimates we have We have proved in theorem 3.3 that A where k + is a positive constant. Since x(t), y(t) and f ε (t) are positive, we have (U 1 (X ε (t)) − U 1 (X(t))) dt, is singular only at double collisions. Now we will regularize the equations of motions at the double collisions by a time scaling, using a kind of Levi-Civita regularization [7]. Introduce the new variables (γ, υ, ζ, Γ, Υ, Z) defined by Notice that (θ, Θ) = (γ, υ, ζ, Γ, Υ, Z) → (X,Ẋ) = (x, y, z,ẋ,ẏ,ż) is a symplectic 8-fold covering outside collisions. Equations of motions in the new variables are still in hamiltonian form. The hamiltonian can be written as and the equations of motion are .
This time scaling will permit us to regularize all the doubles collisions at once. The system (4.2) in the new time paramenter is  Proof. By the symmetry of the problem and the invariance of X under the action of G, we need to check only the collision at t = 0. Notice that our rescaling of time = γ 2 υ 2 ζ 2˙r egularizes simultaneously all double collisions. We know that for a minimizer of A| Ω the total energy h(t) = H(X(t),Ẋ(t)) is constant in each interval without collision, but in a small interval containing t = 0, a priori, the energy could be different for t > 0 and t < 0. We remark that where K 0 (X(t),Ẋ(t)) and U 0 (t) are the continuous function in a neighborhood of t defined in page 12. By proposition 3.3, h x is an absolutely continuous function in a neighborhood of a double collision on the x axis. So, t → h(t) is continuous in a neighborhood of t = 0. Therefore, is constant in (−T /3, T /3), consequently in [0, T ].
The change of variables = γ 2 υ 2 ζ 2˙c an be written as d ds = γ 2 υ 2 ζ 2 d dt , where˙is the derivate with respect to the physical time, t, and is the derivate with respect to the new (i.e. rescaled) time s. The new time can be written as a function of the physical time t as s(t) =  .
This proves that for t > 0, X(t) coincides with the solution obtained through Levi-Civita regularization. For the others double collisions the argument is the same.
Let X = (x, y, z) be a minimizer of A| Ω . By proposition 4.1 X has double collisions regularized, and by proposition 3.1, X is a solution of the 6-body problem. So we can say that X is a collision solution of the symmetric 6-body problem on the axis. To prove the Main Theorem, we just have to prove (ii), because (i), (iii) and (iv) are already satisfied. Notice that, by the symmetry of the problem, we need to prove that t → y(t) is decreasing and t → z(t) is increasing in [0, T /6] is the same thing as prove t → x(t) is decreasing in [2T /3, 5T /6] and increasing in [T /3, T /2], respectively. And, prove that t → y(t) is decreasing and t → z(t) is decreasing in [T /6, T /3] is the same thing as proving that t → x(t) is decreasing in [5T /6, T ] and decreasing in [T /2, 2T /3], respectively. So, we have just study the behavior of t → x(t) in a period. By the equations of motion (2.1), we have