Spatial Central Configurations with Two Twisted Regular 4-Gons

Abstract

We study the configuration formed by two squares in two parallel layers separated by a distance. We picture the two layers horizontally with the 𝑧-axis passing through the centers of the two squares. The masses located on the vertices of each square are equal, but we do not assume that the masses of the top square are equal to the masses of the bottom square. We prove that the above configuration of two squares forms a central configuration if and only if the twist angle is equal to 𝑘𝜋/2 or (𝜋/4+𝑘𝜋/2) (𝑘=1,2,3,4).

1. Introduction and Main Results

This paper uses the same notations as [1]. The Newtonian 𝑁-body problems [2, 3] concern the motions of 𝑁 particles with masses 𝑚_𝑗∈𝑅⁺ and positions 𝑞_𝑗∈𝑅³(𝑗=1,2,…,𝑁). The motion is governed by Newton’s second law and the Universal law: 𝑚_𝑗𝑞̈_𝑗=𝜕𝑈𝑞𝜕𝑞_𝑗, (1) where 𝑞=(𝑞₁,𝑞₂,…,𝑞_𝑁) with the Newtonian potential: 𝑈𝑞=_{1⩽𝑗<𝑘⩽𝑁}𝑚_𝑗𝑚_𝑘𝑞_𝑗−𝑞_𝑘. (2) Consider the space 𝑋=𝑞=𝑞₁,𝑞₂,…,𝑞_𝑁∈𝑅^3𝑁∶^𝑁_𝑗=1𝑚_𝑗𝑞_𝑗=0; (3) that is, suppose that the center of mass is fixed at the origin of the origin of the coordinate axis, because the potential is singular when two particles have the same position. It is natural to assume that the configuration avoids the collision set Δ={𝑞∶𝑞_𝑗=𝑞_𝑘 for some 𝑘≠𝑗}. The set 𝑋⧵Δ is called the configuration space.

Definition 1 (see [2, 3]). A configuration 𝑞=(𝑞₁,𝑞₂,…,𝑞_𝑁)∈𝑋⧵Δ is called a central configuration if there exists a constant 𝜆 such that ^𝑁_{𝑗=1,𝑗≠𝑘}𝑚_𝑗𝑚_𝑘𝑞_𝑗−𝑞_𝑘³𝑞_𝑗−𝑞_𝑘=−𝜆𝑚_𝑘𝑞_𝑘,1⩽𝑘⩽𝑁. (4)
The value of constant 𝜆 in (4) is uniquely determined by 𝑈𝜆=𝐼, (5) where 𝐼=^𝑁_𝑘=1𝑚_𝑘𝑞_𝑘². (6)
Consider the configuration in 𝑅³ consisting of two layer regular 𝑁-gons (𝑁⩾2) with distance ℎ⩾0. It is assumed that the lower layer regular 𝑁-gons lie in horizontal plane, and the upper regular 𝑁-gons parallel the lower one, and 𝑧-axis passes through both centers of two regular 𝑁-gons. Suppose that the lower layer particles have masses 𝑚₁,𝑚₂,…,𝑚_𝑁 and the upper layer particles have masses 𝑚₁,𝑚₂,…,𝑚_𝑁, respectively; then these assumptions can be interpreted more precisely by the following. Let 𝜌_𝑘 denote for all 𝑁 complex roots of unity, that is, 𝜌_𝑘=exp2𝜋𝑘𝑁𝑖. (7) Let 𝜌_𝑘=𝑎𝜌_𝑘𝑒^𝑖𝜑, (8) where 𝑎>0,  𝑖=√−1,  0⩽𝜑⩽2𝜋, and 𝜑 is called twist angle. It is assumed that 𝑚_𝑘(1⩽𝑘⩽𝑁) locates at the vertex 𝑞_𝑘 of the lower layer regular 𝑁-gons; 𝑚_𝑘(1⩽𝑘⩽𝑁) locates on the vertex 𝑞_𝑘 of the upper layer regular 𝑁-gons: 𝑞_𝑘=𝜌_𝑘,𝑞,0_𝑘=𝜌_𝑘,ℎ, (9) where ℎ≥0 is the distance between the two layers. Then the center of masses is 𝑧₀=^𝑁_𝑗=1𝑚_𝑗𝑞_𝑗+𝑚_𝑗𝑞_𝑗𝑀, (10) where 𝑀=^𝑁_𝑗=1𝑚_𝑗+𝑚_𝑗. (11) Let 𝑃_𝐾=𝑞_𝑘−𝑧₀,𝑃_𝐾=𝑞_𝑘−𝑧₀. (12) If 𝑚₁,𝑚₂,…,𝑚_𝑁 and 𝑚₁,𝑚₂,…,𝑚_𝑁 form a central configuration, then there is 𝜆∈𝑅⁺ such that ^𝑁_𝑗≠𝑘𝑚_𝑗𝑃_𝑗−𝑃_𝑘𝑃_𝑗−𝑃_𝑘³+^𝑁_𝑗=1𝑚_𝑗𝑃_𝑗−𝑃_𝑘𝑃_𝑗−𝑃_𝑘³=−𝜆𝑃_𝑘,1⩽𝑘⩽𝑁,_𝑁𝑗=1𝑚_𝑗𝑃_𝑗−𝑃_𝑘𝑃_𝑗−𝑃_𝑘³+^𝑁_𝑗≠𝑘𝑚_𝑗𝑃_𝑗−𝑃_𝑘𝑃_𝑗−𝑃_𝑘³=−𝜆𝑃_𝑘,1⩽𝑘⩽𝑁. (13) Under the case that twist angle 𝜑=0, Moeckel-Simo proved.

Theorem 2 (see [4]). If 𝑁<473, there is a unique pair of spatial central configurations of parallel regular 𝑁-gons. If 𝑁≥473, there is no such central configuration for 𝑏<𝑏₀(𝑁), where b is the mass ratio. At a unique pair bifurcates from the planar central configuration with the smaller masses on the inner polygon. This remains the unique pair of spatial central configurations until 𝑏=1/𝑏₀, where a similar bifurcation occurs in reverse, so that 𝑏>1/𝑏₀, and only the planar central configurations remain.

Xie et al. [5] studied the necessary conditions for the masses of two layer regular polygon central configurations in 𝑅³, and they proved the following theorems.

Theorem 3. Under the assumptions of (9), if 𝑚₁,𝑚₂,…,𝑚_𝑁 and 𝑚₁,𝑚₂,…,𝑚_𝑁 form a central configuration, then(i)𝑀𝜆=𝑁^𝑁_𝑗=11𝑞_𝑗−𝑞_𝑁³, (14)(ii) for 𝑁⩾2 (𝜑≠𝜋/2, when 𝑁=2), 𝑚₁=𝑚₂=⋯=𝑚_𝑁,𝑚₁=𝑚₂=⋯=𝑚_𝑁. (15)Without loss of generality, suppose that 𝑚_𝑗=𝑏𝑚_𝑗 and 𝑁⩾2 (𝜑≠𝜋/2, when 𝑁=2), under the above assumptions, there are four parameters, and ratio 𝑏 of masses ratio, 𝑎 of radius of two regular 𝑁-gons, the distance ℎ between two layers, and the phase difference 𝜑. As for these parameters, Xie et al. [5] proved.

Theorem 4. Under the assumptions of (9) and 𝑚_𝑗=𝑏𝑚_𝑗, then 𝑚₁,𝑚₂,…,𝑚_𝑁 and 𝑚₁,𝑚₂,…,𝑚_𝑁 form a central configuration if and only if the parameters 𝑏, 𝑎, ℎ, and 𝜑 satisfy the following relationships: 𝜆𝑀=1𝑁(1+𝑏)_𝑗≠𝑁1−𝜌_𝑗𝑞_𝑗−𝑞_𝑁³+_𝑗𝑏1−𝑎𝜌_𝑗𝑒^𝑖𝜑𝑞_𝑗−𝑞_𝑁³,𝜆𝑀=1𝑁^𝑁_𝑗=11𝑞_𝑗−𝑞_𝑁³,𝜆𝑀=𝑒^−𝑖𝜑𝑁𝑎(1+𝑏)_𝑗𝑎𝑒^𝑖𝜑−𝜌_𝑗𝑞_𝑗−𝑞_𝑁³+_𝑗≠𝑁𝑏𝑎−𝑎𝜌_𝑗𝑒^𝑖𝜑𝑞_𝑗−𝑞_𝑁³. (16) Zhang and Zhu [1] proved the sufficient conditions for special cases 𝑎=1, 𝑏=1, and 𝜑=𝜋/𝑁.

Theorem 5. Under the assumptions of (9) and (15) and 𝑚_𝑗=𝑏𝑚_𝑗 if 𝑎=1, 𝑏=1, and 𝜑=𝜋/𝑁, then for every 𝑁 there exists a unique central configuration.
When ℎ=0, Bang and Elmabsout [6] study the twist angle.

Theorem 6. Let Π₁ (resp., Π₂) be a regular 𝑁-gon centred around a mass 𝑚₀ at O, 𝑚₁ being at each of its vertices (resp., 𝑚₂). Then 𝑚₀, Π₁, and Π₂ are relative equilibriums if and only if they are homothetic or cursed with an angle equal to (𝜋/𝑁) (and suitable ratio of radii).
A natural interesting problem is that whether (9) form a central configuration for ℎ>0 if and only if twist angle 𝜑=0 or 𝜋/𝑁. In this paper, one will prove the necessary condition of twisted angle 𝜑 for a special case 𝑁=4.

Theorem 7. Under the assumptions of (9) and (15) and 𝑚_𝑗=𝑏𝑚_𝑗 if 𝑁=4, then 𝜑=𝜋/4+(𝜋/2)𝑘 or 𝜑=(𝜋/2)𝑘, (𝑘=0,1,2,3).

2. The Proof of Theorem 7

2.1. Some Lemmas

We need three lemmas.

Lemma 1 (see [7]). For 𝑁≥3 and 𝑚₁=𝑚₂=⋯=𝑚_𝑁, if 𝑚₁,𝑚₂,…,𝑚_𝑁 locate on vertices of a regular polygon, then they form a central configuration.
One has 𝑚^∗₁=𝑚^∗₂=𝑚^∗₃=𝑚^∗₄. Also 𝑚^∗₁, 𝑚^∗₂, 𝑚^∗₃, and 𝑚^∗₄ are located on vertices of a square 𝑆; 𝑞^∗_𝑖 is the position 𝑚^∗_𝑖(𝑖=1,2,3,4) and 𝑞^∗ is the position of 𝑚^∗. Plane 𝑃 contains the square 𝑆. One has 𝐹_∗=∑⁴_𝑖=1𝑚^∗_𝑖𝑚^∗(𝑞^∗_𝑖−𝑞^∗)/|𝑞^∗_𝑖−𝑞^∗|³.

Lemma 2. 𝐹_∗’s projection on 𝑃 is directed toward the center of the square if and only if the point 𝑚^∗ is on a vertical plane of symmetry.

2.2. The Proof of Lemma 2

Without loss of generality, suppose that 𝑚^∗₁=𝑚^∗₂=𝑚^∗₃=𝑚^∗₄=1, 𝑞^∗₁=(1,0,0), 𝑞^∗₂=(0,1,0), 𝑞^∗₃=(−1,0,0), 𝑞^∗₄=(0,−1,0), and 𝑞^∗=(𝜌cos𝜃,𝜌sin𝜃,ℎ), where 𝜌>0. Consider the following: 𝐹_∗=𝑚^∗1−𝜌cos𝜃,−𝜌sin𝜃,−ℎ1+𝜌²+ℎ²−2𝜌cos𝜃^3/2+𝑚^∗−𝜌cos𝜃,1−𝜌sin𝜃,−ℎ1+𝜌²+ℎ²−2𝜌sin𝜃^3/2+𝑚^∗−1−𝜌cos𝜃,−𝜌sin𝜃,−ℎ1+𝜌²+ℎ²+2𝜌cos𝜃^3/2+𝑚^∗−𝜌cos𝜃,−1−𝜌sin𝜃,−ℎ1+𝜌²+ℎ²+2𝜌sin𝜃^3/2. (17)

If 𝐹_∗’s projection on 𝑃 is directed toward the center of the square, there exists 𝜆^∗>0 such that 𝑚^∗1−𝜌cos𝜃1+𝜌²+ℎ²−2𝜌cos𝜃^3/2+𝑚^∗−𝜌cos𝜃1+𝜌²+ℎ²−2𝜌sin𝜃^3/2+𝑚^∗−1−𝜌cos𝜃1+𝜌²+ℎ²+2𝜌cos𝜃^3/2+𝑚^∗−𝜌cos𝜃1+𝜌²+ℎ²+2𝜌sin𝜃^3/2=−𝜆^∗=𝑚𝜌cos𝜃,_∗−𝜌sin𝜃1+𝜌²+ℎ²−2𝜌cos𝜃^3/2+𝑚^∗1−𝜌sin𝜃1+𝜌²+ℎ²−2𝜌sin𝜃^3/2+𝑚^∗−𝜌sin𝜃1+𝜌²+ℎ²+2𝜌cos𝜃^3/2+𝑚^∗−1−𝜌sin𝜃1+𝜌²+ℎ²+2𝜌sin𝜃^3/2=−𝜆^∗𝜌sin𝜃. (18)(19)

If 𝜃=(𝜋/2)𝑘, 𝑚^∗ is on a vertical plane of symmetry.

When 𝜃≠(𝜋/2)𝑘, multiplying both sides of (18) and (19), respectively, by sin𝜃 and cos𝜃, we have sin𝜃−𝜌cos𝜃sin𝜃1+𝜌²+ℎ²−2𝜌cos𝜃^3/2+−𝜌cos𝜃sin𝜃1+𝜌²+ℎ²−2𝜌sin𝜃^3/2+−sin𝜃−𝜌cos𝜃sin𝜃1+𝜌²+ℎ²+2𝜌cos𝜃^3/2+−𝜌cos𝜃sin𝜃1+𝜌²+ℎ²+2𝜌sin𝜃^3/2=−𝜆^∗𝜌cos𝜃sin𝜃,−𝜌cos𝜃sin𝜃1+𝜌²+ℎ²−2𝜌cos𝜃^3/2+cos𝜃−𝜌cos𝜃sin𝜃1+𝜌²+ℎ²−2𝜌sin𝜃^3/2+−𝜌cos𝜃sin𝜃1+𝜌²+ℎ²+2𝜌cos𝜃^3/2+−cos𝜃−𝜌cos𝜃sin𝜃1+𝜌²+ℎ²+2𝜌sin𝜃^3/2=−𝜆^∗𝜌cos𝜃sin𝜃. (20)

By (20) we have sin𝜃1+𝜌²+ℎ²−2𝜌cos𝜃^3/2+−cos𝜃1+𝜌²+ℎ²−2𝜌sin𝜃^3/2+−sin𝜃1+𝜌²+ℎ²+2𝜌cos𝜃^3/2+cos𝜃1+𝜌²+ℎ²+2𝜌sin𝜃^3/2=0. (21)

Then 11sin𝜃1+𝜌²+ℎ²−2𝜌sin𝜃^3/2−11+𝜌²+ℎ²+2𝜌sin𝜃^3/2=11cos𝜃1+𝜌²+ℎ²−2𝜌cos𝜃^3/2−11+𝜌²+ℎ²+2𝜌cos𝜃^3/2. (22)

Let 𝑓(𝑥)=1𝑥11+𝜌²+ℎ²−2𝜌𝑥^3/2−11+𝜌²+ℎ²+2𝜌𝑥^3/2; (23) then the system (22) is equivalent to 𝑓(sin𝜃)=𝑓(cos𝜃). It is obvious that 𝑓(𝑥) is even and 𝑓(𝑥)>0 when 𝑥>0. We will prove that 𝑓(𝑥) is a strictly increasing function when 𝑥>0.

When 0<𝑥≤1 we compute 𝑑𝑓(𝑥)1𝑑𝑥=−𝑥²11+𝜌²+ℎ²−2𝜌𝑥^3/2−11+𝜌²+ℎ²+2𝜌𝑥^3/2+1𝑥3𝜌1+𝜌²+ℎ²−2𝜌𝑥^5/2+3𝜌1+𝜌²+ℎ²+2𝜌𝑥^5/2=1+𝜌²+ℎ²+2𝜌𝑥1+𝜌²+ℎ²−2𝜌𝑥^5/2𝑥²1+𝜌²+ℎ²+2𝜌𝑥^5/21+𝜌²+ℎ²−2𝜌𝑥^5/2−1+𝜌²+ℎ²−2𝜌𝑥1+𝜌²+ℎ²+2𝜌𝑥^5/2𝑥²1+𝜌²+ℎ²+2𝜌𝑥^5/21+𝜌²+ℎ²−2𝜌𝑥^5/2+3𝜌𝑥1+𝜌²+ℎ²+2𝜌𝑥^5/2𝑥²1+𝜌²+ℎ²+2𝜌𝑥^5/21+𝜌²+ℎ²−2𝜌𝑥^5/2+3𝜌𝑥1+𝜌²+ℎ²−2𝜌𝑥^5/2𝑥²1+𝜌²+ℎ²+2𝜌𝑥^5/21+𝜌²+ℎ²−2𝜌𝑥^5/2. (24)

Let𝑔(𝑥)=1+𝜌²+ℎ²×+2𝜌𝑥1+𝜌²+ℎ²−2𝜌𝑥^5/2−1+𝜌²+ℎ²×−2𝜌𝑥1+𝜌²+ℎ²+2𝜌𝑥^5/2+3𝜌𝑥1+𝜌²+ℎ²+2𝜌𝑥^5/2+3𝜌𝑥1+𝜌²+ℎ²−2𝜌𝑥^5/2. (25)

Then 𝑑𝑓(𝑥)=𝑔(𝑥)𝑑𝑥𝑥²1+𝜌²+ℎ²+2𝜌𝑥^5/21+𝜌²+ℎ²−2𝜌𝑥^5/2,𝑔(0)=1+𝜌²+ℎ^27/2−1+𝜌²+ℎ^27/2(𝑥)=0,𝑑𝑔𝑑𝑥=2𝜌1+𝜌²+ℎ²−2𝜌𝑥^5/2−5𝜌1+𝜌²+ℎ²×+2𝜌𝑥1+𝜌²+ℎ²−2𝜌𝑥^3/2+2𝜌1+𝜌²+ℎ²+2𝜌𝑥^5/2−5𝜌1+𝜌²+ℎ²×−2𝜌𝑥1+𝜌²+ℎ²+2𝜌𝑥^3/2+3𝜌1+𝜌²+ℎ²+2𝜌𝑥^5/2+15𝜌²𝑥1+𝜌²+ℎ²+2𝜌𝑥^3/2+3𝜌1+𝜌²+ℎ²−2𝜌𝑥^5/2−15𝜌²𝑥1+𝜌²+ℎ²−2𝜌𝑥^3/2,𝑑𝑔(𝑥)𝑑𝑥_𝑥=0=10𝜌1+𝜌²+ℎ^25/2−10𝜌1+𝜌²+ℎ^25/2𝑑=0,₂𝑔(𝑥)𝑑𝑥²=−10𝜌²1+𝜌²+ℎ²−2𝜌𝑥^3/2−10𝜌²1+𝜌²+ℎ²−2𝜌𝑥^3/2+15𝜌²1+𝜌²+ℎ²×+2𝜌𝑥1+𝜌²+ℎ²−2𝜌𝑥^1/2+10𝜌²1+𝜌²+ℎ²+2𝜌𝑥^3/2+10𝜌²1+𝜌²+ℎ²+2𝜌𝑥^3/2−15𝜌²1+𝜌²+ℎ²×−2𝜌𝑥1+𝜌²+ℎ²+2𝜌𝑥^1/2+15𝜌²1+𝜌²+ℎ²+2𝜌𝑥^3/2+15𝜌²1+𝜌²+ℎ²+2𝜌𝑥^3/2+45𝜌³𝑥1+𝜌²+ℎ²+2𝜌𝑥^1/2−15𝜌²1+𝜌²+ℎ²−2𝜌𝑥^3/2−15𝜌²1+𝜌²+ℎ²−2𝜌𝑥^3/2+45𝜌³𝑥1+𝜌²+ℎ²−2𝜌𝑥^1/2≥45𝜌³𝑥1+𝜌²+ℎ²+2𝜌𝑥^1/2+45𝜌³𝑥1+𝜌²+ℎ²−2𝜌𝑥^1/2>0,∀0<𝑥≤1. (26)(27)(28)(29)(30)

Hence by (26), (27), (29), and (30) we have 𝑔(𝑥)>0,∀0<𝑥≤1. (31)

Hence 𝑑𝑓(𝑥)/𝑑𝑥>0, when 0<𝑥≤1. From 𝑓(sin𝜃)=𝑓(cos𝜃) we obtain |sin𝜃|=|cos𝜃|; that is, 𝜃=(𝜋/4)+(𝜋/2)𝑘, so 𝑚^∗ is on a vertical plane of symmetry.

It is obvious that if the point 𝑚^∗ is on a vertical plane of symmetry, 𝐹_∗’s projection on 𝑃 is directed toward the center of the square.

The proof of Lemma 2 is completed.

2.3. The Proof of Theorem 7

Without loss of generality, suppose that 𝑚₁=𝑚₂=𝑚₃=𝑚₄=𝑚, 𝑚₁=𝑚₂=𝑚₃=𝑚₄=1. We can let 𝑞₁=(1,0,0),𝑞₂=(),𝑞0,1,0₃=(−1,0,0),𝑞₄=(),𝑞0,−1,0₁=,𝑞𝜌cos𝜑,𝜌sin𝜑,ℎ₂=𝜌cos2𝜋4+𝜑,𝜌sin2𝜋4,𝑞+𝜑,ℎ₃=𝜌cos2𝜋4×2+𝜑,𝜌sin2𝜋4,𝑞×2+𝜑,ℎ₄=𝜌cos2𝜋4×3+𝜑,𝜌sin2𝜋4×3+𝜑,ℎ, (32) where 𝜌ℎ>0. It is obvious that 𝑧₀=∑⁴_𝑗=1(𝑚_𝑗𝑞_𝑗+𝑚_𝑗+𝑞_𝑗)/𝑀= (0,0,4𝑚ℎ/(4+4𝑚))=(0,0,𝑚ℎ/(1+𝑚)). By Lemma 1, since 𝑚₁, 𝑚₂, 𝑚₃, and 𝑚₄ locate on vertices of a regular polygon, so they form a central configuration; then there exists a constant 𝜆 (notice that it can be different from 𝜆 in(16)), such that ⁴_𝑗≠𝑘𝑚_𝑗𝑞_𝑗−𝑞_𝑘𝑞_𝑗−𝑞_𝑘=−𝜆𝑞_𝑘−𝑧₀, (33) where 𝑧₀=(∑⁴_𝑗=1𝑞_𝑗𝑚_𝑗)/4𝑚=(0,0,ℎ).

By (1) and (33) we have −𝜆𝑚𝑃_𝑘=⁴_𝑗=1𝑚𝑞_𝑗−𝑞_𝑘𝑞_𝑗−𝑞_𝑘³+⁴_𝑗≠𝑘𝑞𝑚𝑚^𝑗−𝑞_𝑘𝑞_𝑗−𝑞_𝑘³=⁴_𝑗=1𝑚𝑃_𝑗−𝑃_𝑘𝑃_𝑗−𝑃_𝑘³+⁴_𝑗≠𝑘𝑃𝑚𝑚^𝑗−𝑃_𝑘𝑃_𝑗−𝑃_𝑘³=⁴_𝑗=1𝑚𝑞_𝑗−𝑞_𝑘𝑞_𝑗−𝑞_𝑘³−𝜆𝑚𝑞_𝑘−𝑧₀(𝑘=1,2,3,4). (34)

Then ⁴_𝑗=1𝑚𝑞_𝑗−𝑞_𝑘𝑞_𝑗−𝑞_𝑘³=−𝜆𝑚𝑞_𝑘−𝑧₀+𝜆𝑚𝑞_𝑘−𝑧₀=𝑚𝜆−𝜆𝑞_𝑘+𝑚0,0,𝜆𝑚ℎ.1+𝑚−𝑚0,0,𝜆ℎ (35) Letting 𝐹₁=∑⁴_𝑗=1𝑚(𝑞_𝑗−𝑞₁)/|𝑞_𝑗−𝑞₁|³, 𝐹₁ is the force generated by the 𝑚₁, 𝑚₂, 𝑚₃, and 𝑚₄.

By (19), 𝐹₁’s projection on the plane 𝑃₁ is (𝜆−𝜆)(𝜌cos𝜑,𝜌sin𝜑,0), where 𝑃₁ is the plane containing 𝑚₁, 𝑚₂, 𝑚₃, and 𝑚₄. It is obvious that the projection is directed toward the center of the lower layer regular 4-gons. By Lemma 2, we have 𝜑=𝜋/4+(𝜋/2)𝑘 or 𝜑=(𝜋/2)𝑘(𝑘=0,1,2,3).

The proof of Theorem 7 is completed.

Acknowledgments

The authors express their gratitude to Professor Zhang Shiqing for his discussions and helpful suggestions. This work is Supported by NSF of China and Youth Fund of Mianyang Normal University.