2.2. The Proof of Lemma 2
Without loss of generality, suppose that m1*=m2*=m3*=m4*=1, q1*=(1,0,0), q2*=(0,1,0), q3*=(-1,0,0), q4*=(0,-1,0), and q*=(ρcosθ,ρsinθ,h), where ρ>0. Consider the following:
(17)F*=m*(1-ρcosθ,-ρsinθ,-h)|1+ρ2+h2-2ρcosθ|3/2 +m*(-ρcosθ,1-ρsinθ,-h)|1+ρ2+h2-2ρsinθ|3/2 +m*(-1-ρcosθ,-ρsinθ,-h)|1+ρ2+h2+2ρcosθ|3/2 +m*(-ρcosθ,-1-ρsinθ,-h)|1+ρ2+h2+2ρsinθ|3/2.
If F*’s projection on P is directed toward the center of the square, there exists λ*>0 such that
(18)m*(1-ρcosθ)|1+ρ2+h2-2ρcosθ|3/2+m*(-ρcosθ)|1+ρ2+h2-2ρsinθ|3/2 +m*(-1-ρcosθ)|1+ρ2+h2+2ρcosθ|3/2 +m*(-ρcosθ)|1+ρ2+h2+2ρsinθ|3/2 =-λ*ρcosθ,(19) =m*(-ρsinθ)|1+ρ2+h2-2ρcosθ|3/2 +m*(1-ρsinθ)|1+ρ2+h2-2ρsinθ|3/2 +m*(-ρsinθ)|1+ρ2+h2+2ρcosθ|3/2 +m*(-1-ρsinθ)|1+ρ2+h2+2ρsinθ|3/2 =-λ*ρsinθ.
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
(20)sinθ-ρcosθsinθ|1+ρ2+h2-2ρcosθ|3/2+-ρcosθsinθ|1+ρ2+h2-2ρsinθ|3/2 +-sinθ-ρcosθsinθ|1+ρ2+h2+2ρcosθ|3/2 +-ρcosθsinθ|1+ρ2+h2+2ρsinθ|3/2 =-λ*ρcosθsinθ,-ρcosθsinθ|1+ρ2+h2-2ρcosθ|3/2+cosθ-ρcosθsinθ|1+ρ2+h2-2ρsinθ|3/2 +-ρcosθsinθ|1+ρ2+h2+2ρcosθ|3/2 +-cosθ-ρcosθsinθ|1+ρ2+h2+2ρsinθ|3/2 =-λ*ρcosθsinθ.
By (20) we have
(21)sinθ|1+ρ2+h2-2ρcosθ|3/2+-cosθ|1+ρ2+h2-2ρsinθ|3/2 +-sinθ|1+ρ2+h2+2ρcosθ|3/2 +cosθ|1+ρ2+h2+2ρsinθ|3/2=0.
Then
(22)1sinθ(1|1+ρ2+h2-2ρsinθ|3/2 -1|1+ρ2+h2+2ρsinθ|3/2) =1cosθ(1|1+ρ2+h2-2ρcosθ|3/2 -1|1+ρ2+h2+2ρcosθ|3/2).
Let
(23)f(x)=1x(1|1+ρ2+h2-2ρx|3/2 -1|1+ρ2+h2+2ρx|3/2);
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
(24)df(x)dx=-1x2(1|1+ρ2+h2-2ρx|3/2 -1|1+ρ2+h2+2ρx|3/2) +1x(3ρ|1+ρ2+h2-2ρx|5/2 +3ρ|1+ρ2+h2+2ρx|5/2)=(1+ρ2+h2+2ρx)(1+ρ2+h2-2ρx)5/2x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2 -(1+ρ2+h2-2ρx)(1+ρ2+h2+2ρx)5/2x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2 +3ρx(1+ρ2+h2+2ρx)5/2x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2 +3ρx(1+ρ2+h2-2ρx)5/2x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2.
Let(25)g(x)=(1+ρ2+h2+2ρx) ×(1+ρ2+h2-2ρx)5/2 -(1+ρ2+h2-2ρx) ×(1+ρ2+h2+2ρx)5/2 +3ρx(1+ρ2+h2+2ρx)5/2 +3ρx(1+ρ2+h2-2ρx)5/2.
Then
(26)df(x)dx=g(x)x2|1+ρ2+h2+2ρx|5/2|1+ρ2+h2-2ρx|5/2,(27)g(0)=(1+ρ2+h2)7/2-(1+ρ2+h2)7/2=0,(28)dg(x)dx=2ρ(1+ρ2+h2-2ρx)5/2-5ρ(1+ρ2+h2+2ρx)×(1+ρ2+h2-2ρx)3/2+2ρ(1+ρ2+h2+2ρx)5/2-5ρ(1+ρ2+h2-2ρx)×(1+ρ2+h2+2ρx)3/2+3ρ(1+ρ2+h2+2ρx)5/2+15ρ2x(1+ρ2+h2+2ρx)3/2+3ρ(1+ρ2+h2-2ρx)5/2-15ρ2x(1+ρ2+h2-2ρx)3/2,(29)dg(x)dx|x=0=10ρ(1+ρ2+h2)5/2-10ρ(1+ρ2+h2)5/2=0,(30)d2g(x)dx2=-10ρ2(1+ρ2+h2-2ρx)3/2 -10ρ2(1+ρ2+h2-2ρx)3/2 +15ρ2(1+ρ2+h2+2ρx) ×(1+ρ2+h2-2ρx)1/2 +10ρ2(1+ρ2+h2+2ρx)3/2 +10ρ2(1+ρ2+h2+2ρx)3/2 -15ρ2(1+ρ2+h2-2ρx) ×(1+ρ2+h2+2ρx)1/2 +15ρ2(1+ρ2+h2+2ρx)3/2 +15ρ2(1+ρ2+h2+2ρx)3/2 +45ρ3x(1+ρ2+h2+2ρx)1/2 -15ρ2(1+ρ2+h2-2ρx)3/2 -15ρ2(1+ρ2+h2-2ρx)3/2 +45ρ3x(1+ρ2+h2-2ρx)1/2≥45ρ3x(1+ρ2+h2+2ρx)1/2 +45ρ3x(1+ρ2+h2-2ρx)1/2>0, ∀0<x≤1.
Hence by (26), (27), (29), and (30) we have
(31)g(x)>0, ∀0<x≤1.
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 m~1=m~2=m~3=m~4=m, m1=m2=m3=m4=1. We can let
(32)q1=(1,0,0), q2=(0,1,0),q3=(-1,0,0), q4=(0,-1,0),q~1=(ρcosφ,ρsinφ,h),q~2=(ρcos(2π4+φ),ρsin(2π4+φ),h),q~3=(ρcos(2π4×2+φ),ρsin(2π4×2+φ),h),q~4=(ρcos(2π4×3+φ),ρsin(2π4×3+φ),h),
where ρh>0. It is obvious that z0=∑j=14(mjqj+m~j+q~j)/M= (0,0,4mh/(4+4m))=(0,0,mh/(1+m)). By Lemma 1, since m~1, m~2, m~3, and m~4 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
(33)∑j≠k4m~jq~j-q~k|q~j-q~k|=-λ~(q~k-z~0),
where z~0=(∑j=14q~jm~j)/4m=(0,0,h).
By (1) and (33) we have
(34)-λmP~k=∑j=14mqj-q~k|qj-q~k|3+∑j≠k4mmq~j-q~k|q~j-q~k|3=∑j=14mPj-P~k|Pj-P~k|3+∑j≠k4mmP~j-P~k|P~j-P~k|3=∑j=14mqj-q~k|qj-q~k|3-λ~m(q~k-z~0) (k=1,2,3,4).
Then
(35)∑j=14m(qj-q~k)|qj-q~k|3 =-λm(q~k-z0)+λ~m(q~k-z~0) =m(λ~-λ)q~k+m(0,0,λmh1+m)-m(0,0,λ~h).
Letting F1~=∑j=14m(qj-q~1)/|qj-q~1|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.