# Theories and Inequalities on the Satellite System

Authors: Chaobang Gao; Jiajin Wen
Journal: ISRN Mathematical Analysis
 (2011)

---

## Abstract

We define the satellite system without any central, the satellite system with a central, and the satellite system of single point with a central. For the satellite system without any centralS{Γ,ΓN}, we establish the inequality: ∮Γds∮ΓN|PQ|dsQ≤(N|Γ|3/4π2)sin(2π/N). For the satellite system with a central S{O,Γ,ΓN}, we establish the following inequality under the proper hypothesis: ∮ΓMN[t](r)ds≤(|Γ|2/2π)cos(π/N)(∀t∈]-∞,-2]). As an application, we get the inequalities ∮ΓρPds≥∮ΓrPds≥2|D(Γ)|, for the satellite system of single point with a central S{O,Γ,P}. For these results, there are generalized backgrounds in the fields of differential geometry and space science.


---

## Body

## 1. Introduction and the Main Results

In this paper, we will use the following symbols:R and Z to denote the set of real numbers and the set of integers, respectively, a=(a1,a2,…,an), and Rn={a|ai∈R,1≤i≤n}[1, 2].Definition 1.1.
LetΓ=Γ(AB):r=r(t)=x(t)i⃗+y(t)j⃗+z(t)k⃗(a≤t≤b) be a smooth curve in three-space R3. A=r(a) and B=r(b) are called the initial point and terminal point of Γ, respectively. If a<t1<b, a≤t2≤b, t1≠t2, and r(t1)=r(t2), then r(t1) is called a coincident point of Γ; smooth curve without any coincident point is called a simple curve; If a simple curve Γ satisfies A=B, then Γ is said be a simple closed curve. Especially, we say Γ is a Jordan closed curve if Γ is a plane curve in R3. For a Jordan closed curve Γ, D(Γ) denotes the closed region bounded by Γ and its area is written as |D(Γ)|, |Γ| denotes the length of Γ, and we have the following Jordan curve theorem.Theorem 1.2 (Jordan Curve [3]).
An arbitrary Jordan closed curve must divide a plane into two parts, where one part is bounded, and the other is unbounded. The bounded part is called interior and another is outside of the Jordan closed curve.Definition 1.3 (see [4, 5]).
Assume thatΓ(AB) is a smooth space curve with end points A and B, P is a fixed point in R3, and Q∈Γ(AB), where Q is a moving point, then the trail of the line segment PQ is called a bounded cone surface, written as Ω[P,Γ(AB)]. We Also say PQ is the generating line, P is the vertex, and Γ(AB) is the generating curve of the bounded cone surface. |Ω[P,Γ(AB)]| denotes the area of Ω[P,Γ(AB)].
IfΓ is a given smooth curve in R3, then its length is written as |Γ|; Γ(AB) denotes the smooth curve segment with end points A and B, and Γ(AB)⊂Γ.Definition 1.4.
LetΓ be a fixed smooth closed curve in R3, the N-polygon ΓN:A1A2⋯ANA1(N≥3) be inscribed in Γ, and, Ai∈Γ(Ai-1Ai+1)(i=1,2,…,N). If all vertices:A1,A2,…,AN of ΓN move continuously along Γ and keep li:=|Γ(AiAi+1)| invariable, where 0<li<|Γ|/2,i=1,2,…,N, we say that the set S{Γ,ΓN}:={Γ,ΓN} is the satellite system without any central in three-space R3.
ForS{Γ,ΓN}, we write the set of the vertices of ΓN as V(ΓN):={A1,A2,…,AN} and define Ai=Aj⇔i≡j(modN), for all i,j∈Z [6].Definition 1.5.
ForS{Γ,ΓN}, if the following statements are valid:(i)
Γis a fixed closed Jordan curve in R3,(ii)
O is a fixed point in the interior of Γ, and(iii)
ΓN is always a Jordan closed curve and Ω[O,ΓN]=D(ΓN) as all vertices of ΓN move continuously along Γ,then we say the set S{O,Γ,ΓN}:={O,Γ,ΓN} is the satellite system with a central and O is its central.
ForS{O,Γ,ΓN}, ri:=Dis(O,AiAi+1) denotes the distance from O to the line AiAi+1, i=1,2,…,N. And we write r:=(r1,r2,…,rN).Remark 1.6.
TheS{O,Γ,ΓN} may be explained as follows: the point O denotes the central of earth, Γ denotes the trajectory on which satellites move, and the vertexes of ΓN are viewed as N satellites moving on the same trajectory Γ. In order to avoid hitting, they must move by same curve velocity, that is, li:=|Γ(AiAi+1)| is invariable and 0<li<|Γ|/2,i=1,2,…,N.Definition 1.7.
Thetth power mean of the positive real numbers sets a=(a1,a2,…,aN)(N≥2), written as MN[t](a), is defined by [7–15]:
(1.1)MN[t](a)≔{(1N∑i=1Nait)1/t,t∈R,0<|t|<+∞∏i=1NaiN,t=0.Definition 1.8 (see [1, 7, 10, 11, 14]).
LetΓ be a smooth curve in R3. For Riemann integrable function f:Γ→R, M(f,Γ):=(1/|Γ|)∫Γfds is called function means of f on Γ. For the function f:Γ→]0,∞[ and the real number t, if ft(t∈R,t≠0) and lnf are Riemann integrable on Γ, then
(1.2)M[t](f,Γ):={(1|Γ|∫Γftds)1/t,t∈R,t≠0exp⁡(1|Γ|∫Γln⁡⁡fds),t=0
is called the tth power means of f on Γ.Now, we give our main results as follow.Theorem 1.9.
LetS{Γ,ΓN} be a satellite system without any central. For any P∈V(ΓN):={A1,A2,…,AN}, we have inequality
(1.3)∮Γds∮ΓN|PQ|dsQ≤N|Γ|34π2sin⁡2πN,
that is,
(1.4)M(|Ω[P,ΓN]|,Γ)≤N|Γ|28π2sin⁡2πN,
where PQ is the generating line of the bounded cone surface Ω[P,ΓN]. A sufficient condition of equalities in (1.3) and (1.4) is that Γ is a circle and ΓN is always a regular polygon with N sides.Theorem 1.10.
LetS{O,Γ,ΓN} be a satellite system with a central. If N=3 or N≥4, 0<li≤|Γ|/4,i=1,2,…,N, we have inequality, for any the real number t≤-2,
(1.5)∮ΓMN[t](r)ds≤|Γ|22πcos⁡⁡πN,
that is,
(1.6)M(MN[t](r),Γ)≤|Γ|2πcos⁡⁡πN.
The equalities of (1.5) and (1.6) occur if and only if Γ is a circle, O is the central of the circle, and ΓN is a regular polygon with sides N.In Section5, we will give some applications of these results and theories.


## 2. Preliminaries

Lemma 2.1.
For the quadrilateralABCDA in R3, writing a:=|AB⃗|,b:=|BC⃗|,c:=|CD⃗|,d:=|DA⃗|, we obtain
(2.1)|ΔBAD|+|ΔBCD|≤14[(a+d)2-(b-c)2][(b+c)2-(a-d)2],
with equality if and only if ∠BAD+∠BCD=π. If the quadrilateral ABCDA is convex and in plane, equality holds if and only if the quadrilateral ABCDA is inscribed in a circle.Proof.
Writes:=|ΔBAD|+|ΔBCD|,∠BAD=α,∠BCD=β. By the area formula of triangle and cosine theorem, we get
(2.2)s=12(adsin⁡α+bcsin⁡β),(2.3)a2+d2-2adcos⁡⁡α=b2+c2-2bccos⁡⁡β,0≤α,β≤π.
Consider β=β(α) as the implicit function with respect to α. Finding the derivatives of the two sides of (2.2) and (2.3), respectively, we have
(2.4)dsdα=12(adcos⁡⁡α+bccos⁡⁡βdβdα),dβdα=adsin⁡αbcsin⁡β≥0.
Therefore, (ds/dα)=(1/2)adsinα(cotα+cotβ) and β=β(α) is increasing with respect to α. Since
(2.5)|b-c|≤|BD|≤a+d,|a-d|≤|BD|≤b+c,⟹|b-c|≤a+d,|a-d|≤b+c,⟹(a+d)2-(b-c)2≥0,(a-d)2-(b+c)2≤0,⟹-2(ad+bc)≤a2+d2-b2-c2≤2(ad+bc),⟹|a2+d2-b2-c22(ad+bc)|≤1,
we have
(2.6)dsdα=0⟺α+β=π,a2+d2-2adcos⁡⁡α=b2+c2-2bccos⁡⁡β,⟺α=α0:=arccos⁡a2+d2-b2-c22(ad+bc),β=β(α0)=π-α0,0≤α<α0⟹0≤β=β(α)<β(α0)=π-α0⟹0≤β<π-α⟹dsdα>0,α0<α≤π⟹π≥β=β(α)>β(α0)=π-α0⟹π≥β>π-α⟹dsdα<0.
Hence s is max if and only if α+β=π. When s is max, β=π-α,α=α0. Thus,
(2.7)s≤12(adsin⁡α0+bcsin⁡α0)=12(ad+bc)1-[a2+d2-b2-c22(ad+bc)]2=14[(a+d)2-(b-c)2][(b+c)2-(a-d)2].
In other words, (2.1) holds. Equality holds if and only if α+β=π⇔∠BAD+∠BCD=π. This completes the proof.Lemma 2.2.
LetS{Γ,ΓN} be a satellite system without any central. For any P∈V(ΓN), we have inequality:
(2.8)12∮ΓN|PQ|dsQ=|Ω[P,ΓN]|≤|ΓN|24NcotπN,
where PQ is the generating line of the bounded cone surface Ω[P,ΓN]. The second equality occurs if ΓN is a regular polygon with sides N.Proof.
By the definition and geometric meaning of curve integral, we obtain(2.9)12∮ΓN|PQ|dsQ=|Ω[P,ΓN]|.
Now we prove the inequality in (2.8).
IfN=3, (2.8) is known; if N=4, by Lemma 2.1, (2.8) holds. In the following, we suppose that N≥5. First, fix the value of |ΓN|. Without loss of generality, we set P=A1. By the theory of differential geometry, we know a bounded cone surface is a developable surface, which implies that Ω[P,ΓN] can be developed into a bounded cone surface Ω[A1*,ΓN*] in the plane, where ΓN*:A1*A2*⋯AN*A1*(P=A1=A1*) is the developed graph of ΓN in the plane and it is a polygon with sides N (may not be a closed Jordan curve) and satisfies (2.10)ΔAi-1*A1*Ai*∩ΔAi*A1*Ai+1*=A1*Ai*(i=3,4,…,N-1),|ΓN*|=|ΓN|,|Ω[P,ΓN]|=|Ω[A1*,ΓN*]|=∑i=3N|ΔAi-1*A1*Ai*|.
When |Ω[A1*,ΓN*]| is max, for any i∈{2,3,…,N-1}, the quadrilateral Ai-1*Ai*Ai+1*Ai+2*Ai-1* must be convex and the polygon with 5 sides A1*Ai-1*Ai*Ai+1*Ai+2*A1*(AN+1*=A1*) must be also convex by the plane geometry. Now, we prove the four points Ai-1*,Ai*,Ai+1* and Ai+2* are on a common circle. Otherwise, there exists some i∈{2,3,…,N-1} such that Ai-1*,Ai*,Ai+1*,Ai+2* are not on a common circle. Therefore, fix the point Aj*(j∈{1,2,…,N}-{i,i+1}) and modify Ai*,Ai+1* to Ai**,Ai+1** such that Ai-1*,Ai**,Ai+1**,Ai+2* are on a common circle and
(2.11)|Ai-1*Ai**⃗|=|Ai-1*Ai*⃗|,|Ai**Ai+1**⃗|=|Ai*Ai+1*⃗|,|Ai+1**Ai+2*⃗|=|Ai+1*Ai+2*⃗|.
Hence, we obtain a new polygon with N sides and write
(2.12)ΓN**:A1*A2*⋯Ai-1*Ai**Ai+1**Ai+2*⋯AN*A1*.
It follows that
(2.13)|Ω[A1*,ΓN*]|=|Ω[A1*,A1*A2*⋯Ai-1*Ai+2*⋯AN*A1*]|+|D(Ai-1*Ai*Ai+1*Ai+2*Ai-1*)|,|Ω[A1*,ΓN**]|=|Ω[A1*,A1*A2*⋯Ai-1*Ai+2*⋯AN*A1*]|+|D(Ai-1*Ai**Ai+1**Ai+2*Ai-1*)|.
By Lemma 2.1, we have
(2.14)|Ω[A1*,ΓN**]|-|Ω[A1*,ΓN*]|=|D(Ai-1*Ai**Ai+1**Ai+2*Ai-1*)|-|D(Ai-1*Ai*Ai+1*Ai+2*Ai-1*)|>0.
This contradicts the greatest |Ω[A1*,ΓN*]|. Since three points confirm a unique circle and for any i∈{2,3,…,N-1}, we know A1*,A2*,…,AN-1*,AN* are on a common circle if |Ω[A1*,ΓN*]| is greatest, and since
(2.15)ΔAi-1*A1*Ai*∩ΔAi*A1*Ai+1*=A1*Ai*,i=3,4,…,N-1,ΓN* is inscribed in a circle and |Ω[P,ΓN]|=|Ω[A1*,ΓN*]|=|D(ΓN*)|. Otherwise, there exists k∈{2,3,…,N-1} such that ΔAk-1*A1*Ak*∩ΔAk*A1*Ak+1*≠A1*Ak*. When the perimeter of a circle is a fixed value, the area of regular polygon with N sides is greatest in all N-polygons inscribed in the circle [16]. Therefore, ΓN* is a regular polygon with N sides if |Ω[A1*,ΓN*]| is greatest.
Now letΓN* be a regular polygon with N sides. By the plane geometry, we know(2.16)|Ω[A1*,ΓN*]|=|D(ΓN*)|=|ΓN*|24NcotπN=|ΓN|24NcotπN.
Hence, the inequality holds in (2.8). Equality occurs if ΓN* is a regular N-polygon. This completes the proof.Lemma 2.3.
For the quadrilateralABCDA in R3, we have
(2.17)|AC⃗|2+|BD⃗|2≤|BC⃗|2+|DA⃗|2+2|AB⃗|⋅|CD⃗|,
with equality if and only if ∠(AB⃗,CD⃗)=π.Proof.
Write(AB⃗,BC⃗,CD⃗,DA⃗,AC⃗,BD⃗)=(a,b,c,d,e,f), a·b:=〈a,b〉 denotes the inner product of vectors a and b, especially, a2=a·a=|a|2. Thus, (2.17) is expressed as
(2.18)e2+f2≤b2+d2+2|a|⋅|c|.
Since
(2.19){a+b=ec+d=-eb+c=fd+a=-f⟹{2a⋅b=e2-a2-b22c⋅d=e2-c2-d22b⋅c=f2-b2-c22d⋅a=f2-d2-a2
and ∑a:=a+b+c+d=0, which implies that
(2.20)0=(∑a)2=a2+b2+c2+d2+2(a⋅b+c⋅d+b⋅c+d⋅a)+2(a⋅c+b⋅d)=-(a2+b2+c2+d2)+2(e2+f2)+2(a⋅c+b⋅d)=-2(a2+b2+c2+d2)+2(e2+f2)+(a+c)2+(b+d)2=-2(a2+b2+c2+d2)+2(e2+f2)+2(a+c)2=-2(b2+d2)+2(e2+f2)+4a⋅c,
it follows that
(2.21)e2+f2=b2+d2-2a⋅c≤b2+d2+2|a|⋅|c|.(2.18) is proved. Equality of (2.18) holds if and only if ∠(AB⃗,CD⃗)=∠(a,c)=π. This completes the proof.Lemma 2.4.
LetΓN(N≥4) be a polygon with sides N in R3. Setting
(2.22)Sk:=sin⁡kπN,Lk:=[N(2Sk)2]-1∑i=1N|AiAi+k⃗|2,k=1,2,…,N-1,
we get the following inequality:
(2.23)Lk≤(S1Sk)2L1+Sk-1Sk+12Sk2(Lk+1+Lk-1),
for 2≤k≤N-2. A sufficient condition of equality is that ΓN is a plane regular N-polygon in R3.Proof.
Consider the quadrilateralAi-1AiAi-1+kAi+kAi-1. From
(2.24)(Sk-1-1|AiAi-1+k⃗|-Sk+1-1|Ai-1Ai+k⃗|)2≥0,
we obtain
(2.25)2|AiAi-1+k⃗|⋅|Ai-1Ai+k⃗|≤Sk+1Sk-1|AiAi-1+k⃗|2+Sk-1Sk+1|Ai-1Ai+k⃗|2.
It follows from Lemma 2.3 and (2.25) that
(2.26)|AiAi+k⃗|2+|Ai-1Ai-1+k⃗|2≤|Ai-1Ai⃗|2+|Ai-1+kAi+k⃗|2+2|AiAi-1+k⃗||Ai-1Ai+k⃗|≤|Ai-1Ai⃗|2+|Ai-1+kAi+k⃗|2+Sk+1Sk-1|AiAi-1+k⃗|2+Sk-1Sk+1|Ai-1Ai+k⃗|2,
which implies that
(2.27)∑i=1N(|AiAi+k⃗|2+|Ai-1Ai-1+k⃗|2)≤∑i=1N(|Ai-1Ai⃗|2+|Ai-1+kAi+k⃗|2+Sk+1Sk-1|AiAi-1+k⃗|2+Sk-1Sk+1|Ai-1Ai+k⃗|2).
Since
(2.28)∑i=1N|AiAi+k⃗|2=∑i=1N|Ai-1Ai-1+k⃗|2=4NSk2Lk,∑i=1N|Ai-1Ai⃗|2=∑i=1N|Ai-1+kAi+k⃗|2=4NS12L1,∑i=1N|AiAi-1+k⃗|2=4NSk-12Lk-1,∑i=1N|Ai-1Ai+k⃗|2=4NSk+12Lk+1,
the inequality (2.27) is equivalent to
(2.29)8NSk2Lk≤8NS12L1+4NSk+1Sk-1Lk-1+4NSk-1Sk+1Lk+1⟺Lk≤(S1Sk)2L1+Sk-1Sk+12Sk2(Lk+1+Lk-1).
Inequality (2.23) is proved. From this proof and Lemma 2.3, we know that a sufficient condition of equality is that ΓN is a regular polygon with N sides in R3.Remark 2.5.
A sufficient condition of equality of (2.23) is that ΓN is a regular polygon with N sides in R3. This condition is not necessary. For example, when N=4, the equality holds in (2.23) if and only if Γ4 is a parallelogram in R3.Remark 2.6.
IfΓN is a regular N-polygon, Lk defined by Lemma 2.4 is equal to R02, where R0 denotes the circumradius of ΓN. Namely,
(2.30)Lk=R02,k=1,2,…,N-1.Lemma 2.7.
SupposeLk is defined by Lemma 2.4, for any positive integer k,j:k≥2,k+j≤N-1, there exist constants Ck+j,j,Ck-1,j,C1,j which is only related to k,j,N such that
(2.31)Lk≤Ck+j,jLk+j+Ck-1,jLk-1+C1,jL1,Ck+j,j+Ck-1,j+C1,j=1.
A sufficient condition of equality is that ΓN is a regular N-polygon in R3.Proof.
We prove it by mathematical induction with respect toj.
(i) Whenj=1, let Ck+1,1=Ck-1,1=Sk-1Sk+1/(2Sk2)(>0),C1,1=(S1/Sk)2(>0) from Lemma 2.4, we have(2.32)Lk≤Ck+1,1Lk+1+Ck-1,1Lk-1+C1,1L1.
Let ΓN be a regular N-polygon. By Remark 2.6, we know Lk=Lk+1=Lk-1=L1=R02>0; it follows from Lemma 2.4 that the equality of (2.32) holds, thus, Ck+1,1+Ck-1,1+C1,1=1.
(ii) Assume that Lemma2.7 holds for j=n≥1. Now we want to prove that Lemma 2.7 holds for j=n+1. By the hypothesis k,n+1:k≥2,k+n+1≤N-1, we know that k,n:k≥2,k+n≤N-2≤N-1. Thus, from the induction hypothesis, there exist constants Ck+n,n,Ck-1,n,C1,n such that(2.33)Ck+n,n+Ck-1,n+C1,n=1,(2.34)Lk≤Ck+n,nLk+n+Ck-1,nLk-1+C1,nL1.
A sufficient condition of equality of (2.34) is that ΓN is a regular polygon with N sides. Since k+1,n:k+1≥3≥2,(k+1)+n≤N-1, by induction hypothesis, substitute k by k+1 in (2.33) and (2.34), in other words, there exist constants Ck+1+n,n*,Ck,n*,C1,n* such that
(2.35)Ck+1+n,n*+Ck,n*+C1,n*=1,(2.36)Lk+1≤Ck+1+n,n*Lk+1+n+Ck,n*Lk+C1,n*L1.
Substituting (2.36) into (2.32), we get the following inequality:
(2.37)Lk≤Ck+1,1(Ck+1+n,n*Lk+1+n+Ck,n*Lk+C1,n*L1)+Ck-1,1Lk-1+C1,1L1.
Notice 0<Ck+1,1<1,0<Ck,n*<1,1-Ck+1,1Ck,n*>0. Solving inequality (2.37) with respect to Lk, we obtain
(2.38)Lk≤Ck+n+1,n+1**Lk+n+1+Ck-1,n+1**Lk-1+C1,n+1**L1,
where
(2.39)Ck+n+1,n+1**=Ck+1,1Ck+n+1,n*1-Ck+1,1Ck,n*>0,Ck-1,n+1**=Ck-1,11-Ck+1,1Ck,n*>0,C1,n+1**=Ck+1,1C1,n*+C1,11-Ck+1,1Ck,n*>0.
Let ΓN be a regular N-polygon. From Remark 2.6, we know Lk=Lk+n+1=Lk-1=L1=R02>0. It follows from Lemma 2.4 and induction hypothesis that the equality of (2.38) holds. Thus, Ck+n+1,n+1**+Ck-1,n+1**+C1,n+1**=1, that is, Lemma 2.7 holds for j=n+1. This completes the proof.Lemma 2.8.
SupposeLk is defined by Lemma 2.4, we have the inequality as follows:
(2.40)Lk≤L1,k=2,3,…,N-2.
A sufficient condition of equality is that ΓN is a regular polygon with N sides in R3.Proof.
Settingk+j=N-1 in (2.31), we get
(2.41)Lk≤CN-1,N-1-kLN-1+Ck-1,N-1-kLk-1+C1,N-1-kL1.
Since Ai=Aj⇔i≡j(modN),
(2.42)LN-1:=[N(2SN-1)2]-1∑i=1N-1|AiAi+N-1⃗|2=[N(2S1)2]-1∑i=1N-1|AiAi-1⃗|2=L1.
It follows from (2.41) and (2.42) that there exist constants Ck-1,C1:Ck-1+C1=1 such that
(2.43)Lk≤Ck-1Lk-1+C1L1,
where Ck-1=Ck-1,N-1-k>0,C1=CN-1,N-1-k+C1,N-1-k>0, for any k∈{2,3,…,N-2}. Using (2.43) repeatedly, we get
(2.44)Lk≤Ck-1Lk-1+C1L1≤Ck-1(Ck-2*Lk-2+C1*L1)+C1L1=Ck-2**Lk-2+C1**L1≤Ck-3***Lk-3+C1***L1≤⋯≤CL1,
therefore,
(2.45)Lk≤CL1.
Set ΓN is a regular polygon with N sides in R3. By Remark 2.6, we know Lk=L1=R02>0; from Lemma 2.7, we have the equality of (2.45) holds, which implies that C=1. A sufficient condition of equality of (2.40) is that ΓN is a regular N-polygon in R3. This completes the proof.Lemma 2.9.
LetΓ=Γ(AB):r=r(t)=x(t)i⃗+y(t)j⃗+z(t)k⃗(a≤t≤b) be a smooth curve with the end points A and B in R3. If the function f:Γ(AB)→R is Riemann integrable on Γ(AB), considering a partition of Γ(AB) by means of N-1(N≥3) points A=A0,A1,…,Ai-1,Ai,…,AN=B such that |A0A1⃗|=|A1A2⃗|=⋯=|Ai-1Ai⃗|=⋯=|AN-1AN⃗|=|ΓN|/N, where ΓN:A0A1⋯AN-1AN is a broken line, we have
(2.46)M(f,Γ(AB)):=1|Γ(AB)|∫Γ(AB)fds=lim⁡N→∞1N∑i=0N-1f(Ai).Proof.
SinceΓ(AB) is a smooth curve with the end points A and B in R3,
(2.47)|Γ(AB)|=∫abx′2(t)+y′2(t)+z′2(t)dt
exists and |Γ(AB)|>0. And f:Γ(AB)→R is Riemann integrable on Γ(AB), it follows that f:Γ(AB)→R is bounded on Γ(AB), that is, there exists a constant M1>0 such that |f(P)|≤M1 for any P∈Γ(AB). For any i∈{0,1,2,…,N-1},
(2.48)lim⁡N→∞|Γ(AiAi+1)|=lim⁡N→∞|AiAi+1⃗|=lim⁡N→∞|ΓN|N=|Γ(AB)|lim⁡N→∞N=0.
It follows from the definition of the curve integral
(2.49)∫Γ(AB)fds=lim⁡N→∞∑i=0N-1f(Pi)|Γ(AiAi+1)|,(∀Pi∈Γ(AiAi+1))
that
(2.50)∫Γ(AB)fds=lim⁡N→∞∑i=0N-1f(Ai)|Γ(AiAi+1)|.
Since limN→∞(|Γ(AiAi+1)|/|AiAi+1⃗|)=1(∀i∈{0,1,2,…,N-1}), for any ε>0, there exists m∈Z(m>1) such that 1≤|Γ(AiAi+1)|/|AiAi+1⃗|<1+ε, for N>m, therefore
(2.51)0≤|Γ(AiAi+1)|-|AiAi+1⃗|<ε|AiAi+1⃗|=ε|ΓN|N≤ε|Γ(AB)|N.
Thus,
(2.52)|∑i=0N-1f(Ai)|Γ(AiAi+1)|-∑i=0N-1f(Ai)|AiAi+1⃗||=|∑i=0N-1f(Ai)(|Γ(AiAi+1)|-|AiAi+1⃗|)|≤∑i=0N-1|f(Ai)|⋅(|Γ(AiAi+1)|-|AiAi+1⃗|)≤∑i=0N-1M1⋅ε|Γ(AB)|N=M1|Γ(AB)|ε.
It follows that
(2.53)lim⁡N→∞∑i=0N-1f(Ai)|Γ(AiAi+1)|=lim⁡N→∞∑i=0N-1f(Ai)|AiAi+1⃗|.
In view of (2.50), we get
(2.54)M(f,Γ(AB)):=1|Γ(AB)|∫Γfds=1|Γ(AB)|lim⁡N→∞∑i=0N-1f(Ai)|Γ(AiAi+1)|=1|Γ(AB)|lim⁡N→∞∑i=0N-1f(Ai)|AiAi+1⃗|=lim⁡N→∞1|Γ(AB)|∑i=0N-1f(Ai)|ΓN|N=lim⁡N→∞|ΓN||Γ(AB)|⋅lim⁡N→∞1N∑i=0N-1f(Ai)=lim⁡N→∞1N∑i=0N-1f(Ai).
This completes the proof of Lemma 2.9.Lemma 2.10.
LetΓ be a smooth closed curve in R3. If the points A,B move continuously on Γ and keep |Γ(AB)| invariable, where 0<l:=|Γ(AB)|<|Γ|/2, we obtain
(2.55)M[2](|AB⃗|,Γ):=1|Γ|∮Γ|AB⃗|2⋅ds≤|Γ|πsin⁡lπ|Γ|.
A sufficient condition of equality is that Γ is a circle in R3.Proof.
By the theory of real number, there exists an increasing sequence of positive integers{kn} such that
(2.56)lim⁡n→∞kn10n=l|Γ|.
Write k:=kn,N:=10n. Consider a partition of Γ by means of N(N≥10) points A1,A2,…,Ai-1,Ai,…,AN such that |A1A2⃗|=|A2A3⃗|=⋯=|AN-1AN⃗|=|ANA1⃗|=|ΓN|/N, where ΓN:A1A2⋯AN-1ANA1 is a polygon with N sides.
First, we give the following fact: if the pointA moves to some point Ai along Γ, the point B moves to Ai* along Γ, in other words, A=Ai∈Γ,B=Ai*∈Γ,|Γ(AiAi*)|=l, then(2.57)lim⁡n→∞|Ai*Ai+k⃗|=0.
In fact, from limn→∞ΓN(AiAi+k)⊂Γ,|Γ(AiAi*)|=l, we only need to prove
(2.58)lim⁡n→∞|ΓN(AiAi+k)|=l.
Since
(2.59)lim⁡n→∞|ΓN(AiAi+k)|=lim⁡n→∞(k|AiAi+1⃗|)=lim⁡n→∞(k|ΓN|N)=lim⁡n→∞(|ΓN|)lim⁡n→∞kN=|Γ|l|Γ|=l,(2.58) holds, it follows that (2.57) holds.
From (2.57), we know for any ε>0, there exists m∈Z(m>1), when N>m, such that(2.60)||AiAi*⃗|2-|AiAi+k⃗|2|<ε,|1N∑i=1N|AiAi*⃗|2-1N∑i=1N|AiAi+k⃗|2|=1N|∑i=1N(|AiAi*⃗|2-|AiAi+k⃗|2)|≤1N∑i=1N||AiAi*⃗|2-|AiAi+k⃗|2|<1N∑i=1Nε=ε.
By Lemma 2.9, we get
(2.61)1|Γ|∮Γ|AB⃗|2ds=1|Γ|∮Γ|AB⃗|2dsA=lim⁡n→∞1N∑i=1N|AiAi*⃗|2=lim⁡n→∞1N∑i=1N|AiAi+k⃗|2.
It follows from inequality (2.40) that
(2.62)1N∑i=1N|AiAi+k⃗|2≤1N(sin⁡(kπ/N)sin⁡(π/N))2∑i=1N|AiAi+1⃗|2=1N(sin⁡(kπ/N)sin⁡(π/N))2∑i=1N(|ΓN|N)2=1N(sin⁡(kπ/N)sin⁡(π/N))2|ΓN|2N=(sin⁡(kπ/N)Nsin⁡(π/N))2|ΓN|2.
It follows from (2.61) and (2.62) that
(2.63)1|Γ|∮Γ|AB⃗|2ds=lim⁡n→∞1N∑i=1N|AiAi+k⃗|2≤lim⁡n→∞(sin⁡(kπ/N)Nsin⁡(π/N))2|ΓN|2=(sin⁡lim⁡n→∞(kπ/N)lim⁡n→∞Nsin⁡(π/N))2lim⁡n→∞|ΓN|2=(sin⁡(lπ/(|Γ|))π)2|Γ|2=(|Γ|πsin⁡lπ|Γ|)2.
Namely, (2.55) holds. From Lemma 2.8, a sufficient condition of equality of (2.55) is that Γ is a circle in R3. This completes the proof.Lemma 2.11 (Cauchy inequality [1, page 6]).
LetΓ be a smooth closed curve in R3. If the function f:Γ→R and g:Γ→R are smooth, we have the inequality as follows:
(2.64)|∮Γf⋅gds|≤∮Γf2ds⋅∮Γg2ds.Lemma 2.12.
AssumeΓ is a closed Jordan curve in R3 and O∈R3, then D(Γ)⊂Ω[O,Γ].Proof.
LetA,B∈R3,A≠B, AB denote the straight line through the points A and B, and let AB¯ denote the oriented line segment with the initial point A and terminal point B.
For anyP∈D(Γ), if P=O, then P∈Ω[O,Γ]; if P≠O, then (2.65)OP∩D(Γ)=⋃i≡1(mod⁡2),1≤i≤k-1QiQi+1¯,Qi∈Γ,i=1,2,…,k,k≥2,P∈OP∩D(Γ)⟹∃i:i≡1(mod⁡2),1≤i≤k-1,P∈QiQi+1¯.
If O∈QiQi+1¯, then
(2.66)P∈QiQi+1¯=QiO¯∪OQi+1¯⟹P∈QiO¯⊂Ω[O,Γ]∨P∈OQi+1¯⊂Ω[O,Γ]⟹P∈Ω[O,Γ].
If O∉QiQi+1¯, without loss of generality, we set |OQi|≤|OQi+1|, then
(2.67)P∈QiQi+1¯⊂OQi+1¯⊂Ω[O,Γ]⟹P∈Ω[O,Γ].
In all, for any P∈D(Γ)⇒P∈Ω[O,Γ], hence, D(Γ)⊂Ω[O,Γ]. This completes the proof.Lemma 2.13.
ForS{O,Γ,ΓN}, we have the following inequality:
(2.68)MN[-2](r):=(1N∑i=1Nri-2)-1/2≤121N∑i=1N|AiAi+1⃗|2⋅cotπN,
with equality if and only if ΓN is a regular N-polygon in R3 and O is its central.Proof.
By Definition1.5 and Lemma 2.12, for any P∈V(ΓN), we have Ω[O,ΓN]=D(ΓN)⊂Ω[P,ΓN]. Thus,
(2.69)∑i=1Nri|AiAi+1⃗|=2|Ω[O,ΓN]|=2|D(ΓN)|≤2|Ω[P,ΓN]|.
Using Cauchy inequality (∑i=1Naibi)2≤(∑i=1Nai2)(∑i=1Nbi2), (2.69) and (2.8), we get
(2.70)|ΓN|2=(∑i=1N|AiAi+1⃗|)2=(∑i=1Nri|AiAi+1⃗|⋅|AiAi+1⃗|ri)2≤(∑i=1Nri|AiAi+1⃗|)(∑i=1N|AiAi+1⃗|ri)≤(∑i=1Nri|AiAi+1⃗|)∑i=1N|AiAi+1⃗|2∑i=1Nri-2≤2|Ω[P,ΓN]|∑i=1N|AiAi+1⃗|2∑i=1Nri-2≤|ΓN|22NcotπN∑i=1N|AiAi+1⃗|2∑i=1Nri-2=|ΓN|22cotπN1N∑i=1N|AiAi+1⃗|2[MN[-2](r)]-1.
Therefore, (2.68) holds. The equality occurs if and only if ΓN is a regular N-polygon in R3 and O is its central. This completes the proof.Lemma 2.14 (Power Mean Inequality, [1, 7, 10, 11]).
LetΓ be a smooth closed curve in R3. If the function f:Γ→]0,∞[ is smooth, we have the following inequality, for p,q∈R,p<q,
(2.71)M[p](f,Γ)≤M[q](f,Γ).Lemma 2.15.
ForS{Γ,ΓN}, we have
(2.72)1N∑i=1Nsin⁡liπ|Γ|≤sin⁡πN.
If N=3 or N≥4,0<li≤|Γ|/4,i=1,2,…,N, we get the inequality as follows:
(2.73)1N∑i=1N(sin⁡liπ|Γ|)2≤(sin⁡πN)2.
The equalities hold in (2.72) and (2.73) if and only if l1=l2=…=lN=|Γ|/N.Proof.
By Definition1.4, we know ∑i=1Nli=|Γ|. Since sinx is the concave function on ]0,π/2[, it follows from Jensen inequality [17, 18] and 0<li<|Γ|/2(i=1,2,…,N,N≥3) that (2.72) holds. When N=3, l1π/|Γ|,l2π/|Γ|,l3π/|Γ| are three inner angles of a triangle. See [1, page 205], (2.73) holds. Since (sinx)2 is the concave function on ]0,π/4[, from N≥4,0<li≤|Γ|/4,i=1,2,…,N, (2.73) holds. In view of the condition of equality of Jensen inequality, the equalities of (2.72) and (2.73) hold if and only if l1=l2=⋯=lN=|Γ|/N. This completes the proof.


## 3. Proof of Theorem1.9

By (2.8),(2.64),(2.55), and (2.72) in order, we get (3.1)∮Γ|Ω[P,ΓN]|ds≤∮Γ|ΓN|24NcotπNds=14NcotπN∮Γ|ΓN|2ds=14NcotπN∮Γ(∑i=1N|AiAi+1⃗|)2ds=14NcotπN∮Γ(∑i=1N|AiAi+1⃗|2+2∑1≤i<j≤N|AiAi+1⃗||AjAj+1⃗|)ds=14NcotπN(∑i=1N∮Γ|AiAi+1⃗|2ds+2∑1≤i<j≤N∮Γ|AiAi+1⃗||AjAj+1⃗|ds)≤|Γ|4NcotπN(∑i=1N∮Γ|AiAi+1⃗|2ds|Γ|+2∑1≤i<j≤N∮Γ|AiAi+1⃗|2ds|Γ|∮Γ|AjAj+1⃗|2ds|Γ|)≤|Γ|4NcotπN[∑i=1N(|Γ|πsin⁡liπ|Γ|)2+2∑1≤i<j≤N(|Γ|πsin⁡liπ|Γ|)(|Γ|πsin⁡ljπ|Γ|)]=|Γ|34π2NcotπN(∑i=1Nsin⁡liπ|Γ|)2=|Γ|3N4π2cotπN(1N∑i=1Nsin⁡liπ|Γ|)2≤|Γ|3N4π2cotπN(sin⁡πN)2=|Γ|3N8π2sin⁡2πN,which implies that(3.2)M(|Ω[P,ΓN]|,Γ):=1|Γ|∮Γ|Ω[P,ΓN]|ds≤N|Γ|28π2sin⁡2πN,∮Γds∮ΓN|PQ|dsQ=2∮Γ|Ω[P,ΓN]|ds≤N|Γ|34π2sin⁡2πN.
Hence, (1.3) and (1.4) are proved. By this proof, a sufficient condition of equalities is that Γ is a circle and ΓN is always a regular polygon with N sides. This completes the proof of Theorem 1.9.Remark 3.1.
We may extend the concepts of the bounded cone surface and the satellite system without any central toRm(m≥2). And Theorem 1.9 also holds for the satellite system without any central in Rm.


## 4. Proof of Theorem1.10

Since power meanMN[t](r) is increasing with respect to t and by (2.68), we have the following inequalities, for t≤-2,(4.1)MN[t](r)≤MN[-2](r)≤121N∑i=1N|AiAi+1⃗|2⋅cotπN.
Write f:=(1/N)∑i=1N|AiAi+1⃗|2. By (4.1), (2.71), (2.72) and (2.73), we get (4.2)∮ΓMN[t](r)ds≤12cotπN∮Γ1N∑i=1N|AiAi+1⃗|2ds=|Γ|2cotπN1|Γ|∮Γf1/2ds=|Γ|2cotπNM[1/2](f,Γ)≤|Γ|2cotπNM[1](f,Γ)=|Γ|2cotπN1|Γ|∮Γfds=|Γ|2cotπN1|Γ|∮Γ1N∑i=1N|AiAi+1⃗|2ds=|Γ|2cotπN1N∑i=1N1|Γ|∮Γ|AiAi+1⃗|2ds≤|Γ|2cotπN1N∑i=1N(|Γ|π⋅sinliπ|Γ|)2=|Γ|22πcotπN1N∑i=1N(sinliπ|Γ|)2≤|Γ|22πcotπN⋅sinπN=|Γ|22πcos⁡πN.
Therefore, (1.5) and (1.6) are proved. From this proof, the equalities of (1.5) and (1.6) occur if and only if Γ is a circle, O is the central of the circle, and ΓN is a regular polygon with sides N. This completes the proof of Theorem 1.10.


## 5. Applications

In Theorem1.9, if both Γ and ΓN are simple closed curves in R3, and Ω[ΓN] denotes the minimal surface [4, 5, 19–21] bounded by ΓN, then for any P∈V(ΓN), |Ω[ΓN]|≤|Ω[P,ΓN]|, where |Ω[ΓN]| denotes the area of Ω[ΓN]. Thus, we get the theorem as follows.Theorem 5.1.
ForS{Γ,ΓN}, if both the curve Γ and the polygon ΓN are simple closed curves in R3, we have
(5.1)M(|Ω[ΓN]|,Γ):=1|Γ|∮Γ|Ω[ΓN]|ds≤N|Γ|28π2sin2πN,
where Ω[ΓN] is a minimal surface bounded by ΓN and |Ω[ΓN]| denotes its area. A sufficient condition of equality in (5.1) is that Γ is a circle and ΓN is always a regular N-polygon.In Theorem5.1, let V(ΓN)={A1,A2,…,AN} and satisfy |Γ(A1A2)|=|Γ(A2A3)|=⋯=|Γ(AN-1AN)|=|Γ(ANA1)|=(|Γ|)/N. Setting N→∞, we get the theorem as follows.Theorem 5.2.
If the curveΓ is a smooth simple closed curve in R3, we have
(5.2)|Ω[Γ]|≤|Γ|24π,
where |Ω[Γ]| denotes the area of the minimal surface Ω[Γ] bounded by Γ. A sufficient condition of equality is that Γ is a circle.Remark 5.3.
IfΓ is a smooth Jordan closed curve in R3, then Ω[Γ]=D(Γ), thus, (5.2) is the well-known isoperimetric inequality [16, 22–33].Definition 5.4.
IfΓ is a smooth Jordan closed curve in R3, O is a fixed point in the interior of Γ and the point P moves continuously on Γ, then S{O,Γ,P} is called the satellite system of single point with a central.
ForS{O,Γ,P}, write ρP:=|OP⃗|. lP denotes the tangent line of Γ at the point P, and rP:=Dis(O,lP) denotes the distance from the point O to the tangent line lP.Theorem 5.5.
ForS{O,Γ,P}, we have
(5.3)∮ΓρPds≥∮ΓrPds≥2|D(Γ)|,
that is,
(5.4)M(ρP,Γ)≥M(rP,Γ)≥2|D(Γ)||Γ|.
The first equalities in (5.3) and (5.4) hold if and only if Γ is a circle and the point O is the central of the circle; the second equalities hold if and only if Ω[O,Γ]=D(Γ).Proof.
Insert the inscribed polygonΓN:A1A2⋯ANA1(N≥3) such that
(5.5)|A1A2⃗|=|A2A3⃗|=⋯=|AN-1AN⃗|=|ANA1⃗|=|ΓN|N,
then the point O is in the inner of ΓN when N is sufficiently large. In the following, we assume that O is in the inner of ΓN.
According to Definition1.8, we only need to prove (5.3). Since for any P∈Γ, ρP≥rP, the first inequality of (5.3) holds. Let ri:=Dis(O,AiAi+1) denote the distance from the point O to the line AiAi+1. We have(5.6)lim⁡N→∞ri=rAi,i=1,2,3,…,lim⁡N→∞1N∑i=1NrAi=lim⁡N→∞1N∑i=1Nri.
By Lemma 2.12, we know D(ΓN)⊂Ω[O,ΓN]. Thus,
(5.7)|D(ΓN)|≤|Ω[O,ΓN]|=∑i=1N|ΔAiOAi+1|.
Equalities of (5.7) hold if and only if Ω[O,ΓN]=D(ΓN). By (5.6),(5.7), and Lemma 2.9, we get
(5.8)|D(Γ)|=lim⁡N→∞|D(ΓN)|≤lim⁡N→∞∑i=1N|ΔAiOAi+1|=12lim⁡N→∞∑i=1Nri|AiAi+1⃗|=12lim⁡N→∞∑i=1Nri|ΓN|N=lim⁡N→∞|ΓN|2lim⁡N→∞1N∑i=1Nri=|Γ|2lim⁡N→∞1N∑i=1NrAi=|Γ|21|Γ|∮ΓrPds=12∮ΓrPds.
Observe that (5.8) implies
(5.9)∮ΓrPds≥2|D(Γ)|.
The second inequality of (5.3) is proved.
By this proof, the first equalities of (5.3) and (5.4) hold if and only if Γ is a circle and the point O is the central of the circle; the second equalities hold if and only if Ω[O,Γ]=D(Γ). This completes the proof.Definition 5.6.
Theorem5.5 may be explained as follows: the point O denotes the central of earth; P is a satellite and Γ denotes the moving trajectory of satellites; M(rP,Γ) is the function mean of rP on Γ and M(ρP,Γ) is the function mean of ρP on Γ.Definition 5.7.
For theS{O,Γ,P1,P2}, which denotes the satellite system of two points with a central, in other words, there are only two satellites on same trajectory, we will discuss in another  paper.



---

*Source: 909261-2011-05-03.xml*