Isoperimetric inequalities in surround system and space science

The gravity is an essential attribute of any physical matter. Therefore, the study of gravity has great theoretical significance and extensive application value.

The theory of satellite is important in space science. In [1–4], the authors systematically studied the theory of satellite and obtained some interesting results. In particular, in [2], the authors defined the centered 2-surround system, established several geometric inequalities for the centered 2-surround system under the proper hypotheses, and illustrated the background of the centered 2-surround system in space science.

It is well known that the Moon is a satellite of the Earth. In space science, we are concerned with the gravity of the Moon since the gravity may be disastrous causing tsunami and tidal wave, etc.

In this paper, we first define the mean central distance \(\bar{r}_{P}\) of a centered 2-surround system \(S^{(2)} \{ P,\varGamma ,l \}\). Next, we study the boundary curve of the l-central regions and the properties of the asymptotic system and establish several identities and inequalities involving the centered 2-surround system. Next, we prove an isoperimetric inequality in the centered 2-surround system. Finally, we demonstrate the application of our results in space science and obtain the best lower bounds of the mean λ-gravity norm \(\overline {\Vert {\mathbf{F}}_{\lambda} ( \varGamma ,P )\Vert }\).

A large number of algebraic, functional analysis, differential equation, convex geometry, physics, computer, and inequality theories are used in this paper. The proofs of our results are both interesting and difficult, as well as which are depend on our previous work. Some of our proof methods can also be found in the references of this paper, such as [1–3].

We continue to use the notation of the references [1–3].

We begin by recalling some of the basic concepts and preliminary results of [1–3].

We remark here that, for the Jordan closed curve, we have the following Jordan theorem [3]: An arbitrary Jordan closed curve must divide a plane into two regions, and one of the regions is bounded and the another is unbounded. The bounded region is called the interior and the another is called the outside of the Jordan closed curve.

We remark here that if \(l=0\), then \(\bar{r}_{P}=\|A-P\|\). This is another geometrical meaning of \(\bar{r}_{P}\), which has applications in space science; see Section 5.

As pointed out in [13], the theory of inequalities plays an important role in all the fields of mathematics. The concept of mean is the most prominent in the theory, and the p-power mean is the crucial one. The references [8–13] studied the sharp bounds of the p-power mean.

In the convex geometry, a large number of isoperimetric inequalities similar to (4) was obtained [14–16]. Recently, we obtained some new isoperimetric inequalities in the surround system [1–3].

Convexity and concavity are essential attributes of any real-variable function, their research and applications are important topics in mathematics and, in particular, the convex analysis [18].

For the centered surround system \(S^{(2)} \{P,\varGamma \}\), we may think of the point P as the center of the Earth and Γ as the orbit of a satellite A (such as the Moon or an artificial Earth satellite). This is the significance of the centered surround system \(S^{(2)} \{P,\varGamma \}\) in the theory of satellite.

From (15) in Section 3 we know that the centered surround system \(S^{(2)} \{P,\varGamma \}\) exists for any smooth and convex Jordan closed curve Γ.

Theorem 1 implies the following interesting corollary.

In Section 5, we will demonstrate the applications of Corollary 1 in space science and establish an isoperimetric inequality involving the λ-gravity of the Moon to the Earth.

In order to prove Theorem 1, we need some preliminaries involving the centered 2-surround system.

3.1 Boundary curve of the l-central region

In the definition of the centered 2-surround system \(S^{(2)} \{ P,\varGamma ,l \}\), an important assumption is that the l-central region \(D(\varGamma ,l)\) is nonempty, that is, the boundary curve \(\partial D(\varGamma ,l)\) of the l-central region \(D(\varGamma ,l)\) is a Jordan closed curve. Unfortunately, the l-central region \(D(\varGamma ,l)\) may be empty. For example, let Γ be a regular triangle of side length 3, then \(D(\varGamma ,4)=\emptyset\), where ∅ denote the empty set, see [2].

Since

$$ \lim_{l\rightarrow0}{D(\varGamma ,l)}=D(\varGamma )\ne\emptyset, $$

(15)

there exists \(\varepsilon\in(0,|\varGamma |/2)\) such that, for any \(l\in(0,\varepsilon )\), we have \(D(\varGamma ,l)\ne\emptyset\).

On the other hand, in [2], the following statement is proved (see Lemmas 2.1 and 2.3 in [2]): Let Γ be a smooth and convex Jordan closed curve. Then \(D(\varGamma ,l)\neq\emptyset\) for all \(l\in (0,|\varGamma |/2)\) if and only if Γ is a central symmetric curve.

According to this result, we know that if Γ is an ellipse, which is a central symmetric curve, then the l-central region \(D(\varGamma ,l)\) is nonempty. In space science, the orbit of a satellite is an ellipse, and P in \(S^{(2)} \{P,\varGamma ,l \}\) is one of the focuses of the ellipse [4]. Therefore, the centered 2-surround system \(S^{(2)} \{P,\varGamma ,l \}\) is of great application value in the theory of satellite.

Based on the definition of the l-central region \(D(\varGamma ,l)\), we know that the boundary curve \(\partial D(\varGamma ,l)\) of the l-central region \(D(\varGamma ,l)\) is the envelope curve of the family of straight line \(AA_{+}\), that is, for any point \(x\mathbf {i}+y\mathbf{j}\in\partial D(\varGamma ,l)\), there exists a line \(AA_{+}\) such that \(AA_{+}\) is tangent to \(\partial D(\varGamma ,l)\) at the point \(x\mathbf{i}+y\mathbf{j}\). Hence, the point \(x\mathbf{i}+y\mathbf{j}\) must satisfy the equation

$$ \det \left [ \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} x & y & 1 \\ x(t_{A}) & y(t_{A}) & 1 \\ x(t_{A}+l) & y(t_{A}+l) & 1 \end{array}\displaystyle \right ]=0 $$

(16)

and the differential equation

$$ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac {y(t_{A}+l)-y(t_{A})}{x(t_{A}+l)-x(t_{A})}. $$

(17)

Eliminating the parameter \(t_{A}\) from (16) and (17), we can obtain the equation of the boundary curve \(\partial D(\varGamma ,l)\); see the following Propositions 1 and 2.

Proposition 1

Let Γ be a unit circle, that is,

$$\varGamma \triangleq \bigl\{ (x,y)| x=\cos t, y=\sin t, t\in\mathbb{R} \bigr\} , $$

where t is the natural parameter. Then the equation of the boundary curve \(\partial D(\varGamma ,l)\) is

$$ \partial D(\varGamma ,l)\mbox{: }x^{2}+y^{2}=\cos^{2} \frac{l}{2}, \quad \forall l\text{: }0< l< \pi. $$

(18)

Proof

Indeed, (16) and (17) can be rewritten as

$$ \left \{ \textstyle\begin{array}{l} x\cos (t_{A}+\frac{l}{2} )+y\sin (t_{A}+\frac{l}{2} )=\cos\frac{l}{2}, \\ \mathrm{d}x\cos (t_{A}+\frac{l}{2} )+\mathrm{d}y\sin (t_{A}+\frac{l}{2} )=0. \end{array}\displaystyle \right . $$

(19)

Eliminating the parameter \(t_{A}\) from (19), we obtain that

$$ (x\,\mathrm{d}y-y\,\mathrm{d}x )^{2}= \bigl[(\mathrm{d}x)^{2}+( \mathrm{d}y)^{2} \bigr]\cos^{2}\frac{l}{2}. $$

(20)

Setting

$$x=\rho\cos\theta,\qquad y=\rho\sin\theta,\quad \theta\in\mathbb{R}, $$

and substituting them into (20), we get

$$ \rho^{4}= \biggl[ \biggl(\frac{\mathrm{d}\rho}{\mathrm{d}\theta} \biggr)^{2}+ \rho^{2} \biggr] \cos^{2}\frac{l}{2}\quad \Leftrightarrow\quad \rho=\cos\frac{l}{2}\vee\rho=\cos\frac {l}{2} \sec(\theta+C). $$

Since

$$\rho=\cos\frac{l}{2}\sec(\theta+C) $$

is the equation of the family of straight line \(AA_{+}\), the equation of \(\partial D(\varGamma ,l)\) is

$$\rho=\cos\frac{l}{2}\quad \Leftrightarrow\quad x^{2}+y^{2}= \cos^{2}\frac{l}{2}, $$

that is, (18) holds. This ends the proof. □

We can also find the equation of \(\partial D(\varGamma ,l)\) if Γ is piecewise smooth.

Proposition 2

Let \(\varGamma _{N}\triangleq \{ A_{1,}A_{2},\ldots,A_{N} \}\), where \(N\geqslant3\), be a convex polygon [17], and let

$$0< l\leqslant\frac{1}{2}\min_{1\leqslant i\leqslant N} \bigl\{ \|A_{i+1}-A_{i}\| \bigr\} . $$

Then we have

$$ \bigl\vert D(\varGamma _{N})\bigr\vert -\bigl\vert D( \varGamma _{N},l)\bigr\vert =\frac{1}{6}l^{2}\sum _{i=1}^{N}\sin \angle A_{i-1}A_{i}A_{i+1} \leqslant\frac{1}{6}Nl^{2}\sin\frac{2\pi}{N}. $$

(21)

Equality in (21) holds if and only if

$$\angle A_{i-1}A_{i}A_{i+1}=\frac{N-2}{N}\pi,\quad i=1,2,\ldots,N, $$

where \(A_{0}\triangleq A_{N}\) and \(A_{N+1}\triangleq A_{1}\).

Proof

Notice that

$$0< l\leqslant\min_{1\leqslant i\leqslant N}\frac{1}{2} \bigl\{ \Vert A_{i+1}-A_{i}\Vert \bigr\} \quad \Rightarrow\quad 0< l< \frac {|\varGamma _{N}|}{2} \text{ and } D(\varGamma _{N},l) \ne\phi. $$

Consider the regular region \(\widehat{A_{j-1}A_{j}A_{j+1}}\). Let the rays \(A_{i}A_{i-1}\) and \(A_{i}A_{i+1} \) be tangent to \(\partial D(\varGamma _{N},l)\) at the points \(T_{i}\) and \(T_{i}'\), respectively, and let

$$\widetilde{T_{i}T_{i}'}\subset\partial D( \varGamma _{N},l) \quad \text{for } i=1,2,\ldots,N. $$

Then we have

$$\partial D(\varGamma _{N},l)=\sum_{i=1}^{N} \bigl[ T_{i-1}'T_{i} \bigr] +\sum _{i=1}^{N}\widetilde{T_{i}T_{i}'} $$

and

$$ \bigl\vert D(\varGamma _{N})\bigr\vert -\bigl\vert D( \varGamma _{N},l)\bigr\vert =\sum_{i=1}^{N} \bigl\vert D \bigl( [A_{i}T_{i}]+ \bigl[ A_{i}T_{i}' \bigr] +\widetilde{T_{i}T_{i}'} \bigr) \bigr\vert . $$

(22)

For any

$$A\in A_{i}A_{i+1},\qquad B\in A_{i}A_{i-1} \quad \text{and} \quad \angle A_{i-1}A_{i}A_{i+1}=2 \alpha\in(0,\pi), $$

let the corresponding coordinates of A and B be

$$A \biggl( \biggl( \frac{l}{2}-t \biggr) \cos\alpha, \biggl( \frac {l}{2}-t \biggr) \sin\alpha \biggr) \quad \text{and}\quad B \biggl( \biggl( \frac{l}{2}+t \biggr) \cos \alpha ,- \biggl( \frac{l}{2}+t \biggr) \sin\alpha \biggr) , $$

respectively, where

$$-\frac{l}{2}\leqslant t\leqslant \frac{l}{2}; $$

see Figure 1.

Then the equation of the curve \(\widetilde{T_{i}T_{i}'}\) is determined by (16) and (17). Hence,

$$ \widetilde{T_{i}T_{i}'}\mbox{: } \frac{y+t\sin\alpha}{x-l/2\cos\alpha }=\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{l}{2t}\tan\alpha, $$

(23)

where

$$\biggl( \frac{l}{2}\cos\alpha,-t\sin\alpha \biggr) $$

is the midpoint of \([AB]\). Eliminating the parameter \(t_{A}\) from (23), we get

$$ \frac{2y}{l\sin\alpha}-\frac{ \mathrm{d}\frac{2x}{l\cos\alpha} }{\mathrm{d}\frac{2y}{l\sin\alpha}}=\frac{ ( \frac{2x}{l\cos\alpha}-1 ) \,\mathrm{d}\frac{2y}{l\sin\alpha} }{\mathrm{d}\frac{2x}{l\cos \alpha}}. $$

(24)

Set

$$x^{*}=\frac{2x}{l\cos\alpha}, \qquad y^{*}=\frac{2y}{l\sin\alpha} \quad \text{and} \quad u=\frac{\mathrm{d}x^{*}}{\mathrm{d}y^{*}}. $$

Then (24) can be rewritten as

$$ y^{*}-u=u^{-1}\bigl(x^{*}-1\bigr) \quad \text{and}\quad \mathrm{d}x^{*}=u\,\mathrm{d}y^{*}. $$

(25)

From (25) we get

$$\mathrm{d}y^{*}-\mathrm{d}u=\frac{u\,\mathrm{d}x^{*}-(x^{*}-1)\,\mathrm{d}u}{u^{2}}= \mathrm{d}y^{*}- \frac{(x^{*}-1)\,\mathrm{d}u}{u^{2}} $$

and

$$ x^{*}=1+u^{2},\qquad y^{*}=2u. $$

(26)

Eliminating the parameter u in (26), we see that the curve \(\widetilde{T_{i}T_{i}'}\) is a parabola whose equation is

$$ \widetilde{T_{i}T_{i}'}\mbox{: }x^{*}=1+ \biggl(\frac{y^{*}}{2} \biggr)^{2}\quad \Leftrightarrow\quad \frac{2x}{l\cos\alpha}=1+ \biggl( \frac{y}{l\sin \alpha} \biggr) ^{2}, \quad -l\sin \alpha\leqslant y\leqslant l\sin\alpha. $$

(27)

Consequently,

$$\begin{aligned} \bigl\vert D \bigl( [A_{i}T_{i}]+ \bigl[ A_{i}T_{i}' \bigr] +\widetilde{T_{i}T_{i}'} \bigr) \bigr\vert =&2 \int_{0}^{l\sin\alpha} \biggl\{ \frac {l\cos\alpha}{2} \biggl[ 1+ \biggl( \frac{y}{l\sin\alpha} \biggr) ^{2} \biggr] -y\cot\alpha \biggr\} \,\mathrm{d}y \\ =&l^{2}\cos\alpha\sin\alpha \int_{0}^{l\sin\alpha} \biggl( \frac{y}{ l\sin\alpha}-1 \biggr) ^{2}\,\mathrm{d} \biggl( \frac{y}{l\sin\alpha }-1 \biggr) \\ =&{\biggl.l^{2}\cos\alpha\sin\alpha\frac{ ( \frac{y}{l\sin\alpha}-1 ) ^{3}}{3}\biggr| _{0}^{l\sin\alpha}} \\ =&\frac{l^{2}}{6}\sin2\alpha \\ =&\frac{l^{2}}{6}\sin\angle A_{i-1}A_{i}A_{i+1}. \end{aligned}$$

From (22), the formula

$$\sum_{i=1}^{N}\angle A_{i-1}A_{i}A_{i+1}=(N-2) \pi, \quad 0< \angle A_{i-1}A_{i}A_{i+1}< \pi, i=1,2, \ldots,N, $$

and the Jensen inequality [19, 20]

$$\frac{1}{N}\sum_{i=1}^{N}\sin \angle A_{i-1}A_{i}A_{i+1}\leqslant\sin \frac{1}{N}\sum_{i=1}^{N} \angle A_{i-1}A_{i}A_{i+1}=\sin \biggl(\pi- \frac{2\pi}{N} \biggr)=\sin\frac {2\pi}{N} $$

we get

$$\begin{aligned} \bigl\vert D(\varGamma _{N})\bigr\vert -\bigl\vert D( \varGamma _{N},l)\bigr\vert =&\sum_{i=1}^{N} \bigl\vert D \bigl( [A_{i}T_{i}]+ \bigl[ A_{i}T_{i}' \bigr] + \widetilde{T_{i}T_{i}'} \bigr) \bigr\vert \\ =& \sum_{i=1}^{N}\frac{l^{2}}{6}\sin \angle A_{i-1}A_{i}A_{i+1} \\ =& \frac{l^{2}}{6} \sum_{i=1}^{N}\sin \angle A_{i-1}A_{i}A_{i+1} \\ \leqslant& \frac{1}{6}Nl^{2}\sin\frac{2\pi}{N}, \end{aligned}$$

that is, (21) holds.

Based on this proof, we see that the equality in (21) holds if and only if

$$\angle A_{i-1}A_{i}A_{i+1}=\frac{N-2}{N}\pi, \quad i=1,2,\ldots,N. $$

The proof is completed. □

For example, if \(N=4\), \(\varGamma _{4}\) is a square of side length 2 and \(l=1\), then, by Proposition 2, we have

$$D(\varGamma _{N},l) \ne\phi \quad \text{and}\quad \bigl\vert D( \varGamma _{N})\bigr\vert -\bigl\vert D(\varGamma _{N},l)\bigr\vert =\frac{1}{6}Nl^{2}\sin\frac{2\pi}{N}= \frac{2}{3}. $$

3.3 Associated identities and inequalities

In order to prove Theorem 1, we need to establish several identities and inequalities involving the centered 2-surround system as follows.

Lemma 4

Let \(S^{(2)} \{P,\varGamma ,l \}\) be a centered 2-surround system. Then we have the following identity:

$$ \oint_{\varGamma }\angle A_{-}PA_{+}=2l\pi. $$

(31)

Proof

This proof is similar to that of Lemma 2.13 in [2].

We need the following definition:

$$A_{i}=A_{j}\quad \Leftrightarrow\quad i=j\quad ( \operatorname{mod} N_{n}). $$

Consider the asymptotic system \(S^{(2)} \{ P,\varGamma _{N_{n}},\frac {k_{n}}{N_{n}}|\varGamma _{N_{n}}| \}\). By Lemmas 1 and 2 we have that \(P\in D(\varGamma _{N_{n}})\) if n is sufficiently large. By Lemmas 1, 2, and 3 and by the identity

$$\sum_{i=1}^{N_{n}}\angle A_{i}PA_{i+1}=2 \pi $$

we get

$$\begin{aligned} \frac{1}{|\varGamma |} \oint_{\varGamma }\angle A_{-}PA_{+} =&\lim _{n\rightarrow \infty}\frac{1}{N_{n}}\sum_{i=1}^{N_{n}} \angle A_{i}PA_{i+k_{n}} \\ =&\lim_{n\rightarrow \infty}\frac{1}{N_{n}}\sum _{i=1}^{N_{n}}\sum_{j=0}^{k_{n}-1} \angle A_{i+j}PA_{i+1+j} \\ =&\lim_{n\rightarrow \infty}\frac{1}{N_{n}}\sum _{j=0}^{k_{n}-1}\sum_{i=1}^{N_{n}} \angle A_{i+j}PA_{i+1+j} \\ =&\lim_{n\rightarrow \infty}\frac{1}{N_{n}}\sum _{j=0}^{k_{n}-1} 2\pi \\ =&\lim_{n\rightarrow\infty}\frac{2k_{n}\pi}{N_{n}} \\ =&\frac{2l\pi}{|\varGamma |}. \end{aligned}$$

The proof of Lemma 4 is completed. □

Lemma 5

(see Lemma 2.7 in [2])

Let \(S^{(2)} \{P,\varGamma ,l \}\) be a centered 2-surround system. Then we have the following inequality:

$$ \sqrt{\frac{1}{|\varGamma |} \oint_{\varGamma }\|A_{+}-A\|^{2}}\leqslant \frac {|\varGamma |}{\pi}\sin\frac{l\pi}{|\varGamma |}. $$

(32)

Equality in (32) holds if Γ is a circle in \(\mathbb{R}^{2}\).

Lemma 6

Let \(0<\theta<{\pi}/{2}\). Then the inequality

$$ \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta}\,\mathrm{d}t\geqslant1 $$

(33)

holds if and only if \(0<\theta\leqslant{\eta}/{2}\), where \(\eta =1.7571802619873076 \ldots\) is the unique real root of equation (13).

Proof

Using the formula

$$ \int\sqrt{t^{2}+a^{2}}\,\mathrm{d}t=\frac{1}{2} \bigl[t\sqrt{t^{2}+a^{2}}+a^{2}\ln \bigl(t+ \sqrt{t^{2}+a^{2}}\bigr) \bigr]+C, $$

(34)

we get

$$\begin{aligned} \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta}\,\mathrm{d}t =&\frac{1}{2} \bigl[\sqrt {1+\cot^{2}\theta}+ \cot^{2}\theta\ln\bigl(1+\sqrt{1+\cot^{2}\theta}\bigr) \bigr] -\frac {1}{2}\cot^{2}\theta\ln\cot\theta \\ =&\frac{1}{2} \bigl[\csc\theta+\cot^{2}\theta\ln(1+\csc\theta) \bigr] -\frac{1}{2}\cot^{2}\theta\ln\cot\theta \\ =&\frac{1}{2} \biggl(\csc\theta+\cot^{2}\theta\ln \frac{1+\csc\theta}{\cot \theta} \biggr) \\ =&\frac{1}{2} \bigl[\csc\theta+\cot^{2} \theta \ln (\tan\theta+ \sec \theta ) \bigr], \end{aligned}$$

that is,

$$ \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta}\,\mathrm{d}t=\frac{1}{2} \bigl[\csc \theta+\cot^{2} \theta \ln (\tan\theta+\sec \theta ) \bigr]. $$

(35)

Consider the auxiliary function

$$\begin{aligned}& \varphi_{*}: \biggl(0,\frac{\pi}{2} \biggr)\rightarrow \mathbb{R}, \\& \varphi_{*}(\theta)= \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta }\,\mathrm{d}t-1=\frac{1}{2} \bigl[\csc\theta+\cot^{2} \theta\ln(\tan \theta+\sec \theta)\bigr]-1. \end{aligned}$$

The graph of the function \(\varphi_{*}\) is depicted in Figures 2 and 3.

Inequality (33) can be rewritten as

$$\varphi_{*}(\theta)\geqslant0, \quad \theta\in \biggl(0, \frac{\pi}{2} \biggr). $$

By means of the Mathematica software we know that the equation

$$\frac{\mathrm{d}\varphi_{*}}{\mathrm{d} \theta}=0, \quad \theta\in (0, {\pi}/{2} ), $$

has no real roots and \({\mathrm{d}\varphi_{*}}/{\mathrm{d} \theta }<0\). Hence, the function \(\varphi_{*}\) is decreasing. The solution of the inequality

$$\varphi_{*}(\theta)\geqslant0,\quad \theta\in \biggl(0, \frac{\pi}{2} \biggr), $$

is

$$0< \theta\leqslant0.8785901309936538\ldots=\frac{\eta}{2}, $$

where \(\eta=1.7571802619873076 \ldots\) is the unique real root of equation (13). This ends the proof. □

Lemma 7

Let \(S^{(2)} \{P,\varGamma ,l \}\) be a centered 2-surround system. Then the inequality

$$ \bar{r}_{P}\leqslant\frac{\|A_{+}-A\|}{2} \int_{0}^{1}\sqrt{t^{2}+\cot ^{2}\frac{\angle APA_{+}}{2}}\,\mathrm{d}t, \quad \forall A\in \varGamma , $$

(36)

holds if and only if

$$ 0< \angle APA_{+}\leqslant\eta,\quad \forall A\in \varGamma , $$

(37)

where \(\eta=1.7571802619873076 \ldots\) is the unique real root of equation (13). Equality in (36) holds if and only if \(P'\) is the midpoint of the closed straight line segment \([AA_{+}]\).

Proof

The relevant calculations in the proof are dependent on the Mathematica software since these calculations are very complex.

Let

$$A=x_{A} \mathbf{i}, \qquad A_{+}=x_{A_{+}} \mathbf{i},\qquad M=x\mathbf{i},\quad x_{A} \leqslant x \leqslant x_{A_{+}}, \qquad P=r\mathbf{j}, \qquad P'=0\mathbf {i}+0 \mathbf{j}, $$

where \(r\triangleq r_{P}>0\), and let

$$\angle APA_{+}=2\theta,\qquad \alpha\triangleq\angle{APP'}=- \arctan\frac {x_{A}}{r},\qquad \beta\triangleq\angle{A_{+}PP'}= \arctan\frac{x_{A_{+}}}{r}. $$

Then

$$\begin{aligned}& \alpha, \beta\in \biggl(-\frac{\pi}{2},\frac{\pi}{2} \biggr),\qquad \alpha +\beta=2\theta, \quad \theta\in \biggl(0,\frac{\pi}{2} \biggr),\qquad \|M-P\|=\sqrt{x^{2}+r^{2}}, \\& \|A_{+}-A\|=x_{A_{+}}-x_{A},\qquad r= \frac{\|A_{+}-A\|}{\tan\alpha+\tan\beta }=\frac{\|A_{+}-A\|\cos\alpha \cos\beta}{\sin2\theta}, \end{aligned}$$

and

$$\begin{aligned} \bar{r}_{P} =& \frac{1}{\|A_{+}-A\|} \int_{[AA_{+}]}\| M-P\| \\ =& \frac{1}{\|A_{+}-A\|} \int_{x_{A}}^{x_{A_{+}}}\sqrt {x^{2}+r^{2}} \,\mathrm{d}x \\ =& \frac{1}{\|A_{+}-A\|} \int_{-r\tan\alpha}^{r\tan\beta} \sqrt{x^{2}+r^{2}} \,\mathrm{d}x \\ =& \frac{1}{\|A_{+}-A\|} \int_{\frac{\|A_{+}-A\|\sin\alpha \cos\beta}{\sin2\theta}}^{\frac{\|A_{+}-A\|\cos\alpha\sin\beta}{\sin 2\theta}}\sqrt{x^{2}+ \biggl( \frac{\|A_{+}-A\|\cos\alpha\cos\beta}{\sin 2\theta} \biggr)^{2}}\,\mathrm{d}x \\ =& \frac{1}{\|A_{+}-A\|} \biggl(\frac{\|A_{+}-A\|\cos\alpha \cos\beta}{\sin 2\theta} \biggr)^{2} \int_{-\tan\alpha}^{\tan\beta}\sqrt{t^{2}+1}\,\mathrm{d}t \\ =& \frac{\|A_{+}-A\| }{\sin^{2} 2\theta(1+\tan^{2}\alpha)(1+\tan^{2}\beta)} \int_{-\tan \alpha}^{\tan\beta}\sqrt{t^{2}+1}\,\mathrm{d}t, \end{aligned}$$

where

$$x={\frac{\|A_{+}-A\|\cos\alpha\cos{\beta}}{\sin2\theta}}t. $$

Hence,

$$ \bar{r}_{P}= \frac{\|A_{+}-A\| }{\sin^{2} 2\theta(1+\tan^{2}\alpha)(1+\tan^{2}\beta)} \int_{-\tan \alpha}^{\tan\beta}\sqrt{t^{2}+1}\,\mathrm{d}t. $$

(38)

By (38) we see that inequality (36) can be rewritten as

$$ \frac{1 }{(1+\tan^{2}\alpha)(1+\tan^{2}\beta)} \int_{-\tan \alpha}^{\tan\beta}\sqrt{t^{2}+1}\,\mathrm{d}t \leqslant \frac{\sin^{2} 2\theta}{2} \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta}\,\mathrm{d}t. $$

(39)

Equality in (39) holds if \(\alpha=\beta\).

By the symmetry we can further assume that \(\beta\geqslant\alpha\).

For any fixed θ, set

$$\frac{\beta-\alpha}{2}=\omega. $$

Then

$$ \alpha=\alpha(\omega)=\theta-\omega,\qquad \frac{\mathrm{d}\alpha}{\mathrm{d}\omega}=-1, \qquad \beta=\beta(\omega)=\theta+\omega, \qquad \frac{\mathrm{d}\beta}{\mathrm{d}\omega}=1, $$

(40)

and

$$ \omega\in \biggl[0,\frac{\pi}{2}-\theta \biggr)\quad \Leftrightarrow\quad \cos\omega \in (\sin\theta, 1 ]. $$

(41)

Now we define two auxiliary functions:

$$\varphi: \biggl[0,\frac{\pi}{2}-\theta \biggr)\rightarrow\mathbb {R}, \qquad \varphi( \omega)\triangleq\frac{1 }{(1+\tan^{2}\alpha)(1+\tan^{2}\beta )} \int_{-\tan \alpha}^{\tan\beta}\sqrt{t^{2}+1}\,\mathrm{d}t $$

and

$$f: \biggl[0,\frac{\pi}{2}-\theta \biggr)\rightarrow\mathbb{R}, \qquad f(\omega ) \triangleq \int_{-\tan \alpha}^{\tan\beta}\sqrt{t^{2}+1}\,\mathrm{d}t- \bigl(1+\tan^{2}\alpha\bigr) \bigl(1+\tan ^{2}\beta\bigr) \varphi(0). $$

Since equality in (39) holds if \(\alpha=\beta\Leftrightarrow \omega=0\), by (35) we see that

$$ \varphi(0)=\frac{\sin^{2} 2\theta}{2} \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta}\,\mathrm{d}t=\frac{\sin^{2} 2\theta}{4} \bigl[\csc\theta+\cot^{2} \theta \ln (\tan\theta+\sec \theta ) \bigr]. $$

(42)

By (42) we know that inequality (39) can be rewritten as

$$ f(\omega)\leqslant0, \quad \forall \omega\in \biggl[0, \frac{\pi}{2}-\theta \biggr). $$

(43)

Next, we prove that if (37) holds, then (43) holds, that is, (36) holds.

By means of the Mathematica software and (40) we can get

$$\frac{\cos^{2}\alpha+\cos^{2}\beta+\cos\alpha\cos\beta}{\cos\alpha+\cos\beta }=\frac{4\cos^{2}\theta-1}{2\cos\theta}\cos\omega+ \frac{\sin^{2}\theta}{2\cos\theta\cos\omega} $$

and

$$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}\omega} =&\sqrt{\tan^{2}\beta+1} \sec^{2}\beta \frac{\mathrm{d}\beta}{ \mathrm{d}\omega}-\sqrt{\tan^{2}\alpha+1}\bigl(-\sec^{2} \alpha\bigr)\frac {\mathrm{d}\alpha}{\mathrm{d}\omega} \\ &{}- \biggl[ 2\tan\alpha\sec^{2}\alpha\bigl(1+\tan^{2}\beta \bigr)\varphi(0)\frac {\mathrm{d}\alpha}{\mathrm{d}\omega} + 2\tan\beta\sec^{2}\beta\bigl(1+ \tan^{2}\alpha\bigr)\varphi(0)\frac{\mathrm{d}\beta}{\mathrm{d}\omega} \biggr] \\ =&\sec^{3}\beta-\sec^{3}\alpha-2\sec^{2}\alpha \sec^{2}\beta(\tan\beta-\tan\alpha)\varphi(0) \\ =&(\tan\beta-\tan\alpha) \biggl[(\tan\beta+\tan\alpha) \frac {\sec^{3}\beta-\sec^{3}\alpha}{ \sec^{2}\beta-\sec^{2}\alpha}-2 \sec^{2}\alpha \sec^{2}\beta\varphi(0) \biggr] \\ =& \frac{\sin2\omega}{\cos\beta\cos\alpha} \biggl[\frac {\sin2\theta}{\cos\beta\cos\alpha} \frac{\sec^{2}\beta+\sec^{2}\alpha+\sec\beta \sec\alpha}{ \sec \beta+\sec\alpha}-2 \sec^{2}\alpha\sec^{2}\beta\varphi(0) \biggr] \\ =&\sin2\omega\sec^{3}\alpha\sec^{3} \beta \biggl[\sin 2 \theta \frac{\cos^{2}\beta+\cos^{2}\alpha+\cos\beta \cos\alpha}{\cos\beta+\cos\alpha}-2\varphi(0) \biggr] \\ =&\sin2\omega\sec^{3}\alpha\sec^{3} \beta \biggl[\sin 2 \theta \biggl(\frac{4\cos^{2}\theta-1}{2\cos\theta}\cos\omega +\frac{\sin^{2}\theta}{2\cos\theta\cos\omega} \biggr)-2 \varphi(0) \biggr] \\ =&\sin2\omega\sec^{3}\alpha\sec^{3} \beta \biggl\{ \sin \theta \biggl[\bigl(4\cos^{2}\theta-1\bigr) \cos\omega+ \frac{\sin^{2}\theta}{\cos\omega} \biggr]-2\varphi(0) \biggr\} \\ =& \sin2\omega\sec^{3}\alpha\sec^{3} \beta g(\omega), \end{aligned}$$

that is,

$$ \frac{\mathrm{d}f}{\mathrm{d}\omega}=\sin2\omega \sec^{3}\alpha \sec^{3} \beta g(\omega), $$

(44)

where

$$g: \left[0,\frac{\pi}{2}-\theta \right)\rightarrow\mathbb{R},\qquad g(\omega ) \triangleq\sin\theta \biggl[\bigl(4\cos^{2}\theta-1\bigr) \cos\omega+ \frac{\sin^{2}\theta}{\cos\omega} \biggr]-2\varphi(0). $$

Since

$$g_{*}:(\sin\theta,1]\rightarrow\mathbb{R}, \qquad g_{*}( \xi)\triangleq\sin \theta \biggl[\bigl(4\cos^{2}\theta-1\bigr) \xi+ \frac{\sin^{2}\theta}{\xi} \biggr]-2\varphi(0) $$

is a convex function, that is,

$$\frac{\mathrm{d}^{2}g_{*}}{\mathrm{d}\xi^{2}}=\frac{2\sin^{3}\theta }{\xi^{3}}>0,\quad \forall \xi\in(\sin\theta,1] $$

and

$$g(\omega)=g_{*}(\cos\xi), $$

we have

$$ g_{*}(\xi)\leqslant \max\bigl\{ g_{*}(\sin \theta),g_{*}(1)\bigr\} \quad \Leftrightarrow\quad g(\omega)\leqslant \max \biggl\{ g (0),g \biggl(\frac{\pi}{2}-\theta \biggr) \biggr\} . $$

(45)

Now we prove that

$$ g (0)\leqslant 0 \quad \text{and} \quad g \biggl( \frac{\pi}{2}-\theta \biggr)\leqslant 0. $$

(46)

Consider the auxiliary function:

$$g^{*}: \biggl(0, \frac{\pi}{2} \biggr)\rightarrow\mathbb{R}, \qquad g^{*}(\theta)=\csc \theta-2\cot^{2}\theta\ln(\tan \theta+\sec \theta). $$

The graph of the function \(g^{*}\) is depicted in Figures 4 and 5.

By (42) we get

$$\begin{aligned} g(0) =&\sin\theta \bigl[\bigl(4\cos^{2}\theta-1\bigr) + \sin^{2}\theta \bigr]-2\varphi(0) \\ =& 3\sin\theta\cos^{2}\theta-\frac{\sin^{2} 2\theta}{2} \bigl[\csc \theta+ \cot^{2}\theta\ln(\tan\theta+\sec \theta) \bigr] \\ =& 3\sin\theta\cos^{2}\theta-2\sin^{2}\theta \cos^{2}\theta \bigl[\csc \theta+\cot^{2}\theta\ln(\tan \theta+\sec\theta) \bigr] \\ =& \sin^{2}\theta\cos^{2}\theta \bigl[\csc\theta-2 \cot^{2}\theta\ln(\tan \theta+\sec \theta) \bigr] \\ =& \sin^{2}\theta\cos^{2}\theta g^{*}(\theta), \end{aligned}$$

that is,

$$ g (0)=\sin^{2}\theta\cos^{2}\theta g^{*}(\theta). $$

(47)

By means of the Mathematica software we get

$$\begin{aligned} \frac{\mathrm{d}g^{*}}{\mathrm{d} \theta} =&-\cot\theta\csc\theta +4\cot\theta\csc^{2}\theta \ln(\tan \theta+\sec \theta) \\ &{}-\frac{2\cot^{2}\theta(\sec^{2}\theta+\sec\theta\tan \theta)}{\sec\theta+\tan\theta}. \end{aligned}$$

The equation \({\mathrm{d}g^{*}}/{\mathrm{d} \theta}=0\) has no real roots in the interval \((0, {\pi}/{2} )\), and \({\mathrm{d}g^{*}}/{\mathrm{d} \theta}>0\). Hence,

$$g^{*}(\theta)\leqslant0\quad \Leftrightarrow\quad 0< \theta\leqslant \theta _{0} \quad \text{and}\quad g^{*}(\theta)\geqslant0 \quad \Leftrightarrow\quad \theta _{0}\leqslant\theta< \frac{\pi}{2}, $$

where \(\theta_{0}\triangleq0.9212996176628999\ldots\) is the root of the equation

$$g^{*}(\theta)=0, \quad \theta\in \biggl(0,\frac{\pi}{2} \biggr). $$

Since (37) holds, and

$$0< \theta\leqslant\frac{\eta}{2}= 0.8785901309936538\ldots< 0.9212996176628999\ldots=\theta_{0}, $$

by (47) we get

$$ g^{*}(\theta)\leqslant 0\quad \Leftrightarrow\quad g(0)=\sin^{2}\theta\cos^{2}\theta g^{*}(\theta) \leqslant 0. $$

(48)

By (42) and Lemma 6 we get

$$\begin{aligned} g \biggl(\frac{\pi}{2}-\theta \biggr) =&\sin\theta \biggl[\bigl(4 \cos^{2}\theta-1\bigr) \cos\omega+\frac{\sin^{2}\theta}{\cos\omega} \biggr]_{\omega=\frac{\pi }{2}-\theta}-2\varphi(0) \\ =&\sin\theta \bigl[\bigl(4\cos^{2}\theta-1\bigr) \sin\theta+ \sin \theta \bigr]-2 \frac{\sin^{2} 2\theta}{2} \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta}\,\mathrm{d}t \\ =&4\sin^{2}\theta\cos^{2}\theta-\sin^{2}2 \theta \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta}\,\mathrm{d}t \\ =&\sin^{2}2\theta-\sin^{2}2\theta \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta}\,\mathrm{d}t \\ =& \sin^{2} 2\theta \biggl(1- \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta}\,\mathrm{d}t \biggr) \\ \leqslant & 0, \end{aligned}$$

that is,

$$ g \biggl( \frac{\pi}{2}-\theta \biggr)\leqslant0. $$

(49)

Combining (48) and (49), we get (46). Hence, (46) is proved.

From (45) and (46) we get

$$ g(\omega)\leqslant \max \biggl\{ g (0),g\biggl(\frac{\pi}{2}- \theta\biggr) \biggr\} \leqslant0, \quad \forall \omega\in \biggl[0,\frac{\pi}{2}- \theta \biggr). $$

(50)

From (44) and (50) we get

$$\frac{\mathrm{d}f}{\mathrm{d}\omega}= \sin2\omega \sec^{3}\alpha\sec^{3} \beta g(\omega)\leqslant0, \quad \forall \omega \in \left[0,\frac{\pi}{2}-\theta \right). $$

Therefore,

$$f(\omega)\leqslant f(0)=0,\quad \forall \omega\in \left[0,\frac{\pi }{2}-\theta \right), $$

which is just inequality (43). Hence, (43) holds, and (36) is proved.

Next, we prove that if inequality (43) holds (i.e., (36) holds), then (37) holds.

Indeed, inequality (43) is equivalent to inequality (39). Since

$$\begin{aligned}& \frac{1 }{(1+\tan^{2}\alpha)(1+\tan^{2}\beta)} \int_{-\tan \alpha}^{\tan\beta}\sqrt{t^{2}+1}\,\mathrm{d}t \\& \quad = \cos^{2}\alpha\cos^{2}\beta \int_{-\tan \alpha}^{\tan\beta}\sqrt{t^{2}+1}\,\mathrm{d}t \\& \quad = \int_{-\tan \alpha}^{\tan\beta}\sqrt{ (\cos\alpha\cos\beta t )^{2}+ (\cos \alpha\cos\beta )^{2}}\,\mathrm{d} (\cos\alpha \cos\beta t ) \\& \quad = \int_{-\cos\beta\sin \alpha}^{\cos\alpha\sin\beta}\sqrt{t^{2}+(\cos\alpha\cos \beta)^{2}}\,\mathrm{d}t, \end{aligned}$$

we can rewrite inequality (39) as

$$ \frac{2}{\sin^{2} 2\theta} \int_{-\cos\beta\sin \alpha}^{\cos\alpha\sin\beta}\sqrt{t^{2}+(\cos\alpha\cos \beta)^{2}}\,\mathrm{d}t\leqslant \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta}\,\mathrm{d}t. $$

(51)

Set

$$\beta\rightarrow\frac{\pi}{2}\quad \Leftrightarrow\quad \alpha=2\theta- \beta \rightarrow2\theta-\frac{\pi}{2} $$

in (51). Then

$$\begin{aligned}& \frac{2}{\sin^{2} 2\theta} \int_{-\cos\beta\sin \alpha}^{\cos\alpha\sin\beta}\sqrt{t^{2}+(\cos\alpha\cos \beta)^{2}}\,\mathrm{d}t\leqslant \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta}\,\mathrm{d}t \\& \quad \Leftrightarrow\quad \frac{2}{\sin^{2} 2\theta} \int_{0}^{\sin2\theta}t\,\mathrm{d}t\leqslant \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta}\,\mathrm{d}t \\& \quad \Leftrightarrow\quad \frac{1}{\sin^{2} 2\theta}t^{2}|_{0}^{\sin2\theta} \leqslant \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta}\,\mathrm{d}t \\& \quad \Leftrightarrow\quad \int_{0}^{1}\sqrt{t^{2}+\cot^{2} \theta}\,\mathrm{d}t\geqslant1, \end{aligned}$$

that is, inequality (33) holds. By Lemma 6 we have

$$0< \angle APA_{+}=2\theta\leqslant\eta, \quad \forall A\in \varGamma ; $$

hence, (37) holds.

Based on this proof, we know that equality in (36) holds if and only if \(P'\) is the midpoint of the closed straight line segment \([AA_{+}]\). This completes the proof of Lemma 7. □

Lemma 8

(see [19, 20])

Let \(E\subset\mathbb{R}^{m}\) be a bounded and closed region (or curve), and let the functions \(f:E\rightarrow\mathbb{R}\) and \(\phi: f(E)\rightarrow\mathbb{R}\) be integrable, where \(f(E)\) is an interval. If \(\phi: f(E)\rightarrow \mathbb{R}\) is a convex function, then we have the following Jensen inequality:

$$ \frac{\int_{E}\phi(f)}{\int_{E}}\geqslant\phi \biggl(\frac{\int_{E} f}{\int_{E} } \biggr). $$

(52)

Corollary 1 is of great significance in space science.

In the solar system, the gravity of the physical matter X to another physical matter P is \(\mathbf{F} (A,P )\), whereas for another galaxy in the universe, the gravity may be \(\mathbf {F}_{\lambda} (A,P )\), where \(\lambda\in(0, 2)\cup(2, +\infty)\). For example, in the black hole of the universe, we conjecture that the gravity is \(\mathbf{F}_{\lambda} (A,P )\) with \(\lambda \in(0, 2)\), P can be regarded as an atomic nucleus of an atom, A can be regarded as an electron of the atom, and Γ can be regarded as the orbit of the electron.

In this section, we will establish a new isoperimetric inequality involving the λ-gravity.

Corollary 1 implies the following interesting result, which is significant in space science.