In order to prove Theorem 1, we need some preliminaries involving the centered 2-surround system.
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
$$D(\varGamma _{N},l) \ne\phi \quad \text{and}\quad \bigl\vert D( \varGamma _{N})\bigr\vert -\bigl\vert D(\varGamma _{N},l)\bigr\vert =\frac{1}{6}Nl^{2}\sin\frac{2\pi}{N}= \frac{2}{3}. $$
Asymptotic system
In the theory of surround system, one of the important concepts is the asymptotic system.
Definition 1
(see [1–3])
Let \(S^{(2)} \{P,\varGamma ,l \}\) be a centered 2-surround system. Suppose that:
-
(i)
\(\{N_{n}\}_{n=1}^{\infty}\) and \(\{k_{n}\}_{n=1}^{\infty}\) are two positive integer sequences, and
$$N_{n} \geqslant3, \qquad 1\leqslant k_{n}< \frac{N_{n}}{2},\qquad \lim_{n\rightarrow \infty}N_{n}=+\infty, \qquad \lim_{n\rightarrow\infty}\frac{k_{n}}{N_{n}}=\frac {l}{|\varGamma |}; $$
-
(ii)
\(\varGamma _{N_{n}}\triangleq \{A_{1},A_{2},\ldots, A_{N_{n}} \}\) is the inscribed \(N_{n}\)-sided polygonal of the closed curve Γ, and
$$\|A_{2}-A_{1}\|=\|A_{3}-A_{2}\| = \cdots= \|A_{N_{n}}-A_{N_{n}-1}\|=\|A_{1}-A_{N_{n}} \|. $$
Then the set
$$S^{(2)} \biggl\{ P,\varGamma _{N_{n}},\frac{k_{n}}{N_{n}}| \varGamma _{N_{n}}| \biggr\} \triangleq \biggl\{ P,\varGamma _{N_{n}}, \frac{k_{n}}{N_{n}}|\varGamma _{N_{n}}| \biggr\} $$
is called an asymptotic system of the system \(S^{(2)} \{P,\varGamma ,l \}\).
The asymptotic system has the properties as follows.
Lemma 1
(see Lemma 2.4 in [2])
If
\(S^{(2)} \{P,\varGamma ,l \}\)
is a centered 2-surround system, then we have
$$ \lim_{n\rightarrow\infty}S^{(2)} \biggl\{ P, \varGamma _{N_{n}},\frac {k_{n}}{N_{n}}|\varGamma _{N_{n}}| \biggr\} =S^{(2)} \{P,\varGamma ,l \}. $$
(28)
Lemma 2
(see Lemma 2.5 in [2])
If
\(S^{(2)} \{P,\varGamma _{N_{n}},\frac{k_{n}}{N_{n}}|\varGamma _{N_{n}}| \}\)
is an asymptotic system of
\(S^{(2)} \{P,\varGamma ,l \}\), then there exists a sequence
$$\bigl\{ A_{i_{n}}^{(n)} \bigr\} _{n=0}^{\infty} \subseteq \bigl\{ A_{i}^{(n)} \bigr\} _{n=0}^{\infty}, \quad i_{n}\in\{1,2,\ldots,N_{n}\}, $$
such that
$$ \lim_{n\rightarrow\infty} A_{i_{n}}^{(n)}=A, \qquad \lim_{n\rightarrow \infty}A_{i_{n}+k_{n}}^{(n)}=A_{+} \quad \textit{and} \quad \lim_{n\rightarrow \infty}A_{i_{n}-k_{n}}^{(n)}=A_{-}. $$
(29)
Lemma 3
(see Lemma 2.6 in [2])
Let the image
\(\varGamma =\gamma([a,b])\)
of a continuous function
\(\gamma:[a,b]\rightarrow\mathbb{R}^{m}\)
be a smooth curve, and let
\(f:\varGamma \rightarrow\mathbb{R}\)
be a Riemann-integrable function over
Γ. Suppose that
Γ
is partitioned by means of
\(N+1\)
points
$$A_{0},A_{1},\ldots,A_{i-1},A_{i}, \ldots,A_{N}, \quad N\geqslant3, $$
such that
$$\begin{aligned}& A_{i}=\gamma(t_{i}), \quad i=0,1,2,\ldots,N, a=t_{0}< t_{1}< \cdots< t_{N}=b, \\& \lim_{N\rightarrow\infty}(t_{i+1}-t_{i})=0,\quad i=0,1,2,\ldots,N-1, \end{aligned}$$
and
$$\|A_{1}-A_{0}\|=\|A_{2}-A_{1}\|= \cdots=\|A_{i}-A_{i-1}\| =\cdots=\|A_{N}-A_{N-1} \|={|\varGamma _{N}|}/{N}. $$
Then we have
$$ \frac{1}{|\varGamma |} \int_{\varGamma }f=\lim_{N\rightarrow \infty}\frac{1}{N}\sum _{i=0}^{N-1}f(A_{i}). $$
(30)
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)