设$ K $为$ n $维欧氏空间中的闭凸集, 超平面$ H $是$ n $维欧氏空间中的$ n-1 $维子空间.若$ H $与$ K $相交非空, 并把$ K $完全置于一个以$ H $为边界的半空间内, 则称$ H $为$ K $的支撑超平面.当$ K $的任意一对平行支撑超平面间的距离相等时, 称$ K $为常宽凸集, 平行支撑超平面间的距离为$ K $的宽度, 平面常宽凸集的边界为常宽曲线.圆是典型的常宽曲线.

常宽凸集的概念由欧拉首次提出. 1876年, Reuleaux构造了以正三角形三个顶点为圆心, 边长为半径的外接圆弧组成的非圆的常宽曲线, 人们称之为Reuleaux三角形.该方法被推广在奇数边正多边形上得到Reuleaux正多边形.之后, Meissner将Reuleaux三角形推广至三维空间, 构造出两种不同的三维常宽凸集^[1-2]. 1915年, Blaschke与Lebesgue分别证明了平面上等宽度的常宽凸体中Reuleaux三角形面积最小, 这个定理就是著名的Blaschke-Lebesgue定理.对于这个命题, 国内外的数学家们又先后给出了其它巧妙的证明^[1,3-5].数学家十分关注Blaschke-Lebesgue定理的高维情形, 并提出了Blaschke-Lebesgue问题, 即高维等宽度的常宽凸体中谁的体积最小.长久以来, 人们猜测Meissner体是三维Blaschke-Lebesgue问题的解, 但至今未被证明.为此, 数学家不断尝试构造新的常宽凸集并研究其几何性质. 2003年, 潘生亮教授在文献[6]中利用Minkowski支撑函数得到了一类新的光滑的常宽凸集. 2007年, Lachand-Robert与Qudet用数学归纳法给出了一种$ n $维常宽凸集的构造方法^[7]. 2011年, 徐文学博士构造了一类新的偶数边的常宽凸体^[1,8].

本文将讨论如何构造平面常宽曲线.为此, 我们首先定义了一类平面曲线, 称为杠杆轮(见正文中定义3.1), 并利用臂函数(见正文中第3节)给出杠杆轮的参数表示.其次, 我们证明了杠杆轮是平面常宽曲线的一种等价刻画.最后, 我们构造了一类臂函数为三角函数形式的常宽曲线, 并表明广义Reuleaux多边形(即由若干段圆弧围成的平面常宽凸体)是臂函数为分段常函数的一类杠杆轮.在此基础上, 构造了具体的广义Reuleaux多边形, 特别是偶数边的广义Reuleaux多边形.

记$ {{\Bbb R}} $为实数集, $ {{\Bbb R}} ^n $为带有标准内积$ \langle \cdot, \cdot \rangle $的n维欧氏空间, $ S^{n-1} $为$ {{\Bbb R}} ^n $中的单位球面.

定义2.1  设$ K\subset{{\Bbb R}} ^n $是一个有界子集, $ {\bf{u}}\in{S^{n-1}} $, 称

$ H(K, {\bf{u}}) = \left\{{\bf{p}}\mid{{\bf{p}}\in{{\Bbb R}} ^n, \langle{{\bf{u}}, {\bf{p}}}\rangle = \sup\{\langle{{\bf{u}}, {\bf{x}}}\rangle\mid{{\bf{x}}\in{K}}\}}\right\} $

为$ K $在$ {\bf{u}} $方向上的广义支撑超平面.

当$ n = 2 $时, 称$ H(K, {\bf{u}}) $为$ K $在$ {\bf{u}} $方向上的广义支撑线.显然, 当$ K $为有界闭集时

$ H(K, {\bf{u}}) = {\{{\bf{p}}\mid{{\bf{p}}\in{{\Bbb R}} ^n, \langle{{\bf{u}}, {\bf{p}}}\rangle = \max\{\langle{{\bf{u}}, {\bf{x}}}\rangle\mid{{\bf{x}}\in{K}}\}}\}}. $

定义2.2  设$ K\subset{{\Bbb R}} ^n $是一个有界子集.若存在实数$ \omega\geq{0} $, 对任意$ {\bf{u}}\in{S^{n-1}} $, $ H(K, {\bf{u}}) $与$ H(K, -{\bf{u}}) $之间的距离恒为$ \omega $, 则称$ K $是一个广义常宽集, 并称$ \omega $为它的宽度.

一般地, 称$ {{\Bbb R}} ^n $中有非空内部的有界闭凸集为凸体.于是, 若取$ K $为$ {{\Bbb R}} ^n $中的凸体, 则定义2.2就是通常的常宽凸体的定义.

设$ K\subset{{\Bbb R}} ^n $是一个有界子集.若存在同胚映射$ f:{{\Bbb R}} ^n\rightarrow{{\Bbb R}} ^n $使得$ f(K) = B^n $, 其中$ B^n $为$ {{\Bbb R}} ^n $中的单位闭球, 则称$ K $满足球同胚条件.记$ {\cal K}_{B^n} $为满足球同胚条件的有界集的全体.此时, 对$ K $的边界$ \partial K $有$ f(\partial K) = S^{n-1} $.

一般地, 常宽曲线由极值点构成, 即一个常宽曲线与其在任意方向上支撑线的交都为单点集(极值点定义参见文献[9]).因此, 设$ K $是一个平面常宽凸体, 可以构造满射$ \alpha:S^1\rightarrow \partial K $, 使得对给定的$ {\bf{u}}\in S^1 $, $ \{\alpha({\bf{u}})\} = H(K, {\bf{u}})\cap{K} $.记$ c({\bf{{\bf{u}}}}) $为$ S^1 $上以$ {\bf{{\bf{u}}}} $为起点且沿逆时针方向的有向曲线段, 若$ K $是一个Reuleaux多边形, 则上述映射$ \alpha $具有如下两个性质.

(ⅰ) $ \forall {\bf{{\bf{u}}}}\in S^1, \exists c({\bf{{\bf{u}}}}), \lambda\geq{0} $使得对任意$ {\bf{v}}\in c({\bf{{\bf{u}}}}) $, 有

$ \alpha({\bf{v}})-\alpha({\bf{{\bf{u}}}}) = \lambda({\bf{v}}-{\bf{{\bf{u}}}}). $

特别地, 当$ \lambda = 0 $时, 有

$ \alpha\left(c({\bf{{\bf{u}}}})\right) = \{\alpha({\bf{{\bf{u}}}})\}; $

(ⅱ) $ \forall {\bf{{\bf{u}}}}\in S^1 $, $ \omega $为$ K $的宽度, 则$ \alpha({\bf{u}})-\alpha(-{\bf{u}}) = \omega {\bf{u}} $.

下面我们将对局部具有性质(ⅰ)和(ⅱ)的曲线进行研究.

定义3.1  设$ \alpha:S^1\rightarrow{{\Bbb R}} ^2 $是一个连续映射.若$ \alpha $满足

$ \rm(ⅰ) $对任意的$ {\bf{{\bf{u_0}}}}\in S^1 $, 当$ {\bf{u}} $沿着$ c({\bf{{\bf{u_0}}}})\varsubsetneqq S^1 $趋于$ {\bf{u_0}} $时, 极限$ \lim\limits_{{\bf{u}}\rightarrow {\bf{{\bf{u_0}}}}}\frac{\alpha({\bf{u}}) -\alpha({\bf{{\bf{u_0}}}})}{\|{\bf{u}}-{\bf{{\bf{u_0}}}}\|} $存在且

$ \lim\limits_{{\bf{u}}\rightarrow {\bf{{\bf{u_0}}}} }\langle \frac{\alpha({\bf{u}})-\alpha({\bf{{\bf{u_0}}}})}{\|{\bf{u}}-{\bf{{\bf{u_0}}}}\|}, \frac{{\bf{u}}-{\bf{{\bf{u_0}}}}}{\|{\bf{u}}-{\bf{{\bf{u_0}}}}\|}\rangle = \lim\limits_{{\bf{u}}\rightarrow {\bf{{\bf{u_0}}}} }\frac{\|\alpha({\bf{u}})-\alpha({\bf{{\bf{u_0}}}})\|}{\|{\bf{u}}-{\bf{{\bf{u_0}}}}\|}, $

其中$ \| \cdot \| $为矢量的模长;

$ \rm(ⅱ) $$ \exists \omega>0 $使得对任意$ {\bf{u}}\in{S^{1}} $, 有$ \alpha({\bf{u}})-\alpha(-{\bf{u}}) = \omega{\bf{u}} $, 则称$ \alpha(S^{1}) $为杠杆轮, 可以用$ \alpha $表示.称线段$ [\alpha(-{\bf{{\bf{u}}}}), \alpha({\bf{{\bf{u}}}})] $为$ \alpha $在$ {\bf{{\bf{u}}}} $方向上的直径.

注3.1  杠杆轮$ \alpha $上各方向直径长度均为$ \omega $.

设$ \alpha $为杠杆轮, 若任意给定$ {\bf{{\bf{u_0}}}}\in S^1, {\bf{u}}\in c({\bf{{\bf{u_0}}}})\varsubsetneqq S^1 $且$ {\bf{{\bf{u_0}}}}\neq \pm {\bf{u}} $, 记点$ P_{{\bf{{\bf{u_0}}}}, {\bf{u}}} $为过点$ \alpha({\bf{{\bf{u_0}}}}) $, $ \alpha(-{\bf{{\bf{u_0}}}}) $的直线与过点$ \alpha({\bf{u}}) $, $ \alpha(-{\bf{u}}) $的直线的交点, 则$ \exists \lambda_1, \lambda_2\in {{\Bbb R}} $使得

(3.1) $ \begin{equation} P_{{\bf{{\bf{u_0}}}}, {\bf{u}}} = (1-\lambda_1)\alpha({\bf{{\bf{u_0}}}})+\lambda_1\alpha(-{\bf{{\bf{u_0}}}}), P_{{\bf{{\bf{u_0}}}}, {\bf{u}}} = (1-\lambda_2)\alpha({\bf{u}})+\lambda_2\alpha(-{\bf{u}}). \\ \end{equation} $

结合定义3.1中的条件(ⅱ), 可得

(3.2) $ \begin{equation} \alpha({\bf{u}})-\alpha({\bf{{\bf{u_0}}}}) = \omega(\lambda_2 {\bf{u}}-\lambda_1 {\bf{{\bf{u_0}}}}). \end{equation} $

由(3.2)式, 可得

(3.3) $ \begin{equation} \langle \frac{\alpha({\bf{u}})-\alpha({\bf{{\bf{u_0}}}})}{\|{\bf{u}}-{\bf{{\bf{u_0}}}}\|}, \frac{{\bf{u}}-{\bf{{\bf{u_0}}}}}{\|{\bf{u}}-{\bf{{\bf{u_0}}}}\|}\rangle = \frac{\langle \omega(\lambda_2{\bf{u}}-\lambda_1{\bf{{\bf{u_0}}}}), {\bf{u}}-{\bf{{\bf{u_0}}}}\rangle}{\|{\bf{u}}-{\bf{{\bf{u_0}}}}\|^2} = \frac{\omega}{2}(\lambda_1+\lambda_2). \end{equation} $

由$ \alpha $连续和(3.2)式, 当$ {\bf{u}} $沿着$ c({\bf{{\bf{u_0}}}})\varsubsetneqq S^1 $趋于$ {\bf{u_0}} $时, $ \exists \lambda_1({\bf{{\bf{u}}}}), \lambda_2({\bf{{\bf{u}}}}) $使得

$ \begin{eqnarray*} {\bf{0}} & = &\lim\limits_{{\bf{u}}\rightarrow {\bf{{\bf{u_0}}}} }\left(\alpha({\bf{u}}) -\alpha({\bf{{\bf{u_0}}}})\right) = \lim\limits_{{\bf{u}}\rightarrow {\bf{{\bf{u_0}}}}}\omega(\lambda_2({\bf{u}}){\bf{u}} -\lambda_1({\bf{u}}){\bf{{\bf{u_0}}}}) \\ & = &\omega\left(\lim\limits_{{\bf{u}}\rightarrow{\bf{u_0}}}\lambda_2({\bf{{\bf{u}}}}) -\lim\limits_{{\bf{u}}\rightarrow{\bf{u_0}}}\lambda_1({\bf{{\bf{u}}}})\right){\bf{{\bf{u_0}}}}, \end{eqnarray*} $

因此, $ \lim\limits_{{\bf{u}}\rightarrow{\bf{u_0}}}\lambda_1({\bf{{\bf{u}}}}) = \lim\limits_{{\bf{u}}\rightarrow{\bf{u_0}}}\lambda_2({\bf{{\bf{u}}}}) $.

由$ {\bf{u_0}} $的任意性, 对任意的$ {\bf{{\bf{u}}}}\in S^1 $, 当$ {\bf{v}} $沿着$ c({\bf{{\bf{u}}}})\varsubsetneqq S^1 $趋于$ {\bf{u}} $时, 定义

$ \lambda({\bf{{\bf{u}}}}) = \lim\limits_{{\bf{v}}\rightarrow{\bf{u}}}\lambda_i({\bf{{\bf{v}}}}), i = 1, 2 $

为杠杆轮$ \alpha $在$ {\bf{{\bf{u}}}} $方向上的臂函数.一般地, 令$ {\bf{u}}(\theta) = (\cos\theta, \sin\theta), \theta\in[0, 2\pi) $, 记

$ \lambda_{\bf{u}}(\theta) = \lambda({\bf{u}}(\theta)). $

引理3.1  对任意的$ {{\bf{{\bf{u}}}}\in{S^{1}}} $, 臂函数$ \lambda({\bf{{\bf{u}}}}) $具有以下两个性质

$ \rm(ⅰ) $$ \lambda({\bf{{\bf{u}}}})+\lambda(-{\bf{{\bf{u}}}}) = 1; $

$ \rm(ⅱ) $$ \lambda({\bf{{\bf{u}}}})\in[0, 1]. $

证  由定义3.1中的条件(ⅰ)与(3.3)式, 可得

(3.4) $ \begin{equation} \lim\limits_{{\bf{v}}\rightarrow {\bf{{\bf{u}}}} }\frac{\|\alpha({\bf{v}}) -\alpha({\bf{{\bf{u}}}})\|}{\|{\bf{v}}-{\bf{{\bf{u}}}}\|} = \lim\limits_{{\bf{v}}\rightarrow {\bf{{\bf{u}}}} }\langle \frac{\alpha({\bf{v}}) -\alpha({\bf{{\bf{u}}}})}{\|{\bf{v}}-{\bf{{\bf{u}}}}\|}, \frac{{\bf{v}}-{\bf{{\bf{u}}}}}{\|{\bf{v}}-{\bf{{\bf{u}}}}\|}\rangle = \omega\lambda({\bf{{\bf{u}}}}). \end{equation} $

由定义3.1中条件(ⅱ)与(3.4)式, 可得

$ \begin{eqnarray*} \omega\lambda(-{\bf{{\bf{u}}}}) & = &\lim\limits_{-{\bf{v}}\rightarrow -{\bf{{\bf{u}}}}}\langle \frac{\alpha(-{\bf{v}}) -\alpha(-{\bf{{\bf{u}}}})}{\|{\bf{v}}-{\bf{{\bf{u}}}}\|}, \frac{(-{\bf{v}})-(-{\bf{{\bf{u}}}})}{\|{\bf{v}}-{\bf{{\bf{u}}}}\|}\rangle\\ & = &\lim\limits_{{\bf{v}}\rightarrow {\bf{{\bf{u}}}}}\langle \frac{(\alpha({\bf{v}}) -\omega {\bf{v}})-(\alpha({\bf{{\bf{u}}}}) -\omega {\bf{{\bf{u}}}})}{\|{\bf{v}}-{\bf{{\bf{u}}}}\|}, \frac{-({\bf{v}}-{\bf{{\bf{u}}}})}{\|{\bf{v}}-{\bf{{\bf{u}}}}\|}\rangle\\ & = &\lim\limits_{{\bf{v}}\rightarrow {\bf{{\bf{u}}}} } \left(\omega-\langle \frac{\alpha({\bf{v}}) -\alpha({\bf{{\bf{u}}}})}{\|{\bf{v}}-{\bf{{\bf{u}}}}\|}, \frac{{\bf{v}}-{\bf{{\bf{u}}}}}{\|{\bf{v}}-{\bf{{\bf{u}}}}\|}\rangle\right)\\ & = &\omega\left(1-\lambda({\bf{{\bf{u}}}})\right), \end{eqnarray*} $

即$ \lambda({\bf{{\bf{u}}}})+\lambda(-{\bf{{\bf{u}}}}) = 1 $成立.

因为$ \omega>0 $, 由(2.4)式易知$ \lambda({\bf{{\bf{u}}}})\geq0 $.由$ {\bf{{\bf{u}}}} $的任意性, $ \lambda(-{\bf{{\bf{u}}}})\geq0 $, 故$ \lambda({\bf{{\bf{u}}}})\in[0, 1] $.

记$ {\cal L}(\omega) $为平面上直径长度为$ \omega $的杠杆轮的全体.称$ P({\bf{{\bf{u}}}}) = \lim\limits_{{\bf{v}}\rightarrow{\bf{{\bf{u}}}}}P_{{\bf{{\bf{u}}}}, {\bf{v}}} $为杠杆轮$ \alpha $在$ {\bf{{\bf{u}}}} $方向上的支点, 其中$ P_{{\bf{{\bf{u}}}}, {\bf{v}}} $被(2.1)式定义.称线段$ [P({\bf{{\bf{u}}}}), \alpha({\bf{{\bf{u}}}})] $为$ \alpha $在$ {\bf{{\bf{u}}}} $方向上的臂.

注3.2  可以看出, 杠杆轮$ \alpha $在$ {\bf{{\bf{u}}}} $方向上的支点

$ P({\bf{{\bf{u}}}}) = (1-\lambda({\bf{{\bf{u}}}}))\alpha({\bf{{\bf{u}}}}) +\lambda({\bf{{\bf{u}}}})\alpha(-{\bf{{\bf{u}}}}), $

且由引理3.1中的条件(ⅰ)可得$ P({\bf{{\bf{u}}}}) = P(-{\bf{{\bf{u}}}}). $

注3.3  对任意$ {\bf{u}}\in S^1 $, 可见$ \lambda({\bf{{\bf{u}}}}) = \frac{\|\alpha({\bf{{\bf{u}}}}) -P({\bf{{\bf{u}}}})\|}{\|\alpha({\bf{{\bf{u}}}}) -\alpha(-{\bf{{\bf{u}}}})\|} $, 即$ \lambda({\bf{{\bf{u}}}}) $为$ \alpha $在$ {\bf{{\bf{u}}}} $方向上臂与直径的长度之比.

注3.4  设杠杆轮$ \alpha $的直径长度为$ \omega $, $ \alpha $在$ {\bf{{\bf{u}}}} $方向上的臂长$ \|\alpha({\bf{{\bf{u}}}})-P({\bf{{\bf{u}}}})\| = \omega\lambda({\bf{{\bf{u}}}}) $恰好是$ \alpha $在$ \alpha({\bf{{\bf{u}}}}) $处的曲率半径.

定理3.1(杠杆轮的参数方程)  若$ \alpha\in{\cal L}(\omega) $, $ \lambda_{\bf{u}} $为$ \alpha $的臂函数, 则$ \lambda_{\bf{u}} $具有以下两个性质

$ \rm(ⅰ) $$ \lambda_{\bf{u}}(\theta)+\lambda_{\bf{u}}(\theta+\pi) = 1, \theta\in[0, \pi) $;

$ \rm(ⅱ) $$ (\int_{0}^{\pi}\lambda_{\bf{u}}(\theta)\sin\theta{\rm d}\theta, -\int_{0}^{\pi}\lambda_{\bf{u}}(\theta)\cos\theta{\rm d}\theta) = (1, 0) $,

且$ \alpha $的参数方程为

$ \alpha(\theta) = \bigg(-\omega\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi)\sin\varphi{\rm d}\varphi+x_0, \omega\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi)\cos\varphi{\rm d}\varphi+y_0\bigg), $

其中$ \alpha(0) = (x_0, y_0) $.

反过来, 若$ \lambda_{\bf{u}}:[0, 2\pi)\rightarrow [0, 1] $满足性质$ \rm(ⅰ) $和$ \rm(ⅱ) $, 则曲线

$ \alpha(\theta) = \bigg(-\omega\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi)\sin\varphi{\rm d}\varphi, \omega\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi)\cos\varphi{\rm d}\varphi\bigg) $

为杠杆轮.

证   (1) $ \forall \alpha\in{\cal L}(\omega) $, $ \lambda_{\bf{u}} $为$ \alpha $的臂函数.注意到定义3.1中的条件(ⅰ)等价于对任意的$ {\bf{{\bf{u_0}}}}\in S^1, {\bf{u}} $沿着$ c({\bf{{\bf{u_0}}}})\varsubsetneqq S^1 $趋于$ {\bf{u_0}} $, 使得

(3.5) $ \begin{equation} \lim\limits_{{\bf{u}}\rightarrow {\bf{u_0}} }\frac{\alpha({\bf{u}}) -\alpha({\bf{u_0}})}{\|{\bf{u}}-{\bf{u_0}}\|} = \lim\limits_{{\bf{u}}\rightarrow {\bf{u_0}} }\frac{\|\alpha({\bf{u}}) -\alpha({\bf{u_0}})\|}{\|{\bf{u}} -{\bf{u_0}}\|}\lim\limits_{{\bf{u}}\rightarrow {\bf{u_0}}}\frac{{\bf{u}} -{\bf{u_0}}}{\|{\bf{u}}-{\bf{u_0}}\|}. \end{equation} $

结合(3.4)式可得

(3.6) $ \begin{equation} \lim\limits_{{\bf{u}}\rightarrow {\bf{u_0}} }\frac{\alpha({\bf{u}}) -\alpha({\bf{u_0}})}{\|{\bf{u}}-{\bf{u_0}}\|} = \omega\lambda({\bf{u_0}})\lim\limits_{{\bf{u}}\rightarrow {\bf{u_0}} }\frac{{\bf{u}} -{\bf{u_0}}}{\|{\bf{u}}-{\bf{u_0}}\|}. \end{equation} $

令$ {\bf{u_0}} = (\cos\theta, \sin\theta) $, 当$ {\bf{u}} $沿着顺时针方向趋近于$ {\bf{u_0}} $时, 由(3.6)式可得

$ \frac{{\rm d}\alpha(\theta)}{{\rm d}\theta} = (-\omega\lambda_{\bf{u}}(\theta)\sin\theta, \omega\lambda_{\bf{u}}(\theta)\cos\theta). $

当$ \alpha(0) = (x_0, y_0) $时, 由$ {\bf{u_0}} $的任意性可以得到$ \alpha(\theta) $的参数方程

$ \alpha(\theta) = \bigg(-\omega\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi)\sin\varphi{\rm d}\varphi+x_0, \omega\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi)\cos\varphi{\rm d}\varphi+y_0\bigg), $

结合定义3.1中的条件(ⅱ), 则有

$ (1, 0) = \frac{\alpha(0)-\alpha(\pi)}{\omega} = \bigg(\int_{0}^{\pi}\lambda_{\bf{u}}(\varphi)\sin\varphi{\rm d}\varphi, -\int_{0}^{\pi}\lambda_{\bf{u}}(\varphi)\cos\varphi{\rm d}\varphi\bigg). $

结合引理3.1可知臂函数$ \lambda_{\bf{u}} $满足性质(ⅰ)和(ⅱ).

(2) 设函数$ \lambda_{\bf{u}} $满足性质(ⅰ)和(ⅱ), 曲线

$ \alpha(\theta) = \bigg(-\omega\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi)\sin\varphi{\rm d}\varphi, \omega\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi)\cos\varphi{\rm d}\varphi\bigg). $

首先, 证明曲线$ \alpha $连续封闭.

$ \begin{eqnarray*} (1, 0) & = &\bigg(\int_{0}^{\pi}\lambda_{\bf{u}}(\varphi)\sin\varphi{\rm d}\varphi, -\int_{0}^{\pi}\lambda_{\bf{u}}(\varphi)\cos\varphi{\rm d}\varphi\bigg)\\ & = &\int_{\pi}^{0}\lambda_{\bf{u}}(\varphi)(-\sin\varphi, \cos\varphi){\rm d}\varphi \\ & = &\int_{\pi}^{0}(1-\lambda_{\bf{u}}(\varphi+\pi))(-\sin\varphi, \cos\varphi){\rm d}\varphi\\ & = &\int_{2\pi}^{\pi}(1-\lambda_{\bf{u}}(\varphi+\pi))(\sin(\varphi+\pi), -\cos(\varphi+\pi)){\rm d}(\varphi+\pi)\\ & = &(2, 0)-\int_{2\pi}^{\pi}\lambda_{\bf{u}}(\varphi+\pi)(\sin(\varphi+\pi), -\cos(\varphi+\pi)){\rm d}(\varphi+\pi). \end{eqnarray*} $

即

$ (-1, 0) = \bigg(\int_{\pi}^{2\pi}\lambda_{\bf{u}}(\varphi)\sin\varphi{\rm d}\varphi, -\int_{\pi}^{2\pi}\lambda_{\bf{u}}(\varphi)\cos\varphi{\rm d}\varphi\bigg). $

结合性质(ⅱ)可得$ \alpha(0) = \alpha(2\pi) $, 则$ \alpha $是连续闭合的曲线.

其次, 由步骤(1)的证明过程可以看出(3.6)式成立, 进而(3.5)式也成立, 故定义3.1中的条件(ⅰ)成立.下面证明定义3.1中的条件(ⅱ)也是成立的.

任意取定$ \theta\in[0, 2\pi) $, 当$ \theta\in[0, \pi) $时, 有

$ \begin{eqnarray*} (\cos\theta, \sin\theta) & = &-(1, 0)-(\cos(\theta+\pi), \sin(\theta+\pi))+(1, 0)\\ & = &\int_{\theta+\pi}^{\pi}(-\sin\varphi, \cos\varphi){\rm d}\varphi -\int_{0}^{\pi}\lambda_{\bf{u}}(\varphi)(-\sin\varphi, \cos\varphi){\rm d}\varphi\\ & = &\int_{\theta+\pi}^{\pi}(1-\lambda_{\bf{u}}(\varphi))(-\sin\varphi, \cos\varphi){\rm d}\varphi-\int_{0}^{\theta+\pi}\lambda_{\bf{u}}(\varphi) (-\sin\varphi, \cos\varphi){\rm d}\varphi\\ & = &\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi-\pi)(-\sin(\varphi-\pi), \cos(\varphi-\pi)) {\rm d}(\varphi-\pi)\\ &&-\int_{0}^{\theta+\pi}\lambda_{\bf{u}}(\varphi) (-\sin\varphi, \cos\varphi){\rm d}\varphi\\ & = &\int_{\theta+\pi}^{\theta}\lambda_{\bf{u}}(\varphi)(-\sin\varphi, \cos\varphi) {\rm d}\varphi = \frac{\alpha(\theta)-\alpha(\theta+\pi)}{\omega}. \end{eqnarray*} $

当$ \theta\in[\pi, 2\pi) $时, 有

$ (\cos\theta, \sin\theta) = -(\cos(\theta-\pi), \sin(\theta-\pi)) = \frac{\alpha(\theta)-\alpha(\theta-\pi)}{\omega}. $

令$ \bf{u} = (\cos\theta, \sin\theta) $, 则定义3.1中的条件(ⅱ)成立.

上一节通过观察Reuleaux多边形的特征定义了一类平面曲线:杠杆轮.本节将证明杠杆轮是常宽曲线的一种等价刻画.

引理4.1  下列各条款等价

$ \rm(ⅰ) $$ K $为n维常宽凸体;

$ \rm(ⅱ) $$ K\in{\cal K}_{B^n} $且$ K $为广义常宽集;

$ \rm(ⅲ) $设$ K\in{\cal K}_{B^n} $对任意$ {\bf{u}}\in{S^{n-1}} $, 若$ \alpha_1\in H(K, {\bf{u}})\cap{K}, \alpha_2\in H(K, -{\bf{u}})\cap{K} $, 则$ \frac{\alpha_1-\alpha_2}{\|\alpha_1-\alpha_2\|} = {\bf{u}} $.

证   (ⅰ) $ \Rightarrow $ (ⅱ).显然的.

(ⅱ) $ \Rightarrow $ (ⅲ).设$ K\in{\cal K}_{B^n} $是宽度为$ \omega $的广义常宽集.假设$ \exists {\bf{u}}\in{S^{n-1}} $, $ \alpha_1\in H(K, {\bf{u}})\cap{K} $, $ \alpha_2\in H(K, -{\bf{u}})\cap{K} $使得$ \frac{\alpha_1-\alpha_2}{\|\alpha_1-\alpha_2\|}\neq {\bf{u}} $.

令$ {\bf{p}} = \alpha_2+\omega {\bf{u}} $, 易知$ {\bf{p}}\neq\alpha_1 $且

$ \|\alpha_1-\alpha_2\| = \sqrt{\langle \alpha_1-({\bf{p}}-\omega{\bf{u}}), \alpha_1-({\bf{p}}-\omega{\bf{u}})\rangle} = \sqrt{\omega^2+2\omega\langle {\bf{u}}, \alpha_1-{\bf{p}}\rangle+\|\alpha_1-{\bf{p}}\|^2}. $

由$ \langle {\bf{u}} $, $ \alpha_1-{\bf{p}}\rangle = 0 $和$ \|\alpha_1-{\bf{p}}\|>0 $, 可得$ \|\alpha_1-\alpha_2\|>\omega $.

令$ {\bf{u}}^* = \frac{\alpha_1-\alpha_2}{\|\alpha_1-\alpha_2\|} $, 设$ \alpha_1^*\in H(K, {\bf{u}}^*)\cap{K}, \alpha_2^*\in H(K, -{\bf{u}}^*)\cap{K} $, 则

$ \langle {\bf{u}}^*, \alpha_1^*-\alpha_2^*\rangle\geq\langle {\bf{u}}^*, \alpha_1-\alpha_2\rangle = \|\alpha_1-\alpha_2\|>\omega, $

这与定义2.2矛盾, 故(ⅱ) $ \Rightarrow $ (ⅲ)成立.

(ⅲ) $ \Rightarrow $ (ⅱ).当$ K $满足条款(ⅲ)时, 假设$ \exists \alpha_1^*\in H(K, {\bf{u}})\cap{K} $使得$ \alpha_1\neq\alpha_1^* $.由

$ \alpha_1^*-\alpha_2 = \|\alpha_1-\alpha_2\| {\bf{u}}+\alpha_1^*-\alpha_1 $

看出, $ \alpha_1^*-\alpha_2 $与$ {\bf{u}} $不共线, 与$ K $满足条款(ⅲ)矛盾.故对任意$ {\bf{u}}\in{S^{n-1}} $, $ H(K, {\bf{u}})\cap{K} $是一个单点集, 于是, 设映射$ \alpha:S^{n-1}\rightarrow {{\Bbb R}} ^n $满足$ \{\alpha({\bf{u}})\} = H(K, {\bf{u}})\cap{K} $.对任意$ {\bf{u}} $, $ {\bf{v}}\in{S^{n-1}} $, $ {\bf{u}}\neq {\bf{v}} $有

$ \langle\alpha({\bf{u}}), {\bf{u}}\rangle\geq\langle\alpha({\bf{v}}), {\bf{u}}\rangle, \langle\alpha(-{\bf{v}}), {\bf{u}}\rangle\geq\langle\alpha(-{\bf{u}}), {\bf{u}}\rangle. $

同理

$ \langle\alpha({\bf{v}}), {\bf{v}}\rangle\geq\langle\alpha({\bf{u}}), {\bf{v}}\rangle, \langle\alpha(-{\bf{u}}), {\bf{v}}\rangle\geq\langle\alpha(-{\bf{v}}), {\bf{v}}\rangle. $

$ \forall {\bf{u}}\in{S^{n-1}} $, 记$ \omega({\bf{u}}) = \|\alpha({\bf{u}})-\alpha(-{\bf{u}})\| $, 则

$ \alpha({\bf{u}})-\alpha(-{\bf{u}}) = \omega({\bf{u}}){\bf{u}}, \alpha({\bf{v}})-\alpha(-{\bf{v}}) = \omega({\bf{v}}){\bf{v}}. $

因此

$ \begin{eqnarray*} \omega({\bf{u}}) & = &\langle\alpha({\bf{u}})-\alpha(-{\bf{u}}), {\bf{u}}\rangle = \langle\alpha({\bf{u}}), {\bf{u}}\rangle-\langle\alpha(-{\bf{u}}), {\bf{u}}\rangle\\ &\geq&\langle\alpha({\bf{v}}), {\bf{u}}\rangle-\langle\alpha(-{\bf{v}}), {\bf{u}}\rangle = \langle\alpha({\bf{v}})-\alpha(-{\bf{v}}), {\bf{u}}\rangle \\ & = &\omega({\bf{v}})\langle {\bf{u}}, {\bf{v}}\rangle. \end{eqnarray*} $

同理, $ \omega({\bf{v}})\geq \omega({\bf{u}})\langle {\bf{u}} $, $ {\bf{v}}\rangle $.

若将$ S^{n-1} $上从$ {\bf{u}} $至$ {\bf{v}} $的弧长为$ s $的测地线平均分为$ k $段: $ \widehat{{\bf{u_0}}{\bf{u}}_1} $, $ \widehat{{\bf{u}}_1{\bf{u}}_2}, \cdots $, $ \widehat{{\bf{u}}_{k-1}{\bf{u}}_k} $.其中$ {\bf{u}}_0 = {\bf{u}}, {\bf{u}}_k = {\bf{v}} $.于是$ \langle {\bf{u}}_{i-1}, {\bf{u}}_{i}\rangle = \cos\frac{s}{k}, i = 1, 2, \cdots , k $, 则

$ \begin{eqnarray*} \omega({\bf{u}}) & = &\omega({\bf{u}}_0)\in\left[\omega({\bf{u}}_{1})\langle {\bf{u}}_0, {\bf{u}}_{1}\rangle, \frac{\omega({\bf{u}}_{1})}{\langle {\bf{u}}_0, {\bf{u}}_{1}\rangle}\right]\\ &\subset&\left[\omega({\bf{u}}_{2})\langle {\bf{u}}_{1}, {\bf{u}}_{2}\rangle\langle {\bf{u}}_0, {\bf{u}}_{1}\rangle, \frac{\omega({\bf{u}}_{2})}{\langle {\bf{u}}_{1}, {\bf{u}}_{2}\rangle\langle {\bf{u}}_0, {\bf{u}}_{1}\rangle}\right]\\ &\subset&\cdots \subset\left[\omega({\bf{u_k}})\prod\limits_{i = 1}^k\langle {\bf{u}}_{i-1}, {\bf{u}}_{i}\rangle, \frac{\omega({\bf{u_k}})}{\prod\limits_{i = 1}^k\langle {\bf{u}}_{i-1}, {\bf{u}}_{i}\rangle}\right]\\ & = &\left[\omega({\bf{v}})\cos^k\frac{s}{k}, \frac{\omega({\bf{v}})}{\cos^k\frac{s}{k}}\right]. \end{eqnarray*} $

因为

$ \lim\limits_{k\rightarrow \infty } \cos^k\frac{s}{k} = \left(\lim\limits_{k\rightarrow \infty } (1-\sin^2\frac{s}{k})^{\frac{1}{\sin^2\frac{s}{k}}}\right)^{{\frac{s}{2} \lim\limits_{k\rightarrow \infty }\frac{\sin^2\frac{s}{k}}{\frac{s}{k}}}} = \left(\frac{1}{e}\right)^0 = 1, $

故$ \omega({\bf{u}}) = \omega({\bf{v}}) $, 进而$ K $为广义常宽集, 即(ⅲ) $ \Rightarrow $ (ⅱ)成立.

(ⅱ) $ \Rightarrow $ (ⅰ).设$ K $满足条款(ⅱ), 则对任意$ {\bf{u}}\in{S^{n-1}}, H(K, {\bf{u}}) = H(\rm{conv}K, {\bf{u}}) $, 其中$ \rm{conv}K $表示$ K $的凸包.又$ K $为广义常宽集, 故凸体$ \rm{conv}K $也为广义常宽集, 进而$ \rm{conv}K $为常宽凸体.由$ K\subset \rm{conv}K $, 可得

$ H(K, {\bf{u}})\cap K = H(\rm{conv}K, {\bf{u}})\cap K\subset H(\rm{conv}K, {\bf{u}})\cap{\rm{conv}K}. $

对$ \rm{conv}K $应用(ⅲ) $ \Rightarrow $ (ⅱ)的方法, 可以得到$ H(\rm{conv}K, {\bf{u}})\cap{\rm{conv}K} $与$ H(K, {\bf{u}})\cap K $均为单点集, 故

$ H(K, {\bf{u}})\cap K = H(\rm{conv}K, {\bf{u}})\cap{\rm{conv}K}, $

进而

$ \partial{(\rm{conv}K)} = \bigcup\limits_{{\bf{u}}\in{S^{n-1}}}(H(\rm{conv}K, {\bf{u}})\cap{\rm{conv}K}) = \bigcup\limits_{{\bf{u}}\in{S^{n-1}}}(H(K, {\bf{u}})\cap K)\subset\partial K. $

事实上, 若$ \exists {\bf{x}}\in\partial K $, 但$ {\bf{x}}\not\in\partial{(\rm{conv}K)} $, 由$ \partial{(\rm{conv}K)} $的封闭性, 则$ \partial K $有自交点或不连通, 这与$ K\in{\cal K}_{B^n} $矛盾, 故$ \partial{(\rm{conv}K)} = \partial K $.

因此$ K = \rm{conv}K $, 故$ K $为常宽凸体, 即(ⅱ) $ \Rightarrow $ (ⅰ)成立.

引理4.2  设曲线$ \alpha $为杠杆轮, 则对任意$ {\bf{u}}\in S^{1} $有$ H(\alpha, {\bf{u}})\cap{\alpha} = \{\alpha({\bf{u}})\} $.

证  设$ \alpha $是直径长为$ \omega $的杠杆轮.由定理3.1, 存在臂函数$ \lambda_{\bf{u}} $使得

$ \alpha(\theta) = \bigg(-\omega\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi)\sin\varphi{\rm d}\varphi+x_0, \omega\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi)\cos\varphi{\rm d}\varphi+y_0\bigg), x_0, y_0\in{{\Bbb R}} , $

其中$ \alpha(0) = (x_0, y_0) $.设$ {\bf{u}}(\theta) = (\cos\theta, \sin\theta), \theta\in[0, 2\pi) $, 任意取定$ \theta_0\in[0, 2\pi) $, 则

$ \langle{\bf{u}}(\theta_0), \alpha({\bf{u}}(\theta_0))\rangle -\langle{\bf{u}}(\theta_0), \alpha(\theta)\rangle = \omega\int_{\theta}^{\theta_0}\lambda_{\bf{u}} (\varphi)\sin(\theta_0-\varphi){\rm d}\varphi. $

当$ \theta_0-\theta\in(0, \pi) $时, 令$ \phi = \theta_0-\varphi $, 则

$ \omega\int_{\theta}^{\theta_0}\lambda_{\bf{u}}(\varphi) \sin(\theta_0-\varphi){\rm d}\varphi = \omega\int_{0}^{\theta_0-\theta}\lambda_{\bf{u}} (\theta_0-\phi)\sin(\phi){\rm d}\phi\geq 0. $

当$ \theta_0-\theta\in(\pi, 2\pi) $时, 令$ \phi = \theta_0-\varphi $, 由$ \alpha $的封闭性(即$ \alpha(0) = \alpha(2\pi) $)得

$ \omega\int_{\theta}^{\theta_0}\lambda_{\bf{u}} (\varphi)\sin(\theta_0-\varphi){\rm d}\varphi = \omega\int_{\theta_0-\theta}^{2\pi} \lambda_{\bf{u}}(\theta_0-\phi)(-\sin(\phi)){\rm d}\phi\geq 0. $

当$ \theta_0-\theta\in(-2\pi, 0) $时, 令$ \phi = \theta_0-\varphi+2\pi $, 同理可证

$ \omega\int_{\theta}^{\theta_0} \lambda_{\bf{u}}(\varphi)\sin(\theta_0-\varphi){\rm d}\varphi\geq 0. $

故$ \alpha(\theta_0)\in H(\alpha, {\bf{u}}(\theta_0))\cap{\alpha} $.

下面证明$ H(\alpha, {\bf{u}}(\theta_0))\cap{\alpha} $中仅有一点$ \alpha(\theta_0) $.反设存在$ \alpha(\theta)\in H(\alpha, {\bf{u}}(\theta_0))\cap{\alpha} $且$ \alpha(\theta)\neq\alpha(\theta_0) $, 则$ \langle{\bf{u}}(\theta_0), \alpha(\theta_0)-\alpha(\theta)\rangle = 0 $.进而

$ \omega\int_{\theta}^{\theta_0} \lambda_{\bf{u}}(\varphi)\sin(\theta_0-\varphi){\rm d}\varphi = 0. $

令$ m = \min\{\theta_0, \theta\}, M = \max\{\theta_0, \theta\} $.利用$ \lambda_{\bf{u}} $的非负性, 仿照上述的分类讨论, 可得

$ \lambda_{\bf{u}}((m, M)) = \{0\}\ \mbox{ 或}\ \lambda_{\bf{u}}([0, 2\pi)\setminus[m, M]) = \{0\}. $

进而, $ \alpha(\theta) = \alpha(\theta_0) $, 即$ H(\alpha, {\bf{u}}(\theta_0))\cap{\alpha} = \{\alpha(\theta_0)\} $, 由$ \theta_0 $的任意性可得结论成立.

定理4.1(常宽曲线的等价刻画)  $ K\in{\cal K}_{B^2} $, $ K $为常宽凸体当且仅当$ \partial K $为杠杆轮.

证  充分性.设$ \partial K $为杠杆轮, 由引理4.2可知对任意$ {\bf{u}}\in{S^{1}}, $有

$ H(\alpha, {\bf{u}})\cap{\alpha} = \{\alpha({\bf{u}})\}, H(\alpha, -{\bf{u}})\cap{\alpha} = \{\alpha(-{\bf{u}})\}, $

由定义3.1中的条件(ⅱ)可得

$ \langle\alpha({\bf{u}})-\alpha(-{\bf{u}}), {\bf{u}}\rangle = \langle \omega {\bf{u}}, {\bf{u}}\rangle = \omega, $

故杠杆轮为广义常宽集.进一步, 结合已知条件$ K\in{\cal K}_{B^2} $和引理4.1可知$ K $为常宽凸体.

必要性.设$ K $为平面常宽凸体, 由引理4.1可知$ K\in{\cal K}_{B^2} $, 且对任意$ {\bf{u}}\in{S^{1}}, $若$ \alpha_1\in H(K, {\bf{u}})\cap{K} $, $ \alpha_2\in H(K, -{\bf{u}})\cap{K} $, 则

$ \frac{\alpha_1-\alpha_2}{\|\alpha_1-\alpha_2\|} = {\bf{u}}. $

由引理4.1中步骤(ⅲ) $ \Rightarrow $ (ⅱ)的证明过程, 存在一个正常数$ \omega $和一个映射$ \alpha:S^1\rightarrow {{\Bbb R}} ^2 $满足$ \{\alpha({\bf{u}})\} = H(K, {\bf{u}})\cap{K} $, 且

$ \alpha({\bf{u}})-\alpha(-{\bf{u}}) = \|\alpha({\bf{u}})-\alpha(-{\bf{u}})\| {\bf{u}} = \omega {\bf{u}}, $

即$ \alpha $满足定义3.1的条件(ⅱ). $ \forall {\bf{u}}, {\bf{v}}\in S^1 $且$ {\bf{u}}\neq \pm{\bf{v}} $, 令

$ {{\bf{p}}_{{\bf{u}}, {\bf{v}}}} = H(K, {\bf{u}})\cap H(K, {\bf{v}}), $

显然, $ \alpha({\bf{u}})\in[{\bf{p}}_{{\bf{u}}, -{\bf{v}}}, {\bf{p}}_{{\bf{u}}, {\bf{v}}}] $, 其中$ [{\bf{p}}_{{\bf{u}}, -{\bf{v}}}, {\bf{p}}_{{\bf{u}}, {\bf{v}}}] $表示以$ {\bf{p}}_{{\bf{u}}, -{\bf{v}}}, {\bf{p}}_{{\bf{u}}, {\bf{v}}} $为端点的线段.同理, $ \alpha(-{\bf{u}})\in[{\bf{p}}_{-{\bf{u}}, -{\bf{v}}}, {\bf{p}}_{-{\bf{u}}, {\bf{v}}}] $.因$ \alpha({\bf{u}})-\alpha(-{\bf{u}}) = \omega{\bf{u}} $, 故$ \exists \lambda\in[0, 1] $使得

$ \alpha({\bf{u}}) = (1-\lambda){\bf{p}}_{{\bf{u}}, {\bf{v}}} +\lambda({\bf{p}}_{-{\bf{u}}, -{\bf{v}}}+\omega {\bf{u}}), $

同理, $ \exists \mu\in[0, 1] $使得

$ \alpha({\bf{v}}) = (1-\mu){\bf{p}}_{{\bf{u}}, {\bf{v}}}+\mu({\bf{p}}_{-{\bf{u}}, -{\bf{v}}}+\omega {\bf{v}}). $

设$ {\bf{p}}_{{\bf{u}}, {\bf{v}}} = {\bf{p}}_{-{\bf{u}}, -{\bf{v}}}+x \omega {\bf{u}}+y \omega {\bf{v}} $, 其中$ x, y $为实数.由$ {\bf{p}}_{{\bf{u}}, {\bf{v}}}\in H(K, {\bf{u}})\cap H(K, {\bf{v}}) $, 得

$ \left\{ \begin{array}{ll} \langle {\bf{p}}_{{\bf{u}}, {\bf{v}}}-({\bf{p}}_{-{\bf{u}}, -{\bf{v}}}+\omega {\bf{u}}), {\bf{u}}\rangle = 0, \\ \langle {\bf{p}}_{{\bf{u}}, {\bf{v}}}-({\bf{p}}_{-{\bf{u}}, -{\bf{v}}}+\omega {\bf{v}}), {\bf{v}}\rangle = 0. \end{array}\right. $

解得$ x = y = \frac{1}{\langle {\bf{u}}, {\bf{v}}\rangle+1} $.故

$ \alpha({\bf{u}})-\alpha({\bf{v}}) = \frac{\omega}{\langle {\bf{u}}, {\bf{v}}\rangle+1} ((\mu+\lambda\langle {\bf{u}}, {\bf{v}}\rangle) {\bf{u}}-(\lambda+\mu\langle {\bf{u}}, {\bf{v}}\rangle) {\bf{v}}). $

若任意给定$ {\bf{{\bf{u_0}}}}\in S^1 $, 当$ {\bf{u}} $沿$ c({\bf{{\bf{u_0}}}})\varsubsetneqq S^1 $趋近于$ {\bf{u_0}} $时, 常宽曲线$ \alpha $在$ \alpha({\bf{u_0}}) $处的曲率半径$ \lim\limits_{{\bf{u}}\rightarrow {\bf{u_0}} }\frac{\|\alpha({\bf{u}}) -\alpha({\bf{u_0}})\|}{\|{\bf{u}}-{\bf{u_0}}\|} $存在.于是, $ \exists \lambda({\bf{u}}), \mu({\bf{u}})\in[0, 1] $使得

(4.1) $ \begin{equation} \lim\limits_{{\bf{u}}\rightarrow {\bf{u_0}}}\frac{\alpha({\bf{u}}) -\alpha({\bf{u_0}})}{\|{\bf{u}}-{\bf{u_0}}\|} = \frac{\omega}{2}\lim\limits_{{\bf{u}}\rightarrow {\bf{u_0}} }(\mu({\bf{u}})+\lambda({\bf{u}})) \lim\limits_{{\bf{u}}\rightarrow {\bf{u_0}}}\frac{{\bf{u}} -{\bf{u_0}}}{\|{\bf{u}}-{\bf{u_0}}\|}. \end{equation} $

又$ \frac{\omega}{2}(\mu({\bf{u}})+\lambda({\bf{u}}))\geq 0 $, 将(3.1)式等号两边与单位矢量$ \lim\limits_{{\bf{u}}\rightarrow {\bf{u_0}}}\frac{{\bf{u}} -{\bf{u_0}}}{\|{\bf{u}}-{\bf{u_0}}\|} $作内积可以得到$ \alpha $满足定义3.1中的条件(ⅰ).由定义3.1知$ \partial K $为杠杆轮.

下面我们利用定理3.1和定理4.1构造一类平面常宽曲线.设$ [0, \pi) $上的函数

(5.1) $ \begin{equation} \mu(\theta) = \sum\limits_{i = 1}^na_i\cos(i\theta)+\sum\limits_{i = 1}^nb_i\sin(i\theta), \end{equation} $

其中$ a_i, b_i\in{{\Bbb R}} , i = 1, 2, \cdots, n $, 则

$ \int_{0}^{\pi}\mu(\theta)\sin\theta{\rm d}\theta = \sum\limits_{i = 2}^n\frac{1+(-1)^i}{2}\left(\frac{1}{i+1}-\frac{1}{i-1}\right)a_i+\frac{b_1\pi}{2}, $

$ \int_{0}^{\pi}\mu(\theta)\cos\theta{\rm d}\theta = \frac{a_1\pi}{2}+\sum\limits_{i = 2}^n\frac{1+(-1)^i}{2}\left(\frac{1}{i+1}+\frac{1}{i-1}\right)b_i. $

结合定理3.1和定理4.1可以证明, 函数

(5.2) $ \begin{equation} \lambda_{\bf{u}}(\theta) = \left\{\begin{array}{ll} { }\frac{1}{2}+\frac{\mu(\theta)}{2c}, & \theta\in[0, \pi), \\ { }\frac{1}{2}-\frac{\mu(\theta-\pi)}{2c}, {\quad} & \theta\in[\pi, 2\pi) \end{array}\right. \end{equation} $

是杠杆轮(或常宽曲线)的臂函数当且仅当齐次线性方程组

(5.3) $ \begin{equation} \left\{\begin{array}{ll} { } \sum\limits_{i = 2}^n\frac{1+(-1)^i}{2}\left(\frac{1}{i+1}-\frac{1}{i-1}\right)a_i+\frac{b_1\pi}{2} = 0, \\ { } \frac{a_1\pi}{2}+\sum\limits_{i = 2}^n\frac{1+(-1)^i}{2}\left(\frac{1}{i+1}+\frac{1}{i-1}\right)b_i = 0 \end{array}\right. \end{equation} $

成立且非零常数

(5.4) $ \begin{equation} c\geq \max\limits_{\theta\in[0, \pi)}|\mu(\theta)|. \end{equation} $

记$ \alpha $是由臂函数$ \lambda_{\bf{u}}(\theta) $确定的杠杆轮, 令$ \alpha(0) = (x_0, y_0) $, 结合定理3.1可得

(5.5) $ \begin{equation} \alpha(\theta) = \bigg(-\omega\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi)\sin\varphi{\rm d}\varphi+x_0, \omega\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi)\cos\varphi{\rm d}\varphi+y_0\bigg). \end{equation} $

利用(5.1), (5.2)和(5.5)式可以得到$ \alpha $的参数方程$ \alpha(\theta) = (x(\theta), y(\theta)) $

$ \left\{ \begin{array}{rl} x(\theta) = &{ } \omega\bigg(\frac{\cos\theta-1}{2}+a_1\frac{\cos2\theta-1}{8c} +\sum\limits_{i = 2}^n\frac{a_i}{4c}u_i(\theta)\bigg(\frac{\cos(i+1)\theta-1}{i+1} -\frac{\cos(i-1)\theta-1}{i-1}\bigg)\\ &{ } +b_1\frac{\sin 2\theta-2\theta}{8c} +\sum\limits_{i = 2}^n\frac{b_i}{4c}u_i(\theta)\bigg(\frac{\sin(i+1)\theta}{i+1} -\frac{\sin(i-1)\theta}{i-1}\bigg)\bigg)+x_0, \\ y(\theta) = &{ } \omega\bigg(\frac{\sin\theta}{2}+a_1\frac{2\theta+\sin 2\theta}{8c}+\sum\limits_{i = 2}^n\frac{a_i}{4c}u_i(\theta) \bigg(\frac{\sin(i-1)\theta}{i-1} +\frac{\sin(i+1)\theta}{i+1}\bigg)\\ &{ } +b_1\frac{1-\cos2\theta}{8c}+\sum\limits_{i = 2}^n\frac{b_i}{4c}u_i(\theta) \bigg(\frac{1-\cos(i+1)\theta}{i+1} +\frac{1-\cos(i-1)\theta}{i-1}\bigg)\bigg)+y_0, \end{array}\right. $

其中

(5.6) $ \begin{equation} u_i(\theta) = \left\{ \begin{array}{ll} 1, & \theta\in[0, \pi), \\ (-1)^{i+1}, {\quad}& \theta\in[\pi, 2\pi). \end{array}\right. \end{equation} $

例5.1  当$ n = 4 $时, 联立(5.3)和(5.4)式可以求出$ a_1, \cdots, a_4, b_1, \cdots, b_4, c $的解空间, 取其中的一组解$ a_1 = 0 $, $ a_2 = -1 $, $ a_3 = 24 $, $ a_4 = 5 $, $ b_1 = 0 $, $ b_2 = 2 $, $ b_3 = 0 $, $ b_4 = -5 $, $ c = 23 $, 此时臂函数

$ \lambda_{\bf{u}}(\theta) = \frac{1}{2}-\frac{u_2(\theta)\cos(2\theta)}{46}+\frac{12\cos(3\theta)}{23}+\frac{5 u_4(\theta)\cos(4\theta)}{46} +\frac{u_2(\theta)\sin(2\theta)}{23}-\frac{5 u_4(\theta)\sin(4\theta)}{46}, $

其中$ u_i(\theta) $由(5.6)式给出.取$ \omega = 1 $, $ x_0 = 1 $, $ y_0 = 0 $, 确定的杠杆轮记为$ \alpha_1 $, 如图 1.

例5.2  当$ n = 5 $时, 联立(4.3)和(4.4)式可以求出$ a_1, \cdots, a_5, b_1, \cdots, b_5, c $的解空间, 取其中的一组解$ a_1 = \frac{8}{15\pi} $, $ a_2 = -1 $, $ a_3 = 0 $, $ a_4 = 15 $, $ a_5 = -\frac{4}{5} $, $ b_1 = \frac{8}{3\pi} $, $ b_2 = 1 $, $ b_3 = 0 $, $ b_4 = -2 $, $ b_5 = 0 $, $ c = 18 $, 此时臂函数

$ \begin{eqnarray*} \lambda_{\bf{u}}(\theta) & = &\frac{1}{2}+\frac{2 \cos(\theta)}{135\pi}-\frac{u_2(\theta)\cos(2\theta)}{36} +\frac{5 u_4(\theta)\cos(4\theta)}{12}-\frac{\cos(5\theta)}{45}\\ &&+\frac{2\sin(\theta)}{27\pi}+\frac{u_2(\theta)\sin(2\theta)}{36}-\frac{u_4(\theta)\sin(4\theta)}{18}, \end{eqnarray*} $

其中$ u_i(\theta) $由(4.6)给出.取$ \omega = 1 $, $ x_0 = 1 $, $ y_0 = 0 $, 确定的杠杆轮记为$ \alpha_2 $, 如图 2.

我们把包含圆与Reuleaux多边形在内的所有可以分割为若干段圆弧的常宽曲线统称为广义Reuleaux多边形.下面我们讨论广义Reuleaux多边形的构造.由定理3.1和定理4.1可知, 对任意宽度为$ \omega $的广义Reuleaux多边形$ \Gamma $, 存在臂函数$ \lambda_{\bf{u}} $满足定理3.1中的性质(ⅰ)和(ⅱ)且

$ \Gamma(\theta) = \bigg(-\omega\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi)\sin\varphi{\rm d}\varphi+x_0, \omega\int_{0}^{\theta}\lambda_{\bf{u}}(\varphi)\cos\varphi{\rm d}\varphi+y_0\bigg), \ x_0, y_0\in{{\Bbb R}} . $

定理5.1  任意广义$ \rm Reuleaux $多边形的臂函数为分段常函数.反之, 任意臂函数为分段常函数的杠杆轮都是广义$ \rm Reuleaux $多边形.

证  首先, 设$ \Gamma $是宽度为$ \omega $的广义Reuleaux多边形, 则$ \Gamma\in{\cal L}(\omega) $.设$ \gamma $为$ \Gamma $上任意一段圆弧, 其圆心为$ P $半径为$ r $.对任意$ {\bf{x}}\in\gamma $, $ \exists {\bf{u}}\in S^1 $使得$ {\bf{x}} = P+r{\bf{u}} $.由引理4.2可得$ {\bf{x}} = \Gamma({\bf{u}}) $, 故

$ \Gamma({\bf{u}})-P = r{\bf{u}}. $

设$ \Gamma $在$ {\bf{u}} $方向上的支点为$ P({\bf{u}}) $, 因杠杆轮$ \Gamma $在$ {\bf{u}} $方向上的臂长是$ \Gamma $在$ {\bf{x}} $处的曲率半径, 即$ \|\Gamma({\bf{u}})-P({\bf{u}})\| = r $, 故

$ P({\bf{u}}) = \Gamma({\bf{u}})-r{\bf{u}} = P. $

且对任意$ \Gamma({\bf{u}})\in\gamma $有

$ \lambda({\bf{u}}) = \frac{r}{\omega}, $

即在$ \gamma $上对应的$ \lambda({\bf{u}}) $恒为常数.因广义Reuleaux多边形由若干段圆弧构成, 故它的臂函数在$ [0, 2\pi) $上为分段常函数.

其次, 设$ \alpha $是直径为$ \omega $的杠杆轮, 其臂函数$ \lambda_{\bf{u}} $为分段常函数, 故设

$ \lambda_{\bf{u}}(\theta) = \left\{ \begin{array}{ll} \lambda_1, & \theta\in I_1, \\ \lambda_2, & \theta\in I_2, \\ \vdots & \vdots\\ \lambda_k, {\quad} & \theta\in I_k, \end{array}\right. $

其中右开区间$ I_1, I_2, \cdots, I_k $为区间$ [0, 2\pi) $的一组划分, $ \lambda_i\in[0, 1] $是一组常数, $ i = 1, 2, \cdots, k $.结合定理3.1知, $ \lambda_{\bf{u}} $确定的$ \alpha $上的曲线段

$ \gamma(\theta) = \omega\int_{\theta_0}^{\theta}\lambda_{\bf{u}}(\varphi)(-\sin\varphi, \cos\varphi){\rm d}\varphi +\gamma(\theta_0), \theta_0, \theta\in I_i $

是半径为$ \omega\lambda_i $的圆弧, 即$ \alpha $是广义Reuleaux多边形.

事实上, 构造广义Reuleaux多边形就是构造臂函数为分段常函数的杠杆轮.为此, 我们设$ [0, \pi) $上的分段常函数

(5.7) $ \begin{equation} \mu(\theta) = \left\{ \begin{array}{ll} \mu_1, & \theta\in I_1, \\ \mu_2, & \theta\in I_2, \\ \vdots & \vdots\\ \mu_n, {\quad} & \theta\in I_n, \end{array}\right. \end{equation} $

其中右开区间$ I_1, I_2, \cdots, I_n $为区间$ [0, \pi) $的一组划分, $ \mu_i\in{{\Bbb R}} $是一组常数, $ i = 1, 2, \cdots, n $.令

(5.8) $ \begin{equation} S_i = \int_{I_i}\sin\theta{\rm d}\theta, C_i = \int_{I_i}\cos\theta{\rm d}\theta. \end{equation} $

结合定理3.1和定理5.1可以证明, 若$ n $元齐次线性方程组

(5.9) $ \begin{equation} \left\{\begin{array}{ll} { } \sum\limits_{i = 1}^{n}S_i\mu_i = 0, \\ { } \sum\limits_{i = 1}^{n}C_i\mu_i = 0 \end{array}\right. \end{equation} $

成立且非零常数

(5.10) $ \begin{equation} c\geq \max\limits_{1\leq i\leq n}|\mu_i|, \end{equation} $

则分段常函数

(5.11) $ \begin{equation} \lambda_{\bf{u}}(\theta) = \left\{\begin{array}{ll} { } \frac{1}{2}+\frac{\mu(\theta)}{2c}, & \theta\in[0, \pi), \\ { } \frac{1}{2}-\frac{\mu(\theta-\pi)}{2c}, {\quad}& \theta\in[\pi, 2\pi) \end{array}\right. \end{equation} $

是广义Reuleaux多边形的臂函数.反之, 若任意广义Reuleaux多边形的臂函数$ \lambda_{\bf{u}}(\theta) $由(5.7)和(5.11)式给出, 则(5.9)和(5.10)式成立.

设分段常函数

$ \lambda_{\bf{u}}(\theta) = \left\{\begin{array}{ll} \lambda_1, & \theta\in I_1, \\ \lambda_2, & \theta\in I_2, \\ \vdots & \vdots\\ \lambda_n, & \theta\in I_n, \\ 1-\lambda_{\bf{u}}(\theta-\pi), \quad & \theta\in[\pi, 2\pi) \end{array}\right. $

是广义Reuleaux多边形$ \Gamma $的臂函数, 其中右开区间$ I_1, I_2, \cdots, I_n $为区间$ [0, \pi) $的一组划分, $ \lambda_1, \lambda_2, \cdots, \lambda_n $是一组常数, 且任意区间$ I_{i} $与$ I_{i-1} $相邻, 任意常数$ \lambda_{i}\neq\lambda_{i-1}, i = 2, \cdots, n $.

易看出, 当常数$ \lambda_{i} = 0 $时, 区间$ I_{i} $对应$ \Gamma $上的一个奇点.当常数$ \lambda_{i}\neq0 $时, 区间$ I_{i} $对应$ \Gamma $上的一条边.进而, 可以得到广义Reuleaux多边形$ \Gamma $的边数

(5.12) $ \begin{equation} N = 2n-N_0-N_1-N_{\lambda_{1}, \lambda_{n}}, \end{equation} $

其中非负整数$ N_0, N_1 $分别定义为常数组$ \lambda_1, \lambda_2, \cdots, \lambda_n $中$ 0 $和$ 1 $的个数, 且

$ N_{\lambda_{1}, \lambda_{n}} = \left\{\begin{array}{ll} 0, & \lambda_{1}+\lambda_{n}\neq1, \\ 1, & \lambda_{1}+\lambda_{n} = 1, \lambda_{1}\in\{0, 1\}, \\ 2, {\quad} & \lambda_{1}+\lambda_{n} = 1, \lambda_{1}\not\in\{0, 1\}. \end{array}\right. $

综上所述, 将区间$ [0, \pi) $任意划分为一组右开区间$ I_1, I_2, \cdots, I_n $, 通过(4.8)式计算出两组实数$ \{S_i\}, \{C_i\}, i = 1, 2, \cdots, n $, 并代入(4.9)和(4.10)式, 可以求出$ \mu_1, \mu_2, \cdots, \mu_n, c $的解空间.联立(5.7)和(5.11)式得到臂函数$ \lambda_{\bf{u}}(\theta) $, 进而得到对应的广义Reuleaux多边形.

例5.3  将区间$ [0, \pi) $平均分成九个右开区间, 可以求出$ \mu_1, \mu_2, \cdots, \mu_9, c $的解空间, 取其中的一组解$ \mu_1 = \mu_4 = \mu_6 = \mu_9 = \cos\frac{2\pi}{9}-\cos\frac{\pi}{9} $, $ \mu_2 = \mu_8 = 2-2\cos\frac{4\pi}{9}-\cos\frac{2\pi}{9}-\cos\frac{\pi}{9} $, $ \mu_3 = \mu_5 = \mu_7 = \cos\frac{\pi}{9}-\cos\frac{2\pi}{9} $, $ c = \cos\frac{\pi}{9}-\cos\frac{2\pi}{9} $.此时臂函数

$ \lambda_{\bf{u}}(\theta) = \left\{\begin{array}{ll} 0, & { } \theta\in[0, \frac{\pi}{9})\cup[\frac{\pi}{3}, \frac{4\pi}{9})\cup[\frac{5\pi}{9}, \frac{2\pi}{3})\cup[\frac{8\pi}{9}, \pi), \\ { } \frac{1-\cos\frac{4\pi}{9}-\cos\frac{2\pi}{9}} {\cos\frac{\pi}{9}-\cos\frac{2\pi}{9}}, {\quad} &{ } \theta\in[\frac{\pi}{9}, \frac{2\pi}{9})\cup[\frac{7\pi}{9}, \frac{8\pi}{9}), \\ 1, & { } \theta\in[\frac{2\pi}{9}, \frac{\pi}{3})\cup[\frac{4\pi}{9}, \frac{5\pi}{9})\cup[\frac{2\pi}{3}, \frac{7\pi}{9}), \\ 1-\lambda_{\bf{u}}(\theta-\pi), & \theta\in[\pi, 2\pi), \end{array}\right. $

设直径长度$ \omega = 1 $, 利用定理$ \rm3.1 $, 构造出广义$ \rm Reuleaux $十一边形, 如图 3.

例5.4  将区间$ [0, \pi) $划分为$ [0, \theta_1) $, $ [\theta_1, \theta_2) $, $ [\theta_2, \pi) $三个区间, 令$ \frac{\mu_1}{c} = \frac{\mu_3}{c} = -1 $, $ |\frac{\mu_2}{c}|<1 $, 即$ \lambda_1 = \lambda_3 = 0 $, $ \lambda_2\in(0, 1) $, 则(5.12)式中$ N = 4 $, 结合(5.9)式解得$ \theta_1 = \pi-\theta_2\in(0, \frac{\pi}{3}) $.令$ \theta_1 = \frac{\pi}{6} $, 此时臂函数

$ \lambda_{\bf{u}}(\theta) = \left\{\begin{array}{ll} 0, & { } \theta\in[0, \frac{\pi}{6})\cup[\frac{5\pi}{6}, \pi), \\ { } \frac{\sqrt{3}}{3}, & { } \theta\in[\frac{\pi}{6}, \frac{5\pi}{6}), \\ 1-\lambda_{\bf{u}}(\theta-\pi), \quad & \theta\in[\pi, 2\pi), \end{array}\right. $

设直径长度$ \omega = 1 $, 利用定理$ \rm3.1 $, 构造出$ \rm Reuleaux $等腰梯形, 如图 4.

为了构造平面常宽凸体, 我们基于Reuleaux多边形的性质定义了一类平面曲线, 称为杠杆轮, 它是平面常宽曲线的一种等价刻画.我们引入了杠杆轮的臂函数, 用臂函数给出了杠杆轮的参数表示, 从而给出了常宽曲线的一种不同于支撑函数表示的参数方程.特别地, 本文定义的臂函数区别于支撑函数.首先, 臂函数不依赖于原点位置.其次, 可以用臂函数的积分表示曲线参数方程, 不要求臂函数的连续性, 并且比支撑函数更便于构造Reuleaux多边形.

另外, 本文构造了一类臂函数为三角函数形式的常宽曲线, 并且推广了经典的Reuleaux多边形, 得到了广义Reuleaux多边形.