作为曲线的几何不变量, 弧长和曲率在微分几何研究中有着重要的意义. 多年来, 人们对于欧氏空间和Minkowski空间中曲线的弧长参数和曲率进行了广泛而深入的研究, 但是仿射空间中的相关研究结论则比较少. 1979年, Michael Spivak提出应用Cartan定理寻找仿射变换下的不变量这一研究思路^[1]. 2007年, Mehdi Nadjafikhah等人在Spivak工作的基础上结合等价问题得到了仿射不变量, 同时给出了仿射不变量在物理学、计算机视觉和图像处理中的若干应用^{[2, 3]}. 2014年, 刘会立在仿射空间中定义了仿射曲线的中心仿射弧长参数和中心仿射曲率, 并给出了一些分类结果^[4]. 2016年, 邓爽将仿射曲线与欧氏空间中的曲线通过弧长参数联系起来, 找到了中心仿射曲率与欧氏空间中曲线的曲率函数和挠率函数之间的关系^[5].

众所周知, 曲率和挠率是刻画空间曲线的两个重要特征, 曲率刻画了曲线沿主法方向的弯曲程度, 挠率则刻画了曲线沿副法方向的弯曲程度. 一条空间曲线的形状可以由曲率$ \kappa(s) $和挠率$ \tau(s) $完全决定, 如当挠率与曲率的比为非零常数时, 曲线为一般螺线; 当挠率与曲率的比为弧长的线性函数时, 曲线为从切曲线等. 本文主要研究二维以及三维Minkowski空间中类空和类时曲线的仿射性质. 首先研究当类空和类时曲线的弧长与仿射弧长相同时, 二维以及三维类空和类时曲线的中心仿射曲率、中心仿射挠率与曲线的曲率、挠率之间的对应关系, 以及两类曲线的正交标架和仿射标架之间的对应关系. 如二维Minkowski空间中的类空或者类时曲线的中心仿射曲率$ \kappa_{1}(s) $和其曲率$ \kappa(s) $之间满足$ \kappa_{1}(s)=(\log|\kappa(s)|)'. $同时讨论这一条件下, 类空和类时曲线自身的曲率和挠率满足的关系. 然后根据得到的结论, 讨论当类空或者类时曲线的曲率$ \kappa(s) $和挠率$ \tau(s) $满足

$ \tau(s)=a\kappa^\lambda(s)(a\not=0, \lambda\in \mathbb {R}) $

时, $ \lambda $取不同值时曲线的曲率所满足的特殊微分方程, 如当$ \lambda =0 $, 即类空或者类时曲线具有常挠率时, 其曲率满足第二Painlevé方程.

本节主要回顾仿射空间以及Minkowski空间中曲线的基本理论.

设$ \mathbb{A}^{n+1} $是$ n+1 $维仿射空间, 其内积定义为

$ (x, y)_{\delta}=\sum\limits_{i=1}^{n+1}\delta_{i} x_{i} y_{i}, $

这里$ x=\left(x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}\right) $, $ y=\left(y_{1}, y_{2}, \cdots, y_{n}, y_{n+1}\right) \in \mathbb{A}^{n+1} $, $ \delta_{i}=\pm 1. $特别地, 当$ \delta_{i}=1 $时, $ \mathbb{A}^{n+1} $即为$ n+1 $维欧氏空间; 当$ \delta_{i} $中有一个取值为$ -1 $时, $ \mathbb{A}^{n+1} $即为$ n+1 $维Minkowski空间.

设$ \boldsymbol{x(t)} $是$ \mathbb{A}^{n+1} $中的任意一条曲线, 则

$ \mathrm{d} s=\left|\frac{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime \prime}(s)}, \ldots, \boldsymbol{x^{(n)}(s)}, \boldsymbol{x^{(n+1)}(s)}\right]}{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime \prime}(s)}, \ldots, \boldsymbol{x^{(n)}(s)}, \boldsymbol{x(s)}\right]}\right|^{\frac{1}{n+1}} \mathrm{\; d} t $

是$ \boldsymbol{x(t)} $的一个中心仿射不变量, 其中$ \left[ , ..., \right] $, 表示, $ \mathbb{A}^{n+1} $, 中的标准行列式, $ t $为一般参数. 根据此中心仿射不变量, 有如下引理

引理2.1  ^[4] 设$ \boldsymbol{x(s)} $是二维仿射空间$ \mathbb{A}^{2} $中任意一条正则曲线. 若

$ \frac{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime \prime}(s)}\right]}{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x(s)}\right]}=\varepsilon_{0}=\pm1, $

则称$ s $为$ \boldsymbol{x(s)} $的中心仿射弧长参数, 此时曲线$ \boldsymbol{x(s)} $满足如下Frenet公式

$ \left\{\begin{array}{l} \boldsymbol{x^{\prime}(s)}=\boldsymbol{e_{1}(s)}, \\ \boldsymbol{x^{\prime \prime}(s)}=\kappa_{1}(s) \boldsymbol{e_{1}(s)}+\varepsilon_{0}\boldsymbol{x(s)}, \end{array}\right. $

这里$ \kappa_{1}(s) $称为$ \boldsymbol{x(s)} $的中心仿射曲率, 它可以表示为

$ \kappa_{1}(s)=\frac{\left[\boldsymbol{x^{\prime\prime}(s)}, \boldsymbol{x(s)}\right]}{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x(s)}\right]}. $

引理2.2  ^[4] 设$ \boldsymbol{x(s)} $是三维仿射空间$ \mathbb{A}^{3} $中任意一条正则曲线. 若

$ \frac{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime\prime}(s)}, \boldsymbol{x^{\prime\prime\prime}(s)}\right]} {\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime\prime}(s)}, \boldsymbol{x(s)}\right]}=\varepsilon_{0}=\pm1, $

则称$ s $为$ \boldsymbol{x(s)} $的中心仿射弧长参数, 此时曲线$ \boldsymbol{x(s)} $满足如下Frenet公式

$ \left\{\begin{array}{l} \boldsymbol{x^{\prime}(s)}=\boldsymbol{e_{1}(s)}, \\ \boldsymbol{x^{\prime \prime}(s)}=\boldsymbol{e_{2}(s)}, \\ \boldsymbol{x^{\prime \prime \prime}(s)}=\kappa_{1}(s) \boldsymbol{e_{1}(s)}+\kappa_{2}(s) \boldsymbol{e_{2}(s)}+\varepsilon_{0} \boldsymbol{x(s)}, \end{array}\right. $

这里$ \kappa_{1}(s) $称为$ \boldsymbol{x(s)} $的中心仿射曲率, $ \kappa_{2}(s) $称为$ \boldsymbol{x(s)} $的中心仿射挠率, 它们可以分别表示为

$ \kappa_{1}(s)=\frac{\left[\boldsymbol{x^{\prime\prime\prime}(s)}, \boldsymbol{x^{\prime\prime}(s)} , \boldsymbol{x(s)}\right]}{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime \prime}(s)}, \boldsymbol{x(s)}\right]}, \quad \kappa_{2}(s)=\frac{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime \prime\prime}(s)}, \boldsymbol{x(s)}\right]}{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime \prime}(s)}, \boldsymbol{x(s)}\right]}.\quad $

设$ \mathbb{E}^{n+1}_{1} $是$ n+1 $维Minkowski空间, 其内积定义为

$ \langle\cdot, \cdot\rangle=\mathrm{d}x_{1}^{2}+\mathrm{d}x_{2}^{2}+\cdots+\mathrm{d}x_{n}^{2}-\mathrm{d}x_{n+1}^{2}. $

对于 $ \mathbb{E}^{n+1}_{1} $中的非零向量$ \textbf{v} $, 如果$ \langle \textbf{v}, \textbf{v}\rangle\!>\!0 $, 称 $ \textbf{v} $为类空向量; 如果, $ \langle \textbf{v}, \textbf{v}\rangle=0 $, 称 $ \textbf{v} $ 为类光向量; 如果$ \langle \textbf{v}, \textbf{v}\rangle<0 $, 称 $ \textbf{v} $为类时向量. 特别地, 规定零向量为类空向量.

引理2.3  ^[6]  设$ \boldsymbol{x(s)} $是二维Minkowski空间$ \mathbb{E}^{2}_{1} $, 中以$ s $为弧长参数的正则曲线, 即$ |\langle \boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime}(s)}\rangle| =1 $, 则存在正交标架$ \left\{\boldsymbol{\alpha \left( s\right)}, \boldsymbol{\beta \left( s\right)}\right\} $满足如下的Frenet公式

$ \begin{cases}\boldsymbol{x^{\prime}(s)}=\boldsymbol{\alpha \left( s\right)}, &\\ \boldsymbol{\alpha^{\prime}(s)}= \kappa \left( s\right) \boldsymbol{\beta \left( s\right)} , &\\ \boldsymbol{\beta^{\prime}\left( s\right)} = \kappa \left( s\right) \boldsymbol{\alpha \left( s\right)} , &\end{cases} $

这里$ \boldsymbol{\alpha(s)} $和$ \boldsymbol{\beta(s)} $分别表示$ \boldsymbol{x(s)} $的切向量和法向量, $ \kappa(s) $称为$ \boldsymbol{x(s)} $的曲率, 且当$ \langle \boldsymbol{\alpha(s)}, \boldsymbol{\alpha(s)} \rangle\!=\!1 $时, $ \boldsymbol{x(s)} $是类空曲线; 当$ \langle \boldsymbol{\alpha(s)}, \boldsymbol{\alpha(s)} \rangle=-1 $时, $ \boldsymbol{x(s)} $是类时曲线.

引理2.4  ^[6] 设, $ \boldsymbol{x(s)} $是三维Minkowski空间$ \mathbb{E}^{3}_{1} $中以$ s $为弧长参数的类空(类时)曲线, 即$ |\langle \boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime}(s)}\rangle|=1 $, 则存在正交标架$ \left\{ \boldsymbol{\alpha \left( s\right) } , \boldsymbol{\beta \left( s\right) } , \boldsymbol{\gamma(s)}\right\} $满足如下的Frenet公式

$ \begin{equation*} \label{2.2} \begin{cases}\boldsymbol{x^{\prime} s} =\boldsymbol{\alpha s} , &\\ \boldsymbol{\alpha^{\prime} s} = \kappa \left( s\right) \boldsymbol{\beta \left( s\right)} , &\\ \boldsymbol{\beta^{\prime} \left( s\right)} = -\varepsilon\tilde{\varepsilon} \kappa \left( s\right) \boldsymbol{\alpha \left( s\right)} + \tau \left( s\right) \boldsymbol{\gamma \left( s\right)} , &\\ \boldsymbol{\gamma^{\prime} \left( s\right)} = \varepsilon \tau \left( s\right) \boldsymbol{\beta \left( s\right)} , &\end{cases} \end{equation*} $

这里$ \boldsymbol{\alpha(s)} $, $ \boldsymbol{\beta(s)} $, $ \boldsymbol{\gamma(s)} $分别是$ \boldsymbol{x(s)} $的切向量, 主法向量和副法向量, $ \kappa(s) $, $ \tau(s) $分别称为$ \boldsymbol{x(s)} $的曲率和挠率. $ \langle \boldsymbol{\alpha (s)}, \boldsymbol{\alpha (s)}\rangle =\varepsilon =\pm 1 $, $ \langle \boldsymbol{\beta (s)}, \boldsymbol{\beta (s)}\rangle =\tilde{\varepsilon } =\pm 1 $, $ \langle \boldsymbol{\gamma (s)}, \boldsymbol{\gamma (s)}\rangle =-\varepsilon\tilde{\varepsilon } $. 且当$ \varepsilon = \tilde{\varepsilon} = 1 $时, $ \boldsymbol{x(s)} $是第一类类空曲线; 当$ \varepsilon = 1, \tilde{\varepsilon} = -1 $时, $ \boldsymbol{x(s)} $是第二类类空曲线; 当$ \varepsilon =-1, \tilde{\varepsilon} = 1 $时, $ \boldsymbol{x(s)} $是类时曲线.

注2.5   $ \mathbb{E}^{2}_{1} $中的类光曲线退化为直线. $ \mathbb{E}^{3}_{1} $中的第三类类空曲线以及类光曲线, 由于它们的切向量或者法向量(副法向量)中含有类光向量, 其活动标架为渐近正交标架, 本文暂且不对它们进行相关讨论.

注2.6   本文假设第一类(第二类)类空曲线以及类时曲线的中心仿射曲率(曲率), 中心仿射挠率(挠率)均不为零. 这里$ ^\prime $表示仿射空间(Minkowski空间)中的变量关于中心仿射弧长参数(弧长参数)的微分.

定理3.1   设$ \boldsymbol{x(s)} $是$ \mathbb{A}^{2} $中以$ s $为中心仿射弧长的曲线. 如果$ \boldsymbol{x(s)} $同时作为$ \mathbb{E}^{2}_{1} $中的类空（类时）曲线, 且以$ s $为弧长参数, 则曲线$ \boldsymbol{x(s)} $的中心仿射曲率$ \kappa_{1}(s) $和$ \mathbb{E}^{2}_{1} $中的曲率$ \kappa(s) $之间满足

$ \kappa_{1}(s)=(\log|\kappa(s)|)'. $

证  由引理2.1和引理2.3, 曲线$ \boldsymbol{x(s)} $的中心仿射曲率$ \kappa_{1}(s) $, 为

$ \begin{equation} \kappa_{1}(s)=\frac{\left[\boldsymbol{x^{\prime\prime}(s)}, \boldsymbol{x(s)}\right]}{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x(s)}\right]}=\frac{\kappa(s)\left[ \boldsymbol{\beta(s)}, \boldsymbol{x(s)}\right]}{\left[\boldsymbol{\alpha(s)}, \boldsymbol{x(s)}\right]}, \end{equation} $

(3.1)

且同时有

$ \begin{equation} \varepsilon_{0}=\frac{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime \prime}(s)}\right]}{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x(s)}\right]}=\frac{\kappa(s)\left[\boldsymbol{\alpha(s)} , \boldsymbol{\beta(s)}\right]}{\left[\boldsymbol{\alpha(s)}, \boldsymbol{x(s)}\right]}=\pm1. \end{equation} $

(3.2)

由(3.1)式和(3.2)式, 显然有下面的等式

$ \begin{equation} \kappa(s)\left[\boldsymbol{\beta(s)}, \boldsymbol{x(s)} \right]=\kappa_{1}(s)[\boldsymbol{\alpha(s)}, \boldsymbol{x(s)}] \end{equation} $

(3.3)

和

$ \begin{equation} \kappa(s)\left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)} \right]=\varepsilon_{0}[\boldsymbol{\alpha(s)}, \boldsymbol{x(s)}]. \end{equation} $

(3.4)

在(3.4)式两端关于$ s $求导, 可得

$ \begin{equation} \kappa^\prime(s) \left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)}\right]\!+\!\kappa(s)\left[\boldsymbol{\alpha^\prime(s)}, \boldsymbol{\beta(s)} \right]\!+\!\kappa(s)\left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta^\prime(s)} \right]\!\!=\!\!\varepsilon_{0}\{[\boldsymbol{\alpha^\prime(s)}, \boldsymbol{x(s)}]\!+\![\boldsymbol{\alpha(s)}, \boldsymbol{x^\prime(s)}]\}. \end{equation} $

(3.5)

利用引理2.3, (3.5)式可以化简为

$ \begin{equation} \kappa^\prime(s) \left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)} \right ] =\varepsilon_{0}\kappa(s)[\boldsymbol{\beta(s)}, \boldsymbol{x(s)}], \end{equation} $

(3.6)

进一步地, 根据(3.3)和(3.4)式, 我们有

$ \begin{equation} \kappa^\prime(s) \left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)} \right ] =\kappa_{1}(s)\kappa(s)[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)}], \end{equation} $

(3.7)

于是, 中心仿射曲率$ \kappa_{1}(s) $和曲率$ \kappa(s) $满足

$ \begin{equation} \kappa_{1}(s)=\frac{\kappa^\prime(s)}{\kappa(s)}=(\log|\kappa(s)|)'. \end{equation} $

(3.8)

定理3.2   设$ \boldsymbol{x(s)} $是$ \mathbb{A}^{2} $中以$ s $为中心仿射弧长的曲线. 如果$ \boldsymbol{x(s)} $同时作为$ \mathbb{E}^{2}_{1} $中的类空（类时）曲线, 且以$ s $为弧长参数, 则曲线$ \boldsymbol{x(s)} $在$ \mathbb{E}^{2}_{1} $中的曲率$ \kappa(s) $满足

$ \kappa(s)=C_{2} e^{\int \sqrt{\kappa^{2}(s)-2\varepsilon_{0}\log|\kappa(s)|+C_{1}} ds}, (\varepsilon_{0}=\pm1, C_{1}, 0\not= C_{2}\in \mathbb {R}). $

证  根据定理3.1的证明, 在(3.6)式两端关于$ s $求导, 有

$ \ \ \ \begin{equation} \kappa^{\prime\prime}(s) \left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)} \right ]+ \kappa^\prime(s) \left[\boldsymbol{\alpha^\prime(s)}, \boldsymbol{\beta(s)} \right ]+ \kappa^\prime(s) \left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta^\prime(s)} \right ] \end{equation} \\ =\varepsilon_{0}\kappa^\prime(s)[\boldsymbol{\beta(s)}, \boldsymbol{x(s)}]+ \varepsilon_{0}\kappa(s)[\boldsymbol{\beta^\prime(s)}, \boldsymbol{x(s)}]+ \varepsilon_{0}\kappa(s)[\boldsymbol{\beta(s)}, \boldsymbol{x^\prime(s)}]. $

(3.9)

由引理2.3, (3.9)式可以化简为

$ \begin{equation} \kappa^{\prime\prime}(s) \left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)} \right ] =\varepsilon_{0}\kappa^\prime(s)[\boldsymbol{\beta(s)}, \boldsymbol{x(s)}]+ \varepsilon_{0}\kappa^2(s)[\boldsymbol{\alpha(s)}, \boldsymbol{x(s)}]+ \varepsilon_{0}\kappa(s)[\boldsymbol{\beta(s)}, \boldsymbol{\alpha(s)}], \end{equation} $

(3.10)

此外, 根据(3.3)式和(3.4)式, 我们有

$ \begin{equation} \kappa^{\prime\prime}(s) \left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)} \right ] =\frac{(\kappa^\prime(s))^2\left[ \boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)}\right]}{\kappa(s)} +\kappa^3(s)\left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)} \right ] -\varepsilon_{0}\kappa(s)[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)}], \end{equation} $

(3.11)

由(3.11)式, 曲率$ \kappa(s) $满足

$ \kappa^{\prime \prime}(s)=\frac{\left(\kappa^{\prime}(s)\right)^{2}}{\kappa(s)}+\kappa^{3}(s)-\varepsilon_{0} \kappa(s), $

即

$ (\log|\kappa(s)|)''=(\frac{\kappa^{\prime}(s)}{\kappa(s) })^\prime=\kappa^{2}(s)-\varepsilon_{0}, $

解上面的伯努利方程, 得

$ \kappa(s)=C_{2} e^{\int \sqrt{\kappa^{2}(s)-2\varepsilon_{0}\log|\kappa(s)|+C_{1}} ds}, ( C_{1}, 0\not= C_{2}\in \mathbb {R}). $

推论3.3   设$ \boldsymbol{x(s)} $是$ \mathbb{A}^{2} $中以$ s $为中心仿射弧长的曲线. 如果$ \boldsymbol{x(s)} $同时作为$ \mathbb{E}^{2}_{1} $中的类空（类时）曲线, 且以$ s $为弧长参数, 则曲线$ \boldsymbol{x(s)} $的中心仿射标架和Minkowski空间中的正交标架之间满足

$ \left(\begin{array}{l}\boldsymbol{e_{1}(s)}\\ \boldsymbol{x(s)}\end{array}\right)=\left(\begin{array}{ccc}1 & 0 \\ -\varepsilon_{0}(\log|\kappa(s)|)' & \varepsilon_{0} \kappa(s)\end{array}\right)\left(\begin{array}{l}\boldsymbol{\alpha(s)} \\ \boldsymbol{\beta(s)}\end{array}\right). $

证  首先, 由已知条件, 显然$ \boldsymbol{e_{1}(s)}=\boldsymbol{\alpha(s)}. $其次, 由引理2.1, 我们知道

$ \begin{equation*} \varepsilon_{0}\boldsymbol{x(s)} =\boldsymbol{x^{\prime \prime}(s)}-\kappa_{1}(s) \boldsymbol{e_{1}(s)}, \end{equation*} $

由定理3.1, 知$ \kappa_{1}(s)=(\log|\kappa(s)|)'. $于是, 我们有

$ \begin{equation*} \varepsilon_{0} \boldsymbol{x(s)} =\kappa(s)\boldsymbol{\beta(s)}-(\log|\kappa(s)|)'\boldsymbol{\alpha(s)}, \end{equation*} $

即得曲线$ \boldsymbol{x(s)} $的中心仿射标架和Minkowski空间中的正交标架之间的关系.

定理4.1   设$ \boldsymbol{x(s)} $是$ \mathbb{A}^{3} $中以$ s $为中心仿射弧长的曲线. 如果$ \boldsymbol{x(s)} $同时作为$ \mathbb{E}^{3}_{1} $中的第一类(第二类)类空或类时曲线, 且以$ s $为弧长参数, 则曲线$ \boldsymbol{x(s)} $的中心仿射曲率$ \kappa_{1}(s) $、中心仿射挠率$ \kappa_{2}(s) $和$ \mathbb{E}^{3}_{1} $中的曲率$ \kappa(s) $、挠率$ \tau(s) $之间满足

1. $ \kappa_{1}(s)=(\log|\kappa(s)\tau(s)|)'(\log|\frac{\tau(s)}{(\kappa(s)\tau(s))'}|)'-\varepsilon\tilde{\varepsilon}\kappa^2(s)+\varepsilon\tau^2(s); $

2. $ \kappa_{2}(s) =(\log|\kappa^2(s)\tau(s)|)', $

这里, 当$ \boldsymbol{x(s)} $是第一类类空曲线时, $ \varepsilon=\tilde{\varepsilon}=1 $; 当$ \boldsymbol{x(s)} $是第二类类空曲线时, $ \varepsilon=1, \tilde{\varepsilon}=-1 $; 当$ \boldsymbol{x(s)} $是类时曲线时, $ \varepsilon=-1, \tilde{\varepsilon}=1 $.

由于三维Minkowski空间中第一类类空曲线、第二类类空曲线和类时曲线的仿射性质证明过程十分相似, 下面仅以第一类类空曲线的仿射性质为例证明, 另外两种曲线的仿射性质不再另作说明.

证  首先, 由引理2.2和引理2.4, 曲线$ \boldsymbol{x(s)} $的中心仿射曲率$ \kappa_{1}(s) $和中心仿射挠率$ \kappa_{2}(s) $为

$ \begin{equation} \kappa_{1}(s)=\frac{\left[\boldsymbol{x^{\prime\prime\prime}(s)}, \boldsymbol{x^{\prime\prime}(s)} , \boldsymbol{x(s)}\right]}{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime \prime}(s)}, \boldsymbol{x(s)}\right]} =-\kappa^2(s)+\frac{\kappa(s)\tau(s)\left[\boldsymbol{\gamma(s)} , \boldsymbol{\beta(s)}, \boldsymbol{x(s)}\right]}{\left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)}, \boldsymbol{x(s)}\right]} \end{equation} $

(4.1)

和

$ \begin{equation} \kappa_{2}(s)=\frac{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime\prime\prime}(s)} , \boldsymbol{x(s)}\right]}{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime \prime}(s)}, \boldsymbol{x(s)}\right]} =\frac{\kappa^\prime(s)}{\kappa(s)}+\frac{\tau(s)\left[\boldsymbol{\alpha(s)}, \boldsymbol{\gamma(s)} , \boldsymbol{x(s)}\right]}{\left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)}, \boldsymbol{x(s)}\right]}, \end{equation} $

(4.2)

且同时有

$ \begin{equation} \varepsilon_{0}=\frac{\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime\prime}(s)}, \boldsymbol{x^{\prime\prime\prime}(s)}\right]} {\left[\boldsymbol{x^{\prime}(s)}, \boldsymbol{x^{\prime\prime}(s)}, \boldsymbol{x(s)}\right]} =\frac{\kappa(s)\tau(s)\left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s), \boldsymbol{\gamma(s)}}\right]}{\left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)}, \boldsymbol{x(s)}\right]}=\pm1. \end{equation} $

(4.3)

由(4.3)式, 显然有下面的等式

$ \begin{equation} \kappa(s)\tau(s){\left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s), \boldsymbol{\gamma(s)}}\right]}=\varepsilon_{0}[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)}, \boldsymbol{x(s)}], \end{equation} $

(4.4)

在(4.4)式两端关于$ s $求两次导, 有

$ \begin{equation} \begin{aligned} &(\kappa(s)\tau(s))^{\prime}{\left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s), \boldsymbol{\gamma(s)}}\right]}\!\!+\!\! \kappa(s)\tau(s)([\boldsymbol{\alpha^\prime(s)}\!, \!\boldsymbol{\beta(s)}\!, \!\boldsymbol{\gamma(s)}]\!\!+\!\! [\boldsymbol{\alpha(s)}\!, \!\boldsymbol{\beta^\prime(s)}\!, \!\boldsymbol{\gamma(s)}]\!\!+\!\! [\boldsymbol{\alpha(s)}\!, \!\boldsymbol{\beta(s)}\!, \!\boldsymbol{\gamma^\prime(s)}])\\ &=\varepsilon_{0}([\boldsymbol{\alpha^\prime(s)}, \boldsymbol{\beta(s)}, \boldsymbol{x(s)}]+ [\boldsymbol{\alpha(s)}, \boldsymbol{\beta^\prime(s)}, \boldsymbol{x(s)}]+ [\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)}, \boldsymbol{x^\prime(s)}]) \end{aligned} \end{equation} $

(4.5)

和

$ \begin{equation} \begin{aligned} &(\kappa(s)\tau(s))^{\prime\prime}[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)}, \boldsymbol{\gamma(s)}]+ (\kappa(s)\tau(s))^{\prime}\{[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)}, \boldsymbol{\gamma(s)}]\}^{\prime}\\ &=\varepsilon_{0}\tau^\prime(s)[\boldsymbol{\alpha(s)}, \boldsymbol{\gamma(s)}, \boldsymbol{x(s)}]\!+\! \varepsilon_{0}\tau(s)([\boldsymbol{\alpha^\prime(s)}, \boldsymbol{\gamma(s)}, \boldsymbol{x(s)}] \!+\![\boldsymbol{\alpha(s)}, \!\boldsymbol{\gamma^\prime(s)}, \!\boldsymbol{x(s)}]\!+\! [\boldsymbol{\alpha(s)}\!, \!\boldsymbol{\gamma(s)}\!, \!\boldsymbol{x^\prime(s)}]). \end{aligned} \end{equation} $

(4.6)

利用引理2.4, (4.5)式和(4.6)式可以化简为

$ \begin{equation} (\kappa(s)\tau(s))^{\prime}{\left[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s), \boldsymbol{\gamma(s)}}\right]}= \varepsilon_{0}\tau(s)[\boldsymbol{\alpha(s)}, \boldsymbol{\gamma(s)}, \boldsymbol{x(s)}] \end{equation} $

(4.7)

和

$ \begin{equation} \begin{aligned} &(\kappa(s)\tau(s))^{\prime\prime}[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)}, \boldsymbol{\gamma(s)}]\\ &=\varepsilon_{0}\tau^\prime(s)[\boldsymbol{\alpha(s)}, \boldsymbol{\gamma(s)}, \boldsymbol{x(s)}] +\varepsilon_{0}\kappa(s)\tau(s)[\boldsymbol{\beta(s)}, \boldsymbol{\gamma(s)}, \boldsymbol{x(s)}] +\varepsilon_{0}\tau^2(s)[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s)}, \boldsymbol{x(s)}]. \end{aligned} \end{equation} $

(4.8)

把(4.4)式和(4.7)式代入到(4.8)式, 我们有

$ \begin{equation} \varepsilon_{0}[\boldsymbol{\beta(s)}, \boldsymbol{\gamma(s)}, \boldsymbol{x(s)}] =\{\frac{(\kappa(s)\tau(s))^{\prime\prime}}{\kappa(s)\tau(s)} -\frac{\tau^\prime(s)}{\tau(s)}\frac{(\kappa(s)\tau(s))^\prime}{\kappa(s)\tau^(s)} -\tau^2(s)\} [\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s), \boldsymbol{\gamma(s)}}]. \end{equation} $

(4.9)

把(4.4), (4.7)和(4.9)式代入(4.1)和(4.2)式中, 得到

$ \begin{equation} \begin{aligned} \kappa_{1}(s) &=-\left[\frac{(\kappa(s) \tau(s))^{\prime \prime}}{\kappa(s) \tau(s)}-\frac{(\kappa(s) \tau(s))^{\prime}}{\kappa(s) \tau(s)} \frac{\tau(s)^{\prime}}{\tau(s)}+\kappa^{2}(s)-\tau^{2}(s)\right], \\ &=(\log|\kappa(s)\tau(s)|)'(\log|\frac{\tau(s)}{(\kappa(s)\tau(s))'}|)'-\kappa^2(s)+\tau^2(s) \end{aligned} \end{equation} $

(4.10)

和

$ \begin{equation} \kappa_{2}(s) =\frac{\kappa^\prime(s)}{\kappa(s)}+\frac{(\kappa(s)\tau(s))^\prime}{\kappa(s)\tau(s)}=(\log|\kappa^2(s)\tau(s)|)'. \end{equation} $

(4.11)

定理4.2   设$ \boldsymbol{x(s)} $是$ \mathbb{A}^{3} $中以$ s $为中心仿射弧长的曲线. 如果$ \boldsymbol{x(s)} $同时作为$ \mathbb{E}^{3}_{1} $中的第一类(第二类)类空或类时曲线, 且以$ s $为弧长参数, 则曲线$ \boldsymbol{x(s)} $在$ \mathbb{E}^{3}_{1} $中的曲率$ \kappa(s) $, 挠率$ \tau(s) $满足

$ \varepsilon\tilde{\varepsilon}\kappa^2(s)(\log|\kappa(s)\tau(s)|)'+[(\log|\kappa(s)\tau(s)|)' (\log|\frac{(\kappa(s)\tau(s))'}{\tau(s)}|)']'-\varepsilon(\tau^2(s))'=\varepsilon_{0}, (\varepsilon_{0}=\pm1) $

这里, 当$ \boldsymbol{x(s)} $是第一类类空曲线时, $ \varepsilon=\tilde{\varepsilon}=1 $; 当$ \boldsymbol{x(s)} $是第二类类空曲线时, $ \varepsilon=1, \tilde{\varepsilon}=-1 $; 当$ \boldsymbol{x(s)} $是类时曲线时, $ \varepsilon=-1, \tilde{\varepsilon}=1 $.

证  由定理4.1的证明, 在(4.9)式两端关于$ s $求导, 有

$ \begin{equation} \begin{aligned} &\varepsilon_{0}\{[\boldsymbol{\beta^\prime(s)}, \boldsymbol{\gamma(s)}, \boldsymbol{x(s)}]+[\boldsymbol{\beta(s)}, \boldsymbol{\gamma^\prime(s)}, \boldsymbol{x(s)}] +[\boldsymbol{\beta(s)}, \boldsymbol{\gamma(s)}, \boldsymbol{x^\prime(s)}]\}=\\ &\left\{\frac{(\kappa(s)\tau(s))^{\prime\prime}}{\kappa(s)\tau(s)} -\frac{\tau^\prime(s)}{\tau(s)}\frac{(\kappa(s)\tau(s))^\prime}{\kappa(s)\tau^(s)} -\tau^2(s)\right\}^\prime[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s), \boldsymbol{\gamma(s)}}]+\\ &\left\{\frac{(\kappa(s)\tau(s))^{\prime\prime}}{\kappa(s)\tau(s)} -\frac{\tau^\prime(s)}{\tau(s)}\frac{(\kappa(s)\tau(s))^\prime}{\kappa(s)\tau^(s)} -\tau^2(s)\right\}\left\{[\boldsymbol{\alpha(s)}, \boldsymbol{\beta(s), \boldsymbol{\gamma(s)}}]\right\}^{\prime}. \end{aligned} \end{equation} $

(4.12)

利用引理2.4, (4.12)式可以化简为

$ \begin{equation} \begin{aligned} \frac{\kappa(s)}{\tau(s)}(\kappa(s)\tau(s))^{\prime}-[\tau^2(s)]' +\left[\frac{(\kappa(s)\tau(s))^{\prime\prime}}{\kappa(s)\tau(s)} -\frac{\tau^\prime(s)}{\tau(s)}\frac{(\kappa(s)\tau(s))^\prime}{\kappa(s)\tau^(s)} \right]^\prime=\varepsilon_{0}, \end{aligned} \end{equation} $

(4.13)

即

$ \begin{equation} \kappa^2(s)(\log|\kappa(s)\tau(s)|)'+[(\log|\kappa(s)\tau(s)|)'(\log|\frac{(\kappa(s)\tau(s))'}{\tau(s)}|)']'-(\tau^2(s))'=\varepsilon_{0}, \quad (\varepsilon_{0}=\pm1). \end{equation} $

(4.14)

推论4.3   设$ \boldsymbol{x(s)} $是$ \mathbb{A}^{3} $中以$ s $为中心仿射弧长的曲线. 如果$ \boldsymbol{x(s)} $同时作为$ \mathbb{E}^{3}_{1} $中的第一类(第二类)类空或类时曲线, 且以$ s $为弧长参数, 则曲线$ \boldsymbol{x(s)} $的中心仿射标架和Minkowski空间中的正交标架之间满足

$ \left(\begin{array}{l} \!\!\!\boldsymbol{e_{1}(s)}\!\!\!\\ \!\!\!\boldsymbol{e_{2}(s)}\!\!\!\\ \!\!\!\boldsymbol{x(s)}\!\!\! \end{array}\right)\!\!=\!\!\left(\begin{array}{ccc} 1&0&0\!\!\!\\ 0&\kappa&0\!\!\!\\ -\varepsilon_{0}[(\log|\tau(s)|)'(\log|\kappa(s)\tau(s)|)'+\varepsilon\tau^{2}(s)] & -\varepsilon_{0}\frac{(\kappa(s)\tau(s))^{\prime}}{\tau(s)} & \varepsilon_{0}\kappa(s)\tau(s) \end{array}\right) \left(\begin{array}{l} \!\!\boldsymbol{\alpha(s)}\!\!\!\!\\ \!\!\boldsymbol{\beta(s)}\!\!\!\!\\ \!\!\boldsymbol{\gamma(s)}\!\!\!\!\\ \end{array}\right), $

这里$ \kappa(s) $, $ \tau(s) $为曲线$ \boldsymbol{x(s)} $的曲率和挠率. 其中, 当$ \boldsymbol{x(s)} $是第一类（第二类）类空曲线时, $ \varepsilon=1 $; 当$ \boldsymbol{x(s)} $是类时曲线时, $ \varepsilon=-1, \varepsilon_{0}=\pm1. $

证  首先, 由已知条件, 显然$ \boldsymbol{e_{1}(s)}=\boldsymbol{\alpha(s)}. $其次, 由引理2.2, 有

$ \boldsymbol{e_{2}(s)}=\boldsymbol{x''(s)}=\kappa(s)\boldsymbol{\beta(s)}. $

最后, 由引理2.2和定理4.1, 有

$ \begin{aligned} \varepsilon_{0} \boldsymbol{x(s)} &=\boldsymbol{x^{\prime \prime \prime}(s)}-\kappa_{1}(s) \boldsymbol{e_{1}(s)}-\kappa_{2}(s) \boldsymbol{e_{2}(s)}, \\ &=\boldsymbol{x^{\prime \prime\prime}(s)}-\kappa_{1}(s)\boldsymbol{x^{\prime}(s)}-\kappa_{2}(s) \boldsymbol{x^{\prime\prime}(s)}, \\ &=(-\frac{(\tau(s))^\prime}{\tau(s)}\frac{(\kappa(s)\tau(s))^\prime}{\kappa(s)\tau(s)}-\tau^2(s))\boldsymbol{\alpha(s)} -\frac{(\kappa(s)\tau(s))^{\prime}}{\tau(s)}\boldsymbol{\beta(s)}+\kappa(s)\tau(s)\boldsymbol{\gamma(s)}, \\ &=-[(\log|\tau(s)|)'(\log|\kappa(s)\tau(s)|)'+\tau^2(s)]\boldsymbol{\alpha(s)} -\frac{(\kappa(s)\tau(s))^{\prime}}{\tau(s)}\boldsymbol{\beta(s)} +\kappa(s)\tau(s)\boldsymbol{\gamma(s)}, \\ \end{aligned} $

即得曲线, $ \boldsymbol{x(s)} $, 的中心仿射标架和Minkowski空间中正交标架之间满足的关系.

进一步地, 根据定理4.2得到的结论, 讨论当曲线$ \boldsymbol{x(s)} $的曲率和挠率满足

$ \tau(s)=a\kappa^\lambda(s), (a\not=0, \lambda\in \mathbb {R}) $

时, 第一类(第二类)类空或类时曲线的曲率$ \kappa(s) $所满足的微分方程.

定理4.4   设$ \boldsymbol{x(s)} $是$ \mathbb{A}^{3} $中以$ s $为中心仿射弧长的曲线. 如果$ \boldsymbol{x(s)} $同时作为$ \mathbb{E}^{3}_{1} $中的第一类(第二类)类空或类时曲线, 且以$ s $为弧长参数. 当$ \tau(s)=a\kappa^\lambda(s) $, ($ a\not=0, \lambda\in \mathbb R $)时, 曲线$ \boldsymbol{x(s)} $的曲率$ \kappa(s) $满足如下两种情况

1. 当$ \lambda\not=-1 $时, 曲率$ \kappa(s) $满足

$ \begin{eqnarray*} \kappa^{\prime \prime}(s)+\frac{\varepsilon\tilde\varepsilon}{2} \kappa^{3}(s)- \frac{\varepsilon a^{2} }{\lambda+1} \kappa^{2 \lambda+1}(s)-\frac{\varepsilon_{0}}{\lambda+1} s\kappa(s)=0. \end{eqnarray*} $

2. 当$ \lambda=-1 $时, 曲率$ \kappa(s) $满足

$ \begin{equation*} \kappa^{2}(s)=\frac{a^{2}}{s}. \end{equation*} $

其中, 当$ \boldsymbol{x(s)} $是第一类类空曲线时, $ \varepsilon = \tilde{\varepsilon} = 1 $; 当$ \boldsymbol{x(s)} $是第二类类空曲线时, $ \varepsilon = 1, \tilde{\varepsilon} = -1 $; 当$ \boldsymbol{x(s)} $是类时曲线时, $ \varepsilon =-1, \tilde{\varepsilon} = 1 $, $ \varepsilon_{0}=\pm1 $.

证  由定理4.2知道, 当$ \boldsymbol{x(s)} $作为$ \mathbb{E}^{3}_{1} $中的第一类(第二类)类空或类时曲线, 且以$ s $为弧长参数时, 曲线$ \boldsymbol{x(s)} $的曲率$ \kappa(s) $和挠率$ \tau(s) $满足

$ \begin{equation} \varepsilon\tilde{\varepsilon}\kappa^2(s)(\log|\kappa(s)\tau(s)|)'+\left[(\log|\kappa(s)\tau(s)|)'(\log|\frac{(\kappa(s)\tau(s))'}{\tau(s)}|)'\right]'-\varepsilon(\tau^2(s))'=\varepsilon_{0}, \end{equation} $

(4.15)

其中, 当$ \boldsymbol{x(s)} $是第一类类空曲线时, $ \varepsilon = \tilde{\varepsilon} = 1 $; 当$ \boldsymbol{x(s)} $是第二类类空曲线时, $ \varepsilon = 1, \tilde{\varepsilon} = -1 $; 当$ \boldsymbol{x(s)} $是类时曲线时, $ \varepsilon =-1, \tilde{\varepsilon} = 1 $, $ \varepsilon_{0}=\pm1 $.

当$ \tau(s)=a\kappa^\lambda(s) $时, (4.15)式可以进一步整理为

$ \begin{equation} \varepsilon\tilde{\varepsilon}\frac{\lambda+1}{2}\left(\kappa^{2}(s)\right)^{\prime}- a^{2}\varepsilon\left(\kappa^{2 \lambda}(s)\right)^{\prime}+ (\lambda+1)\left(\frac{\kappa^{\prime \prime}(s)}{\kappa(s)}\right)^{\prime}=\varepsilon_{0}. \end{equation} $

(4.16)

在(4.16)式两端关于弧长$ s $积分, 可得

$ \begin{eqnarray*} \begin{aligned} -a^{2}\varepsilon\kappa^{2 \lambda}(s)+ (\lambda+1)\left\{\varepsilon\tilde{\varepsilon}\frac{\kappa^{2}(s)}{2} + \frac{\kappa^{\prime \prime}(s)}{\kappa(s)}\right\}=\varepsilon_{0}s+c, (c\in \mathbb R). \\ \end{aligned} \end{eqnarray*} $

经过适当的平移变换, 即$ \varepsilon_{0} s+c\to \varepsilon_{0} s $, 可得如下两种情况

1. 当$ \lambda\not=-1 $时, 曲率$ \kappa(s) $满足

$ \begin{eqnarray} \kappa^{\prime \prime}(s)+\frac{\varepsilon\tilde\varepsilon}{2} \kappa^{3}(s)- \frac{\varepsilon a^{2} }{\lambda+1} \kappa^{2 \lambda+1}(s)-\frac{\varepsilon_{0}}{\lambda+1} s\kappa(s)=0. \end{eqnarray} $

(4.17)

2. 当$ \lambda=-1 $, 时, 曲率$ \kappa(s) $满足

$ \begin{equation} \kappa^{2}(s)=\frac{a^{2}}{s}. \end{equation} $

(4.18)

推论4.5   设$ \boldsymbol{x(s)} $是$ \mathbb{A}^{3} $中以$ s $为中心仿射弧长的曲线. 如果$ \boldsymbol{x(s)} $同时作为$ \mathbb{E}^{3}_{1} $中的第一类(第二类)类空或类时曲线, 且以$ s $为弧长参数. 当$ \kappa(s)\tau(s)=a $, ($ a\not=0 $)时, 有如下结论

1. $ \boldsymbol{x(s)} $是从切曲线.

2. $ \boldsymbol{x(s)} $的中心仿射曲率$ \kappa_{1}(s) $和中心仿射挠率$ \kappa_{2}(s) $分别为

$ \kappa_{1}(s)=\frac{-\varepsilon\tilde{\varepsilon}a^{2}}{s}+\varepsilon s, \ \kappa_{2}(s)=-\frac{1}{2s}, $

其中, 当$ \boldsymbol{x(s)} $是第一类类空曲线时, $ \varepsilon = \tilde{\varepsilon} = 1 $; 当$ \boldsymbol{x(s)} $是第二类类空曲线时, $ \varepsilon = 1, \tilde{\varepsilon} = -1 $; 当$ \boldsymbol{x(s)} $是类时曲线时, $ \varepsilon =-1, \tilde{\varepsilon} = 1 $.

证  首先, 当$ \kappa(s)\tau(s)=a, $ ($ a\not=0 $)时, 由(4.18)式知曲率$ \kappa(s) $和挠率$ \tau(s) $满足

$ \begin{equation} \kappa(s)=\pm a s^{-\frac{1}{2}}, \tau(s)=\pm s^{\frac{1}{2}}. \end{equation} $

(4.19)

此时, 显然有

$ \frac{\tau(s)}{\kappa(s)}=a^{-1}s, $

由参考文献[7, 8]知曲线$ \boldsymbol{x(s)} $是从切曲线.

其次, 将(4.18)式代入到定理4.1的表达式中, 我们有

$ \begin{equation} \kappa_{1}(s)=(\log|\kappa(s)\tau(s)|)'(\log|\frac{\tau(s)}{(\kappa(s)\tau(s))'}|)'-\varepsilon\tilde{\varepsilon}\kappa^2(s) +\varepsilon\tau^2(s)=-\varepsilon\tilde{\varepsilon}a^{2}s^{-1}+\varepsilon s \end{equation} $

(4.20)

和

$ \begin{equation} \kappa_{2}(s)=(\log|\kappa^2(s)\tau(s)|)'=-\frac{1}{2}s^{-1}, \end{equation} $

(4.21)

其中, 当$ \boldsymbol{x(s)} $是第一类类空曲线时, $ \varepsilon = \tilde{\varepsilon} = 1 $; 当$ \boldsymbol{x(s)} $是第二类类空曲线时, $ \varepsilon= 1, \tilde{\varepsilon} = -1 $; 当$ \boldsymbol{x(s)} $是类时曲线时, $ \varepsilon =-1, \tilde{\varepsilon} = 1 $.

推论4.6   设$ \boldsymbol{x(s)} $是$ \mathbb{A}^{3} $中以$ s $为中心仿射弧长的曲线. 如果$ \boldsymbol{x(s)} $同时作为$ \mathbb{E}^{3}_{1} $中的第一类(第二类)类空或类时曲线, 且以$ s $为弧长参数. 当$ \kappa(s)\tau^{2}(s)=a^{2} $, ($ a\not=0 $)时, 曲率$ \kappa(s) $满足的方程可以转化为第二Painlev é方程

$ \begin{eqnarray*} y^{\prime \prime}(t)=2 y^{3}(t)+ty(t)-\frac{\tilde\varepsilon\varepsilon_{0} a^2}{2}. \end{eqnarray*} $

证  当$ \kappa(s)\tau^{2}(s)=a^{2} $, ($ a\not=0 $)时, (4.17)式即为

$ \begin{eqnarray} \kappa^{\prime \prime}(s)+\frac{\varepsilon\tilde\varepsilon}{2} \kappa^{3}(s)- 2 \varepsilon_{0} s \kappa(s)-2 \varepsilon a^{2}=0. \end{eqnarray} $

(4.22)

对(4.22)式作如下变换

$ \kappa(s)=-2^{\frac{4}{3}} \varepsilon_{0}\tilde{\varepsilon}y(t), \quad s=2^{-\frac{1}{3}}\varepsilon_{0} t . $

即可得第二Painlev é方程^[9], 2.8.1-1.2, p448, 即

$ \begin{eqnarray*} \label{p1} y^{\prime \prime}(t)=2 y^{3}(t)+ty(t)-\frac{\tilde\varepsilon\varepsilon_{0} a^2}{2}, \end{eqnarray*} $

其中, 当$ \boldsymbol{x(s)} $是第一类类空曲线和类时曲线时, $ \tilde{\varepsilon} = 1 $; 当$ \boldsymbol{x(s)} $是第二类类空曲线时, $ \tilde{\varepsilon} = -1 $, $ \varepsilon_{0}=\pm1 $.

推论4.7   设$ \boldsymbol{x(s)} $是$ \mathbb{A}^{3} $中以$ s $为中心仿射弧长的曲线. 如果$ \boldsymbol{x(s)} $同时作为$ \mathbb{E}^{3}_{1} $中的第一类(第二类)类空或类时曲线, 且以$ s $为弧长参数. 当$ \tau(s)=a $, ($ a\not=0 $)时, 曲率$ \kappa(s) $满足的方程可以转化为第二Painlev é方程

$ \begin{eqnarray*} \label{p2} y^{\prime \prime}(t)+2 t^{3}+t y(t)=0. \end{eqnarray*} $

证  当$ \tau(s)=a $, ($ a\not=0 $)时, (4.17)式即为

$ \begin{eqnarray} \kappa^{\prime \prime}(s)+\frac{\varepsilon\tilde\varepsilon}{2} \kappa^{3}(s)- \left(a^{2} \varepsilon+\varepsilon_{0} s\right) \kappa(s)=0. \end{eqnarray} $

(4.23)

对(4.23)式作如下变换

$ \kappa(s)=2{\varepsilon\tilde{\varepsilon}} y(t), \quad t=\varepsilon_{0} s-a^{2} \varepsilon. $

即可得第二Painlev é方程^[9], 2.8.1-1.2, p448, 为

$ y^{\prime \prime}(t)+2 t^3+t y(t)=0 .$

推论4.8   设$ \boldsymbol{x(s)} $是$ \mathbb{A}^{3} $中以$ s $为中心仿射弧长的曲线. 如果$ \boldsymbol{x(s)} $同时作为$ \mathbb{E}^{3}_{1} $中的第一类(第二类)类空或类时曲线, 且以$ s $为弧长参数. 当$ \tau(s)=a\kappa(s) $, ($ a\not=0 $)时, 曲率$ \kappa(s) $满足的方程可以转化为第二Painlev é方程

$ \begin{eqnarray*} \label{p3} \begin{array}{l} \left\{\begin{array}{l} y_{1}^{\prime \prime}(t)=t y_{1}(t) \pm 2 y_{1}^{3}(t), \\ y_{2}^{\prime \prime}(t)=-t y_{2}(t) \pm 2 y_{2}^{3}(t) . \end{array}\right. \end{array} \end{eqnarray*} $

证  当$ \tau(s)=a\kappa(s) $时, (4.17)式即为

$ \begin{eqnarray} \kappa^{\prime\prime}(s)+\frac{\varepsilon}{2}\left(\tilde{\varepsilon}-a^{2} \right) \kappa^{3}(s)- \frac{\varepsilon_{0}}{2} s \kappa(s)=0. \end{eqnarray} $

(4.24)

对(4.24)式作如下变换

$ \kappa(s)=\frac{\mp 2^{\frac{4}{3}} \varepsilon_{0} \varepsilon}{\sqrt{\tilde\varepsilon-a^{2}}} y_{1}(t), \quad s=2^{-\frac{1}{3}} \varepsilon_{0} t . $

$ \kappa(s)=\frac{\mp 2^{\frac{4}{3}} \varepsilon_{0} \varepsilon}{\sqrt{\tilde{\varepsilon}-a^{2}}} y_{2}(t), \quad s=-2^{-\frac{1}{3}} \varepsilon_{0} t. $

即可得第二Painlev é方程^[9], 2.4.2-103, p357, 为

$ \left\{\begin{array}{l} y_1^{\prime \prime}(t)=t y_1(t) \pm 2 y_1^3(t) \\ y_2^{\prime \prime}(t)=-t y_2(t) \pm 2 y_2^3(t) \end{array}\right.$

特别地, 当$ a $=1时, $ \kappa(s)=\tau(s) $, 若$ \boldsymbol{x(s)} $为三维Minkowski空间中的第一类类空曲线或者类时曲线, 则有$ \tilde\varepsilon=1 $, 那么(4.24)式为

$ \begin{eqnarray*} \kappa^{\prime \prime}(s)+\frac{1}{2} s \kappa(s)=0. \end{eqnarray*} $

此时方程的解如下

$ \kappa(s)=\left\{\begin{array}{ll} c_{1} \sqrt{s} J_{\frac{1}{3}}\left(\frac{\sqrt{2}}{3}s^{\frac{3}{2}}\right)+ c_{2} \sqrt{s} Y_{\frac{1}{3}}\left(\frac{\sqrt{2}}{3} s^{\frac{3}{2}}\right). \\ c_{1} \sqrt{s} I_{\frac{1}{3}}\left(\frac{\sqrt{2}}{3} s^{\frac{3}{2}}\right)+ c_{2} \sqrt{s} K_{\frac{1}{3}}\left(\frac{\sqrt{2}}{3} s^{\frac{3}{2}}\right). \end{array}\right. $

这里, $ J_\frac{1}{3}(s) $, $ Y_\frac{1}{3}(s) $, $ I_\frac{1}{3}(s) $, $ K_\frac{1}{3}(s) $分别是第一类, 第二类, 改进后的第一类, 改进后的第二类Bessel函数^[9], 2.1.2-7, p237.