对于给定的复数 $a, b, c$ , 其中 $c\neq 0, -1, -2, \cdots$ , 高斯超几何分布函数定义如下

$ F(a, b;c;z)= _{2}F_1(a, b;c;z)=\sum\limits_{n\geqslant 0}\frac{(a, n)(b, n)}{(c, n)}\frac{z^n}{n!}, |z|<1. $

其中当 $a\neq 0$ 时, $(a, 0)=1$ , 对 $n\in \bf{N}$ , $(a, n)$ 称为移位阶乘或Appell符号, 其定义为 $(a, n)=a(a+1)\cdots(a+n-1).$ 关于高斯超几何分布函数更多的背景、性质和应用可参看文献[1-12], 值得注意的是，许多重要的特殊函数都可以通过超几何分布函数表示出来.例如:第一型和第二型的完全椭圆积分 $K(r)$ 和 $E(r)$ (参见文献[13-14])分别可以表示为

$ K=K(r) = \int_0^{\pi /2} {\frac{1} {{\sqrt {1 - r^2 \sin ^2 t} }}}{\rm d}t =\frac{\pi } {2}F\Big(\frac{1} {2}, \frac{1} {2};1;r^2\Big ), ~~K' = K' (r) = K (\sqrt{1-r^2}) $

和

$ E = E(r) = \int_0^{\pi /2} {\sqrt {1 - r^2 \sin ^2 t} }{\rm d}t = \frac{\pi }{2}F \Big( { - \frac{1}{2}, \frac{1}{2};1;r^2 } \Big), ~~E' = E' (r) = E (\sqrt{1-r^2}). $

接下来, 简单介绍一下Takeuchi给出的广义椭圆积分^[15]定义, Lindqvist^[16]于1995年提出的针对参数 $p>1$ 时的广义三角函数和广义双曲函数, 当 $p=2$ 时, 这些函数即为通常的三角函数(参阅文献[17-18]).一维 $p$ -拉普拉斯问题^[19]

$ -\Delta_p u=-\left(|u'|^{p-2}u'\right)' =\lambda|u|^{p-2}u, \, u(0)=u(1)=0, \ \ \ p>1 $

的特征函数 $\sin_p$ 称为函数 $F:(0, 1)\to \left(0, \frac{\pi_p}{2}\right)$ 的反函数, 其定义如下

$ F(x)={\rm arcsin}_p(x)=\int^x_0(1-t^p)^{-\frac{1}{p}}{\rm d}t, $

并且

$ \pi_p=2{\rm arcsin}_{p}(1)=\frac{2}{p}\int^1_0(1-s)^{-1/p}s^{1/p-1}{\rm d}s=\frac{2} {p}\, B\left(1-\frac{1}{p}, \frac{1}{p}\right)=\frac{2 \pi}{p\, \sin\big(\frac{\pi}{p}\big)}\, , $

其中 $B(\cdot, \cdot)$ 表示经典的贝塔函数.对任意 $p\in (1, \infty), $ $r\in [0, 1)$ , 第一型和第二型广义椭圆积分分别定义为

$ K_p (r) = \int_0^{\pi _p /2} {\left( {1- r^p \sin_{p} ^p t} \right)^{\left( {1/p} \right)- 1} } {\rm d}t, \ K'_p = K'_p (r) = K_p (r') $

和

$ E_p (r) = \int_0^{\pi _p /2} {\left( {1- r^p \sin_{p} ^p t} \right)^{1/p} }{\rm d}t, \ E'_p = E'_p (r) = E_p (r'), $

其中 $p>1, r\in(0, 1)$ 并且 $ r' = \left( {1 - r^p } \right)^{1/p}. $ 依据 $K_p(r)$ 和 $E_p(r)$ 的定义, 它们的超几何函数表示为

$ \begin{equation}\label{eq:1.1} K_p (r) = \frac{{\pi _p }}{2}F\left( {\frac{1}{p}, 1 - \frac{1}{p};1;r^p } \right) \end{equation} $

(1.1)

和

$ \begin{equation}\label{eq:1.2} E_p (r) = \frac{{\pi _p }}{2}F\left( {\frac{1}{p}, - \frac{1}{p};1;r^p } \right). \end{equation} $

(1.2)

关于 $K_p(r)$ 和 $E_p(r)$ 的这些公式和相关性质可参看文献[20-25].

本文主要思想受Alzer和Richards研究^[14]的启发, 他们的主要结果如下.

性质1.1 函数

$ \Delta (r) = \frac{{E - \left( {1 - r^2 } \right)K}}{{r^2 }} - \frac{{E' - r^2 K'}}{{\left( {1 - r^2 } \right)}} $

是从 $(0, 1)$ 映射到 $(\pi/4-1, 1-\pi/4)$ 的严格单增的凸函数, 且对任意 $r\in(0, 1)$ , 有

$ \begin{eqnarray*} \frac{\pi }{4} - 1 + \alpha r < \Delta (r) < \frac{\pi }{4} - 1 + \beta r, \end{eqnarray*} $

其中 $\alpha=0, $ $\beta=2-\frac{\pi}{2}=0.42920\cdots.$

为了将性质1.1推广到广义椭圆积分, 首先, 给出以下定义

$ \Delta _p (r) = \frac{{E_p - \left( {r'} \right)^p K_p }}{{r^p }} - \frac{{E'_p - r^p K'_p }}{{\left( {r'} \right)^p }}. $

对此，我们有如下结论.

定理1.1 当 $p$ 满足条件

(ⅰ) $2 + \frac{1}{p} + \frac{1}{{p^2 }} \le \frac{6}{p} < 3 + \frac{1}{{p^2 }};$

(ⅱ) $\varepsilon(p)>0$ 时, 函数 $\Delta_p$ 为从 $(0, 1)$ 映射到 $ \Big( {\frac{{\big( {1 - \frac{1}{p}} \big)\pi _p }}{2} - 1, 1 - \frac{{\big( {1 - \frac{1}{p}} \big)\pi _p }}{2}} \Big) $ 的严格单调递增的凸函数, 其中

$ 20 - \frac{36}{p} - \frac{{1}}{{p^2 }} + \frac{6}{{p^3}} - \frac{1}{{p^4 }}. $

此外, 对任意 $r\in(0, 1)$ , 有

$ \begin{equation}\label{eq:1.3} {\frac{{\big( {1 - \frac{1}{p}} \big)\pi _p }}{2} - 1} + \alpha (r) < \Delta_p (r) < {\frac{{\big( {1 - \frac{1}{p}} \big)\pi _p }}{2} - 1} + \beta r, \end{equation} $

(1.3)

其中 $\alpha=0, $ $\beta= {2 - \big( {1 - \frac{1}{p}} \big)\pi _p }.$

定理1.2 对任意 $r, s\in(0, 1)$ , 当 $p$ 满足条件(ⅰ)和(ⅱ)时, 有

$ \frac{{\big( {1 - \frac{1}{p}} \big)\pi _p }}{2} - 1 < \Delta _p (rs) - \Delta _p( r) - \Delta _p( s) < 1 - \frac{{\big( {1 - \frac{1}{p}} \big)\pi _p }}{2}. $

引理2.1 记

$ \begin{eqnarray*} H_{p, a} (r) = \frac{{\pi _p }}{{2r^p }}\left[{F\left( {a, -a;1;r^p } \right)-\left( {r'} \right)^p F\left( {a, 1-a;1;r^p } \right)} \right]. \end{eqnarray*} $

当 $p>1, $ $r\in(0, 1)$ 时, 有

$ \begin{eqnarray*} H_{p, a} (r) = \frac{{(1 - a)\pi _p }}{2}F\left( {a, 1 - a;2;r^p } \right). \end{eqnarray*} $

证 由超几何函数的定义可得

$ \begin{eqnarray*} H_{p, a} (r) &=&\frac{{\pi _p }}{{2r^p }}\left[{\sum\limits_{n = 0}^\infty {\frac{{(a)_n (-a)_n }}{{\left( {n!} \right)^2 }}r^{pn} }-\left( {1-r^p } \right)\sum\limits_{n = 0}^\infty {\frac{{(a)_n (1 - a)_n }}{{\left( {n!} \right)^2 }}r^{pn} } } \right] \\ & =&\frac{{\pi _p }}{{2r^p }}\left[{\sum\limits_{n = 0}^\infty {\left( {\frac{{(a)_n (-a)_n }}{{\left( {n!} \right)^2 }}-\frac{{(a)_n (1-a)_n }}{{\left( {n!} \right)^2 }}} \right)r^{pn} } + \sum\limits_{n = 0}^\infty {\frac{{(a)_n (1 - a)_n }}{{\left( {n!} \right)^2 }}r^{p(n + 1)} } } \right] \\ & =&\frac{{\pi _p }}{2}\left[{\sum\limits_{n = 0}^\infty {\left( {\frac{{(a)_{n + 1} (-a)_{n + 1} }}{{\left( {(n + 1)!} \right)^2 }}-\frac{{(a)_{n + 1} (1-a)_{n + 1} }}{{\left( {(n + 1)!} \right)^2 }}} \right)r^{pn} } + \sum\limits_{n = 0}^\infty {\frac{{(a)_n (1 - a)_n }}{{\left( {n!} \right)^2 }}r^{pn} } } \right] \\ &=&\frac{{\pi _p }}{2}\left[{\sum\limits_{n = 0}^\infty {\frac{{(a)_n (1-a)_n }}{{\left( {(n + 1)!} \right)^2 }}\left( {(n + a)(-a)-(a + n)(1 - a + n) + (n + 1)^2 } \right)r^{pn} } } \right] \\ &=&\frac{{(1 - a)\pi _p }}{2}\sum\limits_{n = 0}^\infty {\frac{{(a)_n (1 - a)_n }}{{n!(2)_n }}r^{pn} } \\ & =&\frac{{(1 - a)\pi _p }}{2}F\left( {a, 1 - a;2;r^p } \right). \end{eqnarray*} $

证毕.

引理2.2 当 $p>1, $ $r\in(0, 1)$ 时, 有

$ \begin{equation}\label{eq:2.1} F\left( {2 + \frac{1}{p}, 3 - \frac{1}{p};4;1 - r^p } \right) = \frac{{F\big( {1 + \frac{1}{p}, 2 - \frac{1}{p};4;1 - r^p } \big)}}{{r^p }}. \end{equation} $

(2.1)

证 在文献[14]中的恒等式

$ \begin{eqnarray*} (1 - z)F\left( {a + 1, b + 1;a + b + 1;z} \right) = F\left( {a, b;a + b + 1;z} \right) \end{eqnarray*} $

取 $a=1+\frac{1}{p}, b=2-\frac{1}{p}, z=1-r^p$ , 可得公式(2.1).

引理2.3 当 $p>1, $ $r\in(0, 1)$ 时, 有

$ \begin{eqnarray}\label{eq:2.2} && \left( {2 - \frac{1}{p}} \right)F\left( {1 + \frac{1}{p}, 3 - \frac{1}{p};4;1 - r^p } \right)\\ &=& 3F\left( {1 + \frac{1}{p}, 2 - \frac{1}{p};3;1 - r^p } \right) - \left( {1 + \frac{1}{p}} \right)F\left( {1 + \frac{1}{p}, 2 - \frac{1}{p};4;1 - r^p } \right). \end{eqnarray} $

(2.2)

证 考虑文献[26, 方程26]

$ \begin{eqnarray*} (\sigma - \rho )F\left( {\alpha, \rho ;\sigma + 1;z} \right) = \sigma F\left( {\alpha, \rho ;\sigma ;z} \right) - \rho F\left( {\alpha, \rho + 1;\sigma + 1;z} \right). \end{eqnarray*} $

令 $ \sigma = 3, \alpha = 1 + \frac{1}{p}, \rho = 2 - \frac{1}{p}, z = 1 - r^p $ 可得(2.2)式成立.

引理2.4 以下两式恒成立

$ \begin{eqnarray*} H_{p, 1/p} (0) = \frac{{\big( {1 - \frac{1}{p}} \big)\pi _p }}{2}, H_{p, 1/p} (1) = 1. \end{eqnarray*} $

证 显然

$ H_{p, 1/p} (0) = \frac{{\big( {1 - \frac{1}{p}} \big)\pi _p }}{2}F \left( {\frac{1}{p}, 1 - \frac{1}{p};2;0} \right) = \frac{{\big( {1 - \frac{1}{p}} \big)\pi _p }}{2}. $

利用恒等式(详见文献[27, 第153页])

$ F\left( {\alpha, \beta ;\gamma ;1} \right) = \frac{{\Gamma (\gamma )\Gamma (\gamma - \alpha - \beta )}}{{\Gamma \left( {\gamma - \alpha } \right)\Gamma \left( {\gamma - \beta } \right)}}, $

可得

$ \begin{eqnarray*} H_{p, 1/p} (1) &= &\frac{{\big( {1 - \frac{1}{p}} \big)\pi _p }}{2}F \left( {\frac{1}{p}, 1 - \frac{1}{p};2;1} \right) \\ & = &\frac{{\big( {1 - \frac{1}{p}} \big)\pi _p }}{2} \frac{{\Gamma (2)\Gamma (1)}}{{\Gamma \big( {2 - \frac{1}{p}} \big)\Gamma \big( {1 + \frac{1}{p}} \big)}} \\ &=& \frac{{\big( {1 - \frac{1}{p}} \big)\pi _p }}{2}\frac{1}{{\big( {1 - \frac{1}{p}} \big)\Gamma \big( {1 - \frac{1}{p}} \big) \frac{1}{p}\Gamma \big( {\frac{1}{p}} \big)}} \\ &=& 1. \end{eqnarray*} $

证毕.

定理1.1的证明 由公式(1.1), (1.2)和引理2.1可得

$ \begin{eqnarray*} \Delta _p (r) &= &H_{p, 1/p} (r) - H_{p, 1/p} (r') \\ & = &\frac{{\big( {1 - \frac{1}{p}} \big)\pi _p }}{2}\left[{F\left( {\frac{1}{p}, 1-\frac{1}{p};2;r^p } \right)-F\left( {\frac{1}{p}, 1-\frac{1}{p};2;1 - r^p } \right)} \right]. \end{eqnarray*} $

由

$ \frac{\rm d}{{{\rm d}r}}F\left( {a, b;c;r} \right) = \frac{{ab}}{c}F\left( {a + 1, b + 1;c + 1;r} \right) $

容易得到

$ \Delta '_p (r) = \frac{{\big( {1 - \frac{1}{p}} \big)^2 \pi _p r^{p-1}}}{4}\left[{F\left( {1 + \frac{1}{p}, 2-\frac{1}{p};3;r^p } \right)-F\left( {1 + \frac{1}{p}, 2-\frac{1}{p};3;1 - r^p } \right)} \right] $

和

$ \begin{eqnarray*} &&\frac{4}{{\big( {1 - \frac{1}{p}} \big)^2 \pi _p r^{p - 2} }}\Delta ''_p (r)\\ & =&(p-1)\left\{ {\left[{F\left( {1 + \frac{1}{p}, 2-\frac{1}{p};3;r^p } \right)-F\left( {1 + \frac{1}{p}, 2-\frac{1}{p};3;1 - r^p } \right)} \right]} \right\} \\ && + \frac{{(p + 1)(2p - 1)r^p }}{{3p}}\left[{F\left( {2 + \frac{1}{p}, 3-\frac{1}{p};4;r^p } \right)-F\left( {2 + \frac{1}{p}, 3-\frac{1}{p};4;1 - r^p } \right)} \right]. \end{eqnarray*} $

再由引理2.2可得

$ \begin{eqnarray*} && \frac{4}{{\big( {1 - \frac{1}{p}} \big)^2 \pi _p r^{p - 2}}}\Delta ''_p (r) \\ &=& (p - 1)F\left( {1 + \frac{1}{p}, 2 - \frac{1}{p};3;r^p } \right) - (p - 1)F\left( {1 + \frac{1}{p}, 2 - \frac{1}{p};3;\left( {r'} \right)^p } \right) \\ && + \frac{{p\big( {1 + \frac{1}{p}} \big)\big( {2 - \frac{1}{p}} \big)}} {{\left( {3} \right)}}r^p F\left( {2 + \frac{1}{p}, 3 - \frac{1}{p};4;r^p } \right) - p\left( {2 - \frac{1}{p}} \right)F\left( {1 + \frac{1}{p}, 2 - \frac{1}{p};3;\left( {r'} \right)^p } \right) \\ && + \frac{{p\big( {2 - \frac{1}{p}} \big)^2 }}{{\big( {3 + \frac{1}{p} - \frac{1}{p}} \big)}}F\left( {1 + \frac{1}{p}, 3 - \frac{1}{p};4;\left( {r'} \right)^p } \right) \\ &=& (p - 1)F\left( {1 + \frac{1}{p}, 2 - \frac{1}{p};3;r^p } \right) + \frac{{p\big( {1 + \frac{1}{p}} \big) \big( {2 - \frac{1}{p}} \big)}}{{\left( {3} \right)}}r^p F\left( {2 + \frac{1}{p}, 3 - \frac{1}{p};4;r^p } \right) \\ && - p\left[{p\left( {2-\frac{1}{p}} \right)-(p-1)} \right] F\left( {1 + \frac{1}{p}, 2 - \frac{1}{p};3;\left( {r'} \right)^p } \right) \\ && + \frac{{p\big( {2 - \frac{1}{p}} \big)^2 }}{{ {3 } }}F\left( {1 + \frac{1}{p}, 3 - \frac{1}{p};4;\left( {r'} \right)^p } \right), \end{eqnarray*} $

再应用条件(ⅰ)和(ⅱ)及引理2.3, 得

$ \begin{eqnarray*} && \frac{4}{{\big( {1 - \frac{1}{p}} \big)^2 \pi _p r^{p - 2}}}\Delta ''_p (r) \\ &\ge& (p - 1) - pF\left( {1 + \frac{1}{p}, 2 - \frac{1}{p};3;\left( {r'} \right)^p } \right) + \frac{{p\big( {2 - \frac{1}{p}} \big)^2 }} {{\left( {3 } \right)}}F\left( {1 + \frac{1}{p}, 3 - \frac{1}{p};4; \left( {r'} \right)^p } \right) \\ & =& (p - 1)\left[{1 + \sum\limits_{n = 0}^\infty {\frac{{\big( {1 + \frac{1}{p}} \big)_n \big( {2-\frac{1}{p}} \big)}}{{\left( {3} \right)_{n}\left( {n + 3 } \right)}}\left( {n-\frac{{\frac{1}{p} + \frac{3}{p}-1 - \frac{1}{{p^2 }}}}{{1 - \frac{1}{p}}}} \right)\frac{{\left( {r'} \right)^{pn} }}{{n!}}} } \right] \\ & > &(p - 1)\left[{1 + \sum\limits_{n = 0}^1 {\frac{{\big( {1 + \frac{1}{p}} \big)_n \big( {2-\frac{1}{p}} \big)}} {{\left( {3 } \right)_{n}\left( {n + 3} \right)}}\left( {n- \frac{{\frac{1}{p} + \frac{3}{p}-1 -\frac{1}{{p^2 }}}}{{1 - \frac{1}{p}}}} \right)\frac{{\left( {r'} \right)^{pn} }}{{n!}}} } \right] \\ &=& (p - 1)\left[{\frac{{\varepsilon (p)}} {{12\big( {1-\frac{1}{p}} \big)}} + \frac{{\big( {1 + \frac{1}{p}} \big) \big( {2-\frac{1}{p}} \big)\big( {\frac{5}{p}-2 - \frac{1}{{p^2 }}} \big)r^p }}{{12\big( {1 - \frac{1}{p}} \big)}}} \right] \\ & > &0. \end{eqnarray*} $

由 $ \Delta ''_p (r) > 0$ , 可知 $\Delta '_p (r)$ 是 $(0, 1)$ 上的严格单调递增函数.从而, 有 $\Delta '_p (r)> \Delta '_p (0).$ 利用洛必达法则, 得

$ \begin{eqnarray*} \Delta _p ^\prime (0) &=& \mathop {\lim }\limits_{r \to 0^ + } \frac{{\Delta _p (r) - \Delta _p (0)}}{{r - 0}} \\ &=& \mathop {\lim }\limits_{r \to 0^ + } \frac{{H_{p, 1/p} (r) - H_{p, 1/p} (0)}}{{r - 0}} - \frac{{H_{p, 1/p} (r') - H_{p, 1/p} (1)}}{{r - 0}} \\ &=& H'_{p, 1/p} (0) - \mathop {\lim }\limits_{x \to 1^ + } \frac{{H_{p, 1/p} (x) - 1}}{{\left( {1 - x^p } \right)^{1/p} }} \\ &= &\mathop {\lim }\limits_{x \to 1^ + } H'_{p, 1/p} (x)\frac{{\left( {1 - x^p } \right)^{1 - 1/p} }}{{x^{p - 1} }} \\ &= &0, \end{eqnarray*} $

其中

$ H'_{p, 1/p} (0) = \frac{{\big( {1 - \frac{1}{p}} \big)^2 \pi _p r^{p-1}}}{4}F\left( {1 + \frac{1}{p}, 2 - \frac{1}{p};3;r^p } \right)\left| {_{r = 0} } \right. = 0. $

从而 $\Delta_p (r)$ 是 $(0, 1)$ 上的严格单调递增函数.

定义

$ M_p (r) = \frac{{\Delta _p (r) - \Delta _p (0)}}{{r - 0}}. $

由 $\Delta_p (r)$ 是 $(0, 1)$ 上的严格凸函数可知 $M_p (r)$ 是 $(0, 1)$ 上的严格单调递增函数, 从而有

$ M_p (0)<M_p (r)<M_p (1). $

由引理2.4可知 $M_p (0)=0$ ,

$ M_p (1) = \Delta _p (1) - \Delta _p (0) = 2H_{p, 1/p} (1) - 2H_{p, 1/p} (0) = 2 - \left( {1 - \frac{1}{p}} \right)\pi _p, $

从而公式(1.3)成立.

定理1.2的证明 记

$ \lambda _p (r, s) = \Delta _p (rs) - \Delta _p (r) - \Delta _p (s), $

经简单计算, 得

$ \frac{\partial }{{\partial r}}\lambda _p (r, s) = s\Delta '_p (rs) - \Delta _p ^\prime (r), $

$ \frac{{\partial ^2 }}{{\partial r\partial s}}\lambda _p (r, s) = \Delta '_p (rs) + rs\Delta _p ^{\prime \prime } (rs). $

$ \frac{\partial }{{\partial r}}\lambda _p (r, s) < \frac{\partial }{{\partial r}}\lambda _p (r, s)\left| {_{s = 1} } \right. = 0. $

由定理1.1的结论可知 $\frac{{\partial ^2 }}{{\partial r\partial s}}\lambda _p (r, s)>0$ .从而函数 $\frac{\partial }{{\partial r}}\lambda _p (r, s)$ 是关于变量 $s$ 的严格单调递增函数.由

$ \frac{\partial }{{\partial r}}\lambda _p (r, s) < \frac{\partial }{{\partial r}}\lambda _p (r, s)\left| {_{s = 1} } \right. = 0 $

得 $ r \mapsto \lambda _p (r, s) $ 是严格单调递减的函数, 从而

$ - \Delta _p (1) = \lambda _p (1, s) < \lambda _p (r, s) < \lambda _p (0, s) =-\Delta _p (s)< - \Delta _p (0). $

证毕.