令$ 0<r<1 $,勒让德第一类和第二类完全椭圆积分$ {\cal K}(r) $和$ {\cal E}(r) $^[1]分别定义为

(1.1) $ \begin{equation} {\cal K} = {\cal K}(r) = \int_{0}^{\pi/2}\frac{{\rm d}\theta}{\sqrt{1-r^2\sin^{2}\theta}}, \quad {\cal K}(0^{+}) = \frac{\pi}{2}, \quad {\cal K}(1^{-}) = \infty, \end{equation} $

(1.2) $ \begin{equation} {\cal E} = {\cal E}(r) = \int_{0}^{\pi/2}\sqrt{1-r^2\sin^{2}\theta}{\rm d}\theta, \quad {\cal E}(0^{+}) = \frac{\pi}{2}, \quad {\cal E}(1^{-}) = 1. \end{equation} $

椭圆积分如此命名是因为它们出现在椭圆周长的计算公式.众所周知,完全椭圆积分在数学、物理及其其它的一些学科中具有广泛的应用.例如,圆周率$ \pi $的计算,一些微分方程的解的表达式,电磁场和单摆周期的计算公式等都与两类完全椭圆积分$ {\cal K}(r) $和$ {\cal E}(r) $密切相关.

在过去的三十多年,完全椭圆积分的研究主要集中于它们在几何函数论中的应用.许多共形不变量、拟共形映射中的偏差函数都可以用完全椭圆积分来表示^[2-3].特别地,关于$ {\cal K}(r) $和$ {\cal E}(r) $的许多优美的性质和上下界被证明^[4-21].

令$ p\in(1, \infty) $, $ r\in(0, 1) $, Takeuchi^[22]引入了二类完全$ p $ -椭圆积分$ {K}_{p}(r) $和$ {E}_{p}(r) $,其定义如下

(1.3) $ \begin{equation} {K}_{p} = {K}_{p}(r) = \int_{0}^{\pi_{p}/2}\frac{{\rm d}\theta}{(1-r^p\sin_{p}^{p}\theta)^{1-1/p}}, \quad {K}_{p}(0^{+}) = \frac{\pi_{p}}{2}, \quad {K}_{p}(1^-) = \infty, \end{equation} $

(1.4) $ \begin{equation} {E}_{p} = {E}_{p}(r) = \int_{0}^{\pi_{p}/2}(1-r^p\sin_{p}^{p}\theta)^{1/p}{\rm d}\theta, \quad {E}_{p}(0^{+}) = \frac{\pi_{p}}{2}, \quad {E}_{p}(1^{-}) = 1, \end{equation} $

其中$ \sin_{p}\theta $是广义三角正弦函数,具有反函数

$ {\sin_{p}}^{-1}\theta = \int_{0}^{\theta}\frac{{\rm d}t}{(1-t^p)^{1/p}}, \quad 0\leq \theta \leq 1, $

而

$ \pi_{p} = 2\int_{0}^{1}\frac{{\rm d}t}{(1-t^p)^{1/p}} = \frac{2\pi}{p\sin{(\pi/p)}} $

是广义圆周率.

易见$ \sin_{2}\theta = \sin\theta $, $ \pi_{2} = \pi $,且由(1.1)–(1.4)式可得$ {K}_{2}(r) = {\cal K}(r) $, $ {E}_{2}(r) = {\cal E}(r) $. $ \sin_{p}\theta $及$ \pi_{p} $首先被给出在$ 1 $维的$ p $-Laplacian方程的特征值问题^[23-25].

近几年,完全$ p $ -椭圆积分已经被Takeuchi等从不同的角度进行深入研究. 2016年, Takeuchi^{[22, 26]}证明完全$ p $ -椭圆积分$ K_{p} $和$ E_{p} $满足如下导数公式及其广义Legendre恒等式

$ \frac{{\rm d} K_{p}(r)}{{\rm d}r} = \frac{E_{p}-{r'}^pK_{p}}{r{r'}^p}, \quad \frac{{\rm d} E_{p}(r)}{{\rm d}r} = \frac{E_{p}-K_{p}}{r}, $

$ \frac{{\rm d} (K_{p}-E_{p})}{{\rm d}r} = \frac{r^{p-1}E_{p}(r)}{{r'}^p}, \quad \frac{{\rm d} (E_{p}-{r'}^pK_{p})}{{\rm d}r} = (p-1)r^{p-1}K_{p}(r), $

(1.5) $ \begin{equation} K_{p}(r')E_{p}(r)+K_{p}(r)E_{p}(r')-K_{p}(r)K_{p}(r') = \frac{\pi_{p}}{2}, \end{equation} $

其中对$ r\in(0, 1) $, $ r' = \sqrt[p]{1-r^p} $.利用式(1.5), Takeuchi^{[22, 27]}发现了当$ p = 3 $和$ p = 4 $时$ \pi_{p} $的算法公式.此外,完全$ p $ -椭圆积分与经典的Gauss超几何函数具有如下关系(参见文献[22,命题2.8])

(1.6) $ \begin{equation} K_{p}(r) = \frac{\pi_{p}}{2}F\left(1-\frac{1}{p}, \frac{1}{p};1;r^p\right), \quad E_{p}(r) = \frac{\pi_{p}}{2}F\left(-\frac{1}{p}, \frac{1}{p};1;r^p\right), \end{equation} $

其中

$ \begin{eqnarray*} F(a, b;c;x) = {}_{2}F_{1}(a, b;c;x) = \sum\limits_{n = 0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}}\frac{x^n}{n!}\ \left( |x|<1\right) \end{eqnarray*} $

是Gauss超几何函数^[28-29], $ a, b, c $为实参数且$ c\neq 0, -1, -2, \cdots $, $ (a)_{0} = 1 $$ (a\neq 0) $, $ (a)_{n} = a(a+1)(a+2)(a+3)\cdots(a+n-1) = \Gamma(n+a)/\Gamma(a) $$ (n = 1, 2, 3\cdots) $, $ \Gamma(x) = \int_{0}^{\infty}t^{x-1}e^{-t}{\rm d}t $是Gamma函数^[30-31].

最近,张孝惠^[32]将完全椭圆积分$ {\cal K}(r) $和$ {\cal E}(r) $满足的一些单调性质和不等式推广到了完全$ p $ -椭圆积分$ K_{p}(r) $和$ E_{p}(r) $.

完全椭圆积分的另一种重要的广义形式,称为广义椭圆积分$ {{\cal K}}_{a}(r) $, $ {{\cal E}}_{a}(r) $,已经被Vuorinen教授及其合作者引入并在Ramanujan模方程理论中广泛研究(参见文献[33]).事实上,对$ a\in(0, 1) $, $ r\in(0, 1) $,第一类和第二类广义椭圆积分$ {{\cal K}}_{a}(r) $, $ {{\cal E}}_{a}(r) $分别定义为

(1.7) $ \begin{equation} {{\cal K}}_{a} = {{\cal K}}_{a}(r) = \frac{\pi}{2}F(a, 1-a;1;r^2), \quad {{\cal E}}_{a} = {{\cal E}}_{a}(r) = \frac{\pi}{2}F(a-1, 1-a;1;r^2). \end{equation} $

易见$ {{\cal K}}_{1/2}(r) = {{\cal K}}(r) $, $ {{\cal E}}_{1/2}(r) = {{\cal E}}(r) $.比较(1.6)和(1.7)式,我们便知：令$ p = 1/(1-a) $,

(1.8) $ \begin{equation} {{\cal K}}_{a}(r) = \frac{\pi}{\pi_{p}}{K}_{p}(r^{2/p}), \ {{\cal E}}_{a}(r) = \frac{\pi}{\pi_{p}}{E}_{p}(r^{2/p}). \end{equation} $

更多$ {{\cal K}}_{a}(r) $, $ {{\cal E}}_{a}(r) $的性质及其应用可参见文献[34-40].

对$ \alpha\in{{\Bbb R}} $,两个正数$ x $和$ y $的$ \alpha $阶Hölder平均值定义为

$ \begin{eqnarray*} H_{\alpha}(x, y) = \left\{ \begin{array}{ll} \left( \frac{x^{\alpha}+y^{\alpha}}{2}\right)^{{1}/{\alpha}}, &\alpha\neq0, \\ \sqrt{xy}, &\alpha = 0. \end{array}\right. \end{eqnarray*} $

特别令$ \alpha = -1 $, $ 0 $和$ 1 $,便得到经典的调和平均值$ H(x, y) = H_{-1}(x, y) $,几何平均值$ G(x, y) = H_{0}(x, y) $及其算术平均值$ A(x, y) = H_{1}(x, y) $.众所周知,对固定的$ x, y>0 $且$ x\neq y $, Hölder平均值$ H_{\alpha}(x, y) $关于$ \alpha $在实数域$ {{\Bbb R}} $上是连续且严格单调增加的. $ \!\! $Hölder平均值的一些基本性质及其不等式可参见文献[43].

令$ \alpha, \beta\in{{\Bbb R}} $,区间$ I\subseteq (0, \infty) $,函数$ f:I\mapsto(0, \infty) $连续.若对于所有的$ x, y\in I $都成立如下不等式

(1.9) $ \begin{equation} f(H_{\alpha}(x, y))\leq(\geq)H_{\beta}(f(x), f(y)), \end{equation} $

则称$ f $在$ I $是$ H_{\alpha, \beta} $ -凸的(凹的).而且,若不等式(1.9)成立等号当且仅当$ x = y $,便称$ f $在$ I $是严格$ H_{\alpha, \beta} $ -凸的(凹的).特别当$ \alpha = \beta = 1 $时, $ H_{\alpha, \beta} $ -凸性(凹性)就退化为经典的凸性（凹性）.

2007年,对$ \alpha = 0, -1, 1 $和$ \beta = 0, -1, 1 $, Anderson, Vamanamurthy和Vuorinen^[44]探究了由Maclaurin级数定义的函数的$ H_{\alpha, \beta} $ -凸性和$ H_{\alpha, \beta} $ -凹性.运用该结果, $ \!\! $Gauss超几何函数的$ H_{\alpha, \beta} $ -凸性(凹性)及其一类新的不等式被证得.

Baricz^[45]考察了零平衡超几何函数$ F(a, b;a+b;x)(a, b>0) $的Hölder凸性,发现对$ (\alpha, \beta)\in\{(\alpha, \beta)|\alpha\leq 2, \beta\geq 0\} $,第一类完全椭圆积分$ {\cal K}(r) $是在$ (0, 1) $上严格$ H_{\alpha, \beta} $ -凸的.随后,使得$ {\cal K}(r) $在$ (0, 1) $上严格$ H_{\alpha, \beta} $ -凸的完整区域被王淼坤等^[46]证明.类似地,相同作者在文献[47]中得到了关于第二类完全椭圆积分在$ (0, 1) $满足Hölder凹的充要条件：

定理1.1^[47]  第二类完全椭圆积分$ {\cal E}(r) $在$ (0, 1) $是$ H_{\alpha, \beta} $ -凹的当且仅当

$ (\beta, \alpha)\in \{(\beta, \alpha)|\alpha\leq C(\beta)\}, $

其中$ C(\beta) = \inf\limits_{r\in(0, 1)}\{r^2{\cal E}/[(1-r^2)({\cal K}-{\cal E})+(1-\beta)({\cal K}-{\cal E})/{\cal E}]\} $关于$ \beta $连续,且当$ \beta\leq 5/2 $时, $ C(\beta) = 2 $,而当$ \beta>5/2 $时, $ C(\beta)<2 $.此外,不存在$ \alpha, \beta\in {{\Bbb R}} $使得$ {\cal E}(r) $在$ (0, 1) $是$ H_{\alpha, \beta} $ -凸的.

2012年,周丽明等^[48]考虑了两类广义椭圆积分,给出了$ {\cal K}_{a}(r) $在$ (0, 1) $是$ H_{\alpha, \alpha} $ -凸的及其$ {\cal E}_{a}(r) $ ($ a\in(0, 1/2] $)在$ (0, 1) $是$ H_{\alpha, \alpha} $ -凹的充分必要条件.最近, Baricz和Vuorinen在文献[49-50]中也研究广义三角函数的$ H_{\alpha, \alpha} $ -凸性(凹性).

本文的主要目的是将定理1.1推广到第二类$ p $ -完全椭圆积分和第二类广义椭圆积分的情形.主要结果是如下的定理1.2和推论1.1.

定理1.2  对$ p\in(1, \infty) $,第二类$ p $ -完全椭圆积分$ E_{p}(r) $在$ (0, 1) $是$ H_{\alpha, \beta} $ -凹的,即不等式

$ E_{p}(H_{\alpha}(x, y))\geq H_{\beta}(E_{p}(x), E_{p}(y)) $

对所有的$ x, y\in (0, 1) $恒成立,当且仅当

$ (\beta, \alpha)\in D = \{(\beta, \alpha)|\alpha\leq L(\beta)\}, $

其中$ L(\beta) = \inf\limits_{r\in(0, 1)}\left\{r^{p}E_{p}/[{r'}^{p}(K_{p}-E_{p})]+(1-\beta)(K_{p}-E_{p})/E_{p}\right\} $关于$ \beta $连续,且当$ \beta\leq (p^2+1)/2 $时, $ L(\beta) = p $,而当$ \beta>(p^2+1)/2 $时, $ L(\beta)<p $ (见图 1).此外,不存在$ \alpha, \beta\in {{\Bbb R}} $使得$ E_{p}(r) $在$ (0, 1) $上是$ H_{\alpha, \beta} $ -凸的.

推论1.1  对$ a\in(0, 1) $,第二类广义椭圆积分$ {\cal E}_{a}(r) $在$ (0, 1) $上严格$ H_{\alpha, \beta} $ -凸的,即不等式

$ {\cal E}_{a}(H_{\alpha}(x, y))\geq H_{\beta}({\cal E}_{a}(x), {\cal E}_{a}(y)) $

对所有$ x, y\in (0, 1) $恒成立,当且仅当

$ (\beta, \alpha)\in\{(\beta, \alpha)|\alpha\leq L^*(\beta)\}, $

其中$ L^*(\beta) = \inf\limits_{r\in(0, 1)}2(1-a)\left\{r^{2}{\cal E}_{a}/[(1-r^2)({\cal K}_{a}-{\cal E}_{a})]+(1-\beta)({\cal K}_{a}-{\cal E}_{a})/{\cal E}_{a}\right\} $关于$ \beta $连续,且当$ \beta\leq (a^2-2a+2)/[2(1-a)^2] $时, $ L^*(\beta) = 2 $,而当$ \beta>(a^2-2a+2)/[2(1-a)^2] $时, $ L^*(\beta)<2 $.此外,不存在$ \alpha, \beta\in {{\Bbb R}} $使得$ {\cal E}_{a}(r) $在$ (0, 1) $上是$ H_{\alpha, \beta} $ -凸的.

为证明本文的主要结果,需要在本节中建立几则引理.

引理2.1^[3]  对$ -\infty<a<b<+\infty $,设$ f $和$ g $是两个实值函数,都在$ [a, b] $上连续,在$ (a, b) $上可微,且在$ (a, b) $上$ g'\neq 0 $.如果$ f'/g' $在$ (a, b) $上(严格)单调增加(减小),那么函数

$ \frac{f(x)-f(a)}{g(x)-g(a)}\quad\mbox{和}\quad \frac{f(x)-f(b)}{g(x)-g(b)} $

也在$ (a, b) $上(严格)单调增加(减小).

引理2.2  令$ p\in(1, \infty) $.则

$ (1) $函数$ r\mapsto(E_{p}-{r'}^pK_{p})/r^p $从$ (0, 1) $到$ ({(p-1)\pi_{p}}/{2p}, 1) $严格单调增加;

$ (2) $函数$ r\mapsto(K_{p}-E_{p})/(r^p K_{p}) $从$ (0, 1) $到$ ({1}/{p}, 1) $严格单调增加;

$ (3) $函数$ r\mapsto{r'}^p(K_{p}-E_{p})/(r^pE_{p}) $从$ (0, 1) $到$ (0, 1/p) $严格单调减小;

$ (4) $当且仅当$ c\geq (p-1)/p $,函数$ r\mapsto{r'}^cK_{p} $在$ (0, 1) $上严格单调减小,且此时函数的值域为$ (0, \pi_{p}/2) $;

$ (5) $当且仅当$ c\leq -1/p $,函数$ r\mapsto{r'}^cE_{p} $在$ (0, 1) $上严格单调增加,且此时函数的值域为$ (\pi_{p}/2, \infty) $;

$ (6) $函数$ r\mapsto[(p-1)(K_{p}-E_{p})-(E_{p}-{r'}^pK_{p})]/r^{2p} $具有正的$ \rm Maclaurin $级数,在$ (0, 1) $的值域为$ (\pi_{p}(1-p)^2/(4p^2), \infty) $;

$ (7) $函数$ r\mapsto[E_{p}-{r'}^pK_{p}-(p-1){r'}^p(K_{p}-E_{p})]/[r^{p}(K_{p}-E_{p})] $从$ (0, 1) $到$ (0, (p^2-1)/(2p)) $严格单调减小.

证  (1)–(5)可参见文献[32,引理3.4(1), (4)–(7)].下证明(6),利用Gauss超几何函数的级数展开可得

$ \begin{eqnarray*} \frac{K_{p}-E_{p}}{r^p}& = &\frac{\pi_{p}}{2}\cdot\frac{\sum\limits_{n = 0}^{\infty}\frac{(1/p)_{n}(1-1/p)_{n}}{n!}\frac{r^{np}}{n!}-\sum\limits_{n = 0}^{\infty}\frac{(1/p)_{n}(-1/p)_{n}}{n!}\frac{r^{np}}{n!}}{r^{p}}\\ & = &\frac{\pi_{p}}{2}\cdot\frac{\sum\limits_{n = 1}^{\infty}\frac{(1/p)_{n}(1-1/p)_{n-1}}{n!}(n-1/p+1/p)\frac{r^{np}}{n!}}{r^{p}}\\ & = &\frac{\pi_{p}}{2}\cdot\sum\limits_{n = 0}^{\infty}\frac{(1/p)_{n+1}(1-1/p)_{n}}{n!}\frac{r^{np}}{(n+1)!}, \\ \frac{E_{p}-{r'}^pK_{p}}{r^p}& = &K_{p}-\frac{K_{p}-E_{p}}{r^p}\\ & = &\frac{\pi_{p}}{2}\left(\sum\limits_{n = 0}^{\infty}\frac{(1/p)_{n}(1-1/p)_{n}}{n!}\frac{r^{np}}{n!}-\sum\limits_{n = 0}^{\infty}\frac{(1/p)_{n+1}(1-1/p)_{n}}{n!}\frac{r^{np}}{(n+1)!}\right)\\ & = &\frac{\pi_{p}}{2}\left(1-\frac{1}{p}\right)\sum\limits_{n = 0}^{\infty}\frac{(1/p)_{n}(1-1/p)_{n}}{n!}\frac{r^{np}}{(n+1)!}, \end{eqnarray*} $

及其

(2.1) $ \begin{eqnarray} &&\frac{(p-1)(K_{p}-E_{p})-(E_{p}-{r'}^pK_{p})}{r^{2p}} = \frac{(p-1)(K_{p}-E_{p})/r^p-(E_{p}-{r'}^pK_{p})/r^p}{r^{p}}{}\\ & = &\frac{\pi_{p}}{2}\bigg[(p-1)\sum\limits_{n = 0}^{\infty}\frac{(1/p)_{n+1}(1-1/p)_{n}}{n!}\frac{r^{np}}{(n+1)!} -\big(1-\frac{1}{p}\big)\sum\limits_{n = 0}^{\infty}\frac{(1/p)_{n}(1-1/p)_{n}}{n!}\frac{r^{np}}{(n+1)!}\bigg]\bigg/r^p{}\\ & = &\frac{\pi_{p}}{2}\bigg[(p-1)\sum\limits_{n = 1}^{\infty}\frac{(1/p)_{n+1}(1-1/p)_{n}}{n!}\frac{r^{np}}{(n+1)!} -\big(1-\frac{1}{p}\big)\sum\limits_{n = 1}^{\infty}\frac{(1/p)_{n}(1-1/p)_{n}}{n!}\frac{r^{np}}{(n+1)!}\bigg]\bigg/r^p{}\\ & = &\frac{\pi_{p}}{2}\bigg[(p-1)\sum\limits_{n = 0}^{\infty}\frac{(1/p)_{n+2}(1-1/p)_{n+1}}{(n+1)!}\frac{r^{np}}{(n+2)!} {}\\ && -\big(1-\frac{1}{p}\big)\sum\limits_{n = 0}^{\infty}\frac{(1/p)_{n+1}(1-1/p)_{n+1}}{(n+1)!}\frac{r^{np}}{(n+2)!}\bigg]{}\\ & = &\frac{\pi_{p}}{2}\cdot\sum\limits_{n = 0}^{\infty}\frac{(1/p)_{n+1}(1-1/p)_{n+1}}{(n+1)!}\frac{r^{np}}{(n+2)!} \big[(p-1)(n+1+1/p)-(1-1/p)\big]{}\\ & = &\frac{\pi_{p}}{2}(p-1)\sum\limits_{n = 0}^{\infty}\frac{(1/p)_{n+1}(1-1/p)_{n+1}}{(n+2)!}\frac{r^{np}}{n!}{}\\ & = &\frac{\pi_{p}(1-1/p)^2}{4}F(1+1/p, 2-1/p;3;r^p). \end{eqnarray} $

因此,由等式(2.1)知(6)中定义的函数具有正的Maclaurin级数且值域为

$ (\pi_{p}(1-p)^2/(4p^2), \infty). $

(7)令$ \varphi(r) = E_{p}-{r'}^pK_{p}-(p-1){r'}^p(K_{p}-E_{p}) $, $ \phi(r) = r^p(K_{p}-E_{p}) $.则

$ \varphi(0) = \phi(0) = 0, $

$ \begin{eqnarray*} \varphi'(r)& = &(p-1)r^{p-1}K_{p}-(p-1)(-pr^{p-1})(K_{p}-E_{p})-(p-1)r^{p-1}E_{p}\\ & = &(p^2-1)r^{p-1}(K_{p}-E_{p}), \end{eqnarray*} $

$ \phi'(r) = pr^{p-1}(K_{p}-E_{p})+(r^{2p-1}E_{p})/{r'}^p, $

及其

(2.2) $ \begin{equation} \frac{\varphi'(r)}{\phi'(r)} = \frac{p^2-1}{p+{r}^pE_{p}/[{r'}^p(K_{p}-E_{p})]}. \end{equation} $

式(2.2)和本引理(3)表明$ \varphi'(r)/\phi'(r) $在$ (0, 1) $上严格单调减小

$ \lim\limits_{r\rightarrow 0^+}[\varphi(r)/\phi(r)] = \lim\limits_{r\rightarrow 0^+}[\varphi'(r)/\phi'(r)] = (p^2-1)/(2p). $

应用引理2.1,可推断$ \varphi(r)/\phi(r) $在$ (0, 1) $上也严格单调减小.此外,显然有$ \lim\limits_{r\rightarrow 1^-}[\varphi(r)/\phi(r)] = 0 $.引理2.2证毕.

引理2.3  令$ p\in(1, \infty) $, $ c\in {{\Bbb R}} $和

$ \begin{eqnarray*} f_{c}(r) = \frac{r'^{c}(K_{p}-E_{p})}{r^p}. \end{eqnarray*} $

则以下论断成立

$ (1) $当且仅当$ c\geq (p^2-1)/(2p) $,函数$ f_{c}(r) $从$ (0, 1) $到$ (0, \pi_{p}/(2p)) $严格单调减小;

$ (2) $当且仅当$ c\leq 0 $,函数$ f_{c}(r) $从$ (0, 1) $到$ (\pi_{p}/(2p), \infty) $严格单调增加;

$ (3) $若$ 0<c<(p^2-1)/(2p) $,则存在$ r_{0} = r_{0}(c)\in(0, 1) $使得$ f_{c}(r) $在$ (0, r_{0}) $上严格单调增加,在$ (r_{0}, 1) $严格单调减小.

证  显然当$ c\leq0 $时, $ f_{c}(1^-) = \infty $,而当$ c>0 $时, $ f_{c}(1^-) = 0 $.利用引理2.2(2), $ f_{c}(0^+) = \pi_{p}/(2p) $.对$ f_{c} $求导得

(2.3) $ \begin{eqnarray} f_{c}'(r)& = &-c{r'}^{c-p}r^{p-1}\left(\frac{K_{p}-E_{p}}{r^p}\right)+{r'}^c\left[\frac{r^pE_{p}-p{r'}^p(K_{p}-E_{p})}{r^{p+1}{r'}^p}\right]\\ & = &\frac{{r'}^{c-p}(K_{p}-E_{p})}{r}\left[-c+\frac{E_{p}-{r'}^pK_{p}-(p-1){r'}^p(K_{p}-E_{p})}{r^p(K_{p}-E_{p})}\right]. \end{eqnarray} $

由引理2.2(7)和(2.3)式立即可得第(1)、(2)部分关于$ f_{c} $的单调性论断.此外,若$ 0<c<(p^2-1)/(2p) $,又可由引理2.2(7)和(2.3)式知存在依赖于参数$ c $的$ r_{0} = r_{0}(c)\in(0, 1) $使得当$ r\in(0, r_{0}) $时, $ f_{c}'(r)>0 $,当$ r\in(r_{0}, 1) $时, $ f_{c}'(r)<0 $,故$ f_{c}(r) $在$ (0, r_{0}) $上严格单调增加,在$ (r_{0}, 1) $上严格单调减小.

引理2.4  令$ p\in(1, \infty) $,有

$ f(r) = E_{p}(r)+\frac{{r'}^p(K_{p}-E_{p})^2}{r^pE_{p}(r)}, $

则$ f(r) $从$ (0, 1) $到$ (1, {\pi_{p}}/{2}) $严格单调减小.

证  对$ f $求导得

(2.4) $ \begin{eqnarray} f'(r)& = &\Big\{[-pr^{p-1}(K_{p}-E_{p})^2+2r^{p-1}E_{p}(K_{p}-E_{p})]r^pE_{p}{}\\ &&-{r'}^p(K_{p}-E_{p})^2 \left[pr^{p-1}E_{p}+r^{p-1}(E_{p}-K_{p})\right]\Big\}\Big/ \big\{r^{2p}{E_{p}}^2\big\} +\frac{E_{p}-K_{p}}{r}{}\\ & = &\frac{E_{p}-K_{p}}{r}+\frac{K_{p}-E_{p}}{r^{p+1}{E_{p}}^2}\left[\left(-p(K_{p}-E_{p})+2E_{p}\right)r^{p}E_{p}-{r'}^p(K_{p}-E_{p})(pE_{p}+E_{p}-K_{p})\right]{}\\ & = &\frac{E_{p}-K_{p}}{r}+\frac{K_{p}-E_{p}}{r^{p+1}{E_{p}}^2}\left[-pE_{p}(K_{p}-E_{p})+2r^{p}{E_{p}}^2+{r'}^p(K_{p}-E_{p})^2\right]{}\\ & = &-\frac{K_{p}-E_{p}}{r}\left[1-\frac{-pE_{p}(K_{p}-E_{p})+2r^{p}{E_{p}}^2+{r'}^p(K_{p}-E_{p})^2}{r^p{E_{p}}^2}\right]{}\\ & = &-\frac{K_{p}-E_{p}}{r}\left[-1-\frac{-pE_{p}(K_{p}-E_{p})+{r'}^p(K_{p}-E_{p})K_{p}-{r'}^p(K_{p}-E_{p})E_{p}}{r^p{E_{p}}^2}\right]{}\\ & = &-\frac{K_{p}-E_{p}}{r}\left[-\frac{{r'}^p(K_{p}-E_{p})K_{p}}{r^p{E_{p}}^2}+\frac{{r'}^p(K_{p}-E_{p})}{r^p{E_{p}}}-1-\frac{K_{p}-E_{p}}{r^p{E_{p}}}+\frac{(p+1)(K_{p}-E_{p})}{r^p{E_{p}}}\right]{}\\ & = &-\frac{K_{p}-E_{p}}{r}\left[-\frac{{r'}^p(K_{p}-E_{p})K_{p}}{r^p{E_{p}}^2}-\frac{K_{p}}{E_{p}}+\frac{(p+1)(K_{p}-E_{p})}{r^p{E_{p}}}\right]{}\\ & = &-\frac{(K_{p}-E_{p})K_{p}}{rE_{p}}\left[-\frac{{r'}^p(K_{p}-E_{p})}{r^pE_{p}}-1+(p+1)\frac{K_{p}-E_{p}}{r^pK_{p}}\right]. \end{eqnarray} $

引理2.2(2)和(3)表明函数$ r\mapsto -{r'}^p(K_{p}-E_{p})/(r^pE_{p})-1+(p+1)(K_{p}-E_{p})/(r^pK_{p}) $从$ (0, 1) $到$ (0, p) $严格单调增加.结合式(2.4)可推断对所有$ r\in(0, 1) $,有$ f'(r)<0 $,故$ f(r) $在$ (0, 1) $上严格单调减小.此外,不难计算得$ f(0^+) = E_{p}(0) = \pi_{p}/2 $, $ f(1^-) = E_{p}(1) = 1 $.

引理2.5  令$ p\in(1, +\infty) $,有

$ g(r) = \frac{(K_{p}-E_{p})(E_{p}-{r'}^p K_{p})}{r^{2p}}+E_{p}\frac{(p-1)(K_{p}-E_{p})-(E_{p}-{r'}^pK_{p})}{r^{2p}}. $

则$ g(r) $从$ (0, 1) $到$ ((p^2-1){\pi_{p}}^2/(8p^2), \infty) $严格单调增加.

证  根据引理2.2(1), (2)及(6)得$ \lim\limits_{r\rightarrow 1^-}g(r) = \infty $和

$ \begin{eqnarray*} \lim\limits_{r\rightarrow 0^+}g(r)& = &\lim\limits_{r\rightarrow 0^+}\left(\frac{K_{p}-E_{p}}{r^p}\right)\lim\limits_{r\rightarrow 0^+}\left(\frac{E_{p}-{r'}^pK_{p}}{r^p}\right)\nonumber\\ &&+\lim\limits_{r\rightarrow 0^+}E_{p}\lim\limits_{r\rightarrow 0^+}\left[\frac{(p-1)(K_{p}-E_{p})-(E_{p}-{r'}^pK_{p})}{r^{2p}}\right]\nonumber\\ & = &\frac{(p-1){\pi_{p}}^2}{4p^2}+\frac{(1-p)^2{\pi_{p}}^2}{8p^2} = \frac{{\pi_{p}}^2(p^2-1)}{8p^2}. \end{eqnarray*} $

注意到

$ \frac{{\rm d}[(E_{p}-{r'}^pK_{p})/r^p]}{{\rm d}r} = \frac{(p-1)(K_{p}-E_{p})-(E_{p}-{r'}^pK_{p})}{r^{p+1}}. $

对$ g $求导并结合引理2.2(2), (6)可推断

$ \begin{eqnarray*} g'(r)& = &\frac{{\rm d}[(K_{p}-E_{p})/r^p]}{{\rm d}r}\cdot\left(\frac{E_{p}-{r'}^pK_{p}}{r^p}\right)+\left(\frac{K_{p}-E_{p}}{r^p}\right)\cdot\frac{{\rm d}[(E_{p}-{r'}^pK_{p})/r^p]}{{\rm d}r}\nonumber\\ &&+\frac{{\rm d}E_{p}}{{\rm d}r}\left[\frac{(p-1)(K_{p}-E_{p})-(E_{p}-{r'}^pK_{p})}{r^{2p}}\right]\nonumber\\ &&+E_{p}\frac{{\rm d}\left\{\left[(p-1)(K_{p}-E_{p})-(E_{p}-{r'}^pK_{p})\right]/r^{2p}\right\}}{{\rm d}r}\nonumber\\ & = &\frac{{\rm d}[(K_{p}-E_{p})/r^p]}{{\rm d}r}\cdot\left(\frac{E_{p}-{r'}^pK_{p}}{r^p}\right)+E_{p}\frac{{\rm d}\left\{\left[(p-1)(K_{p}-E_{p})-(E_{p}-{r'}^pK_{p})\right]/r^{2p}\right\}}{{\rm d}r}\\ &>&0. \end{eqnarray*} $

引理2.5成立.

引理2.6  令$ p\in(1, +\infty) $,有

(2.5) $ \begin{equation} h(r) = \frac{r^p{{E}_{p}}^2\left\{({K}_{p}-{E}_{p}) ({E}_{p}-{r'}^p{K}_{p})+{{E}_{p}}\left[(p-1)({K}_{p}-{E}_{p}) -({E}_{p}-{r'}^p{K}_{p})\right]\right\}}{{r'}^p({K}_{p}-{E}_{p})^2\left[{r'}^p ({K}_{p}-{E}_{p})^2+r^p{{E}_{p}}^2\right]}, \end{equation} $

则$ h(r) $从$ (0, 1) $到$ ((p^2-1)/2, \infty) $严格单调增加.

证  将$ h(r) $改写成

$ \begin{eqnarray*} h(r) = \frac{\frac{(K_{p}-E_{p})(E_{p}-{r'}^p K_{p})}{r^{2p}}+E_{p}\frac{(p-1)(K_{p}-E_{p})-(E_{p}-{r'}^pK_{p})}{r^{2p}}} {\left[E_{p}(r)+\frac{{r'}^p(K_{p}-E_{p})^2}{r^pE_{p}(r)}\right]\left[\frac{{r'}^{(p^2-1)/(2p)}(K_{p}-E_{p})}{r^p}\right]^2\left(\frac{{r'}^{1/p}}{E_{p}}\right)}. \end{eqnarray*} $

根据引理2.2(5),引理2.3(1),引理2.4和引理2.5便得引理2.6.

引理2.7  令$ p\in(1, \infty) $, $ \beta\in{{\Bbb R}} $,有

(2.6) $ \begin{equation} G_{\beta}(r) = (1-\beta)\frac{K_{p}-E_{p}}{E_{p}}+\frac{r^pE_{p}}{{r'}^p(K_{p}-E_{p})} \end{equation} $

和$ L(\beta) = \inf\limits_{r\in(0, 1)}G_{\beta}(r) $.则

$ (1) $当且仅当$ \beta\leq (p^2+1)/2 $, $ G_{\beta}(r) $从$ (0, 1) $到$ (p, \infty) $严格单调增加, $ L(\beta) = p $;

$ (2) $若$ \beta>(p^2+1)/2 $,则存在$ \lambda = \lambda(\beta)\in(0, 1) $使得$ G_{\beta}(r) $在$ (0, \lambda) $上严格单调减小,在$ (\lambda, 1) $上严格单调增加. $ G_{\beta}(r) $在$ (0, 1) $上的值域为$ [L(\beta), \infty) $且$ L(\beta)<p $.

证  经计算得

(2.7) $ \begin{equation} \lim\limits_{r\rightarrow 0^+}G_{\beta}(r) = p, \end{equation} $

(2.8) $ \begin{equation} \lim\limits_{r\rightarrow 1^-}G_{\beta}(r) = \lim\limits_{r\rightarrow 1^-}\left(\frac{K_{p}-E_{p}}{E_{p}}\right)\left[1-\beta+\frac{r^p{E_{p}}^2}{{r'}^p(K_{p}-E_{p})^2}\right] = \infty \end{equation} $

及其

(2.9) $ \begin{eqnarray} G_{\beta}'(r)& = &r^{p-1}\frac{{r'}^p(K_{p}-E_{p})(pE_{p}+E_{p}-K_{p}) -r^pE_{p}(E_{p}+pE_{p}-pK_{p})}{{r'}^{2p}(K_{p}-E_{p})^2}\\ &&+(1-\beta)\frac{r^p{E_{p}}^2+{r'}^p(K_{p}-E_{p})^2}{r{r'}^p{E_{p}}^2}\\ & = &r^{p-1}\frac{pE_{p}(K_{p}-E_{p})-{r'}^p(K_{p}-E_{p})K_{p}+[{r'}^p(K_{p}-E_{p})E_{p}-r^p{E_{p}}^2]}{{r'}^{2p}(K_{p}-E_{p})^2}\\ &&+(1-\beta)\frac{r^p{E_{p}}^2+{r'}^p(K_{p}-E_{p})^2}{r{r'}^p{E_{p}}^2}\\ & = &r^{p-1}\frac{E_{p}[(p-1)(K_{p}-E_{p})-(E_{p}-{r'}^pK_{p})]+(K_{p}-E_{p})(E_{p}-{r'}^pK_{p})}{{r'}^{2p}(K_{p}-E_{p})^2}\\ &&+(1-\beta)\frac{r^p{E_{p}}^2+{r'}^p(K_{p}-E_{p})^2}{r{r'}^p{E_{p}}^2}\\ & = &\frac{r^p{E_{p}}^2+{r'}^p(K_{p}-E_{p})^2}{r{r'}^p{E_{p}}^2}\left[1-\beta+h(r)\right], \end{eqnarray} $

其中$ h(r) $是$ (0, 1) $上的严格单调增加函数,其定义见(2.5)式.

由引理2.6和(2.9)式知:当且仅当$ \beta\leq \inf\limits_{r\in(0, 1)}[1+h(r)] = (p^2+1)/2 $, $ {G_{\beta}}'(r)>0 $对所有的$ r\in(0, 1) $恒成立,即函数$ G_{\beta}(r) $在$ (0, 1) $上严格单调增加,进而由(2.7)和(2.8)式可得引理第(1)部分.若$ \beta>(p^2+1)/2 $,则引理2.6及(2.9)式表明存在$ \lambda = \lambda(\beta)\in(0, 1) $使得当$ r\in(0, \lambda) $时, $ G_{\beta}'(r)<0 $,当$ r\in(\lambda, 1) $, $ G_{\beta}'(r)>0 $,故$ G_{\beta}(r) $在$ (0, 1) $分段单调的.结合(2.7)和(2.8)式即得$ L(\beta)<p $, $ G_{\beta}(r) $在$ (0, 1) $上的值域为$ (L(\beta), \infty) $.

引理2.8  令$ p\in(1, +\infty) $, $ \alpha, \beta\in{{\Bbb R}} $,有

$ \begin{eqnarray*} J(r) = {E_{p}}^{{\beta}-1}\frac{K_{p}(r)-E_{p}(r)}{r^{\alpha}}, \end{eqnarray*} $

则以下论断成立

$ (1) $若$ \beta\leq (p^2+1)/2 $,则$ J(r) $在$ (0, 1) $上严格单调增加当且仅当$ \alpha\leq p $,若$ \alpha>p $,则存在$ r_{1} = r_{1}(\alpha, \beta)\in(0, 1) $使得$ J(r) $在$ (0, r_{1}) $上严格单调减小,在$ (r_{1}, 1) $上严格单调增加;

$ (2) $若$ \beta>(p^2+1)/2 $,则$ J(r) $在$ (0, 1) $上严格单调增加当且仅当$ \alpha\leq L(\beta) $,若$ \alpha>L(\beta) $,则$ J(r) $是分段单调的,其中$ L(\beta) $见引理2.7.

证  对数求导得

(2.10) $ \begin{equation} \frac{J'(r)}{J(r)} = (1-\beta)\frac{K_{p}-E_{p}}{rE_{p}}+\frac{r^{p-1}E_{p}}{{r'}^p(K_{p}-E_{p})}-\frac{\alpha}{r} = \frac{1}{r}\left[G_{\beta}(r)-\alpha\right], \end{equation} $

其中$ G_{\beta}(r) $定义见(2.6)式.

下面分四种情形完成引理的证明.

情形Ⅰ  $ \beta\leq (p^2+1)/2 $且$ \alpha\leq p $.则由引理2.7(1)和(2.10)式知：对所有$ r\in(0, 1) $, $ J'(r)\geq 0 $,因此$ J(r) $在$ (0, 1) $上严格单调增加.

情形Ⅱ  $ \beta\leq (p^2+1)/2 $且$ \alpha> p $.则引理2.7(1)及式(2.10)表明存在$ r_{1} = r_{1}(\alpha, \beta)\in(0, 1) $,使得当$ r\in(0, r_{1}) $时, $ J'(r)<0 $,当$ r\in(r_{1}, 1) $时, $ J'(r)>0 $.因此, $ J(r) $在$ (0, r_{1}) $上严格单调减小,在$ (r_{1}, 1) $上严格单调增加.

情形Ⅲ  $ \beta>(p^2+1)/2 $且$ \alpha\leq L(\beta) $.则由引理2.7(1)和(2.10)式知：对所有$ r\in(0, 1) $, $ J'(r)\geq 0 $,因此$ J(r) $在$ (0, 1) $上严格单调增加.

情形Ⅳ  $ \beta>(p^2+1)/2 $且$ \alpha> L(\beta) $.则由引理2.7(2), (2.10)式及其相似于情形II的讨论可得$ J(r) $在$ (0, 1) $上分段单调.

定理1.2的证明  不妨设$ 0<x\leq y<1 $.令$ t = H_{\alpha}(x, y) $,则$ \partial t/\partial x = (x/t)^{\alpha-1}/2 $,且当$ y>x $时, $ t>x $.下面分$ \beta\neq0 $和$ \beta = 0 $两种情形来完成定理1.2的证明.

情形1  $ \beta\neq 0 $.令

(3.1) $ \begin{equation} F(x, y) = E_{p}(H_{\alpha}(x, y))^{\beta}-\frac{{E_{p}(x)}^{\beta}+{E_{q}(y)}^{\beta}}{2}. \end{equation} $

则对$ x $求偏导数得

(3.2) $ \begin{eqnarray} \frac{\partial F}{\partial x}& = &\frac{\beta}{2} {E_{p}(t)}^{\beta-1} \frac{E_{p}(t)-K_{p}(t)}{t}\left(\frac{x}{t}\right)^{\alpha-1}-\frac{\beta}{2}{E_{p}(x)}^{\beta-1}\frac{E_{p}(x)-K_{p}(x)}{x}\\ & = &\frac{\beta}{2}x^{\alpha-1}\left[{E_{p}(x)}^{\beta-1}\frac{K_{p}(x)-E_{p}(x)}{x^{\alpha}}-{E_{p}(t)}^{\beta-1}\frac{K_{p}(t)-E_{p}(t)}{t^{\alpha}}\right]. \end{eqnarray} $

下面再将情形1分成下面四种子情形.

子情形1.1  $ \beta\leq (p^2+1)/2 $且$ \alpha\leq p $.则由引理2.8(1)和(3.2)式知:若$ \beta>0 $, $ \partial F/\partial x<0 $, $ F(x, y)>F(y, y) = 0 $;若$ \beta<0 $, $ \partial F/\partial x>0 $, $ F(x, y)>F(y, y) = 0 $.结合(3.1)式可推得:当$ \alpha\leq p $且$ \beta\leq (p^2+1)/2 $时,始终成立不等式

(3.3) $ \begin{equation} E_{p}\left(H_{\alpha}(x, y)\right)\geq H_{\beta}\left({E_{p}}(x), {E_{p}}(y)\right), \end{equation} $

而且不难验证当且仅当$ x = y $时不等式(3.3)成立等号.

综上可知,当$ (\alpha, \beta)\in\{(\alpha, \beta)|\alpha\leq p, \beta\leq (p^2+1)/2, \beta\neq0\} $时, $ E_{p} $在$ (0, 1) $上严格$ H_{\alpha, \beta} $ -凹的.

子情形1.2  $ \beta\leq (p^2+1)/2 $且$ \alpha>p $.则由(3.1)和(3.2)式,引理2.8(1)及其类似子情形1.1的讨论可知:当$ \alpha>p $且$ \beta\leq (p^2+1)/2 $时,存在$ x_{0}, y_{0}\in(0, r_{1}) $和$ x_{1}, y_{1}\in(r_{1}, 1) $使得$ E_{p}\left(H_{\alpha}(x_{0}, y_{0})\right)\leq H_{\beta}\left({E_{p}}(x_{0}), {E_{p}}(y_{0})\right) $, $ E_{p}\left(H_{\alpha}(x_{1}, y_{1})\right)\geq H_{\beta}\left({E_{p}}(x_{1}), {E_{p}}(y_{1})\right) $.因此, $ E_{p} $在整个$ (0, 1) $区间上既不是$ H_{\alpha, \beta} $ -凹的也不是$ H_{\alpha, \beta} $ -凸的.

子情形1.3  $ \beta>(p^2+1)/2 $且$ \alpha\leq L(\beta) $.则(3.1)和(3.2)式,引理2.8(2)及其类似子情形1.1的讨论表明:当$ (\alpha, \beta)\in\{(\alpha, \beta)|\alpha\leq L(\beta), \beta>(p^2+1)/2, \beta\neq0\} $时, $ E_{p} $在$ (0, 1) $上严格$ H_{\alpha, \beta} $ -凹的.

子情形1.4  $ \beta>(p^2+1)/2 $且$ \alpha>L(\beta) $.则从(3.1)和(3.2)式,引理2.8(2)及其上述子情形1.2的类似讨论可知:$ E_{p} $在整个$ (0, 1) $区间上既不是$ H_{\alpha, \beta} $ -凹的也不是$ H_{\alpha, \beta} $ -凸的.

情形2  $ \beta = 0 $.令

(3.4) $ \begin{equation} G(x, y) = \frac{\left[E_{p}(H_{\alpha}(x, y))\right]^2}{E_{p}(x)E_{p}(y)}, \end{equation} $

则对$ x $求偏导数得

(3.5) $ \begin{equation} \frac{1}{G(x, y)}\frac{\partial G}{\partial x} = x^{\alpha-1}\left[\frac{K_{p}(x)-E_{p}(x)}{x^\alpha E_{p}(x)}-\frac{K_{p}(t)-E_{p}(t)}{t^\alpha E_{p}(t)}\right]. \end{equation} $

下面将情形2的证明分如下两种情形.

子情形2.1  $ \alpha\leq p $.则(3.5)式和引理2.8(1)表明$ \partial G/\partial x<0 $,故$ G(x, y)>G(y, y) = 1 $.于是从(3.4)式便知

$ \begin{eqnarray*} E_{p}\left(H_{\alpha}(x, y)\right)\geq H_{\beta}\left(E_{p}(x), E_{p}(y)\right), \end{eqnarray*} $

且上述不等式成立等式当且仅当$ x = y $.

因此,当$ (\alpha, \beta)\in\{(\alpha, \beta)|\alpha\leq p, \beta = 0\} $时, $ E_{p} $在$ (0, 1) $上严格$ H_{\alpha, \beta} $ -凹的.

子情形2.2  $ \alpha>p $.则利用(3.4)和(3.5)式及其引理2.8(1)推断得：$ E_{p} $在$ (0, 1) $上既不是$ H_{\alpha, \beta} $ -凹的也不是$ H_{\alpha, \beta} $ -凸的.

推论1.1的证明  假设$ \alpha, \beta\in{{\Bbb R}} $, $ x, y\in(0, 1) $.对$ a\in(0, 1) $,令$ p = 1/(1-a) $,则从(1.8)式可推断不等式

$ {\cal E}_{a}(H_{\alpha}(x, y))\geq (\leq) H_{\beta}({\cal E}_{a}(x), {\cal E}_{a}(y)) $

等价于

$ {\cal E}_{a}\left(H_{\alpha}(x^{\frac{p}{2}}, y^{\frac{p}{2}})\right) = {\cal E}_{a}\left(\left[H_{\frac{p\alpha}{2}}(x, y)\right]^{\frac{p}{2}}\right)\geq (\leq) H_{\beta}({{\cal E}}_{a}(x^{\frac{p}{2}}), {{\cal E}}_{a}(y^{\frac{p}{2}})), $

再对不等式两边同时乘以$ \pi_{p}/\pi $,便得

$ E_{p}\left(\left[H_{\frac{p\alpha}{2}}(x, y)\right]\right)\geq (\leq) H_{\beta}(E_{p}(x), E_{p}(y)). $

因此,由定理1.2便得推论1.1.

注3.1  当$ p\in(1, \infty) $时, $ p<(p^2+1)/2 $,故由定理1.2可知:第二类完全$ p $ -椭圆积分$ E_{p}(r) $在$ (0, 1) $上$ H_{\alpha, \alpha} $ -凹的,即不等式

$ E_{p}(H_{\alpha}(x, y))\geq H_{\alpha}(E_{p}(x), E_{p}(y)) $

对所有$ x, y\in (0, 1) $恒成立当且仅当$ \alpha\leq p $,且不存在$ \alpha\in {{\Bbb R}} $使得$ {\cal E}_{a}(r) $在$ (0, 1) $上是$ H_{\alpha, \alpha} $ -凸的.