Monotonicity properties and bounds involving the two-parameter generalized Grötzsch ring function

Abstract

In the article, we present several new monotonicity properties and bounds involving the generalized Grötzsch ring functions $\mu _{a,b}$ in the theory of Ramanujan’s generalized modular equation for $0< a, b<1$. Our results are the variants and extensions of some previously known results.

1 Introduction

Given $x\in (-1,1)$ and real numbers a, b, and c with $c\neq 0,-1,-2,\ldots$ , the Gaussian hypergeometric function $F(a,b;c;x)$ [1–18] is defined by

$$ F(a,b;c;x)={}_{2}F_{1}(a,b;c;x)=\sum _{n=0}^{\infty } \frac{(a,n)(b,n)}{(c,n)} \frac{x^{n}}{n!}, $$

(1.1)

where $(a,0)=1$ for $a\neq 0$ and $(a, n)=a(a+1)(a+2)\cdots (a+n-1)$ for $n=1,2,\ldots$ . $F(a,b;c;x)$ is said to be zero-balanced if $c=a+b$. If $x\rightarrow 1$, then the following asymptotic formulas

$$ \textstyle\begin{cases} F(a,b;c;1)=\frac{\varGamma (c)\varGamma (c-a-b)}{\varGamma (c-a)\varGamma (c-b)},& a+b< c, \\ B(a,b)F(a,b;c;x)+\log (1-x)=R(a,b)+O((1-x)\log (1-x)),&a+b=c, \\ F(a,b;c;x)=(1-x)^{c-a-b}F(c-a,c-b;c;x),&a+b>c, \end{cases} $$

(1.2)

can be found in the literature [19, Theorems 1.19 and 1.48], where $\varGamma (x)=\int _{0}^{\infty }t^{x-1}e^{-t}\,dt$ [20–26] and $B(p,q)=[\varGamma (p)\varGamma (q)]/\varGamma (p+q)$ [27–30] are respectively the classical Euler gamma and beta functions, and

$$ R(a,b)=-\psi (a)-\psi (b)-2\gamma ,\qquad R \biggl(\frac{1}{2}, \frac{1}{2} \biggr)=\log 16, \qquad \psi (x)= \frac{\varGamma '(x)}{\varGamma (x)} $$

and

$$ \gamma =\lim_{n\rightarrow \infty } \Biggl(\sum_{k=1}^{n} \frac{1}{k}- \log n \Biggr)=0.5772156649\cdots $$

is the Euler–Mascheroni constant [31–33].

Ramanujan’s generalized modular equation with order (or degree) $p>0$ is given by

$$ \frac{F(a,b;c;1-s^{2})}{F(a,b;c;s^{2})}=p \frac{F(a,b;c;1-r^{2})}{F(a,b;c;r^{2})}, \quad 0< r< 1. $$

(1.3)

It is well known that equation (1.3) has a unique solution for s if $a,b,c>0$ with $a+b\geq c$ [34, Lemma 4.5].

The two-parameter generalized Grötzsch ring function is defined by

$$ \mu _{a,b}(r)=\frac{B(a,b)}{2} \frac{F(a,b;(a+b+1)/2;1-r^{2})}{F(a,b;(a+b+1)/2;r^{2})}, \quad r\in (0,1) $$

(1.4)

if $a+b\geq 1$.

Our interest is to focus on $c=(a+b+1)/2$, which makes the derivative formula of the two-parameter generalized Grötzsch ring function defined by (1.4) simpler.

Let $0< a,b<1$ with $a+b\geq 1$ and $r\in (0,1)$. Then the two-parameter generalized elliptic integrals of first and second kinds [34, (1.6)–(1.8)] are defined by

$$\begin{aligned} &\mathcal{K}=\mathcal{K}_{a,b}=\mathcal{K}_{a,b}(r)= \frac{B(a,b)}{2}F \biggl(a,b;\frac{a+b+1}{2};r^{2} \biggr), \end{aligned}$$

(1.5)

$$\begin{aligned} &\mathcal{E}=\mathcal{E}_{a,b}=\mathcal{E}_{a,b}(r)= \frac{B(a,b)}{2}F \biggl(a-1,b;\frac{a+b+1}{2};r^{2} \biggr), \end{aligned}$$

(1.6)

$$\begin{aligned} &\mathcal{K}'=\mathcal{K}'_{a,b}= \mathcal{K}_{a,b}\bigl(r'\bigr),\qquad \mathcal{E}'=\mathcal{E}'_{a,b}= \mathcal{E}_{a,b}\bigl(r'\bigr), \end{aligned}$$

(1.7)

where and in what follows $r'=\sqrt{1-r^{2}}$. Moreover, it follows from (1.2) that

$$\begin{aligned} &\mathcal{K}_{a,b}\bigl(0^{+}\bigr)=\mathcal{E}_{a,b} \bigl(0^{+}\bigr)=\frac{B(a,b)}{2}, \\ &\mathcal{K}_{a,b}\bigl(1^{-}\bigr)=\infty , \qquad \mathcal{E}_{a,b}\bigl(1^{-}\bigr)= \frac{B(a,b)B ((a+b+1)/2,(3-a-b)/2 )}{2B ((b-a+3)/2,(a-b+1)/2 )}. \end{aligned}$$

In this paper, we study the two-parameter generalized Grötzsch ring function $\mu _{a,b}(r)$ for $a,b\in (0,1)$, as well as the related functions $\mathcal{K}_{a,b}$, $\mathcal{E}_{a,b}$, and

$$ m_{a,b}(r)=\frac{2}{B(a,b)}r^{\prime 2} \mathcal{K}_{a,b}\mathcal{K}_{a,b}',\quad r \in (0,1). $$

(1.8)

The so-called Legendre $\mathcal{M}$-function introduced in [35] can be used to study the derivative of $m_{a,b}(r)$ and satisfies the formula

$$ \biggl[\frac{B(a,b)}{2} \biggr]^{2}\mathcal{M} \bigl(r^{2}\bigr)=\frac{a+b-1}{2} \mathcal{K} \mathcal{K}'+\frac{b-a+1}{2} \bigl(\mathcal{K} \mathcal{E}'+ \mathcal{K}'\mathcal{E}-\mathcal{K} \mathcal{K}' \bigr) $$

(1.9)

for $r\in (0,1)$. Furthermore, $\mathcal{M}(r)$ can be rewritten as

$$ \mathcal{M}(r)= \frac{\varGamma ((a+b+1)/2)^{2}[r(1-r)]^{(1-a-b)/2}}{\varGamma (a)\varGamma (b)}, $$

(1.10)

and $\mathcal{M}(r)$ becomes a constant if and only if $a+b=1$, in which case $\mathcal{M}(r^{2})$ degenerates to be the generalized Legendre relation.

In the case of $a+b=1$, these functions coincide with the special functions $\mu _{a}(r)$, $\mathcal{K}_{a}(r)$, $\mathcal{E}_{a}(r)$, and $m_{a}(r)$, respectively, which were studied in [36–49]. In particular, if $a=b=1/2$, then these functions reduce to the classical cases denoted by $\mu (r)$, $\mathcal{K}(r)$, $\mathcal{E}(r)$, and $m(r)$, which appeared frequently in the geometric function theory and number theory [50–69].

The main purpose of the article is to find the sub-regions of $\{(a, b)\in {\mathbb{R}}^{2}|0< a, b<1, a+b>1\}$ such that certain quotient functions involving $\mu _{a,b}(r)$, $\mathcal{K}_{a,b}(r)$, $\mathcal{E}_{a,b}(r)$, and $m_{a,b}(r)$ are monotonic on their corresponding sub-regions. As a consequence, several new bounds for $\mu _{a,b}(r)$ and $m_{a,b}(r)$ are discovered, which are the variants and extensions of the results given in [42, Theorems 1.1 and 1.2] for the case of zero-balanced.

2 Notations, formulas, and lemmas

In order to prove our main results, we need several derivative formulas and lemmas, which we present in this section.

2.1 Notations

Throughout the article, we denote $B(a, b)$ by B if no risk for confusion. Let

$$\begin{aligned} &D=\frac{B(\frac{a+b+1}{2},\frac{a+b-1}{2})}{2}, \\ & E= \frac{B(a,b)B(\frac{a+b+1}{2},\frac{1-a-b}{2})}{2B(\frac{a-b+1}{2},\frac{b-a+1}{2})}, \\ &\kappa _{1}(a,b)=a+b+1-2ab(4-a-b), \\ &\kappa _{2}(a,b)=3 +7 (a + b) +2 (a - b)^{2} -6 (a + b)^{3} -5 (a + b)^{4}- (a + b)^{5} \\ &\hphantom{\kappa _{2}(a,b)=}{} +8 a b \bigl[(a + b)^{2} + (a + b)^{3} - (a + b) + 8 a b \bigr], \\ &\kappa _{3}(a,b)=5 + 7 (a + b) - 3 (a + b)^{2} - 7 (a + b)^{3} - 2 (a + b)^{4} \\ &\hphantom{\kappa _{3}(a,b)=}{} + 4 a b \bigl[3 (a + b)^{2}+ 4 (a + b)-3 \bigr], \\ &\kappa _{4}(a,b)=9 + 5 (a + b) - 9 (a + b)^{2} - 5 (a + b)^{3} + 16 a b (a + b), \\ &\kappa _{5}(a,b)=(a+b+1)^{2}-12ab. \end{aligned}$$

For the convenience of readers, we also introduce three sub-regions $\varOmega _{1}$, $\varOmega _{2}$, and $\varOmega _{3}$ of $\{(a, b)\in {\mathbb{R}}^{2}| 0< a, b<1\}$, which are illustrated in Fig. 1.

$$\begin{aligned} &\varOmega _{1}=\bigl\{ (a,b)| 0< a,b< 1,a+b>1,\kappa _{1}(a,b)\geq 0\bigr\} , \\ &\varOmega _{2}=\bigl\{ (a,b)| 0< a,b< 1,a+b>1,\kappa _{2}(a,b)\leq 0,\kappa _{3}(a,b) \leq 0\bigr\} , \\ &\varOmega _{3}=\bigl\{ (a,b)| 0< a,b< 1,a+b>1,\kappa _{5}(a,b)\geq 0\bigr\} . \end{aligned}$$

Figure 1

Visualized sub-regions $\varOmega _{1}$, $\varOmega _{2}$, and $\varOmega _{3}$ of $\{(a, b)\in {\mathbb{R}}^{2}| 0< a, b<1\}$

2.2 Formulas

Let $r\in (0,1)$ and $0< a,b<1$ with $a+b>1$. Then the following derivative formulas

$$\begin{aligned} &\frac{d\mathcal{K}}{dr}=\frac{1}{rr^{\prime 2}} \bigl[2b\bigl( \mathcal{E}-r^{\prime 2} \mathcal{K}\bigr)+(a+b-1) (\mathcal{K}-\mathcal{E}) \bigr], \end{aligned}$$

(2.1)

$$\begin{aligned} &\frac{d\mathcal{E}}{dr}=\frac{2(a-1)}{r}(\mathcal{K}- \mathcal{E}), \end{aligned}$$

(2.2)

$$\begin{aligned} &\frac{d(\mathcal{K}-\mathcal{E})}{dr}=\frac{1}{rr^{\prime 2}} \bigl[2br^{2} \mathcal{K}- \bigl(a+b-1+2(1-a)r^{2} \bigr) (\mathcal{K}-\mathcal{E}) \bigr], \end{aligned}$$

(2.3)

$$\begin{aligned} &\frac{d(\mathcal{E}-r^{\prime 2}\mathcal{K})}{dr}=\frac{1}{r} \bigl[2(1-b)r^{2} \mathcal{K}+(a+b-1) (\mathcal{K}-\mathcal{E}) \bigr] \end{aligned}$$

(2.4)

can be found in [34, Theorem 4.15].

Note that Theorem 1.19(9) of [19] gives the derivative formula

$$ \frac{d\mu _{a,b}(r)}{dr}=- \frac{(a+b-1)B^{2} D}{4r^{a+b}r^{\prime a+b+1}\mathcal{K}^{2}} $$

(2.5)

for $\mu _{a,b}(r)$ if $d=c=(a+b+1)/2$.

From (1.7), (1.9), (1.10), and (2.1) we clearly see that

$$ \frac{d m_{a,b}(r)}{dr}=\frac{4}{Br} \biggl[ \bigl((2b-1)r^{2} \mathcal{K}-(b-a+1) (\mathcal{K}-\mathcal{E}) \bigr) \mathcal{K}'- \frac{(a+b-1)BD}{4(rr')^{a+b-1}} \biggr]. $$

(2.6)

2.3 Lemmas

Lemma 2.1

([70, Theorem 2.1])

Suppose that the power series$f(x)=\sum_{n=0}^{\infty }a_{n}x^{n}$and$g(x)=\sum_{n=0}^{\infty }b_{n}x^{n}$have the radius of convergence$r>0$with$b_{n}>0$for all$n\in \{0,1,2,\ldots \}$. Let$h(x)=f(x)/g(x)$and$H_{f,g}=(f'/g')g-f$. Then the following statements hold true:

• (1) If the non-constant sequence$\{a_{n}/b_{n}\}_{n=0}^{\infty }$is increasing (decreasing) for all$n\geq 0$, then$h(x)$is strictly increasing (decreasing) on$(0,r)$;

• (2) If there exists$n_{0}>0$such that the non-constant sequence$\{a_{n}/b_{n}\}_{n=0}^{\infty }$is increasing (decreasing) for$0\leq n\leq n_{0}$and decreasing (increasing) for$n\geq n_{0}$, then$h(x)$is strictly increasing (decreasing) on$(0,r)$if and only if$H_{f,g}(r^{-})\geq (\leq ) 0$. Moreover, if$H_{f,g}(r^{-})<(>) 0$, then there exists$x_{0}\in (0,r)$such that$h(x)$is strictly increasing (decreasing) on$(0,x_{0})$and strictly decreasing (increasing) on$(x_{0},r)$.

Lemma 2.2

([19, Theorem 1.25])

Suppose that$-\infty < a< b<\infty$, $f, g:[a,b]\rightarrow \mathbb{R}$are continuous on$[a, b]$and differentiable on$(a,b)$, and$g'(x)\neq 0$on$(a,b)$. If$f'(x)/g'(x)$is increasing (decreasing) on$(a,b)$, then so are the functions

$$ \frac{f(x)-f(a)}{g(x)-g(a)} \quad \textit{and}\quad \frac{f(x)-f(b)}{g(x)-g(b)}. $$

If$f'(x)/g'(x)$is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 2.3

Let$0< a,b<1$with$a+b>1$. Then the following assertions are valid:

• (i) The function$(\mathcal{K}-\mathcal{E})/(r^{2}\mathcal{K})$is strictly increasing from$(0,1)$onto$(2b/(a+b+1),1)$;

• (ii) The function$r^{\prime a+b-1}\mathcal{K}$has positive Maclaurin coefficients and maps$(0, 1)$onto$(B/2, D)$;

• (iii) The function$r^{\prime p}\mathcal{K}$is strictly decreasing from$(0,1)$onto$(0,B/2)$if$p\geq 4ab/(a+b+1)$.

Proof

Items (i) and (ii) follow directly from [34, Lemma 4.22]. We only need to prove item (iii).

It follows from (2.1) that

$$\begin{aligned} \frac{d(r^{\prime p}\mathcal{K})}{dr}&=\frac{r^{\prime p-2}}{r} \bigl[(2b-p)r^{2} \mathcal{K}-(b-a+1) (\mathcal{K}-\mathcal{E}) \bigr] \\ &=(b-a+1)rr^{\prime p-2}\mathcal{K} \biggl[ \frac{2b-p}{b-a+1}- \frac{\mathcal{K}-\mathcal{E}}{r^{2}\mathcal{K}} \biggr]. \end{aligned}$$

(2.7)

Lemma 2.3(i) and (2.7) enable us to know that $r^{\prime p}\mathcal{K}$ is strictly decreasing on $(0,1)$ if $(2b-p)/(b-a+1)\leq 2b/(a+b+1)$, that is, $p\geq 4ab/(a+b+1)$.

Note that

$$ r^{\prime p}\mathcal{K}=\frac{B(a,b)}{2}r^{\prime p+1-a-b}F \biggl(\frac{b-a+1}{2}, \frac{a-b+1}{2};\frac{a+b+1}{2};r^{2} \biggr). $$

(2.8)

If $p\geq 4ab/(a+b+1)$, then $p+1-a-b\geq (a-b+1)(b-a+1)/(a+b+1)>0$. This in conjunction with (1.2) and (2.8) gives $\lim_{r\rightarrow 1^{-}}r^{\prime p}\mathcal{K}=0$. □

In the following Lemma 2.4 we provide an asymptotic formula for $\mathcal{K}$ as $r\rightarrow 1$ in the case of $a+b>1$, which is the analog for the zero-balanced hypergeometric function (1.2).

Lemma 2.4

Let$0< a,b<1$with$a+b>1$. Then one has

$$ \mathcal{K}(\sqrt{r}) = D(1-r)^{(1-a-b)/2}+E+o \bigl((1-r)^{ \frac{a+b-1}{2}} \log (1-r) \bigr) $$

as$r\rightarrow 1$.

Proof

It follows from $F(a,b;(a+b+1)/2;r)$ is asymptotic to $2D(1-r)^{(1-a-b)/2}/B$ [19, Theorem 1.19(5)] as $r\rightarrow 1$ for $a+b>1$ and the derivative formula

$$ \frac{dF(a,b;c;r)}{dr}=\frac{ab}{c}F(a+1,b+1;c+1;r) $$

(2.9)

given in [19, (1.16)] for the hypergeometric function together with (1.2), and L’Hôpital’s rule that

$$\begin{aligned} &\lim_{r\rightarrow 1^{-}} \frac{\mathcal{K}(\sqrt{r})-D(1-r)^{(1-a-b)/2}-E}{(1-r)^{(a+b-1)/2}\log (1-r)} \\ &\quad = \lim_{r\rightarrow 1^{-}} \frac{BF (\frac{b-a+1}{2},\frac{a-b+1}{2};\frac{a+b+1}{2};r )-2D-2E(1-r)^{(a+b-1)/2}}{2(1-r)^{a+b-1}\log (1-r)} \\ &\quad =\lim_{r\rightarrow 1^{-}} \frac{[(b-a)^{2}-1]BF (\frac{b-a+3}{2},\frac{a-b+3}{2};\frac{a+b+3}{2};r )-2[(a+b)^{2}-1]E(1-r)^{(a+b-3)/2}}{4(a+b+1)(1-r)^{a+b-2}[(a+b-1)\log (1-r)+1]} \\ &\quad =\lim_{r\rightarrow 1^{-}} \frac{[(b-a)^{2}-1]BF (a,b;\frac{a+b+3}{2};r )-2[(a+b)^{2}-1]E}{4(a+b+1)(1-r)^{(a+b-1)/2}[(a+b-1)\log (1-r)+1]} \\ &\quad =\lim_{r\rightarrow 1^{-}} \frac{ab[1-(b-a)^{2}]B(1-r)^{2-(a+b)}F (\frac{b-a+3}{2},\frac{a-b+3}{2};\frac{a+b+5}{2};r )}{(a+b+3)[(a+b)^{2}-1][(a+b-1)\log (1-r)+3]} \\ &\quad =0. \end{aligned}$$

This completes the proof. □

Lemma 2.4 leads to Corollary 2.5 immediately.

Corollary 2.5

Let$0< a,b<1$and$a+b>1$. Then

$$ Dr^{1-a-b}+E-\mu _{a,b}(r)\rightarrow 0 \quad \textit{and}\quad Dr^{1-a-b}+E-m_{a,b}(r) \rightarrow 0 $$

as$r\rightarrow 0$.

Proof

By replacing r with $1-r^{2}$ in Lemma 2.4, we clearly see that

$$ \mathcal{K}' = Dr^{1-a-b}+E+o \bigl(r^{a+b-1}\log r^{2}\bigr). $$

(2.10)

By definition, it is easy to know that $(\mathcal{K}-B/2)/r\rightarrow 0$ as $r\rightarrow 0$. This in conjunction with (2.10) and $a+b<2$ yields

$$\begin{aligned} &Dr^{1-a-b}+E-\mu _{a,b}(r) \\ &\quad =\frac{B}{2\mathcal{K}} \bigl(Dr^{1-a-b}+E-\mathcal{K}' \bigr)+ \frac{1}{\mathcal{K}} (\mathcal{K}-B/2 ) \bigl(Dr^{1-a-b}+E \bigr)\rightarrow 0 \end{aligned}$$

as $r\rightarrow 0$. The second asymptotic formula can be proved by similar arguments. □

Lemma 2.6

Let$0< a,b<1$with$a+b>1$. Then the following assertions are valid:

• (i) If$\kappa _{1}(a,b)\geq 0$, then$\kappa _{5}(a,b)>0$and$a+b<3/2$;

• (ii) $\kappa _{4}(a,b)<\kappa _{3}(a,b)$;

• (iii) If$\kappa _{5}(a,b)\geq 0$and$a\leq b$, then$3-3a-b>0$.

Proof

(i) We only need to prove that it is not possible for $\kappa _{1}(a,b)\geq 0$ and $\kappa _{5}(a,b)\leq 0$. By calculations, the inequality $\kappa _{1}(a,b)\geq 0$ is equivalent to $0< a\leq 1/2$ and $1-a< b<1$ or $1/2< a<1$ and $1-a< b\leq b_{1}(a)$, where $b_{1}(a)=\frac{1}{4a} [-1 + 8 a - 2 a^{2}-\sqrt{1 - 24 a + 60 a^{2} - 32 a^{3} + 4 a^{4}} ]$ and $\kappa _{5}(a,b)\leq 0$ is equivalent to $1/2<2(2-\sqrt{3})<a<1$ and $b_{2}(a,b)\leq b<1$, where $b_{2}(a)=-1 + 5 a - 2\sqrt{3(2a^{2}-a)}$.

It remains to show that $b_{2}(a)>b_{1}(a)$ for $2(2-\sqrt{3})< a<1$. A simple calculation leads to

$$\begin{aligned} b_{2}(a)-b_{1}(a)={}&\frac{1}{4a} \bigl[1 - 12 a + 22 a^{2}+\sqrt{1 - 24 a + 60 a^{2} - 32 a^{3} + 4 a^{4}} \bigr] \\ &{} -2\sqrt{3a(2a-1)}>0 \end{aligned}$$

if and only if

$$\begin{aligned} & \bigl(1 - 12 a + 22 a^{2}+\sqrt{1 - 24 a + 60 a^{2} - 32 a^{3} + 4 a^{4}} \bigr)^{2}- \bigl(8a \sqrt{3a(2a-1)} \bigr)^{2} \\ &\quad =2 \bigl[\bigl(22a^{2}-12a+1\bigr)\sqrt{1 - 24 a + 60 a^{2} - 32 a^{3} + 4 a^{4}} \\ &\qquad {} +1 - 24 a + 124 a^{2} - 184 a^{3} + 52 a^{4} \bigr]>0, \end{aligned}$$

which is also equivalent to

$$\begin{aligned} & \bigl(22a^{2}-12a+1 \bigr)^{2} \bigl(1 - 24 a + 60 a^{2} - 32 a^{3} + 4 a^{4} \bigr) \\ & \qquad {} - \bigl(1 - 24 a + 124 a^{2} - 184 a^{3} + 52 a^{4} \bigr)^{2} \\ &\quad =64 a^{3} (a+1) (3 - 2 a) (2 a - 1) \bigl(3 a^{2}-3a+1\bigr)>0 \end{aligned}$$

for $1/2< a<1$. On the other hand, as we know, $\kappa _{1}(a,b)$ can be thought of as a quadratic function of b and the parabola opens up. It is easy to verify that $\kappa _{1}(a,1-a)=2 (1 - 3 a + 3 a^{2})>0$ and $\kappa _{1}(a,1)=-(2-a) (2a-1)<0$ for $1/2< a<1$. Combining this with $\kappa _{1}(a,3/2-a)=-5(1-a)(a-1/2)<0$ for $1/2< a<1$, we conclude that $3/2-a< b<1$ makes $\kappa _{1}(a,b)$ negative. This completes the first assertion.

(ii) Observe that $\kappa _{4}(a,b)-\kappa _{3}(a,b)=2(a+b+1)(a+b-1)\mathcal{Q}(b)$, where

$$ \mathcal{Q}(b)=a^{2}+a-2-(4a-1)b+b^{2} $$

(2.11)

is a quadratic function in terms of b. Since the parabola of $\mathcal{Q}(b)$ opens up, it follows from $\mathcal{Q}(1-a)=-6a(1-a)<0$ and $\mathcal{Q}(1)=-a(3-a)<0$ that $\mathcal{Q}(b)<0$ for $0< a<1$ and $1-a< b<1$. This in conjunction with (2.11) yields $\kappa _{4}(a,b)<\kappa _{3}(a,b)$.

(iii) If the conclusion is not true, that is, $3-3a-b\leq 0$, it follows that $b\geq \max \{a,3(1-a)\}$. As we know, $\kappa _{5}(a,b)=b^{2}-2(5a-1)b+(a+1)^{2}$ is a quadratic function of b. We divide the proof into two cases.

Case 1::

$a\geq 3(1-a)$. Then we clearly see that $a\leq b<1$ and $3/4\leq a<1$. Since the symmetric axis $5a-1>1$, $\kappa _{5}(a,b)$ is strictly decreasing for $a< b<1$. This gives $\kappa _{5}(a,b)\leq \kappa _{5}(a,a)=-[8(a-3/4)^{2}+8(a-3/4)+1/2]<0$, which is a contradiction.

Case 2::

$a<3(1-a)$. In other words, $3(1-a)< b<1$ and $2/3< a<3/4$. Similarly, the monotonicity of $\kappa _{5}(a,b)$ gives rise to $\kappa _{5}(a,b)\leq \kappa _{5}(a,3(1-a))=4(2a-1)(5a-4)<0$, which is also a contradiction.

 □

Lemma 2.7

Let$0< a,b<1$with$a+b>1$and$a+b+1\geq 4ab$, and$\varphi (r)$be defined by

$$ \varphi (r)= \frac{1/r^{\prime a+b-1}-1}{B^{2}/ (4r^{\prime a+b+1}\mathcal{K}^{2} )-1}. $$

Then$\varphi (r)$is strictly decreasing from$(0,1)$onto$(0,\frac{(a+b-1)(a+b+1)}{1+2a+2b+a^{2}+b^{2}-6ab} )$.

Proof

Let $\varphi _{1}(r)=1/r^{\prime a+b-1}-1$ and $\varphi _{2}(r)=B^{2}/ (4r^{\prime a+b+1}\mathcal{K}^{2} )-1$. Then $\varphi (r)=\varphi _{1}(r)/\varphi _{2}(r)$ and $\varphi _{1}(0)=\varphi _{2}(0)=0$. Combining this with Lemma 2.2, we clearly see that the monotonicity of $\varphi (r)$ depends on $\varphi '_{1}(r)/\varphi '_{2}(r)$, that is,

$$ \frac{\varphi '_{1}(r)}{\varphi '_{2}(r)}=\frac{4(a+b-1)}{B^{2}} \cdot \bigl(r'\mathcal{K}\bigr)^{2}\cdot \frac{r^{2}\mathcal{K}}{(a+1-3b)r^{2}\mathcal{K}+2(b-a+1)(\mathcal{K}-\mathcal{E})}. $$

(2.12)

It follows from Lemma 2.3(i) that $(a+1-3b)+2(b-a+1)(\mathcal{K}-\mathcal{E})/(r^{2}\mathcal{K})$ is strictly increasing from $(0,1)$ onto $(\frac{1+2a+2b+a^{2}+b^{2}-6ab}{a+b+1},3-a-b)$. Since $a+b+1\geq 4ab$, Lemma 2.3(iii) leads to the conclusion that $r'\mathcal{K}$ is strictly decreasing from $(0,1)$ onto $(0,B/2)$. This in conjunction with (2.12) implies that $\varphi '_{1}(r)/\varphi '_{2}(r)$ is strictly decreasing on $(0,1)$.

On the other hand, it follows from L’Hôpital’s rule and (2.12) that

$$ \varphi \bigl(0^{+}\bigr)=\lim_{r\rightarrow 0^{+}} \frac{\varphi '_{1}(r)}{\varphi '_{2}(r)}= \frac{(a+b-1)(a+b+1)}{1+2a+2b+a^{2}+b^{2}-6ab}, \qquad \varphi \bigl(1^{-} \bigr)=0. $$

 □

Lemma 2.8

Let$(a,b)\in \varOmega _{1}$and$f(r)$be defined by

$$ f(r)= \frac{(1-2b)r^{2}\mathcal{K}+(b-a+1)(\mathcal{K}-\mathcal{E})}{\frac{B^{2}}{4r^{\prime a+b+1}\mathcal{K}^{2}}-1}. $$

Then$f(r)$is strictly decreasing from$(0,1)$onto$(0,\frac{(a+b+1-4ab)B}{1+2a+2b+a^{2}+b^{2}-6ab} )$.

Proof

Let $f_{1}(r)=(1-2b)r^{2}\mathcal{K}+(b-a+1)(\mathcal{K}-\mathcal{E})$ and $f_{2}(r)=B^{2}/ (4r^{\prime a+b+1}\mathcal{K}^{2} )-1$. Then we clearly see that $f(r)=f_{1}(r)/f_{2}(r)$ and $f_{1}(0)=f_{2}(0)=0$.

By calculations, one has

$$ \frac{f'_{1}(r)}{f'_{2}(r)}=\frac{4}{B^{2}}\cdot \bigl[r^{\prime (a+b+1)/3} \mathcal{K} \bigr]^{3}\cdot \widehat{f}(r), $$

(2.13)

where

$$ \widehat{f}(r)= \frac{[\sigma _{1}(a,b)+\sigma _{2}(a,b)r^{2}]r^{2}\mathcal{K}-[\sigma _{3}(a,b)+\sigma _{4}(a,b)r^{2}](\mathcal{K}-\mathcal{E})}{(a+1-3b)r^{2}\mathcal{K}+2(b-a+1)(\mathcal{K}-\mathcal{E})} \triangleq \frac{\widehat{f}_{1}(r)}{\widehat{f}_{2}(r)} $$

(2.14)

and

$$\begin{aligned} &\sigma _{1}(a,b)=2 \bigl(1-b-a b + b^{2}\bigr), \qquad \sigma _{2}(a,b)=2(1-b) (2b-1), \\ &\sigma _{3}(a,b)=(a+b-1) (b-a+1), \qquad \sigma _{4}(a,b)=(b-a+1) (3-2 a-2b). \end{aligned}$$

Let

$$ \widehat{f}_{11}(r)=\sigma _{1}(a,b)-\sigma _{3}(a,b) \frac{\mathcal{K}-\mathcal{E}}{r^{2}\mathcal{K}}, \qquad \widehat{f}_{12}(r)=-r^{2} \biggl[\sigma _{4}(a,b) \frac{\mathcal{K}-\mathcal{E}}{r^{2}\mathcal{K}}-\sigma _{2}(a,b) \biggr]. $$

Then $\widehat{f}_{1}(r)/(r^{2}\mathcal{K})=\widehat{f}_{11}(r)+ \widehat{f}_{12}(r)$.

It follows from Lemma 2.3(i) and $\sigma _{3}(a,b)>0$ that $\widehat{f}_{11}(r)$ is strictly decreasing on $(0,1)$. For $(a,b)\in \varOmega _{1}$, namely $0< a,b<1$, $a+b>1$ and $\kappa _{1}(a,b)\geq 0$, we clearly see from Lemma 2.6(i) that $\kappa _{5}(a,b)=(a+b+1)^{2}-12ab>0$, $a+b<3/2$, and then $\sigma _{4}(a,b)>0$. This in conjunction with Lemma 2.3(i), (iii) implies that $r^{\prime (a+b+1)/3}\mathcal{K}$ is strictly decreasing on $(0,1)$ and

$$ \sigma _{4}(a,b)\frac{\mathcal{K}-\mathcal{E}}{r^{2}\mathcal{K}}- \sigma _{2}(a,b)>\sigma _{4}(a,b)\cdot \frac{2b}{a+b+1}- \sigma _{2}(a,b)= \frac{2 \kappa _{1}(a,b)}{a+b+1}\geq 0. $$

(2.15)

Lemma 2.3(i) and (2.15) enable us to know that $\widehat{f}_{12}(r)$ is strictly decreasing on $(0,1)$. This gives the monotonicity of $\widehat{f}_{1}(r)/(r^{2}\mathcal{K})$. So $\widehat{f}_{1}(r)/(r^{2}\mathcal{K})>\sigma _{1}(a,b)+\sigma _{2}(a,b)- \sigma _{3}(a,b)-\sigma _{4}(a,b)=(2-a-b)(a+b-1)>0$. Moreover, it is easy to verify from Lemma 2.3(i) that $\widehat{f}_{2}(r)/(r^{2}\mathcal{K})$ is strictly increasing from $(0,1)$ onto $(\frac{1+2a+2b+a^{2}+b^{2}-6ab}{a+b+1},3-a-b)$. Combining with (2.14), the monotonicity of $\widehat{f}_{1}(r)/(r^{2}\mathcal{K})$ and $\widehat{f}_{2}(r)/(r^{2}\mathcal{K})$ leads to the conclusion that $\widehat{f}(r)$ is strictly decreasing on $(0,1)$.

Therefore, the monotonicity of $f(r)$ follows from Lemma 2.2 and (2.13) together with the monotonicity of $r^{\prime (a+b+1)/3}\mathcal{K}$ and $\widehat{f}(r)$.

To this end, by L’Hôpital’s rule and (2.13), (2.14),

$$ f\bigl(0^{+}\bigr)=\lim_{r\rightarrow 0^{+}} \frac{f'_{1}(r)}{f'_{2}(r)}= \frac{(a+b+1-4ab)B}{1+2a+2b+a^{2}+b^{2}-6ab}, \qquad f\bigl(1^{-} \bigr)=0. $$

 □

Lemma 2.9

Let$(a,b)\in \varOmega _{2}$and$g(r)$be defined by

$$ g(r)= \frac{B^{2}/ (4r^{\prime a+b+1}\mathcal{K}^{2} )-1}{ [2br^{2}\mathcal{K}+(a-b-1)(\mathcal{K}-\mathcal{E}) ]/r^{\prime 2}}. $$

Then$g(r)$is strictly decreasing from$(0,1)$onto$(0,\frac{1+2a+2b+a^{2}+b^{2}-6ab}{4abB} )$.

Proof

Let $g_{1}(r)=B^{2}/ (4r^{\prime a+b+1}\mathcal{K}^{2} )-1$ and $g_{2}(r)= [2br^{2}\mathcal{K}+(a-b-1)(\mathcal{K}-\mathcal{E}) ]/r^{\prime 2}$. Then $g(r)=g_{1}(r)/g_{2}(r)$ and $g_{1}(0)=g_{2}(0)=0$.

By calculations, one has

$$ \frac{g'_{1}(r)}{g'_{2}(r)}= \frac{B^{2}}{4 [r^{\prime a+b-1}\mathcal{K} ]^{2}}\cdot \frac{\widehat{g}_{1}(r)}{\widehat{g}_{2}(r)}, $$

(2.16)

where

$$\begin{aligned} &\widehat{g}_{1}(r)= \frac{r^{\prime a+b-1} [(a-3b+1)r^{2}\mathcal{K}+2(b-a+1)(\mathcal{K}-\mathcal{E}) ]}{r^{2}\mathcal{K}}, \end{aligned}$$

(2.17)

$$\begin{aligned} &\widehat{g}_{2}(r)= \frac{ [\lambda _{1}(a,b)+\lambda _{2}(a,b)r^{2} ]r^{2}\mathcal{K}+ [\lambda _{3}(a,b)+\lambda _{4}(a,b)r^{2} ](\mathcal{K}-\mathcal{E})}{r^{2}} \end{aligned}$$

(2.18)

and

$$\begin{aligned} \lambda _{1}(a,b)&=2b(a-b+1), \qquad \lambda _{2}(a,b)=4b^{2}, \\ \lambda _{3}(a,b)&=(b-a+1) (a+b-1), \qquad \lambda _{4}(a,b)=-2(a+b) (b-a+1). \end{aligned}$$

By (2.2) and (2.9), we clearly see that

$$\begin{aligned} \frac{\mathcal{K}-\mathcal{E}}{r^{2}}&=\frac{B(a,b)}{4(a-1)r} \frac{dF(a-1,b;(a+b+1)/2;r^{2})}{dr} \\ &=\frac{bB(a,b)}{a+b+1}F\bigl(a,b+1;(a+b+3)/2;r^{2} \bigr). \end{aligned}$$

(2.19)

It follows from (1.2), (1.5), and (2.19) that

$$\begin{aligned} &r^{\prime a+b-1}\mathcal{K}=\frac{B(a,b)}{2}F \biggl( \frac{b-a+1}{2}, \frac{a-b+1}{2};\frac{a+b+1}{2};r^{2} \biggr), \end{aligned}$$

(2.20)

$$\begin{aligned} &r^{\prime a+b-1}\frac{\mathcal{K}-\mathcal{E}}{r^{2}}= \frac{B(a,b)b}{a+b+1}F \biggl(\frac{b-a+3}{2},\frac{a-b+1}{2}; \frac{a+b+3}{2};r^{2} \biggr). \end{aligned}$$

(2.21)

Combining with (2.17), (2.18), (2.20), and (2.21), we rewrite $\widehat{g}_{1}(r)$ and $\widehat{g}_{2}(r)$ in terms of power series:

$$\begin{aligned} &\widehat{g}_{1}(r)= \frac{\sum_{n=0}^{\infty }\frac{(\frac{a-b+1}{2},n)(\frac{b-a+1}{2},n)}{(\frac{a+b+1}{2},n)n!}\xi _{a,b}(n)r^{2n}}{\sum_{n=0}^{\infty }\frac{(a,n)(b,n)}{(\frac{a+b+1}{2},n)n!}r^{2n}}, \end{aligned}$$

(2.22)

$$\begin{aligned} &\widehat{g}_{2}(r)=\frac{B(a,b)}{2}\sum _{n=0}^{\infty }\frac{(a,n-1)(b,n-1)}{(\frac{a+b+1}{2},n+1)n!}\zeta _{a,b}(n)r^{2n}, \end{aligned}$$

(2.23)

where

$$\begin{aligned} &\xi _{a,b}(n)= \frac{1 + 2a+2b+a^{2}+b^{2}-6a b+2(a+b+1)n}{1 + a + b + 2 n}, \\ &\zeta _{a,b}(n)=\bigl[(a + b)^{2}-1\bigr]n^{3}+2 (a + b-1) (a + b + 2 a b-1)n^{2} \\ &\hphantom{\zeta _{a,b}(n)=}{} + \bigl[4 a b (a + b + a b-1) - 3 (a + b)^{2} + 4 (a + b) -1 \bigr]n \\ &\hphantom{\zeta _{a,b}(n)=}{} +4 ab(1-a) (1-b). \end{aligned}$$

We now claim that $\widehat{g}_{1}(r)$ is strictly decreasing on $(0,1)$ and $\widehat{g}_{2}(r)$ is strictly increasing on $(0,1)$; furthermore, $\widehat{g}_{2}(r)$ has positive Maclaurin coefficients.

• Lemma 2.1 and (2.22) enable us to know that the monotonicity of $\widehat{g}_{1}(r)$ depends on the monotonicity of the following sequence:

  $$ \{\alpha _{n}\}_{n\geq 0}= \biggl\{ \frac{(\frac{a-b+1}{2},n)(\frac{b-a+1}{2},n)}{(a,n)(b,n)}\xi _{a,b}(n) \biggr\} _{n\geq 0}. $$

  A simple calculation yields

  $$ \frac{\alpha _{n+1}}{\alpha _{n}}= \frac{(1+b-a +2n ) (1+a-b+2n)\xi _{a,b}(n+1)}{4(b+n)(a+n)\xi _{a,b}(n)} \leq 1 $$

  if and only if

  $$\begin{aligned} \Delta _{a,b}(n)={}&(1+b-a +2n ) (1+a-b+2n)\xi _{a,b}(n+1) \\ &{} -4(a+n) (b+n)\xi _{a,b}(n) \\ ={}& \frac{\widehat{\Delta }_{a,b}(n)}{(1 + a + b + 2 n) (3 + a + b + 2 n)} \leq 0, \end{aligned}$$

  (2.24)

  where

  $$ \widehat{\Delta }_{a,b}(n)= \kappa _{2}(a,b)+4\kappa _{3}(a,b)n+4 \kappa _{4}(a,b)n^{2}-16 \bigl[(a+b)^{2}-1 \bigr]n^{3}. $$

  (2.25)

  For $(a,b)\in \varOmega _{2}$, namely $0< a,b<1$, $a+b>1$, $\kappa _{2}(a,b)\leq 0$, $\kappa _{3}(a,b)\leq 0$, and then $\kappa _{4}(a,b)\leq 0$ by Lemma 2.6(ii). This in conjunction with (2.24) and (2.25) implies that the sequence $\{\alpha _{n}\}_{n\geq 0}$ is decreasing. So the first assertion is valid.

• We mention that the Pochhammer symbol $(a,-1)(b,-1)=\frac{1}{(a-1)(b-1)}>0$ for $0< a,b<1$. It only needs to prove $\zeta _{a,b}(n)>0$ for $n\geq 0$ with $0< a,b<1$ and $a+b>1$.

  Clearly, $\zeta _{a,b}(0)=4 ab(1-a) (1-b)>0$ and $\zeta _{a,b}(1)=4 a b (a + b + 2 a b-1)>0$. Moreover, $\zeta _{a,b}'(n)$ is strictly increasing for $n\geq 0$. This gives $\zeta _{a,b}'(n)\geq \zeta _{a,b}'(1)=4q(b)$ for $n\geq 1$, where $q(b)=(a^{2}+3a+1)b^{2}+(3 a^{2}-a-1)b+a(a-1)$ is regarded as a quadratic function in terms of b and its parabola opens up.

  Observe that

  $$ -\frac{3 a^{2}-a-1}{2(a^{2}+3a+1)}-(1-a)=- \frac{a(1-a)(2a+3)+1}{2(a^{2}+3a+1)}< 0, $$

  that is to say, the symmetric axis of $q(b)$ lies on the left side of the interval $[1-a,1]$. This in conjunction with $q(1-a)=a^{2}(a-1)^{2}>0$ implies that $q(b)>0$ for $1-a< b<1$. So $\zeta _{a,b}(n)$ is strictly increasing for $n\geq 1$ and $\zeta _{a,b}(n)\geq \zeta _{a,b}(1)>0$ for $n\geq 1$. This completes the second assertion.

Therefore, $\widehat{g}_{1}(r)/\widehat{g}_{2}(r)$ is strictly decreasing on $(0,1)$ follows from the above assertions together with $\widehat{g}_{1}(r)>0$ and $\widehat{g}_{2}(r)>0$. Combining this with (2.16), Lemma 2.2 and Lemma 2.3(ii), we conclude that $g(r)$ is strictly decreasing on $(0,1)$.

It remains to compute two end values of $g(r)$. By L’Hôpital’s rule and (2.16) together with Lemma 2.3(i), (ii),

$$ g\bigl(0^{+}\bigr)=\lim_{r\rightarrow 0^{+}} \frac{g'_{1}(r)}{g'_{2}(r)}= \frac{1+2a+2b+a^{2}+b^{2}-6ab}{4abB}, \qquad g\bigl(1^{-} \bigr)=0. $$

 □

Lemma 2.10

Let$0< a\leq b<1$with$a+b>1$and$\kappa _{5}(a,b)\geq 0$, and$h(r)$be defined by

$$ h(r)= \frac{2br^{2}\mathcal{K}-[2(a+b-1)+(3-3a-b)r^{2}](\mathcal{K}-\mathcal{E})}{(a-3b+1)r^{2}\mathcal{K}+2(b-a+1)(\mathcal{K}-\mathcal{E})}. $$

Then$h(r)$is strictly decreasing from$(0,1)$onto$(\frac{a+b-1}{3-a-b},\frac{2b(3-a-b)}{1+2a+2b+a^{2}+b^{2}-6ab} )$.

Proof

We denote by $h_{1}(r)=2b-[2(a+b-1)+(3-3a-b)r^{2}](\mathcal{K}-\mathcal{E})/(r^{2} \mathcal{K})$ and $h_{2}(r)=(a-3b+1)+2(b-a+1)(\mathcal{K}-\mathcal{E})/(r^{2} \mathcal{K})$.

If $0< a\leq b<1$, $a+b>1$, and $\kappa _{5}(a,b)\geq 0$, then $3-3a-b>0$ follows from Lemma 2.6(iii). Combining this with Lemma 2.3(i), we conclude that $h_{1}(r)$ is strictly decreasing from $(0,1)$ onto $(a+b-1,\frac{2b(3-a-b)}{a+b+1} )$ and $h_{2}(r)$ is strictly increasing from $(0,1)$ onto $(\frac{1+2a+2b+a^{2}+b^{2}-6ab}{a+b+1},3-a-b )$. This gives the monotonicity of $h(r)=h_{1}(r)/h_{2}(r)$ together with two limiting values $h(0^{+})$ and $h(1^{-})$. □

3 Main results

Theorem 3.1

Let$(a,b)\in \varOmega _{1}$and$F(r)$be defined on$(0,1)$by

$$ F(r)=\frac{Dr^{1-a-b}+E-m_{a,b}(r)}{Dr^{1-a-b}+E-\mu _{a,b}(r)}. $$

Then$F(r)$is strictly decreasing from$(0,1)$onto$(1,L_{0})$, where

$$ L_{0}= \frac{(a + b + 1) [(a + b - 1)^{2} + 4 ] - 16 a b}{(a + b - 1) (1 + 2 a +2b+ a^{2}+ b^{2}-6ab)}. $$

In particular, the double inequality

$$ m_{a,b}(r)< \mu _{a,b}(r)< \frac{1}{L_{0}}m_{a,b}(r)+ \biggl(1- \frac{1}{L_{0}} \biggr) \bigl(Dr^{1-a-b}+E\bigr) $$

holds for$r\in (0,1)$.

Proof

Let $F_{1}(r)=Dr^{1-a-b}+E-m_{a,b}(r)$ and $F_{2}(r)=Dr^{1-a-b}+E-\mu _{a,b}(r)$. Clearly, $F(r)=F_{1}(r)/F_{2}(r)$ and $F_{1}(0^{+})=F_{2}(0^{+})=0$ follow from Corollary 2.5.

By calculations, one has

$$\begin{aligned} \frac{F'_{1}(r)}{F'_{2}(r)}&= \frac{ (\frac{1}{r^{\prime a+b-1}}-1 )+\frac{4r^{a+b-1}\mathcal{K}'}{(a+b-1)BD} [(1-2b)r^{2}\mathcal{K}+(b-a+1)(\mathcal{K}-\mathcal{E}) ]}{\frac{B^{2}}{4r^{\prime a+b+1}\mathcal{K}^{2}}-1} \\ &=\varphi (r)+\frac{4}{(a+b-1)BD}\cdot r^{a+b-1} \mathcal{K}'\cdot f(r), \end{aligned}$$

(3.1)

where $\varphi (r)$ and $f(r)$ are defined as in Lemma 2.7 and Lemma 2.8, respectively.

Since $r^{a+b-1}\mathcal{K}'$ can be regarded as the composition of $x^{\prime a+b-1}\mathcal{K}(x)$ and $x=r'=\sqrt{1-r^{2}}$, Lemma 2.3(ii) enables us to know that $r^{a+b-1}\mathcal{K}'$ is strictly decreasing from $(0,1)$ onto $(B/2,D)$. This in conjunction with (3.1) together with Lemma 2.2, Lemma 2.7, and Lemma 2.8 gives rise to the monotonicity of $F(r)$ and also, by L’Hôpital’s rule and (3.1),

$$ F\bigl(0^{+}\bigr)=\lim_{r\rightarrow 0^{+}} \frac{F'_{1}(r)}{F'_{2}(r)}= \varphi \bigl(0^{+}\bigr)+\frac{4}{(a+b-1)BD} \cdot D\cdot f\bigl(0^{+}\bigr)=L_{0}, $$

and $F(1^{-})=1$ follows directly from $m_{a,b}(1^{-})=\mu _{a,b}(1^{-})=0$. □

Corollary 3.2

Let$(a,b)\in \varOmega _{1}$and$\widehat{F}(r)$be defined on$(0, 1)$by

$$ \widehat{F}(r)= \frac{m_{a,b}(r)-D(r^{1-a-b}-1)}{\mu _{a,b}(r)-D(r^{1-a-b}-1)}. $$

Then$\widehat{F}(r)$is strictly decreasing from$(0,1)$onto$(0,1)$.

Proof

Let $\widehat{F}_{1}(r)=m_{a,b}(r)-D(r^{1-a-b}-1)$ and $\widehat{F}_{2}(r)=\mu _{a,b}(r)-D(r^{1-a-b}-1)$. Then $\widehat{F}(r)=\widehat{F}_{1}(r)/\widehat{F}_{2}(r)$ and $\widehat{F}_{1}(1^{-})=\widehat{F}_{2}(1^{-})=0$.

Since $\widehat{F}'_{1}(r)/\widehat{F}'_{2}(r)=F'_{1}(r)/F'_{2}(r)$, Lemma 2.2 enables us to know the monotonicity of $\widehat{F}(r)$ depends on that of $F'_{1}(r)/F'_{2}(r)$, which follows from Theorem 3.1. It only remains to compute two limiting values $\widehat{F}(0^{+})$ and $\widehat{F}(1^{-})$.

By Corollary 2.5, it is easy to see that $\widehat{F}(0^{+})=(D+E)/(D+E)=1$. By L’Hôpital’s rule and (3.1) together with Lemma 2.7, Lemma 2.8,

$$ \widehat{F}\bigl(1^{-}\bigr)=\lim_{r\rightarrow 1^{-}} \frac{\widehat{F}'_{1}(r)}{\widehat{F}'_{2}(r)}=\varphi \bigl(1^{-}\bigr)+ \frac{4}{(a+b-1)BD} \cdot \frac{B}{2}\cdot f\bigl(1^{-}\bigr)=0. $$

 □

Theorem 3.3

Let$(a,b)\in \varOmega _{2}$and$G(r)$be defined on$(0,1)$by

$$ G(r)=\frac{Dr^{1-a-b}+E-\mu _{a,b}(r)}{\mathcal{K}-B/2}. $$

Then$G(r)$is strictly decreasing from$(0,1)$onto$(0,\infty )$.

Proof

We denote $G_{1}(r)=Dr^{1-a-b}+E-\mu _{a,b}(r)$ and $G_{2}(r)=\mathcal{K}-B/2$. Then we clearly see that $G(r)=G_{1}(r)/G_{2}(r)$ and $G_{1}(0^{+})=G_{2}(0^{+})=0$.

By taking the derivative of $G_{1}(r)$ and $G_{2}(r)$, one has

$$\begin{aligned} \frac{G'_{1}(r)}{G'_{2}(r)}&=\frac{(a+b-1)D}{r^{a+b-1}}\cdot \frac{B^{2}/ (4r^{\prime a+b+1}\mathcal{K}^{2} )-1}{ [2br^{2}\mathcal{K}+(a-b-1)(\mathcal{K}-\mathcal{E}) ]/r^{\prime 2}} \\ &=\frac{(a+b-1)D}{r^{a+b-1}}\cdot g(r), \end{aligned}$$

(3.2)

where $g(r)$ is defined as in Lemma 2.9.

Therefore, the monotonicity of $G(r)$ follows from Lemma 2.9 and that of $1/r^{a+b-1}$.

To this end, by L’Hôpital’s rule and (3.2),

$$ G\bigl(0^{+}\bigr)=\lim_{r\rightarrow 0^{+}} \frac{G'_{1}(r)}{G'_{2}(r)}=\lim_{r \rightarrow 0^{+}} \frac{(a+b-1)D}{r^{a+b-1}}\cdot g \bigl(0^{+}\bigr)=\infty ,\qquad G\bigl(1^{-}\bigr)=0. $$

 □

Theorem 3.4

Let$(a,b)\in \varOmega _{3}$and$H(r)$be defined on$(0, 1)$by

$$ H(r)=\frac{r^{1-a-b}(B/2-\mathcal{E})}{Dr^{1-a-b}+E-\mu _{a,b}(r)}. $$

Then$H(r)$is strictly decreasing from$(0,1)$onto$(L_{1},L_{2})$, where

$$\begin{aligned} &L_{1}=\frac{2b(1-a)(3-a-b)B}{(a+b-1)(1+2a+2b+a^{2}+b^{2}-6ab)D}, \\ &L_{2}=\frac{(b-a+1)B+2(a+b-1)E}{2(b-a+1)(D+E)}. \end{aligned}$$

As a consequence, the double inequality

$$ r^{1-a-b} \biggl(D-\frac{B}{2L_{1}}+\mathcal{E} \biggr)+E< \mu _{a,b}(r)< r^{1-a-b} \biggl(D-\frac{B}{2L_{2}}+ \mathcal{E} \biggr)+E $$

holds for$r\in (0,1)$.

Proof

Since $H(r)$ is symmetric with respect to a, b, we may assume that $0< a\leq b<1$. Let $H_{1}(r)=r^{1-a-b}(B/2-\mathcal{E})$ and $H_{2}(r)=Dr^{1-a-b}+E-\mu _{a,b}(r)$. Then we clearly see from Corollary 2.5 and $a+b<2$ that $H_{1}(r)=H_{1}(r)/H_{2}(r)$ and $H_{1}(0^{+})=H_{2}(0^{+})=0$.

Moreover,

$$\begin{aligned} \begin{aligned} &\frac{H'_{1}(r)}{H'_{2}(r)}= \frac{r^{-(a+b)} [(1-a-b)B/2+(3a+b-3)\mathcal{E}+2(1-a)\mathcal{K} ]}{(a+b-1)Dr^{-(a+b)} (\frac{B^{2}}{4r^{\prime a+b+1}\mathcal{K}^{2}}-1 )} \\ &\hphantom{\frac{H'_{1}(r)}{H'_{2}(r)}}=\frac{1}{(a+b-1)D}\cdot \frac{(1-a-b)B/2+(3a+b-3)\mathcal{E}+2(1-a)\mathcal{K}}{ (\frac{B^{2}}{4r^{\prime a+b+1}\mathcal{K}^{2}}-1 )} \\ &\hphantom{\frac{H'_{1}(r)}{H'_{2}(r)}}\triangleq \frac{H_{11}(r)}{H_{22}(r)}, \\ &H_{11}\bigl(0^{+}\bigr)=H_{22} \bigl(0^{+}\bigr)=0, \end{aligned} \end{aligned}$$

(3.3)

and

$$ \frac{H'_{11}(r)}{H'_{22}(r)}=\frac{8(1-a)}{(a+b-1)B^{2}D}\cdot \bigl(r^{\prime \frac{a+b+1}{3}} \mathcal{K} \bigr)^{3}\cdot h(r), $$

(3.4)

where $h(r)$ is defined as in Lemma 2.10.

If $(a,b)\in \varOmega _{3}$, in other words, $0< a\leq b<1$, $a+b>1$, and $\kappa _{5}(a,b)=(a+b+1)^{2}-12ab\geq 0$, then it follows from Lemma 2.3(iii) and Lemma 2.10 that $r^{\prime \frac{a+b+1}{3}}\mathcal{K}$ is strictly decreasing on $(0,1)$ and $h(r)$ is strictly decreasing on $(0,1)$. This in conjunction with (3.3), (3.4), and Lemma 2.2 implies that $H(r)$ is strictly decreasing on $(0,1)$. By L’Hôpital’s rule together with Lemma 2.10 and (3.3), (3.4),

$$\begin{aligned} H\bigl(0^{+}\bigr)&=\lim_{r\rightarrow 0^{+}}H(r)=\lim _{r\rightarrow 0^{+}} \frac{H'_{1}(r)}{H'_{2}(r)}=\lim_{r\rightarrow 0^{+}} \frac{H'_{11}(r)}{H'_{22}(r)} \\ &=\frac{8(1-a)}{(a+b-1)B^{2}D}\cdot \biggl(\frac{B}{2} \biggr)^{3} \cdot h\bigl(0^{+}\bigr)=L_{1}, \end{aligned}$$

and $H(1^{-})=L_{2}$ follows easily from $\mu _{a,b}(1^{-})=0$. □

Open Problem

What is the sub-region of $\{(a,b)\in \mathbb{R}^{2}| 0< a,b<1\}$ such that the function

$$ \widehat{G}(r)= \frac{Dr^{1-a-b}+E-\mu _{a,b}(r)}{r^{1-a-b}(\mathcal{K}-B/2)} $$

is strictly decreasing from $(0,1)$ onto $(0,L_{3})$, where

$$ L_{3}=\frac{(a+b-1)(1+2a+2b+a^{2}+b^{2}-6ab)D}{2ab(3-a-b)B}. $$

4 Consequences and discussion

In the article, we study the monotonicity of the functions $F(r)$, $G(r)$, and $H(r)$ related to generalized Grötzsch ring function and generalized elliptic integrals, where $F(r)$, $G(r)$, and $H(r)$ are explicitly given by

$$ F(r)=\frac{Dr^{1-a-b}+E-m_{a,b}(r)}{Dr^{1-a-b}+E-\mu _{a,b}(r)},\qquad G(r)=\frac{Dr^{1-a-b}+E-\mu _{a,b}(r)}{\mathcal{K}-B/2}, $$

and

$$ H(r)=\frac{r^{1-a-b}(B/2-\mathcal{E})}{Dr^{1-a-b}+E-\mu _{a,b}(r)}. $$

5 Conclusion

In the article, we have found the sub-regions of $\{(a, b)\in {\mathbb{R}}^{2}| 0< a, b<1, a+b>1\}$ such that several quotient functions involving $\mu _{a,b}(r)$, $\mathcal{K}_{a,b}(r)$, $\mathcal{E}_{a,b}(r)$, and $m_{a,b}(r)$ are monotonic on their corresponding sub-regions, and established several inequalities for $\mu _{a,b}(r)$ and $m_{a,b}(r)$. Our results are the variants and extensions of the previous results of [42, Theorems 1.1 and 1.2] in the case of zero-balanced.