# 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.

## Introduction

Given $$x\in (-1,1)$$ and real numbers a, b, and c with $$c\neq 0,-1,-2,\ldots$$ , the Gaussian hypergeometric function $$F(a,b;c;x)$$  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 $$\varGamma (x)=\int _{0}^{\infty }t^{x-1}e^{-t}\,dt$$  and $$B(p,q)=[\varGamma (p)\varGamma (q)]/\varGamma (p+q)$$  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 .

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  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 . 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 .

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.

## Notations, formulas, and lemmas

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

### 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}

### 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  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)

### 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. (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. (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:

1. (i)

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

2. (ii)

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

3. (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:

1. (i)

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

2. (ii)

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

3. (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 $$h(1^{-})$$. □

## 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

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

## 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

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

## 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.

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### Acknowledgements

The authors would like to express their sincere thanks to the editor and the anonymous reviewers for their helpful comments and suggestions.

Not applicable.

## Funding

This work was supported by the National Natural Science Foundation of China (Grant Nos. 11971142, 11871202) and the Natural Science Foundation of Zhejiang Province (Grant No. LY19A010012).

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Correspondence to Tie-Hong Zhao.

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