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# On multivariate higher order Lyapunov-type inequalities

Journal of Inequalities and Applications20142014:503

https://doi.org/10.1186/1029-242X-2014-503

• Received: 30 September 2014
• Accepted: 1 December 2014
• Published:

## Abstract

In this paper, by using the best Sobolev constant method, we obtain some new Lyapunov-type inequalities for a class of even-order partial differential equations; the results of this paper are new which generalize and improve some early results in the literature.

## Keywords

• Lyapunov-type inequality
• even-order partial differential equations
• Sobolev constant

## 1 Introduction

It is well known that the Lyapunov inequality for the second-order linear differential equation
${x}^{″}\left(t\right)+q\left(t\right)x\left(t\right)=0$
(1)
states that if $q\in C\left[a,b\right]$, $x\left(t\right)$ is a nonzero solution of (1) such that $x\left(a\right)=x\left(b\right)=0$, then the following inequality holds:
${\int }_{a}^{b}|q\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt>\frac{4}{b-a}$
(2)

and the constant 4 is sharp.

There have been many proofs and generalizations as well as improvements on this inequality. For example, the authors in  generalized the Lyapunov-type inequality to the partial differential equations or systems.

First let us recall some background and notations which are introduced in [1, 2].

Let A be a spherical shell $\subseteq {\mathbb{R}}^{N}$ for $N>1$, i.e. $A=B\left(0,{R}_{2}\right)-\overline{B\left(0,{R}_{1}\right)}$ for $0<{R}_{1}<{R}_{2}$, where $B\left(0,R\right)=\left\{x\in {\mathbb{R}}^{N}:\parallel x\parallel for $R>0$ and $\parallel \cdot \parallel$ is the Euclidean norm. Denote ${S}^{N-1}=\left\{x\in {\mathbb{R}}^{N}:\parallel x\parallel =1\right\}$, the unit sphere in ${\mathbb{R}}^{N}$ with surface area
(3)
where $\mathrm{\Gamma }\left(\cdot \right)$ is the gamma function. Then every $x\in {\mathbb{R}}^{N}-\left\{0\right\}$ has a unique representation of the form $x=r\omega$, where $r=\parallel x\parallel >0$ and $\omega =\frac{x}{r}\in {S}^{N-1}$. Therefore, for any $f\in C\left(\overline{A}\right)$, we have
${\int }_{A}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx={\int }_{{S}^{N-1}}\left({\int }_{{R}_{1}}^{{R}_{2}}f\left(r\omega \right){r}^{N-1}\phantom{\rule{0.2em}{0ex}}dr\right)\phantom{\rule{0.2em}{0ex}}d\omega .$

In , Aktaş obtained the following results.

Theorem A If $f\in {C}^{2n}\left(\overline{A}\right)$ is a nonzero solution of the following even-order partial differential equation:
$\frac{{\partial }^{2n}f\left(x\right)}{\partial {r}^{2n}}+q\left(x\right)f\left(x\right)=0,\phantom{\rule{1em}{0ex}}x\in A,$
(4)
where $n\in \mathbb{N}$ and $q\in C\left(\overline{A}\right)$, with the boundary conditions
$\frac{{\partial }^{2i}f}{\partial {r}^{2i}}\left(\partial B\left(0,{R}_{1}\right)\right)=\frac{{\partial }^{2i}f}{\partial {r}^{2i}}\left(\partial B\left(0,{R}_{2}\right)\right)=0,\phantom{\rule{1em}{0ex}}i=0,1,2,\dots ,n-1,$
(5)
then the following inequality holds:
${\int }_{A}|q\left(x\right)|\phantom{\rule{0.2em}{0ex}}dx>\frac{{2}^{3n-1}}{{\left({R}_{2}-{R}_{1}\right)}^{2n-1}}\frac{2{\pi }^{\frac{N}{2}}}{\mathrm{\Gamma }\left(\frac{N}{2}\right)}{R}_{1}^{N-1}.$
(6)
Theorem B If $f\in {C}^{2n}\left(\overline{A}\right)$ is a nonzero solution of (4) with the boundary conditions
$\frac{{\partial }^{i}f}{\partial {r}^{i}}\left(\partial B\left(0,{R}_{1}\right)\right)=\frac{{\partial }^{i}f}{\partial {r}^{i}}\left(\partial B\left(0,{R}_{2}\right)\right)=0,\phantom{\rule{1em}{0ex}}i=0,1,2,\dots ,n-1,$
(7)
then the following inequality holds:
${\int }_{A}|q\left(x\right)|\phantom{\rule{0.2em}{0ex}}dx>\frac{{4}^{2n-1}\left(2n-1\right){\left[\left(n-1\right)!\right]}^{2}}{{\left({R}_{2}-{R}_{1}\right)}^{2n-1}}\frac{2{\pi }^{\frac{N}{2}}}{\mathrm{\Gamma }\left(\frac{N}{2}\right)}{R}_{1}^{N-1}.$
(8)

In this paper, we generalize Theorem A and Theorem B to a more general class of even order partial differential equations. Moreover, as we shall see by the end of this paper, Theorem 1 improves Theorem A significantly.

## 2 Main results

Let us consider the following even-order partial differential equation:
$\frac{{\partial }^{2n}y\left(x\right)}{\partial {r}^{2n}}+\sum _{k=0}^{n}{p}_{k}\left(x\right)\frac{{\partial }^{k}y\left(x\right)}{\partial {r}^{k}}=0,$
(9)

where $y\left(x\right)\in {C}^{2n}\left(\overline{A}\right)$, ${p}_{k}\left(x\right)\in C\left(\overline{A}\right)$, $k=0,1,2,\dots ,n$, and $x\in {\mathbb{R}}^{N}$.

The main results of this paper are the following theorems.

Theorem 1 If $y\left(x\right)$ is a nonzero solution of (9) satisfying boundary conditions (5), then the following inequality holds:
$\begin{array}{rl}1<& \sqrt{\frac{\left({2}^{2n}-1\right)\zeta \left(2n\right){\left({R}_{2}-{R}_{1}\right)}^{2n-1}\mathrm{\Gamma }\left(\frac{N}{2}\right)}{{2}^{2n}{\pi }^{2n+\frac{N}{2}}{R}_{1}^{N-1}}}{\left({\int }_{A}{p}_{n}^{2}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right)}^{\frac{1}{2}}\\ +\sum _{k=0}^{n-1}\frac{{\left({R}_{2}-{R}_{1}\right)}^{2n-k-1}\mathrm{\Gamma }\left(\frac{N}{2}\right)}{{\left(2\pi \right)}^{2n-k}{R}_{1}^{N-1}{\pi }^{\frac{N}{2}}}\sqrt{\left({2}^{2n}-1\right)\left({2}^{2\left(n-k\right)}-1\right)\zeta \left(2n\right)\zeta \left(2\left(n-k\right)\right)}\\ ×{\int }_{A}|{p}_{k}\left(x\right)|\phantom{\rule{0.2em}{0ex}}dx,\end{array}$
(10)

where $\zeta \left(s\right)={\sum }_{n=1}^{+\mathrm{\infty }}\frac{1}{{n}^{s}}$ is the Riemann zeta function.

Theorem 2 If $y\left(x\right)$ is a nonzero solution of (9) satisfying boundary conditions (7), then the following inequality holds:
$\begin{array}{rl}1<& \frac{1}{\left(n-1\right)!{2}^{2n-1}}\sqrt{\frac{{\left({R}_{2}-{R}_{1}\right)}^{2n-1}\mathrm{\Gamma }\left(\frac{N}{2}\right)}{\left(2n-1\right){R}_{1}^{N-1}2{\pi }^{\frac{N}{2}}}}{\left({\int }_{A}{p}_{n}^{2}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right)}^{\frac{1}{2}}\\ +\sum _{k=0}^{n-1}\frac{{\left({R}_{2}-{R}_{1}\right)}^{2n-k-1}\mathrm{\Gamma }\left(\frac{N}{2}\right)}{\sqrt{\left(2n-1\right)\left(2n-2k-1\right)}\left(n-1\right)!\left(n-k-1\right)!{4}^{2n-k-1}{R}_{1}^{N-1}2{\pi }^{\frac{N}{2}}}{\int }_{A}|{p}_{k}\left(x\right)|\phantom{\rule{0.2em}{0ex}}dx.\end{array}$
(11)

## 3 Proofs of theorems

For the proofs of Theorem 1 and Theorem 2, let us consider first the following ordinary even-order linear ordinary differential equation:
${x}^{\left(2n\right)}\left(t\right)+\sum _{k=0}^{n}{p}_{k}\left(t\right){x}^{\left(k\right)}\left(t\right)=0,$
(12)

where ${p}_{k}\left(t\right)\in C\left[a,b\right]$, $k=0,1,2,\dots ,n$.

Proposition 3 If (12) has a nonzero solution $x\left(t\right)$ satisfying the following boundary value conditions:
${x}^{\left(2i\right)}\left(a\right)={x}^{\left(2i\right)}\left(b\right)=0,\phantom{\rule{1em}{0ex}}i=0,1,2,\dots ,n-1,$
(13)
then the following inequality holds:
$\begin{array}{rl}1<& \sqrt{\frac{\left({2}^{2n}-1\right)\zeta \left(2n\right){\left(b-a\right)}^{2n-1}}{{2}^{2n-1}{\pi }^{2n}}}{\left({\int }_{a}^{b}{p}_{n}^{2}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}\\ +\sum _{k=0}^{n-1}\frac{{\left(b-a\right)}^{2n-k-1}\sqrt{\left({2}^{2n}-1\right)\left({2}^{2n-2k}-1\right)\zeta \left(2n\right)\zeta \left(2\left(n-k\right)\right)}}{{2}^{2n-k-1}{\pi }^{2n-k}}{\int }_{a}^{b}|{p}_{k}\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt,\end{array}$
(14)

where $\zeta \left(s\right)$ is the Riemann zeta function: $\zeta \left(s\right)={\sum }_{k=1}^{+\mathrm{\infty }}\frac{1}{{k}^{s}}$, $s>1$.

Proposition 4 If (12) has a nonzero solution $x\left(t\right)$ satisfying the following boundary value conditions:
${x}^{\left(i\right)}\left(a\right)={x}^{\left(i\right)}\left(b\right)=0,\phantom{\rule{1em}{0ex}}i=0,1,2,\dots ,n-1,$
(15)
then we have the following inequality:
$\begin{array}{rl}1<& \frac{1}{\left(n-1\right)!{2}^{2n-1}}\sqrt{\frac{{\left(b-a\right)}^{2n-1}}{\left(2n-1\right)}}{\left({\int }_{a}^{b}{p}_{n}^{2}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}\\ +\sum _{k=0}^{n-1}\frac{{\left(b-a\right)}^{2n-k-1}}{\left(n-1\right)!\left(n-k-1\right)!{4}^{2n-k-1}\sqrt{\left(2n-1\right)\left(2n-2k-1\right)}}{\int }_{a}^{b}|{p}_{k}\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt.\end{array}$
(16)

In order to prove the above propositions, we need the following lemmas.

Lemma 5 ([, Proposition 2.1])

Let $M\in \mathbb{N}$ and
${H}_{C}=\left\{u|{u}^{\left(M\right)}\in {L}^{2}\left(a,b\right),{u}^{\left(2k\right)}\left(a\right)={u}^{\left(2k\right)}\left(b\right)=0,0\le k\le \left[\left(M-1\right)/2\right]\right\}.$
Then there exists a positive constant C such that, for any $u\in {H}_{C}$, the Sobolev inequality
${\left(\underset{a\le t\le b}{sup}|u\left(t\right)|\right)}^{2}\le C{\int }_{a}^{b}|{u}^{\left(M\right)}\left(t\right){|}^{2}\phantom{\rule{0.2em}{0ex}}dt$
holds. Moreover, the best constant $C=C\left(M\right)$ is as follows:
$C\left(M\right)=\frac{\left({2}^{2M}-1\right)\zeta \left(2M\right){\left(b-a\right)}^{2M-1}}{{2}^{2M-1}{\pi }^{2M}}.$

Lemma 6 ([, Theorem 1.2 and Corollary 1.3])

Let $M\in \mathbb{N}$ and
${H}_{D}=\left\{u|{u}^{\left(M\right)}\in {L}^{2}\left(a,b\right),{u}^{\left(k\right)}\left(a\right)={u}^{\left(k\right)}\left(b\right)=0,0\le k\le M-1\right\}.$
Then there exists a positive constant D such that for any $u\in {H}_{D}$, the Sobolev inequality
${\left(\underset{a\le t\le b}{sup}|u\left(t\right)|\right)}^{2}\le D{\int }_{a}^{b}|{u}^{\left(M\right)}\left(t\right){|}^{2}\phantom{\rule{0.2em}{0ex}}dt$
holds. Moreover, the best constant $D=D\left(M\right)$ is as follows:
$D\left(M\right)=\frac{{\left(b-a\right)}^{2M-1}}{\left(2M-1\right){\left[\left(M-1\right)!\right]}^{2}{4}^{2M-1}}.$
(17)
We give the first seven values of $\zeta \left(2n\right)$, $C\left(n\right)$, and $D\left(n\right)$ in Table 1.

Since the proof of Proposition 4 is similar to that of Proposition 3, we give only the proof of Proposition 3 below.

Proof of Proposition 3 Multiplying both sides of (12) by $x\left(t\right)$ and integrating from a to b by parts and using the boundary value condition (13), we can obtain
${\int }_{a}^{b}{x}^{\left(2n\right)}\left(t\right)x\left(t\right)\phantom{\rule{0.2em}{0ex}}dt={\left(-1\right)}^{n}{\int }_{a}^{b}{\left({x}^{\left(n\right)}\left(t\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt=-\sum _{k=0}^{n}{\int }_{a}^{b}{p}_{k}\left(t\right){x}^{\left(k\right)}\left(t\right)x\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.$
This yields
$\begin{array}{rl}{\int }_{a}^{b}{\left({x}^{\left(n\right)}\left(t\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt& \le \sum _{k=0}^{n}{\int }_{a}^{b}|{p}_{k}\left(t\right)||{x}^{\left(k\right)}\left(t\right)x\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt\\ ={\int }_{a}^{b}|{p}_{n}\left(t\right)||{x}^{\left(n\right)}\left(t\right)x\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt+\sum _{k=0}^{n-1}{\int }_{a}^{b}|{p}_{k}\left(t\right)||{x}^{\left(k\right)}\left(t\right)x\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt.\end{array}$
(18)
Now, by using Lemma 5, we get for any $t\in \left[a,b\right]$, $k=1,2,\dots ,n-1$,
$|x\left(t\right)|\le \sqrt{C\left(n\right)}{\left({\int }_{a}^{b}{\left({x}^{\left(n\right)}\left(t\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}$
(19)
and
$|{x}^{\left(k\right)}\left(t\right)|\le \sqrt{C\left(n-k\right)}{\left({\int }_{a}^{b}{\left({x}^{\left(n\right)}\left(t\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}.$
(20)
Substituting (19) and (20) into (18), we obtain
$\begin{array}{rl}{\int }_{a}^{b}{\left({x}^{\left(n\right)}\left(t\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt\le & \sqrt{C\left(n\right)}{\int }_{a}^{b}|{p}_{n}\left(t\right)||{x}^{\left(n\right)}\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt{\left({\int }_{a}^{b}{\left({x}^{\left(n\right)}\left(t\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}\\ +\sum _{k=0}^{n-1}\sqrt{C\left(n\right)C\left(n-k\right)}{\int }_{a}^{b}|{p}_{k}\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt{\int }_{a}^{b}{\left({x}^{\left(n\right)}\left(t\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt.\end{array}$
(21)
Now by applying Hölder’s inequality, we get
${\int }_{a}^{b}|{p}_{n}\left(t\right){x}^{\left(n\right)}\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt\le {\left({\int }_{a}^{b}{p}_{n}^{2}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}{\left({\int }_{a}^{b}{\left({x}^{\left(n\right)}\left(t\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}.$
(22)
Substituting (22) into (21) and by using the fact that $x\left(t\right)$ is not a constant function, we obtain the following strict inequality:
$\begin{array}{rl}{\int }_{a}^{b}{\left({x}^{\left(n\right)}\left(t\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt<& \sqrt{C\left(n\right)}{\left({\int }_{a}^{b}{p}_{n}^{2}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}{\int }_{a}^{b}{\left({x}^{\left(n\right)}\left(t\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt\\ +\sum _{k=0}^{n-1}\sqrt{C\left(n\right)C\left(n-k\right)}{\int }_{a}^{b}|{p}_{k}\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt{\int }_{a}^{b}{\left({x}^{\left(n\right)}\left(t\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt.\end{array}$
(23)
Dividing both sides of (23) by ${\int }_{a}^{b}{\left({x}^{\left(n\right)}\left(t\right)\right)}^{2}\phantom{\rule{0.2em}{0ex}}dt$, which can be proved to be positive by using the boundary value condition (13) and the assumption that $x\left(t\right)\not\equiv 0$, we obtain
$1<\sqrt{C\left(n\right)}{\left({\int }_{a}^{b}{p}_{n}^{2}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\right)}^{\frac{1}{2}}+\sum _{k=0}^{n-1}\sqrt{C\left(n\right)C\left(n-k\right)}{\int }_{a}^{b}|{p}_{k}\left(t\right)|\phantom{\rule{0.2em}{0ex}}dt.$

This is equivalent to (14). Thus we finished the proof of Proposition 3. □

Lemma 7 For any $f\in C\left(A\right)$, we have
${\int }_{A}|f\left(x\right)|\phantom{\rule{0.2em}{0ex}}dx\ge \frac{{R}_{1}^{N-1}2{\pi }^{\frac{N}{2}}}{\mathrm{\Gamma }\left(\frac{N}{2}\right)}{\int }_{{R}_{1}}^{{R}_{2}}|f\left(r\omega \right)|\phantom{\rule{0.2em}{0ex}}dr.$
(24)
Proof Similar to the proofs given in  and , we have
${\int }_{{R}_{1}}^{{R}_{2}}|f\left(r\omega \right)|\phantom{\rule{0.2em}{0ex}}dr={\int }_{{R}_{1}}^{{R}_{2}}{r}^{1-N}{r}^{N-1}|f\left(r\omega \right)|\phantom{\rule{0.2em}{0ex}}dr\le \left({\int }_{{R}_{1}}^{{R}_{2}}{r}^{N-1}|f\left(r\omega \right)|\phantom{\rule{0.2em}{0ex}}dr\right){R}_{1}^{1-N},$
which implies that
$\begin{array}{rl}{\int }_{A}|f\left(x\right)|\phantom{\rule{0.2em}{0ex}}dx& ={\int }_{{S}^{N-1}}\left({\int }_{{R}_{1}}^{{R}_{2}}{r}^{N-1}|f\left(r\omega \right)|\phantom{\rule{0.2em}{0ex}}dr\right)\phantom{\rule{0.2em}{0ex}}d\omega \\ \ge {\int }_{{S}^{N-1}}\left({R}_{1}^{N-1}{\int }_{{R}_{1}}^{{R}_{2}}|f\left(r\omega \right)|\phantom{\rule{0.2em}{0ex}}dr\right)\phantom{\rule{0.2em}{0ex}}d\omega \\ =\left({\int }_{{R}_{1}}^{{R}_{2}}|f\left(r\omega \right)|\phantom{\rule{0.2em}{0ex}}dr\right)\frac{{R}_{1}^{N-1}2{\pi }^{\frac{N}{2}}}{\mathrm{\Gamma }\left(\frac{N}{2}\right)}.\end{array}$

□

Proof of Theorem 1 It follows from (14) and Lemma 7 that for any fixed $\omega \in {S}^{N-1}$, we have
$\begin{array}{rl}1<& \sqrt{\frac{\left({2}^{2n}-1\right)\zeta \left(2n\right){\left({R}_{2}-{R}_{1}\right)}^{2n-1}}{{2}^{2n-1}{\pi }^{2n}}}{\left({\int }_{{R}_{1}}^{{R}_{2}}{p}_{n}^{2}\left(r\omega \right)\phantom{\rule{0.2em}{0ex}}dr\right)}^{\frac{1}{2}}\\ +\sum _{k=0}^{n-1}\frac{{\left({R}_{2}-{R}_{1}\right)}^{2n-k-1}\sqrt{\left({2}^{2n}-1\right)\left({2}^{2n-2k}-1\right)\zeta \left(2n\right)\zeta \left(2\left(n-k\right)\right)}}{{2}^{2n-k-1}{\pi }^{2n-k}}{\int }_{a}^{b}|{p}_{k}\left(r\omega \right)|\phantom{\rule{0.2em}{0ex}}dr\\ \le & \sqrt{\frac{\left({2}^{2n}-1\right)\zeta \left(2n\right){\left({R}_{2}-{R}_{1}\right)}^{2n-1}\mathrm{\Gamma }\left(\frac{N}{2}\right)}{{2}^{2n}{\pi }^{2n+\frac{N}{2}}{R}_{1}^{N-1}}}{\left({\int }_{A}{p}_{n}^{2}\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\right)}^{\frac{1}{2}}\\ +\sum _{k=0}^{n-1}\frac{{\left({R}_{2}-{R}_{1}\right)}^{2n-k-1}\mathrm{\Gamma }\left(\frac{N}{2}\right)}{{\left(2\pi \right)}^{2n-k}{R}_{1}^{N-1}{\pi }^{\frac{N}{2}}}\sqrt{\left({2}^{2n}-1\right)\left({2}^{2\left(n-k\right)}-1\right)\zeta \left(2n\right)\zeta \left(2\left(n-k\right)\right)}{\int }_{A}|{p}_{k}\left(x\right)|\phantom{\rule{0.2em}{0ex}}dx,\end{array}$

which is (10). This finishes the proof of Theorem 1. □

The proof of Theorem 2 is similar to that of Theorem 1, so we omit it for simplicity.

Let us compare Theorem 1 and Theorem 2 with Theorem A and Theorem B. It is evident that Theorem 2 is a natural generalization of Theorem B. If we let ${p}_{n}\left(x\right)={p}_{n-1}\left(x\right)=\cdots ={p}_{1}\left(x\right)\equiv 0$, ${p}_{0}\left(x\right)=q\left(x\right)$, $\mathrm{\forall }x\in A$, then (10) reduces to the following inequality:
${\int }_{A}|q\left(x\right)|\phantom{\rule{0.2em}{0ex}}dx>\frac{{2}^{2n-1}{\pi }^{2n}}{\left({2}^{2n}-1\right)\zeta \left(2n\right){\left({R}_{2}-{R}_{1}\right)}^{2n-1}}\frac{2{\pi }^{\frac{N}{2}}}{\mathrm{\Gamma }\left(\frac{N}{2}\right)}{R}_{1}^{N-1}.$
(25)
Let us compare the right sides of inequalities (6) and (25): if we denote ${\delta }_{n}=\frac{{2}^{2n-1}{\pi }^{2n}}{\left({2}^{2n}-1\right)\zeta \left(2n\right){2}^{3n-1}}$, then we have
since $\zeta \left(2n\right)\to 1$ as $n\to \mathrm{\infty }$. Table 2 gives the first eight values of ${\delta }_{n}$.

From Table 2 we see that ${\delta }_{n}$ increases very quickly, so Theorem 1 improves Theorem A significantly even in the special case of (4).

## Declarations

### Acknowledgements

The authors thank the anonymous referees for their valuable suggestions and comments on the original manuscript.

## Authors’ Affiliations

(1)
College of Science, Hebei University of Engineering, Handan, 056038, P.R. China

## References 