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An improvement of the Lyapunov inequality for certain higher order differential equations

Journal of Inequalities and Applications20182018:215

https://doi.org/10.1186/s13660-018-1809-5

• Accepted: 7 August 2018
• Published:

Abstract

In this paper, we establish two Lyapunov inequalities for some half-linear higher order differential equations with anti-periodic boundary conditions. Our result improves that obtained by Wang (Appl. Math. Lett. 25:2375–2380, 2012).

Keywords

• Lyapunov inequality
• Half-linear differential equation
• Sobolev inequality
• Riemann zeta function

1 Introduction

In 1907, Lyapunov [2] proved the following remarkable result known as Lyapunov inequality. If u is a solution of
$$u''+q(t)u=0,$$
(1.1)
satisfying $$u(a)=u(b)=0$$ ($$a< b$$) and $$u\neq 0$$, then
$$\int_{a}^{b} \bigl\vert q(t) \bigr\vert \, \mathrm{d}t>\frac{4}{b-a}.$$

Since then, Lyapunov inequality and many of its generalizations have gained a great deal of attention (see [312] and the references therein), because these results have found many applications in the study of various properties of solutions of differential and difference equations such as oscillation theory, disconjugacy, and eigenvalue problems.

In the last twenty years, a lot of efforts have been made to obtain similar results for higher order differential equations (see [1, 1317] and the references therein), and other type integral inequalities (see [1830] and the references therein). In particular, Çakmak [13] considered the following even higher order linear differential equation:
$$u^{(2m)}(t)+r(t)u(t)=0,\quad t\in [a,b],$$
(1.2)
where $$r\in C([a,b],[0,\infty))$$ and u satisfies the following boundary conditions:
$$u^{(2i)}(a)=u^{(2i)}(b)=0, \quad i=0,1,2,\ldots,m-1,$$
(1.3)
and he obtained the following result. If there exists a nontrivial solution u of Eq. (1.2) satisfying (1.3), then one has
$$\int_{a}^{b}r(t)\,\mathrm{d}t>\frac{2^{2m}}{(b-a)^{2m-1}}.$$
(1.4)
Later, Watanabe et al. [14] used a Sobolev inequality to get a new Lyapunov inequality for Eq. (1.2)
$$\int_{a}^{b}r(t)\,\mathrm{d}t>\frac{2^{2m}}{(b-a)^{2m-1}}\cdot \frac{ \pi^{2m}}{{2(2^{2m}-1)\zeta (2m)}},$$
where
$$\zeta (s)=\sum_{n=1}^{+\infty }\frac{1}{n^{s}}, \qquad \operatorname{Re}(s)>1,$$
is the Riemann zeta function. Their result sharpened the result of Çakmak [13].
Recently, Wang [1] considered the following $$(m+1)$$-order half-linear differential equation:
$$\bigl( \bigl\vert u^{(m)}(t) \bigr\vert ^{p-2}u^{(m)}(t) \bigr)'+r(t) \bigl\vert u(t) \bigr\vert ^{p-2}u(t)=0, \quad t\in (a,b),$$
(1.5)
where $$r\in C([a,b],\mathbb{R})$$, $$m\in \mathbb{N}$$, $$p>1$$ is a constant, and u satisfies the following anti-periodic boundary conditions:
$$u^{(i)}(a)+u^{(i)}(b)=0,\quad i=0,1,2,\ldots,m,$$
(1.6)
and he obtained the following result. If there exists a nontrivial solution u of Eq. (1.5) satisfying (1.6), then
$$\int_{a}^{b} \bigl\vert r(t) \bigr\vert \, \mathrm{d}t>2 \biggl( \frac{2}{b-a} \biggr) ^{m(p-1)}.$$
(1.7)

In the present paper, we shall use the Sobolev inequality established in [14] to establish two Lyapunov inequalities for Eq. (1.5) with the anti-periodic boundary conditions (1.6). Our result improves that obtained by Wang [1].

2 Main results

Lemma 2.1

([14])

For $$m\geq 1$$, define the following Sobolev space:
$$H_{m}=\bigl\{ u|u^{(m)}\in L^{2}[a,b], u^{(k)}(a)+u^{(k)}(b)=0, k=0,1,2, \ldots,m-1\bigr\} .$$
There exists a positive constant $$C_{m}$$ such that, for any $$u\in H_{m}$$, the Sobolev inequality
$$\Bigl(\sup_{a\leq t\leq b} \bigl\vert u(t) \bigr\vert \Bigr)^{2}\leq C_{m} \int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{2} \,\mathrm{d}t$$
holds, where
$$C_{m}=\frac{2(2^{2m}-1)(b-a)^{2m-1}\zeta (2m)}{2^{2m}\pi^{2m}},\quad m=1,2, \ldots,$$
and the constants $$\{C_{m}\}$$ are sharp,
$$\zeta (s)=\sum_{n=1}^{+\infty }\frac{1}{n^{s}}, \qquad \operatorname{Re}(s)>1,$$
is the Riemann zeta function.

Remark 2.1

From the definition of $$H_{m}$$, we can easily get if $$u\in H_{m}$$, then $$u^{(m-1)}\in H_{1}$$.

Lemma 2.2

If u is a nontrivial solution of Eq. (1.5) satisfying the anti-periodic boundary conditions (1.6), then the inequalities
$$\bigl\vert u(t) \bigr\vert \leq \frac{(b-a)^{m-1}}{2^{m}} \int_{a}^{b} \bigl\vert u^{(m)}(s) \bigr\vert \,\mathrm{d}s, \quad t\in [a,b],$$
(2.1)
and
$$\bigl\vert u^{(m-1)}(t) \bigr\vert \leq \frac{1}{2} \int_{a}^{b} \bigl\vert u^{(m)}(s) \bigr\vert \,\mathrm{d}s, \quad t\in [a,b]$$
(2.2)
hold.

Proof

Since the nontrivial solution u of Eq. (1.5) satisfies the anti-periodic boundary conditions (1.6), then we have
$$u(t)=\frac{1}{2} \int_{a}^{t}u'(s)\,\mathrm{d}s- \frac{1}{2} \int_{t}^{b}u'(s) \,\mathrm{d}s, \quad t\in [a,b].$$
So,
$$\bigl\vert u(t) \bigr\vert \leq \frac{1}{2} \int_{a}^{t} \bigl\vert u'(s) \bigr\vert \,\mathrm{d}s+\frac{1}{2} \int _{t}^{b} \bigl\vert u'(s) \bigr\vert \,\mathrm{d}s=\frac{1}{2} \int_{a}^{b} \bigl\vert u'(s) \bigr\vert \,\mathrm{d}s, \quad t\in [a,b].$$
(2.3)
Similarly, we have
$$\bigl\vert u^{(k)}(t) \bigr\vert \leq \frac{1}{2} \int_{a}^{b} \bigl\vert u^{(k+1)}(s) \bigr\vert \,\mathrm{d}s, \quad t\in [a,b], k=1,2,\ldots, m-1,$$
(2.4)
then (2.2) holds. It follows from (2.4) that
$$\int_{a}^{b} \bigl\vert u^{(k)}(s) \bigr\vert \,\mathrm{d}s\leq \frac{b-a}{2} \int_{a}^{b} \bigl\vert u ^{(k+1)}(s) \bigr\vert \,\mathrm{d}s, \quad k=1,2,\ldots, m-1.$$
(2.5)
From (2.3) and (2.5), we obtain
$$\bigl\vert u(t) \bigr\vert \leq \frac{b-a}{2^{2}} \int_{a}^{b} \bigl\vert u''(s) \bigr\vert \,\mathrm{d}s\leq \cdots \leq \frac{(b-a)^{m-1}}{2^{m}} \int_{a}^{b} \bigl\vert u^{(m)}(s) \bigr\vert \,\mathrm{d}s, \quad t\in [a,b],$$
(2.6)
i.e., (2.1) holds. □

Theorem 2.1

If u is a nontrivial solution of Eq. (1.5) satisfying the anti-periodic boundary conditions (1.6), then the inequality
$$\int_{a}^{b} \bigl\vert r(t) \bigr\vert \, \mathrm{d}t>2 \biggl( \frac{2}{b-a} \biggr) ^{m(p-1)} \frac{ \pi^{m(p-1)}}{2^{(p-1)/2}(2^{2m}-1)^{(p-1)/2}(\zeta (2m))^{(p-1)/2}}$$
(2.7)
holds, where $$p>2$$.

Proof

Since the nontrivial solution u of Eq. (1.5) satisfies the anti-periodic boundary conditions (1.6), it is easy to see that u is an element of $$H_{m}$$. Multiplying (1.5) by $$u^{(m-1)}(t)$$ and integrating over $$[a, b]$$, yields
$$\int_{a}^{b}\bigl( \bigl\vert u^{(m)}(t) \bigr\vert ^{p-2}u^{(m)}(t)\bigr)'u^{(m-1)}(t) \,\mathrm{d}t+ \int_{a}^{b}r(t) \bigl\vert u(t) \bigr\vert ^{p-2}u(t)u^{(m-1)}(t)\,\mathrm{d}t=0.$$
(2.8)
Using integration by parts to the first integral on the left-hand side of (2.8) and the anti-periodic boundary conditions (1.6), we have
$$\int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{p}\,\mathrm{d}t= \int_{a}^{b}r(t) \bigl\vert u(t) \bigr\vert ^{p-2}u(t)u ^{(m-1)}(t)\,\mathrm{d}t.$$
So
\begin{aligned} \int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{p}\,\mathrm{d}t =& \int_{a}^{b}r(t) \bigl\vert u(t) \bigr\vert ^{p-2}u(t)u ^{(m-1)}(t)\,\mathrm{d}t \\ \leq & \int_{a}^{b} \bigl\vert r(t) \bigr\vert \bigl\vert u(t) \bigr\vert ^{p-1} \bigl\vert u^{(m-1)}(t) \bigr\vert \,\mathrm{d}t \\ \leq & \Bigl(\sup_{a\leq t\leq b} \bigl\vert u(t) \bigr\vert \Bigr)^{p-1}\sup_{a\leq t\leq b} \bigl\vert u ^{(m-1)}(t) \bigr\vert \int_{a}^{b} \bigl\vert r(t) \bigr\vert \, \mathrm{d}t. \end{aligned}
(2.9)
By Lemma 2.1 and Remark 2.1, we obtain
$$\Bigl(\sup_{a\leq t\leq b} \bigl\vert u(t) \bigr\vert \Bigr)^{p-1}\leq C_{m}^{(p-1)/2} \biggl( \int _{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{2}\,\mathrm{d}t \biggr)^{(p-1)/2}$$
(2.10)
and
$$\sup_{a\leq t\leq b} \bigl\vert u^{(m-1)}(t) \bigr\vert \leq C_{1}^{1/2} \biggl( \int_{a}^{b} \bigl\vert u ^{(m)}(t) \bigr\vert ^{2}\,\mathrm{d}t \biggr)^{1/2}.$$
(2.11)
Multiplying (2.10) and (2.11), we have
$$\Bigl(\sup_{a\leq t\leq b} \bigl\vert u(t) \bigr\vert \Bigr)^{p-1}\cdot \sup_{a\leq t\leq b} \bigl\vert u ^{(m-1)}(t) \bigr\vert \leq C_{m}^{(p-1)/2}C_{1}^{1/2} \biggl( \int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{2} \,\mathrm{d}t \biggr)^{p/2}.$$
(2.12)
By using Hölder’s inequality
$$\int_{a}^{b} \bigl\vert f(t)g(t) \bigr\vert \, \mathrm{d}t\leq \biggl( \int_{a}^{b} \bigl\vert f(t) \bigr\vert ^{ \alpha }\,\mathrm{d}t \biggr) ^{1/\alpha } \biggl( \int_{a}^{b} \bigl\vert g(t) \bigr\vert ^{ \beta }\,\mathrm{d}t \biggr) ^{1/\beta }$$
(2.13)
with $$f(t)=1$$, $$g(t)=|u^{(m)}(t)|^{2}$$, $$\alpha =\frac{p}{p-2}$$, and $$\beta =\frac{p}{2}$$, we obtain that
$$\int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{2}\,\mathrm{d}t\leq (b-a)^{(p-2)/p} \biggl( \int _{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{p}\,\mathrm{d}t \biggr)^{2/p}.$$
Thus
$$\biggl( \int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{2}\,\mathrm{d}t \biggr)^{p/2}\leq (b-a)^{(p-2)/2} \int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{p}\,\mathrm{d}t.$$
(2.14)
From (2.12) and (2.14), we have
\begin{aligned}& \Bigl(\sup_{a\leq t\leq b} \bigl\vert u(t) \bigr\vert \Bigr)^{p-1}\cdot \sup_{a\leq t\leq b} \bigl\vert u ^{(m-1)}(t) \bigr\vert \\& \quad \leq C_{m}^{(p-1)/2}C_{1}^{1/2}(b-a)^{(p-2)/2} \int_{a} ^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{p}\,\mathrm{d}t. \end{aligned}
(2.15)
From (2.9) and (2.15), we get
$$\int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{p}\,\mathrm{d}t \leq C_{m}^{(p-1)/2}C_{1}^{1/2}(b-a)^{(p-2)/2} \int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{p}\,\mathrm{d}t \int_{a}^{b} \bigl\vert r(t) \bigr\vert \, \mathrm{d}t.$$
(2.16)
Now, we claim that $$\int_{a}^{b}|u^{(m)}(t)|^{p}\,\mathrm{d}t>0$$. In fact, if the above inequality is not true, we have $$\int_{a}^{b}|u^{(m)}(t)|^{p} \,\mathrm{d}t=0$$, then $$u^{(m)}(t)=0$$ for $$t\in [a, b]$$. By the anti-periodic conditions (1.6), we obtain $$u(t)=0$$ for $$t\in [a, b]$$, which contradicts $$u(t)\not \equiv 0$$, $$t\in [a, b]$$. Thus dividing both sides of (2.16) by $$\int_{a}^{b}|u^{(m)}(t)|^{p}\,\mathrm{d}t$$, we obtain the inequality
$$\int_{a}^{b} \bigl\vert r(t) \bigr\vert \, \mathrm{d}t\geq \frac{1}{C_{m}^{(p-1)/2}C_{1}^{1/2}(b-a)^{(p-2)/2}}.$$
(2.17)
Since
$$C_{m}=\frac{2(2^{2m}-1)(b-a)^{2m-1}\zeta (2m)}{2^{2m}\pi^{2m}} \quad \text{and}\quad \zeta (2)= \frac{\pi^{2}}{6},$$
we get
\begin{aligned} \biggl( \frac{1}{C_{m}} \biggr) ^{(p-1)/2} =& \biggl( \frac{2^{2m}\pi^{2m}}{2(2^{2m}-1)(b-a)^{2m-1} \zeta (2m)} \biggr) ^{(p-1)/2} \\ =&\frac{2^{m(p-1)}\pi^{m(p-1)}}{2^{(p-1)/2}(2^{2m}-1)^{(p-1)/2}(b-a)^{ \frac{(2m-1)(p-1)}{2}}(\zeta (2m))^{(p-1)/2}}, \end{aligned}
and
\begin{aligned} \biggl( \frac{1}{C_{1}} \biggr) ^{(p-1)/2} =& \biggl( \frac{2^{2}\pi^{2}}{2(2^{2}-1)(b-a)^{2-1} \zeta (2)} \biggr) ^{(p-1)/2} \\ =&\frac{2}{(b-a)^{(p-1)/2}}. \end{aligned}
Then
\begin{aligned}& \frac{1}{C_{m}^{(p-1)/2}C_{1}^{1/2}(b-a)^{(p-2)/2}} \\& \quad =\frac{2^{m(p-1)}\pi^{m(p-1)}2}{2^{(p-1)/2}(2^{2m}-1)^{(p-1)/2}(b-a)^{m(p-1)}( \zeta (2m))^{(p-1)/2}} \\& \quad =2 \biggl( \frac{2}{b-a} \biggr) ^{m(p-1)}\frac{\pi^{m(p-1)}}{2^{(p-1)/2}(2^{2m}-1)^{(p-1)/2}( \zeta (2m))^{(p-1)/2}}. \end{aligned}
(2.18)
So, from (2.17) and (2.18), we get (2.7) holds. Moreover, the inequality in (2.7) is strict since u is not a constant. This completes the proof of Theorem 2.1. □

Theorem 2.2

If u is a nontrivial solution of Eq. (1.5) satisfying the anti-periodic boundary conditions (1.6), then the inequality
$$\int_{a}^{b} \bigl\vert r(t) \bigr\vert \, \mathrm{d}t>\frac{2^{m(p-1)+1}}{(b-a)^{m(p-1)}}$$
(2.19)
holds, where $$1< p<2$$.

Proof

As shown in the proof of Theorem 2.1, (2.9) holds, that is,
\begin{aligned} \int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{p}\,\mathrm{d}t \leq & \Bigl(\sup_{a\leq t \leq b} \bigl\vert u(t) \bigr\vert \Bigr)^{p-1}\sup_{a\leq t\leq b} \bigl\vert u^{(m-1)}(t) \bigr\vert \int_{a} ^{b} \bigl\vert r(t) \bigr\vert \, \mathrm{d}t. \end{aligned}
By using Hölder’s inequality (2.13) with $$f(t)=|u^{(m)}(t)|$$, $$g(t)=1$$, $$\alpha =p$$, and $$\beta =\frac{p}{p-1}$$, we obtain that
$$\int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert \,\mathrm{d}t\leq \biggl( \int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{p} \,\mathrm{d}t \biggr)^{1/p}(b-a)^{(p-1)/p}.$$
Thus
$$(b-a)^{1-p} \biggl( \int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert \,\mathrm{d}t \biggr)^{p}\leq \int _{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert ^{p}\,\mathrm{d}t.$$
(2.20)
From (2.1) and (2.2), we have
\begin{aligned}& \Bigl(\sup_{a\leq t\leq b} \bigl\vert u(t) \bigr\vert \Bigr)^{p-1}\sup_{a\leq t\leq b} \bigl\vert u ^{(m-1)}(t) \bigr\vert \\& \quad \leq \frac{(b-a)^{(m-1)(p-1)}}{2^{m(p-1)}} \biggl( \int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert \,\mathrm{d}t \biggr)^{p-1} \frac{1}{2} \int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert \,\mathrm{d}t \\& \quad =\frac{(b-a)^{(m-1)(p-1)}}{2^{m(p-1)+1}} \biggl( \int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert \,\mathrm{d}t \biggr)^{p}. \end{aligned}
(2.21)
Using (2.9), (2.20), and (2.21), we get
\begin{aligned}& (b-a)^{1-p} \biggl( \int_{a}^{b} \bigl\vert u^{(m)}(t) \bigr\vert \,\mathrm{d}t \biggr)^{p} \leq \frac{(b-a)^{(m-1)(p-1)}}{2^{m(p-1)+1}} \biggl( \int_{a}^{b} \bigl\vert u ^{(m)}(t) \bigr\vert \,\mathrm{d}t \biggr)^{p} \int_{a}^{b} \bigl\vert r(t) \bigr\vert \, \mathrm{d}t. \end{aligned}
(2.22)
Now, we claim that $$\int_{a}^{b}|u^{(m)}(t)|\,\mathrm{d}t>0$$. In fact, if the above inequality is not true, we have $$\int_{a}^{b}|u^{(m)}(t)| \,\mathrm{d}t=0$$, then $$u^{(m)}(t)=0$$ for $$t\in [a, b]$$. By the anti-periodic conditions (1.6), we obtain $$u(t)=0$$ for $$t\in [a, b]$$, which contradicts $$u(t)\not \equiv 0$$, $$t\in [a, b]$$. Thus, dividing both sides of (2.22) by $$(\int_{a}^{b}|u^{(m)}(t)|\,\mathrm{d}t )^{p}$$, we obtain the inequality
$$\int_{a}^{b} \bigl\vert r(t) \bigr\vert \, \mathrm{d}t\geq \frac{2^{m(p-1)+1}}{(b-a)^{m(p-1)}}.$$
(2.23)
Moreover, the inequality in (2.23) is strict since u is not a constant. This completes the proof of Theorem 2.2. □

Remark 2.2

Inequality (2.7) improves inequality (1.7) significantly when $$p>2$$ and $$m\geq 2$$. We list the first six values of $$\zeta (2n)$$, $$n = 1,2,\ldots, 6$$, in Table 1.
For any $$m\in \mathbb{N}$$, we have $$\zeta (2m)\leq \zeta (2)<2$$, and then
\begin{aligned} \frac{\pi^{m(p-1)}}{2^{\frac{p-1}{2}}(2^{2m}-1)^{\frac{p-1}{2}}( \zeta (2m))^{\frac{p-1}{2}}} >& \frac{\pi^{m(p-1)}}{2^{(p-1)/2}(2^{2m})^{(p-1)/2}2^{(p-1)/2}} \\ =&\frac{\pi^{m(p-1)}}{2^{p-1}2^{m(p-1)}}= \biggl( \frac{\pi }{2} \biggr) ^{m(p-1)} \frac{1}{2^{p-1}} \\ =& \biggl( \frac{(\frac{\pi }{2})^{m}}{2} \biggr) ^{p-1}. \end{aligned}
(2.24)
Note that $$( \frac{(\frac{\pi }{2})^{m}}{2} ) ^{p-1}>1$$ for $$m>\frac{\ln 2}{\ln \pi -\ln 2}\approx 1.535$$ and $$p>2$$. Then, using (2.24), we get
$$2 \biggl( \frac{2}{b-a} \biggr) ^{m(p-1)}\frac{\pi^{m(p-1)}}{2^{(p-1)/2}(2^{2m}-1)^{(p-1)/2}( \zeta (2m))^{(p-1)/2}}>2 \biggl( \frac{2}{b-a} \biggr) ^{m(p-1)}.$$
So, Theorem 2.1 improves Theorem 2.1 of [1] significantly when $$p>2$$ and $$m\geq 2$$.

3 Application

We give an application of the above Lyapunov inequality for an eigenvalue problem.

Example 3.1

Let λ be an eigenvalue of the following problem:
$$\textstyle\begin{cases} ( \vert u^{(m)}(t) \vert ^{p-2}u^{(m)}(t))'+\lambda r(t) \vert u(t) \vert ^{p-2}u(t)=0, \\ u^{(i)}(a)+u^{(i)}(b)=0, \quad i=0,1,2,\ldots,m, \end{cases}$$
where $$r\in C([a,b],\mathbb{R})$$, $$m\in \mathbb{N}$$, and $$p>2$$ is a constant. Then, from Theorem 2.1, we have
$$\vert \lambda \vert >\frac{2}{\int_{a}^{b} \vert r(t) \vert \,\mathrm{d}t} \biggl( \frac{2}{b-a} \biggr) ^{m(p-1)} \frac{\pi^{m(p-1)}}{2^{(p-1)/2}(2^{2m}-1)^{(p-1)/2}(\zeta (2m))^{(p-1)/2}}.$$

Declarations

Acknowledgements

The author is indebted to the anonymous referees for their valuable suggestions and helpful comments which helped improve the paper significantly.

Funding

This research was supported by A Project of Shandong Province Higher Educational Science and Technology Program (China) (Grant No. J14LI09), and the Natural Science Foundation of Shandong Province (China) (Grant No. ZR2018MA018), and the National Natural Science Foundation of China (Grant No. 11671227).

Authors’ contributions

HDL organized and wrote this paper. Further, he examined all the steps of the proofs in this research. The author read and approved the final manuscript.

Competing interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Authors’ Affiliations

(1)
School of Mathematical Sciences, Qufu Normal University, Qufu, P.R. China

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