Lyapunov inequalities for a class of nonlinear dynamic systems on time scales

Abstract

The purpose of this work is to obtain several Lyapunov inequalities for the nonlinear dynamic systems

$$\left \{ \textstyle\begin{array}{l} x^{\Delta}(t)= -A(t)x(\sigma(t))-B(t)y(t)|\sqrt{B(t)}y(t)|^{p-2}, \\ y^{\Delta}(t)= C(t)x(\sigma(t))|x(\sigma(t))|^{q-2}+A^{T}(t)y(t), \end{array}\displaystyle \right . $$

on a given time scale interval $[a,b]_{\mathbb{T}}$ ($a,b\in{\mathbb{T}}$ with $\sigma(a)< b$), where $p,q\in (1,+\infty)$ satisfy $1/p+1/q=1$, $A(t)$ is a real $n\times n$ matrix-valued function on $[a,b]_{\mathbb{T}}$ such that $I+\mu(t)A(t)$ is invertible, $B(t)$ and $C(t)$ are two real $n\times n$ symmetric matrix-valued functions on $[a,b]_{ \mathbb{T}}$, $B(t)$ is positive definite, and $x(t)$, $y(t)$ are two real n-dimensional vector-valued functions on $[a,b]_{\mathbb{T}}$.

1 Introduction

The theory of dynamic equations on time scales, which follows Hilger’s landmark paper [1], is a new study area of mathematics that has received a lot of attention. For example, we refer the reader to monographs [2, 3] and the references therein. During the last few years, some Lyapunov inequalities for dynamic equations on time scales have been obtained by many authors [4–7].

In 2002, Bohner et al. [8] investigated the second-order Sturm-Liouville dynamic equation

$$ x^{\Delta^{2}}(t)+q(t)x^{\sigma}(t)=0 $$

(1.1)

on time scale ${\mathbb{T}}$ under the conditions $x(a)=x(b)=0$ ($a,b\in{\mathbb{T}}$ with $a< b$) and $q\in C_{\mathrm{rd}}({\mathbb{T}}, (0,\infty))$ and showed that if $x(t)$ is a solution of (1.1) with $\max_{t\in[a,b]_{ \mathbb{T}}}|x(t)|>0$, then

$$\int_{a}^{b}q(t)\Delta t\geq \frac{b-a}{C}, $$

where $[a,b]_{\mathbb{T}}\equiv\{t\in \mathbb{T}:a\leq t\leq b\}$ and $C=\max\{(t-a)(b-t):t\in [a,b]_{\mathbb{T}}\}$.

When ${\mathbb{T}}= \mathbb{R}$, (1.1) reduces to the Hills equation

$$ x''(t) + u(t)x(t) = 0. $$

(1.2)

In 1907, Lyapunov [9] showed that if $u\in C([a, b], {\mathbb{R}})$ and $x(t)$ is a solution of (1.2) satisfying $x(a) = x(b) = 0$ and $\max_{t\in[a,b]}|x(t)|>0$, then the following classical Lyapunov inequality holds:

$$\int_{a}^{b}\bigl\vert u(t)\bigr\vert \, dt> \frac{4}{b-a}. $$

This was later strengthened with $|u(t)|$ replaced by $u^{+}(t)=\max\{u(t),0\}$ by Wintner [10] and thereafter by some other authors:

$$\int_{a}^{b}u^{+}(t)\, dt>\frac{4}{b-a}. $$

Moreover, the last inequality is optimal.

When ${\mathbb{T}}$ is the set ${\mathbb{Z}}$ of the integers, (1.1) reduces to the linear difference equation

$$ \Delta^{2}x(n)+u(n)x(n+1)=0. $$

(1.3)

In 1983, Cheng [11] showed that if $a,b\in{\mathbb{Z}}$ with $0< a< b$ and $x(n)$ is a solution of (1.3) satisfying $x(a)=x(b)=0$ and $\max_{n\in\{a,a+1,\ldots,b\}}|x(n)|>0$, then

$$\sum_{n=a}^{b-2}\bigl\vert u(n)\bigr\vert \geq \left \{ \textstyle\begin{array}{l@{\quad}l} \frac{4(b-a)}{(b-a)^{2}-1} & \mbox{if } b-a-1 \mbox{ is even}, \\ \frac{4}{b-a} & \mbox{if } b-a-1 \mbox{ is odd} . \end{array}\displaystyle \right . $$

The purpose of this paper is to establish several Lyapunov inequalities for the nonlinear dynamic system

$$ \left \{ \textstyle\begin{array}{l} x^{\Delta}(t)= -A(t)x(\sigma(t))-B(t)y(t)|\sqrt{B(t)}y(t)|^{p-2}, \\ y^{\Delta}(t)= C(t)x(\sigma(t))|x(\sigma(t))|^{q-2}+A^{T}(t)y(t), \end{array}\displaystyle \right . $$

(1.4)

on a given time scale interval $[a,b]_{\mathbb{T}}$ ($a,b\in{\mathbb{T}}$ with $\sigma(a)< b$), where $p,q\in (1,+\infty)$ satisfy $1/p+1/q=1$, $A(t)$ is a real $n\times n$ matrix-valued function on $[a,b]_{\mathbb{T}}$ such that $I+\mu(t)A(t)$ is invertible, $B(t)$ and $C(t)$ are two real $n\times n$ symmetric matrix-valued functions on $[a,b]_{ \mathbb{T}}$, $B(t)$ being positive definite, $A^{T}(t)$ is the transpose of $A(t)$, and $x(t)$, $y(t)$ are two real n-dimensional vector-valued functions on $[a,b]_{\mathbb{T}}$.

When $n=1$ and $p=q=2$, (1.4) reduces to

$$ \left \{ \textstyle\begin{array}{l} x^{\Delta}(t)=u(t)x(\sigma(t))+v(t)y(t), \\ y^{\Delta}(t)=-w(t)x(\sigma(t))-u(t)y(t), \end{array}\displaystyle \right . $$

(1.5)

where $u(t)$, $v(t)$, and $w(t)$ are real-valued rd-continuous functions on $\mathbb{T}$ satisfying $v(t)\geq0$ for any $t\in {\mathbb{T}}$.

In 2011, He et al. [12] obtained the following result.

Theorem 1.1

([12])

Let $1-\mu(t)u(t)>0$ for any $t\in\mathbb{T}$ and $a, b \in\mathbb{T}^{k}$ with $\sigma(a)\leq b$. If (1.5) has a real solution $(x(t),y(t))$ such that

$$\begin{aligned}& x(a)=0 \quad \textit{or}\quad x(a)x\bigl(\sigma(a)\bigr)< 0 ; \\& x(b)=0\quad \textit{or}\quad x(b)x\bigl(\sigma(b)\bigr)< 0 ; \qquad \max_{t\in[a,b]_{ \mathbb{T}}} \bigl\vert x(t)\bigr\vert >0, \end{aligned}$$

then we have the following inequality:

$$\int_{a}^{b}\bigl\vert u(t)\bigr\vert \Delta (t)+ \biggl[ \int_{a}^{\sigma (b)}v(t)\Delta (t) \int_{a}^{b}w^{+}(t)\Delta (t) \biggr]^{1/2}\geq 2, $$

where $w^{+}(t)=\max\{w(t),0\}$.

In 2016, Liu et al. [13] obtained the following theorem.

Theorem 1.2

Let $p=q=2$ and $a,b\in \mathbb{T}$ with $\sigma(a)< b$. If (1.4) has a solution $(x(t),y(t))$ such that

$$ x(a)=x(b)=0\quad \textit{and}\quad \max_{t\in[a,b]_{ \mathbb{T}}}x^{T}(t)x(t)>0 , $$

(1.6)

then for any $n\times n$ symmetric matrix-valued function $C_{1}(t)$ with $C_{1}(t)- C(t)\geq0$, we have the following inequalities:

• (1)
  $$\int_{a}^{b}{\frac{[\int_{a}^{\sigma(t)}|B(s)||e_{\Theta A}(\sigma(t),s)|^{2}\Delta s ][\int_{\sigma(t)}^{b}|B(s)||e_{\Theta A}(\sigma(t),s)|^{2}\Delta s]}{\int_{a}^{b}|B(s)||e_{\Theta A}(\sigma(t),s)|^{2}\Delta s }}\bigl\vert C_{1}(t)\bigr\vert \Delta t\geq1, $$
• (2)
  $$\int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \biggl\{ \int_{a}^{b}\bigl\vert B(s)\bigr\vert \bigl\vert e_{\Theta A}\bigl(\sigma(t),s\bigr)\bigr\vert ^{2}\Delta s \biggr\} \Delta t\geq 4, $$
• (3)
  $$\int_{a}^{b}\bigl\vert A(t)\bigr\vert \Delta t+ \biggl( \int_{a}^{b}\bigl\vert \sqrt {B(t)}\bigr\vert ^{2}\Delta t \biggr)^{1/2} \biggl( \int _{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \Delta t \biggr)^{1/2}\geq2. $$

For some other related results on Lyapunov-type inequalities, see, for example, [14–23].

2 Preliminaries and some lemmas

Throughout this paper, we adopt basic definitions and notation of monograph [2]. A time scale $\mathbb{T}$ is a nonempty closed subset of the real numbers $\mathbb{R}$. On a time scale $\mathbb{T}$, the forward jump operator, the backward jump operator, and the graininess function are defined as

$$\sigma(t)=\inf\{s\in\mathbb{T}:s>t\},\qquad \rho(t)=\sup\{s\in \mathbb{ T}:s< t \}, \quad \mbox{and} \quad \mu(t)=\sigma(t)-t, $$

respectively.

The point $t \in\mathbb{T}$ is said to be left-dense (resp. left-scattered) if $\rho(t) = t$ (resp. $\rho(t) < t$). The point $t\in\mathbb{T}$ is said to be right-dense (resp. right-scattered) if $\sigma(t) = t$ (resp. $\sigma(t) > t$). If $\mathbb{T}$ has a left-scattered maximum M, then we define $\mathbb{T}^{k} = \mathbb{T}- \{M\}$, otherwise $\mathbb{T}^{k} = \mathbb{T}$.

A function $f : \mathbb{T}\rightarrow \mathbb{R}$ is said to be rd-continuous if f is continuous at right-dense points and has finite left-sided limits at left-dense points in $\mathbb{T}$. The set of all rd-continuous functions from $\mathbb{T}$ to $\mathbb{R}$ is denoted by $C_{\mathrm{rd}}(\mathbb{T},\mathbb{R})$. For a function $f : \mathbb{T} \rightarrow \mathbb{R}$, the notation $f^{\sigma}$ means the composition $f\circ\sigma$.

For a function $f : \mathbb{T}\rightarrow \mathbb{R}$, the (delta) derivative $f^{\Delta}(t)$ at $t\in\mathbb{T}$ is defined as the number (if it exists) such that for given any $\varepsilon> 0$, there is a neighborhood U of t with

$$\bigl\vert f\bigl(\sigma(t)\bigr)-f(s)-f^{\Delta}(t) \bigl(\sigma(t)-s \bigr)\bigr\vert \leq\varepsilon \bigl\vert \sigma(t)-s\bigr\vert $$

for all $s\in U$. If the (delta) derivative $f^{\Delta}(t)$ exists for every $t\in\mathbb{T}^{k}$, then we say that f is Δ-differentiable on $\mathbb{T}$.

Let $F,f\in C_{\mathrm{rd}}(\mathbb{T}, \mathbb{R})$ satisfy $F^{\Delta }(t)=f(t)$ for all $t\in \mathbb{T}^{k}$. Then, for any $c,d\in\mathbb{T}$, the Cauchy integral of f is defined as

$$\int_{c}^{d}f(t)\Delta t=F(d)-F(c). $$

For any $z\in{\mathbb{R}}^{n}$ and any $S\in{ \mathbb{R}}^{n\times n}$ (the space of real $n\times n$ matrices), write

$$|z|=\sqrt{z^{T}z} \quad \mbox{and} \quad |S|=\max_{z\in\mathbb {R}^{n},z\neq0} \frac{ |Sz|}{|z|}, $$

which are called the Euclidean norm of z and the matrix norm of S, respectively. It is obvious that, for any $z\in{ \mathbb{R}}^{n}$ and $U,V\in{\mathbb{R}}^{n\times n}$,

$$|Uz|\leq|U||z| \quad \mbox{and} \quad |UV|\leq|U||V|. $$

Let $\mathbb{R}^{n\times n}_{s}$ be the set of all symmetric real $n\times n$ matrices. We can show that, for any $U\in{\mathbb{R}}_{s}^{n\times n}$,

$$|U|=\max_{|\lambda I-U|=0}|\lambda| \quad \mbox{and} \quad |U^{2}|=|U|^{2}. $$

A matrix $S\in {\mathbb{R}}_{s}^{n\times n}$ is said to be positive definite (resp. semipositive definite), written as $S> 0$ (resp. $S \geq 0$), if $y^{T}Sy> 0$ (resp. $y^{T}Sy\geq0$) for any $y\in{ \mathbb{R}}^{n}$ with $y\neq0$. If S is positive definite (resp. semipositive definite), then there exists a unique positive definite matrix (resp. semipositive definite matrix), written as $\sqrt{S}$, satisfying $[\sqrt{S}]^{2}=S$.

In this paper, we establish Lyapunov inequalities for (1.4) that has a solution $(x(t),y(t))$ satisfying

$$ x(a)=x(b)=0 \quad \mbox{and}\quad \max_{t\in[a,b]_{ \mathbb{T}}}x^{T}(t)x(t)>0. $$

(2.1)

We first introduce the following lemmas.

Lemma 2.1

([2])

Let $1/p+1/q=1$ ($p,q\in(1,+\infty)$) and $a,b\in{\mathbb{T}}$ ($a< b$). Then, for any $f,g\in C_{\mathrm{rd}}([a,b]_{ \mathbb{T}},{\mathbb{R}})$,

$$\int_{a}^{b}\bigl\vert f(t)g(t)\bigr\vert \Delta t\leq \biggl( \int _{a}^{b}\bigl\vert f(t)\bigr\vert ^{p}\Delta t \biggr)^{\frac{1}{p}} \biggl( \int_{a}^{b}\bigl\vert g(t)\bigr\vert ^{q}\Delta t \biggr)^{\frac{1}{q}}. $$

Lemma 2.2

Let $a,b\in{\mathbb{T}}$ with $a< b$. Suppose that $\alpha,\beta,\gamma,\delta\in \mathbb{R}$ and $p,q\in (1,+\infty)$ with $\alpha/p+\beta/q=\gamma/p+\delta/q=1/p+1/q=1$. Then, for any $f,g\in C_{\mathrm{rd}}([a,b]_{\mathbb{T}},(-\infty,0)\cup(0,\infty))$,

$$\int_{a}^{b}\bigl\vert f(t)g(t)\bigr\vert \Delta t\leq \biggl( \int_{a}^{b}\bigl\vert f(t)\bigr\vert ^{\alpha}\bigl\vert g(t)\bigr\vert ^{\gamma} \Delta t \biggr)^{\frac{1}{p}} \biggl( \int_{a}^{b}\bigl\vert f(t)\bigr\vert ^{\beta}\bigl\vert g(t)\bigr\vert ^{\delta} \Delta t \biggr)^{\frac{1}{q}}. $$

Proof

Let $M(t)=(|f(t)|^{\alpha}|g(t)|^{\gamma})^{\frac{1}{p}}$ and $N(t)=(|f(t)|^{\beta}|g(t)|^{\delta})^{\frac{1}{q}}$. Then by Lemma 2.1 we have

$$\begin{aligned} \int^{b}_{a}\bigl\vert f(t)g(t)\bigr\vert \Delta t =& \int ^{b}_{a}M(t)N(t)\Delta t \\ \leq& \biggl( \int^{b}_{a}M^{p}(t)\Delta t \biggr)^{\frac{1}{p}} \biggl( \int ^{b}_{a}N^{q}(t)\Delta t \biggr)^{\frac{1}{q}} \\ =& \biggl( \int_{a}^{b}\bigl\vert f(t)\bigr\vert ^{\alpha}\bigl\vert g(t)\bigr\vert ^{\gamma} \Delta t \biggr)^{\frac{1}{p}} \biggl( \int_{a}^{b}\bigl\vert f(t)\bigr\vert ^{\beta}\bigl\vert g(t)\bigr\vert ^{\delta} \Delta t \biggr)^{\frac{1}{q}}. \end{aligned}$$

This completes the proof of Lemma 2.2. □

Remark 2.3

Let $\gamma=0$ in Lemma 2.2. Then we obtain that, for any $f,g\in C_{\mathrm{rd}}([a,b]_{ \mathbb{T}},(-\infty, 0)\cup(0,\infty))$,

$$\int_{a}^{b}\bigl\vert f(t)g(t)\bigr\vert \Delta t\leq \Bigl\{ \max_{t\in[a,b]_{ \mathbb{T}}}\bigl\vert f(t)\bigr\vert ^{\beta}\Bigr\} ^{\frac{1}{q}} \biggl( \int _{a}^{b}\bigl\vert f(t)\bigr\vert ^{\alpha} \Delta t \biggr)^{\frac{1}{p}} \biggl( \int_{a}^{b}\bigl\vert g(t)\bigr\vert ^{q}\Delta t \biggr)^{\frac{1}{q}}. $$

Lemma 2.4

([2])

If $A\in C_{\mathrm{rd}}({\mathbb{T}, \mathbb{R}^{n\times n}})$ with invertible $I+\mu(t)A(t)$, $f\in C_{\mathrm{rd}}(\mathbb{T},\mathbb{R}^{n})$, $t_{0} \in\mathbb{T}$, and $a \in\mathbb{R}^{n}$, then

$$x(t)=e_{\Theta A}(t,t_{0})a+ \int_{t_{0}}^{t}e_{\Theta A}(t,\tau)f(\tau)\Delta{ \tau} $$

is the unique solution of the initial value problem

$$\left \{ \textstyle\begin{array}{l} x^{\Delta}(t)= -A(t)x(\sigma(t))+f(t), \\ x(t_{0})= a, \end{array}\displaystyle \right . $$

where $(\Theta A)(t)=-[I+\mu(t)A(t)]^{-1}A(t)$ for any $t\in\mathbb{T}^{k}$, and $e_{\Theta A}(t,t_{0})$ is the unique matrix-valued solution of the initial value problem

$$\left \{ \textstyle\begin{array}{l} Y^{\Delta}(t)= (\Theta A)(t)Y(t), \\ Y(t_{0})= I. \end{array}\displaystyle \right . $$

Lemma 2.5

([2])

Let $A,B\in C_{\mathrm{rd}}({\mathbb{T}, \mathbb{R}^{n\times n}})$ be Δ-differentiable. Then

$$\bigl(A(t)B(t)\bigr)^{\Delta }=A^{\sigma}(t) B^{\Delta }(t)+A^{\Delta }(t) B(t)=A^{\Delta }(t) B^{\sigma}(t)+A(t)B^{\Delta }(t). $$

Lemma 2.6

([13])

If $f_{1}(t),f_{2}(t),\ldots,f_{n}(t)$ are Δ-integrable on ${[a,b]_{\mathbb{T}}}$ and $x(t)=(f_{1}(t),f_{2}(t), \ldots,f_{n}(t))$, then

$$\biggl\vert \int_{a}^{b}x(t)\Delta t\biggr\vert = \Biggl\{ \sum_{i=1}^{n} \biggl( \int _{a}^{b}f_{i}(t)\Delta t \biggr)^{2} \Biggr\} ^{\frac{1}{2}}\leq \int_{a}^{b} \Biggl\{ \sum _{i=1}^{n}f^{2}_{i}(t) \Biggr\} ^{\frac{1}{2}}\Delta t= \int _{a}^{b}\bigl\vert x(t)\bigr\vert \Delta t. $$

Lemma 2.7

([13])

If $A_{1},A_{2}\in \mathbb{R}^{n\times n}_{s}$ and $A_{1}-A_{2}\geq0$, then, for any $x \in\mathbb{R}^{n}$,

$$\bigl(x^{\sigma}\bigr)^{T}A_{2}x^{\sigma}\leq|A_{1}|\bigl\vert x^{\sigma}\bigr\vert ^{2} . $$

3 Main results and proofs

In this section, we assume that $\alpha,\beta\in\mathbb{R}$ and $p,q\in(1,+\infty)$ satisfy

$$\alpha/p+\beta/q=1/p+1/q=1. $$

For any $t,\tau\in [a,b]_{\mathbb{T}}$, write

$$\begin{aligned}& F(t,\tau) = \bigl\vert e_{\Theta A}\bigl(\sigma(t),\tau\bigr)\bigr\vert \bigl\vert \sqrt{B(\tau)}\bigr\vert , \\& G(t) = \bigl\vert \sqrt{B(t)}y(t)\bigr\vert ^{p-2}y^{T}(t)B(t)y(t)= \bigl\vert \sqrt{B(t)}y(t)\bigr\vert ^{p}, \\& \Phi\bigl(\sigma(t)\bigr) = \biggl( \int_{a}^{\sigma(t)}F^{\alpha}(t,s)\Delta s \biggr)^{\frac{q}{p}}, \\& \Psi\bigl(\sigma(t)\bigr) = \biggl( \int_{\sigma(t)}^{b}F^{\alpha}(t,s)\Delta s \biggr)^{\frac{q}{p}}, \\& P(t) = \Phi\bigl(\sigma(t)\bigr)\Psi\bigl(\sigma(t)\bigr)\max _{a\leq\tau\leq \sigma(t)}F^{\beta}(t,\tau) \max_{\sigma(t)\leq\tau\leq b}F^{\beta}(t, \tau), \\& Q(t) = \Phi\bigl(\sigma(t)\bigr)\max_{a\leq\tau\leq \sigma(t)}F^{\beta}(t, \tau)+\Psi\bigl(\sigma(t)\bigr)\max_{\sigma(t)\leq \tau\leq b}F^{\beta}(t, \tau). \end{aligned}$$

Theorem 3.1

Let $a,b\in\mathbb{T}$ with $\sigma(a)< b$ and $C_{1}\in\mathbb{R}^{n\times n}_{s}$ with $C_{1}(t)-C(t)\geq0$. If (1.4) has a solution $(x(t),y(t))$ with $x(t),y(t)\in C_{\mathrm{rd}}(\mathbb{T},\mathbb{R}^{n})$ satisfying (2.1) on the interval $[a,b]_{\mathbb{T}}$, then

$$ \int_{a}^{b}{\frac{P(t)}{Q(t)}}\bigl\vert C_{1}(t)\bigr\vert \Delta t\geq1. $$

(3.1)

Proof

Since $(x(t),y(t))$ is a solution of (1.4), we have

$$ \bigl(y^{T}(t)x(t)\bigr)^{\Delta}=\bigl(x^{\sigma}(t) \bigr)^{T}C(t)x^{\sigma}(t)\bigl\vert x^{\sigma}(t)\bigr\vert ^{q-2}-G(t). $$

(3.2)

Integrating (3.2) from a to b and noting that $x(a)=x(b)=0$, we obtain

$$\int_{a}^{b}G(t)\Delta t= \int_{a}^{b}\bigl\vert x^{\sigma}(t)\bigr\vert ^{q-2}\bigl(x^{\sigma}(t)\bigr)^{T}C(t)x^{\sigma}(t) \Delta t. $$

Noting that $B(t)>0$, we know that $y^{T}(t)B(t)y(t)\geq0$, $t\in [a,b]_{\mathbb{T}}$.

We claim that $y^{T}(t)B(t)y(t)\not\equiv0$ ($t\in [a,b]_{\mathbb{T}}$). Indeed, if $y^{T}(t)B(t)y(t)\equiv0$ ($t\in [a,b]_{\mathbb{T}}$), then

$$\bigl\vert \sqrt{B(t)}y(t)\bigr\vert ^{2}=y^{T}(t)B(t)y(t) \equiv0, $$

which implies $B(t)y(t)\equiv0$ ($t\in[a,b]_{\mathbb{T}}$). Thus, the first equation of (1.4) reduces to

$$x^{\Delta}(t)= -A(t)x\bigl(\sigma(t)\bigr),\qquad x(a)=0. $$

By Lemma 2.4 it follows

$$x(t)=e_{\Theta A}(t,a)\cdot0=0, $$

which is a contradiction to (2.1). Hence, we obtain that

$$ \int_{a}^{b}\bigl\vert x^{\sigma}(t)\bigr\vert ^{q-2}\bigl(x^{\sigma}(t)\bigr)^{T}C(t)x^{\sigma }(t) \Delta t= \int_{a}^{b}G(t)\Delta t>0, $$

(3.3)

and it follows from Lemma 2.4 that, for $t\in[a,b]_{\mathbb{T}}$,

$$\begin{aligned} x(t) =& - \int_{a}^{t}e_{\Theta A}(t,\tau)B(\tau)y(\tau ) \bigl\vert \sqrt{B(\tau)}y(\tau)\bigr\vert ^{p-2}\Delta{\tau} \\ =& - \int_{b}^{t}e_{\Theta A} (t,\tau)B(\tau)y(\tau) \bigl\vert \sqrt{B(\tau)}y(\tau)\bigr\vert ^{p-2}\Delta{\tau}, \end{aligned}$$

which implies that, for $t\in[a,b)_{\mathbb{T}}$,

$$\begin{aligned} x^{\sigma}(t) =&- \int_{a}^{\sigma(t)}e_{\Theta A}\bigl(\sigma (t),\tau \bigr)B(\tau)y(\tau)\bigl\vert \sqrt{B(\tau)}y(\tau)\bigr\vert ^{p-2} \Delta {\tau} \\ =&+ \int_{\sigma(t)}^{b}e_{\Theta A}\bigl(\sigma(t),\tau \bigr)B(\tau)y(\tau)\bigl\vert \sqrt{B(\tau)}y(\tau)\bigr\vert ^{p-2} \Delta {\tau}. \end{aligned}$$

Note that, for $a\leq\sigma(t)\leq b$,

$$\begin{aligned}& \bigl\vert e_{\Theta A}\bigl(\sigma(t),\tau\bigr)B(\tau)y(\tau)\bigr\vert \sqrt {B(\tau)}y(\tau)\bigl\vert ^{p-2}\bigr\vert \\& \quad \leq \bigl\vert e_{\Theta A}\bigl(\sigma(t),\tau\bigr)\bigr\vert \bigl\vert B(\tau)y(\tau)\bigr\vert \bigl\vert \sqrt{B(\tau)}y(\tau)\bigr\vert ^{p-2} \\& \quad \leq F(t,\tau)\bigl\vert \sqrt{B(\tau)}y(\tau)\bigr\vert \bigl\vert \sqrt{B(\tau)}y(\tau)\bigr\vert ^{p-2} \\& \quad = F(t,\tau)G^{\frac{1}{q}}(\tau). \end{aligned}$$

Then by Remark 2.3 and Lemma 2.6 we obtain

$$\begin{aligned} \bigl\vert x^{\sigma}(t)\bigr\vert ^{q} =&\biggl\vert \int_{a}^{\sigma(t)}e_{\Theta A}\bigl(\sigma(t),\tau \bigr)B(\tau)y(\tau)\bigl\vert \sqrt{B(\tau)}y(\tau)\bigr\vert ^{p-2} \Delta {\tau}\biggr\vert ^{q} \\ \leq& \biggl[ \int_{a}^{\sigma(t)}\bigl\vert e_{\Theta A}\bigl( \sigma(t),\tau\bigr)B(\tau )y(\tau)\bigl\vert \sqrt{B(\tau)}y(\tau)\bigr\vert ^{p-2}\bigr\vert \Delta{\tau} \biggr]^{q} \\ \leq& \biggl[ \int_{a}^{\sigma(t)}F(t,\tau)G^{\frac{1}{q}}(\tau)\Delta { \tau} \biggr]^{q} \\ \leq& \biggl( \int_{a}^{\sigma(t)}F^{\alpha}(t,\tau)\Delta { \tau} \biggr)^{\frac{q}{p}} \int_{a}^{\sigma(t)}F^{\beta}(t,\tau)G(\tau )\Delta {\tau} \\ \leq& \max_{a\leq\tau\leq\sigma(t)}F^{\beta}(t,\tau) \biggl( \int _{a}^{\sigma(t)}F^{\alpha}(t,\tau)\Delta { \tau} \biggr)^{\frac{q}{p}} \int_{a}^{\sigma(t)}G(\tau)\Delta {\tau}, \end{aligned}$$

that is,

$$ \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\leq\max _{a\leq\tau\leq\sigma(t)}F^{\beta}(t,\tau)\Phi \bigl(\sigma(t)\bigr) \int_{a}^{\sigma(t)}G(\tau)\Delta{\tau}. $$

(3.4)

Similarly, for $a\leq\sigma(t) \leq b$, we have

$$ \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\leq\max _{\sigma(t)\leq\tau\leq b}F^{\beta}(t,\tau)\Psi \bigl(\sigma(t)\bigr) \int_{\sigma(t)}^{b}G(\tau)\Delta{\tau}. $$

(3.5)

It follows from (3.4) and (3.5) that

$$\bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\leq\frac{P(t)}{Q(t)} \int_{a}^{b}G(\tau)\Delta{\tau}. $$

Then by (3.3) and Lemma 2.7 we have

$$\begin{aligned}& \int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\Delta t \\& \quad \leq \int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \frac{P(t)}{Q(t)}\Delta t \int _{a}^{b}G(t)\Delta t \\& \quad = \int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \frac{P(t)}{Q(t)}\Delta t \int_{a}^{b}\bigl\vert x^{\sigma}(t)\bigr\vert ^{q-2}\bigl(x^{\sigma}(t)\bigr)^{T}C(t)x^{\sigma }(t) \Delta t \\& \quad \leq \int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \frac{P(t)}{Q(t)}\Delta t \int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\Delta t. \end{aligned}$$

Since

$$\int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\Delta t\geq \int_{a}^{b}\bigl\vert x^{\sigma}(t)\bigr\vert ^{q-2}\bigl(x^{\sigma}(t)\bigr)^{T}C(t)x^{\sigma}(t) \Delta t= \int_{a}^{b} G(t)\Delta {t}>0, $$

we get

$$\int_{a}^{b}\frac{P(t)}{Q(t)}\bigl\vert C_{1}(t)\bigr\vert \Delta t\geq1. $$

This completes the proof of Theorem 3.1. □

Corollary 3.2

Let $a,b\in\mathbb{T}$ with $\sigma(a)< b$ and $C_{1}\in\mathbb{R}^{n\times n}_{s}$ with $C_{1}(t)-C(t)\geq0$. If (1.4) has a solution $(x(t),y(t))$ with $x(t),y(t)\in C_{\mathrm{rd}}(\mathbb{T},\mathbb{R}^{n})$ satisfying (2.1) on the interval $[a,b]_{\mathbb{T}}$, then

$$ \int_{a}^{b}Q(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \geq4. $$

(3.6)

Proof

Note that

$$\frac{P(t)}{Q(t)}\leq \frac{Q(t)}{4}. $$

It follows from (3.1) that

$$\int_{a}^{b} \frac{Q(t)}{4}\bigl\vert C_{1}(t)\bigr\vert \Delta t\geq1, $$

that is,

$$\int_{a}^{b}Q(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \geq4. $$

This completes the proof of Corollary 3.2. □

Corollary 3.3

Let $a,b\in\mathbb{T}$ with $\sigma(a)< b$ and $C_{1}\in\mathbb{R}^{n\times n}_{s}$ with $C_{1}(t)-C(t)\geq0$. If (1.4) has a solution $(x(t),y(t))$ with $x(t),y(t)\in C_{\mathrm{rd}}(\mathbb{T},\mathbb{R}^{n})$ satisfying (2.1) on the interval $[a,b]_{\mathbb{T}}$, then

$$ \int_{a}^{b}\sqrt{P(t)}\bigl\vert C_{1}(t)\bigr\vert \Delta t \geq2. $$

(3.7)

Proof

Note that

$$Q(t)\geq2\sqrt{P(t)}. $$

It follows from (3.1) that

$$\int_{a}^{b}\sqrt{P(t)}\bigl\vert C_{1}(t)\bigr\vert \Delta t\geq \int_{a}^{b}2\frac {P(t)}{Q(t)}\bigl\vert C_{1}(t)\bigr\vert \Delta t\geq2. $$

This completes the proof of Corollary 3.3. □

Theorem 3.4

Let $a,b\in\mathbb{T}$ with $\sigma(a)< b$ and $C_{1}\in\mathbb{R}^{n\times n}_{s}$ with $C_{1}(t)-C(t)\geq0$. If (1.4) has a solution $(x(t),y(t))$ with $x(t),y(t)\in C_{\mathrm{rd}}(\mathbb{T},\mathbb{R}^{n})$ satisfying (2.1) on the interval $[a,b]_{\mathbb{T}}$, then there exists $c\in(a,b)$ such that

$$ \left \{ \textstyle\begin{array}{l} {\int}_{a}^{\sigma(c)}\Phi(\sigma(t))\max_{a\leq\tau\leq \sigma(t)}F^{\beta}(t,\tau)|C_{1}(t)|\Delta t\geq1, \\ \int_{c}^{b} \Psi(\sigma(t))\max_{\sigma(t)\leq\tau\leq b}F^{\beta}(t,\tau)|C_{1}(t)|\Delta t\geq1. \end{array}\displaystyle \right . $$

(3.8)

Proof

Set $U(t)=\Phi(\sigma(t))\max_{a\leq \tau\leq\sigma(t)}F^{\beta}(t,\tau)$ and $V(t)= \Psi(\sigma(t))\max_{\sigma(t)\leq\tau\leq b}F^{\beta}(t,\tau)$. Let

$$f(t)= \int_{a}^{t}U(s)\bigl\vert C_{1}(s) \bigr\vert \Delta s- \int_{t}^{b}V(s)\bigl\vert C_{1}(s) \bigr\vert \Delta s. $$

Then we have $f(a)<0$ and $f(b)>0$. Hence, we can choose $c\in(a,b)$ such that $f(c)\leq0$ and $f(\sigma(c))\geq0$, that is,

$$ \int_{a}^{c}U(s)\bigl\vert C_{1}(s) \bigr\vert \Delta s\leq \int _{c}^{b}V(s)\bigl\vert C_{1}(s) \bigr\vert \Delta s $$

(3.9)

and

$$ \int_{a}^{\sigma(c)}U(s)\bigl\vert C_{1}(s) \bigr\vert \Delta s\geq \int_{\sigma(c)}^{b}V(s)\bigl\vert C_{1}(s) \bigr\vert \Delta s. $$

(3.10)

By (3.4) we have that

$$ \bigl\vert C_{1}(t)\bigr\vert \bigl\vert x^{\sigma}(t) \bigr\vert ^{q}\leq U(t)\bigl\vert C_{1}(t)\bigr\vert \int_{a}^{\sigma(t)}G(\tau )\Delta{\tau}. $$

(3.11)

Integrating (3.11) from a to $\sigma(c)$, we obtain

$$\begin{aligned} \int_{a}^{\sigma(c)}\bigl\vert C_{1}(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\Delta t \leq& \int _{a}^{\sigma(c)}U(t)\bigl\vert C_{1}(t) \bigr\vert \biggl( \int_{a}^{\sigma(t)} G(\tau)\Delta{\tau} \biggr) \Delta t \\ \leq& \int_{a}^{c}U(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \int_{a}^{\sigma(c)}G(\tau)\Delta {\tau} \\ &{}+U(c)\bigl\vert C_{1}(c)\bigr\vert \bigl(\sigma(c)-c\bigr) \int_{a}^{\sigma(c)}G(\tau)\Delta {\tau} \\ =& \int_{a}^{\sigma(c)}U(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \int_{a}^{\sigma(c)}G(\tau)\Delta{\tau}. \end{aligned}$$

Similarly, we obtain from (3.4) and (3.10) that

$$\begin{aligned} \int_{\sigma(c)}^{b}\bigl\vert C_{1}(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\Delta t \leq& \int _{\sigma(c)}^{b}V(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \int_{\sigma(c)}^{b}G(\tau )\Delta{\tau} \\ \leq& \int _{a}^{\sigma(c)}U(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \int_{\sigma(c)}^{b}G(\tau )\Delta{\tau}. \end{aligned}$$

This yields

$$\begin{aligned} \int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\Delta t \leq& \int _{a}^{\sigma(c)}U(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \int_{a}^{b}G(t)\Delta t \\ =& \int_{a}^{\sigma (c)}U(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \int_{a}^{b}\bigl\vert x^{\sigma}(t)\bigr\vert ^{q-2}\bigl(x^{\sigma}(t)\bigr)^{T}C(t)x^{\sigma}(t) \Delta t \\ \leq& \int_{a}^{\sigma(c)}U(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\Delta t. \end{aligned}$$

Since

$$\begin{aligned} \int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\Delta t \geq& \int_{a}^{b}\bigl\vert x^{\sigma}(t)\bigr\vert ^{q-2}\bigl(x^{\sigma}(t)\bigr)^{T}C(t)x^{\sigma}(t) \Delta t \\ =& \int _{a}^{b}G(t)\Delta {t}>0, \end{aligned}$$

we have $\int_{a}^{\sigma(c)}U(t)|C_{1}(t)|\Delta t\geq1$.

Next, we obtain from (3.5) that

$$ \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\bigl\vert C_{1}(t)\bigr\vert \leq V(t)\bigl\vert C_{1}(t)\bigr\vert \int_{\sigma(t)}^{b}G(\tau )\Delta{\tau}. $$

(3.12)

Integrating (3.12) from c to b, we have

$$\begin{aligned} \int_{c}^{b}\bigl\vert C_{1}(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\Delta t \leq& \int _{c}^{b}V(t)\bigl\vert C_{1}(t) \bigr\vert \biggl( \int_{\sigma(t)}^{b}G(\tau)\Delta{\tau} \biggr) \Delta t \\ \leq& \int_{c}^{b}V(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \int_{\sigma (c)}^{b}G(\tau)\Delta{\tau}. \end{aligned}$$

Similarly, we obtain

$$\begin{aligned} \begin{aligned} \int_{a}^{c}\bigl\vert C_{1}(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\Delta t&\leq \int _{a}^{c}U(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \int_{a}^{\sigma(c)}G(\tau)\Delta {\tau} \\ &\leq \int _{c}^{b}V(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \int_{a}^{\sigma(c)}G(\tau)\Delta {\tau}. \end{aligned} \end{aligned}$$

This yields

$$\begin{aligned} \int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\Delta t \leq& \int _{c}^{b}V(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \int_{a}^{b}G(t)\Delta t \\ =& \int_{c}^{b}V(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \int_{a}^{b}\bigl\vert x^{\sigma}(t)\bigr\vert ^{q-2}\bigl(x^{\sigma}(t)\bigr)^{T}C(t)x^{\sigma }(t) \Delta t \\ \leq& \int_{c}^{b}V(t)\bigl\vert C_{1}(t) \bigr\vert \Delta t \int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\Delta t. \end{aligned}$$

Thus, we have $\int_{c}^{b}V(t)|C_{1}(t)|\Delta t\geq1$. This completes the proof of Theorem 3.4. □

Theorem 3.5

Let $a,b\in\mathbb{T}$ with $\sigma(a)< b$ and $C_{1}\in\mathbb{R}^{n\times n}_{s}$ with $C_{1}(t)-C(t)\geq0$. If (1.4) has a solution $(x(t),y(t))$ with $x(t),y(t)\in C_{\mathrm{rd}}(\mathbb{T},\mathbb{R}^{n})$ satisfying (2.1) on the interval $[a,b]_{\mathbb{T}}$, then

$$\int_{a}^{b}\bigl\vert A(t)\bigr\vert \Delta t+\Bigl\{ \max_{a\leq t\leq b}\bigl\vert \sqrt {B(t)}\bigr\vert ^{\beta}\Bigr\} ^{\frac{1}{q}} \biggl( \int_{a}^{b}\bigl\vert \sqrt{B(t)}\bigr\vert ^{\alpha} \Delta t \biggr)^{\frac{1}{p}} \biggl( \int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \Delta t \biggr)^{\frac{1}{q}}\geq2. $$

Proof

Since $x(a)=x(b)=0$, we have

$$\int_{a}^{b}G(t)\Delta t= \int_{a}^{b}\bigl\vert x^{\sigma}(t)\bigr\vert ^{q-2}\bigl(x^{\sigma}(t)\bigr)^{T}C(t)x^{\sigma}(t) \Delta t. $$

It follows from the first equation of (1.4) that, for all $a\leq t \leq b$,

$$\begin{aligned} x(t) =& \int_{a}^{t}\bigl(-A(\tau)x^{\sigma}(\tau)-B( \tau)\bigl\vert \sqrt{B(\tau)}y(\tau )\bigr\vert ^{p-2}y(\tau)\bigr) \Delta \tau \\ =& \int_{t}^{b}\bigl(A(\tau)x^{\sigma}(\tau)+B( \tau)\bigl\vert \sqrt{B(\tau )}y(\tau)\bigr\vert ^{p-2}y(\tau)\bigr) \Delta \tau. \end{aligned}$$

Thus, we have

$$\begin{aligned} \bigl\vert x(t)\bigr\vert =&\biggl\vert \int_{a}^{t}\bigl(-A(\tau)x^{\sigma}(\tau)-B( \tau )y(\tau)\bigr)\bigl\vert \sqrt{B(\tau)}y(\tau)\bigr\vert ^{p-2} \Delta \tau\biggr\vert \\ \leq& \int_{a}^{t}\bigl\vert A(\tau)x^{\sigma}( \tau)+B(\tau)y(\tau)\bigr\vert \sqrt{B(\tau )}y(\tau)\bigl\vert ^{p-2} \bigr\vert \Delta \tau \\ \leq& \int_{a}^{t}\bigl\vert A(\tau)x^{\sigma}( \tau)\bigr\vert \Delta \tau+ \int _{a}^{t}\bigl\vert B(\tau)y(\tau)\bigr\vert \bigl\vert \sqrt{B(\tau)}y(\tau)\bigr\vert ^{p-2} \Delta \tau \\ \leq& \int_{a}^{t}\bigl\vert A(\tau)\bigr\vert \bigl\vert x^{\sigma}(\tau)\bigr\vert \Delta \tau+ \int_{a}^{t}\bigl\vert \sqrt{B(\tau)}\bigr\vert G^{\frac{1}{q}}(\tau)\Delta \tau. \end{aligned}$$

Similarly, we have

$$\bigl\vert x(t)\bigr\vert \leq \int_{t}^{b}\bigl\vert A(\tau)\bigr\vert \bigl\vert x^{\sigma}(\tau)\bigr\vert \Delta \tau + \int_{t}^{b}\bigl\vert \sqrt{B(\tau)}\bigr\vert G^{\frac{1}{q}}(\tau)\Delta \tau. $$

Then we obtain

$$\begin{aligned} \bigl\vert x(t)\bigr\vert \leq&\frac{1}{2} \biggl[ \int _{a}^{b}\bigl\vert A(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert \Delta t+ \int_{a}^{b}\bigl\vert \sqrt{B(t )}\bigr\vert G^{\frac{1}{q}}(t) \Delta t \biggr] \\ \leq& \frac{1}{2} \biggl[ \int_{a}^{b}\bigl\vert A(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert \Delta t+\Bigl\{ \max _{a\leq t\leq b}\bigl\vert \sqrt{B(t)}\bigr\vert ^{\beta}\Bigr\} ^{\frac{1}{q}} \\ &{}\times\biggl( \int_{a}^{b} \bigl\vert \sqrt{B(t)}\bigr\vert ^{\alpha} \Delta t \biggr)^{\frac{1}{p}} \biggl( \int _{a}^{b}G(t)\Delta t \biggr)^{\frac{1}{q}} \biggr] \\ =& \frac{1}{2} \biggl[ \int_{a}^{b}\bigl\vert A(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert \Delta t+\Bigl\{ \max _{a\leq t\leq b}\bigl\vert \sqrt{B(t)}\bigr\vert ^{\beta}\Bigr\} ^{\frac{1}{q}} \biggl( \int _{a}^{b}\bigl\vert \sqrt{B(t)}\bigr\vert ^{\alpha} \Delta t \biggr)^{\frac{1}{p}} \\ &{}\times \biggl( \int_{a}^{b}\bigl\vert x^{\sigma}(t)\bigr\vert ^{q-2}\bigl(x^{\sigma }(t)\bigr)^{T}C(t)x^{\sigma}(t) \Delta t \biggr)^{\frac{1}{q}} \biggr] \\ \leq&\frac{1}{2} \biggl[ \int_{a}^{b}\bigl\vert A(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert \Delta t+\Bigl\{ \max _{a\leq t\leq b}\bigl\vert \sqrt{B(t)}\bigr\vert ^{\beta}\Bigr\} ^{\frac{1}{q}} \\ &{}\times\biggl( \int_{a}^{b}\bigl\vert \sqrt {B(t)}\bigr\vert ^{\alpha} \Delta t \biggr)^{\frac{1}{p}} \biggl( \int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \bigl\vert x^{\sigma}(t)\bigr\vert ^{q}\Delta t \biggr)^{\frac{1}{q}} \biggr]. \end{aligned}$$

Denote $M=\max_{a\leq t\leq b}|x(t)|>0$. Then

$$M\leq \frac{1}{2} \biggl[ \int_{a}^{b}\bigl\vert A(t)\bigr\vert M \Delta t+\Bigl\{ \max_{a\leq t\leq b}\bigl\vert \sqrt{B(t)}\bigr\vert ^{\beta}\Bigr\} ^{\frac{1}{q}} \biggl( \int _{a}^{b}\bigl\vert \sqrt{B(t)}\bigr\vert ^{\alpha} \Delta t \biggr)^{\frac{1}{p}} \biggl( \int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert M^{q}\Delta t \biggr)^{\frac{1}{q}} \biggr]. $$

Thus,

$$\int_{a}^{b}\bigl\vert A(t)\bigr\vert \Delta t+\Bigl\{ \max_{a\leq t\leq b}\bigl\vert \sqrt {B(t)}\bigr\vert ^{\beta}\Bigr\} ^{\frac{1}{q}} \biggl( \int_{a}^{b}\bigl\vert \sqrt{B(t)}\bigr\vert ^{\alpha} \Delta t \biggr)^{\frac{1}{p}} \biggl( \int_{a}^{b}\bigl\vert C_{1}(t)\bigr\vert \Delta t \biggr)^{\frac{1}{q}}\geq2. $$

This completes the proof of Theorem 3.5. □