On positive solutions of second-order delayed differential system with indefinite weight

Abstract

In this paper, we study the existence of positive solutions of a second-order delayed differential system, in which the weight functions may change sign. To prove our main results, we apply Krasnosel’skii’s fixed point theorems in cones.

1 Introduction

This paper is mainly concerned with the existence of positive solutions of a second-order two-delay differential system with

$$ \textstyle\begin{cases} x_{1}''(t)+h_{1}(t)f_{1}(x_{1}(t-\tau _{1}),x_{2}(t-\tau _{2}))=0,\quad 0< t< 1, \\ x_{2}''(t)+h_{2}(t)f_{2}(x_{1}(t-\tau _{1}),x_{2}(t-\tau _{2}))=0,\quad 0< t< 1, \\ x_{1}(t)=0,\quad -\tau _{1} \leq t \leq 0, \quad \text{and}\quad x_{1}(1)=0, \\ x_{2}(t)=0,\quad -\tau _{2} \leq t \leq 0, \quad \text{and}\quad x_{2}(1)=0, \end{cases} $$

(1.1)

where $0<\tau _{1}<\tau _{2}<\frac{1}{2}$ are constants, and the weight functions $h_{i}(t)$ may change sign (see ($\mathrm{H}_{1}$)).

Many researchers have been attracted to study the theory, methodology and application of functional differential equations with delays, which have often been put forward as mathematical model to describe various natural phenomena [6, 9]. One of the important aspects of research is that there are many papers devoted to studying nontrivial solutions of boundary-value problems for functional differential equations with delays. For example, J.W. Lee and D. O’Regan established the general existence principle of differential-difference equations with delay type based on the nonlinear alternative (see [10, 11]). Since then, T. Candan [4, 5] applied Krasnosel’skii’s fixed point theorem for the sum of a completely continuous and a contraction mapping to prove the existence of positive periodic solutions for the first- (second-) order neutral differential equation. Y. Liu [12] applied the Mönch fixed point theorem to study the existence and uniqueness of solutions for the nonlinear functional differential equations on infinite interval. D. Bai and Y. Xu [1, 2] employed fixed-point theory to show how the parameters effect the number of positive solution for a two-delay singular differential system (1.1). In addition, to the best of our knowledge, many papers are concerned with the existence of positive solutions of differential equations with indefinite weight functions by using various methods, such as the fixed point theorems, the Leray–Schauder degree theory, bifurcation and so on. We refer the reader to [3, 7, 13, 14, 16–18].

Motivated by the related references, in this paper we mainly apply the Krasnosel’skii fixed point theorems in cones to discuss the existence of positive solutions of problem (1.1), if the functions $h_{i}(t)$ and $f_{i}(x_{1},x_{2})$ satisfy the following assumptions:

($\mathrm{H}_{1}$):

$h_{i}:[0,1]\rightarrow (-\infty ,+\infty )$ are continuous, and there exists a $\xi \in (0,1-\tau _{2})$ such that

$$ \textstyle\begin{cases} h_{i}(t)\geq 0, &\text{if } t\in [0,\xi ], \\ h_{i}(t)\equiv 0, &\text{if } t\in [\xi ,\xi +\tau _{2}], \\ h_{i}(t)\leq 0, &\text{if } t\in [\xi +\tau _{2},1]. \end{cases} $$

Furthermore, $h_{i}$ do not vanish identically on any subinterval of $[0,\xi ]$ and $[\xi +\tau _{2},1]$.

($\mathrm{H}_{2}$):

$f_{i}:\mathbb{R^{+}}^{2}\rightarrow \mathbb{R}^{+}$ are continuous and there exists a $\theta \in (0,1]$ such that

$$ f_{i}(u,v)\geq \theta \phi _{i}(u,v), $$

$\phi _{i}(u,v)=\max \{f_{i}(x_{1},x_{2}):0\leq x_{1}\leq u, 0\leq x_{2} \leq v \}$.

($\mathrm{H}_{3}$):

There exist positive constants $k_{i}$, $\alpha _{i}$ and continuous functions

$$ F_{i}(x_{1},x_{2}):\mathbb{R^{+}}^{2} \rightarrow \mathbb{R}^{+} $$

satisfying

• (i) $F_{i}(x_{1},x_{2})$ are strictly increasing functions with respective to $(x_{1},x_{2})$,

• (ii) $F_{i}(\lambda x_{1},\lambda x_{2})={\lambda }^{\alpha _{i}} F_{i}(x_{1},x_{2})$,

such that

$$ k_{i}F_{i}(x_{1},x_{2}) \leq f_{i}(x_{1},x_{2})\leq F_{i}(x_{1},x_{2}), \quad (x_{1},x_{2})\in \mathbb{R^{+}}^{2}. $$

($\mathrm{H}_{4}$):

There exist $\sigma _{i}$ satisfying $\tau _{2}<\sigma _{i}<\xi$ such that

$$ \theta ^{2} k_{i}(\sigma _{i}-\tau _{2})^{\alpha _{i}} \int ^{\xi }_{ \sigma _{i}} G(t,s)h_{i}^{+}(s) \,ds\geq \xi ^{\alpha _{i}} \int ^{1}_{ \xi +\tau _{2}} G(t,s)h_{i}^{-}(s) \,ds. $$

2 Preliminaries

Let

$$ E_{i}= \bigl\{ x_{i}\in C[-\tau _{i},1]:x_{i}(t)=0, \forall t\in [-\tau _{i},0] \text{ and } x_{i}(1)=0 \bigr\} \quad (i=1,2) $$

be a Banach space with norm $|x_{i}(t)|_{i}=\max_{-\tau _{i}\leq t\leq 1}x_{i}(t)=\max_{0\leq t \leq 1}x_{i}(t)$. Then we can define a Banach space E by $E_{1}\times E_{2}$ with norm $\|x\|=\max \{|x_{1}|_{1},|x_{2}|_{2}\}$, for $x=(x_{1},x_{2}) \in E$. Also, define a subcone $K\subset E$ by

$$ K= \bigl\{ x\in E : x(t) \geq 0,x_{i} \text{ is concave on } [0,\xi ] \text{ and convex on } [\xi ,1] \bigr\} . $$

For $\forall x(t)\in K$, it is obvious that

$$ \vert x_{i} \vert _{i}=\max_{0\leq t\leq \xi }x_{i}(t). $$

For any $\gamma >0$, in the following paragraphs, we set

$$ K_{\gamma }=\bigl\{ x\in K: \Vert x \Vert < \gamma \bigr\} $$

and

$$ \partial K_{\gamma }=\bigl\{ x\in K: \Vert x \Vert =\gamma \bigr\} . $$

Define an operator T by $Tx(t)=(T_{1}x(t),T_{2}x(t))$, where

$$\begin{aligned} T_{i}x(t)= \textstyle\begin{cases} \int ^{1}_{0} G(t,s)h_{i}(s)f_{i}(x_{1}(s-\tau _{1}),x_{2}(s-\tau _{2}))\,ds,& 0< t \leq 1, \\ 0,& {-}\tau _{i} \leq t \leq 0, \end{cases}\displaystyle \end{aligned}$$

where

$$\begin{aligned} G(t,s)= \textstyle\begin{cases} (1-t)s, & 0\leq s\leq t \leq 1, \\ t(1-s), & 0\leq t\leq s \leq 1. \end{cases}\displaystyle \end{aligned}$$

Now solutions of (1.1) can be rewritten as fixed points of T in Banach space E.

Lemma 2.1

Assume that ($\mathrm{H}_{1}$)–($\mathrm{H}_{4}$) hold. Then the operator T is positive and $T:K\rightarrow K$ is completely continuous.

Proof

First of all, we show that the operator $T_{i}$ is positive, namely, for any $x(t)\in K$, we have

$$ \int ^{1}_{0} G(t,s)h_{i}(s)f_{i} \bigl(x_{1}(s-\tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \geq \int ^{\sigma _{i}}_{0} G(t,s)h_{i}^{+}(s)f_{i} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds. $$

For any $x(t)\in K$, we can obtain

$$ x_{i}(t)\geq q(t)x_{i}(\xi ),\quad t\in [0,\xi ] \quad \text{and} \quad x_{i}(t) \leq q(t)x_{i}(\xi ),\quad t\in [ \xi ,1], $$

where $q(t):[0,1]\rightarrow [0,1]$ defined by $q(t)=\min \{\frac{t}{\xi },\frac{1-t}{1-\xi }\}$. Hence, we have

$$ \min_{s\in [\sigma _{1},\xi ]}q(s-\tau _{i})= \frac{\sigma _{1}-\tau _{i}}{\xi },\qquad \max_{s\in [\xi +\tau _{2},1]}q(s- \tau _{i})=\frac{1-\xi -\tau _{2}+\tau _{i}}{1-\xi }. $$

Then, from ($\mathrm{H}_{1}$)–($\mathrm{H}_{4}$), it follows that

$$\begin{aligned}& \int ^{1}_{0} G(t,s)h_{1}(s)f_{1} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds- \int ^{\sigma _{1}}_{0} G(t,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \\& \quad = \int ^{\xi }_{\sigma _{1}} G(t,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds \\& \qquad {}- \int ^{1}_{\xi +\tau _{2}} G(t,s)h_{1}^{-}(s)f_{1} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \\& \quad \geq \theta \int ^{\xi }_{\sigma _{1}} G(t,s)h_{1}^{+}(s) \phi _{1} \bigl(x_{1}(s- \tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds \\& \qquad {}- \int ^{1}_{\xi +\tau _{2}} G(t,s)h_{1}^{-}(s) \phi _{1} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds \\& \quad \geq \theta \int ^{\xi }_{\sigma _{1}} G(t,s)h_{1}^{+}(s) \phi _{1} \bigl(q(s- \tau _{1})x_{1}(\xi ),q(s- \tau _{2})x_{2}(\xi ) \bigr)\,ds \\& \qquad {} - \int ^{1}_{\xi +\tau _{2}} G(t,s)h_{1}^{-}(s) \phi _{1} \bigl(q(s- \tau _{1})x_{1}(\xi ),q(s- \tau _{2})x_{2}(\xi ) \bigr)\,ds \\& \quad \geq \theta \int ^{\xi }_{\sigma _{1}} G(t,s)h_{1}^{+}(s)f_{1} \bigl(q(s- \tau _{1})x_{1}(\xi ),q(s-\tau _{2})x_{2}(\xi ) \bigr)\,ds \\& \qquad {} -\frac{1}{\theta } \int ^{1}_{\xi +\tau _{2}} G(t,s)h_{1}^{-}(s)f_{1} \bigl(q(s- \tau _{1})x_{1}(\xi ),q(s-\tau _{2})x_{2}(\xi ) \bigr)\,ds \\& \quad = \frac{1}{\theta } \biggl[\theta ^{2} \int ^{\xi }_{\sigma _{1}} G(t,s)h_{1}^{+}(s)f_{1} \bigl(q(s- \tau _{1})x_{1}(\xi ),q(s-\tau _{2})x_{2}(\xi ) \bigr)\,ds \\& \qquad {} - \int ^{1}_{\xi +\tau _{2}} G(t,s)h_{1}^{-}(s)f_{1} \bigl(q(s-\tau _{1})x_{1}( \xi ),q(s-\tau _{2})x_{2}(\xi ) \bigr)\,ds \biggr] \\& \quad \geq \frac{1}{\theta } \biggl[\theta ^{2} \int ^{\xi }_{\sigma _{1}} G(t,s)h_{1}^{+}(s)k_{1}F_{1} \bigl(q(s- \tau _{1})x_{1}(\xi ),q(s-\tau _{2})x_{2}(\xi ) \bigr) \,ds \\& \qquad {} - \int ^{1}_{\xi +\tau _{2}} G(t,s)h_{1}^{-}(s)F_{1} \bigl(q(s-\tau _{1})x_{1}( \xi ),q(s-\tau _{2})x_{2}(\xi ) \bigr) \,ds \biggr] \\& \quad \geq \frac{1}{\theta } \biggl[\theta ^{2} \int ^{\xi }_{\sigma _{1}} G(t,s)h_{1}^{+}(s)k_{1}F_{1} \Bigl( \min_{[\sigma _{1},\xi ]}q(s-\tau _{1})x_{1}( \xi ),\min_{[\sigma _{1}, \xi ]}q(s-\tau _{2})x_{2}( \xi ) \Bigr) \,ds \\& \qquad {} - \int ^{1}_{\xi +\tau _{2}} G(t,s)h_{1}^{-}(s)F_{1} \Bigl(\max_{[\xi + \tau _{2},1]}q(s-\tau _{1})x_{1}( \xi ),\max_{[\xi +\tau _{2},1]}q(s- \tau _{2})x_{2}( \xi ) \Bigr) \,ds \biggr] \\& \quad = \frac{1}{\theta } \biggl[\theta ^{2} \int ^{\xi }_{\sigma _{1}} G(t,s)h_{1}^{+}(s) \frac{k_{1}}{\xi ^{\alpha _{1}}}F_{1} \bigl((\sigma _{1}-\tau _{1})x_{1}( \xi ),(\sigma _{1}-\tau _{2})x_{2}(\xi ) \bigr) \,ds \\& \qquad {} - \int ^{1}_{\xi +\tau _{2}} G(t,s)h_{1}^{-}(s) \frac{1}{(1-\xi )^{\alpha _{1}}}F_{1} \bigl((1-\xi -\tau _{2}+\tau _{1})x_{1}( \xi ),(1-\xi )x_{2}(\xi ) \bigr) \,ds \biggr] \\& \quad \geq \frac{1}{\theta } \biggl[\theta ^{2} \int ^{\xi }_{\sigma _{1}} G(t,s)h_{1}^{+}(s) \frac{k_{1}}{\xi ^{\alpha _{1}}}F_{1} \bigl((\sigma _{1}-\tau _{2})x_{1}( \xi ),(\sigma _{1}-\tau _{2})x_{2}(\xi ) \bigr) \,ds \\& \qquad {} - \int ^{1}_{\xi +\tau _{2}} G(t,s)h_{1}^{-}(s) \frac{1}{(1-\xi )^{\alpha _{1}}}F_{1} \bigl((1-\xi )x_{1}(\xi ),(1- \xi )x_{2}( \xi ) \bigr) \,ds \biggr] \\& \quad = \frac{1}{\theta } \biggl[\theta ^{2} \int ^{\xi }_{\sigma _{1}} G(t,s)h_{1}^{+}(s) \frac{k_{1}}{\xi ^{\alpha _{1}}}(\sigma _{1}-\tau _{2})^{\alpha _{1}} \,ds \\& \qquad {} - \int ^{1}_{\xi +\tau _{2}} G(t,s)h_{1}^{-}(s) \frac{1}{(1-\xi )^{\alpha _{1}}}(1-\xi )^{\alpha _{1}} \,ds \biggr]\cdot F_{1} \bigl(x_{1}( \xi ),x_{2}(\xi ) \bigr) \\& \quad \geq 0, \end{aligned}$$

which implies that the operator $T_{1}$ is positive.

In the similar way, we also get

$$ \int ^{1}_{0} G(t,s)h_{2}(s)f_{2} \bigl(x_{1}(s-\tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \geq \int ^{\sigma _{2}}_{0} G(t,s)h_{2}^{+}(s)f_{2} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds, $$

which implies that the operator $T_{2}$ is positive.

Secondly, from the conditions ($\mathrm{H}_{1}$)–($\mathrm{H}_{3}$), we can obtain

$$ (T_{i}x)''(t)\leq 0, \quad t\in [0,\xi ]\quad \text{and}\quad (T_{i}x)''(t) \geq 0,\quad t\in [\xi ,1]. $$

Ultimately, from the standard process, we can prove that $T:K\rightarrow K$ is completely continuous. □

The proofs of this paper are mainly based on the Krasnosel’skii fixed point theorems in cones such as the following.

Lemma 2.2

([8])

Let E be a Banach space, and $K\subset E$ be a cone in E. Assume that $\Omega _{1}$, $\Omega _{2}$ are open subsets of E with $0\in \Omega _{1}$, $\overline{\Omega }_{1}\subset \Omega _{2}$, and let $T:K\cap (\overline{\Omega }_{2}\setminus \Omega _{1})\rightarrow K$ be a completely continuous operator such that either

• (i) $\|Tu\|\leq \|u\|$, $u\in K\cap \partial \Omega _{1}$ and $\|Tu\|\geq \|u\|$, $u\in K\cap \partial \Omega _{2}$; or

• (ii) $\|Tu\|\geq \|u\|$, $u\in K\cap \partial \Omega _{1}$ and $\|Tu\|\leq \|u\|$, $u\in K\cap \partial \Omega _{2}$.

Then T has a fixed point in $K\cap (\overline{\Omega }_{2}\setminus \Omega _{1})$.

Lemma 2.3

([15])

Let $(X,|\cdot|)$ be a normal linear space; $K_{1}$, $K_{2}\subset X$ two cones; $K:K_{1}\times K_{2}$; $r,R\in R_{+}^{2}$ with $0< r_{i}< R_{i}$ for $i=1,2$, $K_{r,R}:=\{u=(u_{1},u_{2})\in K:r_{i}\leq |u_{i}|\leq R_{i},i=1,2\}$, and let $T=(T_{1},T_{2})$ be a compact map. Assume that, for each $i\in \{1,2\}$, one of the following conditions is satisfied in $K_{r,R}$:

• (a) $T_{i}(u)\nless(\geq ) u_{i}$ if $|u_{i}|=r_{i}$, and $T_{i}(u)\ngtr(\leq ) u_{i}$ if $|u_{i}|=R_{i}$;

• (b) $T_{i}(u)\ngtr(\leq ) u_{i}$ if $|u_{i}|=r_{i}$, and $T_{i}(u)\nless(\geq ) u_{i}$ if $|u_{i}|=R_{i}$.

Then T has a fixed point u in K with $r_{i}\leq |u_{i}|\leq R_{i}$.

3 Main results I

Define a function $\delta (t)$ by

$$ \delta (t)=\min \biggl\{ \frac{t}{\xi },\frac{\xi -t}{\xi } \biggr\} ,\quad t \in [0,\xi ]. $$

Then, for $i=1,2$, let

$$\begin{aligned}& \gamma _{i}=\min \Bigl\{ \min_{\frac{\sigma _{i}+\tau _{2}}{2}\leq s \leq \sigma _{i}}\delta (s- \tau _{1}),\min_{ \frac{\sigma _{i}+\tau _{2}}{2}\leq s \leq \sigma _{i}}\delta (s- \tau _{2}) \Bigr\} , \\& A_{i}=k_{i} \gamma _{i}^{\alpha _{i}} \max_{t\in [0,\xi ]}G(t,t) \cdot \int ^{\sigma _{i}}_{\frac{\sigma _{i}+\tau _{2}}{2}} G(s,s)h_{i}^{+}(s) \,ds, \\& B_{i}= \int ^{\tau _{2}}_{\tau _{1}}G(s,s)h_{i}^{+}(s) \,ds, C_{i}= \int ^{\xi }_{\tau _{2}} G(s,s)h_{i}^{+}(s) \,ds. \end{aligned}$$

Theorem 3.1

Assume that ($\mathrm{H}_{1}$)–($\mathrm{H}_{4}$) hold. If there exist two positive constants r, R with $r< R$, satisfying

$$\begin{aligned}& \min \bigl\{ A_{1} F_{1}(r,0),A_{2}F_{2}(0,r) \bigr\} \geq r, \\& \max \bigl\{ B_{1} F_{1}(R,0)+C_{1} F_{1}(R,R), B_{2} F_{2}(R,0)+C_{2} F_{2}(R,R) \bigr\} \leq R, \end{aligned}$$

then problem (1.1) at least has a positive solution.

Proof

On one hand, for any $x\in \partial K_{R}$, we have $0\leq x_{1}$, $x_{2}\leq R$. Then, by the assumptions, we have

$$\begin{aligned} \vert T_{1}x \vert _{1} \leq & \int ^{1}_{0} G(s,s)h_{1}(s)f_{1} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds \\ =& \int ^{\xi }_{0} G(s,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds \\ &{} - \int ^{1}_{\xi }G(s,s)h_{1}^{-}(s)f_{1} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds \\ \leq & \int ^{\xi }_{0} G(s,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds \\ =& \biggl( \int ^{\tau _{2}}_{\tau _{1}}+ \int ^{\xi }_{\tau _{2}} \biggr) G(s,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \\ \leq & \int ^{\tau _{2}}_{\tau _{1}}G(s,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s- \tau _{1}),0 \bigr)\,ds \\ &{} + \int ^{\xi }_{\tau _{2}}G(s,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds \\ \leq & \int ^{\tau _{2}}_{\tau _{1}}G(s,s)h_{1}^{+}(s)F_{1} \bigl(x_{1}(s- \tau _{1}),0 \bigr)\,ds \\ &{}+ \int ^{\xi }_{\tau _{2}} G(s,s)h_{1}^{+}(s)F_{1} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr) \,ds \\ \leq & \int ^{\tau _{2}}_{\tau _{1}}G(s,s)h_{1}^{+}(s)F_{1}(R,0) \,ds \\ &{}+ \int ^{\xi }_{\tau _{2}} G(s,s)h_{1}^{+}(s)F_{1}(R,R) \,ds \\ \leq & R. \end{aligned}$$

In the similar way, we also have

$$\begin{aligned} \vert T_{2}x \vert _{2} \leq & \int ^{1}_{0} G(s,s)h_{2}(s)f_{2} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds \\ \leq & \int ^{\tau _{2}}_{\tau _{1}}G(s,s)h_{2}^{+}(s)F_{2}(R,0) \,ds+ \int ^{\xi }_{\tau _{2}} G(s,s)h_{2}^{+}(s)F_{2}(R,R) \,ds \\ \leq & R. \end{aligned}$$

So, from the above discussions, we get

$$ \Vert Tx \Vert =\max \bigl\{ \vert T_{1}x \vert _{1}, \vert T_{2}x \vert _{2} \bigr\} \leq R= \Vert x \Vert ,\quad \text{for } x \in \partial K_{R}. $$

(3.1)

On the other hand, for any $x(t)\in K$, from the concave property of $x_{i}(t)$ it follows that

$$ x_{i}(t)\geq \delta (t) \bigl\vert x_{i}(t) \bigr\vert _{i},\quad t \in [0,\xi ]. $$

Then, for any $x(t)\in \partial K_{r}$, we have the following.

Case I: if $|x_{1}(t)|_{1}=r$, then $\delta (t) r\leq x_{1}(t)\leq r$ and $0\leq x_{2}(t)\leq r$. By ($\mathrm{H}_{2}$), we have

$$\begin{aligned} \max_{t\in [0,\xi ]}T_{1}x(t) =& \max _{t\in [0,\xi ]} \int ^{1}_{0} G(t,s)h_{1}(s)f_{1} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \\ \geq & \max_{t\in [0,\xi ]} \int ^{\sigma _{1}}_{0} G(t,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \\ \geq & \max_{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{1}}_{0} G(s,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \\ \geq & k_{1}\max_{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{1}}_{ \frac{\sigma _{1}+\tau _{2}}{2}} G(s,s)h_{1}^{+}(s)F_{1} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr) \,ds \\ \geq & k_{1} \max_{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{1}}_{ \frac{\sigma _{1}+\tau _{2}}{2}} G(s,s)h_{1}^{+}(s)F_{1} \bigl(\delta (s- \tau _{1}) \vert x_{1} \vert _{1},\delta (s-\tau _{2}) \vert x_{2} \vert _{2} \bigr) \,ds \\ \geq & k_{1} \max_{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{1}}_{ \frac{\sigma _{1}+\tau _{2}}{2}} G(s,s)h_{1}^{+}(s)F_{1} \bigl( \gamma _{1} \vert x_{1} \vert _{1}, \gamma _{1} \vert x_{2} \vert _{2}\bigr) \,ds \\ =& k_{1} \gamma _{1}^{\alpha _{1}} \max _{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{1}}_{\frac{\sigma _{1}+\tau _{2}}{2}} G(s,s)h_{1}^{+}(s)F_{1} \bigl( \vert x_{1} \vert _{1}, \vert x_{2} \vert _{2}\bigr)\,ds \\ \geq & k_{1} \gamma _{1}^{\alpha _{1}} \max _{t\in [0,\xi ]}G(t,t) \cdot \int ^{\sigma _{1}}_{\frac{\sigma _{1}+\tau _{2}}{2}} G(s,s)h_{1}^{+}(s)F_{1}(r,0) \,ds \\ =& k_{1} \gamma _{1}^{\alpha _{1}} \max _{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{1}}_{\frac{\sigma _{1}+\tau _{2}}{2}} G(s,s)h_{1}^{+}(s) \,ds \cdot F_{1}(r,0) \\ \geq & r. \end{aligned}$$

Case II: if $|x_{1}(t)|_{1}< r$, then $0\leq x_{1}(t)< r$ and $\delta (t) r\leq x_{2}(t)\leq r$. By ($\mathrm{H}_{2}$), we have

$$\begin{aligned} \max_{t\in [0,\xi ]}T_{2}x(t) =& \max _{t\in [0,\xi ]} \int ^{1}_{0} G(t,s)h_{2}(s)f_{2} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \\ \geq & \max_{t\in [0,\xi ]} \int ^{\sigma _{2}}_{0} G(t,s)h_{2}^{+}(s)f_{2} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \\ \geq & \max_{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{2}}_{0} G(s,s)h_{2}^{+}(s)f_{2} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \\ \geq & k_{2}\max_{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{2}}_{ \frac{\sigma _{2}+\tau _{2}}{2}} G(s,s)h_{2}^{+}(s)F_{2} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr) \,ds \\ \geq & k_{2} \max_{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{2}}_{ \frac{\sigma _{2}+\tau _{2}}{2}} G(s,s)h_{1}^{+}(s)F_{2} \bigl(\delta (s- \tau _{1}) \vert x_{1} \vert _{1},\delta (s-\tau _{2}) \vert x_{2} \vert _{2} \bigr) \,ds \\ \geq & k_{2} \max_{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{2}}_{ \frac{\sigma _{2}+\tau _{2}}{2}} G(s,s)h_{1}^{+}(s)F_{2} \bigl( \gamma _{2} \vert x_{1} \vert _{1}, \gamma _{2} \vert x_{2} \vert _{2}\bigr) \,ds \\ =& k_{2} \gamma _{2}^{\alpha _{2}} \max _{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{2}}_{\frac{\sigma _{2}+\tau _{2}}{2}} G(s,s)h_{2}^{+}(s)F_{2} \bigl( \vert x_{1} \vert _{1}, \vert x_{2} \vert _{2}\bigr)\,ds \\ \geq & k_{2} \gamma _{2}^{\alpha _{2}} \max _{t\in [0,\xi ]}G(t,t) \cdot \int ^{\sigma _{2}}_{\frac{\sigma _{2}+\tau _{2}}{2}} G(s,s)h_{2}^{+}(s)F_{2}(0,r) \,ds \\ =& k_{2} \gamma _{2}^{\alpha _{2}} \max _{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{2}}_{\frac{\sigma _{2}+\tau _{2}}{2}} G(s,s)h_{2}^{+}(s) \,ds \cdot F_{2}(0,r) \\ \geq & r. \end{aligned}$$

So, from the above discussions, we get

$$ \Vert Tx \Vert =\max \bigl\{ \vert T_{1}x \vert _{1}, \vert T_{2}x \vert _{2} \bigr\} \geq r= \Vert x \Vert , \quad \text{for } x \in \partial K_{r}. $$

(3.2)

Therefore, from (3.1) and (3.2) and Lemma 2.2, the operator T has a fixed point in $K\cap (\overline{\Omega }_{R}\setminus \Omega _{r})$. □

As by a similar proof to Theorem 3.1, we also get the following.

Theorem 3.2

Assume that ($\mathrm{H}_{1}$)–($\mathrm{H}_{4}$) hold. If there exist two positive constants r, R with $r< R$, satisfying

$$\begin{aligned}& \min \bigl\{ A_{1} F_{1}(R,0),A_{2}F_{2}(0,R) \bigr\} \geq R, \\& \max \bigl\{ B_{1} F_{1}(r,0)+C_{1} F_{1}(r,r), B_{2} F_{2}(r,0)+C_{2} F_{2}(r,r) \bigr\} \leq r, \end{aligned}$$

then problem (1.1) at least has a positive solution.

4 Main results II

Theorem 4.1

Assume that ($\mathrm{H}_{1}$)–($\mathrm{H}_{4}$) hold. If there exist four positive constants $r_{1}$, $r_{2}$, $R_{1}$, $R_{2}$ with $r_{1}< R_{1}$, $r_{2}< R_{2}$, satisfying

$$\begin{aligned}& A_{1} F_{1}(r_{1},r_{2})>r_{1},\qquad A_{2} F_{2}(r_{1},r_{2})>r_{2}, \\& B_{1} F_{1}(R_{1},0)+C_{1}F_{1}(R_{1},R_{2})< R_{1}, \\& B_{2} F_{2}(R_{1},0)+C_{2}F_{2}(R_{1},R_{2})< R_{2}, \end{aligned}$$

then problem (1.1) at least has a positive solution.

Proof of Theorem 4.1

Let

$$ K_{r,R}:= \bigl\{ x=(x_{1},x_{2})\in K:r_{i}\leq \vert x_{i} \vert _{i}\leq R_{i} \bigr\} \quad (i=1,2). $$

Then, for any $x(t)\in K_{r,R}$, we have

$$ x_{i}(t)\geq \delta (t) \bigl\vert x_{i}(t) \bigr\vert _{i},\quad t \in [0,\xi ], $$

where $\delta (t)=\min \{\frac{t}{\xi },\frac{\xi -t}{\xi }\}$, $t\in [0,\xi ]$.

On the one hand, for any $x(t)\in K_{r,R}$, we show that $T_{1}(x)\nless x_{1}$ if $|x_{1}|_{1}=r_{1}$ and $r_{2}\leq |x_{2}|_{2}\leq R_{2}$. On the contrary, if $T_{1}(x)< x_{1}$, then we have

$$\begin{aligned} r_{1} \geq &\max_{t\in [0,\xi ]}x_{1}(t)\geq \max_{t\in [0,\xi ]}T_{1}x(t) \\ =& \max_{t\in [0,\xi ]} \int ^{1}_{0} G(t,s)h_{1}(s)f_{1} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \\ \geq & \max_{t\in [0,\xi ]} \int ^{\sigma _{1}}_{0} G(t,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \\ \geq & \max_{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{1}}_{0} G(s,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \\ \geq & k_{1}\max_{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{1}}_{ \frac{\sigma _{1}+\tau _{2}}{2}} G(s,s)h_{1}^{+}(s)F_{1} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr) \,ds \\ \geq & k_{1} \max_{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{1}}_{ \frac{\sigma _{1}+\tau _{2}}{2}} G(s,s)h_{1}^{+}(s)F_{1} \bigl(\delta (s- \tau _{1}) \vert x_{1} \vert _{1},\delta (s-\tau _{2}) \vert x_{2} \vert _{2} \bigr) \,ds \\ \geq & k_{1} \max_{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{1}}_{ \frac{\sigma _{1}+\tau _{2}}{2}} G(s,s)h_{1}^{+}(s)F_{1} \bigl( \gamma _{1} \vert x_{1} \vert _{1}, \gamma _{1} \vert x_{2} \vert _{2}\bigr) \,ds \\ =& k_{1} \gamma _{1}^{\alpha _{1}} \max _{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{1}}_{\frac{\sigma _{1}+\tau _{2}}{2}} G(s,s)h_{1}^{+}(s)F_{1} \bigl( \vert x_{1} \vert _{1}, \vert x_{2} \vert _{2}\bigr)\,ds \\ \geq & k_{1} \gamma _{1}^{\alpha _{1}} \max _{t\in [0,\xi ]}G(t,t) \cdot \int ^{\sigma _{1}}_{\frac{\sigma _{1}+\tau _{2}}{2}} G(s,s)h_{1}^{+}(s)F_{1}(r_{1},r_{2}) \,ds \\ =& k_{1} \gamma _{1}^{\alpha _{1}} \max _{t\in [0,\xi ]}G(t,t)\cdot \int ^{\sigma _{1}}_{\frac{\sigma _{1}+\tau _{2}}{2}} G(s,s)h_{1}^{+}(s) \,ds \cdot F_{1}(r_{1},r_{2}) \\ >&r_{1}, \end{aligned}$$

which implies a contradiction. In the similar way, for any $x(t)\in K_{r,R}$, we also can obtain $T_{2}(x)\nless x_{2}$ if $|x_{2}|_{2}=r_{2}$ and $r_{1}\leq |x_{1}|_{1}\leq R_{1}$.

On the one hand, for any $x(t)\in K_{r,R}$, we show that $T_{1}(x)\ngtr x_{1}$ if $|x_{1}|_{1}=R_{1}$ and $r_{2}\leq |x_{2}|_{2}\leq R_{2}$. On the contrary, if $T_{1}(x)>x_{1}$, then we have

$$\begin{aligned} x_{1}(t) < &T_{1}x(t) = \int ^{1}_{0} G(t,s)h_{1}(s)f_{1} \bigl(x_{1}(s- \tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \\ =& \int ^{\xi }_{0} G(t,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds \\ &{} - \int ^{1}_{\xi }G(t,s)h_{1}^{-}(s)f_{1} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds \\ \leq & \int ^{\xi }_{0} G(s,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds \\ =& \biggl( \int ^{\tau _{1}}_{0}+ \int ^{\tau _{2}}_{\tau _{1}}+ \int ^{\xi }_{ \tau _{2}} \biggr) G(s,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s-\tau _{1}),x_{2}(s-\tau _{2}) \bigr)\,ds \\ =& \int ^{\tau _{1}}_{0}G(s,s)h_{1}^{+}(s)f_{1}(0,0) \,ds+ \int ^{\tau _{2}}_{ \tau _{1}}G(s,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s-\tau _{1}),0 \bigr)\,ds \\ &{}+ \int ^{\xi }_{\tau _{2}}G(s,s)h_{1}^{+}(s)f_{1} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr)\,ds \\ \leq & \int ^{\tau _{2}}_{\tau _{1}}G(s,s)h_{1}^{+}(s)F_{1} \bigl(x_{1}(s- \tau _{1}),0 \bigr)\,ds \\ &{}+ \int ^{\xi }_{\tau _{2}} G(s,s)h_{1}^{+}(s)F_{1} \bigl(x_{1}(s-\tau _{1}),x_{2}(s- \tau _{2}) \bigr) \,ds \\ \leq & \int ^{\tau _{2}}_{\tau _{1}}G(s,s)h_{1}^{+}(s)F_{1}(R_{1},0) \,ds+ \int ^{\xi }_{\tau _{2}} G(s,s)h_{1}^{+}(s)F_{1}(R_{1},R_{2}) \,ds \\ < &R_{1}, \end{aligned}$$

which implies a contradiction. In the similar way, for any $x(t)\in K_{r,R}$, we also can obtain $T_{2}(x)\ngtr x_{2}$ if $|x_{2}|_{2}=R_{2}$ and $r_{1}\leq |x_{1}|_{1}\leq R_{1}$.

Therefore, from (a) of Lemma 2.3, the operator T has a fixed point x in K with $r_{i}\leq |x_{i}|\leq R_{i}$. □

As in a similar proof to Theorem 4.1, we also get the following.

Theorem 4.2

Assume that ($\mathrm{H}_{1}$)–($\mathrm{H}_{4}$) hold. If there exist four positive constants $r_{1}$, $r_{2}$, $R_{1}$, $R_{2}$ with $r_{1}< R_{1}$, $r_{2}< R_{2}$, satisfying

$$\begin{aligned}& A_{1} F_{1}(R_{1},r_{2})>R_{1},\qquad A_{2} F_{2}(r_{1},R_{2})>R_{2}, \\& B_{1} F_{1}(r_{1},0)+C_{1} F_{1}(r_{1},R_{2})< r_{1}, \\& B_{2} F_{2}(R_{1},0)+C_{2} F_{2}(R_{1},r_{2})< r_{2}, \end{aligned}$$

then problem (1.1) at least has a positive solution.

5 Examples

Example 5.1

Now we consider the following problem:

$$ \textstyle\begin{cases} x_{1}''(t)+h_{1}(t)f_{1}(x_{1}(t-\tau _{1}),x_{2}(t-\tau _{2})), \quad t>0, \\ x_{2}''(t)+h_{2}(t)f_{2}(x_{1}(t-\tau _{1}),x_{2}(t-\tau _{2})),\quad t>0, \\ x_{1}(t)=0,\quad -\tau _{1} \leq t \leq 0,\quad \text{and}\quad x_{1}(1)=0, \\ x_{2}(t)=0,\quad -\tau _{2} \leq t \leq 0, \quad \text{and}\quad x_{2}(1)=0, \end{cases} $$

(5.1)

where $\tau _{1}=\frac{1}{10}$, $\tau _{2}=\frac{1}{5}$,

$$\begin{aligned}& h_{1}(t)=h_{2}(t)= \textstyle\begin{cases} 400(\frac{1}{2}-t), & t\in [0,\frac{1}{2}], \\ 0, & t\in [\frac{1}{2},\frac{7}{10}], \\ -\frac{5}{3}t+\frac{7}{6}, & t\in [\frac{7}{10},1], \end{cases}\displaystyle \\& f_{1}= \frac{(x_{1}+x_{2})^{\frac{1}{2}}}{\frac{3}{2}+\frac{1}{4}\sin ^{2}(x_{1}+x_{2})}, \\& f_{2}= \frac{(x_{1}+x_{2})^{\frac{1}{3}}}{\frac{4}{3}+\frac{1}{3\pi }\arctan (x^{2}_{1}x^{4}_{2})}. \end{aligned}$$

It is obvious that ($\mathrm{H}_{1}$) and ($\mathrm{H}_{2}$) hold. Furthermore, we also have

$$\begin{aligned}& \frac{1}{2}F_{1}(x_{1},x_{2})= \frac{1}{2}(x_{1}+x_{2})^{\frac{1}{2}} \leq f_{1}\leq (x_{1}+x_{2})^{\frac{1}{2}}=F_{1}(x_{1},x_{2}), \\& \frac{1}{2}F_{2}(x_{1},x_{2})= \frac{1}{2}(x_{1}+x_{2})^{\frac{1}{3}} \leq f_{2}\leq (x_{1}+x_{2})^{\frac{1}{3}}=F_{2}(x_{1},x_{2}), \end{aligned}$$

which shows that ($\mathrm{H}_{3}$) holds. Now, we prove that ($\mathrm{H}_{4}$) holds.

Firstly, we show that there exist $\sigma _{i}$ satisfying $\frac{1}{10}<\sigma _{i}<\frac{1}{5}$ such that

$$ \theta ^{2} k_{1}(\sigma _{1}-\tau _{2})^{\frac{1}{2}} \int ^{\xi }_{ \sigma _{1}}G(t,s)h^{+}_{1}(s) \,ds \geq \xi ^{\frac{1}{2}} \int ^{1}_{ \xi +\tau _{2}}G(t,s)h^{-}_{1}(s) \,ds $$

(5.2)

and

$$ \theta ^{2} k_{2}(\sigma _{2}-\tau _{2})^{\frac{1}{3}} \int ^{\xi }_{ \sigma _{2}}G(t,s)h^{+}_{2}(s) \,ds \geq \xi ^{\frac{1}{3}} \int ^{1}_{ \xi +\tau _{2}}G(t,s)h^{-}_{2}(s) \,ds. $$

(5.3)

For fixed $\theta =\frac{1}{2}$, $\sigma _{1}=\sigma _{2}=\frac{2}{5}$, we set

$$\begin{aligned}& M(t)=\theta ^{2} \frac{1}{2}(\sigma _{1}-\tau _{2})^{\frac{1}{2}} \int ^{\xi }_{\sigma _{1}}G(t,s)h^{+}_{1}(s) \,ds= \frac{50}{\sqrt{5}} \int ^{ \frac{1}{2}}_{\frac{2}{5}}G(t,s) \biggl( \frac{1}{2}-s \biggr)\,ds, \\& N(t)=\xi ^{\frac{1}{2}} \int ^{1}_{\xi +\tau _{2}}G(t,s)h^{-}_{1}(s) \,ds= \frac{1}{\sqrt{2}} \int ^{1}_{\frac{7}{10}}G(t,s) \biggl(\frac{5}{3}s- \frac{7}{6} \biggr)\,ds,\quad t\in [0,1]. \end{aligned}$$

For $t\in [0,\frac{2}{5}]$, we have

$$\begin{aligned}& M(t)=\frac{50t}{\sqrt{5}} \int ^{\frac{1}{2}}_{\frac{2}{5}}(1-s) \biggl( \frac{1}{2}-s \biggr)\,ds=\frac{17}{120\sqrt{5}}t, \\& N(t)=\frac{t}{\sqrt{2}} \int ^{1}_{\frac{7}{10}}(1-s) \biggl(\frac{5}{3}s- \frac{7}{6} \biggr)\,ds=\frac{3}{400\sqrt{2}}t. \end{aligned}$$

It is obvious that $M(t)\geq N(t)$, $\forall t\in [0,\frac{2}{5}]$.

For $t\in [\frac{2}{5},\frac{1}{2}]$, we have

$$\begin{aligned}& \begin{aligned} M(t)&=\frac{50}{\sqrt{5}} \biggl[ \int ^{t}_{\frac{2}{5}}(1-t)s \biggl( \frac{1}{2}-s \biggr)\,ds+ \int ^{\frac{1}{2}}_{t}t(1-s) \biggl( \frac{1}{2}-s \biggr)\,ds \biggr] \\ &= \frac{50}{\sqrt{5}} \biggl( \frac{1}{6}t^{3}-\frac{1}{4}t^{2}+ \frac{737}{6000}t-\frac{7}{375} \biggr), \end{aligned} \\& N(t)=\frac{t}{\sqrt{2}} \int ^{1}_{\frac{7}{10}}(1-s) \biggl(\frac{5}{3}s- \frac{7}{6} \biggr)\,ds=\frac{3}{400\sqrt{2}}t. \end{aligned}$$

Via some computations, we have

$$ M(t)_{\mathrm{min}}=M \biggl(\frac{1}{2} \biggr)= \frac{13}{240\sqrt{5}}> \frac{3}{800\sqrt{2}}=N \biggl(\frac{1}{2} \biggr)=N(t)_{\mathrm{max}}, $$

which implies that $M(t)>N(t)$, $\forall t\in [\frac{2}{5},\frac{1}{2}]$.

For $t\in [\frac{1}{2},\frac{7}{10}]$, we have

$$\begin{aligned}& M(t)=\frac{50(1-t)}{\sqrt{5}} \int ^{\frac{1}{2}}_{\frac{2}{5}}s \biggl( \frac{1}{2}-s \biggr)\,ds=\frac{13}{120\sqrt{5}}(1-t), \\& N(t)=\frac{t}{\sqrt{2}} \int ^{1}_{\frac{7}{10}}(1-s) \biggl(\frac{5}{3}s- \frac{7}{6} \biggr)\,ds=\frac{3}{400\sqrt{2}}t. \end{aligned}$$

Via some computations, we have

$$ M(t)_{\mathrm{min}}=M \biggl(\frac{7}{10} \biggr)= \frac{13}{400\sqrt{5}}> \frac{21}{4000\sqrt{2}}=N \biggl(\frac{7}{10} \biggr)=N(t)_{\mathrm{max}}, $$

which implies that $M(t)>N(t)$, $\forall t\in [\frac{1}{2},\frac{7}{10}]$.

For $t\in [\frac{7}{10},1]$, we have

$$\begin{aligned}& M(t)=\frac{50(1-t)}{\sqrt{5}} \int ^{\frac{1}{2}}_{\frac{2}{5}}s \biggl( \frac{1}{2}-s \biggr)\,ds=\frac{13}{120\sqrt{5}}(1-t), \\& \begin{aligned} N(t)&=\frac{1}{\sqrt{2}} \biggl[ \int ^{t}_{\frac{7}{10}}(1-t)s \biggl( \frac{5}{3}s- \frac{7}{6} \biggr)\,ds+ \int ^{1}_{t}t(1-s) \biggl(\frac{5}{3}s- \frac{7}{6} \biggr)\,ds \biggr] \\ &=\frac{1}{\sqrt{2}} \biggl(-\frac{5}{18}t^{3}+ \frac{7}{12}t^{2}- \frac{481}{1200}t+ \frac{343}{3600} \biggr). \end{aligned} \end{aligned}$$

Denote $g(t)=M(t)-N(t)$ and $g(1)=0$. Since

$$ g'(t)=\frac{5}{6\sqrt{2}}t^{2}- \frac{7}{6\sqrt{2}}t+ \frac{481}{1200\sqrt{2}}-\frac{13}{120\sqrt{5}}< 0,\quad t\in \biggl[ \frac{7}{10},1 \biggr], $$

$g(t)$ is strictly decreasing on $[\frac{7}{10},1]$. Then $g(t)\geq 0$, $\forall t\in [\frac{7}{10},1]$, which implies that $M(t)\geq N(t)$, $\forall t\in [\frac{7}{10},1]$.

So from these discussions, we have $M(t)\geq N(t)$, $\forall t\in [0,1]$, which means that (5.2) holds. Hence, in the similar way, we also see that the inequality (5.3) is true.

Secondly, let

$$\begin{aligned}& \Gamma (\rho )=\min \bigl\{ A_{1} F_{1}(\rho ,0),A_{2}F_{2}(0,\rho ) \bigr\} = \min \bigl\{ A_{1}\rho ^{\frac{1}{2}},A_{2}\rho ^{\frac{1}{3}} \bigr\} , \\& \begin{aligned} \Lambda (\rho )&=\max \bigl\{ B_{1} F_{1}(\rho ,0)+C_{1} F_{1}(\rho ,\rho ), B_{2} F_{2}(\rho ,0)+C_{2} F_{2}(\rho ,\rho ) \bigr\} \\ &=\max \bigl\{ B_{1}\rho ^{\frac{1}{2}}+C_{1} (2\rho )^{\frac{1}{2}}, B_{2} \rho ^{\frac{1}{3}}+C_{2} (2 \rho )^{\frac{1}{3}} \bigr\} . \end{aligned} \end{aligned}$$

It is obvious that there exist a sufficiently small constant $r>0$ and a sufficiently large constant $R>0$ such that

$$\begin{aligned}& \min \bigl\{ A_{1}r^{\frac{1}{2}},A_{2}r^{\frac{1}{3}} \bigr\} >r, \\& \max \bigl\{ B_{1}R^{\frac{1}{2}}+C_{1} (2R)^{\frac{1}{2}}, B_{2} R^{\frac{1}{3}}+C_{2} (2R)^{\frac{1}{3}} \bigr\} < R. \end{aligned}$$

Then, by Theorem 3.1, problem (5.1) has a positive solution.

Example 5.2

Now we consider the following problem:

$$ \textstyle\begin{cases} x_{1}''(t)+h_{1}(t)f_{1}(x_{1}(t-\tau _{1}),x_{2}(t-\tau _{2})),\quad t>0, \\ x_{2}''(t)+h_{2}(t)f_{2}(x_{1}(t-\tau _{1}),x_{2}(t-\tau _{2})),\quad t>0, \\ x_{1}(t)=0,\quad -\tau _{1} \leq t \leq 0, \quad \text{and}\quad x_{1}(1)=0, \\ x_{2}(t)=0,\quad -\tau _{2} \leq t \leq 0, \quad \text{and}\quad x_{2}(1)=0, \end{cases} $$

(5.4)

where $\tau _{1}=\frac{1}{10}$, $\tau _{2}=\frac{1}{5}$,

$$\begin{aligned}& h_{1}(t)=h_{2}(t)= \textstyle\begin{cases} \frac{1}{2}-t, & t\in [0,\frac{1}{2}], \\ 0, & t\in [\frac{1}{2}, \frac{7}{10}], \\ 10^{-4}(-\frac{5}{3}t+\frac{7}{6}),& t\in [\frac{7}{10},1], \end{cases}\displaystyle \\& f_{1}=\frac{(x_{1}+x_{2})^{2}}{1+ \vert \sin (x_{1}x_{2}) \vert }, \\& f_{2}=\frac{(x_{1}+x_{2})^{3}}{1+ \vert \sin (x_{1}x_{2}) \vert }. \end{aligned}$$

It is obvious that ($\mathrm{H}_{1}$) and ($\mathrm{H}_{2}$) hold. Moreover, we have

$$\begin{aligned}& \frac{1}{2}F_{1}(x_{1},x_{2})= \frac{1}{2}(x_{1}+x_{2})^{2}\leq f_{1} \leq (x_{1}+x_{2})^{2}=F_{1}(x_{1},x_{2}),\quad k_{1}>0, \\& \frac{1}{2}F_{2}(x_{1},x_{2})= \frac{1}{2}(x_{1}+x_{2})^{3}\leq f_{2} \leq (x_{1}+x_{2})^{3}=F_{2}(x_{1},x_{2}),\quad k_{2}>0, \end{aligned}$$

which shows that ($\mathrm{H}_{3}$) holds. Now, we prove that ($\mathrm{H}_{4}$) holds.

Firstly, we show that there exist $\sigma _{i}$ satisfying $\frac{1}{10}<\sigma _{i}<\frac{1}{5}$ such that

$$ \theta ^{2} k_{1}(\sigma _{1}-\tau _{2})^{2} \int ^{\xi }_{\sigma _{1}}G(t,s)h^{+}_{1}(s) \,ds \geq \xi ^{2} \int ^{1}_{\xi +\tau _{2}}G(t,s)h^{-}_{1}(s) \,ds $$

(5.5)

and

$$ \theta ^{2} k_{2}(\sigma _{2}-\tau _{2})^{3} \int ^{\xi }_{\sigma _{2}}G(t,s)h^{+}_{2}(s) \,ds \geq \xi ^{3} \int ^{1}_{\xi +\tau _{2}}G(t,s)h^{-}_{2}(s) \,ds. $$

(5.6)

For fixed $\theta =\frac{1}{2}$, $\sigma _{1}=\sigma _{2}=\frac{2}{5}$, we set

$$\begin{aligned}& m(t)=\theta ^{2} \frac{1}{2}(\sigma _{1}-\tau _{2})^{2} \int ^{\xi }_{ \sigma _{1}}G(t,s)h^{+}_{1}(s) \,ds= \frac{1}{200} \int ^{\frac{1}{2}}_{ \frac{2}{5}}G(t,s) \biggl( \frac{1}{2}-s \biggr)\,ds, \\& n(t)=\xi ^{2} \int ^{1}_{\xi +\tau _{2}}G(t,s)h^{-}_{1}(s) \,ds= \frac{1}{4\times 10^{4}} \int ^{1}_{\frac{7}{10}}G(t,s) \biggl(\frac{5}{3}s- \frac{7}{6} \biggr)\,ds,\quad t\in [0,1]. \end{aligned}$$

For $t\in [0,\frac{2}{5}]$, we have

$$\begin{aligned}& m(t)=\frac{t}{200} \int ^{\frac{1}{2}}_{\frac{2}{5}}(1-s) \biggl(\frac{1}{2}-s \biggr)\,ds= \frac{17}{12\times 10^{5}}t, \\& n(t)=\frac{t}{4\times 10^{4}} \int ^{1}_{\frac{7}{10}}(1-s) \biggl( \frac{5}{3}s- \frac{7}{6} \biggr)\,ds=\frac{3}{16\times 10^{6}}t. \end{aligned}$$

It is obvious that $m(t)\geq n(t)$, $\forall t\in [0,\frac{2}{5}]$.

For $t\in [\frac{2}{5},\frac{1}{2}]$, we have

$$\begin{aligned}& \begin{aligned} m(t)&=\frac{1}{200} \biggl[ \int ^{t}_{\frac{2}{5}}(1-t)s \biggl( \frac{1}{2}-s \biggr)\,ds+ \int ^{\frac{1}{2}}_{t}t(1-s) \biggl( \frac{1}{2}-s \biggr)\,ds \biggr] \\ &= \frac{1}{200} \biggl( \frac{1}{6}t^{3}-\frac{1}{4}t^{2}+ \frac{737}{6000}t-\frac{7}{375} \biggr), \end{aligned} \\& n(t)=\frac{t}{4\times 10^{4}} \int ^{1}_{\frac{7}{10}}(1-s) \biggl( \frac{5}{3}s- \frac{7}{6} \biggr)\,ds=\frac{3}{16\times 10^{6}}t. \end{aligned}$$

Via some computations, we have

$$ m(t)_{\mathrm{min}}=m \biggl(\frac{1}{2} \biggr)= \frac{13}{24\times 10^{5}}> \frac{3}{32\times 10^{6}}=n \biggl(\frac{1}{2} \biggr)=n(t)_{\mathrm{max}}, $$

which implies that $m(t)>n(t)$, $\forall t\in [\frac{2}{5},\frac{1}{2}]$.

For $t\in [\frac{1}{2},\frac{7}{10}]$, we have

$$\begin{aligned}& m(t)=\frac{1-t}{200} \int ^{\frac{1}{2}}_{\frac{2}{5}}s \biggl(\frac{1}{2}-s \biggr)\,ds= \frac{13}{12\times 10^{5}}(1-t), \\& n(t)=\frac{t}{4\times 10^{4}} \int ^{1}_{\frac{7}{10}}(1-s) \biggl( \frac{5}{3}s- \frac{7}{6} \biggr)\,ds=\frac{3}{16\times 10^{6}}t. \end{aligned}$$

Via some computations, we have

$$ m(t)_{\mathrm{min}}=m \biggl(\frac{7}{10} \biggr)= \frac{13}{4\times 10^{6}}> \frac{21}{16\times 10^{7}}=n \biggl(\frac{7}{10} \biggr)=n(t)_{\mathrm{max}}, $$

which implies that $m(t)>n(t)$, $\forall t\in [\frac{1}{2},\frac{7}{10}]$.

For $t\in [\frac{7}{10},1]$, we have

$$\begin{aligned}& m(t)=\frac{1-t}{200} \int ^{\frac{1}{2}}_{\frac{2}{5}}s \biggl(\frac{1}{2}-s \biggr)\,ds= \frac{13}{12\times 10^{5}}(1-t), \\& \begin{aligned} n(t)&= \frac{1}{4\times 10^{4}} \biggl[ \int ^{t}_{\frac{7}{10}}(1-t)s \biggl( \frac{5}{3}s-\frac{7}{6} \biggr)\,ds+ \int ^{1}_{t}t(1-s) \biggl(\frac{5}{3}s- \frac{7}{6} \biggr)\,ds \biggr] \\ &= \frac{1}{4\times 10^{4}} \biggl(-\frac{5}{18}t^{3}+ \frac{7}{12}t^{2}- \frac{481}{1200}t+ \frac{343}{3600} \biggr). \end{aligned} \end{aligned}$$

Let $\omega (t)=m(t)-n(t)$ satisfying $\omega (1)=0$. Since

$$ \omega '(t)=\frac{5}{24\times 10^{4}}t^{2}- \frac{7}{24\times 10^{4}}t- \frac{13}{16\times 10^{6}}< 0,\quad t\in \biggl[\frac{7}{10},1 \biggr], $$

it means that $\omega (t)$ is strictly decreasing on $[\frac{7}{10},1]$. Then $\omega (t)\geq 0$, $\forall t\in [\frac{7}{10},1]$, namely, $m(t)\geq n(t)$, $\forall t\in [\frac{7}{10},1]$.

So from these discussions, we have $m(t)\geq n(t)$, $\forall t\in [0,1]$, which means that (5.5) holds. Hence, by a similar method, we also can see that the inequality (5.6) is true.

Secondly, let

$$\begin{aligned}& \Gamma (\rho )=\min \bigl\{ A_{1} F_{1}(\rho ,0),A_{2}F_{2}(0,\rho ) \bigr\} = \min \bigl\{ A_{1}\rho ^{2},A_{2}\rho ^{3} \bigr\} , \\& \begin{aligned} \Lambda (\rho )&=\max \bigl\{ B_{1} F_{1}(\rho ,0)+C_{1} F_{1}(\rho ,\rho ), B_{2} F_{2}(\rho ,0)+C_{2} F_{2}(\rho ,\rho ) \bigr\} \\ &=\max \bigl\{ B_{1}\rho ^{2}+C_{1} (2\rho )^{2}, B_{2} \rho ^{3}+C_{2} (2 \rho )^{3} \bigr\} . \end{aligned} \end{aligned}$$

It is obvious that there exist a sufficiently small constant $r>0$ and a sufficiently large constant $R>0$ such that

$$\begin{aligned}& \min \bigl\{ A_{1}R^{2},A_{2}R^{3} \bigr\} >R, \\& \max \bigl\{ B_{1}r^{2}+C_{1} (2r)^{2}, B_{2} r^{3}+C_{2} (2r)^{3} \bigr\} < r. \end{aligned}$$

Then, by Theorem 3.2, problem (5.4) has a positive solution.

Example 5.3

Now we consider the following problem:

$$ \textstyle\begin{cases} x_{1}''(t)+h_{1}(t)f_{1}(x_{1}(t-\tau _{1}),x_{2}(t-\tau _{2})),\quad t>0, \\ x_{2}''(t)+h_{2}(t)f_{2}(x_{1}(t-\tau _{1}),x_{2}(t-\tau _{2})),\quad t>0, \\ x_{1}(t)=0,\quad -\tau _{1} \leq t \leq 0,\quad \text{and}\quad x_{1}(1)=0, \\ x_{2}(t)=0,\quad -\tau _{2} \leq t \leq 0, \quad \text{and}\quad x_{2}(1)=0, \end{cases} $$

(5.7)

where $\tau _{1}=\frac{1}{10}$, $\tau _{2}=\frac{1}{5}$,

$$\begin{aligned}& h_{1}(t)=h_{2}(t)= \textstyle\begin{cases} 400(\frac{1}{2}-t), & t\in [0,\frac{1}{2}], \\ 0, & t\in [\frac{1}{2},\frac{7}{10}], \\ -\frac{5}{3}t+\frac{7}{6}, & t\in [\frac{7}{10},1], \end{cases}\displaystyle \\& f_{1}=(x_{1}+x_{2})^{\frac{1}{2}},\qquad f_{2}=(x_{1}+x_{2})^{\frac{1}{3}}. \end{aligned}$$

Choosing $\theta =\frac{1}{2}$, $k_{1}=k_{2}=\frac{1}{2}$, $\sigma _{1}=\sigma _{2}=\frac{2}{5}$. Then, from Example 5.1, it is obvious that ($\mathrm{H}_{1}$)–($\mathrm{H}_{4}$) hold.

Now, we show that there exist $0< r_{i}< R_{i}$ such that

$$\begin{aligned}& k_{1}\gamma ^{\frac{1}{2}}_{1}\max _{t\in [0,\frac{1}{2}]}G(t,t) \cdot \int ^{\sigma _{1}}_{\frac{\sigma _{1}+\frac{1}{5}}{2}}G(s,s)h^{+}_{1}(s) \,ds \cdot (r_{1}+r_{2})^{\frac{1}{2}}>r_{1}, \end{aligned}$$

(5.8)

$$\begin{aligned}& k_{2}\gamma ^{\frac{1}{3}}_{2}\max _{t\in [0,\frac{1}{2}]}G(t,t) \cdot \int ^{\sigma _{2}}_{\frac{\sigma _{2}+\frac{1}{5}}{2}}G(s,s)h^{+}_{2}(s) \,ds \cdot (r_{1}+r_{2})^{\frac{1}{3}}>r_{2}, \end{aligned}$$

(5.9)

$$\begin{aligned}& \int ^{\frac{1}{5}}_{\frac{1}{10}}G(s,s)h^{+}_{1}(s) \,ds \cdot R_{1}^{ \frac{1}{2}}+ \int ^{\frac{1}{2}}_{\frac{1}{5}}G(s,s)h^{+}_{1}(s) \,ds \cdot (R_{1}+R_{2})^{\frac{1}{2}}< R_{1}, \end{aligned}$$

(5.10)

$$\begin{aligned}& \int ^{\frac{1}{5}}_{\frac{1}{10}}G(s,s)h^{+}_{2}(s) \,ds \cdot R_{1}^{ \frac{1}{3}}+ \int ^{\frac{1}{2}}_{\frac{1}{5}}G(s,s)h^{+}_{2}(s) \,ds \cdot (R_{1}+R_{2})^{\frac{1}{3}}< R_{2}. \end{aligned}$$

(5.11)

Choosing $r_{1}=0.001$, $r_{2}=0.01$, $R_{1}=70$, $R_{2}=140$, we obtain

$$\begin{aligned}& \frac{50}{\sqrt{5}} \int ^{\frac{2}{5}}_{\frac{3}{10}}s(1-s) \biggl( \frac{1}{2}-s \biggr)\,ds\cdot (0.001+0.01)^{\frac{1}{2}}= \frac{0.16875}{\sqrt{5}}\sqrt{0.011}>0.001, \\& \frac{50}{\sqrt[3]{5}} \int ^{\frac{2}{5}}_{\frac{3}{10}}s(1-s) \biggl( \frac{1}{2}-s \biggr)\,ds\cdot (0.001+0.01)^{\frac{1}{3}}= \frac{0.16875}{\sqrt[3]{5}}\sqrt[3]{0.011}>0.01, \\& 400 \biggl[ \int ^{\frac{1}{5}}_{\frac{1}{10}}s(1-s) \biggl( \frac{1}{2}-s \biggr)\,ds\cdot 70^{ \frac{1}{2}}+ \int ^{\frac{1}{2}}_{\frac{1}{5}}s(1-s) \biggl( \frac{1}{2}-s \biggr)\,ds \cdot (70+140)^{\frac{1}{2}} \biggr] \\& \quad =1.75 \sqrt{70}+3.69\sqrt{210}< 70, \\& 400 \biggl[ \int ^{\frac{1}{5}}_{\frac{1}{10}}s(1-s) \biggl( \frac{1}{2}-s \biggr)\,ds\cdot 70^{ \frac{1}{3}}+ \int ^{\frac{1}{2}}_{\frac{1}{5}}s(1-s) \biggl( \frac{1}{2}-s \biggr)\,ds \cdot (70+140)^{\frac{1}{3}} \biggr] \\& \quad =1.75 \sqrt[3]{70}+3.69\sqrt[3]{210}< 140, \end{aligned}$$

which implies that the inequalities (5.8)–(5.11) are true.

Therefore, from Theorem 4.1, one concludes that problem (5.7) has a positive solution.