Existence of positive solutions for a singular semipositone differential system with nonlocal boundary conditions

Abstract

In this paper we consider the existence of at least one positive solution to a class of singular semipositone coupled system of nonlocal boundary value problems. We show that the system possesses at least one positive solution by using fixed point index theory. We remark that to some extent our systems and results generalize and extend some previous works.

1 Introduction

In this paper, we consider the existence of at least one positive solution to the following singular semipositone coupled system of nonlocal boundary value problems:

$$\begin{aligned} \textstyle\begin{cases} -x''=f(t,y(t))+q(t),\quad t\in(0,1), \\ -y''=g(t,x(t)), \quad t\in(0,1), \\ x(0)=H_{1}(\varphi_{1}(y)),\qquad x(1)=0, \\ y(0)=H_{2}(\varphi_{2}(x)),\qquad y(1)=0, \end{cases}\displaystyle \end{aligned}$$

(1.1)

where $f,g:(0,1)\times[0,+\infty)\rightarrow[0,+\infty)$ are continuous and may be singular at $t=0,1$, $q:(0,1)\rightarrow(-\infty,+\infty)$ is Lebesgue integrable, and $q(t)$ may have finitely many singularities in $[0,1]$, $H_{i}:\mathbf{R}\rightarrow\mathbf{R}$ ($\mathbf{R}=(-\infty,+\infty )$) are continuous, and $H_{i}([0,+\infty))\subseteq[0,+\infty)$ and $\varphi_{i}:C([0,1])\rightarrow\mathbf{R}$ $(i=1,2)$ are linear and can be realized as Stieltjes integrals with signed measures. In particular, in the Stieltjes integral representation $\varphi(y)=\int_{[0,1]}y(t) \,d\alpha(t)$ with $\alpha:[0,1]\rightarrow\mathbf{R}$ of bounded variation on $[0,1]$, we no longer assume that α is necessarily monotonically increasing. Thus, in this paper, we allow the map $y\mapsto\varphi(y)$ to be negative even if y is nonnegative.

Recently, the theory of nonlocal and nonlinear boundary value problems and singular semipositone differential systems becomes an important area of investigation because of its wide applicability in control, electrical engineering, physics, chemistry fields, and so on. Equation (1.1) is used to describe chemical reactor theory where the nonlinearity can take negative values. Many works have been done for a kind of nonlinear boundary value problems [1, 2] and nonlinear differential systems [3–6]. However, most investigators only focus on the case where the nonlinearity takes nonnegative values, that is, positone problems. For example, under conditions where $f(t,y)$ and $g(t,x)$ have no any singularities and $q(t)\equiv0$, Agarwal and O’Regan [6], using the Leray-Schauder fixed point theorem, obtained the existence of positive solutions of the following system:

$$\begin{aligned} \textstyle\begin{cases} -x''=f(t,y(t))+q(t),\quad t\in(0,1),\\ -y''=g(t,x(t)), \quad t\in(0,1), \\ x(0)=x(1)=0, \\ \alpha y(0)-\beta y'(0)=\gamma^{2}y(1)+\delta y'(1). \end{cases}\displaystyle \end{aligned}$$

(1.2)

Later, Zhang and Liu [7] obtained the existence of positive solutions of system (1.2) by the Leray-Schauder fixed point theorem under the conditions that $q(t):(0,1)\rightarrow(-\infty,+\infty)$ is Lebesgue integrable, $q(t)$ may have finitely many singularities in $[0,1]$, and $f,g:(0,1)\times[0,+\infty)\rightarrow[0,+\infty)$ are continuous and may be singular at $t=0,1$. The study of semipositone problems has a long history in the literature, the work of Anuradha et al. [8] being an early, classical example. More recent papers include those by Goodrich [9], Graef and Kong [10], and Infante and Webb [11]. Furthermore, recently, there have been many papers on nonlocal BVPs with nonlinear boundary conditions. For example, Anderson [12], Goodrich [9, 13–16], and Infante et al. [17–22]. In this paper, these nonlocal nonlinear boundary conditions have been investigated by Goodrich [13, 14]. For example, in [13], Goodrich investigated the existence of positive solutions of the semipositone boundary value problems with nonlocal nonlinear boundary conditions

$$\begin{aligned} \textstyle\begin{cases} -y''=f(t,y(t)), \quad t\in(0,1), \\ y(0)=H(\varphi(y)), \qquad y(1)=0, \end{cases}\displaystyle \end{aligned}$$

(1.3)

by the fixed point index under the conditions that $f:[0,1]\times \mathbf{R}\rightarrow\mathbf{R}$ and $H:\mathbf{R}\rightarrow\mathbf {R}$ are continuous, $H([0,+\infty))\subseteq[0,+\infty)$, and there is a number $C\geq0$, such that $\lim_{z\rightarrow+\infty}\frac {\vert H(z)-Cz\vert }{z}=0$ ($(H_{3})$ in [13]). The proof of Theorem 3.1 in [13] gives a limiting condition, that is,

$$\begin{aligned} (C_{2}+\varepsilon) \int_{0}^{1} (1-t) \,d\alpha(t)+ \int_{0}^{1} \int _{0}^{1}G(t,s) \biggl[\frac{1}{r_{2}}u(s)+v(s) \biggr] \,d\alpha(t)< 1, \end{aligned}$$

where $C_{2}$ is a constant, $r_{2}\neq0$ satisfies some conditions, ε satisfies $0<\varepsilon<[\int_{0}^{1} (1-t)\,d\alpha (t)]^{-1}-C_{2}$, $v:[0,1]\rightarrow[0,+\infty)$ is continuous, and $u:[0,1]\rightarrow[0,+\infty)$ is not identically zero on any subinterval of $[0,1]$. Goodrich [14] investigated the existence of positive solutions of the coupled system of boundary value problems with nonlocal boundary conditions

$$\begin{aligned} \textstyle\begin{cases} -x''=f(t,y(t)), \quad t\in(0,1), \\ -y''=g(t,x(t)), \quad t\in(0,1), \\ x(0)=H_{1}(\varphi_{1}(y)),\qquad x(1)=0, \\ y(0)=H_{2}(\varphi_{2}(x)),\qquad y(1)=0, \end{cases}\displaystyle \end{aligned}$$

(1.4)

by the Leray-Schauder fixed point theorem under the conditions that $f,g:[0,1]\times[0,+\infty)\rightarrow[0,+\infty)$ are $H_{i}:\mathbf {R}\rightarrow\mathbf{R}$ are continuous, $H_{i}([0,+\infty))\subseteq [0,+\infty)$, and $H_{i}$ satisfies $\lim_{z\rightarrow0^{+}}\frac{H_{i}(z)}{z}=0$ and $\lim_{z\rightarrow +\infty}\frac{H_{i}(z)}{z}=+\infty$, $i=1,2$ (($A_{3}$) in [14]). For example, Goodrich [16] investigated the existence of at least one positive solution of the semipositone boundary value problems with nonlocal, nonlinear boundary conditions

$$\begin{aligned} \textstyle\begin{cases} -y''=\lambda f(t,y(t)),\quad t\in(0,1), \\ y(0)=H(\varphi(y)), \\ y(1)=0, \end{cases}\displaystyle \end{aligned}$$

(1.5)

by the fixed point index, where $\lambda>0$ is a parameter, under the conditions that $f:[0,1]\times\mathbf{R} \rightarrow\mathbf{R}$ are $H:\mathbf{R}\rightarrow\mathbf{R}$ are continuous, $H([0,+\infty ))\subseteq[0,+\infty)$, $\lim_{z\rightarrow0^{+}}\frac{H(z)}{z}=+\infty$ ($(H_{2})$ in [16]), and there is a number $C_{2}\geq0$ such that $\lim_{z\rightarrow+\infty}\frac{\vert H(z)-C_{2}z\vert }{z}=0$ ($(H_{3})$ in [16]).

Motivated by the works mentioned, in this paper, we consider the coupled system (1.1). The main features of this paper are as follows. Firstly, we have more general integral boundary conditions. Secondly, we consider coupled systems rather than a single equation. Finally, we consider f that need not have a lower bound, that is, a semipositone problem. We remark that, to some extent, our systems and results generalize some previous works.

We organize this paper as follows. In Section 2, we first approximate the singular semipositone problem to the singular positone problem by a substitution. Then we present some lemmas to be used later. In Section 3, we state our result and give its proof. In Section 4, we present an example to demonstrate an application of our main results.

2 Preliminaries and lemmas

In this section, we first approximate the singular semipositone problem to the singular positone problem by a substitution. Then we present some lemmas to be used later. We assume that there are four linear functionals $\varphi_{1,1},\varphi _{1,2},\varphi_{2,1},\varphi_{2,2}:C[0,1]\to\mathbf{R}$ such that $\varphi_{1},\varphi_{2}$ satisfy the decompositions

$$\begin{aligned} \varphi_{1}(y) =\varphi_{1,1}(y)+\varphi_{1,2}(y),\qquad \varphi_{2}(y) =\varphi_{2,1}(y)+\varphi_{2,2}(y). \end{aligned}$$

(2.1)

Let $E= C[0,1]$, so that $(E,\Vert \cdot \Vert )$ is a Banach space with usual maximal norm $\Vert y\Vert =\max_{t\in[0,1]}\vert y(t)\vert$. Let

$$\begin{aligned} P=\bigl\{ {y\in E:y(t)\geq0,y(t)\geq t(1-t)\Vert y\Vert , t\in [0,1], \varphi_{1,1}(y)\geq0,\varphi_{2,1}(y)\geq0} \bigr\} . \end{aligned}$$

(2.2)

Clearly, P is a cone in E. We denote $P_{r}:=\{y\in P,\Vert y\Vert < r\}$ for any $r>0$.

Now, for the boundary value problem

$$\begin{aligned} \textstyle\begin{cases} x''(t) =0, \quad t\in(0,1), \\ x(0)=0, \\ x(1)=0, \end{cases}\displaystyle \end{aligned}$$

we denote the Green functions

$$\begin{aligned} \textstyle\begin{cases} t(1-s), &0\leq t\leq s\leq1, \\ s(1-t), &0\leq s\leq t\leq1. \end{cases}\displaystyle \end{aligned}$$

(2.3)

In the rest of the paper, we adopt the following assumptions:

$( H_{1} )$ There exist constants $C_{0},D_{0}>0$ such that $\varphi _{1,2}(y) \geq C_{0}\Vert y\Vert$ and $\varphi_{2,2}(y) \geq D_{0}\Vert y\Vert$ for all $y\in P$.

$( H_{2})$ The functionals described in (2.1) have the form

$$\begin{aligned} &\varphi_{1}(y):= \int_{[0,1]}y(t) \,d\alpha_{1}(t),\qquad \varphi _{1,1}(y):= \int_{[0,1]}y(t) \,d\alpha_{1,1}(t), \\ &\varphi_{1,2}(y):= \int_{[0,1]}y(t) \,d\alpha_{1,2}(t), \\ &\varphi_{2}(y):= \int_{[0,1]}y(t) \,d\alpha_{2}(t),\qquad \varphi _{2,1}(y):= \int_{[0,1]}y(t) \,d\alpha_{2,1}(t),\\ & \varphi_{2,2}(y):= \int_{[0,1]}y(t) \,d\alpha_{2,2}(t), \end{aligned}$$

where all $\alpha_{i},\alpha_{i,j}:C[0,1]\to\mathbf{R}$, $i,j=1,2$, are of bounded variation on $[0,1]$.

$( H_{3} )$ We have

$$\begin{aligned} &\int_{[0,1]}G(t,s) \,d\alpha_{1,1}(t)>0,\qquad \int_{[0,1]}G(t,s) \,d\alpha_{2,1}(t)>0, \quad\forall s\in[0,1], \\ &\int_{[0,1]}(1-t) \,d\alpha_{1,1}(t)>0,\qquad \int_{[0,1]}(1-t) \,d\alpha_{2,1}(t)>0. \end{aligned}$$

$( H_{4} )$ The functions $H_{1} ,H_{2}:\mathbf{R}\to\mathbf{R}$ are continuous with $H_{1}([0,+\infty))$, $H_{2}([0,+\infty))\subseteq [0,+\infty)$.

$( H_{5} )$ $f:(0,1)\times[0,+\infty)\rightarrow[0,+\infty)$ is continuous, and for any $t\in(0,1)$, $f(t,y)$ is nondecreasing in y and satisfies

$$\begin{aligned} f(t,y)\leq p(t)h(y), \end{aligned}$$

(2.4)

where $p:(0,1)\rightarrow[0,+\infty)$ and $h:[0,+\infty)\rightarrow[0,+\infty )$ are continuous, and $\lim_{y\to+\infty} \frac{f(t,y)}{y}=+\infty$ for t uniformly on any closed subinterval of $(0,1)$.

$( H_{6} )$ $g:(0,1)\times[0,+\infty)\rightarrow[0,+\infty)$ is continuous, and $g(t,1)>0$ for all $t\in(0,1)$. Moreover, there exist constants $\lambda_{1} \geq\lambda_{2}>1$ such that, for any $t\in (0,1)$ and $x\in[0,+\infty)$,

$$\begin{aligned} c^{\lambda_{1}}g(t,x)\leq g(t,cx)\leq c^{\lambda _{2}}g(t,x), \quad 0\leq c\leq1, \end{aligned}$$

(2.5)

with $0<\int_{0}^{1}G(t,t)g(t,1) \,dt<\infty$.

$( H_{7} )$ $q(t):(0,1)\rightarrow(-\infty,+\infty)$ is Lebesgue integrable such that $\int_{0}^{1}q_{-}(t)\,dt>0$, where

$$\begin{aligned} q_{-}(t)=\max\bigl\{ -q(t),0\bigr\} ,\quad t\in(0,1). \end{aligned}$$

Remark 2.1

Note that since both $\varphi_{1}$ and $\varphi_{2}$ are linear, there exist constants $C_{1}$ and $D_{1}>0$ such that $\vert \varphi_{1}\vert \leq C_{1}\Vert y\Vert$ and $\vert \varphi_{2}\vert \leq D_{1}\Vert y\Vert$ for all $y\in P$. Henceforth, $C_{1}$ and $D_{1}$ denote these constants.

To state and prove the main result of this paper, we need the following lemmas.

Lemma 2.1

[7]

$q:(0,1)\rightarrow(-\infty,+\infty)$ is Lebesgue integrable, and $q(t)$ may have finitely many singularities.

Lemma 2.2

[7]

For any $c\geq1$ and $(t,x)\in(0,1)\times[0,+\infty )$, we have

$$\begin{aligned} c^{\lambda_{2}}g(t,x)\leq g(t,cx)\leq c^{\lambda _{1}}g(t,x). \end{aligned}$$

(2.6)

Definition 2.1

If $(x,y)\in C[0,1]\cap C^{2}(0,1)\times C[0,1]\cap C^{2}(0,1)$ satisfies (1.1) and $x(t)>0,y(t)>0$ for any $t\in(0,1)$, then we say that $(x,y)$ is a positive solution of system (1.1).

For $u\in E$, let us define the function $[\cdot]^{*}$ by

$$\begin{aligned} &\bigl[u(t)\bigr] ^{*}= \textstyle\begin{cases} u(t), & u(t)\geq0, \\ 0, & u(t)< 0, \end{cases}\displaystyle \\ &H_{i}^{*}(z)=H_{i}\bigl(\max\{0,z\}\bigr),\quad i=1,2. \end{aligned}$$

Clearly, $\omega(t)=\int_{0}^{1}G(t,s)q_{-}(s)\,ds$ is a positive solution of the BVP

$$\begin{aligned} \textstyle\begin{cases} -\omega''(t)=q_{-}(t), \quad t\in(0,1), \\ \omega(0)=\omega(1)=0. \end{cases}\displaystyle \end{aligned}$$

Clearly, $\omega\in P$.

In what follows, we consider the following approximately singular nonlinear system:

$$\begin{aligned} \textstyle\begin{cases} -x''(t)=f(t,y(t))+q_{+}(t), \quad t\in(0,1), \\ -y''(t)=g(t,[x(t)-\omega(t)]^{*}), \quad t\in(0,1), \\ x(0)=H_{1}^{*}(\varphi_{1}(y)),\qquad x(1)=0, \\ y(0)=H_{2}^{*}(\varphi_{2}(x-\omega)), \qquad y(1)=0, \end{cases}\displaystyle \end{aligned}$$

(2.7)

where

$$\begin{aligned} q_{+}(t)=\max\bigl\{ q(t),0\bigr\} ,\quad t\in(0,1). \end{aligned}$$

It is easy to check that $(x,y)$ is a solution of (2.7) if and only if $(x,y)$ is a solution of the following nonlinear integral equation system:

$$\begin{aligned} \textstyle\begin{cases} x(t)=(1-t)H_{1}^{*}(\varphi_{1}(y))+\int _{0}^{1}G(t,s)[f(s,y(s))+q_{+}(s)]\,ds, & t\in[0,1], \\ y(t)=(1-t)H_{2}^{*}(\varphi_{2}(x-\omega))+\int _{0}^{1}G(t,s)g(s,[x(s)-\omega(s)]^{*})\,ds, & t\in[0,1]. \end{cases}\displaystyle \end{aligned}$$

(2.8)

If $x\in P$ and $\omega\in P$, then $y\in P$. In fact,

$$\begin{aligned} y(t)=(1-t)H_{2}^{*}\bigl(\varphi_{2}(x-\omega) \bigr)+ \int _{0}^{1}G(t,s)g\bigl(s,\bigl[x(s)-\omega(s) \bigr]^{*}\bigr)\,ds,\quad t\in[0,1] \end{aligned}$$

and

$$\begin{aligned} y(0)=H_{2}^{*}\bigl(\varphi_{2}(x-\omega) \bigr),\qquad y(1)=0. \end{aligned}$$

If $\Vert y\Vert =y(0)=H_{2}^{*}(\varphi_{2}(x-\omega))$, then we have

$$\begin{aligned} y(t)&\geq(1-t)H_{2}^{*}\bigl(\varphi_{2}(x-\omega) \bigr)\geq t(1-t)H_{2}^{*}\bigl(\varphi_{2}(x- \omega)\bigr)\\ &=t(1-t)\Vert y\Vert , \quad t\in[0,1]. \end{aligned}$$

If $\Vert y\Vert >H_{2}^{*}(\varphi_{2}(x-\omega))$, then there exists $t_{0}\in (0,1)$ such that $\Vert y\Vert =y(t_{0})$. Since $\frac{G(t,s)}{G(t_{0},s)}\geq t(1-t)$, $t,s\in(0,1)\times (0,1)$, we have

$$\begin{aligned} y(t) =&(1-t)H_{2}^{*}\bigl(\varphi _{2}(x-\omega)\bigr)+ \int_{0}^{1}\frac {G(t,s)}{G(t_{0},s)}G(t_{0},s)g \bigl(s,\bigl[x(s)-\omega(s)\bigr]^{*}\bigr) \,ds \\ \geq&t(1-t) \biggl[(1-t_{0})H_{2}^{*}\bigl( \varphi_{2}(x-\omega)\bigr)+ \int _{0}^{1}G(t_{0},s)g \bigl(s, \bigl[x(s)-\omega(s)\bigr]^{*} \bigr)\,ds \biggr] \\ =&t(1-t)y(t_{0}) \\ =&t(1-t)\Vert y\Vert , \quad t\in[0,1]. \end{aligned}$$

If $\Vert y\Vert < H_{2}^{*}(\varphi_{2}(x-\omega))$, then

$$\begin{aligned} y(t) =&(1-t)H_{2}^{*}\bigl(\varphi _{2}(x-\omega)\bigr)+ \int_{0}^{1}\frac {G(t,s)}{G(t_{0},s)}G(t_{0},s)g \bigl(s,\bigl[x(s)-\omega(s)\bigr]^{*}\bigr)\,ds \\ \geq&(1-t)H_{2}^{*}\bigl(\varphi_{2}(x-\omega) \bigr) \\ \geq&t(1-t)H_{2}^{*}\bigl(\varphi_{2}(x-\omega) \bigr) \\ >&t(1-t)\Vert y\Vert , \quad t\in[0,1]. \end{aligned}$$

In other words, we have $x,\omega\in P$, $H_{2}^{*}(\varphi _{2}(x-\omega))\geq0$, and

$$\begin{aligned} \varphi_{i,1}(y)={}& \int_{[0,1]}(1-t)H_{2}^{*}\bigl( \varphi_{2}(x-\omega )\bigr)\,d\alpha_{i,1}(t) \\ &{}+ \int_{[0,1]} \biggl( \int _{0}^{1}G(t,s)g\bigl(s,\bigl[x(s)-\omega(s) \bigr]^{*}\bigr)\,ds \biggr)\,d\alpha_{i,1}(t) \\ ={}&H_{2}^{*}\bigl(\varphi_{2}(x-\omega)\bigr) \int_{0}^{1}(1-t) \,d\alpha _{i,1}(t) \\ &{}+ \int_{0}^{1} \biggl[ \int_{[0,1]}G(t,s)\,d\alpha _{i,1}(t) \biggr]g\bigl(s, \bigl[x(s)-\omega(s)\bigr]^{*}\bigr)\,ds \geq0,\quad i=1,2. \end{aligned}$$

This yields that $y\in P$.

For convenience, we have the following form:

$$\begin{aligned} \varphi_{1}(y) =& \int_{0}^{1} y(t)\,d\alpha_{1}(t) \\ =& \int_{0}^{1} \biggl[(1-t)H_{2}^{*} \bigl(\varphi_{2}(x-\omega)\bigr)+ \int _{0}^{1}G(t,s)g\bigl(s,\bigl[x(s)-\omega(s) \bigr]^{*}\bigr)\,ds \biggr]\,d\alpha_{1}(t). \end{aligned}$$

We define

$$\begin{aligned} D_{x}:=H_{1}^{*}\bigl(\varphi_{1}(y) \bigr). \end{aligned}$$

(2.9)

Obviously, it is a nonnegative number that only depends on x.

We list here more assumptions to be used later.

$( H_{8} )$ We have

$$\begin{aligned} \frac{\frac{C_{1}}{C_{0}}\int_{0}^{1}q_{-}(t)\,dt+1}{\max_{0\leq \tau\leq R}h(\tau)+1}>2 \int_{0}^{1}(1-t)\bigl[p(t)+q_{+}(t) \bigr]\,dt+\frac {D_{x_{0}}}{\max_{0\leq\tau\leq R}h(\tau)+1}, \end{aligned}$$

where

$$\begin{aligned} &r^{*}=\frac{C_{1}}{C_{0}} \int_{0}^{1}q_{-}(t)\,dt+1,\qquad \widetilde {g}= \int_{0}^{1}G(s,s)g(s,1)\,ds, \\ &R=\max\bigl\{ H_{2}^{*}\bigl(\varphi_{2}(x-\omega )\bigr)+\bigl(r^{*}+1\bigr)^{\lambda_{1}}\widetilde{g},x\in \bigl[t(1-t)r^{*},r^{*}\bigr],t\in [0,1]\bigr\} , \\ &D_{x_{0}}=\max\bigl\{ D_{x},x\in\bigl[t(1-t)r^{*},r^{*} \bigr],t\in[0,1]\bigr\} . \end{aligned}$$

$( H_{9} )$ $\lim_{y\to+\infty} \frac{f(t,y)}{y}=+\infty$ for t uniformly on any closed subinterval of $(0,1)$.

As a matter of convenience, we set

$$\begin{aligned} \widetilde{x(t)}=y(t)=(1-t)H_{2}^{*}\bigl( \varphi_{2}(x-\omega )\bigr)+ \int_{0}^{1}G(t,s)g\bigl(s,\bigl[x(s)-\omega(s) \bigr]^{*}\bigr)\,ds,\quad t\in[0,1]. \end{aligned}$$

Then, clearly, the equation system (2.8) is equivalent to the equation

$$\begin{aligned} x(t)=(1-t)D_{x}+ \int_{0}^{1}G(t,s)\bigl[f\bigl(s,\widetilde {x(s)} \bigr)+q_{+}(s)\bigr]\,ds, \quad t\in[0,1]. \end{aligned}$$

(2.10)

Next, let us define the nonlinear operator $F:P\rightarrow C([0,1])$ by

$$\begin{aligned} (Fx) (t)=(1-t)D_{x}+ \int_{0}^{1}G(t,s)\bigl[f\bigl(s,\widetilde {x(s)} \bigr)+q_{+}(s)\bigr]\,ds, \quad t\in[0,1]. \end{aligned}$$

(2.11)

It is well known that the solutions to system (2.7) exist if and only if the solutions to equation (2.10) exist. Therefore, if $x(t)$ is a fixed point of F in P, then system (2.8) has one solution $(u,v)$, which can be written as

$$\begin{aligned} \textstyle\begin{cases} u(t)=x(t), \\ v(t)=(1-t)H_{2}^{*}(\varphi_{2}(x-\omega))+\int _{0}^{1}G(t,s)g(s,[x(s)-\omega(s)]^{*})\,ds, \quad t\in[0,1]. \end{cases}\displaystyle \end{aligned}$$

Lemma 2.3

[23]

Let X be a real Banach space, P be a cone in X, Ω be a bounded open subset of X with $\theta\in\Omega$, and $A:\overline{\Omega}\cap P \rightarrow P$ be a completely continuous operator.

(1) Suppose that

$$\begin{aligned} Au \neq\lambda u, \quad \forall u\in\partial\Omega\cap P , \lambda\geq1. \end{aligned}$$

Then $i(A,\Omega\cap P ,P)=1$.

(2) Suppose that

$$\begin{aligned} Au \nleq u, \quad \forall u \in\partial\Omega\cap P. \end{aligned}$$

Then $i(A,\Omega\cap P ,P)=0$.

Lemma 2.4

[7]

If $g(t,x)$ satisfies $(H_{6})$, then for any $t\in (0,1)$, $g(t,x)$ is increasing in $x\in[0,+\infty)$, and for any $[\alpha,\beta]\subset(0,1)$,

$$\begin{aligned} \lim_{n\rightarrow+\infty}\min_{t\in[\alpha,\beta]}\frac {g(t,x)}{x}=+ \infty. \end{aligned}$$

Lemma 2.5

Suppose that $(u,v)$ with $u(t)\geq\omega(t)$ for any $t\in [0,1]$ is a positive solution of system (2.7) and $\varphi_{2}(u-\omega )\geq0$, $\varphi_{1}(v)\geq0$. Then $(u-\omega,v)$ is a positive solution of the singular semipositone differential system (1.1).

Proof

In fact, if $(u,v)$ is a positive solution of (2.7) and $u(t)\geq\omega(t),\varphi_{2}(u-\omega)\geq0,\varphi _{1}(v)\geq0$ for any $t\in[0,1]$, then by (2.7) and the definition of $[u(t)]^{*}$ we have

$$\begin{aligned} \textstyle\begin{cases} -u''(t)=f(t,v(t))+q_{+}(t), \quad t\in(0,1), \\ -v''(t)=g(t,u(t)-\omega(t)), \quad t\in(0,1),\\ u(0)=H_{1}^{*}(\varphi_{1}(v))=H_{1}(\varphi_{1}(v)),\qquad u(1)=0, \\ v(0)=H_{2}^{*}(\varphi_{2}(u-\omega))=H_{2}(\varphi_{2}(u-\omega )), \qquad v(1)=0. \end{cases}\displaystyle \end{aligned}$$

(2.12)

Let $u_{1}=u-\omega$. Then $u_{1}''=u''-\omega''$, which implies that

$$\begin{aligned} u''(t)=u_{1}''(t)+ \omega''(t)=u_{1}''(t)-q_{-}(t),\quad t\in(0,1). \end{aligned}$$

Thus, (2.12) becomes

$$\begin{aligned} \textstyle\begin{cases} -u_{1}''(t)=f(t,v(t))+q_{+}(t)-q_{-}(t), \quad t\in(0,1), \\ -v''(t)=g(t,u_{1}(t)), \quad t\in(0,1), \\ u_{1}(0)=H_{1}(\varphi_{1}(v)),\qquad u_{1}(1)=0, \\ v(0)=H_{2}(\varphi_{2}(u_{1})),\qquad v(1)=0. \end{cases}\displaystyle \end{aligned}$$

(2.13)

Noticing that $q(t)=q_{+}(t)-q_{-}(t)$, by (2.12) we have that $(u_{1},v)$ is a positive solution of system (1.1), that is, $(u-\omega ,v)$ is a positive solution of system (1.1). This completes the proof of the lemma. □

Lemma 2.6

Assume that $(H_{1})$-$(H_{8})$ hold. Then $F:P \rightarrow P$ is a completely continuous operator.

Proof

For convenience, the proof is divided into the following five steps.

Step 1. We show that $F:P\rightarrow P$ is well defined. For any fixed $x\in P$, choose $0< a<1$ such that $a\Vert x\Vert <1$. Then $a[x(t)-\omega (t)]^{*}\leq a\Vert x\Vert <1$, so by (2.4), (2.6), and Lemma 2.4 we have

$$\begin{aligned} g\bigl(t,\bigl[x(t)-\omega(t)\bigr]^{*}\bigr)\leq \biggl( \frac{1}{a} \biggr)^{\lambda_{1}}g\bigl(t,a\bigl[x(t)-\omega(t) \bigr]^{*}\bigr)\leq a^{\lambda_{2}-\lambda _{1}}\Vert x\Vert ^{\lambda_{2}}g(t,1). \end{aligned}$$

Then

$$\begin{aligned} &\int_{0}^{1}G(s,\tau)g\bigl(\tau,\bigl[x(\tau)- \omega(\tau)\bigr]^{*}\bigr)\,d\tau \\ &\quad\leq a^{\lambda_{2}-\lambda_{1}}\Vert x\Vert ^{\lambda_{2}} \int_{0}^{1}G(\tau,\tau )g(\tau,1)\,d\tau= \widetilde{R_{1}}. \end{aligned}$$

(2.14)

Consequently, for any $t\in[0,1]$, we have

$$\begin{aligned} (Fx) (t) =&(1-t)D_{x}+ \int_{0}^{1}G(t,s)\bigl[f\bigl(s,\widetilde {x(s)} \bigr)+q_{+}(s)\bigr]\,ds \\ \leq&D_{x}+ \int_{0}^{1}G(s,s)\bigl[p(s)h\bigl(\widetilde{x(s)} \bigr)+q_{+}(s)\bigr]\,ds \\ \leq&D_{x}+N \int_{0}^{1}G(s,s)\bigl[p(s)+q_{+}(s) \bigr]\,ds< +\infty, \end{aligned}$$

where

$$\begin{aligned} &N=\max_{0\leq\tau\leq R_{1}}h(\tau)+1, \\ &\max\bigl\{ H_{2}^{*}\bigl(\varphi_{2}(x-\omega)\bigr),\forall x\in P\bigr\} =C< +\infty,\qquad R_{1}=C+\widetilde{R_{1}}. \end{aligned}$$

Thus, $F:P\rightarrow P$ is well defined.

Step 2. We show that $F(P)\subset P$. For any $x\in P$, by the definition of the operator F, we obtain $(Fx)(1)=0,(Fx)(0)=D_{x}$. If $\Vert Fx\Vert =D_{x}$, then we have

$$\begin{aligned} (Fx) (t)\geq t(1-t)D_{x}=t(1-t)\Vert Fx\Vert ,\quad t\in[0,1]. \end{aligned}$$

Then $F(P)\subset P$. If $\Vert Fx\Vert >D_{x}$, then there exists $t_{0}\in(0,1)$ such that $\Vert Fx\Vert =(Fx)(t_{0})$. Since $\frac{G(t,s)}{G(t_{0},s)}\geq t(1-t)$, $t,s\in(0,1)\times(0,1)$, we have

$$\begin{aligned} (Fx) (t) =&(1-t)D_{x}+ \int_{0}^{1}\frac {G(t,s)}{G(t_{0},s)}G(t_{0},s) \bigl[f\bigl(s,\widetilde {x(s)}\bigr)+q_{+}(s)\bigr]\,ds \\ \geq&t(1-t) \biggl[(1-t_{0})D_{x}+ \int_{0}^{1}G(t_{0},s)\bigl[f\bigl(s, \widetilde {x(s)}\bigr)+q_{+}(s)\bigr]\,ds \biggr] \\ \geq&t(1-t) (Fx) (t_{0})=t(1-t)\Vert Fx\Vert , \quad t\in[0,1]. \end{aligned}$$

If $\Vert Fx\Vert < D_{x}$, then

$$\begin{aligned} (Fx) (t) =&(1-t)D_{x}+ \int_{0}^{1}\frac {G(t,s)}{G(t_{0},s)}G(t_{0},s) \bigl[f\bigl(s,\widetilde {x(s)}\bigr)+q_{+}(s)\bigr]\,ds \\ \geq&(1-t)D_{x} > t(1-t)\Vert Fx\Vert ,\quad t\in[0,1]. \end{aligned}$$

We also know that

$$\begin{aligned} \varphi_{i,1}(Fx) =& \int_{[0,1]}(1-t)D_{x}\,d\alpha _{i,1}(t)+ \int_{[0,1]} \biggl( \int_{0}^{1}G(t,s)\bigl[f\bigl(s,\widetilde {x(s)} \bigr)+q_{+}(s)\bigr]\,ds \biggr)\,d\alpha_{i,1}(t) \\ =&D_{x} \int_{0}^{1}(1-t) \,d\alpha_{i,1}(t)+ \int_{0}^{1} \biggl[ \int _{[0,1]}G(t,s)\,d\alpha_{i,1}(t) \biggr] \bigl[f \bigl(s,\widetilde {x(s)}\bigr)+q_{+}(s)\bigr]\,ds \\ \geq& 0, \quad i=1,2. \end{aligned}$$

Thus, $F(P)\subset P$.

Step 3. Let $B\subset P$ be any bounded set. We show that $F(B)$ is uniformly bounded. There exists a constant $L>0$ such that $\Vert u\Vert \leq L$ for any $u\in B$. Moreover, for any $u\in B$ and $s\in[0,1]$, we have $[x(s)-\omega(s)]^{*}\leq x(s)\leq \Vert x\Vert \leq L < L+1$. Then, for any $x\in B$ and $s\in[0,1]$, we have $g(s,[x(s)-\omega(s)]^{*})\leq g(s,L+1)\leq(L+1)^{\lambda _{1}}g(s,1)$, and thus

$$\begin{aligned} (Fx) (t) =&(1-t)D_{x}+ \int _{0}^{1}G(t,s)\bigl[f\bigl(s,\widetilde{x(s)} \bigr)+q_{+}(s)\bigr]\,ds \\ \leq&D_{x}+ \int_{0}^{1}G(s,s)\bigl[p(s)h\bigl(\widetilde{x(s)} \bigr)+q_{+}(s)\bigr]\,ds \\ \leq&D_{x}+M \int_{0}^{1}G(s,s)\bigl[p(s)+q_{+}(s) \bigr]\,ds< +\infty, \end{aligned}$$

where

$$\begin{aligned} &M=\max_{0\leq\tau\leq R_{2}}h(\tau)+1,\qquad \max\bigl\{ H_{2}^{*}\bigl( \varphi_{2}(x-\omega)\bigr),\forall x\in B\subset P\bigr\} =C_{1}< + \infty, \\ &R_{2}=C_{1}+(L+1)^{\lambda_{1}} \int_{0}^{1}G(s,s)g(s,1)\,ds. \end{aligned}$$

Therefore, $F(B)$ is uniformly bounded.

Step 4. Let $B\subset P$ be any bounded set. We show that $F(B)$ is equicontinuous on $[0,1]$. For any $(t,s)\in[0,1]\times[0,1]$, $G(t,s)$ is uniformly continuous. Thus, for any $\varepsilon>0$, there exists a constant $\delta=\frac{\varepsilon}{2D_{x}}$ such that, for any $t,t',\in[0,1]$ such that $\vert t-t'\vert <\delta$, we have

$$\begin{aligned} \bigl\vert G(t,s)-G\bigl(t',s\bigr)\bigr\vert < \frac{\varepsilon}{2M\int_{0}^{1}[p(s)+q_{+}(s)]\,ds}. \end{aligned}$$

On the other hand, for any $x\in B$, we have

$$\begin{aligned} &\bigl\vert (Fx) (t)-(Fx) \bigl(t'\bigr)\bigr\vert \\ &\quad\leq\bigl\vert t-t'\bigr\vert D_{x}+ \int _{0}^{1}\bigl\vert G(t,s)-G \bigl(t',s\bigr)\bigr\vert \bigl[f\bigl(s,\widetilde{x(s)} \bigr)+q_{+}(s)\bigr]\,ds \\ &\quad< \delta D_{x}+\frac{\varepsilon}{2M\int_{0}^{1}[p(s)+q_{+}(s)]\,ds}M \int _{0}^{1}\bigl[p(s)+q_{+}(s)\bigr]\,ds \\ &\quad=\frac{\varepsilon}{2D_{x}}D_{x}+\frac{\varepsilon}{2}=\varepsilon . \end{aligned}$$

Thus, $F(B)$ is equicontinuous on $[0,1]$.

Step 5. We show that $F:P\rightarrow P$ is continuous. Assume that $x_{n},x_{0}\in P$ and $\Vert x_{n}-x_{0}\Vert \rightarrow0, n\rightarrow+\infty$. Then there exists a constant $L_{1}>0$ such that $\Vert x_{n}\Vert \leq L_{1},\Vert x_{0}\Vert \leq L_{1}$ $(n=1,2,\ldots)$. Similarly to (2.15), we have $g(s,[x(s)-\omega(s)]^{*})\leq g(s,L_{1}+1)\leq(L_{1}+1)^{\lambda_{1}}g(s,1)$ $(n=1,2,\ldots)$. Then, we have

$$\begin{aligned} &\bigl\vert (Fx_{n}) (t)-(Fx_{0}) (t) \bigr\vert \\ &\quad\leq\bigl\vert (1-t) (D_{x_{n}}-D_{x_{0}})\bigr\vert + \int_{0}^{1}G(s,s)\bigl\vert f\bigl(s,\widetilde {x_{n}(s)}\bigr)-f(s,\widetilde{x_{0}(s)}\bigr\vert \,ds, \end{aligned}$$

where

$$\begin{aligned} D_{x_{n}}=H_{1}^{*} \biggl( \int_{0}^{1}\bigl[\widetilde {x_{n}(s)} \bigr]\,d\alpha_{1}(s) \biggr),\qquad D_{x_{0}}=H_{1}^{*} \biggl( \int _{0}^{1}\bigl[\widetilde{x_{0}(s)} \bigr]\,d\alpha_{1}(s) \biggr). \end{aligned}$$

Set

$$\begin{aligned} &r_{n}(s)=G(s,s)\vert f\bigl(s,\widetilde{x_{n}(s)}\bigr)-f \bigl(s,\widetilde {x_{0}(s)}\bigr), \\ &F(s)=2M_{1}G(s,s)\bigl[p(s)+q_{+}(s)\bigr], \quad s\in (0,1), \end{aligned}$$

where

$$\begin{aligned} &M_{1}=\max_{0\leq\tau\leq R_{3}}h(\tau)+1,\qquad \max\bigl\{ H_{2}^{*}\bigl(\varphi_{2}(x_{n}-\omega)\bigr), \forall x_{n}\in P\bigr\} =C_{2}< +\infty, \\ &R_{3}'= \int_{0}^{1}G(s,\tau)g\bigl(\tau, \bigl[x_{n}(\tau)-\omega(\tau )\bigr]^{*}\bigr)\,d\tau \leq(L_{1}+1)^{\lambda_{1}} \int_{0}^{1}G(s,s)g(s,1)\,ds, \\ &R_{3}=C_{2}+R_{3}'. \end{aligned}$$

It is clear that $\vert r_{n}(s)\vert \leq F(s), s\in (0,1),n=1,2,\ldots$ , and $\{r_{n}(s)\}$ is a sequence of measurable functions in $(0,1)$. By $(H_{8})$ we have

$$\begin{aligned} 0\leq \int_{0}^{1}F(s)\,ds=2M_{1} \int _{0}^{1}G(s,s)\bigl[p(s)+q_{+}(s) \bigr]\,ds< +\infty. \end{aligned}$$

(2.15)

We assert that $r_{n}(s)\rightarrow0$ $(n\rightarrow+\infty)$ for any fixed $s\in (0,1)$. In fact, for any fixed $s\in (0,1)$, noticing the continuity of $f(s,y)$ in y, we have that $f(s,y)$ is uniformly continuous with respect to y in $[0,R_{3}]$; thus, for any $\varepsilon>0$, there exists a constant $\delta>0$ such that, for any $v_{1},v_{2}\in [0,R_{3}]$ such that $\vert v_{1}-v_{2}\vert <\delta$,

$$\begin{aligned} \bigl\vert f(s,v_{1})-f(s,v_{2})\bigr\vert < \frac{\varepsilon}{G(s,s)}. \end{aligned}$$

(2.16)

On the other hand, in view of the continuity of $g(s,x)$ in x, we obtain that $g(s,x)$ is uniformly continuous in x in $[0,L_{1}]$, so for the above $\delta>0$, there exists a constant $\delta_{1}>0$, such that. for any $u_{1},u_{2}\in[0,L_{1}]$ such that $\vert u_{1}-u_{2}\vert <\delta_{1}$,

$$\begin{aligned} \bigl\vert g(s,u_{1})-g(s,u_{2})\bigr\vert < \frac{\delta}{2G(s,s)} . \end{aligned}$$

Since $x_{n}(s)\rightarrow x_{0}(s)$ $(n\rightarrow+\infty)$, there exists a natural number $N_{0}>0$ such that $\vert x_{n}(s)-x_{0}(s)\vert <\delta_{1}$ for $n>N_{0}$. Noting that

$$\begin{aligned} &\bigl\vert \bigl[x_{n}(s)-\omega(s) \bigr]^{*}-\bigl[x_{0}(s)-\omega(s)\bigr]^{*} \bigr\vert \\ &\quad=\biggl\vert \frac{\vert x_{n}(s)-\omega(s)\vert +x_{n}(s)-\omega(s)}{2}-\frac {\vert x_{0}(s)-\omega(s)\vert +x_{0}(s)-\omega(s)}{2}\biggr\vert \\ &\quad=\biggl\vert \frac{\vert x_{n}(s)-\omega(s)\vert -\vert x_{0}(s)-\omega(s)\vert }{2}+\frac {x_{n}(s)-x_{0}(s)}{2}\biggr\vert \\ &\quad\leq\frac{\vert x_{n}(s)-x_{0}(s)\vert }{2}+\frac {\vert x_{n}(s)-x_{0}(s)\vert }{2} \\ &\quad =\bigl\vert x_{n}(s)-x_{0}(s)\bigr\vert < \delta _{1} \end{aligned}$$

and

$$\begin{aligned} &0\leq\bigl[x_{n}(s)-\omega(s)\bigr]^{*}\leq x_{n}(s)\leq L_{1}, \\ &0\leq\bigl[x_{0}(s)-\omega(s)\bigr]^{*}\leq x_{0}(s)\leq L_{1}, \end{aligned}$$

for $n>N_{0}$, we have

$$\begin{aligned} \bigl\vert g\bigl(s,\bigl[x_{n}(s)-\omega(s)\bigr]^{*} \bigr)-g\bigl(s,\bigl[x_{0}(s)-\omega (s)\bigr]^{*}\bigr) \bigr\vert < \frac{\delta}{2G(s,s)} . \end{aligned}$$

(2.17)

By (2.17) we have

$$\begin{aligned} \biggl\vert \int_{0}^{1}G(s,\tau)g\bigl(\tau, \bigl[x_{n}(\tau)-\omega(\tau )\bigr]^{*}\bigr)\,d\tau- \int_{0}^{1}G(s,\tau)g\bigl(\tau, \bigl[x_{0}(\tau)-\omega(\tau )\bigr]^{*}\bigr)\,d\tau\biggr\vert < \frac{\delta}{2}. \end{aligned}$$

(2.18)

Noting that $H_{2}^{*}$ is continuous, for the above $\delta>0$, there exists $\delta_{2}>0$ such that $\vert z_{1}-z_{2}\vert <\delta_{2}$. Then

$$\begin{aligned} \bigl\vert H_{2}^{*}(z_{1})-H_{2}^{*}(z_{2}) \bigr\vert < \frac{\delta}{2}. \end{aligned}$$

Since

$$\begin{aligned} \varphi_{2}(x_{n}-\omega):= \int_{[0,1]}\bigl(x_{n}(s)-\omega (s)\bigr)\,d \alpha_{2}(s), \qquad \varphi_{2}(x_{0}-\omega):= \int_{[0,1]}\bigl(x_{0}(s)-\omega(s)\bigr)\,d \alpha_{2}(s), \end{aligned}$$

and $x_{n}\rightarrow x_{0},n\rightarrow+\infty$, by the Lebesgue dominated convergence theorem we have

$$\begin{aligned} \varphi_{2}(x_{n}-\omega)\rightarrow\varphi _{2}(x_{0}-\omega),\quad n\rightarrow+\infty. \end{aligned}$$

For the above $\delta_{2} >0$, there exists a natural number $N_{1}>0$ such that, for any $n>N_{1}$, we have

$$\begin{aligned} \bigl\vert H_{2}^{*}\bigl(\varphi_{2}(x_{n}- \omega)\bigr)-H_{2}^{*}\bigl(\varphi _{2}(x_{0}- \omega)\bigr)\bigr\vert < \frac{\delta}{2}. \end{aligned}$$

(2.19)

Then it follows from (2.18) and (2.19) that

$$\begin{aligned} &\biggl\vert (1-s)H_{2}^{*}\bigl( \varphi_{2}(x_{n}-\omega)\bigr)+ \int_{0}^{1}G(s,\tau )g\bigl(\tau, \bigl[x_{n}(\tau)-\omega(\tau)\bigr]^{*}\bigr)\,d\tau \\ &\qquad{}-(1-s)H_{2}^{*}\bigl(\varphi_{2}(x_{0}- \omega)\bigr)+ \int_{0}^{1}G(s,\tau )g\bigl(\tau, \bigl[x_{0}(\tau)-\omega(\tau)\bigr]^{*}\bigr)\,d\tau\biggr\vert \\ &\quad < \delta, \end{aligned}$$

that is,

$$\begin{aligned} \bigl\vert \widetilde{x_{n}(s)}-\widetilde{x_{0}(s)}\bigr\vert < \delta. \end{aligned}$$

(2.20)

By (2.16), choose $N=\max\{N_{0},N_{1}\}$. For $n>N$, we have

$$\begin{aligned} &\biggl\vert f \biggl(s,(1-s)H_{2}^{*} \bigl(\varphi_{2}(x_{n}-\omega)\bigr)+ \int_{0}^{1}G(s,\tau)g\bigl(\tau , \bigl[x_{n}(\tau)-\omega(\tau)\bigr]^{*}\bigr)\,d\tau \biggr) \\ &\qquad{}-f \biggl(s,(1-s)H_{2}^{*}\bigl( \varphi_{2}(x_{0}-\omega)\bigr)+ \int_{0}^{1}G(s,\tau )g\bigl(\tau, \bigl[x_{0}(\tau)-\omega(\tau)\bigr]^{*}\bigr)\,d\tau \biggr) \biggr\vert \\ &\quad< \frac{\varepsilon}{G(s,s)}, \end{aligned}$$

that is,

$$\begin{aligned} \bigl\vert f\bigl(s,\widetilde{x_{n}(s)}\bigr)-f\bigl(s, \widetilde{x_{0}(s)}\bigr)\bigr\vert < \frac {\varepsilon}{G(s,s)}. \end{aligned}$$

Consequently, for any fixed $s\in(0,1)$ and for any $\varepsilon>0$, there exists a natural number $N_{2}>0$ such that, for $n>N_{2}$,

$$\begin{aligned} \bigl\vert r_{n}(s)-0\bigr\vert < \varepsilon, \end{aligned}$$

that is, $r_{n}(s)\rightarrow0$ $(n\rightarrow+\infty)$, $s\in(0,1)$.

Since $H_{1}^{*}$ is continuous, for the above $\varepsilon>0$, there exists $\delta_{3}>\delta>0$ such that if $\vert z_{1}-z_{2}\vert <\delta_{3}$, then

$$\begin{aligned} \bigl\vert H_{1}^{*}(z_{1})-H_{1}^{*}(z_{2}) \bigr\vert < \varepsilon. \end{aligned}$$

So by (2.20) we have

$$\begin{aligned} &\biggl\vert H_{1}^{*} \biggl( \int_{0}^{1} \biggl[(1-s)H_{2}^{*} \bigl(\varphi _{2}(x_{n}-\omega)\bigr)+ \int_{0}^{1}G(s,\tau)g\bigl(\tau, \bigl[x_{n}(\tau)-\omega(\tau )\bigr]^{*}\bigr)\,d\tau \biggr]\,d\alpha_{1}(s) \biggr) \\ &\qquad{}-H_{1}^{*} \biggl( \int_{0}^{1} \biggl[(1-s)H_{2}^{*} \bigl(\varphi _{2}(x_{0}-\omega)\bigr)+ \int_{0}^{1}G(s,\tau)g\bigl(\tau, \bigl[x_{0}(\tau)-\omega(\tau )\bigr]^{*}\bigr)\,d\tau \biggr]\,d\alpha_{1}(s) \biggr)\biggr\vert \\ &\quad< \varepsilon, \end{aligned}$$

that is,

$$\begin{aligned} \vert D_{x_{n}}-D_{x_{0}}\vert < \varepsilon. \end{aligned}$$

By the Lebesgue dominated convergence theorem we have

$$\begin{aligned} \Vert Fx_{n}-Fx_{0}\Vert < \varepsilon+\varepsilon=2 \varepsilon , \quad n\rightarrow+\infty. \end{aligned}$$

Then

$$\begin{aligned} \Vert Fx_{n}-Fx_{0}\Vert \rightarrow0,\quad n \rightarrow+\infty. \end{aligned}$$

Therefore, $F:P\rightarrow P$ is continuous. Thus, $F:P\rightarrow P$ is a completely continuous operator. This completes the proof of the lemma. □

Lemma 2.7

Assume that $(H_{1})$-$(H_{8})$ hold. Then $i(F,P_{r^{*}},P)=1$.

Proof

Assume that there exist $\lambda_{0}\geq1$ and $z_{0}\in\partial P_{r^{*}}$ such that $\lambda_{0}z_{0}=Fz_{0}$. Then $z_{0}=\frac{1}{\lambda_{0}}Fz_{0}$ and $0<\frac{1}{\lambda _{0}}\leq1$. We know that $z_{0}(t)\geq t(1-t)\Vert z_{0}\Vert =t(1-t)r^{*},t\in[0,1]$, and $\omega(t)=\int_{0}^{1}G(t,s)q_{-}(s)\,ds\leq t(1-t)\int_{0}^{1}q_{-}(s)\,ds$. Then, for any $t\in[0,1]$,

$$\begin{aligned} z_{0}(t)-\omega(t) \geq& z_{0}(t)-t(1-t) \int_{0}^{1}q_{-}(s)\,ds \\ \geq& t(1-t)r^{*}-t(1-t) \int_{0}^{1}q_{-}(s)\,ds \\ =&t(1-t) \biggl[r^{*}- \int_{0}^{1}q_{-}(s)\,ds \biggr] \geq0. \end{aligned}$$

Applying $z_{0}=\frac{1}{\lambda_{0}}Fz_{0}$, we obtain $\lambda _{0},z_{0}$ such that

$$\begin{aligned} \textstyle\begin{cases} z_{0}''(t)+\frac{1}{\lambda_{0}}[f(t,\widetilde{z_{0}(t)})+ q_{+}(t)]=0, \\ z_{0}(0)=\frac{1}{\lambda_{0}}H_{1}^{*} (\int_{0}^{1}\widetilde {z_{0}(t)}\,dt )=\frac{1}{\lambda_{0}}D_{z_{0}}, \\ z_{0}(1)=0. \end{cases}\displaystyle \end{aligned}$$

(2.21)

Since $z_{0}''(t)\leq0$ for any $t\in(0,1)$, $z_{0}(t)$ is a concave function on $[0,1]$. By the boundary conditions, if $\Vert z_{0}\Vert =z_{0}(0)$, then $z_{0}'(t)\leq0, t\in(0,1)$, and since $z_{0}(0)=\frac{1}{\lambda_{0}}D_{z_{0}}$ is a nonnegative number depending only on $z_{0}$, we have $z_{0}'(0)=0$. Noting that

$$\begin{aligned} \int_{0}^{1}G(s,\tau)g\bigl(\tau, \bigl[z_{0}(\tau)-\omega(\tau)\bigr]^{*}\bigr)\,d\tau \leq& \int_{0}^{1}G(\tau,\tau)g\bigl(\tau, \bigl[z_{0}(\tau)-\omega(\tau)\bigr]^{*}\bigr)\,d\tau \\ =& \int_{0}^{1}G(\tau,\tau)g\bigl(\tau, \bigl[z_{0}(\tau)-\omega(\tau)\bigr]\bigr)\,d\tau \\ \leq& \int_{0}^{1}G(\tau,\tau)g\bigl(\tau,z_{0}( \tau)\bigr)\,d\tau \\ \leq& \int_{0}^{1}G(\tau,\tau)g\bigl(\tau,r^{*} \bigr)\,d\tau \\ \leq&\bigl(r^{*}+1\bigr)^{\lambda_{1}} \int_{0}^{1}G(\tau,\tau)g(\tau,1)\,d\tau , \end{aligned}$$

we get

$$\begin{aligned} \widetilde {z_{0}(s)} =&(1-s)H_{2}^{*} \bigl(\varphi_{2}(z_{0}-\omega)\bigr)+ \int_{0}^{1}G(s,\tau )g\bigl(\tau, \bigl[z_{0}(\tau)-\omega(\tau)\bigr]^{*}\bigr)\,d\tau \\ \leq&H_{2}^{*}\bigl(\varphi_{2}(z_{0}- \omega)\bigr)+\bigl(r^{*}+1\bigr)^{\lambda_{1}} \int _{0}^{1}G(s,s)g(s,1)\,ds. \end{aligned}$$

Then, choosing $t\in(0,1)$ and integrating (2.21) from 0 to t, we have

$$\begin{aligned} z_{0}'(t) =& \int_{0}^{t} z_{0}''(s)\,ds \geq - \int_{0}^{t}\bigl[f\bigl(s,\widetilde{z_{0}(s)} \bigr)+q_{+}(s)\bigr]\,ds \\ \geq&- \int_{0}^{t}\bigl[p(s)h\bigl(\widetilde{z_{0}(s)} \bigr)+q_{+}(s)\bigr]\,ds \\ \geq&- \Bigl[\max_{0\leq\tau\leq R_{0}}h(\tau)+1 \Bigr] \int _{0}^{t}\bigl[p(s)+q_{+}(s)\bigr]\,ds, \end{aligned}$$

where

$$\begin{aligned} R_{0}=H_{2}^{*}\bigl(\varphi_{2}(z_{0}- \omega )\bigr)+\bigl(r^{*}+1\bigr)^{\lambda_{1}} \int_{0}^{1}G(s,s)g(s,1)\,ds. \end{aligned}$$

(2.22)

By $(H_{8})$ we know that $R_{0}\leq R$. So

$$\begin{aligned} -z_{0}'(t) \leq& \Bigl[\max_{0\leq\tau\leq R_{0}}h( \tau)+1 \Bigr] \int _{0}^{t}\bigl[p(s)+q_{+}(s)\bigr]\,ds \\ \leq& \Bigl[\max_{0\leq\tau\leq R}h(\tau)+1 \Bigr] \int_{0}^{t}\bigl[p(s)+q_{+}(s)\bigr]\,ds. \end{aligned}$$

Next, integrating this inequality from 0 to 1, we obtain

$$\begin{aligned} r^{*} =&z_{0}(0)= \int_{0}^{1}-z_{0}'(s)\,ds \leq \Bigl[\max_{0\leq\tau\leq R}h(\tau)+1 \Bigr] \int_{0}^{1}\,ds \int_{0}^{s}\bigl[p(\xi)+q_{+}(\xi)\bigr]\,d \xi \\ \leq& \Bigl[\max_{0\leq\tau\leq R}h(\tau)+1 \Bigr] \int_{0}^{1}\,d\xi \int_{\xi}^{1}\bigl[p(\xi)+q_{+}(\xi) \bigr]\,ds \\ \leq& \Bigl[\max_{0\leq\tau\leq R}h(\tau)+1 \Bigr] \int_{0}^{1}(1-\xi)\bigl[p(\xi )+q_{+}( \xi)\bigr]\,d\xi. \end{aligned}$$

Hence,

$$\begin{aligned} \frac{r^{*}}{\max_{0\leq\tau\leq R}h(\tau)+1}&\leq \int _{0}^{1}(1-\xi)\bigl[p(\xi)+q_{+}( \xi)\bigr]\,d\xi\\ &\leq2 \int_{0}^{1}(1-\xi)\bigl[p(\xi)+q_{+}(\xi )\bigr]\,d\xi, \end{aligned}$$

which is a contradiction with $(H_{8})$. On the other hand, if $\Vert z_{0}\Vert >z_{0}(0)$, then there exists $t_{0}\in (0,1)$ such that

$$\begin{aligned} \Vert z_{0}\Vert =z_{0}(t_{0});\qquad z_{0}'(t_{0})=0,\qquad z_{0}'(t) \geq 0,\quad t\in(0,t_{0}); \qquad z_{0}'(t)\leq0,\quad t \in(t_{0},1). \end{aligned}$$

If $t\in(0,t_{0})$, integrating (2.21) from t to $t_{0}$, we have

$$\begin{aligned} z_{0}'(t) =& \int_{t}^{t_{0}} -z_{0}''(s)\,ds \\ \leq& \int_{t}^{t_{0}}\bigl[f\bigl(s,\widetilde{z_{0}(s)} \bigr)+q_{+}(s)\bigr]\,ds \\ \leq& \int_{t}^{t_{0}}\bigl[p(s)h\bigl(\widetilde{z_{0}(s)} \bigr)+q_{+}(s)\bigr]\,ds \\ \leq& \Bigl[\max_{0\leq\tau\leq R_{0}}h(\tau)+1 \Bigr] \int _{t}^{t_{0}}\bigl[p(s)+q_{+}(s)\bigr]\,ds, \end{aligned}$$

where $R_{0}$ is defined by (2.22), and by $(H_{8})$ we know that $R_{0}\leq R$. So

$$\begin{aligned} z_{0}'(t) \leq& \Bigl[\max _{0\leq\tau\leq R_{0}}h(\tau)+1 \Bigr] \int _{t}^{t_{0}}\bigl[p(s)+q_{+}(s)\bigr]\,ds \\ \leq& \Bigl[\max_{0\leq\tau\leq R}h(\tau)+1 \Bigr] \int_{t}^{t_{0}}\bigl[p(s)+q_{+}(s)\bigr]\,ds. \end{aligned}$$

Next, integrating this inequality from 0 to $t_{0}$, we obtain

$$\begin{aligned} r^{*} =&z_{0}(t_{0})= \int_{0}^{t_{0}}z_{0}'(s)\,ds+z_{0}(0) \\ \leq& \Bigl[\max_{0\leq\tau\leq R}h(\tau)+1 \Bigr] \int_{0}^{t_{0}}\,ds \int _{s}^{t_{0}}\bigl[p(\xi)+q_{+}(\xi) \bigr]\,d\xi+z_{0}(0) \\ \leq& \Bigl[\max_{0\leq\tau\leq R}h(\tau)+1 \Bigr] \int_{0}^{t_{0}}\,d\xi \int _{0}^{\xi}\bigl[p(\xi)+q_{+}(\xi) \bigr]\,ds+z_{0}(0) \\ \leq& \Bigl[\max_{0\leq\tau\leq R}h(\tau)+1 \Bigr] \int_{0}^{t_{0}}\xi \bigl[p(\xi)+q_{+}(\xi) \bigr]\,d\xi+z_{0}(0) \\ \leq&\frac{\max_{0\leq\tau\leq R}h(\tau)+1}{1-t_{0}} \int_{0}^{1}\xi(1-\xi )\bigl[p(\xi)+q_{+}( \xi)\bigr]\,d\xi+z_{0}(0). \end{aligned}$$

Consequently,

$$\begin{aligned} &\frac{r^{*}(1-t_{0})}{\max_{0\leq\tau\leq R}h(\tau)+1}\\ &\quad\leq \int_{0}^{1}\xi (1-\xi)\bigl[p(\xi)+q_{+}( \xi)\bigr]\,d\xi+ \frac{z_{0}(0)(1-t_{0})}{\max_{0\leq\tau \leq R}h(\tau)+1}. \end{aligned}$$

For $t\in(t_{0},1)$, we have

$$\begin{aligned} z_{0}'(t) =& \int_{t_{0}}^{t} z_{0}''(s)\,ds \geq - \int_{t_{0}}^{t}\bigl[f\bigl(s,\widetilde{z_{0}(s)} \bigr)+q_{+}(s)\bigr]\,ds \\ \geq&- \int_{t_{0}}^{t}\bigl[p(s)h\bigl(\widetilde{z_{0}(s)} \bigr)+q_{+}(s)\bigr]\,ds \\ \geq&- \Bigl[\max_{0\leq\tau\leq R_{0}}h(\tau)+1 \Bigr] \int _{t_{0}}^{t}\bigl[p(s)+q_{+}(s)\bigr]\,ds, \end{aligned}$$

where $R_{0}$ is defined by (2.22), and by $(H_{8})$ we know that $R_{0}\leq R$, so

$$\begin{aligned} -z_{0}'(t) \leq& \Bigl[\max _{0\leq\tau\leq R_{0}}h(\tau)+1 \Bigr] \int _{t_{0}}^{t}\bigl[p(s)+q_{+}(s)\bigr]\,ds \\ \leq& \Bigl[\max_{0\leq\tau\leq R}h(\tau)+1 \Bigr] \int_{t_{0}}^{t}\bigl[p(s)+q_{+}(s)\bigr]\,ds. \end{aligned}$$

Next, integrating this inequality from $t_{0}$ to 1, we obtain

$$\begin{aligned} r^{*} =&z_{0}(t_{0})= \int_{t_{0}}^{1}-z_{0}'(s)\,ds \\ \leq& \Bigl[\max_{0\leq\tau\leq R}h(\tau)+1 \Bigr] \int_{t_{0}}^{1}\,ds \int _{t_{0}}^{s}\bigl[p(\xi)+q_{+}(\xi) \bigr]\,d\xi \\ \leq& \Bigl[\max_{0\leq\tau\leq R}h(\tau)+1 \Bigr] \int_{t_{0}}^{1}\,d\xi \int _{\xi}^{1}\bigl[p(\xi)+q_{+}(\xi) \bigr]\,ds \\ \leq& \Bigl[\max_{0\leq\tau\leq R}h(\tau)+1 \Bigr] \int_{t_{0}}^{1}(1-\xi )\bigl[p(\xi)+q_{+}( \xi)\bigr]\,d\xi \\ \leq&\frac{\max_{0\leq\tau\leq R}h(\tau)+1}{t_{0}} \int_{0}^{1}\xi(1-\xi )\bigl[p(\xi)+q_{+}( \xi)\bigr]\,d\xi. \end{aligned}$$

Then,

$$\begin{aligned} \frac{r^{*}t_{0}}{\max_{0\leq\tau\leq R}h(\tau)+1} \leq& \int_{0}^{1}\xi (1-\xi)\bigl[p(\xi)+q_{+}( \xi)\bigr]\,d\xi. \end{aligned}$$

Thus,

$$\begin{aligned} &\frac{r^{*}}{\max_{0\leq\tau\leq R}h(\tau)+1}\\ &\quad \leq 2 \int_{0}^{1}\xi(1-\xi )\bigl[p(\xi)+q_{+}( \xi)\bigr]\,d\xi+ \frac{z_{0}(0)(1-t_{0})}{\max_{0\leq\tau\leq R}h(\tau)+1} \\ &\quad=2 \int_{0}^{1}\xi(1-\xi)\bigl[p(\xi)+q_{+}( \xi)\bigr]\,d\xi+ \frac{\frac{1}{\lambda _{0}}D_{z_{0}}(1-t_{0})}{\max_{0\leq\tau\leq R}h(\tau)+1} \\ &\quad\leq 2 \int_{0}^{1}(1-\xi)\bigl[p(\xi)+q_{+}(\xi) \bigr]\,d\xi+ \frac{D_{z_{0}}}{\max_{0\leq\tau\leq R}h(\tau)+1}, \end{aligned}$$

which is a contradiction with $(H_{8})$. So, by Lemma 2.3, $i(F,P_{r^{*}},P)=1$. This completes the proof of the lemma. □

Lemma 2.8

Assume that $(H_{1})$-$(H_{7})$ and $(H_{9})$ hold. There exists a constant $R^{*}>r^{*}$ such that $i(F,P_{R^{*}},P)=0$.

Proof

We choose constants $\alpha,\beta$, and L such that

$$\begin{aligned}{} [\alpha,\beta] \subset(0,1),\quad L>2 \biggl[\alpha(1-\beta)\max_{0\leq t \leq1} \int_{\alpha}^{\beta}G(t,s)\,ds \biggr]^{-1}. \end{aligned}$$

By $(H_{9})$ there exists $R_{1}^{*}>2r$ such that

$$\begin{aligned} f(t,y)\geq Ly,\quad t\in[\alpha,\beta],y\in\bigl[R_{1}^{*},+\infty \bigr). \end{aligned}$$

(2.23)

On the other hand, by Lemma 2.4 there exists $R_{2}^{*}>R_{1}^{*}$ such that, for $t\in[\alpha,\beta]$ and $x\in[R_{2}^{*},+\infty)$,

$$\begin{aligned} \frac{g(t,x)}{x}\geq\min_{t\in[\alpha,\beta]}\frac {g(t,x)}{x}\geq \frac{1}{\max_{\alpha\leq s\leq\beta} \int_{\alpha }^{\beta} G(s,\tau)\,d\tau}, \end{aligned}$$

that is,

$$\begin{aligned} g(t,x)\geq\frac{1}{\max_{\alpha\leq s\leq\beta} \int _{\alpha}^{\beta} G(s,\tau)\,d\tau}x,\quad t\in[\alpha,\beta],x\in \bigl[R_{2}^{*},+ \infty\bigr). \end{aligned}$$

(2.24)

Let $R^{*}\geq\frac{2R_{2}^{*}}{\alpha(1-\beta)}$. Obviously, $R^{*}>R_{2}^{*}>R_{1}^{*}>2r^{*}$. Thus, $\frac{r^{*}}{R^{*}}<\frac{1}{2}$.

Now we show that $x\ngeq Fx$, $x\in\partial P_{R^{*}}$. Indeed, otherwise, there exists $x_{1}\in\partial P_{R^{*}}$ such that $x_{1}\geq Fx_{1}$. As in the proof of Lemma 2.7, by the definition of $r^{*}$, for any $t\in[\alpha,\beta]$, we have

$$\begin{aligned} &x_{1}(t)-\omega(t) \\ &\quad\geq x_{1}(t)-t(1-t) \int_{0}^{1}q_{-}(s)\,ds \\ &\quad\geq x_{1}(t)-t(1-t) \biggl[\frac{C_{1}\int_{0}^{1}q_{-}(s)\,ds}{C_{0}}+1 \biggr] \\ &\quad= x_{1}(t)-t(1-t)r^{*} \geq x_{1}(t)- \frac{x_{1}(t)}{\Vert x_{1}\Vert }r^{*} =x_{1}(t)-\frac{r^{*}}{R^{*}}x_{1}(t) \geq\frac{1}{2}x_{1}(t) \\ &\quad\geq \frac{1}{2}t(1-t)\Vert x_{1}\Vert \geq \frac{1}{2}R^{*}\alpha(1-\beta )\geq R_{2}^{*}>0. \end{aligned}$$

So, by (2.24), for any $s\in[\alpha,\beta]$, we have

$$\begin{aligned} & \int_{\alpha}^{\beta}G(s,\tau)g\bigl(\tau, \bigl[x_{1}(\tau)-\omega(\tau )\bigr]^{*}\bigr)\,d\tau \\ &\quad\geq \frac{1}{\max_{\alpha\leq s\leq\beta} \int_{\alpha}^{\beta} G(s,\tau)\,d\tau} \int_{\alpha}^{\beta}G(s,\tau)\bigl[x_{1}(\tau)- \omega(\tau )\bigr]^{*}\,d\tau \\ &\quad\geq\frac{1}{2}R^{*}\alpha(1-\beta)\geq R_{2}^{*}>R_{1}^{*}. \end{aligned}$$

Since f is nondecreasing in y, from the last inequality it follows that

$$\begin{aligned} R^{*} \geq& x_{1}(t)\geq Fx_{1}(t) \geq \int_{0}^{1}G(t,s)\bigl[f\bigl(s, \widetilde{x_{1}(s)}\bigr)+q_{+}(s)\bigr]\,ds \\ \geq& \int_{0}^{1}G(t,s)f\bigl(s,\widetilde{x_{1}(s)} \bigr)\,ds \\ \geq& \int_{\alpha}^{\beta}G(t,s)L\widetilde{x_{1}(s)}\,ds \\ \geq& \int_{\alpha}^{\beta}G(t,s)L \int_{\alpha}^{\beta}G(s,\tau)g\bigl(\tau , \bigl[x_{1}(\tau)-\omega(\tau)\bigr]^{*}\bigr)\,d\tau \,ds \\ \geq&\frac{1}{2}L\alpha(1-\beta)R^{*} \int_{\alpha}^{\beta }G(t,s)\,ds, \quad t\in[0,1]. \end{aligned}$$

Then we have

$$\begin{aligned} 2\bigl[L\alpha(1-\beta)\bigr]^{-1}\geq \int_{\alpha}^{\beta }G(t,s)\,ds,\quad t\in[0,1]. \end{aligned}$$

Taking the maximum in the last inequality, we get

$$\begin{aligned} 2\bigl[L\alpha(1-\beta)\bigr]^{-1}\geq\max_{0\leq t\leq1} \int _{\alpha}^{\beta}G(t,s)\,ds. \end{aligned}$$

Consequently,

$$\begin{aligned} L\leq2 \biggl[\alpha(1-\beta)\max_{0\leq t\leq1} \int_{\alpha }^{\beta}G(t,s)\,ds \biggr]^{-1}. \end{aligned}$$

This contradicts to the choice of L. Thus, by Lemma 2.3, $i(F,P_{R^{*}},P)=0$. The proof is complete. □

3 Main results

In this section, we give our main result.

Theorem 3.1

Suppose that $(H_{1})$-$(H_{9})$ are satisfied. Then system (1.1) has at least one positive solution.

Proof

Applying Lemmas 2.7 and 2.8 and the definition of the fixed point index, we have $i(F,P_{R^{*}}\backslash \overline{P_{r^{*}}},P)=-1$. Thus, F has a fixed point $z_{0}$ in $P_{R^{*}}\backslash\overline{P_{r^{*}}}$ with $r^{*}<\Vert z_{0}\Vert <R^{*}$. Since $r^{*}<\Vert z_{0}\Vert$, we have

$$\begin{aligned} z_{0}(t)-\omega(t) \geq& t(1-t)\Vert z_{0} \Vert - \int _{0}^{1}G(t,s)q_{-}(s)\,ds \geq t(1-t)\Vert z_{0}\Vert -t(1-t) \int_{0}^{1}q_{-}(s)\,ds \\ =&t(1-t) \biggl[\Vert z_{0}\Vert - \int_{0}^{1}q_{-}(s)\,ds \biggr] =kt(1-t) \geq0, \quad t\in[0,1], \end{aligned}$$

where $k=\Vert z_{0}\Vert -\int_{0}^{1}q_{-}(s)\,ds>0$.

Choosing $z_{0}\in P_{R^{*}}\backslash\overline{P_{r^{*}}}$, we have

$$\begin{aligned} \varphi_{i}(z_{0}-\omega)=\varphi_{i}(z_{0})- \varphi_{i}(\omega)\geq C_{0}\Vert z_{0}\Vert -\varphi_{i}(\omega),\quad i=1,2. \end{aligned}$$

Since $\varphi_{i}(\omega)\leq C_{1}\Vert \omega \Vert$ and $\omega\in P$, we have $\omega(t)\geq t(1-t)\Vert \omega \Vert$. Consequently, by the above inequalities and the definition of $\omega(t)$ we have

$$\begin{aligned} 0 \leq& t(1-t)\varphi_{i}(\omega)\leq C_{1}t(1-t)\Vert \omega \Vert \leq C_{1}\omega(t) = C_{1} \int_{0}^{1}G(t,s)q_{-}(s)\,ds \\ \leq&C_{1}t(1-t) \int_{0}^{1}q_{-}(s)\,ds. \end{aligned}$$

Consequently, $\varphi_{i}(\omega)\leq C_{1}\int_{0}^{1}q_{-}(s)\,ds$. Then

$$\begin{aligned} \varphi_{i}(z_{0}-\omega)&\geq C_{0}\Vert z_{0}\Vert -\varphi_{i}(\omega)\geq C_{0} \biggl(\frac{C_{1}\int_{0}^{1}q_{-}(s)\,ds}{C_{0}}+1 \biggr)-C_{1} \int _{0}^{1}q_{-}(s)\,ds\\ &=C_{0}>0,\quad i=1,2. \end{aligned}$$

Then from Lemma 2.5 it follows that

$$\begin{aligned} \textstyle\begin{cases} x(t)=z_{0}(t)-\omega(t), \\ y(t)=(1-t)H_{2}^{*}(\varphi_{2}(z_{0}-\omega))+\int _{0}^{1}G(t,s)g(s,(z_{0}(s)-\omega(s)))\,ds \end{cases}\displaystyle \end{aligned}$$

is a positive solution of system (1.1). Thus, we complete the proof of Theorem 3.1. □

Remark 3.1

In comparison with [13] and [16], we consider coupled systems rather than only a single equation, the nonlinearity $f(t,x)$ may be singular at $t=0,1$, and $q(t)$ can have finitely many singularities in $[0,1]$. Moreover, we do not assume that H satisfies merely an asymptotic condition.

Remark 3.2

In comparison with [14], we also consider the coupled system, but our system is singular semipositone. We consider f that need not have a lower bound, and we do not assume that $H_{i}$ satisfy superlinearity conditions at $t=0$ and $t=+\infty$.

Remark 3.3

In comparison with [7], we have more complex integral boundary conditions. In this paper, $H_{i}$ $(i=1,2)$ are not linear, and $\varphi_{i}:C([0,1])\rightarrow\mathbf{R }$ $(i=1,2)$ are linear Stieltjes integrals with signed measures. Thus, in this paper, we allow the map $y\mapsto\varphi(y)$ to be negative even if y is nonnegative. This is very different from paper [4].

4 Example

Example 4.1

Consider the singular system

$$\begin{aligned} &\textstyle\begin{cases} -x''=\frac{t}{10(\pi+4)}y^{2}\arctan y-\frac{2}{2+3\sqrt [3]{4}} \{\frac{1}{\sqrt{t}}+\frac{1}{\sqrt[3]{(t-\frac {1}{2})^{2}}} \}, \quad t\in(0,1), \\ -y''=\frac{25x^{\frac{3}{2}}}{199^{2}t(1-t)}, \quad t\in (0,1), & \\ x(0)=H_{1}(\varphi_{1}(y)), \qquad x(1)=0, \\ y(0)=H_{2}(\varphi_{2}(x)),\qquad y(1)=0, \end{cases}\displaystyle \end{aligned}$$

(4.1)

$$\begin{aligned} &\varphi_{1}(y)=\frac{1}{2}y \biggl(\frac{1}{3} \biggr)- \frac {1}{5}y \biggl(\frac{2}{3} \biggr)= \underbrace{ \frac{1}{3}y \biggl(\frac{1}{3} \biggr)-\frac{1}{5}y \biggl( \frac {2}{3} \biggr)}_{\varphi_{1,1}(y)}+\underbrace{\frac{1}{6}y \biggl( \frac {1}{3} \biggr)}_{\varphi_{1,2}(y)}, \end{aligned}$$

(4.2)

$$\begin{aligned} &\varphi_{2}(x)=\frac{1}{3}x \biggl(\frac{1}{2} \biggr)- \frac {1}{8}x \biggl(\frac{7}{10} \biggr)= \underbrace{ \frac{1}{6}x \biggl(\frac{1}{2} \biggr)-\frac{1}{8}x \biggl( \frac {7}{10} \biggr)}_{\varphi_{2,1}(x)}+\underbrace{\frac{1}{6}x \biggl( \frac {1}{2} \biggr)}_{\varphi_{2,2}(x)}. \end{aligned}$$

(4.3)

By (4.2) and (4.3) we know that $\varphi_{1},\varphi_{2}$ satisfy $(H_{5})$-$(H_{7})$, and $C_{0}=\frac{1}{27},C_{1}=\frac{7}{10},D_{0}=\frac{1}{24},D_{1}=\frac {11}{24}$. Define $H_{1},H_{2}$ by

$$\begin{aligned} H_{1}(z):=z^{2}+z,\qquad H_{2}(z):=z^{\frac{1}{3}},\quad t \in(-\infty ,+\infty). \end{aligned}$$

We know that $H_{1}$ is not superlinear at $t=0$ and does not satisfy an asymptotic condition and that $H_{2}$ is not superlinear at $t=0$ and $t=+\infty$. Then system (4.1) has at least one positive solution on $C[0,1]\cap C^{2}(0,1)\times C[0,1]\cap C^{2}(0,1)$. Indeed, choose

$$\begin{aligned} &p(t)=\frac{t}{10(\pi+4)},\qquad h(y)=y^{2}\arctan y,\\ & q_{-}(t)= \frac {2}{2+3\sqrt[3]{4}} \biggl\{ \frac{1}{\sqrt{t}}+\frac{1}{\sqrt[3]{(t-\frac {1}{2})^{2}}} \biggr\} , \\ &q_{+}(t)\equiv0,\qquad g(t,x)=\frac{25x^{\frac {3}{2}}}{199^{2}t(1-t)},\qquad \lambda_{1}=2,\qquad \lambda_{2}=\frac{3}{2}. \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} &r^{*}=\frac{C_{1}\int_{0}^{1}q_{-}(t)\,dt}{C_{0}}+1=\frac {194}{5}, \\ & 2 \int_{0}^{1}(1-t)\bigl[p(t)+q_{+}(t) \bigr]\,dt=\frac{1}{30(\pi+4)}\approx 0.0047, \end{aligned} \end{aligned}$$

(4.4)

$$\begin{aligned} &\bigl(r^{*}+1\bigr)^{\lambda_{1}} \int_{0}^{1}G(s,s)g(s,1)\,ds=1, \end{aligned}$$

(4.5)

and thus

$$\begin{aligned} \widetilde{g}= \int_{0}^{1}G(s,s)g(s,1)\,ds= \biggl( \frac {199}{5} \biggr)^{-2}= \biggl(\frac{5}{199} \biggr)^{2}\approx0.0006, \end{aligned}$$

and $0\leq t(1-t)\Vert x\Vert \leq x(t)\leq \Vert x\Vert ,x\in P$. If $x\in [t(1-t)r^{*},r^{*}],t\in[0,1]$, then

$$\begin{aligned} \varphi_{2}(x-\omega)\leq\varphi_{2}(x)\leq D_{1} \Vert x\Vert =\frac{2{,}134}{120}\approx17.7833. \end{aligned}$$

Consequently,

$$\begin{aligned} & H_{2}^{*}\bigl(\varphi_{2}(x-\omega)\bigr)\leq H_{2} \biggl(\frac {2{,}134}{120} \biggr)= \biggl(\frac{2{,}134}{120} \biggr)^{\frac{1}{3}}\approx 2.6102, \\ & y(t)=(1-t)H_{2}^{*}\bigl(\varphi _{2}(x-\omega)\bigr)+ \int_{0}^{1}G(t,s)g\bigl(s,\bigl[x(s)-\omega(s) \bigr]^{*}\bigr)\,ds \\ &\phantom{y(t)}\leq \biggl(\frac{2{,}134}{120} \biggr)^{\frac{1}{3}}+ \int _{0}^{1}G(t,s)g\bigl(s,x(s)\bigr)\,ds \\ &\phantom{y(t)}\leq \biggl(\frac{2{,}134}{120} \biggr)^{\frac{1}{3}}+ \int _{0}^{1}G(t,s)g\bigl(s,r^{*}\bigr)\,ds \\ &\phantom{y(t)}\leq\biggl(\frac{2{,}134}{120} \biggr)^{\frac{1}{3}}+\bigl(r^{*} \bigr)^{\lambda _{1}} \int_{0}^{1}G(t,s)g(s,1)\,ds \\ &\phantom{y(t)}= \biggl(\frac{2{,}134}{120} \biggr)^{\frac{1}{3}}+ \biggl(\frac{194}{5} \biggr)^{2}\cdot \biggl(\frac{5}{199} \biggr)^{2} \approx3.5606. \end{aligned}$$

(4.6)

Then $\varphi_{1}(y)\leq C_{1}\Vert y\Vert \leq\frac{7}{10}\times[(\frac {2{,}134}{120})^{\frac{1}{3}}+(\frac{194}{5})^{2}\cdot(\frac {5}{199})^{2}]\approx2.4924$, and

$$\begin{aligned} H_{1}^{*}\bigl(\varphi_{1}(y) \bigr) \leq &H_{1} \biggl(\frac{7}{10}\times \biggl[ \biggl( \frac{2{,}134}{120} \biggr)^{\frac{1}{3}}+ \biggl(\frac{194}{5} \biggr)^{2}\cdot \biggl(\frac {5}{199} \biggr)^{2} \biggr] \biggr) \\ =& \biggl\{ \frac {7}{10}\times \biggl[ \biggl(\frac{2{,}134}{120} \biggr)^{\frac{1}{3}}+ \biggl(\frac{194}{5} \biggr)^{2}\cdot \biggl(\frac{5}{199} \biggr)^{2} \biggr] \biggr\} ^{2} \\ &{}+\frac{7}{10}\times \biggl[ \biggl(\frac {2{,}134}{120} \biggr)^{\frac{1}{3}}+ \biggl(\frac{194}{5} \biggr)^{2}\cdot \biggl(\frac{5}{199} \biggr)^{2} \biggr] \\ \approx& 8.7045. \end{aligned}$$

By (4.5) and (4.6) we have

$$\begin{aligned} &\max\bigl\{ H_{2}^{*}\bigl( \varphi_{2}(x-\omega )\bigr)+\bigl(r^{*}+1 \bigr)^{\lambda_{1}}\widetilde{g},x\in\bigl[t(1-t)r^{*},r^{*} \bigr],t\in [0,1]\bigr\} \\ &\quad\leq \biggl(\frac{2{,}134}{120} \biggr)^{\frac {1}{3}}+1\approx3.6102. \end{aligned}$$

Then

$$\begin{aligned} &R= \biggl(\frac{2{,}134}{120} \biggr)^{\frac{1}{3}}+1\approx3.6102, \\ &\max_{0\leq\tau\leq R}h(\tau)= \biggl( \biggl(\frac {2{,}134}{120} \biggr)^{\frac{1}{3}}+1 \biggr)^{2}\arctan \biggl( \biggl( \frac {2{,}134}{120} \biggr)^{\frac{1}{3}}+1 \biggr)\approx971.2290, \\ &\frac{r^{*}}{\max_{0\leq\tau\leq R}h(\tau)+1}=\frac{\frac {194}{5}}{((\frac{2{,}134}{120})^{\frac{1}{3}}+1)^{2}\arctan((\frac {2{,}134}{120})^{\frac{1}{3}}+1)+1}\approx0.0399, \\ &\frac{D_{x_{0}}}{\max_{0\leq\tau \leq R}h(\tau)+1} \\ &\quad\leq \frac{\{\frac{7}{10}\times[(\frac {2{,}134}{120})^{\frac{1}{3}}+(\frac{194}{5})^{2}\cdot(\frac {5}{199})^{2}]\}^{2}+\frac{7}{10}\times[(\frac{2{,}134}{120})^{\frac {1}{3}}+(\frac{194}{5})^{2}\cdot(\frac{5}{199})^{2}]}{((\frac {2{,}134}{120})^{\frac{1}{3}}+1)^{2}\arctan((\frac{2{,}134}{120})^{\frac {1}{3}}+1)+1} \\ &\quad\approx 0.0090. \end{aligned}$$

(4.7)

By (4.4), (4.7), and the last inequality we get that condition $(H_{8})$ holds.

Thus, $(H_{1})$-$(H_{8})$ hold. Therefore, by Theorem 3.1 system (4.1) has at least one positive solution on $C[0,1]\cap C^{2}(0,1)\times C[0,1]\cap C^{2}(0,1)$.

Remark 4.1

In Example 4.1, even if we consider only one equation

$$\begin{aligned} \textstyle\begin{cases} -x''=\frac{t}{10(\pi+4)}x^{2}\arctan x-\frac{2}{2+3\sqrt [3]{4}} \{\frac{1}{\sqrt{t}}+\frac{1}{\sqrt[3]{(t-\frac {1}{2})^{2}}} \}, \quad t\in(0,1), \\ x(0)=H(\varphi_{1}(x)),\qquad x(1)=0 \end{cases}\displaystyle \end{aligned}$$

(4.8)

the function $H(z)=z^{2}+z$ does not satisfy the key condition H of [13], that is, there is a number $C_{2}\geq0$ such that

$$\begin{aligned} \lim_{z\rightarrow+\infty}\frac{\vert H(z)-C_{2}z\vert }{z}=0. \end{aligned}$$

So [13] cannot deal with the problem.

Remark 4.2

In Example 4.1, the nonlinearity term f has singularity at $t=0$ and $t=\frac{1}{2}$. Moreover, f can tend to negative infinity as $t\rightarrow0$ or $t\rightarrow\frac{1}{2}$, which implies that f need not have a lower bound. So, Example 4.1 well demonstrates this point. In Example 4.1, if $q(t)\equiv0$, then $f,g:[0,1]\times[0,+\infty)\rightarrow[0,+\infty )$ are continuous. Consider the system

$$\begin{aligned} \textstyle\begin{cases} -x''=\frac{t}{10(\pi+4)}y^{2}\arctan y, \quad t\in(0,1), \\ -y''=\frac{25x^{\frac{3}{2}}}{199^{2}t(1-t)},\quad t\in (0,1), \\ x(0)=H_{1}(\varphi_{1}(y)), \qquad x(1)=0, \\ y(0)=H_{2}(\varphi_{2}(x)), \qquad y(1)=0. \end{cases}\displaystyle \end{aligned}$$

(4.9)

Let $H_{1}(z):=z^{2}+z$ and $H_{2}(z):=z^{\frac{1}{3}}$. We know that $H_{1}$ is not superlinear at $t=0$ and $H_{2}$ is not superlinear at $t=0$ and $t=+\infty$. Then, these do not satisfy the condition for $H_{i}$ $(i=1,2)$ in [14], that is,

$$\begin{aligned} \lim_{z\rightarrow0^{+}}\frac{\vert H_{i}(z)\vert }{z}=0,\quad i=1,2 \end{aligned}$$

and

$$\begin{aligned} \lim_{z\rightarrow+\infty}\frac{\vert H_{i}(z)\vert }{z}=+\infty. \end{aligned}$$

So [14] cannot deal with the problem.