Positive solutions for a system of semipositone coupled fractional boundary value problems

Abstract

We study the existence of positive solutions for a system of nonlinear Riemann-Liouville fractional differential equations with sign-changing nonlinearities, subject to coupled integral boundary conditions.

1 Introduction

Fractional differential equations describe many phenomena in various fields of engineering and scientific disciplines such as physics, biophysics, chemistry, biology, economics, control theory, signal and image processing, aerodynamics, viscoelasticity, electromagnetics, and so on (see [1–6]). Integral boundary conditions arise in thermal conduction problems, semiconductor problems and hydrodynamic problems.

We consider the system of nonlinear fractional differential equations

$$(\mathrm{S})\quad \left \{ \textstyle\begin{array}{l}D_{0+}^{\alpha}u(t)+\lambda f(t,u(t),v(t))=0,\quad t\in (0,1), n-1< \alpha\le n, \\ D_{0+}^{\beta}v(t)+\mu g(t,u(t),v(t))=0,\quad t\in(0,1), m-1< \beta\le m, \end{array}\displaystyle \right . $$

with the coupled integral boundary conditions

$$(\mathrm{BC})\quad \left \{ \textstyle\begin{array}{l}u(0)=u'(0)=\cdots=u^{(n-2)}(0)=0,\qquad u(1)= \int_{0}^{1} v(s)\,dH(s), \\ v(0)=v'(0)=\cdots=v^{(m-2)}(0)=0, \qquad v(1)= \int_{0}^{1} u(s)\,dK(s), \end{array}\displaystyle \right . $$

where $n, m\in\mathbb{N}$, $n, m\ge3$, $D_{0+}^{\alpha}$, and $D_{0+}^{\beta}$ denote the Riemann-Liouville derivatives of orders α and β, respectively, the integrals from (BC) are Riemann-Stieltjes integrals, and f, g are sign-changing continuous functions (that is, we have a so-called system of semipositone boundary value problems). These functions may be nonsingular or singular at $t=0$ and/or $t=1$. The boundary conditions above include multi-point and integral boundary conditions and sum of these in a single framework.

We present intervals for parameters λ and μ such that the above problem (S)-(BC) has at least one positive solution. By a positive solution of problem (S)-(BC) we mean a pair of functions $(u,v)\in C([0,1])\times C([0,1])$ satisfying (S) and (BC) with $u(t)\ge0$, $v(t)\ge0$ for all $t\in[0,1]$ and $u(t)>0$, $v(t)>0$ for all $t\in(0,1)$. In the case when f and g are nonnegative, problem (S)-(BC) has been investigated in [7] by using the Guo-Krasnosel’skii fixed point theorem, and in [8] where $\lambda=\mu=1$ and $f(t,u,v)$ and $g(t,u,v)$ are replaced by $\tilde{f}(t,v)$ and $\tilde{g}(t,u)$, respectively (denoted by (S̃)). In [8], the authors study two cases: f and g are nonsingular and singular functions and they used some theorems from the fixed point index theory and the Guo-Krasnosel’skii fixed point theorem. The systems (S) and (S̃) with uncoupled boundary conditions

$$(\widetilde{\mathrm{BC}})\quad \left \{ \textstyle\begin{array}{l}u(0)=u'(0)=\cdots=u^{(n-2)}(0)=0,\qquad u(1)= \int_{0}^{1} u(s)\,dH(s), \\ v(0)=v'(0)=\cdots=v^{(m-2)}(0)=0, \qquad v(1)= \int_{0}^{1} v(s)\,dK(s), \end{array}\displaystyle \right . $$

were investigated in [9] (problem (S)-(BC̃) with f, g nonnegative), in [10] (problem (S̃)-(BC̃) with f, g nonnegative, singular or not), and in [11] (problem (S)-(BC̃) with f, g sign-changing functions). We also mention paper [12], where the authors studied the existence and multiplicity of positive solutions for system (S) with $\alpha=\beta$, $\lambda=\mu$, and the boundary conditions $u^{(i)}(0)=v^{(i)}(0)=0$, $i=0,\ldots ,n-2$, $u(1)=av(\xi)$, $v(1)=b u(\eta)$, $\xi, \eta\in(0,1)$, with $\xi , \eta\in(0,1)$, $0< ab\xi\eta<1$, and f, g are sign-changing nonsingular or singular functions.

The paper is organized as follows. Section 2 contains some preliminaries and lemmas. The main results are presented in Section 3, and finally in Section 4 some examples are given to support the new results.

2 Auxiliary results

We present here the definitions of Riemann-Liouville fractional integral and Riemann-Liouville fractional derivative and then some auxiliary results that will be used to prove our main results.

Definition 2.1

The (left-sided) fractional integral of order $\alpha>0$ of a function $f:(0,\infty)\to\mathbb{R}$ is given by

$$\bigl(I_{0+}^{\alpha}f\bigr) (t)= \frac{1}{\Gamma(\alpha)} \int_{0}^{t}(t-s)^{\alpha -1}f(s) \,ds,\quad t>0, $$

provided the right-hand side is pointwise defined on $(0,\infty)$, where $\Gamma(\alpha)$ is the Euler gamma function defined by $\Gamma (\alpha)= \int_{0}^{\infty}t^{\alpha-1}e^{-t} \,dt$, $\alpha>0$.

Definition 2.2

The Riemann-Liouville fractional derivative of order $\alpha\ge0$ for a function $f:(0,\infty)\to\mathbb{R}$ is given by

$$\bigl(D_{0+}^{\alpha}f\bigr) (t)= \biggl( \frac{d}{dt} \biggr)^{n} \bigl(I_{0+}^{n-\alpha }f \bigr) (t)= \frac{1}{\Gamma(n-\alpha)} \biggl( \frac{d}{dt} \biggr)^{n} \int_{0}^{t} \frac{f(s)}{(t-s)^{\alpha-n+1}} \,ds, \quad t>0, $$

where $n=\lfloor\alpha\rfloor+1$, provided that the right-hand side is pointwise defined on $(0,\infty)$.

The notation $\lfloor\alpha\rfloor$ stands for the largest integer not greater than α. If $\alpha=m\in\mathbb{N}$ then $D_{0+}^{m}f(t)=f^{(m)}(t)$ for $t>0$, and if $\alpha=0$ then $D_{0+}^{0}f(t)=f(t)$ for $t>0$.

We consider now the fractional differential system

$$ \left \{ \textstyle\begin{array}{l}D_{0+}^{\alpha}u(t)+\tilde{x}(t)=0,\quad t\in(0,1), n-1< \alpha\le n, \\ D_{0+}^{\beta}v(t)+\tilde{y}(t)=0,\quad t\in(0,1), m-1< \beta\le m, \end{array}\displaystyle \right . $$

(1)

with the coupled integral boundary conditions

$$ \left \{ \textstyle\begin{array}{l} u(0)=u'(0)=\cdots=u^{(n-2)}(0)=0,\qquad u(1)= \int_{0}^{1}v(s) \,dH(s), \\ v(0)=v'(0)=\cdots=v^{(m-2)}(0)=0,\qquad v(1)= \int_{0}^{1}u(s) \,dK(s), \end{array}\displaystyle \right . $$

(2)

where $n, m\in\mathbb{N}$, $n, m\ge3$, and $H, K:[0,1]\to\mathbb{R}$ are functions of bounded variation.

Lemma 2.1

([7])

If $H, K:[0,1]\to\mathbb{R}$ are functions of bounded variations, $\Delta=1- (\int_{0}^{1}\tau^{\alpha-1} \,dK(\tau) ) (\int_{0}^{1}\tau^{\beta -1} \,dH(\tau) )\neq0$ and $\tilde{x}, \tilde{y}\in C(0,1)\cap L^{1}(0,1)$, then the pair of functions $(u,v)\in C([0,1])\times C([0,1])$ given by

$$ \left \{ \textstyle\begin{array}{l}u(t)= \int_{0}^{1}G_{1}(t,s)\tilde{x}(s) \,ds+ \int _{0}^{1}G_{2}(t,s)\tilde{y}(s) \,ds,\quad t\in[0,1], \\ v(t)= \int_{0}^{1}G_{3}(t,s)\tilde{y}(s) \,ds+ \int_{0}^{1}G_{4}(t,s)\tilde{x}(s) \,ds, \quad t\in[0,1], \end{array}\displaystyle \right . $$

(3)

where

$$ \left \{ \textstyle\begin{array}{l} G_{1}(t,s)=g_{1}(t,s)+ \frac{t^{\alpha-1}}{\Delta} ( \int_{0}^{1}\tau^{\beta -1}\,dH(\tau) ) ( \int_{0}^{1}g_{1}(\tau,s)\,dK(\tau) ), \\ G_{2}(t,s)= \frac{t^{\alpha-1}}{\Delta} \int_{0}^{1}g_{2}(\tau,s)\,dH(\tau), \\ G_{3}(t,s)=g_{2}(t,s)+ \frac{t^{\beta-1}}{\Delta} ( \int_{0}^{1}\tau^{\alpha -1}\,dK(\tau) ) ( \int_{0}^{1}g_{2}(\tau,s)\,dH(\tau) ), \\ G_{4}(t,s)= \frac{t^{\beta-1}}{\Delta} \int_{0}^{1}g_{1}(\tau,s)\,dK(\tau),\quad \forall t, s\in[0,1] \end{array}\displaystyle \right . $$

(4)

and

$$ \left \{ \textstyle\begin{array}{l} g_{1}(t,s)= \frac{1}{\Gamma(\alpha)} \left\{ \textstyle\begin{array}{l@{\quad}l} t^{\alpha-1}(1-s)^{\alpha-1}-(t-s)^{\alpha-1},& 0\le s\le t\le1,\\ t^{\alpha-1}(1-s)^{\alpha-1},& 0\le t\le s\le1, \end{array}\displaystyle \right . \\ g_{2}(t,s)= \frac{1}{\Gamma(\beta)} \left\{ \textstyle\begin{array}{l@{\quad}l} t^{\beta-1}(1-s)^{\beta-1}-(t-s)^{\beta-1},& 0\le s\le t\le1, \\ t^{\beta-1}(1-s)^{\beta-1},& 0\le t\le s\le1, \end{array}\displaystyle \right . \end{array}\displaystyle \right . $$

(5)

is solution of problem (1)-(2).

Lemma 2.2

The functions $g_{1}$, $g_{2}$ given by (5) have the properties:

(a):

$g_{1}, g_{2}:[0,1]\times[0,1]\to\mathbb{R}_{+}$ are continuous functions, and $g_{1}(t,s)>0$, $g_{2}(t,s)>0$ for all $(t,s)\in(0,1)\times(0,1)$.

(b):

$g_{1}(t,s)\le h_{1}(s)$, $g_{2}(t,s)\le h_{2}(s)$ for all $(t,s)\in [0,1]\times[0,1]$, where $h_{1}(s)=\frac{s(1-s)^{\alpha-1}}{\Gamma(\alpha -1)}$ and $h_{2}(s)=\frac{s(1-s)^{\beta-1}}{\Gamma(\beta-1)}$ for all $s\in[0,1]$.

(c):

$g_{1}(t,s)\ge k_{1}(t)h_{1}(s)$, $g_{2}(t,s)\ge k_{2}(t)h_{2}(s)$ for all $(t,s)\in[0,1]\times[0,1]$, where

$$\begin{aligned}& k_{1}(t)=\min \biggl\{ \frac{(1-t)t^{\alpha-2}}{\alpha -1},\frac{t^{\alpha-1}}{\alpha-1} \biggr\} =\left \{ \textstyle\begin{array}{l@{\quad}l}\frac{t^{\alpha-1}}{\alpha-1},&0\le t\le\frac{1}{2}, \\ \frac{(1-t)t^{\alpha-2}}{\alpha-1},&\frac{1}{2}\le t\le1, \end{array}\displaystyle \right . \\& k_{2}(t)=\min \biggl\{ \frac{(1-t)t^{\beta-2}}{\beta-1},\frac{t^{\beta -1}}{\beta-1} \biggr\} = \left \{ \textstyle\begin{array}{l@{\quad}l}\frac{t^{\beta-1}}{\beta-1},&0\le t\le\frac{1}{2}, \\ \frac{(1-t)t^{\beta-2}}{\beta-1},&\frac{1}{2}\le t\le1. \end{array}\displaystyle \right . \end{aligned}$$

(d):

For any $(t,s)\in[0,1]\times[0,1]$, we have

$$g_{1}(t,s)\le\frac{(1-t)t^{\alpha-1}}{\Gamma(\alpha-1)}\le\frac{t^{\alpha -1}}{\Gamma(\alpha-1)},\qquad g_{2}(t,s)\le\frac{(1-t)t^{\beta-1}}{\Gamma(\beta-1)}\le\frac{t^{\beta -1}}{\Gamma(\beta-1)}. $$

For the proof of Lemma 2.2(a) and (b) see [13], for the proof of Lemma 2.2(c) see [11], and the proof of Lemma 2.2(d) is based on the relations $g_{1}(t,s)=g_{1}(1-s,1-t)$, $g_{2}(t,s)=g_{2}(1-s,1-t)$, and relations (b) above.

Lemma 2.3

([7])

If $H, K:[0,1]\to\mathbb{R}$ are nondecreasing functions, and $\Delta>0$, then $G_{i}$, $i=1,\ldots,4$ given by (4) are continuous functions on $[0,1]\times[0,1]$ and satisfy $G_{i}(t,s)\ge0$ for all $(t,s)\in[0,1]\times[0,1]$, $i=1,\ldots,4$. Moreover, if $\tilde{x}, \tilde{y}\in C(0,1)\cap L^{1}(0,1)$ satisfy $\tilde{x}(t)\ge0$, $\tilde{y}(t)\ge0$ for all $t\in(0,1)$, then the solution $(u,v)$ of problem (1)-(2) given by (3) satisfies $u(t)\ge0$, $v(t)\ge0$ for all $t\in[0,1]$.

Lemma 2.4

Assume that $H, K:[0,1]\to\mathbb{R}$ are nondecreasing functions, $\Delta>0$, $\int_{0}^{1}\tau^{\alpha-1}(1-\tau) \,dK(\tau)>0$, $\int_{0}^{1}\tau^{\beta-1}(1-\tau) \,dH(\tau)>0$. Then the functions $G_{i}$, $i=1,\ldots,4$ satisfy the inequalities:

(a₁):

$G_{1}(t,s)\le\sigma_{1} h_{1}(s)$, $\forall(t,s)\in[0,1]\times [0,1]$, where

$$\sigma_{1}=1+ \frac{1}{\Delta}\bigl(K(1)-K(0)\bigr) \int_{0}^{1}\tau^{\beta-1} \,dH(\tau)>0. $$

(a₂):

$G_{1}(t,s)\le\delta_{1} t^{\alpha-1}$, $\forall(t,s)\in [0,1]\times[0,1]$, where

$$\delta_{1}= \frac{1}{\Gamma(\alpha-1)} \biggl[1+ \frac{1}{\Delta} \biggl( \int _{0}^{1}\tau^{\beta-1} \,dH(\tau) \biggr) \biggl( \int_{0}^{1}(1-\tau)\tau^{\alpha -1} \,dK(\tau) \biggr) \biggr]>0. $$

(a₃):

$G_{1}(t,s)\ge\varrho_{1} t^{\alpha-1} h_{1}(s)$, $(t,s)\in [0,1]\times[0,1]$, where

$$\varrho_{1}= \frac{1}{\Delta} \biggl( \int_{0}^{1}\tau^{\beta-1} \,dH(\tau) \biggr) \biggl( \int_{0}^{1} k_{1}(\tau) \,dK(\tau) \biggr)>0. $$

(b₁):

$G_{2}(t,s)\le\sigma_{2} h_{2}(s)$, $\forall (t,s)\in[0,1]\times [0,1]$, where $\sigma_{2}=\frac{1}{\Delta}(H(1)-H(0))>0$.

(b₂):

$G_{2}(t,s)\le\delta_{2} t^{\alpha-1}$, $\forall(t,s)\in [0,1]\times[0,1]$, where $\delta_{2}=\frac{1}{\Delta\Gamma(\beta-1)}\int _{0}^{1}(1-\tau)\tau^{\beta-1} \,dH(\tau)>0$.

(b₃):

$G_{2}(t,s)\ge\varrho_{2} t^{\alpha-1}h_{2}(s)$, $\forall(t,s)\in [0,1]\times[0,1]$, where $\varrho_{2}=\frac{1}{\Delta}\int_{0}^{1}k_{2}(\tau) \,dH(\tau)>0$.

(c₁):

$G_{3}(t,s)\le\sigma_{3} h_{2}(s)$, $\forall(t,s)\in[0,1]\times [0,1]$, where

$$\sigma_{3}=1+ \frac{1}{\Delta}\bigl(H(1)-H(0)\bigr) \int_{0}^{1}\tau^{\alpha-1} \,dK(\tau)>0. $$

(c₂):

$G_{3}(t,s)\le\delta_{3} t^{\beta-1}$, $\forall(t,s)\in[0,1]\times [0,1]$, where

$$\delta_{3}= \frac{1}{\Gamma(\beta-1)} \biggl[1+ \frac{1}{\Delta} \biggl( \int _{0}^{1}\tau^{\alpha-1} \,dK(\tau) \biggr) \biggl( \int_{0}^{1}(1-\tau)\tau^{\beta -1} \,dH(\tau) \biggr) \biggr]>0. $$

(c₃):

$G_{3}(t,s)\ge\varrho_{3} t^{\beta-1} h_{2}(s)$, $\forall(t,s)\in [0,1]\times[0,1]$, where

$$\varrho_{3}= \frac{1}{\Delta} \biggl( \int_{0}^{1}\tau^{\alpha-1} \,dK(\tau) \biggr) \biggl( \int_{0}^{1} k_{2}(\tau) \,dH(\tau) \biggr)>0. $$

(d₁):

$G_{4}(t,s)\le\sigma_{4} h_{1}(s)$, $\forall(t,s)\in[0,1]\times [0,1]$, where $\sigma_{4}=\frac{1}{\Delta}(K(1)-K(0))>0$.

(d₂):

$G_{4}(t,s)\le\delta_{4} t^{\beta-1}$, $\forall(t,s)\in[0,1]\times [0,1]$, where $\delta_{4}=\frac{1}{\Delta\Gamma(\alpha-1)}\int_{0}^{1}(1-\tau )\tau^{\alpha-1} \,dK(\tau)>0$.

(d₃):

$G_{4}(t,s)\ge\varrho_{4} t^{\beta-1} h_{1}(s)$, $\forall(t,s)\in [0,1]\times[0,1]$, where $\varrho_{4}=\frac{1}{\Delta}\int_{0}^{1} k_{1}(\tau) \,dK(\tau)>0$.

Proof

From the assumptions of this lemma, we obtain

$$\begin{aligned}& \int_{0}^{1}\tau^{\alpha-1} \,dK(\tau)\ge \int_{0}^{1}\tau^{\alpha -1}(1-\tau) \,dK( \tau)>0, \\& \int_{0}^{1}(1-\tau)\tau^{\alpha-2} \,dK(\tau) \ge \int_{0}^{1}(1-\tau)\tau^{\alpha -1} \,dK( \tau)>0, \\& \int_{0}^{1} k_{1}(\tau) \,dK(\tau)\ge \frac{1}{\alpha-1} \int_{0}^{1}\tau^{\alpha -1}(1-\tau) \,dK( \tau)>0, \\& \int_{0}^{1}\tau^{\beta-1} \,dH(\tau)\ge \int_{0}^{1}\tau^{\beta-1}(1-\tau) \,dH( \tau)>0, \\& \int_{0}^{1}(1-\tau)\tau^{\beta-2} \,dH(\tau) \ge \int_{0}^{1}(1-\tau)\tau^{\beta -1} \,dH( \tau)>0, \\& \int_{0}^{1} k_{2}(\tau) \,dH(\tau)\ge \frac{1}{\beta-1} \int_{0}^{1}\tau^{\beta -1}(1-\tau) \,dH( \tau)>0, \\& K(1)-K(0)= \int_{0}^{1} dK(\tau)\ge \int_{0}^{1}\tau^{\alpha-1}(1-\tau) \,dK(\tau )>0, \\& H(1)-H(0)= \int_{0}^{1} dH(\tau)\ge \int_{0}^{1}\tau^{\beta-1}(1-\tau) \,dH( \tau)>0. \end{aligned}$$

By using Lemma 2.2, we deduce, for all $(t,s)\in[0,1]\times[0,1]$:

(a₁)

$$\begin{aligned} G_{1}(t,s) =&g_{1}(t,s)+\frac{t^{\alpha-1}}{\Delta} \biggl( \int_{0}^{1}\tau^{\beta -1} \,dH(\tau) \biggr) \biggl( \int_{0}^{1} g_{1}(\tau,s) \,dK(\tau) \biggr) \\ \le& h_{1}(s)+\frac{1}{\Delta} \biggl( \int_{0}^{1}\tau^{\beta-1} \,dH(\tau) \biggr) \biggl( \int_{0}^{1} h_{1}(s) \,dK(\tau) \biggr) \\ =&h_{1}(s) \biggl[1+\frac{1}{\Delta}\bigl(K(1)-K(0)\bigr) \int_{0}^{1}\tau^{\beta-1} \,dH(\tau ) \biggr]= \sigma_{1} h_{1}(s). \end{aligned}$$

(a₂)

$$\begin{aligned} G_{1}(t,s) \le&\frac{t^{\alpha-1}}{\Gamma(\alpha-1)}+\frac{t^{\alpha -1}}{\Delta} \biggl( \int_{0}^{1}\tau^{\beta-1} \,dH(\tau) \biggr) \biggl( \int _{0}^{1}\frac{(1-\tau)\tau^{\alpha-1}}{\Gamma(\alpha-1)}\,dK(\tau) \biggr) \\ =&t^{\alpha-1}\frac{1}{\Gamma(\alpha-1)} \biggl[1+\frac{1}{\Delta} \biggl( \int_{0}^{1}\tau^{\beta-1} \,dH(\tau) \biggr) \biggl( \int_{0}^{1}(1-\tau)\tau ^{\alpha-1} \,dK(\tau) \biggr) \biggr] =\delta_{1} t^{\alpha-1}. \end{aligned}$$

(a₃)

$$\begin{aligned} G_{1}(t,s) \ge&\frac{t^{\alpha-1}}{\Delta} \biggl( \int_{0}^{1}\tau^{\beta-1} \,dH(\tau) \biggr) \biggl( \int_{0}^{1}k_{1}(\tau)h_{1}(s) \,dK(\tau) \biggr) \\ =&t^{\alpha-1}h_{1}(s)\frac{1}{\Delta} \biggl( \int_{0}^{1}\tau^{\beta-1} \,dH(\tau ) \biggr) \biggl( \int_{0}^{1}k_{1}(\tau) \,dK(\tau) \biggr) =\varrho_{1} t^{\alpha-1}h_{1}(s). \end{aligned}$$

(b₁)

$$\begin{aligned} G_{2}(t,s) =&\frac{t^{\alpha-1}}{\Delta} \int_{0}^{1} g_{2}(\tau,s) \,dH(\tau)\le \frac{1}{\Delta} \int_{0}^{1} h_{2}(s) \,dH(\tau) \\ =&\frac{1}{\Delta}\bigl(H(1)-H(0)\bigr)h_{2}(s)= \sigma_{2} h_{2}(s). \end{aligned}$$

(b₂)

$$G_{2}(t,s)\le\frac{t^{\alpha-1}}{\Delta} \int_{0}^{1}\frac{(1-\tau)\tau^{\beta -1}}{\Gamma(\beta-1)}\,dH(\tau)= \delta_{2} t^{\alpha-1}. $$

(b₃)

$$ G_{2}(t,s)\ge\frac{t^{\alpha-1}}{\Delta} \int_{0}^{1} k_{2}(\tau)h_{2}(s) \,dH(\tau )=\frac{t^{\alpha-1}}{\Delta}h_{2}(s) \int_{0}^{1}k_{2}(\tau) \,dH(\tau)= \varrho_{2} t^{\alpha-1}h_{2}(s). $$

(c₁)

$$\begin{aligned} G_{3}(t,s) =&g_{2}(t,s)+\frac{t^{\beta-1}}{\Delta} \biggl( \int_{0}^{1}\tau^{\alpha -1} \,dK(\tau) \biggr) \biggl( \int_{0}^{1} g_{2}(\tau,s) \,dH(\tau) \biggr) \\ \le& h_{2}(s)+\frac{1}{\Delta} \biggl( \int_{0}^{1}\tau^{\alpha-1} \,dK(\tau) \biggr) \biggl( \int_{0}^{1} h_{2}(s) \,dH(\tau) \biggr) \\ =&h_{2}(s)\biggl[1+\frac{1}{\Delta}\bigl(H(1)-H(0)\bigr) \int_{0}^{1}\tau^{\alpha-1} \,dK(\tau) \biggr]= \sigma_{3} h_{2}(s). \end{aligned}$$

(c₂)

$$\begin{aligned} G_{3}(t,s) \le&\frac{(1-t)t^{\beta-1}}{\Gamma(\beta-1)}+\frac{t^{\beta -1}}{\Delta} \biggl( \int_{0}^{1}\tau^{\alpha-1} \,dK(\tau) \biggr) \biggl( \int _{0}^{1}\frac{(1-\tau)\tau^{\beta-1}}{\Gamma(\beta-1)}\,dH(\tau) \biggr) \\ \le&\frac{t^{\beta-1}}{\Gamma(\beta-1)} \biggl[1+\frac{1}{\Delta} \biggl( \int_{0}^{1}\tau^{\alpha-1}\,dK(\tau) \biggr) \biggl( \int_{0}^{1}(1-\tau)\tau^{\beta -1}\,dH(\tau) \biggr) \biggr]=\delta_{3} t^{\beta-1}. \end{aligned}$$

(c₃)

$$\begin{aligned} G_{3}(t,s) \ge&\frac{t^{\beta-1}}{\Delta} \biggl( \int_{0}^{1}\tau^{\alpha -1}\,dK(\tau) \biggr) \biggl( \int_{0}^{1} k_{2}(\tau)h_{2}(s) \,dH(\tau) \biggr) \\ =&t^{\beta-1}h_{2}(s)\frac{1}{\Delta} \biggl( \int_{0}^{1}\tau^{\alpha-1}\,dK(\tau ) \biggr) \biggl( \int_{0}^{1}k_{2}(\tau)\,dH(\tau) \biggr) =\varrho_{3} t^{\beta-1}h_{2}(s). \end{aligned}$$

(d₁)

$$\begin{aligned} G_{4}(t,s) =&\frac{t^{\beta-1}}{\Delta} \int_{0}^{1} g_{1}(\tau,s) \,dK(\tau)\le \frac{1}{\Delta} \int_{0}^{1}h_{1}(s)\,dK(\tau) \\ =&h_{1}(s)\frac{1}{\Delta}\bigl(K(1)-K(0)\bigr)=\sigma_{4} h_{1}(s). \end{aligned}$$

(d₂)

$$G_{4}(t,s)\le\frac{t^{\beta-1}}{\Delta} \int_{0}^{1}\frac{(1-\tau)\tau^{\alpha -1}}{\Gamma(\alpha-1)}\,dK(\tau)= \delta_{4} t^{\beta-1}. $$

(d₃)

$$G_{4}(t,s)\ge\frac{t^{\beta-1}}{\Delta} \int_{0}^{1}k_{1}(\tau)h_{1}(s) \,dK(\tau )=t^{\beta-1}h_{1}(s)\frac{1}{\Delta} \int_{0}^{1}k_{1}(s)\,dK(\tau) = \varrho_{4} t^{\beta-1}h_{1}(s). $$

 □

Lemma 2.5

Assume that $H, K:[0,1]\to\mathbb{R}$ are nondecreasing functions, $\Delta>0$, $\int_{0}^{1}\tau^{\alpha-1}(1-\tau) \,dK(\tau)>0$, $\int_{0}^{1}\tau^{\beta-1}(1-\tau) \,dH(\tau)>0$, and $\tilde{x}, \tilde{y}\in C(0,1)\cap L^{1}(0,1)$, $\tilde{x}(t)\ge0$, $\tilde{y}(t)\ge0$ for all $t\in(0,1)$. Then the solution $(u,v)$ of problem (1)-(2) given by (3) satisfies the inequalities $u(t)\ge\gamma_{1} t^{\alpha-1}u(t')$, $v(t)\ge\gamma_{2} t^{\beta-1} v(t')$, for all $t, t'\in[0,1]$, where $\gamma_{1}=\min \{\frac{\varrho_{1}}{\sigma_{1}},\frac{\varrho _{2}}{\sigma_{2}} \}>0$, $\gamma_{2}=\min \{\frac{\varrho_{3}}{\sigma _{3}},\frac{\varrho_{4}}{\sigma_{4}} \}>0$.

Proof

By using Lemma 2.4, we obtain

$$\begin{aligned} u(t) =& \int_{0}^{1}G_{1}(t,s)\tilde{x}(s) \,ds+ \int_{0}^{1} G_{2}(t,s)\tilde{y}(s) \,ds \\ \ge& \int_{0}^{1}\varrho_{1} t^{\alpha-1}h_{1}(s)\tilde{x}(s) \,ds+ \int _{0}^{1}\varrho_{2} t^{\alpha-1}h_{2}(s)\tilde{y}(s) \,ds \\ =&t^{\alpha-1} \biggl(\varrho_{1} \int_{0}^{1}h_{1}(s)\tilde{x}(s) \,ds+ \varrho_{2} \int_{0}^{1}h_{2}(s)\tilde{y}(s) \,ds \biggr) \\ \ge& t^{\alpha-1} \biggl( \frac{\varrho_{1}}{\sigma_{1}} \int _{0}^{1}G_{1}\bigl(t',s \bigr)\tilde{x}(s) \,ds+\frac{\varrho_{2}}{\sigma_{2}} \int _{0}^{1}G_{2}\bigl(t',s \bigr)\tilde{y}(s) \,ds \biggr) \\ \ge& t^{\alpha-1}\min \biggl\{ \frac{\varrho_{1}}{\sigma_{1}},\frac{\varrho _{2}}{\sigma_{2}} \biggr\} \biggl( \int_{0}^{1}G_{1}\bigl(t',s \bigr)\tilde{x}(s) \,ds+ \int _{0}^{1}G_{2}\bigl(t',s \bigr)\tilde{y}(s) \,ds \biggr) \\ =&\gamma_{1} t^{\alpha-1}u\bigl(t'\bigr),\quad \forall t, t'\in[0,1], \mbox{where } \gamma _{1}=\min \biggl\{ \frac{\varrho_{1}}{\sigma_{1}},\frac{\varrho_{2}}{\sigma _{2}} \biggr\} >0. \end{aligned}$$

In a similar way, we deduce

$$\begin{aligned} v(t) =& \int_{0}^{1}G_{3}(t,s)\tilde{y}(s) \,ds+ \int_{0}^{1} G_{4}(t,s)\tilde{x}(s) \,ds \\ \ge&\gamma_{2} t^{\beta-1}v\bigl(t'\bigr), \quad \forall t, t'\in[0,1], \mbox{where } \gamma_{2}=\min \biggl\{ \frac{\varrho_{3}}{\sigma_{3}},\frac{\varrho_{4}}{\sigma _{4}} \biggr\} >0. \end{aligned}$$

 □

In the proof of our main results we shall use the nonlinear alternative of Leray-Schauder type and the Guo-Krasnosel’skii fixed point theorem presented below (see [14, 15]).

Theorem 2.1

Let X be a Banach space with $\Omega\subset X$ closed and convex. Assume U is a relatively open subset of Ω with $0\in U$, and let $S:\bar{U}\to\Omega$ be a completely continuous operator (continuous and compact). Then either

• (1) S has a fixed point in Ū, or

• (2) there exist $u\in\partial U$ and $\nu\in(0,1)$ such that $u=\nu Su$.

Theorem 2.2

Let X be a Banach space and let $C\subset X$ be a cone in X. Assume $\Omega_{1}$ and $\Omega_{2}$ are bounded open subsets of X with $0\in \Omega_{1}\subset\bar{\Omega}_{1}\subset\Omega_{2}$ and let ${\mathcal {A}}:C\cap(\bar{\Omega}_{2}\setminus\Omega_{1})\to C$ be a completely continuous operator such that either

• (i) $\|{\mathcal{A}} u\|\le\|u\|$, $u\in C\cap\partial \Omega_{1}$, and $\|{\mathcal{A}} u\|\ge\|u\|$, $u\in C\cap\partial\Omega_{2}$, or

• (ii) $\|{\mathcal{A}} u\|\ge\|u\|$, $u\in C\cap\partial \Omega_{1}$, and $\|{\mathcal{A}} u\|\le\|u\|$, $u\in C\cap\partial \Omega_{2}$.

Then ${\mathcal{A}}$ has a fixed point in $C\cap(\bar{\Omega} _{2}\setminus\Omega_{1})$.

3 Main results

In this section, we investigate the existence and multiplicity of positive solutions for our problem (S)-(BC). We present now the assumptions that we shall use in the sequel.

(H1):

$H, K:[0,1]\to\mathbb{R}$ are nondecreasing functions, $\Delta=1- (\int_{0}^{1}\tau^{\alpha-1} \,dK(\tau) )\times (\int _{0}^{1}\tau^{\beta-1} \,dH(\tau) ) >0$, and $\int_{0}^{1}\tau^{\alpha-1}(1-\tau) \,dK(\tau)>0$, $\int_{0}^{1}\tau^{\beta -1}(1-\tau) \,dH(\tau)>0$.

(H2):

The functions $f, g\in C([0,1]\times[0,\infty)\times [0,\infty), (-\infty,+\infty))$ and there exist functions $p_{1}, p_{2}\in C([0,1],[0,\infty))$ such that $f(t,u,v)\ge-p_{1}(t)$ and $g(t,u,v)\ge -p_{2}(t)$ for any $t\in[0,1]$ and $u, v\in[0,\infty)$.

(H3):

$f(t,0,0)>0$, $g(t,0,0)>0$ for all $t\in[0,1]$.

(H4):

The functions $f, g\in C((0,1)\times[0,\infty)\times [0,\infty),(-\infty,+\infty))$, f, g may be singular at $t=0$ and/or $t=1$, and there exist functions $p_{1}, p_{2}\in C((0,1),[0,\infty))$, $\alpha_{1}, \alpha_{2}\in C((0,1),[0,\infty))$, $\beta_{1}, \beta_{2}\in C([0,1]\times[0,\infty)\times[0,\infty),[0,\infty ))$ such that

$$-p_{1}(t)\le f(t,u,v)\le\alpha_{1}(t)\beta_{1}(t,u,v), \qquad -p_{2}(t)\le g(t,u,v)\le\alpha_{2}(t) \beta_{2}(t,u,v), $$

for all $t\in(0,1)$, $u, v\in[0,\infty)$, with $0<\int_{0}^{1} p_{i}(s) \,ds<\infty$, $0<\int_{0}^{1}\alpha_{i}(s) \,ds<\infty$, $i=1,2$.

(H5):

There exists $c\in(0,1/2)$ such that

$$f_{\infty}= \lim_{u+v\to\infty} \min_{t\in[c,1-c]} \frac {f(t,u,v)}{u+v}=\infty \quad \mbox{or}\quad g_{\infty}= \lim _{u+v\to\infty} \min_{t\in[c,1-c]} \frac {g(t,u,v)}{u+v}=\infty. $$

(H6):

$\beta_{i\infty}= \lim_{u+v\to\infty} \max_{t\in[0,1]} \frac{\beta_{i}(t,u,v)}{u+v}=0$, $i=1,2$.

We consider the system of nonlinear fractional differential equations

$$ \left \{ \textstyle\begin{array}{l} D_{0+}^{\alpha}x(t)+\lambda (f(t,[x(t)-q_{1}(t)]^{*},[y(t)-q_{2}(t)]^{*})+p_{1}(t))=0, \quad 0< t< 1, \\ D_{0+}^{\beta}y(t)+\mu(g(t, [x(t)-q_{1}(t)]^{*},[y(t)-q_{2}(t)]^{*})+p_{2}(t))=0,\quad 0< t< 1, \end{array}\displaystyle \right . $$

(6)

with the integral boundary conditions

$$ \left \{ \textstyle\begin{array}{l} x(0)=x'(0)=\cdots=x^{(n-2)}(0)=0,\qquad x(1)= \int_{0}^{1} y(s)\,dH(s), \\ y(0)=y'(0)=\cdots=y^{(m-2)}(0)=0, \qquad y(1)= \int_{0}^{1}x(s)\,dK(s), \end{array}\displaystyle \right . $$

(7)

where $z(t)^{*}=z(t)$ if $z(t)\ge0$, and $z(t)^{*}=0$ if $z(t)<0$. Here $(q_{1},q_{2})$ with

$$\begin{aligned}& q_{1}(t)=\lambda \int_{0}^{1}G_{1}(t,s)p_{1}(s) \,ds+\mu \int _{0}^{1}G_{2}(t,s)p_{2}(s) \,ds,\quad t\in[0,1], \\& q_{2}(t)=\mu \int_{0}^{1}G_{3}(t,s)p_{2}(s) \,ds+\lambda \int_{0}^{1}G_{4}(t,s)p_{1}(s) \,ds, \quad t\in[0,1], \end{aligned}$$

is solution of the system of fractional differential equations

$$ \left \{ \textstyle\begin{array}{l} D_{0+}^{\alpha}q_{1}(t)+\lambda p_{1}(t)=0,\quad 0< t< 1, \\ D_{0+}^{\beta}q_{2}(t)+\mu p_{2}(t)=0,\quad 0< t< 1, \end{array}\displaystyle \right . $$

(8)

with the integral boundary conditions

$$ \left \{ \textstyle\begin{array}{l} q_{1}(0)=q_{1}'(0)=\cdots=q_{1}^{(n-2)}(0)=0,\qquad q_{1}(1)= \int _{0}^{1} q_{2}(s)\,dH(s), \\ q_{2}(0)=q_{2}'(0)=\cdots=q_{2}^{(m-2)}(0)=0,\qquad q_{2}(1)= \int_{0}^{1}q_{1}(s)\,dK(s). \end{array}\displaystyle \right . $$

(9)

Under the assumptions (H1) and (H2), or (H1) and (H4), we have $q_{1}(t)\ge0$, $q_{2}(t)\ge0$ for all $t\in[0,1]$.

We shall prove that there exists a solution $(x,y)$ for the boundary value problem (6)-(7) with $x(t)\ge q_{1}(t)$ and $y(t)\ge q_{2}(t)$ on $[0,1]$, $x(t)>q_{1}(t)$, $y(t)>q_{2}(t)$ on $(0,1)$. In this case $(u,v)$ with $u(t)=x(t)-q_{1}(t)$ and $v(t)=y(t)-q_{2}(t)$, $t\in[0,1]$ represents a positive solution of boundary value problem (S)-(BC).

By using Lemma 2.1 (relations (3)), a solution of the system

$$\left \{ \textstyle\begin{array}{l} x(t)=\lambda \int _{0}^{1}G_{1}(t,s)(f(s,[x(s)-q_{1}(s)]^{*},[y(s)-q_{2}(s)]^{*})+p_{1}(s)) \,ds \\ \hphantom{x(t)={}}{}+\mu \int_{0}^{1}G_{2}(t,s)(g(s,[x(s)-q_{1}(s)]^{*},[y(s)-q_{2}(s)]^{*})+p_{2}(s)) \,ds,\quad t\in[0,1], \\ y(t)=\mu \int_{0}^{1}G_{3}(t,s)(g(s,[x(s)-q_{1}(s)]^{*},[y(s)-q_{2}(s)]^{*})+p_{2}(s)) \,ds \\ \hphantom{y(t)={}}{}+\lambda \int _{0}^{1}G_{4}(t,s)(f(s,[x(s)-q_{1}(s)]^{*},[y(s)-q_{2}(s)]^{*})+p_{1}(s)) \,ds,\quad t\in[0,1], \end{array}\displaystyle \right . $$

is a solution for problem (6)-(7).

We consider the Banach space $X=C([0,1])$ with the supremum norm $\| \cdot\|$, and the Banach space $Y=X\times X$ with the norm $\|(u,v)\| _{Y}=\|u\|+\|v\|$. We define the cones

$$\begin{aligned}& P_{1}=\bigl\{ x\in X, x(t)\ge\gamma_{1} t^{\alpha-1}\|x \|, \forall t\in[0,1]\bigr\} , \\& P_{2}=\bigl\{ y\in X, y(t)\ge \gamma_{2} t^{\beta-1}\|y\|, \forall t\in[0,1]\bigr\} , \end{aligned}$$

where $\gamma_{1}$, $\gamma_{2}$ are defined in Section 2 (Lemma 2.5), and $P=P_{1}\times P_{2}\subset Y$.

For $\lambda, \mu>0$, we introduce the operators $Q_{1}, Q_{2}:Y\to X$ and ${\mathcal{Q}}:Y\to Y$ defined by ${\mathcal {Q}}(x,y)=(Q_{1}(x,y),Q_{2}(x,y))$, $(x,y)\in Y$ with

$$\begin{aligned}& Q_{1}(x,y) (t) = \lambda \int _{0}^{1}G_{1}(t,s) \bigl(f\bigl(s, \bigl[x(s)-q_{1}(s)\bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*} \bigr)+p_{1}(s)\bigr) \,ds \\& \hphantom{Q_{1}(x,y)(t) ={}}{}+\mu \int_{0}^{1}G_{2}(t,s) \bigl(g\bigl(s, \bigl[x(s)-q_{1}(s)\bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*} \bigr)+p_{2}(s)\bigr) \,ds,\quad t\in[0,1], \\& Q_{2}(x,y) (t)=\mu \int _{0}^{1}G_{3}(t,s) \bigl(g\bigl(s, \bigl[x(s)-q_{1}(s)\bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*} \bigr)+p_{2}(s)\bigr) \,ds \\& \hphantom{Q_{2}(x,y)(t)={}}{}+\lambda \int_{0}^{1}G_{4}(t,s) \bigl(f\bigl(s, \bigl[x(s)-q_{1}(s)\bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*} \bigr)+p_{1}(s)\bigr) \,ds,\quad t\in[0,1]. \end{aligned}$$

It is clear that if $(x,y)$ is a fixed point of operator ${\mathcal {Q}}$, then $(x,y)$ is a solution of problem (6)-(7).

Lemma 3.1

If (H1) and (H2), or (H1) and (H4) hold, then operator ${\mathcal{Q}}:P\to P$ is a completely continuous operator.

Proof

The operators $Q_{1}$ and $Q_{2}$ are well defined. To prove this, let $(x,y)\in P$ be fixed with $\|(x,y)\|_{Y}=\widetilde{L}$. Then we have

$$\begin{aligned}& \bigl[x(s)-q_{1}(s)\bigr]^{*}\le x(s)\le\|x\|\le\bigl\| (x,y) \bigr\| _{Y}=\widetilde{L},\quad \forall s\in[0,1], \\& \bigl[y(s)-q_{2}(s)\bigr]^{*}\le y(s)\le\|y\|\le\bigl\| (x,y) \bigr\| _{Y}=\widetilde{L},\quad \forall s\in[0,1]. \end{aligned}$$

If (H1) and (H2) hold, then we deduce easily that $Q_{1}(x,y)(t)< \infty$ and $Q_{2}(x,y)(t)< \infty$ for all $t\in[0,1]$.

If (H1) and (H4) hold, we deduce, for all $t\in[0,1]$,

$$\begin{aligned}& Q_{1}(x,y) (t)\le\lambda\sigma_{1} \int_{0}^{1}h_{1}(s)\bigl[\alpha _{1}(s)\beta_{1}\bigl(s,\bigl[x(s)-q_{1}(s) \bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*}\bigr)+p_{1}(s)\bigr] \,ds \\& \hphantom{Q_{1}(x,y)(t)\le{}}{}+\mu\sigma_{2} \int_{0}^{1}h_{2}(s)\bigl[ \alpha_{2}(s)\beta _{2}\bigl(s,\bigl[x(s)-q_{1}(s) \bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*}\bigr)+p_{2}(s)\bigr] \,ds \\& \hphantom{Q_{1}(x,y)(t)}\le M \biggl(\lambda\sigma_{1} \int_{0}^{1}h_{1}(s) \bigl( \alpha_{1}(s)+p_{1}(s)\bigr) \,ds+\mu \sigma_{2} \int_{0}^{1}h_{2}(s) \bigl( \alpha_{2}(s)+p_{2}(s)\bigr) \,ds \biggr)< \infty, \\& Q_{2}(x,y) (t)\le\mu\sigma_{3} \int_{0}^{1}h_{2}(s)\bigl[ \alpha_{2}(s)\beta _{2}\bigl(s,\bigl[x(s)-q_{1}(s) \bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*}\bigr)+p_{2}(s)\bigr] \,ds \\& \hphantom{Q_{2}(x,y)(t)\le{}}{}+\lambda\sigma_{4} \int_{0}^{1}h_{1}(s)\bigl[ \alpha_{1}(s)\beta _{1}\bigl(s,\bigl[x(s)-q_{1}(s) \bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*}\bigr)+p_{1}(s)\bigr] \,ds \\& \hphantom{Q_{2}(x,y)(t)}\le M \biggl(\mu\sigma_{3} \int_{0}^{1}h_{2}(s) \bigl( \alpha_{2}(s)+p_{2}(s)\bigr) \,ds+\lambda \sigma_{4} \int_{0}^{1}h_{1}(s) \bigl( \alpha_{1}(s)+p_{1}(s)\bigr) \,ds \biggr)< \infty, \end{aligned}$$

where $M=\max \{ \max_{t\in[0,1], u,v\in[0,\widetilde{L}]}\beta _{1}(t,u,v), \max_{t\in[0,1], u,v\in[0,\widetilde{L}]}\beta _{2}(t,u,v),1 \}$.

Besides, by Lemma 2.5, we conclude that

$$Q_{1}(x,y) (t)\ge\gamma_{1} t^{\alpha-1}\bigl\Vert Q_{1}(x,y)\bigr\Vert ,\qquad Q_{2}(x,y) (t)\ge\gamma _{2} t^{\beta-1}\bigl\Vert Q_{2}(x,y)\bigr\Vert , \quad \forall t\in[0,1], $$

and so $Q_{1}(x,y), Q_{2}(x,y)\in P$.

By using standard arguments, we deduce that operator ${\mathcal {Q}}:P\to P$ is a completely continuous operator (a compact operator, that is, one that maps bounded sets into relatively compact sets and is continuous). □

Theorem 3.1

Assume that (H1)-(H3) hold. Then there exist constants $\lambda_{0}>0$ and $\mu_{0}>0$ such that, for any $\lambda\in(0,\lambda_{0}]$ and $\mu\in(0,\mu_{0}]$, the boundary value problem (S)-(BC) has at least one positive solution.

Proof

Let $\delta\in(0,1)$ be fixed. From (H2) and (H3), there exists $R_{0}\in(0,1]$ such that

$$ f(t,u,v)\ge\delta f(t,0,0),\qquad g(t,u,v)\ge\delta g(t,0,0),\quad \forall t\in [0,1], u, v\in[0,R_{0}]. $$

(10)

We define

$$\begin{aligned}& \bar{f}(R_{0})= \max_{t\in[0,1], u,v\in[0,R_{0}]}\bigl\{ f(t,u,v)+p_{1}(t)\bigr\} \ge \max_{t\in[0,1]}\bigl\{ \delta f(t,0,0)+p_{1}(t)\bigr\} >0, \\& \bar{g}(R_{0})= \max_{t\in[0,1], u,v\in[0,R_{0}]}\bigl\{ g(t,u,v)+p_{2}(t)\bigr\} \ge \max_{t\in[0,1]}\bigl\{ \delta g(t,0,0)+p_{2}(t)\bigr\} >0, \\& c_{1}=\sigma_{1} \int_{0}^{1}h_{1}(s) \,ds,\qquad c_{2}=\sigma_{2} \int_{0}^{1}h_{2}(s) \,ds, \\& c_{3}=\sigma _{3} \int_{0}^{1}h_{2}(s) \,ds, \qquad c_{4}=\sigma_{4} \int_{0}^{1}h_{1}(s) \,ds, \\& \lambda_{0}=\max \biggl\{ \frac{R_{0}}{8c_{1}\bar{f}(R_{0})},\frac{R_{0}}{8c_{4}\bar{f}(R_{0})} \biggr\} ,\qquad \mu_{0}=\max \biggl\{ \frac{R_{0}}{8c_{2}\bar{g}(R_{0})},\frac {R_{0}}{8c_{3}\bar{g}(R_{0})} \biggr\} . \end{aligned}$$

We will show that, for any $\lambda\in(0,\lambda_{0}]$ and $\mu\in(0,\mu _{0}]$, problem (6)-(7) has at least one positive solution.

So, let $\lambda\in(0,\lambda_{0}]$ and $\mu\in(0,\mu_{0}]$ be arbitrary, but fixed for the moment. We define the set $U=\{(x,y)\in P, \|(u,v)\| _{Y}< R_{0}\}$. We suppose that there exist $(x,y)\in\partial U$ ($\|(x,y)\| _{Y}=R_{0}$ or $\|x\|+\|y\|=R_{0}$) and $\nu\in(0,1)$ such that $(x,y)=\nu{\mathcal{Q}}(x,y)$ or $x=\nu Q_{1}(x,y)$, $y=\nu Q_{2}(x,y)$.

We deduce that

$$\begin{aligned}& {\bigl[x(t)-q_{1}(t)\bigr]}^{*}=x(t)-q_{1}(t)\le x(t)\le R_{0},\quad \mbox{if } x(t)-q_{1}(t)\ge 0, \\& {\bigl[x(t)-q_{1}(t)\bigr]}^{*}=0,\quad \mbox{for } x(t)-q_{1}(t)< 0, \forall t\in[0,1], \\& {\bigl[y(t)-q_{2}(t)\bigr]}^{*}=y(t)-q_{2}(t)\le y(t)\le R_{0},\quad \mbox{if } y(t)-q_{2}(t)\ge 0, \\& {\bigl[y(t)-q_{2}(t)\bigr]}^{*}=0, \quad \mbox{for } y(t)-q_{2}(t)< 0, \forall t\in[0,1]. \end{aligned}$$

Then by Lemma 2.4, for all $t\in[0,1]$, we obtain

$$\begin{aligned}& x(t)=\nu Q_{1}(x,y) (t)\le Q_{1}(x,y) (t) \\& \hphantom{x(t)}\le\lambda\sigma_{1} \int_{0}^{1}h_{1}(s)\bar{f}(R_{0}) \,ds+\mu\sigma_{2} \int _{0}^{1}h_{2}(s)\bar{g}(R_{0}) \,ds \\& \hphantom{x(t)}\le\lambda_{0} c_{1}\bar{f}(R_{0})+ \mu_{0} c_{2}\bar{g}(R_{0})\le\frac{R_{0}}{8}+ \frac {R_{0}}{8}=\frac{R_{0}}{4}, \\& y(t)=\nu Q_{2}(x,y) (t)\le Q_{2}(x,y) (t) \\& \hphantom{y(t)}\le\mu\sigma_{3} \int_{0}^{1}h_{2}(s)\bar{g}(R_{0}) \,ds+\lambda\sigma_{4} \int _{0}^{1}h_{1}(s)\bar{f}(R_{0}) \,ds \\& \hphantom{y(t)}\le\mu_{0} c_{3}\bar{g}(R_{0})+ \lambda_{0} c_{4}\bar{f}(R_{0})\le \frac{R_{0}}{8}+\frac {R_{0}}{8}=\frac{R_{0}}{4}. \end{aligned}$$

Hence $\|x\|\le\frac{R_{0}}{4}$ and $\|y\|\le\frac{R_{0}}{4}$. Then $R_{0}=\| (x,y)\|_{Y}=\|x\|+\|y\|\le\frac{R_{0}}{4}+\frac{R_{0}}{4}=\frac{R_{0}}{2}$, which is a contradiction.

Therefore, by Theorem 2.1 (with $\Omega=P$), we deduce that ${\mathcal{Q}}$ has a fixed point $(x_{0},y_{0})\in\bar{U}\cap P$. That is, $(x_{0},y_{0})={\mathcal{Q}}(x_{0},y_{0})$ or $x_{0}=Q_{1}(x_{0},y_{0})$, $y_{0}=Q_{2}(x_{0},y_{0})$, and $\|x_{0}\|+\|y_{0}\|\le R_{0}$ with $x_{0}(t)\ge\gamma _{1} t^{\alpha-1}\|x_{0}\|$ and $y_{0}(t)\ge\gamma_{2} t^{\beta-1}\|y_{0}\|$ for all $t\in[0,1]$.

Moreover, by (10), we conclude

$$\begin{aligned}& x_{0}(t)=Q_{1}(x_{0},y_{0}) (t) \\& \hphantom{x_{0}(t)}\ge \lambda \int_{0}^{1}G_{1}(t,s) \bigl(\delta f(t,0,0)+p_{1}(s)\bigr)\,ds+\mu \int _{0}^{1}G_{2}(t,s) \bigl(\delta g(t,0,0)+p_{2}(s)\bigr)\,ds \\& \hphantom{x_{0}(t)}\ge\lambda \int_{0}^{1}G_{1}(t,s)p_{1}(s) \,ds+\mu \int_{0}^{1}G_{2}(t,s)p_{2}(s) \,ds=q_{1}(t),\quad \forall t\in[0,1], \\& x_{0}(t)>\lambda \int_{0}^{1}G_{1}(t,s)p_{1}(s) \,ds+\mu \int_{0}^{1}G_{2}(t,s)p_{2}(s) \,ds=q_{1}(t), \quad \forall t\in(0,1), \\& y_{0}(t)=Q_{2}(x_{0},y_{0}) (t) \\& \hphantom{y_{0}(t)}\ge \mu \int_{0}^{1}G_{3}(t,s) \bigl(\delta g(t,0,0)+p_{2}(s)\bigr)\,ds+\lambda \int _{0}^{1}G_{4}(t,s) \bigl(\delta f(t,0,0)+p_{1}(s)\bigr)\,ds \\& \hphantom{y_{0}(t)}\ge\mu \int_{0}^{1}G_{3}(t,s)p_{2}(s) \,ds+\lambda \int_{0}^{1}G_{4}(t,s)p_{1}(s) \,ds=q_{2}(t),\quad \forall t\in[0,1], \\& y_{0}(t)>\mu \int_{0}^{1}G_{3}(t,s)p_{2}(s) \,ds+\lambda \int_{0}^{1}G_{4}(t,s)p_{1}(s) \,ds=q_{2}(t),\quad \forall t\in(0,1). \end{aligned}$$

Therefore $x_{0}(t)\ge q_{1}(t)$, $y_{0}(t)\ge q_{2}(t)$ for all $t\in[0,1]$, and $x_{0}(t)>q_{1}(t)$, $y_{0}(t)>q_{2}(t)$ for all $t\in(0,1)$. Let $u_{0}(t)=x_{0}(t)-q_{1}(t)$ and $v_{0}(t)=y_{0}(t)-q_{2}(t)$ for all $t\in[0,1]$. Then $u_{0}(t)\ge0$, $v_{0}(t)\ge0$ for all $t\in[0,1]$, $u_{0}(t)>0$, $v_{0}(t)>0$ for all $t\in(0,1)$. Therefore $(u_{0},v_{0})$ is a positive solution of (S)-(BC). □

Theorem 3.2

Assume that (H1), (H4), and (H5) hold. Then there exist $\lambda^{*}>0$ and $\mu^{*}>0$ such that, for any $\lambda\in(0,\lambda^{*}]$ and $\mu\in(0,\mu^{*}]$, the boundary value problem (S)-(BC) has at least one positive solution.

Proof

We choose a positive number

$$\begin{aligned} R_{1} >&\max\biggl\{ 1, \frac{2}{\gamma_{1}} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr) \,ds, \frac{2}{\gamma_{2}} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr) \,ds, \\ &\frac{2}{\gamma_{1}\gamma_{2}} \biggl( \int_{0}^{1} s^{\beta-1}\,dH(s) \biggr)^{-1} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr) \,ds, \\ &\frac{2}{\gamma_{1}\gamma_{2}} \biggl( \int_{0}^{1} s^{\alpha-1}\,dK(s) \biggr)^{-1} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr) \,ds\biggr\} , \end{aligned}$$

and we define the set $\Omega_{1}=\{(x,y)\in P, \|(x,y)\|_{Y}< R_{1}\}$.

We introduce

$$\begin{aligned}& \lambda^{*}=\min \biggl\{ 1,\frac{R_{1}}{4\sigma_{1} M_{1}} \biggl( \int_{0}^{1}h_{1}(s) \bigl( \alpha_{1}(s)+p_{1}(s)\bigr) \,ds \biggr)^{-1}, \\& \hphantom{\lambda^{*}={}}{} \frac{R_{1}}{4\sigma_{4} M_{1}} \biggl( \int_{0}^{1}h_{1}(s) \bigl( \alpha_{1}(s)+p_{1}(s)\bigr) \,ds \biggr)^{-1} \biggr\} , \\& \mu^{*}=\min \biggl\{ 1,\frac{R_{1}}{4\sigma_{2} M_{2}} \biggl( \int_{0}^{1}h_{2}(s) \bigl(\alpha _{2}(s)+p_{2}(s)\bigr) \,ds \biggr)^{-1}, \\& \hphantom{\mu^{*}={}}{} \frac{R_{1}}{4\sigma_{3} M_{2}} \biggl( \int _{0}^{1}h_{2}(s) \bigl( \alpha_{2}(s)+p_{2}(s)\bigr) \,ds \biggr)^{-1} \biggr\} , \end{aligned}$$

with

$$\begin{aligned}& M_{1}=\max \Bigl\{ \max_{\substack{t\in[0,1]\\u,v\ge0, u+v\le R_{1}}}\beta_{1}(t,u,v),1 \Bigr\} , \\& M_{2}=\max \Bigl\{ \max_{\substack{t\in [0,1]\\u,v\ge0, u+v\le R_{1}}} \beta_{2}(t,u,v),1 \Bigr\} . \end{aligned}$$

Let $\lambda\in(0,\lambda^{*}]$ and $\mu\in(0,\mu^{*}]$. Then, for any $(x,y)\in P\cap\partial\Omega_{1}$ and $s\in[0,1]$, we have

$$\begin{aligned}& \bigl[x(s)-q_{1}(s)\bigr]^{*}\le x(s)\le\|x\|\le R_{1}, \\& \bigl[y(s)-q_{2}(s)\bigr]^{*}\le y(s)\le\| y\|\le R_{1}. \end{aligned}$$

Then, for any $(x,y)\in P\cap\partial\Omega_{1}$, we obtain

$$\begin{aligned}& \bigl\Vert Q_{1}(x,y)\bigr\Vert \le\lambda\sigma_{1} \int_{0}^{1}h_{1}(s)\bigl[ \alpha_{1}(s)\beta _{1}\bigl(s,\bigl[x(s)-q_{1}(s) \bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*}\bigr)+p_{1}(s)\bigr] \,ds \\& \hphantom{\bigl\Vert Q_{1}(x,y)\bigr\Vert \le{}}{}+\mu\sigma_{2} \int_{0}^{1}h_{2}(s)\bigl[ \alpha_{2}(s)\beta _{2}\bigl(s,\bigl[x(s)-q_{1}(s) \bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*}\bigr)+p_{2}(s)\bigr] \,ds \\& \hphantom{\bigl\Vert Q_{1}(x,y)\bigr\Vert }\le\lambda^{*}\sigma_{1} M_{1} \int_{0}^{1}h_{1}(s) \bigl( \alpha_{1}(s)+p_{1}(s)\bigr) \,ds+\mu ^{*}\sigma_{2} M_{2} \int_{0}^{1}h_{2}(s) \bigl( \alpha_{2}(s)+p_{2}(s)\bigr) \,ds \\& \hphantom{\bigl\Vert Q_{1}(x,y)\bigr\Vert }\le\frac{R_{1}}{4}+\frac{R_{1}}{4}=\frac{R_{1}}{2}= \frac{\|(x,y)\|_{Y}}{2}, \\& \bigl\Vert Q_{2}(x,y)\bigr\Vert \le\mu\sigma_{3} \int_{0}^{1}h_{2}(s)\bigl[ \alpha_{2}(s)\beta _{2}\bigl(s,\bigl[x(s)-q_{1}(s) \bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*}\bigr)+p_{2}(s)\bigr] \,ds \\& \hphantom{\bigl\Vert Q_{2}(x,y)\bigr\Vert \le{}}{}+\lambda\sigma_{4} \int_{0}^{1}h_{1}(s)\bigl[ \alpha_{1}(s)\beta _{1}\bigl(s,\bigl[x(s)-q_{1}(s) \bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*}\bigr)+p_{1}(s)\bigr] \,ds \\& \hphantom{\bigl\Vert Q_{2}(x,y)\bigr\Vert }\le\mu^{*}\sigma_{3} M_{2} \int_{0}^{1}h_{2}(s) \bigl( \alpha_{2}(s)+p_{2}(s)\bigr) \,ds+\lambda ^{*} \sigma_{4} M_{1} \int_{0}^{1}h_{1}(s) \bigl( \alpha_{1}(s)+p_{1}(s)\bigr) \,ds \\& \hphantom{\bigl\Vert Q_{2}(x,y)\bigr\Vert }\le\frac{R_{1}}{4}+\frac{R_{1}}{4}=\frac{R_{1}}{2}= \frac{\|(x,y)\|_{Y}}{2}. \end{aligned}$$

Therefore

$$ \bigl\Vert {\mathcal{Q}}(x,y)\bigr\Vert _{Y}=\bigl\Vert Q_{1}(x,y)\bigr\Vert +\bigl\Vert Q_{2}(x,y)\bigr\Vert \le\bigl\Vert (x,y)\bigr\Vert _{Y},\quad \forall(x,y)\in P \cap\partial\Omega_{1}. $$

(11)

On the other hand, we choose a constant $L>0$ such that

$$\begin{aligned}& \lambda L \varrho_{1} \gamma_{1} c^{2(\alpha-1)} \int_{c}^{1-c}h_{1}(s) \,ds\ge 4, \lambda L \varrho_{4} \gamma_{2} c^{2(\beta-1)} \int_{c}^{1-c}h_{1}(s) \,ds\ge 4, \\& \mu L \varrho_{2} \gamma_{1} c^{2(\alpha-1)} \int_{c}^{1-c}h_{2}(s) \,ds\ge4, \mu L \varrho_{3} \gamma_{2} c^{2(\beta-1)} \int_{c}^{1-c}h_{2}(s) \,ds\ge4. \end{aligned}$$

From (H5), we deduce that there exists a constant $M_{0}>0$ such that

$$ f(t,u,v)\ge L(u+v) \quad \mbox{or}\quad g(t,u,v)\ge L(u+v),\quad \forall t\in[c,1-c], u,v\ge0, u+v\ge M_{0}. $$

(12)

Now we define

$$\begin{aligned} R_{2} =&\max \biggl\{ 2R_{1},\frac{4M_{0}}{\gamma_{1} c^{\alpha-1}}, \frac {4M_{0}}{\gamma_{2} c^{\beta-1}},\frac{4}{\gamma_{1}} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr) \,ds, \\ &\frac{4}{\gamma_{2}} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr) \,ds \biggr\} >0, \end{aligned}$$

and let $\Omega_{2}=\{(x,y)\in P, \|(x,y)\|_{Y}< R_{2}\}$.

We suppose that $f_{\infty}=\infty$, that is, $f(t,u,v)\ge L(u+v)$ for all $t\in[c,1-c]$ and $u,v\ge0$, $u+v\ge M_{0}$. Then, for any $(x,y)\in P\cap\partial\Omega_{2}$, we have $\|(x,y)\| _{Y}=R_{2}$ or $\|x\|+\|y\|=R_{2}$. We deduce that $\|x\|\ge\frac{R_{2}}{2}$ or $\|y\|\ge\frac{R_{2}}{2}$.

We suppose that $\|x\|\ge\frac{R_{2}}{2}$. Then, for any $(x,y)\in P\cap \partial\Omega_{2}$, we obtain

$$\begin{aligned} x(t)-q_{1}(t) =&x(t)-\lambda \int_{0}^{1}G_{1}(t,s)p_{1}(s) \,ds-\mu \int_{0}^{1} G_{2}(t,s)p_{2}(s) \,ds \\ \ge& x(t)-t^{\alpha-1} \biggl(\delta_{1} \int_{0}^{1} p_{1}(s) \,ds+ \delta_{2} \int _{0}^{1}p_{2}(s) \,ds \biggr) \\ \ge& x(t)- \frac{x(t)}{\gamma_{1}\|x\|} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr) \,ds \\ =&x(t) \biggl[1- \frac{1}{\gamma_{1}\|x\|} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr) \,ds \biggr] \\ \ge& x(t) \biggl[1- \frac{2}{\gamma_{1} R_{2}} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta _{2} p_{2}(s)\bigr) \,ds \biggr]\ge \frac{1}{2}x(t)\ge0,\quad \forall t\in[0,1]. \end{aligned}$$

Therefore, we conclude

$$\begin{aligned} \bigl[x(t)-q_{1}(t)\bigr]^{*} =&x(t)-q_{1}(t)\ge \frac{1}{2}x(t)\ge\frac{1}{2}\gamma_{1} t^{\alpha-1} \|x\| \\ \ge&\frac{1}{4}\gamma_{1} t^{\alpha-1}R_{2}\ge \frac{1}{4}\gamma_{1} c^{\alpha -1}R_{2}\ge M_{0},\quad \forall t\in[c,1-c]. \end{aligned}$$

Hence

$$ \bigl[x(t)-q_{1}(t)\bigr]^{*}+\bigl[y(t)-q_{2}(t) \bigr]^{*}\ge\bigl[x(t)-q_{1}(t)\bigr]^{*}=x(t)-q_{1}(t)\ge M_{0},\quad \forall t\in[ c,1-c]. $$

(13)

Then, for any $(x,y)\in P\cap\partial\Omega_{2}$ and $t\in[c,1-c]$, by (12) and (13), we deduce

$$\begin{aligned} f\bigl(t,\bigl[x(t)-q_{1}(t)\bigr]^{*},\bigl[y(t)-q_{2}(t) \bigr]^{*}\bigr) \ge& L\bigl(\bigl[x(t)-q_{1}(t)\bigr]^{*}+ \bigl[y(t)-q_{2}(t)\bigr]^{*}\bigr) \\ \ge& L\bigl[x(t)-q_{1}(t)\bigr]^{*}\ge\frac{L}{2}x(t),\quad \forall t\in[c,1-c]. \end{aligned}$$

It follows that, for any $(x,y)\in P\cap\partial\Omega_{2}$, $t\in [c,1-c]$, we obtain

$$\begin{aligned} Q_{1}(x,y) (t) \ge&\lambda \int _{0}^{1}G_{1}(t,s) \bigl(f\bigl(s, \bigl[x(s)-q_{1}(s)\bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*} \bigr)+p_{1}(s)\bigr) \,ds \\ \ge&\lambda \int _{c}^{1-c}G_{1}(t,s) \bigl(f\bigl(s, \bigl[x(s)-q_{1}(s)\bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*} \bigr)+p_{1}(s)\bigr) \,ds \\ \ge&\lambda \int_{c}^{1-c}G_{1}(t,s)L\bigl( \bigl[x(s)-q_{1}(s)\bigr]^{*}\bigr) \,ds \ge\lambda \int_{c}^{1-c}G_{1}(t,s)\frac{1}{4}L \gamma_{1} c^{\alpha-1} R_{2} \,ds \\ \ge&\lambda \int_{c}^{1-c}\varrho_{1} t^{\alpha-1}h_{1}(s)\frac{1}{4}L\gamma _{1} c^{\alpha-1} R_{2} \,ds \\ \ge&\lambda c^{2(\alpha-1)} \frac{1}{4}\varrho_{1} L\gamma_{1} R_{2} \int _{c}^{1-c}h_{1}(s) \,ds\ge R_{2}. \end{aligned}$$

Then $\|Q_{1}(x,y)\|\ge\|(x,y)\|_{Y}$ and

$$ \bigl\Vert {\mathcal{Q}}(x,y)\bigr\Vert _{Y}\ge \bigl\Vert (x,y)\bigr\Vert _{Y},\quad \forall(x,y)\in P\cap \partial\Omega_{2}. $$

(14)

If $\|y\|\ge\frac{R_{2}}{2}$, then by a similar approach, we obtain again relation (14).

We suppose now that $g_{\infty}=\infty$, that is, $g(t,u,v)\ge L(u+v)$, for all $t\in[c,1-c]$ and $u,v\ge0$, $u+v\ge M_{0}$. Then, for any $(x,y)\in P\cap\partial\Omega_{2}$, we have $\|(x,y)\|_{Y}=R_{2}$. Hence $\| x\|\ge\frac{R_{2}}{2}$ or $\|y\|\ge\frac{R_{2}}{2}$.

If $\|x\|\ge\frac{R_{2}}{2}$, then for any $(x,y)\in P\cap\partial \Omega_{2}$ we deduce in a similar manner as above that $x(t)-q_{1}(t)\ge \frac{1}{2}x(t)$ for all $t\in[0,1]$ and

$$\begin{aligned} Q_{1}(x,y) (t) \ge&\mu \int _{0}^{1}G_{2}(t,s) \bigl(g\bigl(s, \bigl[x(s)-q_{1}(s)\bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*} \bigr)+p_{2}(s)\bigr) \,ds \\ \ge&\mu \int _{c}^{1-c}G_{2}(t,s) \bigl(g\bigl(s, \bigl[x(s)-q_{1}(s)\bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*} \bigr)+p_{2}(s)\bigr) \,ds \\ \ge&\mu \int_{c}^{1-c}G_{2}(t,s)L\bigl( \bigl[x(s)-q_{1}(s)\bigr]^{*}\bigr) \,ds \ge\mu \int_{c}^{1-c}G_{2}(t,s)\frac{1}{4}L \gamma_{1} c^{\alpha-1} R_{2} \,ds \\ \ge&\mu \int_{c}^{1-c}\varrho_{2} t^{\alpha-1}h_{2}(s)\frac{1}{4}L\gamma_{1} c^{\alpha-1} R_{2} \,ds \ge\mu c^{2(\alpha-1)}\frac{1}{4} \varrho_{2} L\gamma_{1} R_{2} \int _{c}^{1-c}h_{2}(s) \,ds \\ \ge& R_{2}, \quad \forall t\in[c,1-c]. \end{aligned}$$

Hence we obtain relation (14).

If $\|y\|\ge\frac{R_{2}}{2}$, then in a similar way as above, we deduce again relation (14).

Therefore, by Theorem 2.2, relations (11) and (14), we conclude that ${\mathcal{Q}}$ has a fixed point $(x_{1},y_{1}) \in P\cap(\bar{\Omega}_{2}\setminus\Omega_{1})$, that is, $R_{1}\le\| (x_{1},y_{1})\|_{Y}\le R_{2}$. Since $\|(x_{1},y_{1})\|_{Y}\ge R_{1}$, then $\|x_{1}\|\ge \frac{R_{1}}{2}$ or $\|y_{1}\|\ge\frac{R_{1}}{2}$.

We suppose first that $\|x_{1}\|\ge\frac{R_{1}}{2}$. Then we deduce

$$\begin{aligned} x_{1}(t)-q_{1}(t) =&x_{1}(t)-\lambda \int_{0}^{1}G_{1}(t,s)p_{1}(s) \,ds-\mu \int_{0}^{1} G_{2}(t,s)p_{2}(s) \,ds \\ \ge& x_{1}(t)-t^{\alpha-1} \biggl(\delta_{1} \int_{0}^{1} p_{1}(s) \,ds+ \delta_{2} \int _{0}^{1}p_{2}(s) \,ds \biggr) \\ \ge& x_{1}(t)- \frac{x_{1}(t)}{\gamma_{1}\|x_{1}\|} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr) \,ds \\ \ge& \biggl[1- \frac{2}{\gamma_{1} R_{1}} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr) \,ds \biggr]x_{1}(t) \\ \ge& \biggl[1- \frac{2}{\gamma_{1} R_{1}} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr) \,ds \biggr] \gamma_{1} t^{\alpha-1}\|x_{1}\| \\ \ge& \frac{R_{1}}{2} \biggl[1- \frac{2}{\gamma_{1} R_{1}} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr) \,ds \biggr] \gamma_{1} t^{\alpha-1} \\ =&\Lambda_{1} t^{\alpha-1},\quad \forall t\in[0,1], \end{aligned}$$

and so $x_{1}(t)\ge q_{1}(t)+\Lambda_{1} t^{\alpha-1}$ for all $t\in[0,1]$, where $\Lambda_{1}=\frac{\gamma_{1} R_{1}}{2}-\int_{0}^{1}(\delta_{1} p_{1}(s)+\delta _{2} p_{2}(s)) \,ds>0$.

Then $y_{1}(1)=\int_{0}^{1}x_{1}(s) \,dK(s)\ge\Lambda_{1}\int_{0}^{1}s^{\alpha-1} \,dK(s)>0$ and

$$\begin{aligned} \|y_{1}\| \ge& y_{1}(1)= \int_{0}^{1}x_{1}(s)\,dK(s)\ge \int_{0}^{1}\gamma_{1} s^{\alpha -1} \|x_{1}\|\,dK(s) \\ \ge&\frac{\gamma_{1} R_{1}}{2} \int_{0}^{1}s^{\alpha-1}\,dK(s)>0. \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} y_{1}(t)-q_{2}(t) =&y_{1}(t)-\mu \int_{0}^{1}G_{3}(t,s)p_{2}(s) \,ds-\lambda \int _{0}^{1}G_{4}(t,s)p_{1}(s) \,ds \\ \ge& y_{1}(t)-t^{\beta-1} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr) \,ds \\ \ge& y_{1}(t)- \frac{y_{1}(t)}{\gamma_{2}\|y_{1}\|} \int_{0}^{1}\bigl(\delta _{3}p_{2}(s)+ \delta_{4} p_{1}(s)\bigr) \,ds \\ \ge& y_{1}(t) \biggl[1- \frac{2}{\gamma_{1}\gamma_{2} R_{1}} \biggl( \int_{0}^{1} s^{\alpha-1}\,dK(s) \biggr)^{-1} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr) \,ds \biggr] \\ \ge&\gamma_{2} t^{\beta-1}\|y_{1}\| \biggl[1- \frac{2}{\gamma_{1}\gamma_{2} R_{1}} \biggl( \int_{0}^{1} s^{\alpha-1}\,dK(s) \biggr)^{-1} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr) \,ds \biggr] \\ \ge& \frac{\gamma_{1}\gamma_{2} R_{1}}{2}t^{\beta-1} \int_{0}^{1}s^{\alpha -1}\,dK(s) \\ &{}\times \biggl[1- \frac{2}{\gamma_{1}\gamma_{2} R_{1}} \biggl( \int_{0}^{1} s^{\alpha -1}\,dK(s) \biggr)^{-1} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr)\,ds \biggr] \\ =&\Lambda_{2} t^{\beta-1}, \quad \forall t\in[0,1], \end{aligned}$$

where $\Lambda_{2}=\frac{\gamma_{1}\gamma_{2} R_{1}}{2}\int_{0}^{1}s^{\alpha -1}\,dK(s)-\int_{0}^{1}(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)) \,ds>0$.

Hence $y_{1}(t)\ge q_{2}(t)+\Lambda_{2} t^{\beta-1}$ for all $t\in[0,1]$.

If $\|y_{1}\|\ge\frac{R_{1}}{2}$, then by a similar approach, we deduce that $y_{1}(t)\ge q_{2}(t)+\Lambda_{3} t^{\beta-1}$ and $x_{1}(t)\ge q_{1}(t)+\Lambda_{4} t^{\alpha-1}$ for all $t\in[0,1]$, where $\Lambda_{3}=\frac{\gamma_{2} R_{1}}{2}-\int_{0}^{1}(\delta_{3} p_{2}(s)+\delta _{4} p_{1}(s)) \,ds>0$ and $\Lambda_{4}=\frac{\gamma_{1}\gamma_{2} R_{1}}{2}\int_{0}^{1}s^{\beta-1}\,dH(s)-\int _{0}^{1}(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)) \,ds>0$.

Let $u_{1}(t)=x_{1}(t)-q_{1}(t)$ and $v_{1}(t)=y_{1}(t)-q_{2}(t)$ for all $t\in [0,1]$. Then $(u_{1},v_{1})$ is a positive solution of (S)-(BC) with $u_{1}(t)\ge\Lambda_{5} t^{\alpha-1}$ and $v_{1}(t)\ge\Lambda_{6} t^{\beta -1}$ for all $t\in[0,1]$, where $\Lambda_{5}=\min\{\Lambda_{1},\Lambda_{4}\}$ and $\Lambda_{6}=\min\{\Lambda_{2},\Lambda_{3}\}$. This completes the proof of Theorem 3.2. □

Theorem 3.3

Assume that (H1), (H3), (H5), and

(H4′):

The functions $f, g\in C([0,1]\times[0,\infty)\times [0,\infty),(-\infty,+\infty))$ and there exist functions $p_{1}, p_{2}, \alpha_{1}, \alpha_{2}\in C([0,1],[0,\infty))$, $\beta_{1}, \beta _{2}\in C([0,1]\times[0,\infty)\times[0,\infty),[0,\infty))$ such that

$$-p_{1}(t)\le f(t,u,v)\le\alpha_{1}(t)\beta_{1}(t,u,v), \qquad -p_{2}(t)\le g(t,u,v)\le\alpha_{2}(t) \beta_{2}(t,u,v), $$

for all $t\in[0,1]$, $u, v\in[0,\infty)$, with $\int_{0}^{1} p_{i}(s) \,ds>0$, $i=1,2$,

hold. Then the boundary value problem (S)-(BC) has at least two positive solutions for $\lambda>0$ and $\mu>0$ sufficiently small.

Proof

Because assumption (H4′) implies assumptions (H2) and (H4), we can apply Theorems 3.1 and 3.2. Therefore, we deduce that, for $0<\lambda\le\min\{\lambda_{0},\lambda^{*}\}$ and $0<\mu\le\min\{\mu_{0},\mu^{*}\}$, problem (S)-(BC) has at least two positive solutions $(u_{0},v_{0})$ and $(u_{1},v_{1})$ with $\|(u_{0}+q_{1},v_{0}+q_{2})\|_{Y}\le1$ and $\| (u_{1}+q_{1},v_{1}+q_{2})\|_{Y} >1$. □

Theorem 3.4

Assume that $\lambda=\mu$, and (H1), (H4), and (H6) hold. In addition if

(H7):

there exists $c\in(0,1/2)$ such that

$$f_{\infty}^{i}= \liminf_{\substack{u+v\to\infty\\u,v\ge0}} \min _{t\in [c,1-c]}f(t,u,v)>L_{0} \quad \textit{or}\quad g_{\infty}^{i}= \liminf_{\substack{u+v\to\infty\\u,v\ge0}} \min _{t\in [c,1-c]}g(t,u,v)>L_{0}, $$

where

$$\begin{aligned} L_{0} =&\max\biggl\{ \frac{4}{\gamma_{1}} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr)\,ds, \frac{4}{\gamma_{2}} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr)\,ds, \\ &\frac{4}{\gamma_{1}\gamma_{2}} \biggl( \int_{0}^{1} s^{\alpha-1}\,dK(s) \biggr)^{-1} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr)\,ds, \\ &\frac{4}{\gamma_{1}\gamma_{2}} \biggl( \int_{0}^{1} s^{\beta-1}\,dH(s) \biggr)^{-1} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr)\,ds\biggr\} \\ &{}\times \biggl(\min \biggl\{ c^{\alpha-1}\varrho_{1} \int_{c}^{1-c}h_{1}(s) \,ds,c^{\alpha-1}\varrho_{2} \int_{c}^{1-c}h_{2}(s) \,ds \biggr\} \biggr)^{-1}, \end{aligned}$$

then there exists $\lambda_{*}>0$ such that for any $\lambda\ge\lambda _{*}$ problem (S)-(BC) (with $\lambda=\mu$) has at least one positive solution.

Proof

By (H7) we conclude that there exists $M_{3}>0$ such that

$$f(t,u,v)\ge L_{0}\quad \mbox{or}\quad g(t,u,v)\ge L_{0}, \quad \forall t\in[c,1-c], u,v\ge 0, u+v\ge M_{3}. $$

We define

$$\lambda_{*}=\max \biggl\{ \frac{M_{3}}{c^{\alpha-1}} \biggl( \int_{0}^{1}\bigl(\delta _{1}p_{1}(s)+ \delta_{2}p_{2}(s)\bigr) \,ds \biggr)^{-1}, \frac{M_{3}}{c^{\beta-1}} \biggl( \int_{0}^{1}\bigl(\delta_{3}p_{2}(s)+ \delta_{4}p_{1}(s)\bigr) \,ds \biggr)^{-1} \biggr\} . $$

We assume now $\lambda\ge\lambda_{*}$. Let

$$\begin{aligned} R_{3} =&\max\biggl\{ \frac{4\lambda}{\gamma_{1}} \int_{0}^{1}\bigl(\delta _{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr)\,ds, \frac{4\lambda}{\gamma_{2}} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr)\,ds, \\ &\frac{4\lambda}{\gamma_{1}\gamma_{2}} \biggl( \int_{0}^{1} s^{\alpha-1}\,dK(s) \biggr)^{-1} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr)\,ds, \\ &\frac{4\lambda}{\gamma_{1}\gamma_{2}} \biggl( \int_{0}^{1} s^{\beta -1}\,dH(s) \biggr)^{-1} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr)\,ds\biggr\} , \end{aligned}$$

and $\Omega_{3}=\{(x,y)\in P, \|(x,y)\|_{Y}< R_{3}\}$.

We suppose first that $f_{\infty}^{i}>L_{0}$, that is, $f(t,u,v)\ge L_{0}$ for all $t\in[c,1-c]$ and $u,v\ge0$, $u+v\ge M_{3}$. Let $(x,y)\in P\cap \partial\Omega_{3}$. Then $\|(x,y)\|_{Y}=R_{3}$, so $\|x\|\ge R_{3}/2$ or $\| y\|\ge R_{3}/2$. We assume that $\|x\|\ge R_{3}/2$. Then for all $t\in[0,1]$ we deduce

$$\begin{aligned} x(t)-q_{1}(t) \ge&\gamma_{1} t^{\alpha-1}\|x\|-\lambda t^{\alpha-1}\delta_{1} \int_{0}^{1}p_{1}(s) \,ds-\lambda t^{\alpha-1}\delta_{2} \int_{0}^{1}p_{2}(s) \,ds \\ \ge& t^{\alpha-1} \biggl[ \frac{\gamma_{1} R_{3}}{2}-\lambda \int_{0}^{1}\bigl(\delta _{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr) \,ds \biggr] \\ \ge& t^{\alpha-1} \biggl[2\lambda \int_{0}^{1}\bigl(\delta_{1}p_{1}(s)+ \delta_{2}p_{2}(s)\bigr) \,ds-\lambda \int_{0}^{1}\bigl(\delta_{1}p_{1}(s)+ \delta_{2}p_{2}(s)\bigr) \,ds \biggr] \\ =&t^{\alpha-1}\lambda \int_{0}^{1}\bigl(\delta_{1}p_{1}(s)+ \delta_{2}p_{2}(s)\bigr) \,ds \\ \ge& t^{\alpha-1}\lambda_{*} \int_{0}^{1}\bigl(\delta_{1}p_{1}(s)+ \delta_{2}p_{2}(s)\bigr) \,ds \ge \frac{M_{3}}{c^{\alpha-1}}t^{\alpha-1} \ge0. \end{aligned}$$

Therefore, for any $(x,y)\in P\cap\partial\Omega_{3}$ and $t\in [c,1-c]$, we have

$$ \bigl[x(t)-q_{1}(t)\bigr]^{*}+\bigl[y(t)-q_{2}(t) \bigr]^{*}\ge\bigl[x(t)-q_{1}(t)\bigr]^{*}=x(t)-q_{1}(t)\ge \frac {M_{3}}{c^{\alpha-1}}t^{\alpha-1}\ge M_{3}. $$

(15)

Hence, for any $(x,y)\in P\cap\partial\Omega_{3}$ and $t\in[c,1-c]$, we conclude

$$\begin{aligned} Q_{1}(x,y) (t) \ge&\lambda \int _{0}^{1}G_{1}(t,s)\bigl[f\bigl(s, \bigl[x(s)-q_{1}(s)\bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*} \bigr)+p_{1}(s)\bigr] \,ds \\ \ge&\lambda\varrho_{1}t^{\alpha-1} \int _{c}^{1-c}h_{1}(s)f\bigl(s, \bigl[x(s)-q_{1}(s)\bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*}\bigr) \,ds \\ \ge&\lambda L_{0}\varrho_{1} t^{\alpha-1} \int_{c}^{1-c}h_{1}(s) \,ds\ge\lambda L_{0}\varrho_{1} c^{\alpha-1} \int_{c}^{1-c}h_{1}(s) \,ds\ge R_{3}=\bigl\Vert (x,y)\bigr\Vert _{Y}. \end{aligned}$$

Therefore we obtain $\|Q_{1}(x,y)\|\ge R_{3}$ for all $(x,y)\in P\cap \partial\Omega_{3}$, and so

$$ \bigl\Vert {\mathcal{Q}}(x,y)\bigr\Vert _{Y}\ge R_{3}=\bigl\Vert (x,y)\bigr\Vert _{Y},\quad \forall(x,y) \in P\cap \partial\Omega_{3}. $$

(16)

If $\|y\|\ge R_{3}/2$, then by a similar approach we deduce again relation (16).

We suppose now that $g_{\infty}^{i}>L_{0}$, that is, $g(t,u,v)\ge L_{0}$ for all $t\in[c,1-c]$ and $u,v\ge0$, $u+v\ge M_{3}$. Let $(x,y)\in P\cap \partial\Omega_{3}$. Then $\|(x,y)\|_{Y}=R_{3}$, so $\|x\|\ge R_{3}/2$ or $\| y\|\ge R_{3}/2$. If $\|x\|\ge R_{3}/2$, then we obtain in a similar manner as in the first case above ($f_{\infty}^{i}>L_{0}$) that $x(t)-q_{1}(t)\ge\frac {M_{3}}{c^{\alpha-1}}t^{\alpha-1}\ge0$ for all $t\in[0,1]$.

Therefore, for any $(x,y)\in P\cap\partial\Omega_{3}$ and $t\in [c,1-c]$, we deduce inequalities (15).

Hence, for any $(x,y)\in P\cap\partial\Omega_{3}$ and $t\in[c,1-c]$, we conclude

$$\begin{aligned} Q_{1}(x,y) (t) \ge&\lambda \int _{0}^{1}G_{2}(t,s)\bigl[g\bigl(s, \bigl[x(s)-q_{1}(s)\bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*} \bigr)+p_{2}(s)\bigr] \,ds \\ \ge&\lambda\varrho_{2}t^{\alpha-1} \int _{c}^{1-c}h_{2}(s)g\bigl(s, \bigl[x(s)-q_{1}(s)\bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*}\bigr) \,ds \\ \ge&\lambda L_{0}\varrho_{2} t^{\alpha-1} \int_{c}^{1-c}h_{2}(s) \,ds\ge\lambda L_{0}\varrho_{2} c^{\alpha-1} \int_{c}^{1-c}h_{2}(s) \,ds\ge R_{3}=\bigl\Vert (x,y)\bigr\Vert _{Y}. \end{aligned}$$

Therefore we obtain $\|Q_{1}(x,y)\|\ge R_{3}$, and so $\|{\mathcal {Q}}(x,y)\|_{Y}\ge R_{3}=\|(x,y)\|_{Y}$ for all $(x,y)\in P\cap\partial\Omega _{3}$, that is, we have relation (16).

By a similar approach we obtain relation (16) if $\|y\|\ge R_{3}/2$.

On the other hand, we consider the positive number

$$\begin{aligned} \varepsilon =&\min\biggl\{ \frac{1}{8\lambda\sigma_{1}} \biggl( \int_{0}^{1}h_{1}(s)\alpha_{1}(s) \,ds \biggr)^{-1}, \frac{1}{8\lambda\sigma _{2}} \biggl( \int_{0}^{1}h_{2}(s)\alpha_{2}(s) \,ds \biggr)^{-1}, \\ &\frac{1}{8\lambda\sigma_{3}} \biggl( \int_{0}^{1}h_{2}(s)\alpha_{2}(s) \,ds \biggr)^{-1}, \frac{1}{8\lambda\sigma_{4}} \biggl( \int_{0}^{1}h_{1}(s)\alpha_{1}(s) \,ds \biggr)^{-1}\biggr\} . \end{aligned}$$

Then by (H6) we deduce that there exists $M_{4}>0$ such that

$$\beta_{i}(t,u,v)\le\varepsilon(u+v),\quad \forall t\in[0,1], u, v\ge0, u+v\ge M_{4}, i=1,2. $$

Therefore we obtain

$$\beta_{i}(t,u,v)\le M_{5}+\varepsilon(u+v), \quad \forall t \in[0,1], u, v\ge0, i=1,2, $$

where $M_{5}= \max_{i=1,2} \{ \max_{t\in[0,1], u,v\ge0, u+v\le M_{4}}\beta_{i}(t,u,v) \}$.

We define now

$$\begin{aligned} R_{4} =&\max\biggl\{ 2R_{3},8\lambda\sigma_{1}\max \{M_{5},1\} \int _{0}^{1}h_{1}(s) \bigl( \alpha_{1}(s)+p_{1}(s)\bigr) \,ds, \\ &8\lambda\sigma_{2}\max\{M_{5},1\} \int_{0}^{1}h_{2}(s) \bigl( \alpha_{2}(s)+p_{2}(s)\bigr) \,ds, \\ &8\lambda\sigma_{3}\max\{M_{5},1\} \int_{0}^{1}h_{2}(s) \bigl( \alpha_{2}(s)+p_{2}(s)\bigr) \,ds, \\ &8\lambda\sigma_{4}\max\{M_{5},1\} \int_{0}^{1}h_{1}(s) \bigl( \alpha_{1}(s)+p_{1}(s)\bigr) \,ds\biggr\} , \end{aligned}$$

and let $\Omega_{4}=\{(x,y)\in P, \|(x,y)\|_{Y}< R_{4}\}$.

For any $(x,y)\in P\cap\partial\Omega_{4}$, we have

$$\begin{aligned} Q_{1}(x,y) (t) \le&\lambda \int_{0}^{1}\sigma_{1} h_{1}(s) \bigl[\alpha_{1}(s)\beta _{1}\bigl(s,\bigl[x(s)-q_{1}(s) \bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*}\bigr)+p_{1}(s)\bigr]\,ds \\ &{}+\lambda \int_{0}^{1}\sigma_{2} h_{2}(s) \bigl[\alpha_{2}(s)\beta _{2}\bigl(s,\bigl[x(s)-q_{1}(s) \bigr]^{*},\bigl[y(s)-q_{2}(s)\bigr]^{*}\bigr)+p_{2}(s)\bigr]\,ds \\ \le&\lambda\sigma_{1} \int_{0}^{1}h_{1}(s)\bigl[ \alpha_{1}(s) \bigl(M_{5}+\varepsilon \bigl( \bigl[x(s)-q_{1}(s)\bigr]^{*}+\bigl[y(s)-q_{2}(s)\bigr]^{*}\bigr) \bigr)+p_{1}(s)\bigr] \,ds \\ &{}+\lambda\sigma_{2} \int_{0}^{1}h_{2}(s)\bigl[ \alpha_{2}(s) \bigl(M_{5}+\varepsilon \bigl( \bigl[x(s)-q_{1}(s)\bigr]^{*}+\bigl[y(s)-q_{2}(s)\bigr]^{*}\bigr) \bigr)+p_{2}(s)\bigr] \,ds \\ \le&\lambda\sigma_{1}\max\{M_{5},1\} \int_{0}^{1}h_{1}(s) \bigl( \alpha_{1}(s)+p_{1}(s)\bigr) \,ds+\lambda\sigma_{1} \varepsilon R_{4} \int_{0}^{1}h_{1}(s)\alpha_{1}(s) \,ds \\ &{}+\lambda\sigma_{2}\max\{M_{5},1\} \int_{0}^{1}h_{2}(s) \bigl( \alpha_{2}(s)+p_{2}(s)\bigr) \,ds+\lambda\sigma_{2} \varepsilon R_{4} \int_{0}^{1}h_{2}(s)\alpha_{2}(s) \,ds \\ \le&\frac{R_{4}}{8}+\frac{R_{4}}{8}+\frac{R_{4}}{8}+ \frac{R_{4}}{8}=\frac {R_{4}}{2}=\frac{\|(x,y)\|_{Y}}{2}, \quad \forall t\in[0,1], \end{aligned}$$

and so $\|Q_{1}(x,y)\|\le\frac{\|(x,y)\|_{Y}}{2}$ for all $(x,y)\in P\cap \partial\Omega_{4}$.

In a similar way we obtain $Q_{2}(x,y)(t)\le\frac{\|(x,y)\|_{Y}}{2}$ for all $t\in[0,1]$, and so $\|Q_{2}(x,y)\|\le\frac{\|(x,y)\|_{Y}}{2}$ for all $(x,y)\in P\cap\partial\Omega_{4}$.

Therefore, we deduce

$$ \bigl\Vert {\mathcal{Q}}(x,y)\bigr\Vert _{Y}\le \bigl\Vert (x,y)\bigr\Vert _{Y}, \quad \forall(x,y)\in P\cap \partial \Omega_{4}. $$

(17)

By Theorem 2.2, (16), and (17), we conclude that ${\mathcal{Q}}$ has a fixed point $(x_{1},y_{1})\in P\cap(\bar{\Omega}_{4}\setminus\Omega_{3})$. Since $\|(x_{1},y_{1})\|\ge R_{3}$ then $\|x_{1}\| \ge R_{3}/2$ or $\|y_{1}\|\ge R_{3}/2$.

We suppose that $\|x_{1}\|\ge R_{3}/2$. Then $x_{1}(t)-q_{1}(t)\ge\frac {M_{3}}{c^{\alpha-1}}t^{\alpha-1}$ for all $t\in[0,1]$. Besides

$$y_{1}(1)= \int_{0}^{1}x_{1}(s) \,dK(s)\ge \gamma_{1}\|x_{1}\| \int_{0}^{1} s^{\alpha-1} \,dK(s) \ge \frac{\gamma_{1} R_{3}}{2} \int_{0}^{1}s^{\alpha-1} \,dK(s)>0, $$

and then

$$\|y_{1}\|\ge y_{1}(1)= \int_{0}^{1}x_{1}(s) \,dK(s)\ge \frac{\gamma_{1} R_{3}}{2} \int _{0}^{1} s^{\alpha-1} \,dK(s)>0. $$

Therefore, we deduce that, for all $t\in[0,1]$,

$$\begin{aligned} y_{1}(t)-q_{2}(t) \ge& y_{1}(t)-\lambda \delta_{3} \int_{0}^{1}t^{\beta-1}p_{2}(s) \,ds- \lambda\delta_{4} \int_{0}^{1} t^{\beta-1}p_{1}(s) \,ds \\ \ge&\gamma_{2} t^{\beta-1}\|y_{1}\|-\lambda t^{\beta-1} \int_{0}^{1}\bigl(\delta _{3}p_{2}(s)+ \delta_{4} p_{1}(s)\bigr) \,ds \\ \ge&\frac{\gamma_{1}\gamma_{2} R_{3}}{2} t^{\beta-1} \int_{0}^{1}s^{\alpha-1} \,dK(s)-\lambda t^{\beta-1} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4}p_{1}(s)\bigr) \,ds \\ \ge&\lambda t^{\beta-1} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr) \,ds \\ \ge& \lambda_{*} t^{\beta-1} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr) \,ds\ge \frac{M_{3}}{c^{\beta-1}}t^{\beta-1}. \end{aligned}$$

If $\|y_{1}\|\ge R_{3}/2$, then by a similar approach we conclude again that $x_{1}(t)-q_{1}(t)\ge\frac{M_{3}}{c^{\alpha-1}}t^{\alpha-1}$ and $y_{1}(t)-q_{2}(t)\ge\frac{M_{3}}{c^{\beta-1}}t^{\beta-1}$ for all $t\in[0,1]$.

Let $u_{1}(t)=x_{1}(t)-q_{1}(t)$ and $v_{1}(t)=y_{1}(t)-q_{2}(t)$ for all $t\in [0,1]$. Then $u_{1}(t)\ge\widetilde{\Lambda}_{1} t^{\alpha-1}$ and $v_{1}(t)\ge\widetilde{\Lambda}_{2} t^{\beta-1}$ for all $t\in[0,1]$, where $\widetilde{\Lambda}_{1}=\frac{M_{3}}{c^{\alpha-1}}$, $\widetilde{\Lambda}_{2}=\frac{M_{3}}{c^{\beta-1}}$. Hence we deduce that $(u_{1},v_{1})$ is a positive solution of (S)-(BC), which completes the proof of Theorem 3.4. □

In a similar manner as we proved Theorem 3.4, we obtain the following theorems.

Theorem 3.5

Assume that $\lambda=\mu$, and (H1), (H4), and (H6) hold. In addition if

(H7′):

there exists $c\in(0,1/2)$ such that

$$f_{\infty}^{i}= \liminf_{\substack{u+v\to\infty\\u,v\ge0}} \min _{t\in [c,1-c]}f(t,u,v)>\widetilde{L}_{0} \quad \textit{or} \quad g_{\infty}^{i}= \liminf_{\substack{u+v\to\infty\\u,v\ge0}} \min _{t\in [c,1-c]}g(t,u,v)>\widetilde{L}_{0}, $$

where

$$\begin{aligned} \widetilde{L}_{0} =&\max\biggl\{ \frac{4}{\gamma_{1}} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr)\,ds, \frac{4}{\gamma_{2}} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr)\,ds, \\ &\frac{4}{\gamma_{1}\gamma_{2}} \biggl( \int_{0}^{1} s^{\alpha-1}\,dK(s) \biggr)^{-1} \int_{0}^{1}\bigl(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)\bigr)\,ds, \\ &\frac{4}{\gamma_{1}\gamma_{2}} \biggl( \int_{0}^{1} s^{\beta-1}\,dH(s) \biggr)^{-1} \int_{0}^{1}\bigl(\delta_{1} p_{1}(s)+\delta_{2} p_{2}(s)\bigr)\,ds\biggr\} \\ &{}\times \biggl(\min \biggl\{ c^{\beta-1}\varrho_{3} \int_{c}^{1-c}h_{2}(s) \,ds,c^{\beta-1}\varrho_{4} \int_{c}^{1-c}h_{1}(s) \,ds \biggr\} \biggr)^{-1}, \end{aligned}$$

then there exists $\lambda'_{*}>0$ such that for any $\lambda\ge\lambda '_{*}$ problem (S)-(BC) (with $\lambda=\mu$) has at least one positive solution.

Theorem 3.6

Assume that $\lambda=\mu$, and (H1), (H4), and (H6) hold. In addition if

(H8):

there exists $c\in(0,1/2)$ such that

$$\hat{f}_{\infty}= \lim_{\substack{u+v\to\infty\\u,v\ge0}} \min_{t\in [c,1-c]}f(t,u,v)= \infty \quad \textit{or}\quad \hat{g}_{\infty}= \lim_{\substack{u+v\to\infty\\u,v\ge0}} \min_{t\in [c,1-c]}g(t,u,v)=\infty, $$

then there exists $\tilde{\lambda}_{*}>0$ such that for any $\lambda \ge\tilde{\lambda}_{*}$ problem (S)-(BC) (with $\lambda=\mu$) has at least one positive solution.

4 Examples

Let $\alpha=5/2$ ($n=3$), $\beta=7/3$ ($m=3$), $H(t)=t^{2}$, $K(t)=t^{3}$. Then $\int_{0}^{1}u(s)\,dK(s)=3\int_{0}^{1}s^{2}u(s) \,ds$ and $\int _{0}^{1}v(s)\,dH(s)=2\int_{0}^{1}sv(s) \,ds$.

We consider the system of fractional differential equations

$$(\mathrm{S}_{0})\quad \left \{ \textstyle\begin{array}{l} D_{0+}^{5/2}u(t)+\lambda f(t,u(t),v(t))=0,\quad t\in(0,1), \\ D_{0+}^{7/3}v(t)+\mu g(t,u(t),v(t))=0, \quad t\in(0,1), \end{array}\displaystyle \right . $$

with the boundary conditions

$$(\mathrm{BC}_{0})\quad \left \{ \textstyle\begin{array}{l}u(0)=u'(0)=0,\qquad u(1)=2\int_{0}^{1}sv(s) \,ds, \\ v(0)=v'(0)=0, \qquad v(1)=3\int_{0}^{1}s^{2}u(s) \,ds. \end{array}\displaystyle \right . $$

Then we obtain $\Delta=1- (\int_{0}^{1}s^{\alpha-1}\,dK(s) ) (\int_{0}^{1}s^{\beta-1}\,dH(s) )=\frac{3}{5}>0$, $\int_{0}^{1}\tau^{\alpha -1}(1-\tau) \,dK(\tau)=\frac{4}{33}>0$, $\int_{0}^{1}\tau^{\beta-1}(1-\tau )\,dH(\tau)=\frac{9}{65}>0$. The functions H and K are nondecreasing, and so assumption (H1) is satisfied. Besides, we deduce

$$\begin{aligned}& g_{1}(t,s)= \frac{4}{3\sqrt{\pi}} \left \{ \textstyle\begin{array}{l@{\quad}l} t^{3/2}(1-s)^{3/2}-(t-s)^{3/2},& 0\le s\le t\le1, \\ t^{3/2}(1-s)^{3/2},& 0\le t\le s\le1, \end{array}\displaystyle \right . \\& g_{2}(t,s)= \frac{1}{\Gamma(7/3)}\left \{ \textstyle\begin{array}{l@{\quad}l} t^{4/3}(1-s)^{4/3}-(t-s)^{4/3},& 0\le s\le t\le1, \\ t^{4/3}(1-s)^{4/3},& 0\le t\le s\le1, \end{array}\displaystyle \right . \\& G_{1}(t,s)=g_{1}(t,s)+3t^{3/2} \int_{0}^{1}\tau^{2} g_{1}( \tau,s) \,ds, \qquad G_{2}(t,s)=\frac {10}{3}t^{3/2} \int_{0}^{1}\tau g_{2}(\tau,s)\, d\tau, \\& G_{3}(t,s)=g_{2}(t,s)+ \frac{20}{9}t^{4/3} \int_{0}^{1}\tau g_{2}(\tau,s) d\tau, \qquad G_{4}(t,s)=5t^{4/3} \int_{0}^{1}\tau^{2}g_{1}( \tau,s)\, d\tau. \end{aligned}$$

We also obtain $h_{1}(s)=\frac{2}{\sqrt{\pi}}s(1-s)^{3/2}$, $h_{2}(s)=\frac {1}{\Gamma(4/3)}s(1-s)^{4/3}$,

$$k_{1}(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{2}{3}t^{3/2}, &0\le t\le1/2, \\ \frac{2}{3}(1-t)t^{1/2}, &1/2\le t\le1, \end{array}\displaystyle \right .\qquad k_{2}(t)=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{3}{4}t^{4/3}, &0\le t\le1/2, \\ \frac{3}{4}(1-t)t^{1/3}, &1/2\le t\le1. \end{array}\displaystyle \right . $$

In addition, we have $\sigma_{1}=2$, $\delta_{1}=\frac{74}{33\sqrt{\pi}}$, $\varrho_{1}=\frac{8\sqrt{2}-1}{63\sqrt{2}}$, $\sigma_{2}=\frac{5}{3}$, $\delta_{2}=\frac{3}{13\Gamma(4/3)}$, $\varrho_{2}=\frac{36\sqrt [3]{2}-9}{112\sqrt[3]{2}}$, $\sigma_{3}=\frac{19}{9}$, $\delta_{3}=\frac{15}{13\Gamma(4/3)}$, $\varrho _{3}=\frac{12\sqrt[3]{2}-3}{56\sqrt[3]{2}}$, $\sigma_{4}=\frac{5}{3}$, $\delta_{4}=\frac{40}{99\sqrt{\pi}}$, $\varrho_{4}=\frac{40\sqrt {2}-5}{189\sqrt{2}}$, $\gamma_{1}=\frac{8\sqrt{2}-1}{126\sqrt{2}}\approx 0.0578801$, $\gamma_{2}=\frac{9(12\sqrt[3]{2}-3)}{1064\sqrt[3]{2}}\approx0.08136286$.

Example 1

We consider the functions

$$f(t,u,v)= \frac{(u+v)^{2}}{\sqrt{t(1-t)}}+\ln t,\qquad g(t,u,v)= \frac{2+\sin(u+v)}{\sqrt{t(1-t)}}+ \ln(1-t), \quad t\in(0,1), u,v\ge0. $$

We have $p_{1}(t)=-\ln t$, $p_{2}(t)=-\ln(1-t)$, $\alpha_{1}(t)=\alpha _{2}(t)=\frac{1}{\sqrt{t(1-t)}}$ for all $t\in(0,1)$, $\beta _{1}(t,u,v)=(u+v)^{2}$, $\beta_{2}(t,u,v)=2+\sin(u+v)$ for all $t\in[0,1]$, $u, v\ge0$, $\int_{0}^{1}p_{1}(t) \, dt=1$, $\int_{0}^{1}p_{2}(t)\, dt=1$, $\int_{0}^{1}\alpha_{i}(t)\, dt=\pi$, $i=1,2$. Therefore, assumption (H4) is satisfied. In addition, for $c\in(0,1/2)$ fixed, assumption (H5) is also satisfied ($f_{\infty}=\infty$).

After some computations, we deduce $\int_{0}^{1}(\delta_{1}p_{1}(s)+\delta _{2}p_{2}(s)) \,ds\approx1.52357852$, $\int_{0}^{1}(\delta_{3} p_{2}(s)+\delta_{4} p_{1}(s)) \,ds\approx1.520086$, $\int_{0}^{1}h_{1}(s)(\alpha_{1}(s)+p_{1}(s)) \,ds\approx0.42548534$, $\int_{0}^{1}h_{2}(s)(\alpha_{2}(s)+p_{2}(s)) \,ds\approx 0.44092924$. We choose $R_{1}=1080$, which satisfies the condition from the beginning of the proof of Theorem 3.2. Then $M_{1}=R_{1}^{2}$, $M_{2}=3$, $\lambda^{*}\approx2.7202\cdot10^{-4}$, and $\mu^{*}=1$. By Theorem 3.2, we conclude that ($\mathrm{S}_{0}$)-($\mathrm{BC}_{0}$) has at least one positive solution for any $\lambda\in(0,\lambda^{*}]$ and $\mu \in(0,\mu^{*}]$.

Example 2

We consider the functions

$$f(t,u,v)=(u+v)^{2}+\cos u, \qquad g(t,u,v)=(u+v)^{1/2}+\cos v, \quad t\in[0,1], u, v\ge0. $$

We have $p_{1}(t)=p_{2}(t)=1$ for all $t\in[0,1]$, and then assumption (H2) is satisfied. Besides, assumption (H3) is also satisfied, because $f(t,0,0)=1$ and $g(t,0,0)=1$ for all $t\in[0,1]$.

Let $\delta=\frac{1}{2}<1$ and $R_{0}=1$. Then

$$f(t,u,v)\ge\delta f(t,0,0)=\frac{1}{2}, \qquad g(t,u,v)\ge\delta g(t,0,0)=\frac{1}{2},\quad \forall t\in[0,1], u, v\in[0,1]. $$

In addition,

$$\begin{aligned}& \bar{f}(R_{0})=\bar{f}(1)= \max_{t\in[0,1], u,v\in[0,1]}\bigl\{ f(t,u,v)+p_{1}(t)\bigr\} \approx5.5403023, \\& \bar{g}(R_{0})=\bar{g}(1)= \max_{t\in[0,1], u,v\in[0,1]}\bigl\{ g(t,u,v)+p_{2}(t)\bigr\} \approx3.10479256. \end{aligned}$$

We also obtain $c_{1}\approx0.25791523$, $c_{2}\approx0.23996711$, $c_{3}\approx0.30395834$, $c_{4}\approx0.21492936$, and then $\lambda _{0}=\max \{\frac{R_{0}}{8c_{1}\bar{f}(R_{0})},\frac{R_{0}}{8c_{4}\bar{f}(R_{0})} \}\approx0.10497377$ and $\mu_{0}=\max \{\frac{R_{0}}{8c_{2}\bar{g}(R_{0})},\frac{R_{0}}{8c_{3}\bar{g}(R_{0})} \}\approx 0.1677744$.

By Theorem 3.1, for any $\lambda\in(0,\lambda_{0}]$ and $\mu \in(0,\mu_{0}]$, we deduce that problem ($\mathrm{S}_{0}$)-($\mathrm{BC}_{0}$) has at least one positive solution.

Because assumption (H4′) is satisfied ($\alpha_{1}(t)=\alpha_{2}(t)=1$, $\beta_{1}(t,u,v)=(u+v)^{2}+1$, $\beta_{2}(t,u,v)=(u+v)^{1/2}+1$ for all $t\in[0,1]$, $u,v\ge0$) and assumption (H5) is also satisfied ($f_{\infty}=\infty$), by Theorem 3.3 we conclude that problem ($\mathrm{S}_{0}$)-($\mathrm{BC}_{0}$) has at least two positive solutions for λ and μ sufficiently small.

Example 3

We consider $\lambda=\mu$ and the functions

$$f(t,u,v)= \frac{(u+v)^{a}}{\sqrt[3]{t^{2}(1-t)}}-\frac{1}{\sqrt{t}},\qquad g(t,u,v)= \frac{\ln(1+u+v)}{\sqrt[3]{t(1-t)^{2}}}-\frac{1}{\sqrt{1-t}},\quad t\in(0,1), u,v\ge0, $$

where $a\in(0,1)$.

Here we have $p_{1}(t)=\frac{1}{\sqrt{t}}$, $p_{2}(t)=\frac{1}{\sqrt {1-t}}$, $\alpha_{1}(t)=\frac{1}{\sqrt[3]{t^{2}(1-t)}}$, $\alpha_{2}(t)=\frac {1}{\sqrt[3]{t(1-t)^{2}}}$ for all $t\in(0,1)$, $\beta_{1}(t,u,v)=(u+v)^{a}$, $\beta_{2}(t,u,v)=\ln(1+u+v)$ for all $t\in[0,1]$, $u,v\ge0$. For $c\in (0,1/2)$ fixed, the assumptions (H4), (H6), and (H8) are satisfied ($\beta_{i\infty }=0$ for $i=1,2$ and $\hat{f}_{\infty}=\infty$).

Then by Theorem 3.6, we deduce that there exists $\tilde {\lambda}_{*}>0$ such that for any $\lambda\ge\tilde{\lambda}_{*}$ our problem ($\mathrm{S}_{0}$)-($\mathrm{BC}_{0}$) (with $\lambda=\mu$) has at least one positive solution.