Positive solutions for a system of Riemann–Liouville fractional boundary value problems with p-Laplacian operators

Abstract

We study the existence and nonexistence of positive solutions for a system of Riemann–Liouville fractional differential equations with p-Laplacian operators, nonnegative nonlinearities and positive parameters, subject to coupled nonlocal boundary conditions which contain Riemann–Stieltjes integrals and various fractional derivatives. We use the Guo–Krasnosel’skii fixed point theorem in the proof of the main existence results.

1 Introduction

In this paper we consider the system of nonlinear ordinary fractional differential equations with $\varrho _{1}$-Laplacian and $\varrho _{2}$-Laplacian operators

$$ \textstyle\begin{cases} D_{0+}^{\alpha _{1}}(\varphi _{\varrho _{1}} (D_{0+}^{\beta _{1}} x(t)))+ \lambda f(t,x(t),y(t))=0,\quad t\in (0,1), \\ D_{0+}^{\alpha _{2}}(\varphi _{\varrho _{2}}(D_{0+}^{\beta _{2}}y(t)))+ \mu g(t,x(t),y(t))=0,\quad t\in (0,1), \end{cases} $$

(1)

with the coupled nonlocal boundary conditions

$$ \textstyle\begin{cases} x^{(j)}(0)=0, \quad j=0,\ldots ,n-2; \qquad D_{0+}^{\beta _{1}}x(0)=0, \\ D_{0+}^{\gamma _{0}}x(1)= \sum_{i=1}^{p} \int _{0}^{1} D_{0+}^{\gamma _{i}}y(t) \,dH_{i}(t), \\ y^{(j)}(0)=0, \quad j=0,\ldots ,m-2;\qquad D_{0+}^{\beta _{2}}y(0)=0, \\ D_{0+}^{\delta _{0}}y(1)= \sum_{i=1}^{q} \int _{0}^{1} D_{0+}^{\delta _{i}}x(t) \,dK_{i}(t), \end{cases} $$

(2)

where $\alpha _{1}, \alpha _{2}\in (0,1]$, $\beta _{1}\in (n-1,n]$, $\beta _{2}\in (m-1,m]$, $n, m\in \mathbb{N}$, $n, m\ge 3$, $p,q\in \mathbb{N}$, $\gamma _{i}\in \mathbb{R}$ for all $i=0,1,\ldots ,p$, $0\le \gamma _{1}<\gamma _{2}<\cdots <\gamma _{p}\le \delta _{0}< \beta _{2}-1$, $\delta _{0}\ge 1$, $\delta _{i}\in \mathbb{R}$ for all $i=0,1,\ldots ,q$, $0\le \delta _{1}<\delta _{2}<\cdots <\delta _{q}\le \gamma _{0}< \beta _{1}-1$, $\gamma _{0}\ge 1$, $\varrho _{1}, \varrho _{2}>1$, $\varphi _{\varrho _{i}}(s)=|s|^{\varrho _{i}-2}s$, $\varphi _{\varrho _{i}}^{-1}=\varphi _{\rho _{i}}$, $\frac{1}{\varrho _{i}}+\frac{1}{\rho _{i}}=1$, $i=1,2$, $\lambda , \mu >0$, the functions $f, g\in C([0,1]\times \mathbb{R}_{+}\times \mathbb{R}_{+}, \mathbb{R}_{+})$ ($\mathbb{R}_{+}=[0,\infty )$), the integrals from (2) are Riemann–Stieltjes integrals with $H_{i}$, $i=1,\ldots ,p$, and $K_{i}$, $i=1,\ldots ,q$, functions of bounded variation, and $D_{0+}^{k}$ denotes the Riemann–Liouville derivative of order k (for $k=\alpha _{1}, \beta _{1}, \alpha _{2}, \beta _{2}, \gamma _{i}$ for $i=0,1,\ldots ,p$, $\delta _{i}$ for $i=0,1,\ldots ,q$).

We present sufficient conditions on the functions f and g, and intervals for the parameters λ and μ such that problem (1), (2) has positive solutions. A positive solution of problem (1), (2) is a pair of functions $(x,y)\in (C([0,1],\mathbb{R}_{+}))^{2}$, satisfying (1) and (2) with $x(t)>0$ for all $t\in (0,1]$ or $y(t)>0$ for all $t\in (0,1]$. We also investigate the nonexistence of positive solutions for this problem. The problem (1), (2) is a generalization of the problem studied in [30]. Indeed, if $p=1$, $q=1$, $\gamma _{0}=p_{1}$, $\gamma _{1}=q_{1}$, $\delta _{0}=p_{2}$, $\delta _{1}=q_{2}$, $H_{1}$ is a step function given by $H_{1}(t)=\{0, t\in [0,\xi _{1}); a_{1}, t\in [\xi _{1},\xi _{2}); a_{1}+a_{2}, t\in [\xi _{2},\xi _{3});\ldots ;\sum_{i=1}^{N}a_{i}, t\in [\xi _{N},1]\}$, and $K_{1}$ is a step function given by $K_{1}(t)=\{0, t\in [0,\eta _{1}); b_{1}, t\in [\eta _{1}, \eta _{2}); b_{1}+b_{2}, t\in [\eta _{2},\eta _{3});\ldots ; \sum_{i=1}^{M} b_{i}, t\in [\eta _{M},1]\}$, then the boundary conditions (2) become the multi-point boundary conditions $(BC)$ from [30]. The system (1) with uncoupled multi-point boundary conditions was studied in [29]. Fractional differential equations and systems of fractional differential equations, subject to various multi-point or Riemann–Stieltjes integral boundary conditions, were studied in the last years in [1–8, 17–25, 28, 31, 37, 39–41]. For various applications of the fractional calculus in several scientific and engineering fields, the reader may consult the books [10, 11, 26, 27, 34–36], and the papers [9, 12–15, 32, 33, 38].

In Sect. 2, we study a nonlocal boundary value problem for fractional differential equations with p-Laplacian operators, and we present some properties of the associated Green functions. Section 3 is devoted to the main existence theorems for the positive solutions with respect to a cone for our problem (1), (2), which are based on the Guo–Krasnosel’skii fixed point theorem (see [16]). In Sect. 4, we present some nonexistence results for the positive solutions of (1), (2), and in Sect. 5, we give an example which illustrates our results.

2 Preliminary results

We consider the system of fractional differential equations

$$ \textstyle\begin{cases} D_{0+}^{\alpha _{1}}(\varphi _{\varrho _{1}} (D_{0+}^{\beta _{1}} x(t)))+ \psi (t)=0, \quad t\in (0,1), \\ D_{0+}^{\alpha _{2}}(\varphi _{\varrho _{2}}(D_{0+}^{\beta _{2}}y(t)))+ \chi (t)=0, \quad t\in (0,1), \end{cases} $$

(3)

with the coupled boundary conditions (2), where $\psi , \chi \in C(0,1)\cap L^{1}(0,1)$.

We denote $\varphi _{\varrho _{1}}(D_{0^{+}}^{\beta _{1}}x(t))=u(t)$ and $\varphi _{\varrho _{2}}(D_{0^{+}}^{\beta _{2}}y(t))=v(t)$. Then problem (3), (2) is equivalent to the following three problems:

$$\begin{aligned}& \textstyle\begin{cases} D_{0+}^{\alpha _{1}}u(t)+\psi (t)=0, \quad 0< t< 1, \\ u(0)=0, \end{cases}\displaystyle \end{aligned}$$

(4)

$$\begin{aligned}& \textstyle\begin{cases} D_{0+}^{\alpha _{2}}v(t)+\chi (t)=0,\quad 0< t< 1, \\ v(0)=0, \end{cases}\displaystyle \end{aligned}$$

(5)

and

$$ \textstyle\begin{cases} D_{0+}^{\beta _{1}}x(t)=\varphi _{\rho _{1}}(u(t)),\quad t\in (0,1), \\ D_{0+}^{\beta _{2}}y(t)=\varphi _{\rho _{2}}(v(t)),\quad t\in (0,1), \end{cases} $$

(6)

with the boundary conditions

$$ \textstyle\begin{cases} x^{(j)}(0)=0, \quad j=0,\ldots ,n-2;\qquad D_{0+}^{\gamma _{0}}x(1)= \sum_{i=1}^{p} \int _{0}^{1}D_{0+}^{\gamma _{i}} y(t) \,dH_{i}(t), \\ y^{(j)}(0)=0, \quad j=0,\ldots ,m-2;\qquad D_{0+}^{\delta _{0}}y(1)= \sum_{i=1}^{q} \int _{0}^{1} D_{0+}^{\delta _{i}}x(t) \,dK_{i}(t). \end{cases} $$

(7)

The first two problems (4) and (5) have the unique solutions $u\in C[0,1]$ and $v\in C[0,1]$, respectively, of the form

$$ u(t)=-I_{0+}^{\alpha _{1}}\psi (t)=- \frac{1}{\varGamma (\alpha _{1})} \int _{0}^{t} (t-\nu )^{ \alpha _{1}-1}\psi ( \nu ) \,d\nu ,\quad t\in [0,1], $$

(8)

and

$$ v(t)=-I_{0+}^{\alpha _{2}}\chi (t)=- \frac{1}{\varGamma (\alpha _{2})} \int _{0}^{t} (t-\nu )^{ \alpha _{2}-1}\chi ( \nu ) \,d\nu ,\quad t\in [0,1], $$

(9)

respectively.

For the third problem (6), (7), we denote

$$ \begin{aligned}[b] \Delta ={}& \frac{\varGamma (\beta _{1})\varGamma (\beta _{2})}{\varGamma (\beta _{1}-\gamma _{0})\varGamma (\beta _{2}-\delta _{0})}- \Biggl( \sum_{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1}\tau ^{\beta _{2}-\gamma _{i}-1} \,dH_{i}( \tau ) \Biggr) \\ &{}\times \Biggl( \sum_{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1}\tau ^{\beta _{1}-\delta _{i}-1} \,dK_{i}( \tau ) \Biggr). \end{aligned} $$

(10)

Lemma 2.1

If$\Delta \neq0$, then the unique solution$(x,y)\in C[0,1]\times C[0,1]$of problem (6), (7) is given by

$$\begin{aligned}& x(t)= \frac{1}{\varGamma (\beta _{1})} \int _{0}^{t} (t-\nu )^{\beta _{1}-1}\varphi _{\rho _{1}} \bigl(u(\nu ) \bigr) \,d\nu \\& \hphantom{x(t)= {}}{}+ \frac{t^{\beta _{1}-1}}{\Delta } \Biggl[- \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{1}-\gamma _{0})\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1-\nu )^{\beta _{1}-\gamma _{0}-1} \varphi _{\rho _{1}} \bigl(u(\nu ) \bigr) \,d\nu \\& \hphantom{x(t)= {}}{}+ \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\delta _{0})} \sum_{i=1}^{p} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} \biggl( \int _{0}^{\nu } (\nu -\tau )^{\beta _{2}- \gamma _{i}-1} \varphi _{\rho _{2}} \bigl(v(\tau ) \bigr) \,d\tau \biggr) \,dH_{i}( \nu ) \\& \hphantom{x(t)= {}}{}- \Biggl( \sum_{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} {\nu }^{\beta _{2}-\gamma _{i}-1} \,dH_{i}( \nu ) \Biggr) \\& \hphantom{x(t)= {}}{}\times \biggl( \frac{1}{\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1- \nu )^{\beta _{2}-\delta _{0}-1}\varphi _{\rho _{2}} \bigl(v(\nu ) \bigr) \,d\nu \biggr) \\& \hphantom{x(t)= {}}{}+ \Biggl( \sum_{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} {\nu }^{\beta _{2}-\gamma _{i}-1} \,dH_{i}( \nu ) \Biggr) \\& \hphantom{x(t)= {}}{}\times \Biggl( \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} \biggl( \int _{0}^{\nu } (\nu -\tau )^{ \beta _{1}-\delta _{i}-1} \varphi _{\rho _{1}} \bigl(u(\tau ) \bigr) \,d\tau \biggr) \,dK_{i}( \nu ) \Biggr) \Biggr], \\& \hphantom{x(t)= {}}{} \forall t\in [0,1], \\& y(t)= \frac{1}{\varGamma (\beta _{2})} \int _{0}^{t} (t-\nu )^{\beta _{2}-1}\varphi _{\rho _{2}} \bigl(v(\nu ) \bigr) \,d\nu \\& \hphantom{y(t)={}}{}+ \frac{t^{\beta _{2}-1}}{\Delta } \Biggl[- \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\gamma _{0})\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1-\nu )^{\beta _{2}-\delta _{0}-1} \varphi _{\rho _{2}} \bigl(v(\nu ) \bigr) \,d\nu \\& \hphantom{y(t)={}}{}+ \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\gamma _{0})} \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} \biggl( \int _{0}^{\nu } (\nu -\tau )^{\beta _{1}- \delta _{i}-1} \varphi _{\rho _{1}} \bigl(u(\tau ) \bigr) \,d\tau \biggr) \,dK_{i}( \nu ) \\& \hphantom{y(t)={}}{}- \Biggl( \sum_{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} {\nu }^{\beta _{1}-\delta _{i}-1} \,dK_{i}( \nu ) \Biggr) \\& \hphantom{y(t)={}}{}\times \biggl( \frac{1}{\varGamma (\beta _{1}-\gamma _{0})} \int _{0}^{1}(1- \nu )^{\beta _{1}-\gamma _{0}-1}\varphi _{\rho _{1}} \bigl(u(\nu ) \bigr) \,d\nu \biggr) \\& \hphantom{y(t)={}}{}+ \Biggl( \sum_{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} {\nu }^{\beta _{1}-\delta _{i}-1} \,dK_{i}( \nu ) \Biggr) \\& \hphantom{y(t)={}}{}\times \Biggl( \sum_{i=1}^{p} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} \biggl( \int _{0}^{\nu } (\nu -\tau )^{ \beta _{2}-\gamma _{i}-1} \varphi _{\rho _{2}} \bigl(v(\tau ) \bigr) \,d\tau \biggr) \,dH_{i}( \nu ) \Biggr) \Biggr], \\& \hphantom{y(t)={}}{} \forall t\in [0,1]. \end{aligned}$$

(11)

Proof

The solutions $(x,y)\in (C(0,1)\cap L^{1}(0,1))^{2}$ of system (6) are

$$ \begin{aligned} x(t)=I_{0+}^{\beta _{1}}\varphi _{\rho _{1}} \bigl(u(t) \bigr)+a_{1}t^{\beta _{1}-1}+a_{2} t^{\beta _{1}-2}+\cdots +a_{n} t^{\beta _{1}-n}, \\ y(t)=I_{0+}^{\beta _{2}}\varphi _{\rho _{2}} \bigl(v(t) \bigr)+b_{1}t^{\beta _{2}-1}+b_{2}t^{ \beta _{2}-2}+ \cdots +b_{m} t^{\beta _{2}-m}, \end{aligned} $$

with $a_{1},a_{2},\ldots ,a_{n},b_{1},b_{2},\ldots ,b_{m}\in \mathbb{R}$. By using the boundary conditions $x(0)=x'(0)=\cdots =x^{(n-2)}(0)=0$ and $y(0)=y'(0)=\cdots =y^{(m-2)}=0$, we obtain $a_{2}=\cdots =a_{n}=0$ and $b_{2}=\cdots =b_{m}=0$. So the above solutions become

$$ \begin{aligned} &x(t)=I_{0+}^{\beta _{1}} \varphi _{\rho _{1}} \bigl(u(t) \bigr)+a_{1}t^{\beta _{1}-1}= \frac{1}{\varGamma (\beta _{1})} \int _{0}^{t} (t-\nu )^{\beta _{1}-1} \varphi _{\rho _{1}} \bigl(u(\nu ) \bigr) \,d\nu +a_{1} t^{\beta _{1}-1}, \\ &y(t)=I_{0+}^{\beta _{2}}\varphi _{\rho _{2}} \bigl(v(t) \bigr)+b_{1}t^{\beta _{2}-1}= \frac{1}{\varGamma (\beta _{2})} \int _{0}^{t}(t-\nu )^{\beta _{2}-1} \varphi _{\rho _{2}} \bigl(v(\nu ) \bigr) \,d\nu +b_{1} t^{\beta _{2}-1}. \end{aligned} $$

(12)

For the obtained functions x and y we have

$$\begin{aligned}& D_{0+}^{\gamma _{i}}y(t)=b_{1} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})}t^{\beta _{2}- \gamma _{i}-1}+I_{0+}^{\beta _{1}-\gamma _{i}} \varphi _{\rho _{2}} \bigl(v(t) \bigr),\quad i=1,\ldots ,p, \\& D_{0+}^{\gamma _{0}}x(t)=a_{1} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\gamma _{0})}t^{\beta _{1}- \gamma _{0}-1}+I_{0+}^{\beta _{1}-\gamma _{0}} \varphi _{\rho _{1}} \bigl(u(t) \bigr), \\& D_{0+}^{\delta _{i}}x(t)=a_{1} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})}t^{\beta _{1}- \delta _{i}-1}+I_{0+}^{\beta _{1}-\delta _{i}} \varphi _{\rho _{1}} \bigl(u(t) \bigr),\quad i=1,\ldots ,q, \\& D_{0+}^{\delta _{0}}y(t)=b_{1} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\delta _{0})}t^{\beta _{2}- \delta _{0}-1}+I_{0+}^{\beta _{2}-\delta _{0}} \varphi _{\rho _{2}} \bigl(v(t) \bigr), \\& \begin{aligned} D_{0+}^{\gamma _{0}}x(1)&=a_{1} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\gamma _{0})}+I_{0+}^{ \beta _{1}-\gamma _{0}} \varphi _{\rho _{1}} \bigl(u(t) \bigr)|_{t=1} \\ &=a_{1} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\gamma _{0})}+ \frac{1}{\varGamma (\beta _{1}-\gamma _{0})} \int _{0}^{1}(1-\nu )^{\beta _{1}-\gamma _{0}-1}\varphi _{\rho _{1}} \bigl(u( \nu ) \bigr) \,d\nu, \end{aligned} \\& \begin{aligned} D_{0+}^{\delta _{0}}y(1)&=b_{1} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\delta _{0})}+I_{0+}^{ \beta _{2}-\delta _{0}} \varphi _{\rho _{2}} \bigl(v(t) \bigr)|_{t=1} \\ &=b_{1} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\delta _{0})}+ \frac{1}{\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1-\nu )^{\beta _{2}-\delta _{0}-1}\varphi _{\rho _{2}} \bigl(v( \nu ) \bigr) \,d\nu . \end{aligned} \end{aligned}$$

By imposing the boundary conditions $D_{0+}^{\gamma _{0}}x(1)=\sum_{i=1}^{p}\int _{0}^{1} D_{0+}^{ \gamma _{i}} y(s) \,dH_{i}(s)$ and $D_{0+}^{\delta _{0}}y(1)=\sum_{i=1}^{q}\int _{0}^{1} D_{0+}^{ \delta _{i}} x(s) \,dK_{i}(s)$, we deduce the following system in $a_{1}$ and $b_{1}$:

$$ \textstyle\begin{cases} a_{1} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\gamma _{0})}-b_{1} \sum_{i=1}^{p} \int _{0}^{1} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})}t^{\beta _{2}- \gamma _{i}-1}\,dH_{i}(t) \\ \quad = - \frac{1}{\varGamma (\beta _{1}-\gamma _{0})} \int _{0}^{1}(1-\nu )^{\beta _{1}-\gamma _{0}-1} \varphi _{\rho _{1}}(u(\nu )) \,d\nu \\ \qquad {} + \sum_{i=1}^{p} \int _{0}^{1} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} ( \int _{0}^{t} (t-\nu )^{\beta _{2}-\gamma _{i}-1} \varphi _{\rho _{2}}(v(\nu )) \,d\nu )\,dH_{i}(t), \\ b_{1} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\delta _{0})}-a_{1} \sum_{i=1}^{q} \int _{0}^{1} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})}t^{\beta _{1} -\delta _{i}-1}\,dK_{i}(t) \\ \quad = - \frac{1}{\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1-\nu )^{\beta _{2}-\delta _{0}-1} \varphi _{\rho _{2}}(v(\nu )) \,d\nu \\ \qquad {} + \sum_{i=1}^{q} \int _{0}^{1} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} ( \int _{0}^{t} (t-\nu )^{\beta _{1}-\delta _{i}-1} \varphi _{\rho _{1}}(u(\nu )) \,d\nu )\,dK_{i}(t). \end{cases} $$

(13)

The above system in the unknowns $a_{1}$ and $b_{1}$ has the determinant Δ, which, by the assumption of the lemma, is nonzero. So the system (13) has the unique solution

$$\begin{aligned}& a_{1}=- \frac{\varGamma (\beta _{2})}{\Delta \varGamma (\beta _{1}-\gamma _{0})\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1-\nu )^{\beta _{1}-\gamma _{0}-1} \varphi _{\rho _{1}} \bigl(u(\nu ) \bigr) \,d\nu \\& \hphantom{a_{1}={}}{}+ \frac{\varGamma (\beta _{2})}{\Delta \varGamma (\beta _{2}-\delta _{0})} \sum_{i=1}^{p} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} \biggl( \int _{0}^{t} (t-\nu )^{\beta _{2}-\gamma _{i}-1} \varphi _{\rho _{2}} \bigl(v(\nu ) \bigr) \,d\nu \biggr) \,dH_{i}(t) \\& \hphantom{a_{1}={}}{}- \frac{1}{\Delta } \Biggl( \sum _{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} t^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(t) \Biggr) \\& \hphantom{a_{1}={}}{}\times \biggl( \frac{1}{\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1- \nu )^{\beta _{2}-\delta _{0}-1}\varphi _{\rho _{2}} \bigl(v(\nu ) \bigr) \,d\nu \biggr) \\& \hphantom{a_{1}={}}{}+ \frac{1}{\Delta } \Biggl( \sum _{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} t^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(t) \Biggr) \\& \hphantom{a_{1}={}}{}\times \Biggl( \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} \biggl( \int _{0}^{t}(t-\nu )^{\beta _{1}-\delta _{i}-1} \varphi _{\rho _{1}} \bigl(u(\nu ) \bigr) \,d\nu \biggr)\,dK_{i}(t) \Biggr), \\& b_{1}=- \frac{\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1-\nu )^{\beta _{2}-\delta _{0}-1} \varphi _{\rho _{2}} \bigl(v(\nu ) \bigr) \,d\nu \\& \hphantom{b_{1}={}}{}+ \frac{\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})} \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} \biggl( \int _{0}^{t} (t-\nu )^{\beta _{1}-\delta _{i}-1} \varphi _{\rho _{1}} \bigl(u(\nu ) \bigr) \,d\nu \biggr) \,dK_{i}(t) \\& \hphantom{b_{1}={}}{}- \frac{1}{\Delta } \Biggl( \sum _{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} t^{\beta _{1}-\delta _{i}-1} \,dK_{i}(t) \Biggr) \\& \hphantom{b_{1}={}}{}\times \biggl( \frac{1}{\varGamma (\beta _{1}-\gamma _{0})} \int _{0}^{1}(1- \nu )^{\beta _{1}-\gamma _{0}-1}\varphi _{\rho _{1}} \bigl(u(\nu ) \bigr) \,d\nu \biggr) \\& \hphantom{b_{1}={}}{}+ \frac{1}{\Delta } \Biggl( \sum _{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} t^{\beta _{1}-\delta _{i}-1} \,dK_{i}(t) \Biggr) \\& \hphantom{b_{1}={}}{}\times \Biggl( \sum_{i=1}^{p} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} \biggl( \int _{0}^{t}(t-\nu )^{\beta _{2}-\gamma _{i}-1} \varphi _{\rho _{2}} \bigl(v(\nu ) \bigr) \,d\nu \biggr)\,dH_{i}(t) \Biggr). \end{aligned}$$

(14)

Now replacing the constants $a_{1}$ and $b_{1}$ given by (14) in (12) we find the solution $(x,y)\in C[0,1]\times C[0,1]$ of problem (6), (7) presented in (11). Conversely, we can easily prove that the functions x, y given by (11) satisfy the problem (6), (7). □

Lemma 2.2

If$\Delta \neq0$, then the solution$(x,y)$of problem (6), (7) given by (11) can be written as

$$ \textstyle\begin{cases} x(t)=- \int _{0}^{1}{\mathcal{G}}_{1}(t,\nu )\varphi _{\rho _{1}}(u( \nu )) \,d\nu - \int _{0}^{1} {\mathcal{G}}_{2}(t,\nu ) \varphi _{\rho _{2}}(v(\nu )) \,d\nu , \quad \forall t\in [0,1], \\ y(t)=- \int _{0}^{1}{\mathcal{G}}_{3}(t,\nu )\varphi _{\rho _{1}}(u( \nu )) \,d\nu - \int _{0}^{1} {\mathcal{G}}_{4}(t,\nu ) \varphi _{\rho _{2}}(v(\nu )) \,d\nu ,\quad \forall t\in [0,1], \end{cases} $$

(15)

where

$$ \begin{aligned} &{\mathcal{G}}_{1}(t,\nu )=g_{1}(t,\nu )+ \frac{t^{\beta _{1}-1}}{\Delta } \Biggl( \sum _{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} s^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(s) \Biggr) \\ &\hphantom{{\mathcal{G}}_{1}(t,\nu )={}}{}\times \Biggl( \sum_{i=1}^{q} \int _{0}^{1} g_{1i}(\tau ,\nu ) \,dK_{i}(\tau ) \Biggr), \\ &{\mathcal{G}}_{2}(t,\nu )= \frac{t^{\beta _{1}-1}\varGamma (\beta _{2})}{\Delta \varGamma (\beta _{2}-\delta _{0})} \sum _{i=1}^{p} \int _{0}^{1} g_{2i}(\tau , \nu ) \,dH_{i}(\tau ), \\ &{\mathcal{G}}_{3}(t,\nu )= \frac{t^{\beta _{2}-1}\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})} \sum _{i=1}^{q} \int _{0}^{1} g_{1i}(\tau , \nu ) \,dK_{i}(\tau ), \\ &{\mathcal{G}}_{4}(t,\nu )=g_{2}(t,\nu )+ \frac{t^{\beta _{2}-1}}{\Delta } \Biggl( \sum_{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} s^{\beta _{1}-\delta _{i}-1} \,dK_{i}(s) \Biggr) \\ &\hphantom{{\mathcal{G}}_{4}(t,\nu )={}}{}\times \Biggl( \sum_{i=1}^{p} \int _{0}^{1} g_{2i}(\tau ,\nu ) \,dH_{i}(\tau ) \Biggr), \end{aligned} $$

(16)

for all$(t,\nu )\in [0,1]\times [0,1]$and

$$\begin{aligned}& g_{1}(t,s)= \frac{1}{\varGamma (\beta _{1})} \textstyle\begin{cases} t^{\beta _{1}-1}(1-s)^{\beta _{1}-\gamma _{0}-1}-(t-s)^{\beta _{1}-1},& 0\le s\le t\le 1, \\ t^{\beta _{1}-1}(1-s)^{\beta _{1}-\gamma _{0}-1}, & 0\le t\le s \le 1,\end{cases}\displaystyle \\& g_{1i}(\tau ,s)= \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \textstyle\begin{cases} {\tau }^{\beta _{1}-\delta _{i}-1}(1-s)^{\beta _{1}-\gamma _{0}-1}-( \tau -s)^{\beta _{1}-\delta _{i}-1}, & 0\le s\le \tau \le 1, \\ {\tau }^{\beta _{1}-\delta _{i}-1}(1-s)^{\beta _{1}-\gamma _{0}-1},& 0\le \tau \le s\le 1,\end{cases}\displaystyle \\& g_{2}(t,s)= \frac{1}{\varGamma (\beta _{2})} \textstyle\begin{cases} t^{\beta _{2}-1}(1-s)^{\beta _{2}-\delta _{0}-1}-(t-s)^{\beta _{2}-1}, & 0\le s\le t\le 1, \\ t^{\beta _{2}-1}(1-s)^{\beta _{2}-\delta _{0}-1},& 0\le t\le s \le 1,\end{cases}\displaystyle \\& g_{2j}(\tau ,s)= \frac{1}{\varGamma (\beta _{2}-\gamma _{j})} \textstyle\begin{cases} {\tau }^{\beta _{2}-\gamma _{j}-1}(1-s)^{\beta _{2}-\delta _{0}-1}-( \tau -s)^{\beta _{2}-\gamma _{j}-1}, & 0\le s\le \tau \le 1, \\ {\tau }^{\beta _{2}-\gamma _{j}-1}(1-s)^{\beta _{2}-\delta _{0}-1},& 0\le \tau \le s\le 1,\end{cases}\displaystyle \end{aligned}$$

(17)

for all$i=1,\ldots ,q$and$j=1,\ldots ,p$.

Proof

For $x(t)$ from (11) we have

$$\begin{aligned} x(t) =& \frac{1}{\varGamma (\beta _{1})} \int _{0}^{t} (t-s)^{\beta _{1}-1}\varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}- \frac{t^{\beta _{1}-1}\varGamma (\beta _{2})}{\Delta \varGamma (\beta _{1}-\gamma _{0})\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1-s)^{\beta _{1}-\gamma _{0}-1} \varphi _{ \rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}+ \frac{t^{\beta _{1}}-1}{\Delta } \Biggl( \sum_{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} s^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(s) \Biggr) \\ &{}\times \Biggl( \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} \biggl( \int _{0}^{s}(s-\tau )^{\beta _{1}-\delta _{i}-1} \varphi _{\rho _{1}} \bigl(u(\tau ) \bigr) \,d\tau \biggr) \,dK_{i}(s) \Biggr) \\ &{}+ \frac{t^{\beta _{1}-1}\varGamma (\beta _{2})}{\Delta \varGamma (\beta _{2}-\delta _{0})} \sum_{i=1}^{p} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} \biggl( \int _{0}^{s}(s-\tau )^{\beta _{2}-\gamma _{i}-1} \varphi _{\rho _{2}} \bigl(v(\tau ) \bigr) \,d\tau \biggr) \,dH_{i}(s) \\ &{}- \frac{t^{\beta _{1}-1}}{\Delta } \Biggl( \sum_{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} s^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(s) \Biggr) \\ &{}\times \biggl( \frac{1}{\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1-s)^{ \beta _{2}-\delta _{0}-1}\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \biggr) \\ =&- \frac{1}{\varGamma (\beta _{1})} \int _{0}^{t} \bigl[t^{\beta _{1}-1}(1-s)^{\beta _{1}-\gamma _{0}-1}-(t-s)^{\beta _{1}-1} \bigr]\varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}- \frac{1}{\varGamma (\beta _{1})} \int _{t}^{1} t^{\beta _{1}-1}(1-s)^{\beta _{1}-\gamma _{0}-1} \varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}+ \frac{1}{\varGamma (\beta _{1})} \int _{0}^{1} t^{\beta _{1}-1}(1-s)^{\beta _{1}-\gamma _{0}-1} \varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}- \frac{t^{\beta _{1}-1}\varGamma (\beta _{2})}{\Delta \varGamma (\beta _{1}-\gamma _{0})\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1-s)^{\beta _{1}-\gamma _{0}-1} \varphi _{ \rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}+ \frac{t^{\beta _{1}-1}}{\Delta } \Biggl( \sum_{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} s^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(s) \Biggr) \\ &{}\times \Biggl( \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} \biggl( \int _{0}^{s}(s-\tau )^{\beta _{1}-\delta _{i}-1} \varphi _{\rho _{1}} \bigl(u(\tau ) \bigr) \,d\tau \biggr) \,dK_{i}(s) \Biggr) \\ &{}+ \frac{t^{\beta _{1}-1}\varGamma (\beta _{2})}{\Delta \varGamma (\beta _{2}-\delta _{0})} \sum_{i=1}^{p} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} \biggl( \int _{\tau }^{1} (s-\tau )^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(s) \biggr)\varphi _{\rho _{2}} \bigl(v(\tau ) \bigr) \,d\tau \\ &{}- \frac{t^{\beta _{1}-1}}{\Delta \varGamma (\beta _{2}-\delta _{0})} \sum_{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} \biggl( \int _{0}^{1} \tau ^{ \beta _{2}-\gamma _{i}-1} \,dH_{i}(\tau ) \biggr) \\ &{}\times(1-s)^{\beta _{2}- \delta _{0}-1}\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ =&- \int _{0}^{1} g_{1}(t,s)\varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds+ \frac{1}{\varGamma (\beta _{1})} \int _{0}^{1} t^{\beta _{1}-1}(1-s)^{\beta _{1}-\gamma _{0}-1} \varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}- \frac{t^{\beta _{1}-1}\varGamma (\beta _{2})}{\Delta \varGamma (\beta _{1}-\gamma _{0})\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1-s)^{\beta _{1}-\gamma _{0}-1} \varphi _{ \rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}+ \frac{t^{\beta _{1}-1}}{\Delta } \Biggl( \sum_{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} s^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(s) \Biggr) \\ &{}\times \Biggl( \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} \biggl( \int _{0}^{s}(s-\tau )^{\beta _{1}-\delta _{i}-1} \varphi _{\rho _{1}} \bigl(u(\tau ) \bigr) \,d\tau \biggr) \,dK_{i}(s) \Biggr) \\ &{}- \frac{t^{\beta _{1}-1}\varGamma (\beta _{2})}{\Delta \varGamma (\beta _{2}-\delta _{0})} \sum_{i=1}^{p} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} \biggl[- \int _{0}^{1} \biggl( \int _{s}^{1}(\tau -s)^{\beta _{2}-\gamma _{i}-1} \,dH_{i}( \tau ) \biggr)\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ &{}+ \int _{0}^{1} \biggl( \int _{0}^{1} \tau ^{\beta _{2}-\gamma _{i}-1}(1-s)^{\beta _{2}-\delta _{0}-1} \,dH_{i}( \tau ) \biggr)\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \biggr], \quad \forall t \in [0,1]. \end{aligned}$$

Then we deduce

$$\begin{aligned} x(t) =&- \int _{0}^{1} g_{1}(t,s)\varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}- \frac{1}{\Delta } \Biggl( \sum_{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} s^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(s) \Biggr) \Biggl( \sum_{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} s^{\beta _{1}-\delta _{i}-1} \,dK_{i}(s) \Biggr) \\ &{}\times \frac{1}{\varGamma (\beta _{1})} \int _{0}^{1} t^{\beta _{1}-1}(1-s)^{\beta _{1}-\gamma _{0}-1} \varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}+ \frac{t^{\beta _{1}-1}}{\Delta } \Biggl( \sum_{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} s^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(s) \Biggr) \\ &{}\times \Biggl( \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} \biggl( \int _{0}^{s}(s-\tau )^{\beta _{1}-\delta _{i}-1} \varphi _{\rho _{1}} \bigl(u(\tau ) \bigr) \,d\tau \biggr) \,dK_{i}(s) \Biggr) \\ &{}- \frac{t^{\beta _{1}-1}\varGamma (\beta _{2})}{\Delta \varGamma (\beta _{2}-\delta _{0})} \sum_{i=1}^{p} \int _{0}^{1} \biggl( \int _{0}^{1} g_{2i}(\tau ,s) \,dH_{i}(\tau ) \biggr) \varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ =&- \int _{0}^{1} g_{1}(t,s)\varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds- \frac{t^{\beta _{1}-1}}{\Delta } \Biggl( \sum _{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} s^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(s) \Biggr) \\ &{}\times \Biggl[ \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \biggl( \int _{0}^{1} \tau ^{\beta _{1}-\delta _{i}-1} \,dK_{i}(\tau ) \biggr) \biggl( \int _{0}^{1} (1-s)^{\beta _{1}-\gamma _{0}-1}\varphi _{ \rho _{1}} \bigl(u(s) \bigr) \,ds \biggr) \\ &{}- \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} \biggl( \int _{0}^{s}(s-\tau )^{\beta _{1}-\delta _{i}-1} \varphi _{\rho _{1}} \bigl(u(\tau ) \bigr) \,d\tau \biggr) \,dK_{i}(s) \Biggr] \\ &{}- \int _{0}^{1} \Biggl( \frac{t^{\beta _{1}-1}\varGamma (\beta _{2})}{\Delta \varGamma (\beta _{2}-\delta _{0})} \sum_{i=1}^{p} \int _{0}^{1} g_{2i}(\tau ,s) \,dH_{i}(\tau ) \Biggr)\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ =&- \int _{0}^{1} g_{1}(t,s)\varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds- \frac{t^{\beta _{1}-1}}{\Delta } \Biggl( \sum _{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} s^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(s) \Biggr) \\ &{}\times \Biggl\{ \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \biggl[ \int _{0}^{1} \biggl( \int _{0}^{1} \tau ^{\beta _{1}-\delta _{i}-1}(1-s)^{ \beta _{1}-\gamma _{0}-1} \,dK_{i}(\tau ) \biggr)\varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}- \int _{0}^{1} \biggl( \int _{s}^{1}(\tau -s)^{\beta _{1}-\delta _{i}-1} \,dK_{i}(\tau ) \biggr)\varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \biggr] \Biggr\} \\ &{}- \int _{0}^{1} \Biggl( \frac{t^{\beta _{1}-1}\varGamma (\beta _{2})}{\Delta \varGamma (\beta _{2}-\delta _{0})} \sum_{i=1}^{p} \int _{0}^{1} g_{2i}(\tau ,s) \,dH_{i}(\tau ) \Biggr)\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ =&- \int _{0}^{1} g_{1}(t,s)\varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds- \frac{t^{\beta _{1}-1}}{\Delta } \Biggl( \sum _{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} s^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(s) \Biggr) \\ &{}\times \Biggl[ \sum_{i=1}^{q} \int _{0}^{1} \biggl( \int _{0}^{1} g_{1i}(\tau ,s) \,dK_{i}(\tau ) \biggr)\varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \Biggr] \\ &{}- \int _{0}^{1} \Biggl( \frac{t^{\beta _{1}-1}\varGamma (\beta _{2})}{\Delta \varGamma (\beta _{2}-\delta _{0})} \sum_{i=1}^{p} \int _{0}^{1} g_{2i}(\tau ,s) \,dH_{i}(\tau ) \Biggr)\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ =&- \int _{0}^{1}{\mathcal{G}}_{1}(t,s) \varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds- \int _{0}^{1}{\mathcal{G}}_{2}(t,s) \varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds, \quad \forall t\in [0,1], \end{aligned}$$

where $g_{1}$, $g_{1i}$, $i=1, \ldots ,q$, $g_{2i}$, $i=1,\ldots ,p$, are given by (17), and ${\mathcal{G}}_{1}$ and ${\mathcal{G}}_{2}$ are given by (16).

For $y(t)$ we find

$$\begin{aligned} y(t) =& \frac{t^{\beta _{2}-1}\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})} \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} \biggl( \int _{0}^{s} (s-\tau )^{\beta _{1}-\delta _{i}-1} \varphi _{\rho _{1}} \bigl(u(\tau ) \bigr) \,d\tau \biggr) \,dK_{i}(s) \\ &{}- \frac{t^{\beta _{2}-1}}{\Delta } \Biggl( \sum_{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} s^{\beta _{1}-\delta _{i}-1} \,dK_{i}(s) \Biggr) \\ &{}\times \biggl( \frac{1}{\varGamma (\beta _{1}-\gamma _{0})} \int _{0}^{1} (1-s)^{ \beta _{1}-\gamma _{0}-1}\varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \biggr) \\ &{}+ \frac{1}{\varGamma (\beta _{2})} \int _{0}^{t} (t-s)^{\beta _{2}-1}\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ &{} - \frac{t^{\beta _{2}-1}\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1-s)^{\beta _{2}-\delta _{0}-1}\varphi _{ \rho _{2}} \bigl(v(s) \bigr) \,ds \\ &{}+ \frac{t^{\beta _{2}-1}}{\Delta } \Biggl( \sum_{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} s^{\beta _{1}-\delta _{i}-1} \,dK_{i}(s) \Biggr) \\ &{}\times \Biggl( \sum_{i=1}^{p} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} \biggl( \int _{0}^{s} (s-\tau )^{\beta _{2}-\gamma _{i}-1} \varphi _{\rho _{2}} \bigl(v(\tau ) \bigr) \,d\tau \biggr) \,dH_{i}(s) \Biggr) \\ =& \frac{t^{\beta _{2}-1}\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})} \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} \biggl( \int _{\tau }^{1}(s-\tau )^{\beta _{1}-\delta _{i}-1} \,dK_{i}(s) \biggr)\varphi _{\rho _{1}} \bigl(u(\tau ) \bigr) \,d\tau \\ &{}- \frac{t^{\beta _{2}-1}}{\Delta } \sum_{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})\varGamma (\beta _{1}-\gamma _{0})} \\ &{}\times \int _{0}^{1} \biggl( \int _{0}^{1} \tau ^{ \beta _{1}-\delta _{i}-1}(1-s)^{\beta _{1}-\gamma _{0}-1} \,dK_{i}( \tau ) \biggr)\varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}- \frac{1}{\varGamma (\beta _{2})} \int _{0}^{t} \bigl[t^{ \beta _{2}-1}(1-s)^{\beta _{2}-\delta _{0}-1}-(t-s)^{\beta _{2}-1} \bigr] \varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ &{}- \frac{1}{\varGamma (\beta _{2})} \int _{t}^{1} t^{\beta _{2}-1}(1-s)^{\beta _{2}-\delta _{0}-1} \varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ &{}+ \frac{1}{\varGamma (\beta _{2})} \int _{0}^{1} t^{\beta _{2}-1}(1-s)^{\beta _{2}-\delta _{0}-1} \varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ &{}- \frac{t^{\beta _{2}-1}\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1-s)^{\beta _{2}-\delta _{0}-1}\varphi _{ \rho _{2}} \bigl(v(s) \bigr) \,ds \\ &{}+ \frac{t^{\beta _{2}-1}}{\Delta } \Biggl( \sum_{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} s^{\beta _{1}-\delta _{i}-1} \,dK_{i}(s) \Biggr) \\ &{}\times \Biggl( \sum_{i=1}^{p} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} \biggl( \int _{0}^{s} (s-\tau )^{\beta _{2}-\gamma _{i}-1} \varphi _{\rho _{2}} \bigl(v(\tau ) \bigr) \,d\tau \biggr) \,dH_{i}(s) \Biggr) \\ = &\frac{t^{\beta _{2}-1}\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})} \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} \biggl( \int _{s}^{1} (\tau -s)^{\beta _{1}-\delta _{i}-1} \,dK_{i}( \tau ) \biggr)\varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}- \frac{t^{\beta _{2}-1}\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})} \sum_{i=1}^{q} \frac{1}{\varGamma (\beta _{1}-\delta _{i})} \\ &{}\times \int _{0}^{1} \biggl( \int _{0}^{1} \tau ^{\beta _{1}-\delta _{i}-1}(1-s)^{ \beta _{1}-\gamma _{0}-1} \,dK_{i}( \tau ) \biggr)\varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}- \int _{0}^{1} g_{2}(t,s)\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds+ \frac{1}{\varGamma (\beta _{2})} \int _{0}^{1} t^{\beta _{2}-1}(1-s)^{\beta _{2}-\delta _{0}-1} \varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ &{}- \frac{t^{\beta _{2}-1}\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})\varGamma (\beta _{2}-\delta _{0})} \int _{0}^{1}(1-s)^{\beta _{2}-\delta _{0}-1}\varphi _{ \rho _{2}} \bigl(v(s) \bigr) \,ds \\ &{}+ \frac{t^{\beta _{2}-1}}{\Delta } \Biggl( \sum_{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} s^{\beta _{1}-\delta _{i}-1} \,dK_{i}(s) \Biggr) \\ &{}\times \Biggl( \sum_{i=1}^{p} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} \biggl( \int _{0}^{s} (s-\tau )^{\beta _{2}-\gamma _{i}-1} \varphi _{\rho _{2}} \bigl(v(\tau ) \bigr) \,d\tau \biggr) \,dH_{i}(s) \Biggr),\quad \forall t\in [0,1]. \end{aligned}$$

Hence we obtain

$$\begin{aligned} y(t) =&- \frac{t^{\beta _{2}-1}\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})} \sum_{i=1}^{q} \int _{0}^{1} \biggl( \int _{0}^{1} g_{1i}(\tau ,s) \,dK_{i}(\tau ) \biggr) \varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}- \int _{0}^{1} g_{2}(t,s)\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ &{} - \frac{1}{\Delta } \Biggl( \sum_{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1}s^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(s) \Biggr) \Biggl( \sum_{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} s^{\beta _{1}-\delta _{i}-1} \,dK_{i}(s) \Biggr) \\ &{}\times \frac{1}{\varGamma (\beta _{2})} \int _{0}^{1} t^{\beta _{2}-1}(1-s)^{\beta _{2}-\delta _{0}-1} \varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ &{}+ \frac{t^{\beta _{2}-1}}{\Delta } \Biggl( \sum_{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} s^{\beta _{1}-\delta _{i}-1} \,dK_{i}(s) \Biggr) \\ &{}\times \Biggl( \sum_{i=1}^{p} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} \biggl( \int _{0}^{s} (s-\tau )^{\beta _{2}-\gamma _{i}-1} \varphi _{\rho _{2}} \bigl(v(\tau ) \bigr) \,d\tau \biggr) \,dH_{i}(s) \Biggr) \\ =&- \frac{t^{\beta _{2}-1}\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})} \sum_{i=1}^{q} \int _{0}^{1} \biggl( \int _{0}^{1} g_{1i}(\tau ,s) \,dK_{i}(\tau ) \biggr) \varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}- \int _{0}^{1} g_{2}(t,s)\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds - \frac{t^{\beta _{2}-1}}{\Delta } \Biggl( \sum _{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} s^{\beta _{1}-\delta _{i}-1} \,dK_{i}(s) \Biggr) \\ &{}\times \Biggl[ \sum_{i=1}^{p} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} \biggl( \int _{0}^{1} \tau ^{\beta _{2}-\gamma _{i}-1} (1-s)^{ \beta _{2}-\delta _{0}-1}\,dH_{i}(\tau ) \biggr)\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ &{}- \sum_{i=1}^{p} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} \biggl( \int _{\tau }^{1}(s-\tau )^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(s) \biggr)\varphi _{\rho _{2}} \bigl(v(\tau ) \bigr) \,d\tau \Biggr] \\ =&- \frac{t^{\beta _{2}-1}\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})} \sum_{i=1}^{q} \int _{0}^{1} \biggl( \int _{0}^{1} g_{1i}(\tau ,s) \,dK_{i}(\tau ) \biggr) \varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}- \int _{0}^{1} g_{2}(t,s)\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds - \frac{t^{\beta _{2}-1}}{\Delta } \Biggl( \sum _{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} s^{\beta _{1}-\delta _{i}-1} \,dK_{i}(s) \Biggr) \\ &{}\times \Biggl\{ \sum_{i=1}^{p} \frac{1}{\varGamma (\beta _{2}-\gamma _{i})} \biggl[ \int _{0}^{1} \biggl( \int _{0}^{1} \tau ^{\beta _{2}-\gamma _{i}-1} (1-s)^{\beta _{2}-\delta _{0}-1}\,dH_{i}(\tau ) \biggr) \varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \\ &{}- \int _{0}^{1} \biggl( \int _{s}^{1} (\tau -s)^{\beta _{2}-\gamma _{i}-1} \,dH_{i}( \tau ) \biggr)\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \biggr] \Biggr\} \\ =&- \frac{t^{\beta _{2}-1}\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})} \sum_{i=1}^{q} \int _{0}^{1} \biggl( \int _{0}^{1} g_{1i}(\tau ,s) \,dK_{i}(\tau ) \biggr) \varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds \\ &{}- \int _{0}^{1} g_{2}(t,s)\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds- \frac{t^{\beta _{2}-1}}{\Delta } \Biggl( \sum _{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} s^{\beta _{1}-\delta _{i}-1} \,dK_{i}(s) \Biggr) \\ &{}\times \Biggl[ \sum_{i=1}^{p} \int _{0}^{1} \biggl( \int _{0}^{1} g_{2i}(\tau ,s) \,dH_{i}(\tau ) \biggr)\varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds \Biggr] \\ =&- \int _{0}^{1} {\mathcal{G}}_{3}(t,s) \varphi _{\rho _{1}} \bigl(u(s) \bigr) \,ds- \int _{0}^{1} {\mathcal{G}}_{4}(t,s) \varphi _{\rho _{2}} \bigl(v(s) \bigr) \,ds,\quad \forall t\in [0,1], \end{aligned}$$

where $g_{1i}$, $i=1,\ldots ,q$, $g_{2}$, $g_{2i}$, $i=1,\ldots ,p$, are given by (17), and ${\mathcal{G}}_{3}$ and ${\mathcal{G}}_{4}$ are given by (16). □

Therefore, by (8), (9) and (15) we deduce the following lemma.

Lemma 2.3

If$\Delta \neq0$, then the unique solution$(x,y)\in C[0,1]\times C[0,1]$of problem (3), (2) is given by

$$ \textstyle\begin{cases} x(t)= \int _{0}^{1}{\mathcal{G}}_{1}(t,\nu )\varphi _{\rho _{1}}(I_{0+}^{ \alpha _{1}}\psi (\nu )) \,d\nu + \int _{0}^{1}{\mathcal{G}}_{2}(t, \nu )\varphi _{\rho _{2}}(I_{0+}^{\alpha _{2}}\chi (\nu )) \,d\nu , \quad \forall t\in [0,1], \\ y(t)= \int _{0}^{1} {\mathcal{G}}_{3}(t,\nu )\varphi _{\rho _{1}}(I_{0+}^{ \alpha _{1}}\psi (\nu )) \,d\nu + \int _{0}^{1} {\mathcal{G}}_{4}(t, \nu )\varphi _{\rho _{2}}(I_{0+}^{\alpha _{2}}\chi (\nu )) \,d\nu , \quad \forall t\in [0,1]. \end{cases} $$

(18)

In the next lemma, we present some properties of the functions $g_{1}$, $g_{2}$, $g_{1i}$, $i=1,\ldots ,q$, and $g_{2i}$, $i=1,\ldots ,p$.

Lemma 2.4

([19])

The functions$g_{1}$, $g_{2}$, $g_{1i}$, $i=1,\ldots ,q$, $g_{2i}$, $i=1,\ldots ,p$, given by (17) have the properties:

• (a) The functions$g_{1}, g_{2}, g_{1i}, i=1,\ldots ,q$, $g_{2i}, i=1,\ldots ,p$, are continuous on$[0,1]\times [0,1]$; $g_{1}(t,\nu )\ge 0$, $g_{2}(t,\nu )\ge 0$, $g_{1i}(t,\nu )\ge 0$, $i=1, \ldots ,q$, $g_{2i}(t,\nu )\ge 0$, $i=1,\ldots ,p$, for all$(t,\nu )\in [0,1]\times [0,1]$; $g_{1}(t,\nu )>0$, $g_{2}(t,\nu )>0$, $g_{1i}(t,\nu )>0$, $i=1,\ldots ,q$, $g_{2i}(t,\nu )>0$, $i=1,\ldots ,p$, for all$(t,\nu )\in (0,1)\times (0,1)$;

• (b) $g_{1}(t,\nu )\le h_{1}(\nu )$for all$(t,\nu )\in [0,1]\times [0,1]$, where$h_{1}(\nu )=\frac{1}{\varGamma (\beta _{1})}(1-\nu )^{ \beta _{1}-\gamma _{0}-1}(1-(1-\nu )^{\gamma _{0}})$, $\forall \nu \in [0,1]$;

• (c) $g_{1}(t,\nu )\ge t^{\beta _{1}-1}h_{1}(\nu )$for all$(t,\nu )\in [0,1]\times [0,1]$;

• (d) $g_{2}(t,\nu )\le h_{2}(\nu )$for all$(t,\nu )\in [0,1]\times [0,1]$, where$h_{2}(\nu )=\frac{1}{\varGamma (\beta _{2})}(1-\nu )^{ \beta _{2}-\delta _{0}-1}(1-(1-\nu )^{\delta _{0}})$, $\forall \nu \in [0,1]$;

• (e) $g_{2}(t,\nu )\ge t^{\beta _{2}-1}h_{2}(\nu )$for all$(t,\nu )\in [0,1]\times [0,1]$;

• (f) $g_{1i}(t,\nu )\le \frac{1}{\varGamma (\beta _{1}-\delta _{i})}t^{\beta _{1}- \delta _{i}-1}(1-\nu )^{\beta _{1}-\gamma _{0}-1}$for all$(t,\nu )\in [0,1]\times [0,1]$, $i=1,\ldots ,q$;

• (g) $g_{1i}(t,\nu )\ge t^{\beta _{1}-\delta _{i}-1}h_{1i}(\nu )$for all$(t,\nu )\in [0,1]\times [0,1]$, where$h_{1i}(\nu )=\frac{1}{\varGamma (\beta _{1}-\delta _{i})}(1-\nu )^{ \beta _{1}-\gamma _{0}-1}(1-(1-\nu )^{\gamma _{0}-\delta _{i}})$, $\forall \nu \in [0,1]$, $i=1,\ldots ,q$;

• (h) $g_{2i}(t,\nu )\le \frac{1}{\varGamma (\beta _{2}-\gamma _{i})}t^{\beta _{2}- \gamma _{i}-1}(1-\nu )^{\beta _{2}-\delta _{0}-1}$for all$(t,\nu )\in [0,1]\times [0,1]$, $i=1,\ldots ,p$;

• (i) $g_{2i}(t,\nu )\ge t^{\beta _{2}-\gamma _{i}-1}h_{2i}(\nu )$for all$(t,\nu )\in [0,1]\times [0,1]$, where$h_{2i}(\nu )=\frac{1}{\varGamma (\beta _{2}-\gamma _{i})}(1-\nu )^{ \beta _{2}-\delta _{0}-1}(1-(1-\nu )^{\delta _{0}-\gamma _{i}})$, $\forall \nu \in [0,1]$, $i=1,\ldots ,p$.

By using Lemma 2.4, we obtain the following properties for the functions ${\mathcal{G}}_{i}$, $i=1,\ldots ,4$.

Lemma 2.5

Assume that$\Delta >0$, $H_{i}$, $i=1,\ldots ,p$, $K_{i}$, $i=1,\ldots ,q$, are nondecreasing functions. Then the functions${\mathcal{G}}_{i}$, $i=1,\ldots ,4$, given by (16) have the properties:

• (a) ${\mathcal{G}}_{i}:[0,1]\times [0,1]\to [0,\infty )$, $i=1,\ldots ,4$, are continuous functions;

• (b) ${\mathcal{G}}_{1}(t,\nu )\le {\mathcal{J}}_{1}(\nu )$for all$(t,\nu )\in [0,1]\times [0,1]$, where

  $$\begin{aligned} {\mathcal{J}}_{1}(\nu ) =&h_{1}(\nu )+ \frac{1}{\Delta } \Biggl(\sum_{i=1}^{p} \frac{\varGamma (\beta _{2})}{\varGamma (\beta _{2}-\gamma _{i})} \int _{0}^{1} s^{\beta _{2}-\gamma _{i}-1} \,dH_{i}(s) \Biggr) \\ &{}\times\Biggl(\sum_{i=1}^{q} \int _{0}^{1} g_{1i}(\tau ,\nu ) \,dK_{i}(\tau ) \Biggr), \quad \forall \nu \in [0,1]; \end{aligned}$$

• (c) ${\mathcal{G}}_{1}(t,\nu )\ge t^{\beta _{1}-1}{\mathcal{J}}_{1}(\nu )$for all$(t,\nu )\in [0,1]\times [0,1]$;

• (d) ${\mathcal{G}}_{2}(t,\nu )\le {\mathcal{J}}_{2}(\nu )$, for all$(t,\nu )\in [0,1]\times [0,1]$, where

  $$ {\mathcal{J}}_{2}(\nu )=\frac{\varGamma (\beta _{2})}{\Delta \varGamma (\beta _{2}-\delta _{0})} \sum _{i=1}^{p} \int _{0}^{1} g_{2i}(\tau , \nu ) \,dH_{i}(\tau ), \quad \forall \nu \in [0,1]; $$

• (e) ${\mathcal{G}}_{2}(t,\nu )=t^{\beta _{1}-1}{\mathcal{J}}_{2}(\nu )$for all$(t,\nu )\in [0,1]\times [0,1]$;

• (f) ${\mathcal{G}}_{3}(t,\nu )\le {\mathcal{J}}_{3}(\nu )$for all$(t,\nu )\in [0,1]\times [0,1]$, where

  $$ {\mathcal{J}}_{3}(\nu )=\frac{\varGamma (\beta _{1})}{\Delta \varGamma (\beta _{1}-\gamma _{0})} \sum _{i=1}^{q} \int _{0}^{1} g_{1i}(\tau , \nu ) \,dK_{i}(\tau ), \quad \forall \nu \in [0,1]; $$

• (g) ${\mathcal{G}}_{3}(t,\nu )=t^{\beta _{2}-1}{\mathcal{J}}_{3}(\nu )$for all$(t,\nu )\in [0,1]\times [0,1]$;

• (h) ${\mathcal{G}}_{4}(t,\nu )\le {\mathcal{J}}_{4}(\nu )$for all$(t,\nu )\in [0,1]\times [0,1]$, where

  $$\begin{aligned} {\mathcal{J}}_{4}(\nu ) =&h_{2}(\nu )+ \frac{1}{\Delta } \Biggl(\sum_{i=1}^{q} \frac{\varGamma (\beta _{1})}{\varGamma (\beta _{1}-\delta _{i})} \int _{0}^{1} s^{\beta _{1}-\delta _{i}-1} \,dK_{i}(s) \Biggr) \\ &{}\times\Biggl(\sum_{i=1}^{p} \int _{0}^{1} g_{2i}(\tau ,\nu ) \,dH_{i}(\tau ) \Biggr),\quad \forall \nu \in [0,1]; \end{aligned}$$

• (i) ${\mathcal{G}}_{4}(t,\nu )\ge t^{\beta _{2}-1}{\mathcal{J}}_{4}(\nu )$, for all$(t,\nu )\in [0,1]\times [0,1]$.

3 Existence of positive solutions

In this section we present sufficient conditions for the functions f and g, and intervals for the parameters λ and μ such that problem (1), (2) has at least one positive solution.

We give now the basic assumptions that we will use in the main results.

$(I1)$:

$\alpha _{1}, \alpha _{2}\in (0,1]$, $\beta _{1}\in (n-1,n]$, $\beta _{2}\in (m-1,m]$, $n, m\in \mathbb{N}$, $n, m\ge 3$, $p, q\in \mathbb{N}$, $\gamma _{i}\in \mathbb{R}$ for all $i=0,1,\ldots ,p$, $0\le \gamma _{1}<\gamma _{2}<\cdots <\gamma _{p}\le \delta _{0}< \beta _{2}-1$, $\delta _{0}\ge 1$, $\delta _{i}\in \mathbb{R}$ for all $i=0,1,\ldots ,q$, $0\le \delta _{1}<\delta _{2}<\cdots <\delta _{q}\le \gamma _{0}< \beta _{1}-1$, $\gamma _{0}\ge 1$, $H_{i}:[0,1]\to \mathbb{R}$, $i=1,\ldots ,p$, and $K_{j}:[0,1]\to \mathbb{R}$, $j=1,\ldots ,q$, are nondecreasing functions, $\exists i_{0}\in \{1,\ldots ,p\}$, such that $H_{i_{0}}(1)>H_{i_{0}}(0)$, $\exists j_{0}\in \{1,\ldots ,q\}$, such that $K_{j_{0}}(1)>K_{j_{0}}(0)$, $\lambda >0$, $\mu >0$, $\Delta >0$ (Δ is given by (10)), $\varrho _{i}>1$, $\varphi _{\varrho _{i}}(s)=|s|^{\varrho _{i}-2}s$, $\varphi _{\varrho _{i}}^{-1}=\varphi _{\rho _{i}}$, $\rho _{i}=\frac{\varrho _{i}}{\varrho _{i}-1}$, $i=1,2$.

$(I2)$:

The functions $f, g:[0,1]\times \mathbb{R}_{+}\times \mathbb{R}_{+}\to \mathbb{R}_{+}$ are continuous.

For $[\theta _{1},\theta _{2}]\subset [0,1]$ with $0<\theta _{1}<\theta _{2}\le 1$, we introduce the following extreme limits:

$$\begin{aligned}& {\mathfrak{f}}_{0}^{s}= \limsup_{ \substack{x+y\to 0^{+}\\x,y\ge 0}} \max_{t\in [0,1]} \frac{f(t,x,y)}{\varphi _{\varrho _{1}}(x+y)}, \qquad { \mathfrak{g}}_{0}^{s}= \limsup _{ \substack{x+y\to 0^{+}\\x,y\ge 0}} \max_{t\in [0,1]} \frac{g(t,x,y)}{\varphi _{\varrho _{2}}(x+y)}, \\& {\mathfrak{f}}_{0}^{i}= \liminf_{ \substack{x+y\to 0^{+}\\x,y\ge 0}} \min_{t\in [\theta _{1}, \theta _{2}]} \frac{f(t,x,y)}{\varphi _{\varrho _{1}}(x+y)},\qquad {\mathfrak{g}}_{0}^{i}= \liminf_{\substack{x+y\to 0^{+}\\x,y\ge 0}} \min_{t\in [\theta _{1},\theta _{2}]} \frac{g(t,x,y)}{\varphi _{\varrho _{2}}(x+y)}, \\& {\mathfrak{f}}_{\infty }^{s}= \limsup_{ \substack{x+y\to \infty\\x,y\ge 0}} \max_{t\in [0,1]} \frac{f(t,x,y)}{\varphi _{\varrho _{1}}(x+y)},\qquad { \mathfrak{g}}_{\infty }^{s}= \limsup_{ \substack{x+y\to \infty \\x,y\ge 0}} \max_{t\in [0,1]} \frac{g(t,x,y)}{\varphi _{\varrho _{2}}(x+y)}, \\& {\mathfrak{f}}_{\infty }^{i}= \liminf_{ \substack{x+y\to \infty \\x,y\ge 0}} \min_{t\in [ \theta _{1},\theta _{2}]} \frac{f(t,x,y)}{\varphi _{\varrho _{1}}(x+y)},\qquad {\mathfrak{g}}_{ \infty }^{i}= \liminf_{ \substack{x+y\to \infty \\x,y\ge 0}} \min_{t\in [ \theta _{1},\theta _{2}]} \frac{g(t,x,y)}{\varphi _{\varrho _{2}}(x+y)}. \end{aligned}$$

By using Lemma 2.3 (Eqs. (18)), $(x,y)$ is a solution of problem (1), (2) if and only if $(x,y)$ is a solution of the following nonlinear system of integral equations:

$$ \textstyle\begin{cases} x(t)=\lambda ^{\rho _{1}-1} \int _{0}^{1}{\mathcal{G}}_{1}(t, \nu )\varphi _{\rho _{1}}(I_{0+}^{\alpha _{1}}f(\nu ,x(\nu ),y(\nu ))) \,d\nu \\ \hphantom{x(t)={}}{} +\mu ^{\rho _{2}-1} \int _{0}^{1} {\mathcal{G}}_{2}(t, \nu )\varphi _{\rho _{2}}(I_{0+}^{\alpha _{2}}g(\nu ,x(\nu ),y(\nu ))) \,d\nu ,\quad t\in [0,1], \\ y(t)=\lambda ^{\rho _{1}-1} \int _{0}^{1}{\mathcal{G}}_{3}(t, \nu )\varphi _{\rho _{1}}(I_{0+}^{\alpha _{1}}f(\nu ,x(\nu ),y(\nu ))) \,d\nu \\ \hphantom{y(t)={}}{} +\mu ^{\rho _{2}-1} \int _{0}^{1} {\mathcal{G}}_{4}(t, \nu )\varphi _{\rho _{2}}(I_{0+}^{\alpha _{2}}g(\nu ,x(\nu ),y(\nu ))) \,d\nu , \quad t\in [0,1]. \end{cases} $$

We consider the Banach space ${\mathcal{X}}=C[0,1]$ with the supremum norm

$$ \|x\|=\sup_{t\in [0,1]}\bigl|x(t)\bigr|, $$

and the Banach space ${\mathcal{Y}}={\mathcal{X}}\times {\mathcal{X}}$ with the norm $\|(x,y)\|_{\mathcal{Y}}=\|x\|+\|y\|$. We define the cones

$$\begin{aligned}& {\mathcal{P}}_{1}= \bigl\{ x\in { \mathcal{X}}, x(t)\ge t^{\beta _{1}-1} \Vert x \Vert , \forall t\in [0,1] \bigr\} \subset {\mathcal{X}}, \\& {\mathcal{P}}_{2}= \bigl\{ y\in {\mathcal{X}}, y(t)\ge t^{\beta _{2}-1} \Vert y \Vert , \forall t\in [0,1] \bigr\} \subset { \mathcal{X}}, \end{aligned}$$

and ${\mathcal{P}}={\mathcal{P}}_{1}\times {\mathcal{P}}_{2}\subset {\mathcal{Y}}$.

We also define the operators ${\mathcal{A}}_{1}, {\mathcal{A}}_{2}:{\mathcal{Y}}\to {\mathcal{X}}$ and ${\mathcal{A}}:{\mathcal{Y}}\to {\mathcal{Y}}$ by

$$\begin{aligned}& \begin{aligned} {\mathcal{A}}_{1}(x,y) (t)={}&\lambda ^{\rho _{1}-1} \int _{0}^{1}{ \mathcal{G}}_{1}(t, \nu )\varphi _{\rho _{1}} \bigl(I_{0+}^{\alpha _{1}}f \bigl( \nu ,x( \nu ),y(\nu ) \bigr) \bigr) \,d\nu \\ &{}+\mu ^{\rho _{2}-1} \int _{0}^{1} {\mathcal{G}}_{2}(t, \nu )\varphi _{\rho _{2}} \bigl(I_{0+}^{\alpha _{2}}g \bigl( \nu ,x(\nu ),y(\nu ) \bigr) \bigr) \,d\nu , \quad t\in [0,1], \end{aligned} \\& \begin{aligned} {\mathcal{A}}_{2}(x,y) (t)={}&\lambda ^{\rho _{1}-1} \int _{0}^{1}{ \mathcal{G}}_{3}(t, \nu )\varphi _{\rho _{1}} \bigl(I_{0+}^{\alpha _{1}}f \bigl( \nu ,x( \nu ),y(\nu ) \bigr) \bigr) \,d\nu \\ &{}+\mu ^{\rho _{2}-1} \int _{0}^{1} {\mathcal{G}}_{4}(t, \nu )\varphi _{\rho _{2}} \bigl(I_{0+}^{\alpha _{2}}g \bigl( \nu ,x(\nu ),y(\nu ) \bigr) \bigr) \,d\nu ,\quad t\in [0,1], \end{aligned} \end{aligned}$$

and ${\mathcal{A}}(x,y)=({\mathcal{A}}_{1}(x,y),{\mathcal{A}}_{2}(x,y))$, $(x,y)\in {\mathcal{Y}}$. The pair $(x,y)$ is a solution of problem (1), (2) if and only if $(x,y)$ is a fixed point of the operator ${\mathcal{A}}$.

Lemma 3.1

If$(I1)$and$(I2)$hold, then${\mathcal{A}}:{\mathcal{P}}\to {\mathcal{P}}$is a completely continuous operator.

Proof

Let $(x,y)\in {\mathcal{P}}$ be an arbitrary element. Because ${\mathcal{A}}_{1}(x,y)$ and ${\mathcal{A}}_{2}(x,y)$ satisfy the problem (3), (2) for $\psi (t)=\lambda f(t,x(t),y(t))$ and $\chi (t)=\mu g(t,x(t),y(t))$, $t\in [0,1]$, then by Lemma 2.5 we obtain

$$\begin{aligned}& \begin{aligned} {\mathcal{A}}_{1}(x,y) (t)\le{}& \lambda ^{\rho _{1}-1} \int _{0}^{1}{ \mathcal{J}}_{1}(\nu )\varphi _{\rho _{1}} \bigl(I_{0+}^{\alpha _{1}}f \bigl(\nu ,x( \nu ),y(\nu ) \bigr) \bigr) \,d\nu \\ &{}+\mu ^{\rho _{2}-1} \int _{0}^{1} {\mathcal{J}}_{2}( \nu )\varphi _{\rho _{2}} \bigl(I_{0+}^{\alpha _{2}}g \bigl( \nu ,x(\nu ),y(\nu ) \bigr) \bigr) \,d\nu , \quad \forall t\in [0,1], \end{aligned} \\& \begin{aligned} {\mathcal{A}}_{2}(x,y) (t)\le{}& \lambda ^{\rho _{1}-1} \int _{0}^{1}{ \mathcal{J}}_{3}( \nu )\varphi _{\rho _{1}} \bigl(I_{0+}^{\alpha _{1}}f \bigl( \nu ,x( \nu ),y(\nu ) \bigr) \bigr) \,d\nu \\ &{}+\mu ^{\rho _{2}-1} \int _{0}^{1} {\mathcal{J}}_{4}( \nu )\varphi _{\rho _{2}} \bigl(I_{0+}^{\alpha _{2}}g \bigl( \nu ,x(\nu ),y(\nu ) \bigr) \bigr) \,d\nu , \quad \forall t\in [0,1], \end{aligned} \end{aligned}$$

and then

$$\begin{aligned}& \begin{aligned} \bigl\Vert {\mathcal{A}}_{1}(x,y) \bigr\Vert \le{}& \lambda ^{\rho _{1}-1} \int _{0}^{1}{ \mathcal{J}}_{1}(\nu )\varphi _{\rho _{1}} \bigl(I_{0+}^{\alpha _{1}}f \bigl(\nu ,x( \nu ),y(\nu ) \bigr) \bigr) \,d\nu \\ &{}+\mu ^{\rho _{2}-1} \int _{0}^{1} {\mathcal{J}}_{2}( \nu )\varphi _{\rho _{2}} \bigl(I_{0+}^{\alpha _{2}}g \bigl( \nu ,x(\nu ),y(\nu ) \bigr) \bigr) \,d\nu , \end{aligned} \\& \begin{aligned} \bigl\Vert {\mathcal{A}}_{2}(x,y) \bigr\Vert \le {}&\lambda ^{\rho _{1}-1} \int _{0}^{1}{ \mathcal{J}}_{3}( \nu )\varphi _{\rho _{1}} \bigl(I_{0+}^{\alpha _{1}}f \bigl( \nu ,x( \nu ),y(\nu ) \bigr) \bigr) \,d\nu \\ &{}+\mu ^{\rho _{2}-1} \int _{0}^{1} {\mathcal{J}}_{4}( \nu )\varphi _{\rho _{2}} \bigl(I_{0+}^{\alpha _{2}}g \bigl( \nu ,x(\nu ),y(\nu ) \bigr) \bigr) \,d\nu . \end{aligned} \end{aligned}$$

Hence by using again Lemma 2.5, we deduce

$$\begin{aligned}& \begin{aligned} {\mathcal{A}}_{1}(x,y) (t)\ge{}& \lambda ^{\rho _{1}-1} \int _{0}^{1}t^{ \beta _{1}-1}{ \mathcal{J}}_{1}(\nu )\varphi _{\rho _{1}} \bigl(I_{0+}^{\alpha _{1}}f \bigl( \nu ,x(\nu ),y(\nu ) \bigr) \bigr) \,d\nu \\ &{}+\mu ^{\rho _{2}-1} \int _{0}^{1} t^{\beta _{1}-1}{ \mathcal{J}}_{2}(\nu )\varphi _{\rho _{2}} \bigl(I_{0+}^{\alpha _{2}}g \bigl(\nu ,x( \nu ),y(\nu ) \bigr) \bigr) \,d\nu \\ \ge{}& t^{\beta _{1}-1} \bigl\Vert { \mathcal{A}}_{1}(x,y) \bigr\Vert , \quad \forall t\in [0,1], \end{aligned} \\& \begin{aligned} {\mathcal{A}}_{2}(x,y) (t)\ge{}& \lambda ^{\varrho _{1}-1} \int _{0}^{1}t^{ \beta _{2}-1}{ \mathcal{J}}_{3}(\nu )\varphi _{\rho _{1}} \bigl(I_{0+}^{\alpha _{1}}f \bigl( \nu ,x(\nu ),y(\nu ) \bigr) \bigr) \,d\nu \\ &{}+\mu ^{\rho _{2}-1} \int _{0}^{1} t^{\beta _{2}-1}{ \mathcal{J}}_{4}(\nu )\varphi _{\rho _{2}} \bigl(I_{0+}^{\alpha _{2}}g \bigl(\nu ,x( \nu ),y(\nu ) \bigr) \bigr) \,d\nu \\ \ge{}& t^{\beta _{2}-1} \bigl\Vert {\mathcal{A}}_{2}(x,y) \bigr\Vert ,\quad \forall t\in [0,1]. \end{aligned} \end{aligned}$$

Therefore ${\mathcal{A}}(x,y)=({\mathcal{A}}_{1}(x,y),{\mathcal{A}}_{2}(x,y))\in {\mathcal{P}}$, and then ${\mathcal{A}}({\mathcal{P}})\subset {\mathcal{P}}$. By using the continuity of the functions $f, g, {\mathcal{G}}_{i}, i=1,\ldots,4$, and the Ascoli–Arzela theorem, we can prove that ${\mathcal{A}}_{1}$ and ${\mathcal{A}}_{2}$ are completely continuous operators (compact operators, that is, they map bounded sets into relatively compact sets, and continuous), and then ${\mathcal{A}}$ is a completely continuous operator. □

For $[\theta _{1},\theta _{2}]\subset [0,1]$ with $0<\theta _{1}<\theta _{2}\le 1$, we denote

$$ \begin{aligned} &L_{1}= \frac{1}{(\varGamma (\alpha _{1}+1))^{\rho _{1}-1}} \int _{0}^{1}\nu ^{\alpha _{1}(\rho _{1}-1)}{ \mathcal{J}}_{1}( \nu ) \,d\nu , \\ &L_{2}= \frac{1}{(\varGamma (\alpha _{2}+1))^{\rho _{2}-1}} \int _{0}^{1}\nu ^{\alpha _{2}(\rho _{2}-1)}{ \mathcal{J}}_{2}( \nu ) \,d\nu , \\ &L_{3}= \frac{1}{(\varGamma (\alpha _{1}+1))^{\rho _{1}-1}} \int _{0}^{1}\nu ^{\alpha _{1}(\rho _{1}-1)}{ \mathcal{J}}_{3}( \nu ) \,d\nu , \\ &L_{4}= \frac{1}{(\varGamma (\alpha _{2}+1))^{\rho _{2}-1}} \int _{0}^{1}\nu ^{\alpha _{2}(\rho _{2}-1)}{ \mathcal{J}}_{4}( \nu ) \,d\nu , \\ &\widetilde{L}_{1}= \frac{1}{(\varGamma (\alpha _{1}+1))^{\rho _{1}-1}} \int _{ \theta _{1}}^{\theta _{2}}(\nu -\theta _{1})^{\alpha _{1}(\rho _{1}-1)}{ \mathcal{J}}_{1}(\nu ) \,d\nu , \\ &\widetilde{L}_{2}= \frac{1}{(\varGamma (\alpha _{2}+1))^{\rho _{2}-1}} \int _{ \theta _{1}}^{\theta _{2}}(\nu -\theta _{1})^{\alpha _{2}(\rho _{2}-1)}{ \mathcal{J}}_{2}(\nu ) \,d\nu , \\ &\widetilde{L}_{3}= \frac{1}{(\varGamma (\alpha _{1}+1))^{\rho _{1}-1}} \int _{ \theta _{1}}^{\theta _{2}}(\nu -\theta _{1})^{\alpha _{1}(\rho _{1}-1)}{ \mathcal{J}}_{3}(\nu ) \,d\nu , \\ &\widetilde{L}_{4}= \frac{1}{(\varGamma (\alpha _{2}+1))^{\rho _{2}-1}} \int _{ \theta _{1}}^{\theta _{2}}(\nu -\theta _{1})^{\alpha _{2}(\rho _{2}-1)}{ \mathcal{J}}_{4}(\nu ) \,d\nu , \end{aligned} $$

(19)

where ${\mathcal{J}}_{i}$, $i=1,\ldots,4$, are defined in Lemma 2.5.

For ${\mathfrak{f}}_{0}^{s}, {\mathfrak{g}}_{0}^{s}, {\mathfrak{f}}_{ \infty }^{i}, {\mathfrak{g}}_{\infty }^{i}\in (0,\infty )$ and numbers $c_{1}, c_{2}\in [0,1]$, $c_{3}, c_{4}\in (0,1)$, $a\in [0,1]$ and $b\in (0,1)$, we introduce the numbers

$$\begin{aligned}& A=\max \biggl\{ \frac{1}{{\mathfrak{f}}_{\infty }^{i}} \varphi _{\varrho _{1}} \biggl( \frac{ac_{1}}{\zeta \zeta _{1}\widetilde{L}_{1}} \biggr), \frac{1}{{\mathfrak{f}}_{\infty }^{i}} \varphi _{ \varrho _{1}} \biggl( \frac{(1-a)c_{2}}{\zeta \zeta _{2}\widetilde{L}_{3}} \biggr) \biggr\} , \\& B=\min \biggl\{ \frac{1}{{\mathfrak{f}}_{0}^{s}}\varphi _{ \varrho _{1}} \biggl( \frac{b c_{3}}{L_{1}} \biggr), \frac{1}{{\mathfrak{f}}_{0}^{s}}\varphi _{\varrho _{1}} \biggl( \frac{(1-b)c_{4}}{L_{3}} \biggr) \biggr\} , \\& C=\max \biggl\{ \frac{1}{{\mathfrak{g}}_{\infty }^{i}} \varphi _{\varrho _{2}} \biggl( \frac{a(1-c_{1})}{\zeta \zeta _{1}\widetilde{L}_{2}} \biggr), \frac{1}{{\mathfrak{g}}_{\infty }^{i}}\varphi _{ \varrho _{2}} \biggl( \frac{(1-a)(1-c_{2})}{\zeta \zeta _{2}\widetilde{L}_{4}} \biggr) \biggr\} , \\& D=\min \biggl\{ \frac{1}{{\mathfrak{g}}_{0}^{s}}\varphi _{ \varrho _{2}} \biggl( \frac{b (1-c_{3})}{L_{2}} \biggr), \frac{1}{{\mathfrak{g}}_{0}^{s}}\varphi _{\varrho _{2}} \biggl( \frac{(1-b)(1-c_{4})}{L_{4}} \biggr) \biggr\} , \\& E=\min \biggl\{ \frac{1}{{\mathfrak{f}}_{0}^{s}}\varphi _{ \varrho _{1}} \biggl( \frac{b}{L_{1}} \biggr), \frac{1}{{\mathfrak{f}}_{0}^{s}} \varphi _{\varrho _{1}} \biggl( \frac{1-b}{L_{3}} \biggr) \biggr\} , \\& F=\min \biggl\{ \frac{1}{{\mathfrak{g}}_{0}^{s}}\varphi _{ \varrho _{2}} \biggl( \frac{b}{L_{2}} \biggr), \frac{1}{{\mathfrak{g}}_{0}^{s}} \varphi _{\varrho _{2}} \biggl( \frac{1-b}{L_{4}} \biggr) \biggr\} , \end{aligned}$$

where $\zeta _{1}=\theta _{1}^{\beta _{1}-1}$, $\zeta _{2}=\theta _{1}^{\beta _{2}-1}$, $\zeta =\min \{\zeta _{1},\zeta _{2}\}$.

Theorem 3.1

Assume that$(I1)$and$(I2)$hold, $[\theta _{1},\theta _{2}]\subset [0,1]$with$0<\theta _{1}<\theta _{2}\le 1$, $c_{1}, c_{2}\in [0,1]$, $c_{3}, c_{4}\in (0,1)$, $a\in [0,1]$and$b\in (0,1)$.

• (1) If${\mathfrak{f}}_{0}^{s}, {\mathfrak{g}}_{0}^{s}, {\mathfrak{f}}_{ \infty }^{i}, {\mathfrak{g}}_{\infty }^{i}\in (0,\infty )$, $A< B$and$C< D$, then, for each$\lambda \in (A,B)$and$\mu \in (C,D)$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

• (2) If${\mathfrak{f}}_{0}^{s}=0$, ${\mathfrak{g}}_{0}^{s}, {\mathfrak{f}}_{\infty }^{i}, { \mathfrak{g}}_{\infty }^{i}\in (0,\infty )$and$C< F$, then, for each$\lambda \in (A,\infty )$and$\mu \in (C,F)$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

• (3) If${\mathfrak{g}}_{0}^{s}=0$, ${\mathfrak{f}}_{0}^{s}, {\mathfrak{f}}_{\infty }^{i}, { \mathfrak{g}}_{\infty }^{i}\in (0,\infty )$and$A< E$, then, for each$\lambda \in (A,E)$and$\mu \in (C,\infty )$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

• (4) If${\mathfrak{f}}_{0}^{s}={\mathfrak{g}}_{0}^{s}=0$, ${\mathfrak{f}}_{\infty }^{i}, {\mathfrak{g}}_{\infty }^{i}\in (0, \infty )$, then, for each$\lambda \in (A,\infty )$and$\mu \in (C,\infty )$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

• (5) If${\mathfrak{f}}_{0}^{s}, {\mathfrak{g}}_{0}^{s}\in (0,\infty )$and at least one of${\mathfrak{f}}_{\infty }^{i}$, ${\mathfrak{g}}_{\infty }^{i}$is ∞, then, for each$\lambda \in (0,B)$and$\mu \in (0,D)$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

• (6) If${\mathfrak{f}}_{0}^{s}=0$, ${\mathfrak{g}}_{0}^{s}\in (0,\infty )$and at least one of${\mathfrak{f}}_{\infty }^{i}$, ${\mathfrak{g}}_{\infty }^{i}$is ∞, then, for each$\lambda \in (0,\infty )$and$\mu \in (0,F)$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

• (7) If${\mathfrak{f}}_{0}^{s}\in (0,\infty )$, ${\mathfrak{g}}_{0}^{s}=0$and at least one of${\mathfrak{f}}_{\infty }^{i}$, ${\mathfrak{g}}_{\infty }^{i}$is ∞, then, for each$\lambda \in (0,E)$and$\mu \in (0,\infty )$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

• (8) If${\mathfrak{f}}_{0}^{s}={\mathfrak{g}}_{0}^{s}=0$and at least one of${\mathfrak{f}}_{\infty }^{i}$, ${\mathfrak{g}}_{\infty }^{i}$is ∞, then, for each$\lambda \in (0,\infty )$and$\mu \in (0,\infty )$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

Proof

We consider the cone ${\mathcal{P}}\subset {\mathcal{Y}}$ and the operators ${\mathcal{A}}_{1}$, ${\mathcal{A}}_{2}$ and ${\mathcal{A}}$. The proofs of the above cases are similar, and for this reason we will prove only two cases, namely (1) and (6).

Case (1). We have ${\mathfrak{f}}_{0}^{s}, {\mathfrak{g}}_{0}^{s}, {\mathfrak{f}}_{ \infty }^{i}, {\mathfrak{g}}_{\infty }^{i}\in (0,\infty )$, $A< B$ and $C< D$. Let $\lambda \in (A,B)$ and $\mu \in (C,D)$. We consider $\epsilon >0$ such that $\epsilon <{\mathfrak{f}}_{\infty }^{i}$, $\epsilon <{\mathfrak{g}}_{\infty }^{i}$ and

$$\begin{aligned}& \max \biggl\{ \frac{1}{{\mathfrak{f}}_{\infty }^{i}-\epsilon }\varphi _{\varrho _{1}} \biggl( \frac{a c_{1}}{\zeta \zeta _{1}\widetilde{L}_{1}} \biggr), \frac{1}{{\mathfrak{f}}_{\infty }^{i}-\epsilon } \varphi _{\varrho _{1}} \biggl( \frac{(1-a)c_{2}}{\zeta \zeta _{2}\widetilde{L}_{3}} \biggr) \biggr\} \\& \quad \le \lambda\le \min \biggl\{ \frac{1}{{\mathfrak{f}}_{0}^{s}+\epsilon }\varphi _{\varrho _{1}} \biggl( \frac{bc_{3}}{L_{1}} \biggr), \frac{1}{{\mathfrak{f}}_{0}^{s}+\epsilon }\varphi _{\varrho _{1}} \biggl( \frac{(1-b)c_{4}}{L_{3}} \biggr) \biggr\} , \\& \max \biggl\{ \frac{1}{{\mathfrak{g}}_{\infty }^{i}-\epsilon }\varphi _{\varrho _{2}} \biggl( \frac{a(1-c_{1})}{\zeta \zeta _{1}\widetilde{L}_{2}} \biggr), \frac{1}{{\mathfrak{g}}_{\infty }^{i}-\epsilon } \varphi _{\varrho _{2}} \biggl( \frac{(1-a)(1-c_{2})}{\zeta \zeta _{2}\widetilde{L}_{4}} \biggr) \biggr\} \\& \quad \le \mu\le \min \biggl\{ \frac{1}{{\mathfrak{g}}_{0}^{s}+\epsilon }\varphi _{\varrho _{2}} \biggl( \frac{b(1-c_{3})}{L_{2}} \biggr), \frac{1}{{\mathfrak{g}}_{0}^{s}+\epsilon }\varphi _{\varrho _{2}} \biggl( \frac{(1-b)(1-c_{4})}{L_{4}} \biggr) \biggr\} . \end{aligned}$$

By using $(I2)$ and the definitions of ${\mathfrak{f}}_{0}^{s}$ and ${\mathfrak{g}}_{0}^{s}$, we find that there exists $r_{1}>0$ such that

$$ f(t,x,y)\le \bigl({\mathfrak{f}}_{0}^{s}+\epsilon \bigr)\varphi _{\varrho _{1}}(x+y),\qquad g(t,x,y)\le \bigl({\mathfrak{g}}_{0}^{s}+ \epsilon \bigr)\varphi _{ \varrho _{2}}(x+y), $$

for all $t\in [0,1]$ and $x, y\ge 0$, $x+y\le r_{1}$.

We introduce the set $\varLambda _{1}=\{(x,y)\in {\mathcal{Y}}, \|(x,y)\|_{\mathcal{Y}}< r_{1}\}$. Let $(x,y)\in {\mathcal{P}}\cap \partial \varLambda _{1}$, that is $(x,y)\in {\mathcal{P}}$ with $\|(x,y)\|_{\mathcal{Y}}=r_{1}$ or $\|x\|+\|y\|=r_{1}$. Then $x(t)+y(t)\le r_{1}$ for all $t\in [0,1]$, and by Lemma 2.5, we deduce

$$\begin{aligned}& {\mathcal{A}}_{1}(x,y) (t) \\& \quad \le \lambda ^{\rho _{1}-1} \int _{0}^{1} {\mathcal{J}}_{1}(\nu )\varphi _{\rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{0}^{\nu }(\nu -s)^{ \alpha _{1}-1}f \bigl(s,x(s),y(s) \bigr) \,ds \biggr) \,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1} \int _{0}^{1} {\mathcal{J}}_{2}( \nu ) \varphi _{\rho _{2}} \biggl( \frac{1}{\varGamma (\alpha _{2})} \int _{0}^{\nu }(\nu -s)^{ \alpha _{2}-1}g \bigl(s,x(s),y(s) \bigr) \,ds \biggr) \,d\nu \\& \quad \le \lambda ^{\rho _{1}-1} \int _{0}^{1} {\mathcal{J}}_{1}( \nu )\varphi _{\rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{0}^{\nu } (\nu -s)^{ \alpha _{1}-1} \bigl({ \mathfrak{f}}_{0}^{s}+\epsilon \bigr)\varphi _{\varrho _{1}} \bigl(x(s)+y(s) \bigr) \,ds \biggr)\,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1} \int _{0}^{1}{\mathcal{J}}_{2}(\nu ) \varphi _{\rho _{2}} \biggl( \frac{1}{\varGamma (\alpha _{2})} \int _{0}^{\nu }(\nu -s)^{ \alpha _{2}-1} \bigl({ \mathfrak{g}}_{0}^{s}+\epsilon \bigr)\varphi _{\varrho _{2}} \bigl(x(s)+y(s) \bigr) \,ds \biggr)\,d\nu \\& \quad \le \lambda ^{\rho _{1}-1}\varphi _{\rho _{1}} \bigl({ \mathfrak{f}}_{0}^{s}+ \epsilon \bigr) \int _{0}^{1} {\mathcal{J}}_{1}(\nu )\varphi _{ \rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{0}^{\nu } (\nu -s)^{\alpha _{1}-1}\varphi _{ \varrho _{1}} \bigl( \Vert x \Vert + \Vert y \Vert \bigr) \,ds \biggr)\,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1}\varphi _{\rho _{2}} \bigl({\mathfrak{g}}_{0}^{s}+ \epsilon \bigr) \int _{0}^{1} {\mathcal{J}}_{2}( \nu )\varphi _{ \rho _{2}} \biggl( \frac{1}{\varGamma (\alpha _{2})} \int _{0}^{\nu } (\nu -s)^{\alpha _{2}-1}\varphi _{ \varrho _{2}} \bigl( \Vert x \Vert + \Vert y \Vert \bigr) \,ds \biggr)\,d\nu \\& \quad = \lambda ^{\rho _{1}-1}\varphi _{\rho _{1}} \bigl({ \mathfrak{f}}_{0}^{s}+ \epsilon \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{0}^{1} {\mathcal{J}}_{1}( \nu ) \frac{1}{(\varGamma (\alpha _{1}+1))^{\rho _{1}-1}}{ \nu }^{\alpha _{1}(\rho _{1}-1)} \,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1}\varphi _{\rho _{2}} \bigl({\mathfrak{g}}_{0}^{s}+ \epsilon \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{0}^{1} {\mathcal{J}}_{2}( \nu ) \frac{1}{(\varGamma (\alpha _{2}+1))^{\rho _{2}-1}}{ \nu }^{\alpha _{2}(\rho _{2}-1)} \,d\nu \\& \quad = \bigl[\varphi _{\rho _{1}} \bigl(\lambda \bigl({\mathfrak{f}}_{0}^{s}+ \epsilon \bigr) \bigr)L_{1}+ \varphi _{\rho _{2}} \bigl(\mu \bigl({\mathfrak{g}}_{0}^{s}+\epsilon \bigr) \bigr)L_{2} \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \\& \quad \le \bigl[bc_{3}+b(1-c_{3}) \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}=b \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}},\quad \forall t\in [0,1]. \end{aligned}$$

Therefore $\|{\mathcal{A}}_{1}(x,y)\|\le b\|(x,y)\|_{\mathcal{Y}}$.

In a similar manner we obtain

$$\begin{aligned} {\mathcal{A}}_{2}(x,y) (t) \le& \lambda ^{\rho _{1}-1}\varphi _{\rho _{1}} \bigl({ \mathfrak{f}}_{0}^{s}+ \epsilon \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{0}^{1} {\mathcal{J}}_{3}( \nu ) \frac{1}{(\varGamma (\alpha _{1}+1))^{\rho _{1}-1}}{\nu }^{\alpha _{1}( \rho _{1}-1)} \,d\nu \\ &{}+\mu ^{\rho _{2}-1}\varphi _{\rho _{2}} \bigl({\mathfrak{g}}_{0}^{s}+ \epsilon \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{0}^{1} {\mathcal{J}}_{4}( \nu ) \frac{1}{(\varGamma (\alpha _{2}+1))^{\rho _{2}-1}}{ \nu }^{\alpha _{2}(\rho _{2}-1)} \,d\nu \\ =& \bigl[\varphi _{\rho _{1}} \bigl(\lambda \bigl({\mathfrak{f}}_{0}^{s}+ \epsilon \bigr) \bigr)L_{3}+ \varphi _{\rho _{2}} \bigl(\mu \bigl({\mathfrak{g}}_{0}^{s}+\epsilon \bigr) \bigr)L_{4} \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \\ \le& \bigl[(1-b)c_{4}+(1-b) (1-c_{4}) \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}=(1-b) \bigl\Vert (x,y) \bigr\Vert _{ \mathcal{Y}},\quad \forall t\in [0,1]. \end{aligned}$$

Then $\|{\mathcal{A}}_{2}(x,y)\|\le (1-b)\|(x,y)\|_{\mathcal{Y}}$.

Therefore, for $(x,y)\in {\mathcal{P}}\cap \partial \varLambda _{1}$, we conclude

$$ \bigl\Vert {\mathcal{A}}(x,y) \bigr\Vert _{\mathcal{Y}}= \bigl\Vert {\mathcal{A}}_{1}(x,y) \bigr\Vert + \bigl\Vert {\mathcal{A}}_{2}(x,y) \bigr\Vert \le b \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}+(1-b) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}= \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}. $$

(20)

By the definitions of ${\mathfrak{f}}_{\infty }^{i}$ and ${\mathfrak{g}}_{\infty }^{i}$, we find that there exists $r_{2}'>0$ such that

$$ f(t,x,y)\ge \bigl({\mathfrak{f}}_{\infty }^{i}-\epsilon \bigr)\varphi _{ \varrho _{1}}(x+y),\qquad g(t,x,y)\ge \bigl({\mathfrak{g}}_{\infty }^{i}- \epsilon \bigr)\varphi _{\varrho _{2}}(x+y), $$

for all $t\in [\theta _{1},\theta _{2}]$ and $x, y\ge 0$, $x+y\ge r_{2}'$.

We consider $r_{2}=\max \{2r_{1},r_{2}'/\zeta \}$ and we introduce the set $\varLambda _{2}=\{(x,y)\in {\mathcal{Y}}, \|(x,y)\|_{\mathcal{Y}}< r_{2}\}$. Then, for $(x,y)\in {\mathcal{P}}\cap \partial \varLambda _{2}$, we deduce

$$ \begin{aligned} x(t)+y(t)&\ge \min_{t\in [\theta _{1},\theta _{2}]}t^{ \beta _{1}-1} \Vert x \Vert + \min_{t\in [\theta _{1},\theta _{2}]}t^{ \beta _{2}-1} \Vert y \Vert =\theta _{1}^{\beta _{1}-1} \Vert x \Vert +\theta _{1}^{ \beta _{2}-1} \Vert y \Vert \\ &=\zeta _{1} \Vert x \Vert +\zeta _{2} \Vert y \Vert \ge \zeta \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}=\zeta r_{2} \ge r_{2}', \quad \forall t\in [\theta _{1},\theta _{2}]. \end{aligned} $$

Therefore, by Lemma 2.5, we obtain

$$\begin{aligned}& {\mathcal{A}}_{1}(x,y) (\theta _{1}) \\& \quad \ge \lambda ^{\rho _{1}-1} \int _{0}^{1}\theta _{1}^{\beta _{1}-1}{ \mathcal{J}}_{1}( \nu )\varphi _{\rho _{1}} \bigl(I_{0+}^{\alpha _{1}}f \bigl(\nu ,x(\nu ),y(\nu ) \bigr) \bigr) \,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1} \int _{0}^{1}\theta _{1}^{\beta _{1}-1}{ \mathcal{J}}_{2}(\nu )\varphi _{\rho _{2}} \bigl(I_{0+}^{\alpha _{2}}g \bigl(\nu ,x( \nu ),y(\nu ) \bigr) \bigr) \,d\nu \\& \quad \ge \lambda ^{\rho _{1}-1}\theta _{1}^{\beta _{1}-1} \int _{\theta _{1}}^{\theta _{2}}{\mathcal{J}}_{1}(\nu )\varphi _{\rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{\theta _{1}}^{\nu }(\nu -s)^{\alpha _{1}-1}f \bigl(s,x(s),y(s) \bigr) \,ds \biggr)\,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1}\theta _{1}^{\beta _{1}-1} \int _{ \theta _{1}}^{\theta _{2}}{\mathcal{J}}_{2}( \nu )\varphi _{\rho _{2}} \biggl( \frac{1}{\varGamma (\alpha _{2})} \int _{\theta _{1}}^{\nu }(\nu -s)^{\alpha _{2}-1}g \bigl(s,x(s),y(s) \bigr) \,ds \biggr)\,d\nu \\& \quad \ge \lambda ^{\rho _{1}-1}\theta _{1}^{\beta _{1}-1} \int _{\theta _{1}}^{\theta _{2}}{\mathcal{J}}_{1}(\nu )\varphi _{\rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{\theta _{1}}^{\nu }(\nu -s)^{\alpha _{1}-1} \bigl({ \mathfrak{f}}_{ \infty }^{i}-\epsilon \bigr)\varphi _{\varrho _{1}} \bigl(x(s)+y(s) \bigr) \,ds \biggr)\,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1}\theta _{1}^{\beta _{1}-1} \int _{ \theta _{1}}^{\theta _{2}}{\mathcal{J}}_{2}( \nu )\varphi _{\rho _{2}} \biggl( \frac{1}{\varGamma (\alpha _{2})} \int _{\theta _{1}}^{\nu }(\nu -s)^{\alpha _{2}-1} \bigl({ \mathfrak{g}}_{ \infty }^{i}-\epsilon \bigr)\varphi _{\varrho _{1}} \bigl(x(s)+y(s) \bigr) \,ds \biggr)\,d\nu \\& \quad \ge \lambda ^{\rho _{1}-1}\theta _{1}^{\beta _{1}-1} \int _{\theta _{1}}^{\theta _{2}}{\mathcal{J}}_{1}(\nu )\varphi _{\rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{\theta _{1}}^{\nu }(\nu -s)^{\alpha _{1}-1} \bigl({ \mathfrak{f}}_{ \infty }^{i}-\epsilon \bigr)\varphi _{\varrho _{1}} \bigl(\zeta \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \bigr) \,ds \biggr)\,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1}\theta _{1}^{\beta _{1}-1} \int _{ \theta _{1}}^{\theta _{2}}{\mathcal{J}}_{2}( \nu )\varphi _{\rho _{2}} \biggl( \frac{1}{\varGamma (\alpha _{2})} \int _{\theta _{1}}^{\nu }(\nu -s)^{\alpha _{2}-1} \bigl({ \mathfrak{g}}_{ \infty }^{i}-\epsilon \bigr)\varphi _{\varrho _{1}} \bigl(\zeta \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \bigr) \,ds \biggr)\,d\nu \\& \quad =\zeta \zeta _{1}\varphi _{\rho _{1}} \bigl(\lambda \bigl({ \mathfrak{f}}_{ \infty }^{i}-\epsilon \bigr) \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{ \theta _{1}}^{\theta _{2}}{\mathcal{J}}_{1}(\nu ) \frac{1}{(\varGamma (\alpha _{1}+1))^{\rho _{1}-1}}(\nu -\theta _{1})^{ \alpha _{1}(\rho _{1}-1)}\,d\nu \\& \qquad {}+\zeta \zeta _{1}\varphi _{\rho _{2}} \bigl(\mu \bigl({ \mathfrak{g}}_{\infty }^{i}- \epsilon \bigr) \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{\theta _{1}}^{ \theta _{2}}{\mathcal{J}}_{2}( \nu ) \frac{1}{(\varGamma (\alpha _{2}+1))^{\rho _{2}-1}}(\nu -\theta _{1})^{ \alpha _{2}(\rho _{2}-1)}\,d\nu \\& \quad = \bigl[\zeta \zeta _{1}\varphi _{\rho _{1}} \bigl(\lambda \bigl({\mathfrak{f}}_{ \infty }^{i}-\epsilon \bigr) \bigr) \widetilde{L}_{1}+\zeta \zeta _{1}\varphi _{ \rho _{2}}\bigl(\mu \bigl({\mathfrak{g}}_{\infty }^{i}- \epsilon \bigr)\bigr)\widetilde{L}_{2} \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \\& \quad \ge \bigl[ac_{1}+a(1-c_{1}) \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}=a \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}. \end{aligned}$$

Then $\|{\mathcal{A}}_{1}(x,y)\|\ge {\mathcal{A}}_{1}(x,y)(\theta _{1})\ge a\|(x,y) \|_{\mathcal{Y}}$.

In a similar manner, we deduce

$$\begin{aligned}& {\mathcal{A}}_{2}(x,y) (\theta _{1}) \\& \quad \ge \zeta \zeta _{2}\varphi _{\rho _{1}} \bigl( \lambda \bigl({\mathfrak{f}}_{\infty }^{i}-\epsilon \bigr) \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{\theta _{1}}^{\theta _{2}}{\mathcal{J}}_{3}( \nu ) \frac{1}{(\varGamma (\alpha _{1}+1))^{\rho _{1}-1}}(\nu - \theta _{1})^{\alpha _{1}(\rho _{1}-1)}\,d\nu \\& \qquad {}+\zeta \zeta _{2}\varphi _{\rho _{2}} \bigl(\mu \bigl({ \mathfrak{g}}_{\infty }^{i}- \epsilon \bigr) \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{\theta _{1}}^{ \theta _{2}}{\mathcal{J}}_{4}( \nu ) \frac{1}{(\varGamma (\alpha _{2}+1))^{\rho _{2}-1}}(\nu -\theta _{1})^{ \alpha _{2}(\rho _{2}-1)}\,d\nu \\& \quad = \bigl[\zeta \zeta _{2}\varphi _{\rho _{1}} \bigl(\lambda \bigl({\mathfrak{f}}_{ \infty }^{i}-\epsilon \bigr) \bigr) \widetilde{L}_{3}+\zeta \zeta _{2}\varphi _{ \rho _{2}}\bigl(\mu \bigl({\mathfrak{g}}_{\infty }^{i}- \epsilon \bigr)\bigr)\widetilde{L}_{4} \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \\& \quad \ge \bigl[(1-a)c_{2}+(1-a) (1-c_{2}) \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}=(1-a) \bigl\Vert (x,y) \bigr\Vert _{ \mathcal{Y}}. \end{aligned}$$

Hence $\|{\mathcal{A}}_{2}(x,y)\|\ge {\mathcal{A}}_{2}(x,y)(\theta _{1})\ge (1-a)\|(x,y) \|_{\mathcal{Y}}$.

Therefore for $(x,y)\in {\mathcal{P}}\cap \partial \varLambda _{2}$ we find

$$ \bigl\Vert {\mathcal{A}}(x,y) \bigr\Vert _{\mathcal{Y}}= \bigl\Vert {\mathcal{A}}_{1}(x,y) \bigr\Vert + \bigl\Vert {\mathcal{A}}_{2}(x,y) \bigr\Vert \ge a \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}+(1-a) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}= \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}. $$

(21)

By Lemma 3.1, Eqs. (20), (21) and the Guo–Krasnosel’skii fixed point theorem, we conclude that ${\mathcal{A}}$ has a fixed point $(x,y)\in {\mathcal{P}}\cap (\overline{\varLambda }_{2}\setminus \varLambda _{1})$ such that $r_{1}\le \|x\|+\|y\|\le r_{2}$, $x(t)\ge t^{\beta _{1}-1}\|x\|$, $y(t)\ge t^{\beta _{2}-1}\|y\|$ for all $t\in [0,1]$. If $\|x\|>0$ then $x(t)>0$ for all $t\in (0,1]$ and if $\|y\|>0$ then $y(t)>0$ for all $t\in (0,1]$. Therefore $(x,y)$ is a positive solution for problem (1), (2).

Case (6). We have ${\mathfrak{f}}_{0}^{s}=0$, ${\mathfrak{g}}_{0}^{s}\in (0,\infty )$ and ${\mathfrak{f}}_{\infty }^{i}=\infty$. Let $\lambda \in (0,\infty )$ and $\mu \in (0,F)$. We choose $\widetilde{c}_{3}\in (0,1-\varphi _{\rho _{2}}(\mu {\mathfrak{g}}_{0}^{s}) \frac{L_{2}}{b})$ and $\widetilde{c}_{4}\in (0,1-\varphi _{\rho _{2}}(\mu {\mathfrak{g}}_{0}^{s}) \frac{L_{4}}{1-b})$. The choice of $\widetilde{c}_{3}$ and $\widetilde{c}_{4}$ is possible because $\mu <\frac{1}{{\mathfrak{g}}_{0}^{s}}\varphi _{\varrho _{2}}( \frac{b}{L_{2}})$ and $\mu <\frac{1}{{\mathfrak{g}}_{0}^{s}}\varphi _{\varrho _{2}}( \frac{1-b}{L_{4}})$. Let $\epsilon >0$ be such that

$$ \begin{aligned} &\epsilon \varphi _{\varrho _{1}} \biggl( \frac{1}{\zeta \zeta _{1}\widetilde{L}_{1}} \biggr)\le \lambda \le \min \biggl\{ \frac{1}{\epsilon } \varphi _{\varrho _{1}} \biggl( \frac{b\widetilde{c}_{3}}{L_{1}} \biggr), \frac{1}{\epsilon }\varphi _{\varrho _{1}} \biggl( \frac{(1-b)\widetilde{c}_{4}}{L_{3}} \biggr) \biggr\} , \\ &\mu \le \min \biggl\{ \frac{1}{{\mathfrak{g}}_{0}^{s} +\epsilon }\varphi _{\varrho _{2}} \biggl( \frac{b(1-\widetilde{c}_{3})}{L_{2}} \biggr), \frac{1}{{\mathfrak{g}}_{0}^{s}+\epsilon } \biggl( \frac{(1-b)(1-\widetilde{c}_{4})}{L_{4}} \biggr) \biggr\} . \end{aligned} $$

By using $(I2)$ and the definitions of ${\mathfrak{f}}_{0}^{s}$ and ${\mathfrak{g}}_{0}^{s}$, we find that there exists $r_{1}>0$ such that

$$ f(t,x,y)\le \epsilon \varphi _{\varrho _{1}}(x+y), \qquad g(t,x,y)\le \bigl({ \mathfrak{g}}_{0}^{s}+\epsilon \bigr)\varphi _{\varrho _{2}}(x+y) $$

for all $t\in [0,1]$ and $x, y\ge 0$, $x+y\le r_{1}$.

We introduce the set $\varLambda _{1}=\{(x,y)\in {\mathcal{Y}}, \|(x,y)\|_{\mathcal{Y}}< r_{1}\}$. As in the proof of Case (1), for any $(x,y)\in {\mathcal{P}}\cap \partial \varLambda _{1}$, we deduce

$$\begin{aligned}& \begin{aligned} {\mathcal{A}}_{1}(x,y) (t)&\le \bigl[ \varphi _{\rho _{1}}(\lambda \epsilon )L_{1}+ \varphi _{\rho _{2}} \bigl(\mu \bigl({\mathfrak{g}}_{0}^{s}+ \epsilon \bigr) \bigr)L_{2} \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \\ &\le \bigl[b\widetilde{c}_{3}+b(1-\widetilde{c}_{3}) \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}=b \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}, \quad \forall t\in [0,1], \end{aligned} \\& \begin{aligned} {\mathcal{A}}_{2}(x,y) (t)&\le \bigl[\varphi _{\rho _{1}}( \lambda \epsilon )L_{3}+ \varphi _{\rho _{2}} \bigl(\mu \bigl({ \mathfrak{g}}_{0}^{s}+\epsilon \bigr) \bigr)L_{4} \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \\ &\le \bigl[(1-b)\widetilde{c}_{4}+(1-b) (1-\widetilde{c}_{4}) \bigr] \bigl\Vert (x,y) \bigr\Vert _{ \mathcal{Y}}=(1-b) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}},\quad \forall t\in [0,1], \end{aligned} \end{aligned}$$

and so $\|{\mathcal{A}}(x,y)\|_{\mathcal{Y}}\le \|(x,y)\|_{\mathcal{Y}}$.

For the second part, by the definition of ${\mathfrak{f}}_{\infty }^{i}$, we find that there exists $r_{2}'>0$ such that

$$ f(t,x,y)\ge \frac{1}{\epsilon }\varphi _{\varrho _{1}}(x+y),\quad \forall t\in [ \theta _{1},\theta _{2}], x, y\ge 0, x+y\ge r_{2}'. $$

We consider $r_{2}=\max \{2 r_{1},r_{2}'/\zeta \}$ and we introduce the set $\varLambda _{2}=\{(x,y)\in {\mathcal{Y}}, \|(x,y)\|_{\mathcal{Y}}< r_{2}\}$. Then, for any $(x,y)\in {\mathcal{P}}\cap \partial \varLambda _{2}$, we obtain as in Case (1) that $x(t)+y(t)\ge \zeta \|(x,y)\|_{\mathcal{Y}}=\zeta r_{2}\ge r_{2}'$ for all $t\in [\theta _{1},\theta _{2}]$.

Hence by Lemma 2.5 we deduce

$$\begin{aligned}& {\mathcal{A}}_{1}(x,y) (\theta _{1}) \\& \quad \ge \lambda ^{\rho _{1}-1} \int _{0}^{1}{\theta }_{1}^{\beta _{1}-1}{ \mathcal{J}}_{1}( \nu )\varphi _{\rho _{1}} \bigl(I_{0+}^{\alpha _{1}}f \bigl(\nu ,x(\nu ),y(\nu ) \bigr) \bigr) \,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1} \int _{0}^{1} \theta _{1}^{\beta _{1}-1}{ \mathcal{J}}_{2}(\nu )\varphi _{\rho _{2}} \bigl(I_{0+}^{\alpha _{2}}g \bigl(\nu ,x( \nu ),y(\nu ) \bigr) \bigr) \,d\nu \\& \quad \ge \lambda ^{\rho _{1}-1} \int _{0}^{1} \theta _{1}^{ \beta _{1}-1}{ \mathcal{J}}_{1}(\nu )\varphi _{\rho _{1}} \bigl(I_{0+}^{\alpha _{1}}f \bigl( \nu ,x(\nu ),y(\nu ) \bigr) \bigr) \,d\nu \\& \quad \ge \lambda ^{\rho _{1}-1}\theta _{1}^{\beta _{1}-1} \int _{\theta _{1}}^{\theta _{2}}{\mathcal{J}}_{1}(\nu )\varphi _{\rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{\theta _{1}}^{\nu } (\nu -s)^{\alpha _{1}-1}f \bigl(s,x(s),y(s) \bigr) \,ds \biggr)\,d\nu \\& \quad \ge \lambda ^{\rho _{1}-1}\theta _{1}^{\beta _{1}-1} \int _{\theta _{1}}^{\theta _{2}}{\mathcal{J}}_{1}(\nu )\varphi _{\rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{\theta _{1}}^{\nu } (\nu -s)^{\alpha _{1}-1} \frac{1}{\epsilon }\varphi _{\varrho _{1}} \bigl(x(s)+y(s) \bigr)\,ds \biggr) \,d\nu \\& \quad \ge \lambda ^{\rho _{1}-1}\theta _{1}^{\beta _{1}-1} \int _{\theta _{1}}^{\theta _{2}}{\mathcal{J}}_{1}(\nu )\varphi _{\rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{\theta _{1}}^{\nu } (\nu -s)^{\alpha _{1}-1} \frac{1}{\epsilon }\varphi _{\varrho _{1}} \bigl(\zeta \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \bigr) \,ds \biggr)\,d\nu \\& \quad =\zeta \zeta _{1}\varphi _{\rho _{1}} \biggl( \frac{\lambda }{\epsilon } \biggr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{\theta _{1}}^{\theta _{2}}{\mathcal{J}}_{1}(\nu ) \frac{1}{(\varGamma (\alpha _{1}+1))^{\rho _{1}-1}}(\nu -\theta _{1})^{ \alpha _{1}(\rho _{1}-1)}\,d\nu \\& \quad =\zeta \zeta _{1}\varphi _{\rho _{1}} \biggl( \frac{\lambda }{\epsilon } \biggr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \widetilde{L}_{1} \ge \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}. \end{aligned}$$

Then we conclude that $\|{\mathcal{A}}_{1}(x,y)\|\ge {\mathcal{A}}_{1}(x,y)(\theta _{1})\ge \|(x,y)\|_{ \mathcal{Y}}$ and $\|{\mathcal{A}}(x,y)\|_{\mathcal{Y}}\ge \|{\mathcal{A}}_{1}(x,y)\|\ge \|(x,y)\|_{ \mathcal{Y}}$. Therefore we obtain the conclusion of the theorem. □

Next for ${\mathfrak{f}}_{0}^{i}, {\mathfrak{g}}_{0}^{i}, {\mathfrak{f}}_{ \infty }^{s}, {\mathfrak{g}}_{\infty }^{s}\in (0,\infty )$ and numbers $c_{1}, c_{2}\in [0,1]$, $c_{3}, c_{4}\in (0,1)$, $a\in [0,1]$ and $b\in (0,1)$, we introduce the numbers

$$\begin{aligned}& \widetilde{A}=\max \biggl\{ \frac{1}{{\mathfrak{f}}_{0}^{i}} \varphi _{\varrho _{1}} \biggl( \frac{ac_{1}}{\zeta \zeta _{1}\widetilde{L}_{1}} \biggr), \frac{1}{{\mathfrak{f}}_{0}^{i}}\varphi _{\varrho _{1}} \biggl( \frac{(1-a)c_{2}}{\zeta \zeta _{2}\widetilde{L}_{3}} \biggr) \biggr\} , \\& \widetilde{B}=\min \biggl\{ \frac{1}{{\mathfrak{f}}_{\infty }^{s}}\varphi _{\varrho _{1}} \biggl( \frac{b c_{3}}{L_{1}} \biggr), \frac{1}{{\mathfrak{f}}_{\infty }^{s}}\varphi _{\varrho _{1}} \biggl( \frac{(1-b)c_{4}}{L_{3}} \biggr) \biggr\} , \\& \widetilde{C}=\max \biggl\{ \frac{1}{{\mathfrak{g}}_{0}^{i}}\varphi _{\varrho _{2}} \biggl( \frac{a(1-c_{1})}{\zeta \zeta _{1}\widetilde{L}_{2}} \biggr), \frac{1}{{\mathfrak{g}}_{0}^{i}}\varphi _{ \varrho _{2}} \biggl( \frac{(1-a)(1-c_{2})}{\zeta \zeta _{2}\widetilde{L}_{4}} \biggr) \biggr\} , \\& \widetilde{D}=\min \biggl\{ \frac{1}{{\mathfrak{g}}_{\infty }^{s}}\varphi _{\varrho _{2}} \biggl( \frac{b (1-c_{3})}{L_{2}} \biggr), \frac{1}{{\mathfrak{g}}_{\infty }^{s}}\varphi _{\varrho _{2}} \biggl( \frac{(1-b)(1-c_{4})}{L_{4}} \biggr) \biggr\} , \\& \widetilde{E}=\min \biggl\{ \frac{1}{{\mathfrak{f}}_{\infty }^{s}}\varphi _{\varrho _{1}} \biggl( \frac{b}{L_{1}} \biggr), \frac{1}{{\mathfrak{f}}_{\infty }^{s}} \varphi _{\varrho _{1}} \biggl( \frac{1-b}{L_{3}} \biggr) \biggr\} , \\& \widetilde{F}=\min \biggl\{ \frac{1}{{\mathfrak{g}}_{\infty }^{s}}\varphi _{\varrho _{2}} \biggl( \frac{b}{L_{2}} \biggr), \frac{1}{{\mathfrak{g}}_{\infty }^{s}} \varphi _{\varrho _{2}} \biggl( \frac{1-b}{L_{4}} \biggr) \biggr\} . \end{aligned}$$

Theorem 3.2

Assume that$(I1)$and$(I2)$hold, $[\theta _{1},\theta _{2}]\subset [0,1]$with$0<\theta _{1}<\theta _{2}\le 1$, $c_{1}, c_{2}\in [0,1]$, $c_{3}, c_{4}\in (0,1)$, $a\in [0,1]$and$b\in (0,1)$.

• (1) If${\mathfrak{f}}_{0}^{i}, {\mathfrak{g}}_{0}^{i}, {\mathfrak{f}}_{ \infty }^{s}, {\mathfrak{g}}_{\infty }^{s}\in (0,\infty )$, $\widetilde{A}<\widetilde{B}$and$\widetilde{C}<\widetilde{D}$, then, for each$\lambda \in (\widetilde{A},\widetilde{B})$and$\mu \in (\widetilde{C},\widetilde{D})$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

• (2) If${\mathfrak{f}}_{0}^{i}, {\mathfrak{g}}_{0}^{i}, {\mathfrak{f}}_{ \infty }^{s}\in (0,\infty )$, ${\mathfrak{g}}_{\infty }^{s}=0$and$\widetilde{A}<\widetilde{E}$, then, for each$\lambda \in (\widetilde{A},\widetilde{E})$and$\mu \in (\widetilde{C},\infty )$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

• (3) If${\mathfrak{f}}_{0}^{i}, {\mathfrak{g}}_{0}^{i}, {\mathfrak{g}}_{ \infty }^{s}\in (0,\infty )$, ${\mathfrak{f}}_{\infty }^{s}=0$and$\widetilde{C}<\widetilde{F}$, then, for each$\lambda \in (\widetilde{A},\infty )$and$\mu \in (\widetilde{C},\widetilde{F})$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

• (4) If${\mathfrak{f}}_{0}^{i}, {\mathfrak{g}}_{0}^{i}\in (0,\infty )$, ${\mathfrak{f}}_{\infty }^{s}={\mathfrak{g}}_{\infty }^{s}=0$, then, for each$\lambda \in (\widetilde{A},\infty )$and$\mu \in (\widetilde{C},\infty )$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

• (5) If${\mathfrak{f}}_{\infty }^{s}, {\mathfrak{g}}_{\infty }^{s}\in (0, \infty )$and at least one of${\mathfrak{f}}_{0}^{i}$, ${\mathfrak{g}}_{0}^{i}$is ∞, then, for each$\lambda \in (0,\widetilde{B})$and$\mu \in (0,\widetilde{D})$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

• (6) If${\mathfrak{f}}_{\infty }^{s}\in (0,\infty )$, ${\mathfrak{g}}_{\infty }^{s}=0$and at least one of${\mathfrak{f}}_{0}^{i}$, ${\mathfrak{g}}_{0}^{i}$is ∞, then, for each$\lambda \in (0,\widetilde{E})$and$\mu \in (0,\infty )$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

• (7) If${\mathfrak{f}}_{\infty }^{s}=0$, ${\mathfrak{g}}_{\infty }^{s}\in (0,\infty )$and at least one of${\mathfrak{f}}_{0}^{i}$, ${\mathfrak{g}}_{0}^{i}$is ∞, then, for each$\lambda \in (0,\infty )$and$\mu \in (0,\widetilde{F})$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

• (8) If${\mathfrak{f}}_{\infty }^{s}={\mathfrak{g}}_{\infty }^{s}=0$and at least one of${\mathfrak{f}}_{0}^{i}$, ${\mathfrak{g}}_{0}^{i}$is ∞, then, for each$\lambda \in (0,\infty )$and$\mu \in (0,\infty )$, the problem (1), (2) has at least one positive solution$(x(t),y(t))$, $t\in [0,1]$.

Proof

We consider again the cone ${\mathcal{P}}\subset {\mathcal{Y}}$ and the operators ${\mathcal{A}}_{1}$, ${\mathcal{A}}_{2}$ and ${\mathcal{A}}$. The proofs of the above cases are similar, and for this reason we will prove only two cases, namely (1) and (6).

Case (1). We have ${\mathfrak{f}}_{0}^{i}, {\mathfrak{g}}_{0}^{i}, {\mathfrak{f}}_{ \infty }^{s}, {\mathfrak{g}}_{\infty }^{s}\in (0,\infty )$, $\widetilde{A}<\widetilde{B}$ and $\widetilde{C}<\widetilde{D}$. Let $\lambda \in (\widetilde{A},\widetilde{B})$ and $\mu \in (\widetilde{C},\widetilde{D})$. We consider $\epsilon >0$ such that $\epsilon <{\mathfrak{f}}_{0}^{i}$, $\epsilon <{\mathfrak{g}}_{0}^{i}$ and

$$\begin{aligned}& \max \biggl\{ \frac{1}{{\mathfrak{f}}_{0}^{i}-\epsilon } \varphi _{\varrho _{1}} \biggl( \frac{ac_{1}}{\zeta \zeta _{1}\widetilde{L}_{1}} \biggr), \frac{1}{{\mathfrak{f}}_{0}^{i}-\epsilon } \varphi _{ \varrho _{1}} \biggl( \frac{(1-a)c_{2}}{\zeta \zeta _{2}\widetilde{L}_{3}} \biggr) \biggr\} \\& \quad \le \lambda\le \min \biggl\{ \frac{1}{{\mathfrak{f}}_{\infty }^{s}+\epsilon }\varphi _{\varrho _{1}} \biggl( \frac{bc_{3}}{L_{1}} \biggr), \frac{1}{{\mathfrak{f}}_{\infty }^{s}+\epsilon }\varphi _{\varrho _{1}} \biggl( \frac{(1-b)c_{4}}{L_{3}} \biggr) \biggr\} , \\& \max \biggl\{ \frac{1}{{\mathfrak{g}}_{0}^{i}-\epsilon } \varphi _{\varrho _{2}} \biggl( \frac{a(1-c_{1})}{\zeta \zeta _{1}\widetilde{L}_{2}} \biggr), \frac{1}{{\mathfrak{g}}_{0}^{i}-\epsilon }\varphi _{ \varrho _{2}} \biggl( \frac{(1-a)(1-c_{2})}{\zeta \zeta _{2}\widetilde{L}_{4}} \biggr) \biggr\} \\& \quad \le \mu\le \min \biggl\{ \frac{1}{{\mathfrak{g}}_{\infty }^{s}+\epsilon }\varphi _{\varrho _{2}} \biggl( \frac{b(1-c_{3})}{L_{2}} \biggr), \frac{1}{{\mathfrak{g}}_{\infty }^{s}+\epsilon } \biggl( \frac{(1-b)(1-c_{4})}{L_{4}} \biggr) \biggr\} . \end{aligned}$$

By using $(I2)$ and the definitions of ${\mathfrak{f}}_{0}^{i}$ and ${\mathfrak{g}}_{0}^{i}$ we deduce that there exists $r_{3}>0$ such that

$$ f(t,x,y)\ge \bigl({\mathfrak{f}}_{0}^{i}-\epsilon \bigr)\varphi _{\varrho _{1}}(x+y),\qquad g(t,x,y)\ge \bigl({\mathfrak{g}}_{0}^{i}- \epsilon \bigr)\varphi _{ \varrho _{2}}(x+y), $$

for all $t\in [\theta _{1},\theta _{2}]$, $x, y\ge 0$, $x+y\le r_{3}$.

We introduce the set $\varLambda _{3}=\{(x,y)\in {\mathcal{Y}}, \|(x,y)\|_{\mathcal{Y}}< r_{3}\}$. Let $(x,y)\in {\mathcal{P}}\cap \partial \varLambda _{3}$, that is $(x,y)\in {\mathcal{P}}$ with $\|(x,y)\|_{\mathcal{Y}}=r_{3}$ or $\|x\|+\|y\|=r_{3}$. Then $x(t)+y(t)\le r_{3}$ for all $t\in [0,1]$, and $x(t)+y(t)\ge \zeta \|(x,y)\|_{\mathcal{Y}}$ for all $t\in [\theta _{1},\theta _{2}]$ (see the proof of Case (1) in Theorem 3.1). So, by Lemma 2.5, we obtain

$$\begin{aligned}& {\mathcal{A}}_{1}(x,y) (\theta _{1}) \\& \quad \ge \lambda ^{\rho _{1}-1}\theta _{1}^{ \beta _{1}-1} \int _{\theta _{1}}^{\theta _{2}}{\mathcal{J}}_{1}( \nu )\varphi _{\rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{\theta _{1}}^{\nu } (\nu -s)^{\alpha _{1}-1}f \bigl(s,x(s),y(s) \bigr) \,ds \biggr)\,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1}\theta _{1}^{\beta _{1}-1} \int _{ \theta _{1}}^{\theta _{2}}{\mathcal{J}}_{2}( \nu )\varphi _{\rho _{2}} \biggl( \frac{1}{\varGamma (\alpha _{2})} \int _{\theta _{1}}^{\nu }(\nu -s)^{\alpha _{2}-1}g \bigl(s,x(s),y(s) \bigr) \,ds \biggr)\,d\nu \\& \quad \ge \lambda ^{\rho _{1}-1}\theta _{1}^{\beta _{1}-1} \int _{\theta _{1}}^{\theta _{2}}{\mathcal{J}}_{1}(\nu )\varphi _{\rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{\theta _{1}}^{\nu }(\nu -s)^{\alpha _{1}-1} \bigl({ \mathfrak{f}}_{0}^{i}- \epsilon \bigr)\varphi _{\varrho _{1}} \bigl(x(s)+y(s) \bigr) \,ds \biggr)\,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1}\theta _{1}^{\beta _{1}-1} \int _{ \theta _{1}}^{\theta _{2}}{\mathcal{J}}_{2}( \nu )\varphi _{\rho _{2}} \biggl( \frac{1}{\varGamma (\alpha _{2})} \int _{\theta _{1}}^{\nu } (\nu -s)^{\alpha _{2}-1} \bigl({ \mathfrak{g}}_{0}^{i}- \epsilon \bigr)\varphi _{\varrho _{2}} \bigl(x(s)+y(s) \bigr) \,ds \biggr)\,d\nu \\& \quad \ge \lambda ^{\rho _{1}-1}\theta _{1}^{\beta _{1}-1} \int _{\theta _{1}}^{\theta _{2}}{\mathcal{J}}_{1}(\nu )\varphi _{\rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{\theta _{1}}^{\nu } (\nu -s)^{\alpha _{1}-1} \bigl({ \mathfrak{f}}_{0}^{i}- \epsilon \bigr)\varphi _{\varrho _{1}} \bigl(\zeta \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \bigr) \,ds \biggr)\,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1}\theta _{1}^{\beta _{1}-1} \int _{ \theta _{1}}^{\theta _{2}}{\mathcal{J}}_{2}( \nu )\varphi _{\rho _{2}} \biggl( \frac{1}{\varGamma (\alpha _{2})} \int _{\theta _{1}}^{\nu } (\nu -s)^{\alpha _{2}-1} \bigl({ \mathfrak{g}}_{0}^{i}- \epsilon \bigr)\varphi _{\varrho _{2}} \bigl(\zeta \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \bigr) \,ds \biggr)\,d\nu \\& \quad =\zeta \zeta _{1}\varphi _{\rho _{1}} \bigl(\lambda \bigl({ \mathfrak{f}}_{0}^{i}- \epsilon \bigr) \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{\theta _{1}}^{ \theta _{2}}{\mathcal{J}}_{1}(\nu ) \frac{1}{(\varGamma (\alpha _{1}+1))^{\rho _{1}-1}}(\nu -\theta _{1})^{ \alpha _{1}(\rho _{1}-1)} \,d\nu \\& \qquad {}+\zeta \zeta _{1}\varphi _{\rho _{2}} \bigl(\mu \bigl({ \mathfrak{g}}_{0}^{i}- \epsilon \bigr) \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{\theta _{1}}^{ \theta _{2}}J_{2}(\nu ) \frac{1}{(\varGamma (\alpha _{2}+1))^{\rho _{2}-1}}(\nu -\theta _{1})^{ \alpha _{2}(\rho _{2}-1)} \,d\nu \\& \quad = \bigl[\zeta \zeta _{1}\varphi _{\rho _{1}} \bigl(\lambda \bigl({\mathfrak{f}}_{0}^{i}- \epsilon \bigr) \bigr) \widetilde{L}_{1}+\zeta \zeta _{1}\varphi _{\rho _{2}}\bigl(\mu \bigl({ \mathfrak{g}}_{0}^{i}- \epsilon \bigr)\bigr)\widetilde{L}_{2} \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \\& \quad \ge \bigl[ac_{1}+a(1-c_{1}) \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}=a \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}. \end{aligned}$$

Then $\|{\mathcal{A}}_{1}(x,y)\|\ge {\mathcal{A}}_{1}(x,y)(\theta _{1})\ge a\|(x,y) \|_{\mathcal{Y}}$.

In a similar manner we find

$$\begin{aligned}& {\mathcal{A}}_{2}(x,y) (\theta _{1}) \\& \quad \ge \zeta \zeta _{2}\varphi _{\rho _{1}} \bigl( \lambda \bigl({\mathfrak{f}}_{0}^{i}-\epsilon \bigr) \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{\theta _{1}}^{\theta _{2}}{\mathcal{J}}_{3}( \nu ) \frac{1}{(\varGamma (\alpha _{1}+1))^{\rho _{1}-1}}(\nu - \theta _{1})^{\alpha _{1}(\rho _{1}-1)} \,d\nu \\& \qquad {}+\zeta \zeta _{2}\varphi _{\rho _{2}} \bigl(\mu \bigl({ \mathfrak{g}}_{0}^{i}- \epsilon \bigr) \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{\theta _{1}}^{ \theta _{2}}J_{4}(\nu ) \frac{1}{(\varGamma (\alpha _{2}+1))^{\rho _{2}-1}}(\nu -\theta _{1})^{ \alpha _{2}(\rho _{2}-1)} \,d\nu \\& \quad = \bigl[\zeta \zeta _{2}\varphi _{\rho _{1}} \bigl(\lambda \bigl({\mathfrak{f}}_{0}^{i}- \epsilon \bigr) \bigr) \widetilde{L}_{3}+\zeta \zeta _{2}\varphi _{\rho _{2}}\bigl(\mu \bigl({ \mathfrak{g}}_{0}^{i}- \epsilon \bigr)\bigr)\widetilde{L}_{4} \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \\& \quad \ge \bigl[(1-a)c_{2}+(1-a) (1-c_{2}) \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}=(1-a) \bigl\Vert (x,y) \bigr\Vert _{ \mathcal{Y}}. \end{aligned}$$

So, $\|{\mathcal{A}}_{2}(x,y)\|\ge {\mathcal{A}}_{2}(x,y)(\theta _{1})\ge (1-a)\|(x,y) \|_{\mathcal{Y}}$.

Then, for $(x,y)\in {\mathcal{P}}\cap \partial \varLambda _{3}$, we conclude

$$ \bigl\Vert {\mathcal{A}}(x,y) \bigr\Vert _{\mathcal{Y}}= \bigl\Vert {\mathcal{A}}_{1}(x,y) \bigr\Vert + \bigl\Vert {\mathcal{A}}_{2}(x,y) \bigr\Vert \ge a \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}+(1-a) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}= \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}. $$

(22)

We introduce now the functions ${\mathfrak{f}}^{*}, {\mathfrak{g}}^{*}:[0,1]\times \mathbb{R}_{+} \to \mathbb{R}_{+}$ by ${\mathfrak{f}}^{*}(t,u)=\max_{0\le x+y\le u} f(t,x,y)$, $g^{*}(t,u)=\max_{0\le x+y\le u}g(t,x,y)$ for all $t\in [0,1]$ and $u\in \mathbb{R}_{+}$. Then

$$ f(t,x,y)\le {\mathfrak{f}}^{*}(t,u),\qquad g(t,x,y)\le { \mathfrak{g}}^{*}(t,u),\quad \forall t\in [0,1], x,y\ge 0, x+y\le u. $$

The functions ${\mathfrak{f}}^{*}(t,\cdot )$, ${\mathfrak{g}}^{*}(t,\cdot )$ are nondecreasing for every $t\in [0,1]$ and they satisfy the conditions

$$ \limsup_{u\to \infty }\max_{t\in [0,1]} \frac{{\mathfrak{f}}^{*}(t,u)}{\varphi _{\varrho _{1}}(u)}={ \mathfrak{f}}_{\infty }^{s},\qquad \limsup _{u\to \infty }\max_{t\in [0,1]} \frac{{\mathfrak{g}}^{*}(t,u)}{\varphi _{\varrho _{2}}(u)}={ \mathfrak{g}}_{\infty }^{s}. $$

Then for $\epsilon >0$ there exists $r_{4}'>0$ such that for all $u\ge r_{4}'$ and $t\in [0,1]$ we obtain

$$\begin{aligned}& \frac{{\mathfrak{f}}^{*}(t,u)}{\varphi _{\varrho _{1}}(u)}\le \limsup_{u\to \infty } \max_{t\in [0,1]} \frac{f^{*}(t,u)}{\varphi _{\varrho _{1}}(u)}+\epsilon ={ \mathfrak{f}}_{ \infty }^{s}+\epsilon , \\& \frac{{\mathfrak{g}}^{*}(t,u)}{\varphi _{\varrho _{2}}(u)}\le \limsup_{u\to \infty }\max _{t\in [0,1]} \frac{{\mathfrak{g}}^{*}(t,u)}{\varphi _{\varrho _{2}}(u)}+\epsilon ={ \mathfrak{g}}_{\infty }^{s}+\epsilon , \end{aligned}$$

and hence ${\mathfrak{f}}^{*}(t,u)\le ({\mathfrak{f}}_{\infty }^{s}+\epsilon ) \varphi _{\varrho _{1}}(u)$ and ${\mathfrak{g}}^{*}(t,u)\le ({\mathfrak{g}}_{\infty }^{s}+\epsilon ) \varphi _{\varrho _{2}}(u)$.

We consider $r_{4}=\max \{2r_{3},r_{4}'\}$ and we introduce the set $\varLambda _{4}=\{(x,y)\in {\mathcal{Y}}, \|(x,y)\|_{\mathcal{Y}}< r_{4}\}$. Let $(x,y)\in {\mathcal{P}}\cap \partial \varLambda _{4}$. By the definitions of ${\mathfrak{f}}^{*}$ and ${\mathfrak{g}}^{*}$ we deduce

$$ \begin{aligned} &f \bigl(t,x(t),y(t) \bigr)\le {\mathfrak{f}}^{*} \bigl(t, \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \bigr), \\ & g \bigl(t,x(t),y(t) \bigr) \le {\mathfrak{g}}^{*} \bigl(t, \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \bigr), \quad \forall t\in [0,1]. \end{aligned} $$

(23)

Then for all $t\in [0,1]$ we find

$$\begin{aligned}& {\mathcal{A}}_{1}(x,y) (t) \\& \quad \le \lambda ^{\rho _{1}-1} \int _{0}^{1}{ \mathcal{J}}_{1}(\nu )\varphi _{\rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{0}^{\nu }(\nu -s)^{ \alpha _{1}-1}f \bigl(s,x(s),y(s) \bigr) \,ds \biggr)\,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1} \int _{0}^{1} {\mathcal{J}}_{2}( \nu ) \varphi _{\rho _{2}} \biggl( \frac{1}{\varGamma (\alpha _{2})} \int _{0}^{\nu } (\nu -s)^{ \alpha _{2}-1}g \bigl(s,x(s),y(s) \bigr) \,ds \biggr)\,d\nu \\& \quad \le \lambda ^{\rho _{1}-1} \int _{0}^{1} {\mathcal{J}}_{1}( \nu )\varphi _{\rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{0}^{\nu } (\nu -s)^{ \alpha _{1}-1}{ \mathfrak{f}}^{*} \bigl(s, \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \bigr) \,ds \biggr)\,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1} \int _{0}^{1}{\mathcal{J}}_{2}(\nu ) \varphi _{\rho _{2}} \biggl( \frac{1}{\varGamma (\alpha _{2})} \int _{0}^{\nu }(\nu -s)^{ \alpha _{2}-1}{ \mathfrak{g}}^{*} \bigl(s, \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \bigr) \,ds \biggr)\,d\nu \\& \quad \le \lambda ^{\rho _{1}-1} \int _{0}^{1}{\mathcal{J}}_{1}( \nu )\varphi _{\rho _{1}} \biggl( \frac{1}{\varGamma (\alpha _{1})} \int _{0}^{\nu } (\nu -s)^{ \alpha _{1}-1} \bigl({ \mathfrak{f}}_{\infty }^{s}+\epsilon \bigr)\varphi _{ \varrho _{1}} \bigl( \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \bigr) \,ds \biggr)\,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1} \int _{0}^{1}{\mathcal{J}}_{2}(\nu ) \varphi _{\rho _{2}} \biggl( \frac{1}{\varGamma (\alpha _{2})} \int _{0}^{\nu } (\nu -s)^{ \alpha _{2}-1} \bigl({ \mathfrak{g}}_{\infty }^{s}+\epsilon \bigr)\varphi _{ \varrho _{2}} \bigl( \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \bigr) \,ds \biggr)\,d\nu \\& \quad =\varphi _{\rho _{1}} \bigl(\lambda \bigl({\mathfrak{f}}_{\infty }^{s}+ \epsilon \bigr) \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{0}^{1} \frac{1}{(\varGamma (\alpha _{1}+1))^{\rho _{1}-1}}{\nu }^{\alpha _{1}( \rho _{1}-1)}{\mathcal{J}}_{1}(\nu ) \,d\nu \\& \qquad {}+\varphi _{\rho _{2}} \bigl(\mu \bigl({\mathfrak{g}}_{\infty }^{s}+ \epsilon \bigr) \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{0}^{1} \frac{1}{(\varGamma (\alpha _{2}+1))^{\rho _{2}-1}}{\nu }^{\alpha _{2}( \rho _{2}-1)}{\mathcal{J}}_{2}(\nu ) \,d\nu \\& \quad = \bigl[\varphi _{\rho _{1}} \bigl(\lambda \bigl({\mathfrak{f}}_{\infty }^{s}+ \epsilon \bigr) \bigr)L_{1}+\varphi _{\rho _{2}} \bigl(\mu \bigl({\mathfrak{g}}_{\infty }^{s}+ \epsilon \bigr) \bigr)L_{2} \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \\& \quad \le \bigl[bc_{3}+b(1-c_{3}) \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}=b \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}, \end{aligned}$$

and so $\|{\mathcal{A}}_{1}(x,y)\|\le b\|(x,y)\|_{\mathcal{Y}}$.

In a similar manner, we obtain

$$ \begin{aligned} {\mathcal{A}}_{2}(x,y) (t)\le{}& \varphi _{\rho _{1}} \bigl(\lambda \bigl({\mathfrak{f}}_{ \infty }^{s}+ \epsilon \bigr) \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{0}^{1} \frac{1}{(\varGamma (\alpha _{1}+1))^{\rho _{1}-1}}{\nu }^{ \alpha _{1}(\rho _{1}-1)}{\mathcal{J}}_{3}(\nu ) \,d\nu \\ &{}+\varphi _{\rho _{2}} \bigl(\mu \bigl({\mathfrak{g}}_{\infty }^{s}+ \epsilon \bigr) \bigr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \int _{0}^{1} \frac{1}{(\varGamma (\alpha _{2}+1))^{\rho _{2}-1}}{\nu }^{\alpha _{2}( \rho _{2}-1)}{\mathcal{J}}_{4}(\nu ) \,d\nu \\ ={}& \bigl[\varphi _{\rho _{1}} \bigl(\lambda \bigl({\mathfrak{f}}_{\infty }^{s}+ \epsilon \bigr) \bigr)L_{3}+\varphi _{\rho _{2}} \bigl(\mu \bigl({\mathfrak{g}}_{\infty }^{s}+ \epsilon \bigr) \bigr)L_{4} \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \\ \le{}& \bigl[(1-b)c_{4}+(1-b) (1-c_{4}) \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}=(1-b) \bigl\Vert (x,y) \bigr\Vert _{ \mathcal{Y}}, \end{aligned} $$

and then $\|{\mathcal{A}}_{2}(x,y)\|\le (1-b)\|(x,y)\|_{\mathcal{Y}}$.

Therefore, for $(x,y)\in {\mathcal{P}}\cap \partial \varLambda _{4}$, we deduce

$$ \bigl\Vert {\mathcal{A}}(x,y) \bigr\Vert _{\mathcal{Y}}= \bigl\Vert {\mathcal{A}}_{1}(x,y) \bigr\Vert + \bigl\Vert {\mathcal{A}}_{2}(x,y) \bigr\Vert \le b \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}+(1-b) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}= \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}. $$

(24)

By using Lemma 3.1, Eqs. (23), (24) and the Guo–Krasnosel’skii fixed point theorem, we conclude that ${\mathcal{A}}$ has a fixed point $(x,y)\in {\mathcal{P}}\cap (\overline{\varLambda }_{4}\setminus \varLambda _{3})$, which is a positive solution of problem (1), (2).

Case (6). We have ${\mathfrak{f}}_{\infty }^{s}\in (0,\infty )$, ${\mathfrak{g}}_{\infty }^{s}=0$ and ${\mathfrak{g}}_{0}^{i}=\infty$. Let $\lambda \in (0,\widetilde{E})$ and $\mu \in (0,\infty )$. We choose $\widetilde{c}_{3}\in (\varphi _{\rho _{1}}(\lambda {\mathfrak{f}}_{ \infty }^{s})\frac{L_{1}}{b},1)$ and $\widetilde{c}_{4}\in (\varphi _{\rho _{1}}(\lambda {\mathfrak{f}}_{ \infty }^{s})\frac{L_{3}}{1-b},1)$. The choice of $\widetilde{c}_{3}$ and $\widetilde{c}_{4}$ is possible because $\lambda <\frac{1}{{\mathfrak{f}}_{\infty }^{s}}\varphi _{\varrho _{1}}( \frac{b}{L_{1}})$ and $\lambda <\frac{1}{{\mathfrak{f}}_{\infty }^{s}}\varphi _{\varrho _{1}}( \frac{1-b}{L_{3}})$. Let $\epsilon >0$ be such that

$$\begin{aligned}& \lambda \le \min \biggl\{ \frac{1}{{\mathfrak{f}}_{\infty }^{s}+\epsilon } \varphi _{\varrho _{1}} \biggl( \frac{b\widetilde{c}_{3}}{L_{1}} \biggr), \frac{1}{{\mathfrak{f}}_{\infty }^{s}+\epsilon } \varphi _{\varrho _{1}} \biggl( \frac{(1-b)\widetilde{c}_{4}}{L_{3}} \biggr) \biggr\} , \\& \varepsilon \varphi _{\varrho _{2}} \biggl( \frac{1}{\zeta \zeta _{1}\widetilde{L}_{2}} \biggr)\le \mu \le \min \biggl\{ \frac{1}{\epsilon }\varphi _{\varrho _{2}} \biggl( \frac{b(1-\widetilde{c}_{3})}{L_{2}} \biggr), \frac{1}{\epsilon } \varphi _{\varrho _{2}} \biggl( \frac{(1-b)(1-\widetilde{c}_{4})}{L_{4}} \biggr) \biggr\} . \end{aligned}$$

By $(I2)$ and the definition of ${\mathfrak{g}}_{0}^{i}$ we find that there exists $r_{3}>0$ such that

$$ g(t,x,y)\ge \frac{1}{\epsilon }\varphi _{\varrho _{2}}(x+y), \quad \forall t\in [ \theta _{1},\theta _{2}], x,y\ge 0, x+y\le r_{3}. $$

We define the set $\varLambda _{3}=\{(x,y)\in {\mathcal{Y}}, \|(x,y)\|_{\mathcal{Y}}< r_{3}\}$. Let $(x,y)\in {\mathcal{P}}\cap \partial \varLambda _{3}$, that is $\|x\|+\|y\|=r_{3}$. Because $x(t)+y(t)\le \|x\|+\|y\|=r_{3}$ for all $t\in [0,1]$, then by Lemma 2.5, we find

$$\begin{aligned}& {\mathcal{A}}_{1}(x,y) (\theta _{1}) \\& \quad \ge \lambda ^{\rho _{1}-1} \int _{0}^{1} \theta _{1}^{\beta _{1}-1}{ \mathcal{J}}_{1}( \nu )\varphi _{\rho _{1}} \bigl(I_{0+}^{\alpha _{1}}f \bigl(\nu ,x(\nu ),y(\nu ) \bigr) \bigr) \,d\nu \\& \qquad {}+\mu ^{\rho _{2}-1} \int _{0}^{1} \theta _{1}^{\beta _{1}-1}{ \mathcal{J}}_{2}(\nu )\varphi _{\rho _{2}} \bigl(I_{0+}^{\alpha _{2}}g \bigl(\nu ,x( \nu ),y(\nu ) \bigr) \bigr) \,d\nu \\& \quad \ge \mu ^{\rho _{2}-1}\theta _{1}^{\beta _{1}-1} \int _{ \theta _{1}}^{\theta _{2}}{\mathcal{J}}_{2}(s) \varphi _{\rho _{2}} \biggl( \frac{1}{\varGamma (\alpha _{2})} \int _{ \theta _{1}}^{\nu } (\nu -s)^{\alpha _{2}-1}g \bigl(s,x(s),y(s) \bigr) \,ds \biggr)\,d\nu \\& \quad \ge \mu ^{\rho _{2}-1}\theta _{1}^{\beta _{1}-1} \int _{ \theta _{1}}^{\theta _{2}}{\mathcal{J}}_{2}( \nu )\varphi _{\rho _{2}} \biggl( \frac{1}{\varGamma (\alpha _{2})} \int _{\theta _{1}}^{\nu } (\nu -s)^{\alpha _{2}-1} \frac{1}{\epsilon }\varphi _{\varrho _{2}} \bigl(x(s)+y(s) \bigr) \,ds \biggr)\,d\nu \\& \quad \ge \mu ^{\rho _{2}-1}\theta _{1}^{\beta _{1}-1} \int _{ \theta _{1}}^{\theta _{2}}{\mathcal{J}}_{2}( \nu )\varphi _{\rho _{2}} \biggl(\frac{1}{\varGamma (\alpha _{2})} \int _{\theta _{1}}^{ \nu }(\nu -s)^{\alpha _{2}-1} \frac{1}{\epsilon }\varphi _{\varrho _{2}} \bigl( \zeta \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \bigr) \,ds \biggr)\,d\nu \\& \quad = \zeta \zeta _{1}\varphi _{\rho _{2}} \biggl( \frac{\mu }{\epsilon } \biggr) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \widetilde{L}_{2}\ge \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}. \end{aligned}$$

Hence we deduce $\|{\mathcal{A}}_{1}(x,y)\|\ge {\mathcal{A}}_{1}(x,y)(\theta _{1})\ge \|(x,y)\|_{ \mathcal{Y}}$ and $\|{\mathcal{A}}(x,y)\|_{\mathcal{Y}}\ge \|{\mathcal{A}}_{1}(x,y)\|\ge \|(x,y)\|_{ \mathcal{Y}}$.

For the second part of the proof, we use the functions ${\mathfrak{f}}^{*}$ and ${\mathfrak{g}}^{*}$ from Case (1), which satisfy here the conditions

$$ \limsup_{u\to \infty }\max_{t\in [0,1]} \frac{{\mathfrak{f}}^{*}(t,u)}{\varphi _{\varrho _{1}}(u)}={ \mathfrak{f}}_{\infty }^{s}, \qquad \lim _{u\to \infty } \max_{t\in [0,1]} \frac{{\mathfrak{g}}^{*}(t,u)}{\varphi _{\varrho _{2}}(u)}=0. $$

Then for $\epsilon >0$ there exists $r_{4}'>0$ such that for all $u\ge r_{4}'$ and $t\in [0,1]$ we obtain

$$ \begin{aligned} &\frac{{\mathfrak{f}}^{*}(t,u)}{\varphi _{\varrho _{1}}(u)}\le \limsup_{u\to \infty } \max_{t\in [0,1]} \frac{{\mathfrak{f}}^{*}(t,u)}{\varphi _{\varrho _{1}}(u)}+\epsilon ={ \mathfrak{f}}_{\infty }^{s}+\epsilon , \\ &\frac{{\mathfrak{g}}^{*}(t,u)}{\varphi _{\varrho _{2}}(u)}\le \lim_{u\to \infty }\max _{t\in [0,1]} \frac{{\mathfrak{g}}^{*}(t,u)}{\varphi _{\varrho _{2}}(u)}+\epsilon = \epsilon , \end{aligned} $$

and hence ${\mathfrak{f}}^{*}(t,u)\le ({\mathfrak{f}}_{\infty }^{s}+\epsilon ) \varphi _{\varrho _{1}}(u)$ and $g^{*}(t,u)\le \epsilon \varphi _{\varrho _{2}}(u)$.

We define $r_{4}=\max \{2r_{3},r_{4}'\}$ and we introduce the set $\varLambda _{4}=\{(x,y)\in {\mathcal{Y}}, \|(x,y)\|_{\mathcal{Y}}< r_{4}\}$. Let $(x,y)\in {\mathcal{P}}\cap \partial \varLambda _{4}$. By the definitions of ${\mathfrak{f}}^{*}$ and ${\mathfrak{g}}^{*}$ we deduce Eq. (23). Besides, in a similar manner to that used in the proof of Case (1), we find

$$\begin{aligned}& \begin{aligned} {\mathcal{A}}_{1}(x,y) (t)&\le \varphi _{\rho _{1}} \bigl(\lambda \bigl({\mathfrak{f}}_{ \infty }^{s}+ \epsilon \bigr) \bigr)L_{1} \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}+\varphi _{\rho _{2}}( \mu \epsilon )L_{2} \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \\ &\le \bigl[b \widetilde{c}_{3}+b(1-\widetilde{c}_{3}) \bigr] \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}=b \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}, \quad \forall t\in [0,1], \end{aligned} \\& \begin{aligned} {\mathcal{A}}_{2}(x,y) (t)&\le \varphi _{\rho _{1}} \bigl( \lambda \bigl({\mathfrak{f}}_{ \infty }^{s}+\epsilon \bigr) \bigr)L_{3} \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}+ \varphi _{\rho _{2}}( \mu \epsilon )L_{4} \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}} \\ &\le \bigl[(1-b) \widetilde{c}_{4}+(1-b) (1-\widetilde{c}_{4}) \bigr] \bigl\Vert (x,y) \bigr\Vert _{ \mathcal{Y}}=(1-b) \bigl\Vert (x,y) \bigr\Vert _{\mathcal{Y}}, \quad \forall t\in [0,1], \end{aligned} \end{aligned}$$

and then $\|{\mathcal{A}}(x,y)\|_{\mathcal{Y}}\le b\|(x,y)\|_{\mathcal{Y}}+(1-b)\|(x,y)\|_{ \mathcal{Y}}=\|(x,y)\|_{\mathcal{Y}}$.

Therefore we obtain the conclusion of the theorem. □

4 Nonexistence of the positive solutions

In this section we give intervals for λ and μ for which there exist no positive solutions of problem (1), (2). By using similar arguments to those used in the proofs of Theorems 4.1–4.4 from [30], we obtain the following theorems for problem (1), (2).

Theorem 4.1

Assume that$(I1)$and$(I2)$hold. If there exist positive numbers$T_{1}$, $T_{2}$such that

$$ f(t,x,y)\le T_{1}\varphi _{\varrho _{1}}(x+y),\qquad g(t,x,y)\le T_{2} \varphi _{\varrho _{2}}(x+y),\quad \forall t\in [0,1], x, y\ge 0, $$

(25)

then there exist positive constants$\lambda _{0}$and$\mu _{0}$such that for every$\lambda \in (0,\lambda _{0})$and$\mu \in (0,\mu _{0})$the boundary value problem (1), (2) has no positive solution.

In the proof of Theorem 4.1 we define $\lambda _{0}=\min \{ (T_{1}\varphi _{\varrho _{1}}(4L_{1}))^{-1},(T_{1} \varphi _{\varrho _{1}}(4L_{3}))^{-1} \}$ and $\mu _{0}=\min \{ (T_{2}\varphi _{\varrho _{2}}(4L_{2}))^{-1},(T_{2} \varphi _{\varrho _{2}}(4L_{4}))^{-1} \}$, where $L_{i}$, $i=1,\ldots ,4$, are given by (19).

Remark 4.1

If ${\mathfrak{f}}_{0}^{s}, {\mathfrak{g}}_{0}^{s}, {\mathfrak{f}}_{ \infty }^{s}, {\mathfrak{g}}_{\infty }^{s}<\infty$, then there exist positive constants $T_{1}$, $T_{2}$ such that relation (25) holds, and so we obtain the conclusion of Theorem 4.1.

Theorem 4.2

Assume that$(I1)$and$(I2)$hold. If there exist positive numbers$\theta _{1}$, $\theta _{2}$with$0<\theta _{1}<\theta _{2}\le 1$and$q_{1}>0$such that

$$ f(t,x,y)\ge q_{1}\varphi _{\varrho _{1}}(x+y),\quad \forall t\in [ \theta _{1},\theta _{2}], x, y\ge 0, $$

(26)

then there exists a positive constant$\lambda _{0}'$such that for every$\lambda >\lambda _{0}'$and$\mu >0$, the boundary value problem (1), (2) has no positive solution.

In the proof of Theorem 4.2 we define $\lambda _{0}'=\min \{ (q_{1}\varphi _{\varrho _{1}}(\zeta \zeta _{1}\widetilde{L}_{1}))^{-1},(q_{1}\varphi _{\varrho _{1}}( \zeta \zeta _{2}\widetilde{L}_{3}))^{-1} \}$, where $\widetilde{L}_{1}$ and $\widetilde{L}_{3}$ are given by (19).

Remark 4.2

If for $\theta _{1}$, $\theta _{2}$ with $0<\theta _{1}<\theta _{2}\le 1$, we have ${\mathfrak{f}}_{0}^{i}, {\mathfrak{f}}_{\infty }^{i}>0$, and $f(t,x,y)>0$ for all $t\in [\theta _{1},\theta _{2}]$ and $x,y\ge 0$ with $x+y>0$, then Eq. (26) holds, and therefore we obtain the conclusion of Theorem 4.2.

Theorem 4.3

Assume that$(I1)$and$(I2)$hold. If there exist positive numbers$\theta _{1}$, $\theta _{2}$with$0<\theta _{1}<\theta _{2}\le 1$and$q_{2}>0$such that

$$ g(t,x,y)\ge q_{2}\varphi _{\varrho _{2}}(x+y),\quad \forall t\in [ \theta _{1},\theta _{2}], x, y\ge 0, $$

(27)

then there exists a positive constant$\mu _{0}'$such that for every$\mu >\mu _{0}'$and$\lambda >0$the boundary value problem (1), (2) has no positive solution.

In the proof of Theorem 4.3 we define $\mu _{0}'=\min \{ (q_{2}\varphi _{\varrho _{2}}(\zeta \zeta _{1} \widetilde{L}_{2}))^{-1},(q_{2}\varphi _{\varrho _{2}}(\zeta \zeta _{2} \widetilde{L}_{4}))^{-1} \}$, where $\widetilde{L}_{2}$ and $\widetilde{L}_{4}$ are given by (19).

Remark 4.3

If for $\theta _{1}$, $\theta _{2}$ with $0<\theta _{1}<\theta _{2}\le 1$, we have ${\mathfrak{g}}_{0}^{i}, {\mathfrak{g}}_{\infty }^{i}>0$, and $g(t,x,y)>0$ for all $t\in [\theta _{1},\theta _{2}]$ and $x,y\ge 0$ with $x+y>0$, then Eq. (27) holds, and hence we deduce the conclusion of Theorem 4.3.

Theorem 4.4

Assume that$(I1)$and$(I2)$hold. If there exist positive numbers$\theta _{1}$, $\theta _{2}$with$0<\theta _{1}<\theta _{2}\le 1$and$q_{1}, q_{2}>0$such that

$$ f(t,x,y)\ge q_{1}\varphi _{\varrho _{1}}(x+y),\qquad g(t,x,y)\ge q_{2} \varphi _{\varrho _{2}}(x+y),\quad \forall t\in [ \theta _{1}, \theta _{2}], x, y\ge 0, $$

(28)

then there exist positive constants$\widetilde{\lambda }_{0}$and$\widetilde{\mu }_{0}$such that for every$\lambda >\widetilde{\lambda }_{0}$and$\mu >\widetilde{\mu }_{0}$the boundary value problem (1), (2) has no positive solution.

In the proof of Theorem 4.4 we define $\widetilde{\lambda }_{0}=(q_{1}\varphi _{\varrho _{1}}(2\zeta \zeta _{1} \widetilde{L}_{1}))^{-1}$ and $\widetilde{\mu }_{0}=(q_{2}\varphi _{\varrho _{2}}(2\zeta \zeta _{2} \widetilde{L}_{4}))^{-1}$.

Remark 4.4

If for $\theta _{1}$, $\theta _{2}$ with $0<\theta _{1}<\theta _{2}\le 1$, we have ${\mathfrak{f}}_{0}^{i}, {\mathfrak{f}}_{\infty }^{i}, { \mathfrak{g}}_{0}^{i}, {\mathfrak{g}}_{\infty }^{i}>0$, and $f(t,x,y)>0$ and $g(t,x,y)>0$ for all $t\in [\theta _{1},\theta _{2}]$ and $x,y\ge 0$ with $x+y>0$, then Eq. (28) holds, and so we obtain the conclusion of Theorem 4.4.

5 An example

Let $\alpha _{1}=1/2$, $\alpha _{2}=1/3$, $\beta _{1}=10/3$, $n=4$, $\beta _{2}=17/4$, $m=5$, $p=2$, $q=1$, $\delta _{0}=11/5$, $\gamma _{1}=5/6$, $\gamma _{2}=3/2$, $\gamma _{0}=9/8$, $\delta _{1}=3/7$, $H_{1}(t)=4t$ for all $t\in [0,1]$, $H_{2}(t)=\{1, t\in [0,1/4); 8, t\in [1/4,1]\}$, $K_{1}(t)=\{1, t\in [0,1/2); 3, t\in [1/2,1]\}$, $\varrho _{1}=5$, $\rho _{1}=5/4$, $\varrho _{2}=3$, $\rho _{2}=3/2$, $\varphi _{\varrho _{1}}(s)=|s|^{3}s$, $\varphi _{\rho _{1}}(s)=|s|^{-3/4}s$, $\varphi _{\varrho _{2}}(s)=|s|s$, $\varphi _{\rho _{2}}(s)=|s|^{-1/2}s$.

We consider the system of fractional differential equations

$$ \textstyle\begin{cases} D_{0+}^{1/2} (\varphi _{5} (D_{0+}^{10/3}x(t) ) )+\lambda (3-t)^{a_{0}} (e^{(x(t)+y(t))^{4}}-1 )=0,\quad t\in (0,1), \\ D_{0+}^{1/3} (\varphi _{3} (D_{0+}^{17/4}y(t) ) )+\mu (t+2)^{b_{0}}(x^{3}(t)+y^{3}(t))=0,\quad t\in (0,1), \end{cases} $$

(29)

with the coupled nonlocal boundary conditions

$$ \textstyle\begin{cases} x(0)=x'(0)=x''(0)=0, \qquad D_{0+}^{10/3}x(0)=0, \\ D_{0+}^{9/8}x(1)=4 \int _{0}^{1} D_{0+}^{5/6}y(t) \,dt+7D_{0+}^{3/2}y (\frac{1}{4} ), \\ y(0)=y'(0)=y''(0)=y'''(0)=0, \qquad D_{0+}^{17/4}y(0)=0, \\ D_{0+}^{11/5}y(1)=2D_{0+}^{3/7}x (\frac{1}{2} ), \end{cases} $$

(30)

where $a_{0}, b_{0}>0$.

Here we have $f(t,x,y)=(3-t)^{a_{0}}(e^{(x+y)^{4}}-1)$, $g(t,x,y)=(t+2)^{b_{0}}(x^{3}+y^{3})$ for all $t\in [0,1]$ and $x, y\ge 0$. We obtain $\Delta \approx 15.18283383>0$, and then the assumptions $(I1)$ and $(I2)$ are satisfied. Besides, we find

$$\begin{aligned}& g_{1}(t,s)= \frac{1}{\varGamma (10/3)} \textstyle\begin{cases} t^{7/3}(1-s)^{29/24}-(t-s)^{7/3},& 0\le s\le t\le 1, \\ t^{7/3}(1-s)^{29/24}, &0\le t\le s\le 1, \end{cases}\displaystyle \\& g_{11}(t,s)= \frac{1}{\varGamma (61/21)} \textstyle\begin{cases} t^{40/21}(1-s)^{29/24}-(t-s)^{40/21}, & 0\le s\le t\le 1, \\ t^{40/21}(1-s)^{29/24}, &0\le t\le s\le 1, \end{cases}\displaystyle \\& g_{2}(t,s)= \frac{1}{\varGamma (17/4)} \textstyle\begin{cases} t^{13/4}(1-s)^{21/20}-(t-s)^{13/4},& 0\le s\le t\le 1, \\ t^{13/4}(1-s)^{21/20},& 0\le t\le s\le 1,\end{cases}\displaystyle \\& g_{21}(t,s)= \frac{1}{\varGamma (41/12)} \textstyle\begin{cases} t^{29/12}(1-s)^{21/20}-(t-s)^{29/12}, & 0\le s\le t\le 1, \\ t^{29/12}(1-s)^{21/20}, &0\le t\le s\le 1,\end{cases}\displaystyle \\& g_{22}(t,s)= \frac{1}{\varGamma (11/4)} \textstyle\begin{cases} t^{7/4}(1-s)^{21/20}-(t-s)^{7/4}, & 0\le s\le t\le 1, \\ t^{7/4}(1-s)^{21/20},& 0\le t\le s\le 1,\end{cases}\displaystyle \\& {\mathcal{G}}_{1}(t,s)=g_{1}(t,s)+ \frac{2t^{7/3}}{\Delta } \biggl[\frac{48\varGamma (17/4)}{41\varGamma (41/12)}+7 \biggl(\frac{1}{4} \biggr)^{7/4} \frac{\varGamma (17/4)}{\varGamma (11/4)} \biggr] g_{11} \biggl(\frac{1}{2},s \biggr), \\& {\mathcal{G}}_{2}(t,s)= \frac{t^{7/3}\varGamma (17/4)}{\Delta \varGamma (41/20)} \biggl[4 \int _{0}^{1} g_{21}(\tau ,s) \,d \tau +7 g_{22} \biggl(\frac{1}{4},s \biggr) \biggr], \\& {\mathcal{G}}_{3}(t,s)= \frac{2t^{13/4}\varGamma (10/3)}{\Delta \varGamma (53/24)}g_{11} \biggl( \frac{1}{2},s \biggr), \\& {\mathcal{G}}_{4}(t,s)=g_{2}(t,s)+ \frac{t^{13/4}\varGamma (10/3)}{\Delta \varGamma (61/21)2^{19/21}} \biggl[4 \int _{0}^{1} g_{21}(\tau ,s) \,d \tau +7 g_{22} \biggl(\frac{1}{4},s \biggr) \biggr], \end{aligned}$$

for all $(t,s)\in [0,1]\times [0,1]$. For the functions $h_{1}$, $h_{2}$ and ${\mathcal{J}}_{i}$, $i=1,\ldots ,4$, we obtain

$$\begin{aligned}& h_{1}(s)=\frac{1}{\varGamma (10/3)} \bigl[(1-s)^{29/24}-(1-s)^{7/3} \bigr], \\& h_{2}(s)= \frac{1}{\varGamma (17/4)} \bigl[(1-s)^{21/20}-(1-s)^{13/4} \bigr], \\& \quad \text{for all } s\in [0,1], \\& {\mathcal{J}}_{1}(s)= \textstyle\begin{cases} \frac{1}{\varGamma (10/3)} [(1-s)^{29/24}-(1-s)^{7/3} ]+ \frac{1}{\Delta } [\frac{48\varGamma (17/4)}{41\varGamma (41/12)}+7 (\frac{1}{4} )^{7/4}\frac{\varGamma (17/4)}{\varGamma (11/4)} ] \\ \quad {} \times \frac{2}{\varGamma (61/21)} [ (\frac{1}{2} )^{40/21}(1-s)^{29/24}- (\frac{1}{2}-s )^{40/21} ],& 0\le s < \frac{1}{2}, \\ \frac{1}{\varGamma (10/3)} [(1-s)^{29/24}-(1-s)^{7/3} ]+ \frac{1}{\Delta } [\frac{48\varGamma (17/4)}{41\varGamma (41/12)}+7 (\frac{1}{4} )^{7/4}\frac{\varGamma (17/4)}{\varGamma (11/4)} ] \\ \quad {} \times \frac{2}{\varGamma (61/21)} (\frac{1}{2} )^{40/21}(1-s)^{29/24},& \frac{1}{2}\le s \le 1, \end{cases}\displaystyle \\& {\mathcal{J}}_{2}(s)= \textstyle\begin{cases} \frac{\varGamma (17/4)}{\Delta \varGamma (41/20)} \{ \frac{48}{41 \varGamma (41/12)} [(1-s)^{21/20}-(1-s)^{41/12} ] \\ \quad {} +\frac{7}{\varGamma (11/4)} [ (\frac{1}{4} )^{7/4}(1-s)^{21/20}- (\frac{1}{4}-s )^{7/4} ] \} , &0\le s< \frac{1}{4}, \\ \frac{\varGamma (17/4)}{\Delta \varGamma (41/20)} \{ \frac{48}{41 \varGamma (41/12)} [(1-s)^{21/20}-(1-s)^{41/12} ] \\ \quad {} +\frac{7}{\varGamma (11/4)} (\frac{1}{4} )^{7/4}(1-s)^{21/20} \} , &\frac{1}{4}\le s \le 1, \end{cases}\displaystyle \\& {\mathcal{J}}_{3}(s)= \textstyle\begin{cases} \frac{2\varGamma (10/3)}{\Delta \varGamma (53/24)\varGamma (61/21)} [ (\frac{1}{2} )^{40/21}(1-s)^{29/24}- (\frac{1}{2}-s )^{40/21} ],& 0\le s< \frac{1}{2}, \\ \frac{2\varGamma (10/3)}{\Delta \varGamma (53/24)\varGamma (61/21)} ( \frac{1}{2} )^{40/21}(1-s)^{29/24}, &\frac{1}{2}\le s\le 1, \end{cases}\displaystyle \\& {\mathcal{J}}_{4}(s)= \textstyle\begin{cases} \frac{1}{\varGamma (17/4)} [(1-s)^{21/20}-(1-s)^{13/4} ] \\ \quad {}+ \frac{\varGamma (10/3)}{\Delta \varGamma (61/21)2^{19/21}} \{ \frac{48}{41 \varGamma (41/12)} [(1-s)^{21/20} \\ \quad {} -(1-s)^{41/12} ]+\frac{7}{\varGamma (11/4)} [ (\frac{1}{4} )^{7/4}(1-s)^{21/20}- ( \frac{1}{4}-s )^{7/4} ] \} ,& 0\le s< \frac{1}{4}, \\ \frac{1}{\varGamma (17/4)} [(1-s)^{21/20}-(1-s)^{13/4} ] \\ \quad {}+ \frac{\varGamma (10/3)}{\Delta \varGamma (61/21)2^{19/21}} \{ \frac{48}{41 \varGamma (41/12)} [(1-s)^{21/20} \\ \quad {} -(1-s)^{41/12} ]+\frac{7}{\varGamma (11/4)} (\frac{1}{4} )^{7/4}(1-s)^{21/20} \} ,& \frac{1}{4}\le s\le 1. \end{cases}\displaystyle \end{aligned}$$

We choose $\theta _{1}=1/4$ and $\theta _{2}=3/4$, and hence we obtain $\zeta _{1}=(1/4)^{7/3}$, $\zeta _{2}=(1/4)^{13/4}$ and $\zeta =\zeta _{2}$. Besides, we deduce ${\mathfrak{f}}_{0}^{s}=3^{a_{0}}$, ${\mathfrak{f}}_{\infty }^{i}=\infty$, ${\mathfrak{g}}_{0}^{s}=0$, ${\mathfrak{g}}_{\infty }^{i}=\infty$, $L_{1}\approx 0.08207184$, $L_{3}\approx 0.01229905$, $\widetilde{L}_{1}\approx 0.05185073$, $\widetilde{L}_{3}\approx 0.00775575$.

By Theorem 3.1(7), if we consider $b=1/2$, then, for any $\lambda \in (0,E)$ and $\mu \in (0,\infty )$ with $E=3^{-a_{0}}(2L_{1})^{-4}$, the problem (29), (30) has a positive solution $(x(t),y(t))$, $t\in [0,1]$. For example, if $a_{0}=1$ we find $E\approx 459.179$.

We can also use Theorem 4.2, because $f(t,x,y)\ge q_{1}(x+y)^{4}$ for all $t\in [1/4,3/4]$ and $x, y\ge 0$, with $q_{1}=(9/4)^{a_{0}}$. If $a_{0}=1$, we obtain $\lambda _{0}'=(q_{1}(\zeta \zeta _{1}\widetilde{L}_{1})^{4})^{-1} \approx 1.71714\times 10^{18}$, and then we deduce then, for every $\lambda >\lambda _{0}'$ and $\mu >0$, that the boundary value problem (29), (30) has no positive solution.