Theory and Modern Applications

# 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 [18, 1725, 28, 31, 37, 3941]. For various applications of the fractional calculus in several scientific and engineering fields, the reader may consult the books [10, 11, 26, 27, 3436], and the papers [9, 1215, 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:

1. (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$$(t,\nu )\in (0,1)\times (0,1)$$;

2. (b)

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

3. (c)

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

4. (d)

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

5. (e)

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

6. (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$$(t,\nu )\in [0,1]\times [0,1]$$, $$i=1,\ldots ,q$$;

7. (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$$h_{1i}(\nu )=\frac{1}{\varGamma (\beta _{1}-\delta _{i})}(1-\nu )^{ \beta _{1}-\gamma _{0}-1}(1-(1-\nu )^{\gamma _{0}-\delta _{i}})$$, $$\forall \nu \in [0,1]$$, $$i=1,\ldots ,q$$;

8. (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$$(t,\nu )\in [0,1]\times [0,1]$$, $$i=1,\ldots ,p$$;

9. (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:

1. (a)

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

2. (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}
3. (c)

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

4. (d)

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

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

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

6. (f)

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

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

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

8. (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}
9. (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. (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. (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. (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. (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. (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. (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. (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. (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. (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. (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. (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. (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. (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. (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. (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. (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.

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Alexandru Tudorache, Faculty of Automatic Control and Computer Engineering, Gh. Asachi Technical University, Str. Prof.dr.doc. Dimitrie Mangeron, nr.27, Iasi 700050, Romania. Rodica Luca, Department of Mathematics, Gh. Asachi Technical University, Blvd. Carol I, nr.11, Iasi 700506, Romania.

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Tudorache, A., Luca, R. Positive solutions for a system of Riemann–Liouville fractional boundary value problems with p-Laplacian operators. Adv Differ Equ 2020, 292 (2020). https://doi.org/10.1186/s13662-020-02750-6