The eigenvalue problem for a coupled system of singular p-Laplacian differential equations involving fractional differential-integral conditions

Abstract

In this paper, we deal with a coupled system of singular p-Laplacian differential equations involving fractional differential-integral conditions. By employing Schauder’s fixed point theorem and the upper and lower solution method, we establish an eigenvalue interval for the existence of positive solutions. As an application an example is presented to illustrate the main results.

1 Introduction

In this paper, we consider the following system of nonlinear fractional differential equations with different fractional derivatives:

$$ \left \{ \textstyle\begin{array}{@{}l} -D^{\beta_{1}}(\varphi_{p_{1}}(-D^{\alpha_{1}}u_{1}))(t)=\lambda f_{1}(u_{1}(t),D^{\gamma_{1}}u_{1}(t),D^{\gamma_{2}}u_{2}(t)), \quad 0< t< 1,\\ -D^{\beta_{2}}(\varphi_{p_{2}}(-D^{\alpha_{2}}u_{2}))(t)=f_{2}(t,u_{1}(t)),\\ D^{\alpha_{i}}u_{i}(0)=D^{\alpha_{i}}u_{i}(1)=0, \\ D^{\gamma_{i}}u_{i}(0)=0,\qquad D^{\alpha_{i}-1}u_{i}(1)=\xi_{i}I^{\omega_{i}}(D^{\gamma _{i}}u_{i}(\eta_{i})), \quad i=1,2, \end{array}\displaystyle \right . $$

(1.1)

where $D^{\alpha_{i}},D^{\beta_{i}}, D^{\gamma_{i}}$ ($i=1,2$) are the standard Riemannn-Liouville fractional derivatives, $I^{\omega_{i}}$ is the Riemannn-Liouville fractional integral, $\varphi_{p_{i}}$ is the p-Laplacian operator defined by $\varphi_{p_{i}}(s)=|s|^{p_{i}-2}s,p_{i}>2$ ($i=1,2$), and the nonlinearity $f_{1}(x,y,z)$ may be singular at $x=0,y=0,z=0$.

Throughout this paper, we always suppose that:

(s₀):

$0<\gamma_{i}\leq1<\alpha_{i}<\beta_{i}<2, \alpha_{1}-\gamma _{1}>1, \alpha_{2}-\gamma_{2}>1, \omega_{i}>0, \xi_{i}>0, \eta_{i}\in[0,1]$ ($i=1,2$).

(s₁):

$\Gamma(\alpha_{i}-\gamma_{i}+\omega_{i})>\xi_{i}\eta_{i}^{\alpha _{i}-\gamma_{i}+\omega_{i}-1}$ ($i=1,2$).

(s₂):

Let $q_{i}$ satisfies the relation $\frac{1}{q_{i}}+\frac {1}{p_{i}}=1$, where $p_{i}$ is given by (1.1), then $1< q_{i}<2$.

Fractional calculus provides an excellent tool for describing the hereditary properties of various materials and processes. Concerning the development of theory, method and application of fractional calculus, we refer the reader to the recent papers [1–8].

On the other hand, the study of coupled systems involving fractional differential equations is also important as such systems occur in various problems of applied nature. So considerable work has been done to study the existence result for them nowadays [9–12]. The authors got the existence solutions by the method of the fixed point theorem, the coincidence degree theorem, or Schauder’s fixed point theorem.

The theory of upper and lower solutions is well known to be an effective method to deal with the existence of solutions for the boundary value problems of the fractional differential equations. In [13] the authors used the method of upper and lower solutions and investigated the existence of solutions for initial value problems. By the same method some people got the solutions of boundary value problems for fractional differential equations, such as [14, 15]. To the best of our knowledge, only few papers considered the existence of solutions by using the method of upper and lower solutions for boundary value problems with fractional coupled systems.

The aim of this paper is to deal with the eigenvalue problem for a coupled system of fractional differential equations involving differential-integral conditions. The novelty of this paper is that the nonlinear terms $f_{1}$, $f_{2}$ in the system (1.1) involve different unknown functions $u_{1}(t)$, $u_{2}(t)$ and their Riemann-Liouville fractional derivatives with different orders, and $f_{1}(x,y,z)$ may be singular at $x=0, y=0, z=0$. We establish an eigenvalue interval for the existence of positive solutions by Schauder’s fixed point theorem and the upper and lower solutions method.

2 Preliminaries and lemmas

Lemma 2.1

([16])

Let $h_{i}\in L^{1} (0,1)$, then the problem

$$\left \{ \textstyle\begin{array}{@{}l} -D^{\beta_{i}} v_{i}(t)=h_{i}(t), \quad 0< t< 1,\\ v_{i}(0)=v_{i}(1)=0, \end{array}\displaystyle \right . $$

has the unique solution $v_{i}(t)=\int_{0}^{1}G(\beta_{i},t,s)h_{i}(s)\,ds$ ($i=1,2$), where

$$G(\beta_{i},t,s)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{1}{\Gamma(\beta_{i})}[t(1-s)]^{\beta _{i}-1}, & t\leq s,\\ \frac{1}{\Gamma(\beta_{i})}\{[t(1-s)]^{\beta_{i}-1}-(t-s)^{\beta_{i}-1}\}, & s\leq t. \end{array}\displaystyle \right . $$

Lemma 2.2

([17])

Let $h_{i}\in L^{1} (0,1)$, then the fractional integral boundary value problem

$$\left \{ \textstyle\begin{array}{@{}l} -D^{\alpha_{i}-\gamma_{i}} v_{i}(t)=h_{i}(t), \quad 0< t< 1,\\ v_{i}(0)=0, \qquad D^{\alpha_{i}-\gamma_{i}-1}v_{i}(1)=\xi_{i}I^{\omega_{i}}v_{i}(\eta _{i}),\quad i=1,2, \end{array}\displaystyle \right . $$

has a unique solution $v_{i}(t)=\int_{0}^{1}H_{i}(t,s)h_{i}(s)\,ds$, where

$$H_{i}(t,s)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \frac{[\Gamma(\alpha_{i}-\gamma_{i}+\omega_{i})-\xi_{i}(\eta_{i}-s)^{\alpha _{i}-\gamma_{i}+\omega_{i}-1}]t^{\alpha_{i}-\gamma_{i}-1} - [\Gamma(\alpha_{i}-\gamma_{i}+\omega_{i})-\xi_{i}\eta_{i}^{\alpha_{i}-\gamma _{i}+\omega_{i}-1}](t-s)^{\alpha_{i}-\gamma_{i}-1}}{\Delta_{i}},& s\leq t,s\leq\eta_{i};\\ \frac{[\Gamma(\alpha_{i}-\gamma_{i}+\omega_{i})-\xi_{i}(\eta_{i}-s)^{\alpha _{i}-\gamma_{i}+\omega_{i}-1}]t^{\alpha_{i}-\gamma_{i}-1}}{\Delta_{i}} , & t\leq s\leq\eta_{i};\\ \frac{\Gamma(\alpha_{i}-\gamma_{i}+\omega_{i})[t^{\alpha_{i}-\gamma _{i}-1}-(t-s)^{\alpha_{i}-\gamma_{i}-1}]+\xi_{i}\eta_{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1} (t-s)^{\alpha_{i}-\gamma_{i}-1}}{\Delta_{i}}, & \eta_{i}\leq s\leq t;\\ \frac{\Gamma(\alpha_{i}-\gamma_{i}+\omega_{i})t^{\alpha_{i}-\gamma_{i}-1}}{\Delta_{i}}, & s\geq t,s\geq\eta_{i}, \end{array}\displaystyle \right . $$

and $\Delta_{i}=\Gamma(\alpha_{i}-\gamma_{i})[\Gamma(\alpha_{i}-\gamma_{i}+\omega _{i})-\xi_{i}\eta_{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1}]$.

Lemma 2.3

([17, 18])

The functions $G(\beta_{i},t,s)$ and $H_{i}(t,s)$ have the following properties:

• (1) $G(\beta_{i},t,s)>0$, $H_{i}(t,s)>0$, for $t,s\in(0,1)$.

• (2)
  $$\frac{t^{\beta_{i}-1}(1-t)s(1-s)^{\beta_{i}-1}}{\Gamma(\beta_{i})}\leq G(\beta _{i},t,s)\leq\frac{\beta_{i}-1}{\Gamma(\beta_{i})}t^{\beta_{i}-1}(1-t), \quad \textit{for } t,s\in[0,1]. $$
• (3)
  $$e_{i}t^{\alpha_{i}-\gamma_{i}-1}\bigl[1-(1-s)^{\alpha_{i}-\gamma_{i}-1}\bigr]\leq H_{i}(t,s)\leq d_{i} t^{\alpha_{i}-\gamma_{i}-1}, \quad \textit{for } t,s\in[0,1], $$

  where $d_{i}=\frac{1}{\Delta_{i}}[\Gamma(\alpha_{i}-\gamma_{i}+\omega_{i})+\xi _{i}\eta_{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1}]$, $e_{i}=\Gamma(\alpha_{i}-\gamma_{i})$.

Proof

From [18], we can see that $G(\beta_{i},t,s)>0$ and (2) hold.

In the following, we will prove (3).

For $s\leq t,s\leq\eta_{i}$,

$$\begin{aligned} H_{i}(t,s) =&\frac{1}{\Delta_{i}} \bigl\{ \bigl[\Gamma( \alpha_{i}-\gamma_{i}+\omega _{i})- \xi_{i}(\eta_{i}-s)^{\alpha_{i}-\gamma_{i}+\omega_{i}-1}\bigr]t^{\alpha_{i}-\gamma _{i}-1} \\ &{}- \bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})-\xi_{i}\eta_{i}^{\alpha _{i}-\gamma_{i}+\omega_{i}-1} \bigr](t-s)^{\alpha_{i}-\gamma_{i}-1} \bigr\} \\ \geq&\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})-\xi_{i}\eta _{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1} \bigr] \bigl[t^{\alpha_{i}-\gamma_{i}-1}-(t-s)^{\alpha _{i}-\gamma_{i}-1}\bigr] \\ \geq&\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})-\xi_{i}\eta _{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1} \bigr] \bigl[t^{\alpha_{i}-\gamma _{i}-1}-(t-ts)^{\alpha_{i}-\gamma_{i}-1}\bigr] \\ =&\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})-\xi_{i}\eta _{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1} \bigr]t^{\alpha_{i}-\gamma _{i}-1}\bigl[1-(1-s)^{\alpha_{i}-\gamma_{i}-1}\bigr] \\ =&\Gamma(\alpha_{i}-\gamma_{i})t^{\alpha_{i}-\gamma_{i}-1} \bigl[1-(1-s)^{\alpha _{i}-\gamma_{i}-1}\bigr] \\ =&e_{i}t^{\alpha_{i}-\gamma_{i}-1}\bigl[1-(1-s)^{\alpha_{i}-\gamma_{i}-1}\bigr], \\ H_{i}(t,s) =&\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}- \gamma_{i}+\omega_{i})-\xi _{i}( \eta_{i}-s)^{\alpha_{i}-\gamma_{i}+\omega_{i}-1}\bigr]t^{\alpha_{i}-\gamma_{i}-1} \\ &{} - \bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})-\xi_{i}\eta_{i}^{\alpha _{i}-\gamma_{i}+\omega_{i}-1} \bigr](t-s)^{\alpha_{i}-\gamma_{i}-1} \\ \leq&\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})t^{\alpha _{i}-\gamma_{i}-1}+\xi_{i}\eta_{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1}(t-s)^{\alpha _{i}-\gamma_{i}-1} \bigr] \\ \leq&\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma _{i}+\omega_{i})+\xi_{i}\eta_{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1} \bigr]t^{\alpha _{i}-\gamma_{i}-1} \\ =&d_{i}t^{\alpha_{i}-\gamma_{i}-1}. \end{aligned}$$

For $t\leq s\leq\eta_{i}$,

$$\begin{aligned}& \begin{aligned}[b] H_{i}(t,s)&=\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}- \gamma_{i}+\omega_{i})-\xi _{i}( \eta_{i}-s)^{\alpha_{i}-\gamma_{i}+\omega_{i}-1}\bigr]t^{\alpha_{i}-\gamma_{i}-1} \\ &\geq\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})-\xi_{i}\eta _{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1} \bigr]t^{\alpha_{i}-\gamma_{i}-1} \\ &\geq\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})-\xi_{i}\eta _{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1} \bigr]t^{\alpha_{i}-\gamma _{i}-1}\bigl[1-(1-s)^{\alpha_{i}-\gamma_{i}-1}\bigr] \\ &=\Gamma(\alpha_{i}-\gamma_{i})t^{\alpha_{i}-\gamma_{i}-1} \bigl[1-(1-s)^{\alpha _{i}-\gamma_{i}-1}\bigr] \\ &=e_{i}t^{\alpha_{i}-\gamma_{i}-1}\bigl[1-(1-s)^{\alpha_{i}-\gamma_{i}-1}\bigr], \end{aligned}\\& \begin{aligned}[b] H_{i}(t,s)&=\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}- \gamma_{i}+\omega_{i})-\xi _{i}( \eta_{i}-s)^{\alpha_{i}-\gamma_{i}+\omega_{i}-1}\bigr]t^{\alpha_{i}-\gamma_{i}-1} \\ &\leq\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})t^{\alpha _{i}-\gamma_{i}-1}+\xi_{i}\eta_{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1}t^{\alpha _{i}-\gamma_{i}-1} \bigr] \\ &=\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})+\xi_{i}\eta _{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1} \bigr]t^{\alpha_{i}-\gamma_{i}-1} \\ &=d_{i}t^{\alpha_{i}-\gamma_{i}-1}. \end{aligned} \end{aligned}$$

For $\eta_{i}\leq s\leq t$,

$$\begin{aligned} H_{i}(t,s) =&\frac{1}{\Delta_{i}} \bigl\{ \Gamma(\alpha_{i}- \gamma_{i}+\omega _{i})\bigl[t^{\alpha_{i}-\gamma_{i}-1}-(t-s)^{\alpha_{i}-\gamma_{i}-1} \bigr]+\xi_{i}\eta _{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1} (t-s)^{\alpha_{i}-\gamma_{i}-1} \bigr\} \\ \geq&\frac{1}{\Delta_{i}} \bigl\{ \bigl[\Gamma(\alpha_{i}- \gamma_{i}+\omega_{i})-\xi _{i} \eta_{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1}\bigr]t^{\alpha_{i}-\gamma_{i}-1} \\ &{} - \bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})-\xi_{i}\eta_{i}^{\alpha_{i}-\gamma _{i}+\omega_{i}-1} \bigr](t-s)^{\alpha_{i}-\gamma_{i}-1} \bigr\} \\ =&\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})-\xi_{i}\eta _{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1} \bigr] \bigl[t^{\alpha_{i}-\gamma_{i}-1}-(t-s)^{\alpha _{i}-\gamma_{i}-1}\bigr] \\ \geq&\Gamma(\alpha_{i}-\gamma_{i})t^{\alpha_{i}-\gamma_{i}-1} \bigl[1-(1-s)^{\alpha _{i}-\gamma_{i}-1}\bigr] \\ =&e_{i}t^{\alpha_{i}-\gamma_{i}-1}\bigl[1-(1-s)^{\alpha_{i}-\gamma_{i}-1}\bigr], \\ H_{i}(t,s) =&\frac{1}{\Delta_{i}} \bigl\{ \Gamma(\alpha_{i}- \gamma_{i}+\omega _{i})\bigl[t^{\alpha_{i}-\gamma_{i}-1}-(t-s)^{\alpha_{i}-\gamma_{i}-1} \bigr]+\xi_{i}\eta _{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1} (t-s)^{\alpha_{i}-\gamma_{i}-1} \bigr\} \\ \leq&\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})t^{\alpha _{i}-\gamma_{i}-1}+\xi_{i}\eta_{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1}(t-s)^{\alpha _{i}-\gamma_{i}-1} \bigr] \\ \leq&\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma _{i}+\omega_{i})+\xi_{i}\eta_{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1} \bigr]t^{\alpha _{i}-\gamma_{i}-1} \\ =&d_{i}t^{\alpha_{i}-\gamma_{i}-1}. \end{aligned}$$

For $s\geq t,s\geq\eta_{i}$,

$$\begin{aligned} H_{i}(t,s) =&\frac{1}{\Delta_{i}} \Gamma(\alpha_{i}- \gamma_{i}+\omega_{i})t^{\alpha_{i}-\gamma_{i}-1} \\ \geq&\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})-\xi_{i}\eta _{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1} \bigr]t^{\alpha_{i}-\gamma _{i}-1}\bigl[1-(1-s)^{\alpha_{i}-\gamma_{i}-1}\bigr] \\ =&e_{i}t^{\alpha_{i}-\gamma_{i}-1}\bigl[1-(1-s)^{\alpha_{i}-\gamma_{i}-1}\bigr], \\ H_{i}(t,s) =&\frac{1}{\Delta_{i}}\Gamma(\alpha_{i}- \gamma_{i}+\omega _{i})t^{\alpha_{i}-\gamma_{i}-1} \\ \leq&\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})t^{\alpha _{i}-\gamma_{i}-1}+\xi_{i}\eta_{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1}t^{\alpha _{i}-\gamma_{i}-1} \bigr] \\ =&\frac{1}{\Delta_{i}}\bigl[\Gamma(\alpha_{i}-\gamma_{i}+ \omega_{i})+\xi_{i}\eta _{i}^{\alpha_{i}-\gamma_{i}+\omega_{i}-1} \bigr]t^{\alpha_{i}-\gamma_{i}-1} \\ =&d_{i}t^{\alpha_{i}-\gamma_{i}-1}. \end{aligned}$$

From the above, the proof of (3) is completed. Clearly $H_{i}(t,s)>0$ for $(t,s)\in(0,1)$, since (3) holds. □

Lemma 2.4

([17])

Let $h_{1}\in L^{1} (0,1)$, if (s₀)-(s₂) hold, then the fractional boundary value problem

$$ \left \{ \textstyle\begin{array}{@{}l} -D^{\beta_{1}}(\varphi_{p_{1}}(-D^{\alpha_{1}-\gamma_{1}}v_{1}))(t)=h_{1}(t),\\ D^{\alpha_{1}-\gamma_{1}}v_{1}(0)=D^{\alpha_{1}-\gamma_{1}}v_{1}(1)=0, \\ v_{1}(0)=0,\qquad D^{\alpha_{1}-\gamma_{1}-1}v_{1}(1)=\xi_{1}I^{\omega_{1}}(v_{1}(\eta_{1})), \end{array}\displaystyle \right . $$

(2.1)

has the unique positive solution

$$ v_{1}(t)= \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s, \tau)h_{1}(\tau)\,d\tau \biggr)^{q_{1}-1}\,ds. $$

(2.2)

Now let us consider the following modified problem of the BVP (1.1):

$$ \left \{ \textstyle\begin{array}{@{}l} -D^{\beta_{1}}(\varphi_{p_{1}}(-D^{\alpha_{1}-\gamma_{1}}v_{1}))(t)=\lambda f_{1}(I^{\gamma_{1}}v_{1}(t),v_{1}(t),v_{2}(t)),\\ -D^{\beta_{2}}(\varphi_{p_{2}}(-D^{\alpha_{2}-\gamma _{2}}v_{2}))(t)=f_{2}(t,I^{\gamma_{1}}v_{1}(t)),\\ D^{\alpha_{i}-\gamma_{i}}v_{i}(0)=D^{\alpha_{i}-\gamma_{i}}v_{i}(1)=0, \\ v_{i}(0)=0, \qquad D^{\alpha_{i}-\gamma_{i}-1}v_{i}(1)=\xi_{i}I^{\omega_{i}}(v_{i}(\eta _{i})), \quad i=1,2. \end{array}\displaystyle \right . $$

(2.3)

Lemma 2.5

Let $u_{i}(t)=I^{\gamma_{i}}v_{i}(t)$, $v_{i}(t)\in C[0,1]$ ($i=1,2$). Then (1.1) can be transformed into (2.3). Moveover, if $(v_{1}(t),v_{2}(t))\in C[0,1]\times C[0,1]$ is a positive solution of the problem (2.3), then $(I^{\gamma_{1}}v_{1}(t), I^{\gamma_{2}}v_{2}(t))$ is a positive solution of the problem (1.1).

Proof

Let $u_{i}(t)=I^{\gamma_{i}}v_{i}(t),v_{i}(t)\in C[0,1]$, by the definition of the Riemannn-Liouville fractional derivatives and integrals, we obtain

$$ D^{\alpha_{i}}u_{i}(t)=D^{\alpha_{i}-\gamma_{i}}v_{i}(t), \quad D^{\alpha_{i}+1}u_{i}(t) =D^{\alpha_{i}-\gamma_{i}+1}v_{i}(t), \quad D^{\alpha_{i}-1}u_{i}(t) =D^{\alpha_{i}-\gamma_{i}-1}v_{i}(t). $$

(2.4)

Thus by applying (2.4), the BVP (1.1) reduces to the modified boundary value problem (2.3).

Consequently, if $(v_{1}(t),v_{2}(t))\in C[0,1]\times C[0,1]$ is a positive solution of the problem (2.3), then $(I^{\gamma_{1}}v_{1}(t), I^{\gamma_{2}}v_{2}(t))$ is a positive solution of the problem (1.1).

It is well know that $(v_{1},v_{2})\in C[0,1]\times C[0,1]$ is a solution of system (2.3), if and only if $(v_{1},v_{2})\in C[0,1]\times \in C[0,1]$ is a solution of the following nonlinear integral equation system:

$$ \left \{ \textstyle\begin{array}{@{}l} v_{1}(t)=\lambda^{q_{1}-1}\int_{0}^{1}H_{1}(t,s) (\int_{0}^{1}G(\beta_{1},s,\tau) f_{1}(I^{\gamma_{1}}v_{1}(\tau),v_{1}(\tau),v_{2}(\tau))\,d\tau )^{q_{1}-1}\,ds,\\ v_{2}(t)=\int_{0}^{1}H_{2}(t,s) (\int_{0}^{1}G(\beta_{2},s,\tau)f_{2}(\tau,I^{\gamma _{1}}v_{1}(\tau))\,d\tau )^{q_{2}-1}\,ds. \end{array}\displaystyle \right . $$

(2.5)

Now define an operator

$$(Av_{1}) (t)= \int_{0}^{1}H_{2}(t,s) \biggl( \int_{0}^{1}G(\beta_{2},s, \tau)f_{2}\bigl(\tau ,I^{\gamma_{1}}v_{1}(\tau)\bigr)\,d\tau \biggr)^{q_{2}-1}\,ds. $$

Then the integral system (2.5) is equivalent to the following nonlinear integral-differential equation:

$$ \left \{ \textstyle\begin{array}{@{}l} -D^{\beta_{1}}(\varphi_{p_{1}}(-D^{\alpha_{1}-\gamma_{1}}v_{1}))(t)=\lambda f_{1}(I^{\gamma_{1}}v_{1}(t),v_{1}(t),Av_{1}(t)),\\ D^{\alpha_{1}-\gamma_{1}}v_{1}(0)=D^{\alpha_{1}-\gamma_{1}}v_{1}(1)=0, \\ v_{1}(0)=0,\qquad D^{\alpha_{1}-\gamma_{1}-1}v_{1}(1)=\xi_{1}I^{\omega_{1}}(v_{1}(\eta_{1})), \end{array}\displaystyle \right . $$

(2.6)

i.e. the operator equation

$$v_{1}(t)=\lambda^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s,\tau) f_{1}\bigl(I^{\gamma_{1}}v_{1}(\tau),v_{1}( \tau),(Av_{1}) (t)\bigr)\,d\tau \biggr)^{q_{1}-1}\,ds. $$

 □

Definition 2.1

A continuous function $\Psi(t)$ is called a lower solution of the problem (2.6) if it is satisfies

$$\left \{ \textstyle\begin{array}{@{}l} -D^{\beta_{1}}(\varphi_{p_{1}}(-D^{\alpha_{1}-\gamma_{1}}\Psi))(t)\leq\lambda f_{1}(I^{\gamma_{1}}\Psi(t),\Psi(t),A\Psi(t)),\\ D^{\alpha_{1}-\gamma_{1}}\Psi(0)\geq0,\qquad D^{\alpha_{1}-\gamma_{1}}\Psi(1)\geq0,\\ \Psi(0)\geq0, \qquad D^{\alpha_{1}-\gamma_{1}-1}\Psi(1)\geq\xi_{1}I^{\omega _{1}}(\Psi(\eta_{1})), \end{array}\displaystyle \right . $$

where

$$(A\Psi) (t)= \int_{0}^{1}H_{2}(t,s) \biggl( \int_{0}^{1}G(\beta_{2},s, \tau)f_{2}\bigl(\tau ,I^{\gamma_{1}}\Psi(\tau)\bigr)\,d\tau \biggr)^{q_{2}-1}\,ds. $$

Definition 2.2

A continuous function $\Phi(t)$ is called an upper solution of the problem (2.6) if it is satisfies

$$\left \{ \textstyle\begin{array}{@{}l} -D^{\beta_{1}}(\varphi_{p_{1}}(-D^{\alpha_{1}-\gamma_{1}}\Phi))(t)\geq\lambda f_{1}(I^{\gamma_{1}}\Phi(t),\Phi(t),A\Phi(t)),\\ D^{\alpha_{1}-\gamma_{1}}\Phi(0)\leq0, \qquad D^{\alpha_{1}-\gamma_{1}}\Phi(1)\leq0,\\ \Phi(0)\leq0, \qquad D^{\alpha_{1}-\gamma_{1}-1}\Phi(1)\leq\xi_{1}I^{\omega _{1}}(\Phi(\eta_{1})), \end{array}\displaystyle \right . $$

where

$$(A\Phi) (t)= \int_{0}^{1}H_{2}(t,s) \biggl( \int_{0}^{1}G(\beta_{2},s, \tau)f_{2}\bigl(\tau ,I^{\gamma_{1}}\Phi(\tau)\bigr)\,d\tau \biggr)^{q_{2}-1}\,ds. $$

Lemma 2.6

(Maximal principle)

If $v_{1}\in C([0,1],R)$ satisfies

$$v_{1}(0)=0,\qquad D^{\alpha_{1}-\gamma_{1}-1}v_{1}(1)= \xi_{1}I^{\omega_{1}}\bigl(v_{1}(\eta_{1})\bigr) $$

and $-D^{\alpha_{1}-\gamma_{1}}v_{1}(t)\geq0$ for any $t\in[0,1]$, then $v_{1}(t)\geq0, t\in[0,1]$

Proof

By Lemma 2.3, the conclusion is obvious, we omit the proof. □

3 Main results

To establish the existence of a solution to the boundary value problem (1.1), we need to make the following assumptions.

(H₁):

$f_{1}(x,y,z): (0,+\infty)^{3}\rightarrow[0,+\infty]$ is continuous and non-increasing in $x,y,z>0$, respectively, and for all $r\in(0,1)$, there exists a constant $\varepsilon>0$, such that, for any $(x,y,z)\in(0,+\infty)^{3}$, we have

$$f_{1}(rx,ry,rz)\leq r^{-\varepsilon}f_{1}(x,y,z). $$

(H₂):

$f_{2}(t,x): [0,1]\times[0,+\infty)\rightarrow[0,+\infty]$ is continuous and non-decreasing in $x>0$, and there exists a constant $0<\sigma<\frac{1}{q_{2}-1}$, such that, for any $r\in(0,1)$, $(t,x)\in[0,1]\times[0,+\infty)$, we have

$$f_{2}(t,rx)\geq r^{\sigma}f_{2}(t,x). $$

Remark

For $r\geq1$, and $x,y,z>0$, we have

$$\begin{aligned}& f_{1}(rx,ry,rz)\geq r^{-\varepsilon}f_{1}(x,y,z), \end{aligned}$$

(3.1)

$$\begin{aligned}& f_{2}(t,rx)\leq r^{\sigma}f_{2}(t,x). \end{aligned}$$

(3.2)

Theorem 3.1

Suppose (H₁) and (H₂) hold, and the following condition is satisfied:

(H₃):

$f_{1}(1,1,1)\neq0$, and

$$0< \int_{0}^{1}f_{1}\biggl( \frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha _{1})}t^{\alpha_{1}-1},t^{\alpha_{1}-\gamma_{1}-1}, bt^{\alpha_{2}-\gamma_{2}-1}\biggr)\,dt< + \infty, $$

where

$$\begin{aligned}& \begin{aligned}[b] b={}&e_{2} \int_{0}^{1}\bigl[1-(1-s)^{\alpha_{2}-\gamma_{2}-1} \bigr]s^{(\beta _{2}-1)(q_{2}-1)}(1-s)^{q_{2}-1}\,ds\\ &{}\times \biggl( \frac{1}{\Gamma(\beta_{2})} \int _{0}^{1}\tau(1-\tau)^{\beta_{2}-1}f_{2} \biggl(\tau,\frac{\Gamma(\alpha_{1}-\gamma _{1})}{\Gamma(\alpha_{1})} \tau^{\alpha_{1}-1}\biggr)\,d\tau \biggr)^{q_{2}-1}, \end{aligned}\\& e_{2}=\Gamma(\alpha_{2}-\gamma_{2}). \end{aligned}$$

Then there exists a constant $\lambda^{*}>0$ such that for any $\lambda \in(\lambda^{*},+\infty)$, the BVP (1.1) has at least one positive solution $(u_{1}(t),u_{2}(t))$, and, moreover, there exist two constants $0< l<1$ and $L>1$ such that

$$\begin{aligned}& l\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}t^{\alpha_{1}-1}\leq u_{1}(t)\leq L \frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}t^{\alpha_{1}-1}, \\& bl^{\sigma(q_{2}-1)}\frac{\Gamma(\alpha_{2}-\gamma_{2})}{\Gamma(\alpha _{2})}t^{\alpha_{2}-1}\leq u_{2}(t)\leq L^{\sigma(q_{2}-1)}a\frac{\Gamma(\alpha _{2}-\gamma_{2})}{\Gamma(\alpha_{2})}t^{\alpha_{2}-1}, \end{aligned}$$

where

$$\begin{aligned}& a=d_{2} \biggl(\frac{\beta_{2}-1}{\Gamma(\beta_{2})} \int_{0}^{1}f_{2}\biggl(\tau, \frac {\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}\biggr)\,d\tau \biggr)^{q_{2}-1}, \\& d_{2}=\frac{1}{\Delta_{2}}\bigl[\Gamma(\alpha_{2}- \gamma_{2}+\omega_{2})+\xi_{2}\eta _{2}^{\alpha_{2}-\gamma_{2}+\omega_{2}-1}\bigr], \\& \Delta_{2}=\Gamma(\alpha_{2}-\gamma_{2})\bigl[ \Gamma(\alpha_{2}-\gamma_{2}+\omega _{2})- \xi_{2}\eta_{2}^{\alpha_{2}-\gamma_{2}+\omega_{2}-1}\bigr]. \end{aligned}$$

Proof

Let $E=C[0,1]$, and define a subset P of E as follows: $P=\{v_{1}(t)\in E: \mbox{there exists a}\mbox{ }\mbox{constant }0< l<1 \mbox{ such that } lt^{\alpha_{1}-\gamma_{1}-1}\leq v_{1}(t)\leq l^{-1}t^{\alpha_{1}-\gamma _{1}-1}, t\in[0,1]\}$. Clearly, P is a nonempty set, since $t^{\alpha_{1}-\gamma_{1}-1}\in P$. Also

$$I^{\gamma_{1}}t^{\alpha_{1}-\gamma_{1}-1}=\frac{1}{\Gamma(\gamma_{1})} \int _{0}^{t}(t-s)^{\gamma_{1}-1}s^{\alpha_{1}-\gamma_{1}-1}\,ds= \frac{\Gamma(\alpha _{1}-\gamma_{1})}{\Gamma(\alpha_{1})}t^{\alpha_{1}-1}. $$

Now define the operator $T_{\lambda}$ in E

$$ (T_{\lambda}v_{1}) (t)=\lambda^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta _{1},s, \tau)f_{1}\bigl(I^{\gamma_{1}}v_{1}(\tau),v_{1}( \tau),Av_{1}(\tau)\bigr)\,d\tau \biggr)^{q_{1}-1}\,ds, $$

(3.3)

where

$$Av_{1}(t)= \int_{0}^{1}H_{2}(t,s) \biggl( \int_{0}^{1}G(\beta_{2},s, \tau)f_{2}\bigl(\tau ,I^{\gamma_{1}}v_{1}(\tau)\bigr)\,d\tau \biggr)^{q_{2}-1}\,ds. $$

We assert that $T_{\lambda}$ is well defined and $T_{\lambda}(P)\subset P$. In fact, for any $v_{1}(t)\in P$, there exists a positive number $0< l_{v_{1}}<1$ such that $l_{v_{1}}t^{\alpha_{1}-\gamma_{1}-1}\leq v_{1}(t)\leq l_{v_{1}}^{-1}t^{\alpha_{1}-\gamma_{1}-1}, t\in[0,1]$. It follows from Lemma 2.3 and (H₂) that

$$\begin{aligned}& Av_{1}(t)= \int_{0}^{1}H_{2}(t,s) \biggl( \int_{0}^{1}G(\beta_{2},s, \tau)f_{2}\bigl(\tau ,I^{\gamma_{1}}v_{1}(\tau)\bigr)\,d\tau \biggr)^{q_{2}-1}\,ds, \\& \begin{aligned}[b] &\int_{0}^{1}G(\beta_{2},s, \tau)f_{2}\bigl(\tau,I^{\gamma_{1}}v_{1}(\tau)\bigr)\,d\tau\\ &\quad\leq \int _{0}^{1}G(\beta_{2},s, \tau)f_{2}\biggl(\tau,l_{v_{1}}^{-1}\frac{\Gamma(\alpha_{1}-\gamma _{1})}{\Gamma(\alpha_{1})} \tau^{\alpha_{1}-1}\biggr)\,d\tau \\ &\quad\leq l_{v_{1}}^{-\sigma}\frac{\beta_{2}-1}{\Gamma(\beta_{2})} \int_{0}^{1}s^{\beta _{2}-1}(1-s)f_{2} \biggl(\tau,\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha _{1})}\tau^{\alpha_{1}-1}\biggr)\,d\tau \\ &\quad\leq l_{v_{1}}^{-\sigma}\frac{\beta_{2}-1}{\Gamma(\beta_{2})} \int_{0}^{1}f_{2}\biggl(\tau , \frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{\alpha _{1}-1}\biggr)\,d\tau, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} &\int_{0}^{1}G(\beta_{2},s, \tau)f_{2}\bigl(\tau,I^{\gamma_{1}}v_{1}(\tau)\bigr)\,d\tau\\ &\quad\geq \int_{0}^{1} G(\beta_{2},s, \tau)f_{2}\biggl(\tau,l_{v_{1}}\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma (\alpha_{1})} \tau^{\alpha_{1}-1}\biggr)\,d\tau\\ &\quad\geq l_{v_{1}}^{\sigma}\int_{0}^{1} G(\beta_{2},s, \tau)f_{2}\biggl(\tau,\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha _{1})}\tau^{\alpha_{1}-1}\biggr)\,d\tau\\ &\quad\geq l_{v_{1}}^{\sigma}\frac{s^{\beta_{2}-1}(1-s)}{\Gamma(\beta_{2})} \int _{0}^{1}\tau(1-\tau)^{\beta_{2}-1} f_{2}\biggl(\tau,\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{\alpha _{1}-1}\biggr)\,d\tau. \end{aligned}$$

Then,

$$\begin{aligned} Av_{1}(t) \leq& \int_{0}^{1}H_{2}(t,s) \biggl(l_{v_{1}}^{-\sigma}\frac{\beta _{2}-1}{\Gamma(\beta_{2})} \int_{0}^{1}f_{2}\biggl(\tau, \frac{\Gamma(\alpha_{1}-\gamma_{1})}{ \Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}\biggr)\,d\tau \biggr)^{q_{2}-1}\,ds \\ \leq&d_{2}t^{\alpha_{2}-\gamma_{2}-1}l_{v_{1}}^{-\sigma(q_{2}-1)} \biggl( \frac {\beta_{2}-1}{\Gamma(\beta_{2})} \int_{0}^{1}f_{2}\biggl(\tau, \frac{\Gamma(\alpha _{1}-\gamma_{1})}{ \Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}\biggr)\,d\tau \biggr)^{q_{2}-1} \\ =&al_{v_{1}}^{-\sigma(q_{2}-1)}t^{\alpha_{2}-\gamma_{2}-1}, \end{aligned}$$

(3.4)

where

$$a=d_{2} \biggl(\frac{\beta_{2}-1}{\Gamma(\beta_{2})} \int_{0}^{1}f_{2}\biggl(\tau, \frac {\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}\biggr)\,d\tau \biggr)^{q_{2}-1} $$

and

$$\begin{aligned} Av_{1}(t) \geq& \int_{0}^{1}H_{2}(t,s) \biggl(l_{v_{1}}^{\sigma}\frac{s^{\beta _{2}-1}(1-s)}{\Gamma(\beta_{2})} \int_{0}^{1}\tau(1-\tau)^{\beta_{2}-1} f_{2}\biggl(\tau,\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}t^{\alpha _{1}-1}\biggr)\,d\tau \biggr)^{q_{2}-1}\,ds \\ \geq& e_{2}t^{\alpha_{2}-\gamma_{2}-1}l_{v_{1}}^{\sigma(q_{2}-1)} \int _{0}^{1}s^{(\beta_{2}-1)(q_{2}-1)}(1-s)^{q_{2}-1} \bigl[1-(1-s)^{\alpha_{2}-\gamma_{2}-1}\bigr]\,ds \\ &{} \times \biggl( \int_{0}^{1}\frac{1}{\Gamma(\beta_{2})}\tau(1- \tau)^{\beta_{2}-1} f_{2}\biggl(\tau,\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})} \tau^{\alpha _{1}-1}\biggr)\,d\tau \biggr)^{q_{2}-1} \\ =&bl_{v_{1}}^{\sigma(q_{2}-1)}t^{\alpha_{2}-\gamma_{2}-1}, \end{aligned}$$

(3.5)

where

$$\begin{aligned} b={}&e_{2} \int_{0}^{1}\bigl[1-(1-s)^{\alpha_{2}-\gamma_{2}-1} \bigr]s^{(\beta _{2}-1)(q_{2}-1)}(1-s)^{q_{2}-1}\,ds\\ &{}\times \biggl( \frac{1}{\Gamma(\beta_{2})} \int _{0}^{1}\tau(1-\tau)^{\beta_{2}-1} f_{2}\biggl(\tau,\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{(\alpha _{1}-1)}\biggr)\,d\tau \biggr)^{q_{2}-1}. \end{aligned}$$

Since $0<\sigma<\frac{1}{q_{2}-1}$, and by Lemma 2.3 and (H₁), (H₃), we also have

$$\begin{aligned} (T_{\lambda}v_{1}) (t) =&\lambda^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int _{0}^{1}G(\beta_{1},s,\tau) f_{1}\bigl(I^{\gamma_{1}}v_{1}(\tau),v_{1}( \tau),Av_{1}(\tau)\bigr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ \leq&\lambda^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s,\tau) \\ &{}\times f_{1}\biggl(l_{v_{1}}\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha _{1})} \tau^{\alpha_{1}-1},l_{v_{1}} \tau^{\alpha_{1}-\gamma_{1}-1},bl_{v_{1}}^{\sigma(q_{2}-1)} \tau^{\alpha_{2}-\gamma _{2}-1}\biggr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ \leq&\lambda^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s,\tau) \\ &{}\times f_{1}\biggl(l_{v_{1}}\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha _{1})} \tau^{\alpha_{1}-1},l_{v_{1}} \tau^{\alpha_{1}-\gamma_{1}-1},bl_{v_{1}} \tau^{\alpha_{2}-\gamma_{2}-1}\biggr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ \leq&\lambda^{q_{1}-1}l_{v_{1}}^{-\varepsilon(q_{1}-1)} \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1} \frac{\beta_{1}-1}{\Gamma(\beta_{1})}s^{\beta_{1}-1}(1-s) \\ &{} \times f_{1}\biggl(\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau ^{\alpha_{1}-1}, \tau^{\alpha_{1}-\gamma_{1}-1},b\tau^{\alpha_{2}-\gamma_{2}-1}\biggr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ \leq&\lambda^{q_{1}-1}l_{v_{1}}^{-\varepsilon(q_{1}-1)}d_{1}t^{\alpha_{1}-\gamma _{1}-1} \biggl( \int_{0}^{1}s^{(\beta_{1}-1)(q_{1}-1)}(1-s)^{q_{1}-1}\,ds \biggr) \\ &{}\times \biggl(\frac{\beta_{1}-1}{\Gamma(\beta_{1})} \int_{0}^{1}f_{1}\biggl( \frac {\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}, \tau^{\alpha_{1}-\gamma_{1}-1},b\tau^{\alpha_{2}-\gamma_{2}-1} \biggr)\,d\tau \biggr)^{q_{1}-1} \\ \leq&\lambda^{q_{1}-1}l_{v_{1}}^{-\varepsilon(q_{1}-1)} \\ &{}\times d_{1} \biggl(\frac{\beta _{1}-1}{\Gamma(\beta_{1})} \int_{0}^{1}f_{1}\biggl(\frac{\Gamma(\alpha_{1}-\gamma _{1})}{\Gamma(\alpha_{1})} \tau^{\alpha_{1}-1}, \tau^{\alpha_{1}-\gamma_{1}-1},b\tau^{\alpha_{2}-\gamma_{2}-1}\biggr)\,d\tau \biggr)^{q_{1}-1}t^{\alpha_{1}-\gamma_{1}-1} \\ < &+\infty. \end{aligned}$$

(3.6)

On the other hand, as $0<\sigma<\frac{1}{q_{2}-1}$, from Lemma 2.3 and (3.1), we have

$$\begin{aligned} (T_{\lambda}v_{1}) (t) =&\lambda^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int _{0}^{1}G(\beta_{1},s,\tau) f_{1}\bigl(I^{\gamma_{1}}v_{1}(\tau),v_{1}( \tau),Av_{1}(\tau)\bigr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ \geq&\lambda^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s,\tau) \\ &{}\times f_{1}\biggl(l_{v_{1}}^{-1} \frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha _{1})}\tau^{\alpha_{1}-1},l_{v_{1}}^{-1} \tau^{\alpha_{1}-\gamma_{1}-1},al_{v_{1}}^{{-\sigma}(q_{2}-1)}\tau^{\alpha _{2}-\gamma_{2}-1}\biggr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ \geq&\lambda^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s,\tau) \\ &{} \times f_{1}\biggl(l_{v_{1}}^{-1} \frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha _{1})}\tau^{\alpha_{1}-1},l_{v_{1}}^{-1} \tau^{\alpha_{1}-\gamma_{1}-1},al_{v_{1}}^{-1}\tau^{\alpha_{2}-\gamma_{2}-1}\biggr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ \geq&\lambda^{q_{1}-1}l_{v_{1}}^{\varepsilon(q_{1}-1)} \int_{0}^{1}H_{1}(t,s) \biggl( \frac{s^{\beta_{1}-1}(1-s)}{ \Gamma(\beta_{1})} \int_{0}^{1}\tau(1-\tau)^{\beta_{1}-1} \\ &{}\times f_{1}\biggl(\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau ^{\alpha_{1}-1}, \tau^{\alpha_{1}-\gamma_{1}-1},a\tau^{\alpha_{2}-\gamma_{2}-1}\biggr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ \geq&\lambda^{q_{1}-1}l_{v_{1}}^{\varepsilon(q_{1}-1)}e_{1}t^{\alpha_{1}-\gamma _{1}-1} \int_{0}^{1}\bigl[1-(1-s)^{\alpha_{1}-\gamma_{1}-1}\bigr] s^{(\beta_{1}-1)(q_{1}-1)}(1-s)^{q_{1}-1}\,ds \\ &{}\times \biggl(\frac{1}{ \Gamma(\beta_{1})} \int_{0}^{1}\tau(1-\tau)^{\beta_{1}-1}f_{1} \biggl(\frac{\Gamma(\alpha _{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}, \tau^{\alpha_{1}-\gamma_{1}-1},a \tau^{\alpha_{2}-\gamma_{2}-1}\biggr)\,d\tau \biggr)^{q_{1}-1} \\ =&k\lambda^{q_{1}-1}l_{v_{1}}^{\varepsilon(q_{1}-1)}t^{\alpha_{1}-\gamma _{1}-1} \biggl(\frac{1}{ \Gamma(\beta_{1})} \int_{0}^{1}\tau(1-\tau)^{\beta_{1}-1} \\ &{}\times f_{1}\biggl(\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau ^{\alpha_{1}-1}, \tau^{\alpha_{1}-\gamma_{1}-1},a\tau^{\alpha_{2}-\gamma_{2}-1}\biggr)\,d\tau \biggr)^{q_{1}-1}, \end{aligned}$$

(3.7)

where

$$k= \int_{0}^{1}\bigl[1-(1-s)^{\alpha_{1}-\gamma_{1}-1}\bigr] s^{(\beta_{1}-1)(q_{1}-1)}(1-s)^{q_{1}-1}\,ds. $$

Choose

$$\begin{aligned} \widetilde{l_{v_{1}}} =&\min \biggl\{ \frac{1}{2}, \biggl\{ \lambda ^{q_{1}-1}l_{v_{1}}^{-\varepsilon(q_{1}-1)}d_{1} \biggl(\frac{\beta_{1}-1}{ \Gamma(\beta_{1})} \int_{0}^{1}f_{1}\biggl( \frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma (\alpha_{1})}\tau^{\alpha_{1}-1}, \\ &{}\tau^{\alpha_{1}-\gamma_{1}-1},b\tau^{\alpha_{2}-\gamma_{2}-1} \biggr)\,d\tau \biggr)^{q_{1}-1} \biggr\} ^{-1}, \\ &{} k\lambda^{q_{1}-1}l_{v_{1}}^{\varepsilon(q_{1}-1)} \biggl( \frac{1}{ \Gamma(\beta_{1})} \int_{0}^{1}\tau(1-\tau)^{\beta_{1}-1}f_{1} \biggl(\frac{\Gamma(\alpha _{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}, \\ &{} \tau^{\alpha_{1}-\gamma_{1}-1},a \tau^{\alpha_{2}-\gamma_{2}-1}\biggr)\,d\tau \biggr)^{q_{1}-1} \biggr\} . \end{aligned}$$

(3.8)

Then it follows from (3.4)-(3.8) that

$$\widetilde{l_{v_{1}}}t^{\alpha_{1}-\gamma_{1}-1}\leq T_{\lambda}v_{1}(t)\leq \widetilde{l_{v_{1}}}^{-1}t^{\alpha_{1}-\gamma_{1}-1}. $$

This implies that $T_{\lambda}$ is well defined and $T_{\lambda}(P)\subset P$. Furthermore, comparing (3.3) and (2.2), the right hand side of (3.3) is exactly the same as the right hand of (2.2), if $h_{1}(t)$ in (2.1) is taken as $\lambda f_{1}(I^{\gamma_{1}}v_{1}(t),v_{1}(t),Av_{1}(t))$. Hence as the left hand side of (2.2), i.e. $v_{1}(t)$ satisfies equation (2.1) according to Lemma 2.4, the left hand side of (3.3), i.e. $T_{\lambda}v_{1}(t)$ must also satisfy equation (2.1) with $h_{1}(t)$ replace by $\lambda f_{1}(I^{\gamma_{1}}v_{1}(t),v_{1}(t),Av_{1}(t))$, namely

$$ \left \{ \textstyle\begin{array}{@{}l} -D^{\beta_{1}}(\varphi_{p_{1}}(-D^{\alpha_{1}-\gamma_{1}}(T_{\lambda}v_{1}))(t)=\lambda f_{1}(I^{\gamma_{1}}v_{1}(t),v_{1}(t),Av_{1}(t)),\\ D^{\alpha_{1}-\gamma_{1}}(T_{\lambda}v_{1})(0)=D^{\alpha_{1}-\gamma_{1}}(T_{\lambda}v_{1})(1)=0, \\ (T_{\lambda}v_{1})(0)=0,\qquad D^{\alpha_{1}-\gamma_{1}-1}(T_{\lambda}v_{1})(1)=\xi _{1}I^{\omega_{1}}(T_{\lambda}v_{1})(\eta_{1}), \end{array}\displaystyle \right . $$

(3.9)

where

$$(Av_{1}) (t)= \int_{0}^{1}H_{2}(t,s) \biggl( \int_{0}^{1}G(\beta_{2},s, \tau)f_{2}\bigl(\tau ,I^{\gamma_{1}}v_{1}(\tau)\bigr)\,d\tau \biggr)^{q_{2}-1}\,ds. $$

Next, we shall find the upper and lower solutions of (1.1). First of all, let

$$e(t)= \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s, \tau)f_{1}\biggl(\frac{\Gamma (\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}, \tau^{\alpha_{1}-\gamma_{1}-1},A\tau^{\alpha_{1}-\gamma_{1}-1}\biggr)\,d\tau \biggr)^{q_{2}-1}\,ds, $$

where

$$At^{\alpha_{1}-\gamma_{1}-1}= \int_{0}^{1}H_{2}(t,s) \biggl( \int_{0}^{1}G(\beta_{2},s,\tau )f_{2}\biggl(\tau,\frac{\Gamma (\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}\biggr)\,d\tau \biggr)^{q_{2}-1}\,ds. $$

Similar to (3.4) and (3.5), the following inequalities are also valid:

$$At^{\alpha_{1}-\gamma_{1}-1}\geq bt^{\alpha_{2}-\gamma_{2}-1} $$

and

$$At^{\alpha_{1}-\gamma_{1}-1}\leq at^{\alpha_{2}-\gamma_{2}-1}. $$

By Lemma 2.3, (H₁), and (3.7), we also have

$$\begin{aligned} e(t) \geq& \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s, \tau)f_{1}\biggl(\frac {\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}, \tau^{\alpha_{1}-\gamma_{1}-1},a\tau^{\alpha_{2}-\gamma _{2}-1}\biggr)\,d\tau \biggr)^{q_{2}-1}\,ds \\ \geq&kt^{\alpha_{1}-\gamma_{1}-1} \biggl(\frac{1}{ \Gamma(\beta_{1})} \int_{0}^{1}\tau(1-\tau)^{\beta_{1}-1}f_{1} \biggl(\frac{\Gamma(\alpha _{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}, \\ &{} \tau^{\alpha_{1}-\gamma_{1}-1},a \tau^{\alpha_{2}-\gamma_{2}-1}\biggr)\,d\tau \biggr)^{q_{1}-1}, \end{aligned}$$

and consequently there exists a constant $\lambda_{1}\geq1$ such that

$$ \lambda_{1}^{q_{1}-1}e(t)\geq t^{\alpha_{1}-\gamma_{1}-1},\quad \forall t\in [0,1]. $$

(3.10)

On the other hand by (H₁) and (H₂), we know that A is increasing and $T_{\lambda}$ is decreasing, and thus for $\lambda>\lambda_{1}$, from (3.6) we have

$$\begin{aligned} & \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s,\tau) f_{1}\bigl(I^{\gamma_{1}}\lambda^{q_{1}-1}e(\tau), \lambda^{q_{1}-1}e(\tau ),A\lambda^{q_{1}-1}e(\tau)\bigr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ &\quad\leq \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s,\tau) f_{1}\bigl(I^{\gamma_{1}}\lambda_{1} ^{q_{1}-1}e(\tau), \lambda_{1} ^{q_{1}-1}e(\tau ),A\lambda_{1}^{q_{1}-1}e( \tau)\bigr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ &\quad\leq \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s,\tau) f_{1}\bigl(I^{\gamma_{1}}\tau^{\alpha_{1}-\gamma_{1}-1}, \tau^{\alpha_{1}-\gamma_{1}-1},A \tau^{\alpha_{1}-\gamma_{1}-1}\bigr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ &\quad\leq \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s,\tau) f_{1}\biggl(\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}, \tau^{\alpha_{1}-\gamma_{1}-1},b\tau^{\alpha_{2}-\gamma_{2}-1}\biggr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ &\quad\leq d_{1} \biggl(\frac{\beta_{1}-1}{\Gamma(\beta_{1})} \int_{0}^{1}f_{1}\biggl( \frac{\Gamma (\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}, \tau^{\alpha_{1}-\gamma_{1}-1},b\tau^{\alpha_{2}-\gamma_{2}-1} \biggr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ &\quad< +\infty. \end{aligned}$$

(3.11)

Applying (3.2) and $0<\sigma<\frac{1}{q_{2}-1}$, for any $t\in[0,1]$, we have

$$\begin{aligned} A\bigl(\lambda^{*}\bigr)^{q_{1}-1}e(t) =& \int_{0}^{1}H_{2}(t,s) \biggl( \int_{0}^{1}G(\beta _{2},s, \tau)f_{2}\bigl(\tau,\bigl(\lambda^{*}\bigr) ^{q_{1}-1}I^{\gamma_{1}}e( \tau)\bigr)\,d\tau \biggr)^{q_{2}-1}\,ds \\ \leq&\bigl(\lambda^{*}\bigr)^{\sigma(q_{1}-1)(q_{2}-1)} \int_{0}^{1}H_{2}(t,s) \biggl( \int _{0}^{1}G(\beta_{2},s, \tau)f_{2} \bigl(\tau,I^{\gamma_{1}}e(\tau)\bigr)\,d\tau \biggr)^{q_{2}-1}\,ds \\ \leq&\bigl(\lambda^{*}\bigr)^{(q_{1}-1)} \int_{0}^{1}H_{2}(t,s) \biggl( \int_{0}^{1}G(\beta _{2},s, \tau)f_{2} \bigl(\tau,I^{\gamma_{1}}e(\tau)\bigr)\,d\tau \biggr)^{q_{2}-1}\,ds \\ =&\bigl(\lambda^{*}\bigr)^{(q_{1}-1)}Ae(t). \end{aligned}$$

(3.12)

Let

$$C_{1}=\max_{0\leq t\leq1}e(t),\qquad C_{2}=\max _{0\leq t\leq1} Ae(t),\qquad C=\max\{2,C_{1},C_{2} \}, $$

then we have

$$ I^{\gamma_{1}}e(t)\leq\frac{C_{1}}{\Gamma(\gamma_{1})}\leq C_{1}\leq C, \qquad Ae(t)\leq C_{2}\leq C,\qquad e(t)\leq C_{1}\leq C. $$

(3.13)

Now, take

$$\lambda^{*}> \biggl\{ \lambda_{1}, \biggl[\frac{C^{\varepsilon}}{f_{1}^{q_{1}-1}(1,1,1)k[\frac{1}{\Gamma(\beta_{1})}\int_{0}^{1}\tau(1-\tau) ^{\beta_{1}-1}\,d\tau ]^{q_{1}-1}} \biggr]^{\frac{1}{(q_{1}-1)[1-\varepsilon (q_{1}-1)]}} \biggr\} . $$

Then by (3.12), (3.13), and (H₁), we have

$$\begin{aligned} &\bigl(\lambda^{*}\bigr)^{q_{1}-1} \bigl(f_{1} \bigl(I^{\gamma_{1}}\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e(\tau ),\bigl(\lambda^{*} \bigr) ^{q_{1}-1}e(\tau),A\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e(\tau)\bigr) \bigr)^{q_{1}-1} \\ &\quad\geq\bigl(\lambda^{*}\bigr)^{q_{1}-1} \bigl(f_{1}\bigl(\bigl( \lambda^{*}\bigr)^{q_{1}-1}I^{\gamma_{1}} e(\tau),\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e(\tau),\bigl(\lambda^{*}\bigr) ^{q_{1}-1}Ae(\tau)\bigr) \bigr)^{q_{1}-1} \\ &\quad\geq\bigl(\lambda^{*}\bigr)^{q_{1}-1}\bigl(\lambda^{*} \bigr)^{-\varepsilon(q_{1}-1)^{2}} \bigl(f_{1}\bigl(I^{\gamma_{1}}e(\tau), e( \tau),Ae(\tau)\bigr) \bigr)^{q_{1}-1} \\ &\quad\geq\bigl(\lambda^{*}\bigr)^{(q_{1}-1)[1-\varepsilon(q_{1}-1)]} \bigl(f_{1}(C,C,C) \bigr)^{q_{1}-1} \\ &\quad\geq\bigl(\lambda^{*}\bigr)^{(q_{1}-1)[1-\varepsilon(q_{1}-1)]}C^{-\varepsilon }f_{1}^{q_{1}-1}(1,1,1). \end{aligned}$$

(3.14)

Consequently, (3.7) and (3.14) yield

$$\begin{aligned} &\bigl(\lambda^{*}\bigr)^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s,\tau )f_{1}\bigl(I^{\gamma_{1}}\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e( \tau), \\ &\qquad{}\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e(\tau ),A\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e(\tau)\,d\tau\bigr) \biggr)^{q_{1}-1}\,ds \\ &\quad\geq\bigl(\lambda^{*}\bigr)^{(q_{1}-1)[1-\varepsilon(q_{1}-1)]}C^{-\varepsilon }f_{1}^{q_{1}-1}(1,1,1) \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s,\tau)\,d\tau \biggr)^{q_{1}-1}\,ds \\ &\quad\geq\bigl(\lambda^{*}\bigr)^{(q_{1}-1)[1-\varepsilon(q_{1}-1)]}C^{-\varepsilon }f_{1}^{q_{1}-1}(1,1,1)kt^{\alpha_{1}-\gamma_{1}-1} \biggl[\frac{1}{\Gamma(\beta _{1})} \int_{0}^{1}\tau(1-\tau)^{\beta_{1}-1}\,d\tau \biggr]^{q_{1}-1} \\ &\quad\geq t^{\alpha_{1}-\gamma_{1}-1}. \end{aligned}$$

(3.15)

Let

$$\begin{aligned}& \Phi(t)=\bigl(\lambda^{*}\bigr)^{q_{1}-1}e(t), \\& \begin{aligned}[b] \Psi(t)={}&\bigl(\lambda^{*}\bigr)^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta _{1},s, \tau)f_{1}\bigl(I^{\gamma_{1}}\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e( \tau),\\ &{}\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e(\tau),A\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e(\tau)\bigr)\,d\tau \biggr)^{q_{1}-1}\,ds, \end{aligned} \end{aligned}$$

then

$$ \Phi(t)=T_{\lambda^{*}}\bigl(t^{\alpha_{1}-\gamma_{1}-1}\bigr),\qquad \Psi(t)=T_{\lambda ^{*}}\bigl(\Phi(t)\bigr). $$

(3.16)

It follows from the monotonicity of A, $f_{1}$, and (3.10), (3.15), that for any $t\in[0,1]$

$$\begin{aligned}& \begin{aligned} \Phi(t)={}&\bigl(\lambda^{*}\bigr)^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta _{1},s, \tau)f_{1}\biggl(\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau ^{\alpha_{1}-1},\\ &{} \tau^{\alpha_{1}-\gamma_{1}-1},A \tau^{\alpha_{1}-\gamma_{1}-1}\biggr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ \geq{}&\lambda_{1}e(t)\geq t^{\alpha_{1}-\gamma_{1}-1},\\ \Psi(t)={}&\bigl(\lambda^{*}\bigr)^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta _{1},s, \tau)f_{1}\bigl(I^{\gamma_{1}}\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e( \tau),\\ &{}\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e(\tau),A\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e(\tau)\bigr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ \geq{}& t^{\alpha_{1}-\gamma_{1}-1}. \end{aligned} \end{aligned}$$

(3.17)

Moveover, by (3.9) and (3.16), we know

$$ \begin{aligned} &D^{\alpha_{1}-\gamma_{1}}\Phi(0)=D^{\alpha_{1}-\gamma_{1}}\Phi(1)=0, \qquad \Phi(0)=0, \qquad D^{\alpha_{1}-\gamma_{1}-1}\Phi(1)=\xi_{1}I^{\omega_{1}} \Phi(\eta_{1}), \\ &D^{\alpha_{1}-\gamma_{1}}\Psi(0)=D^{\alpha_{1}-\gamma_{1}}\Psi(1)=0, \qquad \Psi(0)=0, \qquad D^{\alpha_{1}-\gamma_{1}-1}\Psi(1)=\xi_{1}I^{\omega_{1}}\Psi(\eta _{1}). \end{aligned} $$

(3.18)

Proceeding as in (3.6)-(3.8), we get that $\Phi(t),\Psi(t)\in P$. By (3.16) and (3.17), we have

$$ t^{\alpha_{1}-\gamma_{1}-1}\leq\Psi(t)=(T_{\lambda^{*}}\Phi) (t), \qquad t^{\alpha _{1}-\gamma_{1}-1} \leq\Phi(t),\quad \forall t\in[0,1], $$

(3.19)

which implies

$$\begin{aligned} \Psi(t) =&(T_{\lambda^{*}}\Phi) (t) \\ =&\bigl(\lambda^{*}\bigr)^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s,\tau )f_{1}\bigl(I^{\gamma_{1}}\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e( \tau), \\ &{}\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e(\tau ),A\bigl(\lambda^{*}\bigr) ^{q_{1}-1}e(\tau)\bigr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ \leq&\bigl(\lambda^{*}\bigr)^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s,\tau )f_{1}\bigl(I^{\gamma_{1}}\tau^{\alpha_{1}-\gamma_{1}-1}, \tau^{\alpha_{1}-\gamma_{1}-1},A \tau^{\alpha_{1}-\gamma_{1}-1}\bigr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ =&\Phi(t). \end{aligned}$$

(3.20)

Thus, by (3.9), (3.16), (3.17), and (3.20)

$$\begin{aligned}& \begin{aligned}[b] &D^{\beta_{1}}\bigl(\varphi_{p_{1}} \bigl(-D^{\alpha_{1}-\gamma_{1}}\Psi\bigr)\bigr) (t)+\lambda^{*} f_{1} \bigl(I^{\gamma_{1}}\Psi(t),\Psi(t),A\Psi (t)\bigr) \\ &\quad=D^{\beta_{1}}\bigl(\varphi_{p_{1}}\bigl(-D^{\alpha_{1}-\gamma_{1}}(T_{\lambda^{*}} \Phi )\bigr)\bigr) (t)+\lambda^{*} f_{1}(I^{\gamma_{1}}\Psi(t),\Psi(t),A \Psi(t) \\ &\quad\geq-\lambda^{*} f_{1}\bigl(I^{\gamma_{1}}\Phi(t),\Phi(t),A \Phi(t)\bigr)+\lambda^{*} f_{1}\bigl(I^{\gamma_{1}}\Phi(t),\Phi(t),A\Phi (t)\bigr)=0, \end{aligned} \end{aligned}$$

(3.21)

$$\begin{aligned}& \begin{aligned}[b] &D^{\beta_{1}}\bigl(\varphi_{p_{1}} \bigl(-D^{\alpha_{1}-\gamma_{1}}\Phi\bigr)\bigr) (t)+\lambda^{*} f_{1} \bigl(I^{\gamma_{1}}\Phi(t),\Phi(t),A\Phi (t)\bigr) \\ &\quad=D^{\beta_{1}}(\varphi_{p_{1}}\bigl(-D^{\alpha_{1}-\gamma_{1}} \bigl(T_{\lambda ^{*}}\bigl(t^{\alpha_{1}-\gamma_{1}-1}\bigr)\bigr)\bigr)+\lambda^{*} f_{1}\bigl(I^{\gamma_{1}}\Phi(t),\Phi (t),\Phi(t)\bigr) \\ &\quad\leq-\lambda^{*} f_{1}\bigl(I^{\gamma_{1}}t^{\alpha_{1}-\gamma_{1}-1},t^{\alpha _{1}-\gamma_{1}-1},At^{\alpha_{1}-\gamma_{1}-1} \bigr)\\ &\qquad{}+\lambda^{*} f_{1}\bigl(I^{\gamma _{1}}t^{\alpha_{1}-\gamma_{1}-1},t^{\alpha_{1}-\gamma_{1}-1},At^{\alpha _{1}-\gamma_{1}-1} \bigr)=0. \end{aligned} \end{aligned}$$

(3.22)

It follows from (3.18) and (3.21)-(3.22) that $\Psi(t),\Phi(t)$ are upper and lower solutions of BVP (2.6), and that $\Psi(t), \Phi(t)\in P$. Now let us define a function

$$F(v_{1})=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} f_{1}(I^{\gamma_{1}}\Psi(t),\Psi(t),A\Psi(t)), & v_{1}< \Psi(t),\\ f_{1}(I^{\gamma_{1}}v_{1}(t),v_{1}(t),Av_{1}(t)), &\Psi(t)\leq v_{1}\leq\Phi(t),\\ f_{1}(I^{\gamma_{1}}\Phi(t),\Phi(t),A\Phi(t)),& v_{1}>\Phi(t). \end{array}\displaystyle \right . $$

Clearly, $F:[0,+\infty]\rightarrow[0,+\infty]$ is continuous.

We now show that the fractional boundary value problem

$$ \left \{ \textstyle\begin{array}{@{}l} -D^{\beta_{1}}(\varphi_{p_{1}}(-D^{\alpha_{1}-\gamma_{1}}v_{1}))(t)=\lambda^{*} F(v_{1}),\\ D^{\alpha_{1}-\gamma_{1}}v_{1}(0)=D^{\alpha_{1}-\gamma_{1}} v_{1}(1)=0, \\ v_{1}(0)=0, \qquad D^{\alpha_{1}-\gamma_{1}-1} v_{1}(1)=\xi_{1}I^{\omega_{1}}v_{1}(\eta_{1}), \end{array}\displaystyle \right . $$

(3.23)

has a positive solution. Define the operator $D_{\lambda^{*}}$ by

$$D_{\lambda^{*}}v_{1}(t)=\bigl(\lambda^{*}\bigr)^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int _{0}^{1}G(\beta_{1},s,\tau)F \bigl(v_{1}(\tau)\,d\tau\bigr) \biggr)^{q_{1}-1}\,ds. $$

Then $D_{\lambda^{*}}:C[0,1]\rightarrow C[0,1]$, and a fixed point of the operator $D_{\lambda^{*}}$ is a solution of the BVP (3.23). On the other hand, from the definition of F and the fact that the function $f_{1}(x,y,z)$ is non-increasing in $x, y, z$ respectively, and A is non-decreasing, we obtain $f_{1}(I^{\gamma_{1}}\Phi(t),\Phi(t),A\Phi(t)) \leq F(v_{1}(t))\leq f_{1}(I^{\gamma_{1}}\Psi(t),\Psi(t),A\Psi(t))$, provided that $\Psi(t)\leq v_{1}(t)\leq\Phi(t)$, $F(v_{1}(t))=f_{1}(I^{\gamma_{1}}\Psi(t), \Psi(t),A\Psi (t))$, provided that $v_{1}(t)<\Psi(t)$, and $F(v_{1}(t))=f_{1}(I^{\gamma_{1}}\Phi(t),\Phi(t),A\Phi(t))$, provided that $v_{1}(t)>\Phi(t)$. So we have

$$f_{1}\bigl(I^{\gamma_{1}}\Phi(t),\Phi(t),A\Phi(t)\bigr) \leq F \bigl(v_{1}(t)\bigr)\leq f_{1}\bigl(I^{\gamma_{1}}\Psi(t), \Psi(t),A\Psi(t)\bigr),\quad \forall v_{1}(t)\in E. $$

Furthermore, since $\Psi(t)\geq t^{\alpha_{1}-\gamma_{1}-1}$, we have

$$\begin{aligned} f_{1}\bigl(I^{\gamma_{1}}\Phi(t),\Phi(t),A\Phi(t)\bigr) \leq& F \bigl(v_{1}(t)\bigr) \\ \leq& f_{1}\bigl(I^{\gamma_{1}}t^{\alpha_{1}-\gamma_{1}-1},t^{\alpha_{1}-\gamma _{1}-1},At^{\alpha_{1}-\gamma_{1}-1} \bigr), \quad\forall v_{1}(t)\in E. \end{aligned}$$

(3.24)

It follows from (3.11), for any $v_{1}(t)\in E$

$$\begin{aligned} D_{\lambda^{*}}v_{1}(t) =&\bigl(\lambda^{*} \bigr)^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int _{0}^{1}G(\beta_{1},s,\tau)F \bigl(v_{1}(\tau)\,d\tau\bigr) \biggr)^{q_{1}-1}\,ds \\ \leq&\bigl(\lambda^{*}\bigr)^{q_{1}-1} \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G(\beta_{1},s,\tau )f_{1}\bigl(I^{\gamma_{1}}\tau^{\alpha_{1}-\gamma_{1}-1}, \\ &{} \tau^{\alpha_{1}-\gamma_{1}-1},A \tau^{\alpha_{1}-\gamma_{1}-1}\bigr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ \leq&\bigl(\lambda^{*}\bigr)^{q_{1}-1}d_{1} \biggl( \frac{\beta_{1}-1}{\Gamma(\beta_{1})} \int _{0}^{1}f_{1}\biggl( \frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}, \tau^{\alpha_{1}-\gamma_{1}-1},b\tau^{\alpha_{2}-\gamma_{2}-1} \biggr)\,d\tau \biggr)^{q_{1}-1}\,ds \\ < &+\infty, \end{aligned}$$

(3.25)

namely, the operator $D_{\lambda^{*}}$ is uniformly bounded.

On the other hand, let $\Omega\subset E$ be bounded. As the function $H_{1}(t,s), G(\beta_{1},t,s)$ is uniformly continuous on $[0,1]\times[0,1]$, $D_{\lambda^{*}}(\Omega)$ is equicontinuous. By the Arzela-Ascoli theorem, we have $D_{\lambda ^{*}}:E\rightarrow E$ is completely continuous. Thus by using the Schauder fixed point theorem, $D_{\lambda^{*}}$ has at least one fixed point x such the $x=D_{\lambda^{*}}x$.

Now we prove

$$\Psi(t)\leq x(t)\leq\Phi(t), \quad t\in[0,1]. $$

Since x is a fixed point of $D_{\lambda^{*}}$, by (3.18) and (3.23), we have

$$ \begin{aligned} &D^{\alpha_{1}-\gamma_{1}}x(0)=D^{\alpha_{1}-\gamma_{1}}x(1)=0,\qquad x(0)=0,\qquad D^{\alpha_{1}-\gamma_{1}-1}x(1)=\xi_{1}I^{\omega_{1}}\bigl(x( \eta_{1})\bigr), \\ &D^{\alpha_{1}-\gamma_{1}}\Phi(0)=D^{\alpha_{1}-\gamma_{1}}\Phi(1)=0, \qquad\Phi(0)=0,\qquad D^{\alpha_{1}-\gamma_{1}-1} \Phi(1)=\xi_{1}I^{\omega_{1}}\bigl(\Phi (\eta_{1})\bigr). \end{aligned} $$

(3.26)

From (3.9), (3.16), (3.24), and noting that x is a fixed point of $D_{\lambda^{*}}$, we also have

$$\begin{aligned} &D^{\beta_{1}}\bigl(\varphi_{p_{1}}\bigl(-D^{\alpha_{1}-\gamma_{1}}\Phi\bigr) \bigr) (t)-D^{\beta _{1}}\bigl(\varphi_{p_{1}}\bigl(-D^{\alpha_{1}-\gamma_{1}}x \bigr)\bigr) (t) \\ &\quad=-\lambda^{*} f_{1}\bigl(I^{\gamma_{1}}t^{\alpha_{1}-\gamma_{1}-1},t^{\alpha _{1}-\gamma_{1}-1},At^{\alpha_{1}-\gamma_{1}-1} \bigr)+\lambda^{*}F(x(t)\leq0. \end{aligned}$$

Let $z(t)=\varphi_{p_{1}}(-D^{\alpha_{1}-\gamma_{1}}\Phi)(t)-\varphi _{p_{1}}(-D^{\alpha_{1}-\gamma_{1}}x)(t)$, then

$$\begin{aligned}& D^{\beta_{1}}z(t) =D^{\beta_{1}}\bigl(\varphi_{p_{1}} \bigl(-D^{\alpha_{1}-\gamma_{1}}\Phi(t)\bigr)\bigr)-D^{\beta _{1}}\bigl( \varphi_{p_{1}}\bigl(-D^{\alpha_{1}-\gamma_{1}}x(t)\bigr)\bigr) \leq0, \\& z(0)=\varphi_{p_{1}}\bigl(-D^{\alpha_{1}-\gamma_{1}}\Phi(0)\bigr)-\varphi _{p_{1}}\bigl(-D^{\alpha_{1}-\gamma_{1}}x(0)\bigr)=0, \\& z(1)=\varphi_{p_{1}}\bigl(-D^{\alpha_{1}-\gamma_{1}}\Phi(1)\bigr)-\varphi _{p_{1}}\bigl(-D^{\alpha_{1}-\gamma_{1}}x(1)\bigr)=0. \end{aligned}$$

In view of Lemmas 2.1 and 2.3, we obtain

$$z(t)\geq0, $$

i.e.

$$\varphi_{p_{1}}\bigl(-D^{\alpha_{1}-\gamma_{1}}\Phi(t)\bigr)- \varphi_{p_{1}}\bigl(-D^{\alpha _{1}-\gamma_{1}}x(t)\bigr)\geq0, $$

Noticing that $\varphi_{p_{1}}$ is monotone increasing, we have

$$-D^{\alpha_{1}-\gamma_{1}}\Phi(t)\geq-D^{\alpha_{1}-\gamma_{1}}x(t), $$

i.e.

$$-D^{\alpha_{1}-\gamma_{1}}\bigl(\Phi(t)-x(t)\bigr)\geq0. $$

It follows from Lemma 2.6 and (3.26)

$$\Phi(t)-x(t)\geq0. $$

Then we have $x(t)\leq\Phi(t)$ on $[0,1]$. In the same way we also have $x(t)\geq\Psi(t)$ on $[0,1]$. So

$$ \Psi(t)\leq x(t)\leq\Phi(t). $$

(3.27)

Consequently, $F(x(t))=f_{1}(I^{\gamma_{1}}x(t),x(t),Ax(t)), t\in[0,1]$. Hence $x(t)$ is a positive solution of the problem (2.6). Finally, by (3.27) and $\Phi,\Psi\in P$, we have

$$l_{\Psi}t^{\alpha_{1}-\gamma_{1}-1}\leq\Psi(t)\leq x(t)\leq\Phi(t)\leq l_{\Phi}^{-1}t^{\alpha_{1}-\gamma_{1}-1}. $$

Then by Lemmas 2.5

$$\left \{ \textstyle\begin{array}{@{}l} u_{1}(t)=I^{\gamma_{1}}x(t),\\ u_{2}(t)=I^{\gamma_{1}}v_{2}(t), \end{array}\displaystyle \right . $$

where $v_{2}(t)=\int_{0}^{1}H_{2}(t,s) (\int_{0}^{1}G(\beta_{2},s,\tau )f_{2}(\tau,I^{\gamma_{1}}x(\tau))\,d\tau )^{q_{2}-1}\,ds$ is the unique positive solution of system (1.1).

Since the process is similar to (3.4) and (3.5) we obtain

$$\begin{aligned}[b] v_{2}(t)&= \int_{0}^{1}H_{2}(t,s) \biggl( \int_{0}^{1}G(\beta_{2},s, \tau)f_{2}\bigl(\tau ,I^{\gamma_{1}}x(\tau)\bigr)\,d\tau \biggr)^{q_{2}-1}\,ds \\ &\leq \int_{0}^{1}H_{2}(t,s) \biggl( \int_{0}^{1}G(\beta_{2},s, \tau)f_{2}\biggl(\tau,l_{\Phi}^{-1} \frac{\Gamma(\alpha_{1}-\gamma_{1})}{ \Gamma(\alpha_{1})}\tau^{\alpha_{1}-1}\biggr)\,d\tau \biggr)^{q_{2}-1}\,ds \\ &\leq al_{\Phi}^{-\sigma(q_{2}-1)}t^{\alpha_{2}-\gamma_{2}-1} \end{aligned} $$

and

$$\begin{aligned} v_{2}(t) =& \int_{0}^{1}H_{2}(t,s) \biggl( \int_{0}^{1}G(\beta_{2},s, \tau)f_{2}\bigl(\tau ,I^{\gamma_{1}}x(\tau)\bigr)\,d\tau \biggr)^{q_{2}-1}\,ds \\ \geq& \int_{0}^{1}H_{2}(t,s) \biggl( \int_{0}^{1}G(\beta_{2},s, \tau)f_{2}\biggl(\tau,l_{\Psi}\frac{\Gamma(\alpha_{1}-\gamma_{1})}{ \Gamma(\alpha_{1})} \tau^{\alpha_{1}-1}\biggr)\,d\tau \biggr)^{q_{2}-1}\,ds \\ \geq& bl_{\Psi}^{\sigma(q_{2}-1)}t^{\alpha_{2}-\gamma_{2}-1}, \end{aligned}$$

i.e.

$$ bl_{\Psi}^{\sigma(q_{2}-1)}\frac{\Gamma(\alpha_{2}-\gamma_{2})}{ \Gamma(\alpha_{2})}t^{\alpha_{2}-1}\leq u_{2}(t)=I^{\gamma_{1}}v_{2}(t)\leq a l_{\Phi}^{-\sigma(q_{2}-1)} \frac{\Gamma(\alpha_{2}-\gamma_{2})}{ \Gamma(\alpha_{2})}t^{\alpha_{2}-1}. $$

(3.28)

 □

Example

Consider the following boundary value problem:

$$ \left \{ \textstyle\begin{array}{@{}l} -D^{\frac{5}{2}}(\varphi_{3}(-D^{\frac{4}{3}}u_{1}))(t)=\lambda (u_{1}^{-\frac{2}{9}}(t)+[D^{\frac{1}{6}}u_{1}(t)]^{-\frac{1}{2}}+ [D^{\frac{1}{4}}u_{2}(t)]^{-\frac{1}{8}} ),\\ -D^{\frac{11}{4}}(\varphi_{4}(-D^{\frac{3}{2}}u_{2}))(t)=(t^{2}+1)u_{1}^{\frac {1}{7}}(t),\\ D^{\frac{4}{3}}u_{1}(0)=D^{\frac{4}{3}}u_{1}(1)=0, D^{\frac{1}{6}}u_{1}(0)=0, D^{\frac{1}{3}}u_{1}(1)=2I^{\frac {5}{6}}({D^{\frac{1}{6}}}u_{1}(\frac{1}{3})),\\ D^{\frac{3}{2}}u_{2}(0)=D^{\frac{3}{2}}u_{2}(1)=0, D^{\frac{1}{4}}u_{2}(0)=0, D^{\frac{1}{2}}u_{2}(1)=5I^{\frac{7}{4}}(D^{\frac {1}{4}}u_{2}(\frac{1}{2})). \end{array}\displaystyle \right . $$

(3.29)

Let $\alpha_{1}=\frac{4}{3},\alpha_{2}=\frac{3}{2},\beta_{1}=\frac{5}{2},\beta _{2}=\frac{11}{4},\gamma_{1}=\frac{1}{6},\gamma_{2}=\frac{1}{4}, p_{1}=3,p_{2}=4,\omega_{1}=\frac{5}{6},\omega_{2}=\frac{7}{4},\xi_{1}=2,\xi _{2}=5,\eta_{1}=\frac{1}{3},\eta_{2}=\frac{1}{2}$.

First, we have

$$\begin{aligned}& \Gamma(\alpha_{1}-\gamma_{1}+\omega_{1})= \Gamma(2)=1>\xi_{1}\eta_{1}^{\alpha _{1}-\gamma_{1}+\omega_{1}-1}=2\biggl( \frac{1}{3}\biggr), \\& \Gamma(\alpha_{2}-\gamma_{2}+\omega_{2})= \Gamma(3)=2>\xi_{2}\eta_{2}^{\alpha _{2}-\gamma_{2}+\omega_{2}-1}=5\biggl( \frac{1}{2}\biggr)^{2}, \end{aligned}$$

and $q_{1}=\frac{3}{2},q_{2}=\frac{4}{3}$, then $(s_{0}),(s_{1})$, and $(s_{2})$ hold.

Second, let

$$\begin{aligned}& f_{1}(x,y,z)=x^{-\frac{2}{9}}+y^{-\frac{1}{2}}+z^{-\frac{1}{8}}, \qquad f_{2}(t,x)=\bigl(t^{2}+1\bigr)x^{\frac{1}{7}}, \qquad \sigma=\frac{1}{3}< \frac{1}{q_{2}-1}=3, \end{aligned}$$

and for all $r\in(0,1)$, $(x,y,z)\in(0,+\infty)^{3}$, $(t,x)\in(0,1)\times (0,+\infty)$,

$$\begin{aligned}& f_{1}(rx,ry,rz)=r^{-\frac{2}{9}}x^{-\frac{2}{9}}+r^{-\frac{1}{2}}y^{-\frac {1}{2}}+r^{-\frac{1}{8}}z^{-\frac{1}{8}} \leq r^{-\frac{1}{2}}f_{1}(x,y,z), \\& f_{2}(t,rx)=\bigl(t^{2}+1\bigr)r^{\frac{1}{7}}x^{\frac{1}{7}} \geq r^{\frac{1}{3}}f_{2}(t,x), \end{aligned}$$

which implies that (H₁), (H₂) hold. On the other hand, by direct calculation, we have $f_{1}(1,1,1)=3\neq0$, and then

$$\begin{aligned} b =&e_{2} \int_{0}^{1}\bigl[1-(1-s)^{\alpha_{2}-\gamma_{2}-1} \bigr]s^{(\beta _{2}-1)(q_{2}-1)}(1-s)^{q_{2}-1}\,ds \\ &{}\times \biggl( \frac{1}{\Gamma(\beta_{2})} \int_{0}^{1}\tau(1-\tau)^{\beta _{2}-1}f_{2} \biggl(\tau,\frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha_{1})} \tau^{\alpha_{1}-1}\biggr)\,d\tau \biggr)^{q_{2}-1} \\ =&\Gamma\biggl(\frac{5}{4}\biggr) \int_{0}^{1}\bigl[1-(1-s)^{\frac{1}{4}} \bigr]s^{\frac{7}{4}\frac {1}{3}}(1-s)^{\frac{1}{3}}\,ds \\ &{}\times\biggl( \frac{1}{\Gamma(\frac{11}{4})} \int_{0}^{1}\tau(1-\tau)^{\frac {7}{4}}\bigl( \tau^{2}+1\bigr) \biggl[\frac{\Gamma(\frac{7}{6})}{\Gamma(\frac{4}{3})} \biggr]^{\frac{1}{7}} \tau^{\frac{1}{21}}\,d\tau \biggr)^{\frac{1}{3}}>0. \end{aligned}$$

Thus

$$\begin{aligned} & \int_{0}^{1}f_{1}\biggl( \frac{\Gamma(\alpha_{1}-\gamma_{1})}{\Gamma(\alpha _{1})}t^{\alpha_{1}-1},t^{\alpha_{1}-\gamma_{1}-1}, bt^{\alpha_{2}-\gamma_{2}-1}\biggr)\,dt \\ &\quad= \int_{0}^{1} \biggl\{ \biggl[\frac{\Gamma(\frac{7}{6})}{\Gamma(\frac {4}{3})}t^{\frac{1}{3}} \biggr]^{-\frac{2}{9}}+\bigl[t^{\frac{1}{6}}\bigr] ^{-\frac{1}{2}}+ \bigl[bt^{\frac{1}{4}}\bigr]^{-\frac{1}{8}} \biggr\} \,dt \\ &\quad= \int_{0}^{1} \biggl\{ \biggl[\frac{\Gamma(\frac{7}{6})}{\Gamma(\frac {4}{3})} \biggr]^{-\frac{2}{9}}t^{-\frac{2}{27}}+t^{-\frac{1}{12}} +b^{-\frac{1}{8}}t^{-\frac{1}{32}} \biggr\} \,dt< +\infty. \end{aligned}$$

Hence, (H₃) holds. Then by Theorem 3.1 there exists a constant $\lambda^{*}>0$ such that for any $\lambda\in (\lambda^{*},+\infty)$, the BVP (1.1) has at least one positive solution $(u_{1}(t),u_{2}(t))$.