### Lemma 3.1

*Suppose*
\(\epsilon_{i}(q_{i}-1)>1\), \(i=1,2\), *then the equation*

$$ \biggl(\frac{\beta_{1}-1}{\Gamma{(\beta_{1}+2)}}2^{\epsilon_{1}}M \biggr)^{q_{1}-1}x^{\epsilon_{1}(q_{1}-1)-1}+ \biggl(\frac{\beta_{2}-1}{\Gamma{(\beta_{2}+2)} }2^{\epsilon_{2}}M \biggr)^{q_{2}-1}x^{\epsilon_{2}(q_{2}-1)-1}=b^{-1} $$

(3.1)

*has a unique positive solution*
\(r^{*}\)
*in*
\([0,\infty )\).

### Proof

Let

$$\begin{aligned} \varphi(x) =&b^{-1}- \biggl(\frac{\beta _{1}-1}{\Gamma{(\beta_{1}+2)}}2^{\epsilon_{1}}M \biggr) ^{q_{1}-1}x^{\epsilon_{1}(q_{1}-1)-1} \\ &{} - \biggl(\frac{\beta_{2}-1}{\Gamma {(\beta_{2}+2)}}2^{\epsilon_{2}}M \biggr)^{q_{2}-1}x^{\epsilon_{2}(q_{2}-1)-1}, \end{aligned}$$

(3.2)

then \(\varphi(x)\) is continuous in \([0,\infty )\), and

$$\begin{aligned}& \varphi(0)=b^{-1}>0, \end{aligned}$$

(3.3)

$$\begin{aligned}& \varphi \biggl( \biggl(\frac{(\beta _{1}-1)2^{\epsilon_{1}}Mb^{\frac{1}{q_{1}-1}}}{\Gamma{(\beta_{1}+2)}} \biggr)^{-\frac{q_{1}-1}{\epsilon_{1}(q_{1}-1)-1}} \biggr) \\& \quad = - \biggl(\frac{\beta_{2}-1}{\Gamma{(\beta_{2}+2)}}2^{\epsilon_{2}}M \biggr)^{q_{2}-1} \biggl(\frac{(\beta_{1}-1)2^{\epsilon_{1}}Mb^{\frac{1}{q_{1}-1}}}{\Gamma{(\beta_{1}+2)}} \biggr)^{-\frac{(q_{1}-1) (\epsilon_{2}(q_{2}-1)-1)}{\epsilon_{1}(q_{1}-1)-1}} < 0. \end{aligned}$$

(3.4)

On the other hand, we also have

$$\begin{aligned} \varphi'(x) =&-\bigl[\epsilon_{1}(q_{1}-1)-1 \bigr] \biggl(\frac{\beta_{1}-1}{\Gamma{(\beta_{1}+2)}}2^{\epsilon_{1}}M \biggr) ^{q_{1}-1}x^{\epsilon_{1}(q_{1}-1)-2} \\ &{}-\bigl[\epsilon_{2}(q_{2}-1)-1\bigr] \biggl( \frac{\beta_{2}-1}{\Gamma{(\beta _{2}+2)}}2^{\epsilon_{2}}M \biggr)^{q_{2}-1}x^{\epsilon_{2}(q_{2}-1)-2} \\ < &0,\quad x\in(0,\infty). \end{aligned}$$

(3.5)

Thus by (3.2)-(3.5), equation (3.1) has a unique positive solution \(r^{*}\) in \([0,\infty )\). □

### Theorem 3.1

*Suppose conditions* (H0) *and* (H1) *hold*, *and*

$$ \frac{(\beta_{i}-1) f_{i}(0,0)}{\Gamma(\beta_{i}+2)}\le \biggl(\frac{r^{*}}{b} \biggr)^{\frac{1}{q_{i}-1}},\quad i=1,2. $$

(3.6)

*Then system* (1.1) *has the minimal and maximal solutions*, \(x^{*}=(x^{*}_{1},x^{*}_{2})\)
*and*
\(y^{*}=(y^{*}_{1},y^{*}_{2})\), *which are positive*; *and there exist some nonnegative numbers*
\(m_{i}< n_{i}\), \(i=1,2,3,4\), *such that*

$$ \begin{aligned} & m_{1}t^{\alpha-1}\le x^{*}_{1}(t) \le n_{1}t^{\alpha-1},\qquad m_{2}t^{\alpha-1}\le y^{*}_{1}(t) \le n_{2}t^{\alpha-1}, \quad t\in[0,1], \\ &m_{3}t^{\alpha-1}\le x^{*}_{2}(t) \le n_{3}t^{\alpha-1},\qquad m_{4}t^{\alpha-1}\le y^{*}_{2}(t) \le n_{4}t^{\alpha-1}, \quad t\in [0,1]. \end{aligned} $$

(3.7)

*Moreover*, *for initial values*
\(u^{(0)}(t)=(0,0)\), \(w^{(0)}(t)=(r^{*},r^{*})\), *let*
\(\{{u^{(n)}}\}\)
*and*
\(\{{w^{(n)}}\}\)
*be the iterative sequences generated by*

$$ u^{(n)}(t) = T{u^{(n-1)}}(t) = T^{n}u^{(0)}(t),\qquad {w^{(n)}}(t) = T{w^{(n-1)}}(t) = T^{n}w^{(0)}(t), $$

(3.8)

*then*

$$\lim_{n\to+\infty}{u^{(n)}(t)}=x^{*}(t),\qquad \lim _{n\to+\infty }{w^{(n)}(t)}=y^{*}(t) $$

*uniformly for*
\(t\in[0,1]\).

### Proof

Take \(P[0, r^{*}] = \{(x_{1},x_{2}) \in P : 0 \le\|x_{1}\| \le r^{*}, 0 \le\|x_{2}\| \le r^{*}\}\), we firstly prove \(T(P[0, r^{*}]) \subset P[0, r^{*}]\).

In fact, for any \((x_{1},x_{2}) \in P[0, r^{*}]\), if \((x_{1},x_{2})\equiv(0,0)\), it follows from (2.8) that

$$\begin{aligned} \bigl\Vert T_{1}(x_{1},x_{2}) \bigr\Vert =& \max_{t\in[0,1]} \biggl\{ \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G_{\beta_{1}}(s,\tau) f_{1}\bigl(x_{1}(\tau),x_{2}(\tau)\bigr)\,d\tau \biggr)^{q_{1}-1}\,ds \biggr\} \\ \le& b \int_{0}^{1} \biggl( \int_{0}^{1} \frac{\beta_{1}-1}{\Gamma(\beta_{1})}\tau(1-\tau )^{\beta_{1}-1}f_{1}(0,0)\,d\tau \biggr)^{q_{1}-1}\,ds \\ =&b \biggl(\frac{\beta_{1}-1}{\Gamma{(\beta_{1}+2})}f_{1}(0,0) \biggr)^{q_{1}-1}\le r^{*}, \end{aligned}$$

(3.9)

and

$$\begin{aligned} \begin{aligned}[b] \bigl\Vert T_{2}(x_{1},x_{2}) \bigr\Vert & = \max_{t\in [0,1]} \biggl\{ \int_{0}^{1}H_{2}(t,s) \biggl( \int_{0}^{1}G_{\beta_{2}}(s,\tau) f_{2}\bigl(x_{1}(\tau),x_{2}(\tau)\bigr)\,d\tau \biggr)^{q_{2}-1}\,ds \biggr\} \\ &\le b \int_{0}^{1} \biggl( \int_{0}^{1} \frac{\beta_{2}-1}{\Gamma(\beta_{2})}\tau(1-\tau )^{\beta_{2}-1}f_{2}(0,0)\,d\tau \biggr)^{q_{2}-1}\,ds \\ &=b \biggl(\frac{\beta_{2}-1}{\Gamma{(\beta_{2}+2)}}f_{2}(0,0) \biggr)^{q_{2}-1}\le r^{*}. \end{aligned} \end{aligned}$$

(3.10)

Otherwise, for any \(t\in(0,1)\), we have

$$ 0< x_{1}(t)+x_{2}(t) \le\max _{t\in[0,1]}x_{1}(t)+\max_{t\in[0,1]}x_{2}(t) \le2r^{*}. $$

(3.11)

So, by (H1), we get

$$\begin{aligned} \bigl\Vert T_{1}(x_{1},x_{2}) \bigr\Vert =& \max_{t\in[0,1]} \biggl\{ \int_{0}^{1}H_{1}(t,s) \biggl( \int_{0}^{1}G_{\beta_{1}}(s,\tau) f_{1}\bigl(x_{1}(\tau),x_{2}(\tau)\bigr)\,d\tau \biggr)^{q_{1}-1}\,ds \biggr\} \\ \le& b \int_{0}^{1} \biggl( \int_{0}^{1} \frac{\beta_{1}-1}{\Gamma(\beta_{1})}\tau(1-\tau )^{\beta_{1}-1}M\bigl[x_{1}(\tau)+x_{2}(\tau) \bigr]^{\epsilon_{1}}\,d\tau \biggr)^{q_{1}-1}\,ds \\ \le& b \int_{0}^{1} \biggl( \int_{0}^{1} \frac{\beta_{1}-1}{\Gamma(\beta _{1})}\tau(1- \tau)^{\beta_{1}-1}M2^{\epsilon_{1}}\bigl(r^{*}\bigr)^{\epsilon_{1}}\,d\tau \biggr)^{q_{1}-1}\,ds \\ \le& b \biggl(\frac{\beta_{1}-1}{\Gamma{(\beta_{1}+2)}}2^{\epsilon _{1}}M \biggr)^{q_{1}-1}{ \bigl(r^{*}\bigr)}^{\epsilon_{1}(q_{1}-1)} \\ \le& b \biggl(\frac{\beta_{1}-1}{\Gamma{(\beta_{1}+2)}}2^{\epsilon_{1}}M \biggr)^{q_{1}-1}{ \bigl(r^{*}\bigr)}^{\epsilon_{1}(q_{1}-1)} + b \biggl(\frac{\beta_{2}-1}{\Gamma{(\beta_{2}+2)}}2^{\epsilon _{2}}M \biggr)^{q_{2}-1}{\bigl(r^{*}\bigr)}^{\epsilon_{1}(q_{2}-1)} \\ =&r^{*}. \end{aligned}$$

(3.12)

Similar to (3.12), we have

$$\begin{aligned} \bigl\Vert T_{2}(x_{1},x_{2}) \bigr\Vert \le& b \biggl(\frac{\beta _{1}-1}{\Gamma{(\beta_{1}+2)}}2^{\epsilon_{1}}M \biggr)^{q_{1}-1}\bigl(r^{*} \bigr)^{\epsilon_{1}(q_{1}-1)} \\ &{}+b \biggl(\frac{\beta _{2}-1}{\Gamma{(\beta_{2}+2)}}2^{\epsilon_{2}}M \biggr)^{q_{2}-1} \bigl(r^{*}\bigr)^{\epsilon_{1}(q_{2}-1)}=r^{*}, \end{aligned}$$

(3.13)

which implies that \(T(P[0, r^{*}]) \subset P[0, r^{*}]\).

Let \(u^{(0)}(t) =(u^{(0)}_{1}(t),u^{(0)}_{2}(t))=(0,0)\) and

$$\begin{aligned} u^{(1)}(t)&=:\bigl(u^{(1)}_{1}(t),u^{(1)}_{2}(t) \bigr)= \bigl(\bigl(T_{1}\bigl(u^{(0)}_{1},u^{(0)}_{2} \bigr)\bigr) (t),\bigl(T_{2}\bigl(u^{(0)}_{1},u^{(0)}_{2} \bigr)\bigr) (t)\bigr) \\ &=\bigl(\bigl(T_{1}(0,0)\bigr) (t),\bigl(T_{2}(0,0)\bigr) (t) \bigr), \quad t \in[0, 1], \end{aligned}$$

it follows from \(u^{(0)}(t)\in P([0,r^{*}])\) that \(u^{(1)}(t) \in T(P[0, r^{*}])\).

Denote

$$u^{(n+1)}(t) = Tu^{(n)}(t) = T^{n+1} u^{(0)}(t), \quad n = 1, 2, \ldots. $$

It follows from \(T(P[0, r^{*}]) \subset P[0, r^{*}]\) that \(u_{n}(t) \in P[0, r^{*}]\) for \(n\ge1\). Noticing that *T* is compact, we get that \(\{u^{(n)}\}\) is a sequentially compact set.

On the other hand, since \(u^{(1)}(t)\ge0=u^{(0)}(t)\), we have

$$u^{(2)}(t) = \bigl(Tu^{(1)}\bigr) (t) \ge\bigl(Tu^{(0)} \bigr) (t)= u^{(1)}(t),\quad t \in[0, 1]. $$

By induction, we get

$$u^{(n+1)} \ge u^{(n)}, \quad n = 1, 2, \ldots. $$

Consequently, there exists \(x^{*}=(x^{*}_{1},x^{*}_{2}) \in P[0, r]\) such that \(u^{(n)} \to x^{*}\). Letting \(n \to+\infty\), from the continuity of *T* and \(Tu^{(n)} =u^{(n-1)}\), we obtain \(Tx^{*} = x^{*}\), which implies that \(x^{*}=(x^{*}_{1},x^{*}_{2})\) is a nonnegative solution of the nonlinear integral equation (1.1). Since \(x^{*}\in P\), there exist constants \(0< m_{1}< n_{1}\), \(0< m_{2}< n_{2}\) such that

$$0< m_{1}t^{\alpha_{1}-1}\le x_{1}^{*}(t)\le n_{1} t^{\alpha_{1}-1}, 0< m_{2}t^{\alpha _{2}-1}\le x_{2}^{*}(t)\le n_{2} t^{\alpha_{2}-1},\quad t \in(0, 1), $$

and consequently \(x^{*}\) is a positive solution of system (1.1).

Now let \(w^{(0)}(t) =(w^{(0)}_{1}(t),w^{(0)}_{2}(t))=(r^{*},r^{*})\) and

$$w^{(1)}(t)=:\bigl(w^{(1)}_{1}(t),w^{(1)}_{2}(t) \bigr)= \bigl(\bigl(T_{1}\bigl(w^{(0)}_{1},w^{(0)}_{2} \bigr)\bigr) (t),\bigl(T_{2}\bigl(w^{(0)}_{1},w^{(0)}_{2} \bigr) \bigr)(t)\bigr), \quad t \in[0, 1]. $$

Since \(w^{(0)}(t)=(r^{*},r^{*}) \in P[0, r^{*}]\), and then \(w^{(1)}(t)\in P[0, r^{*}]\). Thus denote

$$w^{(n+1)}(t) = Tw^{(n)}(t) = T^{n+1}w^{(0)}(t), \quad n = 1, 2, \ldots. $$

It follows from \(T(P[0, r^{*}]) \subset P[0, r^{*}]\) that

$$w^{(n)}(t) \in P\bigl[0, r^{*}\bigr], \quad n = 0, 1, 2, \ldots. $$

From Lemma 2.5, *T* is compact, consequently \(\{w^{(n)}\}\) is a sequentially compact set.

Now, since \(w^{(1)}(t) \in P[0, r^{*}]\), we get

$$0 \le w^{(1)}_{1}(t) \le \bigl\Vert w^{(1)}_{1} \bigr\Vert \le r = w^{(0)}_{1}(t), \qquad 0 \le w^{(1)}_{2}(t) \le \bigl\Vert w^{(1)}_{2} \bigr\Vert \le r^{*} = w^{(0)}_{2}(t). $$

It follows from (H1) that \(w^{(2)}(t) = Tw^{(1)}(t) \le Tw^{(0)}(t)=w^{(1)}(t)\). By induction, we obtain

$$w^{(n+1)} (t)\le w^{(n)}(t),\quad n = 0, 1, 2, \ldots. $$

Consequently, there exists \(y^{*}(t)\in P[0, r^{*}]\) such that \(w^{(n)}(t) \to y^{*}=(y^{*}_{1},y^{*}_{2})\). Letting \(n \to+\infty\), from the continuity of *T* and \(Tw^{(n)}(t) = w^{(n-1)}(t)\), we have \(Ty^{*} = y^{*}\), which implies that \(y^{*}\) is another nonnegative solution of the boundary value problem (1.1) and \(y^{*}\) also satisfies (3.7) since \(y^{*}\in P\).

In the end, we prove that \(x^{*}\) and \(y^{*}\) are maximum and minimum solutions for system (1.1). Let *x̃* be any positive solution of system (1.1), then \(u^{(0)}=0\le\tilde{x}\le r^{*}=w^{(0)}\), and \(u^{(1)}=Tu^{(0)}\le T\tilde{x}=\tilde{x}\le T(w^{(0)})=w^{(1)}\). By induction, we have \(u^{(n)}\le\tilde{x}\le w^{(n)}\), \(n=1,2,3,\ldots\) . Taking limit, we have \(x^{*}\le\tilde{x}\le y^{*}\). This implies that \(x^{*}\) and \(y^{*}\) are the maximal and minimal solutions of system (1.1), respectively. The proof is completed. □

### Remark

In particular, if \(\alpha_{i}=\beta_{i}=p_{i}=2\), then the nonlocal fractional system (1.1) will reduce to a fourth order classical beam system of ordinary differential equation,

$$ \textstyle\begin{cases} -x^{(4)}_{1}(t)=f_{1}(x_{1}(t), x_{2}(t)), \\ -x_{2}^{(4)}(t)=f_{2}(x_{1}(t), x_{2}(t)) \\ x_{1}(0)=0,\qquad x_{1}''(0)=x_{1}''(1)=0,\qquad x_{1}(1)=\int_{0}^{1}x_{1}(t)\,dA_{1}(t), \\ x_{2}(0)=0,\qquad x_{2}''(0)=x_{2}''(1)=0, \qquad x_{2}(1)=\int_{0}^{1}x_{2}(t)\,dA_{2}(t), \end{cases} $$

(3.14)

we have the following good result.

### Corollary 3.1

*Suppose conditions* (H0) *and* (H1) *hold*, *and*

$$ f_{i}(0,0)\le\frac{6}{b}r^{*},\quad i=1,2, $$

(3.15)

*then system* (3.14) *has the minimal and maximal solutions*, \(x^{*}=(x^{*}_{1},x^{*}_{2})\)
*and*
\(y^{*}=(y^{*}_{1},y^{*}_{2})\), *which are positive*; *and there exist some nonnegative numbers*
\(m_{i}< n_{i}\), \(i=1,2,3,4\), *such that*

$$ \begin{aligned} &m_{1}t\le x^{*}_{1}(t) \le n_{1}t, m_{2}t\le y^{*}_{1}(t) \le n_{2}t,\quad t\in[0,1], \\ &m_{3}t\le x^{*}_{2}(t) \le n_{3}t, m_{4}t\le y^{*}_{2}(t) \le n_{4}t, \quad t\in[0,1]. \end{aligned} $$

(3.16)

*Moreover*, *for initial values*
\(u^{(0)}(t)=(0,0)\), \(w^{(0)}(t)=(r^{*},r^{*})\), *let*
\(\{{u^{(n)}}\}\)
*and*
\(\{{w^{(n)}}\}\)
*be the iterative sequences generated by*

$$ u^{(n)}(t) = T{u^{(n-1)}}(t) = T^{n}u^{(0)}(t),\qquad {w^{(n)}}(t) = T{w^{(n-1)}}(t) = T^{n}w^{(0)}(t), $$

(3.17)

*then*

$$\lim_{n\to+\infty}{u^{(n)}(t)}=x^{*}(t),\qquad \lim _{n\to+\infty }{w^{(n)}(t)}=y^{*}(t) $$

*uniformly for*
\(t\in[0,1]\).

In the end, we know that fractional order integral and derivative operators can describe an important characteristics exhibiting long-memory in time in many complex processes and systems. With this advantage, in many eco-economical systems and diffusive processes with long time memory behavior [2, 3, 15, 39], fractional calculus provides an excellent tool to describe the hereditary properties of them. Here we give a specific example arising from the above complex processes.

### Example

Consider the following nonlocal boundary value problem of the fractional *p*-Laplacian equation:

$$ \textstyle\begin{cases} -\mathscr {D}_{t}^{\frac{4}{3}} (\varphi_{\frac{5}{2}} (-\mathscr {D}_{t}^{\frac{3}{2}}x_{1} ) )(t) \\ \quad =(x_{1}+x_{2})^{2}+{(x_{1}+x_{2})^{2}}(x_{1}+x_{2}+1)\ln ( 1+\frac {1}{1+x_{1}+x_{2}} )+1, \\ -\mathscr {D}_{t}^{\frac{6}{5}} (\varphi_{\frac{3}{2}} (-\mathscr {D}_{t}^{\frac{7}{6}}x_{2} ) )(t) =\ln(2+x_{1}+x_{2}),\quad t\in(0,1), \\ x_{1}(0)=0,\qquad \mathscr {D}_{t}^{\frac{3}{2}}x_{1}(0)=\mathscr {D}_{t}^{\frac{3}{2}}x_{1}(1)=0,\qquad x_{1}(1)=\int^{1}_{0}x_{1}(s)\,dA_{1}(s), \\ x_{2}(0)=0,\qquad \mathscr {D}_{t}^{\frac{7}{6}}x_{2}(0)=\mathscr {D}_{t}^{\frac{7}{6}}x_{2}(1)=0,\qquad x_{2}(1)=\int^{1}_{0}x_{2}(s)\,dA_{2}(s), \end{cases} $$

(3.18)

where

$$A_{1}(t)= \textstyle\begin{cases} 0,& t\in [0,\frac{1}{2} ), \\ 2,& t\in [\frac{1}{2},\frac{3}{4} ), \\ 1,& t\in [\frac{3}{4},1 ], \end{cases}\displaystyle \qquad A_{2}(t)= \textstyle\begin{cases} 0,& t\in [0,\frac{1}{3} ), \\ \frac{3}{2}, &t\in [\frac{1}{3},\frac{2}{3} ), \\ \frac{1}{2}, &t\in [\frac{2}{3},1 ]. \end{cases} $$

Then system (3.18) has the positive minimal and maximal solutions, \(x^{*}=(x^{*}_{1},x^{*}_{2})\) and \(y^{*}=(y^{*}_{1},y^{*}_{2})\); and there exist some nonnegative numbers \(m_{i}\le n_{i}\), \(i=1,2,3,4\), such that

$$ \begin{aligned} &m_{1}t^{\frac{1}{2}}\le x^{*}_{1}(t) \le n_{1}t^{\frac{1}{2}},\qquad m_{2}t^{\frac{1}{2}}\le y^{*}_{1}(t) \le n_{2}t^{\frac{1}{2}}, \quad t\in[0,1], \\ &m_{3}t^{\frac{1}{6}}\le x^{*}_{2}(t) \le n_{3}t^{\frac{1}{6}},\qquad m_{4}t^{\frac{1}{6}}\le y^{*}_{2}(t) \le n_{4}t^{\frac{1}{6}}, \quad t\in[0,1]. \end{aligned} $$

(3.19)

By simple computation, problem (3.18) is equivalent to the following multipoint boundary value problem:

$$ \textstyle\begin{cases} -\mathscr {D}_{t}^{\frac{4}{3}} (\varphi_{\frac{5}{2}} (-\mathscr {D}_{t}^{\frac{3}{2}}x_{1} ) )(t) \\ \quad =(x_{1}+x_{2})^{2} +{(x_{1}+x_{2})^{2}}(x_{1}+x_{2}+1)\ln ( 1+\frac {1}{1+x_{1}+x_{2}} )+1, \\ -\mathscr {D}_{t}^{\frac{6}{5}} (\varphi_{\frac{3}{2}} (-\mathscr {D}_{t}^{\frac{7}{6}}x_{2} ) )(t) =\ln(2+x_{1}+x_{2}), \quad t\in(0,1), \\ x_{1}(0)=0,\qquad \mathscr {D}_{t}^{\frac{3}{2}}x_{1}(0)=\mathscr {D}_{t}^{\frac{3}{2}}x_{1}(1)=0,\qquad x_{1}(1)=2x_{1} (\frac{1}{2} )-x_{1} (\frac{3}{4} ), \\ x_{2}(0)=0,\qquad \mathscr {D}_{t}^{\frac{7}{6}}x_{2}(0)=\mathscr {D}_{t}^{\frac{7}{6}}x_{2}(1)=0, \qquad x_{2}(1)=\frac{3}{2}x (\frac{1}{3} )-x (\frac{2}{3} ). \end{cases} $$

Let

$$\alpha_{1}=\frac{3}{2}, \qquad \alpha_{2}= \frac{7}{6},\qquad \beta_{1}=\frac{4}{3}, \qquad \beta _{2}=\frac{6}{5},\qquad p_{1}=\frac{5}{2}, \qquad p_{2}=\frac{3}{2}, $$

and

$$f_{1}(s,t)={(s+t)^{2}}(s+t+1)\ln \biggl( 1+ \frac{1}{1+s+t} \biggr) +(s+t)^{2}+1, \qquad f_{2}(s,t)= \ln(2+s+t). $$

Firstly, we have

$$\begin{aligned}& \mathcal{A}_{1}= \int_{0}^{1}t^{\alpha_{1}-1}\,dA_{1}(t)=2 \times \biggl(\frac{1}{2} \biggr)^{\frac{1}{2}}- \biggl(\frac{3}{4} \biggr)^{\frac{1}{2}}=0.5482< 1, \\& \mathcal{A}_{2}= \int_{0}^{1}t^{\alpha_{2}-1}\,dA_{2}(t)= \frac{3}{2}\times \biggl(\frac{1}{3} \biggr)^{\frac{1}{6}}- \biggl( \frac{2}{3} \biggr)^{\frac{1}{6}}=0.3053< 1, \end{aligned}$$

and by simple computation, we have \(\mathcal{G}_{A_{i}}(s)\ge0\), \(i=1,2\), and so (H0) holds.

Obviously, \(f_{1},f_{2}: [0,+\infty)\times[0,+\infty)\to[0,+\infty)\) are continuous and nondecreasing in the first variable and second variable and \(f_{i}(0,0)\neq0\), \(i=1,2\). Noticing that

$$\begin{aligned}& \sup_{{s+t\ge1}}\frac{f_{1} ( s,t ) }{(s+t)^{2}}\le\sup_{{s+t\ge1}} \biggl\{ \ln \biggl( 1+\frac{1}{1+s+t} \biggr)^{1+s+t}+2 \biggr\} \le 3, \\& \sup_{{s+t\ge1}}\frac{f_{2} ( s,t ) }{(s+t)}=\sup_{{s+t\ge1}} \frac{\ln(2+s+t) }{(s+t)}\le\ln3\le3, \end{aligned}$$

and then (2.11) is satisfied with \(\epsilon_{1}=2>\frac{1}{q_{1}-1}=\frac{3}{2}\), \(\epsilon_{2}=1>\frac{1}{q_{2}-1}=\frac{1}{2}\), \(M=3\). Thus, by Theorem 3.1, system (3.18) has maximal and minimal solutions which satisfy (3.19).