Existence and multiplicity of positive solutions to a system of fractional difference equations with parameters

Abstract

We consider a fractional difference-sum boundary problem for a system of fractional difference equations with parameters. Using the Banach fixed point theorem, we prove the existence and uniqueness of solutions. We also prove the existence of at least one and two solutions by using the Krasnoselskii’s fixed point theorem for a cone map. Finally, we give some examples to illustrate our results.

1 Introduction

Fractional difference calculus is quite new to researchers. It has been used in mathematical models that explain many real-life situations, for example, economics, electrical networks, and queuing problems (see [1–3] and the references therein). Basic definitions and properties of fractional difference calculus were presented in [4], and discrete fractional boundary value problems have been found in [5–33]. However, the studies of a system of fractional boundary value problems are quite rare (see [34–42]).

For an extension of the research work in this area, in this paper, we study the following system of fractional difference equations with parameters:

$$\begin{aligned} \begin{gathered} -\Delta ^{\alpha _{1}} u_{1}(t) = \lambda _{1}F_{1} \bigl[t+\alpha _{1}-1,t+ \alpha _{2}-1,u_{1}(t+\alpha _{1}-1), u_{2}(t+\alpha _{2}-1) \bigr], \\ -\Delta ^{\alpha _{2}} u_{2}(t) = \lambda _{2}F_{2} \bigl[t+\alpha _{1}-1,t+ \alpha _{2}-1,u_{1}(t+ \alpha _{1}-1), u_{2}(t+\alpha _{2}-1) \bigr], \end{gathered} \end{aligned}$$

(1.1)

subject to nonlocal fractional difference-sum boundary conditions of the form

$$\begin{aligned} \begin{gathered} \Delta ^{-\beta _{1}} u_{1}( \alpha _{1}+\beta _{1}-3) = \Delta ^{ \gamma _{1}} u_{1}(\alpha _{1}-\gamma _{1}-2) = 0, \\ \Delta ^{-\beta _{2}} u_{2}(\alpha _{2}+\beta _{2}-3) = \Delta ^{ \gamma _{2}} u_{2}(\alpha _{2}-\gamma _{2}-2) = 0, \\ u_{1}(T+\alpha _{1}) = \chi _{1} u_{1}(\eta _{1}), \quad \eta _{1} \in \mathbb{N}_{\alpha _{1}-2,T+\alpha _{1}-1}, \\ u_{2}(T+\alpha _{2}) = \chi _{2}u_{2}( \eta _{2}), \quad \eta _{2} \in \mathbb{N}_{\alpha _{2}-2,T+\alpha _{2}-1}, \end{gathered} \end{aligned}$$

(1.2)

where $t\in \mathbb{N}_{0,T}:=\{0,1,\ldots,T\}$, $\alpha _{i}\in (2,3]$, $\beta _{i},\gamma _{i}\in (0,1)$, $i=1,2$. Moreover, we suppose that the following assumptions hold:

$$\begin{aligned} &(A1) \quad F_{i}\in C \bigl(\mathbb{N}_{\alpha _{1}-1,T+\alpha _{1}-1} \times \mathbb{N}_{\alpha _{2}-1,T+\alpha _{2}-1}\times (0,\infty ) \times (0,\infty ), (0,\infty ) \bigr); \\ &\hphantom{(A1)\quad } \text{where }\mathbb{N}_{\alpha _{i}-1,T+\alpha _{i}-1}:=\{ \alpha _{i}-1, \alpha _{i}, \alpha _{i}+1, \ldots, T+ \alpha _{i}-1 \}; \\ &(A2) \quad 0< \chi _{i}\eta _{i}^{\underline{\alpha _{i}-1}} < (T+\alpha _{i})^{ \underline{\alpha _{i}-1}}; \\ &(A3) \quad \lambda _{1},\lambda _{2} \text{ are positive parameters}; \\ &(A4) \quad F_{i} ( t_{1},t_{2},u_{1},u_{2} ) >0 \text{ for } u_{1},u_{2}>0, t_{i} \in \mathbb{N}_{\alpha _{i}-1,T+\alpha _{i}-1}. \end{aligned}$$

For convenience, we use the following notations ($i=1,2$):

$$\begin{aligned}& F_{i}^{0} = \lim_{u_{1},u_{2}\rightarrow 0^{+}} \biggl[ \max _{ t_{i} \in \mathbb{N}_{\alpha _{i}-1,T+\alpha _{i}-1}} \frac{F_{i} ( t_{1},t_{2},u_{1},u_{2} )}{u_{1}+u_{2}} \biggr], \\& F_{i}^{\infty } = \lim_{u_{1},u_{2}\rightarrow \infty } \biggl[ \min _{ t_{i} \in \mathbb{N}_{\alpha _{i}-1,T+\alpha _{i}-1}} \frac{F_{i} ( t_{1},t_{2},u_{1},u_{2} )}{u_{1}+u_{2}} \biggr]. \end{aligned}$$

We organize our paper as follows. In Sect. 2, we recall some definitions and basic lemmas and present some properties of the fractional difference operators. In this section, we also derive a representation for the solution to (1.1)–(1.2) by converting the problem to equivalent summation equations. In Sect. 3, we prove the existence and uniqueness result for problem (1.1)–(1.2) by using the Banach fixed point theorem. In Sect. 4, we prove the existence of at least one and two solutions for problem (1.1)–(1.2) by using the Krasnoselskii fixed point theorem in a cone map. In the last section, we provide some examples to illustrate our results.

Theorem 1.1

([43], Krasnoselskii’s fixed point theorem)

Let E be a Banach space, and let $K\subset E$be a cone. Let $\varOmega _{1}$and $\varOmega _{2}$be open subsets of E such that $0\in \varOmega _{1}$and $\overline{\varOmega }_{1}\subset \varOmega _{2}$, and let

$$ A:K\cap (\overline{\varOmega }_{2}\setminus \varOmega _{1})\longrightarrow K $$

be a completely continuous operator such that

• (i) $\|Au\|\leqslant \|u\|$, $u\in K\cap \partial \varOmega _{1}$, and $\|Au\|\geqslant \|u\|$, $u\in K\cap \partial \varOmega _{2}$, or

• (ii) $\|Au\|\geqslant \|u\|$, $u\in K\cap \partial \varOmega _{1}$, and $\|Au\|\leqslant \|u\|$, $u\in K\cap \partial \varOmega _{2}$.

Then A has a fixed point in $K\cap (\overline{\varOmega }_{2}\setminus \varOmega _{1})$.

Theorem 1.2

([44], Arzelá–Ascoli theorem)

A set of functions in $C[a,b]$with the sup norm is relatively compact if and only it is uniformly bounded and equicontinuous on $[a,b]$.

Theorem 1.3

([44])

If a set is closed and relatively compact, then it is compact.

2 Preliminaries

In this section, we provide some notations, definitions, and lemmas, which are used in the main results.

Definition 2.1

The generalized falling function is defined by $t^{\underline{\alpha }}:= \frac{\varGamma (t+1)}{\varGamma (t+1-\alpha )}$ for any t and α for which the right-hand side is defined. If $t+1-\alpha$ is a pole of the Gamma function and $t+1$ is not a pole, then $t^{\underline{\alpha }}=0$.

Definition 2.2

For $\alpha >0$ and f defined on $\mathbb{N}_{a}$, the α-order fractional sum of f is defined by

$$ \Delta ^{-\alpha }f(t)=\Delta ^{-\alpha }f(t;a):= \frac{1}{\varGamma (\alpha )}\sum_{s=a}^{t-\alpha }\bigl(t- \sigma (s)\bigr)^{ \underline{\alpha -1}}f(s), $$

where $t\in \mathbb{N}_{a+\alpha }$ and $\sigma (s)=s+1$.

Definition 2.3

For $\alpha >0$ and f defined on $\mathbb{N}_{a}$, the αth-order Riemann–Liouville fractional difference of f is defined by

$$ \Delta ^{\alpha }f(t) := \Delta ^{N}\Delta ^{-(N-\alpha )}f(t)= \frac{1}{\varGamma (-\alpha )}\sum_{s=a}^{t+\alpha }\bigl(t- \sigma (s)\bigr)^{ \underline{-\alpha -1}} f(s), $$

where $t \in \mathbb{N}_{a+N-\alpha }$, and $N \in \mathbb{N}$ is chosen so that $0\leq {N-1}<\alpha \leq N$.

Lemma 2.1

([5])

For $0\leq N-1<\alpha \leq N$,

$$ \Delta ^{-\alpha }\Delta ^{\alpha }y(t)=y(t)+C_{1}t^{ \underline{\alpha -1}} +C_{2}t^{\underline{\alpha -2}}+\cdots +C_{N}t^{ \underline{\alpha -N}} $$

for some $C_{i}\in \mathbb{R}$, $1\leq i\leq N$.

The following is a solution of a linear variant of the boundary value problem (1.1).

Lemma 2.2

Suppose that $(A1)$–$(A3)$hold. For $i\in \{1,2\}$, let $\alpha _{i}\in (2, 3]$and $\beta _{i}$, $\gamma _{i}\in (0, 1)$be given constants, and let $h_{i}\in C (\mathbb{N}_{\alpha _{i}-1,T+\alpha _{i}-1}, \mathbb{R}^{+} )$be given functions. The problem

$$\begin{aligned}& -\Delta ^{\alpha _{i}} u_{i}(t) = h_{i}(t+\alpha _{i}-1),\quad t\in \mathbb{N}_{0,T}, \end{aligned}$$

(2.1)

$$\begin{aligned}& \Delta ^{-\beta _{i}}u_{i}(\alpha _{i}+\beta _{i}-3) = \Delta ^{ \gamma _{i}}u_{i}(\alpha _{i}-\gamma _{i}-2) = 0, \end{aligned}$$

(2.2)

$$\begin{aligned}& u_{i}(T+\alpha _{i}) = \chi _{i}u_{i}( \eta _{i}), \eta _{i}\in \mathbb{N}_{\alpha _{i}-2,T+\alpha _{i}-1} \end{aligned}$$

(2.3)

has the unique solution

$$\begin{aligned} u_{i}(t_{i})&= \frac{t_{i}^{\underline{\alpha _{i}-1}}}{ [ (T+\alpha _{i})^{\underline{\alpha _{i}-1}}-\chi _{i} \eta _{i}^{\underline{\alpha _{i}-1}} ] \varGamma (\alpha _{i}) } \Biggl\{ \sum _{s=0}^{T}\bigl(T+\alpha _{i}-\sigma (s) \bigr)^{ \underline{\alpha _{i}-1}} h_{i}(s+\alpha _{i}-1) \\ &\quad {}- \chi _{i} \sum_{s=0}^{\eta _{i}-\alpha _{i} } \bigl(\eta _{i}-\sigma (s)\bigr)^{ \underline{\alpha _{i}-1}} h_{i}(s+ \alpha _{i}-1) \Biggr\} \\ &\quad {}-\frac{1}{\varGamma (\alpha _{i})}\sum_{s=0}^{t_{i}-\alpha _{i}} \bigl(t_{i}- \sigma (s)\bigr)^{\underline{\alpha _{i}-1}} h_{i}(s+\alpha _{i}-1) \end{aligned}$$

(2.4)

for $t_{i}\in \mathbb{N}_{\alpha _{i}-3,T+\alpha _{i}}$.

Proof

Using Lemma 2.1 and the fractional sum of order $\alpha _{i}\in (2,3]$, $i\in \{1,2\}$, for (2.1), we obtain

$$\begin{aligned} u_{i}(t_{i})=C_{1i}t_{i}^{\underline{\alpha _{i}-1}}+C_{2i}t_{i}^{ \underline{\alpha _{i}-2}}+C_{3i}t_{i}^{\underline{\alpha _{i}-3}} - \frac{1}{\varGamma (\alpha _{i})}\sum_{s=0}^{t_{i}-\alpha _{i}} \bigl(t_{i}- \sigma (s)\bigr)^{\underline{\alpha _{i}-1}} h_{i}(s+\alpha _{i}-1) \end{aligned}$$

(2.5)

for $t_{i}\in \mathbb{N}_{\alpha _{i}-3,T+\alpha _{i}}$.

Next, applying the fractional sum of order $\beta _{i}\in (0,1)$, $i\in \{1,2\}$, to (2.5), we have

$$\begin{aligned} \Delta ^{-\beta _{i}}u_{i}(t_{i}) =& \frac{1}{\varGamma (\beta _{i})} \sum_{s=\alpha _{i}-3}^{t_{i}-\beta _{i}} \bigl(t_{i}-\sigma (s)\bigr)^{ \underline{\beta _{i}-1}} \bigl[ C_{1i}s^{\underline{\alpha _{i}-1}}+C_{2i}s^{ \underline{\alpha _{i}-2}}+C_{3i}s^{\underline{\alpha _{i}-3}} \bigr] \\ &{}-\frac{1}{\varGamma (\alpha _{i})\varGamma (\beta _{i})} \sum_{r=\alpha _{i}}^{t_{i}- \beta _{i}}\sum _{s=0}^{r-\alpha _{i}} \bigl(t_{i}-\sigma (r) \bigr)^{ \underline{\beta _{i}-1}} \bigl(r-\sigma (s)\bigr)^{\underline{\alpha _{i}-1}} h_{i}(s+ \alpha _{i}-1) \end{aligned}$$

(2.6)

for $t_{i}\in \mathbb{N}_{\alpha _{i}+\beta _{i}-3,T+\alpha _{i}+\beta _{i}}$.

Taking the fractional difference of order $\gamma _{i}\in (0,1)$, $i\in \{1,2\}$, of (2.5), we obtain

$$\begin{aligned} \Delta ^{\gamma _{i}}u_{i}(t_{i}) =& \frac{1}{\varGamma (-\gamma _{i})} \sum_{s=\alpha _{i}-3}^{t_{i}+\gamma _{i}} \bigl(t_{i}-\sigma (s)\bigr)^{ \underline{-\gamma _{i}-1}} \bigl[ C_{1i}s^{\underline{\alpha _{i}-1}}+C_{2i}s^{ \underline{\alpha _{i}-2}}+C_{3i}s^{\underline{\alpha _{i}-3}} \bigr] \\ &{}-\frac{1}{\varGamma (\alpha _{i})\varGamma (-\gamma _{i})} \sum_{r= \alpha _{i}}^{t_{i}+\gamma _{i}} \sum_{s=0}^{r-\alpha _{i}} \bigl(t_{i}- \sigma (r)\bigr)^{\underline{-\gamma -1}} \bigl(r-\sigma (s)\bigr)^{ \underline{\alpha _{i}-1}} h_{i}(s+\alpha _{i}-1) \end{aligned}$$

(2.7)

for $t_{i}\in \mathbb{N}_{\alpha _{i}-\gamma _{i}-2,T+\alpha _{i}-\gamma _{i}}$.

Using the boundary condition $\Delta ^{-\beta _{i}}u_{i}(\alpha _{i}+\beta _{i}-3)=0$ in (2.2), we find that $C_{3i}=0$.

Then we have

$$\begin{aligned} u_{i}(t_{i})=C_{1i}t_{i}^{\underline{\alpha _{i}-1}}+C_{2i}t_{i}^{ \underline{\alpha _{i}-2}} -\frac{1}{\varGamma (\alpha _{i})}\sum_{s=0}^{t_{i}- \alpha _{i}} \bigl(t_{i}-\sigma (s)\bigr)^{\underline{\alpha _{i}-1}} h_{i}(s+ \alpha _{i}-1). \end{aligned}$$

(2.8)

From the boundary condition $\Delta ^{\gamma _{i}}u_{i}(\alpha _{i}-\gamma _{i}-2)=0$ in (2.2) we have $C_{2i}=0$.

Therefore

$$\begin{aligned} u_{i}(t_{i})=C_{1i}t_{i}^{\underline{\alpha _{i}-1}} - \frac{1}{\varGamma (\alpha _{i})}\sum_{s=0}^{t_{i}-\alpha _{i}} \bigl(t_{i}- \sigma (s)\bigr)^{\underline{\alpha _{i}-1}} h_{i}(s+\alpha _{i}-1). \end{aligned}$$

(2.9)

By using the boundary condition (2.3) we obtain

$$\begin{aligned} C_{1i}& = \frac{1}{ [ (T+\alpha _{i})^{\underline{\alpha _{i}-1}}-\chi _{i} \eta _{i}^{\underline{\alpha _{i}-1}} ] \varGamma (\alpha _{i}) } \Biggl\{ \sum _{s=0}^{T}\bigl(T+\alpha _{i}-\sigma (s) \bigr)^{ \underline{\alpha _{i}-1}} h_{i}(s+\alpha _{i}-1) \\ &\quad {}- \chi _{i} \sum_{s=0}^{\eta _{i}-\alpha _{i} } \bigl(\eta _{i}-\sigma (s)\bigr)^{ \underline{\alpha _{i}-1}} h_{i}(s+ \alpha _{i}-1) \Biggr\} . \end{aligned}$$

(2.10)

Finally, substituting $C_{1i}$ into (2.9), we obtain (2.4). The proof is complete. □

Corollary 2.1

Problem (2.1)–(2.3) has the unique solution, which is of the form

$$\begin{aligned} u_{i}(t_{i}) = \sum _{s=0}^{T} G_{i}(t_{i},s) h_{i}(s+\alpha _{i}-1) \end{aligned}$$

(2.11)

for $t_{i}\in \mathbb{N}_{\alpha _{i}-3,T+\alpha _{i}}$, $i=1,2$, where

$$ G_{i}(t_{i},s):=\frac{1}{\varGamma (\alpha _{i})} \textstyle\begin{cases} g_{1}(t_{i},s), & s\in \mathbb{N}_{0,t_{i}-\alpha _{i}}\cap \mathbb{N}_{0,\eta _{i}-\alpha _{i}}, \\ g_{2}(t_{i},s), & s\in \mathbb{N}_{t_{i}-\alpha _{i}+1,\eta _{i}- \alpha _{i}}, \\ 3_{2}(t_{i},s), & s\in \mathbb{N}_{\eta _{i}-\alpha _{i}+1,t_{i}- \alpha _{i}}, \\ g_{4}(t_{i},s), & s\in \mathbb{N}_{t_{i}-\alpha _{i}+1,T}\cap \mathbb{N}_{\eta _{i}-\alpha _{i}+1,T}, \end{cases}$$

(2.12)

with

$$\begin{aligned} \begin{gathered} g_{1}(t_{i},s) := \frac{t_{i}^{\underline{\alpha _{i}-1}}}{ [ (T+\alpha _{i})^{\underline{\alpha _{i}-1}}-\chi _{i} \eta _{i}^{\underline{\alpha _{i}-1}} ] } \bigl\{ \bigl(T+\alpha _{i}-\sigma (s) \bigr)^{\underline{\alpha _{i}-1}} - \chi _{i}\bigl(\eta _{i}-\sigma (s) \bigr)^{\underline{\alpha _{i}-1}} \bigr\} \\ \hphantom{g_{1}(t_{i},s) :=}{}-\bigl(t_{i}-\sigma (s)\bigr)^{\underline{\alpha _{i}-1}}, \\ g_{2}(t_{i},s) := \frac{t_{i}^{\underline{\alpha _{i}-1}}}{ [ (T+\alpha _{i})^{\underline{\alpha _{i}-1}}-\chi _{i} \eta _{i}^{\underline{\alpha _{i}-1}} ] } \bigl\{ \bigl(T+ \alpha _{i}-\sigma (s)\bigr)^{\underline{\alpha _{i}-1}} - \chi _{i}\bigl( \eta _{i}-\sigma (s)\bigr)^{\underline{\alpha _{i}-1}} \bigr\} , \\ g_{3}(t_{i},s) := \frac{t_{i}^{\underline{\alpha _{i}-1}}}{ [ (T+\alpha _{i})^{\underline{\alpha _{i}-1}}-\chi _{i} \eta _{i}^{\underline{\alpha _{i}-1}} ] } \bigl(T+\alpha _{i}-\sigma (s)\bigr)^{\underline{\alpha _{i}-1}}-\bigl(t_{i}- \sigma (s) \bigr)^{\underline{\alpha _{i}-1}}, \\ g_{4}(t_{i},s) := \frac{t_{i}^{\underline{\alpha _{i}-1}}}{ [ (T+\alpha _{i})^{\underline{\alpha _{i}-1}}-\chi _{i} \eta _{i}^{\underline{\alpha _{i}-1}} ] } \bigl(T+\alpha _{i}-\sigma (s)\bigr)^{\underline{\alpha _{i}-1}}. \end{gathered} \end{aligned}$$

(2.13)

Green’s function (2.12) has the following properties.

Proposition 2.1

([13])

For $i=1,2$, let $G_{i}(t_{i},s)$be Green’s function given in (2.12)–(2.13). Then for all $(t_{i},s)\in \mathbb{N}_{\alpha _{i}-3,T+\alpha _{i}}\times \mathbb{N}_{0,T}$,

$$ G_{i}(t_{i},s)\geq 0. $$

Proposition 2.2

([13])

For $i=1,2$, let $G_{i}(t_{i},s)$be Green’s function given in (2.12)–(2.13). Suppose that for given $\eta _{i}\in \mathbb{N}_{\alpha _{i}-2,T+\alpha _{i}-1}$and $\alpha _{i} \in (2,3]$, $\chi _{i}$satisfies the inequality

$$\begin{aligned} 0\leq \chi _{i}\leq \min_{(t_{i},s)\in \mathbb{N}_{\alpha _{i}-3,T+ \alpha _{i}}\times \mathbb{N}_{0,T}} \biggl\lbrace \frac{(T+\alpha _{i})^{\underline{\alpha _{i}-1}}}{\eta _{i}^{\underline{\alpha _{i}-1}}} - \frac{t_{i}^{\underline{\alpha _{i}-2}}(T+\alpha _{i}-\sigma (s))^{\underline{\alpha _{i}-1}} }{\eta _{i}^{\underline{\alpha _{i}-1}}(t_{i}-\sigma (s))^{\underline{\alpha _{i}-2}} } \biggr\rbrace . \end{aligned}$$

Then

$$\begin{aligned} \max_{(t_{i},s)\in \mathbb{N}_{\alpha _{i}-3,T+\alpha _{i}}\times \mathbb{N}_{0,T}} G_{i}(t_{i},s)=G_{i}(s+ \alpha -1,s). \end{aligned}$$

Proposition 2.3

([13])

For $i=1,2$, let $G_{i}(t_{i},s)$be Green’s function given in (2.12)–(2.13). Then

$$\begin{aligned} \min_{t_{i}\in [\frac{1}{4}(T+\alpha _{i}),\frac{3}{4}(T+\alpha _{i})] } G_{i}(t_{i},s)\geq \theta _{i} \max_{(t_{i},s)\in \mathbb{N}_{ \alpha _{i}-3,T+\alpha _{i}}\times \mathbb{N}_{0,T}}G_{i}(t_{i},s)= \theta _{i} G_{i}(s+\alpha -1,s), \end{aligned}$$

where

$$\begin{aligned} \theta _{i}&:=\min \biggl\lbrace \frac{ ( \frac{1}{4}(T+\alpha _{i}) )^{\underline{\alpha _{i}-1}}}{ ( T+\alpha _{i} )^{\underline{\alpha _{i}-1}}} , \\ &\quad \frac{1}{ ( \frac{3}{4}(T+\alpha _{i}) )^{\underline{\alpha _{i}-1}}} \biggl[ \biggl( \frac{3}{4}(T+\alpha _{i}) \biggr)^{ \underline{\alpha _{i}-1}} - \frac{ ( \frac{3}{4}(T+\alpha _{i}) )^{\underline{\alpha _{i}-1}} ( T+\alpha _{i} )^{\underline{\alpha _{i}-1}}}{ ( T+\alpha _{i} )^{\underline{\alpha _{i}-1}}} \biggr] \biggr\rbrace , \end{aligned}$$

(2.14)

and $\theta _{i}$satisfy the inequality $0<\theta _{i}<1$.

3 Existence and uniqueness of solution

In this section, we apply the Banach fixed point theorem to prove the existence and uniqueness result for problem (1.1)–(1.2). For each $i,j \in \{1,2\}$, we let $E_{i}=C ( \mathbb{N}_{\alpha _{i}-3,T+\alpha _{i}}, \mathbb{R} )$ be the Banach space for all functions on $\mathbb{N}_{\alpha _{i}-3,T+\alpha _{i}}$ with the norm $\|u_{i}\|=\max_{t_{i} \in \mathbb{N}_{\alpha _{i}-3,T+ \alpha _{i}}}|u_{i}(t_{i})|$. The product space $\mathcal{U}=E_{1}\times E_{2}$ is a Banach space with the norm

$$ \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{\mathcal{U}} = \Vert u_{1} \Vert + \Vert u_{2} \Vert . $$

Next, define the operator ${\mathcal{T}}:{\mathcal{U}}\rightarrow {\mathcal{U}}$ by

$$\begin{aligned} \bigl({\mathcal{T}}(u_{1},u_{2}) \bigr) (t_{1},t_{2}) := \bigl( \bigl({ \mathcal{T}}_{1}(u_{1},u_{2}) \bigr) (t_{1},t_{2}) , \bigl({\mathcal{T}}_{2}(u_{1},u_{2}) \bigr) (t_{1},t_{2}) \bigr), \end{aligned}$$

(3.1)

and for $i,j\in \{1,2\}$, $i\neq j$,

$$\begin{aligned} &\bigl({\mathcal{T}}_{i}(u_{1},u_{2}) \bigr) (t_{1},t_{2}) \\ &\quad := \frac{t_{i}^{\underline{\alpha _{i}-1}}}{ [ (T+\alpha _{i})^{\underline{\alpha _{i}-1}}-\chi _{i} \eta _{i}^{\underline{\alpha _{i}-1}} ] \varGamma (\alpha _{i}) } \\ &\qquad {}\times \Biggl\{ \sum_{s=0}^{T} \bigl(T+\alpha _{i}-\sigma (s)\bigr)^{ \underline{\alpha _{i}-1}} F_{i} \bigl[ s+\alpha _{i}-1,t_{j},u_{i}(s+ \alpha _{i}-1),u_{j}(t_{j}) \bigr] \\ &\qquad {}- \chi _{i} \sum_{s=0}^{\eta _{i}-\alpha _{i} } \bigl(\eta _{i}-\sigma (s)\bigr)^{ \underline{\alpha _{i}-1}} F_{i} \bigl[ s+\alpha _{i}-1,t_{j},u_{i}(s+ \alpha _{i}-1),u_{j}(t_{j}) \bigr] \Biggr\} \\ &\qquad {}-\frac{1}{\varGamma (\alpha _{i})}\sum_{s=0}^{t_{i}-\alpha _{1}} \bigl(t_{i}- \sigma (s)\bigr)^{\underline{\alpha _{i}-1}} F_{i} \bigl[ s+ \alpha _{i}-1,t_{j},u_{i}(s+ \alpha _{i}-1),u_{j}(t_{j}) \bigr] \end{aligned}$$

(3.2)

for $t_{i}\in \mathbb{N}_{\alpha _{i}-3,T+\alpha _{i}}$ and $t_{j}\in \mathbb{N}_{\alpha _{j}-1,T+\alpha _{j}-1}$. Obviously, problem (1.1)-(1.2) has a solution if and only if the operator ${\mathcal{T}}$ has a fixed point.

Theorem 3.1

Suppose that $(A1)$–$(A4)$hold. In addition, suppose there exist constants $M_{i},N_{i}>0$for $i=1,2$such that

$$ \bigl\vert F_{i} [ t_{1},t_{2},u_{1},u_{2} ] -F_{i} [ t_{1},t_{2},v_{1},v_{2} ] \bigr\vert \leq M_{i} \vert u_{1}-v_{1} \vert +N_{i} \vert u_{2}-v_{2} \vert $$

(3.3)

for all $t_{i}\in \mathbb{N}_{\alpha _{i}-1,T+\alpha _{i}-1}$and $(u_{1},u_{2}),(v_{1},v_{2})\in \mathcal{U}$. If

$$\begin{aligned} L_{1}\varOmega _{1}+L_{2} \varOmega _{2} < &1, \end{aligned}$$

(3.4)

where

$$\begin{aligned}& L_{i} = \max \{M_{i},N_{i}\}, \\& \varOmega _{i} = \frac{\lambda _{i}}{\varGamma (\alpha _{i}+1)} \biggl[ \biggl\vert \frac{(T+\alpha _{i})^{\underline{\alpha _{i}}}-\chi _{i} \eta _{i}^{\underline{\alpha _{i}}}}{(T+\alpha _{i})^{\underline{\alpha _{i}-1}}-\chi _{i} \eta _{i}^{\underline{\alpha _{i}-1}}} \biggr\vert (T+\alpha _{i})^{\underline{\alpha _{i}-1}} +(T+ \alpha _{i})^{ \underline{\alpha _{i}}} \biggr], \end{aligned}$$

then problem (1.1)–(1.2) has a unique solution.

Proof

Let us prove that ${\mathcal{T}}$ is a contraction mapping. For $i,j\in \{1,2\}$, $i\neq j$, denote

$$\begin{aligned} {\mathcal{F}}_{i} \vert u-v \vert (s,t_{j})&:= \bigl\vert F_{i} \bigl[ s+\alpha _{i}-1,t_{j},u_{i}(s+ \alpha _{i}-1),u_{j}(t_{j}) \bigr] \\ &\quad {} -F_{i} \bigl[ s+\alpha _{i}-1,t_{j},v_{i}(s+ \alpha _{i}-1),v_{j}(t_{j}) \bigr] \bigr\vert . \end{aligned}$$

For $t_{i}\in \mathbb{N}_{\alpha _{i}-3,T+\alpha _{i}}$, $t_{j}\in \mathbb{N}_{\alpha _{j}-1,T+\alpha _{j}-1}$, and $(u_{1},u_{2}),(v_{1},v_{2})\in \mathcal{U}$, we find that

$$\begin{aligned}& \bigl\vert \bigl({\mathcal{T}}_{i}(u_{1},u_{2}) \bigr) (t_{1},t_{2})- \bigl({\mathcal{T}}_{i}(v_{1},v_{2}) \bigr) (t_{1},t_{2}) \bigr\vert \\& \quad \leq \frac{\lambda _{i}t_{i}^{\underline{\alpha _{i}-1}}}{ [ (T+\alpha _{i})^{\underline{\alpha _{i}-1}}-\chi _{i} \eta _{i}^{\underline{\alpha _{i}-1}} ] \varGamma (\alpha _{i}) } \Biggl\vert \sum_{s=0}^{T} \bigl(T+\alpha _{i}-\sigma (s)\bigr)^{ \underline{\alpha _{i}-1}} { \mathcal{F}}_{i} \vert u-v \vert (s,t_{j}) \\& \qquad {}- \chi _{i} \sum_{s=0}^{\eta _{i}-\alpha _{i} } \bigl(\eta _{i}-\sigma (s)\bigr)^{ \underline{\alpha _{i}-1}} {\mathcal{F}}_{i} \vert u-v \vert (s,t_{j}) \Biggr\vert + \frac{\lambda _{i}}{\varGamma (\alpha _{i})}\sum_{s=0}^{t_{i}-\alpha _{1}} \bigl(t_{i}- \sigma (s)\bigr)^{\underline{\alpha _{i}-1}} {\mathcal{F}}_{i} \vert u-v \vert (s,t_{j}) \\& \quad \leq \frac{\lambda _{i} [ M_{i} \vert u_{1}+v_{1} \vert +N_{i} \vert u_{2}-v_{2} \vert ] }{\varGamma (\alpha _{i})} \Biggl[ \frac{(T+\alpha _{i})^{\underline{\alpha _{i}-1}}}{ t_{i}^{\underline{\alpha _{i}-1}}-\chi _{i} \eta _{i}^{\underline{\alpha _{i}-1}} } \\& \qquad {}\times \Biggl\vert \sum_{s=0}^{T} \bigl(T+\alpha _{i}-\sigma (s)\bigr)^{ \underline{\alpha _{i}-1}} - \chi _{i} \sum_{s=0}^{\eta _{i}-\alpha _{i} }\bigl(\eta _{i}-\sigma (s)\bigr)^{\underline{\alpha _{i}-1}} \Biggr\vert + \sum _{s=0}^{t_{i}- \alpha _{1}}\bigl(t_{i}-\sigma (s) \bigr)^{\underline{\alpha _{i}-1}} \Biggr] \\& \quad \leq \frac{\lambda _{i}L_{i} [ \vert u_{1}+v_{1} \vert + \vert u_{2}-v_{2} \vert ] }{\varGamma (\alpha _{i}+1)} \biggl[ \biggl\vert \frac{(T+\alpha _{i})^{\underline{\alpha _{i}}}-\chi _{i} \eta _{i}^{\underline{\alpha _{i}}}}{(T+\alpha _{i})^{\underline{\alpha _{i}-1}}-\chi _{i} \eta _{i}^{\underline{\alpha _{i}-1}}} \biggr\vert (T+\alpha _{i})^{\underline{\alpha _{i}-1}} +(T+\alpha _{i})^{ \underline{\alpha _{i}}} \biggr] \\& \quad \leq L_{i}\varOmega _{i} \bigl\Vert (u_{1},v-1,u_{2}-v_{2}) \bigr\Vert _{ \mathcal{U}}. \end{aligned}$$

Therefore

$$\begin{aligned} \bigl\Vert \bigl({\mathcal{T}}(u_{1},u_{2}) \bigr)- \bigl({\mathcal{T}}(v_{1},v_{2}) \bigr) \bigr\Vert _{ \mathcal{U}} =& \bigl\Vert \bigl( {\mathcal{T}}_{1}(u_{1},u_{2})-{ \mathcal{T}}_{1}(v_{1},v_{2}) , { \mathcal{T}}_{2}(u_{1},u_{2})-{ \mathcal{T}}_{2}(v_{1},v_{2}) \bigr) \bigr\Vert _{ \mathcal{U}} \\ =& \bigl\Vert {\mathcal{T}}_{1}(u_{1},u_{2})-{ \mathcal{T}}_{1}(v_{1},v_{2}) \bigr\Vert + \bigl\Vert {\mathcal{T}}_{2}(u_{1},u_{2})-{ \mathcal{T}}_{2}(v_{1},v_{2}) \bigr\Vert \\ \leq & ( L_{1}\varOmega _{1}+L_{2}\varOmega _{2} ) \bigl\Vert (u_{1},v-1,u_{2}-v_{2}) \bigr\Vert _{ \mathcal{U}}. \end{aligned}$$

By (3.4) we find that ${\mathcal{T}}$ is a contraction. Therefore by the Banach fixed point theorem ${\mathcal{T}}$ has a fixed point, which is a unique solution of problem (1.1)–(1.2). □

4 Existence and multiplicity of positive solutions

In this section, we apply the Krasnoselskii fixed point theorem for a cone map to prove the existence and multiplicity of positive solutions for problem (1.1)–(1.2). The product space $\mathcal{U}$ and the norm $\|(u_{1},u_{2})\|_{\mathcal{U}}$ are defined in Sect. 3. Moreover, we define the cone $\mathcal{P} \subset \mathcal{U}$ as

$$\begin{aligned} \mathcal{P}:= \Bigl\lbrace (u_{1},u_{2})\in \mathcal{U}: u_{1},u_{2} \geq 0 \text{ and } \min_{t_{i}\in \mathbb{N}_{\alpha _{i}-3,T+ \alpha _{i}}} \bigl[ u_{1}(t_{1})+u_{2}(t_{2}) \bigr] \geq \theta \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{\mathcal{U}} \Bigr\rbrace , \end{aligned}$$

where $\theta :=\min \{\theta _{1},\theta _{2}\}$ with $\theta _{1}$, $\theta _{2}$ defined in (2.14).

Next, by Corollary 2.1 we also define the operator ${\mathcal{T}}:{\mathcal{U}}\rightarrow {\mathcal{U}}$ by

$$\begin{aligned} \bigl({\mathcal{T}}(u_{1},u_{2}) \bigr) (t_{1},t_{2}) =& \bigl( \bigl({ \mathcal{T}}_{1}(u_{1},u_{2}) \bigr) (t_{1},t_{2}) , \bigl({\mathcal{T}}_{2}(u_{1},u_{2}) \bigr) (t_{1},t_{2}) \bigr) \end{aligned}$$

(4.1)

and for $i,j\in \{1,2\}$, $i\neq j$,

$$\begin{aligned} \bigl({\mathcal{T}}_{i}(u_{1},u_{2}) \bigr) (t_{1},t_{2}):=& \lambda _{i} \sum _{s=0}^{T}G_{i}(t_{i},s) F_{i} \bigl[ s+\alpha _{i}-1,t_{j},u_{i}(s+ \alpha _{i}-1),u_{j}(t_{j}) \bigr] \end{aligned}$$

(4.2)

for $t_{i}\in \mathbb{N}_{\alpha _{i}-3,T+\alpha _{i}}$ and $t_{j}\in \mathbb{N}_{\alpha _{j}-1,T+\alpha _{j}-1}$. The positive solutions of problem (1.1)–(1.2) and the fixed points of the operator ${\mathcal{T}}$ in the cone $\mathcal{P}$ coincide.

Lemma 4.1

If $(A1)$–$(A4)$hold, then $\mathcal{T}(\mathcal{P})\subset \mathcal{P}$, and $\mathcal{T}:\mathcal{P}\rightarrow \mathcal{P}$is a completely continuous operator.

Proof

The continuity of $\mathcal{T}$ is obvious. To prove $\mathcal{T}(\mathcal{P})\subset \mathcal{P}$, we choose $(u_{1},u_{2}) \in \mathcal{P}$. Since, for $i=1,2$, $G_{i}(t_{i},s)\leq G_{i}(s+\alpha _{i}-1,s)$ for $s \in \mathbb{N}_{0,T}$ and $G_{i}(t_{i},s)\geq \theta _{i} G_{i}(s+\alpha -1,s)$ for $t_{i} \in [ \frac{1}{4}(T+\alpha _{i}),\frac{3}{4}(T+\alpha _{i}) ]$, we have

$$\begin{aligned} &\min_{ t_{i}\in [ \frac{1}{4}(T+\alpha _{i}),\frac{3}{4}(T+ \alpha _{i}) ] } {\mathcal{T}}_{1}(u_{1},u_{2}) (t_{1},t_{2}) \\ &\quad \geq \lambda _{1}\theta _{1} \sum _{s=0}^{T} G_{i}(s+\alpha _{i}-1,s) F_{1} \bigl[s+\alpha _{i}-1,t_{2},u_{1}(s+ \alpha _{i}-1),u_{2}(t_{2}) \bigr] \\ &\quad \geq \theta _{1} \bigl\Vert {\mathcal{T}}_{1}(u_{1},u_{2}) \bigr\Vert . \end{aligned}$$

Similarly, $\min_{ t_{i}\in [ \frac{1}{4}(T+\alpha _{i}), \frac{3}{4}(T+\alpha _{i}) ] } {\mathcal{T}}_{2}(u_{1},u_{2})(t_{1},t_{2}) \geq \theta _{2}\|{\mathcal{T}}_{2}(u_{1},u_{2})\|$.

Thus

$$\begin{aligned} &\min_{ t_{i}\in [ \frac{1}{4}(T+\alpha _{i}),\frac{3}{4}(T+ \alpha _{i}) ] } \bigl( {\mathcal{T}}_{1}(u_{1},u_{2}) (t_{1},t_{2})+{ \mathcal{T}}_{2}(u_{1},u_{2}) (t_{1},t_{2}) \bigr) \\ &\quad \geq \min_{ t_{i}\in [ \frac{1}{4}(T+\alpha _{i}), \frac{3}{4}(T+\alpha _{i}) ] } {\mathcal{T}}_{1}(u_{1},u_{2}) (t_{1},t_{2}) + \min_{ t_{i}\in [ \frac{1}{4}(T+\alpha _{i}),\frac{3}{4}(T+ \alpha _{i}) ] } { \mathcal{T}}_{2}(u_{1},u_{2}) (t_{1},t_{2}) \\ &\quad \geq \theta \bigl\Vert \bigl( {\mathcal{T}}_{1}(u_{1},u_{2}),{ \mathcal{T}}_{2}(u_{1},u_{2}) \bigr) \bigr\Vert _{{\mathcal{U}}}. \end{aligned}$$

Since $G_{i}(t_{i},s)\geq 0$ for all $(t_{i},s)\in \mathbb{N}_{\alpha _{i}-3,T+\alpha _{i}}\times \mathbb{N}_{0,T}$ and (A1)–(A4) hold, we conclude that $\mathcal{T}(\mathcal{P})\subset \mathcal{P}$. It is easy to show that $\mathcal{T}$ is uniformly bounded. By the Arzelá–Ascoli theorem we can conclude that $\mathcal{T}:\mathcal{P}\rightarrow \mathcal{P}$ is a completely continuous operator. □

Theorem 4.1

Suppose assumptions $(A1)$–$(A4)$hold. Then for $\lambda _{1},\lambda _{2}>0$, problem (1.1)–(1.2) has at least one positive solution in the following cases:

$$\begin{aligned} &(a) \quad F_{1}^{0}=F_{2}^{0}=0, \textit{ and either }F_{1}^{\infty }=\infty \textit{ or } F_{2}^{\infty }=\infty \textit{ (superlinear);} \\ &(b) \quad F_{1}^{\infty }=F_{2}^{\infty }=0, \textit{ and either }F_{1}^{0}= \infty \textit{ or } F_{2}^{0}=\infty \textit{ (sublinear)}. \end{aligned}$$

Proof

$(a)$ Since, $F_{i}^{0}=0$, $i=1,2$, we choose $K_{1}>0$ such that $F_{i}[t_{1},t_{2},u_{1},u_{2}]\leq \varepsilon (u_{1}+u_{2})$ for $0< u_{1}+u_{2}\leq K_{1}$ and $t_{i}\in \mathbb{N}_{\alpha _{i}-1,T+\alpha _{i}-1}$, where the constant $\varepsilon >0$ satisfies

$$\begin{aligned} 2\varepsilon \lambda _{i}\sum_{s=0}^{T} G_{i}(s+\alpha _{i}-1,s) \leq 1. \end{aligned}$$

Set $\varOmega _{1}= \lbrace (u_{1},u_{2}) \in \mathcal{U}: \|(u_{1},u_{2}) \|_{\mathcal{U}}< K_{1} \rbrace$. If $(u_{1},u_{2}) \in \mathcal{P}\cap \partial \varOmega _{1}$ and $\|(u_{1},u_{2})\|_{\mathcal{U}}=K_{1}$, then we have

$$\begin{aligned} \mathcal{T}_{1}(u_{1},u_{2}) (t_{1},t_{2}) \leq &\lambda _{1}\sum _{s=0}^{T} G_{1}(t_{1},s) F_{1}\bigl[s-\alpha _{1}-1,t_{2},u_{1}(s- \alpha _{1}-1),u_{2}(t_{2})\bigr] \\ \leq &\varepsilon \lambda _{1}\sum_{s=0}^{T}G_{1}(s+ \alpha _{i}-1,s) (u_{1}+u_{2}) \\ \leq &\varepsilon \lambda _{1} \bigl( \Vert u_{1} \Vert + \Vert u_{2} \Vert \bigr)\sum _{s=0}^{T}G_{1}(s+ \alpha _{i}-1,s) \\ \leq &\frac{1}{2} \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{\mathcal{U}}. \end{aligned}$$

Similarly, $\mathcal{T}_{2}(u_{1},u_{2})(t_{1},t_{2}) \leq \frac{1}{2} \|(u_{1},u_{2}) \|_{\mathcal{U}}$. Therefore

$$ \bigl\Vert \mathcal{T}(u_{1},u_{2}) \bigr\Vert _{\mathcal{U}}= \bigl\Vert \bigl( \mathcal{T}_{1}(u_{1},u_{2}), \mathcal{T}_{2}(u_{1},u_{2}) \bigr) \bigr\Vert _{\mathcal{U}}= \bigl\Vert \mathcal{T}_{1}(u_{1},u_{2}) \bigr\Vert + \bigl\Vert \mathcal{T}_{2}(u_{1},u_{2}) \bigr\Vert \leq \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{\mathcal{U}} $$

for $(u_{1},u_{2}) \in \mathcal{P}\cap \partial \varOmega _{1}$.

If $F_{1}^{\infty }=\infty$, then there exists $\hat{K}>0$ such that $F_{1}[t_{1},t_{2},u_{1},u_{2}] \geq \hat{\varepsilon }(u_{1}+u_{2})$ for $u_{1}+u_{2}\geq \hat{K}_{1}$ and $t_{i}\in \mathbb{N}_{\alpha _{i}-1,T+\alpha _{i}-1}$, where the constant $\hat{\varepsilon }>0$ satisfies

$$\begin{aligned} \hat{\varepsilon }\lambda _{1}\sum_{s=0}^{T} G_{1}(s+\alpha _{1}-1,s) \geq 1. \end{aligned}$$

Let $K_{2}=\max \lbrace 2K_{1},\frac{\hat{K}}{\theta _{1}} \rbrace$ and set $\varOmega _{2}= \lbrace (u_{1},u_{2}) \in \mathcal{U}: \|(u_{1},u_{2}) \|_{\mathcal{U}}< K_{2} \rbrace$. For $(u_{1},u_{2}) \in \mathcal{P}\cap \partial \varOmega _{2}$, we have $\min_{ t_{i}\in [ \frac{1}{4}(T+\alpha _{i}), \frac{3}{4}(T+\alpha _{i}) ] } ( (u_{1}(t_{1})+u_{2}(t_{2}) ) \geq \theta _{i}\|(u_{1},u_{2})\|_{\mathcal{U}}\geq \hat{K}$. For all $t_{i}\in [ \frac{1}{4}(T+\alpha _{i}),\frac{3}{4}(T+\alpha _{i}) ]$, we get

$$\begin{aligned} &\min_{ t_{i}\in [ \frac{1}{4}(T+\alpha _{i}),\frac{3}{4}(T+ \alpha _{i}) ] } \mathcal{T}_{1}(u_{1},u_{2}) (t_{1},t_{2}) \\ &\quad \geq \min_{ t_{i}\in [ \frac{1}{4}(T+\alpha _{i}), \frac{3}{4}(T+\alpha _{i}) ] } \lambda _{1}\sum _{s=0}^{T} G_{1}(t_{1},s) F_{1}\bigl[s-\alpha _{1}-1,t_{2},u_{1}(s- \alpha _{1}-1),u_{2}(t_{2})\bigr] \\ &\quad \geq \hat{\varepsilon }\lambda _{1}\sum _{s=0}^{T}G_{1}(s+\alpha _{i}-1,s) (u_{1}+u_{2}) \\ &\quad \geq \hat{\varepsilon }\lambda _{1}\theta _{1} \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{ \mathcal{U}}\sum _{s=0}^{T}G_{1}(s+\alpha _{i}-1,s) \\ &\quad \geq \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{\mathcal{U}}. \end{aligned}$$

Therefore

$$ \bigl\Vert \mathcal{T}(u_{1},u_{2}) \bigr\Vert _{\mathcal{U}}= \bigl\Vert \bigl( \mathcal{T}_{1}(u_{1},u_{2}), \mathcal{T}_{2}(u_{1},u_{2}) \bigr) \bigr\Vert _{\mathcal{U}}= \bigl\Vert \mathcal{T}_{1}(u_{1},u_{2}) \bigr\Vert + \bigl\Vert \mathcal{T}_{2}(u_{1},u_{2}) \bigr\Vert \geq \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{\mathcal{U}} $$

for $(u_{1},u_{2}) \in \mathcal{P}\cap \partial \varOmega _{2}$. An analogous estimate holds for $F_{2}^{\infty }=\infty$.

By Theorem 1.1(i), $\mathcal{T}$ has a fixed point $(u_{1},u_{2}) \in \mathcal{P}\cap \partial ( \bar{\varOmega }_{2} \setminus \varOmega _{1} )$ such that $K_{1}\leq \|(u_{1},u_{2})\|_{\mathcal{U}}\leq K_{2}$ and problem (1.1)–(1.2) has a positive solution.

$(b)$ If $F_{1}^{0}=\infty$, then we choose $K_{1}>0$ such that $F_{1}[t_{1},t_{2},u_{1},u_{2}] \geq \tilde{\varepsilon } (u_{1}+u_{2})$ for $0< u_{1}+u_{2}\leq K_{2}$ and $t_{i}\in \mathbb{N}_{\alpha _{i}-1,T+\alpha _{i}-1}$, where the constant $\tilde{\varepsilon }>0$ satisfies

$$\begin{aligned} \tilde{\varepsilon }\lambda _{1}\theta _{1}\sum _{s=0}^{T} G_{1}(s+ \alpha _{1}-1,s) \geq 1. \end{aligned}$$

If $(u_{1},u_{2}) \in \mathcal{P}\cap \partial \varOmega _{1}$ and $\|(u_{1},u_{2})\|_{\mathcal{U}}=K_{1}$, then for all $t_{i}\in [ \frac{1}{4}(T+\alpha _{i}),\frac{3}{4}(T+\alpha _{i}) ]$, we have

$$\begin{aligned} &\min_{ t_{i}\in [ \frac{1}{4}(T+\alpha _{i}),\frac{3}{4}(T+ \alpha _{i}) ] } \mathcal{T}_{1}(u_{1},u_{2}) (t_{1},t_{2}) \\ &\quad \geq \min_{ t_{i}\in [ \frac{1}{4}(T+\alpha _{i}), \frac{3}{4}(T+\alpha _{i}) ] } \lambda _{1}\sum _{s=0}^{T} G_{1}(t_{1},s) F_{1}\bigl[s-\alpha _{1}-1,t_{2},u_{1}(s- \alpha _{1}-1),u_{2}(t_{2})\bigr] \\ &\quad \geq \tilde{\varepsilon }\lambda _{1}\sum _{s=0}^{T}G_{1}(s+\alpha _{i}-1,s) (u_{1}+u_{2}) \\ &\quad \geq \tilde{\varepsilon }\lambda _{1}\theta _{1} \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{ \mathcal{U}}\sum _{s=0}^{T}G_{1}(s+\alpha _{i}-1,s) \\ &\quad \geq \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{\mathcal{U}}. \end{aligned}$$

Therefore

$$ \bigl\Vert \mathcal{T}(u_{1},u_{2}) \bigr\Vert _{\mathcal{U}}= \bigl\Vert \bigl( \mathcal{T}_{1}(u_{1},u_{2}), \mathcal{T}_{2}(u_{1},u_{2}) \bigr) \bigr\Vert _{\mathcal{U}}= \bigl\Vert \mathcal{T}_{1}(u_{1},u_{2}) \bigr\Vert + \bigl\Vert \mathcal{T}_{2}(u_{1},u_{2}) \bigr\Vert \geq \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{\mathcal{U}} $$

for $(u_{1},u_{2}) \in \mathcal{P}\cap \partial \varOmega _{2}$. An analogous estimate holds for $F_{2}^{0}=\infty$.

For $i=1,2$, set $F_{i}^{*}(t_{i})=\max_{0\leq u_{1}+u_{2}\leq t_{i}} F_{i}[t_{1},t_{2},u_{1},u_{2}]$. Then $F_{i}^{*}$ are nondecreasing in their respective arguments. In addition, from $F_{i}^{\infty }=0$ we see that $\lim_{t_{i}\rightarrow \infty } \frac{F_{i}^{*}(t_{i})}{t_{i}}=0$. Hence there exist $K_{2}>2K_{1}$ such that $F_{i}^{*}(t_{i})\leq \varepsilon t_{i}$, where the constant $\varepsilon >0$ satisfies

$$\begin{aligned} 2\varepsilon \lambda _{i}\sum_{s=0}^{T} G_{i}(s+\alpha _{i}-1,s) \leq 1. \end{aligned}$$

If $(u_{1},u_{2}) \in \mathcal{P}\cap \partial \varOmega _{2}$ and $\|(u_{1},u_{2})\|_{\mathcal{U}}=K_{2}$, then we have

$$\begin{aligned} \mathcal{T}_{1}(u_{1},u_{2}) (t_{1},t_{2}) \leq &\lambda _{1}\sum _{s=0}^{T} G_{1}(t_{1},s) F_{1}\bigl[s-\alpha _{1}-1,t_{2},u_{1}(s- \alpha _{1}-1),u_{2}(t_{2})\bigr] \\ \leq &\lambda _{1}\sum_{s=0}^{T}G_{1}(s+ \alpha _{i}-1,s) F_{1}^{*}(K_{2}) \\ \leq &\varepsilon \lambda _{1}K_{2} \sum _{s=0}^{T}G_{1}(s+\alpha _{i}-1,s) \\ \leq &\frac{1}{2} \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{\mathcal{U}}. \end{aligned}$$

Similarly, $\mathcal{T}_{2}(u_{1},u_{2})(t_{1},t_{2}) \leq \frac{1}{2} \|(u_{1},u_{2}) \|_{\mathcal{U}}$. Therefore

$$ \bigl\Vert \mathcal{T}(u_{1},u_{2}) \bigr\Vert _{\mathcal{U}}= \bigl\Vert \bigl( \mathcal{T}_{1}(u_{1},u_{2}), \mathcal{T}_{2}(u_{1},u_{2}) \bigr) \bigr\Vert _{\mathcal{U}}= \bigl\Vert \mathcal{T}_{1}(u_{1},u_{2}) \bigr\Vert + \bigl\Vert \mathcal{T}_{2}(u_{1},u_{2}) \bigr\Vert \leq \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{\mathcal{U}} $$

for $(u_{1},u_{2}) \in \mathcal{P}\cap \partial \varOmega _{2}$.

By Theorem 1.1(ii), $\mathcal{T}$ has a fixed point, and thus problem (1.1)–(1.2) has a positive solution $(u_{1},u_{2}) \in \mathcal{P}\cap \partial ( \bar{\varOmega }_{2} \setminus \varOmega _{1} )$. □

Theorem 4.2

Suppose assumptions $(A1)$–$(A4)$hold.

$(a)$:

If $F_{1}^{0}=F_{2}^{0}=F_{1}^{\infty }=F_{2}^{\infty }=0$, then there exists $\delta _{1}>0$such that problem (1.1)–(1.2) has at least two positive solutions for all $\lambda _{1},\lambda _{2} \geq \delta _{1}$.

$(b)$:

If either $F_{1}^{0}=\infty$or $F_{2}^{0}=\infty$and either $F_{1}^{\infty }=0$or $F_{2}^{\infty }=0$, then there exists $\delta _{2}>0$such that problem (1.1)–(1.2) has at least two positive solutions for all $\lambda _{1},\lambda _{2} \leq \delta _{2}$.

Proof

$(a)$ For $(u_{1},u_{2}) \in \mathcal{P}$ such that $\|(u_{1},u_{2})\|_{\mathcal{U}}=\ell$, let

$$\begin{aligned} m(\ell )&=\min \Biggl\lbrace \lambda _{1}\sum _{s=0}^{T} G_{1}(t_{1},s) F_{1}\bigl[s-\alpha _{1}-1,t_{2},u_{1}(s- \alpha _{1}-1),u_{2}(t_{2})\bigr], \\ &\quad \lambda _{2}\sum_{s=0}^{T} G_{2}(t_{1},s) F_{2}\bigl[t_{1},s- \alpha _{2}-1,u_{1}(t_{1}),u_{2}(s- \alpha _{2}-1)\bigr] \Biggr\rbrace . \end{aligned}$$

By assumption $m(\ell )>0$ for $\ell >0$. Choose two numbers $0< K_{3}< K_{4}$, and let

$$\begin{aligned}& \delta _{1} = \max \biggl\lbrace \frac{K_{3}}{2m(K_{3})} , \frac{K_{4}}{2m(K_{4})} \biggr\rbrace , \\& \varOmega _{i} = \bigl\lbrace (u_{1},u_{2}) \in \mathcal{U}: \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{\mathcal{U}}< K_{i} \bigr\rbrace \quad (i=1,2,3,4). \end{aligned}$$

Then, for $\lambda _{1},\lambda _{2} \geq \delta _{1}$ and $(u_{1},u_{2}) \in \mathcal{P}\cap \partial \varOmega _{i}$ ($i=3,4$) such that $\|(u_{1},u_{2})\|_{\mathcal{U}}=K_{i}$, we have

$$\begin{aligned} &\min_{ t_{i}\in [ \frac{1}{4}(T+\alpha _{i}),\frac{3}{4}(T+ \alpha _{i}) ] } \mathcal{T}_{1}(u_{1},u_{2}) (t_{1},t_{2}) \\ &\quad \geq \tilde{\varepsilon }\lambda _{1}\sum _{s=0}^{T} G_{1}(t_{1},s) F_{1}\bigl[s-\alpha _{1}-1,t_{2},u_{1}(s- \alpha _{1}-1),u_{2}(t_{2})\bigr] \\ &\quad \geq \lambda _{1}m(K_{i}) \\ &\quad \geq \frac{K_{i}}{2}\quad (i=3,4). \end{aligned}$$

Similarly, $\min_{ t_{i}\in [ \frac{1}{4}(T+\alpha _{i}), \frac{3}{4}(T+\alpha _{i}) ] } \mathcal{T}_{2}(u_{1},u_{2})(t_{1},t_{2}) \geq \frac{K_{i}}{2}$ ($i=3,4$).

This implies that

$$\begin{aligned} \bigl\Vert \mathcal{T}(u_{1},u_{2}) \bigr\Vert _{\mathcal{U}}= \bigl\Vert \mathcal{T}_{1}(u_{1},u_{2}) \bigr\Vert + \bigl\Vert \mathcal{T}_{2}(u_{1},u_{2}) \bigr\Vert \geq K_{i}= \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{ \mathcal{U}} \end{aligned}$$

for $(u_{1},u_{2}) \in \mathcal{P}\cap \partial \varOmega _{i}$ ($i=3,4$).

Since $F_{1}^{0}=F_{2}^{0}=F_{1}^{\infty }=F_{2}^{\infty }=0$, it follows from the proof of Theorem 4.1(a,b) that we can choose $K_{1}<\frac{K_{3}}{2}$ and $K_{2}>2K_{4}$ such that

$$\begin{aligned} \bigl\Vert \mathcal{T}(u_{1},u_{2}) \bigr\Vert _{\mathcal{U}}\leq \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{ \mathcal{U}} \end{aligned}$$

for $(u_{1},u_{2}) \in \mathcal{P}\cap \partial \varOmega _{i}$ ($i=1,2$).

Applying Theorem 1.1 to $\varOmega _{1}$, $\varOmega _{3}$ and $\varOmega _{3}$, $\varOmega _{4}$, we have a positive solution $(u_{1},u_{2})$ such that $K_{1}\leq \|(u_{1},u_{2})\|_{\mathcal{U}} \leq K_{3}$, and another positive solution $(v_{1},v_{2})$ such that $K_{4}\leq \|(v_{1},v_{2})\|_{\mathcal{U}} \leq K_{2}$. Since $K_{3}< K_{4}$, these two solutions are distinct.

$(b)$ For $(u_{1},u_{2}) \in \mathcal{P}$ and $\|(u_{1},u_{2})\|_{\mathcal{U}}=L$, let

$$\begin{aligned} M(L)&=\max \Biggl\lbrace \lambda _{1}\sum _{s=0}^{T} G_{1}(t_{1},s) F_{1}\bigl[s- \alpha _{1}-1,t_{2},u_{1}(s- \alpha _{1}-1),u_{2}(t_{2})\bigr], \\ &\quad \lambda _{2}\sum_{s=0}^{T} G_{2}(t_{1},s) F_{2}\bigl[t_{1},s- \alpha _{2}-1,u_{1}(t_{1}),u_{2}(s- \alpha _{2}-1)\bigr] \Biggr\rbrace . \end{aligned}$$

By assumption $M(L)>0$ for $L>0$. Choose two numbers $0< K_{3}< K_{4}$, and let

$$\begin{aligned} \delta _{2} =&\max \biggl\lbrace \frac{K_{3}}{2M(K_{3})} , \frac{K_{4}}{2M(K_{4})} \biggr\rbrace . \end{aligned}$$

Then, for $\lambda _{1},\lambda _{2} \leq \delta _{2}$ and $(u_{1},u_{2}) \in \mathcal{P}\cap \partial \varOmega _{i}$ ($i=3,4$) such that $\|(u_{1},u_{2})\|_{\mathcal{U}}=K_{i}$, we have

$$\begin{aligned} \mathcal{T}_{1}(u_{1},u_{2}) (t_{1},t_{2}\leq \lambda _{1}M(H_{i}) \leq \frac{K_{i}}{2} \quad \text{and}\quad \mathcal{T}_{2}(u_{1},u_{2}) (t_{1},t_{2}) \leq \lambda _{2}M(H_{i}) \leq \frac{K_{i}}{2} \quad (i=3,4). \end{aligned}$$

This implies that

$$\begin{aligned} \bigl\Vert \mathcal{T}(u_{1},u_{2}) \bigr\Vert _{\mathcal{U}}\leq K_{i}= \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{\mathcal{U}} \end{aligned}$$

for $(u_{1},u_{2}) \in \mathcal{P}\cap \partial \varOmega _{i}$ ($i=3,4$).

Since either $F_{1}^{0}=\infty$ or $F_{2}^{0}=\infty$ and either $F_{1}^{\infty }=0$ or $F_{2}^{\infty }=0$, it follows from the proof of Theorem 4.1(a,b) that we can choose $K_{1}<\frac{K_{3}}{2}$ and $K_{2}>2K_{4}$ such that

$$\begin{aligned} \bigl\Vert \mathcal{T}(u_{1},u_{2}) \bigr\Vert _{\mathcal{U}}\geq \bigl\Vert (u_{1},u_{2}) \bigr\Vert _{ \mathcal{U}} \end{aligned}$$

for $(u_{1},u_{2}) \in \mathcal{P}\cap \partial \varOmega _{i}$ ($i=1,2$).

Once again, we obtain the existence of two distinct positive solutions. □

By the same arguments as in Theorem 4.2 we obtain the following theorem.

Theorem 4.3

Suppose assumptions $(A1)$–$(A4)$hold.

$(a)$:

If $F_{1}^{0}=F_{2}^{0}=0$or $F_{1}^{\infty }=F_{2}^{\infty }=0$, then there exists $\delta _{3}>0$such that problem (1.1)–(1.2) has at least two positive solutions for all $\lambda _{1},\lambda _{2} \geq \delta _{3}$.

$(b)$:

If $F_{1}^{0}=\infty$or $F_{2}^{0}=\infty$, or if $F_{1}^{\infty }=\infty$or $F_{2}^{\infty }=\infty$, then there exists $\delta _{4}>0$such that problem (1.1)–(1.2) has at least two positive solutions for all $\lambda _{1},\lambda _{2} \leq \delta _{2}$.

5 Examples

In this section, we provide some examples to illustrate our main results.

Example 1

Consider the following system of fractional difference equations with parameters:

$$\begin{aligned} \begin{gathered} -\Delta ^{\frac{5}{2}} u_{1}(t) = \frac{e^{- (t+\frac{4}{3} ) } ( \vert u_{2} \vert +1 ) }{400 (t+\frac{34}{3} )^{2} ( 1+\sin ^{2}u_{2}\pi ) }+ \frac{e^{- (t+\frac{3}{2} )\pi }u_{1} (t+\frac{3}{2} ) }{100e+20\sin ^{2} (t+\frac{3}{2} )\pi }, \\ -\frac{1}{2}\Delta ^{\frac{7}{3}} u_{2}(t) = \frac{e^{- (t+\frac{3}{2} )} ( \vert u_{1} \vert +e^{-\sin ^{2} (t+\frac{3}{2} )\pi } ) }{1000(e^{t+\frac{3}{2}}+10)^{2} ( \vert u_{1} \vert +\cos ^{2} (t+\frac{3}{2} )\pi ) }\\ \hphantom{-\frac{1}{2}\Delta ^{\frac{7}{3}} u_{2}(t) =}{}+ \frac{\arctan (\sin ^{2} (t+\frac{4}{3} )\pi ) u_{2} (t+\frac{4}{3} ) }{100\pi (t+\frac{13}{3} )^{2}} \end{gathered} \end{aligned}$$

(5.1)

for $t\in {\mathbb{N}}_{0,5}$, subject to nonlocal fractional difference-sum conditions

$$\begin{aligned} \begin{gathered} \Delta ^{-\frac{1}{3}}u_{1} \biggl( - \frac{1}{6} \biggr) = \Delta ^{ \frac{1}{5}}u_{1} \biggl( \frac{3}{10} \biggr) = 0, \\ \Delta ^{-\frac{1}{4}}u_{2} \biggl( -\frac{5}{12} \biggr) = \Delta ^{ \frac{2}{5}}u_{2} \biggl( -\frac{1}{15} \biggr) = 0, \\ u_{1} \biggl( \frac{15}{2} \biggr) = \frac{1}{2}u_{1} \biggl( \frac{9}{2} \biggr), \\ u_{2} \biggl( \frac{22}{3} \biggr) = \frac{3}{4}u_{2} \biggl( \frac{13}{3} \biggr). \end{gathered} \end{aligned}$$

(5.2)

Here $\alpha _{1}=\frac{5}{2}$, $\alpha _{2}=\frac{7}{3}$, $\beta _{1}= \frac{1}{3}$, $\beta _{2}=\frac{1}{4}$, $\gamma _{1}=\frac{1}{5}$, $\gamma _{2}= \frac{2}{5}$, $\lambda _{1}=1$, $\lambda _{2}=2$, $\chi _{1}=\frac{1}{2}$, $\chi _{2}=\frac{3}{4}$, $\eta _{1}=\frac{9}{2}$, $\eta _{2}=\frac{13}{3}$, $T=5$, and for $t_{1}\in {\mathbb{N}}_{-\frac{1}{2},\frac{15}{2}}$ and $t_{2}\in {\mathbb{N}}_{-\frac{2}{3},\frac{22}{3}}$,

$$\begin{aligned} &F_{1} [t_{1},t_{2},u_{1} ,u_{2} ] = \frac{e^{-t_{2} } ( \vert u_{2} \vert +1 ) }{400 (t_{2}+10 )^{2} ( 1+\sin ^{2}u_{2}\pi ) }+ \frac{e^{-t_{1}\pi }u_{1} (t_{1} ) }{100e+20\sin ^{2}t_{1}\pi }, \\ &F_{2} [t_{1},t_{2},u_{1},u_{2} ] = \frac{e^{-t_{1}} ( \vert u_{1} \vert +e^{-\sin ^{2}t_{1}\pi } ) }{1000(e^{t_{1}}+10)^{2} ( \vert u_{1} \vert +\cos ^{2}t_{1}\pi ) }+ \frac{\arctan (\sin ^{2}t_{2}\pi ) u_{2} (t_{2} ) }{100\pi (t_{2}+3 )^{2}}. \end{aligned}$$

We get that

$$\begin{aligned} & \bigl\vert F_{1} [t_{1},t_{2},u_{1} ,u_{2} ] -F_{1} [t_{1},t_{2},v_{1} ,v_{2} ] \bigr\vert \\ &\quad \leq \frac{e^{-t_{2}}}{400(t_{2}+10)^{2}} \biggl\vert \frac{ \vert u_{2} \vert +1}{1+\sin ^{2}u_{2}\pi }- \frac{ \vert v_{2} \vert +1}{1+\sin ^{2}v_{2}\pi } \biggr\vert + \frac{e^{-t_{1}\pi }}{100e+20\sin ^{2}t_{1}\pi } \bigl\vert \Delta ^{ \frac{1}{3}}u_{1}-\Delta ^{\frac{1}{3}}v_{1} \bigr\vert \\ &\quad \leq \frac{1}{45\mbox{,}511} \vert u_{2}-v_{2} \vert +\frac{1}{100e} \vert u_{1}-v_{1} \vert , \\ & \bigl\vert F_{2} [t_{1},t_{2},u_{1} ,u_{2} ] -F_{2} [t_{1},t_{2},v_{1} ,v_{2} ] \bigr\vert \\ &\quad \leq \frac{e^{-t_{1}}}{1000(e^{t_{1}}+10)^{2}} \biggl\vert \frac{ \vert u_{1} \vert }{1+ \vert u_{1} \vert }- \frac{ \vert v_{1} \vert }{1+ \vert v_{1} \vert } \biggr\vert + \frac{\arctan (1)}{100\pi (t_{2}+3)^{2}}| u_{2}-v_{2} )| \\ &\quad \leq \frac{1}{12\mbox{,}100} \vert u_{1}-v_{1} \vert +\frac{9}{19\mbox{,}600} \vert u_{2}-v_{2} \vert . \end{aligned}$$

So, (3.3) holds with $M_{1}=0.00002197$, $M_{2}=0.00008264$, $N_{1}=0.00368$, and $N_{2}=0.000459$.

Finally, we find that

$$\begin{aligned}& L_{1}=\max \lbrace M_{1},N_{1} \rbrace =0.00368,\qquad L_{2} = \max \lbrace M_{2},N_{2} \rbrace =0.000459 \quad \text{and} \\& \varOmega _{1}=75.4559, \qquad \varOmega _{2}=136.2638. \end{aligned}$$

Therefore we obtain

$$ L_{1}\varOmega _{1}+L_{2}\varOmega _{2} = 0.3402 < 1. $$

Hence, by Theorem 3.1, problem (5.1)–(5.2) has a unique solution.

Example 2

Consider the following system of fractional difference equations with parameters:

$$\begin{aligned} \begin{gathered} -\Delta ^{\frac{5}{2}} u_{1}(t) = \biggl[ u_{1} \biggl(t+\frac{3}{2} \biggr) + u_{1} \biggl(t+\frac{4}{3} \biggr) \biggr]^{6}, \\ -\frac{1}{2}\Delta ^{\frac{7}{3}} u_{2}(t) = \biggl[u_{1} \biggl(t+ \frac{3}{2} \biggr) + u_{1} \biggl(t+\frac{4}{3} \biggr) \biggr] ^{4}, \quad t \in { \mathbb{N}}_{0,5}, \end{gathered} \end{aligned}$$

(5.3)

for $t\in {\mathbb{N}}_{0,5}$, subject to nonlocal fractional difference-sum conditions (5.2).

For all $t_{1}\in {\mathbb{N}}_{-\frac{1}{2},\frac{15}{2}}$, $t_{2}\in { \mathbb{N}}_{-\frac{2}{3},\frac{22}{3}}$, $u_{1},u_{2}>0$, we have $F_{1}[t_{1},t_{2},u_{1}u_{2}]= [ u_{1} (t_{1} ) + u_{1} (t_{2} ) ]^{6}>0$ and $F_{2}[t_{1},t_{2},u_{1}u_{2}]= [ u_{1} (t_{1} ) + u_{1} (t_{2} ) ]^{4}>0$. We find that

$$\begin{aligned}& 0 < 4.3619 = \chi _{1}\eta _{1}^{\underline{\alpha _{1}-1}} < (T+ \alpha _{1})^{\underline{\alpha _{1}-1}}=19.4922, \\& 0 < 5.0162 = \chi _{2}\eta _{2}^{\underline{\alpha _{2}-1}} < (T+ \alpha _{2})^{\underline{\alpha _{2}-1}}=13.8057. \end{aligned}$$

By direct calculation we have $F_{1}^{0}=F_{2}^{0}=0$ and $F_{1}^{\infty }=F_{2}^{\infty }=\infty$. Then, by Theorem 4.1(a), problem (5.3) and (5.2) has at least one positive solution.

Example 3

Consider the following system of fractional difference equations with parameters:

$$\begin{aligned} \begin{gathered} -\Delta ^{\frac{5}{2}} u_{1}(t) = \biggl[ u_{1} \biggl(t+\frac{3}{2} \biggr) + u_{1} \biggl(t+\frac{4}{3} \biggr) \biggr]^{\frac{1}{3}}, \\ -\frac{1}{2}\Delta ^{\frac{7}{3}} u_{2}(t) = \biggl[u_{1} \biggl(t+ \frac{3}{2} \biggr) + u_{1} \biggl(t+\frac{4}{3} \biggr) \biggr] ^{ \frac{1}{4}}, \quad t\in { \mathbb{N}}_{0,5}, \end{gathered} \end{aligned}$$

(5.4)

for $t\in {\mathbb{N}}_{0,5}$, subject to nonlocal fractional difference-sum conditions (5.2).

For all $t_{1}\in {\mathbb{N}}_{-\frac{1}{2},\frac{15}{2}}$, $t_{2}\in { \mathbb{N}}_{-\frac{2}{3},\frac{22}{3}}$, $u_{1},u_{2}>0$, we have $F_{1}[t_{1},t_{2},u_{1}u_{2}]= [ u_{1} (t_{1} ) + u_{1} (t_{2} ) ]^{\frac{1}{3}}>0$ and $F_{2}[t_{1},t_{2},u_{1}u_{2}]= [ u_{1} (t_{1} ) + u_{1} (t_{2} ) ]^{\frac{1}{4}}>0$.

By the same argument as in Example 2 we have that $\chi _{i}\eta _{i}^{\underline{\alpha _{i}-1}} < (T+\alpha _{i})^{ \underline{\alpha _{i}-1}}$, $i=1,2$.

By direct calculation we have $F_{1}^{0}=F_{2}^{0}=\infty$ and $F_{1}^{\infty }=F_{2}^{\infty }=0$. Then, by Theorem 4.1(b), problem (5.4) and (5.2) has at least one positive solution.