Existence and uniqueness of solutions for a first-order discrete fractional boundary value problem

Abstract

In this paper we use topological degree to establish the existence of nontrivial solutions for a first-order discrete fractional boundary value problem with a sign-changing nonlinearity. Also using the monotone iterative technique we discuss problems with a nonnegative nonlinearity and establish the uniqueness of positive solutions.

1 Introduction

In this paper we study the existence and uniqueness of solutions for the first-order discrete fractional boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} {\Delta _{\nu -1}^\nu y(t)=f(t+\nu -1,y(t+\nu -1)),\quad t\in [0,T]_{{\mathbb {Z}}}},\\ y(\nu -1)=y(\nu +T), \end{array}\right. } \end{aligned}$$

(1.1)

where $\nu$ is a real number with $0<\nu <1$ and $\Delta ^\nu$ is a discrete fractional operator. For the nonlinear term f, we assume that

(H1) $f(t,y)\in C([\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}\times {\mathbb {R}}, {\mathbb {R}})$.

Note that, in this paper we use $[a,b]_{E}$ to stand for $[a,b]\cap E$ for some set E.

In [1,2,3,4], Atici and Eloe developed the fundamental theory of both discrete delta and nabla fractional calculus. For more recent results we refer the reader to [5,6,7,8,9] and the references cited therein. In [5], the authors considered a three-point fractional sum boundary value problem, where the nonlinearity f is either superlinear or sublinear:

$$\begin{aligned} f_0=\lim _{u\rightarrow 0^+}\frac{f(u)}{u}=0\quad \;\text { or }\; \infty ,\quad f_\infty =\lim _{u\rightarrow \infty }\frac{f(u)}{u}=\infty \;\text {or }\; 0; \end{aligned}$$

(1.2)

here $f_0=0,f_\infty =\infty$ is the superlinear case, and $f_0=\infty ,f_\infty =0$ is the sublinear case. In [6,7,8], the authors adopted the growth conditions in (1.2) to study many types of discrete fractional boundary value problems with nonnegative and semipositone nonlinearities. For more details in this direction, we refer the reader to the recent book [9], which summarizes some results on discrete fractional equations.

Motivated by the above, in this paper we use topological degree to study the existence of nontrivial solutions for the discrete fractional boundary value problem (1.1). The novelty is threefold: (1) with the aid of some inequalities of Green’s function, our growth condition on the nonlinearity improves that in (1.2) (see conditions (H3) and (H4) in the following section); (2) nontrivial solutions are obtained using topological degree with a semipositone nonlinearity (this is seldom considered in discrete fractional equations); (3) with a sublinear growth condition, a unique positive solution is obtained from the monotone iterative technique with a nonnegative nonlinearity (in addition the iterative sequences are given).

2 Preliminaries

We first introduce some background materials from discrete fractional calculus, and for more details, we refer the reader to [1,2,3,4, 9].

Definition 2.1

We define $t^{\underline{\nu }}:=\frac{\Gamma (t+1)}{\Gamma (t+1-\nu )}$ for any $t,\nu \in {\mathbb {R}}$ for which the right-hand side is well-defined. We use the convention that if $t+1-\nu$ is a pole of the Gamma function and $t+1$ is not a pole, then $t^{\underline{\nu }}=0$.

Definition 2.2

For $\nu >0$, the $\nu -$th fractional sum of a function f is $\Delta _a ^{-\nu } f(t)=\frac{1}{\Gamma (\nu )}\sum \nolimits _{s=a}^{t-\nu }(t-s-1)^{\underline{\nu -1}} f(s),\text { for } t\in {\mathbb {N}}_{a+N-\nu }$. We also define the $\nu -$th fractional difference for $\nu > 0$ by $\Delta _a ^\nu f(t)=\Delta ^N \Delta _a ^{\nu -N}f(t),$ $\text { for } t\in {\mathbb {N}}_{a+N-\nu }$, where $N\in {\mathbb {N}}$ with $0\le N-1<\nu \le N$.

From [7], we know (1.1) is equivalent to

$$\begin{aligned} y(t)= \sum _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f(s+\nu -1,y(s+\nu -1)),\quad \text { for } t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}, \end{aligned}$$

(2.1)

where

$$\begin{aligned} G(t,s)={\left\{ \begin{array}{ll} \frac{ (\nu +T-s-1) ^{\underline{\nu -1}} t^{\underline{\nu -1}} }{\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}}}+(t-s-1)^{\underline{\nu -1}},&{}\quad 0\le s\le t-\nu \le T,\\ \frac{(\nu +T-s-1) ^{\underline{\nu -1}} t^{\underline{\nu -1}} }{ \Gamma (\nu )-(\nu +T)^{\underline{\nu -1}} },&{}\quad t-\nu <s\le T. \end{array}\right. } \end{aligned}$$

(2.2)

Let $C^*=1+\frac{\Gamma (\nu )-(\nu +T) ^{\underline{\nu -1}} }{(\nu +T-1)^{\underline{\nu -1}}}$, for all $(t,s)\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}\times [0,T]_{{\mathbb {Z}}}$, and note the inequality

$$\begin{aligned} 0< & {} \frac{(\nu +T)^{\underline{\nu -1}}}{\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}}}(\nu +T-s-1) ^{\underline{\nu -1}}\nonumber \\\le & {} G(t,s)\le \frac{C^*\Gamma (\nu )}{\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}}} (\nu +T-s-1) ^{\underline{\nu -1}} . \end{aligned}$$

(2.3)

Let $\phi (t+\nu -1)=(\nu +T-t-1) ^{\underline{\nu -1}}, \ t\in [0,T]_{{\mathbb {Z}}}$ and $\phi ^*(t)=(2\nu +T-t-2) ^{\underline{\nu -1}}, \ t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}$. Then, for all $s\in [0,T]_{{\mathbb {Z}}}$, we have

$$\begin{aligned} \sum _{t=\nu -1}^{\nu +T-1} \frac{ G(t,s)}{\Gamma (\nu )}\phi ^*(t) =\sum _{t=0}^T \frac{ G(t+\nu -1,s)}{\Gamma (\nu )}\phi (t+\nu -1). \end{aligned}$$

(2.4)

Consequently, for all $s\in [0,T]_{{\mathbb {Z}}}$, from (2.3) and (2.4) we have

$$\begin{aligned} \sum _{t=0}^T \frac{ (\nu +T)^{\underline{\nu -1}} \phi (t+\nu -1) }{ \Gamma (\nu ) (\Gamma (\nu ) -(\nu +T)^{\underline{\nu -1}})} \cdot \phi (s+\nu -1)\le & {} \sum _{t=\nu -1}^{\nu +T-1} \frac{ G(t,s)}{\Gamma (\nu )}\phi ^*(t) \nonumber \\\le & {} \sum _{t=0}^T \frac{C^* \phi (t+\nu -1)}{ \Gamma (\nu )-(\nu +T)^{\underline{\nu -1}}} \cdot \phi (s+\nu -1).\nonumber \\ \end{aligned}$$

(2.5)

For convenience, we let

$$\begin{aligned} \kappa _1= & {} \sum \limits _{t=0}^T \frac{(\nu +T)^{\underline{\nu -1}}}{\Gamma (\nu ) (\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}})}\phi (t+\nu -1)\quad \text { and }\\ \kappa _2= & {} \sum \limits _{t=0}^T \frac{C^*}{ \Gamma (\nu )-(\nu +T)^{\underline{\nu -1}}} \phi (t+\nu -1) . \end{aligned}$$

Let E be the collection of all maps from $[\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}$ to ${\mathbb {R}}$, equipped with the max norm, $\Vert \cdot \Vert$. Then E is a Banach space. Let $P\subset E$ be $P=\left\{ y\in E: y(t)\ge 0 ,\ t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}\right\} .$ Then P is a cone in E. Define a linear operator L by $(Ly)(t) = \sum \nolimits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} y(s+\nu -1), \text { for } t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}$. Then we easily have $L(P)\subset P$, i.e., L is a positive operator.

Lemma 2.3

Let r(L) be the spectral radius of L. Then $\kappa _1 \le r(L)\le \kappa _2$.

Proof

For all $n\in {\mathbb {N}}^+$, we have

$$\begin{aligned} \Vert L^n\Vert =\max _{t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}}\sum _{s_1=0}^T\sum _{s_2=0}^T\cdots \sum _{s_n=0}^T\frac{G(t,s_1)}{\Gamma (\nu )}\frac{G(s_1,s_2)}{\Gamma (\nu )} \cdots \frac{G(s_{n-1},s_n)}{\Gamma (\nu )}. \end{aligned}$$

From (2.3), for all $t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}$ we have

$$\begin{aligned} \kappa _1= & {} \sum _{s=0}^T\frac{(\nu +T)^{\underline{\nu -1}}(\nu +T-s-1) ^{\underline{\nu -1}}}{\Gamma (\nu )(\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}})} \le \sum _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )}\nonumber \\\le & {} \sum _{s=0}^T \frac{C^*(\nu +T-s-1) ^{\underline{\nu -1}} }{\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}}} = \kappa _2. \end{aligned}$$

(2.6)

Hence, we have $\kappa _1^n \le \Vert L^n\Vert \le \kappa _2^n$. From Gelfand’s theorem, we have $r(L)=\lim _{n\rightarrow \infty }\root n \of {\Vert L^n\Vert }\in [\kappa _1,\kappa _2]$. This completes the proof. $\square$

Lemma 2.4

Let $P_0=\left\{ y\in P: \sum \nolimits _{t=0}^T\phi (t+\nu -1) y(t+\nu -1)\ge \frac{\kappa _1(\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}})}{C^*} \Vert y\Vert \right\}$. Then $P_0$ is a cone in E, and $L(P)\subset P_0$.

Proof

Note that $\sum \nolimits _{t=0}^T\phi (t+\nu -1) y(t+\nu -1)= \sum \nolimits _{t=\nu -1}^{\nu +T-1}\phi ^*(t) y(t)$. Then we have

$$\begin{aligned} \begin{aligned} \sum _{t=\nu -1}^{\nu +T-1}\phi ^*(t) (Ly)(t)&=\sum _{t=\nu -1}^{\nu +T-1}\phi ^*(t) \sum _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} y(s+\nu -1)\\&\ge \kappa _1 \sum _{s=0}^T \phi (s+\nu -1) y(s+\nu -1)\\&\ge \frac{\kappa _1(\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}})}{C^*} \sum _{s=0}^T \frac{G(\tau ,s)}{\Gamma (\nu )} y(s+\nu -1), \end{aligned} \end{aligned}$$

for all $\tau \in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}.$ This implies that $\sum \nolimits _{t=\nu -1}^{\nu +T-1}\phi ^*(t) (Ly)(t)\ge \frac{\kappa _1(\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}})}{C^*}\Vert Ly\Vert$. This completes the proof. $\square$

We now recall the following three well known results (see [10]).

Lemma 2.5

Let E be a Banach space and $\Omega$ a bounded open set in E. Suppose $A:\overline{\Omega }\rightarrow E$ is a continuous compact operator. If there exists a $\omega _{0}\not =0$ such that $\omega -A\omega \ne \lambda \omega _{0}, \forall \omega \in \partial \Omega , \lambda \ge 0,$ then the topological degree $\deg (I-A,\Omega ,0)=0$.

Lemma 2.6

Let E be a Banach space and $\Omega$ a bounded open set in E with $0\in \Omega$. Suppose $A:\overline{\Omega }\rightarrow E$ is a continuous compact operator. If $\omega -\lambda A\omega \ne 0, \forall \omega \in \partial \Omega , \lambda \in [0,1],$ then $\deg (I-A,\Omega ,0)=1$.

Lemma 2.7

Let E be a partially ordered Banach space, and $x_0 ,y_0 \in E$ with $x_0 \le y_0$, $D = [x_0,y_0]$. Suppose that $A : D \rightarrow E$ satisfies the following conditions

• 1. A is an increasing operator;

• 2. $x_0\le Ax_0$, $y_0\ge Ay_0$, i.e., $x_0$ and $y_0$ is a subsolution and a supersolution of A;

• 3. A is a continuous compact operator.

Then A has the smallest fixed point $x^*$ and the largest fixed point $y^*$ in $[x_0,y_0]$, respectively. Moreover, $x^*=\lim _{n\rightarrow \infty }A^nx_0$ and $y^*=\lim _{n\rightarrow \infty }A^n y_0$.

3 Nontrivial solutions for (1.1)

Let $(Ay)(t)=\sum \nolimits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f(s+\nu -1,y(s+\nu -1)), \text { for } t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}},$ where G is defined in (2.2). Then we note, from the Arzelà–Ascoli theorem, that A is a completely continuous operator on E. Moreover, y is a solution of (1.1) if and only if y is a fixed point of the operator A.

Let $\lambda _1=\kappa _1^{-1},\lambda _2=\kappa _2^{-1}$, and $B_\varrho :=\{x \in E: \Vert x\Vert <\varrho \}$ for $\varrho > 0$. For convenience, we use $c_1,c_2, ...$ to stand for different positive constants. Now we list our assumptions on f.

• (H2) There exists $M\ge 0$ such that $f(t,y)\ge -M$ for all $y\in {\mathbb {R}}$ and $t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}$.

• (H3) $\liminf _{y\rightarrow +\infty }\frac{f(t,y)}{y}> \lambda _1,\ \limsup _{y\rightarrow 0}\left| \frac{f(t,y)}{y}\right| <\lambda _2$ uniformly on $t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}.$

• (H4) $\liminf _{y\rightarrow 0}\frac{f(t,y)}{|y|}> \lambda _1,\ \limsup _{y\rightarrow +\infty }\frac{f(t,y)}{y}<\lambda _2$ uniformly on $t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}.$

Theorem 3.1

Suppose that (H1)–(H3) hold. Then (1.1) has at least one nontrivial solution.

Proof

Let

$$\begin{aligned} \overline{y}(t)=M\sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )}\quad \text { for }t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}. \end{aligned}$$

From the first limit in (H3), there exist $\varepsilon >0$ and $c_1>0$ such that

$$\begin{aligned} f(t,y)\ge (\lambda _1+\varepsilon )y-c_1, \quad \forall y\in {\mathbb {R}},\quad t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}. \end{aligned}$$

(3.1)

We now prove that

$$\begin{aligned} y\not =Ay+\lambda \phi ^{**}, \quad \forall y\in E,\quad \Vert y\Vert =R,\quad \lambda \ge 0, \end{aligned}$$

(3.2)

where $\phi ^{**}\in P_0$ is an arbitrarily fixed element and $R>\Vert \overline{y}\Vert + \frac{C^*(\varepsilon M \kappa _1\kappa _2 +c_1\kappa _1+M\kappa _2-M\kappa _1)}{\varepsilon \kappa _1^2(\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}})}\sum \nolimits _{s=0}^T \phi (s+\nu -1)$.

Suppose there exist $y_0\in E,\Vert y_0\Vert =R$ and $\lambda _0\ge 0$ such that

$$\begin{aligned} \begin{aligned} y_0 (t)&=(Ay_0)(t)+\lambda _0 \phi ^{**}(t) =\sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f(s+\nu -1,y_0(s+\nu -1))\\&\quad +\,\lambda _0 \phi ^{**}(t) , \quad \forall t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}. \end{aligned} \end{aligned}$$

(3.3)

Then we have

$$\begin{aligned} y_0 (t)+ \overline{y}(t)= & {} \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} [f(s+\nu -1,y_0(s+\nu -1))+M]\\&+\,\lambda _0 \phi ^{**}(t),\quad \forall t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}. \end{aligned}$$

Note $\phi ^{**}\in P_0$ and Lemma 2.4, so $y_0+\overline{y}\in P_0$. From (3.1), (2.4) and (2.5), we have

$$\begin{aligned}&\sum _{t=\nu -1}^{\nu +T-1} (Ay_0)(t) \phi ^*(t) - \sum _{t=\nu -1}^{\nu +T-1} y_0 (t)\phi ^*(t) \nonumber \\&\quad = \sum _{t=\nu -1}^{\nu +T-1} \phi ^*(t) \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f(s+\nu -1,y_0(s+\nu -1)) - \sum _{t=\nu -1}^{\nu +T-1} y_0 (t)\phi ^*(t)\nonumber \\&\quad = \sum _{t=\nu -1}^{\nu +T-1} \phi ^*(t) \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} [f(s+\nu -1,y_0(s+\nu -1))+M] \nonumber \\&\qquad - M \sum _{t=\nu -1}^{\nu +T-1} \phi ^*(t) \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )}- \sum _{t=\nu -1}^{\nu +T-1} y_0 (t)\phi ^*(t)\nonumber \\&\quad \ge \sum \limits _{s=0}^T \kappa _1 \phi (s+\nu -1) [(\lambda _1+\varepsilon )y_0(s+\nu -1)-c_1+M]\nonumber \\&\qquad - M\kappa _2 \sum \limits _{s=0}^T \phi (s+\nu -1)-\sum _{t=\nu -1}^{\nu +T-1} y_0 (t)\phi ^*(t) \nonumber \\&\quad = \kappa _1\lambda _1 \sum \limits _{s=0}^T \phi (s+\nu -1) y_0(s+\nu -1)\nonumber \\&\qquad -\,\sum _{t=\nu -1}^{\nu +T-1} y_0 (t)\phi ^*(t) + \varepsilon \kappa _1\sum \limits _{s=0}^T \phi (s+\nu -1) y_0(s+\nu -1)\nonumber \\&\qquad - \,(c_1\kappa _1+M\kappa _2-M\kappa _1)\sum \limits _{s=0}^T \phi (s+\nu -1)\nonumber \\&\quad = \varepsilon \kappa _1\sum _{t=\nu -1}^{\nu +T-1} y_0 (t)\phi ^*(t) -(c_1\kappa _1+M\kappa _2-M\kappa _1) \sum \limits _{s=0}^T \phi (s+\nu -1)\nonumber \\&\quad = \varepsilon \kappa _1\sum _{t=\nu -1}^{\nu +T-1} [y_0 (t)+\overline{y}(t)]\phi ^*(t)- \varepsilon \kappa _1\sum _{t=\nu -1}^{\nu +T-1} \phi ^*(t) M\sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )}\nonumber \\&\qquad -\,(c_1\kappa _1+M\kappa _2-M\kappa _1) \sum \limits _{s=0}^T \phi (s+\nu -1)\nonumber \\&\quad \ge \varepsilon \kappa _1 \frac{\kappa _1(\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}})}{C^*} \Vert y_0+\overline{y}\Vert \nonumber \\&\qquad - \,(\varepsilon M \kappa _1\kappa _2 +c_1\kappa _1+M\kappa _2-M\kappa _1)\sum \limits _{s=0}^T \phi (s+\nu -1)\nonumber \\&\quad \ge \varepsilon \kappa _1 \frac{\kappa _1(\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}})}{C^*} \Vert y_0\Vert - \varepsilon \kappa _1 \frac{\kappa _1(\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}})}{C^*}\Vert \overline{y}\Vert \nonumber \\&\qquad - \,(\varepsilon M \kappa _1\kappa _2 +c_1\kappa _1+M\kappa _2-M\kappa _1)\sum \limits _{s=0}^T \phi (s+\nu -1)\nonumber \\&\quad >0. \end{aligned}$$

(3.4)

Note $\phi ^{**}\in P_0$ and $\lambda _0\ge 0$, so (3.3) enables us to obtain

$$\begin{aligned} \sum _{t=\nu -1}^{\nu +T-1} y_0 (t)\phi ^*(t) -\sum _{t=\nu -1}^{\nu +T-1} (Ay_0)(t) \phi ^*(t) =\sum _{t=\nu -1}^{\nu +T-1} \lambda _0 \phi ^{**}(t) \phi ^*(t) \ge 0. \end{aligned}$$

That is a contradiction with (3.4). As a result, (3.2) holds true. Now Lemma 2.5 implies that

$$\begin{aligned} \deg (I-A,B_R,0)=0. \end{aligned}$$

(3.5)

From the second limit in (H3), there exist $r\in (0,R)$ and $\varepsilon \in (0,\lambda _2)$ such that $|f(t,y)|\le (\lambda _2-\varepsilon ) |y|$, for all $|y|\le r,t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}.$ Next we claim

$$\begin{aligned} Ay\not = \lambda y, \quad \forall y\in \partial B_r, \lambda \ge 1. \end{aligned}$$

(3.6)

If this is false, then there exists a $y_1\in \partial B_r$ and a $\overline{\lambda }_{0}\ge 1$ such that $Ay_1=\overline{\lambda }_{0} y_1$. We may assume that $\overline{\lambda }_{0}>1$ (otherwise we have a fixed point $y_1$). Hence, we have $|Ay_1|=|\overline{\lambda }_{0}y_1|\ge |y_1|$ and

$$\begin{aligned} |y_1(t)|\le & {} \left| \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f(s+\nu -1,y_1(s+\nu -1))\right| \\\le & {} \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} \left| f(s+\nu -1,y_1(s+\nu -1))\right| \\\le & {} \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )}(\lambda _2-\varepsilon ) \left| y_1(s+\nu -1)\right| . \end{aligned}$$

Multiply both sides of the above inequality by $\phi ^*(t)$ and sum from $\nu -1$ to $\nu +T-1$, and from (2.4) and (2.5) we obtain

$$\begin{aligned} \begin{aligned} \sum _{t=\nu -1}^{\nu +T-1} |y_1(t)|\phi ^*(t)&\le \sum _{t=\nu -1}^{\nu +T-1}\phi ^*(t)\sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} (\lambda _2-\varepsilon ) |y_1(s+\nu -1)|\\&= \sum \limits _{s=0}^T \left( \sum _{t=\nu -1}^{\nu +T-1}\frac{G(t,s)}{\Gamma (\nu )} \phi ^*(t)\right) (\lambda _2-\varepsilon ) |y_1(s+\nu -1)|\\&\le \sum \limits _{s=0}^T \kappa _2 \phi (s+\nu -1) (\lambda _2-\varepsilon ) |y_1(s+\nu -1)| \\&\le \frac{\lambda _2-\varepsilon }{\lambda _2}\sum _{t=\nu -1}^{\nu +T-1} |y_1(t)|\phi ^*(t). \end{aligned} \end{aligned}$$

This implies that $\sum \nolimits _{t=\nu -1}^{\nu +T-1} |y_1(t)|\phi ^*(t)=0$ and then $|y_1(t)|\equiv 0$ for $t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}.$ This contradicts $y_1\in \partial B_r$. Consequently, (3.6) is true. Now Lemma 2.6 implies that

$$\begin{aligned} \deg (I-A,B_r,0)=1. \end{aligned}$$

(3.7)

From (3.5) and (3.7), we have that $\deg (I-A,B_R\backslash \overline{B}_r,0)=\deg (I-A,B_R,0)-\deg (I-A,B_r,0)=-1$, which implies that $I-A$ has at least a zero point in $B_R\backslash \overline{B}_r$, i.e., A has at least one fixed point in $B_R\backslash \overline{B}_r$. Thus (1.1) has at least one nontrivial solution. This completes the proof. $\square$

Theorem 3.2

Suppose that (H1), (H2) and (H4) hold. Then (1.1) has at least one nontrivial solution.

Proof

From the first limit in (H4), there exist $\varepsilon >0$ and $r_1>0$ such that

$$\begin{aligned} f(t,y)\ge (\lambda _1+\varepsilon )|y|,\quad \forall |y|\le r_1,\quad t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}. \end{aligned}$$

(3.8)

Therefore, for each $y\in \overline{B}_{r_1}$, we obtain $(Ay)(t)\ge (\lambda _1+\varepsilon ) \sum \nolimits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} |y(s+\nu -1)|$ for $t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}.$ Then $A(\overline{B}_{r_1})\subset P$. For every $y\in \partial B_{r_1}\cap P$, we claim

$$\begin{aligned} y\not =Ay+\lambda \phi ^{***},\quad \lambda \ge 0, \end{aligned}$$

(3.9)

where $\phi ^{***}\in P$ is an arbitrarily fixed element. Suppose there exist $y_0\in \partial B_{r_1}\cap P$ and $\lambda _0\ge 0$ such that $y_0(t) =(Ay_0)(t)+\lambda _0 \phi ^{***}(t)$ for $t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}.$ This implies that

$$\begin{aligned} y_0(t)\ge (Ay_0)(t) \ge (\lambda _1+\varepsilon ) \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} y_0(s+\nu -1). \end{aligned}$$

Multiply both sides of the above inequality by $\phi ^*(t)$ and sum from $\nu -1$ to $\nu +T-1$, and from (2.4) and (2.5) we obtain

$$\begin{aligned} \begin{aligned} \sum _{t=\nu -1}^{\nu +T-1} y_0 (t)\phi ^*(t)&\ge \sum _{t=\nu -1}^{\nu +T-1} \phi ^*(t) \sum \limits _{s=0}^T (\lambda _1+\varepsilon ) \frac{G(t,s)}{\Gamma (\nu )} y_0(s+\nu -1)\\&= (\lambda _1+\varepsilon ) \sum \limits _{s=0}^T \left( \sum _{t=\nu -1}^{\nu +T-1} \frac{G(t,s)}{\Gamma (\nu )} \phi ^*(t) \right) y_0(s+\nu -1)\\&\ge \kappa _1 (\lambda _1+\varepsilon ) \sum \limits _{s=0}^T y_0(s+\nu -1)\phi (s+\nu -1)\\&\ge \frac{\lambda _1+\varepsilon }{\lambda _1}\sum _{t=\nu -1}^{\nu +T-1} y_0 (t)\phi ^*(t), \end{aligned} \end{aligned}$$

and then $y_0(t)\equiv 0$ for $t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}.$ This contradicts $y_0\in \partial B_{r_1}\cap P$. Hence, (3.9) is satisfied. Note that $A(\overline{B}_{r_1})\subset P$. Then from the permanence property of the fixed point index and Lemma 2.5, we obtain

$$\begin{aligned} \deg (I-A,B_{r_1},0)=i(A,B_{r_1}\cap P,P)=0, \end{aligned}$$

(3.10)

where i denotes the fixed point index on P.

Define an operator $\overline{A}y=A(y-\overline{y})+\overline{y}$. Then $\overline{A}:E\rightarrow P$ is a completely continuous operator. From the second limit in (H4), there exist $r_2\in ( r_1+\Vert \overline{y}\Vert ,+\infty )$ and $\sigma \in (0,1)$ such that

$$\begin{aligned} f(t,y)\le \sigma \lambda _2 y, \quad \forall y\ge r_2, t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}. \end{aligned}$$

(3.11)

Let $L_1 y =\sigma \lambda _2 L y$. Then $L_1: E\rightarrow E$ is a bounded linear operator and $L_1(P)\subset P$. Moreover, note $\sigma \in (0,1)$, and Lemma 2.3 enables us to find $r(L_1)=\sigma \lambda _2 r(L)\le \sigma \lambda _2 \kappa _2 =\sigma <1$. Therefore, $(I-L_1)$ has an inverse operator, denoted by $(I-L_1)^{-1}$.

Let $W=\{y\in P: y=\lambda \overline{A}y, 0\le \lambda \le 1 \}$. We show that W is bounded. For $y\in W$, we note that if $y(s+\nu -1)-\overline{y}(s+\nu -1)<0$, then we have

$$\begin{aligned} y(s+\nu -1)-\overline{y}(s+\nu -1)\ge y(s+\nu -1)-\Vert \overline{y}\Vert \ge -\Vert \overline{y}\Vert >-r_2. \end{aligned}$$

Consequently, if $y(s+\nu -1)-\overline{y}(s+\nu -1)<r_2$, we obtain $\Vert y-\overline{y}\Vert \le r_2$ and then by the continuity of f, there exists $M_1>0$ such that

$$\begin{aligned} |f(s+\nu -1,y(s+\nu -1)-\overline{y}(s+\nu -1))|\le M_1,\quad \text { for } s\in [0,T]_{{\mathbb {Z}}}. \end{aligned}$$

Therefore, for all $y\in W$, from (3.11) and (2.3) we have

$$\begin{aligned} y(t)\le & {} (\overline{A}y)(t)=(A(y-\overline{y}))(t)+\overline{y}(t)\\= & {} \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f(s+\nu -1,y(s+\nu -1)-\overline{y}(s+\nu -1))+ M\sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )}\\= & {} \sum \limits _{\{s\in [0,T]_{{\mathbb {Z}}}:y(s+\nu -1)-\overline{y}(s+\nu -1)\ge r_2\}}\frac{G(t,s)}{\Gamma (\nu )} f(s+\nu -1,y(s+\nu -1)-\overline{y}(s+\nu -1))\\&+\sum \limits _{\{s\in [0,T]_{{\mathbb {Z}}}:y(s+\nu -1)-\overline{y}(s+\nu -1)< r_2\}}\frac{G(t,s)}{\Gamma (\nu )} f(s+\nu -1,y(s+\nu -1)-\overline{y}(s+\nu -1))\\&+\,M\sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )}\\\le & {} \sigma \lambda _2 \sum \limits _{\{s\in [0,T]_{{\mathbb {Z}}}:y(s+\nu -1)-\overline{y}(s+\nu -1)\ge r_2\}}\frac{G(t,s)}{\Gamma (\nu )} (y(s+\nu -1)-\overline{y}(s+\nu -1))\\&+\,M_1\sum \limits _{\{s\in [0,T]_{{\mathbb {Z}}}:y(s+\nu -1)-\overline{y}(s+\nu -1)< r_2\}}\frac{G(t,s)}{\Gamma (\nu )}+M\sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )}\\\le & {} \sigma \lambda _2 \sum \limits _{\{s\in [0,T]_{{\mathbb {Z}}}:y(s+\nu -1)-\overline{y}(s+\nu -1)\ge r_2\}}\frac{G(t,s)}{\Gamma (\nu )} y(s+\nu -1)+M_1\sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )}\\&+\,M\sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} \\\le & {} \sigma \lambda _2 \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )}y(s+\nu -1)+(M_1+M)\sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )}\\\le & {} \sigma \lambda _2 \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )}y(s+\nu -1)+\frac{(M_1+M) C^*}{\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}}} \sum \limits _{s=0}^T (\nu +T-s-1) ^{\underline{\nu -1}}\\= & {} (L_1y)(t)+ \frac{(M_1+M) C^*}{\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}}} \sum \limits _{s=0}^T (\nu +T-s-1) ^{\underline{\nu -1}}. \end{aligned}$$

Then $((I-L_1)y)(t)\le \frac{(M_1+M) C^*}{\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}}} \sum \nolimits _{s=0}^T (\nu +T-s-1) ^{\underline{\nu -1}}$ for $t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}.$ Hence,

$$\begin{aligned} y(t)\le (I-L_1)^{-1} \frac{(M_1+M) C^*}{\Gamma (\nu )-(\nu +T)^{\underline{\nu -1}}} \sum \limits _{s=0}^T (\nu +T-s-1) ^{\underline{\nu -1}} \end{aligned}$$

for $t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}$, and so W is bounded.

Let $r_3>\max \{r_2,\sup W+\Vert \overline{y}\Vert \}$. Then $\overline{A}$ has no fixed points on $\partial B_{r_3}$. If this is false, then there is a $y_0\in \partial B_{r_3}$ such that $\overline{A}y_0=y_0$, and thus $y_0\in W$, $\Vert y_0\Vert =r_3>\sup W$, and this is a contradiction. As a result, from the permanence property and the homotopy invariance property of the fixed point index, we obtain $\deg (I-\overline{A},B_{r_3},0)=i(\overline{A},B_{r_3}\cap P,P)=1.$ Next we claim that

$$\begin{aligned} \deg (I-{A},B_{r_3},0)=\deg (I-\overline{A},B_{r_3},0)=1. \end{aligned}$$

(3.12)

Let $H(l,y)=A(y-l\overline{y})+l\overline{y}$ for $(l,y)\in [0,1]\times \partial B_{r_3}$. Then $H(0,y)=Ay, H(1,y)=\overline{A}y$. Suppose that there exists $(l_0,y_1)\in [0,1]\times \partial B_{r_3}$ such that $H(l_0,y_1)=y_1$. Then $A(y_1-l_0\overline{y})+l_0\overline{y}=y_1$, which implies that $A(y_1-l_0\overline{y})=y_1-l_0\overline{y}$, and $\overline{A}(y_1-l_0\overline{y}+\overline{y})=y_1-l_0\overline{y}+\overline{y}$. Therefore, $y_1-l_0\overline{y}+\overline{y}\in W$, and $\Vert y_1-l_0\overline{y}+\overline{y}\Vert \ge \Vert y_1\Vert -(1-l_0)\Vert \overline{y}\Vert =r_3-(1-l_0)\Vert \overline{y}\Vert >\sup W$. This is a contradiction. From the homotopy invariance of the topological degree, we see that (3.12) holds. From (3.10) and (3.12), we have $\deg (I-A,B_{r_3}\backslash \overline{B}_{r_1},0)=\deg (I-A,B_{r_3},0)-\deg (I-A,B_{r_1},0)=1$, which implies that $I-A$ has at least a zero point in $B_{r_3}\backslash \overline{B}_{r_1}$, i.e., A has at least one fixed point in $B_{r_3}{\backslash }\overline{B}_{r_1}$. Thus (1.1) has at least one nontrivial solution. This completes the proof. $\square$

4 Positive solutions for (1.1)

In this section we have the following assumptions on f.

• (H1’) $f(t,y)\in C([\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}\times {\mathbb {R}}^+, {\mathbb {R}}^+)$, and $f(t,y)>0$ for all $(t,y)\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}\times {\mathbb {R}}^+$.

• (H5) There exits $\mu \in (0,1)$ such that $f(t,ky)\ge k^\mu f(t,y)$ for $k\in (0,1)$.

• (H6) f(t, y) is increasing in y, i.e., $f(t,y_1)\le f(t,y_2)$ holds for $y_1\le y_2$.

Theorem 4.1

Suppose that (H1)$'$ and (H5)–(H6) hold. Then there exist $y_0^*, y_1^*\in P\backslash \{0\}$ such that $y_0^*\le Ay_0^*$, $y_1^*\ge Ay_1^*$.

Proof

Let $w(t)=\sum \nolimits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )}$ for $t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}$. Then from (2.6), $w(t)\in [\kappa _1,\kappa _2]$. Therefore, from (H1)$'$, there exist $a_w,b_w>0$ such that $0<a_w\le f(t,w)\le b_w$. This implies that

$$\begin{aligned} a_w w(t)\le \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f(s+\nu -1,w(s+\nu -1)):=w_0(t)\le b_w w(t), \quad \forall t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}.\nonumber \\ \end{aligned}$$

(4.1)

Let $y_0(t)=\delta w_0(t)$ with $0<\delta <\min \{1/b_w, a_w^{\mu /(1-\mu )}\}$. Then choosing $\varepsilon \in (0,\min \{1,a_w\})$, we have

$$\begin{aligned} \begin{aligned} (A\varepsilon y_0)(t)&=\sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f(s+\nu -1,\varepsilon y_0(s+\nu -1))\\&=\sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f\left( s+\nu -1,\frac{\varepsilon y_0(s+\nu -1)}{w(s+\nu -1)}w(s+\nu -1)\right) \\&\ge \varepsilon ^\mu (\delta a_w)^\mu w_0(t) \ge \varepsilon \delta w_0(t)= \varepsilon y_0(t). \end{aligned} \end{aligned}$$

Now, let $y_0^*=\varepsilon y_0$. Then $y_0^*\le Ay_0^*$.

Let $y_1=\xi w_0(t)$ with $\xi >\max \{1/a_w, b_w^{\mu /(1-\mu )}\}$. Taking $\overline{\varepsilon }>\max \{1,b_w\}$, we obtain

$$\begin{aligned} \begin{aligned} \overline{\varepsilon }y_1(t)&\ge \overline{\varepsilon }^\mu \xi w_0(t) =\overline{\varepsilon }^\mu \xi \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f(s+\nu -1,w(s+\nu -1)) \\&\ge \overline{\varepsilon }^\mu \xi \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f\left( s+\nu -1,\frac{w(s+\nu -1)}{\overline{\varepsilon }y_1(s+\nu -1)}\overline{\varepsilon }y_1(s+\nu -1)\right) \\&\ge \overline{\varepsilon }^\mu \xi \overline{\varepsilon }^{-\mu } (\xi b_w)^{-\mu } \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f\left( s+\nu -1,\overline{\varepsilon }y_1(s+\nu -1)\right) \\&\ge \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f\left( s+\nu -1,\overline{\varepsilon }y_1(s+\nu -1)\right) \\&= (A\overline{\varepsilon }y_1)(t). \end{aligned} \end{aligned}$$

Let $y_1^*=\overline{\varepsilon }y_1$. Then $Ay_1^*\le y_1^*$. This completes the proof. $\square$

Theorem 4.2

Let $y_0^*,y_1^*$ be defined in Theorem 4.1. Then (1.1) has a unique positive solution $y^*\in [y_0^*,y_1^*]$. Moreover, for any $y_0\in [y_0^*,y_1^*]$, the sequence $y_n=A^n y_0\rightarrow y^*$ $(n\rightarrow \infty )$ uniformly in $t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}$.

Proof

From (H6), A is an increasing operator. Then from Section 3 and Theorem 4.1, we note that all conditions of Lemma 2.7 are satisfied. Then A has the smallest fixed point $y_0^{**}$ and the largest fixed point $y_1^{**}$ in $[y_0^*,y_1^*]$, respectively. Moreover, $y_0^{**}=\lim _{n\rightarrow \infty }A^ny_0^*$ and $y_1^{**}=\lim _{n\rightarrow \infty }A^n y_1^*$. Next we claim that $y_0^{**}(t)\equiv y_1^{**}(t)$ for $t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}$. Indeed, from (4.1), there exist $b_i\ge a_i>0(i=1,2)$ such that

$$\begin{aligned} a_1 w(t)\le y_0^{**}(t)\le b_1 w(t), \ a_2 w(t)\le y_1^{**}(t)\le b_2 w(t),\quad \forall t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}, \end{aligned}$$

from the fact that $y_0^{**}$ and $y_1^{**}$ are fixed points of A with $y_0^{**}, y_1^{**}\in [y_0^*,y_1^*]$. Therefore, $y_0^{**}\ge \frac{a_1}{b_2}y_1^{**}$. Let $k_0=\sup \{k>0:y_0^{**}\ge k y_1^{**}\}$. Then $k_0>0$ and $y_0^{**}\ge k_0 y_1^{**}$. We now show $k_0\ge 1$. Suppose the contrary. Then $k_0<1$ and

$$\begin{aligned} \begin{aligned} y_0^{**}(t)&=\sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f(s+\nu -1,y_0^{**}(s+\nu -1))\\&\ge \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} f(s+\nu -1,k_0y_1^{**}(s+\nu -1))\\&\ge \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} k_0^\mu f(s+\nu -1,y_1^{**}(s+\nu -1)). \end{aligned} \end{aligned}$$

Let $g(s+\nu -1)=f(s+\nu -1,k_0y_1^{**}(s+\nu -1))-k_0^\mu f(s+\nu -1,y_1^{**}(s+\nu -1))$ for $s\in [0,T]_{{\mathbb {Z}}}$. Then $g\in P\backslash \{0\}$, and from (4.1) there exist $b_3\ge a_3>0$ such that

$$\begin{aligned} a_3 w(t)\le \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} g(s+\nu -1)\le b_3 w(t). \end{aligned}$$

Consequently, we have

$$\begin{aligned} \begin{aligned} y_0^{**}(t)&\ge \sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} g(s+\nu -1) +\sum \limits _{s=0}^T\frac{G(t,s)}{\Gamma (\nu )} k_0^\mu f(s+\nu -1,y_1^{**}(s+\nu -1))\\&\ge \frac{a_3}{b_2} y_1^{**}(t) + k_0^\mu y_1^{**}(t) > k_0y_1^{**}(t), \end{aligned} \end{aligned}$$

contradicting the definition of $k_0$. As a result, $k_0\ge 1$, and $y_0^{**}\ge y_1^{**}$. This implies that $y_0^{**}(t)\equiv y_1^{**}(t)$ for $t\in [\nu -1,\nu +T-1]_{{\mathbb {Z}}_{\nu -1}}$. Hence, (1.1) has a unique positive solution $y^*\in [y_0^*,y_1^*]$. Moreover, for any $y_0\in [y_0^*,y_1^*]$, then we have $y_0^*\le y_0 \le y_1^*$ and $y_0^*\le Ay_0^*\le Ay_0 \le A y_1^*\le y_1^*$. For $n\in {\mathbb {N}}^+$ large enough, we have

$$\begin{aligned} y_0^*\le Ay_0^*\le \cdots \le A^ny_0^*\le A^ny_0 \le A^n y_1^* \le \cdots \le A y_1^*\le y_1^*. \end{aligned}$$

Let $n\rightarrow \infty$. Then $y_n=A^n y_0\rightarrow y^*$ from the fact that $\lim _{n\rightarrow \infty }A^ny_0^*=\lim _{n\rightarrow \infty }A^n y_1^*=y^*$. This completes the proof. $\square$

Remark 4.3

In [11], the authors established the existence of a unique positive solution to the fractional q-difference boundary value problem

$$\begin{aligned} {\left\{ \begin{array}{ll} (D_q^\alpha y)(x)=-f(x,y(x)),&{}\quad 0<x<1,\quad 2<\alpha \le 3,\\ y(0)=(D_q y)(0)=0,&{}\quad (D_q y)(1)=0, \end{array}\right. } \end{aligned}$$

(4.2)

where f is a nonnegative continuous function and one of their assumptions is

(H)$_{\text {Yang}}$ for any $x\in [0,1]$, $y\in {\mathbb {R}}^+$, there exist two constants m, n with $m\le 0\le n<1$ such that

$$\begin{aligned} c^nf(x,y)\le f(x,cy)\le c^m f(x,y),\quad \text { for } 0<c\le 1. \end{aligned}$$

(4.3)

In our argument above we only need the left inequality of (4.3) (see (H5)) to establish a unique positive solution for (1.1).