Positive solutions for eigenvalue problems of fractional q-difference equation with ϕ-Laplacian

Abstract

The aim of this paper is to investigate the boundary value problem of a fractional q-difference equation with ϕ-Laplacian, where ϕ-Laplacian is a generalized p-Laplacian operator. We obtain the existence and nonexistence of positive solutions in terms of different eigenvalue intervals for this problem by means of the Green function and Guo–Krasnoselskii fixed point theorem on cones. Finally, we give some examples to illustrate the use of our results.

1 Introduction

The theory of q-calculus has been developed for more than 100 years; see [1]. As a branch of q-calculus, fractional q-calculus was first proposed by Al-Salam and Agarwal in the 1960s; see [2, 3]. Fractional q-calculus has a wide range of applications in many fields, such as cosmic strings and black holes, quantum theory, aerospace dynamics, biology, economics, control theory, medicine, electricity, signal processing, image processing, biophysics, blood flow phenomenon, and so on; see [4–10] and the references therein. The fractional q-difference equations are very important, and their basic theory has been continuously developed. Recently, as a new research direction, the solvability of boundary value problems (BVPs) of fractional q-difference equations have been widely concerned by scholars at home and abroad, and some conclusions have been obtained; see [11–14]. However, there are a few studies of eigenvalue problems for fractional q-difference equations with ϕ-Laplacian operator, and lots of work should be done.

In 2013, Li et al. [15] studied some positive solutions for a class of nonlinear fractional q-difference equations with parameters involving the Riemann–Liouville fractional derivative by means of a fixed theorem in cones,

$$\begin{aligned} \textstyle\begin{cases} (D_{q}^{\alpha }u)(x)+\lambda f(u(x))=0,\quad 0< x< 1, \\ u(0)=(D_{q}u)(0)=(D_{q}u)(1)=0, \end{cases}\displaystyle \end{aligned}$$

where $2<\alpha \leq 3$, and $f: [0,1]\times R \rightarrow R$ is a nonnegative continuous function.

In 2015, Wang et al. [16] obtained the existence and uniqueness of solutions for a class of singular BVPs of nonlinear fractional q-difference equations by a fixed point theorem in partially ordered sets,

$$\begin{aligned} \textstyle\begin{cases} (D_{q}^{\alpha }u)(t)+f(t,u(t))=0,\quad 0< t< 1, \\ u(0)=u(1)=0,\qquad (D_{q}u)(0)=0, \end{cases}\displaystyle \end{aligned}$$

where $2<\alpha \leq 3$, and $f:[0,1]\times [0,+\infty ) \rightarrow [0,+\infty )$ is continuous with $\lim_{t\rightarrow 0^{+}}f(t,\cdot )=\infty$.

In 2015, Han et al. [17] used the Green function and Guo–Krasnoselskii fixed-point theorem on cones to study solutions for eigenvalue problems of fractional differential equations with generalized p-Laplacian

$$\begin{aligned} \textstyle\begin{cases} D_{0^{+}}^{\beta }(\phi (D_{0^{+}}^{\alpha }u(t)))=\lambda f(u(t)),\quad 0< t< 1, \\ u(0)=u'(0)=u'(1)=0,\qquad \phi (D_{0^{+}}^{\alpha }u(0))=(\phi (D_{0^{+}}^{ \alpha }u(1)))'=0, \end{cases}\displaystyle \end{aligned}$$

(1.1)

where $0< q<1$, $2<\alpha \leq 3$, $1<\beta \leq 2$, $\lambda >0$ is a parameter, and $D_{0^{+}}^{\beta }$ and $D_{0^{+}}^{\alpha }$ are the standard Riemann–Liouville fractional derivatives.

Motivated by the work above, in this paper, we investigate the following BVP of fractional q-difference equation with ϕ-Laplacian:

$$\begin{aligned} \textstyle\begin{cases} D_{q}^{\beta }(\phi (D_{q}^{\alpha }u(t)))=\lambda f(u(t)),\quad 0< t< 1, \\ u(0)=D_{q}u(0)=D_{q}u(1)=0,\qquad \phi (D_{q}^{\alpha }u(0))=D_{q}(\phi (D_{q}^{ \alpha }u(1)))=0, \end{cases}\displaystyle \end{aligned}$$

(1.2)

where $0< q<1$, $2<\alpha \leq 3$, $1<\beta \leq 2$, $\lambda >0$ is a parameter, and $D_{q}^{\beta }$, $D_{q}^{\alpha }$ are the standard Riemann–Liouville fractional q-derivatives. As $q\to 1^{-}$, problem (1.2) reduces to problem (1.1).

In this paper, we always assume that

• (A1) $\phi :R\rightarrow R$ is an odd increasing homeomorphism, and there exist two increasing homeomorphisms $\psi _{1},~\psi _{2}:(0,\infty )\rightarrow (0,\infty )$ such that

  $$\begin{aligned} \psi _{1}(x)\phi (y)\leq \phi (xy)\leq \psi _{2}(x) \phi (y),\quad x,y>0; \end{aligned}$$

• (A2) $f:(0,+\infty )\rightarrow (0,+\infty )$ is a continuous function.

A function ϕ satisfying (A1) is called a generalized p-Laplacian operator. Two important cases are $\phi (u)=u$ and $\phi (u)=|u|^{p-2}u\ (p>1)$; see [18] and the references therein.

We aim to obtain the existence of at least one or two positive solutions in terms of different eigenvalue intervals using the Green function and Guo–Krasnoselskii fixed point theorem on cones. We also consider the nonexistence of positive solutions in terms of the parameter λ. Finally, we give some examples to illustrate our main results.

2 Preliminary results

In this section, we cite some definitions and fundamental results of the q-calculus and fractional q-calculus.

Definition 2.1

([1])

For $0< q<1$, we define the q-derivative of a real-valued function f as

$$\begin{aligned} (D_{q}f) (x)=\frac{f(x)-f(qx)}{(1-q)x},\qquad (D_{q}f) (0)=\lim_{x \rightarrow 0}(D_{q}f) (x). \end{aligned}$$

Note that $\lim_{q\rightarrow 1^{-}}D_{q}f(x)=f'(x)$.

Definition 2.2

([1])

The q-integral of a function f in the interval $[0,b]$ is given by

$$\begin{aligned} (I_{q}f) (x)= \int _{0}^{x}f(t)\,d_{q}t=x(1-q)\sum _{n=0}^{\infty }f\bigl(xq^{n} \bigr)q^{n},\quad x \in [0,b]. \end{aligned}$$

If $a\in [0,b]$ and f is defined in the interval $[0,b]$, then its integral from a to b is defined as

$$\begin{aligned} \int _{a}^{b}f(t)\,d_{q}t= \int _{0}^{b}f(t)\,d_{q}t- \int _{0}^{a}f(t)\,d_{q}t. \end{aligned}$$

More definitions and properties of q-calculus can be found in [1]. In recent years, some results of q-calculus have been obtained; see [18–20] and the references therein.

Definition 2.3

([10])

Let $\alpha \geq 0$, and let f be a function on $[0,1]$. The fractional q-integral of the Riemann–Liouville type is defined by $(I_{q}^{0}f)(x)=f(x)$ and

$$\begin{aligned} \bigl(I_{q}^{\alpha }f\bigr) (x)=\frac{1}{\Gamma _{q}(\alpha )} \int _{0}^{x}(x-qt)^{( \alpha -1)}f(t) \,d_{q}t,\quad \alpha >0, x\in [0,1]. \end{aligned}$$

Definition 2.4

([21])

The fractional q-derivative of the Riemann–Liouville type of order $\alpha \geq 0$ is defined by $(D_{q}^{0}f)(x)=f(x)$ and

$$\begin{aligned} \bigl(D_{q}^{\alpha }f\bigr) (x)=\bigl(D_{q}^{m}I_{q}^{m-\alpha }f \bigr) (x),\quad \alpha >0, \end{aligned}$$

where m is the smallest integer greater than or equal to α.

Lemma 2.1

([10, 21])

Let $\alpha ,\beta \geq 0$, and let f be a continuous differentiable function on $[0,1]$. Then we have the following formulas:

• 1. $(I_{q}^{\beta }I_{q}^{\alpha }f)(x)=(I_{q}^{\alpha +\beta }f)(x)$,

• 2. $(D_{q}^{\alpha }I_{q}^{\alpha }f)(x)=f(x)$.

Lemma 2.2

([22])

Let $f:[0,1]\rightarrow R$ be differentiable, let p be a positive integer, and let $\alpha >p-1$. Then

$$\begin{aligned} \bigl(I_{q}^{\alpha }D_{q}^{p}f \bigr) (x)=\bigl(D_{q}^{p}I_{q}^{\alpha }f \bigr) (x) -\sum_{k=0}^{p-1} \frac{x^{\alpha -p+k}}{\Gamma _{q}(\alpha +k-p+1)}\bigl(D_{q}^{k}f\bigr) (0). \end{aligned}$$

Lemma 2.3

([23])

Let $\alpha \in R^{+}$, $n:=\lceil \alpha \rceil$. Then

$$\begin{aligned} \bigl(I_{q}^{\alpha }D_{q}^{\alpha }f \bigr) (x)=f(x) -\sum_{j=1}^{n}D_{q}^{ \alpha -j}f \bigl(0^{+}\bigr)\frac{x^{\alpha -j}}{\Gamma _{q}(\alpha -j+1)}. \end{aligned}$$

Lemma 2.4

([22])

Let $\alpha \geq 0$. Then we have the following three formulas:

$$\begin{aligned}& \bigl[a(t-s)\bigr]^{(\alpha )}=a^{\alpha }(t-s)^{(\alpha )}, \\& _{t}D_{q}(t-s)^{(\alpha )}=[ \alpha ]_{q}(t-s)^{(\alpha -1)}, \\& {_{x}}D_{q} \int _{0}^{x}f(x,t)\,d_{q}t= \int _{0}^{x}{_{x}}D_{q}f(x,t)\,d_{q}t+f(qx,x). \end{aligned}$$

Remark 2.1

([22])

Note that if $\alpha \geq 0$ and $a\leq b\leq t$, then $(t-a)^{(\alpha )}\geq (t-b)^{(\alpha )}$.

Lemma 2.5

Let $y\in C_{q}[0,1]$, $2<\alpha \leq 3$, $1<\beta \leq 2$. Then the BVP

$$\begin{aligned} \textstyle\begin{cases} D_{q}^{\beta }(\phi (D_{q}^{\alpha }u(t)))=\lambda y(t),\quad 0< t< 1, \\ u(0)=(D_{q}u)(0)=D_{q}u(1)=0,\qquad \phi (D_{q}^{\alpha }u(0))=D_{q}(\phi (D_{q}^{ \alpha }u(1)))=0, \end{cases}\displaystyle \end{aligned}$$

(2.1)

has a unique solution

$$\begin{aligned} u(t)= \int _{0}^{1}G(t,qs)\phi ^{-1} \biggl(\lambda \int _{0}^{1}H(s,q\tau )y( \tau )\,d_{q}\tau \biggr)\,d_{q}s, \end{aligned}$$

where

$$\begin{aligned}& H(s,\tau )=\frac{1}{\Gamma _{q}(\beta )} \textstyle\begin{cases} s^{\beta -1}(1-\tau )^{(\beta -2)}-(s-\tau )^{(\beta -1)},&\tau \leq s, \\ s^{\beta -1}(1-\tau )^{(\beta -2)},&\tau \geq s, \end{cases}\displaystyle \end{aligned}$$

(2.2)

$$\begin{aligned}& G(t,s)=\frac{1}{\Gamma _{q}(\alpha )} \textstyle\begin{cases} t^{\alpha -1}(1-s)^{(\alpha -2)}-(t-s)^{(\alpha -1)},&s\leq t, \\ t^{\alpha -1}(1-s)^{(\alpha -2)},&s\geq t. \end{cases}\displaystyle \end{aligned}$$

(2.3)

Proof

By Lemma 2.3 we have

$$\begin{aligned} \phi \bigl(D_{q}^{\alpha }u(t)\bigr)=C_{1}t^{\beta -1}+C_{2}t^{\beta -2}+ \lambda \int _{0}^{t} \frac{(t-q\tau )^{(\beta -1)}}{\Gamma _{q}(\beta )}y(\tau )\,d_{q}\tau . \end{aligned}$$

Using Lemma 2.4 and the boundary conditions $\phi (D_{q}^{\alpha }u(0))=D_{q}(\phi (D_{q}^{\alpha }u(1)))=0$, we get that

$$\begin{aligned} C_{2}=0,~C_{1}=-\lambda \int _{0}^{1} \frac{(1-q\tau )^{(\beta -2)}}{\Gamma _{q}(\beta )}y(\tau )\,d_{q}\tau . \end{aligned}$$

So we can obtain

$$\begin{aligned} \phi \bigl(D_{q}^{\alpha }u(t)\bigr) =&-\lambda \int _{0}^{1} \frac{t^{\beta -1}(1-q\tau )^{(\beta -2)}}{\Gamma _{q}(\beta )}y( \tau )\,d_{q}\tau +\lambda \int _{0}^{t} \frac{(t-q\tau )^{(\beta -1)}}{\Gamma _{q}(\beta )}y(\tau )\,d_{q}\tau \\ =&-\lambda \int _{0}^{1}H(t,q\tau )y(\tau )\,d_{q}\tau . \end{aligned}$$

Further, from

$$\begin{aligned} D_{q}^{\alpha }u(t)=-\phi ^{-1}\biggl(\lambda \int _{0}^{1}H(t,q\tau )y( \tau )\,d_{q}\tau \biggr) \end{aligned}$$

by Lemma 2.3 we have

$$\begin{aligned} u(t)=C_{3}t^{\alpha -1}+C_{4}t^{\alpha -2}+C_{5}t^{\alpha -3}- \int _{0}^{t} \frac{(t-qs)^{(\alpha -1)}}{\Gamma _{q}(\alpha )} \phi ^{-1}\biggl(\lambda \int _{0}^{1}H(s,q\tau )y(\tau )\,d_{q}\tau \biggr)\,d_{q}s. \end{aligned}$$

Using the boundary condition $u(0)=(D_{q}u)(0)=D_{q}u(1)=0$, we get

$$\begin{aligned} C_{5}=0,\qquad C_{4}=0,\qquad C_{3}= \int _{0}^{1} \frac{(1-qs)^{(\alpha -2)}}{\Gamma _{q}(\alpha )}\phi ^{-1}\biggl(\lambda \int _{0}^{1}H(s,q\tau )y(\tau )\,d_{q}\tau \biggr)\,d_{q}s, \end{aligned}$$

so we have

$$\begin{aligned} u(t) =& \int _{0}^{1} \frac{t^{\alpha -1}(1-qs)^{(\alpha -2)}}{\Gamma _{q}(\alpha )}\phi ^{-1}\biggl( \lambda \int _{0}^{1}H(s,q\tau )y(\tau )\,d_{q}\tau \biggr)\,d_{q}s \\ &{}- \int _{0}^{t}\frac{(t-qs)^{(\alpha -1)}}{\Gamma _{q}(\alpha )} \phi ^{-1}\biggl( \lambda \int _{0}^{1}H(s,q\tau )y(\tau )\,d_{q}\tau \biggr)\,d_{q}s \\ =& \int _{0}^{1}G(t,qs)\phi ^{-1} \biggl(\lambda \int _{0}^{1}H(s,q\tau )y( \tau )\,d_{q}\tau \biggr)\,d_{q}s. \end{aligned}$$

 □

Lemma 2.6

Let $2<\alpha \leq 3$. The functions $G(t,qs)$ and $H(s,q\tau )$ defined by (2.2) and (2.3), respectively, are continuous on $[0,1]\times [0,1]$ and satisfy

• (i) $G(t,qs)\geq 0$, $H(s,q\tau )\geq 0$ for $t,s,\tau \in [0,1]$;

• (ii) $G(t,qs)\leq G(1,qs)$, $H(s,q\tau )\leq H(q\tau ,q\tau )$ for $t,s,\tau \in [0,1]$;

• (iii) $G(t,qs)\geq k(t)G(1,qs)$, $H(s,q\tau )\geq s^{\beta -1} H(1,q\tau )$, where $k(t)=t^{\alpha -1}$, for $t,s,\tau \in [0,1]$.

Proof

(i) Let

$$\begin{aligned}& g_{1}(t,qs)=t^{\alpha -1}(1-qs)^{(\alpha -2)}-(t-qs)^{(\alpha -1)}, \quad s \leq t, \\& g_{2}(t,qs)=t^{\alpha -1}(1-qs)^{(\alpha -2)},\quad s\geq t. \end{aligned}$$

It is clear that $g_{2}(t,qs)\geq 0$ for $t,s\in [0,1]$. If $s\leq t$, then in view of Remark 2.1, for $t\neq 0$,

$$\begin{aligned} g_{1}(t,qs) =&t^{\alpha -1}(1-qs)^{(\alpha -2)}-(t-qs)^{(\alpha -1)} \\ =&t^{\alpha -1}(1-qs)^{(\alpha -2)}-t^{\alpha -1}(1-qs/t)^{(\alpha -1)} \\ \geq &t^{\alpha -1}\bigl[(1-qs)^{(\alpha -2)}-(1-qs)^{(\alpha -1)} \bigr] \\ \geq& 0. \end{aligned}$$

Therefore $G(t,qs)\geq 0$. In the same way, we can obtain that $H(s,q\tau )\geq 0$.

(ii) Fix $s\in [0,1]$. For $t\neq 0$, we have

$$\begin{aligned} _{t}D_{q}g_{1}(t,qs) =&[\alpha -1]_{q}t^{\alpha -2}(1-qs)^{(\alpha -2)}-[ \alpha -1]_{q}(t-qs)^{(\alpha -2)} \\ =&[\alpha -1]_{q}t^{\alpha -2}(1-qs)^{(\alpha -2)}-[\alpha -1]_{q}t^{ \alpha -2}\biggl(1-\frac{qs}{t} \biggr)^{(\alpha -2)} \\ =&[\alpha -1]_{q}t^{\alpha -2}\biggl[(1-qs)^{(\alpha -2)}- \biggl(1-\frac{qs}{t}\biggr)^{( \alpha -2)}\biggr] \geq 0 \end{aligned}$$

and

$$\begin{aligned} _{t}D_{q}g_{2}(t,qs)=[\alpha -1]_{q}t^{\alpha -2}(1-qs)^{(\alpha -2)} \geq 0. \end{aligned}$$

Therefore $g_{1}(t,qs)$, $g_{2}(t,qs)$ are increasing functions of t for $s\in [0,1]$. Thus $G(t,qs)\leq G(1,qs)$. In the same way, we get that $H(s,q\tau )\leq H(q\tau ,q\tau )$.

(iii) Suppose that $s\leq t$. Then

$$\begin{aligned} \frac{G(t,qs)}{G(1,qs)} =& \frac{t^{\alpha -1}(1-qs)^{(\alpha -2)}-(t-qs)^{(\alpha -1)}}{(1-qs)^{(\alpha -2)}-(1-qs)^{(\alpha -1)}} \\ =& \frac{t^{\alpha -1}(1-qs)^{(\alpha -2)}-t^{\alpha -1}(1-\frac{qs}{t})^{(\alpha -1)}}{(1-qs)^{(\alpha -2)}-(1-qs)^{(\alpha -1)}} \\ \geq& t^{\alpha -1}. \end{aligned}$$

For the other circumstance, we also get $G(t,qs)\geq t^{\alpha -1}G(1,qs)$. In the same way, we get that $H(s,q\tau )\geq s^{\beta -1} H(1,q\tau )$. The proof is completed. □

Lemma 2.7

([17])

Assume that (A1) holds. Then

$$\begin{aligned} \psi ^{-1}_{2}(x)y\leq \phi ^{-1}\bigl(x \phi (y)\bigr)\leq \psi ^{-1}_{1}(x)y,\quad x,y \in (0,\infty ). \end{aligned}$$

Theorem 2.1

([24] (Krasnoselskii))

Let E be a Banach space, and let $K\in E$ be a cone in E. Let $\Omega _{1}$ and $\Omega _{2}$ be open subsets of E with $0\in \Omega _{1}$ and $\overline{\Omega }_{1}\subset \Omega _{2}$. Let $T:K\cap (\overline{\Omega }_{2}\backslash \Omega _{1})\rightarrow K$ be a completely continuous operator. In addition, suppose that either

$(\mathrm {H_{1}})$:

$\|Tu\|\leq \|u\|$, $\forall u\in K\cap \partial \Omega _{1}$ and $\|Tu\|\geq \|u\|$, $\forall u\in K\cap \partial \Omega _{2}$ or

$(\mathrm {H_{2}})$:

$\|Tu\|\leq \|u\|$, $\forall u\in K\cap \partial \Omega _{2}$ and $\|Tu\|\geq \|u\|$, $\forall u\in K\cap \partial \Omega _{1}$.

Then T has a fixed point in $K\cap (\overline{\Omega }_{2}\backslash \Omega _{1})$.

3 Main results

In this section, we consider the existence of at least one or two positive solutions or no positive solution for the BVP (1.1).

Let the Banach space $E=C_{q}[0,1]$ be endowed with norm $\|u\|=\max_{0\leq t\leq 1}|u(t)|$. Define the cone

$$\begin{aligned} P=\bigl\{ u\in E\mid u(t)\geq k(t) \Vert u \Vert , t\in [0,1]\bigr\} \subset E . \end{aligned}$$

Let $T_{\lambda }:P\rightarrow P$ be the operator defined by

$$\begin{aligned} T_{\lambda }u(t):= \int _{0}^{1}G(t,qs)\phi ^{-1} \biggl(\lambda \int ^{1}_{0}H(s,q \tau )f\bigl(u(\tau ) \bigr)\,d_{q}\tau \biggr)\,d_{q}s. \end{aligned}$$

Lemma 3.1

Assume that (A2) holds. Then $T_{\lambda }:P\rightarrow P$ is completely continuous.

Proof

By Lemma 2.6 we have

$$\begin{aligned} (T_{\lambda }u) (t) \geq &t^{\alpha -1} \int _{0}^{1}G(1,qs)\phi ^{-1} \biggl( \lambda \int ^{1}_{0}H(s,q\tau )f\bigl(u(\tau ) \bigr)\,d_{q}\tau \biggr)\,d_{q}s = k(t) \bigl\Vert T_{\lambda }u(t) \bigr\Vert . \end{aligned}$$

Thus $T_{\lambda }(P)\subset P$. In view of the nonnegativeness and continuity of $G(t,qs)$, $H(s,q\tau )$, and $f(u(\tau ))$, we have that $T_{\lambda }:P\rightarrow P$ is continuous.

Next, we prove that $T_{\lambda }$ is uniformly bounded.

Let $\Omega \subset P$ be bounded, that is, there exists a positive constant $M>0$ such that $\| u\| \leq M$ for all $u\in \Omega$. Set $L=\max_{0\leq u\leq M}|f(u)|+1$. Then, for $u\in \Omega$ and all $t\in [0,1]$, we have

$$\begin{aligned} \bigl\vert T_{\lambda }u(t) \bigr\vert =& \biggl\vert \int _{0}^{1}G(t,qs)\phi ^{-1} \biggl(\lambda \int ^{1}_{0}H(s,q \tau )f\bigl(u(\tau ) \bigr)\,d_{q}\tau \biggr)\,d_{q}s \biggr\vert \\ \leq& \psi ^{-1}_{1}(\lambda L) \int ^{1}_{0}G(t,qs)\phi ^{-1} \biggl( \int ^{1}_{0}H(s,q \tau )\,d_{q}\tau \biggr)\,d_{q}s \\ \leq& \psi ^{-1}_{1}(\lambda L) \int ^{1}_{0}G(1,qs)\phi ^{-1} \biggl( \int ^{1}_{0}H(q \tau ,q\tau )\,d_{q}\tau \biggr)\,d_{q}s \\ < &+\infty . \end{aligned}$$

Hence $T_{\lambda }(\Omega )$ is uniformly bounded.

On the other hand, we prove that $T_{\lambda }$ is equicontinuous.

Since $G(t,qs)$ is continuous on $[0,1]\times [0,1]$, it is uniformly continuous on $[0,1]\times [0,1]$. Thus, for any $\varepsilon >0$, there exists a constant $\delta >0$ such that $t_{1},t_{2}\in [0,1]$ with $| t_{1}-t_{2}| <\delta$ imply

$$\begin{aligned} \bigl\vert G(t_{2},qs)-G(t_{1},qs) \bigr\vert < \frac{\varepsilon }{\psi ^{-1}_{1}(\lambda L)\phi ^{-1}(\int ^{1}_{0}H(q\tau ,q\tau )\,d_{q}\tau )}. \end{aligned}$$

Then, for all $u\in \Omega$,

$$\begin{aligned} \bigl\vert T_{\lambda }u(t_{2})-T_{\lambda }u(t_{1}) \bigr\vert \leq& \int ^{1}_{0} \bigl\vert G(t_{2},qs)-G(t_{1},qs) \bigr\vert \phi ^{-1}\biggl(\lambda \int ^{1}_{0}H(s,q\tau )f\bigl(u(\tau ) \bigr)\,d_{q}\tau \biggr)\,d_{q}s~ \\ \leq &\psi ^{-1}_{1}(\lambda L) \int ^{1}_{0} \bigl\vert G(t_{2},qs)-G(t_{1},qs) \bigr\vert \phi ^{-1}\biggl( \int ^{1}_{0}H(q\tau ,q\tau )\,d_{q} \tau \biggr)\,d_{q}s \\ =&\psi ^{-1}_{1}(\lambda L)\phi ^{-1}\biggl( \int ^{1}_{0}H(q\tau ,q\tau )\,d_{q} \tau \biggr) \int ^{1}_{0} \bigl\vert G(t_{2},qs)-G(t_{1},qs) \bigr\vert \,d_{q}s \\ < &\varepsilon . \end{aligned}$$

Hence $T_{\lambda }(\Omega )$ is equicontinuous. By the Arzelà–Ascoli theorem we have that $T_{\lambda }:P\rightarrow P$ is completely continuous. The proof is completed. □

For convenience, we denote

$$\begin{aligned}& F_{0}=\lim_{u\rightarrow 0^{+}}\sup \frac{f(u)}{\phi (u)}, \qquad F_{ \infty }=\lim_{u\rightarrow +\infty }\sup \frac{f(u)}{\phi (u)}, \\& f_{0}= \lim_{u\rightarrow 0^{+}}\inf \frac{f(u)}{\phi (u)},\qquad f_{\infty }= \lim_{u\rightarrow +\infty }\inf \frac{f(u)}{\phi (u)}, \\& A_{1}= \int ^{1}_{0}G(1,qs)\psi ^{-1}_{1} \biggl( \int ^{1}_{0}H(q\tau ,q\tau )\,d_{q} \tau \biggr)\,d_{q}s, \\& A_{2}=k(\delta ) \int ^{1}_{0}\psi _{2}^{-1} \bigl(s^{\beta -1}\bigr)G(1,qs)\psi ^{-1}_{2} \biggl( \int ^{1}_{0}\psi _{1}\bigl(\tau ^{\alpha -1}\bigr)H(1,q\tau )\,d_{q}\tau \biggr)\,d_{q}s, \\& A_{3}=k(\delta ) \int ^{1}_{0}\psi _{2}^{-1} \bigl(s^{\beta -1}\bigr)G(1,qs)\psi ^{-1}_{2} \biggl( \int ^{1}_{0}H(1,q\tau )\,d_{q}\tau \biggr)\,d_{q}s. \end{aligned}$$

Theorem 3.1

Assume that (A1), (A2), and $f_{\infty }\psi _{1}(A_{1}^{-1})>F_{0}\psi _{2}(A_{2}^{-1})$ hold. Then for each

$$\begin{aligned} \lambda \in \bigl(\psi _{2}\bigl(A_{2}^{-1} \bigr)f_{\infty }^{-1},\psi _{1} \bigl(A_{1}^{-1}\bigr)F_{0}^{-1} \bigr), \end{aligned}$$

(3.1)

the BVP of fractional q-difference Eq. (1.1) has at least one positive solution. Here we impose that $f_{\infty }^{-1}=0$ if $f_{\infty }=+\infty$ and $F_{0}^{-1}=+\infty$ if $F_{0}=0$.

Proof

Let λ satisfy (3.1), and let $\varepsilon >0$ be such that

$$\begin{aligned} \psi _{2}\bigl(A_{2}^{-1} \bigr) (f_{\infty }-\varepsilon )^{-1}\leq \lambda \leq \psi _{1}\bigl(A_{1}^{-1}\bigr) (F_{0}+\varepsilon )^{-1}. \end{aligned}$$

(3.2)

We separate the proof into two steps.

(1) By the definition of $F_{0}$ there exists $r_{1}>0$ such that

$$\begin{aligned} f(u)\leq (F_{0}+\varepsilon )\phi (u),\quad 0< u< r_{1}. \end{aligned}$$

(3.3)

If $u\in P$ with $\| u\| =r_{1}$, then from (3.2) and (3.3) we obtain

$$\begin{aligned} \bigl\Vert T_{\lambda }u(t) \bigr\Vert \leq& \int _{0}^{1}G(1,qs)\phi ^{-1} \biggl( \lambda \int _{0}^{1}H(q\tau ,q\tau )f\bigl(u(\tau ) \bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ \leq & \int _{0}^{1}G(1,qs)\phi ^{-1} \biggl(\lambda \int _{0}^{1}H(q\tau ,q \tau ) (F_{0}+\varepsilon )\phi (r_{1})\biggr)\,d_{q} \tau )\,d_{q}s \\ \leq& \psi _{1}^{-1}\bigl(\lambda (F_{0}+ \varepsilon )\bigr) \int _{0}^{1}G(1,qs) \phi ^{-1} \biggl( \int _{0}^{1}H(q\tau ,q\tau )\phi (r_{1})\biggr)\,d_{q}\tau )\,d_{q}s \\ \leq& \psi _{1}^{-1}\bigl(\lambda (F_{0}+ \varepsilon )\bigr)r_{1} \int _{0}^{1}G(1,qs) \psi _{1}^{-1} \biggl( \int _{0}^{1}H(q\tau ,q\tau )\,d_{q} \tau \biggr)\,d_{q}s \\ =&\psi _{1}^{-1}\bigl(\lambda (F_{0}+ \varepsilon )\bigr)A_{1}r_{1}\leq r_{1}= \Vert u \Vert . \end{aligned}$$

Let $\Omega _{1}=\{u\in E\mid \| u\| < r_{1}\}$. Then

$$\begin{aligned} \Vert T_{\lambda }u \Vert \leq \Vert u \Vert , \quad u\in P \cap \partial \Omega _{1}. \end{aligned}$$

(3.4)

(2) By the definition of $f_{\infty }$ there exists $r_{3}>0$ such that

$$\begin{aligned} f(u)\geq (f_{\infty }-\varepsilon )\phi (u),\quad u>r_{3}. \end{aligned}$$

(3.5)

If $u\in P$ with $\| u\| =r_{2}=\max \{2r_{1},r_{3}\}$, then from (3.2) and (3.5) we obtain

$$\begin{aligned} \bigl\Vert T_{\lambda }u(t) \bigr\Vert \geq & \int _{0}^{1}k(\delta )G(1,qs) \phi ^{-1}\biggl(\lambda \int _{0}^{1}H(s,q\tau )f\bigl(u(\tau ) \bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ \geq& \int _{0}^{1}k(\delta )G(1,qs)\phi ^{-1}\biggl(\lambda \int _{0}^{1}s^{ \beta -1}H(1,q\tau ) (f_{\infty }-\varepsilon )\phi \bigl(\tau ^{\alpha -1} \Vert u \Vert \bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ \geq& \int _{0}^{1}k(\delta )G(1,qs)\phi ^{-1}\biggl(\lambda \int _{0}^{1}s^{ \beta -1}H(1,q\tau ) (f_{\infty }-\varepsilon )\psi _{1}\bigl(\tau ^{\alpha -1}\bigr) \phi (r_{2})\,d_{q}\tau \biggr)\,d_{q}s \\ \geq& \psi _{2}^{-1}\bigl(\lambda (f_{\infty }- \varepsilon )\bigr)r_{2}k(\delta )\\ &{}\times \int _{0}^{1}\psi _{2}^{-1} \bigl(s^{\beta -1}\bigr)G(1,qs)\psi _{2}^{-1} \biggl( \int _{0}^{1}H(1,q \tau )\psi _{1} \bigl(\tau ^{\alpha -1}\bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ =&\psi _{2}^{-1}\bigl(\lambda (f_{\infty }- \varepsilon )\bigr)A_{2}r_{2}\geq r_{2}= \Vert u \Vert . \end{aligned}$$

Let $\Omega _{2}=\{u\in E\mid \| u\| < r_{2}\}$. Then

$$\begin{aligned} \Vert T_{\lambda }u \Vert \geq \Vert u \Vert , \quad u\in P \cap \partial \Omega _{2}. \end{aligned}$$

(3.6)

From (3.4) and (3.6) and from Theorem 2.1 we have that $T_{\lambda }$ has a fixed point $u\in P\cap (\overline{\Omega }_{2}\setminus \Omega _{1})$ with $r_{1}\leq \| u\| \leq r_{2}$ and that u is a positive solution for the BVP of fractional q-difference Eq. (1.1). The proof is completed. □

Theorem 3.2

Assume that (A1), (A2), and $f_{0}\psi _{1}(A_{1}^{-1})>F_{\infty }\psi _{2}(A_{2}^{-1})$ hold. Then for each

$$\begin{aligned} \lambda \in \bigl(\psi _{2}\bigl(A_{2}^{-1} \bigr)f_{0}^{-1},\psi _{1} \bigl(A_{1}^{-1}\bigr)F_{ \infty }^{-1} \bigr), \end{aligned}$$

(3.7)

the BVP of fractional q-difference Eq. (1.1) has at least one positive solution. Here we impose that $f_{0}^{-1}=0$ if $f_{0}=+\infty$ and $F_{\infty }^{-1}=+\infty$ if $F_{\infty }=0$.

Proof

Let λ satisfy (3.7), and let $\varepsilon >0$ be such that

$$\begin{aligned} \psi _{2}\bigl(A_{2}^{-1} \bigr) (f_{0}-\varepsilon )^{-1}\leq \lambda \leq \psi _{1}\bigl(A_{1}^{-1}\bigr) (F_{ \infty }+\varepsilon )^{-1}. \end{aligned}$$

(3.8)

We separate the proof into two steps.

(1) By the definition of $f_{0}$ there exists $r_{1}>0$ such that

$$\begin{aligned} f(u)\geq (f_{0}-\varepsilon )\phi (u),\quad 0< u \leq r_{1}. \end{aligned}$$

(3.9)

If $u\in P$ with $\| u\| =r_{1}$, then similarly to the second part of the proof of Theorem 3.1, let $\Omega _{1}=\{u\in E\mid \| u\| < r_{1}\}$. Then

$$\begin{aligned} \Vert T_{\lambda }u \Vert \geq \Vert u \Vert , \quad u\in P \cap \partial \Omega _{1}. \end{aligned}$$

(3.10)

(2) By the definition $F_{\infty }$ there exists $R_{1}>0$ such that

$$\begin{aligned} f(u)\leq (F_{\infty }+\varepsilon )\phi (u),\quad u> R_{1}. \end{aligned}$$

(3.11)

We consider two cases:

Case 1: When f is bounded, then there exists $N>0$, such that $|f(u)|\leq N$ for $u\in (0,+\infty )$. If $u\in P$ with $\|u\|=r_{3}$, where $r_{3}=\max \{2r_{1},\phi ^{-1}(\lambda N)A_{1}\}$, then

$$\begin{aligned} \bigl\Vert T_{\lambda }u(t) \bigr\Vert \leq& \int ^{1}_{0}G(1,qs)\phi ^{-1} \biggl(\lambda \int _{0}^{1}H(q \tau ,q\tau )f\bigl(u(\tau ) \bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ \leq& \phi ^{-1}(\lambda N) \int ^{1}_{0}G(1,qs)\psi _{1}^{-1} \biggl( \int _{0}^{1}H(q \tau ,q\tau )\,d_{q}\tau \biggr)\,d_{q}s \\ \leq& \phi ^{-1}(\lambda N)A_{1} \\ \leq& r_{3}= \Vert u \Vert . \end{aligned}$$

So let $\Omega _{3}=\{u\in E\mid \| u\| < r_{3}\}$. Then

$$\begin{aligned} \Vert T_{\lambda }u \Vert \leq \Vert u \Vert , \quad u\in P \cap \partial \Omega _{3}. \end{aligned}$$

(3.12)

Case 2: Suppose f is unbounded. Then there exists $r_{4}>\max \{2r_{1},R_{1}\}$ such that $f(u)\leq f(r_{4})$ for $0< u< r_{4}$. If $u\in P$ with $\| u\| =r_{4}$, then by (3.7) and (3.11) we have

$$\begin{aligned} \bigl\Vert T_{\lambda }u(t) \bigr\Vert \leq& \int _{0}^{1}G(1,qs)\phi ^{-1} \biggl( \lambda \int _{0}^{1}H(q\tau ,q\tau )f\bigl(u(\tau ) \bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ \leq& \int _{0}^{1}G(1,qs)\phi ^{-1} \biggl(\lambda \int _{0}^{1}H(q\tau ,q \tau ) (F_{\infty }+\varepsilon )\phi (r_{4})\,d_{q}\tau \biggr)\,d_{q}s \\ \leq& \psi _{1}^{-1}\bigl(\lambda (F_{\infty }+ \varepsilon )\bigr)A_{1}r_{4} \\ \leq& r_{4}= \Vert u \Vert . \end{aligned}$$

Thus we suppose $\Omega _{4}=\{u\in E\mid \| u\| < r_{4}\}$. Then

$$\begin{aligned} \Vert T_{\lambda }u \Vert \leq \Vert u \Vert , \quad u\in P \cap \partial \Omega _{4}. \end{aligned}$$

(3.13)

In view of Cases 1 and 2, we let $\Omega _{2}=\{u\in E\mid \| u\| < r_{2}\}$, where $r_{2}=\max \{r_{3},r_{4}\}$. Then

$$\begin{aligned} \Vert T_{\lambda }u \Vert \leq \Vert u \Vert , \quad u\in P \cap \partial \Omega _{2}. \end{aligned}$$

(3.14)

From (3.10) and (3.14) and from Theorem 2.1 we obtain that $T_{\lambda }$ has a fixed point $u\in P\cap (\overline{\Omega }_{2}\backslash \Omega _{1})$ with $r_{1}\leq \| u\| \leq r_{2}$. Obviously, u is a positive solution of the BVP of fractional q-difference Eq. (1.1). The proof is completed. □

Theorem 3.3

Assume that (A1) and (A2) hold and there exist $r_{2}>r_{1}>0$ such that

$$\begin{aligned} \lambda \min_{k(\delta )r_{1}\leq u\leq r_{1}}f(u)\geq \phi \biggl( \frac{r_{1}}{A_{3}}\biggr),\qquad \lambda \max_{0\leq u\leq r_{2}}f(u)\leq \phi \biggl(\frac{r_{2}}{A_{1}}\biggr). \end{aligned}$$

Then the BVP of fractional q-difference Eq. (1.1) has a positive solution $u\in P$ with $r_{1}\leq \| u\| \leq r_{2}$.

Proof

Let $\Omega _{1}=\{u\in E|\|u\|< r_{1}\}$. Then for $u\in P\cap \partial \Omega _{1}$, we obtain

$$\begin{aligned} \bigl\Vert T_{\lambda }u(t) \bigr\Vert \geq& T_{\lambda }u( \delta ) \\ =& \int _{0}^{1}G(\delta ,qs)\phi ^{-1}\biggl(\lambda \int _{0}^{1}H(s,q\tau )f\bigl(u( \tau ) \bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ \geq& \int _{0}^{1}k(\delta )G(1,qs)\phi ^{-1}\biggl(\lambda \int _{0}^{1}s^{ \beta -1}H(1,q\tau )\min _{k(\delta )r_{1}\leq u\leq r_{1}}f\bigl(u(\tau )\bigr)\,d_{q} \tau \biggr)\,d_{q}s \\ \geq& \int _{0}^{1}k(\delta )G(1,qs)\psi _{2}^{-1}\bigl(s^{\beta -1}\bigr)\phi ^{-1}\biggl( \lambda \int _{0}^{1}H(1,q\tau )\min _{k(\delta )r_{1}\leq u\leq r_{1}}f\bigl(u( \tau )\bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ \geq& \phi ^{-1}\Bigl(\lambda \min_{k(\delta )r_{1}\leq u\leq r_{1}}f(u) \Bigr)A_{3} \\ \geq& r_{1}= \Vert u \Vert . \end{aligned}$$

Suppose $\Omega _{2}=\{u\in E\mid \| u\| < r_{2}\}$. Then for $u\in P\cap \partial \Omega _{2}$, we have

$$\begin{aligned} \bigl\Vert T_{\lambda }u(t) \bigr\Vert \leq& \int _{0}^{1}G(1,qs)\phi ^{-1} \biggl( \lambda \int _{0}^{1}H(q\tau ,q\tau )f\bigl(u(\tau ) \bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ \leq & \int _{0}^{1}G(1,qs)\phi ^{-1} \biggl(\lambda \int _{0}^{1}H(q\tau ,q \tau )\max _{0\leq u\leq r_{2}}f\bigl(u(\tau )\bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ \leq &\phi ^{-1}\Bigl(\lambda \max_{0\leq u\leq r_{2}}f \bigl(u(\tau )\bigr)\Bigr)A_{1} \\ \leq &r_{2}= \Vert u \Vert . \end{aligned}$$

Thus by Theorem 2.1 the BVP of fractional q-difference Eq. (1.1) has a positive solution $u\in P$ with $r_{1}\leq \| u\| \leq r_{2}$. The proof is completed. □

Theorem 3.4

Assume that (A1) and (A2) hold. Let $\lambda _{1}=\sup_{r>0} \frac{\phi (r)}{\phi (A_{1})\max_{0\leq u\leq r}f(u)}$. If $f_{0}=+\infty$ and $f_{\infty }=+\infty$, then the BVP of fractional q-difference Eq. (1.1) has at least two positive solutions for each $\lambda \in (0,\lambda _{1})$.

Proof

We define $x(r)=\frac{\phi (r)}{\psi _{2}(A_{1})\max_{0\leq u\leq r}f(u)}$. In view of the continuity of $f,~f_{0}=+\infty$, and $f_{\infty }=+\infty$, we obtain that $x(r):(0,+\infty )\rightarrow (0,+\infty )$ is continuous and $\lim_{r\rightarrow 0^{+}}x(r)=\lim_{r\rightarrow +\infty }x(r)=0$. So there exists $r_{0}\in (0,+\infty )$ such that $x(r_{0})=\sup_{r>0}x(r)=\lambda _{1}$. For all $\lambda \in (0,\lambda _{1})$, there exist constants $a_{1},~a_{2}>0$ such that $x(a_{1})=x(a_{2})=\lambda$, where $0< a_{1}< r_{0}< a_{2}<+\infty$. Thus

$$\begin{aligned}& \lambda f(u)\leq \frac{\phi (a_{1})}{\psi _{2}(A_{1})}\leq \phi \biggl( \frac{a_{1}}{A_{1}}\biggr),\quad u\in [0,a_{1}], \end{aligned}$$

(3.15)

$$\begin{aligned}& \lambda f(u)\leq \frac{\phi (a_{2})}{\psi _{2}(A_{1})}\leq \phi \biggl( \frac{a_{2}}{A_{1}}\biggr),\quad u\in [0,a_{2}], \end{aligned}$$

(3.16)

By the conditions $f_{0}=+\infty$ and $f_{\infty }=+\infty$ there exist constants $b_{1},~b_{2}>0$, where $0< b_{1}< a_{1}< r_{0}< a_{2}< b_{2}<+\infty$, such that

$$\begin{aligned} \frac{f(u)}{\phi (u)}\geq \frac{1}{\lambda \psi _{1}(k(\delta ))\phi (A_{3})},\quad u\in (0,b_{1}) \cup \bigl(k(\delta )b_{2},+\infty \bigr), \end{aligned}$$

so that

$$\begin{aligned}& \lambda \min_{k(\delta )b_{1}\leq u\leq b_{1}}f(u)\geq \phi \biggl( \frac{b_{1}}{A_{3}}\biggr), \end{aligned}$$

(3.17)

$$\begin{aligned}& \lambda \min_{k(\delta )b_{2}\leq u\leq b_{2}}f(u)\geq \phi \biggl( \frac{b_{2}}{A_{3}}\biggr). \end{aligned}$$

(3.18)

By (3.15), (3.17), (3.16), and (3.18), combined with Theorems 2.1 and 3.3, the BVP of fractional q-difference Eq. (1.1) has at least two positive solutions for each $\lambda \in (0,\lambda _{1})$. The proof is completed. □

Theorem 3.5

Assume that (A1) and (A2) hold. If $F_{0}<+\infty$ and $F_{\infty }<+\infty$, then there exists $\lambda _{0}>0$ such that for all $0<\lambda <\lambda _{0}$, the BVP of fractional q-difference Eq. (1.1) has no positive solution.

Proof

Since $F_{0}<+\infty$ and $F_{\infty }<+\infty$, there exist $M_{1},M_{2}, r_{1},r_{2}>0$, such that $r_{1}< r_{2}$ and

$$\begin{aligned}& f(u)\leq M_{1}\phi (u),\quad u\in [0,r_{1}], \\& f(u)\leq M_{2}\phi (u),\quad u\in [r_{2},+\infty ). \end{aligned}$$

Let $M_{0}=\max \{M_{1},M_{2},\max_{r_{1}\leq u\leq r_{2}}{ \frac{f(u)}{\phi (u)}}\}$. Then we have

$$\begin{aligned} f(u)\leq M_{0}\phi (u),\quad u\in [0,+\infty ). \end{aligned}$$

Let v be a positive solution of the fractional q-difference equation boundary value problem (1.1). We will show that this leads to a contradiction for $0<\lambda <\lambda _{0}$, where $\lambda _{0}:=M_{0}^{-1}\psi _{1}(A_{1}^{-1})$. Indeed, since $T_{\lambda }v(t)=v(t)$ for $t\in [0,1]$, we have

$$\begin{aligned} \Vert v \Vert =& \Vert T_{\lambda }v \Vert \leq \int _{0}^{1}G(1,qs)\phi ^{-1} \biggl(\lambda \int _{0}^{1}H(q\tau ,q\tau )f\bigl(v(\tau ) \bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ \leq& \int _{0}^{1}G(1,qs)\phi ^{-1} \biggl(\lambda \int _{0}^{1}H(q\tau ,q \tau )M_{0}\phi \bigl(v(\tau )\bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ \leq& \psi _{1}^{-1}(\lambda M_{0}) \Vert v \Vert A_{1}< \Vert v \Vert , \end{aligned}$$

a contradiction. Therefore the BVP of fractional q-difference Eq. (1.1) has no positive solution. The proof is completed. □

Theorem 3.6

Assume that (A1) and (A2) hold. If $f_{0}>0$ and $f_{\infty }>0$, then there exists $\lambda _{0}'>0$ such that for all $\lambda >\lambda _{0}'$, the BVP of fractional q-difference Eq. (1.1) has no positive solution.

Proof

Since $f_{0}>0$ and $f_{\infty }>0$, there exist $m_{1},m_{2},r_{3},r_{4}>0$ such that $r_{3}< r_{4}$ and

$$\begin{aligned}& f(u)\geq m_{1}\phi (u),\quad u\in [0,r_{3}], \\& f(u)\geq m_{2}\phi (u),\quad u\in [r_{4},+\infty ). \end{aligned}$$

Let $m_{0}=\max \{m_{1},m_{2},\max_{r_{3}\leq u\leq r_{4}}{ \frac{f(u)}{\phi (u)}}\}$. Then we have

$$\begin{aligned} f(u)\geq m_{0}\phi (u),\quad u\in [0,+\infty ). \end{aligned}$$

Let v be a positive solution of the fractional q-difference equation BVP (1.1). We will show that this leads to a contradiction for $\lambda >\lambda _{0}'$, where $\lambda _{0}':=m_{0}^{-1}\psi _{2}(A_{2}^{-1})$. Indeed, since $T_{\lambda }v(t)=v(t)$ for $t\in [0,1]$, we have

$$\begin{aligned} \Vert v \Vert =& \Vert T_{\lambda }v \Vert \geq \int _{0}^{1}k(\delta )G(1,qs)\phi ^{-1}\biggl( \lambda \int _{0}^{1}s^{\beta -1}H(1,q\tau )f \bigl(v(\tau )\bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ \geq& \int _{0}^{1}k(\delta )G(1,qs)\phi ^{-1}\biggl(\lambda \int _{0}^{1}s^{ \beta -1}H(1,q\tau )m_{0}\phi \bigl(v(\tau )\bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ \geq & \int _{0}^{1}k(\delta )\psi _{2}^{-1}\bigl(s^{\beta -1}\bigr)G(1,qs)\phi ^{-1}\biggl( \lambda \int _{0}^{1}H(1,q\tau )m_{0}\phi \bigl(\tau ^{\alpha -1} \Vert v \Vert \bigr)\,d_{q}\tau \biggr)\,d_{q}s \\ \geq& \psi _{2}^{-1}(\lambda m_{0}) \Vert v \Vert A_{2}> \Vert v \Vert , \end{aligned}$$

a contradiction. Therefore the BVP of fractional q-difference Eq. (1.1) has no positive solution. The proof is completed. □

4 Some examples of application

Example 4.1

Consider the following fractional q-difference equation BVP:

$$\begin{aligned} \textstyle\begin{cases} D_{q}^{\frac{3}{2}}(D_{q}^{\frac{5}{2}}u(t))=\lambda (7u(t)-6\sin (u(t))),\quad 0< t< 1, \\ u(0)=D_{q}u(0)=D^{q}u(1)=0,\qquad D_{q}^{\frac{5}{2}}u(0)=D_{q}(D_{q}^{ \frac{5}{2}}u(1))=0. \end{cases}\displaystyle \end{aligned}$$

(4.1)

Here $q=\frac{1}{2}$, $\alpha =\frac{5}{2}$, $\beta =\frac{3}{2}$, $\phi (u)=u$, $f(u)=7u-6\sin u$. Take $\psi _{1}(x)=\psi _{2}(x)=x$, $\delta =0.9$.

By a simple calculation we obtain $\Gamma _{\frac{1}{2}}(\frac{5}{2})\approx 1.1906$, $\Gamma _{ \frac{1}{2}}(\frac{3}{2})\approx 0.9209$, $A_{1}\approx 0.05523$, $A_{2} \approx 0.00811$, $F_{0}=1$, $f_{\infty }=7$, $f_{\infty }\psi _{1}(A_{1}^{-1})\approx 126.74271$, $F_{0}\psi _{2}(A_{2}^{-1}) \approx 123.30456$, $\psi _{2}(A_{2}^{-1})f_{\infty }^{-1}\approx 17.6194$ and $\psi _{1}(A_{1}^{-1})F_{0}^{-1})\approx 18.10610$.

Obviously, $f_{\infty }\psi _{1}(A_{1}^{-1})>F_{0}\psi _{2}(A_{2}^{-1})$. By Theorem 3.1 we obtain that BVP (4.1) has at least one positive solution for each $\lambda \in (17.61494, 18.10610)$.

Example 4.2

Consider the following fractional q-difference equation BVP with ϕ-Laplacian:

$$\begin{aligned} \textstyle\begin{cases} D_{q}^{\frac{3}{2}}(\phi (D_{q}^{\frac{5}{2}}u(t)))=\lambda \frac{(u^{3}(t)+ u^{2}(t))(\sin (u(t))+9)}{25u(t)+1},\quad 0< t< 1, \\ u(0)=D_{q}u(0)=D^{q}u(1)=0,\qquad \phi (D_{q}^{\frac{5}{2}}u(0))=D_{q}( \phi (D_{q}^{\frac{5}{2}}u(1)))=0. \end{cases}\displaystyle \end{aligned}$$

(4.2)

Here $q=\frac{1}{2}$, $\alpha =\frac{5}{2}$, $\beta =\frac{3}{2}$, $\phi (u)=|u|u$, and $f(u)=\frac{(u^{3}+u^{2})(\sin u+9)}{25u+1}$.

Take $\psi _{1}(x)=\psi _{2}(x)=x^{2}$, $\delta =0.9$. By calculating we get $A_{1}\approx 0.07088$, $A_{2}\approx 0.01550$, $F_{\infty }=0.4$ and $f_{0}=9$. Thus, $f_{0}\psi _{1}(A_{1}^{-1})>F_{\infty }\psi _{2}(A_{2}^{-1})$. By Theorem 3.1 we obtain that BVP (4.2) has at least one positive solution for each $\lambda \in (462.24260, 497.6280425)$.

Example 4.3

Consider the following fractional q-difference equation BVP:

$$\begin{aligned} \textstyle\begin{cases} D_{q}^{\frac{3}{2}}(\phi (D_{q}^{\frac{5}{2}}u(t)))=\lambda \frac{(20u^{2}(t)+ u(t))(\sin (u(t))+2)}{u(t)+1},\quad 0< t< 1, \\ u(0)=D_{q}u(0)=D^{q}u(1)=0,\qquad D_{q}^{\frac{5}{2}}u(0)=D_{q}(D_{q}^{ \frac{5}{2}}u(1))=0. \end{cases}\displaystyle \end{aligned}$$

(4.3)

Here $q=\frac{1}{2}$, $\alpha =\frac{5}{2}$, $\beta =\frac{3}{2}$, $\phi (u)=u$, and $f(u)=\frac{(20u^{2}+u)(\sin u+2)}{u+1}$. Take $\psi _{1}(x)=\psi _{2}(x)=x$, $\delta =0.9$. By calculating we have $A_{1}\approx 0.05523$, $A_{2}\approx 0.00811$, $F_{0}=f_{0}=2$, $F_{\infty }=60$, $f_{ \infty } =20$, and $u< f(u)<60u$ for $u>0$.

• (i) By Theorem 3.1 we obtain that BVP (4.3) has at least one positive solution for each $\lambda \in (6.1653,9.05305)$.

• (ii) By Theorem 3.5 we obtain that BVP (4.3) has no positive solution for each $\lambda \in (0, 0.30177)$.

• (iii) By Theorem 3.6 we obtain that BVP (4.3) has no positive solution for each $\lambda \in (123.30456, +\infty )$.

Example 4.4

Consider the following fractional q-difference equation BVP with ϕ-Laplacian:

$$\begin{aligned} \textstyle\begin{cases} D_{q}^{\frac{3}{2}}(\phi (D_{q}^{\frac{5}{2}}u(t)))=\lambda \frac{(u^{3}(t)+ u^{2}(t))(\arctan (u(t))+8)}{40u(t)+1},\quad 0< t< 1, \\ u(0)=D_{q}u(0)=D^{q}u(1)=0,\qquad D_{q}^{\frac{5}{2}}u(0)=D_{q}(D_{q}^{ \frac{5}{2}}u(1))=0. \end{cases}\displaystyle \end{aligned}$$

(4.4)

Here $q=\frac{1}{2}$, $\alpha =\frac{5}{2}$, $\beta =\frac{3}{2}$, $\phi (u)=|u|u$, and $f(u)=\frac{(u^{3}+u^{2})(\arctan u+8))}{40u+1}$. Take $\psi _{1}(x)=\psi _{2}(x)=x^{2}$, $\delta =0.9$. Then we get $A_{1}\approx 0.07088$, $A_{2}\approx 0.01550$, $f_{0}=8$, $F_{\infty }=0.23927$, and $\frac{1}{5}\phi (u)< f(u)<9.57080\phi (u)$ for $u>0$.

• (i) By Theorem 3.2 we obtain that BVP (4.4) has at least one positive solution for each $\lambda \in (520.02293, 831.91078)$.

• (ii) By Theorem 3.5 we obtain that BVP (4.3) has no positive solution for each $\lambda \in (0, 20.79777)$.

• (iii) By Theorem 3.6 we obtain that BVP (4.3) has no positive solution for each $\lambda \in (20800.91719, +\infty )$.

5 Conclusions

This research establishes the existence of at least one or two positive solutions in terms of different eigenvalue intervals for the BVP of ϕ-Laplacian fractional q-difference equation, by applying the Green function and Guo–Krasnoselskii fixed point theorem on cones. This enriches the theories for fractional q-difference equations and provides the theoretical guarantee for the application of fractional q-difference equations in such fields as aerodynamics, electrodynamics of complex medium, capacitor theory, electrical circuits, control theory, and so on. At the same time, we also consider the nonexistence of a positive solution in terms of the parameter λ. In the future, we will use bifurcation theory, critical point theory, variational method, and other methods to continue our works in this area.