On nonlocal boundary value problems of nonlinear q-difference equations

Abstract

This paper studies a nonlocal boundary value problem of nonlinear third-order q-difference equations. Our results are based on Leray-Schauder degree theory and some standard fixed point theorems.

MSC 2000: 39A05; 39A13.

1 Introduction

In this paper, we study a nonlocal nonlinear boundary value problem (BVP) of third-order q-difference equations given by

$$\left\{\begin{array}{c}{D}_q^{3}u\left(t\right)=f\left(t,u\left(t\right)\right),t\in I_q,\hfill \\ u\left(0\right)=0,D_{q}u\left(0\right)=0,u\left(1\right)=\alpha u\left(\eta \right),\hfill \end{array}\right.$$

(1.1)

where f ∈ C(I _q × ℝ,ℝ), I _q = {qⁿ: n ∈ ℕ} ∪ {0,1}, q ∈ (0,1) is a fixed constant, η ∈ {qⁿ: n ∈ ℕ} and α ≠ 1/η² is a real number.

The subject of q-difference equations has evolved into a multidisciplinary subject in the last few decades. In fact, it is a truly operational subject and its operational formulas were often used with great success in the theory of classical orthogonal polynomials and Bessel functions [1, 2]. For some pioneer work on q-difference equations, we refer the reader to [1, 3–5], whereas the recent development of the subject can be found in [6–17] and references therein. However, the theory of boundary value problems for nonlinear q-difference equations is still in the initial stages and many aspects of this theory need to be explored. In particular, the study of nonlocal boundary value problems for nonlinear q-difference equations is yet to be initiated.

The aim of our paper is to present some existence results for the problem (1.1). The first result relies on the nonlinear alternative of Leray-Schauder type. In the second result, we apply Banach's contraction principle to prove the uniqueness of the solution of the problem, while the third result is based on Krasnoselskii's fixed point theorem. The methods used are standard; however, their exposition in the framework of problem (1.1) is new. In Sect. 2, we present some basic material that we need in the sequel and Sect. 3 contains main results of the paper. Some illustrative examples are also discussed.

2 Preliminaries

Let us recall some basic concepts of q-calculus [8, 9].

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

$$D_{q}f\left(t\right)=\frac{f\left(t\right)-f\left(qt\right)}{\left(1-q\right)t},t\in I_q-\left\{0\right\},D_{q}f\left(0\right)=\underset{t\to 0}{\text{lim}}{D}_{q}f\left(t\right).$$

Note that

$$\underset{q\to 1^{-}}{\text{lim}}{D}_{q}f\left(t\right)=f^{\prime }\left(t\right).$$

The higher order q-derivatives are defined inductively as

$$D_q^{0}f\left(t\right)=f\left(t\right),D_q^{n}f\left(t\right)=D_q{D}_q^{n-1}f\left(t\right),n\in \mathbb{N}.$$

For example, $D_q\left(t^k\right)={\left[k\right]}_q{t}^{k-1}$, where k is a positive integer and the q-bracket [k] _q = (q^k- 1)/(q - 1). In particular, D _q (t²) = (1 + q)t.

For y ≥ 0, let us set ${\mathcal{J}}_y=\left\{y{q}^n:n\in \mathbb{N}\cup \left\{0\right\}\right\}\cup \left\{0\right\}$ and define the definite q-integral of a function $f:{\mathcal{J}}_y\to ℝ$ by

$$I_{q}f\left(y\right)=\underset{0}{\overset{y}{\int }}f\left(s\right)d_{q}s=\sum _{n=0}^{\infty }y\left(1-q\right)q^{n}f\left(y{q}^n\right)$$

provided that the series converges. For $b_1,b_2\in {\mathcal{J}}_y\left(b_1=y{q}^{n_1},b_2=y{q}^{n_2}\text{for}\text{some}{n}_1,n_2\in \mathbb{N}\right)$, we define

$$\underset{b_1}{\overset{b_2}{\int }}f\left(s\right)d_{q}s=I_{q}f\left(b_2\right)-I_{q}f\left(b_1\right)=\left(1-q\right)\sum _{n=0}^{\infty }{q}^n\left[b_{2}f\left(b_2{q}^n\right)-b_{1}f\left(b_1{q}^n\right)\right].$$

Similarly, we have

$$I_q^{0}f\left(t\right)=f\left(t\right),I_q^{n}f\left(t\right)=I_q{I}_q^{n-1}f\left(t\right),n\in \mathbb{N}.$$

Observe that

$$D_q{I}_{q}f\left(x\right)=f\left(x\right),$$

(2.1)

and if f is continuous at x = 0, then

$$I_q{D}_{q}f\left(x\right)=f\left(x\right)-f\left(0\right).$$

This implies that if D _q f(t) = σ(t), then f(t) = I _q σ(t) + c, where c is an arbitrary constant.

In q-calculus, the product rule and integration by parts formula are

$$D_q\left(gh\right)\left(t\right)=D_{q}g\left(t\right)h\left(t\right)+g\left(qt\right)D_{q}h\left(t\right),$$

(2.2)

$$\underset{0}{\overset{x}{\int }}f\left(t\right)D_{q}g\left(t\right)dqt={\left[f\left(t\right)g\left(t\right)\right]}_0^x-\underset{0}{\overset{x}{\int }}{D}_{q}f\left(t\right)g\left(qt\right)d_{q}t.$$

(2.3)

In the limit q → 1^-, the above results correspond to their counterparts in standard calculus.

For $f,g:{\mathcal{J}}_y\to ℝ$, it is possible to introduce an inner product

$$⟨f,g⟩=\underset{0}{\overset{1}{\int }}f\left(t\right)g\left(t\right)d_{q}t$$

and the resulting Hilbert space is denoted by $L_q^2\left(0,1\right)$.

As argued in [16], we can write the solution of the third-order q-difference equation $D_q^{3}u\left(t\right)=v\left(t\right)$ in the following form:

$$u\left(t\right)=\underset{0}{\overset{t}{\int }}\left({\alpha }_1\left(q\right)t^2+{\alpha }_2\left(q\right)ts+{\alpha }_3\left(q\right)s^2\right)v\left(s\right)d_{q}s+a_0+a_{1}t+a_2{t}^2,$$

(2.4)

where a₀, a₁, a₂ are arbitrary constants and α₁(q), α₂(q), α₃(q) can be fixed appropriately.

Choosing α₁(q) = 1/(1 + q), α₂(q) = -q, α₃(q) = q³/(1 + q) and using (2.1) and (2.2), we find that

$$D_{q}u\left(t\right)=\underset{0}{\overset{t}{\int }}tv\left(s\right)d_{q}s-\underset{0}{\overset{t}{\int }}qsv\left(s\right)d_{q}s,D_q^{2}u\left(t\right)=\underset{0}{\overset{t}{\int }}v\left(s\right)d_{q}s,D_q^{3}u\left(t\right)=v\left(t\right).$$

Thus, the solution (2.4) of $D_q^{3}u\left(t\right)=v\left(t\right)$ takes the form

$$u\left(t\right)=\underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}-qts\right)v\left(s\right)d_{q}s+a_0+a_{1}t+a_2{t}^2.$$

(2.5)

Lemma 2.1 The BVP (1.1) is equivalent to the integral equation

$$\begin{array}{cc}\hfill u\left(t\right)=& \underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}-qts\right)f\left(s,u\left(s\right)\right)d_{q}s+\frac{t_2}{1-\alpha {\eta }^2}\hfill \\ \times \left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta s\right)f\left(s,u\left(s\right)\right)d_{q}s-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)f\left(s,u\left(s\right)\right)d_{q}s\right].\hfill \end{array}$$

(2.6)

Proof. In view of (2.5), the solution of $D_q^{3}u=f\left(t,u\right)$ can be written as

$$u\left(t\right)=\underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}-qts\right)f\left(s,u\left(s\right)\right)d_{q}s+a_0+a_{1}t+a_2{t}^2,$$

(2.7)

where a₁, a₂, a₂ are arbitrary constants. Using the boundary conditions of (1.1) in (2.7), we find that a₀ = 0, a₁ = 0 and

$$a_2=\frac{1}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta s\right)f\left(s,u\left(s\right)\right)d_{q}s-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)f\left(s,u\left(s\right)\right)d_{q}s\right].$$

Substituting the values of a₀, a₁ and a₂ in (2.7), we obtain (2.6). This completes the proof.

We define

$$\begin{array}{ll}\hfill G_1& ={\text{max}}_{t\in I_q}∥\underset{0}{\overset{t}{\int }}\frac{{\left(t-s\right)}^2}{2}{d}_{q}s+\frac{t^2}{1-\alpha {\eta }^2}\left.\left[\alpha \underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^2}{2}{d}_{q}s-\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^2}{2}{d}_{q}s\right]\right|\\ =\text{max}\left\{\frac{\left|\gamma \right|\left(1+q\right)q^2}{{\left(1+q+q^2\right)}^4},\frac{\left|\alpha \right|{\eta }^2\left(1-\eta \right)}{\left|1-\alpha {\eta }^2\right|\left(1+q\right)\left(1+q+q^2\right)}\right\},\end{array}$$

(2.8)

where

$$\gamma =\eta +\frac{1-\eta }{1-\alpha {\eta }^2}.$$

Remark 2.1 For q → 1^-, equation (2.6) takes the form

$$\begin{array}{ll}\hfill u\left(t\right)=& \underset{0}{\overset{t}{\int }}\frac{{\left(t-s\right)}^2}{2}f\left(s,u\left(s\right)\right)ds+\frac{t^2}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^2}{2}f\left(s,u\left(s\right)\right)ds\right.\\ \left.-\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^2}{2}f\left(s,u\left(s\right)\right)ds\right].\end{array}$$

which is equivalent to the solution of a classical third-order nonlocal boundary value problem

$$u^{‴}\left(t\right)=f\left(t,u\left(t\right)\right),u\left(0\right)=0,u^{\prime }\left(0\right)=0,u\left(1\right)=\alpha u\left(\eta \right),0\le t\le 1,0<\eta <1.$$

(2.9)

3 Existence results

Let ${\mathcal{C}}_q=C\left(I_q,ℝ\right)$ denote the Banach space of all continuous functions from I _q → ℝ endowed with the norm defined by ∥x∥ = sup{|x(t)| : t ∈ I _q }.

Theorem 3.1 Assume that there exist constants M₁ ≥ 0 and M₂ > 0 such that M₁G₁ < 1 and |f(t, u)| ≤ M₁|u| + M₂ for all t ∈ I _q , u ∈ ℝ, where G₁ is given by (2.8). Then the problem (1.1) has at least one solution.

Proof. Let $B_R\subset {\mathcal{C}}_q$ be a suitable ball with radius R > 0. Define an operator $Ϝ:{\bar{B}}_R-{\mathcal{C}}_q$ as

$$\begin{array}{ll}\hfill \left[Ϝu\right]\left(t\right)=& \underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}-qts\right)f\left(s,u\left(s\right)\right)d_{q}s\\ +\frac{t^2}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta 2\right)f\left(s,u\left(s\right)\right)\right.d_{q}s\\ \left.-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)f\left(s,u\left(s\right)\right)d_{q}s\right].\end{array}$$

In view of Lemma 2.1, we just need to prove the existence of at least one solution $u\in {\mathcal{C}}_q$ such that $u=Ϝu$. Thus, it is sufficient to show that the operator $Ϝ$ satisfies

$$u\ne \lambda Ϝu,\forall u\in \partial B_R\text{and}\forall \lambda \in \left[0,1\right].$$

(3.1)

Let us define

$$H\left(\lambda ,y\right)=\lambda Ϝu,u\in {\mathcal{C}}_q,\lambda \in \left[0,1\right].$$

Then, by Arzela-Ascoli theorem, $h_{\lambda }\left(u\right)=u-H\left(\lambda ,u\right)=u-\lambda Ϝu$ is completely continuous. If (3.1) is true, then the following Leray-Schauder degrees are well defined and by the homotopy invariance of topological degree, it follows that

$$\begin{array}{ll}\hfill \text{deg}\left(h_{\lambda },B_R,0\right)& =\text{deg}\left(\mathcal{I}-\lambda Ϝ,B_R,0\right)=\text{deg}\left(h_1,B_R,0\right)\\ =\text{deg}\left(h_0,B_R,0\right)=\text{deg}\left(\mathcal{I},B_R,0\right)=1\ne 0,0\in B_R,\end{array}$$

where $\mathcal{I}$ denotes the unit operator. By the nonzero property of Leray-Schauder degree, $h_1\left(t\right)=u-\lambda Ϝu=0$ for at least one u ∈ B _R . Let us set

$$B_R=\left\{u\in {\mathcal{C}}_q:∥u∥<R\right\},$$

where R will be fixed later. In order to prove (3.1), we assume that $u=\lambda Ϝu$ for some λ ∈ [0,1] and for all t ∈ I _q so that

$$\begin{array}{c}\left|u\left(t\right)\right|=\left|\lambda \left[Ϝu\right]\left(t\right)\right|\le \left|\underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}-qts\right)\right.f\left(s,u\left(s\right)\right)d_{q}s\\ +\frac{t^2}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta s\right)f\left(s,u\left(s\right)\right)\right.d_{q}s\\ \left.\left.-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)f\left(s,u\left(s\right)\right)d_{q}s\right]\right|\\ \le \left|\underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}-qts\right)\right.\left(M_1\left|u\left(s\right)\right|+M_2\right)d_{q}s\\ +\frac{t^2}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta s\right)\left(M_1\left|u\left(s\right)\right|+M_2\right)d_{q}s\right.\\ \left.\left.-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)\left(M_1\left|u\left(s\right)\right|+M_2\right)d_{q}s\right]\right|\\ \le \left(M_1∥u∥+M_2\right)\underset{t\in I_q}{\text{max}}\left|\underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}-qts\right)d_{q}s\right.\\ \left.+\frac{t^2}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta s\right)d_{q}s-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)d_{q}s\right]\right|\\ \le \left(M_1∥u∥+M_2\right)G_1,\end{array}$$

which implies that

$$∥u∥\le \frac{M_2{G}_1}{1-M_1{G}_1}.$$

Letting $R=\frac{M_2{G}_1}{1-M_1{G}_1}+1$, (3.1) holds. This completes the proof.

Theorem 3.2 Let f : I _q × ℝ → ℝ be a jointly continuous function satisfying the Lipschitz condition

$$\left|f\left(t,u\right)-f\left(t,v\right)\right|\le L\left|u-v\right|,\forall t\in I_q,u,v\in ℝ,$$

where L is a Lipschitz constant. Then the boundary value problem (1.1) has a unique solution provided L < 1/G₁, where G₁ is given by (2.8).

Proof. Let us define an operator $Ϝ:{\mathcal{C}}_q\to {\mathcal{C}}_q$ by

$$\begin{array}{ll}\hfill \left[Ϝu\right]\left(t\right)=& \underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}-qts\right)f\left(s,u\left(s\right)\right)d_{q}s\\ +\frac{t^2}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta s\right)f\left(s,u\left(s\right)\right)d_{q}s\right.\\ \left.-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)f\left(s,u\left(s\right)\right)d_{q}s\right].\end{array}$$

Let us set ${\text{max}}_{t\in I_q}\left|f\left(t,0\right)\right|=M$ and choose

$$r\ge \frac{M{G}_1}{1-L{G}_1}$$

(3.2)

Then we show that $ϜB_r\subset B_r$, where $B_r=\left\{u\in {\mathcal{C}}_q:∥u∥\le r\right\}$. For u ∈ B _r , we have

$$\begin{array}{ll}\hfill ∥Ϝu∥& ={\text{max}}_{t\in I_q}\left|\underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}-qts\right)f\left(s,u\left(s\right)\right)d_{q}s\right.\\ +\frac{t^2}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta s\right)f\left(s,u\left(s\right)\right)d_{q}s\right.\\ \left.\left.-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)f\left(s,u\left(s\right)\right)d_{q}s\right]\right|\\ ={\text{max}}_{t\in I_q}\left|\underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}-qts\right)\right.\left[\left(f\left(s,u\left(s\right)\right)-f\left(s,0\right)\right)+f\left(s,0\right)\right]d_{q}s\\ +\frac{t^2}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta s\right)\left[\left(f\left(s,u\left(s\right)\right)-f\left(s,0\right)\right)+f\left(s,0\right)\right]d_{q}s\right.\\ \left.\left.-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)\left[\left(f\left(s,u\left(s\right)\right)-f\left(s,0\right)\right)+f\left(s,0\right)\right]d_{q}s\right]\right|\\ \le \left(L∥u∥+\left.M\right|\right){\text{max}}_{t\in I_q}\left|\underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}-qts\right)d_{q}s\right.\\ \left.+\frac{t^2}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta s\right)d_{q}s-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)d_{q}s\right]\right|\\ \le G_1\left(L_r+M\right)\le r.\end{array}$$

where we have used (3.2).

Now, for u, v ∈ ℝ, we obtain

$$\begin{array}{c}∥Ϝu-Ϝv∥\hfill \\ ={\text{max}}_{t\in I_q}\left|\left[Ϝu\right]\left(t\right)-\left[Ϝv\right]\left(t\right)\right|\hfill \\ \le {\text{max}}_{t\in I_q}\left|\underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}-qts\right)f\left(s,u\left(s\right)\right)d_{q}s\right.\hfill \\ +\frac{t^2}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta s\right)f\left(s,u\left(s\right)\right)d_{q}s\right.\hfill \\ \left.\left.-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)f\left(s,u\left(s\right)\right)d_{q}s\right]\right|\hfill \\ \le L{\text{max}}_{t\in I_q}\left|\underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}-qts\right)d_{q}s\right.\hfill \\ \left.+\frac{t^2}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta s\right)d_{q}s-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)d_{q}s\right]\right|∥u-v∥\hfill \\ \le L{G}_1∥u-v∥.\hfill \end{array}$$

As L < 1/G₁, therefore $Ϝ$ is a contraction. Thus, the conclusion of the theorem follows by Banach's contraction mapping principle. This completes the proof.

To prove the next existence result, we need the following known fixed point theorem due to Krasnoselskii [18].

Theorem 3.3 Let $\mathcal{M}$ be a closed convex and nonempty subset of a Banach space X. Let A, B be the operators such that (i) $Ax+By\in \mathcal{M}$ whenever $x,y\in \mathcal{M}$; (ii) A is compact and continuous; (iii) B is a contraction mapping. Then there exists $z\in \mathcal{M}$ such that z = Az + Bz.

Theorem 3.4 Assume that f : I _q × ℝ → ℝ is a continuous function such that

$$\left|f\left(t,u\right)-f\left(t,v\right)\right|\le L\left|u-v\right|,\forall t\in I_q,u,v\in ℝ.$$

(3.3)

Furthermore, |f(t, u)| ≤ μ(t), ∀(t, u) ∈ I _q × ℝ, with μ ∈ C(I _q , ℝ⁺). Then the boundary value problem (1.1) has at least one solution on I _q if

$$\frac{\left|1-\alpha {\eta }^3\right|}{\left|1-\alpha {\eta }^2\right|\left(1+q\right)\left(1+q+q^2\right)}<1.$$

(3.4)

Proof. Letting ${\text{sup}}_{t\in I_q}\left|\mu \left(t\right)\right|=∥\mu ∥$, we fix $\bar{r}\ge ∥\mu ∥G_1$ (G₁ is given by (2.8) and consider $B_{\bar{r}}=\left\{u\in ℝ:∥u∥\le \bar{r}\right\}$. We define the operators ${\mathcal{P}}_1$ and ${\mathcal{P}}_2$ on $B_{\bar{r}}$ as

$$\begin{array}{c}\left[{\mathcal{P}}_{1}u\right]\left(t\right)=\underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}-qts\right)f\left(s,u\left(s\right)\right)d_{q}s,\\ \left[{\mathcal{P}}_{2}u\right]\left(t\right)=\frac{t^2}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta s\right)f\left(s,u\left(s\right)\right)d_{q}s\right.\\ \left.-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)f\left(s,u\left(s\right)\right)d_{q}s\right].\end{array}$$

For $u,v\in B_{\bar{r}}$, we find that

$$\begin{array}{c}∥{\mathcal{P}}_{1}u+{\mathcal{P}}_{2}v∥\hfill \\ \le ∥\mu ∥{\text{max}}_{t\in I}\left|\underset{0}{\overset{t}{\int }}\left(\frac{t^2+q^3{s}^2}{1+q}+qts\right)d_{q}s\right.\hfill \\ \left.+\frac{t^2}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta s\right)d_{q}s-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)d_{q}s\right]\right|\hfill \\ =∥\mu ∥G_1\le \bar{r}.\hfill \end{array}$$

Thus, ${\mathcal{P}}_{1}u+{\mathcal{P}}_{2}v\in B_{\bar{r}}$. It follows from (3.3) and (3.4) that ${\mathcal{P}}_2$ is a contraction mapping. Continuity of f implies that the operator ${\mathcal{P}}_1$ is continuous. Also, ${\mathcal{P}}_1$ is uniformly bounded on $B_{\bar{r}}$ as

$$∥{\mathcal{P}}_{1}u∥\le \frac{1}{\left(1+q\right)\left(1+q+q^2\right)}.$$

Now we prove the compactness of the operator ${\mathcal{P}}_1$.

In view of (H₁), we define ${\text{sup}}_{\left(t,u\right)\in I_q\times B_r}\left|f\left(t,u\right)\right|=\bar{f}$, and consequently we have

$$\begin{array}{c}\left|\left[{\mathcal{P}}_{1}u\right]\left(t_1\right)-\left[{\mathcal{P}}_{1}u\right]\left(t_2\right)\right|\hfill \\ =\left|\underset{0}{\overset{t_1}{\int }}\left(\frac{t_1^2+q^3{s}^2}{1+q}-q{t}_{1}s\right)f\left(s,u\left(s\right)\right)d_{q}s\right.-\underset{0}{\overset{t_2}{\int }}\left(\frac{t_2^2+q^3{s}^2}{1+q}-q{t}_{2}s\right)f\left(s,u\left(s\right)\right)d_{q}s\hfill \\ \left.+\frac{\left(t_1^2+t_2^2\right)}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta 2\right)f\left(s,u\left(s\right)\right)d_{q}s-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)f\left(s,u\left(s\right)\right)d_{q}s\right]\right|\hfill \\ \le \bar{f}\left|\underset{0}{\overset{t_1}{\int }}\frac{\left(t_1-t_2\right)\left[t_1+t_2-q\left(1+q\right)s\right]}{1+q}\right.d_{q}s-\underset{t_1}{\overset{t_2}{\int }}\left(\frac{t_2^2+q^3{s}^2}{1+q}-q{t}_{2}s\right)d_{q}s\hfill \\ \left.+\frac{\left(t_1^2-t_2^2\right)}{1-\alpha {\eta }^2}\left[\alpha \underset{0}{\overset{\eta }{\int }}\left(\frac{{\eta }^2+q^3{s}^2}{1+q}-q\eta s\right)d_{q}s-\underset{0}{\overset{1}{\int }}\left(\frac{1+q^3{s}^2}{1+q}-qs\right)d_{q}s\right]\right|,\hfill \end{array}$$

which is independent of u and tends to zero as t₂ → t₁. So ${\mathcal{P}}_1$ is relatively compact on $B_{\bar{r}}$. Hence, by the Arzelá-Ascoli Theorem, ${\mathcal{P}}_1$ is compact on $B_{\bar{r}}$. Thus all the assump tions of Theorem 3.3 are satisfied. So the conclusion of Theorem 3.3 implies that (1.1) has at least one solution on I _q . This completes the proof.

Remark 3.1 In the limit q → 1^-, our results reduce to the ones for a classical third-order nonlocal nonlinear boundary value problem (2.9).

Example 3.1. Consider the following problem

$$\left\{\begin{array}{c}{D}_{\frac{1}{2}}^{3}u\left(t\right)=\frac{M_1}{\left(2\pi \right)}\text{sin}\left(2\pi u\right)+\frac{\left|u\right|}{1+\left|u\right|}+t^2,t\in {\left[0,1\right]}_{1/2},\hfill \\ u\left(0\right)=0,D_{\frac{1}{2}}u\left(0\right)=0,u\left(1\right)=2u\left(1/2\right).\hfill \end{array}\right.$$

(3.5)

Here q = 1/2 and M₁ will be fixed later. Observe that

$$\left|f\left(t,u\right)\right|=\left|\frac{M_1}{\left(2\pi \right)}\text{sin}\left(2\pi u\right)+\frac{\left|u\right|}{1+\left|u\right|}+t^2\right|\le M_1\left|u\right|+2,$$

and

$$\begin{array}{ll}\hfill G_1& =\text{max}\left\{\frac{\left|\gamma \right|\left(1+q\right)q^2}{{\left(1+q+q^2\right)}^4},\frac{\left|\alpha \right|{\eta }^2\left(1-\eta \right)}{\left|1-\alpha {\eta }^2\right|\left(1+q\right)\left(1+q+q^2\right)}\right\}\\ =\frac{\left|\alpha \right|{\eta }^2\left(1-\eta \right)}{\left|1-\alpha {\eta }^2\right|\left(1+q\right)\left(1+q+q^2\right)}=4/21.\end{array}$$

Clearly M₂ = 2 and we can choose $M_1<\frac{1}{G_1}=21/4$. Thus, Theorem 3.1 applies to the problem (3.5).

Example 3.2. Consider the following problem with unbounded nonlinearity

$$\left\{\begin{array}{c}{D}_{\frac{1}{2}}^{3}u\left(t\right)=5u+cosu+\left(u^2/\left(1+u^2\right)\right),t\in {\left[0,1\right]}_{1/2},\hfill \\ u\left(0\right)=0,D_{\frac{1}{2}}u\left(0\right)=0,u\left(1\right)=2u\left(1/2\right).\hfill \end{array}\right.$$

(3.6)

Clearly

$$\left|f\left(t,u\right)\right|=\left|5u+cosu+\left(u^2/\left(1+u^2\right)\right)\right|\le 5\left|u\right|+2,$$

with M₁ = 5 < 1/G₁ = 21/4 (G₁ is given in Example 3.1) and M₂ = 2. Thus, by the conclusion of Theorem 3.1, the problem (3.6) has a solution.

Example 3.3. Consider

$$\left\{\begin{array}{c}{D}_{\frac{3}{4}}^{3}u\left(t\right)=L\left(\text{cos}t+{\text{tan}}^{-1}u\right),t\in {\left[0,1\right]}_{3/4},\hfill \\ u\left(0\right)=0,D_{\frac{3}{4}}u\left(0\right)=0,u\left(1\right)=u\left(1/4\right).\hfill \end{array}\right.$$

(3.7)

With f(t, u) = L (cos t + tan^-1 u), we find that

$$\left|f\left(t,u\right)-f\left(t,v\right)\right|\le L\left|{\text{tan}}^{-1}u-{\text{tan}}^{-1}v\right|\le L\left|u-v\right|$$

and

$$\begin{array}{ll}\hfill G_1& =\text{max}\left\{\frac{\left|\gamma \right|\left(1+q\right)q^2}{{\left(1+q+q^2\right)}^4},\frac{\left|\alpha \right|{\eta }^2\left(1-\eta \right)}{\left|1-\alpha {\eta }^2\right|\left(1+q\right)\left(1+q+q^2\right)}\right\}\\ =\frac{\left|\gamma \right|\left(1+q\right)q^2}{{\left(1+q+q^2\right)}^4}=\frac{86704128}{2398926080}.\end{array}$$

Fixing $L<\frac{1}{G_1}\approx 27.668$, it follows by Theorem 3.2 that the problem (3.7) has a unique solution.