Existence of solutions for nonlinear second-order q-difference equations with first-order q-derivatives

Abstract

In this paper, we establish the existence of solutions for a boundary value problem with the nonlinear second-order q-difference equation

$$\{\begin{array}{l}{D}_q^{2}u(t)=f(t,u(t),D_{q}u(t)),t\in I,\\ D_{q}u(0)=0,D_{q}u(1)=\alpha u(1).\end{array}$$

The existence and uniqueness of solutions for the problem are proved by means of the Leray-Schauder nonlinear alternative and some standard fixed point theorems. Finally, we give two examples to demonstrate the use of the main results. The nonlinear team f contains $D_{q}u(t)$ in the equation.

1 Introduction

In this paper, we study the existence of solutions for a boundary value problem with nonlinear second-order q-difference equations

$$\{\begin{array}{l}{D}_q^{2}u(t)=f(t,u(t),D_{q}u(t)),t\in I,\\ D_{q}u(0)=0,D_{q}u(1)=\alpha u(1),\end{array}$$

(1.1)

where $f\in C(I\times R^2,R)$, $I=\{q^n:n\in N\}\cup \{0,1\}$, $q\in (0,1)$, and $\alpha \ne 0$ is a fixed real number.

The q-difference equations initiated at the beginning of the twentieth century [1–4] is a very interesting field in difference equations. In the last few decades, it has evolved into a multidisciplinary subject and plays an important role in several fields of physics such as cosmic strings and black holes [5], conformal quantum mechanics [6], nuclear and high energy physics [7]. However, the theory of boundary value problems (BVPs) for nonlinear q-difference equations is still in the initial stages and many aspects of this theory need to be explored. To the best of our knowledge, for the BVPs of nonlinear q-difference equations, a few works were done; see [8–13] and the references therein. In particular, the study of BVPs for nonlinear q-difference equation with first-order q-difference is yet to be initiated.

The main aim of this paper is to develop some existence and uniqueness results for BVP (1.1). Our results are based on a variety of fixed point theorems such as the Banach contraction mapping principle, the Leray-Schauder nonlinear alternative and the Leray-Schauder continuous theorem. Some examples and special cases are also discussed.

2 Preliminary results

In this section, firstly, let us recall some basic concepts of q-calculus [14, 15].

Definition 2.1 For $0<q<1$, we define the q-derivative of a real-value function f as

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

Note that ${lim}_{q\to 1^{-}}{D}_{q}f(t)=f^{\prime }(t)$.

Definition 2.2 The higher-order q-derivatives are defined inductively as

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

For example, $D_q(t^k)={[k]}_q{t}^{k-1}$, where k is a positive integer and the bracket ${[k]}_q=(q^k-1)/(q-1)$. In particular, $D_q(t^2)=(1+q)t$.

Definition 2.3 The q-integral of a function f defined in the interval $[a,b]$ is given by

$${\int }_a^{x}f(t)d_{q}t:=\sum _{n=0}^{\mathrm{\infty }}x(1-q)q^{n}f\left(x{q}^n\right)-af\left(q^{n}a\right),x\in [a,b],$$

and for $a=0$, we denote

$$I_{q}f(x)={\int }_0^{x}f(t)d_{q}t=\sum _{n=0}^{\mathrm{\infty }}x(1-q)q^{n}f\left(x{q}^n\right).$$

Then

$${\int }_a^{b}f(t)d_{q}t={\int }_0^{b}f(t)d_{q}t-{\int }_0^{a}f(t)d_{q}t.$$

Similarly, we have

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

Observe that

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

and if f is continuous at $x=0$, then $I_q{D}_{q}f(x)=f(x)-f(0)$.

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

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

(2.1)

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

(2.2)

Remark 2.4 In the limit $q\to 1^{-}$, the above results correspond to their counterparts in standard calculus.

Definition 2.5 $f:I\times R^2\to R$ is called an S-Carathéodory function if and only if

• (i) for each $(u,v)\in R^2$, $t\mapsto f(t,u,v)$ is measurable on I;

• (ii) for a.e. $t\in I$, $(u,v)\mapsto f(t,u,v)$ is continuous on $R^2$;

• (iii) for each $r>0$, there exists ${\phi }_r(t)\in L^1(I,R^{+})$ with $t{\phi }_r(t)\in L^1(I,R^{+})$ on I such that $max\{|u|,|v|\}\le r$ implies $|f(t,u,v)|\le {\phi }_r(t)$, for a.e. I, where $L^1(I,R^{+})=\{u\in C_q:{\int }_0^{1}u(t)d_{q}t\text{ exists}\}$, and normed by ${\parallel u\parallel }_{L^1}={\int }_0^1|u(t)|d_{q}t$ for all $u\in L^1(I,R^{+})$.

Theorem 2.6 (Nonlinear alternative for single-valued maps [16])

Let E be a Banach space, let C be a closed and convex subset of E, and let U be an open subset of C and $0\in U$. Suppose that $F:\overline{U}\to C$ is a continuous, compact (that is, $F(\overline{U})$ is a relatively compact subset of C) map. Then either

• (i) F has a fixed point in $\overline{U}$, or

• (ii) there is a $u\in \partial U$ (the boundary of U in C) and $\lambda \in (0,1)$ with $u=\lambda F(u)$.

Lemma 2.7 Let $y\in C[0,1]$, then the BVP

$$\{\begin{array}{l}{D}_q^{2}u(t)=y(t),t\in I,\\ D_{q}u(0)=0,D_{q}u(1)=\alpha u(1),\end{array}$$

(2.3)

has a unique solution

$$\begin{array}{rl}u(t)& ={\int }_0^t(t-qs)y(s)d_{q}s+{\int }_0^1(\frac{1}{\alpha }-1+qs)y(s)d_{q}s\\ ={\int }_0^{1}G(t,s;q)y(s)d_{q}s,\end{array}$$

(2.4)

where

$$G(t,s;q)=\frac{1}{\alpha }\{\begin{array}{ll}1-\alpha +\alpha t,& s\le t,\\ 1-\alpha +\alpha qs,& t\le s.\end{array}$$

(2.5)

Proof Integrating the q-difference equation from 0 to t, we get

$$D_{q}u(t)={\int }_0^{t}y(s)d_{q}s+a_1.$$

(2.6)

Integrating (2.6) from 0 to t and changing the order of integration, we have

$$u(t)={\int }_0^t(t-qs)y(s)d_{q}s+a_{1}t+a_0,$$

(2.7)

where $a_1$, $a_0$ are arbitrary constants. Using the boundary conditions $D_{q}u(0)=0,D_{q}u(1)=\alpha u(1)$ in (2.7), we find that $a_1=0$, and

$$a_1={\int }_0^1(\frac{1}{\alpha }-1+qs)y(s)d_{q}s.$$

Substituting the values of $a_0$ and $a_1$ in (2.7), we obtain

$$u(t)={\int }_0^t(t-qs)y(s)d_{q}s+{\int }_0^1(\frac{1}{\alpha }-1+qs)y(s)d_{q}s.$$

This completes the proof. □

Remark 2.8 For $q\to 1$, equation (2.4) takes the form

$$u(t)={\int }_0^t(t-qs)y(s)d_{q}s+a_{1}t+a_0,$$

which is the solution of a classical second-order ordinary differential equation $u^{″}(t)=y(t)$ and the associated form of Green’s function for the classical case is

$$G(t,s)=\frac{1}{\alpha }\{\begin{array}{ll}1-\alpha +\alpha t,& s\le t,\\ 1-\alpha +\alpha s,& t\le s.\end{array}$$

We consider the Banach space $C_q=C(I,R)$ equipped with the standard norm $\parallel u\parallel =max\{{\parallel u\parallel }_{\mathrm{\infty }},{\parallel D_{q}u\parallel }_{\mathrm{\infty }}\}$, and ${\parallel \cdot \parallel }_{\mathrm{\infty }}=sup\{\parallel \cdot \parallel ,t\in I\}$, $u\in C_q$.

Define an integral operator $T:C_p\to C_p$ by

$$\begin{array}{rl}Tu(t)=& {\int }_0^{1}G(t,s;q)f(s,u(s),D_{q}u(s))d_{q}s\\ =& {\int }_0^t(t-qs)f(s,u(s),D_{q}u(s))d_{q}s\\ +{\int }_0^1(\frac{1}{\alpha }-1+qs)f(s,u(s),D_{q}u(s))d_{q}s,t\in I,u\in C_q.\end{array}$$

(2.8)

Obviously, T is well defined and $u\in C_q$ is a solution of BVP (1.1) if and only if u is a fixed point of T.

3 Existence and uniqueness results

In this section, we apply various fixed point theorems to BVP (1.1). First, we give the uniqueness result based on Banach’s contraction principle.

Theorem 3.1 Let $f:I\times R^2\to R$ be a continuous function, and there exists $L_1(t),L_2(t)\in C([0,1],[0,+\mathrm{\infty }))$ such that

$$|f(t,u_1,v_1)-f(t,u_2,v_2)|\le L_1(t)|u_1-u_2|+L_2(t)|v_1-v_2|,t\in I,(u_1,v_1),(u_2,v_2)\in R^2.$$

In addition, suppose either

(${\mathrm{H}}_{\mathrm{1}}$) $\mathrm{\Lambda }<|\alpha |$ for $0<|\alpha |<1$, or

(${\mathrm{H}}_{\mathrm{2}}$) $\mathrm{\Lambda }<1$ for $|\alpha |\ge 1$

holds, where $\mathrm{\Lambda }={max}_{t\in [0,1]}\{L_1(t)+L_2(t)\}$. Then BVP (1.1) has a unique solution.

Proof Case 1: $|\alpha |<1$. Let us set ${sup}_{t\in I}|f(t,0,0)|=M_0$ and choose

$$r\ge \frac{M_0}{|\alpha |(1-\delta )},$$

(3.1)

where δ is such that $\frac{\mathrm{\Lambda }}{|\alpha |}\le \delta \le 1$. Now we show that $T{B}_r\subset B_r$, where $B_r=\{u\in C_q:\parallel u\parallel \le r\}$. For each $u\in B_r$, we have

$$\begin{array}{rl}|Tu(t)|\le & \underset{t\in I}{sup}|{\int }_0^t(t-qs)f(s,u(s),D_{q}u(s))d_{q}s+{\int }_0^1(\frac{1}{\alpha }-1+qs)f(s,u(s),D_{q}u(s))d_{q}s|\\ \le & \underset{t\in I}{sup}|{\int }_0^t(t-qs)(|f(s,u(s),D_{q}u(s))-f(s,0,0)|+|f(s,0,0)|)d_{q}s\\ +{\int }_0^1(\frac{1}{\alpha }-1+qs)(|f(s,u(s),D_{q}u(s))-f(s,0,0)|+|f(s,0,0)|)d_{q}s|\\ \le & \underset{t\in I}{sup}|{\int }_0^t(t-qs)(L_1(s)|u(s)|+L_2(s)|D_{q}u(s)|+|f(s,0,0)|)d_{q}s\\ +{\int }_0^1(\frac{1}{\alpha }-1+qs)(L_1(s)|u(s)|+L_2(s)|D_{q}u(s)|+|f(s,0,0)|)d_{q}s|\\ \le & (\mathrm{\Lambda }\parallel u\parallel +M_0)\underset{t\in I}{sup}|{\int }_0^t(t-qs)d_{q}s+{\int }_0^1(\frac{1}{\alpha }-1+qs)d_{q}s|\\ \le & (\mathrm{\Lambda }\parallel u\parallel +M_0)\underset{t\in I}{sup}\left\{|\frac{t^2}{1+q}+\frac{1}{\alpha }-1+\frac{q}{1+q}|\right\}\\ \le & (\mathrm{\Lambda }\parallel u\parallel +M_0)\frac{1}{|\alpha |}\le (\mathrm{\Lambda }r+M_0)\frac{1}{|\alpha |}\le (\frac{\mathrm{\Lambda }}{|\alpha |}+(1-\delta ))r\le r,\end{array}$$

and

$$\begin{array}{rl}|D_{q}Tu(t)|& \le |D_{q}Tu(t)|\le \underset{t\in I}{sup}\left|{\int }_0^{t}f(s,u(s),D_{q}u(s))d_{q}s\right|\\ \le \underset{t\in I}{sup}{\int }_0^t(|f(s,u(s),D_{q}u(s))-f(s,0,0)|+|f(s,0,0)|)d_{q}s\\ \le \underset{t\in I}{sup}{\int }_0^t(L_1(s)|u(s)|+L_2(s)|D_{q}u(s)|+|f(s,0,0)|)d_{q}s\\ \le (\mathrm{\Lambda }\parallel u\parallel +M_0)\le (\mathrm{\Lambda }r+|\alpha |(1-\delta )r)\le (\frac{\mathrm{\Lambda }}{|\alpha |}+(1-\delta ))r\le r.\end{array}$$

Hence, we obtain that $\parallel Tu\parallel \le r$, so $T{B}_r\subset B_r$.

Now, for $u,v\in C_q$ and for each $t\in I$, we have

$$\begin{array}{rl}|Tu(t)-Tv(t)|\le & \underset{t\in I}{sup}|Tu(t)-Tv(t)|\\ \le & \underset{t\in I}{sup}|{\int }_0^t(t-qs)|f(s,u(s),D_{q}u(s))-f(s,v(s),D_{q}v(s))|d_{q}s\\ +{\int }_0^1(\frac{1}{\alpha }-1+qs)|f(s,u(s),D_{q}u(s))-f(s,v(s),D_{q}v(s))|d_{q}s|\\ \le & \underset{t\in I}{sup}|{\int }_0^t(t-qs)(L_1(s)|u(s)-v(s)|+L_2(s)|D_{q}u(s)-D_{q}v(s)|)d_{q}s\\ +{\int }_0^1(\frac{1}{\alpha }-1+qs)(L_1(s)|u(s)-v(s)|+L_2(s)|D_{q}u(s)-D_{q}v(s)|)d_{q}s|\\ \le & \mathrm{\Lambda }\underset{t\in I}{sup}\left\{|\frac{t^2}{1+q}+\frac{1}{\alpha }-1+\frac{q}{1+q}|\right\}\parallel u-v\parallel \\ \le & \frac{\mathrm{\Lambda }}{|\alpha |}\parallel u-v\parallel <\parallel u-v\parallel ,\end{array}$$

and

$$\begin{array}{rl}|D_{q}Tu(t)-D_{q}Tv(t)|& \le \underset{t\in I}{sup}|D_{q}Tu(t)-D_{q}Tv(t)|\\ \le \underset{t\in I}{sup}\left|{\int }_0^t|f(s,u(s),D_{q}u(s))-f(s,v(s),D_{q}v(s))|d_{q}s\right|\\ \le \underset{t\in I}{sup}\left|{\int }_0^t(L_1(s)|u(s)-v(s)|+L_2(s)|D_{q}u(s)-D_{q}v(s)|)d_{q}s\right|\\ \le \mathrm{\Lambda }\parallel u-v\parallel \le \frac{\mathrm{\Lambda }}{|\alpha |}\parallel u-v\parallel <\parallel u-v\parallel .\end{array}$$

Therefore, we obtain that $\parallel Tu-Tv\parallel <\parallel u-v\parallel$, so T is a contraction. Thus, the conclusion of the theorem follows by Banach’s contraction mapping principle.

Case 2: $|\alpha |\ge 1$. It is similar to the proof of case 1. This completes the proof of Theorem 3.1. □

Corollary 3.2 Assume that $f:I\times R^2\to R$ is a continuous function and there exist two positive constants $L_1$, $L_2$ such that

$$|f(t,u_1,v_1)-f(t,u_2,v_2)|\le L_1|u_1-u_2|+L_2|v_1-v_2|,t\in I,(u_1,v_1),(u_2,v_2)\in R^2.$$

In addition, suppose either

(${\mathrm{H}}_{\mathrm{3}}$) $L_1+L_2<|\alpha |$ for $0<|\alpha |<1$, or

(${\mathrm{H}}_{\mathrm{4}}$) $L_1+L_2<1$ for $|\alpha |\ge 1$

holds. Then BVP (1.1) has a unique solution.

Corollary 3.3 Assume that $f:I\times R^2\to R$ is a continuous function and there exist two functions $L_1(t),L_2(t)\in L^1(I,R^{+})$ such that

$$|f(t,u_1,v_1)-f(t,u_2,v_2)|\le L_1|u_1-u_2|+L_2|v_1-v_2|,t\in I,(u_1,v_1),(u_2,v_2)\in R^2.$$

In addition, suppose either

(${\mathrm{H}}_{\mathrm{5}}$) $A+Bq|\alpha |<|\alpha |$ for $0<|\alpha |<1$, or

(${\mathrm{H}}_{\mathrm{6}}$) $A<1$ for $|\alpha |\ge 1$

holds, where

$$A={\int }_0^1[L_1(s)+L_2(s)]d_{q}s,B={\int }_0^{1}s[L_1(s)+L_2(s)]d_{q}s.$$

Then BVP (1.1) has a unique solution.

Proof It is similar to the proof of Theorem 3.1. □

The next existence result is based on the Leray-Schauder nonlinear alternative theorem.

Lemma 3.4 Let $f:I\times R^2\to R$ be an S-Carathéodory function. Then $T:C_q\to C_q$ is completely continuous.

Proof The proof consists of several steps.

• (i) T maps bounded sets into bounded sets in $C_q$.

Let $B_r=\{u\in C_q:\parallel u\parallel \le r\}$ be a bounded set in $C_q$ and $u\in B_r$. Then we have

$$\begin{array}{rl}|Tu(t)|& \le {\int }_0^t|t-qs|\left|f(s,u(s),D_{q}u(s))\right|d_{q}s+{\int }_0^1|\frac{1}{\alpha }-1+qs|\left|f(s,u(s),D_{q}u(s))\right|d_{q}s\\ \le (1+\frac{1}{|\alpha |}){\int }_0^1{\phi }_r(s)d_{q}s=(1+\frac{1}{|\alpha |}){\parallel {\phi }_r\parallel }_{L^1},\end{array}$$

and

$$|D_{q}Tu(t)|\le {\int }_0^t\left|f(s,u(s),D_{q}u(s))\right|d_{q}s\le {\int }_0^1{\phi }_r(s)d_{q}s={\parallel {\phi }_r\parallel }_{L^1}.$$

Thus $\parallel Tu\parallel \le max\{{\parallel Tu\parallel }_{\mathrm{\infty }},{\parallel D_{q}Tu\parallel }_{\mathrm{\infty }}\}\le (1+\frac{1}{|\alpha |}){\parallel {\phi }_r\parallel }_{L^1}$.

• (ii) T maps bounded sets into equicontinuous sets of $C_q$.

Let $r_1,r_2\in I$, $r_1<r_2$, and let $B_r$ be a bounded set of $C_q$ as before. Then, for $u\in B_r$, we have

$$\begin{array}{rl}|Tu(r_2)-Tu(r_1)|=& |{\int }_0^{r_2}(r_2-qs)f(s,u(s),D_{q}u(s))d_{q}s\\ -{\int }_0^{r_1}(r_1-qs)f(s,u(s),D_{q}u(s))d_{q}s|\\ =& |{\int }_0^{r_1}(r_2-r_1)f(s,u(s),D_{q}u(s))d_{q}s\\ +{\int }_{r_1}^{r_2}(r_2-qs)f(s,u(s),D_{q}u(s))d_{q}s|\\ \le & {\int }_0^{r_1}|r_2-r_1|{\phi }_r(s)d_{q}s+{\int }_{r_1}^{r_2}|r_2-qs|{\phi }_r(s)d_{q}s\to 0\\ (r_2-r_1\to 0),\end{array}$$

and

$$\begin{array}{rl}|d_{q}Tu(r_2)-D_{q}Tu(r_1)|=& |{\int }_0^{r_2}f(s,u(s),D_{q}u(s))d_{q}s-{\int }_0^{r_1}f(s,u(s),D_{q}u(s))d_{q}s|\\ =& \left|{\int }_{r_1}^{r_2}f(s,u(s),D_{q}u(s))d_{q}s\right|\le {\int }_{r_1}^{r_2}{\phi }_r(s)d_{q}s\to 0\\ (r_2-r_1\to 0).\end{array}$$

As a consequence of the Arzelá-Ascoli theorem, we can conclude that $T:C_q\to C_q$ is completely continuous. This proof is completed. □

Theorem 3.5 Let $f:I\times R^2\to R$ be an S-Carathéodory function. Suppose further that there exists a real number $M>0$ such that

$$\frac{|\alpha |M}{(1+|\alpha |){\parallel {\phi }_r\parallel }_{L^1}}>1$$

holds, where

$${\parallel {\phi }_r\parallel }_{L^1}={\int }_0^1{\phi }_r(s)d_{q}s\ne 0.$$

Then BVP (1.1) has at least one solution.

Proof In view of Lemma 3.4, we obtain that $T:C_q\to C_q$ is completely continuous. Let $\lambda \in (0,1)$ and $u=\lambda Tu$. Then, for $t\in I$, we have

$$\begin{array}{rl}|u(t)|=& |\lambda Tu(t)|\\ \le & {\int }_0^t|t-qs|\left|f(s,u(s),D_{q}u(s))\right|d_{q}s\\ +{\int }_0^1|\frac{1}{\alpha }-1+qs|\left|f(s,u(s),D_{q}u(s))\right|d_{q}s\\ \le & (1+\frac{1}{|\alpha |}){\int }_0^1{\phi }_r(s)d_{q}s=(1+\frac{1}{|\alpha |}){\parallel {\phi }_r\parallel }_{L^1},\end{array}$$

and

$$|D_{q}u(t)|=|D_q\lambda Tu(t)|\le {\int }_0^t\left|f(s,u(s),D_{q}u(s))\right|d_{q}s\le {\int }_0^1{\phi }_r(s)d_{q}s={\parallel {\phi }_r\parallel }_{L^1}.$$

Hence, consequently,

$$\frac{|\alpha |\parallel u\parallel }{(1+|\alpha |){\parallel {\phi }_r\parallel }_{L^1}}\le 1.$$

Therefore, there exists $M>0$ such that $\parallel u\parallel \ne M$. Let us set $U=\{u\in C_q:\parallel u\parallel <M\}$. Note that the operator $T:\overline{U}\to C_q$ is completely continuous (which is known to be compact restricted to bounded sets). From the choice of U, there is no $U\in \partial U$ such that $u=\lambda Tu$ for some $\lambda \in (0,1)$. Consequently, by Theorem 2.6, we deduce that T has a fixed point $u\in \overline{U}$ which is a solution of problem (1.1). This completes the proof. □

The next existence result is based on the Leray-Schauder continuation theorem.

Theorem 3.6 Let $f:I\times R^2\to R$ be an S-Carathéodory function. Suppose further that there exist functions $p(t),q(t),r(t)\in L^1(I,R^{+})$ with $tp(t)\in L^1(I,R^{+})$ such that

$$|f(t,u,v)|\le p(t)|u|+q(t)|v|+r(t),\mathit{\text{for a.e.}}t\in I\mathit{\text{and}}(u,v)\in R^2.$$

Then BVP (1.1) has at least one solution provided $(N+1)P+P_1+Q<1$, where $N=max\{\frac{1}{\alpha },1-\frac{1}{\alpha }\}$.

Proof We consider the space

$$P=\{u\in C_q:D_{q}u(0)=0,D_{q}u(1)=\alpha u(1)\}$$

and define the operator $T_1:P\times [0,1]\to P$ by

$$T_1(u,\lambda )=\lambda Tu=\lambda {\int }_0^{1}G(t,s;q)f(s,u(s),D_{q}u(s))d_{q}s,t\in I.$$

(3.2)

Obviously, we can see that $P\subset C_q$. In view of Lemma 3.4, it is easy to know that for each $\lambda \in [0,1]$, $T_1(u,\lambda )$ is completely continuous in P. It is clear that $u\in P$ is a solution of BVP (1.1) if and only if u is a fixed point of $T_1(\cdot ,1)$. Clearly, $T_1(u,0)=0$ for each $u\in P$. If for each $\lambda \in [0,1]$ the fixed points of $T_1(\cdot ,1)$ in P belong to a closed ball of P independent of λ, then the Leray-Schauder continuation theorem completes the proof.

Next we show that the fixed point of $T_1(\cdot ,1)$ has a priori bound M, which is independent of λ. Assume that $u=T_1(u,\lambda )$, and set

$$\begin{array}{c}P={\int }_0^{1}p(s)d_{q}s,P_1={\int }_0^{1}sp(s)d_{q}s,\hfill \\ Q={\int }_0^{1}q(s)d_{q}s,R={\int }_0^{1}r(s)d_{q}s.\hfill \end{array}$$

By (2.5), it is clear that $|G(t,s;q)|\le N$ for each $\alpha \ne 0$. For any $u\in P$, we have

$$\begin{array}{rl}|u(t)|& =|u(1)-{\int }_t^1{D}_{q}u(s)d_{q}s|\le |\frac{1}{\alpha }{D}_{q}u(1)|+|{\int }_t^1{D}_{q}u(s)d_{q}s|\\ \le (N+1-t){\parallel D_{q}u\parallel }_{\mathrm{\infty }}\le (N+1+t){\parallel D_{q}u\parallel }_{\mathrm{\infty }},t\in I,\end{array}$$

and so it holds that

$$\begin{array}{rl}{\parallel D_{q}u\parallel }_{\mathrm{\infty }}& \le {\parallel \lambda f(s,u(s),D_{q}u(s))\parallel }_{L^1}\le {\parallel f(s,u(s),D_{q}u(s))\parallel }_{L^1}\\ \le {\parallel p(t)|u(s)|+q(t)|D_{q}u(s)|+r(s)\parallel }_{L^1}\\ \le ((N+1)P+P_1+Q){\parallel D_{q}u\parallel }_{\mathrm{\infty }}+R;\end{array}$$

therefore,

$${\parallel D_{q}u\parallel }_{\mathrm{\infty }}\le \frac{R}{1-((N+1)P+P_1+Q)}:=M_1.$$

At the same time, we have

$$\begin{array}{rl}|u(t)|& \le \lambda |{\int }_0^{1}G(t,s,q)f(s,u(s),D_{q}u(s))d_{q}s|\\ \le N{\int }_0^1\left|f(s,u(s),D_{q}u(s))\right|d_{q}s\\ \le N{\int }_0^1(p(t)|u(s)|+q(t)|D_{q}u(s)|+r(s))d_{q}s\\ \le NP{\parallel u\parallel }_{\mathrm{\infty }}+NQ{\parallel D_{q}u\parallel }_{\mathrm{\infty }}+NR,t\in I,\end{array}$$

and so

$${\parallel u\parallel }_{\mathrm{\infty }}\le \frac{N(Q{M}_1+R)}{1-NP}:=M_2.$$

Set $M=max\{M_1,M_2\}$, which is independent of λ. So, BVP (1.1) has at least one solution. This completes the proof. □

4 Example

Example 4.1 Consider the following BVP:

$$\{\begin{array}{l}{D}_q^{2}u(t)=e^t+\frac{1}{4}sin(u(t))+\frac{1}{8}{tan}^{-1}(D_{q}u(t)),t\in I,\\ D_{q}u(0)=0,D_{q}u(1)=\frac{1}{2}u(1).\end{array}$$

(4.1)

Here, $f(t,u(t),D_{q}u(t))=e^t+\frac{1}{4}sin(u(t))+\frac{1}{8}{tan}^{-1}(D_{q}u(t))$, $q=\frac{1}{2}$, $\alpha =\frac{1}{2}$. Clearly, $|f(t,u_1,v_1)-f(t,u_2,v_2)|\le \frac{1}{4}|u_1-u_2|+\frac{1}{8}|v_1-v_2|$. Then $L_1=\frac{1}{4}$, $L_2=\frac{1}{8}$ and $L_1+L_2=\frac{3}{8}<\alpha$. By Corollary 3.2, we obtain that BVP (4.1) has a unique solution.

Example 4.2 Consider the following BVP:

$$\{\begin{array}{l}{D}_q^{2}u(t)=\frac{t}{20}sin(u(t))+\frac{1}{(2+\sqrt{2})\sqrt{t}}{(D_{q}u(t))}^{\beta }+\frac{3}{2}t,t\in I,\\ D_{q}u(0)=0,D_{q}u(1)=\frac{1}{2}u(1).\end{array}$$

(4.2)

Here, $q=\frac{1}{2}$, $0<\beta <1$. It is obvious that $|f(t,u,v)|\le \frac{t}{20}|u|+\frac{1}{(2+\sqrt{2})\sqrt{t}}|v|+\frac{3}{2}t$, where $p(t)=\frac{t}{20}$, $q(t)=\frac{1}{(2+\sqrt{2})\sqrt{t}}$, $r(t)=\frac{3}{2}t$. Then $P=\frac{1}{30}$, $P_1=\frac{1}{35}$, $Q=\frac{1}{2}$, $R=1$, so $(N+1)P+P_1+Q=\frac{22}{35}<1$. By Theorem 3.6, we obtain that BVP (4.2) has at least one solution.