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Existence of solutions for fractional q-difference equation with mixed nonlinear boundary conditions

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Abstract

In this paper, we study the boundary value problem for a class of nonlinear fractional q-difference equation with mixed nonlinear conditions involving the fractional q-derivative of Riemann-Liouuville type. By means of the Guo-Krasnosel’skii fixed point theorem on cones, some results concerning the existence of solutions are obtained. Finally, examples are presented to illustrate our main results.

MSC:39A13, 34B18, 34A08.

1 Introduction

Fractional calculus is a generalization of integer order calculus [1, 2]. It has been used by many researchers to adequately describe the evolution of a variety of engineering, economical, physical, and biological processes [3]. There are a large number of papers dealing with the continuous fractional calculus. Among all the topics, boundary value problems for fractional differential equations have attracted considerable attention [46]. However, the discrete fractional calculus has seen slower progress, it is still a relatively new and emerging area of mathematics. Some efforts have also been made to develop the theory of discrete fractional calculus in various directions. For some recent works, see [7]. Of particular note is that Atici and Sengül have shown the usefulness of fractional difference equations in tumor growth modeling in [8].

The early works about q-difference calculus or quantum calculus were first developed by Jackson [9, 10], while basic definitions and properties can be found in the monograph by Kac and Cheung [11]. q-Difference equations have been widely used in mathematical physical problems, dynamical system and quantum models [12], heat and wave equations [13], and sampling theory of signal analysis [14].

As an important part of discrete mathematics, more recently, some researchers devoted their attention to the study of the fractional q-difference calculus, they developed the q-analogs of fractional integral and difference operators properties, the q-Mittag-Leffler function [15], q-Laplace transform, q-Taylor’s formula, etc. [16]. The origin of the fractional q-difference calculus can be traced back to the works in [17, 18] by Al-Salam and Agarwal. A book on this subject by Annaby and Mansour [19] summarizes and organizes much of the q-fractional calculus and equations.

As is well known, the aim of finding solutions to boundary value problems is of main importance in various fields of applied mathematics. Recently, there seems to be a new interest in the study of the boundary value problems for fractional q-difference equations [2027].

In 2012, Liang and Zhang [21] studied the three-point boundary problem of fractional q-differences,

( D q α u ) ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , 2 < α < 3 , u ( 0 ) = ( D q u ) ( 0 ) = 0 , ( D q u ) ( 1 ) = β ( D q u ) ( η ) ,

where 0<β η α 2 <1. By using a fixed point theorem in partially ordered sets, they got some sufficient conditions for the existence and uniqueness of positive solutions to the above boundary problem.

In 2013, Zhou and Liu [22] studied the existence results for fractional q-difference equations with nonlocal q-integral boundary conditions,

( D q α u ) ( t ) + f ( t , u ( t ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = 0 , u ( 1 ) = μ I q β u ( η ) = μ 0 η ( η q s ) ( β 1 ) Γ q ( β ) u ( s ) d q s ,

where μ>0 is a parameter, D q α is the q-derivative of Riemann-Liouville type of order α. By using the generalized Banach contraction principle, the monotone iterative method and Krasnoselskii’s fixed point theorem, some existence results of positive solutions to the above boundary value problems were enunciated.

In 2013, Goodrich [28] proved that the nonlocal boundary value problem with mixed nonlinear boundary conditions

y ( t ) = f ( t , y ( t ) ) , 0 < t < 1 , y ( 0 ) = H 1 ( φ ( y ) ) + E H 2 ( s , y ( s ) ) d s , y ( 1 ) = 0 ,

has at least one positive solution by imposing some relatively mild structural conditions on f, H 1 , H 2 and φ.

To the best of our knowledge, very few authors consider the boundary value problem of fractional q-difference equations with mixed nonlinear boundary conditions. Theories and applications seem to be just initiated. This paper will fill up the gap. Here, motivated by [28], we will consider the boundary value problem of the nonlinear fractional q-difference equations

( D q α u ) (t)+f ( t , u ( t ) ) =0,0<t<1,
(1.1)

subject to the boundary conditions

D q i u(0)=0,i=0,1,,n2, D q u(1)= H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s,
(1.2)

where 0<q<1, n1<αn, n2 are integers, f:[0,1]×[0,)[0,+) is continuous and φ is a linear functional, here E(0,1] is a closed subinterval, H 1 :[0,+)[0,+) and H 2 :[0,1]×[0,+)[0,+) are real-valued, continuous functions. We are interested in the existence of solutions for the boundary value problem (1.1)-(1.2) by utilizing a fixed point theorem on cones.

We should mention that the above boundary conditions are rather general and contain many common cases such as separated boundary conditions, integral boundary conditions, multi-point boundary conditions, etc., by choosing different H 1 , H 2 , and ϕ. Our results generalize and improve some results on the existence of solutions for fractional q-difference equations. Moreover, problems studied in [20] and [27] can be regarded as our special cases.

The paper is organized as follows. In Section 2, we introduce some definitions of q-fractional integral and differential operator together with some basic properties and lemmas to prove our main results. In Section 3, we investigate the existence of solutions for the boundary value problem (1.1)-(1.2) by fixed point theorem on cones. Moreover, some examples are given to illustrate our main results.

2 Preliminaries

In the following section, we collect some definitions and lemmas about fractional q-integral and fractional q-derivative; for an overview of the theory one is referred to [18, 20].

Let q(0,1) and define

[ a ] q = 1 q a 1 q ,aR.

The q-analog of the power function ( a b ) n with n N 0 is

( a b ) 0 =1, ( a b ) n = k = 0 n 1 ( a b q k ) ,nN,a,bR.

More generally, if αR, then

( a b ) ( α ) = a α n = 0 a b q n a b q α + n .

It is easy to see that [ a ( t s ) ] ( α ) = a α ( t s ) ( α ) . Note that, if b=0 then a ( α ) = a α .

The q-gamma function is defined by

Γ q (x)= ( 1 q ) ( x 1 ) ( 1 q ) x 1 ,xR{0,1,2,},

and satisfies Γ q (x+1)= [ x ] q Γ q (x).

The q-derivative of a function f is here defined by

( D q f)(x)= f ( x ) f ( q x ) ( 1 q ) x ,( D q f)(0)= lim x 0 ( D q f)(x)for x0,

and q-derivatives of higher order by

( D q 0 f ) (x)=f(x)and ( D q n f ) (x)= D q ( D q n 1 f ) (x),nN.

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

( I q f)(x)= 0 x f(t) d q t=x(1q) n = 0 f ( x q n ) q n ,x[0,b].

From the definition of q-integral and the properties of series, we can get the following results on q-integral, which are helpful in the proofs of our main results.

Lemma 2.1

  1. (1)

    If f and g are q-integral on the interval [a,b], αR, c[a,b], then

    1. (i)

      a b (f(t)+g(t)) d q t= a b f(t) d q t+ a b g(t) d q t;

    2. (ii)

      a b αf(t) d q t=α b a f(t) d q t;

    3. (iii)

      a b f(t) d q t= a c f(t) d q t+ c b f(t) d q t.

  2. (2)

    If |f| is q-integral on the interval [0,x], then | 0 x f(t) d q t| 0 x |f(t)| d q t.

  3. (3)

    If f and g are q-integral on the interval [0,x], f(t)g(t) for all t[0,x], then 0 x f(t) d q t 0 x g(t) d q t.

Basic properties of q-integral and q-differential operators can be found in the book [11].

We now present out three formulas that will be used later ( D q i denotes the derivative with respect to variable i)

D q t ( t s ) ( α ) = [ α ] q ( t s ) ( α 1 ) , ( x D q 0 x f ( x , t ) d q t ) ( x ) = 0 x x D q f ( x , t ) d q t + f ( q x , x ) .

Remark 2.1 We note that if α>0 and abt, then ( t a ) ( α ) ( t b ) ( α ) .

Definition 2.1 [18]

Let α0 and f be a function defined on [0,b]. The fractional q-integral of the Riemann-Liouville type is defined by ( I q 0 f)(x)=f(x) and

( I q α f ) (x)= 1 Γ q ( α ) 0 x ( x q t ) ( α 1 ) f(t) d q t,α>0,x[0,b].

Definition 2.2 [16]

The fractional q-derivative of the Riemann-Liouville type of order α0 is defined by ( D q 0 f)(x)=f(x) and

( D q α f ) (x)= ( D q p I q p α f ) (x),α>0,

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

Next, we list some properties about q-derivative and q-integral that are already known in the literature.

Lemma 2.2 [16, 18]

Let α,β0 and f be a function defined on [0,1]. Then the following formulas hold:

  1. (i)

    ( I q β I q α f)(x)=( I q α + β f)(x);

  2. (ii)

    ( D q α I q α f)(x)=f(x).

Lemma 2.3 [18, 20]

Let α>0 and p be a positive integer. Then the following equality holds:

( I q α D q p f ) (x)= ( D q p I q α f ) (x) k = 0 p 1 x α p + k Γ q ( α + k p + 1 ) ( D q k f ) (0).

Lemma 2.4 [29]

Let X be a Banach space and PX be a cone. Suppose that Ω 1 and Ω 2 are bounded open sets contained in X such that 0 Ω 1 Ω ¯ 1 Ω 2 . Suppose further that S:P( Ω ¯ 2 Ω 1 )P is a completely continuous operator. If either

  1. (i)

    Suu for uP Ω 1 and Suu for uP Ω 2 , or

  2. (ii)

    Suu for uP Ω 1 and Suu for uP Ω 2 , then S has at least one fixed point in P( Ω ¯ 2 Ω 1 ).

The next result is important in the sequel.

Lemma 2.5 Let hC[0,1] be a given function. Then the boundary value problem

( D q α u ) (t)+h(t)=0,0t1,
(2.1)
D q i u(0)=0,i=0,1,,n2, D q u(1)= H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s,
(2.2)

has a unique solution

u(t)= t α 1 [ α 1 ] q ( H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s ) + 0 1 G(t,qs)h(s) d q s,

where

G(t,qs)= 1 Γ q ( α ) { ( 1 q s ) ( α 2 ) t α 1 ( t q s ) ( α 1 ) , 0 q s t 1 , ( 1 q s ) ( α 2 ) t α 1 , 0 t q s 1 .
(2.3)

Proof In view of Definition 2.2 and Lemma 2.2, we deduce

I q α D q α u(t)= I q α h(t)

and

I q α D q n I q n α u(t)= I q α h(t).

It follows from Lemma 2.3,

u(t)= c 1 t α 1 + c 2 t α 2 ++ c n t α n 1 Γ q ( α ) 0 t ( t q s ) ( α 1 ) h(s) d q s,
(2.4)

where c 1 ,, c n R are some constants to be determined. By the boundary condition u(0)=0, we get c n =0. Now if n2, then differentiating both sides of (2.4) j times for j=1,2,,n2, we obtain

D q j u ( t ) = [ α 1 ] q [ α j ] q c 1 t α j 1 + [ α 2 ] q [ α j 1 ] q c 2 t α j 2 + + [ α + n 1 ] q [ α n j + 2 ] q c n 1 t α n j + 1 1 Γ q ( α ) 0 t [ α 1 ] q [ α j ] q ( t q s ) ( α j 1 ) h ( s ) d q s .

From the boundary conditions D q i u(0)=0, for i=1,2,,n2, it is easy to know c 2 = c 3 == c n 1 =0. Thus, (2.4) reduces to

u(t)= c 1 t α 1 1 Γ q ( α ) 0 t ( t q s ) ( α 1 ) h(s) d q s.
(2.5)

Differentiating both sides of (2.5), we obtain

D q u(1)= [ α 1 ] q c 1 1 Γ q ( α ) 0 1 [ α 1 ] q ( 1 q s ) ( α 2 ) h(s) d q s.

Using the boundary condition D q u(1)= H 1 (φ(u))+ E H 2 (s,u(s)) d q s, we have

c 1 = 1 [ α 1 ] q ( H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s ) + 1 Γ q ( α ) 0 1 ( 1 q s ) ( α 2 ) h(s) d q s.

Hence,

u ( t ) = t α 1 [ α 1 ] q ( H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s ) + t α 1 Γ q ( α ) 0 1 ( 1 q s ) ( α 2 ) h ( s ) d q s 1 Γ q ( α ) 0 t ( t q s ) ( α 1 ) h ( s ) d q s = t α 1 [ α 1 ] q ( H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s ) + 0 1 G ( t , q s ) h ( s ) d q s .

 □

Remark 2.2 [20]

Let 0<τ<1. Then 0< τ α 1 <1 and min t [ τ , 1 ] G(t,qs) τ α 1 G(1,qs) for s[0,1].

The following properties of the Green’s function play important roles in this paper.

Lemma 2.6 [20]

The function G defined as (2.3) satisfies the following properties:

  1. (1)

    G(t,qs)0 and G(t,qs)G(1,qs) for all 0s,t1;

  2. (2)

    G(t,qs)r(t) max t [ 0 , 1 ] G(t,qs)=r(t)G(1,qs) for all 0t,s1 with r(t)= t α 1 .

3 Main results

Let the Banach space B=C[0,1] be endowed with the norm u= max t [ 0 , 1 ] |u(t)|. Let τ be a real constant with 0<τ<1 and define the cone KB by

K= { u C [ 0 , 1 ] | u ( t ) 0 , min t [ τ , 1 ] u ( t ) r u , φ ( u ) 0 } ,

where r =min{ min t [ τ , 1 ] t α 1 max t [ 0 , 1 ] t α 1 , min t [ τ , 1 ] r(t)}= min t [ τ , 1 ] t α 1 , obviously, r (0,1].

Define the operator F:KB by

(Fu)(t)= t α 1 [ α 1 ] q ( H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s ) + 0 1 G(t,qs)f ( s , u ( s ) ) d q s.

In order to get the integrated and rigorous theory, we make the following assumptions.

(H1) There are constants C 1 , C 2 >0 such that the functional φ satisfies the inequality

C 2 uφ(u) C 1 ufor all uC[0,1].
(3.1)

(H2) For each given ε>0, there are C 3 >0 and G ε >0 such that

| H 1 (z) C 3 z|<ε C 3 z,whenever z> G ε .

(H3) There exists a function T:[0,+)[0,+) satisfying the growth condition

T(u) C 4 u
(3.2)

for some C 4 0. For each ε>0 given and x[0,1], there is M ε >0 such that

| H 2 (x,u)T(u)|<εT(u),whenever u> M ε .

The following result plays an important role in the coming discussion.

Lemma 3.1 F:KK is completely continuous.

Proof It is easy to see that the operator F is continuous in view of the continuity of G and f.

By Lemmas 2.1 and 2.6, we have

min t [ τ , 1 ] F u ( t ) = min t [ τ , 1 ] [ t α 1 [ α 1 ] q ( H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s ) + 0 1 G ( t , q s ) f ( s , u ( s ) ) d q s ] = min t [ τ , 1 ] [ t α 1 [ α 1 ] q ( H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s ) ] + min t [ τ , 1 ] 0 1 G ( t , q s ) f ( s , u ( s ) ) d q s r 0 [ H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s ] + min t [ τ , 1 ] 0 1 r ( t ) max t [ 0 , 1 ] G ( t , q s ) f ( s , u ( s ) ) d q s r 0 [ H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s ] + min t [ τ , 1 ] r ( t ) max t [ 0 , 1 ] 0 1 G ( t , q s ) f ( s , u ( s ) ) d q s r F u ,

where r 0 = min t [ τ , 1 ] t α 1 max t [ 0 , 1 ] t α 1 . Thus, F(K)K.

Now let ΩK be bounded, i.e., there exists a positive constant M>0 such that uM for all uΩ. By the continuity of H 1 , H 2 , and φ, we easily see that H 1 and H 2 are bounded, so there exist constants P>0 and Q>0 such that | H 1 (φ(Ω))|P and | H 2 (t,u(t))|Q. Let L= max u M |f(t,u(t))|+1. Then for uΩ, from Lemma 2.6, we have

| F u ( t ) | = | t α 1 [ α 1 ] q ( H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s ) + 0 1 G ( t , q s ) f ( s , u ( s ) ) d q s | | 1 [ α 1 ] q ( H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s ) + 0 1 G ( 1 , q s ) f ( s , u ( s ) ) d q s | 1 [ α 1 ] q ( | H 1 ( φ ( u ) ) | + | E H 2 ( s , u ( s ) ) d q s | ) + L 0 1 G ( 1 , q s ) d q s ,
(3.3)

we rearrange (3.3) as follows:

| F u ( t ) | 1 [ α 1 ] q ( | H 1 ( φ ( u ) ) | + | E H 2 ( s , u ( s ) ) d q s | ) + L 0 1 G ( 1 , q s ) d q s 1 [ α 1 ] q ( | H 1 ( φ ( u ) ) | + E | H 2 ( s , u ( s ) ) | d q s ) + L 0 1 G ( 1 , q s ) d q s 1 [ α 1 ] q ( P + Q m ( E ) ) + L 0 1 G ( 1 , q s ) d q s .

Hence, F(Ω) is bounded.

On the other hand, for any given ε>0, there exists δ>0 small enough, such that |Fu( t 2 )Fu( t 1 )|<ε holds for each uΩ and 0 t 1 t 2 1 with | t 2 t 1 |<δ, that is to say, F(Ω) is equicontinuous. In fact,

| F u ( t 2 ) F u ( t 1 ) | | t 2 α 1 t 1 α 1 | | [ 1 [ α 1 ] q ( H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s ) + 0 1 ( 1 q s ) ( α 1 ) Γ q ( α ) f ( s , u ( s ) ) d q s ] | + 1 Γ q ( α ) | 0 t 2 ( t 2 q s ) ( α 2 ) f ( s , u ( s ) ) d q s 0 t 1 ( t 1 q s ) ( α 2 ) f ( s , u ( s ) ) d q s | | t 2 α 1 t 1 α 1 | [ 1 [ α 1 ] q ( H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s ) + 0 1 ( 1 q s ) ( α 1 ) Γ q ( α ) | f ( s , u ( s ) ) | d q s ] + 1 Γ q ( α ) | 0 1 t 2 α 2 ( 1 q s ) f ( s , u ( s ) ) d q s 1 Γ q ( α ) 0 1 t 1 α 2 ( 1 q s ) f ( s , u ( s ) ) d q s | | t 2 α 1 t 1 α 1 | [ 1 [ α 1 ] q ( P + Q m ( E ) ) + L 0 1 ( 1 q s ) ( α 1 ) Γ q ( α ) d q s ] + | t 2 α 2 t 1 α 2 | L Γ q ( α ) 0 1 ( 1 q s ) d q s .
(3.4)

Now, we estimate t 2 α 1 t 1 α 1 (we discuss t 2 α 2 t 1 α 2 in the same way, the proof here is omitted):

  1. (1)

    for 0 t 1 <δ, δ t 2 <2δ, t 2 α 1 t 1 α 1 t 2 α 1 < ( 2 δ ) α 1 2δ;

  2. (2)

    for 0 t 1 < t 2 δ, t 2 α 1 t 1 α 1 t 2 α 1 < δ α 1 2δ;

  3. (3)

    for δ t 1 < t 2 1, from the mean value theorem of differentiation, we have t 2 α 1 t 1 α 1 (α1)( t 2 t 1 )2δ.

Thus, we have

|Fu( t 2 )Fu( t 1 )|<ε.

By means of the Arzela-Ascoli theorem, F:KK is completely continuous. The proof is complete. □

Theorem 3.1 Assume that the nonlinearity f(t,u) splits in the sense that f(t,u)=a(t)g(u), for continuous functions a:[0,1][0,+) and g:[0,)[0,+) such that lim u 0 + g ( u ) u =+ and lim u + g ( u ) u =0. Suppose conditions (H1)-(H3) and

C 1 C 3 + C 4 m(E)<1
(3.5)

hold, where E(0,1] is a closed subinterval. Then the boundary value problem (1.1)-(1.2) has at least one solution.

Proof Begin by selecting the number η 1 such that

η 1 0 1 γ G(1,qs)a(s) d q s>1.
(3.6)

Now, there exists a number r 1 >0 such that g(u) η 1 u for 0<u< r 1 . Then take the open set

Ω r 1 = { u B : u < r 1 } .

By Remark 2.2, Lemma 2.6, and (3.6), for uK Ω r 1 , we find

( F u ) ( t ) = t α 1 [ α 1 ] q ( H 1 ( φ ( u ) ) + E H ( s , u ( s ) ) d q s ) + 0 1 G ( t , q s ) f ( s , u ( s ) ) d q s 0 1 G ( t , q s ) a ( s ) g ( u ( s ) ) d q s 0 1 G ( t , q s ) a ( s ) η 1 u ( s ) d q s u η 1 0 1 γ G ( 1 , q s ) a ( s ) d q s u .
(3.7)

Hence, Fuu, that is, F is a cone expansion on K Ω r 1 .

On the other hand, we consider two cases:

Case 1. Suppose that g is bounded for u[0,+). We may find r 2 >0 sufficiently large such that

g(u) r 2 for all u0.
(3.8)

Condition (3.5) implies the existence of ε>0 such that

C 1 C 3 ε+ C 1 C 3 +ε+( C 4 + C 4 ε)m(E)1.
(3.9)

Now if u G ε C 2 , then by (H1) we get φ(u) G ε . According to condition (H2), it follows that

| H 1 ( φ ( u ) ) C 3 φ(u)|<ε C 3 φ(u)
(3.10)

for all uC[0,1] with u G ε C 2 .

Next, since E(0,1] is a closed subinterval, we may select 0<τ<1 such that E[τ,1]. Then for each uK,

min t E u(t) min t [ τ , 1 ] u(t) γ u.
(3.11)

Thus combining condition (H3) we see that

| H 2 ( s , u ( s ) ) T ( u ( s ) ) |<εT ( u ( s ) ) for each sE with u M ε γ .
(3.12)

Take r 2 =max{ 2 r 1 γ , r 2 , G ε C 2 , M ε γ }. Set Ω r 2 ={uB:u< r 2 }.

From (3.8) we may assume without loss of generality that

g(u) r 2 0 1 G ( 1 , q s ) a ( s ) d q s ε.
(3.13)

Then by (3.9), (3.10), (3.12), and (3.13), for each uK Ω r 2 , we have

F u H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s + 0 1 G ( t , q s ) f ( s , u ( s ) ) d q s H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s + 0 1 G ( 1 , q s ) a ( s ) g ( u ( s ) ) d q s | H 1 ( φ ( u ) ) C 3 φ ( u ) | + C 3 | φ ( u ) | + E | H 2 ( s , u ( s ) ) T ( u ( s ) ) | d q s + E | T ( u ( s ) ) | d q s + 0 1 r 2 ε d q s ε C 1 C 3 u + C 1 C 3 u + m ( E ) C 4 ( 1 + ε ) u + ε u [ C 1 C 3 ε + ε + C 1 C 3 + m ( E ) C 4 ( 1 + ε ) ] u u .
(3.14)

Case 2. Suppose that g is unbounded at +∞. By condition lim u + g ( u ) u =0, there exists a number r 3 >0 such that for u> r 4 we find that g(u) η 2 u, where η 2 meets

η 2 0 1 G(1,qs)a(s) d q sε.
(3.15)

Noting that g is unbounded at +∞, we may find r 3 >max{ 2 r 1 γ , r 3 , G ε C 2 , M ε γ } such that

g(u)g( r 3 )for 0<u r 3 .
(3.16)

Now, put Ω r 2 ={uB:u< r 3 }. Then for each uK Ω r 3 we find that by (3.9), (3.10), (3.12), (3.15), and (3.16),

F u H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s + 0 1 G ( t , q s ) f ( s , u ( s ) ) d q s H 1 ( φ ( u ) ) + E H 2 ( s , u ( s ) ) d q s + 0 1 G ( 1 , q s ) a ( s ) g ( u ( s ) ) d q s | H 1 ( φ ( u ) ) C 3 φ ( u ) | + C 3 | φ ( u ) | + E | H 2 ( s , u ( s ) ) T ( u ( s ) ) | d q s + E | T ( u ( s ) ) | d q s + 0 1 G ( 1 , q s ) a ( s ) g ( r 3 ) d q s C 3 ε φ ( u ) + C 1 C 3 u + E ( 1 + ε ) T ( u ( s ) ) d q s + 0 1 G ( 1 , q s ) a ( s ) η 2 u d q s ε C 1 C 3 u + C 1 C 3 u + m ( E ) C 4 ( 1 + ε ) u + ε u [ C 1 C 3 ε + ε + C 1 C 3 + m ( E ) C 4 ( 1 + ε ) ] u u .
(3.17)

Hence, Fuu.

To summarize, we conclude from (3.14) and (3.17) that F is a cone compression on K Ω r 2 .

With the help of Lemma 2.4 we can now deduce the existence of function u 0 K Ω ¯ r 2 Ω r 1 such that F u 0 = u 0 . Hence, the problem (1.1)-(1.2) has at least one solution. The proof is completed. □

Next, by choosing suitable forms of H 1 and φ, we present the corresponding boundary value problems with separated boundary conditions, integral boundary conditions and multi-point boundary conditions as corollaries of Theorem 3.1 to illustrate the universality and generalization of our results.

Corollary 3.1 Assume f(t,u) is defined as in Theorem  3.1 and (H1)-(H3) hold. If n=3, H 1 (φ(u))+ E H 2 (s,u(s)) d q s=β0. Then the boundary value problem

( D q α u ) (t)+f ( t , u ( t ) ) =0,0<t<1,2<α3,
(3.18)
u(0)= D q u(0)=0, D q u(1)=β,
(3.19)

has at least one solution.

This existence result of positive solution for the boundary value problem (3.18)-(3.19) has been studied by Ferreira in [20] and Yang et al. in [27].

Corollary 3.2 Assume f(t,u) is defined as in Theorem  3.1, φ(u)= T u(s) d q s. Then φ is linear functional, T=[a,b](0,1] is a closed subinterval, just need m(T) C 1 , where m is Lebesgue measure. If (H 1)-(H 3) hold, in addition

C 1 C 3 + C 4 m(E)<1,

then the boundary value problem (3.20)-(3.21) with integral boundary conditions

( D q α u ) (t)+f ( t , u ( t ) ) =0,0<t<1,n1<αn,
(3.20)
u(0)= D q u(0)=0, D q u(1)= H 1 ( T u ( s ) d q s ) + E H 2 ( s , u ( s ) ) d q s,
(3.21)

has at least one solution.

Corollary 3.3 Assume f(t,u) is defined as in Theorem  3.1, φ(u)= i = 1 m a i u( ξ i ). Then φ is linear functional, just need i = 1 m | a i | C 1 , E(0,1] is a closed subinterval. If (H1)-(H3) hold, moreover,

C 1 C 3 + C 4 m(E)<1,

then the boundary value problem (3.22)-(3.23) with multi-point boundary conditions

( D q α u ) (t)+f ( t , u ( t ) ) =0,0<t<1,n1<αn,
(3.22)
u(0)= D q u(0)=0, D q u(1)= H 1 ( i = 1 m a i u ( ξ i ) ) + E H 2 ( s , u ( s ) ) d q s,
(3.23)

has at least one solution.

4 Example

In this section, we will give an example to expound our main results.

Example 4.1 Consider the following boundary value problem:

( D q α u ) (t)+f(t,u)=0,0<t<1,
(4.1)
u(0)= D q u(0)=0, D q u(1)= j = 1 m 2 a j ξ j + e j = 1 m 2 a j ξ j + E u(s) d q s,
(4.2)

here 2<α3, a j 0 with 1 4 j = 1 m a j 3 4 and ξ j [ 1 100 , 1 10 ] for j=1,2,,m, E(0,1] is a closed subinterval, φ(u)= j = 1 m 2 a j ξ j , H 1 (φ(u))=φ(u)+ e φ ( u ) and E H 2 (s,u(s)) d q s= E u(s) d q s satisfies (H3).

Since a j 0 and ξ j [ 1 100 , 1 10 ], then |φ(u)|= j = 1 m 2 a j ξ j . Choosing C 1 = 3 20 and C 2 = 1 40 , we have

1 200 |φ(u)|= j = 1 m 2 a j ξ j 3 20 .

For each ε>0, setting C 3 =1, it is clear that

lim φ ( u ) | H 1 ( φ ( u ) ) φ(u)|=|φ(u)+ e φ ( u ) φ(u)|=0,

so (H2) holds.

If only the given function f(t,u) satisfies f(t,u)=a(t)g(u), for continuous functions a:[0,1][0,+) and g:R[0,+) such that lim u 0 + g ( u ) u =+ and lim u + g ( u ) u =0, in addition,

C 1 C 3 + C 4 m(E)<1,

then, by Theorem 3.1, the boundary value problem (4.1)-(4.2) has at least one solution.

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Acknowledgements

The authors sincerely thank the reviewers for their valuable suggestions and useful comments, which have led to the present improved version of the original manuscript. This research is supported by the Natural Science Foundation of China (61374074), Shandong Provincial Natural Science Foundation (ZR2012AM009, ZR2013AL003), also supported by Graduate Innovation Foundation of University of Jinan (YCX13013).

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Correspondence to Zhenlai Han.

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Keywords

  • fractional q-difference equations
  • mixed nonlinear conditions
  • fixed point theorem in cones
  • existence of solutions