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Existence results for fractional q-difference equations with nonlocal q-integral boundary conditions

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Abstract

In this paper, we discuss the existence of positive solutions for nonlocal q-integral boundary value problems of fractional q-difference equations. By applying the generalized Banach contraction principle, the monotone iterative method, and Krasnoselskii’s fixed point theorem, some existence results of positive solutions are obtained. In addition, some examples to illustrate our results are given.

MSC:39A13, 34B18, 34A08.

1 Introduction

Studies on q-difference equations appeared already at the beginning of the twentieth century in intensive works especially by Jackson [1], Carmichael [2] and other authors such as Poincare, Picard, Ramanujan. Up to date, q-difference equations have evolved into a multidisciplinary subject; for example, see [36] and the references therein. For some recent work on q-difference equations, we refer the reader to the papers [720], and basic definitions and properties of q-difference calculus can be found in the book [21]. On the other hand, fractional differential equations have gained importance due to their numerous applications in many fields of science and engineering including fluid flow, rheology, diffusive transport akin to diffusion, electrical networks, probability, etc. For details, see [22, 23]. Many researchers studied the existence of solutions to fractional boundary value problems; see, for example, [2433] and the references therein.

The fractional q-difference calculus had its origin in the works by Al-Salam [34] and Agarwal [35]. More recently, perhaps due to the explosion in research within the fractional differential calculus setting, new developments in this theory of fractional q-difference calculus were made, specifically, q-analogues of the integral and differential fractional operators properties such as the Mittag-Leffler function, the q-Laplace transform, and q-Taylor’s formula [12, 21, 36, 37], just to mention some.

However, the theory of boundary value problems for nonlinear q-difference equations is still in the initial stage and many aspects of this theory need to be explored. Recently, there have been some paper considering the existence of solutions to boundary value problems of fractional q-difference equations, for example, [10, 1620, 38] and the references therein.

In [16], Ferreira considered the Dirichlet type nonlinear q-difference boundary value problem

{ D q α u ( t ) + f ( u ( t ) ) = 0 , 0 < t < 1 , 1 < α 2 , u ( 0 ) = u ( 1 ) = 0 .

By applying a fixed point theorem in cones, sufficient conditions for the existence of nontrivial solutions were enunciated.

In [20], Graef and Kong investigated the boundary value problem with fractional q-derivatives

{ ( D q α u ) ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , n 1 < α n , n N , ( D q i u ) ( 0 ) = 0 , i = 0 , , n 2 , b D q u ( 1 ) = j = 1 m a j D q u ( t j ) + λ ,

where λ0 is a parameter, and the uniqueness, existence, and nonexistence of positive solutions are considered in terms of different ranges of λ.

Furthermore, Ahmad, Ntouyas, and Purnaras [10] studied the following nonlinear fractional q-difference equation with nonlocal boundary conditions:

{ ( C D q α u ) ( t ) = f ( t , u ( t ) ) , 0 t 1 , 1 < α 2 , a 1 u ( 0 ) b 1 D q u ( 0 ) = c 1 u ( η 1 ) , a 2 u ( 1 ) + b 2 D q u ( 1 ) = c 2 u ( η 2 ) ,

where D q α C is the fractional q-derivative of the Caputo type and a i , b i , c i , η i R. The existence of solutions for the problem is shown by applying some well-known tools of fixed point theory such as Banach’s contraction principle, Krasnoselskii’s fixed point theorem, and the Leray-Schauder nonlinear alternative.

In this paper, we deal with the following nonlocal q-integral boundary value problem of nonlinear fractional q-derivatives equation:

{ ( 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 ,
(1.1)

where q(0,1), 1<α2, 0<β2, 0<η<1, and μ>0 is a parameter, D q α is the q-derivative of Riemann-Liouville type of order α, f:[0,1]× R + R + is continuous, in which R + =[0,+). To the authors’ knowledge, no one has studied the existence of positive solutions for the fractional q-difference boundary value problem (1.1). In the present work, we gave the corresponding Green’s function of the boundary value problem (1.1) and its properties. 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 are enunciated.

2 Preliminaries on q-calculus and lemmas

For the convenience of the reader, below we recall some known facts on fractional q-calculus. The presentation here can be found in, for example, [1, 21, 36, 37].

Let q(0,1) and define

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

The q-analogue of the power function ( a b ) n with n N 0 :={0,1,2,} is

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

More generally, if γR, then

( a b ) ( γ ) = a γ k = 0 a b q k a b q γ + k ,a0.
(2.1)

Clearly, if b=0, then a ( γ ) = a γ . 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 defined by

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

and the q-derivatives of higher order by

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

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

( I q f)(x)= 0 x f(s) d q s=x(1q) k = 0 f ( x q k ) q k ,x[0,b],

provided the sum converges absolutely.

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

a b f(s) d q s= 0 b f(s) d q s 0 a f(s) d q s.

Obviously, if f(x)g(x) on [a,b], then a b f(x) d q x a b g(x) d q x.

Similar as done for derivatives, an operator I q n is given by

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

The fundamental theorem of calculus applies to these operators I q and D q , i.e.,

( 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).
(2.2)

The following formulas will be used later, namely, the integration by parts formula

0 x f(s)( D q g)(s) d q s= [ f ( s ) g ( s ) ] s = 0 s = x 0 x ( D q f)(s)g(qs) d q s,

and

(2.3)
(2.4)
(2.5)
(2.6)

where D q t denotes the derivative with respect to the variable t.

Definition 2.1 Let α0 and f be a function defined on [0,1]. The fractional q-integral of Riemann-Liouville type is ( I q 0 f)(x)=f(x) and

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

Definition 2.2 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 [ α ] I q [ α ] α f ) (x),α>0,

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

Lemma 2.3 Assume that γ0 and abt, then ( t a ) ( γ ) ( t b ) ( γ ) .

Lemma 2.4 Let α,β0 and f be a function defined on [0,1]. Then the following formulas hold:

  1. (1)

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

  2. (2)

    ( D q α I q α f)(x)=f(x).

Lemma 2.5 ([16])

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

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

Lemma 2.6 ([36])

Let α R + , λ(1,+), the following is valid:

I q α ( ( t a ) ( λ ) ) = Γ q ( λ + 1 ) Γ q ( α + λ + 1 ) ( t a ) ( α + λ ) ,0<a<t<b.

Particularly, for λ=0, a=0, using q-integration by parts, we have

( I q α 1 ) ( t ) = 1 Γ q ( α ) 0 t ( t q s ) ( α 1 ) d q s = 1 Γ q ( α ) 0 t D q s ( ( t s ) ( α ) ) [ α ] q d q s = 1 Γ q ( α + 1 ) 0 t s D q ( ( t s ) ( α ) ) d q s = 1 Γ q ( α + 1 ) t ( α ) .

Obviously, we have 0 t ( t q s ) ( α 1 ) d q s= 1 [ α ] q t ( α ) .

In order to define the solution for the problem (1.1), we need the following lemmas.

Lemma 2.7 Let M:= Γ q (α+β)μ Γ q (α) η α + β 1 >0. Then, for a given yC[0,1], the unique solution of the boundary value problem

( D q α u ) (t)+y(t)=0,t(0,1),1<α2,
(2.7)

subject to the boundary condition

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 , 0 < β 2 , 0 < η < 1 ,
(2.8)

is given by

u(t)= 0 1 G(t,qs)y(s) d q s,
(2.9)

where

(2.10)
(2.11)

and

H(η,s)={ η α + β 1 ( 1 s ) ( α 1 ) ( η s ) ( α + β 1 ) , 0 s η < 1 , η α + β 1 ( 1 s ) ( α 1 ) , 0 < η s 1 .
(2.12)

Proof Since 1<α2, we take n=2. In view of Definition 2.1 and Lemma 2.4, we have

( D q α u ) (t)=y(t) ( I q α D q 2 I q 2 α u ) (t)= ( I q α y ) (t).

Then it follows from Lemma 2.5 that the solution u(t) of (2.7) and (2.8) is given by

u(t)= c 1 t α 1 + c 2 t α 2 0 t ( t q s ) ( α 1 ) Γ q ( α ) y(s) d q s
(2.13)

for some constants c 1 , c 2 R. Since u(0)=0, we have c 2 =0.

Using the Riemann-Liouville integral of order β for (2.13), we have

( I q β u ) ( t ) = 0 t ( t q s ) ( β 1 ) Γ q ( β ) ( c 1 s α 1 0 s ( s q x ) ( α 1 ) Γ q ( α ) y ( x ) d q x ) d q s = 0 t ( t q s ) ( α + β 1 ) Γ q ( α + β ) y ( s ) d q s + c 1 0 t ( t q s ) ( β 1 ) Γ q ( β ) s α 1 d q s = 0 t ( t q s ) ( α + β 1 ) Γ q ( α + β ) y ( s ) d q s + c 1 Γ q ( α ) Γ q ( α + β ) t α + β 1 ,

where we have used Lemma 2.4 and Lemma 2.6. Using the boundary condition u(1)=μ I q β u(η), we get

c 1 = Γ q ( α + β ) M ( 0 1 ( 1 q s ) ( α 1 ) Γ q ( α ) y ( s ) d q s μ 0 η ( η q s ) ( α + β 1 ) Γ q ( α + β ) y ( s ) d q s ) .

Hence, we have

u ( t ) = Γ q ( α + β ) t α 1 M ( 0 1 ( 1 q s ) ( α 1 ) Γ q ( α ) y ( s ) d q s μ 0 η ( η q s ) ( α + β 1 ) Γ q ( α + β ) y ( s ) d q s ) 0 t ( t q s ) ( α 1 ) Γ q ( α ) y ( s ) d q s = 0 1 t α 1 ( 1 q s ) ( α 1 ) Γ q ( α ) y ( s ) d q s 0 t ( t q s ) ( α 1 ) Γ q ( α ) y ( s ) d q s + μ t α 1 M ( 0 1 η α + β 1 ( 1 q s ) ( α 1 ) y ( s ) d q s 0 η ( η q s ) ( α + β 1 ) y ( s ) d q s ) = 0 1 g ( t , q s ) y ( s ) d q s + μ t α 1 M 0 1 H ( η , q s ) d q s = 0 1 G ( t , q s ) y ( s ) d q s .

This completes the proof of the lemma. □

Remark 2.8 For the special case where μ=0, Lemma 2.7 has been obtained by Ferreira [16].

Lemma 2.9 ([16])

The function g(t,s) defined by (2.11) satisfies the following properties:

g(t,qs)0andg(t,qs)g(qs,qs) for all 0t,s1.

Lemma 2.10 The function G(t,qs) defined by (2.10) satisfies the following properties:

  1. (i)

    G is a continuous function and G(t,qs)0 for (t,s)[0,1]×[0,1].

  2. (ii)

    There exists a positive function ρC((0,1),(0,+)) such that

    max 0 t 1 G(t,qs)ρ(s),

where

ρ(s)=g(qs,qs)+ μ M H(η,qs),s(0,1).

Proof It is easy to prove that the statement (i) holds. On the other hand, we note that g(t,s) defined by (2.11) is decreasing with respect to t for st and increasing with respect to t for ts. Hence, we have

max 0 t 1 G ( t , q s ) = max 0 t 1 ( g ( t , q s ) + μ t α 1 M H ( η , q s ) ) g ( q s , q s ) + μ M H ( η , q s ) : = ρ ( s ) , s ( 0 , 1 ) .

The proof is completed. □

3 The main results

Let X=C([0,1]) be a Banach space endowed with the norm u X = max 0 t 1 |u(t)|. Define the cone PX by P={uX:u(t)0,0t1}.

Define the operator T:PX as follows:

(Tu)(t)= 0 1 G(t,qs)f ( s , u ( s ) ) d q s.
(3.1)

It follows from the nonnegativeness and continuity of G and f that the operator T:PX satisfies T(P)P and is completely continuous.

Theorem 3.1 Suppose that f:[0,1]× R + R + is continuous and there exists a function h:[0,1] R + such that

| f ( t , u ) f ( t , v ) | h(t)|uv|,t[0,1],u,v R + .
(3.2)

Then the BVP (1.1) has a unique positive solution provided

0 1 s α 1 ( 1 q s ) ( α 1 ) h(s) d q s< Γ q ( α ) M 2 Γ q ( α + β ) .
(3.3)

Proof We will prove that under the assumptions (3.2) and (3.3), T m is a contraction operator for m sufficiently large.

By (2.10), (2.11), and (2.12), for u,vP, we obtain the estimate

| ( T u ) ( t ) ( T v ) ( t ) | 0 1 G ( t , q s ) | f ( s , u ( s ) ) f ( s , v ( s ) ) | d q s = 0 1 [ g ( t , q s ) + μ t α 1 M H ( η , q s ) ] | f ( s , u ( s ) ) f ( s , v ( s ) ) | d q s 0 1 t α 1 ( 1 q s ) ( α 1 ) Γ q ( α ) h ( s ) d q s u v X + μ η α + β 1 0 1 t α 1 ( 1 q s ) ( α 1 ) M h ( s ) d q s u v X = Γ q ( α + β ) t α 1 u v X Γ q ( α ) M 0 1 ( 1 q s ) ( α 1 ) h ( s ) d q s = Λ 1 Γ q ( α + β ) t α 1 Γ q ( α ) M u v X ,

where Λ 1 = 0 1 ( 1 q s ) ( α 1 ) h(s) d q s.

Consequently,

| ( T 2 u ) ( t ) ( T 2 v ) ( t ) | 0 1 G ( t , q s ) | f ( s , ( T u ) ( s ) ) f ( s , ( T v ) ( s ) ) | d q s Λ 1 Γ q ( α + β ) u v X Γ q ( α ) M 0 1 G ( t , q s ) s α 1 h ( s ) d q s Λ 1 [ Γ q ( α + β ) ] 2 t α 1 u v X [ Γ q ( α ) M ] 2 0 1 s α 1 ( 1 q s ) ( α 1 ) h ( s ) d q s = Λ 1 Λ 2 [ Γ q ( α + β ) ] 2 t α 1 [ Γ q ( α ) M ] 2 u v X ,

where Λ 2 = 0 1 s α 1 ( 1 q s ) ( α 1 ) h(s) d q s.

By introduction, we have

| ( T m u ) ( t ) ( T m v ) ( t ) | Λ 1 Λ 2 m 1 [ Γ q ( α + β ) ] m t α 1 [ Γ q ( α ) M ] m u v X .

According to (3.3), we can choose m sufficiently large such that

Λ 1 Λ 2 m 1 [ Γ q ( α + β ) ] m [ Γ q ( α ) M ] m = Λ 1 Λ 2 [ Λ 2 Γ q ( α + β ) Γ q ( α ) M ] m < ( 1 2 ) m Λ 1 Λ 2 < 1 2 ,

which implies

( T m u ) ( t ) ( T m v ) ( t ) X < 1 2 u v X .

Hence, it follows from the generalized Banach contraction principle that the BVP (1.1) has a unique positive solution. □

Remark 3.2 When h(t) h 0 is a constant, the condition (3.2) reduces to a Lipschitz condition.

For the sake of convenience, we set

where τ 1 = q m 1 , τ 2 = q m 2 with m 1 , m 2 N 0 , m 1 > m 2 .

Theorem 3.3 Suppose that there exists 0< ξ 1 < ξ 2 such that

( H f ) f(t,u):[0,1]×[0, ξ 2 ] R + is continuous and nondecreasing relative to u, and

(3.4)
(3.5)

Then the BVP (1.1) has one positive solution u satisfying

ξ 1 u X ξ 2 and lim m T m u 0 = u , u 0 (t)= ξ 2 .

Proof We denote Ω[ ξ 1 , ξ 2 ]={uP: ξ 1 u ξ 2 }. In what follows, we first show that T:Ω[ ξ 1 , ξ 2 ]Ω[ ξ 1 , ξ 2 ].

Let uΩ[ ξ 1 , ξ 2 ]; then 0< ξ 1 u(t) u X ξ 2 . By assumption ( H f ), we have

Hence, for any uΩ[ ξ 1 , ξ 2 ],

T u X = max 0 t 1 0 1 | G ( t , q s ) f ( s , u ( s ) ) | d q s 0 1 ρ ( s ) f ( s , u ( s ) ) d q s L ξ 2 0 1 ρ ( s ) d q s L ξ 2 0 1 ( q α 1 ( 1 q s ) ( α 1 ) Γ q ( α ) + μ M H ( η , q s ) ) d q s = ξ 2 ,

and

T u X min τ 1 t τ 2 ( 0 1 | g ( t , q s ) f ( s , u ( s ) ) | d q s + μ t α 1 M 0 1 | H ( η , q s ) f ( s , u ( s ) ) | d q s ) μ τ 1 α 1 M ( τ 1 τ 2 H ( η , q s ) d q s ) l ξ 1 = ξ 1 .

Thus, we get T:Ω[ ξ 1 , ξ 2 ]Ω[ ξ 1 , ξ 2 ].

Let u 0 (t)= ξ 2 , 0t1; then u 0 Ω[ ξ 1 , ξ 2 ]. Let u 1 (t)=T u 0 (t); then u 1 Ω[ ξ 1 , ξ 2 ]. We denote u m + 1 =T u m = T m + 1 u 0 , m=0,1,2, . According to T:Ω[ ξ 1 , ξ 2 ]Ω[ ξ 1 , ξ 2 ], we have u n Ω[ ξ 1 , ξ 2 ], n=0,1,2, . Since T is completely continuous, we assert that { u m } m = 1 has a convergent subsequence { u m k } k = 1 and there exists u Ω[ ξ 1 , ξ 2 ] such that lim k + u m k = u . Since u 1 Ω[ ξ 1 , ξ 2 ], then ξ 1 u 1 u 1 X ξ 2 = u 0 . According to the definition of T and ( H f ), we have

u 1 ( t ) = ( T u 0 ) ( t ) = 0 1 G ( t , q s ) f ( s , u 0 ( s ) ) d q s 0 1 ρ ( s ) f ( s , ξ 2 ) d q s L ξ 2 0 1 ( q α 1 ( 1 q s ) ( α 1 ) Γ q ( α ) + μ M H ( η , q s ) ) d q s = ξ 2 = u 0 ( t ) ,

which implies

u 2 (t)=T u 1 (t)T u 0 (t)= u 1 (t),0t1.

By introduction, we have u m + 1 (t) u m (t) for 0t1, m=0,1,2, . Thus, there exists u Ω[ ξ 1 , ξ 2 ] such that lim m + u m = u . From the continuity of T and u m + 1 =T u m , we have u =T u . The proof is completed. □

Our next existence result is based on Krasnoselskii’s fixed point theorem [39].

Lemma 3.4 (Krasnoselskii’s)

Let E be a Banach space, and let PE be a cone. Assume Ω 1 , Ω 2 are open subsets of E with θ Ω 1 Ω ¯ 1 Ω 2 , and let T:P( Ω ¯ 2 Ω 1 )P be a completely continuous operator such that

Tuu,uP Ω 1 andTuu,uP Ω 2 .

Then T has at least one fixed point in P( Ω ¯ 2 Ω 1 ).

Theorem 3.5 Let f(t,u) be a nonnegative continuous function on [0,1]× R + . In addition, we assume that

(H1) There exists a positive constant r 1 such that

f(t,u)κ r 1 , for (t,u)[ τ 3 , τ 4 ]×[0, r 1 ],

where τ 3 = q m 3 , τ 4 = q m 4 with m 3 , m 4 N 0 , m 3 > m 4 , and

κ ( τ 3 τ 4 ( g ( q s , q s ) + μ M H ( η , q s ) ) d q s ) 1 .

(H2) There exists a positive constant r 2 with r 2 > r 1 such that

f(t,u)L r 2 , for (t,u)[0,1]×[0, r 2 ].

Then the BVP (1.1) has at least one positive solution u 0 satisfying 0< r 1 u 0 X r 2 .

Proof By Lemma 2.9, we obtain that max 0 t 1 g(t,qs)=g(qs,qs). Let Ω 1 ={uX: u X < r 1 }. For any uX Ω 1 , according to (H1) and the definitions of τ 3 and τ 4 , we obtain

T u X = max 0 t 1 0 1 g ( t , q s ) f ( s , u ( s ) ) d q s + max 0 t 1 0 1 μ t α 1 M H ( η , q s ) f ( s , u ( s ) ) d q s = 0 1 [ g ( q s , q s ) + μ M H ( η , q s ) ] f ( s , u ( s ) ) d q s τ 3 τ 4 ( g ( q s , q s ) + μ M H ( η , q s ) ) f ( s , u ( s ) ) d q s κ r 1 τ 3 τ 4 ( g ( q s , q s ) + μ M H ( η , q s ) ) d q s u X = r 1 .

Let Ω 2 ={uX: u X < r 2 }. For any uX Ω 2 , by (H2) and Lemma 2.10, we have

T u X = max 0 t 1 0 1 G ( t , q s ) f ( s , u ( s ) ) d q s L r 2 0 1 ρ ( s ) d q s L r 2 0 1 ( q α 1 ( 1 q s ) ( α 1 ) Γ q ( α ) + μ M H ( η , q s ) ) d q s = u X = r 2 .

Now, an application of Lemma 3.4 concludes the proof. □

Theorem 3.6 Assume that there exist m+1 positive numbers 0< r 1 < r 2 << r m + 1 such that

(H3) f(t,u)>κ r 2 j 1 for (t,u)[ τ 3 , τ 4 ]×[0, r 2 j 1 ] and f(t,u)<L r 2 j for (t,u)[0,1]×[0, r 2 j ], j=1,2,,[ m + 1 2 ]; or

(H4) f(t,u)>κ r 2 j for (t,u)[ τ 3 , τ 4 ]×[0, r 2 j ] and f(t,u)<L r 2 j 1 for (t,u)[0,1]×[0, r 2 j 1 ], j=1,2,,[ m + 1 2 ],

where τ 3 , τ 4 , κ are given in (H1).

Then the BVP (1.1) has at least n positive solutions u 1 , u 2 ,, u m with r j < u j X < r j + 1 , j=1,,m.

Proof Suppose that the condition (H3) holds. According to the continuity of f, for every pair ( r j , r j + 1 ), there exists ( a j , b j ) with r j < a j < b j < r j + 1 such that

It follows from Theorem 3.5 that every pair ( a j , b j ) presents a positive solution of the BVP (1.1) such that a j u j X < b j , j=1,2,,m.

When the condition (H4) holds, the proofs are similar to those in the case (H3). The proof is completed. □

4 Examples

Example 4.1

The fractional boundary value problem

{ D 0.5 1.5 u ( t ) + e t ( 1 + u ) 2 ( 1 + e t ) + t sin 2 t + 4 = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = 0.25 I 0.5 0.5 u ( 0.75 )
(4.1)

has a unique positive solution.

Proof In this case, α=1.5, β=q=0.5, μ=0.25, η=0.75. Let

f(t,u)= e t ( 1 + u ) 2 ( 1 + e t ) +t sin 2 t+4,(t,u)[0,1]×[0,+),

and h(t)= e t 2 ( 1 + e t ) . It is easy to prove that

| f ( t , u ) f ( t , v ) | h(t)|uv|,for (t,u),(t,v)[0,1]×[0,+).

By simple calculation, we get

and

0 1 ( 1 q s ) ( α 1 ) h ( s ) d q s = 0 1 ( 1 q s ) ( α 1 ) e s 2 ( 1 + e s ) d q s 1 2 0 1 ( 1 q s ) ( α 1 ) d q s 0.3867 ,

which implies that

0 1 s α 1 ( 1 q s ) ( α 1 ) h(s) d q s< Γ q ( α ) M 2 Γ q ( α + β ) 0.4512.

Obviously, for any m3, we have

Λ 1 Λ 2 m 1 [ Γ q ( α + β ) ] m [ Γ q ( α ) M ] m 0.3867 × Γ q ( α + β ) 2 m 1 Γ q ( α ) M <0.2329< 1 2 .

Thus Theorem 3.1 implies that the boundary value problem (4.1) has a unique positive solution. □

Example 4.2

Consider the following fractional boundary value problem:

{ D 0.5 1.5 u ( t ) + 21 u t 2 10 + t 10 = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = 1.25 I 0.5 0.5 u ( 0.25 ) ,
(4.2)

where α=1.5, β=q=0.5, μ=1.25, η=0.25. Choosing m 1 =2, m 2 =1, then τ 1 =0.25, τ 2 =0.5.

A simple computation showed M0.6786. By Lemma 2.6 and with the aid of a computer, we obtain that

l= ( μ τ 1 α 1 M τ 1 τ 2 H ( η , q s ) d q s ) 1 18.9553,

and

L= ( 0 1 ( q α 1 ( 1 q s ) ( α 1 ) Γ q ( α ) + μ M H ( η , q s ) ) d q s ) 1 1.1770.

Let f(t,u)= t 2 10 + t 10 +21 u . Take ξ 1 =1, ξ 2 =340, then f(t,u) satisfies

  1. (i)

    f(t,u):[0,1]×[0,340] R + is continuous and nondecreasing relative to u;

  2. (ii)

    max 0 t 1 f(t, ξ 2 )=f( 1 2 ,340)387.2459<L ξ 2 400.1800;

  3. (iii)

    min 1 4 t 1 2 f(t, ξ 1 )=f( 1 4 ,1)21.0188>l ξ 1 18.9553.

So, by Theorem 3.3, the problem (4.2) has one positive solution u satisfying

1 u X 340and lim m T m u 0 = u , u 0 (t)=340.

Example 4.3

Consider the following fractional boundary value problem:

{ D 0.5 1.5 u ( t ) + u 2 100 + sin 2 t 20 + t 10 + 1 = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = 0.75 I 0.5 1.5 u ( 0.75 ) ,
(4.3)

where α=β=1.5, q=0.5, μ=η=0.75. Choosing m 3 =3, m 4 =1, then τ 3 =0.125, τ 4 =0.5.

By calculation, we get M1.0661. By Lemma 2.6 and with the aid of a computer, we obtain that

and

L= ( 0 1 ( q α 1 ( 1 q s ) ( α 1 ) Γ q ( α ) + μ M H ( η , q s ) ) d q s ) 1 1.4966.

Let f(t,u)= u 2 100 + sin 2 t 20 + t 10 +1. Take κ=7 and r 1 = 1 7 , r 2 = 9 10 . Then f(t,u) satisfies

  1. (i)

    f(t,u):[0,1]× R + R + is continuous;

  2. (ii)

    f(t,u) u 2 100 + 1 20 + 1 10 +11.1581<L r 2 1.3469, (t,u)[0,1]×[0, 9 10 ];

  3. (iii)

    f(t,u)= u 2 100 + sin 2 t 20 + t 10 +11+ τ 1 10 =1.0125>κ r 1 =1, (t,u)[0.125,0.5]×[0, 1 7 ].

So, by Theorem 3.5, the problem (4.3) has at least one positive solution u 0 with 1 7 u 0 X 9 10 .

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Acknowledgements

Dedicated to Professor Hari M Srivastava.

The authors are highly grateful for the referee’s careful reading and comments on this paper. The research is supported by the National Natural Science Foundation of China (11271372, 11201138); it is also supported by the Hunan Provincial Natural Science Foundation of China (12JJ2004, 13JJ3106), and the Scientific Research Fund of Hunan Provincial Education Department (12B034).

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Correspondence to Yulin Zhao.

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The authors declare that they have no competing interests.

Authors’ contributions

Each of the authors, YZ, HC, and QZ, contributed to each part of this study equally and read and approved the final version of the manuscript.

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Keywords

  • fractional q-difference equation
  • nonlocal boundary value problems
  • existence
  • positive solutions