Open Access

Existence and multiplicity of positive solutions for nonhomogeneous boundary value problems with fractional q-derivatives

Boundary Value Problems20132013:103

DOI: 10.1186/1687-2770-2013-103

Received: 17 January 2013

Accepted: 12 April 2013

Published: 25 April 2013

Abstract

In this paper, we study a class of fractional q-difference equations with nonhomogeneous boundary conditions. By applying the classical tools from functional analysis, sufficient conditions for the existence of single and multiple positive solutions to the boundary value problem are obtained in term of the explicit intervals for the nonhomogeneous term. In addition, some examples to illustrate our results are given.

MSC:34A08, 34B18, 39A13.

Keywords

fractional q-difference equation nonhomogeneous boundary value problem positive solution multiplicity

1 Introduction

Fractional differential equations have attracted considerable interest because of its demonstrated applications in various fields of science and engineering including fluid flow, rheology, diffusive transport akin to diffusion, electrical networks, probability [1, 2]. Many researchers have studied the existence of solutions (or positive solutions) to fractional boundary value problems; for example, see [310] and the references therein.

The early work on q-difference calculus or quantum calculus dates back to Jackson’s papers [11], basic definitions and properties of quantum calculus can be found in the book [12]. For some recent existence results on q-difference equations, we refer to [1315] and the references therein.

The fractional q-difference calculus had its origin in the works by Al-Salam [16] and Agarwal [17]. More recently, there seems to be new interest in the study of this subject and many new developments were made in this theory of fractional q-difference calculus [1822]. Specifically, fractional q-difference equations have attracted the attentions of several researchers. Some recent work on the existence theory of fractional q-difference equations can be found in [20, 2331]. However, the study of boundary value problems for nonlinear fractional q-difference equations is still in the initial stage and many aspects of this topic need to be explored.

By using a fixed-point theorem in a cone, M. El-Shahed and F. Al-Askar [25] were concerned with the existence of positive solutions to nonlinear q-difference equation:
{ D q α C u ( t ) + a ( t ) f ( u ( t ) ) = 0 , 0 < t < 1 , 2 < α 3 , u ( 0 ) = D q 2 u ( 0 ) = 0 , a D q u ( 1 ) + b D q 2 u ( 1 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equa_HTML.gif

where a , b 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq1_HTML.gif and D q α C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq2_HTML.gif is the fractional q-derivatives of the Caputo type.

In [27], 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equb_HTML.gif

where λ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq3_HTML.gif is a parameter, and the uniqueness, existence and nonexistence of positive solutions are considered in terms of different ranges of λ.

By applying the Banach contraction principle, Krasnoselskii’s fixed-point theorem, and the Leray-Schauder nonlinear alternative, Ahmad, Ntouyas and Purnaras [29] studied the existence of solution for 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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equc_HTML.gif

where D q α C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq2_HTML.gif is the fractional q-derivative of the Caputo type, and a i , b i , c i , η i R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq4_HTML.gif.

Recently, in [32], the authors investigate the following singular semipositone integral boundary value problem for fractional q-derivatives equation:
{ ( D q α u ) ( t ) + f ( t , u ( t ) ) = 0 , t ( 0 , 1 ) , 2 < α 3 , u ( 0 ) = ( D q u ) ( 0 ) = 0 , u ( 1 ) = μ 0 1 u ( s ) d q s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equd_HTML.gif

where 0 < μ < [ α ] q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq5_HTML.gif, D q α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq6_HTML.gif is the q-derivative of Riemann-Liouville type of order α, f : [ 0 , 1 ] × ( 0 , + ) ( , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq7_HTML.gif is continuous and semipositone, and may be singular at u = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq8_HTML.gif.

Since finding positive solutions of boundary value problems is interest in various fields of sciences, fractional q-calculus equations has tremendous potential for applications. In this paper, we will deal with the following nonhomogeneous boundary value problem with fractional q-derivatives:
{ ( D q α u ) ( t ) + f ( t , u ( t ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = ( D q u ) ( 0 ) = 0 , γ ( D q u ) ( 1 ) + β ( D q 2 u ) ( 1 ) = λ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ1_HTML.gif
(1.1)

where q ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq9_HTML.gif, 2 < α 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq10_HTML.gif, γ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq11_HTML.gif, β > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq12_HTML.gif, and λ is a parameter, D q α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq6_HTML.gif is the q-derivative of Riemann-Liouville type of order α, f : [ 0 , 1 ] × R R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq13_HTML.gif is continuous. 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 and Krasnoselskii’s fixed-point theorem, the uniqueness, existence, and multiplicity of positive solution to the BVP (1.1) are obtained in term of the explicit intervals for the nonhomogeneous term. Our results are different from those of [25, 27].

2 Preliminaries on q-calculus and lemmas

For the convenience of the reader, below we cite some definitions and fundamental results on q-calculus as well as the fractional q-calculus. The presentation here can be found in, for example, [12, 18, 20, 22].

Let q ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq9_HTML.gif and define
[ a ] q = 1 q a 1 q , a R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Eque_HTML.gif
The q-analogue of the power function ( a b ) n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq14_HTML.gif with n N 0 : = { 0 , 1 , 2 , } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq15_HTML.gif is
( a b ) ( 0 ) = 1 , ( a b ) ( n ) = k = 0 n 1 ( a b q k ) , n N , a , b R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equf_HTML.gif
More generally, if α R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq16_HTML.gif, then
( a b ) ( α ) = a α k = 0 a b q k a b q α + k , a 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ2_HTML.gif
(2.1)
Clearly, if b = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq17_HTML.gif, then a ( α ) = a α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq18_HTML.gif. The q-gamma function is defined by
Γ q ( x ) = ( 1 q ) ( x 1 ) ( 1 q ) x 1 , x R { 0 , 1 , 2 , } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equg_HTML.gif

and satisfies Γ q ( x + 1 ) = [ x ] q Γ q ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq19_HTML.gif.

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 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equh_HTML.gif
and 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 ) , n N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equi_HTML.gif
The q-integral of a function f defined in the interval [ 0 , b ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq20_HTML.gif is given by
( I q f ) ( x ) = 0 x f ( s ) d q s = x ( 1 q ) k = 0 f ( x q k ) q k , x [ 0 , b ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equj_HTML.gif
If a [ 0 , b ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq21_HTML.gif and f is defined in the interval [ 0 , b ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq20_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equk_HTML.gif
Similar to that for derivatives, an operator I q n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq22_HTML.gif is given by
( I q 0 f ) ( x ) = f ( x ) , ( I q n f ) ( x ) = I q ( I q n 1 f ) ( x ) , n N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equl_HTML.gif
The fundamental theorem of calculus applies to these operators I q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq23_HTML.gif and D q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq24_HTML.gif, i.e.,
( D q I q f ) ( x ) = f ( x ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equm_HTML.gif
and if f is continuous at x = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq25_HTML.gif, then
( I q D q f ) ( x ) = f ( x ) f ( 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ3_HTML.gif
(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 ( q s ) d q s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equn_HTML.gif
and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ4_HTML.gif
(2.3)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ5_HTML.gif
(2.4)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ6_HTML.gif
(2.5)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ7_HTML.gif
(2.6)

where D q t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq26_HTML.gif denotes the derivative with respect to the variable t.

Definition 2.1 Let α 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq27_HTML.gif and f be a function defined on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq28_HTML.gif. The fractional q-integral of Riemann-Liouville type is ( I q 0 f ) ( x ) = f ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq29_HTML.gif 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equo_HTML.gif
Definition 2.2 The fractional q-derivative of the Riemann-Liouville type of order α 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq27_HTML.gif is defined by ( D q 0 f ) ( x ) = f ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq30_HTML.gif and
( D q α f ) ( x ) = ( D q [ α ] I q [ α ] α f ) ( x ) , α > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equp_HTML.gif

where [ α ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq31_HTML.gif is the smallest integer greater than or equal to α.

Lemma 2.3 ([20])

Assume that α 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq27_HTML.gif and a b t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq32_HTML.gif, then ( t a ) ( α ) ( t b ) ( α ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq33_HTML.gif.

Lemma 2.4 Let α , β 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq34_HTML.gif and f be a function defined on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq35_HTML.gif. Then the following formulas hold:
  1. (1)

    ( I q β I q α f ) ( x ) = ( I q α + β f ) ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq36_HTML.gif,

     
  2. (2)

    ( D q α I q α f ) ( x ) = f ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq37_HTML.gif.

     

Lemma 2.5 ([20])

Let α > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq38_HTML.gif 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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equq_HTML.gif

Lemma 2.6 ([22])

Let α R + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq39_HTML.gif, λ ( 1 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq40_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equr_HTML.gif
Particularly, for λ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq41_HTML.gif, a = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq42_HTML.gif, 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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equs_HTML.gif
Obviously, we have 0 t ( t q s ) ( α 1 ) d q s = 1 [ α ] q t ( α ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq43_HTML.gif, and
0 t ( 1 q s ) ( α 1 ) d q s = 0 t D q s ( ( 1 s ) ( α ) ) [ α ] q d q s = 1 [ α ] q 0 t s D q ( ( 1 s ) ( α ) ) d q s = 1 [ α ] q [ 1 ( 1 t ) ( α ) ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equt_HTML.gif

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

Lemma 2.7 For given y C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq44_HTML.gif, the unique solution of the boundary value problem
( D q α u ) ( t ) + y ( t ) = 0 , t ( 0 , 1 ) , 2 < α 3 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ8_HTML.gif
(2.7)
subject to the boundary conditions
u ( 0 ) = ( D q u ) ( 0 ) = 0 , γ ( D q u ) ( 1 ) + β ( D q 2 u ) ( 1 ) = λ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ9_HTML.gif
(2.8)
is given by
u ( t ) = 0 1 G ( t , q s ) y ( s ) d q s + λ t α 1 ( γ + [ α 2 ] q β ) [ α 1 ] q , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ10_HTML.gif
(2.9)
where
G ( t , s ) = { γ t α 1 ( 1 s ) ( α 2 ) + [ α 2 ] q β t α 1 ( 1 s ) ( α 3 ) ( γ + [ α 2 ] q β ) Γ q ( α ) ( t s ) ( α 1 ) Γ q ( α ) , 0 s t 1 , γ t α 1 ( 1 s ) ( α 2 ) + [ α 2 ] q β t α 1 ( 1 s ) ( α 3 ) ( γ + [ α 2 ] q β ) Γ q ( α ) , 0 t s 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ11_HTML.gif
(2.10)
Proof Since 2 < α 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq10_HTML.gif, we put n = 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq45_HTML.gif. In view of Definition 2.1 and Lemma 2.4, we see that
( D q α u ) ( t ) = y ( t ) ( I q α D q 3 I q 3 α u ) ( t ) = ( I q α y ) ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equu_HTML.gif
Then it follows from Lemma 2.5 that the solution u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq46_HTML.gif of (2.7) and (2.8) is given by
u ( t ) = c 1 t α 1 + c 2 t α 2 + c 3 t α 3 0 t ( t q s ) ( α 1 ) Γ q ( α ) y ( s ) d q s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ12_HTML.gif
(2.11)

for some constants c 1 , c 2 , c 3 R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq47_HTML.gif. From u ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq48_HTML.gif, we have c 3 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq49_HTML.gif.

Differentiating both sides of (2.11) and with the help of (2.4) and (2.6), we obtain,
( D q u ) ( t ) = [ α 1 ] q c 1 t α 2 + [ α 2 ] q c 2 t α 3 0 t [ α 1 ] q ( t q s ) ( α 2 ) Γ q ( α ) y ( s ) d q s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equv_HTML.gif
and
( D q 2 u ) ( t ) = [ α 1 ] q [ α 2 ] q c 1 t α 3 + [ α 2 ] q [ α 3 ] q c 2 t α 4 0 t [ α 1 ] q [ α 2 ] q ( t q s ) ( α 3 ) Γ q ( α ) y ( s ) d q s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equw_HTML.gif
Then by the boundary condition ( D q u ) ( 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq50_HTML.gif, we get c 2 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq51_HTML.gif. Using the boundary condition γ ( D q u ) ( 1 ) + β ( D q 2 u ) ( 1 ) = λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq52_HTML.gif, we get
c 1 = 1 γ + [ α 2 ] q β × ( γ 0 1 ( 1 q s ) ( α 2 ) Γ q ( α ) y ( s ) d q s + β 0 1 [ α 2 ] q ( 1 q s ) ( α 3 ) Γ q ( α ) y ( s ) d q s + λ [ α 1 ] q ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equx_HTML.gif
Hence, we have
u ( t ) = t α 1 γ + [ α 2 ] q β ( γ 0 1 ( 1 q s ) ( α 2 ) Γ q ( α ) y ( s ) d q s + β 0 1 [ α 2 ] q ( 1 q s ) ( α 3 ) Γ q ( α ) y ( s ) d q s + λ [ α 1 ] q ) 0 t ( t q s ) ( α 1 ) Γ q ( α ) y ( s ) d q s = 0 1 G ( t , q s ) y ( s ) d q s + λ t α 1 ( γ + [ α 2 ] q β ) [ α 1 ] q . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equy_HTML.gif

This completes the proof of the lemma. □

Lemma 2.8 The function G ( t , s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq53_HTML.gif defined by (2.10) satisfies the following conditions:
  1. (i)

    G ( t , q s ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq54_HTML.gif, and G ( t , q s ) G ( 1 , q s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq55_HTML.gif for all 0 t , s 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq56_HTML.gif.

     
  2. (ii)

    G ( t , q s ) t α 1 G ( 1 , q s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq57_HTML.gif for all 0 t , s 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq56_HTML.gif.

     
Proof We start by defining the following two functions:
g 1 ( t , s ) = γ t α 1 ( 1 s ) ( α 2 ) + [ α 2 ] q β t α 1 ( 1 s ) ( α 3 ) ( γ + [ α 2 ] q β ) Γ q ( α ) , 0 t s 1 , g 2 ( t , s ) = g 1 ( t , s ) ( t s ) ( α 1 ) Γ q ( α ) , 0 s t 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equz_HTML.gif
Obviously, g 1 ( t , q s ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq58_HTML.gif. Now g 2 ( 0 , q s ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq59_HTML.gif, and for t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq60_HTML.gif
g 2 ( t , q s ) = γ t α 1 ( 1 q s ) ( α 2 ) + [ α 2 ] q β t α 1 ( 1 q s ) ( α 3 ) ( γ + [ α 2 ] q β ) Γ q ( α ) ( t q s ) ( α 1 ) Γ q ( α ) = t α 1 ( γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ( γ + [ α 2 ] q β ) ( 1 q s t ) ( α 1 ) ) ( γ + [ α 2 ] q β ) Γ q ( α ) t α 1 ( γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ( γ + [ α 2 ] q β ) ( 1 q s ) ( α 1 ) ) ( γ + [ α 2 ] q β ) Γ q ( α ) = t α 1 ( γ ( ( 1 q s ) ( α 2 ) ( 1 q s ) ( α 1 ) ) + [ α 2 ] q β ( ( 1 q s ) ( α 3 ) ( 1 q s ) ( α 1 ) ) ) ( γ + [ α 2 ] q β ) Γ q ( α ) 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equaa_HTML.gif

Therefore, G ( t , q s ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq54_HTML.gif.

Moreover, for s ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq61_HTML.gif, it follows from (2.4) and Lemma 2.3 that
D q t g 2 ( t , q s ) = [ α 1 ] q ( γ + [ α 2 ] q β ) Γ q ( α ) ( γ t α 2 ( 1 q s ) ( α 2 ) + [ α 2 ] q β t α 2 ( 1 q s ) ( α 3 ) ( γ + [ α 2 ] q β ) ( t q s ) ( α 2 ) ) [ α 1 ] q t α 2 ( γ + [ α 2 ] q β ) Γ q ( α ) ( γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ( γ + [ α 2 ] q β ) ( 1 q s ) ( α 2 ) ) = [ α 1 ] q [ α 2 ] q β t α 2 ( γ + [ α 2 ] q β ) Γ q ( α ) ( ( 1 q s ) ( α 3 ) ( 1 q s ) ( α 2 ) ) 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equab_HTML.gif

which implies that g 2 ( t , q s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq62_HTML.gif is an increasing function with respect to t. It is clear that g 1 ( t , q s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq63_HTML.gif is increasing in t. Therefore, G ( t , q s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq64_HTML.gif is an increasing function of t for all s ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq61_HTML.gif, and so G ( t , q s ) G ( 1 , q s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq55_HTML.gif.

When 0 t q s 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq65_HTML.gif, then
G ( t , q s ) = γ t α 1 ( 1 q s ) ( α 2 ) + [ α 2 ] q β t α 1 ( 1 q s ) ( α 3 ) ( γ + [ α 2 ] q β ) Γ q ( α ) G ( q s , q s ) G ( 1 , q s ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equac_HTML.gif
Finally, we prove part (ii). When 0 q s t 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq66_HTML.gif, we have
G ( t , q s ) G ( 1 , q s ) = γ t α 1 ( 1 q s ) ( α 2 ) + [ α 2 ] q β t α 1 ( 1 q s ) ( α 3 ) ( γ + [ α 2 ] q β ) ( t q s ) ( α 1 ) γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ( γ + [ α 2 ] q β ) ( 1 q s ) ( α 1 ) γ t α 1 ( 1 q s ) ( α 2 ) + [ α 2 ] q β t α 1 ( 1 q s ) ( α 3 ) ( γ + [ α 2 ] q β ) t α 1 ( 1 q s ) ( α 1 ) γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ( γ + [ α 2 ] q β ) ( 1 q s ) ( α 1 ) = t α 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equad_HTML.gif
If 0 t q s 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq65_HTML.gif, then we have
G ( t , q s ) G ( 1 , q s ) = γ t α 1 ( 1 q s ) ( α 2 ) + [ α 2 ] q β t α 1 ( 1 q s ) ( α 3 ) γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ( γ + [ α 2 ] q β ) ( 1 q s ) ( α 1 ) t α 1 [ γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ( γ + [ α 2 ] q β ) ( 1 q s ) ( α 1 ) ] γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ( γ + [ α 2 ] q β ) ( 1 q s ) ( α 1 ) = t α 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equae_HTML.gif

which implies that part (ii) holds. This completes the proof of the lemma. □

Remark 2.9 If we let 0 < τ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq67_HTML.gif, then
min t [ τ , 1 ] G ( t , q s ) τ α 1 G ( 1 , q s ) , for  s [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equaf_HTML.gif

According to [20], we may take τ = q n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq68_HTML.gif, n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq69_HTML.gif.

3 The main results

Let X = C ( [ 0 , 1 ] ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq70_HTML.gif be a Banach space endowed with the norm u X = max 0 t 1 | u ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq71_HTML.gif. Define the cone P X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq72_HTML.gif by P = { u X : u ( t ) 0 , 0 t 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq73_HTML.gif.

Define the operator T : P X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq74_HTML.gif as follows:
( T u ) ( t ) = 0 1 G ( t , q s ) f ( s , u ( s ) ) d q s + λ t α 1 ( γ + [ α 2 ] q β ) [ α 1 ] q . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ13_HTML.gif
(3.1)
Theorem 3.1 Assume that f : [ 0 , 1 ] × [ 0 , + ) [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq75_HTML.gif is continuous and there exists a nonnegative function h C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq76_HTML.gif such that
| f ( t , u ) f ( t , v ) | h ( t ) | u v | , t [ 0 , 1 ] , u , v [ 0 , + ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ14_HTML.gif
(3.2)
Then the BVP (1.1) has a unique positive solution for any λ ( 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq77_HTML.gif, provided
0 1 s α 1 ( γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ) h ( s ) d q s < ( γ + [ α 2 ] q β ) Γ q ( α ) 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ15_HTML.gif
(3.3)

If, in addition, f ( t , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq78_HTML.gif on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq35_HTML.gif, then the conclusion is true for λ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq79_HTML.gif.

Proof We will show that under the assumptions (3.2) and (3.3), T m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq80_HTML.gif is a contraction operator for m sufficiently large.

By the definition of G ( t , q s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq64_HTML.gif, for u , v P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq81_HTML.gif, we have
| ( 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 t α 1 ( γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ) ( γ + [ α 2 ] q β ) Γ q ( α ) h ( s ) d q s u v X = t α 1 u v X ( γ + [ α 2 ] q β ) Γ q ( α ) 0 1 ( γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ) h ( s ) d q s = Λ 1 t α 1 ( γ + [ α 2 ] q β ) Γ q ( α ) u v X , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equag_HTML.gif

where Λ 1 = 0 1 ( γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ) h ( s ) d q s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq82_HTML.gif.

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 u v X ( γ + [ α 2 ] q β ) Γ q ( α ) 0 1 G ( t , q s ) s α 1 h ( s ) d q s Λ 1 t α 1 u v X ( γ + [ α 2 ] q β ) 2 [ Γ q ( α ) ] 2 0 1 s α 1 ( γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ) h ( s ) d q s = Λ 1 Λ 2 t α 1 ( γ + [ α 2 ] q β ) 2 [ Γ q ( α ) ] 2 u v X , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equah_HTML.gif

where Λ 2 = 0 1 s α 1 ( γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ) h ( s ) d q s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq83_HTML.gif.

By introduction, we get
| ( T m u ) ( t ) ( T m v ) ( t ) | Λ 1 Λ 2 m 1 t α 1 ( γ + [ α 2 ] q β ) m [ Γ q ( α ) ] m u v X . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equai_HTML.gif
From the condition (3.3), we have
Λ 1 Λ 2 m 1 ( γ + [ α 2 ] q β ) m [ Γ q ( α ) ] m = Λ 1 Λ 2 [ Λ 2 ( γ + [ α 2 ] q β ) Γ q ( α ) ] m Λ 1 Λ 2 ( 1 2 ) m < 1 4 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equaj_HTML.gif
for m sufficiently large. So, we get
( T m u ) ( t ) ( T m v ) ( t ) X < 1 4 u v X . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equak_HTML.gif

Hence, it follows from the generalized Banach contraction principle that the BVP (1.1) has a unique positive solution for any λ ( 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq77_HTML.gif. If λ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq79_HTML.gif, then the condition f ( t , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq84_HTML.gif on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq35_HTML.gif and Lemma 2.8 imply that u ( t ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq85_HTML.gif in ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq86_HTML.gif. This completes the proof of the theorem. □

Remark 3.2 When h ( t ) h 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq87_HTML.gif is a constant, the condition (3.2) reduces to a Lipschitz condition.

Our next existence results is based on Krasnoselskii’s fixed-point theorem [33].

Lemma 3.3 (Krasnoselskii’s)

Let E be a Banach space, and let P E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq88_HTML.gif be a cone. Assume Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq89_HTML.gif, Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq90_HTML.gif are open subsets of E with θ Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq91_HTML.gif, Ω ¯ 1 Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq92_HTML.gif and let T : P ( Ω ¯ 2 Ω 1 ) P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq93_HTML.gif be a completely continuous operator such that, either
  1. (1)

    T u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq94_HTML.gif, u P Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq95_HTML.gif and T u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq96_HTML.gif, u P Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq97_HTML.gif, or

     
  2. (2)

    T u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq96_HTML.gif, u P Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq95_HTML.gif and T u u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq94_HTML.gif, u P Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq97_HTML.gif.

     

Then T has at least one fixed point in P ( Ω ¯ 2 Ω 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq98_HTML.gif.

Define a cone K X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq99_HTML.gif by
K = { u X : u ( t ) 0 , u ( t ) t α 1 u , t [ 0 , 1 ] } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equal_HTML.gif

Obviously, K is a cone of nonnegative functions in X.

Lemma 3.4 The operator T : K K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq100_HTML.gif is completely continuous.

Proof Firstly, we prove that T ( K ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq101_HTML.gif. By (2.9) and Lemma 2.8, we have
T u = max 0 t 1 { 0 1 G ( t , q s ) f ( s , u ( s ) ) d q s + λ t α 1 ( γ + [ α 2 ] q β ) [ α 1 ] q } 0 1 G ( 1 , q s ) f ( s , u ( s ) ) d q s + λ ( γ + [ α 2 ] q β ) [ α 1 ] q . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equam_HTML.gif
On the other hand,
( T u ) ( t ) t α 1 ( 0 1 G ( 1 , q s ) f ( s , u ( s ) ) d q s + λ ( γ + [ α 2 ] q β ) [ α 1 ] q ) t α 1 T u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equan_HTML.gif

Hence, we have T ( K ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq101_HTML.gif.

Next, we show that T is uniformly bounded. For fixed r > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq102_HTML.gif, consider a bounded subset K r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq103_HTML.gif of K defined by K r = { u K : u r , r > 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq104_HTML.gif, and let M = max 0 u r | f ( t , u ) | + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq105_HTML.gif. Then for u K r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq106_HTML.gif, we get
| ( T u ) ( t ) | 0 1 G ( t , q s ) | f ( s , u ( s ) ) | d q s + | λ | t α 1 ( γ + [ α 2 ] q β ) [ α 1 ] q M 0 1 G ( 1 , q s ) d q s + | λ | ( γ + [ α 2 ] q β ) [ α 1 ] q < + , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equao_HTML.gif

which implies that T ( K r ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq107_HTML.gif is bounded.

Finally, we show that T is equicontinuous. For all ε > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq108_HTML.gif, setting
δ = min { ε ω ( α 1 ) , 1 2 ( ε ω ) 1 α 1 } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equap_HTML.gif
where
ω = M ( γ + [ α 2 ] q β ) + | λ | Γ q ( α ) ( γ + [ α 2 ] q β ) Γ q ( α ) [ α 1 ] q . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equaq_HTML.gif
For any u K r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq106_HTML.gif, we can prove that if t 1 , t 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq109_HTML.gif and 0 < t 2 t 1 < δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq110_HTML.gif, then
| ( T u ) ( t 2 ) ( T u ) ( t 2 ) | < ε . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equar_HTML.gif
In fact, we have
| ( T u ) ( t 2 ) ( T u ) ( t 1 ) | 0 1 | G ( t 2 , q s ) G ( t 1 , q s ) | f ( s , u ( s ) ) d q s + | λ | ( t 2 α 1 t 1 α 1 ) ( γ + [ α 2 ] q β ) [ α 1 ] q M 0 1 | G ( t 2 , q s ) G ( t 1 , q s ) | d q s + | λ | ( t 2 α 1 t 1 α 1 ) ( γ + [ α 2 ] q β ) [ α 1 ] q M ( 0 t 1 | G ( t 2 , q s ) G ( t 1 , q s ) | d q s + t 1 t 2 | G ( t 2 , q s ) G ( t 1 , q s ) | d q s + t 2 1 | G ( t 2 , q s ) G ( t 1 , q s ) | d q s ) + | λ | ( t 2 α 1 t 1 α 1 ) ( γ + [ α 2 ] q β ) [ α 1 ] q M γ + [ α 2 ] q β ( 0 t 1 γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) Γ q ( α ) ( t 2 α 1 t 1 α 1 ) d q s + t 1 t 2 γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) Γ q ( α ) ( t 2 α 1 t 1 α 1 ) d q s + t 2 1 γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) Γ q ( α ) ( t 2 α 1 t 1 α 1 ) d q s ) + | λ | ( t 2 α 1 t 1 α 1 ) ( γ + [ α 2 ] q β ) [ α 1 ] q = M ( t 2 α 1 t 1 α 1 ) ( γ + [ α 2 ] q β ) Γ q ( α ) 0 1 ( γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ) d q s + | λ | ( γ + [ α 2 ] q β ) [ α 1 ] q ( t 2 α 1 t 1 α 1 ) = M ( γ + [ α 2 ] q β ) + | λ | Γ q ( α ) ( γ + [ α 2 ] q β ) Γ q ( α ) [ α 1 ] q ( t 2 α 1 t 1 α 1 ) = ω ( t 2 α 1 t 1 α 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equas_HTML.gif
If δ t 1 < t 2 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq111_HTML.gif, then
| ( T u ) ( t 2 ) ( T u ) ( t 1 ) | ω ( t 2 α 1 t 1 α 1 ) < ω ( α 1 ) ( t 2 t 1 ) < ω ( α 1 ) δ ε . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equat_HTML.gif
If 0 t 1 < δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq112_HTML.gif, t 2 < 2 δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq113_HTML.gif, then
| ( T u ) ( t 2 ) ( T u ) ( t 1 ) | ω ( t 2 α 1 t 1 α 1 ) < ω t 2 α 1 < ω ( 2 δ ) α 1 ε . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equau_HTML.gif

By means of Arzela-Ascoli theorem, T : K K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq100_HTML.gif is completely continuous.

For the sake of convenience, we introduce the following weight functions:
ϕ ( r ) = max { f ( t , u ( t ) ) : ( t , u ) [ 0 , 1 ] × [ 0 , r ] } , φ ( r ) = min { f ( t , u ( t ) ) : ( t , u ) [ τ , 1 ] × [ τ α 1 r , r ] } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equav_HTML.gif
and set
l = ( 0 1 G ( 1 , q s ) d q s ) 1 , L = ( τ α 1 τ 1 G ( 1 , q s ) d q s ) 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equaw_HTML.gif

 □

Theorem 3.5 Suppose that there exists two positive numbers ξ 1 < ξ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq114_HTML.gif such that one of the following conditions is satisfied

( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq115_HTML.gif) φ ( ξ 1 ) ξ 1 L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq116_HTML.gif, ϕ ( ξ 2 ) l 2 ξ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq117_HTML.gif;

( H 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq118_HTML.gif) ϕ ( ξ 1 ) l 2 ξ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq119_HTML.gif, φ ( ξ 2 ) ξ 2 L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq120_HTML.gif.

Then the BVP (1.1) has at least one positive solution u K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq121_HTML.gif, such that ξ 1 u ξ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq122_HTML.gif for λ ( 0 , ( γ + [ α 2 ] q β ) [ α 1 ] q 2 ξ 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq123_HTML.gif. If, in addition, f ( t , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq78_HTML.gif on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq35_HTML.gif, then the conclusion is true for λ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq79_HTML.gif.

Proof Because the proofs are similar, we prove only the case ( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq115_HTML.gif). Denote Ω ξ 1 = { u X : u < ξ 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq124_HTML.gif. Then for any u K Ω ξ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq125_HTML.gif, we get u = ξ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq126_HTML.gif, 0 u ( t ) ξ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq127_HTML.gif, 0 t 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq128_HTML.gif, and τ α 1 ξ 1 min τ t 1 t α 1 u u ( t ) u = ξ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq129_HTML.gif, τ t 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq130_HTML.gif. By assumption ( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq115_HTML.gif), we have
f ( t , u ) φ ( ξ 1 ) ξ 1 L , τ t 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equax_HTML.gif
In view of (2.9) and Lemma 2.8, we have
T u min τ t 1 { 0 1 G ( t , q s ) f ( s , u ( s ) ) d q s + λ t α 1 ( γ + [ α 2 ] q β ) [ α 1 ] q } min τ t 1 0 1 G ( t , q s ) f ( s , u ( s ) ) d q s τ α 1 τ 1 G ( 1 , q s ) f ( s , u ( s ) ) d q s τ α 1 τ 1 G ( 1 , q s ) d q s ξ 1 L = ξ 1 = u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equay_HTML.gif
On the other hand, define Ω ξ 2 = { u X : u < ξ 2 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq131_HTML.gif. For any t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq132_HTML.gif and u K Ω ξ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq133_HTML.gif, we have u = ξ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq134_HTML.gif and 0 u ( t ) ξ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq135_HTML.gif, 0 t 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq128_HTML.gif. Thus,
f ( t , u ) ϕ ( ξ 2 ) l 2 ξ 2 , for  0 t 1 , 0 u ξ 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equaz_HTML.gif
It follows
T u = max 0 t 1 { 0 1 G ( t , q s ) f ( s , u ( s ) ) d q s + λ t α 1 ( γ + [ α 2 ] q β ) [ α 1 ] q } 0 1 G ( 1 , q s ) d q s l 2 ξ 2 + λ ( γ + [ α 2 ] q β ) [ α 1 ] q l 1 l 2 ξ 2 + ξ 1 2 < ξ 2 2 + ξ 2 2 = u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equba_HTML.gif

By Lemma 3.3, the operator T has at least one fixed point u K ( Ω ¯ ξ 2 Ω ξ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq136_HTML.gif, and ξ 1 u ξ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq122_HTML.gif. Since u ( t ) t α 1 u ξ 1 t α 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq137_HTML.gif, 0 < t < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq138_HTML.gif, then, the solution u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq139_HTML.gif is positive for λ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq140_HTML.gif. As in the proof of Theorem 3.1, u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq141_HTML.gif is a positive solution for λ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq79_HTML.gif. This completes the proof of the theorem. □

Theorem 3.6 Suppose that there exists three positive numbers ξ 1 < ξ 2 < ξ 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq142_HTML.gif such that one of the following conditions is satisfied

( H 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq143_HTML.gif) φ ( ξ 1 ) ξ 1 L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq116_HTML.gif, ϕ ( ξ 2 ) < l 2 ξ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq144_HTML.gif, φ ( ξ 3 ) ξ 3 L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq145_HTML.gif;

( H 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq146_HTML.gif) ϕ ( ξ 1 ) l 2 ξ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq119_HTML.gif, φ ( ξ 2 ) > ξ 2 L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq147_HTML.gif, ϕ ( ξ 3 ) l 2 ξ 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq148_HTML.gif.

Then the BVP (1.1) has at least two positive solutions u 1 , u 2 K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq149_HTML.gif such that ξ 1 u 1 < ξ 2 u 2 ξ 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq150_HTML.gif for λ ( 0 , ( γ + [ α 2 ] q β ) [ α 1 ] q 2 ξ 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq151_HTML.gif. If, in addition, f ( t , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq78_HTML.gif on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq35_HTML.gif, then the conclusion is true for λ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq79_HTML.gif.

Proof We prove only the case ( H 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq146_HTML.gif). Since φ : [ 0 , + ) [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq152_HTML.gif is continuous and φ ( ξ 2 ) > ξ 2 L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq153_HTML.gif, there exist two positive numbers η 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq154_HTML.gif, η 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq155_HTML.gif such that ξ 1 < η 1 < ξ 2 < η 2 < ξ 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq156_HTML.gif and φ ( η 1 ) η 1 L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq157_HTML.gif, φ ( η 2 ) η 2 L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq158_HTML.gif. Thus, it follows from the assumption ( H 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq146_HTML.gif) that
ϕ ( ξ 1 ) l 2 ξ 1 , φ ( η 1 ) η 1 L , and φ ( η 2 ) η 2 L , ϕ ( ξ 3 ) l 2 ξ 3 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equbb_HTML.gif

From Theorem 3.5, the operator T has two fixed point u 1 K ( Ω ¯ η 1 Ω ξ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq159_HTML.gif, u 2 K ( Ω ¯ ξ 3 Ω η 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq160_HTML.gif with ξ 1 u 1 < ξ 2 < u 2 ξ 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq161_HTML.gif. Therefore, the BVP (1.1) has at least two positive solutions for λ ( 0 , ( γ + [ α 2 ] q β ) [ α 1 ] q 2 ξ 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq162_HTML.gif. As in the proof of Theorem 3.1, u 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq163_HTML.gif, u 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq164_HTML.gif are two positive solutions for λ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq79_HTML.gif. This completes the proof of the theorem. □

Denote the integer part of m by [ m ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq165_HTML.gif. Generally, we have the following theorem.

Theorem 3.7 Suppose that there exists m + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq166_HTML.gif positive numbers ξ 1 < ξ 2 < < ξ m + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq167_HTML.gif such that one of the following conditions is satisfied:

( H 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq168_HTML.gif) φ ( ξ 2 j 1 ) > ξ 2 j 1 L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq169_HTML.gif, ϕ ( ξ 2 j ) < l 2 ξ 2 j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq170_HTML.gif, j = 1 , 2 , , [ m + 2 2 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq171_HTML.gif;

( H 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq172_HTML.gif) ϕ ( ξ 2 j 1 ) < l 2 ξ 2 j 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq173_HTML.gif, φ ( ξ 2 j ) > ξ 2 j L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq174_HTML.gif, j = 1 , 2 , , [ m + 2 2 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq171_HTML.gif.

Then the BVP (1.1) has at least m positive solutions u i K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq175_HTML.gif, i = 1 , 2 , , m https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq176_HTML.gif, such that ξ i < u i < ξ i + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq177_HTML.gif for λ ( 0 , ( γ + [ α 2 ] q β ) [ α 1 ] q 2 ξ 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq123_HTML.gif. If, in addition, f ( t , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq78_HTML.gif on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq35_HTML.gif, then the conclusion is true for λ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq79_HTML.gif.

4 Examples

Example 4.1 The fractional q-difference boundary value problem
{ D 0.5 2.5 u ( t ) + 2 e t 5 ( 1 + e t ) ( tan 1 u + t 2 + t sin 2 t + 1 ) = 0 , 0 < t < 1 , u ( 0 ) = ( D 0.5 u ) ( 0 ) = 0 , 0.25 ( D 0.5 u ) ( 1 ) + 0.75 ( D 0.5 2 u ) ( 1 ) = λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ16_HTML.gif
(4.1)

has a unique positive solution for any λ ( 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq178_HTML.gif.

Proof In this case, α = 2.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq179_HTML.gif, q = 0.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq180_HTML.gif, γ = 0.25 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq181_HTML.gif, β = 0.75 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq182_HTML.gif, λ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq140_HTML.gif. Let
f ( t , u ) = 2 e t 5 ( 1 + e t ) ( tan 1 u + t 2 + t sin 2 t + 1 ) , ( t , u ) [ 0 , 1 ] × ( 0 , + ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equbc_HTML.gif
and h ( t ) = 2 e t 5 ( 1 + e t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq183_HTML.gif. It is easy to prove that
| f ( t , u ) f ( t , v ) | h ( t ) | tan 1 u tan 1 v | h ( t ) | u v | , for  ( t , u ) , ( t , v ) [ 0 , 1 ] × [ 0 , + ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equbd_HTML.gif
A simple computation showed
( γ + [ α 2 ] q β ) Γ q ( α ) = ( 0.25 + [ 0.5 ] 0.5 × 0.75 ) Γ 0.5 ( 2.5 ) 0.9168 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Eqube_HTML.gif
and
Λ 2 = 0 1 s α 1 ( γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ) 2 e s 5 ( 1 + e s ) d q s 2 5 0 1 s α 1 ( γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ) d q s 2 5 0 1 ( γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ) d q s 0.3774 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equbf_HTML.gif
which implies that
0 1 s α 1 ( γ ( 1 q s ) ( α 2 ) + [ α 2 ] q β ( 1 q s ) ( α 3 ) ) h ( s ) d q s < ( γ + [ α 2 ] q β ) Γ q ( α ) 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equbg_HTML.gif
Obviously, for any m 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq184_HTML.gif, we have
Λ 1 Λ 2 m 1 ( γ + [ α 2 ] q β ) m [ Γ q ( α ) ] m 0.3774 0.9168 × 2 m 1 < 0.2058 < 1 4 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equbh_HTML.gif

Thus, Theorem 3.1 implies that the boundary value problem (4.1) has a unique positive solution for any λ ( 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq178_HTML.gif. □

Example 4.2 Consider the following fractional boundary value problem:
{ D 0.5 2.5 u ( t ) + u 2 ( 1 2 cos ( π t π 2 ) + 1 4 ) = 0 , 0 < t < 1 , u ( 0 ) = ( D 0.5 u ) ( 0 ) = 0 , 2 ( D 0.5 u ) ( 1 ) + 7 ( D 0.5 2 u ) ( 1 ) = λ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equ17_HTML.gif
(4.2)

where α = 2.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq179_HTML.gif, q = 0.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq180_HTML.gif, γ = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq185_HTML.gif, β = 7 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq186_HTML.gif. Choosing n = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq187_HTML.gif, then τ = q n = 0.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq188_HTML.gif.

By calculation, we get ( γ + [ α 2 ] q β ) Γ q ( α ) 8.1132 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq189_HTML.gif. By Lemma 2.6, Lemma 2.8 and with the aid of a computer, we obtain that
l = ( 0 1 G ( 1 , 0.5 s ) d 0.5 s ) 1 1.6756 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equbi_HTML.gif
and
L = ( ( 0.5 ) α 1 0.5 1 G ( 1 , 0.5 s ) d 0.5 s ) 1 0.0657 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_Equbj_HTML.gif
Let f ( t , u ) = u 2 ( 1 2 cos ( π t π 2 ) + 1 4 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq190_HTML.gif. Take ξ 1 = 3 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq191_HTML.gif, ξ 2 = 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq192_HTML.gif, then ( γ + [ α 2 ] q β ) [ α 1 ] q 2 ξ 1 2.9578 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq193_HTML.gif, and f ( t , u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq194_HTML.gif satisfies
  1. (i)

    ϕ ( 3 4 ) = max { u 2 ( 1 2 cos ( π t π 2 ) + 1 4 ) : ( t , u ) [ 0 , 1 ] × [ 0 , 3 4 ] } = 27 64 < l 2 ξ 1 0.6283 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq195_HTML.gif;

     
  2. (ii)

    φ ( 3 ) = min { u 2 ( 1 2 cos ( π t π 2 ) + 1 4 ) : ( t , u ) [ 0.5 , 1 ] × [ 1.0608 , 3 ] } 0.2652 > L ξ 2 0.1971 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq196_HTML.gif.

     

So, by Theorem 3.5, the problem (4.2) has one positive solution u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq139_HTML.gif such that 1 u X 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq197_HTML.gif for λ ( 0 , 2.9578 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-103/MediaObjects/13661_2013_Article_357_IEq198_HTML.gif.

Declarations

Acknowledgements

Dedicated to Professor Hari M Srivastava.

The authors are highly grateful for the referees’ careful reading and comments on this paper. The research is supported by the National Natural Science Foundation of China (Grant No. 11271372, 11201138); it is also supported by the Hunan Provincial Natural Science Foundation of China (Grant No. 13JJ3106, 12JJ2004), and the Scientific Research Fund of Hunan Provincial Education Department (Grant No. 12B034).

Authors’ Affiliations

(1)
School of Science, Hunan University of Technology
(2)
Department of Mathematics, Central South University

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