Open Access

Positive solutions of fractional differential equations at resonance on the half-line

Boundary Value Problems20122012:64

DOI: 10.1186/1687-2770-2012-64

Received: 19 January 2012

Accepted: 30 April 2012

Published: 22 June 2012

Abstract

This article deals with the differential equations of fractional order on the half-line. By the recent Leggett-Williams norm-type theorem due to O’Regan and Zima, we present some new results on the existence of positive solutions for the fractional boundary value problems at resonance on unbounded domains.

MSC:26A33, 34A08, 34A34.

Keywords

fractional order half-line coincidence degree at resonance

1 Introduction

In this article, we are concerned with the fractional differential equation
{ D 0 + α u ( t ) = f ( t , u ( t ) ) , t [ 0 , + ) , u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 , D 0 + α 1 u ( 0 ) = lim t + D 0 + α 1 u ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ1_HTML.gif
(1.1)
where D 0 + α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq1_HTML.gif is the Riemann-Liouville fractional derivative, 3 < α < 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq2_HTML.gif, and f : [ 0 , + ) × R R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq3_HTML.gif satisfies the following condition: (H) = f : [ 0 , + ) × R R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq3_HTML.gif is continuous and for each l > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq4_HTML.gif, there exists ϕ l C [ 0 , + ) L 1 [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq5_HTML.gif satisfying sup t 0 | ϕ l ( t ) | < + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq6_HTML.gif and ϕ l ( t ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq7_HTML.gif, t > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq8_HTML.gif such that
| u | < l implies | f ( t , ( 1 + t α 1 ) u ) | ϕ l ( t ) , a.e. t 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equa_HTML.gif
.

The problem (1.1) happens to be at resonance in the sense that the kernel of the linear operator D 0 + α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq1_HTML.gif is not less than one-dimensional under the boundary value conditions.

Fractional calculus is a generalization of the ordinary differentiation and integration. It has played a significant role in science, engineering, economy, and other fields. Some books on fractional calculus and fractional differential equations have appeared recently (see [13]); furthermore, today there is a large number of articles dealing with the fractional differential equations (see [415]) due to their various applications.

In [8], the researchers dealt with the existence of solutions for boundary value problems of fractional order of the form
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equb_HTML.gif

where 1 < α 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq9_HTML.gif and f : [ 0 , + ) × R R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq3_HTML.gif is continuous. The results are based on the fixed point theorem of Schauder combined with the diagonalization method.

In [9], Su and Zhang studied the following fractional differential equations on the half-line using Schauder’s fixed point theorem
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equc_HTML.gif
Employing the Leray-Schauder alternative theorem, in [12], Zhao and Ge considered the fractional boundary value problem
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equd_HTML.gif

However, the articles on the existence of solutions of fractional differential equations on the half-line are still few, and most of them deal with the problems under nonresonance conditions. And as far as we know, recent articles, such as [4, 6, 7], investigating resonant problems are on the finite interval.

Motivated by the articles [1620], in this article we study the differential equations (1.1) under resonance conditions on the unbounded domains. Moreover, we have successfully established the existence theorem by the recent Leggett-Williams norm-type theorem due to O’Regan and Zima. To our best knowledge, there is no article dealing with the resonant problems of fractional order on unbounded domains by the theorem.

The rest of the article is organized as follows. In Section 2, we give the definitions of the fractional integral and fractional derivative, some results about fractional differential equations, and the abstract existence theorem. In Section 3, we obtain the existence result of the solution for the problem (1.1) by the recent Leggett-Williams norm-type theorem. Then, an example is given in Section 4 to demonstrate the application of our result.

2 Preliminaries

First of all, we present some fundamental facts on the fractional calculus theory which we will use in the next section.

Definition 2.1 ([13])

The Riemann-Liouville fractional integral of order ν > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq10_HTML.gif of a function h : ( 0 , ) R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq11_HTML.gif is given by
I 0 + ν h ( t ) = D 0 + ν h ( t ) = 1 Γ ( ν ) 0 t ( t s ) ν 1 h ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ2_HTML.gif
(2.1)

provided that the right-hand side is pointwise defined on ( 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq12_HTML.gif.

Definition 2.2 ([13])

The Riemann-Liouville fractional derivative of order ν > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq10_HTML.gif of a continuous function h : ( 0 , ) R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq11_HTML.gif is given by
D 0 + ν h ( t ) = 1 Γ ( n ν ) ( d d t ) n 0 t ( t s ) n ν 1 h ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ3_HTML.gif
(2.2)

where n = [ ν ] + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq13_HTML.gif, provided that the right-hand side is pointwise defined on ( 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq14_HTML.gif.

Lemma 2.1 ([1, 9])

Assume that h ( t ) L 1 ( 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq15_HTML.gif. If ν 1 , ν 2 , ν > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq16_HTML.gif, then
I 0 + ν 1 I 0 + ν 2 h ( t ) = I 0 + ν 1 + ν 2 h ( t ) , D 0 + ν I 0 + ν h ( t ) = h ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ4_HTML.gif
(2.3)

Lemma 2.2 ([9])

Assume that D 0 + ν h ( t ) L 1 ( 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq17_HTML.gif, ν > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq10_HTML.gif. Then we have
I 0 + ν D 0 + ν h ( t ) = h ( t ) + C 1 t ν 1 + C 2 t ν 2 + + C N t ν N , t > 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ5_HTML.gif
(2.4)

for some C i R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq18_HTML.gif, i = 1 , 2 , , N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq19_HTML.gif, where N is the smallest integer greater than or equal to ν.

Now, let us recall some standard facts and the fixed point theorem due to O’Regan and Zima, and these can be found in [16, 17, 2123].

Let X, Z be real Banach spaces. Consider an operation equation
L u = N u , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Eque_HTML.gif

where L : dom L X Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq20_HTML.gif is a linear operator, N : X Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq21_HTML.gif is a nonlinear operator. If dim Ker L = codim Im L < + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq22_HTML.gif and ImL is closed in Z, then L is called a Fredholm mapping of index zero. And if L is a Fredholm mapping of index zero, there exist linear continuous projectors P : X X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq23_HTML.gif and Q : Z Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq24_HTML.gif such that Ker L = Im P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq25_HTML.gif, Im L = Ker Q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq26_HTML.gif and X = Ker L Ker P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq27_HTML.gif, Z = Im L Im Q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq28_HTML.gif. Then it follows that L P = L | dom L Ker P : dom L Ker P Im L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq29_HTML.gif is invertible. We denote the inverse of this map by K P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq30_HTML.gif. For ImQ is isomorphic to KerL, there exists an isomorphism J : Im Q Ker L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq31_HTML.gif.

It is known that the coincidence equation L u = N u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq32_HTML.gif is equivalent to
u = ( P + J Q N ) u + K P ( I Q ) N u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equf_HTML.gif
A nonempty convex closed set C X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq33_HTML.gif is called a cone if
  1. (i)

    κ x C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq34_HTML.gif for all x C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq35_HTML.gif and κ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq36_HTML.gif;

     
  2. (ii)

    x , x C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq37_HTML.gif implies x = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq38_HTML.gif.

     
Note that C induces a partial order in X by
x y if and only if y x C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equg_HTML.gif

The following lemma is valid for every cone in a Banach space.

Lemma 2.3 ([17, 23])

Let C be a cone in the Banach space X. Then for every u C { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq39_HTML.gif, there exists a positive number σ ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq40_HTML.gifsuch that
x + u σ ( u ) x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equh_HTML.gif

for all x C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq41_HTML.gif.

Let γ : X C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq42_HTML.gif be a retraction, i.e., a continuous mapping such that γ ( x ) = x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq43_HTML.gif for all x C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq41_HTML.gif. Denote
Ψ : = P + J Q N + K P ( I Q ) N , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equi_HTML.gif
and
Ψ γ : = Ψ γ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equj_HTML.gif

Theorem 2.1 ([16, 17])

Let C be a cone in X and let Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq44_HTML.gif, Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq45_HTML.gifbe open bounded subsets of X with Ω ¯ 1 Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq46_HTML.gifand C ( Ω ¯ 2 Ω 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq47_HTML.gif. Assume that: 1 = L is a Fredholm operator of index zero;; 2 = Q N : X Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq48_HTML.gifis continuous and bounded and K P ( I Q ) N : X X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq49_HTML.gifis compact on every bounded subset of X;; 3 = L u λ N u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq50_HTML.giffor all u C Ω 2 dom L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq51_HTML.gifand λ ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq52_HTML.gif;; 4 = γ maps subsets of Ω ¯ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq53_HTML.gifinto bounded subsets of C;; 5 = d B ( [ I ( P + J Q N ) γ ] | Ker L , Ker L Ω 2 , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq54_HTML.gif, where d B https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq55_HTML.gifstands for the Brouwer degree;; 6 = there exists u 0 C { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq56_HTML.gifsuch that u σ ( u 0 ) Ψ u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq57_HTML.giffor u C ( u 0 ) Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq58_HTML.gif, where C ( u 0 ) = { u C : μ u 0 u for some μ > 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq59_HTML.gifand σ ( u 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq60_HTML.gifis such that u + u 0 σ ( u 0 ) u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq61_HTML.giffor every u C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq62_HTML.gif;; 7 = ( P + J Q N ) γ ( Ω 2 ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq63_HTML.gif;; 8 = Ψ γ ( Ω ¯ 2 Ω 1 ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq64_HTML.gif..Then the equation L x = N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq65_HTML.gifhas a solution in the set C ( Ω ¯ 2 Ω 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq66_HTML.gif.

Let
X = { x | x C [ 0 , + ) , lim t + x ( t ) 1 + t α 1 exists } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equk_HTML.gif
with the norm
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equl_HTML.gif
and
Z = { z | z C [ 0 , + ) L 1 [ 0 , + ) , sup t 0 | z ( t ) | < + } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equm_HTML.gif
equipped with the norm
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equn_HTML.gif

Remark 2.1 It is easy for us to prove that https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq67_HTML.gif and https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq68_HTML.gif are Banach spaces.

Set
dom L = { u X | D 0 + α u ( t ) C [ 0 , + ) L 1 [ 0 , + ) , u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 , D 0 + α 1 u ( 0 ) = lim t + D 0 + α 1 u ( t ) } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equo_HTML.gif
Define
L : dom L Z , u D 0 + α u ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ6_HTML.gif
(2.5)
and
N : X Z , u f ( t , u ( t ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ7_HTML.gif
(2.6)
Then the multi-point boundary value problem (1.1) can be written by
L u = N u , u dom L . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equp_HTML.gif

Definition 2.3 u X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq69_HTML.gif is called a solution of the problem (1.1) if u dom L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq70_HTML.gif and u satisfied Equation (1.1).

Next, similar to the compactness criterion in [12, 24], we establish the following criterion, and it can be proved in a similar way.

Lemma 2.4 U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq71_HTML.gifis a relatively compact set in X if and only if the following conditions are satisfied:
  1. (a)

    U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq71_HTML.gif is uniformly bounded, that is, there exists a constant R > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq72_HTML.gif such that for each u U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq73_HTML.gif, https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq74_HTML.gif .

     
  2. (b)
    The functions from U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq71_HTML.gif are equicontinuous on any compact subinterval of [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq75_HTML.gif, that is, let J be a compact subinterval of [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq75_HTML.gif, then ε > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq76_HTML.gif, there exists δ = δ ( ε ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq77_HTML.gif such that for t 1 , t 2 J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq78_HTML.gif, | t 1 t 2 | < δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq79_HTML.gif,
    | u ( t 1 ) 1 + t 1 α 1 u ( t 2 ) 1 + t 2 α 1 | < ε , u U . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equq_HTML.gif
     
  3. (c)
    The functions from U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq71_HTML.gif are equiconvergent, that is, given ε > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq80_HTML.gif, there exists T = T ( ε ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq81_HTML.gif such that
    | u ( s 1 ) 1 + s 1 α 1 u ( s 2 ) 1 + s 2 α 1 | < ε , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equr_HTML.gif
     

for s 1 , s 2 > T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq82_HTML.gif, u U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq83_HTML.gif.

3 Main results

In this section, we will present the existence theorem for the fractional differential equation on the half-line. In order to prove our main result, we need the following lemmas.

Lemma 3.1 Let g Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq84_HTML.gif. Then u X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq85_HTML.gifis the solution of the following fractional differential equation:

{ D 0 + α u ( t ) = g ( t ) , t [ 0 , + ) , u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 , D 0 + α 1 u ( 0 ) = lim t + D 0 + α 1 u ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equs_HTML.gif
if and only if
u ( t ) = c t α 1 + 1 Γ ( α ) 0 t ( t s ) α 1 g ( s ) d s , c R , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equt_HTML.gif
and
0 + g ( t ) d t = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equu_HTML.gif

Proof In view of Lemmas 2.1 and 2.2, we can certify the conclusion easily, so we omit the details here. □

Lemma 3.2 The operator L is a Fredholm mapping of index zero. Moreover,

Ker L = { u | u = c t α 1 , t 0 , c R } X , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ8_HTML.gif
(3.1)
and
Im L = { g Z | 0 + g ( t ) d t = 0 } Z . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ9_HTML.gif
(3.2)

Proof It is obvious that Lemma 3.1 implies (3.1) and (3.2). Now, let us focus our minds on proving that L is a Fredholm mapping of index zero.

Define Q : Z Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq24_HTML.gif
( Q g ) ( t ) = e t 0 + g ( s ) d s , t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ10_HTML.gif
(3.3)
where g Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq86_HTML.gif. Evidently, Ker Q = Im L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq87_HTML.gif, Im Q = { g | g = c e t , t 0 , c R } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq88_HTML.gif, and Q : Z Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq24_HTML.gif is a continuous linear projector. In fact, for an arbitrary g Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq86_HTML.gif, we have
Q 2 g = Q ( Q g ) = Q ( e t 0 + g ( s ) d s ) = Q ( e t ) 0 + g ( s ) d s = e t 0 + g ( s ) d s = Q g , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equv_HTML.gif

that is to say, Q : Z Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq24_HTML.gif is idempotent.

Let g = g Q g + Q g = ( I Q ) g + Q g https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq89_HTML.gif, where g Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq86_HTML.gif is an arbitrary element. Since Q g Im Q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq90_HTML.gif and ( I Q ) g Ker Q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq91_HTML.gif, we obtain that Z = Im Q + Ker Q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq92_HTML.gif. Take z 0 Im Q Ker Q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq93_HTML.gif, then z 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq94_HTML.gif can be written as z 0 = c e t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq95_HTML.gif, c R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq96_HTML.gif , for z 0 Im Q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq97_HTML.gif. Since z 0 Ker Q = Im L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq98_HTML.gif, by (3.2), we get that Q ( z 0 ) = Q ( c e t ) = c Q ( e t ) = c e t = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq99_HTML.gif, which implies that c = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq100_HTML.gif, and then z 0 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq101_HTML.gif. Therefore, Im Q Ker Q = { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq102_HTML.gif, thus, Z = Im Q Ker Q = Im Q Im L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq103_HTML.gif.

Now, dim Ker L = 1 = dim Im Q = codim Ker Q = codim Im L < + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq104_HTML.gif, and observing that ImL is closed in Z, so L is a Fredholm mapping of index zero. □

Let P : X X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq23_HTML.gif be defined by
( P u ) ( t ) = ( 1 Γ ( α ) 0 + e s u ( s ) d s ) t α 1 , t 0 , u X . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ11_HTML.gif
(3.4)
It is clear that P : X X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq23_HTML.gif is a linear continuous projector and
Im P = { u | u = c t α 1 , t 0 , c R } = Ker L . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equw_HTML.gif

Also, proceeding with the proof of Lemma 3.2, we can show that X = Im P Ker P = Ker L Ker P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq105_HTML.gif.

Consider the mapping K P : Im L dom L Ker P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq106_HTML.gif
( K P g ) ( t ) = ( 1 Γ ( α ) 0 + e s g ( s ) d s ) t α 1 + 1 Γ ( α ) 0 t ( t s ) α 1 g ( s ) d s , g Im L . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equx_HTML.gif
Note that
( K P L ) u = K P ( L u ) = u , u dom L Ker P , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equy_HTML.gif
and
( L K P ) g = L ( K P g ) = g , g Im L . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equz_HTML.gif

Thus, K P = ( L P ) 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq107_HTML.gif, where L P = L | dom L Ker P : dom L Ker P Im P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq108_HTML.gif.

Define the linear isomorphism J : Im Q Ker L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq109_HTML.gif as
J ( c e t ) = c t α 1 , t 0 , c R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equaa_HTML.gif
Thus, J Q N + K P ( I Q ) N : X X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq110_HTML.gif is given by
[ J Q N + K P ( I Q ) N ] u ( t ) = t α 1 Γ ( α ) 0 + G ( t , s ) f ( s , u ( s ) ) d s , t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ12_HTML.gif
(3.5)
where
G ( t , s ) = { 0 , t = 0 ; Γ ( α ) + 1 2 e s 0 t ( t τ ) α 1 t α 1 e τ d τ + ( t s ) α 1 t α 1 , t 0 and 0 s t < + ; Γ ( α ) + 1 2 e s 0 t ( t τ ) α 1 t α 1 e τ d τ , 0 < t s < + . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equab_HTML.gif
Then, it is easy to verify that
0 < Γ ( α ) 1 2 G ( t , s ) Γ ( α ) + 3 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ13_HTML.gif
(3.6)

Now, we state the main result on the existence of the positive solutions to the problem (1.1) in the following.

Theorem 3.1 Let f : [ 0 , + ) × R R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq111_HTML.gifsatisfy the condition (H). Assume that there exist six nonnegative functions α i ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq112_HTML.gif ( i = 1 , 2 , 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq113_HTML.gif), β j ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq114_HTML.gif ( j = 1 , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq115_HTML.gif) and μ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq116_HTML.gifsuch that
f ( t , u ) α 1 ( t ) | f ( t , u ) | + α 2 ( t ) u 1 + t α 1 + α 3 ( t ) , t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ14_HTML.gif
(3.7)
and
μ ( t ) u 1 + t α 1 f ( t , u ) β 1 ( t ) u 1 + t α 1 + β 2 ( t ) , t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ15_HTML.gif
(3.8)
where 0 u 1 + t α 1 R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq117_HTML.gif, R > R 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq118_HTML.gif, and R 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq119_HTML.gifis defined by (3.12), α 1 ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq120_HTML.gifis bounded on [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq75_HTML.gif, β 1 ( t ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq121_HTML.gif, t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq122_HTML.gif, α 2 ( t ) , α 3 ( t ) , β 1 ( t ) , β 2 ( t ) L 1 [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq123_HTML.gif,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ16_HTML.gif
(3.9)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ17_HTML.gif
(3.10)
and
0 + μ ( t ) d t < Γ ( α ) + 1 2 ( Γ ( α ) + 1 / 2 ) ( Γ ( α ) + 3 / 2 ) , e t μ ( t ) < 1 + t α 1 Γ ( α ) + 3 / 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ18_HTML.gif
(3.11)

Then the problem (1.1) has at least one positive solution in domL.

Proof For the simplicity of notation, we denote
ε 1 : = Γ ( α ) Γ ( α ) + 1 / 2 + Γ ( α ) + 3 / 2 Γ ( α ) + 1 0 + μ ( s ) d s < 1 , β 0 : = 0 + s α 1 β 1 ( s ) 1 + s α 1 d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equac_HTML.gif
and
R 0 : = max { Γ 0 Γ ( α ) 0 + β 2 ( s ) d s + 2 α 0 Γ ( α ) 0 + α 3 ( s ) d s , 1 β 0 0 + β 2 ( s ) d s } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ19_HTML.gif
(3.12)
Consider the cone
C = { u | u X , u ( t ) 0 , t 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equad_HTML.gif
Set
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equae_HTML.gif

where R 2 ( R 0 , R ) , R 1 ( 0 , R 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq124_HTML.gif, ε 0 ( ε 1 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq125_HTML.gif. Clearly, Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq44_HTML.gif and Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq45_HTML.gif are an open bounded set of X.

Step 1: In view of Lemma 3.2, the condition 1 of Theorem 2.1 is fulfilled.

Step 2: By virtue of Lemma 2.4, we can get that Q N : X Z https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq48_HTML.gif is continuous and bounded and K P ( I Q ) N : X X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq49_HTML.gif is compact on every bounded subset of X, which ensures that the assumption 2 of Theorem 2.1 holds.

Step 3: Suppose that there exist u C Ω 2 dom L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq126_HTML.gif and λ ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq127_HTML.gif such that L u = λ N u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq128_HTML.gif.

Since
u = ( I P ) u + P u = K P L ( I P ) u + P u = K P L u + P u , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equaf_HTML.gif
we have
u ( t ) 1 + t α 1 = 1 Γ ( α ) 0 + e s D 0 + α u ( s ) d s t α 1 1 + t α 1 + 1 Γ ( α ) 0 t ( t s ) α 1 1 + t α 1 D 0 + α u ( s ) d s + 1 Γ ( α ) 0 + e s u ( s ) d s t α 1 1 + t α 1 < 2 Γ ( α ) 0 + | D 0 + α u ( s ) | d s + 1 Γ ( α ) 0 + e s u ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ20_HTML.gif
(3.13)
From (3.7) and (3.8), we get that
D 0 + α u ( t ) = λ f ( t , u ( t ) ) λ α 1 ( t ) | f ( t , u ( t ) ) | + λ α 2 ( t ) u ( t ) 1 + t α 1 + λ α 3 ( t ) α 1 ( t ) | D 0 + α u ( t ) | + α 2 ( t ) u ( t ) 1 + t α 1 + α 3 ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ21_HTML.gif
(3.14)
and
D 0 + α u ( t ) = λ f ( t , u ( t ) ) λ β 1 ( t ) u ( t ) 1 + t α 1 + λ β 2 ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ22_HTML.gif
(3.15)
On account of the fact that
0 + D 0 + α u ( s ) d s = 0 + D ( D 0 + α 1 u ( s ) ) d s = lim t + D 0 + α 1 u ( t ) D 0 + α 1 u ( 0 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equag_HTML.gif
and considering (3.14) and (3.15), we have
0 = 0 + D 0 + α u ( s ) d s 0 + α 1 ( s ) | D 0 + α u ( s ) | d s + 0 + α 2 ( s ) u ( s ) 1 + s α 1 d s + 0 + α 3 ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equah_HTML.gif
and
0 = 0 + D 0 + α u ( s ) d s 0 + λ β 1 ( s ) u ( s ) 1 + s α 1 d s + 0 + λ β 2 ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equai_HTML.gif
Thus,
0 + | D 0 + α u ( s ) | d s 1 α 0 0 + α 2 ( s ) u ( s ) 1 + s α 1 d s + 1 α 0 0 + α 3 ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equaj_HTML.gif
and
0 + β 1 ( s ) u ( s ) 1 + s α 1 d s 0 + β 2 ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equak_HTML.gif
By (3.9), (3.10) and (3.13), we obtain that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equal_HTML.gif

which is a contradiction to u C Ω 2 dom L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq126_HTML.gif. Therefore, 3 is satisfied.

Step 4: Let ( γ u ) ( t ) = | u ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq129_HTML.gif, then we can verify that γ : X C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq42_HTML.gif is a retraction and 4 holds.

Step 5: Let u Ker L Ω ¯ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq130_HTML.gif, then u ( t ) = c t α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq131_HTML.gif t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq122_HTML.gif c R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq132_HTML.gif. Inspired by Aijun and Wang [5], we set
H ( c t α 1 , ρ ) = [ I ρ ( P + J Q N ) γ ] ( c t α 1 ) = ( c ρ | c | ρ 0 + f ( s , | c | s α 1 ) d s ) t α 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equam_HTML.gif

where c [ R 2 , R 2 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq133_HTML.gif and ρ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq134_HTML.gif.

Define homeomorphism J 1 : Ker L Ω ¯ 2 R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq135_HTML.gif by J 1 ( c t α 1 ) = c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq136_HTML.gif, then
d B ( H ( c t α 1 , ρ ) , Ker L Ω 2 , 0 ) = d B ( J 1 H ( J 1 1 c , ρ ) , J 1 ( Ker L Ω 2 ) , J 1 ( 0 ) ) = d B ( J 1 H ( J 1 1 c , ρ ) , J 1 ( Ker L Ω 2 ) , 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equan_HTML.gif

It is obvious that J 1 H ( J 1 1 c , ρ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq137_HTML.gif implies that c 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq138_HTML.gif by (3.8) and (3.11).

Take c 0 J 1 ( Ker L Ω 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq139_HTML.gif, then | c 0 | = R 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq140_HTML.gif. Suppose that J 1 H ( J 1 1 c , ρ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq137_HTML.gif, ρ ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq141_HTML.gif, then we have that c 0 = R 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq142_HTML.gif. Also, in view of (3.8),
R 2 = ρ ( R 2 0 + f ( s , R 2 s α 1 ) d s ) ρ ( R 2 + R 2 0 + β 1 ( s ) s α 1 1 + s α 1 d s + 0 + β 2 ( s ) d s ) < ρ R 2 R 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equao_HTML.gif

It is a contradiction. Besides, if ρ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq143_HTML.gif, then R 2 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq144_HTML.gif, which is impossible. Hence, for c J 1 ( Ker L Ω 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq145_HTML.gif, J 1 H ( J 1 1 c , ρ ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq146_HTML.gif, ρ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq134_HTML.gif.

Therefore,
d B ( [ I ( P + J Q N ) γ ] | Ker L , Ker L Ω 2 , 0 ) = d B ( H ( , 1 ) , Ker L Ω 2 , 0 ) = d B ( J 1 H ( J 1 1 c , 1 ) , J 1 ( Ker L Ω 2 ) , 0 ) = d B ( J 1 H ( J 1 1 c , 0 ) , J 1 ( Ker L Ω 2 ) , 0 ) = d B ( I , J 1 ( Ker L Ω 2 ) , 0 ) = 1 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equap_HTML.gif

which shows that 5 is true.

Step 6: Let u 0 = 1 + t α 1 C { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq147_HTML.gif, then we have
C ( u 0 ) = { u C | inf t 0 u ( t ) 1 + t α 1 > 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equaq_HTML.gif

And we can take σ ( u 0 ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq148_HTML.gif.

Let t 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq149_HTML.gif such that
t 0 α 1 1 + t 0 α 1 > Γ ( α ) + 1 / 2 Γ ( α ) + 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equar_HTML.gif
For u C ( u 0 ) Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq58_HTML.gif, we have that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equas_HTML.gif
Therefore, combining (3.6), (3.8) and (3.11), we get that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equat_HTML.gif

Thus, https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq150_HTML.gif for all u C ( u 0 ) Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq58_HTML.gif. So, 6 holds.

Step 7: For u Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq151_HTML.gif, from (3.8) and (3.11), we have
( P + J Q N ) ( γ u ) ( t ) = ( 1 Γ ( α ) 0 + e s | u ( s ) | d s + 0 + f ( s , | u ( s ) | ) d s ) t α 1 t α 1 Γ ( α ) 0 + [ e s ( 1 + s α 1 ) μ ( s ) ] | u ( s ) | 1 + s α 1 d s 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equau_HTML.gif

which implies that ( P + J Q N ) γ ( Ω 2 ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq63_HTML.gif. Hence, 7 holds.

Step 8: For u Ω ¯ 2 Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq152_HTML.gif, by (3.6), (3.8) and (3.11), we obtain that
Ψ γ u ( t ) = [ P + J Q N + K P ( I Q ) N ] | u ( t ) | = ( 1 Γ ( α ) 0 + e s | u ( s ) | d s + 1 Γ ( α ) 0 + G ( t , s ) f ( s , | u ( s ) | ) d s ) t α 1 t α 1 Γ ( α ) ( 0 + [ e s ( 1 + s α 1 ) G ( t , s ) μ ( s ) ] | u ( s ) | 1 + s α 1 d s ) t α 1 Γ ( α ) ( 0 + [ e s ( 1 + s α 1 ) ( Γ ( α ) + 3 2 ) μ ( s ) ] | u ( s ) | 1 + s α 1 d s ) 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equav_HTML.gif

Thus, Ψ γ ( Ω ¯ 2 Ω 1 ) C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq64_HTML.gif, that is, 8 is satisfied.

Hence, applying Theorem 2.1, the problem (1.1) has a positive solution in the set C ( Ω ¯ 2 Ω 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq66_HTML.gif. □

4 Examples

To illustrate our main result, we will present an example.

Example 4.1
{ D 0 + α u ( t ) = f ( t , u ( t ) ) , t [ 0 , + ) , u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 , D 0 + α 1 u ( 0 ) = lim t + D 0 + α 1 u ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equ23_HTML.gif
(4.1)
where α = 3.5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq153_HTML.gif, and for ( t , u ) R 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq154_HTML.gif,
f ( t , u ) = β 1 ( t ) u 1 + t α 1 + β 2 ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equaw_HTML.gif
and
β 1 ( t ) = 1 40 e t ( 1 + t α 1 ) , β 2 ( t ) = 1 1 + t 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equax_HTML.gif

It is easy for us to certify that f satisfies the condition (H).

Noting that
f ( t , u ) α 1 ( t ) | f ( t , u ) | + α 2 ( t ) u 1 + t α 1 + α 3 ( t ) , t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equay_HTML.gif
and
μ ( t ) u 1 + t α 1 f ( t , u ) β 1 ( t ) u 1 + t α 1 + β 2 ( t ) , t 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equaz_HTML.gif
for u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq155_HTML.gif, where
α 1 ( t ) = 2 , α 2 ( t ) = β 1 ( t ) , α 3 ( t ) = 3 β 2 ( t ) , μ ( t ) = β 1 ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equba_HTML.gif

Evidently, μ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_IEq116_HTML.gif satisfies (3.11).

Meanwhile, by simple computation we can get that
α 0 = 2 , 0 + α 3 ( t ) d t = 3 π 2 , 0 + β 2 ( t ) d t = π 2 , Γ 0 = 41 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-64/MediaObjects/13661_2012_Article_168_Equbb_HTML.gif

Thus, to sum up the points which we have just indicated, by Theorem 3.1, we can conclude that the problem (4.1) has at least one positive solution.

Declarations

Acknowledgement

This project is supported by the Hunan Provincial Innovation Foundation For Postgraduate (NO. CX2011B079) and the National Natural Science Foundation of China (NO. 11171351).

Authors’ Affiliations

(1)
School of Mathematical Sciences and Computing Technology, Central South University

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