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Successive iteration and positive solutions of a fractional boundary value problem on the half-line

Advances in Difference Equations20132013:210

https://doi.org/10.1186/1687-1847-2013-210

  • Received: 17 April 2013
  • Accepted: 28 June 2013
  • Published:

Abstract

In this paper, we study the existence of positive solutions for a nonlinear fractional boundary value problem on the half-line. Based on the monotone iterative technique, we obtain the existence of positive solutions of a fractional boundary value problem and establish iterative schemes for approximating the solutions. As application, an example is presented to illustrate the main results.

MSC:34A08, 34A12, 34B40.

Keywords

  • fractional boundary value problem
  • successive iteration
  • half-line
  • positive solutions

1 Introduction

The initial and boundary value problems for nonlinear fractional differential equations arise from the study of models of viscoelasticity, control, porous media, etc. [1, 2]. In the past few years, the existence and multiplicity of positive solutions for nonlinear fractional boundary value problems have been widely studied by many authors (see [311] and the references therein).

Most of these papers only considered the existence of positive solutions of various boundary value problems. Yet how can we find the solutions when their existence is known? Sun et al. [12] proved the existence of positive solutions for second-order p-Laplacian boundary value problems which are defined on finite intervals via the iterative technique. Liang and Zhang [13] investigated the existence of three positive solutions for the following m-point fractional boundary value problem on an infinite interval by means of the Leggett-Williams fixed point theorems on cones. In [14, 15], by employing the standard fixed point theorems and the monotone iterative technique, the authors deal with the existence of solutions for nonlinear fractional boundary value problems on an unbounded domain. Motivated by all the works above, we investigate the iteration and existence of positive solutions for the following fractional boundary value problems on the half-line:
{ D 0 + α u ( t ) + q ( t ) f ( t , u ( t ) ) = 0 , t ( 0 , ) , u ( 0 ) = 0 , lim t D 0 + α 1 u ( t ) = λ 0 ,
(1.1)

where 1 < α < 2 , D 0 + α is the standard Riemann-Liouville fractional derivative. By applying monotone iterative techniques, we construct some successive iterative schemes to approximate the solutions in this paper. They start off with a simple function and the zero function respectively, which is convenient for application.

Throughout this paper, we assume that the following conditions hold:

(H1) f C ( [ 0 , ) × [ 0 , ) , [ 0 , ) ) , f ( t , 0 ) 0 on any subinterval of [ 0 , ) and when u is bounded, f ( t , ( 1 + t α 1 ) u ) is bounded on [ 0 , ) ;

(H2) q ( t ) : [ 0 , ) [ 0 , ) is not identical zero on any closed subinterval of [ 0 , ) , and 0 < 0 q ( s ) d s < .

2 Preliminaries

We need the following definitions and lemmas that will be used to prove our main results.

Definition 2.1 [1]

The integral
I 0 + α f ( t ) = 1 Γ ( α ) 0 t f ( s ) ( t s ) 1 α d s , t > 0 ,

where α > 0 , is called the Riemann-Liouville fractional integral of order α and Γ ( α ) is the Euler gamma function defined by Γ ( α ) = 0 t α 1 e t d t , α > 0 .

Definition 2.2 [2]

For a function f ( x ) given in the interval [ 0 , ) , the expression
D 0 + α f ( t ) = 1 Γ ( α ) ( d d t ) n 0 t f ( s ) ( t s ) α n + 1 d s ,

where n = [ α ] + 1 , [ α ] denotes the integer part of number α, is called the Riemann-Liouville fractional derivative of order α.

Lemma 2.1 [2]

Let α > 0 and u C ( 0 , ) L ( 0 , ) . Then the fractional differential equation
D 0 + α u ( t ) = 0
has
u ( t ) = c 1 t α 1 + c 2 t α 2 + + c n t α n , c i R , i = 0 , 1 , , n , n = [ α ] + 1

as a unique solution.

Lemma 2.2 [2]

Assume that u ( t ) C ( 0 , ) L ( 0 , ) with a fractional derivative of order α > 0 that belongs to C ( 0 , ) L ( 0 , ) . Then
I 0 + α D 0 + α u ( t ) = u ( t ) + c 1 t α 1 + c 2 t α 2 + + c n t α n ,

for some c i R , i = 0 , 1 , , n , n = [ α ] + 1 .

Lemma 2.3 Let g C [ 0 , ) , then the fractional boundary value problem
{ D 0 + α u ( t ) + g ( t ) = 0 , t ( 0 , ) , 1 < α < 2 , u ( 0 ) = 0 , lim t D 0 + α 1 u ( t ) = λ 0
(2.1)
has a unique solution
u ( t ) = 0 G ( t , s ) g ( s ) d s + λ Γ ( α ) t α 1 ,
where
G ( t , s ) = 1 Γ ( α ) { t α 1 ( t s ) α 1 , 0 s t < , t α 1 , 0 t s < .
(2.2)
Proof By Lemma 2.2, the solution of (2.1) can be written as
u ( t ) = c 1 t α 1 + c 2 t α 2 0 t ( t s ) α 1 Γ ( α ) g ( s ) d s .

From u ( 0 ) = 0 , we know that c 2 = 0 .

Together with lim t D 0 + α 1 u ( t ) = λ 0 , we have
c 1 = λ + 0 g ( s ) d s Γ ( α ) .
Therefore, the unique solution of fractional boundary value problem (2.1) is
u ( t ) = 0 t ( t s ) α 1 Γ ( α ) g ( s ) d s + 0 t α 1 Γ ( α ) g ( s ) d s + λ Γ ( α ) t α 1 = 0 t t α 1 ( t s ) α 1 Γ ( α ) g ( s ) d s + t t α 1 Γ ( α ) g ( s ) d s + λ Γ ( α ) t α 1 = 0 G ( t , s ) g ( s ) d s + λ Γ ( α ) t α 1 .

The proof is complete. □

Lemma 2.4 The function G ( t , s ) defined by (2.2) satisfies the following:
  1. (1)

    G ( t , s ) is a continuous function and G ( t , s ) 0 for ( t , s ) [ 0 , ) × [ 0 , ) ;

     
  2. (2)

    G ( t , s ) t α 1 Γ ( α ) , G ( t , s ) 1 + t α 1 1 Γ ( α ) for ( t , s ) [ 0 , ) × [ 0 , ) .

     

The proof is easy, so we omit it here.

In this paper, we use the following space E, which is denoted by
E = { u C [ 0 , ) : sup 0 t < | u ( t ) | 1 + t α 1 < } ,
to study fractional boundary value problem (1.1). From [16], we know that E is a Banach space equipped with the norm u = sup 0 t < | u ( t ) | 1 + t α 1 < . Define the cone P E by
P = { u E : u ( t ) 0 , t [ 0 , ) } ,
and an integral operator T : P E by
T u ( t ) = 0 G ( t , s ) q ( s ) f ( s , u ( s ) ) d s + λ Γ ( α ) t α 1 , 0 t < ,
(2.3)

where G ( t , s ) is defined by (2.2).

Lemma 2.5 [16]

Let V = { u E : u < l } ( l > 0 ), V 1 = { u ( t ) 1 + t α 1 : u V } . If V 1 is equicontinuous on any compact intervals of [ 0 , ) and equiconvergent at infinity, then V is relatively compact on E.

Remark 2.5 V 1 is called equiconvergent at infinity if and only if for all ε > 0 , there exists v ( ε ) > 0 such that for all u V 1 , t 1 , t 2 v , the following holds:
| u ( t 1 ) 1 + t 1 α 1 u ( t 2 ) 1 + t 2 α 1 | < ε .

Lemma 2.6 Let (H1) and (H2) hold, then T : P P is completely continuous.

Proof First, it is easy to check that T : P P is well defined. Now, we prove that T is continuous and compact respectively. Let u n u as n in P, then there exists r 0 such that sup n N { 0 } u n < r 0 . Let B r 0 = sup { f ( t , ( 1 + t α 1 ) u ) ( t , u ) [ 0 , ) × [ 0 , r 0 ] } . By (H2), we have
0 q ( s ) f ( s , u ( s ) ) d s B r 0 0 q ( s ) d s < .
Then by the Lebesgue dominated convergence theorem and the continuity of f, we can get
0 q ( s ) f ( s , u n ( s ) ) d s 0 q ( s ) f ( s , u ( s ) ) d s as  n .
Therefore, we have
T u n T u = sup 0 t < | 0 G ( t , s ) 1 + t α 1 q ( s ) [ f ( s , u n ( s ) ) f ( s , u ( s ) ) ] d s | 1 Γ ( α ) | 0 q ( s ) f ( s , u n ( s ) ) d s 0 q ( s ) f ( s , u ( s ) ) d s | 0 , n .

Then T is continuous.

Let Ω be any bounded subset of P. Then there exists r > 0 such that u r for any u Ω . Let B r = sup { f ( t , ( 1 + t α 1 ) u ) ( t , u ) [ 0 , ) × [ 0 , r ] } , therefore, from Lemma 2.4, we have
T u = sup 0 t < 1 1 + t α 1 | 0 G ( t , s ) q ( s ) f ( s , u ( s ) ) d s + λ Γ ( α ) t α 1 | B r Γ ( α ) 0 q ( s ) d s + λ Γ ( α ) < .
So, T Ω is bounded. Moreover, for any T ( 0 , ) and t 1 , t 2 [ 0 , T ] , without loss of generality, we may assume that t 2 > t 1 , we have
| T u ( t 1 ) 1 + t 1 α 1 T u ( t 2 ) 1 + t 2 α 1 | 0 | G ( t 1 , s ) 1 + t 1 α 1 G ( t 2 , s ) 1 + t 2 α 1 | q ( s ) f ( s , u ( s ) ) d s + λ Γ ( α ) | t 1 α 1 t 2 α 1 | 0 | G ( t 1 , s ) 1 + t 1 α 1 G ( t 2 , s ) 1 + t 1 α 1 | q ( s ) f ( s , u ( s ) ) d s + 0 | G ( t 2 , s ) 1 + t 1 α 1 G ( t 2 , s ) 1 + t 2 α 1 | q ( s ) f ( s , u ( s ) ) d s + λ Γ ( α ) | t 1 α 1 t 2 α 1 | .
On the other hand, we have
0 | G ( t 1 , s ) 1 + t 1 α 1 G ( t 2 , s ) 1 + t 1 α 1 | q ( s ) f ( s , u ( s ) ) d s 0 t 1 | G ( t 1 , s ) 1 + t 1 α 1 G ( t 2 , s ) 1 + t 1 α 1 | q ( s ) f ( s , u ( s ) ) d s + t 1 t 2 | G ( t 1 , s ) 1 + t 1 α 1 G ( t 2 , s ) 1 + t 1 α 1 | q ( s ) f ( s , u ( s ) ) d s + t 2 | G ( t 1 , s ) 1 + t 1 α 1 G ( t 2 , s ) 1 + t 1 α 1 | q ( s ) f ( s , u ( s ) ) d s B r Γ ( α ) 0 t 1 [ t 2 α 1 t 1 α 1 ] + [ ( t 2 s ) α 1 ( t 1 s ) α 1 ] 1 + t 1 α 1 q ( s ) d s + B r Γ ( α ) t 1 t 2 [ t 2 α 1 t 1 α 1 ] + ( t 2 s ) α 1 1 + t 1 α 1 q ( s ) d s + B r Γ ( α ) t 2 t 2 α 1 t 1 α 1 1 + t 1 α 1 q ( s ) d s .
Similarly, we have
0 | G ( t 2 , s ) 1 + t 1 α 1 G ( t 2 , s ) 1 + t 2 α 1 | q ( s ) f ( s , u ( s ) ) d s 0 uniformly as  t 1 t 2 , λ Γ ( α ) | t 1 α 1 t 2 α 1 | 0 uniformly as  t 1 t 2 .

Hence, T Ω is locally equicontinuous on [ 0 , ) .

Next, we show that T : P E is equiconvergent at ∞. For any u Ω , we have
0 q ( s ) f ( s , u ( s ) ) d s < ,
and
lim t | T u ( t ) 1 + t α 1 | = lim t 1 1 + t α 1 0 G ( t , s ) q ( s ) f ( s , u ( s ) ) d s + lim t λ Γ ( α ) t α 1 1 + t α 1 = λ Γ ( α ) < .

Hence, T Ω is equiconvergent at infinity. By using Lemma 2.5, we obtain that T : P P is completely continuous. The proof is complete. □

3 Main results

We will prove the following existence results.

Theorem 3.1 Assume that (H1) and (H2) hold, and there exists a > 2 λ Γ ( α ) such that:

(H3) f ( t , u 1 ) f ( t , u 2 ) for any 0 t < , 0 u 1 u 2 ;

(H4) f ( t , ( 1 + t α 1 ) u ) a Γ ( α ) 2 0 q ( s ) d s , ( t , u ) [ 0 , ) × [ 0 , a ] .

Then fractional boundary value problem (1.1) has two positive solutions w and v on [ 0 , ) , satisfying 0 w a , 0 v a , 0 < v w , and
lim n w n = lim n T n w 0 = w , lim n v n = lim n T n v 0 ( t ) = v ,
where
w 0 ( t ) = ( a 2 + λ Γ ( α ) ) t α 1 , v 0 ( t ) = 0 , t [ 0 , ) .
Proof By Lemma 2.6, we know that T : P P is completely continuous. For any u 1 , u 2 P with u 1 u 2 , from the definition of T and (H3), we can easily get that T u 1 T u 2 . We denote
P a ¯ = { u P : u a } .

Then, in what follows, we first prove that T : P a ¯ P a ¯ .

If u P a ¯ , then u a , that is, 0 u ( t ) 1 + t α 1 a , 0 t < . Then assumption (H4) implies f ( t , u ( t ) ) a Γ ( α ) 2 0 q ( s ) d s for all ( t , u ) [ 0 , ) × [ 0 , a ] , together with (2.3), (H3), we get
T u = sup 0 t < | 0 G ( t , s ) 1 + t α 1 q ( s ) f ( s , u ( s ) ) d s + λ Γ ( α ) t α 1 1 + t α 1 | 0 q ( s ) d s Γ ( α ) a Γ ( α ) 2 0 q ( s ) d s + λ Γ ( α ) < a 2 + a 2 = a .

Hence, we have proved that T : P a ¯ P a ¯ .

Let w 0 ( t ) = ( a 2 + λ Γ ( α ) ) t α 1 , 0 t < , then w 0 ( t ) P a ¯ . Let w 1 = T w 0 , w 2 = T 2 w 0 ; then by Lemma 2.6, we have that w 1 P a ¯ and w 2 P a ¯ . We denote w n + 1 = T w n = T n w 0 , n = 0 , 1 ,  . Since T : P a ¯ P a ¯ , we have that w n T ( P a ¯ ) P a ¯ , n = 1 , 2 ,  . It follows from the complete continuity of T that { w n } n = 1 is a sequentially compact set.

By (2.3) and (H4), we get
w 1 ( t ) = 0 G ( t , s ) q ( s ) f ( s , w 0 ( s ) ) d s + λ Γ ( α ) t α 1 a Γ ( α ) 2 0 q ( s ) d s 0 t α 1 Γ ( α ) q ( s ) d s + λ Γ ( α ) t α 1 = ( a 2 + λ Γ ( α ) ) t α 1 = w 0 ( t ) .
So, by (H3), we have
w 2 ( t ) = T w 1 ( t ) T w 0 ( t ) w 1 ( t ) , 0 t < .
By induction, we get
w n + 1 w n , 0 t < , n = 0 , 1 , 2 , .
(3.1)

Thus, there exists w P a ¯ such that w n w as n . Applying the continuity of T and w n + 1 = T w n , we get that T w = w .

Let v 0 ( t ) = 0 , 0 t < ; then v 0 ( t ) P a ¯ . Let v 1 = T v 0 , v 2 = T 2 v 0 ; then by Lemma 2.6, we know that v 1 P a ¯ and v 2 P a ¯ . We denote v n + 1 = T v n = T n v 0 , n = 0 , 1 , 2 ,  . Since T : P a ¯ P a ¯ , we have v n T ( P a ¯ ) P a ¯ , n = 1 , 2 ,  . It follows from the complete continuity of T that { v n } n = 1 is a sequentially compact set.

Since v 1 = T v 0 P a ¯ , we have
v 1 ( t ) = ( T v 0 ) ( t ) = ( T 0 ) ( t ) 0 , 0 t < .
So, we have
v 2 ( t ) = ( T v 1 ) ( t ) ( T v 0 ) ( t ) = v 1 ( t ) , 0 t < .
By induction, we get
v n + 1 v n , 0 t < , n = 0 , 1 , 2 , .
(3.2)

Thus, there exists v P a ¯ such that v n v as n . Applying the continuity of T and v n + 1 = T v n , we get T v = v .

Since w 0 ( t ) = ( a 2 + λ Γ ( α ) ) t α 1 v 0 = 0 , and the operator T is increasing, then
w 1 = T w 0 T v 0 = v 1 ,
by induction, we have
w n v n , 0 t < , n = 0 , 1 , 2 , .
Together with (3.1), (3.2), we obtain
v 0 v 1 v n v n + 1 w n + 1 w n w 0 .

If f ( t , 0 ) 0 , 0 t < , then the zero function is not the solution of fractional boundary value problem (1.1). Thus, v is a positive solution of fractional boundary value problem (1.1).

It is well known that each fixed point of T in P is a solution of fractional boundary value problem (1.1). Hence, w and v are two positive solutions on [ 0 , ) of fractional boundary value problem (1.1), satisfying 0 < T v 0 v w . □

4 An example

Example 4.1 Consider the boundary value problem of the fractional differential equation.
{ D 0 + 3 2 u ( t ) + e t f ( t , u ( t ) ) = 0 , t ( 0 , ) , u ( 0 ) = 0 , lim t D 0 + 1 2 u ( t ) = 0 ,
(4.1)
where
f ( t , u ( t ) ) = { | cos t | + ( u 1 + t ) 3 , 0 t < , 0 u 2 ; | cos t | + ( 2 1 + t ) 3 , 0 t < , u 2 .
In this case, α = 3 2 , q ( t ) = e t , λ = 0 , Γ ( 3 2 ) 0.886 . It is clear that (H1)-(H3) hold. Select a = 30 , by direct calculation, we can obtain
0 q ( s ) d s = 0 e s d s = 1 , f ( t , ( 1 + t 1 2 ) u ) 1 + 2 3 = 9 < a Γ ( α ) 2 0 q ( s ) d s 30 × 0.886 2 f ( t , ( 1 + t 1 2 ) u ) = 13.29 , t [ 0 , ) , u [ 0 , 30 ] .

The conditions in Theorem 3.1 are all satisfied. Therefore, fractional boundary value problem (4.1) has two positive solutions.

Declarations

Acknowledgements

This work is supported by the National Nature Science Foundation of P.R. China (10871063), partially supported by the National Nature Science Foundation of P.R. China (61170320), the Foundation of Guangdong Natural Science (S2011040002981), and the Nature Science Foundation of Guangdong Medical College (B2012053).

Authors’ Affiliations

(1)
Department of Mathematics, Hunan Normal University, Changsha, Hunan, 410081, China
(2)
Department of Information Engineering, Guangdong Medical College, Dongguan, Guangdong, 523808, China

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© Xie et al.; licensee Springer 2013

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