Open Access

Unbounded solution for a fractional boundary value problem

Advances in Difference Equations20142014:154

https://doi.org/10.1186/1687-1847-2014-154

Received: 7 January 2014

Accepted: 7 May 2014

Published: 23 May 2014

Abstract

This paper concerns the existence of unbounded positive solutions of a fractional boundary value problem on the half line. By means of the properties of the Green function and the compression and expansion fixed point theorem (Kwong in Fixed Point Theory Appl. 2008:164537, 2008), sufficient conditions are obtained to guarantee the existence of a solution to the posed problem.

MSC:26A33, 34B15, 34B27.

Keywords

integral condition Caputo derivative unbounded solution existence of solution Leray-Schauder nonlinear alternative Guo-Krasnosel’skii fixed point theorem

1 Introduction

In this paper, we investigate the existence of solutions for a fractional boundary value problem (P) on the half line:
{ D 0 + q c u ( t ) = f ( t , u ( t ) , u ( t ) ) , t > 0 , u ( 0 ) = u ( 0 ) = 0 , lim t c D 0 + q 1 u ( t ) = α u ( 1 ) ,
(P)

where f : [ 0 , [ × R × R R is a given function, 2 < q < 3 , α > 0 , D 0 + q c denotes the Caputo fractional derivative. Note that few papers in the literature dealing with fractional differential equations considered the nonlinearity f in (P) depending on the derivative of u.

Since many problems in the natural sciences require a notion of positivity (only non-negative densities, population sizes or probabilities make sense in real life), in the present study we discuss the existence of positive solutions for the problem (P). The proofs of the main results are based on the properties of the associated Green function, Leray-Schauder nonlinear alternative and Guo-Krasnosel’skii fixed point theorem on cone. Different methods are applied to investigate such boundary value problems, we can cite fixed point theory, topological degree methods, Mawhin theory, upper and lower solutions…; see [113].

Fractional boundary value problems on infinite intervals often appear in applied mathematics and physics. They can model some physical phenomena, such as the models of gas pressure in a semi-infinite porous medium; see [13]. The population growth model can also be characterized by a nonlinear fractional Volterra integrodifferential equation on the half line [14]. For more results on fractional differential equations in science and engineering and their applications we refer to [1517].

Fractional boundary value problems on infinite intervals have been investigated by many authors; see [1, 2, 4, 5, 8, 1113]. In [12], the authors proved the existence of unbounded solutions for the following nonlinear fractional boundary value problem:
{ D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , t > 0 , u ( 0 ) = 0 , lim t D 0 + α 1 u ( t ) = α u ( η ) ,

by using Leray-Schauder nonlinear alternative. Here 1 < α 2 , and D 0 + q denotes the Riemann-Liouville fractional derivative.

In [5] by means of fixed point theorem on cone, the authors discussed the existence of multiple positive solutions for m-point fractional boundary value problems with p-Laplacian operator on infinite interval.

Further, in [13], the authors studied the following second order nonlinear differential equation on the half line:
{ u ( t ) + q ( t ) f ( t , u ( t ) , u ( t ) ) = 0 , t > 0 , α u ( 0 ) β u ( 0 ) = 0 , u ( ) = u 0 .

Applying a fixed point theorem and the monotone iterative technique, they proved the existence of positive solution.

The organization of this paper is as follows. In Section 2, we provide necessary background and properties of the Green function. The existence result is established under some sufficient conditions on the nonlinear term f. Section 3 is devoted to the existence of positive solutions on a cone. We conclude the paper with some examples.

2 Existence results

For the convenience of the readers, we first present some useful definitions and fundamental facts of fractional calculus theory, which can be found in [15, 18].

Definition 1 The Riemann-Liouville fractional integral of order α of a function g is defined by
I a + α g ( t ) = 1 Γ ( α ) a t g ( s ) ( t s ) 1 α d s ,

where Γ ( α ) = 0 + e t t α 1 d t is the Gamma function, α > 0 .

Definition 2 The Caputo fractional derivative of order q of a function g is defined by
D a + q c g ( t ) = 1 Γ ( n q ) a t g ( n ) ( s ) ( t s ) q n + 1 d s ,

where n = [ q ] + 1 ( [ q ] is the entire part of q).

Lemma 3 For q > 0 , g C ( [ 0 , 1 ] ) , the homogeneous fractional differential equation D a + q c g ( t ) = 0 has a solution
g ( t ) = c 1 + c 2 t + c 3 t 2 + + c n t n 1 ,

where c i R , i = 0 , , n , and n = [ q ] + 1 .

Lemma 4 Let p , q 0 , f L 1 [ a , b ] . Then I 0 + p I 0 + q f ( t ) = I 0 + p + q f ( t ) = I 0 + q I 0 + p f ( t ) and D a + q c I 0 + q f ( t ) = f ( t ) , for all t [ a , b ] .

Lemma 5 Let α , β > 0 and n = [ α ] + 1 , then the following relations hold:
D a + α c t β 1 = Γ ( β ) Γ ( β α ) t β α 1 , β > n , D a + α c t k = 0 , k = 0 , 1 , 2 , , n 1 .

To prove the main results of this paper we need the following lemma.

Lemma 6 Let y C ( R + , R ) with 0 y ( s ) d s < , the linear nonhomogeneous boundary value problem
D a + q c u ( t ) = y ( t ) , t > 0 , u ( 0 ) = u ( 0 ) = 0 , lim t c D 0 + q 1 u ( t ) = α u ( 1 ) ,
has a unique solution
u ( t ) = 0 G ( t , s ) y ( s ) d s ,
where
G ( t , s ) = { ( t s ) q 1 Γ ( q ) + t α t Γ ( q ) ( 1 s ) q 1 , s min ( t , 1 ) , ( t s ) q 1 Γ ( q ) + t α , 1 s t , t α t Γ ( q ) ( 1 s ) q 1 , t s 1 , t α , s max ( t , 1 ) .
Proof By Lemmas 3 and 4, we obtain
u ( t ) = I 0 + q y ( t ) + a + b t + c t 2 .
The boundary conditions u ( 0 ) = u ( 0 ) = 0 , imply that
u ( t ) = I 0 + q y ( t ) + b t ;
applying Lemma 5 and the condition lim t c D 0 + q 1 u ( t ) = α u ( 1 ) , we obtain
lim t c D 0 + q 1 u ( t ) = lim t ( I 0 + y ( t ) + c D 0 + q 1 ( b t ) ) = lim t I 0 + y ( t ) = 0 y ( s ) d s , u ( 1 ) = I 0 + q y ( 1 ) + b ,
consequently
b = 1 α 0 y ( s ) d s I 0 + q y ( 1 ) = 1 α 0 y ( s ) d s 1 Γ ( q ) 0 1 ( 1 s ) q 1 y ( s ) d s ,
substituting b by its value, it yields
u ( t ) = 1 Γ ( q ) [ 0 t ( t s ) q 1 y ( s ) d s t 0 1 ( 1 s ) q 1 y ( s ) d s ] + t α 0 y ( s ) d s .

The proof is complete. □

Lemma 7 Assume that 0 < α Γ ( q ) , then for all s , t 0 we have
0 G ( t , s ) 1 + t q 1 2 α , 0 G t ( t , s ) 1 + t q 1 4 α Γ ( q 1 ) .
Proof Simple computations give
G t ( t , s ) = { ( t s ) q 2 Γ ( q 1 ) + 1 α 1 Γ ( q ) ( 1 s ) q 1 , s min ( t , 1 ) , ( t s ) q 2 Γ ( q 1 ) + 1 α , 1 s t , 1 α 1 Γ ( q ) ( 1 s ) q 1 , t s 1 , 1 α , s max ( t , 1 ) .
Let us consider the case s min ( t , 1 ) , then we get
G ( t , s ) t α t Γ ( q ) ( 1 s ) q 1 t ( Γ ( q ) α α Γ ( q ) ) 0 , G t ( t , s ) 1 α 1 Γ ( q ) ( 1 s ) q 1 0 .
Firstly if s t 1 , then
G ( t , s ) 1 + t q 1 ( t s ) q 1 + t α ( 1 + t q 1 ) 2 α , G t ( t , s ) 1 + t q 1 α ( t s ) q 2 + Γ ( q 1 ) α ( 1 + t q 1 ) Γ ( q 1 ) α + Γ ( q 1 ) α Γ ( q 1 ) ( 1 + t q 1 ) 4 α Γ ( q 1 ) .
Secondly if s 1 t , then
G ( t , s ) 1 + t q 1 ( t s ) q 1 + t α ( 1 + t q 1 ) t q 1 + t α ( 1 + t q 1 ) 2 t q 1 α ( 1 + t q 1 ) 2 α , G t ( t , s ) 1 + t q 1 α ( t s ) q 2 + Γ ( q 1 ) α ( 1 + t q 1 ) Γ ( q 1 ) 4 α Γ ( q 1 ) .

Applying the same techniques to the other cases, the conclusion follows. □

In this paper, we will use the Banach space E defined by
E = { u C 1 ( R + , R ) , lim t u ( t ) 1 + t q 1 < , lim t u ( t ) 1 + t q 1 < }

and equipped with the norm u = max ( u , u ) , where u = sup t 0 | u ( t ) | 1 + t q 1 and R + = [ 0 , [ .

Define the integral operator T : E E by
T u ( t ) = 0 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s ,

so we have transformed the problem (P) to a Hammerstein integral equation by using the Green function.

Lemma 8 The function u E is solution of the boundary value problem (P) if and only if T u ( t ) = u ( t ) , for all t R + .

From this we see that to solve the problem (P) it remains to prove that the map T has a fixed point in E. Since the Arzela-Ascoli theorem cannot be applied in this situation, then, to prove that T is completely continuous, we need the following compactness criterion:

Lemma 9 [19]

Let V = { u C , u < l ,  where  l > 0 } , V ( t ) = { u ( t ) 1 + t q 1 , u V } , V ( t ) = { u ( t ) 1 + t q 1 , u V } . V is relatively compact in E, if V ( t ) and V ( t ) are both equicontinuous on any finite subinterval of R + and equiconvergent at ∞, that is for any ε > 0 , there exists η = η ( ε ) > 0 such that
| u ( t 1 ) 1 + t 1 q 1 u ( t 2 ) 1 + t 2 q 1 | < ε , | u ( t 1 ) 1 + t 1 q 1 u ( t 2 ) 1 + t 2 q 1 | < ε ,

u V , t 1 , t 2 η (uniformly according to u).

We recall that a continuous mapping F from a subset M of a normed space X into another normed space Y is called completely continuous iff F maps bounded subset of M into relatively compact subset of Y.

Lemma 10 Assume that f C ( R + × R × R , R ) , f ( t , 0 , 0 ) 0 on any subinterval of R + and there exist non-negative functions h , k L 1 ( R + , R + ) and ψ 1 , ψ 2 C ( R , R + ) nondecreasing on R + , such that
| f ( t , ( 1 + t q 1 ) x , ( 1 + t q 1 ) x ¯ ) | k ( t ) ψ 1 ( | x | ) + h ( t ) ψ 2 ( | x ¯ | ) , ( t , x , x ¯ ) R + × R 2 ,
(2.1)

then T is completely continuous. (Here R + = ] 0 , [ .)

Proof The proof will be done in some steps.

Step 1: T is continuous. Let ( u n ) n N E be a convergent sequence to u in E. Let r 1 > max ( u , sup u n ) and r 2 > max ( u , sup u n ) , then we obtain with the help of Lemma 7, hypothesis (2.1) and some elementary inequalities
0 G ( t , s ) 1 + t q 1 | f ( s , u ( s ) , u ( s ) ) f ( s , u n ( s ) , u n ( s ) ) | d s 2 α 0 | f ( s , ( 1 + t q 1 ) u ( s ) 1 + t q 1 , ( 1 + t q 1 ) u ( s ) 1 + t q 1 ) | | f ( s , ( 1 + t q 1 ) u n ( s ) 1 + t q 1 , ( 1 + t q 1 ) u n ( s ) 1 + t q 1 ) | d s 2 α 0 ( | f ( s , ( 1 + t q 1 ) u ( s ) 1 + t q 1 , ( 1 + t q 1 ) u ( s ) 1 + t q 1 ) | + | f ( s , ( 1 + t q 1 ) u n ( s ) 1 + t q 1 , ( 1 + t q 1 ) u n ( s ) 1 + t q 1 ) | ) d s 2 α 0 k ( s ) ψ 1 ( u ( s ) 1 + t q 1 ) + h ( s ) ψ 2 ( u ( s ) 1 + t q 1 ) d s + 2 α 0 k ( s ) ψ 1 ( u n ( s ) 1 + t q 1 ) + h ( s ) ψ 2 ( u n ( s ) 1 + t q 1 ) d s 4 α ( ψ 1 ( r 1 ) 0 k ( s ) d s + ψ 2 ( r 2 ) 0 h ( s ) d s ) < .
Using similar techniques we prove that
0 G t ( t , s ) 1 + t q 1 | f ( s , u ( s ) , u ( s ) ) f ( s , u n ( s ) , u n ( s ) ) | d s < ,
hence the integrals are convergent. With the help of Lebesgue dominated convergence theorem and the fact that f is continuous we get
T u n T u = sup t 0 0 G ( t , s ) 1 + t q 1 | f ( s , u ( s ) , u ( s ) ) f ( s , u n ( s ) , u n ( s ) ) | d s 0 as  n , T u n T u = sup t 0 0 G t ( t , s ) 1 + t q 1 | f ( s , u ( s ) , u ( s ) ) f ( s , u n ( s ) , u n ( s ) ) | d s 0 as  n ,
therefore
T u n T u 0 , as  n .
Step 2: T is relatively compact. Let B r = { u E , u < r } , first let us show that T B r is uniformly bounded. Let u B r , taking (2.1) into account and the fact that ψ 1 and ψ 2 are nondecreasing on R + , it yields
| T u ( t ) | 1 + t q 1 2 α 0 | f ( s , u ( s ) , u ( s ) ) | d s = 2 α 0 | f ( s , ( 1 + s q 1 ) u ( s ) 1 + s q 1 , ( 1 + s q 1 ) u ( s ) 1 + s q 1 ) | d s 2 α ( ψ 1 ( r ) 0 k ( s ) d s + ψ 2 ( r ) 0 h ( s ) d s ) .
Consequently
T u 2 α ( ψ 1 ( r ) 0 k ( s ) d s + ψ 2 ( r ) 0 h ( s ) d s ) .
(2.2)
Similarly, we prove that
T u 4 α Γ ( q 1 ) ( ψ 1 ( r ) 0 k ( s ) d s + ψ 2 ( r ) 0 h ( s ) d s ) ;
(2.3)
this along with (2.3) yields
T u 4 α Γ ( q 1 ) ( ψ 1 ( r ) 0 k ( s ) d s + ψ 2 ( r ) 0 h ( s ) d s )
(2.4)

thus T B r is uniformly bounded.

Next, we show that T B r is equicontinuous on any compact interval of R + . Let u B r , t 1 , t 2 [ a , b ] , 0 a < b < , t 1 t 2 we have
| T u ( t 2 ) 1 + t 2 q 1 T u ( t 1 ) 1 + t 1 q 1 | 0 | G ( t 2 , s ) 1 + t 2 q 1 G ( t 1 , s ) 1 + t 1 q 1 | | f ( s , u ( s ) , u ( s ) ) | d s 0 | G ( t 2 , s ) G ( t 1 , s ) | 1 + t 2 q 1 | f ( s , u ( s ) , u ( s ) ) | d s + 0 G ( t 1 , s ) ( t 2 q 1 t 1 q 1 ) ( 1 + t 1 q 1 ) ( 1 + t 2 q 1 ) | f ( s , u ( s ) , u ( s ) ) | d s 2 α ( ( t 2 t 1 ) 1 + t 2 q 1 + ( t 2 q 1 t 1 q 1 ) ( 1 + t 1 q 1 ) ( 1 + t 2 q 1 ) ) ( ψ 1 ( r ) 0 k ( s ) d s + ψ 2 ( r ) 0 h ( s ) d s ) ,
which approaches zero uniformly when t 1 t 2 . On the other hand we have
| T u ( t 2 ) 1 + t 2 q 1 T u ( t 1 ) 1 + t 1 q 1 | 0 | G t ( t 2 , s ) 1 + t 2 q 1 G t ( t 1 , s ) 1 + t 1 q 1 | | f ( s , u ( s ) , u ( s ) ) | d s 0 | G t ( t 2 , s ) G t ( t 1 , s ) | 1 + t 2 q 1 | f ( s , u ( s ) , u ( s ) ) | d s + 0 G t ( t 1 , s ) ( t 2 q 1 t 1 q 1 ) ( 1 + t 1 q 1 ) ( 1 + t 2 q 1 ) | f ( s , u ( s ) , u ( s ) ) | d s .
(2.5)
Let us estimate the second integral on the right hand side of the inequality (2.5):
0 G t ( t 1 , s ) ( t 2 q 1 t 1 q 1 ) ( 1 + t 1 q 1 ) ( 1 + t 2 q 1 ) | f ( s , u ( s ) , u ( s ) ) | d s 4 α Γ ( q 1 ) ( t 2 q 1 t 1 q 1 ) ( 1 + t 1 q 1 ) ( 1 + t 2 q 1 ) ( ψ 1 ( r ) 0 k ( s ) d s + ψ 2 ( r ) 0 h ( s ) d s ) ,

which approaches zero uniformly when t 1 t 2 . Now we analyze the first integral on the right hand side of inequality (2.5) in different cases when the compact [ a , b ] contains 1 or not.

If t 1 < t 2 1 , then
0 | G t ( t 2 , s ) G t ( t 1 , s ) | 1 + t 2 q 1 | f ( s , u ( s ) , u ( s ) ) | d s = 0 t 1 | ( t 2 s ) q 2 ( t 1 s ) q 2 | Γ ( q 1 ) ( 1 + t 2 q 1 ) | f ( s , u ( s ) , u ( s ) ) | d s + t 1 t 2 ( t 2 s ) q 2 Γ ( q 1 ) ( 1 + t 2 q 1 ) | f ( s , u ( s ) , u ( s ) ) | d s ( q 2 ) ( t 2 t 1 ) Γ ( q 1 ) ( 1 + t 2 q 1 ) ( ψ 1 ( r ) 0 t 1 k ( s ) d s + ψ 2 ( r ) 0 t 1 h ( s ) d s ) + 1 Γ ( q 1 ) ( 1 + t 2 q 1 ) ( ψ 1 ( r ) t 1 t 2 k ( s ) d s + ψ 2 ( r ) t 1 t 2 h ( s ) d s ) ,

which approaches zero uniformly when t 1 t 2 .

If t 1 < 1 t 2 , then
0 | G t ( t 2 , s ) G t ( t 1 , s ) | 1 + t 2 q 1 | f ( s , u ( s ) , u ( s ) ) | d s = 0 t 1 ( t 2 s ) q 2 ( t 1 s ) q 2 Γ ( q 1 ) ( 1 + t 2 q 1 ) | f ( s , u ( s ) , u ( s ) ) | d s + t 1 t 2 ( t 2 s ) q 2 Γ ( q 1 ) ( 1 + t 2 q 1 ) | f ( s , u ( s ) , u ( s ) ) | d s ( t 2 t 1 ) q 2 Γ ( q 1 ) ( 1 + t 2 q 1 ) ( ψ 1 ( r ) 0 t 1 k ( s ) d s + ψ 2 ( r ) 0 t 1 h ( s ) d s ) + ( t 2 ) q 2 Γ ( q 1 ) ( 1 + t 2 q 1 ) ( ψ 1 ( r ) t 1 t 2 k ( s ) d s + ψ 2 ( r ) t 1 t 2 h ( s ) d s ) 0 , uniformly as  t 1 t 2 .
If 1 t 1 < t 2 , then
0 | G t ( t 2 , s ) G t ( t 1 , s ) | 1 + t 2 q 1 | f ( s , u ( s ) , u ( s ) ) | d s = 0 t 1 ( t 2 s ) q 2 ( t 1 s ) q 2 Γ ( q 1 ) ( 1 + t 2 q 1 ) | f ( s , u ( s ) , u ( s ) ) | d s + t 1 t 2 ( t 2 s ) q 2 Γ ( q 1 ) ( 1 + t 2 q 1 ) | f ( s , u ( s ) , u ( s ) ) | d s ( t 2 t 1 ) q 2 Γ ( q 1 ) ( 1 + t 2 q 1 ) ( ψ 1 ( r ) 0 t 1 k ( s ) d s + ψ 2 ( r ) 0 t 1 h ( s ) d s ) + t 2 q 2 Γ ( q 1 ) ( 1 + t 2 q 1 ) ( ψ 1 ( r ) t 1 t 2 k ( s ) d s + ψ 2 ( r ) t 1 t 2 h ( s ) d s ) 0 , uniformly as  t 1 t 2 .

Thus T is equicontinuous on the compact [ a , b ] .

Step 3: T is equiconvergent at ∞. Since
0 | f ( s , u ( s ) , u ( s ) ) | d s ( ψ 1 ( r ) 0 k ( s ) d s + ψ 2 ( r ) 0 h ( s ) d s ) <
we have
lim t + | T u ( t ) 1 + t q 1 | 1 Γ ( q ) 0 | f ( s , u ( s ) , u ( s ) ) | d s < , lim t + | T u ( t ) 1 + t q 1 | = 0 ,

consequently T is equiconvergent at ∞. The proof is complete. □

Now, we can give an existence result.

Theorem 11 Assume that the hypotheses of Lemma  10 hold and that there exists r > 0 , such that
4 α Γ ( q 1 ) ( ψ 1 ( r ) 0 k ( s ) d s + ψ 2 ( r ) 0 h ( s ) d s ) < r .
(2.6)

Then the fractional boundary value problem (P) has at least one nontrivial solution u E .

To prove this theorem, we apply the Leray-Schauder nonlinear alternative.

Lemma 12 [20]

Let F be a Banach space and Ω a bounded open subset of F, 0 Ω . Let T : Ω ¯ F be a completely continuous operator. Then either there exist x Ω , λ > 1 , such that T ( x ) = λ x , or there exists a fixed point x Ω ¯ of T.

Proof of Theorem 11 From the proof of Lemma 10, we know that T is a completely continuous operator. Now we apply the nonlinear alternative of Leray-Schauder to prove that T has at least one nontrivial solution in E. Let u B r , such that u = λ T u , 0 < λ < 1 ; we get with the help of (2.4):
u = λ T u T u 4 α Γ ( q 1 ) ( ψ 1 ( r ) 0 k ( s ) d s + ψ 2 ( r ) 0 h ( s ) d s ) .
(2.7)
This together with (2.6) implies
r = u 4 α Γ ( q 1 ) ( ψ 1 ( r ) 0 k ( s ) d s + ψ 2 ( r ) 0 h ( s ) d s ) < r ,

which contradicts the fact that u B r . Lemma 12 allows one to conclude that the operator T has a fixed point u B ¯ r and then the fractional boundary value problem (P) has a nontrivial solution u E . The proof is complete. □

3 Positive solutions

To study the existence of positive solution of the problem (P), first, we will introduce a positive cone constituted of continuous positive functions or some suitable subset of it. Second, we will impose suitable assumptions on the nonlinear terms such that the hypotheses of the cone theorem are satisfied. Third, we will apply a fixed point theorem to conclude the existence of a positive solution in the annular region.

Definition 13 A function u is called positive solution of the problem (P) if u ( t ) 0 , t R + , and it satisfies the boundary conditions in (P).

Definition 14 A nonempty subset P of a Banach space E is called a cone if P is convex, closed, and satisfies the conditions
  1. (i)

    α x P for all x P and α R + ;

     
  2. (ii)

    x ; x P imply x = 0 .

     
Lemma 15 Assume that 0 < α < Γ ( q ) , then for 0 < τ 1 t τ 2 and s > 0 we have
G ( t , s ) 1 + t q 1 γ τ 1 1 + τ 2 q 1 and G t ( t , s ) 1 + t q 1 γ 1 + τ 2 q 1 ,

where γ = 1 α 1 Γ ( q ) > 0 .

Proof The proof is easy, we omit it. □

Let us make the following hypotheses on the nonlinear term f:
  1. (H)

    f C ( R + 3 , R + ) , 0 f ( t , u , v ) d t < , f ( t , u , v ) = a ( t ) g ( t , u , v ) , where a L 1 ( R + , R + ) , g C ( R + 3 , R + ) and 0 < τ 1 τ 2 a ( t ) d t < .

     
Define the cone K by
K = { u E , u ( t ) 0 , u ( t ) 0 , t 0 , min t [ τ 1 , τ 2 ] u ( t ) + u ( t ) 1 + t q 1 γ 1 u } ,

where γ 1 = γ α ( 1 + τ 1 ) Γ ( q 1 ) 4 ( 1 + τ 2 2 ) .

Lemma 16 We have T K K .

Proof Taking Lemma 7 into account, we get
T u ( t ) 1 + t q 1 2 α 0 f ( s , u ( s ) , u ( s ) ) d s , T u ( t ) 1 + t q 1 4 α Γ ( q 1 ) 0 f ( s , u ( s ) , u ( s ) ) d s ,
thus
T u 4 α Γ ( q 1 ) 0 f ( s , u ( s ) , u ( s ) ) d s .
Lemma 15 implies for all t [ τ 1 , τ 2 ]
T u ( t ) 1 + t q 1 γ τ 1 1 + τ 2 q 1 0 f ( s , u ( s ) , u ( s ) ) d s γ α τ 1 Γ ( q 1 ) 4 ( 1 + τ 2 q 1 ) T u , T u ( t ) 1 + t q 1 γ 1 + τ 2 q 1 0 f ( s , u ( s ) , u ( s ) ) d s γ α Γ ( q 1 ) 4 ( 1 + τ 2 q 1 ) T u .
Therefore,
min t [ τ 1 , τ 2 ] T u ( t ) + T u ( t ) 1 + t q 1 γ α ( 1 + τ 1 ) Γ ( q 1 ) 4 ( 1 + τ 2 q 1 ) T u .

 □

Let us introduce the following notation:
A δ = lim u + v δ sup t 0 g ( t , ( 1 + t q 1 ) u , ( 1 + t q 1 ) v ) u + v , A δ = lim u + v δ inf t 0 g ( t , ( 1 + t q 1 ) u , ( 1 + t q 1 ) v ) u + v ( δ = 0 +  or  + ) .

Theorem 17 Under the hypothesis (H) and if 0 < α < Γ ( q ) , then the fractional boundary value problem (P) has at least one positive solution in the case A 0 = 0 and A = .

To prove Theorem 17, we apply the well-known Guo-Krasnosel’skii fixed point theorem on cone.

Theorem 18 [21, 22]

Let E be a Banach space, and let K E , be a cone. Assume Ω 1 and Ω 2 are open subsets of E with 0 Ω 1 , Ω ¯ 1 Ω 2 and let
A : K ( Ω ¯ 2 Ω 1 ) K ,
be a completely continuous operator such that
  1. (i)

    (Expansive form) A u u , u K Ω 1 , and A u u , u K Ω 2 ; or

     
  2. (ii)

    (Compressive form) A u u , u K Ω 1 , and A u u , u K Ω 2 .

     

Then A has a fixed point in K ( Ω ¯ 2 Ω 1 ) .

Proof of Theorem 17 From A 0 = 0 , we deduce that for any ε > 0 , there exists R 1 > 0 , such that if 0 < u + v R 1 , then g ( t , ( 1 + t q 1 ) u , ( 1 + t q 1 ) v ) ε ( u + v ) , t 0 . Let Ω 1 = { u E , u < R 1 2 } and u K Ω 1 , by Lemma 7 we get
T u ( t ) 1 + t q 1 2 α 0 a ( s ) g ( s , u ( s ) , u ( s ) ) d s = 2 α 0 a ( s ) g ( s , ( 1 + s q 1 ) u ( s ) 1 + s q 1 , ( 1 + s q 1 ) u ( s ) 1 + s q 1 ) d s 2 ε α 0 a ( s ) ( u ( s ) 1 + s q 1 + u ( s ) 1 + s q 1 ) d s 4 ε α u 0 a ( s ) d s .
Similarly we obtain
T u ( t ) 1 + t q 1 8 ε α Γ ( q 1 ) u 0 a ( s ) d s ,
therefore,
T u 8 ε α Γ ( q 1 ) u 0 a ( s ) d s .

Choosing ε α Γ ( q 1 ) 8 0 a ( s ) d s , it yields T u u , for any u K Ω 1 .

Now, since A = , then for any M > 0 , there exists R > 0 , such that t 0
g ( t , ( 1 + t q 1 ) u , ( 1 + t q 1 ) v ) M ( u + v )
for u + v R . Let R 2 > max { R 1 2 , R γ 1 } and denote by Ω 2 = { u E : u < R 2 } . For u K Ω 2 and t [ τ 1 , τ 2 ] , we obtain
u ( t ) + u ( t ) 1 + t q 1 min [ τ 1 , τ 2 ] u ( t ) + u ( t ) 1 + t q 1 γ 1 u = γ 1 R 2 > R .
Using Lemma 15 and the fact that u K , we obtain for all t [ τ 1 , τ 2 ]
T u ( t ) 1 + t q 1 γ τ 1 1 + τ 2 q 1 0 a ( s ) g ( s , u ( s ) , u ( s ) ) d s = γ τ 1 1 + τ 2 q 1 0 a ( s ) g ( s , ( 1 + s q 1 ) u ( s ) 1 + s q 1 , ( 1 + s q 1 ) u ( s ) 1 + s q 1 ) d s γ τ 1 1 + τ 2 q 1 M 0 a ( s ) ( u ( s ) 1 + s q 1 + u ( s ) 1 + s q 1 ) d s γ τ 1 1 + τ 2 q 1 M τ 1 τ 2 a ( s ) u ( s ) + u ( s ) 1 + s q 1 d s γ τ 1 1 + τ 2 q 1 M min t [ τ 1 , τ 2 ] u ( t ) + u ( t ) 1 + t q 1 τ 1 τ 2 a ( s ) d s γ γ 1 τ 1 1 + τ 2 q 1 M u τ 1 τ 2 a ( s ) d s ,
similarly, we get
T u ( t ) 1 + t q 1 γ γ 1 1 + τ 2 q 1 M u τ 1 τ 2 a ( s ) d s .
Thus
T u max ( 1 , τ 1 ) γ γ 1 1 + τ 2 q 1 M u τ 1 τ 2 a ( s ) d s .
Let us choose M such that
M 1 + τ 2 q 1 γ γ 1 max ( 1 , τ 1 ) τ 1 τ 2 a ( s ) d s ,
then we obtain
T u u , u K Ω 2 .

The first statement of Theorem 18 implies that T has a fixed point in K ( Ω ¯ 2 Ω 1 ) . The proof is complete. □

Now define the function
A : R + R + , A ( r ) = max { sup t 0 g ( t , ( 1 + t q 1 ) u , ( 1 + t q 1 ) v ) , ( u + v ) [ 0 , r ] } , A 0 = lim r 0 + A ( r ) r , A = lim r + A ( r ) r .

It is proved in [9] that:

Lemma 19 If g is continuous then A 0 = A 0 and A = A .

Theorem 20 Under the hypothesis (H) and if 0 < α < Γ ( q ) and g is decreasing according to the both variables, then the problem (P) has at least one nontrivial positive solution in the cone K, in the case A 0 = + and A = 0 .

Proof Since A 0 = + , then for M 1 + τ 2 q 1 4 τ 1 γ 0 a ( s ) d s > 0 , there exists r 1 > 0 , such that if 0 < u + v r 1 , then for all t 0 , we have
g ( t , ( 1 + t q 1 ) u , ( 1 + t q 1 ) v ) M ( u + v ) .
Let Ω 1 = { u E , u < r 1 2 } , we should prove the second statement of Theorem 18. Suppose u 1 P Ω 1 , then
T u 1 T u 1 ( t ) ( 1 + t q 1 ) γ τ 1 1 + τ 2 q 1 0 a ( s ) g ( s , ( 1 + s q 1 ) u 1 ( s ) ( 1 + s q 1 ) , ( 1 + s q 1 ) u 1 ( s ) ( 1 + s q 1 ) ) d s .
Now from the fact that g is decreasing, we get
T u 1 γ τ 1 1 + τ 2 q 1 0 a ( s ) g ( s , ( 1 + s q 1 ) u 1 , ( 1 + s q 1 ) u 1 ) d s 2 γ τ 1 u 1 1 + τ 2 q 1 M 0 a ( s ) d s .
Thus
T u 1 u 1 on  P Ω 1 .
From A = 0 and Lemma 19, we get A = 0 , so for ε α Γ ( q 1 ) 8 0 a ( s ) d s , there exists R > 0 , such that if r R , then A ( r ) ε r . Let Ω 2 = { u E , u < r 2 } , where r 2 > max ( r 1 2 , R 2 ) , then Ω ¯ 1 Ω 2 . Suppose that u 2 P Ω 2 , then it yields
T u 2 ( t ) 1 + t q 1 2 α 0 a ( s ) g ( s , ( 1 + s q 1 ) u 2 ( s ) ( 1 + s q 1 ) , ( 1 + s q 1 ) u 2 ( s ) ( 1 + s q 1 ) ) d s 2 A ( 2 r 2 ) α 0 a ( s ) d s 4 r 2 ε α 0 a ( s ) d s .
On the other hand we get
T u 2 ( t ) 1 + t q 1 4 α Γ ( q 1 ) 0 a ( s ) g ( s , ( 1 + s q 1 ) u 2 ( s ) ( 1 + s q 1 ) , ( 1 + s q 1 ) u 2 ( s ) ( 1 + s q 1 ) ) d s 4 A ( 2 r 2 ) α Γ ( q 1 ) 0 a ( s ) d s 8 r 2 ε α Γ ( q 1 ) 0 a ( s ) d s .
Therefore
T u 2 8 r 2 ε α Γ ( q 1 ) 0 a ( s ) d s < r 2 = u 2

then from the second statement of Theorem 18, T has a fixed point in P ( Ω ¯ 2 Ω 1 ) . The proof is complete. □

Remark If 0 f ( t , u ( t ) , v ( t ) ) d t < , then every positive solution of the problem (P) is unbounded. Indeed
u ( t ) = 1 Γ ( q ) [ 0 t ( t s ) q 1 f ( s , u ( s ) , u ( s ) ) d s t 0 1 ( 1 s ) q 1 f ( s , u ( s ) , u ( s ) ) d s ] + t α 0 f ( s , u ( s ) , u ( s ) ) d s t ( 1 α 0 f ( s , u ( s ) , u ( s ) ) d s 1 Γ ( q ) 0 1 ( 1 s ) q 1 f ( s , u ( s ) , u ( s ) ) d s ) t ( 1 α 1 Γ ( q ) ) 0 f ( s , u ( s ) , u ( s ) ) d s ,

therefore our conclusion follows.

Example 21 Let us consider the problem (P) with
g ( t , u , v ) = ( u + v ) 2 ( 1 + t q 1 ) 2 , a ( t ) = 1 1 + t 2 , q = 5 2 , α = 1 ,

by direct calculation we obtain Γ ( 5 2 ) = 1.3293 > α , τ 1 τ 2 1 1 + s 2 d s = arctan τ 2 arctan τ 1 > 0 , A 0 = 0 and A = . Clearly hypothesis (H) is satisfied, so by Theorem 17 there exists at least one nontrivial positive solution in the cone K.

Example 22 Let us reconsider the above example with
g ( t , u , v ) = 1 + t q 1 1 + u + v .

Easily we check the hypothesis (H) and find that g is decreasing with respect to u and v. Furthermore we have the case A 0 = + and A = 0 . Thus by Theorem 20 there exists at least one nontrivial positive solution in the cone K.

Declarations

Acknowledgements

The authors express their sincere thanks to the referees for the careful and details reading of the manuscript and very helpful suggestions. This joint work was done when the first author visited University Putra Malaysia as a visiting scientist during 21st December-30th December 2013. Thus the authors gratefully acknowledge that this research was partially supported by the University Putra Malaysia under the GP-IBT Grant Scheme having project number GP-IBT/2013/9420100.

Authors’ Affiliations

(1)
Mathematics Department, Faculty of Sciences, Badji Mokhtar Annaba University
(2)
Department of Mathematics and Institute of Mathematical Research, Universiti Putra Malaysia (UPM)

References

  1. Agarwal RP, O’Regan D: Infinite Interval Problems for Differential, Difference and Integral Equations. Kluwer Academic, Dordrecht; 2001.View ArticleMATHGoogle Scholar
  2. Agarwal RP, O’Regan D: Boundary value problems of nonsingular type on the semi-infinity interval. Tohoku Math. J. 1999, 51: 391–397. 10.2748/tmj/1178224769MathSciNetView ArticleMATHGoogle Scholar
  3. Guezane-Lakoud A, Khaldi R: Solvability of a three-point fractional nonlinear boundary value problem. Differ. Equ. Dyn. Syst. 2012, 20(4):395–403. 10.1007/s12591-012-0125-7MathSciNetView ArticleMATHGoogle Scholar
  4. Liang S, Zhang J: Existence of three positive solutions of m -point boundary value problems for some nonlinear fractional differential equations on an infinite interval. Comput. Math. Appl. 2011, 61: 3343–3354. 10.1016/j.camwa.2011.04.018MathSciNetView ArticleMATHGoogle Scholar
  5. Liang S, Shi S: Existence of multiple positive solutions for m -point fractional boundary value problems with p -Laplacian operator on infinite interval. J. Appl. Math. Comput. 2012, 38: 687–707. 10.1007/s12190-011-0505-0MathSciNetView ArticleMATHGoogle Scholar
  6. Liu Y, Ahmad B, Agarwal RP: Existence of solutions for a coupled system of nonlinear fractional differential equations with fractional boundary conditions on the half-line. Adv. Differ. Equ. 2013., 2013: Article ID 46Google Scholar
  7. Nyamoradi N, Baleanu D, Agarwal RP: On a multipoint boundary value problem for a fractional order differential inclusion on an infinite interval. Adv. Math. Phys. 2013., 2013: Article ID 823961Google Scholar
  8. Su X, Zhang S: Unbounded solutions to a boundary value problem of fractional order on the half-line. Comput. Math. Appl. 2011, 61: 1079–1087. 10.1016/j.camwa.2010.12.058MathSciNetView ArticleMATHGoogle Scholar
  9. Wang H: On the numbers of positive solutions of a nonlinear systems. J. Math. Anal. Appl. 2003, 281: 287–306. 10.1016/S0022-247X(03)00100-8MathSciNetView ArticleGoogle Scholar
  10. Webb JRL, Infante G: Positive solutions of nonlocal boundary value problems involving integral conditions. NoDEA Nonlinear Differ. Equ. Appl. 2008, 15(1–2):45–67. 10.1007/s00030-007-4067-7MathSciNetView ArticleMATHGoogle Scholar
  11. Yan BQ, O’Regan D, Agarwal RP: Unbounded solutions for singular boundary value problems on the semi-infinite interval: upper and lower solutions and multiplicity. J. Comput. Appl. Math. 2006, 197: 365–386. 10.1016/j.cam.2005.11.010MathSciNetView ArticleMATHGoogle Scholar
  12. Zhao X, Ge W: Unbounded solutions for a fractional boundary value problems on the infinite interval. Acta Appl. Math. 2010, 109: 495–505. 10.1007/s10440-008-9329-9MathSciNetView ArticleMATHGoogle Scholar
  13. Zhang X: Existence of positive solutions for boundary value problems for second order nonlinear differential equations on the half line. Electron. J. Differ. Equ. 2009, 141: 1–10.MathSciNetGoogle Scholar
  14. Xu H: Analytical approximations for a population growth model with fractional order. Commun. Nonlinear Sci. Numer. Simul. 2009, 14(5):1978–1983. 10.1016/j.cnsns.2008.07.006View ArticleMATHGoogle Scholar
  15. Kilbas AA, Srivastava HM, Trujillo JJ North-Holland Mathematics Studies 204. In Theory and Applications of Fractional Differential Equations. Elsevier, Amsterdam; 2006.Google Scholar
  16. Mainardi F: Fractals and Fractional Calculus in Continuum Mechanics. Springer, New York; 1997.MATHGoogle Scholar
  17. Podlubny I: Fractional Differential Equations Mathematics in Sciences and Engineering. Academic Press, New York; 1999.MATHGoogle Scholar
  18. Lakshmikantham V, Vatsala AS: Basic theory of fractional differential equations. Nonlinear Anal. 2008, 69(8):2677–2682. 10.1016/j.na.2007.08.042MathSciNetView ArticleMATHGoogle Scholar
  19. Corduneanu C: Integral Equations and Stability of Feedback Systems. Academic Press, New York; 1973.MATHGoogle Scholar
  20. Deimling K: Nonlinear Functional Analysis. Springer, Berlin; 1985.View ArticleMATHGoogle Scholar
  21. Guo DJ, Lakshmikantham V Notes and Reports in Mathematics in Science and Engineering 5. In Nonlinear Problems in Abstract Cones. Academic Press, Boston; 1988.Google Scholar
  22. Kwong MK: On Krasnoselskii’s cone fixed point theorem. Fixed Point Theory Appl. 2008., 2008: Article ID 164537Google Scholar

Copyright

© Guezane-Lakoud and Kılıçman; licensee Springer. 2014

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.