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Positive solution for a class of singular semipositone fractional differential equations with integral boundary conditions

Abstract

In this article, by employing a fixed point theorem in cones, we investigate the existence of a positive solution for a class of singular semipositone fractional differential equations with integral boundary conditions. We also obtain some relations between the solution and Green’s function.

MSC: 26A33, 34B15, 34B16, 34G20.

1 Introduction

In this article, we consider the existence of a positive solution for the following singular semipositone fractional differential equations:

{ D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = λ 0 η u ( s ) d s ,
(1)

where fC[(0,1)×[0,+),(,+)], 3<α4, 0<η1, 0 λ η α α <1, D 0 + α is the standard Riemann-Liouville derivative, f(t,u) may be singular at t=0 and/or t=1. Since the nonlinearity f(t,x) may change sign, the problem studied in this paper is called the semipositone problem in the literature which arises naturally in chemical reactor theory. Up to now, much attention has been attached to the existence of positive solutions for semipositone differential equations and the system of differential equations; see [111] and references therein to name a few.

Boundary value problems with integral boundary conditions for ordinary differential equations arise in different fields of applied mathematics and physics such as heat conduction, chemical engineering, underground water flow, thermo-elasticity, and plasma physics. Moreover, boundary value problems with integral conditions constitute a very interesting and important class of problems. They include two-point, three-point, multi-point, and nonlocal boundary value problems as special cases, which have received much attention from many authors. For boundary value problems with integral boundary conditions and comments on their importance, we refer the reader to the papers by Gallardo [12], Karakostas and Tsamatos [13], Lomtatidze and Malaguti [14], and the references therein.

On the other hand, fractional differential equations have been of great interest for many researchers recently. This is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in various fields of science and engineering such as control, porous media, electromagnetic, and other fields. For an extensive collection of such results, we refer the readers to the monographs by Samko et al.[15], Podlubny [16] and Kilbas et al.[17]. For the case where α is an integer, a lot of work has been done dealing with local and nonlocal boundary value problems. For example, in [18] Webb studied the n th-order nonlocal BVP

{ u ( n ) ( t ) + a ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d A ( s ) ,

where a(t) can have singularities, and the nonlinearity f satisfies Carathéodory conditions. Under weak assumptions, Webb obtained sharp results on the existence of positive solutions under a suitable condition on f. In [19] Hao et al. consider the n th-order singular nonlocal BVP

{ u ( n ) ( t ) + λ a ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d A ( s ) ,

where λ>0 is a parameter, a may be singular at t=0 and/or t=1, f(t,x) may also have singularity at x=0.

In two recent papers [20] and [21], by means of the fixed point theory and fixed point index theory, the authors investigated the existence and multiplicity of positive solutions for the following two kinds of fractional differential equations with integral boundary value problems:

{ D 0 + α u ( t ) + q ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d A ( s ) ,

and

{ D α C u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = λ 0 1 u ( s ) d s ,

where 0<λ<2, D 0 + α and D α C are the standard Riemann-Liouville derivative and the Caputo fractional derivative, respectively.

To the author’s knowledge, there are few papers in the literature to consider fractional differential equations with integral boundary value conditions. Motivated by above papers, the purpose of this article is to investigate the existence of positive solutions for the more general fractional differential equations BVP (1). Firstly, we derive corresponding Green’s function known as fractional Green’s function and argue its positivity. Then a fixed point theorem is used to obtain the existence of positive solutions for BVP (1). We also obtain some relations between the solution and Green’s function. From the example given in Section 4, we know that λ in this article may be greater than 2 and η may take the value 1. Therefore, compared with that in [21], BVP (1) considered in this article has a more general form.

The rest of this article is organized as follows. In Section 2, we give some preliminaries and lemmas. The main result is formulated in Section 3, and an example is worked out in Section 4 to illustrate how to use our main result.

2 Preliminaries and several lemmas

Let E=C[0,1], u= max 0 t 1 |u(t)|, then (E,) is a Banach space. Denote I=[0,1], J=(0,1), R + =[0,+).

For the reader’s convenience, we present some necessary definitions from fractional calculus theory and lemmas. They can be found in the recent literature; see [1417].

Definition 2.1 The Riemann-Liouville fractional integral of order α>0 of a function y:(0,)R is given by

I 0 + α y(t)= 1 Γ ( α ) 0 t ( t s ) α 1 y(s)ds

provided the right-hand side is pointwise defined on (0,).

Definition 2.2 The Riemann-Liouville fractional derivative of order α>0 of a continuous function y:(0,)R is given by

D 0 + α y(t)= 1 Γ ( n α ) ( d d t ) n 0 t y ( s ) ( t s ) α n + 1 y(s)ds,

where n=[α]+1, [α] denotes the integer part of the number α, provided that the right-hand side is pointwise defined on (0,).

From the definition of the Riemann-Liouville derivative, we can obtain the statement.

Lemma 2.1 ([17])

Letα>0. If we assumeuC(0,1)L(0,1), then the fractional differential equation

D 0 + α u(t)=0

hasu(t)= C 1 t α 1 + C 2 t α 2 ++ C N t α N , C i R, i=1,2,,N, as unique solutions, where N is the smallest integer greater than or equal to α.

Lemma 2.2 ([17])

Assume thatuC(0,1)L(0,1)with a fractional derivative of orderα>0that belongs toC(0,1)L(0,1).

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=1,2,,N, where N is the smallest integer greater than or equal to α.

In the following, we present Green’s function of the fractional differential equation boundary value problem.

Lemma 2.3 GivenyC[0,1], the problem

{ D 0 + α u ( t ) + y ( t ) = 0 , u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = λ 0 η u ( s ) d s ,
(2)

where0<t<1, 3<α4, 0<η1, 0 λ η α α <1, is equivalent to

u(t)= 0 1 G(t,s)y(s)ds,

where

G(t,s)={ t α 1 ( 1 s ) α 1 λ α ( η s ) α t α 1 ( 1 λ α η α ) ( t s ) α 1 p ( 0 ) Γ ( α ) , 0 s t 1 , s η ; t α 1 ( 1 s ) α 1 ( 1 λ α η α ) ( t s ) α 1 p ( 0 ) Γ ( α ) , 0 η s t 1 ; t α 1 ( 1 s ) α 1 λ α ( η s ) α t α 1 p ( 0 ) Γ ( α ) , 0 t s η 1 ; t α 1 ( 1 s ) α 1 p ( 0 ) Γ ( α ) , 0 t s 1 , η s .
(3)

Here, p(s):=1 λ η α α (1s), G(t,s)is called the Green function of BVP (2). Obviously, G(t,s)is continuous on[0,1]×[0,1].

Proof We may apply Lemma 2.2 to reduce (2) to an equivalent integral equation

u(t)= I 0 + α y(t)+ C 1 t α 1 + C 2 t α 2 + C 3 t α 3 + C 4 t α 4 ,

for some C 1 , C 2 , C 3 , C 4 R. Consequently, the general solution of (2) is

u(t)= 1 Γ ( α ) 0 t ( t s ) α 1 y(s)ds+ C 1 t α 1 C 2 t α 2 + C 3 t α 3 + C 4 t α 4 .

By u(0)= u (0)= u (0)=0, one gets that C 2 = C 3 = C 4 =0. On the other hand, u(1)=λ 0 η u(s)ds combining with

yields

C 1 = 0 1 ( 1 s ) α 1 Γ ( α ) ( 1 λ η α α ) y(s)dsλ 0 η ( η s ) α α Γ ( α ) ( 1 λ η α α ) y(s)ds.

Therefore, the unique solution of the problem (2) is

u ( t ) = 0 t ( t s ) α 1 Γ ( α ) y ( s ) d s + 1 ( 1 λ η α α ) 0 1 ( 1 s ) α 1 t α 1 Γ ( α ) y ( s ) d s 1 ( 1 λ η α α ) 0 η λ α ( η s ) α t α 1 Γ ( α ) y ( s ) d s .

For tη, one has

u ( t ) = 0 t ( t s ) α 1 Γ ( α ) y ( s ) d s + 1 ( 1 λ α η α ) [ ( 0 t + t η + η 1 ) ( 1 s ) α 1 t α 1 Γ ( α ) y ( s ) d s ] λ ( 1 λ α η α ) [ ( 0 t + t η ) 1 α ( η s ) α t α 1 Γ ( α ) y ( s ) d s ] = 0 t t α 1 ( 1 s ) α 1 λ α ( η s ) α t α 1 ( 1 λ α η α ) ( t s ) α 1 ( 1 λ α η α ) Γ ( α ) y ( s ) d s + t η t α 1 ( 1 s ) α 1 λ α ( η s ) α t α 1 ( 1 λ α η α ) Γ ( α ) y ( s ) d s + η 1 t α 1 ( 1 s ) α 1 ( 1 λ α η α ) Γ ( α ) y ( s ) d s = 0 1 G ( t , s ) y ( s ) d s .

For tη, one has

u ( t ) = ( 0 η + η t ) ( t s ) α 1 Γ ( α ) y ( s ) d s + 1 ( 1 λ α η α ) [ ( 0 η + η t + t 1 ) ( 1 s ) α 1 t α 1 Γ ( α ) y ( s ) d s ] λ ( 1 λ α η α ) 0 η 1 α ( η s ) α t α 1 Γ ( α ) y ( s ) d s = 0 η t α 1 ( 1 s ) α 1 λ α ( η s ) α t α 1 ( 1 λ α η α ) ( t s ) α 1 ( 1 λ α η α ) Γ ( α ) y ( s ) d s + η t t α 1 ( 1 s ) α 1 ( 1 λ α η α ) ( t s ) α 1 ( 1 λ α η α ) Γ ( α ) y ( s ) d s + t 1 t α 1 ( 1 s ) α 1 ( 1 λ α η α ) Γ ( α ) y ( s ) d s = 0 1 G ( t , s ) y ( s ) d s .

The proof is complete. □

Lemma 2.4 The functionG(t,s)defined by (3) satisfies

(a1) G(t,s) m 1 e(t)s ( 1 s ) α 1 , t,s[0,1];

(a2) G(t,s) M 1 e(t) ( 1 s ) α 1 , t,s[0,1];

(a3) G(t,s) M 1 s ( 1 s ) α 1 , t,s[0,1];

(a4) p(s)>0, andp(s)is not decreasing on[0,1];

(a5) G(t,s)>0, t,s(0,1),

where m 1 = 1 p ( 0 ) Γ ( α ) p ( 0 ) , M 1 = α 1 Γ ( α ) + 4 λ α η α 1 p ( 0 ) Γ ( α ) , e(t)= t α 1 .

Proof For st, sη,

For ηst,

For tsη,

For ηs, ts,

From above, (a1), (a2), (a3), (a5) are complete. Clearly, (a4) is true. The proof is complete. □

Throughout this article, we adopt the following conditions.

(H1) fC[(0,1)×[0,+),(,+)] and there exist a,bC[J, R + ], hC[ R + , R + ], g(t,u)[J× R + , R + ] such that

g(t,u)b(t)f(t,u)a(t)h(u),tJ,u R + ;

(H2) There exists [a,b]I such that lim u + g ( t , u ) u =+ uniformly for t[a,b];

(H3) There exists r>0 such that

0 1 ( 1 s ) α 1 b ( s ) d s < m 1 M 1 2 r , M 1 0 1 ( 1 s ) α 1 ( a ( s ) + b ( s ) ) d s < r max 0 u r { h ( u ) , 1 } .

Let

(4)

Obviously, Q is a cone in a Banach space E and (E,Q) is an ordering Banach space.

Let

x 0 (t)= 0 1 G(t,s)b(s)ds,tI,
(5)

where b(t) is defined as that in (H1). It follows from Lemma 2.4 and (H3) that

0 x 0 (t) M 1 e(t) 0 1 ( 1 s ) α 1 b(s)ds<+,tI.
(6)

So, x 0 P and it satisfies

{ D 0 + α x 0 ( t ) + x 0 ( t ) = 0 , 0 < t < 1 , x 0 ( 0 ) = x 0 ( 0 ) = x 0 ( 0 ) = 0 , x 0 ( 1 ) = λ 0 η x 0 ( s ) d s .
(7)

For any uQ{θ}, u(t) m 1 M 1 ue(t), tI. Consequently, by (6) and Lemma 2.4, we have

x 0 ( t ) M 1 e ( t ) 0 1 ( 1 s ) α 1 b ( s ) d s M 1 u m 1 M 1 e ( t ) 1 u m 1 M 1 0 1 ( 1 s ) α 1 b ( s ) d s u ( t ) u M 1 2 m 1 0 1 ( 1 s ) α 1 b ( s ) d s .
(8)

For any kE, denote

[ k ( t ) ] + ={ k ( t ) , k ( t ) 0 , 0 , k ( t ) < 0 .

We define an operator A as follows:

(Au)(t)= 0 1 G(t,s) ( f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ) ds,uP.
(9)

Lemma 2.5 Suppose that ( H 1 )-( H 3 ) hold. ThenA:QQis completely continuous.

Proof For any uP, it is clear that [ u ( s ) x 0 ( s ) ] + u(s)u. By (H1), we get

f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) a ( s ) h ( [ u ( s ) x 0 ( s ) ] + ) + b ( s ) max 0 r u { h ( r ) , 1 } ( a ( s ) + b ( s ) ) , s J .
(10)

By (10) and Lemma 2.4, we have

(11)

which together with (H3) means that operator A defined by (9) is well defined.

Now, we show that A:QQ.

For any uQ, by (H1) we have by (9) and Lemma 2.4 that

(Au)(t) M 1 0 1 s ( 1 s ) α 1 [ f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ] ds,tI,

which means that

Au M 1 0 1 s ( 1 s ) α 1 [ f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ] ds.
(12)

It follows from (12) and Lemma 2.4 that

( A u ) ( t ) = 0 1 G ( t , s ) ( f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ) d s m 1 e ( t ) 0 1 s ( 1 s ) α 1 [ f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ] d s = m 1 M 1 e ( t ) 0 1 M 1 s ( 1 s ) α 1 [ f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ] d s m 1 M 1 e ( t ) A u .

Thus, A maps Q into Q.

Finally, we prove that A maps Q into Q is completely continuous.

Let DQ be any bounded set. Then there exists a constant L 1 >0 such that u L 1 for any uD. Notice that [ u ( s ) x 0 ( s ) ] + u(s) L 1 , for any uD, sI, by (H3) and (11), we have

| ( A u ) ( t ) | M 1 0 1 ( 1 s ) α 1 ( a ( s ) + b ( s ) ) ds max 0 r L 1 { h ( r ) , 1 } <+.

Therefore, A(D) is uniformly bounded.

On the other hand, since G(t,s) is continuous on [0,1]×[0,1], it is uniformly continuous on [0,1]×[0,1] as well. Thus, for fixed sI and for any ε>0, there exists a constant δ>0 such that for any t 1 , t 2 [0,1] and | t 1 t 2 |<δ,

| G ( t 1 , s ) G ( t 2 , s ) | < ε max 0 r L 1 { h ( r ) , 1 } 0 1 ( a ( s ) + b ( s ) ) d s .
(13)

Therefore, for any xD, we get by (10) and (13)

| ( A u ) ( t 1 ) ( A u ) ( t 2 ) | 0 1 | G ( t 1 , s ) G ( t 2 , s ) | ( f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ) d s ε max 0 r L 1 { h ( r ) , 1 } 0 1 ( a ( s ) + b ( s ) ) d s 0 1 ( a ( s ) + b ( s ) ) d s max 0 r L 1 { h ( r ) , 1 } = ε ,

which implies that the operator A is equicontinuous. Thus, the Ascoli-Arzela theorem guarantees that A(D) is a relatively compact set.

Let u n , u 0 Q, u n u 0 (n). Then { u n } is bounded. Let L 2 =sup{ u n ,n=0,1,2,}, by (10), we get

f ( s , [ u n ( s ) x 0 ( s ) ] + ) +b(s) ( max 0 r L 2 { h ( r ) , 1 } ) ( a ( s ) + b ( s ) ) ,n=0,1,2,.
(14)

By (9), we have

(15)

It follows from (14), (15), (H1), (H3), and the Lebesgue dominated convergence theorem that A is continuous. Thus, we have proved the continuity of the operator A. This completes the complete continuity of A. □

To prove the main result, we need the following well-known fixed point theorem.

Lemma 2.6 (Fixed point theorem of cone expansion and compression of norm type [22])

Let Ω 1 and Ω 2 be two bounded open sets in a Banach space E such thatθ Ω 1 and Ω ¯ 1 Ω 2 ,A:P( Ω ¯ 2 Ω 1 )Pbe a completely continuous operator, where θ denotes the zero element of E and P a cone of E. Suppose that one of the two conditions holds:

  1. (i)

    Auu, uP Ω 1 ; Auu, uP Ω 2 ;

  2. (ii)

    Auu, uP Ω 1 ; Auu, uP Ω 2 .

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

3 Main result

Theorem 3.1 Assume that conditions ( H 1 )-( H 3 ) are satisfied. Then the singular semipositone BVP (1) has at least one positive solutionω(t). Furthermore, there exist two constantsM>m>0such that

me(t)ω(t)Me(t),tI.
(16)

Proof Firstly, we show that the operator A has a fixed point in Q. Let

Ω r = { u E | u < r } ,

where r is the same as that defined in (H3). For any u Ω r Q, by (10) and (12), we have that

( A u ) ( t ) = 0 1 G ( t , s ) ( f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ) d s M 1 0 1 s ( 1 s ) α 1 ( a ( s ) + b ( s ) ) d s max 0 u r { h ( u ) , 1 } < + , t J .

Therefore,

Au M 1 0 1 s ( 1 s ) α 1 ( a ( s ) + b ( s ) ) ds max 0 u r { h ( u ) , 1 } <r,

which together with (H3) implies that

Aur=u,u Ω r Q.
(17)

For [a,b] in (H2), it is clear that

e(t) a α 1 ,t[a,b].
(18)

By (H3), we know that there exists a natural number m 0 big enough such that

0 1 ( 1 s ) α 1 b(s)ds< m 0 m 0 + 1 m 1 M 1 2 r.
(19)

Choose

M> M 1 ( m 0 + 1 ) m 1 a α 1 min t [ a , b ] 0 1 G ( t , s ) d s .
(20)

By (H2), we know there exists R 1 >r such that

g(t,u)Mu,t[a,b],u R 1 .
(21)

Take

R>max { r , 1 , R 1 ( m 0 + 1 ) M 1 a α 1 m 1 } .
(22)

In the following, we are in a position to show that

Auu,u Ω R Q.
(23)

For any u Ω R Q, by (8) we get

x 0 (t) u ( t ) u M 1 2 m 1 0 1 ( 1 s ) α 1 b(s)ds,

which together with (18), (19), (22), and (H3) implies that

(24)

For u Ω R Q, t[a,b], it follows from (H1), (20), (21), (22), and (24) that

( A u ) ( t ) = 0 1 G ( t , s ) ( f ( s , [ u ( s ) x 0 ( s ) ] + ) + b ( s ) ) d s 0 1 G ( t , s ) g ( s , [ u ( s ) x 0 ( s ) ] + ) d s 0 1 G ( t , s ) M ( u ( s ) x 0 ( s ) ) d s M R m 0 + 1 m 1 M 1 a α 1 min t [ a , b ] 0 1 G ( t , s ) d s > R , t [ a , b ] .
(25)

By (25), we know that (23) holds. So, (17), (23), and Lemma 2.6 guarantee that A has at least one fixed point z 0 (t) in ( Ω ¯ R Ω r )Q and r z 0 R. Furthermore,

z 0 (t)= 0 1 G(t,s) ( f ( s , [ z 0 ( s ) x 0 ( s ) ] + ) + b ( s ) ) ds,tI.
(26)

By simple computation, we have that

{ D 0 + α z 0 ( t ) + f ( t , [ z 0 ( t ) x 0 ( t ) ] + ) + b ( t ) = 0 , 0 < t < 1 , z 0 ( 0 ) = z 0 ( 0 ) = z 0 ( 0 ) = 0 , z 0 ( 1 ) = λ 0 η z 0 ( s ) d s .
(27)

Secondly, we show BVP (1) has a positive solution. It follows from (8) and the fact z 0 r that

x 0 (t) z 0 ( t ) z 0 M 1 2 m 1 0 1 ( 1 s ) α 1 b(s)ds z 0 ( t ) r M 1 2 m 1 0 1 ( 1 s ) α 1 b(s)ds,

which combined with (19) implies that

z 0 (t) x 0 (t) ( 1 1 r M 1 2 m 1 0 1 ( 1 s ) α 1 b ( s ) d s ) z 0 (t) 1 m 0 + 1 z 0 (t)0.
(28)

By (27) and (28), we have

{ D 0 + α z 0 ( t ) + f ( t , [ z 0 ( t ) x 0 ( t ) ] ) + b ( t ) = 0 , 0 < t < 1 , z 0 ( 0 ) = z 0 ( 0 ) = z 0 ( 0 ) = 0 , z 0 ( 1 ) = λ 0 η z 0 ( s ) d s .
(29)

Let ω(t)= z 0 (t) x 0 (t), tI. It follows from (28) and z 0 Q that

ω(t) 1 m 0 + 1 z 0 (t) m 1 r ( m 0 + 1 ) M 1 e(t)>0,tJ.
(30)

By (7), (29), and (30), we obtain

{ D 0 + α ω ( t ) + f ( t , ω ( t ) ) = 0 , 0 < t < 1 , ω ( 0 ) = ω ( 0 ) = ω ( 0 ) = 0 , ω ( 1 ) = λ 0 η ω ( s ) d s .

Thus, we have proved that ω(t) is a positive solution for BVP (1).

Finally, we show that (16) holds. From (26) and Lemma 2.4, we know that

z 0 (t) M 1 e(t) 0 1 ( 1 s ) α 1 ( f ( s , [ z 0 ( s ) x 0 ( s ) ] + ) + b ( s ) ) ds,tI.
(31)

Since ω(t) z 0 (t), (30), and (31) mean that (16) holds for m= m 1 r ( m 0 + 1 ) M 1 and M= M 1 0 1 ( 1 s ) α 1 (f(s, [ z 0 ( s ) x 0 ( s ) ] + )+b(s))ds holds. This completes the proof of Theorem 3.1. □

4 Example

Consider the following singular semipositone fractional differential equations:

{ D 0 + 7 2 u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = 16 2 0 1 2 u ( s ) d s ,
(32)

where f(t,u)= t 2 8 1 t u 5 2 + t 1 4 20 ( 1 t ) (u u 1 2 3 4 ). It is clear (32) has the form of (1), where α= 7 2 , λ=16 2 , η= 1 2 . By simple computation, we know that 0< λ η α α 0.5714<1, p(0)0.4286. Let

a ( t ) = t 2 8 1 t + t 1 4 20 ( 1 t ) , b ( t ) = t 1 4 20 ( 1 t ) , g ( t , u ) = t 2 8 1 t u 5 2 , h ( u ) = u 5 2 + u u 1 2 + 1 4 .

Notice that

1u u 1 2 3 4 u u 1 2 + 1 4 ,

we have

t 1 4 20 ( 1 t ) t 1 4 20 ( 1 t ) ( u u 1 2 3 4 ) t 1 4 20 ( 1 t ) ( u u 1 2 + 1 4 ) .
(33)

It follows from the left side of (33) that

b(t)f(t,u)g(t,u).
(34)

Considering u u 1 2 + 1 4 0, we get

f(t,u) t 2 8 1 t u 5 2 + t 1 4 20 ( 1 t ) ( u u 1 2 + 1 4 ) a(t)h(u).
(35)

By (34) and (35) we know (H1) holds. Obviously, (H2) holds for [a,b]=[ 1 4 , 3 4 ].

Now, we check (H3). By simple computation, we have m 1 =0.4012, M 1 =3.9619, m 1 M 1 2 =0.0256, 0 1 ( 1 s ) α 1 b(s)ds0.0136, M 1 0 1 ( 1 s ) α 1 (a(s)+b(s))ds0.1245. Take r=1, then max 0 u r {h(u),1}1.2500, r max 0 u r { h ( u ) , 1 } 0.8000. Thus, (H3) is valid. It follows from Theorem 3.1 that BVP (32) has at least one positive solution.

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Acknowledgements

The author thanks the referee for his/her careful reading of the manuscript and useful suggestions. The project is supported financially by the Foundation for Outstanding Middle-Aged and Young Scientists of Shandong Province (Grant No. BS2010SF004), a Project of Shandong Province Higher Educational Science and Technology Program (Grant No. J10LA53, No. J11LA02), the China Postdoctoral Science Foundation (Grant No. 20110491154) and the National Natural Science Foundation of China (Grant No. 10971179).

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Correspondence to Xingqiu Zhang.

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Keywords

  • fractional differential equations
  • integral boundary value problem
  • positive solution
  • semipositone
  • cone