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Positive solutions to boundary value problems of a high-order fractional differential equation in a Banach space

Abstract

In this paper, by using the fixed-point theorem in the cone of strict-set-contraction operators, we study a class of higher-order boundary value problems of nonlinear fractional differential equation in a Banach space. The sufficient conditions for the existence of at least two positive solutions is obtained. In addition, an example to illustrate the main results is given.

1 Introduction

Fractional calculus is a generalization of ordinary differentiation and integration to arbitrary noninteger order. The fractional differential equations have been of great interest recently. This is because of both the intensive development of the theory of fractional calculus itself and its numerous applications in various fields of science and engineering including fluid flow, rheology, control, electrochemistry, electromagnetic, porous media and probability, etc. (see [14]).

In recent years, the existence and uniqueness of solutions of the initial and boundary value problems for fractional equations have been extensively studied (see [416] and the references therein). But there are few works that deal with the existence of solutions of nonlinear fractional differential equations in Banach spaces; see [1724]. In [19], Hussein investigated the existence of pseudo solutions for the following nonlinear m-point boundary value problem of fractional type:

{ D 0 + q u ( t ) + a ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , u ( 1 ) = i = 1 m 2 ζ i u ( η i )

in a reflexive Banach space E, where D 0 + q is the pseudo fractional differential operator of order n1<qn, n2.

In [22], by the monotone iterative technique and the Mönch fixed point theorem, Lv et al. investigated the existence of a solution to the following Cauchy problem for the differential equation with fractional order in a real Banach space E:

D q C u(t)=f ( t , u ( t ) ) ,u(0)= u 0 ,

where D q C u(t) is the Caputo derivative of order 0<q<1.

By means of Darbo’s fixed point theorem, Su [23] established the existence result of solutions to the following boundary value problem of fractional differential equation on unbounded domain [0,+):

{ D 0 + q u ( t ) = f ( t , u ( t ) ) , t [ 0 , + ) , 1 < q 2 , u ( 0 ) = θ , D 0 + q 1 u ( ) = u

in a Banach space E. D 0 + q is the Riemann-Liouville fractional derivative.

Being directly inspired by [9, 15, 19, 22] but taking quite a different method from that in [15, 1923], we discuss in this paper the following high-order boundary value problem (BVP for short) in a Banach space E:

{ D 0 + q u ( t ) + f ( t , u , u , , u ( n 2 ) ) = θ , t ( 0 , 1 ) , q ( n 1 , n ] , u ( i ) ( 0 ) = 0 , 0 i n 3 , α u ( n 2 ) ( 0 ) β u ( n 1 ) ( 0 ) = θ , γ u ( n 2 ) ( 1 ) + δ u ( n 1 ) ( 1 ) = θ ,
(1.1)

where θ is the zero element of E, n2, α, β, γ and δ are nonnegative constants satisfying ρ 1 =αγ+αδ+βγ>0, and D 0 + q is the Caputo fractional derivative. Note that the nonlinear term f depends on u and its derivatives u , u ,, u ( n 2 ) .

The paper is organized as follows. In Section 2 we give some basic definitions in Riemann-Liouville fractional calculus and the Kuratowski noncompactness. In Section 3 we present the expression and properties of Green’s function associated with BVP (1.1), and by using the fixed-point theorem for strict-set-contraction operators and introducing a new cone Ω, we obtain the existence of at least two positive solutions for BVP (1.1) under certain conditions on the nonlinearity. Moreover, an example illustrating our main result is given in Section 4.

2 Preliminaries and lemmas

For convenience of the reader, we present here some definitions and preliminaries which are used throughout the paper. These definitions and lemmas can be found in the recent literature such as [1, 5].

Definition 2.1 ([1])

The Riemann-Liouville fractional integral of order q>0 of a function y(t) is given by

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

provided that the right-hand side is defined pointwise.

Definition 2.2 ([1])

The fractional derivative of order q>0 of a function y: R 0 + R is given by

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

where n=[q]+1, [q] denotes the integer part of number q, provided that the right-hand side is defined pointwise. In particular, for q=n, D 0 + q y(t)= y ( n ) (t).

Lemma 2.3 ([1])

Let q>0. Then the fractional differential equation

D 0 + q y(t)=0

has the unique solution y(t)= c 1 t q 1 + c 2 t q 2 ++ c n t q n , c i R, i=1,2,,n, here n1<qn.

Lemma 2.4 ([5])

Let q>0. Then the following equality holds for yL(0,1), D 0 + q yL(0,1):

I 0 + q D 0 + q y(t)=y(t)+ c 1 t q 1 + c 2 t q 2 ++ c N t q N

for some c i R, i=1,2,,N, here N is the smallest integer greater than or equal to q.

Let the real Banach space E with the norm be partially ordered by a cone P of E, i.e., uv if and only if vuP, and P is said to be normal if there exists a positive constant N such that θuv implies uNv, where the smallest N is called the normal constant of P. For details on cone theory, see [25].

The basic space used in this paper is C[I,E]. For any uC[I,E], evidently, (C[I,E], C ) is a Banach space with the norm u C = sup t I |u(t)|, and P={uC[I,E]:u(t)θ for tI} is a cone of the Banach space C[I,E]. We use α, α C to denote the Kuratowski noncompactness measure of bounded sets in the spaces E, C(I,E), respectively. As for the definition of the Kuratowski noncompactness measure, we refer to Ref. [25].

Definition 2.5 ([25], Strict-set contraction operator)

Let E 1 , E 2 be real Banach spaces, S E 1 . T:S E 2 is a continuous and bounded operator. If there exists a constant k such that α(T(S))kα(S), then T is called a k-set contraction operator. When k<1, T is called a strict-set contraction operator.

Lemma 2.6 ([25])

If DC[I,E] is bounded and equicontinuous, then α(D(t)) is continuous on I and

α C (D)= max t I α ( D ( t ) ) ,α ( { I u ( t ) d t : u D } ) I α ( D ( t ) ) dt,

where D(t)={u(t):uD,tI}.

Lemma 2.7 ([25])

Let K be a cone in a Banach space E. Assume that Ω 1 , Ω 2 are open subsets of E with θ Ω 1 , Ω ¯ 1 Ω 2 . If T:K( Ω ¯ 2 Ω 1 )K is a strict-set contraction operator such that either

  1. (i)

    Txx, xK Ω 1 and Txx, xK Ω 2 , or

  2. (ii)

    Txx, xK Ω 1 and Txx, xK Ω 2 ,

then T has a fixed point in K( Ω ¯ 2 Ω 1 ).

In the paper, we always assume that the following three assumptions hold:

  • (H0) 0< ρ :=ρ 0 1 φ(s)ds<+, where φ(s) is defined as

    φ(s)= 1 Γ ( q n + 2 ) (β+αs) [ γ ( 1 s ) + δ ( q n + 1 ) ] ( 1 s ) q n .
  • (H1) There exist aC[I, R + ] and hC[ R + n 1 , R + ] such that

    f ( t , u 1 , , u n 1 ) a ( t ) h ( u 1 , , u n 1 ) , t I , u k P , k = 1 , , n 1 .
    (2.1)
  • (H2) f:I× P r n 1 P for any r>0, f is uniformly continuous on I× P r n 1 and there exist nonnegative constants L k , k=1,,n1, with

    2 ρ ( k = 1 n 2 L k ( n 2 k ) ! + L n 1 ) <1
    (2.2)

    such that

    α ( f ( t , D 1 , D 2 , , D n 1 ) ) k = 1 n 1 L k α( D k ),tI, bounded  D k P r ,
    (2.3)

    where P r ={uP:ur}.

3 Main results

Lemma 3.1 Given yC[I,E], then the unique solution of

{ D 0 + q n + 2 x ( t ) + y ( t ) = 0 , 0 < t < 1 , n 1 < q n , n > 2 , α x ( 0 ) β x ( 0 ) = 0 , γ x ( 1 ) + δ x ( 1 ) = 0
(3.1)

is

x(t)= 0 1 G(t,s)y(s)ds,
(3.2)

where

G(t,s)={ G 1 ( t , s ) = ( t s ) q n + 1 Γ ( q n + 2 ) + G 0 ( t , s ) , 0 s t 1 , G 0 ( t , s ) , 0 t s 1 ,
(3.3)

and

G 0 (t,s)= ρ ( β + α t ) Γ ( q n + 2 ) [ γ ( 1 s ) + δ ( q n + 1 ) ] ( 1 s ) q n .

Proof Deduced from Lemma 2.4, we have

x(t)= I 0 + q n + 2 y(t)+ c 1 + c 2 t

for some c 1 , c 2 R. Then we get

x ( t ) = 1 Γ ( q n + 2 ) 0 t ( t s ) q n + 1 y ( s ) d s + c 1 + c 2 t , x ( t ) = 1 Γ ( q n + 1 ) 0 t ( t s ) q n y ( s ) d s + c 2 .

By boundary conditions αx(0)β x (0)=0, γx(1)+δ x (1)=0, and noting that Γ(qn+2)=(qn+1)Γ(qn+1), we have

c 1 = ρ β Γ ( q n + 2 ) 0 1 [ γ ( 1 s ) + δ ( q n + 1 ) ] ( 1 s ) q n y(s)ds

and

c 2 = ρ α Γ ( q n + 2 ) 0 1 [ γ ( 1 s ) + δ ( q n + 1 ) ] ( 1 s ) q n y(s)ds.

Thus

x ( t ) = 0 t ( t s ) q n + 1 Γ ( q n + 2 ) y ( s ) d s + ρ β Γ ( q n + 2 ) 0 1 [ γ ( 1 s ) + δ ( q n + 1 ) ] ( 1 s ) q n y ( s ) d s + ρ α t Γ ( q n + 2 ) 0 1 [ γ ( 1 s ) + δ ( q n + 1 ) ] ( 1 s ) q n y ( s ) d s = 0 t ( t s ) q n + 1 Γ ( q n + 2 ) y ( s ) d s + ρ ( β + α t ) 0 1 [ γ ( 1 s ) + δ ( q n + 1 ) ] ( 1 s ) q n Γ ( q n + 2 ) y ( s ) d s = 0 1 G ( t , s ) y ( s ) d s ,

where G(t,s) is Green’s function defined by (3.3). This completes the proof. □

Moreover, there is one paper [10] in which the following statement was shown.

Lemma 3.2 ([10])

The function G(t,s) defined by Lemma  2.3 has the following properties:

  1. (i)

    G(t,s) is continuous on [0,1]×[0,1];

  2. (ii)

    if β> n q q n + 1 α, then 0<G(t,s)G(s,s) for any t,s[0,1].

Lemma 3.3 If β> n q q n + 1 α, then the function G(t,s) satisfies:

λφ(s)G(t,s)ρφ(s)fort,s[0,1],

where

λ : = min { 4 ρ α γ δ [ β ( q n + 1 ) + α ( q n ) ] ζ , 4 ρ α β γ δ ( q n + 1 ) ζ } , ζ : = [ α γ β γ + α δ ( q n + 1 ) ] 2 + 4 α β γ [ γ + δ ( q n + 1 ) ] .

Proof According to (3.3) and Lemma 3.2, we have

G (s)G(t,s)ρφ(s),

where

G (s)={ G 1 ( 1 , s ) , 0 s < β γ α δ ( q n ) α δ + β γ , G 0 ( 0 , s ) , β γ α δ ( q n ) α δ + β γ s < 1 .

Since

inf 0 < s < 1 G 1 ( 1 , s ) φ ( s ) = inf 0 < s < 1 ρ 1 ( 1 s ) + ( β + α ) [ γ ( 1 s ) + δ ( q n + 1 ) ] ρ 1 ( β + α s ) [ γ ( 1 s ) + δ ( q n + 1 ) ] ρ δ [ β ( q n + 1 ) + α ( q n ) ] ( β + α s ) [ γ ( 1 s ) + δ ( q n + 1 ) ] 4 ρ α γ δ [ β ( q n + 1 ) + α ( q n ) ] [ α γ β γ + α δ ( q n + 1 ) ] 2 + 4 α β γ [ γ + δ ( q n + 1 ) ] = 4 ρ α γ δ [ β ( q n + 1 ) + α ( q n ) ] ζ ,

and

inf 0 < s < 1 G 0 ( 0 , s ) φ ( s ) = inf 0 < s < 1 ρ [ β γ ( 1 s ) + β δ ( q n + 1 ) ] ( β + α s ) [ γ ( 1 s ) + δ ( q n + 1 ) ] 4 ρ 1 α β γ δ ( q n + 1 ) [ α γ β γ + α δ ( q n + 1 ) ] 2 + 4 α β γ [ γ + δ ( q n + 1 ) ] = 4 ρ α β γ δ ( q n + 1 ) ζ ,

then we get

λφ(s)G(t,s)G(s,s)ρφ(s).

 □

Lemma 3.4 Let u(t)= I 0 + n 2 x(t), xC[I,E]. Then problem (1.1) can be transformed into the following modified problem:

{ D 0 + q n + 2 x ( t ) + f ( s , I 0 + n 2 x ( s ) , , I 0 + 1 x ( s ) , x ( s ) ) = θ , α x ( 0 ) β x ( 0 ) = θ , γ x ( 1 ) + δ x ( 1 ) = θ ,
(3.4)

where 0<t<1, n1<qn, n2. Moreover, if xC[I,E] is a solution of problem (3.3) and xθ, xθ, then the function u(t)= I 0 + n 2 x(s) is a positive solution of (1.1).

To obtain a positive solution, we construct a cone Ω by

Ω= { x ( t ) P : x ( t ) λ ρ x ( s ) , t , s I } ,
(3.5)

where P={xC[I,E]:x(t)θ,tI}.

Define an integral operator T:ΩE by

(Tx)(t)= 0 1 G(t,s)f ( s , I 0 + n 2 x ( s ) , I 0 + n 3 x ( s ) , , I 0 + 1 x ( s ) , x ( s ) ) ds,0t1.
(3.6)

Lemma 3.5 Assume that (H0)-(H2) hold. Then T:ΩΩ is a strict-set contraction operator.

Proof From Lemma 3.3 and (3.6), we obtain

(Tx)(t)λ 0 1 φ(s)f ( s , I 0 + n 2 x j ( s ) , , I 0 + 1 x j ( s ) , x j ( s ) ) ds.

On the other hand,

( T x ) ( s ) = 0 1 G ( s , s ) f ( s , I 0 + n 2 x j ( s ) , , I 0 + 1 x j ( s ) , x j ( s ) ) d s ρ 0 1 φ ( s ) f ( s , I 0 + n 2 x j ( s ) , , I 0 + 1 x j ( s ) , x j ( s ) ) d s = ρ λ 0 1 λ φ ( s ) f ( s , I 0 + n 2 x j ( s ) , , I 0 + 1 x j ( s ) , x j ( s ) ) d s ρ λ ( T x ) ( t ) .

Then (Tx)(t) λ ρ (Tx)(s), which implies (Tx)Ω, i.e., T(Ω)Ω.

Next we prove that T is continuous on Ω. Let { x j },{x}Ω and x j x Ω 0 (j). Hence { x j } is a bounded subset of Ω. Thus, there exists r>0 such that r= sup j x j Ω < and x Ω r.

According to the properties of f, for ε>0, there exists J>0 such that

f ( s , I 0 + n 2 x j ( s ) , , I 0 + 1 x j ( s ) , x j ( s ) ) f ( s , I 0 + n 2 x ( s ) , , I 0 + 1 x ( s ) , x ( s ) ) ε ρ

for jJ, tI.

Then

( T x j ) ( t ) ( T x ) ( t ) 0 1 G ( t , s ) f ( s , I 0 + n 2 x j ( s ) , , I 0 + 1 x j ( s ) , x j ( s ) ) f ( s , I 0 + n 2 x ( s ) , , I 0 + 1 x ( s ) , x ( s ) ) d s ρ 0 1 φ ( s ) d s f ( s , I 0 + n 2 x j ( s ) , , I 0 + 1 x j ( s ) , x j ( s ) ) f ( s , I 0 + n 2 x ( s ) , , I 0 + 1 x ( s ) , x ( s ) ) < ε .

Therefore, ε>0 for any tI and jJ. We get

( T x j ) ( t ) ( T x ) ( t ) 0.

This implies T is continuous on Ω.

By the properties of continuity of G(t,s), it is easy to see that T is equicontinuous on I.

Finally, we are going to show that T is a strict-set contraction operator. Let DΩ be bounded. Then, by condition (H1), Lemma 3.1 implies α C (TD)= max t I α((TD)(t)). It follows from (3.6) that

α ( ( T D ) ( t ) ) = α ( { 0 1 G ( t , s ) f ( s , ( T 1 x ) ( s ) , , ( T n 2 x ) ( s ) , x ( s ) ) d s D } ) 0 1 α ( { G ( t , s ) f ( s , ( T 1 x ) ( s ) , , ( T n 2 x ) ( s ) , x ( s ) ) d s D } ) ρ 0 1 φ ( s ) α ( f ( I × ( I 0 + n 2 D ) ( I ) × × ( I 0 + 1 D ) ( I ) × D ( I ) ) ) d s ρ ( k = 1 n 2 L k α ( ( I 0 + n 1 k D ) ( I ) ) + L n 1 α ( D ( I ) ) ) ,

which implies

α C (TD) ρ ( k = 1 n 2 L k α ( ( I 0 + n 1 k D ) ( I ) ) + L n 1 α ( D ( I ) ) ) .
(3.7)

Obviously,

α ( I 0 + n 1 k D ) ( I ) = α ( { 0 s ( s ν ) n 2 k ( n 2 k ) ! x ( s ) d s : ν [ 0 , s ] , s I , k = 1 , , n 2 } ) 1 ( n 2 k ) ! α ( D ( I ) ) .
(3.8)

Using a similar method as in the proof of Theorem 2.1.1 in [25], we have

α ( D ( I ) ) 2 α C (D).
(3.9)

Therefore, it follows from (3.7), (3.8) and (3.9) that

α C (TD) 2 ρ ( k = 1 n 2 L k ( n 2 k ) ! + L n 1 ) α C (D).

Noticing that (2.3), we obtain that T is a strict-set contraction operator. This completes the proof. □

Now we are in a position to give the main result of this work.

Theorem 3.6 Let the cone P be normal and conditions (H0)-(H2) hold. In addition, assume that the following conditions are satisfied:

  • (H3) There exist u P{θ}, c 1 C[I, R + ] and h 1 C[ P n 1 , R + ] such that

    f(t, u 1 ,, u n 1 ) c 1 (t) h 1 ( u 1 ,, u n 1 ) u ,tI, u k P,

    and

    h 1 = lim k = 1 n 1 u k h 1 ( u 1 , , u n 1 ) k = 1 n 1 u k > 1 ,as u k P,

    where := N 2 ρ 1 ( λ 2 0 1 φ(s) c 1 (s)ds u ).

  • (H4) There exist u P{θ}, c 2 C[I, R + ] and h 2 C[ P n 1 , R + ] such that

    f(t, u 1 ,, u n 2 ) c 2 (t) h 2 ( u 1 ,, u n 1 ) u ,tI, u k P,

    and

    h 1 0 = lim k = 1 n 1 u k 0 h 2 ( u 1 , , u n 1 ) k = 1 n 1 u i > 1 ,as u k P,

    where := N 2 ρ 1 ( λ 2 0 1 φ(s) c 2 (s)ds u ).

  • (H5) There exists ξ>0 such that

    ρN M ξ 0 1 φ(s)a(s)ds<ξ,

    where M ξ = max u k P ξ {h( u 1 ,, u n 1 )}.

Then problem (1.1) has at least two different positive solutions.

Proof Consider condition (H3), there exists r 1 >0 such that

h 1 ( u 1 ,, u n 1 ) ( h 1 ε 1 ) k = 1 n 1 u k , u k P, k = 1 n 1 u k r 1 ,

where ε 1 >0 satisfies ( h 1 ε 1 )1.

Therefore,

f(t, u 1 ,, u n 1 ) ( h 1 ε 1 ) k = 1 n 1 u k c 1 (t) u , u k P, k = 1 n 1 u k r 1 .

Take

R>max { N ρ λ 1 r 1 , ξ } .

Then, for tI, xΩ, x Ω =R, we have by (3.5)

x ( t ) λ ( ρ N ) 1 x Ω λ ( ρ N ) 1 R> r 1 .

Thus

( T x ) ( t ) = 0 1 G ( t , s ) f ( s , I 0 + n 2 x ( s ) , , I 0 + 1 x ( s ) , x ( s ) ) d s 0 1 λ φ ( s ) f ( s , I 0 + n 2 x ( s ) , , I 0 + 1 x ( s ) , x ( s ) ) d s ( h 1 ε 1 ) λ 0 u 1 φ ( s ) ( k = 1 n 2 I 0 + n 1 k x ( s ) + x ( s ) ) c 1 ( s ) d s u ( h 1 ε 1 ) λ 0 1 φ ( s ) c 1 ( s ) x ( s ) d s u ( h 1 ε 1 ) λ 2 ( ρ N ) 1 x Ω ( 0 1 φ ( s ) c 1 ( s ) d s ) u = ( h 1 ε 1 ) λ 2 ρ N 2 ( 0 1 φ ( s ) c 1 ( s ) d s u ) N x Ω u u N x Ω u u ,
(3.10)

and consequently,

T x Ω x Ω ,xΩ, x Ω =R.
(3.11)

Similarly, by condition (H4), there exists r 2 >0 such that

h 2 ( u 1 ,, u n 1 ) ( h 1 0 ε 2 ) k = 1 n 1 u k , u k P,0< k = 1 n 1 u k r 2 ,

where ε 2 >0 satisfies ( h 1 0 ε 2 )1.

Therefore,

f(t, u 1 ,, u n 1 ) ( h 1 0 ε 2 ) k = 1 n 1 u k c 2 (t) u , u k P,0< k = 1 n 1 u k r 2 .

Choose

0<r<min { ( k = 0 n 2 1 k ! ) 1 r 2 , ξ } .

Then, for t I τ , xΩ, x Ω =r, similar to (3.10), we have

( T x ) ( t ) ( h 1 0 ε 2 ) λ 0 1 φ ( s ) c 2 ( s ) x ( s ) d s u ( h 1 0 ε 2 ) λ 2 ( ρ N ) 1 x Ω ( 0 1 φ ( s ) c 2 ( s ) d s ) u = ( h 1 0 ε 2 ) λ 2 ρ N 2 ( 0 1 φ ( s ) c 2 ( s ) d s u ) N x Ω u u N x Ω u u ,

which implies

T x Ω x ( s ) Ω ,xΩ, x Ω =r.
(3.12)

On the other hand, according to Lemma 3.3 and (3.6), we get

(Tx)(t)ρ 0 1 φ(s)f ( s , I 0 + n 2 x ( s ) , , I 0 + 1 x ( s ) , x ( s ) ) ds.
(3.13)

By condition (H1), for tI, xΩ, x Ω =ξ, we have

f ( t , u 1 , , u n 1 ) a(t)h ( u 1 , , u n 1 ) M ξ a(t).

Therefore,

( T x ) ( t ) Ω ρN M ξ 0 1 φ(s)a(s)ds<ξ= x Ω .
(3.14)

Applying Lemma 2.7 to (3.12), (3.13) and (3.14) yields that T has a fixed point x 1 Ω ¯ r , ξ , r x 1 ξ, and a fixed point x 2 Ω ¯ ξ , R , ξ x 2 R. Noticing (3.13), we get x 1 ξ and x 2 ξ. This and Lemma 3.4 complete the proof. □

Theorem 3.7 Let the cone P be normal and conditions (H0) (H3) hold. In addition, assume that the following condition is satisfied:

(H6)

h ( u 1 , , u n 1 ) k = 1 n 1 u k 0,as u k P, k = 1 n 1 u k 0 + .
(3.15)

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

Proof By (H3), we can choose R>Nρ λ 1 r 1 . As in the proof of Theorem 3.6, it is easy to see that (3.11) holds. On the other hand, considering (3.15), there exists r 3 >0 such that

h ( u 1 , , u n 1 ) ε 3 k = 1 n 1 u k ,for tI, u k P,0< k = 1 n 1 u k r 3 ,

where ε 3 >0 satisfies

ε 3 = ( N ρ k = 1 n 1 1 k ! 0 1 φ ( s ) a ( s ) d s ) 1 .

Choose 0< r <min{ ( k = 0 n 2 1 k ! ) 1 r 3 ,R}. For tI, xΩ, x Ω = r , it follows from (3.5) that

0 < ( ρ N ) 1 λ r x < r 3 , 0 < ( ρ N ) 1 λ r k = 1 n 1 I 0 + n 1 k x ( t ) k = 1 n 1 1 k ! x < r 3 .
(3.16)

Then, for tI, xΩ, x Ω = r , we have

( T x ) ( t ) ρ 0 1 φ ( s ) f ( s , I 0 + n 2 x ( s ) , , I 0 + 1 x ( s ) , x ( s ) ) d s ρ 0 1 φ ( s ) a ( s ) h ( I 0 + n 2 x ( s ) , , I 0 + 1 x ( s ) , x ( s ) ) d s ε 3 ρ 0 1 φ ( s ) a ( s ) ( k = 1 n 2 I 0 + n 1 k x ( s ) + x ( s ) ) d s N ε 3 ρ k = 0 n 1 1 k ! r 0 1 φ ( s ) a ( s ) d s = r ,

and consequently,

( T x ) ( t ) Ω x Ω ,xΩ,x r .
(3.17)

Since 0< r <R, applying Lemma 2.7 to (3.11) and (3.17) yields that T has a fixed point x 0 Ω ¯ r , R , r x 0 R. This and Lemma 3.5 complete the proof. □

4 An example

Consider the following system of scalar differential equations of fractional order:

{ D 5 2 u k ( t ) = 1 24 u k ( t ) + 1 + t 324 k 2 { [ 2 u k + 2 ( t ) + 2 u k + 1 ( t ) + j = 1 u j ( t ) + j = 1 u 2 j ( t ) ] 3 D 5 2 u k ( t ) = + 3 u 2 k ( t ) + 3 u k + 1 ( t ) + j = 1 u j ( t ) + j = 1 u j ( t ) } , t I , u k ( 0 ) = u k ( 0 ) = 0 , 1 4 u k ( 0 ) 1 2 u k ( 0 ) = 0 , u k ( 1 ) + 1 2 u k ( 1 ) = 0 , k = 1 , 2 , 3 , .
(4.1)

Conclusion Problem (4.1) has at least two positive solutions.

Proof Let E= l 1 ={u=( u 1 , u 2 ,, u k ,): k = 1 | u k |<} with the norm u= k = 1 | u k |, and P={( u 1 ,, u k ,): u k 0,k=1,2,3,}. Then P is a normal cone in E with a normal constant N=1, and system (4.1) can be regarded as a boundary value problem of the form (1.1). In this situation, q= 5 2 , n=3, α= 1 4 , β= 1 2 , γ=1, δ= 1 2 , u=( u 1 ,, u n ,), f=( f 1 , f 2 ,, f n ,), in which

f k ( t , u k , u k ) = 1 24 u k ( t ) + 1 + t 324 k 2 { [ 2 u k + 2 ( t ) + 2 u k + 1 ( t ) + j = 1 u j ( t ) + j = 1 u 2 j ( t ) ] 3 + 3 u 2 k ( t ) + 3 u k + 1 ( t ) + j = 1 u j ( t ) + j = 1 u j ( t ) } .
(4.2)

By calculating, we have ρ= 4 5 , λ0.1515, and

1 2 =β> n q q n + 1 α= 1 4 ,

and

ρ =ρ 0 1 φ(s)ds= 1 5 π 0 1 (2+s) ( 1 s + 1 4 1 s ) ds= 44 75 π 0.3310,

which implies that condition (H0) is satisfied. Observing the inequality k = 1 1 k 2 <2, we get, by (4.2),

f ( t , u , v ) = k = 1 | f k ( t , u k , v k ) | 1 + t 3 ( 1 8 u + 1 2 ( u + v ) 3 + 1 27 u + v ) .
(4.3)

Hence (H1) is satisfied for a(t)= 1 + t 3 , and

h(x,y)= 1 8 x+ 1 2 ( x + y ) 3 + 1 27 x + y .

Now, we check condition (H2). Obviously, f:I× P r 2 P for any r>0, and f is uniformly continuous on I× P r 2 . Let f= f ( 1 ) + f ( 2 ) , where f ( 1 ) =( f 1 ( 1 ) ,, f k ( 1 ) ,) and f ( 2 ) =( f 1 ( 2 ) ,, f k ( 2 ) ,), in which

f k ( 1 ) ( t , u , v ) = 1 + t 324 k 2 { [ 2 u k + 2 ( t ) + 2 u k + 1 ( t ) + j = 1 u j ( t ) + j = 1 u 2 j ( t ) ] 3 + 3 u 2 k ( t ) + 3 u k + 1 ( t ) + j = 1 u j ( t ) + j = 1 u j ( t ) } ( k = 1 , 2 , 3 , ) ,
(4.4)

and

f k ( 2 ) (t,u,v)= 1 24 u k (t)(k=1,2,3,).
(4.5)

For any tI and bounded subsets D 1 , D 2 E, by (4.4), (4.5), we know

α ( f ( 2 ) ( I , D 1 , D 2 ) ) 1 24 α( D 1 ),tI, D 1 , D 2 E,
(4.6)

and

0 f ( 1 ) ( t , u , v ) = k = 1 | f k ( 1 ) ( t , u k , v k ) | ( u + v ) 3 + 2 27 u + v , t I , u , v E .

Similar to the proof of [[25], Example 2.12], we have

α ( f ( 1 ) ( t , D 1 , D 2 ) ) =0,tI, bounded sets  D 1 , D 2 E.
(4.7)

It follows from (4.6) and (4.7) that

α ( f ( I , D 1 , D 2 ) ) 1 24 α( D 1 ),tI, D 1 , D 2 E,

and

2 ρ ( k = 1 n 2 L k ( n 2 k ) ! + L n 1 ) = 24 29 π 0.2518<1,

i.e., condition (H2) holds for L 1 = 1 24 , L 2 =0.

On the other hand, by (4.2), we have

f k (t,u,v) 1 + t 324 k 2 ( u + v ) 3 ,t I τ ,u,vP(k=1,2,3,),

and

f k (t,u,v) 1 + t 324 k 2 u + v ,t I τ ,u,vP(k=1,2,3,).

Hence condition (H3) is satisfied for

c 1 (t)= 1 + t 324 k 2 , h 1 , k (u,v)= ( u + v ) 3 ,and u = ( 1 , , 1 k 2 , ) ,

in this situation,

h 1 , k = lim u + v ( u + v ) 3 u + v => 1 .

And condition (H4) is also satisfied for

c 2 (t)= 1 + t 324 k 2 , h 2 , k (u,v)= u + v ,and u = ( 1 , , 1 k 2 , ) ,

in this situation,

h 2 , k = lim u + v 0 u + v u + v => 1 .

Finally, choose ξ=1. It is easy to check that condition (H5) is satisfied. In this case, M ξ 4.1774, and so

ρN M ξ 0 1 φ(s)a(s)ds0.9604<ξ=1.

From Theorem 3.6, the conclusion follows and the proof is complete. □

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Acknowledgements

The authors are highly grateful for the referees’s careful reading and comments on this paper. The first author is supported financially by Hunan Provincial Natural Science Foundation of China (Grant No: 13JJ3106).

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Zhao, Y., Yang, L. & Chen, P. Positive solutions to boundary value problems of a high-order fractional differential equation in a Banach space. Adv Differ Equ 2013, 344 (2013). https://doi.org/10.1186/1687-1847-2013-344

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Keywords

  • fractional differential equation
  • boundary value problem
  • measure of noncompactness
  • strict-set-contraction operators
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