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Existence of positive solutions for a fractional high-order three-point boundary value problem

Abstract

In this paper, the authors consider the following fractional high-order three-point boundary value problem: D 0 + α u(t)+f(t,u(t))=0, t(0,1), u(0)= u (0)== u ( n 2 ) (0)=0, D 0 + α 1 u(η)=k D 0 + α 1 u(1), where k>1, η(0,1), n1<αn, n3, D 0 + α is the standard Riemann-Liouville derivative of order α, and f:[0,1]×[0,+)[0,+) is continuous. By using some fixed point index theorems on a cone for differentiable operators, the authors obtain the existence of positive solutions to the above boundary value problem.

MSC:34A08, 34B15.

1 Introduction

In this paper, we investigate the existence of solutions for the following fractional high-order equation:

D 0 + α u(t)+f ( t , u ( t ) ) =0,t(0,1)
(1.1)

with the three-point boundary value conditions

u(0)= u (0)== u ( n 2 ) (0)=0, D 0 + α 1 u(η)=k D 0 + α 1 u(1),
(1.2)

where k>1, η(0,1), n1<αn, n3, D 0 + α is the standard Riemann-Liouville derivative of order α and f:[0,1]×[0,+)[0,+) is continuous.

Differential equations with fractional order are a generalization of the ordinary differential equations to non-integer order. This generalization is not a mere mathematical curiosity but rather has interesting applications in many areas of science and engineering such as electrochemistry, control, porous media, electromagnetism, etc. (see [15]). There has been a significant development in the study of fractional differential equations in recent years; see for example [627]. Furthermore, several kinds of the high-order boundary value problems of fractional equations have been studied; see [610, 2831] for example. In [28], using the Guo-Krasnosel’skii fixed point theorem, Goodrich discussed the existence of positive solutions for the following fractional boundary value problem:

{ D 0 + v u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , n 1 v n , u ( i ) ( 0 ) = 0 , 0 i n 2 , D 0 + α u ( t ) | t = 1 = 0 , 1 α n 2 ,

where fC([0,1]×[0,+),[0,+)), n>3.

Moreover, Goodrich [29] investigated the existence of a positive solution to system of fractional boundary value problems and extended his previous study in [28].

Recently, motivated by the above work of Goodrich, Xu et al. [6] investigated the existence and uniqueness of positive solution for the following fractional boundary value problem:

{ D 0 + v u ( t ) + h ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , n 1 < v n , u ( i ) ( 0 ) = 0 , 0 i n 2 , D 0 + α u ( t ) | t = 1 = 0 , 1 α n 2 ,

where fC([0,1]×[0,+),[0,+)), hC(0,1)L(0,1) and n>3.

More recently, by the method of upper and lower solution together with Schauder fixed position theorem, Vong [7] studied the existence of positive solutions of the nonlocal boundary value problem for fractional equation:

{ D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , u ( 0 ) = = u ( n 1 ) ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d μ ( s ) ,

where n2, α(n1,n), and μ(s) is a function of bounded variation.

Moreover, Waug et al. [30], El-Shahed and Shammakh [31], Yang et al. [8], Zhang and Han [9], Wu et al. [10] also studied similar problems.

It is worth pointing out that the fixed point index theorems on cone for differentiable operators are the effective tools to investigate positive solutions of fractional equation. However, to the author’s knowledge, such theorems are rarely used in the literature. Different from the literature mentioned above, in the present paper, the authors apply some fixed point theorems for differentiable operators to establish the existence results on positive solutions to the fractional nonlocal boundary value problem (1.1)-(1.2). That is one of the features of this paper. Another feature of this paper is that some spectral properties of a correlative linear integral operator are introduced to obtain some positive eigenvector.

The rest of this paper is organized as follows. In Section 2, we present some necessary definitions and preliminary results that will be used to prove our main results. In Section 3, we put forward and prove our main results. Finally, we will give two examples to demonstrate our main results.

2 Preliminaries

In this section, we introduce some preliminary facts which are used throughout this paper.

Let be the set of positive integers, be the set of real numbers. R + be the set of real positive numbers. Denote by C([a,b],R) the Banach space endowed with the norm u= max t [ a , b ] |u(t)|. Let E=C([0,1],R).

Definition 2.1 ([3])

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

I a + α y(t)= 1 Γ ( α ) a t ( t s ) α 1 y(s)ds,t(a,b].

Definition 2.2 ([3])

The Riemann-Liouville fractional derivative of order α>0 of function y:(a,b]R is given by

D a + α y(t)= 1 Γ ( n α ) ( d d t ) n a t y ( s ) d s ( t s ) α n + 1 ,t(a,b],

where n=[α]+1, [α] denotes the integer part of α.

Lemma 2.1 ([32])

Let α>0. If uC(0,1)L(0,1) with a fractional derivative of order α that belongs to C(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, n1<αn.

To study the existence of solutions to the boundary value problems (BVPs, for short), we first consider the following auxiliary BVP:

{ D 0 + α u ( t ) + h ( t ) = 0 , u ( 0 ) = u ( 0 ) = = u ( n 2 ) ( 0 ) = 0 , D 0 + α 1 u ( η ) = k D 0 + α 1 u ( 1 ) ,
(2.1)

where hE, and α, η, k, n are given in (1.2).

We have the following lemma.

Lemma 2.2 For hE, the BVP (2.1) has a unique solution given by

u(t)= 0 1 G(t,s)h(s)ds,t[0,1],

where

G(t,s)= 1 Γ ( α ) { t α 1 ( t s ) α 1 , 0 s min { η , t } , t α 1 , t s η , k k 1 t α 1 ( t s ) α 1 , η < s t , k k 1 t α 1 , max { η , t } < s 1 .
(2.2)

Proof By Lemma 2.1, it follows from hE that there exist some constants c k (1kn) such that

u(t)= I 0 + α h(t)+ c 1 t α 1 + c 2 t α 2 ++ c n t α n ,t[0,1].

The boundary value conditions u(0)= u (0)== u ( n 2 ) (0)=0 in (2.1) imply that c 2 = c 3 == c n =0, and so

u(t)= I 0 + α h(t)+ c 1 t α 1 ,t[0,1].
(2.3)

Hence

D 0 + α 1 u(t)= 0 t h(s)ds+ c 1 Γ(α),t[0,1].

Thus,

D 0 + α 1 u(η)= 0 η h(s)ds+ c 1 Γ(α),
(2.4)
D 0 + α 1 u(1)= 0 1 h(s)ds+ c 1 Γ(α).
(2.5)

The condition D 0 + α 1 u(η)=k D 0 + α 1 u(1) together with (2.4) and (2.5) yields

c 1 = ( k 1 ) 0 η h ( s ) d s + k η 1 h ( s ) d s ( k 1 ) Γ ( α )

and so

u(t)= I 0 + α h(t)+ ( k 1 ) 0 η h ( s ) d s + k η 1 h ( s ) d s ( k 1 ) Γ ( α ) t α 1 ,t[0,1]

from (2.3).

  1. (1)

    If 0tη, then

    u ( t ) = 1 Γ ( α ) [ 0 t ( t α 1 ( t s ) α 1 ) h ( s ) d s + t η t α 1 h ( s ) d s + k k 1 η 1 t α 1 h ( s ) d s ] = . T 1 h ( t ) .
    (2.6)
  2. (2)

    If tη, then

    u ( t ) = 1 Γ ( α ) [ 0 η ( t α 1 ( t s ) α 1 ) h ( s ) d s + η t ( k k 1 t α 1 ( t s ) α 1 ) h ( s ) d s + k k 1 t 1 t α 1 h ( s ) d s ] = . T 2 h ( t ) .
    (2.7)

So, we always have

u(t)= 0 1 G(t,s)h(s)ds,t[0,1],

where Green’s function G(t,s) is given by (2.2). The proof is complete. □

Now, we give some properties of G(t,s).

Lemma 2.3 The Green function G has the following properties:

  1. (1)

    G(t,s)>0 for all t,s(0,1];

  2. (2)

    min t [ η , 1 ] G(t,s) ω 0 G(τ,s) for all s,τ[0,1], where ω 0 = η α 1 1 + k 0 and k 0 = 1 ( k 1 ) ( 1 ( 1 η ) α 1 ) .

Proof It is easy to see that the conclusion (1) of Lemma 2.3 is true from the expression of G(t,s) in (2.2). So, it remains to show that the conclusion (2) of Lemma 2.3 is true. Our proof is divided into two steps.

Step 1. In this step, we show that

min t [ η , 1 ] G(t,s) 1 Γ ( α ) η α 1 ( 1 ( 1 s ) α 1 ) ,s[0,1].
(2.8)

Let ϕ(t,s)=Γ(α)G(t,s), t[η,1], s[0,1]. Then we have the following cases to consider.

  1. (i)

    If s[0,η], then

    ϕ ( t , s ) = t α 1 ( t s ) α 1 = t α 1 [ 1 ( 1 s t ) α 1 ] t α 1 [ 1 ( 1 s ) α 1 ] η α 1 [ 1 ( 1 s ) α 1 ] .
  2. (ii)

    If s(η,t], then

    ϕ(t,s)= k k 1 t α 1 ( t s ) α 1 > t α 1 ( t s ) α 1 t α 1 [ 1 ( 1 s ) α 1 ] η α 1 [ 1 ( 1 s ) α 1 ] ,

    because k>1

  3. (iii)

    If s(t,1], then

    ϕ(t,s)= k k 1 t α 1 > k k 1 η α 1 > k k 1 η α 1 [ 1 ( 1 s ) α 1 ] > η α 1 [ 1 ( 1 s ) α 1 ] .

So, from the above analysis, we know that the inequality (2.8) holds.

Step 2. Now, we show that

G(τ,s) 1 Γ ( α ) (1+ k 0 ) [ 1 ( 1 s ) α 1 ] ,τ,s[0,1].
(2.9)

Let ϕ(τ,s)=Γ(α)G(τ,s), τ,s[0,1]. Similar to the proof in Step 1, we deduce the relation (2.9).

  1. (i)

    If 0smin{η,τ}, then ϕ(τ,s)= τ α 1 ( τ s ) α 1 . Owing to that α>2, it is easy to know that

    τ α 1 ( τ s ) α 1 1 ( 1 s ) α 1 ,

and so

ϕ(τ,s)1 ( 1 s ) α 1 .
  1. (ii)

    If max{η,τ}<s, then

    ϕ(τ,s)= k k 1 τ α 1 k k 1 s α 1 < k k 1 [ 1 ( 1 s ) α 1 ] (1+ k 0 ) [ 1 ( 1 s ) α 1 ]

    noting that s α 1 + ( 1 s ) α 1 1 for s(0,1] and k 0 = 1 ( k 1 ) ( 1 ( 1 η ) α 1 ) .

  2. (iii)

    If τsη, then

    ϕ(τ,s)= τ α 1 s α 1 1 ( 1 s ) α 1

noting that s α 1 + ( 1 s ) α 1 1 for s[0,1).

  1. (iv)

    If η<sτ, then

    ϕ(τ,s)= k k 1 τ α 1 ( τ s ) α 1 .

From

ϕ τ (τ,s)=(α1) [ k k 1 τ α 2 ( τ s ) α 2 ] 0,

we have

ϕ ( τ , s ) ϕ ( 1 , s ) = k k 1 ( 1 s ) α 1 = 1 ( 1 s ) α 1 + 1 k 1 = 1 ( 1 s ) α 1 + k 0 ( 1 ( 1 η ) α 1 ) < ( 1 + k 0 ) [ 1 ( 1 s ) α 1 ]

noting that k 0 = 1 ( k 1 ) ( 1 ( 1 η ) α 1 ) .

Summing up the above discussion, we know that the inequality (2.9) holds.

Now, from (2.8) and (2.9), the conclusion (2) of Lemma 2.3 follows. The proof is complete. □

We introduce a cone PE as follows:

P= { u E : u ( t ) 0 , t [ 0 , 1 ] ; u ( t ) ω 0 u  for all  t [ η , 1 ] } .

Define an operator T:EE by

Th= 0 1 G(t,s)h(s)ds,for hE.
(2.10)

We establish the following lemma, which will be used in the next section.

Lemma 2.4 T:PP is completely continuous.

Proof Let B be an arbitrary bounded set in E. Then there exists M>0 such that hM for any hB. First, we show that the set G={w|w=Th,hB} is equicontinuous on [0,1].

In fact, for an arbitrary ε>0 and any hB as well as t 1 , t 2 [0,1] with t 1 < t 2 , there are three cases to consider.

Case 1. If t 1 < t 2 η, then from (2.6) and (2.10), it follows that

| T h ( t 2 ) T h ( t 1 ) | = | T 1 h ( t 2 ) T 1 h ( t 1 ) | 1 Γ ( α ) [ | 0 t 2 ( t 2 α 1 ( t 2 s ) α 1 ) h ( s ) d s 0 t 1 ( t 1 α 1 ( t 1 s ) α 1 ) h ( s ) d s | + | t 2 η t 2 α 1 h ( s ) d s t 1 η t 1 α 1 h ( s ) d s | + k k 1 | η 1 ( t 2 α 1 t 1 α 1 ) h ( s ) d s | ] M Γ ( α ) [ 3 k 2 k 1 ( t 2 α 1 t 1 α 1 ) + 2 ( t 2 t 1 ) ] .
(2.11)

Therefore, there is a δ 1 >0 such that

| T h ( t 2 ) T h ( t 1 ) | < ε 2 ,when 0< t 2 t 1 <δ, t 1 < t 2 η.
(2.12)

Case 2. If η t 1 < t 2 , then by a similar argument to (2.11), from (2.7) and (2.10), we have

| T h ( t 2 ) T h ( t 1 ) | = | T 2 h ( t 2 ) T 2 h ( t 1 ) | M Γ ( α ) [ 3 k 1 k 1 ( t 2 α 1 t 1 α 1 ) + 2 k k 1 ( t 2 t 1 ) ] .

Thus, there exists a δ 2 >0 such that

| T h ( t 2 ) T h ( t 1 ) | < ε 2 ,when 0< t 2 t 1 < δ 2 ,η t 1 < t 2 .
(2.13)

Case 3. If t 1 <η< t 2 with 0< t 2 t 1 <δ=min{ δ 1 , δ 2 }, then from (2.12)-(2.13), it follows that

| T h ( t 2 ) T h ( t 1 ) | | T h ( t 2 ) T h ( η ) | + | T h ( η ) T h ( t 1 ) | <ε

noting that T 1 h(η)= T 2 h(η).

Summing up the above analysis on Cases 1-3, we conclude that |Th( t 2 )Th( t 1 )|<ε when 0< t 2 t 1 <δ=min{ δ 1 , δ 2 }, for t 1 , t 2 [0,1], that is, G is equicontinuous on [0,1].

Now, we show that G is bounded in E.

In fact, from the fact that 0G(t,s) k k 1 , t,s[0,1], we immediately have

| T h ( t ) | 0 1 G(t,s) | h ( s ) | ds k k 1 M,t[0,1], for all hE.
(2.14)

So, by the Arzela-Ascoli theorem, we know that T:EE is a compact operator. Again, because T is a bounded operator on E owing to (2.14), T:EE is continuous, and therefore T is completely continuous on E.

Finally, we apply the Lemma 2.3 to obtain

Th(t)= 0 1 G(t,s)h(s)ds0,t[0,1]

and

min t [ η , 1 ] Th(t)= min t [ η , 1 ] 0 1 G(t,s)h(s)ds ω 0 0 1 G(τ,s)h(s)ds= ω 0 Th(τ),τ[0,1]

for any hP. So, min t [ η , 1 ] (Th)(t)Th, that is, T:PP. The proof is complete. □

For the remainder of this section, we introduce the following lemmas, which will be used to obtain our main result in the next section.

Lemma 2.5 ([33])

Let P be a cone in a Banach space E, A:PP be completely continuous, and Aθ=θ. Suppose that A is differentiable at θ along P and 1 is not an eigenvalue of A + (θ) corresponding to a positive eigenvector. Moreover, if A + (θ) has no positive eigenvectors corresponding to an eigenvalue greater than one, then there exists r 0 >0 such that

i(A, P r ,P)=1,0<r r 0 ,

where P r ={xP:u<r}.

Lemma 2.6 ([33])

Let P be a cone in a Banach space E, A:PP be completely continuous. Suppose that A is differentiable atalong P and 1 is not an eigenvalue of A + () corresponding to a positive eigenvector. Moreover, if A + () has no positive eigenvectors corresponding to an eigenvalue greater than one, then there exists R 0 >0 such that

i(A, Ω R ,P)=1, R 0 R,

where Ω R ={xP:u<R}.

Lemma 2.7 ([34])

Let k(x,y) be nonnegative on [a,b]×[a,b], and let the operator K:C[a,b]C[a,b] be completely continuous, where K is defined as K ϕ = [ a , b ] k(x,y)ϕ(y)dy. If the spectral radius r(A)0, then K has a positive eigenfunction ϕ 1 corresponding to its first eigenvalue λ 1 = ( r ( A ) ) 1 , i.e. there exists ϕ 1 (t)C[a,b] with λ 1 K ϕ 1 = ϕ 1 , ϕ 1 (t)0, ϕ 1 (t)0, t[a,b].

3 Main results

Let E 0 =C([η,1],R), u 0 = max t [ η , 1 ] |u(t)|. Define an operator T 0 on E 0 as

T 0 u(t)= η 1 G(t,s)u(s)dt,t[η,1],

for u E 0 , where G(t,s) is the Green function (2.2), whose domain is restricted on [η,1]×[0,1].

Let P 0 ={u E 0 :u(t) ω 0 u 0 ,t[η,1]}, where ω 0 is given in Lemma 2.3. Obviously, P 0 is a cone in E 0 . We have the following lemma.

Lemma 3.1 T 0 : E 0 E 0 is completely continuous. Moreover, the spectral radius r( T 0 )>0.

Proof Since the proof of the complete continuity of T 0 is similar to that in Lemma 2.4, we omit it. Here, we only show that r( T 0 )>0.

Let ϕ(t)= η 1 G(t,s)ds, t[η,1]. Because

ϕ(t)= k k 1 t α 1 (1η) 1 α ( t η ) α ,t[η,1],

and

ϕ ( t ) = k k 1 ( α 1 ) t α 2 ( 1 η ) ( t η ) α 1 > t α 1 ( 1 η ) ( t η ) α 1 > 0 , t [ η , 1 ] ,

we have

min t [ η , 1 ] ϕ(t)=ϕ(η)= k k 1 η α 1 (1η) = Δ γ 0 >0.

For any u P 0 {θ}, from

( T 0 u ) ( t ) = η 1 G ( t , s ) u ( s ) d s ω 0 u 0 η 1 G ( t , s ) d s ω 0 γ 0 u 0 ω 0 γ 0 u ( t ) , t [ η , 1 ] ,

we have

( T 0 2 u ) (t)= T 0 ( T 0 u) ω 0 γ 0 T 0 u ( ω 0 γ 0 ) 2 u(t),t[η,1],

and so

T 0 n u 0 ( T 0 n u ) (t) ( ω 0 γ 0 ) n u(t),t[η,1].

Hence, T 0 n u 0 ( ω 0 γ 0 ) n u 0 , and so T 0 n u 0 u 0 ( ω 0 γ 0 ) n , which implies that T 0 n ( ω 0 γ 0 ) n . So, we obtain

r( T 0 )= lim ¯ n T n 1 n ω 0 γ 0 >0.

The proof is complete. □

Let us list the following assumptions, which will be used later.

(H1) fC([0,1]× R + , R + ).

(H2) f(t,0)=0, f x (t,0)0, t[0,1], f x (t,0)0, t[η,1], where the partial derivative f x (t,0)= f x (t,x) | ( t , 0 ) . Moreover, b>0 such that f x (t,x) is continuous on [0,1]×[0,b].

(H3) 0 1 (1 ( 1 s ) α 1 ) f x (s,0)ds< Γ ( α ) 1 + k 0 , where k 0 = 1 ( k 1 ) ( 1 ( 1 η ) α 1 ) , f x (s,0)= f x (s,x) | ( s , 0 ) .

(H4) There exists ϕC([0,1], R + ) such that lim x + f ( t , x ) x =ϕ(t) holds uniformly on [0,1] with respect to t. Moreover, ϕ(t)0, t[η,1].

(H5) 0 1 (1 ( 1 s ) α 1 )ϕ(s)ds< Γ ( α ) 1 + k 0 , where k 0 = 1 ( k 1 ) ( 1 ( 1 η ) α 1 ) .

Define an operator A on P as

(Au)(t)= 0 1 G(t,s)f ( s , u ( s ) ) ds,t[0,1],for uP.

Obviously, the following lemma is true in view of Lemma 2.4.

Lemma 3.2 Let (H1) hold. Then A:PP is completely continuous.

We need the following two lemmas, which will play an important role to obtain the existence results.

Lemma 3.3 Let (H1)-(H2) hold. Then the operator A is differentiable at θ along P, and Aθ=θ, A + (θ)h=Bh, hP, where

Bh= 0 1 G(t,s) f x (s,0)h(s)ds,hP.

Proof For any (s,x)[0,1]×[0,b], by the mean value theorem, there exists ξ(0,x) such that

f(s,x)=f(s,x)f(s,0)= f x (s,ξ)x,

that is,

f(s,x) f x (s,ξ)x=0.
(3.1)

Again, due to the fact that f x (s,x) is uniformly continuous on [0,1]×[0,b], for arbitrary ε>0, there exists δ(0,b) such that

| f x ( s , x ) f x ( s , 0 ) | < α Γ ( α ) ( α 1 ) ( 1 + k 0 ) ε,
(3.2)

when 0<x<δ, for all s[0,1].

So, from (3.1)-(3.2), it follows that

| f ( s , x ) f x ( s , 0 ) x | | f ( s , x ) f x ( s , ξ ) x | + | f x ( s , ξ ) f x ( s , 0 ) | x < α Γ ( α ) ( α 1 ) ( 1 + k 0 ) ε x ,
(3.3)

when 0<x<δ, for all s[0,1].

Consequently, for any hP with h<δ, from (3.3) and (2.9), it follows that

| ( A h ) ( t ) ( B h ) ( t ) | 0 1 G ( t , s ) | f ( s , h ( s ) ) f x ( s , 0 ) h ( s ) | d s α Γ ( α ) ( α 1 ) ( 1 + k 0 ) ε h 0 1 G ( t , s ) d s α α 1 ε h 0 1 ( 1 ( 1 s ) α 1 ) d s = ε h .

So, AhBhεh, that is, A + (θ)h=Bh, hP. The proof is complete. □

Lemma 3.4 Let (H1)-(H3) hold. Then A + (θ) has no positive eigenvectors corresponding to an eigenvalue greater than or equal to one.

Proof If not, then there exist a λ 0 1 and h 0 P{θ} with λ 0 h 0 = A + (θ) h 0 , and so

h 0 ( t ) λ 0 h 0 ( t ) = ( A + ( θ ) h ) ( t ) = 0 1 G ( t , s ) f x ( s , 0 ) h 0 ( s ) d s 1 + k 0 Γ ( α ) 0 1 ( 1 ( 1 s ) α 1 ) f x ( s , 0 ) h 0 ( s ) d s .

Thus,

( 1 ( 1 t ) α 1 ) f x ( t , 0 ) h 0 ( t ) 1 + k 0 Γ ( α ) ( 1 ( 1 t ) α 1 ) f x ( t , 0 ) × 0 1 ( 1 ( 1 s ) α 1 ) f x ( s , 0 ) h 0 ( s ) d s ,

and so

0 1 ( 1 ( 1 t ) α 1 ) f x ( t , 0 ) h 0 ( t ) d t 1 + k 0 Γ ( α ) 0 1 ( 1 ( 1 t ) α 1 ) f x ( t , 0 ) d t × 0 1 ( 1 ( 1 s ) α 1 ) f x ( s , 0 ) h 0 ( s ) d s .
(3.4)

Because h 0 P{θ}, f x (t,0)0, t[0,1], f x (t,0)0, t[η,1] and f x (t,0) is continuous on [0,1], the following inequality:

0 1 ( 1 ( 1 t ) α 1 ) f x (t,0) h 0 (t)dt>0

holds. Immediately, from (3.4) it follows that 1 + k 0 Γ ( α ) 0 1 (1 ( 1 t ) α 1 ) f x (t,0)dt1, which contradicts (H3). So, the conclusion of Lemma 3.4 is true. The proof is complete. □

Lemma 3.5 Let (H1), (H4), and (H5) hold. Then A is differentiable atalong P and A + ()h=Gh, hP, where

Gh= 0 1 G(t,s)ϕ(s)h(s)ds,hP.

Proof From (H4), it follows that, for arbitrary ε>0, there exists R>0 such that

| f ( t , x ) x ϕ(t)|<ε,t[0,1],

when x>R, and so

| f ( t , x ) ϕ ( t ) x | <εx,t[0,1],

when x>R.

Let M= sup ( t , x ) [ 0 , 1 ] × [ 0 , R ] |f(t,x)ϕ(t)x|. Then

| f ( t , x ) ϕ ( t ) x | <M+εx,t[0,1]

holds for any x R + .

Now, for any hP, by the above inequality, we have

| ( A h ) ( t ) ( G h ) ( t ) | 0 1 G ( t , s ) | f ( s , h ( s ) ) ϕ ( s ) h ( s ) | d s ( M + ε h ) 0 1 G ( t , s ) d s ( M + ε h ) L , t [ 0 , 1 ] ,

where L= max t [ 0 , 1 ] 0 1 G(t,s)ds. Thus A h G h h M L h +Lε. So, A h G h h <2Lε when h> M ε , that is, A + ()h=Gh. The proof is complete. □

Lemma 3.6 Let (H1), (H4), and (H5) hold. Then A + () has no positive eigenvectors corresponding to an eigenvalue greater than or equal to one.

Proof The proof is similar to the proof of Lemma 3.4. In fact, if not, then there exist a λ 0 1 and h 0 P{θ} such that λ 0 h 0 = A + () h 0 , and so

h 0 ( t ) λ 0 h 0 ( t ) = ( A + ( ) h 0 ) ( t ) = 0 1 G ( t , s ) ϕ ( s ) h 0 ( s ) d s 1 + k 0 Γ ( α ) 0 1 ( 1 ( 1 s ) α 1 ) ϕ ( s ) h 0 ( s ) d s .

Thus,

( 1 ( 1 t ) α 1 ) ϕ(t) h 0 (t) 1 + k 0 Γ ( α ) ( 1 ( 1 t ) α 1 ) ϕ(t) 0 1 ( 1 ( 1 s ) α 1 ) ϕ(s) h 0 (s)ds,

and so

0 1 ( 1 ( 1 t ) α 1 ) ϕ ( t ) h 0 ( t ) d t 1 + k 0 Γ ( α ) 0 1 ( 1 ( 1 t ) α 1 ) ϕ ( t ) d t × 0 1 ( 1 ( 1 s ) α 1 ) ϕ ( s ) h 0 ( s ) d s .

Because h 0 P{θ} and ϕ(t)0, t[η,1], ϕC([0,1], R + ), the relation

0 1 ( 1 ( 1 t ) α 1 ) ϕ(t) h 0 (t)dt>0

holds, and therefore 1 + k 0 Γ ( α ) 0 1 (1 ( 1 t ) α 1 )ϕ(t)dt1, which contradicts (H5). The proof is complete. □

Let f = lim x inf min t [ η , 1 ] f ( t , x ) x . We are in a position to state our main result in the present paper.

Theorem 3.1 Let (H1)-(H3) hold. If f > r 1 ( T 0 ), then BVP (1.1)-(1.2) has a positive solution.

Proof In view of Lemmata 3.2-3.4 and by applying Lemma 2.5, we conclude that there exist a r 0 >0 such that

i(A, Ω r 0 ,P)=1,
(3.5)

where Ω r =P B r , B r ={xE:u<r} and P is defined as before.

By Lemma 2.7 and Lemma 3.1, we know that there exists a ϕ 1 C[η,1] with ϕ 1 (t)0, ϕ 1 (t)0, t[η,1] satisfying

r 1 ( T 0 ) η 1 G(t,s) ϕ 1 (s)ds= ϕ 1 (t),t[η,1].

So, from Lemma 2.3, it follows that

ϕ 1 (t) ω 0 r 1 ( T 0 ) η 1 G(τ,s) ϕ 1 (s)ds= ω 0 ϕ 1 (τ),t,τ[η,1].

Thus, ϕ 1 (t) ω 0 ϕ 1 0 , t[η,1], that is, ϕ 1 P 0 .

Let

ϕ 0 (t)= { ϕ 1 ( η ) , t [ 0 , η ] , ϕ 1 ( t ) , t [ η , 1 ] .

It is easy to see that ϕ 1 P.

On the other hand, by f > r 1 ( T 0 ), there exists a R>0 such that

f(t,x)> r 1 ( T 0 )x,t[η,1]

when xR. Take R 0 max{ ω 0 1 R, r 0 }. Set Ω R 0 ={uP:u< R 0 }. Then for any u Ω R 0 , the inequality u(t) ω 0 u= ω 0 R 0 R, t[η,1] implies

f ( t , u ( t ) ) > r 1 ( T 0 )u(t),t[η,1].
(3.6)

We show that the following relation holds:

u(t)(Au)(t)λ ϕ 0 (t),λ0,u Ω R 0 ,t[0,1].
(3.7)

In fact, if not, then there exist a u 0 Ω R 0 and a λ 0 0 such that

u 0 (t)=(A u 0 )(t)+ λ 0 ϕ 0 (t),t[0,1].
(3.8)

Obviously, we can assume that λ 0 >0. From (3.8), it follows that

u 0 (t) λ 0 ϕ 0 (t),t[0,1]

because APP.

Let λ =sup{λ| u 0 (t)λ ϕ 0 (t),t[η,1]}. Then 0< λ <+ and

u 0 (t) λ ϕ 0 (t),t[η,1].
(3.9)

Again, from (3.8), (3.6), and (3.9), for t[η,1], we have

u 0 ( t ) = ( A u 0 ) ( t ) + λ 0 ϕ 0 ( t ) = 0 1 G ( t , s ) f ( s , u 0 ( s ) ) d s + λ 0 ϕ 0 ( t ) η 1 G ( t , s ) f ( s , u 0 ( s ) ) d s + λ 0 ϕ 0 ( t ) > r 1 ( T 0 ) η 1 G ( t , s ) u 0 ( s ) d s + λ 0 ϕ 0 ( t ) r 1 ( T 0 ) λ η 1 G ( t , s ) ϕ 0 ( s ) d s + λ 0 ϕ 0 ( t ) = r 1 ( T 0 ) λ ( T 0 ϕ 0 ) ( t ) + λ 0 ϕ 0 ( t ) = ( λ + λ 0 ) ϕ 0 ( t ) , t [ η , 1 ] ,

which contradicts the definition of λ . Hence, the relation (3.7) holds. So, in terms of the fixed point index theorem on a cone, we have

i(A, Ω R 0 ,P)=0.
(3.10)

Thus, (3.5) and (3.10) imply that

i(A, Ω R 0 Ω ¯ r 0 ,P)=1.

So, A has a fixed point u ¯ Ω R 0 Ω ¯ r 0 , that is, u ¯ is a positive solution of BVP (1.1)-(1.2). The proof is complete. □

Let μ 0 = Γ ( α ) ω 0 k 1 k η 1 α ( 1 η ) 1 , where ω 0 is given as in Lemma 2.3. We state another result in this paper.

Theorem 3.2 Let (H1), (H4), and (H5) hold. Assume that there exists r 0 such that

f(t,x) μ 0 x,x[0, r 0 ],t[0,1].
(3.11)

Then BVP (1.1)-(1.2) has a positive solution.

Proof We show that

Auu,u Ω r 0 ,
(3.12)

where Ω r 0 ={uP:u< r 0 }.

In fact, for any u Ω r 0 , from (3.11) we have

f ( t , u ( t ) ) μ 0 u(t),t[0,1]

owing to

0u(t) r 0 ,t[0,1].

Thus, by Lemma 2.3,

A u ( A u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s μ 0 0 1 G ( t , s ) u ( s ) d s μ 0 ω 0 u 0 1 G ( τ , s ) d s μ 0 ω 0 u η 1 G ( τ , s ) d s , τ [ 0 , 1 ] , t [ η , 1 ] .

So,

Au μ 0 ω 0 u η 1 G(η,s)ds=u,

because

ω 0 η 1 G(η,s)ds= ω 0 Γ ( α ) k k 1 η α 1 (1η)= μ 0 1 .

Therefore, the relation (3.12) holds. Consequently, applying the fixed point index theorem, we get

i(A, Ω r 0 ,P)=0.
(3.13)

On the other hand, by Lemma 3.2, Lemma 3.5, Lemma 3.6, and Lemma 2.6, we know that there exists R 0 > r 0 such that

i(A, Ω R 0 ,P)=1.
(3.14)

So, by (3.13) and (3.14), we have

i(A, Ω R 0 Ω ¯ r 0 ,P)=1.

Therefore, A has a fixed point u ¯ Ω R 0 Ω ¯ r 0 , that is, u ¯ is a positive solution of BVP (1.1)-(1.2). The proof is complete. □

Example 3.1 Consider the following boundary value problem:

{ D 0 + 5 2 u ( t ) = ln ( 1 + t 2 ) sin 2 u ( t ) + t 2 u ( t ) e u ( t ) 1 + u 2 ( t ) , t ( 0 , 1 ) , u ( 0 ) = u ( 0 ) = 0 , D 0 + 3 2 u ( 1 2 ) = 2 D 0 + 3 2 u ( 1 ) .
(3.15)

To obtain the existence result, we will apply Theorem 3.1 with α= 5 2 , n=3, η= 1 2 , k=2 and function f(t,x)=ln(1+ t 2 ) sin 2 x+ t 2 x e x 1 + x 2 , t[0,1], x[0,+). Clearly, the function f(t,x) satisfies fC([0,1]× R + , R + ), f(t,0)=0, f x (t,x)=ln(1+ t 2 )sin2x+ t 2 e x ( 1 + x x 2 + x 3 ) ( 1 + x 2 ) 2 and f x (t,0)= t 2 >0, t[ 1 2 ,1]. Further, 0 1 (1 ( 1 s ) 3 2 ) f x (s,0)ds= 89 315 < 3 ( 15 2 2 ) 124 π = Γ ( α ) 1 + k 0 and f =+. So, all the assumptions of Theorem 3.1 are satisfied and therefore BVP (3.15) has at least one positive solution.

Example 3.2 Consider the following boundary value problem:

{ D 0 + 5 2 u ( t ) = 15 42 e t ( 1 + u ( t ) ) + ln ( 2 + t ) | cos u ( t ) | , t ( 0 , 1 ) , u ( 0 ) = u ( 0 ) = 0 , D 0 + 3 2 u ( 1 2 ) = 2 D 0 + 3 2 u ( 1 ) .
(3.16)

To obtain the existence result, we will apply Theorem 3.2 with α= 5 2 , n=3, η= 1 2 , k=2 and function f(t,x)= 15 42 e t (1+x)+ln(2+t)|cosx|, t[0,1], x[0,+). Clearly, the function f(t,x) satisfies fC([0,1]× R + , R + ), lim x + f ( t , x ) x ϕ(t)= 15 42 e t holds uniformly on [0,1] with respect to t. Further, 0 1 (1 ( 1 s ) 3 2 )ϕ(s)ds< 9 14 < 3 ( 15 2 2 ) 124 π = Γ ( α ) 1 + k 0 . On the other hand, it is easy to see that exists a r 0 (0, π 2 ) such that f(t,x) μ 0 x, x[0, r 0 ], t[0,1] noting that f(t,x)>(ln2)cosx, x[0, π 2 ], t[0,1] and ((ln2)cosx μ 0 x) | x = 0 =ln2, ( ( ln 2 ) cos x μ 0 x ) <0, x[0, π 2 ], t[0,1]. So, all the assumptions of Theorem 3.2 are satisfied and therefore BVP (3.16) has at least one positive solution.

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Acknowledgements

The authors sincerely thank the anonymous referees for their valuable suggestions and comments, which have greatly helped improve this article. This work is supported by the Natural Science Foundation of Hubei Provincial Education Department (Q20132505).

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Chai, G., Hu, S. Existence of positive solutions for a fractional high-order three-point boundary value problem. Adv Differ Equ 2014, 90 (2014). https://doi.org/10.1186/1687-1847-2014-90

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Keywords

  • fractional differential equations
  • three-point boundary value problems
  • existence results
  • fixed point index theorem for differentiable operators