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Positive solutions for fractional differential equations with three-point multi-term fractional integral boundary conditions

Abstract

We are concerned with the existence of at least one, two or three positive solutions for the boundary value problem with three-point multi-term fractional integral boundary conditions:

{ D q u ( t ) + f ( t , u ( t ) ) = 0 , 1 < q 2 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = i = 1 m α i ( I p i u ) ( η ) , 0 < η < 1 ,

where D q is the standard Riemann-Liouville fractional derivative. Our analysis relies on the Krasnoselskii fixed point theorem and the Leggett-Williams fixed point theorem. Some examples are also given to illustrate the main results.

MSC:26A33, 34A08, 34B18.

1 Introduction

In recent years, the interest in the study of fractional differential equations has been growing rapidly. Fractional differential equations have arisen in mathematical models of systems and processes in various fields such as aerodynamics, acoustics, mechanics, electromagnetism, signal processing, control theory, robotics, population dynamics, finance, etc.

We refer a reader interested in the systematic development of the topic to the books [17]. A variety of results on initial and boundary value problems of fractional differential equations and inclusions can easily be found in the literature on the topic. For some recent results, we can refer to [819] and references cited therein.

In this paper, we concentrate on the study of positive solutions to the boundary value problems of fractional differential equations. More precisely, we consider the nonlinear fractional differential equation

D q u(t)+f ( t , u ( t ) ) =0,1<q2,0<t<1,
(1.1)

subject to three-point multi-term fractional integral boundary conditions

u(0)=0,u(1)= i = 1 m α i ( I p i u ) (η),0<η<1,
(1.2)

where D q is the standard Riemann-Liouville fractional derivative of order q, I p i is the Riemann-Liouville fractional integral of order p i >0, i=1,2,,m, f:[0,1]×[0,)[0,) and α i 0, i=1,2,,m, are real constants such that i = 1 m α i η p i + q 1 Γ ( q ) Γ ( p i + q ) <1.

We mention that integral boundary conditions are encountered in various applications such as population dynamics, blood flow models, chemical engineering, cellular systems, heat transmission, plasma physics, thermoelasticity, etc. Nonlocal conditions come up when values of the function on the boundary is connected to values inside the domain.

One of the most frequently used tools for proving the existence of positive solutions to the integral equations and boundary value problems is the Krasnoselskii theorem on cone expansion and compression and its norm-type version due to Guo and Lakshmikantham [20]. The main idea is to construct a cone in a Banach space and a completely continuous operator defined on this cone based on the corresponding Green’s function and then find fixed points of the operator. See [9, 18] and references therein for recent development.

The rest of this paper is organized as follows. In Section 2 we present some necessary basic knowledge and definitions for fractional calculus theory and give the corresponding Green’s function of boundary value problem (1.1)-(1.2). Moreover, some properties of the Green’s function are also proved. In Section 3 we use the properties of the corresponding Green’s function and the Guo-Krasnoselskii fixed point theorem to show the existence of at least one or two positive solutions of (1.1)-(1.2) under the condition that the nonlinear f is either sublinear or superlinear. In Section 4 we prove the existence of at least three positive solutions via the Leggett-Williams fixed point theorem. Finally, illustrative examples are presented in Section 5.

2 Preliminaries

In this section, we introduce some notations and definitions of fractional calculus [3, 4] and present preliminary results needed in our proofs later.

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

I α g(t)= 0 t ( t s ) α 1 Γ ( α ) g(s)ds,

provided the right-hand side is point-wise defined on (0,), where Γ is the gamma function.

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

D α g(t)= 1 Γ ( n α ) ( d d t ) n 0 t g ( s ) ( t s ) α n + 1 ds,n1<α<n,

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

From the definition of the Riemann-Liouville fractional derivative, we can obtain the following lemmas.

Lemma 2.1 (see [4])

Let α>0 and yC(0,1)L(0,1). Then the fractional differential equation D α y(t)=0 has a unique solution

y(t)= c 1 t α 1 + c 2 t α 2 ++ c n t α n ,

where c i R, i=1,2,,n, and n1<α<n.

Lemma 2.2 (see [4])

Let α>0. Then for yC(0,1)L(0,1) it holds

I α D α y(t)=y(t)+ c 1 t α 1 + c 2 t α 2 ++ c n t α n ,

where c i R, i=1,2,,n, and n1<α<n.

The following property (Dirichlet’s formula) of the fractional calculus is well known [1]:

I γ I β u(t)= I γ + β u(t),t[0,1],uL(0,1),γ+β1,

which has the form

0 t ( t s ) γ 1 ( 0 s ( s τ ) β 1 u ( τ ) d τ ) ds= Γ ( γ ) Γ ( β ) Γ ( γ + β ) 0 t ( t s ) γ + β 1 u(s)ds.

For convenience we put

Ω:=1 i = 1 m α i η p i + q 1 Γ ( q ) Γ ( p i + q ) .
(2.1)

Lemma 2.3 Let i = 1 m α i η p i + q 1 Γ ( q ) Γ ( p i + q ) <1, α i 0, p i >0, i=1,2,,m, and hC([0,1],R). The unique solution uAC([0,1],R) of the boundary value problem

D q u(t)+h(t)=0,t(0,1),q(1,2],
(2.2)
u(0)=0,u(1)= i = 1 m α i ( I p i u ) (η),0<η<1,
(2.3)

is the integral equation

u(t)= 0 1 G(t,s)h(s)ds,
(2.4)

where G(t,s) is the Green’s function given by

G(t,s)=g(t,s)+ i = 1 m α i t q 1 Ω Γ ( p i + q ) g i (η,s),
(2.5)

where

g(t,s)= { t q 1 ( 1 s ) q 1 ( t s ) q 1 Γ ( q ) ; 0 s t 1 , t q 1 ( 1 s ) q 1 Γ ( q ) ; 0 t s 1
(2.6)

and

g i (η,s)= { η p i + q 1 ( 1 s ) q 1 ( η s ) p i + q 1 ; 0 s η < 1 , η p i + q 1 ( 1 s ) q 1 ; 0 < η s 1 .
(2.7)

Proof Using Lemmas 2.1-2.2, problem (2.2)-(2.3) can be expressed as an equivalent integral equation

u(t)= c 1 t q 1 + c 2 t q 2 0 t ( t s ) q 1 Γ ( q ) h(s)ds
(2.8)

for c 1 , c 2 R. The first condition of (2.3) implies that c 2 =0. Taking the Riemann-Liouville fractional integral of order p i >0 for (2.8) and using Dirichlet’s formula, we get that

( I p i u ) ( t ) = 0 t ( t s ) p i 1 Γ ( p i ) ( c 1 s q 1 0 s ( s r ) q 1 Γ ( q ) h ( r ) d r ) d s = c 1 0 t ( t s ) p i 1 s q 1 Γ ( p i ) d s 0 t ( t s ) p i 1 Γ ( p i ) 0 s ( s r ) q 1 Γ ( q ) h ( r ) d s d r = c 1 t p i + q 1 Γ ( q ) Γ ( p i + q ) 1 Γ ( p i + q ) 0 t ( t s ) p i + q 1 h ( s ) d s .

The second condition of (2.3) yields

c 1 0 1 ( 1 s ) q 1 Γ ( q ) h ( s ) d s = c 1 i = 1 m α i η p i + q 1 Γ ( q ) Γ ( p i + q ) i = 1 m α i Γ ( p i + q ) 0 η ( η s ) p i + q 1 h ( s ) d s .

Then we have that

c 1 = 1 Ω [ 0 1 ( 1 s ) q 1 Γ ( q ) h ( s ) d s i = 1 m α i Γ ( p i + q ) 0 η ( η s ) p i + q 1 h ( s ) d s ] .

Therefore, the unique solution of boundary value problem (2.2)-(2.3) is written as

u ( t ) = 0 t ( t s ) q 1 Γ ( q ) h ( s ) d s + 1 Ω Γ ( q ) 0 1 ( 1 s ) q 1 t q 1 h ( s ) d s 1 Ω i = 1 m α i Γ ( p i + q ) 0 η ( η s ) p i + q 1 t q 1 h ( s ) d s .

Hence, by taking into account (2.1), we have

u ( t ) = 0 t ( t s ) q 1 Γ ( q ) h ( s ) d s + 1 Ω Γ ( q ) 0 1 ( 1 s ) q 1 t q 1 h ( s ) d s 1 Ω i = 1 m α i Γ ( p i + q ) 0 η ( η s ) p i + q 1 t q 1 h ( s ) d s + 0 1 ( 1 s ) q 1 t q 1 Γ ( q ) h ( s ) d s 0 1 ( 1 s ) q 1 t q 1 Γ ( q ) h ( s ) d s = 0 1 ( 1 s ) q 1 t q 1 Γ ( q ) h ( s ) d s 0 t ( t s ) q 1 Γ ( q ) h ( s ) d s + i = 1 m α i t q 1 Ω Γ ( p i + q ) ( 0 1 η p i + q 1 ( 1 s ) q 1 h ( s ) d s 0 η ( η s ) p i + q 1 h ( s ) d s ) = 0 1 g ( t , s ) h ( s ) d s + 0 1 i = 1 m α i t q 1 Ω Γ ( p i + q ) g i ( η , s ) h ( s ) d s = 0 1 G ( t , s ) h ( s ) d s .

The proof is completed. □

Lemma 2.4 The Green’s function G(t,s) in (2.5) satisfies the following conditions:

(P1) G(t,s) is continuous on [0,1]×[0,1];

(P2) G(t,s)0 for all 0s,t1;

(P3) G(t,s) max 0 t 1 G(t,s)g(s,s)+ i = 1 m α i Ω Γ ( p i + q ) g i (η,s) for all s,t[0,1];

(P4) 0 1 max 0 t 1 G(t,s)ds Γ ( q ) Γ ( 2 q ) + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) );

(P5) min η t 1 G(t,s) i = 1 m α i η q 1 Ω Γ ( p i + q ) g i (η,s) for s[0,1].

Proof It is easy to check that (P1) holds. To prove (P2), we will show that g(t,s)0 and g i (η,s)0, i=1,2,,m, for all 0s,t1.

Let k 1 (t,s)= t q 1 ( 1 s ) q 1 ( t s ) q 1 for 0st1, then we have

k 1 (t,s)= ( t s t ) q 1 ( t s ) q 1 ( t s ) q 1 ( t s ) q 1 =0.

Let k 2 (t,s)= t q 1 ( 1 s ) q 1 for 0ts1, then we get k 2 (t,s)0. Therefore, g(t,s)0 for all 0s,t1. Now, let k 3 i (η,s)= η p i + q 1 ( 1 s ) q 1 ( η s ) p i + q 1 for 0sη<1, then we get

k 3 i ( η , s ) = η p i + q 1 ( 1 s ) q 1 η p i + q 1 ( 1 s η ) p i + q 1 > η p i + q 1 ( 1 s ) q 1 η p i + q 1 ( 1 s ) p i + q 1 = η p i + q 1 [ ( 1 s ) q 1 ( 1 s ) p i + q 1 ] > 0 .

Let k 4 i (η,s)= η p i + q 1 ( 1 s ) q 1 for 0<ηs1, then we have k 4 i (η,s)0. Therefore, k 3 i (η,s), k 4 i (η,s)0, i=1,2,,m, which implies that g i (η,s)0, i=1,2,,m, for all 0s1.

To prove (P3), we will show that g(t,s)g(s,s) for s,t[0,1]. For 0st1, by the definition of k 1 (t,s), we have

t k 1 ( t , s ) = ( q 1 ) t q 2 ( 1 s ) q 1 ( q 1 ) ( t s ) q 2 = ( q 1 ) ( t t s ) q 2 ( 1 s ) ( q 1 ) ( t s ) q 2 ( q 1 ) ( t s ) q 2 ( 1 s ) ( q 1 ) ( t s ) q 2 = s ( q 1 ) ( t s ) q 2 0 .

Hence, k 1 (t,s) is decreasing with respect to t. Then we have g(t,s)g(s,s) for 0st1. For 0ts1, by the definition of k 2 (t,s), we have that k 2 (t,s) is increasing with respect to t. Thus g(t,s)g(s,s) for 0ts1. Therefore, g(t,s)g(s,s) for 0s,t1.

From the above analysis, we have for 0s1 that

G ( t , s ) max 0 t 1 G ( t , s ) = max 0 t 1 ( g ( t , s ) + i = 1 m α i t q 1 Ω Γ ( p i + q ) g i ( η , s ) ) g ( s , s ) + i = 1 m α i Ω Γ ( p i + q ) g i ( η , s ) .

To prove (P4), by direct integration, we have

0 1 max 0 t 1 G ( t , s ) d s 0 1 ( g ( s , s ) + i = 1 m α i Ω Γ ( p i + q ) g i ( η , s ) ) d s = 0 1 s q 1 ( 1 s ) q 1 Γ ( q ) d s + i = 1 m α i Ω Γ ( p i + q ) ( η 1 η p i + q 1 ( 1 s ) q 1 d s + 0 η [ η p i + q 1 ( 1 s ) q 1 ( η s ) p i + q 1 ] d s ) = Γ ( q ) Γ ( 2 q ) + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) .

To prove (P5), from g(t,s)0 and g i (η,s)0, i=1,2,,m, for all 0s,t1, we have

min η t 1 G ( t , s ) = min η t 1 ( g ( t , s ) + i = 1 m α i t q 1 Ω Γ ( p i + q ) g i ( η , s ) ) min η t 1 g ( t , s ) + min η t 1 i = 1 m α i t q 1 Ω Γ ( p i + q ) g i ( η , s ) i = 1 m α i η q 1 Ω Γ ( p i + q ) g i ( η , s )

for 0s1. This completes the proof. □

Let E=C([0,1],R) be the Banach space endowed with the supremum norm . Define the cone PE by

P= { u E : u ( t ) 0 } ,

and the operator A:PE by

Au(t):= 0 1 G(t,s)f ( s , u ( s ) ) ds.
(2.9)

In view of Lemma 2.3, the positive solutions of problem (1.1)-(1.2) are given by the operator equation u(t)=Au(t).

Lemma 2.5 Suppose that f:[0,1]×[0,)[0,) is continuous. The operator A:PP is completely continuous.

Proof Since G(t,s)0 for s,t[0,1], we have Au(t)0 for all uP. Hence, A:PP.

For a constant R>0, we define Φ={uP:u<R}.

Let

L= max 0 t 1 , 0 u R |f(t,u)|.
(2.10)

Then, for uΦ, from Lemma 2.4, one has

| A u ( t ) | = | 0 1 G ( t , s ) f ( s , u ( s ) ) d s | L 0 1 G ( t , s ) d s L 0 1 ( g ( s , s ) + i = 1 m α i Ω Γ ( p i + q ) g i ( η , s ) ) d s = L [ Γ ( q ) Γ ( 2 q ) + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) ] : = M .

Therefore, AuM, and so A(Φ) is uniformly bounded.

Now, we shall show that A(Φ) is equicontinuous. For uΦ, t 1 , t 2 [0,1], t 1 < t 2 , we have

|Au( t 2 )Au( t 1 )|L 0 1 |G( t 2 ,s)G( t 1 ,s)|ds,

where L is defined by (2.10). Since G(t,s) is continuous on [0,1]×[0,1], therefore G(t,s) is uniformly continuous on [0,1]×[0,1]. Hence, for any ε>0, there exists a positive constant

δ= 1 2 { ε Γ ( q ) L ( 1 1 q + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) ) } 1 q 1 >0,

whenever | t 2 t 1 |<δ, we have the following two cases.

Case 1. δ t 1 < t 2 <1.

Therefore,

| A u ( t 2 ) A u ( t 1 ) | L 0 1 | G ( t 2 , s ) G ( t 1 , s ) | d s = L [ 0 t 1 | G ( t 2 , s ) G ( t 1 , s ) | d s + t 1 t 2 | G ( t 2 , s ) G ( t 1 , s ) | d s + t 2 1 | G ( t 2 , s ) G ( t 1 , s ) | d s ] < ( t 2 q 1 t 1 q 1 ) L Γ ( q ) [ 0 1 ( 1 s ) q 1 d s + i = 1 m α i Ω Γ ( p i + q ) 0 1 g i ( η , s ) d s ] = ( t 2 q 1 t 1 q 1 ) L Γ ( q ) [ 1 q + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) ] ( q 1 ) δ q 1 L Γ ( q ) [ 1 q + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) ] < ε .

Case 2. 0 t 1 <δ, t 2 <2δ.

Therefore,

| A u ( t 2 ) A u ( t 1 ) | L 0 1 | G ( t 2 , s ) G ( t 1 , s ) | d s < ( t 2 q 1 t 1 q 1 ) L Γ ( q ) [ 1 q + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) ] t 2 q 1 L Γ ( q ) [ 1 q + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) ] < ( 2 δ ) q 1 L Γ ( q ) [ 1 q + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) ] = ε .

Thus, A(Φ) is equicontinuous. In view of the Arzelá-Ascoli theorem, we have that A ( Φ ) ¯ is compact, i.e., A:PP is a completely continuous operator. This completes the proof. □

For convenience, we set

Λ 1 = i = 1 m α i η p i + 2 ( q 1 ) Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) , Λ 2 = Γ ( q ) Γ ( 2 q ) + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) , Λ 3 = i = 1 m α i η p i + 2 ( q 1 ) ( 1 η ) q Ω Γ ( p i + q ) q .

3 Existence of at least one or two positive solutions

For the main results of this section, we use the well-known Guo-Krasnoselskii fixed point theorem.

Theorem 3.1 ([20])

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

  1. (i)

    Tuu, uP Ω 1 , and Tuu, uP Ω 2 ; or

  2. (ii)

    Tuu, uP Ω 1 , and Tuu, uP Ω 2 .

Then T has a fixed point in P( Ω ¯ 2 Ω 1 ).

Theorem 3.2 Let f:[0,1]×[0,)[0,) be a continuous function. Assume that there exist constants r 2 > r 1 >0, ρ 1 ( Λ 1 1 ,) and ρ 2 (0, Λ 2 1 ) such that:

(H1) f(t,u) ρ 1 r 1 , for (t,u)[0,1]×[0, r 1 ];

(H2) f(t,u) ρ 2 r 2 , for (t,u)[0,1]×[0, r 2 ].

Then boundary value problem (1.1)-(1.2) has at least one positive solution u such that

r 1 <u< r 2 .

Proof We shall show that the first part of Theorem 3.1 is satisfied. By Lemma 2.5, the operator A:PP is completely continuous.

Let Φ 1 ={uE:u< r 1 }, then for any uP Φ 1 , we have 0u(t) r 1 for all t[0,1]. From (H1), it follows for t[η,1] that

( A u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s 0 1 min η t 1 G ( t , s ) f ( s , u ( s ) ) d s ρ 1 r 1 i = 1 m α i η q 1 Ω Γ ( p i + q ) 0 1 g i ( η , s ) d s = ρ 1 r 1 i = 1 m α i η q 1 Ω Γ ( p i + q ) ( η 1 η p i + q 1 ( 1 s ) q 1 d s + 0 η [ η p i + q 1 ( 1 s ) q 1 ( η s ) p i + q 1 ] d s ) = ρ 1 r 1 [ i = 1 m α i η p i + 2 ( q 1 ) Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) ] r 1 = u ,

which yields

Auufor uP Φ 1 .
(3.1)

Let Φ 2 ={uE:u< r 2 }, then for any uP Φ 2 , we have 0u(t) r 2 for all t[0,1]. For t[0,1], assumption (H2) yields

( A u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s 0 1 G ( s , s ) f ( s , u ( s ) ) d s ρ 2 r 2 0 1 ( g ( s , s ) + i = 1 m α i Ω Γ ( p i + q ) g i ( η , s ) ) d s = ρ 2 r 2 [ Γ ( q ) Γ ( 2 q ) + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) ] r 2 = u ,

one has

Auufor uP Φ 2 .
(3.2)

Therefore, from (3.1), (3.2) and the first part of Theorem 3.1, it follows that A has a fixed point in P( Φ ¯ 2 Φ 1 ) which is a positive solution of boundary value problem (1.1)-(1.2). Hence, problem (1.1)-(1.2) has at least one positive solution u such that

r 1 <u< r 2 .

The proof is complete. □

Theorem 3.3 Let all the assumptions of Theorem  3.2 hold. In addition, assume that

(H3) lim u max t [ 0 , 1 ] f ( t , u ) u =.

Then boundary value problem (1.1)-(1.2) has at least two positive solutions u 1 and u 2 such that

0< r 1 < u 1 < r 2 < u 2 .

Proof It follows from Theorem 3.2 that there exists a positive solution u 1 such that r 1 < u 1 < r 2 . From (H3), there exists r > r 2 such that for any t[0,1] and for any M ( Λ 1 1 ,),

f(t,u) M ufor u r .

Let Φ 3 ={uE:u< r }. Then, for any uP Φ 3 and for t[η,1], we have

( A u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s 0 1 min η t 1 G ( t , s ) f ( s , u ( s ) ) d s M r i = 1 m α i η q 1 Ω Γ ( p i + q ) 0 1 g i ( η , s ) d s = M r [ i = 1 m α i η p i + 2 ( q 1 ) Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) ] r = u .

This implies that

Auufor uP Φ 3 .
(3.3)

It follows from (3.2), (3.3) and the second part of Theorem 3.1 that A has a fixed point in P( Φ ¯ 3 Φ 2 ).

Therefore, we conclude that boundary value problem (1.1)-(1.2) has at least two positive solutions such that

0< r 1 < u 1 < r 2 < u 2 .

 □

Similarly to the previous theorems, we can prove the following.

Theorem 3.4 Let f:[0,1]×[0,)[0,) be a continuous function. Assume that there exist constants 0< r 1 < r 2 and ρ 1 ( Λ 1 1 ,), ρ 2 (0, Λ 2 1 ) such that:

(H4) f(t,u) ρ 2 r 1 for (t,u)[0,1]×[0, r 1 ];

(H5) f(t,u) ρ 1 r 2 for (t,u)[0,1]×[0, r 2 ];

(H6) lim u max t [ 0 , 1 ] f ( t , u ) u =0.

Then boundary value problem (1.1)-(1.2) has at least two positive solutions u 1 and u 2 such that

0< r 1 < u 1 < r 2 < u 2 .

Corollary 3.1 Assume that conditions (H4)-(H5) are satisfied. Then boundary value problem (1.1)-(1.2) has at least one positive solution u such that

r 1 <u< r 2 .

4 Existence of at least three positive solutions

In this section we use the Leggett-Williams fixed point theorem to prove the existence of at least three positive solutions.

Definition 4.1 A continuous mapping θ:P[0,+) is said to be a nonnegative continuous concave functional on the cone P of a real Banach space E provided that

θ ( λ u + ( 1 λ ) v ) λθ(u)+(1λ)θ(v)

for all u,vP and λ[0,1].

Let a,b,d>0 be constants. We define P d ={uP:u<d}, P ¯ d ={uP:ud} and P(θ,a,b)={uP:θ(u)a,ub}.

Theorem 4.1 ([21])

Let P be a cone in the real Banach space E and c>0 be a constant. Assume that there exists a concave nonnegative continuous functional θ on P with θ(u)u for all u P ¯ c . Let A: P ¯ c P ¯ c be a completely continuous operator. Suppose that there exist constants 0<a<b<dc such that the following conditions hold:

  1. (i)

    {uP(θ,b,d):θ(u)>b} and θ(Au)>b for uP(θ,b,d);

  2. (ii)

    Au<a for ua;

  3. (iii)

    θ(Au)>b for uP(θ,b,c) with Au>d.

Then A has at least three fixed points u 1 , u 2 and u 3 in P ¯ c .

Furthermore, u 1 <a, b<θ( u 2 ), a< u 3 with θ( u 3 )<b.

We now prove the following result.

Theorem 4.2 Let f:[0,1]×[0,)[0,) be a continuous function. Suppose that there exist constants 0<a<b<c such that the following assumptions hold:

(H7) f(t,u)< Λ 2 1 a for (t,u)[0,1]×[0,a];

(H8) f(t,u)> Λ 3 1 b for (t,u)[η,1]×[b,c];

(H9) f(t,u) Λ 2 1 c for (t,u)[0,1]×[0,c].

Then boundary value problem (1.1)-(1.2) has at least three positive solutions u 1 , u 2 and u 3 with

u 1 <a,b< min η t 1 u 2 (t)

and

a< u 3 with min η t 1 u 3 (t)<b.

Proof We will show that all the conditions of the Leggett-Williams fixed point theorem are satisfied for the operator A defined by (2.9).

For u P ¯ c , we have uc. From condition (H9), we have f(t,u(t)) Λ 2 1 c for t[0,1]. Therefore,

( A u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) ) d s 0 1 G ( s , s ) f ( s , u ( s ) ) d s Λ 2 1 c 0 1 ( g ( s , s ) + i = 1 m α i Ω Γ ( p i + q ) g i ( η , s ) ) d s = Λ 2 1 c [ Γ ( q ) Γ ( 2 q ) + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) ] = c ,

which implies Auc. Hence, A: P ¯ c P ¯ c .

If u P ¯ a , then condition (H7) yields

( A u ) ( t ) < Λ 2 1 a 0 1 ( g ( s , s ) + i = 1 m α i Ω Γ ( p i + q ) g i ( η , s ) ) d s = Λ 2 1 a [ Γ ( q ) Γ ( 2 q ) + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) ] = a .

Thus Au<a. Therefore, condition (ii) of Theorem 4.1 holds.

Define a concave nonnegative continuous functional θ on P by θ(u)= min t [ η , 1 ] |u(t)|. To check condition (i) of Theorem 4.1, we choose u(t)=(b+c)/2 for t[0,1]. It is easy to see that u(t)=(b+c)/2P(θ,b,c) and θ(u)=θ((b+c)/2)>b; consequently, the set {uP(θ,b,c):θ(u)>b}. Hence, if uP(θ,b,c), then bu(t)c for t[η,1]. From condition (H8), we have

θ ( A u ) = min η t 1 | ( A u ) ( t ) | η 1 min η t 1 G ( t , s ) f ( s , u ( s ) ) d s > Λ 3 1 b i = 1 m α i η q 1 Ω Γ ( p i + q ) η 1 g i ( η , s ) d s = Λ 3 1 b [ i = 1 m α i η p i + 2 ( q 1 ) ( 1 η ) q Ω Γ ( p i + q ) q ] = b .

Thus θ(Au)>b for all uP(θ,b,c). This shows that condition (i) of Theorem 4.1 is also satisfied.

We finally show that condition (iii) of Theorem 4.1 also holds. Assume that uP(θ,b,c) with Au>d, then we have bu(t)c for all t[η,1]. From (H8) and Lemma 2.4, one has

θ ( A u ) = min η t 1 | ( A u ) ( t ) | > Λ 3 1 b i = 1 m α i η q 1 Ω Γ ( p i + q ) η 1 g i ( η , s ) d s = Λ 3 1 b [ i = 1 m α i η p i + 2 ( q 1 ) ( 1 η ) q Ω Γ ( p i + q ) q ] = b .

So, condition (iii) of Theorem 4.1 is satisfied. Therefore, an application of Theorem 4.1 implies that boundary value problem (1.1)-(1.2) has at least three positive solutions u 1 , u 2 and u 3 such that

u 1 <a,b< min η t 1 | u 2 (t)|anda< u 3 with min η t 1 | u 3 (t)|<b.

The proof is complete. □

5 Examples

In this section, we present some examples to illustrate our results.

Example 5.1 Consider the following three-point three-term fractional integral boundary value problem:

D 3 2 u(t)+f(t,u)=0,t(0,1),
(5.1)
u(0)=0,u(1)= 3 ( I π 2 u ) ( 1 4 ) + π 2 ( I 3 2 u ) ( 1 4 ) +2 ( I 2 3 u ) ( 1 4 ) ,
(5.2)

where

f(t,u)= { u ( 1 u ) + 6 ( 1 + t ) ; 0 t 1 ; 0 u 1 , 6 ( 1 + t ) e 1 u + sin 2 ( u π ) ; 0 t 1 ; 1 u < .

Set m=3, η=1/4, q=3/2, p 1 =π/2, p 2 =3/2, p 3 =2/3, α 1 = 3 , α 2 =π/2, α 3 =2, and we can show that

Ω=1 i = 1 m α i η p i + q 1 Γ ( q ) Γ ( p i + q ) 0.590859.

Through a simple calculation we can get

Λ 1 = i = 1 m α i η p i + 2 ( q 1 ) Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) 0.218023 , Λ 2 = Γ ( q ) Γ ( 2 q ) + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) 0.879159 .

Choose r 1 =1, r 2 =14, ρ 1 =5 and ρ 2 =1, then f(t,u) satisfies

f(t,u)65= ρ 1 r 1 ,(t,u)[0,1]×[0,1]

and

f(t,u)1314= ρ 2 r 2 ,(t,u)[0,1]×[0,14].

Thus, (H1) and (H2) hold. By Theorem 3.2, we have that boundary value problem (5.1)-(5.2) has at least one positive solution u such that 1<u<14.

Example 5.2 Consider the following three-point four-term fractional integral boundary value problem:

D 3 2 u(t)+f(t,u)=0,t(0,1),
(5.3)
{ u ( 0 ) = 0 , u ( 1 ) = 1 6 ( I 1 2 u ) ( 2 3 ) + 3 2 ( I 2 u ) ( 2 3 ) + 1 2 ( I π u ) ( 2 3 ) + π 6 ( I 5 6 u ) ( 2 3 ) ,
(5.4)

where

f(t,u)= { u 2 ( 1 2 u ) + 2 ( 1 + t 2 ) ; 0 t 1 ; 0 u 1 / 2 , 4 ( 1 + t 2 ) cos 2 ( u π 2 ) + 1 95 ( u 1 2 ) 2 ; 0 t 1 ; 1 / 2 u < .

Here m=4, η=2/3, q=3/2, p 1 =1/2, p 2 = 2 , p 3 = π , p 4 =5/6, α 1 =1/6, α 2 = 3 /2, α 3 =1/2, α 4 = π /6, and we can show that

Ω=1 i = 1 m α i η p i + q 1 Γ ( q ) Γ ( p i + q ) 0.515010.

Through a simple calculation we can get

Λ 1 = i = 1 m α i η p i + 2 ( q 1 ) Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) 0.351492 , Λ 2 = Γ ( q ) Γ ( 2 q ) + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) 0.873602 .

Choose r 1 =1/2, r 2 =10, ρ 1 =3 and ρ 2 =1, then f(t,u) satisfies

f(t,u)2 3 2 = ρ 1 r 1 ,(t,u)[0,1]×[0,1/2]

and

f(t,u)910= ρ 2 r 2 ,(t,u)[0,1]×[0,10],

and

lim u max t [ 0 , 1 ] f ( t , u ) u = lim u max t [ 0 , 1 ] [ 4 ( 1 + t 2 ) cos 2 ( u π / 2 ) + ( 1 / 95 ) ( u 1 / 2 ) 2 u ] =.

Thus, (H1), (H2) and (H3) hold. By Theorem 3.3, we have that boundary value problem (5.3)-(5.4) has at least two positive solutions u 1 and u 2 such that 0<1/2< u 1 <10< u 2 .

Example 5.3 Consider the following three-point five-term fractional integral boundary value problem:

D 3 2 u(t)+f(t,u)=0,t(0,1),
(5.5)
{ u ( 0 ) = 0 , u ( 1 ) = 1 2 ( I 1 2 u ) ( 1 2 ) + 4 3 ( I 3 2 u ) ( 1 2 ) + π 4 ( I 5 2 u ) ( 1 2 ) u ( 1 ) = + 5 6 ( I 7 2 u ) ( 1 2 ) + 3 2 ( I 9 2 u ) ( 1 2 ) ,
(5.6)

where

f(t,u)= { u ( 3 4 u ) + 1 4 ( t + 1 ) ; 0 t 1 ; 0 u 3 / 4 , 1 2 ( t + 1 ) cos 2 ( u π ) + 43 ( u 3 4 ) 2 ; 0 t 1 ; 3 / 4 u 5 / 4 , 1 4 ( t + 44 ) + sin 2 ( u 5 4 ) π ; 0 t 1 ; 5 / 4 u < .

Set m=5, η=1/2, q=3/2, p 1 =1/2, p 2 =3/2, p 3 =5/2, p 4 =7/2, p 5 =9/2, α 1 =1/2, α 2 =4/3, α 3 = π /4, α 4 =5/6, α 5 = 3 /2, and we can show that

Ω=1 i = 1 m α i η p i + q 1 Γ ( q ) Γ ( p i + q ) 0.620434.

Through a simple calculation we can get

Λ 2 = Γ ( q ) Γ ( 2 q ) + i = 1 m α i η p i + q 1 Ω Γ ( p i + q ) ( p i + q ( 1 η ) q ( p i + q ) ) 0.755575 , Λ 3 = i = 1 m α i η p i + 2 ( q 1 ) ( 1 η ) q Ω Γ ( p i + q ) q 0.115052 .

Choose a=3/4, b=5/4 and c=10, then f(t,u) satisfies

f(t,u) 41 64 <0.992622 Λ 2 1 a,(t,u)[0,1]×[0,3/4]

and

f(t,u) 89 8 >10.864653 Λ 3 1 b,(t,u)[1/2,1]×[5/4,10],

and

f(t,u) 49 4 13.234954= Λ 2 1 c,(t,u)[0,1]×[0,10].

Thus, (H7), (H8) and (H9) hold. By Theorem 4.2, we have that boundary value problem (5.5)-(5.6) has at least three positive solutions u 1 , u 2 and u 3 such that u 1 <3/4, 5/4< min ( 1 / 2 ) t 1 u 2 (t) and 3/4< u 3 with min ( 1 / 2 ) t 1 u 3 (t)<5/4.

Authors’ information

Sotiris K Ntouyas is a member of Nonlinear Analysis and Applied Mathematics (NAAM) - Research Group at King Abdulaziz University, Jeddah, Saudi Arabia.

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Acknowledgements

We would like to thank the reviewers for their valuable comments and suggestions on the manuscript. This research of J Tariboon and W Sudsutad is supported by King Mongkut’s University of Technology North Bangkok, Thailand.

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Tariboon, J., Ntouyas, S.K. & Sudsutad, W. Positive solutions for fractional differential equations with three-point multi-term fractional integral boundary conditions. Adv Differ Equ 2014, 28 (2014). https://doi.org/10.1186/1687-1847-2014-28

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Keywords

  • fractional differential equations
  • nonlocal boundary conditions
  • positive solutions
  • fixed point theorem