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Theory and Modern Applications

Nonlinear fractional Caputo-Langevin equation with nonlocal Riemann-Liouville fractional integral conditions

Abstract

In this paper, we study the existence and uniqueness of solution for a problem consisting of a sequential nonlinear fractional Caputo-Langevin equation with nonlocal Riemann-Liouville fractional integral conditions. A variety of fixed point theorems, such as Banach’s fixed point theorem, Krasnoselskii’s fixed point theorem, Leray-Schauder’s nonlinear alternative and Leray-Schauder degree theory, are used. Examples illustrating the obtained results are also presented.

MSC:26A33, 34A08, 34B10.

1 Introduction

In this paper, we concentrate on the study of existence and uniqueness of solution for the following nonlinear fractional Caputo-Langevin equation with nonlocal Riemann-Liouville fractional integral conditions:

D p ( D q + λ ) x ( t ) = f ( t , x ( t ) ) , t [ 0 , T ] , i = 1 m μ i I α i x ( η i ) = σ 1 , j = 1 n ν j I β j x ( ξ j ) = σ 2 ,
(1.1)

where 0<p,q1, 1<p+q2, D q and D p are the Caputo fractional derivatives of order q and p, respectively, I ϕ is the Riemann-Liouville fractional integral of order ϕ, where ϕ= α i , β j >0, η i , ξ j (0,T) are given points, μ i , ν j ,λ, σ 1 , σ 2 R, i=1,2,,m, j=1,2,,n, and f:[0,T]×RR is a continuous function.

The significance of studying problem (1.1) is that the nonlocal conditions are very general and include many conditions as special cases. In particular, if α i = β j =1, for all i=1,2,,m, j=1,2,,n, then the nonlocal condition of (1.1) reduces to

{ μ 1 0 η 1 x ( s ) d s + μ 2 0 η 2 x ( s ) d s + + μ m 0 η m x ( s ) d s = σ 1 , ν 1 0 ξ 1 x ( s ) d s + ν 2 0 ξ 2 x ( s ) d s + + ν n 0 ξ n x ( s ) d s = σ 2 ,
(1.2)

and if σ 1 = σ 2 =0, m=n=2, μ 2 , ν 2 0, then (1.2) is reduced to

ε 1 0 η 1 x(s)ds= 0 η 2 x(s)ds, ε 2 0 ξ 1 x(s)ds= 0 ξ 2 x(s)ds,
(1.3)

where ε 1 =( μ 1 / μ 2 ) and ε 2 =( ν 1 / ν 2 ). Note that the nonlocal conditions (1.2) and (1.3) do not contain values of an unknown function x on the left-hand side and the right-hand side of boundary points t=0 and t=T, respectively.

Fractional differential equations have been shown to be very useful in the study of models of many phenomena in various fields of science and engineering, such as physics, chemistry, biology, signal and image processing, biophysics, blood flow phenomena, control theory, economics, aerodynamics and fitting of experimental data. For examples and recent development of the topic, see [113] and the references cited therein.

The Langevin equation (first formulated by Langevin in 1908) is found to be an effective tool to describe the evolution of physical phenomena in fluctuating environments [14]. For some new developments on the fractional Langevin equation, see, for example, [1524].

In the present paper several new existence and uniqueness results are proved by using a variety of fixed point theorems (such as Banach’s contraction principle, Krasnoselskii’s fixed point theorem, Leray-Schauder’s nonlinear alternative and Leray-Schauder’s degree theory).

The rest of the paper is organized as follows. In Section 2 we recall some preliminary facts that we need in the sequel. In Section 3 we present our existence and uniqueness results. Examples illustrating the obtained results are presented in Section 4.

2 Preliminaries

In this section, we introduce some notations and definitions of fractional calculus [2, 3] and present preliminary results needed in our proofs later.

Definition 2.1 For an at least n-times differentiable function g:[0,)R, the Caputo derivative of fractional order q is defined as

D q c g(t)= 1 Γ ( n q ) 0 t ( t s ) n q 1 g ( n ) (s)ds,n1<q<n,n=[q]+1,

where [q] denotes the integer part of the real number q.

Definition 2.2 The Riemann-Liouville fractional integral of order q is defined as

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

provided the integral exists.

Lemma 2.1 For q>0, the general solution of the fractional differential equation D q c u(t)=0 is given by

u(t)= c 0 + c 1 t++ c n 1 t n 1 ,

where c i R, i=1,2,,n1 (n=[q]+1).

In view of Lemma 2.1, it follows that

I q c D q u(t)=u(t)+ c 0 + c 1 t++ c n 1 t n 1

for some c i R, i=1,2,,n1 (n=[q]+1).

In the following, for the sake of convenience, we set constants

Ω 1 = i = 1 m μ i η i q + α i Γ ( α i + q + 1 ) , Ψ 1 = i = 1 m μ i η i α i Γ ( α i + 1 ) , Ω 2 = j = 1 n ν j ξ j q + β j Γ ( β j + q + 1 ) , Ψ 2 = j = 1 n ν j ξ j β i Γ ( β j + 1 ) ,

and Δ= Ω 1 Ψ 2 Ω 2 Ψ 1 .

Lemma 2.2 Let Δ0, 0<p,q1, 1<p+q2, α i , β j >0, μ i , ν j ,λ, σ 1 , σ 2 R, η i , ξ j (0,T), i=1,2,,m, j=1,2,,n, and yC([0,T],R). Then the nonlinear fractional Caputo-Langevin equation

D p ( D q + λ ) x(t)=y(t),
(2.1)

subject to the nonlocal Riemann-Liouville fractional integral conditions

i = 1 m μ i I α i x( η i )= σ 1 , j = 1 n ν j I β j x( ξ j )= σ 2 ,
(2.2)

has a unique solution given by

x ( t ) = I q + p y ( t ) λ I q x ( t ) + Ψ 2 t q Ω 2 Γ ( q + 1 ) Δ Γ ( q + 1 ) ( σ 1 i = 1 m μ i I α i + q + p y ( η i ) + λ i = 1 m μ i I α i + q x ( η i ) ) Ψ 1 t q Ω 1 Γ ( q + 1 ) Δ Γ ( q + 1 ) ( σ 2 j = 1 n ν j I β j + q + p y ( ξ j ) + λ j = 1 n ν j I β j + q x ( ξ j ) ) .
(2.3)

Proof The general solution of equation (2.1) is expressed as the following integral equation:

x(t)= I q + p y(t)λ I q x(t)+ c 0 t q Γ ( q + 1 ) + c 1 ,
(2.4)

where c 0 and c 1 are arbitrary constants. By taking the Riemann-Liouville fractional integral of order α i >0 for (2.4), we get

I α i x(t)= I α i + q + p y(t)λ I α i + q x(t)+ c 0 ( t α i + q Γ ( α i + q + 1 ) ) + c 1 ( t α i Γ ( α i + 1 ) ) .

In particular, for t= η i , we have

I α i x( η i )= I α i + q + p y( η i )λ I α i + q x( η i )+ c 0 ( η i α i + q Γ ( α i + q + 1 ) ) + c 1 ( η i α i Γ ( α i + 1 ) ) .

Repeating the above process for the Riemann-Liouville fractional integral of order β j >0, substituting t= ξ j and applying the nonlocal condition (2.2), we obtain the following system of linear equations:

c 0 Ω 1 + c 1 Ψ 1 = σ 1 i = 1 m μ i I α i + q + p y ( η i ) + λ i = 1 m μ i I α i + q x ( η i ) , c 0 Ω 2 + c 1 Ψ 2 = σ 2 j = 1 n ν j I β j + q + p y ( ξ j ) + λ j = 1 n ν j I β j + q x ( ξ j ) .
(2.5)

Solving the linear system of equations in (2.5) for constants c 0 , c 1 , we have

c 0 = Ψ 2 Δ ( σ 1 i = 1 m μ i I α i + q + p y ( η i ) + λ i = 1 m μ i I α i + q x ( η i ) ) c 0 = Ψ 1 Δ ( σ 2 j = 1 n ν j I β j + q + p y ( ξ j ) + λ j = 1 n ν j I β j + q x ( ξ j ) ) , c 1 = Ω 1 Δ ( σ 2 j = 1 n ν j I β j + q + p y ( ξ j ) + λ j = 1 n ν j I β j + q x ( ξ j ) ) c 1 = Ω 2 Δ ( σ 1 i = 1 m μ i I α i + q + p y ( η i ) + λ i = 1 m μ i I α i + q x ( η i ) ) .

Substituting c 0 and c 1 into (2.4), we obtain solution (2.3). □

3 Main results

Throughout this paper, for convenience, the expression I x ϕ(y) means

I x ϕ(y)= 1 Γ ( x ) 0 y ( y s ) x 1 ϕ(s)dsfor y[0,T].

Let C=C([0,T],R) denote the Banach space of all continuous functions from [0,T] to endowed with the norm defined by u= sup t [ 0 , T ] |u(t)|. As in Lemma 2.2, we define an operator K:CC by

K x ( t ) = I q + p f ( s , x ( s ) ) ( t ) λ I q x ( t ) + Ψ 2 t q Ω 2 Γ ( q + 1 ) Δ Γ ( q + 1 ) ( σ 1 i = 1 m μ i I α i + q + p f ( s , x ( s ) ) ( η i ) + λ i = 1 m μ i I α i + q x ( η i ) ) Ψ 1 t q Ω 1 Γ ( q + 1 ) Δ Γ ( q + 1 ) ( σ 2 j = 1 n ν j I β j + q + p f ( s , x ( s ) ) ( ξ j ) + λ j = 1 n ν j I β j + q x ( ξ j ) ) .
(3.1)

It should be noticed that problem (1.1) has solutions if and only if the operator K has fixed points.

In the following subsections, we prove existence, as well as existence and uniqueness results, for problem (1.1) by using a variety of fixed point theorems.

3.1 Existence and uniqueness result via Banach’s fixed point theorem

Theorem 3.1 Let f:[0,T]×RR be a continuous function. Assume that

(H1) there exists a constant L>0 such that |f(t,x)f(t,y)|L|xy| for each t[0,T] and x,yR.

If

L Λ 1 + Λ 2 <1,
(3.2)

where constants Λ 1 , Λ 2 are defined by

Λ 1 : = T q + p Γ ( q + p + 1 ) + ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) i = 1 m | μ i | η i α i + q + p Γ ( α i + q + p + 1 ) Λ 1 : = + ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) j = 1 n | ν j | ξ j β j + q + p Γ ( β j + q + p + 1 ) ,
(3.3)
Λ 2 : = | λ | ( T q Γ ( q + 1 ) + ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) i = 1 m | μ i | η i α i + q Γ ( α i + q + 1 ) Λ 2 : = + ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) j = 1 n | ν j | ξ j β j + q Γ ( β j + q + 1 ) ) ,
(3.4)

then problem (1.1) has a unique solution on [0,T].

Proof Problem (1.1) is equivalent to a fixed point problem by defining the operator K as in (3.1), which yields x=Kx. Using the Banach contraction mapping principle, we will show that problem (1.1) has a unique solution. Setting sup t [ 0 , T ] |f(t,0)|=M<, we define a set B r ={xC:xr},

r M Λ 1 + Φ 1 ( L Λ 1 + Λ 2 ) ,

where

Φ:=| σ 1 | ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) +| σ 2 | ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) .
(3.5)

For any x B r , we have

| K x ( t ) | I q + p | f ( s , x ( s ) ) | ( t ) + | λ | I q | x ( s ) | ( t ) + | Ψ 2 | t q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( | σ 1 | + i = 1 m | μ i | I α i + q + p | f ( s , x ( s ) ) | ( η i ) + | λ | i = 1 m | μ i | I α i + q | x ( s ) | ( η i ) ) + | Ψ 1 | t q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( | σ 2 | + j = 1 n | ν j | I β j + q + p | f ( s , x ( s ) ) | ( ξ j ) + | λ | j = 1 n | ν j | I β j + q | x ( s ) | ( ξ j ) ) I q + p ( | f ( s , x ( s ) ) f ( s , 0 ) | + | f ( s , 0 ) | ) ( t ) + | λ | I q | x ( s ) | ( t ) + | Ψ 2 | t q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( | σ 1 | + i = 1 m | μ i | I α i + q + p ( | f ( s , x ( s ) ) f ( s , 0 ) | + | f ( s , 0 ) | ) ( η i ) + | λ | i = 1 m | μ i | I α i + q | x ( s ) | ( η i ) ) + | Ψ 1 | t q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( | σ 2 | + j = 1 n | ν j | I β j + q + p ( | f ( s , x ( s ) ) f ( s , 0 ) | + | f ( s , 0 ) | ) ( ξ j ) + | λ | j = 1 n | ν j | I β j + q | x ( s ) | ( ξ j ) ) ( L r + M ) [ T q + p Γ ( q + p + 1 ) + ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) i = 1 m | μ i | η i α i + q + p Γ ( α i + q + p + 1 ) + ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) j = 1 n | ν j | ξ j β j + q + p Γ ( β j + q + p + 1 ) ] + r | λ | [ T q Γ ( q + 1 ) + ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) i = 1 m | μ i | η i α i + q Γ ( α i + q + 1 ) + ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) j = 1 n | ν j | ξ j β j + q Γ ( β j + q + 1 ) ] + | σ 1 | ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) + | σ 2 | ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) = ( L r + M ) Λ 1 + r Λ 2 + Φ r ,

which implies that K B r B r . Next, we need to show that K is a contraction mapping. Let x,yC. Then, for t[0,T], we have

| K x ( t ) K y ( t ) | I q + p | f ( s , x ( s ) ) f ( s , y ( s ) ) | ( t ) + | λ | I q | x ( s ) y ( s ) | ( t ) + | Ψ 2 | t q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( i = 1 m | μ i | I α i + q + p | f ( s , x ( s ) ) f ( s , y ( s ) ) | ( η i ) + | λ | i = 1 m | μ i | I α i + q | x ( s ) y ( s ) | ( η i ) ) + | Ψ 1 | t q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( j = 1 n | ν j | I β j + q + p | f ( s , x ( s ) ) f ( s , y ( s ) ) | ( ξ j ) + | λ | j = 1 n | ν j | I β j + q | x ( s ) y ( s ) | ( ξ j ) ) x y [ L T p + q Γ ( 1 + p + q ) + | λ | T q Γ ( 1 + q ) + | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( i = 1 m L | μ i | η i α i + q + p Γ ( α i + p + q + 1 ) + | λ | i = 1 m | μ i | η i α i + q Γ ( α i + q + 1 ) ) + | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( j = 1 n L | ν j | ξ j β j + q + p Γ ( β j + q + p + 1 ) + | λ | j = 1 n | ν j | ξ j β j + q Γ ( β j + q + 1 ) ) ] ( L Λ 1 + Λ 2 ) x y ,

which leads to KxKy(L Λ 1 + Λ 2 )xy. Since (L Λ 1 + Λ 2 )<1, K is a contraction mapping. Therefore K has only one fixed point, which implies that problem (1.1) has a unique solution. □

3.2 Existence and uniqueness result via Banach’s fixed point theorem and Hölder’s inequality

Now we give another existence and uniqueness result for problem (1.1) by using Banach’s fixed point theorem and Hölder’s inequality. For σ(0,1), we set

Λ 3 : = [ ( 1 σ q + p σ ) 1 σ T q + p σ Γ ( q + p ) + | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( i = 1 m ( 1 σ α i + q + p σ ) 1 σ | μ i | η i α i + q + p Γ ( α i + q + p ) ) + | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( j = 1 n ( 1 σ β j + q + p σ ) 1 σ | ν j | ξ j β j + q + p Γ ( β j + q + p ) ) ] .
(3.6)

Theorem 3.2 Let f:[0,T]×RR be a continuous function. In addition we assume that

(H2) |f(t,x)f(t,y)|δ(t)|xy| for each t[0,T], x,yR, where δ L σ ([0,T], R + ), σ(0,1).

Denote δ= ( 0 T δ 1 σ ( s ) d s ) σ .

If

Λ 3 δ+ Λ 2 <1,
(3.7)

where Λ 2 and Λ 3 are defined by (3.4) and (3.6), respectively, then problem (1.1) has a unique solution.

Proof For x,yC and each t[0,T], by Hölder’s inequality, we have

| K x ( t ) K y ( t ) | I q + p δ ( s ) | x ( s ) y ( s ) | ( t ) + | λ | I q | x ( s ) y ( s ) | ( t ) + | Ψ 2 | t q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( i = 1 m | μ i | I α i + q + p δ ( s ) | x ( s ) y ( s ) | ( η i ) + | λ | i = 1 m | μ i | I α i + q | x ( s ) y ( s ) | ( η i ) ) + | Ψ 1 | t q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( j = 1 n | ν j | I β j + q + p δ ( s ) | x ( s ) y ( s ) | ( ξ j ) + | λ | j = 1 n | ν j | I β j + q | x ( s ) y ( s ) | ( ξ j ) ) x y [ 0 t ( t s ) q + p 1 δ ( s ) d s + | λ | t q Γ ( q + 1 ) + | Ψ 2 | t q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( i = 1 m | μ i | 0 η i ( η i s ) α i + q + p 1 δ ( s ) d s + | λ | i = 1 m | μ i | η i α i + q Γ ( α i + q + 1 ) ) + | Ψ 1 | t q + | Ω 1 | Γ ( q + 1 ) | Δ | ( j = 1 n | ν j | 0 ξ j ( ξ j s ) β j + q + p 1 δ ( s ) d s + | λ | j = 1 n | ν j | ξ j β j + q Γ ( β j + q + 1 ) ) ] x y [ ( 0 t ( t s ) q + p 1 1 σ d s ) 1 σ ( 0 t δ ( s ) 1 σ d s ) σ + | λ | t q Γ ( q + 1 ) + | Ψ 2 | t q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( i = 1 m | μ i | ( 0 η i ( η i s ) α i + q + p 1 1 σ d s ) 1 σ × ( 0 η i δ ( s ) 1 σ d s ) σ + | λ | i = 1 m | μ i | η i α i + q γ ( α i + q + 1 ) ) + | Ψ 1 | t q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( j = 1 n | ν j | ( 0 ξ j ( ξ j s ) β j + q + p 1 1 σ d s ) 1 σ × ( 0 ξ j δ ( s ) 1 σ d s ) σ + | λ | j = 1 n | ν j | ξ j β j + q Γ ( β j + q + 1 ) ) ] ( Λ 3 δ + Λ 2 ) x y .

Therefore,

KxKy ( Λ 3 δ + Λ 2 ) xy.

Hence, from (3.7), K is a contraction mapping. Banach’s fixed point theorem implies that K has a unique fixed point, which is the unique solution of problem (1.1). This completes the proof. □

3.3 Existence result via Krasnoselskii’s fixed point theorem

Lemma 3.1 (Krasnoselskii’s fixed point theorem [25])

Let M be a closed, bounded, convex and nonempty subset of a Banach space X. Let A, B be operators such that (a) Ax+BxM whenever x,yM; (b) A is compact and continuous; (c) B is a contraction mapping. Then there exists zM such that z=Az+Bz.

Theorem 3.3 Let f:[0,T]×RR be a continuous function. Moreover, we assume that

(H3) |f(t,x)|ϕ(t), (t,x)[0,T]×R and ϕC([0,T], R + ).

Then problem (1.1) has at least one solution on [0,T] if

Λ 2 <1,
(3.8)

where Λ 2 is defined by (3.4).

Proof We define the operators A and on B r by

A x ( t ) = I q + p f ( s , x ( s ) ) ( t ) + Ψ 2 t q Ω 2 Γ ( q + 1 ) Δ Γ ( q + 1 ) ( σ 1 i = 1 m μ i I α i + q + p f ( s , x ( s ) ) ( η i ) ) A x ( t ) = Ψ 1 t q Ω 1 Γ ( q + 1 ) Δ Γ ( q + 1 ) ( σ 2 j = 1 n ν j I β j + q + p f ( s , x ( s ) ) ( ξ j ) ) , B x ( t ) = λ I q + p x ( s ) ( t ) + Ψ 2 t q Ω 2 Γ ( q + 1 ) Δ Γ ( q + 1 ) ( λ i = 1 m μ i I α i + q x ( s ) ( η i ) ) B x ( t ) = Ψ 1 t q Ω 1 Γ ( q + 1 ) Δ Γ ( q + 1 ) ( λ j = 1 n ν j I β j + q x ( s ) ( ξ j ) ) ,

where the ball B r is defined by B r ={xC,xr} for some suitable r such that

r Λ 1 ϕ + Φ 1 Λ 2 ,

with ϕ= sup t [ 0 , T ] |ϕ(t)| and Λ 1 , Λ 2 and Φ are defined by (3.3), (3.4) and (3.5), respectively. To show that Ax+By B r , we let x,y B r . Then we have

| A x ( t ) + B y ( t ) | I q + p | f ( s , x ( s ) ) | ( t ) + | Ψ 2 | t q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) × ( | σ 1 | + i = 1 m | μ i | I α i + q + p | f ( s , x ( s ) ) | ( η i ) ) + | Ψ 1 | t q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( | σ 2 | + j = 1 n | ν j | I β j + q + p | f ( s , x ( s ) ) | ( ξ j ) ) + | λ | I q + p | y ( s ) | ( t ) + | Ψ 2 | t q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) × ( | λ | i = 1 m | μ i | I α i + q | y ( s ) | ( η i ) ) + | Ψ 1 | t q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( | λ | j = 1 n | ν j | I β j + q | y ( s ) | ( ξ j ) ) Λ 1 ϕ + Φ + r Λ 2 r .

It follows that Ax+By B r , and thus condition (a) of Lemma 3.1 is satisfied. For x,yC, we have BxBy Λ 2 xy. Since Λ 2 <1, the operator is a contraction mapping. Therefore, condition (c) of Lemma 3.1 is satisfied.

The continuity of f implies that the operator A is continuous. For x B r , we obtain

Ax Λ 1 ϕ+Φ.

This means that the operator A is uniformly bounded on B r . Next we show that A is equicontinuous. We set sup t [ 0 , T ] f(t,x(t))= f ¯ , and consequently we get

| A x ( t 2 ) A x ( t 1 ) | 1 Γ ( q + p + 1 ) | 0 t 1 [ ( t 2 s ) q + p 1 ( t 1 s ) q + p 1 ] f ( s , x ( s ) ) d s + t 1 t 2 ( t 2 s ) q + p 1 f ( s , x ( s ) ) d s | + | Ψ 2 | | t 2 q t 1 q | | Δ | Γ ( q + 1 ) ( | σ 1 | + i = 1 m | μ i | I α i + q + p | f ( s , x ( s ) ) | ( η i ) ) + | Ψ 1 | | t 2 q t 1 q | | Δ | Γ ( q + 1 ) ( | σ 2 | + j = 1 n | ν j | I β j + q + p | f ( s , x ( s ) ) | ( ξ j ) ) f ¯ Γ ( q + p + 1 ) | t 2 q + p t 1 q + p | + f ¯ | Ψ 2 | | t 2 q t 1 q | | Δ | ( | σ 1 | + i = 1 m | μ i | η i α i + q + p Γ ( α i + q + p + 1 ) ) + f ¯ | Ψ 1 | | t 2 q t 1 q | | Δ | Γ ( q + 1 ) ( | σ 2 | + j = 1 n | ν j | ξ j β j + q + p Γ ( β j + q + p + 1 ) ) ,

which is independent of x and tends to zero as t 1 t 2 . Then A is equicontinuous. So A is relatively compact on B r , and by the Arzelá-Ascoli theorem, A is compact on B r . Thus condition (b) of Lemma 3.1 is satisfied. Hence the operators A and satisfy the hypotheses of Krasnoselskii’s fixed point theorem; and consequently, problem (1.1) has at least one solution on [0,T]. □

3.4 Existence result via Leray-Schauder’s nonlinear alternative

Theorem 3.4 (Nonlinear alternative for single-valued maps [26])

Let E be a Banach space, C be a closed, convex subset of E, U be an open subset of C and 0U. Suppose that A: U ¯ C is a continuous, compact (that is, F( U ¯ ) is a relatively compact subset of C) map. Then either

  1. (i)

    A has a fixed point in U ¯ , or

  2. (ii)

    there is xU (the boundary of U in C) and λ(0,1) with x=λA(x).

Theorem 3.5 Let f:[0,T]×RR be a continuous function. Assume that

(H4) there exists a continuous nondecreasing function denoted by ψ:[0,)(0,) and a function gC([0,T], R + ) such that

|f(t,u)|g(t)ψ ( x ) for each (t,x)[0,T]×R;

(H5) there exists a constant M>0 such that

M ψ ( M ) g Λ 1 + M Λ 2 + Φ >1,

where Λ 1 , Λ 2 and Φ are defined by (3.3), (3.4) and (3.5), respectively.

Then problem (1.1) has at least one solution on [0,T].

Proof Let the operator K be defined by (3.1). Firstly, we shall show that K maps bounded sets (balls) into bounded sets in C. For a number r>0, let B r ={xC:xr} be a bounded ball in C. Then, for t[0,T], we have

| K x ( t ) | I q + p | f ( s , x ( s ) ) | ( t ) + | λ | I q | x ( s ) | ( t ) + | Ψ 2 | t q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( | σ 1 | + i = 1 m | μ i | I α i + q + p | f ( s , x ( s ) ) | ( η i ) + | λ | i = 1 m | μ i | I α i + q | x ( s ) | ( η i ) ) + | Ψ 1 | t q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( | σ 2 | + j = 1 n | ν j | I β j + q + p | f ( s , x ( s ) ) | ( ξ j ) + | λ | j = 1 n | ν j | I β j + q | x ( s ) | ( ξ j ) ) ψ ( r ) g Λ 1 + r Λ 2 + Φ ,

and consequently,

Kxψ(r)g Λ 1 +r Λ 2 +Φ.

Next, we will show that K maps bounded sets into equicontinuous sets of C. Let t 1 , t 2 [0,T] with t 1 < t 2 and x B r . Then we have

| K x ( t 2 ) K x ( t 1 ) | 1 Γ ( q + p + 1 ) | 0 t 1 [ ( t 2 s ) q + p 1 ( t 1 s ) q + p 1 ] f ( s , x ( s ) ) d s + t 1 t 2 ( t 2 s ) q + p 1 f ( s , x ( s ) ) d s | + | Ψ 2 | | t 2 q t 1 q | | Δ | Γ ( q + 1 ) ( | σ 1 | + i = 1 m | μ i | I α i + q + p | f ( s , x ( s ) ) | ( η i ) + | λ | i = 1 m | μ i | I α i + q | x ( s ) | ( η i ) ) + | Ψ 1 | | t 2 q t 1 q | | Δ | Γ ( q + 1 ) ( | σ 2 | + j = 1 n | ν j | I β j + q + p | f ( s , x ( s ) ) | ( ξ j ) + | λ | j = 1 n | ν j | I β j + q | x ( s ) | ( ξ j ) ) ψ ( r ) g Γ ( 1 + q + p ) | t 2 q + p t 1 q + p | + | Ψ 2 | t 2 q t 1 q | | Δ | Γ ( q + 1 ) ( | σ 1 | + i = 1 m ψ ( r ) g | μ i | η i α i + q + p Γ ( α i + q + p + 1 ) + i = 1 m r | μ i | η i α i + q Γ ( α i + q + 1 ) ) + | Ψ 1 | | t 2 q t 1 q | | Δ | Γ ( q + 1 ) ( | σ 2 | + j = 1 n ψ ( r ) g | ν j | ξ j β j + q + p Γ ( β j + q + p + 1 ) + j = 1 n r | ν j | x i j β j + q Γ ( β j + q + 1 ) ) .

As t 2 t 1 0, the right-hand side of the above inequality tends to zero independently of x B r . Therefore, by the Arzelá-Ascoli theorem, the operator K:CC is completely continuous.

Let x be a solution. Then, for t[0,T], and following similar computations as in the first step, we have

|x(t)|ψ ( x ) g Λ 1 +x Λ 2 +Φ,

which leads to

x ψ ( x ) g Λ 1 + x Λ 2 + Φ 1.

By (H5) there is M such that xM. Let us set

U= { x C : x < M } .

We see that the operator K: U ¯ C is continuous and completely continuous. From the choice of U, there is no xU such that x=νKx for some ν(0,1). Consequently, by the nonlinear alternative of Leray-Schauder type, we deduce that K has a fixed point x U ¯ which is a solution of problem (1.1). This completes the proof. □

3.5 Existence result via Leray-Schauder’s degree theory

Theorem 3.6 Let f:[0,T]×RR be a continuous function. Suppose that

(H6) there exist constants 0γ<(1 Λ 2 ) Λ 1 1 and M>0 such that

|f(t,x)|γ|x|+Mfor all (t,x)[0,T]×R,

where Λ 1 , Λ 2 are defined by (3.3) and (3.4), respectively.

Then problem (1.1) has at least one solution on [0,T].

Proof We define an operator K:CC as in (3.1) and consider the fixed point equation

x=Kx.

We shall prove that there exists a fixed point xC satisfying (1.1).

Set a ball B r C as

B r = { x C : sup t [ 0 , T ] | x ( t ) | < r } ,

where a constant radius r>0. Hence, we show that K: B ¯ r C satisfies the condition

xθKx,x B r ,θ[0,1].
(3.9)

We define

H(θ,x)=θKx,xC,θ[0,1].

As shown in Theorem 3.5, the operator K is continuous, uniformly bounded and equicontinuous. Then, by the Arzelá-Ascoli theorem, a continuous map h θ defined by h θ (x)=xH(θ,x)=xθKx is completely continuous. If (3.9) holds, then the following Leray-Schauder degrees are well defined, and by the homotopy invariance of topological degree, it follows that

deg ( h θ , B r , 0 ) = deg ( I θ K , B r , 0 ) = deg ( h 1 , B r , 0 ) = deg ( h 0 , B r , 0 ) = deg ( I , B r , 0 ) = 1 0 , 0 B r ,

where I denotes the unit operator. By the nonzero property of Leray-Schauder degree, h 1 (x)=xKx=0 for at least one x B r . Let us assume that x=θKx for some θ[0,1] and for all t[0,T] so that

| x ( t ) | = | θ ( K x ) ( t ) | I q + p | f ( s , x ( s ) ) | ( t ) + | λ | I q | x ( s ) | ( t ) + | Ψ 2 | t q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( | σ 1 | + i = 1 m | μ i | I α i + q + p | f ( s , x ( s ) ) | ( η i ) + | λ | i = 1 m | μ i | I α i + q | x ( s ) | ( η i ) ) + | Ψ 1 | t q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( | σ 2 | + j = 1 n | ν j | I β j + q + p | f ( s , x ( s ) ) | ( ξ j ) + | λ | j = 1 n | ν j | I β j + q | x ( s ) | ( ξ j ) ) ( γ | x ( t ) | + M ) ( t q + p Γ ( q + p + 1 ) + ( | Ψ 2 | t q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) i = 1 m | μ i | η i α i + q + p Γ ( α i + q + p + 1 ) + ( | Ψ 1 | t q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) j = 1 n | ν j | ξ j β j + q + p Γ ( β j + q + p + 1 ) ) + | λ | | x ( t ) | ( t q Γ ( q + 1 ) + ( | Ψ 2 | t q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) i = 1 m | μ i | η i α i + q Γ ( α i + q + 1 ) + ( | Ψ 1 | t q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) j = 1 n | ν j | ξ j β j + q Γ ( β j + q + 1 ) ) + | σ 1 | ( | Ψ 2 | t q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) + | σ 2 | ( | Ψ 1 | t q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) .

Taking norm sup t [ 0 , T ] |x(t)|=x, we get

x ( γ x + M ) Λ 1 +x Λ 2 +Φ.

Solving the above inequality for x yields

x M Λ 1 + Φ 1 γ Λ 1 Λ 2 .

If r= M Λ 1 + Φ 1 γ Λ 1 Λ 2 +1, then inequality (3.9) holds. This completes the proof. □

4 Examples

Example 4.1 Consider the following fractional Caputo-Langevin equation with Riemann-Liouville fractional integral conditions:

{ D 7 10 ( D 2 5 + 1 10 ) x ( t ) = 1 e t 4 ( t + 1 ) 2 | x | 2 | x | + 3 + 2 3 , t ( 0 , 1 ) , 1 2 I 9 10 x ( 1 10 ) + 3 10 I 4 5 x ( 3 20 ) + 1 5 I 7 10 x ( 1 5 ) = 5 , 2 5 I 3 10 x ( 4 5 ) + 2 5 I 1 5 x ( 17 20 ) + 1 5 I 1 10 x ( 9 10 ) = 20 .
(4.1)

Here p=7/10, q=2/5, λ=1/10, T=1, m=3, n=3, μ 1 =1/2, α 1 =9/10, η 1 =1/10, μ 2 =3/10, α 2 =4/5, η 2 =3/20, μ 3 =1/5, α 3 =7/10, η 3 =1/5, σ 1 =5, ν 1 =2/5, β 1 =3/10, ξ 1 =4/5, ν 2 =2/5, β 2 =1/5, ξ 2 =17/20, ν 3 =1/5, β 3 =1/10, ξ 3 =9/10, σ 2 =20 and f(t,x)=((1 e t )|x|/(4 ( t + 1 ) 2 (2|x|+3)))+2/3. Since |f(t,x)f(t,y)|(1/4)|xy|, then (H1) is satisfied with L=1/4. We can find that

Λ 1 = T q + p Γ ( q + p + 1 ) + ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) i = 1 m | μ i | η i α i + q + p Γ ( α i + q + p + 1 ) Λ 1 = + ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) j = 1 n | ν j | ξ j β j + q + p Γ ( β j + q + p + 1 ) 2.849024 , Λ 2 = | λ | ( T q Γ ( q + 1 ) + ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) i = 1 m | μ i | η i α i + q Γ ( α i + q + 1 ) Λ 2 = + ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) j = 1 n | ν j | ξ j β j + q Γ ( β j + q + 1 ) ) 0.052040 .

Therefore, we have

L Λ 1 + Λ 2 0.764296<1.

Hence, by Theorem 3.1, problem (4.1) has a unique solution on [0,1].

Example 4.2 Consider the following fractional Caputo-Langevin equation with Riemann-Liouville fractional integral conditions:

{ D 3 5 ( D 3 7 + 1 50 ) x ( t ) = x e 10 t sin 1 2 ( t / 2 ) x 4 t + 1 ( 4 + t ) ( | x | + 1 ) + 5 , t ( 0 , π ) , 7 2 I 5 2 x ( 3 π 5 ) 1 5 I 14 11 x ( 4 π 15 ) 3 7 I 17 12 x ( 3 π 8 ) 2 7 I 4 3 x ( 2 π 9 ) + 11 15 I 4 5 x ( π 5 ) = 3 7 , 5 3 I 10 3 x ( 3 π 11 ) 5 2 I 19 10 x ( π 5 ) + 6 5 I 13 18 x ( π 3 ) 17 13 I 13 7 x ( π 4 ) + 3 17 I 6 5 x ( π 16 ) = 5 6 .
(4.2)

Here p=3/5, q=3/7, λ=1/50, T=π, m=5, n=5, μ 1 =7/2, α 1 =5/2, η 1 =3π/5, μ 2 =1/5, α 2 =14/11, η 2 =4π/15, μ 3 =3/7, α 3 =17/12, η 3 =3π/8, μ 4 =2/7, α 4 =4/3, η 4 =2π/9, μ 5 =11/15, α 5 =4/5, η 5 =π/5, σ 1 =3/7, ν 1 =5/3, β 1 =10/3, ξ 1 =3π/11, ν 2 =5/2, β 2 =19/10, ξ 2 =π/5, ν 3 =6/5, β 3 =13/18, ξ 3 =π/3, ν 4 =17/13, β 4 =13/7, ξ 4 =π/4, ν 5 =3/17, β 5 =6/5, ξ 5 =π/16, σ 2 =5/6 and f(t,x)=((x e 10 t sin 1 2 (t/2)x)/(4t))+(1/((4+t)(|x|+1)))+5. Since |f(t,x)f(t,y)| e 10 t sin 1 2 (t/2)|xy|, then (H2) is satisfied with δ(t)= e 10 t sin 1 2 (t/2) such that δ L 1 / 2 ([0,π], R + ). We can find that

Λ 2 = | λ | ( T q Γ ( q + 1 ) + ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) i = 1 m | μ i | η i α i + q Γ ( α i + q + 1 ) Λ 2 = + ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) j = 1 n | ν j | ξ j β j + q Γ ( β j + q + 1 ) ) 0.350711 , Λ 3 = [ ( 1 σ q + p σ ) 1 σ T q + p σ Γ ( q + p ) Λ 3 = + | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( i = 1 m ( 1 σ α i + q + p σ ) 1 σ | μ i | η i α i + q + p Γ ( α i + q + p ) ) Λ 3 = + | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ( j = 1 n ( 1 σ β j + q + p σ ) 1 σ | ν j | ξ j β j + q + p Γ ( β j + q + p ) ) ] Λ 3 17.417544 ,

and δ0.035344. Therefore, we have

Λ 3 δ+ Λ 2 0966321<1.

Hence, by Theorem 3.2, problem (4.2) has a unique solution on [0,π].

Example 4.3 Consider the following fractional Caputo-Langevin equation with Riemann-Liouville fractional integral conditions:

{ D 2 3 ( D 3 4 3 35 ) x ( t ) = ( t 2 + ( 1 + | x | ) t + | x | t + | x | + 1 ) ln ( t + 1 ) , t ( 0 , e 1 ) , 3 11 I 1 6 x ( 1 10 ) + 1 5 I 1 5 x ( 10 e 11 10 ) + 3 7 I 3 8 x ( 2 e 9 ) + 4 3 I 1 7 x ( 2 e 5 ) = 15 , 2 5 I 1 2 x ( 3 8 ) 1 5 I 6 25 x ( 3 e 2 4 ) + 1 10 I 1 4 x ( 5 e 6 5 ) = 5 .
(4.3)

Here p=2/3, q=3/4, λ=3/35, T=e1, m=4, n=3, μ 1 =3/11, α 1 =1/6, η 1 =1/10, μ 2 =1/5, α 2 =1/5, η 2 =(10e11)/10, μ 3 =3/7, α 3 =3/8, η 3 =2e/9, μ 4 =4/3, α 4 =1/7, η 4 = 2 e/5, σ 1 =15, ν 1 =2/5, β 1 =1/2, ξ 1 =3/8, ν 2 =1/5, β 2 =6/25, ξ 2 =(3e2)/4, ν 3 =1/10, β 3 =1/4, ξ 3 =(5e6)/5, σ 2 =5 and f(t,x)=(( t 2 +(1+|x|)t+|x|)/(t+|x|+1))ln(t+1).

Since |f(t,x)|ln(t+1)+e, then (H3) is satisfied. We can find that

Λ 2 = | λ | ( T q Γ ( q + 1 ) + ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) i = 1 m | μ i | η i α i + q Γ ( α i + q + 1 ) + ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) j = 1 n | ν j | ξ j β j + q Γ ( β j + q + 1 ) ) 0.980502 .

This means that Λ 2 <1. By Theorem 3.3 problem (4.3) has as least one solution on [0,e1].

Example 4.4 Consider the following fractional Caputo-Langevin equation with Riemann-Liouville fractional integral conditions:

{ D 5 6 ( D 2 3 1 50 ) x ( t ) = e t x 2 + 40 sin t 40 | x | + 60 t ( sin ( 2 t / 3 ) ( t + 1 ) 2 ) , t ( 0 , π 2 ) , 2 2 I 3 8 x ( π 18 ) 2 I 1 10 x ( 4 π 9 ) + 3 I 1 4 x ( 1 ) I 2 3 x ( 3 4 ) = 1 , 2 3 I 2 7 x ( π 6 ) 3 I 3 10 x ( π 3 ) + 2 2 7 I 11 20 x ( 5 4 ) 2 5 I 7 9 x ( 1 2 ) = 0 .
(4.4)

Here p=5/6, q=2/3, λ=1/50, m=4, n=4, μ 1 = 2 /2, α 1 =3/8, η 1 =π/18, μ 2 =2, α 2 =1/10, η 2 =4π/9, μ 3 = 3 , α 3 =1/4, η 3 =1, μ 4 =1, α 4 =2/3, η 4 =3/4, σ 1 =1, ν 1 = 2 / 3 , β 1 =2/7, ξ 1 =π/6, ν 2 =3, β 3 =3/10, ξ 2 =π/3, ν 3 =2 2 /7, β 3 =11/20, ξ 3 =5/4, ν 4 =2/5, β 4 =7/9, ξ 4 =1/2, σ 2 =0, and f(t,x)=(( e t x 2 +40sint)/(40|x|+60t))(sin(2t/3)/ ( t + 1 ) 2 ). Then we can find that

Λ 1 = T q + p Γ ( q + p + 1 ) + ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) i = 1 m | μ i | η i α i + q + p Γ ( α i + q + p + 1 ) Λ 1 = + ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) j = 1 n | ν j | ξ j β j + q + p Γ ( β j + q + p + 1 ) 18.473830 , Λ 2 = | λ | ( T q Γ ( q + 1 ) + ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) i = 1 m | μ i | η i α i + q Γ ( α i + q + 1 ) Λ 2 = + ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) j = 1 n | ν j | ξ j β j + q Γ ( β j + q + 1 ) ) 0.531365 , Φ = | σ 1 | ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) + | σ 2 | ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) 3.515304 .

Clearly,

|f(t,x)|=| e t x 2 ( t ) + 40 sin t 40 | x | + 60 t ( sin ( 2 t / 3 ) ( t + 1 ) 2 ) | 1 40 ( | x | + 1 ) |sin(2t/3)|.

By choosing ψ(|x|)=|x|+1 and g(t)=|sin(2t/3)|/40, we can show that

M ψ ( M ) g Λ 1 + M λ 2 + Φ >1,

which implies M>57.019704. By Theorem 3.5, problem (4.4) has at least one solution on [0,π/2].

Example 4.5 Consider the following fractional Caputo-Langevin equation with Riemann-Liouville fractional integral conditions:

{ D 4 5 ( D 3 10 + 1 80 ) x ( t ) = t 2 e 2 t 2 4 π sin ( x + π 3 ) , t ( 0 , 2 π ) , 3 2 I 2 5 x ( 8 5 ) 1 3 I 3 10 x ( 4 3 ) 2 11 I 2 15 x ( 9 2 ) + 1 7 I 1 3 x ( 3 2 ) = 1 , 5 2 I 1 6 x ( 6 ) + 1 2 I 1 20 x ( 5 2 ) 4 5 I 2 9 x ( 3 20 ) 4 3 I 2 7 x ( 1 2 ) 2 17 I 3 5 x ( 1 16 ) = 1 .
(4.5)

Here p=4/5, q=3/10, λ=1/80, m=4, n=5, μ 1 =3/2, α 1 =2/5, η 1 =8/5, μ 2 =1/3, α 2 =3/10, η 2 =4/3, μ 3 =2/11, α 3 =2/15, η 3 =9/2, μ 4 =1/7, α 4 =1/3, η 4 =3/2, σ 1 =1, ν 1 =5/2, β 1 =1/6, ξ 1 =6, ν 2 =1/2, β 2 =1/20, ξ 2 =5/2, ν 3 =4/5, β 3 =2/9, ξ 3 =3/20, ν 4 =4/3, β 4 =2/7, ξ 4 =1/2, ν 5 =2/17, β 5 =3/5, ξ 5 =1/16, σ 2 =1, and f(t,x)= t 2 e 2 t 2 sin(x+π/3)/4π. By a direct computation, we have

Λ 1 = T q + p Γ ( q + p + 1 ) + ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) i = 1 m | μ i | η i α i + q + p Γ ( α i + q + p + 1 ) Λ 1 = + ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) j = 1 n | ν j | ξ j β j + q + p Γ ( β j + q + p + 1 ) 39.248431 , Λ 2 = | λ | ( T q Γ ( q + 1 ) + ( | Ψ 2 | T q + | Ω 2 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) i = 1 m | μ i | η i α i + q Γ ( α i + q + 1 ) Λ 2 = + ( | Ψ 1 | T q + | Ω 1 | Γ ( q + 1 ) | Δ | Γ ( q + 1 ) ) j = 1 n | ν j | ξ j β j + q Γ ( β j + q + 1 ) ) 0.214034 .

Choosing γ=0.02<(1 Λ 2 ) Λ 1 1 0.020025 and M=0.016, we can show that

| f ( t , x ( t ) ) | | t 2 e 2 t 2 4 π | | sin ( x + π 3 ) | ( t 2 e 2 t 2 4 π ) | x | + ( t 2 e 2 t 2 12 ) 0.014637 | x | + 0.015328 γ | x | + M ,

which satisfies (H6). By Theorem 3.6, problem (4.5) has at least one solution on [0,2π].

References

  1. Samko SG, Kilbas AA, Marichev OI: Fractional Integrals and Derivatives: Theory and Applications. Gordon & Breach, Yverdon; 1993.

    Google Scholar 

  2. Podlubny I: Fractional Differential Equations. Academic Press, San Diego; 1999.

    Google Scholar 

  3. Kilbas AA, Srivastava HM, Trujillo JJ North-Holland Mathematics Studies 204. In Theory and Applications of Fractional Differential Equations. Elsevier, Amsterdam; 2006.

    Google Scholar 

  4. Baleanu D, Diethelm K, Scalas E, Trujillo JJ Series on Complexity, Nonlinearity and Chaos. In Fractional Calculus Models and Numerical Methods. World Scientific, Boston; 2012.

    Google Scholar 

  5. Agarwal RP, Zhou Y, He Y: Existence of fractional neutral functional differential equations. Comput. Math. Appl. 2010, 59: 1095-1100. 10.1016/j.camwa.2009.05.010

    Article  MathSciNet  Google Scholar 

  6. Baleanu D, Mustafa OG, Agarwal RP:On L p -solutions for a class of sequential fractional differential equations. Appl. Math. Comput. 2011, 218: 2074-2081. 10.1016/j.amc.2011.07.024

    Article  MathSciNet  Google Scholar 

  7. Ahmad B, Nieto JJ: Riemann-Liouville fractional integro-differential equations with fractional nonlocal integral boundary conditions. Bound. Value Probl. 2011., 2011: Article ID 36

    Google Scholar 

  8. Ahmad B, Ntouyas SK, Alsaedi A: New existence results for nonlinear fractional differential equations with three-point integral boundary conditions. Adv. Differ. Equ. 2011., 2011: Article ID 107384

    Google Scholar 

  9. O’Regan D, Stanek S: Fractional boundary value problems with singularities in space variables. Nonlinear Dyn. 2013, 71: 641-652. 10.1007/s11071-012-0443-x

    Article  MathSciNet  Google Scholar 

  10. Ahmad B, Ntouyas SK, Alsaedi A: A study of nonlinear fractional differential equations of arbitrary order with Riemann-Liouville type multistrip boundary conditions. Math. Probl. Eng. 2013., 2013: Article ID 320415

    Google Scholar 

  11. Ahmad B, Nieto JJ: Boundary value problems for a class of sequential integrodifferential equations of fractional order. J. Funct. Spaces Appl. 2013., 2013: Article ID 149659

    Google Scholar 

  12. Zhang L, Ahmad B, Wang G, Agarwal RP: Nonlinear fractional integro-differential equations on unbounded domains in a Banach space. J. Comput. Appl. Math. 2013, 249: 51-56.

    Article  MathSciNet  Google Scholar 

  13. Liu X, Jia M, Ge W: Multiple solutions of a p -Laplacian model involving a fractional derivative. Adv. Differ. Equ. 2013., 2013: Article ID 126

    Google Scholar 

  14. Coffey WT, Kalmykov YP, Waldron JT: The Langevin Equation. 2nd edition. World Scientific, Singapore; 2004.

    Google Scholar 

  15. Lim SC, Li M, Teo LP: Langevin equation with two fractional orders. Phys. Lett. A 2008, 372: 6309-6320. 10.1016/j.physleta.2008.08.045

    Article  MathSciNet  Google Scholar 

  16. Lim SC, Teo LP: The fractional oscillator process with two indices. J. Phys. A, Math. Theor. 2009., 42: Article ID 065208

    Google Scholar 

  17. Uranagase M, Munakata T: Generalized Langevin equation revisited: mechanical random force and self-consistent structure. J. Phys. A, Math. Theor. 2010., 43: Article ID 455003

    Google Scholar 

  18. Denisov SI, Kantz H, Hänggi P: Langevin equation with super-heavy-tailed noise. J. Phys. A, Math. Theor. 2010., 43: Article ID 285004

    Google Scholar 

  19. Lozinski A, Owens RG, Phillips TN: The Langevin and Fokker-Planck equations in polymer rheology. 16. Handbook of Numerical Analysis 2011, 211-303.

    Google Scholar 

  20. Lizana L, Ambjörnsson T, Taloni A, Barkai E, Lomholt MA: Foundation of fractional Langevin equation: harmonization of a many-body problem. Phys. Rev. E 2010., 81: Article ID 051118

    Google Scholar 

  21. Gambo YY, Jarad F, Baleanu D, Abdeljawad T: On Caputo modification of the Hadamard fractional derivative. Adv. Differ. Equ. 2014., 2014: Article ID 10

    Google Scholar 

  22. Ahmad B, Eloe PW: A nonlocal boundary value problem for a nonlinear fractional differential equation with two indices. Commun. Appl. Nonlinear Anal. 2010, 17: 69-80.

    MathSciNet  Google Scholar 

  23. Ahmad B, Nieto JJ, Alsaedi A, El-Shahed M: A study of nonlinear Langevin equation involving two fractional orders in different intervals. Nonlinear Anal., Real World Appl. 2012, 13: 599-606. 10.1016/j.nonrwa.2011.07.052

    Article  MathSciNet  Google Scholar 

  24. Sudsutad W, Tariboon J: Nonlinear fractional integro-differential Langevin equation involving two fractional orders with three-point multi-term fractional integral boundary conditions. J. Appl. Math. Comput. 2013, 43: 507-522. 10.1007/s12190-013-0676-y

    Article  MathSciNet  Google Scholar 

  25. Krasnoselskii MA: Two remarks on the method of successive approximations. Usp. Mat. Nauk 1955, 10: 123-127.

    MathSciNet  Google Scholar 

  26. Granas A, Dugundji J: Fixed Point Theory. Springer, New York; 2003.

    Book  Google Scholar 

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Acknowledgements

The research of JT is supported by King Mongkut’s University of Technology North Bangkok, Thailand.

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Yukunthorn, W., Ntouyas, S.K. & Tariboon, J. Nonlinear fractional Caputo-Langevin equation with nonlocal Riemann-Liouville fractional integral conditions. Adv Differ Equ 2014, 315 (2014). https://doi.org/10.1186/1687-1847-2014-315

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