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Research Article

Solvability for integral boundary value problems of fractional differential equation on infinite intervals

Changlong Yu, Jufang Wang, Yanping Guo

College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018, Hebei, P. R. China.

Communicated by Mohamed Jleli

Abstract

In this paper, we establish the solvability for integral boundary value problems of fractional differential equation with the nonlinear term dependent in a fractional derivative of lower order on infinite intervals.

The existence and uniqueness of solutions for the boundary value problem are proved by means of the Schauder’s fixed point theorem and Banach’s contraction mapping principle. Finally, we give two examples to demonstrate the use of the main results. c2016 All rights reserved.

Keywords: Integral boundary value problem, fractional differential equation, infinite interval, Fixed point theorem.

2010 MSC: 34A08, 34B40, 34B15.

1. Introduction

Boundary value problems on infinite intervals appear often in applied mathematics and physics. More examples and a collection of works on the existence of solutions of Boundary value problems on infinite intervals for differential, difference and integral equations may be found in the monographs [1, 15]. For some works and various techniques dealing with such boundary value problems, see [2, 5, 7, 13, 21, 22, 23, 26]

and the references therein.

The fractional differential equation has emerged as a new branch in the field of differential equations for their deep back grounds. For an extensive collection of such results, we refer the readers to the monographs [9, 11, 16, 17]. There has been a significant development in nonlocal problems for fractional differential equations or inclusions, see [3, 4, 8, 10, 12, 14, 19, 20, 24, 27] and the references therein.

Corresponding author

Email addresses: [email protected](Changlong Yu),[email protected](Jufang Wang), [email protected](Yanping Guo)

Received 2015-03-23

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Boundary value problems for fractional differential equations on infinite intervals have been considered widely and there are some excellent results on the existence of solutions, see [6, 18, 25] and the references therein. However, to our knowledge, it is rare for works to be done on the solutions for integral boundary value problems (IBVPs) of fractional differential equations on infinite interval.

Recently, in [18], X. Su and S. Zhang considered the BVP:

Dα0+u(t) =f(t, u(t), Dα−10+ u(t)), t∈J := [0,+∞), u(0) = 0, D0α−1+ u(∞) =u, u∈R,

where 1 < α ≤ 2, f ∈ C(J ×R×R, R),Dα0+ and Dα−10+ are the standard Riemann-Liouville fractional derivatives. The existence of unbounded positive solutions was obtained by the Schauder’s fixed point theorem on unbounded domain.

Motivated by the work above, in this paper, we will discuss the following IBVP:

( D0α+u(t) =f(t, u(t), D0α−1+ u(t)), t∈J, u(0) = 0, Dα−10+ u(∞) =R+∞

η g(t)u(t)dt, (1.1)

whereJ = [0,+∞),1< α≤2, f ∈C(J×R×R, R), η≥0, g(t)∈L1[0,+∞) andR+∞

η g(t)tα−1dt <Γ(α), D0α+ andDα−10+ are the standard Riemann-Liouville fractional derivatives andDα−10+ u(∞) = lim

t→+∞Dα−10+ u(t).

We deal with the existence and uniqueness of solutions for BVP (1.1) by using the Schauder’s fixed point theorem and Banach’s contraction mapping principle and obtain multiplicity results which extend and improve the known results.

2. Preliminary results

In this section, we introduce definitions and preliminary facts which are used throughout paper. Follow- ing, let us recall some basic concepts of fractional calculus, see [9, 16, 17] and the references therein.

Definition 2.1. The Riemann-Liouville fractional integral of order δ >0 of a functionf(t) is defined by Iaδ+f(t) = 1

Γ(δ) Z t

a

(t−s)δ−1f(s)ds, t > a, provided that the right-hand side is pointwise defined.

It is well known that Iaδ+f(a) = 0, for f(t)∈C[a, b],δ >0 and Iaδ+ :C[a, b]→C[a, b] forδ >0.

Definition 2.2. The Riemann-Liouville fractional derivative of order δ >0 of a function f(t) is defined by Daδ+f(t) =

d dt

n

Ian−δ+ f(t) = 1 Γ(n−δ)

d dt

nZ t a

(t−s)n−δ−1f(s)ds, t > a,

where n is the smallest integer greater than or equal to δ, provided that the right-hand side is pointwise defined. In particular, forδ =n,Dan+f(t) =f(n)(t).

Lemma 2.3 ([9]). In this work, we need the following composition relations:

(a) Daδ+Iaδ+f(t) =f(t), δ >0, f(t)∈L1[0,+∞);

(b) Daδ+Iaγ+f(t) =Iaγ−δ+ f(t), γ > δ >0, f(t)∈L1[0,+∞).

Definition 2.4 ([13]). It holds thatf : [0,∞)×R2→R is called an S-Carath´eodory function if and only if (i) for each (u, v)∈R2, t∈f(t, u, v) is measurable on [0,∞);

(ii) for almost every t∈[0,∞), (u, v)7→f(t, u, v) is continuous on R2;

(iii) for eachr >0, there existϕr(t)∈L1[0,∞), ϕr(t)>0 on [0,∞) such that max{|u|,|v|} ≤r implies

|f(t, u, v)| ≤ϕr(t), f or a.e. t∈[0,∞).

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Lemma 2.5 ([18]). For δ > 0, the equation Daδ+x(t) = 0 is valid if and only if, x(t) = Σnj=1cj(t−a)δ−j, where cj ∈R, j = 1,2,· · ·, n are arbitrary constants and n is the smallest integer greater than or equal to δ.

Lemma 2.6. Let y(t)∈L1[0,+∞) and R+∞

η g(t)tα−1dt6= Γ(α), then IBVP ( D0α+u(t) =y(t), t∈J,

u(0) = 0, D0α−1+ u(∞) =R+∞

η g(t)u(t)dt, (2.1)

has a unique soltuion

u(t) = Z +∞

0

G(t, s)y(s)ds, where

G(t, s) = 1

Γ(α)(Γ(α)−R+∞

η g(t)tα−1dt) (

[H(η, s)−Γ(α)]tα−1+ (Γ(α)−R+∞

η g(t)tα−1dt)(t−s)α−1, s≤t,

[H(η, s)−Γ(α)]tα−1, s≥t,

and

H(η, s) =

( R+∞

η g(t)(t−s)α−1dt, s≤η, R+∞

s g(t)(t−s)α−1dt, s≥η.

Proof. We may apply Lemma 2.5 to reduce the differential equation in (2.1) to the integral equation u(t) =c1tα−1+c2tα−2+ 1

Γ(α) Z t

0

(t−s)α−1y(s)ds.

In accordance with Lemma 2.3, Lemma 2.5 and the relationDα−10+ tα−1 = Γ(α), we have Dα−10+ u(t) =c1Γ(α) +

Z t 0

y(s)ds.

The boundary condition in (2.1) imply thatc2= 0 and c1=

R+∞

0 H(η, s)y(s)ds−Γ(α)R+∞

0 y(s)ds Γ(α)(Γ(α)−βR+∞

η g(t)tα−1dt) , where

H(η, s) =

( R+∞

η g(t)(t−s)α−1dt, s≤η, R+∞

s g(t)(t−s)α−1dt, s≥η.

Hence,

u(t) = R+∞

0 H(η, s)y(s)ds−Γ(α)R+∞

0 y(s)ds Γ(α)(Γ(α)−R+∞

η g(t)tα−1dt) tα−1+ 1 Γ(α)

Z t 0

(t−s)α−1y(s)ds

= Z +∞

0

G(t, s)y(s)ds.

This proof is complete.

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Lemma 2.7. Let R+∞

η g(t)tα−1dt <Γ(α),then

|G(t, s)| ≤ 2tα−1 Γ(α)−R+∞

η g(t)tα−1dt. (2.2)

Proof. Obviously, it holds

0≤H(η, s)≤ Z +∞

η

g(t)tα−1dt, therefore, whens≤t, then

|G(t, s)| ≤ (R+∞

η g(t)tα−1dt+ Γ(α))tα−1+ (Γ(α)−R+∞

η g(t)tα−1dt)tα−1 Γ(α)(Γ(α)−R+∞

η g(t)tα−1dt)

≤ 2tα−1 Γ(α)−R+∞

η g(t)tα−1dt. Obviously, when s≥t, (2.2) holds. This proof is complete.

Remark 2.8. By Lemma 2.6, it is easy to get Dα−10+ u(t) =

Z +∞

0

G1(t, s)y(s)ds, where

G1(t, s) = 1

Γ(α)−R+∞

η g(t)tα−1dt

( H(η, s)−R+∞

η g(t)tα−1dt, s≤t,

H(η, s)−Γ(α), s≥t,

and if R+∞

η g(t)tα−1dt <Γ(α), then

|G1(t, s)| ≤ 2Γ(α) Γ(α)−R+∞

η g(t)tα−1dt. (2.3)

Define the spaces by

X ={u(t)∈C(J, R) : sup

t∈J

|u(t)|

1 +tα−1 <+∞}, C={u(t)∈X:D0α−1+ u(t)∈C(J, R), sup

t∈J

|D0α−1+ u(t)|<+∞}, and the norm kukX = sup

t∈J

|u(t)|

1+tα−1,kuk= max{sup

t∈J

|u(t)|

1+tα−1,sup

t∈J

|Dα−10+ u(t)|}.

Lemma 2.9 ([18]). (X,k · kX) and (C,k · k) are Banach spaces.

Define operator T :C→C, T u(t) :=

Z +∞

0

G(t, s)f(s, u(s), Dα−10+ u(s))ds, (2.4) and

Dα−10+ T u(t) = R+∞

0 H(η, s)f(s, u(s), D0α−1+ u(s))ds−Γ(α)R+∞

0 f(s, u(s), D0α−1+ u(s))ds Γ(α)−R+∞

η g(t)tα−1dt +

Z t 0

f(s, u(s), D0α−1+ u(s))ds

= Z +∞

0

G1(t, s)f(s, u(s), D0α−1+ u(s))ds.

(2.5)

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In this paper, our basic space isC. Note that the Arzela-Ascoli theorem fails to work inC. Therefore, we need the following compactness criterion.

Lemma 2.10 ([18]). Let Z ⊆ Y be a bounded set, then Z is relatively compact in Y if the following conditions hold:

(i) For any u(t)∈Z, 1+tu(t)α−1 and Dα−10+ u(t) are equicontinuous on any compact interval of J;

(ii) Givenε >0, there exists a constant T =T(ε)>0 such that

u(t1)

1 +tα−11 − u(t2) 1 +tα−12

< ε and |D0α−1+ u(t1)−D0α−1+ u(t2)|< ε for anyt1, t2 ≥T and u(t)∈Z.

3. Main results

In this section, we apply fixed point theorem to IBVP (1.1). First, we give the uniqueness result based on Banach’s contraction mapping principle.

Theorem 3.1. Let f :J×R2→R be an S-Caratheodory function and there existL1(t), L2(t)∈L1[0,+∞) such that

|f(t, u1, v1)−f(t, u2, v2)| ≤L1(t)|u1−u2|+L2(t)|v1−v2|, t∈I, (u1, v1),(u2, v2)∈R2. In addition, suppose thatΛ<1 holds, where

Λ = 2

Γ(α)−R+∞

η g(t)tα−1dt Z +∞

0

[(1 +tα−1)L1(t) +L2(t)]dt.

Then IBVP (1.1) has a unique solution.

Proof. Let us choose

r≥ 2R+∞

0 ϕr(t)dt Γ(α)−R+∞

η g(t)tα−1dt−2R+∞

0 [(1 +tα−1)L1(t) +L2(t)]dt. Now, we show that T Br ⊂Br, where Br={u∈C:kuk ≤r}.

For eachu∈Br, we have

|T u(t)|

1 +tα−1 = Z +∞

0

|G(t, s)|

1 +tα−1 |f(s, u(s), D0α−1+ u(s))|ds

≤ 2

Γ(α)−R+∞

η g(t)tα−1dt Z +∞

0

|f(s, u(s), Dα−10+ u(s))|ds

≤ 2

Γ(α)−R+∞

η g(t)tα−1dt Z +∞

0

|f(s, u(s), D0α−1+ u(s))−f(s,0,0)|+|f(s,0,0)|

ds

≤ 2

Γ(α)−R+∞

η g(t)tα−1dt

||u||

Z +∞

0

[(1 +tα−1)L1(t) +L2(t)]dt+ Z +∞

0

ϕr(t)dt

≤r, and

|D0α−1+ T u(t)| ≤ 2Γ(α) Γ(α)−R+∞

η g(t)tα−1dt Z +∞

0

|f(s, u(s), Dα−10+ u(s))|ds≤Γ(α)r≤r.

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Hence, we obtain that||T u|| ≤r, so T Br⊂Br. Next, for u, v∈C and for eacht∈J, we have

T u(t)

1 +tα−1 − T v(t) 1 +tα−1

= 1

1 +tα−1

Z +∞

0

G(t, s)(f(s, u(s), Dα−10+ u(s))−f(s, v(s), D0α−1+ v(s)))ds

≤ 2

Γ(α)−R+∞

η g(t)tα−1dt Z +∞

0

|f(s, u(s), Dα−10+ u(s))−f(s, v(s), D0α−1+ v(s))|ds

≤ 2

Γ(α)−R+∞

η g(t)tα−1dt Z +∞

0

L1(s)|u1(s)−u2(s)|+L2(s)|v1(s)−v2(s)|

ds

≤ 2

Γ(α)−R+∞

η g(t)tα−1dt Z +∞

0

[(1 +tα−1)L1(t) +L2(t)]dt||u−v||

≤Λ||u−v||<||u−v||, and

|D0α−1+ T u(t)−D0α−1+ T v(t)|=

Z +∞

0

G1(t, s)(f(s, u(s), D0α−1+ u(s))−f(s, v(s), Dα−10+ v(s)))ds

≤ 2Γ(α)

Γ(α)−R+∞

η g(t)tα−1dt Z +∞

0

|f(s, u(s), Dα−10+ u(s))−f(s, v(s), D0α−1+ v(s))|ds

<Γ(α)||u−v|| ≤ ||u−v||.

Therefore, we obtain that ||T u−T v|| < ||u −v||, T is a contraction map. Thus, the conclusion of the theorem follows by Bananch’s contraction mapping principle.

The next existence result is based on the Schauder’s fixed-point theorem.

Theorem 3.2. Assume that there exists nonnegative functions p(t), q(t), r(t) ∈L1(J, R+) with tα−1p(t) ∈ L1(J, R+) such that

|f(t, u, v)| ≤p(t)|u|+q(t)|v|+r(t), t∈J, (u, v)∈R2. (3.1) Then BVP (1.1)has at least one solution provided

Z +∞

0

(1 +tα−1)p(t) +q(t)

dt < Γ(α)−R+∞

η g(t)tα−1dt

2 .

Proof. First of all, by the continuity off, we can conclude thatT u(t) andDα−10+ T u(t) are continuous onJ. In what follows, we divide the proof into several steps.

Setp 1. Choose

R≥ 2R+∞

0 r(s)ds Γ(α)−R+∞

η g(t)tα−1dt−2R+∞

0 [(1 +sα−1)p(s) +q(s)]ds, and let

U ={u(t)∈C:||u|| ≤R}.

Then,T :U →U.

Indeed, for any u(t)∈U, by (2.2)-(2.5) and the condition of Theorem 3.2, we can get

|T u(t)|

1 +tα−1 ≤ 2

Γ(α)−R+∞

η g(t)tα−1dt Z +∞

0

|f(s, u(s), Dα−10+ u(s))|ds

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≤ 2 Γ(α)−R+∞

η g(t)tα−1dt Z +∞

0

p(s)|u(s)|+q(s)|Dα−10+ u(s)|+r(s)

ds

≤ 2

Γ(α)−R+∞

η g(t)tα−1dt

||u||

Z +∞

0

[(1 +sα−1)p(s) +q(s)]ds+ Z +∞

0

r(s)ds

≤R, and

|Dα−10+ T u(t)| ≤ 2Γ(α) Γ(α)−R+∞

η g(t)tα−1dt Z

0

|f(s, u(s), D0α−1+ u(s))|ds

≤ 2Γ(α)

Γ(α)−R+∞

η g(t)tα−1dt

Z +∞

0

[(1 +sα−1)p(s) +q(s)]ds+ Z +∞

0

r(s)ds

≤Γ(α)R≤R.

Hence,||T u|| ≤R and this show that T :U →U. Setp 2. T :U →U is continuous operator.

Letun, u∈U, n= 1,2,· · ·,and ||un−u|| →0 as n→ ∞. Then by by (2.2)-(2.5) and the condition of Theorem 3.2, we obtain that

T un(t)

1 +tα−1 − T u(t) 1 +tα−1

= 1

1 +tα−1

Z +∞

0

G(t, s)(f(s, un(s), Dα−10+ un(s))−f(s, u(s), Dα−10+ u(s)))ds

≤ 2

Γ(α)−R+∞

η g(t)tα−1dt Z +∞

0

|f(s, un(s), Dα−10+ un(s))−f(s, u(s), Dα−10+ u(s))|ds

≤ 2

Γ(α)−R+∞

η g(t)tα−1dt

Z +∞

0

|f(s, un(s), Dα−10+ un(s))|ds+ Z +∞

0

|f(s, u(s), D0α−1+ u(s))|ds

≤ 2

Γ(α)−R+∞

η g(t)tα−1dt

||un||

Z +∞

0

[(1 +sα−1)p(s) +q(s)]ds+ Z +∞

0

r(s)ds

+||u||

Z +∞

0

[(1 +sα−1)p(s) +q(s)]ds+ Z +∞

0

r(s)ds

≤ 4RR+∞

0 [(1 +sα−1)p(s) +q(s)]ds+ 4R+∞

0 r(s)ds Γ(α)−R+∞

η g(t)tα−1dt ,

and

|D0α−1+ T un(t)−Dα−10+ T u(t)|

≤ 2Γ(α)

Γ(α)−R+∞

η g(t)tα−1dt Z +∞

0

|f(s, un(s), D0α−1+ un(s))−f(s, u(s), D0α−1+ u(s))|ds

≤ 4Γ(α)

Γ(α)−R+∞

η g(t)tα−1dt

R Z +∞

0

[(1 +sα−1)p(s) +q(s)]ds+ Z +∞

0

r(s)ds

.

Therefore, by Lebesgue’s dominated convergence theorem, we have||T un−T u|| →0 as n→ ∞. Hence,T is continuous.

Setp 3. Let V be a subset of U. We apply Lemma 2.10 to verify that T V is relatively compact.

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Let I ∈ J be a compact interval, t1, t2 ∈ I and t1 < t2. Then for any u(t) ∈ V, it is easy to know f(t, u(t), Dα−10+ u(t)) is bounded onI, so we can obtain that

T u(t2)

1 +tα−12 − T u(t1) 1 +tα−11

R+∞

0 H(η, s)f(s, u(s), Dα−10+ u(s))ds−Γ(α)R+∞

0 f(s, u(s), D0α−1+ u(s))ds Γ(α)

Γ(α)−R+∞

η g(t)tα−1dt

tα−12

1 +tα−12 − tα−11 1 +tα−11

+ 1

Γ(α)

Z t2

0

(t2−s)α−1

1 +tα−12 f(s, u(s), Dα−10+ u(s))ds− Z t1

0

(t1−s)α−1

1 +tα−11 f(s, u(s), D0α−1+ u(s))ds

≤ R+∞

0 H(η, s)|f(s, u(s), Dα−10+ u(s))|ds+ Γ(α)R+∞

0 |f(s, u(s), Dα−10+ u(s))|ds Γ(α)

Γ(α)−R+∞

η g(t)tα−1dt

tα−12

1 +tα−12 − tα−11 1 +tα−11

+ 1

Γ(α) Z t1

0

(t2−s)α−1

1 +tα−12 −(t1−s)α−1 1 +tα−11

|f(s, u(s), Dα−10+ u(s))|ds

+ 1

Γ(α) Z t2

t1

(t2−s)α−1

1 +tα−12 |f(s, u(s), Dα−10+ u(s))ds

≤ 2

RR+∞

0 [(1 +sα−1)p(s) +q(s)]ds+R+∞

0 r(s)ds

Γ(α)−R+∞

η g(t)tα−1dt

tα−12

1 +tα−12 − tα−11 1 +tα−11

+max|f(s, u(s), D0α−1+ u(s))|

Γ(α)

Z t2

t1

(t2−s)α−1 1 +tα−12 ds+

Z t1

0

(t2−s)α−1

1 +tα−12 −(t1−s)α−1 1 +tα−11

ds

,

→0, uniformly as t1 →t2, and

|Dα−10+ T u(t2)−Dα−10+ T u(t1)| ≤ Z t2

t1

|f(t, u(t), D0α−1+ u(t))|ds→0,uniformly as t1→t2. Then it is easy to see that 1+tT u(t)α−1 and Dα−10+ T u(t) are equicontinuous onI.

Now, we show that for any u(t) ∈ V, 1+tT u(t)α−1 and D0α−1+ T u(t) satisfy the condition (ii) of Lemma 2.10.

Observing that by the condition of Theorem 3.2, we have Z +∞

0

|f(t, u(t), Dα−10+ u(t))|dt≤ ||u||

Z +∞

0

[(1 +tα−1)p(t) +q(t)]dt+ Z +∞

0

r(t)dt

≤ 2

Γ(α)−R+∞

η g(t)tα−1dt

||u||

Z +∞

0

[(1 +tα−1)p(t) +q(t)]dt+ Z +∞

0

r(t)dt

≤R,

we know that for givenε >0, there exists a constant L >0, such that Z +∞

L

|f(t, u(t), Dα−10+ u(t))|dt < ε. (3.2) On the other hand, since lim

t→+∞

tα−1

1+tα−1 = 1,there exists a constant T1 >0,such that for any t1, t2 ≥T1, we

have

tα−11

1 +tα−11 − tα−12 1 +tα−12

< ε. (3.3)

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Similarly, lim

t→+∞

(t−L)α−1

1+tα−1 = 1 and thus there exists a constant T2 > L >0,such that for any t1, t2≥T2 and 0≤s≤L,

(t1−s)α−1

1 +tα−11 −(t2−s)α−1 1 +tα−12

< ε. (3.4)

Now, choose T0 >max{T1, T2}, then for any t1, t2≥T0, by (3.2)-(3.4), we can obtain that

T u(t2)

1 +tα−12 − T u(t1) 1 +tα−11

≤ 2

RR+∞

0 [(1 +sα−1)p(s) +q(s)]ds+R+∞

0 r(s)ds

Γ(α)−R+∞

η g(t)tα−1dt

tα−12

1 +tα−12 − tα−11 1 +tα−11

+ 1

Γ(α) Z L

0

(t2−s)α−1

1 +tα−12 −(t1−s)α−1 1 +tα−11

|f(s, u(s), Dα−10+ u(s))|ds

+ 1

Γ(α)

Z +∞

L

|f(s, u(s), Dα−1

0+ u(s))|ds+ Z +∞

L

|f(s, u(s), D0α−1+ u(s))|ds

≤ 2

RR+∞

0 [(1 +sα−1)p(s) +q(s)]ds+R+∞

0 r(s)ds

Γ(α)−R+∞

η g(t)tα−1dt ε

+

t∈[0,L],u∈Vmax |f(s, u(s), D0α−1+ u(s))|

Γ(α) Lε+ 2

Γ(α)ε, and

|D0α−1+ T u(t2)−D0α−1+ T u(t1)| ≤ Z t2

t1

|f(t, u(t), D0α−1+ u(t))|ds

≤ Z +∞

L

|f(t, u(t), D0α−1+ u(t))|ds < ε.

Consequently, lemma 2.10 yields thatT V is relatively compact.

Therefore, by Schauder’s fixed point theorem, we conclude that the BVP (1.1) has at least one solutions inU and the proof is finished.

4. Example

Example 4.1. Consider the following IBVP for fractional differential equation on infinite intervals:

 D

3 2

o+u(t) =e−t+ 1

10(1+t2)(1+

t)sin(u(t)) + 4(1+e1 t)arctan(D

1 2

0+u(t)), t∈J, u(0) = 0, D

1 2

0+u(+∞) =R+∞

1 1

10t12e−tu(t)dt.

(4.1)

Here, α = 32, η = 1, f(t, u, v) = e−t + 1

10(1+t2)(1+

t)sin(u) + 4(1+e1 t)arctan(v), L1(t) = 1

10(1+t2)(1+ t), L2(t) = 4(1+e1 t) and g(t) = 101t12e−t. With the aid of simple computation, we can obtain that

f(t, u1, v1)−f(t, u2, v2)

≤ 1

10(1 +t2)(1 +√

t)|u1−u2|+ 1

4(1 +et)|v1−v2|, R+∞

η g(t)tα−1dt=R+∞

1 1

10e−tdt <Γ(32) and Λ≈0.77785<1. In the view of Theorem 3.1, then IBVP (4.1) has a unique solution.

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Example 4.2. Consider the following IBVP for fractional differential equation on infinite intervals:

 D

5 4

o+u(t) = ln(1+|u(t)|) 10(1+t2)(1+4

t) +10e1t sin(|D

1 4

0+u(t)|) +te−t2, t∈J, u(0) = 0, D

1 4

0+u(+∞) =R+∞

1 1

2t34e−tu(t)dt.

(4.2)

Here,α= 54,η = 1, f(t, u, v) = ln(1+|u|)

10(1+t2)(1+4

t) +10e1t sin(|v|) +te−t2 and g(t) = 12e−t. Obviously,

|f(t, u, v)| ≤ |u|

10(1 +t2)(1 +√4

t) + 1

10et|v|+te−t2, With the aid of simple computation, we find that

Z +∞

0

(1 +t14) 1

10(1 +t2) + 1 10et

dt= π

20 + 1

10 ≈0.25708, and

Γ(54)−12R+∞

1 te−tdt

2 ≈ 0.906402−12 ×0.73576

2 = 0.26926,

Hence, the conditions of Theorem 3.2 we satisfied, so IBVP (4.2) has at least one solution.

Acknowledgements:

This paper is supported by the Natural Science Foundation of China (11201112), the Natural Science Foundation of Hebei Province (A2013208147), (A2014208152), and (A2015208114), and the Foundation of Hebei Education Department (Z2014062) and (QN2015175).

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