Applied Mathematics and Computation 244 (2014) 422–431

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Applied Mathematics and Computation

journal homepage: www.elsevier .com/ locate /amc

Solvability of periodic boundary value problem for fractionalp-Laplacian equation q

http://dx.doi.org/10.1016/j.amc.2014.06.1050096-3003/� 2014 Elsevier Inc. All rights reserved.

q This work was supported by the Fundamental Research Funds for the Central Universities (2012QNA50) and the National Natural Science FounChina (11271364).⇑ Corresponding author.

E-mail addresses: taiyongchen@cumt.edu.cn (T. Chen), wblium@163.com (W. Liu), relinaliu@163.com (J. Liu).

Taiyong Chen ⇑, Wenbin Liu, Jiaying LiuDepartment of Mathematics, China University of Mining and Technology, Xuzhou 221116, PR China

a r t i c l e i n f o

Keywords:Fractional differential equationp-Laplacian operatorPeriodic boundary conditionsContinuation theorem

a b s t r a c t

In this paper, we establish a continuation theorem for the fractional p-Laplacian operatorwith periodic boundary conditions. And then, we apply this continuation theorem to thestudy of periodic boundary value problem for fractional p-Laplacian equation. Undercertain nonlinear growth condition of the nonlinearity, we obtain a new result on theexistence of solutions.

� 2014 Elsevier Inc. All rights reserved.

1. Introduction

Fractional calculus is a generalization of ordinary differentiation and integration on an arbitrary order that can be non-integer. Fractional differential equations appear naturally in a number of fields such as physics, polymer rheology, regularvariation in thermodynamics, biophysics, blood flow phenomena, aerodynamics, electro-dynamics of complex medium, vis-coelasticity, Bode analysis of feedback amplifiers, capacitor theory, electrical circuits, electro-analytical chemistry, biology,control theory, fitting of experimental data, etc. (see [1–4]).

Recently, fractional differential equations have been of great interest due to the intensive development of the theory offractional calculus itself and its applications. For example, for fractional initial value problems, the existence and multiplicityof solutions were discussed in [5–8]. In addition, for fractional boundary value problems, Agarwal et al. (see [9]) considered atwo-point boundary value problem at nonresonance, and Bai (see [10]) considered a m-point boundary value problem atresonance. For more papers on fractional boundary value problems, see [11–19] and the references therein.

The turbulent flow in a porous medium is a fundamental mechanics problem. For studying this type of problems,Leibenson (see [20]) introduced the p-Laplacian equation as follows

ð/pðx0ðtÞÞÞ0 ¼ f ðt; xðtÞ; x0ðtÞÞ; ð1:1Þ

where /pðsÞ ¼ jsjp�2s; p > 1. Obviously, /p is invertible and its inverse operator is /q, where q > 1 is a constant such that

1=pþ 1=q ¼ 1.In the past few decades, many important results on Eq. (1.1) with certain boundary conditions had been obtained. We

refer the readers to [21–28] and the references cited therein. However, to the best of our knowledge, there are relativelyfew results on periodic boundary value problems (PBVPs for short) for fractional p-Laplacian equations.

dation of

T. Chen et al. / Applied Mathematics and Computation 244 (2014) 422–431 423

Our aim in this paper is to study the existence of solutions for PBVP of fractional p-Laplacian equation of the form

Db0þ/pðDa

0þxðtÞÞ ¼ f ðt; xðtÞ;Da0þxðtÞÞ; t 2 ½0; T�;

xð0Þ ¼ xðTÞ; Da0þxð0Þ ¼ Da

0þxðTÞ;

(ð1:2Þ

where 0 < a; b � 1; Da0þ is a Caputo fractional derivative, and f : ½0; T� � R2 ! R is continuous. Note that, the nonlinear oper-

ator Db0þ/pðDa

0þ Þ is reduced to the linear operator Db0þDa

0þ when p ¼ 2 and the additive index law

Db0þDa

0þuðtÞ ¼ Daþb0þ uðtÞ

holds under some reasonable constraints on the function uðtÞ (see [29]).The choice of periodic boundary conditions (PBCs for short) is motivated by the supplementary difficulty they introduce

with respect to anti-periodic boundary conditions xð0Þ ¼ �xðTÞ; Da0þxð0Þ ¼ �Da

0þxðTÞ (see [12]). Indeed, it can be shown thatthe anti-periodic boundary value problem

Db0þ/pðDa

0þxðtÞÞ ¼ hðtÞ; t 2 ½0; T�;xð0Þ ¼ �xðTÞ; Da

0þxð0Þ ¼ �Da0þxðTÞ

(

is uniquely solvable for each h 2 Cð½0; T�;RÞ and the solution define a nice nonlinear operator T : Cð½0; T�;RÞ ! Cð½0; T�;RÞ.Consequently, the corresponding nonlinear problem

Db0þ/pðDa

0þxðtÞÞ ¼ f ðt; xðtÞÞ; t 2 ½0; T�;xð0Þ ¼ �xðTÞ; Da

0þxð0Þ ¼ �Da0þxðTÞ

(

is equivalent to the fixed point problem in Cð½0; T�;RÞ

x ¼ ðT � Nf Þx;

where Nf is the Nemytskii operator defined by Nf xðtÞ ¼ f ðt; xðtÞÞ; 8t 2 ½0; T�. Moreover, one can show that T � Nf is a com-pletely continuous operator, which allows the use of Leray–Schauder theory.

However, PBVP

Db0þ/pðDa

0þxðtÞÞ ¼ hðtÞ; t 2 ½0; T�;xð0Þ ¼ xðTÞ; Da

0þxð0Þ ¼ Da0þxðTÞ

(ð1:3Þ

is not solvable for each h 2 Cð½0; T�;RÞ, and, when solvable, has no unique solution because xðtÞ þ c; 8c 2 R is a solutiontogether with xðtÞ. In this case, a trivial necessary condition for the solvability of PBVP (1.3) is that

�h :¼ b

Tb

Z T

0ðT � sÞb�1hðsÞds ¼ 0:

Furthermore, since Db0þ/pðDa

0þ Þ is a nonlinear operator, the coincidence degree theory for linear differential operators withPBCs is invalid in the direct application to it.

The rest of this paper is organized as follows. Section 2 contains some necessary notations, definitions and lemmas. InSection 3, we construct a mechanism which allows us to deal with PBVP (1.3) of forced fractional p-Laplacian equation. InSection 4, based on the previous results, we establish a continuation theorem for the fractional p-Laplacian operator withPBCs. In Section 5, we apply this continuation theorem to the study of PBVP (1.2) for fractional p-Laplacian equation andobtain a new existence result under the nonlinear growth restriction of f. Finally, in Section 6, an example is given to illus-trate the main result.

2. Preliminaries

For convenience of the readers, we present here some necessary basic knowledge and definitions about fractional calculustheory, which can be found, for instance, in [2,4].

Definition 2.1. The Riemann–Liouville fractional integral operator of order a > 0 of a function u : ð0;þ1Þ ! R is given by

Ia0þuðtÞ ¼ 1CðaÞ

Z t

0ðt � sÞa�1uðsÞds;

provided that the right side integral is pointwise defined on ð0;þ1Þ.

Definition 2.2. The Caputo fractional derivative of order a > 0 of a continuous function u : ð0;þ1Þ ! R is given by

Da0þuðtÞ ¼ In�a

0þdnuðtÞ

dtn ¼ 1Cðn� aÞ

Z t

0ðt � sÞn�a�1uðnÞðsÞds;

424 T. Chen et al. / Applied Mathematics and Computation 244 (2014) 422–431

where n is the smallest integer greater than or equal to a, provided that the right side integral is pointwise defined onð0;þ1Þ.

Lemma 2.1 (see [1]). Let a > 0. Assume that u;Da0þu 2 Lð½0; T�;RÞ. Then the following equality holds

Ia0þDa0þuðtÞ ¼ uðtÞ þ c0 þ c1t þ . . .þ cn�1tn�1;

where ci 2 R; i ¼ 0;1; . . . ;n� 1, here n is the smallest integer greater than or equal to a.

Lemma 2.2 (see [30]). For any u; v P 0, then

/pðuþ vÞ 6 /pðuÞ þ /pðvÞ; if p < 2;

/pðuþ vÞ 6 2p�2ð/pðuÞ þ /pðvÞÞ; if p P 2:

Now, we give some necessary notations. Let X; Y be real Banach spaces, L : domL � X ! Y and P : X ! X; Q : Y ! Y beprojectors such that

ImP ¼ KerL; KerQ ¼ ImL;

X ¼ KerL� KerP; Y ¼ ImL� ImQ :

It follows that

LjdomL\KerP : domL \ KerP ! ImL

is invertible.In this paper, we take Y ¼ Cð½0; T�;RÞ with the norm kyk0 ¼maxt2½0;T�jyðtÞj, and X ¼ xjx;Da

0þx 2 Y� �

; XT ¼ x 2 Xjxð0Þf¼ xðTÞ; Da

0þxð0Þ ¼ Da0þxðTÞg with the norm kxkX ¼max kxk0; kD

a0þxk0

� �. By means of the linear functional analysis theory,

we can prove that X; XT are Banach spaces.Define the operator L : domL � X ! Y by

Lx ¼ Db0þ/pðDa

0þxÞ; ð2:1Þ

where domL ¼ x 2 XT jDb0þ/pðDa

0þxÞ 2 Yn o

. Let Nf : X ! Y be the Nemytskii operator defined by

Nf xðtÞ ¼ f ðt; xðtÞ;Da0þxðtÞÞ; 8t 2 ½0; T�: ð2:2Þ

Then PBVP (1.2) is equivalent to the operator equation

Lx ¼ Nf x; x 2 domL:

3. Forced fractional p-Laplacian equation with PBCs

In this section, a mechanism will be constructed which allows us to deal with PBVP (1.3) of forced fractional p-Laplacianequation.

Let us consider PBVP (1.3), where h 2 Y such that �h ¼ 0, and let x be a solution of PBVP (1.3). From Lemma 2.1 we have

/pðDa0þxðtÞÞ ¼ aþ Ib0þhðtÞ ¼ aþ 1

CðbÞ

Z t

0ðt � sÞb�1hðsÞds; 8a 2 R; ð3:1Þ

which together with the boundary condition xð0Þ ¼ xðTÞ yields that

Z T

0ðT � sÞa�1/qðaþ Ib0þhðsÞÞds ¼ 0:

For any fixed l 2 Y , define the function GlðaÞ : R! R by

GlðaÞ ¼aTa

Z T

0ðT � sÞa�1/qðaþ lðsÞÞds: ð3:2Þ

Lemma 3.1. The function GlðaÞ, defined by (3.2), has the following properties

(1) for any fixed l 2 Y, the equation

GlðaÞ ¼ 0

has a unique solution ~aðlÞ;(2) the function ~a : Y ! R, defined in (1), is continuous and sends bounded sets into bounded sets.

T. Chen et al. / Applied Mathematics and Computation 244 (2014) 422–431 425

The proof of Lemma 3.1 is similar to the proof of Proposition 2.2 in [31], so we omit the details.Let a : Y ! R be defined by

aðhÞ ¼ ~aðIb0þhÞ:

Then, basing on Lemma 3.1, a is a completely continuous mapping. Furthermore, by (3.1) and Lemma 2.1, we obtain that

xðtÞ ¼ xð0Þ þ Ia0þ/qðaðhÞ þ Ib0þhÞðtÞ; ð3:3Þ

where /q is understood as the operator /q : Y ! Y defined by /qðxÞðtÞ ¼ /qðxðtÞÞ.

Lemma 3.2. Let L be defined by (2.1), then

KerL ¼ x 2 XjxðtÞ ¼ c; 8t 2 ½0; T�; c 2 Rf g; ð3:4Þ

ImL ¼ y 2 Y jZ T

0ðT � sÞb�1yðsÞds ¼ 0

� �: ð3:5Þ

Proof. From Lemma 2.1, Db0þ/pðDa

0þxðtÞÞ ¼ 0 has the following solutions

xðtÞ ¼ c þ/qðbÞ

Cðaþ 1Þ ta; 8b; c 2 R:

Thus, by the boundary condition xð0Þ ¼ xðTÞ, one has that (3.4) holds.If y 2 ImL, then there exists a function x 2 domL such that yðtÞ ¼ Db

0þ/pðDa0þxðtÞÞ. Hence, basing on Lemma 2.1, we have

/pðDa0þxðtÞÞ ¼ dþ 1

CðbÞ

Z t

0ðt � sÞb�1yðsÞds; 8d 2 R:

By condition Da0þxð0Þ ¼ Da

0þxðTÞ, one has

Z T

0ðT � sÞb�1yðsÞds ¼ 0: ð3:6Þ

On the other hand, suppose y 2 Y and satisfies (3.6). Let

xðtÞ ¼ Ia0þ/qðaðyÞ þ Ib0þyÞðtÞ:

Then Da0þxð0Þ ¼ Da

0þxðTÞ and, from the definition of mapping a, we have

xð0Þ ¼ 0 ¼ Ia0þ/qðaðyÞ þ Ib0þyÞðTÞ ¼ xðTÞ:

Hence x 2 domL. Moreover

LxðtÞ ¼ Db0þ/p½Da

0þ Ia0þ/qðaðyÞ þ Ib0þyÞ�ðtÞ ¼ yðtÞ:

So that, y 2 ImL. Thus, one has that (3.5) holds. The proof is complete. h

Let us define the projectors P : X ! X and Q : Y ! Y by

PxðtÞ ¼ xð0Þ; 8t 2 ½0; T�;

QyðtÞ ¼ b

Tb

Z T

0ðT � sÞb�1yðsÞds; 8t 2 ½0; T�: ð3:7Þ

Thus, from (3.3), we obtain that if x 2 XT is a solution of PBVP (1.3), then x satisfies the following abstract equation

x ¼ Pxþ QhþKh; ð3:8Þ

where the operator K : Y ! XT is given by

KhðtÞ ¼ Ia0þ/q½aððI � QÞhÞ þ Ib0þ ðI � QÞh�ðtÞ; 8t 2 ½0; T�: ð3:9Þ

On the other hand, from the proof of Lemma 3.2, we can see that if x satisfies Eq. (3.8), then x is a solution of PBVP (1.3).Note that, since að0Þ ¼ ~að0Þ ¼ 0, we get Kð0Þ ¼ 0.

Lemma 3.3. The operator K, defined by (3.9), is a completely continuous operator.

Proof. By the definition of K, we obtain that

Da0þKhðtÞ ¼ /q½aððI � QÞhÞ þ Ib0þ ðI � QÞh�ðtÞ:

426 T. Chen et al. / Applied Mathematics and Computation 244 (2014) 422–431

Obviously, the operators K; Da0þK are also a composition of continuous operators. SoK; Da

0þK are continuous in Y. Hence,K isa continuous operator.

Let X � Y be an open bounded set, then Kð�XÞ and Da0þKð�XÞ are bounded. Thus, in view of the Arzel�a–Ascoli theorem, we

need only to prove that Kð�XÞ � XT is equicontinuous.From the conclusion (2) of Lemma 3.1, we know that there exists a constant M > 0 such that

j½aððI � QÞhÞ þ Ib0þ ðI � QÞh�ðtÞj 6 M; 8h 2 �X; t 2 ½0; T�. For 0 6 t1 < t2 6 T; h 2 �X, we have

jKhðt2Þ � Khðt1Þj ¼1

CðaÞ

Z t2

0ðt2 � sÞa�1/q½aððI � QÞhÞ þ Ib0þ ðI � QÞh�ðsÞds�

Z t1

0ðt1 � sÞa�1/q½aððI � QÞhÞ

����þIb0þ ðI � QÞh�ðsÞds

�� 6 Mq�1

CðaÞ

Z t1

0½ðt1 � sÞa�1 � ðt2 � sÞa�1�dsþ

Z t2

t1

ðt2 � sÞa�1ds� �

¼ Mq�1

Cðaþ 1Þ ½ta1 � ta

2 þ 2ðt2 � t1Þa�:

Since ta is uniformly continuous in ½0; T�, we can obtain that Kð�XÞ � Y is equicontinuous. Similar proof can show that½aðI � QÞ þ Ib0þ ðI � QÞ�ð�XÞ � Y is equicontinuous, where aðI � QÞh is understood as aððI � QÞhÞ. This, together with the uni-formly continuity of /qðsÞ on ½�M;M�, yields that Da

0þKð�XÞ � Y is also equicontinuous. Thus, we get that K : Y ! XT is com-pact. The proof is complete. h

4. A continuation theorem for fractional p-Laplacian operator with PBCs

In this section, based on the results of Section 3, a continuation theorem for the fractional p-Laplacian operator with PBCswill be established, which is an extension of the coincidence degree theory for linear differential operators with PBCs.

Theorem 4.1. Let f : ½0; T� � R2 ! R be continuous, L; Nf ; Q be defined respectively by (2.1), (2.2), (3.7), and X be an openbounded subset of XT such that domL \ �X – ;. Assume that the following conditions are satisfied

(C1) for each k 2 ð0;1Þ, the equation

Lx ¼ kNf x ð4:1Þ

has no solution on ðdomL n KerLÞ \ @X;(C2) the equation

QNf x ¼ 0

has no solution on KerL \ @X;(C3) the Brouwer degree

degðQNf jKerL;X \ KerL; 0Þ– 0:

Then the equation Lx ¼ Nf x has at least one solution in domL \ �X.

Proof. Let us consider the homotopic equation of Lx ¼ Nf x as follows

Lx ¼ kNf xþ ð1� kÞQNf x; x 2 domL: ð4:2Þ

That is

Db0þ/pðDa

0þxðtÞÞ ¼ kf ðt; xðtÞ;Da0þxðtÞÞ þ ð1� kÞ b

Tb

R T0 ðT � sÞb�1f ðs; xðsÞ;Da

0þxðsÞÞds;

xð0Þ ¼ xðTÞ; Da0þxð0Þ ¼ Da

0þxðTÞ:

(

It is clear that, for k 2 ð0;1�, if x is a solution of Eq. (4.1) or Eq. (4.2), then we also have the necessary condition

QNf xðtÞ ¼ b

Tb

Z T

0ðT � sÞb�1f ðs; xðsÞ;Da

0þxðsÞÞds ¼ 0:

It follows that Eq. (4.1) and Eq. (4.2) have the same solutions. Furthermore, Eq. (4.2) can be written in the following equiv-alent form

x ¼ Gf ðx; kÞ; ð4:3Þ

where Gf : XT � ½0;1� ! XT is defined by

Gf ðx; kÞ ¼ Pxþ QNf xþ ½K � ðkNf þ ð1� kÞQNf Þ�x ¼ Pxþ QNf xþ ½K � ðkðI � QÞNf Þ�x:

By the continuity of f and Lemma 3.3, one has that Gf is a completely continuous operator.

T. Chen et al. / Applied Mathematics and Computation 244 (2014) 422–431 427

We assume that, for k ¼ 1, Eq. (4.3) has no solution on @X because otherwise we are done with the proof. It follows that,from hypothesis (C1), Eq. (4.3) has no solution for ðx; kÞ 2 @X� ð0;1�. For k ¼ 0, Eq. (4.2) is reduced to the following PBVP

Db0þ/pðDa

0þxðtÞÞ ¼ bTb

R T0 ðT � sÞb�1f ðs; xðsÞ;Da

0þxðsÞÞds;

xð0Þ ¼ xðTÞ; Da0þxð0Þ ¼ Da

0þxðTÞ:

(

Hence, if x is a solution of this PBVP, we must have

b

Tb

Z T

0ðT � sÞb�1f ðs; xðsÞ;Da

0þxðsÞÞds ¼ 0:

So, by (3.4), one has xðtÞ ¼ c 2 KerL; 8c 2 R. Thus, we obtain that

ðQNf jKerLÞxðtÞ ¼b

Tb

Z T

0ðT � sÞb�1f ðs; c;0Þds ¼ 0;

which together with hypothesis (C2) yields that x ¼ c R @X. Therefore, the Eq. (4.3) has no solution for ðx; kÞ 2 @X� ½0;1�.Then we get that, for each k 2 ½0;1�, the Leray–Schauder degree degðI � Gf ð; kÞ;X;0Þ is well defined and, by the homotopyproperty of degree, that

degðI � Gf ð;1Þ;X;0Þ ¼ degðI � Gf ð;0Þ;X; 0Þ: ð4:4Þ

It is clear that equation x ¼ Gf ðx;1Þ is equivalent to the equation Lx ¼ Nf x, and (4.4) tell us it will have a solution if we canshow that degðI � Gf ð;0Þ;X;0Þ– 0. Next we will do this. From the definition of Gf , we get

Gf ðx;0Þ ¼ Pxþ QNf xþKð0Þ ¼ Pxþ QNf x:

Thus, we have x ¼ Gf ðx;0Þ ¼ c; 8c 2 R, which implies that

x� Gf ðx;0Þ ¼ �b

Tb

Z T

0ðT � sÞb�1f ðs; c; 0Þds:

That is

I � Gf ð;0Þ ¼ �QNf jKerL:

Therefore, basing on the properties of the Leray–Schauder degree, we obtain that

degðI � Gf ð;0Þ;X;0Þ ¼ �degðQNf jKerL;X \ KerL;0Þ;

where the right side degree is the Brouwer degree. Hence, by hypothesis (C3), the equation Lx ¼ Nf x has at least one solutionin domL \ �X. The proof is complete. h

5. Solutions of PBVP for fractional p-Laplacian equation

In this section, by using the continuation theorem given in Section 4, a new existence result on PBVP (1.2) for fractionalp-Laplacian equation will be stated and proved under the nonlinear growth restriction of f.

Theorem 5.1. Let f : ½0; T� � R2 ! R be continuous. Assume that

ðH1Þ there exist nonnegative functions a; b; c 2 Y such that

jf ðt;u;vÞj 6 aðtÞ þ bðtÞjujp�1 þ cðtÞjv jp�1; 8t 2 ½0; T�; ðu; vÞ 2 R2;

ðH2Þ there exists a constant A > 0 such that either

uf ðt;u;vÞ > 0; 8t 2 ½0; T�; v 2 R; juj > A ð5:1Þ

or

uf ðt;u;vÞ < 0; 8t 2 ½0; T�; v 2 R; juj > A: ð5:2Þ

Then PBVP (1.2) has at least one solution, provided that

c1 :¼ 2Tb

Cðbþ 1Þð2TaÞp�1kbk0

ðCðaþ 1ÞÞp�1 þ kck0

" #< 1; if p < 2; ð5:3Þ

c2 :¼ 2Tb

Cðbþ 1Þ2p�2ð2TaÞp�1kbk0

ðCðaþ 1ÞÞp�1 þ kck0

" #< 1; if p P 2:

428 T. Chen et al. / Applied Mathematics and Computation 244 (2014) 422–431

Proof. Set

X1 ¼ x 2 domL n KerLjLx ¼ kNf x; k 2 ð0;1Þ� �

:

For x 2 X1, we get Nf x 2 ImL. Thus, by (3.5), we have

Z T

0ðT � sÞb�1f ðs; xðsÞ;Da

0þxðsÞÞds ¼ 0:

Then, basing on the integral mean value theorem, there exists a constant n 2 ð0; TÞ such that f ðn; xðnÞ;Da0þxðnÞÞ ¼ 0. So, from

the hypothesis ðH2Þ, one has jxðnÞj 6 A.By Lemma 2.1, we get

xðtÞ ¼ xðnÞ � Ia0þDa0þxðnÞ þ Ia0þDa

0þxðtÞ;

which together with

jIa0þDa0þxðtÞj ¼ 1

CðaÞ

Z t

0ðt � sÞa�1Da

0þxðsÞds����

���� 6 1CðaÞ kD

a0þxk0

1a

ta 6Ta

Cðaþ 1Þ kDa0þxk0; 8t 2 ½0; T�

and jxðnÞj 6 A yields that

kxk0 6 Aþ 2Ta

Cðaþ 1Þ kDa0þxk0: ð5:4Þ

Then, from the hypothesis ðH1Þ and (5.4), we obtain that

jIb0þNf xðtÞj ¼1

CðbÞ

Z t

0ðt � sÞb�1f ðs; xðsÞ;Da

0þxðsÞÞds����

���� 6 1CðbÞ

Z t

0ðt � sÞb�1ðaðsÞ þ bðsÞjxðsÞjp�1 þ cðsÞjDa

0þxðsÞjp�1Þds

61

CðbÞ ðkak0 þ kbk0kxkp�10 þ kck0kD

a0þxkp�1

0 Þ 1b

tb

6Tb

Cðbþ 1Þ kak0 þ kck0kDa0þxkp�1

0 þ kbk0 Aþ 2Ta

Cðaþ 1Þ kDa0þxk0

� �p�1" #

; 8t 2 ½0; T�: ð5:5Þ

By Lx ¼ kNf x and Lemma 2.1, one has

xðtÞ ¼ d2 þ Ia0þ/qðd1 þ kIb0þNf xÞðtÞ; 8d1; d2 2 R:

Then, from xð0Þ ¼ xðTÞ, we get

1CðaÞ

Z T

0ðT � sÞa�1/qðd1 þ kIb0þNf xðsÞÞds ¼ 0:

Thus, there exists a constant g 2 ð0; TÞ such that /qðd1 þ kIb0þNf xðgÞÞ ¼ 0, which implies d1 ¼ �kIb0þNf xðgÞ. So, we have

/pðDa0þxðtÞÞ ¼ �kIb0þNf xðgÞ þ kIb0þNf xðtÞ;

which, together with j/pðDa0þxðtÞÞj ¼ jDa

0þxðtÞjp�1 and (5.5), yields that

kDa0þxkp�1

0 62Tb

Cðbþ 1Þ kak0 þ kck0kDa0þxkp�1

0 þ kbk0 Aþ 2Ta

Cðaþ 1Þ kDa0þxk0

� �p�1" #

: ð5:6Þ

If p < 2, based on (5.6) and Lemma 2.2, one gets

kDa0þxkp�1

0 62Tb

Cðbþ 1Þ kak0 þ Ap�1kbk0 þð2TaÞp�1kbk0

ðCðaþ 1ÞÞp�1 þ kck0

!kDa

0þxkp�10

" #;

which together with (5.3) implies that

kDa0þxk0 6

2Tb kak0 þ Ap�1kbk0

ð1� c1ÞCðbþ 1Þ

24

35

q�1

:¼ M1: ð5:7Þ

Thus, from (5.4), we have

kxk0 6 Aþ 2TaM1

Cðaþ 1Þ : ð5:8Þ

T. Chen et al. / Applied Mathematics and Computation 244 (2014) 422–431 429

Similarly, if p P 2, we can obtain that

kDa0þxk0 6

2Tb kak0 þ 2p�2Ap�1kbk0

ð1� c2ÞCðbþ 1Þ

24

35

q�1

:¼ M2; ð5:9Þ

kxk0 6 Aþ 2TaM2

Cðaþ 1Þ : ð5:10Þ

Therefore, combining (5.7) and (5.9) with (5.8) and (5.10), we have

kxkX ¼max kxk0; kDa0þxk0

� �6max M1;M2;Aþ

2TaM1

Cðaþ 1Þ ;Aþ2TaM2

Cðaþ 1Þ

� �:¼ M:

It follows that X1 is bounded.Let

X2 ¼ x 2 KerLjQNf x ¼ 0� �

:

For x 2 X2, we have xðtÞ ¼ d; 8d 2 R. Then we get

Z T

0ðT � sÞb�1f ðs;d;0Þds ¼ 0;

which together with hypothesis ðH2Þ implies that jdj 6 A. Thus, one has

kxkX 6 maxfA;0g ¼ A:

Hence, X2 is bounded.If (5.1) holds, set

X3 ¼ x 2 KerLjkIxþ ð1� kÞQNf x ¼ 0; k 2 ½0;1�� �

:

For x 2 X3, we have xðtÞ ¼ d; 8d 2 R and

kdþ ð1� kÞ b

Tb

Z T

0ðT � sÞb�1f ðs;d; 0Þds ¼ 0: ð5:11Þ

If k ¼ 0, then jdj 6 A because of (5.1). If k 2 ð0;1�, we can also show that jdj 6 A. Otherwise, if jdj > A, in view of (5.1), one has

kd2 þ ð1� kÞ b

Tb

Z T

0ðT � sÞb�1df ðs;d;0Þds > 0;

which contradicts to (5.11). It follows that, X3 is bounded.If (5.2) holds, let

X03 ¼ x 2 KerLj � kIxþ ð1� kÞQNf x ¼ 0; k 2 ½0;1�� �

:

Similar to the above argument, we can show that X03 is also bounded.In the following, we shall prove that all assumptions of Theorem 4.1 are satisfied. Set

X ¼ x 2 XT jkxkX < maxfM;Ag þ 1� �

:

Obviously, ðX1 [X2 [X3Þ � X (or ðX1 [X2 [X03Þ � X). It follows from the above arguments that the following two condi-tions are satisfied

(C1) for each k 2 ð0;1Þ, the equation

Lx ¼ kNf x

has no solution on ðdomL n KerLÞ \ @X;(C2) the equation

QNf x ¼ 0

has no solution on KerL \ @X.

Now, the condition (C3) of Theorem 4.1 remains to be verified. To do that, let us define the homotopy

Hðx; kÞ ¼ kIxþ ð1� kÞQNf x:

According to the above argument, we know

Hðx; kÞ– 0; 8x 2 @X \ KerL:

430 T. Chen et al. / Applied Mathematics and Computation 244 (2014) 422–431

Thus, by the homotopy property of degree, we have

degðQNf jKerL;X \ KerL; 0Þ ¼ degðHð;0Þ;X \ KerL;0Þ ¼ degðHð;1Þ;X \ KerL;0Þ ¼ degðI;X \ KerL;0Þ– 0:

So that, the condition (C3) of Theorem 4.1 is satisfied.Consequently, by using Theorem 4.1, the operator equation Lx ¼ Nf x has at least one solution in domL \ �X. Namely, PBVP

(1.2) has at least one solution in XT . The proof is complete. h

6. An example

In this section, we will give an example to illustrate our main result.

Example 6.1. Consider the following PBVP for fractional p-Laplacian equation

D340þ/3ðD

120þxðtÞÞ ¼ � 8

3þ 124 x2ðtÞ þ te

� D120þ

xðtÞ 2

; t 2 ½0;1�;

xð0Þ ¼ xð1Þ; D120þxð0Þ ¼ D

120þxð1Þ:

8><>: ð6:1Þ

Corresponding to PBVP (1.2), we get that p ¼ 3; a ¼ 1=2; b ¼ 3=4; T ¼ 1 and

f ðt;u; vÞ ¼ �83þ 1

24u2 þ te�v2

:

Choose aðtÞ ¼ 4; bðtÞ ¼ 1=24; cðtÞ ¼ 0; A ¼ 8. By a simple calculation, we can obtain that kbk0 ¼ 1=24; kck0 ¼ 0 and

uf ðt;u;vÞ ¼ u1

24ðu2 � 64Þ þ te�v2

� �> 0 ðor < 0Þ; 8t 2 ½0;1�; v 2 R; juj > 8;

c2 ¼2

C 34þ 1 � 2� 22

C 12þ 1 � �2 �

124þ 0

" #< 1:

Obviously, PBVP (6.1) satisfies all assumptions of Theorem 5.1. Hence, PBVP (6.1) has at least one solution.

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