Solvability of fractional boundary value problemwithp-laplacian operator at resonance

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1 Shen et al. Advances in Difference Equations 23 23:295 R E S E A R C H Open Access Solvability of fractional boundary value problemwithp-laplacian operator at resonance Tengfei Shen Wenbin Liu * and Xiaohui Shen * Correspondence: wblium@63.com College of Sciences China University of Mining and Technology Xuzhou 226 P.R. China Abstract In this paper a class of multi-point boundary value problems for nonlinear fractional differential equations at resonance with p-laplacian operator is considered. By using the extension of Mawhin s continuation theorem due to Ge the existence of solutions is obtained which enriches previous results. MSC: 34A8; 34B5 Keywords: fractional differential equation; boundary value problem; p-laplacian operator; coincidence degree theory; resonance Introduction In the recent years fractional differential equations played an important role in many fields such as physics electrical circuits biology control theory etc. see [ 7]. Thus many scholars have paid more attention to fractional differential equations and gained some achievements see [8 22].For examplewang [2] considered a class of fractional multipoint boundary value problems at resonance by Mawhin s continuation theorem see [23]: D +ut=f t ut D + ut t a.e. t u = + u = m a i + uξ i D 2 + u = m b id 2 + uη i. where < 2 < ξ < ξ 2 < < ξ m <<η < η 2 < < η n < m a i = n b i = n b iη i =D + is the standard fractional derivative f :[] R2 R satisfies the Carathéodory condition. But Mawhin s continuation theorem is not suitable for quasi-linear operators. In [24] Ge and Ren had extended Mawhin s continuation theorem which was used to deal with more general abstract operator equations. In [25] Pang et al. considered a higher order nonlinear differential equation with a p-laplacianoperatorat resonance: ϕ p u n t = f t ut...u n t + et t u i = i =2...n u = us dgs.2 23 Shen et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License which permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited.

2 Shen et al. Advances in Difference Equations 23 23:295 Page 2 of 3 where ϕ p s= s p 2 s p >f :[] R n R n and e :[] R are continuous n 2is an integer. g :[] R is a nondecreasing function with dgs=theintegralinthe second partof.2 is meant in the Riemann-Stieltjes sense. However there are few articles which consider the fractional multi-point boundary value problem at resonance with p-laplacian operator and dim Ker M =2because p-laplacian operator is a nonlinear operator and it is hard to construct suitable continuous projectors. In this paper we will improve and generalize some known results. Motivated by the work above our article is to investigate the multi-point boundary value problem at resonance for a class of Riemanne-Liouville fractional differential equations with p-laplacian operator and dim Ker M = 2 by constructing suitable continuous projectors and using the extension of Mawhin s continuation theorem: D +ϕ pd +ut = f t ut D 2 + ut D + ut D +ut t u = D +u = u = m a iuξ i + u = m b i + uη i.3 where 2 < 3 < 3 < + 4 < ξ < ξ 2 < < ξ m <<η < η 2 < < η m < a i R b i R <m m N m a iξi = m a iξi 2 = m b i =ϕ p s = s p 2 s ϕ p = < p/p+/q =ϕ p is invertible and its inverse operator is ϕ q D + is Riemann- Liouville standard fractional derivative f :[] R 4 R is continuous. In order to investigate the problem we need to suppose that the following conditions hold: = where = Ɣq Ɣq + q q +2 Ɣ + q Ɣq + q q +2 a i ξ q+q q + i 2 = Ɣ q ƔƔq + q 2q +3 Ɣ + q Ɣq + q 2q +3 Ɣ q 3 = Ɣ + q q + q q +2 a i ξ q+q 2q +2 i b i η q+q q +2 i Ɣ q 4 = b Ɣ + q i η q+q 2q +3 i. q + q 2q +3 The rest of this article is organized as follows: In Section 2 wegivesomenotations definitions and lemmas. In Section 3 based on the extension of Mawhin s continuation theorem due to Ge we establish a theorem on existence of solutions for BVP.3. 2 Preliminaries For the convenience of the reader we present here some necessary basic knowledge and definitions for fractional calculus theory that can be found in the recent literature see [ ].

3 Shen et al. Advances in Difference Equations 23 23:295 Page 3 of 3 Let X and Y be two Banach spaces with norms X and u Y respectively. A continuous operator M dom M X : X dom M Y is said to be quasi-linear if i Im M := MX dom M isaclosedsubsetofy ii Ker M := {u X dom M : Mu =} is linearly homeomorphic to R n n <. Let X = Ker M and X 2 be the complement space of X in X thenx = X X 2.Onthe other hand suppose that Y is a subspace of Y andy 2 is the complement space of Y in Y so that Y = Y Y 2.LetP: X X be a projector and Q : Y Y a semi-projector and X an open and bounded set with origin θ. θ is the origin of a linear space. Suppose that N λ : Y λ [ ] is a continuous operator. Denote N by N.Let λ = {u : Mu = N λ u}. N λ is said to be M-compact in if there is an Y Y with dim Y = dim X and an operator R : [ ] X continuous and compact such that for λ [ ] I QN λ Im M I QY 2. QN λ x = θ λ QNx = θ 2.2 R λ λ =I P λ 2.3 and R is the zero operator M [ P + R λ ] =I QN λ. 2.4 Lemma 2. Ge-Mawhin s continuation theorem [24] Let X X and Y Y be two Banach spaces and X an open and bounded nonempty set. Suppose that M : X dom M Y is a quasi-linear operator N λ : Y λ [ ] is M-compact in. In addition if i Lu N λ u u λ dom M ii degjqn Ker M where J : Im Q Ker M is a homeomorphism with Jθ=θ and N = N then the equation Mu = Nu has at least one solution in dom M. Definition 2. The Riemann-Liouville fractional integral of order >ofafunctionu is given by I +ut= Ɣ t s us ds provided the right-hand side integral is pointwise almost everywhere defined on +. Definition 2.2 The Riemann-Liouville fractional derivative of order > of a function u is given by D +ut= d n us ds Ɣn dt t s n+

4 Shen et al. Advances in Difference Equations 23 23:295 Page 4 of 3 provided the right-hand side integral is pointwise everywhere defined on + where n =[]+. Definition 2.3 Let X be a Banach space and X X is a subspace. A mapping Q : X X is a semi-projector if Q satisfies i Q 2 x = Qx x X ii Qμx=μQx x X μ R. Lemma 2.2 Assume that u C L with a fractional derivative of order > that belongs to C L. Then I + D + ut=ut+c t + c 2 t c N t N for some c i R i =2...N where N =[]+. Lemma 2.3 Assume that ut C[ ] p q then D q +Ip +ut=ip q + ut. Lemma 2.4 Assume that then: i If λ > λ i i =2...[]+ we have that D +tλ = Ɣλ + Ɣλ + tλ. ii D +t i = i =2...[]+. In this paper we take X = {u u D 2 + u D + u D +u C[ ]} with the norm u X = max{ u D 2 + u + u D +u } where u = max t [] ut and Y = C[ ] with the norm y Y = y. By means of the linear functional analysis theory it is easy to prove that X and Y are Banach spaces so we omit it. Define the operator M : dom M Y by Mu = D +ϕ p D +ut 2.5 { dom M = u X D +ϕ p D +u Y u = D +u = } m u = a i uξ i + u = b i + uη i. 2.6 Based on the definition of dom M itiseasytofindthatdom M such as ut=ct dom M c R. Define the operator N λ : X Y λ [ ] N λ ut=λf t ut D 2 + ut D + ut D +ut t [ ]. Then BVP.3 is equivalent to the operator equation Mu = NuwhereN = N.

5 Shen et al. Advances in Difference Equations 23 23:295 Page 5 of 3 3 Main result In this section a theorem on existence of solutions for BVP.3 will be given. Define operators T j : Y Y j = 2 as follows: T y = T 2 y = s ϕ q Ɣ a i ξi ϕ q Ɣ b i ηi ξ i s ϕ q Ɣ s τ yτ dτ ds s τ yτ dτ ϕ q Ɣ s τ yτ dτ ds ds s τ yτ dτ ds. Let us make some assumptions which will be used in the sequel. H There exist nonnegative functions r d e h k Y such that for all t [ ] u v w z R 4 f t u v w z rt+dt u p + et v p + ht w p + kt z p. H2 There exists a constant A >such that for u dom Mif + ut > A for all t [ ]then or sgn { + ut} 4 T Nut 3 T 2 Nut > sgn { + ut} 4 T Nut 3 T 2 Nut <. H3 There exists a constant B >such that for u dom Mif D 2 + ut > B for all t [ ]then or sgn { D 2 + ut} 2 T Nut+ T 2 Nut > sgn { D 2 + ut} 2 T Nut+ T 2 Nut <. Theorem 3. Let f :[] R 4 R be continuous and condition H-H3 hold then BVP.3 has at least one solution provided that A d + e + h + k < 3. Ɣ + where A = Ɣ+ + 2 Ɣ + 7 2Ɣ. InordertoproveTheorem3. we need to prove some lemmas below.

6 Shen et al. Advances in Difference Equations 23 23:295 Page 6 of 3 Lemma 3. The operator M : dom M X Yisquasi-linear. Ker M = { u X ut=c t + c 2 t 2 c c 2 R } 3.2 Im M = {y Y T j y =j =2}. 3.3 Proof Suppose that ut dom MbyD +ϕ pd +ut = we have D +ut=ϕ q c t. Based on D +u = one has ut=c t + c 2 t 2 + c 3 t 3 which together with u = yields that Ker M = { u X ut=c t + c 2 t 2 c c 2 R }. It is clear that dim Ker M =2.SoKer M is linearly homeomorphic to R 2. If y Im M then there exists a function u dom M such that yt =D +ϕ pd +ut. Based on Lemmas 2.2 and 2.3wehave ut=i +ϕ q I +ys + c t + c 2 t 2 + ut=d + I +ϕ q I +ys + c Ɣ which together with m a iξi = m a iξi 2 =and m b i =yieldsthatt j yt= j =2. On the other hand suppose that y Y and satisfies 3.3 and let ut =I +ϕ qi +yt then u dom M and Mut=D +ϕ pd +ut = ysoy Im M and Im M := Mdom Misa closed subset of Y.ThusM is a quasi-linear operator. Lemma 3.2 Let Xbeanopenandboundedset then N λ is M-compact in. Proof Define the continuous projector P : X X by Put= Ɣ D + ut + Ɣ D 2 + ut 2 t [ ]. Define the continuous projector Q : Y Y by Qyt= Q yt t + Q 2 yt t 2 t [ ] where Q yt=ϕ p 4 T yt 3 T 2 yt Q 2 yt=ϕ p 2 T yt+ T 2 yt.

7 Shen et al. Advances in Difference Equations 23 23:295 Page 7 of 3 Obviously X = Ker M = Im P and Y =ImQ. Thuswehavedim Y = dim X = 2. For any y Y wehave Q Q ytt = ϕ p 4 T Q ytt 3 T 2 Q ytt = Q ytϕ p = Q yt. Similarly we can get Q Q2 ytt 2 = Q 2 Q ytt = Q 2 Q2 ytt 2 = Q 2 yt. Hence the map Q is idempotent. Similarly we can get Qμy=μQy for all y Y μ R. Thus Q is a semi-projector. For any y Im M wecangetthatqy =andy Ker Q conversely if y Ker Q we can obtain that Qy =thatistosayy Im M.ThusKer Q = Im M. Let X be an open and bounded set with θ foreachu wecanget Q[I QN λ u] =. Thus I QN λ u Im M = Ker Q.Takeanyy Im M in the type y = y Qy+QysinceQy =wecangety I QY.So2.holds.Itiseasytoverify2.2. Define R : [ ] X 2 by Ru λt= Ɣ t s ϕ q Ɣ s τ I QN λ uτ dτ ds. By the continuity of f itiseasytogetthatru λ is continuous on [ ]. Moreover for all u there exists a constant T >suchthat I +I QN λuτ Tsowecaneasily obtain that R λ D 2 + R λ D + R λ andd +R λ are uniformly bounded. By Arzela-Ascoli theorem we just need to prove that R : [ ] X 2 is equicontinuous. For u <t < t 2 2 < 3 < 3 < + 4 we have Ru λt2 Ru λt = 2 Ɣ t 2 s ϕ q I + I QNλ uτ ds t s ϕ q I + I QNλ uτ ds ϕ ql Ɣ t2 s t s ds + 2 t t 2 s ds = ϕ qt t Ɣ + 2 t D 2 + Ru λt 2 D 2 + Ru λt 2 = t sϕ q I + I QNλ uτ ds t sϕ q I + I QNλ uτ ds 2 ϕ q T t 2 s t s ds + t 2 s ds = ϕ qt t t 2 t

8 Shen et al. Advances in Difference Equations 23 23:295 Page 8 of 3 and D + Ru λt 2 + Ru λt 2 = ϕ q I + I QNλ uτ ds ϕ q Tt 2 t. ϕ q I + I QNλ uτ ds Since t is uniformly continuous on [ ] so R λ D 2 + R λ andd + R λ are equicontinuous. Similarly we can get that I +I QN λuτ C[ ] is equicontinuous. Considering that ϕ q s isuniformlycontinuouson[ T T] we have that D +R λ = I +I QN λ is also equicontinuous. So we can obtain that R : [ ] X 2 is compact. For each u λ wehaved +ϕ pd +ut = N λut Im M.Thus Ru λt= Ɣ = Ɣ t s ϕ q Ɣ t s ϕ q Ɣ which together with D +u = u = yields that s τ I QN λ uτ dτ s τ D +ϕ p D +uτ dτ ds ds Ru λt=ut Ɣ D + ut Ɣ D 2 + ut 2 =I Put. It is easy to verify that Rut is the zero operator. So 2.3 holds. Besides for any u M [ Pu + Ru λ ] t [ t = M t s ϕ q Ɣ Ɣ + Ɣ D + ut + =I QN λ ut s τ I QN λ uτ dτ ] Ɣ D 2 + ut 2 ds which implies 2.4. So N λ is M-compact in. Lemma 3.3 Suppose that H H2hold then the set = { u dom M \ Ker M Mu = N λ u λ } is bounded. Proof By Lemma 2.2foreachu dom Mwehave ut=i +D +ut+c t + c 2 t 2 + c 3 t 3.

9 Shen et al. Advances in Difference Equations 23 23:295 Page 9 of 3 Combined with u = we get c 3 =.Thus ut=i +D +ut+c t + c 2 t 2 + ut=i +D +ut+c Ɣ D 2 + ut=i2 +D +ut+c Ɣt + c 2 Ɣ. By simple calculation we get c = Ɣ c 2 = Ɣ + ut D +us ds D 2 + ut t s + us ds + ut D t +us ds. Take any u thennu Im M = Ker Q and QNu =. It follows from H2 and H3 that there exist ε ε 2 [ ] such that + uε A D 2 + uε 2 B.Thus + ut=d + uε + D +ut dt ε D 2 + ut=d 2 + uε 2+ + u A + D +u + ut dt ε 2 D 2 + u B + D + u A + B + D +u. So we get c Ɣ + u + D +u A +2D Ɣ +u D 2 c 2 Ɣ + u + 3 D 2 + u + D +u Ɣ u A D +u + B 5A 2 + B + 7 D 2 +u where A = Ɣ+ + 2 Ɣ + 7 2Ɣ B = Based on D +u = we have ϕ p D +ut = λi +Nut. A Ɣ + 5A 2Ɣ + B Ɣ. From H and λ we have ϕp D +ut Ɣ Ɣ t s f s us D 2 + ut D + us D +us ds t s rs+ds us p + es D 2 + us p

10 Shen et al. Advances in Difference Equations 23 23:295 Page of 3 + hs D + us p + ks D +us p ds Ɣ + r + d u p + h + u p + k + e D 2 + u p D +u p which together with ϕ p D +ut = D +ut p wecanget D +u p Ɣ + r + d u p + h + u p + k + e D 2 + u p D +u p r + d A D Ɣ + +u + B + e 2A + D +u p + h A + D +u p + k D +u p In view of 3. we can obtain that there exists a constant M >suchthat D +u M + u A + M := M 2 D 2 + u 2A + M := M 3 u A M + B := M 4. Thus we have. u X = max { u D 2 + u D + u D +u } max{m M 2 M 3 M 4 } := M. So is bounded. Lemma 3.4 Suppose that H2 holds then the set 2 = {u u Ker M Nu Im M} is bounded. Proof For each u 2 wehavethatut =c t + c 2 t 2 c c 2 R and QNu =.It follows from H2 and H3 that there exists an ε ε 2 [ ] such that + uε A D 2 + uε 2 A which implies that c A Ɣ and c 2 A+B Ɣ.So 2 is bounded. Define the isomorphism J : Ker M Im Q by J c t + c 2 t 2 =c t + c 2 t 2 c c 2 R t [ ]. In fact for each c c 2 R supposethatq yt Q 2 yt = c c 2 we have 4 T yt 3 T 2 yt= ϕ q c := c 2 T yt+ T 2 yt= ϕ q c 2 := c where c c 2 R by the condition there exists a unique solution for 3.4 which is T yt T 2 yt = m m 2 m m 2 R. Now we will prove that there exists y Y such that T yt T 2 yt = m m 2. Based on yt C[ ] we choose yt=d +yt

11 Shen et al. Advances in Difference Equations 23 23:295 Page of 3 where yt=ϕ p l t + q +l 2 t + 2q l = Ɣ q Ɣ+ q l 2 = 2Ɣ q Ɣ+ q = m 4 m = m 2 m 3 t [ ] which together with 2 < 3 < 3 < + 4 q > we have + q >and + 2q >.Sowehavey =. Thus we can obtain that I +yt=i +D +yt=yt+ct where c R which together with y = we have c =.Sowehave T yt= s ϕ q ys ds m = = m T 2 yt= ϕ qys ds m b i a i ξi ξ i s ϕ q ys ds ηi ϕ qys ds = = m 2. Thus there exists y Y such that T yt T 2 yt = m m 2. So J : Ker L Im Q is well defined. Lemma 3.5 Suppose that the first part of H3 holds then the set 3 = { u Ker M λj u + λqnu =λ [ ] } is bounded. Proof For each u 3 wecangetthatut=c t + c 2 t 2 c c 2 R. By the definition of the set 3 we can obtain that λ c t + c 2 t 2 + λ Q N c t + c 2 t 2 t + Q 2 N c t + c 2 t 2 t 2 =. Thus λc + λϕ p 4 T N c t + c 2 t 2 3 T 2 N c t + c 2 t 2 = 3.5 λc 2 + λϕ p 2 T N c t + c 2 t 2 + T 2 N c t + c 2 t 2 =. 3.6 If λ =then 3 is bounded because of the first part of H2 and H3. If λ =weget c = c 2 =obviously 3 is bounded. If λ by the first part of H2 and 3.5 we can obtain that c A Ɣ by the first part of H3 and 3.6 we have c 2 A+B Ɣ.So 3 is bounded. Remark 3. If the second part of H3 holds then the set 3 = { u Ker M λj u + λqnu =λ [ ] } is bounded.

12 Shen et al. Advances in Difference Equations 23 23:295 Page 2 of 3 Proof of Theorem 3. Assume that is a bounded open set of X with 3 i 3. By Lemmas 3. and 3.2 we can obtain that M : dom M X Y is quasi-linear and N λ is M-compact on. By the definition of wehave Lu N λ u u λ dom M Hu λ=±λj u+ λqnu Ker M [ ]. Thus by the homotopic property of degree we can get degjqn Ker M=deg H Ker M = deg H Ker M = deg±i Ker M. So Lemma 2. is satisfied and Mu = Nu has at least one solution in dom M. Namely BVP.3 have at least one solution in the space X. Competing interests The authors declare that they have no competing interests. Authors contributions The authors contributed equally in this article. All authors read and approved the final manuscript. Acknowledgements The authors are grateful to those who gave useful suggestions about the original manuscript. This research is supported by the National Natural Science Foundation of China No and the Fundamental Research Funds for the Central Universities 2LKSX9. Received: January 23 Accepted: 2 September 23 Published: 7 Nov 23 References. Kilbas AA Srivastava HM Trujillo JJ: Theory and Applications of Fractional Differential Equations. Elsevier Amsterdam Sabatier J Agrawal OP Machado JAT: Advances in Fractional Calculus: Theoretical Developments and Applications in Physics and Engineering. Springer Dordrecht Samko SG Kilbas AA Marichev OI: Fractional Integrals and Derivatives: Theory and Applications. Gordon & Breach New York Magin R: Fractional calculus models of complex dynamics in biological tissues. Comput. Math. Appl Metzler R Klafter J: Boundary value problems for fractional diffusion equations. Physica A Mainardi F: Fractional diffusive waves in viscoelastic solids. In: Wegner JL Norwood FR eds. Nonlinear Waves in Solids pp ASME Fairfield Bai J Feng X: Fractional-order anisotropic diffusion for image denoising. IEEE Trans. Image Process Agarwal RP O Regan D Stanek S: Positive solutions for Dirichlet problems of singular nonlinear fractional differential equations. J. Math. Anal. Appl Kosmatov N: A boundary value problem of fractional order at resonance. Electron. J. Differ. Equ Jafari H Gejji VD: Positive solutions of nonlinear fractional boundary value problems using Adomian decomposition method.appl.math.comput Bai Z: Solvability fora class of fractional m-point boundary value problem at resonance. Comput. Math. Appl Bai Z: On solutions of some fractional m-point boundary value problems at resonance. Electron. J. Qual. Theory Differ. Equ Bai Z Zhang Y: Solvability of fractional three-point boundary value problems with nonlinear growth. Appl. Math. Comput Zhang Y Bai Z: Existence of positive solutions for nonlinear fractional three-point boundary value problem at resonance. Appl. Math. Comput Hu Z Liu W: Solvability for fractional order boundary value problem at resonance. Bound. Value Probl Rui W: Existence of solutions of nonlinear fractional differential equations at resonance. Electron. J. Qual. Theory Differ. Equ

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