Solvability of Neumann boundary value problem for fractional p-laplacian equation

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1 Zhang Advances in Difference Equations 215) 215:76 DOI /s R E S E A R C H Open Access Solvability of Neumann boundary value problem for fractional p-laplacian equation Bo Zhang * * Correspondence: zhangbohuaibei@163.com School of Mathematical Sciences, Huaibei Normal University, Huaibei, 235, P.R. China Abstract We consider the existence of solutions for a Neumann boundary value problem for the fractional p-laplacian equation. Under certain nonlinear growth conditions of the nonlinearity, we obtain a new result on the existence of solutions by using the continuation theorem of coincidence degree theory. MSC: 34A8; 34B15 Keywords: Neumann boundary value problem; fractional differential equation; p-laplacian operator; continuation theorem 1 Introduction The purpose of this paper is to establish the existence of solutions for the following Neumann boundary value problem NBVP for short) for a fractional p-laplacian equation: { D β +φ pd +xt)) = gt, xt)), t [, T], D +x) = D +xt)=, 1.1) where <, β 1, D + is a Caputo fractional derivative, φ ps)= s p 2 s p >1),T >isa given constant and g :[,T] R R is continuous. Obviously, φ p is invertible and its inverse operator is φ q,whereq > 1 is a constant such that 1/p +1/q =1. The fractional calculus is a generalization of the ordinary differentiation and integration on an arbitrary order that can be noninteger. Fractional differential equations appear in a number of fields such as physics, polymer rheology, regular variation in thermodynamics, biophysics, blood flow phenomena, aerodynamics, electro-dynamics of a complex medium, viscoelasticity, Bode analysis of feedback amplifiers, capacitor theory, electrical circuits, electro-analytical chemistry, biology, control theory, fitting of experimental data, etc. see [1 4]). In recent years, because of the intensive development of the fractional calculus theory itself and its applications, fractional differential equations have been of great interest. For example, Agarwal et al. see [5]) considered a two-point boundary value problem at nonresonance, and Bai see [6]) considered a m-point boundary value problem at resonance. For more papers on fractional boundary value problems, see [7 15] andthe references therein. In [7], by using the coincidence degree theory for Fredholm operators, the authors studied the existence of solutions for the following NBVP: { D β +φ pd +xt)) = f t, xt), D +xt)), t [, 1], D +x) = D +x1) =. 215 Zhang; 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 credited.

2 ZhangAdvances in Difference Equations 215) 215:76 Page 2 of 1 Notice that D β +φ pd +) is a nonlinear operator, so it is not a Fredholm operator. Hence, there is a bug in the proof of the main result. 2 Preliminaries In this section, for convenience of the reader, we will present here some necessary basic knowledge and definitions as regards the fractional calculus theory, which can be found, for instance, in [2, 4]. Definition 2.1 The Riemann-Liouville fractional integral operator of order >ofa function u :,+ ) R is given by I +ut)= 1 Ɣ) t t s) 1 us) ds, provided that the right-side integral is pointwise defined on, + ). Definition 2.2 The Caputo fractional derivative of order > of a continuous function u :,+ ) R is given by D d n ut) +ut)=in + dt n = 1 Ɣn ) t t s) n 1 u n) s) ds, where n is the smallest integer greater than or equal to, provided that the right-side integral is pointwise defined on, + ). Lemma 2.1 see [1]) Let >.Assume that u, D +u L[, T], R). Then the following equality holds: I +D +ut)=ut)+c + c 1 t + + c n 1 t n 1, where c i R, i =,1,...,n 1,here n is the smallest integer greater than or equal to. Lemma 2.2 see [16]) For any u, v, then φ p u + v) φ p u)+φ p v), if p <2; φ p u + v) 2 p 2 φ p u)+φ p v) ), if p 2. Now we briefly recall some notations and an abstract existence result, which can be found in [17]. Let X, Y be real Banach spaces, L : dom L X Y be a Fredholm operator with index zero, and P : X X, Q : Y Y be projectors such that Im P = Ker L, Ker Q = Im L, X = Ker L Ker P, Y = Im L Im Q.

3 ZhangAdvances in Difference Equations 215) 215:76 Page 3 of 1 It follows that L dom L Ker P : dom L Ker P Im L is invertible. We denote the inverse by K P. If is an open bounded subset of X such that dom L, then the map N : X Y will be called L-compact on if QN ) is bounded and K P I Q)N : X is compact. Lemma 2.3 see [17]) Let X and Y be two Banach spaces, L : dom L X YbeaFredholm operator with index zero, Xbeanopenboundedset, and N : YbeL-compact on. Suppose that all of the following conditions hold: 1) Lx λnx, x dom L, λ, 1); 2) QNx, x Ker L; 3) degjqn, Ker L,), where J : Im Q Ker L is an isomorphism map. Then the equation Lx = Nx has at least one solution on dom L. 3 Main result In this section, we will give the main result on the existence of solutions for NBVP 1.1). Theorem 3.1 Let g :[,T] R R be continuous. Assume that C 1 ) there exists a constant d >such that 1) i ugt, u)> i =1,2), t [, T], u > d; C 2 ) there exist nonnegative functions a, b C[, T] such that gt, u) at) u p 1 + bt), t [, T], u R. Then NBVP 1.1) has at least one solution, provided that γ 1 := 2p 1 T β+p a <1, if p <2; Ɣβ +1)Ɣ +1)) p 1 γ 2 := 22p 3 T β+p a <1, if p 2. Ɣβ +1)Ɣ +1)) p 1 3.1) For making use of the continuation theorem to study the existence of solutions for NBVP 1.1), we consider the following system: D +x 1t)=φ q x 2 t)), D β +x 2t)=gt, x 1 t)), 3.2) x 2 ) = x 2 T)=. Clearly, if x )=x 1 ), x 2 )) T is a solution of NBVP 3.2), then x 1 ) must be a solution of NBVP 1.1). Hence, to prove that NBVP 1.1) has solutions, it suffices to show that NBVP 3.2)hassolutions. In this paper, we take X = {x =x 1, x 2 ) T x 1, x 2 C[, T]} with the norm x = max{ x 1, x 2 },where x i = max t [,T] x i t) i = 1, 2). By means of the linear functional analysis theory, we can prove X is a Banach space.

4 ZhangAdvances in Difference Equations 215) 215:76 Page 4 of 1 Define the operator L : dom L X X by where D Lx = +x 1 D β +x, 3.3) 2 dom L = { x X D +x 1, D β +x 2 C[, T], x 2 ) = x 2 T)= }. Let N : X X be the Nemytskii operator φ q x 2 t)) Nxt)=, t [, T]. 3.4) gt, x 1 t)) Then NBVP 3.2) is equivalent to the operator equation as follows: Lx = Nx, x dom L. Next we will give some lemmas which are useful in the proof of Theorem 3.1. Lemma 3.1 Let L be defined by 3.3), then Ker L = { x X xt)=c R 2, t [, T] }, 3.5) { } Im L = y X y 1 ) =, T s) β 1 y 2 s) ds =. 3.6) Proof Obviously, from Lemma 2.1,3.5)holds. If y Im L, then there exists x dom L such that y = Lx.Thatis,y 1 t)=d +x 1t), y 2 t)= D β +x 2t). By Lemma 2.1,wehave x 2 t)=c + 1 Ɣβ) t t s) β 1 y 2 s) ds, c R. From the boundary value conditions D +x 1) = x 2 ) = x 2 T)=,weobtain y 1 ) =, T s) β 1 y 2 s) ds =. 3.7) So we get 3.6). On the other hand, suppose y X which satisfies 3.7). Let x 1 t) =I +y 1t), x 2 t) = I β +y 2t). Clearly x 2 ) = x 2 T)=.Hencex =x 1, x 2 ) T dom L and Lx = y. Thusy Im L. The proof is completed. Lemma 3.2 Let L be defined by 3.3), then L is a Fredholm operator of index zero. The linear projectors P : X XandQ: X Xcanbedefinedas Pxt)=x), t [, T], ) y 1 ) Qy) Qyt)= β 1 t) T T β T :=, t [, T]. s)β 1 y 2 s) ds Qy) 2 t)

5 ZhangAdvances in Difference Equations 215) 215:76 Page 5 of 1 Furthermore, the operator K P : Im L dom L Ker P can be written by I K P y = +y 1 I β +y. 2 Proof For any y X,wehave Q 2 y 1 ) yt)=q Qy) 2 t) ) y 1 ) = β Qy) 2 t) T β T = Qyt). s)β 1 ds Let y = y Qy,thenwegety 1 ) = and = T s) β 1 y 2 s) ds T s) β 1 y 2 s) ds = T β Qy)2 t) Q 2 y ) β 2 t)) =. T s) β 1 Qy) 2 s) ds So y Im L. ThusX = Im L + Im Q. SinceIm L Im Q = {}, wehavex = Im L Im Q. Hence dim Ker L = dim Im Q = codim Im L =2. This means that L is a Fredholm operator of index zero. For y Im L, from the definition of K P,wehave D LK P y = +I +y 1 D β +Iβ +y = y. 2 On the other hand, for x dom L Ker P,wegetx 1 ) = x 2 ) =. By Lemma 2.1,weobtain x 1 x 1 ) K P Lx = = x. x 2 x 2 ) So we know that K P =L dom L Ker P ) 1. The proof is completed. Lemma 3.3 Let N be defined by 3.4). Assume X is an open bounded subset such that dom L, then N is L-compact on. Proof By the continuity of φ q and g,wefindthatqn )andk P I Q)N ) are bounded. Moreover, there exists a constant A >suchthat I Q)Nx A, x, t [, T]. Hence, in view of the Arzelà-Ascoli theorem, we need only to prove that K P I Q)N ) X is equicontinuous.

6 ZhangAdvances in Difference Equations 215) 215:76 Page 6 of 1 For t 1 < t 2 T, x,wehave K P I Q)Nxt 2 ) K P I Q)Nxt 1 ) ) I = +I Q)Nx) 1t 2 ) I +I Q)Nx) 1t 1 ) I β +I Q)Nx) 2t 2 ) I β +I Q)Nx). 2t 1 ) From I Q)Nx A, x, t [, T], we can see that I + I Q)Nx 1 t 2) I + I Q)Nx 1 t 1) = 1 t2 Ɣ) t 2 s) 1 I Q)Nx ) s) ds 1 = t1 A Ɣ) t 1 s) 1 I Q)Nx ) 1 s) ds { t1 [ t1 s) 1 t 2 s) 1] ds + t2 A [ t Ɣ +1) 1 t2 +2t 2 t 1 ) ]. t 1 } t 2 s) 1 ds Since t is uniformly continuous on [, T],wecanobtainthatK P I Q)N )) 1 C[, T] is equicontinuous. A similar proof can show that K P I Q)N )) 2 C[, T] isalso equicontinuous. Hence, we find that K P I Q)N : X is compact. The proof is completed. Lemma 3.4 Suppose C 1 ), C 2 ) hold, then the set 1 = { x dom L Lx = λnx, λ, 1) } is bounded. Proof For x 1,wehaveNx Im L.Thus,from3.6), we obtain T s) β 1 g s, x 1 s) ) ds =. Then, by the integral mean value theorem, there exists a constant ξ, T) suchthat gξ, x 1 ξ)) =. So, from C 1 ), we get x 1 ξ) d. By Lemma 2.1,wehave x 1 t)=x 1 ξ) I +D +x 1ξ)+I +D +x 1t), which together with I +D +x 1t) 1 t = Ɣ) t s) 1 D +x 1s) ds 1 D Ɣ) +x 1 1 t T D Ɣ +1) +x 1, t [, T],

7 ZhangAdvances in Difference Equations 215) 215:76 Page 7 of 1 and x 1 ξ) d yields x 1 d + 2T D Ɣ +1) +x ) By Lx = λnx,wehave { D +x 1t)=λφ q x 2 t)), D β +x 2t)=λgt, x 1 t)). 3.9) From the first equation of 3.9), we get x 2 t)=λ 1 p φ p D +x 1t)). Then, by substituting it to the second equation of 3.9), we get D β +φ p D +x 1 t) ) = λ p gt, x 1 ):=λ p N g x 1 t). Thus, from Lemma 2.1 and the boundary value condition x 2 ) =, we obtain φ p D +x 1 t) ) = λ p I β +N gx 1 t). So, from C 2 ), we have φp D +x 1 t) ) 1 Ɣβ) 1 Ɣβ) t t t s) β 1 g s, x1 s) ) ds t s) β 1 as) x1 s) p 1 + bs) ) ds T β a x 1 p 1 ) + b, t [, T], Ɣβ +1) which together with φ p D +x 1t)) = D +x 1t) p 1 and 3.8)yields D +x p 1 1 T β [ b + a d + 2T ) D Ɣβ +1) Ɣ +1) +x p 1 1 ]. If p < 2, by Lemma 2.2,weget D +x 1 p 1 T β [ b + a d p 1 + 2T ) p 1 )] D Ɣβ +1) Ɣ +1)) p 1 +x 1 p 1 where A 1 = = A 1 + 2p 1 T β+p a Ɣβ +1)Ɣ +1)) p 1 D +x 1 p 1, Tβ Ɣβ+1) b + d p 1 a ). Then, from 3.1), we have q 1 D A1 +x 1 := B 1. 1 γ 1 Thus, from 3.8), we get x 1 d + 2T Ɣ +1) B )

8 ZhangAdvances in Difference Equations 215) 215:76 Page 8 of 1 If p 2,similartotheaboveargument,weletA 2 = x 1 d + Tβ Ɣβ+1) b +2 p 2 d p 1 a ), we obtain 2T Ɣ +1) B 2, 3.11) where B 2 = A 2 1 γ 2 ) q 1.Hence,combining3.1)with3.11), we have x 1 max {d + 2T Ɣ +1) B 1, d + 2T } Ɣ +1) B 2 := B. 3.12) From the second equation of 3.9), Lemma 2.1,andx 2 ) =, we have x 2 t)=λi β +N gx 1 t). So we have x 2 T β Ɣβ +1) G B, where G B = max{ gt, x) t [, T], x B}.Thus,from3.12), we obtain x = max { x 1, x 2 } max {B, T β } Ɣβ +1) G B := M. Hence, 1 is bounded. The proof is completed. Lemma 3.5 Suppose C 1 ) holds, then the set 2 = {x Ker L QNx =} is bounded. Proof For x 2,wehavex 1 t)=c 1, x 2 t)=c 2, t [, T], c 1, c 2 R,and φ q c 2 )=, 3.13) T s) β 1 gs, c 1 ) ds =. 3.14) From 3.13), we get c 2 =.From3.14)andC 1 ), we get c 1 d.thus,wehave x = max { c 1, c 2 } d. Hence, 2 is bounded. The proof is completed. Lemma 3.6 Suppose C 1 ) holds, then the set 3 = { x Ker L μx +1 μ)jqnx =,μ [, 1] }

9 ZhangAdvances in Difference Equations 215) 215:76 Page 9 of 1 is bounded, where J : Im Q Ker Ldefinedby Jx 1, x 2 ) T = 1) i x 2, x 1 ) T is an isomorphism map. Proof For x 3,wehavex 1 t)=c 1, x 2 t)=c 2, t [, T], c 1, c 2 R,and μc 1 + 1) i 1 μ) β T s) β 1 gs, c T β 1 ) ds =, 3.15) μc 2 +1 μ)φ q c 2 )=. 3.16) From 3.16), we get c 2 =becausec 2 and φ q c 2 ) have the same sign. From 3.15), if μ =, we get c 1 d because of C 1 ). If μ, 1], we can also get c 1 d. Infact,if c 1 > d, in view of C 1 ), one has μc μ) β T s) β 1 1) i c T β 1 gs, c 1 ) ds >, which contradicts to 3.15). So x d. Hence, 3 is bounded. The proof is completed. Proof of Theorem 3.1 Set = { x X x < max{m, d} +1 }. Obviously ). It follows from Lemmas 3.2 and 3.3 that L defined by 3.3)) is a Fredholm operator of index zero and N defined by 3.4)) is L-compact on. Moreover, by Lemmas 3.4 and 3.5, the conditions 1) and 2) of Lemma 2.3 are satisfied. Hence, it remains to verify the condition 3) of Lemma 2.3. Define the operator H : [, 1] X by Hx, μ)=μx +1 μ)jqnx. Then, from Lemma 3.6,wehave Hx, μ), x, μ) Ker L) [, 1]. Thus, by the homotopy property of the degree, we have degjqn, Ker L, θ)=deg H,), Ker L, θ ) = deg H,1), Ker L, θ ) = degi, Ker L, θ), where θ is the zero element of X. So the condition 3) of Lemma 2.3 is satisfied. Consequently, by Lemma 2.3, the operator equation Lx = Nx has at least one solution x )=x 1 ), x 2 )) T on dom L. Namely,NBVP1.1) has at least one solution x 1 ). The proof is completed.

10 ZhangAdvances in Difference Equations 215) 215:76 Page 1 of 1 Competing interests The author declares that he has no competing interests. Acknowledgements The author is grateful for the valuable comments and suggestions of the referees. Received: 2 October 214 Accepted: 19 December 214 References 1. Kilbas, AA, Srivastava, HM, Trujillo, JJ: Theory and Applications of Fractional Differential Equations. Elsevier, Amsterdam 26) 2. Podlubny, I: Fractional Differential Equations. Academic Press, San Diego 1999) 3. Sabatier, J, Agrawal, OP, Machado, JAT eds.): Advances in Fractional Calculus: Theoretical Developments and Applications in Physics and Engineering. Springer, Dordrecht 27) 4. Samko, SG, Kilbas, AA, Marichev, OI: Fractional Integrals and Derivatives: Theory and Applications. Gordon & Breach, Yverdon 1993) 5. Agarwal, RP, O Regan, D, Stanek, S: Positive solutions for Dirichlet problems of singular nonlinear fractional differential equations. J. Math. Anal. Appl. 371, ) 6. Bai, Z: On solutions of some fractional m-point boundary value problems at resonance. Electron. J. Qual. Theory Differ. Equ. 21, 37 21) 7. Chen, T, Liu, W, Hu, Z: A boundary value problem for fractional differential equation with p-laplacian operator at resonance. Nonlinear Anal. TMA 75, ) 8. Chen, T, Liu, W: An anti-periodic boundary value problem for the fractional differential equation with a p-laplacian operator. Appl. Math. Lett. 25, ) 9. Benchohra, M, Hamani, S, Ntouyas, SK: Boundary value problems for differential equations with fractional order and nonlocal conditions. Nonlinear Anal. TMA 71, ) 1. Bai, Z, Lü, H: Positive solutions for boundary value problem of nonlinear fractional differential equation. J. Math. Anal. Appl. 311, ) 11. Darwish, MA, Ntouyas, SK: On initial and boundary value problems for fractional order mixed type functional differential inclusions. Comput. Math. Appl. 59, ) 12. El-Shahed, M, Nieto, JJ: Nontrivial solutions for a nonlinear multi-point boundary value problem of fractional order. Comput. Math. Appl. 59, ) 13. Jiang, W: The existence of solutions to boundary value problems of fractional differential equations at resonance. Nonlinear Anal. TMA 74, ) 14. Kosmatov, N: A boundary value problem of fractional order at resonance. Electron. J. Differ. Equ. 21, ) 15. Su, X: Boundary value problem for a coupled system of nonlinear fractional differential equations. Appl. Math. Lett. 22, ) 16. Lian, H: Boundary Value Problems for Nonlinear Ordinary Differential Equations on Infinite Intervals. Doctoral thesis 27) 17. Gaines, R, Mawhin, J: Coincidence Degree and Nonlinear Differential Equations. Springer, Berlin 1977)