Shen et al. Boundary Value Problems 5 5:4 DOI.86/s366-5-59-z R E S E A R C H Open Access On the existence of solution to a boundary value problem of fractional differential equation on the infinite interval Chunfang Shen,HuiZhou Liu Yang,* * Correspondence: xjiangfeng@63.com College of Mathematics Statistics, Hefei Normal University, Hefei, 36, P.R. China College of Mathematical Science, University of Science Technology of China, Hefei, 3, P.R. China Abstract This work deals with a boundary value problem for a nonlinear multi-point fractional differential equation on the infinite interval. By constructing the proper function spaces the norm, we overcome the difficulty following from the noncompactness of [,. By using the Schauder fixed point theorem, we show the existence of one solution with suitable growth conditions imposed on the nonlinear term. MSC: 34B; 34B5 Keywords: fractional differential equation; boundary value problem; infinite interval; fixed point theorem Introduction In this paper, we consider the existence of solution of boundary value problem for a nonlinear multi-point fractional differential equation, D α ut=f t, ut, D ut, t J := [,,. m u =, u =, D u = β i uξ i,. where < α 3isarealnumber,f CJ R R, R. Due to the intensive development of the theory of fractional calculus itself as well as its applications, such as in the fields of physics, chemistry, aerodynamics, polymer rheology, etc., many papers books on fractional calculus, fractional differential equations have appeared see [ 6]. For example,bai [] established the existence results of positive solutions for the problem D α utf t, ut =, t, u =, u = βuη, η,. 5 Shen et al. This article is distributed under the terms of the Creative Commons Attribution 4. International License http://creativecommons.org/licenses/by/4./, which permits unrestricted use, distribution, reproduction in any medium, provided you give appropriate credit to the original authors the source, provide a link to the Creative Commons license, indicate if changes were made.
Shen et al. Boundary Value Problems 5 5:4 Page of 3 In [3], the authors considered the three-point boundary value problem of a coupled system of the nonlinear fractional differential equation D α ut=f t, vt, D p vt, t, D β vt=f t, ut, D q ut, t, u = v =, u = γ uη, v = γ uη, under the conditions < γη <,<γη β <. By using the Schauder fixed point theorem, they obtained at least one solution of this problem. The theory of boundary value problems on infinite intervals arises naturally has many applications; see [7].The existence multiplicityofsolutionstoboundary value problems of fractional differential equations on the infinite interval have been investigated in recent years [8 ]. Agarwal et al. [] established existence results of solutions for a class of boundary value problems involving the Riemann-Liouville fractional derivative on the half line by using the nonlinear alternative of Leray-Schauder type combined with the diagonalization process. Arara et al. [3] considered boundary value problems involving the Caputo fractional derivative on the half line, c D α ut=f t, ut, t J := [,, u = u, u is bounded on J. By using fixed point theorem combined with the diagonalization process, they obtained the existence of solutions. Liang Zhang [4] consider the m-point boundary value problem of fractional differential equation on the infinite interval D α utatf t, ut =, <t <, m u =, u =, D u = β i uξ i, where < α 3, D α is the stard Riemann-Liouville derivative. Using a fixed point theorem for operators on a cone, sufficient conditions for the existence of multiple positive solutions were established. We point out that the nonlinear term of the equation does not depend on the lower order derivative of the unknown function. In this paper, by constructing the proper function spaces the norm to overcome the difficulty of the noncompactness of [, using the Schauder fixed point theorem, we show the existence of one solution with suitable growth conditions imposed on the nonlinear term. Our method is different from [, 3]inessence. Preliminaries lemmas For convenience of the reader, we present the necessary definitions from fractional calculustheory[].
Shen et al. Boundary Value Problems 5 5:4 Page 3 of 3 Definition. The Riemann-Liouville fractional integral of order α >ofafunctionut: R R is given by I α ut= t s us ds provided the right side is point-wise defined on,. Definition. The fractional derivative of order α > of a continuous function ut : R R is given by D α ut= d n us ds Ɣn α dt t s α n where n =[α], provided that the right side is point-wise defined on,. Lemma. Assume that u C, L,, D α C, L,. Then I α Dα ut=utc C t α C N t α N, for some C i R, i =,,...,N, where N is the smallest integer greater than or equal to α. Lemma. Given yt L[,. Theproblem D α ut=yt, < t <,<α <3, u = u =, m D u = β i uξ i, is equivalent to ut= t s ys ds β i ξ i s ys ds. m Proof By Lemma.,wehave ut= t s ys ds c c t α c 3 t α 3. ys ds The boundary condition u = u = implies that c = c 3 =. Considering the boundary condition D u = β i uξ i, we have c = ys ds β i ξ i s ys ds. The proof is completed.
Shen et al. Boundary Value Problems 5 5:4 Page 4 of 3 Define the function spaces { } ut X = ut CJ, R:sup < t J t with the norm u X = sup t J ut { Y = ut X : u t, D u t ut CJ, R, sup <, sup t J tα D ut } < t J with the norm { u Y = max sup t J ut, sup t t J u t t Lemma.3 X, X is a Banach space., sup α D ut }. t J Proof Let {u n } n= be a Cauchy sequence in the space X, X, then ε >, N >such that u n t t u mt < ε for any t J n, m > N. Thus,{u n } n= converges uniformly to a function vt we can verify easily that vt X.ThenX, X isabanachspace. Lemma.4 Y, Y is a Banach space. Proof Let {u n } n= be a Cauchy sequence in the space Y, Y, then {u n } n= isalsoacauchy sequence in X, X. Thus there exists a function vt X such that lim n Moreover, lim n sup t J u n t ut = t t. u n t t vt t vt =, lim α tα n D u n = wt, <, sup α D ut <. t J It is easy to check that v = u t. Next we need to ensure that w = D ut.
Shen et al. Boundary Value Problems 5 5:4 Page 5 of 3 In view of the Lebesgue dominated convergence theorem the uniform convergence of {D u n t} n=, there exists a positive constant M >suchthat u nt M, n =,,... Then wt= lim n D u n t= together with t s α s α u ns s α M Ɣ α t s α s ds = M [t α τ α dτ t d lim n dt t s α s α u ns s α ds ] τ τ α dτ = M α t α Bα, αmt ensures that w = D ut. Thus Y Y isabanachspace. Because the Arzela-Ascoli theorem fails to work in Y, we need a modified compactness criterion to prove the compactness of the operator. Lemma.5 Let Z Y be a bounded set the following conditions hold: ut u i for any ut Z,, t D ut are equicontinuous on any compact t α interval of J; ii given ε >, there exists a constant T = Tε>such that ut ut t D ut D ut < ε < ε, u t t α u t t α < ε, for any t, t > T ut Z. Then Z is relatively compact in Y. Proof We need to prove that Z is totally bounded. First we consider the case t [, T]. Define Z [,T] = { ut:ut Z, t [, T] }. It is easy to check that Z [,T] with the norm u = sup t [,T] ut is a Banach space. Then condition i combined with the Arzela-Ascoli theorem indicates that Z [,T] is relatively compact. Thus for any positive number ε, there exist finitely many balls B ε u i such that Z [,T] n B ε u i, where { B ε u i = ut Z [,T] : u u i = sup ut t u it t [,T] }. < ε
Shen et al. Boundary Value Problems 5 5:4 Page 6 of 3 Similarly, the space Z [,T] = { u t:ut Z, t [, T] } with the norm u = u t t α Z [,T] = { D ut:ut Z, t [, T] } with the norm D u = sup D ut t [,T] are Banach spaces. Then m Z[,T] B ε v j, Z j= k [,T] B ε D w p, p= where { B ε v j = u t Z[,T] : u v j < ε }, B ε D w p = { D u w Z α [,T] : D u D w j < ε }. Next we define Z ijp = { ut Z, u [,T] B ε u i, u [,T] B ε v j, D u [,T] B ε D w p }. Now we take u ijp Z ijp.thenz canbecoveredbytheballsb 5ε u ijp, i =,,...,n, j =,,...,m, p =,,...,k,where B 5ε u ijp = { ut Z : u u ijp Y <5ε }. In fact, for t [, T], ut t u ijpt ut t u it < ε ε ε =3ε, u t t u ijp t α t α u t t u i t α < ε ε ε =3ε, t α u i t t u ijt u ij t t u ijpt u i t t u ij t α t α u ij t t u ijp t α t α
Shen et al. Boundary Value Problems 5 5:4 Page 7 of 3 D ut D u ijp t D ut D u ijp t D u i t D u ij t D u ij t D u ijp t < ε ε ε =3ε. For t [T, ], we have ut t u ijpt ut ut t ut t u ijpt u ijp T t u ijpt < ε ε 3ε =5ε, u t t u ijp t α t α u t t u T α t α u T t u ijp T α t α u ijp T t u ijp t α t α < ε ε 3ε =5ε, D ut D u ijp t D ut D ut D ut D u ijp T D u ijp T D u ijp t < ε ε 3ε =5ε. These ensure that ut uijp t Y <5ε. 3 Main results Define the operator T by Tut= t s f s, us, D us ds f s, us, D us ds β i ξ i s f s, us, D us ds. Theorem 3. Assume that f : J R R R is continuous. Then problem.-. has at least one solution under the assumption that H there exist nonnegative functions at, bt, ct L J, such that f t, x, y at x bt y ct, where ct dt <.
Shen et al. Boundary Value Problems 5 5:4 Page 8 of 3 Proof First of all, in view of Tu t= D Tut = t s α f s, u, D u ds f s, u, D u ds β i f s, us, D us ds ξ i s f s, us, D us ds β i ξ i s f s, u, D u ds α t α, f s, us, D us ds, together with the continuity of f,weseethatt utd TutarecontinuousonJ. In the following we divide the proof into several steps. Step Choose the positive number where R > max{r, R, R 3 }, R = R = R 3 = ct dt atbt dt β i ξ i t ct dt β i ξ i t atbt dt ct dt ξi ct dt β i ξ i t ct dt ct dt β i ξ i t atbt dt atbt dt ct dt atbt dt = β i ξ i. Let set β i ξ i t ct dt β i ξ i t atbt dt, atbt dt, atbt dt ct dt, atbt dt U = { ut Y : ut Y R }. Then, A : U U.Infact,foranyut U,wehave Tut = t s f s, us, D us ds f s, us, D us ds β i as us bs D us cs ds as us bs D us cs ds ξ i s f s, us, D us ds
Shen et al. Boundary Value Problems 5 5:4 Page 9 of 3 u Y R, T ut t α = β i u Y u Y ξ i s as us bs D us cs ds atbt dt ct dt atbt dt m ξ i s β i t s α t α f s, us, D us ds ct dt atbt dt β i f s, us, D us ds β i ξ i s ct dt ξ i s f s, us, D us ds α tα t α as us bs D us cs ds α as us bs D us cs ds β ξi i ξ i s α as us bs D us cs ds u Y α u Y atbt dt ct dt atbt dt α ct dt α u m Y ξ i s β i atbt dt α β ξi i ξ i s ct dt R, D Tut f s, us, D us t ds m f s, us, D k= β us ds iξi β ξi i ξ m i s f s, us, D us ds R atbt dt ct dt R atbt dt ct dt
Shen et al. Boundary Value Problems 5 5:4 Page of 3 β i R R. ξ i s atbt dt β i ξ i s ct dt Hence, Tut Y R, which shows that A : U U. Step LetV be a nonempty subset of U. We will show that TV is relative compact. Let I J be a compact interval, t, t I t < t. Then for any ut V,wehave Tut t Tut t = t s t f s, u, D u ds f s, u, D u ds β i t s t f s, u, D u ds f s, u, D u ds β i ξ i s f s, u, D u ds ξ i s f s, u, D u ds t s t t s t f s, u, D u ds t s t f s, u, D u ds f s, us, D us ds β i t t t, T ut t T ut α t α t s α t α f s, u, D u ds α f s, u, D u ds β i t t D Tut D Tut f s, us, D us ds f s, us, D us ds. t ξ i s t s α t α ξ i s f s, us, D us ds f s, u, D u ds f s, u, D u ds f s, u, D u ds
Shen et al. Boundary Value Problems 5 5:4 Page of 3 Note that for any ut V,wehaveft, ut, D ut is bounded on I. Thenitiseasyto see that Tut, T ut,d Tut are equicontinuous on I. t α Considering the condition H,forgivenε >, thereexistsaconstant L >suchthat L f t, ut, D ut < ε. On the other h, since lim t =, there exists a constant T >suchthatt, t T, t t < ε. Similarly, in view of lim t t L =, there exists a constant T > L >suchthatt, t T s L, t s t s < ε. In view of lim t t L α t α =, there exists a constant T 3 > L >suchthatt, t T 3, s L, t s α t α t s α t α < ε. Now choose T > max{t, T, T 3 }.Thenfort, t T,wehave Tut Tut t max t [,L],u V f t, u, D u Lε ε T ut T ut ξ i s f s, u, D u ds β i t max t [,L],u V f t, u, D u Lε ε α f s, u, D u ds ε, ξ i s f s, u, D u ds β i D Tut D Tut f s, us, D us ds < ε. t Consequently, Lemma.5 shows that TV is relative compact. Step 3 T : U U is a continuous operator. f s, u, D u ds ε,
Shen et al. Boundary Value Problems 5 5:4 Page of 3 Let u n, u U, n =,,..., u n u Y asn. Then, we have Tu n t Tut t t s f s, un s, D u n s f s, u n s, D u n s ds m β i ξ i s f s, un s, D u n s f s, u n s, D u n s ds f s, un s, D u n s f s, u n s, D u n s ds 4 f s, un s, D u n s f s, u n s, D u n s ds 4 8 [ R t atbt ] dt 4 8 ct dt, T u n t t T ut α t α t s α f s, un s, D u n s f s, u n s, D u n s ds α tα m β t α i ξi s f s, un s, D u n s f s, u n s, D u n s ds α tα f s, un s, D u t α n s f s, u n s, D u n s ds 4α f s, un s, D u n s f s, u n s, D u n s ds 4 8α [ R t atbt ] dt 4 8α ct dt, D Tu n t D Tut f s, u n s, D u n s f s, u n s, D u n s ds m ξ i s β i f s, u n s, D u n s f s, u n s, D u n s ds k= f s, u n s, D u n s f s, u n s, D u n s ds f s, u n s, D u n s f s, u n s, D u n s ds
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