EXISTENCE OF SOLUTIONS TO FRACTIONAL DIFFERENTIAL EQUATIONS WITH MULTI-POINT BOUNDARY CONDITIONS AT RESONANCE IN HILBERT SPACES

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1 Electronic Journal of Differential Equations, Vol. 216 (216), No. 61, pp ISSN: URL: or ftp ejde.math.txstate.edu EXISTENCE OF SOLUTIONS TO FRACTIONAL DIFFERENTIAL EQUATIONS WITH MULTI-POINT BOUNDARY CONDITIONS AT RESONANCE IN HILBERT SPACES HUA-CHENG ZHOU, FU-DONG GE, CHUN-HAI KOU Abstract. This article is devoted to investigating the existence of solutions to fractional multi-point boundary-value problems at resonance in a Hilbert space. More precisely, the dimension of the kernel of the fractional differential operator with the boundary conditions be any positive integer. We point out that the problem is new even when the system under consideration is reduced to a second-order ordinary differential system with resonant boundary conditions. We show that the considered system admits at least a solution by applying coincidence degree theory first introduced by Mawhin. An example is presented to illustrate our results. 1. Introduction In this article, we are concerned with the existence of solutions to the following fractional multi-point boundary value problems(bvps) at resonance D α x(t) = f(t, x(t), Dα 1 x(t)), 1 < α 2, t (, 1), I 2 α x(t) (1.1) t= = θ, x(1) = Ax(ξ), where D α and I α are the Riemann-Liouville differentiation and integration, respectively; θ is the zero vector in l 2 := {x = (x 1, x 2,...,.) : i=1 x i 2 < }; A : l 2 l 2 is a bounded linear operator satisfying 1 dim ker(i Aξ α 1 ) < ; ξ (, 1) is a fixed constant; f : [, 1] l 2 l 2 l 2 is a Carathéodory function; that is, (i) for each (u, v) l 2 l 2, t f(t, u, v) is measurable on [, 1]; (ii) for a.e. t [, 1], (u, v) f(t, u, v) is continuous on l 2 l 2 ; (iii) for every bounded set Ω l 2 l 2, the set {f(t, u, v) : (u, v) Ω} is a relatively compact set in l 2. Moreover, the function ϕ Ω (t) = sup{ f(t, u, v) l 2 : (u, v) Ω} L 1 [, 1], where x l 2 = i=1 x i 2 is the norm of x = (x 1, x 2,..., ) in l 2. System (1.1) is said to be at resonance in l 2 if dim ker(i Aξ α 1 ) 1, otherwise, it is said to be non-resonant. In the past three decades, the existence of solutions for the fractional differential equations with the boundary value conditions have been 21 Mathematics Subject Classification. 34A8, 34B1, 34B4. Key words and phrases. Fractional differential equations; resonance; coincidence degree. c 216 Texas State University. Submitted August 22, 215. Published February 29,

2 2 H.-C. ZHOU, F.-D. GE, C.-H. KOU EJDE-216/61 given considerable attention by many mathematical researchers. The attempts on dim ker(i Aξ α 1 ) =, non-resonance case, for fractional differential equations are available in [1, 2, 1, 11, 17, 21, 22, 23], and the attempts on 1 dim ker(i Aξ α 1 ) 2, resonance case, can be found in [3, 4, 8, 9, 13, 14, 18, 2]. However, to the best of our knowledge, all results derived in these papers are for one equation with dim ker L = or 1 and for two equations with dim ker L = 2. Recently, the authors in [16] investigated the following second differential system u (t) = f(t, u(t), u (t)), < t < 1, u () = θ, u(1) = Au(η) (1.2) where f : [, 1] R n R n R n is a Carathéodory function and the square matrix A satisfies certain condition. Moreover, fractional order α (1, 2] case was investigated in [7], where the results for second order ordinary differential equation in [16] was generalized to fractional order case. However, these considered problems were investigated in finite dimensional space. Therefore, it is more natural to ask whether it exists a solution when such kind of boundary value problem considered in a infinite dimensional space. Recently, in [24], the author discussed the existence of solution for fractional boundary value problem with non-resonant conditions in an arbitrary Banach space which, of course, can be in the infinite dimensional space. However, it is still open for the equation in infinite dimensional space with resonance conditions. It deservers to point out that the problem new even when α = 2 the system (1.1) becomes second order ordinary differential system with resonant boundary conditions. In this paper, we investigate the existence of solution for fractional differential equation in l 2. There is remarkable difference that any bounded closed set is compact in finite dimensional space, while bounded closed set may be not compact in the infinite dimensional, for instance, {x l 2 : x 1} l 2 is non-compact in l 2. Therefore, compactness criterion of the infinite dimensional space is more complicated, the problem we considered is in the infinite dimensional setting. To apply the coincidence degree theory of Mawhin [15], we suppose additionally that A satisfies 1 dim ker(i Aξ α 1 ) < and one of the following conditions ](A1)] Aξ α 1 is idempotent, that is, A 2 ξ 2α 2 = Aξ α 1, or; ](A2)] A 2 ξ 2α 2 = I, where I stands for the identity operator from l 2 to l 2. The requirement 1 dim ker(i Aξ α 1 ) is to make the problem to be resonant and the requirement dim ker(i Aξ α 1 ) < is to make the kernel operator to be a Fredholm operator which is a basic requirement in applying the Mawhin theorem. It is also obvious that dim ker(i Aξ α 1 ) can take any integer n N for suitable A, which can be regards as a generalization of the previous efforts [3, 4, 8, 9, 13, 14, 18, 2]. However, we point out that without the above assumptions (A1) or (A2), it will be difficult to construct the projector Q as (3.1) below. Actually, the assumptions (A1) or (A2) play a key role in the process of the proof. This is the reason why we only choose the two special cases of A. Without such an assumption, i.e., the general A satisfying dim ker(i Aξ α 1 ) <, (1.1) may be a challenge problem, which we will study in the future. In particular, when A = ξ 1 α I, it is clear that A satisfies (A2) but with dim ker(i Aξ α 1 ) =, which leads to the kernel operator not to be Fredholm operator. Thus, such operator is excluded. Unlike the case in R n, the operator A is allowed to be identity operator ξ α 1 I. Let A = diag(ξ 1 α I k, B) with

3 EJDE-216/61 EXISTENCE OF SOLUTIONS 3 dim ker(i Bξ α 1 ) = and B satisfying (A1) or (A2), where I k is the identity matrix in R k. It is seen that dim ker(i Aξ α 1 ) = k, ker L = {(c 1, c 2,..., c k,,,..., ) t α 1 : c i R, i = 1, 2,..., k} and dim ker L = k, where L is defined by (2.2) below. So under this boundary condition, the system (1.1) is at resonance. The goal of this paper is to study the existence of solutions for the fractional differential equations with boundary value conditions at resonance in Hilbert space l 2. We proceed as follows: In Section 2, we give some necessary background and some preparations for our consideration. The proof for the main results is presented in Section 3 by applying the coincidence degree theory of Mawhin. In Section 4, an example is given to illustrate the main result. 2. Preliminaries In this section, we introduce some necessary definitions and lemmas which will be used later. For more details, we refer the reader to [5, 12, 15] and the references therein. Definition 2.1 ([12]). The fractional integral of order α > of a function x : (, ) R is given by I α x(t) = 1 t (t s) α 1 x(s)ds, provided that the right-hand side is pointwise defined on (, ). Remark 2.2. The notation I α x(t) t= means that the limit is taken at almost all points of the right-sided neighborhood (, ε)(ε > ) of as follows: I α x(t) t= = lim t Iα x(t). Generally, I α x(t) t= is not necessarily to be zero. For instance, let α (, 1), x(t) = t α. Then I α 1 t α t= = lim t t (t s) α 1 s α ds = lim Γ(1 α) = Γ(1 α). t Definition 2.3 ([12]). The fractional derivative of order α > of a function x : (, ) R is given by D α x(t) = 1 d ) n t x(s) ds, Γ(n α)( dt (t s) α n1 where n = [α] 1, provided that the right side is pointwise defined on (, ). Lemma 2.4 ([12]). Assume that x C(, ) L loc (, ) with a fractional derivative of order α > belonging to C(, ) L loc (, ). Then I α Dα x(t) = x(t) c 1t α 1 c 2 t α 2 c n t α n, for some c i R, i = 1,..., n, where n = [α] 1. For any x(t) = (x 1 (t), x 2 (t),... ) l 2, the fractional derivative of order α > of x is defined by D α x(t) = (Dα x 1(t), D α x 2(t),... ) l 2.

4 4 H.-C. ZHOU, F.-D. GE, C.-H. KOU EJDE-216/61 The following definitions and the coincidence degree theory are fundamental in the proof of our main result. We refer the reader to [5, 15]. Definition 2.5. Let X and Y be normed spaces. A linear operator L : dom(l) X Y is said to be a Fredholm operator of index zero provided that (i) im L is a closed subset of Y, and (ii) dim ker L = codim im L <. It follows from definition 2.5 that there exist continuous projectors P : X X and Q : Y Y such that im P = ker L, ker Q = im L, X = ker L ker P, Y = im L im Q and the mapping L dom L ker P : dom L ker P im L is invertible. We denote the inverse of L dom L ker P by K P : im L dom L ker P. The generalized inverse of L denoted by K P,Q : Y dom L ker P is defined by K P,Q = K P (I Q). Furthermore, for every isomorphism J : im Q ker L, we can obtain that the mapping K P,Q JQ : Y dom L is also an isomorphism and for all x dom L, we know that (K P,Q JQ) 1 x = (L J 1 P )x. (2.1) Definition 2.6. Let L be a Fredholm operator of index zero, let Ω X be a bounded subset and dom L Ω. Then the operator N : Ω Y is called to be L-compact in Ω if (i) the mapping QN : Ω Y is continuous and QN(Ω) Y is bounded, and (ii) the mapping K P,Q N : Ω X is completely continuous. Assume that L is defined by Definition 2.6 and N : Ω Y is L-compact. For any x Ω, by (2.1), we shall see that Lx = (K P,Q JQ) 1 x J 1 P x = (K P,Q JQ) 1 [ Ix K P,Q J 1 P x JQJ 1 P x ] = (K P,Q JQ) 1 (Ix P x). Then we can equivalently transform the existence problem of the equation Lx = Nx, x Ω into a fixed point problem of the operator P (K P,Q JQ)N in Ω. This can be guaranteed by the following lemma, which is also the main tool in this paper. Lemma 2.7 ([15]). Let Ω X be bounded, L be a Fredholm mapping of index zero and N be L-compact on Ω. Suppose that the following conditions are satisfied: (i) Lx λnx for every (x, λ) ((dom L\ ker L) Ω) (, 1); (ii) Nx im L for every x ker L Ω; (iii) deg(jqn ker L Ω, Ω ker L, ), with Q : Y Y a continuous projector such that ker Q = im L and J : im Q ker L is an isomorphism. Then the equation Lx = Nx has at least one solution in dom L Ω. In this paper, we use spaces X, Y introduced as X = { x(t) l 2 : x(t) = I α 1 u(t), u C([, 1]; l2 ), t [, 1] } with the norm x X = max{ x C([,1];l2 ), D α 1 x C([,1];l 2 )} and Y = L 1 ([, 1]; l 2 ) with the norm y L 1 ([,1];l 2 ) = y(s) l 2ds, respectively, where x C([,1];l 2 ) = sup t [,1] x(t) l 2.

5 EJDE-216/61 EXISTENCE OF SOLUTIONS 5 We have the following compactness criterion on subset F of X which is a slight modification of [19, Lemma 2.2] (see also the Ascoli-Arzela theorem [6, Theorem 1.2.5, p. 15]). Lemma 2.8. F X is a sequentially compact set if and only if F (t) is a relatively compact set and equicontinuous which are understood in the following sense: (1) for any t [, 1], F (t) := {x(t) x F } is a relatively compact set in l 2 ; (2) for any given ε >, there exists a δ > such that x(t 1 ) x(t 2 ) l 2 < ε, D α 1 x(t 1 ) D α 1 x(t 2 ) l 2 < ε, for t 1, t 2 [, 1], t 1 t 2 < δ, for all x F. Now we define the linear operator L : dom L X Y by Lx := D α x, (2.2) where dom L = {x X : D α x Y, x() = θ, x(1) = Ax(ξ)}. Define N : X Y by Nx(t) := f(t, x(t), D α 1 x(t)), t [, 1]. (2.3) Then the problem can be equivalently rewritten as Lx = Nx. The next lemma plays a vital role in estimating the boundedness of some sets. Lemma 2.9. Let z 1, z 2, γ 1, γ 2 [, 1) and λ i, µ i, i = 1, 2, 3, and the following two inequalities hold, z 1 λ 1 z γ1 1 λ 2z 2 λ 3, z 2 µ 1 z 1 µ 2 z γ2 2 µ 3 Then z 1, z 2 is bounded if λ 2 µ 1 < 1. Proof. From (2.4), we have z 1 λ 1z γ1 1 λ 2µ 2 z γ2 2 λ 2µ 3 λ 3 1 λ 2 µ 1, z 2 λ 1µ 1 z γ1 1 µ 2z γ2 2 λ 3µ 1 µ 3 1 λ 2 µ 1. (2.4) (2.5) Let z = max{z 1, z 2 }, κ 1 = max{λ 1, λ 1 µ 1 } and κ 2 = max{λ 2 µ 2, µ 2 }. It follows from (2.5) that z κ 1z γ1 κ 2 z γ2 λ 2 µ 3 λ 3 µ 1 λ 3 µ 3 1 λ 2 µ 1. This, together with γ 1, γ 2 [, 1), yields that z is bounded. Lemma 2.1. The operator L, defined by (2.2), is a Fredholm operator of index zero. Proof. For any x dom L, by Lemma 2.4 and x() = θ, we obtain x(t) = I α Lx(t) ct α 1, c l 2, t [, 1], (2.6) which, together with x(1) = Ax(ξ), yields Now we claim that ker L = {x X : x(t) = ct α 1, t [, 1], c ker(i Aξ α 1 )} ker(i Aξ α 1 )t α 1. (2.7) im L = {y Y : h(y) im(i Aξ α 1 )}, (2.8)

6 6 H.-C. ZHOU, F.-D. GE, C.-H. KOU EJDE-216/61 where h : Y l 2 is a continuous linear operator defined by h(y) := A ξ (ξ s) α 1 y(s)ds I (1 s) α 1 y(s)ds. (2.9) Actually, for any y im L, there exists a function x dom L such that y = Lx. It follows from (2.6) that x(t) = I α y(t)ct α 1. From this equality and x(1) = Ax(ξ), we obtain A ξ (ξ s) α 1 y(s)ds I (1 s) α 1 y(s)ds = (I Aξ α 1 )c, c l 2, which means that h(y) im(i Aξ α 1 ). On the other hand, for any y Y satisfying h(y) im(i Aξ α 1 ), there exists a constant c such that h(y) = (I Aξ α 1 )c. Let x (t) = I α y(t) c t α 1. A straightforward computation shows that x () = θ and x (1) = Ax (ξ). Hence, x dom L and y(t) = D α x (t), which implies that y im L. Next, put ρ A = κ(i Aξ α 1 ), where { 1, if (A1) holds, i.e., A 2 ξ 2α 2 = Aξ α 1 ; κ = 1 2, if (A2) holds, i.e., A2 ξ 2α 2 (2.1) = I. For A 2 ξ 2α 2 = Aξ α 1, we have ρ 2 A = (I Aξ α 1 ) 2 = I 2Aξ α 1 A 2 ξ 2α 2 = I Aξ α 1 = ρ A, (I ρ A )(ξ 2α 1 A I) = Aξ α 1 (ξ 2α 1 A I) = ξ 3α 2 A 2 Aξ α 1 For A 2 ξ 2α 2 = I, we have = (ξ α )Aξ α 1 = (ξ α )(I ρ A ). ρ 2 A = 1 4 (I Aξα 1 ) 2 = 1 4 (I 2Aξα 1 A 2 ξ 2α 2 ) = 1 2 (I Aξα 1 ) = ρ A, (I ρ A )(ξ 2α 1 A I) (2.11) = 1 2 (I Aξα 1 )(ξ 2α 1 A I) (2.12) = 1 2 [ξ2α 1 A I ξ 3α 2 A 2 Aξ α 1 ] = 1 2 (ξα )(I Aξ α 1 ) = (ξ α )(I ρ A ). It follows from (2.11) and (2.12) that ρ A satisfies the following properties ρ 2 A = ρ A, (I ρ A )(ξ 2α 1 A I) = (ξ α )(I ρ A ). (2.13) Furthermore, we note that if y = ct α 1, c l 2, then h(y) = A ξ = (ξ2α 1 A I)c. Γ(2α) (ξ s) α 1 y(s)ds I Define the continuous linear mapping Q : Y Y by (1 s) α 1 y(s)ds (2.14) Qy(t) := Γ(2α) ξ α (I ρ A )h(y)t α 1, t [, 1], y Y. (2.15)

7 EJDE-216/61 EXISTENCE OF SOLUTIONS 7 By the first identity in (2.13), we obtain (I ρ A ) 2 = (I ρ A ), which together with (2.13) implies Q 2 y(t) = Γ(2α) ξ α (I ρ A )h(qy(t))t α 1 = Γ(2α) ξ α (I ρ A ) (ξ2α 1 A I) Γ(2α) = Γ(2α) ξ α (I ρ A ) 2 h(y)t α 1 = Qy(t); Γ(2α) ξ α (I ρ A )h(y)t α 1 that is, Q is a projection operator. The equality ker Q = im L follows from the trivial fact that y ker Q h(y) ker(i ρ A ) h(y) im ρ A h(y) im(i Aξ α 1 ) y im L. Therefore, we get Y = ker Q im Q = im L im Q. Finally, we shall prove that im Q = ker L. Indeed, for any z im Q, let z = Qy, y Y. By (2.13), we have k(i Aξ α 1 )z(t) = ρ A z(t) = ρ A Qy(t) = Γ(2α) ξ α ρ A (I ρ A )g(y)t α 1 = θ, which implies z ker L. Conversely, for each z ker L, there exists a constant c ker(i Aξ α 1 ) such that z = c t α 1 for t [, 1]. By (2.13) and (2.14), we derive Qz(t) = Γ(2α) ξ α (I ρ A )h(c t α 1 )t α 1 = c t α 1 = z(t), t [, 1], which implies that z im Q. Hence we know that im Q = ker L, i.e., the operator L is a Fredholm operator of index zero. The proof is complete. Define the operator P : X X as follows P x(t) = 1 (I ρ A)D α 1 x()tα 1. (2.16) Lemma The mapping P : X X, defined by (2.16), is a continuous projector such that im P = ker L, X = ker L ker P and the linear operator K P : im L dom L ker P can be written as also K P y(t) = I α y(t), K P = (L dom L ker P ) 1, K P y X 1/ y L 1 ([,1];l 2 ). Proof. By (2.16), one can see that P is a continuous operator. From the first identity of (2.13), we have (I ρ A ) 2 = (I ρ A ), which implies that the mapping P is a projector. Moreover, if v im P, there exists a x X such that v = P x. By the first identity of (2.13) again, we see that 1 (I Aξα 1 )(I ρ A )D α 1 x() = 1 k ρ A(I ρ A )D α 1 x() =,

8 8 H.-C. ZHOU, F.-D. GE, C.-H. KOU EJDE-216/61 which gives us v ker L. Conversely, if v ker L, then v(t) = c t α 1 for some c ker(i Aξ α 1 ), and we deduce that P v(t) = 1 (I ρ A)D α 1 v()tα 1 = (I ρ A )c t α 1 = c t α 1 = v(t), t [, 1], which gives us v im P. Thus, we get that ker L = im P and consequently X = kerl ker P. Moreover, let y im L. There exists x dom L such that y = Lx, and we obtain K P y(t) = x(t) ct α 1 where c l 2 satisfies c = ξ α 1 Ac. It is easy to see that K P y dom L and K P y ker P. Therefore, K P is well defined. Further, for y im L, we have L(K P y(t)) = D α (K P y(t)) = y(t) and for x dom L ker P, we obtain that K P (Lx(t)) = x(t) c 1 t α 1 c 2 t α 1, for some c 1, c 2 l 2. In view of x dom L ker P, we know that c 1 = c 2 = θ. Therefore, (K P L)x(t) = x(t). This shows that K P = (L dom L ker P ) 1. Finally, by the definition of K P, we derive and D α 1 K P y C([,1];l 2 ) = K P y C([,1];l 2 ) = 1 It follows from (2.17) and (2.18) that y(s)ds C([,1];l 2 ) y L 1 ([,1];l 2 ) (2.17) ( s) α 1 y(s)ds C([,1];l 2 ) 1 y L 1 ([,1];l 2 ). K P y X = max{ D α 1 K P y C([,1];l2 ), K P y C([,1];l2 )} max { 1 y L1 ([,1];l 2 ), y } L 1 ([,1];l 2 ) This completes of the proof. = 1 y L 1 ([,1];l 2 ). (2.18) (2.19) Lemma Let f be a Carathéodory function. Then N, defined by (2.3)), is L-compact. Proof. Let Ω be a bounded subset in X. By hypothesis (iii) on the function f, there exists a function ϕ Ω (t) L 1 [, 1] such that for all x Ω, f(t, x(t), D α 1 x(t)) l 2 ϕ Ω (t), a.e. t [, 1], (2.2) which, along with (2.9) implies h(nx(t)) l 2 = A I ξ A 1 ϕ Ω L1 [,1]. (ξ s) α 1 f(s, x(s), D α 1 x(s))ds (1 s) α 1 f(s, x(s), D α 1 x(s))ds l 2 (2.21)

9 EJDE-216/61 EXISTENCE OF SOLUTIONS 9 Thus, from (2.15) and (2.21) it follows that QNx L1 ([,1];l 2 ) = Γ(2α) ξ α (I ρ A )h(nx) l 2 s α 1 ds Γ(2α)( A 1) I ρ A 1 ξ α ϕ Ω L 1 [,1] <. (2.22) This shows that QN(Ω) Y is bounded. The continuity of QN follows from the hypothesis on f and the Lebesgue dominated convergence theorem. Next, we shall show that K P,Q N is completely continuous. First, for any x Ω, we have and D α 1 K P,Q Nx(t) = K P (I Q)Nx(t) = K P Nx(t) K P QNx(t) = INx(t) α Γ(2α) ξ α (I ρ A )h(nx(t))i α t α 1. (2.23) K P,Q Nx(t) = INx(t) 1 Γ(2α) ξ α (I ρ A )h(nx(t))i 1 t α 1. (2.24) By the hypothesis on f and the Lebesgue dominated convergence theorem, it is easy to see that K P,Q N is continuous. Since f is a Carathéodory function, for every bounded set Ω l 2 l 2, the set {f(t, u, v) : (u, v) Ω } is relatively compact set in l 2. Therefore, for almost all t [, 1], {K P,Q Nx(t) : x Ω} and {D α 1 K P,Q Nx(t) : x Ω} are relatively compact in l 2. From (2.21), (2.23) and (2.24), we derive that and K P,Q Nx C([,1];l 2 ) = I α Nx(t) Γ(2α) ξ α (I ρ A )h(nx(t))i α t α 1 C([,1];l 2 ) 1 ϕ Ω L1 (,1) Γ(2α) I ρ A ξ α h(nx(t)) l 2 1 ϕ Ω L 1 (,1) Γ(2α) I ρ A ( A 1) ξ α ϕ Ω L 1 (,1) <, D α 1 K P,Q Nx C([,1];l2 ) = I 1 Nx(t) Γ(2α) ξ α (I ρ A )h(nx(t))i 1 t α 1 C([,1];l2 ) ϕ Ω L 1 (,1) Γ(2α) I ρ A ξ α h(nx(t)) l 2 ϕ Ω L1 (,1) Γ(2α) I ρ A ( A 1) ξ α ϕ Ω L1 (,1) <, which shows that K P,Q NΩ is uniformly bounded in X. Noting that b p a p (b a) p for any b a >, < p 1. (2.25) for any t 1, t 2 [, 1] with t 1 < t 2, we shall see that K P,Q Nx(t 2 ) K P,Q Nx(t 1 ) l 2

10 1 H.-C. ZHOU, F.-D. GE, C.-H. KOU EJDE-216/61 and = 1 t1 t2 [(t 2 s) α 1 (t 1 s) α 1 ]Nx(s)ds Γ(2α) ξ α (I ρ A )h(nx(t))[i α t α 1 2 It α α 1 1 ] 1 D α 1 t2 = t1 (t 2 t 1 ) α 1 ϕ Ω (s)ds 1 t2 t 1 ϕ Ω (s)ds t 1 l 2 (t 2 s) α 1 Nx(s)ds Γ2 (α) I ρ A ( A 1) ξ α ϕ Ω L 1 (,1) t2 2α 1 t1 2α 1 as t 2 t 1 K P,Q Nx(t 2 ) D α 1 K P,Q Nx(t 1 ) l 2 t 1 t2 Nx(s)ds l 2 Γ(2α) ξ α (I ρ A )h(nx(t)) ϕ Ω (s)ds Γ(2α) I ρ A ( A 1) t 1 ξ α t2 t 1 s α 1 ds l 2 ϕ Ω L1 (,1) t α 2 t α 1 as t 2 t 1. Then K P,Q NΩ is equicontinuous in X. By Lemma 2.8, K P,Q NΩ X is relatively compact. Thus we can conclude that the operator N is L-compact in Ω. The proof is complete. 3. Main results Theorem 3.1. Let f be a Carathéodory function and the following conditions hold: (H1) There exist five nonnegative functions a 1, a 2, b 1, b 2, c L 1 [, 1] and constants γ 1, γ 2 [, 1) such that for all t [, 1], u, v l 2, f(t, u, v) l 2 a 1 (t) u l 2 b 1 (t) v l 2 a 2 (t) u γ1 l 2 b 2 (t) v γ2 l 2 c(t). (H2) There exists a constant A 1 > such that for x dom L, if D α 1 x(t) l 2 > A 1 for all t [, 1], then A ξ I (ξ s) α 1 f(s, x(s), D α 1 x(s))ds (1 s) α 1 f(s, x(s), D α 1 x(s))ds / im(i Aξ α 1 ). (H3) There exists a constant A 2 > such that for any e = {(e i )} l 2 satisfying e = ξ α 1 Ae and e l 2 > A 2, either e, QNe l 2 or e, QNe l 2, where, l 2 is the inner product in l 2. Then (1.1) has at least one solution in space X provided that > max { ( I ρ A 1) a 1 L 1 (,1), ( I ρ A 1) b 1 L 1 (,1)}, ( I ρ A 1) 2 a 1 L 1 (,1) b 1 L 1 (,1) ( ( I ρ A 1) a 1 L1 (,1))( ( I ρ A 1) b 1 L1 (,1)) < 1. (3.1) Proof. We shall construct an open bounded subset Ω in X satisfying all assumption of Lemma 2.7. Let Ω 1 = { x dom L\ ker L : Lx = λnx for some λ [, 1]}. (3.2)

11 EJDE-216/61 EXISTENCE OF SOLUTIONS 11 For any x Ω 1, x / ker L, we have λ. Since Nx im L = ker Q, by (2.8), we have h(nx) im(i Aξ α 1 ), where h(nx) = A ξ I (ξ s) α 1 f(s, x(s), D α 1 x(s))ds (1 s) α 1 f(s, x(s), D α 1 x(s))ds. (3.3) From (H2) there exists t [, 1] such that D α 1 x(t ) l 2 A 1. Then from the equality D α 1 x() = Dα 1 x(t ) t Dα x(s)ds, we deduce that D α 1 x() l 2 A 1 D α x L1 (,1;l 2 ) = A 1 Lx 1 A 1 Nx L1 (,1;l 2 ), which implies P x X = 1 (I ρ A)D α 1 x()tα 1 X I ρ A (A 1 Nx L 1 (,1;l 2 )). (3.4) Further, for x Ω 1, since im P = ker L, X = ker L ker P, we have (I P )x dom L ker P and LP x = θ. Then (I P )x X = K P L(I P )x X K P Lx X 1 Lx L 1 (,1;l 2 ) 1 Nx (3.5) L 1 (,1;l 2 ). From (3.4) and (3.5), we conclude that x X = P x (I P )x X P x X (I P )x X I ρ A A 1 I ρ A 1 Nx L 1 (,1;l 2 ). Moreover, by the definition of N and (H1), one has Nx L 1 (,1;l 2 ) = Thus, f(s, x(s), D α 1 x(s)) l 2dt a 1 L1 (,1) x C([,1];l2 ) b 1 L1 (,1) D α 1 x C([,1];l2 ) a 2 L1 (,1) x γ1 C([,1];l 2 ) b 2 L1 (,1) D α 1 x γ2 C([,1];l 2 ) c L 1 (,1). x X I ρ A A 1 I ρ A 1 ( a 1 L1 (,1) x C([,1];l2 ) b 1 L 1 (,1) D α 1 x C([,1];l 2 ) ( ) I ρ A 1 a 2 L1 (,1) x γ1 C([,1];l 2 ) b 2 L1 (,1) D α 1 x γ2 ) c L1 (,1). C([,1];l 2 ) (3.6) (3.7) (3.8) It follows from (3.1), (3.8), x C([,1];l2 ) x X, D α 1 x C([,1];l2 ) x X and Lemma 2.9 that there exists M > such that max{ x C([,1];l 2 ), D α 1 x C([,1];l 2 )} M, that is to say Ω 1 is bounded.

12 12 H.-C. ZHOU, F.-D. GE, C.-H. KOU EJDE-216/61 Let Ω 2 = {x ker L : Nx im L}. (3.9) For any x Ω 2, it follows from x ker L that x = et α 1 for some e ker(i Aξ α 1 ) l 2, and it follows from Nx im L that h(nx) im(i Aξ α 1 ), where h(nx) is defined by (3.3). By hypothesis (H 2 ), we arrive at D α 1 x(t ) l = e l 2 A 1. Thus we obtain x e l A 1. That is, Ω 2 is bounded in X. If the first part of (H 3 ) holds, denote Ω 3 = {x ker L : λx (1 λ)qnx = θ, λ [, 1]}, then for any x Ω 3, we know that x = et α 1 with e ker(i Aξ α 1 ) and λx = (1 λ)qnx. If λ =, we have Nx ker Q = im L, then x Ω 2, by the argument above, we get that x A 1. Moreover, if λ (, 1] and if e l 2 > A 2, by (H3), we deduce that < λ e 2 l 2 = λ e, e l 2 = (1 λ) e, QNe l 2, which is a contradiction. Then x X = et α 1 X max{ e l 2, e l 2}. That is to say, Ω 3 is bounded. If the other part of (H3) holds, we take Ω 3 = {x ker L : λx (1 λ)qnx = θ, λ [, 1]}. By using the same arguments as above, we can conclude that Ω 3 is also bounded. Next, we show that all conditions of Lemma 2.7 are satisfied. Assume that Ω is a bounded open subset of X such that 3 i=1 Ω i Ω. It follows from Lemmas 2.1 and 2.12 that L is a Fredholm operator of index zero and N is L-compact on Ω. By the definition of Ω and the argument above, to complete the theorem, we only need to prove that condition (iii) of Lemma 2.7 is satisfied. For this purpose, let H(x, λ) = ±λx (1 λ)qnx, (3.1) where we let the isomorphism the J : im Q ker L be the identical operator. Since Ω 3 Ω, H(x, λ) for (x, λ) ker L Ω [, 1], then by homotopy property of degree, we obtain deg (JQN ker L Ω, Ω ker L, ) = deg (H (, ), Ω ker L, ) = deg (H (, 1), Ω ker L, ) = deg (±Id, Ω ker L, ) = ±1. Thus (H3) of Lemma 2.7 is fulfilled and Theorem 3.1 is proved. 4. Example In this section, we shall present an example to illustrate our main result in l 2 with dim ker L = 2k, which surely generalize the previous results [3, 4, 8, 9, 13, 14, 18, 2], where the dimension of dim ker L is only 1 or 2.

13 EJDE-216/61 EXISTENCE OF SOLUTIONS 13 Consider the following system with dim ker L = 2k, k = 1, 2, 3,... in l 2. x 1 (t) x 2 (t) x 3 (t) x 4 (t). { 1, if D 1/2 x(t) l 2 < 1 D 1/2 x = 1 1 (t) [D 1/2 x 1 (t)] 1, if D 1/2 x(t) l 2 1 ( x2 (t) D 1/2 x 3 (t) ) /2 ( 1 x3 (t) D 1/2 x 3 (t) ) /2 ( 2 x4 (t) D 1/2 x 4 (t) ) /2 3. D 3/2 x i () =, i = 1, 2,... x(1) = Ax(1/9). (4.1) Let α = 3/2, ξ = 1/9. For all t [, 1], let u = (x 1, x 2, x 3,... ), v = (y 1, y 2, y 3,... ) l 2 and f = (f 1, f 2,... ) T with f 1 (t, u, v) = { 1/1, if v l 2 < 1, (y 1 y1 1 )/1, if v l 2 1, f 2 (t, u, v) = (x 2 y 3 )/2 and f i (t, u, v) = 1 x iy i 5 2, i = 3, 4,.... Moreover, i B 1... B A = B k with B i = 3 6, (4.2) 3 i = 1, 2,..., k, k N. Obviously, we see that Bi 2 = 9I 3 and dim ker(i 3 ξ α 1 B i ) = dim ker(i 3 B i /3) = 2, i = 1, 2,..., where I 3 is the 3 3 identity matrix. Then A 2 ξ 2α 2 = I, dim ker(i Aξ α 1 ) = 2k, k N and the problem (4.1), with A and f defined above, has one solution if and only if problem (1.1) admits one solution. Checking (H1) of Theorem 3.1: For some r R, Ω = {(u, v) l 2 l 2 : u l 2 r, v l 2 r}, let ϕ Ω (t) = 1 1 [(r 1/r 1)2 4r2 3 ]1/2 L 1 [, 1]. Since A l 2 9 k, letting a 1 (t) = b 1 (t) = 1 5 3, a 2(t) = b 2 (t) =, c(t) = r 1/r 1. (4.3) 1 condition (H1) is satisfied.

14 14 H.-C. ZHOU, F.-D. GE, C.-H. KOU EJDE-216/61 Checking (H2) of Theorem 3.1: From the definition of f it follows that f 1 > 1/1 > when D 1/2 x(t) l 2 > 1. This, (B 1 ξ α I) f 1 f 2 = 8/9 1/9 2/9 f 8f f 2 = f 3 8/9 f 3 and im(i Aξ α 1 ) = {(,, τ 3,,, τ 6,...,,, τ 3i,... ) : τ 3i R, i = 1, 2,... } implies that condition (H2) is satisfied. Checking (H3) of Theorem 3.1: Since dim ker(i Aξ α 1 ) = dim ker(i A/3) = 2k, k N, for any e l 2 satisfying e = Ae, e can be expressed as e = e 1 e 2 e k, with e i = σ i1 ε 3i 2 σ i2 (ε 3i 1 ε 3i ), σ ij R, i = 1, 2,..., k, j = 1, 2, where ε j = (,,..., 1j th,,,... ) l 2 is a vector with all elements equaling to except the j-th equaling to 1, j = 1, 2,.... In addition, for any y Y, by (2.15) and ρ A = 1 2 (I A/3), we have Qy(t) = Γ(2α) ξ α (I ρ A )h(y)t α 1 = 27 π (I A/3)h(y)t α 1, (4.4) 52 where h(y) = A ξ (ξ s) α 1 y(s)ds I By (2.3), let d = t 1/2 π 2, we have ( 1, dσ12 2, dσ12 2, dσ21 2 N(et 1/2 ) = 1 1 dσ i1 2 3i 3, dσi2 2 3i 2, dσi2 2, dσ22 3 2, dσ22 ) 4 2,... 3i 1, if σ 11 < 1, 2 i k; ( σ 11 1 σ 11, dσ12 dσ i1 2, dσi2 3i 3 2, dσi2 3i 2 2,... 3i 1, if σ 11 1, 2 i k. 2 5,..., 2, dσ12 2, dσ21 2 2, dσ22 3 2, dσ22 ) 4 (1 s) α 1 y(s)ds. (4.5) 2 5,..., (4.6) Suppose that σ 11 > 1, σ 12, and let A 2 = 1, d = 27π 28 π 648 <. From (4.4) and (4.6) it follows that Q(Net 1/2 ) = 27 π 52 (I A/3)h(Net1/2 )t 1/2 and = 27 πt 1/ , ( π (σ 11 1 σ 11 ), dσ , dσ , dσ , dσ , dσ , dσ i1 2 3i 3, dσi2 2 3i 1, dσi2 2 3i 1,... ) T e, QNet 1/2 = 27 πt 1/2 [ π ((σ 11 /2) 2 3/4) d ( 2σ 2 )] σ σ σ2 i1 2 3i 3 2σ2 i2 2 3i 1... >.

15 EJDE-216/61 EXISTENCE OF SOLUTIONS 15 Therefore, (4.1) admits at least one solution. Acknowledgements. This work was supported by Chinese Universities Scientific Fund No.CUSF-DH-D-21461, the Natural Science Foundation of Shanghai (No.15ZR148) and by the National Natural Science Foundation of China (No ). References [1] B. Ahmad, P. Eloe; A nonlocal boundary value problem for a nonlinear fractional differential equation with two indices, Comm. Appl. Nonlinear Anal., 17 (21), [2] Z. B. Bai, Y. H. Zhang; Solvability of fractional three-point boundary value problems with nonlinear growth, Appl. Math. Comput., 218 (211), [3] Z. B. Bai; On solutions of some fractional m-point boundary value problems at resonance. J. Qual. Theory Differ. Equ. 21, No. 37, [4] Y. Chen, X. H. Tang; Solvability of sequential fractional order multi-point boundary value problems at resonance, Appl. Math. Comput., 218 (212), [5] R. E. Gaines, J. Mawhin; Coincidence degree and nonlinear differential equations, Lecture Notes in Math., vol. 568, Springer-Verlag, Berlin, [6] D. J. Guo, V. Lakshmikantham, X. Z. Liu; Nonlinear Integral Equations in Abstract Spaces, Kluwer Academic, Dordrecht, [7] F. D. Ge, H. C. Zhou; Existence of solutions for fractional differential equations with threepoint boundary conditions at resonance in R n, J. Qual. Theory Differ. Equ. 214, No. 68, [8] W. H. Jiang; Solvability for a coupled system of fractional differential equations at resonance, Nonlinear Anal. Real World Appl., 13 (212), [9] W.H. Jiang, The existence of solutions for boundary value problems of fractional differential equations at resonance, Nonlinear Anal., 74 (211) [1] M. Jia, X. P. Liu; Multiplicity of solutions for integral boundary value problems of fractional differential equations with upper and lower solutions, Appl. Math. Comput., 232 (214), [11] F. Jiao, Y. Zhou; Existence of solutions for a class of fractional boundary value problems via critical point theory, Comput. Math. Appl., 62 (211), [12] A. A. Kilbas, H. M. Srivastava, J. J. Trujillo; Theory and Applications of Fractional Differential Equations, Elsevier, 26. [13] N. Kosmatov; Multi-point boundary value problems on an unbounded domain at resonance, Nonlinear Anal., 68 (28), [14] N. Kosmatov; A boundary value problem of fractional order at resonance, Electron. J. Differ. Equ., 135(21) 1-1. [15] J. Mawhin; NSFCBMS Regional Conference Series in Mathematics, American Mathematical Society, Providence, RI, [16] P. D. Phung, L. X. Truong; On the existence of a three point boundary value problem at resonance in R n, J. Math. Anal. Appl., 416 (214), [17] M. Rehman, P. Eloe; Existence and uniqueness of solutions for impulsive fractional differential equations. Appl. Math. Comput., 224 (213), [18] Y. H. Zhang, Z. B. Bai, T. T. Feng; Existence results for a coupled system of nonlinear fractional three-point boundary value problems at resonance, Comput. Math. Appl., 61 (211), [19] Y. Zhang, Z. Bai; Existence of solutions for nonlinear fractional three-point boundary value problems at resonance. J. Appl. Math. Comput. 36 (211), [2] H. C. Zhou, C. H. Kou, F. Xie; Existence of solutions for fractional differential equations with multi-point boundary conditions at resonance on a half-line, J. Qual. Theory Differ. Equ. 211, No. 27, [21] C. B. Zhai, L. Xu; Properties of positive solutions to a class of four-point boundary value problem of Caputo fractional differential equations with a parameter, Commun. Nonlinear Sci. Numer. Simul., 19 (214),

16 16 H.-C. ZHOU, F.-D. GE, C.-H. KOU EJDE-216/61 [22] X. Q. Zhang, L. Wang, Q. Sun; Existence of positive solutions for a class of nonlinear fractional differential equations with integral boundary conditions and a parameter, Appl. Math. Comput., 226 (214), [23] C. X. Zhu, X. Z. Zhang, Z. Q. Wu; Solvability for a coupled system of fractional differential equations with integral boundary conditions, Taiwanese J. Math., 17 (213), [24] X. W. Su; Solutions to boundary value problem of fractional order on unbounded domains in a Banach space, Nonlinear Anal., 74 (211), Hua-Cheng Zhou (corresponding author) Academy of Mathematics and Systems Science, Academia Sinica, Beijing 119, China address: hczhou@amss.ac.cn Fu-Dong Ge College of Information Science and Technology, Donghua University, Shanghai 2162, China address: gefd211@gmail.com Chun-Hai Kou Department of Applied Mathematics, Donghua University, Shanghai 2162, China address: kouchunhai@dhu.edu.cn