Existence of solutions for fractional differential equations with three-point boundary conditions at resonance in R n

In this paper, by applying the coincidence degree theory which was first introduced by Mawhin, we obtain an existence result for the fractional three-point boundary value problems in Rn, where the dimension of the kernel of fractional differential operator with the boundary conditions can take any value in {1, 2, . . . , n}. This is our novelty. Several examples are presented to illustrate the result.


Introduction
In this paper, we are concerned with the existence of solutions for the following fractional three-point boundary value problems (BVPs) at resonance in R n : where D α 0 + and I α 0 + are the Riemann-Liouville differentiation and integration; θ is the zero vector in R n ; A is a square matrix of order n satisfying rank(I − A) < n; ξ ∈ (0, 1) is a fixed constant; f : (iii) for every compact set Ω ⊆ R n × R n , the function ϕ Ω (t) = sup{| f (t, u, v)| : (u, v) ∈ Ω} ∈ L 1 [0, 1], where |x| = max{|x i |, i = 1, 2, . . ., n}, the norm of x = (x 1 , x 2 , . . ., x n ) in R n .
Corresponding author.Email: hczhou@amss.ac.cn 2 F. D. Ge and H. C. Zhou The system (1.1) is said to be at resonance in R n if det(I − A) = 0, i.e., dim ker(I − A) ≥ 1, otherwise, it is said to be non-resonant.In the past three decades, many authors investigated the existence of solutions for the fractional differential equations with the boundary value conditions.The attempts on det(I − A) = 0, non-resonance case, for fractional differential equations are available in [1-3, 10, 11, 17, 21-23], and the attempts on det(I − A) = 0 and n ≤ 2, resonance case, can be found in [4-6, 8, 9, 13, 14, 18-20], and the references therein.However, to the best of our knowledge, almost all results derived in these papers are for the case n = 1 with dim ker L = 0 or 1 and for the case n = 2 with dim ker L = 2.It is still open for the case n ≥ 3.So we study this issue in this paper.
For instance, when n = 1, consider the following problems where 1 < α ≤ 2, ξ i ∈ (0, 1), It follows from the argument above that (1.2) is resonant when ∑ m−2 i=1 β i = 1 and it is non-resonant when ∑ m−2 i=1 β i = 1.Further, in order to apply the coincidence degree theory of Mawhin [15], we suppose additionally that A satisfies rank(I − A) < n and one of the following conditions (a 1 ) A is idempotent, that is, A 2 = A, or; (a 2 ) A 2 = I, where I stands for the identity matrix of size n.
It is also obvious that dim ker(I − A) can take any value in {1, 2 . . ., n} for suitable A, which is surely generalize the previous efforts.However, we point out that without the above assumptions, it will be difficult to construct the projector Q as (3.1) below.This is the reason why we only choose the two special cases of A. Removing such an assumption, for the general A satisfying rank(I − A) < n, the problem (1.1) may be a challenging problem, which is also an issue of our further research.
In paper [16], the authors investigated the following second differential system where Carathéodory function and the square matrix A satisfies the condition (a 1 ) or (a 2 ).Therefore, it is more natural to ask whether there exists a solution when the order of the derivative is fractional.In this paper, we offer an answer by considering the system (1.1).The goal of this paper is to study the existence of solutions for the fractional differential equations with boundary value conditions when n ≥ 3. The layout of this paper will be as follows: in Section 2, we give some necessary background and some preparations for our consideration.The statement and the proof of our main result will be given in Section 3 by the coincidence degree theory of Mawhin [15].In Section 4, we present several examples to illustrate the main result.

Background materials and preliminaries
In this section, we introduce some necessary definitions and lemmas which will be used later.For more details, we refer the reader to [7,12,15], and the references therein.Definition 2.1 ([12]).The fractional integral of order α > 0 of a function x : (0, ∞) → R is given by provided the right-hand side is pointwise defined on (0, ∞).
For any x(t) = (x 1 (t), x 2 (t), . . ., x n (t)) ∈ R n , the fractional derivative of order α > 0 of x is defined by The following definitions and coincidence degree theory are fundamental in the proof of our main result.One can refer to [7,15].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 Ge and H. C. Zhou and the mapping L| dom L∩kerP : 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 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.
In this paper, we utilize spaces X, Y introduced as We have the following compactness criterion on subset F of X (see, e.g., [19]).
Lemma 2.8.F ⊂ X is a sequentially compact set if and only if F is uniformly bounded and equicontinuous which are understood in the following sense: (1) there exists an M > 0 such that for every x ∈ F, x ≤ M; (2) for any given ε > 0, there exists a δ > 0 such that

Now we define the linear operator
where dom L = x ∈ X : Then the problem can be equivalently rewritten as Lx = Nx.
Lemma 2.9.The operator L defined above is a Fredholm operator of index zero.
Proof.For any x ∈ dom L, by Lemma 2.4 and x(0) = θ, we obtain which, together with (2.5) where g : Y → R n is a continuous linear operator defined by Actually, for any y ∈ im L, there exists a function which means that g(y) ∈ im(I − A).
On the other hand, for any y ∈ Y satisfying g(y) ∈ im(I − A), there exists a constant c * such that g(y) (2.8) For A 2 = A, we have (2.9) For A 2 = I, we have It follows from (2.9) and (2.10) that ρ A satisfies the following properties By (2.11), it is easy to verify Q 2 y = Qy, that is, Q is a projection operator.The equality ker Q = im L follows from the trivial fact that Therefore, we get Finally, we shall prove that im Q = ker L. Indeed, for any z ∈ im Q, let z = Qy, y ∈ Y.By (2.11), we have which implies z ∈ ker L. Conversely, for each z ∈ ker L, there exists a constant c * ∈ ker(I − A) such that z = c * t α−1 for t ∈ [0, 1].By (2.11) and (2.12), we derive which implies that z ∈ im Q. Hence we know that im Q = ker L. Then the operator L is a Fredholm operator of index zero.The proof is complete.
Define the operator P : X → X as follows: Lemma 2.10.The mapping P : X → X defined as above 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 K P y(t) = I α 0+ y(t), also K P = (L| dom L∩ker P ) −1 and K P y ≤ 1/Γ(α) y 1 .
Proof.The continuity of P is obvious.By the first identity of (2.11), 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 an x ∈ X such that v = Px.By the first identity of (2.11) again, we see that and we deduce that which gives us v ∈ im P. Thus, we get that ker L = im P and consequently X = ker L ⊕ ker P.Moreover, let y ∈ im L. There exists x ∈ dom L such that y = Lx, and we obtain where c ∈ R n satisfies c = 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 and for x ∈ dom L ∩ ker P, we obtain that for some c 1 , c 2 ∈ R n .In view of x ∈ dom L ∩ ker P, we know that c 1 = c 2 = 0. Therefore, (K P L)x(t) = x(t).This shows that K P = (L| domL∩kerP ) −1 .Finally, by the definition of K P , we derive and It follows from (2.15) and (2.16) that (2.17) Then Lemma 2.10 is proved.
Lemma 2.13.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 the hypothesis (iii) on the function f , there exists a function ϕ which, along with (2.7) and (2.13), implies 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 By Lemma 2.8, it is easy to know that K P,Q N is continuous.From (2.19) and (2.7), we derive that (2.21) From (2.21), we obtain for any t 1 , t 2 ∈ [0, 1] with t 1 < t 2 , we shall see that Then we get that 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 continuous in Ω.The proof is complete.

Main result
In this section, we shall present and prove our main result.
Theorem 3.1.Let f be a Carathéodory function and the following conditions hold: (H 1 ) There exist three nonnegative functions a, b, c (H 3 ) There exists a constant A 2 > 0 such that for any e ∈ R n satisfying e = Ae and min 1≤i≤n {|e i |} > A 2 , either e, QNe ≤ 0, min or else e, QNe ≥ 0, min where •, • is the scalar product in R n .
Then the BVPs (1.1) have at least one solution in space X provided that Proof.We shall construct an open bounded subset Ω in X satisfying all assumption of Lemma 2.7. Let For any x ∈ Ω 1 , x / ∈ kerL, we get that λ = 0. Since Nx ∈ im L = ker Q, by the definition of im L, we have g(Nx) ∈ im(I − A), where (3. 3) It follows from hypothesis (H 2 ) and (3.3) that there exists Further, again for x ∈ Ω 1 , since im P = ker L, X = ker L ⊕ ker P, we have (I − P)x ∈ dom L ∩ ker P and LPx = θ.Then From (3.4) and (3.5), we can conclude that Moreover, by the definition of N and the hypothesis (H 1 ), we see that From (3.8) and x ∞ ≤ x , we derive which, together with D α−1 0 + x ∞ ≤ x , (3.8) and (3.9), gives us That is, Ω 2 is bounded in X.If the first part of (H 3 ) holds, denote then for any x ∈ Ω 3 , we know that x = et α−1 with e ∈ ker(I − A) and λx = (1 − λ)QNx.
In the sequel, we will 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 2.9 and 2.13 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, in order to complete the theorem, we only need to prove that the condition (iii) of Lemma 2.7 is also satisfied.For this purpose, let where we let the isomorphism J : im Q → ker L be the identical operator.Since Ω 3 ⊆ Ω, H(x, λ) = 0 for (x, λ) ∈ ker L ∩ ∂Ω × [0, 1], then by the homotopy property of degree, we obtain Thus (H 3 ) of Lemma 2.7 is fulfilled and Theorem 3.1 is proved.The proof is complete.

Examples
In this section, we shall present three examples to illustrate our main result in R 3 with dim ker L = 1, dim ker L = 2, dim ker L = 3, respectively, which surely generalize the previous results [4-6, 8, 9, 13, 14, 18-20], where the dimension of dim ker L is only 1 or 2.
Example 4.3.Consider the following system with dim ker L = 3 in R 3 .

Concluding remarks
In this paper, we consider fractional BVPs at resonance in R n .The dimension of the kernel of fractional differential operator with the boundary conditions can take any value in {1, 2 . . ., n}, which generalizes the existing literature [4-6, 8, 9, 13, 14, 18-20], where the dim ker L = 1 for n = 1, or dim ker L = 2 for n = 2.The illustrative examples validate the applicability of Theorem (2.7).Note that only the two particular cases: A 2 = A, A 2 = I are considered.For the general A satisfying rank(I − A) < n, the system is still resonant.However, we do not know for the system if there are solutions for (1.1) due to some difficulty in constructing the projector Q.We shall investigate this problem in our forthcoming paper.Finally, our result can also be easily generalized to other fractional BVPs, for instance,    D α 0 + x(t) = f (t, x(t), D α−1 0 + x(t)), 1 < α ≤ 2, t ∈ (0, 1), I 2−α 0+ x(t)| t=0 = θ, x(1) = Ax(ξ), (5.1) where the matrix A satisfies rank(I − Aξ α−1 ) < n which means this system is resonant.Particularly, when α = 2, the system (1.1) and (5.1) become a system of second order differential equations, which can be regarded as a generalization of the results in [16], where a system of second order differential equations was considered.

F
. D. Ge and H. C. Zhou

.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 in Definition 2.6 and N : Ω → Y is L-compact.For any x ∈ Ω, by (2.1), we shall see that Then problem (4.1) has one solution if and only if problem (1.1), with A and f defined by (4.2), (4.3), has one solution.Hence we need only to verify that the conditions of Theorem 3.1 are satisfied.Check (H 1 ) of Theorem 3.1: for some r ∈ R, where |u| Consider the following system with dim ker L .10)It is not difficult to see that A 2 = I and dim ker(I − A) = 2. Then problem (4.8), with A and f defined by (4.10) and (4.9), has one solution if and only if problem (1.1) has one solution.