On a New Class of Antiperiodic Fractional Boundary Value Problems

and Applied Analysis 3 where g(t, s) is g (t, s) = {{{{{{{{{{{{{{ {{{{{{{{{{{{{{ { (t − s) q−1 − (1/2) (T − s) q−1 Γ (q) + (T − 2t) (T − s) q−2 4Γ (q − 1) + t (T − t) (T − s) q−3 4Γ (q − 2) , s ≤ t, − (T − s) q−1 2Γ (q) + (T − 2t) (T − s) q−2 4Γ (q − 1) + t (T − t) (T − s) q−3 4Γ (q − 2) , t < s. (14) If we let p → 1− in (7), we obtain G T (t, s) 󵄨󵄨󵄨󵄨p→1− = {{{{{{{{{{{{{{{{{{{ {{{{{{{{{{{{{{{{{{{ { (t − s) q−1 − (1/2) (T − s) q−1 Γ (q) + (T − 2t) (T − s) q−2 2Γ (q − 1) + (−2t 2 − T 2 + 4tT) (T − s) q−3 4Γ (q − 2) , s ≤ t, − (T − s) q−1 2Γ (q) + (T − 2t) (T − s) q−2 2Γ (q − 1) + (−2t 2 − T 2 + 4tT) (T − s) q−3 4Γ (q − 2) , t < s. (15) We note that the solutions given by (14) and (15) are different. As amatter of fact, (15) contains an additional term: (−t2−T2+ 3tT)(T − s) q−3 /4Γ(q − 2). Therefore the fractional boundary conditions introduced in (2) give rise to a new class of problems. Remark 5. When the phenomenon of antiperiodicity occurs at an intermediate point η ∈ (0, T), the parametric-type antiperiodic fractional boundary value problem takes the form c D q x (t) = f (t, x (t)) , t ∈ [0, T] , 2 < q ≤ 3, x (0) = −x (η) , c D p (0) = − c D p (η) , c D p+1 (0) = − c D p+1 (η) , (16) whose solution is x (t) = ∫ T 0 G η (t, s) f (s, x (s)) ds, (17) where G η (t, s) is given by (7). Notice that G η (t, s) → G T (t, s) when η → T. 3. Existence Results LetC = C([0, T], R) denotes a Banach space of all continuous functions defined on [0, T] into R endowed with the usual supremum norm. In relation to (2), we define an operatorF : C → C as

Definition 1.The Riemann-Liouville fractional integral of order  for a continuous function  : [0, +∞) → R is defined as provided the integral exists.
Definition 2. For (−1) times absolutely continuous function  : [0, +∞) → R, the Caputo derivative of fractional order  is defined as where [] denotes the integer part of the real number . ( is where   (, ) is Green's function (depending on  and ) given by Proof.We know that the general solution of equation    () = (), 2 <  ≤ 3 can be written as [3] for some constants  0 ,  1 , and  2 ∈ R. Using the facts     = 0 ( is a constant), we get Applying the boundary conditions for the problem (5), we find that Substituting the values of  0 ,  1 , and  2 in (8), we get the solution ( 6).This completes the proof.
Remark 4. For  = 1, the solution of the antiperiodic problem is given by [18] where (, ) is If we let  → 1 − in ( 7), we obtain We note that the solutions given by ( 14) and ( 15) are different.
As a matter of fact, (15) contains an additional term: . Therefore the fractional boundary conditions introduced in (2) give rise to a new class of problems.

Existence Results
Let C = ([0, ], ) denotes a Banach space of all continuous functions defined on [0, ] into  endowed with the usual supremum norm.
In relation to (2), we define an operator F : C → C as Observe that the problem (2) has a solution if and only if the operator F has a fixed point.
For the sequel, we need the following known fixed point theorems.
Theorem 7 (see [25]).Let  be a Banach space.Assume that Ω is an open bounded subset of  with  ∈ Ω and let  : Ω →  be a completely continuous operator such that ‖‖ ≤ ‖‖ , ∀ ∈ Ω. ( Then  has a fixed point in Ω.
Now we are in a position to present the main results of the paper.
which implies that ‖(F)()‖ ≤ This implies that F is equicontinuous on [0, ], by the Arzela-Ascoli theorem, the operator F : C → C is completely continuous.
Next, we consider the set 23) and show that the set  is bounded.Let  ∈ , then  = F, 0 <  < 1.For any  ∈ [0, ], we have Thus, ‖‖ ≤  3 for any  ∈ [0, ].So, the set  is bounded.Thus, by the conclusion of Theorem 6, the operator F has at least one fixed point, which implies that (2) has at least one solution.
Theorem 9. Let there exists a positive constant  such that |(, )| ≤ || with 0 < || < , where  is a positive constant satisfying Then the problem (2) has at least one solution.
Proof.Define Ω 1 = { ∈ C : ‖‖ < } and take  ∈ C such that ‖‖ = ; that is,  ∈ Ω.As before, it can be shown that F is completely continuous and that for  ∈ Ω, where we have used (25).Therefore, by Theorem 7, the operator F has at least one fixed point which in turn implies that the problem (2) has at least one solution.
with  < 1, where Then the problem (2) has a unique solution.
Example 11.Consider the following antiperiodic fractional boundary value problem: