On a New Class of Antiperiodic Fractional Boundary Value Problems

Definition 1. The Riemann-Liouville fractional integral of order q for a continuous function g : [0, +∞) → ℝ is defined as

provided the integral exists.

Definition 2. For (n − 1) times absolutely continuous function g : [0, +∞) → ℝ, the Caputo derivative of fractional order q is defined as

where [q] denotes the integer part of the real number q.

Lemma 3. For any y ∈ C[0,1], the unique solution of the linear fractional boundary value problem

is where G_T(t, s) is Green’s function (depending on q and p) given by

Proof. We know that the general solution of equation  ^cD^qx(t) = y(t), 2 < q ≤ 3 can be written as [3]

for some constants b₀, b₁, and b₂ ∈ ℝ. Using the facts   ^cD^pb = 0 (b is a constant), we get Applying the boundary conditions for the problem (5), we find that Substituting the values of b₀, b₁, and b₂ in (8), we get the solution (6). This completes the proof.

Remark 4. For p = 1, the solution of the antiperiodic problem

is given by [18] where g(t, s) is

If we let p → 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: (−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

whose solution is where G_η(t, s) is given by (7). Notice that G_η(t, s) → G_T(t, s) when η → T⁻.

Theorem 6 (see [25].)Let X be a Banach space. Assume that T : X → X is a completely continuous operator and the set V = {u ∈ X∣u = μTu, 0 < μ < 1} is bounded. Then T has a fixed point in X.

Theorem 7 (see [25].)Let X be a Banach space. Assume that Ω is an open bounded subset of X with θ ∈ Ω and let $$ be a completely continuous operator such that

Then T has a fixed point in $$.

Theorem 8. Assume that there exists a positive constant L₁ such that |f(t, x(t))| ≤ L₁ for t ∈ [0, T], x ∈ ℭ. Then the problem (2) has at least one solution.

Proof. First, we show that the operator ℱ is completely continuous. Clearly continuity of the operator ℱ follows from the continuity of f. Let Ω ⊂ ℭ be bounded. Then, for all x ∈ Ω together with the assumption |f(t, x(t))| ≤ L₁, we get

which implies that ∥(ℱx)(t)∥≤M₁.

Furthermore,

Hence, for t₁, t₂ ∈ [0, T], we have

This implies that ℱ is equicontinuous on [0, T], by the Arzela-Ascoli theorem, the operator ℱ : ℭ → ℭ is completely continuous.

Next, we consider the set

and show that the set V is bounded. Let x ∈ V, then x = μℱx, 0 < μ < 1. For any t ∈ [0, T], we have Thus, ∥x∥≤M₃ for any t ∈ [0, T]. So, the set V 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 r such that |f(t, x)| ≤ δ | x| with 0<|x | < r, where δ is a positive constant satisfying

Then the problem (2) has at least one solution.

Proof. Define Ω₁ = {x ∈ ℭ : ∥x∥<r} and take x ∈ ℭ such that ∥x∥ = r; that is, x ∈ ∂Ω. As before, it can be shown that ℱ is completely continuous and that

for x ∈ ∂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.

Theorem 10. Assume that f : [0, T] × ℛ → ℛ is a continuous function satisfying the condition

with Lκ < 1, where Then the problem (2) has a unique solution.

Proof. Let us fix sup _t∈[0,T] | f(t, 0)| = M < ∞ and select

where κ is given by (28). Then we show that ℱB_r ⊂ B_r, where B_r = {x ∈ ℭ : ∥x∥≤r}. For x ∈ B_r, we have Thus we get ℱx ∈ B_r. Now, for x, y ∈ ℭ and for each t ∈ [0, T], we obtain which, in view of the condition κL < 1 (κ is given by (28)), implies that the operator ℱ is a contraction. Hence, by Banach′s contraction mapping principle, the problem (2) has a unique solution.

Example 11. Consider the following antiperiodic fractional boundary value problem:

Example 12. Consider the following antiperiodic fractional boundary value problem:

where q = 5/2, p = 3/4, $$, and T = 2. Clearly, where we have used the fact that $$. Further, With L < 1/κ, all the assumptions of Theorem 10 are satisfied. Hence, the fractional boundary value problem (33) has a unique solution on [0,2].