A Note on Fractional Equations of Volterra Type with Nonlocal Boundary Condition

Definition 1. For α > 0, the integral

is called the Riemann-Liouville fractional integral of order α.

Definition 2. The Riemann-Liouville derivative of order α (n − 1 < α ≤ n) can be written as

Lemma 3 (see [5].)Let n − 1 < α ≤ n. If f(t) ∈ L(0, T) and $$, then one has the following equality:

Lemma 4. Let (H1) hold. x ∈ C_1−α(J) and x is a solution of the following problem:

if and only if x(t) is a solution of the following integral equation:

Proof. Assume that x(t) satisfies (8). From the first equation of (8) and Lemma 3, we have

Conversely, assume that x(t) satisfies (9). Applying the operator D^α to both sides of (9), we have In addition, by calculation, we can conclude $$. The proof is completed.

Theorem 5. Let (H1), (H2) hold, f ∈ C(J × R², R),   and  k ∈ C(Δ, R). Then problem (2) has a unique solution.

Proof. Define the operator N : C_1−α(J) → C_1−α(J) by

It is easy to check that the operator N is well defined on C_1−α(J). Next we show that N is a contradiction operator on C_1−α(J). For convenience, let and choose where λ is a positive constant defined in the norm of the space C_1−α(J).

Then, for any x, y ∈ C_1−α(J), we have from (H1), (H2), and the Hölder inequality

According to λ > ρ^q and the Banach fixed point theorem, the problem (2) has a unique solution. The proof is completed.

Remark 6. Theorem 5 is an essential improvement of [3, Theorem 1].

Lemma 7. Let 0 < α < 1, M, N ∈ C(J), |M(t)| ≤ M₁, |N(t)| ≤ N₁. Suppose that

and p ∈ C_1−α(J) satisfies the problem Then p(t) ≤ 0 for all t ∈ (0, T].

Proof. Suppose that the inequality p(t) ≤ 0, for all t ∈ (0, T], is not true. Therefore, there exists at least a t_* ∈ (0, T] such that $$. Without loss of generality, we assume $$.

We obtain that

Let t = t_*; we have So This is a contradiction. Hence p(t) ≤ 0 for all t ∈ (0, T]. The proof is completed.

Definition 8. We say that x₀ ∈ C_1−α(J) is called a lower solution of problem (2) if

We say that y₀ ∈ C_1−α(J) is called an upper solution of problem (2) if

In the following discussion, we need the following assumptions.

• (H3)

  Assume that g : C_1−α(J) → R is a nondecreasing continuous function, f(t, β₁, β₂) ∈ C_1−α(J) for all t ∈ J, x₀ ≤ β₁ ≤ y₀, $$. x₀ and y₀ are lower and upper solutions of problem (2), respectively, and x₀ ≤ y₀.

• (H4)

  Consider

  $$ ()

•  

  where x₀ ≤ u₁ ≤ v₁ ≤ y₀, $$. M, N ∈ C(J).

Let $$.

Theorem 9. Let inequality (17), (H2)–(H4) hold. Then there exist monotone sequences {x_n},{y_n}⊂[x₀, y₀] which converge uniformly to the extremal solutions of (2) in [x₀, y₀], respectively.

Proof. This proof consists of the following three steps.

Step 1. Construct the sequences {x_n}, {y_n}.

For any η ∈ [x₀, y₀], we consider the following linear problem:

By Theorem 5, (25) has a unique solution which satisfies Define an operator A : [x₀, y₀]→[x₀, y₀] by x = Aη. It is easy to check that the operator A is well defined on [x₀, y₀]. Let η₁, η₂ ∈ [x₀, y₀] with η₁ ≤ η₂.

Setting p(t) = z₁(t) − z₂(t), z₁(t) = Aη₁(t), and z₂(t) = Aη₂(t), by (26), we obtain

Besides, By Lemma 7, we know p(t) ≤ 0, t ∈ (0, T]. It means that A is nondecreasing. Obviously, we can easily get that A is a continuous map. Let x_n = Ax_n−1, y_n = Ay_n−1, n = 1,2, ….

Step 2. The sequences {t^1−αx_n}, {t^1−αy_n} converge uniformly to t^1−αx^*, t^1−αy^*, respectively.

In fact, x_n, y_n satisfy the following relation:

Setting p(t) = x₀(t) − x₁(t) and x₀(t) is a lower solution of problem (2): Besides, By Lemma 7, we can obtain that x₀ ≤ x₁ for all t ∈ (0, T]. Similarly, we can show that y₁ ≤ y₀ for all t ∈ (0, T]. Applying the operator A to both sides of x₀ ≤ x₁, y₁ ≤ y₀, and x₀ ≤ y₀, we can easily get (29). Obviously, the sequences {t^1−αx_n}, {t^1−αy_n} are uniformly bounded and equicontinuous. Then by using the Ascoli-Arzela criterion, we can conclude that the sequences {t^1−αx_n}, {t^1−αy_n} converge uniformly on (0, T] with lim _n→∞t^1−αx_n = t^1−αx^*, lim _n→∞t^1−αy_n = t^1−αy^* uniformly on (0, T].

Step 3. x^*, y^* are extremal solutions of (1).

x^*, y^* are solutions of (1) on [x₀, y₀], because of the continuity of operator A. Let z ∈ [x₀, y₀] be any solution of (1). That is,

Suppose that there exists a positive integer n such that x_n(t) ≤ z(t) ≤ y_n(t) on (0, T]. Let p(t) = x_n+1(t) − z(t); we have By Lemma 7, we know that p(t) ≤ 0 on (0, T], which implies x_n+1(t) ≤ z(t) on (0, T]. Similarly, we obtain that z(t) ≤ y_n+1(t) on (0, T]. Since x₀(t) ≤ z(t) ≤ y₀(t) on (0, T], by induction we get that x_n(t) ≤ z(t) ≤ y_n(t) on (0, T] for every n. Therefore, x^*(t) ≤ z(t) ≤ y^*(t) on (0, T] by taking n → ∞. Thus, we completed this proof.

Example 1. Consider the following problem:

Obviously, T = 1, α = 1/2, k(t, s) = ts, and f(t, v₁, v₂) = t + (1/60)v₁ + (1/30)v₂.

Let w = 1, L₁ = 1/60, L₂ = 1/30, and L₃ = 1/12.

It is easy to check that

So, (H1) and (H2) are satisfied. By the choice of p = 3/2, q = 3, we can get that λ > ρ³ and ρ ≡ (4/(11 × 3^1/3)){(1/20Γ(1/2))[Γ(1/4) ²/Γ(1/2)] ^2/3 + (1/15Γ(1/2))[Γ(1/4)Γ(7/4)] ^2/3}. According to Theorem 5, the problem (34) has a unique solution.

Consider the same equation as (34), taking x₀(t) = 0, y₀(t) = t^−1/2 + 6, and then we have $$.

Moreover,

On the other hand, it is easy to check that (H3) holds. And let M(t) = 1/(t − 1), N(t) = cos  t/30, and then we have where x₀ ≤ u₁ ≤ v₁ ≤ y₀, $$ . So (H4) is satisfied. Obviously, M₁ = 1/30, N₁ = 1, and then we can get Inequality (17) holds. All conditions of Theorem 9 are satisfied, so problem (34) has extremal solutions.