We are in a position to establish our main results. In the following subsections we prove existence as well as existence and uniqueness results for the impulsive fractional BVP (1.1) by using a variety of fixed-point theorems.
3.1 Existence and uniqueness results via Banach’s fixed-point theorem
In this subsection we give first an existence and uniqueness result for the impulsive fractional BVP (1.1) by using Banach’s fixed-point theorem.
For convenience, we set
(3.1)
(3.2)
(3.3)
Theorem 3.1 Assume the following.
(H1) There exists a constant such that , for each and .
(H2) There exists a constant such that for each , .
If
(3.4)
then impulsive fractional boundary-value problem (1.1) has a unique solution in .
Proof We transform the problem (1.1) into a fixed-point problem, , where the operator A is defined by equation (2.4). Using Banach’s contraction principle, we shall show that A has a fixed point.
Setting , and choosing , we show that , where . For , we have
which proves that .
Now let . Then, for , we have
Therefore,
As follows from equation (3.4), A is a contraction. As a consequence of Banach’s fixed-point theorem, we have A has a fixed point which is a unique solution of the impulsive fractional boundary-value problem (1.1). This completes the proof. □
Now we give another existence and uniqueness result for impulsive fractional BVP (1.1) by using Banach’s fixed-point theorem and Hölder’s inequality. In addition, for , we set
(3.5)
(3.6)
(3.7)
Theorem 3.2 Assume that the following conditions hold:
(H3) , for each , , where , .
(H4) , for each , with constants , .
Denote .
If
then the impulsive fractional boundary-value problem (1.1) has a unique solution.
Proof For and for each , by Hölder’s inequality, we get
Therefore,
It follows that A is a contraction mapping. Hence Banach’s fixed-point theorem implies that A has a unique fixed point, which is the unique solution of the impulsive fractional boundary-value problem (1.1). This completes the proof. □
3.2 Existence result via Krasnoselskii’s fixed-point theorem
Lemma 3.1 (Krasnoselskii’s fixed point theorem) [29]
Let M be a closed, bounded, convex and nonempty subset of a Banach space X. Let A, B be the operators such that (a) whenever ; (b) A is compact and continuous; (c) B is a contraction mapping. Then there exists such that .
Theorem 3.3 Let be a continuous function and let (H2) holds. In addition, we assume that:
(H5) , , and .
(H6) There exists a constant such that , , for .
Then the impulsive fractional boundary-value problem (1.1) has at least one solution on if
where Φ is defined by equation (3.2).
Proof We define and choose a suitable constant as
where Ω and Ψ are defined by equations (3.1) and (3.3), respectively. We define the operators and on as
For , we find that
Thus, . It follows from the assumption (H2) together with (3.8) that is a contraction mapping. Continuity of f implies that the operator is continuous. Also, is uniformly bounded on as
Now we prove the compactness of the operator .
We define , with and consequently we have
which is independent of x and tends to zero as . Thus, is equicontinuous. So is relatively compact on . Hence, by the Arzelá-Ascoli theorem, is compact on . Thus all the assumptions of Lemma 3.1 are satisfied. So the conclusion of Lemma 3.1 implies that the impulsive fractional boundary-value problem (1.1) has at least one solution on . The proof is completed. □
3.3 Existence result via Leray-Schauder’s Nonlinear Alternative
Lemma 3.2 (Nonlinear alternative for single valued maps) [30]
Let E be a Banach space, C a closed, convex subset of E, U an open subset of C and . Suppose that is a continuous, compact (that is, is a relatively compact subset of C) map. Then either
-
(i)
F has a fixed point in , or
-
(ii)
there is a (the boundary of U in C) and with .
Theorem 3.4 Assume the following.
(H7) There exist a continuous nondecreasing function and a function such that
(H8) There exists a continuous nondecreasing function such that
(H9) There exists a constant such that
Then the impulsive fractional boundary-value problem (1.1) has at least one solution on .
Proof We show that A maps bounded sets (balls) into bounded sets in . For a positive number r, let be a bounded ball in . Then for we have
Consequently
Next we show that A maps bounded sets into equicontinuous sets of . Let , with , , , and . Then we have
Obviously the right-hand side of the above inequality tends to zero independently of as . As A satisfies the above assumptions, therefore it follows by the Arzelá-Ascoli theorem that is completely continuous.
Let x be a solution. Then, for , and following the similar computations as in the first step, we have
Consequently, we have
In view of (H9), there exists such that . Let us set
Note that the operator is continuous and completely continuous. From the choice of U, there is no such that for some . Consequently, by the nonlinear alternative of Leray-Schauder type (Lemma 3.2), we deduce that A has a fixed point which is a solution of the problem (1.1). This completes the proof. □
3.4 Existence result via Leray-Schauder degree
Theorem 3.5 Assume the following.
(H10) There exist constants and such that
(H11) There exist constants and such that
where Ω and Φ are given by equations (3.1) and (3.2), respectively.
Then the impulsive fractional boundary-value problem (1.1) has at least one solution on .
Proof We define an operator as in equation (2.4) and consider the fixed-point problem
We are going to prove that there exists a fixed point satisfying equation (3.9). It is sufficient to show that satisfies
(3.10)
where . We define
As shown in Theorem 3.4, we find that the operator A is continuous, uniformly bounded, and equicontinuous. Then, by the Arzelá-Ascoli theorem, a continuous map defined by is completely continuous. If equation (3.10) is true, then the following Leray-Schauder degrees are well defined and by the homotopy invariance of topological degree, it follows that
(3.11)
where I denotes the identity operator. By the nonzero property of the Leray-Schauder degree, for at least one . In order to prove equation (3.10), we assume that for some . Then
Computing directly for , we have
If , inequality (3.10) holds. This completes the proof. □