with its q
-derivative existing on
is said to be a solution of (1.1) if u
satisfies the equation
and satisfies the conditions
Now, we define the operator
Clearly, the fixed points of the operator T are the solutions of problem (1.1). To begin with, we need the following assumptions to prove the existence and uniqueness of a solution of the integral equation (2.3) which satisfies BVP (1.1):
(A1) is a continuous bounded linear operator and there exists a constant such that for all ;
(A2) The function is continuous and there exists a constant such that ;
(A3) are continuous and there exist constants and such that for each and ;
(A4) There exist constants and are continuous functions such that , ;
(A5) There exists a constant
, and ;
is continuous and there exists a constant
for all ;
(A7) There exist constants , such that , for each and ;
(A8) There exist constants such that , .
The following are the main results of this paper. Our first result relies on Schauder’s fixed point theorem which gives an existence result for solutions of BVP (1.1).
Theorem 1 Assume that the assumptions (A1)-(A4) hold. Then BVP (1.1) has at least one solution on J.
Proof In order to show the existence of a solution of BVP (1.1), we need to transform BVP (1.1) to a fixed point problem by using the operator T in (3.1). Now, we shall use Schauder’s fixed point theorem to prove T has a fixed point which is then a solution of BVP (1.1). First, let us define for any . Then it is clear that the set is a closed, bounded and convex. The proof will be given in several steps.
Step 1: T is continuous.
be a sequence such that
Since A is a continuous operator and f, g, I, are continuous functions, we have as .
Step 2: T maps bounded sets into bounded sets.
Now, it is enough to show that there exists a positive constant l
. Then we have, for each
Then it follows that .
Step 3: T maps bounded sets into equicontinuous sets.
be a bounded set of
as in Step 2, and let
. Then, letting
, we have
Hence, is equicontinuous on all the subintervals , . Then we can deduce that is completely continuous as a result of the Arzela-Ascoli theorem together with Steps 1 to 3.
As a consequence of Schauder’s fixed point theorem, we conclude that T has a fixed point. That is, BVP (1.1) has at least one solution. The proof is complete. □
Our second result is about the uniqueness of the solution of BVP (1.1). And it depends on Banach’s fixed point theorem.
Theorem 2 Assume that
(A1)-(A8) hold with
First, we show that
. Indeed, in order to do this, it is adequate to replace l
in Step 2 in Theorem 1. Thus, T
into itself. Now, define the mapping
. Then, for each
, we have
Observing the inequality
which implies that
Therefore, by (3.2), the operator T is a contraction. As a consequence of Banach’s fixed point theorem, we deduce that T has a fixed point which is a unique solution of BVP (1.1). □
Consider the following boundary value problem for impulsive integrodifferential evolution equation of fractional order:
where , , and , are given positive constants with and .
and by (2.5), it can be found that
Therefore, due to the fact that all the assumptions of Theorem 2 hold, BVP (3.3) has a unique solution. Besides, one can easily check the result of Theorem (1) for BVP (3.3).