Let = {: is continuous everywhere except for some at which and exist and , }. is a Banach space with the norm.
From Lemma 2.1, we define an operator by
(3.1)
where constants , , , , and Ω are defined as in Lemma 2.1. Itshould be noticed that problem (1.4) has solutions if and only if the operator
has fixed points.
Our first result is an existence and uniqueness result for the impulsive boundaryvalue problem (1.4) by using the Banach contraction mapping principle.
For convenience, we set
(3.2)
and
(3.3)
Theorem 3.1 Assume that the following conditions hold:
(H1) is a continuous function and there exists a constantsuch that
for eachand.
(H2) The functionsare continuous and there exist constantssuch that
for each, .
If
whereis defined by (3.2), then the impulsive-difference Langevin boundary value problem(1.4) has a unique solution on J.
Proof Firstly, we transform the impulsive -difference Langevin boundary value problem (1.4) intoa fixed point problem, , where the operator
is defined by (3.1).Applying the Banach contraction mapping principle, we shall show that
has a fixed point which is the unique solution of theboundary value problem (1.4).
Let , , and be nonnegative constants such that, , and . We choose a suitable constant ρ by
where and defined by (3.3). Now, we will show that, where a set is defined as . For , we have
which implies that .
For any and for each , we have
which implies that . As ,
is a contraction.Therefore, by the Banach contraction mapping principle, we find that
has a fixed point which is the unique solution of problem(1.4). □
The second existence result is based on Schaefer’s fixed point theorem.
Theorem 3.2 Assume that the following conditions hold:
(H3) is a continuous function and there exists a constantsuch that
for eachand all.
(H4) The functionsare continuous and there exist constantssuch that
for all, .
If
(3.5)
then the impulsive-difference Langevin boundary value problem(1.4) has at least one solution on J.
Proof We shall use Schaefer’s fixed point theorem to prove that theoperator
defined by (3.1) has a fixed point. We divide the proof intofour steps.
Step 1: Continuity of
.
Let be a sequence such that in . Since f is a continuous function on and , are continuous functions on ℝ for, we have
for , as .
Then, for each , we get
which gives as . This means that
is continuous.
Step 2:
maps bounded sets into bounded sets in.
Let us prove that for any , there exists a positive constant σsuch that for each , we have . For any , we have
Hence, we deduce that .
Step 3:
maps bounded sets into equicontinuous sets of.
Let for some , , be a bounded set of as in Step 2, and let . Then we have
The right-hand side of the above inequality is independent of x and tendsto zero as . As a consequence of Steps 1 to 3, together with theArzelá-Ascoli theorem, we deduce that is completely continuous.
Step 4: We show that the set
is bounded.
Let . Then for some . Thus, for each , we have
This implies by (H3) and (H4) that for each, we have
Setting
we have
which yields
This shows that the set E is bounded. As a consequence of Schaefer’sfixed point theorem, we conclude that
has a fixed point whichis a solution of the impulsive -difference Langevin boundary value problem(1.4). □