In view of Lemma 2.1, we define an operator by
(3.1)
It should be noticed that problem (1.1) has solutions if and only if the operator has fixed points.
Our first result is an existence and uniqueness result for the impulsive boundary value problem (1.1) by using the Banach contraction mapping principle.
Let . Further, for convenience we set
(3.2)
and
(3.3)
Theorem 3.1 Assume that:
(H1) The function is continuous and there exist constants such that
for each and , .
(H2) The functions are continuous and there exist constants such that
for each , .
If
where ω is defined by (3.2), and , then the boundary value problem (1.1) has a unique solution on J.
Proof We transform the boundary value problem (1.1) into a fixed point problem, , where the operator is defined by (3.1). By using the Banach contraction mapping principle, we shall show that has a fixed point which is the unique solution of the boundary value problem (1.1).
Let , , and be nonnegative constants such that , , and . By choosing a constant R as
where and defined by (3.3), we will show that , where a ball is defined by . 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 (1.1). This completes the proof. □
The second existence result is based on Schaefer’s fixed point theorem.
Theorem 3.2 Assume that:
(H3) is a continuous function and there exists a constant such that
for each and all , .
(H4) The functions are continuous and there exist constants such that
for all , .
Then the boundary value problem (1.1) has at least one solution on J.
Proof We will use Schaefer’s fixed point theorem to prove that , defined by (3.1), has a fixed point. We divide the proof into four 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
and , for , as .
Then, for each , we get
which gives as . This means that is continuous.
Step 2: maps bounded sets into bounded sets in .
So, let us prove that for any , there exists a positive constant ρ such 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
As , the right-hand side of the above inequality (which is independent of x) tends to zero. As a consequence of Steps 1 to 3, together with the Arzelá-Ascoli theorem, we deduce that is completely continuous.
Step 4: We show that the set
is bounded.
Let . Then for some . Thus, for each , by using the computations of Step 2, we have that
This shows that the set E is bounded. As a consequence of Schaefer’s fixed point theorem, we conclude that has a fixed point which is a solution of the impulsive -integro-difference boundary value problem (1.1). □
The third existence result for the impulsive boundary value problem (1.1) is based on Krasnoselskii’s fixed point theorem.
Lemma 3.3 (Krasnoselskii’s fixed point theorem) [19]
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 a compact and continuous;
-
(c)
B is a contraction mapping.
Then there exists such that .
For convenience we put
(3.5)
and
(3.6)
Theorem 3.4 Let be a continuous function. Assume that:
(A1) , and .
(A2) There exist constants such that and , , for .
(A3) There exist constants such that and , , for .
If
(3.7)
then boundary value problem (1.1) has at least one solution on J.
Proof We define and choose a suitable constant ρ as
where Λ and are defined by (3.5) and (3.6), respectively. We define the operators Φ and Ψ on as
and
For , we have
Thus, .
For , from (A3), we have
which implies, by (3.7), that Ψ is a contraction mapping.
Continuity of f implies that the operator Φ is continuous. Also, Φ is uniformly bounded on as
Now we prove the compactness of the operator Φ.
We define , for some with and consequently we get
which is independent of x and tends to zero as . Thus, Φ is equicontinuous. So Φ is relatively compact on . Hence, by the Arzelá-Ascoli theorem, Φ is compact on . Thus all the assumptions of Lemma 3.3 are satisfied. So the boundary value problem (1.1) has at least one solution on J. The proof is completed. □
Our final, fourth existence result is based on the Leray-Schauder Nonlinear Alternative.
Lemma 3.5 (Nonlinear alternative for single valued maps) [20]
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 3.6 Assume that:
(A4) There exist a continuous nondecreasing function and a continuous function such that
(A5) There exist continuous nondecreasing functions such that
for all , .
(A6) There exists a constant such that
where , and
Then the impulsive boundary value problem (1.1) has at least one solution on J.
Proof First we show that maps bounded sets (balls) into bounded sets in . For a positive number , let be a bounded ball in . Then for we have
Hence, we deduce that .
Next we show that maps bounded sets into equicontinuous sets of .
Let for some , , be a bounded set of as in the previous step, and let . Then we have
The right-hand side of the above inequality, which is independent of x, tends to zero as . From all above and by the Arzelá-Ascoli theorem is completely continuous.
The result will follow from the Leray-Schauder nonlinear alternative (Lemma 3.5) once we have proved the boundedness of the set of all solutions to the equations for some .
Let x be a solution. Thus, for each , we have