Research Article | Open Access

# Multiple Positive Solutions for First-Order Impulsive Integrodifferential Equations on the Half Line in Banach Spaces

**Academic Editor:**A. M. El-Sayed

#### Abstract

The author discusses the multiple positive solutions for an infinite boundary value problem of first-order impulsive superlinear integrodifferential equations on the half line in a Banach space by means of the fixed point theorem of cone expansion and compression with norm type.

#### 1. Introduction

Let be a real Banach space and a cone in which defines a partial ordering in by if and only if . is said to be normal if there exists a positive constant such that implies , where denotes the zero element of and the smallest is called the normal constant of . If and , we write . For details on cone theory, see [1].

In paper [2], we considered the infinite boundary value problem (IBVP) for first-order impulsive nonlinear integrodifferential equation of mixed type on the half line in : where , ,, , , (), , , and , , denotes the set of all nonnegative numbers. denotes the jump of at , that is, where and represent the right and left limits of at , respectively. By using the fixed point index theory, we discussed the multiple positive solutions of IBVP(1.1). But the discussion dealt with sublinear equations, that is, we assume that there exists such that uniformly for (see condition in [2]).

Now, in this paper, we discuss the multiple positive solutions of an infinite three-point boundary value problem (which includes IBVP(1.1) as a special case) for superlinear case by means of different method, that is, by using the fixed point theorem of cone expansion and compression with norm type, which was established by the author in [3] (see also [1]), and the key point is to introduce a new cone .

Consider the infinite three-point boundary value problem for first-order impulsive nonlinear integrodifferential equation of mixed type on the half line in : where , and (for some ). It is clear that IBVP(1.5) includes IBVP(1.1) as a special case when .

Let is a map from into such that is continuous at , left continuous at , and exists, and . It is clear that is a Banach space with norm Let and . Obviously, and are two cones in space and . is called a positive solution of IBVP(1.5) if for and satisfies (1.5).

#### 2. Several Lemmas

Let us list some conditions.(), and In this case, let ()There exist and such that ()There exist () and such that ()For any and and () are relatively compact in , where .

*Remark 2.1. *Obviously, condition () is satisfied automatically when is finite dimensional.

*Remark 2.2. *It is clear that if condition is satisfied, then the operators and defined by (1.2) are bounded linear operators from into and ; moreover, we have and .

We shall reduce IBVP(1.5) to an impulsive integral equation. To this end, we consider the operator defined by In what follows, we write ().

Lemma 2.3. *If conditions ()–() are satisfied, then operator defined by (2.5) is a completely continuous (i.e., continuous and compact) operator from into .*

*Proof. *Let be given. Let
For , we see that by virtue of condition and (2.6),
which implies the convergence of the infinite integral
On the other hand, condition and (2.7) imply the convergence of the infinite series
It follows from (2.5), (2.10), and (2.12) that
which implies that and
Moreover, by (2.5), we have
It is clear that
so, (2.16) and (2.17) imply
It follows from (2.15) and (2.18) that
Hence, . That is, maps into .

Now, we are going to show that is continuous. Let (). Then and . Similar to (2.14), it is easy to get
It is clear that
Moreover, we see from (2.8) that
It follows from (2.21), (2.22) and the dominated convergence theorem that
On the other hand, for any , we can choose a positive integer such that
And then, choose a positive integer such that
From (2.24) and (2.25), we get
hence
It follows from (2.20), (2.23), and (2.51) that , and the continuity of is proved.

Finally, we prove that is compact. Let be bounded and (). Consider for any fixed . By (2.5) and (2.8), we have
which implies that the functions () defined by
( denotes the right limit of at ) are equicontinuous on . On the other hand, for any , choose a sufficiently large and a sufficiently large positive integer such that
We have, by (2.29), (2.5), (2.8), (2.30), and condition ,
It follows from (2.31), (2.32), (2.33), (2.8), and [4, Theorem 1.2.3] that
where , , , and denotes the Kuratowski measure of noncompactness of bounded set (see [4, Section 1.2]). Since for , where , we see that, by condition (),
It follows from (2.34)–(2.36) that
which implies by virtue of the arbitrariness of that for .

By Ascoli-Arzela theorem (see [4, Theorem 1.2.5]), we conclude that is relatively compact in ; hence, has a subsequence which is convergent uniformly on , so, has a subsequence which is convergent uniformly on . Since may be any positive integer, so, by diagonal method, we can choose a subsequence of such that is convergent uniformly on each (). Let
It is clear that . By (2.14), we have
which implies that and
Let be arbitrarily given and choose a sufficiently large positive number such that
For any , we have, by (2.5),
which implies by virtue of (2.8), condition and (2.41) that
Letting in (2.43), we get
On the other hand, since converges uniformly to on as , there exists a positive integer such that
It follows from (2.43)–(2.45) that
By (2.45) and (2.46), we have
hence as , and the compactness of is proved.

Lemma 2.4. *Let conditions ()–() be satisfied. Then is a solution of IBVP(1.5) if and only if is a solution of the following impulsive integral equation:
**
that is, is a fixed point of operator defined by (2.5) in .*

*Proof. *For , it is easy to get the following formula:

Let be a solution of IBVP(1.5). By (1.5) and (2.49), we have
We have shown in the proof of Lemma 2.3 that the infinite integral (2.9) and the infinite series (2.11) are convergent, so, by taking limits as in both sides of (2.50), we get
On the other hand, by (1.5) and (2.50), we have
It follows from (2.51)–(2.53) that
and, substituting it into (2.50), we see that satisfies (2.48), that is, . Since by virtue of Lemma 2.3, we conclude that .

Conversely, assume that is a solution of (2.48). We have, by (2.48),
Moreover, by taking limits as in (2.33), we see that exists and
It follows from (2.55)–(2.57) that
On the other hand, direct differentiation of (2.48) gives
and, it is clear, by (2.48),
Hence, and satisfies (1.5).

Corollary 2.5. *Let cone be normal. If is a fixed point of operator defined by (1.5) in and , then for , so, is a positive solution of IBVP(1.5).*

*Proof. *For , we have
so,
where denotes the normal constant of . Since , (2.61) and (2.62) imply that for .

Lemma 2.6 (Fixed point theorem of cone expansion and compression with norm type, see [3, Theorem 3] or [1, Theorem 2.3.4]). * Let be a cone in real Banach space and two bounded open sets in such that , , where denotes the zero element of and denotes the closure of . Let operator be completely continuous. Suppose that one of the following two conditions is satisfied: *(a)*
where denotes the boundary of ().*(b)*Then has at least one fixed point in .*

#### 3. Main Theorems

Let us list more conditions.() There exist , and such that

*Remark 3.1. *Condition () means that is superlinear with respect to .() There exist , and such that

Theorem 3.2. *Let cone be normal and conditions ()–() satisfied. Assume that there exists a such that
**
where
**
(for and , see conditions () and ()). Then IBVP(1.5) has at least two positive solutions such that .*

*Proof. *By Lemmas 2.3, 2.4, and Corollary 2.5, operator defined by (2.5) is completely continuous from into , and we need to prove that has two fixed points and in such that .

By condition (), there exists an such that
where denotes the normal constant of , so,
Choose
For ; we have by (2.62) and (3.7),
so, (2.5), (3.8), (3.6), and (2.62) imply
and consequently,
Similarly, by condition (), there exists such that
so,
Choose
For , we have by (3.13) and (2.62),
so, similar to (3.9), we get by (2.5), (3.12), and (3.14)
hence
On the other hand, for , , by condition (), condition (), (3.4), we have
It is clear that
It follows from (3.17)–(3.19) that
Thus, (3.20) and (3.3) imply
From (3.7) and (3.13), we know ; hence, (3.10), (3.16), (3.21), and Lemma 2.6 imply that has two fixed points such that . The proof is complete.

Theorem 3.3. *Let cone be normal and conditions ()–() satisfied. Assume that
**
(for and , see conditions () and ()). Then IBVP(1.5) has at least one positive solution .*

*Proof. *As in the proof of Theorem 3.2, we can choose such that (3.10) holds (in this case, we only choose instead of (3.7)). On the other hand, by (3.22), there exists such that
where
Choose
For , we have by (2.62) and (3.25),
so, (3.23) imply
It follows from (3.19), condition , condition , (3.27), and (3.24) that
and consequently,
Since by virtue of (3.25), we conclude from (3.10), (3.29), and Lemma 2.6 that has a fixed point such that . The theorem is proved.

*Example 3.4. *Consider the infinite system of scalar first-order impulsive integrodifferential equations of mixed type on the half line:
Evidently, is the trivial solution of infinite system (3.30).

*Conclusion. *Infinite system (3.30) has at least two positive solutions () and () such that

*Proof. *Let with norm and . Then is a normal cone in with normal constant , and infinite system (3.30) can be regarded as an infinite three-point boundary value problem of form (1.5). In this situation, , , , , , , , , , , and , in which
It is easy to see that , (