Abstract
In this paper, the author discusses the multiple positive solutions for an infinite threepoint boundary value problem of firstorder impulsive superlinear singular integrodifferential equations on the half line in a Banach space by means of the fixedpoint theorem of cone expansion and compression with norm type.
MSC: 45J05, 34G20, 47H10.
Keywords:
impulsive singular integrodifferential equation in a Banach space; infinite threepoint boundary value problem; fixedpoint theorem of cone expansion and compression with norm type1 Introduction
In recent years, multiple solutions of boundary value problems for impulsive differential equations in scalar spaces had been extensively studied (see, for example, [13]). In recent papers [4] and [5], Professor D. Guo discussed two infinite boundary value problems for nthorder impulsive nonlinear singular integrodifferential equations of mixed type on the half line in a Banach space. By constructing a bounded closed convex set, apart from the singularities, and using the Schauder fixedpoint theorem, he obtained the existence of positive solutions for the infinite boundary value problems. But such equations are sublinear, and there are no results on existence of two positive solutions. Now, in this paper, we shall discuss the existence of two positive solutions for firstorder superlinear singular equations by means of a different method, i.e., by using the fixedpoint theorem of cone expansion and compression with norm type (see [6,7]), and the key point is to introduce a new cone Q.
Let E be a real Banach space and P be a cone in E, which defines a partial ordering in E by if and only if . P is said to be normal if there exists a positive constant N such that implies , where θ denotes the zero element of E, and the smallest N is called the normal constant of P. If and , we write . Let , i.e., . For details on cone theory, see [7].
Consider the infinite threepoint boundary value problem for a firstorder impulsive nonlinear singular integrodifferential equation of mixed type on the half line in E:
where , , , , , , (), , , (for some m), and
, , , denotes the set of all nonnegative numbers. denotes the jump of at , i.e.,
where and represent the right and left limits of at , respectively. In the following, we always assume that
and
(where means , ), i.e., is singular at and . We also assume that
Let = { is a map from J into E 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 Q are two cones in space and . is called a positive solution of the infinite threepoint boundary value problem (1) if for and satisfies (1). Let and for .
2 Several lemmas
Let us list some conditions.
In this case, let
and
() There exist () and such that
and
() For any and , and () are relatively compact in E, where and .
Remark Obviously, condition () is satisfied automatically when E is finite dimensional.
Remark It is clear: If condition () is satisfied, then the operators T and S defined by (2) are bounded linear operators from into and , ; moreover, we have and .
We shall reduce the infinite threepoint boundary value problem (1) to an impulsive integral equation. To this end, we consider the operator A defined by
In what follows, we write , ().
Lemma 1Let conePbe normal and conditions ()() be satisfied. Then operatorAdefined by (6) is a continuous operator fromintoQ; moreover, for any, is relatively compact.
so,
where N denotes the normal constant of cone P, and consequently,
By condition () and (8), we have
where
which implies the convergence of the infinite integral
and
On the other hand, by condition () and (8), we have
where
which implies the convergence of the infinite series
and
It follows from (6), (11), and (14) that
Moreover, by (6), we have
and
It is clear,
so, (17) and (18) imply
It follows from (16) and (19) that
Now, we are going to show that A is continuous. Let , (). Write () and we may assume that
So, by (7),
and
Similar to (15), it is easy to get
It is clear that
and, similar to (9) and observing (21) and (22), we have
where
It follows from (24), (25), and the dominated convergence theorem that
On the other hand, for any , we can choose a positive integer j such that
where
And then, choose an positive integer such that
From (27), (28), and observing condition () and (21), (22), we get
hence,
It follows from (23), (26), and (29) that as , and the continuity of A is proved.
Finally, we prove that is relatively compact, where are arbitrarily given. Let (). Then, by (7),
Similar to (9), (12), (15), and observing (30), we have
and
where
and
Consider for any fixed i. By (6) and (31), 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 (35), (6), (31), (32), and (36),
and
It follows from (37), (38), (39), and ([8], Theorem 1.2.3) that
where , , , and denotes the Kuratowski measure of noncompactness of bounded set (see [[8], Section 1.2]). Since and for , where and , we see that, by condition (),
and
It follows from (40) to (42) that
which implies by virtue of the arbitrariness of ϵ that for .
By the AscoliArzela theorem (see [[8], 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 i 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 (33), we have
Let be arbitrarily given and choose a sufficiently large positive number τ such that
which implies by virtue of (31), (32), and (43) that
On the other hand, since converges uniformly to on as , there exists a positive integer such that
It follows from (44) to (46) that
By (46) and (47), we have
hence, as , and the relative compactness of is proved. □
Lemma 2Let conePbe normal and conditions ()() be satisfied. Thenis a positive solution of the infinite threepoint boundary value problem (1) if and only ifis a solution of the following impulsive integral equation:
i.e., uis a fixed point of operatorAdefined by (6) in.
Proof For , it is easy to get the following formula:
Let be a positive solution of the infinite threepoint boundary value problem (1). By (1) and (49), we have
We have shown in the proof of Lemma 1 that the infinite integral (10) and the infinite series (13) are convergent, so, by taking limits as in both sides of (50), we get
On the other hand, by (1) and (50), we have
and
It follows from (51) to (53) that
and, substituting it into (50), we see that satisfies equation (48), i.e., .
Conversely, assume that is a solution of Equation (48). We have, by (48),
and
Moreover, by taking limits as in (48), we see that exists and
It follows from (54) to (56) that
On the other hand, direct differentiation of (48) gives
and, it is clear, by (48),
Hence, and satisfies (1). Since , so (7) holds and , hence for . Consequently, is a positive solution of the infinite threepoint boundary value problem (1). □
Lemma 3 (The fixedpoint theorem of cone expansion and compression with norm type; see [[6], Theorem 3] or [[7], Theorem 2.3.4])
LetPbe a cone in real Banach spaceEand, be two bounded open sets inEsuch that, , and operatorbe completely continuous, whereθdenotes the zero element ofEanddenotes the closure of (). Suppose that one of the following two conditions is satisfied:
wheredenotes the boundary of ().
ThenAhas at least one fixed point in.
Remark 1 Lemma 3 is different from the Krasnoselskii fixedpoint theorem of cone expansion and compression (see [[9], Theorem 44.1]). In Krasnoselskii’s theorem, the condition corresponding to (a) is
It is clear, conditions (a) and (a′) are independent each other. On the other hand, in Krasnoselskii’s theorem, and are balls with center θ.
3 Main theorems
Let us list more conditions.
() There exist , and such that
and
and
Remark 2 Condition () means that is superlinear with respect to u.
() There exist , and such that
and
and
Theorem 1Let conePbe normal and conditions ()() be satisfied. Assume that there exists asuch that
whereNdenotes the normal constant ofP, and
(for, , and; see conditions () and ()). Then the infinite threepoint boundary value problem (1) has at least two positive solutionssuch that.
Proof By Lemma 1 and Lemma 2, operator A defined by (6) is continuous from into Q and we need to prove that A has two fixed points and in such that .
By condition (), there exists a such that
so,
Choose
For , , we have by (7) and (62),
so, (6), (63), (61), and (7) imply
and consequently,
By condition (), there exists such that
so,
Choose
For , , we have by (68) and (7),
so, we get by (6), (69), and (67),
which implies
and consequently,
On the other hand, for , , by condition (), condition (), (58), and (59), we have
and
It is clear, by (17),
It follows from (71) to (73) that
Thus, (74) and (57) imply
From (62) and (68), we know , and by Lemma 1, is completely continuous, where , hence, (65), (70), (75), and Lemma 3 imply that A has two fixed points such that . The proof is complete. □
Theorem 2Let conePbe normal and conditions ()() and () be satisfied. Assume that
(forand, see conditions () and ()). Then the infinite threepoint boundary value problem (1) has at least one positive solution.
Proof As in the proof of Theorem 1, we can choose such that (70) holds (in this case, we put in (66) and (68)). On the other hand, by (76) and (77), there exists such that
and
where
Choose
For , , we have by (7) and (81),
so, (78) and (79) imply
and
It follows from (73), conditions (), condition (), (82), (83), and (80) that
and consequently,
Since by virtue of (81), we conclude from (70), (84), and Lemma 3 that A has a fixed point such that . The theorem is proved. □
Example 1
Consider the infinite system of scalar firstorder impulsive singular integrodifferential equations of mixed type on the half line:
Conclusion Infinite system (85) has at least two positive solutions () and () such that
Proof Let with norm and . Then P is a normal cone in E with normal constant , and infinite system (85) can be regarded as an infinite threepoint boundary value problem of form (1). In this situation, , , , (), , , , , , and , in which
and
It is easy to see that , () and condition () is satisfied and , . We have, by (86),
so, observing the inequality , we get
which implies that condition () is satisfied for
and
with
By (87), we have
so,
which implies that condition () is satisfied for and
On the other hand, (86) implies
and
so, we see that condition () is satisfied for (), and and condition () is satisfied for (), and . In addition, from (90), we have
which implies that (3) and (4) hold, i.e., is singular at and . Moreover, from (87), we get
and so,
which implies that (5) holds, i.e., () are singular at . Now, we check that condition () is satisfied. Let and be fixed, and be any sequence in , where . Then, we have, by (86) and (88),
So, is bounded, and, by diagonal method, we can choose a subsequence such that
which implies by virtue of (91) that
Consequently, . Let be given. Choose a positive integer such that
By (92), we see that there exists a positive integer such that
It follows from (91) to (95) that
hence, in E as . Thus, we have proved that is relatively compact in E. Similarly, by using (89), we can prove that is relatively compact in E. Hence, condition () is satisfied. Finally, we check that inequality (57) is satisfied for . In this case,
and
so,
i.e., inequality (57) is satisfied for . Hence, our conclusion follows from Theorem 1. □
Example 2
Consider the infinite system of scalar first order impulsive singular integrodifferential equations of mixed type on the half line:
Conclusion Infinite system (96) has at least one positive solution () such that
Proof Let with norm and . Then P is a normal cone in E with normal constant , and infinite system (96) can be regarded as an infinite threepoint boundary value problem of form (1) in E. In this situation, , , , (), , , , , , and , in which
and
It is clear that , () and condition () is satisfied and , . We have, by (97) and (98),
and
so, observing
we get
and
which imply that conditions () is satisfied for
and
with
and () is satisfied for () and
By (97), we have
so, condition () is satisfied for
so, (3) and (4) are satisfied, i.e., is singular at and . Similarly, (98) implies
so, (5) is satisfied, i.e., () are singular at . Similar to the discussion in Example 1, we can prove that and (for fixed and ; ) are relatively compact in , so, condition () is satisfied. On the other hand, we have
so, (76) is satisfied. Moreover, it is clear that (77) is satisfied. Hence, our conclusion follows from Theorem 2. □
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors typed, read, and approved the final manuscript.
Acknowledgements
The author would like to thank Professor D. Guo for his valuable suggestions.
References

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