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Fixed-point theorems for nonlinear operators with singular perturbations and applications
Journal of Inequalities and Applications volume 2014, Article number: 37 (2014)
Abstract
In this paper, using fixed-point index theory and approximation techniques, we consider the existence and multiplicity of fixed points of some nonlinear operators with singular perturbation. As an application we consider the existence and multiplicity of positive solutions of singular systems of multi-point boundary value problems, which improve the results in the literature.
1 Introduction
In this paper we consider the problem
where A is continuous and compact and B is a singular continuous and compact operator (defined in Section 2).
In the study of nonlinear phenomena many models give rise to singular boundary value problems (singular in the dependent variable) (see [1–3]). In [4], Taliaferro showed that the singular boundary value problem
has a solution; here , with on and . For more recent work we refer the reader to [5–14] and the references therein.
In this paper we consider abstract singular operators (defined in Section 2) and we consider the existence and multiplicity of fixed points of some nonlinear operators with singular perturbations. As an application we discuss the existence and multiplicity of positive solutions of singular systems of multi-point boundary value problems.
2 Fixed-point theorems
Let E be a Banach space, P a cone of E, bounded and open. The following theorems are needed in our paper.
Theorem 2.1 ([8])
Suppose , is continuous and compact and
Then
Theorem 2.2 ([8])
Assume that is continuous and compact. If there exists a compact and continuous operator such that
-
(1)
;
-
(2)
, , ,
then
Now we give a new definition.
Definition 2.1 If is continuous with
and is relatively compact, for any , then is called a singular continuous and compact operator.
Remark Consider
where and or equivalently
where
Set
where . For , let
It is easy to see that is a singular continuous and compact operator (see [7, 14]).
Theorem 2.3 Suppose that , is continuous and compact and is singular continuous and compact. Assume that
Then there exists a such that, for any , there exist with
Proof Choose , and define
Set
Now we claim that
If , there exists such that
First, we show is bounded.
To see this suppose is unbounded. Without loss of generality, we assume that . Then
and this is a contradiction.
Next, we show that there exists a such that
The boundedness of means that has a convergent subsequence. Without loss of generality, we assume that . Since is bounded and A is continuous and compact, has a convergent subsequence with . From (2.3), we have
which implies that
Then
Let . Clearly, and
which contradicts (2.1).
Let
Now we claim that
To see this suppose that
Without loss of generality, assume that
which implies that there exists a sequence such that
For all , we have
Since, for any , is relatively compact, (2.5) guarantees that there exists a subsequence such that
Thus
which implies that . This contradicts . Hence, (2.4) holds.
Set
and
For , , , we have
Theorem 2.1 guarantees that
Note (2.6) guarantees that, for any , there exists a such that
Now we show that
which implies that
To see this suppose that
Then there exists a such that
and so
Thus
The compactness of A guarantees that has a convergent subsequence. Without loss of generality, we assume that . From (2.10), we have
which contradicts (2.9).
Now (2.8) guarantees that
Then has a convergent subsequence. Without loss of generality, we assume that
Then
Now (2.8) guarantees that . Letting in (2.7), and we have
and . The proof is complete. □
Corollary 2.1 Suppose that , is continuous and compact and is singular continuous and compact. Assume that
or
Then there exists a such that, for any , there exist with
It is easy to see that (2.12) or (2.13) guarantees that (2.1) holds (see [8]).
Theorem 2.4 Suppose that , are bounded open sets and , are continuous and compact and is singular continuous and compact. Assume that
(C1) , , ;
(C2) ;
(C3) , , .
Then there exists a such that, for any , there exist and with
Proof Choose , and define
Set
and
We claim that
An argument similar to that in (2.2) shows that
Now we show that
To see this suppose that . Then there exists such that
Now since
we have is bounded, which means that has a convergent subsequence. Without loss of generality, we assume that
Since is bounded and A and K are compact, and have convergent subsequences and with and . Now
which implies that
Let . Clearly, . Now
which contradicts condition (C3).
Consequently, (2.16) is true, which together with (2.15) yields (2.14).
Let . Obviously, .
Let
An argument similar to that in (2.4) shows that
Let
and
For , , , we have
which guarantees that
and for , , we have
which guarantees that
Thus
Now (2.18) and (2.19) guarantee that there exist and such that
An argument similar to that in (2.11) shows that there exist and with
The proof is complete. □
Corollary 2.2 Suppose that , are bounded open sets and , is continuous and compact and is singular continuous and compact. Assume that
(C4)
or
(C5)
or such that
or
Then there exists a such that, for any , there exist and with
and
It is easy to see that (C4) and (C5) guarantee that (C1)-(C3) hold (see [8]).
3 Applications for singular systems of multi-point boundary value problems
In [9], Henderson and Luca considered the system of nonlinear second-order ordinary differential equations
with multi-point boundary conditions
The following conditions come from [9]:
(H1) , , , , , , , , , ,
(H2) we have the functions and , for all ,
(H3) there exists a positive constant such that
-
(1)
;
-
(2)
,
(H4) there exists a such that
-
(1)
;
-
(2)
,
(H5)
-
(1)
;
-
(2)
,
(H6) for each , and are nondecreasing with respect to u, and there exists a constant such that
where and , are defined in [9].
Theorem 3.1 ([9])
Assume that (H1)-(H2) and (H4)-(H5) hold. Then Problem (3.1), (3.2) has at least one positive solution , .
Theorem 3.2 ([9])
Assume that (H1)-(H3) and (H5)-(H6) hold. Then Problem (3.1), (3.2) has at least two positive solutions , , .
Here we consider
with multi-point boundary conditions (3.2), where .
Let with norm
Obviously, is a Banach space. Let
where and γ are defined in Section 2 in [9].
For , define an operator
and for , define an operator
where and are defined in [9].
It is easy to see that is a singular continuous and compact operator (see [9, 11]).
Theorem 3.3 Assume that (H1)-(H2) and (H4) hold. Then there exists a such that Problem (3.3), (3.2) has at least one positive solution , for all .
Proof Let be defined as that in Theorem 3.2 of [9]. From the proof in [9], it is easy to see that
Now Corollary 2.1 guarantees that there exists a such that, for any , there exist such that
Let
Then is a positive solution for (3.3), (3.2). The proof is complete. □
Theorem 3.4 Assume that (H1)-(H3) and (H6) hold. Then Problem (3.3), (3.2) has at least two positive solutions , , .
Proof Let and be defined as in Theorem 3.3 of [9]. From the proof in [9], it is easy to see that
Now Corollary 2.2 guarantees that there exists a , for any , such that there exist and with
and
Let
and
Then and are two positive solutions for (3.3), (3.2). The proof is complete. □
Remark Note that f and g have no singularity at and in [9, 10], so Theorems 3.3 and 3.4 improve the results in [9, 10].
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Acknowledgements
This research is supported by Young Award of Shandong Province (ZR2013AQ008).
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Yan, B., O’Regan, D. & Agarwal, R.P. Fixed-point theorems for nonlinear operators with singular perturbations and applications. J Inequal Appl 2014, 37 (2014). https://doi.org/10.1186/1029-242X-2014-37
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DOI: https://doi.org/10.1186/1029-242X-2014-37