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Positive Solutions of Singular Initial-Boundary Value Problems to Second-Order Functional Differential Equations
Boundary Value Problems volume 2008, Article number: 457028 (2008)
Abstract
Positive solutions to the singular initial-boundary value problems are obtained by applying the Schauder fixed-point theorem, where
on
and
may be singular at
and
. As an application, an example is given to demonstrate our result.
1. Introduction
Recently, in [1–4], Erbe, Kong, Jiang, Wang, and Weng considered the following singular functional differential equations:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ1_HTML.gif)
where and the existence of positive solutions to (1.1) is obtained. When
in (1.1), Agarwal and O'Regan in [5], Lin and Xu in [6] discussed the existence of positive solutions to (1.1) also. We notice that the nonlinearities
in all the above-mentioned references depend on
.
The more difficult case is that the term depends on
for second-order functional differential equations with delay. When
has no singularity at
and
, there are many results on the following (1.2) (see [7–9] and references therein). Up to now, to our knowledge, there are fewer results on (1.2) when the term
is allowed to possess singularity for the term
at
and
, which is of more actual significance.
In this paper, motivated by above results, we consider the second-order initial-boundary value problems:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ2_HTML.gif)
where . By Leray-Schauder fixed-point theorem, the existence of positive solutions to (1.2) is obtained when
is singular at
and
.
For and
, let
and
. Then,
and
are Banach spaces. Let
and
. Obviously,
and
are cones in
and
respectively. Now, we give a new definition.
Deffinition.
is said to be singular at
for
when
satisfies
for
and
is said to be singular at
for
when
satisfies
for
.
And one defines some functions which one has to use in this paper.
Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ3_HTML.gif)
where is a Green's function. It is clear that
for
and
on
We now introduce the definition of a solution to IBVP(1.2).
Deffinition.
A function is said to be a solution to IBVP(1.2) if it satisfies the following conditions:
(1) is continuous and nonnegative on
;
(2);
(3) and
exist on
;
(4) is Lebesgue integrable on
;
(5) for
.
Furthermore, a solution is said to be positive if
on
.
Let be a solution to IBVP(1.2). Then, it can be represented as
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ4_HTML.gif)
It is clear that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ5_HTML.gif)
for all solutions, , to IBVP(1.2), where
. For
, let
on
throughout this paper. Obviously,
and
for all
.
Throughout this paper, we assume the following hypotheses hold.
(H1) is continuous on
.
(H2) There exists , such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ6_HTML.gif)
Lemma 1.3.
Assume that (H 1 )-(H 2 ) hold, then there exists a , such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ7_HTML.gif)
for all solutions, , to (1.2).
Proof.
Suppose that the claim is false. (1.5) guarantees that there exists a sequence of solutions to IBVP(1.2) such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ8_HTML.gif)
Without loss of generality, we may assume that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ9_HTML.gif)
From and (1.5), it follows that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ10_HTML.gif)
which contradicts the assumption that and hence the claim is true provided
is suitably small.
Remark 1.4.
The following inequality
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ11_HTML.gif)
holds provided that is sufficiently small, where
is in Lemma 1.3.
There exist a nonnegative continuous function
defined on (0,1) and two nonnegative continuous functions
defined on, respectively,
, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ12_HTML.gif)
where and
) satisfy
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ13_HTML.gif)
Furthermore, is nonincreasing and
is nondecreasing, that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ14_HTML.gif)
Lemma 1.5 (see [7]).
Let be the Banach space and let X be any nonempty, convex, closed, and bounded subset of
. If
is a continuous mapping of
into itself and
is relatively compact, then the mapping
has at least one fixed point (i.e., there exists an
with
) .
Using Lemma 1.5, we present the existence of at least one positive solution to (1.2) when is singular at
and
(notice the new Definition 1.1). To some extent, our paper complements and generalizes these in [1–6, 8–10].
2. Main Results
Theorem 2.1.
Assume that (H 1 )–(H 3 ) hold. Then, the IBVP( 1.2 ) has at least one positive solution.
Proof.
Since , we can choose an
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ15_HTML.gif)
where the positive number satisfies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ16_HTML.gif)
Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ17_HTML.gif)
For each , we define
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ18_HTML.gif)
It is obvious that satisfies the hypotheses
and
.
We now consider the modified initial-boundary value problem:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ19_HTML.gif)
We claim that for all solutions, , to IBVP(2.5),
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ20_HTML.gif)
Suppose that the claim is false. Then there exists such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ21_HTML.gif)
Since on
, there are the following three cases.
Case 1.
for all
.
The solution of IBVP(2.5) can be represented as (notice Remark 1.4)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ22_HTML.gif)
which contradicts (2.7).
Case 2.
There exists a such that
and
.
In this case, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ23_HTML.gif)
which contradicts (2.7).
Case 3.
There exists a such that
and
.
From (1.5), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ24_HTML.gif)
which contradicts (2.7).
So we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ25_HTML.gif)
To prove the existence of positive solutions to IBVP(2.5), we seek to transform (2.5) into an integral equation via the use of Green's function and then find a positive solution by using Lemma 1.5.
Define a nonempty convex and closed subset of by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ26_HTML.gif)
Then, we define an operator by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ27_HTML.gif)
From and the definition of
, we have, for every
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ28_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ29_HTML.gif)
Together with the definition of , we get
.
Also,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ30_HTML.gif)
is continuous in (0,1), and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ31_HTML.gif)
From and (2.15), we can get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ32_HTML.gif)
which implies that is integrable on
.
Now, we claim that is equicontinuous on
. We will prove the claim. For any
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ33_HTML.gif)
Since is continuous on
and
, then for any
, there is a
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ34_HTML.gif)
By (2.6), we have, for
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ35_HTML.gif)
where is a constant number.
Put , then for
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ36_HTML.gif)
Set . Then for
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ37_HTML.gif)
Since on
, the above inequality holds for
.
Thus, is a relative compact subset of
. That is,
is a compact operator.
We are now going to prove that the mapping is continuous on
.
Let be arbitrarily chosen and let
converge to
uniformly on
as
. Now, we claim that
converge to
uniformly as
. From the definition of
, we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ38_HTML.gif)
Thus,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ39_HTML.gif)
that is, the claim is true.
Since is continuous with respect to
for
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ40_HTML.gif)
for each fixed . From the definition of
and
, we know that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ41_HTML.gif)
and hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ42_HTML.gif)
where is a Lebesgue integrable function defined on
because of
. Consequently, we apply the dominated convergence theorem to get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ43_HTML.gif)
which shows that the mapping is continuous on
.
Then from Lemma 1.5, we get that there exists at least one positive solution, , to IBVP(2.5) in
. The solution can be represented by (1.4), where
is replaced with
. So, (2.6) holds. Furthermore, from the definition of
, we can get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ44_HTML.gif)
Thus, the solution of IBVP(2.5) is also the one of (1.2). The proof is complete.
3. Application
Example 3.1.
Consider the singular IBVP(3.1):
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ45_HTML.gif)
where .
4. Conclusion
Equation (3.1) has at least one positive solution.
Now, we will check that hold in (3.1).
In IBVP(3.1), . It is clear that
is continuous and singular at
and
. For
we choose
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ46_HTML.gif)
when ; by simple computation, we can get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ47_HTML.gif)
It is obvious that is nonincreasing and
is nondecreasing.
Now, we check . For any
,
(notice the definition of
), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ48_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ49_HTML.gif)
We define
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ50_HTML.gif)
Now, we will prove that there exists such that
is decreasing on
.
Obviously,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ51_HTML.gif)
Put , then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ52_HTML.gif)
From the continuity of , we can find
such that
on
. Then,
on
. That is,
is decreasing on
.
Furthermore, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ53_HTML.gif)
Thus,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2008%2F457028/MediaObjects/13661_2007_Article_806_Equ54_HTML.gif)
which implies that holds.
So, from Theorem 2.1, IBVP(3.1) has at least one positive solution.
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Acknowledgments
The research was supported by NNSF of China (10571111) and the fund of Shandong Education Committee (J07WH08).
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Jin, F., Yan, B. Positive Solutions of Singular Initial-Boundary Value Problems to Second-Order Functional Differential Equations. Bound Value Probl 2008, 457028 (2008). https://doi.org/10.1155/2008/457028
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DOI: https://doi.org/10.1155/2008/457028