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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:

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:

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

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

It is clear that

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

Lemma 1.3.
Assume that (H 1 )-(H 2 ) hold, then there exists a , such that

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

Without loss of generality, we may assume that

From and (1.5), it follows that

which contradicts the assumption that and hence the claim is true provided
is suitably small.
Remark 1.4.
The following inequality

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

where and
) satisfy

Furthermore, is nonincreasing and
is nondecreasing, that is,

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

where the positive number satisfies

Let

For each , we define
by

It is obvious that satisfies the hypotheses
and
.
We now consider the modified initial-boundary value problem:

We claim that for all solutions, , to IBVP(2.5),

Suppose that the claim is false. Then there exists such that

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)

which contradicts (2.7).
Case 2.
There exists a such that
and
.
In this case, we have

which contradicts (2.7).
Case 3.
There exists a such that
and
.
From (1.5), we get

which contradicts (2.7).
So we have

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

Then, we define an operator by

From and the definition of
, we have, for every


Together with the definition of , we get
.
Also,

is continuous in (0,1), and

From and (2.15), we can get

which implies that is integrable on
.
Now, we claim that is equicontinuous on
. We will prove the claim. For any
, we have

Since is continuous on
and
, then for any
, there is a
such that

By (2.6), we have, for

where is a constant number.
Put , then for

Set . Then for

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

Thus,

that is, the claim is true.
Since is continuous with respect to
for
, we have

for each fixed . From the definition of
and
, we know that

and hence

where is a Lebesgue integrable function defined on
because of
. Consequently, we apply the dominated convergence theorem to get

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

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):

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

when ; by simple computation, we can get

It is obvious that is nonincreasing and
is nondecreasing.
Now, we check . For any
,
(notice the definition of
), we have


We define

Now, we will prove that there exists such that
is decreasing on
.
Obviously,

Put , then

From the continuity of , we can find
such that
on
. Then,
on
. That is,
is decreasing on
.
Furthermore, we have

Thus,

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
Keywords
- Differential Equation
- Continuous Function
- Integral Equation
- Partial Differential Equation
- Ordinary Differential Equation