## Boundary Value Problems

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# Existence of Positive Solutions of a Singular Nonlinear Boundary Value Problem

Boundary Value Problems20102010:458015

https://doi.org/10.1155/2010/458015

Accepted: 11 August 2010

Published: 18 August 2010

## Abstract

We are concerned with the existence of positive solutions of singular second-order boundary value problem , , , which is not necessarily linearizable. Here, nonlinearity is allowed to have singularities at . The proof of our main result is based upon topological degree theory and global bifurcation techniques.

## 1. Introduction

Existence and multiplicity of solutions of singular problem
(1.1)
where is allowed to have singularities at and , have been studied by several authors, see Asakawa [1], Agarwal and O Regan [2], O Regan [3], Habets and Zanolin [4], Xu and Ma [5], Yang [6], and the references therein. The main tools in [16] are the method of lower and upper solutions, Leray-Schauder continuation theorem, and the fixed point index theory in cones. Recently, Ma [7] studied the existence of nodal solutions of the singular boundary value problem
(1.2)

by applying Rabinowitz's global bifurcation theorem, where is allowed to have singularities at and is linearizable at as well as at . It is the purpose of this paper to study the existence of positive solutions of (1.1), which is not necessarily linearizable.

Let be Banach space defined by
(1.3)
with the norm
(1.4)
Let
(1.5)

Definition 1.1.

A function is said to be an -Carathéodory function if it satisfies the following:

(i)for each , is measurable;

(ii)for a.e. , is continuous;

(iii)for any , there exists such that

(1.6)

In this paper, we will prove the existence of positive solutions of (1.1) by using the global bifurcation techniques under the following assumptions.

(H1) Let be an -Carathéodory function and there exist functions , , , and such that

(1.7)
for some -Carathéodory functions defined on with
(1.8)
uniformly for a.e. , and
(1.9)
for some -Carathéodory functions defined on with
(1.10)

uniformly for a.e. .

(H2) for a.e. and .

(H3) There exists function such that
(1.11)

Remark 1.2.

If , , , and , then (1.8) implies that
(1.12)
and (1.10) implies that
(1.13)

The main tool we will use is the following global bifurcation theorem for problem which is not necessarily linearizable.

Theorem A (Rabinowitz, [8]).

Let be a real reflexive Banach space. Let be completely continuous, such that , . Let , such that is an isolated solution of the following equation:
(1.14)
for and , where , are not bifurcation points of (1.14). Furthermore, assume that
(1.15)
where is an isolating neighborhood of the trivial solution. Let
(1.16)

then there exists a continuum (i.e., a closed connected set) of containing , and either

(i) is unbounded in , or

(ii) .

To state our main results, we need the following.

Lemma 1.3 (see [1, Proposition ]).

Let , then the eigenvalue problem
(1.17)
has a sequence of eigenvalues as follows:
(1.18)

Moreover, for each , is simple and its eigenfunction has exactly zeros in .

Remark 1.4.

Note that and for each . Therefore, there exist constants , such that
(1.19)

Our main result is the following.

Theorem 1.5.

Let (H1)–(H3) hold. Assume that either
(1.20)
or
(1.21)

then (1.1) has at least one positive solution.

Remark 1.6.

For other references related to this topic, see [914] and the references therein.

## 2. Preliminary Results

Lemma 2.1 (see [15, Proposition ]).

For any , the linear problem
(2.1)
has a unique solution and , such that
(2.2)
where
(2.3)
Furthermore, if , then
(2.4)
Let be the Banach space with the norm , and
(2.5)
Let be an operator defined by
(2.6)
where
(2.7)

Then, from Lemma 2.1, is well defined.

Lemma 2.2.

Let and be the first eigenfunction of (1.17). Then for all , one has
(2.8)

Proof.

For any , integrating by parts, we have
(2.9)
Since and , then
(2.10)
Therefore, we only need to prove that
(2.11)
Let us deal with the first equality, the second one can be treated by the same way. Note that , then
(2.12)
which implies that . Then is bounded on . Now, we claim that
(2.13)
Suppose on the contrary that , then for small enough, we have
(2.14)
Therefore,
(2.15)
which is a contradiction. Combining (1.19) with (2.13), we have
(2.16)

This completes the proof.

Remark 2.3.

Under the conditions of Lemma 2.2, for the later convenience, (2.8) is equivalent to
(2.17)

Lemma 2.4 (see [1, Lemma ]).

For every , the subset defined by
(2.18)

is precompact in .

Let be the closure of the set of positive solutions of the problem
(2.19)
We extend the function to an -Carathéodory function defined on by
(2.20)
Then for and a.e. . For , let be an arbitrary solution of the problem
(2.21)

Since for a.e. , Lemma 2.2 yields for . Thus, is a nonnegative solution of (2.19), and the closure of the set of nontrivial solutions of (2.21) in is exactly .

Let be an -Carathéodory function. Let be the Nemytskii operator associated with the function as follows:
(2.22)

Lemma 2.5.

Let on . Let be such that in , . Then,
(2.23)

Moreover, , whenever .

Let be the Nemytskii operator associated with the function as follows:
(2.24)
Then (2.21), with , is equivalent to the operator equation
(2.25)
that is,
(2.26)

Lemma 2.6.

Let (H1) and (H2) hold. Then the operator is completely continuous.

Proof.

From (1.10) in (H1), there exists , such that, for a.e. and ,
(2.27)
Since is an -Carathéodory function, then there exists , such that, for a.e. and , . Therefore, for a.e. and , we have
(2.28)
For convenience, let . We first show that is continuous. Suppose that in as . Clearly, as for a.e. and there exists such that for every . It is easy to see that
(2.29)

By the Lebesgue dominated convergence theorem, we have that in as . Thus, is continuous.

Let be a bounded set in . Lemma 2.4 together with (2.28) shows that is precompact in . Therefore, is completely continuous.

In the following, we will apply the Leray-Schauder degree theory mainly to the mapping ,
(2.30)

For , let , let denote the degree of on with respect to .

Lemma 2.7.

Let be a compact interval with , then there exists a number with the property
(2.31)

Proof.

Suppose to the contrary that there exist sequences and in in , such that for all , then, in .

Set . Then and . Now, from condition (H1), we have the following:
(2.32)
and accordingly
(2.33)
Let and denote the nonnegative eigenfunctions corresponding to and , respectively, then we have from the first inequality in (2.33) that
(2.34)
From Lemma 2.2, we have that
(2.35)
Since in , from (1.12), we have that
(2.36)
By the fact that , we conclude that in . Thus,
(2.37)
Combining this and (2.35) and letting in (2.34), it follows that
(2.38)
and consequently
(2.39)
Similarly, we deduce from second inequality in (2.33) that
(2.40)

Corollary 2.8.

For and , .

Proof.

Lemma 2.7, applied to the interval , guarantees the existence of such that for
(2.41)
This together with Lemma 2.6 implies that for any ,
(2.42)

which ends the proof.

Lemma 2.9.

Suppose , then there exists such that with , ,
(2.43)

where is the nonnegative eigenfunction corresponding to .

Proof.

Suppose on the contrary that there exist and a sequence with and in such that for all . As
(2.44)
and in , it concludes from Lemma 2.2 that
(2.45)
Notice that has a unique decomposition
(2.46)

where and . Since on and , we have from (2.46) that .

Choose such that
(2.47)
By (H1), there exists , such that
(2.48)
Therefore, for a.e. ,
(2.49)
Since , there exists , such that
(2.50)
and consequently
(2.51)
Applying (2.51), it follows that
(2.52)
Thus,
(2.53)

Corollary 2.10.

For and , .

Proof.

Let , where is the number asserted in Lemma 2.9. As is bounded in , there exists such that , . By Lemma 2.9, one has
(2.54)
This together with Lemma 2.6 implies that
(2.55)

Now, using Theorem A, we may prove the following.

Proposition 2.11.

is a bifurcation interval from the trivial solution for (2.30). There exists an unbounded component of positive solutions of (2.30) which meets . Moreover,
(2.56)

Proof.

For fixed with , let us take that , and . It is easy to check that, for , all of the conditions of Theorem A are satisfied. So there exists a connected component of solutions of (2.30) containing , and either

(i) is unbounded, or

(ii) .

By Lemma 2.7, the case (ii) can not occur. Thus, is unbounded bifurcated from in . Furthermore, we have from Lemma 2.7 that for any closed interval , if , then in is impossible. So must be bifurcated from in .

## 3. Proof of the Main Results

Proof of Theorem 1.5.

It is clear that any solution of (2.30) of the form yields solutions of (1.1). We will show that crosses the hyperplane in . To do this, it is enough to show that joins to . Let satisfy
(3.1)

We note that for all since is the only solution of (2.30) for and .

Case 1.

consider the following:
(3.2)
In this case, we show that the interval
(3.3)

We divide the proof into two steps.

Step 1.

We show that is bounded.

Since , . From (H3), we have
(3.4)

Let denote the nonnegative eigenfunction corresponding to .

From (3.4), we have
(3.5)
By Lemma 2.2, we have
(3.6)
Thus,
(3.7)

Step 2.

We show that joins to .

From (3.1) and (3.7), we have that . Notice that (2.30) is equivalent to the integral equation
(3.8)
which implies that
(3.9)
We divide the both sides of (3.9) by and set . Since is bounded in , there exist a subsequence of and with and on , such that
(3.10)
relabeling if necessary. Thus, (3.9) yields that
(3.11)
Let and denote the nonnegative eigenfunctions corresponding to and , respectively, then it follows from the second inequality in (3.11) that
(3.12)
and consequently
(3.13)
Similarly, we deduce from the first inequality in (3.11) that
(3.14)
Thus,
(3.15)

So joins to .

Case 2.

.

In this case, if is such that
(3.16)
then
(3.17)
and moreover,
(3.18)
Assume that is bounded, applying a similar argument to that used in Step 2 of Case 1, after taking a subsequence and relabeling if necessary, it follows that
(3.19)

Again joins to and the result follows.

Remark 3.1.

Lomtatidze [13, Theorem ] proved the existence of solutions of singular two-point boundary value problems as follows:
(3.20)

under the following assumptions:

(A1)
(3.21)
where satisfies the following condition:
(3.22)
?(A2) For , let be the solution of singular IVPs
(3.23)

satisfying has at least one zero in and has no zeros in .

It is worth remarking that (A1)-(A2) imply Condition (1.21) in Theorem 1.5. However, Condition (1.21) is easier to be verified than (A1)-(A2) since and are easily estimated by Rayleigh's Quotient.

The language of eigenvalue of singular linear eigenvalue problem did not occur until Asakawa [1] in 2001. The first part of Theorem 1.5 is new.

## Declarations

### Acknowledgments

The authors are very grateful to the anonymous referees for their valuable suggestions. This work was supported by the NSFC 11061030, the Fundamental Research Funds for the Gansu Universities.

## Authors’ Affiliations

(1)
College of Mathematics and Information Science, Northwest Normal University
(2)
The School of Mathematics, Physics & Software Engineering, Lanzhou Jiaotong University

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