We are concerned with the existence of positive solutions of singular secondorder 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
where is allowed to have singularities at and , have been studied by several authors, see Asakawa [1], Agarwal and ORegan [2], ORegan [3], Habets and Zanolin [4], Xu and Ma [5], Yang [6], and the references therein. The main tools in [1–6] are the method of lower and upper solutions, LeraySchauder 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
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
with the norm
Let
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
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
for some Carathéodory functions defined on with
uniformly for a.e. , and
for some Carathéodory functions defined on with
uniformly for a.e. .
(H2) for a.e. and .
(H3) There exists function such that
Remark 1.2.
If , , , and , then (1.8) implies that
and (1.10) implies that
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:
for and , where , are not bifurcation points of (1.14). Furthermore, assume that
where is an isolating neighborhood of the trivial solution. Let
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
has a sequence of eigenvalues as follows:
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
Our main result is the following.
Theorem 1.5.
Let (H1)–(H3) hold. Assume that either
or
then (1.1) has at least one positive solution.
Remark 1.6.
For other references related to this topic, see [9–14] and the references therein.
2. Preliminary Results
Lemma 2.1 (see [15, Proposition ]).
For any , the linear problem
has a unique solution and , such that
where
Furthermore, if , then
Let be the Banach space with the norm , and
Let be an operator defined by
where
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
Proof.
For any , integrating by parts, we have
Since and , then
Therefore, we only need to prove that
Let us deal with the first equality, the second one can be treated by the same way. Note that , then
which implies that . Then is bounded on . Now, we claim that
Suppose on the contrary that , then for small enough, we have
Therefore,
which is a contradiction. Combining (1.19) with (2.13), we have
This completes the proof.
Remark 2.3.
Under the conditions of Lemma 2.2, for the later convenience, (2.8) is equivalent to
Lemma 2.4 (see [1, Lemma ]).
For every , the subset defined by
is precompact in .
Let be the closure of the set of positive solutions of the problem
We extend the function to an Carathéodory function defined on by
Then for and a.e. . For , let be an arbitrary solution of the problem
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:
Lemma 2.5.
Let on . Let be such that in , . Then,
Moreover, , whenever .
Let be the Nemytskii operator associated with the function as follows:
Then (2.21), with , is equivalent to the operator equation
that is,
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 ,
Since is an Carathéodory function, then there exists , such that, for a.e. and , . Therefore, for a.e. and , we have
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
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 LeraySchauder degree theory mainly to the mapping ,
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
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:
and accordingly
Let and denote the nonnegative eigenfunctions corresponding to and , respectively, then we have from the first inequality in (2.33) that
From Lemma 2.2, we have that
Since in , from (1.12), we have that
By the fact that , we conclude that in . Thus,
Combining this and (2.35) and letting in (2.34), it follows that
and consequently
Similarly, we deduce from second inequality in (2.33) that
Thus, . This contradicts .
Corollary 2.8.
For and , .
Proof.
Lemma 2.7, applied to the interval , guarantees the existence of such that for
This together with Lemma 2.6 implies that for any ,
which ends the proof.
Lemma 2.9.
Suppose , then there exists such that with , ,
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
and in , it concludes from Lemma 2.2 that
Notice that has a unique decomposition
where and . Since on and , we have from (2.46) that .
Choose such that
By (H1), there exists , such that
Therefore, for a.e. ,
Since , there exists , such that
and consequently
Applying (2.51), it follows that
Thus,
This contradicts (2.47).
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
This together with Lemma 2.6 implies that
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,
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
We note that for all since is the only solution of (2.30) for and .
Case 1.
consider the following:
In this case, we show that the interval
We divide the proof into two steps.
Step 1.
We show that is bounded.
Since , . From (H3), we have
Let denote the nonnegative eigenfunction corresponding to .
From (3.4), we have
By Lemma 2.2, we have
Thus,
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
which implies that
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
relabeling if necessary. Thus, (3.9) yields that
Let and denote the nonnegative eigenfunctions corresponding to and , respectively, then it follows from the second inequality in (3.11) that
and consequently
Similarly, we deduce from the first inequality in (3.11) that
Thus,
So joins to .
Case 2.
.
In this case, if is such that
then
and moreover,
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
Again joins to and the result follows.
Remark 3.1.
Lomtatidze [13, Theorem ] proved the existence of solutions of singular twopoint boundary value problems as follows:
under the following assumptions:
(A1)
where satisfies the following condition:
(A2) For , let be the solution of singular IVPs
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.
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.
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