Abstract
We are interested in the following singular boundary value problem:
where is a parameter and is the Stieltjes integral. The function and w may be singular at and/or , and . Some a priori estimates and the existence, multiplicity and nonexistence of positive solutions are obtained. Our proofs are based on the method of global continuous theorem, the lowerupper solutions methods and fixed point index theory. Furthermore, we also discuss the interval of parameter μ such that the problem has a positive solution.
Keywords:
singularity; global continuous theorem; solution of boundedness; fixed point index; positive solution1 Introduction
We are concerned with the second order nonlocal boundary value problem:
where is a parameter and is a Stieltjes integral. The function and w may be singular at and/or , and .
Integral boundary conditions and multipoint boundary conditions for differential equations come from many areas of applied mathematics and physics [17]. Recently, singular boundary value problems have been extensively considered in a lot of literature [1,2,5,8], since they model many physical phenomena including gas diffusion through porous media, nonlinear diffusion generated by nonlinear sources, chemically reacting systems as well as concentration in chemical or biological problems. In all these problems, positive solutions are very meaningful.
In [1,2], Webb and Infante considered the existence of positive solutions of nonlinear boundary value problem:
where hg are nonnegative functions, subjected to the nonlocal boundary conditions
Here is a linear functional on given by
involving a Stieltjes integral with a signed measure, that is, A has bounded variation. They dealt with many boundary conditions given in the literature in a unified way by utilizing the fixed point index theory in cones.
Recently, many researchers were interested in the global structure of positive solutions for the nonlinear boundary value problem (see, e.g., [3,6,7]). In 2009, Ma and An [3] considered the problem (1.1). Assume that
(A0) is nondecreasing and is not a constant on , with , and for (for the definition of , see (2.1) below).
(A1) is continuous and on any subinterval of , and , where , .
They obtained the following main result:
Theorem 1.1 ([3], Theorem 4.1])
Assume that (A0)(A3) hold. Then there exists a componentin ∑ which joinsto, and
for some. Moreover, there existssuch that (1.1) has at least two positive solutions for. Herejoinstosuch that
Here, ∑ is the closure of the set of positive solutions of (1.1) on , , and the component of a set M is a maximal connected subset of M.
A natural problem arises: How can we consider the global structure of positive solutions for the case ?
In this paper, we first obtain the global structure of positive solutions by the use of global continuous theorem, and some a priori estimates. Applying the analysis technique, we construct the lower and upper solutions. Those combined with the fixed point index theory, the existence, multiplicity and nonexistence of positive solutions to (1.1) in the case are investigated. Finally, we discuss the interval of parameter μ such that the problem (1.1) has positive solutions. The proof of the method which is based on the construction of some bounds of the solution together with global continuous theorem and fixed point index is of independent interest, and is different from the other papers.
This paper is arranged as follows. We will give some hypotheses and lemmas in Section 2. In Section 3, new criteria of the existence, multiplicity and nonexistence of a positive solution are obtained. Moreover, an example is given to illustrate our result.
2 Preliminaries and lemmas
Let X denote the Banach space with the maximum norm
Define , then K is a cone. Let
Denote
Throughout this paper, we suppose that the following conditions hold:
(H0) is nondecreasing, on and with .
(H1) and w may be singular at and/or 1, satisfying
Remark 2.1 It is easy to see from (H0) that . Therefore, if we assume that , then (2.3) obviously holds.
Define an operator as follows:
Assume that the conditions (H0)(H2) hold, then it is easy to verify that is well defined and completely continuous.
Lemma 2.1 ([9] Global continuation theorem)
LetXbe a Banach space and letKbe an order cone inX. Consider the equation
whereand. Suppose thatis completely continuous andfor all, then, the component of the solution set of (2.4) containingis unbounded.
Remark 2.2
(1) We note that u is a positive solution of the problem (1.1) if and only if on K.
(2) If for and for all , then we get from Lemma 2.1 that there exists an unbounded continuum emanating from in the closure of the set of positive solutions (1.1) in .
Lemma 2.2 ([10])
LetXbe a Banach space, Kan order cone inXandan open bounded set inXwith. Suppose thatis a completely continuous operator. If, for alland all, then.
3 Main results
Lemma 3.1Let (H0)(H3) hold and letwith. Then there exists a constantsuch that for alland all possible positive solutionsof (1.1), the inequalityholds.
Proof Suppose on the contrary that there exist a sequence and a sequence of the positive solutions of (1.1) corresponding to such that
Denote . From the concavity of , it follows that
Choose , where . Then we find from (H3) that there exists a constant such that
This, together with (3.1) and (3.2), implies
Put , for . Hence, we know from (3.3) that
On the other hand, multiplying (1.1) by ψ and integrating by parts, we obtain that
leads to
i.e.,
This is a contradiction. □
Lemma 3.2Assume that the hypotheses (H0)(H3) hold. Then there existssuch that if the problem (1.1) has a positive solution for parameterμ, then.
Proof Let u be a positive solution of (1.1) corresponding to μ. From the hypotheses (H2) and (H3), it follows that there exists such that for all . Consequently, we have
Let be the first eigenvalue of
and let be the positive eigenfunction corresponding to (see [8]). It is easy to see that . Multiplying (3.4) by and integrating by parts, we get that
implies
This completes the proof. □
Lemma 3.3Assume that (H0)(H3) hold. Then we have
Proof Suppose this fails, that is, there exists such that
where is a positive constant. Since , we get that for all
where
Hence, we find that
Thus, it implies that . This is contradiction. □
Theorem 3.1Assume that the conditions (H0)(H3) hold and. Then there exists a constantsuch that the problem (1.1) has at least two positive solutions for, and at least one positive solution for, and no positive solution for.
Proof Define
From Lemma 2.1 and Remark 2.2, we can find that there exists an unbounded continuum emanating from in the closure of the set of positive solutions in and for all . Meanwhile, Lemma 3.1 and Lemma 3.3 respectively imply that is bounded (, ) and unbounded (, and ). Therefore, we conclude that the set of (3.5) is nonempty. Those combined with Lemma 3.2 follows that is well defined and . From the definition of , it is easy to see that the problem (1.1) has at least two positive solutions for . Again, since the continuum is a compact connected set and T is a completely continuous operator, the problem (1.1) has at least one positive solution at .
Next, we only show that the problem (1.1) has no positive solution for any . Suppose on the contrary that there exists some () such that the problem (1.1) has a positive solution corresponding to . Then we will prove that the problem (1.1) has at least two positive solutions for any which contradicts the definition of (3.5).
For the sake of obtaining the contradiction, we divide the proof into four steps.
Step 1. Constructing a modified boundary value problem.
Choose arbitrarily a constant . Since f is uniformly continuous on , there exists a constant such that
where
Denote , for . Then we claim that
and
Indeed, using (3.6), we have
Define a set . Then the set is bounded and open in . Now, we construct the modified secondorder boundary value problem:
Step 2. We will show that if u is a positive solution of (3.9), then .
Let u be a positive solution of (3.9), then we claim that
Suppose this fails, that is, . Clearly, we only show that , for . Comparing the boundary conditions (3.8) and (3.9), the only following three cases need to be considered. Case I. There exists such that and , for and some ; Case II. There exists such that and for . Case III. There exists such that , , and . See the three Figures 1, 2 and 3.
Case I. From (3.7), it implies that there exists a constant such that
Since f is uniformly continuous on , there exists a such that if and , then we get
where . From the assumption of Case I, it follows that there exists a subinterval with such that
and
This together with (3.11) and (3.12) leads to
This is a contradiction.
Case II. Let , for . Then we have that for
Obviously, we have . From (3.13), it implies that , for any . Hence, the function is strictly increasing in . From (3.8) and the boundary condition (3.9), we find
We obtain a contradiction. In particular, if , then we obtain
This contradicts (3.13).
Case III. Since , for , we have that , for . Again since , we know from the maximum principle that , for . This contradicts the assumption of Case III.
Therefore, we conclude that the claim (3.10) holds.
Step 3. (the definition of see below).
Using (2.2), we define an operator by
Then is completely continuous and u is a positive solution of (3.9) if and only if on K. From the definition of , it implies that there exists such that , for all . Consequently, we get from Lemma 2.2 that
where . Applying the conclusion of Step 2 and the excision property of fixed point index, we find that
Step 4. We conclude that the problem (1.1) has at least two positive solutions corresponding to μ.
Since the problem (1.1) is equivalent to the problem (3.9) on , we get that the problem (1.1) has a positive solution in . Without loss of generality, we may suppose that T has no fixed point on (otherwise the proof is completed). Then is well defined and (3.15) implies
On the other hand, from Lemma 3.2, we choose such that the problem (1.1) has no positive solution in K. By apriori estimate in , there exists such that for all possible positive solutions of (1.1) with , we know that . Define by
Then it is easy to verify that is completely continuous on and for all . From the property of homotopy invariance, it follows that
Hence, by the additivity property and (3.16), we have
Then we conclude that the problem (1.1) has at least two positive solutions corresponding to μ. □
Remark 3.1
(i) From the hypotheses (H2) and (H3), it implies that there exists such that
Let f attain its maximum at the point of . If , then adopting the similar method as in [11], Theorem 1], we get that for , the problem (1.1) has at least two positive solutions and such that by the use of compression of conical shells in [12], Corollary 20.1]. Consequently, we know that .
(ii) If u is a positive solution of the equation (1.1) corresponding to μ, then we have
i.e., from (3.17),
Corollary 3.1Assume that (H1)(H3) hold. Consider the followingmpoint boundary value problem
whereμis a positive parameter, , , and. Then there exists a constantsuch that the problem (3.18) has at least two positive solutions for, and at least one positive solution for, and no positive solution for.
Proof In the boundary condition of (1.1), if we let
where is the characteristic function on , i.e.,
Then the boundary condition of (1.1) reduces to the mpoint boundary condition of (3.18). Applying the method of Theorem 3.1, we get the conclusion. □
Example 3.1 We consider the following singular boundary value problem:
Computing yields
Thus, it implies that (2.3) holds. It is easy to verify that the conditions (H2) and (H3) hold. Therefore, by Theorem 3.1, we obtain that there exists a constant such that the problem (3.19) has at least two positive solutions for , and at least one positive solution for , and no positive solution for .
Competing interests
The author declares that they have no competing interests.
Author’s contributions
The author typed, read and approved the final manuscript.
Acknowledgement
The author would like to thank the anonymous referees very much for helpful comments and suggestions which led to the improvement of presentation and quality of the work. The work was supported partly by NSCF of Tianyuan Youth Foundation (No. 11126125), K. C. Wong Magna Fund of Ningbo University, Subject Foundation of Ningbo University (No. xkl11044) and Hulan’s Excellent Doctor Foundation of Ningbo University.
References

Webb, JRL, Infante, G: Positive solutions of nonlocal boundary value problems: a unified approach. J. Lond. Math. Soc.. 74, 673–693 (2006). Publisher Full Text

Webb, JRL, Infante, G: Positive solutions of nonlocal boundary value problems involving integral conditions. Nonlinear Differ. Equ. Appl.. 15, 45–67 (2008). Publisher Full Text

Ma, RY, An, YL: Global structure of positive solutions for nonlocal boundary value problems involving integral conditions. Nonlinear Anal. TMA. 71, 4364–4376 (2009). Publisher Full Text

Anderson, DR, Smyrlis, G: Solvability for a thirdorder threepoint boundary value problem on time scales. Math. Comput. Model.. 49, 1994–2001 (2009). Publisher Full Text

Lan, KQ, Webb, JRL: Positive solutions of semilinear differential equations with singularities. J. Differ. Equ.. 148, 407–421 (1998). Publisher Full Text

Li, HY, Sun, JX: Positive solutions of superlinear semipositone nonlinear boundary value problems. Comput. Math. Appl.. 61, 2806–2815 (2011). Publisher Full Text

Rynne, B: Spectral properties and nodal solutions for secondorder, mpoint, boundary value problems. Nonlinear Anal. TMA. 67, 3318–3327 (2007). Publisher Full Text

Asakawa, H: On nonresonant singular twopoint boundary value problems. Nonlinear Anal.. 47, 4849–4860 (2001). Publisher Full Text

Zeidler, E: Nonlinear Functional Analysis and Its Applications, I. FixedPoint Theorems, Springer, New York (1986)

Guo, DJ, Lakshmikantham, V: Nonlinear Problems in Abstract Cones, Academic Press, New York (1988)

Sánchez, J: Multiple positive solutions of singular eigenvalue type problems involving the onedimensional pLaplacian. J. Math. Anal. Appl.. 292, 401–414 (2004). Publisher Full Text

Deimling, K: Nonlinear Functional Analysis, Springer, Berlin (1985)