On positive solutions for nonhomogeneous m-point boundary value problems with two parameters
© Wang and An; licensee Springer 2012
Received: 14 May 2012
Accepted: 27 July 2012
Published: 6 August 2012
This paper is concerned with the existence, multiplicity, and nonexistence of positive solutions for nonhomogeneous m-point boundary value problems with two parameters. The proof is based on the fixed-point theorem, the upper-lower solutions method, and the fixed-point index.
where , (), , , may be singular at and/or . They showed that there exists a positive number such that the problem has at least two positive solutions for , at least one positive solution for and no solution for by using the Krasnosel’skii-Guo fixed-point theorem, the upper-lower solutions method, and the topological degree theory.
where λ, μ are positive parameters, , . The main result of the present paper is summarized as follows.
Theorem 1.1 Assume the following conditions hold:
(H1) are nonnegative parameters;
(H2) is continuous, does not vanish identically on any subinterval of and , where is given in Sect. 2;
And either or ;
(H4) There exist constants such that and , respectively, for all ;
(H5) , .
Then on , the function is continuous and nonincreasing, that is, if , we have .
For the proof of Theorem 1.1, we also need the following lemmas.
Lemma 1.2 
Lemma 1.3 
Lemma 2.1 
for , and is continuous on ;
for all .
where . Then we have
Lemma 2.3 If (H 1)-(H 3) hold, then is completely continuous.
The proof procedure of Lemma 2.3 is standard, so we omit it.
Lemma 2.4 Let , be lower and upper solutions, respectively, of (1) such that . Then (1) has a nonnegative solution satisfying for .
It is clear to see that is a bounded, convex and closed subset in Banach space E. Now we can prove that .
From above inequalities, we obtain that .
Therefore, by Schauder’s fixed theorem, the operator T has a fixed point , which is the solution of (1). □
3 Proof of Theorem 1.1
Lemma 3.1 Assume (H 1)-(H 5) hold and Σ be a compact subset of . Then there exists a constant such that for all and all possible positive solutions of (1) at , one has .
and at least or .
This is a contradiction.
This is a contradiction. □
Lemma 3.2 Assume (H 1)-(H 4) hold. If (1) has a positive solution at , then Eq. (1) has a positive solution at for all .
Proof Let be the solution of Eq. (1) at , then be the upper solution of (1) at with . Since or , is not a solution of (1), but it is the lower solution of (1) at . Therefore, by Lemma 2.4, we obtain the result. □
Lemma 3.3 Assume (H 1)-(H 5) hold. Then there exists such that Eq. (1) has a positive solution for all .
which implies that is an upper solution of (3) at . On the other hand, 0 is a lower solution of (1) and . By (H3), 0 is not a solution of (1). Hence, (1) has a positive solution at , Lemma 3.2 now implies the conclusion of Lemma 3.3. □
Then it follows from Lemma 3.3 that and is a partially ordered set.
Lemma 3.4 Assume (H 1)-(H 5) hold. Then is bounded above.
Lemma 3.5 Assume (H 1)-(H 5) hold. Then every chain in S has a unique supremum in S.
Lemma 3.6 Assume (H 1)-(H 5) hold. Then there exists a such (1) has a positive solution at for all , no solution at for all . Similarly, there exists a such that (1) has a positive solution at for all , and no solution at for all .
for all and .
From above inequalities, it is clear to see that , is an upper solution of (1) at for all . □
Hence, T has at least one fixed point in and another one in ; this shows that in , (1) has at least two positive solution. □
where , , and .
The authors would like to thank the referees for valuable comments and suggestions for improving this paper. The first author is supported financially by the Fundamental Research Funds for the Central Universities.
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