# Existence of Positive Solutions of Fourth-Order Problems with Integral Boundary Conditions

- Ruyun Ma
^{1}Email author and - Tianlan Chen
^{1}

**Received: **5 May 2010

**Accepted: **7 July 2010

**Published: **3 August 2010

## Abstract

## Keywords

## 1. Introduction

where is continuous; see Gupta [1, 2]. In the past twenty more years, the existence of solutions and positive solutions of these kinds of problems and the Lidstone problem has been extensively studied; see [3–9] and the references therein. In [3], Ma was concerned with the existence of positive solutions of (1.1) and (1.2) under the assumptions:

Ma proved the following.

Theorem A (see [3, Theorem ]).

Then (1.1) and (1.2) have at least one positive solution.

where may be singular at and (or) ; is continuous, and are nonnegative.

under the assumption

(H4) are nonnegative, and . The main result of this paper is the following.

Theorem 1.1.

Then (1.9) has at least one positive solution.

Remark 1.2.

Theorem 1.1 generalizes [3, Theorem ] where the special case and was treated.

Remark 1.3.

In this case, and the corresponding eigenfunction is . However, (1.15) and (1.16) has no positive solution. (In fact, suppose on the contrary that (1.15) and (1.16) has a positive solution . Multiplying (1.15) with and integrating from to , we get a desired contradiction!).

*positive*if . It is said to be

*-completely continuous*if is continuous and maps bounded subsets of to precompact subset of . Finally, a positive linear operator on is said to be a

*linear minorant*for if for . If is a continuous linear operator on , denote the spectral radius of . Define

The following lemma will play a very important role in the proof of our main results, which is essentially a consequence of Dancer [14, Theorem ].

Lemma 1.4.

- (i)
- (ii)

where is a strongly positive linear compact operator on with the spectral radius , satisfies as locally uniformly in .

Proof.

Since is a strongly positive compact endomorphism of and has nonempty interior, we have from Amann [15, Theorem ] that the set in [14, Theorem ] reduces to a single point . Now the desired result is a consequence of Dancer [14, Theorem ].

The rest of the paper is arranged as follows. In Section 2, we state and prove some preliminary results about the spectrum of (1.12)–(1.14). Finally, in Section 3, we proved our main result.

## 2. Generalized Eigenvalues

Lemma 2.1 (see [10]).

Lemma 2.2 (see [10]).

Lemma 2.3 (see [10]).

Let

(H5) be two given constants with .

Definition 2.4.

if (2.8) and (2.9) have nontrivial solutions.

Let with the norm . Let with the norm .

We claim that is a Banach space.

Then the cone is normal and nonempty interior and .

In fact, for any , it follows from the definition of that

Lemma 2.5.

Proof.

From , we have that , and so , and accordingly .

which implies that , and consequently .

Theorem 2.6.

Assume that (H4) and (H5) hold. Let be the spectral radius of . Then (2.8) and (2.9) has an algebraically simple eigenvalue, , with a positive eigenfunction . Moreover, there is no other eigenvalue with a positive eigenfunction.

Remark 2.7.

and the corresponding eigenfunction .

Proof of Theorem 2.6.

where , it follows that , and accordingly .

Now, since , and is compactly embedded in , we have that is compact.

Next, we show that is positive.

This together with (2.9) and (H4) imply on .

Therefore, it follows from (2.43) and (2.45) that .

Now, by the Krein-Rutman theorem ([16, Theorem C]; [17, Theorem ]), has an algebraically simple eigenvalue with an eigenfunction . Moreover, there is no other eigenvalue with a positive eigenfunction. Correspondingly, with a positive eigenfunction of , is a simple eigenvalue of (2.8) and (2.9). Moreover, for (2.8) and (2.9), there is no other eigenvalue with a positive eigenfunction.

## 3. The Proof of the Main Result

It is easy to check that is compact.

Proof of Theorem 1.1.

we note that for all since is the only solution of (3.8) for and .

We divide the proof into two steps.

Step 1.

This together with Theorem 2.6 imply that . Therefore, joins to .

Step 2.

We show that there exists a constant be such that for all .

By Lemma 1.4, we only need to show that has a linear minorant and there exists a such that and .

Combining this with (2.39), we conclude that , here, . Therefore, we have that from Lemma 1.4 that .

## Declarations

### Acknowledgments

The authors are very grateful to the anonymous referees for their valuable suggestions. This paper was supported by the NSFC (no. 11061030), the Fundamental Research Funds for the Gansu Universities.

## Authors’ Affiliations

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