1 Introduction

The periodic boundary value problems

(1)

where f is a continuous or L1-Caratheodory type function have been extensively studied. A very popular technique to obtain the existence and multiplicity of positive solutions to the problem is Krasnosel'skii's fixed point theorem of cone expansion/compression type, see for example [1–4], and the references contained therein. In those papers, the following condition is an essential assumptions:

(A) The Green function G(t, s) associated with problem (1) is positive for all (t, s) ∈ [0, T] × [0, T].

Under condition (A), Torres get in [4] some existence results for (1) with jumping nonlinearities as well as (1) with a repulsive or attractive singularity, and the authors in [3] obtained the multiplicity results to (1) when f(t, u) has a repulsive singularity near x = 0 and f(t, u) is super-linear near x = +∞. In [2], a special case, a(t) ≡ m2 and , was considered, the multiplicity results to (1) are obtained when the nonlinear term f(t, u) is singular at u = 0 and is super-linear at u = ∞.

Recently, in [5], the hypothesis (A) is weakened as

(B) The Green function G(t, s) associated with problem (1) is nonnegative for all (t, s) ∈ [0, T] × [0, T] but vanish at some interior points.

By defining a new cone, in order to apply Krasnosel'skii's fixed point theorem, the authors get an existence result when and is sub-linear at u = 0 and u = ∞ or is super-linear at u = 0 and u = ∞ with is convex and nondecreasing.

In [6], the author improve the result of [5] and prove the existence results of at least two positive solutions under conditions weaker than sub- and super-linearity.

In [7], the author study (1) with f(t, u) = λb(t)f(u) under the following condition:

(C) The Green function G(t, s) associated with problem (1) changes sign and where G - is the negative part of G.

Inspired by those papers, here we study the problem:

(2)

where is a constant and the associated Green's function may changes sign. The aim is to prove the existence of positive solutions to the problem.

2 Preliminaries

Consider the periodic boundary value problem

(3)

where and e(t) is a continuous function on [0, T]. It is well known that the solutions of (3) can be expressed in the following forms

where G(t, s) is Green's function associated to (3) and it can be explicitly expressed

By direct computation, we get

and

for when , and

where G+ and G- are the positive and negative parts of G.

We denote

and

Let E denote the Banach space C[0, T] with the norm ||u|| = maxt∈[0,T]|u(t)|.

Define the cone K in E by

We know that and therefore K ≠ ∅. For r > 0, let K r = {u ∈ K : ||u|| < r}, and ∂K r = {u ∈ K : ||u|| = r}, which is the relative boundary of K r in K.

To prove our result, we need the following fixed point index theorem of cone mapping.

Lemma 1 (Guo and Lakshmikantham [8]). Let E be a Banach space and let K ⊂ E be a closed convex cone in E. Let L : K → K be a completely continuous operator and let i(L, K r , K) denote the fixed point index of operator L.

(i) If μLu ≠ u for any u ∈ ∂K r and 0 < μ ≤ 1, then

(ii) If and μLu ≠ u for any u ∈ ∂K r and μ ≥ 1, then

3 Existence result

We make the following assumptions: (H 1) f : [0, +∞) → [0, +∞) is continuous;

(H 2) 0 ≤ m = inf u∈[0,+ ∞]f (u) and M = supu∈[0,+ ∞)f (u) ≤ +∞;

(H 3) , when m = 0 we define .

To be convenience, we introduce the notations:

and suppose that f0, f∞ ∈ [0, ∞].

Define a mapping L : K → E by

It can be easily verified that u ∈ K is a fixed point of L if and only if u is a positive solution of (2).

Lemma 2. Suppose that (H1), (H2) and (H3) hold, then L : E → E is completely continuous and L(K) ⊆ K.

Proof Let u ∈ K, then in case of γ = +∞, since G(t, s) ≥ 0, we have Lu(t) ≥ 0 on [0, T]; in case of γ < +∞, we have

On the other hand,

and

for t ∈ [0, T]. Thus,

i.e., L(K) ⊆ K. A standard argument can be used to show that L : E → E is completely continuous.

Now we give and prove our existence theorem:

Theorem 3. Assume that (H1), (H2) and (H3) hold. Furthermore, suppose that f0 > ρ2 and f∞ < ρ2 in case of γ = +∞. Then problem (2) has at least one positive solution.

Proof Since f0 > ρ2, there exist ε > 0 and ξ > 0 such that

(4)

Let r ∈ (0, ξ), then for every u ∈ ∂K r , we have

Hence, . Next, we show that μLu ≠ u for any u ∈ ∂K r and μ ≥ 1. In fact, if there exist u0 ∈ ∂K r and μ0 ≥ 1 such that μ0Lu0 = u0, then u0(t) satisfies

(5)

Integrating the first equation in (5) from 0 to T and using the periodicity of u0(t) and (4), we have

Since , we see that ρ2 ≥ (ρ2 + ε), which is a contradiction. Hence, by Lemma 1, we have

(6)

On the other hand, since f∞ < ρ2, there exist ε ∈ (0, ρ2) and ζ > 0 such that

Set C = max0≤u≤ζ|f (u) - (ρ2 - ε)u| + 1, it is clear that

(7)

If there exist u0 ∈ K and 0 < μ0 ≤ 1 such that μ0Lu0 = u0, then (5) is valid.

Integrating again the first equation in (5) from 0 to T, and from (7), we have

Therefore, we obtain that

i.e.,

(8)

Let , then μLu ≠ u for any u ∈ ∂K R and 0 < μ ≤ 1. Therefore, by Lemma 1, we get

(9)

From (6) and (9) it follows that

Hence, L has a fixed point in , which is the positive solution of (2).

Remark 4. Theorem 3 contains the partial results of [4–7] obtained in case of positive Green's function, vanishing Green's function and sign-changing Green's function, respectively.

4 An example

Let 0 ≠ q < 1 be a constant, h be the function:

and let

By the direct calculation, we get m = 1 and M = γ, and f0 = ∞ and f∞ = 0 in case of γ = +∞. Consider the following problem

(10)

where is a constant. We know that the conditions of Theorem 3 hold for the problem (10) and therefore, (10) have at least one positive solution from Theorem 3.