Open Access

Existence of positive solutions to periodic boundary value problems with sign-changing Green's function

Boundary Value Problems20112011:8

DOI: 10.1186/1687-2770-2011-8

Received: 27 January 2011

Accepted: 27 July 2011

Published: 27 July 2011

Abstract

This paper deals with the periodic boundary value problems

https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equa_HTML.gif

where https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq1_HTML.gif is a constant and in which case the associated Green's function may changes sign. The existence result of positive solutions is established by using the fixed point index theory of cone mapping.

Keywords

periodic boundary value problem positive solution sign-changing Green's function cone fixed point theorem

1 Introduction

The periodic boundary value problems
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equ1_HTML.gif
(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 [14], 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq2_HTML.gif , 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq3_HTML.gif and https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq4_HTML.gif is sub-linear at u = 0 and u = ∞ or https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq4_HTML.gif is super-linear at u = 0 and u = ∞ with https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq4_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq5_HTML.gif where G - is the negative part of G.

Inspired by those papers, here we study the problem:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equ2_HTML.gif
(2)

where https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq1_HTML.gif 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
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equ3_HTML.gif
(3)
where https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq1_HTML.gif 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
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equb_HTML.gif
where G(t, s) is Green's function associated to (3) and it can be explicitly expressed
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equc_HTML.gif
By direct computation, we get
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equd_HTML.gif
and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Eque_HTML.gif
for https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq6_HTML.gif when https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq7_HTML.gif , and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equf_HTML.gif
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equg_HTML.gif

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

We denote
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equh_HTML.gif
and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equi_HTML.gif

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
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equj_HTML.gif

We know that https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq8_HTML.gif 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 : KK be a completely continuous operator and let i(L, K r , K) denote the fixed point index of operator L.

(i) If μLuu for any u K r and 0 < μ ≤ 1, then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equk_HTML.gif
(ii) If https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq9_HTML.gif and μLuu for any u K r and μ ≥ 1, then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equl_HTML.gif

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) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq10_HTML.gif , when m = 0 we define https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq11_HTML.gif .

To be convenience, we introduce the notations:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equm_HTML.gif

and suppose that f0, f [0, ∞].

Define a mapping L : KE by
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equn_HTML.gif

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 : EE 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
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equo_HTML.gif
On the other hand,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equp_HTML.gif
and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equq_HTML.gif
for t [0, T]. Thus,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equr_HTML.gif

i.e., L(K) K. A standard argument can be used to show that L : EE 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
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equ4_HTML.gif
(4)
Let r (0, ξ), then for every u K r , we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equs_HTML.gif
Hence, https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq9_HTML.gif . Next, we show that μLuu 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
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equ5_HTML.gif
(5)
Integrating the first equation in (5) from 0 to T and using the periodicity of u0(t) and (4), we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equt_HTML.gif
Since https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq12_HTML.gif , we see that ρ2 ≥ (ρ2 + ε), which is a contradiction. Hence, by Lemma 1, we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equ6_HTML.gif
(6)
On the other hand, since f < ρ2, there exist ε (0, ρ2) and ζ > 0 such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equu_HTML.gif
Set C = max0≤uζ|f (u) - (ρ2 - ε)u| + 1, it is clear that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equ7_HTML.gif
(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
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equv_HTML.gif
Therefore, we obtain that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equw_HTML.gif
i.e.,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equ8_HTML.gif
(8)
Let https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq13_HTML.gif , then μLuu for any u K R and 0 < μ ≤ 1. Therefore, by Lemma 1, we get
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equ9_HTML.gif
(9)
From (6) and (9) it follows that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equx_HTML.gif

Hence, L has a fixed point in https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq14_HTML.gif , which is the positive solution of (2).

Remark 4. Theorem 3 contains the partial results of [47] 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:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equy_HTML.gif
and let
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equz_HTML.gif
By the direct calculation, we get m = 1 and M = γ, and f0 = ∞ and f = 0 in case of γ = +∞. Consider the following problem
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_Equ10_HTML.gif
(10)

where https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-8/MediaObjects/13661_2011_Article_8_IEq1_HTML.gif 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.

Declarations

Acknowledgements

The authors are very grateful to the anonymous referee whose careful reading of the manuscript and valuable comments enhanced presentation of the manuscript.

Authors’ Affiliations

(1)
Department of Engineering Technology, Wuwei Occupational College Wuwei
(2)
Department of Mathematics, Nanjing University of Aeronautics and Astronautics Nanjing

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Copyright

© Zhong and An; licensee Springer. 2011

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.