Existence of positive solutions for fourth-order semipositone multi-point boundary value problems with a sign-changing nonlinear term

Boundary Value Problems20122012:12

DOI: 10.1186/1687-2770-2012-12

Received: 23 July 2011

Accepted: 9 February 2012

Published: 9 February 2012

Abstract

In this article, some new sufficient conditions are obtained by making use of fixed point index theory in cone and constructing some available integral operators together with approximating technique. They guarantee the existence of at least one positive solution for nonlinear fourth-order semipositone multi-point boundary value problems. The interesting point is that the nonlinear term f not only involve with the first-order and the second-order derivatives explicitly, but also may be allowed to change sign and may be singular at t = 0 and/or t = 1. Moreover, some stronger conditions that common nonlinear term f ≥ 0 will be modified. Finally, two examples are given to demonstrate the validity of our main results.

2000 Mathematics Subject Classification: 34B10; 34B18; 47N20.

Keywords

semipositone positive solutions multi-point boundary value problems

1 Introduction

In this article, we consider the existence of positive solutions to the following nonlinear fourth-order semipositone multi-point boundary value problems with derivatives
y ( 4 ) ( t ) + λ f ( t , y ( t ) , y ( t ) , y ( t ) ) = 0 , 0 < t < 1 , y ( 0 ) = y ( 0 ) = 0 , y ( 1 ) = i = 1 m - 2 α i y ( ξ i ) , y ( 0 ) = i = 1 m - 2 β i y ( ξ i ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ1_HTML.gif
(1.1)

where fC((0, 1) × R×R×R, R) satisfies f(t, y1 y2, y3) ≥ -p(t), pL1 ((0,1), (0, +∞)). λ > 0, ξ i ∈ (0, 1) with 0 < ξ1 < ξ2 < ⋯ < ξm-2< 1, α i , β i ∈ [0, +∞), i = 1, 2,... , m-2, are given constants satisfying 0 < i = 1 m - 2 α i < 1 , 0 < i = 1 m - 2 β i < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq1_HTML.gif. Here, by a positive solution of the problem (1.1) we mean a function y*(t) which is positive on (0, 1) and satisfies the problem (1.1).

The existence of positive solutions for multi-point boundary value problems has been widely studied in recent years. For details, see [115] and references therein. We note that the existence of n solutions and/or positive solutions to the following semipositone elastic beam equation boundary value problem
u ( 4 ) ( t ) = f ( t , u ( t ) , u ( t ) ) , t ( 0 , 1 ) , u ( 0 ) = u ( 1 ) = u ( 0 ) = u ( 1 ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equa_HTML.gif
was obtained by Yao [13] in a Banach space setting. Gupta [3] proved the existence of positive solutions for more general multi-point boundary value problems
x ( t ) = g ( t , x ( t ) , x ( t ) ) + e ( t ) , a . e . t ( 0 , 1 ) x ( 0 ) = i = 1 m - 2 h i x ( τ i ) , x ( 1 ) = i = 1 m - 2 k i x ( ξ i ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equb_HTML.gif

For further background information of multi-point boundary value problems we refer the reader to [11, 12, 16]. However, in previous work, the positivity which imposed on nonlinear term plays an important role for boundary value problems. Naturally, one is interested in establishing the existence of positive solutions for multi-point boundary value problems under the relaxed conditions.

Inspired and motivated greatly by the above mentioned works, the present work may be viewed as a direct attempt to extend the results of [3, 13] to a broader class of nonlinear boundary value problems in a general Banach spaces. When the nonlinearity is negative, such kinds of the problems are called semipositone problems, which occur in chemical rector theory, combustion and management of natural resources, see [11, 1316]. To our best knowledge, few results were obtained for the problem (1.1).

The purpose of the article is to establish some new criteria for the existence of positive solutions to the problem (1.1). The nonlinear term f may take negative values and the nonlinearity may be sign-changing. Firstly, we employ a exchange technique and construct an integral operator for the corresponding second-order multi-point boundary value problem. Then we establish a special cone associated with concavity of functions. Finally, the existence of positive solutions for the problem (1.1) is obtained by applying fixed-point index theory. The common restriction on f ≥ 0 is modified.

The plan of the article is as follows. Section 2 contains a number of lemmas useful to the derivation of the main results. The proof of the main results will be stated in Section 3. A class of examples are given to show that our main result is applicable to many problems in Section 4.

2 Preliminaries and lemmas

In this section, we shall state some necessary definitions and preliminaries.

Definition 2.1. Let E be a real Banach space. A nonempty closed convex set KE is called a cone if it satisfies the following two conditions:

(1) xK, λ > 0 implies λxK;

(2) xK, -xK implies x = 0.

Definition 2.2. An operator T is called completely continuous if it is continuous and maps bounded sets into precompact sets.

For convenience, we list the following assumptions:

(H1) For i ∈ {1,2, ⋯, m - 2}, ξ i ∈ (0, 1), 0 < ξ1 < ξ2 < ⋯ < ξm-2< 1 and α i , β i ∈ [0, +∞) satisfying 0 < i = 1 m - 2 α i < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq2_HTML.gif, 0 < i = 1 m - 2 β i < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq3_HTML.gif and 0 < i = 1 m - 2 α i ξ i < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq4_HTML.gif.

(H2) fC((0, 1) × R × R × R, R) and there exist functions p, qL1((0, 1), (0, +∞)), gC(R × R × R, (0, +∞)) such that
- p ( t ) f ( t , x 1 , x 2 , x 3 ) q ( t ) g ( x 1 , x 2 , x 3 ) for  ( t , x 1 , x 2 , x 3 ) ( 0 , 1 ) × R × R × R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equc_HTML.gif

(H3) lim ( | x 1 | + | x 2 | + | x 3 | ) + f ( t , x 1 , x 2 , x 3 ) | x 1 | + | x 2 | + | x 3 | = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq5_HTML.gif for t uniformly on [0,1].

Remark 2.1. From (H2) we know that for given points t1, t2,..., t m on [0,1], the functions p, q = (0, 1)\{t i , i = 1, 2,..., m} → (0, +∞) are continuous and integrable, that is 0 < 0 1 ( p ( t ) + q ( t ) ) d t < + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq6_HTML.gif. The condition (H2) also implies that f may have finitely singularities at t1, t2,..., t m on [0,1].

Lemma 2.1. Suppose that (H1) and (H2) hold. Then the problem (1.1) has a positive solution if and only if the following nonlinear second-order integro-differential equation
x ( t ) + λ f t , 0 t ( t - u ) x ( u ) d u , 0 t x ( u ) d u , x ( t ) = 0 , 0 < t < 1 , x ( 0 ) = i = 1 m - 2 β i x ( ξ i ) , x ( 1 ) = i = 1 m - 2 α i x ( ξ i ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ2_HTML.gif
(2.1)

has a positive solution.

Proof. Let y(t) be a positive solution of the problem (1.1) and let x(t) = y''(t). Then it follows from the problem (1.1) and combining with exchanging the integral sequence we know that
y ( t ) = 0 t ( t - u ) x ( u ) d u , y ( t ) = 0 t x ( u ) d u . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equd_HTML.gif

Thus x(t) = y''(t) is a positive solution of the second-order integro-differential equation multi-point boundary value problem (2.1).

Conversely, let x(t) be a positive solution of the problem (2.1), then y ( t ) = 0 t ( t - u ) x ( u ) d u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq7_HTML.gif is a positive solution of the problem (1.1). In fact, y ( t ) = 0 t x ( u ) d u , y ( t ) = x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq8_HTML.gif, which implies that y(0) = 0, y'(0) = 0. The proof is complete.   □

Now, let X = C[0,1]. Then X is a real Banach space with norm x = max t [ 0 , 1 ] | x ( t ) | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq9_HTML.gif for xC[0, 1]. Let
C + [ 0 , 1 ] = { x C [ 0 , 1 ] : x ( t ) 0 , t [ 0 , 1 ] } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Eque_HTML.gif
Lemma 2.2. Suppose that (H1) holds. In addition, assume that u(t) ∈ L1(0, 1) and u(t) ≥ 0. Then the following problem
x ( t ) + u ( t ) = 0 , 0 < t < 1 , x ( 0 ) = i = 1 m - 2 β i x ( ξ i ) , x ( 1 ) = i = 1 m - 2 α i x ( ξ i ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ3_HTML.gif
(2.2)
has a unique positive solution
x ( t ) = - 0 t ( t - s ) u ( s ) d s + i = 1 m - 2 β i 0 ξ i u ( s ) d s i = 1 m - 2 β i - 1 t + 1 1 - i = 1 m - 2 α i 0 1 ( 1 - s ) u ( s ) d s - 1 1 - i = 1 m - 2 α i i = 1 m - 2 α i 0 ξ i ( ξ i - s ) u ( s ) d s + i = 1 m - 2 β i 0 ξ i u ( s ) d s i = 1 m - 2 β i - 1 1 - i = 1 m - 2 α i ξ i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ4_HTML.gif
(2.3)
satisfies x(t) ≥ 0, t ∈ [0,1] and
min t [ 0 , 1 ] x ( t ) w x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ5_HTML.gif
(2.4)
where
ω = i = 1 m - 2 α i ( 1 - ξ i ) 1 - i = 1 m - 2 α i ξ i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ6_HTML.gif
(2.5)
Proof. From (2.2), we have x'' (t) = -u(t), 0 < t < 1. For t ∈ [0,1], integrating from 0 to t we get
x ( t ) = x ( 0 ) - 0 t u ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ7_HTML.gif
(2.6)
Thus
x ( 0 ) = i = 1 m - 2 β i x ( ξ i ) = i = 1 m - 2 β i i = 1 m - 2 β i - 1 0 ξ i u ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ8_HTML.gif
(2.7)
For t ∈ [0, 1], integrating (2.6) from t to 1 yields
x ( 1 ) - x ( t ) = i = 1 m - 2 β i ( 1 - t ) i = 1 m - 2 β i - 1 0 ξ i u ( s ) d s - 0 t ( s - t ) u ( s ) d s - 0 1 ( 1 - s ) u ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ9_HTML.gif
(2.8)
which means that
- x ( t ) = - i = 1 m - 2 α i x ( ξ i ) + i = 1 m - 2 β i 0 ξ i u ( s ) d s i = 1 m - 2 β i - 1 - i = 1 m - 2 β i 0 ξ i u ( s ) d s i = 1 m - 2 β i - 1 t + 0 t ( t - s ) u ( s ) d s - 0 1 ( 1 - s ) u ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ10_HTML.gif
(2.9)
From (2.9), we have
x ( ξ i ) = 1 1 - i = 1 m - 2 α i 0 1 ( 1 - s ) u ( s ) d s - i = 1 m - 2 β i ( 1 - ξ i ) 0 ξ i u ( s ) d s i = 1 m - 2 β i - 1 - 0 ξ i ( ξ i - s ) u ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ11_HTML.gif
(2.10)
It follows from (2.9) and (2.10) that
x ( t ) = - 0 t ( t - s ) u ( s ) d s + i = 1 m - 2 β i 0 ξ i u ( s ) d s i = 1 m - 2 β i - 1 t + 0 1 ( 1 - s ) u ( s ) d s + 1 1 - i = 1 m - 2 α i i = 1 m - 2 α i 0 1 ( 1 - s ) u ( s ) d s - i = 1 m - 2 α i 0 ξ i ( ξ i - s ) u ( s ) d s - i = 1 m - 2 β i 0 ξ i u ( s ) d s i = 1 m - 2 β i - 1 1 - i = 1 m - 2 α i ξ i = - 0 t ( t - s ) u ( s ) d s + i = 1 m - 2 β i 0 ξ i u ( s ) d s i = 1 m - 2 β i - 1 t + 1 1 - i = 1 m - 2 α i 0 1 ( 1 - s ) u ( s ) d s - 1 1 - i = 1 m - 2 α i i = 1 m - 2 α i 0 ξ i ( ξ i - s ) u ( s ) d s + i = 1 m - 2 β i 0 ξ i u ( s ) d s i = 1 m - 2 β i - 1 1 - i = 1 m - 2 α i ξ i . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ12_HTML.gif
(2.11)
Combining (2.11) with (H1) we know that
x ( 0 ) = 1 1 - i = 1 m - 2 α i 0 1 ( 1 - s ) u ( s ) d s - i = 1 m - 2 α i 0 ξ i ( ξ i - s ) u ( s ) d s - i = 1 m - 2 β i 0 ξ i u ( s ) d s i = 1 m - 2 β i - 1 1 - i = 1 m - 2 α i ξ i 1 1 - i = 1 m - 2 α i i = 1 m - 2 α i 0 1 ( 1 - s ) u ( s ) d s - i = 1 m - 2 α i 0 ξ i ( ξ i - s ) u ( s ) d s - i = 1 m - 2 β i 0 ξ i u ( s ) d s i = 1 m - 2 β i - 1 1 - i = 1 m - 2 α i ξ i i = 1 m - 2 α i 1 - i = 1 m - 2 α i ξ i 1 ( 1 - s ) u ( s ) d s + i = 1 m - 2 β i 0 ξ i u ( s ) d s 1 - i = 1 m - 2 α i ξ i 1 - i = 1 m - 2 β i 1 - i = 1 m - 2 α i 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ13_HTML.gif
(2.12)

From the fact that x'' (t) = -u(t) ≤ 0, we know that the graph of x(t) is concave on [0,1].

Thus

If x(1) ≥ 0, we know that x(t) ≥ 0 for all t ∈[0,1].

If x(1) < 0, from the concavity of x once again we know that
x ( ξ i ) ξ i x ( 1 ) 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equf_HTML.gif
for i ∈ {1, 2, ..., m - 2}. This implies
x ( 1 ) = i = 1 m - 2 α i x ( ξ i ) i = 1 m - 2 α i ξ i x ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equg_HTML.gif

which contracts with the fact 0 < i = 1 m - 2 α i ξ i < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq10_HTML.gif. Thus we know that (2.4) holds.

Again from x''(t) = -u(t) ≤ 0, we see that x'(t) is non-increasing on (0, 1). Combining the condition 0 < i = 1 m - 2 β i < 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq11_HTML.gif we have x' (0) ≤ 0 and x ( t ) = x ( 0 ) - 0 t u ( s ) d s 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq12_HTML.gif for t ∈ (0, 1). Hence x(t) is non-increasing on (0, 1). By making use of the concavity of x(t) on (0, 1) we get ||x|| = x(0) and min t [ 0 , 1 ] x ( t ) = x ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq13_HTML.gif. Therefore, for all i = 1, 2, ⋯, m - 2, we obtain
x = x ( 0 ) x ( 1 ) + x ( ξ i ) - x ( 1 ) 1 - ξ i ( 1 - 0 ) = x ( 1 ) + x ( ξ i ) 1 - ξ i - x ( 1 ) 1 - ξ i = x ( ξ i ) - x ( 1 ) ξ i 1 - ξ i = 1 - i = 1 m - 2 α i ξ i i = 1 m - 2 α i ( 1 - ξ i ) x ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equh_HTML.gif
which implies that
min t [ 0 , 1 ] x ( t ) i = 1 m - 2 α i ( 1 - ξ i ) 1 - i = 1 m - 2 α i ξ i x = ω x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equi_HTML.gif

where ω is given by (2.5). This completes the proof.   □

Lemma 2.3. Suppose that (H1) holds. In addition, assume that pL1((0, 1), (0, +∞)) .

Then the following boundary value problem
x ( t ) + λ p ( t ) = 0 , 0 < t < 1 , x ( 0 ) = i = 1 m - 2 β i x ( ξ i ) , x ( 1 ) = i = 1 m - 2 α i x ( ξ i ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ14_HTML.gif
(2.13)
has a unique positive solution z satisfying z(t) ≥ 0, t ∈ [0,1], min t [ 0 , 1 ] z ( t ) ω z http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq14_HTML.gif and
z ( t ) λ G ω 0 1 p ( s ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ15_HTML.gif
(2.14)
where
G = 1 1 - i = 1 m - 2 α i + i = 1 m - 2 β i 1 - i = 1 m - 2 α i ξ i 1 - i = 1 m - 2 α i 1 - i = 1 m - 2 β i ω - 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equj_HTML.gif

ω is given by (2.5).

Proof. From Lemma 2.2. we have z(t) ≥ 0 and min t [ 0 , 1 ] z ( t ) ω z , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq15_HTML.gif t ∈ [0,1]. By making use of (2.3) we get
z ( t ) λ 1 i = 1 m 2 α i ( 0 1 ( 1 s ) p ( s ) d s i = 1 m 2 α i 0 ξ i ( ξ i s ) p ( s ) d s + i = 1 m 2 β i 0 ξ i p ( s ) d s 1 i = 1 m 2 β i ) ( 1 i = 1 m 2 α i ξ i ) λ 1 i = 1 m 2 α i ( 0 1 p ( s ) d s + i = 1 m 2 β i 0 1 p ( s ) d s 1 i = 1 m 2 β i ( 1 i = 1 m 2 α i ξ i ) ) = λ 0 1 p ( s ) d s ( 1 1 i = 1 m 2 α i + i = 1 m 2 β i ( 1 i = 1 m 2 α i ξ i ) ( 1 i = 1 m 2 α i ) ( 1 i = 1 m 2 β i ) ) = λ G ω 0 1 p ( s ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equk_HTML.gif

The proof is complete.   □

Let
[ v ( t ) ] * = v ( t ) , v ( t ) 0 , 0 , v ( t ) < 0 , 0 < t < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equl_HTML.gif
and
F ( t , [ x ( t ) z ( t ) ] * ) = f ( t , 0 t ( t u ) [ x ( u ) z ( u ) ] * d u , 0 t [ x ( u ) z ( u ) ] * d u , [ x ( t ) z ( t ) ] * ) + p ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equm_HTML.gif
Lemma 2.4. Suppose that (H1) and (H2) hold. Then the following nonlinear second-order integro-differential equation boundary value problem
{ x ( t ) + λ [ f ( t , 0 t ( t u ) [ x ( u ) z ( u ) ] * d u , 0 t [ x ( u ) z ( u ) ] * d u , [ x ( t ) z ( t ) ] * ) + p ( t ) ] = 0 , 0 < t < 1 , x ( 0 ) = i = 1 m 2 β i x ( ξ i ) , x ( 1 ) = i = 1 m 2 α i x ( ξ i ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ16_HTML.gif
(2.15)

has a positive solution x(t) with x(t) ≥ z(t) for t ∈ [0, 1] if and only if y(t) = x(t) - z(t) is a nonnegative solution (positive on (0, 1)) of the problem (2.1).

Proof. Assume that y(t) = x(t) - z(t) is a nonnegative solution (positive on (0,1)) of the problem (2.1). Then we know that x(t) ≥ z(t) and
y ( t ) + λ f t , 0 t ( t - u ) y ( u ) d u , 0 t y ( u ) d u , y ( t ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equn_HTML.gif
Noticing that z is a positive solution of the problem (2.13). Thus we get
x ( t ) = λ [ f ( t , 0 t ( t u ) [ x ( u ) z ( u ) ] * d u , 0 t [ x ( u ) z ( u ) ] * d u , [ x ( t ) z ( t ) ] * ) + p ( t ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equo_HTML.gif

Therefore x(t) is a positive solution of the problem (2.15) with x(t) ≥ z(t) for t ∈ [0,1].

Conversely, we assume that x(t) and z(t) are positive solutions of the problem (2.15) and the problem (2.13), respectively, and it implies that the boundary conditions of the problem (2.13) are also satisfied. Thus y(t) = x(t) - z(t) is a nonnegative solution (positive on (0, 1)) of the problem (2.1). The proof is complete.   □

Remark 2.2. Combining Lemma 2.4. with Lemma 2.1. we know that if the problem (2.15) has a positive solution, then the fourth-order multi-point boundary value problem (1.1) has a positive solution. So, we need only to study the problem (2.15).

Remark 2.3. For any fixed xC+[0,1], let L = max t [ 0 , 1 ] x ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq16_HTML.gif. Noticing that [x(u) - z(u)]*x(u) ≤ L and 0 s ( s - u ) [ x ( u ) - z ( u ) ] * d u 0 1 L d u = L http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq17_HTML.gif, by virtue of (H2), we obtain
0 1 F ( s , [ x ( s ) z ( s ) ] * ) d s 0 1 [ q ( s ) g ( 0 s ( s u ) [ x ( u ) z ( u ) ] * d u , 0 s [ x ( u ) z ( u ) ] * d u , [ x ( s ) z ( s ) ] * ) + p ( s ) ] d s 0 1 ( p ( s ) + M q ( s ) ) d s M 0 1 ( p ( s ) + q ( s ) ) d s < + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ17_HTML.gif
(2.16)
where
M = max ( | u 1 | , | u 2 | , | u 3 | ) [ 0 , L ] × [ 0 , L ] × [ 0 , L ] g ( | u 1 | , | u 2 | , | u 3 | ) + 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ18_HTML.gif
(2.17)
We introduce an integral mapping T : C+[0, 1] → C+[0, 1] defined by
( T x ) ( t ) = λ ( 0 t ( t s ) F ( s , [ x ( s ) z ( s ) ] * ) d s t i = 1 m 2 β i 0 ξ i F ( s , [ x ( s ) z ( s ) ] * ) d s 1 i = 1 m 2 β i ) + λ 1 i = 1 m 2 α i ( 0 1 ( 1 s ) F ( s , [ x ( s ) z ( s ) ] * ) d s i = 1 m 2 α i 0 ξ i ( ξ i s ) F ( s , [ x ( s ) z ( s ) ] * ) d s + i = 1 m 2 β i 0 ξ i F ( s , [ x ( s ) z ( s ) ] * ) d s 1 i = 1 m 2 β i ( 1 i = 1 m 2 α i ξ i ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ19_HTML.gif
(2.18)
Denote
K = { y C + [ 0 , 1 ] : min 0 t 1 y ( t ) ω | | y | | } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equp_HTML.gif

where ω is given by the problem (2.5). It is obvious that K is a positive cone of C[0,1].

Lemma 2.5. Suppose that (H1) - (H3) hold. Then T : KK is a completely continuous operator.

Proof. It follows from Lemma 2.2. we see T (K) ⊂ K. Combining (H1) with (2.18) we know that T (K) is equicontinuous and uniformly bounded. In fact, let DC+[0,1] be a bounded set. Then there exists M0 > 0 such that ||x|| ≤ M0 for all xD. By virtue of (H2) we obtain
| ( T x ) ( t ) | = | λ 0 t ( t s ) F ( s , [ x ( s ) z ( s ) ] * ) d s λ i = 1 m 2 β i 0 ξ i F ( s , [ x ( s ) z ( s ) ] * ) d s 1 i = 1 m 2 β i t + λ 1 i = 1 m 2 α i ( 0 1 ( 1 s ) F ( s , [ x ( s ) z ( s ) ] * ) d s i = 1 m 2 α i 0 ξ i ( ξ i s ) F ( s , [ x ( s ) z ( s ) ] * ) d s i = 1 m 2 β i 0 ξ i F ( s , [ x ( s ) z ( s ) ] * ) d s i = 1 m 2 β i 1 ( 1 i = 1 m 2 α i ξ i ) ) | | λ 1 i = 1 m 2 α i ( 0 1 F ( s , [ x ( s ) z ( s ) ] * ) d s + i = 1 m 2 β i ( 1 i = 1 m 2 α i ξ i ) 1 i = 1 m 2 β i 0 1 F ( s , [ x ( s ) z ( s ) ] * ) d s ) | = λ 1 i = 1 m 2 α i ( 1 + i = 1 m 2 β i ( 1 i = 1 m 2 α i ξ i ) 1 i = 1 m 2 β i ) 0 1 F ( s , [ x ( s ) z ( s ) ] * ) d s λ M ( 1 i = 1 m 2 β i i = 1 m 2 α i ξ i ) ( 1 i = 1 m 2 α i ) ( 1 i = 1 m 2 β i ) 0 1 ( q ( s ) + p ( s ) ) d s < + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equq_HTML.gif

which implies that T (K) is uniformly bounded.

On the other hand, for all xD, once again from (H2) we have
0 | ( T x ) ( t ) | = λ | 0 t F ( s , [ x ( s ) z ( s ) ] * ) d s i = 1 m 2 β i 0 ξ i F ( s , [ x ( s ) z ( s ) ] * ) d s 1 i = 1 m 2 β i | λ | 0 1 F ( s , [ x ( s ) z ( s ) ] * ) d s + i = 1 m 2 β i 1 i = 1 m 2 β i 0 1 F ( s , [ x ( s ) z ( s ) ] * ) d s | = λ 1 i = 1 m 2 β i 0 1 F ( s , [ x ( s ) z ( s ) ] * ) d s λ M 1 i = 1 m 2 β i 0 1 [ q ( s ) + p ( s ) ] d s = λ M M * 1 i = 1 m 2 β i < + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equr_HTML.gif
here M * = 0 1 [ p ( s ) + q ( s ) ] d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq18_HTML.gif. So, for any 0 ≤ t1 < t2 ≤ 1, and for all xD, we get
| ( T x ) ( t 1 ) - ( T x ) ( t 2 ) | = t 1 t 2 ( T x ) ( t ) d t λ M M * 1 - i = 1 m - 2 β i | t 2 - t 1 | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equs_HTML.gif

By the absolutely continuous of integral, we know that T(K) is equicontinuous on [0,1]. Thus, an application of the Ascoli-Arzela theorem we know that T(K) is a relatively compact set.

Now we show that T is continuous. Let x n x* (n → ∞), x n , x*C+[0,1]. It follows from the Lebesgue control convergence theorem that we obtain
( T x n ) ( t ) - ( T x * ) ( t ) 0 ( n ) , t [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equt_HTML.gif

Therefore T : KK is a completely continuous operator. The proof is complete.   □

Lemma 2.6. [17] Let X = (X, ||·||) be a Banach space and KX be a cone. For r > 0 define K r = {uK : ||u|| < r}. Assume that T : K ¯ r K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq19_HTML.gif is a completely continuous operator, such that Tuu for u ∈ ∂K r = {uK : ||u|| = r}, and

(1) If ||Tu|| ≥ ||u|| for u ∈ ∂K r , then i(T, K r , K) = 0,

(2) If ||Tu|| ≤ ||u|| for u ∈ ∂K r , then i(T, K r , K) = 1.

3 Main results

In this section, we shall apply Lemma 2.6. to establish the existence of at least one positive solutions of the problem (1.1).

Theorem 3.1. Suppose that (H1)-(H3) hold. Then there exists sufficiently small λ* > 0 such that the problem (1.1) has at least one positive solution for any λ ∈ (0, λ*) .

Proof. Let r > 1 and λ ∈ (0, λ*) be fixed, where
λ * = min r M 1 0 1 ( p ( s ) + q ( s ) ) d s 1 - i = 1 m - 2 β i i = 1 m - 2 α i ξ i 1 - i = 1 m - 2 α i 1 - i = 1 m - 2 β i - 1 , r G 0 1 p ( s ) d s , 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ20_HTML.gif
(3.1)
here M 1 = max ( | u 1 | , | u 2 | , | u 3 | ) [ 0 , r ] × [ 0 , r ] × [ 0 , r ] { g ( | u 1 | , | u 2 | , | u 3 | ) } + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq20_HTML.gif. Choose Ω r = {xC+ [0,1]: ||x|| < r}. If there is a fixed point on ∂Ω r , we complete the proof. Without loss of generality, we may assume that there is no fixed point on ∂Ω r . Thus, for any xK ∩∂Ω r , from (3.1) we get
( T x ) ( t ) λ 1 i = 1 m 2 α i 0 1 ( 1 s ) F ( s , [ x ( s ) z ( s ) ] * ) d s + λ i = 1 m 2 β i 0 ξ i F ( s , [ x ( s ) z ( s ) ] * ) d s ( 1 i = 1 m 2 α i ) ( 1 i = 1 m 2 β i ) ( 1 i = 1 m 2 α i ξ i ) λ M 1 ( 1 i = 1 m 2 β i i = 1 m 2 α i ξ i ) 0 1 ( p ( s ) + q ( s ) ) d s ( 1 i = 1 m 2 α i ) ( 1 i = 1 m 2 β i ) < r = x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equu_HTML.gif
So
T x < | | x for all  x K Ω r . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equv_HTML.gif
It follows from Lemma 2.6. we know that
i ( T , Ω r , Ω ) = 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ21_HTML.gif
(3.2)
Let d be a real number such that
1 2 λ d w r . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ22_HTML.gif
(3.3)
Choose R > max{r, λdω + 1} such that if φ > 1 r R ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq21_HTML.gif, then
F ( t , φ ) φ d  for  t [ 0 , 1 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ23_HTML.gif
(3.4)
and
1 - λ G p ( s ) R 0 1 p ( s ) d s 1 r . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ24_HTML.gif
(3.5)
Let z = p ( s ) z ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq22_HTML.gif, where z ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq23_HTML.gif is the unique solution of the problem (2.13). Denote
Ω R = { x C + [ 0 , 1 ] : x < R } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ25_HTML.gif
(3.6)
From (2.14) we have
z ( s ) = p ( s ) z ̃ ( s ) λ p ( s ) G w 0 1 p ( s ) d s λ p ( s ) G 0 1 p ( s ) d s x ( s ) x , x K Ω R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equw_HTML.gif
Thus
x ( s ) - z ( s ) 1 - λ G p ( s ) 0 1 p ( s ) d s R x ( s ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ26_HTML.gif
(3.7)
Combining (3.7) with (3.5) and by making use of Lemma 2.2. we know that
x ( s ) - z ( s ) 1 r x ( s ) 1 r min s [ 0 , 1 ] x ( s ) 1 r x w 1 r R w , s [ 0 , 1 ] , x K Ω R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ27_HTML.gif
(3.8)
From (3.8) together with (3.4), we see that
F ( s , ( x - z ) ) d ( x - z ) R d w r , s [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equx_HTML.gif
Then
0 s ( s u ) [ x ( u ) z ( u ) ] * d u + 0 s [ x ( u ) z ( u ) ] * d u + [ x ( s ) z ( s ) ] * = 0 s ( s u ) [ x ( u ) z ( u ) ] d u + 0 s [ x ( u ) z ( u ) ] d u + [ x ( s ) z ( s ) ] > [ x ( s ) z ( s ) ] 1 r R w , s [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equy_HTML.gif
Thus, for any xK ∩ ∂Ω R , it follows from (H3) we know that
f ( s , 0 s ( s u ) [ x ( u ) z ( u ) ] * d u , 0 s [ x ( u ) z ( u ) ] * d u , [ x ( s ) z ( s ) ] * ) d ( 0 s ( s u ) [ x ( u ) z ( u ) ] * d u + 0 s [ x ( u ) z ( u ) ] * d u + [ x ( s ) z ( s ) ] * ) + p ( s ) d [ x ( s ) z ( s ) ] * d [ x ( s ) z ( s ) ] d R w r , s [ 0 , 1 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ28_HTML.gif
(3.9)
Therefore, in view of (2.18) and (3.9) together with (3.3) we get
( T x ) ( 0 ) = λ 1 i = 1 m 2 α i ( 0 1 ( 1 s ) F ( s , [ x ( s ) z ( s ) ] * ) d s i = 1 m 2 α i 0 ξ i ( ξ i s ) F ( s , [ x ( s ) z ( s ) ] * ) d s i = 1 m 2 β i 0 ξ i F ( s , [ x ( s ) z ( s ) ] * ) d s i = 1 m 2 β i 1 ( 1 i = 1 m 2 α i ξ i ) ) λ 1 i = 1 m 2 α i [ 0 1 ( 1 s ) F ( s , [ x ( s ) z ( s ) ] * ) d s i = 1 m 2 α i 0 1 ( ξ i s ) F ( s , [ x ( s ) z ( s ) ] * ) d s ] = λ 0 1 ( 1 s ) F ( s , [ x ( s ) z ( s ) ] * ) d s = λ 0 1 ( 1 s ) [ f ( s , 0 s ( s u ) [ x ( u ) z ( u ) ] * d u , 0 s [ x ( u ) z ( u ) ] * d u , [ x ( s ) z ( s ) ] * ) + p ( s ) ] d s λ d R w 2 r R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equz_HTML.gif
which implies that ||Tx|| ≥ || x || for all xK ∩ ∂Ω R . It follows from Lemma 2.6 that
i ( T , K Ω R , K ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ29_HTML.gif
(3.10)
Combining (3.2) with (3.10) and the additivity of fixed point index, we know that
i ( T , K ( Ω ¯ R \ Ω r ) , K ) = i ( T , K Ω ¯ R , K ) - i ( T , K Ω r , K ) = - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equaa_HTML.gif
As a result, there exists x * K ( Ω ¯ R \ Ω r ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq24_HTML.gif satisfying Tx* = x* and r ≤ ||x*|| ≤ R. From (3.1) we have
x * ( t ) - z ( t ) w x * - λ G w 0 1 p ( s ) d s w x * - r w > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equab_HTML.gif

Let y(t) = x*(t) - z(t). Then y(t) is a positive solution of the problem (2.1). By virtue of Lemma 2.1. we know that y ( t ) = 0 t ( t - s ) x * ( s ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq25_HTML.gif is a positive solution of the fourth-order multi-point boundary value problem (1.1). This completes the proof.   □

Remark 3.1. In the case, when f = f(t, u) and f has lower bound i. e. f(t, u) + M ≥ 0 for some M > 0, we can study the second-order multi-point boundary value problem under suitable condition by making use of the similar method. In particular, if p(t) = M, the conclusion of Theorem 3.1. is still valid.

Remark 3.2. The constant λ in problem (1.1) is usually called the Thiele modulus, in ap-plications, one is interested in showing the existence of positive solutions for semipositone problems for small enough λ > 0.

4 Examples

Example 4.1. Consider the following singular fourth-order semipositone boundary value problem:
{ y ( 4 ) ( t ) + 4 λ 3 ( 1 t ) 2 3 [ sin 8 ( | y ( t ) | + | y ( t ) | ) + e | y ( t ) | + | y ( t ) | + | y ( t ) | + ( | y ( t ) | + | y ( t ) | + | y ( t ) | ) 1 3 ] 2 t = 0 , t ( 0 , 1 ) y ( 0 ) = y ( 0 ) = 0 , y ( 1 ) = 1 3 y ( 1 2 ) , y ( 0 ) = 1 4 y ( 1 2 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ30_HTML.gif
(4.1)
Proof. Let
f ( t , u 1 , u 2 , u 3 ) = 4 3 ( 1 t ) 2 3 [ sin 8 ( | u 1 | + | u 2 | ) + e | u 1 | + | u 2 | + | u 3 | + ( | u 1 | + | u 2 | + | u 3 | ) 1 3 ] 2 t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equac_HTML.gif
Then
- p ( t ) f ( t , u 1 , u 2 , u 3 ) q ( t ) g ( u 1 , u 2 , u 3 ) and lim ( | u 1 | + | u 2 | + | u 3 | ) + f ( t , u 1 , u 2 , u 3 ) | u 1 | + | u 2 | + | u 3 | = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equad_HTML.gif
where p ( t ) = 1 t , q ( t ) = 4 3 ( 1 - t ) 2 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq26_HTML.gif, g ( u 1 , u 2 , u 3 ) = sin 8 ( | u 1 | + | u 2 | ) + e | u 1 | + | u 2 | + | u 3 | + ( | u 1 | + | u 2 | + | u 3 | ) 1 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq27_HTML.gif, which implies that (H1)-(H3) hold. Since α = 1 3 , β = 1 4 , ξ = 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq28_HTML.gif, then we know that
0 1 ( p ( s ) + q ( s ) ) d s = 0 1 2 s + 4 3 ( 1 - s ) 2 3 d s = 8 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equae_HTML.gif
Take r = 2, then
M 1 = max ( | u 1 | , | u 2 | , | u 3 | ) [ 0 , 2 ] × [ 0 , 2 ] × [ 0 , 2 ] { g ( | u 1 | , | u 2 | , | u 3 | ) } + 1 = max ( | u 1 | , | u 2 | , | u 3 | ) [ 0 , 2 ] × [ 0 , 2 ] × [ 0 , 2 ] { [ sin 8 ( | u 1 | + | u 2 | ) + e | u 1 | + | u 2 | + | u 3 | + ( | u 1 | + | u 2 | + | u 3 | ) 1 3 ] } + 1 = 2 + e 6 + 6 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equaf_HTML.gif
G = ( 1 1 i = 1 m 2 α i + i = 1 m 2 β i ( 1 i = 1 m 2 α i ξ i ) ( 1 i = 1 m 2 α i ) ( 1 i = 1 m 2 β i ) ) ( 1 i = 1 m 2 α i ξ i ) i = 1 m 2 α i ( 1 ξ i ) = 115 12 . r G 0 1 p ( s ) d s = 2 G 0 1 2 s d s = 6 115 . r [ M 1 0 1 ( p ( s ) + q ( s ) ) d s ( 1 i = 1 m 2 β i i = 1 m 2 α i ξ i ) ( 1 i = 1 m 2 α i ) ( 1 i = 1 m 2 β i ) ] 1 = 3 23 ( 2 + e 6 + 6 3 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equag_HTML.gif

It follows from Theorem 3.1 that the problem (4.1) has at least one positive solution for any λ * 0 , 3 23 ( 2 + e 6 + 6 3 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq29_HTML.gif.   □

Example 4.2. Consider the following singular fourth-order semipositone boundary value problem:
{ y ( 4 ) ( t ) + 8 λ π t ( 1 t ) [ sin 19 ( | y ( t ) | + | y ( t ) | ) + 28 e | y ( t ) | + | y ( t ) | + | y ( t ) | + ( | y ( t ) | + | y ( t ) | + | y ( t ) | ) 1 8 ] 3 1 t = 0 , t ( 0 , 1 ) y ( 0 ) = y ( 0 ) = 0 , y ( 1 ) = 1 3 y ( 1 2 ) + 1 6 y ( 2 3 ) + 1 10 y ( 5 6 ) , y ( 0 ) = 1 6 y ( 1 2 ) + 1 4 y ( 2 3 ) + 1 10 y ( 5 6 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equ31_HTML.gif
(4.2)
Proof. Let
f ( t , u 1 , u 2 , u 3 ) = 8 π t ( 1 t ) [ sin 19 ( | u 1 | + | u 2 | ) + 28 e | u 1 | + | u 2 | + | u 3 | + ( | u 1 | + | u 2 | + | u 3 | ) 1 8 ] 3 1 t http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equah_HTML.gif
Then
- p ( t ) f ( t , u 1 , u 2 , u 3 ) q ( t ) g ( u 1 , u 2 , u 3 ) and lim ( | u 1 | + | u 2 | + | u 3 | ) + f ( t , u 1 , u 2 , u 3 ) | u 1 | + | u 2 | + | u 3 | = + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equai_HTML.gif
where p ( t ) = 3 1 - t , q ( t ) = 8 π t ( 1 - t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq30_HTML.gif, g ( u 1 , u 2 , u 3 ) = sin 19 ( | u 1 | + | u 2 | ) + 28 e | u 1 | + | u 2 | + | u 3 | + ( | u 1 | + | u 2 | + | u 3 | ) 1 8 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq31_HTML.gif, which implies that (H1)-(H3) hold. Since α 1 = 1 3 , α 2 = 1 2 , α 3 = 1 5 , ξ 1 = 1 2 , ξ 2 = 2 3 , ξ 3 = 5 6 , β 1 = 1 6 , β 2 = 1 4 , β 3 = 1 10 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq32_HTML.gif, then we know that
0 1 ( p ( s ) + q ( s ) ) d s = 0 1 3 1 - s + 8 π s ( 1 - s ) d s = 6 + 8 = 14 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equaj_HTML.gif
Take r = 3, then
M 1 = max ( | u 1 | , | u 2 | , | u 3 | ) [ 0 , 3 ] × [ 0 , 3 ] × [ 0 , 3 ] { g ( | u 1 | , | u 2 | , | u 3 | ) } + 1 = max ( | u 1 | , | u 2 | , | u 3 | ) [ 0 , 3 ] × [ 0 , 3 ] × [ 0 , 3 ] { [ sin 19 ( | u 1 | + | u 2 | ) + 28 e | u 1 | + | u 2 | + | u 3 | + ( | u 1 | + | u 2 | + | u 3 | ) 1 8 ] } + 1 = 2 + 28 e 9 + 3 4 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equak_HTML.gif
G = 1 1 - i = 1 m - 2 α i + i = 1 m - 2 β i 1 - i = 1 m - 2 α i ξ i 1 - i = 1 m - 2 α i 1 - i = 1 m - 2 β i 1 - i = 1 m - 2 α i ξ i i = 1 m - 2 α i ( 1 - ξ i ) = 1 2 5 + 31 60 × 23 36 2 5 × 29 60 × 23 36 43 180 = 1010275 89784 . r G 0 1 p ( s ) d s = 3 1010275 89784 × 6 = 44892 1010275 . r M 1 0 1 ( p ( s ) + q ( s ) ) d s 1 - i = 1 m - 2 β i i = 1 m - 2 α i ξ i 1 - i = 1 m - 2 α i 1 - i = 1 m - 2 β i - 1 = 3 × ( 2 + 28 e 9 + 3 4 ) × 14 2 5 × 1 - 31 60 × 13 36 29 60 - 1 = 3132 61495 ( 2 + 28 e 9 + 3 4 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_Equal_HTML.gif

It follows from Theorem 3.1 that the problem (4.2) has at least one positive solution for any λ * 0 , 3132 61495 ( 2 + 28 e 9 + 3 4 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-12/MediaObjects/13661_2011_Article_124_IEq33_HTML.gif.   □

Declarations

Acknowledgements

The author is very grateful to Editor of the Journal and the anonymous referees for their carefully reading of the first draft of the manuscript and making many valuable suggestions and comments. The author was supported financially by the Foundation of Shanghai Municipal Education Commission (Grant Nos. DZL803, 10YZ77, and DYL201105).

Authors’ Affiliations

(1)
Department of Mathematics, Shanghai Normal University

References

  1. Feng W: On a m -point nonlinear boundary value problem. Nonlinear Anal 1997, 30(6):5369–5374.MathSciNetView Article
  2. Feng W, Webb JRL: Solvability of a m -point boundary value problems with non-linear growth. J Math Anal Appl 1997, 212: 467–480. 10.1006/jmaa.1997.5520MathSciNetView Article
  3. Gupta CP: A generalized multi-point boundary value problem for second order or-dinary differential equation. Appl Math Comput 1998, 89: 133–146. 10.1016/S0096-3003(97)81653-0MathSciNetView Article
  4. Gupta CP: A Second order m-point boundary value problem at resonance. Nonlinear Anal 1995, 24(10):1483–1489. 10.1016/0362-546X(94)00204-UMathSciNetView Article
  5. Gupta CP, Ntouyas SK, Tsamatos P: On an m -point boundary value problem for second-order ordinary differential equations. Nonlinear Anal 1994, 23: 1427–1436. 10.1016/0362-546X(94)90137-6MathSciNetView Article
  6. Gupta CP, Ntouyas SK, Tsamatos P: Solvability of an m -point boundary value problem for second order ordinary differential equations. J Math Anal Appl 1995, 189: 575–584. 10.1006/jmaa.1995.1036MathSciNetView Article
  7. Gupta CP: Solvability of a three-point nonlinear boundary value problems for a second order ordinary differential equation. J Math Anal Appl 1992, 168: 540–551. 10.1016/0022-247X(92)90179-HMathSciNetView Article
  8. Gupta CP: A sharp condition for the solvability of a three-point second order boundary value problem. J Math Anal Appl 1997, 205: 579–586.View Article
  9. Gupta CP, Trofimchuk S: Existence of a solution to a three-point boundary value problem and the spectral radius of a related linear operator. Nonlinear Anal 1998, 34: 498–507.MathSciNetView Article
  10. Il'in VA, Moiseev EI: Nonlocal boundary-value problem of the first kind for a Sturm-liouville operator in its differential and finite difference aspects. Diff Equ 1987, 23: 803–810.
  11. Krasnoselskii MA: Positive Solutions of Operator Equation. Noordhoff, Groningen; 1964.
  12. Erbe L, Wang H: On the existence of positive solutions of ordinary differential equations. Proc Am Math Soc 1994, 120: 743–748. 10.1090/S0002-9939-1994-1204373-9MathSciNetView Article
  13. Yao Q: Existence of n solutions and/or positive solutions to a semipositone elastic. Nonlinear Anal 2007, 66: 138–150. 10.1016/j.na.2005.11.016MathSciNetView Article
  14. Yao Q: Existence and multiplicity of positive solutions to a singular elastic beam equation rigidly fixed at both ends. Nonlinear Anal 2008, 69: 2683–2694. 10.1016/j.na.2007.08.043MathSciNetView Article
  15. Yao Q: Positive solutions of a nonlinear elastic beam equation rigidly fastened on the left and simply supported on the right. Nonlinear Anal 2008, 69: 1570–1580. 10.1016/j.na.2007.07.002MathSciNetView Article
  16. Anuradha V, Hai DD, Shivaji R: Existence results for superlinear semipositone BVPs. Proc Am Math Soc 1996, 124: 757–746. 10.1090/S0002-9939-96-03256-XMathSciNetView Article
  17. Guo D, Lakshmikantham V: Nonlinear Problems in Abstract Cone. Academic Press, New York; 1988.

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