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Existence of positive solutions for fourth-order semipositone multi-point boundary value problems with a sign-changing nonlinear term Boundary Value Problems 2012, 2012:12 doi:10.1186/16

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Existence of positive solutions for fourth-order semipositone multi-point

boundary value problems with a sign-changing nonlinear term

Boundary Value Problems 2012, 2012:12 doi:10.1186/1687-2770-2012-12

Yan Sun (ysun@shnu.edu.cn)

ISSN 1687-2770

Article type Research

Submission date 23 July 2011

Acceptance date 9 February 2012

Publication date 9 February 2012

Article URL http://www.boundaryvalueproblems.com/content/2012/1/12

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semipositone multi-point boundary value problems

with a sign-changing nonlinear term

Yan Sun

Department of Mathematics, Shanghai Normal University,Shanghai 200234, People’s Republic of ChinaEmail addresses: ysun@shnu.edu.cn; ysun881@sina.com.cn

Abstract

In this article, some new sufficient conditions are obtained by making use of fixedpoint index theory in cone and constructing some available integral operators to-gether with approximating technique They guarantee the existence of at least onepositive 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 exam-

ples are given to demonstrate the validity of our main results

1

Keywords: semipositone; positive solutions; multi-point boundary value problems.

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

β i < 1 Here, by a positive solution

problem (1.1)

The existence of positive solutions for multi-point boundary value problems has beenwidely studied in recent years For details, see [1–15] 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

positive solutions for more general multi-point boundary value problems

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 nonlinearboundary value problems in a general Banach spaces When the nonlinearity is negative,such kinds of the problems are called semipositone problems, which occur in chemicalrector theory, combustion and management of natural resources, see [11, 13–16] To ourbest 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 andconstruct an integral operator for the corresponding second-order multi-point boundaryvalue 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 inSection 3 A class of examples are given to show that our main result is applicable tomany 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 K ⊂ E is

called a cone if it satisfies the following two conditions:

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:

that is 0 < 01(p(t) + q(t))dt < +∞ The condition (H2) also implies that f may have finitely singularities at t1, t2, , t m on [0, 1].

solution if and only if the following nonlinear second-order integro-differential equation

has a positive solution.

follows from the problem (1.1) and combining with exchanging the integral sequence weknow that

multi-point boundary value problem (2.1)

u)x(u)du is a positive solution of the problem (1.1) In fact, y 0 (t) = R0t x(u)du, y 00 (t) =

Now, let X = C[0, 1] Then X is a real Banach space with norm kxk = max

u(t) ≥ 0 Then the following problem

which means that

Combining (2.11) with (H1) we know that

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

α i ξ i < 1 Thus we know that (2.4) holds.

i=1

β i < 1 we have x 0 (0) ≤ 0 and x 0 (t) = x 0 (0) −R0t u(s)ds ≤ 0 for

t ∈ (0, 1) Hence x(t) is non-increasing on (0, 1) By making use of the concavity of x(t) on

t∈[0,1] x(t) = x(1) Therefore, for all i = 1, 2, · · · , m − 2,

Then the following boundary value problem

Proof From Lemma 2.2 we have z(t) ≥ 0 and min

second-order integro-differential equation boundary value problem

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

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

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)

t∈[0,1] x(t) Noticing that [x(u)−z(u)] ∗ ≤

x(u) ≤ L and¯ 0s (s − u)[x(u) − z(u)] ∗ du ¯ ≤ 1

0 Ldu = L, by virtue of (H2), we obtain

(|u1|,|u2|,|u3|)∈[0,L]×[0,L]×[0,L] g(|u1|, |u2|, |u3|) + 1. (2.17)

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

continuous operator.

On the other hand, for all x ∈ D, once again from (H2) we have

follows from the Lebesgue control convergence theorem that we obtain

k(T x n )(t) − (T x ∗ )(t)k −→ 0 (n → ∞), t ∈ [0, 1].

Lemma 2.6 [17] Let X = (X, k · k) be a Banach space and K ⊂ X be a cone For r > 0

(1) If kT uk ≥ kuk for u ∈ ∂K r , then i(T, K r , K) = 0,

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

such that the problem (1.1) has at least one positive solution for any λ ∈ (0, λ ∗ ).

(|u1|,|u2|,|u3|)∈[0,r]×[0,r]×[0,r] {g(|u1|, |u2|, |u3|)} + 1 Choose Ω r = {x ∈ C+[0, 1] :

kxk < r} If there is a fixed point on ∂Ω r, we complete the proof Without loss of

From (3.8) together with (3.4), we see that

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.

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.

+ (|y(t)| + |y 0 (t)| + |y 00 (t)|)13

i

− √2

t = 0, t ∈ (0, 1) y(0) = y 0 (0) = 0, y 00(1) = 1

i

− √2t

Then

−p(t) ≤ f (t, u1, u2, u3) ≤ q(t)g(u1, u2, u3) and lim

(|u1|+|u2|+|u3|)→+∞

(1−t)2, g(u1, u2, u3) = sin8(|u1| + |u2|) + e |u1|+|u2|+|u3| +(|u1| +

|u2| + |u3|)13, which implies that (H1)−(H3) hold Since α = 1

1 − t

Then

−p(t) ≤ f (t, u1, u2, u3) ≤ q(t)g(u1, u2, u3) and lim

(|u1|+|u2|+|u3|)→+∞

t(1−t) , g(u1, u2, u3) = sin19(|u1| + |u2|) + 28e |u1|+|u2|+|u3| +

(|u1| + |u2| + |u3|)1, which implies that (H1)−(H3) hold Since α1 = 1

2 5

+

31

60 × 23 36 2

5 ×29 60

¶

×

23 36 43 180

31

60× 13 36 29 60

and comments The author was supported financially by the Foundation of ShanghaiMunicipal Education Commission (Grant Nos DZL803, 10YZ77, and DYL201105).

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