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Existence of positive solutions for nonlocal second-order boundary value problem with variable parameter in Banach spaces

  • Peiguo Zhang
Open Access
Research

Abstract

By obtaining intervals of the parameter λ, this article investigates the existence of a positive solution for a class of nonlinear boundary value problems of second-order differential equations with integral boundary conditions in abstract spaces. The arguments are based upon a specially constructed cone and the fixed point theory in cone for a strict set contraction operator.

MSC: 34B15; 34B16.

Keywords

boundary value problem positive solution fixed point theorem measure of noncompactness 

1 Introduction

The existence of positive solutions for second-order boundary value problems has been studied by many authors using various methods (see [1, 2, 3, 4, 5, 6]). Recently, the integral boundary value problems have been studied extensively. Zhang et al. [7] investigated the existence and multiplicity of symmetric positive solutions for a class of p-Laplacian fourth-order differential equations with integral boundary conditions. By using Mawhin's continuation theorem, some sufficient conditions for the existence of solution for a class of second-order differential equations with integral boundary conditions at resonance are established in [8]. Feng et al. [9] considered the boundary value problems with one-dimensional (1D) p-Laplacian and impulse effects subject to the integral boundary condition. This study in this article is motivated by Feng and Ge [1], who applied a fixed point theorem [10] in cone to the second-order differential equations.
x ( t ) + f ( t , x ( t ) ) = θ , 0 < t < 1 , x ( 0 ) = 0 1 g ( t ) x ( t ) d t , x ( 1 ) = θ . Open image in new window
Let E be a real Banach space with norm || · || and PE be a cone of E. The purpose of this article is to investigate the existence of positive solutions of the following second-order integral boundary value problem:
- x ( t ) + q ( t ) x ( t ) = λ f ( t , x ) , 0 < t < 1 , x ( 0 ) = 0 1 g ( t ) x ( t ) d t , x ( 1 ) = θ , Open image in new window
(1.1)

where qC[0, 1], λ > 0 is a parameter, f(t, x) ∈ C([0, 1] × P, P), and gL1[0, 1] is nonnegative, θ is the zero element of E.

The main features of this article are as follows. First, the author discusses the existence results in the case qC[0, 1], not q(t) = 0 as in [1]. Second, comparing with [1], let us consider the existence results in the case λ > 0, not λ = 1 as in [1]. To our knowledge, no article has considered problem (1.1) in abstract spaces.

The organization of this article is as follows. In Section 2, the author provides some necessary background. In particular, the author states some properties of the Green function associated with problem (1.1). In Section 3, the main results will be stated and proved.

Basic facts about ordered Banach space E can be found in [10, 11]. In this article, let me just recall a few of them. The cone P in E induces a partial order on E, i.e., xy if and only if y - xP. P is said to be normal if there exists a positive constant N such that θxy implies ||x|| ≤ N||y||. Without loss of generality, let us suppose that, in the present article, the normal constant N = 1.

Now let us consider problem (1.1) in C[I, E], in which I = [0, 1]. Evidently, (C[I, E], || · || c ) is a Banach space with norm ||x|| c = max t∈I ||x(t)|| for xC[I, E]. In the following, xC[I, E] is called a solution of (1.1) if it satisfies (1.1). x is a positive solution of (1.1) if, in addition, x(t) > θ for t ∈ (0, 1).

In the following, the author denotes Kuratowski's measure of noncompactness by α(·).

Lemma 1.1[10]Let K be a cone of Banach space E and K r, R = {xK, r ≤ ||x|| ≤ R}, R > r > 0. Suppose that A:K r, R K is a strict set contraction such that one of the following two conditions is satisfied:
a | | A x | | | | x | | , x K , | | x | | = r ; | | A x | | | | x | | , x K , | | x | | = R . Open image in new window
b | | A x | | | | x | | , x K , | | x | | = r ; | | A x | | | | x | | , x K , | | x | | = R . Open image in new window

Then, A has a fixed point xK r, R such that r ≤ ||x|| ≤ R.

2 Preliminaries

To establish the existence and nonexistence of positive solutions in C[I, P] of (1.1), let us list the following assumptions, which will hold throughout this article:
  1. (H)

    m ( t ) = 0 t q ( s ) d s , 0 1 e m ( x ) d x = c R Open image in new window, inf t∈I {m(t)} = d > -∞, and for any r > 0, f is uniformly continuous on I × P r . f(t, P r ) is relatively compact, and there exist a, bL(I, R +), and wC(R +, R +), such that ||f(t, x)|| ≤ a(t) + b(t)w(||x||), a.e. tI, xP, where P r = PT r .

     

In the case of main results of this study, let us make use of the following lemmas.

Lemma 2.1 Assume that (H) holds, then x is a nonnegative solution of (1.1) if and only if x is a fixed point of the following integral operator:
( T x ) ( t ) = λ 0 1 H ( t , s ) f ( s , x ( s ) ) d s , Open image in new window
(2.1)
where
( T x ) ( t ) = λ c e m ( t ) - 0 t e - m ( s ) f ( s , x ( s ) ) 0 s e m ( x ) d x d s + t 1 e - m ( s ) f ( s , x ( s ) ) s 1 e m ( x ) d x d s (1) - λ 0 1 1 c ( 1 - σ ) e m ( t ) 0 1 g ( τ ) G ( τ , s ) d τ f ( s , x ( s ) ) d s , (2) ( T x ) ( t ) = m ( t ) ( T x ) ( t ) - λ f ( t , x ( t ) ) . (3) (4)  Open image in new window
Proof. By
T x ( t ) = λ 0 1 H ( t , s ) f ( s , x ( s ) ) d s = λ 0 1 G ( t , s ) f ( s , x ( s ) ) d s + λ 0 1 1 c ( 1 - σ ) t 1 e m ( x ) d x 0 1 g ( τ ) G ( τ , s ) d τ f ( s , x ( s ) ) d s = λ c 0 t f ( s , x ( s ) ) e m ( s ) 0 s e m ( x ) d x t 1 e m ( x ) d x d s + t 1 f ( s , x ( s ) ) e m ( s ) 0 t e m ( x ) d x s 1 e m ( x ) d x d s + λ 0 1 1 c ( 1 - σ ) t 1 e m ( x ) d x 0 1 g ( τ ) G ( τ , s ) d τ f ( s , x ( s ) ) d s , Open image in new window
we get
( T x ) ( t ) = λ c e m ( t ) - 0 t e - m ( s ) f ( s , x ( s ) ) 0 s e m ( x ) d x d s + t 1 e - m ( s ) f ( s , x ( s ) ) s 1 e m ( x ) d x d s - λ 0 1 1 c ( 1 - σ ) e m ( t ) 0 1 g ( τ ) G ( τ , s ) d τ f ( s , x ( s ) ) d s , ( T x ) ( t ) = m ( t ) ( T x ) ( t ) - λ f ( t , x ( t ) ) . Open image in new window
Therefore,
- ( T x ) ( t ) + m ( t ) ( T x ) ( t ) = λ f ( t , x ( t ) ) , t ( 0 , 1 ) . Open image in new window

Moreover, by G(0, s) = G(1, s) = 0, it is easy to verify that T x ( 0 ) = 0 1 g ( s ) T x ( s ) d s Open image in new window, Tx(1) = θ. The lemma is proved.

For convenience, let us define
k = sup t ( 0 , 1 ) 1 c e - m ( t ) 0 t e m ( x ) d x t 1 e m ( x ) d x , e ( t ) = 1 c t 1 e m ( x ) d x , h ( t ) = G ( t , t ) + 1 1 - σ 0 1 g ( x ) G ( x , t ) d x , k 0 = k + k 1 - σ 0 1 g ( x ) d x . Open image in new window

For the Green's function G(t, s), it is easy to prove that it has the following two properties.

Proposition 2.1 For t, sI, we have 0 ≤ H(t, s) ≤ h(s) ≤ k0.

Proposition 2.2 For t, w, sI, we have H(t, t) ≥ e(s)H(w, s).

To obtain a positive solution, let us construct a cone K by
K = { x Q : x ( t ) e ( t ) x ( s ) , t , s I } Open image in new window
(2.2)

where Q = {xC[I, E]: x(t) ≥ θ, tI}.

It is easy to see that K is a cone of C[I, E] and K r, R = {xK: r ≤ ||x|| ≤ R} ⊂K, KQ.

In the following, let B l = {xC[I, E]: ||x|| c l}, l > 0.

Lemma 2.2[10]Let H be a countable set of strongly measurable function x: JE such that there exists a ML[I, R+] such that ||x(t)|| ≤ M(t) a.e. tI for all xH. Then α(H(t)) ∈ L[I, R+] and
α J x ( t ) d t : x H 2 J α ( H ( t ) ) d t . Open image in new window

Lemma 2.3 Suppose that (H) holds. Then T(K) ⊂K and T: K r, R K is a strict set contraction.

Proof. Observing H(t, s) ∈ C(I × I) and fC(I × P, P), we can get TuC(I, E). For any uK, we have

T u ( t ) = λ 0 1 H ( t , s ) f ( s , u ( s ) ) d s λ e ( t ) 0 1 H ( w , s ) f ( s , u ( s ) ) d s = e ( t ) T u ( w ) , t , w ( 0 , 1 ) Open image in new window, thus, T: KK. Therefore, by (H), it is easily seen that TC(K, K). On the other hand, let V = { u n } n = 1 Open image in new window, be a bounded sequence, ||u n || c r, let M r = {w(v): 0 ≤ vr}, be (H), then we have
f ( t , u n ( t ) ) a ( t ) + b ( t ) M r , u n V , a . e . t I . Open image in new window
Then
α ( T u n ( t ) : u n V ) = α λ 0 1 H ( t , s ) f ( s , u n ( s ) ) d s : u n V 2 λ k 0 0 1 α ( f ( s , u n ( s ) ) : u n V ) d s = 0 . Open image in new window

Hence, T: K r, R K is a strict set contraction. The proof is complete.

3 Main results

Definition 3.1 Let P be a cone of real Banach space E. If P* = {φE* |φ(x) ≥ 0, xP}, then P* is a dual cone of cone P. Write
f β = limsup x β max t I f ( t , x ) x , ( φ f ) β = liminf x β min t I φ ( f ( t , x ) ) x , A = max t I 0 1 e ( s ) H ( t , s ) d s , B = max t I 0 1 H ( t , s ) d s , Open image in new window

where β denotes 0 or ∞, φP*, and ||φ|| = 1.

In this section, let us apply Lemma 1.1 to establish the existence of a positive solution for problem (1.1).

Theorem 3.1 Assume that (H) holds, P is normal and for any xP, A(φf) > Bf0. Then problem (1.1) has at least one positive solution in K provided
1 A ( φ f ) < λ < 1 B f 0 . Open image in new window

Proof. Let T be a cone preserving, strict set contraction that was defined by (2.1).

According to (3.1), there exists ε > 0 such that
1 A [ ( φ f ) - ε ] < λ < 1 B ( f 0 + ε ) . Open image in new window

Considering f0 < ∞, there exists r1 > 0 such that ||f(t, x)|| ≤ (f0 + ε)||x||, for ||x|| ≤ r1, xP, and tI.

Therefore, for tI, xK, ||x|| c = r1, we have
T x ( t ) = λ 0 1 H ( t , s ) f ( s , x ( s ) ) d s λ ( f 0 + ε ) 0 1 H ( t , s ) x ( s ) d s λ ( f 0 + ε ) x c 0 1 H ( t , s ) d s λ ( f 0 + ε ) x c B x c . Open image in new window
Therefore,
T x c x c , t I , x K , x c = r 1 . Open image in new window
Next, turning to (φf) > 0, there exists r2 > r1, such that φ(f(t, x(t))) ≥ [(φf) - ε] ||x||, for ||x|| ≥ r2, xP, tI. Then, for tI, xK, ||x|| c = r2, we have by Proposition 2.2 and (2.8),
T x ( t ) φ ( ( T u ) ( t ) ) = λ 0 1 H ( t , s ) φ ( f ( s , x ( s ) ) ) d s λ 0 1 H ( t , s ) ( ( φ f ) - ε ) x ( s ) d s λ ( ( φ f ) - ε ) 0 1 H ( t , s ) e ( s ) x c d s λ ( ( φ f ) - ε ) x c A x c . Open image in new window
Therefore,
| | T x | | c | | x | | c , t I , x K , | | x | | c = r 2 . Open image in new window

Applying (b) of Lemma 1.1 to (3.3) and (3.4) yields that T has a fixed point x * K r 1 , r 2 , r 1 x * c r 2 Open image in new window and x*(t) ≤ e(t)x*(s) > θ, tI, sI.

The proof is complete.

Similar to the proof of Theorem 3.1, we can prove the following results.

Theorem 3.2 Assume that (H) holds, P is normal and for any xP, A(φf)0 > Bf. Then problem (1.1) has at least one positive solution in K provided
1 A ( φ f ) 0 < λ < 1 B f . Open image in new window

Proof. Considering (φf)0 > 0, there exists r3 > 0 such that φ(f(t, x)) ≥ [(φf)0 - ε]||x||, for ||x|| ≤ r3, xP, tI.

Therefore, for tI, xK, ||x|| c = r3, similar to (3.3), we have
T x ( t ) φ ( ( T u ) ( t ) ) λ [ ( φ f ) 0 - ε ] x c A x c . Open image in new window
Therefore,
T x c x c , t I , x K , x c = r 3 . Open image in new window
Using a similar method, we can get r4 > r3, such that
| | T x | | c | | x | | c , t I , x K , | | x | | c = r 4 . Open image in new window

Applying (a) of Lemma 1.1 to (3.3) and (3.4) yields that T has a fixed point x * K r 3 , r 4 , r 3 x * c r 4 Open image in new window and x*(t) ≤ e(t)x*(s) > θ, tI, sI.

The proof is complete.

Theorem 3.3 Assume that (H) holds, P is normal and for any ||f(t, x)|| ≤ ||x||, ||x|| > 0. Then problem (1.1) has no positive solution in K provided λB < 1.

Proof. Assume to the contrary that x(t) is a positive solution of the problem (1.1). Then xK, ||x|| c > 0 for tI, and
x ( t ) = λ 0 1 H ( t , s ) f ( s , x ( s ) ) d s λ 0 1 H ( t , s ) x ( s ) d s λ x c 0 1 H ( t , s ) d s λ B x c x c , Open image in new window

which is a contradiction, and completes the proof.

Similarly, we have the following results.

Theorem 3.4 Assume that (H) holds, P is normal and for any ||f(t, x)|| ≥ ||x||, ||x|| > 0 Then problem (1.1) has no positive solution in K provided λA > 1.

Remark 3.1 When q(t) ≡ 0, λ = 1, the problem (1.1) reduces to the problem studied in[1], and so our results generalize and include some results in[1].

Notes

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Copyright information

© Zhang; 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.

Authors and Affiliations

  1. 1.Department of Elementary EducationHeze UniversityHezePeople's Republic of China

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