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Existence of positive solutions to a non-positive elastic beam equation with both ends fixed

Haixia Lu1, Li Sun23* and Jingxian Sun2

Author Affiliations

1 Department of Teachers Education, Suqian College, Suqian, Jiangsu 223800, P. R. China

2 Department of Mathematics, Jiangsu Normal University, Xuzhou, Jiangsu 221116, P. R. China

3 School of Mechanics and Civil Engineering, China University of Mining & Technology, Xuzhou, Jiangsu 221008, P. R. China

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Boundary Value Problems 2012, 2012:56  doi:10.1186/1687-2770-2012-56

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2012/1/56

Received:30 November 2011
Accepted:14 May 2012
Published:14 May 2012

© 2012 Lu et al; licensee Springer.

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.


This article is concerned with the existence of nontrivial solutions for a non-positive fourth-order two-point boundary value problem (BVP) and the existence of positive solutions for a semipositive fourth-order two-point BVP. In mechanics, the problem describes the deflection of an elastic beam rigidly fixed at both ends. The method to show our main results is the topological degree and fixed point theory of nonlinear operator on lattice.

Mathematics Subject Classification 2010: 34B18; 34B16; 34B15.

lattice; topological degree; fixed point; nontrivial solutions and positive solutions; elastic beam equations

1 Introduction

The purpose of this article is to investigate the existence of nontrivial solutions and positive solutions of the following nonlinear fourth-order two-point boundary value problem (for short, BVP)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M1">View MathML</a>


where λ is a positive parameter, f : [0,1] × R1 R1 is continuous.

Fourth-order two-point BVPs are useful for material mechanics because the problems usually characterize the deflection of an elastic beam. The problem (P) describes the deflection of an elastic beam with both ends rigidly fixed. The existence and multiplicity of positive solutions for the elastic beam equations has been studied extensively when the non-linear term f : [0,1] × [0, +∞) → [0, +∞) is continuous, see for example [1-10] and references therein. Agarwal and Chow [1] investigated problem (P) by using of contraction mapping and iterative methods. Bai [3] applied upper and lower solution method and Yao [9] used Guo-Krasnosel'skii fixed point theorem of cone expansion-compression type. However, there are only a few articles concerned with the nonpositive or semipositive elastic beam equations. Yao [11] considered the existence of positive solutions of semipositive elastic beam equations by constructing control functions and a special cone and using fixed point theorem of cone expansion-compression type. In this article, we assume that f : [0,1] × R1 R1, which implies the problem (P) is nonpositive (or semipositive particularly). By the topological degree and fixed point theory of superlinear operator on lattice (the definition of lattice will be given in Section 2), we obtain the existence of nontrivial solutions for the non-positive BVP (P) and the existence of positive solutions for the semipositive BVP (P).

2 Preliminaries

Let E be an ordered Banach space in which the partial ordering ≤ is induced by a cone P E, θ denote the zero element of E. P is called solid if int P ≠ ∅, i.e., P has nonempty interior. P is called a generating cone if E = P - P . For the concepts and properties about the cones we refer to [12-14].

We call E a lattice in the partial ordering ≤, if for arbitrary x, y E, sup{x, y} and inf{x, y} exist. For x E, let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M2">View MathML</a>

which are called the positive and the negative part of x, respectively. Take |x| = x+ + x-, then |x| ∈ P , and |x| is called the module of x. One can see [15] for the definition and the properties about the lattice.

For convenience, we use the following notations:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M3">View MathML</a>



<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M4">View MathML</a>

Remark 2.1 If E is a lattice, then P is a generating cone.

Definition 2.1 [[16], Definition 3.2, p. 929]. Let D E and F : D E be a nonlinear operator. F is said to be quasi-additive on lattice, if there exists y0 E such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M5">View MathML</a>


where x+ and x- are defined by (2.1).

Remark 2.2 We point out that the condition (2.2) appears naturally in the applications of nonlinear differential equations and integral equations.


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M6">View MathML</a>

and f : [a, b] × R1 R1. Consider the Nemytskii operator

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M7">View MathML</a>

Set P = {x C[a, b] | x(t) ≥ 0}, then E = C[a, b] is a lattice in the partial ordering which is induced by P . For any x C[a, b], it is evident that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M8">View MathML</a>

and hence |x|(t) = |x(t)|. By Remark 3.1 in [16], we know that there exists y0 C[a, b] such that Fx = Fx+ + Fx- + y0, ∀x C[a, b].

Suppose that B is a linear operator and A = BF . It follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M9">View MathML</a>

which means that A is quasi-additive on lattice.

Definition 2.2 [[17], Definition, p. 261]. Let B : E E be a linear operator. B is said to be a u0-bounded linear operator, if there exists u0 P\{θ}, such that for any x P\{θ}, there exist a natural number n and real numbers ζ, η > 0, such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M10">View MathML</a>

Lemma 2.1 [[18], Theorem 4.2.2, p. 122]. Let P be a generating cone and B a u0-bounded completely continuous linear operator. Then the spectral radius r(B) ≠ 0 and r-1(B) is the only positive eigenvalue corresponding to positive eigenvectors and B has no other eigenvectors except those corresponding to r-1(B).

Let B : E E be a positive completely continuous linear operator, r(B) a spectral radius of B, B* the conjugated operator of B, and P* the conjugated cone of P. Since P E is a generating cone, according to the famous Krein-Rutman theorem (see [14]), if r(B) ≠ 0, then there exist <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M11">View MathML</a>, and g* ∈ P*\{θ}, such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M12">View MathML</a>


Fix <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M13">View MathML</a>, g* ∈ P*\{θ} such that (2.3) holds. For δ > 0, let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M14">View MathML</a>

then P (g*, δ) is also a cone in E.

Definition 2.3 [[19], Definition, p. 528]. Let B be a positive linear operator. B is said to satisfy H condition, if there exist <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M15">View MathML</a>, g* ∈ P*\{θ} and δ > 0 such that (2.3) holds, and B maps P into P(g*, δ).

Remark 2.3 Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M16">View MathML</a>, where k(x, y) ∈ C([a, b] × [a, b]), k(x, y) ≥ 0, φ C[a, b]. Suppose that (2.3) holds and there exists v(x) ∈ P\{θ} such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M17">View MathML</a>

and v(x)g*(x) ≢ 0, then B satisfies H condition (see [19]).

Lemma 2.2 [[16], Theorem 3.1, p. 929]. Let P be a solid cone, A : E E be a completely continuous operator satisfying A = BF, where F is quasi-additive on lattice, B is a positive bounded operator satisfying H condition. Suppose that

(i) there exist a1 > r-1(B), y1 P such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M18">View MathML</a>


(ii) there exist 0 ≤ a2 r-1(B), y2 P such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M19">View MathML</a>


Then there exists R0 > 0 such that deg(I - A, TR , θ) = 0 for any R > R0 , where TR = {x C[0, 1] : ||x|| < R}.

Lemma 2.3 [[16], Theorem 3.3, p. 932]. Let Ω be a bounded open subset of E, θ ∈ Ω, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M20">View MathML</a> a completely continuous operator. Suppose that A has no fixed point on ∂Ω. If

(i) there exists a positive bounded linear operator B such that |Ax| ≤ B|x|, for all x ∈ ∂Ω;

(ii) r(B) ≤ 1.

Then deg(I - A, Ω, θ) = 1.

3 Existence of nontrivial solutions for the non-positive BVP (P)

In the sequel we always take E = C[0,1] with the norm ||u|| = max0≤t≤1 |u(t)| and P = {u C[0, 1] | u(t) ≥ 0, 0 ≤ t ≤ 1}. Then P is a solid cone in E and E is a lattice under the partial ordering ≤ induced by P.

A solution of BVP (P) is a fourth differentiable function u : [0,1] → R such that u satisfies (P). u is said to be a positive solution of BVP (P) if u(t) > 0, 0 < t < 1. Let G(t, s) be Green's function of homogeneous linear problem u(4)(t) = 0, u(0) = u(1) = u'(0) = u'(1) = 0. From Yao [11] we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M21">View MathML</a>


(G1) G(t, s) ≥ 0, 0 ≤ t, s ≤ 1;

(G2) G(t, s) = G(s, t);

(G3) G(t, s) ≥ p(t)G(τ; s), 0 ≤ t, s, τ ≤ 1, where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M22">View MathML</a>.

Lemma 3.1 Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M23">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M24">View MathML</a>. Then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M25">View MathML</a>


Proof. Since G(0, s) = G(1, s) = 0, 0 ≤ s ≤ 1, H(0) = H(1) = 0, then q(s)H(t) = G(t, s) = H(t) holds for t = 0 and t = 1. If 0 < t s ≤ 1 and t < 1, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M26">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M27">View MathML</a>

Similarly, (3.1) holds for 0 ≤ s t < 1 and t > 0. The proof is complete.    □

It is well known that the problem (P) is equivalent to the integral equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M28">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M29">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M30">View MathML</a>


Lemma 3.2 Let B be defined by (3.3). Then B is a u0-bounded linear operator.

Proof. Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M31">View MathML</a>, t ∈ [0,1]. For any u P\{θ}, by Lemma 3.1

we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M32">View MathML</a>

Take arbitrarily <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M33">View MathML</a>, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M34">View MathML</a>

Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M35">View MathML</a>, η = ||u|| > 0. Then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M36">View MathML</a>

This indicates that B : E E is a u0-bounded linear operator. □

From Lemma 2.1 we have r(B) ≠ 0 and r-1(B) is the only eigenvalue of B. Denote λ1 = r-1(B).

Now let us list the following conditions which will be used in this article:

(H1) there exist constants α and β with α > β ≥ 0 satisfying

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M37">View MathML</a>


(H2) there exists a constant γ ≥ 0 satisfying

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M38">View MathML</a>


(H3) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M39">View MathML</a>.

Theorem 3.1 Suppose that (H1) and (H2) hold. Then for any <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M40">View MathML</a>, BVP (P) has at least one nontrivial solution, where λ1 = r-1(B) is the only eigenvalue of B, B is denoted by (3.3), ι = max{β, γ}.

Proof. Let (Fu)(t) = f(t, u(t)). Then A = BF, where A is denoted by (3.2). By Remark 2.2, F is quasi-additive on lattice. Applying the Arzela-Ascoli theorem and a standard argument, we can prove that A : E E is a completely continuous operator.

Now we show that λA = λBF has at least one nontrivial fixed point, which is the nontrivial solution of BVP (P).

On account of (G3) we have that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M41">View MathML</a> such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M42">View MathML</a>

Notice that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M43">View MathML</a>, where G(t, s) ≥ 0, G(t, s) ∈ C([0,1] × [0,1]). From Lemma 3.2 B is a u0-bounded linear operator. By Lemma 2.1 we have r(B) ≠ 0 and λ1 = r-1(B) is the only eigenvalue of B. Then there exist <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M44">View MathML</a> and g* ∈ P*\{θ} such that (2.3) holds. Notice that λ > 0, from Remark 2.3, λB satisfies H condition.

By (3.4) and (3.5), there exist r > 0, M > 0 and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M45">View MathML</a> such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M46">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M47">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M48">View MathML</a>


By (3.6) and (3.7), we have (2.4) and (2.5) hold, where a1 = α - ε, a2 = β + ε.

Let B1 = λB. Then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M49">View MathML</a>. Obviously, for any <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M50">View MathML</a>, a1 > r-1(B1), a2 < r-1(B1). From Lemma 2.2 there exists R0 > 0 such that for any R > max{R0, r},

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M51">View MathML</a>


Let B2 = λ(γ + ε)B. From (3.8) we have |λAu| ≤ B2|u|, also <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M52">View MathML</a>. Without loss of generality we assume that λA has no fixed point on ∂Tr, where Tr = {u C[0,1] | ||u|| < r}. By Lemma 2.3 we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M53">View MathML</a>


It is easy to see from (3.9) and (3.10) that λA has at least one nontrivial fixed point. Thus problem (P) has at least one nontrivial solution. □

Remark 3.1 If α = +∞, β = γ = 0, then for any λ > 0 problem (P) has at least one nontrivial solution.

Theorem 3.2 Suppose that (H1) holds. Assume f(t, 0) ≡ 0, ∀t ∈ [0,1] and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M54">View MathML</a>


Then for any <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M55">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M56">View MathML</a>, BVP (P) has at least one nontrivial solution.

Proof. Since f(t, 0) ≡ 0, ∀t ∈ [0,1], then = θ. By (3.11) we have that the Frechet derivative <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M57">View MathML</a> of A at θ exists and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M58">View MathML</a>

Notice that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M59">View MathML</a>, then 1 is not an eigenvalue of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M60">View MathML</a>. By the famous Leray-Schauder theorem there exists r > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M61">View MathML</a>


where κ is the sum of algebraic multiplicities for all eigenvalues of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M62">View MathML</a> lying in the interval (0, 1). From the proof of Theorem 3.1 we have that (3.9) holds for any <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M63">View MathML</a>. By (3.9) and (3.12), λA has at least one nontrivial fixed point. Thus problem (P) has at least one nontrivial solution. □

4 Existence of positive solutions for the semipositive BVP (P)

In many real problems, the positive solution is more significant. In this section we will study this kind of question.

Lemma 4.1 [[20], Theorem 1, p. 90]. Let D = [a, b]. Suppose

(i) G(t, s) is a symmetric kernel. And there exist D0 D, mesD0 ≠ 0 and δ > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M64">View MathML</a>

(ii) f(t, u) is bounded from below and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M65">View MathML</a> uniformly holds for t D0. Then for any λ* > 0, there exists R = R(λ*) > 0 such that if 0 < λ0 ≤ λ*, ||φ0|| ≥ R and φ0 = λ00, then φ0(x) ≥ 0, where A is denoted by (3.2).

Theorem 4.1 Suppose that (H3) holds. Then there exists λ* > 0 such that for any 0 < λ < λ* BVP (P) has at least one positive solution.

Proof. By (H3) there exists a constant b > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M66">View MathML</a>



<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M67">View MathML</a>

From (4.1) f1 is bounded from below. Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M68">View MathML</a>

Then A1 : E E is a completely continuous operator.

From the proof of Theorems 2.7.3 and 2.7.4 in Sun [18], there exists R0 > 0 such that for any R > R0,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M69','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M69">View MathML</a>


Take 0 < r < R0. Let m = max0≤t≤1,|u|<r|f1(t, u)|, G = max0≤s,t≤1 G(t, s), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M70','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M70">View MathML</a>. For any <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M71','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M71">View MathML</a>, u ∈ ∂Tr, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M95','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M95">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M72','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M72">View MathML</a>


From (4.2) and (4.3), we have that for any <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M73','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M73">View MathML</a>, there exist uλ, C[0,1], ||uλ|| > r such that uλ = λA1uλ. In order to apply Lemma 4.1 we claim that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M97','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M97">View MathML</a>


In fact, if (4.4) doesn't hold, then there exist λn > 0, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M74','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M74">View MathML</a> such that λn → 0, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M75">View MathML</a> (c > 0 is a constant) and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M76','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M76">View MathML</a>


Since A1 is completely continuous, then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M77">View MathML</a> has a subsequence converging to u* ∈ C[0,1]. Assume, without loss of generality, that it is <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M77">View MathML</a>. Taking n → +∞ in (4.5), we have u* = θ, which is a contradiction to <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M79','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M79">View MathML</a>. Hence (4.4) holds.

Let D = [0,1], D0 = [t1, t2] ⊂ (0, 1) ⊂ D, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M80','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M80">View MathML</a>. By (G3)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M81','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M81">View MathML</a>

From (H3) and the definition of f1, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M82','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M82">View MathML</a>

By Lemma 4.1 there exists <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M83','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M83">View MathML</a> such that if <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M84','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M84">View MathML</a>, ||uλ|| ≥ R and uλ = λA1uλ, then uλ(t) ≥ 0. By (4.4), there exists <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M85','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M85">View MathML</a> such that if 0 < λ ≤ λ*, ||uλ|| ≥ r and uλ = λA1uλ, then ||uλ|| ≥ R. Thus uλ(t) ≥ 0. By the definitions of A1 and f1 we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M86">View MathML</a>

And so uλ(t) is a positive solution of problem (P). □

Remark 4.1 In Theorem 4.1, without assuming that f(t, u) ≥ 0 when u ≥ 0, we obtain the existence of positive solutions for the semipositive BVP (P).

Remark 4.2 Noticing that, in this article, we only study the existence of positive solutions for BVP (P), which is irrelevant to the value of f(t, u) when u ≤ 0, we only need to suppose that f(t, u) is bounded from below when u ≥ 0. In fact, f(t, u) may be unbounded from below when u ≤ 0.

5 Two examples

In order to illustrate possible applications of Theorems 3.2 and 4.1, we give two examples.

Example 5.1 Consider the fourth-order BVP

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M87','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M87">View MathML</a>


In this example, f(t, u) = sinu + uarctanu + πu, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M88','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M88">View MathML</a>


<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M89','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M89">View MathML</a>


Take α = 3π/2, β = π/2, ρ = π +1. Then (5.1), (5.2) indicate (H1), (3.11) hold, repectively. Notice that α > ρ > β > 0 and f(t, 0) ≡ 0, ∀t ∈ [0,1], by Theorem 3.2 for any <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M90','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M90">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M91','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M91">View MathML</a>, BVP (P1) has at least one nontrivial solution.

Example 5.2 Consider the fourth-order BVP

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M92','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M92">View MathML</a>


In this example, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M93','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M93">View MathML</a>, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M94','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/56/mathml/M94">View MathML</a>


(5.3) means (H3) holds. By Theorem 4.1 there exists λ* > 0 such that for any 0 < λ < λ* BVP (P2) has at least one positive solution.

Competing interests

The authors declare that they have no competing interests.

Authors' contributions

All authors typed, read and approved the final manuscript.


The authors would like to thank the referees for carefully reading this article and making valuable comments and suggestions. This work is supported by the Foundation items: NSFC (10971179, 11026203), NSF (BK2011202) of the Jiangsu Province, NSF (09XLR04) of the Xuzhou Normal University and NSF (2010KY07) of the Suqian College.


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