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Positive solutions for the third-order boundary value problems with the second derivatives

Yanping Guo1, Yujing Liu2 and Yonhchun Liang1*

Author affiliations

1 College of Electrical Engineering and Information, Hebei University of Science and Technology, Shijiazhuang, 050018, Hebei, PR China

2 College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018, Hebei, PR China

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Citation and License

Boundary Value Problems 2012, 2012:34  doi:10.1186/1687-2770-2012-34

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2012/1/34


Received:3 November 2011
Accepted:26 March 2012
Published:26 March 2012

© 2012 Guo 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.

Abstract

By using the fixed-point index theory in a cone and defining a linear operator, we obtain the existence of at least one positive solution for the third-order boundary value problem with integral boundary conditions

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M1">View MathML</a>

where f : [0, 1] × R+ × R- → R+ is a nonnegative function. The associated Green's function for the above problem is also used, and a new reproducing cone also used.

Keywords:
fixed-point index theory; Green's function; positive solution; boundary value problem

1 Introduction

By eigenvalue criteria, Webb [1] obtained the existence of multiple positive solutions of a Hammerstein integral equation of the form

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M2">View MathML</a>

where k can have discontinuities and g L1. Then, some articles have studied different BVPs by this way (see [2-5]). Webb [4] introduced an unified method to study existence of at least one nonzero solution for higher order boundary value problems

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M3">View MathML</a>

In 2010, Hao [5] considered the existence of positive solutions of the nth-order BVP

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M4">View MathML</a>

Guo [6] studied the existence of positive solutions for the there-point boundary problem with the first-order derivative.

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M5">View MathML</a>

where f is a nonnegative continuous function. In 2011, Zhao [7] studied third-order differential equations:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M6">View MathML</a>

subject to integral boundary condition of the form

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M7">View MathML</a>

where f C([0, 1] × P, P).

In this article, we study the existence of positive solutions for the following boundary value problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M8">View MathML</a>

(1.1)

The results are proved by applying the fixed point index theory in a cone and spectral radius of a linear operator. Unlike reference [7], the nonlinear part f involves the second-order derivative and just satisfies Caratheodory conditions.

The following conditions are satisfied throughout this article:

(H1) f : [0, 1] × R+ × R- R+ satisfies Caratheodory conditions, that is, f(·,u, v) is measurable for each fixed u R+, v R-, and f(t, ·,·) is continuous for a.e. t ∈ [0, 1]. For any r, r' > 0, there exists <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M9">View MathML</a>, such that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M10">View MathML</a>, where (u, v) ∈ [0, r] × [-r', 0], a.e. t ∈ [0, 1];

(H2) g L[0, 1] is nonnegative, b ∈ [0, 1), where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M11">View MathML</a>.

2 Preliminaries

Lemma 2.1 [7]. Let y L1[0, 1] and y ≥ 0, the problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M12">View MathML</a>

(2.1)

has a unique solution

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M13">View MathML</a>

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M14">View MathML</a>,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M15">View MathML</a>

Lemma 2.2. Let y L1[0, 1] and y ≥ 0, the unique solution of the boundary value problem (2.1) satisfies the following conditions: u(t) ≥ 0, u"(t) ≤ 0, for t ∈ [0, 1].

Proof. By Lemma 2.1, u(t) ≥ 0. By differential equations u'"(t) + y(t) = 0, t ∈ (0, 1), we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M16">View MathML</a>

Let X = C2[0, 1] with <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M17">View MathML</a>. Obviously, (X, ||·||) is a Banach space. Define the cone P X by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M18">View MathML</a>

Obviously P is a cone in X, and Pris a bounded open subset in P.

Definition 2.1 [1]. Let P be a cone in a Banach space X. If for any x X and x+, x- P, writing x = x+ + x- shows that P is a reproducing cone.

Lemma 2.3. P is a reproducing cone in X.

Proof. Suppose u X, so u" ∈ C[0, 1] and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M19">View MathML</a>

(2.2)

where u- = min{u"(t), 0}, u+ = min{-u"(t), 0}. Obviously u+,u- C[0, 1] and u+ ≤ 0,u- ≤ 0. For (2.2), we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M20">View MathML</a>

If u(0) ≥ 0, u'(0)t ≥ 0, let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M21">View MathML</a>

So u1 ≥ 0, u2 ≥ 0, then u1, u2 P and u = u1 - u2.

If u(0) ≤ 0, u'(0)t ≤ 0, let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M22">View MathML</a>

So u1 ≥ 0, u2 ≥ 0, then u1, u2 P and u = u1 - u2.

If u(0) ≥ 0, u'(0)t ≤ 0, let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M23">View MathML</a>

So u1 ≥ 0, u2 ≥ 0, then u1, u2 P and u = u1 - u2.

If u(0) ≤ 0, u'(0)t ≥ 0, let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M24">View MathML</a>

So u1 ≥ 0, u2 ≥ 0, then u1, u2 P and u = u1 - u2.

Then P is a reproducing cone in X.

Lemma 2.4 (Krein-Rutman) [8]. Let K be a reproducing cone in a real Banach space X and let L : K K be a compact linear operator with L(K) ⊂ K. r(L) is the spectral radius of L. If r(L) > 0, then there is φ1 K\{0} such that 1 = r(L)φ1.

Lemma 2.5 [9]. Let X be a Banach space, P be a cone in X and Ω(P) be a bounded open subset in P. Suppose that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M25">View MathML</a> is a completely continuous operator. Then the following results hold

(1) If there exists u0 P\{0} such that u Au + λu0, for any u Ω(P), λ ≥ 0, then the fixed-point index i(A, Ω(P), P) = 0.

(2) If 0 ∈ Ω(P), Au ≠ λu, for any u Ω(P), λ ≥ 1, then the fixed-point index i(A, Ω(P), P) = 1.

Define the operator A: X X, L: X X, by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M26">View MathML</a>

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M27">View MathML</a>

So A : P P is completely continuous operator; L : P → P is a compact linear operator.

Lemma 2.6 [7]. Assume that (H2) holds, then choose <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M28">View MathML</a>, for all t ∈ [δ, 1 - δ],v, s ∈ [0, 1], we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M29">View MathML</a>

where ρ = 4δ2(1 - δ).

Note: r(L) is the spectral radius of L. <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M30">View MathML</a>, where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M28">View MathML</a>. By Lemma 2.6, obviously h > 0.

Lemma 2.7. Suppose conditions (H1), (H2) hold, then r(L) > 0.

Proof. Take u(t) ≡ 1, then u"(t) = 0, for any t ∈ [δ, 1 - δ] we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M31">View MathML</a>

Repeating the process gives

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M32">View MathML</a>

So, we get <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M33">View MathML</a>. The proof is completed.

By Lemma 2.4, then there is φ1 P\{0} such that 1 = r(L)φ1.

3 Main results

In the following, we use the notation:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M34">View MathML</a>

where E is a fixed subset of [0, 1] of measure zero, d > 0.

Lemma 3.1. Suppose

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M35">View MathML</a>

(3.1)

where μ = 1/r(L), then there exists R0 > 0 such that i(A, Pr, P) = 1 for each r > R0.

Proof. Let ε > 0 satisfy fμ - ε, then there exist r1 > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M36">View MathML</a>

for all u > r1 or v < -r1 and a.e. t ∈ [0, 1].

By (H1), there exists <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M37">View MathML</a> such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M38">View MathML</a>

for all (u, v) ∈ [0, r1] × [-r1, 0] and a.e. t ∈ [0, 1]. Hence, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M39">View MathML</a>

(3.2)

for all u ∈ R+, v R- and a.e. t ∈ [0, 1].

Since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M40">View MathML</a> is the spectrum radius of L. It follows from <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M41">View MathML</a>, (I/(μ - ε) - L)-l exists, let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M42">View MathML</a>

For r > R0, by Lemma 2.5 we will prove

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M43">View MathML</a>

for each u ∈ ∂Prand λ ≥ 1.

In fact, if not, there exist u0 ∂Prand λ0 ≥ 1 such that Au0 = λ0u0.

Together with (3.2) implies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M44">View MathML</a>

So

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M45">View MathML</a>

Then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M46">View MathML</a>

So

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M47">View MathML</a>

Then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M48">View MathML</a>

This is a contradiction. By Lemma 2.5 (2), we get that i(A, Pr, P) = 1 for each r > R0. The proof is completed.

Lemma 3.2. Suppose there exists d > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M49">View MathML</a>

(3.3)

Then there exists ρ0 > 0 and d ρ0 such that for each ρ∈ (0, ρ0], if u Au for u ∈ ∂, then i(A, Pρ, P) = 0.

Proof. Let ε > 0 satisfy <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M50">View MathML</a>, there exist d ρ0 > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M51">View MathML</a>

(3.4)

for u ∈ [0, ρ0],v ∈ [-ρ0,0] and a.e. t ∈ [0, 1].

Let ρ ∈ (0,ρ0], by Lemma 2.5 (1), we prove that: u Au + λφ1 for all u ∂Pρ, λ > 0, where φ1 P\{0} is the eigenfunction of L corresponding to the eigenvalue <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M40">View MathML</a>. In fact, if not, there exist u0 ∂Pρ, λ0 > 0 such that u0 = Au0 + λ0φ1. This implies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M52">View MathML</a>

Let: <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M53">View MathML</a>.

So 0 < λ0 < λ* < ∞ and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M54">View MathML</a>. Then, u0 - λ*φ1 P.

For L(P) ⊂ P, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M55">View MathML</a>

By (3.4), we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M56">View MathML</a>

So, we know

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M57">View MathML</a>

which contradicts the definition of λ*.

Lemma 3.3. Suppose there is ρ1 > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M58">View MathML</a>

(3.5)

for u ∈ [0, ρ1] and v ∈ [-ρ1, 0] a.e. t ∈ [0, 1], where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M59">View MathML</a>, if Au u for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M60">View MathML</a>, then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M61">View MathML</a>.

Proof. Suppose <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M60">View MathML</a>, by Lemma 2.2, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M62">View MathML</a>

That is Au ≠ λu for each <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M60">View MathML</a>, λ > 1. If Au u for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M60">View MathML</a>, by Lemma 2.5, then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M61">View MathML</a>.

Lemma 3.4. Suppose there is ρ2 > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M63">View MathML</a>

(3.6)

for u ∈ [0, ρ2] and v ∈ [-ρ2, 0] a.e. t ∈ [0, 1], where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M64">View MathML</a>. If Au u for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M65">View MathML</a>, then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M66">View MathML</a>.

Proof. For <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M65">View MathML</a>, t ∈ [δ, 1 - δ], by Lemma 2.2, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M67">View MathML</a>

This implies that u Au + λφ for each <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M65">View MathML</a>, λ > 0, where φ P\{0} is the eigenfunction of L corresponding to r(L). Suppose u Au for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M65">View MathML</a>, by Lemma 2.5, then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M66">View MathML</a>.

Theorem 3.1. The boundary value problem (1.1) has at least one positive solution if one of the following conditions holds.

(C1) There exists d > 0 such that (3.3) and (3.1) hold.

(C2) There exists d > 0, ρ1 > 0 such that (3.3) and (3.5) hold.

(C3) There exists ρ2 > 0 such that (3.6) and (3.1) hold.

(C4) There exists ρ1, ρ2 > 0 with 0 < ρ2 < ρ1d1/d2 such that (3.5) and (3.6) hold.

Proof. When condition (C1) holds, by Lemma 3.1 and 0 ≤ f< μ, we get that there exists r > 0 such that i(A, Pr, P) = 1. It follows from Lemma 3.2 and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/34/mathml/M68">View MathML</a>, then there exists 0 < ρ < min{r, d} such that either there exists u ∂Pρthat i(A, Pρ, P) = 0 or u = Au. So BVP (1.1) has at least one positive solution u P with ρ ≤ ||u|| < r.

When one of other conditions holds, the results can be proved similarly.

Competing interests

The authors declare that they have no competing interests.

Authors' contributions

The authors declare that the study was realized in collaboration with same responsibility. All authors read and approved the final manuscript.

Acknowledgements

The project is supported by the Natural Science Foundation of China (10971045) and the Natural Science Foundation of Hebei Province (A2009000664, A2011208012). The research item financed by the talent training project funds of Hebei Province. The authors would like to thank the referee for helpful comments and suggestions.

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