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Existence of nodal solutions of a nonlinear fourth-order two-point boundary value problem

Wenguo Shen

Author affiliations

School of Mathematics and Statistics, Lanzhou University, Lanzhou 730000, People's Republic of China

Department of Basic Courses, Lanzhou Polytechnic College, Lanzhou 730050, People's Republic of China

Citation and License

Boundary Value Problems 2012, 2012:31  doi:10.1186/1687-2770-2012-31


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2012/1/31


Received:15 August 2011
Accepted:20 March 2012
Published:20 March 2012

© 2012 Shen; 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

In this article, we give conditions on parameters k, l that the generalized eigenvalue problem x″″ + kx″ + lx = λh(t)x, 0 < t < 1, x(0) = x(1) = x′(0) = x′(1) = 0 possesses an infinite number of simple positive eigenvalues <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M1">View MathML</a> and to each eigenvalue there corresponds an essential unique eigenfunction ψk which has exactly k - 1 simple zeros in (0,1) and is positive near 0. It follows that we consider the fourth-order two-point boundary value problem x″″ + kx″ + lx = f(t,x), 0 < t < 1, x(0) = x(1) = x′(0) = x′(1) = 0, where f(t, x) ∈ C([0,1] × ℝ, ℝ) satisfies f(t, x)x > 0 for all x ≠ 0, t ∈ [0,1] and lim|x|→0 f(t,x)/x = a(t), lim|x|→+∞ f(t,x)/x = b(t) or limx→-∞ f(t,x)/x = 0 and limx→+∞f(t,x)/x = c(t) for some a(t), b(t), c(t) ∈ C([0,1], (0,+∞)) and t ∈ [0,1]. Furthermore, we obtain the existence and multiplicity results of nodal solutions for the above problem. The proofs of our main results are based upon disconjugate operator theory and the global bifurcation techniques.

MSC (2000): 34B15.

Keywords:
disconjugacy theory; bifurcation; nodal solutions; eigenvalue

1 Introduction

The deformations of an elastic beam in equilibrium state with fixed both endpoints can be described by the fourth-order boundary value problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M2">View MathML</a>

(1.1)

where f: ℝ ℝ is continuous, λ ∈ ℝ is a parameter and l is a given constant. Since the problem (1.1) cannot transform into a system of second-order equation, the treatment method of second-order system does not apply to the problem (1.1). Thus, existing literature on the problem (1.1) is limited. Recently, when l = 0, the existence and multiplicity of positive solutions of the problem (1.1) has been studied by several authors, see Agarwal and Chow [1], Ma and Wu [2], Yao [3,4] and Korman [5]. Especially, when l ≠ 0, l satisfying (H1) and h(t) satisfying (H2), Xu and Han [6] studied the existence of nodal solutions of the problem (1.1) by applying bifurcation techniques, where

(H1) l ∈ (-π4, π4/64) is given constant.

(H2) h C([0,1], [0, ∞)) with h(t) ≢ 0 on any subinterval of [0,1].

Motivated by [6], we consider the existence of nodal solutions of general fourth-order boundary value problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M3">View MathML</a>

(1.2)

and under the assumptions:

(A1) One of following conditions holds

(i) k, l satisfying <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M4">View MathML</a> are given constants with

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M5">View MathML</a>

(1.3)

(ii) k, l satisfying <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M6">View MathML</a> are given constants with

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M7">View MathML</a>

(1.4)

(A2) f(t, x) ∈ C([0,1] × ℝ, ℝ) satisfies f(t, x)x > 0 for all x ≠ 0 and t ∈ [0,1].

(A3) There exists a(t) ∈ C([0,1], (0, ∞)) such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M8">View MathML</a>

(1.5)

(A4) There exists b(t) ∈ C([0,1], (0, ∞)) such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M9">View MathML</a>

(1.6)

(A5) There exists c(t) ∈ C([0,1], (0, ∞)) such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M10">View MathML</a>

(1.7)

However, in order to use bifurcation technique to study the nodal solutions of the problem (1.2), we first prove that the generalized eigenvalue problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M11">View MathML</a>

(1.8)

(where h satisfies (H2)) has an infinite number of positive eigenvalues

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M12">View MathML</a>

(1.9)

and each eigenvalue corresponding an essential unique eigenfunction ψk which has exactly k - 1 simple zeros in (0,1) and is positive near 0. Fortunately, Elias [7] developed a theory on the eigenvalue problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M13">View MathML</a>

(1.10)

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M14">View MathML</a>

(1.11)

and ρi Cn-i[a, b] with ρi > 0 (i = 0,1,..., n) on [a, b]. 0y,...., n-1y are called the quasi-derivatives of y(t). To apply Elias's theory, we have to prove that (1.8) can be rewritten to the form of (1.10), that is, the linear operator

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M15">View MathML</a>

(1.12)

has a factorization of the form

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M16">View MathML</a>

(1.13)

on [0,1], where li C4-i[0,1] with li > 0 (i = 0, 1, 2, 3, 4) on [0, 1], and x(0) = x(1) = x′(0) = x′(1) = 0 if and only if

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M17">View MathML</a>

(1.14)

This can be achieved under (A1) by using the disconjugacy theory in [8].

The rest of the article is arranged as follows: In Section 2, we state some disconjugacy theory which can be used in this article, and then show that (A1) implies the equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M18">View MathML</a>

(1.15)

is disconjugate on [0, 1], and establish some preliminary properties on the eigenvalues and eigenfunctions of the generalized eigenvalue problem (1.8). Finally in Section 3, we state and prove our main results (Theorems 3.1 and 3.2 ).

Remark 1.1. If we let k = 0, then the condition (A1) reduces to (H1) in [6].

Remark 1.2. Since the function f(t, x) is more general than the function h(t)f(x) in [6], then the problem considered in this article is more general than the problem in [6].

Remark 1.3. If we let k = 0 and f(t, x) = λh(t)f(x), then Theorem 3.2 reduces to [[6], Theorem 3.1].

Remark 14. For other results on the existence and multiplicity of positive solutions and nodal solutions for the boundary value problems of fourth-order ordinary differential equations based on bifurcation techniques, see [9-14]s and their references.

2 Preliminary results

Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M19">View MathML</a>

(2.1)

be nth-order linear differential equation whose coefficients pk(⋅) (k = 1,..., n) are continuous on an interval I.

Definition 2.1 [[8], Definition 0.2, p. 2]. Equation (2.1) is said to be disconjugate on an interval I if no nontrivial solution has n zeros on I, multiple zeros being counted according to their multiplicity.

Lemma 2.2 [[8], Theorem 0.7, p. 3]. Equation (2.1) is disconjugate on a compact interval I if and only if there exists a basis of solutions y0, ...,yn-1 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M20">View MathML</a>

(2.2)

on I. A disconjugate operator L[y] = y(n) + p1(t)y(n-1)+ ⋯ + pn(t)y can be written as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M21">View MathML</a>

(2.3)

where ρk Cn-k(I) (k = 0,1,..., n) and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M22">View MathML</a>

(2.4)

and ρ0ρ1ρn 1.

Lemma 2.3 [[8], Theorem 0.13, p. 9]. Green's function G(t,s) of the disconjugate equation (2.3) and the two-point boundary value conditions

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M23">View MathML</a>

(2.5)

satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M24">View MathML</a>

(2.6)

Now using Lemmas 2.2 and 2.3, we will prove some preliminary results.

Theorem 2.4. Let (A1) hold. Then

(i) L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M25">View MathML</a>

(2.7)

where ρk C4-k[0,1] with ρk > 0 (k = 0, 1, 2, 3, 4).

(ii) x(0) = x(1) = x′(0) = x′(1) = 0 if and only if

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M26">View MathML</a>

(2.8)

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M27">View MathML</a>

(2.9)

Proof of Theorem 2.4. We divide the proof into nine cases.

Case 1. <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M28">View MathML</a>

In the case, we have corresponding L[x] = x″″ + kx″ + lx = 0 that the equation λ4 + kλ2 + lx = 0 has 4 roots λ1 = m1 + m2i, λ2 = m1- m2i, λ3 = -m1 + m2i, and λ4 = -m1- m2i, where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M29">View MathML</a>

(2.10)

Combining <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M30">View MathML</a> with (2.10), we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M31">View MathML</a>. Thus, we get that either the following (1) or (2) holds:

(1) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M32">View MathML</a>, for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M33">View MathML</a>;

(2) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M34">View MathML</a>, for <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M35">View MathML</a>.

Furthermore, it is easy to check that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M36">View MathML</a>

(2.11)

Take

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M37">View MathML</a>

(2.12)

It is easy to check that x0(t), x1(t), x2(t), and x3(t) form a basis of solutions of L[x] = 0. By simple computation, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M38">View MathML</a>

(2.13)

This together with (2.11) implies that wi > 0(i=1, 2, 3, 4) on [0,1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M39">View MathML</a>

(2.14)

and accordingly

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M40">View MathML</a>

(2.15)

Using (2.15), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

Case 2. k ∈ (-, 0) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M41">View MathML</a>.

In the case, applying the similar method used in Case 1, we take

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M42">View MathML</a>

(2.16)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M43">View MathML</a>.

It is easy to check that x0(t), x1(t), x2(t), and x3(t) form a basis of solutions of L[x] = 0.

By simple computation, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M44">View MathML</a>

(2.17)

Clearly, wi > 0 (i = 1, 2, 3, 4) on [0,1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M45">View MathML</a>

(2.18)

and accordingly

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M46">View MathML</a>

(2.19)

Using (2.19), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

Case 3.k ∈ (-, 0) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M47">View MathML</a>.

In the case, we take

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M48">View MathML</a>

(2.20)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M49">View MathML</a>,

It is easy to check that x0(t), x1(t), x2(t), and x3(t) form a basis of solutions of L[x] = 0.

By simple computation, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M50">View MathML</a>

(2.21)

Clearly, wi > 0 (i = 1, 2, 3, 4) on [0, 1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0, 1], and L[x] has a factorization

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M51">View MathML</a>

(2.22)

and accordingly

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M52">View MathML</a>

(2.23)

Using (2.23), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

Case 4. k ∈ (-,0), l = 0.

In the case, we take

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M53">View MathML</a>

(2.24)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M54">View MathML</a>.

It is easy to check that x0(t), x1(t), x2(t), and x3(t) form a basis of solutions of L[x] = 0.

By simple computation, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M55">View MathML</a>

(2.25)

Clearly, wi > 0 (i = 1, 2, 3, 4) on [0, 1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M56">View MathML</a>

(2.26)

and accordingly

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M57">View MathML</a>

(2.27)

Using (2.27), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

Case 5. k = 0, l = 0. The case is obvious.

Case 6. k ∈ (0,π2), l = 0.

In the case, we take

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M58">View MathML</a>

(2.28)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M59">View MathML</a>, σ is a positive constant. Clearly, m ∈ (0,π) and then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M60">View MathML</a>

(2.29)

It is easy to check that x0(t), x1(t), x2(t), and x3(t) form a basis of solutions of L[x] = 0.

By simple computation, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M61">View MathML</a>

(2.30)

Clearly, wi > 0 (i = 1, 2, 3, 4) on [0, 1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M62">View MathML</a>

(2.31)

and accordingly

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M63">View MathML</a>

(2.32)

Using (2.32), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

Case 7. π2 (k - π2) < l < 0.

In the case, we take

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M64">View MathML</a>

(2.33)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M65">View MathML</a>, σ is a positive constant. Clearly, m2 ∈ (0,π) and then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M66">View MathML</a>

(2.34)

It is easy to check that x0(t), x1(t), x2(t), and x3(t) form a basis of solutions of L[x] = 0. By simple computation, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M67">View MathML</a>

(2.35)

Clearly, wi > 0 (i = 1, 2, 3, 4) on [0, 1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M68">View MathML</a>

(2.36)

and accordingly

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M69','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M69">View MathML</a>

(2.37)

Using (2.37), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

Case 8. <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M70','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M70">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M71','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M71">View MathML</a>

In the case, we take

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M72','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M72">View MathML</a>

(2.38)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M73','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M73">View MathML</a>, σ is a positive constant. Clearly, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M74','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M74">View MathML</a> and then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M75">View MathML</a>

(2.39)

It is easy to check that x0(t), x1(t), x2(t), and x3(t) form a basis of solutions of L[x] = 0. By simple computation, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M76','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M76">View MathML</a>

(2.40)

Clearly, wi > 0 (i = 1, 2, 3, 4) on [0, 1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M77">View MathML</a>

(2.41)

and accordingly

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M78','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M78">View MathML</a>

(2.42)

Using (2.42), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

Case 9. <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M79','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M79">View MathML</a>

In the case, we take

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M80','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M80">View MathML</a>

(2.43)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M81','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M81">View MathML</a> Clearly, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M82','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M82">View MathML</a>, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M83','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M83">View MathML</a>, m1 < m2 and then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M84','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M84">View MathML</a>

(2.44)

It is easy to check that x0(t), x1(t), x2(t), and x3(t) form a basis of solutions of L[x] = 0.

By simple computation, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M85','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M85">View MathML</a>

(2.45)

Clearly, wi > 0 (i = 1, 2, 3, 4) on [0, 1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M86">View MathML</a>

(2.46)

and accordingly

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M87','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M87">View MathML</a>

(2.47)

Using (2.47), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

This completes the proof of Theorem 2.4.

Remark 2.5. If condition (A1) does not hold, the results of Theorem 2.4 cannot be obtained. For example, in the case of L[x] = 0 with <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M88','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M88">View MathML</a>, we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M89','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M89">View MathML</a>. Applying the similar method to prove case 1 in Theorem 2.4, we conclude that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M90','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M90">View MathML</a>

form a basis of solutions of L[x] = 0. By simple computation, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M91','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M91">View MathML</a>

From <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M92','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M92">View MathML</a>, we easily get that tan <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M93','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M93">View MathML</a>. Furthermore, w3 < 0. Thus, Theorem 2.4 does not hold in this case.

Remark 2.6. In the following, consider L[x] = 0, for k, l are given constants, by the similar method in Remark 2.5, we may gain the location of (k,l) in the (k, l)-plane and the results of w3 or w1 corresponding <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M94','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M94">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M95','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M95">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M96','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M96">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M97','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M97">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M98','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M98">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M99','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M99">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M100','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M100">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M101','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M101">View MathML</a> and ω3 < 0. Furthermore, it follows that the conclusion of Theorem 2.4 cannot be yielded in the cases.

Theorem 2.7. Let (A1) hold and h satisfy (H2). Then

(i) The problem (1.8) has an infinite number of positive eigenvalue

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M102','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M102">View MathML</a>

(ii) λk(h) as k → ∞.

(iii) To each eigenvalue λk(h) there corresponds an essential unique eigenfunction ψk which has exactly k - 1 simple zeros in (0,1) and is positive near 0.

(iv) Given an arbitrary subinterval of [0,1], then an eigenfunction which belongs to a sufficiently large eigenvalue change its sign in that subinterval.

(v) For each k ∈ ℕ, the geometric multiplicity of λk(h) is 1.

Proof of Theorem 2.7. (i)-(iv) are immediate consequences of Elias [[7], Theorem 1-5] and Theorem 2.4, we only prove (υ).

Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M103','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M103">View MathML</a>

(2.48)

with

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M104','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M104">View MathML</a>

To show (υ), it is enough to prove

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M105','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M105">View MathML</a>

Clearly

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M106','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M106">View MathML</a>

(2.49)

Suppose on the contrary that the geometric multiplicity of λk(h) is greater than 1. Then there exists <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M107','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M107">View MathML</a> and subsequently

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M108','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M108">View MathML</a>

(2.50)

for some γ ≠ 0. Multiplying both sides of (2.50) by ψk(t) and integrating from 0 to 1, we deduce that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M109','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M109">View MathML</a>

(2.51)

which is a contradiction !

Theorem 2.8 (Maximum principle). Let (A1) hold. Let e C[0,1] with e ≥ 0 on [0,1] and e ≢ 0 on any compact subinterval in [0,1]. If x C4[0,1] satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M110','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M110">View MathML</a>

(2.52)

Then x > 0 on (0,1).

Proof. When (A1) holds, the homogeneous problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M111','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M111">View MathML</a>

(2.53)

has only trivial solution. So the boundary value problem (2.52) has a unique solution which may be represented in the form

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M112','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M112">View MathML</a>

(2.54)

where G(t, s) is Green's function. By Theorem 2.4 and Lemma 2.3 (take n = 4, k = 2), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M113','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M113">View MathML</a>

(2.55)

that is, G(t, s) > 0, for all (t, s) (0,1) × (0,1).

Using (2.54), when e ≥ 0 on [0,1] and e ≢ 0 on any compact subinterval in [0,1], then x > 0 on (0,1).

3 Main results

Theorem 3.1. Let (A1), (A2), (A3) and (A4) hold. Assume that either (i) or (ii) holds for some k ∈ ℕ and j ∈ {0} ∪ ℕ:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M114','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M114">View MathML</a>

(3.1)

Then the problem (1.2) has 2(j + 1) solutions <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M115','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M115">View MathML</a>has exactly k+i-1 zeros in (0,1) and is positive near t = 0, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M116','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M116">View MathML</a>has exactly k + i - 1 zeros in (0,1) and is negative near t = 0.

Theorem 3.2 Let (A1), (A2), (A3) and (A5) hold. Assume that for some k ∈ ℕ,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M117','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M117">View MathML</a>

(3.2)

Then there are at least 2k - 1 nontrivial solutions of the problem (1.2). In fact, there exist solutions ω1,...,ωk, such that for 1 ≤ j ≤ k, ωj has exactly j - 1 simple zeros on the open interval (0,1) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M118','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M118">View MathML</a>and there exist solutions z2,...,zk, such that for 2 ≤ j ≤ k, zj has exactly j - 1 simple zeros on the open interval (0,1) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M119','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M119">View MathML</a>.

Let Y = C[0,1] with the norm

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M120','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M120">View MathML</a>

(3.3)

Let E = {x C2[0, 1]|x(0) = x(1) = x′(0) = x′(1) = 0} with the norm

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M121','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M121">View MathML</a>

(3.4)

Then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M122','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M122">View MathML</a> is completely continuous. Here <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M123','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M123">View MathML</a> is given as in (2.48).

Let ζ(,⋅), ξ1(,⋅), ξ2(,⋅) ∈C([0,1] ×ℝ,ℝ) be such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M124','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M124">View MathML</a>

(3.5)

Here x+ = max{x,0}.

Clearly,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M125','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M125">View MathML</a>

(3.6)

uniformly for t ∈ [0,1].

Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M126','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M126">View MathML</a>

(3.7)

then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M127','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M127">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M128','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M128">View MathML</a> are nondecreasing and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M129','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M129">View MathML</a>

(3.8)

Let us consider

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M130','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M130">View MathML</a>

(3.9)

as a bifurcation problem from the trivial solution x ≡ 0.

Equation (3.9) can be converted to the equivalent equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M131','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M131">View MathML</a>

(3.10)

Clearly, the compactness of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M132','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M132">View MathML</a> together with (3.6) imply that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M133','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M133">View MathML</a>

Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M134','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M134">View MathML</a> denotes the set of functions in E which have exactly k - 1 interior nodal (i.e., non-degenerate) zeros in (0,1) and are positive near t = 0, set <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M135','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M135">View MathML</a>, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M136','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M136">View MathML</a>. They are disjoint and open sets in E. Finally, let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M137','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M137">View MathML</a> and Φk = ℝ × Sk.

The results of Rabinowitz [15] for (3.9) can be stated as follows: For each integer k ≥ 1 and each ν = {+, -}, there exists a continuum <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M138','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M138">View MathML</a> of solution of (3.9), joining (λk(a), 0) to infinity in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M139','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M139">View MathML</a>. Moreover, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M140','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M140">View MathML</a>.

Notice that we have used the fact that if x is a nontrivial solution of (3.9), then all zeros of x on (0, 1) are simple under (A1), (A2), (A3), and (A4).

In fact, (3.9) can be rewritten to

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M141','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M141">View MathML</a>

(3.11)

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M142','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M142">View MathML</a>

Clearly â(t) satisfies (H2). So Theorem 2.7 (iii) yields that all zeros of x on (0,1) are simple.

Proof of Theorem 3.1. We first prove the theorem when j = 0.

It is clear that any solution of (3.9) of the form (1, x) yields solutions x of (1.2). We will show that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M143','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M143">View MathML</a> crosses the hyperplane {1} × E in ℝ × E. To do this, it is enough to show that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M143','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M143">View MathML</a> joins (λk(a),0) to (λk(b),∞). Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M144','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M144">View MathML</a> satisfy

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M145','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M145">View MathML</a>

(3.12)

We note that µn > 0 for all n ∈ ℕ since (0, 0) is the only solution of (3.9) for λ = 0 and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M146','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M146">View MathML</a>.

Case 1. λk(b) < 1 < λk(a).

In this case, we show that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M147','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M147">View MathML</a>

We divide the proof into two steps.

Step 1. We show that if there exists a constant number M > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M148','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M148">View MathML</a>

(3.13)

then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M143','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M143">View MathML</a> joins (λk(a),0) to (λk(b),∞).

In this case ║xnE ∞. We divide the equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M149','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M149">View MathML</a>

(3.14)

by ║xnE and set <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M150','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M150">View MathML</a>. Since yn is bounded in C2[0,1], choosing a subsequence and relabeling if necessary, we have that yn → y for some y E with ║y║E = 1. Moreover, from (3.8) and the fact that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M127','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M127">View MathML</a> is nondecreasing, we have that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M151','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M151">View MathML</a>

(3.15)

since

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M152','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M152">View MathML</a>

(3.16)

Thus

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M153','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M153">View MathML</a>

(3.17)

where µ: = limn→∞μn, again choosing a subsequence and relabeling if necessary. Thus

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M154','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M154">View MathML</a>

(3.18)

We claim that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M155','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M155">View MathML</a>

(3.19)

Suppose, to the contrary, that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M156','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M156">View MathML</a>. Since y ≠ 0 is a solution of (3.18), all zeros of y in [0,1] are simple. It follows that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M157','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M157">View MathML</a> for some h ∈ ℝ and l ∈ {+, -}. By the openness of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M158','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M158">View MathML</a> we know that there exists a neighborhood U(y,ρ0) such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M159','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M159">View MathML</a>

which, together with the fact yn → y, implies that exists n0 ∈ ℕ such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M160','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M160">View MathML</a>

However, this contradicts the fact that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M161','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M161">View MathML</a> Therefore, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M162','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M162">View MathML</a>

Now, by Theorem 2.7, we obtain µ = λk(b).

Thus <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M143','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M143">View MathML</a> joins (λk(a),0) to (λk(b),∞).

Step 2. We show that there exists a constant number M > 0 such that µn ∈ (0, M], for all n.

Suppose there is no such M. Choosing a subsequence and relabeling if necessary, it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M163','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M163">View MathML</a>

(3.20)

Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M164','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M164">View MathML</a>

denotes the zeros of xn. Then there exists a subsequence {τ(1, nm)} ⊆ {τ(1, n)} such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M165','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M165">View MathML</a>

Clearly

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M166','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M166">View MathML</a>

We claim that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M167','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M167">View MathML</a>

(3.21)

Suppose, to the contrary, that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M168','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M168">View MathML</a>

(3.22)

Define a function p: [0,1] × [0, ∞) ℝ by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M169','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M169">View MathML</a>

(3.23)

Then, by (A2), (A3), and (A4), there exist two positive numbers ρl and ρ2, such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M170','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M170">View MathML</a>

(3.24)

Using (3.22), (3.24), and the fact that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M171','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M171">View MathML</a>, we conclude that there exists a closed interval I1 ⊂ (τ(0, ∞), τ(1, ∞)) such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M172','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M172">View MathML</a>

uniformly for t I1.

However, since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M173','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M173">View MathML</a> satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M174','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M174">View MathML</a>

the proof of Lemma 4 in [7] (see also the remarks in the final paragraph in [[7], p. 43]), shows that for all n sufficiently large, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M173','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M173">View MathML</a> must change sign on I1. However, this contradicts the fact that for all m sufficiently large we have I1 ⊂ (τ(0, nm),τ(1,nm)) and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M175','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M175">View MathML</a>

Thus, (3.21) holds.

Next, we work with (τ(1, nm), τ(2, nm)). It is easy to see that there is a subsequence <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M176','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M176">View MathML</a> such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M177','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M177">View MathML</a>

Clearly

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M178','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M178">View MathML</a>

(3.25)

We claim that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M179','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M179">View MathML</a>

(3.26)

Suppose, to the contrary, that τ(1,) < τ(2,). Then, from (3.23), (3.24), and the fact that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M180','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M180">View MathML</a>, there exists a closed interval I2 ⊂ (τ(1, ), τ(2, )) such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M181','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M181">View MathML</a>

uniformly for t I2.

This implies the solution <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M182','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M182">View MathML</a> of the equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M183','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M183">View MathML</a>

must change sign on I2. However, this contradicts the fact that for all j sufficiently large we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M184','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M184">View MathML</a> and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M185','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M185">View MathML</a>

Therefore, (3.26) holds.

By a similar argument to obtain (3.21) and (3.26), we can show that for each l ∈ {2,...,k-1},

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M186','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M186">View MathML</a>

(3.27)

Taking a subsequence and relabeling it as {(µn, xn)}, if necessary, it follows that for each l ∈ {0,..., k-1},

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M187','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M187">View MathML</a>

(3.28)

But this is impossible since

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M188','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M188">View MathML</a>

(3.29)

for all n. Therefore,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M189','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M189">View MathML</a>

for some constant number M > 0, independent of n ∈ ℕ.

Case 2. λk(a) < 1< λk(b).

In this case, if <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M144','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M144">View MathML</a> is such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M190','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M190">View MathML</a>

(3.30)

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M191','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M191">View MathML</a>

(3.31)

then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M192','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M192">View MathML</a>

(3.32)

and, moreover, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M193','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M193">View MathML</a>.

Assume that there exists M > 0 such that for all n ∈ ℕ,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M194','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M194">View MathML</a>

(3.33)

Applying a similar argument to that used in step 1 of Case 1, after taking a subsequence and relabeling if necessary, it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M195','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M195">View MathML</a>

(3.34)

Again <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M143','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M143">View MathML</a> joins (λk(a), 0) to (λk(b), ∞) and the result follows.

Finally, let j ∈ ℕ. By repeating the arguments used in the proof of the case j = 0, we see that for each ν ∈ {+, -} and each i ∈ {k, k + 1,..., k + j},

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M196','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M196">View MathML</a>

(3.35)

The result follows.

Proof of Theorem 3.2.

We only need to show that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M197','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M197">View MathML</a>

Suppose on the contrary that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M198','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M198">View MathML</a>

(3.36)

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M199','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M199">View MathML</a>

(3.37)

Since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M200','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M200">View MathML</a> joins (λi,(a), 0) to infinity in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M201','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M201">View MathML</a> and (λ, x) = (0, 0) is the unique solution of (3.9)λ = 0 in E, there exists a sequence <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M202','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M202">View MathML</a> such that µn ∈ (0,1) and ║xnE → ∞ as n → ∞. We may assume that µn→µ ∈ [0, 1] as n → ∞. Let yn = xn/║xnE, n ≥ 1. From the fact

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M203','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M203">View MathML</a>

(3.38)

We have that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M204','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M204">View MathML</a>

(3.39)

Furthermore, since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M205','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M205">View MathML</a> is completely continuous, we may assume that there exists y E with ║y║E = 1 such that ║yn - y║E → 0 as n . Since

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M206','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M206">View MathML</a>

(3.40)

uniformly for t ∈ [0,1], we have from (3.39) and (3.8) that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M207','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M207">View MathML</a>

(3.41)

that is,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M208','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M208">View MathML</a>

(3.42)

By (A1), (A5), and (3.42) and the fact that ║y║E = 1, we conclude that µc(t)y+ ≢ 0 on any compact subinterval in [0,1], and consequently

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M209','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M209">View MathML</a>

(3.43)

By Theorem 2.8, we know that y(t) > 0 in (0,1). This means µ is the first eigenvalue of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M210','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M210">View MathML</a> and y is the corresponding eigenfunction. Hence <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M211','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M211">View MathML</a> and therefore, since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M212','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M212">View MathML</a> is open and ║yn - y║E 0, we have that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M213','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M213">View MathML</a> for n large. But this contradicts the assumption that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M214','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/31/mathml/M214">View MathML</a> and (i,l) ∈ Γ, so (3.36) is wrong, which completes the proof.

Competing interests

The author declares that they have no competing interests.

Acknowledgements

Thanks are given to Professor R.Y. Ma for his valuable suggestion. The author is also grateful to the anonymous referee for his/her valuable suggestions. This study was supported by: the NSFC (No. 11031003); the Scientific Research Foundation of the Education department of Gansu Province (No. 1114-04).

References

  1. Agarwal, RP, Chow, YM: Iterative methods for a fourth-order boundary value problem. J Comput Appl Math. 10(2), 203–217 (1984). Publisher Full Text OpenURL

  2. Ma, R, Wu, HP: Positive solutions of a fourth-order two-point boundary value problem. Acta Math Sci A. 22(2), 244–249 (2002)

  3. Yao, Q: Positive solutions for eigenvalue problems of fourth-order elastic beam equations. Appl Math Lett. 17(2), 237–243 (2004). Publisher Full Text OpenURL

  4. Yao, Q: Solvability of an elastic beam equation with Caratheodory function. Math Appl. 17(3), 389–392 (2004)

  5. Korman, P: Uniqueness and exact multiplicity of solutions for a class of fourth-order semilinear problems. Proc Royal Soc Edinburgh A. 134(1), 179–190 (2004). Publisher Full Text OpenURL

  6. Xu, J, Han, X: Nodal solutions for a fourth-order two-point boundary value problem. Boundary Value Problem. 2010, 11 Article ID 570932 (2010)

  7. Elias, U: Eigenvalue problems for the equations Ly + λp(x)y = 0. J Diff Equ. 29(1), 28–57 (1978). Publisher Full Text OpenURL

  8. Elias, U: Oscillation Theory of Two-Term Differential Equations, vol. 396 of Mathematics and Its Applications, Kluwer Academic Publishers, Dordrecht, The Netherlands (1997)

  9. Ma, R: Nodal Solutions for a fourth-order two-point boundary value problem. J Math Anal Appl. 314(1), 254–265 (2006). Publisher Full Text OpenURL

  10. Ma, R: Nodal Solutions of boundary value problem of fourth-order ordinary differential equations. J Math Anal Appl. 319(2), 424–434 (2006). Publisher Full Text OpenURL

  11. Ma, R, Xu, J: Bifurcation from interval and positive solutions of a fourth-order boundary value problem. Nonlinear Anal Theory Methods Appl. 72(1), 113–122 (2010). Publisher Full Text OpenURL

  12. Ma, R, Wang, H: On the existence of positive solutions of a fourth-order ordinary differential equations. Appl Anal. 59(1-4), 225–231 (1995). Publisher Full Text OpenURL

  13. Ma, R: Existence of positive solutions of a fourth-order boundary value problem. Appl Math Comput. 168(2), 1219–1231 (2005). Publisher Full Text OpenURL

  14. Ma, R, Thompson, B: Nodal solutions for a nonlinear fourth-order eigenvalue problem. Acta Math Sin (Engl Ser). 24(1), 27–34 (2008). Publisher Full Text OpenURL

  15. Rabinowitz, PH: Some global results for nonlinear eigenvalue problems. J Funct Anal. 7(32), 487–513 (1971)