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Nodal solutions of second-order two-point boundary value problems

Ruyun Ma, Bianxia Yang and Guowei Dai*

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Department of Mathematics, Northwest Normal University, Lanzhou 730070, P. R. China

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Boundary Value Problems 2012, 2012:13  doi:10.1186/1687-2770-2012-13


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2012/1/13


Received:16 August 2011
Accepted:10 February 2012
Published:10 February 2012

© 2012 Ma 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

We shall study the existence and multiplicity of nodal solutions of the nonlinear second-order two-point boundary value problems,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M1">View MathML</a>

The proof of our main results is based upon bifurcation techniques.

Mathematics Subject Classifications: 34B07; 34C10; 34C23.

Keywords:
nodal solutions; bifurcation

1 Introduction

In [1], Ma and Thompson were considered with determining interval of μ, in which there exist nodal solutions for the boundary value problem (BVP)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M2">View MathML</a>

(1.1)

under the assumptions:

(C1) w(·) ∈ C([0, 1], [0, ∞)) and does not vanish identically on any subinterval of [0, 1];

(C2) f C(ℝ, ℝ) with sf(s) > 0 for s ≠ 0;

(C3) there exist f0, f∞ ∈ (0, ∞) such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M3">View MathML</a>

It is well known that under (C1) assumption, the eigenvalue problem

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M4">View MathML</a>

(1.2)

has a countable number of simple eigenvalues μk, k = 1, 2,..., which satisfy

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M5">View MathML</a>

and let μk be the kth eigenvalue of (1.2) and φk be an eigenfunction corresponding to μk, then φk has exactly k -- 1 simple zeros in (0,1) (see, e.g., [2]).

Using Rabinowitz bifurcation theorem, they established the following interesting results:

Theorem A (Ma and Thompson [[1], Theorem 1.1]). Let (C1)-(C3) hold. Assume that for some k ∈ ℕ, either <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M6">View MathML</a>or <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M7">View MathML</a>. Then BVP (1.1) has two solutions <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M8">View MathML</a>and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M9">View MathML</a>such that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M8">View MathML</a>has exactly k -- 1 zeros in (0, 1) and is positive near 0, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M9">View MathML</a>has exactly k -- 1 zeros in (0,1) and is negative near 0.

In [3], Ma and Thompson studied the existence and multiplicity of nodal solutions for BVP

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M10">View MathML</a>

(1.3)

They gave conditions on the ratio <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M11">View MathML</a> at infinity and zero that guarantee the existence of solutions with prescribed nodal properties.

Using Rabinowitz bifurcation theorem also, they established the following two main results:

Theorem B (Ma and Thompson [[1], Theorem 2]). Let (C1)-(C3) hold. Assume that either (i) or (ii) holds for some k ∈ ℕ and j ∈ {0} ∪ ℕ;

(i) f0 < μk < ⋯ < μk+j < f;

(ii) f< μk < ⋯ < μk+j < f0,

where μk denotes the kth eigenvalue of (1.2). Then BVP (1.3) has 2(j + 1) solutions <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M12">View MathML</a>, such that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M13">View MathML</a>has exactly k + i -- 1 zeros in (0, 1) and are positive near 0, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M14">View MathML</a>has exactly k + i -- 1 zeros in (0,1) and are negative near 0.

Theorem C (Ma and Thompson [[1], Theorem 3]). Let (C1)-(C3) hold. Assume that there exists an integer k ∈ ℕ such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M15">View MathML</a>

where μk denotes the kth eigenvalue of (1.2). Then BVP (1.3) has no nontrivial solution.

From above literature, we can see that the existence and multiplicity results are largely based on the assumption that t and u are separated in nonlinearity term. It is interesting to know what will happen if t and u are not separated in nonlinearity term? We shall give a confirm answer for this question.

In this article, we consider the existence and multiplicity of nodal solutions for the nonlinear BVP

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M16">View MathML</a>

(1.4)

under the following assumptions:

(H1) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M17">View MathML</a>uniformly on [0, 1], and the inequality is strict on some subset of positive measure in (0,1), where λk denotes the kth eigenvalue of

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M18">View MathML</a>

(1.5)

(H2) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M19">View MathML</a>uniformly on [0, 1], and all the inequalities are strict on some subset of positive measure in (0, 1), where λk denotes the kth eigenvalue of (1.5);

(H3) f(t, s)s > 0 for t ∈ (0, 1) and s ≠ 0.

Remark 1.1. From (H1)-(H3), we can see that there exist a positive constant ϱ and a subinterval [α, β] of [0, 1] such that α < β and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M20">View MathML</a> for all r ∈ [α, β] and s ≠ 0.

In the celebrated study [4], Rabinowitz established Rabinowitz's global bifurcation theory [[4], Theorems 1.27 and 1.40]. However, as pointed out by Dancer [5,6] and López-Gómez [7], the proofs of these theorems contain gaps, the original statement of Theorem 1.40 of [4] is not correct, the original statement of Theorem 1.27 of [4] is stronger than what one can actually prove so far. Although there exist some gaps in the proofs of Rabinowitz's Theorems 1.27, 1.40, and 1.27 has been used several times in the literature to analyze the global behavior of the component of nodal solutions emanating from u = 0 in wide classes of boundary value problems for equations and systems [1,2,8,9]. Fortunately, López-Gómez gave a corrected version of unilateral bifurcation theorem in [7].

By applying the bifurcation theorem of López-Gómez [[7], Theorem 6.4.3], we shall establish the following:

Theorem 1.1. Suppose that f(t, u) satisfies (H1), (H2), and (H3), then problem (1.4) possesses two solutions <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M8">View MathML</a>and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M9">View MathML</a>, such that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M8">View MathML</a>has exactly k -- 1 zeros in (0, 1) and is positive near 0, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M9">View MathML</a>has exactly k -- 1 zeros in (0,1) and is negative near 0.

Similarly, we also have the following:

Theorem 1.2. Suppose that f(t, u) satisfies (H3) and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M21">View MathML</a>uniformly on [0, 1], and all the inequalities are strict on some subset of positive measure in (0, 1), where λk denotes the kth eigenvalue of (1.5);

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M22">View MathML</a>uniformly on [0, 1], and the inequality is strict on some subset of positive measure in (0, 1), where λk denotes the kth eigenvalue of (1.5), then problem (1.4) possesses two solutions <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M8">View MathML</a>and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M9">View MathML</a>, such that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M8">View MathML</a> has exactly k -- 1 zeros in (0,1) and is positive near 0, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M9">View MathML</a>has exactly k -- 1 zeros in (0,1) and is negative near 0.

Remark 1.2. We would like to point out that the assumptions (H1) and (H2) are weaker than the corresponding conditions of Theorem A. In fact, if we let f(t, s) ≡ μw(t)f(s), then we can get <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M23">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M24">View MathML</a>. By the strict decreasing of μk(f) with respect to weight function f (see [10]), where μk(f) denotes the kth eigenvalue of (1.2) corresponding to weight function f, we can show that our condition c(t) ≤ λk a(t) is equivalent to the condition <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M6">View MathML</a>. Similarly, our condition c (t) ≥ λk a (t) is equivalent to the condition <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M7">View MathML</a>. Therefore, Theorem A is the corollary of Theorems 1.1 and 1.2.

Using the similar proof with the proof Theorems 1.1 and 1.2, we can obtain the more general results as follows.

Theorem 1.3. Suppose that (H3) holds, and either (i) or (ii) holds for some k ∈ ℕ and j ∈ {0} ∪ ℕ:

(i) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M25">View MathML</a>uniformly on [0, 1], and the inequalities are strict on some subset of positive measure in (0,1), where λk denotes the kth eigenvalue of (1.5);

(ii) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M26">View MathML</a>uniformly on [0, 1], and the inequality is strict on some subset of positive measure in (0, 1), where λk denotes the kth eigenvalue of (1.5).

Then BVP (1.4) has 2(j + 1) solutions <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M12">View MathML</a>, such that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M13">View MathML</a>has exactly k + i -- 1 zeros in (0,1) and are positive near 0, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M14">View MathML</a>has exactly k + i -- 1 zeros in (0,1) and are negative near 0.

Using Sturm Comparison Theorem, we also can get a non-existence result when f satisfies a non-resonance condition.

Theorem 1.4. Let (H3) hold. Assume that there exists an integer k ∈ ℕ such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M27">View MathML</a>

(1.6)

for any t ∈ [0, 1], where λk denotes the kth eigenvalue of (1.5). Then BVP (1.4) has no nontrivial solution.

Remark 1.3. Similarly to Remark 1.2, we note that the assumptions (i) and (ii) are weaker than the corresponding conditions of Theorem B. In fact, if we let f(t, s) ≡ w(t) f(s), then we can get <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M28">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M29">View MathML</a>. By the strict decreasing of μk(f) with respect to weight function f (see [11]), where μk(f) denotes the kth eigenvalue of (1.2) corresponding to weight function f, we can show that our condition c(t) ≤ λk < ⋯ < λk+j a(t) is equivalent to the condition f0 < μk < · · · < μk+j < f. Similarly, our condition a(t) ≤ λk < · · · < λk+j c(t) is equivalent to the condition f< μk < ⋯ < μk+j < f0. Therefore, Theorem B is the corollary of Theorem 1.3. Similar, we get Theorem C is also the corollary of Theorem 1.4.

2 Preliminary results

To show the nodal solutions of the BVP (1.4), we need only consider an operator equation of the following form

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M30">View MathML</a>

(2.1)

Equations of the form (2.1) are usually called nonlinear eigenvalue problems. López-Gómez [7] studied a nonlinear eigenvalue problem of the form

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M31">View MathML</a>

(2.2)

where r ∈ ℝ is a parameter, u X, X is a Banach space, θ is the zero element of X, and G: <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M32">View MathML</a> is completely continuous. In addition, G(r, u) = rTu + H(r, u), where H(r, u) = o(||u||) as ||u|| → 0 uniformly on bounded r interval, and T is a linear completely continuous operator on X. A solution of (2.2) is a pair <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M33">View MathML</a>, which satisfies the equation (2.2). The closure of the set nontrivial solutions of (2.2) is denoted by ℂ, let Σ(T) denote the set of eigenvalues of linear operator T. López-Gómez [7] established the following results:

Lemma 2.1 [[7], Theorem 6.4.3]. Assume Σ(T) is discrete. Let λ0 ∈ Σ(T) such that ind(0, λ0T) changes sign as λ crosses λ0, then each of the components <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M34">View MathML</a>satisfies <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M35">View MathML</a>, and either

(i) meets infinity in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M36">View MathML</a>,

(ii) meets (τ, θ), where τ λ0 ∈ Σ(T) or

(iii) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M34">View MathML</a>contains a point

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M37">View MathML</a>

where V is the complement of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M38">View MathML</a>denotes the eigenfunction corresponding to eigenvalue λ0.

Lemma 2.2 [[7], Theorem 6.5.1]. Under the assumptions:

(A) X is an order Banach space, whose positive cone, denoted by P, is normal and has a nonempty interior;

(B) The family ϒ(r) has the special form

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M39">View MathML</a>

where T is a compact strongly positive operator, i.e., T(P\{0}) ⊂ int P;

(C) The solutions of u = rTu + H(r, u) satisfy the strong maximum principle.

Then the following assertions are true:

(1) Spr (T) is a simple eigenvalue of T, having a positive eigenfunction denoted by ψ0 > 0, i.e., ψ0 ∈ int P, and there is no other eigenvalue of T with a positive eigenfunction;

(2) For every y ∈ int P, the equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M40">View MathML</a>

has exactly one positive solution if <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M41">View MathML</a>, whereas it does not admit a positive solution if <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M42">View MathML</a>.

Lemma 2.3 [[10], Theorem 2.5]. Assume T : X X is a completely continuous linear operator, and 1 is not an eigenvalue of T, then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M43">View MathML</a>

where β is the sum of the algebraic multiplicities of the eigenvalues of T large than 1, and β = 0 if T has no eigenvalue of this kind.

Let Y = C[0, 1] with the norm <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M44">View MathML</a>. Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M45">View MathML</a>

with the norm

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M46">View MathML</a>

Define L: D(L) → Y by setting

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M47">View MathML</a>

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M48">View MathML</a>

Then L-1: Y E is compact. Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M49">View MathML</a> under the product topology. For any C1 function u, if u(x0) = 0, then x0 is a simple zero of u, if u'(x0) ≠ 0. For any integer k ∈ ℕ and ν ∈ {+, --}, define <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M50">View MathML</a> consisting of functions u C1 [0, 1] satisfying the following conditions:

(i) u(0) = 0, νu'(0) > 0;

(ii) u has only simple zeros in [0, 1] and exactly n -- 1 zeros in (0,1).

Then sets <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M51">View MathML</a> are disjoint and open in E. Finally, let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M52">View MathML</a>.

Furthermore, let ζ ∈ C[0, 1] × ℝ) be such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M53">View MathML</a>

with

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M54">View MathML</a>

(2.3)

Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M55">View MathML</a>

then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M56">View MathML</a> is nondecreasing with respect to u and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M57">View MathML</a>

If u ∈ E, it follows from (2.3) that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M58">View MathML</a>

uniformly for t ∈ [0, 1].

Let us study

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M59">View MathML</a>

(2.4)

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

Equation (2.4) can be converted to the equivalent equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M60">View MathML</a>

Further we note that ||L-1[ζ(t, u(t))] ||E = o(||u||E) for u near 0 in E.

Lemma 2.4. For each k ∈ ℕ and ν ∈ {+. -- }, there exists a continuum <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M61">View MathML</a> of solutions of (2.4) with the properties:

(i) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M62">View MathML</a>;

(ii) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M63">View MathML</a>;

(iii) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M64">View MathML</a>is unbounded in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M65">View MathML</a>, where λk denotes the kth eigenvalue of (1.5).

Proof. It is easy to see that the problem (2.4) is of the form considered in [7], and satisfies the general hypotheses imposed in that article.

Combining Lemma 2.1 with Lemma 2.3, we know that there exists a continuum <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M66">View MathML</a> of solutions of (2.4) such that:

(a) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M67">View MathML</a> is unbounded and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M68">View MathML</a>;

(b) or <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M69','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M69">View MathML</a>, where j ∈ ℕ, λj is another eigenvalue of (1.5) and different from λk;

(c) or <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M64">View MathML</a> contains a point

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M37">View MathML</a>

where V is the complement of span{φk}, φk denotes the eigenfunction corresponding to eigenvalue λk.

We finally prove that the first choice of the (a) is the only possibility.

In fact, all functions belong to the continuum sets <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M67">View MathML</a> have exactly k -- 1 simple zeros, this implies that it is impossible to exist <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M70','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M70">View MathML</a>.

Next, we shall prove (c) is impossible, suppose (c) occurs, then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M67">View MathML</a> is bounded and without loss of generality, suppose there exists a point <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M71','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M71">View MathML</a>. Moreover, it follows from Lemma 2.1 that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M72','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M72">View MathML</a>

Note that as the complement V of span{φk} in E, we can take

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M73','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M73">View MathML</a>

Thus, for this choice of V, the component <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M74','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M74">View MathML</a> cannot contain a point

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M75">View MathML</a>

Indeed, if

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M75">View MathML</a>

then y > 0 in (0, a0), where a0 denotes the first zero point of y, and there exists u E for which

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M76','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M76">View MathML</a>

Thus, for each sufficiently large α > 0, we have that u + αφk >> 0 in (0, a0) and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M77">View MathML</a>

Define

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M78','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M78">View MathML</a>

Hence, according to Lemma 2.2

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M79','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M79">View MathML</a>

which is impossible since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M80','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M80">View MathML</a>.

Lemma 2.5. If <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M81','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M81">View MathML</a>is a non-trivial solution of (2.4), then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M82','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M82">View MathML</a>for ν and some k ∈ ℕ.

Proof. Taking into account Lemma 2.4, we only need to prove that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M83','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M83">View MathML</a>.

Suppose <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M84','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M84">View MathML</a>. Then there exists <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M85','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M85">View MathML</a> such that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M86">View MathML</a>, and (μj,uj) → (μ*, u) with <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M87','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M87">View MathML</a>. Since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M88','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M88">View MathML</a>, so u ≡ 0. Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M89','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M89">View MathML</a>, then cj should be a solution of problem,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M90','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M90">View MathML</a>

(2.5)

By (2.3), (2.5) and the compactness of L-1, we obtain that for some convenient subsequence cj c0 ≠ 0 as j → + ∞. Now c0 verifies the equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M91','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M91">View MathML</a>

and ||c0||E = 1. Hence μ* = λi, for some i k, i ∈ ℕ. Therefore, (μj, uj) → (λi, θ) with <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M92','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M92">View MathML</a>. This contradicts to Lemma 2.3.

3 Proof of main results

Proof of Theorems 1.1 and 1.2. We only prove Theorem 1.1 since the proof of Theorem 1.2 is similar. It is clear that any solution of (2.4) of the form (1, u) yields a solution u of (1.4). We shall show <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M67">View MathML</a> crosses the hyperplane {1} × E in ℝ × E.

By the strict decreasing of μk(c(t)) with respect to c(t) (see [11]), where μk(c(t)) is the kth eigenvalue of (1.2) corresponding to the weight function c(t), we have μk(c(t)) > μk(λk) = 1.

Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M93','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M93">View MathML</a> with <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M94','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M94">View MathML</a> satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M95','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M95">View MathML</a>

We note that μj > 0 for all j ∈ ℕ, since (0,0) is the only solution of (2.4) for μ = 0 and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M96','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M96">View MathML</a>.

Step 1: We show that if there exists a constant M > 0, such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M97','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M97">View MathML</a>

for j ∈ ℕ large enough, then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M67">View MathML</a> crosses the hyperplane {1} × E in ℝ × E.

In this case it follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M98','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M98">View MathML</a>

Let ξ ∈ C([0, 1] × ℝ) be such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M99','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M99">View MathML</a>

with

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M100','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M100">View MathML</a>

(3.1)

We divide the equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M101','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M101">View MathML</a>

(3.2)

set <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M102','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M102">View MathML</a>. Since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M103','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M103">View MathML</a> is bounded in C2 [0, 1], after taking a subsequence if necessary, we have that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M104','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M104">View MathML</a> for some <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M105','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M105">View MathML</a> with ||u||E = 1. By (3.1), using the similar proof of (2.3), we have that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M106','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M106">View MathML</a>

By the compactness of L we obtain

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M107','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M107">View MathML</a>

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M108','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M108">View MathML</a>, again choosing a subsequence and relabeling if necessary.

It is clear that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M109','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M109">View MathML</a> since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M110','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M110">View MathML</a> is closed in ℝ × E. Therefore, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M111','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M111">View MathML</a> is the kth eigenvalue of

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M112','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M112">View MathML</a>

By the strict decreasing of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M111','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M111">View MathML</a> with respect to a(t) (see [11]), where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M111','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M111">View MathML</a> is the kth eigenvalue of (1.2) corresponding to the weight function a(t), we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M113','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M113">View MathML</a>. Therefore, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M110','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M110">View MathML</a> crosses the hyperplane {1} × E in ℝ × E.

Step 2: We show that there exists a constant M such that μj ∈ (0, M] for j ∈ ℕ large enough.

On the contrary, we suppose that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M114','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M114">View MathML</a>

On the other hand, we note that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M115','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M115">View MathML</a>

In view of Remark 1.1, we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M116','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M116">View MathML</a> on [α, β] and for j large enough and all t ∈ [0, 1]. By Lemma 3.2 of [12], we get uj must change its sign more than k times on [α, β] for j large enough, which contradicts the act that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M117','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M117">View MathML</a>.

Therefore,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M118','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M118">View MathML</a>

for some constant number M > 0 and j ∈ ℕ sufficiently large.

Proof of Theorem 1.3. Repeating the arguments used in the proof of Theorems 1.1 and 1.2, we see that for ν ∈ {+, --} and each i ∈ {k, k + 1,..., k + j}

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M119','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M119">View MathML</a>

The results follows.

Proof of Theorem 1.4. Assume to the contrary that BVP (1.4) has a solution u E, we see that u satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M120','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M120">View MathML</a>

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M121','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M121">View MathML</a>.

Note that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M122','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M122">View MathML</a> and hence <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M123','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/13/mathml/M123">View MathML</a> can be regarded as a continuous function on ℝ. Thus we get b(·) ∈ C[0, 1]. Also, notice that a nontrivial solution of (1.4) has a finite number of zeros. From (2.8) and the above fact λk < b(t) < λk+1 for all t ∈ [0, 1].

We know that the eigenfunction φk corresponding to λk has exactly k -- 1 zeros in [0, 1]. Applying Lemma 2.4 of [13] to φk and u, we see that u has at least k zeros in I. By Lemma 2.4 of [13] again to u and φk+1, we get that φk+1 has at least k + 1 zeros in [0, 1]. This is a contradiction.

Competing interests

The authors declare that they have no competing interests.

Authors' contributions

GD conceived of the study, and participated in its design and coordination and helped to draft the manuscript. BY drafted the manuscript. RM participated in the design of the study. All authors read and approved the final manuscript.

Acknowledgements

The authors were very grateful to the anonymous referees for their valuable suggestions. This study was supported by the NSFC (No. 11061030, No. 10971087) and NWNU-LKQN-10-21.

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