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Solvability for fractional order boundary value problems at resonance

Zhigang Hu* and Wenbin Liu

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Department of Mathematics, China University of Mining and Technology, Xuzhou 221008, People's Republic of China

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Citation and License

Boundary Value Problems 2011, 2011:20  doi:10.1186/1687-2770-2011-20


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2011/1/20


Received:10 May 2011
Accepted:5 September 2011
Published:5 September 2011

© 2011 Hu and Liu; 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 paper, by using the coincidence degree theory, we consider the following boundary value problem for fractional differential equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M1">View MathML</a>

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M2">View MathML</a> denotes the Caputo fractional differential operator of order α, 2 < α ≤ 3. A new result on the existence of solutions for above fractional boundary value problem is obtained.

Mathematics Subject Classification (2000): 34A08, 34B15.

Keywords:
Fractional differential equations; boundary value problems; resonance; coincidence degree theory

1 Introduction

Fractional calculus is a generalization of ordinary differentiation and integration on an arbitrary order that can be noninteger. This subject, as old as the problem of ordinary differential calculus, can go back to the times when Leibniz and Newton invented differential calculus. As is known to all, the problem for fractional derivative was originally raised by Leibniz in a letter, dated September 30, 1695.

In recent years, the fractional differential equations have received more and more attention. The fractional derivative has been occurring in many physical applications such as a non-Markovian diffusion process with memory [1], charge transport in amorphous semiconductors [2], propagations of mechanical waves in viscoelastic media [3], etc. Phenomena in electromagnetics, acoustics, viscoelasticity, electrochemistry, and material science are also described by differential equations of fractional order (see [4-9]).

Recently, boundary value problems (BVPs for short) for fractional differential equations at nonresonance have been studied in many papers (see [10-16]). Moreover, Kosmatov studied the BVPs for fractional differential equations at resonance (see [17]). Motivated by the work above, in this paper, we consider the following BVP of fractional equation at resonance

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M3">View MathML</a>

(1.1)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M2">View MathML</a> denotes the Caputo fractional differential operator of order α, 2 < α ≤ 3. f : [0, 1] × ℝ3 → ×ℝ is continuous.

The rest of this paper is organized as follows. Section 2 contains some necessary notations, definitions, and lemmas. In Section 3, we establish a theorem on existence of solutions for BVP (1.1) under nonlinear growth restriction of f, basing on the coincidence degree theory due to Mawhin (see [18]). Finally, in Section 4, an example is given to illustrate the main result.

2 Preliminaries

In this section, we will introduce notations, definitions, and preliminary facts that are used throughout this paper.

Let X and Y be real Banach spaces and let L : domL X Y be a Fredholm operator with index zero, and P : X X, Q : Y Y be projectors such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M4">View MathML</a>

It follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M5">View MathML</a>

is invertible. We denote the inverse by KP.

If Ω is an open bounded subset of X, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M6">View MathML</a>, the map N : X Y will be called L-compact on <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M7">View MathML</a> if <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M8">View MathML</a> is bounded and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M9">View MathML</a> is compact. Where I is identity operator.

Lemma 2.1. ([18]) If Ω is an open bounded set, let L : domL X Y be a Fredholm operator of index zero and N : X Y L-compact on <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M7">View MathML</a>. Assume that the following conditions are satisfied

(1) Lx λNx for every (x, λ) ∈ [(domL\KerL)] ∩ ∂Ω × (0, 1);

(2) Nx ∉ ImL for every x ∈ KerL ∩ ∂Ω;

(3) deg(QN|KerL, KerL ∩ Ω, 0) ≠ 0, where Q : Y Y is a projection such that ImL = KerQ.

Then the equation Lx = Nx has at least one solution in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M10">View MathML</a>.

Definition 2.1. The Riemann-Liouville fractional integral operator of order α > 0 of a function x is given by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M11">View MathML</a>

provided that the right side integral is pointwise defined on (0, +∞).

Definition 2.2. The Caputo fractional derivative of order α > 0 of a continuous function x is given by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M12">View MathML</a>

where n is the smallest integer greater than or equal to α, provided that the right side integral is pointwise defined on (0, +∞).

Lemma 2.2. ([19]) For α > 0, the general solution of the Caputo fractional differential equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M13">View MathML</a>

is given by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M14">View MathML</a>

where ci ∈ ℝ, i = 0, 1, 2, . . ., n - 1; here, n is the smallest integer greater than or equal to α.

Lemma 2.3. ([19]) Assume that x C(0, 1) ∩ L(0, 1) with a Caputo fractional derivative of order α > 0 that belongs to C(0, 1) ∩ L(0, 1). Then,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M15">View MathML</a>

where ci ∈ ℝ, i = 0, 1, 2, . . ., n - 1; here, n is the smallest integer greater than or equal to α.

In this paper, we denote X = C2[0, 1] with the norm ||x||X = max{||x||, ||x'||, ||x"||} and Y = C[0, 1] with the norm ||y||Y = ||y||, where ||x||= maxt∈[0, 1] |x(t)|. Obviously, both X and Y are Banach spaces.

Define the operator L : domL X Y by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M16">View MathML</a>

(2.1)

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M17">View MathML</a>

Let N : X Y be the Nemytski operator

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M18">View MathML</a>

Then, BVP (1.1) is equivalent to the operator equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M19">View MathML</a>

3 Main result

In this section, a theorem on existence of solutions for BVP (1.1) will be given.

Theorem 3.1. Let f : [0, 1] × ℝ3 → ℝ be continuous. Assume that

(H1) there exist nonnegative functions p, q, r, s C[0, 1] with Γ(α - 1) - q1 - r1 - s1 > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M20">View MathML</a>

where p1 = ||p||, q1 = ||q||, r1 = ||r||, s1 = ||s||.

(H2) there exists a constant B > 0 such that for all u ∈ ℝ with |u| > B either

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M21">View MathML</a>

or

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M22">View MathML</a>

Then, BVP (1.1) has at leat one solution in X.

Now, we begin with some lemmas below.

Lemma 3.1. Let L be defined by (2.1), then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M23">View MathML</a>

(3.1)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M24">View MathML</a>

(3.2)

Proof. By Lemma 2.2, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M25">View MathML</a> has solution

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M26">View MathML</a>

Combining with the boundary value condition of BVP (1.1), one has (3.1) hold.

For y ∈ ImL, there exists x ∈ domL such that y = Lx Y. By Lemma 2.3, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M27">View MathML</a>

Then, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M28">View MathML</a>

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M29">View MathML</a>

By conditions of BVP (1.1), we can get that y satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M30">View MathML</a>

Thus, we get (3.2). On the other hand, suppose y Y and satisfies <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M31">View MathML</a>. Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M32">View MathML</a>, then x ∈ domL and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M33">View MathML</a>. So that, y ∈ ImL. The proof is complete.

Lemma 3.2. Let L be defined by (2.1), then L is a Fredholm operator of index zero, and the linear continuous projector operators P : X X and Q : Y Y can be defined as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M34">View MathML</a>

Furthermore, the operator KP : ImL → domL ∩ KerP can be written by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M35">View MathML</a>

Proof. Obviously, ImP = KerL and P2x = Px. It follows from x = (x - Px) + Px that X = KerP + KerL. By simple calculation, we can get that KerL ∩ KerP = {0}. Then, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M36">View MathML</a>

For y Y, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M37">View MathML</a>

Let y = (y - Qy) + Qy, where y - Qy ∈ KerQ = ImL, Qy ∈ ImQ. It follows from KerQ = ImL and Q2y = Qy that ImQ ∩ ImL = {0}. Then, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M38">View MathML</a>

Thus,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M39">View MathML</a>

This means that L is a Fredholm operator of index zero.

From the definitions of P, KP, it is easy to see that the generalized inverse of L is KP. In fact, for y ∈ ImL, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M40">View MathML</a>

(3.3)

Moreover, for x ∈ domL ∩ KerP, we get x(0) = x'(0) = x"(0) = 0. By Lemma 2.3, we obtain that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M41">View MathML</a>

which together with x(0) = x'(0) = x"(0) = 0 yields that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M42">View MathML</a>

(3.4)

Combining (3.3) with (3.4), we know that KP = (L|domL∩KerP)-1. The proof is complete.

Lemma 3.3. Assume Ω ⊂ X is an open bounded subset such that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M6">View MathML</a>, then N is L-compact on <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M7">View MathML</a>.

Proof. By the continuity of f, we can get that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M8">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M43">View MathML</a> are bounded. So, in view of the Arzelà -Ascoli theorem, we need only prove that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M44">View MathML</a> is equicontinuous.

From the continuity of f, there exists constant A > 0 such that |(I - Q)Nx| ≤ A, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M45">View MathML</a>, t ∈ [0, 1]. Furthermore, denote KP,Q = KP(I - Q)N and for 0 ≤ t1 < t2 ≤ 1, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M46">View MathML</a>, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M47">View MathML</a>

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M48">View MathML</a>

Since tα, tα-1 and tα-2 are uniformly continuous on [0, 1], we can get that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M49">View MathML</a>, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M50">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M51">View MathML</a> are equicontinuous. Thus, we get that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M52">View MathML</a> is compact. The proof is completed.

Lemma 3.4. Suppose (H1), (H2) hold, then the set

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M53">View MathML</a>

is bounded.

Proof. Take x ∈ Ω1, then Nx ∈ ImL. By (3.2), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M54">View MathML</a>

Then, by the integral mean value theorem, there exists a constant ξ ∈ (0, 1) such that f(ξ, x(ξ), x'(ξ), x"(ξ)) = 0. Then from (H2), we have |x(ξ)| ≤ B.

Then, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M55">View MathML</a>

That is

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M56">View MathML</a>

(3.5)

From x ∈ domL, we get x'(0) = 0. Therefore,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M57">View MathML</a>

That is

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M58">View MathML</a>

(3.6)

By Lx = λNx and x ∈ domL, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M59">View MathML</a>

Then we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M60">View MathML</a>

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M61">View MathML</a>

From (3.5),(3.6), and (H1), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M62">View MathML</a>

Thus, from Γ(α - 1) - q1 - r1 - s1 > 0, we obtain that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M63">View MathML</a>

Thus, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M64">View MathML</a>

and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M65">View MathML</a>

Therefore,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M66">View MathML</a>

So Ω1 is bounded. The proof is complete.

Lemma 3.5. Suppose (H2) holds, then the set

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M67">View MathML</a>

is bounded.

Proof. For x ∈ Ω2, we have x(t) = c, c ∈ ℝ, and Nx ∈ ImL. Then, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M68">View MathML</a>

which together with (H2) implies |c| ≤ B. Thus, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M69','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M69">View MathML</a>

Hence, Ω2 is bounded. The proof is complete.

Lemma 3.6. Suppose the first part of (H2) holds, then the set

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M70','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M70">View MathML</a>

is bounded.

Proof. For x ∈ Ω3, we have x(t) = c, c ∈ ℝ, and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M71','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M71">View MathML</a>

(3.7)

If λ = 0, then |c| ≤ B because of the first part of (H2). If λ ∈ (0, 1], we can also obtain |c| ≤ B. Otherwise, if |c| > B, in view of the first part of (H2), one has

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M72','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M72">View MathML</a>

which contradicts to (3.7).

Therefore, Ω3 is bounded. The proof is complete.

Remark 3.1. Suppose the second part of (H2) hold, then the set

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M73','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M73">View MathML</a>

is bounded.

The proof of Theorem 3.1. Set Ω = {x X | ||x||X < max{M1, B, B + M1} + 1}. It follows from Lemma 3.2 and 3.3 that L is a Fredholm operator of index zero and N is L-compact on <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M7">View MathML</a>. By Lemma 3.4 and 3.5, we get that the following two conditions are satisfied

(1) Lx λNx for every (x, λ) ∈ [(domL\KerL) ∩ ∂Ω] × (0, 1);

(2) Nx ∉ ImL for every x ∈ KerL ∩ ∂Ω.

Take

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M74','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M74">View MathML</a>

According to Lemma 3.6 (or Remark 3.1), we know that H(x, λ) ≠ 0 for x ∈ KerL ∩ ∂Ω. Therefore,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M75">View MathML</a>

So that, the condition (3) of Lemma 2.1 is satisfied. By Lemma 2.1, we can get that Lx = Nx has at least one solution in <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M10">View MathML</a>. Therefore, BVP (1.1) has at least one solution. The proof is complete.

4 An example

Example 4.1. Consider the following BVP

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M76','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M76">View MathML</a>

(4.1)

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M77">View MathML</a>

Choose <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M78','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M78">View MathML</a>, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M79','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M79">View MathML</a>, r(t) = 0, s(t) = 0, B = 10. We can get that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M80','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M80">View MathML</a>, r1 = 0, s1 = 0 and

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M81','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2011/1/20/mathml/M81">View MathML</a>

Then, all conditions of Theorem 3.1 hold, so BVP (4.1) has at least one solution.

Competing interests

The authors declare that they have no competing interests.

Authors' contributions

All authors typed, read and approved the final manuscript.

Acknowledgements

The authors would like to thank the referees very much for their helpful comments and suggestions. This research was supported by the Fundamental Research Funds for the Central Universities (2010LKSX09) and the Science Foundation of China University of Mining and Technology (2008A037).

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