Open Access Research

Existence and uniqueness of nonlinear deflections of an infinite beam resting on a non-uniform nonlinear elastic foundation

Sung Woo Choi1 and Taek Soo Jang2*

Author Affiliations

1 Department of Mathematics, Duksung Womens's University, Seoul 132-714, Republic of Korea

2 Department of Naval Architecture and Ocean Engineering, Pusan National University, Busan 609-735, Republic of Korea

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Boundary Value Problems 2012, 2012:5  doi:10.1186/1687-2770-2012-5


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2012/1/5


Received:29 June 2011
Accepted:17 January 2012
Published:17 January 2012

© 2012 Choi and Jang; 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 consider the static deflection of an infinite beam resting on a nonlinear and non-uniform elastic foundation. The governing equation is a fourth-order nonlinear ordinary differential equation. Using the Green's function for the well-analyzed linear version of the equation, we formulate a new integral equation which is equivalent to the original nonlinear equation. We find a function space on which the corresponding nonlinear integral operator is a contraction, and prove the existence and the uniqueness of the deflection in this function space by using Banach fixed point theorem.

2010 Mathematics Subject Classification: 34A12; 34A34; 45G10; 74K10.

Keywords:
Infinite beam; elastic foundation; nonlinear; non-uniform; fourth-order ordinary differential equation; Banach fixed point theorem; contraction.

1 Introduction

The topic of the problem of finite or infinite beams which rest on an elastic foundation has received increased attention in a wide range of fields of engineering, because of its practical design applications, say, to highways and railways. The analysis of the problem is thus of interest to many mechanical, civil engineers and, so on: a number of researchers have made their contributions to the problem. For example, from a very early time, the problem of a linear elastic beam resting on a linear elastic foundation and subjected to lateral forces, was investigated by many techniques [1-8].

In contrast to the problem of beams on linear foundation, Beaufait and Hoadley [9] analyzed elastic beams on "nonlinear" foundations. They organized the midpoint difference method for solving the basic differential equation for the elastic deformation of a beam supported on an elastic, nonlinear foundation. Kuo et al. [10] obtained an asymptotic solution depending on a small parameter by applying the perturbation technique to elastic beams on nonlinear foundations.

Recently, Galewski [11] used a variational approach to investigate the nonlinear elastic simply supported beam equation, and Grossinho et al. [12] studied the solvability of an elastic beam equation in presence of a sign-type Nagumo control. With regard to the beam equation, Alves et al. [13] discussed about iterative solutions for a nonlinear fourth-order ordinary differential equation. Jang et al. [14] proposed a new method for the nonlinear deflection analysis of an infinite beam resting on a nonlinear elastic foundation under localized external loads. Although their method appears powerful as a mathematical procedure for beam deflections on nonlinear elastic foundation, in practice, it has a limited applicability: it cannot be applied to a "non-uniform" elastic foundation. Also, their analysis is limited to compact intervals.

Motivated by these limitations, we herein extend the previous study [14] to propose an original method for determining the infinite beam deflection on nonlinear elastic foundation which is no longer uniform in space. In fact, although there are a large number of studies of beams on nonlinear elastic foundation [10,15], most of them are concerned with the uniform foundation; that is, little is known about the non-uniform foundation analysis. This is because the solution procedure for a nonlinear fourth-order ordinary differential equation has not been fully developed. The method proposed in this article does not depend on a small parameter and therefore can overcome the disadvantages and limitations of perturbation expansions with respect to the small parameter. In this article, we derive a new, nonlinear integral equation for the deflection, which is equivalent to the original nonlinear and non-uniform differential equation, and suggest an iterative procedure for its solution: a similar iterative technique was previously proposed to obtain the nonlinear Stokes waves [14,16-19]. Our basic tool is Banach fixed point theorem [20], which has many applications in diverse areas. One difficulty here is that the integral operator concerning the iterative procedure is not a contraction in general for the case of infinite beam. We overcome this by finding out a suitable subspace inside the whole function space, wherein our integral operator becomes a contraction. Inside this subspace, we then prove the existence and the uniqueness of the deflection of an infinite beam resting on a both non-uniform and nonlinear elastic foundation by means of Banach fixed point theorem. In fact, this restriction on the candidate space for solutions is justified by physical considerations.

The rest of the article is organized as follows: in Section 2, we describe our problem in detail, and formulate an integral equation equivalent to the nonlinear and non-uniform beam equation. The properties of the nonlinear, non-uniform elastic foundation are analyzed in Section 3, and a close investigation on the basic integral operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1">View MathML</a>, which has an important role in both linear and nonlinear beam equations, is performed in Section 4. In Section 5, we define the subspace on which our integral operator Ψ becomes a contraction, and show the existence and the uniqueness of the solution in this space. Finally, Section 6 recapitulates the overall procedure of the article, and explains some of the intuitions behind our formulation for the reader.

2 Definition of the problem

We deal with the question of existence and uniqueness of solutions of nonlinear deflections for an infinitely long beam resting on a nonlinear elastic foundation which is non-uniform in x. Figure 1 shows that the vertical deflection of the beam u(x) results from the net load distribution p(x):

thumbnailFigure 1. Infinite beam on nonlinear and non-uniform elastic foundation.

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M2','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M2">View MathML</a>

(1)

In (1), the two variable function f(u, x) is the nonlinear spring force upward, which depends not only on the beam deflection u but also on the position x, and w(x) denotes the applied loading downward. For simplicity, the weight of the beam is neglected. In fact, the weight of the beam could be incorporated in our static beam deflection problem by adding m(x)g to the loading w(x), where m(x) is the lengthwise mass density of the beam in x-coordinate, and g is the gravitational acceleration. The term m(x)g also plays an important role in the dynamic beam problem, since the second-order time derivative term of deflection must be included as d/dt(m(x)du/dt) in the motion equation. Denoting by EI the flexural rigidity of the beam (E and I are Young's modulus and the mass moment of inertia, respectively), the vertical deflection u(x), according to the classical Euler beam theory, is governed by a fourth-order ordinary differential equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M3','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M3">View MathML</a>

which, in turn, becomes the following nonlinear differential equation for the deflection u by (1):

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M4','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M4">View MathML</a>

(2)

The boundary condition that we consider is

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M5','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M5">View MathML</a>

(3)

Note that (2) and (3) together form a well-defined boundary value problem.

We shall attempt to seek a nonlinear integral equation, which is equivalent to the nonlinear differential equation (2). We start with a simple modification made on (2) by introducing an artificial linear spring constant k: (2) is rewritten as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M6','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M6">View MathML</a>

(4)

where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M7','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M7">View MathML</a>

or

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M8','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M8">View MathML</a>

(5)

The exact determination of k out of the function f(u, x) will be given in Section 3. The modified differential equation (5) is a starting point to the formulation of a nonlinear integral equation equivalent to the original equation (2). For this, we first recall that the linear solution of (2), which corresponds to the case N(u, x) ≡ 0 in (4), was derived by Timoshenko [21], Kenney [8], Saito et al. [22], Fryba [23]. They used the Fourier and Laplace transforms to obtain a closed-form solution:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M9','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M9">View MathML</a>

(6)

expressed in terms of the following Green's function G:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M10','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M10">View MathML</a>

(7)

where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M11','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M11">View MathML</a>. A localized loading condition was assumed in the derivation of (6): u, u', u", and u'" all tend toward zero as |x| → ∞. Green's functions such as (7) play a crucial role in the solution of linear differential equations, and are a key component to the development of integral equation methods. We utilize the Green's function (7) and the solution (6) as a framework for setting up the following nonlinear relations for the case of N(u, x) ≠ 0:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M12','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M12">View MathML</a>

(8)

With the substitution of (5), (8) immediately reveals the following nonlinear Fredholm integral equation for u:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M13','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M13">View MathML</a>

(9)

Physically, the term <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M14">View MathML</a> in (9) amounts to the linear deflection of an infinite beam on a linear elastic foundation having the artificial linear spring constant k, which is uniform in x. The term <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M15','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M15">View MathML</a> in (9) corresponds to the difference between the exact nonlinear solution u and the linear deflection <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M14','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M14">View MathML</a>. We define the nonlinear operator Ψ by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M16','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M16">View MathML</a>

(10)

for functions u : ℝ → ℝ. Then the integral equation (9) becomes just Ψ[u] = u, which is the equation for fixed points of the operator Ψ. We will show in exact sense the equivalence between (2) and (9) in Lemma 7 in Section 5.

3 Assumptions on f and the operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M17">View MathML</a>

Denote ∥u= supx∈ℝ |u(x)| for u : ℝ → ℝ, and let L(ℝ) be the space of all functions u : ℝ → ℝ such that ∥u< ∞. Let C0(ℝ) be the space of all continuous functions vanishing at infinity. It is well known [24] that C0(ℝ) and L(ℝ) are Banach spaces with the norm ∥·∥, and C0(ℝ) ⊂ L(ℝ). For q = 0, 1, 2, ..., let Cq(ℝ) be the space of q times differentiable functions from ℝ to ℝ. Here, C0(ℝ) is just the space of continuous functions C(ℝ).

We have a few assumptions on f(u, x) and w(x). There are four assumptions F1, F2, F3, F4 on f, and two W1, W2 on w. As one can find out soon, they are general enough, and have natural physical meanings. In this section, we list the assumptions on f. Those on w will appear in Section 5.1.

(F1) f(u, x) is sufficiently differentiable, so that f(u(x), x) ∈ Cq(ℝ) if u Cq(ℝ) for q = 0, 1, 2, ....

(F2) f(u, x) · u ≥ 0, and fu(u, x) ≥ 0 for every u, x ∈ ℝ.

(F3) For every <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M18','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M18">View MathML</a> for q = 0, 1, 2.

(F4) infx∈ℝ fu(0, x) > η0 supx∈ℝ fu(0, x), where

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M19','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M19">View MathML</a>

Note first that F1 will free us of any unnecessary consideration for differentiability, and in fact, f(u, x) is usually infinitely differentiable in most applications. F2 means that the elastic force of the elastic foundation, represented by f(u, x), is restoring, and increases in magnitude as does the amount of the deflection u. F3 also makes sense physically: The case q = 0 implies that, within the same amount of deflection u < |υ|, the restoring force f(u, x), though non-uniform, cannot become arbitrarily large. Note that fu(u, x) ≥ 0 is the linear approximation of the spring constant (infinitesimal with respect to x) of the elastic foundation at (u, x). Hence, the case q = 1 means that this non-uniform spring constant fu(u, x) be bounded within a finite deflection |u| < υ. Although the case q = 2 of F3 does not have obvious physical interpretation, we can check later that it is in fact satisfied in usual situations.

Especially, F3 enables us to define the constant k:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M20','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M20">View MathML</a>

(11)

We justifiably rule out the case k = 0; hence, we assuming k > 0 for the rest of the article. Define

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M21','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M21">View MathML</a>

(12)

which is the nonlinear and non-uniform part of the restoring force f(u, x) = ku + N(u, x). Finally, F4 implies that, for any x ∈ ℝ, the spring constant fu(0, x) at (0, x) cannot become smaller than about 12.3% of the maximum spring constant k = supx∈ℝ fu(0, x). This restriction, which is realistic, comes from the unfortunate fact that the operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1">View MathML</a> in Section 4 is not a contraction. The constant η0 is related to another constant τ, which will be introduced later in (41) in Section 4, by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M22','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M22">View MathML</a>

(13)

We define a parameter η which measures the non-uniformity of the elastic foundation:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M23','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M23">View MathML</a>

(14)

Then, by F4, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M24','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M24">View MathML</a>

(15)

A uniform elastic foundation corresponds to the extreme case η = 1, and the non-uniformity increases as η becomes smaller. In order for our current method to work, the condition F4 limits the non-uniformity η by η0 ≈ 0.123.

Using the function N, we define the operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M17">View MathML</a> by <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M25','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M25">View MathML</a> for functions u : ℝ → ℝ. Note that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M17">View MathML</a> is nonlinear in general.

Lemma 1. (a) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M26">View MathML</a> for every u C0(ℝ).

(b) For every u, v L(ℝ), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M27','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M27">View MathML</a>

for some strictly increasing continuous function ρ : [0, ∞) → [0, ∞), such that ρ(0) = 0.

Proof. Suppose u C0(ℝ). <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M28','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M28">View MathML</a> is continuous by F1. Let ϵ > 0. Then there exists M > 0 such that |u(x)| < ϵ if |x| > M, since limx→±∞ u(x) = 0. By the mean value theorem, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M29','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M29">View MathML</a>

for some μ between 0 and u(x), and hence |μ| ≤ |u(x)| < ϵ if |x| > M. Hence, for |x| > M, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M30','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M30">View MathML</a>

(16)

Note that (16) can be made arbitrarily small as M gets larger, since supx∈ℝ, |μ|≤ϵ fu(μ, x) < ∞ by F3. Thus, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M26">View MathML</a>, which proves (a).

By the mean value theorem, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M31','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M31">View MathML</a>

for some μ between u and v, and hence |μ| ≤ max{|u|, |v|}. Hence,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M32','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M32">View MathML</a>

Now suppose u, v L(ℝ). Then

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M33','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M33">View MathML</a>

(17)

Put

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M34','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M34">View MathML</a>

(18)

Note that (18) is well-defined by F3, since we have Nu(μ, x) = fu(μ, x) - k from (12). Clearly, ρ1 is non-negative and non-decreasing.

We want to show ρ1 is continuous. Fix t0 ≥ 0. We first show the left-continuity of ρ1 at t0. Let <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M35">View MathML</a> be a sequence in [0, t0) such that tn t0. Suppose there exists t' < t0 such that ρ1(t') = ρ1(t0). Then, since ρ1 is non-decreasing, it becomes constant on [t', t0], and hence ρ1 is clearly left-continuous at t0. So we assume that ρ1(t') < ρ1(t0) for every t' < t0. It follows that there exists a sequence <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M36">View MathML</a> in [-t0, t0] × ℝ, such that |μn| = tn and |Nu(μn, xn)| → ρ1(t0) as n → ∞, since |Nu(u, x)| is continuous. Thus, we have ρ1(tn) → ρ1(t0) as n → ∞, since |Nu(μn, xn)| ≤ ρ1(tn) ≤ ρ1(t0) for n = 1, 2, .... This shows that ρ1 is left-continuous at t0.

Suppose ρ1 is not right-continuous at t0. Then there exist ϵ > 0 and a sequence <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M35','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M35">View MathML</a> in (t0, ∞), such that tn t0 and ρ1(tn) - ρ1(t0) ≥ ϵ for n = 1, 2, .... Suppose there exists t' > t0 such that ρ1(t') = ρ1(t0). Then ρ1 becomes constant on [t0, t'], so that ρ1 is right-continuous at t0. So we assume that ρ1(t') > ρ1(t0) for every t' > t0. It follows that there exists a sequence <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M36','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M36">View MathML</a> in {(t0, ∞) ∪ (-∞, -t0)} × ℝ, such that t0 < |μn| ≤ tn and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M37','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M37">View MathML</a> for n = 1, 2, ..., since |Nu(u, x)| is continuous. With no loss of generality, we can assume μn > 0. By the mean value theorem, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M38','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M38">View MathML</a>

for some μ between t0 and μn, and so we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M39','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M39">View MathML</a>

(19)

for n = 1, 2, .... By F3, (19) goes to 0 as n → ∞, since

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M40','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M40">View MathML</a>

This is a contradiction. It follows that ρ1 is right-continuous, and thus, is continuous.

By (11) and (14), we have ηk fu(0, x) ≤ k, and so -(1 - η) k fu(0, x) - k ≤ 0 for every x ∈ ℝ. It follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M41','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M41">View MathML</a>

Put ρ2(t): = ρ1(t) - ρ1(0). Then ρ2 is a nondecreasing continuous function such that ρ2(0) = 0. By Lemma 2 below, there exists a strictly increasing continuous function ρ such that ρ(0) = 0, and ρ(t) ≥ ρ2(t) for t ≥ 0. Thus, we have a desired function ρ, since

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M42','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M42">View MathML</a>

where the first inequality is from (17) and (18). This proves (b), and the proof is complete.

Lemma 2. Let g : [0, ∞) → [0, ∞) be a non-decreasing continuous function such that g(0) = 0. Then there exists a strictly increasing continuous function <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M43">View MathML</a> such that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M44">View MathML</a>, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M45">View MathML</a> for t ≥ 0.

Proof. Note that, for every s ∈ [0, ∞), g-1(s)is a compact connected subset of [0, ∞), since g is continuous and non-decreasing. It follows that g-1(s) is either a point or a closed interval in [0, ∞) for every s ∈ [0, ∞). Let A be the set of all points in [0, ∞) at which g is locally constant, i.e.,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M46','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M46">View MathML</a>

Define <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M43','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M43">View MathML</a> by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M47','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M47">View MathML</a>

where l(B) is the Lebesque measure, and hence the length in our case, of the set B ⊂ [0, ∞). From the definition of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M48">View MathML</a>, it is clear that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M44','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M44">View MathML</a>, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M45','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M45">View MathML</a> for t ≥ 0. We omit the proof that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M48','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M48">View MathML</a> is continuous and strictly increasing, which is an easy exercise.

Example 1. Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M49','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M49">View MathML</a>

Then,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M50','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M50">View MathML</a>

and hence,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M51','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M51">View MathML</a>

We also have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M52','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M52">View MathML</a>

Thus, we can take ρ(t) = ρ2(t) = 2(2n + 1)λt2n.

Example 2. Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M53','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M53">View MathML</a>

Then,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M54','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M54">View MathML</a>

and hence,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M55','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M55">View MathML</a>

We also have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M56','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M56">View MathML</a>

Thus, we can take ρ(t) = ρ2(t) = 2{exp(at) - 1}.

Example 3. As an extreme case, we take f(u, x) = ku, for which the original differential equation (2) becomes linear. Clearly, η = 1. Since N(u, x) = Nu(u, x) ≡ 0, we have ρ2(t) ≡ 0. The function ρ taken according to Lemma 2 would be ρ(t) = t. However, a better choice is

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M57','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M57">View MathML</a>

(20)

as we will check in Section 5.1, where the constant σ is defined as well.

4 The Operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1">View MathML</a>

Let

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M58','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M58">View MathML</a>

so that G(x, ξ) = K (|ξ - x|) for G in (7). Using the function K, we define the linear operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1">View MathML</a> by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M59','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M59">View MathML</a>

for functions u : ℝ → ℝ. With this notation, we can rewrite the solution u in (6) of the following linear differential equation:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M60','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M60">View MathML</a>

(21)

which is just the linear case of (2), as <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M61','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M61">View MathML</a>. In fact, understanding the properties of the operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1">View MathML</a> is important not only for the linear case (21), but also for the general nonlinear non-uniform case (2).

Lemma 3.

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M62','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M62">View MathML</a>

Proof. We use induction on i. Note first that the case i = 0 is trivially true. Suppose that the statement is true for some i ≥ 0. Using the following trigonometric equality

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M63','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M63">View MathML</a>

we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M64','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M64">View MathML</a>

which shows that the statement is true for i + 1. Thus, we have the proof.

Using Lemma 3, we can obtain more detailed information on the derivatives of <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M65','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M65">View MathML</a>. Note that, for every u L(ℝ),

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M66','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M66">View MathML</a>

(22)

Lemma 4. (a) Let u C(ℝ) ∩ L(ℝ). Then we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M67','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M67">View MathML</a>

Consequently, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M68','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M68">View MathML</a>for every u C(ℝ) ∩ L(ℝ).

(b) Let q = 0, 1, 2, .... Suppose u Cq(ℝ) and u(i) L(ℝ) for i = 0, 1, ..., q. Then we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M69','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M69">View MathML</a>.

Proof. Let u C(ℝ) ∩ L(ℝ). Then there exists a function U Cl(ℝ) such that U' = u. Since u L(ℝ), U has at most linear growth, and hence by Lemma 3,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M70','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M70">View MathML</a>

(23)

for i = 0, 1, 2, .... Using integration by parts, (22) becomes

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M71','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M71">View MathML</a>

by (23), and hence we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M72','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M72">View MathML</a>

(24)

By (23) and integration by parts again, (24) becomes

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M73','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M73">View MathML</a>

since K'(0) = 0 by Lemma 3. Hence,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M74','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M74">View MathML</a>

(25)

Again by (23) and integration by parts, (25) becomes

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M75','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M75">View MathML</a>

and hence,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M76','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M76">View MathML</a>

(26)

Once more by (23) and integration by parts, (26) becomes

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M77','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M77">View MathML</a>

and hence, by (22),

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M78','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M78">View MathML</a>

(27)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M79','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M79">View MathML</a>

(28)

since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M80','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M80">View MathML</a> and K(4)(y) = -α4K(y) by Lemma 3. Thus (a) follows from (24), (25), (26), (27), and (28).

From (22), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M81','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M81">View MathML</a>

(29)

for every u Cl(ℝ) with u, u' L(ℝ). Suppose now u Cq(ℝ) and u(i) L(ℝ) for i = 0, 1, ..., q. Then, by successively applying (29), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M82','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M82">View MathML</a>

and hence, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M83','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M83">View MathML</a> by applying (22) to u(q). This proves (b), and the proof is complete.

Lemma 5. For every u C0(ℝ), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M84','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M84">View MathML</a> for i = 0, 1, 2, 3, 4.

Proof. Suppose u C0(ℝ). Since C0(ℝ) ⊂ C(ℝ) ∩ L(ℝ), we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M85','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M85">View MathML</a> for i = 0, 1, 2, 3, 4 by Lemma 4 (a). So it is sufficient to show that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M86','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M86">View MathML</a> for i = 0, 1, 2, 3, 4. We first consider the case i = 0, 1, 2, 3. Let ϵ > 0 be arbitrary. Since u C0(ℝ), there exists M > 0 such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M87','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M87">View MathML</a>

(30)

for every |x| ≥ M/2. Moreover, we can assume M is large enough so that it also satisfies

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M88','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M88">View MathML</a>

(31)

Suppose x > M. By (22) and Lemma 4 (a), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M89','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M89">View MathML</a>

(32)

for i = 0, 1, 2, 3. Consider the second term in (32). If y ≥ 0, then x + y M > M/2, and so <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M90','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M90">View MathML</a> by (30). Hence,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M91','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M91">View MathML</a>

(33)

since

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M92','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M92">View MathML</a>

(34)

by Lemma 3, and hence

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M93','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M93">View MathML</a>

(35)

Note that the first term in (32) is

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M94','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M94">View MathML</a>

(36)

If 0 ≤ y x - M/2, then x - y M/2, and hence <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M95','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M95">View MathML</a> by (30). If y x + M/2, then x - y ≤ - M/2, and we also have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M95','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M95">View MathML</a> by (30). Thus, by (35), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M96','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M96">View MathML</a>

(37)

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M97','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M97">View MathML</a>

(38)

By (31) and (34), the remaining term in (36) becomes

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M98','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M98">View MathML</a>

(39)

since x > M. Combining (32), (33), (36), (37), (38), and (39), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M99','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M99">View MathML</a>

for every x > M. This implies <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M100','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M100">View MathML</a> for i = 0, 1, 2, 3. We omit the similar proof that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M100','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M100">View MathML</a>. Thus we conclude that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M101','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M101">View MathML</a> for i = 0, 1, 2, 3. It follows that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M102','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M102">View MathML</a>, since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M103','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M103">View MathML</a> by Lemma 4 (a).

In what follows, we put τ to be the following constant:

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M104','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M104">View MathML</a>

(40)

The exact value of τ can be determined by elementary calculation, which we omit. It turns out that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M105','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M105">View MathML</a>

(41)

Lemma 6. (a) <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M106','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M106">View MathML</a> for every u L(ℝ). Thus, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M107','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M107">View MathML</a>for every u L(ℝ).

(b) For every u C(ℝ) ∩ L(ℝ), we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M108','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M108">View MathML</a> for i = 1, 2, 3, and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M109','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M109">View MathML</a>.

Proof. By (22) and (40), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M110','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M110">View MathML</a>

for every u L(ℝ). This shows (a).

Suppose u C(ℝ) ∩ L(ℝ). By Lemma 4 (a),

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M111','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M111">View MathML</a>

(42)

for i = 1, 2, 3. By Lemma 3 and with the substitution <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M112','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M112">View MathML</a>, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M113','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M113">View MathML</a>

which, together with (40) and (42), gives

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M114','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M114">View MathML</a>

for i = 1, 2, 3. This proves (b) for i = 1, 2, 3.

Finally, by Lemma 4 (a) and the above result (a),

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M115','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M115">View MathML</a>

and the proof is therefore complete.

5 Main result

Using the operators <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M17','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M17">View MathML</a> and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1">View MathML</a> in Sections 3 and 4, the nonlinear integral operator Ψ defined in (10) can be expressed in abstract notation as

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M116','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M116">View MathML</a>

for u : ℝ → ℝ. We will show that Ψ is a contraction when it is restricted to an appropriate function space X C0(ℝ) which will be defined later in this section.

5.1 Assumptions on w and the space X

Here, we introduce two assumptions W1 and W2 on the function w in (2):

(W1) w C0(ℝ).

(W2) ∥w< sup0≤sσk {ρ-1(s) · (σk - s)}, where σ is defined by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M117','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M117">View MathML</a>

(43)

W1 means that the loading w should be sufficiently localized, which was also assumed for the linear solution (6) of (21). Nonetheless, w need not be compactly supported, and it is sufficient that limx→±∞ w(x) = 0. Note that the constant σ is positive by (13), (14), and (15). The function ρ is taken to satisfy Lemma 1 (b). Since ρ is continuous and strictly increasing, the inverse function ρ-1 : ρ([0, ∞)) → [0, ∞) is well defined, and is also a strictly increasing continuous function with ρ-1(0) = 0. It is easy to see that the range ρ([0, ∞)) of ρ, which is the domain of ρ-1, is always of the form <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M118','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M118">View MathML</a> for some <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M119','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M119">View MathML</a>. In fact, the supremum in W2 should be meant to be taken in the range <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M120','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M120">View MathML</a>. Note that the set <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M121','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M121">View MathML</a> should be connected, and hence of the form <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M122','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M122">View MathML</a> or <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M123','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M123">View MathML</a> for some <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M124','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M124">View MathML</a>, since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M125','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M125">View MathML</a> is connected and ρ-1(s) · (σk - s) is continuous. In fact, we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M126','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M126">View MathML</a>. It follows from W2 that there exists <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M127','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M127">View MathML</a> such that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M128','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M128">View MathML</a>

(44)

We remark that the trivial case ∥w∥= 0 is safely excluded in this article. The physical meaning of W2 is that the size ∥w∥of the loading w cannot be arbitrarily large, and its upper limit is closely related to the nonlinearity and the non-uniformity of the given elastic foundation.

Now define the subset X of C0(ℝ) by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M129','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M129">View MathML</a>

(45)

We view X as a metric space with the metric ∥· - ·∥. Note that X is a complete metric space, since it is a closed set in C0(ℝ) which itself is a complete metric space. Note that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M130','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M130">View MathML</a>

since 0 < s* < σk. It follows that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M131','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M131">View MathML</a>

(46)

In our system described by the differential equation (2), it is physically clear that the size ∥uof the output deflection u cannot be too large compared to the size ∥wof the input loading w. In fact, Lemma 6 (a) describes this relationship quantitatively in the linear case (21). Thus, (46) implies that the space X, though it is not the whole of C0(ℝ), is big enough in some sense.

Example 4. Consider the case

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M132','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M132">View MathML</a>

in Example 1. Then we have ρ(t) = 2(2n + 1)λt2n, and hence <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M133','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M133">View MathML</a>. Put ϕ(s) = ρ-1(s) · (σk - s). Since

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M134','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M134">View MathML</a>

ϕ is strictly increasing on <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M135','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M135">View MathML</a>, and strictly decreasing on <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M136','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M136">View MathML</a>. Note also that ϕ(0) = ϕ(σk) = 0. Thus,

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M137','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M137">View MathML</a>

There are exactly two solutions in (0, σk) of the equation ρ-1(s) · (σk - s) = ∥w, or equivalently, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M138','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M138">View MathML</a>. Note that we have bigger X, if we take s* to be the larger among them.

Example 5. Consider the case

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M139','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M139">View MathML</a>

in Example 2. Then we have ρ(t) = 2{exp(at) - 1}, and hence <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M140','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M140">View MathML</a>. Putting ϕ(s) = ρ-1(s) · (σk - s), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M141','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M141">View MathML</a>

It follows that ϕ is strictly increasing on <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M142','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M142">View MathML</a>, and strictly decreasing on <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M143','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M143">View MathML</a>, and hence, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M144','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M144">View MathML</a>, where <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M145','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M145">View MathML</a> is the unique solution in (0, σk) of the equation

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M146','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M146">View MathML</a>

Again, there are exactly two solutions in (0, σk) of the equation ρ-1(s) · (σk - s) = ||w||. Among them, we take s* to be preferably the larger.

Example 6. In Example 3, we took ρ as in (20), rather than ρ(t) = t, for the case f(u, x) = ku. Then we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M147','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M147">View MathML</a>

Let ϕ(s) = ρ-1(s) · (σk - s). We can easily check that ϕ is strictly increasing on [0, σk), ϕ(0) = 0, and limsσk- ϕ(s) = ∞. Thus, we have sup0≤sσk {ρ-1(s) · (σk - s)} = ∞. This implies that we have no restriction on the upper bound of ∥w, which indeed is expected with the linear equation (21). Note, however, this observation could not have been possible to be made, if we took ρ(t) = t. The equation ϕ(s) = ∥w, which is equivalent to s2 - σk(2 + σkw)s + σ3k3w= 0, has the unique solution

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M148','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M148">View MathML</a>

in (0, σk).

5.2 Contractiveness of the operator Ψ

Suppose u C0(ℝ). Then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M26','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M26">View MathML</a> by Lemma 1 (a), and again, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M149','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M149">View MathML</a> by Lemma 5. We also have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M150','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M150">View MathML</a> by W1 and Lemma 5. Thus, we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M151','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M151">View MathML</a> for every u C0(ℝ). In short, the operator Ψ is a well-defined map from C0(ℝ) into C0(ℝ). The next lemma confirms that the solutions of (2) are the fixed points of Ψ in C0(ℝ).

Lemma 7. Suppose u C4(ℝ) ∩ C0(ℝ) and u(i) L(ℝ) for i = 1, 2, 3, 4. Then u is a solution of the differential equation (2), if and only if Ψ[u] = u.

Proof. Suppose u satisfies Ψ[u] = u. By Lemma 4 (a), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M152','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M152">View MathML</a>

and hence, u is a solution of (2) by (12).

Conversely, suppose u is a solution of (2), so that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M153','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M153">View MathML</a> by (12).

Applying the operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1">View MathML</a>, we get

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M154','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M154">View MathML</a>

and hence

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M155','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M155">View MathML</a>

by Lemma 4 (a), and (b), and the proof is complete.

Unfortunately, Ψ is not a contraction on the whole of C0(ℝ). Nevertheless, if we restrict Ψ to the subset X of C0(ℝ) defined in (45), then we can show that Ψ is a contraction from X into X. This enables us to use the usual argument of the Banach fixed point theorem, and to prove the existence and the uniqueness of the fixed point of Ψ, which is the solution of the differential equation (2), at least in X.

Lemma 8. Ψ[u] ∈ X for every u X. Moreover, Ψ: X X is a contraction, i.e., ∥Ψ[u] - Ψ[v]∥L · ∥u - vfor every u, v X for some constant L < 1.

Proof. Suppose u X. Note that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M156','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M156">View MathML</a> by F2, if we denote the zero function by 0(x) ≡ 0. Hence, by Lemma 1 (b) and Lemma 6 (a), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M157','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M157">View MathML</a>

where ρ is taken as in Lemma 1 (b). Hence, by (44) and (43), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M158','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M158">View MathML</a>

which shows Ψ[u] ∈ X.

Now suppose u, v X. Again by Lemma 6 (a) and Lemma 1 (b), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M159','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M159">View MathML</a>

Since 0 < s* < σk, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M160','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M160">View MathML</a>

by (43). Thus, we have the desired inequality by taking <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M161','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M161">View MathML</a>, and the proof is complete.

Proposition 1. (Banach Fixed Point Theorem [20]) Let Y be a complete metric space with the metric d(·, ·), and suppose the map Φ: Y Y satisfies d(Φ(y1), Φ(y2)) ≤ L · d(y1, y2) for every y1, y2 Y for some constant L < 1. Then Φ has a unique fixed point in Y. Moreover, for any y0 Y, the sequence <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M162','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M162">View MathML</a>, defined by yn = Φ(yn-1), n = 1, 2, ..., converges to this unique fixed point.

Lemma 9. Ψ has a unique fixed point in X. Moreover, this fixed point, denoted by u*, is in C4(ℝ), and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M163','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M163">View MathML</a> for i = 1, 2, 3, 4.

Proof. The fact that Ψ has a unique fixed point in X is immediate from Proposition 1 and Lemma 8, since X is a complete metric space with the metric ∥· - ·∥. Let u* be this unique fixed point.

Take any u0 in X, and define un = Ψ[un-1], n = 1, 2, .... By Proposition 1, the sequence of functions <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M164','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M164">View MathML</a> in X converges uniformly to the fixed point u* X. We assume u0 C4(ℝ) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M165','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M165">View MathML</a> for i = 1, 2, 3, 4, which can always be achieved: For example, we could take u0 to be the zero function. Suppose, for some n, un-1 C4(ℝ) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M166','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M166">View MathML</a> for i = 1, 2, 3, 4. Then <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M167','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M167">View MathML</a> by Lemma 1 (a), since un-1 C0(ℝ). Hence, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M168','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M168">View MathML</a> by Lemma 4 (a), and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M169','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M169">View MathML</a>, i = 1, 2, 3, 4 by Lemma 5. Since w C0(ℝ) by W1, we also have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M170','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M170">View MathML</a> by Lemma 4 (a), and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M171','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M171">View MathML</a> for i = 1, 2, 3, 4 by Lemma 5. Hence, we have un C4(ℝ) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M172','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M172">View MathML</a> for i = 1, 2, 3, 4, since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M173','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M173">View MathML</a>. Thus, by induction on n, we have un C4(ℝ) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M172','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M172">View MathML</a> for i = 1, 2, 3, 4, for every n = 0, 1, 2, ....

By Lemma 6 (b), we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M174','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M174">View MathML</a>, i = 1, 2, 3, 4, for every u C0(ℝ) ⊂ C(ℝ) ∩ L(ℝ), where we put

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M175','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M175">View MathML</a>

Hence, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M176','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M176">View MathML</a>

(47)

for every n = 0, 1, 2, ..., since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M177','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M177">View MathML</a>. Since un-1, un X L(ℝ), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M178','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M178">View MathML</a>

(48)

by Lemma 1 (b) and (44), (47). Since

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M179','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M179">View MathML</a>

by Lemma 8, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M180','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M180">View MathML</a>

(49)

Combining (48) and (49), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M181','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M181">View MathML</a>

(50)

Let ϵ > 0. Since 0 ≤ L < 1, we can take N large enough so that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M182','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M182">View MathML</a>

(51)

Let m, n > N. Assume m > n with no loss of generality. Then by (50) and (51), we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M183','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M183">View MathML</a>

This implies that, for every i = 1, 2, 3, 4, the sequence <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M184','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M184">View MathML</a> is Cauchy in C0(ℝ) with respect to the metric ∥· - ·∥, and hence, converges uniformly to a function vi C0(R). So by Lemma 10 below, u* C1(ℝ) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M185','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M185">View MathML</a>, since un converges uniformly to u* and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M186','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M186">View MathML</a> converges uniformly to v1. Applying Lemma 10 again to <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M186','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M186">View MathML</a>, we see that v1 C1(ℝ) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M187','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M187">View MathML</a>. By repeating the same argument, we see that v2 C1(ℝ), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M188','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M188">View MathML</a>, and v3 C1(ℝ), <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M189','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M189">View MathML</a>. Thus, we have u* C4(ℝ) and <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M190','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M190">View MathML</a> for i = 1, 2, 3, 4. Hence, the proof is complete.

Lemma 10. Suppose a sequence of functions <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M191','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M191">View MathML</a> in C1(ℝ) converges uniformly to a function g. Suppose also <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M192','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M192">View MathML</a> converges uniformly to a function h. Then g C1(ℝ) and g' = h.

Proof. Fix x0 ∈ ℝ, and define hn : ℝ → ℝ, n = 1, 2, ... by

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M193','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M193">View MathML</a>

which is continuous since gn C1(ℝ). Note that <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M194','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M194">View MathML</a> as n → ∞. For x x0, we have

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M195','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M195">View MathML</a>

for some ξ between x0 and x by the mean value theorem for gm - gn. Thus, we have <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M196','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M196">View MathML</a> for any m, n. It follows that hn converges uniformly to a continuous function, since <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M192','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M192">View MathML</a> converges uniformly. Note that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M197','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M197">View MathML</a>

(52)

Since hn converges uniformly, we can change the order of the limit in (52), so that

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M198','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M198">View MathML</a>

Hence, <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M199','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M199">View MathML</a> exists, and is equal to h(x0). Thus, the proof is complete, since x0 is arbitrary.

Now the following main result of the article is immediate from Lemmas 9 and 7.

Theorem Suppose the functions f(u, x) and w(x) satisfy the conditions F1, F2, F3, F4, and W1, W2. Then the differential equation (2) has a unique solution in

<a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M200','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M200">View MathML</a>

where k, σ, s* are as defined in (11), (43), (44) respectively. Moreover, the unique solution, denoted by u*, satisfies <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M201','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M201">View MathML</a>for i = 1, 2, 3, 4.

6 Concluding remarks

It is intuitively clear that the nature of the resulting beam deflection depends on both the nonlinearity and the non-uniformity of the given elastic foundation. In this study, we introduced a physical parameter η in (14) measuring the non-uniformity, and a function ρ in Lemma 1 which mainly measures the nonlinearity. Accordingly, the pair (η, ρ) may be considered as a systematic encoding of the non-uniformity and the nonlinearity of the given foundation. Together with the maximal linear spring constant k in (11) at the equilibrium state u ≡ 0, η and ρ capture the dominating mechanical properties of the present beam problem represented by the differential equation (2).

We transformed the original nonlinear differential equation into an equivalent nonlinear integral equation Ψ[u] = u, thereby positioning our problem into the realm of the fixed point theory. However, the integral operator Ψ is not a contraction in the whole function space C0(ℝ) equipped with the usual sup-norm ∥·∥. The reason for this is twofold: first, the nonlinearity of the elastic foundation, encoded in the function ρ, makes Ψ expansive for functions with large norms, which can be seen from Lemma 1 (b). Second, the value of the constant τ which gives the L-norm of the operator <a onClick="popup('http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1','MathML',630,470);return false;" target="_blank" href="http://www.boundaryvalueproblems.com/content/2012/1/5/mathml/M1">View MathML</a> in Lemma 6 (a), is greater than 1. Because of this, too much non-uniformity of the elastic foundation, encoded in the parameter η, can also contribute to the non-contractiveness of Ψ.

Thus to resort to the Banach fixed point theorem, it is necessary to find a subspace smaller than C0(ℝ), where the operator Ψ is contractive. This "shrinking the space" idea also conforms with the physical intuition that the norm of the resulting beam deflection cannot be too large compared to that of the input loading w. Meanwhile, the nonlinearity and the non-uniformity of the system suggest that the norm of the loading w itself should also be bounded. All these heuristic ideas were materialized into the actual construction of the upper-bound in W2 and the subspace X, which, besides the input loading w, depend only on the three main attributes k, η, and ρ of the given mechanical system.

The subspace we are looking for should satisfy two conditions other than completeness: first, it should be invariant under the operator Ψ. Second, the restriction of Ψ to it should be contractive. Once we proved in Lemma 8 that the function space X actually satisfies these conditions, the existence and the uniqueness of the solution in Lemma 9 follows immediately from the Banach fixed point theorem. Note carefully that Lemma 7 establishes the equivalence between the original differential equation (2) and our integral equation (9), only for solutions satisfying the regularity condition to be in C4(ℝ). In this respect, what Lemma 9 is really up to are the regularity of the unique solution thus found, and its behavior at infinity. Consequently, the main theorem in Section 5.2 states that the unique solution u* has enough regularity for the differential equation (2), and satisfies our boundary condition (3), and hence, is the solution of the present nonlinear boundary value problem.

Competing interests

The authors declare that they have no competing interests.

Authors' contributions

TSJ formulated the integral equation (9) which is equivalent to the original nonlinear beam equation (2), and introduced the overall problem to SWC. SWC found the subspace (45) and proved that the integral operator (10) is a contraction on that subspace. All authors read and approved the final manuscript.

Acknowledgements

T.S. Jang was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant no.: 2011-0010090). The same author was also supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant no.: K20903002030-11E0100-04610). Finally, the authors would like to thank the anonymous reviewers for their valuable comments, and appreciate their time and effort to review our manuscript and to make suggestions and constructive criticism, which we believe have improved the quality of the article substantially.

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