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

  • Sung Woo Choi1 and

    Affiliated with

    • Taek Soo Jang2Email author

      Affiliated with

      Boundary Value Problems20122012:5

      DOI: 10.1186/1687-2770-2012-5

      Received: 29 June 2011

      Accepted: 17 January 2012

      Published: 17 January 2012

      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 [18].

      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, 1619]. 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 K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq1_HTML.gif, 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):
      http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Fig1_HTML.jpg
      Figure 1

      Infinite beam on nonlinear and non-uniform elastic foundation.

      p ( x ) = w ( x ) - f ( u , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ1_HTML.gif
      (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
      E I d 4 u d x 4 = p ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equa_HTML.gif
      which, in turn, becomes the following nonlinear differential equation for the deflection u by (1):
      E I d 4 u d x 4 + f ( u , x ) = w ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ2_HTML.gif
      (2)
      The boundary condition that we consider is
      lim x ± u ( x ) = lim x ± u ( x ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ3_HTML.gif
      (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
      E I d 4 u d x 4 + k u + N ( u , x ) = w ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ4_HTML.gif
      (4)
      where
      f ( u , x ) = k u + N ( u , x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equb_HTML.gif
      or
      E I d 4 u d x 4 + k u = w ( x ) - N ( u , x ) Φ ( u , x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ5_HTML.gif
      (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:
      u ( x ) = - G ( x , ξ ) w ( ξ ) d ξ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ6_HTML.gif
      (6)
      expressed in terms of the following Green's function G:
      G ( x , ξ ) = α 2 k exp - α | ξ - x | 2 sin α | ξ - x | 2 + π 4 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ7_HTML.gif
      (7)
      where α = k / E I 4 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq2_HTML.gif. 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:
      u ( x ) = - G ( x , ξ ) Φ ( u ( ξ ) , ξ ) d ξ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ8_HTML.gif
      (8)
      With the substitution of (5), (8) immediately reveals the following nonlinear Fredholm integral equation for u:
      u ( x ) = - G ( x , ξ ) w ( x ) d ξ - - G ( x , ξ ) N ( u ( ξ ) , ξ ) d ξ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ9_HTML.gif
      (9)
      Physically, the term - G ( x , ξ ) w ( x ) d ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq3_HTML.gif 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 - - G ( x , ξ ) N ( u ( ξ ) , ξ ) d ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq4_HTML.gif in (9) corresponds to the difference between the exact nonlinear solution u and the linear deflection - G ( x , ξ ) w ( x ) d ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq3_HTML.gif. We define the nonlinear operator Ψ by
      Ψ [ u ] ( x ) : = - G ( x , ξ ) w ( x ) d ξ - - G ( x , ξ ) N ( u ( ξ ) , ξ ) d ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ10_HTML.gif
      (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 N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq5_HTML.gif

      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 C q (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) ∈ C q (ℝ) if uC q (ℝ) for q = 0, 1, 2, ....

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

      (F3) For every υ 0 , sup x , | u | υ q f u q ( u , x ) < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq6_HTML.gif for q = 0, 1, 2.

      (F4) infx∈ℝf u (0, x) > η0 supx∈ℝf u (0, x), where
      η 0 = 2 exp - 3 π 4 1 - exp ( - π ) + 2 exp - 3 π 4 0 . 123 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equc_HTML.gif

      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 f u (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 f u (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:
      k : = sup x f u ( 0 , x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ11_HTML.gif
      (11)
      We justifiably rule out the case k = 0; hence, we assuming k > 0 for the rest of the article. Define
      N ( u , x ) : = f ( u , x ) - k u , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ12_HTML.gif
      (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 f u (0, x) at (0, x) cannot become smaller than about 12.3% of the maximum spring constant k = supx∈ℝf u (0, x). This restriction, which is realistic, comes from the unfortunate fact that the operator K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq1_HTML.gif 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
      η 0 = τ - 1 τ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ13_HTML.gif
      (13)
      We define a parameter η which measures the non-uniformity of the elastic foundation:
      η : = inf x f u ( 0 , x ) sup x f u ( 0 , x ) = inf x f u ( 0 , x ) k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ14_HTML.gif
      (14)
      Then, by F4, we have
      η 0 < η 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ15_HTML.gif
      (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 N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq5_HTML.gif by N [ u ] ( x ) : = N ( u ( x ) , x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq7_HTML.gif for functions u : ℝ → ℝ. Note that N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq5_HTML.gif is nonlinear in general.

      Lemma 1. (a) N [ u ] C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq8_HTML.giffor every uC0(ℝ).
      1. (b)
        For every u, vL (ℝ), we have
        | | N [ u ] - N [ v ] | | { ( 1 - η ) k + ρ ( max { | | u | | , | | v | | } ) } | | u - v | | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equd_HTML.gif
         

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

      Proof. Suppose uC0(ℝ). N [ u ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq9_HTML.gif 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
      N [ u ] ( x ) = N ( u ( x ) , x ) = f ( u ( x ) , x ) - k u ( x ) = f u ( μ , x ) { u ( x ) - 0 } - k u ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Eque_HTML.gif
      for some μ between 0 and u(x), and hence |μ| ≤ |u(x)| < ϵ if |x| > M. Hence, for |x| > M, we have
      | N [ u ] ( x ) | = | f u ( μ , x ) u ( x ) - k u ( x ) | { f u ( μ , x ) + k } | u ( x ) | k + sup x , | μ | ε f u ( μ , x ) ε . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ16_HTML.gif
      (16)

      Note that (16) can be made arbitrarily small as M gets larger, since supx∈ℝ, |μ|≤ϵf u (μ, x) < ∞ by F3. Thus, N [ u ] C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq8_HTML.gif, which proves (a).

      By the mean value theorem, we have
      N ( u , x ) - N ( v , x ) = N u ( μ , x ) ( u - v ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equf_HTML.gif
      for some μ between u and v, and hence |μ| ≤ max{|u|, |v|}. Hence,
      | N ( u , x ) - N ( v , x ) | sup | μ | max { | u | , | v | } | N u ( μ , x ) | | u - v | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equg_HTML.gif
      Now suppose u, vL(ℝ). Then
      | | N [ u ] - N [ v ] | | = sup x | N ( u ( x ) , x ) - N ( v ( x ) , x ) | sup x sup | μ | max { | u ( x ) | , | v ( x ) | } | N u ( μ , x ) | | u ( x ) - v ( x ) | sup x sup | μ | max { | u ( x ) | , | v ( x ) | } | N u ( μ , x ) | sup x | u ( x ) - v ( x ) | sup x , | μ | max { | | u | | , | | v | } | N u ( μ , x ) | | | u - v | | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ17_HTML.gif
      (17)
      Put
      ρ 1 ( t ) : = sup x , | μ | t | N u ( μ , x ) | , t 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ18_HTML.gif
      (18)

      Note that (18) is well-defined by F3, since we have N u (μ, x) = f u (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 { t n } n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq10_HTML.gif be a sequence in [0, t0) such that t n 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 { ( μ n , x n ) } n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq11_HTML.gif in [-t0, t0] × ℝ, such that |μ n | = t n and |N u (μ n , x n )| → ρ1(t0) as n → ∞, since |N u (u, x)| is continuous. Thus, we have ρ1(t n ) → ρ1(t0) as n → ∞, since |N u (μ n , x n )| ≤ ρ1(t n ) ≤ 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 { t n } n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq10_HTML.gif in (t0, ∞), such that t n t0 and ρ1(t n ) - 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 { ( μ n , x n ) } n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq11_HTML.gif in {(t0, ∞) ∪ (-∞, -t0)} × ℝ, such that t0 < |μ n | ≤ t n and | N u ( μ n , x n ) | > ρ 1 ( t n ) - ε 2 n > ρ 1 ( t 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq12_HTML.gif for n = 1, 2, ..., since |N u (u, x)| is continuous. With no loss of generality, we can assume μ n > 0. By the mean value theorem, we have
      N u ( μ n , x n ) - N u ( t 0 , x n ) = N u u ( μ , x n ) ( μ n - t 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equh_HTML.gif
      for some μ between t0 and μ n , and so we have
      ε ρ 1 ( t n ) - ρ 1 ( t 0 ) = { ρ 1 ( t n ) - | N u ( μ n , x n ) | } + { | N u ( μ n , x n ) | - | N u ( t 0 , x n ) | } + { | N u ( t 0 , x n ) | - ρ 1 ( t 0 ) } { ρ 1 ( t n ) - | N u ( μ n , x n ) | } + { | N u ( μ n , x n ) | - | N u ( t 0 , x n ) | } { ρ 1 ( t n ) - | N u ( μ n , x n ) | } + | N u ( μ n , x n ) - N u ( t 0 , x n ) | < ε 2 n + | N u u ( μ , x n ) | | μ n - t 0 | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ19_HTML.gif
      (19)
      for n = 1, 2, .... By F3, (19) goes to 0 as n → ∞, since
      | N u u ( μ , x n ) | | μ n - t 0 | sup x , | μ | t 1 | N u u ( μ , x ) | | t n - t 0 | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equi_HTML.gif

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

      By (11) and (14), we have ηkf u (0, x) ≤ k, and so -(1 - η) kf u (0, x) - k ≤ 0 for every x ∈ ℝ. It follows that
      ρ 1 ( 0 ) = sup x , | μ | 0 | N u ( μ , x ) | = sup x | N u ( 0 , x ) | = sup x | f u ( 0 , x ) - k | ( 1 - η ) k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equj_HTML.gif
      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
      | | N [ u ] - N [ v ] | | ρ 1 ( max { | | u | | , | | v | | } ) | | u - v | | { ρ 1 ( 0 ) + ρ 2 ( max { | | u | | , | | v | | } ) } | | u - v | | { ( 1 - η ) k + ρ ( max { | | u | | , | | v | | } ) } | | u - v | | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equk_HTML.gif

      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 g ̃ : [ 0 , ) [ 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq13_HTML.gifsuch that g ̃ ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq14_HTML.gif, and g ̃ ( t ) g ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq15_HTML.giffor 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 = { t [ 0 , ) | g - 1 ( g ( t ) ) is an interval with non-zero length } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equl_HTML.gif
      Define g ̃ : [ 0 , ) [ 0 , ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq13_HTML.gif by
      g ̃ ( t ) : = g ( t ) + l ( A [ 0 , t ] ) , t 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equm_HTML.gif

      where l(B) is the Lebesque measure, and hence the length in our case, of the set B ⊂ [0, ∞). From the definition of g ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq16_HTML.gif, it is clear that g ̃ ( 0 ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq14_HTML.gif, and g ̃ ( t ) g ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq15_HTML.gif for t ≥ 0. We omit the proof that g ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq16_HTML.gif is continuous and strictly increasing, which is an easy exercise.

      Example 1. Let
      f ( u , x ) = ( 1 + ε cos x ) k 1 + ε u + λ u 2 n + 1 , 0 ε 1 2 , n 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equn_HTML.gif
      Then,
      f u ( u , x ) = 1 + ε cos x 1 + ε k + λ ( 2 n + 1 ) ( 1 + ε cos x ) u 2 n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equo_HTML.gif
      and hence,
      1 - ε 1 + ε k f u ( 0 , x ) k , η = 1 - ε 1 + ε . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equp_HTML.gif
      We also have
      N ( u , x ) = f ( u , x ) - k u = - k ε 1 + ε ( 1 - cos x ) u + λ ( 1 + ε cos x ) u 2 n + 1 , N u ( u , x ) = - k ε 1 + ε ( 1 - cos x ) + λ ( 2 n + 1 ) ( 1 + ε cos x ) u 2 n , | N u ( u , x ) | 2 ε 1 + ε k + λ ( 2 n + 1 ) | 1 + ε cos x | | u | 2 n ( 1 - η ) k + 2 ( 2 n + 1 ) λ | u | 2 n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equq_HTML.gif

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

      Example 2. Let
      f ( u , x ) = ( 1 + ε cos x ) k 1 + ε u + λ { exp ( a u ) - 1 - a u } , 0 ε 1 2 , a > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equr_HTML.gif
      Then,
      f u ( u , x ) = 1 + ε cos x 1 + ε k + a λ ( 1 + ε cos x ) { exp ( a u ) - 1 } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equs_HTML.gif
      and hence,
      1 - ε 1 + ε k f u ( 0 , x ) k , η = 1 - ε 1 + ε . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equt_HTML.gif
      We also have
      N ( u , x ) = f ( u , x ) - k u = - k ε 1 + ε ( 1 - cos x ) u + λ ( 1 + ε cos x ) { exp ( a u ) - 1 - a u } , N u ( u , x ) = - k ε 1 + ε ( 1 - cos x ) + a λ ( 1 + ε cos x ) { exp ( a u ) - 1 } , | N u ( u , x ) | 2 ε 1 + ε k + a λ ( 1 + ε ) { exp ( a u ) - 1 } ( 1 - η ) k + 2 a λ { exp ( a u ) - 1 } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equu_HTML.gif

      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) = N u (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
      ρ ( t ) = σ k 1 - 1 1 + σ 2 k 2 t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ20_HTML.gif
      (20)

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

      4 The Operator K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq1_HTML.gif

      Let
      K ( y ) : = α 2 k exp - α 2 y sin α 2 y + π 4 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equv_HTML.gif
      so that G(x, ξ) = K (|ξ - x|) for G in (7). Using the function K, we define the linear operator K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq1_HTML.gif by
      K [ u ] ( x ) : = - K ( | x - ξ | ) u ( ξ ) d ξ = - G ( x , ξ ) u ( ξ ) d ξ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equw_HTML.gif
      for functions u : ℝ → ℝ. With this notation, we can rewrite the solution u in (6) of the following linear differential equation:
      E I d 4 u ( x ) d x 4 + k u ( x ) = w ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ21_HTML.gif
      (21)

      which is just the linear case of (2), as u = K [ w ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq17_HTML.gif. In fact, understanding the properties of the operator K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq1_HTML.gif is important not only for the linear case (21), but also for the general nonlinear non-uniform case (2).

      Lemma 3.
      K ( i ) ( y ) = α i + 1 2 k exp - α 2 y sin α 2 y + ( 3 i + 1 ) π 4 , i = 0 , 1 , 2 , . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equx_HTML.gif
      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
      - sin t + cos t = 2 sin t cos 3 π 4 + cos t sin 3 π 4 = 2 sin t + 3 π 4 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equy_HTML.gif
      we have
      K ( i + 1 ) ( y ) = d d y K ( i ) ( y ) = α i + 1 2 k exp - α 2 - α 2 sin α 2 y + ( 3 i + 1 ) π 4 + α i + 1 2 k exp - α 2 y cos α 2 y + ( 3 i + 1 ) π 4 α 2 = α i + 2 2 2 k exp - α 2 y - sin α 2 y + ( 3 i + 1 ) 4 π + cos α 2 y + ( 3 i + 1 ) 4 π = α i + 2 2 2 k exp - α 2 y 2 sin α 2 y + ( 3 i + 1 ) π 4 + 3 π 4 = α ( i + 1 ) + 1 2 k exp - α 2 y sin α 2 y + { 3 ( i + 1 ) + 1 } π 4 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equz_HTML.gif

      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 K [ u ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq18_HTML.gif. Note that, for every uL(ℝ),
      K [ u ] ( x ) = - x K ( x - ξ ) u ( ξ ) d ξ + x K ( ξ - x ) u ( ξ ) d ξ = - 0 K ( y ) u ( x - y ) d y + 0 K ( y ) u ( x + y ) d y = 0 K ( y ) { u ( x - y ) + u ( x + y ) } d y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ22_HTML.gif
      (22)
      Lemma 4. (a) Let uC(ℝ) ∩ L(ℝ). Then we have
      K [ u ] ( i ) ( x ) = 0 K ( i ) ( y ) { u ( x - y ) + ( - 1 ) i u ( x + y ) } d y , i = 1 , 2 , 3 , K [ u ] ( 4 ) ( x ) = 0 K ( 4 ) ( y ) { u ( x - y ) + u ( x + y ) } d y + 2 K ( 3 ) ( 0 ) u ( x ) = - α 4 K [ u ] ( x ) + α 4 k u ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equaa_HTML.gif
      Consequently, K [ u ] C 4 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq19_HTML.giffor every uC(ℝ) ∩ L(ℝ).
      1. (b)

        Let q = 0, 1, 2, .... Suppose uC q (ℝ) and u (i)L (ℝ) for i = 0, 1, ..., q. Then we have K [ u ( q ) ] = K [ u ] ( q ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq20_HTML.gif.

         
      Proof. Let uC(ℝ) ∩ L(ℝ). Then there exists a function UCl(ℝ) such that U' = u. Since uL(ℝ), U has at most linear growth, and hence by Lemma 3,
      lim y K ( i ) ( y ) U ( x - y ) = lim y K ( i ) ( y ) U ( x + y ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ23_HTML.gif
      (23)
      for i = 0, 1, 2, .... Using integration by parts, (22) becomes
      K [ u ] ( x ) = [ K ( y ) { - U ( x - y ) + U ( x + y ) } ] 0 - 0 K ( y ) { - U ( x - y ) + U ( x + y ) } d y = 0 K ( y ) { U ( x - y ) - U ( x + y ) } d y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equab_HTML.gif
      by (23), and hence we have
      K [ u ] ( x ) = d d x 0 K ( y ) { U ( x - y ) - U ( x + y ) } d y = 0 K ( y ) { u ( x - y ) - u ( x + y ) } d y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ24_HTML.gif
      (24)
      By (23) and integration by parts again, (24) becomes
      K [ u ] ( x ) = [ K ( y ) { - U ( x - y ) - U ( x + y ) } ] 0 - 0 K ( y ) { - U ( x - y ) - U ( x + y ) } d y = 2 K ( 0 ) U ( x ) + 0 K ( y ) { U ( x - y ) + U ( x + y ) } d y = 0 K ( y ) { U ( x - y ) + U ( x + y ) } d y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equac_HTML.gif
      since K'(0) = 0 by Lemma 3. Hence,
      K [ u ] ( x ) = d d x 0 K ( y ) { U ( x - y ) + U ( x + y ) } d y = 0 K ( y ) { u ( x - y ) + u ( x + y ) } d y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ25_HTML.gif
      (25)
      Again by (23) and integration by parts, (25) becomes
      K [ u ] ( x ) = [ K ( y ) { - U ( x - y ) + U ( x + y ) } ] 0 - 0 K ( 3 ) ( y ) { - U ( x - y ) + U ( x + y ) } d y = 0 K ( 3 ) ( y ) { U ( x - y ) - U ( x + y ) } d y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equad_HTML.gif
      and hence,
      K [ u ] ( 3 ) ( x ) = d d x 0 K ( 3 ) ( y ) { U ( x - y ) - U ( x + y ) } d y = 0 K ( 3 ) ( y ) { u ( x - y ) - u ( x + y ) } d y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ26_HTML.gif
      (26)
      Once more by (23) and integration by parts, (26) becomes
      K [ u ] ( 3 ) ( x ) = [ K ( 3 ) ( y ) { - U ( x - y ) - U ( x + y ) } ] 0 - 0 K ( 4 ) ( y ) { - U ( x - y ) - U ( x + y ) } d y = 2 K ( 3 ) ( 0 ) U ( x ) + 0 K ( 4 ) ( y ) { U ( x - y ) + U ( x + y ) } d y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equae_HTML.gif
      and hence, by (22),
      K [ u ] ( 4 ) ( x ) = d d x 0 K ( 4 ) ( y ) { U ( x - y ) + U ( x + y ) } d y + 2 K ( 3 ) ( 0 ) U ( x ) = 0 K ( 4 ) ( y ) { u ( x - y ) + u ( x + y ) } d y + 2 K ( 3 ) ( 0 ) u ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ27_HTML.gif
      (27)
      = - α 4 K [ u ] ( x ) + α 4 k u ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ28_HTML.gif
      (28)

      since K ( 3 ) ( 0 ) = α 4 2 k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq21_HTML.gif and K(4)(y) = -α4K(y) by Lemma 3. Thus (a) follows from (24), (25), (26), (27), and (28).

      From (22), we have
      K [ u ] ( x ) = d d x 0 K ( y ) { u ( x - y ) + u ( x + y ) } d y = 0 K ( y ) { u ( x - y ) + u ( x + y ) } d y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ29_HTML.gif
      (29)
      for every uCl(ℝ) with u, u'L(ℝ). Suppose now uC q (ℝ) and u(i)L(ℝ) for i = 0, 1, ..., q. Then, by successively applying (29), we have
      K [ u ] ( q ) ( x ) = 0 K ( y ) { u ( q ) ( x - y ) + u ( q ) ( x + y ) } d y , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equaf_HTML.gif

      and hence, K [ u ] ( q ) ( x ) = K [ u ( q ) ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq22_HTML.gif by applying (22) to u(q). This proves (b), and the proof is complete.

      Lemma 5. For every uC0(ℝ), K [ u ] ( i ) C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq23_HTML.giffor i = 0, 1, 2, 3, 4.

      Proof. Suppose uC0(ℝ). Since C0(ℝ) ⊂ C(ℝ) ∩ L(ℝ), we have K [ u ] ( i ) C ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq24_HTML.gif for i = 0, 1, 2, 3, 4 by Lemma 4 (a). So it is sufficient to show that lim x ± K [ u ] ( i ) ( x ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq25_HTML.gif for i = 0, 1, 2, 3, 4. We first consider the case i = 0, 1, 2, 3. Let ϵ > 0 be arbitrary. Since uC0(ℝ), there exists M > 0 such that
      | u ( x ) | < 2 k 6 α i ε , i = 0 , 1 , 2 , 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ30_HTML.gif
      (30)
      for every |x| ≥ M/2. Moreover, we can assume M is large enough so that it also satisfies
      α i 2 k | | u | | exp - α 2 M 2 < ε 2 , i = 0 , 1 , 2 , 3 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ31_HTML.gif
      (31)
      Suppose x > M. By (22) and Lemma 4 (a), we have
      | K [ u ] ( i ) ( x ) | 0 K ( i ) ( y ) u ( x - y ) d y + ( - 1 ) i 0 K ( i ) ( y ) u ( x + y ) d y 0 | K ( i ) ( y ) | | u ( x - y ) | d y + 0 | K ( i ) ( y ) | | u ( x + y ) | d y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ32_HTML.gif
      (32)
      for i = 0, 1, 2, 3. Consider the second term in (32). If y ≥ 0, then x + yM > M/2, and so | u ( x + y ) | < 2 k 6 α i ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq26_HTML.gif by (30). Hence,
      0 | K ( i ) ( y ) | | u ( x + y ) | d y 2 k 6 α i ε 0 | K ( i ) ( y ) | d y 2 k 6 α i ε α i 2 k = ε 6 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ33_HTML.gif
      (33)
      since
      | K ( i ) ( y ) | α i + 1 2 k exp - α 2 y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ34_HTML.gif
      (34)
      by Lemma 3, and hence
      0 | K ( i ) ( y ) | d y α i + 1 2 k - 2 α exp - α 2 y 0 = α i 2 k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ35_HTML.gif
      (35)
      Note that the first term in (32) is
      0 | K ( i ) ( y ) | | u ( x - y ) | d y = 0 x - M / 2 | K ( i ) ( y ) | | u ( x - y ) | d y + x - M / 2 x + M / 2 | K ( i ) ( y ) | | u ( x - y ) | d y + x + M / 2 | K ( i ) ( y ) | | u ( x - y ) | d y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ36_HTML.gif
      (36)
      If 0 ≤ yx - M/2, then x - yM/2, and hence | u ( x - y ) | < 2 k 6 α i ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq27_HTML.gif by (30). If yx + M/2, then x - y ≤ - M/2, and we also have | u ( x - y ) | < 2 k 6 α i ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq27_HTML.gif by (30). Thus, by (35), we have
      0 x - M / 2 | K ( i ) ( y ) | | u ( x - y ) | d y 2 k 6 α i ε 0 x - M / 2 | K ( i ) ( y ) | d y 2 k 6 α i ε 0 | K ( i ) ( y ) | d y = 2 k 6 α i ε α i 2 k = ε 6 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ37_HTML.gif
      (37)
      x + M / 2 | K ( i ) ( y ) | | u ( x - y ) | d y 2 k 6 α i ε x + M / 2 | K ( i ) ( y ) | d y 2 k 6 α i ε 0 | K ( i ) ( y ) | d y = 2 k 6 α i ε α i 2 k = ε 6 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ38_HTML.gif
      (38)
      By (31) and (34), the remaining term in (36) becomes
      x - M / 2 x + M / 2 | K ( i ) ( y ) | | u ( x - y ) | d y | | u | | x - M / 2 x + M / 2 | K ( i ) ( y ) | d y | | u | | x - M / 2 | K ( i ) ( y ) | d y | | u | | x - M / 2 α i + 1 2 k exp - α 2 y d y = | | u | | α i + 1 2 k - 2 α exp - α 2 y x - M / 2 = | | u | | α i 2 k exp - α 2 ( x - M / 2 ) - 0 < | | u | | α i 2 k exp - α 2 M 2 < ε 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ39_HTML.gif
      (39)
      since x > M. Combining (32), (33), (36), (37), (38), and (39), we have
      | K ( i ) [ u ] ( x ) | < ε 6 + ε 2 + ε 6 + ε 6 = ε http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equag_HTML.gif

      for every x > M. This implies lim x | K ( i ) [ u ] ( x ) | = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq28_HTML.gif for i = 0, 1, 2, 3. We omit the similar proof that lim x | K ( i ) [ u ] ( x ) | = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq28_HTML.gif. Thus we conclude that K ( i ) [ u ] C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq29_HTML.gif for i = 0, 1, 2, 3. It follows that K ( 4 ) [ u ] C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq30_HTML.gif, since K ( 4 ) [ u ] = - α 4 K [ u ] + α 4 k u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq31_HTML.gif by Lemma 4 (a).

      In what follows, we put τ to be the following constant:
      τ : = 2 k 0 | K ( y ) | d y . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ40_HTML.gif
      (40)
      The exact value of τ can be determined by elementary calculation, which we omit. It turns out that
      τ = 1 + 2 exp - 3 π 4 1 - exp ( - π ) 1 . 140 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ41_HTML.gif
      (41)
      Lemma 6. (a) | | K [ u ] | | < ( τ / k ) | | u | | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq32_HTML.giffor every uL(ℝ). Thus, K [ u ] L ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq33_HTML.giffor every uL(ℝ).
      1. (b)

        For every uC(ℝ) ∩ L (ℝ), we have | | K [ u ] ( i ) | | ( τ α i / k ) exp ( 3 i π / 4 ) | | u | | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq34_HTML.gif for i = 1, 2, 3, and | | K [ u ] ( 4 ) | | < ( ( τ + 1 ) α 4 / k ) | | u | | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq35_HTML.gif.

         
      Proof. By (22) and (40), we have
      | | K [ u ] | | sup x 0 | K ( y ) | | u ( x - y ) + u ( x + y ) | d y 0 | K ( y ) | sup x | u ( x - y ) + u ( x + y ) | d y 2 | | u | | 0 | K ( y ) | d y = τ k | | u | | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equah_HTML.gif

      for every uL(ℝ). This shows (a).

      Suppose uC(ℝ) ∩ L(ℝ). By Lemma 4 (a),
      | | K [ u ] ( i ) | | sup x 0 | K ( i ) ( y ) | | u ( x - y ) + ( - 1 ) i u ( x + y ) | d y 0 | K ( i ) ( y ) | sup x | u ( x - y ) + ( - 1 ) i u ( x + y ) | d y 0 | K ( i ) ( y ) | sup x | u ( x - y ) | + sup x | u ( x + y ) | d y 2 | | u | | 0 | K ( i ) ( y ) | d y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ42_HTML.gif
      (42)
      for i = 1, 2, 3. By Lemma 3 and with the substitution z = y + 3 i π 2 2 α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq36_HTML.gif, we have
      0 | K ( i ) ( y ) | d y = 0 α i + 1 2 k exp - α 2 y sin α 2 y + π 4 + 3 i π 4 d y = 3 i π / ( 2 2 α ) α i + 1 2 k exp - α 2 z + 3 i π 4 sin α 2 z + π 4 d z α i exp 3 i π 4 0 α 2 k exp - α 2 z sin α 2 z + π 4 d z = α i exp 3 i π 4 0 | K ( z ) | d z , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equai_HTML.gif
      which, together with (40) and (42), gives
      | | K [ u ] ( i ) | | 2 | | u | | α i exp 3 i π 4 τ 2 k = τ α i k exp 3 i π 4 | | u | | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equaj_HTML.gif

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

      Finally, by Lemma 4 (a) and the above result (a),
      | | K ( 4 ) [ u ] | | = - α 4 K [ u ] + α 4 k u α 4 | | K [ u ] | | + α 4 k | | u | | α 4 τ k | | u | | + α 4 k | | u | | = ( τ + 1 ) α 4 k | | u | | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equak_HTML.gif

      and the proof is therefore complete.

      5 Main result

      Using the operators N http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq5_HTML.gif and K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq1_HTML.gif in Sections 3 and 4, the nonlinear integral operator Ψ defined in (10) can be expressed in abstract notation as
      Ψ [ u ] = K [ w ] - K [ N [ u ] ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equal_HTML.gif

      for u : ℝ → ℝ. We will show that Ψ is a contraction when it is restricted to an appropriate function space XC0(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) wC0(ℝ).

      (W2) ∥w < sup0≤sσk{ρ-1(s) · (σk - s)}, where σ is defined by
      σ : = 1 - τ τ + η . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ43_HTML.gif
      (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 [ 0 , s ̄ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq37_HTML.gif for some 0 < s ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq38_HTML.gif. In fact, the supremum in W2 should be meant to be taken in the range s [ 0 , σ k ] [ 0 , s ̄ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq39_HTML.gif. Note that the set { ρ - 1 ( s ) ( σ k - s ) | s [ 0 , σ k ] [ 0 , s ̄ ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq40_HTML.gif should be connected, and hence of the form [ 0 , c ̄ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq41_HTML.gif or [ 0 , c ̄ ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq42_HTML.gif for some 0 < c ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq43_HTML.gif, since [ 0 , σ k ] [ 0 , s ̄ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq44_HTML.gif is connected and ρ-1(s) · (σk - s) is continuous. In fact, we have c ̄ = sup 0 s σ k { ρ - 1 ( s ) ( σ k - s ) } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq45_HTML.gif. It follows from W2 that there exists s * ( 0 , σ k ) ( 0 , s ̄ ) ( 0 , σ k ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq46_HTML.gif such that
      | | w | | = ρ - 1 ( s * ) { σ k - s * } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ44_HTML.gif
      (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
      X : = u C 0 ( ) | | u | | | | w | | σ k - s * . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ45_HTML.gif
      (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
      1 σ k w < w σ k - s * , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equam_HTML.gif
      since 0 < s* < σk. It follows that
      u C 0 ( ) | | u | | 1 σ k | | w | | X . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equ46_HTML.gif
      (46)

      In our system described by the differential equation (2), it is physically clear that the size ∥u of the output deflection u cannot be too large compared to the size ∥w of 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
      f ( u , x ) = ( 1 + ε cos x ) k 1 + ε u + λ u 2 n + 1 , 0 ε 1 2 , n 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equan_HTML.gif
      in Example 1. Then we have ρ(t) = 2(2n + 1)λt2n, and hence ρ - 1 ( s ) = s 2 ( 2 n + 1 ) λ 1 2 n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq47_HTML.gif. Put ϕ(s) = ρ-1(s) · (σk - s). Since
      ϕ ( s ) = d d s s 2 ( 2 n + 1 ) λ 1 2 n ( σ k - s ) = 1 2 ( 2 n + 1 ) λ 2 n 1 2 n s 1 2 n - 1 ( σ k - s ) - s 1 2 n = s 1 2 n - 1 2 n 2 ( 2 n + 1 ) λ 2 n { ( σ k - s ) - 2 n s } = ( 2 n + 1 ) s 1 2 n - 1 2 n 2 ( 2 n + 1 ) λ 2 n σ k 2 n + 1 - s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equao_HTML.gif
      ϕ is strictly increasing on 0 , σ k 2 n + 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq48_HTML.gif, and strictly decreasing on σ k 2 n + 1 , σ k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq49_HTML.gif. Note also that ϕ(0) = ϕ(σk) = 0. Thus,
      sup 0 s σ k ρ - 1 ( s ) ( σ k - s ) = ρ - 1 σ k 2 n + 1 σ k - σ k 2 n + 1 = σ k 2 ( 2 n + 1 ) 2 λ 1 2 n 2 n 2 n + 1 σ k = 2 n ( 2 n + 1 ) { 2 ( 2 n + 1 ) 2 λ } 1 2 n ( σ k ) 1 + 1 2 n < . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equap_HTML.gif

      There are exactly two solutions in (0, σk) of the equation ρ-1(s) · (σk - s) = ∥w, or equivalently, s ( s - σ k ) 2 n - 2 ( 2 n + 1 ) λ | | w | | 2 n = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq50_HTML.gif. Note that we have bigger X, if we take s* to be the larger among them.

      Example 5. Consider the case
      f ( u , x ) = ( 1 + ε cos x ) k 1 + ε u + λ { exp ( a u ) - 1 - a u } , 0 ε 1 2 , a > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equaq_HTML.gif
      in Example 2. Then we have ρ(t) = 2 {exp(at) - 1}, and hence ρ - 1 ( s ) = 1 a ln 1 + s 2 a λ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq51_HTML.gif. Putting ϕ(s) = ρ-1(s) · (σk - s), we have
      ϕ ( s ) = d d s 1 a ln 1 + s 2 a λ ( σ k - s ) = 1 a 1 2 a λ 1 + s 2 a λ ( σ k - s ) - ln 1 + s 2 a λ = 1 2 a 2 λ 1 + s 2 a λ ( σ k - s ) - 2 a λ 1 + s 2 a λ ln 1 + s 2 a λ = 1 2 a 2 λ 1 + s 2 a λ σ k - s + 2 a λ 1 + s 2 a λ ln 1 + s 2 a λ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equar_HTML.gif
      It follows that ϕ is strictly increasing on [ 0 , s ̃ ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq52_HTML.gif, and strictly decreasing on [ s ̃ , σ k ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq53_HTML.gif, and hence, sup 0 s σ k { ρ - 1 ( s ) ( σ k - s ) } = ϕ ( s ̃ ) < http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq54_HTML.gif, where s ̃ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq55_HTML.gif is the unique solution in (0, σk) of the equation
      σ k - s + 2 a λ 1 + s 2 a λ ln 1 + s 2 a λ = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equas_HTML.gif

      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
      ρ - 1 ( s ) = 1 ( σ k - s ) 2 - 1 σ 2 k 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equat_HTML.gif
      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
      s * = σ k 1 + σ k | | w | | 2 - 1 + σ k | | w | | 2 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equau_HTML.gif

      in (0, σk).

      5.2 Contractiveness of the operator Ψ

      Suppose uC0(ℝ). Then N [ u ] C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq8_HTML.gif by Lemma 1 (a), and again, K [ N [ u ] ] C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq56_HTML.gif by Lemma 5. We also have K [ w ] C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq57_HTML.gif by W1 and Lemma 5. Thus, we have Ψ [ u ] = K [ u ] - K [ N [ u ] ] C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq58_HTML.gif for every uC0(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 uC4(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
      u ( 4 ) = Ψ [ u ] ( 4 ) = { K [ w ] - K [ N [ u ] ] } ( 4 ) = K [ w ] ( 4 ) - K [ N [ u ] ] ( 4 ) = - α 4 K [ w ] + α 4 k w - - α 4 K [ N [ u ] ] + α 4 k N [ u ] = α 4 k { w - N [ u ] } - α 4 { K [ w ] - K [ N [ u ] ] } = α 4 k { w - N [ u ] } - α 4 Ψ [ u ] = α 4 k { w - N [ u ] } - α 4 u , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equav_HTML.gif

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

      Conversely, suppose u is a solution of (2), so that u ( 4 ) + α 4 u + α 4 k N [ u ] = α 4 k w http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq59_HTML.gif by (12).

      Applying the operator K http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq1_HTML.gif, we get
      K [ u ( 4 ) ] + α 4 K [ u ] + α 4 k K [ N [ u ] ] = α 4 k K [ w ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equaw_HTML.gif
      and hence
      Ψ [ u ] = K [ w ] - K [ N [ u ] ] = k α 4 K [ u ( 4 ) ] + k K [ u ] = k α 4 K [ u ] ( 4 ) + k K [ u ] = k α 4 - α 4 K [ u ] + α 4 k u + k K [ u ] = u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equax_HTML.gif

      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 uX. Moreover, Ψ: XX is a contraction, i.e., ∥Ψ[u] - Ψ[v]∥L · ∥u - vfor every u, vX for some constant L < 1.

      Proof. Suppose uX. Note that N [ 0 ] = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq60_HTML.gif by F2, if we denote the zero function by 0(x) ≡ 0. Hence, by Lemma 1 (b) and Lemma 6 (a), we have
      | | Ψ [ u ] | | = | | K [ w ] - K [ N [ u ] ] | | | | K [ w ] | | + | | K [ N [ u ] ] | | τ k | | w | | + τ k | | N [ u ] | | τ k | | w | | + τ k | | N [ u ] - N [ 0 ] | | τ k | | w | | + τ k { ( 1 - η ) k + ρ ( | | u | | ) } | | u | | , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equay_HTML.gif
      where ρ is taken as in Lemma 1 (b). Hence, by (44) and (43), we have
      | | Ψ [ u ] | | τ k | | w | | + τ k ( 1 - η ) k + ρ | | w | | σ k - s * | | w | | σ k - s * = τ k | | w | | + τ k { ( 1 - η ) k + ρ ( ρ - 1 ( s * ) ) } | | w | | σ k - s * = τ k 1 + ( 1 - η ) k + s * σ k - s * | | w | | = τ ( σ + 1 - η ) σ k - s * | | w | | = τ 1 - τ τ + η + 1 - η σ k - s * | | w | | = | | w | | σ k - s * , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equaz_HTML.gif

      which shows Ψ[u] ∈ X.

      Now suppose u, vX. Again by Lemma 6 (a) and Lemma 1 (b), we have
      | | Ψ [ u ] - Ψ [ v ] | | = | | { K [ w ] - K [ N [ u ] ] } - { K [ w ] - K [ N [ v ] ] } | | = | | K [ N [ u ] ] - K [ N [ v ] ] | | = | | K [ N [ u ] - N [ v ] ] | | 2 τ k | | N [ u ] - N [ v ] | | τ k { ( 1 - η ) k + ρ ( max { | | u | | , | | v | | } ) } | | u - v | | τ k ( 1 - η ) k + ρ | | w | | σ k - s * | | u - v | | = τ k { ( 1 - η ) k + ρ ( ρ - 1 ( s * ) ) } | | u - v | | = τ 1 + η + s * k | | u - v | | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equba_HTML.gif
      Since 0 < s* < σk, we have
      τ 1 - η + s * k < τ 1 - η + σ k k = τ ( 1 - η + σ ) = τ 1 - η + 1 - τ τ + η = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equbb_HTML.gif

      by (43). Thus, we have the desired inequality by taking L = τ 1 - η + s * k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq61_HTML.gif, 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: YY satisfies d(Φ(y1), Φ(y2)) ≤ L · d(y1, y2) for every y1, y 2 Y for some constant L < 1. Then Φ has a unique fixed point in Y. Moreover, for any y0Y, the sequence { y n } n = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq62_HTML.gif, defined by y n = 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 u * ( i ) C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq63_HTML.giffor 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 u n = Ψ[un-1], n = 1, 2, .... By Proposition 1, the sequence of functions { u n } n = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq64_HTML.gif in X converges uniformly to the fixed point u*X. We assume u0C4(ℝ) and u 0 ( i ) C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq65_HTML.gif 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-1C4(ℝ) and u n - 1 ( i ) C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq66_HTML.gif for i = 1, 2, 3, 4. Then N [ u n - 1 ] C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq67_HTML.gif by Lemma 1 (a), since un-1C0(ℝ). Hence, K [ N [ u n - 1 ] ] C 4 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq68_HTML.gif by Lemma 4 (a), and K [ N [ u n - 1 ] ] ( i ) C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq69_HTML.gif, i = 1, 2, 3, 4 by Lemma 5. Since wC0(ℝ) by W1, we also have K [ w ] C 4 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq70_HTML.gif by Lemma 4 (a), and K [ w ] ( i ) C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq71_HTML.gif for i = 1, 2, 3, 4 by Lemma 5. Hence, we have u n C4(ℝ) and u n ( i ) C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq72_HTML.gif for i = 1, 2, 3, 4, since u n = Ψ [ u n - 1 ] = K [ w ] - K [ N [ u n - 1 ] ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq73_HTML.gif. Thus, by induction on n, we have u n C4(ℝ) and u n ( i ) C 0 ( ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq72_HTML.gif for i = 1, 2, 3, 4, for every n = 0, 1, 2, ....

      By Lemma 6 (b), we have | | K [ u ] ( i ) | | A | | u | | http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_IEq74_HTML.gif, i = 1, 2, 3, 4, for every uC0(ℝ) ⊂ C(ℝ) ∩ L(ℝ), where we put
      A = max max i = 1 , 2 , 3 τ α i k exp 3 i π 4 , ( τ + 1 ) α 4 k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-5/MediaObjects/13661_2011_Article_130_Equbc_HTML.gif
      Hence, we have
      u n + 1 ( i ) - u n ( i ) = Ψ [ u n ] ( i ) - Ψ [ u n - 1 ] ( i ) = { K [ w ] - K [ N [ u n ] ] } ( i ) - { K [ w ] - K [ N