Abstract
We consider the static deflection of an infinite beam resting on a nonlinear and nonuniform elastic foundation. The governing equation is a fourthorder nonlinear ordinary differential equation. Using the Green's function for the wellanalyzed 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; nonuniform; fourthorder 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 signtype Nagumo control. With regard to the beam equation, Alves et al. [13] discussed about iterative solutions for a nonlinear fourthorder 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 "nonuniform" 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 nonuniform foundation analysis. This is because the solution procedure for a nonlinear fourthorder 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 nonuniform 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 nonuniform 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 nonuniform beam equation. The properties of the nonlinear, nonuniform elastic foundation are analyzed in Section 3, and a close investigation on the basic integral operator , 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 nonuniform in x. Figure 1 shows that the vertical deflection of the beam u(x) results from the net load distribution p(x):
Figure 1. Infinite beam on nonlinear and nonuniform elastic foundation.
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 xcoordinate, and g is the gravitational acceleration. The term m(x)g also plays an important role in the dynamic beam problem, since the secondorder 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 fourthorder ordinary differential equation
which, in turn, becomes the following nonlinear differential equation for the deflection u by (1):
The boundary condition that we consider is
Note that (2) and (3) together form a welldefined 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
where
or
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 closedform solution:
expressed in terms of the following Green's function G:
where . 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:
With the substitution of (5), (8) immediately reveals the following nonlinear Fredholm integral equation for u:
Physically, the term 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 in (9) corresponds to the difference between the exact nonlinear solution u and the linear deflection . We define the nonlinear operator Ψ by
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
Denote ∥u∥_{∞ }= sup_{x∈ℝ }u(x) for u : ℝ → ℝ, and let L^{∞}(ℝ) be the space of all functions u : ℝ → ℝ such that ∥u∥_{∞ }< ∞. Let C_{0}(ℝ) be the space of all continuous functions vanishing at infinity. It is well known [24] that C_{0}(ℝ) and L^{∞}(ℝ) are Banach spaces with the norm ∥·∥_{∞}, and C_{0}(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, C^{0}(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 u ∈ C^{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 for q = 0, 1, 2.
(F4) inf_{x∈ℝ }f_{u}(0, x) > η_{0 }sup_{x∈ℝ }f_{u}(0, x), where
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 nonuniform, 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 nonuniform 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:
We justifiably rule out the case k = 0; hence, we assuming k > 0 for the rest of the article. Define
which is the nonlinear and nonuniform 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 = sup_{x∈ℝ }f_{u}(0, x). This restriction, which is realistic, comes from the unfortunate fact that the operator in Section 4 is not a contraction. The constant η_{0 }is related to another constant τ, which will be introduced later in (41) in Section 4, by
We define a parameter η which measures the nonuniformity of the elastic foundation:
Then, by F4, we have
A uniform elastic foundation corresponds to the extreme case η = 1, and the nonuniformity increases as η becomes smaller. In order for our current method to work, the condition F4 limits the nonuniformity η by η_{0 }≈ 0.123.
Using the function N, we define the operator by for functions u : ℝ → ℝ. Note that is nonlinear in general.
Lemma 1. (a) for every u ∈ C_{0}(ℝ).
(b) For every u, v ∈ L^{∞}(ℝ), we have
for some strictly increasing continuous function ρ : [0, ∞) → [0, ∞), such that ρ(0) = 0.
Proof. Suppose u ∈ C_{0}(ℝ). is continuous by F1. Let ϵ > 0. Then there exists M > 0 such that u(x) < ϵ if x > M, since lim_{x→±∞ }u(x) = 0. By the mean value theorem, we have
for some μ between 0 and u(x), and hence μ ≤ u(x) < ϵ if x > M. Hence, for x > M, we have
Note that (16) can be made arbitrarily small as M gets larger, since sup_{x∈ℝ, μ≤ϵ }f_{u}(μ, x) < ∞ by F3. Thus, , which proves (a).
By the mean value theorem, we have
for some μ between u and v, and hence μ ≤ max{u, v}. Hence,
Now suppose u, v ∈ L^{∞}(ℝ). Then
Put
Note that (18) is welldefined by F3, since we have N_{u}(μ, x) = f_{u}(μ, x)  k from (12). Clearly, ρ_{1 }is nonnegative and nondecreasing.
We want to show ρ_{1 }is continuous. Fix t_{0 }≥ 0. We first show the leftcontinuity of ρ_{1 }at t_{0}. Let be a sequence in [0, t_{0}) such that t_{n }↗ t_{0}. Suppose there exists t' < t_{0 }such that ρ_{1}(t') = ρ_{1}(t_{0}). Then, since ρ_{1 }is nondecreasing, it becomes constant on [t', t_{0}], and hence ρ_{1 }is clearly leftcontinuous at t_{0}. So we assume that ρ_{1}(t') < ρ_{1}(t_{0}) for every t' < t_{0}. It follows that there exists a sequence in [t_{0}, t_{0}] × ℝ, such that μ_{n} = t_{n }and N_{u}(μ_{n}, x_{n}) → ρ_{1}(t_{0}) as n → ∞, since N_{u}(u, x) is continuous. Thus, we have ρ_{1}(t_{n}) → ρ_{1}(t_{0}) as n → ∞, since N_{u}(μ_{n}, x_{n}) ≤ ρ_{1}(t_{n}) ≤ ρ_{1}(t_{0}) for n = 1, 2, .... This shows that ρ_{1 }is leftcontinuous at t_{0}.
Suppose ρ_{1 }is not rightcontinuous at t_{0}. Then there exist ϵ > 0 and a sequence in (t_{0}, ∞), such that t_{n }↘ t_{0 }and ρ_{1}(t_{n})  ρ_{1}(t_{0}) ≥ ϵ for n = 1, 2, .... Suppose there exists t' > t_{0 }such that ρ_{1}(t') = ρ_{1}(t_{0}). Then ρ_{1 }becomes constant on [t_{0}, t'], so that ρ_{1 }is rightcontinuous at t_{0}. So we assume that ρ_{1}(t') > ρ_{1}(t_{0}) for every t' > t_{0}. It follows that there exists a sequence in {(t_{0}, ∞) ∪ (∞, t_{0})} × ℝ, such that t_{0 }< μ_{n} ≤ t_{n }and 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
for some μ between t_{0 }and μ_{n}, and so we have
for n = 1, 2, .... By F3, (19) goes to 0 as n → ∞, since
This is a contradiction. It follows that ρ_{1 }is rightcontinuous, and thus, is continuous.
By (11) and (14), we have ηk ≤ f_{u}(0, x) ≤ k, and so (1  η) k ≤ f_{u}(0, x)  k ≤ 0 for every x ∈ ℝ. It follows that
Put ρ_{2}(t): = ρ_{1}(t)  ρ_{1}(0). Then ρ_{2 }is a nondecreasing continuous function such that ρ_{2}(0) = 0. By Lemma 2 below, there exists a strictly increasing continuous function ρ such that ρ(0) = 0, and ρ(t) ≥ ρ_{2}(t) for t ≥ 0. Thus, we have a desired function ρ, since
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 nondecreasing continuous function such that g(0) = 0. Then there exists a strictly increasing continuous function such that , and 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 nondecreasing. 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.,
where l(B) is the Lebesque measure, and hence the length in our case, of the set B ⊂ [0, ∞). From the definition of , it is clear that , and for t ≥ 0. We omit the proof that is continuous and strictly increasing, which is an easy exercise.
Example 1. Let
Then,
and hence,
We also have
Thus, we can take ρ(t) = ρ_{2}(t) = 2(2n + 1)λt^{2n}.
Example 2. Let
Then,
and hence,
We also have
Thus, we can take ρ(t) = ρ_{2}(t) = 2aλ {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 ρ_{2}(t) ≡ 0. The function ρ taken according to Lemma 2 would be ρ(t) = t. However, a better choice is
as we will check in Section 5.1, where the constant σ is defined as well.
4 The Operator
Let
so that G(x, ξ) = K (ξ  x) for G in (7). Using the function K, we define the linear operator by
for functions u : ℝ → ℝ. With this notation, we can rewrite the solution u in (6) of the following linear differential equation:
which is just the linear case of (2), as . In fact, understanding the properties of the operator is important not only for the linear case (21), but also for the general nonlinear nonuniform case (2).
Lemma 3.
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
we have
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 . Note that, for every u ∈ L^{∞}(ℝ),
Lemma 4. (a) Let u ∈ C(ℝ) ∩ L^{∞}(ℝ). Then we have
Consequently, for every u ∈ C(ℝ) ∩ L^{∞}(ℝ).
(b) Let q = 0, 1, 2, .... Suppose u ∈ C^{q}(ℝ) and u^{(i) }∈ L^{∞}(ℝ) for i = 0, 1, ..., q. Then we have .
Proof. Let u ∈ C(ℝ) ∩ L^{∞}(ℝ). Then there exists a function U ∈ C^{l}(ℝ) such that U' = u. Since u ∈ L^{∞}(ℝ), U has at most linear growth, and hence by Lemma 3,
for i = 0, 1, 2, .... Using integration by parts, (22) becomes
by (23), and hence we have
By (23) and integration by parts again, (24) becomes
since K'(0) = 0 by Lemma 3. Hence,
Again by (23) and integration by parts, (25) becomes
and hence,
Once more by (23) and integration by parts, (26) becomes
and hence, by (22),
since and K^{(4)}(y) = α^{4}K(y) by Lemma 3. Thus (a) follows from (24), (25), (26), (27), and (28).
From (22), we have
for every u ∈ C^{l}(ℝ) with u, u' ∈ L^{∞}(ℝ). Suppose now u ∈ C^{q}(ℝ) and u^{(i) }∈ L^{∞}(ℝ) for i = 0, 1, ..., q. Then, by successively applying (29), we have
and hence, by applying (22) to u^{(q)}. This proves (b), and the proof is complete.
Lemma 5. For every u ∈ C_{0}(ℝ), for i = 0, 1, 2, 3, 4.
Proof. Suppose u ∈ C_{0}(ℝ). Since C_{0}(ℝ) ⊂ C(ℝ) ∩ L^{∞}(ℝ), we have for i = 0, 1, 2, 3, 4 by Lemma 4 (a). So it is sufficient to show that for i = 0, 1, 2, 3, 4. We first consider the case i = 0, 1, 2, 3. Let ϵ > 0 be arbitrary. Since u ∈ C_{0}(ℝ), there exists M > 0 such that
for every x ≥ M/2. Moreover, we can assume M is large enough so that it also satisfies
Suppose x > M. By (22) and Lemma 4 (a), we have
for i = 0, 1, 2, 3. Consider the second term in (32). If y ≥ 0, then x + y ≥ M > M/2, and so by (30). Hence,
since
by Lemma 3, and hence
Note that the first term in (32) is
If 0 ≤ y ≤ x  M/2, then x  y ≥ M/2, and hence by (30). If y ≥ x + M/2, then x  y ≤  M/2, and we also have by (30). Thus, by (35), we have
By (31) and (34), the remaining term in (36) becomes
since x > M. Combining (32), (33), (36), (37), (38), and (39), we have
for every x > M. This implies for i = 0, 1, 2, 3. We omit the similar proof that . Thus we conclude that for i = 0, 1, 2, 3. It follows that , since by Lemma 4 (a).
In what follows, we put τ to be the following constant:
The exact value of τ can be determined by elementary calculation, which we omit. It turns out that
Lemma 6. (a) for every u ∈ L^{∞}(ℝ). Thus, for every u ∈ L^{∞}(ℝ).
(b) For every u ∈ C(ℝ) ∩ L^{∞}(ℝ), we have for i = 1, 2, 3, and .
Proof. By (22) and (40), we have
for every u ∈ L^{∞}(ℝ). This shows (a).
Suppose u ∈ C(ℝ) ∩ L^{∞}(ℝ). By Lemma 4 (a),
for i = 1, 2, 3. By Lemma 3 and with the substitution , we have
which, together with (40) and (42), gives
for i = 1, 2, 3. This proves (b) for i = 1, 2, 3.
Finally, by Lemma 4 (a) and the above result (a),
and the proof is therefore complete.
5 Main result
Using the operators and in Sections 3 and 4, the nonlinear integral operator Ψ defined in (10) can be expressed in abstract notation as
for u : ℝ → ℝ. We will show that Ψ is a contraction when it is restricted to an appropriate function space X ⊂ C_{0}(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 ∈ C_{0}(ℝ).
(W2) ∥w∥_{∞ }< sup_{0≤s≤σk }{ρ^{1}(s) · (σk  s)}, where σ is defined by
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 lim_{x→±∞ }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 for some . In fact, the supremum in W2 should be meant to be taken in the range . Note that the set should be connected, and hence of the form or for some , since is connected and ρ^{1}(s) · (σk  s) is continuous. In fact, we have . It follows from W2 that there exists such that
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 nonuniformity of the given elastic foundation.
Now define the subset X of C_{0}(ℝ) by
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 C_{0}(ℝ) which itself is a complete metric space. Note that
since 0 < s_{* }< σk. It follows that
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 C_{0}(ℝ), is big enough in some sense.
Example 4. Consider the case
in Example 1. Then we have ρ(t) = 2(2n + 1)λt^{2n}, and hence . Put ϕ(s) = ρ^{1}(s) · (σk  s). Since
ϕ is strictly increasing on , and strictly decreasing on . Note also that ϕ(0) = ϕ(σk) = 0. Thus,
There are exactly two solutions in (0, σk) of the equation ρ^{1}(s) · (σk  s) = ∥w∥_{∞}, or equivalently, . Note that we have bigger X, if we take s_{* }to be the larger among them.
Example 5. Consider the case
in Example 2. Then we have ρ(t) = 2aλ {exp(at)  1}, and hence . Putting ϕ(s) = ρ^{1}(s) · (σk  s), we have
It follows that ϕ is strictly increasing on , and strictly decreasing on , and hence, , where is the unique solution in (0, σk) of the equation
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
Let ϕ(s) = ρ^{1}(s) · (σk  s). We can easily check that ϕ is strictly increasing on [0, σk), ϕ(0) = 0, and lim_{s→σk }ϕ(s) = ∞. Thus, we have sup_{0≤s≤σk }{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 s^{2 } σk(2 + σk∥w∥_{∞})s + σ^{3}k^{3}∥w∥_{∞ }= 0, has the unique solution
in (0, σk).
5.2 Contractiveness of the operator Ψ
Suppose u ∈ C_{0}(ℝ). Then by Lemma 1 (a), and again, by Lemma 5. We also have by W1 and Lemma 5. Thus, we have for every u ∈ C_{0}(ℝ). In short, the operator Ψ is a welldefined map from C_{0}(ℝ) into C_{0}(ℝ). The next lemma confirms that the solutions of (2) are the fixed points of Ψ in C_{0}(ℝ).
Lemma 7. Suppose u ∈ C^{4}(ℝ) ∩ C_{0}(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
and hence, u is a solution of (2) by (12).
Conversely, suppose u is a solution of (2), so that by (12).
Applying the operator , we get
and hence
by Lemma 4 (a), and (b), and the proof is complete.
Unfortunately, Ψ is not a contraction on the whole of C_{0}(ℝ). Nevertheless, if we restrict Ψ to the subset X of C_{0}(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  v∥_{∞ }for every u, v ∈ X for some constant L < 1.
Proof. Suppose u ∈ X. Note that by F2, if we denote the zero function by 0(x) ≡ 0. Hence, by Lemma 1 (b) and Lemma 6 (a), we have
where ρ is taken as in Lemma 1 (b). Hence, by (44) and (43), we have
which shows Ψ[u] ∈ X.
Now suppose u, v ∈ X. Again by Lemma 6 (a) and Lemma 1 (b), we have
Since 0 < s_{* }< σk, we have
by (43). Thus, we have the desired inequality by taking , 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(y_{1}), Φ(y_{2})) ≤ L · d(y_{1}, y_{2}) for every y_{1}, y_{2 }∈ Y for some constant L < 1. Then Φ has a unique fixed point in Y. Moreover, for any y_{0 }∈ Y, the sequence , defined by y_{n }= Φ(y_{n1}), 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 C^{4}(ℝ), and 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 u_{0 }in X, and define u_{n }= Ψ[u_{n1}], n = 1, 2, .... By Proposition 1, the sequence of functions in X converges uniformly to the fixed point u_{* }∈ X. We assume u_{0 }∈ C^{4}(ℝ) and for i = 1, 2, 3, 4, which can always be achieved: For example, we could take u_{0 }to be the zero function. Suppose, for some n, u_{n1 }∈ C^{4}(ℝ) and for i = 1, 2, 3, 4. Then by Lemma 1 (a), since u_{n1 }∈ C_{0}(ℝ). Hence, by Lemma 4 (a), and , i = 1, 2, 3, 4 by Lemma 5. Since w ∈ C_{0}(ℝ) by W1, we also have by Lemma 4 (a), and for i = 1, 2, 3, 4 by Lemma 5. Hence, we have u_{n }∈ C^{4}(ℝ) and for i = 1, 2, 3, 4, since . Thus, by induction on n, we have u_{n }∈ C^{4}(ℝ) and for i = 1, 2, 3, 4, for every n = 0, 1, 2, ....
By Lemma 6 (b), we have , i = 1, 2, 3, 4, for every u ∈ C_{0}(ℝ) ⊂ C(ℝ) ∩ L^{∞}(ℝ), where we put
Hence, we have
for every n = 0, 1, 2, ..., since . Since u_{n1}, u_{n }∈ X ⊂ L^{∞}(ℝ), we have
by Lemma 1 (b) and (44), (47). Since
by Lemma 8, we have
Combining (48) and (49), we have
Let ϵ > 0. Since 0 ≤ L < 1, we can take N large enough so that
Let m, n > N. Assume m > n with no loss of generality. Then by (50) and (51), we have
This implies that, for every i = 1, 2, 3, 4, the sequence is Cauchy in C_{0}(ℝ) with respect to the metric ∥·  ·∥_{∞}, and hence, converges uniformly to a function v_{i }∈ C_{0}(R). So by Lemma 10 below, u_{* }∈ C^{1}(ℝ) and , since u_{n }converges uniformly to u_{* }and converges uniformly to v_{1}. Applying Lemma 10 again to , we see that v_{1 }∈ C^{1}(ℝ) and . By repeating the same argument, we see that v_{2 }∈ C^{1}(ℝ), , and v_{3 }∈ C^{1}(ℝ), . Thus, we have u_{* }∈ C^{4}(ℝ) and for i = 1, 2, 3, 4. Hence, the proof is complete.
Lemma 10. Suppose a sequence of functions in C^{1}(ℝ) converges uniformly to a function g. Suppose also converges uniformly to a function h. Then g ∈ C^{1}(ℝ) and g' = h.
Proof. Fix x_{0 }∈ ℝ, and define h_{n }: ℝ → ℝ, n = 1, 2, ... by
which is continuous since g_{n }∈ C^{1}(ℝ). Note that as n → ∞. For x ≠ x_{0}, we have
for some ξ between x_{0 }and x by the mean value theorem for g_{m } g_{n}. Thus, we have for any m, n. It follows that h_{n }converges uniformly to a continuous function, since converges uniformly. Note that
Since h_{n }converges uniformly, we can change the order of the limit in (52), so that
Hence, exists, and is equal to h(x_{0}). Thus, the proof is complete, since x_{0 }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
where k, σ, s_{* }are as defined in (11), (43), (44) respectively. Moreover, the unique solution, denoted by u*, satisfies 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 nonuniformity of the given elastic foundation. In this study, we introduced a physical parameter η in (14) measuring the nonuniformity, 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 nonuniformity 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 C_{0}(ℝ) equipped with the usual supnorm ∥·∥_{∞}. 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 in Lemma 6 (a), is greater than 1. Because of this, too much nonuniformity of the elastic foundation, encoded in the parameter η, can also contribute to the noncontractiveness of Ψ.
Thus to resort to the Banach fixed point theorem, it is necessary to find a subspace smaller than C_{0}(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 nonuniformity 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 upperbound 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 C^{4}(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.: 20110010090). 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.: K2090300203011E010004610). 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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