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

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.

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

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

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

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

The rest of the article is organized as follows: in Section 2, we describe our problem in detail, and formulate an integral equation equivalent to the nonlinear and non-uniform beam equation. The properties of the nonlinear, non-uniform elastic foundation are analyzed in Section 3, and a close investigation on the basic integral operator K , 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.

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):

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

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 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

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 closed-form solution:

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

where α = k / E I 4 . 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 ∫ - ∞ ∞ G ( x , ξ ) w ( x ) d ξ 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 in (9) corresponds to the difference between the exact nonlinear solution u and the linear deflection ∫ - ∞ ∞ G ( x , ξ ) w ( x ) d ξ . 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.

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 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 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 υ ≥ 0 , sup x ∈ ℝ , | u | ≤ υ ∂ q f ∂ u q ( u , x ) < ∞ for q = 0, 1, 2.

(F4) inf_x∈ℝ f_u(0, x) > η₀ 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 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:

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 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 = sup_x∈ℝ f_u(0, x). This restriction, which is realistic, comes from the unfortunate fact that the operator K 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

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

Then, by F4, we have

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 by N [ u ] ( x ) : = N ( u ( x ) , x ) for functions u : ℝ → ℝ. Note that N is nonlinear in general.

Lemma 1. (a) N [ u ] ∈ C 0 ( ℝ ) for every u ∈ C₀(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 ∈ C₀(ℝ). N [ u ] 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, N [ u ] ∈ C 0 ( ℝ ) , 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 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 ∞ 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 ∞ 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 ∞ 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 ∞ 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 ) 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 t₀ 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 ρ₁ is right-continuous, 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 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

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 , ∞ ) such that g ̃ ( 0 ) = 0 , and g ̃ ( t ) ≥ g ( t ) for t ≥ 0.

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

Define g ̃ : [ 0 , ∞ ) → [ 0 , ∞ ) by

where l(B) is the Lebesque measure, and hence the length in our case, of the set B ⊂ [0, ∞). From the definition of g ̃ , it is clear that g ̃ ( 0 ) = 0 , and g ̃ ( t ) ≥ g ( t ) for t ≥ 0. We omit the proof that g ̃ 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) = ρ₂(t) = 2(2n + 1)λt²ⁿ.

Example 2. Let

Then,

and hence,

We also have

Thus, we can take ρ(t) = ρ₂(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 ρ₂(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.

Let

so that G(x, ξ) = K (|ξ - x|) for G in (7). Using the function K, we define the linear operator K 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 u = K [ w ] . In fact, understanding the properties of the operator K is important not only for the linear case (21), but also for the general nonlinear non-uniform 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 K [ u ] . Note that, for every u ∈ L^∞(ℝ),

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

Consequently, K [ u ] ∈ C 4 ( ℝ ) 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 K [ u ( q ) ] = K [ u ] ( q ) .

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 K ( 3 ) ( 0 ) = α 4 2 k and K⁽⁴⁾(y) = -α⁴K(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, K [ u ] ( q ) ( x ) = K [ u ( q ) ] by applying (22) to u^(q). This proves (b), and the proof is complete.

Lemma 5. For every u ∈ C₀(ℝ), K [ u ] ( i ) ∈ C 0 ( ℝ ) for i = 0, 1, 2, 3, 4.

Proof. Suppose u ∈ C₀(ℝ). Since C₀(ℝ) ⊂ C(ℝ) ∩ L^∞(ℝ), we have K [ u ] ( i ) ∈ C ( ℝ ) 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 for i = 0, 1, 2, 3, 4. We first consider the case i = 0, 1, 2, 3. Let ϵ > 0 be arbitrary. Since u ∈ C₀(ℝ), there exists M > 0 such that

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 | u ( x + y ) | < 2 k 6 α i ε 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 | u ( x - y ) | < 2 k 6 α i ε by (30). If y ≥ x + M/2, then x - y ≤ - M/2, and we also have | u ( x - y ) | < 2 k 6 α i ε 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 lim x → ∞ | K ( i ) [ u ] ( x ) | = 0 for i = 0, 1, 2, 3. We omit the similar proof that lim x → ∞ | K ( i ) [ u ] ( x ) | = 0 . Thus we conclude that K ( i ) [ u ] ∈ C 0 ( ℝ ) for i = 0, 1, 2, 3. It follows that K ( 4 ) [ u ] ∈ C 0 ( ℝ ) , since K ( 4 ) [ u ] = - α 4 K [ u ] + α 4 k u 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) | | K [ u ] | | ∞ < ( τ / k ) ⋅ | | u | | ∞ for every u ∈ L^∞(ℝ). Thus, K [ u ] ∈ L ∞ ( ℝ ) for every u ∈ L^∞(ℝ).

(b) For every u ∈ C(ℝ) ∩ L^∞(ℝ), we have | | K [ u ] ( i ) | | ∞ ≤ ( τ α i / k ) exp ( 3 i π / 4 ) ⋅ | | u | | ∞ for i = 1, 2, 3, and | | K [ u ] ( 4 ) | | ∞ < ( ( τ + 1 ) α 4 / k ) ⋅ | | u | | ∞ .

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 z = y + 3 i π 2 2 α , 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.