### Abstract

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

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

##### Keywords:

Infinite beam; elastic foundation; nonlinear; non-uniform; fourth-order ordinary differential equation; Banach fixed point theorem; contraction.### 1 Introduction

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

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

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

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

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

### 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*):

**Figure 1.** **Infinite beam on nonlinear and non-uniform 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 *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/d*t*(*m*(*x*)d*u*/d*t*) 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
*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
*k*, which is uniform in *x*. The term
*u *and the linear deflection

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

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
*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 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
*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 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 *η*_{0 }≈ 0.123.

Using the function *N*, we define the operator
*u *: ℝ → ℝ. Note that

**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).
**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,

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, *ρ*_{1 }is non-negative and non-decreasing.

We want to show *ρ*_{1 }is continuous. Fix *t*_{0 }≥ 0. We first show the left-continuity of *ρ*_{1 }at *t*_{0}. Let
*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 non-decreasing, it becomes constant on [*t'*, *t*_{0}], and hence *ρ*_{1 }is clearly left-continuous 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
*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 left-continuous at *t*_{0}.

Suppose *ρ*_{1 }is not right-continuous at *t*_{0}. Then there exist *ϵ *> 0 and a sequence
*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 right-continuous 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
*t*_{0}, ∞) ∪ (-∞, -*t*_{0})} × ℝ, such that *t*_{0 }< |*μ*_{n}| ≤ *t*_{n }and
*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 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 *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 non-decreasing 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 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

where *l*(*B*) is the Lebesque measure, and hence the length in our case, of the set *B *⊂ [0, ∞). From the definition of
*t *≥ 0. We omit the proof that

*Example *1. Let

Then,

and hence,

We also have

Thus, we can take *ρ*(*t*) = *ρ*_{2}(*t*) = 2(2*n *+ 1)*λt*^{2n}.

*Example *2. Let

Then,

and hence,

We also have

Thus, we can take *ρ*(*t*) = *ρ*_{2}(*t*) = 2*aλ *{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
K

Let

so that *G*(*x*, *ξ*) = *K *(|*ξ *- *x*|) for *G *in (7). Using the function *K*, we define the linear operator

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

**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
*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
*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,
*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
*i *= 0, 1, 2, 3, 4 by Lemma 4 (a). So it is sufficient to show that
*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

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
*y *≥ *x *+ *M*/2, then *x *- *y *≤ - *M*/2, and we also 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
*i *= 0, 1, 2, 3. We omit the similar proof that
*i *= 0, 1, 2, 3. It follows that

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

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

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
**W2 **should be meant to be taken in the range
*ρ*^{-1}(*s*) · (*σk *- *s*) is continuous. In fact, we have
**W2 **that there exists

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 *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(2*n *+ 1)*λt*^{2n}, and hence
*ϕ*(*s*) = *ρ*^{-1}(*s*) · (*σk *- *s*). Since

*ϕ *is strictly increasing on
*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,
*X*, if we take *s*_{* }to be the larger among them.

*Example *5. Consider the case

in Example 2. Then we have *ρ*(*t*) = 2*aλ *{exp(*at*) - 1}, and hence
*ϕ*(*s*) = *ρ*^{-1}(*s*) · (*σk *- *s*), we have

It follows that *ϕ *is strictly increasing on
*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
**W1 **and Lemma 5. Thus, we have
*u *∈ *C*_{0}(ℝ). In short, the operator Ψ is a well-defined 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

Applying the operator

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
**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