Electronic Journal of Qualitative Theory of Differential Equations 2003, No. 10, 1-9; http://www.math.u-szeged.hu/ejqtde/

THE GENERALIZED METHOD OF QUASILINEARIZATION AND NONLINEAR BOUNDARY VALUE PROBLEMS WITH INTEGRAL

BOUNDARY CONDITIONS

RAHMAT ALI KHAN

Abstract. The generalized method of quasilinearization is applied to obtain a monotone sequence of iterates converging uniformly and rapidly to a solution of second order nonlinear boundary value problem with nonlinear integral boundary conditions.

1. Introduction

In this paper, we shall study the method of quasilinearization for the nonlinear boundary value problem with integral boundary conditions

\begin{matrix}x′′(t) = f(t, x), t ∈ J = [0, 1], \\ x(0) − k_{1}x′(0) = ∫_{0}^{1} h_{1}(x(s))ds, \\ x(1) + k_{2}x′(1) = ∫_{0}^{1} h_{2}(x(s))ds, \end{matrix} ; (1.1)

where f : J × ℝ → ℝ and h_{i} : ℝ → ℝ (i = 1, 2) are continuous functions and k_{i} are nonnegative constants. Boundary value problems with integral boundary conditions constitute a very interesting and important class of problems. They include two, three, multipoint and nonlocal boundary value problems as special cases. For boundary value problems with integral boundary conditions and comments on their importance, we refer the reader to the papers [13, 14, 15] and the references therein. Moreover, boundary value problems with integral boundary conditions have been studied by a number of authors, for example [11, 12, 16, 17].

The purpose of this paper is to develop the method of quasilinearization for the boundary value problem (1.1). The main idea of the method of quasilinearization as developed by Bellman and Kalaba [1] and generalized by Lakshmikantham [4, 5] has been applied to a variety of problems [3, 6, 7]. Recently, Eloe and Gao [8], Ahmad, Khan and Eloe [9] have developed the quasilinearization method for three point boundary value problems. More recently, Khan, Ahmad [10] developed the method to treat first order problems

EJQTDE, 2003 No. 10, p. 1 with integral boundary conditions

2000 Mathematics Subject Classification. Primary 34A45, 34B15, secondary .

Key words and phrases. Quasilinearization, integral boundary value problem, upper and lower solutions, rapid convergence Financial support by the MoST Govt. of Pakistan is gratefully acknowledged.

\begin{matrix}x′(t) = f(t, x(t)), t ∈ [0, T] \\ x(0) = ax(T) + ∫_{0}^{T} b(s)x(s)ds + k = Bx + k. \end{matrix}

In the present paper we extend the method of generalized quasilinearization to the boundary value problem (1.1) and we obtain a sequence of solutions converging uniformly and rapidly to a solution of the problem.

2. Preliminaries

We know that the homogeneous problem

\begin{matrix}x′′(t) = 0, t ∈ J = [0, 1], \\ x(0) − k_{1}x′(0) = 0, x(1) + k_{2}x′(1) = 0, \end{matrix}

has only the trivial solution. Consequently, for any σ(t), ρ_{1}(t), ρ_{2}(t) ∈ C[0, 1], the corresponding nonhomogeneous linear problem

\begin{matrix}x′′(t) = σ(t), t ∈ J = [0, 1], \\ x(0) − k_{1}x′(0) = ∫_{0}^{1} ρ_{1}(s)ds, x(1) + k_{2}x′(1) = ∫_{0}^{1} ρ_{2}(s)ds, \end{matrix}

has a unique solution x ∈ C^{2}[0, 1],

x(t) = P (t) + ∫_{0}^{1} G(t, s)σ(s)ds,

where

P (t) = \frac{1}{1 + k_{1} + k_{2}} \{(1 − t + k_{2}) ∫_{0}^{1} ρ_{1}(s)ds + (k_{1} + t) ∫_{0}^{1} ρ_{2}(s)ds\}

is the unique solution of the problem

x′′(t) = 0, t ∈ J = [0, 1],
x(0) − k_{1}x′(0) = ∫_{0}^{1} ρ_{1}(s)ds,
x(1) + k_{2}x′(1) = ∫_{0}^{1} ρ_{2}(s)ds,

and

G(t, s) = \frac{−1}{k_{1} + k_{2} + 1} \begin{cases}(k_{1} + t)(1 − s + k_{2}), & 0 ≤ t < s ≤ 1 \\ (k_{1} + s)(1 − t + k_{2}), & 0 ≤ s < t ≤ 1 \end{cases}

is the Green’s function of the problem. We note that G(t, s) < 0 on (0, 1) × (0, 1). EJQTDE, 2003 No. 10, p. 2 Definition 2.1. Let α, β ∈ C^{2}[0, 1]. We say that α is a lower solution of (1.1) if

α′′ ≥ f(t, α(t)), t ∈ [0, 1]
α(0) − k_{1}α′(0) ≤ ∫_{0}^{1} h_{1}(α(s))ds,
α(1) + k_{2}α′(1) ≤ ∫_{0}^{1} h_{2}(α(s))ds.

Similarly, β is an upper solution of the BVP (1.1), if β satisfies similar inequalities in the reverse direction.

Now, we state and prove the existence and uniqueness of solutions in an ordered interval generated by the lower and upper solutions of the boundary value problem.

Theorem 2.2. Assume that α and β are respectively lower and upper solutions of (1.1) such that α(t) ≤ β(t), t ∈ [0, 1]. If f : [0, 1] × ℝ → ℝ and h_{i} : ℝ → ℝ (i = 1, 2) are continuous and h′_{i}(x) ≥ 0, then there exists a solution x(t) of the boundary value problem (1.1) such that

α(t) ≤ x(t) ≤ β(t), t ∈ [0, 1].

Proof. Define the following modifications of f(t, x) and h_{i}(x)(i = 1, 2)

F (t, x) = \begin{cases}f(t, β(t)) + \frac{x − β(t)}{1 + |x − β|}, & if x > β, \\ f(t, x), & if α ≤ x ≤ β, \\ f(t, α) + \frac{x − α(t)}{1 + |x − α|}, & if x < α. \end{cases}

and

H_{i}(x) = \begin{cases}h_{i}(β(t)), & x > β(t), \\ h_{i}(x), & α(t) ≤ x ≤ β(t), \\ h_{i}(α(t)), & x < α(t). \end{cases}

Consider the modified problem

\begin{matrix}x′′(t) = F (t, x), t ∈ J = [0, 1], \\ x(0) − k_{1}x′(0) = ∫_{0}^{1} H_{1}(x(s)) ds, x(1) + k_{2}x′(1) = ∫_{0}^{1} H_{2}(x(s)) ds. \end{matrix} ; (2.1)

Since F (t, x) : [0, 1] × ℝ → ℝ and H_{i} : ℝ → ℝ are continuous and bounded, it follows that the boundary value problem (2.1) has a solution. Further, note that

α′′(t) ≥ f(t, α(t)) = F (t, α(t)), t ∈ [0, 1]

α(0) − k_{1}α′(0) ≤ ∫_{0}^{1} h_{1}(α(s))ds = ∫_{0}^{1} H_{1}(α(s))ds,
α(1) + k_{2}α′(1) ≤ ∫_{0}^{1} h_{2}(α(s))ds = ∫_{0}^{1} H_{2}(α(s))ds

and

β′′(t) ≤ f(t, β(t)) = F (t, β(t)), t ∈ [0, 1]
β(0) − k_{1}β′(0) ≥ ∫_{0}^{1} h_{1}(β(s))ds = ∫_{0}^{1} H_{1}(β(s))ds,
β(1) + k_{2}β′(1) ≥ ∫_{0}^{1} h_{2}(β(s))ds = ∫_{0}^{1} H_{2}(β(s))ds

which imply that α and β are respectively lower and upper solutions of (2.1). Also, we note that any solution of (2.1) which lies between α and β, is a solution of (1.1). Thus, we only need to show that any solution x(t) of (2.1) is such that α(t) ≤ x(t) ≤ β(t), t ∈ [0, 1]. Assume that α(t) ≤ x(t) is not true on [0, 1]. Then the function k(t) = α(t) − x(t) has a positive maximum at some t = t_{0} ∈ [0, 1]. If t_{0} ∈ (0, 1), then

k(t_{0}) > 0, k′(t_{0}) = 0, k′′(t_{0}) ≤ 0

and hence

0 ≥ k′′(t_{0}) = α′′(t_{0}) − x′′(t_{0}) ≥ f(t_{0}, α(t_{0})) − (f(t_{0}, α(t_{0})) + \frac{x(t_{0}) − α(t_{0})}{1 + |x(t_{0}) − α(t_{0})|}) > 0,

a contradiction. If t_{0} = 0, then k(0) > 0 and k′(0) ≤ 0, but then the boundary conditions and the nondecreasing property of h_{i} gives

\begin{aligned}k(0) & ≤ k_{1}k′(0) + ∫_{0}^{1}[h_{1}(α(s)) − H_{1}(x(s))]ds \\ & ≤ ∫_{0}^{1}[h_{1}(α(s)) − H_{1}(x(s))]ds. \end{aligned}

If x < α(t), then H_{1}(x(s)) = h_{1}(α(s)) and hence k(0) ≤ 0, a contradiction. If x > β(t), then H_{1}(x(s)) = h_{1}(β(s)) ≥ h_{1}(α(s)) which implies k(0) ≤ 0, a contradiction. Hence α(t) ≤ x(t) ≤ β(t) and H_{1}(x(s)) = h_{1}(x(s)) ≥ h_{1}(α(s)) and again k(0) ≤ 0, another contradiction. Similarly, if t_{0} = 1, we get a contradiction. Thus α(t) ≤ x(t), t ∈ J. Similarly, we can show that x(t) ≤ β(t), t ∈ [0, 1]. □

Theorem 2.3. Assume that α and β are lower and upper solutions of the boundary value problem (1.1) respectively. If f : [0, 1] × ℝ → ℝ and h_{i} : ℝ → ℝ are continuous, f_{x}(t, x) > 0 for t ∈ [0, 1], x ∈ ℝ and 0 ≤ h′_{i}(x) < 1. Then α(t) ≤ β(t), t ∈ [0, 1].

EJQTDE, 2003 No. 10, p. 4 Proof. Define m(t) = α(t) − β(t), t ∈ [0, 1], then m(t) ∈ C^{2}[0, 1] and

\begin{matrix}m(0) − k_{1}m′(0) ≤ ∫_{0}^{1}[h_{1}(α(s)) − h_{1}(β(s))]ds \\ m(1) + k_{2}m′(1) ≤ ∫_{0}^{1}[h_{2}(α(s)) − h_{2}(β(s))]ds. \end{matrix} ; (2.2)

Assume that m(t) ≤ 0 is not true for t ∈ [0, 1]. Then m(t) has a positive maximum at some t_{0} ∈ [0, 1]. If t_{0} ∈ (0, 1), then m(t_{0}) > 0, m′(t_{0}) = 0 and m′′(t_{0}) ≤ 0. Using the increasing property of the function f(t, x) in x, we obtain

f(t_{0}, α(t_{0})) ≤ α′′(t_{0}) ≤ β′′(t_{0}) ≤ f(t_{0}, β(t_{0})) < f(t_{0}, α(t_{0})),

a contradiction. If t_{0} = 0, then m(0) > 0 and m′(0) ≤ 0. On the other hand, using the boundary conditions (2.2) and the assumption 0 ≤ h′_{1}(x) < 1, we have

\begin{aligned}m(0) ≤ m(0) − k_{1}m′(0) & ≤ ∫_{0}^{1}[h_{1}(α(s)) − h_{1}(β(s))]ds ≤ ∫_{0}^{1} h′_{1}(c)m(s)ds \\ & ≤ h′_{1}(c) max_{t∈[0,1]} m(t) = h′_{1}(c)m(0) < m(0), \end{aligned} ; (2.3)

a contradiction. If t_{0} = 1, then m(1) > 0 and m′(1) ≥ 0. But again, the boundary conditions (2.2) and the assumption 0 ≤ h′_{2}(x) < 1, gives

\begin{aligned}m(1) ≤ m(0) + k_{2}m′(1) & ≤ ∫_{0}^{1}[h_{2}(α(s)) − h_{2}(β(s))]ds ≤ ∫_{0}^{1} h′_{2}(c)m(s)ds \\ & ≤ h′_{2}(c) max_{t∈[0,1]} m(t) = h′_{2}(c)m(1) < m(1), \end{aligned} ; (2.4)

a contradiction. Hence

α(t) ≤ β(t), t ∈ [0, 1]
□

As a consequence of the theorem (2.3), we have

Corollary 2.4. Assume that α and β are lower and upper solutions of the boundary value problem (1.1) respectively. If f : [0, 1] × ℝ → ℝ and h : ℝ → ℝ are continuous, f_{x}(t, x) > 0 and 0 ≤ h′(x) < 1, for t ∈ [0, 1], x ∈ ℝ. Then the solution of the boundary value problem (1.1) is unique.

3. Quasilinearization Technique

Theorem 3.1. Assume that (A_{1}) α and β ∈ C^{2}[0, 1] are respectively lower and upper solutions of (1.1) such that

α(t) ≤ β(t), t ∈ [0, 1]. EJQTDE, 2003 No. 10, p. 5 (A_{2}) f(t, x) ∈ C^{2}[0, 1] × ℝ is such that f_{x}(t, x) > 0 and f_{xx}(t, x) + ϕ_{xx}(t, x) ≤ 0, where

ϕ(t, x) ∈ C^{2}[0, 1] × ℝ and ϕ_{xx}(t, x) ≤ 0. (A_{3}) h_{i} ∈ C^{2}(ℝ) (i = 1, 2) are nondecreasing, 0 ≤ h′_{i}(x) < 1 and h′′_{i}(x) ≥ 0. Then, there exists a monotone sequence {w_{n}} of solutions converging uniformly and quadratically to the unique solution of the problem.

Proof. Define, F : [0, 1] × ℝ → ℝ by F (t, x) = f(t, x) + ϕ(t, x). Then in view of (A_{2}), we note that F ∈ C^{2}[0, 1] × ℝ and

(3.1)  F_{xx}(t, x) ≤ 0.

For any t ∈ [0, 1], using Taylor’s theorem, (3.1) and (A_{3}), we have

(3.2)  f(t, x) ≤ F (t, y) + F_{x}(t, y)(x − y) − ϕ(t, x)

and

(3.3)  h_{i}(x) ≥ h_{i}(y) + h′_{i}(y)(x − y),

where x, y ∈ ℝ. Again applying Taylor’s theorem to ϕ(t, x), we can find ξ ∈ ℝ with y ≤ ξ ≤ x such that

(3.4)  ϕ(t, x) = ϕ(t, y) + ϕ_{x}(t, y)(x − y) + \frac{1}{2}ϕ_{xx}(t, ξ)(x − y)^{2},

which in view of (A_{2}) implies that

(3.5)  ϕ(t, x) ≤ ϕ(t, y) + ϕ_{x}(t, y)(x − y)

and

(3.6)  ϕ(t, x) ≥ ϕ(t, y) + ϕ_{x}(t, y)(x − y) − \frac{1}{2}|ϕ_{xx}(t, ξ)|∥x − y∥^{2},

where ∥x − y∥ = max_{t∈[0,1]}{|x(t) − y(t)|} denotes the supremum norm in the space of continuous functions on [0, 1]. Using (3.6) in (3.2), we obtain

(3.7)  f(t, x) ≤ f(t, y) + f_{x}(t, y)(x − y) + \frac{1}{2}|ϕ_{xx}(t, ξ)|∥x − y∥^{2}.

Let Ω = {(t, x) : t ∈ [0, 1], x ∈ [α, β]} and define on Ω

(3.8)  g(t, x, y) = f(t, y) + f_{x}(t, y)(x − y) + \frac{1}{2}|ϕ_{xx}(t, ξ)|∥x − y∥^{2}

and

(3.9)  H_{i}(x, y) = h_{i}(y) + h′_{i}(y)(x − y).

EJQTDE, 2003 No. 10, p. 6 Note that g(t, x, y) and H_{i}(x, y) are continuous, bounded and are such that g_{x}(t, x, y) = f_{x}(t, y) > 0 and 0 ≤ \frac{∂}{∂x}H_{i}(x, y) < 1. Further, from \{(3.7), (3.8)\} and \{(3.3), (3.9)\}, we have the relations

(3.10)  \begin{cases}f(t, x) ≤ g(t, x, y) \\ f(t, x) = g(t, x, x) \end{cases}

and

(3.11)  \begin{cases}h_{i}(x) ≥ H_{i}(x, y) \\ h_{i}(x) = H_{i}(x, x) \end{cases}

Now, set w_{0} = α and consider the linear problem

\begin{matrix}x′′(t) = g(t, x, w_{0}), t ∈ [0, 1], \\ x(0) − k_{1}x′(0) = ∫_{0}^{1} H_{1}(x(s), w_{0}(s))ds, \\ x(1) + k_{2}x′(1) = ∫_{0}^{1} H_{2}(x(s), w_{0}(s))ds. \end{matrix} ; (3.12)

Using (A_{1}), (3.10) and (3.11), we obtain

w′′_{0}(t) ≥ f(t, w_{0}) = g(t, w_{0}, w_{0}), t ∈ [0, 1],
w_{0}(0) − k_{1}w′_{0}(0) ≤ ∫_{0}^{1} h_{1}(w_{0}(s))ds = ∫_{0}^{1} H_{1}(w_{0}(s), w_{0}(s))ds,
w_{0}(1) + k_{2}w′_{0}(1) ≤ ∫_{0}^{1} h_{2}(w_{0}(s))ds = ∫_{0}^{1} H_{2}(w_{0}(s), w_{0}(s))ds

and

β′′(t) ≤ f(t, β) ≤ g(t, β, w_{0}), t ∈ [0, 1],
β(0) − k_{1}β′(0) ≥ ∫_{0}^{1} h_{1}(β(s))ds ≥ ∫_{0}^{1} H_{1}(β(s), w_{0}(s))ds,
β(1) + k_{2}β′(1) ≥ ∫_{0}^{1} h_{2}(β(s))ds ≥ ∫_{0}^{1} H_{2}(β(s), w_{0}(s))ds,

which imply that w_{0} and β are respectively lower and upper solutions of (3.12). It follows by theorems 2.2 and 2.3 that there exists a unique solution w_{1} of (3.12) such that

w_{0}(t) ≤ w_{1}(t) ≤ β(t), t ∈ [0, 1].

In view of (3.10), (3.11) and the fact that w_{1} is a solution of (3.12), we note that w_{1} is a lower solution of (1.1).

EJQTDE, 2003 No. 10, p. 7 Now consider the problem

\begin{matrix}x′′(t) = g(t, x, w_{1}), t ∈ [0, 1], \\ x(0) − k_{1}x′(0) = ∫_{0}^{1} H_{1}(x(s), w_{1}(s))ds, \\ x(1) + k_{2}x′(1) = ∫_{0}^{1} H_{2}(x(s), w_{1}(s))ds. \end{matrix} ; (3.13)

Again we can show that w_{1} and β are lower and upper solutions of (3.13) and hence by theorems (2.2, 2.3), there exists a unique solution w_{2} of (3.13) such that

w_{1}(t) ≤ w_{2}(t) ≤ β(t), t ∈ [0, 1].

Continuing this process, we obtain a monotone sequence {w_{n}} of solutions satisfying

w_{0}(t) ≤ w_{1}(t) ≤ w_{2}(t) ≤ ...w_{n}(t) ≤ β(t), t ∈ [0, 1]

where, the element w_{n} of the sequence {w_{n}} is a solution of the boundary value problem

x′′(t) =g(t, x, w_{n−1}), t ∈ [0, 1],
x(0) − k_{1}x′(0) = ∫_{0}^{1} H_{1}(x(s), w_{n−1}(s))ds,
x(1) + k_{2}x′(1) = ∫_{0}^{1} H_{2}(x(s), w_{n−1}(s))ds

and

(3.14)  w_{n}(t) = P_{n}(t) + ∫_{0}^{1} G(t, s)g(s, w_{n}(s), w_{n−1}(s))ds,

where

(3.15) P_{n}(t) = \frac{1}{1 + k_{1} + k_{2}} \{(1 − t + k_{2}) ∫_{0}^{1} H_{1}(w_{n}(s), w_{n−1}(s)) ds

+ (k_{1} + t) ∫_{0}^{1} H_{2}(w_{n}(s), w_{n−1}(s)) ds}.

Employing the standard arguments [2], it follows that the convergence of the sequence is uniform. If x(t) is the limit point of the sequence, passing to the limit as n → ∞, (3.14) gives

x(t) = P (t) + ∫_{0}^{1} G(t, s)f(s, x(s))ds,

where

P (t) = \frac{1}{1 + k_{1} + k_{2}} \{(1 − t + k_{2}) ∫_{0}^{1} h_{1}(x(s))ds + (k_{1} + t) ∫_{0}^{1} h_{2}(x(s))ds\};

that is, x(t) is a solution of the boundary value problem (1.1).

Now, we show that the convergence of the sequence is quadratic. For that, set e_{n}(t) = x(t) − w_{n}(t), t ∈ [0, 1]. Note that, e_{n}(t) ≥ 0, t ∈ [0, 1]. Using Taylor’s theorem and (3.9), we obtain

\begin{aligned}e_{n}(0) − k_{1}e′_{n}(0) & = ∫_{0}^{1}[h_{1}(x(s)) − H_{1}(w_{n}(s), w_{n−1}(s)]ds \\ & = ∫_{0}^{1}[h′_{1}(w_{n−1}(s))e_{n}(s) + \frac{1}{2}h′′_{1}(ξ_{1})e_{n−1}^{2}(s)]ds \end{aligned}

and

\begin{aligned}e_{n}(1) + k_{2}e′_{n}(1) & = ∫_{0}^{1}[h_{2}(x(s)) − H_{2}(w_{n}(s), w_{n−1}(s))]ds \\ & = ∫_{0}^{1}[h′_{2}(w_{n−1}(s))e_{n}(s) + \frac{1}{2}h′′_{2}(ξ_{2})e_{n−1}^{2}(s)]ds \end{aligned}

where, w_{n−1} ≤ ξ_{1}, ξ_{2} ≤ x. In view of (A_{3}), there exist λ_{i} < 1 and C_{i} ≥ 0 such that h′_{i}(w_{n−1}(s)) ≤ λ_{i} and \frac{1}{2}h′′_{i}(ξ_{1}) ≤ C_{i} (i = 1, 2). Let λ(< 1) = max\{λ_{1}, λ_{2}\} and C(≥ 0) = max{C_{1}, C_{2}}, then

\begin{matrix}e_{n}(0) − k_{1}e′_{n}(0) ≤ λ ∫_{0}^{1} e_{n}(s)ds + C ∫_{0}^{1} e_{n−1}^{2}(s)ds ≤ λ ∫_{0}^{1} e_{n}(s)ds + C∥e_{n−1}∥^{2} \\ e_{n}(1) + k_{2}e′_{n}(1) ≤ λ ∫_{0}^{1} e_{n}(s)ds + C ∫_{0}^{1} e_{n−1}^{2}(s)ds ≤ λ ∫_{0}^{1} e_{n}(s)ds + C∥e_{n−1}∥^{2}. \end{matrix} ; (3.16)

Further, using Taylor’s theorem, (A_{2}), (3.5) and (3.9), we obtain

\begin{aligned}e′′_{n}(t) & = x′′(t) − w′′_{n}(t) = (F (t, x) − ϕ(t, x)) \\ & − [f(t, w_{n−1}) + f_{x}(t, w_{n−1})(w_{n} − w_{n−1}) + \frac{1}{2}|ϕ_{xx}(t, ξ)|∥w_{n} − w_{n−1}∥^{2}] \\ & ≥ f_{x}(t, w_{n−1})e_{n}(t) + F_{xx}(t, ξ_{3})\frac{e_{n−1}^{2}(t)}{2!} − \frac{1}{2}|ϕ_{xx}(t, ξ)|∥e_{n−1}∥^{2} \\ & ≥ −\frac{1}{2}(|F_{xx}(t, ξ_{3})| + |ϕ_{xx}(t, ξ)|)∥e_{n−1}∥^{2} \\ & ≥ −M∥e_{n−1}∥^{2}, \end{aligned} ; (3.17)

where w_{n−1} ≤ ξ_{3} ≤ x, w_{n−1} ≤ ξ ≤ w_{n}, |F_{xx}| ≤ M_{1}, |ϕ_{xx}| ≤ M_{2} and 2M = M_{1} + M_{2}. From (3.16) and (3.17), it follows that

e_{n}(t) ≤ r(t) on [0, 1],

EJQTDE, 2003 No. 10, p. 9 where, r(t) ≥ 0 is the unique solution of the boundary value problem

r′′(t) = −M∥e_{n−1}∥^{2}, t ∈ [0, 1]
r(0) − k_{1}r′(0) = λ ∫_{0}^{1} r(s)ds + C∥e_{n−1}∥^{2}
r(1) + k_{2}r′(1) = λ ∫_{0}^{1} r(s)ds + C∥e_{n−1}∥^{2}.

Thus, r(t) =

= \frac{1}{1 + k_{1} + k_{2}} [(1 − t + k_{2})(λ ∫_{0}^{1} r(s)ds + C∥e_{n−1}∥^{2})

+ (t + k_{1})(λ ∫_{0}^{1} r(s)ds + C∥e_{n−1}∥^{2})] − M ∫_{0}^{1} G(t, s)∥e_{n−1}∥^{2}ds

≤ \frac{1}{1 + k_{1} + k_{2}} [λ\{(1 − t + k_{2}) + (t + k_{1})\}∥r∥ + C\{(1 − t + k_{2}) + (t + k_{1})\}∥e_{n−1}∥^{2})

+ M∥e_{n−1}∥^{2} ∫_{0}^{1} |G(t, s)|ds

= λ∥r∥ + C∥e_{n−1}∥^{2} + Ml∥e_{n−1}∥^{2} = λ∥r∥ + L∥e_{n−1}∥^{2},

where l is a bound for ∫_{0}^{1} |G(t, s)|ds and L = C + lM. Taking the maximum over [0, 1], we get

∥r∥ ≤ δ∥e_{n−1}∥^{2},

where, δ = \frac{L}{1−λ}. □

4. Rapid convergence

Theorem 4.1. Assume that (B_{1}) α, β ∈ C^{2}[0, 1] are lower and upper solutions of (1.1) respectively such that α(t) ≤

β(t), t ∈ [0, 1]. (B_{2}) f(t, x) ∈ C^{k}[[0, 1] × ℝ] such that \frac{∂^{j}}{∂x^{j}} f(t, x) ≥ 0 (j = 1, 2, 3..., k − 1), and

\frac{∂^{k}}{∂x^{k}} (f(t, x) + ϕ(t, x)) ≤ 0, where, ϕ ∈ C^{k}[[0, 1] × ℝ] and \frac{∂^{k}}{∂x^{k}} ϕ(t, x) ≤ 0, (B_{3}) h_{j}(x) ∈ C^{k}[ℝ] such that \frac{d^{i}}{dx^{i}} h_{j}(x) ≤ \frac{M}{(β−α)^{i−1}} (i = 1, 2, ..., k − 1) and \frac{d^{k}}{dx^{k}} h_{j}(x) ≥ 0,

where M < 1/3 and j = 1, 2. Then, there exists a monotone sequence {w_{n}} of solutions converging uniformly to the unique solution of the problem. Moreover the rate of convergence is of order k ≥ 2.

Proof. Define, F : [0, 1] × ℝ → ℝ by F (t, x) = f(t, x) + ϕ(t, x), t ∈ [0, 1], then in view of (B_{2}), we note that F ∈ C^{k}[[0, 1] × ℝ] and

(4.1)  \frac{∂^{k}}{∂x^{k}} F (t, x) ≤ 0.

EJQTDE, 2003 No. 10, p. 10 Using (B_{3}), Taylor’s theorem and (4.1), we have

(4.2)  f(t, x) ≤ ∑_{i=0}^{k−1} \frac{∂^{i}}{∂x^{i}} F (t, y)\frac{(x − y)^{i}}{i!} − ϕ(t, x)

and

(4.3)  h_{j}(x) ≥ ∑_{i=0}^{k−1} \frac{d^{i}}{dx^{i}} h_{j}(y)\frac{(x − y)^{i}}{i!} .

Expanding ϕ(t, x) about (t, y) by Taylor’s theorem, we can find y ≤ ξ ≤ x, such that

(4.4)  ϕ(t, x) = ∑_{i=0}^{k−1} \frac{∂^{i}}{∂x^{i}} ϕ(t, y)\frac{(x − y)^{i}}{i!} + \frac{∂^{k}}{∂x^{k}} ϕ(t, ξ)\frac{(x − y)^{k}}{k!} ,

which in view of (B_{2}) implies that

(4.5)  ϕ(t, x) ≤ ∑_{i=0}^{k−1} \frac{∂^{i}}{∂x^{i}} ϕ(t, y)\frac{(x − y)^{i}}{i!} .

Using (4.4) in (4.2), we obtain

(4.6)  f(t, x) ≤ ∑_{i=0}^{k−1} \frac{∂^{i}}{∂x^{i}} f(t, y)\frac{(x − y)^{i}}{i!} − \frac{∂^{k}}{∂x^{k}} ϕ(t, ξ)\frac{(x − y)^{k}}{k!} .

Let Ω = {(t, x) : t ∈ [0, 1], x ∈ [α, β]} and define on Ω the functions

(4.7)  g^{∗}(t, x, y) = ∑_{i=0}^{k−1} \frac{∂^{i}}{∂x^{i}} f(t, y)\frac{(x − y)^{i}}{i!} − \frac{∂^{k}}{∂x^{k}} ϕ(t, ξ)\frac{(x − y)^{k}}{k!}

and

(4.8)  H_{j}^{∗}(x, y) = ∑_{i=0}^{k−1} \frac{d^{i}}{dx^{i}} h_{j}(y)\frac{(x − y)^{i}}{i!} .

Then, we note that g^{∗}(t, x, y) and H_{j}^{∗}(x, y) are continuous, bounded and are such that

g_{x}^{∗}(t, x, y) = ∑_{i=1}^{k−1} \frac{∂^{i}}{∂x^{i}} f(t, y)\frac{(x − y)^{i−1}}{(i − 1)!} − \frac{∂^{k}}{∂x^{k}} ϕ(t, ξ)\frac{(x − y)^{k−1}}{(k − 1)!} ≥ 0

and

\frac{∂}{∂x}H_{j}^{∗}(x, y) = ∑_{i=1}^{k−1} \frac{d^{i}}{dx^{i}} h_{j}(y)\frac{(x − y)^{i−1}}{(i − 1)!}

≤ ∑_{i=1}^{k−1} \frac{M}{(β − α)^{i−1}} \frac{(β − α)^{i−1}}{(i − 1)!} ≤ M(3 − \frac{1}{2^{k−2}} ) < 1.

Further, from {(4.6),(4.7) } and {(4.3),(4.8) }, we have the relations

(4.9)  \begin{cases}f(t, x) ≤ g^{∗}(t, x, y), \\ f(t, x) = g^{∗}(t, x, x) \end{cases}

and

(4.10) \begin{cases}h_{j}(x) ≥ H_{j}^{∗}(x, y), \\ h_{j}(x) = H_{j}^{∗}(x, x). \end{cases}

Now, set α = w_{0} and consider the linear problem

\begin{matrix}x′′(t) = g^{∗}(t, x, w_{0}), t ∈ [0, 1], \\ x(0) − k_{1}x′(0) = ∫_{0}^{1} H_{1}^{∗}(x(s), w_{0}(s))ds, \\ x(1) + k_{2}x′(1) = ∫_{0}^{1} H_{2}^{∗}(x(s), w_{0}(s))ds. \end{matrix} ; (4.11)

The assumption (B_{1}) and the expressions (4.9), (4.10) yields w′′_{0}(t) ≥ f(t, w_{0}) = g^{∗}(t, w_{0}, w_{0}), t ∈ [0, 1],

w_{0}(0) − k_{1}w′_{0}(0) ≤ ∫_{0}^{1} h_{1}(w_{0}(s))ds = ∫_{0}^{1} H_{1}^{∗}(w_{0}(s), w_{0}(s))ds,
w_{0}(1) + k_{2}w′_{0}(1) ≤ ∫_{0}^{1} H_{2}^{∗}(w_{0}(s))ds = ∫_{0}^{1} H_{2}^{∗}(w_{0}(s), w_{0}(s))ds

and

β′′(t) ≤ f(t, β) ≤ g(t, β, w_{0}), t ∈ [0, 1],
β(0) − k_{1}β′(0) ≥ ∫_{0}^{1} h_{1}(β(s))ds ≥ ∫_{0}^{1} H_{1}^{∗}(β(s), w_{0}(s))ds,
β(1) + k_{2}β′(1) ≥ ∫_{0}^{1} h_{2}(β(s))ds ≥ ∫_{0}^{1} H_{2}^{∗}(β(s), w_{0}(s))ds,

imply that w_{0} and β are respectively lower and upper solutions of (4.11). Hence by theorems (2.2, 2.3), there exists a unique solution w_{1} of (4.11) such that

w_{0}(t) ≤ w_{1}(t) ≤ β(t), t ∈ [0, 1].

Continuing this process, we obtain a monotone sequence {w_{n}} of solutions satisfying

w_{0}(t) ≤ w_{1}(t) ≤ w_{2}(t) ≤ ...w_{n}(t) ≤ β(t), t ∈ [0, 1],

where the element w_{n} of the sequence {w_{n}} is a solution of the boundary value problem

x′′(t) =g^{∗}(t, x, w_{n−1}), t ∈ [0, 1],
x(0) − k_{1}x′(0) = ∫_{0}^{1} H_{1}^{∗}(x(s), w_{n−1}(s))ds,
x(1) + k_{2}x′(1) = ∫_{0}^{1} H_{2}^{∗}(x(s), w_{n−1}(s))ds.

EJQTDE, 2003 No. 10, p. 12 By the same process as in theorem (3.1), we can show that the sequence converges uniformly to the unique solution of (1.1). Now, we show that the convergence of the sequence is of order k ≥ 2. For that, set

e_{n}(t) = x(t) − w_{n}(t) and a_{n}(t) = w_{n+1}(t) − w_{n}(t), t ∈ [0, 1].

Then,

e_{n}(t) ≥ 0, a_{n} ≥ 0, e_{n+1} = e_{n} − a_{n}, e_{n}^{i} ≥ a_{n}^{i} (i = 1, 2...)

and

\begin{matrix}e_{n}(0) − k_{1}e′_{n}(0) = ∫_{0}^{1}[h_{1}(x(s)) − H_{1}^{∗}(w_{n}(s), w_{n−1}(s)]ds \\ e_{n}(1) + k_{2}e′_{n}(1) = ∫_{0}^{1}[h_{2}(x(s)) − H_{2}^{∗}(w_{n}(s), w_{n−1}(s)]ds. \end{matrix} ; (4.12)

Using Taylor’s theorem and (4.8), we obtain

h_{j}(x(s)) − H_{j}^{∗}(w_{n}(s), w_{n−1}(s) = ∑_{i=0}^{k−1} \frac{d^{i}}{dx^{i}} h_{j}(w_{n−1})\frac{(x − w_{n−1})^{i}}{i!} + \frac{d^{k}}{dx^{k}} h_{j}(c)\frac{(x − w_{n−1})^{k}}{k!}

− ∑_{i=0}^{k−1} \frac{d^{i}}{dx^{i}} h_{j}(w_{n−1})\frac{(w_{n} − w_{n−1})^{i}}{i!}

= (∑_{i=1}^{k−1} \frac{d^{i}}{dx^{i}} h_{j}(w_{n−1})\frac{1}{i!} ∑_{l=0}^{i−1} e_{n−1}^{i−1−l}a_{n−1}^{l})e_{n} + \frac{d^{k}}{dx^{k}} h_{j}(c)\frac{e_{n−1}^{k}}{k!}

≤ p_{j}(t)e_{n}(t) + \frac{M}{γ^{k−1}} \frac{e_{n−1}^{k}}{k!} ≤ p_{j}(t)e_{n}(t) + \frac{M}{γ^{k−1}} \frac{∥e_{n−1}^{k}∥}{k!},

where p_{j}(t) = ∑_{i=1}^{k−1} \frac{d^{i}}{dx^{i}} h_{j}(w_{n−1})\frac{1}{i!} ∑_{l=0}^{i−1} e_{n−1}^{i−1−l}a_{n−1}^{l} and γ = max_{t∈[0,1]} β(t)−min_{t∈[0,1]} α(t). In view of (B_{3}), we have

p_{j}(t) ≤ ∑_{i=1}^{k−1} \frac{M}{(β − α)^{i−1}} \frac{1}{i!} ∑_{l=0}^{i−1} e_{n−1}^{i−1} ≤ ∑_{i=1}^{k−1} \frac{M}{(β − α)^{i−1}} \frac{1}{(i − 1)!}(β − α)^{i−1} < 1.

It follows that, we can find λ < 1 such that p_{j}(t) ≤ λ, t ∈ [0, 1], (j = 1, 2) and hence

\begin{matrix}e_{n}(0) − k_{1}e′_{n}(0) ≤ λ ∫_{0}^{1} e_{n}(s)ds + \frac{M}{γ^{k−1}k!}∥e_{n−1}∥^{k} \\ e_{n}(1) + k_{2}e′_{n}(1) ≤ λ ∫_{0}^{1} e_{n}(s)ds + \frac{M}{γ^{k−1}k!}∥e_{n−1}∥^{k}. \end{matrix} ; (4.13)

Now, using Taylor’s theorem and (4.5), we obtain

\begin{aligned}e′′_{n}(t) & = x′′(t) − w′′_{n}(t) \\ & = [F (t, x) − ϕ(t, x)] − [∑_{i=0}^{k−1} \frac{∂^{i}}{∂x^{i}} f(t, w_{n−1})\frac{a_{n−1}^{i}}{i!} − \frac{∂^{k}}{∂x^{k}} ϕ(t, ξ)\frac{a_{n−1}^{k}}{k!} ] \\ & = ∑_{i=1}^{k−1} \frac{∂^{i}}{∂x^{i}} f(t, w_{n−1})\frac{(e_{n−1}^{i} − a_{n−1}^{i})}{i!} + \frac{∂^{k}}{∂x^{k}} F (t, c_{1})\frac{e_{n−1}^{k}}{k!} + \frac{∂^{k}}{∂x^{k}} ϕ(t, ξ))\frac{a_{n−1}^{k}}{k!} \\ & ≥ ∑_{i=1}^{k−1} \frac{∂^{i}}{∂x^{i}} f(t, w_{n−1})\frac{∑_{l=0}^{i−1} e_{n−1}^{i−1−l}a_{n−1}^{l}}{i!} e_{n} + \frac{e_{n−1}^{k}}{k!} (\frac{∂^{k}}{∂x^{k}} F (t, c_{1}) + \frac{∂^{k}}{∂x^{k}} ϕ(t, ξ)) \\ & ≥ −N \frac{∥e_{n−1}∥^{k}}{k!} , \end{aligned} ; (4.14)

where −N_{1} ≤ \frac{∂^{k}}{∂x^{k}} F (t, x) ≤ 0, −N_{2} ≤ \frac{∂^{k}}{∂x^{k}} ϕ(t, x) ≤ 0 and N = max\{N_{1}, N_{2}\}. From (4.13) and (4.14), it follows that 0 ≤ e_{n}(t) ≤ r(t), t ∈ [0, 1], where r(t) is the unique solution of the problem

r′′(t) = −N \frac{e_{n−1}^{k}}{k!} , t ∈ [0, 1]

r(0) − k_{1}r′(0) = λ ∫_{0}^{1} r(s)ds + \frac{M}{γ^{k−1}k!}∥e_{n−1}∥^{k}

r(1) + k_{2}r′(1) = λ ∫_{0}^{1} r(s)ds + \frac{M}{γ^{k−1}k!}∥e_{n−1}∥^{k}

and

r(t) = \frac{1}{1 + k_{1} + k_{2}} [(1 − t + k_{2})(λ ∫_{0}^{1} r(s) ds + \frac{M}{γ^{k−1}k!}∥e_{n−1}∥^{k}) + (t + k_{1})(λ ∫_{0}^{1} r(s) ds

+ \frac{M}{γ^{k−1}k!}∥e_{n−1}∥^{k}] − N ∫_{0}^{1} G(t, s)\frac{∥e_{n−1}∥^{k}}{k!} ds

≤ \frac{1}{1 + k_{1} + k_{2}} [λ\{(1 − t + k_{2}) + (t + k_{1})\}∥r∥

+ \{(1 − t + k_{2}) + (t + k_{1})\}\frac{M}{γ^{k−1}k!}∥e_{n−1}∥^{k}] + N∥e_{n−1}∥^{2} ∫_{0}^{1} |G(t, s)| ds

= λ∥r∥ + C′∥e_{n−1}∥^{k},

where L is a bound for ∫_{0}^{1} |G(t, s)|ds and C′ = \frac{M}{γ^{k−1}k!} + NL. Taking the maximum over [0, 1], we get

∥r∥ ≤ δ∥e_{n−1}∥^{k},

where, δ = \frac{C′}{1−λ}. □ EJQTDE, 2003 No. 10, p. 14

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Department of Mathematics, University of Glasgow, Glasgow G12 8QW, UK

E-mail address: rak@maths.gla.ac.uk