Boundary Value Problems
Volume 2010 (2010), Article ID 459754, 19 pages
doi:10.1155/2010/459754
Research Article

Existence and Numerical Method for Nonlinear Third-Order Boundary Value Problem in the Reproducing Kernel Space

Xueqin Lü1,2 and Minggen Cui1

1Department of Mathematics, Harbin Institute of Technology, Weihai, Shandong 264209, China
2School of Mathematics and Sciences, Harbin Normal University, Harbin, Heilongjiang 150025, China

Received 14 July 2009; Revised 12 November 2009; Accepted 11 January 2010

Academic Editor: Irena Rachůnkova

Copyright © 2010 Xueqin Lü and Minggen Cui. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We are concerned with general third-order nonlinear boundary value problems. An existence theorem of solution is given under weaker conditions. In the meantime, an iterative algorithm with global convergence is presented. The higher order derivatives of approximate solution is obtained by using this method can approximate the corresponding derivatives of exact solution well.

1. Introduction

The third-order boundary value problems arise in the study of draining and coating flows. In the past few years, third-order boundary value problems with various types of boundary conditions have been studied in many literatures. For details, see [1–6] and their references therein. In the last decade or so, several papers [7–12] have been devoted to the study of third-order differential equations with two point boundary value conditions. In [13], the authors discussed the solvability of a class of particular third-order nonlinear boundary value problems:

In this paper, we consider third-order nonlinear equations subject to the other boundary value conditions:

In order to derive the existence theorems for solution of (1.2), we make the following assumptions H:

( H 1 ) 𝑓 ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ∈ [ 0 , 1 ] × 𝑅 3 is completely continuous function; ( H 2 ) 𝑓 ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) , 𝑓 𝑥 ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) , 𝑓 𝑦 ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) , 𝑓 𝑧 ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) and 𝑓 𝑤 ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) are bounded; ( H 3 ) 𝑓 ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) > 0 on [ 0 , 1 ] × 𝑅 3 ,

where 𝑓 ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ∈ 𝑊 1 [ 0 , 1 ] as 𝑦 = 𝑦 ( 𝑥 ) ∈ 𝑊 1 [ 0 , 1 ] , 𝑧 = 𝑧 ( 𝑥 ) ∈ 𝑊 1 [ 0 , 1 ] , 𝑤 = 𝑤 ( 𝑥 ) ∈ 𝑊 1 [ 0 , 1 ] , ( 0 ≤ 𝑥 ≤ 1 , − ∞ < 𝑦 , 𝑧 , 𝑤 < ∞ ) .

Under weaker conditions, we give an existence theorem. In the meantime, we give an iterative method of obtaining the solution of (1.2). The method has following features: firstly, the method is an iterative method in a wide range, that is, the convergence of iterative sequences is independent of the choice of initial values; secondly, the approximate solution obtained by using this method can approximate the higher order derivatives of exact solution well; thirdly, we consider the problem in the reproducing kernel Hilbert space and use sufficiently the good properties of the space. When we choose an appropriate reproducing kernel Hilbert space, we can discuss higher-order boundary value problems in a similar manner.

2. Several Reproducing Kernel Spaces

2.1. The Reproducing Kernel Space 𝑊 1 [ 0 , 1 ]

The inner product space 𝑊 1 [ 0 , 1 ] is defined by 𝑊 1 [ 0 , 1 ] = { 𝑢 ( 𝑥 ) ∣ 𝑢 ( 𝑥 ) is absolutely continuous real value function in [ 0 , 1 ] , 𝑢 ( 𝑥 ) ∈ 𝐿 2 [ 0 , 1 ] } . The inner product and norm in 𝑊 1 [ 0 , 1 ] are given, respectively, by

In [14, 15], the author had proved that 𝑊 1 [ 0 , 1 ] is a complete reproducing kernel space. That is, there exists a reproducing kernel function 𝑄 𝑥 ( 𝑦 ) ∈ 𝑊 1 [ 0 , 1 ] , 𝑦 ∈ [ 0 , 1 ] , for each fixed 𝑥 ∈ [ 0 , 1 ] and any 𝑢 ( 𝑦 ) ∈ 𝑊 1 [ 0 , 1 ] , such that ⟨ 𝑢 ( 𝑦 ) , 𝑄 𝑥 ( 𝑦 ) ⟩ 𝑊 1 = 𝑢 ( 𝑥 ) . The reproducing kernel 𝑄 𝑥 ( 𝑦 ) can be denoted by

2.2. The Reproducing Kernel Space 𝑊 2 [ 0 , 1 ]

The inner product space 𝑊 2 [ 0 , 1 ] is defined by 𝑊 2 [ 0 , 1 ] = { 𝑢 ( 𝑥 ) ∣ 𝑢 ( 3 ) are absolutely continuous real value functions, 𝑢 ( 4 ) ∈ 𝐿 2 [ 0 , 1 ] , 𝑢 ( 1 ) = 0 , 𝑢 ′ ( 0 ) = 0 , 𝑢 ′ ( 1 ) = 0 } . The inner product and norm in 𝑊 2 [ 0 , 1 ] are given, respectively, by

where 𝑢 , 𝑣 ∈ 𝑊 2 [ 0 , 1 ] .

Theorem 2.1. The space 𝑊 2 [ 0 , 1 ] is a reproducing kernel space. That is, for any fixed 𝑥 ∈ [ 0 , 1 ] , there exists 𝑇 𝑥 ( 𝑦 ) ∈ 𝑊 2 [ 0 , 1 ] , such that for any 𝑢 ( 𝑥 ) ∈ 𝑊 2 [ 0 , 1 ] , 𝑢 ( 𝑥 ) = ⟨ 𝑢 ( 𝑦 ) , 𝑇 𝑥 ( 𝑦 ) ⟩ 𝑊 2 . The reproducing kernel 𝑇 𝑥 ( 𝑦 ) can be denoted by

2.3. The Reproducing Kernel Space 𝑊 3 [ 0 , 1 ]

The inner product space 𝑊 3 [ 0 , 1 ] is defined by 𝑊 3 [ 0 , 1 ] = { 𝑢 ( 𝑥 ) ∣ 𝑢 ( 3 ) are absolutely continuous real value functions, 𝑢 ( 4 ) ∈ 𝐿 2 [ 0 , 1 ] , 𝑢 ( 0 ) = 0 , 𝑢 ′ ( 0 ) = 0 , 𝑢 ′ ( 1 ) = 0 } . The definitions of inner product and norm in 𝑊 3 [ 0 , 1 ] are the same as (2.3).

Theorem 2.2. The space 𝑊 3 [ 0 , 1 ] is a reproducing kernel space. That is, for any fixed 𝑥 ∈ [ 0 , 1 ] , there exists 𝑅 𝑥 ( 𝑦 ) ∈ 𝑊 3 [ 0 , 1 ] , such that for any 𝑢 ( 𝑥 ) ∈ 𝑊 3 [ 0 , 1 ] , 𝑢 ( 𝑥 ) = ⟨ 𝑢 ( 𝑦 ) , 𝑅 𝑥 ( 𝑦 ) ⟩ 𝑊 3 . The reproducing kernel 𝑅 𝑥 ( 𝑦 ) can be denoted by

The method of obtaining the reproducing kernel 𝑇 𝑥 ( 𝑦 ) , 𝑅 𝑥 ( 𝑦 ) and the proofs of Theorem 2.1, Theorem 2.2 are given in [16].

3. Preliminaries of Constructing Iterative Method

For the convenience of discussion, let 𝐿 ∶ 𝑊 2 → 𝑊 1 , 𝐿 𝑢 = 𝑢 ( 3 ) , then (1.2) can be converted into the form as follows:

Now, we fix ̂ 𝑢 ∈ 𝑊 2 [ 0 , 1 ] and denote 𝑓 ( 𝑥 , ̂ 𝑢 , ̂ 𝑢 , ̂ 𝑢 ) by 𝑓 ( ̂ 𝑢 ) . 𝑓 ( ̂ 𝑢 ) ∈ 𝑊 1 [ 0 , 1 ] and 𝑓 ( ̂ 𝑢 ) is a known function. In order to construct the present iterative method, we will consider the following equation:

We convert (3.1) and (3.2) into the following (3.3) and (3.4), respectively,

Now, we construct an orthogonal system of functions firstly.

Let { 𝑥 𝑖 } ∞ 𝑖 = 1 , where [ 0 , 1 ] is dense on 𝜑 0 = 𝑓 ( ̂ 𝑢 ) . Let { 𝜑 𝑖 ( 𝑥 ) } ∞ 𝑖 = 0 . The orthogonal system of functions { 𝜑 𝑖 ( 𝑥 ) } ∞ 𝑖 = 0 can be derived from Gram-Schmidt orthogonalization process of 𝜑 𝑖 ( 𝑥 ) = 𝑖 𝑘 = 0 𝛼 𝑖 𝑘 𝜑 𝑘 ( 𝑥 ) , ( 3 . 5 ) :

where 𝑓 ( ̂ 𝑢 ) In order to emphasize the fact that the first function of orthogonal system of functions is 𝜑 𝑖 , we denote 𝜑 𝑖 ( 𝑓 ( ̂ 𝑢 ) ) by 𝜓 𝑖 ( 𝑥 ) = 𝐿 ∗ 𝜑 𝑖 ( 𝑥 ) ∈ 𝑊 2 , 𝑖 = 0 , 1 , 2 , … , . Put 𝐿 ∗ where 𝐿 is the conjugate operator of 𝜓 𝑖 . Similarly, we denote 𝜓 𝑖 ( 𝑓 ( ̂ 𝑢 ) ) by .

Now, we introduce some notations:

Ψ 𝑛 ( 𝑓 ( ̂ 𝑢 ) ) = S p a n { 𝜓 1 ( 𝑓 ( ̂ 𝑢 ) ) , 𝜓 2 ( 𝑓 ( ̂ 𝑢 ) ) , … , 𝜓 𝑛 ( 𝑓 ( ̂ 𝑢 ) ) } ; Ψ ( 𝑓 ( ̂ 𝑢 ) ) = S p a n { 𝜓 1 ( 𝑓 ( ̂ 𝑢 ) ) , 𝜓 2 ( 𝑓 ( ̂ 𝑢 ) ) , … } ; 𝑃 𝑛 ∶ 𝑊 2 → Ψ 𝑛 ( 𝑓 ( ̂ 𝑢 ) ) is a orthogonal projector; 𝑃 ∶ 𝑊 2 → Ψ ( 𝑓 ( ̂ 𝑢 ) ) is a orthogonal projector; Ψ ⟂ 𝑛 ( 𝑓 ( ̂ 𝑢 ) ) Ψ 𝑛 ( 𝑓 ( ̂ 𝑢 ) ) is the orthogonal complement space of ; Ψ ⟂ ( 𝑓 ( ̂ 𝑢 ) ) Ψ ( 𝑓 ( ̂ 𝑢 ) ) is the orthogonal complement space of ; 𝑔 𝑛 ( 𝑓 ( ̂ 𝑢 ) ) Ψ ⟂ 𝑛 ( 𝑓 ( ̂ 𝑢 ) ) denotes the element in 𝑔 𝑛 ( 𝑓 ( ̂ 𝑢 ) ) = 𝐿 ∗ 𝑓 ( ̂ 𝑢 ) − 𝑃 𝑛 𝐿 ∗ 𝑓 ( ̂ 𝑢 ) , pick ; 𝑔 ( 𝑓 ( ̂ 𝑢 ) ) Ψ ⟂ ( 𝑓 ( ̂ 𝑢 ) ) denotes the element in 𝑔 ( 𝑓 ( ̂ 𝑢 ) ) = 𝐿 ∗ 𝑓 ( ̂ 𝑢 ) − 𝑃 𝐿 ∗ 𝑓 ( ̂ 𝑢 ) , pick { 𝑥 𝑖 } ∞ 𝑖 = 1 .

It is easy to obtain the following lemmas.

Lemma 3.1. Suppose that [ 0 , 1 ] is dense on , then ( 1 0 ) { 𝜑 𝑖 ( 𝑥 ) } ∞ 𝑖 = 0 𝑊 1 is a complete system of ; ( 2 0 ) { 𝜑 𝑖 ( 𝑥 ) } ∞ 𝑖 = 1 𝑊 1 is not a complete system of ; ( 3 0 ) { 𝜓 𝑖 ( 𝑥 ) } ∞ 𝑖 = 0 𝑊 2 is a complete system of ; ( 4 0 ) { 𝜓 𝑖 ( 𝑥 ) } ∞ 𝑖 = 1 𝑊 2 is not a complete system of 𝑢 .

Theorem 3.2. ⟨ 𝑢 , 𝜓 0 ( 𝑓 ( ̂ 𝑢 ) ) ⟩ 𝑊 2 = ‖ 𝑓 ( ̂ 𝑢 ) ‖ 2 𝑊 1 , ⟨ 𝑢 , 𝜓 𝑖 ( 𝑓 ( ̂ 𝑢 ) ) ⟩ 𝑊 2 = 0 , 𝑖 = 1 , 2 , … . ( 3 . 6 ) is the solution of (3.4) if and only if

Proof. (Sufficiency)
From 𝐿 𝑢 , 𝜑 0 𝑊 1 = 𝑢 , 𝐿 ∗ 𝜑 0 𝑊 2 = ⟨ 𝑢 , 𝜓 0 ⟩ 𝑊 2 ‖ = ‖ 𝑓 ( ̂ 𝑢 ) 2 𝑊 1 = ⟨ 𝑓 ( ̂ 𝑢 ) , 𝑓 ( ̂ 𝑢 ) ⟩ 𝑊 1 = 𝑓 ( ̂ 𝑢 ) , 𝜑 0 𝑊 1 . ( 3 . 7 ) , it follows that From ⟨ 𝜑 0 , 𝜑 𝑖 ⟩ 𝑊 1 = 0 , 𝑖 ≠ 0 and 𝐿 𝑢 , 𝜑 𝑖 𝑊 1 = 𝑢 , 𝐿 ∗ 𝜑 𝑖 𝑊 2 = ⟨ 𝑢 , 𝜓 𝑖 ⟩ 𝑊 2 = 0 = 𝜑 0 , 𝜑 𝑖 𝑊 1 = 𝑓 ( ̂ 𝑢 ) , 𝜑 𝑖 𝑊 1 . ( 3 . 8 ) , we have By the completeness of 𝐿 𝑢 = 𝑓 ( ̂ 𝑢 ) and (3.7), (3.8), one obtains 𝑢 . (Necessity)
If 𝐿 𝑢 − 𝑓 ( ̂ 𝑢 ) , 𝜑 𝑖 𝑊 1 = 0 , 𝑖 = 0 , 1 , 2 , … . ( 3 . 9 ) is the solution of (3.4), then When ⟨ 𝑢 , 𝜓 0 ⟩ 𝑊 2 = 𝐿 𝑢 , 𝜑 0 𝑊 1 = 𝑓 ( ̂ 𝑢 ) , 𝜑 0 𝑊 1 = ‖ 𝑓 ( ̂ 𝑢 ) ‖ 2 𝑊 1 . ( 3 . 1 0 ) , we have When ⟨ 𝑢 , 𝜓 𝑖 ⟩ 𝑊 2 = 𝐿 𝑢 , 𝜑 𝑖 𝑊 1 = 𝜑 0 , 𝜑 𝑖 𝑊 1 = 0 . ( 3 . 1 1 ) , we have (3.10) and (3.11) yield (3.6).

In the same way, we can prove the following corollary.

Corollary 3.3. 𝑔 ‖ 𝑢 = ( 𝑓 ( ̂ 𝑢 ) ) ‖ 𝑓 ( ̂ 𝑢 ) 2 𝑊 1 ⟨ 𝜓 0 ( 𝑓 ( ̂ 𝑢 ) ) , 𝑔 ( 𝑓 ( ̂ 𝑢 ) ) ⟩ 𝑊 2 . ( 3 . 1 2 ) is the solution of (3.4) if and only if

4. Iterative Formulas

4.1. Iterative Formula I

Pick arbitrarily initial function 𝑢 𝑛 + 1 in (3.4) and denote the solution of (3.4) by 𝑢 1 = 𝑔 𝑓 𝑢 0 ‖ ‖ 𝑓 𝑢 0 ‖ ‖ 2 𝑊 1 𝜓 0 𝑓 𝑢 0 𝑓 𝑢 , 𝑔 0 𝑊 2 , 𝑢 2 = 𝑔 𝑓 𝑢 1 ‖ ‖ 𝑓 𝑢 1 ‖ ‖ 2 𝑊 1 𝜓 0 𝑓 𝑢 1 𝑓 𝑢 , 𝑔 1 𝑊 2 , ⋮ 𝑢 𝑛 + 1 = 𝑔 𝑓 𝑢 𝑛 ‖ ‖ 𝑓 𝑢 𝑛 ‖ ‖ 2 𝑊 1 𝜓 0 𝑓 𝑢 𝑛 𝑓 𝑢 , 𝑔 𝑛 𝑊 2 , ( 4 . 1 ) . Then, applying Corollary 3.3, one sees that

where

This is the iterative formula I. However, 𝑔 cannot be computed through finite steps. Thus, we will revise the formula.

4.2. Iterative Formula II

Replacing 𝑔 𝑛 with 𝑢 𝑛 + 1 = 𝑔 𝑛 𝑓 𝑢 𝑛 ‖ ‖ 𝑓 𝑢 𝑛 ‖ ‖ 2 𝑊 1 𝜓 0 𝑓 𝑢 𝑛 , 𝑔 𝑛 𝑓 𝑢 𝑛 𝑊 2 , ( 4 . 3 ) in iterative formula I, we derive iterative formula II, that is

where

Lemma 4.1. Let 𝛽 = i n f 𝑛 ‖ ‖ 𝜓 0 𝑓 𝑢 𝑛 − 𝑃 𝜓 0 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 2 = ‖ ‖ 𝜓 0 𝑓 ∗ − 𝑃 𝜓 0 𝑓 ∗ ‖ ‖ 𝑊 2 , ( 4 . 5 ) and then ‖ ‖ 𝜓 0 𝑓 𝑢 𝑛 − 𝑃 𝑛 𝜓 0 𝑓 𝑢 𝑛 ‖ ‖ 2 𝑊 2 = ‖ ‖ 𝜓 0 𝑓 𝑢 𝑛 − 𝑃 𝜓 0 𝑓 𝑢 𝑛 + 𝑃 𝜓 0 𝑓 𝑢 𝑛 − 𝑃 𝑛 𝜓 0 𝑓 𝑢 𝑛 ‖ ‖ 2 𝑊 2 , ( 4 . 6 ) .

Proof. Note that since this gives us Thus, (4.6) leads to However, Because 𝛽 2 ≤ 𝜓 0 𝑓 𝑢 𝑛 , 𝑔 𝑛 𝑓 𝑢 𝑛 𝑊 2 . ( 4 . 1 1 ) , hence
Now, we prove 𝛽 = 0 . Suppose that ‖ ‖ 𝜓 0 = 0 𝑓 ∗ − 𝑃 𝜓 0 𝑓 ∗ ‖ ‖ 2 𝑊 2 = 𝜓 0 𝑓 ∗ − 𝑃 𝜓 0 𝑓 ∗ , 𝜓 0 𝑓 ∗ − 𝑃 𝜓 0 𝑓 ∗ 𝑊 2 = 𝜓 0 𝑓 ∗ − 𝑃 𝜓 0 𝑓 ∗ 𝑓 , 𝑔 ∗ 𝑊 2 = 𝜓 0 𝑓 ∗ 𝑓 , 𝑔 ∗ 𝑊 2 − 𝑃 𝜓 0 𝑓 ∗ 𝑓 , 𝑔 ∗ 𝑊 2 . ( 4 . 1 2 ) , then (4.5) gives us Since 𝑃 𝜓 0 ( 𝑓 ∗ ) = 𝑃 𝐿 ∗ 𝑓 ∗ ∈ Ψ ( 𝑓 ∗ ) and ⟨ 𝑃 𝜓 0 ( 𝑓 ∗ ) , 𝑔 ( 𝑓 ∗ ) ⟩ 𝑊 2 = 0 , hence ⟨ 𝜓 0 ( 𝑓 ∗ ) , 𝑔 ( 𝑓 ∗ ) ⟩ 𝑊 2 = 0 , , furthermore, 𝜓 0 ( 𝑓 ∗ ) ⟂ 𝑔 ( 𝑓 ∗ ) namely, 𝑔 ( 𝑓 ∗ ) ∈ Ψ ⟂ ( 𝑓 ∗ ) . Note here that 𝜓 0 𝑓 ∗ 𝑓 ∈ Ψ ∗ 𝜓 = S p a n 1 𝑓 ∗ , 𝜓 2 𝑓 ∗ , … . ( 4 . 1 3 ) , one has From ( { 𝜓 𝑖 ( 𝑥 ) } ∞ 𝑖 = 0 ) of Lemma 3.1, we know that 𝑊 2 [ 0 , 1 ] is a complete system of 𝜓 0 ( 𝑓 ∗ ) ∈ Ψ ( 𝑓 ∗ ) . On the other hand, by { 𝜓 𝑖 ( 𝑥 ) } ∞ 𝑖 = 1 , we obtain 𝑊 2 [ 0 , 1 ] is a complete system of 4 0 . This is a contradiction to ( 𝛽 > 0 ) of Lemma 3.1, thus, ̂ 𝑢 = 𝑢 𝑛 .

From Lemma 4.1 we know that iterative formula is significative.

5. Existence of Solution for (3.3)

Put 𝐿 𝑢 = 𝑓 ( 𝑢 𝑛 ) in (3.4), then by Corollary 3.3, 𝑔 𝑓 𝑢 𝑢 = 𝑛 ‖ ‖ 𝑓 𝑢 𝑛 ‖ ‖ 2 𝜓 0 𝑓 𝑢 𝑛 𝑓 𝑢 , 𝑔 𝑛 𝑊 2 , ( 5 . 1 ) can be written by

Replacing 𝑣 𝑛 with 𝐿 𝑣 𝑛 𝑢 = 𝑓 𝑛 𝑣 , ( 5 . 2 ) 𝑛 = 𝑔 𝑓 𝑢 𝑛 ‖ ‖ 𝑓 𝑢 𝑛 ‖ ‖ 2 𝜓 0 𝑓 𝑢 𝑛 𝑓 𝑢 , 𝑔 𝑛 𝑊 2 . ( 5 . 3 ) , then

Obviously, iterative formula II can be converted into the form as follows:

where

Note that by Lemma 4.1 we know

in (5.5). Now suppose 𝛼 𝑛 ( 𝑥 ) , if there exists an 𝛼 𝑛 𝜓 ( 𝑥 ) = 0 𝑓 𝑢 𝑛 𝑓 𝑢 , 𝑔 𝑛 𝑊 2 𝜓 0 𝑓 𝑢 𝑛 , 𝑔 𝑛 𝑓 𝑢 𝑛 𝑊 2 ⋅ 𝑔 𝑛 𝑓 𝑢 𝑛 + 𝑀 𝑔 𝑓 𝑢 𝑛 + 𝑀 , ( 5 . 8 ) , such that 𝑀 then we will change definition of 𝑔 ( 𝑓 ( 𝑢 𝑛 ) ) + 𝑀 ≥ 𝛼 > 0 :

where 𝐶 [ 0 , 1 ] satisfies 𝑢 ∈ 𝑊 2 , 𝑣 ∈ 𝑊 1 .

Lemma 5.1. A bounded set in | | 𝑢 ( 𝑖 ) | | ( 𝑥 ) ≤ 𝜇 𝑖 ‖ 𝑢 ‖ 𝑊 2 | | | | , 𝑖 = 0 , 1 , 2 , 3 , 𝑣 ( 𝑥 ) ≤ 𝜇 ‖ 𝑣 ‖ 𝑊 1 . ( 5 . 9 ) is a compacted set in 𝑊 2 .

Lemma 5.2. For 𝐶 3 [ 0 , 1 ] , one has

Lemma 5.3. A bounded set in ‖ 𝛼 𝑛 − 1 ‖ 𝐶 3 → 0 , ( 𝑛 → ∞ ) . is a compacted set in ‖ ‖ 𝜓 0 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 2 = ‖ ‖ ‖ ‖ 𝜓 0 𝑓 𝑢 𝑛 ‖ ‖ 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 1 ‖ ‖ ‖ ‖ 𝑊 2 = 1 ‖ ‖ 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 1 ‖ ‖ 𝜓 0 ( 𝑓 ( 𝑢 𝑛 ‖ ‖ ) ) 𝑊 2 = 1 ‖ ‖ 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 1 ‖ ‖ 𝐿 ∗ 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 2 ≤ 1 ‖ ‖ 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 1 ‖ ‖ 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 1 ‖ ‖ 𝑔 ‖ 𝐿 ‖ = ‖ 𝐿 ‖ , ( 5 . 1 0 ) 𝑓 𝑢 𝑛 − 𝑔 𝑛 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 2 = 1 ‖ ‖ 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 1 ‖ ‖ 𝑔 ( 𝑓 ( 𝑢 𝑛 ) ) − 𝑔 𝑛 ( 𝑓 ( 𝑢 𝑛 ‖ ‖ ) ) 𝑊 2 = 1 ‖ ‖ 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 1 ‖ 𝑃 𝐿 ∗ 𝑓 𝑢 𝑛 − 𝑃 𝑛 𝐿 ∗ 𝑓 𝑢 𝑛 ‖ 𝑊 2 ≤ 1 ‖ ‖ 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 1 ‖ ‖ 𝑃 − 𝑃 𝑛 ‖ ‖ ‖ ‖ 𝐿 ∗ 𝑓 𝑢 𝑛 ‖ ‖ ≤ 1 ‖ ‖ 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 1 ‖ ‖ 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 1 ‖ ‖ ‖ 𝐿 ‖ 𝑃 − 𝑃 𝑛 ‖ ‖ ⟶ 0 , ( 𝑛 ⟶ ∞ ) , ( 5 . 1 1 ) .

Lemma 5.4. Let | | | | | 𝜓 0 𝑓 𝑢 𝑛 𝑓 𝑢 , 𝑔 𝑛 𝑊 2 𝜓 0 𝑓 𝑢 𝑛 , 𝑔 𝑛 𝑓 𝑢 𝑛 𝑊 2 | | | | | = | | | | | | | 𝜓 − 1 0 𝑓 𝑢 𝑛 , 𝑔 𝑓 𝑢 𝑛 𝑊 2 𝜓 0 𝑓 𝑢 𝑛 , 𝑔 𝑛 𝑓 𝑢 𝑛 𝑊 2 | | | | | | | = | | | | | | | 𝜓 − 1 0 𝑓 𝑢 𝑛 , 𝑔 𝑓 𝑢 𝑛 − 𝑔 𝑛 𝑓 𝑢 𝑛 𝑊 2 𝜓 0 𝑓 𝑢 𝑛 , 𝑔 𝑛 𝑓 𝑢 𝑛 𝑊 2 | | | | | | | ≤ 1 𝛽 2 | | | | 𝜓 0 𝑓 𝑢 𝑛 , 𝑔 𝑓 𝑢 𝑛 − 𝑔 𝑛 𝑓 𝑢 𝑛 𝑊 2 | | | | ≤ 1 , ( 𝛽 > 0 ) 𝛽 2 ‖ ‖ 𝜓 0 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 2 ‖ ‖ 𝑔 𝑓 𝑢 𝑛 − 𝑔 𝑛 𝑓 𝑢 𝑛 ‖ ‖ 𝑊 2 ⟶ 0 , ( 𝑛 ⟶ ∞ ) , ( 5 . 1 2 ) , then 𝜓 0 𝑓 𝑢 𝑛 𝑓 𝑢 , 𝑔 𝑛 𝑊 2 𝜓 0 𝑓 𝑢 𝑛 , 𝑔 𝑛 𝑓 𝑢 𝑛 𝑊 2 ⟶ 1 ( 𝑛 ⟶ ∞ ) . ( 5 . 1 3 )

Proof. (1) Since hence, by Lemma 4.1, one gets namely,
𝑔 𝑛 𝑘 𝑗 𝑓 𝑢 𝑛 𝑘 𝑗 𝐶 3 ⟶ 𝑔 ∗ ( 𝑥 ) ≠ 0 . ( 5 . 1 6 ) Note that thus, by Lemma 5.3, there exists l i m 𝑛 ‖ ‖ ‖ ‖ 𝑔 𝑛 𝑓 𝑢 𝑛 𝑔 𝑓 𝑢 𝑛 ‖ ‖ ‖ ‖ − 1 𝐶 3 = 0 . ( 5 . 1 9 ) of l i m 𝑛 ‖ ‖ ‖ ‖ 𝑔 𝑛 𝑓 𝑢 𝑛 𝑔 𝑓 𝑢 𝑛 ‖ ‖ ‖ ‖ − 1 𝐶 3 = 0 . ( 5 . 2 0 ) , such that
By (5.11), it follows that
Therefore, In the similar manner, one gets Combining the above argument, we obtain
The lemma is complete from (5.13) and (5.20).

Lemma 5.5. 𝑣 𝑛 ( 𝜃 𝑛 ) = 0

Lemma 5.6. If 𝜃 𝑛 satisfies condition H, then 𝜃 ∈ ( 0 , 1 ) , 𝜃 𝑛 < 𝜃 < 1 are bounded. (Technique of proof is similar to the Lemma 7.1 in appendix).

Lemma 5.7. If 𝑣 𝑛 ( 0 ) = 0 and 𝑣 𝑛 ( 1 ) = 0 , then 𝜃 𝑛 ∈ ( 0 , 1 ) does not infinitely go closer to 1, as there exists 𝑣 𝑛 ( 𝜃 𝑛 ) = 0 , such that 𝜃 𝑛 → 1 .

Proof. Since 𝑛 → ∞ and 𝑣 ( 𝑥 ) , there exists ‖ 𝑣 𝑛 − 𝑣 ‖ 𝐶 3 → 0 , such that | 𝑣 𝑛 ( 𝑥 ) − 𝑣 ( 𝑥 ) | → 0 .
Suppose | 𝑣 𝑛 ( 𝑥 ) − 𝑣 ( 𝑥 ) | → 0 , as | 𝑣 𝑛 ( 𝑥 ) − 𝑣 ( 𝑥 ) | → 0 . By Lemmas 5.3 and 5.6, there exists 𝑣 𝑛 ( 0 ) = 0 such that 𝑣 𝑛 ( 1 ) = 0 ; thus 𝑣 𝑛 ( 𝜃 𝑛 ) = 0 , 𝑣 ( 0 ) = 0 , 𝑣 ( 1 ) = 0 . Since 𝑣 ( 1 ) = 0 , 𝜂 ∈ ( 0 , 1 ) , 𝑣 ( 𝜂 ) = 0 , it follows that 𝜉 ∈ ( 𝜂 , 1 ) , 𝑣 ( 3 ) ( 𝜉 ) = 0 , 𝐿 𝑣 𝑛 ( 𝑥 ) = 𝑓 ( 𝑥 , 𝑢 𝑛 ( 𝑥 ) , 𝑢 𝑛 ( 𝑥 ) , 𝑢 𝑛 ( 𝑥 ) ) . Hence there exists an 𝑛 , such that 𝐿 𝑣 ( 𝑥 ) = 𝑓 𝑥 , 𝑢 ( 𝑥 ) , 𝑢 ( 𝑥 ) , 𝑢 ( 𝑥 ) , ( 5 . 2 1 ) , thus there exists an 𝑓 ( 𝜉 , 𝑢 ( 𝜉 ) , 𝑢 ( 𝜉 ) , 𝑢 ( 𝜉 ) ) = 𝑣 ( 3 ) ( 𝜉 ) = 0 , such that 𝑓 ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) . By (5.4), ‖ 𝑢 𝑛 + 1 − 𝑢 𝑛 ‖ 𝐶 2 → 0 , taking limit for 𝑛 → ∞ on both sides, we have thus 𝑊 2 [ 0 , 1 ] ; this is a contradiction to (H3).

Lemma 5.8. If ‖ 𝑢 𝑛 ‖ 𝑊 3 ≤ 𝑀 satisfies condition H, then { 𝑢 𝑛 ( 𝑥 ) } as 𝐶 3 [ 0 , 1 ] . (see Lemma 7.3 in the appendix)

Theorem 5.9 (Existence). If { 𝑢 𝑛 𝑘 ( 𝑥 ) } satisfies condition H, then the solution of (1.2) exists in { 𝑢 𝑛 ( 𝑥 ) } .

Proof. From Lemma 5.6, ‖ 𝑢 𝑛 𝑘 − 𝑢 ‖ 𝐶 3 → 0 , so by Lemma 5.3, 𝑣 ( 𝑥 ) is compact in ‖ 𝑣 𝑛 𝑘 − 𝑣 ‖ 𝐶 3 → 0 , then there exists convergent subsequence | 𝑢 𝑛 𝑘 ( 𝑥 ) − 𝑢 ( 𝑥 ) | → 0 of | 𝑣 𝑛 𝑘 ( 𝑥 ) − 𝑣 ( 𝑥 ) | → 0 , such that 𝑘 → ∞ , in the similar manner, there exists | 𝛼 𝑛 𝑘 − 1 | → 0 , such that 𝑘 → ∞ , thus | 𝑢 𝑛 𝑘 + 1 ( 𝑥 ) − 𝑢 𝑛 𝑘 ( 𝑥 ) | → 0 , | 𝑢 𝑛 𝑘 + 1 ( 𝑥 ) − 𝑢 ( 𝑥 ) | ≤ | 𝑢 𝑛 𝑘 + 1 ( 𝑥 ) − 𝑢 𝑛 𝑘 ( 𝑥 ) | + | 𝑢 𝑛 𝑘 ( 𝑥 ) − 𝑢 ( 𝑥 ) | → 0 , as 𝑣 ( 𝑥 ) = l i m 𝑘 → ∞ 𝑣 𝑛 𝑘 ( 𝑥 ) = l i m 𝑘 → ∞ 1 𝛼 𝑛 𝑘 𝑢 𝑛 𝑘 + 1 = l i m 𝑘 → ∞ 1 𝛼 𝑛 𝑘 ⋅ l i m 𝑘 → ∞ 𝑢 𝑛 𝑘 + 1 = 1 ⋅ l i m 𝑘 → ∞ 𝑢 𝑛 𝑘 = 𝑢 ( 𝑥 ) . ( 5 . 2 2 ) .
By Lemma 5.4, 𝐿 𝑢 ( 𝑥 ) = 𝑓 ( 𝑥 , 𝑢 ( 𝑥 ) , 𝑢 ( 𝑥 ) , 𝑢 ( 𝑥 ) ) , as f ( 𝑥 , 𝑢 ( 𝑥 ) , 𝑢 ( 𝑥 ) , 𝑢 ( 𝑥 ) ) ∈ 𝑊 1 [ 0 , 1 ] and by Lemma 5.8, 𝑢 ( 𝑥 ) ∈ 𝑊 2 [ 0 , 1 ] , then 𝑓 ( 𝑥 , 𝑦 , 𝑧 , 𝑤 ) ; thus
From (5.4) and (5.22), { 𝑢 𝑛 ( 𝑥 ) } ∞ 𝑛 = 1 , the solution of (1.2) exists. Since 𝑢 ( 𝑥 ) , thus ‖ 𝑢 𝑛 ( 𝑘 ) ( 𝑥 ) − 𝑢 ( 𝑘 ) ( 𝑥 ) ‖ 𝐶 → 0 , 𝑘 = 0 , 1 , 2 , 3 . .

Theorem 5.10. If { 𝑢 𝑛 ( 𝑥 ) } ∞ 𝑛 = 1 satisfies condition H and the solution 𝑢 ( 𝑥 ) of (1.2) is unique, then the sequence ‖ 𝑢 𝑛 ‖ 𝑊 2 is convergent to { 𝑢 𝑛 𝑘 ( 𝑥 ) } ∞ 𝑘 = 1 , and { 𝑢 𝑛 𝑗 ( 𝑥 ) } ∞ 𝑗 = 1

Proof. Suppose that { 𝑢 𝑛 ( 𝑥 ) } ∞ 𝑛 = 1 is not convergent to 𝑢 𝑛 𝑘 ( 𝑥 ) → 𝑢 1 ( 𝑥 ) , since 𝑢 𝑛 𝑗 ( 𝑥 ) → 𝑢 2 ( 𝑥 ) is bounded, we can choose two subsequences 𝑢 1 ( 𝑥 ) ≠ 𝑢 2 ( 𝑥 ) , 𝑢 1 ( 𝑥 ) of 𝑢 2 ( 𝑥 ) such that ‖ 𝑢 𝑛 − 𝑢 ‖ 𝐶 → 0 . , ‖ 𝑢 𝑛 ( 𝑘 ) − 𝑢 ( 𝑘 ) ‖ 𝐶 → 0 , 𝑛 → ∞ , 𝑘 = 1 , 2 , 3 . and 𝑊 2 [ 0 , 1 ] , From Theorem 5.9, 5 . 0 and 𝑢 3 ( 𝑥 ) = 𝑥 𝑢 ( 𝑥 ) + 𝑢 ( 𝑥 ) + 𝑥 𝑢 ( 𝑥 ) + 𝑢 2 ( 𝑥 ) , 𝑢 ( 1 ) = 𝑢 ( 0 ) = 𝑢 ( 1 ) = 0 , ( 6 . 1 ) are the solutions of (1.2), this contradicts the uniqueness of solution of (1.2). consequently, 𝑥 ∈ [ 0 , 1 ]
In the same way, we can verify 𝑢 ( 𝑥 ) = 𝑥 ( 𝑥 − 1 )

6. Numerical Example

Now we present a numerical example for solving (1.2) in the reproducing kernel space