Abstract and Applied Analysis

Volume 2013 (2013), Article ID 925192, 6 pages

http://dx.doi.org/10.1155/2013/925192

Research Article

Solving a Class of Singular Fifth-Order Boundary Value Problems Using Reproducing Kernel Hilbert Space Method

Yulan Wang,¹ Shuai Lu,¹ Fugui Tan,² Mingjing Du,¹ and Hao Yu¹

¹Department of Mathematics, Inner Mongolia University of Technology, Hohhot 010051, China

²Jining Teachers College, Inner Mongolia, Wulanchabu 012000, China

Received 4 May 2013; Accepted 13 June 2013

Academic Editor: Rodrigo Lopez Pouso

Copyright © 2013 Yulan Wang et al. 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 use the reproducing kernel Hilbert space method to solve the fifth-order boundary value problems. The exact solution to the fifth-order boundary value problems is obtained in reproducing kernel space. The approximate solution is given by using an iterative method and the finite section method. The present method reveals to be more effective and convenient compared with the other methods.

1. Introduction

The reproducing kernel Hilbert space method has been shown [1–7] to solve effectively, easily, and accurately a large class of linear and nonlinear, ordinary, partial differential equations. However, in [1–7], it cannot be used directly boundary value problems with mixed boundary conditions, since it is very difficult to obtain a reproducing kernel function satisfying mixed nonlinear boundary conditions. The aim of this work is to fill this gap. In [8], we give a new reproducing kernel Hilbert space for solving singular linear fourth-order boundary value problems with mixed boundary conditions. In this paper, we use the new reproducing kernel Hilbert function space method to solve the nonlinear fifth-order boundary value problems.

Singular fifth-order boundary value problems arise in the fields of gas dynamics, Newtonian fluid mechanics, fluid mechanics, fluid dynamics, elasticity, reaction-diffusion processes, chemical kinetics, and other branches of applied mathematics.

Let us consider the following class of singular fifth-order mixed boundary value problems:

where () are known functions. () are linear conditions. We assume that (1) has a unique solution which belongs to , where is a reproducing kernel space.

Remark 1. If (), then (1) is an initial value problem. If (), then (1) is a multipoint problem and so on. That is, problem (1) has a rather general form.

Let and , .

Consider

It is easy to prove that is a bounded linear operator. On the other hand, we suppose that the linear conditions can also always be homogenized; after homogenization of these conditions, we put these conditions into the reproducing kernel space constructed in the following section. Equation (1) can be transformed into the following form in :

2. Reproducing Kernel Hilbert Space

Definition 2. Let be a real Hilbert space of functions . A function is called reproducing kernel for if

(i) for all , (ii) for all and all .

Definition 3. A real Hilbert space of functions on a set is called a reproducing kernel Hilbert space if there exists a reproducing kernel of .

One defines that the inner product space are absolutely continuous  function, .

The inner product in is given by

Theorem 4 (see [8]). The space is a reproducing kernel space, and its reproducing kernel is

For studying the solution of (1) in the homogenized form, we give space (6) as follows:

is equipped with the same inner product . In the following, we construct a reproducing kernel for the space , and we give Lemmas 5 and 6.

Lemma 5. Let be a bounded linear operator; function is the reproducing kernel of space . Let ; then , where denotes any reproducing kernel space of functions over , the symbol indicates that the operator applies to functions of the variable , and the symbol indicates that the operator applies to function of the variable and .

Proof. Consider

Lemma 6. If , , and are defined as in Lemma 5, let ; consider the space , then, is the reproducing kernel of space .

Proof. For any , next, we will prove .

Consider

Let , , , , and , where the symbol () indicates that the operator () applies to functions of the variable . Using Lemma 6, we get Theorem 7.

Theorem 7. The space is a reproducing kernel space, and its reproducing kernel is

3. Analytical Solution

Let , . Via Gram-Schmidt orthonormalization for , we get

where the are the coefficients resulting from Gram-Schmid orthonormalization.

Lemma 8. If are distinct points dense in and is existent, then is a complete function system in .

Proof. For each fixed , if , then

Taking into account the density of , it results in . It follows that from the existence of .

Theorem 9. If are distinct points dense in and is existent, then

is an analytical solution of (3).

Proof. can be expanded to Fourier series in terms of normal orthogonal basis in as follows:

4. Numerical Solution

We define an approximate solution by

Theorem 10. Let , where and are given by (12) and (14); then the sequence of real numbers is monotonously decreasing and .

Proof. We have

Clearly, and consequently is monotonously decreasing in the sense of . By Theorem 9, we know that is convergent in the norm of ; then we have

Hence, .

Theorem 11 (convergence analysis). and are uniformly convergent to and , , where and are given by (12) and (14).

Proof. For any , ,

Then there exists such that

The numerical solution to (3) can be obtained using the following method:

where the coefficients , , are determined by the equation

Using (19) and (20), we have , . So, is the approximation solution of (3).

5. Numerical Experiment

In this section, two numerical examples are studied to demonstrate the accuracy of the present method.

Example 12. Consider the following fifth-order boundary value problem with nonclassical side condition (the right-hand side of this problem has a singularity at , ):

where . The exact solution is . The numerical results are presented in Table 1.

Table 1: The numerical results of Example 12.

0
0
0
0
0
0
0

0.08
−0.00150556
−0.00149854

0.16
−0.00300419
−0.00299716

0.24
−0.000739712
−0.000738437

0.32
0.00926282
0.00925363

0.4
0.0316161
0.0315926

0.48
0.0717954
0.0717548

0.56
0.136337
0.136276

0.64
0.23304
0.232959

0.72
0.371189
0.371084

0.8
0.561801
0.561673

0.88
0.817902
0.817739

0.96
1.15484
1.15458

Example 13 (see [9]). Consider the following fifth-order boundary value problem (the right-hand side of this problem has a singularity at ):

where the exact solution is . By the homogeneous process of the boundary condition, letting , the problem can be transformed into the equivalent form

The numerical results are presented in Tables 2, 3, and 4.

Table 2: Comparison of the absolute error of Example 13.

Solution
Absolute error

Reference [9]

Reference [9]

0.0
0.0
0.0
0.0
0.0
0.0
0.0

0.1249
0.0000752
0.0000754
0.0000754

0.2431
0.0013039
0.0013043
0.0013037

0.3806
0.0080242
0.0080249
0.0080244

0.4195
0.0116531
0.0116538
0.0116533

0.5
0.0220970
0.0220978
0.0220972

0.6923
0.0588207
0.0588201
0.0588209

0.7854
0.0723723
0.0723726
0.0723724

0.8917
0.0646361
0.0646363
0.0646366

1.0
0.0
0.0
0.0
0.0
0.0
0.0

Table 3: The numerical results of Example 13.

0.0
0
0
0
0
0
0

0.1
0.0012491
0.00125306

0.0419792
0.0420291

0.2
0.0121642
0.0121721

0.193196
0.193223

0.3
0.0421473
0.0421562

0.410381
0.410375

0.4
0.0930975
0.0931043

0.591978
0.591944

0.5
0.15468
0.154682

0.596621
0.59657

0.6
0.200775
0.200773

0.250969
0.250915

0.7
0.186533
0.186526

−0.645692
−0.645732

0.8
0.0457947
0.0457844

−2.31836
−2.31836

0.9
−0.311216
−0.311224

−5.01403
−5.01398

1.0
−1
−1

−9
−8.99988

Table 4: The numerical results of Example 13.

0.1
0.971215
0.971117

11.8289
11.8244

0.2
1.97221
1.9719

7.04361
7.0429

0.3
2.19979
2.19947

−3.23499
−3.2345

0.4
1.19534
1.19511

−17.4321
−17.431

0.5
−1.39212
−1.39222

−34.8029
−34.8014

0.6
−5.85595
−5.8559

−54.8995
−54.8978

0.7
−12.4526
−12.4524

−77.4172
−77.4153

0.8
−21.4126
−21.4122

−102.132
−102.13

0.9
−32.9466
−32.9459

−128.873
−128.871

1.0
−47.25
−47.2491

−157.5
−157.498

6. Conclusions and Remarks

In this paper, a new reproducing kernel space satisfying mixed boundary value conditions is constructed skillfully. This makes it easy to solve such kind of problems. Furthermore, the exact solution of the problem can be expressed in series form. The numerical results demonstrate that the new method is quite accurate and efficient for singular problems of fifth-order ordinary differential equations. All computations have been performed using the Mathematica 7.0 software package.

Acknowledgments

The authors would like to express their thanks to the unknown referees for their careful reading and helpful comments. This paper is supported by the Natural Science Foundation of Inner Mongolia (2013MS0109) and Project Application Technology Research and Development Foundation of Inner Mongolia.

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