Journal of Applied Mathematics

Volume 2012 (2012), Article ID 175292, 13 pages

doi:10.1155/2012/175292

Research Article

Calculation of the Reproducing Kernel on the Reproducing Kernel Space with Weighted Integral

Er Gao,1,2 Songhe Song,1,2 and Xinjian Zhang¹

¹Department of Mathematics and Systems Science, College of Science, National University of Defense Technology, Changsha 410073, China

²State key Laboratory of High Performance Computing, National University of Defense Technology, Changsha 410073, China

Received 28 April 2012; Accepted 19 July 2012

Academic Editor: Livija Cveticanin

Copyright © 2012 Er Gao 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 provide a new definition for reproducing kernel space with weighted integral and present a method to construct and calculate the reproducing kernel for the space. The new reproducing kernel space is an enlarged reproducing kernel space, which contains the traditional reproducing kernel space. The proposed method of this paper is a universal method and is suitable for the case of that the weight is variable. Obviously, this new method will generalize a number of applications of reproducing kernel theory to many areas.

1. Introduction

A reproducing kernel is a basic tool for studying the spline interpolation of differential operators and is also the base of the reproducing kernel method, which were widely used in numerical analysis, genetic models, pattern analysis, and so forth. The concept of reproducing kernel is derived from the study of the integration equation, and paper [1] studied specially the reproducing kernels and presented its primary theory. From then on, the reproducing kernel theory and the reproducing kernel method have been studied by many authors [2–7].

Let denote the function space on a finite interval , = , and this space becomes a reproducing kernel Hilbert space (RKHS) if we endow it with some inner product. This kind of the reproducing kernel space is the most popular space for solving the boundary value problems using reproducing kernel method. But in paper [8], the author firstly considered the reproducing kernel space with weighted integral is an absolute continuous real-valued function on , and used it solving Volterra integral equation with weakly singular kernel. It is obvious that and will be more widely applied.

In this paper, we are concerned with the reproducing kernel space with weighted integral are absolute continuous real-valued functions on , where is a constant and satisfies (when , is ). The method for computing the corresponding reproducing kernel is given.

2. Preliminaries

In order to get the main results of the paper, we introduce the method of Zhang for calculating the reproducing kernel of in a nutshell in this section.

The method of Zhang has very powerful system modeling capability. The idea is coming from the relationship between the Green function with reproducing kernel.

Set , where and = .

Definition 2.1. are the basis in . The -th row of Wronskian matrix is , and the last line of its inverse matrix is . Call are the adjunct functions of .

Lemma 2.2. Assume

and is a system of linear independent functions in and satisfies

where are the dual basis of relative to , and are the adjunct functions of . Then for any functions , they satisfy the form

where is defined below

and the expression is exclusive.

Lemma 2.3. is a linear differential operator. Assume are linear independent functions in , satisfying (2.2). Then is a Hilbert space if the inner product is defined by the following form:

Lemma 2.4. Under the assumptions of Lemma 2.2 and the inner product (2.5), the Hilbert space is reproducing kernel Hilbert space with the reproducing kernel can be denoted by

3. The New Method for Computing the Reproducing Kernel

It is known that the reproducing kernel of a reproducing kernel Hilbert space is existence and uniqueness. The reproducing kernel of a Hilbert space completely determines the space .

This section discusses the method of calculating reproducing kernels for the following two cases. The first case is when the weight is constant. In the second case we deal with the general space , where , and the result of this part is the main result of this paper.

3.1. Case 1: The Weight Is Constant

For general space , let be the linear differential operator of order , and let be the linear independent functions on , where is defined by . Let be the dual basis of relative to . That means

Let be the Green's function of and satisfy

By the Lemma 2.4, is a reproducing kernel Hilbert space if the inner product is defined by the following form:

and the reproducing kernel is

Let

where both and are positive real numbers. The following proposition holds.

Theorem 3.1. Using the above hypothesis, is a reproducing kernel Hilbert space if it has been endowed with the inner product (3.5) and the reproducing kernel is

Proof. Let , and .

It is obvious that are also the linear independent functions on . From Lemma 2.3, we have that is a Hilbert space if the inner product is defined by

Next, we will proof is the reproducing kernel of the space with the inner product .

is the reproducing kernel of the space with the inner product . In particular, is contained in . So is also contained in .

For any ,

From (3.1) and (3.2), we have

Similarly, from (3.1) and (3.2), we obtain

So holds.

The proof is complete.

3.2. Case 2: The Weight Is Variable

In this case, we construct the inner product of the space , and calculate the corresponding reproducing kernel.

Define . From the definition of the space , we know that (mapping the space to the square integrable space on ).

Under the hypothesis of Case 1, are also the linear independent functions on , where is defined by , and is also the dual basis of relative to . are the adjunct functions of .

Similar to Case 1, define an algorithm as the following form

Theorem 3.2. Under the above assumption, is the inner product of the space .

If act in accordance with the four basic rules of the inner product, the proof of this proposition is easy. So one overlaps the proof.

Divide the space into two parts and , where is the linear space of order . From the results in [9], one has the following proposition.

Proposition 3.3. Under the above assumption, is the reproducing kernel Hilbert space with the inner product below

and the corresponding reproducing kernel is

Let

It is obvious that .

The following theorem holds.

Theorem 3.4. is the Green's function of and for any , , satisfies

Proof. For any , we have

From (3.2),

Then from the results in [9, 10], for any , we obtain

So

The proof is complete.

Remark 3.5. If acting in accordance with the process of the paper [9], we have

But is not the reproducing kernel of , since .

Now, we will give an important property of the arbitrary element of .

Theorem 3.6. For any , . Then there are some real constant , satisfying

and the expression is exclusive.

Proof. is a linear mapping, and is a homomorphic mapping. For any , we have a function satisfies

Because , holds. So is a surjective homomorphism.

is a linear system and the dimension of the system is . So from the knowledge of the group homomorphism, we have

is isomorphic.

On the one hand for any , there exists the exclusive , satisfying . On the other hand for the , satisfies and . So for , that holds

For any , holds, where and . At the same time, the decomposition is exclusive because of the orthogonality between and .

Furthermore, , so , where are real numbers, and the expression is exclusive.

So for any , we have

where are real numbers, and the the expression is exclusive.

The proof is complete.

Then similar to the Theorem 3.4, we have the following theorem.

Theorem 3.7. is the Green's function of and for any , , satisfies

Proof. For any , ,

From Theorem 3.6, we have

Thus, we know that (3.26) is true.

The proof is complete.

Theorem 3.8. Under the above hypothesis and the inner product , is the Hilbert space.

Proof. The norm of the space is denoted by , where .

Suppose that is a Cauchy sequence in , that is,

From Theorems 3.6 and 3.1, we have

By the completeness of and , there exist a real number , and a real function , such that

Set . It follows that and

So is complete. Namely, is Hilbert space.

The proof is complete.

Theorem 3.9. Under the above hypothesis and the inner product , is the reproducing kernel Hilbert space, and the reproducing kernel is

Proof. From Theorem 3.8 and Proposition 3.3, we only need to demonstrate that

is the reproducing kernel of , where the inner product is defined by

From Theorem 3.7, , so .

For any ,

From Theorem 3.7,

Furthermore, from the definition of the , we have

Finally, from the Theorem 3.6,

So

The proof is complete.

4. Example

Example 4.1. We consider the space mentioned in the introduction is an absolute continuous real-valued function on , . Let , , and . Using Theorems 3.8 and 3.9, is endowed with the inner product:

and the corresponding reproducing kernel is

This result is in accord with Theorem 2.1 of [8].

If and and the inner product of is given by

using the method of this paper, the reproducing kernel of the this space is

Example 4.2. We consider the space , is an absolute continuous real-valued function on , . Let , , , and . The inner product is given by

Similar to Example 4.1, we can compute the reproducing kernel of the reproducing kernel space is

where

5. Conclusion

In this paper, we have proposed a method to compute the reproducing kernel on the reproducing kernel space with weighted integral. Theorems 3.8 and 3.9 are the most important theorems of the paper. To our best knowledge, Theorem 3.6 is the first results about the component of the space . From the example, we know that the reproducing kernel space of [8] is just one space of the , and the proposed method of this paper is a universal method.

Acknowledgment

The work is supported by NSF of China under Grant no. 10971226.

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