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

Authors: Er Gao; Songhe Song; Xinjian Zhang
Journal: Journal of Applied Mathematics
 (2012)

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## 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.


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## Body

## 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].LetW2m[0,T] denote the function space on a finite interval [0,T],  W2m[0,T] = {f(t),t∈[0,T],f(m-1)(t)∈L2[0,T]}, 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 W2,ρ2[0,T]={u(t),t∈[0,T],u'(t) is an absolute continuous real-valued function on  [0,T], ∫0Tt(u′′(t))2dt<+∞} and used it solving Volterra integral equation with weakly singular kernel. It is obvious that W22[0,T]⊆W2,ρ2[0,T] and W2,ρ2[0,T] will be more widely applied.In this paper, we are concerned with the reproducing kernel space with weighted integralW2,αm[0,T]={u(t),t∈[0,T],u'(t),…,u(m-1) are absolute continuous real-valued functions on [0,T],∫0T‍tα(u(m)(t))2dt<+∞}, where α is a constant and satisfies  0≤α<1 (when α=0,  W2,αm[0,T] is W2m[0,T]). 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 ofW2m[a,b] 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.SetL=Dm+am-1Dm-1+⋯+a1D+a0(t),t∈[a,b], where aj(t)∈Cj[a,b] and KerL = {f∈W2m[a,b]:Lf=0}.Definition 2.1.
φ
1
(
t
)
,
…
,
φ
m
(
t
) are the basis in KerL. The i-th row of Wronskian matrix is (φ1(i-1)(t),…,φm(i-1)(t)), and the last line of its inverse matrix is (φ~1(t),…,φ~m(t)). Call φ~1(t),…,φ~m(t) are the adjunct functions of φ1(t),…,φm(t).Lemma 2.2.
Assume(2.1)g(t,τ)=∑i=1mei(t)e~i(τ)(t-τ)+0
and γ1,…,γm is a system of linear independent functions in KerL and satisfies
(2.2)γk∫abg(·,τ)u(τ)dτ=∫abγkg(·,τ)u(τ)dτ,
where e1(t),…,em(t) are the dual basis of KerL relative to γ1,…,γm, and e~1(t),…,e~m(t) are the adjunct functions of e1(t),…,em(t). Then for any functions f∈W2m[a,b], they satisfy the form
(2.3)f(t)=∑i=1m(γif)ei(t)+∫ab‍G(t,τ)·Lf(τ)dτ,
where G(t,τ) is defined below
(2.4)G(t,τ)=g(t,τ)-∑i=1m‍(γig(·,τ))ei(t)
and the expression is exclusive.Lemma 2.3.
L is a linear differential operator. Assume γ1,…,γm are linear independent functions in KerL, satisfying (2.2). Then W2m[a,b] is a Hilbert space if the inner product is defined by the following form:
(2.5)(f,h)=∑i=1m(γif)(γih)¯+∫ab‍Lf(t)Lh(t)¯dt,f,h∈W2m[a,b].Lemma 2.4.
Under the assumptions of Lemma2.2 and the inner product (2.5), the Hilbert space W2m[a,b] is reproducing kernel Hilbert space with the reproducing kernel can be denoted by
(2.6)K(t,τ)=∑i=1mei(t)ei(τ)¯+∫ab‍G(t,x)G(τ,x)¯dx.


## 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 kernelK of a Hilbert space H completely determines the space H.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 spaceW2,αm[0,T], where 0≤α<1, and the result of this part is the main result of this paper.


### 3.1. Case 1: The Weight Is Constant

For general spaceW2m[0,T], let L=Dm be the linear differential operator of order m, and let λ1,λ2,…,λm be the linear independent functions on  KerL, where  KerL is defined by  KerL={f∈W2m[0,T],Lf=0}. Let  e1,e2,…,em be the dual basis of  KerL relative to  λ1,λ2,…,λm. That means
(3.1)Lei=0,λiej=δij,i,j=1,2,…,m.
Let  G be the Green's function of  L and satisfy
(3.2)LtG(t,s)=δ(t-s),λiG(·,s)=0,i=1,…,m.
By the Lemma 2.4, W2m[0,T] is a reproducing kernel Hilbert space if the inner product is defined by the following form:
(3.3)(f,h)1=∑i=1m‍(λif)(λih)¯+∫0T‍Lf(t)Lh(t)¯dt,f,h∈W2m[0,T]
and the reproducing kernel is
(3.4)K1(t,τ)=∑i=1m‍ei(t)ei(τ)¯+∫0T‍G(t,x)G(τ,x)¯dx.Let(3.5)(f,h)2=∑i=1m‍ai(λif)(λih)¯+∫0T‍bLf(t)Lh(t)¯dt,f,h∈W2m[0,T],
where both a1,a2,…,am and b are positive real numbers. The following proposition holds.Theorem 3.1.
Using the above hypothesis,W2m[0,T] is a reproducing kernel Hilbert space if it has been endowed with the inner product (3.5) and the reproducing kernel is
(3.6)K2(t,τ)=∑i=1mei(t)ei(τ)¯ai+∫0T‍1bG(t,x)G(τ,x)¯dx.Proof.
LetL~=bL=bDm, and  λ~if=aiλif.
It is obvious thatλ~1,λ~2,…,λ~m are also the linear independent functions on KerL~. From Lemma 2.3, we have that W2m[0,T] is a Hilbert space if the inner product is defined by
(3.7)(f,h)L~=∑i=1m(λ~if)(λ~ih)¯+∫0T‍L~f(t)L~h(t)¯dt=∑i=1m‍ai(λif)(λih)¯+∫0T‍bLf(t)Lh(t)¯dt=(f,h)2.
Next, we will proof K2(t,s) is the reproducing kernel of the space W2m[0,T] with the inner product (·,·)2.
K
1
(
t
,
s
) is the reproducing kernel of the space W2m[0,T] with the inner product (·,·)1. In particular, K1(t,s) is contained in W2m[0,T]. So K2(t,s) is also contained in W2m[0,T].
For anyf∈W2m[0,T],
(3.8)(f(s),K2(t,s))2=∑i=1m‍ai(λif)(λiK2(t,·))¯+∫0T‍bf(m)(s)∂m∂smK2(t,s)¯ds.
From (3.1) and (3.2), we have
(3.9)(f(τ),K2(t,τ))2f(t)=∑i=1m‍ai(λif)λi(∑i=1m‍ei(t)ei(τ)¯ai)¯+∫0T‍bf(m)(τ)∂m∂τm∫0T‍1bG(t,x)G(τ,x)¯dx¯dτf(t)=∑i=1m(λif)λi(∑i=1m‍ei(t)ei(τ)¯)¯+∫0T‍f(m)(τ)∂m∂τm∫0T‍G(t,x)G(τ,x)¯dx¯dτ.
Similarly, from (3.1) and (3.2), we obtain
(3.10)f(t)=(f(τ),K1(t,τ))1=∑i=1m(λif)λi(∑i=1m‍ei(t)ei(τ)¯)¯+∫0T‍f(m)(s)∂m∂τm∫0T‍G(t,x)G(τ,x)¯dx¯dτ.
So f(t)=(f(τ),K2(t,τ))2 holds.
The proof is complete.


### 3.2. Case 2: The Weight Is Variable

In this case, we construct the inner product of the spaceW2,αm[0,T], and calculate the corresponding reproducing kernel.DefineL1=tα/2Dm. From the definition of the space W2,αm[0,T], we know that L1:W2,αm[0,T]→L2[0,T] (mapping the space W2,αm[0,T] to the square integrable space on [0,T]).Under the hypothesis of Case 1,λ1,λ2,…,λm are also the linear independent functions on  KerL1, where  KerL1 is defined by  KerL1={f∈W2,αm[0,T],L1f=0}, and e1,e2,…,em is also the dual basis of  KerL1 relative to  λ1,λ2,…,λm. e~1(t),…,e~m(t) are the adjunct functions of e1(t),…,em(t).Similar to Case 1, define an algorithm(·,·)3 as the following form
(3.11)(f,h)3=∑i=1m(λif)(λih)¯+∫0T‍L1f(t)L1h(t)¯dt,f,h∈W2,αm[0,T]Theorem 3.2.
Under the above assumption,(·,·)3 is the inner product of the space W2,αm[0,T].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 spaceW2,αm[0,T] into two parts KerL1 and (KerL1)⊥, where KerL1 is the linear space of order m. From the results in [9], one has the following proposition.Proposition 3.3.
Under the above assumption,KerL1 is the reproducing kernel Hilbert space with the inner product below
(3.12)(f,h)4=∑i=1m(λif)(λih)¯f,h∈KerL1
and the corresponding reproducing kernel is
(3.13)K4(t,τ)=∑i=1mei(t)e¯i(τ).Let(3.14)g1(t,τ)=1τα/2∑i=1mei(t)e~i(τ)(t-τ)+0={1τα/2∑i=1mei(t)e~i(τ),t≥τ,0,t<τ,G1(t,τ)=g1(t,τ)-∑i=1m(λig1(·,τ))ei(t).It is obvious thatG1(t,τ)=(1/τα/2)G(t,τ).The following theorem holds.Theorem 3.4.
G
1 is the Green's function of L1 and for any f∈W2m[0,T], u(t)=L1f(t), satisfies
(3.15)L1t∫0TG1(t,τ)u(τ)dτ=u(t),λiG1(·,τ)=0,i=1,2,…,m.Proof.
For anyi=1,2,·,m, we have
(3.16)λiG1(·,τ)=τα/2λiG(·,τ).
From (3.2),
(3.17)λiG1(·,τ)=τα/2λiG(·,τ)=0,i=1,2,·,m.
Then from the results in [9, 10], for any f∈W2m[0,T], we obtain
(3.18)Lt∫0TG(t,τ)Lf(τ)dτ=Lf.
So
(3.19)L1t∫0TG1(t,τ)u(τ)dτ=tα/2Lt∫0T‍1τα/2G(t,τ)τα/2Lf(τ)dτ=tα/2Lf(t)=L1f(t)=u(t).
The proof is complete.Remark 3.5.
If acting in accordance with the process of the paper [9], we have
(3.20)K3(t,τ)=∑i=1mei(t)ei(τ)¯+∫0T‍G1(t,x)G1(τ,x)¯dx.
But K3(t,τ) is not the reproducing kernel of W2m[0,T], since K3(t,τ)∉W2m[0,T].Now, we will give an important property of the arbitrary element ofW2,αm[0,T].Theorem 3.6.
For anyf∈W2,αm[0,T], L1f(t)=u(t). Then there are some real constant c1,c2,…,cm, satisfying
(3.21)f(t)=∑i=1m‍ciei(t)+∫0T(∑i=1mei(t)e~i(τ))u(τ)dτ,
and the expression is exclusive.Proof.
L
1
=
t
α
/
2
D
m is a linear mapping, and L1:W2,αm[0,T]→L2[0,T] is a homomorphic mapping. For any u(t)∈L2[0,T], we have a function h(x) satisfies
(3.22)h(t)=∫0T‍(∑i=1mei(t)e~i(τ))u(τ)dτ.
Because L1h(t)=u(t)∈L2[0,T], h(t)∈W2,αm[0,T] holds. So L1:W2,αm[0,T]→L2[0,T] is a surjective homomorphism.
Ker
L
1
=
Span
{
e
1
,
e
2
,
…
,
e
m
} is a linear system and the dimension of the system is m. So from the knowledge of the group homomorphism, we have
(3.23)L1:W2,αm[0,T]/KerL1→L2[0,T]
is isomorphic.
On the one hand for anyh(t)∈W2,αm[0,T]/KerL1, there exists the exclusive u(t), satisfying L1h(t)=u(t). On the other hand for the u(t)∈L2[0,T], h0(t)=∫0T‍(∑i=1mei(t)e~i(τ))u(τ)dτ satisfies h0(t)∈W2,αm[0,T]/KerL1 and L1h0(t)=u(t). So for h(t)∈W2,αm[0,T]/KerL1, that holds
(3.24)h(t)=∫0T(∑i=1m‍ei(t)e~i(τ))L1h(τ)dτ.
For anyf(t)∈W2,αm[0,T], f(t)=f0(t)+h(t) holds, where f0(t)∈KerL1 and h(t)∈W2,αm[0,T]/KerL1. At the same time, the decomposition is exclusive because of the orthogonality between KerL1 and W2,αm[0,T]/KerL1.
Furthermore,KerL1=Span{e1,e2,…,em}, so f0(t)=∑i=1mciei(t), where c1,c2,…,cm are real numbers, and the expression is exclusive.
So for anyf(t)∈W2,αm[0,T], we have
(3.25)f(t)=f0(t)+h(t)=∑i=1mciei(t)+∫0T‍(∑i=1mei(t)e~i(τ))u(τ)dτ,
wherec1,c2,…,cm are real numbers, and the the expression is exclusive.
The proof is complete.Then similar to the Theorem3.4, we have the following theorem.Theorem 3.7.
G
1 is the Green's function of L1 and for any f∈W2,αm[0,T], L1f(t)=u(t), satisfies
(3.26)L1t∫0TG1(t,τ)u(τ)dτ=u(t).Proof.
For anyf∈W2,αm[0,T], L1f(t)=u(t),
(3.27)L1t∫0T‍G1(t,τ)u(τ)dτ=tα/2Lt∫0T1τα/2G(t,τ)τα/2Lf(τ)dτ=tα/2Lt∫0TG(t,τ)Lf(τ)dτ.
From Theorem3.6, we have
(3.28)Lf(t)=Lt∫0TG(t,τ)Lf(τ)dτ.
Thus, we know that (3.26) is true.
The proof is complete.Theorem 3.8.
Under the above hypothesis and the inner product(·,·)3, W2,αm[0,T] is the Hilbert space.Proof.
The norm of the space is denoted by∥u∥3=(u,u)3, where u∈W2,αm[0,T].
Suppose that{fn}n=1∞ is a Cauchy sequence in W2,αm[0,T], that is,
(3.29)∥fn+p-fn∥32→0(n→∞).
From Theorems 3.6 and 3.1, we have
(3.30)∥fn+p-fn∥32=∑i=1m‍(λifn+p-λifn)2+∫0T[L1(fn+p(t)-fn(t))]2dt→0(n→∞).
By the completeness of KerL1 and L2[0,T], there exist a real number ri, (i=1,2,…,m) and a real function h∈L2[0,T], such that
(3.31)limn→∞λifn=ri,i=1,2,…,m,limn→∞∫0T[L1fn(t)-h(t)]2=0.
Setf0(t)=∑i=1m‍riei(t)+∫0T‍G1(t,τ)h(τ)dτ. It follows that f0∈W2,αm[0,T] and
(3.32)∥fn-f0∥32→0(n→∞).
So W2,αm[0,T] is complete. Namely, W2,αm[0,T] is Hilbert space.
The proof is complete.Theorem 3.9.
Under the above hypothesis and the inner product(·,·)3, W2,αm[0,T] is the reproducing kernel Hilbert space, and the reproducing kernel is
(3.33)K3(t,τ)=∑i=1mei(t)ei(τ)¯+∫0TG1(t,x)G1(τ,x)¯dx.Proof.
From Theorem3.8 and Proposition 3.3, we only need to demonstrate that
(3.34)K5(t,τ)=∫0T‍G1(t,x)G1(τ,x)¯dx.
is the reproducing kernel of (KerL1)⊥, where the inner product is defined by
(3.35)(f,h)5=∫0T‍L1f(t)L1h(t)¯dt,f,h∈(KerL1)⊥.
From Theorem 3.7, L1K5(t,τ)≠0, so K5(t,τ)∈(KerL1)⊥.
For anyh∈(KerL1)⊥,
(3.36)(h(τ),K5(t,τ))5=∫0TL1h(τ)L1∫0TG1(t,x)G1(τ,x)¯dx¯dτ.
From Theorem 3.7,
(3.37)(h(τ),K5(t,τ))5=∫0TL1h(τ)G1(t,τ)¯dτ.
Furthermore, from the definition of the (KerL1)⊥, we have
(3.38)∫0TL1h(τ)G1(t,τ)¯dτ=∑i=1m‍(λih)ei(t)+∫0TL1h(τ)G1(t,τ)¯dτ.
Finally, from the Theorem 3.6,
(3.39)h(t)=∑i=1m(λih)ei(t)+∫0TL1h(τ)G1(t,τ)¯dτ.
So
(3.40)(h(τ),K5(t,τ))5=h(t).
The proof is complete.


## 3.1. Case 1: The Weight Is Constant

For general spaceW2m[0,T], let L=Dm be the linear differential operator of order m, and let λ1,λ2,…,λm be the linear independent functions on  KerL, where  KerL is defined by  KerL={f∈W2m[0,T],Lf=0}. Let  e1,e2,…,em be the dual basis of  KerL relative to  λ1,λ2,…,λm. That means
(3.1)Lei=0,λiej=δij,i,j=1,2,…,m.
Let  G be the Green's function of  L and satisfy
(3.2)LtG(t,s)=δ(t-s),λiG(·,s)=0,i=1,…,m.
By the Lemma 2.4, W2m[0,T] is a reproducing kernel Hilbert space if the inner product is defined by the following form:
(3.3)(f,h)1=∑i=1m‍(λif)(λih)¯+∫0T‍Lf(t)Lh(t)¯dt,f,h∈W2m[0,T]
and the reproducing kernel is
(3.4)K1(t,τ)=∑i=1m‍ei(t)ei(τ)¯+∫0T‍G(t,x)G(τ,x)¯dx.Let(3.5)(f,h)2=∑i=1m‍ai(λif)(λih)¯+∫0T‍bLf(t)Lh(t)¯dt,f,h∈W2m[0,T],
where both a1,a2,…,am and b are positive real numbers. The following proposition holds.Theorem 3.1.
Using the above hypothesis,W2m[0,T] is a reproducing kernel Hilbert space if it has been endowed with the inner product (3.5) and the reproducing kernel is
(3.6)K2(t,τ)=∑i=1mei(t)ei(τ)¯ai+∫0T‍1bG(t,x)G(τ,x)¯dx.Proof.
LetL~=bL=bDm, and  λ~if=aiλif.
It is obvious thatλ~1,λ~2,…,λ~m are also the linear independent functions on KerL~. From Lemma 2.3, we have that W2m[0,T] is a Hilbert space if the inner product is defined by
(3.7)(f,h)L~=∑i=1m(λ~if)(λ~ih)¯+∫0T‍L~f(t)L~h(t)¯dt=∑i=1m‍ai(λif)(λih)¯+∫0T‍bLf(t)Lh(t)¯dt=(f,h)2.
Next, we will proof K2(t,s) is the reproducing kernel of the space W2m[0,T] with the inner product (·,·)2.
K
1
(
t
,
s
) is the reproducing kernel of the space W2m[0,T] with the inner product (·,·)1. In particular, K1(t,s) is contained in W2m[0,T]. So K2(t,s) is also contained in W2m[0,T].
For anyf∈W2m[0,T],
(3.8)(f(s),K2(t,s))2=∑i=1m‍ai(λif)(λiK2(t,·))¯+∫0T‍bf(m)(s)∂m∂smK2(t,s)¯ds.
From (3.1) and (3.2), we have
(3.9)(f(τ),K2(t,τ))2f(t)=∑i=1m‍ai(λif)λi(∑i=1m‍ei(t)ei(τ)¯ai)¯+∫0T‍bf(m)(τ)∂m∂τm∫0T‍1bG(t,x)G(τ,x)¯dx¯dτf(t)=∑i=1m(λif)λi(∑i=1m‍ei(t)ei(τ)¯)¯+∫0T‍f(m)(τ)∂m∂τm∫0T‍G(t,x)G(τ,x)¯dx¯dτ.
Similarly, from (3.1) and (3.2), we obtain
(3.10)f(t)=(f(τ),K1(t,τ))1=∑i=1m(λif)λi(∑i=1m‍ei(t)ei(τ)¯)¯+∫0T‍f(m)(s)∂m∂τm∫0T‍G(t,x)G(τ,x)¯dx¯dτ.
So f(t)=(f(τ),K2(t,τ))2 holds.
The proof is complete.


## 3.2. Case 2: The Weight Is Variable

In this case, we construct the inner product of the spaceW2,αm[0,T], and calculate the corresponding reproducing kernel.DefineL1=tα/2Dm. From the definition of the space W2,αm[0,T], we know that L1:W2,αm[0,T]→L2[0,T] (mapping the space W2,αm[0,T] to the square integrable space on [0,T]).Under the hypothesis of Case 1,λ1,λ2,…,λm are also the linear independent functions on  KerL1, where  KerL1 is defined by  KerL1={f∈W2,αm[0,T],L1f=0}, and e1,e2,…,em is also the dual basis of  KerL1 relative to  λ1,λ2,…,λm. e~1(t),…,e~m(t) are the adjunct functions of e1(t),…,em(t).Similar to Case 1, define an algorithm(·,·)3 as the following form
(3.11)(f,h)3=∑i=1m(λif)(λih)¯+∫0T‍L1f(t)L1h(t)¯dt,f,h∈W2,αm[0,T]Theorem 3.2.
Under the above assumption,(·,·)3 is the inner product of the space W2,αm[0,T].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 spaceW2,αm[0,T] into two parts KerL1 and (KerL1)⊥, where KerL1 is the linear space of order m. From the results in [9], one has the following proposition.Proposition 3.3.
Under the above assumption,KerL1 is the reproducing kernel Hilbert space with the inner product below
(3.12)(f,h)4=∑i=1m(λif)(λih)¯f,h∈KerL1
and the corresponding reproducing kernel is
(3.13)K4(t,τ)=∑i=1mei(t)e¯i(τ).Let(3.14)g1(t,τ)=1τα/2∑i=1mei(t)e~i(τ)(t-τ)+0={1τα/2∑i=1mei(t)e~i(τ),t≥τ,0,t<τ,G1(t,τ)=g1(t,τ)-∑i=1m(λig1(·,τ))ei(t).It is obvious thatG1(t,τ)=(1/τα/2)G(t,τ).The following theorem holds.Theorem 3.4.
G
1 is the Green's function of L1 and for any f∈W2m[0,T], u(t)=L1f(t), satisfies
(3.15)L1t∫0TG1(t,τ)u(τ)dτ=u(t),λiG1(·,τ)=0,i=1,2,…,m.Proof.
For anyi=1,2,·,m, we have
(3.16)λiG1(·,τ)=τα/2λiG(·,τ).
From (3.2),
(3.17)λiG1(·,τ)=τα/2λiG(·,τ)=0,i=1,2,·,m.
Then from the results in [9, 10], for any f∈W2m[0,T], we obtain
(3.18)Lt∫0TG(t,τ)Lf(τ)dτ=Lf.
So
(3.19)L1t∫0TG1(t,τ)u(τ)dτ=tα/2Lt∫0T‍1τα/2G(t,τ)τα/2Lf(τ)dτ=tα/2Lf(t)=L1f(t)=u(t).
The proof is complete.Remark 3.5.
If acting in accordance with the process of the paper [9], we have
(3.20)K3(t,τ)=∑i=1mei(t)ei(τ)¯+∫0T‍G1(t,x)G1(τ,x)¯dx.
But K3(t,τ) is not the reproducing kernel of W2m[0,T], since K3(t,τ)∉W2m[0,T].Now, we will give an important property of the arbitrary element ofW2,αm[0,T].Theorem 3.6.
For anyf∈W2,αm[0,T], L1f(t)=u(t). Then there are some real constant c1,c2,…,cm, satisfying
(3.21)f(t)=∑i=1m‍ciei(t)+∫0T(∑i=1mei(t)e~i(τ))u(τ)dτ,
and the expression is exclusive.Proof.
L
1
=
t
α
/
2
D
m is a linear mapping, and L1:W2,αm[0,T]→L2[0,T] is a homomorphic mapping. For any u(t)∈L2[0,T], we have a function h(x) satisfies
(3.22)h(t)=∫0T‍(∑i=1mei(t)e~i(τ))u(τ)dτ.
Because L1h(t)=u(t)∈L2[0,T], h(t)∈W2,αm[0,T] holds. So L1:W2,αm[0,T]→L2[0,T] is a surjective homomorphism.
Ker
L
1
=
Span
{
e
1
,
e
2
,
…
,
e
m
} is a linear system and the dimension of the system is m. So from the knowledge of the group homomorphism, we have
(3.23)L1:W2,αm[0,T]/KerL1→L2[0,T]
is isomorphic.
On the one hand for anyh(t)∈W2,αm[0,T]/KerL1, there exists the exclusive u(t), satisfying L1h(t)=u(t). On the other hand for the u(t)∈L2[0,T], h0(t)=∫0T‍(∑i=1mei(t)e~i(τ))u(τ)dτ satisfies h0(t)∈W2,αm[0,T]/KerL1 and L1h0(t)=u(t). So for h(t)∈W2,αm[0,T]/KerL1, that holds
(3.24)h(t)=∫0T(∑i=1m‍ei(t)e~i(τ))L1h(τ)dτ.
For anyf(t)∈W2,αm[0,T], f(t)=f0(t)+h(t) holds, where f0(t)∈KerL1 and h(t)∈W2,αm[0,T]/KerL1. At the same time, the decomposition is exclusive because of the orthogonality between KerL1 and W2,αm[0,T]/KerL1.
Furthermore,KerL1=Span{e1,e2,…,em}, so f0(t)=∑i=1mciei(t), where c1,c2,…,cm are real numbers, and the expression is exclusive.
So for anyf(t)∈W2,αm[0,T], we have
(3.25)f(t)=f0(t)+h(t)=∑i=1mciei(t)+∫0T‍(∑i=1mei(t)e~i(τ))u(τ)dτ,
wherec1,c2,…,cm are real numbers, and the the expression is exclusive.
The proof is complete.Then similar to the Theorem3.4, we have the following theorem.Theorem 3.7.
G
1 is the Green's function of L1 and for any f∈W2,αm[0,T], L1f(t)=u(t), satisfies
(3.26)L1t∫0TG1(t,τ)u(τ)dτ=u(t).Proof.
For anyf∈W2,αm[0,T], L1f(t)=u(t),
(3.27)L1t∫0T‍G1(t,τ)u(τ)dτ=tα/2Lt∫0T1τα/2G(t,τ)τα/2Lf(τ)dτ=tα/2Lt∫0TG(t,τ)Lf(τ)dτ.
From Theorem3.6, we have
(3.28)Lf(t)=Lt∫0TG(t,τ)Lf(τ)dτ.
Thus, we know that (3.26) is true.
The proof is complete.Theorem 3.8.
Under the above hypothesis and the inner product(·,·)3, W2,αm[0,T] is the Hilbert space.Proof.
The norm of the space is denoted by∥u∥3=(u,u)3, where u∈W2,αm[0,T].
Suppose that{fn}n=1∞ is a Cauchy sequence in W2,αm[0,T], that is,
(3.29)∥fn+p-fn∥32→0(n→∞).
From Theorems 3.6 and 3.1, we have
(3.30)∥fn+p-fn∥32=∑i=1m‍(λifn+p-λifn)2+∫0T[L1(fn+p(t)-fn(t))]2dt→0(n→∞).
By the completeness of KerL1 and L2[0,T], there exist a real number ri, (i=1,2,…,m) and a real function h∈L2[0,T], such that
(3.31)limn→∞λifn=ri,i=1,2,…,m,limn→∞∫0T[L1fn(t)-h(t)]2=0.
Setf0(t)=∑i=1m‍riei(t)+∫0T‍G1(t,τ)h(τ)dτ. It follows that f0∈W2,αm[0,T] and
(3.32)∥fn-f0∥32→0(n→∞).
So W2,αm[0,T] is complete. Namely, W2,αm[0,T] is Hilbert space.
The proof is complete.Theorem 3.9.
Under the above hypothesis and the inner product(·,·)3, W2,αm[0,T] is the reproducing kernel Hilbert space, and the reproducing kernel is
(3.33)K3(t,τ)=∑i=1mei(t)ei(τ)¯+∫0TG1(t,x)G1(τ,x)¯dx.Proof.
From Theorem3.8 and Proposition 3.3, we only need to demonstrate that
(3.34)K5(t,τ)=∫0T‍G1(t,x)G1(τ,x)¯dx.
is the reproducing kernel of (KerL1)⊥, where the inner product is defined by
(3.35)(f,h)5=∫0T‍L1f(t)L1h(t)¯dt,f,h∈(KerL1)⊥.
From Theorem 3.7, L1K5(t,τ)≠0, so K5(t,τ)∈(KerL1)⊥.
For anyh∈(KerL1)⊥,
(3.36)(h(τ),K5(t,τ))5=∫0TL1h(τ)L1∫0TG1(t,x)G1(τ,x)¯dx¯dτ.
From Theorem 3.7,
(3.37)(h(τ),K5(t,τ))5=∫0TL1h(τ)G1(t,τ)¯dτ.
Furthermore, from the definition of the (KerL1)⊥, we have
(3.38)∫0TL1h(τ)G1(t,τ)¯dτ=∑i=1m‍(λih)ei(t)+∫0TL1h(τ)G1(t,τ)¯dτ.
Finally, from the Theorem 3.6,
(3.39)h(t)=∑i=1m(λih)ei(t)+∫0TL1h(τ)G1(t,τ)¯dτ.
So
(3.40)(h(τ),K5(t,τ))5=h(t).
The proof is complete.


## 4. Example

Example 4.1.
We consider the space mentioned in the introductionW2,1/22[0,1]={u(t),t∈[0,1],u'(t) is an absolute continuous real-valued function on [0,1], ∫01t(u′′(t))2dt<+∞}. Let L=D2, λ1u=u(0), and λ2u=u′(0). Using Theorems 3.8 and 3.9, W2,1/22[0,1] is endowed with the inner product:
(4.1)(u,v)3=u(0)v(0)+u′(0)v′(0)+∫01xu′′(x)v′′(x)dx
and the corresponding reproducing kernel is
(4.2)K(x,y)={1+xy+4xy3/23-4y5/215,y≤x,1+xy+4x3/2y3-4x5/215,y>x.
This result is in accord with Theorem 2.1 of [8].
Ifλ1u=u(0) and λ2u=u(1) and the inner product of W2,1/22[0,1] is given by
(4.3)(u,v)3=u(0)v(0)+u(1)v(1)+∫01xu′′(x)v′′(x)dx,
using the method of this paper, the reproducing kernel of the this space is
(4.4)K(x,y)=1-x-y+46xy15+4(y-5)xy3/215+4(x-5)x3/2y15=-{4y3/2(y-5x)15,y≤x,4x3/2(x-5y)15,y>x.Example 4.2.
We consider the spaceW2,α4[0,1]={u(t),t∈[0,1], u(3)(t) is an absolute continuous real-valued function on [0,1], ∫01‍tα(u(4)(t))2dt<+∞}. Let L=D4, λ1u=u(0), λ2u=u(1), λ3u=u'(0) and λ4u=u'(1). The inner product is given by
(4.5)(u,v)3=u(0)v(0)+u(1)v(1)+u'(0)v'(0)+u'(1)v'(1)+∫01xαu(4)(x)v(4)(x)dx
Similar to Example 4.1, we can compute the reproducing kernel of the reproducing kernel space  W2,α4[0,1] is
(4.6)K(x,y)=(x-1)2x(-1+y)2y+x2(-3+2x)y2(-3+2y)=+(-1+x)x2(-1+y)y2+(-1+x)2(1+2x)(-1+y)2(1+2y)=-(2(-1+x)x2(3+3α(-1+y)-8y)y2)(720-1764α+1624α2-735α3+175α4-21α5+α6)=-(10x2(-3+2x)(1+a(-1+y)-3y)y2)(-5040+13068α-13132α2+6769α3-1960α4+322α5-28α6+α7)=-r(x,y)-r(y,x)+{R(x,y),y≤x,R(y,x),y>x,
where
(4.7)r(x,y)=x(4-α)y2(-3(-42+55α-14α2+α3)x-3(-14+23α-10α2+α3)x2(-2+y))6(-7+α)(-2+α)(-1+α)(360-342α+119α2-18α3+α4)+(x(4-α)y2((-210+107α-18α2+α3)y+(-6+11α-6α2+α3)x3(-3+2y)))(6(-7+α)(-2+α)(-1+α)(360-342α+119α2-18α3+α4)),R(x,y)=y(4-α)((-7+α)(-6+α)(-5+α)x3-3(-7+α)(-6+α)(-1+α)x2y)6(-7+α)(-6+α)(-5+α)(-4+α)(-3+α)(-2+α)(-1+α)+y(4-α)(3(-7+α)(-2+α)(-1+α)xy2-(-3+α)(-2+α)(-1+α)y3)6(-7+α)(-6+α)(-5+α)(-4+α)(-3+α)(-2+α)(-1+α).


## 5. Conclusion

In this paper, we have proposed a method to compute the reproducing kernel on the reproducing kernel space with weighted integral. Theorems3.8 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 W2,αm[0,T]. From the example, we know that the reproducing kernel space of [8] is just one space of the W2,αm[0,T], and the proposed method of this paper is a universal method.



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*Source: 175292-2012-08-29.xml*