# A Novel Approach to Calculation of Reproducing Kernel on Infinite Interval and Applications to Boundary Value Problems

Authors: Jing Niu; Yingzhen Lin; Minggen Cui
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

A new analytical method for the computation of reproducing kernel is proposed and tested on some examples. The expression of
reproducing kernel on infinite interval is obtained concisely in polynomial form for the first time. Furthermore, as a particular effective application of this method, we give an explicit representation formula for calculation of reproducing kernel in reproducing kernel space with boundary value conditions.


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## Body

## 1. Introduction

It is well known that reproducing kernel theory has been used in many research fields such as complex analysis, dilation of linear operators, stochastic processes [1–5], and solution of various differential and integral equations [6–10]. Recently, there are a rather small number of methods for calculating the reproducing kernel expression. One is using Green’s function for a differential operator to construct a reproducing kernel [11, 12]. Another very standard method involves boundary value conditions depending on the property of δ function [13, 14]. The disadvantages of the above two approaches are clear. In the first method, the expression of reproducing kernel including integral form is complicated as a result of Green’s function. Because δ function is highly abstractive in the second approach, it is quite difficult to calculate the kernel in the procedure.The purpose of this paper is to avoid the complex operation of Green’s function andδ function and simply give representation of reproducing kernel in polynomial form. The principal step of the procedure consists of classifying and discussing the infinite interval to satisfy the reproducing property. Therefore, our approach has the advantages that no additional condition is required in order to solve the kernel and a much simpler formalism which is in contrast to the previous two methods. In effect, the universal formula can be obtained in the case of finite interval.Numerically solving an initial and boundary value problem for a differential equation by the reproducing kernel method can be described as follows: construct reproducing kernel spaces which can absorb initial or boundary value conditions, and then, transfer the initial and boundary value problem into an operator equation in the reproducing kernel space where the exact solution to the initial and boundary value problem is expressed by the reproducing kernel, and at last solve the operator equation by approximation. It is obvious that constructing reproducing kernel space which satisfies the initial or boundary conditions and effectively solving for the reproducing kernel become the key to apply reproducing kernel method for initial and boundary value problems.In this work, in order to apply the new approach to solving differential equations with multiform boundary value problems, the explicit formula for calculation of reproducing kernel in the appropriate reproducing kernel space is provided successfully by using the orthogonal decomposition property.The rest of the paper is organized as follows. In Section2, a new reproducing kernel space on infinite interval is presented. Section 3 shows for representation of class one how reproducing kernel can be expressed in polynomial form and gives some examples. Then, these basic ideas are shown to extend to cases involving reproducing kernel space with boundary value condition in Section 4. Finally, in Section 5, we give a brief conclusion and discuss extensions and generalizations of the present work.


## 2. A New Reproducing Kernel Space on Infinite Interval

Definition 1 (see [15]).
LetH be a Hilbert function space on a set D. H is called a reproducing kernel space if and only if for any x∈D, there exists a unique function Rx(y)∈H, such that 〈f,Rx〉=f(x) for any f∈H. Meanwhile, R(x,y)≜Rx(y) is called a reproducing kernel function.Definition 2.
H
m
(
-
∞
,
+
∞
)
=
{
f
(
x
)
∣
f
(
m
-
1
)
(
x
)
is
an
absolutely
continuous
real
value
function
in
(
-
∞
,
+
∞
)
,
f
(
m
)
(
x
)
∈
L
2
(
-
∞
,
+
∞
)
}. The inner product and norm are given, respectively, by
(1)〈f(x),g(x)〉H=∑i=0m-1f(i)(0)g(i)(0)+∫-∞+∞‍f(m)(x)g(m)(x)dx,∥f∥H=〈f(x),f(x)〉H.Theorem 3.
H
m
(
-
∞
,
+
∞
) is a Hilbert reproducing kernel space; namely, it is complete, and for any fixed x∈(-∞,+∞) and f(x)∈Hm(-∞,+∞), there exists a cx>0, such that |f(x)|≤cx∥f∥H.Proof.
Suppose thatfn(x) is a Cauchy sequence in Hm(-∞,+∞). Then it holds that
(2)∥fn+p-fn∥H2=∑i=0m-1‍|fn+p(i)(0)-fn(i)(0)|2+∫-∞+∞‍|fn+p(m)(x)-fn(m)(x)|2dx⟶0.
Therefore, we have(3)fn+p(i)(0)-fn(i)(0)⟶0,i=0,1,…,m-1,∫-∞+∞‍[fn+p(m)(x)-fn(m)(x)]2dx⟶0,
which indicate that the sequences fn(i)(0)(0≤i≤m-1) and fn(m)(x) are Cauchy sequences, respectively, in R and L2(-∞,+∞). So, there exist unique real number λi(i=0,1,…,m-1) and function h(x)∈L2(-∞,+∞), such that
(4)limn→∞fn(i)(0)=λi,(i=0,1,…,m-1),limn→∞[fn(m)(x)-h(x)]2=L20.
Let(5)g(x)=∑i=0m-1λii!xi+1(m-1)!∫0x(x-t)m-1h(t)dt.
One easily sees that(6)g(i)(0)=λi,(i=0,1,…,m-1),g(m-1)(x)=λm-1+∫0xh(t)dt.
Becauseh(t)∈L2(-∞,+∞), we get g(x)∈Hm(-∞,+∞). Moreover, it holds
(7)∥fn-g∥H2=∑i=0m-1‍|fn(i)(0)-g(i)(0)|2+∫-∞+∞|fn(m)(x)-g(m)(x)|2dx=∑i=0m-1|fn(i)(0)-λi|2+∫-∞+∞‍|fn(m)(x)-h(x)|2dx⟶0,
which means that Hm(-∞,+∞) is a Hilbert space.
Meanwhile, we can introduce that(8)|f(m-1)(x)|=f(m-1)(0)+∫0x‍f(m)(t)dt≤|f(m-1)(0)|+∫0x‍|f(m)(t)|dt≤∥f∥H+x∫0x‍|f(m)(t)|2dt≤∥f∥H+x(∑i=0m-1‍|f(i)(0)|2+∫-∞+∞‍|f(m)(x)|2dx)1/2≤(1+x)∥f∥H=ax∥f∥H,|f(m-2)(x)|≤|f(m-2)(0)|+∫0x‍|f(m-1)(t)|dt≤∥f∥H+∫0x‍at∥f∥Hdt≤bx∥f∥H,
where ax and bx are positive numbers.
Then, we also introduce thatf(x)≤cx∥f∥H. This completes the proof of Theorem 3.


## 3. Calculation of Reproducing Kernel on Infinite Interval

In this section, we will give a novel method to calculate the reproducing kernel ofHm(-∞,+∞). Suppose R(x,y) is the reproducing kernel function. According to the definition of inner, for any fixed f(y)∈Hm(-∞,+∞), we have
(9)〈f(y),R(x,y)〉H=∑i=0m-1‍f(i)(0)∂yiR(x,0)+∫-∞+∞‍f(m)(y)∂ymR(x,y)dy=f(0)R(x,0)+f′(0)∂yR(x,0)+⋯+f(m-1)(0)∂y(m-1)R(x,0)+{(∫-∞0‍+∫0x‍+∫x+∞‍)f(m)×(y)∂ymR(x,y)dy,x≥0,(∫-∞x‍+∫x0‍+∫0+∞‍)f(m)×(y)∂ymR(x,y)dy,x<0.According to the formula(10)f(x)=f(0)+f′(0)x+f′′(0)2x2+⋯+1m!∫0x‍f(m+1)(t)(x-t)mdt,
it follows that if
(11)R(x,0)=1,∂yR(x,0)=x,∂y2R(x,0)=12x2,⋮∂ymR(x,y)={1(m-1)!Sign(x)(x-y)m-1,y∈[0,x]or[x,0],0,otherwise,
then 〈f(y),R(x,y)〉H=f(x) which indicates that R(x,y) is the reproducing kernel function.It is now clear how to computeR(x,y) by solving (11). Moreover, the resulting function is represented locally by polynomials.Example 4.
H
1
(
-
∞
,
+
∞
)
=
{
f
(
x
)
∣
f
(
x
)
is
an
absolutely
continuous
real
value
function
in
(
-
∞
,
+
∞
)
,
f
′
(
x
)
∈
L
2
(
-
∞
,
+
∞
)
} endowed inner product
(12)〈f(x),g(x)〉H1=f(0)g(0)+∫-∞+∞f′(x)g′(x)dx.
By applying the procedure of the previous statement, we can construct the equations:(13)R(x,0)=1,∂yR(x,y)={Sign(x),y∈[0,x]or[x,0],0,otherwise.Then, the reproducing kernel ofH1(-∞,+∞) is given by the following.(1)
Whenx<0(14)R(x,y)={1,0≤y,-y+1,x<y≤0,-x+1,y≤x.(2)
Whenx=0, R(x,y)=1.(3)
Whenx>0(15)R(x,y)={1,y≤0,y+1,0<y≤x,x+1,x≤y.It can be obviously obtained that for anyx∈R, we have
(16)R(x,y)=12(|y|+|x|-|y-x|)+1.The graph of theR(x,y) of H1(-∞,+∞) up to y=1,-1 is presented in Figures 1 and 2, and it shows the curvilinear figure of R(x,y).The curves are, respectively,R(x,1) and R(x,-1) for Example 4, where x∈[-2,2].
(a)
(b)Figure 2
Image ofR(x,y) for Example 4.Example 5.
H
3
[
0
,
+
∞
)
=
{
f
(
x
)
∣
f
′′
(
x
)
is
an
absolutely
continuous
real
value
function
in
[
0
,
+
∞
)
, f′′′(x)∈L2[0,+∞)} endowed inner product
(17)〈f(x),g(x)〉H3=f(0)g(0)+f′(0)g′(0)+f′′(0)g′′(0)+∫0+∞‍f′′′(x)g′′′(x)dx.
Similarly, theR(x,y) of H3[0,+∞) satisfying the following:
(18)R(x,0)=1,∂yR(x,0)=x,∂y2R(x,0)=12x2,∂y3R(x,y)={12(x-y)2,y≤x,0,y>x
can be obtained; namely,
(19)R(x,y)={1+xy+14x2y2+112x2y3-124xy4+1120y5,kkkkkkkkkkkkkkkkkkkkkkkkkkkkky≤x,1+xy+14x2y2+112x3y2-124x4y+1120x5,kkkkkkkkkkkkkkkkkkkkkkkkkkkkky>x.


## 4. A Concrete Application to Boundary Value Problems

### 4.1.  Application to Initial Value Problems

Actually, when we apply reproducing kernel theory to solve problems with boundary value conditions [16, 17], it is important to find the representation of reproducing kernel in the appropriate reproducing kernel space. In this section, we show how to express reproducing kernel function in terms of reproducing kernel space with boundary value condition on infinite interval.SetD=(-∞,+∞). Let H(D) be a reproducing kernel Hilbert space and R(x,y) its reproducing kernel function. As is well known,
(20)H0(D)={f(x)∣f(x0)=0,x0∈D,f∈H(D)}
is the closed subspace of H(D).Lemma 6.
IfH0(D) is the proper subspace of H(D), then for any y∈D,R(x,y)∉H0(D); namely, R(x0,y)≠0.Proof.
The proof is by contradiction. Suppose thatR(x,y)∈H0(D). Then for a fixed x0 it holds R(x0,y)=0. Choosing f(x)∈H(D), we have
(21)f(x0)=〈f(y),R(x0,y)〉H=〈f(y),0〉H=0,
which infers f(x)∈H0(D). It is conflict about proper subspace.Lemma 7.
The orthogonal complement space ofH0(D) about H(D) is(22)H0⊥(D)={g(x)∣g(x)=λR(x,x0),λ∈C}.Proof.
Due to the only restriction ofH0(D), it means that H0⊥ is one dimensional space. Moreover, for any f(x)∈H0(D) we get
(23)〈f(x),R(x,x0)〉H0=f(x0)=0;
Then, it indicates R(x,x0)⊥H0(D). Therefore, we have R(x,x0)∈H0⊥(D), and H0⊥(D) satisfies (22).By virtue of the orthogonal decomposition property of reproducing kernel, the reproducing kernel functionK(x,y) of H0(D) can be represented as
(24)R(x,y)=K(x,y)+λR(x,x0).So under the assumption thatK(x0,y)=0, we can introduce
(25)λ=R(x0,y)R(x0,x0).Here, according to Lemma6, we know that R(x0,x0)≠0.Now, the formula of reproducing kernel function ofH0(D) can be obtained by
(26)K(x,y)=R(x,y)-R(x0,y)R(x,x0)R(x0,x0).Example 8.
This problem corresponds to the closed subspace ofH3[0,+∞) in Example 5.
H
3
0
[
0
,
+
∞
)
=
{
f
(
x
)
∣
f
′′
(
x
)
is
an
absolutely
continuous
real
value
function
in
[
0
,
+
∞
)
, f′′′(x)∈L2[0,+∞)andf(10)=0}.
Here, we simply use formula (26) to find the reproducing kernel function K(x,y) of H30[0,+∞)(27)K(x,y)=R(10,10)R(x,y)-R(10,y)R(x,10)R(10,10),
where R(x,y) is determined by (19).


### 4.2. Application to Multipoint Boundary Value Problem

To see that the previous results can be generalized, letH(D) be a reproducing kernel space with kernel R(x,y). For any finite point set {x1,x2,…,xm}⊂D, we get
(28)H0(D)={f(x)∣f(xi)=0,i=1,2,…,m,f∈H(D)},
which is the closed subspace of H(D). Because, for any f∈H0(D),
(29)〈f(·),R(xi,·)〉H0=f(xi)=0i=1,2,…m,
we know that R(xi,x)∈H0⊥(D), where H0⊥(D) is m dimensional orthogonal complement space of H(D).Lemma 9.
If, for anyn∈N and differential points x1,x2,…,xn, the valuation functional x1*,x2*,…,xn* of reproducing kernel space H(D) are linearly independent, then the reproducing kernel function R(x,y) is positive definite.Proof.
Suppose thatR(x,y) is not positive definite. Then there exist differential points x1,x2,…,xn and nonzero vector u=(u1,u2,…,un)∈Cn such that
(30)∑i,jui¯ujR(xi,xj)=〈∑iui¯R(xi,·),∑juj¯R(xj,·)〉H=0.
It follows∑i‍ui¯R(xi,·)≡0. Therefore, for any f∈H(D), we have
(31)(∑i‍uixj*)(f)=∑i‍uif(xi)=∑i‍ui〈f,R(xi,·)〉H=〈f,∑i‍ui¯R(xi,·)〉H=0.
It indicates thatx1*,x2*,…,xn* are linearly dependent, which can be a contradiction.Theorem 10.
The reproducing kernel functionK(x,y) of H0(D) has the form
(32)K(x,y)=R(x,y)+∑i=1mλiR(x,xi).Proof.
It needs to be proven thatR(x,xi)(i=1,2,…,m) is the basis of H0⊥(D); namely, R(x,xi) are linearly independent.
Let∑i=1muiR(x,xi)=0. Then ∑i=1muiR(xi,xj)=0(j=1,2,…,m). Applying Lemma 9, we get that matrix (R(xi,xj))m×m is positive definite. So, ui=0(i=1,2,…,m), and formula (32) holds. Meanwhile, according to K(x,y)∈H0(D), we can obtain the uniquely defined λi.Example 11.
Here we apply the proposed method to solve the following simplified model of propagation of nonadiabatic flames inside long tubes [18–22]:
(33)-x′′(t)+cx′(t)+λx(t)=f(t,x(t),x′(t)),kkkkkkkkkkkkkkkkkkkkkkkkkkkkt∈(0,∞),x(0)-αx(η)=a0,limt→+∞x′(t)ert=b0.
Firstly, due to the complex boundary conditions of (33), the following reproducing kernel space is constructed.
H
=
{
x
(
t
)
∣
x
′
′
(
t
)
is
absolutely
continuous
in
[
0
,
+
∞
)
and
lim
t
→
∞
(
x
′
(
t
)
/
e
r
t
)
=
0
, ∫0+∞‍e-rt[x(3)(t)]2dt<+∞}. The inner product is defined, respectively, by
(34)〈x(t),y(t)〉H=∑i=02‍x(i)(0)y(i)(0)+∫0+∞‍e-rtx(3)(t)y(3)(t)dt.
Suppose thatR(t,s) is the reproducing kernel of H. According to the previous method for computation of reproducing kernel, we can get
(35)R(t,s)={r1(t,s)=ert∑i=13‍a1i(s)ti-1+∑i=46‍a1i(s)ti-4,t≤s,r2(t,s)=ert∑i=13‍a2i(s)ti-1+∑i=46‍a2i(s)ti-4s≤t.
Forr=1, the concrete expression of R(t,s) is given as follows:
(36)R(t,s)=12[es(12+6t+t2-2(3+t)s)kkkkk+et(12-6t+t2-2(t-3)s+s2)kkkkk-10-6t-t2-6skkkkk-2ts-t2s-s2-ts2kkkkk+e(1/2)(t+s+|t-s|)(6(t-s)-12-(t-s)2)].
Secondly, we need to establish the subspace ofH:
(37)H0={x(t)∣x(t)∈H,x(0)-αx(η)=0}.
Now, we use the formula (32) to get the reproducing kernel of H0:
(38)K(t,s)=R(t,s)-f(t)f(s)f(0)-αf(η),
where f(t)=R(t,0)-αR(t,η) and R(t,s) is the reproducing kernel function of H.
Thirdly, we will prove thatK(t,s) in (38) is the reproducing kernel of H0. For any x(t)∈H0, there holds that
(39)〈x(t),K(t,s)〉H0=〈x(t),R(t,s)-f(t)f(s)f(0)-αf(η)〉H0=x(s)-f(s)f(0)-αf(η)〈x(t),R(t,0)-αR(t,η)〉H0=x(s)-f(s)f(0)-αf(η)[x(0)-αx(η)]=x(s).
Forr=1,α=2, and η=1, the concrete expression of K(t,s) is given as follows
(40)K(t,s)={k1(t,s)s≤t≤η,k2(t,s)s≤η≤t,k3(t,s)η≤s≤t,k4(t,s)η≤t≤s,k5(t,s)t≤η≤s,k6(t,s)t≤s≤η;k1(t,s)=110(-52+5s2(-1+t)-5es(12-6s+s2-t)kkkkkk×(-1+t)+61t+60est-5t2kkkkkk-5ets(19-8t+t2)+s(61-68t+5t2));k2(t,s)=110(-52+5s2(-1+t)-5es(12-6s+s2-t)kkkkkkk×(-1+t)+61t-5t2-5es(7-8t+t2)kkkkkkk+s(61-68t+5t2));k3(t,s)=110(-52+5s2(-1+t)kkkkkk+61t-5t2+s(61-68t+5t2)kkkkkk+5es(12+s2+6t+t2-2s(3+t))kkkkkk-5e(7t+s2t+s(7-4t+t2)));k4(t,s)=k3(s,t);k5(t,s)=k2(s,t);k6(t,s)=k1(s,t).
Compared with the procedure for computation of the reproducing kernel in [22], we can see that our method is easier to implement, and it avoids the complexity of δ function.


## 4.1.  Application to Initial Value Problems

Actually, when we apply reproducing kernel theory to solve problems with boundary value conditions [16, 17], it is important to find the representation of reproducing kernel in the appropriate reproducing kernel space. In this section, we show how to express reproducing kernel function in terms of reproducing kernel space with boundary value condition on infinite interval.SetD=(-∞,+∞). Let H(D) be a reproducing kernel Hilbert space and R(x,y) its reproducing kernel function. As is well known,
(20)H0(D)={f(x)∣f(x0)=0,x0∈D,f∈H(D)}
is the closed subspace of H(D).Lemma 6.
IfH0(D) is the proper subspace of H(D), then for any y∈D,R(x,y)∉H0(D); namely, R(x0,y)≠0.Proof.
The proof is by contradiction. Suppose thatR(x,y)∈H0(D). Then for a fixed x0 it holds R(x0,y)=0. Choosing f(x)∈H(D), we have
(21)f(x0)=〈f(y),R(x0,y)〉H=〈f(y),0〉H=0,
which infers f(x)∈H0(D). It is conflict about proper subspace.Lemma 7.
The orthogonal complement space ofH0(D) about H(D) is(22)H0⊥(D)={g(x)∣g(x)=λR(x,x0),λ∈C}.Proof.
Due to the only restriction ofH0(D), it means that H0⊥ is one dimensional space. Moreover, for any f(x)∈H0(D) we get
(23)〈f(x),R(x,x0)〉H0=f(x0)=0;
Then, it indicates R(x,x0)⊥H0(D). Therefore, we have R(x,x0)∈H0⊥(D), and H0⊥(D) satisfies (22).By virtue of the orthogonal decomposition property of reproducing kernel, the reproducing kernel functionK(x,y) of H0(D) can be represented as
(24)R(x,y)=K(x,y)+λR(x,x0).So under the assumption thatK(x0,y)=0, we can introduce
(25)λ=R(x0,y)R(x0,x0).Here, according to Lemma6, we know that R(x0,x0)≠0.Now, the formula of reproducing kernel function ofH0(D) can be obtained by
(26)K(x,y)=R(x,y)-R(x0,y)R(x,x0)R(x0,x0).Example 8.
This problem corresponds to the closed subspace ofH3[0,+∞) in Example 5.
H
3
0
[
0
,
+
∞
)
=
{
f
(
x
)
∣
f
′′
(
x
)
is
an
absolutely
continuous
real
value
function
in
[
0
,
+
∞
)
, f′′′(x)∈L2[0,+∞)andf(10)=0}.
Here, we simply use formula (26) to find the reproducing kernel function K(x,y) of H30[0,+∞)(27)K(x,y)=R(10,10)R(x,y)-R(10,y)R(x,10)R(10,10),
where R(x,y) is determined by (19).


## 4.2. Application to Multipoint Boundary Value Problem

To see that the previous results can be generalized, letH(D) be a reproducing kernel space with kernel R(x,y). For any finite point set {x1,x2,…,xm}⊂D, we get
(28)H0(D)={f(x)∣f(xi)=0,i=1,2,…,m,f∈H(D)},
which is the closed subspace of H(D). Because, for any f∈H0(D),
(29)〈f(·),R(xi,·)〉H0=f(xi)=0i=1,2,…m,
we know that R(xi,x)∈H0⊥(D), where H0⊥(D) is m dimensional orthogonal complement space of H(D).Lemma 9.
If, for anyn∈N and differential points x1,x2,…,xn, the valuation functional x1*,x2*,…,xn* of reproducing kernel space H(D) are linearly independent, then the reproducing kernel function R(x,y) is positive definite.Proof.
Suppose thatR(x,y) is not positive definite. Then there exist differential points x1,x2,…,xn and nonzero vector u=(u1,u2,…,un)∈Cn such that
(30)∑i,jui¯ujR(xi,xj)=〈∑iui¯R(xi,·),∑juj¯R(xj,·)〉H=0.
It follows∑i‍ui¯R(xi,·)≡0. Therefore, for any f∈H(D), we have
(31)(∑i‍uixj*)(f)=∑i‍uif(xi)=∑i‍ui〈f,R(xi,·)〉H=〈f,∑i‍ui¯R(xi,·)〉H=0.
It indicates thatx1*,x2*,…,xn* are linearly dependent, which can be a contradiction.Theorem 10.
The reproducing kernel functionK(x,y) of H0(D) has the form
(32)K(x,y)=R(x,y)+∑i=1mλiR(x,xi).Proof.
It needs to be proven thatR(x,xi)(i=1,2,…,m) is the basis of H0⊥(D); namely, R(x,xi) are linearly independent.
Let∑i=1muiR(x,xi)=0. Then ∑i=1muiR(xi,xj)=0(j=1,2,…,m). Applying Lemma 9, we get that matrix (R(xi,xj))m×m is positive definite. So, ui=0(i=1,2,…,m), and formula (32) holds. Meanwhile, according to K(x,y)∈H0(D), we can obtain the uniquely defined λi.Example 11.
Here we apply the proposed method to solve the following simplified model of propagation of nonadiabatic flames inside long tubes [18–22]:
(33)-x′′(t)+cx′(t)+λx(t)=f(t,x(t),x′(t)),kkkkkkkkkkkkkkkkkkkkkkkkkkkkt∈(0,∞),x(0)-αx(η)=a0,limt→+∞x′(t)ert=b0.
Firstly, due to the complex boundary conditions of (33), the following reproducing kernel space is constructed.
H
=
{
x
(
t
)
∣
x
′
′
(
t
)
is
absolutely
continuous
in
[
0
,
+
∞
)
and
lim
t
→
∞
(
x
′
(
t
)
/
e
r
t
)
=
0
, ∫0+∞‍e-rt[x(3)(t)]2dt<+∞}. The inner product is defined, respectively, by
(34)〈x(t),y(t)〉H=∑i=02‍x(i)(0)y(i)(0)+∫0+∞‍e-rtx(3)(t)y(3)(t)dt.
Suppose thatR(t,s) is the reproducing kernel of H. According to the previous method for computation of reproducing kernel, we can get
(35)R(t,s)={r1(t,s)=ert∑i=13‍a1i(s)ti-1+∑i=46‍a1i(s)ti-4,t≤s,r2(t,s)=ert∑i=13‍a2i(s)ti-1+∑i=46‍a2i(s)ti-4s≤t.
Forr=1, the concrete expression of R(t,s) is given as follows:
(36)R(t,s)=12[es(12+6t+t2-2(3+t)s)kkkkk+et(12-6t+t2-2(t-3)s+s2)kkkkk-10-6t-t2-6skkkkk-2ts-t2s-s2-ts2kkkkk+e(1/2)(t+s+|t-s|)(6(t-s)-12-(t-s)2)].
Secondly, we need to establish the subspace ofH:
(37)H0={x(t)∣x(t)∈H,x(0)-αx(η)=0}.
Now, we use the formula (32) to get the reproducing kernel of H0:
(38)K(t,s)=R(t,s)-f(t)f(s)f(0)-αf(η),
where f(t)=R(t,0)-αR(t,η) and R(t,s) is the reproducing kernel function of H.
Thirdly, we will prove thatK(t,s) in (38) is the reproducing kernel of H0. For any x(t)∈H0, there holds that
(39)〈x(t),K(t,s)〉H0=〈x(t),R(t,s)-f(t)f(s)f(0)-αf(η)〉H0=x(s)-f(s)f(0)-αf(η)〈x(t),R(t,0)-αR(t,η)〉H0=x(s)-f(s)f(0)-αf(η)[x(0)-αx(η)]=x(s).
Forr=1,α=2, and η=1, the concrete expression of K(t,s) is given as follows
(40)K(t,s)={k1(t,s)s≤t≤η,k2(t,s)s≤η≤t,k3(t,s)η≤s≤t,k4(t,s)η≤t≤s,k5(t,s)t≤η≤s,k6(t,s)t≤s≤η;k1(t,s)=110(-52+5s2(-1+t)-5es(12-6s+s2-t)kkkkkk×(-1+t)+61t+60est-5t2kkkkkk-5ets(19-8t+t2)+s(61-68t+5t2));k2(t,s)=110(-52+5s2(-1+t)-5es(12-6s+s2-t)kkkkkkk×(-1+t)+61t-5t2-5es(7-8t+t2)kkkkkkk+s(61-68t+5t2));k3(t,s)=110(-52+5s2(-1+t)kkkkkk+61t-5t2+s(61-68t+5t2)kkkkkk+5es(12+s2+6t+t2-2s(3+t))kkkkkk-5e(7t+s2t+s(7-4t+t2)));k4(t,s)=k3(s,t);k5(t,s)=k2(s,t);k6(t,s)=k1(s,t).
Compared with the procedure for computation of the reproducing kernel in [22], we can see that our method is easier to implement, and it avoids the complexity of δ function.


## 5. Conclusions and Future Work

To summarize, in this paper, a new method for the calculation of reproducing kernel on infinite interval was introduced, and the representation in polynomial form was obtained for the first time. The scheme was then used to generate the formula for the reproducing kernel in reproducing kernel space with boundary value conditions. We end this paper by mentioning the following applications. On one hand, the approach detailed here can be readily adapted to the case of reproducing kernel on the finite interval. According to the former method in [15], which cannot represent reproducing kernel on the infinite interval in polynomial form, the advantages of the present approach are that we use theory of elementary to avoid the complex operation and numerical algorithm will be much more timesaving. On the other hand, the formula in Section 4 can be used to solve multipoint boundary value problems on the positive half-line, such as in [23–25].



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*Source: 959346-2013-09-19.xml*