# A Unified Reproducing Kernel Method and Error Estimation for Solving Linear Differential Equation with Functional Constraints

Authors: Xinjian Zhang; Xiongwei Liu
Journal: Mathematical Problems in Engineering
 (2015)

---

## Abstract

A unified reproducing kernel method for solving linear differential equations with functional constraint is provided. We use a specified inner product to obtain a class of piecewise polynomial reproducing kernels which have a simple unified description. Arbitrary order linear differential operator is proved to be bounded about the special inner product. Based on space decomposition, we present the expressions of exact solution and approximate solution of linear differential equation by the polynomial reproducing kernel. Error estimation of approximate solution is investigated. Since the approximate solution can be described by polynomials, it is very suitable for numerical calculation.


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## Body

## 1. Introduction

The reproducing kernel method of solving differential equation is one of the important topics of reproducing kernel numerical methods. In recent years, reproducing kernel theories are widely applied to solve all kinds of the differential equations, which provide new description forms for exact solutions and approximate solutions of many classical equations. For example, Zhang and Cui [1] and Li [2] obtained the analytical solutions described as series for the steady state convecting equations and boundary value problems, separately. Wang et al. [3] not only got the exact solution described as series for a class of boundary value problems of ordinary differential equations, but also proposed iterative method for the approximate solutions. Gao et al. [4] and Li and Wu [5] investigated the reproducing kernel methods to solve singular second-order initial/boundary value problems and multipoint boundary value problems. Zhang [6] and Wu and Lin [7] specially introduced the reproducing kernel methods of solving linear differential equation based on reproducing kernel theory. Shi et al. [8] and Du and Zhang [9] obtained the analytical solutions and the corresponding approximate solutions of initial value problems of linear ordinary differential equations and error estimation was considered. Li and Wu [10] considered the error estimation of solving boundary value problem of linear equations. Castro et al. [11] introduced a new method to solve general initial value problems by means of reproducing kernel theory.However, these studies were generally concentrated on constructing the corresponding reproducing kernels for solving specific equations. There was no unified description for these methods. Therefore, how to construct a common method of reproducing kernel for a class of differential equations is worthy of further study about theories and methods. In this paper, for a class of frequently used reproducing kernel spaceW
2
s
[
0,1
], a class of piecewise polynomial reproducing kernels is presented by a specific inner product. These reproducing kernels have a simple unified expression for all W
2
s
[
0,1
]. They are suitable for numerical calculation and independent of the concrete form of the operator L. According to the given inner product, the arbitrary kth-order linear differential operator is proved to be bounded operator of W
2
n
[
0,1
]
→
W
2
m
[
0,1
]  (if only k
=
n
-
m). Based on space decomposition, {
ψ
i
} (defined as (1)) were proved to always be complete in a subspace. Finally, we give the error estimation of approximate solution. These results provide a unified reproducing kernel theory and method for solving arbitrary linear differential equation with arbitrary linear functional constraint.


## 2. Preliminary

For the convenience of description, we first give a brief description of the problem concerned in this paper. LetH
1, H
2 be reproducing kernel Hilbert spaces composed of functions defined on a certain Ω. The inner products are ·
,
·
H
1, ·
,
·
H
2 and reproducing kernels are K
H
1
(
t
,
τ
), K
H
2
(
t
,
τ
), separately. L denotes a linear differential operator from H
1 to H
2. For the given u
∈
H
2, under certain conditions, we need to find f
∈
H
1 such that L
f
(
t
)
=
u
(
t
). The main idea of solving this kind of equation by reproducing kernel method is presented as follows:(i)
For supposingL
∗
:
H
2
→
H
1 to be conjugate operator of L, prove L to be a linear bounded operator from H
1 to H
2.
(ii)
For the specified dense sequence{
x
i
}
⊂
Ω, which is composed of distinct numbers, let (1)
σ
i
t
=
K
H
2
t
,
x
i
,
ψ
i
t
=
L
∗
σ
i
t
,
i
=
1,2
,
…
,
and  then prove{
ψ
i
} to be complete about H
1.
(iii)
Based on (ii), orthonormalize{
ψ
i
} to obtain {
ψ
~
i
} and suppose ψ
~
i
=
∑
j
=
1
i
β
i
j
ψ
j; then, from (1) and properties of reproducing kernel and conjugate operator, we can get the solution of L
f
(
t
)
=
u
(
t
) as(2)
f
t
=
∑
i
=
1
∞
f
,
ψ
~
i
H
1
ψ
~
i
t
=
∑
i
=
1
∞
∑
j
=
1
i
β
i
j
L
f
,
K
H
2
·
,
x
i
H
2
ψ
~
i
t
=
∑
i
=
1
∞
∑
j
=
1
i
β
i
j
L
f
x
i
ψ
~
i
t
=
∑
i
=
1
∞
∑
j
=
1
i
β
i
j
u
x
i
ψ
~
i
t
.From (2), we can obtain the exact expression of the solution and can do numerical calculation by values of the given u
(
x
i
).According to the above process, we can see that if we want to solve differential equation, (i) and (ii) must be true. To verify these two conditions, the frequently used method is to firstly design an inner product and corresponding reproducing kernel according to the specified differential operatorL. Then, the boundedness of L is considered and, for ensuring {
ψ
i
} to be complete, L has to be invertible. For each specific differential equation, we have the same process. Moreover, operator L usually is not invertible. The forms of reproducing kernels obtained by solving high-order differential equation are complicated and have no unified structural properties. These also make it difficult to verify the boundedness of operator L and the completeness of {
ψ
i
}.


## 3. Polynomial Reproducing Kernel inW
2
s
[
0,1
]

For any positive integers, let(3)
W
2
s
0,1
=
f
t
,
t
∈
0,1
:
f
s
-
1
t
is  absolutely  continuous  on
0,1
and
f
s
t
∈
L
2
0,1
,where L
2
[
0,1
] is a square integrable function space. According to [6], every f
(
t
)
∈
W
2
s
[
0,1
] can be written as(4)
f
t
=
∑
i
=
1
s
f
i
-
1
0
e
i
t
+
∫
0
1
g
s
t
,
τ
f
s
τ
d
τ
,where(5)
e
i
t
=
t
i
-
1
i
-
1
!
,
g
s
t
,
τ
=
1
s
-
1
!
t
-
τ
+
s
-
1
.For any u
∈
L
2
[
0,1
], we have (6)
d
i
d
t
i
∫
0
1
g
s
t
,
τ
u
τ
d
τ
=
∫
0
1
∂
i
∂
t
i
g
t
,
τ
u
τ
d
τ
,
0
≤
i
≤
s
-
1
,
d
s
d
t
s
∫
0
1
g
s
t
,
τ
u
τ
d
τ
=
u
t
.
W
2
s
[
0,1
] is a reproducing kernel Hilbert space with respect to inner product (7)
f
,
h
s
=
∑
i
=
1
s
f
i
-
1
0
·
h
i
-
1
0
+
∫
0
1
f
s
t
·
h
s
t
d
tand has a reproducing kernel (8)
K
s
t
,
τ
=
∑
i
=
1
s
e
i
t
e
i
τ
+
∫
0
1
g
s
t
,
x
·
g
s
τ
,
x
d
x
.For the convenience of numerical calculation, we explore more concrete form of reproducing kernelK
s
(
t
,
τ
). Suppose that (9)
E
s
t
,
τ
=
∫
0
1
g
s
t
,
x
·
g
s
τ
,
x
d
x
.The key of calculating reproducing kernel K
s
(
t
,
τ
) is to calculate E
s
(
t
,
τ
). Let (10)
e
^
i
t
=
-
1
s
-
i
t
s
-
i
s
-
i
!
,
i
=
1,2
,
…
,
s
;then we can obtain (11)
g
s
t
,
x
=
1
s
-
1
!
t
-
x
+
s
-
1
=
∑
i
=
1
s
e
i
t
·
e
^
i
x
t
-
x
+
0
,where e
i
(
t
) is as the definition in (5).It follows that whent
≥
τ, (12)
E
s
t
,
τ
=
∫
0
τ
∑
i
=
1
s
e
i
t
e
^
i
x
·
∑
j
=
1
s
e
j
τ
e
^
j
x
d
x
=
∑
i
=
1
s
e
i
t
∫
0
τ
∑
j
=
1
s
e
j
τ
e
^
j
x
·
e
^
i
x
d
x
=
1
s
-
1
!
∑
i
=
1
s
e
i
t
∫
0
τ
τ
-
x
s
-
1
·
e
^
i
x
d
x
,
∫
0
τ
τ
-
x
s
-
1
·
e
^
i
x
d
x
=
-
1
s
-
i
s
-
i
!
∫
0
τ
τ
-
x
s
-
1
·
x
s
-
i
d
x
=
-
1
s
-
i
s
-
i
!
τ
2
s
-
i
·
B
s
+
1
-
i
,
s
,where B
(
a
,
b
)
=
∫
0
1
‍
x
a
-
1
(
1
-
x
)
b
-
1
d
x is a Beta function. As a consequence, we have (13)
E
s
t
,
τ
=
1
s
-
1
!
∑
i
=
1
s
-
1
s
-
i
B
s
+
1
-
i
,
s
s
-
i
!
i
-
1
!
t
i
-
1
τ
2
s
-
i
,
t
≥
τ
.When t
<
τ, we can get E
s
(
t
,
τ
)
=
E
s
(
τ
,
t
) by symmetry. By Beta function, we have(14)
E
s
t
,
τ
=
∑
i
=
1
s
-
1
s
-
i
2
s
-
i
!
i
-
1
!
t
i
-
1
τ
2
s
-
i
,
t
≥
τ
,
∑
i
=
1
s
-
1
s
-
i
2
s
-
i
!
i
-
1
!
t
2
s
-
i
τ
i
-
1
,
t
<
τ
.Then, from (8), we can get(15)
K
s
t
,
τ
=
∑
i
=
1
s
t
i
-
1
τ
i
-
1
i
-
1
!
2
+
E
s
t
,
τ
.Obviously, K
s
(
t
,
τ
) is a piecewise polynomial of two variables.Lemma 1.
For any dense sequence{
x
i
} composed of distinct numbers in [
0,1
], function sequence (16)
ϕ
i
t
=
K
s
t
,
x
i
,
i
=
1,2
,
…
,is a linearly independent complete system in W
2
s
[
0,1
].Proof.
For any positive integerN, if there are constants c
1
,
c
2
,
…
,
c
N such that (17)
ϕ
=
c
1
ϕ
1
+
c
2
ϕ
2
+
⋯
+
c
N
ϕ
N
=
θ
s
θ
s
represents  the  zero  element  of
W
2
s
0,1
,namely, ϕ
s
=
0, then, for any f
∈
W
2
s
[
0,1
], it can follow that (18)
0
=
f
,
ϕ
s
=
c
1
f
x
1
+
c
2
f
x
2
+
⋯
+
c
N
f
x
N
.Take f in the above equation as 1
,
x
,
…
,
x
N
-
1 in turn; we can obtain c
1
=
c
2
=
⋯
=
c
N
=
0. Therefore, ϕ
1
,
ϕ
2
,
…
,
ϕ
N are linearly independent.
For anyf
∈
W
2
s
[
0,1
], if f
,
ϕ
i
s
=
0
(
i
=
1,2
,
…
), that is, f
(
x
i
)
=
f
,
K
s
(
·
,
x
i
)
s
=
0, since {
x
i
} are dense and f is continuous in [
0,1
], then f
(
t
)
=
0 and ϕ
1
,
ϕ
2
,
… are complete.
This completes the proof of Lemma1.Remark 2.
The above proof is independent of the inner product ofW
2
s
[
0,1
] and the concrete form of reproducing kernel. Therefore, Lemma 1 is always true for arbitrary inner product of W
2
s
[
0,1
] and the corresponding reproducing kernel.


## 4. Reproducing Kernel Method for Solution of Linear Differential Equation

Suppose thatkth-order linear differential operator is (19)
L
=
D
k
+
a
k
-
1
t
D
k
-
1
+
⋯
+
a
1
t
D
+
a
0
t
,
t
∈
0,1
,where a
j
(
t
)
∈
C
j
[
0,1
] and integers m and n satisfy 0
≤
m
<
n and k
=
n
-
m. It is obvious that L is a linear surjective map from W
2
n
[
0,1
] to W
2
m
[
0,1
]. The null space Ker
⁡
L is a k-dimensional subspace of W
2
n
[
0,1
].Whenm
=
0, then k
=
n. Denote W
2
0
[
0,1
] by L
2
[
0,1
]; that is, W
2
0
[
0,1
]
=
L
2
[
0,1
]. The corresponding inner product is (20)
f
,
h
0
=
∫
0
1
f
t
·
h
t
d
t
=
f
,
h
L
2
.Theorem 3.
Suppose thata
i
(
m
)
(
t
)
(
0
≤
i
≤
k
-
1
) exist and are bounded on [
0,1
]; then the differential operator L is a bounded linear operator from W
2
n
[
0,1
] to W
2
m
[
0,1
].Proof.
We only need to prove thatL is bounded. For any f
∈
W
2
n
[
0,1
] and 0
≤
i
≤
n
-
1, there is (21)
f
i
t
2
≤
2
f
i
0
2
+
2
∫
0
t
f
i
+
1
τ
d
τ
2
≤
2
f
i
0
2
+
2
∫
0
1
f
i
+
1
t
2
d
t
.It follows that(22)
∫
0
1
f
i
t
2
d
t
≤
2
f
i
0
2
+
2
∫
0
1
f
i
+
1
t
2
d
t
.According to the above inequality, for i
≤
n, we have(23)
∫
0
1
f
i
t
2
d
t
≤
2
f
i
0
2
+
2
2
f
i
+
1
0
2
+
⋯
+
2
n
-
i
f
n
-
1
0
2
+
2
n
-
i
∫
0
1
f
n
t
2
d
t
≤
2
n
-
i
f
i
0
2
+
f
i
+
1
0
2
+
⋯
+
f
n
-
1
0
2
+
∫
0
1
f
n
t
2
d
t
≤
2
n
-
i
∑
i
=
1
n
f
i
-
1
0
2
+
∫
0
1
f
n
t
2
d
t
=
2
n
-
i
f
n
2
.
Whenm
≥
1, the norm of L
f in W
2
m
[
0,1
] is (24)
L
f
m
2
=
∑
i
=
1
m
L
f
i
-
1
0
2
+
∫
0
1
L
f
m
t
2
d
t
.Since(25)
L
f
i
2
=
f
k
+
a
k
-
1
f
k
-
1
+
a
k
-
2
f
k
-
2
+
⋯
+
a
1
f
1
+
a
0
f
i
2
=
f
k
+
i
+
∑
j
=
0
i
C
i
j
a
k
-
1
j
f
k
-
1
+
i
-
j
+
a
k
-
2
j
f
k
-
2
+
i
-
j
+
⋯
+
a
0
j
f
i
-
j
2
≤
2
f
k
+
i
2
+
2
∑
j
=
0
i
C
i
j
a
k
-
1
j
f
k
-
1
+
i
-
j
+
a
k
-
2
j
f
k
-
2
+
i
-
j
+
⋯
+
a
0
j
f
i
-
j
2
,it is easily known that, for arbitrary n numbers c
1
,
c
2
,
…
,
c
n, we have (26)
c
1
+
c
2
+
⋯
+
c
n
2
≤
2
n
-
1
c
1
2
+
c
2
2
+
⋯
+
c
n
2and C
i
j
≤
m
! for 0
≤
i
≤
m. Because a
i
(
m
)
(
t
)
(
0
≤
i
≤
k
-
1
) are bounded, there exists a constant A
>
0 (we might as well put A
>
1) such that (27)
a
i
j
t
≤
A
0
≤
i
≤
k
-
1
,
0
≤
j
≤
m
.Then, we can get(28)
∑
j
=
0
i
C
i
j
a
k
-
1
j
f
k
-
1
+
i
-
j
+
a
k
-
2
j
f
k
-
2
+
i
-
j
+
⋯
+
a
0
j
f
i
-
j
2
≤
m
!
2
2
i
∑
j
=
0
i
∑
l
=
1
k
a
k
-
l
j
f
k
-
l
+
i
-
j
2
≤
m
!
2
2
m
2
k
-
1
∑
j
=
0
i
∑
l
=
1
k
a
k
-
l
j
f
k
-
l
+
i
-
j
2
≤
m
!
2
2
n
-
1
A
∑
j
=
0
i
∑
l
=
1
k
f
k
+
i
-
l
-
j
2
≤
m
!
2
2
n
-
1
A
·
i
+
1
f
k
+
i
-
1
2
+
f
k
+
i
-
2
2
+
⋯
+
f
1
2
+
f
2
.Let C
1
=
(
m
!
)
2
2
n
(
m
+
1
)
A; we have(29)
L
f
i
2
≤
C
1
f
k
+
i
2
+
f
k
+
i
-
1
2
+
⋯
+
f
1
2
+
f
2
≤
C
1
f
n
2
+
f
n
-
1
2
+
⋯
+
f
1
2
+
f
2
,
0
≤
i
≤
m
.Note that C
1 is independent of i, so, from (29) and (23), we can obtain (30)
∑
i
=
1
m
L
f
i
-
1
0
2
≤
m
C
1
f
n
0
2
+
f
n
-
1
0
2
+
⋯
+
f
2
0
,
∫
0
1
L
f
m
t
2
d
t
≤
C
1
∫
0
1
f
n
t
2
+
f
n
-
1
t
2
+
⋯
+
f
2
t
d
t
≤
C
1
1
+
2
+
⋯
+
2
n
f
n
2
≤
2
n
+
1
C
1
f
n
2
.Substituting the above two formulas into (24) and letting C
=
[
2
n
+
1
+
m
]
C
1, we can get (31)
L
f
m
2
≤
m
C
1
∑
i
=
0
n
-
1
f
i
0
2
+
2
n
+
1
C
1
f
n
2
≤
C
f
n
2
.It shows that L is a bounded operator from W
2
n
[
0,1
] to W
2
m
[
0,1
].
Whenm
=
0, then k
=
n, and the norm of L
f in W
2
0
[
0,1
] is L
f
0
2
=
∫
0
1
‍
[
L
f
(
t
)
]
2
d
t. According to the conditions, we can know that a
i
(
t
) is bounded on [
0,1
]. Therefore, there exists A
0
>
0 (we might as well put A
0
>
1) such that a
i
(
t
)
≤
A
0
(
0
≤
i
≤
n
-
1
).
From (29) and (23), we can obtain(32)
∫
0
1
L
f
t
2
d
t
≤
2
n
A
0
∫
0
1
f
n
t
2
+
f
n
-
1
t
2
+
⋯
+
f
1
t
2
+
f
t
2
d
t
≤
2
n
A
0
1
+
2
+
⋯
+
2
n
f
n
2
≤
2
n
+
k
+
1
A
0
f
n
2
.This shows that L is a bounded operator from W
2
n
[
0,1
] to W
2
0
[
0,1
].
This completes the proof of Theorem3.The null spaceKer
⁡
L of L is k-dimensional subspace of W
2
n
[
0,1
]. Suppose that λ
1
,
λ
2
,
…
,
λ
k are continuous linear functional and linearly independent on Ker
⁡
L and suppose that the values of r
1
,
r
2
,
…
,
r
k are given. Now we consider the following definite solutions problem.For a givenu
∈
W
2
m
[
0,1
], find f
∈
W
2
n
[
0,1
] such that it satisfies(33)
L
f
t
=
u
t
,
λ
i
f
=
r
i
i
=
1,2
,
…
,
k
.Denoted by Λ
=
(
λ
1
,
λ
2
,
…
,
λ
k
), Λ is a vector-valued linear functional on W
2
n
[
0,1
]; namely, Λ
f
=
(
λ
1
f
,
…
,
λ
k
f
).Lemma 4.
W
2
n
[
0,1
] has direct sum decomposition: (34)
W
2
n
0,1
=
Ker
⁡
L
+
Ker
⁡
Λ
,
Ker
⁡
L
∩
Ker
⁡
Λ
=
0
.Proof.
Becauseλ
1
,
λ
2
,
…
,
λ
k are linearly independent on Ker
⁡
L, there is a basis φ
1
,
φ
2
,
…
,
φ
k of Ker
⁡
L which is dual to λ
1
,
λ
2
,
…
,
λ
k; namely, (35)
L
φ
i
t
=
0
,
λ
i
φ
j
=
δ
i
j
,
1
≤
i
,
j
≤
k
.
SinceKer
⁡
L is a closed subspace of W
2
n
[
0,1
], every f
∈
W
2
n
[
0,1
] has a unique decomposition form: (36)
f
t
=
f
1
t
+
f
2
t
,
f
1
∈
Ker
⁡
L
,
f
2
∈
Ker
⁡
L
⊥
.From (35), we can obtain f
1
(
t
)
=
∑
i
=
1
k
(
λ
i
f
1
)
φ
i
(
t
); then (37)
f
2
t
=
f
t
-
∑
i
=
1
k
λ
i
f
1
φ
i
t
.Let (38)
f
L
t
=
∑
i
=
1
k
λ
i
f
φ
i
t
,
f
Λ
t
=
f
2
t
-
∑
i
=
1
k
λ
i
f
2
φ
i
t
;we have (39)
f
t
=
∑
i
=
1
k
λ
i
f
φ
i
t
+
f
Λ
t
=
f
L
t
+
f
Λ
t
.From (35) again, we can get f
L
∈
Ker
⁡
L and f
Λ
∈
Ker
⁡
Λ.
Iff
∈
Ker
⁡
L
∩
Ker
⁡
Λ, there is f
(
t
)
=
∑
i
=
1
k
(
λ
i
f
)
φ
i
(
t
) and λ
i
f
=
0
(
1
≤
i
≤
k
). It follows that f
(
t
)
=
0. Therefore, Ker
⁡
L
∩
Ker
⁡
Λ
=
{
0
}.
This completes the proof of Lemma4.According to the proof of Lemma4, we have (40)
L
f
Λ
t
=
L
f
t
,
λ
i
f
L
=
λ
i
f
,
i
=
1,2
,
…
,
k
.Since L
:
Ker
⁡
Λ
→
W
2
m
[
0,1
] is a bounded linear operator, there exists a conjugate operator L
∗
:
W
2
m
[
0,1
]
→
Ker
⁡
Λ. For any dense sequence {
x
i
} composed of distinct numbers in [
0,1
], when m
≥
1, suppose that(41)
σ
i
t
=
K
m
t
,
x
i
,
ψ
i
t
=
L
∗
σ
i
t
,
i
=
1,2
,
…
;then we have the following lemma.Lemma 5.
As in Lemma4, suppose that f
(
t
)
=
f
L
(
t
)
+
f
Λ
(
t
); then(i)
for anyf
∈
W
2
n
[
0,1
], f
,
ψ
i
n
=
0
(
i
=
1,2
,
…
)
⇔
f
=
f
L;
(ii)
for anyf
Λ
∈
Ker
⁡
Λ, f
Λ
,
ψ
i
n
=
0
(
i
=
1,2
,
…
)
⇔
f
Λ
=
0.Proof.
Suppose thatf
∈
W
2
n
[
0,1
]. If f
,
ψ
i
n
=
0
(
i
=
1,2
,
…
), then, from (41), we have (42)
L
f
Λ
,
K
m
·
,
x
i
m
=
L
f
,
K
m
·
,
x
i
m
=
f
,
ψ
i
n
=
0
i
=
1,2
,
…
.Since L
f
Λ
∈
W
2
m
[
0,1
] and from Lemma 1, {
K
m
(
·
,
x
i
)
} are complete in W
2
m
[
0,1
], so L
f
Λ
=
0. As f
Λ satisfies λ
i
f
Λ
=
0
(
1
≤
i
≤
k
), therefore, f
Λ
=
0; that is to say, f
=
f
L. By contrast, if f
=
f
L, namely, L
f
=
0, then f
,
ψ
i
n
=
L
f
,
K
m
(
·
,
x
i
)
m
=
0
(
i
=
1,2
,
…
). It follows that (i) is true. Meanwhile, according to the proof, we can also know that (ii) is true.
Thus, we finish the proof of Lemma5.Theorem 6.
Suppose thatH
n
=
Span
⁡
{
ψ
i
}
+
Ker
⁡
Λ; then {
ψ
i
} are linearly independent and complete in H
n.Proof.
Clearly,{
ψ
i
}
⊂
H
n. We first prove that {
ψ
i
} are linearly independent. Suppose that there are a positive integer N and constants c
1
,
c
2
,
…
,
c
N such that ψ
=
c
1
ψ
1
+
c
2
ψ
2
+
⋯
+
c
N
ψ
N
=
θ
n; namely, (43)
L
∗
c
1
σ
1
+
c
2
σ
2
+
⋯
+
c
N
σ
N
=
θ
n
,
c
1
σ
1
+
c
2
σ
2
+
⋯
+
c
N
σ
N
∈
Ker
⁡
L
∗
.Since Ker
⁡
L
∗
=
(
R
a
n
g
L
)
⊥ and L is surjective, (
R
a
n
g
L
)
⊥
=
(
W
2
m
[
0,1
]
)
⊥
=
θ
m, we have (44)
c
1
σ
1
+
c
2
σ
2
+
⋯
+
c
N
σ
N
=
θ
m
.From Lemma 1, σ
1
,
σ
2
,
…
,
σ
N are linearly independent; then c
1
=
c
2
=
⋯
=
c
N
=
0. Therefore, {
ψ
i
} are linearly independent.
Next, we prove that{
ψ
i
} is a complete system of H
n.
Suppose thatf
∈
H
n satisfies f
,
ψ
i
n
=
0
(
i
=
1,2
,
…
). When f
∈
S
p
a
n
{
ψ
i
}, it is obvious that f
=
0. From Lemma 5, it is easily obtained that when f
=
f
Λ, f
=
0 and when f
=
h
+
f
Λ
(
h
∈
S
p
a
n
{
ψ
i
}
), f
=
f
L; namely, f
=
h
=
h
L. Then, f
∈
S
p
a
n
{
ψ
i
} and f
,
ψ
i
n
=
0
(
i
=
1,2
,
…
). As a result, f
=
0. Therefore, {
ψ
i
} is a complete system of H
n.
This completes the proof of Theorem6.Theorem 7.
Denote{
ψ
i
} orthonormalized in W
2
n
[
0,1
] by {
ψ
~
i
} and suppose that (45)
ψ
~
i
t
=
∑
j
=
1
i
β
i
j
ψ
j
t
i
=
1,2
,
…
;then, the solution f
-
(
t
) of (33) can be expressed as (46)
f
-
t
=
∑
i
=
1
k
r
i
φ
i
t
+
∑
i
=
1
∞
∑
j
=
1
i
β
i
j
u
x
j
ψ
~
i
t
.Proof.
From Lemma4 and (39), f
-
(
t
) can be expressed as (47)
f
-
t
=
∑
i
=
1
k
r
i
φ
i
t
+
f
-
Λ
t
.Since f
-
Λ
∈
H
n, from Theorem 6, we suppose that f
-
Λ
(
t
)
=
∑
i
=
1
∞
f
-
Λ
,
ψ
~
i
n
ψ
~
i
(
t
); according to the properties of conjugate operator and reproducing kernel, it follows that(48)
f
-
Λ
t
=
∑
i
=
1
∞
f
-
Λ
,
ψ
~
i
n
ψ
~
i
t
=
∑
i
=
1
∞
∑
j
=
1
i
β
i
j
f
-
Λ
,
ψ
j
n
ψ
~
i
t
=
∑
i
=
1
∞
∑
j
=
1
i
β
i
j
f
-
Λ
,
L
∗
K
m
·
,
x
j
n
ψ
~
i
t
=
∑
i
=
1
∞
∑
j
=
1
i
β
i
j
L
f
-
,
K
m
·
,
x
j
m
ψ
~
i
t
=
∑
i
=
1
∞
∑
j
=
1
i
β
i
j
u
,
K
m
·
,
x
j
m
ψ
~
i
t
=
∑
i
=
1
∞
∑
j
=
1
i
β
i
j
u
x
j
ψ
~
i
t
.So we obtain (46).
Thus, the proof of Theorem7 is completed.From (48), we can obtain the approximation of f
-
Λ
(
t
) as(49)
f
-
Λ
N
t
=
∑
i
=
1
N
∑
j
=
1
i
β
i
j
u
x
j
ψ
~
i
t
.Remark 8.
Solving (33) by using (48) and (49) depends on the orthonormalization of {
ψ
i
}. This can sometimes be troublesome. Therefore, we hope that (
f
-
Λ
)
N
(
t
) can directly be written as a linear combination of {
ψ
i
}.
Note that whenj
>
N, ψ
~
j
,
ψ
i
n
=
0
(
1
≤
i
≤
N
), so when j
≤
N, there is(50)
f
-
Λ
N
,
ψ
j
n
=
f
-
Λ
,
ψ
j
n
=
L
f
-
Λ
,
K
m
·
,
x
j
m
=
u
x
j
.From (49) and Abel transformation of the partial sum of series, we have (51)
f
-
Λ
N
t
=
∑
i
=
1
N
f
-
Λ
,
ψ
~
i
n
ψ
~
i
t
=
∑
i
=
1
N
∑
j
=
i
N
f
-
Λ
,
ψ
~
j
n
β
j
i
ψ
i
t
≜
∑
i
=
1
N
a
i
ψ
i
t
.
Let two sides of the above equation do inner product withψ
j
(
1
≤
j
≤
N
); by (49), we can obtain the system of equations (52)
∑
i
=
1
N
ψ
i
,
ψ
j
n
a
i
=
u
x
j
,
j
=
1,2
,
…
,
N
,where the coefficient matrix A
=
ψ
i
,
ψ
j
n
N
×
N is symmetric and positive definite. Then, there exists only a set of solutions a
1
,
a
2
,
…
,
a
N. By this solution, we can obtain the approximate solution (
f
-
Λ
)
N
(
t
).
In order to obtain matrixA, we must calculate ψ
i
,
ψ
j
n
(
1
≤
i
,
j
≤
N
). For this purpose, we first deduce the concrete form of ψ
i. According to the properties of conjugate operator and reproducing kernel, we have(53)
ψ
i
t
=
ψ
i
,
K
n
·
,
t
n
=
L
∗
K
m
·
,
x
i
,
K
n
·
,
t
n
=
K
m
·
,
x
i
,
L
K
n
·
,
t
m
=
L
τ
K
n
τ
,
t
τ
=
x
i
.Using the reproducing kernel expression (14) and (15), we obtain the concrete form of ψ
i as(54)
ψ
i
t
=
∑
j
=
1
n
q
i
,
j
-
1
t
j
-
1
j
-
1
!
2
+
∑
j
=
1
n
-
1
n
-
j
q
i
,
j
-
1
2
n
-
j
!
j
-
1
!
t
2
n
-
j
,
t
≤
x
i
,
∑
j
=
1
n
-
1
n
-
j
q
i
,
2
n
-
j
2
n
-
j
!
j
-
1
!
t
j
-
1
,
t
>
x
i
,where(55)
q
i
j
=
L
t
j
τ
=
x
i
,
j
=
0,1
,
…
,
2
n
-
1
.
In order to calculateψ
i
,
ψ
j
n
(
1
≤
i
,
j
≤
N
), from (54), we have (56)
ψ
i
l
-
1
0
=
q
i
,
l
-
1
l
-
1
!
,
l
=
1,2
,
…
,
n
,
ψ
i
n
t
=
∑
j
=
1
n
-
1
j
-
1
q
i
,
n
-
j
n
-
j
!
j
-
1
!
t
j
-
1
,
t
≤
x
i
,
0
,
t
>
x
i
,
∑
l
=
1
n
ψ
i
l
-
1
0
·
ψ
j
l
-
1
0
=
∑
l
=
1
n
q
i
,
l
-
1
l
-
1
!
·
q
j
,
l
-
1
l
-
1
!
.When x
i
≤
x
j, it can be obtained that(57)
∫
0
1
ψ
i
n
t
·
ψ
j
n
t
d
t
=
∑
α
=
1
n
∑
β
=
1
n
-
1
α
+
β
q
i
,
n
-
α
q
j
,
n
-
β
n
-
α
!
α
-
1
!
n
-
β
!
β
-
1
!
∫
0
x
i
t
α
+
β
-
2
d
t
=
∑
α
=
1
n
∑
β
=
1
n
-
1
α
+
β
q
i
,
n
-
α
q
j
,
n
-
β
n
-
α
!
α
-
1
!
n
-
β
!
β
-
1
!
·
x
i
α
+
β
-
1
α
+
β
-
1
.When x
i
>
x
j, by symmetry, there is(58)
∫
0
1
ψ
i
n
t
·
ψ
j
n
t
d
t
=
∑
α
=
1
n
∑
β
=
1
n
-
1
α
+
β
q
i
,
n
-
α
q
j
,
n
-
β
n
-
α
!
α
-
1
!
n
-
β
!
β
-
1
!
·
x
j
α
+
β
-
1
α
+
β
-
1
.
Thus, substituting (56)–(58) into (7), we can obtain ψ
i
,
ψ
j
n
(
1
≤
i
,
j
≤
N
).


## 5. Convergence and Error Estimation

In this section, we will consider the convergence and error estimation of the approximate solution given by (49).Lemma 9.
f
-
N
(
t
) is convergent uniformly to f
-
(
t
) on [
0,1
].Proof.
Since{
ψ
~
i
} is a complete orthonormal system of W
2
n
[
0,1
], it is obvious that (59)
f
-
-
f
-
N
n
2
=
f
-
Λ
-
f
-
Λ
N
n
2
=
∑
i
=
N
+
1
∞
∑
j
=
1
i
β
i
j
u
x
j
2
⟶
0
,
N
⟶
∞
.From (14) and (15), it can be obtained that (60)
K
n
t
,
t
=
∑
i
=
1
n
t
2
i
-
2
i
-
1
!
2
+
∑
i
=
1
n
-
1
n
-
i
2
n
-
i
!
i
-
1
!
t
2
n
-
1
.Since K
n
(
t
,
t
) is continuous on [
0,1
], there exists a constant M such that 0
≤
K
n
(
t
,
t
)
≤
M. Moreover,(61)
f
-
t
-
f
-
N
t
2
=
f
-
-
f
-
N
,
K
n
·
,
t
n
2
≤
f
-
-
f
-
N
n
2
K
n
·
,
t
n
2
=
f
-
f
N
n
2
K
n
t
,
t
≤
M
f
-
f
N
n
2
.Hence, f
-
N
(
t
) is convergent uniformly to f
-
(
t
) on [
0,1
].
This completes the proof of Lemma9.Theorem 10.
The exact solution and approximate solution of (33) satisfy (62)
f
-
t
-
f
-
N
t
2
≤
f
-
Λ
n
2
·
K
n
t
,
t
-
∑
i
=
1
N
ψ
~
i
t
2
.Proof.
From (48) and (49), we have (63)
f
-
Λ
N
t
=
∑
i
=
1
N
f
-
Λ
,
ψ
~
i
n
ψ
~
i
t
=
∑
i
=
1
N
ψ
~
i
t
ψ
~
i
,
f
-
Λ
n
.It follows that(64)
f
-
t
-
f
-
N
t
2
=
f
-
Λ
t
-
f
-
Λ
N
t
2
=
K
n
·
,
t
-
∑
i
=
1
N
ψ
~
i
t
ψ
~
i
,
f
-
Λ
n
2
≤
f
-
Λ
n
2
K
n
·
,
t
-
∑
i
=
1
N
ψ
~
i
t
ψ
~
i
n
2
.Since ∑
i
=
1
N
ψ
~
i
(
t
)
ψ
~
i
=
∑
i
=
1
N
K
n
(
·
,
t
)
,
ψ
~
i
ψ
~
i, then(65)
K
n
·
,
t
-
∑
i
=
1
N
ψ
~
i
t
ψ
~
i
⊥
∑
i
=
1
N
ψ
~
i
t
ψ
~
i
.Therefore,(66)
K
n
·
,
t
-
∑
i
=
1
N
ψ
~
i
t
ψ
~
i
n
2
=
K
n
·
,
t
,
K
n
·
,
t
n
-
∑
i
=
1
N
ψ
~
i
t
K
n
·
,
t
,
ψ
~
i
n
=
K
n
t
,
t
-
∑
i
=
1
N
ψ
~
i
t
2
.Thus, we prove that (62) is true.
This completes the proof of Theorem10.Remark 11.
When we use Theorem10 to estimate f
-
(
t
)
-
f
-
N
(
t
), the value of f
-
Λ
n
2 has to be evaluated. It is usually difficult in practical problems. We will provide a new method of estimation in the following.As in Lemma4, suppose that φ
1
,
φ
2
,
…
,
φ
k is a basis of Ker
⁡
L and is dual to λ
1
,
λ
2
,
…
,
λ
k. According to [6, 7], every f
∈
W
2
n
[
0,1
] has the decomposition form as (67)
f
t
=
∑
i
=
1
k
λ
i
f
φ
i
t
+
∫
0
1
G
L
t
,
τ
·
L
f
τ
d
τ
,where G
L
(
t
,
τ
) is Green function of differential operator L and satisfies (68)
λ
i
∫
0
1
G
L
t
,
τ
·
u
τ
d
τ
=
0
,
i
=
1,2
,
…
,
k
,
L
∫
0
1
G
L
t
,
τ
·
u
τ
d
τ
=
u
t
,
∀
u
∈
L
2
0,1
.Since G
L
(
t
,
τ
) is continuous, there exists a constant C such that G
L
(
t
,
τ
)
≤
C.Theorem 12.
Suppose thatu
i
=
∑
j
=
1
i
β
i
j
[
u
(
x
j
)
]; then the exact solution f
-
(
t
) and approximate solution f
-
N
(
t
) of (33) satisfy (69)
f
-
t
-
f
-
N
t
≤
C
∫
0
1
∑
i
=
N
+
1
∞
∑
l
=
1
i
u
i
β
i
l
L
ψ
l
t
d
t
.The right-hand side of (69) converges to zero as N
→
∞.Proof.
From (39), (48), and (67), it follows that, for any f
Λ
∈
H
n, there is (70)
f
Λ
t
=
∫
0
1
G
L
t
,
τ
·
L
f
τ
d
τ
=
∑
i
=
1
∞
∑
j
=
1
i
β
i
j
L
f
x
j
ψ
~
i
t
.Then, we have (71)
f
-
Λ
t
-
f
-
Λ
N
t
≤
C
∫
0
1
L
f
-
Λ
τ
-
f
-
Λ
N
τ
d
τ
.From Lemmas 1 and 9, we know that the right-hand side of (71) converges to zero as N
→
∞. From (49), we can obtain(72)
L
f
-
Λ
τ
-
f
-
Λ
N
τ
=
∑
i
=
N
+
1
∞
u
i
·
L
ψ
~
i
t
=
∑
i
=
N
+
1
∞
∑
l
=
1
i
u
i
β
i
l
L
ψ
l
t
.Substituting this equation into (71) and f
-
(
t
)
-
f
-
N
(
t
)
=
f
-
Λ
(
t
)
-
(
f
-
Λ
)
N
(
t
), we can get (69) and assert that the right-hand side of (69) converges to zero as N
→
∞.
This completes the proof of Theorem12.Remark 13.
From (54), L
ψ
l
(
t
) can be easily obtained as the piecewise polynomial.


## 6. Conclusion

When we use the reproducing kernel method to solve differential equation, we always hope that the reproducing kernel is polynomial reproducing kernel. In general, to take the reproducing kernel ofW
2
s
[
0,1
] as polynomial reproducing kernel, the inner product taken must be as (7). But if we take this fixed inner product, it is difficult to prove the boundedness of operator and the completeness of {
ψ
i
(
t
)
}. How do these two aspects integrate effectively with each other? For the linear differential operator equation with functional constraints, this paper investigates this topic and some meaningful results are presented. Moreover, the proofs of some results do not depend on whether operator L is differential operator. Therefore, these processes can be extended for general linear bounded operator. In this paper, some conclusions are based on m
≥
1; they need to be further explored for case m
=
0.



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*Source: 823264-2015-09-10.xml*