Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 541935, 13 pages
http://dx.doi.org/10.1155/2013/541935
Research Article
A New Method Based on the RKHSM for Solving Systems of
Nonlinear IDDEs with Proportional Delays
Changbo He,1 Xueqin Lv,2 and Jing Niu2
1
2
School of Sports and Sciences, Harbin Normal University, Harbin, Heilongjiang 150025, China
School of Mathematics and Sciences, Harbin Normal University, Harbin, Heilongjiang 150025, China
Correspondence should be addressed to Xueqin Lv; hashidalvxueqin@126.com
Received 10 November 2012; Revised 1 March 2013; Accepted 25 March 2013
Academic Editor: Zhenya Yan
Copyright © 2013 Changbo He 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.
An efficient computational method is given in order to solve the systems of nonlinear infinite-delay-differential equations (IDDEs)
with proportional delays. Representation of the solution and an iterative method are established in the reproducing kernel space.
Some examples are displayed to demonstrate the computation efficiency of the method.
1. Introduction
In functional-differential equations (FDEs), there is a class
of infinite-delay-differential equations (IDDEs) with proportional delays such systems are often encountered in many
scientific fields such as electric mechanics, quantum mechanics, and optics. In view of this, developing the research for
this class of IDDEs possesses great significance on theory
and practice, for this attracts constant interest of researchers.
Ones have found that there exist very different mathematical
challenges between FDEs with proportional delays and those
with constant delays. Some researches on the numerical
solutions and the corresponding analysis for the linear FDEs
with proportional delays have been presented by several
authors. In the last few years, there has been a growing interest
in studying the existence of solutions of functional differential
equations with state dependent delay [1–7]. Initial-value
problem for neutral functional-differential equations with
proportional time delays had been studied in [8–11]; in
the literature [11] authors had discussed the existence and
uniqueness of analytic solution of linear proportional delays
equations.
Ishiwata et al. used the rational approximation method
and the collocation method to compute numerical solutions
of delay-differential equations with proportional delays in
[12, 13]. At [14–17], Hu et al. gave the numerical method
to compute numerical solutions of neutral delay differential equations. For neutral delay differential equations
with proportional delays, Chen and Wang proposed the
variational iteration method [18] and the homotopy perturbation method [19]. Recently, time-delay systems become
interested in applications like population growth models,
transportation, communications, and agricultural models so
those systems were widely studied both in a theoretical aspect
and in that of related applications [20–22].
We consider the following nonlinear infinite delaydifferential equation (IDDE) with proportional delay [23]:
𝑢󸀠 (𝑥) = 𝑔 (𝑥, 𝑢 (𝑥) , 𝑢 (𝑝𝑥)) ,
𝑢 (0) = 𝜂,
(1)
where 𝑝 ∈ (0, 1), 𝜂 is a given initial value, 𝑢(𝑥) ∈ 𝑊2 [0, +∞),
and for 𝑥 ∈ [0, +∞), 𝑦, 𝑧 ∈ (−∞, +∞), 𝑔(𝑥, 𝑦, 𝑧) is a
continuous function; 𝑔(𝑥, 𝑦, 𝑧) ∈ 𝑊1 [0, +∞) as 𝑦 = 𝑦(𝑥),
𝑧 = 𝑧(𝑥) ∈ 𝑊1 [0, +∞).
Next, the following system of nonlinear infinite-delaydifferential equations (IDDEs) with proportional delays will
be studied:
𝑢󸀠 (𝑥) = 𝐹 (𝑥, 𝑢 (𝑥) , V (𝑝𝑥)) ,
V󸀠 (𝑥) = 𝐺 (𝑥, V (𝑥) , 𝑢 (𝑞𝑥)) ,
𝑢 (0) = 0,
(2)
V (0) = 0,
where 𝑝, 𝑞 ∈ (0, 1), for 𝑥 ∈ [0, +∞), 𝑦, 𝑧 ∈ (−∞, +∞),
𝑓(𝑥, 𝑦, 𝑧), 𝑔(𝑥, 𝑦, 𝑧) are continuous bounded function, and
2
Abstract and Applied Analysis
𝑓(𝑥, 𝑦, 𝑧) ∈ 𝑊1 [0, +∞), 𝑔(𝑥, 𝑦, 𝑧) ∈ 𝑊1 [0, +∞) as 𝑦 = 𝑦(𝑥),
𝑧 = 𝑧(𝑥). 𝑊2 [0, +∞) and 𝑊1 [0, ∞) are reproducing kernel
spaces. Equations (2) are obtained through homogenization
of initial condition for model in [24]. In this study, existence and a new iterative algorithm are established for the
nonlinear infinite-delay-differential equations (IDDEs) with
proportional delays in the reproducing kernel spaces.
The paper is organized as follows. In Section 2, some
definitions of the reproducing kernel spaces are introduced.
In Section 3, main results and the structure of the solution
for operator equation are discussed. Existence of the solution
to (2) and an iterative method are developed for the kind
of problems in the reproducing kernel space. We verify that
the approximate solution converges to the exact solution
uniformly. In Section 4, some experiments are given to
demonstrate the computation efficiency of the algorithm. The
conclusion is given in Section 5.
2. Preliminaries
Definition 1 (see [25] (reproducing kernel)). Let 𝑀 be a
nonempty abstract set. A function 𝐾 : 𝑀 × 𝑀 → 𝐶 is a
reproducing kernel of the Hilbert space 𝐻 if and only if
(a) ∀𝑦 ∈ 𝑀, 𝐾(⋅, 𝑥) ∈ 𝐻,
+∞
0
(𝑢V + 𝑢󸀠 V󸀠 ) 𝑑𝑥,
(5)
‖𝑢‖𝑊1 = √⟨𝑢, 𝑢⟩𝑊1 ,
respectively, where 𝑢(𝑥), V(𝑥) ∈ 𝑊1 [0, +∞). It has been
proved that 𝑊1 [0, +∞) is a complete reproducing kernel
space and its reproducing kernel is as follows [27]:
1 −𝑥−𝑦
(1 + 𝑒2𝑥 ) ,
{
{2𝑒
𝑄𝑥 (𝑦) = {
{ 1 −𝑥−𝑦
(1 + 𝑒2𝑦 ) ,
𝑒
{2
𝑥 < 𝑦,
(6)
𝑥 ≥ 𝑦.
3. Statements of the Main Results
In this section, the implementation method of obtaining the
solution of (2) is proposed in the reproducing kernel space
𝑊2 [0, +∞).
Put the differential operator L = 𝑑/𝑑𝑥; then we can
convert (2) into the following form:
where 𝑥 ∈ [0, +∞), and
𝐹 (𝑥, 𝑢 (𝑥) , V (𝑝𝑥)) = 𝑁 (𝑢 (𝑥) , V (𝑝𝑥)) + 𝑓 (𝑥) ,
Definition 2. 𝑊2 [0, +∞) = {𝑢(𝑥) | 𝑢󸀠 is absolutely
continuous real value functions, 𝑢󸀠󸀠 ∈ 𝐿2 [0, +∞), 𝑢(0) = 0}.
𝑊2 [0, +∞) is a Hilbert space, for 𝑢(𝑥), V(𝑥) ∈ 𝑊2 [0, +∞);
the inner product and norm in 𝑊2 [0, +∞) are given by
(4𝑢V + 5𝑢󸀠 V󸀠 + 𝑢󸀠󸀠 V󸀠󸀠 ) 𝑑𝑥,
(3)
‖𝑢‖𝑊2 = √⟨𝑢, 𝑢⟩𝑊2 ,
respectively.
Theorem 3. The space 𝑊2 [0, +∞) is a reproducing kernel
space that is, for any 𝑢(𝑦) ∈ 𝑊2 [0, +∞) and each fixed 𝑥 ∈
[0, +∞), there exists 𝑅𝑥 (𝑦) ∈ 𝑊2 [0, +∞), 𝑦 ∈ [0, +∞), such
that 𝑢(𝑥) = ⟨𝑅𝑥 (𝑦), 𝑢(𝑦)⟩. And the corresponding reproducing
kernel can be represented as follows [26]:
1 −2(𝑥+𝑦)
{
(−1+𝑒2𝑦 ) (1+𝑒2𝑦 −2𝑒𝑥+𝑦 ) , 𝑦 ≤ 𝑥,
{
{− 12 𝑒
𝑅𝑥 (𝑦) = {
{
{− 1 𝑒−2(𝑥+𝑦) (−1+𝑒2𝑥 ) (1+𝑒2𝑥 −2𝑒𝑥+𝑦 ) , 𝑦 > 𝑥.
{ 12
(4)
Definition 4. 𝑊1 [0, +∞) = {𝑢(𝑥) | 𝑢 is absolutely continuous
real-valued function, 𝑢󸀠 ∈ 𝐿2 [0, +∞)}.
(7)
𝑢 (0) = V (0) = 0,
def
Next, two reproducing kernel spaces are given.
0
⟨𝑢 (𝑥) , V (𝑥)⟩𝑊1 = ∫
LV (𝑥) = 𝐺 (𝑥, V (𝑥) , 𝑢 (𝑞𝑥)) ,
The condition (b) is called “the reproducing property”; a
Hilbert space which possesses a reproducing kernel is called
a Reproducing Kernel Hilbert Space (RKHS).
+∞
The inner product and norm in 𝑊1 [0, +∞) can be defined
L𝑢 (𝑥) = 𝐹 (𝑥, 𝑢 (𝑥) , V (𝑝𝑥)) ,
(b) ∀𝑦 ∈ 𝑀, ∀𝑢 ∈ 𝐻, ⟨𝑢(⋅), 𝐾(⋅, 𝑦)⟩ = 𝑢(𝑦).
⟨𝑢 (𝑥) , V (𝑥)⟩𝑊2 = ∫
by
def
𝐺 (𝑥, V (𝑥) , 𝑢 (𝑞𝑥)) = 𝐷 (V (𝑥) , 𝑢 (𝑞𝑥)) + 𝑔 (𝑥) .
(8)
(9)
It is clear that L : 𝑊2 [0, +∞) → 𝑊1 [0, +∞) is a
bounded linear operator. Let 𝜑𝑖 (𝑥) = 𝑄𝑥𝑖 (𝑥), where {𝑥𝑖 }∞
𝑖=1
is dense in the interval [0, +∞), and 𝜓𝑖 (𝑥) = L∗ 𝜑𝑖 (𝑥),
where L∗ is the conjugate operator of L. Define the normal
orthogonal system 𝜓𝑖 (𝑥) in 𝑊2 [0, +∞), which derives from
Gram-Schmidt orthogonalization process of {𝜓𝑖 (𝑥)}∞
𝑖=1 ,
𝑖
𝜓𝑖 (𝑥) = ∑ 𝛽𝑖𝑘 𝜓𝑘 (𝑥) ,
𝑘=1
𝛽𝑖𝑖 > 0,
𝑖 = 1, 2, . . .
(10)
Lemma 5. Assume that {𝑥𝑖 }∞
𝑖=1 are dense in [0, +∞); then
{𝜓𝑖 (𝑥)}∞
𝑖=1 is a complete system in 𝑊2 [0, +∞) and {𝜓𝑖 (𝑥)} =
(𝑑/𝑑𝑦)𝑅𝑥 (𝑦)|𝑦=𝑥𝑖 .
Proof. Ones have
𝜓𝑖 (𝑥) = ⟨𝜓𝑖 (𝑦) , 𝑅𝑥 (𝑦)⟩𝑊2 = ⟨L∗ 𝜑𝑖 (𝑦) , 𝑅𝑥 (𝑦)⟩𝑊2
= ⟨𝜑𝑖 (𝑦) , L𝑅𝑥 (𝑦)⟩𝑊1 = L𝑅𝑥 (𝑥𝑖 )
󵄨󵄨
𝑑
󵄨
=
𝑅𝑥 (𝑦)󵄨󵄨󵄨 .
󵄨󵄨𝑦=𝑥
𝑑𝑦
𝑖
Clearly, 𝜓𝑖 (𝑥) ∈ 𝑊2 [0, +∞).
(11)
Abstract and Applied Analysis
3
For any 𝑢(𝑥) ∈ 𝑊2 [0, +∞), let ⟨𝑢(𝑥), 𝜓𝑖 (𝑥)⟩ = 0, 𝑖 =
1, 2, . . ., which means that
⟨𝑢 (𝑥) , (L∗ 𝜑𝑖 (𝑥) ⟩ = ⟨L𝑢 (⋅) , 𝜑𝑖 (⋅)⟩ = (L𝑢) (𝑥𝑖 ) = 0.
(12)
Proof. Note that ⟨𝑢(𝑥), 𝜑𝑖 (𝑥)⟩ = 𝑢(𝑥𝑖 ) and {𝜓𝑖 (𝑥)}∞
𝑖=1 is
an orthonormal basis of 𝑊2 [0, +∞); hence according to
Lemma 5 we have
∞
𝑀𝑛 (𝑥) = ∑ ⟨𝑀𝑛 (𝑥) , 𝜓𝑖 (𝑥)⟩ 𝜓𝑖 (𝑥)
𝑖=1
∞
𝑖
𝑖=1
𝑘=1
= ∑ ⟨𝑀𝑛 (𝑥) , ∑ 𝛽𝑖𝑘 𝜓𝑘 (𝑥)⟩ 𝜓𝑖 (𝑥)
Note that {𝑥𝑖 }∞
𝑖=1 is dense in [0, +∞); therefore, L𝑢(𝑥) = 0. It
follows that 𝑢(𝑥) ≡ 0 from the existence of L−1 .
∞
𝑖
= ∑ ∑ 𝛽𝑖𝑘 ⟨𝑀𝑛 (𝑥) , 𝜓𝑘 (𝑥)⟩ 𝜓𝑖 (𝑥)
(17)
𝑖=1 𝑘=1
3.1. Construction the of Iterative Sequence 𝑢̃𝑛 (𝑥) and ̃V𝑛 (𝑥).
Next we construct the iterative sequence 𝑢̃𝑛 (𝑥) and ̃V𝑛 (𝑥),
putting
∞
𝑖
= ∑ ∑ 𝛽𝑖𝑘 ⟨L𝑀𝑛 (𝑥) , 𝜑𝑘 (𝑥)⟩ 𝜓𝑖 (𝑥)
𝑖=1 𝑘=1
∞
L𝑀𝑛 (𝑥) = 𝐹 (𝑥, 𝑢̃𝑛−1 (𝑥) , ̃V𝑛−1 (𝑝𝑥)) ,
𝑖=1 𝑘=1
𝑢̃𝑛 (𝑥) = 𝑃𝑛 𝑀𝑛 (𝑥) ,
L𝑁𝑛 (𝑥) = 𝐺 (𝑥, ̃V𝑛−1 (𝑥) , 𝑢̃𝑛−1 (𝑞𝑥)) ,
𝑖
= ∑ ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢𝑛−1 (𝑥𝑘 ) , V𝑛−1 (𝑝𝑥𝑘 )) 𝜓𝑖 (𝑥) .
(13)
In the same manner
∞
𝑖
𝑁𝑛 (𝑥) = ∑ ∑ 𝛽𝑖𝑘 𝐺 (𝑥𝑘 , V𝑛−1 (𝑥𝑘 ) , 𝑢𝑛−1 (𝑞𝑥𝑘 )) 𝜓𝑖 (𝑥) . (18)
̃V𝑛 (𝑥) = 𝑃𝑛 𝑁𝑛 (𝑥) ,
𝑖=1 𝑘=1
where 𝑀𝑛 (𝑥), 𝑁𝑛 (𝑥) ∈ 𝑊2 [0, +∞) and 𝑃𝑛 : 𝑊2 →
Span{𝜓1 , 𝜓2 , . . . , 𝜓𝑛 } is a orthogonal projection operator.
Then by (13) it followed that:
Take 𝑢0 (𝑥) = V0 (𝑥) = 0; define the iterative sequence
𝑢𝑛 (𝑥) = 𝑃𝑛 𝑀𝑛 (𝑥)
𝑛
𝑖
= ∑ ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢𝑛−1 (𝑥𝑘 ) , V𝑛−1 (𝑝𝑥𝑘 )) 𝜓𝑖 (𝑥) ,
𝑖=1 𝑘=1
𝑥
𝑀𝑛 (𝑥) = ∫ 𝐹 (𝜇, 𝑢̃𝑛−1 (𝜇) , ̃V𝑛−1 (𝑝𝜇)) 𝑑𝜇,
0
𝑀𝑛󸀠 (𝑥) = 𝐹 (𝑥, 𝑢̃𝑛−1 (𝑥) , ̃V𝑛−1 (𝑝𝑥)) ,
𝑀𝑛󸀠󸀠 (𝑥) =
𝑛
(14)
3.2. The Boundedness of Sequence 𝑢𝑛 (𝑥) and V𝑛 (𝑥). From
𝑥
𝜕𝑦3 𝑅𝑥 (𝑥 + 0) − 𝜕𝑦3 𝑅𝑥 (𝑥 − 0) = 1,
𝑁𝑛 (𝑥) = ∫ 𝐺 (𝜇, ̃V𝑛−1 (𝜇) , 𝑢̃𝑛−1 (𝑞𝜇)) 𝑑𝜇,
0
(15)
𝑑
𝐺 (𝑥, ̃V𝑛−1 (𝑥) , 𝑢̃𝑛−1 (𝑞𝑥)) .
(𝑥) =
𝑑𝑥
𝑁𝑛󸀠󸀠
𝑖
𝑖=1 𝑘=1
∞
𝜕𝑦4 𝑅𝑥 (𝑦) = 𝑔𝑥 (𝑦) + (𝜕𝑦3 𝑅𝑥 (𝑥 + 0) −𝜕𝑦3 𝑅𝑥 (𝑥 − 0))
𝑖
𝑁𝑛 (𝑥) = ∑ ∑ 𝛽𝑖𝑘 𝐺 (𝑥𝑘 , ̃V𝑛−1 (𝑥𝑘 ) , 𝑢̃𝑛−1 (𝑞𝑥𝑘 )) 𝜓𝑖 (𝑥) .
𝑖=1 𝑘=1
(16)
(21)
= 𝑔𝑥 (𝑦) + 𝛿 (𝑥 − 𝑦) ,
where
1
{𝑔𝑥 , 𝑦 < 𝑥,
{
{
{ 2
𝑔𝑥 (𝑦) = {𝑔𝑥 , 𝑦 > 𝑥,
{
{
{
𝑦 = 𝑥,
{0,
𝑀𝑛 (𝑥) = ∑ ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢̃𝑛−1 (𝑥𝑘 ) , ̃V𝑛−1 (𝑝𝑥𝑘 )) 𝜓𝑖 (𝑥) ,
(20)
we have
× 𝛿 (𝑥 − 𝑦)
Lemma 6. Let {𝑥𝑖 }∞
𝑖=1 be dense on [0, +∞); if the solution of
(2) is unique, then the solution satisfies the form
∞
𝑖
= ∑ ∑ 𝛽𝑖𝑘 𝐺 (𝑥𝑘 , V𝑛−1 (𝑥𝑘 ) , 𝑢𝑛−1 (𝑞𝑥𝑘 )) 𝜓𝑖 (𝑥) .
𝑖=1 𝑘=1
𝑑
𝐹 (𝑥, 𝑢̃𝑛−1 (𝑥) , ̃V𝑛−1 (𝑝𝑥)) ,
𝑑𝑥
𝑁𝑛󸀠 (𝑥) = 𝐺 (𝑥, ̃V𝑛−1 (𝑥) , 𝑢̃𝑛−1 (𝑞𝑥)) ,
(19)
V𝑛 (𝑥) = 𝑃𝑛 𝑁𝑛 (𝑥)
(22)
𝑔𝑥1 (𝑦) = 𝜕𝑦4 (𝑎1 𝑒𝑦 + 𝑎2 𝑒−𝑦 + 𝑎3 𝑒2𝑦 + 𝑎4 𝑒−2𝑦 ); 𝑔𝑥2 (𝑦) = 𝜕𝑦4 (𝑏1 𝑒𝑦 +
𝑏2 𝑒−𝑦 + 𝑏3 𝑒2𝑦 + 𝑏4 𝑒−2𝑦 ); and 𝑔𝑥1 (𝑦), 𝑔𝑥2 (𝑦), 𝜕𝑦 𝑔𝑥1 (𝑦), 𝜕𝑦 𝑔𝑥2 (𝑦) are
bounded functions with respect to 𝑥, 𝑦, respectively.
4
Abstract and Applied Analysis
= 𝑚 (𝑥) − 𝑀𝑛󸀠 (𝑥) (𝑔𝑥1 (𝑥) − 𝑔𝑥2 (𝑥))
Lemma 7. For 𝑥 ∈ [0, +∞), 𝑦, 𝑧 ∈ (−∞, +∞), and 𝐹(𝑥, 𝑦, 𝑧),
𝐺(𝑥, 𝑦, 𝑧) are continuous bounded functions on [0, +∞), we
have that 𝑀𝑛(𝑖) (𝑥) and 𝑁𝑛(𝑖) (𝑥) (𝑖 = 0, 1, 2) are bounded.
+ (𝑀𝑛󸀠 (0) 𝑔𝑥1 (0) − 𝑀𝑛󸀠 (𝑀) 𝑔𝑥2 (𝑀))
Proof. By the expression of 𝑀𝑛 (𝑥), 𝑀𝑛󸀠 (𝑥) and the assumptions, we know that 𝑀𝑛 (𝑥) and 𝑀𝑛󸀠 (𝑥) are bounded. In the
following, we will discuss the boundedness of 𝑀𝑛󸀠󸀠 (𝑥).
2
Since the function which is in 𝐶[0,+∞)
is dense in
𝑊2 [0, +∞), without loss of generality, we assume that 𝑀𝑛󸀠󸀠 (𝑥)
is continuous.
Note that
4𝑅𝑥 (𝑦) − 5𝜕𝑦2 𝑅𝑥 (𝑦) + 𝜕𝑦4 𝑅𝑥 (𝑦) = 𝛿 (𝑥 − 𝑦) .
(23)
𝑥
+ ∫ 𝑀𝑛󸀠 (𝑦) 𝜕𝑦 𝑔𝑥1 (𝑦) 𝑑𝑦
0
∞
+ ∫ 𝑀𝑛󸀠 (𝑦) 𝜕𝑦 𝑔𝑥2 (𝑦) 𝑑𝑦.
𝑥
(25)
So
𝑀𝑛󸀠 (𝑥) (𝑔𝑥1 (𝑥) − 𝑔𝑥2 (𝑥))
One gets
∞
= 𝑚 (𝑥) + (𝑀𝑛󸀠 (0) 𝑔𝑥1 (0) − 𝑀𝑛󸀠 (𝑀) 𝑔𝑥2 (𝑀))
𝑀𝑛󸀠󸀠 (𝑥) = ∫ 𝑀𝑛󸀠󸀠 (𝑦) 𝛿 (𝑥 − 𝑦) 𝑑𝑦
0
=∫
∞
0
=∫
𝑥
𝑀𝑛󸀠󸀠
∞
0
(𝑦) [4𝑅𝑥 (𝑦)−5𝜕𝑦2 𝑅𝑥
(𝑦)+𝜕𝑦4 𝑅𝑥
(26)
+ ∫ 𝑀𝑛󸀠 (𝑦) 𝜕𝑦 𝑔𝑥1 (𝑦)
(𝑦)] 𝑑𝑦
0
∞
𝑀𝑛󸀠󸀠
(𝑦) [4𝑅𝑥 (𝑦) −
5𝜕𝑦2 𝑅𝑥
+ ∫ 𝑀𝑛󸀠 (𝑦) 𝜕𝑦 𝑔𝑥2 (𝑦) 𝑑𝑦.
(𝑦)] 𝑑𝑦
𝑥
∞
+ ∫ 𝑀𝑛󸀠󸀠 (𝑦) 𝜕𝑦4 𝑅𝑥 (𝑦) 𝑑𝑦
Furthermore, we see that
0
󵄨󵄨∞
󵄨
= 𝑀𝑛󸀠 (𝑦) [4𝑅𝑥 (𝑦) − 5𝜕𝑦2 𝑅𝑥 (𝑦)] 󵄨󵄨󵄨
󵄨󵄨0
−∫
+∞
0
𝑀𝑛󸀠󸀠 (𝑥) =
1
𝑔𝑥1 (𝑥) − 𝑔𝑥2 (𝑥)
× {𝜕𝑥 [𝑚 (𝑥) + (𝑀𝑛󸀠 (0) 𝑔𝑥1 (0) − 𝑀𝑛󸀠 (𝑀) 𝑔𝑥2 (𝑀))
𝑀𝑛󸀠 (𝑦) [4𝜕𝑦 𝑅𝑥 (𝑦) − 5𝜕𝑦3 𝑅𝑥 (𝑦)] 𝑑𝑦
∞
𝑥
+ ∫ 𝑀𝑛󸀠󸀠 (𝑦) [𝑔𝑥 (𝑦) + 𝛿 (𝑥 − 𝑦)] 𝑑𝑦
+ ∫ 𝑀𝑛󸀠 (𝑦) 𝜕𝑦 𝑔𝑥1 (𝑦)
0
0
∞
∞
= 𝑚 (𝑥) + ∫ 𝑀𝑛󸀠󸀠 (𝑦) [𝑔𝑥 (𝑦) + 𝛿 (𝑥 − 𝑦)] 𝑑𝑦
+ ∫ 𝑀𝑛󸀠 (𝑦) 𝜕𝑦 𝑔𝑥2 (𝑦) 𝑑𝑦]
0
𝑥
∞
= 𝑚 (𝑥) + ∫ 𝑀𝑛󸀠󸀠 (𝑦) 𝑔𝑥 (𝑦) 𝑑𝑦 + 𝑀𝑛󸀠󸀠 (𝑥) ,
0
where 𝑚(𝑥)
=
∞
−𝑀𝑛󸀠 (𝑥) 𝜕𝑥 (𝑔𝑥1 (𝑥) − 𝑔𝑥2 (𝑥)) } .
(24)
𝑀𝑛󸀠 (𝑦)[4𝑅𝑥 (𝑦) − 5𝜕𝑦2 𝑅𝑥 (𝑦)]|∞
−
0
∫0 𝑀𝑛󸀠 (𝑦)[4𝜕𝑦 𝑅𝑥 (𝑦) − 5𝜕𝑦3 𝑅𝑥 (𝑦)]𝑑𝑦. Thus, we have
∞
0 = 𝑚 (𝑥) − ∫ 𝑀𝑛󸀠󸀠 (𝑦) 𝑔𝑥 (𝑦) 𝑑𝑦
0
𝑥
∞
0
𝑥
= 𝑚 (𝑥) − ∫ 𝑀𝑛󸀠󸀠 (𝑦) 𝑔𝑥1 (𝑦) 𝑑𝑦 − ∫ 𝑀𝑛󸀠󸀠 (𝑦) 𝑔𝑥2 (𝑦) 𝑑𝑦
󵄨󵄨𝑥
𝑥
󵄨
= 𝑚 (𝑥) − 𝑀𝑛󸀠 (𝑦) 𝑔𝑥1 (𝑦) 󵄨󵄨󵄨 + ∫ 𝑀𝑛󸀠 (𝑦) 𝜕𝑦 𝑔𝑥1 (𝑦) 𝑑𝑦
󵄨󵄨0
0
∞
󵄨󵄨󵄨𝑀
− 𝑀𝑛󸀠 (𝑦) 𝑔𝑥2 (𝑦) 󵄨󵄨󵄨 + ∫ 𝑀𝑛󸀠 (𝑦) 𝜕𝑦 𝑔𝑥2 (𝑦) 𝑑𝑦
󵄨󵄨𝑥
𝑥
(27)
In view of the expression of 𝑅𝑥 (𝑦), we know that 𝜕𝑦𝑚 𝑅𝑥 (𝑦) are
bounded for 𝑚 = 1, 2, 3. It follows that 𝑀𝑛󸀠󸀠 (𝑥) is bounded
from the boundedness of 𝑔𝑥1 (𝑦), 𝑔𝑥2 (𝑦), 𝜕𝑦 𝑔𝑥1 (𝑦), 𝜕𝑦 𝑔𝑥2 (𝑦),
and 𝑀𝑛󸀠 (𝑥). In the same way, 𝑁𝑛 (𝑥), 𝑁𝑛󸀠 (𝑥), and 𝑁𝑛󸀠󸀠 (𝑥) are
bounded.
Lemma 8. Assume that 𝐹(𝑥, 𝑦, 𝑧), 𝐺(𝑥, 𝑦, 𝑧) are continuous
bounded functions for 𝑥 ∈ [0, +∞), 𝑦, 𝑧 ∈ (−∞, +∞), and
𝐹(𝑥, 𝑦, 𝑧), 𝐺(𝑥, 𝑦, 𝑧) ∈ 𝑊1 [0, +∞) as 𝑦 = 𝑦(𝑥) ∈ 𝑊2 , 𝑧 =
𝑧(𝑥) ∈ 𝑊2 , then
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑢𝑛 󵄩󵄩𝑊2 ≤ 𝐶,
󵄩󵄩 󵄩󵄩
󵄩󵄩V𝑛 󵄩󵄩𝑊2 ≤ 𝐷,
(𝐶, 𝐷 are constants) . (28)
Abstract and Applied Analysis
5
Proof.
∞
󵄩󵄩 󵄩󵄩2
2
󸀠2
󸀠󸀠2
󵄩󵄩𝑀𝑛 󵄩󵄩𝑊2 = ⟨𝑀𝑛 , 𝑀𝑛 ⟩ = ∫ (4𝑀𝑛 + 5𝑀𝑛 + 𝑀𝑛 ) 𝑑𝑥.
0
(29)
where 𝐴 𝑖 = ∑𝑖𝑘=1 𝛽𝑖𝑘 𝐹(𝑥𝑘 , 𝑢(𝑥𝑘 ), V(𝑝𝑥𝑘 )) and 𝐵𝑖 =
∑𝑖𝑘=1 𝛽𝑖𝑘 𝐺(𝑥𝑘 , V(𝑥𝑘 ), 𝑢(𝑞𝑥𝑘 )). In fact, 𝐴 𝑖 and 𝐵𝑖 are unknown;
we will approximate 𝐴 𝑖 and 𝐵𝑖 by using the known 𝐴𝑖 and 𝐵𝑖 .
We take 𝑢1 (𝑥) = 0, V1 (𝑥) = 0 and define the following
iterative sequence
Note that (14) and the assumptions, by Lemma 7, |𝑀𝑛(𝑖) (𝑥)| ≤
𝐶𝑖 , 𝑖 = 0, 1, 2, thus ‖𝑀𝑛 ‖ ≤ 𝐶, where 𝐶 = max{𝐶0 , 𝐶1 , 𝐶2 },
from (19), we have ‖𝑢𝑛 ‖𝑊2 ≤ ‖𝑀𝑛 ‖𝑊2 ≤ 𝐶. In the same way,
we obtain ‖V𝑛 ‖𝑊2 ≤ 𝐷.
𝑛−1
𝑢𝑛 (𝑥) = ∑ 𝐴𝑖 𝜓𝑖 (𝑥) ,
𝑖=1
V𝑛 (𝑥) = ∑ 𝐵𝑖 𝜓𝑖 (𝑥) ,
𝑖=1
3.3. Construction the of Another Iterative Sequence
𝑢𝑛 (𝑥) and V𝑛 (𝑥)
where
{𝑥𝑖 }∞
𝑖=1
be dense on [0, +∞); if the solution of
Theorem 9. Let
(2) exists and unique, then the solution satisfies the form
∞
𝐴1 = 𝛽11 𝐹 (𝑥1 , 𝑢1 (𝑥1 ) , V1 (𝑝𝑥1 ))
2
𝑖
𝐴2 = ∑ 𝛽2𝑘 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 ))
𝑢 (𝑥) = ∑ ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢 (𝑥𝑘 ) , V (𝑝𝑥𝑘 )) 𝜓𝑖 (𝑥) ,
𝑖=1 𝑘=1
∞
𝑘=1
(30)
𝑖
(36)
..
.
V (𝑥) = ∑ ∑ 𝛽𝑖𝑘 𝐺 (𝑥𝑘 , V (𝑥𝑘 ) , 𝑢 (𝑞𝑥𝑘 )) 𝜓𝑖 (𝑥) .
𝑖=1 𝑘=1
𝑛−1
Proof. Note that ⟨𝑢(𝑥), 𝜑𝑖 (𝑥)⟩ = 𝑢(𝑥𝑖 ) and {𝜓𝑖 (𝑥)}∞
𝑖=1 is an
orthonormal basis of 𝑊2 [0, +∞); hence we have
𝐴𝑛−1 = ∑ 𝛽𝑛𝑘 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 )) ,
𝑘=1
∞
𝐵1 = 𝛽11 𝐺 (𝑥1 , V1 (𝑥1 ) , 𝑢1 (𝑞𝑥1 ))
𝑢 (𝑥) = ∑ ⟨𝑢 (𝑥) , 𝜓𝑖 (𝑥)⟩ 𝜓𝑖 (𝑥)
𝑖=1
∞
𝑖
𝑖=1
𝑘=1
2
𝐵2 = ∑ 𝛽2𝑘 𝐺 (𝑥𝑘 , V𝑘 (𝑥𝑘 ) , 𝑢𝑘 (𝑞𝑥𝑘 ))
= ∑ ⟨𝑢 (𝑥) , ∑ 𝛽𝑖𝑘 𝜓𝑘 (𝑥)⟩ 𝜓𝑖 (𝑥)
∞
𝑘=1
(37)
..
.
𝑖
= ∑ ∑ 𝛽i𝑘 ⟨𝑢 (𝑥) , 𝜓𝑘 (𝑥)⟩ 𝜓𝑖 (𝑥)
(31)
𝑖=1 𝑘=1
∞
𝑛−1
𝑖
𝐵𝑛−1 = ∑ 𝛽𝑛𝑘 𝐺 (𝑥𝑘 , V𝑘 (𝑥𝑘 ) , 𝑢𝑘 (𝑞𝑥𝑘 )) .
= ∑ ∑ 𝛽𝑖𝑘 ⟨L𝑢 (𝑥) , 𝜑𝑘 (𝑥)⟩ 𝜓𝑖 (𝑥)
𝑘=1
𝑖=1 𝑘=1
∞
𝑖
Next, lemmas are given.
= ∑ ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢 (𝑥𝑘 ) , V (𝑝𝑥𝑘 )) 𝜓𝑖 (𝑥) .
Lemma 10. The following iterative sequences
𝑖=1 𝑘=1
In the same manner
∞
(35)
𝑛−1
𝑛−1
𝑢𝑛 (𝑥) = ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 )) 𝜓𝑖 (𝑥) ,
𝑖
V (𝑥) = ∑ ∑ 𝛽𝑖𝑘 𝐺 (𝑥𝑘 , V (𝑥𝑘 ) , 𝑢 (𝑞𝑥𝑘 )) 𝜓𝑖 (𝑥) .
𝑖=1
(32)
𝑖=1 𝑘=1
(38)
𝑛−1
V𝑛 (𝑥) = ∑ 𝛽𝑖𝑘 𝐺 (𝑥𝑘 , V𝑘 (𝑥𝑘 ) , 𝑢𝑘 (𝑞𝑥𝑘 )) 𝜓𝑖 (𝑥)
𝑖=1
In the following, a new method of solving (2) is presented.
Equations (31) and (32) can be denoted by
∞
𝑢 (𝑥) = ∑𝐴 𝑖 𝜓𝑖 (𝑥) ,
(33)
𝑖=1
𝑖=1
L𝑢𝑛 (𝑥𝑗 ) = 𝐹 (𝑥𝑗 , 𝑢𝑗 (𝑥𝑗 ) , V𝑗 (𝑝𝑥𝑗 )) ,
LV𝑛 (𝑥𝑗 ) = 𝐺 (𝑥𝑗 , V𝑗 (𝑥𝑗 ) , 𝑢𝑗 (𝑞𝑥𝑗 )) ,
∞
V (𝑥) = ∑𝐵𝑖 𝜓𝑖 (𝑥) ,
satisfy
(34)
respectively.
𝑗 ≤ 𝑛 − 1,
(39)
6
Abstract and Applied Analysis
Proof. If 𝑗 = 1,
Theorem 11. The iterative form
𝑛−1
L𝑢𝑛 (𝑥1 ) = ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 )) L𝜓𝑖 (𝑥1 )
𝑖=1
𝑛
𝑖
𝑢̃𝑛 (𝑥) = ∑ ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢̃𝑛−1 (𝑥𝑘 ) , ̃V𝑛−1 (𝑝𝑥𝑘 )) 𝜓𝑖 (𝑥) ,
𝑖=1 𝑘=1
𝑛−1
𝑛
= ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 ))
𝑖
̃V𝑛 (𝑥) = ∑ ∑ 𝛽𝑖𝑘 𝐺 (𝑥𝑘 , ̃V𝑛−1 (𝑥𝑘 ) , 𝑢̃𝑛−1 (𝑞𝑥𝑘 )) 𝜓𝑖 (𝑥) ,
𝑖=1
(46)
𝑖=1 𝑘=1
× ⟨L𝜓𝑖 (𝑥) , 𝜑1 (𝑥)⟩
𝑢̃1 (0) = ̃V1 (0) = 0
𝑛−1
= ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 ))
and the iterative form
𝑖=1
× ⟨𝜓𝑖 (𝑥) , 𝜓1 (𝑥)⟩ ,
𝑛−1 𝑖
𝑢𝑛 (𝑥) = ∑ ∑ 𝛽i𝑘 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 )) 𝜓𝑖 (𝑥) ,
𝑛−1
𝑖=1 𝑘=1
𝛽11 L𝑢𝑛 (𝑥1 ) = ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 ))
𝑛−1 𝑖
𝑖=1
V𝑛 (𝑥) = ∑ ∑ 𝛽𝑖𝑘 𝐺 (𝑥𝑘 , V𝑘 (𝑥𝑘 ) , 𝑢𝑘 (𝑞𝑥𝑘 )) 𝜓𝑖 (𝑥) ,
× ⟨𝜓𝑖 (𝑥) , 𝜓1 (𝑥)⟩
(47)
𝑖=1 𝑘=1
𝑢1 (0) = V1 (0) = 0
= 𝛽11 𝐹 (𝑥1 , 𝑢1 (𝑥1 ) , V1 (𝑝𝑥1 )) ,
(40)
are the same.
so,
L𝑢𝑛 (𝑥1 ) = 𝐹 (𝑥1 , 𝑢1 (𝑥1 ) , V1 (𝑝𝑥1 )) .
(41)
Proof. In Lemma 10, let 𝑛 − 1 = 𝑘; then
If 𝑗 = 2,
𝑛−1
𝑖=1
𝑛−1
= ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 )) ⟨𝜓𝑖 (𝑥) , 𝜓2 (𝑥)⟩ ,
𝑖=1
(42)
L𝑢𝑛 (𝑥𝑘 ) = 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 )) ,
(49)
thus
𝑛−1 𝑖
𝛽21 × (40) +𝛽22 × (42), we have
𝑢𝑛 (𝑥) = ∑ ∑ 𝛽𝑖𝑘 𝐿𝑢𝑘+1 (𝑥𝑘 ) 𝜓𝑖 (𝑥)
𝑖=1 𝑘=1
𝛽21 L𝑢𝑛 (𝑥1 ) + 𝛽22 L𝑢𝑛 (𝑥2 )
𝑛−1 𝑖
𝑛−1
= ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 )) ⟨𝜓𝑖 (𝑥) , 𝜓2 (𝑥)⟩
𝑖=1
= ∑ ∑ 𝛽𝑖𝑘 L𝑢𝑛 (𝑥𝑘 ) 𝜓𝑖 (𝑥)
𝑖=1 𝑘=1
𝑛−1 𝑖
= ∑ ∑ 𝛽𝑖𝑘 ⟨L𝑢𝑛 (𝑥) , 𝜑𝑘 (𝑥)⟩ 𝜓𝑖 (𝑥)
= 𝛽21 𝐹 (𝑥1 , 𝑢1 (𝑥1 ) , V1 (𝑝𝑥1 ))
(50)
𝑖=1 𝑘=1
+ 𝛽22 𝐹 (𝑥2 , 𝑢2 (𝑥2 ) , V2 (𝑝𝑥2 )) ,
(43)
by (41), L𝑢𝑛 (𝑥2 ) = 𝐹(𝑥2 , 𝑢2 (𝑥2 ), V2 (𝑝𝑥2 )). In the same way,
we have
𝑗 ≤ 𝑛 − 1.
(44)
𝑛−1 𝑖
= ∑ ∑ 𝛽𝑖𝑘 ⟨𝑢𝑛 (𝑥) , 𝜓𝑘 (𝑥)⟩ 𝜓𝑖 (𝑥)
𝑖=1 𝑘=1
𝑛−1
= ∑ ⟨𝑢𝑛 (𝑥) , 𝜓𝑖 (𝑥)⟩ 𝜓𝑖 (𝑥) .
𝑖=1
Equation (46) can be written as
Similarly,
LV𝑛 (𝑥𝑗 ) = 𝐹 (𝑥𝑗 , V𝑗 (𝑥𝑗 ) , 𝑢𝑗 (𝑞𝑥𝑗 )) ,
(48)
but
L𝑢𝑛 (𝑥2 ) = ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 )) L𝜓𝑖 (𝑥2 )
L𝑢𝑛 (𝑥𝑗 ) = 𝐹 (𝑥𝑗 , 𝑢𝑗 (𝑥𝑗 ) , V𝑗 (𝑝𝑥𝑗 )) ,
L𝑢𝑘+1 (𝑥𝑘 ) = 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 )) ,
𝑗 ≤ 𝑛 − 1.
(45)
L𝑀𝑛 (𝑥) = 𝐹 (𝑥, 𝑢̃𝑛−1 (𝑥) , ̃V𝑛−1 (𝑝𝑥)) ,
𝑢̃𝑛 (𝑥) = 𝑃𝑛 𝑀𝑛 (𝑥) .
(51)
Abstract and Applied Analysis
7
𝑖
̃𝑛−1 (𝑥𝑘 ), ̃V𝑛−1 (𝑝𝑥𝑘 ))𝜓𝑖 (𝑥);
In fact, 𝑀𝑛 (𝑥) = ∑∞
𝑖=1 ∑𝑘=1 𝛽𝑖𝑘 𝐹(𝑥𝑘 , 𝑢
then
L̃
𝑢𝑛 (𝑥𝑗 ) = L𝑃𝑛 𝑀𝑛 (𝑥𝑗 )
‖⋅‖𝑊2
= ⟨𝑀𝑛 (𝑥) , 𝜓𝑗 (𝑥)⟩
(52)
󳨀→ 𝐹 (𝑦, 𝑢 (𝑦) , V (𝑝𝑦))
as 𝑛 󳨀→ ∞,
󳨀→ 𝐺 (𝑦, V (𝑦) , 𝑢 (𝑞𝑦))
= L𝑀𝑛 (𝑥𝑗 ) ,
(56)
as 𝑛 󳨀→ ∞.
{𝑥𝑖 }∞
𝑖=1
Theorem 15. Let
be dense in [0, +∞), and 𝐹(𝑥, 𝑦, 𝑧)
and 𝐺(𝑥, 𝑦, 𝑧) satisfy the conditions of Lemma 8, then the nterm approximate solutions 𝑢𝑛 (𝑥) and V𝑛 (𝑥) in (35) converge to
the exact solution u(𝑥) and V(𝑥) of (2), respectively, and 𝑢(𝑥) =
∞
∑∞
𝑖=1 𝐴𝑖 𝜓𝑖 (𝑥), V(𝑥) = ∑𝑖=1 𝐵𝑖 𝜓𝑖 (𝑥), where 𝐴𝑖 and 𝐵𝑖 are given,
respectively, by (36), and (37).
by (51) and (52), we have
𝑛−1 𝑖
𝑢̃𝑛 (𝑥) = ∑ ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢̃𝑛−1 (𝑥𝑘 ) , ̃V𝑛−1 (𝑝𝑥𝑘 )) 𝜓𝑖 (𝑥)
𝑖=1 𝑘=1
𝑛−1 𝑖
Proof. (1) Firstly, we will prove the convergence of
𝑢𝑛 (𝑥), V𝑛 (𝑥).
By (35), we infer that
= ∑ ∑ 𝛽𝑖𝑘 L𝑀𝑛 (𝑥𝑘 ) 𝜓𝑖 (𝑥)
𝑖=1 𝑘=1
𝑛−1 𝑖
𝑢𝑛 (𝑥𝑘 ) 𝜓𝑖 (𝑥)
= ∑ ∑ 𝛽𝑖𝑘 L̃
𝑛−1 𝑖
𝐹 (𝑥𝑛 , 𝑢𝑛−1 (𝑥𝑛 ) , V𝑛−1 (𝑝𝑥𝑛 ))
𝐺 (𝑥𝑛 , V𝑛−1 (𝑥𝑛 ) , 𝑢𝑛−1 (𝑞𝑥𝑛 ))
= ⟨L𝑀𝑛 (𝑥) , 𝜑𝑗 (𝑥)⟩
𝑖=1 𝑘=1
‖⋅‖𝑊2
Lemma 14. If 𝑢𝑛 (𝑥) 󳨀→ 𝑢(𝑥)(𝑛 → ∞), V𝑛 (𝑥) 󳨀→
V(𝑥)(𝑛 → ∞), 𝑥𝑛 → 𝑦(𝑛 → ∞), and 𝐹(𝑥, 𝑦, 𝑧) and
𝐺(𝑥, 𝑦, 𝑧) satisfy the conditions of Lemma 8, then
= ⟨L𝑃𝑛 𝑀𝑛 (𝑥) , 𝜑𝑗 (𝑥)⟩
= ⟨𝑃𝑛 𝑀𝑛 (𝑥) , 𝜓𝑗 (𝑥)⟩
By Lemma 13 and Theorem 12, it is easy to obtain the
following Lemma 14.
(53)
𝑢𝑛 (𝑥) , 𝜑𝑘 (𝑥)⟩ 𝜓𝑖 (𝑥)
= ∑ ∑ 𝛽𝑖𝑘 ⟨L̃
𝑢𝑛+1 (𝑥) = 𝑢𝑛 (𝑥) + 𝐴𝑛 𝜓𝑛 (𝑥) ,
(57)
V𝑛+1 (𝑥) = V𝑛 (𝑥) + 𝐵𝑛 𝜓𝑛 (𝑥) .
(58)
From the orthogonality of {𝜓𝑖 (𝑥)}∞
𝑖=1 , it follows that
𝑖=1 𝑘=1
2
󵄩2
󵄩 󵄩2
󵄩󵄩
󵄩󵄩𝑢𝑛+1 󵄩󵄩󵄩𝑊2 = 󵄩󵄩󵄩𝑢𝑛 󵄩󵄩󵄩𝑊2 + (𝐴𝑛 )
𝑛−1 𝑖
= ∑ ∑ 𝛽𝑖𝑘 ⟨̃
𝑢𝑛 (𝑥) , 𝜓𝑘 (𝑥)⟩ 𝜓𝑖 (𝑥)
2
2
󵄩2
󵄩
= 󵄩󵄩󵄩𝑢𝑛−1 󵄩󵄩󵄩𝑊2 + (𝐴𝑛−1 ) + (𝐴𝑛 )
𝑖=1 𝑘=1
𝑛−1
(59)
..
.
= ∑ ⟨̃
𝑢𝑛 (𝑥) , 𝜓𝑖 (𝑥)⟩ 𝜓𝑖 (𝑥) .
𝑖=1
𝑛
Equation (53) is the same as (50). We may prove for ̃V𝑛 (𝑥)
similarly.
So, by Theorem 11 and Lemma 8, we have the following
Theorem.
Theorem 12. Under the conditions of Lemma 8,
𝑛−1
𝑢𝑛 (𝑥) = ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 )) 𝜓𝑖 (𝑥) ,
(54)
𝑖=1
𝑖=1
2
󵄩󵄩
󵄩2
󵄩 󵄩2
󵄩󵄩V𝑛+1 󵄩󵄩󵄩𝑊2 = 󵄩󵄩󵄩V𝑛 󵄩󵄩󵄩 𝑊 + (𝐵𝑛 )
2
2
2
󵄩2
󵄩
= 󵄩󵄩󵄩V𝑛−1 󵄩󵄩󵄩𝑊2 + (𝐵𝑛−1 ) + (𝐵𝑛 )
(60)
..
.
𝑛
2
󵄩 󵄩2
= 󵄩󵄩󵄩V1 󵄩󵄩󵄩𝑊2 + ∑(𝐵𝑖 ) .
𝑛−1
V𝑛 (𝑥) = ∑ 𝛽𝑖𝑘 𝐺 (𝑥𝑘 , V𝑘 (𝑥𝑘 ) , 𝑢𝑘 (𝑞𝑥𝑘 )) 𝜓𝑖 (𝑥)
2
󵄩 󵄩2
= 󵄩󵄩󵄩𝑢1 󵄩󵄩󵄩𝑊2 + ∑(𝐴𝑖 ) ,
(55)
𝑖=1
satisfy ‖𝑢𝑛 (𝑥)‖ ≤ 𝐶, ‖V𝑛 (𝑥)‖ ≤ 𝐷.
Lemma 13. If 𝑢(𝑥) and V(𝑥) ∈ 𝑊2 [0, +∞), then there exists
𝑀1 , 𝑀2 > 0, such that |𝑢(𝑥)| ≤ 𝑀1 ‖𝑢(𝑥)‖𝑊2 and |V(𝑥)| ≤
𝑀2 ‖V(𝑥)‖𝑊2 .
Proof. It is easy to obtain from the properties in the
reproducing kernel space.
𝑖=1
From (59) and (60), we know sequence ‖𝑢𝑛 ‖𝑊2 and ‖V𝑛 ‖𝑊2 is
increasing. By Theorem 12, ‖𝑢𝑛 ‖𝑊2 and ‖V𝑛 ‖ 𝑊 are bounded;
2
hence ‖𝑢𝑛 ‖𝑊2 and ‖V𝑛 ‖𝑊2 are convergent such that
∞
2
∑(𝐴𝑖 ) < ∞,
𝑖=1
∞
(61)
2
∑(𝐵𝑖 ) < ∞.
𝑖=1
8
Abstract and Applied Analysis
This implies that
Since
∞
𝑖
𝐴𝑖 = ∑ 𝛽𝑖𝑘 𝐹 (𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ) , V𝑘 (𝑝𝑥𝑘 )) ∈ 𝑙2
(L𝑢) (𝑥𝑗 ) = ∑𝐴𝑖 ⟨L𝜓𝑖 , 𝜑𝑗 ⟩
(𝑖 = 1, 2 , . . .) ,
𝑖=1
𝑘=1
(62)
𝑖
𝐵𝑖 = ∑ 𝛽𝑖𝑘 𝐺 (𝑥𝑘 , V𝑘 (𝑥𝑘 ) , 𝑢𝑘 (𝑞𝑥𝑘 )) ∈ 𝑙
2
∞
= ∑𝐴𝑖 ⟨𝜓𝑖 , L∗ 𝜑𝑗 ⟩
(𝑖 = 1, 2, . . .) .
∞
𝑘=1
= ∑𝐴𝑖 ⟨𝜓𝑖 , 𝜓𝑗 ⟩ ,
(63)
Without loss of generality, assume 𝑚 > 𝑛; we have
𝑖=1
in the same way, we have
󵄩2
󵄩
󵄩󵄩
󵄩󵄩𝑢𝑚 (𝑥) − 𝑢𝑛 (𝑥)󵄩󵄩󵄩𝑊2 = 󵄩󵄩󵄩𝑢𝑚 (𝑥) − 𝑢𝑚−1 (𝑥)
∞
(LV) (𝑥𝑗 ) = ∑𝐵𝑖 ⟨𝜓𝑖 , 𝜓𝑗 ⟩ ,
+ 𝑢𝑚−1 (𝑥) − 𝑢𝑚−2 (𝑥)
it follows that
󵄩2
󵄩
≤ 󵄩󵄩󵄩𝑢𝑚 (𝑥) − 𝑢𝑚−1 (𝑥)󵄩󵄩󵄩𝑊2
2
𝑛
∞
𝑛
𝑗=1
𝑖=1
𝑗=1
∑𝛽𝑛𝑗 (L𝑢) (𝑥𝑗 ) = ∑𝐴𝑖 ⟨𝜓𝑖 , ∑𝛽𝑛𝑗 𝜓𝑗 ⟩
󵄩2
󵄩
+ ⋅ ⋅ ⋅ + 󵄩󵄩󵄩𝑢𝑛+1 (𝑥) − 𝑢𝑛 (𝑥)󵄩󵄩󵄩𝑊2
= ∑ (𝐴𝑖 ) 󳨀→ 0,
∞
= ∑𝐴𝑖 ⟨𝜓𝑖 , 𝜓𝑛 ⟩
(𝑛 󳨀→ ∞) ,
𝑖=1
𝑖=𝑛+1
= 𝐴𝑛 ,
(64)
󵄩2
󵄩󵄩
󵄩
󵄩󵄩V𝑚 (𝑥) − V𝑛 (𝑥)󵄩󵄩󵄩𝑊2 = 󵄩󵄩󵄩V𝑚 (𝑥) − V𝑚−1 (𝑥)
𝑛
∞
𝑛
𝑗=1
𝑖=1
𝑗=1
󵄩2
+ ⋅ ⋅ ⋅ + V𝑛+1 (𝑥) − V𝑛 (𝑥)󵄩󵄩󵄩 𝑊
2
∞
= ∑𝐵𝑖 ⟨𝜓𝑖 , 𝜓𝑛 ⟩
󵄩
󵄩2
≤ 󵄩󵄩󵄩V𝑚 (𝑥) − V𝑚−1 (𝑥)󵄩󵄩󵄩 𝑊
𝑖=1
2
= 𝐵𝑛 .
󵄩
󵄩2
+ . . . + 󵄩󵄩󵄩V𝑛+1 (𝑥) − V𝑛 (𝑥)󵄩󵄩󵄩 𝑊
2
2
= ∑ (𝐵i ) 󳨀→ 0,
If 𝑛 = 1, then
(𝑛 󳨀→ ∞) .
𝑖=𝑛+1
Considering the completeness of 𝑊2 [0, +∞), there exist 𝑢(𝑥)
and V(𝑥) in 𝑊2 [0, +∞) such that
‖⋅‖
𝑢𝑛 (𝑥) 󳨀→ 𝑢 (𝑥)
as 𝑛 󳨀→ ∞,
‖⋅‖
(65)
V𝑛 (𝑥) 󳨀→ V (𝑥) as 𝑛 󳨀→ ∞.
(2) Secondly, we will prove that 𝑢(𝑥) and V(𝑥) are the
solutions of (2).
By Lemma 14 and the proof of (1), we may know that
𝑢𝑛 (𝑥), and V𝑛 (𝑥) respectively, converge uniformly to 𝑢(𝑥) and
V(𝑥) (𝑛 → ∞). Taking limits in (35), we have
∞
𝑢 (𝑥) = ∑𝐴𝑖 𝜓𝑖 (𝑥) ,
𝑖=1
∞
V (𝑥) = ∑𝐵𝑖 𝜓𝑖 (𝑥) .
𝑖=1
(69)
∑ 𝛽𝑛𝑗 (LV) (𝑥𝑗 ) = ∑𝐵𝑖 ⟨𝜓𝑖 , ∑𝛽𝑛𝑗 𝜓𝑗 ⟩
+ V𝑚−1 (𝑥) − V𝑚−2 (𝑥)
𝑚
(68)
𝑖=1
󵄩2
+ ⋅ ⋅ ⋅ +𝑢𝑛+1 (𝑥) − 𝑢𝑛 (𝑥)󵄩󵄩󵄩𝑊2
𝑚
(67)
𝑖=1
(L𝑢) (𝑥1 ) = 𝐹 (𝑥1 , 𝑢1 (𝑥1 ) , V1 (𝑝𝑥1 )) ,
(70)
(LV) (𝑥1 ) = 𝐺 (𝑥1 , V1 (𝑥1 ) , 𝑢1 (𝑞𝑥1 )) .
(71)
If 𝑛 = 2, then
𝛽21 (L𝑢) (𝑥1 )+𝛽22 (L𝑢) (𝑥2 )
= 𝛽21 𝐹 (𝑥1 , 𝑢1 (𝑥1 ) , V1 (𝑝𝑥1 ))
+ 𝛽22 𝐹 (𝑥2 , 𝑢2 (𝑥2 ) , V2 (𝑝𝑥2 )) ,
𝛽21 (LV) (𝑥1 )+𝛽22 (LV) (𝑥2 )
(72)
= 𝛽21 𝐺 (𝑥1 , V1 (𝑥1 ) , 𝑢1 (𝑞𝑥1 ))
+ 𝛽22 𝐺 (𝑥2 , V2 (𝑥2 ) , 𝑢2 (𝑞𝑥2 )) .
From (70) and (71), it is clear that
(66)
(L𝑢) (𝑥2 ) = 𝐹 (𝑥2 , 𝑢2 (𝑥2 ) , V2 (𝑝𝑥2 )) ,
(LV) (𝑥2 ) = 𝐺 (𝑥2 , V2 (𝑥2 ) , 𝑢2 (𝑞𝑥2 )) .
(73)
Abstract and Applied Analysis
9
4. Numerical Examples
Furthermore, it is easy to see by induction that
(L𝑢) (𝑥𝑗 ) = 𝐹 (𝑥𝑗 , 𝑢𝑗 (𝑥𝑗 ) , V𝑗 (𝑝𝑥𝑗 )) ,
𝑗 = 1, 2, . . . , (74)
(LV) (𝑥𝑗 ) = 𝐺 (𝑥𝑗 , V𝑗 (𝑥𝑗 ) , 𝑢𝑗 (𝑞𝑥𝑗 )) ,
𝑗 = 1, 2, . . . . (75)
Since {𝑥𝑖 }∞
𝑖=1 is dense in [0, +∞), for any 𝑦 ∈ [0, +∞), there
exists subsequence {𝑥𝑛𝑗 } such that
𝑥𝑛𝑗 󳨀→ 𝑦
as 𝑗 󳨀→ ∞.
(76)
Hence, let 𝑗 → ∞ in (74), and (75); by Lemma 14 and the
continuity of 𝐹(𝑥, 𝑢(𝑥), V(𝑝𝑥)) and 𝐺(𝑥, V(𝑥), 𝑢(𝑞𝑥)), we have
(L𝑢) (𝑦) = 𝐹 (𝑦, 𝑢 (𝑦) , V (𝑝𝑦)) ,
(77)
(LV) (𝑦) = 𝐺 (𝑦, V (𝑦) , 𝑢 (𝑞𝑦)) .
(78)
That is, 𝑢(𝑥) and V(𝑥) are the solutions of (2) and
∞
𝑢 (𝑥) = ∑𝐴𝑖 𝜓𝑖 (𝑥) ,
(79)
𝑖=1
In order to demonstrate the efficiency of our algorithm
for solving (2), we will present two numerical examples
in the reproducing kernel space 𝑊2 [0, +∞). Let 𝑛 be the
number of discrete points in [0, +∞). Denote 𝐸𝑛𝑢 (𝑥)
V (𝑥) = ∑𝐵𝑖 𝜓𝑖 (𝑥) ,
=
def
𝐸𝑛V (𝑥) =
|(V(𝑥) − V𝑛 (𝑥))/V(𝑥)|. All
|(𝑢(𝑥) − 𝑢𝑛 (𝑥))/𝑢(𝑥)|,
computations are performed by the Mathematica 5.0 software
package. Results obtained by the method are compared with
the exact solution of each example and are found to be in good
agreement with each other.
Example 18. In this example we consider the problem
𝑢󸀠 (𝑥) =
1
3
+ 2𝑥 cos (V ( 𝑥)) + 𝑓 (𝑥) ,
2
3
1 + 𝑢(𝑥)
1
V󸀠 (𝑥) = 2𝑒−V(𝑥) + 4𝑥3 sin (𝑢 ( 𝑥)) + 𝑔 (𝑥) ,
2
𝑢 (0) = 0,
∞
def
(83)
V (0) = 0,
(80)
𝑖=1
where 𝐴𝑖 = ∑𝑖𝑘=1 𝛽𝑖𝑘 𝐹(𝑥𝑘 , 𝑢𝑘 (𝑥𝑘 ), V𝑘 (𝑝𝑥𝑘 )) and 𝐵𝑖
∑𝑖𝑘=1 𝛽𝑖𝑘 𝐺(𝑥𝑘 , V𝑘 (𝑥𝑘 ), 𝑢𝑘 (𝑞𝑥𝑘 )).
=
In the proof of the convergence in Theorem 15 we only use
‖𝑢𝑛 ‖ ≤ 𝐶, ‖V𝑛 ‖ ≤ 𝐷; thus we obtain the following corollary.
with exact solution 𝑢(𝑥) = 𝑥𝑒−𝑥 , V(𝑥) = sin 𝑥, and
𝑓(𝑥) = 𝑒−𝑥 − 𝑒−𝑥 𝑥 − (3/(1 + 𝑒−2 𝑥2 )) − 2𝑥 cos(sin(𝑥/3)),
𝑔(𝑥) = −2𝑒− sin 𝑥 + cos 𝑥 − 4𝑥3 sin((1/2)𝑥𝑒−𝑥/2 ). Applying the
presented method in Section 3, we calculate the approximate
solution 𝑢50 (𝑥) and V50 (𝑥) in [0, 1] as follows.
Corollary 16. Suppose ‖𝑢𝑛 ‖ and ‖V𝑛 ‖ are bounded; then the
iterative sequence (35) is convergent to the exact solution of (2).
Step 1. By the method of the appendix, the corresponding
reproducing kernel functions can be obtained.
Theorem 17. Assume 𝑢(𝑥) and V(𝑥) are the solutions of (2),
𝑟𝑛1 = ‖𝑢(𝑥) − 𝑢𝑛 (𝑥)‖𝑊2 and 𝑟𝑛2 = ‖V(𝑥) − V𝑛 (𝑥)‖𝑊2 are the
approximate errors of 𝑢𝑛 (𝑥), V𝑛 (𝑥), where 𝑢𝑛 (𝑥) and V𝑛 (𝑥) are
given by (35). Then the errors 𝑟𝑛1 , 𝑟𝑛2 are monotone decreasing
in the sense of ‖ ⋅ ‖𝑊2 .
Step 2. Choosing a dense subset in [0, 1], then we get the
orthogonalization coefficients 𝛽𝑖𝑘 .
Step 4. Selecting the initial value 𝑢0 (𝑥) = V0 (𝑥) = 0, we
obtain 𝑢1 (𝑥), 𝑢2 (𝑥), . . . , 𝑢50 (𝑥) and V1 (𝑥), V2 (𝑥), . . . , V50 (𝑥) by
(19) developed in the paper.
Proof. From (35), and (79), it follows that
󵄩󵄩 ∞
󵄩󵄩2
󵄩
󵄩
󵄩󵄩 1 󵄩󵄩2
󵄩󵄩𝑟𝑛 (𝑥)󵄩󵄩 = 󵄩󵄩󵄩󵄩 ∑ 𝐴𝑖 𝜓𝑖 (𝑥)󵄩󵄩󵄩󵄩
󵄩
󵄩𝑊2 󵄩󵄩
󵄩󵄩
󵄩𝑖=𝑛+1
󵄩𝑊2
∞
(81)
2
= ∑ (𝐴𝑖 ) .
𝑖=𝑛+1
In the same way, we obtain from (35), (80)
󵄩󵄩 ∞
󵄩󵄩2
󵄩
󵄩
󵄩󵄩 2 󵄩󵄩2
󵄩󵄩𝑟𝑛 (𝑥)󵄩󵄩 = 󵄩󵄩󵄩󵄩 ∑ 𝐵𝑖 𝜓𝑖 (𝑥)󵄩󵄩󵄩󵄩
󵄩
󵄩𝑊2 󵄩󵄩
󵄩󵄩
󵄩𝑖=𝑛+1
󵄩𝑊2
∞
(82)
= ∑ (𝐵𝑖 ) .
𝑖=𝑛+1
2
The graphs of the superimposed image emerge in Figure 1.
At the same time, we have computed the approximate
solutions 𝑢𝑛 (𝑥) and V𝑛 (𝑥) (𝑛 = 300, 900) in [0, 3] and also
calculated the relative errors 𝐸𝑛𝑢 and 𝐸𝑛V in Table 1. The root
mean square errors (RMSE) about 𝑢(𝑥) with 𝑢𝑛 and V(𝑥) with
V𝑛 are shown in Table 2.
Example 19. Considering equations
2
Equations (81) and (82) show that the errors
monotone decreasing in the sense of ‖ ⋅ ‖ 𝑊 .
Step 3. According to (10), we can get the normal orthogonal
systems {𝜓𝑖 (𝑥)}50
𝑖=1 .
𝑟𝑛1
and
𝑟𝑛2
are
1
𝑢󸀠 (𝑥) = 3 cos (𝑢 (𝑥)) + 2𝑥 sin (V ( 𝑥)) + 𝑓 (𝑥) ,
5
1
V󸀠 (𝑥) = 2 sin (V (𝑥)) + 4𝑥3 cos (𝑢 ( 𝑥)) + 𝑔 (𝑥) ,
3
𝑢 (0) = 0,
V (0) = 0.
(84)
10
Abstract and Applied Analysis
0.35
0.8
0.3
0.6
0.25
0.2
0.4
0.15
0.1
0.2
0.05
0.2
0.4
0.6
0.8
1
0.2
0.4
(a)
0.6
0.8
1
(b)
Figure 1: The left is the superimposed image of 𝑢(𝑥) with 𝑢50 (𝑥) in [0, 1]. The right is the superimposed image of V(𝑥) with V50 (𝑥) in [0, 1].
Table 3: Relative errors in [0, 4] for Example 19.
Table 1: Relative errors in [0, 3] for Example 18.
Node
0.3
0.6
0.9
1.2
1.5
1.8
2.1
2.4
2.7
3.0
𝑢
𝐸300
4.97833𝑒 − 5
4.43080𝑒 − 5
3.86801𝑒 − 5
3.41875𝑒 − 5
3.09088𝑒 − 5
2.87391𝑒 − 5
2.75649𝑒 − 5
2.7288𝑒 − 5
2.7829𝑒 − 5
2.90670𝑒 − 5
V
𝐸300
8.4527𝑒 − 5
7.3330𝑒 − 5
6.6731𝑒 − 5
6.1845𝑒 − 5
5.7095𝑒 − 5
5.1303𝑒 − 5
4.3153𝑒 − 5
3.0422𝑒 − 5
6.7566𝑒 − 5
7.5017𝑒 − 5
𝑢
𝐸900
5.53163𝑒 − 6
4.92288𝑒 − 6
4.29733𝑒 − 6
3.79797𝑒 − 6
3.43349𝑒 − 6
3.19221𝑒 − 6
3.06149𝑒 − 6
3.03034𝑒 − 6
3.09008𝑒 − 6
3.23206𝑒 − 6
V
𝐸300
9.39806𝑒 − 6
8.15664𝑒 − 6
7.42512𝑒 − 6
6.88400𝑒 − 6
6.35872𝑒 − 6
5.71898𝑒 − 6
4.81984𝑒 − 6
3.41736𝑒 − 6
8.12936𝑒 − 7
8.72219𝑒 − 6
Node
0.4
0.8
1.2
1.6
2.0
2.4
2.8
3.2
3.6
4.0
Table 2: The RMS errors in [0, 3] for Example 18.
√∑𝑛𝑖=1 (𝑢𝑛 (𝑥𝑖 ) − 𝑢(𝑥𝑖 ))2 /𝑛
√∑𝑛𝑖=1 (V𝑛 (𝑥𝑖 ) − V(𝑥𝑖 ))2 /𝑛
300
9.95194𝑒 − 6
4.00007𝑒 − 5
900
1.10551𝑒 − 6
4.45554𝑒 − 6
𝑛
The true solutions are 𝑢(𝑥) = 5𝑥, V(𝑥) = 4 tan(𝑥), 𝑓(𝑥) =
5 − 3 cos(5𝑥) − 2𝑥 sin(4 tan(1/5)), 𝑔(𝑥) = −4𝑥3 cos(5𝑥/3) +
4sec𝑥2 − 2 sin(4 tan 𝑥). The numerical results are given in
Figure 2 and Tables 3 and 4. The figures and tables illustrate
that the method given in the paper is efficient.
𝑢
𝐸800
8.06470𝑒 − 5
1.07473𝑒 − 4
2.56877𝑒 − 4
1.19110𝑒 − 4
8.56122𝑒 − 5
2.27843𝑒 − 4
1.38237𝑒 − 4
7.26035𝑒 − 5
1.86684𝑒 − 5
1.50042𝑒 − 4
𝑢
𝐸1200
3.58334𝑒 − 5
4.77632𝑒 − 5
1.14199𝑒 − 4
5.29170𝑒 − 5
3.80383𝑒 − 5
1.01280𝑒 − 4
6.13912𝑒 − 5
3.22429𝑒 − 5
8.29627𝑒 − 5
6.66348𝑒 − 5
𝑢
𝐸1200
2.20033𝑒 − 5
2.34475𝑒 − 5
2.37040𝑒 − 5
2.23098𝑒 − 5
1.93810𝑒 − 5
1.56705𝑒 − 5
1.21676𝑒 − 5
9.56745𝑒 − 6
8.01449𝑒 − 6
7.14957𝑒 − 6
Table 4: The RMS errors in [0, 4] for Example 19.
√∑𝑛𝑖=1 (𝑢𝑛 (𝑥𝑖 ) − 𝑢(𝑥𝑖 ))2 /𝑛
√∑𝑛𝑖=1 (V𝑛 (𝑥𝑖 ) − V(𝑥𝑖 ))2 /𝑛
800
1.87807𝑒 − 3
7.85856𝑒 − 5
1200
8.34276𝑒 − 4
3.48826𝑒 − 5
𝑛
Appendix
The Reproducing Kernel Space 𝑊2 [0,𝑚]
𝑊2 [0, 𝑚] is defined as 𝑊2 [0, 𝑚] = {𝑢(𝑥) | 𝑢󸀠 are absolutely
continuous real value functions, 𝑢󸀠󸀠 ∈ 𝐿2 [0, 𝑚], 𝑢(0) = 0}.
The inner product in 𝑊2 [0, 𝑚] is given by
5. Conclusion
In this paper, RKHSM has been successfully applied to
find the solutions of systems of nonlinear IDDEs with
proportional delays. The efficiency and accuracy of the
proposed decomposition method were demonstrated by two
test problems. It is concluded from above tables and figures
that the RKHSM is an accurate and efficient method to solve
IDDEs with proportional delays. Moreover, the method is
also effective for solving some nonlinear initial-boundary
value problems and nonlocal boundary value problems.
V
𝐸800
4.94886𝑒 − 5
5.27211𝑒 − 5
5.32904𝑒 − 5
5.01608𝑒 − 5
4.35915𝑒 − 5
3.52666𝑒 − 5
2.74018𝑒 − 5
2.15572𝑒 − 5
1.80587𝑒 − 5
1.64230𝑒 − 5
𝑚
⟨𝑢 (𝑦) , V (𝑦)⟩𝑊2 = ∫ (4𝑢V + 5𝑢󸀠 V󸀠 + 𝑢󸀠󸀠 V󸀠󸀠 ) 𝑑𝑦,
0
(A.1)
where 𝑢, V ∈ 𝑊2 [0, 𝑚] and the norm ‖𝑢‖𝑊2 is denoted by
‖𝑢‖𝑊2 = √⟨𝑢, 𝑢⟩𝑊2 .
Theorem A.1. The space 𝑊2 [0, 𝑚] is a reproducing kernel
space; that is, for any 𝑢(𝑦) ∈ 𝑊2 [0, 𝑚] and each fixed 𝑥 ∈
[0, 𝑚], there exists 𝑅𝑥 (𝑦) ∈ 𝑊2 [0, 𝑚], 𝑦 ∈ [0, 𝑚], such that
Abstract and Applied Analysis
11
5
6
5
4
4
3
3
2
2
1
1
0.2
0.4
0.6
0.8
1
0.2
0.4
(a)
0.6
0.8
1
(b)
Figure 2: The left is the superimposed image of 𝑢(𝑥) with 𝑢50 (𝑥) in [0, 1]. The right is the superimposed image of V(𝑥) with V50 (𝑥) in [0, 1].
⟨𝑢(𝑦), 𝑅𝑥 (𝑦)⟩𝑊2 [0,𝑚] = 𝑢(𝑥). The reproducing kernel 𝑅𝑥 (𝑦) can
be denoted by
𝑅𝑥 (𝑦) = {
𝑐1 𝑒𝑦 + 𝑐2 𝑒−𝑦 + 𝑐3 𝑒2𝑦 + 𝑐4 𝑒−2𝑦 ,
𝑦 ≤ 𝑥,
(A.2)
𝑑1 𝑒𝑦 + 𝑑2 𝑒−𝑦 + 𝑑3 𝑒2𝑦 + 𝑑4 𝑒−2𝑦 , 𝑦 > 𝑥.
Proof. Applying to the integrations by parts for (A.1), we
have
⟨𝑢 (𝑦) , 𝑅𝑥 (𝑦) ⟩𝑊2 [0,𝑚]
𝑅𝑥 (𝑦) = {
0
𝑅𝑥(𝑘) (𝑥 + 0) = 𝑅𝑥(𝑘) (𝑥 − 0)
𝑅𝑥(3) (𝑥 + 0) − 𝑅𝑥(3) (𝑥 − 0) = 1,
𝑅𝑥󸀠󸀠 (𝑚) = 0,
𝑅𝑥󸀠󸀠 (0) = 0,
For 𝑢(𝑦) ∈ 𝑊2 [0, 𝑚], thus, 𝑢(0) = 0.
Suppose that 𝑅𝑥 (𝑦) satisfies the following generalized
differential equations:
𝑅𝑥 (0) = 0,
𝑅𝑥󸀠󸀠
(A.5)
𝑐1 =
𝑒−2𝑥 (4𝑒3𝑚 −7𝑒3𝑥 − 9𝑒2𝑚+𝑥 +7𝑒6𝑚+𝑥 +9𝑒4𝑚+3𝑥 −4𝑒3𝑚+4𝑥 )
(0) = 0.
𝑚
(A.7)
from which, the unknown coefficients of (A.6) can be
obtained:
4𝑅𝑥 (𝑦) − 5𝑅𝑥󸀠󸀠 (𝑦) + 𝑅𝑥(4) (𝑦) = 𝛿 (𝑦 − 𝑥) ,
𝑅𝑥󸀠󸀠 (𝑚) = 0,
(𝑘 = 0, 1, 2) ,
𝑅𝑥(3) (𝑚) − 5𝑅𝑥󸀠 (𝑚) = 0,
(A.4)
5𝑅𝑥󸀠 (𝑚) − 𝑅𝑥(3) (𝑚) = 0,
𝑦 ≤ 𝑥,
(A.6)
𝑦 > 𝑥.
(A.3)
Since 𝑅𝑥 (𝑦) ∈ 𝑊2 [0, 𝑚], it follows that
𝑅𝑥 (0) = 0.
𝑐1 𝑒𝑦 + 𝑐2 𝑒−𝑦 + 𝑐3 𝑒2𝑦 + 𝑐4 𝑒−2𝑦 ,
𝑑1 𝑒𝑦 + d2 𝑒−𝑦 + 𝑑3 𝑒2𝑦 + 𝑑4 𝑒−2𝑦 ,
By the definition of space 𝑊2 [0, 𝑚], coefficients 𝑐1 , . . . , 𝑐4 ,
𝑑1 , . . . , 𝑑4 satisfy
𝑚
= ∫ 𝑢 (𝑦) [4𝑅𝑥 (𝑦) − 5𝑅𝑥󸀠󸀠 (𝑦) + 𝑅𝑥(4) (𝑦)] 𝑑𝑦
󵄨󵄨𝑚
󵄨󵄨󵄨𝑚
󵄨
+ 5𝑢 (𝑦) 𝑅𝑥󸀠 (𝑦) 󵄨󵄨󵄨 + 𝑢󸀠 (𝑦) 𝑅𝑥󸀠󸀠 (𝑦) 󵄨󵄨󵄨
󵄨󵄨0
󵄨󵄨0
󵄨󵄨𝑚
󵄨
− 𝑢 (𝑦) 𝑅𝑥(3) (𝑦) 󵄨󵄨󵄨 .
󵄨󵄨0
The characteristic equation of 𝑅𝑥 (𝑦) − 5𝑅𝑥󸀠󸀠 (𝑦) + 𝑅𝑥(4) (𝑦) =
𝛿(𝑦 − 𝑥) is given by 𝜆4 − 5𝜆2 + 4 = 0, and the characteristic
roots are 𝜆 1,2 = ±1, 𝜆 3,4 = ±2.
We denote 𝑅𝑥 (𝑦) by
Then ⟨𝑢(𝑦), 𝑅𝑥 (𝑦)⟩𝑊2 [0,𝑚] = ∫0 𝑢(𝑦)𝛿(𝑦 − 𝑥)𝑑𝑦 = 𝑢(𝑥).
Hence, 𝑅𝑥 (𝑦) is the reproducing kernel of space 𝑊2 [0, 𝑚].
In the following, we will get the expression of the
reproducing kernel 𝑅𝑥 (𝑦).
6 (−7 − 9𝑒2𝑚 +9𝑒4𝑚 +7𝑒6𝑚 )
𝑐2 =
𝑒−2𝑥 (−4𝑒3𝑚 +7𝑒3𝑥 +9𝑒2𝑚+𝑥 −7𝑒6𝑚+𝑥 −9𝑒4𝑚+3𝑥 +4𝑒3𝑚+4𝑥 )
6 (−7 − 9𝑒2𝑚 + 9𝑒4𝑚 + 7𝑒6 m )
12
Abstract and Applied Analysis
𝑐3 =
𝑒−2𝑥 (−9𝑒4𝑚 −7𝑒6𝑚 +7𝑒4𝑥 +8𝑒3𝑚+𝑥 −8𝑒3(𝑚+𝑥) +9𝑒2𝑚+4𝑥 )
12 (−7 − 9𝑒2𝑚 + 9𝑒4𝑚 + 7𝑒6𝑚 )
𝑐4 =
−𝑒−2𝑥 (−9𝑒4𝑚 −7𝑒6𝑚 +7𝑒4𝑥 +8𝑒3𝑚+𝑥 −8𝑒3(𝑚+𝑥) +9𝑒2𝑚+4𝑥 )
12 (−7 − 9𝑒2𝑚 + 9𝑒4𝑚 + 7𝑒6𝑚 )
𝑑1 =
−
𝑒−2𝑥 (−1 + 𝑒2𝑥 ) (4𝑒3𝑚 + 7𝑒𝑥 − 9𝑒4𝑚+𝑥 + 4𝑒3𝑚+2𝑥 )
6 (−7 − 9𝑒2𝑚 + 9𝑒4𝑚 + 7𝑒6𝑚 )
𝑑2 =
𝑒2𝑚−2𝑥 (−1 + 𝑒2𝑥 ) (4𝑒𝑚 + 7𝑒4𝑚+𝑥 − 9𝑒𝑥 + 4𝑒𝑚+2𝑥 )
6 (−7 − 9𝑒2𝑚 + 9𝑒4𝑚 + 7𝑒6𝑚 )
𝑑3 =
𝑒−2𝑥 (−1 + 𝑒2𝑥 ) (7 + 9𝑒2𝑚 + 7𝑒2𝑥 + 9𝑒2(𝑚+𝑥) − 8𝑒3𝑚+𝑥 )
12 (−7 − 9𝑒2𝑚 + 9𝑒4𝑚 + 7𝑒6𝑚 )
𝑑4 =
−
𝑒3𝑚−2𝑥 (−1+𝑒2𝑥 ) (9𝑒𝑚 +7𝑒3𝑚 +9𝑒𝑚+2𝑥 −8𝑒𝑥 +7𝑒3𝑚+2𝑥 )
12 (−7 − 9𝑒2𝑚 + 9𝑒4𝑚 + 7𝑒6𝑚 )
.
(A.8)
Acknowledgments
This research is supported by the National Natural Science
Foundation of China (61071181), the Educational Department
Scientific Technology Program of Heilongjiang Province
(12531180, 12512133, and 12521148), and the Academic Foundation for Youth of Harbin Normal University (KGB201226).
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