\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{graphicx}

\AtBeginDocument{{\noindent\small \emph{
Electronic Journal of Differential Equations}, 
Vol. 2008(2008), No. 29, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or
 http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/29\hfil Reproducing kernel methods]
{Reproducing kernel methods for solving linear
initial-boundary-value problems}

\author[L.-H. Yang, Y. Lin\hfil EJDE-2008/29\hfilneg]
{Li-Hong Yang, Yingzhen Lin} 

\address{Li-Hong Yang \newline
College of Science, Harbin Engineering University, 150001, China}
\email{lihongyang@hrbeu.edu.cn}

\address{Yingzhen Lin \newline
 Department of Mathematics, Harbin Institute of Technology (WEIHAI),
 264209, China}
\email{liliy55@163.com }

\thanks{Submitted January 26, 2008. Published February 28, 2008.}
\thanks{Supported by grants 60572125 from the National Natural Science
Foundation of China, and \hfill\break\indent HEUF707061 from the
Basic Scientific Research Foundation of
 Harbin Engineering University.}
\subjclass[2000]{35A35, 35A45, 35G05, 65N99} 
\keywords{Hyperbolic equation; linear initial-boundary conditions; 
\hfill\break\indent reproducing kernel space}

\begin{abstract}
 In this paper, a  reproducing kernel with polynomial
 form is used for finding analytical and approximate solutions
 of a second-order hyperbolic equation with linear initial-boundary
 conditions. The analytical solution is represented as a series
 in the reproducing kernel space, and the approximate solution
 is obtained as an n-term summation. Error estimates are proved to
 converge to zero in  the sense of the space norm, and a numerical
 example is given to illustrate the method.
\end{abstract}

\maketitle 
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{property}[theorem]{Property}

\section{Introduction}

A reproducing kernel Hilbert space is a useful framework for
constructing approximate solutions of partial differential equations
(PDE). Many numerical methods have been proposed for solving linear
and nonlinear PDEs, but we did not find a method that uses
reproducing kernels. In this paper, we focus on the exact and
approximate solutions to PDEs with linear initial-boundary
conditions. A reproducing kernel with polynomial form in the
corresponding Hilbert space is given and the space completion is
proved.

We consider the following second-order one-dimensional hyperbolic
equation in a reproducing kernel space:
\begin{equation} \label{e1.1}
\frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial
x^2}+f(x,t),\quad 0<x<1,\; t\geq 0
\end{equation}
subject to the mixed boundary conditions
\begin{gather} \label{e1.2}
\frac{\partial u(0,t)}{\partial x}+w_1 u(0,t)=0,\quad w_1\in
 \mathbb{R}, \; t \geq 0, \\
 \label{e1.3}
\frac{\partial u(1,t)}{\partial x}+w_2 u(1,t)=0,\quad w_2\in
\mathbb{R},\; t \geq 0
\end{gather}
and the initial conditions
\begin{gather} \label{e1.4}
u(x,0)=g_1(x)\quad 0<x<1, \\
 \label{e1.5}
\frac{\partial u(x,0)}{\partial t}=g_2(x)\quad 0<x<1
\end{gather}
 Although the focus is on homogeneous mixed boundary conditions, we
can study problems with non-homogenous mixed boundary conditions by
the homogenization methods. Equation \eqref{e1.1} with conditions
 \eqref{e1.2}-\eqref{e1.5} has been applied to many problems in
 physics,
engineering, fluid mechanics, and so on. It is well known that the
finite difference method \cite{George}, spline method  \cite{Jatoe},
and collocation method \cite{Russell} can be used to solve this
equation. Now, employing the reproducing property of the kernel, we
give an efficient method for solving \eqref{e1.1}. The analytical
solution is represented in the form of series in the reproducing
kernel space, and the approximate solution is obtained by the n-term
intercept of the analytical solution, and error is proved to
converge to zero in the sense of the space norm. In Section 2, the
reproducing kernel function with polynomial form is obtained, and
one-dimensional and two-dimensional reproducing kernel spaces needed
in this paper are defined. After that, we devote Section 3 to solve
\eqref{e1.1} with initial-boundary value conditions
\eqref{e1.2}-\eqref{e1.5}. Finally, a numerical example is discussed
to demonstrate the accuracy of the presented method in Section 4.

\section{The reproducing kernel space}

In this section, several reproducing kernel spaces are introduced.
Throughout this paper, we  discuss problems on the domain
$\Omega=[a,b]\times[c,d]$.

\subsection*{One-dimensional reproducing kernel space}
The reproducing kernel space  $W_2^m[a,b]$  is defined as the set of
functions  such that $u^{(m-1)}(x)$is  absolutely continuous
 on $[a,b]$, and $u^{(m)}(x)\in L^2[a,b]$, for $x\in[a,b]$,
where  $m$ is a positive integer. This space is equipped with the
inner product
\begin{equation} \label{e2.1}
\langle u(x),v(x)\rangle
=\sum_{i=0}^{m-1}u^{(i)}(a)v^{(i)}(a)+\int_a^bu^{(m)}(x)v^{(m)}(x)dx,
\end{equation}
for $u(x),v(x)\in W_2^m[a,b]$. The the norm is
\begin{equation} \label{e2.2}
\|u\|=\sqrt{\langle u(x),u(x)\rangle },\quad u(x)\in W_2^m[a,b]\,.
\end{equation}

\begin{theorem} \label{thm2.1}
The space  $W_2^m[a,b]$  equipped with the norm \eqref{e2.2}, is a
 Hilbert
space.
\end{theorem}

\begin{proof}
Suppose that  $\{f_n(x)\}_{i=1}^{\infty}$  is a Cauchy sequence of
the space  $W_2^m[a,b]$, that is, as  $n\to \infty$, it follows that
$$
\|f_{n+p}-f_n\|^2=\sum_{i=0}^{m-1}(f_{n+p}^{(i)}(a)-f_{n}^{(i)}(a))^2
+\int_a^b(f_{n+p}^{(m)}(x)-f_{n}^{(m)}(x))^2dx\to  0\,.
$$
So, we have
\begin{gather*}
f_{n+p}^{(i)}(a)-f_{n}^{(i)}(a)\to
0,\quad i=0,1,\dots,m-1, \\
\int_a^b(f_{n+p}^{(m)}(x)-f_{n}^{(m)}(x))^2dx\to  0\,.
\end{gather*}
The above formulas show that, for  $0\leq i\leq m-1$,
 $f_n^{(i)}(a)$ ($n=1,2,\dots$)  are  Cauchy sequences and
 $f_n^{(m)}(x)$ ($n=1,2,\dots$)  is a Cauchy sequence in
 $L^2[a,b]$. Hence, there exist the unique real number
 $\lambda_i$ ($i=0,1,\dots,m-1$)  and the unique function
$h(x)\in L^2[a,b]$. It holds that
\begin{gather*}
\lim_{n\to \infty}f_n^{(i)}(a)=\lambda_i,\quad \,(i=0,1,\dots,m-1),\\
\lim_{n\to \infty}\int_a^b(f_{n}^{(m)}(x)-h(x))^2dx=0\,.
\end{gather*}
Let
\[
g(x)=\sum_{k=0}^{m-1}\frac{\lambda_k}{k!}(x-a)^k+
\underbrace{\,\int_a^x\dots\!\int_a^x\!\!}_m\;h(x)dx^m,
\]
 from $h(x)\in L^2[a,b]$, we obtain that
 $g^{(m-1)}(x)=\lambda_{m-1}+\int_a^xh(x)dx$  is absolutely
continuous on the interval  $[a,b]$, and  $g^{(m)}(x)$  is almost
equal to  $h(x)$  on the interval  $[a,b]$. Hence,  $g(x)\in
W_2^m[a,b]$, and  $g^{(i)}(a)=\lambda_i$ ($0\leq i\leq m-1$). Then
\begin{align*}
\|f_n(x)-g(x)\|^2 & =\sum_{i=0}^{m-1}(f_{n}^{(i)}(a)-g^{(i)}(a))^2
+\int_a^b(f_{n}^{(m)}(x)-g^{(m)}(x))^2dx \\
& =\sum_{i=0}^{m-1}(f_{n}^{(i)}(a)-\lambda_i)^2
+\int_a^b(f_{n}^{(m)}(x)-h(x))^2dx \to  0\,.
\end{align*}
Hence,Space  $W_2^m[a,b]$  equipped with the norm \eqref{e2.2}, is a
Hilbert space.
\end{proof}

\begin{theorem} \label{thm2.2}
Hilbert space  $W_2^m[a,b]$  is a reproducing kernel space, that is,
for  for all $f(y)\in W_2^m[a,b]$  and fixed  $x\in [a,b]$ , there
exists  $R_m(x,y)\in W_2^m[a,b]$  such that
\begin{equation} \label{e2.3}
\langle f(x),R_m(x,y)\rangle =f(y)
\end{equation}
and  $R_m(x,y)$  is called the reproducing kernel function of space
 $W_2^m[a,b]$.
\end{theorem}

\begin{proof}
Let  $R_m(x,y)$  be the reproducing kernel function. By \eqref{e2.1}
and \eqref{e2.2}, we have
\begin{equation} \label{e2.4}
\langle f(x),R_m(x,y)\rangle
=\sum_{i=0}^{m-1}f^{(i)}(a)\frac{\partial^i R_m(a,y)}{\partial
x^i}+\int_a^bf^{(m)}(x)\frac{\partial^m R_m(a,y)}{\partial x^m}dx\,.
\end{equation}
Applying the integration by parts for the second scheme of the
right-hand of \eqref{e2.4}, we obtain
\begin{equation} \label{e2.5}
\begin{aligned}
&\int_a^bf(x)\frac{\partial^m R_m(a,y)}{\partial x^m}dx\\
&=\sum_{i=0}^{m-1}(-1)^if^{(m-i-1)}(x)\frac{\partial^{m+i}
R_m(a,y)}{\partial
x^{m+i}}|_{x=a}^b+(-1)^m\int_a^bf(x)\frac{\partial^{2m}
R_m(a,y)}{\partial x^{m+i}}dx\,.
\end{aligned}
\end{equation}
Let  $j=m-i-1$, the first term of the right-hand side of the above
formula is rewritten as
\begin{equation} \label{e2.6}
\begin{aligned}
&\sum_{i=0}^{m-1}(-1)^if^{(m-i-1)}(x)\frac{\partial^{m+i}
R_m(a,y)}{\partial x^{m+i}}|_{x=a}^b\\
&=\sum_{j=0}^{m-1}(-1)^{m-j-1}f^{j}(x)
\frac{\partial^{2m-j-1}R_m(a,y)}{\partial x^{2m-j-1}}|_{x=a}^b.
\end{aligned}
\end{equation}
Let  $i=j$. Then substituting the two expressions \eqref{e2.5} and
 \eqref{e2.6} into \eqref{e2.4} yields
\begin{align*}
&\langle f(x),R_m(x,y)\rangle\\
&=\sum_{i=0}^{m-1}f^{(i)}(a)(\frac{\partial^i R_m(a,y)}{\partial
x^i}-(-1)^{m-i-1}\frac{\partial^{2m-i-1}
R_m(a,y)}{\partial x^{2m-i-1}})\\
&\quad +\sum_{i=0}^{m-1}(-1)^{m-i-1}f^{i}(b)\frac{\partial^{2m-i-1}
R_m(a,y)}{\partial x^{2m-i-1}}+(-1)^m\int_a^bf(x)\frac{\partial^{2m}
R_m(x,y)}{\partial x^{2m}}dx\,.
\end{align*}
Suppose that  $R_m(x,y)$  satisfies the following generalized
differential equations
\begin{equation} \label{e2.8}
\begin{gathered}
 (-1)^m\frac{\partial^{2m} R_m(x,y)}{\partial
x^{2m}}=\delta(x-y) \\
\frac{\partial^{i} R_m(a,y)}{\partial
x^{i}}-(-1)^{m-i-1}\frac{\partial^{2m-i-1} R_m(a,y)}{\partial
x^{2m-i-1}}=0,\quad i=0,1,\dots,m-1 \\
\frac{\partial^{2m-i-1} R_m(b,y)}{\partial x^{2m-i-1}}=0,\quad
i=0,1,\dots,m-1\,.
\end{gathered}
\end{equation}
Then $ \langle f(x),R_m(x,y)\rangle
=\int_a^bf(x)\delta(x-y)dx=f(y)$. Hence,  $R_m(x,y)$  is the
reproducing kernel of space  $W_2^m[a,b]$.
\end{proof}

Next, we  give the expression of the reproducing kernel  $R_m(x,y)$.
The characteristic equation of \eqref{e2.8} is $\lambda^{2m}=0$, and
the characteristic roots are  $\lambda_i=0$, $i=1,\dots,2m$. So we
write the reproducing kernel as
\begin{equation} \label{e2.9}
R_m(x,y)=\begin{cases}
lR_m(x,y)=\sum_{i=1}^{2m}c_i(y)x^{i-1},& x<y\\
rR_m(x,y)=\sum_{i=1}^{2m}d_i(y)x^{i-1}, &x>y\,.
\end{cases}
\end{equation}
By the definition of   $W_2^m[a,b]$  and \eqref{e2.8}, the
coefficients  $c_i,d_i$, $i=1,\dots,2m$  satisfy
\begin{equation} \label{e2.10}
\begin{gathered}
 \frac{\partial^i lR_m(y,y)}{\partial x^i}
 =\frac{\partial^i rR_m(y,y)}{\partial x^i},\quad i=0,1,\dots,2m-2\\
(-1)^m(\frac{\partial^{2m-1} rR_m(y+,y)}{\partial
x^{2m-1}}-\frac{\partial^{2m-1} lR_m(y-,y)}{\partial
x^{2m-1}})=1\\
 \frac{\partial^{i} R_m(a,y)}{\partial
x^{i}}-(-1)^{m-i-1}\frac{\partial^{2m-i-1} R_m(a,y)}{\partial
x^{2m-i-1}}=0,\quad i=0,1,\dots,m-1\\
\frac{\partial^{2m-i-1} R_m(b,y)}{\partial x^{2m-i-1}}=0,\quad
i=0,1,\dots,m-1\,.
\end{gathered}
\end{equation}
Then the solution of \eqref{e2.10} yields the expression of the
 reproducing
kernel  $R_m(x,y)$.


In this paper we consider  the case
 $m=3$  and  $[a,b]=[0,1]$, the corresponding kernel space
is defined as
 $W_2^3[0,1]$ are function $f$ such that $f^{(2)}(x)$ is absolutely
continuous  on $[0,1]$ and $f^{(3)}(x)\in L^2[0,1],x\in[0,1]$, and
the reproducing kernel as
$$
 R_3(x,y)=\begin{cases}
\frac{1}{120}(120+x^5+120xy-5x^4y+30x^2y^2+10x^3y^2), &x<y\\
1+\frac{y^5}{120}+\frac{1}{12}x^2y^2(3+y)+x(y-\frac{y^4}{24}), &
x>y)
\end{cases}
$$
Other reproducing kernel spaces needed in this paper are described
similarly.

The space  $W_{2,1}^3[0,1]$  is a subspace of  $W_2^3[0,1]$  with
$f(0)=f'(0)=0$, and the reproducing kernel is
$$
R_{31}(x,y)=\begin{cases}
\frac{1}{120}x^2(x^3-5x^2y+30y^2+10xy^2), & x<y,\\
\frac{1}{120}y^2(y^3-5xy^2+10x^2(3+y)), & x>y.
\end{cases}
$$
The space  $W_{2,2}^3[0,1]$  is a subspace of  $W_2^3[0,1]$ with
   $f(0)+w_1 f'(0)=0$, $f(1)+w_2 f'(1)=0$, and the reproducing kernel
 is
\begin{align*}
&R_{31}(x,y)\\
&=\begin{cases}
\frac{1}{8400}\Big(-350x^4y+x^5(46-48y+30y^2+10y^3-y^5)
+24(46+92y\\
+30y^2+10y^3-y^5) +48x(46+92y+30y^2+10y^3-y^5)
+30x^2(24+48y\\
+40y^2-10y^3+y^5) +10x^3(24+48y+40y^2-10y^3+y^5)\Big),\\
\text{ if } x<y,\\[3pt]
\frac{1}{8400}\Big(1104+2208y+720y^2+240y^3
+x(2208+4416y+1440y^2+480y^3\\
-350y^4-48y^5) +10x^3(24+48y-30y^2-10y^3+-y^5)
-x^5(24+48y\\
-30y^2-10y^3+y^5) +10x^2(72+144y+120y^2+40y^3+3y^5)\Big),\\
\text{ if } x>y.
\end{cases}
\end{align*}


\subsection*{Two-dimensional reproducing kernel space}

We construct the two-dimen\-sional reproducing kernel spaces as in
\cite{Yang, Aronszajn}. Let
\begin{align*}
P(\Omega)& = \overline{W_{2,1}^3[0,1]\otimes W_{2,2}^3[0,1]} &
=\{\sum_{i,j=1}^{\infty}c_{i,j}g_i^{(1)}(x)g_j^{(2)}(t):
\sum_{i,j=1}^{\infty}|c_{i,j}|^2<\infty\}
\end{align*}
where  $\{g_i^{(k)}\}$  is a complete orthonormal sequence in the
space  $W_{2,k}^3$, $k=1,2$, and endowed with the inner product
% \label{e2.12}
\begin{align*}
(u(x,t),v(x,t))_{P}
&=(\sum_{k,l=1}^{\infty}c_{k,l}^{(1)}g^{(1)}_k(x)g^{(2)}_l(t),
\sum_{p,q=1}^{\infty}c^{(2)}_{p,q}g^{(1)}_p(x)g^{(2)}_q(t))\\
&=\sum_{k,l=1}^{\infty}c_{k,l}^{(1)}
  \sum\lrcorner_{p,q=1}^{\infty}c^{(2)}_{p,q}(g^{(1)}_k(x)
  g^{(2)}_l(t),g^{(1)}_p(x)g^{(2)}_q(t))\\
&=\sum_{k,l=1}^{\infty}c^{(1)}_{k,l}c^{(2)}_{k,l}\\
\end{align*}
and the norm
\[ % \label{e2.12}
\|u\|_{P}=\sqrt{(u,u)_{P}}=(\sum_{k,l=1}^{\infty}c_{k,l}^2)^{1/2}
\]
According to \cite{Russell}, the space  $P(\Omega)$ is a Hilbert
space
 with
the norm  $\| \cdot \|_{P}$, and possesses the reproducing kernel
\[ %\label{e2.13}
\bar{R}((\xi,\eta),(x,t))=R_{31}(\xi,x)\cdot R_{32}(\eta,t).
\]
It is easy to prove that the following properties hold.

\begin{property}
For  $u(x)\in W_{3,1}[0,1],\,v(y)\in W_{32}[0,1]$, it follows that
 $u(x)\cdot v(y)\in P(\Omega)$ .
\end{property}

\begin{property}
 $(u_1(x)\cdot v_1(y),\,u_2(x)\cdot
v_2(y))_P=\langle u_1(x),\,u_2(x)\rangle _{W_{31}}\cdot \langle
v_1(y),\,v_2(y)\rangle_{W_{32}}$  holds for any $u_1,\,u_2\in
W_{3,1}[0,1],\,v_1,v_2\in W_{32}[0,1]$.
\end{property}

Similarly, the other two-dimensional reproducing kernel space
 can be defined as
\begin{align*}
P_1(\Omega)& = \overline{W_{2}^1[0,1]\otimes W_{2}^1[0,1]}\\
& =\{\sum_{i,j=1}^{\infty}c_{i,j}g_i(x)g_j(t):
\sum_{i,j=1}^{\infty}|c_{i,j}|^2<\infty\},
\end{align*}
where  $\{g_i\}$  is a complete orthonormal sequence in the space
 $W_{2}^1$. Its reproducing kernel function
 $\tilde{R}((\xi,\eta),(x,t))$  can be obtain from the reproducing
kernel function  $R_1(x,y)$  of the space  $W_2^1[0,1]$, that is,
 $ \tilde{R}((\xi,\eta),(x,t))=R_{1}(\xi,x)\cdot R_{1}(\eta,t)$.

\section{Solution of Equation \eqref{e1.1}}

In this section, we consider the second-order one-dimensional
hyperbolic equation \eqref{e1.1} with initial-value conditions
\eqref{e1.4}--\eqref{e1.5} and mixed boundary-value conditions
\eqref{e1.2}--\eqref{e1.3}. Without the loss of generality, we
discuss equation \eqref{e1.1} with homogeneous
 conditions,
that is,
\begin{equation} \label{e3.1}
\begin{gathered}
\frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial
x^2}+f(x,t),\quad 0<x<1,\; t\geq 0
\\
\frac{\partial u(0,t)}{\partial x}+w_1 u(0,t)=0,\quad w_1\in
 R, \,\, t \geq 0\\
\frac{\partial u(1,t)}{\partial x}+w_2 u(1,t)=0,\quad w_2\in
R,\; t \geq 0\\
u(x,0)=0,\quad 0<x<1\\
\frac{\partial u(x,0)}{\partial t}=0,\quad 0<x<1\\
\end{gathered}
\end{equation}
Through the homogenization, we can complete the equivalence
transformation. Hence, we can solve  \eqref{e3.1} in the same way as
\eqref{e1.1}.

Define the linear operator  $L$  from the reproducing kernel space
$P(\Omega)$  to the reproducing kernel space
 $P_1(\Omega)$:
\begin{equation} \label{e3.2}
(Lu)(x,t)=\frac{\partial^2 u}{\partial t^2}-\frac{\partial^2
u}{\partial x^2}\,.
\end{equation}

\begin{theorem} \label{thm3.1}
The operator  $L:P(\Omega)\to  P_1(\Omega)$ is a bounded operator.
\end{theorem}

\begin{proof} Note that
\begin{gather*}
\|(Lu)(x,t)\|^2=\|u_{tt}-u_{xx}\|^2\leq \|u_{tt}\|^2+\|u_{xx}\|^2,\\
u(x,t)=(u(\xi,\eta),\,R_{31}(\xi,x)R_{32}(\eta,t))\\
u_{tt}(x,t)=(u(\xi,\eta),R_{31}(\xi,x)
 \frac{\partial^2}{\partial t^2}R_{32}(\eta,t))\\
u_{xx}(x,t)=(u(\xi,\eta),
\frac{\partial^2}{\partial x^2}R_{31}(\xi,x)R_{32}(\eta,t))\\
|u_{tt}(x,t)|
 \leq\|u\|\,\|R_{31}(\xi,x)\|\,\|\frac{\partial^2}{\partial
t^2}R_{32}(\eta,t)\|\\
|u_{xx}(x,t)| \leq\|u\|\,\|\frac{\partial^2}{\partial
x^2}R_{31}(\xi,x)\|\,\|R_{32}(\eta,t)\|\,.
\end{gather*}
Also note that
\[
\|R_{31}(\xi,x)\|=\sqrt{\langle
R_{31}(\xi,x),\,R_{31}(\xi,x)\rangle} =\sqrt{R_{31}(x,x)},\quad
\|R_{32}(\eta,t))\|=\sqrt{R_{32}(t,t)}
\]
are continuous functions on the interval  $[0,1]$; that is, it holds
that  $\|R_{31}(\xi,x)\|\leq M_1$, $\|R_{32}(\eta,t))\|\leq M_2$.
Meanwhile, setting  $\|\frac{\partial^2}{\partial
x^2}R_{31}(\xi,x)\|=M_3$, $\|\frac{\partial^2}{\partial
t^2}R_{32}(\eta,t)\|=M_4$, we have
$$
|u_{tt}(x,t)|\leq \|u\|M_1M_4, \quad |u_{xx}(x,t)|\leq \|u\|M_2M_3
$$
Hence,
$$
\|(Lu)(x,t)\|^2\leq \|u\|^2(M_1^2M_4^2+M_2^2M_3^2)
$$
The proof is complete.
\end{proof}

For a fix  countable dense subset
 $\{M_i=(x_i,y_i)\}_{i=1}^{\infty}$  of the domain  $\Omega$,
we put
\begin{equation} \label{e3.3}
\varphi_i(x,y)=\tilde{R}((x_i,t_i),\,(x,t))=R_1(x_i,x)\cdot
R_1(t_i,t),
\end{equation}
where  $R_1(x_i,x)$ is the reproducing kernel of
 $W_2^1[0,1]$. Let  $\psi_i(x,t)=(L^*\varphi_i)(x,t)$, where
 $L^*$  denotes the adjoint operator of  $L$.
By the definitions of adjoint operator and the reproducing property,
the following Lemmas hold.

\begin{lemma} \label{lem3.1}
 $\psi_i(x,t)=(LR_{31}(\bullet,x)\cdot
R_{32}(\star,t))(x_i,t_i)$
\end{lemma}

\begin{proof}
Let  $\bullet,\,\star$  denote the variables corresponding to
functions respectively. Then
\begin{equation} \label{e3.4}
\begin{aligned}
\psi_i(x,t)&=(L^*\varphi_i)(x,t)\\
&=((L^*\varphi_i)(\bullet,\star),R_{31}(\bullet,x)\cdot
R_{32}(\star,t))_{P(\Omega)}\\
&=(\varphi_i(\circ),(LR_{31}(\bullet,x)\cdot
 R_{32}(\star,t))(\circ))_{P(\Omega)}\\
&=(LR_{31}(\bullet,x)\cdot R_{32}(\star,t))(x_i,t_i)\,.
\end{aligned}
\end{equation}
 From the definition of $L$, we have
\begin{equation} \label{e3.5}
\psi_i(x,t)=R_{31}(x_i,x)\cdot \frac{\partial^2}{\partial
t_i^2}R_{32}(t_i,t) -\frac{\partial^2}{\partial
x_i^2}R_{31}(x_i,x)\cdot R_{32}(t_i,t)
\end{equation}
\end{proof}

\begin{lemma} \label{lem3.2}
If  $\{M_i\}_{i=1}^\infty$  is dense on  $P(\Omega)$ , then
 $\{\psi_i(x,t)\}_{i=1}^\infty$  is a complete system of
 $P(\omega)$.
\end{lemma}

\begin{proof}
For each fixed  $u(x,t)\in P(\Omega)$, let
 $(u(x,t),\psi_i(x,t))=0$ ($i=1,2,\dots$), which implies
\begin{equation} \label{e3.6}
(u(x,t),(L^*\varphi_i)(x,t))=(Lu(x,t),\varphi_i(x,t))=(Lu)(x_i,t_i)=0\,.
\end{equation}
Since  $\{M_i\}_{i=1}^\infty$  is dense on  $P(\Omega)$, we have
 $(Lu)(x,t)=0$. It follows that  $u\equiv 0$  from the existence of
 $L^{-1}$ .
\end{proof}

 Consequently, we employ Gram-Schmidt process to orthonormalize the
sequence  $\{\psi_i\}_{i=1}^\infty$ in the reproducing kernel space
$P(\Omega)$. Denote by
 $\{\overline{\psi}_i\}_{i=1}^\infty$  the orthonormalized sequence;
that is,
\begin{equation} \label{e3.7}
\overline{\psi}_i(x,t)=\sum_{k=1}^i\beta_{ik}\psi_k(x,t),\quad
i=1,2,\dots
\end{equation}
where  $\beta_{ik}$  are orthoganal coefficients.

\begin{theorem} \label{thm3.2}
If   $\{M_i\}_{i=1}^\infty$  is dense on  $P(\Omega)$  and the
solution of \eqref{e3.1} is unique, then the solution of
\eqref{e3.1} has the form
\begin{equation} \label{e3.8}
u(x,t)=\sum_{i=1}^\infty(\sum_{k=1}^i\beta_{ik}f(x_k,t_k))
\overline{\psi}_i(x,t)
\end{equation}
\end{theorem}

\begin{proof}
Note that the Lemma \ref{lem3.1} and the orthonormal system
 $\{\overline{\psi}_i(x,t)\}_{i=1}^\infty$  of  $P(\Omega)$, we
have
\begin{align*}
u(x,t)&
=\sum_{i=1}^\infty(u(x,t),\,\overline{\psi}_i(x,t))\overline{\psi}_i(x,t)\\
&=\sum_{i=1}^\infty(u(x,t),\,\sum_{k=1}^i\beta_{ik}\psi_k(x,t))
 \overline{\psi}_i(x,t)\\
&=\sum_{i=1}^\infty\sum_{k=1}^i\beta_{ik}(u(x,t),\,(L^*\varphi_k)(x,t))
 \overline{\psi}_i(x,t)\\
&=\sum_{i=1}^\infty\sum_{k=1}^i\beta_{ik}(Lu(x,t),\,\varphi_k(x,t))
 \overline{\psi}_i(x,t)\\
&=\sum_{i=1}^\infty(\sum_{k=1}^i\beta_{ik}f(x_k,t_k))
 \overline{\psi}_i(x,t)
\end{align*}
\end{proof}

 We denote the approximate solution of $u(x,t)$ by
\begin{equation} \label{e3.9}
u_{n}(x,t)=\sum_{i=1}^{n}\sum_{k=1}^{i}
\beta_{ik}f(x_{k},t_k)\overline{\psi}_{i}(x,t)\,.
\end{equation}

\begin{theorem} \label{thm3.3}
For each  $u(x,t)\in  P(\Omega)$, let
$\varepsilon_{n}^2=\|u(x,t)-u_{n}(x,t)\|^{2}$,
 then  sequence  $\{\varepsilon_{n}\}$  is  monotone  decreasing  and
 $\varepsilon_{n}\to  0$ ($n\to  \infty$).
\end{theorem}

\begin{proof}
Because
\begin{align*}
 \varepsilon_{n}^2&=\|u(x,t)-u_{n}(x,t)\|^{2}\\
&=\|\sum_{i=n+1}^{\infty}(u(x,t),\overline{\psi}_{i}(x,t))
 \overline{\psi}_{i}(x,t)\|^{2}\\
&=\sum_{i=n+1}^{\infty}((u(x,t),\overline{\psi}_{i}(x,t)))^2,
\end{align*}
we have
\begin{align*}
\varepsilon_{n-1}^2
&=\|u(x,t)-u_{n-1}(x,t)\|^{2}\\
&=\|\sum_{i=n}^{\infty}(u(x,t),\overline{\psi}_{i}(x,t))
 \overline{\psi}_{i}(x,t)\|^{2}\\
&=\sum_{i=n}^{\infty}((u(x,t),\overline{\psi}_{i}(x,t)))^2.
\end{align*}
Clearly $\varepsilon_{n-1}\geq\varepsilon_{n}$. By Theorem
\ref{thm3.2}, $\{\varepsilon_{n}\}$ is monotone decreasing
 and
$\varepsilon_{n}\to  0$ ($n\to  \infty$).
\end{proof}

\section{Numerical Example}

In this Section, we employ the method introduced in Section 3 to
solve \eqref{e3.1} through symbolic and numerical computations are
performed by using Mathematica 5.0. The results obtained by the
method are compared with the analytical solution and are found to be
in good agreement with each other.

\begin{figure}[ht]
 \begin{center}
\includegraphics[width=0.7\textwidth]{fig1} %  error.eps
\end{center}
\caption{Error $u(x,t)-u_n(x,t)$ }
\end{figure}


As an example, we solve  the equation
\begin{gather*}
\frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial
x^2}+f(x,t),\quad 0<x<1,\; t\geq 0 \\
\frac{\partial u(0,t)}{\partial x}+w_1 u(0,t)=0,\quad w_1\in
 R, \; t \geq 0\\
\frac{\partial u(1,t)}{\partial x}+w_2 u(1,t)=0,\quad w_2\in
R,\; t \geq 0\\
u(x,0)=g_1(x),\quad  0<x<1\\
\frac{\partial u(x,0)}{\partial t}=g_2(x),\quad 0<x<1
\end{gather*}
 where
 $f(x,t)=e^{-t}(-te^{x+t}-(1+4t^2)e^{x+t+t^2}+x-2x(1+2t^2)e^{t+t^2})$ ,
 $w_1=-2,\,w_2=-1$ , $g_1(x)=x+e^x,\,\,g_2(x)=-(x+e^x)$.
The true solution is  $u(x,t)=(x+e^x)e^{-t}$. The numerical results
and error are given by table and figure, respectively.

\begin{table}
\caption{The error in approximating $u(x,t)$} \centering
\begin{small}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline $(x,t)$& $u(x,t)$ & $u_n(x,t)$ &relative error
&$(x,t)$& $u(x,t)$ & $u_n(x,t)$ &relative error\\
\hline ($\frac{1}{21},\frac{1}{21}$)&1.0454&1.04539&0.0000109747&
($\frac{1}{3},\frac{1}{21}$)&1.64854& 1.64852& 0.0000125776\\
($\frac{2}{3},\frac{2}{21}$)&2.3769& 2.3768& 0.0000425437&
($\frac{20}{21},\frac{2}{21}$)&3.22228& 3.22214& 0.0000443688\\
($\frac{1}{21},\frac{1}{7}$)&0.950436& 0.950381&0.0000583054&
($\frac{1}{3},\frac{1}{7}$)&1.49878&1.49868&0.0000678766\\
($\frac{2}{3},\frac{1}{7}$)&2.26637& 2.26619&0.0000782034&
 ($\frac{20}{21},\frac{1}{7}$)&3.07244& 3.07219&0.0000815501\\
($\frac{1}{21},\frac{5}{21}$)&0.864095& 0.86399&0.000122309&
($\frac{1}{3},\frac{5}{21}$)&1.36263& 1.36242&0.000150798\\
($\frac{1}{20},\frac{2}{7}$)&0.823912& 0.823783& 0.000156987&
($\frac{1}{3},\frac{2}{7}$)&1.29926& 1.299&0.000203236\\
($\frac{2}{3},\frac{2}{7}$)&1.96466& 1.96421& 0.000229523&
($\frac{20}{21},\frac{2}{7}$)&2.66343& 2.66279& 0.000239147\\
($\frac{1}{21},\frac{8}{21}$)&0.749065& 0.748897& 0.000223804&
($\frac{1}{3},\frac{8}{21}$)&1.18123& 1.18084& 0.000333805\\
 ($\frac{1}{21},\frac{3}{7}$)&0.714231& 0.714051&0.000253051&
($\frac{1}{3},\frac{3}{7}$)&1.1263& 1.12584&0.000414213\\
($\frac{2}{3},\frac{3}{7}$)&1.70312& 1.70234&0.000460057&
 ($\frac{20}{21},\frac{3}{7}$)&2.30887& 2.30776&0.000477361\\
($\frac{2}{3},\frac{10}{21}$)&1.62392& 1.62301&0.000562991&
($\frac{20}{21},\frac{10}{21}$)&2.42147& 2.42054& 0.000386727\\
($\frac{1}{21},\frac{4}{7}$)&0.619151& 0.618958& 0.000313067&
($\frac{1}{3},\frac{4}{7}$)&0.976367& 0.975648& 0.000736244\\
($\frac{2}{3},\frac{4}{7}$)&1.4764& 1.47517&0.000830423&
($\frac{20}{21},\frac{4}{7}$)&2.0015& 1.99979&0.000854846\\
($\frac{1}{21},\frac{13}{21}$)&0.590359& 0.590168&0.000324107&
($\frac{1}{3},\frac{13}{21}$)&0.930963& 0.930144& 0.000879517\\

 \hline
\end{tabular}\end{small}
\end{table}


\subsection*{Conclusion}
 In this paper, we employed a reproducing kernel and its conjugate
operator to construct the complete orthonormal basis in the
reproducing kernel space. By adding the initial and boundary
conditions to the reproducing kernel space, we obtain the analytic
solution of equation \eqref{e1.1}. Numerical example illustrates the
accuracy and validity of the algorithm. Meanwhile, constructing the
new form of the reproducing kernel function, we can obtain the
analytic solution for the multi-dimensional equations because it
reduces the computational complexity. In a future article will study
the nonlinear problem with mixed boundary conditions by using this
method.


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\end{document}