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\AtBeginDocument{{\noindent\small
Eighth Mississippi State - UAB Conference on Differential Equations and
Computational Simulations.
{\em Electronic Journal of Differential Equations},
Conf. 19 (2010), pp. 75--83.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document} \setcounter{page}{75}
\title[\hfilneg EJDE-2010/Conf/19/\hfil Numerical solution]
{Numerical solution for nonlocal Sobolev-type
differential equations}
\author[S. A. Dubey\hfil EJDE/Conf/19 \hfilneg]
{Shruti A. Dubey}
\address{Shruti A. Dubey \newline
Laboratoire d'Analyse et d'Architecture des Systemes du
CNRS (LAAS-CNRS),\newline
Toulouse, France}
\email{shrurkd@gmail.com}
\thanks{Published September 25, 2010.}
\subjclass[2000]{35L65, 65M06, 65M12}
\keywords{Sobolev-type differential equations; Laplace
transform; \hfill\break\indent
numerical inversion; nonlocal conditions}
\begin{abstract}
We present a numerical approximate solution to Sobolev-type
differential equation subject to nonlocal initial boundary
conditions. A Laplace transform method is described for the
solution of considered equation. Following Laplace transform of
the original problem, an appropriate method of solving
differential equations is used to solve the resultant
time-independent modified equation and solution is inverted
numerically back into the time domain. Numerical results are
provided to show the accuracy of the proposed method.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks
\section{Introduction}
This work is concerned with the Sobolev-type partial differential
equation
\begin{equation}
\frac{\partial}{\partial t}\Big(u(x,t)-\frac{\partial^2}{\partial
x^2}u(x,t)\Big)-\frac{\partial ^2}{\partial x^2}u(x,t)=f(x,t),\quad
x\in [0,1],\;t>0,\label{eq1}
\end{equation}
subject to the conditions
\begin{equation}
\begin{gathered}
u(x,0)=u_0(x),\quad x\in [0,1], \\
u(0,t)=\int^1_0 a(x)u(x,t)dx + p(t),\quad t>0,\\
u(1,t)=\int^1_0 b(x)u(x,t)dx + q(t),\quad t>0.
\end{gathered} \label{eq2}
\end{equation}
where, $f$, $u_0$, $a,\;b,\;p,\;q$ are
prescribed continuous functions and $u(x,t)$ is an unknown
function which is a solution of \eqref{eq1} and satisfies
conditions \eqref{eq2} at the same time.
Sobolev-type equation appears in a variety of physical problems
such as flow of fluid through fissured rocks, thermodynamics and
propagation of long waves of small amplitude. There is an
extensive literature in which Sobolev type of equations are
investigated, in the abstract framework, see for instance
\cite{ab,bk,kt,lr,sh2}. Some other models of nonlocal boundary
conditions are numerically solved by Dehghan \cite{d1,d2}. The
first results on Sobolev-type equation were obtained by Hilbert
space methods \cite{ss}. Subsequently several authors studied and
discussed same type of problem subject to classical as well as
nonlocal conditions. Very strong and complete results are known
concerning existence,uniqueness and properties of solutions. Brill
\cite{b} and Showalter \cite{sh1} established the existence of
solutions of semilinear evolution equations of Sobolev type in
Banach space. Balachandran and Park \cite{bp} investigated
integrodifferential equation of Sobolev type with nonlocal
condition and proved the existence of mild and strong solutions
using semigroup theory and Schauder fixed point theorem. Recently,
the existence of solution to semilinear Sobolev type equation with
integral conditions is studied by Bouziani and Merazga \cite{bm}
using Rothe time discretization method.
So far not much seems to be done for obtaining an explicit
solution of Sobolev type partial differential equations, however
the solvability of these equations have been theoretically studied
in terms of the existence and uniqueness of a solution. The
purpose of present article is to give a method of solution to
Equation \eqref{eq1}, \eqref{eq2} using Laplace transform
technique. In recent years, Laplace transform method has been used
to approximate the solution of different class of linear partial
differential equations \cite{a,bs,kw,ms}. Suying et al
\cite{smzw}, established a numerical method based on Laplace
transform for solving initial problem of nonlinear dynamic
differential equations. The main difficulty in using Laplace
transform method consists in finding its inverse. Numerical
inversion methods are then used to overcome this difficulty. There
are many numerical techniques available in literature to invert
Laplace transform. In this paper we focus exclusively on the
Stehfest inversion algorithm \cite{s} in order to efficiently and
accurately invert the Laplace transform (which cannot be done
analytically).
The plan of the paper is as follows. In section 2, we develop a
method of solution and find out a solution of the problem in
Laplace domain. Obtained solution is then converted in to real
domain by the means of numerical inversion algorithm. Some
examples are given in Section 3 to demonstrate the competence of
the method, followed by conclusion in Section 4.
\section{Method based on Laplace transform}
\subsection{Laplace transform technique}
Laplace transform is widely used in the area of engineering
technology and mathematical science. There are many problems whose
solution may be found in terms of a Laplace transform. In fact, it
is an efficient method for solving various differential equations.
Laplace transform of a function $w(x,t)$ with respect to $t$ can
be expressed as
$$
W(x;s)=\mathcal{L}\{w(x,t);t\to s\}
=\int_0^{\infty} w(x,t)e^{-st}dt,
$$
where, $s$ is known as Laplace variable. A capital letter $W$
represents Laplace transform of a function $w$, i.e., $W$ is a
function in Laplace domain. Starting with this approach and taking
Laplace transform on both sides of \eqref{eq1}, \eqref{eq2} with
respect to $t$, we get
\begin{gather*}
\frac{d^2}{dx^2}U(x;s)-\frac{s}{(1+s)}U(x;s)=-\frac{1}{(1+s)}
\Big[F(x;s)+U(x;0)-\frac{d^2}{dx^2}U(x;0)\Big], \\
U(x;0)=u_0(x), \\
U(0;s)=\int^1_0a(x)U(x;s)dx+P(s), \\
U(1;s)=\int^1_0b(x)U(x;s)dx+Q(s).
\end{gather*}
Using the initial condition into the differential equation, we obtain
\begin{equation}
\begin{gathered}
\frac{d^2}{dx^2}U(x;s)-\frac{s}{(1+s)}U(x;s)=-\frac{1}{(1+s)}
\Big[F(x;s)+u_0(x)-\frac{d^2}{dx^2}u_0(x)\Big], \\
U(0;s)=\int^1_0a(x)U(x;s)dx+P(s), \\
U(1;s)=\int^1_0b(x)U(x;s)dx+Q(s),
\end{gathered} \label{eq3}
\end{equation}
where, $U(x;s)=\mathcal{L}\{u(x,t);t\to s\}$,
$F(x;s)=\mathcal{L}\{f(x,t);t\to s\}$,
$P(s)=\mathcal{L}\{p(t);t\to s\}$ and
$Q(s)=\mathcal{L}\{q(t);t\to s\}$.
Thus, considered equation is reduced in boundary value problem governed by a second order inhomogeneous ordinary differential equation. On solving it, we obtain a general solution of \eqref{eq3} as
\begin{equation}
\begin{aligned}
U(x;s)&=\frac{-1}{(1+s)}\sqrt{\frac{(s+1)}{s}}\int
_0^x[F(\tau;s)+u_0(\tau)-\frac{d^2}{d\tau^2}u_0(\tau)]\\
&\quad\times \sinh\Big(\sqrt{\frac{s}{s+1}}(x-\tau)\Big)d\tau
+C_1(s)e^{-\sqrt{\frac{s}{s+1}}x}
+C_2(s)e^{\sqrt{\frac{s}{s+1}}x},
\end{aligned}\label{eq4}
\end{equation}
where, $C_1$ and $C_2$ are arbitrary functions of $s$.
Substituting \eqref{eq4} in to the boundary conditions, we have
\begin{align*}
&C_1(s)\Big[1-\int_0^1a(x)e^{-\sqrt{\frac{s}{s+1}}x}dx\Big]
+C_2(s)\Big[1-\int_0^1a(x)e^{\sqrt{\frac{s}{s+1}}x}dx\Big]\\
&= \frac{-1}{(s+1)}\sqrt{\frac{s+1}{s}}\int_0^1
\Big[[F(\tau;s)+u_0(\tau)-\frac{d^2}{d\tau^2}u_0(\tau)] \\
&\quad\times \int_\tau^1a(x)\sinh\Big(\sqrt{\frac{s}{s+1}}(x-\tau)
\Big)dx\Big]d\tau+P(s),
\end{align*}
\begin{align*}
&C_1(s)\Big[e^{-\sqrt{\frac{s}{s+1}}}-\int_0^1b(x)
e^{-\sqrt{\frac{s}{s+1}}x}dx\Big]
+C_2(s)\Big[e^{\sqrt{\frac{s}{s+1}}}
-\int_0^1b(x)e^{\sqrt{\frac{s}{s+1}}x}dx\Big]\\
&=\frac{-1}{(s+1)}\sqrt{\frac{s+1}{s}}\int_0^1
\Big[[F(\tau;s)+u_0(\tau)-\frac{d^2}{d\tau^2}u_0(\tau)]\\
&\quad\times \int_\tau^1b(x)\sinh
\Big(\sqrt{\frac{s}{s+1}}(x-\tau)\Big)dx\Big]d\tau+Q(s)\\
&\quad+\frac{1}{(s+1)}\sqrt{\frac{s+1}{s}}
\int_0^1[F(\tau;s)+u_0(\tau)-\frac{d^2}{d\tau^2}u_0(\tau)]\sinh
\Big(\sqrt{\frac{s}{s+1}}(1-\tau)\Big)d\tau.
\end{align*}
It gives
\begin{equation}
\begin{pmatrix}
C_1(s) \\
C_2(s)
\end{pmatrix}
=\begin{pmatrix} a_{11}(s)\quad a_{12}(s)\\
a_{21}(s)\quad a_{22}(s)
\end{pmatrix}^{-1}
\begin{pmatrix} b_1(s)\\b_2(s)
\end{pmatrix},\label{eq5}
\end{equation}
where,
\begin{equation}
\begin{gathered} \label{eq6}
a_{11}(s)= 1-\int_0^1a(x)e^{-\sqrt{\frac{s}{s+1}}x}dx, \quad
a_{12}(s)=1-\int_0^1a(x)e^{\sqrt{\frac{s}{s+1}}x}dx, \\
a_{21}(s)=e^{-\sqrt{\frac{s}{s+1}}}
-\int_0^1b(x)e^{-\sqrt{\frac{s}{s+1}}x}dx, \quad
a_{22}(s)=e^{\sqrt{\frac{s}{s+1}}}
-\int_0^1b(x)e^{\sqrt{\frac{s}{s+1}}x}dx,\\
\begin{aligned}
b_1(s)&=-\frac{1}{(s+1)}\sqrt{\frac{s+1}{s}}\int_0^1
\Big[[F(\tau;s)+u_0(\tau)\\
&\quad -\frac{d^2}{d\tau^2}u_0(\tau)
\int_\tau^1a(x)\sinh\Big(\sqrt{\frac{s}{s+1}}(x-\tau)\Big)dx\Big]
d\tau+P(s),
\end{aligned} \\
\begin{aligned}
b_2(s)&=\frac{-1}{(s+1)}\sqrt{\frac{s+1}{s}}\int_0^1
\Big[[F(\tau;s)+u_0(\tau)-\frac{d^2}{d\tau^2}u_0(\tau)] \\
&\quad \times \int_\tau^1b(x)
\sinh\Big(\sqrt{\frac{s}{s+1}}(x-\tau)\Big)dx\Big]d\tau+Q(s)
+\frac{1}{(s+1)}\sqrt{\frac{s+1}{s}}\\
&\quad \times \int_0^1[F(\tau;s)
+u_0(\tau)-\frac{d^2}{d\tau^2}u_0(\tau)]
\sinh\Big(\sqrt{\frac{s}{s+1}}(1-\tau)\Big)d\tau .
\end{aligned}
\end{gathered}
\end{equation}
Thus, to find out the solution in Laplace domain one has to evaluate
all the integrals appear in \eqref{eq4} and \eqref{eq6}.
This can be done for known functions $F,P,Q,a,b,u_0$, however,
in many cases, the resulting function is not
easy to integrate exactly. Therefore, there is need for numerical
approximation of the integrals. A well known Gaussian Quadrature
formula exists for computing integrals numerically (see \cite{as}).
Using this formula we have the following
approximations of the above integrals:
\begin{gather*}
\int_0^1\begin{pmatrix}a(x)\\b(x)
\end{pmatrix} e^{\pm \sqrt{\frac{s}{s+1}}x}\,dx
\simeq \frac{1}{2}\sum_{i=1}^{n}w_i
\begin{pmatrix}
a(\frac{1}{2}(x_i+1))\\
b(\frac{1}{2}(x_i+1))
\end{pmatrix} )e^{\pm
\sqrt{\frac{s}{s+1}}(\frac{1}{2}(x_i+1))},\\
\begin{aligned}
&\int_0^x\Big[F(\tau;s)+u_0(\tau)-\frac{d^2}{d\tau^2}u_0(\tau)\Big]
\sinh\Big(\sqrt{\frac{s}{s+1}}(x-\tau)\Big)d\tau \\
&\simeq \frac{x}{2}\sum_{i=1}^nw_i\Big[F\left(\frac{x}{2}
(x_i+1);s\right)+u_0\left(\frac{x}{2}(x_i+1)\right)
-\frac{d^2}{d\tilde{\tau}^2}u_0\left(\frac{x}{2}(x_i+1)\right)\Big]\\
&\quad \times \sinh\Big(\sqrt{\frac{s}{s+1}}
\left(x-\frac{x}{2}(x_i+1)\right)\Big),
\end{aligned}
\end{gather*}
where $\tilde{\tau}=\frac{x}{2}(x_i+1)$.
\begin{align*}
&\int_0^1\Big[\Big(F(\tau;s)+u_0(\tau)-\frac{d^2}{d\tau^2}u_0(\tau)
\Big)\int_\tau^1 \begin{pmatrix}a(x)\\b(x) \end{pmatrix}
\sinh\Big(\sqrt{\frac{s}{s+1}}(x-\tau)\Big)dx\Big]d\tau \\
&\simeq \frac{1}{2}\sum_{i=1}^nw_i
\Big[F\left(\frac{1}{2}(x_i+1);s\right)+u_0\left(\frac{1}{2}(x_i+1)
\right)-\frac{d^2}{d\tilde{\tau}^2}u_0\left(\frac{1}{2}(x_i+1)\right)
\Big]\\
&\quad \times\left(\frac{1-\frac{1}{2}(x_i+1)}{2}\right)
\sum_{j=1}^{n}w_j \begin{pmatrix}
a\big(\frac{1-\frac{1}{2}(x_i+1)}{2}x_j
+\frac{1+\frac{1}{2}(x_i+1)}{2}\big)\\
b\big(\frac{1-\frac{1}{2}(x_i+1)}{2}x_j
+\frac{1+\frac{1}{2}(x_i+1)}{2}\big)\\
\end{pmatrix}\\
&\quad\times \sinh\Big(\sqrt{\frac{s}{s+1}}\Big(
\big(\frac{1-\frac{1}{2}(x_i+1)}{2}\big)x_j
+\frac{1+\frac{1}{2}(x_i+1)}{2}-\frac{1}{2}(x_i+1)\Big)\Big),
\end{align*}\\
where, $\tilde{\tau}=\frac{1}{2}(x_i+1)$.
$x_i$ and $w_i$ are defined by
$$
x_i: i^{th} \text{ zero of } P_n(x),\quad
w_i=2/(1-x_i^2)[P'_n(x_i)]^2
$$
and known as abscissas and weights respectively.
Their tabulated values can be found in \cite{as} for different
values of $n$.
\subsection{Numerical inversion of Laplace transform}
Now, we have a solution in Laplace transform domain as given
in \eqref{eq4}. So we expect to obtain a solution of original
problem by means of inverting the Laplace transform.
Simple transforms can often be inverted using readily available
table. More complex functions can be analytically inverted
through the complex inversion formula
$$
g(t)=\frac{1}{2\pi j}\int_{c-j\infty}^{c+j\infty}
e^{st}G(s)ds,
$$
where, $c$ is a positive real number such that all the poles
of the function $G(s)$ lie at the left of the line
$\operatorname{Re}(s)=c$.
Sometimes analytical inversion of a Laplace domain solution is
difficult to obtain, therefore, a numerical inversion method
must be used. A variety of different methods for numerically
inverting the Laplace transform are available that can be employed.
There exists no universal method but different types of methods
work well for different classes of functions. A nice comparison
of four frequently used numerical Laplace inversion algorithms
is given by Hasan et al. \cite{hd}. We use the Stehfest
algorithm \cite{s} in the work as it is easy to implement and
leads to result of sufficient accuracy throughout the time range.
This numerical technique was first introduced by Graver \cite{g}
and its algorithm then offered by Stehfest. Stehfest's algorithm
approximates the time domain solution as
$$
u(x,t)\approx \frac{ \ln 2}{t}\sum_{k=1}^{2m}\beta_kU\Big(x;
\frac{\ln 2}{t}k\Big),
$$
where, $m$ is the positive integer and
$$
\beta_k=(-1)^{m+k}\sum_{l=\left[\frac{k+1}{2}\right]}^{\min(k,m)}
\frac{l^{m}(2l)!}{(m-l)!\;l!\;(l-1)!(k-l)!(2l-k)!}.
$$
Here $[r]$ denotes the integer part of $r$. The parameter $m$ is a free
parameter that should be optimized by trial and error. It was seen
that with increasing $m$ accuracy of result increases up to a point
and then owing to the rounding errors it decreases \cite{s}. Thus, for
choosing optimum $m$, it is beneficial to apply an algorithm
repeatedly for different values of $m$ and study its effect on the
solution. The other way to choose optimal value of $m$ could be, to
apply the Stehfest's algorithm for inverting the Laplace transform of
some elementry functions which are known.
\noindent\textbf{Remarks:}
* Stehfest's method gives accurate results for many problems
including diffusion problem, fractional functions in the Laplace domain.
However, it fails to predict $e^t$ type functions or those with
oscillatory behavior such as sine and wave functions (see \cite{hd}).
* Note that more than one numerical inversion algorithm can also
be performed to check the accuracy of the result.
\section{Numerical results}
This section illustrates some of the numerical examples that
we carried out in order to test the reliability of the method
proposed. We consider the problem \eqref{eq1}, \eqref{eq2}
with different choice of functions as follows:
\begin{example} \label{exa1} \rm
Here, we take
\begin{gather*}
f(x,t)=\Big(x(x-1)-\frac{1}{7}\Big)e^{-t},\quad
u_0(x)=x(1-x)+\frac{1}{7},\\
a(x)=b(x)=6/13,\quad p(t)=q(t)=0.
\end{gather*}
In this case exact solution is
$$
u(x,t)=\Big(x(1-x)+\frac{1}{7}\Big)e^{-t}.
$$
We use developed method for finding the numerical solution.
It is found that the best choice of the parameter $m$ for this
and subsequent example is 5. Thus we take m=5, n=8 and follow
exactly the same steps described in the previous section for
the method of solution. Obtained numerical result is compared
with exact solution in Fig. \ref{fig1} and Fig. \ref{fig2}.
\end{example}
Fig. \ref{fig1} provides the contrast of the numerical result
and the exact solution for $t=0.5 $ and $x\in [0,1]$.
In Fig. \ref{fig2}, the comparison between these solutions
are given for a fixed value of $x=0.2 $ and for $t\in[0.1,1]$.
Results show that in both cases, approximate solution agrees
well with the exact solution.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig1} % graph1_x.eps
\end{center}
\caption{Comparison between numerical and exact solutions
for $t=0.5$, $x\in[0,1]$}
\label{fig1}
\end{figure}
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig2} % graph1_t.eps
\end{center}
\caption{Comparison between numerical and exact solutions for
$x=0.2$, $t\in[0.1, 1]$}
\label{fig2}
\end{figure}
\begin{example} \label{exa2} \rm
Here, we take
\begin{gather*}
f(x,t)=\frac{-2(x^2+t+1)}{(t+3)^3},\quad
u_0(x)=x^2/9, \quad
a(x)=b(x)=x,\\
p(t)=-\frac{1}{4(t+3)^2},\quad q(t)=\frac{3}{4(t+3)^2}.
\end{gather*}
Its exact solution is
$$
\frac{x^2}{(t+3)^2}.
$$
\end{example}
To apply the numerical method proposed, we need the Laplace
transform of the above mentioned functions
$f(x,t)$, $p(t)$, $q(t)$ with respect to $t$ which can be given by
\begin{gather*}
F(x;s)=\int_0^\infty \frac{-2s(sx^2+\xi+s)}{(\xi+3s)^3}e^{-\xi}d\xi,\\
P(s)=\int_0^\infty\frac{-s}{4(\xi+3s)^2}e^{-\xi}d\xi,\\
Q(s)=\int_0^\infty\frac{3s}{4(\xi+3s)^2}e^{-\xi}d\xi.
\end{gather*}
Integrals involved here are evaluated with the help of numerical
approximation formula \cite[(25.45)]{as}.
Proceeding in the similar manner as of the previous example
and choosing $m=5$ and $n=8$, a comparison of numerical and
exact solutions is done. In Fig. \ref{fig3} graphs of the
approximate and exact solution are given for $t=0.4$
and $x\in [0,1]$. For $x=0.6$ and $t\in[0.1, 1]$ results are
drawn in Fig. \ref{fig4}. Presented graphs clearly show that
the approximate and exact solutions are almost superposed.
\noindent\textbf{Remarks:}
* The numerical solution matches with exact solution up to at
least 4 significant places of decimal.
* The simulation results show that our proposed method achieves
good performance.
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig3} % graph2_x.eps
\end{center}
\caption{Comparison between numerical and exact solutions for $t=0.4$,
$x\in[0,1]$}
\label{fig3}
\end{figure}
\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig4} % graph2_t.eps
\end{center}
\caption{Comparison between numerical and exact solutions for
$x=0.6$, $t\in[0.1,1]$}
\label{fig4}
\end{figure}
\subsection{Conclusion} A method of solution based on Laplace
transform to the considered nonlocal problem is described. The
benefit of the presented method is that it gives explicitly a
numerical approximate solution of the problem, however, several
theoretical approaches to existence and uniqueness of solution to
Sobolev-type equations are summarized in literature since last few
decades. Numerical results show that the time domain solution
evaluated by presented method is comparable with the exact
solution.
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