\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 48, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/48\hfil Initial-boundary value problems]
{Initial-boundary value problems for the \\ wave equation}

\author[T. Sh. Kalmenov, D. Suragan \hfil EJDE-2014/48\hfilneg]
{Tynysbek Sh. Kalmenov, Durvudkhan Suragan}  % in alphabetical order

\address{Tynysbek Sh. Kalmenov \newline
Institute of Mathematics and Mathematical Modeling,
28 Shevchenko str.,
050010 Almaty, Kazakhstan}
\email{kalmenov.t@mail.ru}

\address{Durvudkhan Suragan \newline
Institute of Mathematics and Mathematical Modeling,
28 Shevchenko str.,
050010 Almaty, Kazakhstan}
\email{suragan@list.ru}

\thanks{Submitted January 9, 2014. Published February 19, 2014.}
\subjclass[2000]{35M10}
\keywords{Hyperbolic equation;  wave potential;
\hfill\break\indent  initial boundary value problems}

\begin{abstract}
 In this work we consider an initial-boundary value problem
 for the one-dimensional wave equation.
 We prove the uniqueness of the solution  and show that the solution
 coincides with the wave potential.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\allowdisplaybreaks

\section{Introduction}

In $\Omega=(0,1)$ consider the  one-dimensional potential
\begin{equation}
 u(x)=\int_0^1-\frac{1}{2}|x-y|f(y)dy, \label{eq1}
\end{equation}
where $f$ is an integrable function in $(0,1)$.
The kernel of the potential is a fundamental solution of the second order 
differential equation
\begin{equation}
 -\varepsilon''(x-y)=\delta(x-y),\label{eq2}
 \end{equation}
where  $\varepsilon(x-y)=-\frac{1}{2}|x-y|$ and $\delta$ is the Dirac delta function.
Hence the potential \eqref{eq1} satisfies the equation
 \begin{equation}
 -u''(x)=f(x), x\in\Omega.\label{eq3}
 \end{equation}
On the other hand, integrating by part, we obtain
\begin{align*}
u(x)&=\int_0^1-\frac{1}{2}|x-y|f(y)dy=\int_0^1\frac{1}{2}|x-y|u''(y)dy\\
&=\int_0^{x}\frac{1}{2}(x-y)u''(y)dy+\int_{x}^1\frac{1}{2}(y-x)u''(y)dy\\
&=u(x)-x\frac{u'(0)+u'(1)}{2}-\frac{-u'(1)+u(0)+u(1)}{2}, \quad \forall x\in (0,1);
\end{align*}
i.e.,
$$
x(u'(0)+u'(1))+(-u'(1)+u(0)+u(1))=0, \quad \forall x\in (0,1).
$$
Therefore, the self-adjoint boundary conditions for the potential \eqref{eq1}  are
\begin{equation}
u'(0)+u'(1)=0,\quad -u'(1)+u(0)+u(1)=0.\label{eq4}
\end{equation}

Hence if we solve the equation \eqref{eq3} with the boundary conditions 
\eqref{eq4}, then we find an unique solution of this boundary value problem 
in the form \eqref{eq1}.

The simple method finds equivalent boundary value problems of ODE for one 
dimensional potential integrals.
However this task becomes tedious for PDE, and we obtained boundary conditions 
of the volume potentials for elliptic equations and showed some their 
applications in works \cite{k1,k2,k3}. 
In particular in \cite{k1}, by using a new non-local boundary 
value problem, which is equivalent to the Newton potential, we found 
explicitly all eigenvalues and eigenfunctions of the Newton potential 
in the 2-disk and the 3-ball.
The aim of this paper is to give an analogy of the boundary value 
problem \eqref{eq3}-\eqref{eq4} for the wave potential. Unlike elliptic 
and parabolic cases, where we obtained non-local boundary conditions 
for the corresponding volume potentials, and some other nonclassic non-local 
boundary initial boundary value problems of hyperbolic equations 
(see, for example, \cite{p1,p2}) we get a local initial boundary value problem 
for the wave potential.

\section{Main result and their proof}

 In the bounded domain  $\Omega\equiv\{(x,t):(0,l)\times(0,T)\}$ 
we consider the  wave potential
\begin{equation}
 u(x,t)=\int_{\Omega}\varepsilon(x-\xi,t-\tau)f(\xi,\tau)d\xi d\tau,\label{e1}
 \end{equation}
where
$\varepsilon(x-\xi,t-\tau)=\frac{1}{2}\theta(t-\tau-|x-\xi|)$ is a
fundamental solution of Cauchy problem for the wave equation;
i.e.,
\begin{gather*}
\frac{\partial^{2}\varepsilon(x-\xi,t-\tau)}{\partial
t^{2}}-\frac{\partial^{2}\varepsilon(x-\xi,t-\tau)}{\partial
x^{2}}=\delta(x-\xi,t-\tau),
\\
\frac{\partial^{2}\varepsilon(x-\xi,t-\tau)}{\partial
\tau^{2}}-\frac{\partial^{2}\varepsilon(x-\xi,t-\tau)}{\partial
\xi^{2}}=\delta(x-\xi,t-\tau),
\\
\varepsilon(x-\xi,t-\tau)|_{\tau=t}=\frac{\partial\varepsilon(x-\xi,t-\tau)}{\partial
t}|_{\tau=t}=
\frac{\partial\varepsilon(x-\xi,t-\tau)}{\partial\tau}|_{\tau=t}=0
\end{gather*}
if $f(x,t)\in L_{2}(\Omega)$ then $u(x,t)\in W_{2}^1(\Omega)\cap
W_{2}^1(\partial\Omega)$ and the wave potential  \eqref{e1}  satisfies
to the  equation \cite{v1}
 \begin{equation}
\frac{\partial^{2}u(x,t)}{\partial
t^{2}}-\frac{\partial^{2}u(x,t)}{\partial x^{2}}=f(x,t),
\quad (x,t)\in\Omega, \label{e2}
 \end{equation}
with the initial conditions
\begin{equation}
 u(x,0)=u_{t}(x,0)=0,\quad  0<x<l.\label{e3}
 \end{equation}

The wave potential \eqref{e1} is widely used to solve various initial-boundary 
problems for the wave equation.
Here we find the lateral boundary conditions of the wave potential \eqref{e1}.
Main result of this article reads as follows.

 \begin{theorem} \label{thm2.1}
If $f(x,t)\in L_{2}(\Omega)$ then  the wave potential \eqref{e1}
satisfies the lateral boundary conditions
\begin{gather}
 (u_{x}-u_{t})(0,t)=0, \quad x=0,\; 0<t<T; \label{e4}\\
 (u_{x}+u_{t})(l,t)=0, \quad x=l,\;  0<t<T\,. \label{e5}
  \end{gather}
Conversely, if  a function $u(x,t)\in W_{2}^1(\Omega)\cap
W_{2}^1(\partial\Omega)$ satisfies the equation \eqref{e2},
the initial conditions \eqref{e3}, and the lateral boundary conditions
\eqref{e4}-\eqref{e5}, then the function $u(x,t)$ uniquely defines 
the wave potential \eqref{e1}.
 \end{theorem}


\begin{proof} 
We use techniques from \cite{k3}.
 Consider the one-dimensional wave potential in the bounded
domain $\Omega\equiv\{(x,t):(0,l)\times(0,T)\}$ with the boundary $S$,
$$
u(x,t)=\int_{\Omega}\varepsilon(x-\xi,t-\tau)f(\xi,\tau)d\xi
d\tau.
$$

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig1}
\end{center}
\caption{Domain of integration} \label{Fig1}
\end{figure}


Since $\tau=t-x+\xi$ and $\tau=t+x-\xi$ are characteristics, 
the integral vanishes outside of characteristic domain.
Therefore, we integrate by ABCDF (Figure \ref{Fig1}). 
Assuming that $u(x,t)\in W_{2}^1(\Omega)$, taking into account properties 
of the fundamental solution, and integrating by part, we calculate
\begin{align*}
u(x,t)&=\int_{\Omega}\varepsilon(x-\xi,t-\tau)f(\xi,\tau)d\xi d\tau\\
&= \int_{ABCDF}\frac{1}{2}[u_{\tau\tau}(\xi,\tau)-u_{\xi\xi}(\xi,\tau)]d\xi d\tau\\
&=\frac{1}{2}\int_0^{x}d \xi\int_0^{t+\xi-x}
 \frac{\partial^{2}u(\xi,\tau)}{\partial\tau^{2}}d \tau+\frac{1}{2}\int_{x}^{l}d
\xi\int_0^{t-\xi+x}\frac{\partial^{2}u(\xi,\tau)}{\partial\tau^{2}}d
\tau\\
&\quad -\frac{1}{2}\int_0^{t-x}d
\tau\int_0^{l}\frac{\partial^{2}u(\xi,\tau)}{\partial\xi^{2}}d
\xi- \frac{1}{2}\int_{t-\xi}^{t-l+x}d
\tau\int_{\tau-t+x}^{l}\frac{\partial^{2}u(\xi,\tau)}{\partial\xi^{2}}d\xi
\\
&\quad-\frac{1}{2}\int_{t-l+x}^{t}d\tau
\int_{\tau-t+x}^{t-\tau+x}\frac{\partial^{2}u(\xi,\tau)}{\partial\xi^{2}}d\xi\\
&=\frac{1}{2}\int_0^{x}[\frac{\partial
u(\xi,t+\xi-x)}{\partial\tau}- \frac{\partial
u(\xi,0)}{\partial\tau}]d \xi\\
&\quad +\frac{1}{2}\int_{x}^{l}[\frac{\partial
u(\xi,t-\xi+x)}{\partial\tau}- \frac{\partial u(\xi,0)}{\partial\tau}]d \xi
-\frac{1}{2}\int_0^{t-x}[\frac{\partial
u(l,\tau)}{\partial\xi}- \frac{\partial u(0,\tau)}{\partial\xi}]d \tau
\\
&\quad -\frac{1}{2}\int_{t-x}^{t-l+x}[\frac{\partial
u(l,\tau)}{\partial\xi}-\frac{\partial
u(\tau-t+x,\tau)}{\partial\xi}]d
\tau-\frac{1}{2}\int_{t-l+x}^{t}[\frac{\partial
u(t-\tau+x,\tau)}{\partial\xi}\\
&\quad - \frac{\partial u(\tau-t+x,\tau)}{\partial\xi}]d \tau
\\
&=\frac{1}{2}\int_0^{x}\frac{\partial u(\xi,t+\xi-x)}{\partial\tau}d \xi
 +\frac{1}{2}\int_{x}^{l}\frac{\partial u(\xi,t-\xi+x)}{\partial\tau}d \xi
 +\frac{1}{2}\int_0^{t-x}\frac{\partial u(0,\tau)}{\partial\xi}d \tau
\\
&\quad  -\frac{1}{2}\int_0^{t-l+x}\frac{\partial u(l,\tau)}{\partial\xi}d \tau
 +\frac{1}{2}\int_{t-x}^{t}\frac{\partial
u(\tau-t+x,\tau)}{\partial\xi}d \tau\\
&\quad-\frac{1}{2}\int_{t-l+x}^{t}\frac{\partial
u(t-\tau+x,\tau)}{\partial\xi}d \tau, \quad  \forall(x,t)\in \Omega.
\end{align*}
Using the total differential formula, we have
\begin{align*}
u(x,t)
&=\frac{1}{2}\int_0^{t-x}\frac{\partial u(0,\tau)}{\partial\xi}d
\tau-\frac{1}{2}\int_0^{t-l+x}\frac{\partial
u(l,\tau)}{\partial\xi}d
\tau+\frac{1}{2}\int_0^{x}[\frac{\partial
u(\xi,t+\xi-x)}{\partial\tau}
\\
&\quad +\frac{\partial u(\xi,t+\xi-x)}{\partial\xi}]d \xi
+\frac{1}{2}\int_{x}^{l}[\frac{\partial
u(\xi,t-\xi+x)}{\partial\tau}-\frac{\partial
u(\xi,t-\xi+x)}{\partial\xi}]d \xi
\\
&=\frac{1}{2}\int_0^{t-x}\frac{\partial u(0,\tau)}{\partial\xi}d
\tau-\frac{1}{2}\int_0^{t-l+x}\frac{\partial
u(l,\tau)}{\partial\xi}d \tau
\\
&\quad +\frac{1}{2}\int_0^{x}\frac{d u(\xi,t+\xi-x)}{d\xi}d \xi
-\frac{1}{2}\int_{x}^{l}\frac{d u(\xi,t-\xi+x)}{d\xi}d \xi.
\end{align*}
Thus, we obtain the identity
\begin{align*}
u(x,t)
&=u(x,t)+\frac{1}{2}\int_0^{t-x}\frac{\partial u(0,\tau)}{\partial\xi}d \tau
-\frac{1}{2}\int_0^{t-l+x}\frac{\partial u(l,\tau)}{\partial\xi}d \tau\\
&\quad -\frac{[u(0,t-x)+u(l,t-l+x)]}{2}, \forall(x,t)\in \Omega\,;
\end{align*}
i.e.,
\begin{equation}
\begin{aligned}
&\frac{1}{2}\int_0^{t-x}\frac{\partial u(0,\tau)}{\partial\xi}d\tau
-\frac{1}{2}\int_0^{t-l+x}\frac{\partial u(l,\tau)}{\partial\xi}d \tau\\
&-\frac{[u(0,t-x)+u(l,t-l+x)]}{2}=0, \quad \forall(x,t)\in \Omega.
\end{aligned} \label{e6}
\end{equation}

Now we consider the identity \eqref{e6} when $(x,t)\to S$.
Taking the limit  as $x\to 0$, we have
$$
(u_{x}-u_{t})(0,t)=0,\quad  x=0,\; 0<t<T,
$$
and similarly
$$
(u_{x}+u_{t})(l,t)=0,\quad x=l,\, 0<t<T,
$$
as $x\to l$.
Hence, the one-dimensional wave potential \eqref{e1} satisfies to
the lateral boundary conditions \eqref{e4}-\eqref{e5}.

Conversely, if a solution of the equation \eqref{e2} satisfies to the initial 
conditions \eqref{e3} and the lateral boundary conditions \eqref{e4}-\eqref{e5}, 
then it is determined only by the formula \eqref{e1},
in the other words it coincides the one-dimensional wave potential \eqref{e1}.

Indeed, if a function $u_1$ satisfies to the equation \eqref{e2}, 
the initial conditions \eqref{e3} and the lateral boundary 
conditions \eqref{e4}-\eqref{e5}, then
$u_1\equiv u$, where $u$ is the wave
potential \eqref{e1}. If it is not so, then the function
$\vartheta(x,t)=u_1(x,t)-u(x,t)$ satisfies
\begin{gather*}
\vartheta_{tt}(x,t)-\vartheta_{xx}(x,t)=0,\quad (x,t)\in\Omega,\\
\vartheta(x,0)=\vartheta_{t}(x,0)=0,\quad 0<x<1\\
(\vartheta_{x}-\vartheta_{t})(0,t)=0,\quad  x=0,\; 0<t<T,\\
(\vartheta_{x}+\vartheta_{t})(l,t)=0,\quad x=l,\; 0<t<T\,.
\end{gather*}
Since
$$
0=\int_{\Omega}\varepsilon(x-\xi,t-\tau)0d \xi d \tau
= \int_{\Omega}\varepsilon(x-\xi,t-\tau)[\vartheta_{\tau\tau}(\xi,\tau)
-\vartheta_{\xi\xi}(\xi,\tau)]d \xi d \tau,
$$
 a similar calculation as above shows that
\begin{align*}
0&=\int_{\Omega}\varepsilon(x-\xi,t-\tau)[\vartheta_{\tau\tau}(\xi,\tau)-
\vartheta_{\xi\xi}(\xi,\tau)]d\xi d\tau
=\vartheta(x,t)+\frac{1}{2}\int_0^{t-x}\frac{\partial
\vartheta(0,\tau)}{\partial\xi}d
\tau\\
&\quad -\frac{1}{2}\int_0^{t-l+x}\frac{\partial
\vartheta(l,\tau)}{\partial\xi}d \tau
-\frac{[\vartheta(0,t-x)+\vartheta(l,t-l+x)]}{2}, \quad\forall(x,t)\in \Omega.
\end{align*}
Denoting,
$$
I_{\vartheta}(x,t):=\frac{1}{2}\int_0^{t-x}\frac{\partial
\vartheta(0,\tau)}{\partial\xi}d
\tau-\frac{1}{2}\int_0^{t-l+x}\frac{\partial
\vartheta(l,\tau)}{\partial\xi}d \tau
-\frac{[\vartheta(0,t-x)+\vartheta(l,t-l+x)]}{2}
$$
And when $(x,t)\to S$, we obtain
$$
\vartheta(x,t)|_{x=0}=-I_{\vartheta}(x,t)|_{x=0}=0,\quad
 \vartheta(x,t)|_{x=l}=-I_{\vartheta}(x,t)|_{x=l}=0.
$$
Thus, the function $\vartheta(x,t)$ satisfies
\begin{equation}
\begin{gathered}
\vartheta_{tt}(x,t)-\vartheta_{xx}(x,t)=0,\quad (x,t)\in\Omega,\\
\vartheta(x,t)=\vartheta_{t}(x,t)=0, \quad 0<x<l,\\
\vartheta(0,t)=0,\quad 
\vartheta(l,t)=0.
\end{gathered}\label{e7}
\end{equation}
Let us define the function
$$
E(t)=\int_0^{l}[(\vartheta_{t}(x,t))^{2}+(\vartheta_{x}(x,t))^{2}]dx
$$
which we call an energy integral. From the physical point of
view, it is a total energy up to a constant, for instance, energy of
oscillating string.

It is obvious that the function $E(t)$  is differentiable because
of our conditions on the function  $\vartheta(x,t)$. Consequently,
its derivative is calculated as
$$
E'(t)=\int_0^{l}[2\vartheta_{t}(x,t)\vartheta_{tt}(x,t)
+2\vartheta_{x}(x,t)\vartheta_{xt}(x,t)]dx
$$
Integrating by parts, one writes the second term in the
form
$$
E'(t)=\int_0^{l}[2\vartheta_{t}(x,t)\{\vartheta_{tt}(x,t)-
\vartheta_{xx}(x,t)\}]dx+
2\vartheta_{x}(x,t)\vartheta_{t}(x,t)|_0^{l}.
$$
Note that the integrand function is equal to zero identically
since $\vartheta(x,t)$ is a solution of the homogeneous wave equation. 
Then by differentiating with respect to $t$  boundary conditions, we
have  $\vartheta_{t}(0,t)\equiv0\equiv\vartheta_{t}(l,t)$.
 It follows that the non-integral term also vanishes. So,
$E'(t)\equiv0$, or
$$
E(t)=\int_0^{l}[(\vartheta_{t}(x,t))^{2}+(\vartheta_{x}(x,t))^{2}]dx\equiv 
{\rm const.}
$$
Actually, we just obtain another law of energy conservation in
a closed system which is described by the initial-boundary value 
problem \eqref{e7} where amount
of energy is permanent. Obviously,
$$
E(t)=E(0)=\int_0^{l}[(\vartheta_{t}(x,0))^{2}+(\vartheta_{x}(x,0))^{2}]dx
$$
From the initial conditions we obtain
$\vartheta_{t}(x,0)=\vartheta_{x}(x,0)=0, 0\leq x\leq l$.
Hence
$$
E(0)=0\;\Rightarrow\; E(t)\equiv 0\,.
$$
Because the integrand functions are nonnegative we have
$\vartheta_{t}(x,t)=\vartheta_{x}(x,t)=0$. It follows that
$\vartheta(x,t)=$const, and from the initial conditions it follows
that $\vartheta(x,t)\equiv0$; i.e., $u_1\equiv u$, $u_1$ coincides 
with the wave potential potential \eqref{e1}.

To summarize, the initial-boundary value problem \eqref{e2}-\eqref{e5} 
has an unique solution and the solution coincides with the wave potential \eqref{e1}.
This completes the proof.
\end{proof}


\subsection*{Acknowledgements}
Authors are grateful to Professor M. Sadybekov for his useful suggestions, 
which made the paper more readable.


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\end{document}