\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 277, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/277\hfil Nonlinear singular hyperbolic problems]
{Nonlinear singular hyperbolic initial boundary value problems}

\author[S. Mesloub, I. Bachar \hfil EJDE-2017/277\hfilneg]
{Said Mesloub, Imed Bachar}

\address{Said Mesloub \newline
King Saud University,
College of sciences Department of Mathematics,
 P.O. Box 2455, Riyadh 11451, Saudi Arabia}
\email{mesloubs@yahoo.com}

\address{Imed Bachar \newline
King Saud University,
College of sciences Department of Mathematics,
 P.O. Box 2455, Riyadh 11451, Saudi Arabia}
\email{abachar@ksu.edu.sa}

\dedicatory{Communicated by Mokhtar Kirane}

\thanks{Submitted May 10 2017. Published November 8, 2017.}
\subjclass[2010]{35L10, 35L20, 35L70}
\keywords{Fixed point theorem; a priori bound; existence and uniqueness;
\hfill\break\indent integro-differential equation}

\begin{abstract}
In this article, we aim to prove the well posedness of a one point
initial boundary value problem for a nonlinear hyperbolic singular
integro-differential partial differential equation. Our proofs are
mainly based on a fixed point theorem and some a priori bounds.
The solvability of the problem is proved when the given data are
related to a weighted Sobolev space. An additional result may allow us
to study the regularity of the solution.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks


\section{Introduction}

A big number of physical phenomena can be modeled and formulated by some
nice partial differential equations. It happens sometimes that such
phenomena cannot be modeled by classical partial differential equations
while describing the system as a function at a given time; it fails to take
into account the effect of past history, such as in thermo-elasticity and
heat diffusion. In such situations, a memory term on the form of an integral
should be included the equation. This integral term can be considered as a
damping term in the equation. A range of elliptic integro-differential
equations were studied in the works of numerous researchers, including for
example Bokalo and Dmytriv \cite{b6}, Barles and Imbert \cite{b3},
Caffarelli and Silvestre \cite{c1}, Chipot and Guesmia \cite{c3},
Balachandran and Park \cite{b2}, Bakalo
and Dmytriv \cite{b6}. Integro-differential equations of hyperbolic and parabolic
type are studied by many authors see for example Adolfsson \cite{a1},
Baker \cite{b1}, Sloan and Thomee \cite{s2}, Berrimi and Messaoudi
\cite{b4}, Cannarsa and Sforza \cite{c2}
and Weiking \cite{x1}. In the present work, we consider a hyperbolic
integro-differential equation with an unknown function that enters both into
the differential part and into the integral part of the equation. It can
occur in visco-elasticity see Renardi and Hrusa \cite{r1,r2} and other areas
as fluid dynamics, biological  models, and chemical kinetics.
Well-posedness of problem \eqref{e1.1}--\eqref{e1.3} is studied, in fact, we prove the
existence and uniqueness of solutions for a posed nonlinear hyperbolic
integro-differential equation with a Neumann and Dirichlet conditions.
The used tool is a variant of fixed point theorems.

In the rectangle $Q_{T}=(0,1)\times (0,T)$, where $0<T<\infty $, we consider
the nonlinear singular second order hyperbolic integro-differential equation
\begin{equation}
\begin{aligned}
\mathcal{L}v
&=\frac{\partial ^2v}{\partial t^2}-\frac{1}{x}
\frac{\partial }{\partial x}( x\frac{\partial v}{\partial x})
-\frac{d}{dt}\max \Big( \int_0^x\xi v(\xi ,t)d\xi ,0\Big)
\label{e1.1} \\
&=f(x,t),
\end{aligned}
\end{equation}
With \eqref{e1.1}, we associate the following initial conditions
\begin{equation}
v(x,0) =Z_1(x),\quad v_{t}(x,0)=Z_2(x),\quad x\in (0,1),
\label{e1.2}
\end{equation}
and the boundary conditions
\begin{equation}
v_{x}(1,t) =0,\quad v(1,t)=0,\quad t\in (0,T),  \label{e1.3}
\end{equation}
where $Z_1(x)$, $Z_2(x)$ and $f(x,t)$ are given functions which will be
specified later on.

In Section 2, we shall give some function spaces and tools
which will be used repeatedly below.
Section 3 is devoted to the proof of
uniqueness of solution of the given problem in the classical Sobolev space
$W_{\gamma ,2}^{1,1}(Q_{T})$. In section 4, we established the existence of
solution and the proof was mainly based on the Schauder fixed point theorem.
In the last section a priori bound is obtained and can help to establish
some regularity results for the solution of problem \eqref{e1.1}--\eqref{e1.3}

\section{Preliminaries}


 For the investigation of problem \eqref{e1.1}--\eqref{e1.3}, we need the following
functions spaces: Let $L_{\gamma }^2(Q_{T})$, $L_{\sigma }^2(Q_{T})$,
$L_{\rho }^2(Q_{T})$ be the weighted Hilbert spaces of square integrable
functions on $Q_{T}$ with $\gamma =x+x^2$, $\sigma =x^2$, and $\rho =x$.
The inner products in $L_{\gamma }^2(Q_{T})$, $L_{\sigma }^2(Q_{T})$,
$L_{\rho }^2(Q_{T})$ are respectively denoted by
 $(\cdot,\cdot)_{L_{\gamma}^2(Q_{T})}$, $(\cdot,\cdot)_{L_{\sigma }^2(Q_{T})}$,
$(\cdot,\cdot)_{L_{\rho}^2(Q_{T})}$ such that
\begin{equation}
(u,v)_{L_{A}^2(Q_{T})} =\int_{0}^{1}A(x)uvdx,\quad
A(x)=\gamma,\sigma ,\rho .  \label{e2.1}
\end{equation}
Let $L^2( 0,T;\text{ }W_{\gamma ,2}^{1,1}(0,1)) $ be
the space of (classes) of function $u$ measurable on $\left[ 0,T\right] $
for the Lebesgue measure and having their values in $W_{\gamma
,2}^{1,1}(0,1),$and such that
\begin{equation*}
\| u\| _{L^2( 0,T;W_{\gamma ,2}^{1,1}(0,1))
}=\Big( \int_{0}^{T}\| u( \cdot ,t) \|
_{W_{\gamma ,2}^{1,1}(0,1)}^2dt\Big) ^{1/2}<\infty
\end{equation*}
The space $L^2( 0,T;\text{ }W_{\gamma ,2}^{1,1}(0,1)) $
is a Hilbert space having the inner product
\begin{equation}
(u,v)_{L^2( 0,T;\text{ }W_{\gamma ,2}^{1,1}(0,1)) }
=\int_0^T(u,v)_{W_{\gamma ,2}^{1,1}(0,1)}dt, \label{e2.2}
\end{equation}
where
\begin{equation}
(u,v)_{W_{\gamma ,2}^{1,1}(0,1)}
=\int_{0}^{1}(x+x^2)(uv+u_{x}v_{x}+u_{t}v_{t})dx.  \label{e2.3}
\end{equation}
The space $W_{\gamma ,2}^{1,1}(Q_{T})$, is the set of functions
$u\in L_{\gamma }^2(Q_{T})$ such that $u_{x},u_{t}$ $\in L_{\gamma }^2(Q_{T})$.
In general the elements of $W_{\gamma ,2}^{m,n}(Q_{T})$, with $m,n$
nonnegative integers are functions having $x-$derivatives up to $m^{th}$
order in $L_{\gamma }^2(Q_{T})$, and $t-$derivative up to $n^{th}$ order
in $L_{\gamma }^2(Q_{T})$. We also use weighted spaces in $(0,1)$ such as:
$L_{\gamma }^2(0,1)$, $L_{\sigma }^2(0,1)$, $L_{\rho }^2(0,1)$, $
W_{\gamma ,2}^{1,1}(0,1)$ and $W_{\gamma ,2}^{1}(0,1)$.

The following inequalities are needed:

\noindent(1) Cauchy $\varepsilon$-inequality which holds for all $\varepsilon >0$
and for arbitrary $A$ and $B$.
\begin{equation}
AB \leq \frac{\varepsilon }{2}A^2+\frac{1}{2\varepsilon }B^2,
\label{e2.4}
\end{equation}

\noindent(2) A Poincar\'{e} type inequality (see \cite{m1}).
\begin{equation}
\begin{gathered}
\| \Lambda _{x}(\xi u)\| _{L^2( 0,1) }^2\leq
\frac{1}{2}\| u\| _{L_{\rho }^2( Q_{T}) }^2,
\\
\| \Lambda _{x}^2(\xi u)\| _{L^2( 0,1)
}^2\leq \frac{1}{2}\| \Lambda _{x}(\xi u)\| _{L^2(
Q_{T}) }^2,
\end{gathered}   \label{e2.5}
\end{equation}
where
\begin{equation*}
\Lambda _{x}(\xi u)=\int_{0}^{x}\xi u(\xi ,t)d\xi ,\text{ \ }\Lambda
_{x}^2(\xi u)=\int_{0}^{x}\int_{0}^{\xi }\eta u(\eta
,t)d\eta d\xi .
\end{equation*}

\noindent(3) Gronwall's inequality (see \cite[Lemma 4.1]{m1}).

\section{Uniqueness of solution}

\begin{theorem} \label{thm1}
Let $Z_1(x)\in W_{\gamma ,2}^{1}((0,1))$, $Z_2(x)$
$\in L_{\gamma }^2((0,1)$ and $\ f(x,t)\in L_{\gamma }^2(Q_{T})$. Then
the posed problem \eqref{e1.1}--\eqref{e1.3} has at most one solution in
$W_{\gamma,2}^{1,1}(Q_{T})$, if it exists.
\end{theorem}

\begin{proof}
 Let $v_1$ and $v_2$ be two solutions of problem \eqref{e1.1}--\eqref{e1.3} and
let $\eta (x,t)=V_1(x,t)-V_2(x,t)$, where
\begin{equation}
V_{j}(x,t) =\int_{0}^{t}v_{j}(x,\tau )d\tau ,\quad j=1,2,
\label{e3.1}
\end{equation}
then the function $\eta (x,t)$ satisfies
\begin{gather}
\mathcal{L}\eta =\frac{\partial ^2\eta }{\partial t^2}-\frac{1}{x}
\frac{\partial }{\partial x}\big(x\frac{\partial \eta }{\partial x}\big)
=\gamma _1(x,t)-\gamma _2(x,t),  \label{e3.2} \\
\eta _{x}(1,t) =0,\quad \eta (1,t)=0,\quad t\in (0,T),  \label{e3.3}\\
\eta (x,0) =0,\quad \eta _{t}(x,0)=0,\quad x\in (0,1),  \label{e3.4}
\end{gather}
where
\begin{equation}
\gamma _{j}(x,t) =\max ( \Lambda _{x}(\xi v_{j}) ,0),\quad j=1,2. \label{e3.5}
\end{equation}
Consider the identity
\begin{equation}
\begin{aligned}
&\Big( \frac{\partial \eta }{\partial t},\frac{\partial ^2\eta }{
\partial t^2}-\frac{1}{x}\frac{\partial }{\partial x}( x\frac{
\partial \eta }{\partial x}) \Big) _{L_{\rho }^2(Q_{T})}
+2\Big( \frac{\partial \eta }{\partial t},\frac{\partial ^2\eta }{\partial t^2}-
\frac{1}{x}\frac{\partial }{\partial x}( x\frac{\partial \eta }{
\partial x}) \Big) _{L_{\sigma }^2(Q_{T})}   \\
&\quad +\Big( \frac{\partial \eta }{\partial x},\frac{\partial ^2\eta }{
\partial t^2}-\frac{1}{x}\frac{\partial }{\partial x}( x\frac{
\partial \eta }{\partial x}) \Big) _{L_{\sigma }^2(Q_{T})}
\\
&=\Big( \frac{\partial \eta }{\partial t},\gamma _1-\gamma _2\Big)
_{L_{\rho }^2(Q_{T})}+2\Big( \frac{\partial \eta }{\partial t},\gamma
_1-\gamma _2\Big) _{L_{\sigma }^2(Q_{T})}
+\Big( \frac{\partial \eta  }{\partial x},
 \gamma _1-\gamma _2\Big) _{L_{\sigma }^2(Q_{T})}.
\end{aligned} \label{e3.6}
\end{equation}
Integrating by parts, and taking into account conditions \eqref{e3.3} and \eqref{e3.4},
we deduce the following expressions for the terms on the left-hand side
of \eqref{e3.6}
\begin{gather}
\Big( \frac{\partial \eta }{\partial t},\frac{\partial ^2\eta }{\partial
t^2}\Big) _{L_{\rho }^2(Q_{T})}
=\frac{1}{2}\| \eta_{t}(x,T)\| _{L_{\rho }^2((0,1))}^2,  \label{e3.7}
\\
-\Big( \frac{\partial \eta }{\partial t},\frac{1}{x}\frac{\partial }{
\partial x}\big( x\frac{\partial \eta }{\partial x}\big) \Big)
_{L_{\rho }^2(Q_{T})}
=\frac{1}{2}\| \eta _{x}(x,T)\|
_{L_{\rho }^2((0,1))}^2,  \label{e3.8}
\\
2\Big( \frac{\partial \eta }{\partial t},\frac{\partial ^2\eta }{\partial
t^2}\Big) _{L_{\sigma }^2(Q_{T})}
=\| \eta _{t}(x,T)\| _{L_{\sigma }^2((0,1))}^2,  \label{e3.9}
\\
-2\Big( \frac{\partial \eta }{\partial t},\frac{1}{x}\frac{\partial }{
\partial x}\big( x\frac{\partial \eta }{\partial x}\big) \Big)
_{L_{\sigma }^2(Q_{T})}
=\| \eta _{x}(x,T)\|_{L_{\sigma }^2((0,1))}^2
+2( \frac{\partial \eta }{\partial t},
\frac{\partial \eta }{\partial x}) _{L_{\rho }^2(Q_{T})},
\label{e3.10}
\\
\Big( \frac{\partial \eta }{\partial x},\frac{\partial ^2\eta }{\partial
t^2}\Big) _{L_{\sigma }^2(Q_{T})}
=\| \eta _{t}\|
_{L_{\rho }^2(Q_{T})}^2+\int_0^1 x^2\eta
_{x}(x,T)\eta _{t}(x,T)dx,  \label{e3.11}
\\
-\Big( \frac{\partial \eta }{\partial x},\frac{1}{x}\frac{\partial }{
\partial x}\big( x\frac{\partial \eta }{\partial x}\big) \Big)
_{L_{\sigma }^2(Q_{T})} =0.  \label{e3.12}
\end{gather}
Substituting formulas \eqref{e3.7}--\eqref{e3.12} into \eqref{e3.6}, we obtain
\begin{equation}
\begin{aligned}
&\frac{1}{2}\| \eta _{t}(x,T)\| _{L_{\rho
}^2((0,1))}^2+\frac{1}{2}\| \eta _{x}(x,T)\| _{L_{\rho
}^2((0,1))}^2+\| \eta _{t}(x,T)\| _{L_{\sigma
}^2((0,1))}^2   \\
&+\| \eta _{x}(x,T)\| _{L_{\sigma
}^2((0,1))}^2+\| \eta _{t}\| _{L_{\rho }^2(Q_{T})}^2
 \\
&=\Big( \frac{\partial \eta }{\partial t},\gamma _1-\gamma _2\Big)
_{L_{\rho }^2(Q_{T})}
+2\Big( \frac{\partial \eta }{\partial t},\gamma
_1-\gamma _2\Big) _{L_{\sigma }^2(Q_{T})}
+\Big( \frac{\partial \eta
}{\partial x},\gamma _1-\gamma _2\Big) _{L_{\sigma }^2(Q_{T})}
 \\
&\quad -\int_0^1 x^2\eta _{x}(x,T)\eta
_{t}(x,T)dx-2( \frac{\partial \eta }{\partial t},
\frac{\partial \eta }{\partial x}) _{L_{\rho }^2(Q_{T})}.
\end{aligned}\label{e3.13}
\end{equation}
By using Cauchy $\varepsilon -$inequality and the Poincar\'{e} type
inequality \eqref{e2.5}, we estimate the terms on the right-hand side of \eqref{e3.13} as
follows
\begin{gather}
\begin{aligned}
&-\int_0^1 x^2\eta _{x}(x,T)\eta _{t}(x,T)dx \\
&\leq \frac{\varepsilon _1}{2}\| \eta _{t}(x,T)\| _{L_{\sigma}^2((0,1))}^2
 +\frac{1}{2\varepsilon _1}\| \eta _{x}(x,T)\|
_{L_{\sigma }^2((0,1))}^2,
\end{aligned}  \label{e3.14}
\\
-2\Big( \frac{\partial \eta }{\partial t},\frac{\partial \eta }{\partial x}
\Big) _{L_{\rho }^2(Q_{T})}
\leq \varepsilon _2\| \eta _{x}\| _{L_{\rho }^2(Q_{T})}^2
+\frac{1}{\varepsilon _2}\| \eta _{t}\| _{L_{\rho }^2(Q_{T})}^2,  \label{e3.15}
\\
\Big( \frac{\partial \eta }{\partial t},\gamma _1-\gamma _2\Big)
_{L_{\rho }^2(Q_{T})}
\leq \frac{\varepsilon _{3}}{2}\|
\eta_{t}\| _{L_{\rho }^2(Q_{T})}^2+\frac{1}{4\varepsilon _{3}}
\| \eta _{t}\| _{L_{\rho }^2(Q_{T})}^2,  \label{e3.16}
\\
2\Big( \frac{\partial \eta }{\partial t},\gamma _1-\gamma _2\Big)
_{L_{\sigma }^2(Q_{T})}
\leq \varepsilon _{4}\| \eta_{t}\| _{L_{\sigma }^2(Q_{T})}^2
+\frac{1}{\varepsilon _{4}}\| \eta _{t}\| _{L_{\sigma }^2(Q_{T})}^2,  \label{e3.17}
\\
\Big( \frac{\partial \eta }{\partial x},\gamma _1-\gamma _2\Big)
_{L_{\sigma }^2(Q_{T})}
\leq \frac{\varepsilon _{5}}{2}\| \eta
_{t}\| _{L_{\sigma }^2(Q_{T})}^2+\frac{1}{2\varepsilon _{5}}
\| \eta _{x}\| _{L_{\sigma }^2(Q_{T})}^2.  \label{e3.18}
\end{gather}
Let $\varepsilon _1=1$, $\varepsilon _2=8$, $\varepsilon _{3}=1$,
$\varepsilon _{4}=1$, $\varepsilon _{5}=1$, and combine
\eqref{e3.14}--\eqref{e3.18} and \eqref{e3.13}, we obtain
\begin{equation}
\begin{aligned}
&\| \eta _{t}(x,T)\| _{L_{\rho}^2((0,1))}^2
 +\| \eta _{x}(x,T)\| _{L_{\rho}^2((0,1))}^2
 +\| \eta _{t}(x,T)\| _{L_{\sigma}^2((0,1))}^2   \\
&+\| \eta _{x}(x,T)\| _{L_{\sigma
}^2((0,1))}^2+\| \eta _{t}\| _{L_{\rho }^2(Q_{T})}^2
 \\
&\leq 64\Big( \| \eta _{t}\| _{L_{\rho
}^2(Q_{T})}^2+\| \eta _{x}\| _{L_{\rho
}^2(Q_{T})}^2+\| \eta _{t}\| _{L_{\sigma
}^2(Q_{T})}^2+\| \eta _{x}\| _{L_{\sigma
}^2(Q_{T})}^2\Big) .
\end{aligned} \label{e3.19}
\end{equation}
Application of Gronwall's lemma \cite[Lemma 4.1]{m1} to \eqref{e3.19}, yields
\begin{equation*}
\| \eta _{t}\| _{L_{\rho }^2(Q_{T})}^2=0\Rightarrow
\eta _{t}=v_1-v_2=0\quad \text{for all }(x,t)\in Q_{T},
\end{equation*}
hence, we conclude the uniqueness of the solution.
\end{proof}

\section{Existence of the solution}


\begin{theorem} \label{thm2}
 Let $Z_1(x)\in W_{\gamma ,2}^{1}((0,1))$,
$Z_2(x)\in L_{\gamma }^2((0,1)$ and
$f(x,t)\in L_{\gamma }^2(Q_{T})$ be given
and satisfy
\begin{equation}
\| Z_1\| _{W_{\gamma ,2}^{1}((0,1))}^2+\|
Z_2\| _{L_{\gamma }^2((0,1))}^2+\| f\|
_{L_{\gamma }^2(Q_{T})}^2 \leq C_2^2,  \label{e4.1}
\end{equation}
for $C_2>0$ small enough, and that
\begin{equation}
\frac{\partial Z_1(1)}{\partial x} =0,\quad
Z_1(1)=0,\quad \frac{\partial Z_2(1)}{\partial x}=0,\quad
Z_2(1)=0,  \label{e4.2}
\end{equation}
Then problem \eqref{e1.1}--\eqref{e1.3} has a unique solution
$v\in W_{\gamma,2}^{1,1}(Q_{T})$.
\end{theorem}

\begin{proof} Consider the class of functions
\begin{equation}
K(A) =\big\{ v\in L_{\gamma }^2(Q_{T}):\| v\|
_{L^2(0,T;W_{2,\gamma }^{1,1}(0,1))}\leq A,\;
\| v_{t}\| _{L_{\gamma }^2(Q_{T})}\leq A\},
  \label{e4.3}
\end{equation}
satisfying conditions \eqref{e1.2} and \eqref{e1.3}, where $A$ is a positive constant
which will be specified later on. 
For any $u\in K(A)$,  problem
\begin{gather}
\frac{\partial ^2v}{\partial t^2}-\frac{1}{x}\frac{\partial }{\partial x}
\big( x\frac{\partial v}{\partial x}\big)
=\mathcal{I}u+f(x,t),  \label{e4.4} \\
v(x,0) =Z_1(x),\quad v_{t}(x,0)=Z_2(x),\quad x\in (0,1) \label{e4.5} \\
v_{x}(1,t) =0,\quad v(1,t)=0,\quad t\in (0,T),  \label{e4.6}
\end{gather}
where
\begin{equation}
\mathcal{I}u =\frac{d}{dt}\max ( \Lambda _{x}(\xi u),0) ,
\label{e4.7}
\end{equation}
has a unique solution in $W_{\gamma ,2}^{1,1}(Q_{T})$.

Define a mapping $\mathcal{M}$ such that $v=\mathcal{M}u$. If we show that
the mapping $\mathcal{M}$ admits a fixed point $v$ in the closed bounded
convex subset $K(A)$, then $v$ will be our solution. Let us first show that
the mapping $\mathcal{M}$ maps the set $K(A)$ to $K(A)$. Let $v=W+S$, such
that $W$ solves
\begin{gather}
\frac{\partial ^2W}{\partial t^2}-\frac{1}{x}\frac{\partial }{\partial x}
\big( x\frac{\partial W}{\partial x}\big) =\mathcal{I}u,\quad (x,t)\in Q_{T},
\label{e4.8} \\
W(x,0) =0,\quad W_{t}(x,0)=0,\quad x\in (0,1),  \label{e4.9} \\
W_{x}(1,t) =0,\quad W(1,t)=0,\quad t\in (0,T),  \label{e4.10}
\end{gather}
and $S$ solves
\begin{gather}
\frac{\partial ^2S}{\partial t^2}-\frac{1}{x}\frac{\partial }{\partial x}
\big( x\frac{\partial S}{\partial x}\big) =f(x,t),\quad (x,t)\in
Q_{T},  \label{e4.11} \\
S(x,0) =Z_1(x),\quad S_{t}(x,0)=Z_2(x),\quad x\in (0,1)  \label{e4.12} \\
S_{x}(1,t) =0,\quad S(1,t)=0,\quad t\in (0,T)  \label{e4.13} 
\end{gather}
We now consider the following inner products in $L^2(Q_{y})$,
$y\in (0,T)$
\begin{gather}
( \mathcal{L}W,M_1W) _{L^2(Q_{y})}
=( \mathcal{I}u,M_1W) _{L^2(Q_{y})},  \label{e4.14} \\
( \mathcal{L}S,M_2S) _{L^2(Q_{y})}
=( f,M_2S)_{L^2(Q_{y})},  \label{e4.15}
\end{gather}
where
\begin{equation*}
M_1W=2x^2W_{t}+x^2W_{x}+2xW_{t}, \quad
M_2S=2x^2S_{t}+x^2S_{x}+2xS_{t}.
\end{equation*}
By integrating by parts in \eqref{e4.14} and evoking conditions \eqref{e4.9} and \eqref{e4.10},
we have
\begin{equation}
\begin{aligned}
&\| W_{t}(x,y)\| _{L_{\sigma }^2((0,1))}^2+\|
W_{x}(x,y)\| _{L_{\sigma }^2((0,1))}^2+\|
W_{t}(x,y)\| _{L_{\rho }^2((0,1))}^2   \\
&+\| W_{x}(x,y)\| _{L_{\rho }^2((0,1))}^2
+\|W_{t}\| _{L_{\rho }^2(Q_{y})}^2   \\
&=-\int_0^1 x^2W_{x}(x,y)W_{t}(x,y)dx-2(
W_{t},W_{x}) _{L_{\rho }^2(Q_{y})}+2( W_{t},\mathcal{I}u)
_{L_{\sigma }^2(Q_{y})}   \\
&\quad +( W_{x},\mathcal{I}u) _{L_{\sigma }^2(Q_{y})}+2( W_{t},
\mathcal{I}u) _{L_{\rho }^2(Q_{y})}.
\end{aligned} \label{e4.16}
\end{equation}
Thanks to Cauchy $\varepsilon$-inequality, \eqref{e4.16} reduces to
\begin{equation}
\begin{aligned}
&\| W_{t}(x,y)\| _{L_{\sigma }^2((0,1))}^2+\|
W_{x}(x,y)\| _{L_{\sigma }^2((0,1))}^2+\frac{1}{2}\|
W_{t}(x,y)\| _{L_{\rho }^2((0,1))}^2   \\
&+\frac{1}{2}\| W_{x}(x,y)\| _{L_{\rho }^2((0,1))}^2
 \\
&\leq \| W_{t}\| _{L_{\rho }^2(Q_{y})}^2+\|
W_{x}\| _{L_{\rho }^2(Q_{y})}^2+\| W_{t}\|
_{L_{\sigma }^2(Q_{y})}^2+\frac{1}{2}\| W_{x}\|
_{L_{\sigma }^2(Q_{y})}^2   \\
&\quad +\| \mathcal{I}u\| _{L_{\rho }^2(Q_{y})}^2+\frac{3}{2}
\| \mathcal{I}u\| _{L_{\sigma }^2(Q_{y})}^2.
\end{aligned} \label{e4.17}
\end{equation}
Consider the following two elementary inequalities
\begin{gather}
\| W(x,y)\| _{L_{\sigma }^2((0,1))}^2
\leq \|W\| _{L_{\sigma }^2(Q_{y})}^2+\| W_{t}\|
_{L_{\sigma }^2(Q_{y})}^2,  \label{e4.18} \\
\| W(x,y)\| _{L_{\rho }^2((0,1))}^2
\leq \|W\| _{L_{\rho }^2(Q_{y})}^2+\| W_{t}\|
_{L_{\rho }^2(Q_{y})}^2.  \label{e4.19}
\end{gather}
Inequalities \eqref{e4.17}--\eqref{e4.19}, yield
\begin{equation}
\begin{aligned}
&\| W(x,y)\| _{L_{\gamma }^2((0,1))}^2+\|
W_{t}(x,y)\| _{L_{\gamma }^2((0,1))}^2+\|
W_{x}(x,y)\| _{L_{\gamma }^2((0,1))}^2   \\
&\leq 3\Big( \| W\| _{L_{\gamma}^2(Q_{y})}^2
 +\| W_{t}\| _{L_{\gamma}^2(Q_{y})}^2
 +\| W_{x}\| _{L_{\gamma}^2(Q_{y})}^2
 +\| \mathcal{I}u\| _{L_{\gamma}^2(Q_{y})}^2\Big) .
\end{aligned}  \label{e4.20}
\end{equation}
Application of Gronwall's lemma to \eqref{e4.20} gives
\begin{equation}
\begin{aligned}
&\| W(x,y)\| _{L_{\gamma }^2((0,1))}^2+\|
W_{t}(x,y)\| _{L_{\gamma }^2((0,1))}^2+\|
W_{x}(x,y)\| _{L_{\gamma }^2((0,1))}^2   \\
&\leq  3e^{3T}\| \mathcal{I}u\| _{L_{\gamma}^2(Q_{T})}^2.
\end{aligned} \label{e4.21}
\end{equation}
Integrating of both sides of \eqref{e4.21} over $(0,T)$ gives
\begin{equation}
\begin{aligned}
&\| W(x,t)\| _{L^2(0,T;W_{2,\gamma }^{1,1}(0,1))}^2
&\leq 3Te^{3T}\| \mathcal{I}u\| _{L_{\gamma}^2(Q_{T})}^2.
\end{aligned} \label{e4.22}
\end{equation}
We now consider \eqref{e4.15} and use conditions \eqref{e4.12}
and \eqref{e4.13}, to obtain
\begin{equation}
\begin{aligned}
&\| S_{t}(x,y)\| _{L_{\sigma }^2((0,1))}^2
 +\|S_{x}(x,y)\| _{L_{\sigma }^2((0,1))}^2
 +\|S_{t}(x,y)\| _{L_{\rho }^2((0,1))}^2   \\
&+\|S_{t}(x,y)\| _{L_{\rho }^2((0,1))}^2
 +\|S_{t}\| _{L_{\rho }^2(Q_{y})}^2   \\
&=-\int_0^1 x^2S_{x}(x,y)S_{t}(x,y)dx
 +2(S_{t},S_{x}) _{L_{\rho }^2(Q_{y})}+2( S_{t},f)_{L_{\sigma }^2(Q_{y})}   \\
&\quad +2( S_{x},f) _{L_{\sigma }^2(Q_{y})}+2( S_{t},f)
_{L_{\rho }^2(Q_{y})}+\| \frac{\partial Z_1}{\partial x}
\| _{L_{\gamma }^2((0,1))}^2   \\
&\quad +\| (Z_2)^2\| _{L_{\gamma }^2((0,1))}^2
+\int_0^1 x^2\frac{\partial Z_1}{\partial x}Z_2dx.
\end{aligned} \label{e4.23}
\end{equation}
By using the Cauchy $\varepsilon$-inequality, we can transform \eqref{e4.23} into
\begin{equation}
\begin{aligned}
&\| S_{t}(x,y)\| _{L_{\gamma }^2((0,1))}^2
+\|S_{x}(x,y)\| _{L_{\gamma }^2((0,1))}^2   \\
&\leq  4\Big( \| S_{t}\| _{L_{\gamma
}^2(Q_{y})}^2+\| S_{x}\| _{L_{\gamma
}^2(Q_{y})}^2\Big)   \\
&\quad +4\Big( \| (Z_1)_{x}^2\| _{L_{\gamma
}^2((0,1))}^2+\| Z_2\| _{L_{\gamma
}^2((0,1))}^2+\| f\| _{L_{\gamma
}^2(Q_{y})}^2\Big) .
\end{aligned}  \label{e4.24}
\end{equation}
Consider the elementary inequality
\begin{equation}
\| S(x,y)\| _{L_{\gamma }^2((0,1))}^2
\leq \|S\| _{L_{\gamma }^2(Q_{y})}^2+\| S_{t}\|
_{L_{\gamma }^2(Q_{y})}^2+\| Z_1\| _{L_{\gamma
}^2((0,1))}^2.
 \label{e4.25}
\end{equation}
Combination of \eqref{e4.24} and \eqref{e4.25} leads to
\begin{equation}
\begin{aligned}
&\| S(x,y)\| _{W_{2,\gamma }^{1,1}(0,1)}^2 \\
&\leq 4\Big(
\| S\| _{W_{2,\gamma }^{1,1}(Q_{y})}^2+\|
Z_1\| _{W_{2,\gamma }^{1}((0,1))}^2+\| Z_2\|
_{L_{\gamma }^2((0,1))}^2+\| f\| _{L_{\gamma}^2(Q_{y})}^2\Big).
\end{aligned}  \label{e4.26}
\end{equation}
Applying of Gronwall's lemma to inequality \eqref{e4.26} and then integrating
over $(0,T)$ gives
\begin{equation}
\begin{aligned}
&\| S(x,y)\| _{L^2(0,T;W_{2,\gamma }^{1,1}(0,1))}^2\\
&\leq 4Te^{4T}\Big( \| Z_1\| _{W_{2,\gamma
}^{1}((0,1))}^2+\| Z_2\| _{L_{\gamma
}^2((0,1))}^2+\| f\| _{L_{\gamma
}^2(Q_{y})}^2\Big) .
\end{aligned}  \label{e4.27}
\end{equation}
It is obvious that
\begin{equation}
\| \mathcal{I}u\| _{L_{\gamma }^2(Q_{T})}^2
\leq C_1^2,     \label{e4.28}
\end{equation}
where $C_1>0$.

Inequalities \eqref{e4.22}, \eqref{e4.27} and \eqref{e4.28} yield
\begin{equation}
\begin{aligned}
\| v\| _{L^2(0,T;W_{2,\gamma }^{1,1}(0,1))}^2
&\leq 2\| W\| _{L^2(0,T;W_{2,\gamma
}^{1,1}(0,1))}^2+2\| S\| _{L^2(0,T;W_{2,\gamma
}^{1,1}(0,1))}^2   \\
&\leq 6Te^{3T}C_1^2+8Te^{4T}C_2^2.
\end{aligned} \label{e4.29}
\end{equation}
and
\begin{equation}
\| v_{t}\| _{L_{\gamma }^2(Q_{T})}^2
\leq 2\|W_{t}\| _{L_{\gamma }^2(Q_{T})}^2+2\| S_{t}\|
_{L_{\gamma }^2(Q_{T})}^2   
\leq 6Te^{3T}C_1^2+8Te^{4T}C_2^2 \label{e4.30}
\end{equation}
If we take $A\geq 6Te^{3T}C_1^2+8Te^{4T}C_2^2$, we then deduce from
the inequalities \eqref{e4.29} and \eqref{e4.30} that
\begin{equation}
\| v\| _{L^2(0,T;W_{2,\gamma }^{1,1}(0,1))}^2 \leq A,\quad
\| v_{t}\| _{L_{\gamma }^2(Q_{T})}^2\leq A.
\label{e4.31}
\end{equation}
Hence $v\in K(A)$ and consequently $\mathcal{M}$ maps $K(A)$ into itself.

We will now show that the mapping $\mathcal{M}:K(A)\to K(A)$ is
continuous. Let $v_1$, $v_2\in K(A)$ and let $V_1=\mathcal{M}(v_1)$,
and $V_2=\mathcal{M}(v_1)$.

We observe that $V=V_1-V_2$ satisfies
\begin{gather}
\frac{\partial ^2V}{\partial t^2}-\frac{1}{x}\frac{\partial }{\partial x}
\big( x\frac{\partial V}{\partial x}\big) 
=\frac{d}{dt}\max (\Lambda _{x}(\xi v_1),0)
 -\frac{d}{dt}\max ( \Lambda _{x}(\xi v_2),0),  \label{e4.32}\\
V(1,t) =0,\quad V_{x}(1,t)=0,  \label{e4.33} \\
V(x,0) =0,\quad V_{t}(x,0)=0  \label{e4.34}
\end{gather}
Define the function
\begin{equation}
\wp (x,t) =\int_{0}^{t}V(x,\tau )d\tau ,  \label{e4.35}
\end{equation}
then it follows from \eqref{e4.32}--\eqref{e4.35} that $\wp (x,t)$ satisfies
\begin{gather}
\frac{\partial ^2\wp }{\partial t^2}-\frac{1}{x}\frac{\partial }{\partial x}
\big( x\frac{\partial \wp }{\partial x}\big)
=\max (\Lambda _{x}(\xi v_1),0) -\max ( \Lambda _{x}(\xi
v_2),0) ,    \label{e4.36} \\
\wp (1,t) =0,\quad \wp _{x}(1,t)=0,  \label{e4.37} \\
\wp (x,0) =0,\quad \wp _{t}(x,0)=0.  \label{e4.38}
\end{gather}
It is obvious that
\begin{equation}
\| \mathcal{H}(x,t)\| _{L_{\gamma }^2(Q_{T})}^2
\leq \| v_1-v_2\| _{L_{\gamma }^2(Q_{T})}^2,
\label{e4.39}
\end{equation}
where
\begin{equation}
\mathcal{H}(x,t) =\max ( \Lambda _{x}(\xi v_1),0)
-\max ( \Lambda _{x}(\xi v_2),0)  \label{e4.40}
\end{equation}
Then we have
\begin{gather}
\| V\| _{L_{\gamma }^2(Q_{T})}^2 \leq
d\|v_1-v_2\| _{L_{\gamma }^2(Q_{T})}^2,  \label{e4.41} \\
\| V_1-V_2\| _{L_{\gamma }^2(Q_{T})}
=\| \mathcal{M}(v_1)-\mathcal{M}(v_1)\| _{L_{\gamma }^2(Q_{T})}  
\leq \sqrt{d}\| v_1-v_2\| _{L_{\gamma }^2(Q_{T})}. \label{e4.42}
\end{gather}
Consequently he mapping $\mathcal{M}:K(A)\to K(A)$ is continuous.
The set $\overline{K(A)}$ is compact, because of the following result.
\end{proof}

\begin{theorem} \label{thm3}
Let $B_{0}\subset B\subset B_1$ with compact embedding
(reflexive Banach spaces) (see \cite{l1,s1}). Assume that
$\alpha ,\lambda \in (0,\infty )$ and $T>0$. Then
\begin{equation*}
W=\{ \theta :\theta \in L^{\alpha }(0,T;B), \theta _{t}\in
L^{\lambda }(0,T;B_1)\}
\end{equation*}
is compactly embedded in $L^{\alpha }(0,T;B)$, that is the bounded sets are
relatively compact in $L^{\alpha }(0,T;B)$.
\end{theorem}

Observe that $L^2(0,T;L_{\gamma }^2(0,1))=L_{\gamma }^2(Q_{T})$,
$\mathcal{M}(K(A))\subset K(A)\subset L_{\gamma }^2(Q_{T})$. Then by
Schauder fixed point theorem the mapping $\mathcal{M}$ admits a fixed point
$v\in K(A)$.

\section{A priori bound for the solution}

We now obtain a priori bound for the solution of  problem
\eqref{e1.1}--\eqref{e1.3}. This a priori bound can be used to establish
some regularity results. We may expect the solution of \eqref{e1.1}--\eqref{e1.3}
to be in the function space $L^2(0,T;W_{q,\gamma }^{1,1}(0,1))$ with $q\leq \infty$.

\begin{theorem} \label{thm4}
 Let $u\in L^2(0,T;W_{2,\gamma }^{1,1}(0,1))$, be a
solution of problem \eqref{e1.1}--\eqref{e1.3}, then following a priori bound holds
\begin{equation}
\begin{aligned}
&\| u\| _{L^2(0,T;W_{2,\gamma }^{1,1}(0,1))}^2\\
&\leq 8Te^{8T}\Big( \| Z_1\| _{W_{2,\gamma
}^{1}((0,1))}^2+\| Z_2\| _{L_{\gamma
}^2((0,1))}^2+\| f\| _{L_{\gamma
}^2(Q_{T})}^2\Big)
\end{aligned} \label{e5.1}
\end{equation}
\end{theorem}

\begin{proof} Note that
\begin{equation}
\begin{aligned}
&\int_0^s \frac{d}{dt}\int_0^1 x^2u_{t}^2dx   \\
&=2\int_0^s \int_0^1 x^2u_{t}u_{tt}\,dx\,dt
=2\int_0^s \int_0^1 x^2u_{t}\Big[ \frac{1}{x}\frac{\partial }{\partial x}( x
\frac{\partial u}{\partial x}) +\mathcal{I}u\Big] \,dx\,dt   \\
&=2\int_0^s x^2(
u_{t}(1,t)u_{x}(1,t)-x^2u_{t}(0,t)u_{x}(0,t)) dt
-2\int_0^s \int_0^1 x^2u_{tx}u_{x}\,dx\,dt
 \\
&\quad -2\int_0^s \int_0^1  xu_{t}u_{x}\,dx\,dt
+2\int_0^s \int_0^1 x^2u_{t}\mathcal{I}u\,dx\,dt
+2\int_0^s \int_0^1 x^2u_{t}f\,dx\,dt
\end{aligned} \label{e5.2}
\end{equation}
By using boundary and initial conditions \eqref{e3.2} and \eqref{e3.3},
then from \eqref{e5.2} one has
\begin{equation}
\begin{aligned}
&\int_0^1 x^2u_{t}^2(x,s)dx+\overset{1}{\underset
{0}{\int }}x^2u_{x}^2(x,s)dx   \\
&=\int_0^1 x^2Z_2^2(x)dx
+\int_0^1 x^2( \frac{\partial Z_1}{\partial x}) ^2dx-2
\int_0^s \int_0^1 xu_{t}u_{x}\,dx\,dt   \\
&\quad +2\int_0^s \int_0^1  x^2u_{t}\mathcal{I}u\,dx\,dt
+2\int_0^s \int_1^1 x^2u_{t}f\,dx\,dt.
\end{aligned}  \label{e5.3}
\end{equation}
On the other hand,
\begin{equation}
\begin{aligned}
&\int_0^s \int_0^1 \Big[
x^2u_{x}u_{tt}-\frac{\partial }{\partial x}( x\frac{\partial u}{
\partial x}) \big(x\frac{\partial u}{\partial x}\big) \Big] \,dx\,dt
 \\
&=\int_0^1 x^2u_{x}(x,s)u_{t}(x,s)dx
 +\int_0^s\int_0^1 xu_{t}^2\,dx\,dt   \\
&=\int_0^s \int_0^1 x^2u_{x} \mathcal{I}u\,dx\,dt
 +\int_0^s \int_1^0 x^2u_{x}f\,dx\,dt,
\end{aligned}\label{e5.4}
\end{equation}
and
\begin{equation}
\begin{aligned}
&2\int_)^s \int_0^1 \Big[
xu_{t}u_{tt}-\frac{\partial }{\partial x}
( x\frac{\partial u}{\partial x}) u_{t}\Big] dx   \\
&=\int_0^1 xu_{t}^2(x,T)dx+\int_0^1 xu_{x}^2(x,T)dx   \\
&=2\int_0^s \int_0^1 xu_{t} \mathcal{I}u\,dx\,dt
+2\int_0^s \int_0^1 xu_{t}f\,dx\,dt
+\int_0^1 xZ_2^2(x)dx   \\
&\quad +\int_0^1 x\big( \frac{\partial Z_1}{\partial x}\big) ^2dx.
\end{aligned}  \label{e5.5}
\end{equation}
Equalities \eqref{e5.3}--\eqref{e5.5} and the elementary equality
\begin{equation}
\begin{aligned}
&\int_0^1 (x+x^2)u^2(x,s)dx \\
&\leq \int_0^s \int_0^1 (x+x^2)u^2\,dx\,dt
+ \int_0^s \int_0^1 (x+x^2)u_{t}^2\,dx\,dt   \\
&\quad +\int_0^1 (x+x^2)Z_1^2(x)dx,
\end{aligned}  \label{e5.6}
\end{equation}
lead to
\begin{align*}
&\int_0^1 x^2u_{t}^2(x,s)dx+\int_0^1 x^2u_{x}^2(x,s)dx
+\int_0^s \int_0^1 xu_{t}^2\,dx\,dt   \\
&+\int_0^1 xu_{t}^2(x,s)dx+\int_0^1 xu_{x}^2(x,s)dx   \\
&=\int_0^1 (x^2+x)Z_2^2(x)dx
+\int_0^1 (x^2+x)\big( \frac{\partial Z_1}{\partial x}\big)^2dx
 -2\int_0^s \int_0^1 xu_{t}u_{x}\,dx\,dt   \\
&\quad +2\int_0^s \int_0^1x^2u_{t}\mathcal{I}u\,dx\,dt
+2\int_0^s \int_0^1 x^2u_{t}f\,dx\,dt
+\int_0^1(x+x^2)Z_1^2(x)dx   \\
&\quad+\int_0^s \int_0^1 x^2u_{x} \mathcal{I}u\,dx\,dt
 +\int_0^s \int_0^1 x^2u_{x}f\,dx\,dt
 +\int_0^1x^2u_{x}(x,s)u_{t}(x,s)dx   \\
&\quad+2\int_0^s \int_0^1 xu_{t} \mathcal{I}u\,dx\,dt
+2\int_0^s \int_0^1 xu_{t}f\,dx\,dt.
\end{align*}
Young's inequality and a Poincar\'{e} type of inequality \cite{s1}, transforms
the above inequality  into
\begin{align*}
&\int_0^1 (x+x^2)u^2(x,s)dx
+\int_0^1 (x+x^2)u_{t}^2(x,s)dx
+\int_0^1 (x+x^2)u_{x}^2(x,s)dx   \\
&\leq 8\Big( \int_0^s \int_0^1 (x+x^2)u^2\,dx\,dt
+\int_0^s \int_0^1 (x+x^2)u_{t}^2\,dx\,dt
+\int_0^s \int_0^1 (x+x^2)u_{x}^2\,dx\,dt\Big)  \\
&\quad +8\Big( \int_0^1 (x^2+x)Z_2^2(x)dx
+\int_0^1 (x+x^2)\Big( Z_1^2(x)
+\Big( \frac{\partial Z_1}{\partial x}\Big) ^2\Big) dx \\
&\quad +\int_0^s \int_0^1 (x+x^2)f^2\,dx\,dt\Big) .
\end{align*}
Using Gronwall's lemma \cite[Lemma 4.1]{m1} this implies
\[
\| u(x,s)\| _{W_{2,\gamma }^{1,1}(0,1))}^2
\leq 8Te^{8T}\Big( \| Z_1\| _{W_{2,\gamma
}^{1}((0,1))}^2+\| Z_2\| _{L_{\gamma
}^2((0,1))}^2+\| f\| _{L_{\gamma
}^2(Q_{T})}^2\Big)  \label{e5.9}
\]
Integrating with respect to $s$ over the interval $[0,T]$ gives the 
a priori bound \eqref{e5.1}.
\end{proof}

\subsection*{Conclusion}
 The well posedness of a one point initial boundary
value problem for a nonlinear hyperbolic singular integro-differential
equation is proved. The Schauder fixed point theorem is the main tool used
to establish the existence of solution. The solvability of the problem is
proved when the given data are related to a weighted Sobolev space. A priori
bound for the solution may allow to study the regularity of the solution is
obtained.

\subsection*{Acknowledgements}
 The authors would like to extend their sincere
appreciation to the Deanship of Scientific Research at King Saud
University for its funding this Research group NO (RG-1435-043).

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\end{document}
