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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 78, 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}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2013/78\hfil Delay differential equations]
{Delay differential equations with homogeneous integral conditions}

\author[A. Raheem, D. Bahuguna \hfil EJDE-2013/78\hfilneg]
{Abdur Raheem, Dhirendra Bahuguna}  % in alphabetical order

\address{Abdur Raheem \newline
Department of Mathematics and Statistics,
Indian Institute of Technology Kanpur,
 Kanpur -208016, India}
\email{araheem@iitk.ac.in}

\address{Dhirendra Bahuguna \newline
Department of Mathematics and Statistics,
Indian Institute of Technology Kanpur,
 Kanpur -208016, India}
\email{dhiren@iitk.ac.in}

\thanks{Submitted April 6, 2012. Published March 22, 2013.}
\subjclass[2000]{35D35, 34K05, 34K07, 34K10, 34K30}
\keywords{Method of semidiscretization; delay differential equation;
\hfill\break\indent strong solution}

\begin{abstract}
 In this article we prove the existence and uniqueness of a strong solution
 of a delay differential equation with homogenous integral conditions using
 the method of semidiscretization in time. As an application,
 we include an example that illustrates the main result.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

This article concerns the  delay differential
equation having homogeneous integral conditions,
\begin{gather}
\frac{\partial u}{\partial t}- \frac{\partial^2 u}{\partial
x^2}-\lambda \frac{\partial^3 u}{\partial x^2 \partial
t}= F(x,t,u_t) \quad \text{on }  (0,1)\times (0,T],\label{eq1}\\
u(x,t)=\phi (x,t) \quad \text{on }  (0,1)\times [-T,0],\label{eq2}
\end{gather}
with integral conditions
\begin{gather}
\int_{0}^{1}u(x,t)\,dx=0, \quad t \in [0,T], \label{eq3} \\
\int_{0}^{1}xu(x,t)\,dx=0,\quad t \in [0,T], \label{eq4}
\end{gather}
where $0<T< \infty$, the unknown function $u:[-T,T] \to L^2(0,1)$,
the history $\phi:[-T,0] \to L^2(0,1)$  and the
nonlinear map $F:(0,T] \times \mathcal{C}_0 \to B(0,1)$, are
defined by $u(t)(x)=u(x,t)$, $\phi(t)(x)=\phi(x,t)$ and
$F(t,u_t)(x)=F(x,t,u_t)$, respectively. Here $L^2(0,1)$ is the real
Hilbert of all square integrable real valued functions on $(0,1)$
with the standard inner product, and $B(0,1)$ is the completion of
$C_0(0,1)$, the space of all continuous functions on $(0,1)$ having
compact support in $(0,1)$, with the inner product defined by
$$
(u,v)_B=\int_{0}^{1}\Im_x u(x)\Im_x v(x)\,dx,
$$
where
$\Im_x u(x)=\int_{0}^{x}u(\xi)d\xi$. We recall that, if
$\|\cdot\|_B$ denote the corresponding norm; that is,
$$
\|\psi\|_B=\sqrt {(\psi,\psi)_B},
$$
then it follows that
$$
\|\psi\|^2_B \leq \frac{1}{2}\|\psi\|^2.
$$
Also for $t \in (0,T]$, $\mathcal{C}_t:= C([-T,t];B(0,1))$  is the
Banach space of all continuous functions from $[-T,t]$ into $B(0,1)$
endowed with the supremum norm
$$
\|\psi\|_t=\sup_{-T \leq\theta \leq t} \|\psi(\theta)\|_B.
$$
For $\psi \in \mathcal{C}_T$, we denote $\psi_t \in \mathcal{C}_0$
given by $\psi_t(\theta)=\psi(t+\theta),\theta \in [-T,0]$.

For the consideration of integral conditions, we use the space
$V$ introduce by  Merazga and Bouziani \cite{ab1},
$$
V=\big\{\phi \in L^2(0,1): \int_{0}^{1}\phi(x)\,dx=\int_{0}^{1}x\phi (x)\,dx=0 \big\}.
$$
Note that $V$ is a Hilbert space with respect to the standard inner product.

Since 1930, various classical types of initial boundary value
problems have been investigated by many authors using the method of
semidiscretization, see for instance, \cite{jk1,kr1,kr2,kr3} and
references therein.

In this paper our aim is to extend  the application of the method of
semidiscretization in time to delay differential equations with homogeneous
integral conditions and to establish the existence and uniqueness of a
strong solution for a delay differential equation with homogeneous
integral conditions given by \eqref{eq1}-\eqref{eq4}.

The method of semidiscretization in time is a constructive method
and has a strong numerical aspect. For the application of the method
of semidiscretization to integrodifferential equations with
nonclassical boundary conditions, we refer the readers to
\cite{b2,b3,ab2,la1,la2} and references therein.

Dubey \cite{sd1}  established the existence and uniqueness of a
strong solution for the following nonlinear differential equation
in a reflexive Banach space with a nonlocal history condition
using the method of semidiscretization in time
\begin{gather*}
u'(t)+Au(t)=f(t,u(t),u_t), \quad t \in (0,T], \\
h(u_0)=\phi \quad \text{on} \quad [-\tau,0],
\end{gather*}
where $0<T<\infty$, $\phi \in \mathcal{C}_0:=C([-\tau,0];X),\tau >0$,
the nonlinear operator $A$ is single-valued and $m$-accretive defined
from the domain $D(A)\subset X$ into $X$, the nonlinear map $f$ is
defined from $[0,T]\times X \times \mathcal{C}_0:=C([-\tau,0];X)$
into $X$, the map $h$ is defined from $\mathcal{C}_0$ and
$\mathcal{C}_0$.
Bahuguna,  Abbas, and Dabas \cite{b1}  applied the method of
semidiscretization to a semilinear functional partial differential
equation with an integral condition.

Our problem is motivated by the work of Lakoud and Belakroum \cite{la3}
and Dubey \cite{sd1}. Lakoud and Balakroum \cite{la3}\
 established the existence and uniqueness of a weak solution for
the  integro-differential evolution with a memory term,
\[
\frac{\partial v}{\partial t}- \frac{\partial^2 v}{\partial
x^2}-\lambda \frac{\partial^3 v}{\partial x^2 \partial
t}=g(x,t)+\int_{0}^{t}a(t-s)k(s,v(x,s))\,ds
\quad \text{on }(0,1)\times (0,T],
\]
subject to the initial conditions
\[
v(x,0)=V_0(x),
\]
and integral conditions
\begin{gather*}
\int_{0}^{1}v(x,t)\,dx=E(t),  \\
\int_{0}^{1}xv(x,t)\,dx=G(t),
\end{gather*}
where $f,V_0,G,E$ are given functions and $T, \lambda$ are positive
constants.

The plan of the paper is as follows. In section 2, we state all the
assumptions and preliminaries. In section 3, we state the main
result. In section 4, we state and prove all the lemmas that are
required to prove the main result and at the end of this section, we
prove the main result. In the last section, we give an application
of the main result.

Throughout the paper we denote a generic constant by $C$. This
constant may have different values in the same discussion.

\section{Preliminaries}

We will use the following assumptions:
\begin{itemize}
\item[(H1)] The nonlinear map $F:(0,T] \times \mathcal{C}_0
\to B(0,1)$ satisfies a local Lipschitz condition
$$
\|F(t_1,\psi_1)-F(t_2,\psi_2)\|_B \leq L_F(r)[|t_1-t_2|+\|\psi_1-\psi_2\|_0],
$$
for all  $t_1,t_2 \in (0,T]$ and $\psi_1,\psi_2 \in \mathcal{C}_0$
with $\|\psi_i-\phi(0)\|_0 \leq r$, $i=1,2$ and $L_F(r)$ is a
nondecreasing function of $r>0$.

\item[(H2)] The history function $\phi :[-T,0] \to L^2(0,1)$ is
uniformly Lipschitz continuous with Lipschitz constant
$K>0$; i.e., $\|\phi(t)-\phi(s)\| \leq K |t-s|$.

\item[(H3)] $\int_{0}^{1} \phi(x,0) \,dx=0$, $\int_{0}^{1} x \phi(x,0)\,dx=0$.
\end{itemize}

\begin{lemma} \label{lm1}
If $-A$ is the infinitesimal generator of a $C_0$-semigroup of contractions
in a Banach space $X$ then $A$ is m-accretive; i.e.,
$$
(Au,J(u))\geq 0, \quad \text{for }u\in D(A),
$$
where $J$ is the duality mapping and $R(I+\lambda A)=X$ for
$\lambda >0$, $I$ is the identity operator on $X$ and $R(\cdot)$ is the range of
an operator.
\end{lemma}

The proof of the above lemma follows from Lumer Phillips theorem
\cite[Thm. 1.4.3]{pz}.

\section{Main result}

\begin{theorem}  \label{t1}
Suppose that assumptions {\rm (H1)--(H3)} are
satisfied. Then problem \eqref{eq1}-\eqref{eq4} has a unique strong solution
on the interval $[-T,T]$.
\end{theorem}

\section{Discretization and a priori estimates}

 To apply the method of semidiscretization we divide the interval $[0,T]$ into
the subintervals of length $h_n=\frac{T}{n}$. We set $u_0^n=\phi(0)$
for all $n\in \mathbb{N}$ and define $\{u_j^n\}$ successively as the
unique solution of the  problem
\begin{gather}
\delta u^n_j-\frac{\partial^2 u^n_j}{\partial x^2}-\lambda
\frac{\partial^2 \delta u^n_j}{\partial x^2}
=F(t_j^n,\tilde{u}_{j-1}^n), \label{eq5}\\
\int_{0}^{1}u^n_j\,dx = 0, \label{eq6}\\
\int_{0}^{1}xu^n_j\,dx = 0, \label{eq7}
\end{gather}
where $\tilde{u}^n_0=\phi(t)$ for $t\in [-T,0]$
and $2 \leq j \leq n$,
\[
\tilde{u}^n_{j-1}(t)
= \begin{cases} \phi(t^n_{j-1}+t), & \text{if } t\in [-T,-t^n_{j-1}] \\
 u^n_{i-1}+(t^n_{j-1}+t-t^n_{i-1})\delta u^n_i, & \text{if }
 t\in[-t^n_{j-i},-t_{j-i-1}^n],\;
 1\leq i \leq j-1,
\end{cases}
\]
and
\[
\delta u^n_j=\frac{u^n_j-u^n_{j-1}}{h_n}.
\]
Let $w_j^n=u^n_j+\lambda \delta u^n_j$. This implies that
$\delta u^n_j=\frac{1}{h_n+\lambda}w^n_j-\frac{1}{h_n+\lambda}u^n_{j-1}$.
So \eqref{eq5}-\eqref{eq7} will reduce to
\begin{gather}
-\frac{\partial^2w^n_j}{\partial x^2} +\frac{1}{h_n+\lambda}w_j^n = f^n_j,
  \label{eq8}\\
\int_{0}^{1}w^n_j\,dx=0 \label{eq9},\\
\int_{0}^{1}xw^n_j\,dx=0 \label{eq10},
\end{gather}
where
$$
f^n_j=\frac{1}{h_n+\lambda}u^n_{j-1}+F(t^n_j,\tilde{u}^n_{j-1}).
$$
Now we show the existence and uniqueness of functions $w^n_j$ satisfying
\eqref{eq8}-\eqref{eq10}. For this consider $H=L^2(0,1)$, the
Hilbert space of all real valued square integrable functions on the
interval $(0,1)$. Let the linear operator $A$ be defined by
$$
D(A):=\{u \in H: u'' \in H, \int_{0}^{1}u(x)\,dx=\int_{0}^{1}xu(x)\,dx=0\},\quad
Au=-u''.
$$
Then we know that $-A$ is the infinitesimal generator of a
$C_0$-semigroup $S(t),t\geq 0$ of contractions in $H$.

The existence of unique $w^n_j$ satisfying equations
\eqref{eq8}-\eqref{eq10} is a consequence of Lemma \ref{lm1}.
As
$$
u^n_j=\frac{h_n}{\lambda+h_n}w^n_j+\frac{\lambda}{\lambda+h_n}u^n_{j-1},
$$
there exist unique $u^n_j \in D(A)$ satisfying
\eqref{eq5}--\eqref{eq7}. Now we define
\begin{equation}
U^n(t)= \begin{cases} \phi(t), & \text{if } t\in [-T,0] \\
u^n_{j-1}+(t-t^n_{j-1})\frac{u^n_j-u^n_{j-1}}{h_n},
&\text{if } t\in (t^n_{j-1},t^n_j].
\end{cases}\label{eq11}
\end{equation}

\begin{lemma}  \label{lem4.1}
For $n \in \mathbb{N}$ and $j=1,2,\cdots n$,
$$
\|u^n_j-\phi(0)\| \leq C,
$$
where $C$ is a generic constant independent of $n,j,h_n$.
\end{lemma}

\begin{proof}
Now for any $\psi \in V$, from \eqref{eq5} we have
\begin{equation}
(\delta u^n_j,\psi)_B-\Big(\frac{\partial^2 u^n_j}{\partial
x^2},\psi \Big)_B-\lambda \Big(\frac{\partial^2 \delta
u^n_j}{\partial x^2},\psi \Big)_B=(F^n_j,\psi)_B, \label{eq12}
\end{equation}
where $F^n_j=F(t^n_j,\tilde{u}^n_{j-1})$.
By the definition of the
inner product $(,)_B$, we have
\begin{eqnarray}
\Big(\frac{\partial^2 u^n_j}{\partial x^2},\psi
\Big)_B=-\int_{0}^{1}u^n_j \psi \,dx=-(u^n_j,\psi).\label{eq13}
\end{eqnarray}
So \eqref{eq12} reduces to
\begin{equation}
(\delta  u^n_j,\psi)_B+(u^n_j,\psi)+\lambda(\delta u^n_j,\psi)=(F^n_j,\psi)_B.
 \label{eq14}
\end{equation}
Taking $j=1,\psi=u_1^n-u^n_0$ in \eqref{eq14},
\begin{align*}
&(u^n_1-u^n_0,u_1^n-u^n_0)_B+h_n(u^n_1,u_1^n-u^n_0)
 +\lambda( u^n_1-u^n_0,u_1^n-u^n_0) \\
&=h_n(F^n_1,u_1^n-u^n_0)_B .
\end{align*}
Now using \eqref{eq13}, we obtain
\begin{align*}
&(u^n_1-u^n_0,u_1^n-u^n_0)_B+h_n(u^n_1-u^n_0,u_1^n-u^n_0)
 +\lambda( u^n_1-u^n_0,u_1^n-u^n_0)\\
&=h_n\left(F^n_1+\frac{d^2u^n_0}{dx^2},u_1^n-u^n_0\right)_B.
\end{align*}
Now, we obtain
\[
\|u^n_1-u^n_0\|^2_B+ h_n \|u^n_1-u^n_0\|^2+ \lambda
\|u^n_1-u^n_0\|^2  \leq h_n \big[\|F^n_1\|_B+
\|\frac{d^2u^n_0}{dx^2} \|_B\big]\|u^n_1-u^n_0\|_B.
\]
By ignoring first two terms on the left hand side, we obtain
\[
\lambda \|u^n_1-u^n_0\|^2 \leq h_n \big[\|F^n_1\|_B+h_n
\|\frac{d^2u^n_0}{dx^2} \|_B\big]\|u^n_1-u^n_0\|_B.
\]
As $\|u^n_1-u^n_0 \|_B \leq \frac{1}{\sqrt{2}}\|u^n_1-u^n_0\|$, we
have
\[
\|u^n_1-u^n_0\| \leq \frac{h_n}{\lambda \sqrt{2}}\big[\|F^n_1\|_B+
\|\frac{d^2u^n_0}{dx^2}\|_B\big].
\]
By using assumption (H1), and the inequality $h_n \leq T$, we obtain
\[
\|u^n_1-u^n_0\| \leq \frac{T}{\lambda \sqrt{2}}
\big[L_F(r)(T+r)+\|F(0,\phi(0))\|_B+
\|\frac{d^2u^n_0}{dx^2}\|_B\big]
\leq C.
\]
By putting $\psi =u^n_j-u^n_0$ in  \eqref{eq14}, we obtain
\begin{align*}
&(u^n_j-u^n_{j-1},u^n_j-u^n_0)_B+h_n(u^n_j,u^n_j-u^n_0)
 +\lambda(u^n_j-u^n_{j-1},u^n_j-u^n_0)\\
&=h_n(F^n_j,u^n_j-u^n_0)_B. \hspace{6.7cm}
\end{align*}
Using \eqref{eq13}, we obtain
\begin{align*}
&\big(u^n_j-u^n_0,u^n_j-u^n_0\big)_B+h_n\big(u^n_j-u^n_0,u^n_j-u^n_0\big)
 +\lambda\big(u^n_j-u^n_0,u^n_j-u^n_0\big)  \\
&=h_n\big(F^n_j,u^n_j-u^n_0\big)_B
 +\big(u^n_{j-1}-u^n_0,u^n_j-u^n_0\big)_B+h_n
 \big(\frac{d^2u^n_0}{dx^2},u^n_j-u^n_0 \big)_B\\
&\quad  +\lambda \big(u^n_{j-1}-u^n_0,u^n_j-u^n_0\big).
\end{align*}
By ignoring the first two terms on the left hand side, we obtain
\begin{align*}
\lambda\|u^n_j-u^n_0\|^2
& \leq h_n \|F^n_j\|_B \|u^n_j-u^n_0\|_B
 +\|u^n_{j-1}-u^n_0\|_B \|u^n_j-u^n_0\|_B  \\
&\quad +h_n \| \frac{d^2u^n_0}{dx^2}\|_B\|u^n_j-u^n_0\|_B+\lambda
\|u^n_{j-1}-u^n_0\| \|u^n_j-u^n_0\|.
\end{align*}
As $\|u^n_j-u^n_0\|_B \leq \frac{1}{\sqrt{2}} \|u^n_j-u^n_0\|$, we
have
\begin{equation}
\|u^n_j-u^n_0\| \leq \frac{1}{\lambda}\big[\frac{h_n}{\sqrt{2}}(\|F^n_j\|_B
+ \|
\frac{d^2u^n_0}{dx^2}\|_B)+(\frac{1}{2}+\lambda)\|u^n_{j-1}-u^n_0\|\big].
\label{eq15}
\end{equation}
By assumption (H1), we have
\begin{equation}
\begin{aligned}
\|F^n_j\|_B&= \|F(t^n_j,\tilde{u}^n_{j-1})\|_B \\
           &=  \|F(t^n_j,\tilde{u}^n_{j-1})-F(0,\phi(0))\|_B
 +\|F(0,\phi(0))\|_B \\
           &\leq  L_F(r)[|t^n_j|+\|\tilde{u}^n_{j-1}-\phi(0)\|_0]
 +\|F(0,\phi(0))\|_B\\
           &\leq  L_F(r)[T+r]+\|F(0,\phi(0))\|_B.
\end{aligned} \label{eq16}
\end{equation}
Using \eqref{eq16} in \eqref{eq15}, we obtain
\begin{gather*}
\begin{aligned}
\|u^n_j-u^n_0\|
&\leq  \frac{h_n}{\lambda \sqrt{ 2}}\big[L_F(r)(T+r)+\|F(0,\phi(0))\|_B
+\| \frac{d^2u^n_0}{dx^2}\|_B \big]\\
&\quad +\Big(1+\frac{1}{2\lambda}\Big)\|u^n_{j-1}-u^n_0\|
\end{aligned} \\
\|u^n_j-u^n_0\| \leq h_n K+\Big(1+\frac{1}{2\lambda}\Big)\|u^n_{j-1}-u^n_0\|,
\end{gather*}
where $K=\frac{1}{\lambda \sqrt{2}}\big[L_F(r)(T+r)+\|F(0,\phi(0))\|_B+\|
\frac{d^2u^n_0}{dx^2}\|_B\big]$.
Repeating the above procedure, we obtain
$$
\|u^n_j-u^n_0\| \leq C.
$$
This completes the proof.
\end{proof}

\begin{lemma}\label{lm2}
 For $j=1,2, \cdots , n$,
$$
\|\frac{u^n_j-u^n_{j-1}}{h_n} \| \leq C.
$$
\end{lemma}

\begin{proof}
By putting $j=1$, $\psi=u^n_j-u^n_0$ in \eqref{eq14} and using
\eqref{eq13}, we obtain
\begin{align*}
&\Big(\frac{u^n_1-u^n_0}{h_n},u^n_1-u^n_0 \Big)_B
+ (u^n_1-u^n_0,u^n_1-u^n_0)+ \lambda
\Big(\frac{u^n_1-u^n_0}{h_n},u^n_1-u^n_0 \Big) \\
&=(F_1^n,u^n_1-u^n_0)_B+ \Big(\frac{d^2u^n_0}{dx^2},u^n_1-u^n_0\Big)_B.
\end{align*}
By ignoring the first two terms on the left hand sides, we obtain
\begin{eqnarray}
\frac{\lambda}{h_n}\|u^n_1-u^n_0\|^2
\leq  \Big(F^n_1+\frac{d^2u^n_0}{dx^2},u^n_1-u^n_0 \Big)_B.
\end{eqnarray}
As $\|u^n_j-u^n_0\|_B \leq \frac{1}{\sqrt{2}} \|u^n_j-u^n_0\|$, we have
\[
\|\frac{u^n_1-u^n_0}{h_n}\| \leq \frac{1}{\lambda\sqrt{2}}
\big[\|F^n_1 \|_B + \| \frac {d^2u^n_0}{dx^2} \|_B\big].
\]
Using assumption (H1), we obtain
\[
\|\frac{u^n_1-u^n_0}{h_n} \|
\leq \frac{1}{\lambda \sqrt{2}}\big[L_F(r)(T+r)+\|F(0,\phi(0))\|_B
+ \| \frac {d^2u^n_0}{dx^2} \|_B \big]
\leq  C.
\]
Subtracting \eqref{eq14} written for $j$, from the same identity for
$j-1$ and then putting $\psi=u^n_j-u^n_{j-1}$, we obtain
\begin{align*}
&(\delta u^n_j,u^n_j-u^n_{j-1})_B + (u^n_j-u^n_{j-1},u^n_j-u^n_{j-1})
+ \lambda(\delta u^n_j,u^n_j-u^n_{j-1})  \\
&=(F^n_j-F^n_{j-1},u^n_j-u^n_{j-1})_B+ (\delta
u^n_{j-1},u^n_j-u^n_{j-1})_B +\lambda \left(\delta
u^n_{j-1},u^n_j-u^n_{j-1} \right ).
\end{align*}
Ignoring the first two terms on the left hand side,
\begin{align*}
\frac{\lambda}{h_n}\| u^n_j-u^n_{j-1}\| ^2
& \leq \|F^n_j-F^n_{j-1}\|_B
\|u^n_j-u^n_{j-1}\|_B+\|\frac{u^n_{j-1}-u^n_{j-2}}{h_n} \|_B
\| u^n_j-u^n_{j-1}\| _B  \\
&\quad +\lambda\| \frac{u^n_{j-1}-u^n_{j-2}}{h_n} \|\| u^n_j-u^n_{j-1}\|.
\end{align*}
As $\|\psi\|_B \leq \|\psi\|/\sqrt{2}$,  we have
\begin{eqnarray}
\|\frac{u^n_j-u^n_{j-1}}{h_n} \| \leq \frac{1}{\lambda
\sqrt{2}}\|F^n_j-F^n_{j-1}\|_B
+(1+\frac{1}{2\lambda})\|\frac{u^n_{j-1}-u^n_{j-2}}{h_n} \|. \label{eq17}
\end{eqnarray}
Now using assumption (H1),
\begin{align*}
\|F^n_j-F^n_{j-1}\|_B
&= \|F(t^n_j,\tilde{u}^n_{j-1})-F(t^n_{j-1},\tilde{u}^n_{j-2})\|_B \\
&\leq L_F(r)[|t^n_j-t^n_{j-1}|+\|\tilde{u}^n_{j-1}-\tilde{u}^n_{j-2}\|_0] \\
&\leq L_F(r) [T+2r].
\end{align*}
Using the above inequality in \eqref{eq17}, we obtain
\begin{eqnarray}
\|\frac{u^n_j-u^n_{j-1}}{h_n} \| \leq \frac{1}{\lambda
\sqrt{2}}L_F(r) [T+2r]
+(1+\frac{1}{2\lambda})\|\frac{u^n_{j-1}-u^n_{j-2}}{h_n} \|.
\end{eqnarray}
Repeating the above procedure, we finally obtain
\begin{eqnarray}
\|\frac{u^n_j-u^n_{j-1}}{h_n} \| \leq C.
\end{eqnarray}
This completes the proof.
\end{proof}

Now we introduce a sequence of step functions $\{X^n(t)\}$ defined by
\begin{equation}
X^n(t)= \begin{cases}
\phi(0), & \text{if }  t=0 \\
 u^n_j, & \text{if } t\in (t^n_{j-i},t_j^n].
\end{cases}\label{eq18}
\end{equation}

\begin{remark} \label{rk1} \rm
From Lemma \ref{lm2} it follows that the functions $U^n$ are
uniformly Lipschitz continuous on $[-T,T]$ and
$U^n(t)-X^n(t)\to 0$, as $n \to \infty$ on $[0,T]$.
\end{remark}

Let $F^n(t)=F(t^n_j,\tilde{u}^n_{j-1})$.
By assumption (H1) and remark \ref{rk1}, we  see that
$F^n(t) \to F(t,u_t)$.
Using \eqref{eq11} and \eqref{eq18}, in \eqref{eq5}, we obtain
\begin{eqnarray}
\frac{d^-}{dt}U^n(t)-\frac{\partial^2}{\partial x^2}X^n(t)-\lambda \frac{\partial^3}{\partial x^2 \partial t}X^n(t)=F^n(t). \label{eq19}
\end{eqnarray}
Integrating with respect to $t$, we obtain
\begin{equation}
-\int_{0}^{t}\big[\frac{\partial^2}{\partial x^2}X^n(s)+\lambda
\frac{\partial^3}{\partial x^2 \partial
s}X^n(s)\big]\,ds=\phi(0)-U^n(t)+\int_{0}^{t}F^n(s)\,ds.\label{eq20}
\end{equation}

\begin{lemma} \label{lem4.4}
There exists $u \in C([-T,T];B(0,1))$ such that
$U^n(t) \to u(t)$ uniformly on $[-T,T]$. Moreover $u(t)$ is Lipschitz continuous
on $[-T,T]$.
\end{lemma}

\begin{proof}
From  \eqref{eq19}, we have
\begin{align*}
&\Big(\frac{d^-}{dt}U^n(t)-\frac{d^-}{dt}U^k(t),U^n(t)-U^k(t)
\Big)_B+(X^n(t)-X^k(t),U^n(t)-U^k(t)) \hspace{.2cm}\hspace{.5cm}\\
&+\lambda \Big(\frac{\partial}{\partial t}X^n(t)-\frac{\partial}{\partial
t}X^k(t),U^n(t)-U^k(t)\Big)\\
&=(F^n(t)-F^k(t),U^n(t)-U^k(t))_B.
\end{align*}
Now,
\begin{align*}
&\frac{1}{2}\frac{d^-}{dt}\|U^n(t)-U^k(t)\|^2_B+\|X^n(t)-X^k(t)\|^2
+\frac {\lambda}{2} \frac{\partial}{\partial t}\|X^n(t)-X^k(t)\|^2\\
&=(X^n(t)-X^k(t),X^n(t)-X^k(t)-U^n(t)+U^k(t))\\
&\quad + \lambda \Big(\frac{\partial}{\partial t}(X^n(t)-X^k(t)),
X^n(t)-X^k(t)-U^n(t)+U^k(t)\Big)\\
&\quad +(F^n(t)-F^k(t),U^n(t)-U^k(t))_B.
\end{align*}
By ignoring the last two terms on the left hand side, we obtain
\[
\frac{1}{2}\frac{d^-}{dt}\|U^n(t)-U^k(t)\|^2_B
\leq \delta_{nk}(t) + \|F^n(t)-F^k(t)\|_B \|U^n(t)-U^k(t)\|_B.
\]
where
\begin{align*}
\delta_{nk} (t)
&=\|X^n(t)-X^k(t)\|[\|X^n(t)-U^n(t)\|+\|X^k(t)-U^k(t)\|] \\
&\quad +\frac{\lambda}{2} \|\frac{\partial}{\partial t}(X^n(t)-X^k(t))
\| \big[\|X^n(t)-U^n(t)\|+\|X^k(t)-U^k(t)\|\big].
\end{align*}
By Remark~\ref{rk1}, it is clear that $\delta_{nk} (t) \to 0$ as
$n,k \to \infty$ uniformly on the interval $[0,T]$.
Now by assumption (H1), we have
\begin{eqnarray}
\|F^n(t)-F^k(t)\|_B
&= \|F(t^n_j,\tilde{u}^n_{j-1})-F(t^k_l,\tilde{u}^k_{l-1})\|_B  \\
&\leq  \delta'_{nk}(t) + L_F(r) \|U^n(t)-U^k(t)\|_B,
\end{eqnarray}
where
$$
\delta'_{nk}(t)
=L_F(r)[|t^n_j-t^k_l|+\|U^n(t)-\tilde{u}^n_{j-1}\|_0
+\|U^k(t)-\tilde{u}^k_{l-1}\|_0].
$$
Clearly $\delta'_{nk}(t) \to 0$ as
$ n,k \to \infty$ uniformly on $[0,T]$. This implies that for a.e.
$t \in [0,T]$,
\begin{eqnarray}
\frac{1}{2}\frac{d^-}{dt}\|U^n(t)-U^k(t)\|^2_B
\leq \delta'_{nk} (t) + L_F(r)\|U^n(t)-U^k(t)\|^2_B,
\end{eqnarray}
Integrating the above inequality over $(0,t)$ with $0 \leq t\leq T$,
we obtain
\[
\|U^n(t)-U^k(t)\|^2_B \leq 2\delta'_{nk}T +
2L_F(r)\int_{0}^{t}\|U^n(s)-U^k(s)\|^2_B\,ds.
\]
Applying Gronwall's inequality, we obtain that $U^n \to u$ in
$C([-\tau,T],B(0,1))$. As each $U^n$ is uniformly Lipschitz
continuous, and by assumption (H2), $u$ is Lipschitz continuous.
This completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{t1}]
Taking limits as $n \to \infty$ in \eqref{eq20}, we obtain
\[
-\int_{0}^{t}\big[\frac{\partial^2}{\partial x^2}u(t)
+\lambda \frac{\partial^3}{\partial x^2 \partial s}u(t)\big]\,ds
=\phi(0)-u(t)+\int_{0}^{t}F(s,u_t)\,ds.
\]
This implies that
\[
\frac{\partial u(t)}{\partial t}- \frac{\partial^2 u(t)}{\partial
x^2}-\lambda \frac{\partial^3 u(t)}{\partial x^2 \partial
t}= F(t,u_t),\quad\text{a.e. }t\in [0,T].
\]
Clearly $u(t)$ is differentiable with $u(t) \in V$
a.e. on $[0,T]$ and $u(t)=\phi(t)$,  $t \in [-T,0]$. This
implies that $u(t)(x)=u(x,t)$ is a strong solution of
\eqref{eq1}--\eqref{eq4}. Now we show the uniqueness of the strong solution.
To do this, suppose that $u_1,u_2$ are two strong solutions of
\eqref{eq1}-\eqref{eq4}.

 Let $u=u_1-u_2$, then for $\psi \in V$,  we have
\[
\Big(\frac{\partial u}{\partial t}, \psi \Big)_B+ (u, \psi) +
\lambda \Big(\frac{\partial u}{\partial t}, \psi \Big)
=\Big(
F(t,(u_1)_t)-F(t,(u_2)_t), \psi \Big)_B.
\]
Putting $\psi=u$ and ignoring last two terms in the left hand side,
we obtain
\[
\Big(\frac{\partial u}{\partial t}, u \Big)_B
\leq \Big(F(t,(u_1)_t)-F(t,(u_2)_t), u \Big)_B.
\]
By using assumption (H1), we obtain
\begin{align*}
\frac{1}{2} \frac{\partial}{\partial t} \|u(t)\|^2_B
 &\leq  L_F(r) \|(u_1)_t-(u_2)_t\|_B \|u(t)\|_B  \\
 &\leq  L_F(r) \sup_{-T \leq t+\theta \leq t}
\|u_t(\theta)\|_B\sup_{-T \leq \theta \leq t}
\|u(\theta)\|_B.
\end{align*}
Integrating between $0$ and $t$, we obtain
\begin{gather*}
\sup_{-T \leq\theta \leq t}\|u(\theta)\|^2_B
 \leq  2 L_F(r) \int_{0}^{t} \|u\|^2_s \,ds  \\
\|u\|^2_t
 \leq  2 L_F(r) \int_{0}^{t} \|u\|^2_s \,ds.
\end{gather*}
Applying Gronwall's inequality, we obtain $u=0$ on $[-T,T]$.
Hence we obtain a unique strong solution of
 problem \eqref{eq1}-\eqref{eq4} on the interval $[-T,T]$.
\end{proof}

\section{Application}
Consider the  partial differential equation
\begin{equation}
\begin{gathered}
\frac{\partial v}{\partial t}- \frac{\partial^2 v}{\partial
x^2}-\lambda \frac{\partial^3 v}{\partial x^2 \partial
t}= g(x,t)+\int_{0}^{t}a(t-s)k(s,v(x,s))\,ds \quad
\text{on } (0,1)\times (0,T],  \\
v(x,t)= \phi (x,t) \quad \text{on} \quad (0,1)\times
[-T,0],
\end{gathered} \label{eq21}
\end{equation}
with the integral conditions
\begin{gather}
\int_{0}^{1}v(x,t)\,dx=0,\label{eq22}  \\
\int_{0}^{1}xv(x,t)\,dx=0.\label{eq23}
\end{gather}
In the above problem, we identify the unknown function
$v:(0,T] \to B(0,1)$, by $v(t)(x)=v(x,t)$,
$g: (0,T] \to B(0,1)$ by $g(t)(x)=g(x,t)$, $k:(0,T]\times \mathbb{R} \to B(0,1)$
by $k(t,v(x,t))=k(t,v(t))(x)$ and the history
function $\phi:[-T,0] \to B(0,1)$ by $\phi(t)(x)=\phi(x,t)$. Also we take
$$
V=\big\{\phi \in L^2(0,1):\int_{0}^{1}\phi(x)\,dx=\int_{0}^{1}x\phi (x)\,dx=0
\big\}.
$$
Putting $t=s-\eta$ in the integral term, problem \eqref{eq21}-\eqref{eq23}
reduces to
\begin{equation}
\begin{gathered}
\frac{\partial v}{\partial t}- \frac{\partial^2 v}{\partial
x^2}-\lambda \frac{\partial^3 v}{\partial x^2 \partial
t}= g(t)+\int_{-t}^{0}a(-\eta)k(t+\eta,v(t+\eta))d\eta \quad
\text{on }  (0,T],  \\
v(t)= \phi (t), \quad t \in [-T,0].
\end{gathered}\label{eq24}
\end{equation}
Now we consider the following assumptions:
\begin{itemize}
\item[(i)]  There exists $k_1>0$, such that for all $t,s \in (0,T]$
$$
\|g(t)-g(s)\|_B \leq k_1 |t-s|.
$$
\item[(ii)] There exists  $k_2>0$, such that for all $t,s \in (0,T]$
and $\psi_1,\psi_2 \in \mathcal{C}_0$,
$$
\|k(t,\psi_1(t))-k(s,\psi_2(s))\|_B \leq
k_2[|t-s|+\|\psi_1-\psi_2\|_0].
$$
\item[(iii)] Also there exist $M>0$,
such that
$$
\|a(t)\|_B \leq M, \quad t \in [-T,0].
$$
\end{itemize}
Now we define $G :(0,T]\times \mathcal{C}_0 \to B(0,1)$ by
$$
G(t,\psi)=g(t)+\int_{-t}^{0}a(-\eta)k(t+\eta,\psi)d\eta.
$$
Thus \eqref{eq24} reduces to
\begin{equation}
\begin{gathered}
\frac{\partial v}{\partial t}- \frac{\partial^2 v}{\partial
x^2}-\lambda \frac{\partial^3 v}{\partial x^2 \partial t}
= G(t,v_t) \quad \text{on } (0,T],  \\
v(t)= \phi (t), \quad t \in [-T,0].
\end{gathered}\label{eq25}
\end{equation}
Now we show that $G$ satisfies assumption (H1). For this
take $t,s \in (0,T]$ and $\psi_1, \psi_2 \in \mathcal{C}_0$
\begin{align*}
&\|G(t,\psi_1)-G(s,\psi_2)\|_B\\
&\leq \|g(t)-g(s)\|_B
+  \big\|\int_{-t}^{0}a(-\eta)k(t+\eta,\psi_1(\eta))d\eta
-\int_{-s}^{0}a(-\eta)k(s+\eta,\psi_2(\eta))d\eta \big \|_B.
\end{align*}
Using the given conditions on $g,a$ and $k$, we obtain
\begin{align*}
\|G(t,\psi_1)-G(s,\psi_2)\|_B
&\leq k_1 |t-s|+Mk_2\int_{-t}^{0} \{|t-s|+\|\psi_1-\psi_2\|_0\}d\eta  \\
&\quad  +M\int_{-s}^{-t}\|k(s+\eta,\psi_2(\eta))\|_Bd\eta.
\end{align*}
After some simplifications, we obtain
\begin{align*}
\|G(t,\psi_1)-G(s,\psi_2)\|_B
&\leq  (k_1+Mk_2T+MK)|t-s|+Mk_2T \|\psi_1-\psi_2\|_0  \\
&\leq  L [|t-s|+\|\psi_1-\psi_2\|_0],
\end{align*}
where
$$
L = \max\{(k_1+Mk_2T+MK),Mk_2T\},\quad
\|k(s+\eta,\psi_2(\eta))\|_B \leq K.
$$
As $G$ satisfies a Lipschitz like condition, we  apply the result
of Theorem \ref{t1} to ensure the existence and uniqueness of a
strong solution of \eqref{eq21}-\eqref{eq23} .

\subsection*{Acknowledgements}
The authors want to thank the anonymous referees for their valuable suggestions.
The first author acknowledges the sponsorship from CSIR, India, under
research grant 09/092 (0652) /2008-EMR-1.
The second author acknowledges the
financial help from the Department of Science and Technology, New
Delhi, under  research project SR/S4/MS:796/12.


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