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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 176, pp. 1--14.\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 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/176\hfil Integral inequalities]
{Integral inequalities similar to Gronwall inequality}

\author[M. Denche, H. Khellaf\hfil EJDE-2007/176\hfilneg]
{Mohamed Denche, Hassane Khellaf}  % in alphabetical order

\address{Mohamed Denche \newline
University of Mentouri, Faculty of Science, Department of
Mathematics, Constantine 25000, Algeria}
\email{denech@wissal.dz}

\address{Hassane Khellaf \newline
University of  Mentouri, Faculty of Science, Department of Mathematics,
Constantine 25000, Algeria}
\email{khellafhassane@yahoo.fr}

\thanks{Submitted November 15, 2007. Published December 12, 2007.}
\subjclass[2000]{26D10, 26D20, 26D15}
\keywords{Integral inequality;  subadditive and submultiplicative function;
\hfill\break\indent nonlinear partial differential equation}

\begin{abstract}
 In the present paper, we establish some nonlinear integral inequalities
 for functions of one variable, with a further generalization functions
 with $n$ independent variables. We apply our results to a system of
 nonlinear differential equations for functions of one variable and
 to the nonlinear hyperbolic partial integrodifferential equation
 in $n$-independent variables. These results extend the Gronwall type
 inequalities obtained by  Pachpatte \cite{p1} and  Oguntuase \cite{o1}.
\end{abstract}

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

\section{Introduction}

Integral inequalities play a big role in the study of differential
integral equation and partial differential equations. They were introduced
for  by Gronwall in 1919 \cite{gr}, who gave their
applications in the study of some problems concerning ordinary differential
equation. One of the most useful inequalities with one variable of Gronwall
type is stated as follows.

\begin{lemma} \label{l1}
Let $u,\Psi $ and $g$ be real continuous functions
defined in $[a,b]$, $g(t)\geq 0$ for $t\in [a,b]$.
Suppose that on $[a,b]$ we have the inequality
\begin{equation} \label{e1.1}
u(t)\leq \Psi (t)+\int_{a}^{t}g(s)u(s)ds.
\end{equation}
Then
\begin{equation} \label{e1.2}
u(t)\leq \Psi (t)+\int_{a}^{t}g(s)\Psi (s)\exp\big[\int_{a}^{s}g(\sigma
)d\sigma \big]ds.
\end{equation}
\end{lemma}

Since that time, the theory of these inequalities knew a fast growth and a
great number of monographs were devoted to this subject \cite{b1,gr2,kh,y1}.
The applications of the integral inequalities were developed in a remarkable
way in the study of the existence, the uniqueness, the comparison, the
stability and continuous dependence of the solution in respect to data.
In the last few years, a series of generalizations of those inequalities
appeared. Among these generalization, we can quote Pachpatte's work \cite{p1}.

In the present paper we establish some new nonlinear integral inequalities
for functions of one variable, with a further generalization of these
inequalities to  function with $n$ independent variables.
These results extend the Gronwall type inequalities obtained by
Pachpatte \cite{p1} and  Oguntuase \cite{o1}.

\section{Mains Results}

Our main results are given in the following theorems:

\begin{theorem} \label{tb10}
Let $u(t)$, $f(t)$ be nonnegative
continuous functions in a real interval $I=[a,b]$. Suppose
that $k(t,s)$ and its partial derivatives $k_{t}(t,s)$ exist and
are nonnegative continuous functions for almost
every $t,s\in I$. Let $\Phi (u(t))$ be real-valued, positive, continuous,
strictly non-decreasing, subadditive, and submultiplicative function for
$u(t)\geq 0$ and let $W(u(t))$ be real-valued, positive, continuous, and
non-decreasing function defined for $t\in I$. Assume that $a(t)$ is a
positive continuous function and nondecreasing for $t\in I$. If
\begin{equation} \label{3.74}
u(t)\leq a(t)+\int_{a}^{t}f(s)u(s)
ds+\int_{a}^{t}f(s)W(\int_{a}^{s}k(s,\tau )
\Phi (u(\tau ))d\tau )ds,
\end{equation}
for $a \leq \tau \leq s\leq t\leq b$,
then for $a\leq t\leq t_{1}$,
\begin{equation}\label{3.75}
\begin{aligned}
u(t) &\leq p(t)\Big\{ a(t)+\int_{a}^{t}f(s)\Psi ^{-1}\Big(
\Psi (\zeta )    \\
&\quad +\int_{a}^{s}k(s,\tau )\Phi (p(\tau ))\Phi
(\int_{a}^{\tau }f(\sigma )d\sigma )d\tau \Big)ds\Big\} ,
\end{aligned}
\end{equation}
where
\begin{gather}
p(t) =1+\int_{a}^{t}f(s)\exp (\int_{a}^{s}f(\sigma )d\sigma )
ds,  \label{3.75'} \\
\zeta  =\int_{a}^{b}k(b,s)\Phi (p(s)a(s))ds,
\label{3.76'} \\
\Psi (x) =\int_{x_{0}}^{x}\frac{ds}{\Phi (W(s))},\;\;\;x\geq x_{0}>0.
\label{3.77'}
\end{gather}
Here $\Psi ^{-1}$ is the inverse of $\Psi $ and $t_{1}$is chosen so that
\[
\Psi (\zeta )+\int_{a}^{s}k(s,\tau )\Phi (p(\tau ))\Phi
(\int_{a}^{\tau }f(\sigma )d\sigma )d\tau \in \mathop{\rm Dom}(\Psi ^{-1}).
\]
\end{theorem}

\begin{proof}
Define a function $z(t)$ by
\begin{equation}
z(t)=a(t)+\int_{a}^{t}f(s)W\Big(\int_{a}^{s}k(s,\tau
)\Phi (u(\tau ))d\tau \Big)ds,  \label{3.76}
\end{equation}
then \eqref{3.76} can be restated as
\begin{equation}
u(t)\leq z(t)+\int_{a}^{t}f(s)u(s)ds.  \label{3.77}
\end{equation}
Clearly $z(t)$ is nonnegative and continuous in $t\in I$, using
lemma \ref{l1} to \eqref{3.77}, we get
\begin{equation}
u(t)\leq z(t)+\int_{a}^{t}f(s)z(s)\exp \Big(\int_{a}^{s}f(\sigma )d\sigma
\Big)ds;  \label{3.78}
\end{equation}
moreover if $z(t)$ is nondecreasing in $t\in I$, we obtain
\begin{equation}
u(t)\leq z(t)p(t),  \label{3.79}
\end{equation}
where $p(t)$ is defined by \ref{3.75'}.
From \eqref{3.76}, we have
\begin{equation}
z(t)\leq a(t)+\int_{a}^{t}f(s)W(v(s))ds,  \label{3.80}
\end{equation}
where
\begin{equation}
v(t)=\int_{a}^{t}k(t,s)\Phi (u(s))ds.  \label{3.81}
\end{equation}
 From (\ref{3.79}) we observe that
\begin{equation} \label{3.82}
\begin{aligned}
v(t) &\leq \int_{a}^{t}k(t,s)\Phi \Big[p(s)
\Big(a(s)+\int_{a}^{s}f(\tau )W(v(\tau ))d\tau \Big)\Big]ds   \\
&\leq \int_{a}^{t}k(t,s)\Phi (p(s)
a(s))ds+\int_{a}^{t}k(t,s)\Phi \Big(p(s)
\int_{a}^{s}f(\tau )W(v(\tau ))d\tau \Big)ds   \\
&\leq \int_{a}^{b}k(b,s)\Phi (p(s)
a(s))ds+\int_{a}^{t}k(t,s)\Phi \Big(p(s)
\int_{a}^{s}f(\tau )d\tau \Big)\Phi (W(v(s)))ds \\
&\leq \zeta +\int_{a}^{t}k(t,s)\Phi \Big(p(s)
\int_{a}^{s}f(\tau )d\tau \Big)\Phi (W(v(s)))ds.
\end{aligned}
\end{equation}
Where $\zeta $ is defined by (\ref{3.76'}).

Since $\Phi $ is a subadditive and a submultiplicative function, $W$ and
$v(t)$ are nondecreasing. Define $r(t)$ as the right side of (\ref{3.82}),
then $r(a)=\zeta $ and $v(t)\leq r(t)$, $r(t)$ is positive nondecreasing in
$t\in I$ and
\begin{equation}
\begin{aligned}
r'(t)
&=k(t,t)\Phi \Big(p(t)\int_{a}^{t}f(\tau )d\tau \Big)\Phi (W(v(t))) \\
&\quad +\int_{a}^{t}k_{t}(t,s)\Phi (p(s)
\int_{a}^{s}f(\tau )d\tau )\Phi (W(v(s)))ds,   \\
&\leq \Phi (W(r(t)))\Big[k(t,t)\Phi \Big(
p(t)\int_{a}^{t}f(\tau )d\tau \Big)   \\
&\quad +\int_{a}^{t}k_{t}(t,s)\Phi \Big(p(s)
\int_{a}^{s}f(\tau )d\tau \Big)ds\Big],
\end{aligned} \label{3.84}
 \end{equation}
dividing both sides of (\ref{3.84}) by $\Phi (W(r(t)))$ we obtain
\begin{equation}
\frac{r'(t)}{\Phi (W(r(t)))}\leq \Big[\int_{a}^{t}k(
t,s)\Phi (p(s)\int_{a}^{s}f(\tau )d\tau )ds\Big]'\,.  \label{3.85}
\end{equation}
Note that for
\[
\Psi (x)=\int_{x_{0}}^{x}\frac{ds}{\Phi (W(s))},\quad
x\geq x_{0}>0,
\]
it follows that
\begin{equation}
[\Psi (r(t))]'=\frac{r'(t)}{\Phi (W(r(t)))}\text{.}
\label{m}
\end{equation}
From (\ref{m}) and (\ref{3.85}), we have
\begin{equation}
[\Psi (r(t))]'\leq \Big[\int_{a}^{t}k(t,s)
\Phi (p(s)\int_{a}^{s}f(\tau )d\tau )ds\Big]',
\label{3.86}
\end{equation}
integrate (\ref{3.86}) from $a$ to $t$, leads to
\[
\Psi (r(t))\leq \Psi (\zeta )+\int_{a}^{t}k(t,s)\Phi
(p(s)\int_{a}^{s}f(\tau )d\tau )ds,
\]
then
\begin{equation}
r(t)\leq \Psi ^{-1}\Big(\Psi (\zeta )+\int_{a}^{t}k(t,s)\Phi
(p(s))\Phi (\int_{a}^{s}f(\tau )d\tau )ds\Big).  \label{3.87}
\end{equation}
By (\ref{3.87}), (\ref{3.82}), \eqref{3.80} and (\ref{3.79}) we have the
desired result
\end{proof}

The preceding theorem is a generalization of the result obtained by
Pachpatte in \cite[Theorem 2.1]{p1}.


\begin{theorem} \label{tb11}
Let $u(t),f(t),b(t)$, $h(t)$ be nonnegative continuous functions in
a real interval $I=[a,b]$. Suppose that
$h(t)\in C^{1}(I,\mathbb{R}^{+})$ is
nondecreasing. Let $\Phi (u(t)),W(u(t))$ and $a(t)$ be as defined in Theorem
\ref{tb10}. If
\[
u(t)\leq a(t)+\int_{a}^{t}f(s)u(s)
ds+\int_{a}^{t}f(s)h(s)W\Big(\int_{a}^{s}b(\tau )
\Phi (u(\tau ))d\tau \Big)ds,
\]
for $a\leq \tau \leq s\leq t\leq b$,
then for $a\leq t\leq t_{2}$,
\begin{align*}
u(t)&\leq p(t)\Big\{ a(t)+\int_{a}^{t}f(s)h(s)\Psi ^{-1}\Big(
\Psi (\vartheta ) \\
&\quad  +\int_{a}^{s}b(\tau )\Phi \Big(p(\tau )
\int_{a}^{\tau }f(\sigma )h(\sigma )d\sigma \Big)d\tau
\Big)ds\Big\} .
\end{align*}
Where $p(t)$ is defined by \eqref{3.75'}, $\Psi $ is defined by
\eqref{3.77'} and
\[
\vartheta =\int_{a}^{b}b(s)\Phi (p(s)a(s))ds,
\]
the  $t_{2}$ is chosen so that
$\Psi (\vartheta )+\int_{a}^{s}b(\tau )\Phi
(p(\tau )\int_{a}^{\tau }f(\sigma )h(\sigma
)d\sigma )d\tau $ is in $ \mathop{\rm Dom}(\Psi ^{-1})$.
\end{theorem}

The proof of the above theorem follows similar arguments as the proof of
Theorem \ref{tb10}; So we omit it.

The preceding theorem is a generalization of the result obtained by
Oguntuase in \cite[Theorem 2.3, 2.9]{o1}.

In this section we use the following class of function. A
function $g:\mathbb{R}_{+}\to \mathbb{R}_{+}$ is said to belong to the
 class $S$ if it satisfies the following conditions,
\begin{enumerate}
\item  $g(u)$ is positive, nondecreasing and continuous for $u\geq 0$ and

\item  $(1/v)g(u)\leq g(u/v)$, $u>0$, $v\geq 1$.
\end{enumerate}

\begin{theorem}\label{tb12}
Let $u(t)$, $f(t),a(t),k(t,s),\Phi $ and $W$ be as defined in
Theorem \ref{tb10}, let $g\in S$. If
\begin{equation} \label{2.1}
u(t)\leq a(t)+\int_{a}^{t}f(s)g(u(s)
)ds+\int_{a}^{t}f(s)W\Big(\int_{a}^{s}k(s,\tau )
\Phi (u(\tau ))d\tau \Big)ds,
\end{equation}
for $a \leq \tau \leq s\leq t\leq b$,
 then for $a\leq t\leq t_{3}$,
\begin{equation}
\begin{aligned}
u(t) &\leq \overline{p}(t)\Big\{ a(t)+\int_{a}^{t}f(s)\Psi
^{-1}\Big(\Psi (\overline{\zeta }) \\
&\quad  +\int_{a}^{s}k(s,\tau )\Phi (\overline{p}(\tau
))\Phi (\int_{a}^{\tau }f(\sigma )d\sigma )d\tau \Big)ds\Big\},
\end{aligned} \label{2.1'}
\end{equation}
where
\begin{gather}
\overline{p}(t) =\Omega ^{-1}\Big(\Omega (1)+\int_{a}^{t}f(s)ds\Big),
\label{2.2} \\
\overline{\zeta } =\int_{a}^{b}k(b,s)\Phi (\overline{p}(
s)a(s))ds,  \label{2.3} \\
\Omega (\delta ) =\int_{\varepsilon }^{\delta }\frac{ds}{g(s)},\quad
\delta \geq \varepsilon >0.  \label{2.4}
\end{gather}
Here $\Omega ^{-1}$ is the inverse function of $\Omega $, and
$\Psi ,\Psi ^{-1}$ are defined in theorem \ref{tb10}, $t_{3}$
is chosen so that $\Omega (1)+\int_{a}^{t}f(s)ds$ is in the domain
of $\Omega ^{-1}$, and
\[
\Psi (\overline{\zeta })+\int_{a}^{s}k(s,\tau )
\Phi (\overline{p}(\tau ))\Phi \big(\int_{a}^{\tau }f(\sigma )d\sigma \big)d\tau ,
\]
is in the domain of $\Psi ^{-1}$.
\end{theorem}

\begin{proof}
Define the function
\begin{equation}
z(t)=a(t)+\int_{a}^{t}f(s)W\Big(\int_{a}^{s}k(s,\tau
)\Phi (u(\tau ))d\tau \Big)ds.  \label{2.5'}
\end{equation}
Then \eqref{2.1} can be restated as
\begin{equation}
u(t)\leq z(t)+\int_{a}^{t}f(s)g(u(s))ds.\label{2.5}
\end{equation}
When $z(x)$ is a positive, continuous, nondecreasing in $x\in I$ and
$g\in S$, then it can be restated as
\begin{equation}
\frac{u(t)}{z(t)}\leq 1+\int_{a}^{t}f(s)g(\frac{u(
s)}{z(s)})ds.  \label{2.6}
\end{equation}
The inequality (\ref{2.6}) may be treated as one-dimensional Bihari-La Salle
inequality (see \cite{b1}), which implies
\begin{equation}
u(t)\leq \overline{p}(t)z(t),  \label{2.7}
\end{equation}
where $\overline{p}(t)$ is defined by (\ref{2.2}). By (\ref{2.5'})
and (\ref{2.7}) we get
\[
u(t)\leq \overline{p}(t)\Big[a(t)+\int_{a}^{t}f(s)
W(v(s))ds\Big],
\]
where
\[
v(s)=\int_{a}^{s}k(s,\tau )\Phi (u(\tau ))d\tau .
\]
Now, by following the  argument as in the proof of Theorem \ref{tb10},
we obtain the desired inequality in (\ref{2.1'}).
\end{proof}

\begin{theorem} \label{tb13}
Let $u(t)$, $f(t)$, $b(t)$, $h(t)$, $\Phi (u(t))$, $W(u(t))$
and $a(t)$ be as defined in Theorem \ref{tb11}, let $g\in S$.
If
\[
u(t)\leq a(t)+\int_{a}^{t}f(s)g(u(s)
)ds+\int_{a}^{t}f(s)h(s)W(\int_{a}^{s}b(\tau
)\Phi (u(\tau ))d\tau )ds,
\]
for $a \leq \tau \leq s\leq t\leq b$,
 then for $a\leq t\leq t_{4}$,
\begin{align*}
u(t) &\leq \overline{p}(t)\Big\{ a(t)+\int_{a}^{t}f(s)
h(s)\Psi ^{-1}\Big(\Psi (\overline{\vartheta })   \\
&\quad +\int_{a}^{s}b(\tau )\Phi (\overline{p}(\tau )
\int_{a}^{\tau }f(\sigma )h(\sigma )d\sigma )d\tau \Big)
ds\Big\}.
\end{align*}
Here $\overline{p}(t)$ is defined by \eqref{2.2}, $\Psi $ is defined by
\eqref{3.77'} and
\[
\overline{\vartheta }=\int_{a}^{b}b(s)\Phi (\overline{p}(s)
a(s))ds,
\]
the value $t_{4}$ is chosen so that
$\Psi (\overline{\vartheta })+\int_{a}^{s}b(\tau
)\Phi (\overline{p}(\tau )\int_{a}^{\tau }f(\sigma
)h(\sigma )d\sigma )d\tau $ $\in \mathop{\rm Dom}(\Psi ^{-1})$.
\end{theorem}

The proof of the above theorem follows similar arguments as in the proof
 of Theorem \ref{tb12}, we omit it.

\section{Integral Inequalities in several variables}

In what follows we denote by $\mathbb{R}$ the set of real numbers,
and $\mathbb{R}_{+}=[0,\infty )$. All the functions
which appear in the inequalities are assumed to be real valued of $n$
variables which are nonnegative and continuous. All integrals are assumed
to exist on their domains of definitions.

Throughout this paper, we assume that $\mathbb{I}=[a;b]$ in any bounded open
set in the dimensional Euclidean space $\mathbb{R}^{n}$ and that our
integrals are on $\mathbb{R}^{n}(n\geq 1)$, where
$a=(a_{1},a_{2},\dots ,a_{n})$,
$b=(b_{1},b_{2},\dots ,b_{n})\in \mathbb{R}_{+}^{n}$.
For $x=(x_{1},x_{2},\dots x_{n})$,
$t=(t_{1},t_{2},\dots t_{n})\in \mathbb{I}$, we
shall denote
\[
\int_{a}^{x} = \int_{a_{1}}^{x_{1}}\int_{a_{2}}^{x_{2}}\dots
\int_{a_{n}}^{x_{n}}\dots dt_{n}\dots dt_{1}\,.
\]
Furthermore, for $x,t\in \mathbb{R}^{n}$, we shall write $t\leq x$
whenever $t_{i}\leq x_{i}$, $i=1,2,\dots ,n$
and $0\leq a\leq x\leq b$, for $x\in \mathbb{I}$,
and $D=D_{1}D_{2}\dots D_{n}$, where
$D_{i}=\frac{\partial }{\partial x_{i}}$ for  $i=1,2,\dots ,n$.
Let $C(\mathbb{I},\mathbb{R}_{+})$ denote the class of continuous functions
from $\mathbb{I}$ to $\mathbb{R}_{+}$.

The following theorem deals with $n$-independent variables versions of the
inequalities established in Pachpatte \cite[Theorem 2.3]{p1}.

\begin{theorem}\label{t2}
Let $u(x),f(x),a(x)$ be in $C(\mathbb{I},\mathbb{R}_{+})$ and let $K(x,t)$,
$D_{i}k(x,t)$ be in $C(\mathbb{I\times I},\mathbb{R}_{+})$ for all
$i=1,2,\dots ,n$, and let $c$ be a nonnegative constant.
(1) If
\begin{equation}
u(x)\leq c+\int_{a}^{x}f(s)\Big[u(s)
+\int_{a}^{s}k(s,\tau )u(\tau )d\tau \Big]ds,
\label{3.1}
\end{equation}
for $x\in \mathbb{I}$ and $a\leq \tau \leq s\leq b$, then
\begin{equation}
u(x)\leq c\Big[1+\int_{a}^{x}f(t)\exp \Big(
\int_{a}^{t}(f(s)+k(b,s))ds\Big)dt\Big] \label{3.2}
\end{equation}
(2) If
\begin{equation}
u(x)\leq a(x)+\int_{a}^{x}f(s)\Big[u(
s)+\int_{a}^{s}k(s,\tau )u(\tau )d\tau
\Big]ds,  \label{3.3}
\end{equation}
for $x\in \mathbb{I}$ and $a\leq \tau \leq s\leq b$, then
\begin{equation}
u(x)\leq a(x)+e(x)\Big[1+\int_{a}^{x}f(t)\exp \Big(
\int_{a}^{t}(f(s)+k(b,s))ds\Big)dt\Big],  \label{3.4}
\end{equation}
where
\begin{equation}
e(x)=\int_{a}^{x}f(s)\Big[a(s)+\int_{a}^{s}k(s,\tau
)a(\tau )d\tau \Big]ds.  \label{3.5}
\end{equation}
\end{theorem}

\begin{proof} (1)
The inequality (\ref{3.1}) implies the estimate
\[
u(x)\leq c+\int_{a}^{x}f(s)\Big[[u(s)
+\int_{a}^{s}k(b,\tau )u(\tau )d\tau \Big]ds.
\]
We define the function
\[
z(x)=c+\int_{a}^{x}f(s)\Big[u(s)
+\int_{a}^{s}k(b,\tau )u(\tau )d\tau \Big]ds.
\]
Then $z(a_{1},x_{2},\dots ,x_{n})=c$, $u(x)\leq z(x)$ and
\begin{align*}
Dz(x) &=f(x)\Big[u(x)+\int_{a}^{x}k(b,s)u(s)ds\Big], \\
&\leq f(x)\Big[z(x)+\int_{x^{0}}^{x}k(b,s)z(s)ds\Big].
\end{align*}
Define the function
\[
v(x)=z(x)+\int_{a}^{x}k(b,s)z(s)ds,
\]
then $z(a_{1},x_{2},\dots ,x_{n})=v(a_{1},x_{2},\dots ,x_{n})=c$,
$Dz(x)\leq f(x)v(x)$ and $z(x)\leq v(x)$, and we have
\begin{equation}
Dv(x) =Dz(x)+k(b,x)z(x)\leq (f(x)+k(b,x))v(x).  \label{3.10}
 \end{equation}
Clearly $v(x)$ is positive for all $x\in \mathbb{I}$, hence the inequality
(\ref{3.10}) implies the estimate
\[
\frac{v(x)Dv(x)}{v^{2}(x)}\leq f(x)+k(b,x);
\]
that is
\[
\frac{v(x)Dv(x)}{v^{2}(x)}\leq f(x)+k(b,x)
+\frac{(D_{n}v(x))(D_{1}D_{2}\dots D_{n-1}v(x))}{v^{2}(x)};
\]
hence
\[
D_{n}\Big(\frac{D_{1}D_{2}\dots D_{n-1}v(x)}{v(x)}\Big)\leq f(x)+k(b,x).
\]
Integrating with respect to $x_{n}$ from $a_{n}$ to $x_{n}$, we have
\[
\frac{D_{1}D_{2}\dots D_{n-1}v(x)}{v(x)}\leq \int_{a_{n}}^{x_{n}}[
f(x_{1},\dots ,x_{n-1},t_{n})+k(b,x_{1},\dots ,x_{n-1},t_{n})]dt_{n};
\]
thus
\begin{align*}
\frac{v(x)D_{1}D_{2}\dots D_{n-1}v(x)}{v^{2}(x)}
&\leq \int_{a_{n}}^{x_{n}}[
f(x_{1},\dots ,x_{n-1},t_{n})+k(b,x_{1},\dots ,x_{n-1},t_{n})]dt_{n} \\
&\quad +\frac{(D_{n-1}v(x))(D_{1}D_{2}\dots D_{n-2}v(x))}{v^{2}(x)}.
\end{align*}
That is,
\[
D_{n-1}\big(\frac{D_{1}D_{2}\dots D_{n-2}v(x)}{v(x)}\big)
\leq \int_{a_{n}}^{x_{n}}[
f(x_{1},\dots ,x_{n-1},t_{n})+k(b,x_{1},\dots ,x_{n-1},t_{n})]dt_{n},
\]
integrating with respect to $x_{n-1}$ from $a_{n-1}$ to $x_{n-1}$, we have
\begin{align*}
\frac{D_{1}D_{2}\dots D_{n-2}v(x)}{v(x)}
&\leq \int_{a_{n-1}}^{x_{n-1}}\int_{a_{n}}^{x_{n}}\big[
f(x_{1},\dots ,x_{n-2},t_{n-1},t_{n})\\
&\quad +k(b,x_{1},\dots ,x_{n-2},t_{n-1},t_{n})
\big]dt_{n}dt_{n-1}.
\end{align*}
Continuing this process, we obtain
\[
\frac{D_{1}v(x)}{v(x)}\leq \int_{a_{2}}^{x_{2}}\dots \int_{a_{n}}^{x_{n}}[
f(x_{1},t_{2},t_{3},\dots ,t_{n})+k(b,x_{1},t_{2},t_{3},\dots ,t_{n})]
dt_{n}\dots dt_{2}\,.
\]
Integrating with respect to $x_{1}$ from $a_{1}$ to $x_{1}$, we have
\[
\log \frac{v(x)}{v(a_{1},x_{2},\dots ,x_{n})}\leq \int_{a}^{x}[
f(t)+k(b,t)]dt;
\]
that is,
\begin{equation}
v(x)\leq c\exp \Big(\int_{a}^{x}[f(t)+k(b,t)]dt\Big).
\label{3.11}
\end{equation}
Substituting (\ref{3.11}) into $Dz(x)\leq f(x)v(x)$, we have
\begin{equation}
Dz(x)\leq cf(x)\exp \Big(\int_{a}^{x}[f(t)+k(b,t)]dt\Big),
\label{3.12}
\end{equation}
integrating (\ref{3.12}) with respect to the $x_{n}$ component from $a_{n}$
to $x_{n}$, then with respect to the $a_{n-1}$ to $x_{n-1}$, and continuing
until finally $a_{1}$ to $x_{1}$, and noting that
$z(a_{1},x_{2},\dots ,x_{n})=c $, we have
\[
z(x)\leq c\Big[1+\int_{a}^{x}f(t)\exp \Big(\int_{a}^{t}[
f(s)+k(b,s)]ds\Big)dt\Big].
\]
This completes the proof of the first part.

(2) Define a function $z(x)$ by
\begin{equation}
z(x)=\int_{a}^{x}f(s)\Big[u(s)
+\int_{a}^{s}k(s,\tau )u(\tau )d\tau \Big]ds.
\label{3.13}
\end{equation}
Then from (\ref{3.3}), $u(x)\leq a(x)+z(x)$ and using this in (\ref{3.13}),
we get
\begin{equation}
\begin{aligned}
z(x) &\leq \int_{a}^{x}f(s)\Big[a(s)
+z(s)+\int_{a}^{s}k(s,\tau )[a(\tau )+z(\tau
)]d\tau \Big]ds,   \\
&\leq e(x)+\int_{a}^{x}f(s)\Big[z(s)+\int_{a}^{s}k(
s,\tau )z(\tau )d\tau \Big]ds,
\end{aligned} \label{3.14}
\end{equation}
where $e(x)$ is defined by (\ref{3.5}). Clearly $e(x)$ is positive,
continuous an nondecreasing for all $x\in \mathbb{I}$. From (\ref{3.14}) it
is easy to observe that
\[
\frac{z(x)}{e(x)}\leq 1+\int_{a}^{x}f(s)\Big[\frac{z(s)}{e(s)}%
+\int_{a}^{s}k(s,\tau )\frac{z(\tau )}{e(\tau )}%
d\tau \Big]ds.
\]
Now, by applying the inequality in part 1, we
have
\begin{equation}
z(x)\leq e(x)\Big[1+\int_{a}^{x}f(t)\exp \Big(
\int_{a}^{t}(f(s)+k(b,s))ds\Big)dt\Big].  \label{3.15}
\end{equation}
The desired inequality in (\ref{3.4}) follows from (\ref{3.15}) and the fact
that $u(x)\leq a(x)+z(x)$.
\end{proof}

The following theorem deals with $n$-independent variables versions of the
inequalities established in Theorem \ref{tb12}.
We need the inequalities in the following lemma (see \cite{kh}).

\begin{lemma} \label{l2}
Let $u(x)$ and $b(x)$ be nonnegative continuous functions, defined
for $x\in \mathbb{I}$, and let $g\in S$. Assume that $a(x)$ is positive,
continuous function, nondecreasing in each of the variables
$x\in \mathbb{I}$. Suppose that
\begin{equation}
u(x)\leq c+\int_{a}^{x}b(t)g(u(t))dt,  \label{3.16}
\end{equation}
holds for all $x\in \mathbb{I}$ with $x\geq a$, then
\begin{equation}
u(x)\leq G^{-1}\Big[G(c)+\int_{a}^{x}b(t)dt\Big],  \label{3.17}
\end{equation}
for all $x\in \mathbb{I}$ such that $G(c)+\int_{a}^{x}b(t)dt\in
\mathop{\rm Dom}(G^{-1})$,
where $G(u)=\int_{u_{0}}^{u}dz/g(z)$, $u>0(u_{0}>0)$.
\end{lemma}

\begin{theorem}\label{t3}
Let $u(x),f(x),a(x)$ and $k(x,t)$ be as defined in Theorem
\ref{t2}. Let $\Phi (u(x))$ be real-valued, positive, continuous, strictly
non-decreasing, subadditive and submultiplicative function for $u(x)\geq 0$
and let $W(u(x))$ be real-valued, positive, continuous and non-decreasing
function defined for $x\in \mathbb{I}$. Assume that $a(x)$ is positive
continuous function and nondecreasing for $x\in \mathbb{I}$. If
\begin{equation}
u(x)\leq a(x)+\int_{a}^{x}f(t)g(u(t)
)dt+\int_{a}^{x}f(t)W\Big(\int_{a}^{t}k(t,s)\Phi
(u(s))ds\Big)dt, \label{3.18}
\end{equation}
for $a\leq s\leq t\leq x\leq b$,
then for $a\leq x\leq x^{*}$,
\begin{equation}
\begin{aligned}
u(x)&\leq \beta (x)\Big\{ a(x)+\int_{a}^{x}f(t)
 W\Big[\Psi ^{-1}\Big(\Psi (\eta )\\
&\quad +\int_{a}^{t}k(b,s)\Phi [\beta (s)\int_{a}^{s}f(\tau )d\tau ]
ds\Big)\Big]dt\Big\},
\end{aligned}\label{3.19}
\end{equation}
where
\begin{gather}
\beta (x)=G^{-1}(G(1)+\int_{a}^{x}f(s)ds),  \label{3.20} \\
\eta =\int_{a}^{b}k(b,s)\Phi (\beta (s)a(s))ds, \label{3.22} \\
G(u)=\int_{u_{0}}^{u}1/g(z)\,dz,\quad u>0(u_{0}>0),  \label{3.21} \\
\Psi (x)=\int_{x_{0}}^{x}\frac{ds}{\Phi (W(s))},\quad x\geq x_{0}>0.
\label{3.22'}
\end{gather}
Here $G^{-1}$ is the inverse function of $G$, and $\Psi $ is the inverse
function of $\Psi ^{-1}$, $x^{*}$is chosen so that
$G(1)+\int_{a}^{x}f(s)ds $ is in the domain of $G^{-1}$, and
\[
\Psi (\eta )+\int_{a}^{t}k(b,s)\Phi \big[\beta (s)
\int_{a}^{s}f(\tau )d\tau \big]ds,
\]
is in the domain of $\Psi ^{-1}$.
\end{theorem}

\begin{proof}
Define the function
\begin{equation}
z(x)=a(x)+\int_{a}^{x}f(t)W\Big(\int_{a}^{t}k(
t,s)\Phi (u(s))ds\Big)dt.  \label{3.23}
\end{equation}
Then \ref{3.18} can be restated as
\[
u(x)\leq z(x)+\int_{a}^{x}f(t)g(u(t))dt.
\]
We have $z(x)$ is a positive, continuous, nondecreasing in $x\in \mathbb{I}$
and $g\in S$. Then the above inequality can be restated as
\begin{equation}
\frac{u(x)}{z(x)}\leq 1+\int_{a}^{x}f(t)g\big(\frac{u(
t)}{z(t)}\big)dt.  \label{3.24}
\end{equation}
By Lemma \ref{l2} we have
\begin{equation}
u(x)\leq z(x)\beta (x),  \label{3.25}
\end{equation}
where $\beta (x)$ is defined by (\ref{3.20}).
By (\ref{3.23}), we have
\begin{equation}
z(x) = a(x)+\int_{a}^{x}f(t)W(v(t))dt,  \label{3.26}
\end{equation}
where
\begin{equation}
v(x)=\int_{a}^{x}k(x,t)\Phi (u(t))dt.  \label{3.27}
\end{equation}
By (\ref{3.27}) and (\ref{3.25}) , we observe that
\begin{equation}
\begin{aligned}
v(x)&\leq \int_{a}^{x}k(b,t)\Phi \Big[\beta (t)
\Big(a(t)+\int_{a}^{t}f(s)W(v(s))ds\Big)\Big]dt \\
&\leq \int_{a}^{x}k(b,s)\Phi (\beta (s) a(s))ds\\
&\quad +\int_{a}^{t}k(b,s)\Phi (\beta (s)
\int_{a}^{s}f(\tau )W(v(\tau ))d\tau )ds,   \\
&\leq \eta +\int_{a}^{x}k(b,s)\Phi [\beta (s)
\int_{a}^{s}f(\tau )d\tau ]\Phi (W(v(s)))ds.
\end{aligned}\label{3.28}
\end{equation}
Where $\eta $ is defined by (\ref{3.22}). Since $\Phi $ is subadditive and
submultiplicative function, $W$ and $v(x)$ are nondecreasing for all
$x\in \mathbb{I}$. Define $r(x)$ as the right side of (\ref{3.28}), then
$r(a_{1},x_{2},\dots ,x_{n})=\eta $ and $v(x)\leq r(x)$, $r(x)$ is positive and
nondecreasing in each of the variables $x_{1},x_{2},x_{3},\dots x_{n}$ . Hence
\[
\frac{Dr(x)}{\Phi (W(r(x)))}\leq k(b,x)\Phi [\beta (
x)\int_{a}^{x}f(s)ds].
\]
Since
\[
D_{n}\big(\frac{D_{1}\dots D_{n-1}r(x)}{\Phi (W(r(x)))}\big)
=\frac{Dr(x)}{\Phi (W(r(x)))}-\frac{D_{n}\Phi (W(r(x)))D_{1}
\dots D_{n-1}r(x)}{\Phi ^{2}(W(r(x)))},
\]
the above inequality implies
\[
D_{n}\big(\frac{D_{1}\dots D_{n-1}r(x)}{\Phi (W(r(x)))}\big)
\leq \frac{Dr(x)}{\Phi (W(r(x)))},
\]
and
\[
D_{n}\big(\frac{D_{1}\dots D_{n-1}r(x)}{\Phi (W(r(x)))}\big)
\leq k(b,x)\Phi [\theta (x)],
\]
where $\theta (x)=\beta (x)\int_{a}^{x}f(s)ds$.
Integrating with respect to $x_{n}$ from $a_{n}$ to $x_{n}$, we have
\[
\frac{D_{1}\dots D_{n-1}r(x)}{\Phi (W(r(x)))}\leq \int_{a_{n}}^{x_{n}}k(
b,x_{1},x_{2},\dots ,x_{n-1},s_{n})\Phi [\theta
(x_{1},x_{2},\dots ,x_{n-1},s_{n})]ds_{n}.
\]
Repeating this argument, we find that
\begin{align*}
&\frac{D_{1}r(x)}{\Phi (W(r(x)))}\\
&\leq
\int_{a_{2}}^{x_{2}}\dots \int_{a_{n-1}}^{x_{n-1}}\int_{a_{n}}^{x_{n}}k(
b,x_{1},s_{2},\dots ,s_{n})\Phi [\theta
(x_{1},s_{2},\dots ,s_{n})]ds_{n}ds_{n-1}\dots ds_{2}.
\end{align*}
Integrating both sides of the above inequality with respect to $x_{1}$
from $a_{1}$ to $x_{1}$, we have
\[
\Psi (r(x))-\Psi (\eta )\leq \int_{a}^{x}k(b,s)\Phi [\theta (s)]
ds,
\]
and
\[
r(x)\leq \Psi ^{-1}\Big(\Psi (\eta )+\int_{a}^{x}k(b,s)\Phi
\Big[\beta (s)\int_{a}^{s}f(\tau )d\tau \Big]ds\Big).
\]
From this we obtain
\begin{equation}
v(x)\leq r(x)\leq \Psi ^{-1}\Big(\Psi (\eta
)+\int_{a}^{x}k(b,s)\Phi [\beta (s)\int_{a}^{s}f(
\tau )d\tau ]ds\Big).  \label{3.29}
\end{equation}
By (\ref{3.25}), (\ref{3.26}) and (\ref{3.29}) we obtain the desired
inequality in (\ref{3.19}).
\end{proof}

\section{Some applications}

In this section, our results are applied to the qualitative
analysis of two application. The first is the system of nonlinear
differential equations for one variable functions. The second is a
nonlinear hyperbolic partial integrodifferential equation of
$n$-independent variables.

First we consider the system of nonlinear differential equations
\begin{equation}
\frac{du}{dt}=F_{1}\Big(t,u(t),\int_{x_{0}}^{t}K_{1}(t,u(s))ds\Big),
\label{4.1}
\end{equation}
for $t\in I=[t_{0},t_{\infty }]\subset R_{+}$, where
$u\in C(I,\mathbb{R}^{n})$,
$F_{1}\in C(I\times \mathbb{R}^{n}\times \mathbb{R}^{n},
\mathbb{R}^{n})$ and $K_{1}\in C(I\times \mathbb{R}^{n},\mathbb{R}^{n})$.

In what follows, we shall assume that the Cauchy problem
\begin{equation}
\begin{gathered}
\frac{du}{dt}=F_{1}(t,u(t),\int_{t_{0}}^{t}K_{1}(t,u(s))ds),\quad
x\in I, \\
u(t_{0})=u_{0}\in \mathbb{R}^{n},
\end{gathered}  \label{4.2}
\end{equation}
has a unique solution, for every $t_{0}\in I$ and $u_{0}\in \mathbb{R}^{n}$.
We shall denote this solution by $u(.,t_{0},u_{0})$. The following theorem
deals the estimate on the solution of the nonlinear Cauchy problem
(\ref{4.2}).

\begin{theorem}\label{a1}
Assume that the functions $F_{1}$ and $K_{1}$ in \eqref{4.2}
satisfy the conditions
 \begin{gather}
\| K_{1}(t,u)\| \leq h(t)\Phi (\| u\| ),\quad t\in I, \label{4.3} \\
\| F_{1}(t,u,v)\| \leq \| u\| +\| v\|,\quad u,v\in \mathbb{R}^{n},  \label{4.4}
\end{gather}
where $h$ and $\Phi $ are as defined in Theorem \ref{tb11}.
Then we have the estimate, for $t_{0}\leq t\leq t_{2}$,
\begin{equation}
\| u(t,t_{0},u_{0})\| \leq e^{t-t_{0}}\Big(\|
u_{0}\| +\int_{t_{0}}^{t}h(s)E_{1}(s,\| u_{0}\| )ds\Big),
\label{4.5}
\end{equation}
where
\begin{gather}
E_{1}(t,\| u_{0}\| ) = \Psi ^{-1}\Big(\Psi (\vartheta
)+\int_{t_{0}}^{t}\Phi \big(e^{\tau -x_{0}}\int_{t_{0}}^{\tau }h(\sigma
)d\sigma \big)d\tau \Big),  \label{4.6} \\
\Psi (t) =\int_{a}^{t}\frac{ds}{\Phi (s)},\quad t\geq a>0,  \label{4.7} \\
\vartheta =\int_{t_{0}}^{t_{\infty }}\| u_{0}\| \Phi (
e^{s-t_{0}})ds,  \label{4.8}
\end{gather}
and $t_{2}$ is chosen so that $\Psi (\vartheta )+\int_{x_{0}}^{s}\Phi (
e^{\tau -t_{0}}\int_{t_{0}}^{\tau }h(\sigma )d\sigma )d\tau $ is in
$\mathop{\rm Dom}(\Psi ^{-1})$
\end{theorem}

\begin{proof}
Let $t_{0}\in I$, $u_{0}\in \mathbb{R}^{n}$ and $u(.,t_{0},u_{0})$ be the
solution of the Cauchy problem (\ref{4.2}). Then we have
\begin{equation}
u(t,t_{0},u_{0})=u_{0}+\int_{t_{0}}^{t}F_{1}\Big(
s,u(s,t_{0},u_{0}),\int_{t_{0}}^{s}K_{1}(s,u(\tau ,t_{0},u_{0}))d\tau
\Big)ds.  \label{4.9}
\end{equation}
Using (\ref{4.3}) and (\ref{4.4}) in (\ref{4.9}), we have
\begin{equation}
\begin{aligned}
\| u(t,t_{0},u_{0})\|
&\leq \| u_{0}\| +\int_{t_{0}}^{t}f(s)\Big[\| u(s,t_{0},u_{0})\|
+\int_{t_{0}}^{s}\| K_{1}(s,u(\tau ,t_{0},u_{0}))\| d\tau \Big]ds,   \\
&\leq \| u_{0}\| +\int_{t_{0}}^{t}f(s)\Big(\|
u(s,t_{0},u_{0})\| +h(s)\int_{t_{0}}^{s}\Phi (\| u(\tau
,t_{0},u_{0})\| )d\tau \Big)ds.
\end{aligned}  \label{4.10}
 \end{equation}
Now, a suitable application of Theorem \ref{tb11} with
$a(t)=\|u_{0}\|$, $f(t)=b(t)=1$ and $W(u)=u$ to (\ref{4.10})
yields (\ref{4.5}).
\end{proof}

If, in addition, we assume that the function $F_{1}$ satisfies the
general condition
\begin{equation}
\| F_{1}(t,u,v)\| \leq f(t)(g(\| u\|
)+W(\| v\| )),  \label{4.11'}
\end{equation}
where $f$ $,g$ and $W$ are as defined in Theorem \ref{tb13}, we obtain
an estimate for $u(.,t_{0},u_{0})$, and from any particular
conditions of (\ref{4.11'}) and (\ref{4.3}), we can get some useful
results similar to Theorem \ref{a1}.

Secondly, we shall demonstrate the usefulness of the inequality established
in Theorem \ref{t3} by obtaining pointwise bounds on the solutions of a
certain class of nonlinear equation in $n$-independent variables. We
consider the nonlinear hyperbolic partial integrodifferential equation
\begin{equation}
\frac{\partial ^{n}u(x)}{\partial x_{1}\partial x_{2}\dots \partial x_{n}}
=F\Big(x,u(x),\int_{x^{0}}^{x}K(x,s,u(s))ds\Big)
+G(x,u(x)) \label{4.12}
\end{equation}
for all $x\in \mathbb{I}=[x^{0};x^{\infty }]\subset \mathbb{R}_{+}^{n}$,
where $x=(x_{1},x_{2},\dots ,x_{n})$,
$x^{0}=(x_{1}^{0},x_{2}^{0},\dots ,x_{n}^{0})$,
$x^{\infty }=(x_{1}^{\infty},x_{2}^{\infty },\dots ,x_{n}^{\infty })$
are in $\mathbb{R}_{+}^{n}$ and
$u\in C(\mathbb{I},\mathbb{R})$,
$F\in C(\mathbb{I}\times \mathbb{R}\times \mathbb{R}, \mathbb{R})$,
$K\in C(\mathbb{I}\times \mathbb{I\times R},\mathbb{R})$ and
$G\in C(\mathbb{I}\times \mathbb{R},\mathbb{R})$. With suitable boundary
conditions, the solution of (\ref{4.12}) is of the form
\begin{equation}
u(x)=l(x)+\int_{x^{0}}^{x}F\Big(s,u(s),\int_{x^{0}}^{s}K(
s,t,u(t))dt\Big)ds+\int_{x^{0}}^{x}G(s,u(s))ds.
\label{4.13}
\end{equation}

The following theorem gives the bound of the solution of (\ref{4.12}).

\begin{theorem}
Assume that the functions $l,F$, $K$ and $G$ in \eqref{4.12} satisfy the
conditions
\begin{gather}
| K(s,t,u(t))| \leq k(s,t)\Phi (| u(t)|),\quad t,s\in \mathbb{I},\;
 u\in \mathbb{R},  \label{4.14}
\\
| F(t,u,v)| \leq \frac{1}{2}| u| +| v|,\quad u,v\in \mathbb{R},\;
t\in \mathbb{I},  \label{4.15}
\\
| G(s,u)| \leq \frac{1}{2}| u|,\quad s\in \mathbb{I},\; u\in \mathbb{R},
\label{4.16} \\
| l(x)| \leq a(x),\quad x\in \mathbb{I}, \label{4.17}
\end{gather}
where $a,f,k$ and $\Phi $ are as defined in Theorem \ref{tb11}, with
$f(x)=b(x)+e(x)$ for all $x\in \mathbb{I}$ where
$b,e\in C(\mathbb{I},\mathbb{R}_{+})$, then we have the estimate,
 for $x^{0}\leq x\leq x^{*}$,
\begin{equation}
| u(x)| \leq \exp \Big(\prod_{i=1}^{n}(x_{i}-x_{i}^{0})\Big)
\Big(a(x)+\int_{a}^{x}E(t)dt\Big).  \label{4.18}
\end{equation}
Here
\begin{gather}
E(t) =\Psi ^{-1}\Big(\Psi (\eta )+\int_{a}^{t}k(x^{\infty },s)\Phi
\big[\exp \big(\prod_{i=1}^{n}(s_{i}-x_{i}^{0})\big)
\int_{a}^{s}f(\tau )d\tau \big]ds\Big),  \label{4.19} \\
\eta =\int_{x^{0}}^{x^{\infty }}k(x^{\infty },s)\Phi \big(
a(s)\exp \big(\prod_{i=1}^{n}(s_{i}-x_{i}^{0})\big)\big)ds,
\label{4.20}\\
\Psi (x)=\int_{x^{0}}^{x}\frac{ds}{\Phi (s)},\;\;\;x\geq x^{0}>0,
\label{4.21}
\end{gather}
where $x^{*}$is chosen so that $\Psi (\eta )+\int_{a}^{t}k(x^{\infty
},s)\Phi [\exp (\prod_{i=1}^{n}(s_{i}-x_{i}^{0}))
\int_{a}^{s}f(\tau )d\tau ]ds$, is in the domain of $\Psi^{-1}$.
\end{theorem}

\begin{proof}
Using the conditions (\ref{4.14}), (\ref{4.17}) in (\ref{4.13}), we have
\begin{equation}
 \begin{aligned}
| u(x)| &\leq a(x)+\int_{x^{0}}^{x}| G(s,u(s))|
ds+\int_{x^{0}}^{x}f(s)\big[| u(s)| +\int_{x^{0}}^{s}|
K(s,t,u(t))| dt\big]ds,   \\
&\leq a(x)+\int_{x^{0}}^{x}\Big(| u(s)|
+\int_{x^{0}}^{s}k(s,t)\Phi (| u(t)| )dt\Big)ds.
\end{aligned} \label{4.23}
\end{equation}
Now, a suitable application of Theorem \ref{t3} with $f(s)=1$, $g(u)=u$ and
$W(u)=u$ to (\ref{4.23}) yields (\ref{4.18}).
\end{proof}

\subsection*{Remarks}
If we assume that the functions $F$ and $G$ satisfy the general conditions
\begin{gather}
| F(t,u,v)| \leq f(t)(g(| u| )+W(| v| )),  \label{4.24} \\
| G(t,u)| \leq f(t)g(| u| ),\quad\text{for }t\in \mathbb{I},\;
u\in \mathbb{R},  \label{4.25}
\end{gather}
we can obtain an estimation of $u(x)$.

From the particular conditions of (\ref{4.14}), (\ref{4.24}) and (\ref{4.25}),
we can obtain some results similar to Theorem \ref{t3}. To save
space, we omit the details here.

Under some suitable conditions, the uniqueness and continuous dependence of
the solutions of (\ref{4.1}) and (\ref{4.12}) can also be discussed using
our results.


\subsection*{Acknowledgments}
The authors are grateful to the anonymous referee for his/her suggestions
on the original manuscript. The second author would
like to thank Prof. Julio G. Dix for his help,  to my
advisor Prof. M. Denche for his assistance.

\begin{thebibliography}{9}

\bibitem{b1}  D. Bainov and P. Simeonov;
\emph{Integral Inequalities and Applications},
Kluwer Academic Publishers, Dordrecht, (1992).

\bibitem{gr}  T. H. Gronwall;
 \emph{Note on the derivatives with respect to a
parameter of solutions of a system of differential equations},
Ann. of Math. ,0 (1919) 4, pp. 292-296.

\bibitem{gr2}  I. Gyori;
\emph{A Generalization of Bellman's Inequality for
Stieltjes Integrals and a Uniqueness Theorem}, Studia Sci. Math. Hungar, 6
(1971), pp. 137-145.

\bibitem{kh}  H. Khellaf;
\emph{On integral inequalities for functions of several
independent variables}, Electron. J. Diff. Eqns. , Vol. 2003 (2003),
No.123, pp. 1--12.

\bibitem{o1}  J. A. Oguntuase; \emph{On an inequality of Gronwall},
J. Ineq. Pure and Appl. Math., 2(1) (2001), No. 9.

\bibitem{p1}  B. G. Pachpatte;
\emph{Bounds on Certain Integral Inequalities}, J.
Ineq. Pure and Appl. Math., 3(3) (2002), Article No. 47.

\bibitem{y1}  C. C. Yeh and M-H. Shim;
\emph{The Gronwall-Bellman Inequality in Several Variables},
J. Math. Anal. Appl., 86 (1982), No. 1, pp. 157-167.

\end{thebibliography}

\end{document}