1. Introduction
The Gronwall type integral inequalities provide a necessary tool for the study of the theory of differential equations, integral equations and inequalities of the various types (please, see Gronwall [1] and Guiliano [2] ). Some applications of this result to the study of stability of the solution of linear and nonlinear differential equations may be found in Bellman [3] . Numerous applications to the existence and uniqueness theory of differential equations may be found in Nemyckii-Stepanov [4] , Bihari [5] , and Langenhop [6] . During the past few years several authors (see references below and some of the references cited therein) have established several Gronwall type integral inequalities in one or two independent real variables [1] -[14] . Of course, such results have application in the theory of partial differential equations and Volterra integral equations.
In [14] , Pachpatte investigated the following inequality:
Lemma: Let
be nonnegative continuous functions defined on
and
be nonnegative constant. Let
be a nonnegative continuous function defined for
, and monotonic nondecreasing with respect to
for ant fixed
. If

For
, then

where
is the maximal solution of

For
.
2. Main Results
Theorem 2.1: Let
and
be nonnegative continuous functions defined on
. Let
be a positive continuous and nondecreasing functions in both variables and defined for
. If
(2.1)
Then

where
and
, (2.2)
and
is the maximal solution of
(2.3)
Proof: Define a function
by the right-hand side of (2.1).Then
(2.4)
By using (2.4) in (2.1), we get
(2.5)
Since
is a positive continuous and nondecreasing function, then
(2.6)
Let
(2.7)
From (2.6) and (2.7), we observe that
(2.8)
And
(2.9)
Differentiating both sides of (2.8) with respect to
and
, we get

(2.10)
By keeping first
fixed in (2.10) and set
and integrate from 0 to
then again keeping
fixed, set
and integrate from 0 to
respectively and using (2.9), we get
(2.11)
From (2.7) and (2.11), it is clear that

From (2.4), it can be restated as

where
is the maximal solution of

This completes the proof.
Theorem 2.2: Let
be defined as in Theorem 2.1. If
(2.12)
Then

where
and
, (2.13)
where
is the maximal solution of

Proof: Define a function
by the right-hand side of (2.12).Then
(2.14)
By using (2.14) in (2.12), we get
(2.15)
By following the same steps of Theorem 2.1 from (2.5)-(2.11), we get
(2.16)
From (2.7), (2.14) and (2.16), we observe that

where
is the maximal solution of

This completes the proof.
Theorem 2.3: Let
be defined as in Theorem 2.1. If
(2.17)
Then

where
and
, (2.18)
where
is the maximal solution of

Proof: Define a function
by the right-hand side of (2.17).Then
(2.19)
where 
By using (2.19) in the above equation, we get
(2.20)
and
(2.21)
Since
is a positive continuous and nondecreasing function, then from (2.20),
(2.22)
Let
(2.23)
where
(2.24)
And
(2.25)
Differentiating both sides of (2.24) with respect to
and
, and from (2.23), we get

(2.26)
By keeping first x fixed in (2.26) and set
and integrate from 0 to y then again keeping y fixed, set
and integrate from 0 to x respectively and using (2.25), we get
(2.27)
From (2.19), (2.23) and (2.27), it is clear that

where
is the maximal solution of

This completes the proof.
Application: As an application, let us consider the bound on the solution of a nonlinear hyperbolic partial differential equation of the form
(2.28)
with the given boundary conditions
(2.29)
where
and
such that
(2.30)
(2.31)
where
and
be nonnegative continuous functions defined on a domain D
. The Equation (2.28) with (2.29) is equivalent to the integral equation
(2.32)
Let
be any solution of (2.28) with (2.29) and taking absolute values of both sides, we get
(2.33)
Using (2.30)-(2.32) in (2.33) and assuming that
, where
be a positive continuous and nondecreasing function defined in the respective domain, we have

Then 
where
, and
is the maximal solution of
The remaining proof will be the same as the proof of Theorem 2.1 with suitable modifications.
We note that Theorem 2.1 can be used to study the stability, boundedness and continuous dependence of the solutions of (2.28).