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Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2007, Article ID 32949, 11 pages
doi:10.1155/2007/32949
Research Article
Nonlinear Integral Inequalities in Two Independent Variables
and Their Applications
Kelong Zheng, Yu Wu, and Shengfu Deng
Received 10 June 2007; Accepted 27 July 2007
Recommended by Wing-Sum Cheung
This paper generalizes results of Cheung and Ma (2005) to more general inequalities with
more than one distinct nonlinear term. From our results, some results of Cheung and
Ma (2005) can be deduced as some special cases. Our results are also applied to show the
boundedness of the solutions of a partial differential equation.
Copyright © 2007 Kelong Zheng et a l. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1. Introduction
The integral inequalities play a fundamental role in the study of existence, uniqueness,
boundedness, stability, invariant manifolds, and other qualitative properties of solutions
of the theory of differential and integral e quations. There are a lot of papers investigating
them such as [1–8]. In particular, Pachpatte [2] discovered some new integral inequalities
involving functions of two variables. These inequalities are applied to study the bound-
edness and uniqueness of the solutions of the following terminal value problem for the
hyperbolic partial differential equation (1.1) with conditions (1.2):
D
1
D
2
u(x, y) = h
x, y,u(x, y)
+ r(x, y), (1.1)
u(x,
∞) = σ
∞
(x), u(∞, y) = τ
∞
(y), u(∞,∞) =k. (1.2)
Cheung [9], and Dragomir and Kim [10, 11] established additional Gronwall-Ou-Iang
type integral inequalities involving functions of two independent variables. Meng and Li
[12] generalized the results of Pachpatte [2] to certain new integrals. Recently, Cheung
2 Journal of Inequalities and Applications
and Ma[13] discussed the following inequalities
u(x, y) ≤ a(x, y)+c(x, y)
x
0
∞
y
d(s,t)w
u(s,t)
dtds,
u(x, y)
≤ a(x, y)+c(x, y)
∞
x
∞
y
d(s,t)w
u(s,t)
dtds,
(1.3)
where a(x, y)andc(x, y) have certain monotonicity.
Our main a im here, motivated by the work of Cheung and Ma [13], is to discuss more
general integr al inequalities with n nonlinear terms:
u(x, y)
≤ a(x, y)+
n
i=1
x
0
∞
y
d
i
(x, y,s,t)w
i
u(s,t)
dtds, (1.4)
u(x, y)
≤ a(x, y)+
n
i=1
∞
x
∞
y
d
i
(x, y,s,t)w
i
u(s,t)
dtds, (1.5)
where we do not require the monotonicity of a(x, y)andd
i
(x, y,s,t). Furthermore, we
also show that some results of Cheung and Ma [13] can b e deduced from our results as
some special cases. Our results are also applied to show the boundedness of the solutions
of a partial differential equation.
2. Main results
Let
R
=
(−∞,∞)andR
+
= [0,∞). D
1
z(x, y)andD
2
z(x, y) denote the first-order partial
derivatives of z(x, y) with respect to x and y, respectively.
As in [1, 5, 6], we define w
1
∝ w
2
for w
1
,w
2
: A ⊂ R → R\{0} if w
2
/w
1
is nondecreasing
on A. This concept helps us compare monotonicity of different functions. Suppose that
(C
1
) w
i
(u)(i = 1, ,n) is a nonnegative, nondecreasing, and continuous function for
u
∈ R
+
with w
i
(u) > 0foru>0suchthatw
1
∝ w
2
∝···∝w
n
;
(C
2
) a(x, y) is a nonnegative and continuous function for x, y ∈ R
+
;
(C
3
) d
i
(x, y,s,t)(i = 1, ,n) is a continuous and nonnegative function for x, y, s,t ∈
R
+
.
Take the notation W
i
(u):=
u
u
i
(dz/w
i
(z)), for u ≥ u
i
,whereu
i
> 0 is a given constant.
Clearly, W
i
is strictly increasing, so its inverse W
−1
i
is well defined, continuous, and in-
creasing in its corresponding domain.
Theorem 2.1. In addition to the assumpt ions (C
1
), (C
2
), and (C
3
), suppos e that a(x, y)
and d
i
(x, y,s,t) are bounded in y ∈ R
+
for each fixed x,s,t ∈ R
+
.Ifu(x, y) is a continuous
and nonnegative function satisfying (1.4)forx, y
∈ R
+
, then
u(x, y)
≤ W
−1
n
W
n
b
n
(x, y)
+
x
0
∞
y
d
n
(x, y,s,t)dtds
(2.1)
Kelong Zheng et a l. 3
for all 0
≤ x ≤ x
1
, y
1
≤ y<∞,whereb
n
(x, y) is determined recursively by
b
1
(x, y) =
a(x, y),
b
i+1
(x, y) = W
−1
i
W
i
b
i
(x, y)
+
x
0
∞
y
d
i
(x, y,s,t)dtds
,
a(x, y) = sup
0≤τ≤x
sup
y≤μ<∞
a(τ,μ),
d
i
(x, y,s,t) = sup
0≤τ≤x
sup
y≤μ<∞
d
i
(τ,μ,s,t),
(2.2)
W
1
(0) := 0,andx
1
, y
1
∈ R
+
are chosen such that
W
i
b
i
x
1
, y
1
+
x
1
0
∞
y
1
d
i
(x, y,s,t)dtds ≤
∞
u
i
dz
w
i
(z)
(2.3)
for i
= 1, ,n.
Remark 2.2. x
1
and y
1
are confined by (2.3). In par ticular, (2.1)istrueforallx, y ∈ R
+
when all w
i
(i = 1, ,n) satisfy
∞
u
i
(dz/w
i
(z)) =∞.
Remark 2.3. As in [6, 5, 1], different choices of u
i
in W
i
do not affect our results.
Proof of Theorem 2.1. From the assumptions, we know that
a(x, y)and
d
i
(x, y,s,t)are
well defined. Moreover,
a(x, y)and
d
i
(x, y,s,t) are nonnegative, nondecreasing in x,non-
increasing in y; and satisfy
a(x, y) ≥ a(x, y)and
d
i
(x, y,s,t) ≥ d
i
(x, y,s,t)foreachi =
1, ,n.
We first discuss the case that a(x, y) > 0forallx, y
∈ R
+
.Thus,b
1
(x, y) is positive,
nondecreasing in x, nonincreasing in y; and satisfies b
1
(x, y) ≥ a(x, y)forallx, y ∈ R
+
.
From (1.4), we have
u(x, y)
≤ b
1
(x, y)+
n
i=1
x
0
∞
y
d
i
(x, y,s,t)w
i
u(s,t)
dtds. (2.4)
Choose arbit rary
x
1
, y
1
such that 0 ≤ x
1
≤ x
1
, y
1
≤ y
1
< ∞.From(2.4), we obtain
u(x, y)
≤ b
1
x
1
, y
1
+
n
i=1
x
0
∞
y
d
i
x
1
, y
1
,s,t
w
i
u(s,t)
dtds (2.5)
for all 0
≤ x ≤ x
1
≤ x
1
, y
1
≤ y
1
≤ y<∞.
Having (2.5), we claim
u(x, y)
≤ W
−1
n
W
n
b
n
x
1
, y
1
,x, y
+
x
0
∞
y
d
n
x
1
, y
1
,s,t
dtds
(2.6)
for all 0
≤ x ≤ min{x
1
,x
2
},max{y
1
, y
2
}≤y<∞,where
b
1
x
1
, y
1
,x, y
=
b
1
x
1
, y
1
,
b
i+1
x
1
, y
1
,x, y
=
W
−1
i
W
i
b
i
x
1
, y
1
,x, y
+
x
0
∞
y
d
i
x
1
, y
1
,s,t
dtds
(2.7)
4 Journal of Inequalities and Applications
for i
= 1, ,n−1andx
2
, y
2
∈ R
+
are chosen such that
W
i
b
i
x
1
, y
1
,x
2
, y
2
+
x
2
0
∞
y
2
d
i
x
1
, y
1
,s,t
dtds ≤
∞
u
i
dz
w
i
(z)
(2.8)
for i
= 1, ,n.
Note that we may take x
2
= x
1
and y
2
= y
1
.Infact,
b
i
(x
1
, y
1
,x, y)and
d
i
(x
1
, y
1
,x, y)are
nondecreasing in
x
1
, nonincreasing in y
1
for fixed x, y.Furthermore,itiseasytocheck
that
b
i
(x
1
, y
1
, x
1
, y
1
) = b
i
(x
1
, y
1
)fori = 1, ,n.Ifx
2
, y
2
are replaced by x
1
, y
1
on the left
side of (2.8), we have from ( 2.3)
W
i
b
i
x
1
, y
1
,x
1
, y
1
+
x
1
0
∞
y
1
d
i
x
1
, y
1
,s,t
dtds
≤ W
i
b
i
x
1
, y
1
,x
1
, y
1
+
x
1
0
∞
y
1
d
i
x
1
, y
1
,s,t
dtds
= W
i
b
i
x
1
, y
1
+
x
1
0
∞
y
1
d
i
x
1
, y
1
,s,t
dtds ≤
∞
u
i
dz
w
i
(z)
.
(2.9)
Thus, it means that we can take x
2
= x
1
, y
2
= y
1
.
In the following, we will use mathematical induction to prove (2.6).
For n
= 1, let
z(x, y)
=
x
0
∞
y
d
1
x
1
, y
1
,s,t
w
1
u(s,t)
dtds.
(2.10)
Then z(x, y)isdifferentiable, nonnegative, nondecreasing for x
∈ [0, x
1
], and nonincreas-
ing for y
∈ [y
1
,∞)andz(0, y) = z(x,∞) =0. From (2.5), we have the following:
u(x, y)
≤ b
1
x
1
, y
1
+ z(x, y),
D
1
z(x, y) =
∞
y
d
1
x
1
, y
1
,x,t
w
1
u(x,t)
dt
≤
∞
y
d
1
x
1
, y
1
,x,t
w
1
b
1
x
1
, y
1
+ z(x,t)
dt
≤ w
1
b
1
x
1
, y
1
+ z(x, y)
∞
y
d
1
x
1
, y
1
,x,t
dt.
(2.11)
Since w
1
is nondecreasing and b
1
(x
1
, y
1
)+z(x, y) > 0, we get
D
1
b
1
x
1
, y
1
+ z(x, y)
w
1
b
1
x
1
, y
1
+ z(x, y)
=
D
1
z(x, y)
w
1
b
1
x
1
, y
1
+ z(x, y)
≤
w
1
b
1
x
1
, y
1
+ z(x, y)
∞
y
d
1
x
1
, y
1
,x,t
dt
w
1
b
1
x
1
, y
1
+ z(x, y)
=
∞
y
d
1
x
1
, y
1
,x,t
dt.
(2.12)
Kelong Zheng et a l. 5
Integrating both sides of the above inequality from 0 to x,weobtain
W
1
b
1
x
1
, y
1
+ z(x, y)
≤
W
1
b
1
x
1
, y
1
+ z(0, y)
+
x
0
∞
y
d
1
x
1
, y
1
,s,t
dtds
= W
1
b
1
x
1
, y
1
+
x
0
∞
y
d
1
x
1
, y
1
,s,t
dtds.
(2.13)
Thus the monotonicity of W
−1
1
implies
u(x, y)
≤ b
1
x
1
, y
1
+ z(x, y) ≤ W
−1
1
W
1
b
1
x
1
, y
1
+
x
0
∞
y
d
1
x
1
, y
1
,s,t
dtds
,
(2.14)
that is, (2.6)istrueforn
= 1.
Assume that (2.6)istrueforn
= m. Consider
u(x, y)
≤ b
1
x
1
, y
1
)+
m+1
i=1
x
0
∞
y
d
i
x
1
, y
1
,s,t
w
i
u(s,t)
dtds (2.15)
for all 0
≤ x ≤ x
1
, y
1
≤ y<∞.Let
z(x, y)
=
m+1
i=1
x
0
∞
y
d
i
x
1
, y
1
,s,t
w
i
u(s,t)
dtds. (2.16)
Then z(x, y)isdifferentiable, nonnegative, nondecreasing for x
∈ [0, x
1
], and nonincreas-
ing for y
∈ [y
1
,∞). Obviously, z(0, y) = z(x, ∞) = 0andu(x, y) ≤ b
1
(x
1
, y
1
)+z(x, y).
Since w
1
is nondecreasing and b
1
(x
1
, y
1
)+z(x, y) > 0, we have
D
1
b
1
x
1
, y
1
+ z(x, y)
w
1
b
1
x
1
, y
1
+ z(x, y)
≤
m+1
i
=1
∞
y
d
i
x
1
, y
1
,x,t
w
i
u(x,t)
dt
w
1
b
1
x
1
, y
1
+ z(x, y)
≤
m+1
i
=1
∞
y
d
i
x
1
, y
1
,x,t
w
i
b
1
(x
1
, y
1
+ z(x,t)
dt
w
1
b
1
x
1
, y
1
+ z(x, y)
≤
∞
y
d
1
x
1
, y
1
,x,t
dt +
m+1
i=2
∞
y
d
i
x
1
, y
1
,x,t
φ
i
b
1
x
1
, y
1
+ z(x,t)
dt
≤
∞
y
d
1
x
1
, y
1
,x,t
dt +
m
i=1
∞
y
d
i+1
x
1
, y
1
,x,t
φ
i+1
b
1
x
1
, y
1
+ z(x,t)
dt,
(2.17)
6 Journal of Inequalities and Applications
where φ
i+1
(u) = w
i+1
(u)/w
1
(u), i = 1, ,m. Integrating the above inequality from 0 to x,
we obtain
W
1
b
1
x
1
, y
1
+ z(x, y)
≤
W
1
b
1
x
1
, y
1
+
x
0
∞
y
d
1
x
1
, y
1
,s,t
dtds
+
m
i=1
x
0
∞
y
d
i+1
x
1
, y
1
,s,t
φ
i+1
b
1
(x
1
, y
1
+ z(s,t)
dtds,
(2.18)
or
ξ(x, y)
≤ c
1
(x, y)+
m
i=1
x
0
∞
y
d
i+1
x
1
, y
1
,s,t
φ
i+1
W
−1
1
ξ(s,t)
dtds (2.19)
for 0
≤ x ≤ x
1
and y
1
≤ y<∞, the same as (2.6)forn = m,whereξ(x, y)=W
1
(b
1
(x
1
, y
1
)+
z(x, y)) and c
1
(x, y) = W
1
(b
1
(x
1
, y
1
)) +
x
0
∞
y
d
1
(x
1
, y
1
,s,t)dtds.
From the assumption (C
1
), each φ
i+1
(W
−1
1
(u)), i = 1, ,m, is continuous and non-
decreasing for u.Moreover,φ
2
(W
−1
1
) ∝ φ
3
(W
−1
1
) ∝···∝φ
m+1
(W
−1
1
). By the inductive
assumption, we have
ξ(x, y) ≤ Φ
−1
m+1
Φ
m+1
c
m
(x, y)
+
x
0
∞
y
d
m+1
x
1
, y
1
,s,t
dtds
(2.20)
for all 0
≤ x ≤ min{x
1
,x
3
},max{y
1
, y
3
}≤y<∞,whereΦ
i+1
(u) =
u
u
i+1
(dz/φ
i+1
(W
−1
1
(z))) ,
u>0,
u
i+1
= W
1
(u
i+1
), Φ
−1
i+1
is the inverse of Φ
i+1
, i = 1, ,m,
c
i+1
(x, y) = Φ
−1
i+1
Φ
i+1
c
i
(x, y)
+
x
0
∞
y
d
i+1
x
1
, y
1
,s,t
dtds
, i = 1, ,m, (2.21)
and x
3
, y
3
∈ R
+
are chosen such that
Φ
i+1
c
i
x
3
, y
3
+
x
3
0
∞
y
3
d
i+1
x
1
, y
1
,s,t
dtds ≤
W
1
(∞)
u
i+1
dz
φ
i+1
W
−1
1
(z)
(2.22)
for i
= 1, ,m.
Note that
Φ
i
(u) =
u
u
i
dz
φ
i
W
−1
1
(z)
=
u
W
1
(u
i
)
w
1
W
−1
1
(z)
dz
w
i
W
−1
1
(z)
=
W
−1
1
(u)
u
i
dz
w
i
(z)
= W
i
◦W
−1
1
(u), i = 2, ,m +1.
(2.23)
From (2.20), we have
u(x, y)
≤ b
1
x
1
, y
1
+ z(x, y) = W
−1
1
ξ(x, y)
≤
W
−1
m+1
W
m+1
W
−1
1
c
m
(x, y)
+
x
0
∞
y
d
m+1
x
1
, y
1
,s,t
dtds
(2.24)
Kelong Zheng et a l. 7
for all 0
≤ x ≤ min{x
1
,x
3
},max{y
1
, y
3
}≤y<∞.Letc
i
(x, y) = W
−1
1
(c
i
(x, y)). Then,
c
1
(x, y) = W
−1
1
c
1
(x, y)
=
W
−1
1
W
1
b
1
x
1
, y
1
+
x
0
∞
y
d
1
x
1
, y
1
,s,t
dtds
=
b
2
x
1
, y
1
,x, y
.
(2.25)
Moreover, with the assumption that
c
m
(x, y) =
b
m+1
(x
1
, y
1
,x, y), we have
c
m+1
(x, y) = W
−1
1
Φ
−1
m+1
Φ
m+1
c
m
(x, y)
+
x
0
∞
y
d
m+1
x
1
, y
1
,s,t
dtds
=
W
−1
m+1
W
m+1
W
−1
1
c
m
(x, y)
+
x
0
∞
y
d
m+1
x
1
, y
1
,s,t
dtds
=
W
−1
m+1
W
m+1
c
m
(x, y)
+
x
0
∞
y
d
m+1
x
1
, y
1
,s,t
dtds
=
W
−1
m+1
W
m+1
b
m+1
x
1
, y
1
,x, y
+
x
0
∞
y
d
m+1
x
1
, y
1
,s,t
dtds
=
b
m+2
x
1
, y
1
,x, y
.
(2.26)
This proves that
c
i
(x, y) =
b
i+1
x
1
, y
1
,x, y
, i = 1, ,m. (2.27)
Therefore, (2.22)becomes
W
i+1
b
i+1
x
1
, y
1
,x
3
, y
3
+
x
3
0
∞
y
3
d
i+1
x
1
, y
1
,s,t
dtds
≤
W
1
(∞)
u
i+1
dz
φ
i+1
W
−1
1
(z)
=
∞
u
i+1
dz
w
i+1
(z)
, i
= 1, ,m.
(2.28)
The above inequalities and (2.8) imply that we may t ake x
2
= x
3
, y
2
= y
3
.From(2.24),
we get
u(x, y)
≤ W
−1
m+1
W
m+1
b
m+1
x
1
, y
1
,x, y
+
x
0
∞
y
d
m+1
x
1
, y
1
,s,t
dtds
(2.29)
for all 0
≤ x ≤ x
1
≤ x
2
, y
2
≤ y
1
≤ y<∞. This proves (2.6) by mathematical induction.
Taking x
=
x
1
, y =
y
1
, x
2
= x
1
,andy
2
= y
1
,wehave
u
x
1
, y
1
≤
W
−1
n
W
n
b
n
x
1
, y
1
, x
1
, y
1
+
x
1
0
∞
y
1
d
n
x
1
, y
1
,s,t
dtds
(2.30)
8 Journal of Inequalities and Applications
for 0
≤ x
1
≤ x
1
, y
1
≤ y
1
< ∞.Itiseasytoverify
b
n
(x
1
, y
1
, x
1
, y
1
) = b
n
(x
1
, y
1
). Thus, (2.30)
can be written as
u
x
1
, y
1
≤
W
−1
n
W
n
b
n
x
1
, y
1
+
x
1
0
∞
y
1
d
n
x
1
, y
1
,s,t
dtds
. (2.31)
Since
x
1
, y
1
are arbitrary, replace x
1
and y
1
by x and y respectively and we have
u(x, y)
≤ W
−1
n
W
n
b
n
(x, y)
+
x
0
∞
y
d
n
(x, y,s,t)dtds
(2.32)
for all 0
≤ x ≤ x
1
, y
1
≤ y<∞.
In case a(x, y)
= 0forsomex, y ∈ R
+
.Letb
1,
(x, y):= b
1
(x, y)+ for all x, y ∈ R
+
,
where
> 0 is arbitrary, and then b
1,
(x, y) > 0. Using the same arguments as above, where
b
1
(x, y)isreplacedwithb
1,
(x, y) > 0,weget
u(x, y)
≤ W
−1
n
W
n
b
n,
(x, y)
+
x
0
∞
y
d
n
(x, y,s,t)dtds
. (2.33)
Letting
→
0
+
,weobtain(2.1) by the continuity of b
1,
in and the continuity of W
i
and
W
−1
i
under the notation W
1
(0) := 0.
Theorem 2.4. In addition to the assumpt ions (C
1
), (C
2
), and (C
3
), suppos e that a(x, y)
and d
i
(x, y,s,t) are bounded in x, y ∈R
+
for each fixed s,t ∈ R
+
.Ifu(x, y) is a continuous
and nonnegative function satisfying (1.5)forx, y
∈ R
+
, then
u(x, y)
≤ W
−1
n
W
n
b
n
(x, y)
+
∞
x
∞
y
d
n
(x, y,s,t)dtds
(2.34)
for all x
4
≤ x<∞, y
4
≤ y<∞,whereb
n
(x, y) is determined recursively by
b
1
(x, y) =
a(x, y),
b
i+1
(x, y) = W
−1
i
W
i
b
i
(x, y)
+
∞
x
∞
y
d
i
(x, y,s,t)dtds
,
(2.35)
a(x, y) = sup
x≤τ<∞
sup
y≤μ<∞
a(τ,μ),
d
i
(x, y,s,t) = sup
x≤τ<∞
sup
y≤μ<∞
d
i
(τ,μ,s,t),
(2.36)
W
1
(0) := 0,andx
4
, y
4
∈ R
+
are chosen such that
W
i
b
i
x
4
, y
4
+
∞
x
4
∞
y
4
d
i
(x, y,s,t)dtds ≤
∞
u
i
dz
w
i
(z)
(2.37)
for i
= 1, ,n.
The proof is similar to the argument in the proof of Theorem 2.1 with suitable modi-
fication. We omit the details here.
Kelong Zheng et a l. 9
Remark 2.5. Take d
1
(x, y,s,t) = c(x, y)d(s,t)andn = 1in(1.4). Suppose that a(x, y)and
c(x, y) are continuous, nonnegative, nondecreasing in x and nonincreasing in y;and
d(s,t) is nonnegative and continuous. We note that
b
1
(x, y) = a(x, y),
d
1
(x, y,s,t) = c(x, y)d(s,t). (2.38)
From Theorem 2.1,weget
u(x, y)
≤ W
−1
1
W
1
a(x, y)
+ c(x, y)
x
0
∞
y
d(s,t)dtds
, (2.39)
which is exactly (2.6) of Lemma 2.2 in [13].
Remark 2.6. Take d
1
(x, y,s,t) = c(x, y)d(s,t)andn = 1in(1.5). Suppose that a(x, y)and
c(x, y) are continuous, nonnegative, nonincreasing in x, y;andd(s,t) is nonnegative and
continuous. It is easy to check that
b
1
(x, y) = a(x, y),
d
1
(x, y,s,t) = c(x, y)d(s,t). (2.40)
From Theorem 2.4,weget
u(x, y)
≤ W
−1
1
W
1
a(x, y)
+ c(x, y)
∞
x
∞
y
d(s,t)dtds
(2.41)
which is (2.10) of Lemma 2.2 in [13].
3. Applications
Consider the partial differential equation
D
1
D
2
v(x, y) =
1
(x +1)
2
(y +1)
2
+exp(−x)exp(−y)
v(x, y)
+1
+ xexp (
−x)exp(−y)Tv(x, y),
(3.1)
v(x,
∞) = σ(x),v(0, y) = τ(y), v(0,∞) = k (3.2)
for x, y
∈ R
+
,whereσ,τ ∈ C(R
+
,R), σ(x) is n ondecreasing in x, τ(y) is nonincreasing
in y, k is a real constant, and
T is a continuous operator on C(R
+
×R
+
,R)suchthat
|Tv|≤c
0
|v| for a constant c
0
> 0. Integrating (3.1) with respect to x and y and using the
initial conditions (3.2), we get
v(x, y)
= σ( x)+τ(y) −k −
x
(x +1)(y +1)
−
x
0
∞
y
exp(−s)exp(−t)
v(s,t)
+1dtds
−
x
0
∞
y
sexp(−s)exp(−t)Tv(s,t)dtds.
(3.3)
10 Journal of Inequalities and Applications
Thus,
v(x, y)
≤
σ(x)+τ(y) −k
+
x
(x +1)(y +1)
+
x
0
∞
y
exp(−s)exp(−t)
v(s,t)
+1dtds
+
x
0
∞
y
sexp(−s)exp(−t)c
0
v(s,t)
dtds.
(3.4)
Letting u(x, y)
=|v(x, y)|,wehave
u(x, y)
≤ a(x, y)+
x
0
∞
y
d
1
(x, y,s,t)w
1
(u)dtds+
x
0
∞
y
d
2
(x, y,s,t)w
2
(u)dtds, (3.5)
where a(x, y)
=|σ(x)+τ(y) −k|+ x/(x +1)(y +1),w
1
(u) =
√
u +1,w
2
(u) = c
0
u, d
1
(x, y,
s,t)
= exp (−s)exp(−t), d
2
(x, y,s,t) = sexp(−s)exp(−t). Clearly, w
2
(u)/w
1
(u) = c
0
(u/
√
u +1)isnondecreasingforu>0, that is, w
1
∝ w
2
.Thenforu
1
,u
2
> 0,
b
1
(x, y) = a(x, y),
d
1
(x, y,s,t) = d
1
(x, y,s,t),
d
2
(x, y,s,t) = d
2
(x, y,s,t),
W
1
(u) =
u
u
1
dz
√
z +1
= 2
√
u +1−
u
1
+1
, W
−1
1
(u) =
u
2
+
u
1
+1
2
−1,
W
2
(u) =
u
u
2
dz
c
0
z
=
1
c
0
ln
u
u
2
, W
−1
2
(u) = u
2
exp
c
0
u
,
b
2
(x, y) = W
−1
1
W
1
b
1
(x, y)
+
x
0
∞
y
d
1
(x, y,s,t)dtds
=
W
−1
1
2
b
1
(x, y)+1−
u
1
+1
+
1 −exp (−x)
exp(−y)
=
b
1
(x, y)+1+
1 −exp (−x)
2
exp(
−y)
2
−1.
(3.6)
By Theorem 2.1,wehave
v(x, y)
≤
W
−1
2
W
2
b
2
(x, y)
+
x
0
∞
y
d
2
(x, y,s,t)dtds
=
W
−1
2
1
c
0
ln
b
2
(x, y)
u
2
+
1 −(x +1)exp(−x)
exp(−y)
=
u
2
exp
c
0
1
c
0
ln
b
2
(x, y)
u
2
+
1 −(x +1)exp(−x)
exp(−y)
=
b
2
(x, y)exp
c
0
1 −(x +1)exp(−x)
exp(−y)
=
σ(x)+τ(y) −k
+
x
(x +1)(y +1)
+1+
1
−exp(−x)
2
exp(
−y)
2
−1
×
exp
c
0
1 −(x +1)exp(−x)
exp(−y)
.
(3.7)
This implies that the solution of ( 3.1) is bounded for x, y
∈ R
+
provided that σ(x)+
τ(y)
−k is bounded for all x, y ∈ R
+
.
Kelong Zheng et al. 11
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Kelong Zheng: College of Science, Southwest University of Science and Technology,
Mianyang, Sichuan 621010, China
Email address:
Yu Wu: Yibin University, Yibin, Sichuan 644007, China
Email address:
Shengfu Deng: Department of Mathematics, Virginia Polytechnical Institute and State University,
Blacksburg, VA 24061, USA
Email address:
