SOME RETARDED NONLINEAR INTEGRAL
INEQUALITIES IN TWO VARIABLES AND
APPLICATIONS
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
KELONG ZHENG
School of Science
Southwest University of Science and Technology
Mianyang, Sichuan 621010, P. R. China
vol. 9, iss. 2, art. 57, 2008
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Contents
EMail: zhengkelong@swust.edu.cn
Received:
23 October, 2007
Accepted:
19 March, 2008
Communicated by:
W.S. Cheung
2000 AMS Sub. Class.:
26D10, 26D15.
Key words:
Integral inequality, Nonlinear, Two variables, Retarded.
Abstract:
In this paper, some retarded nonlinear integral inequalities in two variables with
more than one distinct nonlinear term are established. Our results are also applied
to show the boundedness of the solutions of certain partial differential equations.
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Contents
1
Introduction
3
2
Statement of Main Results
4
3
Proof of Theorem 2.1
8
4
Applications
18
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
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1.
Introduction
The Gronwall-Bellman integral inequality plays an important role in the qualitative
analysis of the solutions of differential and integral equations. During the past few
years, many retarded inequalities have been discovered (see in [1, 2, 4, 5, 6, 10, 11]).
Lipovan [4] investigated the following retarded inequality
Z b(t)
(1.1)
u(t) ≤ a +
f (s)w(u(s))ds,
t0 ≤ t ≤ t1 ,
b(t0 )
Kelong Zheng
and Agarwal et al. [6] generalized (1.1) to a more general case as follows
n Z bi (t)
X
(1.2)
u(t) ≤ a(t) +
fi (s)wi (u(s))ds,
t0 ≤ t ≤ t1 .
i=1
bi (t0 )
Recently, many people such as Wang [10], Cheung [9] and Dragomir [8] established
some new integral inequalities involving functions of two independent variables and
Zhao et al. [11] also established advanced integral inequalities.
The purpose of this paper, motivated by the works of Agarwal [6], Cheung [9] and
Zhao [11], is to discuss more general integral inequalities with n nonlinear terms
n Z αi (x) Z ∞
X
(1.3)
u(x, y) ≤ a(x, y) +
fi (x, y, s, t)wi (u(s, t))dtds
i=1
αi (0)
u(x, y) ≤ a(x, y) +
n Z
X
i=1
vol. 9, iss. 2, art. 57, 2008
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βi (y)
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and
(1.4)
Retarded Nonlinear Integral
Inequalities
∞
αi (x)
Z
Close
∞
fi (x, y, s, t)wi (u(s, t))dtds.
βi (y)
Our results can be used more effectively to study the boundedness and uniqueness
of the solutions of certain partial differential equations. Moreover, at the end of this
paper, an example is presented to show the applications of our results.
2.
Statement of Main Results
Let R = (−∞, ∞) and R+ = [0, ∞). D1 z(x, y) and D2 z(x, y) denote the first-order
partial derivatives of z(x, y) with respect to x and y respectively.
w2
As in [6], define w1 ∝ w2 for w1 , w2 : A ⊂ R → R\{0} if
is nondecreasing
w1
on A. Assume that
(B1 ) wi (u) (i = 1, . . . , n) is a nonnegative, nondecreasing and continuous function
for u ∈ R+ with wi (u) > 0 for u > 0 such that w1 ∝ w2 ∝ · · · ∝ wn ;
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
(B2 ) a(x, y) is a nonnegative and continuous function for x, y ∈ R+ ;
(B3 ) fi (x, y, s, t) (i = 1, . . . , n) is a continuous and nonnegative function for x, y, s, t ∈
R+ .
Ru
Take the notation Wi (u) := ui wdz
for u ≥ ui , where ui > 0 is a given constant.
i (z)
Clearly, Wi is strictly increasing, so its inverse Wi−1 is well defined, continuous and
increasing in its corresponding domain.
Theorem 2.1. Under the assumptions (B1 ), (B2 ) and (B3 ), suppose a(x, y) and
fi (x, y, s, t) are bounded in y ∈ R+ . Let αi (x), βi (y) be nonnegative, continuously
differentiable and nondecreasing functions with αi (x) ≤ x and βi (y) ≥ y on R+ for
i = 1, 2, . . . , n. If u(x, y) is a continuous and nonnegative function satisfying (1.3),
then
"
#
Z αn (x) Z ∞
(2.1)
u(x, y) ≤ Wn−1 Wn (bn (x, y)) +
f˜n (x, y, s, t)dtds
αn (0)
βn (y)
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for all 0 ≤ x ≤ x1 , y1 ≤ y < ∞, where bn (x, y) is determined recursively by
b1 (x, y) = sup
sup a(τ, µ),
"
0≤τ ≤x y≤µ<∞
(2.2)
bi+1 (x, y) = Wi−1 Wi (bi (x, y)) +
Z
αi (x)
αi (0)
f˜i (x, y, s, t) = sup
Z
#
∞
f˜i (x, y, s, t)dtds ,
βi (y)
Retarded Nonlinear Integral
Inequalities
sup fi (τ, µ, s, t),
0≤τ ≤x y≤µ<∞
Kelong Zheng
W1 (0) := 0, and x1 , y1 ∈ R+ are chosen such that
Z αi (x1 ) Z ∞
Z
˜
(2.3)
Wi (bi (x1 , y1 )) +
fi (x, y, s, t)dtds ≤
αi (0)
βi (y1 )
vol. 9, iss. 2, art. 57, 2008
∞
ui
dz
wi (z)
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Contents
for i = 1, . . . , n.
The proof of Theorem 2.1 will be given in the next section.
Remark 1. As in [6], different
R ∞ choices of ui in Wi do not affect our results. If all
wi (i = 1, . . . , n) satisfy ui wdz
= ∞, then (2.1) is true for all x, y ∈ R+ .
i (z)
Remark 2. As in [10], if wi (u) (i = 1, . . . , n) are continuous functions on R+ and
positive on (0, ∞) but the sequence of {wi (u)} does not satisfy w1 ∝ w2 ∝ · · · ∝
wn , we can use a technique of monotonization of the sequence of functions wi (u),
calculated by
(2.4)
w̃1 (u) := max w1 (θ),
θ∈[0,u]
wi+1 (θ)
w̃i (u),
w̃i+1 (u) := max
θ∈[0,u]
w̃i (θ)
i = 1, . . . , n − 1.
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Clearly, w̃i (u) ≥ wi (u) (i = 1, . . . , n). (1.3) and (1.4) can also become
n Z αi (x) Z ∞
X
(2.5)
u(x, y) ≤ a(x, y) +
fi (x, y, s, t)w̃i (u(s, t))dtds
i=1
αi (0)
βi (y)
and
(2.6)
u(x, y) ≤ a(x, y) +
n Z
X
i=1
∞
αi (x)
Z
Retarded Nonlinear Integral
Inequalities
∞
fi (x, y, s, t)w̃i (u(s, t))dtds,
Kelong Zheng
βi (y)
vol. 9, iss. 2, art. 57, 2008
where the function sequence {w̃i (u)} satisfies the assumption (B1 ).
Theorem 2.2. Under the assumptions (B1 ), (B2 ) and (B3 ), suppose a(x, y) and
fi (x, y, s, t) are bounded in x, y ∈ R+ . Let αi (x), βi (y) be nonnegative, continuously differentiable and nondecreasing functions with αi (x) ≥ x and βi (y) ≥ y on
R+ for i = 1, 2, . . . , n. If u(x, y) is a continuous and nonnegative function satisfying
(1.4), then
Z ∞ Z ∞
−1
ˆ
(2.7)
u(x, y) ≤ Wn Wn (bn (x, y)) +
fn (x, y, s, t)dtds
αn (x)
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βn (y)
for all x̂1 ≤ x < ∞, ŷ1 ≤ y < ∞, where bn (x, y) is determined recursively by
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b1 (x, y) = sup
sup a(τ, µ),
x≤τ <∞ y≤µ<∞
Z
−1
bi+1 (x, y) = Wi
Wi (bi (x, y)) +
Close
∞
Z
αi (x)
(2.8)
fˆi (x, y, s, t) = sup
sup fi (τ, µ, s, t),
x≤τ <∞ y≤µ<∞
∞
βi (y)
fˆi (x, y, s, t)dtds ,
W1 (0) := 0, and x̂1 , ŷ1 ∈ R+ are chosen such that
Z ∞ Z ∞
Z
ˆ
(2.9)
Wi (bi (x̂1 , ŷ1 )) +
fi (x, y, s, t)dtds ≤
αi (x̂1 )
βi (ŷ1 )
∞
ui
dz
wi (z)
for i = 1, . . . , n.
The proof is similar to the argument in the proof of Theorem 2.1 with suitable
modifications. In the next section, we omit its proof.
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
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3.
Proof of Theorem 2.1
From the assumptions, we know that b1 (x, y) and f˜i (x, y, s, t) are well defined.
Moreover, ã(x, y) and f˜i (x, y, s, t) are nonnegative, nondecreasing in x and nonincreasing in y and satisfy b1 (x, y) ≥ a(x, y) and f˜i (x, y, s, t) ≥ fi (x, y, s, t) for
each i = 1, . . . , n.
We first discuss the case a(x, y) > 0 for all x, y ∈ R+ . From (1.3), we have
n Z αi (x) Z ∞
X
(3.1)
u(x, y) ≤ b1 (x, y) +
f˜i (x, y, s, t)wi (u(s, t))dtds.
i=1
αi (0)
βi (y)
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
Choose arbitrary x̃1 , ỹ1 such that 0 ≤ x̃1 ≤ x1 , y1 ≤ ỹ1 < ∞. From (3.1), we obtain
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(3.2)
u(x, y) ≤ b1 (x̃1 , ỹ1 ) +
n Z αi (x)
X
i=1
Z
αi (0)
∞
βi (y)
for all 0 ≤ x ≤ x̃1 ≤ x1 , y1 ≤ ỹ1 ≤ y < ∞.
We claim that
"
(3.3) u(x, y) ≤
Wn−1
Z
αn (x)
Z
#
∞
f˜n (x̃1 , ỹ1 , s, t)dtds
Wn (b̃n (x̃1 , ỹ1 , x, y)) +
αn (0)
βn (y)
for all 0 ≤ x ≤ min{x̃1 , x2 }, max{ỹ1 , y2 } ≤ y < ∞, where
b̃i+1 (x̃1 , ỹ1 , x, y)
"
=
Wi−1
Z
αi (x)
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#
∞
Wi (b̃i (x̃1 , ỹ1 , x, y)) +
αi (0)
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b̃1 (x̃1 , ỹ1 , x, y) = b1 (x̃1 , ỹ1 ),
(3.4)
Contents
f˜i (x̃1 , ỹ1 , s, t)wi (u(s, t))dtds
βi (y)
f˜i (x̃1 , ỹ1 , s, t)dtds
for i = 1, . . . , n − 1 and x2 , y2 ∈ R+ are chosen such that
Z αi (x2 ) Z ∞
Z
˜
(3.5)
Wi (b̃i (x̃1 , ỹ1 , x2 , y2 )) +
fi (x̃1 , ỹ1 , s, t)dtds ≤
αi (0)
βi (y2 )
∞
ui
dz
wi (z)
for i = 1, . . . , n.
Note that we may take x2 = x1 and y2 = y1 . In fact, b̃i (x̃1 , ỹ1 , x, y) and
˜
fi (x̃1 , ỹ1 , x, y) are nondecreasing in x̃1 and nonincreasing in ỹ1 for fixed x, y. Furthermore, it is easy to check that b̃i (x̃1 , ỹ1 , x̃1 , ỹ1 ) = bi (x̃1 , ỹ1 ) for i = 1, . . . , n. If x2
and y2 are replaced by x1 and y1 on the left side of (3.5) respectively, from (2.3) we
have
Z αi (x1 ) Z ∞
Wi (b̃i (x̃1 ,ỹ1 , x1 , y1 )) +
f˜i (x̃1 , ỹ1 , s, t)dtds
αi (0)
vol. 9, iss. 2, art. 57, 2008
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αi (x1 )
Z
∞
≤ Wi (b̃i (x1 , y1 , x1 , y1 )) +
Z
αi (0)
αi (x1 ) Z ∞
αi (0)
f˜i (x1 , y1 , s, t)dtds
βi (y1 )
f˜i (x1 , y1 , s, t)dtds
βi (y1 )
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Thus, we can take x2 = x1 , y2 = y1 .
In the following, we will use mathematical induction to prove (3.3).
For n = 1, let
Z α1 (x) Z ∞
z(x, y) = b1 (x̃1 , ỹ1 ) +
f˜1 (x̃1 , ỹ1 , s, t)w1 (u(s, t))dtds.
α1 (0)
Kelong Zheng
βi (y1 )
Z
= Wi (bi (x1 , y1 )) +
Z ∞
dz
.
≤
ui wi (z)
Retarded Nonlinear Integral
Inequalities
β1 (y)
Close
Then z(x, y) is differentiable, nonnegative, nondecreasing for x ∈ [0, x̃1 ] and nonincreasing for y ∈ [ỹ1 , ∞] and z(0, y) = z(x, ∞) = b1 (x̃1 , ỹ1 ). From (3.2), we have
u(x, y) ≤ z(x, y).
(3.6)
0
Considering α1 (x) ≤ x and α1 (x) ≥ 0 for x ∈ R+ , we have
Z ∞
0
D1 z(x, y) =
f˜1 (x̃1 , ỹ1 , α1 (x), t)w1 (u(α1 (x), t))dtα1 (x)
β (y)
Z 1∞
0
≤
f˜1 (x̃1 , ỹ1 , α1 (x), t)w1 (z(α1 (x), t))dtα1 (x)
β1 (y)
Z ∞
0
(3.7)
≤ w1 (z(x, y))
f˜1 (x̃1 , ỹ1 , α1 (x), t)dtα1 (x).
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
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β1 (y)
Since w1 is nondecreasing and z(x, y) > 0, we get
Z ∞
D1 (z(x, y))
0
≤
f˜1 (x̃1 , ỹ1 , α1 (x), t)dtα1 (x).
(3.8)
w1 (z(x, y))
β1 (y)
Integrating both sides of the above inequality from 0 to x, we obtain
Z xZ ∞
0
W1 (z(x, y)) ≤ W1 (z(0, y)) +
f˜1 (x̃1 , ỹ1 , α1 (s), t)α1 (s)dtds
0
Z
(3.9)
β1 (y)
α1 (x)
Z
∞
= W1 (b1 (x̃1 , ỹ1 )) +
α1 (0)
β1 (y)
f˜1 (x̃1 , ỹ1 , s, t)dtds.
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Thus the monotonicity of W1−1 and (3.5) imply
u(x, y) ≤ z(x, y)
"
≤ W1−1 W1 (b1 (x̃1 , ỹ1 )) +
Z
α1 (x)
α1 (0)
Z
∞
#
f˜1 (x̃1 , ỹ1 , s, t)dtds ,
β1 (y)
namely, (3.3) is true for n = 1.
Assume that (3.3) is true for n = m. Consider
u(x, y) ≤ b1 (x̃1 , ỹ1 ) +
m+1
XZ
i=1
αi (x) Z
αi (0)
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
∞
f˜i (x̃1 , ỹ1 , s, t)wi (u(s, t))dtds
βi (y)
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for all 0 ≤ x ≤ x̃1 , ỹ1 ≤ y < ∞. Let
z(x, y) = b1 (x̃1 , ỹ1 ) +
m+1
X Z αi (x)
i=1
αi (0)
Contents
Z
∞
f˜i (x̃1 , ỹ1 , s, t)wi (u(s, t))dtds.
βi (y)
Then z(x, y) is differentiable, nonnegative, nondecreasing for x ∈ [0, x̃1 ] and nonincreasing for y ∈ [ỹ1 , ∞]. Obviously, z(0, y) = z(x, 0) = b1 (x̃1 , ỹ1 ) and u(x, y) ≤
z(x, y). Since w1 is nondecreasing and z(x, y) > 0, noting that αi (x) ≤ x and
0
αi (x) ≥ 0 for x ∈ R+ , we have
Pm+1 R ∞ ˜
0
f (x̃ , ỹ , αi (x), t)wi (u(αi (x), t))dtαi (x)
D1 (z(x, y))
i=1
βi (y) i 1 1
≤
w1 (z(x, y))
w1 (z(x, y))
Pm+1 R ∞ ˜
0
f (x̃ , ỹ , αi (x), t)wi (z(αi (x), t))dtαi (x)
i=1
βi (y) i 1 1
≤
w1 (z(x, y))
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Z
∞
0
f˜1 (x̃1 , ỹ1 , α1 (x), t)dtα1 (x)
≤
β1 (y)
+
m+1
XZ ∞
βi (y)
i=2
Z
∞
0
f˜i (x̃1 , ỹ1 , αi (x), t)φi (z(αi (x), t))dtαi (x)
0
f˜1 (x̃1 , ỹ1 , α1 (x), t)dtα1 (x)
≤
β1 (y)
+
m Z
X
0
f˜i+1 (x̃1 , ỹ1 , αi+1 (x), t)φi+1 (z(αi+1 (x), t))dtαi+1 (x),
βi+1 (y)
i=1
where φi+1 (u) =
x, we obtain
∞
wi+1 (u)
,
w1 (u)
i = 1, . . . , m. Integrating the above inequality from 0 to
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
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W1 (z(x, y))
x
Z
Z
∞
0
f˜1 (x̃1 , ỹ1 , α1 (s), t)α1 (s)dtds
≤ W1 (b1 (x̃1 , ỹ1 )) +
0
+
m Z
X
i=1
0
x
Z
∞
β1 (y)
0
f˜i+1 (x̃1 , ỹ1 , αi+1 (s), t)φi+1 (z(αi+1 (s), t))αi+1 (s)dtds
βi+1 (y)
α1 (x) Z
Z
∞
≤ W1 (b1 (x̃1 , ỹ1 )) +
α1 (0)
+
m Z αi+1 (x)
X
i=1
αi+1 (0)
Z
f˜1 (x̃1 , ỹ1 , s, t)dtds
β1 (y)
∞
βi+1 (y)
f˜i+1 (x̃1 , ỹ1 , s, t)φi+1 (z(s, t))dtds,
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or
ξ(x, y) ≤ c1 (x, y) +
m Z
X
i=1
αi+1 (x)
αi+1 (0)
Z
∞
f˜i+1 (x̃1 , ỹ1 , s, t)φi+1 (W1−1 (ξ(s, t)))dtds
βi+1 (y)
for 0 ≤ x ≤ x̃1 , ỹ1 ≤ y < ∞. This is the same as (3.3) for n = m, where
ξ(x, y) = W1 (z(x, y)) and
Z α1 (x) Z ∞
c1 (x, y) = W1 (b1 (x̃1 , ỹ1 )) +
f˜1 (x̃1 , ỹ1 , s, t)dtds.
α1 (0)
(3.10) ξ(x, y)
≤ Φ−1
m+1 Φm+1 (cm (x, y)) +
Z
αm+1 (x) Z
αm+1 (0)
#
∞
f˜m+1 (x̃1 , ỹ1 , s, t)dtds
βm+1 (y)
for all 0 ≤ x ≤ min{x̃1 , x3 }, max{ỹ1 , y3 } ≤ y < ∞, where
Z u
dz
Φi+1 (u) =
,
−1
ũi+1 φi+1 (W1 (z))
u > 0, ũi+1 = W1 (ui+1 ), Φ−1
i+1 is the inverse of Φi+1 , i = 1, . . . , m,
#
"
Z αi+1 (x) Z ∞
ci+1 (x, y) = Φ−1
f˜i+1 (x̃1 , ỹ1 , s, t)dtds , i = 1, . . . , m,
i+1 Φi+1 (ci (x, y)) +
αi+1 (0)
βi+1 (y)
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
β1 (y)
From the assumption (B1 ), each φi+1 (W1−1 (u)) (i = 1, . . . , m) is continuous and
nondecreasing for u. Moreover, φ2 (W1−1 ) ∝ φ3 (W1−1 ) ∝ · · · ∝ φm+1 (W1−1 ). By
the inductive assumption, we have
"
Retarded Nonlinear Integral
Inequalities
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and x3 , y3 ∈ R+ are chosen such that
Z
(3.11)
αi+1 (x3 )
Z
∞
f˜i+1 (x̃1 , ỹ1 , s, t)dtds
Φi+1 (ci (x3 , y3 )) +
αi+1 (0)
βi+1 (y3 )
Z
W1 (∞)
≤
ũi+1
dz
φi+1 (W1−1 (z))
for i = 1, . . . , m.
Note that
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
Z
u
dz
−1
ũ φi (W1 (z))
Z iu
w1 (W1−1 (z))dz
=
−1
W1 (ui ) wi (W1 (z))
Z W1−1 (u)
dz
=
= Wi ◦ W1−1 (u),
wi (z)
ui
Φi (u) =
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i = 2, . . . , m + 1.
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Wm+1 (W1−1 (cm (x, y)))
Z
(3.12)
J
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u(x, y) ≤ z(x, y) = W1−1 (ξ(x, y))
"
≤
II
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From (3.10), we have
−1
Wm+1
JJ
αm+1 (x)
Z
#
∞)
+
αm+1 (0)
Close
f˜m+1 (x̃1 , ỹ1 , s, t)dtds
βm+1 (y)
for all 0 ≤ x ≤ min{x̃1 , x3 }, max{ỹ1 , y3 } ≤ y < ∞. Let c̃i (x, y) = W1−1 (ci (x, y)).
Then,
c̃1 (x, y) = W1−1 (c1 (x, y))
"
α1 (x)
Z
= W1−1 W1 (b1 (x̃1 , ỹ1 )) +
α1 (0)
#
∞
Z
f˜1 (x̃1 , ỹ1 , s, t)dtds
β1 (y)
= b̃2 (x̃1 , ỹ1 , x, y).
Retarded Nonlinear Integral
Inequalities
Moreover, with the assumption that c̃m (x, y) = b̃m+1 (x̃1 , ỹ1 , x, y), we have
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
c̃m+1 (x, y)
"
= W1−1 Φ−1
m+1 (Φm+1 (cm (x, y)) +
Z
αm+1 (x)
αm+1 (0)
"
−1
Wm+1 (W1−1 (cm (x, y))) +
= Wm+1
Z
αm+1 (0)
"
−1
= Wm+1
Wm+1 (c̃m (x, y)) +
Z
αm+1 (x)
Z
αm+1 (0)
"
=
−1
Wm+1
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f˜m+1 (x̃1 , ỹ1 , s, t)dtds)
βm+1 (y)
αm+1 (x)
Z
#
∞
Z
Contents
#
∞
f˜m+1 (x̃1 , ỹ1 , s, t)dtds
βm+1 (y)
#
∞
f˜m+1 (x̃1 , ỹ1 , s, t)dtds
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βm+1 (y)
Z
αm+1 (x)
Z
Wm+1 (b̃m+1 (x̃1 , ỹ1 , x, y)) +
αm+1 (0)
#
∞
f˜m+1 (x̃1 , ỹ1 , s, t)dtds
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βm+1 (y)
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= b̃m+2 (x̃1 , ỹ1 , x, y).
This proves that
c̃i (x, y) = b̃i+1 (x̃1 , ỹ1 , x, y),
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i = 1, . . . , m.
Therefore, (3.11) becomes
Z
αi+1 (x3 )
Z
∞
f˜i+1 (x̃1 , ỹ1 , s, t)dtds
Wi+1 (b̃i+1 (x̃1 ,ỹ1 , x3 , y3 )) +
αi+1 (0)
Z
W1 (∞)
≤
Z
βi+1 (y3 )
ũi+1
∞
=
ui+1
dz
φi+1 (W1−1 (z))
dz
, i = 1, . . . , m.
wi+1 (z)
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
The above inequalities and (3.5) imply that we may take x2 = x3 , y2 = y3 . From
(3.12) we get
"
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Contents
−1
u(x, y) ≤ Wm+1
Wm+1 (b̃m+1 (x̃1 , ỹ1 , x, y))
Z
αm+1 (x)
Z
#
∞
f˜m+1 (x̃1 , ỹ1 , s, t)dtds
+
αm+1 (0)
βm+1 (y)
(3.13)
u(x̃1 , ỹ1 ) ≤
Wn (b̃n (x̃1 , ỹ1 , x̃1 , ỹ1 ))
Z
αn (x̃1 )
Z
αn (0)
#
∞
+
βn (ỹ1 )
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for all 0 ≤ x ≤ x̃1 ≤ x2 , y2 ≤ ỹ1 ≤ y < ∞. This proves (3.3) by mathematical
induction.
Taking x = x̃1 , y = ỹ1 , x2 = x1 and y2 = y1 , we have
"
Wn−1
JJ
f˜n (x̃1 , ỹ1 , s, t)dtds
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for 0 ≤ x̃1 ≤ x1 , y1 ≤ ỹ1 < ∞. It is easy to verify that b̃n (x̃1 , ỹ1 , x̃1 , ỹ1 ) =
bn (x̃1 , ỹ1 ). Thus, (3.13) can be written as
#
"
Z αn (x̃1 ) Z ∞
u(x̃1 , ỹ1 ) ≤ Wn−1 Wn (bn (x̃1 , ỹ1 )) +
f˜n (x̃1 , ỹ1 , s, t)dtds .
αn (0)
βn (ỹ1 )
Since x̃1 , ỹ1 are arbitrary, replace x̃1 and ỹ1 by x and y respectively and we have
"
#
Z αn (x) Z ∞
u(x, y) ≤ W −1 Wn (bn (x, y)) +
f˜n (x, y, s, t)dtds
n
αn (0)
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
βn (y)
for all 0 ≤ x ≤ x1 , y1 ≤ y < ∞.
In case a(x, y) = 0 for some x, y ∈ R+ . Let b1, (x, y) := b1 (x, y) + for all
x, y ∈ R+ , where > 0 is arbitrary, and then b1, (x, y) > 0. Using the same
arguments as above, where b1 (x, y) is replaced with b1, (x, y) > 0, we get
"
#
Z αn (x) Z ∞
d˜n (x, y, s, t)dtds .
u(x, y) ≤ W −1 Wn (bn, (x, y)) +
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n
αn (0)
βn (y)
Letting → 0+ , we obtain (2.1) by the continuity of b1, in and the continuity of
Wi and Wi−1 under the notation W1 (0) := 0.
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4.
Applications
Consider the retarded partial differential equation
p
1
+
exp
(−x)
exp
(−y)
|v(x, y)|
(x + 1)2 (y + 1)2
x
x
3
+ x exp −
exp (−3y)v
, 3y ,
4
2
2
v(x, ∞) = σ(x), v(0, y) = τ (y), v(0, ∞) = k,
D1 D2 v(x, y) =
(4.1)
(4.2)
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
for x, y ∈ R+ , where σ, τ ∈ C(R+ , R), σ(x) is nondecreasing in x, τ (y) is nonincreasing in y, and k is a real constant. Integrating (4.1) with respect to x and y and
using the initial conditions (4.2), we get
x
v(x, y) = σ(x) + τ (y) − k −
(x + 1)(y + 1)
Z xZ ∞
p
−
exp (−s) exp (−t) |v(s, t)|dtds
0
Z yZ
s
s
3 x ∞
−
s exp − exp (−3t)v( , 3t)dtds
4 0 y
2
2
x
= σ(x) + τ (y) − k −
(x + 1)(y + 1)
Z xZ ∞
p
−
exp (−s) exp (−t) |v(s, t)|dtds
0
Z
−
Z
∞
s exp (−s) exp (−t)v(s, t)dtds.
0
Thus,
y
x
2
3y
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x
|v(x, y)| ≤ |σ(x) + τ (y) − k| +
(x + 1)(y + 1)
Z xZ ∞
p
+
exp (−s) exp (−t) |v(s, t)|dtds
0
y
x
2
Z
Z
∞
+
s exp (−s) exp (−t)|v(s, t)|dtds.
0
Letting u(x, y) = |v(x, y)|, we have
Z α1 (x) Z
u(x, y) ≤ a(x, y) +
α1 (0)
3y
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
∞
vol. 9, iss. 2, art. 57, 2008
f1 (x, y, s, t)w1 (u)dtds
β1 (y)
Z
α2 (x)
Z
f2 (x, y, s, t)w2 (u)dtds,
α2 (0)
where
x
,
2
f1 (x, y, s, t) = exp (−s) exp (−t),
β1 (y) = y,
w2 (u)
Clearly, w
=
1 (u)
u1 , u2 > 0
√u
u
√
x
,
(x + 1)(y + 1)
β2 (y) = 3y,
u
W1 (u) =
u1
w1 (u) =
√
u,
w2 (u) = u,
f2 (x, y, s, t) = s exp (−s) exp (−t).
u is nondecreasing for u > 0, that is, w1 ∝ w2 . Then for
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f˜1 (x, y, s, t) = f1 (x, y, s, t),
b1 (x, y) = a(x, y),
Z
=
α2 (x) =
Contents
β2 (y)
a(x, y) = |σ(x) + τ (y) − k| +
α1 (x) = x,
Title Page
∞
+
√
√ dz
√ = 2 u − u1 ,
z
f˜2 (x, y, s, t) = f2 (x, y, s, t),
W1−1 (u)
=
u
2
√ 2
+ u1 ,
Z
u
dz
u
= ln ,
W2−1 (u) = u2 exp(u),
z
u
2
u2
Z xZ ∞
−1
b2 (x, y) = W1 W1 (b1 (x, y)) +
f˜1 (x, y, s, t)dtds
0
y
h p
i
√ = W1−1 2
b1 (x, y) − u1 + (1 − exp (−x)) exp (−y)
2
p
1
=
b1 (x, y) + (1 − exp (−x)) exp (−y) .
2
W2 (u) =
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
By Theorem 2.1, we have
"
|v(x, y)| ≤ W2−1 W2 (b2 (x, y)) +
Z
0
x
2
Z
#
∞
d˜2 (x, y, s, t)dtds
Title Page
3y
x
x b2 (x, y) −1
= W2 ln
+ 1−
+ 1 exp −
exp (−3y)
u2
2
2
x
x b2 (x, y) + 1−
+ 1 exp −
exp (−3y)
= u2 exp ln
u2
2
2
h
x
x i
= b2 (x, y) exp 1 −
+ 1 exp −
exp (−3y)
2
2
r
2
x
1
|σ(x) + τ (y) − k| +
+ (1 − exp (−x)) exp (−y)
=
(x + 1)(y + 1) 2
h
x
x i
× exp 1 −
+ 1 exp −
exp (−3y) .
2
2
This implies that the solution of (4.1) is bounded for x, y ∈ R+ provided that σ(x) +
τ (y) − k is bounded for all x, y ∈ R+ .
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References
[1] B.G. PACHPATTE, On some new nonlinear retarded integral inequalities, J.
Inequal. Pure Appl. Math., 5(3) (2004), Art. 80. [ONLINE: http://jipam.
vu.edu.au/article.php?sid=436].
[2] F.C. JIANG AND F.W. MENG, Explicit bounds on some new nonlinear integral
inequalities with delay, J. Comput. Appl. Math., 205 (2007), 479–486.
[3] M. PINTO, Integral inequalities of Bihari-type and applications, Funkcial. Ekvac., 33 (1990), 387–403.
[4] O. LIPOVAN, A retarded integral inequality and its applications, J. Math. Anal.
Appl., 285 (2003), 436–443.
[5] Q.H. MA AND E.H. YANG, On some new nonlinear delay integral inequalities,
J. Math. Anal. Appl., 252 (2000), 864–878.
[6] R.P. AGARWAL, S.F. DENG AND W.N. ZHANG, Generalization of a retarded
Gronwall-like inequality and its applications, Appl. Math. Comput., 165 (2005),
599–612.
[7] S.K. CHOI, S.F. DENG, N.J. KOO AND W.N. ZHANG, Nonlinear integral
inequalities of Bihari-type without class H, Math. Inequal. Appl., 8 (2005),
643–654.
[8] S.S. DRAGOMIR AND Y.H. KIM, Some integral inequalities for functions of
two variables, Electron. J. Differential Equations, 2003 (2003), Art.10.
[9] W.S. CHEUNG AND Q.H. MA, On certain new Gronwall-Ou-Ing type integral inequalities in two variables and their applications, J. Inequal. Appl., 2005
(2005), 347–361.
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
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[10] W.S. WANG, A generalized retarded Gronwall-like inequality in two variables
and applications to BVP, Appl. Math. Comput., 191 (2007), 144–154.
[11] X.Q. ZHAO AND F.W. MENG, On some advanced integral inequalities and
their applications, J. Inequal. Pure Appl. Math., 6(3) (2005), Art. 60. [ONLINE:
http://jipam.vu.edu.au/article.php?sid=533].
[12] Y.H. KIM, On some new integral inequalities for functions in one and two
variables, Acta Math. Sinica, 2(2) (2005), 423–434.
Retarded Nonlinear Integral
Inequalities
Kelong Zheng
vol. 9, iss. 2, art. 57, 2008
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