Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2013, Article ID 853476, 10 pages
http://dx.doi.org/10.1155/2013/853476
Research Article
A Generalized Nonlinear Gronwall-Bellman Inequality with
Maxima in Two Variables
Yong Yan
Department of Mathematics, Sichuan University for Nationalities, Kangding, Sichuan 626001, China
Correspondence should be addressed to Yong Yan; kdyan698@163.com
Received 15 November 2012; Accepted 20 January 2013
Academic Editor: Jitao Sun
Copyright © 2013 Yong Yan. 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.
This paper deals with a generalized form of nonlinear retarded Gronwall-Bellman type integral inequality in which the maximum
of the unknown function of two variables is involved. This form includes both a nonconstant term outside the integrals and more
than one distinct nonlinear integrals. Requiring neither monotonicity nor separability of given functions, we apply a technique of
monotonization to estimate the unknown function. Our result can be used to weaken conditions for some known results. We apply
our result to a boundary value problem of a partial differential equation with maxima for uniqueness.
1. Introduction
The Gronwall-Bellman inequality [1, 2] plays an important
role in the study of existence, uniqueness, boundedness,
stability, invariant manifolds, and other qualitative properties
of solutions of differential equations and integral equations.
There can be found a lot of its generalizations in various cases
from literatures (see, e.g., [3–18]). In 1956, Bihari [3] discussed
the integral inequality
𝑡
𝑢 (𝑡) ≤ 𝑐 + ∫ 𝑓 (𝑠) 𝜔 (𝑢 (𝑠)) 𝑑𝑠,
0
𝑡 ≥ 0,
Their results were further generalized by Agarwal et al. [5] to
the inequality
𝑛
𝑖=1 𝑏𝑖 (𝑡0 )
𝜔1 ∝ 𝜔2 ∝ ⋅ ⋅ ⋅ ∝ 𝜔𝑛 ,
𝑢𝑝 (𝑥, 𝑦) ≤ 𝑎 (𝑥, 𝑦)
+∫
𝑏(𝑡)
𝑏(𝑡0 )
(2)
𝑔 (𝑠) 𝜔 (𝑢 (𝑠)) 𝑑𝑠,
𝑏𝑖 (𝑥)
∫
𝑐𝑖 (𝑦)
𝑖=1 𝑏𝑖 (𝑥0 ) 𝑐𝑖 (𝑦0 )
𝑡0
𝑡0 ≤ 𝑡 < 𝑡1 .
(4)
that is, each ratio 𝜔𝑖+1 /𝜔𝑖 is also nondecreasing on 𝐴 ⊆ R\{0},
called in [6] that 𝜔𝑖+1 is stronger nondecreasing than 𝜔𝑖 . On
the basis of this work, Wang [7] considered the inequality of
two variables
𝑛
𝑢 (𝑡) ≤ 𝑐 + ∫ 𝑓 (𝑠) 𝜔 (𝑢 (𝑠)) 𝑑𝑠
𝑡0 ≤ 𝑡 < 𝑡1 ,
where the constant 𝑐 is replaced with a function 𝑎(𝑡), 𝑏𝑖 ’s
are continuously differentiable and nondecreasing functions,
and 𝜔𝑖 ’s are continuous and nondecreasing positive functions
such that
+ ∑∫
𝑡
𝑓𝑖 (𝑡, 𝑠) 𝜔𝑖 (𝑢 (𝑠)) 𝑑𝑠,
(3)
(1)
where 𝑐 > 0 is a constant, 𝑓 is a continuous and nonnegative
function, and 𝜔 is a continuous and nondecreasing positive
function. Replacing 𝑡 by a function 𝑏(𝑡) in (1), Lipovan [4]
investigated the retarded integral inequality
𝑏𝑖 (𝑡)
𝑢 (𝑡) ≤ 𝑎 (𝑡) + ∑ ∫
𝑓𝑖 (𝑥, 𝑦, 𝑠, 𝑡) 𝜔𝑖 (𝑢 (𝑠, 𝑡)) 𝑑𝑡 𝑑𝑠,
(5)
where the functions 𝑎, 𝑓𝑖 , and 𝜔𝑖 are not required to be
monotone, and those 𝜔𝑖 ’s are not required to be stronger
2
Journal of Applied Mathematics
monotone than the one after the next as shown in (4). This
inequality belongs to both the case of multivariables, to which
great attentions [7–11] have been paid, and to the case that
the left-hand side is a composition of the unknown function
with a known function, in which Ou-Iang’s idea [19] was
applied [11–14]. He applied a technique of monotonization
to construct a sequence of functions, made each function
possess stronger monotonization than the previous one, and
gave an estimate for the unknown function 𝑢.
On the other aspect, many problems in the control theory
can be modeled in the form of differential equations with
the maxima of the unknown function [20–22]. In connection
with the development of the theory of differential equations
with maxima (see, e.g., [20, 21, 23]) and partial differential
equations with maxima [24, 25], a new type of integral
inequalities with maxima is required, respectively. There have
been given some results for integral inequalities containing
the maxima of the unknown function [23, 26–28]. Concretely,
in 2012, Bohner et al. [26] discussed the following system of
integral inequalities:
𝑚
𝜑 (𝑢 (𝑡)) ≤ 𝑎 (𝑡)+ ∑ ∫
𝛼𝑖 (𝑡)
𝑝
𝑖=1 𝛼𝑖 (𝑡0 )
𝑚+𝑛
+ ∑ ∫
𝛼𝑗 (𝑡)
𝑗=𝑚+1 𝛼𝑗 (𝑡0 )
𝑓𝑖 (𝑠) 𝑢 (𝑠) 𝜔𝑖 (𝑢 (𝑠)) 𝑑𝑠
𝑓𝑗 (𝑠) 𝑢𝑝 (𝑠)
(6)
× 𝜔𝑗 ( max 𝑔 (𝑢 (𝜉))) 𝑑𝑠,
𝜉∈[𝑠−ℎ,𝑠]
𝑢 (𝑡) ≤ 𝜓 (𝑡) ,
where 𝑎, 𝑓𝑖 ’s, 𝜔𝑖 ’s, 𝜑, and 𝜓 are nonnegative continuous
functions and 𝛼𝑖 ’s are nonnegative continuously differentiable
and nondecreasing functions. They required that 𝑎(𝑡) ≥ 1,
𝜑 is 𝐶1 on R+ := [0, +∞) and increasing such that 𝜑(𝑡𝑥) ≥
𝑡𝜑(𝑥) for 0 ≤ 𝑡 ≤ 1, and 𝜔𝑖 satisfies the following: (i) 𝜔𝑖 ∈
𝐶1 (R+ , R+ ) is an increasing function, and (ii) 𝜔𝑖 (𝑡𝑥) ≥ 𝑡𝜔𝑖 (𝑥)
for all 0 ≤ 𝑡 ≤ 1 and 𝑥 > 0. Bainov and Hristova [23]
considered the following system:
𝑥
𝑦
𝑥0
𝑦0
𝑢 (𝑥, 𝑦) ≤ 𝑎 (𝑥, 𝑦) + ∫ ∫ 𝑓 (𝑠, 𝑡) 𝑢𝑝 (𝑠, 𝑡) 𝑑𝑡 𝑑𝑠
+∫
𝛼(𝑥)
𝑦
𝜑 (𝑢 (𝑥, 𝑦))
≤ 𝑎 (𝑥, 𝑦)
𝑛
+ ∑∫
𝛼𝑖 (𝑥)
∫
𝛽𝑖 (𝑦)
𝑖=1 𝛼𝑖 (𝑥0 ) 𝛽𝑖 (𝑦0 )
𝑛+𝑚
𝛼𝑗 (𝑥)
+ ∑ ∫
𝑓𝑖 (𝑥, 𝑦, 𝑠, 𝑡) 𝜔𝑖 (𝑢 (𝑠, 𝑡)) 𝑑𝑡 𝑑𝑠
𝛽𝑗 (𝑦)
∫
𝑗=𝑛+1 𝛼𝑗 (𝑥0 ) 𝛽𝑗 (𝑦0 )
𝑓𝑗 (𝑥, 𝑦, 𝑠, 𝑡)
× 𝜔𝑗 ( max 𝑔 (𝑢 (𝜉, 𝑡))) 𝑑𝑡 𝑑𝑠,
𝜉∈[𝑠−ℎ,𝑠]
(𝑥, 𝑦) ∈ [𝑥0 , 𝑥1 ) × [𝑦0 , 𝑦1 ) ,
𝑢 (𝑥, 𝑦) ≤ 𝜓 (𝑥, 𝑦) ,
(𝑥, 𝑦) ∈ [𝛼∗ (𝑥0 )−ℎ, 𝑥0 ]×[𝑦0 , 𝑦1 ) ,
(8)
where 𝑎, 𝑓𝑖 ’s, 𝜔𝑖 ’s, and 𝑔 are continuous and nonnegative
functions, 𝛼𝑖 ’s and 𝛽𝑖 ’s are nonnegative continuously differentiable and nondecreasing functions, and 𝛼∗ (𝑥0 ) :=
min1≤𝑖≤𝑚+𝑛 𝛼𝑖 (𝑥0 ). As required in previous works [27–29], we
suppose that 0 ≤ 𝛼𝑖 (𝑡) ≤ 𝑡, 0 ≤ 𝛽𝑖 (𝑡) ≤ 𝑡, ℎ > 0 is
constant. In this paper, we require neither monotonicity of
𝑎, 𝜔𝑖 ’s, 𝑓𝑖 ’s, and 𝑔 nor 𝑎(𝑥, 𝑦) ≥ 1. We monotonize those 𝜔𝑖 ’s
to make a sequence of functions in which each one possesses
stronger monotonicity than the previous one so as to give an
estimation for the unknown function. We can use our result
to discuss inequalities (6) and (7), giving the stronger results
under weaker conditions. We finally apply the obtained result
to a boundary value problem of a partial differential equation
with maxima for uniqueness.
2. Main Result
Consider system (8) of integral inequalities with 𝑥0 < 𝑥1 and
𝑦0 < 𝑦1 in R+ := [0, ∞). Let Λ := [𝑥0 , 𝑥1 ) × [𝑦0 , 𝑦1 ), Ω :=
[𝛼∗ (𝑥0 ) − ℎ, 𝑥0 ] × [𝑦0 , 𝑦1 ). Suppose that
(H1 ) 𝛼𝑖 : [𝑥0 , 𝑥1 ) → R+ (𝑖 = 1, 2, . . . , 𝑚 + 𝑛) and
𝛽𝑖 : [𝑦0 , 𝑦1 ) → [𝑦0 , 𝑦1 ), 𝑖 = 1, 2, . . . , 𝑚 +
𝑛, are nondecreasing such that 𝛼𝑖 (𝑥) ≤ 𝑥 on
[𝑥0 , 𝑥1 ), 𝛽𝑖 (𝑦) ≤ 𝑦 on [𝑦0 , 𝑦1 ) and 𝛽𝑖 (𝑦0 ) = 𝑦0 ;
(H2 ) all 𝑓𝑖 ’s (𝑖 = 1, 2, . . . , 𝑚 + 𝑛) are continuous and
nonnegative functions on Λ × [𝛼∗ (𝑥0 ), 𝑥1 ) × [𝑦0 , 𝑦1 );
𝑝
∫ ℎ (𝑠, 𝑡) ( max 𝑢 (𝜉, 𝑡)) 𝑑𝑡 𝑑𝑠,
𝛼(𝑥0 ) 𝑦0
In this paper, we consider the system of integral inequalities as follows:
𝜉∈[𝑠−ℎ,𝑠]
𝑢 (𝑥, 𝑦) ≤ 𝜓 (𝑥, 𝑦) ,
(7)
where 𝑎(𝑥, 𝑦) is nonnegative and nondecreasing in both of
its arguments, 𝑓, ℎ, and 𝜓 are continuous and nonnegative
functions, and 𝑝 ∈ (0, 1].
(H3 ) 𝑔, 𝜑 : R+ → R+ and 𝜓 : [𝛼∗ (𝑥0 ) − ℎ, 𝑥1 ) → R+
are continuous, and 𝜑 is strictly increasing such that
lim𝑡 → ∞ 𝜑(𝑡) = +∞;
(H4 ) all 𝜔𝑖 ’s (𝑖 = 1, 2, . . . , 𝑚 + 𝑛) are continuous on R+ and
positive on (0, +∞);
(H5 ) 𝑎(𝑥, 𝑦) is a continuous and nonnegative function on
Λ.
Journal of Applied Mathematics
3
̃ 𝑖 (𝑡), 𝑖 = 1, . . . , 𝑚 + 𝑛,
For those 𝜔𝑖 ’s given in (H4 ), define 𝜔
inductively by
for 𝑖 = 1, 2, . . . , 𝑚 + 𝑛 − 1, and 𝑋1 ∈ [𝑥0 , 𝑥1 ), 𝑌1 ∈ [𝑦0 , 𝑦1 ) are
chosen such that
̃ 1 (𝑡) := max {𝜔1 (𝜏)} ,
𝜔
𝜏∈[0,𝑡]
(9)
𝜔 (𝜏)
̃ 𝑖 (𝑡) ,
̃ 𝑖+1 (𝑡) := max { 𝑖+1
}𝜔
𝜔
𝜏∈[0,𝑡] 𝜔
̃ 𝑖 (𝜏) + 𝜖𝑖
𝜏∈[0,𝑡]
̂ 𝑚+1 (max𝑠∈[0,𝜏] {𝑔 (𝑠)})
𝜔
̃ 𝑚 (𝑡) ,
}𝜔
̃ 𝑚 (𝜏) + 𝜖𝑚
𝜔
̃ 𝑗+1 (𝑡) := max {
𝜔
̂ 𝑗+1 (max𝑠∈[0,𝜏] {𝑔 (𝑠)})
𝜔
̃ 𝑗 (𝜏) + 𝜖𝑗
𝜔
𝜏∈[0,𝑡]
̃ 𝑗 (𝑡) ,
}𝜔
(10)
̂ 𝑗 (𝑡) := max𝜏∈[0,𝑡] {𝜔𝑗 (𝜏)} for
for 𝑗 = 𝑚 + 1, . . . , 𝑚 + 𝑛, where 𝜔
̃ 𝑖 (0) = 0 or := 0 if 𝜔
̃ 𝑖 (0) ≠ 0
𝑗 = 𝑚 + 1, . . . , 𝑚 + 𝑛, 𝜖𝑖 := 𝜀1 if 𝜔
for 𝑖 = 1, 2, . . . , 𝑚 + 𝑛 − 1, and 𝜀1 > 0 be a given very small
constant.
hold,
Theorem
1. Suppose
that
(H1 )–(H5 )
max𝑠∈[𝛼∗ (𝑥0 )−ℎ,𝑥0 ] 𝜓(𝑠, 𝑦) ≤ 𝜑−1 (𝑎(𝑥0 , 𝑦)) for all 𝑦 ∈ [𝑦0 , 𝑦1 )
and 𝑢 ∈ 𝐶(Ω, R+ ) satisfies the system (8) of integral
inequalities. Then,
𝑢𝑖
−1
(Ω𝑚+𝑛 (𝑥, 𝑦))) ,
𝑢 (𝑥, 𝑦) ≤ 𝜑−1 (𝑊𝑚+𝑛
𝑥
𝑦
0
0
+ 4∫
√𝑥
𝑦
∫ 𝑡𝑠 ( max 𝑢 (𝜉, 𝑡) + 1) 𝑑𝑡 𝑑𝑠,
𝛽𝑖 (𝑦)
(𝑥, 𝑦) ∈ [0, 𝑥1 ) × [0, 𝑦1 ) ,
𝑑𝑠
,
̃ 𝑖 (𝜑−1 (𝑠))
𝜔
+ 4∫
𝑢 ≥ 𝑢𝑖 , 𝑖 = 1, 2, . . . , 𝑚 + 𝑛,
(13)
max
𝑟𝑖+1 (𝑥, 𝑦) =
0
0
√𝑥
(Ω𝑖 (𝑥, 𝑦)) ,
(14)
𝑦
∫ 𝑡𝑠 ( max 𝑢 (𝜉, 𝑡) + 1) 𝑑𝑡 𝑑𝑠, (17)
𝜉∈[𝑠−ℎ,𝑠]
0
(𝑥, 𝑦) ∈ [0, 𝑥1 ) × [0, 𝑦1 ) ,
(𝑥, 𝑦) ∈ [−ℎ, 0] × [0, 𝑦1 ) ,
𝑢 (𝑥, 𝑦) ≤ 𝑥 + 3,
by enlarging √𝑠 + 1 to 𝑠 + 1. Applying Theorem 1, we obtain
𝑎 (𝜄, 𝜉) ,
(𝜄,𝜉)∈[𝑥0 ,𝑥]×[𝑦0 ,𝑦]
𝑊𝑖−1
𝑦
0
̃ 𝑖 is defined just before the theorem,
𝑢𝑖 > 0 is a given constant, 𝜔
and 𝑟𝑖 (𝑥, 𝑦) is defined recursively by
𝑟1 (𝑥, 𝑦) =
𝑥
𝑢 (𝑥, 𝑦) ≤ 3 + 4 ∫ ∫ 𝑡𝑠 (𝑢 (𝑠, 𝑡) + 1) 𝑑𝑡 𝑑𝑠
𝑊𝑖−1 is the inverse of the function
𝑢𝑖
(𝑥, 𝑦) ∈ [−ℎ, 0] × [0, 𝑦1 ) ,
𝑓𝑖 (𝜄, 𝜉, 𝑠, 𝑡) 𝑑𝑡 𝑑𝑠,
max
(12)
𝑢
(16)
implies that
𝛼𝑖 (𝑥0 ) 𝛽𝑖 (𝑦0 ) (𝜄,𝜉)∈[𝑥0 ,𝑥]×[𝑦0 ,𝑦]
𝑊𝑖 (𝑢) := ∫
𝜉∈[𝑠−ℎ,𝑠]
0
𝑢 (𝑥, 𝑦) ≤ 𝑥 + 3,
Ω𝑖 (𝑥, 𝑦) := 𝑊𝑖 (𝑟𝑖 (𝑥, 𝑦))
∫
(15)
𝑢 (𝑥, 𝑦) ≤ 3 + 4 ∫ ∫ 𝑡𝑠√𝑢 (𝑠, 𝑡) + 1𝑑𝑡 𝑑𝑠
(11)
for all (𝑥, 𝑦) ∈ [𝑥0 , 𝑋1 ) × [𝑦0 , 𝑌1 ), where
𝛼𝑖 (𝑥)
𝑑𝑠
,
̃ 𝑖 (𝜑−1 (𝑠))
𝜔
For the special choice that 𝑛 = 𝑚 = 1, 𝜔1 (𝑠) = 𝑠𝑝 , 𝜔2 (𝑠) =
𝑠, 𝑓1 (𝑥, 𝑦, 𝑠, 𝑡) = 𝑓(𝑠, 𝑡), 𝑓2 (𝑥, 𝑦, 𝑠, 𝑡) = ℎ(𝑠, 𝑡), 𝛼1 (𝑠) = 𝑠,
𝛼2 (𝑠) = 𝛼(𝑠), 𝑔(𝑠) = 𝑠𝑝 , and 𝛽1 (𝑠) = 𝛽2 (𝑠) = 𝑠, where 𝛼 is
a nonnegative continuously differentiable and nondecreasing
function, Theorem 1 gives an estimate for the unknown 𝑢 in
the system (7). we require neither the monotonicity of 𝑎 nor
the monotonicity of 𝜔𝑖 . Obviously, Lemma 2 and Theorem 1
are applicable to more general forms than Corollary 2.3.4 in
[23]. Even if 𝜔𝑖 (𝑠) is enlarged to max1≤𝑖≤𝑚+𝑛 𝜔𝑖 (𝑠) such that (8)
is changed into the form of (2.1) in [29], where 𝑚 = 𝑛 = 1,
our theorem gives a better estimate. For example, the system
of inequalities
0
+∫
∞
for 𝑖 = 1, 2, . . . , 𝑚 + 𝑛.
for 𝑖 = 1, 2, . . . , 𝑚 − 1 and
̃ 𝑚+1 (𝑡) := max {
𝜔
Ω𝑖 (𝑋1 , 𝑌1 ) ≤ ∫
𝑢 (𝑥, 𝑦) ≤
(𝑥2 𝑦2 + 4)
4
2
2
𝑒𝑥𝑦 ,
(𝑥, 𝑦) ∈ [0, 𝑥1 ) × [0, 𝑦1 ) .
(18)
4
Journal of Applied Mathematics
On the other hand, Theorem 2.2 of [29] gives from (17) that
2 2
2
𝑢 (𝑡) ≤ 4𝑒(𝑥 𝑦 +𝑥𝑦 ) ,
(𝑥, 𝑦) ∈ [0, 𝑥1 ) × [0, 𝑦1 ) .
for all (𝑥, 𝑦) ∈ [𝑥0 , 𝑋∗ ) × [𝑦0 , 𝑌∗ ), where 𝐻𝑖−1 is the inverse of
the function
(19)
Clearly, (18) is sharper than (19) for large 𝑥 and 𝑦.
In order to prove Theorem 1, we need the following
lemma.
Lemma 2. Suppose that
(C1) 𝛼𝑖 : [𝑥0 , 𝑥1 ) → R+ (𝑖 = 1, 2, . . . , 𝑚 + 𝑛) and 𝛽𝑖 :
[𝑦0 , 𝑦1 ) → R+ (𝑖 = 1, 2, . . . , 𝑚+𝑛) are nondecreasing
such that 𝛼𝑖 (𝑥) ≤ 𝑥 on [𝛼∗ (𝑥0 ), 𝑥1 ) and 𝛽𝑖 (𝑦) ≤ 𝑦 on
[𝑦0 , 𝑦1 ) and 𝛽𝑖 (𝑦0 ) = 𝑦0 ;
(C2) 𝜓 ∈ 𝐶([𝛼∗ (𝑥0 ) − ℎ, 𝑥1 ) × [𝑦0 , 𝑦1 ), R+ ), 𝑏𝑖 ∈ 𝐶([𝛼∗ (𝑥0 ),
𝑥1 ) × [𝑦0 , 𝑦1 ), R+ ) for 𝑖 = 1, 2, . . . , 𝑚 + 𝑛;
(C3) all ℎ𝑖 ’s (𝑖 = 1, 2, . . . , 𝑚 + 𝑛) are continuous and nondecreasing on R+ and positive on (0, +∞) such that
ℎ1 ∝ ℎ2 ∝ . . . ∝ ℎ𝑚+𝑛 ;
(C4) 𝑎(𝑥, 𝑦) is continuously differentiable in 𝑥 and
𝑦, nonnegative on [𝛼∗ (𝑥0 ), 𝑥1 ) × [𝑦0 , 𝑦1 ), and
max𝑠∈[𝛼∗ (𝑥0 )−ℎ,𝑥1 ] 𝜓(𝑠, 𝑦) ≤ 𝑎(𝑥0 , 𝑦) for all 𝑦 ∈ [𝑦0 , 𝑦1 ).
If 𝑢 ∈ 𝐶([𝛼∗ (𝑥0 ) − ℎ, 𝑥1 ) × [𝑦0 , 𝑦1 ), R+ ) satisfies the system
of inequalities as follows:
𝑢
𝑑𝑥
, 𝑢 ≥ 𝑢𝑖 , 𝑖 = 1, 2, . . . , 𝑚 + n, (22)
𝑢𝑖 𝜔𝑖 (𝑥)
𝑢𝑖 > 0 is a given constant, and 𝛾𝑖 (𝑥, 𝑦) is defined recursively by
𝐻𝑖 (𝑢) := ∫
𝑥
𝑦
0
0
󵄨
󵄨
𝛾1 (𝑥, 𝑦) := 𝑎 (𝑥0 , 𝑦0 ) + ∫ 󵄨󵄨󵄨𝑎𝑥 (𝑡, 𝑦)󵄨󵄨󵄨 𝑑𝑡 + ∫
𝑥
𝑦
𝛾𝑖+1 (𝑥, 𝑦) := 𝐻𝑖−1 (𝐻𝑖 (𝛾𝑖 (𝑥, 𝑦))+∫
𝛼𝑖 (𝑡)
∫
𝛼𝑖 (𝑡0 ) 𝛽𝑖 (𝑡0 )
𝑚
𝛼𝑖 (𝑥)
∫
𝛽𝑖 (𝑦)
𝑖=1 𝛼𝑖 (𝑥0 ) 𝛽𝑖 (𝑦0 )
𝑚+𝑛
𝛼𝑗 (𝑥)
+ ∑ ∫
∫
𝛽𝑗 (𝑦)
𝑗=𝑚+1 𝛼𝑗 (𝑥0 ) 𝛽𝑗 (𝑦0 )
𝑏𝑖 (𝑠, 𝑡) ℎ𝑖 (𝑢 (𝑠, 𝑡)) 𝑑𝑡 𝑑𝑠
𝜉∈[𝑠−ℎ,𝑠]
(𝑥, 𝑦) ∈ Λ,
∗
∗
𝐻𝑖 (𝛾𝑖 (𝑋 , 𝑌 )) + ∫
𝛼𝑖 (𝑋∗ )
𝛼𝑖 (𝑥0 )
∫
𝛽𝑖 (𝑌∗ )
𝛽𝑖 (𝑦0 )
𝑏𝑖 (𝑠, 𝑡) 𝑑𝑡 𝑑𝑠 ≤ ∫
𝑢𝑖
𝑚
𝑢 (𝑥, 𝑦) ≤ 𝛾1 (𝑥, 𝑦) + ∑ ∫
𝛼𝑖 (𝑥)
∫
𝛽𝑖 (𝑦)
𝑖=1 𝛼𝑖 (𝑥0 ) 𝛽𝑖 (𝑦0 )
then
𝑢 (𝑥, 𝑦) ≤
𝑚+𝑛
+ ∑ ∫
(𝐻𝑚+𝑛 (𝛾𝑚+𝑛 (𝑥, 𝑦))
𝛼𝑚+𝑛 (𝑥)
+∫
𝛽𝑚+𝑛 (𝑦)
∫
𝛼𝑚+𝑛 (𝑥0 ) 𝛽𝑚+𝑛 (𝑦0 )
∞
𝑑𝑠
,
ℎ𝑖 (𝑠)
(24)
Proof. From (23), we see that 𝛾1 (𝑥, 𝑦) is nondecreasing on Λ,
𝑎(𝑥, 𝑦) ≤ 𝛾1 (𝑥, 𝑦), and max𝑠∈[𝛼∗ (𝑥0 )−ℎ,𝑥1 ] 𝜓(𝑠, 𝑦) ≤ 𝑎(𝑥0 , 𝑦) ≤
𝛾1 (𝑥0 , 𝑦) for 𝑦 ∈ [𝑦0 , 𝑦1 ). It implies from (20) that
(𝑥, 𝑦) ∈ Ω,
(20)
−1
𝐻𝑚+𝑛
(23)
for 𝑖 = 1, 2, . . . , 𝑚 + 𝑛.
ℎ𝑗 (𝑠, 𝑡) ℎ𝑗 ( max 𝑢 (𝜉, 𝑡)) 𝑑𝑡 𝑑𝑠,
𝑢 (𝑥, 𝑦) ≤ 𝜓 (𝑥, 𝑦) ,
𝑏𝑖 (𝑠, 𝑡) 𝑑𝑡𝑑𝑠)
for 𝑖 = 1, 2, . . . , 𝑚 + 𝑛 − 1, and 𝑥0 ≤ 𝑋∗ < 𝑥1 , 𝑦0 ≤ 𝑌∗ < 𝑦1
are chosen such that
𝑢 (𝑥, 𝑦)
≤ 𝑎 (𝑥, 𝑦) + ∑ ∫
𝛽𝑖 (𝑡)
󵄨
󵄨󵄨
󵄨󵄨𝑎𝑦 (𝑥, 𝑠)󵄨󵄨󵄨 𝑑𝑠,
󵄨
󵄨
𝛼𝑗 (𝑥)
∫
𝛽𝑗 (𝑦)
𝑗=𝑚+1 𝛼𝑗 (𝑥0 ) 𝛽𝑗 (𝑦0 )
𝑏𝑖 (𝑠, 𝑡) ℎ𝑖 (𝑢 (𝑠, 𝑡)) 𝑑𝑡 𝑑𝑠
𝑏𝑗 (𝑠, 𝑡)
× ℎ𝑗 ( max 𝑢 (𝜉, 𝑡)) 𝑑𝑡 𝑑𝑠
ℎ𝑚+𝑛 (𝑠, 𝑡) 𝑑𝑡𝑑𝑠) ,
𝜉∈[𝑠−ℎ,𝑠]
(25)
(21)
𝑚
𝛼 (𝑥)
for all (𝑥, 𝑦) ∈ Λ. Let
𝛽 (𝑦)
𝑖
𝑖
{
{
𝑏𝑖 (𝑠, 𝑡) ℎ𝑖 (𝑢 (𝑠, 𝑡)) 𝑑𝑡 𝑑𝑠
(𝑥,
𝑦)
+
𝛾
∫
∑
∫
{
1
{
{
𝛼𝑖 (𝑥0 ) 𝛽𝑖 (𝑦0 )
{
𝑖=1
{
{
{
𝑚+𝑛
𝛼𝑗 (𝑥)
𝛽𝑖 (𝑦)
𝑧 (𝑥, 𝑦) := {
+
𝑏𝑗 (𝑠, 𝑡) ℎ𝑗 ( max 𝑢 (𝜉, 𝑡)) 𝑑𝑡 𝑑𝑠,
∫
∑
∫
{
{
{
𝜉∈[𝑠−ℎ,𝑠]
{
𝑗=𝑚+1 𝛼𝑗 (𝑥0 ) 𝛽𝑖 (𝑦0 )
{
{
{
{
{𝛾1 (𝑥0 , 𝑦) ,
(𝑥, 𝑦) ∈ Λ,
(𝑥, 𝑦) ∈ Ω.
(26)
Journal of Applied Mathematics
5
Clearly, 𝑧(𝑥, 𝑦) is nondecreasing in 𝑥. Then, we have
𝑢 (𝑥, 𝑦) ≤ 𝑧 (𝑥, 𝑦) ,
(𝑥, 𝑦) ∈ [𝛼∗ (𝑥0 ) − ℎ, 𝜉) × [𝑦0 , 𝜂) ,
𝜉∈[𝑠−ℎ,𝑠]
𝜉∈[𝑠−ℎ,𝑠]
(33)
𝜉∈[𝑠−ℎ,𝑠]
(𝑠, 𝑦) ∈ [𝛼∗ (𝑥0 ) , 𝑥1 ) × [𝑦0 , 𝑦1 ) .
(27)
From (25), (27), and (28) and the definition of 𝑧(𝑥, 𝑦) on Λ,
we get
𝑚
𝑧 (𝑥, 𝑦) ≤ 𝛾1 (𝑥, 𝑦) + ∑ ∫
𝛼𝑖 (𝑥)
𝛽𝑖 (𝑦)
∫
𝑖=1 𝛼𝑖 (𝑥0 ) 𝛽𝑖 (𝑦0 )
𝑚+𝑛
̃ (𝑢 (𝜉, 𝑦)) ≤ 𝑔
̃ ( max 𝑢 (𝜉, 𝑦))
max 𝑔 (𝑢 (𝜉, 𝑦)) ≤ max 𝑔
𝜉∈[𝑠−ℎ,𝑠]
max 𝑢 (𝜉, 𝑦) ≤ max 𝑧 (𝜉, 𝑦) = 𝑧 (𝑠, 𝑦) ,
𝜉∈[𝑠−ℎ,𝑠]
which is nondecreasing in 𝑥 and 𝑦 for each fixed 𝑡 and 𝑠 and
̃ (𝑥, 𝑦, 𝑠, 𝑡) ≥ 𝑓 (𝑥, 𝑦, 𝑠, 𝑡) ≥ 0 for all 𝑖 = 1, 2, . . . , 𝑚 +
satisfies 𝑓
𝑖
𝑖
̃ implies that
𝑛. The monotonicity of 𝑔
+ ∑ ∫
𝛼𝑗 (𝑥)
∫
𝛽𝑖 (𝑦)
𝑗=𝑚+1 𝛼𝑗 (𝑥0 ) 𝛽𝑖 (𝑦0 )
max
𝑡 ≥ 0,
{𝑎 (𝜏, 𝜉)} ,
̂ 𝑖 (̃
̃ 𝑖 (𝑡) ,
𝜔
𝑔 (𝑡)) ≤ 𝜔
𝛼𝑗 (𝑥)
∫
𝛽𝑗 (𝑦)
̃ (𝑥, 𝑦, 𝑠, 𝑡)
𝑓
𝑗
× 𝜔𝑗 ( max 𝑔 (𝑢 (𝜉, 𝑡))) 𝑑𝑡 𝑑𝑠,
𝜉∈[𝑠−ℎ,𝑠]
𝑢 (𝑥, 𝑦) ≤ 𝜓 (𝑥, 𝑦) ,
(𝑥, 𝑦) ∈ Λ,
(𝑥, 𝑦) ∈ Ω.
(34)
Concerning (34), we consider the auxiliary system of inequalities
𝜑 (𝑢 (𝑥, 𝑦)) ≤ 𝑎̃ (𝑋, 𝑌)
𝑛
+ ∑∫
𝛼𝑖 (𝑥)
∫
𝛽𝑖 (𝑦)
̃ (𝑋, 𝑌, 𝑠, 𝑡)
𝑓
𝑖
× 𝜔𝑖 (𝑢 (𝑠, 𝑡)) 𝑑𝑡 𝑑𝑠
(29)
𝑚
+∑∫
𝛼𝑗 (𝑥)
∫
𝛽𝑖 (𝑦)
𝑗=1 𝛼𝑗 (𝑥0 ) 𝛽𝑖 (𝑦0 )
̃ (𝑋, 𝑌, 𝑠, 𝑡)
𝑓
𝑗
(35)
× 𝜔𝑗 ( max 𝑔 (𝑢 (𝜉, 𝑡))) 𝑑𝑡 𝑑𝑠,
𝜉∈[𝑠−ℎ,𝑠]
(𝑥, 𝑦) ∈ [𝑥0 , 𝑋) × [𝑦0 , 𝑌) ,
𝑢 (𝑥, 𝑦) ≤ 𝜓 (𝑥, 𝑦) ,
(𝑥, 𝑦) ∈ Ω,
where 𝑥0 ≤ 𝑋 ≤ 𝑋1 and 𝑦0 ≤ 𝑌 ≤ 𝑌1 are chosen arbitrarily,
and claim
𝑖 = 1, 2, . . . , 𝑚,
𝑖 = 𝑚 + 1, . . . , 𝑚 + 𝑛,
̃ (𝑥, 𝑦, 𝑠, 𝑡)
𝑓
𝑖
𝑖=1 𝛼𝑖 (𝑥0 ) 𝛽𝑖 (𝑦0 )
From (13), we see that the function 𝑊𝑖 is strictly increasing,
and therefore its inverse 𝑊𝑖−1 is well defined, continuous, and
increasing in its domain. The sequence {̃
𝜔𝑖 (𝑡)}, defined by
𝜔𝑖 (𝑠), consists of nondecreasing nonnegative functions on R+
and satisfies
̂ 𝑖 (𝑡) ,
𝜔𝑖 (𝑡) ≤ 𝜔
𝛽𝑖 (𝑦)
𝑗=𝑛+1 𝛼𝑗 (𝑥0 ) 𝛽𝑗 (𝑦0 )
(𝑥, 𝑦) ∈ [𝑥0 , 𝑥1 ) × [𝑦0 , 𝑦1 ) .
̃ 𝑖 (𝑡) ,
𝜔𝑖 (𝑡) ≤ 𝜔
∫
× 𝜔𝑖 (𝑢 (𝑠, 𝑡)) 𝑑𝑡𝑑𝑠
+ ∑ ∫
Proof of Theorem 1. First of all, we monotonize some given
functions 𝑓𝑖 , 𝜔𝑖 , 𝑔, and 𝑎 in the system (8) of integral
inequalities. Let
(𝜏,𝜉)∈[𝑥0 ,𝑥]×[𝑦0 ,𝑦]
𝛼𝑖 (𝑥)
𝑛+𝑚
Applying Theorem 1 of [7] to the case that 𝑓𝑖 (𝑥, 𝑦, 𝑠, 𝑡) =
𝑏𝑖 (𝑠, 𝑡), 𝑎(𝑥, 𝑦) = 𝛾1 (𝑥, 𝑦), 𝑝 = 1, and 𝜔𝑖 (𝑡) = ℎ𝑖 (𝑡), 𝑖 =
1, 2, . . . , 𝑚 + 𝑛, we obtain (21) from (28). This completes the
proof.
𝑎̃ (𝑥, 𝑦) :=
𝑛
+ ∑∫
𝑏𝑖 (𝑠, 𝑡) ℎ𝑖 (𝑢 (𝑠, 𝑡)) 𝑑𝑡 𝑑𝑠
(28)
𝜏∈[0,𝑡]
𝜑 (𝑢 (𝑥, 𝑦)) ≤ 𝑎̃ (𝑥, 𝑦)
𝑖=1 𝛼𝑖 (𝑥0 ) 𝛽𝑖 (𝑦0 )
𝑏𝑗 (𝑠, 𝑡) ℎ𝑗 (𝑧 (𝑠, 𝑡)) 𝑑𝑡 𝑑𝑠.
̃ (𝑡) := max {𝑔 (𝜏)} ,
𝑔
for (𝑠, 𝑦) ∈ [𝛼∗ (𝑥0 ) − ℎ, 𝑥1 ) × [𝑦0 , 𝑦1 ). From (8) and the
̃ (𝑥, 𝑦, 𝑠, 𝑡), we obtain
definition of 𝑓
𝑖
(30)
𝑖 = 𝑚 + 1, . . . , 𝑚 + 𝑛.
−1
̃ 𝑖 (𝑋, 𝑌, 𝑥, 𝑦))) ,
(Ω
𝑢 (𝑥, 𝑦) ≤ 𝜑−1 (𝑊𝑚+𝑛
(36)
for all 𝑥0 ≤ 𝑥 ≤ min{𝑋, 𝑋2 }, 𝑦0 ≤ 𝑦 ≤ min{𝑌, 𝑌2 }, where
̃ 𝑖 (𝑋, 𝑌, 𝑥, 𝑦) := 𝑊𝑖 (̃𝑟𝑖 (𝑋, 𝑌, 𝑥, 𝑦))
Ω
Moreover,
̃ 𝑖+1 ,
̃𝑖 ∝ 𝜔
𝜔
𝑖 = 1, 2, . . . , 𝑚 + 𝑛,
(31)
̃ 𝑖+1 /̃
because the ratios 𝜔
𝜔𝑖 , 𝑖 = 1, 2, . . . , 𝑚 + 𝑛, are all nondecreasing. Furthermore, let
̃ (𝑥, 𝑦, 𝑠, 𝑡) :=
𝑓
𝑖
max
𝑓𝑖 (𝜄, 𝜉, 𝑠, 𝑡) ,
(𝜄,𝜉)∈[𝑥0 ,𝑥]×[𝑦0 ,𝑦]
(32)
+∫
𝛼𝑖 (𝑥)
∫
𝛽𝑖 (𝑦)
𝛼𝑖 (𝑥0 ) 𝛽𝑖 (𝑦0 )
̃ (𝑋, 𝑌, 𝑠, 𝑡) 𝑑𝑡 𝑑𝑠,
𝑓
(37)
𝑖 = 1, 2, . . . , 𝑚 + 𝑛, ̃𝑟𝑖 (𝑋, 𝑌, 𝑥, 𝑦) is defined inductively by
̃𝑟1 (𝑋, 𝑌, 𝑥, 𝑦) := 𝑎̃ (𝑋, 𝑌) ,
̃ 𝑖 (𝑋, 𝑌, 𝑥, 𝑦))) ,
̃𝑟𝑖+1 (𝑋, 𝑌, 𝑥, 𝑦) := 𝑊𝑖−1 (𝑊𝑖 (Ω
(38)
6
Journal of Applied Mathematics
for 𝑖 = 1, 2, . . . , 𝑚 + 𝑛 − 1, and 𝑋2 , 𝑌2 are chosen such that
̃ 𝑖 (𝑋, 𝑌, 𝑋2 , 𝑌2 ) ≤ ∫
Ω
∞
𝑢𝑖
𝑑𝑠
,
̃ 𝑖 (𝜑−1 (𝑠))
𝜔
(39)
for 𝑖 = 1, 2, . . . , 𝑚 + 𝑛.
Notice that we may take 𝑋2 = 𝑋1 and 𝑌2 = 𝑌1 . In
̃ (𝑋, 𝑌, 𝑥, 𝑦) are
fact, the monotonicity that ̃𝑟𝑖 (𝑋, 𝑌, 𝑥, 𝑦) and 𝑓
𝑖
both nondecreasing in 𝑋 and 𝑌 for fixed 𝑥, 𝑦. Furthermore,
it is easy to check that ̃𝑟𝑖+1 (𝑋, 𝑌, 𝑋, 𝑌) = 𝑟𝑖 (𝑋, 𝑌), for 𝑖 =
1, 2, . . . , 𝑚 + 𝑛. If 𝑋2 , 𝑌2 are replaced with 𝑋1 , 𝑌1 , respectively,
on the left side of (39), we get from (15) that
Now, we prove (36) by induction. From (33), (35), and the
̂ 𝑖 (𝑡), we obtain
̃ (𝑡), 𝜔
̃ 𝑖 (𝑡), and 𝜔
definitions of 𝑔
𝜑 (𝑢 (𝑥, 𝑦))
≤ 𝑎̃ (𝑋, 𝑌)
𝑚
+ ∑∫
𝛼𝑖 (𝑋1 )
𝛼𝑖 (𝑥0 )
∫
𝛽𝑖 (𝑌1 )
𝛽𝑖 (𝑦0 )
𝑚+𝑛
+ ∑ ∫
+∫
𝛼𝑖 (𝑥0 )
∫
𝛽𝑖 (𝑌1 )
𝛽𝑖 (𝑦0 )
𝛼𝑖 (𝑋1 )
𝛼𝑖 (𝑥0 )
𝑦
̃ (𝑋, 𝑌, 𝑠, 𝑡)
∫ 𝑓
𝑗
𝑗=𝑚+1 𝛼𝑗 (𝑥0 ) 𝑦0
̂ 𝑗 (̃
×𝜔
𝑔 ( max 𝑢 (𝜉, 𝑡))) 𝑑𝑡 𝑑𝑠
𝜉∈[𝑠−ℎ,𝑠]
≤ 𝑎̃ (𝑋, 𝑌)
𝑚
+ ∑∫
̃ (X , 𝑌 , 𝑠, 𝑡) 𝑑𝑡 𝑑𝑠
𝑓
1 1
𝛼𝑖 (𝑥)
𝛽𝑖 (𝑦)
∫
𝑖=1 𝛼𝑖 (𝑥0 ) 𝛽𝑖 (𝑦0 )
𝑚+𝑛
+ ∑ ∫
= 𝑟𝑖 (𝑋1 , 𝑌1 )
+∫
𝛼𝑗 (𝑥)
̃ (𝑋, 𝑌, 𝑠, 𝑡) 𝑑𝑡 𝑑𝑠
𝑓
≤ 𝑊𝑖 (̃𝑟𝑖 (𝑋1 , 𝑌1 , 𝑋1 , 𝑌1 ))
𝛼𝑖 (𝑋1 )
𝑦
̃ (𝑋, 𝑌, 𝑠, 𝑡) 𝜔 (𝑢 (𝑠, 𝑡))
∫ 𝑓
𝑖
𝑖
𝑖=1 𝛼𝑖 (𝑥0 ) 𝑦0
̃ 𝑖 (𝑋, 𝑌, 𝑋1 , 𝑌1 ) = 𝑊𝑖 (̃𝑟𝑖 (𝑋, 𝑌, 𝑋1 , 𝑌1 ))
Ω
+∫
𝛼𝑖 (𝑥)
𝛼𝑗 (𝑥)
∫
̃ (𝑋, 𝑌, 𝑠, 𝑡) 𝜔
̃ 𝑖 (𝑢 (𝑠)) 𝑑𝑠
𝑓
𝑖
𝛽𝑗 (𝑦)
𝑗=𝑚+1 𝛼𝑗 (𝑥0 ) 𝛽𝑗 (𝑦0 )
∫
𝛽𝑖 (𝑌1 )
𝛽𝑖 (𝑦0 )
̃ (𝑋 , 𝑌 , 𝑠, 𝑡) 𝑑𝑡 𝑑𝑠
𝑓
1 1
̃ (𝑋, 𝑌, 𝑠, 𝑡)
𝑓
𝑗
̃ 𝑗 ( max 𝑢 (𝜉)) 𝑑𝑡 𝑑𝑠,
×𝜔
𝜉∈[𝑠−ℎ,𝑠]
(41)
= Ω𝑖 (𝑋1 , 𝑌1 )
≤∫
∞
𝑢𝑖
𝑑𝑠
.
̃ 𝑖 (𝜑−1 (𝑠))
𝜔
(40)
Thus, it means that we can take 𝑋2 = 𝑋1 , 𝑌2 = 𝑌1 .
for all (𝑥, 𝑦) ∈ [𝑥0 , 𝑋) × [𝑦0 , 𝑌), where 𝑥0 ≤ 𝑋 ≤ 𝑋1 and 𝑦0 ≤
𝑌 ≤ 𝑌1 are chosen arbitrarily. Since max𝑠∈[𝛼∗ (𝑥0 )−ℎ,𝑥0 ] 𝜓(𝑠, 𝑦) ≤
𝜑−1 (𝑎(𝑥0 , 𝑦)) and 𝑎(𝑥0 , 𝑦) ≤ 𝑎̃(𝑥0 , 𝑦) ≤ 𝑎̃(𝑋, 𝑌), we have
max𝑠∈[𝛼∗ (𝑥0 )−ℎ,𝑥0 ] 𝜓(𝑠, 𝑦) ≤ 𝜑−1 (̃𝑎(𝑋, 𝑌)). Define a function
𝑧(𝑥, 𝑦) : [𝛼∗ (𝑥0 ) − ℎ, 𝑋) × [𝑦0 , 𝑌) → R+ by
𝑧 (𝑥, 𝑦)
𝑚
𝛼𝑖 (𝑥)
𝛽𝑖 (𝑦)
̃ (𝑋, 𝑌, 𝑠, 𝑡) 𝜔
{
̃ 𝑖 (𝑢 (𝑠, 𝑡)) 𝑑𝑡 𝑑𝑠
𝑓
𝑎̃ (𝑋, 𝑌) + ∑ ∫
∫
{
𝑖
{
{
𝛼𝑖 (𝑥0 ) 𝛽𝑖 (𝑦0 )
{
𝑖=1
{
{
{
{
{
{
𝑚+𝑛
𝛼𝑗 (𝑥)
𝛽𝑖 (𝑦)
={
̃ (𝑠, 𝑡) 𝜔
{
̃ 𝑗 ( max 𝑢 (𝜉, 𝑡)) 𝑑𝑡 𝑑𝑠,
𝑓
+
∫
∑
∫
{
𝑗
{
𝜉∈[𝑠−ℎ,𝑠]
𝛼
𝛽
𝑥
𝑦
{
(
)
(
)
𝑗
0
𝑖
0
𝑗=𝑚+1
{
{
{
{
{
{
{𝑎̃ (𝑋, 𝑌) ,
Clearly, 𝑧(𝑥, 𝑦) is nondecreasing in 𝑥. By (41) and the
definition of 𝑧(𝑥, 𝑦), we have
(𝑥, 𝑦) ∈ [𝛼∗ (𝑥0 ) − ℎ, 𝑋) × [𝑦0 , 𝑌) .
(𝑥, 𝑦) ∈ [𝑥0 , 𝑋) × [𝑦0 , 𝑌) ,
(𝑥, 𝑦) ∈ [𝛼∗ (𝑥0 ) − ℎ, 𝑥0 ] × [𝑦0 , 𝑌) .
increasing, from (43), we obtain
max 𝑢 (𝜉, 𝑦) ≤ max 𝜑−1 (𝑧 (𝜉, 𝑦)) ≤ max 𝜑−1 (𝑧 (𝑠, 𝑦))
𝜉∈[𝑠−ℎ,𝑠]
−1
𝑢 (𝑥, 𝑦) ≤ 𝜑 (𝑧 (𝑥, 𝑦)) ,
(42)
(43)
Then noting that 𝑧(𝑥, 𝑦) is nondecreasing and 𝜑(𝑡) is strictly
𝜉∈[𝑠−ℎ,𝑠]
𝜉∈[𝑠−ℎ,s]
≤ 𝜑−1 ( max 𝑧 (𝜉, 𝑦)) ,
𝜉∈[𝑠−ℎ,𝑠]
(𝑠, 𝑦) ∈ [𝛼∗ (𝑥0 ) , 𝑋) × [𝑦0 , 𝑌) .
(44)
Journal of Applied Mathematics
7
It follows from (43), (44), and the definition of 𝑧(𝑥, 𝑦) that
𝑧 (𝑥, 𝑦) ≤ 𝑎̃ (𝑋, 𝑌)
𝑚
+ ∑∫
Taking 𝑥 = 𝑋, 𝑦 = 𝑌, 𝑋2 = 𝑋1 , and 𝑌2 = 𝑌1 in (36), we
have
𝑢 (𝑋, 𝑌)
𝛼𝑖 (𝑥)
∫
𝛽𝑖 (𝑦)
𝑖=1 𝛼𝑖 (𝑥0 ) 𝛽𝑖 (𝑦0 )
̃ (𝑋, 𝑌, 𝑠, 𝑡)
𝑓
𝑖
−1
(𝑊𝑚+𝑛 (̃𝑟𝑚+𝑛 (𝑋, 𝑌, 𝑋, 𝑌))
≤ 𝜑−1 (𝑊𝑚+𝑛
̃ 𝑖 (𝜑−1 (𝑧 (𝑠, 𝑡))) 𝑑𝑡 𝑑𝑠
×𝜔
𝑚+𝑛
𝛼𝑗 (𝑥)
+ ∑ ∫
∫
𝛽𝑖 (𝑦)
𝑗=𝑚+1 𝛼𝑗 (𝑥0 ) 𝛽𝑖 (𝑦0 )
+∫
̃ (𝑋, 𝑌, 𝑠, 𝑡)
𝑓
𝑖
(𝑥, 𝑦) ∈ [𝑥0 , 𝑋) × [𝑦0 , 𝑌] ,
+∫
𝛼𝑚+𝑛 (𝑋)
𝛽𝑚+𝑛 (𝑌)
̃
𝑓
𝑚+𝑛 (𝑋, 𝑌, 𝑠, 𝑡) 𝑑𝑡 𝑑𝑠) ,
(49)
𝑢 (𝑥, 𝑦)
−1
≤ 𝜑−1 (𝑊𝑚+𝑛
(𝑊𝑚+𝑛 (𝑟𝑚+𝑛 (𝑥, 𝑦))
𝛽𝑚+𝑛 (𝑦)
∫
𝛼𝑚+𝑛 (𝑥0 ) 𝛽𝑚+𝑛 (𝑦0 )
−1
(𝑊 (̃𝑟𝑚+𝑛 (𝑋, 𝑌, 𝑥, 𝑦))
𝑧 (𝑥, 𝑦) ≤ 𝑊𝑚+𝑛
̃
𝑓
𝑚+𝑛 (𝑋, 𝑌, 𝑠, 𝑡) 𝑑𝑡 𝑑𝑠)) .
Since 𝑋, 𝑌 are arbitrary, replacing 𝑋 and 𝑌 with 𝑥 and 𝑦,
respectively, we get
𝛼𝑚+𝑛 (𝑥)
∫
∫
𝛼𝑚+𝑛 (𝑥0 ) 𝛽𝑚+𝑛 (𝑦0 )
+∫
𝛼𝑚+𝑛 (𝑥0 ) 𝛽𝑚+𝑛 (𝑦0 )
(48)
−1
(𝑊𝑚+𝑛 (𝑟𝑚+𝑛 (𝑋, 𝑌))
≤ 𝜑−1 (𝑊𝑚+𝑛
In order to demonstrate the basic condition of monotonicity,
let ℎ(𝑡) := 𝜑−1 (𝑡), which is clearly a continuous and nondẽ 𝑖 (ℎ(𝑡)) is continuous
creasing function on R+ . Thus, each 𝜔
̃ 𝑖 (ℎ(𝑡)) > 0 for
and nondecreasing on R+ and satisfies 𝜔
̃ 𝑖+1 (𝑡), 𝜔
̃ 𝑖+1 (ℎ(𝑡))/̃
̃ 𝑖 (𝑡) ∝ 𝜔
𝜔𝑖 (ℎ(𝑡))
𝑡 > 0. Moreover, since 𝜔
is also continuous and nondecreasing on R+ and positive
̃ 𝑖+1 (ℎ(𝑡)), for 𝑖 =
̃ 𝑖 (ℎ(𝑡)) ∝ 𝜔
on (0, +∞), implying that 𝜔
1, 2, . . . , 𝑚 + 𝑛 − 1. By Lemma 2 and (45),
+∫
̃
𝑓
𝑚+𝑛 (𝑋, 𝑌, 𝑠, 𝑡) 𝑑𝑡 𝑑𝑠)) ,
𝑢 (𝑋, 𝑌)
(𝑥, 𝑦) ∈ [𝛼∗ (𝑥0 ) − ℎ, 𝑥0 ] × [𝑦0 , 𝑌] .
(45)
𝛽𝑚+𝑛 (𝑦)
𝛽𝑚+𝑛 (𝑌)
for all 𝑥0 ≤ 𝑋 < 𝑋1 , 𝑦0 ≤ 𝑌 < 𝑌1 . It is easy to verify
̃𝑟𝑚+𝑛 (𝑋, 𝑌, 𝑋, 𝑌) = 𝑟𝑚+𝑛 (𝑋, 𝑌). Thus, (48) can be written
𝜉∈[𝑠−ℎ,𝑠]
𝛼𝑚+𝑛 (𝑥)
∫
𝛼𝑚+𝑛 (𝑥0 ) 𝛽𝑚+𝑛 (𝑦0 )
̃ 𝑗 (𝜑−1 ( max 𝑧 (𝜉, 𝑡))) 𝑑𝑡𝑑𝑠,
×𝜔
𝑧 (𝑥, 𝑦) ≤ 𝑎̃ (𝑋, 𝑌) ,
𝛼𝑚+𝑛 (𝑋)
̃
𝑓
𝑚+𝑛 (𝑥, 𝑦, 𝑠, 𝑡) 𝑑𝑡 𝑑𝑠)) ,
(50)
for all (𝑥, 𝑦) ∈ [𝑥0 , 𝑋1 ) × [𝑦0 , 𝑌1 ). This completes the proof.
3. Applications
(46)
for 𝑥0 ≤ 𝑥 < 𝑋2 and 𝑦0 ≤ 𝑦 < 𝑌2 . It follows from (43) and
(46) that
In this section, we apply our result to prove the boundedness
of solutions for a differential equation with the maxima.
Consider a system of partial differential equations with
maxima
𝜕2 𝑧 (𝑥, 𝑦)
= 𝐹 (𝑥, 𝑦, 𝑧 (𝑥, 𝑦) , max 𝜔 (𝑧 (𝑠, 𝑦))) ,
𝜕𝑥𝜕𝑦
𝑠∈[𝛼(𝑥),𝛽(𝑥)]
𝑢 (𝑥, 𝑦)
−1
(𝑊 (̃𝑟𝑚+𝑛 (𝑋, 𝑌, 𝑥, 𝑦))
≤ 𝜑−1 (𝑊𝑚+𝑛
+∫
𝛼𝑚+𝑛 (𝑥)
𝛽𝑚+𝑛 (𝑦)
∫
(𝑥, 𝑦) ∈ [𝑥0 , 𝑥1 ) × [𝑦0 , 𝑦1 )
̃
𝑓
𝑚+𝑛 (𝑋, 𝑌, 𝑠, 𝑡) 𝑑𝑡 𝑑𝑠)) ,
𝑧 (𝑥, 𝑦) = 𝜓 (𝑥, 𝑦) ,
𝑧 (𝑥, 𝑦0 ) = 𝑓 (𝑥) ,
(𝑥, 𝑦) ∈ [𝛽 (𝑥0 ) − ℎ, 𝑥0 ] × [𝑦0 , 𝑦1 ) ,
(47)
𝑥 ≥ 𝑥0 , 𝑦 ≥ 𝑦0 ,
(51)
for 𝑥0 ≤ 𝑥 < 𝑋2 and 𝑦0 ≤ 𝑦 < 𝑌2 . This proves the claimed
(36).
where 𝐹 ∈ 𝐶([𝑥0 , 𝑥1 ) × [𝑦0 , 𝑦1 ) × R2 , R), 𝜔 ∈ 𝐶([0, ∞), R+ ),
𝛼, 𝛽 ∈ 𝐶1 ([𝑥0 , 𝑥1 ), R+ ) are nondecreasing such that 𝛼(𝑥) ≤ 𝑥,
𝛼𝑚+𝑛 (𝑥0 ) 𝛽𝑚+𝑛 (𝑦0 )
𝑧 (𝑥0 , 𝑦) = 𝑔 (𝑦) ,
8
Journal of Applied Mathematics
𝛽(𝑥) ≤ 𝑥, and 0 < 𝛽(𝑥) − 𝛼(𝑥) ≤ ℎ (ℎ is a positive constant)
for 𝑥 ∈ [𝑥0 , 𝑥1 ], 𝜓 ∈ 𝐶([𝛽(𝑥0 ) − ℎ, 𝑥0 ] × [𝑦0 , 𝑦1 ), and 𝑓 ∈
𝐶([𝑥0 , 𝑥1 ), R), 𝑔 ∈ 𝐶([𝑦0 , 𝑦1 ), R) satisfy that 𝑓(𝑥0 ) = 𝑔(𝑦0 )
and 𝑔(𝑦) = 𝜓(𝑥0 , 𝑦), for all 𝑦 ∈ [𝑦0 , 𝑦1 ).
Equation (51) is more general than the equation considered in Section 2.4 of [23]. The following result gives an
estimate for its solutions.
Corollary 3. Suppose that functions 𝐹 and 𝜓 in (51) satisfy
󵄨
󵄨󵄨
󵄨󵄨𝐹 (𝑥, 𝑦, 𝑠, 𝑡)󵄨󵄨󵄨 ≤ ℎ1 (𝑥, 𝑦) 𝜇1 (|𝑠|) + ℎ2 (𝑥, 𝑦) 𝜇2 (|𝑡|) ,
󵄨
󵄨 󵄨
󵄨󵄨
󵄨󵄨𝜓 (𝑥, 𝑦)󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨𝑔 (𝑦)󵄨󵄨󵄨 , ∀ (𝑥, 𝑦) ∈ [𝛽 (𝑥0 ) − ℎ, 𝑥0 ] × [𝑦0 , 𝑦1 ) ,
(52)
where ℎ𝑖 ∈ 𝐶([𝑥0 , 𝑥1 ) × [𝑦0 , 𝑦1 ), R+ ) and 𝜇𝑖 ∈ 𝐶(R+ , (0, ∞)),
𝑖 = 1, 2. Then, any solution 𝑧(𝑥, 𝑦) of (51) has the estimate
𝑥
𝑦
0
0
󵄨
󵄨󵄨
−1
󵄨󵄨𝑧 (𝑥, 𝑦)󵄨󵄨󵄨 ≤ 𝑄2 (𝑄2 (𝛾 (𝑥, 𝑦)) + ∫ ∫ ℎ2 (𝑠, 𝑡) 𝑑𝑡 𝑑𝑠) ,
𝑥 𝑦
for all (𝑥, 𝑦) ∈ [𝑥0 , 𝑋1 ) × [𝑦0 , 𝑌1 ), where
𝑦
𝑥0
𝑦0
𝑑𝑠
𝑢2 {max𝜏∈[0,𝑠]{̃
𝜇2 (̃
𝜔(𝜏))/max𝜏1 ∈[0,𝜏]{𝜇1 (𝜏1 )}}max𝜏∈[0,𝑠]{𝜇1 (𝜏)}}
𝑢
𝑑𝑠
𝑢1
max𝜏∈[0,𝑠] {𝜇1 (𝜏)}
̃ (𝑡) := max {𝜔 (𝑠)} ,
𝜔
𝑠∈[0,𝑡]
,
𝑠∈[0,𝑡]
𝑋1 , 𝑌1 are given as in Theorem 1, and constants 𝑢1 > 0, 𝑢2 > 0
are given arbitrarily.
Proof. From (51), we obtain
𝑦0
𝑥
𝑦
𝑥0
𝑦0
𝑥
𝑦
𝑥0
𝑦0
(56)
+ ∫ ∫ ℎ1 (𝑠, 𝑡) 𝜇1 (|𝑧 (𝑠, 𝑡)|) 𝑑𝑡𝑑𝑠
̃2 (
+∫ ∫ ℎ2 (𝑠, 𝑡) 𝜇
max
𝜉∈[𝛼(𝑠),𝛽(𝑠)]
̃ (󵄨󵄨󵄨󵄨𝑧 (𝜉, 𝑡)󵄨󵄨󵄨󵄨) 𝑑𝑡𝑑𝑠,
𝜔
󵄨
󵄨 󵄨
󵄨󵄨
󵄨󵄨𝑧 (𝑥, 𝑦)󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨𝜓 (𝑥, 𝑦)󵄨󵄨󵄨 ,
𝑥
𝑦
+∫
𝛽(𝑥)
𝑥0
𝑦0
𝑦
󸀠
̃2
∫ ℎ2 (𝛽−1 (𝜂) , 𝑡) (𝛽−1 (𝜂)) 𝜇
𝛽(𝑥0 ) 𝑦0
󵄨
󵄨
V (𝑥, 𝑦) ≤ 󵄨󵄨󵄨𝜓 (𝑥, 𝑦)󵄨󵄨󵄨 ,
(𝑥, 𝑦) ∈ [𝛽 (𝑥0 ) − ℎ, 𝑥0 ] × [𝑦0 , 𝑦1 ) .
(57)
Applying Theorem 1 to specified 𝑚 = 𝑛 = 1, 𝜑(𝑢) = 𝑢,
𝑓1 (𝑥, 𝑦, 𝑠, 𝑡) = ℎ1 (𝑠, 𝑡), 𝛼1 (𝑡) = 𝑡, and 𝛼2 (𝑡) = 𝛽(𝑡), 𝛽𝑖 (𝑡) =
󸀠
𝑡, 𝑖 = 1, 2, 𝑓2 (𝑠, 𝑡) = ℎ2 (𝛽−1 (𝑠), 𝑡)(𝛽−1 (𝑠)) , and 𝜔𝑖 (𝑢) =
𝜇𝑖 (𝑢), 𝑖 = 1, 2, we obtain (53) from (57).
Corollary 4. Suppose that 𝑔(𝑠) = 𝑠 and
= 𝑓 (𝑥) + 𝑔 (𝑦) − 𝑓 (𝑥0 )
𝑥0
𝑦
Next, we discuss the uniqueness of solutions for system
(51).
𝑧 (𝑥, 𝑦)
𝑦
󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨
+ ∫ ∫ 𝜇2 (󵄨󵄨 max 𝜔 (𝑧 (𝜉, 𝑡))󵄨󵄨󵄨) 𝑑𝑡 𝑑𝑠
󵄨󵄨
󵄨
𝑥0 𝑦0
󵄨󵄨𝜉∈[𝛼(𝑠),𝛽(𝑠)]
󵄨
󵄨󵄨
󵄨󵄨
≤ 󵄨󵄨𝑓 (𝑥) + 𝑔 (𝑦) − 𝑓 (𝑥0 )󵄨󵄨
𝑥
̃ (V (𝜉, 𝑡)) 𝑑𝑡 𝑑𝜂
× ( max 𝜔
𝜉∈[𝜂−ℎ,𝜂]
(54)
𝑥
𝑦0
,
̃ 2 (𝑡) := max {𝜇2 (𝑠)} , and
𝜇
+ ∫ ∫ 𝐹 (𝑠, 𝑡, 𝑧 (𝑠, 𝑡) ,
𝑥0
V (𝑥, 𝑦) ≤ 𝛾1 (𝑥, 𝑦) + ∫ ∫ ℎ1 (𝑠, 𝑡) 𝜇1 (V (𝑠, 𝑡)) 𝑑𝑡 𝑑𝑠
𝑄2 (𝑢) :=
𝑄1 (𝑢) := ∫
𝑦
Set V(𝑥, 𝑦) = |𝑧(𝑥, 𝑦)| for (𝑥, 𝑦) ∈ [𝛽(𝑥0 ) − ℎ, 𝑥1 ) × [𝑦0 , 𝑦1 ).
Noting that max𝜉∈[𝛼(𝑠),𝛽(𝑠)] 𝑧(𝜉, 𝑦) ≤ max𝜉∈[𝛽(𝑠)−ℎ,𝛽(𝑠)] 𝑧(𝜉, 𝑦),
from (56), we get
󵄨
󵄨
max
(󵄨󵄨󵄨𝑓 (𝜉) + 𝑔 (𝜂) − 𝑓 (𝑥0 )󵄨󵄨󵄨) ,
(𝜉,𝜂)∈[𝑥0 ,𝑥]×[𝑦0 ,𝑦]
𝑢
𝑥
+ ∫ ∫ ℎ1 (𝑠, 𝑡) 𝜇1 (|𝑧 (𝑠, 𝑡)|) 𝑑𝑡 𝑑𝑠
(𝑥, 𝑦) ∈ [𝛽 (𝑥0 ) − ℎ, 𝑥0 ] × [𝑦0 , 𝑦1 ) .
𝑥
𝛾 (𝑥, 𝑦) := 𝑄1−1 (𝑄1 (𝛾1 (𝑥, 𝑦)) + ∫ ∫ ℎ1 (𝑠, 𝑡) 𝑑𝑡 𝑑𝑠) ,
∫
󵄨
󵄨󵄨
󵄨󵄨𝑧 (𝑥, 𝑦)󵄨󵄨󵄨
󵄨
󵄨
≤ 󵄨󵄨󵄨𝑓 (𝑥) + 𝑔 (𝑦) − 𝑓 (𝑥0 )󵄨󵄨󵄨
(𝑥, 𝑦) ∈ Λ,
(53)
𝛾1 (𝑥, 𝑦) :=
From (52) and (55), we get
max
𝜉∈[𝛼(𝑠),𝛽(𝑠)]
𝜔 (𝑧 (𝜉, 𝑡))) 𝑑𝑡 𝑑𝑠,
(𝑥, 𝑦) ∈ Λ,
𝑧 (𝑥, 𝑦) = 𝜓 (𝑥, 𝑦) ,
(𝑥, 𝑦) ∈ (𝑥, 𝑦) ∈ [𝛽 (𝑥0 ) − ℎ, 𝑥0 ] × [𝑦0 , 𝑦1 ) .
(55)
󵄨󵄨
󵄨
󵄨󵄨𝐹 (𝑥, 𝑦, 𝑠1 , 𝑡1 ) − 𝐹 (𝑥, 𝑦, 𝑠2 , 𝑡2 )󵄨󵄨󵄨
󵄨
󵄨
󵄨
󵄨
≤ ℎ1 (𝑥, 𝑦) 𝜇1 (󵄨󵄨󵄨𝑥1 − 𝑥2 󵄨󵄨󵄨) + ℎ2 (𝑥, 𝑦) 𝜇2 (󵄨󵄨󵄨𝑦1 − 𝑦2 󵄨󵄨󵄨) ,
(58)
for all (𝑥, 𝑦) ∈ [𝑥0 , 𝑥1 ] × [𝑦0 , 𝑦1 ] and all 𝑥𝑖 , 𝑦𝑖 ∈ R (𝑖 = 1, 2),
where ℎ𝑖 ∈ 𝐶([𝑥0 , 𝑥1 ] × [𝑦0 , 𝑦1 ], R+ ) and 𝜇𝑖 ∈ 𝐶([R, R+ ) are
Journal of Applied Mathematics
9
both nondecreasing such that 𝜇𝑖 (0) = 0, 𝜇𝑖 (𝑢) > 0 for 𝑢 > 0,
1
𝜇2 /𝜇1 is also nondecreasing, and ∫0 𝑑𝑠/𝜇𝑖 (𝑠) = +∞, 𝑖 = 1, 2.
Then, system (51) has at most one solution on [𝑥0 , 𝑥1 )×[𝑦0 , 𝑦1 ).
Hence,
󵄨
󵄨󵄨
󵄨󵄨𝑢 (𝑥, 𝑦) − V (𝑥, 𝑦)󵄨󵄨󵄨
Proof. 𝑔(𝑠) = 𝑠. From (51), we get
𝑦
𝑥0
𝑦0
𝑥
𝑦
𝑥0
𝑦0
+ ∫ ∫ ℎ2 (𝑠, 𝑡) 𝜇2 (
𝜕2 𝑧 (𝑥, 𝑦)
= 𝐹 (𝑥, 𝑦, 𝑧 (𝑥, 𝑦) , max 𝑧 (𝑠, 𝑦)) ,
𝜕𝑥 𝜕𝑦
𝑠∈[𝛼(𝑥),𝛽(𝑥)]
Let
𝑥 ≥ 𝑥0 , 𝑦 ≥ 𝑦0 .
(59)
Assume that (59) has two different solutions 𝑢(𝑥, 𝑦) and
V(𝑥, 𝑦). From the equivalent integral equation system (55), we
have
𝑦
𝑥0
𝑦0
Because max𝜉∈[𝛼(𝑠),𝛽(𝑠)] 𝑢(𝜉, 𝑦) ≤ max𝜉∈[𝛽(𝑠)−ℎ,𝛽(𝑠)] 𝑢(𝜉, 𝑦), from
(62), we obtain
𝑦0
𝑦
𝑥0
𝑦0
𝑦0
𝛽(𝑥)
𝑦
∫ ℎ2 (𝛽−1 (𝜂) , 𝑡) (𝛽−1 (𝜂))
(64)
(𝑥, 𝑦) ∈ [𝑥0 , 𝑥1 ) × [𝑦0 , 𝑦1 ) ,
(𝑥, 𝑦) ∈ [𝛽 (𝑥0 ) − ℎ, 𝑥0 ] × [𝑦0 , 𝑦1 ) .
𝜙 (𝑥, 𝑦) ≤ 0,
Applying Theorem 1 to specified 𝑚 = 𝑛 = 1, 𝜑(𝑢) = 𝑢,
𝑓1 (𝑠, 𝑡) = ℎ1 (𝑠, 𝑡), 𝛼1 (𝑡) = 𝑡, 𝛼2 (𝑡) = 𝛽(𝑡), 𝑓2 (𝑠, 𝑡) =
󸀠
ℎ2 (𝛽−1 (𝑡), 𝑠)(𝛽−1 (𝑡)) , 𝑔(𝑡) = 𝑡, 𝑎(𝑥, 𝑦) = 0, and 𝜔𝑖 (𝑡) =
𝜇i (𝑡), 𝑖 = 1, 2, from (64), we obtain
−1
max
𝜉∈[𝛼(𝑠),𝛽(𝑠)]
𝑥
𝑦
𝑥0
𝑦0
̂ 2 (𝛾 (𝑠, 𝑡)) + ∫ ∫ ℎ2 (𝑠, 𝑡) 𝑑𝑡 𝑑𝑠) ,
̂ (𝑄
𝜙 (𝑥, 𝑦) ≤ 𝑄
2
2
+ ∫ ∫ ℎ2 (𝑠, 𝑡) 𝜇2
×(
󸀠
× 𝜇2 ( max 𝜙 (𝜉, 𝑡)) 𝑑𝑡 𝑑𝜂,
𝜉∈[𝜂−ℎ,𝜂]
≤ ∫ ∫ ℎ1 (𝑠, 𝑡) 𝜇1 (|𝑢 (𝑠) − V (𝑠)|) 𝑑𝑡 𝑑𝑠
𝑥
𝑥0
𝛽(𝑥0 ) 𝑦0
󵄨󵄨
󵄨󵄨
− max V (𝜉, 𝑡)󵄨󵄨󵄨) 𝑑𝑡 𝑑𝑠
󵄨󵄨
𝜉∈[𝛼(𝑠),𝛽(𝑠)]
󵄨
𝑥0
𝑦
+∫
󵄨󵄨
𝑥 𝑦
󵄨󵄨
+ ∫ ∫ ℎ2 (𝑠, 𝑡) 𝜇2 (󵄨󵄨󵄨 max 𝑢 (𝜉, 𝑡)
󵄨󵄨𝜉∈[𝛼(𝑠),𝛽(𝑠)]
𝑥0 𝑦0
󵄨
𝑦
𝑥
𝜙 (𝑥, 𝑦) ≤ ∫ ∫ ℎ1 (𝑠, 𝑡) 𝜇1 (𝜙 (𝑠, 𝑡)) 𝑑𝑡 𝑑𝑠
≤ ∫ ∫ ℎ1 (𝑠, 𝑡) 𝜇1 (|𝑢 (𝑠, 𝑡) − V (𝑠, 𝑡)|) 𝑑𝑡 𝑑𝑠
𝑥
(63)
(𝑥, 𝑦) ∈ [𝛽 (𝑥0 ) − ℎ, 𝑥1 ] × [𝑦0 , 𝑦1 ] .
󵄨
󵄨󵄨
󵄨󵄨𝑢 (𝑥, 𝑦) − V (𝑥, 𝑦)󵄨󵄨󵄨
𝑥
󵄨
󵄨󵄨
󵄨󵄨𝑢 (𝜉, 𝑡) − V (𝜉, 𝑡)󵄨󵄨󵄨) 𝑑𝑠.
󵄨
󵄨
𝜙 (𝑥, 𝑦) := 󵄨󵄨󵄨𝑢 (𝑥, 𝑦) − V (𝑥, 𝑦)󵄨󵄨󵄨 ,
(𝑥, 𝑦) ∈ [𝛼 (𝑥0 ) − ℎ, 𝑥0 ] × [𝑦0 , 𝑦1 ] ,
𝑧 (𝑥, 𝑦0 ) = 𝑓 (𝑥) , 𝑧 (𝑥0 , 𝑦) = 𝑔 (𝑦) ,
max
𝜉∈[𝛼(𝑠),𝛽(𝑠)]
(62)
(𝑥, 𝑦) ∈ [𝑥0 , 𝑥1 ) × [𝑦0 , 𝑦1 ) ,
𝑧 (𝑥, 𝑦) = 𝜓 (𝑥, 𝑦) ,
𝑥
󵄨
󵄨
≤ ∫ ∫ ℎ1 (𝑠, 𝑡) 𝜇1 (󵄨󵄨󵄨𝑢 (𝑠, 𝑦) − V (𝑠, 𝑦)󵄨󵄨󵄨) 𝑑𝑡 𝑑𝑠
(65)
for all (𝑥, 𝑦) ∈ [𝑥0 , 𝑋1 ] × [𝑦0 , 𝑌1 ], where
󵄨
󵄨󵄨
󵄨󵄨𝑢 (𝜉, 𝑡) − V (𝜉, 𝑡)󵄨󵄨󵄨) 𝑑𝑠,
(60)
for all (𝑥, 𝑦) ∈ [𝑥0 , 𝑥1 ] × [𝑦0 , 𝑦1 ]. The continuity of the
function 𝑢(𝑥, 𝑦) implies that for any fixed points 𝑠 ∈ [𝑥0 , 𝑥]
and 𝑡 ∈ [𝑦0 , 𝑦] there exists a point 𝜂 ∈ [𝛼(𝑠), 𝛽(𝑠)] such
that the inequality max𝜉∈[𝛼(𝑠),𝛽(𝑠)] 𝑢(𝜉, 𝑡) = 𝑢(𝜂, 𝑡) holds, and
therefore
󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨
󵄨󵄨 max 𝑢 (𝜉, 𝑡) − max V (𝜉, 𝑡)󵄨󵄨󵄨
󵄨󵄨𝜉∈ 𝛼(𝑠),𝛽(𝑠)
󵄨󵄨
𝜉∈[𝛼(𝑠),𝛽(𝑠)]
󵄨󵄨 [
󵄨󵄨
]
󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨󵄨
= 󵄨󵄨󵄨𝑢 (𝜂, 𝑡) − max V (𝜉, 𝑡)󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
𝜉∈
𝛼(𝑠),𝛽(𝑠)
[
]
󵄨
󵄨
󵄨󵄨
󵄨󵄨
󵄨
󵄨
≤ 󵄨󵄨𝑢 (𝜂, 𝑡) − V (𝜂, 𝑡)󵄨󵄨 ≤ max 󵄨󵄨󵄨𝑢 (𝜉, 𝑡) − V (𝜉, 𝑡)󵄨󵄨󵄨 .
𝜉∈[𝛼(𝑠),𝛽(𝑠)]
(61)
̂ 1 (𝑢) := ∫
𝑄
𝑢
1
𝑑𝑠
,
𝜇1 (𝑠)
̂ 2 (𝑢) := ∫
𝑄
𝑢
1
𝑑𝑠
,
𝜇2 (𝜏)
𝑟1 (𝑥, 𝑦) := 0,
(66)
(67)
𝑥
𝑦
𝑥0
𝑦0
̂ −1 (𝑄
̂ 1 (𝑟1 (𝑥, 𝑦)) + ∫ ∫ ℎ1 (𝑠, 𝑡) 𝑑𝑡 𝑑𝑠) .
𝑟2 (𝑥, 𝑦) := 𝑄
1
(68)
̂ 𝑖 and properties of 𝜇𝑖 , noting that
By the definition of 𝑄
1
∫0 𝑑𝑠/𝜇𝑖 (𝑠) = +∞ (𝑖 = 1, 2), we obtain
̂ 𝑖 (𝑢) = −∞,
lim 𝑄
𝑢 → 0+
𝑥
̂ −1 (𝑢) = 0,
lim 𝑄
𝑖
𝑢 → −∞
𝑖 = 1, 2.
(69)
𝑦
Since ∫𝑥 ∫𝑦 ℎ1 (𝑠, 𝑡)𝑑𝑡𝑑𝑠 is finite on a finite interval, [𝑥0 , 𝑥1 ]
0
0
and [𝑦0 , 𝑦1 ], by (67), we obtain
𝑥
𝑦
𝑥0
𝑦0
̂ 1 (𝑟1 (𝑥, 𝑦)) + ∫ ∫ ℎ1 (𝑠, 𝑡) 𝑑𝑡 𝑑𝑠 = −∞.
𝑄
(70)
10
Journal of Applied Mathematics
Thus, we obtain 𝛾2 (𝑥, 𝑦) = 0 from (68), (69), and (70)
𝑥 𝑦
immediately. Similarly, noting that ∫𝑥 ∫𝑦 ℎ2 (𝑠, 𝑡)𝑑𝑡𝑑𝑠 is finite
0
0
on finite interval, [𝑥0 , 𝑥1 ] and [𝑦0 , 𝑦1 ], from (69), we obtain
𝑥
𝑦
𝑥0
𝑦0
̂ 2 (𝑟2 (𝑥, 𝑦)) + ∫ ∫ ℎ2 (𝑠, 𝑡) 𝑑𝑡 𝑑𝑠 = −∞.
𝑄
(71)
Thus, we conclude from (65), (69), and (71) that |𝑢(𝑥, 𝑦) −
V(𝑥, 𝑦)| ≤ 0, which implies that 𝑢(𝑥, 𝑦) = V(𝑥, 𝑦), for all
(𝑥, 𝑦) ∈ [𝑥0 , 𝑋1 ) × [𝑦0 , 𝑌1 ), where 𝑋1 , 𝑌1 are given as in
Theorem 1. The uniqueness is proved.
Acknowledgments
The author is grateful to the reviewer for his valuable suggestions. This research was supported by Scientific Research
Fund of Sichuan Provincial Education Department of China.
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Applied Mathematics
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Function Spaces
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Optimization
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014