Volume 9 (2008), Issue 2, Article 57, 11 pp.
SOME RETARDED NONLINEAR INTEGRAL INEQUALITIES IN TWO
VARIABLES AND APPLICATIONS
KELONG ZHENG
S CHOOL OF S CIENCE
S OUTHWEST U NIVERSITY OF S CIENCE AND T ECHNOLOGY
M IANYANG , S ICHUAN 621010, P. R. C HINA
zhengkelong@swust.edu.cn
Received 23 October, 2007; accepted 19 March, 2008
Communicated by W.S. Cheung
A BSTRACT. 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.
Key words and phrases: Integral inequality, Nonlinear, Two variables, Retarded.
2000 Mathematics Subject Classification. 26D10, 26D15.
1. I NTRODUCTION
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 )
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
fi (x, y, s, t)wi (u(s, t))dtds
(1.3)
u(x, y) ≤ a(x, y) +
i=1
322-07
αi (0)
βi (y)
2
K ELONG Z HENG
and
(1.4)
u(x, y) ≤ a(x, y) +
n Z
X
i=1
∞
Z
∞
fi (x, y, s, t)wi (u(s, t))dtds.
αi (x)
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. S TATEMENT OF M AIN R ESULTS
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 on A.
w1
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 ;
(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. Clearly,
i (z)
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)
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)
sup fi (τ, µ, s, t),
0≤τ ≤x y≤µ<∞
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 )
∞
ui
dz
wi (z)
for i = 1, . . . , n.
The proof of Theorem 2.1 will be given in the next section.
Remark 1. As in
R ∞[6], different 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)
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 57, 11 pp.
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R ETARDED N ONLINEAR I NTEGRAL I NEQUALITIES
3
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
w̃1 (u) := max w1 (θ),
θ∈[0,u]
wi+1 (θ)
w̃i+1 (u) := max
w̃i (u),
θ∈[0,u]
w̃i (θ)
(2.4)
i = 1, . . . , n − 1.
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
∞
Z
∞
fi (x, y, s, t)w̃i (u(s, t))dtds,
αi (x)
βi (y)
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)
βn (y)
for all x̂1 ≤ x < ∞, ŷ1 ≤ y < ∞, where bn (x, y) is determined recursively by
b1 (x, y) = sup
sup a(τ, µ),
Z
−1
bi+1 (x, y) = Wi
Wi (bi (x, y)) +
x≤τ <∞ y≤µ<∞
∞
Z
αi (x)
(2.8)
fˆi (x, y, s, t) = sup
∞
fˆi (x, y, s, t)dtds ,
βi (y)
sup fi (τ, µ, s, t),
x≤τ <∞ y≤µ<∞
W1 (0) := 0, and x̂1 , ŷ1 ∈ R+ are chosen such that
Z ∞ Z ∞
Z
(2.9)
Wi (bi (x̂1 , ŷ1 )) +
fˆi (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.
3. P ROOF OF T HEOREM 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)
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 57, 11 pp.
βi (y)
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4
K ELONG Z HENG
Choose arbitrary x̃1 , ỹ1 such that 0 ≤ x̃1 ≤ x1 , y1 ≤ ỹ1 < ∞. From (3.1), we obtain
(3.2)
u(x, y) ≤ b1 (x̃1 , ỹ1 ) +
n Z
X
i=1
αi (x)
Z
αi (0)
∞
f˜i (x̃1 , ỹ1 , s, t)wi (u(s, t))dtds
βi (y)
for all 0 ≤ x ≤ x̃1 ≤ x1 , y1 ≤ ỹ1 ≤ y < ∞.
We claim that
"
(3.3)
u(x, y) ≤ Wn−1 Wn (b̃n (x̃1 , ỹ1 , x, y)) +
αn (x)
Z
Z
αn (0)
#
∞
f˜n (x̃1 , ỹ1 , s, t)dtds
βn (y)
for all 0 ≤ x ≤ min{x̃1 , x2 }, max{ỹ1 , y2 } ≤ y < ∞, where
b̃1 (x̃1 , ỹ1 , x, y) = b1 (x̃1 , ỹ1 ),
"
(3.4)
b̃i+1 (x̃1 , ỹ1 , x, y) = Wi−1 Wi (b̃i (x̃1 , ỹ1 , x, y)) +
Z
αi (x)
αi (0)
Z
#
∞
f˜i (x̃1 , ỹ1 , s, t)dtds
βi (y)
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 f˜i (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 ∞
f˜i (x̃1 , ỹ1 , s, t)dtds
Wi (b̃i (x̃1 ,ỹ1 , x1 , y1 )) +
αi (0)
βi (y1 )
Z
αi (x1 )
Z
∞
≤ Wi (b̃i (x1 , y1 , x1 , y1 )) +
αi (0)
Z
= Wi (bi (x1 , y1 )) +
Z ∞
dz
≤
.
ui wi (z)
αi (x1 )
αi (0)
Z
f˜i (x1 , y1 , s, t)dtds
βi (y1 )
∞
f˜i (x1 , y1 , s, t)dtds
βi (y1 )
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)
β1 (y)
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
(3.6)
u(x, y) ≤ z(x, y).
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 57, 11 pp.
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R ETARDED N ONLINEAR I NTEGRAL I NEQUALITIES
5
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
≤ w1 (z(x, y))
(3.7)
f˜1 (x̃1 , ỹ1 , α1 (x), t)dtα1 (x).
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
β1 (y)
α1 (x)
Z
(3.9)
Z
∞
f˜1 (x̃1 , ỹ1 , s, t)dtds.
= W1 (b1 (x̃1 , ỹ1 )) +
α1 (0)
β1 (y)
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
X Z αi (x)
Z
αi (0)
i=1
∞
f˜i (x̃1 , ỹ1 , s, t)wi (u(s, t))dtds
βi (y)
for all 0 ≤ x ≤ x̃1 , ỹ1 ≤ y < ∞. Let
z(x, y) = b1 (x̃1 , ỹ1 ) +
m+1
X Z αi (x)
i=1
αi (0)
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
0
nondecreasing and z(x, y) > 0, noting that αi (x) ≤ x and 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))
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 57, 11 pp.
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6
K ELONG Z HENG
∞
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
wi+1 (u)
,
w1 (u)
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) =
∞
i = 1, . . . , m. Integrating the above inequality from 0 to x, we obtain
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
∞
f˜1 (x̃1 , ỹ1 , s, t)dtds
≤ W1 (b1 (x̃1 , ỹ1 )) +
α1 (0)
+
m Z
X
i=1
αi+1 (x)
Z
αi+1 (0)
β1 (y)
∞
f˜i+1 (x̃1 , ỹ1 , s, t)φi+1 (z(s, t))dtds,
βi+1 (y)
or
ξ(x, y) ≤ c1 (x, y) +
m Z
X
i=1
αi+1 (x)
Z
αi+1 (0)
∞
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 ∞
f˜1 (x̃1 , ỹ1 , s, t)dtds.
c1 (x, y) = W1 (b1 (x̃1 , ỹ1 )) +
α1 (0)
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
"
#
Z αm+1 (x) Z ∞
(3.10)
ξ(x, y) ≤ Φ−1
f˜m+1 (x̃1 , ỹ1 , s, t)dtds
m+1 Φm+1 (cm (x, y)) +
αm+1 (0)
βm+1 (y)
for all 0 ≤ x ≤ min{x̃1 , x3 }, max{ỹ1 , y3 } ≤ y < ∞, where Φi+1 (u) =
Ru
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 Φi+1 (ci (x, y)) +
f˜i+1 (x̃1 , ỹ1 , s, t)dtds ,
i+1
αi+1 (0)
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 57, 11 pp.
dz
,
ũi+1 φi+1 (W1−1 (z))
i = 1, . . . , m,
βi+1 (y)
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R ETARDED N ONLINEAR I NTEGRAL I NEQUALITIES
and x3 , y3 ∈ R+ are chosen such that
Z αi+1 (x3 ) Z
(3.11)
Φi+1 (ci (x3 , y3 )) +
αi+1 (0)
7
∞
f˜i+1 (x̃1 , ỹ1 , s, t)dtds
βi+1 (y3 )
Z
W1 (∞)
≤
ũi+1
dz
φi+1 (W1−1 (z))
for i = 1, . . . , m.
Note that
Z
u
dz
−1
ũi φi (W1 (z))
Z u
w1 (W1−1 (z))dz
=
−1
W1 (ui ) wi (W1 (z))
Z W1−1 (u)
dz
=
= Wi ◦ W1−1 (u),
w
(z)
i
ui
Φi (u) =
i = 2, . . . , m + 1.
From (3.10), we have
u(x, y)
≤ z(x, y) = W1−1 (ξ(x, y))
"
(3.12)
−1
≤ Wm+1
Wm+1 (W1−1 (cm (x, y))) +
αm+1 (x)
Z
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 < ∞. 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).
Moreover, with the assumption that c̃m (x, y) = b̃m+1 (x̃1 , ỹ1 , x, y), we have
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
Wm+1 (W1−1 (cm (x, y))) +
Z
f˜m+1 (x̃1 , ỹ1 , s, t)dtds)
βm+1 (y)
αm+1 (x)
Z
αm+1 (0)
"
−1
= Wm+1
Wm+1 (c̃m (x, y)) +
Z
αm+1 (x) Z
αm+1 (0)
"
−1
= Wm+1
Wm+1 (b̃m+1 (x̃1 , ỹ1 , x, y)) +
#
∞
Z
#
∞
f˜m+1 (x̃1 , ỹ1 , s, t)dtds
βm+1 (y)
#
∞
f˜m+1 (x̃1 , ỹ1 , s, t)dtds
βm+1 (y)
Z
αm+1 (x)
αm+1 (0)
Z
#
∞
f˜m+1 (x̃1 , ỹ1 , s, t)dtds
βm+1 (y)
= b̃m+2 (x̃1 , ỹ1 , x, y).
This proves that
c̃i (x, y) = b̃i+1 (x̃1 , ỹ1 , x, y),
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 57, 11 pp.
i = 1, . . . , m.
http://jipam.vu.edu.au/
8
K ELONG Z HENG
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)
The above inequalities and (3.5) imply that we may take x2 = x3 , y2 = y3 . From (3.12) we get
"
#
Z αm+1 (x) Z ∞
−1
u(x, y) ≤ Wm+1
Wm+1 (b̃m+1 (x̃1 , ỹ1 , x, y)) +
f˜m+1 (x̃1 , ỹ1 , s, t)dtds
αm+1 (0)
βm+1 (y)
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
"
#
Z αn (x̃1 ) Z ∞
−1
(3.13)
u(x̃1 , ỹ1 ) ≤ W
Wn (b̃n (x̃1 , ỹ1 , x̃1 , ỹ1 )) +
f˜n (x̃1 , ỹ1 , s, t)dtds
n
αn (0)
βn (ỹ1 )
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 ∞
f˜n (x̃1 , ỹ1 , s, t)dtds .
u(x̃1 , ỹ1 ) ≤ W −1 Wn (bn (x̃1 , ỹ1 )) +
n
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 ∞
f˜n (x, y, s, t)dtds
u(x, y) ≤ W −1 Wn (bn (x, y)) +
n
αn (0)
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 ∞
u(x, y) ≤ W −1 Wn (bn, (x, y)) +
d˜n (x, y, s, t)dtds .
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.
4. A PPLICATIONS
Consider the retarded partial differential equation
D1 D2 v(x, y) =
(4.1)
(4.2)
1
(x +
1)2 (y
+
1)2
p
+ exp (−x) exp (−y) |v(x, y)|
x
x
3
+ x exp −
exp (−3y)v
, 3y ,
4
2
2
v(x, ∞) = σ(x), v(0, y) = τ (y), v(0, ∞) = k,
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 57, 11 pp.
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R ETARDED N ONLINEAR I NTEGRAL I NEQUALITIES
9
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
y
Z xZ ∞
s
s
3
−
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
y
x
2
Z
−
Z
∞
s exp (−s) exp (−t)v(s, t)dtds.
0
3y
Thus,
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
3y
Letting u(x, y) = |v(x, y)|, we have
Z
α1 (x)
Z
∞
u(x, y) ≤ a(x, y) +
f1 (x, y, s, t)w1 (u)dtds
α1 (0)
β1 (y)
Z
α2 (x)
Z
∞
+
f2 (x, y, s, t)w2 (u)dtds,
α2 (0)
where
a(x, y) = |σ(x) + τ (y) − k| +
β2 (y)
x
,
(x + 1)(y + 1)
√
x
, β2 (y) = 3y, w1 (u) = u, w2 (u) = u,
2
f1 (x, y, s, t) = exp (−s) exp (−t), f2 (x, y, s, t) = s exp (−s) exp (−t).
√
w2 (u)
√u =
Clearly, w
=
u is nondecreasing for u > 0, that is, w1 ∝ w2 . Then for u1 , u2 > 0
u
1 (u)
α1 (x) = x,
β1 (y) = y,
α2 (x) =
b1 (x, y) = a(x, y),
f˜1 (x, y, s, t) = f1 (x, y, s, t),
f˜2 (x, y, s, t) = f2 (x, y, s, t),
Z u
u √ 2
√
√ dz
√ = 2 u − u1 ,
W1 (u) =
W1−1 (u) =
+ u1 ,
2
z
u1
Z u
dz
u
W2 (u) =
= ln ,
W2−1 (u) = u2 exp(u),
u2
u2 z
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 57, 11 pp.
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10
K ELONG Z HENG
Z xZ ∞
˜
b2 (x, y) =
W1 (b1 (x, y)) +
f1 (x, y, s, t)dtds
0
y
h p
i
√ −1
= W1 2
b1 (x, y) − u1 + (1 − exp (−x)) exp (−y)
2
p
1
=
b1 (x, y) + (1 − exp (−x)) exp (−y) .
2
W1−1
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
3y
x
x b2 (x, y) −1
= W2 ln
+ 1−
+ 1 exp −
exp (−3y)
u2
2
2
x
x b2 (x, y) = u2 exp ln
+ 1−
+ 1 exp −
exp (−3y)
u2
2
2
h
x
x i
= b2 (x, y) exp 1 −
+ 1 exp −
exp (−3y)
2
2
2
r
x
1
|σ(x) + τ (y) − k| +
+ (1 − exp (−x)) exp (−y)
=
(x + 1)(y + 1) 2
h
x
x i
+ 1 exp −
exp (−3y) .
× exp 1 −
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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