Journal of Inequalities in Pure and
Applied Mathematics
ESTIMATION FOR BOUNDED SOLUTIONS OF INTEGRAL
INEQUALITIES INVOLVING INFINITE INTEGRATION LIMITS
volume 7, issue 5, article 189,
2006.
MAN-CHUN TAN AND EN-HAO YANG
Department of Mathematics
Jinan University
Guangzhou 510632
People’s Republic of China
Received 02 April, 2006;
accepted 31 July, 2006.
Communicated by: W.S. Cheung
EMail: tanmc@jnu.edu.cn
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
101-06
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Abstract
Some integral inequalities with infinite integration limits are established as generalizations of a known result due to B.G. Pachpatte.
2000 Mathematics Subject Classification: 26D10, 45J05.
Key words: Nonlinear integral inequality, Infinite integration limit, Explicit bound on
solutions.
Project supported by the National Natural Science Foundation of P.R. China
(No.50578064), and the Natural Science Foundation of Guangdong Province of P.R.
China (No.06025219).
Estimation for Bounded
Solutions of Integral
Inequalities Involving Infinite
Integration Limits
Man-chun Tan and En-hao Yang
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Linear Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3
Nonlinear Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
4
Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
References
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1.
Introduction
As well known, various differential and integral inequalities have played a dominant role in the development of the theories of differential, functional-differential
as well as integral equations. The most powerful integral inequalities applied
frequently in the literature are the famous Gronwall-Bellman inequality [1] and
its first nonlinear generalization due to Bihari (cf., [2]). A large number of
generalizations and their applications of the Gronwall-Bellman inequality have
been obtained by many authors (cf., [4] – [7], [3], [5]). Pachpatte [6, p. 28]
proved the following interesting variant of the Gronwall-Bellman inequality
which contains an infinite integration limit:
Theorem A. LetRf be a nonnegative continuous function defined for t ∈ R+ =
∞
[0, ∞) such that 0 f (s)ds < ∞ and c(t) > 0 be a continuous and decreasing
function defined for t ∈ R+ . If u(t) ≥ 0 is a bounded continuous function
defined for t ∈ R+ and satisfies
Z ∞
u(t) ≤ c(t) +
f (s)u(s)ds,
t ∈R+ ,
t
then
Z
u(t) ≤ c(t) exp
∞
f (s)ds ,
t ∈ R+ .
t
We note that, the condition above on c(t) can be relaxed to only require
that, it is nonnegative, continuous and nonincreasing on R+ . The importance
of the last result was indicated in [6] by the fact that, it can be used to derive
the Rodrigues’ inequality [8] that played a crucial role in the study of many
perturbed linear delay differential equations.
Estimation for Bounded
Solutions of Integral
Inequalities Involving Infinite
Integration Limits
Man-chun Tan and En-hao Yang
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The aim of the present paper is to establish some new linear and nonlinear
generalizations of Theorem A. In the sequel, we denote by C(S, M ) the class
of continuous functions defined on set S with range contained in set M .
Estimation for Bounded
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Integration Limits
Man-chun Tan and En-hao Yang
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2.
Linear Generalizations
Firstly we show that an inversed version of Theorem A is valid:
R∞
Theorem 2.1. Let f ∈ C(R+ , R+ ) satisfy the condition 0 f (s)ds < ∞ and
m ∈ C(R+ , (0, ∞)) be nondecreasing. If x ∈ C(R+ , R+ ) is bounded and
satisfies the inequality
Z ∞
(2.1)
x(t) ≥ m(t) +
f (s)x(s)ds, t ∈R+ ,
Estimation for Bounded
Solutions of Integral
Inequalities Involving Infinite
Integration Limits
t
then
Z
(2.2)
∞
x(t) ≥ m(t) exp
f (s)ds,
t ∈R+ .
Man-chun Tan and En-hao Yang
t
Proof. From (2.1) we derive
(2.3)
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x(t)
1
>1+
m(t) − ε
m(t) − ε
Z ∞
≥1+
f (s)
t
Z
Contents
∞
f (s)x(s)ds
t
x(s)
ds,
m(s) − ε
t ∈R+ ,
x(t) > [m(t) − ε]V (t),
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where ε > 0 is an arbitrary number satisfying m(0) − ε > 0. Define a positive
and nonincreasing function V ∈ C(R+ , R+ ) by the right member of (2.3). Then
we have V (∞) = 1 and
(2.4)
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By differentiation we obtain
dV (t)
x(t)
= −f (t)
< −f (t)V (t),
dt
m(t) − ε
Rewrite the last relation in the form
dV (t)
< −f (t),
V (t)dt
t ∈ R+ .
t ∈ R+ ,
and integrating its both sides from t to ∞,then we have
Z ∞
ln V (∞) − ln V (t) < −
f (s)ds,
t ∈R+ ,
t
i.e.,
Man-chun Tan and En-hao Yang
∞
Z
V (t) > exp
Estimation for Bounded
Solutions of Integral
Inequalities Involving Infinite
Integration Limits
f (s)ds ,
t ∈R+ .
t
Substituting the last relation into (2.4), and letting ε → 0, the desired inequality (2.2) follows.
From Theorem A and Theorem 2.1, we obtain the following
R∞
Corollary 2.2. Let f ∈ C(R+ , R+ ) satisfy 0 f (s)ds < ∞. Let c ≥ 0 be a
constant. Then the linear integral equation
Z ∞
(2.5)
x(t) = c +
f (s)x(s)ds,
t ∈R+ ,
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has an unique bounded continuous solution represented by
Z ∞
(2.6)
x(t) = c exp
f (s)ds,
t ∈R+ .
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Proof. If c > 0, by letting c(t) ≡ c and
A and
R ∞m(t) ≡ c respectively in Theorem
R∞
Theorem 2.1, we have x(t) ≤ c exp t f (s)ds and x(t) ≥ c exp t f (s)ds.
Hence (2.6) is the unique bounded continuous solution of the equation (2.5).
By the continuous dependence on c of x(t) given by (2.6), the conclusion holds
also when c = 0.
The next result is a new generalization of Pachpatte’s inequality in the case
when an iterated integral functional is involved.
Theorem 2.3. Let n ∈ C(R+ , R
R +∞)be nonincreasing. Let f, h ∈ C(R+ , R+ ),g ∈
0
C(R+ , R+ ) with g (t) ≥ 0 and 0 [f (s) + g(s)h(s)]ds < ∞. If x ∈ C(R+ , R+ )
is bounded and satisfies the inequality
Z ∞
Z ∞
(2.7) x(t) ≤ n(t) +
f (s) x(s) + g(s)
h(k)x(k)dk ds, t ∈R+ ,
t
Estimation for Bounded
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Inequalities Involving Infinite
Integration Limits
Man-chun Tan and En-hao Yang
s
then
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Z ∞
Z ∞
(2.8) x(t) ≤ n(t) 1+ f (s) exp
[f (k)+g(k)h(k)]dk ds , t ∈ R+ .
Contents
t
s
Proof. From (2.7) we have
(2.9)
x(t)
n(t) + ε
Z ∞
Z ∞
1
< 1+
f (s) x(s) + g(s)
h(k)x(k)dk ds
n(t) + ε t
s
Z ∞
Z ∞
x(s)
x(k)
+ g(s)
h(k)
dk ds, t ∈R+ ,
< 1+
f (s)
n(s) + ε
n(k) + ε
t
s
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where ε > 0 is an arbitrary positive number. Define a function V ∈ C(R+ , R+ )
by the right member of inequality (2.9). Then V (t) is positive and nonincreasing
with V (∞) = 1, and by (2.9) we have
(2.10)
x(t) < [n(t) + ε] V (t),
t ∈ R+ .
By differentiation we obtain
Z ∞
x(t)
x(k)
dV (t)
= −f (t)
+ g(t)
h(k)
dk
dt
n(t) + ε
n(k) + ε
t
Z ∞
≥ −f (t) V (t) + g(t)
h(k)V (k)dk , t ∈ R+ .
t
Now we define
Man-chun Tan and En-hao Yang
Z
W (t) = V (t) + g(t)
∞
h(k)V (k)dk.
t
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Then W (t) ∈ C(R+ , R+ ) is positive, W (∞) = V (∞) = 1, and we have
(2.11)
W (t) ≥ V (t),
t ∈ R+ ,
and
dV (t)
≥ −f (t) W (t),
t ∈ R+ .
dt
By differentiation we derive
Z ∞
dV (t)
dW (t)
0
(2.13)
=
+ g (t)
h(k)V (k)dk − g(t)h(t)V (t)
dt
dt
t
≥ − [f (t) + g(t)h(t)] W (t), t ∈ R+ ,
(2.12)
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here (2.11) and (2.12) are used. Rewrite the last relation in the form
dW (t)
≥ − [f (t) + g(t)h(t)],
t ∈ R+ ,
W (t)dt
and then integrating both sides from t to ∞, we obtain
Z ∞
ln W (∞) − ln W (t) ≥ −
[f (k) + g(k)h(k)]dk,
t
or
∞
Z
W (t) ≤ exp
t ∈ R+ .
[f (k) + g(k)h(k)]dk ,
t
Substituting the last inequality into (2.12) and then integrating both sides from
t to ∞, we have
Z ∞
Z ∞
V (∞) − V (t) ≥ −
f (s) exp
[f (k) + g(k)h(k)]dk ds,
t
s
V (t) ≤ 1 +
∞
Z
f (s) exp
t
∞
Contents
t ∈R+ .
[f (k) + g(k)h(k)]dk ds,
s
From inequality (2.10) we obtain
Z
Z ∞
x(t) < [n(t)+ε] 1 +
f (s) exp
t
∞
Man-chun Tan and En-hao Yang
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i.e.,
Z
Estimation for Bounded
Solutions of Integral
Inequalities Involving Infinite
Integration Limits
[f (k) + g(k)h(k)]dk ds , t ∈ R+ .
s
Hence, by letting ε → 0 the desired inequality (2.8) follows from the last
relation directly.
Note that, if g(t) ≡ 0 or h(t) ≡ 0, then from Theorem 2.3 we derive Theorem A.
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3.
Nonlinear Extensions
R∞
Theorem 3.1. Let f ∈ C(R+ , R+ ) satisfy the condition 0 f (s)ds < ∞ and
c is a nonnegative number. Let ϕ, ψ ∈ C(R+ , R+ ) be strictly increasing and
ϕ−1 denote the inverse of ϕ. If x ∈ C(R+ , R+ ) is bounded and satisfies the
inequality
Z ∞
(3.1)
ϕ[x(t)] ≤ c +
f (s)ψ[x(s)]ds,
t ∈R+ ,
t
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Inequalities Involving Infinite
Integration Limits
then for t ∈ (T, ∞) we have
x(t) ≤ ϕ
(3.2)
−1
◦
G−1
c
Z
∞
f (s)ds ,
Man-chun Tan and En-hao Yang
t
where G−1
c is the inverse of Gc and
Z z
ds
,
(3.3)
Gc (z) :=
−1
c ψ ◦ ϕ (s)
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z ≥ c,
and T > 0 is the smallest number satisfying the condition
Z ∞
(3.4)
f (s)ds ∈ Dom(G−1 ), as long as t ∈ (T, ∞).
t
Proof. Without loss of generality we may assume c > 0. Otherwise we may
replace it by an arbitrary positive number ε and then let ε → 0 in (3.1) and (3.2)
to complete the proof.
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Define a nonincreasing and differentiable function H ∈ C(R+ , [c, ∞)) by
the right member of (3.1), then we have
(3.5)
x(t) ≤ ϕ−1 [H(t)],
t ∈ R+ ,
and H(∞) = c holds. By differentiation we obtain
dH(t)
= −f (t)ψ[x(t)] ≥ −f (t)ψ ◦ ϕ−1 [H(t)],
dt
where we used inequality (3.5). Rewrite this relation as
dH(t)
≥ −f (t),
ψ ◦ ϕ−1 [H(t)]dt
t ∈ R+ ,
t ∈ R+ .
Integrating both sides from t to ∞, we derive
Z ∞
Gc (H(∞)) − Gc (H(t)) ≥ −
f (s)ds,
Man-chun Tan and En-hao Yang
t ∈R+ ,
t
i.e.,
Z
Gc (H(t)) ≤ Gc (c) +
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∞
f (s)ds,
Estimation for Bounded
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Inequalities Involving Infinite
Integration Limits
t ∈R+ ,
t
where the function Gc is defined by (3.3). Since Gc (c) = 0, in view of the
choice of T in (3.4), the last relation implies
Z ∞
−1
H(t) ≤ Gc
f (s)ds ,
t ∈ (T, ∞).
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Finally, substituting the last inequality into (3.5), the desired inequality (3.2)
follows immediately.
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Remark 1. In the case when c = 0 and ϕ(0) = ψ(0) = 0 hold, to ensure
the correct definition of the function G(z), an additional condition is needed,
namely,
Z 1
ds
lim
= M < ∞.
δ→0 δ ψ ◦ ϕ−1 (s)
Theorem 3.2. Let p, q be positive numbers and c ∈ C(R+ , RR+ ) be positive and
∞
nonincreasing. Let f ∈ C(R+ , R+ ) satisfy the condition 0 f (s)ds < ∞. If
x ∈ C(R+ , R+ ) is bounded and satisfies the inequality
Z ∞
p
(3.6)
[x(t)] ≤ c(t) +
f (s)[x(s)]q ds,
t ∈R+ ,
t
Man-chun Tan and En-hao Yang
the following conclusions are true:
(I) If p > q,
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1
p−q
Z
p−q ∞
(q−p)/p
1/p
c(s)
f (s)ds
,
(3.7) x(t) ≤ c (t) 1 +
q
t
t ∈ R+ ;
(II) If p = q,
(3.8)
Estimation for Bounded
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Inequalities Involving Infinite
Integration Limits
x(t) ≤ c
1/p
Z ∞
1
(t) exp
f (s)ds , t ∈ R+ ;
p t
(III) If p < q,
(3.9) x(t) ≤ c
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1/p
1
p−q
Z
p−q ∞
(q−p)/p
c(s)
f (s)ds
, t ∈ (T, ∞),
(t) 1+
p
t
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where T is the smallest non-negative number that satisfies
Z ∞
p
c(s)(q−p)/p f (s)ds ≤
.
q−p
T
Proof. (I) If p > q holds, from inequality (3.6) we obtain
Z ∞
p
y (t) ≤ 1 +
c(s)(q−p)/p f (s)y q (s)ds,
t ∈ R+ ,
t
x(t)
where y(t) := c1/p
. The last integral inequality is a special case of (3.1) when
(t)
p
ϕ(ξ) = ξ , ψ(η) = η q . By (3.3) we derive
Z z
p
G1 (z) =
s−q/p ds =
(z (p−q)/p − 1),
p
−
q
1
and hence,
Estimation for Bounded
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p
p−q
p−q
−1
G1 (v) =
v+1
.
p
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Since G−1
1 (v) ⊃ [0, ∞) holds, from (3.2) we derive that
Z ∞
x(t)
−1
−1
(q−p)/p
≤ ϕ ◦ G1
c(s)
f (s)ds
c1/p (t)
t
Z ∞
p1
−1
(q−p)/p
= G1
c(s)
f (s)ds
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t
1
p−q
Z
p−q ∞
(q−p)/p
= 1+
c(s)
f (s)ds
,
p
t
II
I
t ∈ R+ .
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The desired inequality (3.7) follows from the last relation directly.
h
ip
x(t)
(II) If p = q holds, letting z(t) = c1/p
,from (3.6) we derive
(t)
∞
Z
(3.10)
z(t) ≤ 1 +
f (s)z(s)ds,
t ∈ R+ .
t
Define a positive, nonincreasing and differentiable function V (t) by the right
member of (3.10), then z(t) ≤ V (t) and V (∞) = 1 hold. Since c(t), f (t), z(t)
are nonnegative, by differentiation we obtain from (3.9)
V 0 (t) = −f (t)z(t) ≥ −f (t)V (t),
t ∈ R+ ,
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Man-chun Tan and En-hao Yang
i.e.,
V 0 (t)
≥ −f (t),
V (t)
t ∈ R+ .
Title Page
Integrating both sides of the last relation from t to ∞, then we have
Z ∞
ln V (∞) − ln V (t) ≥ −
f (s)ds,
t ∈R+ ,
t
or
Z
ln V (t) ≤ ln V (∞) +
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∞
f (s)ds,
t ∈R+ .
Close
t
Hence we obtain
p
Z ∞
x(t)
= z(t) ≤ V (t) ≤ exp
f (s)ds ,
c1/p (t)
t
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t ∈ R+ .
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This relation implies the desired inequality (3.8) immediately.
(III) If p < q holds, similar to the process of (I), we can get
p
G1 (z) =
(z (p−q)/p − 1),
p−q
Since
G−1
1 (v)
∞
Z
c(s)(q−p)/p f (s)ds =
T
p
p−q
p−q
=
v+1
.
p
p
,
q−p
we can derive
(3.11)
p−q
1+
p
Z
∞
c(s)(q−p)/p f (s)ds > 0,
for t ∈ (T, ∞).
Estimation for Bounded
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Integration Limits
Man-chun Tan and En-hao Yang
t
R∞
(q−p)/p
Inequality (3.11) ensures that G−1
c(s)
f
(s)ds
exists for t ∈
1
t
(T, ∞). Then we get the desired inequality (3.9).
Note that, Theorem A is a special case of Theorem 3.2 (II), when p = q = 1.
Some similar integral inequalities without infinite integration limits had been
established by Yang [8, 9].
Corollary 3.3. Let p,Rq be positive numbers with p ≤ q. Let f ∈ C(R+ , R+ )
∞
satisfy the condition 0 f (s)ds < ∞. Then x(t) ≡ 0 (t ∈ R+ ) is the unique
bounded continuous and nonnegative solution of inequality
Z ∞
p
(3.12)
[x(t)] ≤
f (s)[x(s)]q ds, t ∈R+ .
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Proof. Let x ∈ C(R+ , R+ ) be any bounded function satisfying (3.12). We
obtain
Z ∞
p
(3.13)
[x(t)] ≤ ε +
f (s)[x(s)]q ds, t ∈R+ ,
t
where ε is an arbitrary positive number.
When p < q and ε is small enough, the inequality
Z ∞
p
ε(q−p)/p f (s)ds <
q−p
t
holds for all t ∈ R+ .
A suitable application of Theorem 3.2 to (3.13) yields that, for t ∈ R+
x(t) ≤

h

1/p

1+
 ε
p−q
p
R∞
t
ε
(q−p)/p
i
h


 ε1/p exp 1 R ∞ f (s)ds ,
p t
f (s)ds
1
i p−q
, p < q;
Estimation for Bounded
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Man-chun Tan and En-hao Yang
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p = q.
Finally, letting ε → 0, from the last relation we obtain x(t) ≡ 0, t ∈ R+ .
If the condition p ≤ q is replaced by p > q, the result x(t) ≡ 0R cannot be
∞
derived directly from Theorem 3.2. In fact, if p > q and M (t) := t f (s)ds,
then
1
1
p−q
p−q
Z ∞
p
−
q
p
−
q
lim ε1/p 1 +
ε(q−p)/p f (s)ds
=
M (t)
.
ε→0
p
p
t
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4.
Examples
Example 4.1. Let x ∈ C(R+ , R+ ) be bounded and satisfy the integral inequality
Z
∞
s e−3s x(s)ds,
x(t) ≥ 1 +
t ∈R+ .
t
Then by Theorem 2.1, we have
Z ∞
3t + 1 −3t
−3s
e
,
x(t) ≥ exp
se ds = exp
9
t
t ∈R+ .
Example 4.2. Let x ∈ C(R+ , R+ ) be a bounded function satisfying the inequality
Z ∞
Z ∞
Z ∞
−s
−s
x(t) ≤ 1 +
e x(s)ds +
e
e−k x(k)dk ds, t ∈R+ .
t
t
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s
Then by Theorem 2.3 ,we easily establish
Z
Z ∞
−s
x(t) ≤ 1 +
e exp
t
Estimation for Bounded
Solutions of Integral
Inequalities Involving Infinite
Integration Limits
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∞
2e−k dk ds
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s
1
=
1 + exp 2e−t ,
2
t ∈ R+ .
Example 4.3. Let x ∈ C(R+ , R+ ) be a bounded function satisfying the inequality
Z
∞
e−3s x(s)ds,
x1/2 (t) ≤ 1 +
t
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t ∈R+ .
J. Ineq. Pure and Appl. Math. 7(5) Art. 189, 2006
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Since
dom
G−1
1
Z
∞
−3s
e
t
ds
= dom
3
3 − e−3t
⊃ R+
holds, referring to the proof of Theorem 3.2, we obtain
2
3
, t ∈ R+ .
x(t) ≤
3 − e−3t
Estimation for Bounded
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Inequalities Involving Infinite
Integration Limits
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References
[1] R. BELLMAN AND K.L. COOKE, Differential-difference Equations,
Acad. Press, New York, 1963.
[2] I. BIHARI, A generalization of a lemma of Bellman and its application
to uniqueness problems of differential equations, Acta Math. Acad. Sci.
Hungar., 7 (1956), 71–94.
[3] H. EL-OWAIDY, A. RAGAB AND A. ABDELDAIM, On some new integral inequalities of Growall Bellman type, Appl. Math. Comput., 106(2-3)
(1999), 289–303.
[4] V. LAKSHIMIKANTHAM AND S. LEELA, Differential and Integral Inequalities: Theory and Applications, Academic Press, New
York/London,1969.
[5] F.W. MENG AND W.N. LI, On some new integral inequalities and their
applications, Appl. Math. Comput., 148 (2004), 381–392.
[6] B.G. PACHPATTE, Inequalities for Differential and Integral Equations,
Academic Press, San Diego, 1998.
[7] B.G. PACHPATTE, Explicit bounds on certain integral inequalities, J.
Math. Anal. and Applics., 267(1) (2002), 48–61.
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Inequalities Involving Infinite
Integration Limits
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[8] H.M. RODRIGUES, On growth and decay of solutions of perturbed retarded linear equations, Tohoku Math. J., 32 (1980), 593–605.
[9] EN-HAO YANG, Generalizations of Pachpatte’s integral and discrete inequalities, An. Diff. Eqs., 13(2) (1997), 180–188.
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[10] EN-HAO YANG, On some nonlinear integral and discrete inequalities
related to Ou-Iang’s inequality, Acta Math. Sinica., New Series, 14(3)
(1998), 353–360.
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Solutions of Integral
Inequalities Involving Infinite
Integration Limits
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