Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 5, Article 189, 2006
ESTIMATION FOR BOUNDED SOLUTIONS OF INTEGRAL INEQUALITIES
INVOLVING INFINITE INTEGRATION LIMITS
MAN-CHUN TAN AND EN-HAO YANG
D EPARTMENT OF M ATHEMATICS
J INAN U NIVERSITY
G UANGZHOU 510632
P EOPLE ’ S R EPUBLIC OF C HINA
tanmc@jnu.edu.cn
Received 02 April, 2006; accepted 31 July, 2006
Communicated by W.S. Cheung
A BSTRACT. Some integral inequalities with infinite integration limits are established as generalizations of a known result due to B.G. Pachpatte.
Key words and phrases: Nonlinear integral inequality, Infinite integration limit, Explicit bound on solutions.
2000 Mathematics Subject Classification. 26D10, 45J05.
1. I NTRODUCTION
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:
A. Let f be a nonnegative continuous function defined for t ∈ R+ = [0, ∞) such that
RTheorem
∞
f
(s)ds
< ∞ and c(t) > 0 be a continuous and decreasing function defined for t ∈ R+ . If
0
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
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
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).
101-06
2
M AN - CHUN TAN AND E N - HAO YANG
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.
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 .
2. L INEAR G ENERALIZATIONS
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+ ,
t
then
Z
(2.2)
∞
x(t) ≥ m(t) exp
f (s)ds,
t ∈R+ .
t
Proof. From (2.1) we derive
Z ∞
1
x(t)
(2.3)
>1+
f (s)x(s)ds
m(t) − ε
m(t) − ε t
Z ∞
x(s)
≥1+
f (s)
ds,
t ∈R+ ,
m(s) − ε
t
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)
x(t) > [m(t) − ε]V (t),
t ∈ R+ .
By differentiation we obtain
dV (t)
x(t)
= −f (t)
< −f (t)V (t),
t ∈ R+ .
dt
m(t) − ε
Rewrite the last relation in the form
dV (t)
< −f (t),
t ∈ R+ ,
V (t)dt
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.,
Z
V (t) > exp
∞
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
J. Inequal. Pure and Appl. Math., 7(5) Art. 189, 2006
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I NTEGRAL I NEQUALITIES I NVOLVING I NFINITE I NTEGRATION L IMITS
3
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+ ,
t
has an unique bounded continuous solution represented by
Z ∞
(2.6)
x(t) = c exp
f (s)ds,
t ∈R+ .
t
Proof. If c > 0, by letting
in Theorem A and Theorem 2.1,
R ∞ c(t) ≡ c and m(t) ≡ c respectively
R∞
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 nR ∈ C(R+ , R+ )be nonincreasing. Let f, h ∈ C(R+ , R+ ),g ∈ C(R+ , R+ )
∞
with g 0 (t) ≥ 0 and 0 [f (s) + g(s)h(s)]ds < ∞. If x ∈ C(R+ , R+ ) is bounded and satisfies
the inequality
Z ∞
Z ∞
h(k)x(k)dk ds, t ∈R+ ,
f (s) x(s) + g(s)
(2.7)
x(t) ≤ n(t) +
s
t
then
(2.8)
∞
Z
x(t) ≤ n(t) 1 +
Z
∞
[f (k) + g(k)h(k)]dk ds ,
f (s) exp
t ∈ R+ .
s
t
Proof. From (2.7) we have
x(t)
(2.9)
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,
<1+
f (s)
n(s) + ε
n(k) + ε
s
t
t ∈R+ ,
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
Z
W (t) = V (t) + g(t)
∞
h(k)V (k)dk.
t
Then W (t) ∈ C(R+ , R+ ) is positive, W (∞) = V (∞) = 1, and we have
(2.11)
W (t) ≥ V (t),
J. Inequal. Pure and Appl. Math., 7(5) Art. 189, 2006
t ∈ R+ ,
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4
M AN - CHUN TAN AND E N - HAO YANG
and
dV (t)
≥ −f (t) W (t),
dt
(2.12)
t ∈ R+ .
By differentiation we derive
Z ∞
dW (t)
dV (t)
0
=
+ 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+ ,
(2.13)
here (2.11) and (2.12) are used. Rewrite the last relation in the form
dW (t)
≥ − [f (t) + g(t)h(t)],
W (t)dt
t ∈ R+ ,
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
i.e.,
Z
∞
V (t) ≤ 1 +
s
∞
Z
f (s) exp
t ∈R+ .
[f (k) + g(k)h(k)]dk ds,
t
s
From inequality (2.10) we obtain
Z
Z ∞
x(t) < [n(t) + ε] 1 +
f (s) exp
t
∞
[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.
3. N ONLINEAR E XTENSIONS
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
then for t ∈ (T, ∞) we have
x(t) ≤ ϕ
(3.2)
−1
◦
G−1
c
Z
∞
f (s)ds ,
t
where
G−1
c
is the inverse of Gc and
Z
(3.3)
Gc (z) :=
c
J. Inequal. Pure and Appl. Math., 7(5) Art. 189, 2006
z
ds
,
ψ ◦ ϕ−1 (s)
z ≥ c,
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I NTEGRAL I NEQUALITIES I NVOLVING I NFINITE I NTEGRATION L IMITS
5
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.
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,
t ∈R+ ,
t
i.e.,
Z
Gc (H(t)) ≤ Gc (c) +
∞
f (s)ds,
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, ∞).
t
Finally, substituting the last inequality into (3.5), the desired inequality (3.2) follows immediately.
Remark 3.2. 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 3.3. Let p, q be positive numbersR and c ∈ C(R+ , R+ ) 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
the following conclusions are true:
(I) If p > q,
1
p−q
Z
p−q ∞
(q−p)/p
1/p
(3.7)
x(t) ≤ c (t) 1 +
c(s)
f (s)ds
,
q
t
J. Inequal. Pure and Appl. Math., 7(5) Art. 189, 2006
t ∈ R+ ;
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6
M AN - CHUN TAN AND E N - HAO YANG
(II) If p = q,
x(t) ≤ c
(3.8)
1/p
Z ∞
1
(t) exp
f (s)ds ,
p t
t ∈ R+ ;
(III) If p < q,
(3.9)
1
p−q
Z
p−q ∞
(q−p)/p
x(t) ≤ c (t) 1 +
c(s)
f (s)ds
,
p
t
where T is the smallest non-negative number that satisfies
Z ∞
p
c(s)(q−p)/p f (s)ds ≤
.
q−p
T
1/p
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 ∈ (T, ∞),
t ∈ R+ ,
t
x(t)
. The last integral inequality is a special case of (3.1) when ϕ(ξ) = ξ p ,
where y(t) := c1/p
(t)
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,
p
p−q
p−q
−1
G1 (v) =
v+1
.
p
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
t
1
p−q
Z
p−q ∞
= 1+
c(s)(q−p)/p f (s)ds
,
p
t
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 (t) ,from (3.6) we derive
Z ∞
(3.10)
z(t) ≤ 1 +
f (s)z(s)ds,
t ∈ R+ .
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+ ,
i.e.,
V 0 (t)
≥ −f (t),
t ∈ R+ .
V (t)
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
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I NTEGRAL I NEQUALITIES I NVOLVING I NFINITE I NTEGRATION L IMITS
or
Z
7
∞
ln V (t) ≤ ln V (∞) +
t ∈R+ .
f (s)ds,
t
Hence we obtain
x(t)
1/p
c (t)
p
Z
∞
= z(t) ≤ V (t) ≤ exp
f (s)ds ,
t ∈ R+ .
t
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
Z
G−1
1 (v)
∞
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, ∞).
t
Inequality (3.11) ensures that G−1
1
get the desired inequality (3.9).
R∞
t
c(s)(q−p)/p f (s)ds exists for t ∈ (T, ∞). Then we
Note that, Theorem A is a special case of Theorem 3.3 (II), when p = q = 1. Some similar
integral inequalities without infinite integration limits had been established by Yang [8, 9].
Corollary
R ∞ 3.4. Let p, q 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+ .
t
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 ∞
ε(q−p)/p f (s)ds <
t
p
q−p
holds for all t ∈ R+ .
A suitable application of Theorem 3.3 to (3.13) yields that, for t ∈ R+
x(t) ≤

h

1/p

ε
1+

p−q
p
R∞
t
1
i p−q
ε(q−p)/p f (s)ds
, p < q;
h
i


 ε1/p exp 1 R ∞ f (s)ds ,
p t
p = q.
Finally, letting ε → 0, from the last relation we obtain x(t) ≡ 0, t ∈ R+ .
J. Inequal. Pure and Appl. Math., 7(5) Art. 189, 2006
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8
M AN - CHUN TAN AND E N - HAO YANG
If the condition p ≤ q is replaced by p > q, the
R ∞result x(t) ≡ 0 cannot be derived directly
from Theorem 3.3. 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
4. E XAMPLES
Example 4.1. Let x ∈ C(R+ , R+ ) be bounded and satisfy the integral inequality
Z ∞
s e−3s x(s)ds, t ∈R+ .
x(t) ≥ 1 +
t
Then by Theorem 2.1, we have
Z
x(t) ≥ exp
∞
se
−3s
t
3t + 1 −3t
ds = exp
e
,
9
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
s
Then by Theorem 2.3 ,we easily establish
Z
Z ∞
−s
x(t) ≤ 1 +
e exp
t
∞
−k
2e dk ds
s
1
1 + exp 2e−t ,
t ∈ R+ .
2
Example 4.3. Let x ∈ C(R+ , R+ ) be a bounded function satisfying the inequality
Z ∞
1/2
x (t) ≤ 1 +
e−3s x(s)ds, t ∈R+ .
=
t
Since
dom
G−1
1
Z
∞
−3s
e
ds
= dom
t
3
3 − e−3t
⊃ R+
holds, referring to the proof of Theorem 3.3, we obtain
2
3
x(t) ≤
, t ∈ R+ .
3 − e−3t
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I NTEGRAL I NEQUALITIES I NVOLVING I NFINITE I NTEGRATION L IMITS
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