Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 3, Article 60, 2005
ON SOME ADVANCED INTEGRAL INEQUALITIES AND THEIR APPLICATIONS
XUEQIN ZHAO AND FANWEI MENG
xqzhao1972@126.com
D EPARTMENT OF M ATHEMATICS
Q UFU N ORMAL U NIVERSITY
Q UFU 273165
P EOPLE ’ S R EPUBLIC OF C HINA
fwmeng@qfnu.edu.cn
Received 08 November, 2004; accepted 03 April, 2005
Communicated by S.S. Dragomir
A BSTRACT. In this paper, we obtain a generalization of advanced integral inequality and by
means of examples we show the usefulness of our results.
Key words and phrases: Advanced integral inequality; Integral equation.
2000 Mathematics Subject Classification. 26D15, 26D10.
1. I NTRODUCTION
Integral inequalities play an important role in the qualitative analysis of the solutions to differential and integral equations. Many retarded inequalities have been discovered (see [2], [3],
[5], [7]). However, we almost neglect the importance of advanced inequalities. After all, it does
great benefit to solve the bound of certain integral equations, which help us to fulfill a diversity
of desired goals. In this paper we establish two advanced integral inequalities and an application
of our results is also given.
2. P RELIMINARIES AND L EMMAS
In this paper, we assume throughout that R+ = [0, ∞), is a subset of the set of real numbers
R. The following lemmas play an important role in this paper.
Lemma 2.1. Let ϕ ∈ C(R+ , R+ ) be an increasing function with ϕ(∞) = ∞. Let ψ ∈
C(R+ , R+ ) be a nondecreasing function and let c be a nonnegative constant. Let α ∈ C 1 (R+ , R+ )
be nondecreasing with α(t) ≥ t on R+ . If u, f ∈ C(R+ , R+ ) and
Z ∞
(2.1)
ϕ(u(t)) ≤ c +
f (s)ψ(u(s))ds,
t ∈ R+ ,
α(t)
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
Project supported by a grant from NECC and NSF of Shandong Province, China (Y2001A03).
214-04
2
X UEQIN Z HAO AND FANWEI M ENG
then for 0 ≤ T ≤ t < ∞,
(2.2)
u(t) ≤ ϕ
−1
Z
−1
G [G(c) +
∞
f (s)ds] ,
α(t)
Rz
ds
where G(z) = z0 ψ[ϕ−1
, z ≥ z0 > 0, ϕ−1 , G−1 are respectively the inverse of ϕ and G,
(s)]
T ∈ R+ is chosen so that
Z ∞
(2.3a)
G(c) +
f (s)ds ∈ Dom(G−1 ),
t ∈ [T, ∞).
α(t)
Z ∞
−1
(2.3b)
G
G(c) +
f (s)ds ∈ Dom(ϕ−1 ),
t ∈ [T, ∞).
α(t)
Proof. Define the nonincreasing positive function z(t) and make
Z ∞
(2.4)
z(t) = c + ε +
f (s)ψ(u(s))ds,
t ∈ R+ ,
α(t)
where ε is an arbitrary small positive number. From inequality (2.1), we have
u(t) ≤ ϕ−1 [z(t)].
(2.5)
Differentiating (2.4) and using (2.5) and the monotonicity of ϕ−1 , ψ, we deduce that
z 0 (t) = −f α(t) ψ u α(t) α0 (t)
≥ −f α(t) ψ ϕ−1 z(α(t)) α0 (t)
≥ −f α(t) ψ ϕ−1 z(t) α0 (t).
For
ψ[ϕ−1 (z(t))] ≥ ψ[ϕ−1 (z(∞))] = ψ[ϕ−1 (c + ε)] > 0,
from the definition of G, the above relation gives
0
d
z 0 (t)
G(z(t)) =
≥
−f
α(t)
α (t).
dt
ψ[ϕ−1 (z(t))]
Setting t = s, and integrating it from t to ∞ and letting ε → 0 yields
Z ∞
G z(t) ≤ G(c) +
f (s)ds,
t ∈ R+ .
α(t)
From (2.3), (2.5) and the above relation, we obtain the inequality (2.2).
In fact, we can regard Lemma 2.1 as a generalized form of an Ou-Iang type inequality with
advanced argument.
Lemma 2.2. Let u f and g be nonnegative continuous functions defined on R+ , and let ϕ ∈
C(R+ , R+ ) be an increasing function with ϕ(∞) = ∞ and let c be a nonnegative constant.
Moreover, let w1 , w2 ∈ C(R+ , R+ ) be nondecreasing functions with wi (u) > 0 (i = 1, 2) on
(0, ∞), α ∈ C 1 (R+ , R+ ) be nondecreasing with α(t) ≥ t on R+ . If
Z ∞
Z ∞
(2.6)
ϕ(u(t)) ≤ c +
f (s)w1 (u(s))ds +
g(s)w2 (u(s))ds,
t ∈ R+ ,
α(t)
t
then for 0 ≤ T ≤ t < ∞,
(i) For the case w2 (u) ≤ w1 (u),
Z
−1
−1
(2.7)
u(t) ≤ ϕ
G1 G1 (c) +
∞
α(t)
J. Inequal. Pure and Appl. Math., 6(3) Art. 60, 2005
Z
f (s)ds +
∞
g(s)ds
.
t
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O N S OME A DVANCED I NTEGRAL I NEQUALITIES AND THEIR A PPLICATIONS
(ii) For the case w1 (u) ≤ w2 (u),
Z
−1
−1
(2.8)
u(t) ≤ ϕ
G2 G2 (c) +
∞
f (s)ds +
Z
r
Gi (r) =
r0
ds
,
wi (ϕ−1 (s))
g(s)ds
α(t)
where
∞
Z
3
,
t
r ≥ r0 > 0,
(i = 1, 2)
and ϕ−1 , G−1
i (i = 1, 2) are respectively the inverse of ϕ, Gi , T ∈ R+ is chosen so that
Z
(2.9)
∞
Z
Gi (c) +
∞
g(s)ds ∈ Dom(G−1
i ),
f (s)ds +
α(t)
t ∈ [T, ∞).
(i = 1, 2),
t
Proof. Define the nonincreasing positive function z(t) and make
Z ∞
Z ∞
(2.10) z(t) = c + ε +
f (s)w1 (u(s))ds +
g(s)w2 (u(s))ds,
α(t)
0 ≤ T ≤ t < ∞,
t
where ε is an arbitrary small positive number. From inequality (2.6), we have
u(t) ≤ ϕ−1 [z(t)],
(2.11)
t ∈ R+ .
Differentiating (2.10) and using (2.11) and the monotonicity of ϕ−1 , w1 , w2 , we deduce that
z 0 (t) = −f α(t) w1 u α(t) α0 (t) − g(t)w2 u(t) ,
≥ −f α(t) w1 ϕ−1 z(α(t)) α0 (t) − g(t)w2 ϕ−1 z(t) ,
≥ −f α(t) w1 ϕ−1 z(t) α0 (t) − g(t)w2 ϕ−1 z(t) .
(i) When w2 (u) ≤ w1 (u)
z 0 (t) ≥ −f α(t) w1 ϕ−1 z(t) α0 (t) − g(t)w1 ϕ−1 z(t) ,
t ∈ R+ .
For
w1 [ϕ−1 (z(t))] ≥ w1 ϕ−1 (z(∞)) = w1 [ϕ−1 (c + ε)] > 0,
from the definition of G1 (r), the above relation gives
0
d
z 0 (t)
G1 (z(t)) =
≥
−f
α(t)
α (t) − g(t),
dt
w1 [ϕ−1 (z(t))]
Setting t = s and integrating it from t to ∞ and let ε → 0 yields
Z ∞
Z ∞
G1 z(t) ≤ G1 (c) +
f (s)ds +
g(s)ds,
α(t)
t ∈ R+ .
t ∈ R+ ,
t
so,
z(t) ≤
G−1
1
Z
G1 (c) +
∞
Z
∞
f (s)ds +
α(t)
Using (2.11), we have
Z
−1
−1
u(t) ≤ ϕ
G1 G1 (c) +
∞
α(t)
Z
f (s)ds +
∞
g(s)ds
,
0 ≤ T ≤ t < ∞.
t
(ii) When w1 (u) ≤ w2 (u), the proof can be completed similarly.
J. Inequal. Pure and Appl. Math., 6(3) Art. 60, 2005
0 ≤ T ≤ t < ∞.
g(s)ds ,
t
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4
X UEQIN Z HAO AND FANWEI M ENG
3. M AIN R ESULTS
In this section, we obtain our main results as follows:
Theorem 3.1. Let u, f and g be nonnegative continuous functions defined on R+ and let c
be a nonnegative constant. Moreover, let ϕ ∈ C(R+ , R+ ) be an increasing function with
ϕ(∞) = ∞, ψ ∈ C(R+ , R+ ) be a nondecreasing function with ψ(u) > 0 on (0, ∞) and
α ∈ C 1 (R+ , R+ ) be nondecreasing with α(t) ≥ t on R+ . If
Z ∞
(3.1)
ϕ(u(t)) ≤ c +
[f (s)u(s)ψ(u(s)) + g(s)u(s)]ds,
t ∈ R+
α(t)
then for 0 ≤ T ≤ t < ∞,
Z
−1
−1
−1
(3.2)
u(t) ≤ ϕ
Ω
G
G[Ω(c) +
∞
Z
∞
g(s)ds] +
α(t)
f (s)ds
,
α(t)
where
Z
r
Ω(r) =
r0
ds
,
−1
ϕ (s)
z
Z
r ≥ r0 > 0,
G(z) =
ds
ψ{ϕ−1 [Ω−1 (s)]}
z0
,
z ≥ z0 > 0,
Ω−1 , ϕ−1 , G−1 are respectively the inverse of Ω, ϕ, G and T ∈ R+ is chosen so that
Z ∞
Z ∞
G Ω(c) +
g(s)ds +
f (s)ds ∈ Dom(G−1 )
α(t)
and
G
−1
Z
G Ω(c) +
α(t)
∞
∞
Z
g(s)ds +
α(t)
f (s)ds
∈ Dom(Ω−1 )
α(t)
for t ∈ [T, ∞).
Proof. Let us first assume that c > 0. Define the nonincreasing positive function z(t) by the
right-hand side of (3.1). Then z(∞) = c, u(t) ≤ ϕ−1 [z(t)] and
z 0 (t) = − f α(t) u α(t) ψ u α(t) − g α(t) u α(t) α0 (t)
≥ − f α(t) ϕ−1 z(α(t)) ψ ϕ−1 z(α(t)) − g α(t) ϕ−1 z(α(t)) α0 (t)
≥ − f α(t) ϕ−1 z(t) ψ ϕ−1 z(α(t)) − g α(t) ϕ−1 z(t) α0 (t).
Since ϕ−1 (z(t)) ≥ ϕ−1 (c) > 0,
z 0 (t)
≥ − f α(t) ψ ϕ−1 z(α(t)) + g α(t) α0 (t).
ϕ−1 z(t)
Setting t = s and integrating it from t to ∞ yields
Z ∞
Z
Ω(z(t)) ≤ Ω(c) +
g(s)ds +
α(t)
∞
f (s)ψ[ϕ−1 (z(s))]ds.
α(t)
R∞
Let T ≤ T1 be an arbitrary number. We denote p(t) = Ω(c) + α(t) g(s)ds. From the above
relation, we deduce that
Z ∞
Ω(z(t)) ≤ p(T1 ) +
f (s)ψ[ϕ−1 (z(s))]ds,
T1 ≤ t < ∞.
α(t)
Now an application of Lemma 2.1 gives
Z
−1
−1
z(t) ≤ Ω
G
G(p(T1 )) +
∞
f (s)ds ,
T1 ≤ t < ∞,
α(t)
J. Inequal. Pure and Appl. Math., 6(3) Art. 60, 2005
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O N S OME A DVANCED I NTEGRAL I NEQUALITIES AND THEIR A PPLICATIONS
5
so,
u(t) ≤ ϕ
−1
Z
−1
−1
Ω
G
G(p(T1 )) +
∞
f (s)ds
,
T1 ≤ t < ∞.
α(t)
Taking t = T1 in the above inequality, since T1 is arbitrary, we can prove the desired inequality
(3.2).
If c = 0 we carry out the above procedure with ε > 0 instead of c and subsequently let
ε → 0.
Corollary 3.2. Let u, f and g be nonnegative continuous functions defined on R+ and let c
be a nonnegative constant. Moreover, let ψ ∈ C(R+ , R+ ) be a nondecreasing function with
ψ(u) > 0 on (0, ∞) and α ∈ C 1 (R+ , R+ ) be nondecreasing with α(t) ≥ t on R+ . If
Z ∞
2
2
u (t) ≤ c +
[f (s)u(s)ψ(u(s)) + g(s)u(s)]ds,
t ∈ R+ ,
α(t)
then for 0 ≤ T ≤ t < ∞,
u(t) ≤ Ω
−1
Z
Z
1 ∞
1 ∞
g(s)ds +
f (s)ds ,
Ω c+
2 α(t)
2 α(t)
where
Z
r
ds
,
r > 0,
1 ψ(s)
Ω−1 is the inverse of Ω, and T ∈ R+ is chosen so that
Z
Z
1 ∞
1 ∞
Ω c+
g(s)ds +
f (s)ds ∈ Dom(Ω−1 )
2 α(t)
2 α(t)
Ω(r) =
for all t ∈ [T, ∞).
Corollary 3.3. Let u, f and g be nonnegative continuous functions defined on R+ and let c be
a nonnegative constant. Moreover, let α ∈ C 1 (R+ , R+ ) be nondecreasing with α(t) ≥ t on R+ .
If
Z
∞
u2 (t) ≤ c2 +
[f (s)u2 (s) + g(s)u(s)]ds,
t ≥ 0,
α(t)
then
u(t) ≤
1
c+
2
Z ∞
1
g(s)ds exp
f (s)ds ,
2 α(t)
α(t)
Z
∞
t ≥ 0.
Corollary 3.4. Let u, f and g be nonnegative continuous functions defined on R+ and let c
be a nonnegative constant. Moreover, let p, q be positive constants with p ≥ q, p 6= 1. Let
α ∈ C 1 (R+ , R+ ) be nondecreasing with α(t) ≥ t on R+ . If
Z ∞
p
u (t) ≤ c +
[f (s)uq (s) + g(s)u(s)]ds,
t ∈ R+ ,
α(t)
then for t ∈ R+ ,
 p
p−1
h R
i
R

(1− p1 )
p−1 ∞
1 ∞

c
+
g(s)ds
exp
f
(s)ds
,
when p = q,

p
p α(t)
α(t)


u(t) ≤
1
p−q
p−q

p−1

R
R
1

∞
∞
(1−
)

+ p−q
f (s)ds
, when p > q.
c p + p−1
g(s)ds

p
p
α(t)
α(t)
J. Inequal. Pure and Appl. Math., 6(3) Art. 60, 2005
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6
X UEQIN Z HAO AND FANWEI M ENG
Theorem 3.5. Let u, f and g be nonnegative continuous functions defined on R+ , and let
ϕ ∈ C(R+ , R+ ) be an increasing function with ϕ(∞) = ∞ and let c be a nonnegative constant.
Moreover, let w1 , w2 ∈ C(R+ , R+ ) be nondecreasing functions with wi (u) > 0 (i = 1, 2) on
(0, ∞) and α ∈ C 1 (R+ , R+ ) be nondecreasing with α(t) ≥ t on R+ . If
Z ∞
Z ∞
f (s)u(s)w1 (u(s))ds +
g(s)u(s)w2 (u(s))ds,
(3.3)
ϕ(u(t)) ≤ c +
α(t)
t
then for 0 ≤ T ≤ t < ∞,
(i) For the case w2 (u) ≤ w1 (u),
Z
−1
−1
−1
(3.4)
u(t) ≤ ϕ
Ω
G1 G1 (Ω(c)) +
∞
Z
g(s)ds
,
∞
g(s)ds
,
f (s)ds +
α(t)
(ii) For the case w1 (u) ≤ w2 (u),
Z
−1
−1
−1
(3.5)
u(t) ≤ ϕ
Ω
G2 G2 (Ω(c)) +
∞
t
∞
Z
f (s)ds +
α(t)
t
where
Z
r
Ω(r) =
Zr0z
Gi (z) =
z0
ds
ϕ−1 (s)
,
r ≥ r0 > 0,
ds
wi
{ϕ−1 [Ω−1 (s)]}
z ≥ z0 > 0
,
(i = 1, 2),
Ω−1 , ϕ−1 , G−1 are respectively the inverse of Ω, ϕ, G, and T ∈ R+ is chosen so that
Z ∞
Z ∞
Gi Ω(c) +
f (s)ds +
g(s)ds ∈ Dom(G−1
i ),
α(t)
t
Z ∞
Z ∞
−1
Gi Gi Ω(c) +
f (s)ds +
g(s)ds
∈ Dom(Ω−1 ),
α(t)
t
for all t ∈ [T, ∞).
Proof. Let c > 0, define the nonincreasing positive function z(t) and make
Z ∞
Z ∞
(3.6)
z(t) = c +
f (s)u(s)w1 (u(s))ds +
g(s)u(s)w2 (u(s))ds.
α(t)
t
From inequality (3.3), we have
u(t) ≤ ϕ−1 [z(t)].
(3.7)
Differentiating (3.6) and using (3.7) and the monotonicity of ϕ−1 , w1 , w2 , we deduce that
z 0 (t) = −f α(t) u α(t) w1 u α(t) α0 (t) − g(t)u(t)w2 u(t) ,
≥ −f α(t) ϕ−1 z(α(t)) w1 ϕ−1 z(α(t)) α0 (t) − g(t)ϕ−1 z(t) w2 ϕ−1 z(t) ,
≥ −f α(t) ϕ−1 z(t) w1 ϕ−1 z(t) α0 (t) − g(t)ϕ−1 z(t) w2 ϕ−1 z(t) .
(i) When w2 (u) ≤ w1 (u)
z 0 (t)
≥ −f α(t) w1 ϕ−1 z(t) α0 (t) − g(t)w1 ϕ−1 z(t) .
ϕ−1 z(t)
For
w1 [ϕ−1 (z(t))] ≥ w1 [ϕ−1 (z(∞))] = w1 [ϕ−1 (c + ε)] > 0,
J. Inequal. Pure and Appl. Math., 6(3) Art. 60, 2005
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O N S OME A DVANCED I NTEGRAL I NEQUALITIES AND THEIR A PPLICATIONS
setting t = s and integrating from t to ∞ yields
Z
Z ∞
−1
Ω(z(t)) ≤ Ω(c) +
f (s)w1 ϕ z(t) ds +
α(t)
7
∞
g(s)w1 ϕ−1 z(t) ds.
t
From Lemma 2.2, we obtain
Z
−1
−1
z(t) ≤ Ω
G1 G1 (Ω(c)) +
∞
Z
∞
f (s)ds +
g(s)ds
α(t)
0 ≤ T ≤ t < ∞.
,
t
Using u(t) ≤ ϕ−1 [z(t)], we get the inequality in (3.4)
If c = 0, we can carry out the above procedure with ε > 0 instead of c and subsequently let
ε → 0.
(ii) When w1 (u) ≤ w2 (u), the proof can be completed similarly.
4. A N A PPLICATION
We consider an integral equation
(4.1)
Z
p
x (t) = a(t) +
∞
F [s, x(s), x(φ(s))]ds.
t
Assume that:
|F (x, y, u)| ≤ f (x)|u|q + g(x)|u|
(4.2)
and
(4.3)
|a(t)| ≤ c, c > 0 p ≥ q > 0, p 6= 1,
where f, g are nonnegative continuous real-valued functions, and φ ∈ C 1 (R+ , R+ ) is nondecreasing with φ(t) ≥ t on R+ . From (4.1), (4.2) and (4.3) we have
Z ∞
p
|x(t)| ≤ c +
f (s)|x(φ(s))|q + g(s)|x(φ(s))|ds.
t
Making the change of variables from the above inequality and taking
M = sup
t∈R+
1
φ0 (t)
,
we have
p
Z
∞
|x(t)| ≤ c + M
f¯(s)|x(s)|q + ḡ(s)|x(s)|ds,
φ(t)
in which f¯(s) = f (φ−1 (s)), ḡ(s) = g(φ−1 (s)). From Corollary 3.4, we obtain
 p
p−1
h R
i
∞ ¯

M (p−1) R ∞
(1− p1 )
M

+
ḡ(s)ds
exp
f
(s)ds
,
when
c

p
p φ(t)
φ(t)


|x(t)| ≤
1
p−q
p−q

p−1

R
R
1

∞
∞

ḡ(s)ds
+ M (p−q)
f¯(s)ds
, when
c(1− p ) + M (p−1)

p
p
φ(t)
φ(t)
p=q
p > q.
If the integrals of f (s), g(s) are bounded, then we have the bound of the solution of (4.1).
J. Inequal. Pure and Appl. Math., 6(3) Art. 60, 2005
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8
X UEQIN Z HAO AND FANWEI M ENG
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J. Inequal. Pure and Appl. Math., 6(3) Art. 60, 2005
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