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Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2007, Article ID 24385, 18 pages
doi:10.1155/2007/24385
Research Article
New Integral Inequalities for Iterated Integrals with Applications
Ravi P. Agarwal, Cheon Seoung Ryoo, and Young-Ho Kim
Received 20 September 2007; Accepted 15 November 2007
Recommended by Sever S. Dragomir
Some new nonlinear retarded integral inequalities of Gronwall type are established. These
inequalities can be used as basic tools in the study of certain classes of integrodifferential
equations.
Copyright © 2007 Ravi P. Agarwal et al. 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.
1. Introduction
The Gronwall-type integr al inequalities provide necessary tools in the study of the theory
of differential equations, integral equations, and inequalities of various types. Some such
inequalities can be found in the works of Agarwal, Deng et al. [1]. The result has been used
in the study of global existence of solutions of a retarded di fferential equations and esti-
mation of solution of function differential equation, Cheung [2]. The result has been used
in the study of certain initial boundary value problem for hyperbolic partial differential
equations, Cheung and Ma [3]. The result has been used in the study of global existence
of solutions for a partial differential equations, Pachpatte [4–9]. The results have been ap-
plied in the study of certain properties of solutions for the integrodifferential equations,
partial integrodifferential equations, retarded Volterra-Fredholm integral equations, re-
tarded nonself–adjoint hyperbolic partial differential equations, Ye et al. [10]. The result
has been used in the study of the Riemann-Liouville fractional integral equations, Zhao
and Meng [11]. The result has been used in the study of integral equations. During the
past few years, several authors (see [12–19] and some of the references cited therein) have
established many other very useful Gronwall—like integral inequalities. Recently, in [16]
a new interesting Gronwall—like integral inequality involving iterated integrals has been
established.
2 Journal of Inequalities and Applications
Theorem 1.1. Let u(t) be nonnegative continuous function in J
= [α, β] and let a(t) be
positive nondecreasing continuous function in J, and let f
i
(t,s), i = 1, ,n, be nonne gative
continuous functions for α
≤ s ≤ t ≤ β which are nondecreasing in t for fixed s ∈ J.If
u(t)
≤ a(t)+
t
α
f
1
t,t
1
t
1
α
f
2
t
1
,t
2
···
t
n−1
α
f
n
t
n−1
,t
n
u
p
t
n
dt
n
···
dt
1
(1.1)
for t
∈ J,wherep ≥ 0, p =1, is a constant. Then u(t) ≤ Y
1
(t,t),whereY
1
(T,t) can be suc-
cessively determined from the formulas
Y
n
(T,t) = exp
t
α
n−1
i=1
f
i
(T,s)ds
×
a
q
(T)+q
t
α
f
n
(T,s)exp
−
q
s
α
n
−1
i=1
f
i
(T,τ)dτ
ds
1/q
(1.2)
for t
∈ [α,β
1
),withq = 1 − p and β
1
is chosen so that the expression between [···] is posi-
tive in the subinterval [α,β
1
), and
Y
k
(T,t) = E
k
(T,t)
a(T)+
t
α
f
k
(T,s)
Y
k+1
(T,s)
E
k
(T,s)
ds
,
E
k
(T,t) = exp
t
α
k−1
i=1
f
i
(T,τ) − f
k
(T,τ)
dτ
,
(1.3)
for k
= n − 1, ,1,α ≤ t ≤ T ≤ β.
The main aim of the present paper is to establish some nonlinear retarded inequalities,
which extend the above theorem and other results appeared in [16]. We will also illustrate
the usefulness of our results.
2. Gronwall-type inequalities
First we int roduce some notation,
R denotes the set of real numbers and R
+
= [0,∞),
J
= [α,β]isthegivensubsetofR. Denote by C
i
(M,N)theclassofalli-times continuously
differentiable functions defined on the set M to the set N for i
= 1,2, ,andC
0
(M,N) =
C(M,N).
Theorem 2.1. Let u(t) and a(t) be nonnegative continuous functions in J
= [α, β] with
a(t) nondecreasing in J,andlet f
i
(t,s), i = 1, ,n, be nonnegative continuous functions
for α
≤ s ≤ t ≤ β which are nondec reasing in t for fixed s ∈ J.Supposethatφ ∈ C
1
(J,J) is
nondecreasing w ith φ(t)
≤ t on J, g(u) is a nondecreasing continuous function for u ∈ R
+
with g(u) > 0 for u>0,andϕ ∈ C(R
+
,R
+
) is an increasing function with ϕ(∞) =∞.If
ϕ
u(t)
≤
a(t)+
φ(t)
φ(α)
f
1
t,t
1
φ(t
1
)
φ(α)
f
2
t
1
,t
2
···
φ(t
n−1
)
φ(α)
f
n
t
n−1
,t
n
g
u
t
n
dt
n
···
dt
1
(2.1)
Ravi P. Agarwal et al. 3
for t
∈ [α,β],thenfort ∈ [α, T
1
],
u(t)
≤ ϕ
−1
G
−1
G
a(t)
+
n
i=1
φ(t)
φ(α)
f
i
(t,s)ds
, (2.2)
where
G(r)
=
r
r
0
ds
s + g(s)
, r
≥ r
0
> 0, (2.3)
G
−1
denotes the inverse function of G,andT
1
∈J is chosen so that (G(a(t))+
n
i
=1
φ(t)
φ(α)
f
i
(t,
s)ds)
∈ Dom(G
−1
).
Proof. Let us first assume that a(t) > 0. Fix T
∈ (α,β]. For α ≤ t ≤ T,weobtainfrom(2.1)
ϕ
u(t)
≤
a(T)+
φ(t)
φ(α)
f
1
T,t
1
φ(t
1
)
φ(α)
f
2
T,t
2
···
φ(t
n−1
)
φ(α)
f
n
T,t
n
g
u
t
n
dt
n
···
dt
1
.
(2.4)
Now we introduce the functions
m
1
(t)= a(T)+
φ(t)
φ(α)
f
1
T,t
1
φ(t
1
)
φ(α)
f
2
T,t
2
···
φ(t
n−1
)
φ(α)
f
n
T,t
n
g
u
t
n
dt
n
···
dt
1
,
m
k
(t) = m
k−1
(t)+
φ(t)
φ(α)
f
k
T,t
k
φ(t
k
)
φ(α)
f
k+1
T,t
k+1
···
×
φ(t
n−1
)
φ(α)
f
n
T,t
n
g
m
k−1
t
n
dt
n
···
dt
k
,
(2.5)
for t
∈ [α,T]andk = 2, ,n.Thenwehavem
k
(α) = a(T)fork = 1, , n,andm
1
(t) ≤
m
2
(t) ≤ ··· ≤ m
n
(t), t ∈ [α,T]. From the inequality (2.4), we obtain u(t) ≤ ϕ
−1
(m
1
(t))
or u(t)
≤ ϕ
−1
(m
n
(t)), t ∈ [α,T]. Moreover the function m
1
(t) is nondecreasing. Differ-
entiating m
1
(t), we get
m
1
(t) = f
1
T,φ(t)
φ(φ(t))
φ(α)
f
2
T,t
2
···
φ(t
n−1
)
φ(α)
f
n
T,t
n
g
u
t
n
dt
n
···
dt
2
φ
(t),
≤
− f
1
T,φ(t)
φ
(t)m
1
(t)+ f
1
T,φ(t)
φ
(t)m
2
(t)
.
(2.6)
Thus, induction with respect to k gives
m
k
(t) ≤
k−1
i=1
f
i
T,φ(t)
−
f
k
T,φ(t)
φ
(t)m
k
(t)+ f
k
T,φ(t)
φ
(t)m
k+1
(t), (2.7)
4 Journal of Inequalities and Applications
for t
∈ [α,T], k = 1,2, ,n − 1. From the definition of the function m
n
(t) and inequality
(2.7), we have
m
n
(t) = m
n−1
(t)+ f
n
T,φ(t)
g
m
n−1
φ(t)
φ
(t)
≤
n−2
i=1
f
i
T,φ(t)
m
n−1
(t)+ f
n−1
T,φ(t)
m
n
(t)+ f
n
T,φ(t)
g
m
n
(t)
φ
(t)
≤
n−1
i=1
f
i
T,φ(t)
m
n
(t)+ f
n
T,φ(t)
g
m
n
(t)
φ
(t)
≤
n
i=1
f
i
T,φ(t)
φ
(t)
m
n
(t)+g
m
n
(t)
.
(2.8)
That is,
m
n
(t)
m
n
(t)+g
m
n
(t)
≤
n
i=1
f
i
T,φ(t)
φ
(t).
(2.9)
Tak ing t
= s in (2.9) and then integrating it from α to any t ∈ [α,β], changing the variable
and using the definition of the function G,wefind
G
m
n
(t)
≤
G
m
n
(α)
+
n
i=1
φ(t)
φ(α)
f
i
(T,s)ds, (2.10)
or
m
n
(t) ≤ G
−1
G
m
n
(α)
+
n
i=1
φ(t)
φ(α)
f
i
(T,s)ds
(2.11)
for α ≤ t ≤ T ≤ β.Now,acombinationofu(t) ≤ ϕ
−1
(m
n
(t)) and the last inequality gives
the required inequality in (2.2)forT
= t.Ifa(t) = 0, we replace a(t)bysomeε>0and
subsequently let ε
→0. This completes the proof.
For the special case g(u) = u
p
(p>0 is a constant), Theorem 2.1 gives the following
retarded integral inequality for iterated integrals.
Corollary 2.2. Let u(t), a(t) , f
i
(t,s), φ(t),andϕ(u) be as in Theorem 2.1.Andletp>0
be a constant. Suppose that
ϕ
u(t)
≤
a(t)+
φ(t)
φ(α)
f
1
t,t
1
φ(t
1
)
φ(α)
f
2
t
1
,t
2
···
φ(t
n−1
)
φ(α)
f
n
t
n−1
,t
n
u
p
t
n
dt
n
···
dt
1
(2.12)
for any t
∈ [α,β].Then,foranyt ∈ [α, T
1
],
u(t)
≤ ϕ
−1
G
−1
1
G
1
a(t)
+
n
i=1
φ(t)
φ(α)
f
i
(t,s)ds
, (2.13)
Ravi P. Agarwal et al. 5
where
G
1
(r) =
r
r
0
ds
s + s
p
, r ≥ r
0
> 0, (2.14)
G
−1
1
denotes the inverse function of G
1
,andT
1
∈J is chosen so that (G
1
(a(t))+
n
i
=1
φ(t)
φ(α)
f
i
(t,
s)ds)
∈ Dom (G
−1
1
).
Remark 2.3. (i) When ϕ(u)
= u and g(u) = u,formTheorem 2.1, we derive the following
retarded integr al inequality:
u(t) ≤ a(t)exp
2
n
i=1
φ(t)
φ(α)
f
i
(t,s)ds
. (2.15)
(ii) When ϕ(u)
= u,inTheorem 2.1, we obtain the following retarded integral in-
equality:
u(t)
≤ G
−1
G
a(t)
+
n
i=1
φ(t)
φ(α)
f
i
(t,s)ds
. (2.16)
(iii) When ϕ(u)
= u
p
(p>0 is a constant) in Theorem 2.1, we have the following
retarded integr al inequality:
u(t)
≤
G
−1
G
a(t)
+
n
i=1
φ(t)
φ(α)
f
i
(t,s)ds
1/p
. (2.17)
Now we introduce the following notation. For α<β,letJ
i
={(t
1
,t
2
, ,t
i
) ∈ R
i
: α ≤
t
i
≤ ··· ≤ t
1
≤ β} for i = 1, ,n.
Theorem 2.4. Let u(t) and a(t) be nonnegative continuous functions in J
= [α,β] with a(t)
nondecreasing in J,andletp
i
(t), i = 1, ,n, be nonnegative continuous functions for α ≤
t ≤ β.Supposethatφ ∈ C
1
(J,J) is nondecreasing with φ( t) ≤ t on J, g(u) is a nondecreasing
continuous function for u
∈ R
+
with g(u) > 0 for u>0,andϕ ∈ C(R
+
,R
+
) is an increasing
function with ϕ(
∞) =∞. If
ϕ
u(t)
≤
a(t)+
φ(t)
φ(α)
p
1
t
1
g
u
t
1
dt
1
+
n
i=2
φ(t)
φ(α)
p
1
t
1
φ(t
1
)
φ(α)
p
2
t
2
···
φ(t
i−2
)
φ(α)
p
i−1
t
i−1
×
φ(t
i−1
)
φ(α)
p
i
t
i
g
u
t
i
dt
i
dt
i−1
···
dt
2
dt
1
,
(2.18)
for any t
∈ J, then
u(t)
≤ ϕ
−1
G
−1
G(a(t
+ F(t)
(2.19)
6 Journal of Inequalities and Applications
for t
∈ [α,T
2
],whereT
2
∈ I is chosen so that (G(a(t)) + F(t)) ∈ Dom (G
−1
),
G(r)
=
r
r
0
ds
g
ϕ
−1
(s)
, r ≥ r
0
> 0, (2.20)
G
−1
denotes the inverse function of G,and
F(t)
=
φ(t)
φ(α)
p
1
t
1
dt
1
+
n
i=2
φ(t)
φ(α)
p
1
t
1
φ(t
1
)
φ(α)
p
2
t
2
···
φ(t
i−2
)
φ(α)
p
i−1
t
i−1
×
φ(t
i−1
)
φ(α)
p
i
t
i
dt
i
dt
i−1
···
dt
2
dt
1
,
(2.21)
for any t
∈ I.
Proof. Let the function a(t) be positive. Define a function v(t) by the right side of (2.18).
Clearly, v(t) is nondecreasing continuous, u(t)
≤ ϕ
−1
(v(t)) for t ∈ I and v(α) = a(α).
Differentiating v(t) and rew riting, we have
v
(t) − a
(t)
φ
(t)p
1
φ(t)
−
g
u
φ(t)
≤
v
1
(t), (2.22)
where
v
1
(t) =
φ(t)
φ(α)
p
2
t
2
g
u
t
2
dt
2
+
n
i=3
φ(t)
φ(α)
p
2
t
2
φ(t
2
)
φ(α)
p
3
t
3
···
φ(t
i−2
)
φ(α)
p
i−1
t
i−1
×
φ(t
i−1
)
φ(α)
p
i
t
i
g
u
t
i
dt
i
dt
i−1
···
dt
3
dt
2
.
(2.23)
Now differentiating v
1
(t) and rewriting, we get
v
1
(t)
φ
(t)p
2
φ(t)
−
g
u
φ(t)
≤
v
2
(t), (2.24)
where
v
2
(t)=
φ(t)
φ(α)
p
3
t
3
g
u
t
3
dt
3
+
n
i=4
φ(t)
φ(α)
p
3
t
3
φ(t
3
)
φ(α)
p
4
t
4
···
φ(t
i−2
)
φ(α)
p
i−1
t
i−1
×
φ(t
i−1
)
φ(α)
p
i
t
i
g
u
t
i
dt
i
dt
i−1
···
dt
4
dt
3
.
(2.25)
Ravi P. Agarwal et al. 7
Continuing in this way, we obtain
v
n−2
(t)
φ
(t)p
n−1
φ(t)
−
g
u
φ(t)
≤
v
n−1
(t), (2.26)
where
v
n−1
(t) =
φ(t)
φ(α)
p
n
t
n
g
u
t
n
dt
n
.
(2.27)
From the definition of v
n−1
(t) and the inequality u(t) ≤ ϕ
−1
(v(t)), we find
v
n−1
(t)
g
ϕ
−1
v(t)
≤
φ
(t)p
n
φ(t)
. (2.28)
Integrating the inequality (2.28), we get
t
α
v
n−1
(s)
g
ϕ
−1
v(s)
ds ≤
φ(t)
φ(α)
p
n
(s)ds. (2.29)
Now integrating by parts the left–hand side of (2.29), we obtain
t
α
v
n−1
(s)
g
ϕ
−1
v(s)
ds =
v
n−1
(t)
g
ϕ
−1
v(t)
+
t
α
v
n−1
g
ϕ
−1
(v)
g
2
ϕ
−1
(v)
v
ϕ
ϕ
−1
(v)
ds
≥
v
n−1
(t)
g
ϕ
−1
v(t)
.
(2.30)
From the i nequalities (2.29)and(2.30), we have
v
n−1
(t)
g
ϕ
−1
v(t)
≤
φ(t)
φ(α)
p
n
(s)ds. (2.31)
Next from the inequality (2.26), we observe that
v
n−2
(t) ≤ φ
(t)p
n−1
φ(t)
g
u
φ(t)
+ φ
(t)p
n−1
φ(t)
v
n−1
(t), (2.32)
Thus, it follows that
v
n−2
(t)
g
ϕ
−1
v(t)
≤
φ
(t)p
n−1
φ(t)
g
u
φ(t)
g
ϕ
−1
v(t)
+ φ
(t)p
n−1
φ(t)
v
n−1
(t)
g
ϕ
−1
v(t)
≤
φ
(t)p
n−1
φ(t)
+ φ
(t)p
n−1
φ(t)
v
n−1
(t)
g
ϕ
−1
v(t)
.
(2.33)
Using the same procedure from (2.29)to(2.31) to the inequality (2.33), we get
v
n−2
(t)
g
ϕ
−1
v(t)
≤
φ(t)
φ(α)
p
n−1
t
1
dt
1
+
φ(t)
φ(α)
p
n−1
t
1
v
n−1
t
1
g
ϕ
−1
v
t
1
dt
1
. (2.34)
8 Journal of Inequalities and Applications
Now combining the inequalities (2.31)and(2.34), we find
v
n−2
(t)
g
ϕ
−1
v(t)
≤
φ(t)
φ(α)
p
n−1
t
1
dt
1
+
φ(t)
φ(α)
p
n−1
t
1
φ(t
1
)
φ(α)
p
n
t
2
dt
2
dt
1
. (2.35)
Proceeding in this way we arrive at
v
1
(t)
g
ϕ
−1
v(t)
≤
φ(t)
φ(α)
p
2
t
1
dt
1
+ ···
+
φ(t)
φ(α)
p
2
t
1
φ(t
1
)
φ(α)
p
3
t
2
···
φ(t
n−2
)
φ(α)
p
n
t
n−1
dt
n−1
···
dt
2
dt
1
.
(2.36)
On the other hand, from the inequality (2.22), we have
v
(t) − a
(t) ≤ φ
(t)p
1
φ(t)
g
u
φ(t)
+ φ
(t)p
1
φ(t)
v
1
(t), (2.37)
or
v
(t) − a
(t)
g
ϕ
−1
v(t)
≤
φ
(t)p
1
φ(t)
g
u
φ(t)
g
ϕ
−1
v(t)
+ φ
(t)p
1
φ(t)
v
1
(t)
g
ϕ
−1
v(t)
≤
φ
(t)p
1
φ(t)
+ φ
(t)p
1
φ(t)
v
1
(t)
g
ϕ
−1
v(t)
,
(2.38)
that is,
v
(t)
g
ϕ
−1
v(t)
−
a
(t)
g
ϕ
−1
a(t)
≤
φ
(t)p
1
φ(t)
+ φ
(t)p
1
φ(t)
v
1
(t)
g
ϕ
−1
v(t)
. (2.39)
Setting t
= t
1
and integrating from α to t, and using the definition of G,weobtain
G
v(t)
≤
G
a(t)
+
φ(t)
φ(α)
p
1
t
1
dt
1
+
φ(t)
φ(α)
p
1
t
1
v
1
t
1
g
ϕ
−1
v(t
1
)
dt
1
. (2.40)
Consequently, using (2.36) to the inequality (2.40), we get
v(t)
≤ G
−1
G
a(t)
+ F(t)
, (2.41)
where the function F(t)isdefinedin(2.21). Now, the desired inequality in (2.24)fol-
lows by the inequalit y u(t)
≤ ϕ
−1
(v(t)). If a(t) = 0, we replace a(t)bysomeε>0and
subsequently let ε
→0. This completes the proof.
Ravi P. Agarwal et al. 9
For the special case ϕ(u)
= u
p
(p>1 is a constant), Theorem 2.4 gives the fol lowing
retarded integral inequality for iterated integrals.
Corollary 2.5. Let u(t), a(t), p
i
(t), φ(t),andg(u) be as in Theorem 2.4.Andletp>0 be
a constant. If
u
p
(t) ≤ a(t)+
φ(t)
φ(α)
p
1
t
1
g
u
t
1
dt
1
+
n
i=2
φ(t)
φ(α)
p
1
t
1
φ(t
1
)
φ(α)
p
2
t
2
···
φ(t
i−2
)
φ(α)
p
i−1
t
i−1
×
φ(t
i−1
)
φ(α)
p
i
t
i
g
u
t
i
dt
i
dt
i−1
···
dt
2
dt
1
,
(2.42)
for any t
∈ J, then
u(t)
≤
G
−1
G
a(t)
+ F(t)
1/p
(2.43)
for t
∈ [α,T
3
],whereT
3
∈ I is chosen so that (G
1
(a(t)) + F(t)) ∈ Dom (G
−1
1
),
G
1
(r) =
r
r
0
ds
g
v
1/p
(s)
, r ≥ r
0
> 0, (2.44)
G
−1
denotes the inverse function of G,andthefunctionF(t) is defined in (2.21)foranyt ∈ I.
Theorem 2.6. Let u(t) and a(t) be nonnegative continuous functions in J
= [α,β] with a(t)
nondecreasing in J,andlet f
i
(t) and p
i
(t), i = 1, ,n, be nonnegative continuous functions
for α
≤ t ≤ β.Supposethatφ ∈ C
1
(J,J) is nondec reasing with φ(t) ≤ t on J, g(u) is a non-
decreasing continuous function for u
∈ R
+
with g(u) > 0 for u>0,andϕ ∈ C(R
+
,R
+
) is an
increasing function with ϕ(
∞) =∞.If
ϕ
u(t)
≤
a(t)+
φ(t)
φ(α)
p
1
t
1
f
1
t
1
u
t
1
g
u
t
1
dt
1
+
n
i=2
φ(t)
φ(α)
p
1
t
1
φ(t
1
)
φ(α)
p
2
t
2
···
φ(t
i−2
)
φ(α)
p
i−1
t
i−1
×
φ(t
i−1
)
φ(α)
p
i
t
i
f
i
t
i
u
t
i
g
u
t
i
dt
i
dt
i−1
···
dt
2
dt
1
,
(2.45)
for any t
∈ J, then
u(t)
≤ ϕ
−1
Φ
−1
G
−1
2
G
2
Φ
a(t)
+ F
1
(t)
(2.46)
10 Journal of Inequalities and Applications
for t
∈ [α,T
4
],whereT
4
∈ I is chosen so that (G
2
[Φ(a(t))] + F
1
(t)) ∈ Dom (G
−1
2
),
[G
−1
2
(G
2
[Φ(a(t))] + F
1
(t))] ∈ Dom (Φ
−1
),
G
2
(r) =
r
r
0
ds
g
ϕ
−1
Φ
−1
(s)
, r ≥ r
0
> 0,
Φ(r)
=
r
r
0
ds
ϕ
−1
(s)
, r
≥ r
0
> 0,
(2.47)
G
−1
2
denotes the inverse function of G
2
,and
F
1
(t) =
φ(t)
φ(α)
p
1
t
1
f
1
(t)dt
1
+
n
i=2
φ(t)
φ(α)
p
1
t
1
φ(t
1
)
φ(α)
p
2
t
2
×
···
φ(t
i−2
)
φ(α)
p
i−1
t
i−1
×
φ(t
i−1
)
φ(α)
p
i
t
i
f
i
t
i
dt
i
dt
i−1
···
dt
2
dt
1
,
(2.48)
for any t
∈ I.
Proof. Let the function a(t) be positive. Define a function w(t) by the right side of (2.45).
Clearly, w(t) is a nondecreasing continuous function, u(t)
≤ ϕ
−1
(w(t)) for t ∈ I and
w(α)
= a(α). Differentiating w(t) and rewriting, we have
w
(t) − a
(t)
φ
(t)p
1
φ(t)
−
f
1
φ(t)
u
φ(t)
g
u
φ(t)
≤
w
1
(t), (2.49)
where
w
1
(t) =
φ(t)
φ(α)
p
2
t
2
f
2
t
2
u
t
2
g
u
t
2
dt
2
+
n
i=3
φ(t)
φ(α)
p
2
t
2
φ(t
2
)
φ(α)
p
3
t
3
···
φ(t
i−2
)
φ(α)
p
i−1
t
i−1
×
φ(t
i−1
)
φ(α)
p
i
t
i
f
i
t
i
u
t
i
g
u
t
i
dt
i
dt
i−1
···
dt
3
dt
2
.
(2.50)
Now differentiating v
1
(t) and rewriting, we get
w
1
(t)
φ
(t)p
2
φ(t)
−
f
2
φ(t)
u
φ(t)
g
u
φ(t)
≤
w
2
(t), (2.51)
Ravi P. Agarwal et al. 11
where
w
2
(t) =
φ(t)
φ(α)
p
3
t
3
f
3
t
3
u
t
3
g
u
t
3
dt
3
+
n
i=4
φ(t)
φ(α)
p
3
t
3
φ(t
3
)
φ(α)
p
4
t
4
···
φ(t
i−2
)
φ(α)
p
i−1
t
i−1
×
φ(t
i−1
)
φ(α)
p
i
t
i
f
i
t
i
u
t
i
g
u
t
i
dt
i
dt
i−1
···
dt
4
dt
3
.
(2.52)
Continuing in this way, we obtain
w
n−2
(t)
φ
(t)p
n−1
φ(t)
−
f
n−1
φ(t)
u
φ(t)
g
u
φ(t)
≤
w
n−1
(t), (2.53)
where
w
n−1
(t) =
φ(t)
φ(α)
p
n
t
n
f
n
t
n
u
t
n
g
u
t
n
dt
n
.
(2.54)
From the definition of w
n−1
(t) and the inequality u(t) ≤ ϕ
−1
(w(t)), we get
w
n−1
(t)
ϕ
−1
w(t)
≤
φ
(t)p
n
φ(t)
f
n
φ(t)
g
ϕ
−1
w
φ(t)
. (2.55)
Integrating the inequality (2.55), we have
t
α
w
n−1
(s)
ϕ
−1
w(s)
ds ≤
φ(t)
φ(α)
p
n
(s) f
n
(s)g
ϕ
−1
w(s)
ds. (2.56)
Next integrating by par ts the left–hand side of (2.56), we obtain
t
α
w
n−1
(s)
ϕ
−1
w(s)
ds =
w
n−1
(t)
ϕ
−1
w(t)
+
t
α
w
n−1
ϕ
−1
(w)
2
w
ϕ
ϕ
−1
(w)
ds
≥
w
n−1
(t)
ϕ
−1
w(t)
.
(2.57)
From the i nequalities (2.56)and(2.57), we get
w
n−1
(t)
ϕ
−1
w(t)
≤
φ(t)
φ(α)
p
n
(s) f
n
(s)g
ϕ
−1
w(s)
ds. (2.58)
Now from the inequality (2.53), we observe that
w
n−2
(t) ≤ φ
(t)p
n−1
φ(t)
w
n−1
(t)
+ φ
(t)p
n−1
φ(t)
f
n−1
φ(t)
ϕ
−1
w(t)
g
ϕ
−1
w(t)
.
(2.59)
12 Journal of Inequalities and Applications
Also, from the inequality (2.59), we have
w
n−2
(t)
ϕ
−1
w(t)
≤
φ
(t)p
n−1
φ(t)
w
n−1
(t)
ϕ
−1
(w(t))
+ φ
(t)p
n−1
φ(t)) f
n−1
φ(t)
g
ϕ
−1
w(t)
.
(2.60)
Using the same procedure from (2.56)to(2.58) to the inequality (2.60), we find
w
n−2
(t)
ϕ
−1
w(t)
≤
φ(t)
φ(α)
p
n−1
t
1
w
n−1
t
1
ϕ
−1
w
t
1
dt
1
+
φ(t)
φ(α)
p
n−1
t
1
f
n−1
t
1
g
ϕ
−1
w
t
1
dt
1
.
(2.61)
Next using (2.58) in the inequality (2.61), we get
w
n−2
(t)
ϕ
−1
w(t)
≤
φ(t)
φ(α)
p
n−1
t
1
φ(t
1
)
φ(α)
p
n
(s) f
n
(s)g
ϕ
−1
w(s)
ds dt
1
+
φ(t)
φ(α)
p
n−1
t
1
f
n−1
t
1
g
ϕ
−1
w
t
1
dt
1
.
(2.62)
Proceeding in this way, we arrive at
w
1
(t)
ϕ
−1
w(t)
=
φ(t)
φ(α)
p
2
t
1
f
2
t
1
g
ϕ
−1
w
t
1
dt
1
+ ···
+
φ(t)
φ(α)
p
2
t
1
···
φ(t
n−2
)
φ(α)
p
n
t
s
f
n
t
s
g
ϕ
−1
w
t
s
ds···
dt
1
.
(2.63)
On the other hand, from the inequality (2.49), we have
w
(t) − a
(t) ≤ φ
(t)p
1
φ(t)
f
1
φ(t)
ϕ
−1
w(t)
g
ϕ
−1
w(t)
+ φ
(t)p
1
φ(t)
w
1
(t),
(2.64)
or
w
(t) − a
(t)
ϕ
−1
w(t)
≤
φ
(t)p
1
φ(t)
w
1
(t)
ϕ
−1
w(t)
+ φ
(t)p
1
φ(t)
f
1
φ(t)
g
ϕ
−1
w(t)
.
(2.65)
Now, theleft-hand side of the inequality (2.65) implies that
w
(t)
ϕ
−1
w(t)
−
a
(t)
ϕ
−1
a(t)
≤
w
(t) − a
(t)
ϕ
−1
w(t)
. (2.66)
Ravi P. Agarwal et al. 13
In the inequalities (2.65)and(2.66), setting t
= t
1
, integrating from α to t, and using the
definition of Φ,weobtain
Φ
w(t)
≤
Φ
a(t)
+
φ(t)
φ(α)
p
1
t
1
w
1
t
1
ϕ
−1
w
t
1
dt
1
+
φ(t)
φ(α)
p
1
t
1
f
1
t
1
g
ϕ
−1
w(t)
dt
1
.
(2.67)
Consequently, from the inequalities (2.63)and(2.67), we get
w(t)
≤ Φ
−1
[k(t)], (2.68)
where the function k(t)isdefinedby
k(t)
= Φ
a(T)
+
φ(t)
φ(α)
p
1
t
1
f
1
t
1
g
ϕ
−1
w
t
1
dt
1
+
n
i=2
φ(t)
φ(α)
p
1
t
1
φ(t
1
)
φ(α)
p
2
t
2
×
···
φ(t
i−1
)
φ(α)
p
i
t
i
f
i
t
i
g
ϕ
−1
w
t
i
dt
i
···
dt
2
dt
1
,
(2.69)
for some fixed T, t
≤ T ≤ β. Clearly, k(t) is a nondecreasing continuous function and
k(α)
= Φ(a(T)). Differentiating k(t) and rewriting, we have
k
(t)
φ
(t)p
1
φ(t)
−
f
1
φ(t)
g
ϕ
−1
w
φ(t)
≤
k
1
(t), (2.70)
where
k
1
(t) =
φ(t)
φ(α)
p
2
t
2
f
2
t
2
g
ϕ
−1
w
t
2
dt
2
+
n
i=3
φ(t)
φ(α)
p
2
t
2
φ(t
2
)
φ(α)
p
3
t
3
···
φ(t
i−2
)
φ(α)
p
i−1
t
i−1
×
φ(t
i−1
)
φ(α)
p
i
t
i
f
i
t
i
g
ϕ
−1
w
t
i
dt
i
dt
i−1
···
dt
3
dt
2
.
(2.71)
14 Journal of Inequalities and Applications
Using the same procedures from (2.51)to(2.65) to the equality (2.71), we have
k
1
(t)
g
ϕ
−1
w(t)
≤
φ(t)
φ(α)
p
2
t
2
f
2
t
2
dt
2
+
n
i=3
φ(t)
φ(α)
p
2
t
2
φ(t
2
)
φ(α)
p
3
t
3
···
φ(t
i−1
)
φ(α)
p
i
t
i
f
i
t
i
dt
i
···
dt
3
dt
2
,
(2.72)
k
(t)
g
ϕ
−1
Φ
−1
k(t)
≤
φ
(t)p
1
φ(t)
k
1
(t)
ϕ
−1
w(t)
+ φ
(t)p
1
φ(t)
f
1
φ(t)
.
(2.73)
In the inequality (2.73), setting t
= s and integrating from α to t, using the definition of
G
2
,weobtain
G
2
k(t)
≤
G
2
k(α)
+
φ(t)
φ(α)
p
1
(s)
k
1
(s)
g
ϕ
−1
w(s)
ds+
φ(t)
φ(α)
p
1
(s) f
1
(s)ds. (2.74)
Finally, from the inequalities (2.72)and(2.74), we get
k(t)
≤ G
−1
2
G
2
Φ
a(T)
+ F
1
(t)
, (2.75)
where the function F
1
(t)isdefinedin(2.46). In particular, for T = t, we find that the de-
sired inequality (2.46) follows by the inequalities u(t)
≤ ϕ
−1
(w(t)) and w(t) ≤ Φ
−1
(k(t)).
This completes the proof.
When ϕ(u) = u
p
(p>1 is a constant) in Theorem 2.6, we get the following Ou-Iang
type-retarded integral inequality with iterated integrals.
Corollary 2.7. Let u, a, f
i
, p
i
, φ, and g(u) be as in Theorem 2.6 and let p>1 beacon-
stant. If
u
p
(t) ≤ a(t)+
φ(t)
φ(α)
p
1
t
1
f
1
t
1
u
t
1
g
u
t
1
dt
1
+
n
i=2
φ(t)
φ(α)
p
1
t
1
φ(t
1
)
φ(α)
p
2
t
2
···
φ(t
i−2
)
φ(α)
p
i−1
t
i−1
×
φ(t
i−1
)
φ(α)
p
i
t
i
f
i
t
i
u
t
i
g
u
t
i
dt
i
dt
i−1
···
dt
2
dt
1
,
(2.76)
for any t ∈ J, then
u(t)
≤
G
−1
3
G
3
a
(p−1)/p
(t)
+
p
− 1
p
F
1
(t)
1/(p−1)
(2.77)
Ravi P. Agarwal et al. 15
for t
∈[α,T
3
],whereT
3
∈I is chosen so that [G
3
(a
(p−1)/p
(t)) + (( p − 1)/p)F
1
(t)]∈Dom (G
−1
3
),
G
3
(r) =
r
r
0
ds
g
v
1/(p−1)
(s)
, r ≥ r
0
> 0, (2.78)
G
−1
3
denotes the inverse function of G
3
,andF
1
(t) is defined in (2.48)foranyt ∈ I.
3. Applications
In this section, we will show that our results are directly useful in proving the global exis-
tence of solutions to certain integrodifferential equations. First we consider the following
integrodifferential equation:
px
p−1
(t)x
(t) = F
t,x
t − τ(t)
,
t−τ(t)
α
G
t
1
,x
t
1
− τ
t
1
dt
1
(3.1)
for t
∈ I,wherep>1 is constant, let F ∈ C(I × R
2
,R), G ∈ C
1
(I × R,R), and τ ∈ C
1
(I,I)
be nonincreasing with t
− τ(t) ≥ 0, t − τ(t) ∈ C
1
(I,I), τ
(t) < 1, and τ(α) = 0. The fol-
lowing result provides a bound on the solutions of (3.1).
Theorem 3.1. Assume that F : I
× R
2
→R is a continuous function, and there exist contin-
uous nonnegative functions b
i
(t), i = 1,2, such that
F(t,u,v)
≤
b
1
(t)g
|
u|
+ b
1
(t)|v|,
G(s,w)
≤
b
2
(s)g
|
w|
,
(3.2)
where the function g is the same as in Theorem 2.4.LetM
= max
x∈I
(1/(1 − τ
(x))).Ifx(η)
is any solution of the problem (3.1), then
x(η)
≤
G
−1
1
G
1
x(α)
p
+ B
Mb
1
η
1
,M
2
b
1
η
1
b
2
η
2
1/p
, (3.3)
where the functions G
1
,G
−1
1
are as in Corollary 2.5, η
1
= η
1
+ τ(t
1
), η
2
= η
2
+ τ(t
2
),for
η
1
, η
2
∈ I,and
B
Mb
1
η
1
,M
2
b
1
η
1
b
2
η
2
=
φ(η)
φ(α)
Mb
1
η
1
dη
1
+
φ(η)
φ(α)
φ(η
1
)
φ(α)
M
2
b
2
η
1
b
3
η
2
dη
2
dη
1
,
(3.4)
where φ(γ)
= γ − τ(γ) for γ ∈ I.
Proof. It is easy to see that the solution x(η)oftheproblem(3.1) satisfies the equivalent
integral equation
x
p
(η) = x
p
(α)+
η
α
F
t
1
,x
t
1
−
τ
t
1
,
t
1
−τ(t
1
)
α
G
t
2
,x
t
2
− τ
t
2
dt
2
dt
1
. (3.5)
16 Journal of Inequalities and Applications
From (3.2), and making the change of variables, we have
x(η)
p
≤
x(α)
p
+
η
α
b
1
t
1
g
x
t
1
− τ
t
1
dt
1
+
η
α
t
1
−τ(t
1
)
α
b
1
t
1
b
2
t
2
g
x
t
2
− τ
t
2
dt
2
dt
1
≤
x(α)
p
+
η−τ(η)
α
Mb
1
η
1
g
x
η
1
dη
1
+
η−τ(η)
α
η
1
−τ(η
1
)
α
M
2
b
1
η
1
b
2
η
2
g
x
η
2
dη
2
dη
1
,
(3.6)
where η
1
= η
1
+ τ(t
1
), η
2
= η
2
+ τ(t
2
), for η
1
,η
2
∈ I. Now an immediate application of
the inequality established in Corollary 2.5 to (3.6) yields the desired result.
We next consider the following integ rodifferential equation:
h(t)x
(t)
= F
t,x
t − τ(t)
,
t
α
G
t
1
,x
t
1
− τ
t
1
dt
1
(3.7)
for t
∈ I, F ∈ C(I × R
2
,R), G ∈ C
1
(I × R,R), and h is positive and continuous in I.The
following theorem provides an upper bound on the solutions of (3.7).
Theorem 3.2. Assume that F : I
× R
2
→R is a continuous function, and there exist contin-
uous nonnegative functions f
i
(t), i = 2,3, such that
F(t,u,v)
≤
f
2
(t)
u|g
|
u|
+ |v|
,
G(s,w)
≤
f
3
(s)|w|g
|
w|
,
(3.8)
where function g is the same as in Theorem 2.6.Ifx(t) is any solution of the problem (3.7),
then
x(t)
≤
exp
G
−1
e
G
e
(lna)+C
f
2
s
2
, f
2
s
2
f
3
s
3
−
1, (3.9)
where G
e
(r) =
r
r
0
(ds/g(e
s
)) for r ≥ r
0
> 0, a = 1+|x(α)| + h(α)|x
(α)|
φ(t)
φ(α)
1
h(s
1
)
ds
1
, and
C
f
2
, f
2
f
3
=
φ(t)
φ(α)
1
h
s
1
φ(t
1
)
φ(α)
f
2
s
2
M
2
ds
2
ds
1
+
φ(t)
φ(α)
1
h
s
1
φ(t
1
)
φ(α)
f
2
s
2
φ(t
2
)
φ(α)
f
3
s
3
M
3
ds
3
ds
2
ds
1
,
(3.10)
s
1
= s
1
+ τ(t
1
), s
2
= s
2
+ τ(t
2
),ands
3
= s
3
+ τ(t
2
) for s
1
,s
2
,s
3
∈ I.
Ravi P. Agarwal et al. 17
Proof. It is easy to see that the solution x(t)oftheproblem(3.7) satisfies the equivalent
integral equation
x(t)
= x(α)+h(α)x
(α)
t
α
1
h
t
1
dt
1
+
t
α
1
h
t
1
t
1
α
F
t
2
,x
t
2
− τ
t
2
,
t
2
α
G
t
3
,x
t
3
− τ
t
3
dt
3
dt
2
dt
1
.
(3.11)
From (3.8), and making the change of variables, we have
x(t)
+1≤ 1+x(α)+h(α)
x
(α)
φ(t)
φ(α)
1
h
s
1
ds
1
+
φ(t)
φ(α)
1
h
s
1
φ(t
1
)
φ(α)
f
2
s
2
M
2
x
s
2
g
x
s
2
ds
2
ds
1
+
φ(t)
φ(α)
1
h
s
1
φ(t
1
)
φ(α)
f
2
s
2
φ(t
2
)
φ(α)
f
3
s
3
M
3
x
s
3
g
x
s
3
ds
3
ds
2
ds
1
,
(3.12)
where s
i
= t
i
− τ(t
i
), s
i
= s
i
+ τ(t
i
), s
i
∈ I for i = 1,2,3. Now w hen ϕ(u) = u and f
1
= f
4
=
··· =
f
n
= 0, a suitable application of the inequality given in Theorem 2.6 to (3.12)yields
the desired result.
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Ravi P. Agarwal: Department of Mathematical Sciences, Florida Institute of Technology,
150 West University Boulevard, Melbourne, FL 32901-6975, USA
Email address: agarwal@fit.edu
Cheon Seoung Ryoo: Department of Mathematics, Hannam University, Daejeon 306-791,
South Korea
Email address:
Young-Ho Kim: Department of Applied Mathematics, Changwon National University, Changwon,
Kyungnam 641-773, South Korea
Email address:
