RESEARC H Open Access
Some new nonlinear integral inequalities and
their applications in the qualitative analysis of
differential equations
Bin Zheng
1*
and Qinghua Feng
1,2
* Correspondence:

1
School of Science, Shandong
University of Technology, Zibo,
Shandong 255049, China
Full list of author information is
available at the end of the article
Abstract
In this paper, some new nonlinear integral inequalities are established, which provide
a handy tool for analyzing the global existence and boundedness of solutions of
differential and integral equations. The established results generalize the main results
in Sun (J. Math. Anal. Appl. 301, 265-275, 2005), Ferreira and Torres (Appl. Math. Lett.
22, 876-881, 2009), Xu and Sun (Appl. Math. Comput. 182, 1260-1266, 2006) and Li
et al. (J. Math. Anal. Appl. 372, 339-349 2010).
MSC 2010: 26D15; 26D10
Keywords: integral inequality, global existence, integral equation, differential equa-
tion, bounded
1 Introduction
During the past decades, with the development of the theory of differential and integral
equations, a lot of integral inequalities, for example [1-12], have been discovered,
which play an important role in the research of boundedness, global existence, stability
of solutions of differential and integral equations.

α
(
t
)
0
[f (s)u
n
(s)ω(u(s)) + g(s)u
n
(s)] ds, t ∈ R
+
,
then for t Î [0, ξ]
u(t ) ≤{Ω
−1
[Ω(c +

α(t)
0
g(s)ds)+

α(t)
0
f (s)ds]}
1
m−n
,
where
Ω
(r)=

Zheng and Feng Journal of Inequalities and Applications 2011, 2011:20
/>© 2011 Zheng and Feng; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons
Attribution License ( which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited .
Theorem B: Under the hypothesis of Theorem B, if
u
m
(t) ≤ c
m
m−n
+
m
m − n

α(t)
0
f (s)u
n
(s)ω(u(s ))ds+
m
m − n

t
0
g(s)u
n
(s)ω(u(s ))ds, t ∈ R
+
,
then for t Î [0, ξ]

+
0
, R
+
0
)
be nonde-
creasing in t for every s fixed. Moreover, let
φ ∈ C(R
+
0
, R
+
0
)
beastrictlyincreasingfunc-
tion such that
l
im
x
→∞
φ
(
x
)
=
∞
and suppose that
c ∈ C(R
+

(R
+
0
, R
+
0
)
is nondecreasing with a(t) ≤ t.If
u ∈ C(R
+
0
, R
+
0
)
satisfies
φ(u(t)) ≤ c(t)+

α(t)
0
[f (t, s)η(u(s))ω(u(s)) + g(t, s)η(u(s))]ds, t ∈ R
+
0
then there exists τ Î R
+
so that for all t Î [0, τ] we have
ψ(p(t)) +

α
(

(s))
ds
.
with x ≥ c(0) >x
0
>0if

x
0
1
η
(
φ
−1
(
s
))
ds =
∞
and x ≥ c(0) >x
0
≥ 0if

x
0
1
η(φ
−1
(s))
ds < ∞.

+
, R
+
)
be nondecreasing with ω( u) >0on(0,∞)anda, b, Î C
1
(R
+
, R
+
) be nondecreasing with
a(x) ≤ x, b(y) ≤ y on R
+
. m, n are constants, and m>n>0. If
u
m
(x, y) ≤a(x)+b(y)+
m
m − n

α
(
x
)
0

β
(
y
)

f (t, s)dsdt]}
1
m−n
,
where
p(x, y)=[a(0) + b(y)]
m−n
m
+
m − n
m

x
0
a

(t )
[a(t)+b(0)]
n
m
d
t
+

α(x)
0

β(y)
0
g( t, s) dsdt,

0
f (t, s)u
n
(t , s)ω(u(t, s))dsd
t
+
m
m − n

x
0

y
0
g(t, s)u
n
(t , s)ω(u(t, s))dsdt, x, y ∈ R
+
,
then
u
(x, y) ≤{Ω
−1
[Ω(q(x, y)) +

α(x)
0

β(y)
0

)
]
n
m
dt
.
In this paper, motivated by the above work, we will prove more general theorems
and establish some new integral inequalities. Also we will give some examples so as to
illustrate the validity of the present integral inequalities.
2 Main results
In the rest of the paper we denote the set of real numbers as R,andR
+
=[0,∞)isa
subset of R. Dom(f)andIm(f) denote the definition domain and the image of f,
respectively.
Theorem 2.1: Assume that x, a Î C(R
+
, R
+
)anda(t) is nondecreasing. f
i
, g
i
, h
i
, ∂
t
f
i
,

p
(t ) ≤ a(t)+

α
(
t
)
0
[f
1
(s, t)x
q
(s)ω(x(s)) + g
1
(s, t)x
q
(s)+

s
0
h
1
(τ , t)x
q
(τ )dτ ] d
s
+

t
0

{Ω(H(t)) +
p − q
p
[

α(t)
0
f
1
(s, t)+

t
0
f
2
(s, t)ds]}}
1
p−q
,
(2)
Zheng and Feng Journal of Inequalities and Applications 2011, 2011:20
/>Page 3 of 15
where
H(t)=a
p
−q
p
(t )+
p − q
p

(r)=

r
1
1
ω
(
s
1
p−q
)
d
s
, r >0.Ω
-1
is the inverse of Ω, and Ω (∞)=∞.
Proof: The proof for the existence of
t
can be referred to Remark 1 in [10]. We
notice (3) obviously holds for t =0.Nowgivenanarbitrarynumber
T ∈
(
0, t
]
,fort Î
(0, T], we have
x
p
(t ) ≤a(T)+


(s, t)x
q
(s)+

s
0
h
2
(τ , t)x
q
(τ )dτ ] ds.
(4)
Let the right-hand side of (4) be z(t), then x
p
(t) ≤ z(t) and x
p
(a(t)) ≤ z(a (t)) ≤ z(t). So
z

(t)=[f
1
(α(t), t)x
q
(α(t))ω(x(α(t))) + g
1
(α(t), t)x
q
(α(t)) +

α(t)

s
0
h
1
(τ , t)x
q
(τ )dτ
∂t
] ds
+[f
2
(t, t)x
q
(t)ω(x(t)) + g
2
(t, t)x
q
(t)+

t
0
h
2
(τ , t)x
q
(τ )dτ]
+

t
0

(α(t), t)+

α(t)
0
h
1
(τ , t)dτ]α

(t)
+

α(t)
0
[
∂f
1
(s, t)
∂t
ω(x(s)) +
∂g
1
(s, t)
∂t
+
∂

s
0
h
1


s
0
h
2
(τ , t)dτ
∂t
]ds}z
q
p
(t)
Then
z

(t )
z
q
p
(t )
≤
d

α(t)
0
[f
1
(s, t)ω(x(s)) + g
1
(s, t)+


p−q
p
(T)+
p − q
p
{

α(t)
0
[f
1
(s, t)ω(x(s)) + g
1
(s, t)+

s
0
h
1
(τ , t)dτ ] d
s
+

t
0
[f
2
(s, t)ω(x(s)) + g
2
(s, t)+

2
(s, t)ω(z
1
p
(s))ds]
.
(7)
Let the right-hand side of ( 7) be y(t). Then we have
z
p−q
p
(
t
)
≤ y
(
t
)
,
z
p
−q
p
(
α
(
t
))
≤ y
(

(s, t)
∂t
ω(z
1
p
(s)) d
s
+ f
2
(t , t) ω(z
1
p
(t )) +

t
0
∂f
2
(s, t)
∂t
ω(z
1
p
(s))ds]
≤
p − q
p
d[

α(t)

p
d[

α(t)
0
f
1
(s, t)+

t
0
f
2
(s, t)ds]
dt
.
(9)
Integrating (9) from 0 to t, considering y(0) = H(T), it follows
Ω
(y(t)) − Ω(H(T)) ≤
p − q
p
[

α(t)
0
f
1
(s, t)+


f
2
(s, t)ds]}}
1
p−q
, t ∈ (0, t]
.
(11)
Taking t = T in (11), then
x(T) ≤{Ω
−1
{Ω(H(T)) +
p − q
p
[

α(T)
0
f
1
(s, T)+

T
0
f
2
(s, T)ds]}}
1
p−q
,

i
, n
i
, l
i
Î C
1
(R
+
, R
+
), i =1,2.If
x
p
(t ) ≤ a(t)+

α(t)
0
[m
1
(t ) f (s)x
q
(s)ω(x(s)) + n
1
(t ) g
1
(s)x
q
(s)
+

0
l
2
(t ) h
2
(τ )x
q
(τ )dτ ]ds, t ∈ R
+
,
(12)
Zheng and Feng Journal of Inequalities and Applications 2011, 2011:20
/>Page 5 of 15
then we can find some
t ∈ R
+
such that for
t ∈
[
0, t
]
x(t) ≤{Ω
−1
{Ω(H(t)) +
p − q
p
[

α(t)
0

(t ) g
1
(s)+

s
0
l
1
(t ) h
1
(τ )dτ ] d
s
+

t
0
[n
2
(t ) g
2
(s)+

s
0
l
2
(t ) h
2
(τ )dτ ] ds}.
(14)

p−q
, m
1
(t) ≡
1, g
1
(t) ≡ 0, l
1
(t) ≡ 0, m
2
(t) ≡ 1, n
2
(t)=l
2
(t) ≡ 0fort Î R
+
, then Corollary 2.1 reduce s
to Theorem B [9, Theorem 2.2].
Corollary 2:2:Assumethatx, a, a, ω, Ω are defined as in Theorem 2.1. f, g, h, ∂
t
f,
∂
t
g, ∂
t
h Î C(R
+
× R
+
, R

p − q
p

α
(
t
)
0
f (s, t)ds]}
1
p−q
,
(16)
where
H(t)=a
p−q
p
(t )+
p − q
p
{

α(t)
0
[g(s, t)+

s
0
h(τ , t)dτ ] ds}
.

(τ )dτ ]ds, t ∈ R
+
,
(18)
then for
t ∈
[
0, t
]
x(t) ≤{Ω
−1
[Ω(H(t)) +
p − q
p

α(t)
0
m(t ) f (s)ds]}
1
p−q
,
(19)
where
H(t)=a
p−q
p
(t )+
p − q
p
{

φ(x)=
∞
. ψ, ω Î C(R
+
, R
+
) are nondecreasing with ψ (x) >0, ω(x) >0forx Î
(0, ∞)and

∞
t
0
1
ψ(φ
−1
(s))
ds =
∞
, a(t), a(t) are defined as in Theorem 2.1, and a(0) >
t
0
>0. If x Î C(R
+
, R
+
) satisfies the following integral inequality containing multiple
integrals
φ(x(t)) ≤ a(t)+

α

Y(H( t)) +

α
2
(t)
0
g(s, t)ds ∈ Dom(Y
−1
)
,
and
x(t) ≤ φ
−1
{J
−1
[Y
−1
(Y(H ( t)) +

α
2
(
t
)
0
g(s, t)ds)]}
,
(22)
where
H(t)=J(a(t)) +


t
t
1
1
ω(φ
−1
(J
−1
(s)))
ds, t
1
> 0, t > 0
.
(24)
Proof: The proof for the existence of
t
can be referred to Remark 1 in [10]. We
notice (22) obviously holds for t = 0. Now given an arbitrary number T>0,
T ∈
(
0, t
]
.
Define
d(t)=a(T)+

α
1
(

(
d
(
t
)),
(25)
and
d

(t )=f (α
1
(t ), T)ψ(x(α
1
(t )))α

1
(t )+g(α
2
(t ), T)ψ(x(α
2
(t )))ω(x(α
2
(t )))α

2
(t
)
+ α

3

(t )+α

3
(t ) ψ (φ
−1
(d(α
3
(t ))))

α
3
(t)
0
h(τ , T)dτ
≤ f (α
1
(t ), T)ψ(φ
−1
(d(t)))α

1
(t )+g( α
2
(t ), T)ψ(φ
−1
(d(t)))
ω(φ
−1
(d(α
2

(t )+g(α
2
(t ), T)ω(φ
−1
(d(α
2
(t ))))α

2
(t
)
+ α

3
(t )

α
3
(t)
0
h(τ , T)dτ .
(27)
Integrating (27) from 0 to t, considering J is increasing, we can obtain
d(t) ≤ J
−1
[J(a(T)) +

α
1
(

2
(t)
0
g(s, T)ω(φ
−1
(d(s)))ds], t ∈ (0, T].
(28)
Define
G(t )=H( T)+

α
2
(t)
0
g(s, T)ω(φ
−1
(d(s)))d
s
, then
d
(
t
)
≤ J
−1
(
G
(
t
))

≤ g
(
α
2
(
t
)
, T
)
ω
(
φ
−1
(
J
−1
(
G
(
t
)))
α

2
(
t
)
,
(30)
that is,

T
)
and Y is increasing, it follows
G(t ) ≤ Y
−1
[Y( H(T)) +

α
2
(t)
0
g(s, T)ds], t ∈ (0, T]
.
(32)
Combining (25), (29) and (32) we have
x(t) ≤ φ
−1
{J
−1
[Y
−1
(Y(H ( T)) +

α
2
(t)
0
g(s, T)ds)]}, t ∈ (0, T]
.
(33)

i
(x, y), g
i
(x, y), h
i
(x, y) Î C(R
+
× R
+
, R
+
), i =1,2,andj Î C
(R
+
, R
+
) is a strictly increasing function with
lim
x
→
∞
φ(x)=
∞
. a(x, y) Î C(R
+
× R
+
, R
+
)is

× R
+
, R
+
) satisfies the following integral inequality containing multiple
integrals
φ(u(x, y)) ≤ a(x, y)+

β(y)
0

α(x)
0
[f
1
(s, t)ψ(u( s , t )) + g
1
(s, t)ψ(u( s , t ))ω(u(s, t)
)
+

t
0

s
0
h
1
(ξ, τ)ψ(u(ξ, τ))dξdτ ]dsdt
+

so that for all
x ∈
[
0, x
]
,
y
∈
[
0,
y]
Y(

H( x , y)) +

β(y)
0

α(x)
0
g
1
(s, t)dsdt +

y
0

x
0
g

t
+

y
0

x
0
g
2
(s, t)dsdt)]}
(35)
where J, Y are defined as in Theorem 2.2, and

H(x, y)=J(a(x , y)) +

β(y)
0

α(x)
0
[f
1
(s, t)+

t
0

s
0

0
∈
(
0, x
]
,
y
0
∈
(
0, y
]
,and
x Î (0, x
0
], y Î (0, y
0
]. Let
z
(x, y)=a(x, y
0
)+

β(y)
0

α(x)
0
[f
1


s
0
h
2
(ξ, τ)ψ(u(ξ, τ))dξdτ ]dsdt.
(36)
Considering a(x, y) is nondecreasing, we have u(x, y) ≤ j
-1
(z(x, y)) ≤ j
-1
(z(x
0
, y)).
Moreover,
Zheng and Feng Journal of Inequalities and Applications 2011, 2011:20
/>Page 9 of 15
z
y
(x
0
, y)=β

(y)

α(x
0
)
0
[f

s
0
h
2
(ξ, τ )ψ(u(ξ ,τ ))dξdτ ]ds
≤{β

(y)

α(x
0
)
0
[f
1
(s, t)+g
1
(s, β(y))ω(u(s, β(y))) +

β(y)
0

s
0
h
1
(ξ, τ )dξdτ ] ds
+

x

(z(x
0
, y)))
≤ β

(y)

α(x
0
)
0
[f
1
(s, β(y)) + g
1
(s, β(y))ω(φ
−1
(z(s, β(y))))
+

β(y)
0

s
0
h
1
(ξ, τ )dξdτ ]ds +

x

β(y)
0

α(x
0
)
0
[f
1
(s, t)+g
1
(s, t)ω(φ
−1
(z(s, t))) +

t
0

s
0
h
1
(ξ, τ )dξdτ ]dsd
t
+

y
0

x


α(x
0
)
0
[f
1
(s, t)+g
1
(s, t)ω(φ
−1
(z(s, t)))
+

t
0

s
0
h
1
(ξ, τ )dξdτ ] dsdt +

y
0

x
0
0
[f


H(x
0
, y
0
)+

β(y)
0

α(x
0
)
0
g
1
(s, t)ω(φ
−1
(z(s, t)))dsd
t
+

y
0

x
0
0
g
2

(z(s, t))) dsd
t
+

y
0

x
0
0
g
2
(s, t)ω(φ
−1
(z(s, t))) dsdt.
Zheng and Feng Journal of Inequalities and Applications 2011, 2011:20
/>Page 10 of 15
Then
z(
x
0
, y
)
≤ J
−
1
[v
(
x
0

g
2
(s, y)ω(φ
−1
(z(s, y))) ds
≤ [β

(y)

α(x
0
)
0
g
1
(s, β(y))ds +

x
0
0
g
2
(s, y)ds]ω(φ
−1
(J
−1
(v(x
0
, y))))
,

(s, β(y))ds +

x
0
0
g
2
(s, y)ds
.
(42)
Integrating (42) from 0 to y, considering
v
(
x
0
,0
)
=

H
(
x
0
, y
0
)
we have
Y(v(x
0
, y)) − Y(

v(x
0
, y) ≤ Y
−1
[Y(

H( x
0
, y
0
)) +

β
(
y
)
0

α
(
x
0
)
0
g
1
(s, t)dsdt +

y
0

+

β(y)
0

α(x
0
)
0
g
1
(s, t)dsdt +

y
0

x
0
0
g
2
(s, t)dsdt)]}.
(43)
Take x = x
0
, y = y
0
and we have
u
(x

+

y
0
0

x
0
0
g
2
(s, t)dsdt)]}.
(44)
Since
x
0
∈
(
0, x
]
,
y
0
∈
(
0, y
]
are arbitrary, substitute x
0
, y

0
[f (s, t)ψ(u(s, t)) + g(s, t)ψ( u( s , t))ω(u(s, t)
)
+

t
0

s
0
h(ξ, τ)ψ(u(ξ, τ))dξdτ ] dsdt,
then we can find some
x
>
0
,
y
>
0
such that for all
x ∈
[
0, x
]
,
y
∈
[
0,
y]

0

α(x)
0
g(s, t)dsdt)]}
,
where

H(x, y)=J(a(x, y)) +

β(y)
0

α(x)
0
[f (s, t)+

t
0

s
0
h(ξ, τ)dξdτ ] dsdt]
.
Remark 5:Ifwetakeh(x, y) ≡ 0, ψ (u(x, y)) = u
n
(x, y), j(x, y)=u
m
(x, y), m>n>0,
then Corollary 2.4 reduces to Theorem D [11, Theorem 2.1].

)
0
[f
1
(s, t)ψ (u(s, t)) + g
1
(s, t)ψ (u(s, t))ω(u(s, t))]dsd
t
+

y
0

x
0
[f
2
(s, t)ψ (u(s, t)) + g
2
(s, t)ψ (u(s, t))ω(u(s, t))]dsdt,
then we can find some
x
>
0
,
y
>
0
such that for all
x ∈


x
0
g
2
(s, t) dsdt ∈ Dom(Y
−1
)
,
and
u
(x, y) ≤φ
−1
{J
−1
[Y
−1
(Y(

H(x, y)) +

β(y)
0

α(x)
0
g
1
(s, t) dsd
t

(s, t) dsdt
.
Remark 6:Ifwetakef
1
(x, y)=f
2
(x, y) ≡ 0, ψ(u(x, y)) = u
n
(x, y), j(x, y)=u
m
(x, y), m
>n, then Corollary 8 reduces to Theorem E [11, Theorem 2.2].
3 Applications
In this section, we will present two examples in order to illustrate the validity of the
above results. In the first example, we will try to prove the global existence of the solu-
tions of a delay differential equation, while in the second example, we will obtain the
bound of the solutions of an integral equation.
For the sake of proving the global existence of solutions of differential equations, we
first recall some basic facts. Consider the following equation

X

(t )=H( t, X(t), X(α(t))
)
X(0) = X
0
(45)
with X
0
Î R

⎧
⎨
⎩
(x
p
(t ))

= y
p + q
2
(t )+F(x(t), t
)
(y
p
(t ))

= G(t, x(α(t)))
(46)
where p is an ev en number. a(t) is a nond ecreasing function, a(t) Î C
1
(R
+
, R
+
), a(t)
≤ t, ∀t ≥ 0. p>q>0. Assume
|
F
(
x

1,
˜
f
1
(
t
))
,

f
3
(
t
)
= max
(
1,
˜
f
1
(
t
))
,where

f
1
,

f

p
(
t
))

=
(
x
p
(
t
))

+
(
y
p
(
t
))

= y
p + q
2
(
t
)
+ F
(
x

s
≤

t
0
[|u
p + q
2
(s)| +

f
1
(s)|x
q
(s)|v(|x|)+
˜
f
2
(s)|x
q
(α(s))|] ds
≤

t
0
[|u
p + q
2
(s)| +


f
2
(s)|u
q
(α(s))|] d
s
≤|u
p
(0)| +

t
0
[

f
3
(s)|u
q
(s)|ω( |u|)+

f
2
(s)|u
q
(α(s))|] ds
= |u
p
(0)| +

t

(
|u|
)
+ |u
p
− q
2
|
. From Theorem 2.1 we have
|u(t)|≤{Ω
−1
[Ω(|u
p−q
(0)| +
p − q
p

α(t)
0
˜
f
2
(α
−1
r)
α

(α
−1
r)

t
0
˜
f
3
(s)ds]}
1
p − q
,0 ≤ t < T.
Obviouslywehave{|x(t)|, |y(t)|} ≤ |u(t)|. So x(t), y(t) do not blow up in finite time.
Then T = ∞, and the solutions of (46) are global.
Zheng and Feng Journal of Inequalities and Applications 2011, 2011:20
/>Page 13 of 15
Example 2: Considering the following integral equation
u
(x, y) ln(u(x, y)+1)=a( x , y)+

β
(
y
)
0

α
(
x
)
0
[F(s, t, u(x, y)) + G(s, t, u(x, y))] dsdt
,


β
(
y
)
0

α
(
x
)
0
g(s, t)dsdt]}, x ∈ [0, x], y ∈ [0, y]
,
(49)
where
x
,
y
are determined similar to the process in Theorem 2.3, and

H(x, y)=J(a(x, y)) +

β
(
y
)
0

α

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doi:10.1186/1029-242X-2011-20
Cite this article as: Zheng and Feng: Some new nonlinear integral inequalities and their applications in the
qualitative analysis of differential equations. Journal of Inequalities and Applications 2011 2011:20.
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