Open Access

A class of retarded nonlinear integral inequalities and its application in nonlinear differential-integral equation

Journal of Inequalities and Applications20122012:154

https://doi.org/10.1186/1029-242X-2012-154

Received: 12 March 2012

Accepted: 22 June 2012

Published: 5 July 2012

Abstract

In this paper, we discuss a class of retarded nonlinear integral inequalities and give an upper bound estimation of an unknown function by the integral inequality technique. This estimation can be used as a tool in the study of differential-integral equations with the initial conditions.

MSC:26D10, 26D15, 26D20, 34A12, 34A40.

Keywords

integral inequalityanalysis techniqueretarded differential-integral equationestimation

1 Introduction

Gronwall-Bellman inequalities [1, 2] can be used as important tools in the study of existence, uniqueness, boundedness, stability, and other qualitative properties of solutions of differential equations, integral equations, and integral-differential equations. There can be found a lot of generalizations of Gronwall-Bellman inequalities in various cases from literature (e.g., [313]).

Lemma 1 (Abdeldaim and Yakout [4])

We assume that u ( t ) and f ( t ) are nonnegative real-valued continuous functions defined on I = [ 0 , ) and they satisfy the inequality
u p + 1 ( t ) u 0 + ( 0 t f ( s ) u p ( s ) d s ) 2 + 2 0 t f ( s ) u p ( s ) [ u p ( s ) + 0 s f ( λ ) u p ( λ ) d λ ] d s ,
(1.1)
for all t I , where u 0 > 0 and p ( 0 , 1 ) are constants. Then
u ( t ) u 0 1 p + 1 + 2 p + 1 0 t f ( s ) D 2 ( s ) d s , t I ,
(1.2)
where
D 2 ( t ) = β ( t ) [ u 0 1 p 1 + p + 2 1 p 1 + p 0 t f ( s ) exp ( 2 1 p p 0 s f ( λ ) d λ ) d s ] p 1 p ,
(1.3)

and β ( t ) = exp ( 2 0 t f ( s ) d s ) , for all t I .

In this paper, we discuss a class of retarded nonlinear integral inequalities and give an upper bound estimation of an unknown function by the integral inequality technique.

2 Main result

In this section, we discuss some retarded integral inequalities of Gronwall-Bellman type. Throughout this paper, let I = [ 0 , ) .

Theorem 1 Suppose α C 1 ( I , I ) is increasing function with α ( t ) t , α ( 0 ) = 0 , t I . We assume that u ( t ) and f ( t ) are nonnegative real-valued continuous functions defined on I, and they satisfy the inequality
u p + 1 ( t ) u 0 + ( 0 α ( t ) f ( s ) u p ( s ) d s ) 2 + 2 0 α ( t ) f ( s ) u p ( s ) [ u p ( s ) + 0 s f ( λ ) u p ( λ ) d λ ] d s , t I ,
(2.1)
where u 0 > 0 and p ( 0 , 1 ) are constants. Then
u ( t ) u 0 1 p + 1 + 2 p + 1 0 α ( t ) f ( s ) θ 1 ( α 1 ( s ) ) d s , t I ,
(2.2)
where
θ 1 ( t ) = β 1 ( t ) [ u 0 1 p 1 + p + 2 1 p 1 + p 0 α ( t ) f ( s ) exp ( 2 1 p p 0 s f ( λ ) d λ ) d s ] p 1 p ,
(2.3)

and β 1 ( t ) = exp ( 2 0 α ( t ) f ( s ) d s ) , for all t I .

Remark 1 If α ( t ) = t , then Theorem 1 reduces Lemma 1.

Proof Let z 1 p + 1 ( t ) denote the function on the right-hand side of (2.1), which is a positive and nondecreasing function on I with z 1 ( 0 ) = u 0 1 p + 1 . Then (2.1) is equivalent to
u ( t ) z 1 ( t ) , u ( α ( t ) ) z 1 ( α ( t ) ) , t I .
(2.4)
Differentiating z 1 p + 1 ( t ) with respect to t, using (2.4) we have
( p + 1 ) z 1 p ( t ) d z 1 ( t ) d t = 2 α ( t ) f ( α ( t ) ) u p ( α ( t ) ) 0 α ( t ) f ( s ) u p ( s ) d s + 2 α ( t ) f ( α ( t ) ) u p ( α ( t ) ) [ u p ( α ( t ) ) + 0 α ( t ) f ( λ ) u p ( λ ) d λ ] 2 α ( t ) f ( α ( t ) ) z 1 p ( t ) [ z 1 p ( t ) + 2 0 α ( t ) f ( λ ) z 1 p ( λ ) d λ ] , t I .
(2.5)
Since z 1 p ( t ) > 0 , from (2.5) we have
d z 1 ( t ) d t 2 p + 1 α ( t ) f ( α ( t ) ) Y 1 ( t ) , t I ,
(2.6)
where
Y 1 ( t ) : = z 1 p ( t ) + 2 0 α ( t ) f ( λ ) z 1 p ( λ ) d λ , t I .
(2.7)
Then Y 1 ( t ) is a positive and nondecreasing function on I with Y 1 ( 0 ) = u 0 p / ( p + 1 ) and
z 1 ( t ) Y 1 ( t ) 1 / p .
(2.8)
Differentiating Y 1 ( t ) with respect to t, and using (2.6), (2.7) and (2.8), we get
d Y 1 ( t ) d t 2 p p + 1 α ( t ) f ( α ( t ) ) z 1 p 1 ( t ) Y 1 ( t ) + 2 α ( t ) f ( α ( t ) ) z 1 p ( α ( t ) ) 2 p p + 1 α ( t ) f ( α ( t ) ) Y 1 2 p 1 p ( t ) + 2 α ( t ) f ( α ( t ) ) Y 1 ( t ) , t I .
(2.9)
From (2.9), we have
Y 1 1 2 p p ( t ) d Y 1 ( t ) d t 2 α ( t ) f ( α ( t ) ) Y 1 1 p p ( t ) 2 p p + 1 α ( t ) f ( α ( t ) ) , t I .
(2.10)
Let S 1 ( t ) = Y 1 1 p p ( t ) , then S 1 ( 0 ) = u 0 1 p p + 1 , from (2.10) we obtain
d S 1 ( t ) d t 2 1 p p α ( t ) f ( α ( t ) ) S 1 ( t ) 2 1 p p + 1 α ( t ) f ( α ( t ) ) , t I .
(2.11)
Consider the ordinary differential equation
{ d S 2 ( t ) d t 2 1 p p α ( t ) f ( α ( t ) ) S 2 ( t ) = 2 1 p p + 1 α ( t ) f ( α ( t ) ) , t I , S 2 ( 0 ) = u 0 1 p p + 1 .
(2.12)
The solution of Equation (2.12) is
S 2 ( t ) = exp ( 0 α ( t ) 2 1 p p f ( s ) d s ) ( u 0 1 p p + 1 + 0 α ( t ) 2 1 p p + 1 f ( s ) exp ( 0 s 2 1 p p f ( τ ) d τ ) d s ) ,
(2.13)
for all t I . By (2.11), (2.12) and (2.13), we obtain
Y 1 ( t ) = S 1 p 1 p ( t ) S 2 p 1 p ( t ) = θ 1 ( t ) , t I ,
(2.14)
where θ 1 ( t ) as defined in (2.3). From (2.6) and (2.14), we have
d z 1 ( t ) d t 2 p + 1 α ( t ) f ( α ( t ) ) θ 1 ( t ) , t I .
By taking t = s in the above inequality and integrating it from 0 to t, we get
u ( t ) z 1 ( t ) u 0 1 p + 1 + 2 p + 1 0 α ( t ) f ( s ) θ 1 ( α 1 ( s ) ) d s , t I .

The estimation (2.2) of the unknown function in the inequality (2.1) is obtained. □

Theorem 2 Suppose α C 1 ( I , I ) is increasing function with α ( t ) t , α ( 0 ) = 0 , t I . We assume that u ( t ) and f ( t ) are nonnegative real-valued continuous functions defined on I and satisfy the inequality
u p + 1 ( t ) u 0 + ( 0 α ( t ) f ( s ) u p ( s ) d s ) 2 + 2 0 α ( t ) f ( s ) u p ( s ) [ u ( s ) + 0 s f ( λ ) u ( λ ) d λ ] d s , t I ,
(2.15)
where u 0 > 0 and p ( 0 , 1 ) are constants. Then
u ( t ) u 0 1 p + 1 + 2 p + 1 0 α ( t ) f ( s ) θ 2 ( α 1 ( s ) ) d s , t I ,
(2.16)
where
θ 2 ( t ) = β 2 ( t ) [ u 0 1 p p + 1 + 0 α ( t ) ( 1 p ) f ( s ) exp ( 0 s ( 1 p ) ( p + 3 ) p + 1 f ( τ ) d τ ) d s ] 1 1 p ,
(2.17)

and β 2 ( t ) = exp ( 0 α ( t ) p + 3 p + 1 f ( s ) d s ) , for all t I .

Proof Let z 2 p + 1 ( t ) denote the function on the right-hand side of (2.15), which is a positive and nondecreasing function on I with z 2 ( 0 ) = u 0 1 p + 1 . Then (2.15) is equivalent to
u ( t ) z 2 ( t ) , u ( α ( t ) ) z 2 ( α ( t ) ) , t I .
(2.18)
Differentiating z 2 p + 1 ( t ) with respect to t, using (2.18) we have
( p + 1 ) z 2 p ( t ) d z 2 ( t ) d t = 2 α ( t ) f ( α ( t ) ) u p ( α ( t ) ) 0 α ( t ) f ( s ) u p ( s ) d s + 2 α ( t ) f ( α ( t ) ) u p ( α ( t ) ) [ u ( α ( t ) ) + 0 α ( t ) f ( λ ) u ( λ ) d λ ] 2 α ( t ) f ( α ( t ) ) z 2 p ( t ) [ z 2 ( t ) + 0 α ( t ) f ( λ ) z 2 ( λ ) d λ + 0 α ( t ) f ( λ ) z 2 p ( λ ) d λ ] , t I .
(2.19)
Since z 2 p ( t ) > 0 , we have
d z 2 ( t ) d t 2 p + 1 α ( t ) f ( α ( t ) ) Y 2 ( t ) , t I ,
(2.20)
where
Y 2 ( t ) : = z 2 ( t ) + 0 α ( t ) f ( λ ) z 2 ( λ ) d λ + 0 α ( t ) f ( λ ) z 2 p ( λ ) d λ , t I .
(2.21)
Then Y 2 ( t ) is a positive and nondecreasing function on I with Y 2 ( 0 ) = z 2 ( 0 ) = u 0 1 / ( p + 1 ) and
z 2 ( t ) Y 2 ( t ) .
(2.22)
Differentiating Y 2 ( t ) with respect to t, and using (2.20), (2.21) and (2.22), we get
d Y 2 ( t ) d t 2 p + 1 α ( t ) f ( α ( t ) ) Y 2 ( t ) + α ( t ) f ( α ( t ) ) z 2 ( α ( t ) ) + α ( t ) f ( α ( t ) ) z 2 p ( α ( t ) ) p + 3 p + 1 α ( t ) f ( α ( t ) ) Y 2 ( t ) + α ( t ) f ( α ( t ) ) Y 2 p ( t ) , t I .
(2.23)
From (2.23), we have
Y 2 p ( t ) d Y 2 ( t ) d t p + 3 p + 1 α ( t ) f ( α ( t ) ) Y 2 1 p ( t ) α ( t ) f ( α ( t ) ) , t I .
(2.24)
Let S 3 ( t ) = Y 2 1 p ( t ) , then S 3 ( 0 ) = u 0 1 p p + 1 , from (2.24) we obtain
d S 3 ( t ) d t ( 1 p ) ( p + 3 ) p + 1 α ( t ) f ( α ( t ) ) S 3 ( t ) ( 1 p ) α ( t ) f ( α ( t ) ) , t I .
(2.25)
Consider the ordinary differential equation
{ d S 4 ( t ) d t ( 1 p ) ( p + 3 ) p + 1 α ( t ) f ( α ( t ) ) S 4 ( t ) = ( 1 p ) α ( t ) f ( α ( t ) ) , t I , S 4 ( 0 ) = u 0 1 p p + 1 .
(2.26)
The solution of Equation (2.26) is
S 4 ( t ) = exp ( 0 α ( t ) ( 1 p ) ( p + 3 ) p + 1 f ( s ) d s ) ( u 0 1 p p + 1 + 0 α ( t ) ( 1 p ) f ( s ) exp ( 0 s ( 1 p ) ( p + 3 ) p + 1 f ( τ ) d τ ) d s ) ,
(2.27)
for all t I . By (2.25), (2.26) and (2.27), we obtain
Y 2 ( t ) = S 3 1 1 p ( t ) S 4 1 1 p ( t ) = θ 2 ( t ) , t I ,
(2.28)
where θ 2 ( t ) as defined in (2.17). From (2.20) and (2.28), we have
d z 2 ( t ) d t 2 p + 1 α ( t ) f ( α ( t ) ) θ 2 ( t ) , t I .
By taking t = s in the above inequality and integrating it from 0 to t, we get
u ( t ) z 2 ( t ) u 0 1 p + 1 + 2 p + 1 0 α ( t ) f ( s ) θ 2 ( α 1 ( s ) ) d s , t I .

The estimation (2.16) of the unknown function in the inequality (2.15) is obtained. □

Theorem 3 Suppose ϕ 1 , ϕ 2 , ϕ 2 / ϕ 1 , α C 1 ( I , I ) are increasing functions with α ( t ) t , ϕ i ( t ) > 0 , t > 0 , i = 1 , 2 , α ( 0 ) = 0 . We assume that u ( t ) and f ( t ) are nonnegative real-valued continuous functions defined on I and satisfy the inequality
u ( t ) u 0 + ( 0 α ( t ) f ( s ) ϕ 1 ( u ( s ) ) d s ) 2 + 2 0 α ( t ) f ( s ) ϕ 1 ( u ( s ) ) [ u ( s ) + 0 s g ( λ ) ϕ 2 ( u ( λ ) ) d λ ] d s , t I ,
(2.29)
where u 0 > 0 is a constant. Then
u ( t ) Φ 1 1 [ Φ 2 1 ( Φ 2 ( Φ 1 ( u 0 ) + 0 α ( t ) g ( s ) d s ) + 0 α ( t ) 2 f ( s ) d s ) ] , t < T 1 ,
(2.30)
where
Φ 1 ( r ) : = 1 r d t ϕ 2 ( t ) , r > 0 ,
(2.31)
Φ 2 ( r ) : = 1 r ϕ 2 ( Φ 1 1 ( s ) ) d s ϕ 1 ( Φ 1 1 ( s ) ) ( Φ 1 1 ( s ) + 1 ) , r > 0 ,
(2.32)
and T 1 is the largest number such that
Φ 2 ( Φ 1 ( u 0 ) + 0 α ( t ) g ( s ) d s ) + 0 α ( t ) 2 f ( s ) d s 1 ϕ 2 ( Φ 1 1 ( s ) ) d s ϕ 1 ( Φ 1 1 ( s ) ) ( Φ 1 1 ( s ) + 1 ) , Φ 2 1 ( Φ 2 ( Φ 1 ( u 0 ) + 0 α ( t ) g ( s ) d s ) + 0 α ( t ) 2 f ( s ) d s ) 1 d t ϕ 2 ( t )

for all t T 1 .

Proof Let z 3 ( t ) denote the function on the right-hand side of (2.29), which is a positive and nondecreasing function on I with z 3 ( 0 ) = u 0 . Then (2.29) is equivalent to
u ( t ) z 3 ( t ) , u ( α ( t ) ) z 3 ( t ) , t I .
(2.33)
Differentiating z 3 ( t ) with respect to t, using (2.33) we have
d z 3 ( t ) d t 2 α ( t ) f ( α ( t ) ) ϕ 1 ( z 3 ( t ) ) 0 α ( t ) f ( s ) ϕ 1 ( z 3 ( s ) ) d s + 2 α ( t ) f ( α ( t ) ) ϕ 1 ( z 3 ( t ) ) [ z 3 ( t ) + 0 α ( t ) g ( λ ) ϕ 2 ( z 3 ( λ ) ) d λ ] d s 2 α ( t ) f ( α ( t ) ) ϕ 1 ( z 3 ( t ) ) [ z 3 ( t ) + 0 α ( t ) f ( s ) ϕ 1 ( z 3 ( s ) ) d s + 0 α ( t ) g ( λ ) ϕ 2 ( z 3 ( λ ) ) d λ ] , t I .
(2.34)
Let
Y 3 ( t ) : = z 3 ( t ) + 0 α ( t ) f ( s ) ϕ 1 ( z 3 ( s ) ) d s + 0 α ( t ) g ( λ ) ϕ 2 ( z 3 ( λ ) ) d λ , t I .
(2.35)
Then Y 3 ( t ) is a positive and nondecreasing function on I with Y 3 ( 0 ) = z 3 ( 0 ) = u 0 and
z 3 ( t ) Y 3 ( t ) .
(2.36)
Differentiating Y 3 ( t ) with respect to t, and using (2.34), (2.35) and (2.36), we get
d Y 3 ( t ) d t 2 α ( t ) f ( α ( t ) ) ϕ 1 ( z 3 ( t ) ) Y 3 ( t ) + α ( t ) f ( α ( t ) ) ϕ 1 ( z 3 ( t ) ) + α ( t ) g ( α ( t ) ) ϕ 2 ( z 3 ( t ) ) 2 α ( t ) f ( α ( t ) ) ϕ 1 ( Y 3 ( t ) ) ( Y 3 ( t ) + 1 ) + α ( t ) g ( α ( t ) ) ϕ 2 ( Y 3 ( t ) ) ,
(2.37)
for all t I . Since ϕ 2 ( Y 3 ( t ) ) > 0 , t > 0 , from (2.37) we have
d Y 3 ( t ) ϕ 2 ( Y 3 ( t ) ) 2 α ( t ) f ( α ( t ) ) ϕ 1 ( Y 3 ( t ) ) ( Y 3 ( t ) + 1 ) ϕ 2 ( Y 3 ( t ) ) + α ( t ) g ( α ( t ) ) , t I .
By taking t = s in the above inequality and integrating it from 0 to t, we get
Φ 1 ( Y 3 ( t ) ) Φ 1 ( Y 3 ( 0 ) ) + 0 t 2 α ( s ) f ( α ( s ) ) ϕ 1 ( Y 3 ( s ) ) ( Y 3 ( s ) + 1 ) ϕ 2 ( Y 3 ( s ) ) d s + 0 t α ( s ) g ( α ( s ) ) d s ,
(2.38)
for all t I , where Φ 1 is defined by (2.31). From (2.38), we have
Φ 1 ( Y 3 ( t ) ) Φ 1 ( Y 3 ( 0 ) ) + 0 T α ( s ) g ( α ( s ) ) d s + 0 t 2 α ( s ) f ( α ( s ) ) ϕ 1 ( Y 3 ( s ) ) ( Y 3 ( s ) + 1 ) ϕ 2 ( Y 3 ( s ) ) d s ,
(2.39)
for all t < T , where 0 < T < T 1 is chosen arbitrarily. Let Y 4 ( t ) denote the function on the right-hand side of (2.39), which is a positive and nondecreasing function on I with Y 4 ( 0 ) = Φ 1 ( u 0 ) + 0 T α ( s ) g ( α ( s ) ) d s and
Y 3 ( t ) Φ 1 1 ( Y 4 ( t ) ) , t < T .
(2.40)
Differentiating Y 4 ( t ) with respect to t, using the hypothesis on ϕ 2 / ϕ 1 , from (2.40) we have
d Y 4 ( t ) d t 2 α ( t ) f ( α ( t ) ) ϕ 1 ( Y 3 ( t ) ) ( Y 3 ( t ) + 1 ) ϕ 2 ( Y 3 ( t ) ) 2 α ( t ) f ( α ( t ) ) ϕ 1 ( Φ 1 1 ( Y 4 ( t ) ) ) ( Φ 1 1 ( Y 4 ( t ) ) + 1 ) ϕ 2 ( Φ 1 1 ( Y 4 ( t ) ) ) , t < T .
(2.41)
By the definition of Φ 2 in (2.32), from (2.41) we obtain
Φ 2 ( Y 4 ( t ) ) Φ 2 ( Y 4 ( 0 ) ) + 0 t 2 α ( s ) f ( α ( s ) ) d s Φ 2 ( Φ 1 ( u 0 ) + 0 α ( T ) g ( s ) d s ) + 0 α ( t ) 2 f ( s ) d s , t < T .
(2.42)
Let t = T , from (2.42) we have
Φ 2 ( Y 4 ( T ) ) Φ 2 ( Φ 1 ( u 0 ) + 0 α ( T ) g ( s ) d s ) + 0 α ( T ) 2 f ( s ) d s .
(2.43)
Since 0 < T < T 1 is chosen arbitrarily, from (2.33), (2.36), (2.40) and (2.43), we have
u ( t ) Φ 1 1 [ Φ 2 1 ( Φ 2 ( Φ 1 ( u 0 ) + 0 α ( t ) g ( s ) d s ) + 0 α ( t ) 2 f ( s ) d s ) ] , t < T 1 .

This proves (2.30). □

3 Application

In this section, we apply our Theorem 3 to the following differential-integral equation
{ d x ( t ) d t = F ( t , x ( α ( t ) ) ) + H ( t , x ( α ( t ) ) ) , t I , x ( 0 ) = x 0 ,
(3.1)
where F C ( I × I , R ) , H C ( I 3 , R ) , | x 0 | > 0 is a constant satisfying the following conditions
| F ( t , x ( t ) ) | f 2 ( t ) ϕ 1 2 ( | x ( t ) | ) ,
(3.2)
| H ( t , x ( t ) ) | 2 f ( t ) ϕ 1 ( | x ( t ) | ) ( | x ( t ) | + 0 t g ( s ) ϕ 2 ( | x ( s ) | ) d s ) ,
(3.3)

where f, g is nonnegative real-valued continuous function defined on I.

Corollary 1 Consider the nonlinear system (3.1) and suppose that F, H satisfy the conditions (3.2) and (3.3), and ϕ 1 , ϕ 2 , ϕ 2 / ϕ 1 , α C 1 ( I , I ) are increasing functions with α ( t ) t , ϕ i ( t ) > 0 , t > 0 , i = 1 , 2 , α ( 0 ) = 0 . Then all solutions of Equation (3.1) exist on I and satisfy the following estimation
| x ( t ) | Φ 1 1 [ Φ 2 1 ( Φ 2 ( Φ 1 ( | x 0 | ) + 0 α ( t ) g ( α 1 ( s ) ) α ( α 1 ( s ) ) d s ) + 0 α ( t ) 2 f ( α 1 ( s ) ) α ( α 1 ( s ) ) d s ) ] ,
(3.4)
for all t < T 2 , where
Φ 1 ( r ) : = 1 r d t ϕ 2 ( t ) , r > 0 , Φ 2 ( r ) : = 1 r ϕ 2 ( Φ 1 1 ( s ) ) d s ϕ 1 ( Φ 1 1 ( s ) ) ( Φ 1 1 ( s ) + 1 ) , r > 0 ,
and T 2 is the largest number such that
Φ 2 ( Φ 1 ( | x 0 | ) + 0 α ( t ) g ( α 1 ( s ) ) α ( α 1 ( s ) ) d s ) + 0 α ( t ) 2 f ( α 1 ( s ) ) α ( α 1 ( s ) ) d s 1 ϕ 2 ( Φ 1 1 ( s ) ) d s ϕ 1 ( Φ 1 1 ( s ) ) ( Φ 1 1 ( s ) + 1 ) , Φ 2 1 ( Φ 2 ( Φ 1 ( | x 0 | ) + 0 α ( t ) g ( α 1 ( s ) ) α ( α 1 ( s ) ) d s ) + 0 α ( t ) 2 f ( α 1 ( s ) ) α ( α 1 ( s ) ) d s ) 1 d t ϕ 2 ( t )

for all t T 2 .

Proof Integrating both sides of Equation (3.1) from 0 to t, we get
x ( t ) = x 0 + 0 t F ( s , x ( α ( s ) ) ) d s + 0 t H ( s , x ( α ( s ) ) ) d s , t I .
(3.5)
Using the conditions (3.2) and (3.3), from (3.5) we obtain
| x ( t ) | | x 0 | + 0 t f 2 ( s ) ϕ 1 2 ( | x ( α ( s ) ) | ) d s + 2 0 t f ( s ) ϕ 1 ( | x ( α ( s ) ) | ) ( | x ( α ( s ) ) | + 0 s g ( τ ) ϕ 2 ( | x ( α ( τ ) ) | ) d τ ) d s | x 0 | + ( 0 α ( t ) f ( α 1 ( s ) ) α ( α 1 ( s ) ) ϕ 1 ( | x ( s ) | ) d s ) 2 + 2 0 α ( t ) f ( α 1 ( s ) ) α ( α 1 ( s ) ) ϕ 1 ( | x ( s ) | ) ( | x ( s ) | + 0 s g ( α 1 ( τ ) ) α ( α 1 ( τ ) ) ϕ 2 ( | x ( τ ) | ) d τ ) d s ,
(3.6)

for all t I . Applying Theorem 3 to (3.6), we get the estimation (3.4). This completes the proof of the Corollary 1. □

Declarations

Acknowledgement

The author is very grateful to the editor and the referees for their helpful comments and valuable suggestions. This research was supported by National Natural Science Foundation of China (Project No. 11161018), Guangxi Natural Science Foundation (Project No. 0991265 and 2012GXNSFAA053009), Scientific Research Foundation of the Education Department of Guangxi Province of China (Project No. 201106LX599), and the Key Discipline of Applied Mathematics of Hechi University of China (200725).

Authors’ Affiliations

(1)
Department of Mathematics, Hechi University

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© Wang; licensee Springer 2012

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