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Some new generalized retarded nonlinear integral inequalities with iterated integrals and their applications

Journal of Inequalities and Applications20122012:236

https://doi.org/10.1186/1029-242X-2012-236

Received: 19 February 2012

Accepted: 3 October 2012

Published: 17 October 2012

Abstract

Some new generalized retarded nonlinear integral inequalities with iterated integrals are discussed and upper bound estimations of unknown functions are given by analysis technique. These estimations can be used as tools in the study of differential-integral equations with the initial conditions.

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

Keywords

integral inequalityiterated integralsanalysis 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 and integral equations. There can be found a lot of generalizations of Gronwall-Bellman inequalities in various cases from literature (e.g., [315]).

Agarwal et al. [5] studied the inequality
u ( t ) a ( t ) + i = 1 n b i ( t 0 ) b i ( t ) g i ( t , s ) w i ( u ( s ) ) d s , t 0 t < t 1 .
Agarwal et al. [6] obtained the explicit bound to the unknown function of the following retarded integral inequality:
φ ( u ( t ) ) c + i = 1 n α i ( t 0 ) α i ( t ) u q ( s ) [ f i ( s ) φ 1 ( u ( s ) ) + g i ( s ) φ 2 ( log ( u ( s ) ) ) ] d s .
In 2011, Abdeldaim and Yakout [4] studied the following integral inequalities:

However, the bound given on such an inequality in [4] is not directly applicable in the study of certain retarded nonlinear differential and integral equations. It is desirable to establish new inequalities of the above type, which can be used more effectively in the study of certain classes of retarded nonlinear differential and integral equations.

In this paper, we discuss some new retarded nonlinear integral inequalities with iterated integrals
(1.1)
(1.2)
(1.3)
(1.4)

where u 0 is a positive constant, and give upper bound estimation of the unknown function by integral inequality technique. Furthermore, we apply our result to differential-integral equations for estimation.

2 Main result

In this section, we discuss some retarded integral inequalities with iterated integrals. Throughout this paper, let I = [ 0 , ) .

Lemma 1 (Abdeldaim and Yakout [4])

We assume that u ( t ) , f ( t ) and g ( t ) are nonnegative real-valued continuous functions defined on I = [ 0 , ) and satisfy the inequality
u ( t ) u 0 + 0 t f ( s ) u ( s ) [ u ( s ) + 0 s g ( λ ) u ( λ ) d λ ] p d s
for all t I , where u 0 and p are positive constants. Then
u ( t ) u 0 exp ( 0 t f ( s ) B 1 ( s ) ) d s , t I ,
where
B 1 ( t ) = u 0 p exp ( p 0 t g ( s ) d s ) 1 p u 0 p 0 t f ( s ) exp ( p 0 s g ( τ ) d τ ) d s ,

such that p u 0 p 0 t f ( s ) exp ( p 0 s g ( τ ) d τ ) d s 1 for all t I .

Lemma 2 (Abdeldaim and Yakout [4])

We assume that u ( t ) , f ( t ) and g ( t ) are nonnegative real-valued continuous functions defined on I = [ 0 , ) and satisfy the inequality
u ( t ) u 0 + 0 t ( f ( s ) u ( s ) + q ( s ) ) d s + 0 t f ( s ) u ( s ) [ u ( s ) + 0 s g ( λ ) u ( λ ) d λ ] d s ,
(2.1)
for all t I , where u 0 is a positive constant. Then
u ( t ) ( u 0 + 0 t q ( s ) exp ( A ( s ) ) d s ) exp ( A ( t ) ) , t I ,
(2.2)
where A ( t ) = 0 t ( f ( s ) + f ( s ) Q 1 ( s ) ) d s , and
Q 1 ( t ) = L ( t ) ( L ( 0 ) u 0 ) exp ( A 1 ( t ) ) 1 + ( L ( 0 ) u 0 ) 0 t f ( s ) exp ( A 1 ( s ) ) d s ,
where A 1 ( t ) = 0 t 2 ( f ( s ) L ( s ) + f ( s ) + g ( s ) ) d s , and L ( t ) is the maximal solution of the differential equation
d L ( t ) d t = q ( t ) + f ( t ) L 2 ( t ) + ( f ( t ) + g ( t ) ) L ( t ) , t I ,

such that L ( 0 ) > u 0 .

Lemma 3 Suppose that φ ( t ) is a positive and increasing function on I with φ ( 0 ) = 0 , f i C ( I , I ) , i = 1 , 2 ; u ( t ) is a nonnegative real-valued continuous function defined on I with u ( 0 ) = u 0 > 0 and satisfies the inequality
d u ( t ) d t f 1 ( t ) + f 2 ( t ) φ ( u ( t ) ) , t I ,
(2.3)
then u ( t ) has the following estimation:
u ( t ) Φ 1 ( Φ ( u 0 + 0 t f 1 ( s ) d s ) + 0 t f 2 ( s ) d s ) , t [ 0 , T 1 ] ,
(2.4)
where
Φ ( r ) : = 1 r d s φ ( s ) , r > 0 ,
(2.5)
and T 1 is the largest number such that
Φ ( u 0 + 0 T 1 f 1 ( s ) d s ) + 0 T 1 f 2 ( s ) d s 1 d s φ ( s ) .
(2.6)

Remark 1 Φ ( t ) = ln ( t ) when φ ( t ) = t .

Proof Integrating both sides of (2.3) from 0 to t,
u ( t ) u 0 + 0 t f 1 ( s ) d s + 0 t f 2 ( s ) φ ( u ( s ) ) d s u 0 + 0 T f 1 ( s ) d s + 0 t f 2 ( s ) φ ( u ( s ) ) d s , t [ 0 , T ] ,
(2.7)
where T [ 0 , T 1 ] is a positive constant chosen arbitrarily, T 1 is defined by (2.6). Let
R 1 ( t ) = u 0 + 0 T f 1 ( s ) d s + 0 t f 2 ( s ) φ ( u ( s ) ) d s , t [ 0 , T ] ,
(2.8)
then R 1 ( t ) is a nonnegative and nondecreasing function on I with R 1 ( 0 ) = u 0 + 0 T f 1 ( s ) d s . Then (2.7) is equivalent to
u ( t ) R 1 ( t ) , t [ 0 , T ] .
(2.9)
Differentiating R 1 ( t ) with respect to t, from (2.8) and (2.9), we have
d R 1 ( t ) d t = f 2 ( t ) φ ( u ( t ) ) f 2 ( t ) φ ( R 1 ( t ) ) , t [ 0 , T ] .
(2.10)
Since R 1 ( t ) > 0 , from (2.10) we have
d R 1 ( t ) φ ( R 1 ( t ) ) d t f 2 ( t ) , t [ 0 , T ] .
(2.11)
By taking t = s in (2.11) and integrating it from 0 to t, we get
R 1 ( t ) Φ 1 ( Φ ( R 1 ( 0 ) ) + 0 t f 2 ( s ) d s ) , t [ 0 , T ] ,
(2.12)
where Φ is defined by (2.5). From (2.9) we have
u ( t ) Φ 1 ( Φ ( u 0 + 0 T f 1 ( s ) d s ) + 0 t f 2 ( s ) d s ) , t [ 0 , T ] .
(2.13)
Letting t = T , from (2.13) we get
u ( T ) Φ 1 ( Φ ( u 0 + 0 T f 1 ( s ) d s ) + 0 T f 2 ( s ) d s ) .

Because T [ 0 , T 1 ] is chosen arbitrarily, this proves (2.4). □

Lemma 4 Let f i C ( I , I ) , i = 1 , 2 , 3 ; we assume that u ( t ) is a nonnegative real-valued continuous function defined on I with u ( 0 ) = u 0 > 0 and satisfies the inequality
d u ( t ) d t f 1 ( t ) + f 2 ( t ) u ( t ) + f 3 ( t ) u 2 ( t ) , t I ,
(2.14)
then u ( t ) has the following estimation:
u ( t ) exp ( 0 t f 2 ( s ) d s ) ( ( u 0 + 0 t f 1 ( s ) d s ) 1 0 t f 3 ( s ) exp ( 0 s f 2 ( τ ) d τ ) d s ) 1 ,
(2.15)
for all t [ 0 , T 2 ] , where T 2 is the largest number such that
( u 0 + 0 t f 1 ( s ) d s ) 1 0 t f 3 ( s ) exp ( 0 s f 2 ( τ ) d τ ) d s > 0 , t [ 0 , T 2 ] .
(2.16)
Proof Integrating both sides of (2.14) from 0 to t, we get
u ( t ) u 0 + 0 t f 1 ( s ) d s + 0 t f 2 ( s ) u ( s ) d s + 0 t f 3 ( s ) u 2 ( s ) d s u 0 + 0 T f 1 ( s ) d s + 0 t f 2 ( s ) u ( s ) d s + 0 t f 3 ( s ) u 2 ( s ) d s , t [ 0 , T ] ,
(2.17)
where T [ 0 , T 2 ] is a positive constant chosen arbitrarily, T 2 is defined by (2.16). Let
R 2 ( t ) = u 0 + 0 T f 1 ( s ) d s + 0 t f 2 ( s ) u ( s ) d s + 0 t f 3 ( s ) u 2 ( s ) d s , t [ 0 , T ] ,
(2.18)
then R 2 ( t ) is a nonnegative and nondecreasing function on I with R 2 ( 0 ) = u 0 + 0 T f 1 ( s ) d s . Then (2.17) is equivalent to
u ( t ) R 2 ( t ) , t [ 0 , T ] .
(2.19)
Differentiating R 2 ( t ) with respect to t, from (2.18) and (2.19), we have
d R 2 ( t ) d t = f 2 ( t ) u ( t ) + f 3 ( t ) u 2 ( t ) f 2 ( t ) R 2 ( t ) + f 3 ( t ) R 2 2 ( t ) , t [ 0 , T ] .
(2.20)
Since R 2 ( t ) > 0 , from (2.20) we have
R 2 2 ( t ) d R 2 ( t ) d t f 2 ( t ) R 2 1 ( t ) + f 3 ( t ) , t [ 0 , T ] .
(2.21)
Let S 1 ( t ) = R 2 1 ( t ) , then S 1 ( 0 ) = ( u 0 + 0 T f 1 ( s ) d s ) 1 , from (2.21) we obtain
d S 1 ( t ) d t + f 2 ( t ) S 1 ( t ) f 3 ( t ) , t [ 0 , T ] .
(2.22)
Consider the ordinary differential equation
{ d S 2 ( t ) d t + f 2 ( t ) S 2 ( t ) = f 3 ( t ) , t [ 0 , T ] , S 2 ( 0 ) = ( u 0 + 0 T f 1 ( s ) d s ) 1 .
(2.23)
The solution of equation (2.23) is
S 2 ( t ) = exp ( 0 t f 2 ( s ) d s ) ( ( u 0 + 0 T f 1 ( s ) d s ) 1 0 t f 3 ( s ) exp ( 0 s f 2 ( τ ) d τ ) d s )
(2.24)
for all t [ 0 , T ] . Letting t = T in (2.24), from (2.19), (2.22), (2.23) and (2.24), we obtain
u ( T ) R 2 ( T ) = 1 S 1 ( T ) 1 S 2 ( T ) exp ( 0 T f 2 ( s ) d s ) ( ( u 0 + 0 T f 1 ( s ) d s ) 1 0 T f 3 ( s ) exp ( 0 s f 2 ( τ ) d τ ) d s ) 1 .
(2.25)

Because T [ 0 , T 2 ] is chosen arbitrarily, this proves (2.15). □

Lemma 5 Suppose that φ ( t ) , φ ( t ) / t are positive and increasing functions on I, f i C ( I , I ) , i = 1 , 2 , 3 , 4 ; u ( t ) is a nonnegative real-valued continuous function defined on I with u ( 0 ) = u 0 > 0 and satisfies the inequality
d u ( t ) d t f 1 ( t ) + f 2 ( t ) u ( t ) + f 3 ( t ) u ( t ) φ ( u ( t ) ) + f 4 ( t ) φ ( u ( t ) ) , u ( 0 ) = u 0 , t I ,
(2.26)
then u ( t ) has the following estimation:
u ( t ) exp ( Φ 1 1 ( Φ 1 ( ln ( u 0 + 0 t f 1 ( s ) d s ) + 0 t f 2 ( s ) d s ) + 0 t f 3 ( s ) d s + 0 t f 4 ( s ) d s exp ( ln ( u 0 + 0 t f 1 ( τ ) d τ ) + 0 t f 2 ( τ ) d τ ) ) ) ,
(2.27)
for all t [ 0 , T 3 ] , where
Φ 1 ( r ) : = 1 r d s φ ( exp ( s ) ) , r > 0 ,
(2.28)
and T 3 is the largest number such that
(2.29)
Proof Integrating both sides of (2.26) from 0 to t, we get
u ( t ) u 0 + 0 t f 1 ( s ) d s + 0 t f 2 ( s ) u ( s ) d s + 0 t f 3 ( s ) u ( s ) φ ( u ( s ) ) d s + 0 t f 4 ( s ) φ ( u ( s ) ) d s u 0 + 0 T f 1 ( s ) d s + 0 t f 2 ( s ) u ( s ) d s + 0 t f 3 ( s ) u ( s ) φ ( u ( s ) ) d s + 0 t f 4 ( s ) φ ( u ( s ) ) d s ,
(2.30)
for all t [ 0 , T ] , where T [ 0 , T 3 ] is a positive constant chosen arbitrarily, T 3 is defined by (2.29). Let R 3 ( t ) denote the function on the right-hand side of (2.30), which is a positive and nondecreasing function on [ 0 , T ] with
R 3 ( 0 ) = u 0 + 0 T f 1 ( s ) d s .
(2.31)
Then (2.26) is equivalent to
u ( t ) R 3 ( t ) , t [ 0 , T ] .
(2.32)
Differentiating R 3 ( t ) with respect to t and using (2.32), we have
d R 3 ( t ) d t f 2 ( t ) R 3 ( t ) + f 3 ( t ) R 3 ( t ) φ ( R 3 ( t ) ) + f 4 ( t ) φ ( R 3 ( t ) ) , t [ 0 , T ] .
(2.33)
From (2.33) we get
R 3 1 ( t ) d R 3 ( t ) d t f 2 ( t ) + f 3 ( t ) φ ( R 3 ( t ) ) + f 4 ( t ) φ ( R 3 ( t ) ) R 3 1 ( t ) , t [ 0 , T ] .
(2.34)
Integrating both sides of (2.34) from 0 to t and using (2.31), we get
R 3 ( t ) exp ( ln R 3 ( 0 ) + 0 t f 2 ( s ) d s + 0 t f 3 ( s ) φ ( R 3 ( s ) ) d s + 0 t f 4 ( s ) φ ( R 3 ( s ) ) R 3 1 ( s ) d s ) exp ( ln ( u 0 + 0 T f 1 ( s ) d s ) + 0 T f 2 ( s ) d s + 0 t f 3 ( s ) φ ( R 3 ( s ) ) d s + 0 t f 4 ( s ) φ ( R 3 ( s ) ) R 3 1 ( s ) d s ) , t [ 0 , T ] ,
(2.35)
here we use the monotonicity of φ ( t ) and φ ( t ) / t . Let
R 4 ( t ) = ln ( u 0 + 0 T f 1 ( s ) d s ) + 0 T f 2 ( s ) d s + 0 t f 3 ( s ) φ ( R 3 ( s ) ) d s + 0 t f 4 ( s ) φ ( R 3 ( s ) ) R 3 1 ( s ) d s
(2.36)
for all t [ 0 , T ] , then R 4 ( t ) is a positive and nondecreasing function on [ 0 , T ] with
R 4 ( 0 ) = ln ( u 0 + 0 T f 1 ( s ) d s ) + 0 T f 2 ( s ) d s .
(2.37)
(2.36) is equivalent to
R 3 ( t ) exp ( R 4 ( t ) ) , t [ 0 , T ] .
(2.38)
Differentiating R 4 ( t ) with respect to t and using (2.38), we have
d R 4 ( t ) d t = f 3 ( t ) φ ( R 3 ( t ) ) + f 4 ( t ) φ ( R 3 ( t ) ) R 3 1 ( t ) f 3 ( t ) φ ( exp ( R 4 ( t ) ) ) + f 4 ( t ) φ ( exp ( R 4 ( t ) ) ) ( exp ( R 4 ( t ) ) ) 1 , t [ 0 , T ] .
(2.39)
From (2.39) we get
φ 1 ( exp ( R 4 ( t ) ) ) d R 4 ( t ) d t f 3 ( t ) + f 4 ( t ) ( exp ( R 4 ( t ) ) ) 1 , t [ 0 , T ] .
(2.40)
Integrating both sides of (2.40) from 0 to t,
Φ 1 ( R 4 ( t ) ) Φ 1 ( R 4 ( 0 ) ) + 0 t f 3 ( s ) d s + 0 t f 4 ( s ) ( exp ( R 4 ( 0 ) ) ) 1 d s Φ 1 ( ln ( u 0 + 0 T f 1 ( s ) d s ) + 0 T f 2 ( s ) d s ) + 0 t f 3 ( s ) d s + 0 t f 4 ( s ) d s exp ( ln ( u 0 + 0 T f 1 ( τ ) d τ ) + 0 T f 2 ( τ ) d τ )
(2.41)
for all t [ 0 , T ] , where Φ 1 is defined by (2.28). From (2.32), (2.38) and (2.41), we have
u ( t ) R 3 ( t ) exp ( R 4 ( t ) ) exp ( Φ 1 1 ( Φ 1 ( ln ( u 0 + 0 T f 1 ( s ) d s ) + 0 T f 2 ( s ) d s ) + 0 t f 3 ( s ) d s + 0 t f 4 ( s ) d s exp ( ln ( u 0 + 0 T f 1 ( τ ) d τ ) + 0 T f 2 ( τ ) d τ ) ) ) ,
(2.42)
for all t [ 0 , T ] . Letting t = T , from (2.42) we get
u ( T ) exp ( Φ 1 1 ( Φ 1 ( ln ( u 0 + 0 T f 1 ( s ) d s ) + 0 T f 2 ( s ) d s ) + 0 T f 3 ( s ) d s + 0 T f 4 ( s ) d s exp ( ln ( u 0 + 0 T f 1 ( τ ) d τ ) + 0 T f 2 ( τ ) d τ ) ) ) .

Because T [ 0 , T 3 ] is chosen arbitrarily, this gives the estimation (2.27) of the unknown function in the inequality (2.26). □

Theorem 1 Suppose that α C 1 ( I , I ) is an increasing function with α ( t ) t , α ( 0 ) = 0 , t I ; u ( t ) , f ( t ) , q ( t ) and g ( t ) are nonnegative real-valued continuous functions defined on I = [ 0 , ) and satisfy the inequality (1.1). Then
u ( t ) exp ( ln ( u 0 + 0 α ( t ) q ( s ) d s ) + 0 α ( t ) ( f ( s ) + f ( s ) L 1 ( α 1 ( s ) ) ) d s )
(2.43)
for all t [ 0 , T 4 ] , where
L 1 ( t ) = exp ( 0 α ( t ) ( f ( s ) + g ( s ) ) d s ) ( ( u 0 + 0 α ( t ) q ( s ) d s ) 1 0 α ( t ) f ( s ) exp ( 0 s ( f ( τ ) + g ( τ ) ) d τ ) d s ) 1 ,
(2.44)
T 4 is the largest number such that
( u 0 + 0 α ( t ) q ( s ) d s ) 1 0 α ( t ) f ( s ) exp ( 0 s ( f ( τ ) + g ( τ ) ) d τ ) d s > 0 .

Remark 2 Theorem 1 gives the explicit estimation (2.43) for the inequality (1.1) which is just the inequality (2.1) when α ( t ) = t . Lemma 1 gives the implicit estimation (2.2) for the inequality (2.1).

Proof Let z 1 ( t ) denote the function on the right-hand side of (1.1), which is a positive and nondecreasing function on I with z 1 ( 0 ) = u 0 . Then (1.1) is equivalent to
u ( t ) z 1 ( t ) , u ( α ( t ) ) z 1 ( α ( t ) ) z 1 ( t ) , t I .
(2.45)
Differentiating z 1 ( t ) with respect to t and using (2.45), we have
d z 1 ( t ) d t = α ( t ) f ( α ( t ) ) u ( α ( t ) ) + α ( t ) q ( α ( t ) ) + α ( t ) f ( α ( t ) ) u ( α ( t ) ) [ u ( α ( t ) ) + 0 α ( t ) g ( λ ) u ( λ ) d λ ] α ( t ) q ( α ( t ) ) + [ α ( t ) f ( α ( t ) ) + α ( t ) f ( α ( t ) ) Y 1 ( t ) ] z 1 ( t ) , t I ,
(2.46)
where
Y 1 ( t ) : = z 1 ( t ) + 0 α ( t ) g ( λ ) z 1 ( λ ) d λ , t I .
(2.47)
Then Y 1 ( t ) is a positive and nondecreasing function on I with Y 1 ( 0 ) = z 1 ( 0 ) = u 0 and
z 1 ( t ) Y 1 ( t ) .
(2.48)
Differentiating Y 1 ( t ) with respect to t and using (2.46), (2.47) and (2.48), we get
d Y 1 ( t ) d t α ( t ) q ( α ( t ) ) + [ α ( t ) f ( α ( t ) ) + α ( t ) f ( α ( t ) ) Y 1 ( t ) ] z 1 ( t ) + α ( t ) g ( α ( t ) ) z 1 ( t ) α ( t ) q ( α ( t ) ) + [ α ( t ) f ( α ( t ) ) + α ( t ) g ( α ( t ) ) ] Y 1 ( t ) + α ( t ) f ( α ( t ) ) Y 1 2 ( t )
(2.49)
for all t I . Applying Lemma 4 to (2.49), we obtain
Y 1 ( t ) L 1 ( t ) , t [ 0 , T 4 ] .
(2.50)
From (2.46) and (2.50), we get
d z 1 ( t ) d t α ( t ) q ( α ( t ) ) + [ α ( t ) f ( α ( t ) ) + α ( t ) f ( α ( t ) ) L 1 ( t ) ] z 1 ( t ) , t [ 0 , T 4 ] .
(2.51)
Applying Lemma 3 to (2.51) and using Remark 1, we obtain
z 1 ( t ) exp ( ln ( u 0 + 0 α ( t ) q ( s ) d s ) + 0 α ( t ) ( f ( s ) + f ( s ) L 1 ( α 1 ( s ) ) ) d s ) , t [ 0 , T 4 ] .

From (2.45), the estimation (2.43) of the unknown function in the inequality (1.1) is obtained. □

Theorem 2 Suppose φ C ( I , I ) and α C 1 ( I , I ) are increasing functions with α ( t ) t , α ( 0 ) = 0 , t I . We assume that u ( t ) , f ( t ) , q ( t ) and g ( t ) are nonnegative real-valued continuous functions defined on I and satisfy the inequality (1.2). Then
u ( t ) Φ 1 ( Φ ( u 0 + 0 α ( t ) q ( s ) d s ) + 0 α ( t ) ( f ( s ) + f ( s ) L 2 ( α 1 ( s ) ) ) d s ) , t [ 0 , T 4 ] ,
(2.52)
where Φ is defined by (2.5),
L 2 ( t ) = exp ( Φ 1 1 ( Φ 1 ( ln ( u 0 + 0 α ( t ) q ( s ) d s ) + 0 α ( t ) g ( s ) d s ) + 0 α ( t ) f ( s ) d s + 0 α ( t ) f ( s ) d s exp ( ln ( u 0 + 0 α ( t ) q ( s ) d s ) + 0 α ( t ) g ( s ) d s ) ) ) , t [ 0 , T 4 ] ,
(2.53)
Φ 1 is defined by (2.28), T 4 is the largest number such that
(2.54)
Proof Let z 2 ( t ) denote the function on the right-hand side of (1.2), which is a positive and nondecreasing function on I with z 2 ( 0 ) = u 0 . Then (1.2) is equivalent to
u ( t ) z 2 ( t ) , u ( α ( t ) ) z 2 ( α ( t ) ) z 2 ( t ) , t I .
(2.55)
Differentiating z 2 ( t ) with respect to t and using (2.55), we have
d z 2 ( t ) d t = α ( t ) f ( α ( t ) ) φ ( u ( α ( t ) ) ) + α ( t ) q ( α ( t ) ) + α ( t ) f ( α ( t ) ) φ ( u ( α ( t ) ) ) [ u ( α ( t ) ) + 0 α ( t ) g ( λ ) u ( λ ) d λ ] α ( t ) q ( α ( t ) ) + [ α ( t ) f ( α ( t ) ) + α ( t ) f ( α ( t ) ) Y 2 ( t ) ] φ ( z 2 ( t ) ) , t I ,
(2.56)
where
Y 2 ( t ) : = z 2 ( t ) + 0 α ( t ) g ( λ ) z 2 ( λ ) d λ , t I .
(2.57)
Then Y 2 ( t ) is a positive and nondecreasing function on I with Y 2 ( 0 ) = z 2 ( 0 ) = u 0 and
z 2 ( t ) Y 2 ( t ) .
(2.58)
Differentiating Y 2 ( t ) with respect to t and using (2.56), (2.57) and (2.58), we get
d Y 2 ( t ) d t α ( t ) q ( α ( t ) ) + [ α ( t ) f ( α ( t ) ) + α ( t ) f ( α ( t ) ) Y 2 ( t ) ] φ ( z 2 ( t ) ) + α ( t ) g ( α ( t ) ) z 2 ( t ) α ( t ) q ( α ( t ) ) + α ( t ) g ( α ( t ) ) Y 2 ( t ) + α ( t ) f ( α ( t ) ) Y 2 ( t ) φ ( Y 2 ( t ) ) + α ( t ) f ( α ( t ) ) φ ( Y 2 ( t ) )
(2.59)
for all t I . Applying Lemma 5 to (2.59), we obtain
Y 2 ( t ) exp ( Φ 1 1 ( Φ 1 ( ln ( u 0 + 0 α ( t ) q ( s ) d s ) + 0 α ( t ) g ( s ) d s ) + 0 α ( t ) f ( s ) d s + 0 α ( t ) f ( s ) d s exp ( ln ( u 0 + 0 α ( t ) q ( s ) d s ) + 0 α ( t ) g ( s ) d s ) ) ) = L 2 ( t ) , t [ 0 , T 4 ] ,
(2.60)
where T 4 and L 2 are defined by (2.53) and (2.54) respectively. From (2.56) and (2.60), we get
d z 2 ( t ) d t α ( t ) q ( α ( t ) ) + [ α ( t ) f ( α ( t ) ) + α ( t ) f ( α ( t ) ) L 2 ( t ) ] φ ( z 2 ( t ) ) , t [ 0 , T 4 ] .
(2.61)
Applying Lemma 3 to (2.61), we obtain
z 2 ( t ) Φ 1 ( Φ ( u 0 + 0 α ( t ) q ( s ) d s ) + 0 α ( t ) ( f ( s ) + f ( s ) L 2 ( α 1 ( s ) ) ) d s ) , t [ 0 , T 4 ] ,

where Φ is defined by (2.5). From (2.55), the estimation (2.52) of the unknown function in the inequality (1.2) is obtained. □

Theorem 3 Suppose α C 1 ( I , I ) are increasing functions with α ( t ) t , α ( 0 ) = 0 , t I . We assume that u ( t ) , f ( t ) and g ( t ) are nonnegative real-valued continuous functions defined on I and satisfy the inequality (1.3). Then
u ( t ) u 0 exp ( 0 t f ( s ) B 2 ( s ) ) d s , t I ,
(2.62)
where
B 2 ( t ) = u 0 p exp ( p 0 α ( t ) g ( s ) d s ) 1 p u 0 p 0 α ( t ) f ( s ) exp ( p 0 s g ( τ ) d τ ) d s ,
(2.63)

such that p u 0 p 0 α ( t ) f ( s ) exp ( p 0 s g ( τ ) d τ ) d s 1 for all t I .

Remark 3 If α ( t ) = t , then Theorem 3 reduces Lemma 1.

Proof Let z 3 ( t ) denote the function on the right-hand side of (1.3), which is a positive and nondecreasing function on I with z 3 ( 0 ) = u 0 . Then (1.3) is equivalent to
u ( t ) z 3 ( t ) , u ( α ( t ) ) z 3 ( α ( t ) ) z 3 ( t ) , t I .
(2.64)
Differentiating z 3 ( t ) with respect to t and using (2.64), we have
d z 3 ( t ) d t = α ( t ) f ( α ( t ) ) u ( α ( t ) ) [ u ( α ( t ) ) + 0 α ( t ) g ( λ ) u ( λ ) d λ ] p α ( t ) f ( α ( t ) ) z 3 ( t ) [ z 3 ( t ) + 0 α ( t ) g ( λ ) z 3 ( λ ) d λ ] p = α ( t ) f ( α ( t ) ) z 3 ( t ) Y 3 p ( t ) , t I ,
(2.65)
where
Y 3 ( t ) : = z 3 ( t ) + 0 α ( t ) g ( λ ) z 3 ( λ ) d λ , t I .
(2.66)
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 ) .
(2.67)
Differentiating Y 3 ( t ) with respect to t and using (2.65), (2.66) and (2.67), we get
d Y 3 ( t ) d t α ( t ) f ( α ( t ) ) z 3 ( t ) Y 3 p ( t ) + α ( t ) g ( α ( t ) ) z 3 ( t ) α ( t ) f ( α ( t ) ) Y 3 1 + p ( t ) + α ( t ) g ( α ( t ) ) Y 3 ( t ) , t I .
(2.68)
From (2.68) we have
Y 3 ( 1 + p ) ( t ) d Y 3 ( t ) d t α ( t ) g ( α ( t ) ) Y 3 p ( t ) α ( t ) f ( α ( t ) ) , t I .
(2.69)
Let S 3 ( t ) = Y 3 p ( t ) , then S 3 ( 0 ) = u 0 p . From (2.69) we obtain
d S 3 ( t ) d t + p α ( t ) g ( α ( t ) ) S 3 ( t ) p α ( t ) f ( α ( t ) ) , t I .
(2.70)
Consider the ordinary differential equation
{ d S 4 ( t ) d t + p α ( t ) g ( α ( t ) ) S 4 ( t ) = p α ( t ) f ( α ( t ) ) , t I , S 4 ( 0 ) = u 0 p .
(2.71)
The solution of equation (2.71) is
S 4 ( t ) = exp ( 0 α ( t ) p g ( s ) d s ) ( u 0 p 0 α ( t ) p f ( s ) exp ( 0 s p g ( τ ) d τ ) d s )
(2.72)
for all t I . By (2.70), (2.71) and (2.72), we obtain
Y 3 p ( t ) = S 3 1 ( t ) S 4 1 ( t ) = B 2 ( t ) , t I ,
(2.73)
where B 2 ( t ) is as defined in (2.63). From (2.65) and (2.73), we have
d z 3 ( t ) d t α ( t ) f ( α ( t ) ) z 3 ( t ) B 2 ( t ) , t I .
By taking t = s in the above inequality and integrating it from 0 to t, from (2.64) we get
u ( t ) z 3 ( t ) u 0 exp ( 0 α ( t ) f ( s ) B 2 ( s ) ) d s , t I .

The estimation (2.62) of the unknown function in the inequality (1.3) is obtained. □

Theorem 4 Suppose φ 1 , φ 2 , α C 1 ( I , I ) are increasing functions with φ i ( t ) > 0 , α ( t ) t , t > 0 , i = 1 , 2 , α ( 0 ) = 0 . We assume that u ( t ) , f ( t ) and g ( t ) are nonnegative real-valued continuous functions defined on I = [ 0 , ) and satisfy the inequality (1.4). Then
u ( t ) Φ 2 1 [ Φ 3 1 ( Φ 3 ( Φ 2 ( u 0 ) + 0 α ( t ) g ( s ) d s ) + 0 α ( t ) f ( s ) d s ) ] , t < T 5 ,
(2.74)
where
(2.75)
(2.76)
and T 5 is the largest number such that
Proof Let z 4 ( t ) denote the function on the right-hand side of (1.4), which is a positive and nondecreasing function on I with z 4 ( 0 ) = u 0 . Then (1.4) is equivalent to
u ( t ) z 4 ( t ) , u ( α ( t ) ) z 4 ( α ( t ) ) z 4 ( t ) , t I .
(2.77)
Differentiating z 4 ( t ) with respect to t and using (2.77), we have
d z 4 ( t ) d t = α ( t ) f ( α ( t ) ) φ 1 ( u ( α ( t ) ) ) [ u ( α ( t ) ) + 0 α ( t ) g ( λ ) φ 2 ( u ( λ ) ) d λ ] p α ( t ) f ( α ( t ) ) φ 1 ( z 4 ( t ) ) [ z 4 ( t ) + 0 α ( t ) g ( λ ) φ 2 ( z 4 ( λ ) ) d λ ] p = α ( t ) f ( α ( t ) ) φ 1 ( z 4 ( t ) ) Y 4 p ( t ) , t I ,
(2.78)
where
Y 4 ( t ) : = z 4 ( t ) + 0 α ( t ) g ( λ ) φ 2 ( z 4 ( λ ) ) d λ , t I .
(2.79)
Then Y 4 ( t ) is a positive and nondecreasing function on I with Y 4 ( 0 ) = z 4 ( 0 ) = u 0 and
z 4 ( t ) Y 4 ( t ) .
(2.80)
Differentiating Y 4 ( t ) with respect to t and using (2.78), (2.79) and (2.80), we get
d Y 4 ( t ) d t α ( t ) f ( α ( t ) ) φ 1 ( z 4 ( t ) ) Y 4 p ( t ) + α ( t ) g ( α ( t ) ) φ 2 ( z 4 ( t ) ) α ( t ) f ( α ( t ) ) φ 1 ( Y 4 ( t ) ) Y 4 p ( t ) + α ( t ) g ( α ( t ) ) φ 2 ( Y 4 ( t ) ) , t I .
(2.81)
Since φ 2 ( Y 4 ( t ) ) > 0 , t > 0 , from (2.81) we have
d Y 4 ( t ) φ 2 ( Y 4 ( t ) ) α ( t ) f ( α ( t ) ) φ 1 ( Y 4 ( t ) ) Y 4 p ( t ) φ 2 ( Y 4 ( t ) ) d t + α ( t ) g ( α ( t ) ) d t , t I .
By taking t = s in the above inequality and integrating it from 0 to t, we get
Φ 2 ( Y 4 ( t ) ) Φ 2 ( Y 4 ( 0 ) ) + 0 t α ( s ) f ( α ( s ) ) φ 1 ( Y 4 ( s ) ) Y 4 p ( s ) φ 2 ( Y 4 ( s ) ) d s + 0 t α ( s ) g ( α ( s ) ) d s ,
(2.82)
for all t I , where Φ 2 is defined by (2.75). From (2.82) we have
Φ 2 ( Y 4 ( t ) ) Φ 2 ( Y 4 ( 0 ) ) + 0 T α ( s ) g ( α ( s ) ) d s + 0 t α ( s ) f ( α ( s ) ) φ 1 ( Y 4 ( s ) ) Y 4 p ( s ) φ 2 ( Y 4 ( s ) ) d s
(2.83)
for all t < T , where 0 < T < T 5 is chosen arbitrarily. Let Y 5 ( t ) denote the function on the right-hand side of (2.83), which is a positive and nondecreasing function on I with Y 5 ( 0 ) = Φ 2 ( u 0 ) + 0 T α ( s ) g ( α ( s ) ) d s and
Y 4 ( t ) Φ 2 1 ( Y 5 ( t ) ) , t < T .
(2.84)
Differentiating Y 5 ( t ) with respect to t and using the hypothesis on φ 2 / φ 1 , from (2.84) we have
d Y 5 ( t ) d t α ( t ) f ( α ( t ) ) φ 1 ( Y 4 ( t ) ) Y 4 p ( t ) φ 2 ( Y 4 ( t ) ) α ( t ) f ( α ( t ) ) φ 1 ( Φ 2 1 ( Y 5 ( t ) ) ) ( Φ 2 1 ( Y 5 ( t ) ) ) p φ 2 ( Φ 2 1 ( Y 5 ( t ) ) ) , t < T .
(2.85)
By the definition of Φ 3 in (2.76), from (2.85) we obtain
Φ 3 ( Y 5 ( t ) ) Φ 3 ( Y 5 ( 0 ) ) + 0 t α ( s ) f ( α ( s ) ) d s Φ 3 ( Φ 2 ( u 0 ) + 0 α ( T ) g ( s ) d s ) + 0 α ( t ) f ( s ) d s , t < T .
(2.86)
Let t = T , from (2.86) we have
Φ 3 ( Y 5 ( T ) ) Φ 3 ( Φ 2 ( u 0 ) + 0 α ( T ) g ( s ) d s ) + 0 α ( T ) f ( s ) d s , t < T .
(2.87)
Since 0 < T < T 5 is chosen arbitrarily, from (2.77), (2.80), (2.84) and (2.87), we have
u ( t ) Φ 2 1 [ Φ 3 1 ( Φ 3 ( Φ 2 ( u 0 ) + 0 α ( t ) g ( s ) d s ) + 0 α ( t ) f ( s ) d s ) ] , t < T 5 .

This proves (2.74). □

3 Application

In this section we apply our Theorem 4 to the following differential-integral equation:
{ d x ( t ) d t = H ( t , x ( α ( t ) ) , 0 t K ( s , x ( α ( s ) ) ) d s ) , t I , x ( 0 ) = x 0 ,
(3.88)
where K C ( R × R , R ) , H C ( R 3 , R ) , | x 0 | > 0 is a constant, satisfy the following conditions:
(3.89)
(3.90)

where f, g are nonnegative real-valued continuous functions defined on I.

Corollary 1 Consider the nonlinear system (3.88) and suppose that K, H satisfy the conditions (3.89) and (3.90), and ψ 1 , ψ 2 , ψ 2 / ψ 1 , α C 1 ( I , I ) are increasing functions with α ( t ) t , ψ i ( t ) > 0 , t > 0 , i = 1 , 2 , α ( 0 ) = 0 . Then all the solutions of equation (3.88) exist on I and satisfy the following estimation:
| x ( t ) | Ψ 1 1 [ Ψ 2 1 ( Ψ 2 ( Ψ 1 ( | x 0 | ) + 0 α ( t ) g ( α 1 ( s ) ) α ( α 1 ( s ) ) d s ) + 0 α ( t ) f ( α 1 ( s ) ) α ( α 1 ( s ) ) d s ) ]
(3.91)
for all t < T 6 , where
and T 6 is the largest number such that
Proof Integrating both sides of equation (3.88) from 0 to t, we get
x ( t ) = x 0 + 0 t H ( s , x ( α ( s ) ) , 0 s K ( τ , x ( α ( τ ) ) ) d τ ) d s , t I .
(3.92)
Using the conditions (3.89) and (3.90), from (3.92) we obtain
| x ( t ) | | x 0 | + 0 t f ( s ) φ 1 ( | x ( α ( s ) ) | ) ( | x ( α ( s ) ) | + 0 s g ( τ ) φ 2 ( | x ( α ( τ ) ) | ) d τ ) p d s | x 0 | + 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 ) p d s
(3.93)

for all t I . Applying Theorem 4 to (3.93), we get the estimation (3.91). This completes the proof of Corollary 1. □

Declarations

Acknowledgements

The author is very grateful to the editor and the referees for their helpful comments and valuable suggestions. This research was supported by the 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, Yizhou, P.R. China

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© Wang; licensee Springer 2012

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