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

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.

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 thatu(t)andf(t)are nonnegative real-valued continuous functions defined onI=[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 ] ds,
(1.1)

for alltI, where u 0 >0andp(0,1)are constants. Then

u(t) u 0 1 p + 1 + 2 p + 1 0 t f(s) D 2 (s)ds,tI,
(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)

andβ(t)=exp(2 0 t f(s)ds), for alltI.

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 Supposeα C 1 (I,I)is increasing function withα(t)t, α(0)=0, tI. We assume thatu(t)andf(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 >0andp(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 ) ) ds,tI,
(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)ds), for alltI.

Remark 1 Ifα(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 ) ) ,tI.
(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),tI,
(2.6)

where

Y 1 (t):= z 1 p (t)+2 0 α ( t ) f(λ) z 1 p (λ)dλ,tI.
(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 ) ) ,tI.
(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 ) ) ,tI.
(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 tI. 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),tI,
(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),tI.

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 ) ) ds,tI.

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

Theorem 2 Supposeα C 1 (I,I)is increasing function withα(t)t, α(0)=0, tI. We assume thatu(t)andf(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 >0andp(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 ) ) ds,tI,
(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)ds), for alltI.

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 ) ) ,tI.
(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),tI,
(2.20)

where

Y 2 (t):= z 2 (t)+ 0 α ( t ) f(λ) z 2 (λ)dλ+ 0 α ( t ) f(λ) z 2 p (λ)dλ,tI.
(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 ) ) ,tI.
(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)(1p) α (t)f ( α ( t ) ) ,tI.
(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 tI. 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),tI,
(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),tI.

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 ) ) ds,tI.

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 withα(t)t, ϕ i (t)>0, t>0, i=1,2, α(0)=0. We assume thatu(t)andf(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 >0is 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 allt 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),tI.
(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 ) ) ds+ 0 α ( t ) g(λ) ϕ 2 ( z 3 ( λ ) ) dλ,tI.
(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 tI. 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 ) ) ,tI.

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 tI, 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))ds 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 ) 2f(s)ds.
(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 FC(I×I,R), HC( 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 ) ) | 2f(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 withα(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 allt< 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 allt 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 ) ) ) ds+ 0 t H ( s , x ( α ( s ) ) ) ds,tI.
(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 tI. Applying Theorem 3 to (3.6), we get the estimation (3.4). This completes the proof of the Corollary 1. □

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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).

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Wang, WS. A class of retarded nonlinear integral inequalities and its application in nonlinear differential-integral equation. J Inequal Appl 2012, 154 (2012). https://doi.org/10.1186/1029-242X-2012-154

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Keywords

  • integral inequality
  • analysis technique
  • retarded differential-integral equation
  • estimation