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Volume 2008, Article ID 312395, 8 pagesdoi:10.1155/2008/312395 Research Article On Some New Impulsive Integral Inequalities Jianli Li Department of Mathematics, Hunan Normal University,

Volume 2008, Article ID 312395, 8 pages

doi:10.1155/2008/312395

Research Article

On Some New Impulsive Integral Inequalities

Jianli Li

Department of Mathematics, Hunan Normal University, Changsha, Hunan 410081, China

Correspondence should be addressed to Jianli Li,ljianli@sina.com

Received 4 June 2008; Accepted 21 July 2008

Recommended by Wing-Sum Cheung

We establish some new impulsive integral inequalities related to certain integral inequalities arising in the theory of differential equalities The inequalities obtained here can be used as handy tools in the theory of some classes of impulsive differential and integral equations

Copyrightq 2008 Jianli Li 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

Differential and integral inequalities play a fundamental role in global existence,uniqueness, stability, and other properties of the solutions of various nonlinear differential equations; see1 4 A great deal of attention has been given to differential and integral inequalities; see 1, 2, 5 8 and the references given therein Motivated by the results in 1,5, 7, the main purpose of this paper is to establish some new impulsive integral inequalities similar to Bihari’s inequalities

Let 0≤ t0 < t1 < t2 < · · · , lim k→∞t k  ∞, R  0, ∞, and I ⊂ R, then we introduce

the following spaces of function:

P CR, I   {u : R → I, u is continuous for t / t k , u0, ut

k , and ut−

k exist, and

u t−

k   ut k , k  1, 2, },

P C1R, I   {u ∈ PCR, I  : u is continuously differentiable for t / t k , u0, ut

k,

and ut−

k  exist, and ut−

k   ut k , k  1, 2, }.

To prove our main results, we need the following resultsee 1, Theorem 1.4.1

Lemma 1.1 Assume that

A0 the sequence {t k } satisfies 0 ≤ t0< t1 < t2 < · · · , with lim k→∞t k  ∞;

A1 m ∈ PC1R, R and mt is left-continuous at t k , k  1, 2, ;

A2 for k  1, 2, , t ≥ t0,

mt ≤ ptmt  qt, t / t k ,

m

tk

where q, p ∈ PCR, R, d k ≥ 0, and b k are constants.

m t ≤ mt0 

t0<t k <t

d kexp

 t

t0

p sds

t

t0



s<t k <t

d kexp

 t s

p σdσq sds

t0<t k <t

 

t k <t j <t

d j

 exp

 t

t k

p sds



b k , t ≥ t0.

1.2

2 Main results

In this section, we will state and prove our results

Theorem 2.1 Let u, f ∈ PCR,R, b k ≥ 1, and c ≥ 0 be constants If

u2t ≤ c2 2

t

0

f susds  

0<t k <t



b2k− 1u2t k , 2.1

for t∈ R, then

u t ≤ c 

0<t k <t

b k





t

0

 

0<t k <t

b k



for t∈ R.

Proof Define a function z t by

z t  c  ε2 2

t

0

f susds  

0<t k <t



b2k− 1u2t k , 2.3

where ε > 0 is an arbitrary small constant For t /  t k, differentiating 2.3 and then using the

fact that ut ≤ z t, we have

and so

d

z t

dt  zt

2

For t  t k , we have zt

k  − zt k   b2

k − 1u2t k  ≤ b2

k − 1zt k ; thus zt

k  ≤ b2

k z t k Let

z t  xt; it follows that

xt ≤ ft, t / t k , t ≥ 0,

x

tk

x t ≤ x0 

0<t k <t

b k





t

0

 

s<t k <t

b k



f sds ≤ c  ε 

0<t k <t

b k





t

0

 

s<t k <t

b k



f sds.

2.7

Now by using the fact that ut ≤ z t  xt in 2.7 and then letting ε → 0, we get the

desired inequality in2.2 This proof is complete

Theorem 2.2 Let u, f ∈ PCR,R and b k ≥ 1 be constants, and let c be a nonnegative constant.

If

u2t ≤ c2 2

t

0

f su2s  hsusds 

0<t k <t



b2k− 1u2t k , 2.8

for t∈ R, then

u t ≤ c 

0<t k <t

b k

 exp

 t

0

f sds





t

0

 

s<t k <t

b k

 exp

 t

s

f τdτ



for t∈ R.

Proof This proof is similar to that ofTheorem 2.1; thus we omit the details here

Theorem 2.3 Let u, f, g, h ∈ PCR,R, c ≥ 0, and b k ≥ 1 be constants If

u2t ≤ c2 2

t

0

f sus



u s 

s

0

g τuτdτ



 hsus



ds 

0<t k <t



b2

k− 1u2t k ,

2.10

for t∈ R, then

u t ≤ c 

0<t k <t

b k





t

0

 

s<t k <t

b k



for t∈ R, where

a t  c 

0<t k <t

b k

 exp

 t

0

fτ  gτdτ





t

0

 

s<t k <t

b k

 exp

 t

s

fτ  gτdτ



h sds.

2.12

Proof Let ε > 0 be an arbitrary small constant, and define a function z t by

z t  c  ε2 2

t

0

f sus



u s 

s

0

g τuτdτ



 hsus



ds 

0<t k <t



b2k− 1u2t k .

2.13 Let

z t  xt; similar to the proof ofTheorem 2.1, we have

xt ≤ ft



x t 

t

0

g sxsds



 ht, t / t k ,

x

tk

≤ b k x t k , k  1, 2,

2.14

Set vt  xtt

0g sxsds; then vt ≥ xt, and so from 2.14 we get that xt ≤ ftvt

h t Thus, for t / t k,

vt  xt  gtxt ≤ ftvt  ht  gtxt ≤ ft  gtvt  ht, 2.15

and for t  t k,

v

tk

− vt k   xtk

− xt k  ≤ b k − 1xt k  ≤ b k − 1vt k , 2.16

and so vt

k  ≤ b k v t k ByLemma 1.1, we have

v t≤cε 

0<t k <t

b k

 exp

 t

0

fτgτdτ





t

0

 

s<t k <t

b k

 exp

 t

s

fτgτdτ



h sds.

2.17

Let ε→ 0, then we obtain

where at is defined in 2.12 Substituting 2.18 into 2.14, we have

xt ≤ ftat  ht, t / t k ,

x

tk

ApplyingLemma 1.1again, we obtain

x t ≤ c  ε 

0<t k <t

b k





t

0

 

s<t k <t

b k



Now using ut ≤ xt and letting ε → 0, we get the desired inequality in 2.11

Theorem 2.4 Let u, f, g, h ∈ PCR,R, c ≥ 0, and b k ≥ 1 be constants If

u2t ≤ c2 2

t

0

f sus

 s

0

g τuτdτ



 hsus



ds 

0<t k <t

b2

k − 1u2t k , 2.21

for t∈ R, then

u t ≤ c 

0<t k <t

b k

 exp

 t

0

f s

 s

0

g τdτ



ds





t

0

 

s<t k <t

b k

 exp

 t

s

f τ

 τ

0

g ωdω



dτ



h sds,

2.22

for t∈ R.

Proof Set

z t  c  ε2 2

t

0

f sus

 s

0

g τuτdτ



 hsus



ds 

0<t k <t



b2k− 1u2t k , 2.23

where ε is an arbitrary small constant; then zt is nondecreasing Let xt  z t, then it follows for t /  t kthat

xt ≤ ft

t

0

g sxsds  ht ≤



f t

t

0

g sds



since xt is nondecreasing Also, for t  t k , we have xt

k  ≤ b k x t k ApplyingLemma 1.1,

we obtain

x t ≤ c  εc 

0<t k <t

b k

 exp

 t

0

f s

 s

0

g τdτ



ds





t

0

 

s<t k <t

b k

 exp

 t

s

f τ

 τ

0

g ωdω



dτ



h sds.

2.25

Now by using the fact that ut ≤ xt in 2.25 and letting ε → 0, we get the inequality

2.22

Remark 2.5 If b k ≡ 1, then 2.1, 2.8, 2.10, and 2.21 have no impulses In this case, it is clear that Theorems2.2-2.3improve the corresponding results of5, Theorem 1

Theorem 2.6 Let u, f ∈ PCR,R, ht, s ∈ CR2

,R, for 0 ≤ s ≤ t < ∞, c ≥ 0, b k ≥ 1, and

p > 1 be constants Let g ∈ PCR,R be a nondecreasing function with gu > 0, for u > 0, and

g λu ≥ μλgu, for λ > 0, u ∈ R; here μλ > 0, for λ > 0 If

u p t ≤ c 

t

0

f sgus 

s

0

h s, σguσdσ



ds 

0<t k <t

b k − 1u p t k , 2.26

for t∈ R, then for 0 ≤ t < T,

u t ≤ G−1



G



c

0<t k <t

b k





t

0



s<t k <t

b k

μ

b 1/p k psds

1/p

where

p t  ft 

t

0

G r 

r

r0

ds

g

T  sup



t≥ 0 : G



c

0<t k <t

b k





t

0



s<t k <t

b k

μ

b k 1/p psds



∈ dom G−1

Proof We first assume that c > 0 and define a function z t by the right-hand side of 2.26

Then, zt > 0, z0  c, ut ≤ zt 1/p , and z t is nondecreasing For t / t k,

zt  ftgut 

t

0

h t, σguσdσ

≤ ftgzt 1/p



t

0

h t, σgzσ 1/p

dσ

≤ gzt 1/p f t 

t

0

h t, σdσ



,

2.31

and for t  t k , z t

k  ≤ b k z t k  As t ∈ 0, t1, from 2.31 we have

G zt − Gz0 

z t

z0

ds

g

s 1/p ≤

t

0

and so

z t ≤ G−1

G c 

t

0

p sds



Now assume that for 0≤ t ≤ t n, we have

z t ≤ G−1

G



c

0<t k <t

b k





t

0



0<t k <t

b k

μ

b k 1/p psds



Then, for t ∈ t n , t n1, it follows from 2.32 that Gzt ≤ Gzt

n t

t n p sds Using zt

k ≤

b k z t k, we arrive at

G zt ≤ Gb n z t n 

t

t n

From the supposition of g, we see that

G λu − Gλv 

λu

0

ds

g

s 1/p −

λv

0

ds

g

s 1/p ≤ λ

μ

λ 1/p Gu − Gv, for u ≥ v, λ > 0.

2.36

If Gzt n  ≤ Gcn−1

k1b k, then

G zt ≤ Gb n z t n 

t

t n

p sds ≤ G



c

n



k1

b k





t

0



s<t k <t

b k

μ

b 1/p k psds. 2.37

Otherwise, we have

G b n z t n  − G



c

0<t k <t

b k



μ

b n 1/p G zt n  − G



c

n−1



k1

b k



This implies, by induction hypothesis, that

G b n z t n  − G



c

0<t k <t

b k



μ

b 1/p n 

t n

0



s<t k <t n

b k

μ

b 1/p k psds 

t n

0



s<t k <t

b k

μ

b 1/p k psds.

2.39 Thus,2.35 and 2.39 yield, for 0 < t ≤ t n1,

G zt ≤ G



c

0<t k <t

b k





t

0



s<t k <t

b k

μ

and so

z t ≤ G−1 G



c

0<t k <t

b k





t

0



s<t k <t

b k

μ

b k 1/p psds



Using2.41 in ut ≤ zt 1/p, we have the required inequality in2.27

If c is nonnegative, we carry out the above procedure with c  ε instead of c, where

ε > 0 is an arbitrary small constant, and by letting ε → 0, we obtain 2.27 The proof is complete

Remark 2.7 If∞

r0ds/gs 1/p   ∞, then G∞  ∞ and the inequality in 2.27 is true for

t∈ R

An interesting and useful special version ofTheorem 2.6is given in what follows

Corollary 2.8 Let u, f, h, c, p, and b k be as in Theorem 2.6 If

u p t ≤ c 

t

0

f sus 

s

0

h s, σuσdσ



ds 

0<t k <t

b k − 1u p t k , 2.42

for t∈ R, then

0<t k <t

b k

p−1/p

p− 1

p

t

0



s<t k <t

b k p−1/p p sds

p/ p−1

for t∈ R, where p t is defined by 2.28.

Proof Let g u  u inTheorem 2.6 Then,2.26 reduces to 2.42 and

G r  p

p− 1r p−1/p − r

p−1/p

0 ,

G−1r  p− 1

p r  r0p−1/p

p/ p−1

.

2.44

Consequently, byTheorem 2.6, we have

0<t k <t

b k

p−1/p

p− 1

p

t

0



s<t k <t

b k p−1/p p sds

p/ p−1

This proof is complete

3 Application

Example 3.1 Consider the integrodifferential equations

xt − F



t, x t,

t

0

K t, s, xsds



 ht,

x

tk

 b k x t k , k  1, 2, ,

x 0  x0,

3.1

where 0  t0 < t1 < t2 < · · · with limk→∞t k  ∞; h : R → R and K : R2

 × R → R are

continuous; F : R × R2 → R is continuous at t / t k; limt →t

k F t, ·, · and lim t →t−

k F t, ·, · exist

and limt →t−

k F t, ·, ·  Ft, ·, ·; b k are constants with|b k | ≥ 1 k  1, 2,  Here, we assume that the solution xt of 3.1 exists on R Multiplying both sides of3.1 by xt and then integrating them from 0 to t, we obtain

x2t  x2

0 2

t

0

x sF



s, x s,

s

0

K s, τ, xτdτ



 hsxs



ds 

0<t k <t



b2k− 1x2t k .

3.2

We assume that

|Kt, s, xs| ≤ ftgs|xs|, |Ft, xt, v| ≤ ft|xt|  |v|, 3.3

where f, g ∈ CR,R From 3.2 and 3.3, we obtain

|xt|2≤|x0|22

t

0

f s|xs|



|xs|

s

0

g τ|xτ|dτ



|hs||xs|



ds

0<t k <t



|b k|2−1|xt k|2.

3.4 Now applyingTheorem 2.3, we have

|xt| ≤ |x0| 

0<t k <t

|b k|





t

0

 

s<t k <t

|b k|



where

a t|x0| 

0<t k <t

|b k|

 exp

t

0

fτgτdτ





t

0

 

s<t k <t

|b k|

 exp

t s

fτ  gτdτ



h sds,

3.6

for all t∈ R The inequality3.5 gives the bound on the solution xt of 3.1

Acknowledgments

This work is supported by the National Natural Science Foundation of ChinaGrants nos

10571050 and 60671066 The project is supported by Scientific Research Fund of Hunan Provincial Education Department 07B041 and Program for Young Excellent Talents at Hunan Normal University

References

1 V Lakshmikantham, D D Ba˘ınov, and P S Simeonov, Theory of Impulsive Differential Equations, vol 6

of Series in Modern Applied Mathematics, World Scientific, Singapore, 1989.

2 D D Ba˘ınov and P Simeonov, Integral Inequalities and Applications, vol 57 of Mathematics and Its Applications (East European Series), Kluwer Academic Publishers, Dordrecht, The Netherlands, 1992.

3 X Liu and Q Wang, “The method of Lyapunov functionals and exponential stability of impulsive

systems with time delay,” Nonlinear Analysis: Theory, Methods & Applications, vol 66, no 7, pp 1465–

1484, 2007

4 J Li and J Shen, “Periodic boundary value problems for delay differential equations with impulses,”

Journal of Computational and Applied Mathematics, vol 193, no 2, pp 563–573, 2006.

5 B G Pachpatte, “On some new inequalities related to certain inequalities in the theory of differential

equations,” Journal of Mathematical Analysis and Applications, vol 189, no 1, pp 128–144, 1995.

6 B G Pachpatte, “On some new inequalities related to a certain inequality arising in the theory of

differential equations,” Journal of Mathematical Analysis and Applications, vol 251, no 2, pp 736–751,

2000

7 B G Pachpatte, “Integral inequalities of the Bihari type,” Mathematical Inequalities & Applications, vol.

5, no 4, pp 649–657, 2002

8 N.-E Tatar, “An impulsive nonlinear singular version of the Gronwall-Bihari inequality,” Journal of Inequalities and Applications, vol 2006, Article ID 84561, 12 pages, 2006.

g sds



since xt is nondecreasing Also, for t  t k ,... ft  gtvt  ht, 2.15

and for t  t k,

v...



,

2.31

and for t  t k , z t

k