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Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2010, Article ID 430512, 18 pages
doi:10.1155/2010/430512
Research Article
A Note on the Integral Inequalities with
Two Dependent Limits
Allaberen Ashyralyev,
1, 2
Emine Misirli,
3
and Ozlem Mogol
3
1
Department of Mathematics, Fatih University, Buyukcekmece, 34500 Istanbul, Turkey
2
Department of Mathematics, ITTU, 74200 Ashgabat, Turkmenistan
3
Department of Mathematics, Ege University, 35100 Bornova-Izmir, Turkey
Correspondence should be addressed to Emine Misirli,
Received 4 October 2009; Revised 28 April 2010; Accepted 5 July 2010
Academic Editor: Martin Bohner
Copyright q 2010 Allaberen Ashyralyev et al. 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.
The theorem on the Gronwall’s type integral inequalities with two dependent limits is established.
In application, the boundedness of the solutions of nonlinear differential equations is presented.
1. Introduction
Integral inequalities play a significant role in the study of qualitative properties of solutions
of integral, differential and integro-differential equations see, e.g., 1–4 and the references
given therein. One of the most useful inequalities in the development of the theory of

t
0
f

s

ds
1.2
for all t ≥ 0.
2 Journal of Inequalities and Applications
Note that the generalization of this integral inequality and its discrete analogies are
given in papers 5–8. In paper 9 the following useful inequality with two dependent limits
was established.
Lemma 1.2. Let ut be a real-valued nonnegative continuous function defined on −T, T and let c
and a be nonnegative constants. Then the inequality
u

t

≤ c  sgn

t


t
−t
au

s



t


t
−t
m

s

u

s

ds
2.1
for t ∈ R, then
u

t

≤ c  sgn

t


t
−t
m



u
p

s

 h

s

u

s


ds
2.3
Journal of Inequalities and Applications 3
for t ∈ R, then
u

t



a

t

 b


× sgn

t


t
−t

a

s


g

s


1
p
h

s



p − 1
p
h

dr

ds

1/p
2.4
for all t ∈ R.
(iii) Let ct be a real-valued positive continuous and nondecreasing function defined on R and
p>1 be a real constant. If
u
p

t

≤ c
p

t

 b

t

sgn

t


t
−t


1  b

t

exp

sgn

t


t
−t
b

r


g

r

 h

r

c
1−p


−sgn

s


s
−s
b

r


g

r

 h

r

c
1−p

r

p

dr

ds


g

s

u
p

s

 h

s

u

s


ds
2.7
for t ∈ R, then
u

t



a


p
h

r


dr

×

sgn

t


t
0
sgn

s


s
−s

∂s
k

s, r



s


s
−s
k

r, r

b

r


g

r


1
p
h

r


dr

4 Journal of Inequalities and Applications



1
p
h

y


dy dr

ds
 sgn

t


t
−t
k

s, s

exp

−sgn

s





g

s


1
p
h

s



p − 1
p
h

s


B
k

t, s

ds



k−

t, s















exp


t
s

r
−r

∂r
k


2.10
B
k

t, s















exp


t
s

r
−r

∂r


, −t ≤ s ≤ 0.
2.11
Proof. i Define a function vt by
v

t

 c
2
 2sgn

t


t
−t
m

s

u

s

ds.
2.12
Note that vt is a nonnegative function and v0c
2
. Then 2.1 can be rewritten as

 2

t
−t
m

s

u

s

ds.
2.14
Differentiating 2.14 and using 2.13,weget
v


t

≤ 2m

t


v

t

 2m

−t

.
2.16
Integrating the last inequality from 0 to t,weget

v

t

≤ c 

t
0
m

s

ds 

t
0
m

−s

ds  c  sgn

t


Differentiating 2.18 and using 2.13,weget
−v


t

≤ 2m

t


v

t

 2m

−t


v

t

.
2.19
Dividing both sides of 2.19 by 2

vt,weget


t
m

s

ds 

0
t
m

−s

ds  c  sgn

t


t
−t
m

s

ds.
2.21
6 Journal of Inequalities and Applications
Finally, using 2.17 and 2.21,weobtain

v

g

s

u
p

s

 h

s

u

s


ds.
2.23
It is evident that vt is an even and n onnegative function. We have that
u
p

t

≤ a

t


u

t


a

t

 b

t

v

t

p

p − 1
p
.
2.25
Let t ≥ 0. Then
v

t





u
p

t

 h

t

u

t

 g

−t

u
p

−t

 h

−t

u

−t

 b

−t


g

−t


1
p
h

−t


 a

t


g

t


1
p
h


−t

.
2.28
Denoting
B

t

 b

t


g

t


1
p
h

t


,A

t


t

− v

t

B

t

 B

−t

≤ A

t

 A

−t

. 2.30
Journal of Inequalities and Applications 7
From that it follows that
exp


t


≤ exp


t
s

B

r

 B

−r

dr


A

s

 A

−s

2.31
for any s ≤ t. Integrating the last inequality from 0 to t and using v00, we get
v



ds. 2.32
It is easy to see that

t
s

B

r

 B

−r

dr 

t
−t
B

r

dr −

s
−s
B

r


exp



s
−s
B

r

dr

ds. 2.34
Since 0 ≤ s ≤ t, we have that
v

t

≤ exp

sgn

t


t
−t
B



s

exp



−s
s
B

r

dr

ds

 exp

sgn

t


t
−t
B

r


A

s

exp

−sgn

s


s
−s
B

r

dr

ds

 exp

sgn

t


t
−t

dr

.
2.35
8 Journal of Inequalities and Applications
Applying 2.24,weobtain
u

t



a

t

 b

t

exp

sgn

t


t
−t
B


ds

1/p
.
2.36
From 2.36,and2.29 it follows 2.4 for t ≥ 0. Let t ≤ 0; then
v

t

 −

t
−t

g

s

u
p

s

 h

s

u


u
p

−t

 h

−t

u

−t

.
2.37
Using 2.24 and 2.25,weget
−v


t

≤ v

t

B

t



v


s

 v

s

B

s

 B

−s


≤ exp


s
t

B

r

 B


−s

exp


s
t

B

r

 B

−r

dr

ds.
2.40
It is easy to see that

s
t

B

r


−t
t
B

r

dr


0
t

A

s

 A

−s

exp



−s
s
B

r


s

 A

−s

exp



−s
s
B

r

dr

ds
 exp

sgn

t


t
−t
B


−s

exp


s
−s
B

r

dr

ds

 exp

sgn

t


t
−t
B

r

dr



−s
s
B

r

dr

ds

 exp

sgn

t


t
−t
B

r

dr


−t
t
A


dr

sgn

t


t
−t
A

s

exp

−sgn

s


s
−s
B

r

dr

ds.

p
 h

s

c
1−p

s

u

s

c

s


ds.
2.44
Now the application of the inequality proven in ii yields the desired result in 2.6.
iv We define a function vt by
v

t

 sgn

t

We have that
u
p

t

≤ a

t

 b

t

v

t

,u

t



a

t

 b



s

 h

s

u

s


ds.
2.47
Differentiating 2.47,weget
v


t

 k

t, t


g

t

u


u

−t




t
−t

∂t
k

t, s


g

s

u
p

s

 h

s




1
p
h

t


 k

t, −t

b

−t


g

−t


1
p
h

−t




t, t


g

t

a

t

 h

t


1
p
a

t


p − 1
p

 k

t, −t

∂t
k

t, s


g

s

a

s

 h

s


1
p
a

s


p − 1
p

ds.

t
−t

∂t
k

t, s

B

s

ds

 k

t, t

A

t

 k

t, −t

A

−t


A

s

 k

s, −s

A

−s



s
−s

∂s
k

s, r

A

r

dr

× exp


B

y

dy

dr

.
2.51
Journal of Inequalities and Applications 11
Since kt, sk−t, s, we have that
v

t



t
0

k

s, s

A

s

 k


r, r

B

r

 k

−r, −r

B

−r



r
−r

∂r
k

r, y

B

y

dy


s
−s

∂s
k

s, r

A

r

dr
× exp



s
−s
k

r, r

B

r

dr 


exp



s
−s
k

r, r

B

r

dr 

t
s

r
−r

∂r
k

r, y

B

y

dr ds 

t
s

r
−r

∂r
k

r, y

B

y

dy dr

ds

.
2.53
Since 0 ≤ s ≤ t, we have that
v

t

≤ exp



s
−s

∂s
k

s, r

A

r

dr
× exp



s
−s
k

r, r

B

r

dr 


exp



s
−s
k

r, r

B

r

dr 

t
s

r
−r

∂r
k

r, y

B

y

dr 

t
−s

r
−r

∂r
k

r, y

B

y

dy dr

ds

.
2.54
12 Journal of Inequalities and Applications
Using 2.9 and 2.11,weget
v

t

≤ exp



s
−s

∂s
k

s, r

A

r

dr
× exp

−sgn

s


s
−s
k

r, r

B



t
0
k

s, s

A

s

exp

−sgn

s


s
−s
k

r, r

B

r

dr



B

r

dr

B
k

t, s

ds

 exp

sgn

t


t
−t
k

r, r

B

r

× exp

−sgn

s


s
−s
k

r, r

B

r

dr  sgn

t


t
s
sgn

r


r



s
−s
k

r, r

B

r

dr

A

s

B
k

t, s

ds

.
2.55
Let t ≤ 0. Then
v


2.56
Differentiating 2.56,weget
−v


t

 k

t, t


g

t

u
p

t

 h

t

u

t




t, s


g

s

u
p

s

 h

s

u

s


ds.
2.57
Journal of Inequalities and Applications 13
Using 2.46 and Young’s inequality, we obtain that
−v


t


b

−t


g

−t


1
p
h

−t




t
−t

∂t
k

t, s

b


 h

t


1
p
a

t


p − 1
p

 k

t, −t


g

−t

a

−t

 h


s

 h

s


1
p
h

s


p − 1
p

ds.
2.58
Using 2.29,weget
−v


t

≤ v

t



ds

 k

t, t

A

t

 k

t, −t

A

−t



−t
t

∂t
k

t, s

A




−s
s

∂s
k

s, r

A

r

dr

× exp


s
t

k

r, r

B

r



t



0
t

k

s, s

A

s

 k

−s, −s

A

−s



−s
s

∂s

−r



−r
r

∂r
k

r, y

B

y

dy

dr

ds.
2.61
14 Journal of Inequalities and Applications
Using 2.41,weget
v

t

≤ exp



dr
× exp



−s
s
k

r, r

B

r

dr 

s
t

−r
r

∂r
k

r, y

B

r

dr 

s
t

−r
r

∂r
k

r, y

B

y

dy dr

ds


0
t
k

−s, −s



r, y

B

y

dy dr ds

.
2.62
Since t ≤ s ≤ 0, we have that
v

t

≤ exp

sgn

t


t
−t
k

r, r

B


dr
× exp

−sgn

s


s
−s
k

r, r

B

r

dr 

s
t

−r
r

∂r
k



B

r

dr 

s
t

−r
r

∂r
k

r, y

B

y

dy dr

ds


−t
0
k

k

r, y

B

y

dy dr

ds

.
2.63
Using 2.9 and 2.10,weget
v

t

≤ exp

sgn

t


t
−t
k


A

r

dr
× exp

−sgn

s


s
−s
k

r, r

B

r

dr  sgn

t


t
s
sgn


exp

−sgn

s


s
−s
k

r, r

B

r

dr

B
k−

t, s

ds


−t
0

t, s

ds

Journal of Inequalities and Applications 15
 exp

sgn

t


t
−t
k

r, r

B

r

dr

×

sgn

t


k

r, r

B

r

dr  sgn

t


t
s
sgn

r

×

r
−r

∂r
k

r, y

B

B

r

dr

A

s

B
k

t, s

ds

.
2.64
The inequality 2.8 follows from 2.29, 2.55,and2.64. Theorem 2.1 is proved.
3. An Application
In this section, we indicate an application of Theorem 2.1 part ii to obtain the explicit
bound on the solution of the following boundary value problem for one dimensional partial
differential equations:
v
p
tt

t, x


 v

t, l

,v
x

t, 0

 v
x

t, l

,t∈ R,
v

0,x

 ϕ

x

,v
t

0,x

 ψ


2p

t, x

dx

1/2
 h

t



l
0
v
2

t, x

dx

1/2
3.2
for all t ∈ R. Here gt and ht are real-valued nonnegative continuous functions defined on
R.
This allows us to reduce the nonlocal boundary-value 3.1 to the initial-value problem
v
p
tt

the formula Aux−axu
x
x
x
 δux, with the domain DA{ux : u

x ∈
L
2
0,l,u0ul,u

0u

l} see, e.g., 15, 16.
Let us give a corollary of Theorem 2.1.
Theorem 3.1. The solution of problem 3.1 satisfies the estimates


l
0
v
2p

t, x

dx

1/2p




dr

× sgn

t


t
−t

M

g

s


1
p
l
1−1/p
h

s



p − 1
p


r


dr

ds

1/p
3.4
for all t ∈ R. Here M 

l
0
ϕ
2p
xdx
1/2
p/δ

l
0
ϕ
2p−1
xψ
2
xdx
1/2
.
Proof. It is known thatthe formula see, e.g., 15, 16 


t − s

F

s, v

s

ds
3.5
gives a solution of problem 3.3. Here
c

t


e
itA
1/2
 e
−itA
1/2
2
,s

t

 A
−1/2

−1/2



H →H

1

δ
,
3.7
we get

v
p
t

H


v
p
0

H

1

δ


s


vs

H

ds.
3.8
Journal of Inequalities and Applications 17
Since

v
p
0

H

1

δ



v
p


0


x

dx

1/2
,

vs

H
≤ l
1−1/p

v
p
s

1/p
H
3.9
we have that

v
p
t

H
≤ M 
1


1/p
H

ds.
3.10
Denote that utv
p
t
1/p
H
. Then
u
p

t

≤ M 
1

δ
sgn

t


t
−t

g


δ
exp

sgn

t


t
−t
1

δ

g

r


1
p
l
1−1/p
h

r


dr



s


× exp

−sgn

s


s
−s
1

δ

g

r


1
p
l
1−1/p
h

r



dx

1/2p
.
3.13
Therefore, the inequality 3.4 follows from the last inequality. Theorem 3.1 is proved.
Acknowledgments
The authors thank professor O. Celebi Turkey,professorR.P.AgarwalUSA,and
anonymous reviewers for their valuable comments.
18 Journal of Inequalities and Applications
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3 D. S. Mitrinovi
´
c, Analytic Inequalities, Springer, New York, NY, USA, 1970.
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