Volume 2010, Article ID 430512, 18 pagesdoi:10.1155/2010/430512 Research Article A Note on the Integral Inequalities with Two Dependent Limits 1 Department of Mathematics, Fatih Universi
Trang 1Volume 2010, Article ID 430512, 18 pages
doi:10.1155/2010/430512
Research Article
A Note on the Integral Inequalities with
Two Dependent Limits
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,emine.misirli@gmail.com
Received 4 October 2009; Revised 28 April 2010; Accepted 5 July 2010
Academic Editor: Martin Bohner
Copyrightq 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 differential equations is given in the following lemma see 5
Lemma 1.1 Let ut and ft be real-valued nonnegative continuous functions for all t ≥ 0 If
u2t ≤ c2 2
t
0
f susds 1.1
for all t ≥ 0, where c ≥ 0 is a real constant, then
u t ≤ c
t
0
for all t ≥ 0.
Trang 22 Journal of Inequalities and Applications
Note that the generalization of this integral inequality and its discrete analogies are given in papers5 8 In paper 9 the following useful inequality with two dependent limits was established
Lemma 1.2 Let ut 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 sgnt
t
−t au sds, −T ≤ t ≤ T 1.3
implies that
u t ≤ ce 2a|t| , −T ≤ t ≤ T. 1.4
The theory of integral inequalities with several dependent limits and its applications
to differential equations has been investigated in 10–14
The present study involves some Gronwall’s type integral inequalities with two dependent limits.Section 2includes some new integral inequalities with two dependent lim-its and relevant proofs Subsequently,Section 3includes an application on the boundedness
of the solutions of nonlinear differential equations
2 A Main Statement
Our main statement is given by the following theorem
Theorem 2.1 Let ut, at, bt, gt, ht, and mt be real-valued nonnegative continuous
(i) Let c be a nonnegative constant If
u2t ≤ c2 2 sgnt
t
−t m susds 2.1
for t ∈ R, then
u t ≤ c sgnt
t
−t m sds 2.2
for all t ∈ R.
(ii) Let p > 1 be a real constant If
u p t ≤ at bt sgnt
t
−t
g su p s hsusds 2.3
Trang 3for t ∈ R, then
u t ≤
a t bt exp
sgnt
t
−t b r
g r 1
p h r
dr
× sgnt
t
−t a s
g s 1
p h s
p− 1
× exp − sgns
s
−s b r
g r 1
p h r
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 bt sgnt
t
−t
g su p s hsusds 2.5
for t ∈ R, then
u t ≤ ct
1 bt exp
sgnt
t
−t b r g r hr c1−pr
p
dr
× sgnt
t
−t
g s hsc1−ps
× exp
− sgns
s
−s b r g r hr c1−pr
p
dr
ds
1/p
2.6
for all t ∈ R.
(iv) Let k t, s and its partial derivative ∂kt, s/∂t be real-valued nonnegative continuous
functions on −∞ < s ≤ t < ∞ and let kt, s be even function in t If
u p t ≤ at bt sgnt
t
−t k t, sg su p s hsusds 2.7
for t ∈ R, then
u t ≤
a t bt exp
sgnt
t
−t k r, rbr
g r 1
p h r
dr
×
sgnt
t
0
sgns
s
−s
∂
∂s k s, r
a r
g r 1
p h r
p− 1
dr
× exp − sgns
s
−s k r, rbr
g r 1
p h r
dr
Trang 4
4 Journal of Inequalities and Applications
× exp sgnt
t
s
sgnr
r
−r
∂
∂r k
r, y
b
g
y
1
p h
dy dr
ds
sgnt
t
−t k s, s exp
− sgns
s
−s k r, rbr
g r 1
p h r
dr
×
a s
g s 1
p h s
p− 1
B k t, sds
1/p
.
2.8
for all t ∈ R Here
B k t, s
⎧
⎪
⎪
B kt, s, t ≥ 0, s ∈ R,
B k−t, s, t ≤ 0, s ∈ R, 2.9
where
B k−t, s
⎧
⎪
⎪
⎪
⎪
exp
t s
r
−r
∂
∂r k
r, y
B
y
dy dr
, t ≤ s ≤ 0,
exp
t
−s
r
−r
∂
∂r k
r, y
B
y
dy dr
, 0≤ s ≤ −t,
2.10
B kt, s
⎧
⎪
⎪
⎪
⎪
exp
t s
r
−r
∂
∂r k
r, y
B
y
dy dr
, 0≤ s ≤ t,
exp
t
−s
r
−r
∂
∂r k
r, y
B
y
dy dr
, −t ≤ s ≤ 0.
2.11
v t c2 2 sgnt
t
−t m susds. 2.12
Note that vt is a nonnegative function and v0 c2 Then2.1 can be rewritten as
u2t ≤ vt, u t ≤v t. 2.13
It is easy to see that vt is an even function.
Trang 5First, let t≥ 0; then 2.12 can be rewritten as
v t c2 2
t
−t m susds. 2.14 Differentiating 2.14 and using 2.13, we get
vt ≤ 2mtv t 2m−tv t. 2.15
Dividing both sides of2.15 by 2v t, we get
vt
2
v t ≤ mt m−t. 2.16
Integrating the last inequality from 0 to t, we get
v t ≤ c
t
0
m sds
t
0
m −sds c sgnt
t
−t m sds. 2.17
Second, let t≤ 0 Then, 2.12 can be written as
v t c2− 2
t
−t m susds. 2.18 Differentiating 2.18 and using 2.13, we get
−vt ≤ 2mtv t 2m−tv t. 2.19 Dividing both sides of2.19 by 2v t, we get
− vt
2
v t ≤ mt m−t. 2.20 Integrating2.20 from t to 0, we get
v t ≤ c
0
t
m sds
0
t
m −sds c sgnt
t
−t m sds. 2.21
Trang 66 Journal of Inequalities and Applications
Finally, using2.17 and 2.21, we obtain
v t ≤ c sgnt
t
−t m sds. 2.22
The inequality2.2 follows from 2.13 and 2.22
ii Define a function vt by
v t sgnt
t
−t
g su p s hsusds. 2.23
It is evident that vt is an even and nonnegative function We have that
u p t ≤ at btvt, u t ≤ at btvt 1/p 2.24 Using Young’s inequalitysee, e.g., 2, we obtain that
u t ≤ a t btvt
Let t≥ 0 Then
v t
t
−t g su p s hsusds. 2.26 Differentiating 2.26, we get
vt gtu p t htut g−tu p −t h−tu−t. 2.27 Using2.24 and 2.25, we get
vt ≤ vt b t
g t 1
p h t
b−t
g −t 1
p h −t
at
g t 1
p h t
a−t
g −t 1
p h −t
p− 1
p ht h−t.
2.28
Denoting
B t bt
g t 1
p h t
, A t at
g t 1
p h t
p− 1
p h t, 2.29
we get
vt − vtBt B−t ≤ At A−t. 2.30
Trang 7From that it follows that
exp
t s
Br B−rdr
vs − vsBs B−s
≤ exp t
s
Br B−rdr
As A−s
2.31
for any s ≤ t Integrating the last inequality from 0 to t and using v0 0, we get
v t ≤
t
0
As A−s exp t
s
Br B−rdr
ds. 2.32
It is easy to see that
t
s
Br B−rdr
t
−t B rdr −
s
−s B rdr. 2.33 Then
v t ≤ exp t
−t B rdr
t
0
As A−s exp
−
s
−s B rdr
ds. 2.34
Since 0≤ s ≤ t, we have that
v t ≤ exp sgnt
t
−t B rdr
×
t
0
A s exp
−
s
−s B rdr
0
−t A s exp
−
−s
s
B rdr
ds
exp sgnt
t
−t B rdr
×
t
0
A s exp
− sgns
s
−s B rdr
0
−t A s exp
− sgns
s
−s B rdr
ds
exp sgnt
t
−t B rdr
sgnt
t
−t A s exp
− sgns
s
−s B rdr
.
2.35
Trang 88 Journal of Inequalities and Applications
Applying2.24, we obtain
u t ≤
a t bt exp sgnt
t
−t B rdr
× sgnt
t
−t A s exp
− sgns
s
−s B rdr
ds
1/p
.
2.36
From2.36, and 2.29 it follows 2.4 for t ≥ 0 Let t ≤ 0; then
v t −
t
−t
g su p s hsusds,
−vt gtu p t htut g−tu p −t h−tu−t.
2.37
Using2.24 and 2.25, we get
−vt ≤ vtBt B−t At A−t. 2.38 From that it follows that
− exp
s
t
Br B−rdr
vs vsBs B−s
≤ exps
t
Br B−rdr
As A−s
2.39
for any t ≤ s Integrating the last inequality from t to 0 and using v0 0, we get
v t ≤
0
t
As A−s exp
s t
Br B−rdr
ds. 2.40
It is easy to see that
s
t
Br B−rdr
−t
t
B rdr −
−s
s
B rdr. 2.41
Then
v t ≤ exp −t
t
B rdr
0
t
As A−s exp
−
−s
s
B rdr
ds. 2.42
Trang 9Since t ≤ s ≤ 0, we have that
v t ≤ exp sgnt
t
−t B rdr
0
t
As A−s exp
−
−s
s
B rdr
ds
exp sgnt
t
−t B rdr
×
0
t
A s exp
s
−s B rdr
0
t
A −s exp
s
−s B rdr
ds
exp sgnt
t
−t B rdr
×
0
t
A s exp
s
−s B rdr
−t
0
A s exp
−s
s
B rdr
ds
exp sgnt
t
−t B rdr
−t
t
A s exp
− sgns
s
−s B rdrds
exp sgnt
t
−t B rdr
sgnt
t
−t A s exp
− sgns
s
−s B rdr
ds.
2.43
Applying2.43 and 2.24, we obtain 2.36 for t ≤ 0 Then from 2.36 and 2.29, 2.4
follows for t ≤ 0.
iii Since ct is a positive, continuous, and nondecreasing function for t ∈ R, we have
that
u t
c t
p
≤ 1 bt sgnt
t
−t g s
u s
c s
p
hsc1−ps u s
c s
ds. 2.44
Now the application of the inequality proven inii yields the desired result in 2.6
iv We define a function vt by
v t sgnt
t
−t k t, sg su p s hsusds. 2.45
Evidently, the function vt is a nonnegative, monotonic, and nondecreasing in t and v0 0.
We have that
u p t ≤ at btvt, u t ≤ at btvt 1/p 2.46
Trang 1010 Journal of Inequalities and Applications
Let t≥ 0 Then
v t
t
−t k t, sg su p s hsusds. 2.47 Differentiating 2.47, we get
vt kt, tg tu p t htut kt, −tg −tu p −t h−tu−t
t
−t
∂
∂t k t, sg su p s hsusds.
2.48
Using2.46 and Young’s inequality, we obtain that
vt ≤ vt k t, tbt
g t 1
p h t
kt, −tb−t
g −t 1
p h −t
t
−t
∂
∂t k t, sbs
g s 1
p h s
ds
kt, t g tat ht
1
p a t p− 1
p
kt, −t g −ta−t h−t
1
p a −t p− 1
p
t
−t
∂
∂t k t, s g sas hs
1
p a s p− 1
p
ds.
2.49
Using2.29, we get
vt ≤ vt k t, tBt kt, −tB−t
t
−t
∂
∂t k t, sBsds
kt, tAt kt, −tA−t
t
−t
∂
∂t k t, sAsds.
2.50
Applying the differential inequality, we get
v t ≤
t
0
k s, sAs ks, −sA−s
s
−s
∂
∂s k s, rArdr
× exp
t s
k r, rBr kr, −rB−r
r
−r
∂
∂r k
r, y
B
y
dy
dr
.
2.51
Trang 11Since kt, s k−t, s, we have that
v t ≤
t
0
k s, sAs k−s, −sA−s
s
−s
∂
∂s k s, rArdr
× exp
t s
k r, rBr k−r, −rB−r
r
−r
∂
∂r k
r, y
B
y
dy
dr
ds.
2.52
Using2.33, we get
v t ≤ exp t
−t k r, rBrdr
×
t
0
s
−s
∂
∂s k s, rArdr
× exp −
s
−s k r, rBrdr
t
s
r
−r
∂
∂r k
r, y
B
y
dy dr
ds
t
0
k s, sAs exp −
s
−s k r, rBrdr
t
s
r
−r
∂
∂r k
r, y
B
y
dy dr
ds
t
0
k −s, −sA−sexp −
s
−s k r, rBrdr 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 sgnt
t
−t k r, rBrdr
×
sgnt
t
0
sgns
s
−s
∂
∂s k s, rArdr
× exp −
s
−s k r, rBrdr
t
s
r
−r
∂
∂r k
r, y
B
y
dy dr
ds
t
0
k s, sAs exp −
s
−s k r, rBrdr
t
s
r
−r
∂
∂r k
r, y
B
y
dy dr
ds
0
−t k s, sAs exp −
−s
s
k r, rBrdr
t
−s
r
−r
∂
∂r k
r, y
B
y
dy dr
ds
.
2.54
Trang 1212 Journal of Inequalities and Applications
Using2.9 and 2.11, we get
v t ≤ exp sgnt
t
−t k r, rBrdr
×
sgnt
t
0
sgns
s
−s
∂
∂s k s, rArdr
× exp − sgns
s
−s k r, rBrdr sgnt
t
s
sgnr
r
−r
∂
∂r k
r, y
B
y
dy dr
ds
t
0
k s, sAs exp
− sgns
s
−s k r, rBrdr
B kt, sds
0
−t k s, sAs exp
− sgns
s
−s k r, rBrdr
B kt, sds
exp sgnt
t
−t k r, rBrdr
×
sgnt
t
0
sgns
s
−s
∂
∂s k s, rArdr
× exp − sgns
s
−s k r, rBrdr sgnt
t
s
sgnr
r
−r
∂
∂r k
r, y
B
y
dy dr
ds
sgnt
0
−t k s, s exp
− sgns
s
−s k r, rBrdr
A sB k t, sds
.
2.55
Let t≤ 0 Then
v t −
t
−t k t, sg su p s hsusds. 2.56 Differentiating 2.56, we get
−vt kt, tg tu p t htut kt, −tg −tu p −t h−tu−t
t
−t
∂
∂t k t, sg su p s hsusds.
2.57
Trang 13Using2.46 and Young’s inequality, we obtain that
−vt ≤ vt k t, tbt
g t 1
p h t
kt, −tb−t
g −t 1
p h −t
t
−t
∂
∂t k t, sbs
g s 1
p h s
ds
kt, t g tat ht
1
p a t p− 1
p
kt, −t g −ta−t h−t
1
p a −t p− 1
p
−t
t
∂
∂t k t, s g sas hs
1
p h s p− 1
p
ds.
2.58
Using2.29, we get
−vt ≤ vt k t, tBt kt, −tB−t
−t
t
∂
∂t k t, sBsds
kt, tAt kt, −tA−t
−t
t
∂
∂t k t, sAsds.
2.59
Applying the differential inequality, we get
v t ≤
0
t
k s, sAs ks, −sA−s
−s
s
∂
∂s k s, rArdr
× exp s
t
k r, rBr kr, −rB−r
−r
r
∂
∂r k
r, y
B
y
dy
dr
ds.
2.60
Since kt, s k−t, s, we have that
v t ≤
0
t
k s, sAs k−s, −sA−s
−s
s
∂
∂s k s, rArdr
× exp s
t
k r, rBr k−r, −rB−r
−r
r
∂
∂r k
r, y
B
y
dy
dr
ds.
2.61
Trang 1414 Journal of Inequalities and Applications
Using2.41, we get
v t ≤ exp −t
t
k r, rBrdr
×
0
t
−s
s
∂
∂s k s, rArdr
× exp
−
−s
s
k r, rBrdr
s
t
−r
r
∂
∂r k
r, y
B
y
dy dr
ds
0
t
k s, sAs exp
−
−s
s
k r, rBrdr
s
t
−r
r
∂
∂r k
r, y
B
y
dy dr
ds
0
t
k −s, −sA−s exp
−
−s
s
k r, rBrdr
ds
s
t
−r
r
∂
∂r k
r, y
B
y
dy dr ds
.
2.62
Since t ≤ s ≤ 0, we have that
v t ≤ exp sgnt
t
−t k r, rBrdr
×
sgnt
t
0
sgns
s
−s
∂
∂s k s, rArdr
× exp
− sgns
s
−s k r, rBrdr
s
t
−r
r
∂
∂r k
r, y
B
y
dy dr
ds
0
t
k s, sAs exp
−
−s
s
k r, rBrdr
s
t
−r
r
∂
∂r k
r, y
B
y
dy dr
ds
−t
0
k s, sAs exp
−
s
−s k r, rBrdr
−s
t
−r
r
∂
∂r k
r, y
B
y
dy dr
ds
.
2.63 Using2.9 and 2.10, we get
v t ≤ exp sgnt
t
−t k r, rBrdr
×
sgnt
t
0
sgns
s
−s
∂
∂s k s, rArdr
× exp − sgns
s
−s k r, rBrdr sgnt
t
s
sgnr
r
−r
∂
∂r k
r, y
B
y
dy dr
ds
0
t
k s, sAs exp
− sgns
s
−s k r, rBrdr
B k−t, sds
−t
0
k s, sAs exp
− sgns
s
−s k r, rBrdr
B k−t, sds
Trang 15
exp sgnt
t
−t k r, rBrdr
×
sgnt
t
0
sgns
s
−s
∂
∂s k s, rArdr
× exp − sgns
s
−s k r, rBrdr sgnt
t
s
sgnr
×
r
−r
∂
∂r k
r, y
B
y
dy dr
ds
sgnt
−t
0
k s, s exp
− sgns
s
−s k r, rBrdr
A sB k t, sds
.
2.64 The inequality2.8 follows from 2.29, 2.55, and 2.64.Theorem 2.1is proved
3 An Application
In this section, we indicate an application of Theorem 2.1part ii to obtain the explicit bound on the solution of the following boundary value problem for one dimensional partial differential equations:
v tt p t, x −a xv p
x t, x
x δv p t, x Ft, x; vt, x, t ∈ R, 0 < x < l,
v t, 0 vt, l, v x t, 0 v x t, l, t ∈ R,
v 0, x ϕx, v t 0, x ψx, 0 ≤ x ≤ l,
3.1
where p > 1 is a fixed real number and δ const > 0 Let Ft, x; vt, x, t ∈ R, x ∈ 0, l,
a x ≥ a > 0, x ∈ 0, l, ϕx, ψx, x ∈ 0, l be smooth functions and problem 3.1 has a
unique smooth solution vt, x Assume that
l
0
F2t, x; vt, xdx
1/2
≤ gt l
0
v 2p t, xdx
1/2
ht l
0
v2t, xdx
1/2
3.2
for all t ∈ R Here gt and ht are real-valued nonnegative continuous functions defined on
R
This allows us to reduce the nonlocal boundary-value3.1 to the initial-value problem
v tt p t Av p t Ft, vt, t ∈ R,
v 0 ϕ, v t 0 ψ 3.3
...−t m sds. 2.21
Trang 66 Journal of Inequalities and Applications
Finally,...
2.54
Trang 1212 Journal of Inequalities and Applications
Using2.9 and 2.11, we get
v...
ds.
2.61
Trang 1414 Journal of Inequalities and Applications
Using2.41, we get
v