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 Mogol3
1
Department of Mathematics, Fatih University, Buyukcekmece, 34500 Istanbul, Turkey
Department of Mathematics, ITTU, 74200 Ashgabat, Turkmenistan
3
Department of Mathematics, Ege University, 35100 Bornova-Izmir, Turkey
2
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
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
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
u2 t ≤ c2 2
t
fsusds
1.1
0
for all t ≥ 0, where c ≥ 0 is a real constant, then
ut ≤ c t
fsds
0
for all t ≥ 0.
1.2
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 ut be a real-valued nonnegative continuous function defined on −T, T and let c
and a be nonnegative constants. Then the inequality
ut ≤ c sgnt
t
−t
ausds,
−T ≤ t ≤ T
1.3
implies that
ut ≤ ce2a|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 2 includes some new integral inequalities with two dependent limits and relevant proofs. Subsequently, Section 3 includes 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
functions defined on R −∞, ∞.
(i) Let c be a nonnegative constant. If
u2 t ≤ c2 2 sgnt
t
−t
msusds
2.1
msds
2.2
for t ∈ R, then
ut ≤ c sgnt
t
−t
for all t ∈ R.
(ii) Let p > 1 be a real constant. If
u t ≤ at bt sgnt
p
t
−t
gsup s hsus ds
2.3
Journal of Inequalities and Applications
3
for t ∈ R, then
ut ≤
1
at bt exp sgnt
br gr hr dr
p
−t
t
t p−1
1
hs
× sgnt
as gs hs p
p
−t
s
1/p
1
× exp − sgns
br gr hr dr ds
p
−s
2.4
for all t ∈ R.
(iii) Let ct be a real-valued positive continuous and nondecreasing function defined on R and
p > 1 be a real constant. If
up t ≤ cp t bt sgnt
t
−t
gsup s hsus ds
2.5
for t ∈ R, then
t
c1−p r
ut ≤ ct 1 bt exp sgnt
br gr hr
p
−t
dr
t gs hsc1−p s
× sgnt
2.6
−t
s
c1−p r
× exp − sgns
br gr hr
p
−s
1/p
dr ds
for all t ∈ R.
(iv) Let kt, 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 t ≤ at bt sgnt
p
t
−t
kt, s gsup s hsus ds
for t ∈ R, then
ut ≤
1
at bt exp sgnt
kr, rbr gr hr dr
p
−t
s
t
p−1
∂
1
ks, r ar gr hr hr dr
× sgnt sgns
p
p
0
−s ∂s
s
1
× exp − sgns
kr, rbr gr hr dr
p
−s
t
2.7
4
× exp sgnt
t
r
sgnr
s
t
−r
Journal of Inequalities and Applications
∂ 1 k r, y b y g y h y dy dr ds
∂r
p
s
1
sgnt
ks, s exp − sgns
kr, rbr gr hr dr
p
−t
−s
1/p
p−1
1
× as gs hs hs Bk t, sds
.
p
p
2.8
for all t ∈ R. Here
Bk t, s ⎧
⎪
⎨Bk t, s,
t ≥ 0, s ∈ R,
⎪
⎩Bk− t, s,
t ≤ 0, s ∈ R,
2.9
where
t r
⎧
⎪
∂ ⎪
⎪
k r, y B y dy dr ,
exp
⎪
⎪
⎨
s −r ∂r
Bk− t, s t r
⎪
⎪
∂
⎪
⎪exp
⎪
k r, y B y dy dr ,
⎩
−s −r ∂r
t r
⎧
⎪
∂ ⎪
⎪
k r, y B y dy dr ,
exp
⎪
⎪
⎨
s −r ∂r
Bk t, s t r
⎪
⎪
∂ ⎪
⎪
⎪
k r, y B y dy dr ,
⎩exp
−s −r ∂r
t ≤ s ≤ 0,
2.10
0 ≤ s ≤ −t,
0 ≤ s ≤ t,
2.11
−t ≤ s ≤ 0.
Proof. i Define a function vt by
vt c2 2 sgnt
t
−t
msusds.
2.12
Note that vt is a nonnegative function and v0 c2 . Then 2.1 can be rewritten as
u2 t ≤ vt,
It is easy to see that vt is an even function.
ut ≤
vt.
2.13
Journal of Inequalities and Applications
5
First, let t ≥ 0; then 2.12 can be rewritten as
vt c 2
2
t
−t
2.14
msusds.
Differentiating 2.14 and using 2.13, we get
v t ≤ 2mt vt 2m−t vt.
2.15
Dividing both sides of 2.15 by 2 vt, we get
v t
≤ mt m−t.
2 vt
2.16
Integrating the last inequality from 0 to t, we get
vt ≤ c t
msds 0
t
m−sds c sgnt
0
t
−t
msds.
2.17
Second, let t ≤ 0. Then, 2.12 can be written as
vt c2 − 2
t
−t
2.18
msusds.
Differentiating 2.18 and using 2.13, we get
−v t ≤ 2mt vt 2m−t vt.
2.19
Dividing both sides of 2.19 by 2 vt, we get
v t
− ≤ mt m−t.
2 vt
2.20
Integrating 2.20 from t to 0, we get
vt ≤ c 0
t
msds 0
t
m−sds c sgnt
t
−t
msds.
2.21
6
Journal of Inequalities and Applications
Finally, using 2.17 and 2.21, we obtain
t
vt ≤ c sgnt
msds.
−t
2.22
The inequality 2.2 follows from 2.13 and 2.22.
ii Define a function vt by
vt sgnt
t
−t
gsup s hsus ds.
2.23
It is evident that vt is an even and nonnegative function. We have that
up t ≤ at btvt,
ut ≤ at btvt1/p .
2.24
Using Young’s inequality see, e.g., 2, we obtain that
at btvt p − 1
.
p
p
ut ≤
2.25
Let t ≥ 0. Then
vt t
−t
gsup s hsusds.
2.26
Differentiating 2.26, we get
v t gtup t htut g−tup −t h−tu−t.
2.27
Using 2.24 and 2.25, we get
1
1
v t ≤ vt bt gt ht b−t g−t h−t
p
p
p−1
1
1
at gt ht a−t g−t h−t ht h−t.
p
p
p
2.28
Denoting
1
Bt bt gt ht ,
p
p−1
1
At at gt ht ht,
p
p
2.29
we get
v t − vtBt B−t ≤ At A−t.
2.30
Journal of Inequalities and Applications
7
From that it follows that
t
exp
Br B−rdr
v s − vsBs B−s
s
≤ exp
t
2.31
Br B−rdr As A−s
s
for any s ≤ t. Integrating the last inequality from 0 to t and using v0 0, we get
vt ≤
t
t
As A−s exp
0
Br B−rdr ds.
2.32
s
It is easy to see that
t
Br B−rdr t
−t
s
Brdr −
s
−s
2.33
Brdr.
Then
t
vt ≤ exp
t
−t
Brdr
s
Brdr ds.
As A−s exp −
2.34
−s
0
Since 0 ≤ s ≤ t, we have that
vt ≤ exp sgnt
×
t
t
−t
Brdr
0
s
−s
As exp −
Brdr ds As exp −
Brdr ds
−s
0
exp sgnt
×
t
−t
−t
Brdr
As exp − sgns
0
exp sgnt
s
t
−s
Brdr ds t
−t
s
Brdr
t
sgnt
−t
0
−t
As exp − sgns
As exp − sgns
s
−s
s
−s
Brdr ds
Brdr .
2.35
8
Journal of Inequalities and Applications
Applying 2.24, we obtain
ut ≤ at bt exp sgnt
× sgnt
t
−t
t
−t
Brdr
s
1/p
As exp − sgns
Brdr ds
.
2.36
−s
From 2.36, and 2.29 it follows 2.4 for t ≥ 0. Let t ≤ 0; then
vt −
t
−t
gsup s hsus ds,
2.37
−v t gtu t htut g−tu −t h−tu−t.
p
p
Using 2.24 and 2.25, we get
−v t ≤ vtBt B−t At A−t.
2.38
From that it follows that
− exp
s
Br B−rdr
t
≤ exp
s
v s vsBs B−s
2.39
Br B−rdr As A−s
t
for any t ≤ s. Integrating the last inequality from t to 0 and using v0 0, we get
vt ≤
0
As A−s exp
s
t
Br B−rdr ds.
2.40
t
It is easy to see that
s
Br B−rdr t
−t
t
Brdr −
−s
Brdr.
2.41
s
Then
−t
vt ≤ exp
0
Brdr
t
t
−s
Brdr ds.
As A−s exp −
s
2.42
Journal of Inequalities and Applications
9
Since t ≤ s ≤ 0, we have that
vt ≤ exp sgnt
exp sgnt
×
0
−t
t
−t
s
t
Brdr
−s
t
exp sgnt
×
Brdr
−s
Brdr ds
As A−s exp −
0
s
s
As exp
Brdr ds A−s exp
Brdr ds
0
0
t
−t
−s
t
t
2.43
Brdr
−t
s
−s
As exp
Brdr ds As exp
Brdr ds
−s
t
exp sgnt
exp sgnt
t
−t
Brdr
As exp − sgns
Brdr
s
t
t
−t
0
−t
t
sgnt
−t
s
−s
Brdrds
s
As exp − sgns
Brdr ds.
−s
Applying 2.43 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
ut
ct
p
≤ 1 bt sgnt
t −t
us
us p
1−p
ds.
hsc s
gs
cs
cs
2.44
Now the application of the inequality proven in ii yields the desired result in 2.6.
iv We define a function vt by
vt sgnt
t
−t
kt, s gsup s hsus ds.
2.45
Evidently, the function vt is a nonnegative, monotonic, and nondecreasing in t and v0 0.
We have that
up t ≤ at btvt,
ut ≤ at btvt1/p .
2.46
10
Journal of Inequalities and Applications
Let t ≥ 0. Then
vt t
−t
kt, s gsup s hsus ds.
2.47
Differentiating 2.47, we get
v t kt, t gtup t htut kt, −t g−tup −t h−tu−t
t
∂
kt, s gsup s hsus ds.
−t ∂t
2.48
Using 2.46 and Young’s inequality, we obtain that
1
1
v t ≤ vt kt, tbt gt ht kt, −tb−t g−t h−t
p
p
t
1
∂
kt, sbs gs hs ds
p
−t ∂t
p−1
1
kt, t gtat ht
at p
p
p−1
1
kt, −t g−ta−t h−t
a−t p
p
t
p−1
1
∂
kt, s gsas hs
as ds.
p
p
−t ∂t
2.49
Using 2.29, we get
v t ≤ vt kt, tBt kt, −tB−t kt, tAt kt, −tA−t t
−t
t
−t
∂
kt, sBsds
∂t
2.50
∂
kt, sAsds.
∂t
Applying the differential inequality, we get
vt ≤
t
s
∂
ks, rArdr
ks, sAs ks, −sA−s 0
−s ∂s
t r
∂ × exp
k r, y B y dy dr .
kr, rBr kr, −rB−r s
−r ∂r
2.51
Journal of Inequalities and Applications
11
Since kt, s k−t, s, we have that
vt ≤
t
s
∂
ks, rArdr
0
−s ∂s
t r
∂ × exp
k r, y B y dy dr ds.
kr, rBr k−r, −rB−r s
−r ∂r
ks, sAs k−s, −sA−s 2.52
Using 2.33, we get
vt ≤ exp
×
t
−t
kr, rBrdr
t s
−s
0
∂
ks, rArdr
∂s
s
× exp −
−s
kr, rBrdr t r
s
−r
∂ k r, y B y dy dr ds
∂r
s
t r
∂ ks, sAs exp −
k r, y B y dy dr ds
kr, rBrdr 0
−s
s −r ∂r
s
t
t r
∂ k−s, −sA−s exp −
k r, y B y dy dr ds .
kr, rBrdr ds 0
−s
s −r ∂r
2.53
t
Since 0 ≤ s ≤ t, we have that
vt ≤ exp sgnt
× sgnt
t
−t
kr, rBrdr
s
t
sgns
0
−s
∂
ks, rArdr
∂s
s
t r
∂ × exp −
k r, y B y dy dr ds
kr, rBrdr −s
s −r ∂r
s
t
t r
∂ ks, sAs exp −
k r, y B y dy dr ds
kr, rBrdr 0
−s
s −r ∂r
−s
0
t r
∂ k r, y B y dy dr ds .
ks, sAs exp −
kr, rBrdr −t
s
−s −r ∂r
2.54
12
Journal of Inequalities and Applications
Using 2.9 and 2.11, we get
vt ≤ exp sgnt
× sgnt
t
−t
kr, rBrdr
t
s
sgns
−s
0
× exp − sgns
t
∂
ks, rArdr
∂s
s
−s
kr, rBrdr sgnt
r
sgnr
s
−r
∂ k r, y B y dy dr ds
∂r
s
ks, sAs exp − sgns
kr, rBrdr Bk t, sds
−s
0
t
0
−t
s
ks, sAs exp − sgns
kr, rBrdr Bk t, sds
−s
exp sgnt
× sgnt
t
−t
kr, rBrdr
t
s
sgns
0
× exp − sgns
sgnt
0
−t
−s
∂
ks, rArdr
∂s
s
−s
kr, rBrdr sgnt
t
r
sgnr
s
−r
∂ k r, y B y dy dr ds
∂r
s
ks, s exp − sgns
kr, rBrdr AsBk t, sds .
−s
2.55
Let t ≤ 0. Then
vt −
t
−t
kt, s gsup s hsus ds.
2.56
Differentiating 2.56, we get
−v t kt, t gtup t htut kt, −t g−tup −t h−tu−t
t
−t
∂
kt, s gsup s hsus ds.
∂t
2.57
Journal of Inequalities and Applications
13
Using 2.46 and Young’s inequality, we obtain that
1
1
−v t ≤ vt kt, tbt gt ht kt, −tb−t g−t h−t
p
p
t
−t
1
∂
kt, sbs gs hs ds
∂t
p
p−1
1
at kt, t gtat ht
p
p
2.58
p−1
1
kt, −t g−ta−t h−t
a−t p
p
−t
t
p−1
1
∂
kt, s gsas hs
hs ds.
∂t
p
p
Using 2.29, we get
−v t ≤ vt kt, tBt kt, −tB−t −t
t
kt, tAt kt, −tA−t −t
t
∂
kt, sBsds
∂t
2.59
∂
kt, sAsds.
∂t
Applying the differential inequality, we get
0
−s
∂
ks, rArdr
vt ≤
ks, sAs ks, −sA−s t
s ∂s
× exp
s kr, rBr kr, −rB−r t
−r
r
∂ k r, y B y dy dr ds.
∂r
2.60
Since kt, s k−t, s, we have that
vt ≤
0
t
ks, sAs k−s, −sA−s −s
s
∂
ks, rArdr
∂s
−r
s ∂ × exp
k r, y B y dy dr ds.
kr, rBr k−r, −rB−r t
r ∂r
2.61
14
Journal of Inequalities and Applications
Using 2.41, we get
−t
vt ≤ exp
kr, rBrdr
t
×
0 −s
∂
ks, rArdr
t s ∂s
s −r
−s
∂ k r, y B y dy dr ds
kr, rBrdr × exp −
s
t r ∂r
0
s −r
−s
∂ ks, sAs exp −
k r, y B y dy dr ds
kr, rBrdr t
s
t r ∂r
0
s −r
−s
∂ k−s, −sA−s exp −
k r, y B y dy dr ds .
kr, rBrdr ds t
s
t r ∂r
2.62
Since t ≤ s ≤ 0, we have that
t
vt ≤ exp sgnt
t
−t
kr, rBrdr
s
∂
ks, rArdr
∂s
0
−s
s −r
s
∂ k r, y B y dy dr ds
kr, rBrdr × exp − sgns
−s
t r ∂r
0
s −r
−s
∂ ks, sAs exp −
k r, y B y dy dr ds
kr, rBrdr t
s
t r ∂r
−t
−s −r
s
∂ k r, y B y dy dr ds .
ks, sAs exp −
kr, rBrdr 0
−s
t
r ∂r
2.63
× sgnt
sgns
Using 2.9 and 2.10, we get
t
vt ≤ exp sgnt
× sgnt
t
−t
kr, rBrdr
s
sgns
0
× exp − sgns
0
t
−s
∂
ks, rArdr
∂s
s
−s
kr, rBrdr sgnt
t
r
sgnr
s
−r
∂ k r, y B y dy dr ds
∂r
s
ks, sAs exp − sgns
kr, rBrdr Bk− t, sds
−t
0
ks, sAs exp − sgns
−s
s
−s
kr, rBrdr Bk− t, sds
Journal of Inequalities and Applications
t
exp sgnt
× sgnt
−t
15
kr, rBrdr
s
t
sgns
−s
0
∂
ks, rArdr
∂s
× exp − sgns
s
−s
kr, rBrdr sgnt
t
sgnr
s
r
∂ k r, y B y dy dr ds
−r ∂r
−t
s
sgnt
ks, s exp − sgns
kr, rBrdr AsBk t, sds .
×
−s
0
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:
p
p
vtt t, x − axvx t, x δvp t, x Ft, x; vt, x,
x
vx t, 0 vx t, l,
vt, 0 vt, l,
v0, x ϕx,
vt 0, x ψx,
t ∈ R, 0 < x < l,
3.1
t ∈ R,
0 ≤ x ≤ l,
where p > 1 is a fixed real number and δ const > 0. Let Ft, x; vt, x, t ∈ R, x ∈ 0, l,
ax ≥ 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
1/2
l
≤ gt
F t, x; vt, xdx
2
0
1/2
ht
v t, xdx
2p
0
l
1/2
v t, xdx
2
3.2
0
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-value 3.1 to the initial-value problem
p
vtt t Avp t Ft, vt,
v0 ϕ,
t ∈ R,
vt 0 ψ
3.3
16
Journal of Inequalities and Applications
in a Hilbert space H L2 0, l with a self-adjoint positive definite operator A defined by
the formula Aux −axux xx δux, with the domain DA {ux : u x ∈
L2 0, l, u0 ul, 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
1/2p
v t, xdx
2p
≤
0
t
1
1
1 1−1/p
M √ exp sgnt
hr dr
√ gr l
p
δ
δ
−t
× sgnt
t −t
p − 1 1−1/p
1 1−1/p
l
hs hs
M gs l
p
p
3.4
s
1/p
1
1 1−1/p
hr dr ds
× exp − sgns
√ gr l
p
δ
−s
l
l
for all t ∈ R. Here M 0 ϕ2p xdx1/2 p/δ 0 ϕ2p−1 xψ 2 xdx1/2 .
Proof. It is known thatthe formula see, e.g., 15, 16
vp t ctvp 0 stvp 0 t
st − sFs, vsds
3.5
0
gives a solution of problem 3.3. Here
ct eitA
1/2
e−itA
2
1/2
st A−1/2
,
eitA
1/2
− e−itA
2i
1/2
.
3.6
Applying the triangle inequality, condition 3.2, formula 3.5, and estimates see, e.g., 17
ctH → H ≤ 1,
1/2
A st
H →H
≤ 1,
−1/2 A
H →H
1
≤√ ,
δ
3.7
we get
v tH
p
1 1
≤ v 0H √ vp 0H √
δ
δ
p
t
0
gsvp sH hsvsH ds.
3.8
Journal of Inequalities and Applications
17
Since
1 v 0H √ vp 0H δ
p
l
1/2
ϕ xdx
2p
0
l
1/2
p
2p−1
2
ϕ
,
xψ xdx
δ 0
3.9
1/p
vsH ≤ l1−1/p vp sH
we have that
1
vp tH ≤ M √ sgnt
δ
t −t
1/p
gsvp sH l1−1/p hsvp sH
ds.
3.10
1/p
Denote that ut vp tH . Then
1
u t ≤ M √ sgnt
δ
p
t gsup s l1−1/p hsus ds
−t
3.11
for t ∈ R. Applying the integral inequality 2.4, we get
ut ≤
t
1
1
1 1−1/p
M √ exp sgnt
hr dr
√ gr l
p
δ
δ
−t
× sgnt
t p − 1 1−1/p
1
l
hs
M gs l1−1/p hs p
p
−t
3.12
s
1/p
1
1 1−1/p
hr dr ds
.
× exp − sgns
√ gr l
p
δ
−s
We have that
ut v
p
1/p
tH
l
1/2p
v t, xdx
2p
.
3.13
0
Therefore, the inequality 3.4 follows from the last inequality. Theorem 3.1 is proved.
Acknowledgments
The authors thank professor O. Celebi Turkey, professor R. P. Agarwal USA, and
anonymous reviewers for their valuable comments.
18
Journal of Inequalities and Applications
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