Volume 2010, Article ID 430512,18pages doi:10.1155/2010/430512
Research Article
A Note on the Integral Inequalities with Two Dependent Limits
Allaberen Ashyralyev,
1, 2Emine Misirli,
3and Ozlem Mogol
31Department of Mathematics, Fatih University, Buyukcekmece, 34500 Istanbul, Turkey
2Department of Mathematics, ITTU, 74200 Ashgabat, Turkmenistan
3Department of Mathematics, Ege University, 35100 Bornova-Izmir, Turkey
Correspondence should be addressed to Emine Misirli,[email protected] Received 4 October 2009; Revised 28 April 2010; Accepted 5 July 2010
Academic Editor: Martin Bohner
Copyrightq2010 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 equationssee, e.g.,1–4and the references given therein. One of the most useful inequalities in the development of the theory of differential equations is given in the following lemmasee5.
Lemma 1.1. Letutandftbe real-valued nonnegative continuous functions for allt≥0. If
u2t≤c22 t
0
fsusds 1.1
for allt≥0, wherec≥0 is a real constant, then
ut≤c t
0
fsds 1.2
for allt≥0.
Note that the generalization of this integral inequality and its discrete analogies are given in papers5–8. In paper9the following useful inequality with two dependent limits was established.
Lemma 1.2. Letutbe a real-valued nonnegative continuous function defined on−T, Tand letc andabe nonnegative constants. Then the inequality
ut≤csgnt t
−tausds, −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 in10–14.
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 functions defined onR −∞,∞.
(i) Letcbe a nonnegative constant. If
u2t≤c22 sgnt t
−tmsusds 2.1
fort∈R,then
ut≤csgnt t
−tmsds 2.2
for allt∈R.
(ii) Letp >1 be a real constant. If
upt≤at btsgnt t
−t
gsups hsus
ds 2.3
fort∈R,then
ut≤
at btexp
sgnt t
−tbr
gr 1 phr
dr
×sgnt t
−t
as
gs 1 phs
p−1
p hs
×exp
−sgns s
−sbr
gr 1 phr
dr
ds
1/p
2.4
for allt∈R.
(iii) Letctbe a real-valued positive continuous and nondecreasing function defined onRand p >1 be a real constant. If
upt≤cpt btsgnt t
−t
gsups hsus
ds 2.5
fort∈R,then
ut≤ct
1btexp
sgnt t
−tbr
gr hrc1−pr p
dr
×sgnt t
−t
gs hsc1−ps
×exp
−sgns s
−sbr
gr hrc1−pr p
dr ds
1/p
2.6
for allt∈R.
(iv) Letkt, sand its partial derivative∂kt, s/∂t be real-valued nonnegative continuous functions on−∞< s≤t <∞and letkt, sbe even function int.If
upt≤at btsgnt t
−tkt, s
gsups hsus
ds 2.7
fort∈R,then
ut≤
at btexp
sgnt t
−tkr, rbr
gr 1 phr
dr
×
sgnt t
0
sgns s
−s
∂
∂sks, r
ar
gr 1 phr
p−1
p hr
dr
×exp
−sgns s
−skr, rbr
gr 1 phr
dr
×exp
sgnt t
s
sgnr r
−r
∂
∂rk r, y
b
y
g y
1 ph
y
dy dr
ds
sgnt t
−tks, sexp
−sgns s
−skr, rbr
gr 1 phr
dr
×
as
gs 1 phs
p−1
p hs
Bkt, sds 1/p
.
2.8 for allt∈R. Here
Bkt, s
⎧⎪
⎨
⎪⎩
Bkt, s, t≥0, s∈R,
Bk−t, s, t≤0, s∈R, 2.9
where
Bk−t, s
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩ exp
t
s
r
−r
∂
∂rk r, y
B y
dy dr
, t≤s≤0,
exp t
−s
r
−r
∂
∂rk r, y
B y
dy dr
, 0≤s≤ −t,
2.10
Bkt, s
⎧⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎩ exp
t
s
r
−r
∂
∂rk r, y
B y
dy dr
, 0≤s≤t,
exp t
−s
r
−r
∂
∂rk r, y
B y
dy dr
, −t≤s≤0.
2.11
Proof. iDefine a functionvtby
vt c22 sgnt t
−tmsusds. 2.12
Note thatvtis a nonnegative function andv0 c2.Then2.1can be rewritten as
u2t≤vt, ut≤
vt. 2.13
It is easy to see thatvtis an even function.
First, lett≥0; then2.12can be rewritten as
vt c22 t
−tmsusds. 2.14
Differentiating2.14and using2.13, we get
vt≤2mt
vt 2m−t
vt. 2.15
Dividing both sides of2.15by 2
vt, we get
vt 2
vt ≤mt m−t. 2.16
Integrating the last inequality from 0 tot, we get vt≤c
t
0
msds
t
0
m−sdscsgnt t
−tmsds. 2.17
Second, lett≤0. Then,2.12can be written as
vt c2−2 t
−tmsusds. 2.18
Differentiating2.18and using2.13, we get
−vt≤2mt
vt 2m−t
vt. 2.19
Dividing both sides of2.19by 2
vt, we get
− vt 2
vt ≤mt m−t. 2.20
Integrating2.20fromtto 0, we get vt≤c
0
t
msds
0
t
m−sdscsgnt t
−tmsds. 2.21
Finally, using2.17and2.21, we obtain
vt≤csgnt t
−tmsds. 2.22
The inequality2.2follows from2.13and2.22.
iiDefine a functionvtby
vt sgnt t
−t
gsups hsus
ds. 2.23
It is evident thatvtis an even and nonnegative function. We have that
upt≤at btvt, ut≤at btvt1/p. 2.24
Using Young’s inequalitysee, e.g.,2, we obtain that
ut≤ at btvt
p p−1
p . 2.25
Lett≥0. Then
vt t
−tgsups hsusds. 2.26
Differentiating2.26, we get
vt gtupt htut g−tup−t h−tu−t. 2.27
Using2.24and2.25, we get
vt≤vt
bt
gt 1 pht
b−t
g−t 1 ph−t
at
gt 1 pht
a−t
g−t 1 ph−t
p−1
p ht h−t.
2.28
Denoting
Bt bt
gt 1 pht
, At at
gt 1 pht
p−1
p ht, 2.29
we get
vt−vtBt B−t≤At A−t. 2.30
From 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 anys≤t.Integrating the last inequality from 0 totand usingv0 0, we get
vt≤ t
0
As A−sexp t
s
Br B−rdr
ds. 2.32
It is easy to see that t
s
Br B−rdr t
−tBrdr− s
−sBrdr. 2.33
Then
vt≤exp t
−tBrdr t
0
As A−sexp
− s
−sBrdr
ds. 2.34
Since 0≤s≤t, we have that
vt≤exp
sgnt t
−tBrdr
× t
0
Asexp
− s
−sBrdr
ds 0
−tAsexp
− −s
s
Brdr
ds
exp
sgnt t
−tBrdr
× t
0
Asexp
−sgns s
−sBrdr
ds 0
−tAsexp
−sgns s
−sBrdr
ds
exp
sgnt t
−tBrdr
sgnt t
−tAsexp
−sgns s
−sBrdr
.
2.35
Applying2.24, we obtain
ut≤
at btexp
sgnt t
−tBrdr
×sgnt t
−tAsexp
−sgns s
−sBrdr
ds
1/p
.
2.36
From2.36, and2.29it follows2.4fort≥0.Lett≤0; then
vt − t
−t
gsups hsus ds,
−vt gtupt htut g−tup−t h−tu−t.
2.37
Using2.24and2.25, 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
≤exp s
t
Br B−rdr
As A−s
2.39
for anyt≤s.Integrating the last inequality fromtto 0 and usingv0 0, we get
vt≤ 0
t
As A−sexp s
t
Br B−rdr
ds. 2.40
It is easy to see that s
t
Br B−rdr
−t
t
Brdr− −s
s
Brdr. 2.41
Then
vt≤exp −t
t
Brdr 0
t
As A−sexp
− −s
s
Brdr
ds. 2.42
Sincet≤s≤0, we have that
vt≤exp
sgnt t
−tBrdr 0
t
As A−sexp
− −s
s
Brdr
ds
exp
sgnt t
−tBrdr
× 0
t
Asexp s
−sBrdr
ds 0
t
A−sexp s
−sBrdr
ds
exp
sgnt t
−tBrdr
× 0
t
Asexp s
−sBrdr
ds −t
0
Asexp −s
s
Brdr
ds
exp
sgnt t
−tBrdr −t
t
Asexp
−sgns s
−sBrdrds
exp
sgnt t
−tBrdr
sgnt t
−tAsexp
−sgns s
−sBrdr
ds.
2.43
Applying2.43 and 2.24, we obtain2.36for t ≤ 0. Then from 2.36 and 2.29,2.4 follows fort≤0.
iiiSincectis a positive, continuous, and nondecreasing function fort∈R, we have that
ut ct
p
≤1btsgnt t
−t
gs
us cs
p
hsc1−psus cs
ds. 2.44
Now the application of the inequality proven iniiyields the desired result in2.6.
ivWe define a functionvtby
vt sgnt t
−tkt, s
gsups hsus
ds. 2.45
Evidently, the functionvtis a nonnegative, monotonic, and nondecreasing intandv0 0.
We have that
upt≤at btvt, ut≤at btvt1/p. 2.46
Lett≥0. Then
vt t
−tkt, s
gsups hsus
ds. 2.47
Differentiating2.47, we get
vt kt, t
gtupt htut
kt,−t
g−tup−t h−tu−t
t
−t
∂
∂tkt, s
gsups hsus ds.
2.48
Using2.46and Young’s inequality, we obtain that
vt≤vt
kt, tbt
gt 1 pht
kt,−tb−t
g−t 1 ph−t
t
−t
∂
∂tkt, sbs
gs 1 phs
ds
kt, t
gtat ht
1
pat p−1 p
kt,−t
g−ta−t h−t
1
pa−t p−1 p
t
−t
∂
∂tkt, s
gsas hs
1
pas p−1 p
ds.
2.49
Using2.29, we get
vt≤vt
kt, tBt kt,−tB−t t
−t
∂
∂tkt, sBsds
kt, tAt kt,−tA−t t
−t
∂
∂tkt, sAsds.
2.50
Applying the differential inequality, we get
vt≤ t
0
ks, sAs ks,−sA−s s
−s
∂
∂sks, rArdr
×exp t
s
kr, rBr kr,−rB−r r
−r
∂
∂rk r, y
B y
dy
dr .
2.51
Sincekt, s k−t, s, we have that
vt≤ t
0
ks, sAs k−s,−sA−s s
−s
∂
∂sks, rArdr
×exp t
s
kr, rBr k−r,−rB−r r
−r
∂
∂rk r, y
B y
dy
dr ds.
2.52
Using2.33, we get
vt≤exp t
−tkr, rBrdr
× t
0
s
−s
∂
∂sks, rArdr
×exp
− s
−skr, rBrdr t
s
r
−r
∂
∂rk r, y
B y
dy dr
ds
t
0
ks, sAsexp
− s
−skr, rBrdr t
s
r
−r
∂
∂rk r, y
B y
dy dr
ds
t
0
k−s,−sA−sexp
− s
−skr, rBrdr ds t
s
r
−r
∂
∂rk r, y
B y
dy dr
ds . 2.53
Since 0≤s≤t, we have that
vt≤exp
sgnt t
−tkr, rBrdr
×
sgnt t
0
sgns s
−s
∂
∂sks, rArdr
×exp
− s
−skr, rBrdr t
s
r
−r
∂
∂rk r, y
B y
dy dr
ds
t
0
ks, sAsexp
− s
−skr, rBrdr t
s
r
−r
∂
∂rk r, y
B y
dy dr
ds
0
−tks, sAsexp
− −s
s
kr, rBrdr t
−s
r
−r
∂
∂rk r, y
B y
dy dr
ds . 2.54
Using2.9and2.11, we get
vt≤exp
sgnt t
−tkr, rBrdr
×
sgnt t
0
sgns s
−s
∂
∂sks, rArdr
×exp
−sgns s
−skr, rBrdrsgnt t
s
sgnr r
−r
∂
∂rk r, y
B y
dy dr
ds
t
0
ks, sAsexp
−sgns s
−skr, rBrdr
Bkt, sds
0
−tks, sAsexp
−sgns s
−skr, rBrdr
Bkt, sds
exp
sgnt t
−tkr, rBrdr
×
sgnt t
0
sgns s
−s
∂
∂sks, rArdr
×exp
−sgns s
−skr, rBrdrsgnt t
s
sgnr r
−r
∂
∂rk r, y
B y
dy dr
ds
sgnt 0
−tks, sexp
−sgns s
−skr, rBrdr
AsBkt, sds .
2.55
Lett≤0. Then
vt − t
−tkt, s
gsups hsus
ds. 2.56
Differentiating2.56, we get
−vt kt, t
gtupt htut
kt,−t
g−tup−t h−tu−t
t
−t
∂
∂tkt, s
gsups hsus ds.
2.57
Using2.46and Young’s inequality, we obtain that
−vt≤vt
kt, tbt
gt 1 pht
kt,−tb−t
g−t 1 ph−t
t
−t
∂
∂tkt, sbs
gs 1 phs
ds
kt, t
gtat ht
1
pat p−1 p
kt,−t
g−ta−t h−t
1
pa−t p−1 p
−t
t
∂
∂tkt, s
gsas hs
1
phs p−1 p
ds.
2.58
Using2.29, we get
−vt≤vt
kt, tBt kt,−tB−t −t
t
∂
∂tkt, sBsds
kt, tAt kt,−tA−t −t
t
∂
∂tkt, sAsds.
2.59
Applying the differential inequality, we get
vt≤ 0
t
ks, sAs ks,−sA−s −s
s
∂
∂sks, rArdr
×exp s
t
kr, rBr kr,−rB−r −r
r
∂
∂rk r, y
B y
dy
dr
ds.
2.60
Sincekt, s k−t, s, we have that
vt≤ 0
t
ks, sAs k−s,−sA−s −s
s
∂
∂sks, rArdr
×exp s
t
kr, rBr k−r,−rB−r −r
r
∂
∂rk r, y
B y
dy
dr
ds.
2.61
Using2.41, we get vt≤exp
−t
t
kr, rBrdr
× 0
t
−s
s
∂
∂sks, rArdr
×exp
− −s
s
kr, rBrdr s
t
−r
r
∂
∂rk r, y
B y
dy dr
ds
0
t
ks, sAsexp
− −s
s
kr, rBrdr s
t
−r
r
∂
∂rk r, y
B y
dy dr
ds
0
t
k−s,−sA−sexp
− −s
s
kr, rBrdr
ds s
t
−r
r
∂
∂rk r, y
B y
dy dr ds
. 2.62
Sincet≤s≤0, we have that vt≤exp
sgnt
t
−tkr, rBrdr
×
sgnt t
0
sgns s
−s
∂
∂sks, rArdr
×exp
−sgns s
−skr, rBrdr s
t
−r
r
∂
∂rk r, y
B y
dy dr
ds
0
t
ks, sAsexp
− −s
s
kr, rBrdr s
t
−r
r
∂
∂rk r, y
B y
dy dr
ds
−t
0
ks, sAsexp
− s
−skr, rBrdr −s
t
−r
r
∂
∂rk r, y
B y
dy dr
ds . 2.63
Using2.9and2.10, we get vt≤exp
sgnt
t
−tkr, rBrdr
×
sgnt t
0
sgns s
−s
∂
∂sks, rArdr
×exp
−sgns s
−skr, rBrdrsgnt t
s
sgnr r
−r
∂
∂rk r, y
B y
dy dr
ds
0
t
ks, sAsexp
−sgns s
−skr, rBrdr
Bk−t, sds
−t
0
ks, sAsexp
−sgns s
−skr, rBrdr
Bk−t, sds
exp
sgnt t
−tkr, rBrdr
×
sgnt t
0
sgns s
−s
∂
∂sks, rArdr
×exp
−sgns s
−skr, rBrdrsgnt t
s
sgnr
× r
−r
∂
∂rk r, y
B y
dy dr
ds
sgnt −t
0
ks, sexp
−sgns s
−skr, rBrdr
AsBkt, sds .
2.64
The inequality2.8follows from2.29,2.55, and2.64.Theorem 2.1is proved.
3. An Application
In this section, we indicate an application of Theorem 2.1part iito obtain the explicit bound on the solution of the following boundary value problem for one dimensional partial differential equations:
vttpt, x−
axvpxt, x
xδvpt, x Ft, x;vt, x, t∈R,0< x < l, vt,0 vt, l, vxt,0 vxt, l, t∈R,
v0, x ϕx, vt0, x ψx, 0≤x≤l,
3.1
wherep > 1 is a fixed real number andδ const > 0. LetFt, x;vt, x,t ∈ R,x ∈ 0, l, ax ≥ a > 0,x ∈ 0, l, ϕx, ψx,x ∈ 0, lbe smooth functions and problem3.1has a unique smooth solutionvt, x.Assume that
l
0
F2t, x;vt, xdx 1/2
≤gt l
0
v2pt, xdx 1/2
ht l
0
v2t, xdx 1/2
3.2
for allt∈R.Here gtandhtare real-valued nonnegative continuous functions defined on R.
This allows us to reduce the nonlocal boundary-value3.1to the initial-value problem
vttpt Avpt Ft, vt, t∈R,
v0 ϕ, vt0 ψ 3.3
in a Hilbert spaceH L20, l with a self-adjoint positive definite operatorAdefined by the formula Aux −axuxxx δux, with the domain DA {ux : ux ∈ L20, l, u0 ul, u0 ul}see, e.g.,15,16.
Let us give a corollary ofTheorem 2.1.
Theorem 3.1. The solution of problem3.1satisfies the estimates
l
0
v2pt, xdx 1/2p
≤
M 1
√δexp
sgnt t
−t
√1 δ
gr 1
pl1−1/phr
dr
×sgnt t
−t
M
gs 1
pl1−1/phs
p−1
p l1−1/phs
×exp
−sgns s
−s
√1 δ
gr 1
pl1−1/phr
dr
ds 1/p
3.4
for allt∈R.HereM l
0ϕ2pxdx1/2 p/δl
0ϕ2p−1xψ2xdx1/2. Proof. It is known thatthe formulasee, e.g.,15,16
vpt ctvp0 stvp0 t
0
st−sFs, vsds 3.5
gives a solution of problem3.3. Here
ct eitA1/2e−itA1/2
2 , st A−1/2eitA1/2−e−itA1/2
2i . 3.6
Applying the triangle inequality, condition3.2, formula3.5, and estimatessee, e.g.,17
ctH→H ≤1, A1/2st
H→H ≤1, A−1/2
H→H≤ 1
√δ, 3.7
we get
vptH≤ vp0H 1
√δvp0
H 1
√δ t
0
gsvpsHhsvsH
ds. 3.8
Since
vp0H 1
√δvp0
H
l
0
ϕ2pxdx 1/2
p
δ l
0
ϕ2p−1xψ2xdx 1/2
,
vsH≤l1−1/pvps1/pH
3.9
we have that
vptH≤M 1
√δsgnt t
−t
gsvpsHl1−1/phsvps1/pH
ds. 3.10
Denote thatut vpt1/pH .Then
upt≤M 1
√δsgnt t
−t
gsups l1−1/phsus
ds 3.11
fort∈R.Applying the integral inequality2.4, we get
ut≤
M 1
√δexp
sgnt t
−t
√1 δ
gr 1
pl1−1/phr
dr
×sgnt t
−t
M
gs 1
pl1−1/phs
p−1
p l1−1/phs
×exp
−sgns s
−s
√1 δ
gr 1
pl1−1/phr
dr
ds 1/p
.
3.12
We have that
ut vpt1/pH l
0
v2pt, xdx 1/2p
. 3.13
Therefore, the inequality3.4follows from the last inequality.Theorem 3.1is proved.
Acknowledgments
The authors thank professor O. Celebi Turkey, professor R. P. Agarwal USA, and anonymous reviewers for their valuable comments.
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![The quadratic integral equation can be very often encountered in many applications (see[1]-[4] and [8]-[11](data:image/gif;base64,R0lGODlhAQABAIAAAP///wAAACH5BAEAAAAALAAAAAABAAEAAAICRAEAOw==)